[cut: only the first 33554432 bytes of this file were read]
<?xml version="1.0" encoding="utf-8"?>
<posts>
  <row Id="1" PostTypeId="1" AcceptedAnswerId="9" CreationDate="2010-07-20T19:09:27.200" Score="21" ViewCount="589" Body="&lt;p&gt;Can someone explain to me how there can be different kinds of infinities?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was reading &quot;the man who loved only numbers&quot; and came across the concept of countable and uncountable infinities, but they're only words to me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any help would be appreciated.&lt;/p&gt;&#xA;" OwnerUserId="10" LastEditorUserId="126" LastEditorDisplayName="" LastEditDate="2010-07-22T19:27:38.740" LastActivityDate="2010-11-02T17:28:02.230" Title="Different kinds of infinities?" Tags="&lt;set-theory&gt;&lt;intuition&gt;" AnswerCount="8" CommentCount="1" FavoriteCount="6" />
  <row Id="2" PostTypeId="1" AcceptedAnswerId="19" CreationDate="2010-07-20T19:09:44.937" Score="2" ViewCount="382" Body="&lt;p&gt;This &lt;code&gt;limit&lt;/code&gt; thing keeps coming up in my calculus textbook.  What is it?&lt;/p&gt;&#xA;" OwnerUserId="6" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-07-20T20:21:40.707" LastActivityDate="2010-08-21T20:25:07.867" Title="What is a limit?" Tags="&lt;calculus&gt;&lt;limit&gt;&lt;definition&gt;" AnswerCount="7" CommentCount="4" FavoriteCount="1" />
  <row Id="3" PostTypeId="1" CreationDate="2010-07-20T19:12:14.353" Score="8" ViewCount="255" Body="&lt;p&gt;&lt;a href=&quot;http://mathfactor.uark.edu/&quot; rel=&quot;nofollow&quot;&gt;mathfactor&lt;/a&gt; is one I listen to.  Does anyone else have a recommendation?&lt;/p&gt;&#xA;" OwnerUserId="29" LastEditorUserId="126" LastEditorDisplayName="" LastEditDate="2010-07-21T01:52:17.563" LastActivityDate="2010-07-28T17:39:29.080" CommunityOwnedDate="2010-07-20T20:35:23.263" Title="List of interesting math podcasts?" Tags="&lt;big-list&gt;&lt;soft-question&gt;" AnswerCount="4" CommentCount="3" FavoriteCount="6" />
  <row Id="4" PostTypeId="2" ParentId="3" CreationDate="2010-07-20T19:14:10.603" Score="2" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://www.bbc.co.uk/podcasts/series/moreorless&quot; rel=&quot;nofollow&quot;&gt;More or Less&lt;/a&gt; is a BBC Radio 4 programme about maths and statistics in the news, and there is a free podcast. It's presented by &lt;a href=&quot;http://timharford.com/&quot; rel=&quot;nofollow&quot;&gt;Tim Harford&lt;/a&gt;, the Undercover Economist from the &lt;a href=&quot;http://www.ft.com/home/uk&quot; rel=&quot;nofollow&quot;&gt;Financial Times&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="31" LastActivityDate="2010-07-20T19:14:10.603" CommunityOwnedDate="2010-07-20T20:35:23.263" CommentCount="1" />
  <row Id="5" PostTypeId="1" AcceptedAnswerId="7" CreationDate="2010-07-20T19:18:01.250" Score="11" ViewCount="676" Body="&lt;p&gt;I have read a few proofs that $\sqrt{2}$ is irrational.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have never, however, been able to really grasp what they were talking about.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a simplified proof that $\sqrt{2}$ is irrational?&lt;/p&gt;&#xA;" OwnerUserId="38" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2011-03-04T22:17:05.217" LastActivityDate="2011-04-03T00:28:55.917" Title="How can you prove that the square root of two is irrational?" Tags="&lt;number-theory&gt;&lt;irrational-numbers&gt;" AnswerCount="6" CommentCount="4" FavoriteCount="3" />
  <row Id="6" PostTypeId="1" AcceptedAnswerId="281" CreationDate="2010-07-20T19:20:00.543" Score="7" ViewCount="203" Body="&lt;p&gt;I'm looking for a nice, quick online graphing tool. The ability to link to, or embed the output would be handy, too.&lt;/p&gt;&#xA;" OwnerUserId="41" LastEditorUserId="126" LastEditorDisplayName="" LastEditDate="2010-07-21T07:34:23.613" LastActivityDate="2011-04-03T09:08:04.437" CommunityOwnedDate="2010-07-20T19:20:00.543" Title="What is your favorite online graphing tool?" Tags="&lt;math-software&gt;" AnswerCount="4" CommentCount="2" FavoriteCount="5" />
  <row Id="7" PostTypeId="2" ParentId="5" CreationDate="2010-07-20T19:21:52.240" Score="29" ViewCount="" Body="&lt;p&gt;You use a proof by contradiction. Basically, you suppose that $\sqrt{2}$ can be written as $p/q$. Then you know that $2q^2 = p^2$. However, both $q^2$ and $p^2$ have an even number of factors of two, so $2q^2$ has an odd number of factors of 2, which means it can't be equal to $p^2$.&lt;/p&gt;&#xA;" OwnerUserId="45" LastActivityDate="2010-07-20T19:21:52.240" CommentCount="4" />
  <row Id="8" PostTypeId="1" AcceptedAnswerId="17" CreationDate="2010-07-20T19:22:12.190" Score="8" ViewCount="290" Body="&lt;p&gt;I was reading up on the Fibonacci Sequence when I've noticed some were able to calculate specific numbers. So far I've only figured out creating an array and counting to the value, which is incredibly simple, but I reckon I can't find any formula for calculating a Fibonacci number based on it's position.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a way to do this? If so, how are we able to apply these formulas to arrays?&lt;/p&gt;&#xA;" OwnerUserId="40" LastEditorUserId="126" LastEditorDisplayName="" LastEditDate="2010-07-21T17:01:20.937" LastActivityDate="2010-07-28T09:52:46.637" Title="How are we able to calculate specific numbers in the Fibonacci Sequence?" Tags="&lt;linear-algebra&gt;&lt;combinatorics&gt;&lt;generating-functions&gt;" AnswerCount="6" FavoriteCount="1" />
  <row Id="9" PostTypeId="2" ParentId="1" CreationDate="2010-07-20T19:22:20.193" Score="40" ViewCount="" Body="&lt;p&gt;Suppose no one every taught you the names for ordinary numbers.  Then suppose that you and I agreed that we would trade one bushel of corn for each of my sheep.  But there's a problem, we don't know how to count the bushels or the sheep!  So what do we do?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We form a &quot;bijection&quot; between the two sets.  That's just fancy language for saying you pair things up by putting one bushel next to each of the sheep.  When we're done we swap.  We've just proved that the number of sheep is the same as the number of bushels without actually counting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can try doing the same thing with infinite sets.  So suppose you have the set of positive numbers and I have the set of rational numbers and you want to trade me one positive number for each of my rationals.  Can you do so in a way that gets all of my rational numbers?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps surprisingly the answer is yes!  You make the rational numbers into a big square grid with the numerator and denominators as the two coordinates.  Then you start placing your &quot;bushels&quot; along diagonals of increasing size, &lt;a href=&quot;http://en.wikipedia.org/wiki/File%3aPairing_natural.svg&quot; rel=&quot;nofollow&quot;&gt;see wikipedia&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This says that the rational numbers are &quot;countable&quot; that is you can find a clever way to count them off in the above fashion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The remarkable fact is that for the real numbers there's &lt;em&gt;no way at all&lt;/em&gt; to count them off in this way.  No matter how clever you are you won't be able to scam me out of all of my real numbers by placing a natural number next to each of them.  The proof of that is Cantor's clever &quot;&lt;a href=&quot;http://en.wikipedia.org/wiki/Cantor%27s_diagonal_argument&quot; rel=&quot;nofollow&quot;&gt;diagonal argument&lt;/a&gt;.&quot;&lt;/p&gt;&#xA;" OwnerUserId="8" LastActivityDate="2010-07-20T19:22:20.193" CommentCount="3" />
  <row Id="11" PostTypeId="1" AcceptedAnswerId="49" CreationDate="2010-07-20T19:23:09.117" Score="11" ViewCount="1848" Body="&lt;p&gt;I'm told by smart people that &lt;code&gt;0.999... = 1&lt;/code&gt; and I believe them but is there a proof that explains why?&lt;/p&gt;&#xA;" OwnerUserId="41" LastEditorUserId="171" LastEditorDisplayName="" LastEditDate="2010-08-16T19:08:51.070" LastActivityDate="2011-05-04T05:58:42.093" CommunityOwnedDate="2010-07-28T06:53:43.987" Title="Does .99999... = 1?" Tags="&lt;algebra-precalculus&gt;&lt;decimal-expansion&gt;" AnswerCount="19" CommentCount="11" FavoriteCount="8" />
  <row Id="13" PostTypeId="2" ParentId="1" CreationDate="2010-07-20T19:23:17.570" Score="4" ViewCount="" Body="&lt;p&gt;The basic concept is thus:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;A 'countable' infinity is one where you can give each item in the set an integer and 'count' them (even though there are an infinite number of them)&lt;/li&gt;&#xA;&lt;li&gt;An 'uncountable' infinity defies this. You cannot assign an integer to each item in the set because you will miss items.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The key to seeing this is using the 'diagonal slash' argument as originally put forward by Cantor. With a countable infinity, you can create a list of all the items in the set and assign each one a different natural number. This can be done with the naturals (obviously) and the complete range of integers (including negative numbers) and even the rational numbers (so including fractions). It cannot be done with the reals due to the diagonal slash argument:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Create your list of all real numbers and assign each one an integer&lt;/li&gt;&#xA;&lt;li&gt;Create a real number with the rule that the first digit after the decimal point is different from the first digit of your first number, the second digit is different from the second digit of your second number, and so on for all digits&lt;/li&gt;&#xA;&lt;li&gt;Try and place this number in your list of all numbers... it can't be the first number, or the second or the third... and so on down the list. &lt;/li&gt;&#xA;&lt;li&gt;Reductio Ad Absurdium, your number does not exist in your countable list of all real numbers and must be added on to create a new list. The same process can then be done again to show the list still isn't complete.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;This shows a difference between two obviously infinite sets and leads to the somewhat scary conclusion that there are (at least) 2 different forms of infinity.&lt;/p&gt;&#xA;" OwnerUserId="35" LastActivityDate="2010-07-20T19:23:17.570" />
  <row Id="14" PostTypeId="2" ParentId="1" CreationDate="2010-07-20T19:23:20.317" Score="2" ViewCount="" Body="&lt;p&gt;How many Natural (Integers) Numbers could you count?&#xA;There are infinitely many, yet you can count them.&#xA;It's called &lt;a href=&quot;http://en.wikipedia.org/wiki/Countable_set&quot; rel=&quot;nofollow&quot;&gt;Countable Set&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How many Real Numbers are there?&#xA;Infinitely as well (Since at least every Natural Number is a Real Number).&#xA;Yet you won't be able to count them (Intuitively, Let's say you name a number the first, then find the second, I can, for sure, find a number in between, their average which is Real Number as well)..&#xA;It's called &lt;a href=&quot;http://en.wikipedia.org/wiki/Uncountable_set&quot; rel=&quot;nofollow&quot;&gt;Uncountable Set&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you are after is how we define how big is a given set.&#xA;Then you should look for &lt;a href=&quot;http://en.wikipedia.org/wiki/Cardinality&quot; rel=&quot;nofollow&quot;&gt;Cardinality&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="33" LastEditorUserId="33" LastEditorDisplayName="" LastEditDate="2010-11-02T17:28:02.230" LastActivityDate="2010-11-02T17:28:02.230" CommentCount="3" />
  <row Id="15" PostTypeId="2" ParentId="6" CreationDate="2010-07-20T19:23:21.220" Score="6" ViewCount="" Body="&lt;p&gt;Well, I am not sure where you want to embed the graphs, but &lt;a href=&quot;http://www.wolframalpha.com/input/?i=sin%28x%29&quot; rel=&quot;nofollow&quot;&gt;Wolfram Alpha&lt;/a&gt; is pretty handy for graphing. It has most of the features of Mathematica, can handle 3D functions, and fancy scaling and such. I highly recommend it.&lt;/p&gt;&#xA;" OwnerUserId="16" LastEditorUserId="41" LastEditorDisplayName="" LastEditDate="2010-07-20T19:26:37.047" LastActivityDate="2010-07-20T19:26:37.047" CommunityOwnedDate="2010-07-20T19:23:21.220" />
  <row Id="16" PostTypeId="2" ParentId="5" CreationDate="2010-07-20T19:23:50.563" Score="11" ViewCount="" Body="&lt;p&gt;Consider this proof by contradiction:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume that $\sqrt{2}$ is rational. Then there exists some rational $R=\sqrt{2}=\frac{Q}{D}$, where $Q$ and $D$ are positive integers and relatively prime (since $R$ can be expressed in simplified form).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now consider $R^2 = 2 = \frac{Q^2}{D^2}$. Since $Q$ and $D$ are relatively prime, this means that only $Q^2$ can have $2$ in its prime decomposition, and the exponent must be one. Thus, $Q^2 = 2^1 x$, for some odd integer $x$. But $Q^2$ is a square, and thus the exponents for all of its prime factors must be even. Here we have a contradiction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, $\sqrt{2}$ must be irrational.&lt;/p&gt;&#xA;" OwnerUserId="48" LastActivityDate="2010-07-20T19:23:50.563" />
  <row Id="17" PostTypeId="2" ParentId="8" CreationDate="2010-07-20T19:23:54.230" Score="11" ViewCount="" Body="&lt;p&gt;Wikipedia has a closed-form function called &quot;Binet's formula&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Fibonacci_number#Relation_to_the_golden_ratio&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Fibonacci_number#Relation_to_the_golden_ratio&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$F\left(n\right) = {{\varphi^n-(1-\varphi)^n} \over {\sqrt 5}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is based on the Golden Ratio.&lt;/p&gt;&#xA;" OwnerUserId="38" LastEditorUserId="38" LastEditorDisplayName="" LastEditDate="2010-07-22T18:38:42.820" LastActivityDate="2010-07-22T18:38:42.820" CommentCount="5" />
  <row Id="18" PostTypeId="1" CreationDate="2010-07-20T19:24:34.823" Score="3" ViewCount="173" Body="&lt;p&gt;Given the semimajor axis and a flattening factor, is it possible to calculate the semiminor axis?&lt;/p&gt;&#xA;" OwnerUserId="17" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-24T16:08:54.593" LastActivityDate="2010-12-24T16:08:54.593" Title="How do you calculate the semiminor axis for an ellipsoid?" Tags="&lt;geometry&gt;" AnswerCount="2" CommentCount="7" />
  <row Id="19" PostTypeId="2" ParentId="2" CreationDate="2010-07-20T19:24:52.200" Score="7" ViewCount="" Body="&lt;p&gt;For many functions there are undefined values. For example in the function $$f(x) = \frac{x^2 - 1}{x - 1}$$ there is no value of $f(x)$ for $x = 1$, since that involves a division by 0.&#xA;In this instance though, as x gets closer and closer to 1, $f(x)$ gets closer and closer to 2, so we define that to be the limit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In calculus limits come up when getting the slope of a linear function as I'm sure you've seen, the problem is that getting the slope at a certain point, using the slope formula we all know, involves dividing by zero, so we use limits to get the answer. More specifically we create a formula for the slope between the point $(x, f(x))$ and the point $(x+h, f(x+h))$ and find the slope as $h \to 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Limits as $x \to \infty$ are also used often, since even if a function is well defined for all $x$ it is still impossible to simply plug in infinite and work it out, in this way it can be used for approximation purposes. Calculating the runtime of computer programs on large inputs is often done by deriving a formula for the run-time given input of a certain length, then finding the limit as $x \to \infty$&lt;/p&gt;&#xA;" OwnerUserId="34" LastEditorUserId="171" LastEditorDisplayName="" LastEditDate="2010-08-21T20:25:07.867" LastActivityDate="2010-08-21T20:25:07.867" />
  <row Id="20" PostTypeId="1" CreationDate="2010-07-20T19:25:09.563" Score="3" ViewCount="562" Body="&lt;p&gt;In mathematics there seem to be a lot of different types of number. What exactly are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Real numbers&lt;/li&gt;&#xA;&lt;li&gt;Integers&lt;/li&gt;&#xA;&lt;li&gt;Rationals numbers&lt;/li&gt;&#xA;&lt;li&gt;Decimals&lt;/li&gt;&#xA;&lt;li&gt;Complex numbers&lt;/li&gt;&#xA;&lt;li&gt;Natural numbers&lt;/li&gt;&#xA;&lt;li&gt;Cardinals&lt;/li&gt;&#xA;&lt;li&gt;Ordinals&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;And as workmad3 points out, some more advanced categories (I'd never heard of)&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Hyper-reals&lt;/li&gt;&#xA;&lt;li&gt;Quaternions&lt;/li&gt;&#xA;&lt;li&gt;Imaginary numbers&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Or any other type of classification of a number I missed.&lt;/p&gt;&#xA;" OwnerUserId="54" LastEditorUserId="126" LastEditorDisplayName="" LastEditDate="2010-07-21T23:12:36.630" LastActivityDate="2010-10-13T06:01:26.353" Title="Can you explain what a real number is (also rational, decimal, integer, natural, cardinal, ordinal...)" Tags="&lt;terminology&gt;&lt;definition&gt;" AnswerCount="8" CommentCount="4" FavoriteCount="1" />
  <row Id="22" PostTypeId="1" AcceptedAnswerId="53" CreationDate="2010-07-20T19:25:58.483" Score="6" ViewCount="324" Body="&lt;p&gt;By matrix-defined, I mean&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&#xA; x  =&#xA;&#xA;    | i j k |&#xA;det(| a b c |)&#xA;    | d e f |&#xA;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;(instead of the definition of the product of the magnitudes multiplied by the sign of their angle, in the direction orthogonal)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I try cross producting two vectors with no &lt;code&gt;k&lt;/code&gt; component, I get one with only &lt;code&gt;k&lt;/code&gt;, which is expected.  But why?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As has been pointed out, I am asking why the algebraic definition lines up with the geometric definition.&lt;/p&gt;&#xA;" OwnerUserId="51" LastEditorUserId="126" LastEditorDisplayName="" LastEditDate="2010-07-21T01:42:23.140" LastActivityDate="2010-07-28T02:03:15.647" Title="Why is the matrix-defined Cross Product of two 3D vectors always orthogonal?" Tags="&lt;linear-algebra&gt;" AnswerCount="4" />
  <row Id="23" PostTypeId="2" ParentId="11" CreationDate="2010-07-20T19:26:13.353" Score="-3" ViewCount="" Body="&lt;p&gt;well the simplest is &lt;code&gt;3x(1/3) = 1&lt;/code&gt; where &lt;code&gt;1/3 = 0.33·&lt;/code&gt;&lt;/p&gt;&#xA;" OwnerUserId="22" LastActivityDate="2010-07-20T19:26:13.353" CommunityOwnedDate="2010-07-28T06:53:43.987" CommentCount="1" />
  <row Id="24" PostTypeId="2" ParentId="8" CreationDate="2010-07-20T19:26:16.643" Score="2" ViewCount="" Body="&lt;p&gt;The closed form calculation for Fibonacci sequences is known as &lt;a href=&quot;http://mathworld.wolfram.com/BinetsFibonacciNumberFormula.html&quot; rel=&quot;nofollow&quot;&gt;Binet's Formula&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="53" LastActivityDate="2010-07-20T19:26:16.643" />
  <row Id="25" PostTypeId="2" ParentId="11" CreationDate="2010-07-20T19:26:36.643" Score="0" ViewCount="" Body="&lt;p&gt;The proof I've always seen was that if 1/3 == 0.333... then 0.333... x 3 must be equal to 1, but at the same time, calculating it on a digit level gives 0.999...&lt;/p&gt;&#xA;" OwnerUserId="17" LastActivityDate="2010-07-20T19:26:36.643" CommunityOwnedDate="2010-07-28T06:53:43.987" CommentCount="8" />
  <row Id="26" PostTypeId="2" ParentId="8" CreationDate="2010-07-20T19:26:38.187" Score="1" ViewCount="" Body="&lt;p&gt;You can use Binet's formula, described at &lt;a href=&quot;http://mathworld.wolfram.com/BinetsFibonacciNumberFormula.html&quot; rel=&quot;nofollow&quot;&gt;http://mathworld.wolfram.com/BinetsFibonacciNumberFormula.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(see also Wikipedia for a proof: &lt;a href=&quot;http://en.wikipedia.org/wiki/Binet_formula#Closed_form_expression&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Binet_formula#Closed_form_expression&lt;/a&gt; )&lt;/p&gt;&#xA;" OwnerUserId="55" LastActivityDate="2010-07-20T19:26:38.187" />
  <row Id="27" PostTypeId="2" ParentId="2" CreationDate="2010-07-20T19:26:38.670" Score="0" ViewCount="" Body="&lt;p&gt;A limit is the term used when you examine a function as it approaches some value. For example, the function f(x) = 1/x. You can take a limit of this function as x approaches infinity and as x approaches 0 and see what occurs, and get an idea of the behaviour of the function in these regions even though the function cannot be defined at the exact values you have taken limits of. (1/0 is undefined, and 1/infinity cannot be done without adding infinity to the number system - see the hyperreals for this)&lt;/p&gt;&#xA;" OwnerUserId="35" LastActivityDate="2010-07-20T19:26:38.670" />
  <row Id="29" PostTypeId="1" CreationDate="2010-07-20T19:27:09.700" Score="9" ViewCount="445" Body="&lt;p&gt;I'm looking for an online or software calculator that can show me the history of items I typed in, much like an expensive Ti calculator. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you recommend any?&lt;/p&gt;&#xA;" OwnerUserId="54" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-07-28T01:04:25.440" LastActivityDate="2010-10-08T02:42:51.233" CommunityOwnedDate="2010-07-28T01:32:30.370" Title="Can you recommend a decent online or software calculator?" Tags="&lt;soft-question&gt;&lt;big-list&gt;&lt;math-software&gt;&lt;computer-algebra-systems&gt;" AnswerCount="18" FavoriteCount="5" />
  <row Id="30" PostTypeId="2" ParentId="11" CreationDate="2010-07-20T19:27:38.963" Score="-3" ViewCount="" Body="&lt;p&gt;Given (by long division):&lt;br&gt;&#xA;$\frac{1}{3} = 0.\bar{3}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Multiply by 3:&lt;br&gt;&#xA;$3\times \left( \frac{1}{3} \right) = \left( 0.\bar{3} \right) \times 3$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore:&lt;br&gt;&#xA;$\frac{3}{3} = 0.\bar{9}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;QED.&lt;/p&gt;&#xA;" OwnerUserId="38" LastEditorUserId="38" LastEditorDisplayName="" LastEditDate="2011-02-21T02:52:26.000" LastActivityDate="2011-02-21T02:52:26.000" CommunityOwnedDate="2010-07-28T06:53:43.987" />
  <row Id="31" PostTypeId="2" ParentId="11" CreationDate="2010-07-20T19:27:39.433" Score="-1" ViewCount="" Body="&lt;p&gt;Indeed this is true. The underlying reason is that decimal numbers are not unique representations of the reals. (Technically, there does not exist a bijection between the set of all decimal numbers and the reals.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a very simple proof:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;1 / 3 = 0.333... (by long division)&#xA;&#xA;=&amp;gt; 0.333 * 3 = 0.999... (multiplying each digit by 3)&#xA;&#xA;But then we already know 0.333... * 3 = 1&#xA;&#xA;Therefore 0.999... = 1&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="56" LastEditorUserId="56" LastEditorDisplayName="" LastEditDate="2010-07-21T07:21:11.583" LastActivityDate="2010-07-21T07:21:11.583" CommunityOwnedDate="2010-07-28T06:53:43.987" CommentCount="7" />
  <row Id="35" PostTypeId="2" ParentId="1" CreationDate="2010-07-20T19:29:54.777" Score="6" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Hilbert%27s_paradox_of_the_Grand_Hotel&quot; rel=&quot;nofollow&quot;&gt;Hilbert's Hotel&lt;/a&gt; is a classic demonstration.&lt;/p&gt;&#xA;" OwnerUserId="49" LastActivityDate="2010-07-20T19:29:54.777" CommentCount="2" />
  <row Id="36" PostTypeId="2" ParentId="2" CreationDate="2010-07-20T19:30:11.283" Score="2" ViewCount="" Body="&lt;p&gt;What others have provided is a pretty good intuitive/more general definition of a limit. If you want something a bit more technical and rigorous, have a look at the &lt;a href=&quot;http://en.wikipedia.org/wiki/%28%CE%B5,_%CE%B4%29-definition_of_limit&quot; rel=&quot;nofollow&quot;&gt;(ε, δ)-definition of limit&lt;/a&gt;, which is the one used in mathematical analysis.&lt;/p&gt;&#xA;" OwnerUserId="56" LastActivityDate="2010-07-20T19:30:11.283" />
  <row Id="38" PostTypeId="2" ParentId="29" CreationDate="2010-07-20T19:30:50.313" Score="4" ViewCount="" Body="&lt;p&gt;Lots of people like to use &lt;a href=&quot;http://instacalc.com/&quot; rel=&quot;nofollow&quot;&gt;Instacalc&lt;/a&gt; which lets you do unit conversions and store intermediate calculations in variables.&lt;/p&gt;&#xA;" OwnerUserId="45" LastActivityDate="2010-07-20T19:30:50.313" CommunityOwnedDate="2010-07-28T01:32:30.370" />
  <row Id="40" PostTypeId="2" ParentId="20" CreationDate="2010-07-20T19:32:21.590" Score="6" ViewCount="" Body="&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Natural numbers&lt;br&gt;&#xA;The &quot;counting&quot; numbers.  (That is, all integers, that are one or greater).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Whole numbers&lt;br&gt;&#xA;The Natural numbers, and zero.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Integers&lt;br&gt;&#xA;The Whole numbers, and the negatives of the Natural numbers.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Rational numbers&lt;br&gt;&#xA;Any number that may be expressed by any integer A divided by any integer B, where B is not zero.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Irrational numbers&lt;br&gt;&#xA;Any number that cannot be expressed as a rational number, but is not imaginary.  All irrational numbers have an infinite decimal representation.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Real numbers&lt;br&gt;&#xA;All of the Rational and Irrational numbers.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Imaginary numbers&lt;br&gt;&#xA;All Real numbers, multiplied by the square root of negative one.  Imaginary numbers are signified by the letter &lt;code&gt;i&lt;/code&gt;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Complex numbers&lt;br&gt;&#xA;Numbers composed of the sum of a Real and an Imaginary number.  This includes all Real and all Imaginary numbers.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="38" LastEditorUserId="38" LastEditorDisplayName="" LastEditDate="2010-07-20T19:37:58.457" LastActivityDate="2010-07-20T19:37:58.457" CommentCount="15" />
  <row Id="41" PostTypeId="2" ParentId="11" CreationDate="2010-07-20T19:32:44.553" Score="5" ViewCount="" Body="&lt;p&gt;&lt;code&gt;.999...  = 1&lt;/code&gt; because &lt;code&gt;.999...&lt;/code&gt; is a concise symbolic representation of &quot;the limit of some variable as it approaches one.&quot; Therefore, &lt;code&gt;.999... = 1&lt;/code&gt; for the same reason the limit of x as x approaches 1 equals 1.&lt;/p&gt;&#xA;" OwnerUserId="19" LastEditorUserId="19" LastEditorDisplayName="" LastEditDate="2010-07-23T02:13:49.417" LastActivityDate="2010-07-23T02:13:49.417" CommunityOwnedDate="2010-07-28T06:53:43.987" />
  <row Id="42" PostTypeId="2" ParentId="22" CreationDate="2010-07-20T19:33:50.870" Score="-1" ViewCount="" Body="&lt;p&gt;The obvious but slightly trite answer is &quot;because that's just how the cross-product works as an operation&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're looking for an intuitive reason, the cross-product by definition produces a vector that is orthogonal to the two operand (input) vectors. You know that the base vectors i, j, and k are all orthogonal, thus if your two input vectors lie on the (x, y) plane (i.e. only i and j components), you know that any orthogonal vector must have only a component in the z direction (multiple of k).&lt;/p&gt;&#xA;" OwnerUserId="56" LastActivityDate="2010-07-20T19:33:50.870" CommentCount="7" />
  <row Id="43" PostTypeId="2" ParentId="11" CreationDate="2010-07-20T19:34:26.857" Score="0" ViewCount="" Body="&lt;p&gt;You can visualise it by thinking about it in infinitesimals. The more 9's you have on the end of 0.999, the closer you get to 1. When you add an infinite number of 9's to the decimal expansion, you are infinitely close to 1 (or an infinitesimal distance away).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And this isn't a rigorous proof, just an aid to visualisation of the result.&lt;/p&gt;&#xA;" OwnerUserId="35" LastActivityDate="2010-07-20T19:34:26.857" CommunityOwnedDate="2010-07-28T06:53:43.987" />
  <row Id="44" PostTypeId="2" ParentId="11" CreationDate="2010-07-20T19:35:09.227" Score="11" ViewCount="" Body="&lt;p&gt;What I really don't like about all the above answers, is the underlying assumption that &lt;code&gt;1/3=0.3333...&lt;/code&gt;, How do you know that?. It seems to me like assuming the something which is already known.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A proof I really like is:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;0.9999... × 10 = 9.999999...&#xA;&#xA;0.9999... × (9+1) = 9.999999...  [by distribution rule:]&#xA;&#xA;0.9999... × 9 + 0.99999.... × 1 = 9.99999....&#xA;&#xA;0.9999... × 9 = 9.9999....-0.999999 = 9&#xA;&#xA;0.9999... × 9 = 9&#xA;&#xA;0.9999... = 1&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The only things I need to assume is, that &lt;code&gt;9.999.... - 0.9999... = 9&lt;/code&gt; and that &lt;code&gt;0.999... x 10 = 9.999...&lt;/code&gt;. These seems to me intuitive enough to take for granted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The proof is from an old highschool level math book of the Open University in Israel.&lt;/p&gt;&#xA;" OwnerUserId="52" LastEditorUserId="52" LastEditorDisplayName="" LastEditDate="2011-05-04T05:58:42.093" LastActivityDate="2011-05-04T05:58:42.093" CommunityOwnedDate="2010-07-28T06:53:43.987" CommentCount="6" />
  <row Id="46" PostTypeId="2" ParentId="11" CreationDate="2010-07-20T19:36:43.670" Score="-1" ViewCount="" Body="&lt;p&gt;Here's a collection of proofs:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://premium.uploadit.org/AlphaNumeric/Recurring.pdf&quot; rel=&quot;nofollow&quot;&gt;http://premium.uploadit.org/AlphaNumeric/Recurring.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="67" LastActivityDate="2010-07-20T19:36:43.670" CommunityOwnedDate="2010-07-28T06:53:43.987" />
  <row Id="47" PostTypeId="2" ParentId="29" CreationDate="2010-07-20T19:37:42.700" Score="6" ViewCount="" Body="&lt;p&gt;Google's calculator is very powerful: &#xA;&lt;a href=&quot;http://www.googleguide.com/help/calculator.html&quot; rel=&quot;nofollow&quot;&gt;http://www.googleguide.com/help/calculator.html&lt;/a&gt; and your use history will be stored in your browser history.&lt;/p&gt;&#xA;" OwnerUserId="19" LastActivityDate="2010-07-20T19:37:42.700" CommunityOwnedDate="2010-07-28T01:32:30.370" />
  <row Id="48" PostTypeId="2" ParentId="20" CreationDate="2010-07-20T19:37:45.070" Score="1" ViewCount="" Body="&lt;p&gt;You might as well start at the 'lowest':&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Integers are all the whole numbers,&lt;/li&gt;&#xA;&lt;li&gt;Natural numbers (&lt;code&gt;N&lt;/code&gt;) are the set of positive integers, &lt;code&gt;{1, 2, ...}&lt;/code&gt; (0 optional),&lt;/li&gt;&#xA;&lt;li&gt;Rational numbers (&lt;code&gt;Q&lt;/code&gt;) are any number that can be represented &lt;code&gt;a/b&lt;/code&gt; with a and b being Integers (&lt;code&gt;|b| &amp;lt; 0&lt;/code&gt;.&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Real numbers are all the Rational numbers and all the others.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Cardinality is the number of elements in a set.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;An Ordinal is a well ordered set.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="22" LastEditorUserId="22" LastEditorDisplayName="" LastEditDate="2010-07-29T12:16:54.263" LastActivityDate="2010-07-29T12:16:54.263" CommentCount="1" />
  <row Id="49" PostTypeId="2" ParentId="11" CreationDate="2010-07-20T19:38:21.590" Score="42" ViewCount="" Body="&lt;p&gt;What does it mean when you refer to $.99999...$?  Symbols don't mean anything in particular until you've &lt;em&gt;defined what you mean by them&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case the definition is that you're taking the limit of $.9$, $.99$, $.999$, $.9999$, etc.  What does it mean to say that limit is $1$?  Well, it means that no matter how small a number $x$ you pick, I can show you a point in that sequence such that all further numbers in the sequence are within distance $x$ of $1$.  But certainly whatever number you chose your number is bigger than $10^{-k}$ for some $k$.  So I can just pick my point to be the $k$th spot in the sequence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A more intuitive way of explaining the above argument is that the reason $.99999... = 1$ is that their difference is zero.  So let's subtract $1.0000... -.99999... = .00000... = 0$.  That is, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$1.0 -.9 = .1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$1.00-.99 = .01$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$1.000-.999=.001$, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$...$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$1.000... -.99999... = .000... = 0$&lt;/p&gt;&#xA;" OwnerUserId="8" LastEditorUserId="2736" LastEditorDisplayName="" LastEditDate="2011-02-21T05:55:02.470" LastActivityDate="2011-02-21T05:55:02.470" CommunityOwnedDate="2010-07-28T06:53:43.987" CommentCount="11" />
  <row Id="50" PostTypeId="2" ParentId="11" CreationDate="2010-07-20T19:38:49.920" Score="-6" ViewCount="" Body="&lt;p&gt;If you want to go the calculus route, then basically you're gonna follow this pattern and use mathy magic and limits to follow it up to its logical conclusion:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;0.9&#xA;0.99&#xA;0.999&#xA;0.9999&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And so on. This pattern is a sum of 9*10^-i from i=1 to i=infinity. Every iteration, the difference between the pattern and 1 is a bit less. So at i=infinity, the difference is so small that it is literally zero.&lt;/p&gt;&#xA;" OwnerUserId="62" LastActivityDate="2010-07-20T19:38:49.920" CommunityOwnedDate="2010-07-28T06:53:43.987" />
  <row Id="51" PostTypeId="1" AcceptedAnswerId="109" CreationDate="2010-07-20T19:39:11.557" Score="5" ViewCount="264" Body="&lt;p&gt;What length of rope should be used to tie a cow to an &lt;strong&gt;exterior fence post&lt;/strong&gt; of a &lt;em&gt;circular&lt;/em&gt; field so that the cow can only graze half of the grass within that field?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&lt;strong&gt;updated:&lt;/em&gt;&lt;/strong&gt; To be clear: the cow should be tied to a post on the exterior of the field, not a post at the center of the field. &lt;/p&gt;&#xA;" OwnerUserId="59" LastEditorUserId="126" LastEditorDisplayName="" LastEditDate="2010-07-21T01:55:02.220" LastActivityDate="2010-08-05T19:02:22.920" Title="The cow in the field problem (intersecting circular areas)" Tags="&lt;geometry&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="4" />
  <row Id="52" PostTypeId="2" ParentId="1" CreationDate="2010-07-20T19:40:38.467" Score="4" ViewCount="" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;A &lt;em&gt;countably infinite&lt;/em&gt; set is a set for which you can list the elements a&lt;sub&gt;1&lt;/sub&gt;,a&lt;sub&gt;2&lt;/sub&gt;,a&lt;sub&gt;3&lt;/sub&gt;,...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For example, the set of all integers is countably infinite since I can list its elements as follows 0,1,-1,2,-2,3,-3,... So is the set of rational numbers, but this is more difficult to see. Let's start with the positive rationals. Can you see the pattern in this listing?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1/1,1/2,2/1,1/3,2/2,3/1,1/4,2/3,3/2,4/1,1/5,2/4,...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Hint: Add the numerator and denominator to see a different pattern.) This listing has lots of repeats, e.g. 1/1, 2/2 and 1/2, 2/4. That's ok since I can condense the listing by skipping over any repeats.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1/1,1/2,2/1,1/3,3/1,1/4,2/3,3/2,4/1,1/5,...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's write q&lt;sub&gt;n&lt;/sub&gt; for the n-th element of this list. Then 0,q&lt;sub&gt;1&lt;/sub&gt;,-q&lt;sub&gt;1&lt;/sub&gt;,q&lt;sub&gt;2&lt;/sub&gt;,-q&lt;sub&gt;2&lt;/sub&gt;,q&lt;sub&gt;3&lt;/sub&gt;,-q&lt;sub&gt;3&lt;/sub&gt;,... is a listing of all rational numbers.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A &lt;em&gt;countable set&lt;/em&gt; is a set which is either finite or countably infinite; an &lt;em&gt;uncountable set&lt;/em&gt; is a set which is not countable.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Thus, an uncountable set is an infinite set which has no listing of all of its elements (as in the definition of countably infinite set).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An example of an uncountable set is the set of all real numbers. To see this, you can use the &lt;em&gt;diagonal method&lt;/em&gt;. Ask another question to see how this works...&lt;/p&gt;&#xA;" OwnerUserId="25" LastActivityDate="2010-07-20T19:40:38.467" />
  <row Id="53" PostTypeId="2" ParentId="22" CreationDate="2010-07-20T19:40:39.277" Score="4" ViewCount="" Body="&lt;p&gt;Assuming you know the definition of orthogonal as &quot;a is orthogonal to b iff $a\cdot b=0$ then we could calculate $(a \times b)\cdot a = a_1(a_2b_3-a_3b_2)-a_2(a_1b_3-a_3b_1)-a_3(a_1b_2-a_2b_1)=0$ and  $(a \times b)\cdot b-0$, so the cross product is orthogonal to both. As Nold mentioned, if the two vectors a and b lie in the x,y plane, then the orthogonal vectors must be purely in the z direction.&lt;/p&gt;&#xA;" OwnerUserId="16" LastActivityDate="2010-07-20T19:40:39.277" />
  <row Id="54" PostTypeId="2" ParentId="20" CreationDate="2010-07-20T19:41:05.063" Score="4" ViewCount="" Body="&lt;h1&gt;Real numbers&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;Real numbers are any numbers you can locate (even approximately) on an infinite number line.  This is a theoretical number line with infinite &quot;resolution&quot; that extends infinitely in both positive and negative directions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://upload.wikimedia.org/wikipedia/commons/0/09/Number-line.gif&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One neat property about real numbers is that they are orderable -- that is, given any two real numbers, you can tell which one is &quot;higher&quot; and which one is &quot;lower&quot; than the other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Real numbers are closed under multiplication, addition, and subtraction.  That is, if you perform any of these operations on two real numbers, their result will always be real as well.  They are &lt;em&gt;almost&lt;/em&gt; closed under division, except for the whole divide-by-zero issue.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Not real numbers:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;infinity&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;the square root of -1&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;1/0&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;h1&gt;Decimals&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;There isn't a rigorous definition of &quot;decimals&quot;, because depending on where you use it, you'll get different definitions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the elementary sense, it means any number that has a &quot;decimal part&quot;; or a part after the radix (decimal point, etc.).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a more advance sense, it means any number written in &lt;a href=&quot;http://en.wikipedia.org/wiki/Base_10&quot; rel=&quot;nofollow&quot;&gt;Base 10&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Natural numbers&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;Natural numbers are often also called &quot;counting numbers&quot;, because they are the numbers you count with.  (0,) 1, 2, 3, 4, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is some disagreement in the mathematics community over whether or not 0 is a natural number.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Cardinals&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;In linguistics, this means the natural &quot;numbers&quot; themselves (1, 2, 3, etc.)  But you probably don't want to know about linguistics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In Set Theory, two sets have the same &lt;strong&gt;cardinality&lt;/strong&gt; if each element could be paired up with an element of the other set.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;{1,2,3} and {4,5,6} share the same cardinality because you can pair up 1&amp;amp;4, 2&amp;amp;5, 3&amp;amp;6.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Ordinals&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;In linguistics, this means 1st, 2nd, 3rd, etc.  But you probably don't want to know about linguistics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In Set Theory, an ordinal is a well-ordered set.&lt;/p&gt;&#xA;" OwnerUserId="51" LastEditorUserId="51" LastEditorDisplayName="" LastEditDate="2010-07-29T23:06:54.147" LastActivityDate="2010-07-29T23:06:54.147" CommentCount="5" />
  <row Id="55" PostTypeId="2" ParentId="29" CreationDate="2010-07-20T19:41:44.267" Score="18" ViewCount="" Body="&lt;p&gt;Essentially the most helpful is &lt;a href=&quot;http://www.wolframalpha.com&quot; rel=&quot;nofollow&quot;&gt;WolframAlpha&lt;/a&gt;, as Ami said, you can use your browser history here too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;WolframAlpha can carry out complex equations can comparisons much like a TI Calculator. Additionally they have some areas where you can see the simplification of an equation paired with charts and graphs where possible.&lt;/p&gt;&#xA;" OwnerUserId="40" LastActivityDate="2010-07-20T19:41:44.267" CommunityOwnedDate="2010-07-28T01:32:30.370" CommentCount="2" />
  <row Id="56" PostTypeId="1" CreationDate="2010-07-20T19:41:55.687" Score="7" ViewCount="269" Body="&lt;p&gt;As I've understood it, what I've learned as the dot product is just one of many possible &quot;inner product spaces&quot;. Can someone explain this concept? When is it useful to define it as something other than the dot product?&lt;/p&gt;&#xA;" OwnerUserId="63" LastEditorUserId="126" LastEditorDisplayName="" LastEditDate="2010-07-21T01:47:25.767" LastActivityDate="2010-07-23T06:43:35.283" Title="What is an inner product space?" Tags="&lt;linear-algebra&gt;" AnswerCount="2" FavoriteCount="2" />
  <row Id="57" PostTypeId="1" AcceptedAnswerId="134" CreationDate="2010-07-20T19:42:33.500" Score="3" ViewCount="332" Body="&lt;p&gt;This is simply not clicking for me. I'm currently learning math during the summer vacation and I'm on the chapter for relations and functions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are five properties for a relation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reflexive - R -&gt; R&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Symmetrical - R -&gt; S ; S -&gt; R&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Antisymmetrical - R -&gt; S &amp;amp;&amp;amp; (R -&gt; R || S -&gt; S)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Asymmetrical - R -&gt; S &amp;amp;&amp;amp; !(R -&gt; R || S -&gt; S)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Transitive - if R -&gt; S &amp;amp;&amp;amp; S -&gt; T, then R -&gt; T&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If that's not what you call the properties in English, then please let me know- I have to study it in German, unfortunately, and these are my rough translations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Continuing on, I just don't know what to do with this information practically. The examples of the book are horrible:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) &quot;Is the same age as&quot; is apparently reflexive, symmetrical and transitive. &#xA;2) &quot;Is related to&quot; is also apparently reflexive, symmetrical and transitive.&#xA;3) &quot;Is older than&quot; is asymmetric, antisymetric and transitive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are more useless examples like this. I have no idea how it comes to these conclusions because we're talking about a literal statement. I was hoping perhaps for some real mathematical examples, but the book falls short on those.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would greatly appreciate it if somebody could explain the above example and perhaps give me a better use for Relations other than... that. Also, how can a relation be a- and antisymmetrical at the same time? Don't they cancel each other out? &lt;/p&gt;&#xA;" OwnerUserId="44" LastEditorUserId="38" LastEditorDisplayName="" LastEditDate="2010-07-25T23:39:24.247" LastActivityDate="2010-07-28T21:12:19.983" Title="How do the Properties of Relations work?" Tags="&lt;set-theory&gt;&lt;logic&gt;" AnswerCount="6" CommentCount="2" FavoriteCount="2" />
  <row Id="58" PostTypeId="1" AcceptedAnswerId="70" CreationDate="2010-07-20T19:43:20.457" Score="10" ViewCount="478" Body="&lt;p&gt;I recently discovered Benford's Law. I find it very fascinating. I'm wondering what are some of the real life uses of Benford's law. Specific examples would be great.&lt;/p&gt;&#xA;" OwnerUserId="5" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-07-24T18:41:55.237" LastActivityDate="2010-12-28T01:14:13.617" Title="Real life usage of Benford's Law" Tags="&lt;soft-question&gt;&lt;big-list&gt;&lt;statistics&gt;&lt;applications&gt;" AnswerCount="7" CommentCount="2" FavoriteCount="4" />
  <row Id="59" PostTypeId="1" AcceptedAnswerId="65" CreationDate="2010-07-20T19:43:30.097" Score="2" ViewCount="210" Body="&lt;p&gt;Given two points, around an origin (0,0), in 2D space how would you calculate an angle from p1 to p2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How would this change in 3D space?&lt;/p&gt;&#xA;" OwnerUserId="24" LastEditorUserId="126" LastEditorDisplayName="" LastEditDate="2010-07-21T01:54:27.277" LastActivityDate="2010-07-28T09:43:00.053" Title="Calculating an Angle from 2 points in space" Tags="&lt;linear-algebra&gt;" AnswerCount="3" CommentCount="3" />
  <row Id="60" PostTypeId="2" ParentId="11" CreationDate="2010-07-20T19:45:14.633" Score="22" ViewCount="" Body="&lt;p&gt;Suppose this was not the case, i.e. &lt;code&gt;0.9999... != 1&lt;/code&gt;. Then &lt;code&gt;0.9999... &amp;lt; 1&lt;/code&gt; (I hope we agree on that). But between two distinct real numbers, there's always another one (say &lt;code&gt;x&lt;/code&gt;) in between, hence &lt;code&gt;0.9999... &amp;lt; x &amp;lt; 1&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The decimal representation of &lt;code&gt;x&lt;/code&gt; must have a digit somewhere that is not &lt;code&gt;9&lt;/code&gt; (otherwise &lt;code&gt;x == 0.9999...&lt;/code&gt;). But that means it's actually smaller &amp;ndash; &lt;code&gt;x &amp;lt; 0.9999...&lt;/code&gt;, contradicting the definition of &lt;code&gt;x&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, the assumption that there's a number between &lt;code&gt;0.9999...&lt;/code&gt; and &lt;code&gt;1&lt;/code&gt; is false, hence they're equal.&lt;/p&gt;&#xA;" OwnerUserId="3" LastActivityDate="2010-07-20T19:45:14.633" CommunityOwnedDate="2010-07-28T06:53:43.987" CommentCount="1" />
  <row Id="62" PostTypeId="2" ParentId="51" CreationDate="2010-07-20T19:46:35.563" Score="0" ViewCount="" Body="&lt;p&gt;So, the area of the field is pi*r^2 and you want the cow to be able to graze an area equal to half of that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All you need to do is set up the equation (r[1] is the radius of the field, r[2] is the length of the rope desired):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(pi*r[1]^2) / 2 = pi*r[2]^2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can then simplify it down:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;r[1]^2 / 2 = r[2]^2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and then taking roots:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;r[2] = r[1] / sqrt(2)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you need a rope that is equal to the radius divided by the square root of 2, and the post can be no closer than this distance to the edge of the field.&lt;/p&gt;&#xA;" OwnerUserId="35" LastActivityDate="2010-07-20T19:46:35.563" />
  <row Id="63" PostTypeId="2" ParentId="58" CreationDate="2010-07-20T19:46:38.047" Score="1" ViewCount="" Body="&lt;p&gt;I've heard of it being as a rough check to see if accounting numbers were being made up&lt;/p&gt;&#xA;" OwnerUserId="37" LastActivityDate="2010-07-20T19:46:38.047" CommentCount="1" />
  <row Id="64" PostTypeId="2" ParentId="51" CreationDate="2010-07-20T19:47:08.217" Score="0" ViewCount="" Body="&lt;p&gt;Let the total area of the field = $A$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know $A = \pi R^2$ where $R$ = the radius of the field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We want the cow to be able to graze half the area, so we solve for a length of rope $r$ such that $\pi r^2 = A / 2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This gives: $\pi r^2 = \pi R^2 / 2$, hence $r = R / \sqrt(2)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In words, the length of the cow's rope should be the radius of the field divided by sqrt(2).&lt;/p&gt;&#xA;" OwnerUserId="56" LastActivityDate="2010-07-20T19:47:08.217" CommentCount="4" />
  <row Id="65" PostTypeId="2" ParentId="59" CreationDate="2010-07-20T19:47:26.950" Score="4" ViewCount="" Body="&lt;p&gt;Assuming this is relative to the origin (as John pointed out): Given two position vectors p&lt;sub&gt;1&lt;/sub&gt; and p&lt;sub&gt;2&lt;/sub&gt; the dot product is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;p&lt;sub&gt;1&lt;/sub&gt; * p&lt;sub&gt;2&lt;/sub&gt; = |p&lt;sub&gt;1&lt;/sub&gt;| *&#xA;  |p&lt;sub&gt;2&lt;/sub&gt;| * cos(v)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Solving for v, we get:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;v = arccos(p&lt;sub&gt;1&lt;/sub&gt; * p&lt;sub&gt;2&lt;/sub&gt; / (|p&lt;sub&gt;1&lt;/sub&gt;| * |p&lt;sub&gt;2&lt;/sub&gt;|))&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For 2d space this equals:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;v = arccos((x&lt;sub&gt;1&lt;/sub&gt; * x&lt;sub&gt;2&lt;/sub&gt; + y&lt;sub&gt;1&lt;/sub&gt; * y&lt;sub&gt;2&lt;/sub&gt;) / sqrt((x&lt;sub&gt;1&lt;/sub&gt;&lt;sup&gt;2&lt;/sup&gt; + y&lt;sub&gt;1&lt;/sub&gt;&lt;sup&gt;2&lt;/sup&gt;) * (x&lt;sub&gt;2&lt;/sub&gt;&lt;sup&gt;2&lt;/sup&gt; + y&lt;sub&gt;2&lt;/sub&gt;&lt;sup&gt;2&lt;/sup&gt;)))&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And extended for 3d space:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;v = arccos((x&lt;sub&gt;1&lt;/sub&gt; * x&lt;sub&gt;2&lt;/sub&gt; + y&lt;sub&gt;1&lt;/sub&gt; * y&lt;sub&gt;2&lt;/sub&gt; + z&lt;sub&gt;1&lt;/sub&gt; * z&lt;sub&gt;2&lt;/sub&gt;) / sqrt((x&lt;sub&gt;1&lt;/sub&gt;&lt;sup&gt;2&lt;/sup&gt; + y&lt;sub&gt;1&lt;/sub&gt;&lt;sup&gt;2&lt;/sup&gt; + z&lt;sub&gt;1&lt;/sub&gt;&lt;sup&gt;2&lt;/sup&gt;) * (x&lt;sub&gt;2&lt;/sub&gt;&lt;sup&gt;2&lt;/sup&gt; + y&lt;sub&gt;2&lt;/sub&gt;&lt;sup&gt;2&lt;/sup&gt; + z&lt;sub&gt;2&lt;/sub&gt;&lt;sup&gt;2&lt;/sup&gt;)))&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="63" LastEditorUserId="63" LastEditorDisplayName="" LastEditDate="2010-07-20T19:52:47.110" LastActivityDate="2010-07-20T19:52:47.110" />
  <row Id="66" PostTypeId="2" ParentId="57" CreationDate="2010-07-20T19:49:03.750" Score="2" ViewCount="" Body="&lt;p&gt;There are several important kinds of relations, each of which satisfy a different collection of properties:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Equivalence relations:&#xA;These are reflexive, symmetric, and transitive.  Essentially they're relations that &quot;behave like equality.&quot;  The most important elementary one is &quot;equivalence modulo m,&quot; where say 1 = 6 = 11 modulo 5.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Partial orderings:&#xA;These are reflexive, transitive, and (anti-symmetric or maybe asymmetric, I'm having trouble parsing your logical statements).  Essentially they're relations that &quot;behave like less than or equal to.&quot;  An important elementary example is &quot;divides&quot; where we say a|b if the ratio b/a is an integer.  Note that 2|6 but 6 does not divide 2.  However if 2|4 and 4|8 certainly 2|8.&lt;/p&gt;&#xA;" OwnerUserId="8" LastActivityDate="2010-07-20T19:49:03.750" CommentCount="1" />
  <row Id="67" PostTypeId="1" AcceptedAnswerId="72" CreationDate="2010-07-20T19:49:29.490" Score="18" ViewCount="359" Body="&lt;p&gt;I hear a lot about Elliptic Curve Cryptography these days, but I'm still not quite sure what they are or how they relate to crypto...&lt;/p&gt;&#xA;" OwnerUserId="48" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-08-06T06:16:23.970" LastActivityDate="2010-09-05T10:22:57.960" Title="What is an elliptic curve, and how are they used in cryptography?" Tags="&lt;cryptography&gt;&lt;elliptic-curves&gt;" AnswerCount="4" FavoriteCount="6" />
  <row Id="68" PostTypeId="2" ParentId="59" CreationDate="2010-07-20T19:49:50.033" Score="1" ViewCount="" Body="&lt;p&gt;I will assume that you mean the angle of the line from p1 to p2 &lt;strong&gt;with respect to the x-axis&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the best I can do given the information you have provided.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In any case, the official mathsy way would be to find the dot product between the two, and divide by the magnitude of p1-p2 and take the arccossine.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;v = (normalized vector from p1 to p2)&#xA;theta = arccos( v * &amp;lt;1,0&amp;gt; )    (dot product)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;You can normalize a vector by dividing every term by the magnitude (length) of the entire vector.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For 3D, the same thing applies:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;theta = arccos( v * &amp;lt;1,0,0&amp;gt; )   (dot product)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;You could also possibly mean the angle between the line from the origin to p1 and the line from the origin to p2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can do this with dot products, as well; but both vectors must be normalized.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;theta = arccos( a * b ) (dot product)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;where &lt;code&gt;a&lt;/code&gt; is the normalized vector from the origin to p1 and &lt;code&gt;b&lt;/code&gt; is the normalized vector from the origin to p2.&lt;/p&gt;&#xA;" OwnerUserId="51" LastActivityDate="2010-07-20T19:49:50.033" />
  <row Id="69" PostTypeId="2" ParentId="1" CreationDate="2010-07-20T19:49:51.500" Score="3" ViewCount="" Body="&lt;p&gt;You can see that there are infinitely many natural numbers 1, 2, 3, ..., and infinitely many real numbers, such as 0, 1, pi, etc. But are these two infinities the same?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, suppose you have two sets of objects, e.g. people and horses, and you want to know if the number of objects in one set is the same as in the other. The simplest way is to find a way of corresponding the objects one-to-one. For instance, if you see a parade of people riding horses, you will know that there are as many people as there are horses, because there is such a one-to-one correspondence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We say that an set with infinitely many things is &quot;countable,&quot; if we can find a one-to-one correspondence between the things in this set and the natural numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;E.g., the integers are countable: 1 &amp;lt;-&gt; 0, 2 &amp;lt;-&gt; -1, 3 &amp;lt;-&gt; 1, 4 &amp;lt;-&gt; -2, 5 &amp;lt;-&gt; 2, etc, gives such a correspondence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the set of real numbers is NOT countable! This was proven for the first time by Georg Cantor. Here is a proof using the so-called &lt;a href=&quot;http://en.wikipedia.org/wiki/Cantor%27s_diagonal_argument&quot; rel=&quot;nofollow&quot;&gt;diagonal argument&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="39" LastActivityDate="2010-07-20T19:49:51.500" />
  <row Id="70" PostTypeId="2" ParentId="58" CreationDate="2010-07-20T19:51:24.680" Score="10" ViewCount="" Body="&lt;p&gt;Forensic accountancy is a popular use, and is actually admissible as evidence in the USA.&lt;/p&gt;&#xA;" OwnerUserId="35" LastActivityDate="2010-07-20T19:51:24.680" />
  <row Id="71" PostTypeId="1" AcceptedAnswerId="78" CreationDate="2010-07-20T19:52:14.927" Score="7" ViewCount="377" Body="&lt;p&gt;I've recently started reading about Quaternions, and I keep reading that for example they're used in computer graphics and mechanics calculations to calculate movement and rotation, but without real explanations of the benefits of using them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm wondering what exactly can be done with Quaternions that can't be done as easily (or easier) using more tradition approaches, such as with Vectors?&lt;/p&gt;&#xA;" OwnerUserId="34" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-07-24T18:49:40.137" LastActivityDate="2010-07-24T18:49:40.137" Title="Real world uses of Quaternions?" Tags="&lt;soft-question&gt;&lt;big-list&gt;&lt;linear-algebra&gt;&lt;applications&gt;&lt;quaternions&gt;" AnswerCount="4" FavoriteCount="2" />
  <row Id="72" PostTypeId="2" ParentId="67" CreationDate="2010-07-20T19:52:31.587" Score="15" ViewCount="" Body="&lt;p&gt;Here is a super nice powerpoint on the subject!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.math.brown.edu/~jhs/Presentations/WyomingEllipticCurve.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.math.brown.edu/~jhs/Presentations/WyomingEllipticCurve.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="16" LastActivityDate="2010-07-20T19:52:31.587" CommentCount="1" />
  <row Id="73" PostTypeId="2" ParentId="56" CreationDate="2010-07-20T19:54:30.663" Score="5" ViewCount="" Body="&lt;p&gt;An inner product space is a vector space for which the inner product is defined. The inner product is also known as the 'dot product' for 2D or 3D Euclidean space. An arbitrary number of inner products can be defined according to &lt;a href=&quot;http://en.wikipedia.org/wiki/Inner_product_space#Definition&quot; rel=&quot;nofollow&quot;&gt;three rules&lt;/a&gt;, though most are a lot less intuitive/practical than the Euclidean (dot) product.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Side note:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It may seem slightly esoteric, but as a physicist the obvious application of inner product spaces are Hilbert spaces used in quantum mechanics. The inner product of an eigenfunction with a wavefunction in Hilbert space gives the corresponding eigenvalue.&lt;/p&gt;&#xA;" OwnerUserId="56" LastActivityDate="2010-07-20T19:54:30.663" />
  <row Id="74" PostTypeId="1" AcceptedAnswerId="1273" CreationDate="2010-07-20T19:54:33.893" Score="3" ViewCount="345" Body="&lt;p&gt;Let's say I know 'X' is a Gaussian Variable.&lt;br&gt;&#xA;Moreover, I know 'Y' is a Gaussian Variable and Y=X+Z.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's X and Z are Independent.&#xA;How can I prove Y is a Gaussian Random Variable if and only if Z is a Gaussian R.V.?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's easy to show the other way around (X, Z Orthogonal and Normal hence create a Gaussian Vector hence any Linear Combination of the two is a Gaussian Variable).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="33" LastEditorUserId="33" LastEditorDisplayName="" LastEditDate="2010-07-21T05:28:28.740" LastActivityDate="2010-07-30T22:04:55.433" Title="Sum of Gaussian Variables" Tags="&lt;probability-theory&gt;" AnswerCount="1" CommentCount="6" FavoriteCount="1" />
  <row Id="75" PostTypeId="1" AcceptedAnswerId="80" CreationDate="2010-07-20T19:55:05.293" Score="4" ViewCount="414" Body="&lt;p&gt;Rings, groups, and fields all feel similar. What are the differences between them, both in definition and in how they are used?&lt;/p&gt;&#xA;" OwnerUserId="37" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-07-31T17:50:08.467" LastActivityDate="2010-07-31T17:50:08.467" Title="differences between rings, groups, and fields" Tags="&lt;terminology&gt;&lt;abstract-algebra&gt;" AnswerCount="4" CommentCount="0" />
  <row Id="76" PostTypeId="2" ParentId="57" CreationDate="2010-07-20T19:55:53.373" Score="1" ViewCount="" Body="&lt;p&gt;Asymmetric means simply &quot;not symmetric&quot;. So in the binary case, it is NOT the case that if a is related to b, b is related to a.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Antisymmetric means that if a is related to b, and b is related to a, a = b.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To explain your third example:&#xA;&quot;is older than&quot; is asymmetric because if Alice is older than Bob, Bob is NOT older than Alice. &quot;is older than&quot; is antisymmetric since if Alice is older than Bob, and Bob is older than Alice, Alice must be Bob because someone must be older (and if this is not the case, Alice simply has two names..). &quot;is older than&quot; is transitive since if Alice is older than Bob, and Bob is older than Charlie, Alice is also older than Charlie.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So asymmetric and antisymmetric don't cancel out because the first means it's sort of a one-way relation, whereas the second means, loosely, that if it you reverse the operands and both statements are true, the operands must be the same.&lt;/p&gt;&#xA;" OwnerUserId="50" LastActivityDate="2010-07-20T19:55:53.373" CommentCount="2" />
  <row Id="77" PostTypeId="1" AcceptedAnswerId="85" CreationDate="2010-07-20T19:57:06.617" Score="12" ViewCount="377" Body="&lt;p&gt;What purposes do the Dot and Cross products serve? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do you have any clear examples of when you would use them?&lt;/p&gt;&#xA;" OwnerUserId="24" LastEditorUserId="126" LastEditorDisplayName="" LastEditDate="2010-07-31T15:15:30.327" LastActivityDate="2011-02-02T17:22:41.567" Title="Understanding Dot and Cross Product" Tags="&lt;linear-algebra&gt;" AnswerCount="4" FavoriteCount="3" />
  <row Id="78" PostTypeId="2" ParentId="71" CreationDate="2010-07-20T19:57:51.043" Score="5" ViewCount="" Body="&lt;p&gt;I believe they are used in quantum physics as well, because rotation with quaternions models Spinors extremely well (due to the lovely property that you need to rotate a point in quaternionic space around 2 full revolutions to get back to your 'original', which is exactly what happens with spin-1/2 particles).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They are also, as you said, used in computer graphics a lot for several reasons:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;they are much more space efficient to store than rotation matrices (4 floats rather than 16)&lt;/li&gt;&#xA;&lt;li&gt;They are much easier to interpolate than euler angle rotations (spherical interpolation or normalised liner interpolation)&lt;/li&gt;&#xA;&lt;li&gt;They avoid gimbal lock&lt;/li&gt;&#xA;&lt;li&gt;It's much cooler to say that your rotation is described as 'a great circle on the surface of a unit 4 dimensional hypersphere' :)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I think there are other uses, but a lot of them have been superseded by more general Vectors.&lt;/p&gt;&#xA;" OwnerUserId="35" LastActivityDate="2010-07-20T19:57:51.043" CommentCount="5" />
  <row Id="79" PostTypeId="2" ParentId="20" CreationDate="2010-07-20T19:58:12.743" Score="9" ViewCount="" Body="&lt;p&gt;The &lt;strong&gt;natural numbers&lt;/strong&gt; can be defined by &lt;a href=&quot;http://mathworld.wolfram.com/PeanosAxioms.html&quot; rel=&quot;nofollow&quot;&gt;Peano's Axioms&lt;/a&gt; (sometimes called the Peano Postulates):&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Zero is a number.&lt;/li&gt;&#xA;&lt;li&gt;If n is a number, the successor of n is a number.&lt;/li&gt;&#xA;&lt;li&gt;zero is not the successor of a number.&lt;/li&gt;&#xA;&lt;li&gt;Two numbers of which the successors are equal are themselves equal.&lt;/li&gt;&#xA;&lt;li&gt;(induction axiom.) If a set S of numbers contains zero and also the successor of every number in S, then every number is in S.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;(This definition includes 0 in the natural numbers; altering rules 1, 3, and 5 to refer to one instead of zero excludes 0 from the natural numbers.  Whether or not 0 is a natural number varies in various texts.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;strong&gt;whole numbers&lt;/strong&gt; are the natural numbers with the additive identity element called 0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;strong&gt;integers&lt;/strong&gt; are the whole numbers and their additive inverses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;strong&gt;rational numbers&lt;/strong&gt; are numbers that can be expressed as a ratio of an integer to a non-zero integer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;strong&gt;real numbers&lt;/strong&gt; are the set of numbers that are limits of &lt;a href=&quot;http://mathworld.wolfram.com/CauchySequence.html&quot; rel=&quot;nofollow&quot;&gt;Cauchy sequences&lt;/a&gt; of rational numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;strong&gt;irrational numbers&lt;/strong&gt; are the real numbers that are not rational numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;strong&gt;complex numbers&lt;/strong&gt; are the numbers that can be expressed as &lt;em&gt;a&lt;/em&gt; + &lt;em&gt;b&lt;/em&gt; * &lt;em&gt;i&lt;/em&gt; where &lt;em&gt;a&lt;/em&gt; and &lt;em&gt;b&lt;/em&gt; are real numbers and &lt;em&gt;i&lt;/em&gt; behaves like a real number under addition/multiplication/distribution/etc., with the added rule that &lt;em&gt;i&lt;/em&gt;&lt;sup&gt;2&lt;/sup&gt; = -1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;strong&gt;imaginary numbers&lt;/strong&gt; are sometimes defined to be the &quot;pure imaginary&quot; numbers--complex numbers for which the &quot;real part&quot; &lt;em&gt;a&lt;/em&gt; = 0, sometimes with the added restriction that &lt;em&gt;b&lt;/em&gt; is not zero--and are sometimes defined to be the non-real complex numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;strong&gt;algebraic numbers&lt;/strong&gt; are numbers that are solutions to polynomial equations with integer coefficients.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;strong&gt;transcendental numbers&lt;/strong&gt; are complex numbers (sometimes limited to real numbers) that are not algebraic.&lt;/p&gt;&#xA;" OwnerUserId="72" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-07-24T09:13:19.853" LastActivityDate="2010-07-24T09:13:19.853" CommentCount="3" />
  <row Id="80" PostTypeId="2" ParentId="75" CreationDate="2010-07-20T19:58:48.127" Score="3" ViewCount="" Body="&lt;p&gt;They should feel similar! In fact, every ring is a group, and every field is a ring. A ring is a group with an additional operation, where the second operation is associative and the distributive properties make the two operations &quot;compatible&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A field is a ring such that the second operation also satisfies all the group properties (after throwing out the additive identity); i.e. it has multiplicative inverses, multiplicative identity, and is commutative.&lt;/p&gt;&#xA;" OwnerUserId="16" LastEditorUserId="16" LastEditorDisplayName="" LastEditDate="2010-07-20T20:10:50.810" LastActivityDate="2010-07-20T20:10:50.810" CommentCount="2" />
  <row Id="81" PostTypeId="1" CreationDate="2010-07-20T20:01:55.013" Score="18" ViewCount="914" Body="&lt;p&gt;I have the one or other interesting Math blog in my feedreader that I follow. It would be interesting to compile a list of Math blogs that are interesting to read, and do not require research-level math skills.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll start with my entries:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Division By Zero, &lt;a href=&quot;http://divisbyzero.com/&quot; rel=&quot;nofollow&quot;&gt;http://divisbyzero.com/&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;Tanya Khovanova’s Math Blog, &lt;a href=&quot;http://blog.tanyakhovanova.com/&quot; rel=&quot;nofollow&quot;&gt;http://blog.tanyakhovanova.com/&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="79" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-07-21T16:26:58.500" LastActivityDate="2011-04-23T20:23:45.707" CommunityOwnedDate="2010-07-20T20:01:55.013" Title="List of Interesting Math Blogs" Tags="&lt;soft-question&gt;&lt;big-list&gt;" AnswerCount="15" CommentCount="4" FavoriteCount="16" />
  <row Id="82" PostTypeId="2" ParentId="81" CreationDate="2010-07-20T20:04:19.563" Score="4" ViewCount="" Body="&lt;p&gt;Gil Kalai's &lt;a href=&quot;http://gilkalai.wordpress.com/&quot; rel=&quot;nofollow&quot;&gt;blog&lt;/a&gt; is pretty awesome. Mostly covers combinatorics.&lt;/p&gt;&#xA;" OwnerUserId="50" LastActivityDate="2010-07-20T20:04:19.563" CommunityOwnedDate="2010-07-20T20:04:19.563" />
  <row Id="83" PostTypeId="1" AcceptedAnswerId="88" CreationDate="2010-07-20T20:04:20.733" Score="31" ViewCount="791" Body="&lt;p&gt;I'm talking in the range of 10-12 years old, but this question isn't limited to only that range.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do you have any advice on cool things to show kids that might spark their interest in spending more time with math? The difficulty for some to learn math can be pretty overwhelming. Do you have any teaching techniques that you find valuable?&lt;/p&gt;&#xA;" OwnerUserId="24" LastEditorUserId="100" LastEditorDisplayName="" LastEditDate="2010-07-22T10:06:42.220" LastActivityDate="2010-10-16T20:14:28.650" CommunityOwnedDate="2010-08-07T22:32:44.343" Title="What are some good ways to get children excited about math?" Tags="&lt;big-list&gt;&lt;teaching&gt;" AnswerCount="16" CommentCount="4" FavoriteCount="16" />
  <row Id="84" PostTypeId="2" ParentId="77" CreationDate="2010-07-20T20:05:03.633" Score="1" ViewCount="" Body="&lt;p&gt;The dot product can be used to find the length of a vector or the angle between two vectors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The cross product is used to find a vector which is perpendicular to the plane spanned by two vectors.&lt;/p&gt;&#xA;" OwnerUserId="11" LastActivityDate="2010-07-20T20:05:03.633" />
  <row Id="85" PostTypeId="2" ParentId="77" CreationDate="2010-07-20T20:05:31.633" Score="21" ViewCount="" Body="&lt;p&gt;When you deal with vectors, sometimes you say to yourself, &quot;Darn I wish there was a function that...&quot;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;was zero when two vectors are perpendicular, letting me test perpendicularness.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Dot Product&lt;/strong&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;would let me find the angle between two vectors.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Dot Product&lt;/strong&gt; (actually gives the cosine of the angle between two normalized vectors)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;would let me 'project' one vector onto another, or give the length of one vector in the direction of another.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Dot Product&lt;/strong&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;could tell me how much force is actually helping the object move, when pushing at an angle.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Dot Product&lt;/strong&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;could tell me how much a vector field is 'spreading out'.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Dot Product&lt;/strong&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;could give me a vector that is perpendicular to two other vectors.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Cross Product&lt;/strong&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;could tell me how much torque a force was applying to a rotating system.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Cross Product&lt;/strong&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;could tell me how much this vector field is 'curling' up.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Cross Product&lt;/strong&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;There are actually a lot more uses, but the more I study vectors, the more and more I run into a situation where I need a function to do &lt;em&gt;exactly&lt;/em&gt; something, and then realize that the cross/dot products already did exactly what I needed!&lt;/p&gt;&#xA;" OwnerUserId="51" LastActivityDate="2010-07-20T20:05:31.633" />
  <row Id="86" PostTypeId="2" ParentId="56" CreationDate="2010-07-20T20:09:05.667" Score="15" ViewCount="" Body="&lt;p&gt;As for the utility of inner product spaces: They're vector spaces where notions like the length of a vector and the angle between two vectors are available. In this way, they generalize R^n but preserve some of its additional structure that comes on top of it being a vector space. Familiar friends like Cauchy-Schwartz, the parallelogram rule, and orthogonality all work in inner product spaces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Note that there is a more general class of spaces, normed spaces, where notions of length make sense always, but an inner product cannot necessarily be defined.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The dot product is the standard inner product on R^n. In general, any symmetric, positive definite matrix will give you an inner product on C^n. And you can have inner products on infinite dimensional vector spaces, like &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ \langle \, f, \, g \, \rangle = \int_a^b \ f(x)\overline{g(x)} \, dx$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for f, g square-integrable functions on [a,b].&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This becomes useful, for example, in applications like Fourier series where you want a basis of orthonormal functions for some function space (it's not just the trigonometric functions that work).&lt;/p&gt;&#xA;" OwnerUserId="76" LastEditorUserId="76" LastEditorDisplayName="" LastEditDate="2010-07-23T06:43:35.283" LastActivityDate="2010-07-23T06:43:35.283" CommentCount="4" />
  <row Id="87" PostTypeId="2" ParentId="11" CreationDate="2010-07-20T20:10:04.057" Score="1" ViewCount="" Body="&lt;pre&gt;&lt;code&gt;x = 0.999...&#xA;10x = 9.999...&#xA;10x - x = 9.999... - 0.999... = 9&#xA;-&amp;gt; 9x = 9&#xA;-&amp;gt; x = 1 thus, 0.999... = 1&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="84" LastActivityDate="2010-07-20T20:10:04.057" CommunityOwnedDate="2010-07-28T06:53:43.987" CommentCount="3" />
  <row Id="88" PostTypeId="2" ParentId="83" CreationDate="2010-07-20T20:10:38.563" Score="16" ViewCount="" Body="&lt;p&gt;Graph theory! It's essentially connecting the dots, but with theorems working wonders behind the scenes for when they're old enough. Simple exercises like asking how many colors you need to color the faces or vertices of a graph are often fun (so I hear). (Also, most people won't believe the 4-color theorem.)&lt;/p&gt;&#xA;" OwnerUserId="50" LastActivityDate="2010-07-20T20:10:38.563" CommunityOwnedDate="2010-08-07T22:32:44.343" CommentCount="1" />
  <row Id="89" PostTypeId="2" ParentId="22" CreationDate="2010-07-20T20:11:12.370" Score="2" ViewCount="" Body="&lt;p&gt;Note that if you replace i, j, and k with m, n, and p, the determinant becomes the dot-product of the vector (m, n, p) with the cross-product of the two original vectors.  If (m, n, p) = (a, b, c) or (m, n, p) = (d, e, f), the determinant is zero (any matrix with two identical rows has determinant zero), so the dot product of (a, b, c) or (d, e, f) with the cross-product is zero.  Hence, (a, b, c) and (d, e, f) are orthogonal to their cross-product.&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-07-20T20:11:12.370" />
  <row Id="90" PostTypeId="1" CreationDate="2010-07-20T20:11:37.703" Score="15" ViewCount="424" Body="&lt;p&gt;Not sure if this is the place for it, but there are similar posts for podcasts and blogs, so I'll post this one. I'd be interested in seeing a list of online resources for mathematics learning.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As someone doing a non-maths degree in college I'd be interested in finding some resources for learning more maths online, most resources I know of tend to either assume a working knowledge of maths beyond secondary school level, or only provide a brief summary of the topic at hand.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll start off by posting &lt;a href=&quot;http://ocw.mit.edu/courses/mathematics/&quot; rel=&quot;nofollow&quot;&gt;MIT Open Courseware&lt;/a&gt;, which is a large collection of lecture notes, assignments and multimedia for the MIT mathematics courses, although in many places it's quite incomplete.&lt;/p&gt;&#xA;" OwnerUserId="34" LastEditorUserId="167" LastEditorDisplayName="" LastEditDate="2010-07-21T10:47:30.240" LastActivityDate="2011-01-24T14:45:18.877" CommunityOwnedDate="2010-07-20T20:11:37.703" Title="Online resources for learning Mathematics." Tags="&lt;books&gt;&lt;reference-request&gt;" AnswerCount="8" FavoriteCount="16" />
  <row Id="91" PostTypeId="2" ParentId="75" CreationDate="2010-07-20T20:13:11.210" Score="4" ViewCount="" Body="&lt;p&gt;You're right to think that the definitions are very similar. The main difference between groups and rings is that rings have two binary operations (usually called addition and multiplication) instead of just one binary operation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you forget about multiplication, then a ring becomes a group with respect to addition (the identity is 0 and inverses are negatives). This group is always commutative!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you forget about addition, then a ring does not become a group with respect to multiplication. The binary operation of multiplication is associative and it does have an identity 1, but some elements like 0 do not have inverses. (This structure is called a monoid.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A commutative ring is a field when all nonzero elements have a multiplicative inverses. In this case, if you forget about addition and remove 0, the remaining elements do form a group under multiplication. This group is again commutative.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A division ring is a (not necessarily commutative) ring in which all nonzero elements have multiplicative inverses. Again, if you forget about addition and remove 0, the remaining elements do form a group under multiplication. This group is not necessarily commutative. An example of a division ring which is not a field are the quaternions.&lt;/p&gt;&#xA;" OwnerUserId="25" LastActivityDate="2010-07-20T20:13:11.210" />
  <row Id="92" PostTypeId="2" ParentId="90" CreationDate="2010-07-20T20:15:09.723" Score="1" ViewCount="" Body="&lt;p&gt;Two good general references:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Portal%3aMathematics&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://mathworld.wolfram.com/&quot; rel=&quot;nofollow&quot;&gt;MathWorld&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="37" LastActivityDate="2010-07-20T20:15:09.723" CommunityOwnedDate="2010-07-20T20:15:09.723" />
  <row Id="93" PostTypeId="2" ParentId="18" CreationDate="2010-07-20T20:16:01.453" Score="1" ViewCount="" Body="&lt;p&gt;Possibly something like this. Correct me if I'm wrong. :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;j = semi-major&lt;br /&gt;&#xA;n = semi-minor&lt;br /&gt;&#xA;e = eccentricity&lt;/p&gt;&#xA;&#xA;&lt;p&gt;n = sqrt((j * sqrt(1 - e^2)) * (j * (1 - e^2)))&lt;/p&gt;&#xA;" OwnerUserId="66" LastActivityDate="2010-07-20T20:16:01.453" />
  <row Id="94" PostTypeId="2" ParentId="90" CreationDate="2010-07-20T20:16:09.050" Score="4" ViewCount="" Body="&lt;ul&gt;&#xA;&lt;li&gt;Kahn Academy, &lt;a href=&quot;http://www.khanacademy.org/&quot; rel=&quot;nofollow&quot;&gt;http://www.khanacademy.org/&lt;/a&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;You'll find tons of explanatory videos from various branches of mathematics; plus, each subject is explained pretty good, and the videos are easy to follow&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="79" LastEditorUserId="79" LastEditorDisplayName="" LastEditDate="2010-07-20T20:21:11.893" LastActivityDate="2010-07-20T20:21:11.893" CommunityOwnedDate="2010-07-20T20:16:09.050" CommentCount="2" />
  <row Id="95" PostTypeId="2" ParentId="90" CreationDate="2010-07-20T20:17:42.790" Score="2" ViewCount="" Body="&lt;p&gt;A useful one for undergraduate level maths is &lt;a href=&quot;http://www.mathscentre.ac.uk/&quot; rel=&quot;nofollow&quot;&gt;Mathcentre&lt;/a&gt;.  It has useful background material for people studying maths, or who need some maths background for other courses.&lt;/p&gt;&#xA;" OwnerUserId="49" LastActivityDate="2010-07-20T20:17:42.790" CommunityOwnedDate="2010-07-20T20:17:42.790" />
  <row Id="96" PostTypeId="2" ParentId="81" CreationDate="2010-07-20T20:18:49.590" Score="12" ViewCount="" Body="&lt;p&gt;Not always pure math, but I think John Baez' &lt;a href=&quot;http://math.ucr.edu/home/baez/this.week.html&quot; rel=&quot;nofollow&quot;&gt;This Week in Mathematical Physics&lt;/a&gt; contains a lot of really interesting math reads.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I should add Terry Tao's &lt;a href=&quot;http://terrytao.wordpress.com/&quot; rel=&quot;nofollow&quot;&gt;What's new&lt;/a&gt;. It's a very active math blog (both in posts and comments) and definitely covers some cutting edge math, even if it can be way over my head.&lt;/p&gt;&#xA;" OwnerUserId="82" LastEditorUserId="82" LastEditorDisplayName="" LastEditDate="2010-07-21T15:37:45.827" LastActivityDate="2010-07-21T15:37:45.827" CommunityOwnedDate="2010-07-20T20:18:49.590" />
  <row Id="97" PostTypeId="1" AcceptedAnswerId="1224" CreationDate="2010-07-20T20:20:22.223" Score="2" ViewCount="389" Body="&lt;p&gt;I am looking for an accurate algorithm to calculate the &lt;a href=&quot;http://en.wikipedia.org/wiki/Error_function&quot; rel=&quot;nofollow&quot;&gt;error function&lt;/a&gt;. I have tried using &lt;a href=&quot;http://stackoverflow.com/questions/457408/is-there-an-easily-available-implementation-of-erf-for-python/457805#457805&quot;&gt;this formula&lt;/a&gt; (&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0486612724&quot; rel=&quot;nofollow&quot;&gt;Handbook of Mathematical Functions&lt;/a&gt;, formula 7.1.26), but the results are not accurate enough for the application.&lt;/p&gt;&#xA;" OwnerUserId="92" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-08-27T23:56:18.930" LastActivityDate="2011-05-04T04:59:57.150" Title="How to accurately calculate erf(x) with a computer?" Tags="&lt;algorithms&gt;&lt;statistics&gt;&lt;special-functions&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="98" PostTypeId="1" AcceptedAnswerId="106" CreationDate="2010-07-20T20:21:33.407" Score="8" ViewCount="157" Body="&lt;p&gt;I am looking for a reference (book or article) that poses a problem that seems to be a classic, in that I've heard it posed many times, but that I've never seen written anywhere: that of the possibility of a man in a circular pen with a lion, each with some maximum speed, avoiding capture by that lion. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;References to these sprts of pursuit problems in general would also be appreciated, and the original source of this problem.&lt;/p&gt;&#xA;" OwnerUserId="76" LastActivityDate="2010-10-14T08:01:40.487" Title="Chased By a Lion and other Pursuit Problems" Tags="&lt;reference-request&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="1" />
  <row Id="99" PostTypeId="2" ParentId="81" CreationDate="2010-07-20T20:27:41.723" Score="2" ViewCount="" Body="&lt;p&gt;Not a pure math blog, but it's one of the most fascinating blogs in my RSS. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Futility Closet&#xA;&lt;a href=&quot;http://www.futilitycloset.com/&quot; rel=&quot;nofollow&quot;&gt;http://www.futilitycloset.com/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5" LastActivityDate="2010-07-20T20:27:41.723" CommunityOwnedDate="2010-07-20T20:27:41.723" />
  <row Id="100" PostTypeId="2" ParentId="83" CreationDate="2010-07-20T20:29:10.767" Score="4" ViewCount="" Body="&lt;p&gt;If they're fairly mathematically inclined anyway, then try to get them solving interesting problems with an obvious mathematical content, if they're less mathematically inclined try to find problems where the usage of maths isn't as explicit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Problems with a very mathematical bent can be found at places like &lt;a href=&quot;http://nrich.maths.org/public/&quot; rel=&quot;nofollow&quot;&gt;NRich&lt;/a&gt;, they update their problems monthly, the Stage 2 and 3 problems cover that age range. Other sources for problems could be video games (resource management based games require mathematical thinking), code breaking or programming a game (a simple driving game in Flash requires a lot of maths based problem solving).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If they've convinced themselves that they can't do Maths then you basically need to smuggle the maths into your chosen activity without them realising that they're doing maths to solve the problem.&lt;/p&gt;&#xA;" OwnerUserId="86" LastActivityDate="2010-07-20T20:29:10.767" CommunityOwnedDate="2010-08-07T22:32:44.343" CommentCount="1" />
  <row Id="101" PostTypeId="1" AcceptedAnswerId="105" CreationDate="2010-07-20T20:30:39.280" Score="2" ViewCount="87" Body="&lt;p&gt;This is in relation to the Euler Problem 13 from www.ProjectEuler.net.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Work out the first ten digits of the sum of the following one-hundred 50-digit numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;37107287533902102798797998220837590246510135740250&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, this was my thinking:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can freely discard the last fourty digits and leave the last ten.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;0135740250&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And then simply sum those. This would be large enough to be stored in a 64-bit data-type and a lot easier to compute. However, my answer isn't being accepted, so I'm forced to question my logic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I don't see a problem. The last fourty digits will never make a difference because they are at least a magnitude of 10 larger than the preceding values and therefore never carry backwards into smaller areas. Is this not correct?&lt;/p&gt;&#xA;" OwnerUserId="44" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-07-24T18:42:42.350" LastActivityDate="2010-07-24T18:42:42.350" Title="Faulty logic when summing large integers?" Tags="&lt;arithmetic&gt;&lt;project-euler&gt;" AnswerCount="2" CommentCount="5" />
  <row Id="102" PostTypeId="2" ParentId="83" CreationDate="2010-07-20T20:34:32.283" Score="4" ViewCount="" Body="&lt;p&gt;This really depend on how smart the kid is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I lean toward discrete math, elementary number theory related topics when talking to non-math people about math. They requires little background knowledge. There are some fun problems in discrete math, especially combinatorics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Simple probability is also nice. So are logic problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both topics can be used to formulate some simple puzzles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A simple number theory puzzle&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;How many zeros are there in 20!&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I assume a bright 10 year old can solve it.&lt;/p&gt;&#xA;" OwnerUserId="96" LastActivityDate="2010-07-20T20:34:32.283" CommunityOwnedDate="2010-08-07T22:32:44.343" CommentCount="2" />
  <row Id="103" PostTypeId="2" ParentId="71" CreationDate="2010-07-20T20:37:52.477" Score="4" ViewCount="" Body="&lt;p&gt;One place they are frequently used is in computer games when you want to smoothly transition from one rotation to another. An artist might have said &quot;at this time I want the head oriented like this and at that time I want it like this&quot;. The computer needs to work out what happens in-between these poses. It's quite easy to find in-between poses using quaternions. If the two poses are reasonably similar, then you can get a half-way orientation simply by taking the average of the quaternions. You can find out more &lt;a href=&quot;http://en.wikipedia.org/wiki/Slerp&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="80" LastActivityDate="2010-07-20T20:37:52.477" />
  <row Id="104" PostTypeId="2" ParentId="101" CreationDate="2010-07-20T20:37:56.830" Score="2" ViewCount="" Body="&lt;p&gt;First you are doing it in the wrong end, second, the statement in general is still not correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for example:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;9999999999&#xA;1000000001&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Say if you want the first 2 digits,&#xA;you will get 10 if you discard the last 2 digit and do the sum.&#xA;The right answer is 11&lt;/p&gt;&#xA;" OwnerUserId="96" LastActivityDate="2010-07-20T20:37:56.830" />
  <row Id="105" PostTypeId="2" ParentId="101" CreationDate="2010-07-20T20:38:04.910" Score="5" ViewCount="" Body="&lt;p&gt;If you were supposed to find the &lt;em&gt;last&lt;/em&gt; ten digits, you could just ignore the first 40 digits of each number. However you're supposed to find the &lt;em&gt;first&lt;/em&gt; ten digits, so that doesn't work. And you can't just ignore the last digits of each number either because those &lt;em&gt;can&lt;/em&gt; carry over.&lt;/p&gt;&#xA;" OwnerUserId="11" LastActivityDate="2010-07-20T20:38:04.910" />
  <row Id="106" PostTypeId="2" ParentId="98" CreationDate="2010-07-20T20:38:30.527" Score="3" ViewCount="" Body="&lt;p&gt;Since you're asking for a reference, perhaps this will do?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wolfram Mathworld says the problem was listed by Rado in 1925. The reference is on the problem description page, &lt;a href=&quot;http://mathworld.wolfram.com/LionandManProblem.html&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="50" LastActivityDate="2010-07-20T20:38:30.527" />
  <row Id="107" PostTypeId="1" AcceptedAnswerId="1159" CreationDate="2010-07-20T20:38:57.700" Score="14" ViewCount="1065" Body="&lt;p&gt;It is 10 o'clock, and I have a box. Inside the box is a ball marked 1.&lt;br&gt;&#xA;At 10:30, I will remove the ball marked 1, and add two balls, labeled 2 and 3. At 10:45, I will remove the balls labeled 2 and 3, and add 4 balls, marked 4, 5, 6, and 7. 7.5 minutes before 11, I will remove the balls labeled 4, 5, and 6, and add 8 balls, labeled 8, 9, 10, 11, 12, 13, 14, and 15. This pattern continues. Each time I reach the halfway point between my previous action and 11 o'clock, I add some balls, and remove some other balls. Each time I remove one more ball than I removed last time, but add twice as many balls as I added last time.&lt;br&gt;&#xA;The result is that as it gets closer and closer to 11, the number of balls in the box continues to increase. Yet every ball that I put in was eventually removed. So just how many balls will be in the box when the clock strikes 11? 0, or infinitely many? What's going on here? &lt;/p&gt;&#xA;" OwnerUserId="73" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2010-11-27T14:01:59.213" LastActivityDate="2011-05-25T14:23:30.573" Title="paradox: increasing sequence that goes to 0?" Tags="&lt;soft-question&gt;&lt;paradox&gt;" AnswerCount="11" CommentCount="6" FavoriteCount="4" />
  <row Id="108" PostTypeId="2" ParentId="71" CreationDate="2010-07-20T20:41:27.650" Score="1" ViewCount="" Body="&lt;p&gt;You can view a real-world example of quaternions in computer graphics with the open source program known as NASA WorldWind (http://worldwind.arc.nasa.gov/java/).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It uses a Quaternion object to represent rotation of various geometries. The class definition itself is located in the src/gov/nasa/worldwind/geom/Quaternion.java file.&lt;/p&gt;&#xA;" OwnerUserId="66" LastActivityDate="2010-07-20T20:41:27.650" />
  <row Id="109" PostTypeId="2" ParentId="51" CreationDate="2010-07-20T20:48:24.463" Score="6" ViewCount="" Body="&lt;p&gt;&lt;img src=&quot;http://www.imgftw.net/img/797940734.jpg&quot; alt=&quot;diagram&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The field is the smaller/left circle, centered at A.  The cow is tied to the post at E.  The larger/right circle is the grazing radius.  Let the radius of the field be R and the length of the rope be L.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The grazable area is the union of a segment of the circular field and a segment of the circle defined by the rope length.  (A segment of a circle is a sector of a circle less the triangle defined by the center of the circle and the endpoints of the arc.)  The area of a segment of a circle of radius r with central angle &amp;theta; is $\frac{1}{2}R^2(\theta-\sin\theta)$, where &amp;theta; is measured in radians.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to express the grazable area in terms of L and R, we need expressions for the measures of &amp;ang;CED and &amp;ang;CAD (which define the segments of the circles; call these &amp;alpha; and &amp;beta; for convenience).  By using the Law of Cosines on the blue triangle in the diagram, $\alpha=\mathrm{m}\angle CED=2\arccos\left(\frac{L}{2R}\right)$ and $\beta=\mathrm{m}\angle CAD=2\arccos\left(\frac{2R^2-L^2}{2R^2}\right)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The grazable area is $\frac{1}{2}L^2(\alpha-\sin\alpha)+\frac{1}{2}R^2(\beta-\sin\beta)$.  We want that to be equal to half the area of the field, $\frac{1}{2}\pi R^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is, the relationship between L and R is $$\begin{align}&amp;amp;\frac{L^2}{2}\left(2\arccos\left(\frac{L}{2R}\right)-\sin\left(2\arccos\left(\frac{L}{2R}\right)\right)\right) \\ &amp;amp;\quad\quad\quad+\frac{R^2}{2}\left(2\arccos\left(\frac{2R^2-L^2}{2R^2}\right)-\sin\left(2\arccos\left(\frac{2R^2-L^2}{2R^2}\right)\right)\right) \\ &amp;amp;=\frac{\pi R^2}{2}.\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is no algebraic method of solving this equation for L in terms of R (Mathematica cannot &lt;code&gt;Solve[]&lt;/code&gt; or &lt;code&gt;Reduce[]&lt;/code&gt; it).&lt;/p&gt;&#xA;" OwnerUserId="72" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-08-05T19:02:22.920" LastActivityDate="2010-08-05T19:02:22.920" />
  <row Id="110" PostTypeId="2" ParentId="11" CreationDate="2010-07-20T20:48:39.627" Score="-2" ViewCount="" Body="&lt;p&gt;The argument against this is that 0.99999999... is &quot;somewhat&quot; less than 1. How much exactly?&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;      1 - 0.999999... = ε              (0)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;If the above is true, the following also must be true:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;9 × (1 - 0.999999...) = ε × 9&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Let's calculate:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;0.999... ×&#xA;9        =&#xA;───────────&#xA;8.1&#xA;  81&#xA;   81&#xA;     .&#xA;      .&#xA;       .&#xA;&#xA;───────────&#xA;8.999...&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Thus:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;     9 - 8.999999... = 9ε              (1)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;But:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;         8.999999... = 8 + 0.99999...  (2)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Indeed:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;8.00000000... +&#xA;0.99999999... =&#xA;────────────────&#xA;8.99999999...&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Now let's see what we can deduce from &lt;code&gt;(0)&lt;/code&gt;, &lt;code&gt;(1)&lt;/code&gt; and &lt;code&gt;(2)&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;9 - 8.999999... = 9ε                      because of (2)&#xA;9 - 8.999999... = 9 - (8 + 0.99999...) =  because of (1)&#xA;                = 9 -  8 - (1 - ε)        because of (0)&#xA;                =   1    -  1 + ε         &#xA;                =               ε.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Thus:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;9ε = ε&#xA;&#xA;10ε = 0&#xA;&#xA;ε = 0&#xA;&#xA;1 - 0.999999... = ε = 0&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Quod erat dimostrandum. Pardon my unicode.&lt;/p&gt;&#xA;" OwnerUserId="92" LastEditorUserId="92" LastEditorDisplayName="" LastEditDate="2010-07-20T21:12:35.550" LastActivityDate="2010-07-20T21:12:35.550" CommunityOwnedDate="2010-07-28T06:53:43.987" CommentCount="4" />
  <row Id="111" PostTypeId="2" ParentId="107" CreationDate="2010-07-20T20:50:35.817" Score="-2" ViewCount="" Body="&lt;p&gt;This is a problem of &lt;a href=&quot;http://en.wikipedia.org/wiki/Zeno%27s_paradoxes&quot; rel=&quot;nofollow&quot;&gt;Zeno's Paradox&lt;/a&gt;. If you continue to subdivide the time ad infinitum, you'll never reach 11 o'clock so the answer is undefined.&lt;/p&gt;&#xA;" OwnerUserId="36" LastActivityDate="2010-07-20T20:50:35.817" CommentCount="3" />
  <row Id="112" PostTypeId="1" AcceptedAnswerId="115" CreationDate="2010-07-20T20:51:07.517" Score="4" ViewCount="105" Body="&lt;p&gt;I keep seeing this symbol ∇ around and I know enough to understand that it represents the term &quot;gradient.&quot; But what is a gradient? When would I want to use one mathematically?&lt;/p&gt;&#xA;" OwnerUserId="82" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-07-24T18:44:02.677" LastActivityDate="2010-07-24T18:44:02.677" Title="What are gradients and how would I use them?" Tags="&lt;calculus&gt;&lt;terminology&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="113" PostTypeId="1" CreationDate="2010-07-20T20:56:33.853" Score="4" ViewCount="215" Body="&lt;p&gt;If you know it, also try to include the precise reason why the proof is fallacious. To start this off, let me post the one that most people know already:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt; Let a = b.&#xA;&lt;li&gt; Then a^2 = ab&#xA;&lt;li&gt; a^2 - b^2 = ab - b^2&#xA;&lt;li&gt; Factor to (a-b)(a+b) = b(a-b)&#xA;&lt;li&gt; Then divide out (a-b) to get a+b = b&#xA;&lt;li&gt; Since a = b, then b+b = b&#xA;&lt;li&gt; Therefore 2b = b&#xA;&lt;li&gt; Reduce to 2 = 1&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;As @jan-gorzny pointed out, in this case, line 5 is wrong since a = b implies a-b = 0, and so you can't divide out (a-b).&lt;/p&gt;&#xA;" OwnerUserId="66" LastEditorUserId="126" LastEditorDisplayName="" LastEditDate="2010-07-21T05:09:26.823" LastActivityDate="2010-07-21T05:09:26.823" CommunityOwnedDate="2010-07-20T21:00:04.093" Title="What are some classic fallacious proofs?" Tags="&lt;big-list&gt;&lt;soft-question&gt;" AnswerCount="2" CommentCount="6" FavoriteCount="1" />
  <row Id="114" PostTypeId="2" ParentId="107" CreationDate="2010-07-20T20:57:00.700" Score="3" ViewCount="" Body="&lt;p&gt;During one math class I had the same problem. the professor told us the answer can be a arbitrary integer, depend on how you remove the balls.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since you said every ball you put in was eventually removed, I will make it simple and assume you first number the balls to natural numbers, and then add and remove the balls with the smallest numbers first.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case, the bin will be empty--every ball put in have been removed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you define the sequence in the way such that a_n = amount of balls when you do the nth move, then this sequence never get to zero, because there is no n correspond to 11 o'clock.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The increasing sequence doesn't define how many balls in the bin at 11. Therefore there is no paradox.&lt;/p&gt;&#xA;" OwnerUserId="96" LastActivityDate="2010-07-20T20:57:00.700" CommentCount="1" />
  <row Id="115" PostTypeId="2" ParentId="112" CreationDate="2010-07-20T21:02:32.183" Score="7" ViewCount="" Body="&lt;p&gt;The ∇ (pronounced &quot;del&quot;) is an operator, more technically.  In 3D, it (more or less) means the vector&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;&amp;lt; df/dx, df/dy, df/dz &amp;gt;&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;So, if &lt;code&gt;f(x,y,z) = x^2 + y^3*z + sin(z)&lt;/code&gt;, &lt;code&gt;∇f = &amp;lt; 2x, 3y^2*z, y^3 + cos(z) &amp;gt;&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's actually a bit more subtle than that; technically it means&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;&amp;lt; d/dx, d/dy, d/dz &amp;gt;&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;And when you do ∇f, it's sort of like a &quot;multiplication&quot; of ∇ and f;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;&amp;lt; d/dx, d/dy, d/dz &amp;gt; f = &amp;lt; d/dx f, d/dy f, d/dz f &amp;gt;&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Only, not multiplication, but &lt;em&gt;operation&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are some neat properties about the del operator.  Here are a couple:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;The most famous is that &lt;code&gt;∇f&lt;/code&gt; yields the &lt;strong&gt;gradient&lt;/strong&gt; of &lt;code&gt;f&lt;/code&gt;.  That is, at any point &lt;code&gt;(x,y,z)&lt;/code&gt;, &lt;code&gt;∇f(x,y,z)&lt;/code&gt; is the vector pointing in the direction where it is &lt;strong&gt;most increasing&lt;/strong&gt;.  The magnitude of it is the magnitude of the increase.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is easier to understand with, say, a 2D &lt;code&gt;f(x,y)&lt;/code&gt;.  If &lt;code&gt;f(x,y)&lt;/code&gt; represents the height of a point at &lt;code&gt;(x,y)&lt;/code&gt;, then &lt;code&gt;∇f(x,y)&lt;/code&gt; represents the steepest incline from that point.  Or rather, if you placed a ball on that point, it would &lt;em&gt;start rolling in the opposite direction of the gradient vector&lt;/em&gt;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Normally, for multi-dimensional functions, it is easiest to find the derivative along an axis (x, y, z, etc.).  With ∇, you can find the derivative along any arbitrary direction by using ∇f * u, where * is the dot product and u is the unit vector along the direction you are calculating.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;∇ is also used to calculate divergence (amount that vectors are &quot;spreading out&quot;) and curl (amount that vectors are &quot;curling up&quot;) of a vector field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Divergence is ∇ * f (dot product), and curl is ∇ x f (cross product)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They aren't truly &quot;products&quot; in the sense.  Rather, when you are calculating divergence and curl and you must do &lt;code&gt;d/dx * (something)&lt;/code&gt;, you are actually doing &lt;code&gt;d/dx (something)&lt;/code&gt; or &lt;code&gt;d(something)/dx&lt;/code&gt;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="51" LastEditorUserId="51" LastEditorDisplayName="" LastEditDate="2010-07-20T21:24:14.510" LastActivityDate="2010-07-20T21:24:14.510" />
  <row Id="116" PostTypeId="2" ParentId="11" CreationDate="2010-07-20T21:02:52.433" Score="10" ViewCount="" Body="&lt;p&gt;Assuming:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;infinite decimals are series where the terms are the digits divided by the proper power of the base&lt;/li&gt;&#xA;&lt;li&gt;the infinite geometric series a + a * r + a * r^2 + a * r^3 + ... has sum a/(1 - r) as long as |r|&amp;lt;1&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;0.99999... = 9/10 + 9/10^2 + 9/10^3 + ...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the infinite geometric series with first term a = 9/10 and common ratio r = 1/10, so it has sum (9/10) / (1 - 1/10) = (9/10) / (9/10) = 1.&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-07-20T21:02:52.433" CommunityOwnedDate="2010-07-28T06:53:43.987" CommentCount="1" />
  <row Id="117" PostTypeId="2" ParentId="98" CreationDate="2010-07-20T21:06:00.833" Score="4" ViewCount="" Body="&lt;p&gt;Here is a book on this type of problem&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://press.princeton.edu/titles/8371.html&quot; rel=&quot;nofollow&quot;&gt;http://press.princeton.edu/titles/8371.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;it is also briefly mentioned in his other book &quot;Euler's Fabulous Formula&quot;.&lt;/p&gt;&#xA;" OwnerUserId="16" LastActivityDate="2010-07-20T21:06:00.833" CommentCount="1" />
  <row Id="118" PostTypeId="1" AcceptedAnswerId="125" CreationDate="2010-07-20T21:06:33.610" Score="9" ViewCount="561" Body="&lt;p&gt;I keep hearing about this weird type of math called calculus. I only have experience with geometry and algebra. Can you try to explain what it is to me?&lt;/p&gt;&#xA;" OwnerUserId="104" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-07-22T03:05:43.040" LastActivityDate="2011-01-14T05:57:44.497" Title="How would you describe calculus in simple terms?" Tags="&lt;soft-question&gt;&lt;calculus&gt;" AnswerCount="7" CommentCount="1" FavoriteCount="2" />
  <row Id="119" PostTypeId="2" ParentId="90" CreationDate="2010-07-20T21:08:28.333" Score="5" ViewCount="" Body="&lt;p&gt;The following reddit post has a decent list of math resources:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.reddit.com/r/math/comments/bqbex/lets_list_all_the_useful_free_online_math/&quot; rel=&quot;nofollow&quot;&gt;http://www.reddit.com/r/math/comments/bqbex/lets_list_all_the_useful_free_online_math/&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;One site I did not see it their list that I've found very helpful:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://betterexplained.com/&quot; rel=&quot;nofollow&quot;&gt;http://betterexplained.com/&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="105" LastActivityDate="2010-07-20T21:08:28.333" CommunityOwnedDate="2010-07-20T21:08:28.333" />
  <row Id="120" PostTypeId="1" AcceptedAnswerId="122" CreationDate="2010-07-20T21:08:29.207" Score="17" ViewCount="792" Body="&lt;p&gt;Is 1 classified as a prime number?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And if so, why? If not, why not?&lt;/p&gt;&#xA;" OwnerUserId="106" LastEditorUserId="126" LastEditorDisplayName="" LastEditDate="2010-07-21T01:37:46.723" LastActivityDate="2010-12-07T05:25:50.567" Title="Is 1 a prime number?" Tags="&lt;number-theory&gt;&lt;terminology&gt;" AnswerCount="5" FavoriteCount="3" />
  <row Id="122" PostTypeId="2" ParentId="120" CreationDate="2010-07-20T21:11:24.393" Score="22" ViewCount="" Body="&lt;p&gt;One of the whole &quot;points&quot; of defining primes is to be able to uniquely and finitely prime factorize every natural number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If 1 was prime, then this would be more or less impossible.&lt;/p&gt;&#xA;" OwnerUserId="51" LastEditorUserId="51" LastEditorDisplayName="" LastEditDate="2010-07-25T21:59:33.263" LastActivityDate="2010-07-25T21:59:33.263" CommentCount="11" />
  <row Id="123" PostTypeId="1" CreationDate="2010-07-20T21:14:07.827" Score="8" ViewCount="3049" Body="&lt;p&gt;I covered hyperbolic trigonometric functions in a recent maths course. However I was never presented with any reasons as to &lt;em&gt;why&lt;/em&gt; (or even if) they are useful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any good examples of their uses outside academia?&lt;/p&gt;&#xA;" OwnerUserId="108" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-07-24T18:52:17.790" LastActivityDate="2010-12-29T11:10:41.143" Title="Real world uses of hyperbolic trigonometric functions" Tags="&lt;soft-question&gt;&lt;big-list&gt;&lt;applications&gt;&lt;trigonometry&gt;" AnswerCount="9" CommentCount="4" FavoriteCount="2" />
  <row Id="124" PostTypeId="2" ParentId="123" CreationDate="2010-07-20T21:18:58.203" Score="20" ViewCount="" Body="&lt;p&gt;If you take a rope, fix the two ends, and let it hang under the force of gravity, it will naturally form a hyperbolic cosine curve.&lt;/p&gt;&#xA;" OwnerUserId="30" LastActivityDate="2010-07-20T21:18:58.203" CommentCount="2" />
  <row Id="125" PostTypeId="2" ParentId="118" CreationDate="2010-07-20T21:19:03.680" Score="14" ViewCount="" Body="&lt;p&gt;There came a time in mathematics when people encountered situations where they had to deal with really, really, really small things.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Not just small like 0.01; but small as in &lt;em&gt;infinitesimally small&lt;/em&gt;.  Think of &quot;the smallest positive number that is still greater than zero&quot; and you'll realize what sort of problems mathematicians began encountering.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Soon, this problem became more than just theoretical or abstract.  It became very, very real.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, velocity.  We know that average velocity is the change in position per change in time (i.e., 5 miles per hour).  But what about velocity &lt;strong&gt;at a point in time&lt;/strong&gt;?  What does it mean to be going 5 mph &lt;strong&gt;at this moment&lt;/strong&gt;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One solution someone came up with was to say &quot;it's the change in position divided by the change in time, where the change in time is an infinitesimally small amount of time&quot;.  But how would you handle/calculate that?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another problem came about trying to find the area under a curve.  The current accepted solution was to divide the curve into rectangles, and add together the area of the rectangles.  However, in order to find the &lt;strong&gt;exact&lt;/strong&gt; area under the curve, you'd need to divide it into rectangles that were &lt;strong&gt;infinitesimally tiny&lt;/strong&gt;, and, therefore, add up an infinite amount of tiny rectangles -- to something that was finite (area).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Calculus came about as the system of math dedicated to studying these infinitesimally small changes.  In fact, I do believe some people describe calculus as &quot;the study of continuous changes&quot;.&lt;/p&gt;&#xA;" OwnerUserId="51" LastActivityDate="2010-07-20T21:19:03.680" CommentCount="2" />
  <row Id="126" PostTypeId="2" ParentId="123" CreationDate="2010-07-20T21:19:10.890" Score="4" ViewCount="" Body="&lt;p&gt;An equation for a catenary curve can be given in terms of hyperbolic cosine.  Catenary curves appear in many places, such as the Gateway Arch in St. Louis, MO.&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-07-20T21:19:10.890" />
  <row Id="127" PostTypeId="2" ParentId="118" CreationDate="2010-07-20T21:19:52.887" Score="3" ViewCount="" Body="&lt;p&gt;Calculus is a field which deals with two seemingly unrelated things.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) the area beneath a graph and the x-axis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) the slope (or gradient) of a curve at different points.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Part (1) is also called 'integration' and 'anti-differentiation', and part (2) is called 'differentiation'.&lt;/p&gt;&#xA;" OwnerUserId="106" LastActivityDate="2010-07-20T21:19:52.887" CommentCount="4" />
  <row Id="128" PostTypeId="1" AcceptedAnswerId="1163" CreationDate="2010-07-20T21:20:16.380" Score="6" ViewCount="287" Body="&lt;p&gt;The Weyl equidistribution theorem states that the sequence of fractional parts ${n \xi}$, $n = 0, 1, 2, \dots$ is uniformly distributed for $\xi$ irrational. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This can be proved using a bit of ergodic theory, specifically the fact that an irrational rotation is uniquely ergodic with respect to Lebesgue measure. It can also be proved by simply playing with trigonometric polynomials (i.e., polynomials in $e^{2\pi i k x}$ for $k$ an integer) and using the fact they are dense in the space of all continuous functions with period 1.  In particular, one shows that if $f(x)$ is a continuous function with period 1, then for any $t$,  $\int_0^1 f(x) dx = \lim \frac{1}{N} \sum_{i=0}^{N-1} f(t+i \xi)$. One shows this by checking this (directly) for trigonometric polynomials via the geometric series.  This is a very elementary and nice proof.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The general form of Weyl's theorem states that if $p$ is a monic integer-valued polynomial, then the sequence ${p(n \xi)}$ for $\xi$ irrational is uniformly distributed modulo 1.  I believe this can be proved using extensions of these ergodic theory techniques -- it's an exercise in Katok and Hasselblatt.  I'd like to see an elementary proof.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can the general form of Weyl's theorem be proved using the same elementary techniques as in the basic version?&lt;/p&gt;&#xA;" OwnerUserId="536" LastEditorUserId="126" LastEditorDisplayName="" LastEditDate="2010-07-21T01:35:36.433" LastActivityDate="2010-07-29T16:10:13.567" Title="How do you prove that $p(n \xi)$ for $\xi$ irrational and $p$ a polynomial is uniformly distributed modulo 1?" Tags="&lt;number-theory&gt;" AnswerCount="1" CommentCount="7" FavoriteCount="1" />
  <row Id="129" PostTypeId="2" ParentId="118" CreationDate="2010-07-20T21:20:59.237" Score="2" ViewCount="" Body="&lt;p&gt;To be very brief and succinct:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Calculus is the study of how quantities change&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Slightly more technically, it a subject based on &lt;a href=&quot;http://en.wikipedia.org/wiki/Infinitesimal&quot; rel=&quot;nofollow&quot;&gt;infinitesimals&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It may be pointing out the obvious, but the &lt;a href=&quot;http://en.wikipedia.org/wiki/Calculus&quot; rel=&quot;nofollow&quot;&gt;Wikipedia article&lt;/a&gt; does actually provide a pretty decent beginners introduction to the subject. You'll generally want to start with &lt;em&gt;differential calculus&lt;/em&gt; and move on quickly to &lt;em&gt;integral calculus&lt;/em&gt;, followed by linking up the two (fundamental theorem of calculus) and moving on from there.&lt;/p&gt;&#xA;" OwnerUserId="56" LastActivityDate="2010-07-20T21:20:59.237" />
  <row Id="130" PostTypeId="2" ParentId="113" CreationDate="2010-07-20T21:25:30.033" Score="8" ViewCount="" Body="&lt;p&gt;Wikipedia has a long list of these:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Mathematical_fallacy&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Mathematical_fallacy&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="30" LastActivityDate="2010-07-20T21:25:30.033" CommunityOwnedDate="2010-07-20T21:25:30.033" />
  <row Id="132" PostTypeId="2" ParentId="83" CreationDate="2010-07-20T21:31:08.557" Score="13" ViewCount="" Body="&lt;p&gt;I recently taught a once-a-week geometry class to grades 7-10 where we did some graph theory, some surfaces, some spherical geometry, a lot of complex numbers, and basic ideas of homeomorphismm and homotopy. Most couldn't work with complex numbers at the beginning: showing their use in proofs of analytic geometry results had the benefit of not boring the kids who had already seen and worked with complex numbers, and giving the kids who hadn't a reason to learn the technicalities (getting under the hood of these pretty picures).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Euler's formula and the relation of planar graphs to polyhedra showed them the basics of a connection between geometric and algebraic intuition, outside the coordinate geometry many were used to, and got them thinking about ways to define familiar things mathematically (space), and what we can learn about familiar things through mathematics (surprising things like orientability through the Mobius strip, or sphere eversion).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general I've found geometry to be a very good place to start with people with a professed fear or disinterest in mathematics. For me, it's the quickest way to show the distance between what mathematicians play with when they're doing math and what was taught in high school, with the two column proof nonsense and the treatment of math as a branch of formal logic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, I suspect that I find this path the easiest because it is one of the ones I am most excited about in mathematics, and I further imagine that in many cases, the best thing to use to get people excited about math is something you yourself are really excited about, as long as you can think to translate it well.&lt;/p&gt;&#xA;" OwnerUserId="76" LastActivityDate="2010-07-20T21:31:08.557" CommunityOwnedDate="2010-08-07T22:32:44.343" CommentCount="1" />
  <row Id="133" PostTypeId="2" ParentId="3" CreationDate="2010-07-20T21:31:44.030" Score="1" ViewCount="" Body="&lt;p&gt;I listen to &lt;a href=&quot;http://www.aracnet.com/~eseligma/mm/&quot; rel=&quot;nofollow&quot;&gt;Math Mutation Podcast&lt;/a&gt;. The topics are interesting and understandable by a layman.&lt;/p&gt;&#xA;" OwnerUserId="96" LastActivityDate="2010-07-20T21:31:44.030" CommunityOwnedDate="2010-07-20T21:31:44.030" />
  <row Id="134" PostTypeId="2" ParentId="57" CreationDate="2010-07-20T21:34:01.537" Score="5" ViewCount="" Body="&lt;p&gt;I'd like to change the notation of your definitions, since R, S and T would usually be used to stand for the relations themselves (and x, y and z would be more commonly chosen for the objects that might bear the relation to each other).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Reflexive&lt;/em&gt; - For all x: xRx&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example reflexive relation: xRy stands for 'x is a factor of y' (in the set of natural numbers)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Symmetric&lt;/em&gt; - For all x,y: if xRy then yRx&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example symmetric relation: xRy stands for 'x and y are 2 metres apart' (in the set of all people in a particular room)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Antisymmetric&lt;/em&gt; - For all x,y: if xRy and yRx then x = y&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example antisymmetric relation: xRy stands for 'x is a factor of y' (in the set of natural numbers)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Asymmetric&lt;/em&gt; - For all x,y: if xRy then not yRx&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example asymmetric relation: xRy stands for 'x is taller than y' (in the set of all people)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Transitive&lt;/em&gt; - For all x,y,z: if xRy and yRz then xRz&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example transitive relation: xRy stands for 'x is taller than y' (in the set of all people)&lt;/p&gt;&#xA;" OwnerUserId="106" LastActivityDate="2010-07-20T21:34:01.537" />
  <row Id="135" PostTypeId="1" AcceptedAnswerId="138" CreationDate="2010-07-20T21:34:29.273" Score="10" ViewCount="375" Body="&lt;p&gt;Why is any number (other than zero) to the power of zero equal to one? Please include in your answer an explanation of why 0^0 should be undefined.&lt;/p&gt;&#xA;" OwnerUserId="106" LastEditorUserId="126" LastEditorDisplayName="" LastEditDate="2010-07-21T16:59:46.510" LastActivityDate="2010-07-23T11:41:17.373" Title="Why is x^0 = 1 except when x = 0?" Tags="&lt;notation&gt;&lt;convention&gt;" AnswerCount="6" CommentCount="7" />
  <row Id="136" PostTypeId="1" AcceptedAnswerId="183" CreationDate="2010-07-20T21:36:40.990" Score="3" ViewCount="244" Body="&lt;p&gt;I was playing around with the squares and saw an interesting pattern in their differences.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&#xA;0^2 = 0&#xA;          + 1&#xA;1^2 = 1&#xA;          + 3&#xA;2^2 = 4&#xA;          + 5&#xA;3^2 = 9&#xA;          + 7&#xA;4^2 = 16&#xA;          + 9&#xA;5^2 = 25&#xA;          + 11&#xA;6^2 = 36&#xA;          etc.&#xA;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;(Also, in a very related question, which major Math Research Journal should I contact to publish my groundbreaking find in?)&lt;/p&gt;&#xA;" OwnerUserId="51" LastEditorUserId="126" LastEditorDisplayName="" LastEditDate="2010-07-21T01:36:39.207" LastActivityDate="2011-03-18T08:53:26.810" Title="Why are the differences between consecutive squares equal to the sequence of odd numbers?" Tags="&lt;number-theory&gt;" AnswerCount="4" CommentCount="2" />
  <row Id="137" PostTypeId="2" ParentId="67" CreationDate="2010-07-20T21:37:18.823" Score="7" ViewCount="" Body="&lt;p&gt;The technical definition is a nonsingular projective curve of genus 1, which is an abelian variety under the group law: basially, this means that you   draw the line through two points on the curve -- which can be embedded in the projective plane -- and find where that line intersects the curve again (and call that the negative of the sum).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can always put elliptic curves in the (projectivization of the) form $y^2 = x^3 - Ax + B$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the meaning of &quot;abelian variety&quot; is that you can add points on the elliptic curve, which is really useful; there isn't a way to do this for most objects in algebraic geometry.  Then one can study things like the torsion points on an elliptic curve, with respect to this abelian group structure: it's a theorem that there are $m^2$ torsion points of order $m$ for instance, if you 're working in an algebraically closed field. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, one way to think of this is that an elliptic curve is really--algebraically and topologically--a torus if you are working over the complex numbers, and the torsion points in the torus are easily determined.  (Namely, a torus is algebraically the product of two copies of the unit circle.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This also yields the theorem about torsion points for algebraically closed fields of characteristic zero via the &quot;Lefschetz principle.&quot;  (For characteristic p, you need a different argument.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other things one can consider include the group of points with coordinates in, say, the rational numbers (assuming the curve is defined by rational coefficients). One of the central theorems is that this group is finitely generated.  The point is that the geometry of the elliptic curve leads to a rich algebraic structure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's a bit about elliptic curves; I know nothing about cryptography and can't comment on that.&lt;/p&gt;&#xA;" OwnerUserId="536" LastEditorUserId="536" LastEditorDisplayName="" LastEditDate="2010-07-20T21:57:10.940" LastActivityDate="2010-07-20T21:57:10.940" CommentCount="1" />
  <row Id="138" PostTypeId="2" ParentId="135" CreationDate="2010-07-20T21:38:34.720" Score="3" ViewCount="" Body="&lt;p&gt;For non-zero bases and exponents, the relation x^a * x^b = x^(a+b) holds. For this to make sense with an exponent of 0, x^0 needs to equal one. This gives you x^a * 1 = x^a * x^0 = x^(a+0) = x^a.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the base is also zero, it's not possible to define a value for 0^0 because there is no value that is consistent with all the necessary constraints. For example, 0^x = 0 and x^0 = 1 for all positive x, and 0^0 can't be consistent with both of these.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another way to see that 0^0 can't have a reasonable definition is to look at the graph of f(x,y) = x^y which is discontinuous around (0,0). No chosen value for 0^0 will avoid this discontinuity.&lt;/p&gt;&#xA;" OwnerUserId="113" LastEditorUserId="113" LastEditorDisplayName="" LastEditDate="2010-07-22T13:28:18.537" LastActivityDate="2010-07-22T13:28:18.537" CommentCount="7" />
  <row Id="139" PostTypeId="1" AcceptedAnswerId="176" CreationDate="2010-07-20T21:38:38.840" Score="2" ViewCount="205" Body="&lt;p&gt;I have a square that's 10m x 10m. I want to cut it in half so that I have a square with half the area. But if I cut it from top to bottom or left to right, I don't get a square, I get a rectangle!&lt;br&gt;&#xA;I know the area of the small square is supposed to be 50, so I can use my calculator to find out how long a side should be: it's 7.07106781. But my teacher said I should be able to do this without a calculator. How am I supposed to get that number by hand?&lt;/p&gt;&#xA;" OwnerUserId="73" LastActivityDate="2010-07-20T23:55:32.657" Title="how do i cut a square in half?" Tags="&lt;geometry&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="1" />
  <row Id="140" PostTypeId="2" ParentId="135" CreationDate="2010-07-20T21:39:08.137" Score="5" ViewCount="" Body="&lt;p&gt;0^x = 0, x^0 = 1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;both are true when x &gt; 0 &lt;/p&gt;&#xA;&#xA;&lt;p&gt;what happens when x=0? undefined, because there is no way to chose one definition over the other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some people define 0^0 = 1 in their books, like Knuth, because 0^x is less 'useful' than x^0.&lt;/p&gt;&#xA;" OwnerUserId="96" LastEditorUserId="96" LastEditorDisplayName="" LastEditDate="2010-07-21T22:09:52.383" LastActivityDate="2010-07-21T22:09:52.383" CommentCount="3" />
  <row Id="141" PostTypeId="2" ParentId="135" CreationDate="2010-07-20T21:40:17.747" Score="10" ViewCount="" Body="&lt;p&gt;This is a question of definition, the question is &quot;why does it make sense to define x^0=1 except when x=0?&quot; or &quot;How is this definition better than other definitions?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer is that $x^a * x^b = x^{a+b}$ is an excellent formula that makes a lot of sense (multiplying a times and then multiplying b times is the same as multiplying a+b times) and which you can prove for a and b positive integers.  So any sensible definition of x^a for numbers a which aren't positive integers &lt;em&gt;should still satisfy this identity&lt;/em&gt;.  In particular, $x^0 * x^b = x^{0+b} = x^b$; now &lt;em&gt;if x is not zero&lt;/em&gt; then you can cancel x^b from both sides and get that x^0 = 1.  But if x=0 then x^b is zero and so this argument doesn't tell you anything about what you should define x^0 to be.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A similar argument should convince you that when x is not zero then x^-a should be defined as 1/x^a.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An argument using the related identity (x^a)^b = x^(ab) should convince you that x^{1/n} is taking the nth root.&lt;/p&gt;&#xA;" OwnerUserId="8" LastActivityDate="2010-07-20T21:40:17.747" CommentCount="4" />
  <row Id="142" PostTypeId="2" ParentId="123" CreationDate="2010-07-20T21:40:22.333" Score="3" ViewCount="" Body="&lt;p&gt;Apart from occuring naturally in the form of the catenary mentioned, hyperbolic trig functions are also fundamentally important in calculus. I hope you'll agree that calculus has many, many real world applications!&lt;/p&gt;&#xA;" OwnerUserId="67" LastActivityDate="2010-07-20T21:40:22.333" CommentCount="1" />
  <row Id="143" PostTypeId="2" ParentId="136" CreationDate="2010-07-20T21:41:21.523" Score="11" ViewCount="" Body="&lt;p&gt;A nice thing to notice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's basically because&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;(n + 1) ^ 2 - n ^ 2 = 2 * n + 1&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="88" LastEditorUserId="88" LastEditorDisplayName="" LastEditDate="2010-07-20T21:50:48.713" LastActivityDate="2010-07-20T21:50:48.713" CommentCount="1" />
  <row Id="144" PostTypeId="2" ParentId="135" CreationDate="2010-07-20T21:42:02.833" Score="3" ViewCount="" Body="&lt;p&gt;Exponents are only &quot;basically&quot; defined under the natural numbers above zero.  By this I mean, defined as &quot;iterated multiplication&quot; the same way multiplication is defined as iterated addition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The property &lt;code&gt;a^0 = 1&lt;/code&gt; only arises when we look at generalizing multiplication to the integers.  We do this by:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;a^4 / a^3 = (a*a*a*a)/(a*a*a) = a^1&#xA;a^4 / a^3 = a^(4-3)           = a^1&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;And using this, we can say:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;a^2 / a^3 = a^(-1) = 1/a&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;and also:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;a^2 / a^2                 = 1&#xA;a^2 / a^2 = a^(2-2) = a^0 = 1&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;So we say &lt;code&gt;a^0 = 1&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, notice that these proofs don't have any meaning when &lt;code&gt;a=0&lt;/code&gt;, because the whole concept/idea involves fractions, and you cannot have zero be in the denominator.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When we say &lt;code&gt;2^0 = 1&lt;/code&gt;, we really mean:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;2^(1-1) = 2^1 / 2^1 = 2/2 = 1&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;But we cannot say the same for &lt;code&gt;0^0&lt;/code&gt;:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;0^(1-1) = 0^1/0^1 = 0/0 = UNDEFINED&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="51" LastActivityDate="2010-07-20T21:42:02.833" />
  <row Id="145" PostTypeId="2" ParentId="139" CreationDate="2010-07-20T21:42:38.077" Score="0" ViewCount="" Body="&lt;p&gt;A simple approximation&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;x^2 = 50&#xA;7^2 = 49&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;so the number is around 7.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A even better approximation&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;(7+x)^2 = 50&#xA;14x + x^2 = 1 &#xA;(14+x)x = 1&#xA;14+x = 1/x&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This basically shows x is pretty small, and around 1/14, without a calculator, it's not likely you can find what exactly 1/14 is, but at least you know it be smaller than 1/10. 7.1 is a good approximation. (off by less than 0.03)&lt;/p&gt;&#xA;" OwnerUserId="96" LastEditorUserId="96" LastEditorDisplayName="" LastEditDate="2010-07-20T21:48:27.607" LastActivityDate="2010-07-20T21:48:27.607" CommentCount="1" />
  <row Id="146" PostTypeId="2" ParentId="139" CreationDate="2010-07-20T21:44:00.230" Score="5" ViewCount="" Body="&lt;p&gt;Take a pair of compasses and draw an arc between two opposite corners, centred at another corner, and then draw a diagonal across the arc. If you now draw two lines from the point of intersection, parallel to the sides of the square, the biggest of the resulting squares will have half the area of the original square.&lt;/p&gt;&#xA;" OwnerUserId="67" LastEditorUserId="67" LastEditorDisplayName="" LastEditDate="2010-07-20T23:55:32.657" LastActivityDate="2010-07-20T23:55:32.657" CommentCount="6" />
  <row Id="147" PostTypeId="2" ParentId="90" CreationDate="2010-07-20T21:45:47.093" Score="3" ViewCount="" Body="&lt;p&gt;Andrea Feretti's &lt;a href=&quot;http://mathonline.andreaferretti.it/&quot; rel=&quot;nofollow&quot;&gt;MathOnline&lt;/a&gt; page.&lt;/p&gt;&#xA;" OwnerUserId="536" LastActivityDate="2010-07-20T21:45:47.093" CommunityOwnedDate="2010-07-20T21:45:47.093" />
  <row Id="148" PostTypeId="2" ParentId="118" CreationDate="2010-07-20T21:55:14.763" Score="4" ViewCount="" Body="&lt;p&gt;One of the greatest achievements of human civilization is Newton's laws of motions.  The first law says that unless a force is acting then the &lt;em&gt;velocity&lt;/em&gt; (not the position!) of objects stay constant, while the second law says that forces act by causing an &lt;em&gt;acceleration&lt;/em&gt; (though heavy objects require more force to accellerate).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However to make sense of those laws and to apply them to real life you need to understand how to move between the following three notions:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Position&lt;/li&gt;&#xA;&lt;li&gt;Velocity (that is the rate of change in position)&lt;/li&gt;&#xA;&lt;li&gt;Acceleration (that is the rate of change of the velocity)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Moving down that list is called &quot;taking the derivative&quot; while moving up that list is called &quot;taking the integral.&quot;  Calculus is the study of derivatives and integerals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, if you want to figure out how objects move under some force you need to be able to integrate twice.  This requires understanding a lot of calculus!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a first semester class you usually learn about derivatives and integrals of functions of one variable, that is what you need to understand physics in 1-dimension!  To understand the actual physics of the world you need to understand derivatives and integrals in 3-dimensions which requires several courses.&lt;/p&gt;&#xA;" OwnerUserId="8" LastActivityDate="2010-07-20T21:55:14.763" />
  <row Id="149" PostTypeId="2" ParentId="120" CreationDate="2010-07-20T21:59:54.497" Score="17" ViewCount="" Body="&lt;p&gt;The main point of talking about prime numbers is Euclid's theorem that every positive integer can be written uniquely as a product of primes.  As Justin remarks, this would break horribly if 1 were considered prime, for example we could factor 2 as 2*1*1*1*1*1.  Instead we say that 1 is not a prime, but it is the &lt;em&gt;product of zero primes&lt;/em&gt; (see &lt;a href=&quot;http://math.stackexchange.com/questions/135/why-is-x0-1-except-when-x-0&quot; rel=&quot;nofollow&quot;&gt;this question&lt;/a&gt; to understand why any prime multiplied by itself 0 times is 1) so Euclid's theorem works out nicely!&lt;/p&gt;&#xA;" OwnerUserId="8" LastActivityDate="2010-07-20T21:59:54.497" CommentCount="1" />
  <row Id="150" PostTypeId="1" AcceptedAnswerId="151" CreationDate="2010-07-20T22:04:21.640" Score="6" ViewCount="494" Body="&lt;p&gt;It seems like functions that are continuous always seem to be differentiable, to me.  I can't imagine one that is not.  Are there any examples of functions that are continuous, yet not differentiable?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other way around seems a bit simpler -- a differentiable function is obviously always going to be continuous.  But are there any that do not satisfy this?&lt;/p&gt;&#xA;" OwnerUserId="51" LastEditorUserId="126" LastEditorDisplayName="" LastEditDate="2010-07-21T04:42:28.373" LastActivityDate="2011-02-23T06:27:37.147" Title="Are there any functions that are (always) continuous yet not differentiable? Or vice-versa?" Tags="&lt;real-analysis&gt;" AnswerCount="4" CommentCount="1" FavoriteCount="0" />
  <row Id="151" PostTypeId="2" ParentId="150" CreationDate="2010-07-20T22:07:02.073" Score="12" ViewCount="" Body="&lt;p&gt;It's easy to find a function which is continuous but not differentiable at a single point, e.g. f(x) = |x| is continuous but not differentiable at 0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moreover, there are functions which are continuous but nowhere differentiable, such as the &lt;a href=&quot;http://en.wikipedia.org/wiki/Nowhere_differentiable&quot; rel=&quot;nofollow&quot;&gt;Weierstrass function&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, continuity follows from differentiability, so there are no differentiable functions which aren't also continuous. If a function is differentiable at $x$, then the limit $(f(x+h)-f(x))/h$ must exist (and be finite) as $h$ tends to 0, which means $f(x+h)$ must tend to $f(x)$ as $h$ tends to 0, which means $f$ is continuous at $x$.&lt;/p&gt;&#xA;" OwnerUserId="31" LastEditorUserId="31" LastEditorDisplayName="" LastEditDate="2010-07-20T22:22:36.287" LastActivityDate="2010-07-20T22:22:36.287" CommentCount="5" />
  <row Id="152" PostTypeId="2" ParentId="71" CreationDate="2010-07-20T22:08:08.390" Score="3" ViewCount="" Body="&lt;p&gt;To understand the benefits of using quaternions you have to consider different ways to represent rotations. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are few ways with a summary of the pros and cons:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Euler angles&lt;/li&gt;&#xA;&lt;li&gt;Rotation matrices&lt;/li&gt;&#xA;&lt;li&gt;Axis angle&lt;/li&gt;&#xA;&lt;li&gt;Quaternions  &lt;/li&gt;&#xA;&lt;li&gt;Rotors (normalized Spinors)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Euler angles are the best choice if you want a user to specify an orientation in a intuitive way. They are are also space efficient (three numbers). However, it is more difficult to linear interpolate values. Consider the case where you want to interpolate between 359 and 0 degrees. Linearly interpolating would cause a large rotation, even though the two orientations are almost the same. Writing shortest path interpolation, is easy for one axis, but non-trivial when considering the three Euler angles(for instance the shortest route between (240, 57, 145) and (35, -233, -270) is not immediately clear).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rotation matrices specify a new frame of reference using three normalized and orthogonal vectors (Right, Up, Out, which when multiplied become the new x, y, z). Rotation matrices are useful for operations like strafing (side ways movement), which only requires translating along the Right vector of the camera's rotation matrix. However, there is no clear method of interpolating between them. The are also expensive to normalize which is necessary to prevent scaling from being introduced.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Axis angle, as the name suggests, are a way of specifying a rotation axis and angle to rotate around that axis. You can think of Euler angles, as three axis angle rotations, where the axises are the x, y, z axis respectively. Linearly interpolating the angle in a axis angle is pretty straight forward (if you remember to take the shortest path), however linearly interpolating between different axises is not. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Quaternions are a way of specifying a rotation through a axis and the cosine of half the angle. They main advantage is I can pick any two quaternions and smoothly interpolate between them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rotors are another way to perform rotations. Rotors are basically quaternions, but instead of thinking of them as 4D complex numbers, rotors are thought of as real 3D multivectors. This makes their visualization much more understandable (compared to quaternions), but requires fluency in geometric algebra to grasp their significance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Okay with that as the background I can discuss a real world example. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Say you are writing a computer game&#xA;  where the characters are animated in&#xA;  3ds Max. You need to export a&#xA;  animation of the character to play in&#xA;  your game, but cannot faithfully&#xA;  represent the interpolation used by&#xA;  the animation program, and thus have&#xA;  to sample. The animation is going to&#xA;  be represented as a list of rotations&#xA;  for each joint. How should we store&#xA;  the rotations? &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;If I am going to sample every frame,&#xA;  not interpolate, and space is not an&#xA;  issue, I would probably store the&#xA;  rotations as rotation matrices. If&#xA;  space was issue, then Euler angles.&#xA;  That would also let me do things like&#xA;  only store one angle for joints like&#xA;  the knee that have only one degree of&#xA;  freedom.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;If I only sampled every 4 frames, and&#xA;  need to interpolate it depends on&#xA;  whether I am sure the frame-rate will&#xA;  hold. If I am positive that the&#xA;  frame-rate will hold I can use axis&#xA;  angle relative rotations to perform&#xA;  the interpolation. This is atypical.&#xA;  In most games the frame rate can drop&#xA;  past my sampling interval, which would&#xA;  require skipping an element in the&#xA;  list to maintain the correct playback&#xA;  speed. If I am unsure of what two&#xA;  orientations I need to interpolate&#xA;  between, then I would use quaternions&#xA;  or rotors.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="99" LastEditorUserId="99" LastEditorDisplayName="" LastEditDate="2010-07-20T23:34:33.507" LastActivityDate="2010-07-20T23:34:33.507" />
  <row Id="153" PostTypeId="1" AcceptedAnswerId="277" CreationDate="2010-07-20T22:17:25.933" Score="12" ViewCount="1314" Body="&lt;p&gt;I have a non-professional interest in math and I would like to be able to be curious and play around with some math that I'm learning about (both here and from calculus-and-higher level math classes).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mathematica and MatLab seem like the best ways to play with math. I've seen demos of both and they both seem exceptionally powerful and complex. I'd like to start learning more about the syntax of one or the other so I can develop ideas and evaluate them, but I don't know which is better suited for me. Which one should I use?&lt;/p&gt;&#xA;" OwnerUserId="82" LastEditorUserId="82" LastEditorDisplayName="" LastEditDate="2010-07-21T17:14:33.037" LastActivityDate="2010-09-13T18:56:55.120" Title="What should I learn first, Mathematica or MatLab?" Tags="&lt;math-software&gt;&lt;mathematica&gt;&lt;matlab&gt;" AnswerCount="8" CommentCount="6" FavoriteCount="7" />
  <row Id="154" PostTypeId="1" AcceptedAnswerId="168" CreationDate="2010-07-20T22:17:36.010" Score="95" ViewCount="8071" Body="&lt;p&gt;Complex numbers involve the square root of negative one, and most non-mathematicians find it hard to accept that such a number is meaningful. In contrast, they feel that real numbers have an obvious and intuitive meaning. What's the best way to explain &lt;strong&gt;to a non-mathematician&lt;/strong&gt; that complex numbers are necessary and meaningful, in the same way that real numbers are?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is not a Platonic question about the reality of mathematics, or whether abstractions are as real as physical entities, but an attempt to bridge a comprehension gap that many people experience when encountering complex numbers for the first time. The wording, although provocative, is deliberately designed to match the way that many people do in fact ask this question.&lt;/p&gt;&#xA;" OwnerUserId="113" LastActivityDate="2011-04-15T20:45:29.080" CommunityOwnedDate="2011-04-15T20:45:29.080" Title="Do complex numbers really exist?" Tags="&lt;soft-question&gt;&lt;teaching&gt;&lt;philosophy&gt;" AnswerCount="31" CommentCount="29" FavoriteCount="41" />
  <row Id="155" PostTypeId="1" AcceptedAnswerId="163" CreationDate="2010-07-20T22:17:38.477" Score="23" ViewCount="1024" Body="&lt;p&gt;I've come across statements in the past along the lines of &quot;function f has no closed form integral&quot;, which I assume means that there is no combination of the operations:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;addition/subtraction&lt;/li&gt;&#xA;&lt;li&gt;multiplication/division&lt;/li&gt;&#xA;&lt;li&gt;raising to powers and roots&lt;/li&gt;&#xA;&lt;li&gt;trigonometric functions&lt;/li&gt;&#xA;&lt;li&gt;exponential functions&lt;/li&gt;&#xA;&lt;li&gt;logarithmic functions&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;which when differentiated gives the function f. I've heard this said about the function f(x) = x^x for example.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What sort of techniques are used to prove statements like this? What is this branch of mathematics called?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Merged with &quot;&lt;a href=&quot;http://math.stackexchange.com/questions/2328/&quot;&gt;How to prove that some functions don’t have a primitive&lt;/a&gt;&quot; by &lt;a href=&quot;http://math.stackexchange.com/users/918/ismael&quot;&gt;Ismael&lt;/a&gt;:&lt;br&gt;&#xA;Sometimes we are told that some functions like $\sin(x)/x$ don't have a indefinite integral, or that it can't be expressed in term of other simple functions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I do wonder how we can prove that kind of assertions.&lt;/p&gt;&#xA;" OwnerUserId="31" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-09-29T18:41:54.267" LastActivityDate="2011-04-24T09:46:53.990" Title="How can you prove that a function has no closed form integral?" Tags="&lt;calculus&gt;&lt;real-analysis&gt;&lt;differential-algebra&gt;" AnswerCount="6" CommentCount="4" FavoriteCount="17" />
  <row Id="156" PostTypeId="2" ParentId="154" CreationDate="2010-07-20T22:21:34.883" Score="8" ViewCount="" Body="&lt;p&gt;Quantum mechanics, and hence physics and everything around us, fundamentally involves complex numbers.&lt;/p&gt;&#xA;" OwnerUserId="8" LastActivityDate="2010-07-20T22:21:34.883" CommunityOwnedDate="2011-04-15T20:45:29.080" CommentCount="2" />
  <row Id="157" PostTypeId="2" ParentId="154" CreationDate="2010-07-20T22:21:45.087" Score="3" ViewCount="" Body="&lt;p&gt;The argument isn't worth having, as you disagree about what it means for something to 'exist'. There are many interesting mathematical objects which don't have an obvious physical counterpart. What does it mean for the Monster group to exist?&lt;/p&gt;&#xA;" OwnerUserId="31" LastActivityDate="2010-07-20T22:21:45.087" CommunityOwnedDate="2011-04-15T20:45:29.080" CommentCount="3" />
  <row Id="158" PostTypeId="2" ParentId="153" CreationDate="2010-07-20T22:22:04.197" Score="5" ViewCount="" Body="&lt;p&gt;Conventional wisdom is that Matlab is the natural choice for numerical computation, whereas Mathematica is better for symbolic calculation and manipulation. Both can do both, and it's probably a matter of taste and the convention in the fields you're interested in. Matlab is much more widely used across the sciences and engineering, so if you've any interest in that direction, it might be more helpful for you to know.&lt;/p&gt;&#xA;" OwnerUserId="76" LastActivityDate="2010-07-20T22:22:04.197" />
  <row Id="159" PostTypeId="2" ParentId="154" CreationDate="2010-07-20T22:23:02.310" Score="7" ViewCount="" Body="&lt;p&gt;I'll start by pointing out that a whole host of things that people think of as 'real' are on shakier ground than imaginary numbers. Given that quantum mechanics predicts a fundamental limit to how granular reality is, the whole concept of real numbers is on very shakey ground, yet people accept those as fine. I'd therefore suggest that it is merely a case of familiarity - people are less familiar with complex numbers than with some other mathematical constructs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for an actual existence outside the realms of pure maths... your best bet is to look at quantum mechanics again. This area has some fascinating results that are only possible through the use of imaginary numbers. Incidentally, fundamental particles are the place in nature that gave a 'physicality' to negative numbers (the charge of an electron is negative) well after they were accepted as normal by most people. &lt;/p&gt;&#xA;" OwnerUserId="35" LastActivityDate="2010-07-20T22:23:02.310" CommunityOwnedDate="2011-04-15T20:45:29.080" CommentCount="9" />
  <row Id="160" PostTypeId="2" ParentId="154" CreationDate="2010-07-20T22:24:43.197" Score="55" ViewCount="" Body="&lt;p&gt;The concept of mathematical numbers and &quot;existing&quot; is a tricky one.  What actually &quot;exists&quot;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do negative numbers exist?  Of course they do not.  You can't have a negative amount of apples.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yet, the beauty of negative numbers is that when we define them (rigorously), then all of a sudden we can use them to solve problems we were never ever been able to solve before, or to solve them in a much simpler way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine trying to do simple physics without the idea of negative numbers!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But are they &quot;real&quot;?  Do they &quot;exist&quot;?  No, they don't.  But they are just tools that help us solve real life problems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To go back to your question about complex numbers, I would say that the idea that they exist or not has no bearing onto whether or not they are actually useful in solving the problem of every day life, or making them many, many, many times more easy to solve.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The math that makes your computer run and involves the tool that is complex numbers, for instance.&lt;/p&gt;&#xA;" OwnerUserId="51" LastActivityDate="2010-07-20T22:24:43.197" CommunityOwnedDate="2011-04-15T20:45:29.080" CommentCount="7" />
  <row Id="161" PostTypeId="2" ParentId="2" CreationDate="2010-07-20T22:25:54.740" Score="-3" ViewCount="" Body="&lt;p&gt;Conceptually, to me at least, a limit is a way approximate something, by realizing the long run behavior is predictable and more important then the short term behavior. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's say your baby sitting a rambunctious kid. You could frustrate yourself trying to quite him, or just realize, in the limit he will tire himself out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A lot of systems seem complicated in the short term, but are simple when viewed in the long term. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The limit is way to take advantage of that, and relate a complex system to a simpler approximation. &lt;/p&gt;&#xA;" OwnerUserId="99" LastEditorUserId="99" LastEditorDisplayName="" LastEditDate="2010-07-20T23:38:39.223" LastActivityDate="2010-07-20T23:38:39.223" CommentCount="2" />
  <row Id="162" PostTypeId="1" AcceptedAnswerId="171" CreationDate="2010-07-20T22:34:36.637" Score="11" ViewCount="795" Body="&lt;p&gt;I've heard of some other paradoxes involving sets (ie, &quot;the set of all sets that do not contain themselves&quot;) and I understand how paradoxes arise from them.  But this one I do not understand.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is &quot;the set of all sets&quot; a paradox?  It seems like it would be fine, to me.  There is nothing paradoxical about a set containing itself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it something that arises from the &quot;rules of sets&quot; that are involved in more rigorous set theory?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: I don't really have much of a background with Set Theory, so an answer that deals with common english would be much appreciated =)&lt;/p&gt;&#xA;" OwnerUserId="51" LastEditorUserId="100" LastEditorDisplayName="" LastEditDate="2010-07-21T11:06:27.840" LastActivityDate="2010-08-19T19:57:08.697" Title="Why is &quot;the set of all sets&quot; a paradox?" Tags="&lt;paradox&gt;&lt;logic&gt;&lt;set-theory&gt;" AnswerCount="7" CommentCount="1" FavoriteCount="5" />
  <row Id="163" PostTypeId="2" ParentId="155" CreationDate="2010-07-20T22:34:58.367" Score="16" ViewCount="" Body="&lt;p&gt;It is a theorem of Liouville, reproven later with purely algebraic methods, that for rational functions f and g, g nonconstant, the antiderivative&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int [f(x)\exp(g(x))] \mathrm dx$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;can be expressed in terms of elementary functions if and only if there exists some rational function h such that it is a solution to the differential equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f = h&amp;#39; + hg$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\int e^{x^2} dx$ is another classic example of such a function with no elementary antiderivative.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know how much math you've had, but some of this paper might be comprehensible in its broad strokes: &lt;a href=&quot;http://www.sci.ccny.cuny.edu/~ksda/PostedPapers/liouv06.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.sci.ccny.cuny.edu/~ksda/PostedPapers/liouv06.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Liouville's original paper: Liouville, J. &quot;&lt;em&gt;Sur la classification des Transcendantes et sur l'impossibilité d'exprimer les racines des certaines équations en fonction finie explicite des coefficients&lt;/em&gt;. Part 2.&quot; J. Math. Pure Appl. 3, 523-547, 1838. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Michael Spivak's book on Calculus also has a section with a discussion of this.&lt;/p&gt;&#xA;" OwnerUserId="76" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-25T23:10:30.527" LastActivityDate="2010-11-25T23:10:30.527" CommentCount="1" />
  <row Id="164" PostTypeId="1" AcceptedAnswerId="181" CreationDate="2010-07-20T22:36:56.010" Score="25" ViewCount="1158" Body="&lt;p&gt;I learned that the volume of a sphere is $\frac{4}{3}\pi r^3$, but why? The $\pi$ kind of makes sense because its round like a circle, and the $r^3$ because it's 3-D, but $\frac{4}{3}$ is so random! How could somebody guess something like this for the formula?&lt;/p&gt;&#xA;" OwnerUserId="73" LastActivityDate="2011-03-27T12:04:31.650" Title="Why is the volume of a sphere $\frac{4}{3}\pi r^3$?" Tags="&lt;geometry&gt;" AnswerCount="10" CommentCount="3" FavoriteCount="8" />
  <row Id="165" PostTypeId="2" ParentId="153" CreationDate="2010-07-20T22:37:23.013" Score="1" ViewCount="" Body="&lt;p&gt;MatLab is a numerical computing environment. Mathematica is about symbolic manipulations. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mathematica helped me with calculus and some number theory problems. Almost everything I need is built into Mathematica, i can't say the same for MatLab. The math professors in my school uses maple(closer to mathematica than matlab). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can have the good of both worlds, there are mathematica toolbox for matlab.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you find the problems on &lt;a href=&quot;http://projecteuler.net/&quot; rel=&quot;nofollow&quot;&gt;project euler&lt;/a&gt; interesting, mathematica are something you want to look into. &lt;/p&gt;&#xA;" OwnerUserId="96" LastActivityDate="2010-07-20T22:37:23.013" CommentCount="3" />
  <row Id="166" PostTypeId="2" ParentId="97" CreationDate="2010-07-20T22:38:47.287" Score="4" ViewCount="" Body="&lt;p&gt;You can use a Taylor polynomial of sufficient degree to guarantee the accuracy that you need.  (The Taylor series for erf(x) is given on the Wikipedia page to which you linked.)  The &lt;a href=&quot;http://mathworld.wolfram.com/LagrangeRemainder.html&quot; rel=&quot;nofollow&quot;&gt;Lagrange Remainder term&lt;/a&gt; can be used to bound the error in the Taylor series approximation.&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-07-20T22:38:47.287" />
  <row Id="167" PostTypeId="2" ParentId="154" CreationDate="2010-07-20T22:40:39.890" Score="4" ViewCount="" Body="&lt;p&gt;There are geometric interpretations of imaginary numbers where they are thought of as parallelograms with a front and back, or oriented parallelograms. That interpretation requires geometric algebra but only uses real numbers. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a link:&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Geometric_algebra#Complex_numbers&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Geometric_algebra#Complex_numbers&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That doesn't have any pictures so it is admittedly not intuitive, but the answer is yes. Whether you think of imaginary numbers as square root of negative 1 or as parallelogram with a front and back, they exist.&lt;/p&gt;&#xA;" OwnerUserId="99" LastEditorUserId="99" LastEditorDisplayName="" LastEditDate="2010-07-20T22:52:52.997" LastActivityDate="2010-07-20T22:52:52.997" CommunityOwnedDate="2011-04-15T20:45:29.080" CommentCount="3" />
  <row Id="168" PostTypeId="2" ParentId="154" CreationDate="2010-07-20T22:42:01.277" Score="102" ViewCount="" Body="&lt;p&gt;There are a few good answers to this question, depending on the audience. I've used all of these on occasion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;A way to solve polynomials&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We came up with equations like &lt;code&gt;x - 5 = 0, what is x?&lt;/code&gt;, and the naturals solved them (easily). Then we asked, &quot;wait, what about &lt;code&gt;x + 5 = 0&lt;/code&gt;?&quot; So we invented negative numbers. Then we asked &quot;wait, what about &lt;code&gt;2x = 1&lt;/code&gt;?&quot; So we invented rational numbers. Then we asked &quot;wait, what about &lt;code&gt;x^2 = 2&lt;/code&gt;?&quot; so we invented irrational numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, we asked, &quot;wait, what about &lt;code&gt;x^2 = -1&lt;/code&gt;?&quot; This is the only question that was left, so we decided to invent the &quot;imaginary&quot; numbers to solve it. All the other numbers, at some point, didn't exist and didn't seem &quot;real&quot;, but now they're fine. Now that we have imaginary numbers, we can solve every polynomial, so it makes sense that that's the last place to stop.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Pairs of numbers&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This explanation goes the route of redefinition. Tell the listener to forget everything he knows about imaginary numbers. You're defining a new number system, only now there are always pairs of numbers. Why? For fun. Then go through explaining how addition/multiplication work. Try and find a good &quot;realistic&quot; use of pairs of numbers (many exist).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, show that in this system,  &lt;code&gt;(0,1) * (0,1) = (-1,0)&lt;/code&gt;, in other words, we've defined a new system, under which it makes sense to say that &lt;code&gt;sqrt(-1) = i&lt;/code&gt;, when &lt;code&gt;i=(0,1)&lt;/code&gt;. And that's really all there is to imaginary numbers: a definition of a new number system, which makes sense to use in most places. And under that system, there &lt;strong&gt;is&lt;/strong&gt; an answer to &lt;code&gt;sqrt(-1)&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The historical explanation&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Explain the history of the imaginary numbers. Showing that mathematicians &lt;em&gt;also&lt;/em&gt; fought against them for a long time helps people understand the mathematical process, i.e., that it's all definitions in the end.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm a little rusty, but I think there were certain equations that kept having parts of them which used &lt;code&gt;sqrt(-1)&lt;/code&gt;, and the mathematicians kept  throwing out the equations since there is no such thing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, one mathematician decided to just &quot;roll with it&quot;, and kept working, and found out that all those square roots cancelled each other out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Amazingly, the answer that was left was the &lt;strong&gt;correct answer&lt;/strong&gt; (he was working on finding roots of polynomials, I think). Which lead him to think that there &lt;em&gt;was&lt;/em&gt; a valid reason to use &lt;code&gt;sqrt(-1)&lt;/code&gt;, even if it took a long time to understand it.&lt;/p&gt;&#xA;" OwnerUserId="122" LastActivityDate="2010-07-20T22:42:01.277" CommunityOwnedDate="2011-04-15T20:45:29.080" CommentCount="9" />
  <row Id="169" PostTypeId="2" ParentId="164" CreationDate="2010-07-20T22:45:12.100" Score="6" ViewCount="" Body="&lt;p&gt;This was first &quot;guessed at&quot; by Archimedes, by what fraction of the volume of a cylinder a sphere is.  That is, a sphere that is contained within that cylinder.  From that, he figured out 4/3.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can do it these days with the tools of calculus.  One way would to use the &lt;a href=&quot;http://library.thinkquest.org/3616/Calc/S3/TDM.html&quot; rel=&quot;nofollow&quot;&gt;Disk Method&lt;/a&gt;, over the graph of a semicircle.  Another would be to use spherical coordinates and calculate the 3D Integral.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All of these methods should give you 4/3rds pi * r^3&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that it's not just r^3 because it's &quot;3D&quot;; but also because it's the only way the dimensional analysis would work; that is, the units would have to be in distance^3.  But that might be what you said, but just clarifying.&lt;/p&gt;&#xA;" OwnerUserId="51" LastActivityDate="2010-07-20T22:45:12.100" />
  <row Id="170" PostTypeId="2" ParentId="120" CreationDate="2010-07-20T22:46:09.323" Score="13" ViewCount="" Body="&lt;p&gt;It's important to understand that this is not something that can be &lt;strong&gt;proved&lt;/strong&gt;: it's a &lt;strong&gt;definition&lt;/strong&gt;. We &lt;em&gt;choose&lt;/em&gt; not to regard 1 as a prime number, simply because it makes writing lots of theorems much easier.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Noah gives the best example in his answer: Euclid's theorem that every positive integer can be written uniquely as a product of primes. If 1 &lt;em&gt;is&lt;/em&gt; defined to be a prime number, then we'd have to change that theorem to: &quot;every positive integer can be written uniquely as a product of primes, except for infinite multiplications by 1&quot;. So we choose to go with the easier path of defining 1 to not be a prime.&lt;/p&gt;&#xA;" OwnerUserId="122" LastActivityDate="2010-07-20T22:46:09.323" />
  <row Id="171" PostTypeId="2" ParentId="162" CreationDate="2010-07-20T22:47:54.220" Score="19" ViewCount="" Body="&lt;p&gt;Let |S| be the cardinality of S&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$|S| &amp;lt; |2^S|$ can be proven with &lt;a href=&quot;http://en.wikipedia.org/wiki/Cantor%27s_diagonal_argument&quot; rel=&quot;nofollow&quot;&gt;generalized Cantor's diagonal argument&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proof of the set of all sets doesn't exist:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let S be the set of all sets,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$|S| &amp;lt; |2^S|$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but $2^S$ is a subset of S, because every set in $2^S$ is in S. Therefore&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$|2^S| \leq |S|$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A contradiction. Therefore the set of all sets doesn't exist.&lt;/p&gt;&#xA;" OwnerUserId="96" LastEditorUserId="123" LastEditorDisplayName="" LastEditDate="2010-08-04T03:37:04.360" LastActivityDate="2010-08-04T03:37:04.360" CommentCount="9" />
  <row Id="173" PostTypeId="2" ParentId="154" CreationDate="2010-07-20T22:56:03.757" Score="9" ViewCount="" Body="&lt;p&gt;You may be interested to read the MathOverflow question &quot;Demystifying Complex Numbers,&quot; &lt;a href=&quot;http://mathoverflow.net/questions/30156/demystifying-complex-numbers&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;. A teacher is asking how to motivate complex numbers to students taking complex analysis.&lt;/p&gt;&#xA;" OwnerUserId="76" LastActivityDate="2010-07-20T22:56:03.757" CommunityOwnedDate="2011-04-15T20:45:29.080" CommentCount="1" />
  <row Id="174" PostTypeId="2" ParentId="164" CreationDate="2010-07-20T22:57:01.727" Score="5" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Pappus%27s_centroid_theorem&quot; rel=&quot;nofollow&quot;&gt;Pappus's centroid theorem&lt;/a&gt; (second theorem) says that the volume of a solid formed by revolving a region about an axis is the product of the area of the region and the distance traveled by the centroid of the region when it is revolved.  A sphere can be formed by revolving a semicircle about is diameter edge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The area of the semicircle is 1/2 &amp;pi; r^2.  The centroid of the semicircle can be found by intersecting two lines that both divide the area of the semicircle into two equal parts.  One such line is perpendicular to the diameter edge through the center of the semicircle (this is a line of symmetry of the semicircle).  Another such line is parallel to the diameter edge, (4r)/(3&amp;pi;) away from it (verification of this is left as an exercise for the reader).  When revolved about the diameter edge of the semicircle, the centroid travels 2&amp;pi; * (4r)/(3&amp;pi;) = 8/3 * r, so the volume of the sphere is 1/2 &amp;pi; r^2 * 8/3 * r = 4/3 &amp;pi; r^3.&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-07-20T22:57:01.727" />
  <row Id="175" PostTypeId="2" ParentId="118" CreationDate="2010-07-20T22:58:42.847" Score="2" ViewCount="" Body="&lt;p&gt;Calculus is the mathematics of change. In algebra, almost nothing ever changes. Here's a comparison of some algebra vs. calc problems:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;algebra: car A is driving at 50 kph. How far has it gone after 6 hours?&#xA;calc: car B starts at 10 mph and begins accelerating at the rate of 10 kph^2&#xA;(kilometers per hour per hour). How far has car B gone after 6 hours?&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Note how the algebra problem nothing changes, where in the calc problem, the speed of the car is constantly changing.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;calc: If a ball is rolling in a straight line at 10 fps with a diameter of 1 foot&#xA;and Q is a the point at the top of the ball when t=0, how fast is point Q moving &#xA;at time t=4 relative to the ground?&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The speed of the point in relation to the ground is never the same (its zero when its at the bottom, 20fps when it's at the top. Calculus lets you figure out how fast it's going exactly at a specific moment. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are two main branches of calculus, differential and integral. These problems pertain to differential calculus as they concern how something is changing. Integral calculus deals with how much something has changed, the opposite of differential calculus. &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;The number of cubic feet of oxygen circulated by Jeff lungs per hour at time t &#xA;follows the equation f(t) = t^3 + sin(t) How many cubic feet of oxygen do &#xA;Jeff's lungs cycle per day?&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;To find out how much something has changed when its rate of change isn't constant requires integral calculus. (the equation is purely hypothetical unless Jeff happens to be the size of a beluga whale).&lt;/p&gt;&#xA;" OwnerUserId="104" LastActivityDate="2010-07-20T22:58:42.847" CommunityOwnedDate="2010-07-20T22:58:42.847" />
  <row Id="176" PostTypeId="2" ParentId="139" CreationDate="2010-07-20T23:09:57.103" Score="9" ViewCount="" Body="&lt;p&gt;Does this give you any ideas?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://farm5.static.flickr.com/4119/4813793520_0d7fa53b99_o.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="120" LastActivityDate="2010-07-20T23:09:57.103" CommentCount="3" />
  <row Id="177" PostTypeId="1" CreationDate="2010-07-20T23:10:36.897" Score="3" ViewCount="308" Body="&lt;p&gt;Consider the task of generating random points uniformly distributed within a circle of a given radius $r$ that is centered at the origin. Assume that we are given a random number generator $R$ that generates a floating point number uniformly distributed in the range $[0, 1)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the following procedure:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Generate a random point $p = (x, y)$ within a square of side $2r$ centered at the origin. This can be easily achieved by:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a. Using the random number generator $R$ to generate two random numbers $x$ and $y$, where $x, y \in [0, 1)$, and then transforming $x$ and $y$ to the range $[0, r)$ (by multiplying each by $r$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;b. Flipping a fair coin to decide whether to reflect $p$ around the $x$-axis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;c. Flipping another fair coin to decide whether to reflect $p$ around the $y$-axis.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Now, if $p$ happens to fall outside the given circle, discard $p$ and generate another point. Repeat the procedure until $p$ falls within the circle.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Is the previous procedure correct? That is, are the random points generated by it &lt;strong&gt;uniformly distributed&lt;/strong&gt; within the given circle? How can one formally [dis]prove it?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Background Info&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The task was actually given in &lt;a href=&quot;http://rubyquiz.strd6.com/quizzes/234-random-points-within-a-circle&quot; rel=&quot;nofollow&quot;&gt;Ruby Quiz - Random Points within a Circle (#234)&lt;/a&gt;. If you're interested, you can &lt;a href=&quot;http://gist.github.com/447554&quot; rel=&quot;nofollow&quot;&gt;check my solution&lt;/a&gt; in which I've implemented the procedure described above. I would like to know whether the procedure is mathematically correct or not, but I couldn't figure out how to formally [dis]prove it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the actual task was to generate random points uniformly distributed within a circle of a given radius &lt;em&gt;and position&lt;/em&gt;, but I intentionally left that out in the question because the generated points can be easily translated to their correct positions relative to the given center.&lt;/p&gt;&#xA;" OwnerUserId="114" LastEditorUserId="126" LastEditorDisplayName="" LastEditDate="2010-07-21T01:42:00.333" LastActivityDate="2010-07-21T17:39:38.927" Title="Will this Procedure Generate Random Points Uniformly Distributed within a Given Circle? Proof?" Tags="&lt;algorithms&gt;&lt;probability-theory&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="178" PostTypeId="2" ParentId="154" CreationDate="2010-07-20T23:11:01.500" Score="1" ViewCount="" Body="&lt;p&gt;We will will first consider the most common definition of i, as the square root of -1. When you first hear this, it sounds crazy. 0 squared is 0; a positive times a positive is positive and a negative times a negative is positive too. So there doesn't actually appear to be any number that we can square to get -1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A mathematician would collectively term 0, negative numbers and positive numbers as the real numbers. They would also define the term complex numbers as a group of numbers that includes these real numbers. So while we have shown that no real number can square to get -1, we haven't even defined complex numbers at this point, so we can't rule out that one might have this property.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At this point, it makes sense to ask what does a mathematician mean by a number? It certainly isn't what most people associate it with - as an abstract representation some kind of real world quantity. We need to understand that it isn't uncommon for one word to have different meanings for different groups of people - after all words mean whatever we make them mean. Most people only need to real world quantities, so they find it convenient to call those numbers. On the other hand, mathematicians explore a variety of different number systems. Indeed some, such as complex numbers, are useful for solving problems that are actually about real numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So mathematicians define i as a number that obeys most of the normal algebraic laws. They also defined i*i to equal -1. From this we can derive all of the standard results about &lt;a href=&quot;http://en.wikipedia.org/wiki/Complex_number&quot; rel=&quot;nofollow&quot;&gt;complex number&lt;/a&gt;s.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for whether they are real - it depends on what you want to know. Obviously, they don't correspond to quantities of physical objects. On the other hand, complex numbers can useful for representing &lt;a href=&quot;http://en.wikipedia.org/wiki/Electrical_impedance&quot; rel=&quot;nofollow&quot;&gt;resistance in an electric circuit&lt;/a&gt;. Ultimately, they are an idea and while ideas don't exist physically, saying they don't exist at all is inaccurate.&lt;/p&gt;&#xA;" OwnerUserId="123" LastActivityDate="2010-07-20T23:11:01.500" CommunityOwnedDate="2011-04-15T20:45:29.080" CommentCount="1" />
  <row Id="179" PostTypeId="2" ParentId="162" CreationDate="2010-07-20T23:13:32.773" Score="2" ViewCount="" Body="&lt;p&gt;An informal explanation is Russel's Paradox. The &lt;a href=&quot;http://en.wikipedia.org/wiki/Russel%27s_paradox&quot; rel=&quot;nofollow&quot;&gt;wiki page&lt;/a&gt; is informative, here's the relevant quote:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Let us call a set &quot;abnormal&quot; if it is&#xA;  a member of itself, and &quot;normal&quot;&#xA;  otherwise. For example, take the set&#xA;  of all squares. That set is not itself&#xA;  a square, and therefore is not a&#xA;  member of the set of all squares. So&#xA;  it is &quot;normal&quot;. On the other hand, if&#xA;  we take the complementary set that&#xA;  contains all non-squares, that set is&#xA;  itself not a square and so should be&#xA;  one of its own members. It is&#xA;  &quot;abnormal&quot;. &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Now we consider the set of&#xA;  all normal sets, R. Attempting to&#xA;  determine whether R is normal or&#xA;  abnormal is impossible: If R were a&#xA;  normal set, it would be contained in&#xA;  the set of normal sets (itself), and&#xA;  therefore be abnormal; and if it were&#xA;  abnormal, it would not be contained in&#xA;  the set of normal sets (itself), and&#xA;  therefore be normal. This leads to the&#xA;  conclusion that R is both normal and&#xA;  abnormal: Russell's paradox.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="122" LastActivityDate="2010-07-20T23:13:32.773" CommentCount="3" />
  <row Id="180" PostTypeId="2" ParentId="75" CreationDate="2010-07-20T23:14:16.800" Score="4" ViewCount="" Body="&lt;p&gt;I won't explain what a ring or a group is, because that's already been done, but I'll add something else.  One reason groups and rings feel similar is that they are both &quot;algebraic structures&quot; in the sense of universal algebra. So for instance, the operation of quotienting via a normal subgroup (for a group) and a two-sided ideal (for a ring) are basically instances of quotienting via an invariant equivalence relation in universal algebra.  A field, by contrast, is not really a construction of universal algebra (because the operation $x \to x^{-1}$ is not everywhere defined) -- which is why free fields don't exist, for instance -- though they are a special case of rings.  &lt;/p&gt;&#xA;" OwnerUserId="536" LastActivityDate="2010-07-20T23:14:16.800" />
  <row Id="181" PostTypeId="2" ParentId="164" CreationDate="2010-07-20T23:15:55.393" Score="27" ViewCount="" Body="&lt;p&gt;In addition to the methods of calculus, Pappus, and Archimedes already mentioned, &lt;a href=&quot;http://en.wikipedia.org/wiki/Cavalieri%27s_principle&quot; rel=&quot;nofollow&quot;&gt;Cavalieri's Principle&lt;/a&gt; can be useful for these kinds of problems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose you have two solid figures lined up next to each other, each fitting between the same two parallel planes. (E.g., two stacks of pennies lying on the table, of the same height). Then, consider cutting the two solids by a plane parallel to the given two and in between them. If the cross-sectional area thus formed is the same for each of the solids for any such plane, the volumes of the solids are the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're willing to accept that you know the volume of a cone is 1/3 that of the cylinder with the same base and height, you can use Cavalieri, comparing a hemisphere to a cylinder with an inscribed cone, to get the volume of the sphere. This diagram (from Wikipedia) illustrates the construction: &lt;a href=&quot;http://upload.wikimedia.org/wikipedia/en/5/58/Sphere_cavalieri.svg&quot; rel=&quot;nofollow&quot;&gt;look here&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider a cylinder of radius R and height R, with, inside it, an inverted cone, with base of radius R coinciding with the top of the cylinder, and again height R. Put next to it a hemisphere of radius R. Now consider the cross section of each at height y above the base. For the cylinder/cone system, the area of the cross-section is pi*(R^2-y^2). It's the same for the hemisphere cross-section, as you can see by doing the Pythagorean theorem with any vector from the sphere center to a point on the sphere at height y to get the radius of the cross section (which is circular).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the cylinder/cone and hemisphere have the same height, by Cavalieri's Principle the volumes of the two are equal. The cylinder volume is Pi*R^3, the cone is a third that, so the hemisphere volume is (2/3)*pi*R^3. Thus the sphere of radius R has volume (4/3)*pi*R^3. &lt;/p&gt;&#xA;" OwnerUserId="76" LastEditorUserId="76" LastEditorDisplayName="" LastEditDate="2010-07-21T02:20:34.283" LastActivityDate="2010-07-21T02:20:34.283" CommentCount="2" />
  <row Id="183" PostTypeId="2" ParentId="136" CreationDate="2010-07-20T23:27:10.833" Score="9" ViewCount="" Body="&lt;p&gt;&lt;img src=&quot;http://www.education.vic.gov.au/images/content/studentlearning/mathscontinuum/SquareNos_sml.gif&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://math.stackexchange.com/questions/136/why-are-the-differences-between-consecutive-squares-equal-to-the-sequence-of-odd/143#143&quot; rel=&quot;nofollow&quot;&gt;Dan's&lt;/a&gt; algebraic justification is correct, but you may get more intuition about why this is happening from the above picture. Each time you want to enlarge the square by one unit, you have to add an extra row, an extra column, and one more square to fill in the corner. These correspond directly to the $n+n+1=2n+1$ that Dan mentioned. And of course, $2n+1$ is how odd numbers look.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Looking at it from the other direction, you can use the same idea to convince yourself of Noah's claim that $1+3+5+...+(2n-1) = n^2$. Imagine the left-hand side as representing the upper sequence of green and orange squares. You add on first 1 tile, then 3, then 5, and so on to construct larger and larger squares. At each step, you are adding one row (n) and one column (another n), then removing that one tile in the top right corner where they overlap (for a total of 2n-1). &lt;/p&gt;&#xA;" OwnerUserId="73" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-07-21T01:10:36.993" LastActivityDate="2010-07-21T01:10:36.993" />
  <row Id="184" PostTypeId="2" ParentId="83" CreationDate="2010-07-20T23:30:03.247" Score="2" ViewCount="" Body="&lt;p&gt;I have found most people liked math at some point, but something happened in their learning process that made them feel so stupid, they became disenfranchised with mathematics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What tends to happen is students are presented with some mathematically result they are expected to memorize by route, which takes all the joy out of mathematics and prevents them from approaching mathematics intuitively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I would first try to zero in on what they don't like and what parts of mathematics they have had to take on faith. You might not have to excite them if you help them learn mathematics intuitively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As per the suggestions, I would show them how mathematically equations make pretty shapes in Processing.&lt;/p&gt;&#xA;" OwnerUserId="99" LastEditorUserId="99" LastEditorDisplayName="" LastEditDate="2010-07-21T02:43:14.273" LastActivityDate="2010-07-21T02:43:14.273" CommunityOwnedDate="2010-08-07T22:32:44.343" />
  <row Id="185" PostTypeId="1" AcceptedAnswerId="193" CreationDate="2010-07-20T23:34:22.990" Score="1" ViewCount="66" Body="&lt;p&gt;I have a collection of 3D points in the standard x, y, z vector space. Now I pick one of the points &lt;code&gt;p&lt;/code&gt; as a new origin and two other points &lt;code&gt;a&lt;/code&gt; and &lt;code&gt;b&lt;/code&gt; such that &lt;code&gt;a - p&lt;/code&gt; and &lt;code&gt;b - p&lt;/code&gt; form two vectors of a new vector space. The third vector of the space I will call &lt;code&gt;x&lt;/code&gt; and calculate that as the cross product of the first two vectors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I would like to recast or reevaluate each of the points in my collection in terms of the new vector space. How do I do that?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Also, if 'recasting' not the right term here, please correct me.)&lt;/p&gt;&#xA;" OwnerUserId="124" LastEditorUserId="76" LastEditorDisplayName="" LastEditDate="2010-07-20T23:51:38.390" LastActivityDate="2010-07-20T23:51:38.390" Title="Recasting points from one vector space to another" Tags="&lt;linear-algebra&gt;&lt;vector-space&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="187" PostTypeId="1" AcceptedAnswerId="1492" CreationDate="2010-07-20T23:38:48.037" Score="4" ViewCount="197" Body="&lt;p&gt;Basic definitions: a tiling of d-dimensional Euclidean space is a decomposition of that space into polyhedra such that there is no overlap between their interiors, and every point in the space is contained in some one of the polyhedra.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A vertex-uniform tiling is a tiling such that each vertex figure is the same: each vertex is contained in the same number of k-faces, etc: the view of the tiling is the same from every vertex.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A vertex-transitive tiling is one such that for every two vertices in the tiling, there exists an element of the symmetry group taking one to the other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clearly all vertex-transitive tilings are vertex-uniform. For n=2, these notions coincide. However, Grunbaum, in his book on tilings, mentions but does not explain that for n &gt;= 3, there exist vertex uniform tilings that are not vertex transitive. Can someone provide an example of such a tiling, or a reference that explains this?&lt;/p&gt;&#xA;" OwnerUserId="76" LastActivityDate="2010-08-28T20:57:48.987" Title="Euclidean Tilings that are Uniform but not Vertex-Transitive" Tags="&lt;discrete-geometry&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="1" />
  <row Id="188" PostTypeId="2" ParentId="154" CreationDate="2010-07-20T23:39:39.317" Score="30" ViewCount="" Body="&lt;p&gt;No number does &quot;really exist&quot; the way trees or atoms exist.&#xA;In physics people however have found use for complex numbers just as they have found use for real numbers.&lt;/p&gt;&#xA;" OwnerUserId="2263" LastEditorUserId="2263" LastEditorDisplayName="" LastEditDate="2011-02-19T14:12:37.030" LastActivityDate="2011-02-19T14:12:37.030" CommunityOwnedDate="2011-04-15T20:45:29.080" CommentCount="10" />
  <row Id="190" PostTypeId="2" ParentId="11" CreationDate="2010-07-20T23:44:38.403" Score="0" ViewCount="" Body="&lt;p&gt;If you take two real numbers &lt;code&gt;x&lt;/code&gt; and &lt;code&gt;y&lt;/code&gt; then there per definition of the real number &lt;code&gt;z&lt;/code&gt; for which &lt;code&gt;x &amp;lt; z &amp;lt; y&lt;/code&gt; or &lt;code&gt;x &amp;gt; z &amp;gt; y&lt;/code&gt; is true.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For &lt;code&gt;x = 0.99999...&lt;/code&gt; and &lt;code&gt;y = 1&lt;/code&gt; you can't find a &lt;code&gt;z&lt;/code&gt; and therefore &lt;code&gt;0.99999... = 1&lt;/code&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2263" LastEditorUserId="2263" LastEditorDisplayName="" LastEditDate="2010-07-20T23:50:15.053" LastActivityDate="2010-07-20T23:50:15.053" CommunityOwnedDate="2010-07-28T06:53:43.987" />
  <row Id="191" PostTypeId="2" ParentId="177" CreationDate="2010-07-20T23:46:23.517" Score="0" ViewCount="" Body="&lt;p&gt;I'm going to use $P(a,b)$ for probability of x = a, and y = b.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$P(a,b) = P[x = |a|]P[y = |b|] * 1/4$ for any a, b inside the square.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore you are picking a coordinate uniformly inside the square.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the problem reduced to the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pick a element uniformly in a set A, then discard ones that are not B (B is a subset of A). Is it the same as pick a element uniformly in the set $A \cap B = B$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think this is true, I don't know if the following is a formal proof(haven't done probability in a while). please point out my mistakes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P(B) is the event that a element in B is picked. P(A) is the event that a element in A is picked.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pick a x uniformly, we have b \in B is P(x=b|B), it's obviously uniform.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The conditional probability P(B|A) = P(A \cap B)/P(A) = P(B)/P(A) = P(B) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;P(A) = 1, w/e you pick are in A.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P(x=b| (B|A)) = P(x=b|B), which is uniform.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Answer to your question: It is uniformly distributed. &lt;/p&gt;&#xA;" OwnerUserId="96" LastActivityDate="2010-07-20T23:46:23.517" />
  <row Id="193" PostTypeId="2" ParentId="185" CreationDate="2010-07-20T23:48:51.560" Score="2" ViewCount="" Body="&lt;p&gt;You are changing your basis vectors, call your new ones i, j, and k where i is defined from a-p, j from b-p, and k the cross product. Now recall that your basis vectors should be unit, so take the length of your three vectors and divide the vectors by their length, making i,j, and k unit. Now a = d(a,p)*i, b =d(b,p)*j.&lt;/p&gt;&#xA;" OwnerUserId="16" LastActivityDate="2010-07-20T23:48:51.560" />
  <row Id="194" PostTypeId="1" AcceptedAnswerId="199" CreationDate="2010-07-20T23:49:17.660" Score="6" ViewCount="109" Body="&lt;p&gt;Do primes become more or less frequent as you go further out on the number line?  That is, are there more or fewer primes between 1 and 1,000,000 than between 1,000,000 and 2,000,000?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A proof or pointer to a proof would be appreciated.&lt;/p&gt;&#xA;" OwnerUserId="129" LastEditorUserId="126" LastEditorDisplayName="" LastEditDate="2010-07-21T01:37:11.750" LastActivityDate="2010-07-21T03:59:29.613" Title="Distribution of primes?" Tags="&lt;number-theory&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="195" PostTypeId="2" ParentId="185" CreationDate="2010-07-20T23:49:40.717" Score="1" ViewCount="" Body="&lt;p&gt;What you are describing is an &lt;a href=&quot;http://en.wikipedia.org/wiki/Affine_transformation&quot; rel=&quot;nofollow&quot;&gt;Affine Transformation&lt;/a&gt;, which is a linear transformation followed by a translation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know this because any straight lines in your original vector space is also going to be a straight line in your transformed vector space.&lt;/p&gt;&#xA;" OwnerUserId="51" LastActivityDate="2010-07-20T23:49:40.717" />
  <row Id="196" PostTypeId="1" AcceptedAnswerId="229" CreationDate="2010-07-20T23:50:41.403" Score="6" ViewCount="179" Body="&lt;p&gt;I keep looking at this picture and its driving me crazy. How can the smaller circle travel the same distance when its circumference is less than the entire wheel? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://mathworld.wolfram.com/images/gifs/AristotlesWheel.gif&quot; alt=&quot;demonstration image&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="104" LastEditorUserId="126" LastEditorDisplayName="" LastEditDate="2010-07-21T01:40:25.533" LastActivityDate="2010-07-21T10:33:49.587" Title="How does the wheel paradox work?" Tags="&lt;geometry&gt;" AnswerCount="3" FavoriteCount="3" />
  <row Id="197" PostTypeId="2" ParentId="196" CreationDate="2010-07-20T23:53:59.960" Score="7" ViewCount="" Body="&lt;p&gt;If the two circles are fixed, then they will be traveling the same difference, but at different velocities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, the ratio of the radii is equal to the ratio of the velocities a point on either circle will be traveling.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you tried to repeat this by putting two different-sized circles on a track and making them spin to come out to be the same distance with the same angular velocity, you will notice that one of the circles will have to slide/slip along the track in order to keep them at the same pace.&lt;/p&gt;&#xA;" OwnerUserId="51" LastActivityDate="2010-07-20T23:53:59.960" />
  <row Id="198" PostTypeId="1" CreationDate="2010-07-20T23:54:20.210" Score="3" ViewCount="703" Body="&lt;p&gt;Since constructive mathematics allows us to avoid things like Russell's Paradox, then why don't they replace traditional proofs?  How do we know the &quot;regular&quot; kind of mathematics are free of paradox without a proof construction?&lt;/p&gt;&#xA;" OwnerUserId="117" LastEditorUserId="126" LastEditorDisplayName="" LastEditDate="2010-07-21T05:16:00.523" LastActivityDate="2010-08-17T06:34:19.387" Title="Aren't constructive math proofs more &quot;sound&quot;?" Tags="&lt;proof-theory&gt;&lt;constructive-mathematics&gt;" AnswerCount="6" CommentCount="2" FavoriteCount="2" />
  <row Id="199" PostTypeId="2" ParentId="194" CreationDate="2010-07-20T23:55:54.543" Score="9" ViewCount="" Body="&lt;p&gt;From the &lt;a href=&quot;http://en.wikipedia.org/wiki/Prime_number_theorem&quot; rel=&quot;nofollow&quot;&gt;Wikipedia article about the prime number theorem&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Roughly speaking, the prime number&#xA;  theorem states that if a random number&#xA;  nearby some large number N is&#xA;  selected, the chance of it being prime&#xA;  is about 1 / ln(N), where ln(N)&#xA;  denotes the natural logarithm of N.&#xA;  For example, near N = 10,000, about&#xA;  one in nine numbers is prime, whereas&#xA;  near N = 1,000,000,000, only one in&#xA;  every 21 numbers is prime. In other&#xA;  words, the average gap between prime&#xA;  numbers near N is roughly ln(N).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="114" LastActivityDate="2010-07-20T23:55:54.543" />
  <row Id="200" PostTypeId="2" ParentId="162" CreationDate="2010-07-20T23:58:51.917" Score="7" ViewCount="" Body="&lt;p&gt;The &quot;set of all sets&quot; is not so much a paradox in itself as something  that inevitably leads to a contradiction, namely the well-known (and referenced in the question) Russell's paradox.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given any set and a predicate applying to sets, the set of all things satisfying the predicate should be a subset of the original set.  If the &quot;set of all sets&quot; were to exist, because self-containment and non-self-containment are valid predicates, the set of all sets not containing themselves would have to exist as a set in order for our set theory to be consistent.  But this &quot;set of all sets&quot; cannot exist in a consistent set theory because of the Russel paradox.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the non-existence of the &quot;set of all sets&quot; is a consequence of the fact that presuming it's existence would lead to the contradiction described by Russel's paradox.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is in fact the origin of Russel's paradox. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In his work &quot;The Basic Laws of Arithmetic&quot;, Gottlob Frege had taken as a postulate the existence of this &quot;set of all sets&quot;.  In a letter to Frege, Bertrand Russell essentially blew away the basis of Frege's entire work by describing the paradox and proving that this postulate could not be a part of a consistent set theory.&lt;/p&gt;&#xA;" OwnerUserId="128" LastActivityDate="2010-07-20T23:58:51.917" CommentCount="2" />
  <row Id="201" PostTypeId="1" AcceptedAnswerId="211" CreationDate="2010-07-21T00:01:59.303" Score="9" ViewCount="1494" Body="&lt;p&gt;Prime numbers are numbers with no factors other than one and itself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Factors of a number are always lower or equal to than a given number; so, the larger the number is, the larger the pool of &quot;possible factors&quot; that number might have.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the larger the number, it seems like the less likely the number is to be a prime.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Surely there must be a number where, simply, every number above it has some other factors.  A &quot;critical point&quot; where every number larger than it simply will always have some factors other than one and itself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Has there been any research as to finding this critical point, or has it been proven not to exist?  That for any &lt;code&gt;n&lt;/code&gt; there is always guaranteed to be a number higher than &lt;code&gt;n&lt;/code&gt; that has no factors other than one and itself?&lt;/p&gt;&#xA;" OwnerUserId="51" LastEditorUserId="1778" LastEditorDisplayName="" LastEditDate="2010-11-04T04:08:49.177" LastActivityDate="2010-11-04T04:08:49.177" Title="Is there possibly a largest prime number?" Tags="&lt;number-theory&gt;&lt;prime-numbers&gt;" AnswerCount="8" CommentCount="8" FavoriteCount="0" />
  <row Id="202" PostTypeId="2" ParentId="155" CreationDate="2010-07-21T00:02:33.280" Score="6" ViewCount="" Body="&lt;p&gt;Surprisingly, there is a procedure for determining this, called the &lt;a href=&quot;http://en.wikipedia.org/wiki/Risch_algorithm&quot; rel=&quot;nofollow&quot;&gt;Risch Algorithm&lt;/a&gt;. However, it is apparently very complicated to implement, and furthermore it is not a true algorithm: if the input &lt;em&gt;does&lt;/em&gt; have an anti-derivative, it can find it in a finite amount of time, but demonstrating there is no anti-derivative requires solving the &lt;a href=&quot;http://en.wikipedia.org/wiki/Constant_problem&quot; rel=&quot;nofollow&quot;&gt;constant problem&lt;/a&gt;, which can fail to halt. (In that case, probabilistic methods can be used.)&lt;/p&gt;&#xA;" OwnerUserId="130" LastActivityDate="2010-07-21T00:02:33.280" CommentCount="1" />
  <row Id="203" PostTypeId="2" ParentId="198" CreationDate="2010-07-21T00:04:39.220" Score="3" ViewCount="" Body="&lt;p&gt;The distinction between constructive mathematics and traditional mathematics has nothing to do with Russell's Paradox.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Constructive mathematics simply requires working with one less basic postulate that many mathematicians have believed to be sensible and on which some proofs are based, namely the &lt;a href=&quot;http://en.wikipedia.org/wiki/Axiom_of_choice&quot; rel=&quot;nofollow&quot;&gt;Axiom of Choice&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="128" LastActivityDate="2010-07-21T00:04:39.220" CommentCount="4" />
  <row Id="204" PostTypeId="2" ParentId="201" CreationDate="2010-07-21T00:06:12.167" Score="12" ViewCount="" Body="&lt;p&gt;No there is not, here is a collection of proofs;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://math.mit.edu/~ssam/writings/primes.pdf&quot; rel=&quot;nofollow&quot;&gt;http://math.mit.edu/~ssam/writings/primes.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="16" LastActivityDate="2010-07-21T00:06:12.167" CommentCount="5" />
  <row Id="205" PostTypeId="2" ParentId="29" CreationDate="2010-07-21T00:12:28.473" Score="7" ViewCount="" Body="&lt;p&gt;Personally, I always use &lt;a href=&quot;http://www.python.org/download/&quot; rel=&quot;nofollow&quot;&gt;Python&lt;/a&gt;'s console.&lt;br&gt;&#xA;It has history and allows all kinds of math operations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://img6.imageshack.us/img6/5497/pythony.png&quot; alt=&quot;Screenshot&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is available for Linux, Windows, Mac, and others.&lt;/p&gt;&#xA;" OwnerUserId="135" LastEditorUserId="135" LastEditorDisplayName="" LastEditDate="2010-07-28T01:02:54.477" LastActivityDate="2010-07-28T01:02:54.477" CommunityOwnedDate="2010-07-28T01:32:30.370" CommentCount="5" />
  <row Id="206" PostTypeId="1" AcceptedAnswerId="248" CreationDate="2010-07-21T00:12:32.950" Score="10" ViewCount="892" Body="&lt;p&gt;Often in my studies (economics) the assumption of a &quot;well-behaved&quot; function will be invoked. I don't exactly know what that entails (I think twice continuously differentiability is one of the requirements), nor do I know why this is necessary (though I imagine the why will depend on each case).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can someone explain it to me, and if there is an explanation of the why as well, I would be grateful. Thanks!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;p&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;: To give one example where the term appears, see this Wikipedia entry for utility functions, which says at one point:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In order to simplify calculations,&#xA;  various assumptions have been made of&#xA;  utility functions.     &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;CES (constant elasticity of substitution, or&#xA;  isoelastic) utility&lt;br&gt;&#xA;  Exponential utility&lt;br&gt;&#xA;  Quasilinear utility&lt;br&gt;&#xA;  Homothetic preferences &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Most utility functions&#xA;  used in modeling or theory are&#xA;  &lt;strong&gt;well-behaved&lt;/strong&gt;. They are usually&#xA;  monotonic, quasi-concave, continuous&#xA;  and globally non-satiated.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I might be wrong, but I don't think &quot;well-behaved&quot; means monotonic, quasi-concave, continuous and globally non-satiated. What about twice differentiable?&lt;/p&gt;&#xA;" OwnerUserId="85" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-08-04T16:26:27.393" LastActivityDate="2010-08-10T21:44:15.657" Title="What exactly does it mean for a function to be &quot;well-behaved&quot;?" Tags="&lt;terminology&gt;&lt;analysis&gt;&lt;economics&gt;" AnswerCount="4" CommentCount="13" FavoriteCount="3" />
  <row Id="207" PostTypeId="2" ParentId="153" CreationDate="2010-07-21T00:13:19.343" Score="11" ViewCount="" Body="&lt;p&gt;As a free/libre alternative to Mathematica, you may want to have a look at &lt;a href=&quot;http://www.sagemath.org/&quot; rel=&quot;nofollow&quot;&gt;Sage&lt;/a&gt; for &lt;a href=&quot;http://en.wikipedia.org/wiki/Python_%28programming_language%29&quot; rel=&quot;nofollow&quot;&gt;Python&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a free/libre alternative to MATLAB, which duplicates almost all of its functionality, have a look at &lt;a href=&quot;http://en.wikipedia.org/wiki/GNU_Octave&quot; rel=&quot;nofollow&quot;&gt;GNU Octave&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That said, &lt;a href=&quot;http://www.wolfram.com/products/mathematica/index.html&quot; rel=&quot;nofollow&quot;&gt;Mathematica&lt;/a&gt; is a beautiful system that has no equal.&lt;/p&gt;&#xA;" OwnerUserId="132" LastEditorUserId="132" LastEditorDisplayName="" LastEditDate="2010-08-15T21:20:11.747" LastActivityDate="2010-08-15T21:20:11.747" CommentCount="2" />
  <row Id="209" PostTypeId="2" ParentId="29" CreationDate="2010-07-21T00:15:05.363" Score="5" ViewCount="" Body="&lt;p&gt;I use &lt;a href=&quot;http://www.r-project.org/&quot; rel=&quot;nofollow&quot;&gt;R&lt;/a&gt; these days. It was build to be a calculator and does the job well. It's syntax might be a bit strange but it allows you to do a lot with little typing.&lt;/p&gt;&#xA;" OwnerUserId="2263" LastEditorUserId="2263" LastEditorDisplayName="" LastEditDate="2010-07-21T00:43:07.147" LastActivityDate="2010-07-21T00:43:07.147" CommunityOwnedDate="2010-07-28T01:32:30.370" />
  <row Id="210" PostTypeId="2" ParentId="198" CreationDate="2010-07-21T00:20:29.157" Score="7" ViewCount="" Body="&lt;p&gt;A whole bunch of things in mathematics are inherently nonconstructive. For instance, invariant theory--recall the famous quote by Gordan that Hilbert's mathematics was &quot;theology.&quot; (A quote which, I believe, was in jest.) &#xA;The Hahn-Banach theorem, a fundamental tool in functional analysis (and a great tool for proving all sorts of results, like approximation results--Runge's theorem, the Stone-Weierstrass theorem, and more) relies on the axiom of choice, and is consequently nonconstructive.  The fact that any proper ideal in a ring is contained in a maximal ideal is frequently used in algebra, and yet it needs the axiom of choice.  The use of ultraproducts in logic (or the construction of hyperreal numbers) is inherently nonconstructive: you can't just &lt;em&gt;exhibit&lt;/em&gt; a nonprincipal ultrafilter on the natural numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically, a lot of mathematics just doesn't work without Zorn's lemma, and this is equivalent to the axiom of choice.  &lt;/p&gt;&#xA;" OwnerUserId="536" LastEditorUserId="536" LastEditorDisplayName="" LastEditDate="2010-08-17T06:34:19.387" LastActivityDate="2010-08-17T06:34:19.387" CommentCount="4" />
  <row Id="211" PostTypeId="2" ParentId="201" CreationDate="2010-07-21T00:24:29.570" Score="27" ViewCount="" Body="&lt;p&gt;Euclid's famous proof is as follows: Suppose there is a finite number of primes.  Let X be the product of all of these primes.  Then look at X+1.  It is clear that X is coprime to X+1.  Therefore, no nontrivial factor of X is a factor of X+1, but every prime is a factor of X.  By the fundamental theorem of arithmetic, X+1 admits a prime factorization, and by the above remark, none of these prime factors can be a factor of X, but X is the product of all primes.  This is a contradiction.&lt;/p&gt;&#xA;" OwnerUserId="126" LastActivityDate="2010-07-21T00:24:29.570" CommentCount="6" />
  <row Id="212" PostTypeId="2" ParentId="201" CreationDate="2010-07-21T00:32:50.413" Score="3" ViewCount="" Body="&lt;p&gt;According to &lt;a href=&quot;http://xkcd.com/622/&quot; rel=&quot;nofollow&quot;&gt;XKCD&lt;/a&gt;, we have the following Haiku:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Top Prime's Divisors'&lt;br&gt;&#xA;Product (Plus one)'s factors are...?&lt;br&gt;&#xA;Q.E.D B@%&amp;amp;$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wonder if we can edit it to make it correct&lt;/p&gt;&#xA;" OwnerUserId="123" LastEditorUserId="171" LastEditorDisplayName="" LastEditDate="2010-08-17T14:34:23.607" LastActivityDate="2010-08-17T14:34:23.607" CommentCount="14" />
  <row Id="213" PostTypeId="1" AcceptedAnswerId="498" CreationDate="2010-07-21T00:34:22.937" Score="17" ViewCount="1223" Body="&lt;p&gt;I am learning geometric algebra, and it is incredible how much it helps me understand other branches of mathematics. I wish I had been exposed to it earlier. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Additionally I feel the same way about enumerative combinatorics. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are some less popular mathematical subjects that you think should be more popular?&lt;/p&gt;&#xA;" OwnerUserId="99" LastEditorUserId="171" LastEditorDisplayName="" LastEditDate="2010-08-14T06:29:05.127" LastActivityDate="2011-02-05T00:09:46.027" CommunityOwnedDate="2010-07-21T00:34:22.937" Title="Mathematical subjects you wish you learned earlier" Tags="&lt;soft-question&gt;&lt;learning&gt;" AnswerCount="16" CommentCount="5" FavoriteCount="16" />
  <row Id="214" PostTypeId="2" ParentId="162" CreationDate="2010-07-21T00:36:42.743" Score="8" ViewCount="" Body="&lt;p&gt;As others have explained, the existence of universal set is incompatible with the &lt;a href=&quot;http://en.wikipedia.org/wiki/Zermelo%E2%80%93Fraenkel_set_theory&quot; rel=&quot;nofollow&quot;&gt;Zermelo&amp;ndash;Fraenkel axioms&lt;/a&gt; of set theory. However, there are alternative set theories which admit a universal set. One such theory is &lt;a href=&quot;http://en.wikipedia.org/wiki/New_Foundations&quot; rel=&quot;nofollow&quot;&gt;Quine's New Foundations&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="25" LastActivityDate="2010-07-21T00:36:42.743" />
  <row Id="215" PostTypeId="2" ParentId="206" CreationDate="2010-07-21T00:40:08.307" Score="6" ViewCount="" Body="&lt;p&gt;The short answer is that there is no &quot;exact&quot; meaning. Ideally, additional axioms are introduced to ensure that a certain function (or any mathematical object, for that matter) is &quot;well-behaved&quot; which, in effect, makes analysis easier. So, the meaning of &quot;well-behaved&quot; should be derived from those specific additional axioms.&lt;/p&gt;&#xA;" OwnerUserId="114" LastActivityDate="2010-07-21T00:40:08.307" CommentCount="6" />
  <row Id="216" PostTypeId="2" ParentId="83" CreationDate="2010-07-21T00:41:36.323" Score="4" ViewCount="" Body="&lt;p&gt;I think ideas like cardinality are suitable for getting kids interested in math -- the notion of a one-to-one correspondence is easy to grasp, and the proof of the uncountability of the reals is great fun.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, elementary number theory can be useful.  This is the sort of thing that would be especially useful for, say, contest math, which is where a lot of kids get interested in math.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, popular math books can be a good source of ideas.&lt;/p&gt;&#xA;" OwnerUserId="536" LastActivityDate="2010-07-21T00:41:36.323" CommunityOwnedDate="2010-08-07T22:32:44.343" CommentCount="1" />
  <row Id="217" PostTypeId="2" ParentId="155" CreationDate="2010-07-21T00:42:47.973" Score="1" ViewCount="" Body="&lt;p&gt;Brian Conrad explains this in the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.claymath.org/programs/outreach/academy/LectureNotes05/Conrad.pdf&quot; rel=&quot;nofollow&quot;&gt;link&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="126" LastEditorUserId="8246" LastEditorDisplayName="" LastEditDate="2011-04-24T09:46:53.990" LastActivityDate="2011-04-24T09:46:53.990" CommentCount="1" />
  <row Id="218" PostTypeId="2" ParentId="177" CreationDate="2010-07-21T00:50:34.200" Score="7" ViewCount="" Body="&lt;p&gt;Yes this will work; it's called &lt;a href=&quot;http://en.wikipedia.org/wiki/Rejection_sampling&quot; rel=&quot;nofollow&quot;&gt;rejection sampling&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even better is to generate a point in &lt;a href=&quot;http://mathworld.wolfram.com/DiskPointPicking.html&quot; rel=&quot;nofollow&quot;&gt;polar coordinates&lt;/a&gt; though:  pick θ from [0, 2π) and r&lt;sup&gt;2&lt;/sup&gt; from [0, R&lt;sup&gt;2&lt;/sup&gt;] (ie. multiply R by the square-root of a random number in [0, 1] - without the square-root it is non-uniform).&lt;/p&gt;&#xA;" OwnerUserId="136" LastEditorUserId="136" LastEditorDisplayName="" LastEditDate="2010-07-21T17:39:38.927" LastActivityDate="2010-07-21T17:39:38.927" CommentCount="1" />
  <row Id="219" PostTypeId="2" ParentId="83" CreationDate="2010-07-21T00:53:34.957" Score="5" ViewCount="" Body="&lt;p&gt;Besides all of these mathematics topics, I would contend that it is equally as important to show them the more human side of the subject.  Mathematics has a history and a culture that is accessible and exciting to some people.  Of course there are the exciting lives of Pascal and Galois, but there is also the story of Bourbaki (and the individual stories of their various members).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, introducing children to actual mathematicians/engineers/scientists is a great way to motivate their interest in the subject.  Let them see that mathematics has not been borne from the head of Zeus, that it is a living and breathing subject.  &lt;/p&gt;&#xA;" OwnerUserId="121" LastActivityDate="2010-07-21T00:53:34.957" CommunityOwnedDate="2010-08-07T22:32:44.343" CommentCount="4" />
  <row Id="220" PostTypeId="1" CreationDate="2010-07-21T00:56:14.780" Score="3" ViewCount="226" Body="&lt;p&gt;Okay, so hopefully this isn't too hard or off-topic. Let's say I have a very simple lowpass filter (something that smooths out a signal), and the filter object has a position variable and a cutoff variable (between 0 and 1). In every step, a value is put into the following bit of pseudocode as &quot;input&quot;: &lt;code&gt;position = position*(1-c)+input*c&lt;/code&gt;, or more mathematically, &lt;code&gt;f(n) = f(n-1)*(1-c)+x[n]*c&lt;/code&gt;. The output is the value of &quot;position.&quot; Basically, it moves a percentage of the distance between the current position and then input value, stores this value internally, and returns it as output. It's intentionally simplistic, since the project I'm using this for is going to have way too many of these in sequence processing audio in real time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given the filter design, how do I construct a function that takes input frequency (where 1 means a sine wave with a wavelength of 2 samples, and .5 means a sine wave with wavelength 4 samples, and 0 is a flat line), and cutoff value (between 1 and 0, as shown above) and outputs the amplitude of the resulting sine wave? Sine wave comes in, sine wave comes out, I just want to be able to figure out how much quieter it is at any input and cutoff frequency combination.&lt;/p&gt;&#xA;" OwnerUserId="62" LastEditorUserId="62" LastEditorDisplayName="" LastEditDate="2010-08-01T07:24:06.283" LastActivityDate="2010-08-01T07:24:06.283" Title="Simple lowpass frequency response" Tags="&lt;applications&gt;&lt;signal-processing&gt;" AnswerCount="2" CommentCount="7" />
  <row Id="221" PostTypeId="2" ParentId="213" CreationDate="2010-07-21T00:57:50.690" Score="10" ViewCount="" Body="&lt;p&gt;Though I'm sure it's not unpopular, I don't think many people learn it early: Group Theory. It's a real nice area with a lot of cool math and some neat applications (like cryptography).&lt;/p&gt;&#xA;" OwnerUserId="50" LastActivityDate="2010-07-21T00:57:50.690" CommunityOwnedDate="2010-07-21T00:57:50.690" CommentCount="1" />
  <row Id="222" PostTypeId="2" ParentId="213" CreationDate="2010-07-21T00:59:39.780" Score="2" ViewCount="" Body="&lt;p&gt;I don't really think that &lt;strong&gt;graph theory&lt;/strong&gt; is a &quot;less popular mathematical subject,&quot; but I certainly wish I had been exposed to it earlier.&lt;/p&gt;&#xA;" OwnerUserId="114" LastActivityDate="2010-07-21T00:59:39.780" CommunityOwnedDate="2010-07-21T00:59:39.780" />
  <row Id="223" PostTypeId="2" ParentId="213" CreationDate="2010-07-21T01:01:56.937" Score="3" ViewCount="" Body="&lt;p&gt;Theory of computation, information theory and logic/foundation of mathematics are very interesting topics. I wish I knew them earlier. They are not unpopular(almost every university have a bunch of ToC people in CS depatment...) , but many math major I know have never touched them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They show you the limits of mathematics, computation and communication.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Logic shows there are things can't be proved from a set of axioms even if it's true--Godel's incompleteness theorem. There are other interesting theorems in foundation of mathematics. Like the independence of continuum hypothesis to ZFC.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Theory of computation showed me things that's not computable. Problems that takes exponential time, exponential space, no matter what kind of algorithm you come up with.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Information theory proves the minimum amount of information required to reconstruct some other information. It pops up in unexpected places. There is a proof of there are infinite number of primes by information theory (Sorry I can't find it, I can only tell you it exists. I might find it later).&lt;/p&gt;&#xA;" OwnerUserId="96" LastEditorUserId="76" LastEditorDisplayName="" LastEditDate="2010-07-21T07:05:16.320" LastActivityDate="2010-07-21T07:05:16.320" CommunityOwnedDate="2010-07-21T01:01:56.937" CommentCount="4" />
  <row Id="224" PostTypeId="2" ParentId="154" CreationDate="2010-07-21T01:02:26.140" Score="3" ViewCount="" Body="&lt;p&gt;Are real numbers &lt;em&gt;&quot;real&quot;&lt;/em&gt;?  It's not even computationally possible to compare two real numbers for equality!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Interestingly enough, it is shown in Abstract Algebra courses that the idea of complex numbers arises naturally from the idea of real numbers - you could not say, for instance, that the real numbers are &lt;em&gt;valid&lt;/em&gt; but the complex numbers aren't (whatever your definition of &lt;em&gt;valid&lt;/em&gt; is...)&lt;/p&gt;&#xA;" OwnerUserId="136" LastActivityDate="2010-07-21T01:02:26.140" CommunityOwnedDate="2011-04-15T20:45:29.080" CommentCount="1" />
  <row Id="225" PostTypeId="2" ParentId="213" CreationDate="2010-07-21T01:10:42.140" Score="5" ViewCount="" Body="&lt;p&gt;I don't know what level of mathematics you are referring to, but here's my opinion after recently finishing my university's undergraduate curriculum.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Firstly, I would like to second Jan Gorzny's reply of Group Theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second, I wish that I had learned linear algebra earlier.  The topic usually has two semesters: matrix algebra and then an early proof-based introduction to vector spaces and linear transformations.  The real work in this topic can't begin until after both of these classes are completed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also wish I had been exposed to topology earlier than I was.  Of course there are two &quot;standard&quot; approaches here, and I suppose the approach that introduces general topology before advanced calculus would have better suited my tastes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a good book that may give some more insight into the heavily debated area where your question lies: Thomas Garrity, &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0521792851&quot; rel=&quot;nofollow&quot;&gt;All the Mathematics You Missed But Need to Know for Graduate School&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="121" LastActivityDate="2010-07-21T01:10:42.140" CommunityOwnedDate="2010-07-21T01:10:42.140" CommentCount="6" />
  <row Id="226" PostTypeId="2" ParentId="206" CreationDate="2010-07-21T01:23:22.517" Score="5" ViewCount="" Body="&lt;p&gt;In general, we think of well-behaved functions as &lt;strong&gt;simpler&lt;/strong&gt;, somehow. In any field, we might want to limit ourselves to considering only well-behaved functions in order to avoid having to deal with nasty edge cases. And in each of these domains, the community is free to choose whatever definition of 'well-behaved' makes sense for them. A quick look at the &lt;a href=&quot;http://en.wikipedia.org/wiki/Well-behaved&quot; rel=&quot;nofollow&quot;&gt;wiki link&lt;/a&gt; that &lt;a href=&quot;http://math.stackexchange.com/questions/206/what-exactly-does-it-mean-for-a-function-to-be-well-behaved/215#215&quot; rel=&quot;nofollow&quot;&gt;e.James posted&lt;/a&gt; will show you the diversity in ideas of what it means to be well-behaved. I am no economist, so I will take for granted that the definition you put forth in your question is the one in common use. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can see &lt;strong&gt;twice-differentiable&lt;/strong&gt; as a reasonable requirement for a utility function to be 'well-behaved' is because the derivative of the utility function is marginal utility, and economists often care about the derivative of marginal utility. For example, if the second derivative of utility is negative, this means that the marginal utility has a negative derivative in other words, additional quantity of the good or service does not add utility as quickly. Also commonly referred to as diminishing returns.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we want to be able to take the derivative even once, of course, we need the function to be &lt;strong&gt;continuous&lt;/strong&gt;. You probably don't need to worry about the formal definition here. Pencil test should work fine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The requirement for utility to be &lt;strong&gt;monotonic&lt;/strong&gt; means that it is either always increasing or always decreasing. In other words, that a particular good or service is either desirable or not. If 10 widgets were good, 20 must be better. Of course, as mentioned above, maybe not &lt;em&gt;that&lt;/em&gt; much better.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Monotonic means that it will also be &lt;a href=&quot;http://en.wikipedia.org/wiki/Concave_function#Properties&quot; rel=&quot;nofollow&quot;&gt;&lt;strong&gt;quasi-concave&lt;/strong&gt;&lt;/a&gt;. (Except for weird stuff like a flat function) That is, they have at most one local maximum. We would prefer that functions be quasiconcave because we wish to avoid cases like the one below. It just makes it so much easier to optimize when you only have one possible maximum to worry about.&lt;br&gt;&#xA;&lt;img src=&quot;http://img.sparknotes.com/figures/E/eeb0fa64da5c2a1115cdf73c3ef2f0fc/derivapps1.gif&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Globally non-satiated&lt;/strong&gt; someone else can talk about. I don't know enough to be sure I won't just be misleading you further.&lt;/p&gt;&#xA;" OwnerUserId="73" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-07-21T02:31:15.863" LastActivityDate="2010-07-21T02:31:15.863" CommentCount="5" />
  <row Id="227" PostTypeId="1" AcceptedAnswerId="231" CreationDate="2010-07-21T01:24:02.017" Score="5" ViewCount="760" Body="&lt;p&gt;I am doing a PhD in economics and I have the chance of taking one subject a semester in the maths department (I would like to do more, but &quot;unfortunately&quot; I have to work on my thesis). I want to have a plan so that I start with subjects I might know a bit but making sure I have all the basis, while at the same time ensuring I get the chance to sit in more advanced courses. My plan, for now, is the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I am skipping calculus I and II, because even though I don't know all of it, I know enough I think to be able to do the next course without trouble)&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Linear Algebra  &lt;/li&gt;&#xA;&lt;li&gt;Real Analysis     &lt;/li&gt;&#xA;&lt;li&gt;Discrete Maths and Operations Research &lt;em&gt;or&lt;/em&gt;&lt;br&gt;&#xA;Engineering Maths &lt;em&gt;or&lt;/em&gt;&lt;br&gt;&#xA;Group Theory and Linear Algebra  &lt;/li&gt;&#xA;&lt;li&gt;Numerical &amp;amp; Symbolic Mathematics &lt;em&gt;or&lt;/em&gt;&lt;br&gt;&#xA;Vector Calculus&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Can you anyone give me a hand on making the decision or giving suggestions? In economics we use software such as matlab, lots of differentiation, Euler equations, difference equations, dynamic programming, statistical analysis (but for that I plan on taking Probability and Stochastic Modelling). &lt;/p&gt;&#xA;" OwnerUserId="85" LastEditorUserId="126" LastEditorDisplayName="" LastEditDate="2010-07-21T02:00:53.190" LastActivityDate="2010-11-22T22:32:04.050" CommunityOwnedDate="2010-08-12T14:58:43.903" Title="Which 4 maths courses to take as an Economics PhD student?" Tags="&lt;economics&gt;&lt;soft-question&gt;&lt;education&gt;" AnswerCount="9" CommentCount="5" FavoriteCount="2" />
  <row Id="228" PostTypeId="1" CreationDate="2010-07-21T01:27:27.687" Score="1" ViewCount="308" Body="&lt;p&gt;&lt;i&gt;The sum of two Gaussian variables is another Gaussian.&lt;/i&gt;&lt;br&gt;&#xA;It seems natural, but I could not find a proof using Google.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What's a short way to prove this?&lt;br&gt;&#xA;Thanks!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: Provided the two variables are independent.&lt;/p&gt;&#xA;" OwnerUserId="135" LastEditorUserId="135" LastEditorDisplayName="" LastEditDate="2010-07-28T00:53:59.567" LastActivityDate="2010-07-28T00:53:59.567" Title="Proof that the sum of two Gaussian variables is another Gaussian" Tags="&lt;probability-theory&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="229" PostTypeId="2" ParentId="196" CreationDate="2010-07-21T01:34:37.700" Score="9" ViewCount="" Body="&lt;p&gt;That picture confuses things by making it look as though the red line is being &quot;unwound&quot; from the circle like paper towel being unwound from a roll. Our brains pick up on that, since it is a real-world example.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both circles complete a single revolution, and both travel the same distance from left to right. If these really were rolls of paper towel, the smaller roll would have to spin faster (and therefore complete more than one full revolution) in order to lay out the same length of paper towel as the larger roll. Alternatively, if the two rolls &lt;em&gt;were&lt;/em&gt; spinning at the same rate, the free end of the strip of towel left behind by the smaller roll would also move to the right.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In short, the image is a kind of optical/mental illusion, and you're not going crazy &lt;code&gt;:)&lt;/code&gt;&lt;/p&gt;&#xA;" OwnerUserId="59" LastActivityDate="2010-07-21T01:34:37.700" />
  <row Id="230" PostTypeId="2" ParentId="227" CreationDate="2010-07-21T01:35:04.157" Score="3" ViewCount="" Body="&lt;p&gt;You may consider focusing on Operations Research, and anything numerical.  I think you can pick up the tools of vector calculus as you go (which for your specialty will mostly be partial-differentiation, I think).  Maybe you can take two semesters of OR and a semester of something like Numerical Solutions to PDE, and one of Numerical Linear Algebra or Numerical Analysis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And, I suppose all of this depends on what you plan to do with Econ.&lt;/p&gt;&#xA;" OwnerUserId="121" LastActivityDate="2010-07-21T01:35:04.157" CommunityOwnedDate="2010-08-12T14:58:43.903" CommentCount="2" />
  <row Id="231" PostTypeId="2" ParentId="227" CreationDate="2010-07-21T01:37:29.953" Score="7" ViewCount="" Body="&lt;p&gt;If you do take vector calculus, it's probably best taken before real analysis (where you'll learn many of the results from vector calc in a more general setting). If it were me, I'd take discrete math and operations research over the other two in (3), because they'll be useful for economics and because those subjects are the most different from linear algebra and real analysis. The group theory wouldn't, as far as I know, be tremendously applicable to economics except as it relates to analysis, and you'll get a sense of an algebraic way of thinking through linear. Numerical and symbolic math, if it includes numerical analysis, would be very relevant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A main criterion I am using here is that you'll want to learn more math as you go on in econ, and so the most important thing, for your ability to learn various things on your own later and for developing an appreciation for different kinds of math, is to take courses that cover many different kinds of mathematical thinking (including that which is more directly relevant to econ and other applied areas).&lt;/p&gt;&#xA;" OwnerUserId="76" LastActivityDate="2010-07-21T01:37:29.953" CommunityOwnedDate="2010-08-12T14:58:43.903" CommentCount="3" />
  <row Id="232" PostTypeId="2" ParentId="228" CreationDate="2010-07-21T01:55:04.623" Score="1" ViewCount="" Body="&lt;p&gt;I don't know how I missed that one, indeed:&lt;br&gt;&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Sum_of_normally_distributed_random_variables&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Sum_of_normally_distributed_random_variables&lt;/a&gt;&lt;br&gt;&#xA;Thanks Kaestur Hakarl!&lt;/p&gt;&#xA;" OwnerUserId="135" LastActivityDate="2010-07-21T01:55:04.623" CommentCount="1" />
  <row Id="233" PostTypeId="2" ParentId="123" CreationDate="2010-07-21T01:58:47.687" Score="10" ViewCount="" Body="&lt;p&gt;On a map using the Mercator projection, the relationship between the latitude &lt;em&gt;L&lt;/em&gt; of a point and its &lt;em&gt;y&lt;/em&gt; coordinate on the map is given by &lt;em&gt;y&lt;/em&gt; = atanh(sin(&lt;em&gt;L&lt;/em&gt;)), where atanh is the inverse of the hyperbolic tangent function.&lt;/p&gt;&#xA;" OwnerUserId="115" LastActivityDate="2010-07-21T01:58:47.687" />
  <row Id="234" PostTypeId="1" AcceptedAnswerId="510" CreationDate="2010-07-21T02:11:36.817" Score="6" ViewCount="301" Body="&lt;p&gt;Suppose there are two face down cards each with a positive real number and with one twice the other. Each card has value equal to its number. You are given one of the cards (with value x) and after you have seen it, the dealer offers you an opportunity to swap without anyone having looked at the other card.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you choose to swap, your expected value should be the same, as you still have a 50% chance of getting the higher card and 50% of getting the lower card.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the other card has a 50% chance of being 0.5x and a 50% chance of being 2x. If we keep the card, our expected value is x, while if we swap it, then our expected value if 05.*0.5x+0.5*2x=1.25x, so it seems like it is better to swap. Can anyone explain this apparent contradiction.&lt;/p&gt;&#xA;" OwnerUserId="123" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-07-21T04:29:25.300" LastActivityDate="2010-07-22T23:44:39.753" Title="Card doubling paradox" Tags="&lt;paradox&gt;&lt;probability-theory&gt;" AnswerCount="3" CommentCount="7" FavoriteCount="3" />
  <row Id="235" PostTypeId="2" ParentId="154" CreationDate="2010-07-21T02:16:46.523" Score="13" ViewCount="" Body="&lt;p&gt;To the degree that anything actually &quot;exists&quot; in math, yes complex number exist.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you accept that groups, rings and fields exist, and that isomorphism of rings makes sense, complex numbers can be recognized as (isomorphic to) the subring (which happens to be a field) of the ring of 2 by 2 real matrices.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Generators of this subring are the matrices&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&#xA; |1  0|&#xA; |0  1|&#xA;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&#xA; |0 -1|&#xA; |1  0|&#xA;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;which correspond to 1 and i in the normal notation of complex numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As people tend to accept that matrices exist, this may be a convincing argument. ;-)&lt;/p&gt;&#xA;" OwnerUserId="128" LastActivityDate="2010-07-21T02:16:46.523" CommunityOwnedDate="2011-04-15T20:45:29.080" CommentCount="3" />
  <row Id="236" PostTypeId="2" ParentId="227" CreationDate="2010-07-21T02:24:29.700" Score="1" ViewCount="" Body="&lt;p&gt;During my graduate studies, I had a friend who was extremely interested in differential topology and focused on that in his work toward an MS in math.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In his PhD work, he switched to Economics, but was still essentially doing Differential Topology.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Economics can get just as abstract as math!&lt;/p&gt;&#xA;" OwnerUserId="128" LastActivityDate="2010-07-21T02:24:29.700" CommunityOwnedDate="2010-08-12T14:58:43.903" />
  <row Id="237" PostTypeId="1" AcceptedAnswerId="238" CreationDate="2010-07-21T02:31:40.650" Score="5" ViewCount="300" Body="&lt;p&gt;Given a point's coordinates (x,y), what is the procedure for determining if it lies within a polygon whose vertices are (x1,y1), (x2,y2), ..., (xn,yn)?&lt;/p&gt;&#xA;" OwnerUserId="106" LastEditorUserId="126" LastEditorDisplayName="" LastEditDate="2010-07-21T02:33:05.423" LastActivityDate="2010-07-21T09:24:09.863" Title="How do you determine if a point sits inside a polygon?" Tags="&lt;geometry&gt;&lt;algorithms&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="2" />
  <row Id="238" PostTypeId="2" ParentId="237" CreationDate="2010-07-21T02:43:53.087" Score="5" ViewCount="" Body="&lt;p&gt;I will assume the polygon has no intersections between the edges except at corners. Call the point (x0, y0). First we determine whether we are on a line - this is simple using substitution and range checking. For the range checking, suppose we have a segment (x1,y1), (x2,y2). We check that x1&amp;lt;=x0&amp;lt;=x2 or x2&amp;lt;=x0&amp;lt;=x1 and do the same for y.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if we aren't on a line, we draw a ray from the point and count how many lines you intersect. If the line doesn't intersect a corner or contain a non-degenerate segment of an edge, then an odd number of intersections should mean you are inside and an even number mean outside.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we intersect with a corner, it is more difficult. Clearly, the ray could leave with it only intersecting a corner, but it could also pass through the corner without entering the polygon. One solution is to pick the ray so that it doesn't intersect a corner or go along a line - this should always be possible. We can also use cross product to determine whether we are crossing the lines at their corner or not (just check if the adjacent vertices are on the same side of the line or not).&lt;/p&gt;&#xA;" OwnerUserId="123" LastEditorUserId="123" LastEditorDisplayName="" LastEditDate="2010-07-21T02:52:26.640" LastActivityDate="2010-07-21T02:52:26.640" CommentCount="2" />
  <row Id="239" PostTypeId="2" ParentId="83" CreationDate="2010-07-21T03:07:37.587" Score="2" ViewCount="" Body="&lt;p&gt;When I was that age, I discovered Raymond Smullyan's classic logic puzzle books in the library (such as &lt;em&gt;What is the name of this book?&lt;/em&gt;), and really got into it. I remember my amazement when I first understood how a complicated logic puzzle could become trivial, just symbolic manipulation really, with the right notation.&lt;/p&gt;&#xA;" OwnerUserId="115" LastActivityDate="2010-07-21T03:07:37.587" CommunityOwnedDate="2010-08-07T22:32:44.343" />
  <row Id="240" PostTypeId="1" AcceptedAnswerId="1688" CreationDate="2010-07-21T03:19:46.880" Score="6" ViewCount="545" Body="&lt;p&gt;I remember hearing several times the advice that, we should avoid using a proof by contradiction, if it is simple to convert to a direct proof or a proof by contrapositive.  Could you explain the reason?  Do logicians think that proofs by contradiction are somewhat weaker than direct proofs?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: To clarify the question.  I am wondering, if there is any reason that one would still continue looking for a direct proof of some theorem, although a proof by contradiction has already been found.  I don't mean improvements in terms of elegance or exposition, I am asking about logical reasons.  For example, in the case of axiom of choice, there is obviously reason to look for a proof that does not use the axiom of choice.  Is there a similar case for proofs by contradiction?&lt;/p&gt;&#xA;" OwnerUserId="145" LastEditorUserId="126" LastEditorDisplayName="" LastEditDate="2010-07-21T05:10:26.290" LastActivityDate="2010-08-06T01:54:45.447" Title="Are the &quot;proofs by contradiction&quot; weaker than other proofs?" Tags="&lt;logic&gt;&lt;proof-theory&gt;" AnswerCount="7" CommentCount="2" FavoriteCount="1" />
  <row Id="241" PostTypeId="2" ParentId="240" CreationDate="2010-07-21T03:22:18.083" Score="0" ViewCount="" Body="&lt;p&gt;Nearly always the direct proof is easier to understand, shorter, and more helpful!&lt;/p&gt;&#xA;" OwnerUserId="28" LastActivityDate="2010-07-21T03:22:18.083" CommentCount="6" />
  <row Id="242" PostTypeId="2" ParentId="240" CreationDate="2010-07-21T03:28:42.423" Score="5" ViewCount="" Body="&lt;p&gt;Sometimes you might want to know not just that there &lt;em&gt;exists&lt;/em&gt; something, you might want to know how to actually go about finding it (and related questions like how quickly you can find it).  Proofs by contradiction are non-constructive, while direct proofs are typically constructive in the sense that they actually construct an answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, the proof that there are infinitely many primes usually proceeds by contradiction.  However, you can make it a direct proof which gives the stronger result that the nth prime is less than e^{e^n}.  (This is a good exercise to work out for yourself, but you can also find it as Prop 1.1.3 in my &lt;a href=&quot;http://math.columbia.edu/~nsnyder/thesismain.pdf&quot; rel=&quot;nofollow&quot;&gt;senior thesis&lt;/a&gt; and probably many other places as well.)&lt;/p&gt;&#xA;" OwnerUserId="8" LastActivityDate="2010-07-21T03:28:42.423" />
  <row Id="243" PostTypeId="2" ParentId="240" CreationDate="2010-07-21T03:31:45.497" Score="7" ViewCount="" Body="&lt;p&gt;Most logicians consider proofs by contradiction to be equally valid, however some people are &lt;a href=&quot;http://en.wikipedia.org/wiki/Constructivism_%28mathematics%29&quot; rel=&quot;nofollow&quot;&gt;constructivists/intuitionists&lt;/a&gt; and don't consider them valid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One possible reason that the constructivist point of view makes a certain amount of sense is that statements like the continuum hypothesis are independent of the axioms, so it's a bit weird to claim that it's either true or false, in a certain sense it's neither.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nonetheless constructivism is a relatively uncommon position among mathematicians/logicians.  However, it's not considered totally nutty or beyond the pale.  Fortunately, in practice most proofs by contradiction can be translated into constructivist terms and actual constructivists are rather adept at doing so.  So the rest of us mostly don't bother worrying about this issue, figuring it's the constructivists problem.&lt;/p&gt;&#xA;" OwnerUserId="8" LastEditorUserId="8" LastEditorDisplayName="" LastEditDate="2010-07-21T03:44:03.817" LastActivityDate="2010-07-21T03:44:03.817" CommentCount="3" />
  <row Id="244" PostTypeId="2" ParentId="83" CreationDate="2010-07-21T03:31:49.727" Score="0" ViewCount="" Body="&lt;p&gt;Speaking from my own experience with elementary mathematics, yours is a very hard question to answer because there is little in the elementary math curriculum worth getting excited about. Before students are going to get excited about math, the math curriculum has to be changed so that the process of doing mathematics is made to be engaging and thought provoking - not tedious. I think every teacher of elementary mathematics needs to watch &lt;a href=&quot;http://www.ted.com/talks/lang/eng/dan_meyer_math_curriculum_makeover.html&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; video about how math education can and should be reformed.&lt;/p&gt;&#xA;" OwnerUserId="19" LastActivityDate="2010-07-21T03:31:49.727" CommunityOwnedDate="2010-08-07T22:32:44.343" />
  <row Id="245" PostTypeId="2" ParentId="240" CreationDate="2010-07-21T03:39:26.747" Score="0" ViewCount="" Body="&lt;p&gt;At first this seems like a silly question - after all isn't the point of a mathematical proof to be a proof and hence to be beyond question. But of course, to prove anything we need assumptions and &lt;a href=&quot;http://en.wikipedia.org/wiki/Constructivism_%28mathematics%29&quot; rel=&quot;nofollow&quot;&gt;some people&lt;/a&gt; do disagree with the axioms commonly used by mathematicians. I don't have much knowledge of this view, but I am sure they have theorems that show that contradiction like proofs are valid (given their axioms) within certain conditions. I would recommend going along with what everyone else does and treating &quot;proofs by contradiction&quot; as equally valid, unless you have investigated the Constructivism view and you decide that they are correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As to whether they are clearer, that will depend on the actual proof. Sometimes the clearest way to make a proof is to start from the assumptions and see what they are really saying and why that is going to lead to a contradiction. The most illustrative proof depends on the circumstances.&lt;/p&gt;&#xA;" OwnerUserId="123" LastActivityDate="2010-07-21T03:39:26.747" />
  <row Id="246" PostTypeId="2" ParentId="150" CreationDate="2010-07-21T03:42:05.877" Score="2" ViewCount="" Body="&lt;p&gt;Actually, in some sense, almost all of the continuous functions are nowhere differentiable:&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Weierstrass_function#Density_of_nowhere-differentiable_functions&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Weierstrass_function#Density_of_nowhere-differentiable_functions&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="145" LastActivityDate="2010-07-21T03:42:05.877" />
  <row Id="247" PostTypeId="2" ParentId="194" CreationDate="2010-07-21T03:59:29.613" Score="1" ViewCount="" Body="&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Sieve_of_Eratosthenes&quot; rel=&quot;nofollow&quot;&gt;Sieve of Eratosthenes&lt;/a&gt; is a very intuitive visual representation of why the frequency of prime numbers goes down as you go further out on the number line.&lt;/p&gt;&#xA;" OwnerUserId="19" LastActivityDate="2010-07-21T03:59:29.613" CommentCount="3" />
  <row Id="248" PostTypeId="2" ParentId="206" CreationDate="2010-07-21T04:19:29.520" Score="15" ViewCount="" Body="&lt;p&gt;In the sciences (as opposed to in mathematics) people are often a bit vague about exactly what assumptions they are making about how &quot;well-behaved&quot; things are.  The reason for this is that ultimately these theories are made to be put to the test, so why bother worrying about exactly which properties you're assuming when what you care about is functions coming up in real life which are probably going to satisfy all of your assumptions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is particularly ubiquitous in physics where it is extremely common to make heuristic assumptions about well-behavior.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even in mathematics we do this sometimes.  When people say something is true for n sufficiently large, they often won't bother writing down exactly how large is sufficiently large as long as it's clear from context how to work it out.  Similarly, in an economics paper you could read through the argument and figure out exactly what assumptions they need, but it makes it easier to read to just say &quot;well-behaved.&quot;&lt;/p&gt;&#xA;" OwnerUserId="8" LastEditorUserId="8" LastEditorDisplayName="" LastEditDate="2010-07-21T04:24:47.783" LastActivityDate="2010-07-21T04:24:47.783" />
  <row Id="250" PostTypeId="1" CreationDate="2010-07-21T04:24:20.250" Score="47" ViewCount="2512" Body="&lt;p&gt;The following is a quote from &lt;em&gt;Surely you're joking, Mr. Feynman&lt;/em&gt; .  The question is: are there any interesting theorems that you think would be a good example to tell Richard Feynman, as an answer to his challenge?  Theorems should be totally counter-intuitive, and be easily translatable to everyday language.  (Apparently Banach-Tarski paradox was not a good example.)&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Then I got an idea.  I challenged&#xA;  them: &quot;I bet there isn't a single&#xA;  theorem that you can tell me - what&#xA;  the assumptions are and what the&#xA;  theorem is in terms I can understand -&#xA;  where I can't tell you right away&#xA;  whether it's true or false.&quot;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;It often went like this: They would&#xA;  explain to me, &quot;You've got an orange,&#xA;  OK? Now you cut the orange into a&#xA;  finite number of pieces, put it back&#xA;  together, and it's as big as the sun.&#xA;  True or false?&quot;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&quot;No holes.&quot;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&quot;Impossible!&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&quot;Ha! Everybody gather around! It's&#xA;  So-and-so's theorem of immeasurable&#xA;  measure!&quot;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Just when they think they've got&#xA;  me, I remind them, &quot;But you said an&#xA;  orange! You can't cut the orange peel&#xA;  any thinner than the atoms.&quot;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&quot;But we have the condition of&#xA;  continuity: We can keep on cutting!&quot;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&quot;No, you said an orange, so I&#xA;  assumed that you meant a real orange.&quot;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;So I always won. If I guessed it&#xA;  right, great. If I guessed it wrong,&#xA;  there was always something I could&#xA;  find in their simplification that they&#xA;  left out. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="145" LastEditorUserId="122" LastEditorDisplayName="" LastEditDate="2010-07-21T06:58:27.397" LastActivityDate="2010-11-26T21:43:39.813" CommunityOwnedDate="2010-07-21T04:26:00.670" Title="A challenge by R. P. Feynman: give counter-intuitive theorems that can be translated into everyday language" Tags="&lt;soft-question&gt;&lt;big-list&gt;&lt;examples&gt;" AnswerCount="24" CommentCount="3" FavoriteCount="33" />
  <row Id="251" PostTypeId="2" ParentId="250" CreationDate="2010-07-21T04:26:55.160" Score="4" ViewCount="" Body="&lt;p&gt;Now that Wiles has done the job, I think that Fermat's Last Theorem may suffice. I find it a bit surprising still.&lt;/p&gt;&#xA;" OwnerUserId="16" LastActivityDate="2010-07-21T04:26:55.160" CommunityOwnedDate="2010-07-21T04:26:55.160" CommentCount="3" />
  <row Id="252" PostTypeId="2" ParentId="250" CreationDate="2010-07-21T04:32:06.670" Score="0" ViewCount="" Body="&lt;p&gt;An infinite amount of coaches, each containing an infinite amount of people can be accommodated at &lt;a href=&quot;http://en.wikipedia.org/wiki/Hilbert%27s_paradox_of_the_Grand_Hotel&quot; rel=&quot;nofollow&quot;&gt;Hilbert's Grand Hotel&lt;/a&gt;. Visual demonstration &lt;a href=&quot;http://demonstrations.wolfram.com/TheHilbertHotel/&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="19" LastActivityDate="2010-07-21T04:32:06.670" CommunityOwnedDate="2010-07-21T04:32:06.670" CommentCount="3" />
  <row Id="253" PostTypeId="2" ParentId="250" CreationDate="2010-07-21T04:33:03.923" Score="6" ViewCount="" Body="&lt;p&gt;All of three dimensional space can be filled up with an infinite &lt;a href=&quot;http://en.wikipedia.org/wiki/Space_filling_curve&quot; rel=&quot;nofollow&quot;&gt;curve&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="19" LastEditorUserId="19" LastEditorDisplayName="" LastEditDate="2010-07-21T05:02:27.473" LastActivityDate="2010-07-21T05:02:27.473" CommunityOwnedDate="2010-07-21T04:33:03.923" CommentCount="4" />
  <row Id="254" PostTypeId="2" ParentId="250" CreationDate="2010-07-21T04:34:05.297" Score="8" ViewCount="" Body="&lt;p&gt;My first thought is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Ham_sandwich_theorem&quot; rel=&quot;nofollow&quot;&gt;ham sandwich theorem&lt;/a&gt;--given a sandwich formed by two pieces of bread and one piece of ham (these pieces can be of any reasonable/well-behaved shape) in any positions you choose, it is possible to cut this &quot;sandwich&quot; exactly in half, that is divide each of the three objects exactly in half by volume, with a single &quot;cut&quot; (meaning a single plane).&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-07-21T04:34:05.297" CommunityOwnedDate="2010-07-21T04:34:05.297" />
  <row Id="255" PostTypeId="1" AcceptedAnswerId="259" CreationDate="2010-07-21T05:00:09.020" Score="17" ViewCount="319" Body="&lt;p&gt;Can someone give a simple explanation for why the series 1 + 1/2 + 1/3 + ... doesn't converge, but just grows very slowly? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd prefer an easily comprehensible explanation rather than a rigorous proof of the type I could get from an undergraduate text book.&lt;/p&gt;&#xA;" OwnerUserId="106" LastEditorUserId="121" LastEditorDisplayName="" LastEditDate="2010-07-21T05:23:55.037" LastActivityDate="2010-07-21T05:28:34.980" Title="Why does the series 1/1 + 1/2 + 1/3 + ... not converge?" Tags="&lt;calculus&gt;" AnswerCount="2" FavoriteCount="2" />
  <row Id="256" PostTypeId="2" ParentId="113" CreationDate="2010-07-21T05:06:14.363" Score="0" ViewCount="" Body="&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Two_envelopes_problem&quot; rel=&quot;nofollow&quot;&gt;two envelopes problem&lt;/a&gt; is a good one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See also:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://math.stackexchange.com/questions/234/card-doubling-paradox&quot; rel=&quot;nofollow&quot;&gt;http://math.stackexchange.com/questions/234/card-doubling-paradox&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://mathoverflow.net/questions/9037&quot; rel=&quot;nofollow&quot;&gt;http://mathoverflow.net/questions/9037&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="19" LastActivityDate="2010-07-21T05:06:14.363" CommunityOwnedDate="2010-07-21T05:06:14.363" />
  <row Id="258" PostTypeId="2" ParentId="250" CreationDate="2010-07-21T05:12:52.277" Score="24" ViewCount="" Body="&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Monty_Hall_problem&quot; rel=&quot;nofollow&quot;&gt;Monty Hall problem&lt;/a&gt; fits the bill pretty well. Almost everyone, including most mathematicians, answered it wrong on their first try, and some took a lot of convincing before they agreed with the correct answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's also very easy to explain it to people.&lt;/p&gt;&#xA;" OwnerUserId="122" LastActivityDate="2010-07-21T05:12:52.277" CommunityOwnedDate="2010-07-21T05:12:52.277" CommentCount="5" />
  <row Id="259" PostTypeId="2" ParentId="255" CreationDate="2010-07-21T05:13:12.447" Score="21" ViewCount="" Body="&lt;p&gt;Let's group the terms as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;group 1: 1/1                       (1 term) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;group 2: 1/2 + 1/3                 (2 terms)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;group 3: 1/4 + 1/5 + 1/6 + 1/7     (4 terms) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;group 4: 1/8 + 1/9 + ... + 1/15    (16 terms) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, group n contains 2^n terms.  But also, notice that the smallest element in group n is larger than 1/2^(n+1).  For example all elements in group 2 are larger than 1/2^2.  So the sum of the terms in each group is larger than 2^n * 1/2^(n+1) =  1/2.  Since there are infinitely many groups, and the sum in each group is larger than 1/2, it follows that the total sum is infinite.  &lt;/p&gt;&#xA;" OwnerUserId="145" LastActivityDate="2010-07-21T05:13:12.447" CommentCount="4" />
  <row Id="260" PostTypeId="2" ParentId="250" CreationDate="2010-07-21T05:23:28.717" Score="12" ViewCount="" Body="&lt;p&gt;There are true statements in arithmetic which are unprovable.  Even more remarkably &lt;em&gt;there are explicit polynomial equations&lt;/em&gt; where it's unprovable whether or not they have integer solutions with ZFC! (We need ZFC + consistency of ZFC)&lt;/p&gt;&#xA;" OwnerUserId="31" LastEditorUserId="123" LastEditorDisplayName="" LastEditDate="2010-08-04T03:52:24.753" LastActivityDate="2010-08-04T03:52:24.753" CommunityOwnedDate="2010-07-21T05:23:28.717" CommentCount="6" />
  <row Id="261" PostTypeId="2" ParentId="255" CreationDate="2010-07-21T05:28:34.980" Score="7" ViewCount="" Body="&lt;p&gt;This is not as good an answer as AgCl's, nonetheless people may find it interesting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're used to calculus then you might notice that the sum 1+1/2+1/3+...+1/n is very close to the integral from 1 to n of 1/x.  This definite integral is ln(n), so you should expect 1+1/2+1/3+...+1/n to grow like ln(n).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Although this argument can be made rigorous, it's still unsatisfying because it depends on the fact that the derivative of ln(x) is 1/x, which is probably harder than the original question.  Nonetheless it does illustrate a good general heuristic for quickly determining how sums behave if you already know calculus.&lt;/p&gt;&#xA;" OwnerUserId="8" LastActivityDate="2010-07-21T05:28:34.980" CommentCount="2" />
  <row Id="262" PostTypeId="1" CreationDate="2010-07-21T05:42:01.433" Score="18" ViewCount="1839" Body="&lt;p&gt;If you could go back in time and tell yourself to read a specific book at the beginning of your career as a mathematician, which book would it be?&lt;/p&gt;&#xA;" OwnerUserId="102" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-07-21T06:22:24.537" LastActivityDate="2011-02-15T05:15:58.593" CommunityOwnedDate="2010-07-21T05:42:01.433" Title="What is the single most influential book every mathematician should read?" Tags="&lt;soft-question&gt;&lt;big-list&gt;&lt;books&gt;" AnswerCount="20" CommentCount="1" FavoriteCount="14" />
  <row Id="263" PostTypeId="2" ParentId="262" CreationDate="2010-07-21T05:47:22.827" Score="4" ViewCount="" Body="&lt;p&gt;William Dunham's &quot;Journey through Genius.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, rather that is the book I read that made me want to be a mathematician.&lt;/p&gt;&#xA;" OwnerUserId="8" LastActivityDate="2010-07-21T05:47:22.827" CommunityOwnedDate="2010-07-21T05:47:22.827" CommentCount="3" />
  <row Id="264" PostTypeId="2" ParentId="262" CreationDate="2010-07-21T05:57:33.337" Score="2" ViewCount="" Body="&lt;p&gt;Following Noah's lead I will mention;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;The Man Who Loved Only Numbers&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;How to Read and Do Proofs&quot;&lt;/p&gt;&#xA;" OwnerUserId="16" LastActivityDate="2010-07-21T05:57:33.337" CommunityOwnedDate="2010-07-21T05:57:33.337" />
  <row Id="265" PostTypeId="2" ParentId="262" CreationDate="2010-07-21T05:57:57.783" Score="10" ViewCount="" Body="&lt;p&gt;Polya's &quot;How To Solve It&quot;&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-07-21T05:57:57.783" CommunityOwnedDate="2010-07-21T05:57:57.783" />
  <row Id="266" PostTypeId="2" ParentId="250" CreationDate="2010-07-21T06:12:49.547" Score="60" ViewCount="" Body="&lt;p&gt;Every simple closed curve that you can draw by hand will pass through the corners of some square.  The question was asked by Toeplitz in 1911, and has only been partially answered in 1989 by Stromquist.  As of now, the answer is only known to be positive, for the curves that can be drawn by hand. (i.e. the curves that are piecewise the graph of a continuous function) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I find the result beyond my intuition. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/gl7Ge.gif&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For details, see &lt;a href=&quot;http://www.webpages.uidaho.edu/~markn/squares/&quot; rel=&quot;nofollow&quot;&gt;http://www.webpages.uidaho.edu/~markn/squares/&lt;/a&gt;  (the figure is also borrowed from this site)&lt;/p&gt;&#xA;" OwnerUserId="145" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-09-11T01:39:38.820" LastActivityDate="2010-09-11T01:39:38.820" CommunityOwnedDate="2010-07-21T06:12:49.547" CommentCount="4" />
  <row Id="267" PostTypeId="2" ParentId="262" CreationDate="2010-07-21T06:13:34.803" Score="1" ViewCount="" Body="&lt;p&gt;Probability Theory the Logic of Science.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or anything by Edwin T Jaynes.&lt;/p&gt;&#xA;" OwnerUserId="99" LastActivityDate="2010-07-21T06:13:34.803" CommunityOwnedDate="2010-07-21T06:13:34.803" />
  <row Id="270" PostTypeId="2" ParentId="262" CreationDate="2010-07-21T06:28:48.563" Score="3" ViewCount="" Body="&lt;p&gt;This is an extremely broad question, especially given the wide variety of mathy people here, but I'll bite.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;HSM Coxeter's &lt;em&gt;Introduction to Geometry&lt;/em&gt; is a book that was very important to the development of my interest in mathematics and inclination towards its geometric aspects.&lt;/p&gt;&#xA;" OwnerUserId="76" LastActivityDate="2010-07-21T06:28:48.563" CommunityOwnedDate="2010-07-21T06:28:48.563" CommentCount="1" />
  <row Id="271" PostTypeId="1" AcceptedAnswerId="273" CreationDate="2010-07-21T06:29:43.507" Score="12" ViewCount="191" Body="&lt;p&gt;Is there a way of taking a number known to limited precision (e.g. 1.644934) and finding out an &quot;interesting&quot; real number (e.g. $\pi^2/6$) that's close to it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm thinking of something like Sloane's Online Encyclopedia of Integer Sequences, only for real numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The intended use would be: write a program to calculate an approximation to $\sum_{i=0}^\infty 1/n^2$, look up the answer (&quot;looks close to $\pi^2/6$&quot;) and then use the likely answer to help find a proof that the sum really is $\pi^2/6$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does such a thing exist?&lt;/p&gt;&#xA;" OwnerUserId="31" LastEditorUserId="31" LastEditorDisplayName="" LastEditDate="2010-07-26T19:57:37.337" LastActivityDate="2010-07-26T19:57:37.337" Title="Is there a real number lookup algorithm or service?" Tags="&lt;math-software&gt;&lt;experimental-mathematics&gt;" AnswerCount="3" />
  <row Id="272" PostTypeId="2" ParentId="262" CreationDate="2010-07-21T06:32:05.717" Score="3" ViewCount="" Body="&lt;p&gt;Nicolas Bourbaki's Éléments de mathématique (specifically Topologie Générale and Algèbre).  &lt;/p&gt;&#xA;" OwnerUserId="126" LastActivityDate="2010-07-21T06:32:05.717" CommunityOwnedDate="2010-07-21T06:32:05.717" />
  <row Id="273" PostTypeId="2" ParentId="271" CreationDate="2010-07-21T06:34:58.427" Score="18" ViewCount="" Body="&lt;p&gt;Try &lt;a href=&quot;http://www.wolframalpha.com/input/?i=1.644934&quot; rel=&quot;nofollow&quot;&gt;Wolfram Alpha&lt;/a&gt;. It actually does sequences as well.&lt;/p&gt;&#xA;" OwnerUserId="123" LastActivityDate="2010-07-21T06:34:58.427" CommentCount="2" />
  <row Id="274" PostTypeId="2" ParentId="29" CreationDate="2010-07-21T06:57:37.633" Score="2" ViewCount="" Body="&lt;p&gt;For desktop software, I use &lt;a href=&quot;http://www.speedcrunch.org/en_US/index.html&quot; rel=&quot;nofollow&quot;&gt;SpeedCrunch&lt;/a&gt;. Has a history, lots of mathematical functions, supports variable, etc.&lt;/p&gt;&#xA;" OwnerUserId="122" LastActivityDate="2010-07-21T06:57:37.633" CommunityOwnedDate="2010-07-28T01:32:30.370" />
  <row Id="275" PostTypeId="1" CreationDate="2010-07-21T07:12:50.247" Score="5" ViewCount="981" Body="&lt;p&gt;I'm not a real Mathematician, just an enthusiast. I'm often in the situation where I want to learn some interesting Maths through a good book, but not through an actual Maths textbook. I'm also often trying to give people good Maths books to get them &quot;hooked&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the question: What is a good book, for laymen, which teaches interesting Mathematics, but actually does it in a &quot;real&quot; way. For example, &quot;Fermat's Last Enigma&quot; doesn't count, since it doesn't actually feature any Maths, just a story, and most textbook don't count, since they don't feature a story.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My favorite example of this is &quot;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/014014739X&quot; rel=&quot;nofollow&quot;&gt;Journey Through Genius&lt;/a&gt;&quot;, which is a brilliant combination of interesting storytelling and large amounts of actual Mathematics. It took my love of Maths to a whole other level.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A few more details on what I'm looking for.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The audience of &quot;laymen&quot; should be anyone who has the ability (and desire) to understand actual mathematics, but does not want to learn from a textbook. Obviously I'm thinking about myself here, as a programmer who loves mathematics, I love being exposed to real maths, but I'm not going to get into it seriously. That's why books that show actual maths, but give a lot more exposition (and much clearer explanations, especially of what the intuition should be) are great.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I say &quot;real maths&quot;, I'm talking about actual proofs, formulas, or other mathematical theories. Specifically, I'm not talking about philosophy, nor am I talking about books which &lt;strong&gt;only&lt;/strong&gt; talk about the history of mats (Simon Singh style), since they only talk &lt;em&gt;about&lt;/em&gt; maths, they don't actually show anything. William Dunham's books and Paul J. Nahin's books are good examples.&lt;/p&gt;&#xA;" OwnerUserId="122" LastEditorUserId="122" LastEditorDisplayName="" LastEditDate="2010-07-21T22:25:14.570" LastActivityDate="2010-11-30T17:31:47.517" CommunityOwnedDate="2010-07-21T07:12:50.247" Title="Best Maths books for non-mathematicians" Tags="&lt;soft-question&gt;&lt;big-list&gt;&lt;books&gt;" AnswerCount="18" CommentCount="2" FavoriteCount="19" />
  <row Id="276" PostTypeId="2" ParentId="275" CreationDate="2010-07-21T07:14:25.783" Score="6" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/014014739X&quot; rel=&quot;nofollow&quot;&gt;Journey Through Genius&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://ecx.images-amazon.com/images/I/519AarC3x0L.jpg&quot; alt=&quot;cover&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A brilliant combination of interesting storytelling and large amounts of actual Mathematics. It took my love of Maths to a whole other level.&lt;/p&gt;&#xA;" OwnerUserId="122" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-07T14:45:32.043" LastActivityDate="2010-11-07T14:45:32.043" CommunityOwnedDate="2010-07-21T07:14:25.783" CommentCount="1" />
  <row Id="277" PostTypeId="2" ParentId="153" CreationDate="2010-07-21T07:20:11.040" Score="16" ViewCount="" Body="&lt;p&gt;From &lt;a href=&quot;http://www.helium.com/items/895074-comparing-matlab-and-mathematica&quot; rel=&quot;nofollow&quot;&gt;helium.com&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The comparison between Matlab and&#xA;  Mathematica is NOT numeric versus&#xA;  symbolic as some claim. It is 'high&#xA;  level complete system' versus 'low&#xA;  level core with specialist toolboxes'.&#xA;  Mathematica can do all the numeric&#xA;  math and matrix work that Matlab can&#xA;  do, as fast and as accurately. Matlab&#xA;  is almost strictly a subset of the&#xA;  functionality of Mathematica.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Much of the wider functionality for&#xA;  Matlab is sold as additional&#xA;  toolboxes. e.g. Matlab has toolboxes&#xA;  for curve fitting, statistics,&#xA;  optimization, symbolic math, PDEs,&#xA;  Image acquisition etc but ALL of these&#xA;  areas of functionality are built in to&#xA;  Mathematica already. Aside from the&#xA;  cost and inconvenience of making you&#xA;  buy every function that you need&#xA;  separately, the Matlab design means&#xA;  that the core is inherently low level&#xA;  (since it cannot call on any of the&#xA;  toolbox functionality in case you have&#xA;  not purchased that toolbox), and many&#xA;  of these functions do not fit well&#xA;  together (e.g. no toolbox function can&#xA;  be used with symbolic parameters as&#xA;  they are all written using the&#xA;  numeric-only Matlab core).&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Matlab's only comparative strength&#xA;  lies in some of its specialist&#xA;  toolboxes (e.g. control theory where&#xA;  it is dominant) but if you are not in&#xA;  that specialism, those are of no use.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Some specific differences between the&#xA;  core systems:&lt;/p&gt;&#xA;  &#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;Matlab is machine precision numeric only with some support for complex&#xA;  arithmetic - all Mathematica functions&#xA;  support machine precision and&#xA;  arbitrary precision real and complex&#xA;  arithmetic, exact arithmetic and&#xA;  symbolic computation.&lt;/li&gt;&#xA;  &lt;li&gt;Matlab basic data model is 2D (matrices) with a pointer mechanism to&#xA;  emulate higher dimensions -&#xA;  Mathematica basic data model handles&#xA;  any dimensionality without extra work.&lt;/li&gt;&#xA;  &lt;li&gt;Many of Mathematica's functions switch automatically between different&#xA;  algorithms (with a manual overide) -&#xA;  in Matlab you must always choose the&#xA;  algorithm.&lt;/li&gt;&#xA;  &lt;li&gt;Matlab requires you to manually linearize ODEs to a system of first&#xA;  order equations before it can solve&#xA;  them - Mathematica accepts any order&#xA;  ODE.&lt;/li&gt;&#xA;  &lt;li&gt;Matlab can do integration up to only three dimensions (with&#xA;  triplequad) Mathematica handles&#xA;  n-dimensional integration for any n.&lt;/li&gt;&#xA;  &lt;li&gt;Matlab's language is procedural only (rather like higher level&#xA;  FORTRAN) - Mathematica allows&#xA;  programming in procedural, functional,&#xA;  object oriented and rule based styles.&lt;/li&gt;&#xA;  &lt;li&gt;Matlab has text interface and a basic home grown IDE - Mathematica has&#xA;  text interface, document interface,&#xA;  GUI tools, and full Eclipse based IDE.&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="97" LastEditorUserId="97" LastEditorDisplayName="" LastEditDate="2010-08-16T07:10:58.270" LastActivityDate="2010-08-16T07:10:58.270" CommentCount="7" />
  <row Id="278" PostTypeId="2" ParentId="275" CreationDate="2010-07-21T07:20:16.393" Score="5" ViewCount="" Body="&lt;p&gt;I've been successful using Courant and Robbins, &lt;em&gt;What is Mathematics&lt;/em&gt;, on adults who have not had a math class fora few decades, but are open to the idea of learning more about math. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some sections are too advanced for someone with only high school math, and many more will appear that way to the person at first, but do not actually rely on anything beyond high school math.&lt;/p&gt;&#xA;" OwnerUserId="76" LastActivityDate="2010-07-21T07:20:16.393" CommunityOwnedDate="2010-07-21T07:20:16.393" />
  <row Id="279" PostTypeId="2" ParentId="271" CreationDate="2010-07-21T07:26:04.723" Score="2" ViewCount="" Body="&lt;p&gt;Sometimes the decimal digits of numbers will appear in Sloane's On-Line Encyclopedia of Integer Sequences &lt;a href=&quot;http://www.research.att.com/~njas/sequences/&quot; rel=&quot;nofollow&quot;&gt;OIES&lt;/a&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;E.g. &lt;a href=&quot;http://www.research.att.com/~njas/sequences/A000796&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; is the decimal expansion of pi.&lt;/p&gt;&#xA;" OwnerUserId="139" LastActivityDate="2010-07-21T07:26:04.723" />
  <row Id="280" PostTypeId="2" ParentId="262" CreationDate="2010-07-21T07:32:14.413" Score="1" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Euclid%27s_Elements&quot; rel=&quot;nofollow&quot;&gt;Euclid's Elements&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Principia_mathematica&quot; rel=&quot;nofollow&quot;&gt;Newton's Principia Mathematica&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ideally in the original languages of Ancient Greek and Latin respectively! No, just kidding. But they are &lt;em&gt;true&lt;/em&gt; classics that any accomplished mathematician should read at some point during their career. Not because they'll teach you something you don't already know, but they provide a unique insight into the mind of these giants.&lt;/p&gt;&#xA;" OwnerUserId="56" LastEditorUserId="56" LastEditorDisplayName="" LastEditDate="2010-07-28T15:50:09.213" LastActivityDate="2010-07-28T15:50:09.213" CommunityOwnedDate="2010-07-21T07:32:14.413" CommentCount="6" />
  <row Id="281" PostTypeId="2" ParentId="6" CreationDate="2010-07-21T07:32:58.407" Score="4" ViewCount="" Body="&lt;p&gt;Some good options:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://fooplot.com/&quot; rel=&quot;nofollow&quot;&gt;FooPlot&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.onlinecharttool.com/graph.php&quot; rel=&quot;nofollow&quot;&gt;Online Chart Tool&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://graph.seriesmathstudy.com/&quot; rel=&quot;nofollow&quot;&gt;GraphFunc Tool&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="97" LastActivityDate="2010-07-21T07:32:58.407" CommunityOwnedDate="2010-07-21T07:32:58.407" />
  <row Id="282" PostTypeId="1" CreationDate="2010-07-21T07:34:33.270" Score="-1" ViewCount="241" Body="&lt;p&gt;What are the latest major discoveries in math? What areas and problems are considered important for overall advance of math?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, given that most of those topics are most likely very advanced - what is the best learning path to approach them?&lt;/p&gt;&#xA;" OwnerUserId="68" LastEditorUserId="126" LastEditorDisplayName="" LastEditDate="2010-07-21T07:36:43.060" LastActivityDate="2010-07-21T07:54:15.483" CommunityOwnedDate="2010-07-21T07:55:49.940" ClosedDate="2010-07-21T07:55:54.620" Title="What are the cutting-edge topics in modern math?" Tags="&lt;soft-question&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="283" PostTypeId="1" AcceptedAnswerId="293" CreationDate="2010-07-21T07:42:33.560" Score="4" ViewCount="320" Body="&lt;p&gt;Is there a consensus in the mathematical community, or some accepted authority, to determine whether zero should be classified as a natural number?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems as though formerly 0 was considered in the set of natural numbers but now it seems more common to see definitions saying that the natural numbers are precisely the positive integers.  &lt;/p&gt;&#xA;" OwnerUserId="106" LastEditorUserId="92" LastEditorDisplayName="" LastEditDate="2010-07-21T12:15:31.773" LastActivityDate="2010-07-23T14:53:20.897" Title="Is 0 a natural number?" Tags="&lt;terminology&gt;&lt;natural-number&gt;" AnswerCount="5" CommentCount="4" FavoriteCount="2" />
  <row Id="284" PostTypeId="2" ParentId="90" CreationDate="2010-07-21T07:42:48.740" Score="2" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://www.purplemath.com/lessons.htm&quot; rel=&quot;nofollow&quot;&gt;&lt;strong&gt;Purplemath&lt;/strong&gt;&lt;/a&gt; has a list for math lessons and tutoring. It's a list with various links and short reviews referring to tutoring and instructional resources.&lt;/p&gt;&#xA;" OwnerUserId="97" LastActivityDate="2010-07-21T07:42:48.740" CommunityOwnedDate="2010-07-21T07:42:48.740" />
  <row Id="285" PostTypeId="2" ParentId="282" CreationDate="2010-07-21T07:45:02.323" Score="2" ViewCount="" Body="&lt;p&gt;Higher category theory and Brave new algebraic geometry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to approach these subjects, you should take a look at (roughly in this order):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Algebra by Lang&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Categories for the working mathematician by Mac Lane&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Categories and Sheaves by Kashiwara-Schapira&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basic Concepts of Enriched Category Theory by Kelly&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Model Categories by Hovey&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Model Categories and their Localizations by Hirschhorn&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Simplicial Homotopy theory by Goerss-Jardine&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notes on Quasicategories by Joyal&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Higher Topos Theory by Lurie&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Higher Operads, Higher Categories by Leinster&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Introduction to commutative algebra by Atiyah-MacDonald&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Math 614 notes by Hochster&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Math 615 notes (2010) by Hochster&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Derived Algebraic Geometry (thesis) by Lurie&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Homotopical Algebraic Geometry (I&amp;amp;II) by Toen-Vezzosi&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In all seriousness though, this question is stupid.&lt;/p&gt;&#xA;" OwnerUserId="126" LastActivityDate="2010-07-21T07:45:02.323" CommunityOwnedDate="2010-07-21T07:45:02.323" />
  <row Id="286" PostTypeId="2" ParentId="81" CreationDate="2010-07-21T07:47:29.243" Score="4" ViewCount="" Body="&lt;p&gt;It isn't quite a blog, but &lt;a href=&quot;http://opinionator.blogs.nytimes.com/category/steven-strogatz/&quot; rel=&quot;nofollow&quot;&gt;Steven Strogatz's 15 part series&lt;/a&gt; for the New York Times was excellent.&lt;/p&gt;&#xA;" OwnerUserId="123" LastActivityDate="2010-07-21T07:47:29.243" CommunityOwnedDate="2010-07-21T07:47:29.243" />
  <row Id="287" PostTypeId="1" AcceptedAnswerId="290" CreationDate="2010-07-21T07:53:17.747" Score="17" ViewCount="1044" Body="&lt;p&gt;Below is a visual proof (!) that 32.5 = 31.5. How could that be?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://farm1.static.flickr.com/48/152036443_ca28c8d2a1_o.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="97" LastEditorUserId="6296" LastEditorDisplayName="" LastEditDate="2011-03-07T11:26:10.013" LastActivityDate="2011-06-04T13:56:07.523" Title="How come 32.5 = 31.5?" Tags="&lt;geometry&gt;&lt;recreational-mathematics&gt;" AnswerCount="5" CommentCount="22" FavoriteCount="6" />
  <row Id="288" PostTypeId="2" ParentId="282" CreationDate="2010-07-21T07:54:15.483" Score="2" ViewCount="" Body="&lt;p&gt;There are a few places on the web that have lists of open problems in specific subfields. For example,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://garden.irmacs.sfu.ca/&quot; rel=&quot;nofollow&quot;&gt;The Open Problems Garden&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(this is not an endorsement, I don't know much about the site besides its existence)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://maven.smith.edu/~orourke/TOPP/&quot; rel=&quot;nofollow&quot;&gt;The Open Problems Project&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is specific to discrete and computational geometry. The latter is a favorite of some people looking for nice research projects for undergraduates or high school students, as the problems tend to have statements, at least, accessible without too much background.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The broader question you ask is one I won't address, by dint of intractability.&lt;/p&gt;&#xA;" OwnerUserId="76" LastActivityDate="2010-07-21T07:54:15.483" />
  <row Id="289" PostTypeId="2" ParentId="287" CreationDate="2010-07-21T07:57:42.277" Score="10" ViewCount="" Body="&lt;p&gt;The red and blue triangles are not similar (the ratios of the sides are 3/8 = 0.375 and 2/5 = 0.4 respectively), so the &quot;hypotenuse&quot; of your big triangle is not a straight line.&lt;/p&gt;&#xA;" OwnerUserId="31" LastActivityDate="2010-07-21T07:57:42.277" />
  <row Id="290" PostTypeId="2" ParentId="287" CreationDate="2010-07-21T07:58:04.337" Score="40" ViewCount="" Body="&lt;p&gt;It's an optical illusion - neither the first nor second set of blocks actually describes a triangle. The diagonal edge of the first is slightly concave and that of the second is slightly convex.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To see clearly, look at the gradients of the hypotenuses of the red and blue triangles - they're not 'similar'.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;gradient of blue triangle hypotenuse = 2/5&lt;br&gt;&#xA;gradient of red triangle hypotenuse = 3/8  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since these gradients are different, combining them in the ways shown in the diagram does not produce an overall straight (diagonal) line.&lt;/p&gt;&#xA;" OwnerUserId="56" LastEditorUserId="56" LastEditorDisplayName="" LastEditDate="2010-07-21T12:36:30.200" LastActivityDate="2010-07-21T12:36:30.200" CommentCount="1" />
  <row Id="291" PostTypeId="2" ParentId="81" CreationDate="2010-07-21T07:58:35.410" Score="2" ViewCount="" Body="&lt;p&gt;John D Cook writes &lt;a href=&quot;http://www.johndcook.com/blog/&quot; rel=&quot;nofollow&quot;&gt;The Endeavor&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of the MathWorks blogs: &lt;a href=&quot;http://blogs.mathworks.com/loren/&quot; rel=&quot;nofollow&quot;&gt;Loren on the Art of Matlab&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;... a few more:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://unimodular.net/blog/?p=185&quot; rel=&quot;nofollow&quot;&gt;eon&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://cameroncounts.wordpress.com/&quot; rel=&quot;nofollow&quot;&gt;Peter Cameron's Blog&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.walkingrandomly.com/&quot; rel=&quot;nofollow&quot;&gt;Walking Randomly&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://topologicalmusings.wordpress.com/&quot; rel=&quot;nofollow&quot;&gt;Todd and Vishal's Blog&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Check their blogrolls for more)&lt;/p&gt;&#xA;" OwnerUserId="121" LastActivityDate="2010-07-21T07:58:35.410" CommunityOwnedDate="2010-07-21T07:58:35.410" />
  <row Id="292" PostTypeId="2" ParentId="287" CreationDate="2010-07-21T07:59:12.960" Score="8" ViewCount="" Body="&lt;p&gt;No, this is not a proof of your statement. If you look very closely, you will see that the hypotenuse of the triangles aren't straight. You can also verify this algebraically by calculating the internal angles of the triangle using trigonometry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other way to demonstrate the non-straightness of the hypotenuse is in the fact that the red and blue triangles aren't similar, which can be seen by the difference in the ratios of their width and height. For the blue triangle, this is 2/5 and for the red triangle this is 3/8. As they are both right angled triangles, these ratios would be the same for similar triangles and the fact that they aren't means the interior angles are different. &lt;/p&gt;&#xA;" OwnerUserId="35" LastActivityDate="2010-07-21T07:59:12.960" CommentCount="1" />
  <row Id="293" PostTypeId="2" ParentId="283" CreationDate="2010-07-21T08:02:01.160" Score="5" ViewCount="" Body="&lt;p&gt;Simple answer: sometimes yes, sometimes no, it's usually stated (or impled by notation). From the &lt;a href=&quot;http://en.wikipedia.org/wiki/Natural_number&quot; rel=&quot;nofollow&quot;&gt;Wikipedia article&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In mathematics, there are two&#xA;  conventions for the set of natural&#xA;  numbers: it is either the set of&#xA;  positive integers  {1, 2, 3, ...}&#xA;  according to the traditional&#xA;  definition; or the set of non-negative&#xA;  integers {0, 1, 2, ...} according to a&#xA;  definition first appearing in the&#xA;  nineteenth century.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Saying that, more often than not I've seen the natural numbers only representing the 'counting numbers' (i.e. excluding zero). This was the traditional historical definition, and makes more sense to me. Zero is in many ways the 'odd one out' - indeed, historically it was not discovered (described?) until some time after the natural numbers.&lt;/p&gt;&#xA;" OwnerUserId="56" LastActivityDate="2010-07-21T08:02:01.160" CommentCount="5" />
  <row Id="294" PostTypeId="2" ParentId="283" CreationDate="2010-07-21T08:03:58.067" Score="3" ViewCount="" Body="&lt;p&gt;There are the two definitions, as you say. However the set of strictly positive numbers being the natural numbers is actually the older definition. Inclusion of 0 in the natural numbers is a definition for them that first occurred in the 19th century. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Peano Axioms for natural numbers take 0 to be one though, so if you are working with these axioms (and a lot of natural number theory does) then you take 0 to be a natural number.&lt;/p&gt;&#xA;" OwnerUserId="35" LastEditorUserId="35" LastEditorDisplayName="" LastEditDate="2010-07-21T11:58:57.843" LastActivityDate="2010-07-21T11:58:57.843" CommentCount="2" />
  <row Id="295" PostTypeId="2" ParentId="11" CreationDate="2010-07-21T08:20:33.500" Score="-5" ViewCount="" Body="&lt;p&gt;For any digit n = 0,1,2,3,4,5,6,7,8,9 the fraction&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;n/9 = 0.nnnnnnnn... (periodically).&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Therefore 9/9 is equal to 0.999999999.... but also since 9 divided by 9 equals one,&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;9/9 = 0.9999999.... = 1&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="163" LastActivityDate="2010-07-21T08:20:33.500" CommunityOwnedDate="2010-07-28T06:53:43.987" CommentCount="2" />
  <row Id="296" PostTypeId="2" ParentId="283" CreationDate="2010-07-21T08:25:25.423" Score="8" ViewCount="" Body="&lt;p&gt;There is no &quot;official rule&quot;, it depends from what you want to do with natural numbers. Originally they started from 1 because 0 was not given the status of number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nowadays if you see N&lt;sup&gt;+&lt;/sup&gt; (it's a superscript) you may be assured we are talking about numbers from 1 above; N is usually for numbers from 0 above.&lt;/p&gt;&#xA;" OwnerUserId="89" LastActivityDate="2010-07-21T08:25:25.423" />
  <row Id="297" PostTypeId="1" AcceptedAnswerId="315" CreationDate="2010-07-21T08:41:58.600" Score="4" ViewCount="577" Body="&lt;p&gt;Are there any simple methods for calculating the digits of Pi? Computers are able to calculate billions of digits, so there must be an algorithm for computing them. Is there a simple algorithm that can be computed by hand in order to compute the first few digits?&lt;/p&gt;&#xA;" OwnerUserId="59" LastEditorUserId="126" LastEditorDisplayName="" LastEditDate="2010-07-21T19:37:32.223" LastActivityDate="2010-09-04T01:48:27.217" Title="Simple numerical methods for calculating the digits of Pi" Tags="&lt;approximation&gt;&lt;numerical-methods&gt;&lt;pi&gt;" AnswerCount="6" CommentCount="5" FavoriteCount="2" />
  <row Id="299" PostTypeId="2" ParentId="120" CreationDate="2010-07-21T08:47:03.843" Score="8" ViewCount="" Body="&lt;p&gt;actually 1 was considered a prime number until the beginning of 20th century. Unique factorization was a driving force beneath its changing of status, since it's formulation is quickier if 1 is not considered a prime; but I think that group theory was the other force. Indeed I prefer to describe numbers as primes, composites and &lt;strong&gt;unities&lt;/strong&gt;, that is numbers whose inverse exists (so if we take the set of integer numbers Z, we have that 1 and -1 are unities and we still have unique factorization up to unities).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can always amend the defition of a prime number and say it is a number with exactly two divisors: in this way 1 is not a prime by definition :-)&lt;/p&gt;&#xA;" OwnerUserId="89" LastEditorUserId="89" LastEditorDisplayName="" LastEditDate="2010-07-21T10:15:01.427" LastActivityDate="2010-07-21T10:15:01.427" />
  <row Id="300" PostTypeId="2" ParentId="198" CreationDate="2010-07-21T08:54:52.967" Score="1" ViewCount="" Body="&lt;p&gt;when dealing with infinite sets, many propositions cannot be proved mithout using non-constructive proofs. Axiom of Choice (AC, in brief) or an equivalent proposition is required. Russell's paradox is not a problem per se, you just rule out certain collections of things as sets; Banach-Tarski paradox (you may take a ball, &quot;divide&quot; it in a finite number of parts, translate and rotate them and obtain &lt;strong&gt;two&lt;/strong&gt; balls equal to the first) may be worse indeed. But few mathematicians would prefer not to do a lot of maths because AC is not allowed!&lt;/p&gt;&#xA;" OwnerUserId="89" LastActivityDate="2010-07-21T08:54:52.967" />
  <row Id="301" PostTypeId="1" AcceptedAnswerId="312" CreationDate="2010-07-21T08:59:10.337" Score="6" ViewCount="144" Body="&lt;p&gt;The arithmetic hierarchy defines the &amp;Pi;_1 formulae of arithmetic to be formulae that are provably equivalent to a formula in &lt;a href=&quot;http://en.wikipedia.org/wiki/Prenex_normal_form&quot; rel=&quot;nofollow&quot;&gt;prenex normal form&lt;/a&gt; that only has universal quantifiers, and &amp;Sigma;_1 if it is provably equivalent to a prenex normal form with only existential quantifiers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A formula is &amp;Delta;_1 if it is both &amp;Pi;_1 and &amp;Sigma;_1.  These formulae are often called recursive: why?&lt;/p&gt;&#xA;" OwnerUserId="100" LastEditorUserId="630" LastEditorDisplayName="" LastEditDate="2011-05-03T10:53:25.877" LastActivityDate="2011-05-03T10:53:25.877" Title="Why are Delta-1 sentences of arithmetic called recursive?" Tags="&lt;logic&gt;&lt;computability&gt;&lt;proof-theory&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="302" PostTypeId="2" ParentId="237" CreationDate="2010-07-21T09:24:09.863" Score="1" ViewCount="" Body="&lt;p&gt;Representing a polygon by its edge path might not be the most useful, especially if you want to ask about inclusion for many points.  Consider &lt;a href=&quot;http://en.wikipedia.org/wiki/Polygon_triangulation&quot; rel=&quot;nofollow&quot;&gt;triangulating the polygon&lt;/a&gt;, which is trivial for convex polygons, and not difficult to find O(n log(n)) for hairier cases.  Then determining whether the point is in the polygon reduces to whether it is in anyone of the triangles, which is easy.&lt;/p&gt;&#xA;" OwnerUserId="100" LastActivityDate="2010-07-21T09:24:09.863" CommentCount="1" />
  <row Id="303" PostTypeId="1" AcceptedAnswerId="310" CreationDate="2010-07-21T09:33:20.273" Score="1" ViewCount="129" Body="&lt;p&gt;Any homomorphism φ between the rings Z_18 and Z_15 is completely defined by φ(1). So from&lt;/p&gt;&#xA;&#xA;&lt;p&gt;0 = φ(0) = φ(18) = φ(18 * 1) = 18 * φ(1) = 15 * φ(1) + 3 * φ(1) = 3 * φ(1)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;we get that φ(1) is either 5 or 10. But how can I prove or disprove that these 2 are valid homomorphisms?&lt;/p&gt;&#xA;" OwnerUserId="26" LastEditorUserId="38" LastEditorDisplayName="" LastEditDate="2010-07-23T18:28:06.873" LastActivityDate="2010-07-23T18:28:06.873" Title="What are all the homomorphisms between the rings Z_18 and Z_15?" Tags="&lt;commutative-algebra&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="304" PostTypeId="2" ParentId="283" CreationDate="2010-07-21T09:52:00.850" Score="0" ViewCount="" Body="&lt;p&gt;I remember all of my courses at University using only positive integers (not including 0) for the Natural Numbers. It's possible that they had come to an agreement amongst the Maths Faculty, but during at least two courses we generated the set of natural numbers in ways that wouldn't make sense if 0 was included.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One involved the cardinality of Sets of Sets, the other defined the natural numbers in terms of the number 1 and addition only (0 and Negative Integers come into the picture later when you define an inverse to addition).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a result when teaching the difference between Integers and Natural Numbers I always define 0 as an integer that isn't a Natural Number.&lt;/p&gt;&#xA;" OwnerUserId="86" LastActivityDate="2010-07-21T09:52:00.850" CommentCount="1" />
  <row Id="305" PostTypeId="2" ParentId="90" CreationDate="2010-07-21T10:20:17.880" Score="3" ViewCount="" Body="&lt;p&gt;For basic mathematics &lt;a href=&quot;http://www.themathpage.com/Arith/arithmetic.htm&quot; rel=&quot;nofollow&quot;&gt;mathpage&lt;/a&gt; has a long course, including the list below. Don't be put off by the elementary school style lesson names, it does go into some depth with each one.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&#xA;Lesson 1    Reading and Writing Whole Numbers&#xA;Lesson 2    The Meaning of Decimals&#xA;Lesson 3    Multiplying and Dividing&#xA;Lesson 4    More Elementary Addition&#xA;Lesson 5    Adding Whole Numbers and Decimals&#xA;Lesson 6    The Meaning of Subtraction&#xA;Lesson 7    Subtracting Whole Numbers and&#xA;Lesson 8    The Meaning of Multiplication&#xA;Lesson 9     Multiplying Whole Numbers&#xA;Lesson 10    The Meaning of Division&#xA;Lesson 11    Short Division&#xA;Lesson 12    Dividing Decimals  &#xA;Lesson 13    Percent with a Calculator  &#xA;Lesson 14    Parts of Natural Numbers 1  &#xA;Lesson 15    Parts of Natural Numbers 2  &#xA;Lesson 16    Ratio and Proportion  1  &#xA;Lesson 17    Ratio and Proportion  2  &#xA;Lesson 18    Proportionality  &#xA;Lesson 19    Proper Fractions, Mixed Numbers&#xA;Lesson 20    Unit Fractions  &#xA;Lesson 21    Equivalent Fractions  &#xA;Lesson 22    Lowest Common Multiple  &#xA;Lesson 23    Fractions into Decimals  &#xA;Lesson 24    Adding and Subtracting Fractions &#xA;Lesson 25    Multiplying Fractions &#xA;Lesson 26    The Meaning of Multiplying Fractions&#xA;Lesson 27    Percents are Ratios  &#xA;Lesson 28    Percent of a Number  &#xA;Lesson 29    What Percent?&#xA;Lesson 30    Percent Increase or Decrease  &#xA;Lesson 31    Prime Numbers  &#xA;Lesson 32    Greatest Common Divisor&#xA;&lt;/pre&gt;&#xA;" OwnerUserId="54" LastActivityDate="2010-07-21T10:20:17.880" CommunityOwnedDate="2010-07-21T10:20:17.880" />
  <row Id="306" PostTypeId="2" ParentId="196" CreationDate="2010-07-21T10:33:49.587" Score="6" ViewCount="" Body="&lt;p&gt;the smaller wheel does not just rotate, but also slides. If you had cogwheels instead of smooth wheels, you'd notice that movement is not possible.&lt;/p&gt;&#xA;" OwnerUserId="89" LastActivityDate="2010-07-21T10:33:49.587" />
  <row Id="307" PostTypeId="1" AcceptedAnswerId="308" CreationDate="2010-07-21T11:23:38.983" Score="3" ViewCount="913" Body="&lt;p&gt;How can I show that (n-1)! is congruent to -1 (mod n) iff n is prime?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="26" LastActivityDate="2010-10-15T15:51:28.870" Title="Prove that (n-1)! = -1(mod n) iff n is prime" Tags="&lt;number-theory&gt;&lt;group-theory&gt;" AnswerCount="4" />
  <row Id="308" PostTypeId="2" ParentId="307" CreationDate="2010-07-21T11:28:20.957" Score="10" ViewCount="" Body="&lt;p&gt;$$n\text{ is prime if }(n-1)! \equiv -1 \pmod n$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This direction is easy. If $n$ is composite, then there exists $k|n$ and $k\lt n$. So $k|(n-1)!$ and $k \equiv 1 \pmod n$. This means $k$ needs to divide $1$. So $n$ must be prime (or $1$, but we can eliminate this by substitution).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(n-1)! \equiv -1\text{ if }n\text{ is prime}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wikipedia contains &lt;a href=&quot;http://en.wikipedia.org/wiki/Wilson%27s_theorem#Proofs&quot; rel=&quot;nofollow&quot;&gt;two proofs&lt;/a&gt; of this result known as Wilson's theorem. The first proof only uses basic abstract algebra and so should be understandable with a good knowledge of modular arithmetic. Just in case, I prove below that each element &lt;code&gt;1, 2, ... n-1&lt;/code&gt; has a unique inverse &lt;code&gt;mod n&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They use the fact that integers &lt;code&gt;mod p&lt;/code&gt; form a group and hence that each element &lt;code&gt;x not congruent 0&lt;/code&gt; has a multiplicative inverse (a number y such that &lt;code&gt;xy congruent 1 (mod n)&lt;/code&gt;.&#xA;We show this as follows. Suppose &lt;code&gt;n not | x&lt;/code&gt;, for &lt;code&gt;n&lt;/code&gt; prime. From the uniqueness of prime factorisations, &lt;code&gt;xn&lt;/code&gt; is the first product of &lt;code&gt;x&lt;/code&gt;, after &lt;code&gt;0x&lt;/code&gt;, divisible by &lt;code&gt;n&lt;/code&gt; (use prime factorisation theorem). If we look at the series &lt;code&gt;kn&lt;/code&gt; in &lt;code&gt;mod n&lt;/code&gt;, this cycles and must have cycle length &lt;code&gt;n&lt;/code&gt;. Therefore, each element &lt;code&gt;x, 2x,... nx&lt;/code&gt; must be different modulo &lt;code&gt;n&lt;/code&gt;, including one, &lt;code&gt;y&lt;/code&gt;, with &lt;code&gt;xy congruent 1 mod n&lt;/code&gt;. Furthermore, due to the cycle length being &lt;code&gt;n&lt;/code&gt;, each only one of those elements will be an inverse. So every element has a unique inverse (although 1 and -1 are their own inverses).&lt;/p&gt;&#xA;" OwnerUserId="123" LastActivityDate="2010-10-15T15:51:28.870" CommentCount="1" />
  <row Id="309" PostTypeId="1" AcceptedAnswerId="327" CreationDate="2010-07-21T12:16:11.880" Score="6" ViewCount="214" Body="&lt;p&gt;In category theory, a subobject of X is defined as an object Y with a monomorphism, from Y to X. If A is a subobject of B, and B a subobject of A, are they isomorphic? It is not true in general that having monomorphisms going both ways between two objects is sufficient for isomorphy, so it would seem the answer is no.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I ask because I'm working through the exercises in Geroch's Mathematical Physics, and one of them asks you to prove that the relation &quot;is a subobject of&quot; is reflexive, transitive and antisymmetric. But it can't be antisymmetric if I'm right...&lt;/p&gt;&#xA;" OwnerUserId="172" LastActivityDate="2011-02-13T16:33:38.740" Title="If A is a subobject of B, and B a subobject of A, are they isomorphic?" Tags="&lt;category-theory&gt;" AnswerCount="3" CommentCount="5" FavoriteCount="2" />
  <row Id="310" PostTypeId="2" ParentId="303" CreationDate="2010-07-21T12:27:04.667" Score="8" ViewCount="" Body="&lt;p&gt;If one has a homomorphism of two rings $R, S$, and $R$ has an identity, then the identity must be mapped to an idempotent element of $S$, because the equation $x^2=x$ is preserved under homomorphisms.  Now $5$ is not an idempotent element in $Z_{15}$, so the map generated by $1 \to 5$ is not a homomorphism.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, 10 is an idempotent element of $Z_{15}$. In particular, the subring $T \subset Z_{15}$ generated by 10 has unit 10.  Since it is annihilated by 3, and consequently by 18, there is a unital homomorphism $Z_{18} \to T$ (i.e., mapping 1 to 10).  So your second map is a legitimate homomorphism of rings (composing with the injection $T \to Z_{15}$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically, the point of this answer is to check that one of your maps preserves the relations of the two rings, while the other doesn't.&lt;/p&gt;&#xA;" OwnerUserId="536" LastEditorUserId="536" LastEditorDisplayName="" LastEditDate="2010-07-21T13:29:30.147" LastActivityDate="2010-07-21T13:29:30.147" CommentCount="7" />
  <row Id="311" PostTypeId="1" AcceptedAnswerId="325" CreationDate="2010-07-21T12:30:36.140" Score="5" ViewCount="140" Body="&lt;p&gt;I've read about about &lt;a href=&quot;http://en.wikipedia.org/wiki/Higher_order_logic&quot; rel=&quot;nofollow&quot;&gt;higher-order logics&lt;/a&gt; (i.e. those that build on first-order predicate logic) but am not too clear on their applications. While they are capable of expressing a greater range of proofs (though never &lt;em&gt;all&lt;/em&gt;, by Godel's Incompleteness theorem), they are often said to be less &quot;well-behaved&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mathematicians generally seem to stay clear of such logics when possible, yet they are certainly necessary for prooving some more complicated concepts/theorems, as I understand. (For example, it seems the reals can only be constructed using at least 2nd order logic.) Why is this, what makes them less well-behaved or less useful with respect to logic/proof theory/other fields?&lt;/p&gt;&#xA;" OwnerUserId="56" LastActivityDate="2010-07-27T23:31:34.267" Title="Higher Order Logics" Tags="&lt;logic&gt;&lt;higher-order-logic&gt;" AnswerCount="3" CommentCount="4" FavoriteCount="3" />
  <row Id="312" PostTypeId="2" ParentId="301" CreationDate="2010-07-21T12:31:56.217" Score="5" ViewCount="" Body="&lt;p&gt;If a formula $\phi(x_1, ... x_n)$ is in both $\Sigma_1, \Prod_1$, then one can define a Turing machine to determine whether it is true or false. Namely, in parallel, search for a collection of parameters that makes true the existential formula, and search for a collection of formulas that makes false the universal formula. If the first happens, return true; if the second happens, return false.  One of these must exist, so the Turing machine always halts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(The set of $x_1,...x_n$ such that $\phi(x_1, ... , x_n)$ is valid if $\phi$ belongs to  $\Sigma_1$ is, by contrast, is only recursively enumerable.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By contrast, since the action of any Turing machine is simulable by existential formulas in first-order logic (i.e. there exists a number $k$ such that $M$ halts in $k$ steps), any language which is recursively enumerable can be expressed by existential formulas. Any language whose &lt;em&gt;complement&lt;/em&gt; is recursively enumerable can similarly be expressed by universal formulas (by the analog of deMorgan's laws). So any recursive language (i.e., one which is both recursively enumerable and whose complement is r.e.), can be expressed in both ways.&lt;/p&gt;&#xA;" OwnerUserId="536" LastEditorUserId="536" LastEditorDisplayName="" LastEditDate="2010-07-22T11:34:02.793" LastActivityDate="2010-07-22T11:34:02.793" CommentCount="2" />
  <row Id="313" PostTypeId="2" ParentId="311" CreationDate="2010-07-21T12:35:23.993" Score="2" ViewCount="" Body="&lt;p&gt;Second-order logic does not satisfy the completeness and compactness theorems. Here is a proof that the compactness theorem fails (which itself implies that the completeness theorem fails, because completeness implies compactness).  Namely, there is a second-order way (call it $F$) of expressing that a set is finite: every injective function on the set is surjective. This means that if $P$ is a relation on the product set $S \times S$ such that $P$ satisfies the conditions to be a function, and $P(x,y), P(z,y)$ imply $x=z$, then for all $w$ there is $q$ with $P(w,q)$. This statement is quantified over $P$ as well as the variables $x,y,z,w,q$ so is second-order and clearly expresses finiteness.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, we also have the statement $T_n$ &quot;there exist distinct $x_0, \dots x_n$ in the set,&quot; which is even first order, and states that the set has cardinality at least $n$.&#xA;So the conjunction of $F$ and any finite collection of the $T_n$ is satisfiable, but all of them together cannot be satisfied.&lt;/p&gt;&#xA;" OwnerUserId="536" LastActivityDate="2010-07-21T12:35:23.993" CommentCount="2" />
  <row Id="314" PostTypeId="2" ParentId="271" CreationDate="2010-07-21T12:42:23.590" Score="13" ViewCount="" Body="&lt;p&gt;I've long used Simon Plouffe's &lt;a href=&quot;http://oldweb.cecm.sfu.ca/projects/ISC/ISCmain.html&quot; rel=&quot;nofollow&quot;&gt;inverse symbolic calculator&lt;/a&gt; for this purpose.  It is essentially a searchable list of &quot;interesting&quot; numbers.&lt;/p&gt;&#xA;" OwnerUserId="173" LastActivityDate="2010-07-21T12:42:23.590" CommentCount="1" />
  <row Id="315" PostTypeId="2" ParentId="297" CreationDate="2010-07-21T12:46:15.347" Score="8" ViewCount="" Body="&lt;p&gt;One of the simplest is the series for arctangent:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;tan&lt;sup&gt;-1&lt;/sup&gt; x = x - x&lt;sup&gt;3&lt;/sup&gt;/3 + x&lt;sup&gt;5&lt;/sup&gt;/5 - x&lt;sup&gt;7&lt;/sup&gt;/7 ...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&amp;pi;/4 = tan&lt;sup&gt;-1&lt;/sup&gt; 1, but that converges slowly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can use the &lt;a href=&quot;http://mathworld.wolfram.com/TrigonometricAdditionFormulas.html&quot; rel=&quot;nofollow&quot;&gt;addition formula for the tangent&lt;/a&gt; (tan (a+b) = (tan a + tan b) / (1 - tan a tan b)) to break down &amp;pi;/4  to the sum of two angles and repeat; this can be used to come up with values for the arctangent that are smaller than 1 and therefore converge faster.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;exercise for the reader: show that &amp;pi;/4 = 2 tan&lt;sup&gt;-1&lt;/sup&gt; 2/11 + 3 tan&lt;sup&gt;-1&lt;/sup&gt; 1/7&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This then means that  &amp;pi; = 8 tan&lt;sup&gt;-1&lt;/sup&gt; 2/11 + 12 tan&lt;sup&gt;-1&lt;/sup&gt; 1/7&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Use the arctangent formula with only a few terms and you will have pi to quite a few decimal places. (3 terms yields 3.14160, 4 terms yields 3.1415924, 5 terms yields 3.1415926592)&lt;/p&gt;&#xA;" OwnerUserId="120" LastActivityDate="2010-07-21T12:46:15.347" CommentCount="1" />
  <row Id="316" PostTypeId="1" AcceptedAnswerId="345" CreationDate="2010-07-21T12:47:23.143" Score="8" ViewCount="1910" Body="&lt;p&gt;When I tried to approximate $\int_0^1 (1-x^7)^{1/5} - (1-x^5)^{1/7} dx$, I kept getting answers that were really close to 0, so I think it might be true. But why? When I &lt;a href=&quot;http://integrals.wolfram.com/index.jsp?expr=%281-x%5E7%29%5E%281%2F5%29+-+%281-x%5E5%29%5E%281%2F7%29&amp;amp;random=false&quot; rel=&quot;nofollow&quot;&gt;ask Mathematica&lt;/a&gt;, I get a bunch of symbols I don't understand! &lt;/p&gt;&#xA;" OwnerUserId="73" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-07-22T11:06:49.250" LastActivityDate="2010-08-03T09:08:01.803" Title="Why is $\int_0^1 (1-x^7)^{1/5} - (1-x^5)^{1/7} dx=0$?" Tags="&lt;calculus&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="3" />
  <row Id="318" PostTypeId="2" ParentId="283" CreationDate="2010-07-21T12:56:10.100" Score="2" ViewCount="" Body="&lt;p&gt;Unambiguous names for the set of nonnegative integers (0, 1, 2, 3, ...) are &lt;strong&gt;whole numbers&lt;/strong&gt;&#xA;Unambiguous names for the set of positive integers (1, 2, 3, ...) are &lt;strong&gt;counting numbers&lt;/strong&gt; and $\mathbb{N}^+$.&lt;br&gt;&#xA;As other answers (and your own question) indicate, &lt;strong&gt;natural numbers&lt;/strong&gt;, or $\mathbb{N}$, can have different meanings, depending on context. If which usage you mean is not clear from context, you are better off using one of the other terms.&lt;/p&gt;&#xA;" OwnerUserId="73" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-07-23T14:53:20.897" LastActivityDate="2010-07-23T14:53:20.897" CommentCount="3" />
  <row Id="319" PostTypeId="2" ParentId="8" CreationDate="2010-07-21T12:59:08.067" Score="11" ViewCount="" Body="&lt;p&gt;A lot of people have mentioned Binet's formula.  But I suspect this is not the most practical way to compute the nth Fibonacci number for large n, because it requires either having a very accurate value of sqrt(5) and carrying around lots of decimal places (if you want to do floating-point arithmetic) or expanding large powers of (1+sqrt(5)) using the binomial formula.  The latter comes out to writing the Fibonacci number as a sum of binomial coefficients.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The following formulas hold, though:&#xA;F(2n-1) = F(n)^2 + F(n-1)^2&#xA;F(2n) = (2*F(n-1) + F(n)) F(n)&#xA;which you can find derivations of in &lt;a href=&quot;http://en.wikipedia.org/wiki/Fibonacci_number&quot; rel=&quot;nofollow&quot;&gt;the Wikipedia article on Fibonacci numbers&lt;/a&gt;.  This lets you find F(k), for any k even or odd, in terms of two Fibonacci numbers with approximately half the index.  The result is faster than Binet's formula.&lt;/p&gt;&#xA;" OwnerUserId="173" LastActivityDate="2010-07-21T12:59:08.067" />
  <row Id="320" PostTypeId="2" ParentId="307" CreationDate="2010-07-21T13:00:49.560" Score="2" ViewCount="" Body="&lt;p&gt;to show that (p-1)! &amp;equiv; -1 (mod p) without explicitly use group theory, maybe the simplest path is: (the following assumes p is odd, but if p=2 then the result is immediate)&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;given n &amp;ne; 0, all values n, 2n, ... (p-1)n are different mod p. Otherwise, if hn &amp;equiv; kn (mod p) then (h-k)n &amp;equiv; 0 (mod p) against the hyphotesis that p is prime.&lt;/li&gt;&#xA;&lt;li&gt;this means that each n has an inverse mod p, that is for each n there is a m such that mn &amp;equiv; 1 (mod p).&lt;/li&gt;&#xA;&lt;li&gt;the equation x&lt;sup&gt;2&lt;/sup&gt; &amp;equiv; 1 (mod p) may be written as (x+1)(x-1) &amp;equiv; 0 (mod p); therefore its only solutions are x &amp;equiv; 1 (mod p) and x &amp;equiv; -1 (mod p). For each other number n, an inverse m must exist (because of the pigeonhole principle) but m &amp;ne; n.&lt;/li&gt;&#xA;&lt;li&gt;we are nearly done. Let's couple every number from 2 to p-2 with its own inverse. Their product is 1 (mod p), so they don't count in the overall total. 1 does not count either; it remains just p-1, that is -1 (mod p) as requested.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="89" LastActivityDate="2010-07-21T13:00:49.560" CommentCount="2" />
  <row Id="321" PostTypeId="2" ParentId="303" CreationDate="2010-07-21T13:01:37.873" Score="0" ViewCount="" Body="&lt;p&gt;I will quote &lt;a href=&quot;http://en.wikipedia.org/wiki/Ring_homomorphism&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt; on the definition of a ring homomorphism.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;More precisely, if R and S are rings,&#xA;  then a ring homomorphism is a function&#xA;  f : R → S such that&lt;a href=&quot;http://en.wikipedia.org/wiki/Ring_homomorphism&quot; rel=&quot;nofollow&quot;&gt;1&lt;/a&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;f(a + b) = f(a) + f(b) for all a and b in R&lt;/li&gt;&#xA;  &lt;li&gt;f(ab) = f(a) f(b) for all a and b in R&lt;/li&gt;&#xA;  &lt;li&gt;f(1) = 1&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The last requirement is being relaxed. We can then sub in our &lt;code&gt;f&lt;/code&gt;, in this case &lt;code&gt;f(a1)=5a&lt;/code&gt; or &lt;code&gt;f(a1)=10a&lt;/code&gt; and see if these relations hold. Given that these are small, finite and well understood groups, the problem is easy to solve from here.&lt;/p&gt;&#xA;" OwnerUserId="123" LastActivityDate="2010-07-21T13:01:37.873" />
  <row Id="322" PostTypeId="2" ParentId="262" CreationDate="2010-07-21T13:03:12.423" Score="2" ViewCount="" Body="&lt;p&gt;T.W.Körner, &lt;em&gt;The Pleasures of Counting&lt;/em&gt;. It shows how mathematics is alive.&lt;/p&gt;&#xA;" OwnerUserId="89" LastActivityDate="2010-07-21T13:03:12.423" CommunityOwnedDate="2010-07-21T13:03:12.423" />
  <row Id="323" PostTypeId="2" ParentId="275" CreationDate="2010-07-21T13:07:21.883" Score="7" ViewCount="" Body="&lt;p&gt;I think that a non-mathematician could appreciate T.W.Körner's book &lt;em&gt;The Pleasures of Counting&lt;/em&gt;; but I still believe that the collection of &quot;Mathematical Games&quot; columns from Martin Gardner are the very best thing.&lt;/p&gt;&#xA;" OwnerUserId="89" LastEditorUserId="4079" LastEditorDisplayName="" LastEditDate="2010-11-30T17:31:47.517" LastActivityDate="2010-11-30T17:31:47.517" CommunityOwnedDate="2010-07-21T13:07:21.883" CommentCount="1" />
  <row Id="324" PostTypeId="1" AcceptedAnswerId="330" CreationDate="2010-07-21T13:09:57.197" Score="1" ViewCount="76" Body="&lt;p&gt;What is a unital homomorphism? Why are they important?&lt;/p&gt;&#xA;" OwnerUserId="123" LastEditorUserId="38" LastEditorDisplayName="" LastEditDate="2010-07-23T18:28:26.733" LastActivityDate="2010-07-23T18:28:26.733" Title="Unital homomorphism" Tags="&lt;terminology&gt;&lt;definition&gt;&lt;commutative-algebra&gt;" AnswerCount="2" />
  <row Id="325" PostTypeId="2" ParentId="311" CreationDate="2010-07-21T13:10:04.170" Score="3" ViewCount="" Body="&lt;p&gt;Unfortunately the term is ambiguous: there two kinds of semantics of higher-order languages, and only one is problematic.  Consider the language of second-order arithmetic, where there are quantifiers over both natural numbers and sets of natural numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First is what Quine called &quot;set-theory in sheep's clothing&quot;: this is where  quantification over sets of natural numbers is defined to be over all the sets of number that can be posited.  It's the theory we use when we prove that there can be only one complete, totally ordered field.  It isn't really a logic, there's no complete notion of proof formalisation for it.  Wikipedia calls this &quot;standard semantics&quot;; I'm not sure if there is an authority for this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then there's Henkin semantics, which uses the rule analogous to the lambda-calculus to define a  semantics for second-order quantifiers.  This can be seen as still in the realm of first-order logic, in that sense that the second-order system can be translated into a first-order system conserving provability.  This is how second-order arithmetic is defined.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All the theorems of the Henkin semantics will be &quot;theorems&quot; in the first.&lt;/p&gt;&#xA;" OwnerUserId="100" LastActivityDate="2010-07-21T13:10:04.170" CommentCount="5" />
  <row Id="326" PostTypeId="1" AcceptedAnswerId="328" CreationDate="2010-07-21T13:17:16.040" Score="2" ViewCount="274" Body="&lt;p&gt;So you have 2 dice and you want to get at least a 1 or a 5 (on the dice not added).  How do you go about calculating the answer for this question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This question comes from the game farkle.&lt;/p&gt;&#xA;" OwnerUserId="84" LastEditorUserId="84" LastEditorDisplayName="" LastEditDate="2010-07-21T18:43:07.317" LastActivityDate="2010-07-21T18:43:07.317" Title="Calculating the probability of two dice getting at least a 1 or a 5" Tags="&lt;probability-theory&gt;" AnswerCount="3" CommentCount="4" />
  <row Id="327" PostTypeId="2" ParentId="309" CreationDate="2010-07-21T13:20:59.917" Score="5" ViewCount="" Body="&lt;p&gt;I don't think this can be true in general.  What if we just take the category consisting of two objects A, B and take maps f:A --&gt; B, g: B--&gt; A with no relations between the maps, but forcing associativity.  Then certainly f and g are monomorphisms but A and B are not isomorphic (since there are no relations between maps). &lt;/p&gt;&#xA;" OwnerUserId="9" LastActivityDate="2010-07-21T13:20:59.917" CommentCount="8" />
  <row Id="328" PostTypeId="2" ParentId="326" CreationDate="2010-07-21T13:23:13.453" Score="11" ViewCount="" Body="&lt;p&gt;Go backwards: Calculate the probability that &lt;em&gt;neither&lt;/em&gt; of them shows a 1 or 5. That means both show a 2, 3, 4, or 6. Thats $(4/6)^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence the probability that &lt;em&gt;at least one&lt;/em&gt; shows a 1 or 5 is $1-(2/3)^2=5/9$.&lt;/p&gt;&#xA;" OwnerUserId="3" LastActivityDate="2010-07-21T13:23:13.453" CommentCount="5" />
  <row Id="329" PostTypeId="1" AcceptedAnswerId="483" CreationDate="2010-07-21T13:28:20.443" Score="17" ViewCount="1845" Body="&lt;p&gt;Which is the single best book for Number Theory that everyone who loves Mathematics should read?&lt;/p&gt;&#xA;" OwnerUserId="149" LastEditorUserId="4290" LastEditorDisplayName="" LastEditDate="2011-04-26T16:22:55.610" LastActivityDate="2011-05-31T15:35:10.310" CommunityOwnedDate="2010-07-21T14:53:06.763" Title="Best ever book on Number Theory" Tags="&lt;number-theory&gt;&lt;soft-question&gt;&lt;reference-request&gt;&lt;big-list&gt;&lt;books&gt;" AnswerCount="16" CommentCount="6" FavoriteCount="13" />
  <row Id="330" PostTypeId="2" ParentId="324" CreationDate="2010-07-21T13:28:39.320" Score="0" ViewCount="" Body="&lt;p&gt;A unital homomorphism between rings R and S is a ring homomorphism that sends the identity element of R to the identity element of S.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Homomorphisms (between objects in any algebraic category like groups, rings, vector spaces, etc.) preserve the algebraic structure, and if you want a map between rings with an identity element, it is natural to require this to preserve this element (since it satisfies unique properties).&lt;/p&gt;&#xA;" OwnerUserId="9" LastActivityDate="2010-07-21T13:28:39.320" />
  <row Id="331" PostTypeId="2" ParentId="329" CreationDate="2010-07-21T13:31:47.427" Score="16" ViewCount="" Body="&lt;p&gt;I would still stick with Hardy and Wright, even if it is quite old.&lt;/p&gt;&#xA;" OwnerUserId="89" LastActivityDate="2010-07-21T13:31:47.427" CommentCount="1" />
  <row Id="332" PostTypeId="2" ParentId="324" CreationDate="2010-07-21T13:32:48.360" Score="2" ViewCount="" Body="&lt;p&gt;A lot of results about rings just won't work otherwise: for instance, a unital homomorphism of rings sends units to units.  A nonunital homomorphism doesn't have to do that.  Nonunital homomorphisms can be very degenerate, e.g. the zero homomorphism.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another reason you want homomorphisms to preserve the unit is that this is how you get a map $\spec S \to \spec R$ from a ring-homomorphism $R \to S$.  &lt;/p&gt;&#xA;" OwnerUserId="536" LastActivityDate="2010-07-21T13:32:48.360" />
  <row Id="333" PostTypeId="2" ParentId="326" CreationDate="2010-07-21T13:38:29.280" Score="3" ViewCount="" Body="&lt;p&gt;To visually see the answer given by balpha above, you could write out the entire set of dice rolls&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;[1, 1], [1, 2], [1, 3], [1, 4], [1, 5], [1, 6]&#xA;[2, 1], [2, 2], [2, 3], [2, 4], [2, 5], [2, 6]&#xA;[3, 1], [3, 2], [3, 3], [3, 4], [3, 5], [3, 6]&#xA;[4, 1], [4, 2], [4, 3], [4, 4], [4, 5], [4, 6]&#xA;[5, 1], [5, 2], [5, 3], [5, 4], [5, 5], [5, 6]&#xA;[6, 1], [6, 2], [6, 3], [6, 4], [6, 5], [6, 6]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Total number of possible dice rolls: 36&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dice rolls that contain 1 or a 5: 20&lt;/p&gt;&#xA;&#xA;&lt;p&gt;20/36 = 5/9&lt;/p&gt;&#xA;" OwnerUserId="177" LastActivityDate="2010-07-21T13:38:29.280" />
  <row Id="334" PostTypeId="2" ParentId="275" CreationDate="2010-07-21T13:49:51.470" Score="0" ViewCount="" Body="&lt;p&gt;I think any book by John Allen Paulos would be something any Math enthusiast could enjoy and learn from.&lt;/p&gt;&#xA;" OwnerUserId="177" LastActivityDate="2010-07-21T13:49:51.470" CommunityOwnedDate="2010-07-21T13:49:51.470" />
  <row Id="335" PostTypeId="2" ParentId="326" CreationDate="2010-07-21T13:57:25.867" Score="6" ViewCount="" Body="&lt;p&gt;The other way to visualise this would be to draw a probability tree like so:&#xA;&lt;img src=&quot;http://img.skitch.com/20100721-xwruwx7qnntx1pjmkjq8gxpifs.gif&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(apologies for my poor standard of drawing :) )&lt;/p&gt;&#xA;" OwnerUserId="35" LastEditorUserId="35" LastEditorDisplayName="" LastEditDate="2010-07-21T16:36:15.153" LastActivityDate="2010-07-21T16:36:15.153" />
  <row Id="336" PostTypeId="1" AcceptedAnswerId="338" CreationDate="2010-07-21T13:58:48.657" Score="7" ViewCount="522" Body="&lt;p&gt;&lt;strong&gt;Background:&lt;/strong&gt; Many (if not all) of the transformation matrices used in 3D computer graphics are 4x4, including the three values for &lt;code&gt;x&lt;/code&gt;, &lt;code&gt;y&lt;/code&gt; and &lt;code&gt;z&lt;/code&gt;, plus an additional term which usually has a value of 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given the extra computing effort required to multiply 4x4 matrices instead of 3x3 matrices, there must be a substantial benefit to including that extra fourth term, even though 3x3 matrices &lt;em&gt;should&lt;/em&gt; (?) be sufficient to describe points and transformations in 3D space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; Why is the inclusion of a fourth term beneficial? I can guess that it makes the computations easier in some manner, but I would really like to know &lt;em&gt;why&lt;/em&gt; that is the case.&lt;/p&gt;&#xA;" OwnerUserId="59" LastEditorUserId="126" LastEditorDisplayName="" LastEditDate="2010-07-21T16:56:59.480" LastActivityDate="2010-07-21T16:56:59.480" Title="Why are 3D transformation matrices [4]x[4] instead of [3]x[3]?" Tags="&lt;geometry&gt;&lt;linear-algebra&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="3" />
  <row Id="337" PostTypeId="2" ParentId="329" CreationDate="2010-07-21T14:03:38.800" Score="5" ViewCount="" Body="&lt;p&gt;&lt;em&gt;A concise introduction to the theory of numbers&lt;/em&gt; by Alan Baker (1970 Fields medalist) covers a lot of ground in less than 100 pages, and does so in a fluid way that never feels rushed. I love this little book.&lt;/p&gt;&#xA;" OwnerUserId="115" LastActivityDate="2010-07-21T14:03:38.800" />
  <row Id="338" PostTypeId="2" ParentId="336" CreationDate="2010-07-21T14:04:20.653" Score="10" ViewCount="" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;even though 3x3 matrices should (?) be sufficient to describe points and transformations in 3D space.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No, they aren't enough! Suppose you represent points in space using 3D vectors. You can transform these using 3x3 matrices. But if you examine the definition of matrix multiplication you should see immediately that multiplying a zero 3D vector by a 3x3 matrix gives you another zero vector. So simply multiplying by a 3x3 matrix can never move the origin. But translations and rotations do need to move the origin. So 3x3 matrices are not enough.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I haven't tried to explain exactly how 4x4 matrices are used. But I hope I've convinced you that 3x3 matrices aren't up to the task and that something more is needed.&lt;/p&gt;&#xA;" OwnerUserId="80" LastActivityDate="2010-07-21T14:04:20.653" CommentCount="2" />
  <row Id="339" PostTypeId="2" ParentId="275" CreationDate="2010-07-21T14:12:45.597" Score="6" ViewCount="" Body="&lt;p&gt;John Derbyshire's &lt;em&gt;Prime Obsession&lt;/em&gt; is about Riemann's hypothesis. One of the stated goals of the author is to explain what &quot;all non-trivial zeros of the zeta function have real part one-half&quot; means to readers who have no background in calculus. Odd-numbered chapters tell the story of how Riemann came to his hypothesis, and even-numbered chapters are more mathematical in nature.&lt;/p&gt;&#xA;" OwnerUserId="115" LastActivityDate="2010-07-21T14:12:45.597" CommunityOwnedDate="2010-07-21T14:12:45.597" CommentCount="2" />
  <row Id="340" PostTypeId="2" ParentId="336" CreationDate="2010-07-21T14:20:30.210" Score="7" ViewCount="" Body="&lt;p&gt;To follow up user80's answer, you want to get transformations of the form v --&gt; Av + b, where A is a 3 by 3 matrix (the linear part of transformation) and b is a 3-vector.  We can encode this transformation in a 4 x 4 matrix by putting A in the top left with three 0's below it and making the last column be (b,1).  Multiplying the 4-vector (v,1) with this matrix will give you (Av + b, 1).&lt;/p&gt;&#xA;" OwnerUserId="9" LastEditorUserId="9" LastEditorDisplayName="" LastEditDate="2010-07-21T16:12:21.147" LastActivityDate="2010-07-21T16:12:21.147" CommentCount="1" />
  <row Id="341" PostTypeId="2" ParentId="153" CreationDate="2010-07-21T14:24:58.623" Score="1" ViewCount="" Body="&lt;p&gt;On a slightly different angle, Matlab is widely used in engineering, while Mathematica is more used among mathematicians. If I had to learn just one, I'd go with the one most used in my field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mathematica is also much cheaper for personal use (I think they have a licence for a few hundred dollars for home use, and even cheaper versions for students).&lt;/p&gt;&#xA;" OwnerUserId="64" LastActivityDate="2010-07-21T14:24:58.623" CommentCount="2" />
  <row Id="342" PostTypeId="2" ParentId="329" CreationDate="2010-07-21T14:25:23.817" Score="8" ViewCount="" Body="&lt;p&gt;I like Niven and Zuckerman, &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0471625469&quot; rel=&quot;nofollow&quot;&gt;Introduction to the Theory of Numbers&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="120" LastActivityDate="2010-07-21T14:25:23.817" CommentCount="1" />
  <row Id="343" PostTypeId="2" ParentId="336" CreationDate="2010-07-21T14:31:54.567" Score="8" ViewCount="" Body="&lt;p&gt;I'm going to copy &lt;a href=&quot;http://stackoverflow.com/questions/2465116/understanding-opengl-matrices/2465290#2465290&quot;&gt;my answer from Stack Overflow&lt;/a&gt;, which also shows why 4-component vectors (and hence 4&amp;times;4 matrices) are used instead of 3-component ones.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;In most 3D graphics a point is represented by a 4-component vector (x, y, z, w), where w = 1. Usual operations applied on a point include translation, scaling, rotation, reflection, skewing and combination of these. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;These transformations can be represented by a mathematical object called &quot;matrix&quot;. A matrix applies on a vector like this:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;[ a b c tx ] [ x ]   [ a*x + b*y + c*z + tx*w ]&#xA;| d e f ty | | y | = | d*x + e*y + f*z + ty*w |&#xA;| g h i tz | | z |   | g*x + h*y + i*z + tz*w |&#xA;[ p q r s  ] [ w ]   [ p*x + q*y + r*z +  s*w ]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;For example, scaling is represented as&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;[ 2 . . . ] [ x ]   [ 2x ]&#xA;| . 2 . . | | y | = | 2y |&#xA;| . . 2 . | | z |   | 2z |&#xA;[ . . . 1 ] [ 1 ]   [ 1  ]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;and translation as&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;[ 1 . . dx ] [ x ]   [ x + dx ]&#xA;| . 1 . dy | | y | = | y + dy |&#xA;| . . 1 dz | | z |   | z + dz |&#xA;[ . . . 1  ] [ 1 ]   [   1    ]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&lt;strong&gt;One of the reason for the 4th component is to make a translation representable by a matrix.&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The advantage of using a matrix is that multiple transformations can be combined into one via matrix multiplication.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if the purpose is simply to bring translation on the table, then I'd say (x, y, z, 1) instead of (x, y, z, w) and make the last row of the matrix always &lt;code&gt;[0 0 0 1]&lt;/code&gt;, as done usually for 2D graphics. In fact, the 4-component vector will be mapped back to the normal 3-vector vector via this formula:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;[ x(3D) ]   [ x / w ]&#xA;| y(3D) ] = | y / w |&#xA;[ z(3D) ]   [ z / w ]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This is called &lt;a href=&quot;http://en.wikipedia.org/wiki/Homogeneous_coordinates#Use_in_computer_graphics&quot; rel=&quot;nofollow&quot;&gt;homogeneous coordinates&lt;/a&gt;. &lt;strong&gt;&lt;em&gt;Allowing this makes the perspective projection expressible with a matrix too,&lt;/em&gt;&lt;/strong&gt; which can again combine with all other transformations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, since objects farther away should be smaller on screen, we transform the 3D coordinates into 2D using formula&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;x(2D) = x(3D) / (10 * z(3D))&#xA;y(2D) = y(3D) / (10 * z(3D))&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Now if we apply the projection matrix&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;[ 1 . .  . ] [ x ]   [  x   ]&#xA;| . 1 .  . | | y | = |  y   |&#xA;| . . 1  . | | z |   |  z   |&#xA;[ . . 10 . ] [ 1 ]   [ 10*z ]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;then the real 3D coordinates would become&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;x(3D) := x/w = x/10z&#xA;y(3D) := y/w = y/10z&#xA;z(3D) := z/w = 0.1&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;so we just need to chop the z-coordinate out to project to 2D.&lt;/p&gt;&#xA;" OwnerUserId="171" LastActivityDate="2010-07-21T14:31:54.567" CommunityOwnedDate="2010-07-21T14:31:54.567" />
  <row Id="344" PostTypeId="2" ParentId="329" CreationDate="2010-07-21T14:35:54.197" Score="13" ViewCount="" Body="&lt;p&gt;Serre's &quot;A course in Arithmetic&quot; is pretty phenomenal.&lt;/p&gt;&#xA;" OwnerUserId="16" LastActivityDate="2010-07-21T14:35:54.197" />
  <row Id="345" PostTypeId="2" ParentId="316" CreationDate="2010-07-21T14:51:01.790" Score="26" ViewCount="" Body="&lt;p&gt;Note that if&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ y = \left(1 - x^7\right)^{1/5} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;then &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \left(1 - y^5\right)^{1/7} = x $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means $(1-x^7)^{1/5}$ is the inverse function of $(1-x^5)^{1/7}$. In the graph, one will be the same as the other when reflected along the diagonal line y = x.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, both functions &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;share the same range [0, 1] and domain [0, 1] and&lt;/li&gt;&#xA;&lt;li&gt;monotonically decreasing,&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Therefore, the area under the graph in [0, 1] will be the same for both functions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \int_0^1 \left(1-x^7\right)^{1/5} dx = \int_0^1 \left(1-y^5\right)^{1/7} dy $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Grouping the two integrals yield the equation in the title.&lt;/p&gt;&#xA;" OwnerUserId="171" LastEditorUserId="171" LastEditorDisplayName="" LastEditDate="2010-08-03T09:08:01.803" LastActivityDate="2010-08-03T09:08:01.803" CommentCount="1" />
  <row Id="346" PostTypeId="2" ParentId="250" CreationDate="2010-07-21T14:52:45.437" Score="8" ViewCount="" Body="&lt;p&gt;Similar to the Monty Hall problem, but trickier: at the latest Gathering 4 Gardner, Gary Foshee asked &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I have two children. One is a boy born on a Tuesday. What is the probability I have two boys?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We are assuming that births are equally distributed during the week, that every child is a boy or girl with probability 1/2, and that there is no dependence relation between sex and day of birth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Answer: 13/27. This was in the news a lot recently, see for instance &lt;a href=&quot;http://news.bbc.co.uk/2/hi/programmes/more_or_less/8735812.stm&quot; rel=&quot;nofollow&quot;&gt;BBC News&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="115" LastActivityDate="2010-07-21T14:52:45.437" CommunityOwnedDate="2010-07-21T14:52:45.437" CommentCount="4" />
  <row Id="347" PostTypeId="2" ParentId="250" CreationDate="2010-07-21T15:33:16.183" Score="19" ViewCount="" Body="&lt;p&gt;You have two identical pieces of paper with the same picture printed on them. You put one flat on a table and the other one you crumple up (without tearing it) and place it on top of the first one. &lt;a href=&quot;http://en.wikipedia.org/wiki/Brouwer_fixed_point_theorem&quot; rel=&quot;nofollow&quot;&gt;Brouwer's fixed point theorem&lt;/a&gt; states that there is some point in the picture on the crumpled-up page that is directly above the same point on the bottom page. It doesn't matter how you place the pages, or how you deform the top one.&lt;/p&gt;&#xA;" OwnerUserId="26" LastActivityDate="2010-07-21T15:33:16.183" CommunityOwnedDate="2010-07-21T15:33:16.183" CommentCount="5" />
  <row Id="348" PostTypeId="2" ParentId="57" CreationDate="2010-07-21T15:41:46.653" Score="1" ViewCount="" Body="&lt;p&gt;It's not clear what the question is asking. Is it asking for logical relations among the properties? e.g. a relation that is transitive and symmetric MUST be reflexive (unless the relation holds between no two objects).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or is it asking for mathematical relations that have these properties? &quot;&gt;&quot; is transitive, asymmetric and total or complete (for all distinct x,y either xRy or yRx). It is a strict total order. &quot;greater or equal&quot; is transitive, complete, symmetric, reflexive. This is a weak total order.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;x is a subset of y&quot; is reflexive and antisymmetric but not necessarily transitive, symmetric (although it can be either of those things if you restrict yourself to the right sort of sets). &quot;x is a proper subset of y&quot; is asymmetric and irreflexive (for all x, it is not the case that xRx), and can be transitive if you restrict yourself to the right sort of sets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that in both these cases, one can define one of each pair of relations in terms of the other (plus a notion of identity or non-identity).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In logic the notion of &quot;x entails y&quot; is transitive, reflexive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The relation &quot;x has the same integer part as y&quot; over the real numbers is transitive, reflexive, symmetric. This is called an equivalence relation.&lt;/p&gt;&#xA;" OwnerUserId="172" LastActivityDate="2010-07-21T15:41:46.653" />
  <row Id="349" PostTypeId="2" ParentId="81" CreationDate="2010-07-21T15:49:18.933" Score="2" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://math-blog.com/&quot; rel=&quot;nofollow&quot;&gt;Math-blog&lt;/a&gt; is one I have in my google reader and I just found this one through google reader but it's a little over my head:&lt;a href=&quot;http://unapologetic.wordpress.com/2010/07/20/upper-and-lower-ordinate-sets/&quot; rel=&quot;nofollow&quot;&gt;The Unapologetic Mathematician&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="180" LastEditorUserId="180" LastEditorDisplayName="" LastEditDate="2010-07-22T14:00:08.583" LastActivityDate="2010-07-22T14:00:08.583" CommunityOwnedDate="2010-07-21T15:49:18.933" CommentCount="1" />
  <row Id="350" PostTypeId="1" AcceptedAnswerId="354" CreationDate="2010-07-21T15:58:51.407" Score="2" ViewCount="206" Body="&lt;p&gt;I'm looking for course notes and assignments  and hopefully some example exams for Discrete Math, I'm taking a placement exam in the subject after having taken it 4 years ago. &lt;/p&gt;&#xA;" OwnerUserId="180" LastEditorUserId="126" LastEditorDisplayName="" LastEditDate="2010-07-21T16:56:08.763" LastActivityDate="2010-09-13T18:42:45.433" CommunityOwnedDate="2010-09-13T20:09:02.853" Title="Where can I find a review of discrete math" Tags="&lt;soft-question&gt;&lt;reference-request&gt;&lt;education&gt;&lt;discrete-mathematics&gt;" AnswerCount="5" CommentCount="2" FavoriteCount="1" />
  <row Id="351" PostTypeId="2" ParentId="135" CreationDate="2010-07-21T16:05:57.817" Score="3" ViewCount="" Body="&lt;p&gt;If $a$ and $b$ are natural numbers, then $a^b$ is the number of ways you can make a sequence of length $b$ where each element in the sequence is chosen from a set of size $a$. You're allowed replacements. For example $2^3$ is the number of 3 digit sequences where each digit is zero or 1: 000, 001, 010, ..., 111.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is precisely one way to make a zero length sequence: one. So you'd expect $0^0=1$.&lt;/p&gt;&#xA;" OwnerUserId="80" LastActivityDate="2010-07-21T16:05:57.817" CommentCount="1" />
  <row Id="352" PostTypeId="2" ParentId="57" CreationDate="2010-07-21T16:09:07.313" Score="1" ViewCount="" Body="&lt;p&gt;I think you're thinking about this in the wrong way. Properties don't &quot;work&quot;, properties are things that are true for the given relation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, you say that the a relation has the reflexive property if it satisfies the condition that all elements are related to themselves. Similarly, it has the symmetric property if for all a and b, if a is related to b then b is related to a.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could simply say that &quot;∀a : a R a&quot; each time, but because this is something that happens often, this particular property has been given a name, i.e. reflexivity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Certain types of specific relations are also given names. For instance, a partial order is reflexive, antisymmetric and transitive, and equivalence relations are reflexive, symmetric and transitive. Again, these are just names that aren't strictly needed, but they make it more convenient talking about these kinds of things.&lt;/p&gt;&#xA;" OwnerUserId="157" LastEditorUserId="157" LastEditorDisplayName="" LastEditDate="2010-07-21T16:22:11.827" LastActivityDate="2010-07-21T16:22:11.827" />
  <row Id="353" PostTypeId="2" ParentId="297" CreationDate="2010-07-21T16:13:22.677" Score="7" ViewCount="" Body="&lt;p&gt;Some of the easiest algorithms to use are the &lt;a href=&quot;http://en.wikipedia.org/wiki/Bailey-Borwein-Plouffe_formula&quot; rel=&quot;nofollow&quot;&gt;spigot&lt;/a&gt; algorithms. They allow you to jump rapidly to the nth digit without computing the n-1 digits before. The catch is that they only really work in binary. &lt;a href=&quot;http://en.literateprograms.org/Pi_with_the_BBP_formula_%28Python%29&quot; rel=&quot;nofollow&quot;&gt;Here&lt;/a&gt; are some implementations in Python. You can see how short they are.&lt;/p&gt;&#xA;" OwnerUserId="80" LastActivityDate="2010-07-21T16:13:22.677" CommentCount="1" />
  <row Id="354" PostTypeId="2" ParentId="350" CreationDate="2010-07-21T16:14:24.860" Score="1" ViewCount="" Body="&lt;p&gt;This is the discrete math course one in my school. It contain lecture notes, homework and previous exams. &lt;a href=&quot;http://www.cs.sunysb.edu/~cse547/&quot; rel=&quot;nofollow&quot;&gt;http://www.cs.sunysb.edu/~cse547/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="96" LastEditorUserId="96" LastEditorDisplayName="" LastEditDate="2010-07-21T16:57:39.213" LastActivityDate="2010-07-21T16:57:39.213" CommunityOwnedDate="2010-09-13T20:09:02.853" />
  <row Id="355" PostTypeId="2" ParentId="275" CreationDate="2010-07-21T16:18:15.213" Score="6" ViewCount="" Body="&lt;p&gt;One's that were suggested to me by my Calculus teacher in High School. Even my wife liked them and she hates math now:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/1589880331&quot; rel=&quot;nofollow&quot;&gt;The Education of T.C. Mits: What&#xA;modern mathematics means to you&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/1589880366&quot; rel=&quot;nofollow&quot;&gt;Infinity: Beyond the Beyond the&#xA;Beyond&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Written and illustrated(Pictures are great ;p) by a couple: Lillian R. Lieber, and Hugh Gray Lieber. These books were hard to find before because they went out of print but I have this new version and like it a lot. The books explains profound topics in a way that is graspable by anyone without being dumbed down. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/1589880331&quot; rel=&quot;nofollow&quot;&gt;Godel's proof&lt;/a&gt; is one I enjoyed. It's was a little hard to understand but there is nothing in this book that makes it inaccessible to someone without a strong math background.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Keeping with Godel in the title, &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0465026567&quot; rel=&quot;nofollow&quot;&gt;Godel, Escher, Bach: An Eternal Golden Braid&lt;/a&gt; while not just about math was a good read (a bit long ;p).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0060935588&quot; rel=&quot;nofollow&quot;&gt;The Music of the Primes: Searching to Solve the Greatest Mystery in Mathematics&lt;/a&gt;  It describes the Riemann Hypothesis and people who were involved with it somehow. My favorite part was learning about the people who attempted to solve it. Many I never heard off before this book. (Side not: I'll have to read pguertin suggestion, sounds in like a similar but more profound book).&lt;/p&gt;&#xA;" OwnerUserId="180" LastEditorUserId="4122" LastEditorDisplayName="" LastEditDate="2010-11-30T16:31:08.247" LastActivityDate="2010-11-30T16:31:08.247" CommunityOwnedDate="2010-07-21T16:18:15.213" CommentCount="1" />
  <row Id="356" PostTypeId="1" CreationDate="2010-07-21T16:42:55.237" Score="4" ViewCount="512" Body="&lt;p&gt;Hi everyone,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm very interested in Computer Science (computational complexity, etc.). I've already finished a University course in the subject (using Sipser's &quot;Introduction to the Theory of Computation&quot;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know the basics, i.e. Turing Machines, Computability (Halting problem and related reductions), Complexity classes (time and space, P/NP, L/NL, a little about BPP).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, I'm looking for a good book to learn about some more advanced concepts. Any ideas?&lt;/p&gt;&#xA;" OwnerUserId="122" LastEditorUserId="100" LastEditorDisplayName="" LastEditDate="2010-07-22T07:52:26.070" LastActivityDate="2011-05-09T21:28:57.997" CommunityOwnedDate="2010-07-21T19:11:11.363" Title="Best Intermediate/Advanced Computer Science book" Tags="&lt;soft-question&gt;&lt;big-list&gt;&lt;books&gt;&lt;reference-request&gt;" AnswerCount="6" CommentCount="12" FavoriteCount="3" />
  <row Id="357" PostTypeId="2" ParentId="329" CreationDate="2010-07-21T16:49:45.870" Score="2" ViewCount="" Body="&lt;p&gt;Stewart&amp;amp;Tall's &quot;Algebraic Number Theory&quot; is great.&lt;/p&gt;&#xA;" OwnerUserId="8" LastActivityDate="2010-07-21T16:49:45.870" CommunityOwnedDate="2010-07-21T16:49:45.870" />
  <row Id="358" PostTypeId="2" ParentId="275" CreationDate="2010-07-21T16:50:55.913" Score="0" ViewCount="" Body="&lt;p&gt;Possibly this may not really qualify as presenting much interesting maths, but I think Hardy's &lt;a href=&quot;http://math.boisestate.edu/~holmes/holmes/A%20Mathematician%27s%20Apology.pdf&quot; rel=&quot;nofollow&quot;&gt;A Mathematician's Apology&lt;/a&gt; should be on the must-read list.&lt;/p&gt;&#xA;" OwnerUserId="144" LastActivityDate="2010-07-21T16:50:55.913" CommunityOwnedDate="2010-07-21T16:50:55.913" CommentCount="2" />
  <row Id="359" PostTypeId="2" ParentId="356" CreationDate="2010-07-21T16:56:49.387" Score="4" ViewCount="" Body="&lt;p&gt;Papadimitriou's &lt;em&gt;Computational Complexity&lt;/em&gt; covers complexity theory at a higher level than Sipser, but has essentially no prerequisites.&lt;/p&gt;&#xA;" OwnerUserId="536" LastActivityDate="2010-07-21T16:56:49.387" CommunityOwnedDate="2010-07-21T19:50:37.107" />
  <row Id="360" PostTypeId="2" ParentId="262" CreationDate="2010-07-21T17:07:54.750" Score="6" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/A_Mathematician%27s_Apology&quot; rel=&quot;nofollow&quot;&gt;A Mathematician's Apology&lt;/a&gt; by G H Hardy. I did in fact read this in high school, and it raised my view of mathematics from a thing of utility to a thing of beauty and wonder. It inspired me to go on to study mathematics at Cambridge myself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's a pity that the &quot;introduction&quot; by C P Snow is longer than the original and contains a rather depressing view of Hardy's later life. I would recommend readers to skip the introduction altogether and concentrate on Hardy's own words.&lt;/p&gt;&#xA;" OwnerUserId="113" LastActivityDate="2010-07-21T17:07:54.750" CommunityOwnedDate="2010-07-21T17:07:54.750" CommentCount="2" />
  <row Id="361" PostTypeId="2" ParentId="153" CreationDate="2010-07-21T17:09:43.997" Score="1" ViewCount="" Body="&lt;p&gt;If you have prior programming experience, Matlab will probably seem more intuitive. Mathematica's notions about structure, layout and order of execution can often be vexing to the coder mindset -- as can its set of operators, which we might charitably describe as &lt;em&gt;rich&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, the notion that it is helpful for programs to resemble what a mathematician might scrawl on a piece of parchment, with a quill pen, is one you will either love or hate...&lt;/p&gt;&#xA;" OwnerUserId="144" LastActivityDate="2010-07-21T17:09:43.997" />
  <row Id="362" PostTypeId="1" AcceptedAnswerId="363" CreationDate="2010-07-21T17:12:50.780" Score="13" ViewCount="493" Body="&lt;p&gt;I am vaguely familiar with the broad strokes of the development of group theory, first when ideas of geometric symmetries were studied in concrete settings without the abstract notion of a group available, and later as it was formalized by Cayley, Lagrange, etc (and later, infinite groups being well-developed). In any case, it's intuitively easy for me to imagine that there was substantial lay, scientific, and artistic interest in several of the concepts well-encoded by a theory of groups.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know a few of the corresponding names for who developed the abstract formulation of rings initially (Wedderburn etc.), but I'm less aware of the ideas and problems that might have given rise to interest in ring structures. Of course, now they're terribly useful in lots of math, and $\mathbb{Z}$ is a natural model for elementary properties of commutative rings, and I'll wager number theorists had an interest in developing the concept. And if I wanted noncommutative models, matrices are a good place to start looking. But I'm not even familiar with what the state of knowledge and formalization of things like matrices/linear operators was at the time rings were developed, so maybe these aren't actually good examples for how rings might have been motivated. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone outline or point me to some basics on the history of the development of basic algebraic structures besides groups? &lt;/p&gt;&#xA;" OwnerUserId="76" LastActivityDate="2010-08-07T20:34:33.857" Title="History of the Concept of a Ring" Tags="&lt;math-history&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="2" />
  <row Id="363" PostTypeId="2" ParentId="362" CreationDate="2010-07-21T17:22:25.990" Score="8" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; Bill Dubuque has pointed out that much of this answer (specifically, the part about FLT) is essentially a mathematical urban legend, albeit a pervasive one. I cannot delete an accepted answer, so here is a &lt;a href=&quot;http://mathoverflow.net/questions/26613/papers-that-debunk-common-myths-in-the-history-of-mathematics/30272#30272&quot; rel=&quot;nofollow&quot;&gt;link&lt;/a&gt; to an answer of his on MO explaining it.&lt;br&gt;&#xA;Here is also a link to a related &lt;a href=&quot;http://mathoverflow.net/questions/34806/what-was-the-relative-importance-of-flt-vs-higher-reciprocity-laws-in-kummers-i&quot; rel=&quot;nofollow&quot;&gt;question&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's some of the history here in Bourbaki's &lt;em&gt;Commutative Algebra,&lt;/em&gt; in the appendix.  Basically, a fair bit of ring theory was developed for algebraic number theory.  This in turn was because people were trying to prove Fermat's last theorem. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why's this? Let $p$ be a prime. Then the equation $x^p + y^p = z^p$ can be written as $\prod (x+\zeta_p^iy) = z^p$ for $\zeta_p$ a primitive $p$th root of unity.  All these quantities are elements of the ring $Z[\zeta_p]$.  So if $p&gt;3$ and there is unique factorization in the ring $Z[\zeta_p]$, it isn't terribly hard to show that this is impossible at least in the basic case where $p $ does not divide $xyz$ (and can be found, for instance, in Borevich-Shafarevich's book on number theory).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lame actually thought he had a proof of FLT via this argument. But he was wrong: these rings generally don't admit unique factorization.  So, it became a problem to study these &quot;generalized integers&quot; $Z[\zeta_p]$, which of course are basic examples of rings.  It wasn't until Dedekind that the right notion of unique factorization -- namely, factorization of ideals -- was found. In fact, the case of FLT I just mentioned generalizes to the case where $p$ does not divide the class number of $Z[\zeta_p]$ (the class number is the invariant that measures how far it is from being a UFD).  And, according to this &lt;a href=&quot;http://www-gap.dcs.st-and.ac.uk/~history/HistTopics/Ring_theory.html&quot; rel=&quot;nofollow&quot;&gt;article&lt;/a&gt;, Dedekind was the first to define a ring.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The article I linked to, incidentally, has a fair bit of additional interesting history.&lt;/p&gt;&#xA;" OwnerUserId="536" LastEditorUserId="536" LastEditorDisplayName="" LastEditDate="2010-08-07T20:34:33.857" LastActivityDate="2010-08-07T20:34:33.857" CommentCount="11" />
  <row Id="364" PostTypeId="2" ParentId="275" CreationDate="2010-07-21T17:51:22.493" Score="4" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://www.amazon.com/Paul-J.-Nahin/e/B001HCS1XI/ref=sr_ntt_srch_lnk_1?_encoding=UTF8&amp;amp;qid=1279734093&amp;amp;sr=8-1&quot; rel=&quot;nofollow&quot;&gt;Paul Nahin&lt;/a&gt; has a number of accessible mathematics books written for non-mathematicians, the most famous being&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0691146004&quot; rel=&quot;nofollow&quot;&gt;An Imaginary Tale: The Story of $\sqrt{-1}$&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0691118221&quot; rel=&quot;nofollow&quot;&gt;Dr. Euler's Fabulous Formula (Cures Many Mathematical Ills!)&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.amazon.com/Ian-Stewart/e/B000APQ9NM/ref=ntt_athr_dp_pel_1&quot; rel=&quot;nofollow&quot;&gt;Professor Ian Stewart&lt;/a&gt; also has many books which each give laymen overviews of various fields or surprising mathematical results&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0465013023&quot; rel=&quot;nofollow&quot;&gt;Professor Stewart's Cabinet of Mathematical Curiosities&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0465017754&quot; rel=&quot;nofollow&quot;&gt;Professor Stewart's Hoard of Mathematical Treasures&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0199562075&quot; rel=&quot;nofollow&quot;&gt;Cows in the Maze: And Other Mathematical Explorations&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0631232516&quot; rel=&quot;nofollow&quot;&gt;Does God Play Dice? The New Mathematics of Chaos&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="136" LastEditorUserId="136" LastEditorDisplayName="" LastEditDate="2010-07-21T19:17:49.260" LastActivityDate="2010-07-21T19:17:49.260" CommunityOwnedDate="2010-07-21T17:51:22.493" CommentCount="2" />
  <row Id="365" PostTypeId="1" CreationDate="2010-07-21T18:11:33.560" Score="8" ViewCount="230" Body="&lt;p&gt;I'm looking to find out if there's any easy way to calculate the number of ways to tile a $3 \times 2n$ rectangle with dominoes. I was able to do it with the two codependent recurrences&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;f(0) = g(0) = 1&#xA;f(n) = f(n-1) + 2g(n-1)&#xA;g(n) = f(n) + g(n-1)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;where $f(n)$ is the actual answer and $g(n)$ is a helper function that represents the number of ways to tile a $3 \times 2n$ rectangle with two extra squares on the end (the same as a $3 \times 2n+1$ rectangle missing one square).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By combining these and doing some algebra, I was able to reduce this to&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;f(n) = 4f(n-1) - f(n-2)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;which shows up as sequence &lt;a href=&quot;http://www.research.att.com/~njas/sequences/A001835&quot; rel=&quot;nofollow&quot;&gt;A001835&lt;/a&gt;, confirming that this is the correct recurrence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The number of ways to tile a $2 \times n$ rectangle is the Fibonacci numbers because every rectangle ends with either a verticle domino or two horizontal ones, which gives the exact recurrence that Fibonacci numbers do. My question is, &lt;strong&gt;is there a similar simple explanation for this recurrence for tiling a $3 \times 2n$ rectangle&lt;/strong&gt;?&lt;/p&gt;&#xA;" OwnerUserId="45" LastEditorUserId="45" LastEditorDisplayName="" LastEditDate="2010-07-21T18:22:13.133" LastActivityDate="2010-07-21T18:40:01.863" Title="Tiling a 3 by 2n rectangle with dominoes" Tags="&lt;combinatorics&gt;&lt;tiling&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="366" PostTypeId="2" ParentId="350" CreationDate="2010-07-21T18:12:48.237" Score="2" ViewCount="" Body="&lt;p&gt;When wanting to know about a particular mathematics subject, I often find that starting with the &quot;&lt;a href=&quot;http://en.wikipedia.org/wiki/Discrete_mathematics#Further_reading&quot; rel=&quot;nofollow&quot;&gt;further reading&lt;/a&gt;&quot; section of the relevant wikipedia page is a good way in.&lt;/p&gt;&#xA;" OwnerUserId="172" LastActivityDate="2010-07-21T18:12:48.237" CommunityOwnedDate="2010-09-13T20:09:02.853" />
  <row Id="367" PostTypeId="2" ParentId="365" CreationDate="2010-07-21T18:40:01.863" Score="5" ViewCount="" Body="&lt;p&gt;Here's my best shot at the sort of explanation you're asking for, although it's not nearly as clear as the 2-by-n case.  The negative sign makes combinatorial proofs difficult, so let's rearrange this as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;f(n) + f(n-2) = 4f(n-1).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then you want to show that the number of n-tilings, plus the number of (n-2)-tilings, is four times the number of (n-1)-tilings. (An &quot;n-tiling&quot; is a tiling of a 3 by 2n rectangle by dominoes.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In bijective terms, then, we want a bijection between the set of n-tilings and (n-2)-tilings and the set of (n-1)-tilings, where the (n-1)-tilings are each tagged with the number 1, 2, 3, or 4.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given an (n-1)-tiling, there are three &quot;obvious&quot; ways to obtain an n-tiling from it, namely by adding one of the three 1-tilings on the right end.  These generate tilings which have a vertical line passing all the way through,  two units from the right end; call these &quot;faulted&quot; tilings, and those which don't have a vertical line in that position &quot;faultless&quot;.&lt;br&gt;&#xA;So it suffices to show that the number of faultless n-tilings, plus the number of (n-2)-tilings, is the number of (n-1)-tilings.  It's easy to see that the number of faultless n-tilings is 2g(n-2); a faultless tilings must have a horizontal domino covering the second and third squares from the right in some row, and this assumption forces the placement of some other dominoes.  So we need 2g(n-2) + f(n-2) = f(n-1).  Shifting the indices, we need 2g(n-1) + f(n-1) = f(n), which you have already said is true.&lt;/p&gt;&#xA;" OwnerUserId="173" LastActivityDate="2010-07-21T18:40:01.863" />
  <row Id="368" PostTypeId="1" AcceptedAnswerId="374" CreationDate="2010-07-21T19:18:25.310" Score="7" ViewCount="427" Body="&lt;p&gt;I know that the Fibonacci sequence can be described via the Binet's formula.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I was wondering if there was a similar formula for $n!$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Is this possible?  If not, why not?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="38" LastActivityDate="2010-09-11T05:43:04.290" Title="Is there a closed-form equation for $n!$? If not, why not?" Tags="&lt;combinatorics&gt;&lt;algorithms&gt;" AnswerCount="4" CommentCount="6" FavoriteCount="1" />
  <row Id="369" PostTypeId="2" ParentId="368" CreationDate="2010-07-21T19:24:55.437" Score="6" ViewCount="" Body="&lt;p&gt;The relative error of &lt;a href=&quot;http://en.wikipedia.org/wiki/Stirling%27s_approximation&quot; rel=&quot;nofollow&quot;&gt;Stirling's approximation&lt;/a&gt; gets arbitrarily small as n gets larger.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://upload.wikimedia.org/math/4/6/c/46c33d496f99c39d9886c66fa12e655d.png&quot; alt=&quot;(picture)&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, it is only an approximation, not a closed-form of $n!$&lt;/p&gt;&#xA;" OwnerUserId="136" LastEditorUserId="136" LastEditorDisplayName="" LastEditDate="2010-07-21T20:01:30.527" LastActivityDate="2010-07-21T20:01:30.527" CommentCount="3" />
  <row Id="370" PostTypeId="1" AcceptedAnswerId="377" CreationDate="2010-07-21T19:27:09.313" Score="10" ViewCount="1459" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;what is the best book or lecture notes on category theory?&lt;/p&gt;&#xA;" OwnerUserId="179" LastEditorUserId="179" LastEditorDisplayName="" LastEditDate="2010-07-24T10:25:04.913" LastActivityDate="2011-03-28T17:47:44.587" CommunityOwnedDate="2010-07-24T10:25:04.913" Title="Good book/lecture notes about category theory" Tags="&lt;soft-question&gt;&lt;books&gt;&lt;reference-request&gt;&lt;category-theory&gt;" AnswerCount="18" CommentCount="5" FavoriteCount="15" />
  <row Id="371" PostTypeId="2" ParentId="350" CreationDate="2010-07-21T19:34:46.737" Score="1" ViewCount="" Body="&lt;p&gt;if you don't mind shell out a good amount of money, &lt;em&gt;Concrete Mathematics&lt;/em&gt; by Graham, Knuth and Patashnik could be nice (I own the first edition)&lt;/p&gt;&#xA;" OwnerUserId="89" LastActivityDate="2010-07-21T19:34:46.737" CommunityOwnedDate="2010-09-13T20:09:02.853" CommentCount="2" />
  <row Id="372" PostTypeId="2" ParentId="297" CreationDate="2010-07-21T19:35:23.600" Score="4" ViewCount="" Body="&lt;p&gt;There is also the option of approximating $\pi$ using &lt;a href=&quot;http://en.wikipedia.org/wiki/Monte_Carlo_integration&quot; rel=&quot;nofollow&quot;&gt;Monte Carlo integration&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The idea is this: If we agree that the area of a circle is $\pi r^2$, for simplicity we build a circle of area $\pi$ by setting $r=1$.  Placing this circle wholly inside of another region of known area, preferably by inscribing it in a square of side length 2, then we have a ratio of the circle's area to the total area of the square  (in this example that ratio is $\frac{\pi}{4}$).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Monte Carlo method works by approximating areas based on the ratio of the number of sample points lying within our region of interest and the total number of sample points we choose to try.  If we spread a uniformly distributed sequence of $N$ points over our square of area 4, and call the number of points that land inside of the inscribed circle $p$, then we can say $\frac{\pi}{4} = \frac{p}{N}$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My implementation of this in Matlab requires tens of thousands of test points to achieve 3.14159xxxx, but I have not tried it for low-discrepancy sequences, or any other uniformly distributed point sets.    &lt;/p&gt;&#xA;" OwnerUserId="121" LastActivityDate="2010-07-21T19:35:23.600" />
  <row Id="373" PostTypeId="2" ParentId="370" CreationDate="2010-07-21T19:35:57.110" Score="11" ViewCount="" Body="&lt;p&gt;Categories for the Working mathematician by Mac Lane&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Categories and Sheaves by Kashiwara and Schapira&lt;/p&gt;&#xA;" OwnerUserId="126" LastActivityDate="2010-07-21T19:35:57.110" CommunityOwnedDate="2010-08-04T15:59:49.803" CommentCount="1" />
  <row Id="374" PostTypeId="2" ParentId="368" CreationDate="2010-07-21T19:36:10.433" Score="6" ViewCount="" Body="&lt;p&gt;If you're willing to accept an integral as an answer, then $n! = \int_0^\infty t^n e^{-t} \: dt$.&lt;/p&gt;&#xA;" OwnerUserId="173" LastActivityDate="2010-07-21T19:36:10.433" CommentCount="13" />
  <row Id="375" PostTypeId="1" AcceptedAnswerId="376" CreationDate="2010-07-21T19:43:27.840" Score="4" ViewCount="330" Body="&lt;p&gt;Context: Rings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Arguably three questions in one, but since they all are properties of the multiplication, I'll &#xA;try my luck...&lt;/p&gt;&#xA;" OwnerUserId="174" LastEditorUserId="174" LastEditorDisplayName="" LastEditDate="2010-07-22T22:38:08.670" LastActivityDate="2011-05-07T07:13:52.223" Title="Are x*0 = 0 and x*1 = x and -(-x) = x axioms?" Tags="&lt;algebra&gt;&lt;definition&gt;" AnswerCount="4" CommentCount="2" />
  <row Id="376" PostTypeId="2" ParentId="375" CreationDate="2010-07-21T19:48:29.717" Score="11" ViewCount="" Body="&lt;p&gt;I will assume this is in the context of rings (e.g. real numbers, integers, etc).  In this case, the axiom defining 0 is that x + 0 = x for all x.  x*0 = 0 is a result of this since we have x*0 = x*(0+0) = x*0 + x*0 which implies x*0 = 0 (canceling one of the x*0's).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am guessing that for the second one you mean x*1 = x.  This is a definition (axiom).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The third one is a consequence of the definition of -x being the element such that x + (-x) = 0.  For then we have (-x) + x is also zero so that x is the negative of -x.&lt;/p&gt;&#xA;" OwnerUserId="9" LastActivityDate="2010-07-21T19:48:29.717" CommentCount="4" />
  <row Id="377" PostTypeId="2" ParentId="370" CreationDate="2010-07-21T20:18:39.363" Score="11" ViewCount="" Body="&lt;p&gt;Lang's &lt;em&gt;Algebra&lt;/em&gt; contains a lot of introductory material on categories, which is really nice since it's done with constant motivation from algebra (e.g. coproducts are introduced right before the free product of groups is discussed). &lt;/p&gt;&#xA;" OwnerUserId="536" LastActivityDate="2010-07-21T20:18:39.363" CommunityOwnedDate="2010-08-04T15:59:49.803" CommentCount="2" />
  <row Id="378" PostTypeId="2" ParentId="350" CreationDate="2010-07-21T20:24:35.730" Score="1" ViewCount="" Body="&lt;p&gt;When it comes to textbooks, the Kenneth Rosen text &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0073229725&quot; rel=&quot;nofollow&quot;&gt;Discrete Mathematics and its Applications&lt;/a&gt; is highly recommended. I was first introduced to it at my university, but I've seen it cited in several places.&lt;/p&gt;&#xA;" OwnerUserId="91" LastActivityDate="2010-07-21T20:24:35.730" CommunityOwnedDate="2010-09-13T20:09:02.853" CommentCount="2" />
  <row Id="379" PostTypeId="2" ParentId="375" CreationDate="2010-07-21T20:28:20.760" Score="6" ViewCount="" Body="&lt;p&gt;The question is more profound than is initially seems, and is really about algebraic structures. The first question you have to ask yourself is &lt;em&gt;where you're working&lt;/em&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, addition and multiplication are defined on a &lt;em&gt;structure&lt;/em&gt;, which in this case  is a &lt;em&gt;set&lt;/em&gt; (basically a collection of &quot;things&quot;) with two &lt;em&gt;operators&lt;/em&gt; we call &lt;em&gt;addition&lt;/em&gt; (marked +) and &lt;em&gt;multiplication&lt;/em&gt; (marked · or × or * or whatever). If this structure holds some properties, which are sometimes called &lt;em&gt;axioms&lt;/em&gt;, then it is called a &lt;a href=&quot;http://en.wikipedia.org/wiki/Unit_ring&quot; rel=&quot;nofollow&quot;&gt;unit ring&lt;/a&gt;. The properties are:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The set is closed under the operator +. That is, if a and b are in R, then a+b is also in R.&lt;/li&gt;&#xA;&lt;li&gt;The set has a member which we mark as 0. It has the properties that for every a in R, a+0 = 0 and 0+a = 0.&lt;/li&gt;&#xA;&lt;li&gt;The operation + is commutative: a+b = b+a.&lt;/li&gt;&#xA;&lt;li&gt;The operation + is associative: (a+b)+c = a+(b+c).&lt;/li&gt;&#xA;&lt;li&gt;Every member has an additive inverse: for every a in R there is some b in R such that a+b = 0 (we mark b as -a).&lt;/li&gt;&#xA;&lt;li&gt;The set is closed under the operator *. That is, if a and b are in R, then a*b is also in R.&lt;/li&gt;&#xA;&lt;li&gt;The set has a member which we mark as 1. It has the properties that for every a in R, a*1 = a and 1*a = a.&lt;/li&gt;&#xA;&lt;li&gt;The operation * is associative: (a*b) * c = a * (b*c).&lt;/li&gt;&#xA;&lt;li&gt;Multiplication is distributive over addition: a * (b+c) = a*b + a*c and (a+b) * c = a*b + a*c. &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;While this is a long list, and introduces the operator + which is not even explicitly mentioned in the question, these properties are quite natural. For example, the integers {..., -2, -1, 0, 1, 2, ...} we all know and love indeed form a ring. The real numbers also form a ring  (in fact they form a &lt;a href=&quot;http://en.wikipedia.org/wiki/Field_%28mathematics%29&quot; rel=&quot;nofollow&quot;&gt;field&lt;/a&gt;, which means they hold even more properties).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In regard to your question, the identity x * 1 = x (I assume that's what you meant) is in fact an axiom - it is axiom 7. However, the other two identities are results of the other axioms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;First identity:&lt;/strong&gt; We use axioms 2 and 9 to get  0 * x = (0+0) * x = 0*x + 0*x  and then by adding -(0*x) (the additive inverse of 0*x, from axiom 5) to both sides,  0 = 0*x&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Second identity:&lt;/strong&gt; As stated in axiom 5, -(-x) is just a notation used which means &quot;the additive inverse of -x&quot;. To show that -(-x) = x we need to show that x is in fact the additive inverse of -x, or in other words that x + -x = 0 and -x + x = 0. But that's just what axiom 5 says, so we're done.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Last point: You might be wondering why did we have to go and introduce addition to answer a question about multiplication? Well, it so happens that without addition the other two identities are simply not true. For example, if we look at the positive integers {1, 2, 3, ...} with only multiplication, then there is no 0 there! Simply put, this is because the positive integers do not form a ring.&lt;/p&gt;&#xA;" OwnerUserId="26" LastEditorUserId="26" LastEditorDisplayName="" LastEditDate="2010-07-22T10:17:50.120" LastActivityDate="2010-07-22T10:17:50.120" />
  <row Id="380" PostTypeId="2" ParentId="370" CreationDate="2010-07-21T20:35:21.180" Score="2" ViewCount="" Body="&lt;p&gt;Arbib, &lt;em&gt;Arrows, Structures, and Functors: The Categorical Imperative&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More elementary than MacLane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know very much about this, but some stripes of computer scientist have taken an interest in category theory recently, and there are lecture notes floating around with that orientation.&lt;/p&gt;&#xA;" OwnerUserId="76" LastActivityDate="2010-07-21T20:35:21.180" CommunityOwnedDate="2010-08-04T15:59:49.803" />
  <row Id="381" PostTypeId="1" AcceptedAnswerId="449" CreationDate="2010-07-21T20:40:41.727" Score="9" ViewCount="562" Body="&lt;p&gt;The fibonacci sequence is very well known, and is often explained with a story about how many rabbits there are after &lt;code&gt;n&lt;/code&gt; generations if they each produce a new pair every generation. Is there any &lt;em&gt;other&lt;/em&gt; reason you would care about the fibonacci sequence?&lt;/p&gt;&#xA;" OwnerUserId="104" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-07-28T19:16:39.697" LastActivityDate="2010-07-29T13:55:45.977" CommunityOwnedDate="2010-08-03T18:46:03.740" Title="Applications of the fibonacci sequence" Tags="&lt;big-list&gt;&lt;combinatorics&gt;&lt;applications&gt;" AnswerCount="15" CommentCount="2" FavoriteCount="1" />
  <row Id="382" PostTypeId="2" ParentId="381" CreationDate="2010-07-21T20:50:30.707" Score="6" ViewCount="" Body="&lt;p&gt;Yep, the solution by Matiyasevich to Hilbert's tenth problem relies heavily on showing exponentiation is Diophantine.  In other words, the relation R given by $R(x,y,z)$ if and only if $x^y = z$ can be expressed by a polynomial equation $T(x,y,z,w)=0$, such that $R(x,y,z)$ holds if and only if $T(x,y,z,w)=0$ has a solution with $w$ integral.  This is a very difficult and technical result, as well as rather counterintuitive. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Recall that the Fibonacci numbers grow essentially exponentially. One of the lemmas in showing that exponentiation is Diophantine is to show that the Fibonacci sequence (and its variants) are Diophantine.  See the book by Matiyasevich on Hilbert's 10th problem.&lt;/p&gt;&#xA;" OwnerUserId="536" LastActivityDate="2010-07-21T20:50:30.707" CommunityOwnedDate="2010-08-03T18:46:03.740" />
  <row Id="383" PostTypeId="2" ParentId="381" CreationDate="2010-07-21T20:50:50.440" Score="0" ViewCount="" Body="&lt;p&gt;They are sometimes used or occur in financial applications, e.g. &lt;a href=&quot;http://en.wikipedia.org/wiki/Elliot_wave&quot; rel=&quot;nofollow&quot;&gt;Elliot Wave Theory&lt;/a&gt;.  This seems more like magic than math to me, but I am not a trader.&lt;/p&gt;&#xA;" OwnerUserId="129" LastActivityDate="2010-07-21T20:50:50.440" CommunityOwnedDate="2010-08-03T18:46:03.740" />
  <row Id="384" PostTypeId="2" ParentId="381" CreationDate="2010-07-21T20:51:59.050" Score="5" ViewCount="" Body="&lt;p&gt;It isn't exactly an &lt;em&gt;application&lt;/em&gt; as such, but the upper bound of the size of a subtree in a Fibonacci heap whose root is a node with degree k is F_(k+2) where F_n is the n'th Fibonacci number.&lt;/p&gt;&#xA;" OwnerUserId="157" LastEditorUserId="157" LastEditorDisplayName="" LastEditDate="2010-07-21T22:15:15.967" LastActivityDate="2010-07-21T22:15:15.967" CommunityOwnedDate="2010-08-03T18:46:03.740" CommentCount="1" />
  <row Id="385" PostTypeId="2" ParentId="375" CreationDate="2010-07-21T20:55:51.783" Score="0" ViewCount="" Body="&lt;p&gt;Answering what you asked, the question is poorly formed because you didn't specify the theory you are talking about. To do that, well, you must define symbols, deduction rules... and axioms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Answering what you probably meant to ask, no. Those are properties that - and × have in the Z set, not axioms. They can be deduced from the operations themselves.&lt;/p&gt;&#xA;" OwnerUserId="92" LastActivityDate="2010-07-21T20:55:51.783" />
  <row Id="386" PostTypeId="2" ParentId="381" CreationDate="2010-07-21T20:56:34.280" Score="1" ViewCount="" Body="&lt;p&gt;They're useful in determining the amortized run time of the appropriately named Fibonacci Heap in computer science.&lt;/p&gt;&#xA;" OwnerUserId="50" LastActivityDate="2010-07-21T20:56:34.280" CommunityOwnedDate="2010-08-03T18:46:03.740" />
  <row Id="387" PostTypeId="1" AcceptedAnswerId="388" CreationDate="2010-07-21T20:57:35.510" Score="7" ViewCount="275" Body="&lt;p&gt;I know that the harmonic series 1 + 1/2 + 1/3 + 1/4 + ... diverges. I also know that the sum of the inverse of prime numbers 1/2 + 1/3 + 1/5 + 1/7 + 1/11 + ... diverges too, even if really slowly since it's O(log log n).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I think I read that if we consider the numbers whose decimal representation do not have a certain digit (say, 7) and sum the inverse of these numbers, the sum is finite (usually between 19 and 20, it depends from the missing digit). Does anybody know the result, and some way to prove that the sum is finite?&lt;/p&gt;&#xA;" OwnerUserId="89" LastEditorUserId="89" LastEditorDisplayName="" LastEditDate="2010-07-21T21:08:50.333" LastActivityDate="2010-07-21T21:35:25.143" Title="sum of inverse of numbers with certain terms omitted" Tags="&lt;sequences-and-series&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="5" />
  <row Id="388" PostTypeId="2" ParentId="387" CreationDate="2010-07-21T21:06:16.283" Score="4" ViewCount="" Body="&lt;p&gt;EDIT: &lt;a href=&quot;http://mathworld.wolfram.com/KempnerSeries.html&quot; rel=&quot;nofollow&quot;&gt;This&lt;/a&gt; might be what you're looking for. Found it from looking at the source below. They're called Kempner series.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An article &lt;a href=&quot;http://www.jstor.org/stable/2974352&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; (and cited below) says that one Dr. Kempner proved in 1914 that the series 1+ 1/2 + 1/3 + ..., with any term that has a 9 in the denominator removed, is convergent (though he doesn't say what it converges to in the introductory paragraph). The article goes on to generalize the result.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A Curious Convergent Series&#xA;Frank Irwin&#xA;The American Mathematical Monthly, Vol. 23, No. 5 (May, 1916), pp. 149-152&#xA;Published by: Mathematical Association of America&#xA;Stable URL: &lt;a href=&quot;http://www.jstor.org/stable/2974352&quot; rel=&quot;nofollow&quot;&gt;http://www.jstor.org/stable/2974352&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="50" LastActivityDate="2010-07-21T21:06:16.283" CommentCount="2" />
  <row Id="389" PostTypeId="2" ParentId="381" CreationDate="2010-07-21T21:07:51.460" Score="6" ViewCount="" Body="&lt;p&gt;it pops up every now and then. For example, positions (F&lt;sub&gt;2n&lt;/sub&gt;, F&lt;sub&gt;2n+1&lt;/sub&gt;) are winning ones in &lt;a href=&quot;http://en.wikipedia.org/wiki/Wythoff%27s_game&quot; rel=&quot;nofollow&quot;&gt;Wythoff's Game&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="89" LastActivityDate="2010-07-21T21:07:51.460" CommunityOwnedDate="2010-08-03T18:46:03.740" />
  <row Id="392" PostTypeId="1" AcceptedAnswerId="1093" CreationDate="2010-07-21T21:18:40.193" Score="11" ViewCount="1521" Body="&lt;p&gt;One of the first things ever taught in a differential calculus class:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The derivative of sin(x) is cos(x)&lt;/li&gt;&#xA;&lt;li&gt;The derivative of cos(x) is -sin(x)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This leads to a rather neat (and convenient?) chain of derivatives:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&#xA;sin(x)&#xA;cos(x)&#xA;-sin(x)&#xA;-cos(x)&#xA;sin(x)&#xA;...&#xA;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;An analysis of the shape of their graphs confirms &lt;em&gt;some&lt;/em&gt; points; for example, when sin(x) is at a maximum, cos(x) is zero and moving downwards; when cos(x) is at a maximum, sin(x) is zero and moving upwards.  But these &quot;matching points&quot; only work for multiples of pi/4.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let us move back towards the original definition(s) of sine and cosine:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the most basic level, sin(x) is defined as -- for a right triangle with internal angle x -- the length of the side opposite of the angle divided by the hypotenuse of the triangle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To generalize this to the domain of all real numbers, sin(x) was then defined as the y coordinate of a point on the unit circle that is an angle x from the positive x axis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The definition of cos(x) was then made the same way, but with adj/hyp and the x coordinate, as we all know.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there anything about this &lt;strong&gt;basic&lt;/strong&gt; definition that allows someone to look at these definitions, alone, and think, &quot;Hey, the derivative of the sine function with respect to angle is the cosine function!&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is, from &lt;strong&gt;the unit circle definition alone&lt;/strong&gt;.  Or, even more amazingly, the &lt;strong&gt;right triangle definition alone&lt;/strong&gt;.  Ignoring graphical analysis of their plot.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In essence, I am asking, essentially, &quot;Intuitively &lt;em&gt;why&lt;/em&gt; is the derivative of the sine the cosine?&quot;&lt;/p&gt;&#xA;" OwnerUserId="51" LastEditorUserId="51" LastEditorDisplayName="" LastEditDate="2010-07-21T22:19:59.910" LastActivityDate="2011-02-04T15:01:22.633" Title="Intuitive understanding of the derivatives of sin(x) and cos(x)" Tags="&lt;calculus&gt;&lt;trigonometry&gt;" AnswerCount="14" CommentCount="12" FavoriteCount="9" />
  <row Id="393" PostTypeId="2" ParentId="392" CreationDate="2010-07-21T21:27:01.883" Score="0" ViewCount="" Body="&lt;p&gt;This isn't exactly what you asked, but look at the &lt;a href=&quot;http://en.wikipedia.org/wiki/Taylor_series&quot; rel=&quot;nofollow&quot;&gt;Taylor series&lt;/a&gt; for the polynomials:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;\sin x = \sum^{\infin}_{n=0} \frac{(-1)^n}{(2n+1)!} x^{2n+1} = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots\text{ for all } x\!&#xA;&#xA;\cos x = \sum^{\infin}_{n=0} \frac{(-1)^n}{(2n)!} x^{2n} = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \cdots\text{ for all } x\! &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The relationships between the derivatives are clear from this.&lt;/p&gt;&#xA;" OwnerUserId="123" LastActivityDate="2010-07-21T21:27:01.883" CommentCount="6" />
  <row Id="394" PostTypeId="2" ParentId="387" CreationDate="2010-07-21T21:35:25.143" Score="6" ViewCount="" Body="&lt;p&gt;It is not very surprising that the sum is finite, since numbers without a 7 (or any other digit) get rarer and rarer as the number of digits increases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a proof.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let &lt;em&gt;S&lt;/em&gt; be the harmonic series with all terms whose denominator contains the digit &lt;em&gt;k&lt;/em&gt; removed. We can write &lt;em&gt;S&lt;/em&gt; =&lt;em&gt;S&lt;/em&gt;_1 + &lt;em&gt;S&lt;/em&gt;_2 + &lt;em&gt;S&lt;/em&gt;_3 + ..., where &lt;em&gt;S&lt;/em&gt;_i is the sum of all terms whose denominator contains exactly &lt;em&gt;i&lt;/em&gt; digits, all different from &lt;em&gt;k&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the number of &lt;em&gt;i&lt;/em&gt;-digit numbers that do not contain the digit k is 8 x 9^(&lt;em&gt;i&lt;/em&gt;-1) (there are 8 choices for the first digit, excluding 0 and k, and 9 choices for the other digits). [Well, if k=0 there are 9 choices for the first digit, but the proof still works.] So there are 8 x 9^(&lt;em&gt;i&lt;/em&gt;-1) numbers in the sum &lt;em&gt;S&lt;/em&gt;_i.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now each number in &lt;em&gt;S&lt;/em&gt;_i is of the form 1/&lt;em&gt;a&lt;/em&gt;, where &lt;em&gt;a&lt;/em&gt; is an &lt;em&gt;i&lt;/em&gt;-digit number. So &lt;em&gt;a&lt;/em&gt; &gt;= 10^(&lt;em&gt;i&lt;/em&gt;-1), which implies 1/&lt;em&gt;a&lt;/em&gt; &amp;lt;= 1/10^(&lt;em&gt;i&lt;/em&gt;-1).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore &lt;em&gt;S&lt;/em&gt;_i &amp;lt;= 8 x 9^(&lt;em&gt;i&lt;/em&gt;-1) / 10^(&lt;em&gt;i&lt;/em&gt;-1) = 8 x (9/10)^(&lt;em&gt;i&lt;/em&gt;-1).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So &lt;em&gt;S&lt;/em&gt; = sum of &lt;em&gt;S&lt;/em&gt;_i &amp;lt;= sum of 8 x (9/10)^(&lt;em&gt;i&lt;/em&gt;-1)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is a  geometric series of ratio 9/10 &amp;lt; 1, which converges. Since &lt;em&gt;S&lt;/em&gt; is a positive series bounded above by a converging series, &lt;em&gt;S&lt;/em&gt; converges. &lt;/p&gt;&#xA;" OwnerUserId="115" LastActivityDate="2010-07-21T21:35:25.143" />
  <row Id="395" PostTypeId="2" ParentId="392" CreationDate="2010-07-21T21:56:46.540" Score="3" ViewCount="" Body="&lt;p&gt;As a Physics Major, I would like to propose an answer that comes from my understanding of seeing sine and cosine in the real world.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In doing this, I will examine uniform circular motion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because of the point-on-a-unit-circle definition of sine and cosine, we can say that:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;r(t) = &amp;lt; cos(t), sin(t) &amp;gt;&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Is a proper parametric function to describe a point moving along the unit circle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let us consider what the first derivate, in a physical context, should be.  The first derivative of position &lt;em&gt;should&lt;/em&gt; represent, ideally, the &lt;em&gt;velocity&lt;/em&gt; of the point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a physical context, we would expect the velocity to be the line tangent to the direction of motion at a given time &lt;code&gt;t&lt;/code&gt;.  Following from this, it would be tangent to the circle at angle &lt;code&gt;t&lt;/code&gt;.  Also, because the angular velocity is constant, the magnitude of the velocity should be constant as well.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;r'(t)     = &amp;lt; -sin(t), cos(t) &amp;gt;&#xA;|r'(t)|^2 = (-sin(t))^2 + cos(t)^2&#xA;|r'(t)|^2 = sin(t)^2 + cos(t)^2&#xA;|r'(t)|^2 = 1&#xA;|r'(t)|   = 1&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;As expected, the velocity is constant, so the derivatives of sine and cosine are behaving as they should.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can also think about what the &lt;strong&gt;direction&lt;/strong&gt; of the velocity would be, as well, compared to the position vector.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure if this is &quot;cheating&quot; by the bounds of the question, but by visualizing the graph we can see that the velocity, by nature of being tangent to the circle, must be perpendicular to the position vector.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If this is true, then position * velocity = 0 (dot product).&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;                              r(t) * r'(t) = 0&#xA;  &amp;lt; cos(t), sin(t) &amp;gt; * &amp;lt; -sin(t), cos(t) &amp;gt; = 0&#xA;( cos(t) * -sin(t) ) + ( sin(t) * cos(t) ) = 0&#xA;              -sin(t)cos(t) + sin(t)cos(t) = 0&#xA;                                         0 = 0&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Life is good.  If we assume that the definition of cos(t) is -sin(t) and that the definition of sin(t) is cos(t), we find physical behavior &lt;strong&gt;exactly&lt;/strong&gt; like expected: a constant velocity that is always perpendicular to the position vector.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can take this further and look at the acceleration.  In Physics, we would call this the restoring force.  In a circle, what acceleration would have to exist in order to keep a point moving in a circle?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More specifically, in what direction would this acceleration have to be?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It takes little thought to arrive at the idea that acceleration would have to be center-seeking, and pointing towards the center.  So, if we can find that acceleration is &lt;strong&gt;in the opposite direction&lt;/strong&gt; as the position vector, the we can be almost certain about the derivatives of sine and cosine.  That is, their internal angle should be &lt;code&gt;pi&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;                            r(t) * r''(t) = |r(t)| * |r''(t)| * cos(pi)&#xA;                            r(t) * r''(t) = |r(t)| * |r''(t)| * -1&#xA;&amp;lt; cos(t), sin(t) &amp;gt; * &amp;lt; -cos(t), -sin(t) &amp;gt; = |&amp;lt;cos(t),sin(t)&amp;gt;| * |&amp;lt;-cos(t),-sin(t)&amp;gt;| * -1&#xA;                    -cos(t)^2 + -sin(t)^2 = 1 * 1 * -1&#xA;               -1 * (cos(t)^2 + sin(t)^2) = -1&#xA;                                   -1 * 1 = -1&#xA;                                       -1 = -1&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="51" LastEditorUserId="51" LastEditorDisplayName="" LastEditDate="2010-07-21T22:06:58.390" LastActivityDate="2010-07-21T22:06:58.390" />
  <row Id="396" PostTypeId="2" ParentId="381" CreationDate="2010-07-21T22:05:25.537" Score="6" ViewCount="" Body="&lt;p&gt;Suppose you're writing a computer program to search a sorted array for a particular value. Usually the best method to use is a binary search. But binary search assumes it's the same cost to read from anywhere in the array. If it costs something to move from one array element to another, and the cost is proportional to how many array elements you need to skip over to get from one element you read to the next, then &lt;a href=&quot;http://en.wikipedia.org/wiki/Fibonacci_search_technique&quot; rel=&quot;nofollow&quot;&gt;Fibonacci search&lt;/a&gt; works better. This can apply to situations like searching through arrays that don't fit entirely in your computer's cache so it's generally cheaper to read nearby elements that distant ones.&lt;/p&gt;&#xA;" OwnerUserId="80" LastActivityDate="2010-07-21T22:05:25.537" CommunityOwnedDate="2010-08-03T18:46:03.740" />
  <row Id="397" PostTypeId="2" ParentId="392" CreationDate="2010-07-21T22:11:32.083" Score="0" ViewCount="" Body="&lt;p&gt;I don't think you can get an intuitive feel for the derivatives without looking at the plots personally. When you consider that a derivative is a rate of change, you need to be looking at a function that is varying, which implies you are looking at the plot/graph of the function. When you further consider that a derivative (by definition of it being a rate of change) is a gradient function, the intuitive answer is that cos is the gradient function of sin, and -sin is the gradient function of cos (and so on). So if you calculate the gradient of the sin curve at any point, the value you get will be the cosine value for that point.&lt;/p&gt;&#xA;" OwnerUserId="35" LastActivityDate="2010-07-21T22:11:32.083" />
  <row Id="398" PostTypeId="1" AcceptedAnswerId="406" CreationDate="2010-07-21T22:16:34.997" Score="5" ViewCount="368" Body="&lt;p&gt;We will call the set of all positive even numbers &lt;code&gt;E&lt;/code&gt; and the set of all positive integers &lt;code&gt;N&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At first glance, it seems obvious that &lt;code&gt;E&lt;/code&gt; is smaller than &lt;code&gt;N&lt;/code&gt;, because for &lt;code&gt;E&lt;/code&gt; is basically &lt;code&gt;N&lt;/code&gt; with half of its terms taken out.  The size of &lt;code&gt;E&lt;/code&gt; is the size of &lt;code&gt;N&lt;/code&gt; divided by two.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could see this as, for every item in &lt;code&gt;E&lt;/code&gt;, two items in &lt;code&gt;N&lt;/code&gt; could be matched (the item x and x-1).  This implies that &lt;code&gt;N&lt;/code&gt; is &lt;strong&gt;twice as large as &lt;code&gt;E&lt;/code&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On second glance though, it seems less obvious.  Each item in &lt;code&gt;N&lt;/code&gt; could be mapped with &lt;em&gt;one&lt;/em&gt; item in &lt;code&gt;E&lt;/code&gt; (the item x*2).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which is larger, then?  Or are they both equal in size?  Why?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(My background in Set theory is quite extremely scant)&lt;/p&gt;&#xA;" OwnerUserId="51" LastActivityDate="2010-08-02T17:06:16.650" Title="What is larger -- the set of all positive even numbers, or the set of all positive integers?" Tags="&lt;set-theory&gt;&lt;infinity&gt;" AnswerCount="8" CommentCount="2" FavoriteCount="1" />
  <row Id="399" PostTypeId="2" ParentId="398" CreationDate="2010-07-21T22:19:51.873" Score="5" ViewCount="" Body="&lt;p&gt;They are both the same size, the size being 'countable infinity' or 'aleph-null'. The reasoning behind it is exactly that which you have already identified - you can assign each item in E to a single value in N. This is true for the Natural numbers, the Integers, the Rationals but not the Reals (see the &lt;a href=&quot;http://en.wikipedia.org/wiki/Cantor%27s_diagonal_argument&quot; rel=&quot;nofollow&quot;&gt;Diagonal Slash&lt;/a&gt; argument for details on this result).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-- Added explanation from comment --&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first reasoning is invalid because the cardinality of infinite sets doesn't follow 'normal' multiplication rules. If you multiply a set with cardinality of aleph-0 by 2, you still have aleph-0. The same is true if you divide it, add to it, subtract from it by any finite amount. &lt;/p&gt;&#xA;" OwnerUserId="35" LastEditorUserId="35" LastEditorDisplayName="" LastEditDate="2010-07-21T22:36:17.057" LastActivityDate="2010-07-21T22:36:17.057" CommentCount="6" />
  <row Id="400" PostTypeId="2" ParentId="398" CreationDate="2010-07-21T22:20:00.300" Score="1" ViewCount="" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Each item in N could be mapped with one item in E  (the item x*2).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes. Both sets have &lt;a href=&quot;http://en.wikipedia.org/wiki/Cardinality&quot; rel=&quot;nofollow&quot;&gt;cardinality&lt;/a&gt; aleph-0.&lt;/p&gt;&#xA;" OwnerUserId="120" LastActivityDate="2010-07-21T22:20:00.300" CommentCount="4" />
  <row Id="401" PostTypeId="2" ParentId="398" CreationDate="2010-07-21T22:40:29.437" Score="5" ViewCount="" Body="&lt;p&gt;The word 'size' doesn't have a intuitive meaning for set of infinite items. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mathematicians defined cardinality by one-to-one correspondence(bijection), and it's generally what it means by 'size'. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there &lt;strong&gt;exist&lt;/strong&gt; a bijection between A and B, then the two sets have the same cardinality. You have shown the existence of a bijection, therefore E and N have the same cardinality.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might mean the 'density' of N is twice as large as E. density of A(a subset of natural number) is limit of |{a&amp;lt;=n|a\in A}|/n as n goes to infinity.&lt;/p&gt;&#xA;" OwnerUserId="96" LastActivityDate="2010-07-21T22:40:29.437" />
  <row Id="402" PostTypeId="2" ParentId="275" CreationDate="2010-07-21T22:47:58.727" Score="1" ViewCount="" Body="&lt;p&gt;As an undergrad, I read a fair number of pop math books. The best by far was Ash and Gross' &quot;Fearless Symmetry&quot;. This book is very beautiful. It sustains a nice level of rigor while being approachable by those who aren't professionals. Additionally, it weaves the tale of one of the most beautiful recent stories in mathematics. Everyone I know of who have read the book have found it wonderful.&lt;/p&gt;&#xA;" OwnerUserId="16" LastActivityDate="2010-07-21T22:47:58.727" CommunityOwnedDate="2010-07-21T22:47:58.727" CommentCount="1" />
  <row Id="403" PostTypeId="2" ParentId="392" CreationDate="2010-07-21T22:48:13.517" Score="5" ViewCount="" Body="&lt;p&gt;This interesting pattern of derivatives involving sine and cosine is related to the fact that e^x is its own derivative and that e^(ix) = cos(x) + i*sin(x) (&lt;a href=&quot;http://en.wikipedia.org/wiki/Euler%27s_formula&quot; rel=&quot;nofollow&quot;&gt;Euler's Formula&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These two facts are in some sense the math hiding behind Justin L's more physical explanation, which you might well find more intuitive.  &lt;/p&gt;&#xA;" OwnerUserId="128" LastActivityDate="2010-07-21T22:48:13.517" CommentCount="3" />
  <row Id="404" PostTypeId="2" ParentId="392" CreationDate="2010-07-21T23:03:36.737" Score="4" ViewCount="" Body="&lt;p&gt;One of the main ways that sine and cosine come up is as two solutions to the differential equation y'' = -y.  Why is this an important differential equation?  Well, by Newton's second law it's the differential equation that says &quot;the force is proportional and opposite to the position.&quot;  For example, this is what happens with a spring!  Now that's a degree 2 diff eq so it has a 2-dimensional space of solutions.  How to pick a nice basis of that space?  Well one way would be to pick f and g such that f' = i f and g' = -i g.  However, that involves too many imaginary numbers, so another option is f' = -g, and g' = -f.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus if you're trying to find two functions which explain oscillatory motion you're naturally lead to picking functions that have f' = g, g' = -f, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(On the other hand it's totally unclear from this point of view why Sine and Cosine should have anything to do with triangles...)&lt;/p&gt;&#xA;" OwnerUserId="8" LastActivityDate="2010-07-21T23:03:36.737" />
  <row Id="405" PostTypeId="1" AcceptedAnswerId="408" CreationDate="2010-07-21T23:25:43.533" Score="15" ViewCount="593" Body="&lt;p&gt;I have been learning some functional programming recently and I so I have come across monads. I understand what they are in programming terms, but I would like to understand what they are mathematically. Can anyone explain what a monad is using as little category theory as possible?&lt;/p&gt;&#xA;" OwnerUserId="123" LastEditorUserId="123" LastEditorDisplayName="" LastEditDate="2010-07-31T10:14:29.723" LastActivityDate="2010-08-19T12:28:32.053" Title="Simple explanation of a monad" Tags="&lt;intuition&gt;&lt;category-theory&gt;&lt;universal-algebra&gt;" AnswerCount="4" CommentCount="7" FavoriteCount="7" />
  <row Id="406" PostTypeId="2" ParentId="398" CreationDate="2010-07-21T23:32:07.050" Score="20" ViewCount="" Body="&lt;p&gt;Mathematics is the art of clever forgetting.  The first mathematical breakthrough, numbers, came about when people realized that if you just forgot about whether it was 5 cows + 3 cows, or 5 rocks + 3 rocks or whatever you always got 8.  Numbers are what you get when you look at collections of objects and &lt;em&gt;forget what kind of object they are&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you say &quot;as sets&quot; you mean you're forgetting a lot of information, in particular you don't care about what the names of the elements in that set are or what properties those elements have.  As sets the positive numbers and the positive even numbers are &quot;the same&quot; (that is are in bijection) because you can take 1,2,3,...  and just &lt;em&gt;rename&lt;/em&gt; 1 to 2, and 2 to 4, and n to 2n, and you've just renamed all the elements and got the even numbers!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if you want to remember more about these sets, for example that they're not arbitrary sets they're both subsets of the natural numbers, then they become distinguishable.  Depending on how you want to measure &quot;size of a subset of the natural numbers&quot; they might be different sizes.  For example, a common way to measure &quot;size of a subset of the natural numbers&quot; is by its &quot;density.&quot;  That is you look at the first N numbers and calculate what fraction of them are in your set, and then take the limit as N goes to infinity (warning for sufficiently complicated sets this limit might not exist).  So for your two examples, one has density 1 and the other has density 1/2, which is one way to make precise the intuition that the former is bigger &lt;em&gt;as a subset of the natural numbers&lt;/em&gt; (though not &lt;em&gt;as a set&lt;/em&gt;) than the latter.&lt;/p&gt;&#xA;" OwnerUserId="8" LastActivityDate="2010-07-21T23:32:07.050" CommentCount="1" />
  <row Id="407" PostTypeId="2" ParentId="398" CreationDate="2010-07-21T23:36:51.333" Score="1" ViewCount="" Body="&lt;p&gt;Your example is an example of a countable sets: &lt;a href=&quot;http://en.wikipedia.org/wiki/Countable_set&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Countable_set&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As others have explained, if you find a bijection between these two sets then they are of the same cardinality or have the same number of elements.(I hope that statement is true :)) But in this case you have found one with N which is what makes it a countable set. f(x)=2x would be an example of a bijection between N and 2N. Since for every x in the set of natural numbers there is a corresponding number in the set of 2N (only one) then you can say that they have the same cardinality. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A cooler example would be the interval of the real numbers (0, 1) and R itself, those are not countable but their cardinality is the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other notes, if you are wondering what aleph is: &lt;a href=&quot;http://en.wikipedia.org/wiki/Aleph_number&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Aleph_number&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="180" LastEditorUserId="180" LastEditorDisplayName="" LastEditDate="2010-07-22T01:51:50.317" LastActivityDate="2010-07-22T01:51:50.317" />
  <row Id="408" PostTypeId="2" ParentId="405" CreationDate="2010-07-21T23:42:48.640" Score="2" ViewCount="" Body="&lt;p&gt;If you want to avoid too much category theory, you can first read this link to understand the definition of monads in &lt;a href=&quot;http://www.haskell.org/tutorial/monads.html&quot; rel=&quot;nofollow&quot;&gt;Haskell&lt;/a&gt;. Then look at &lt;a href=&quot;http://en.wikibooks.org/wiki/Haskell/Category_theory&quot; rel=&quot;nofollow&quot;&gt;Wikibooks&lt;/a&gt; for a more mathematical look (thanks Jonathan Fischoff).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are two descriptions that I know of. The first can easily be found by looking at wiki under Monad or consulting Harry's nice summary. The second is more interesting in my opinion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I will assume that you don't know the definition of a monoidal action, if you do, just skip ahead.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A monoidal action is a functor from a monoid to the category of endofunctors on a category satisfying two coherence relations. These two coherence relations simply verify that your monoidal product is the same as composition in the target, and that the identity object behaves with the action. The relations are normally written as diagrams, but without latex implement, I wont type them here. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To get an idea of a monoidal action, consider a group action, and formulate it a little more categorically, by writing the two axioms as diagrams. These diagrams, when converted to the language of monoidal categories, are exactly those of a monoidal action.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the best part is once you have monoidal action, monads on a category are simply the category of monoidal actions from the trivial monoidal category to your category. Note here that the trivial monoidal category will be the monoidal category with one object one morphism and all the other monoidal data is trivially determined. The monadic coherence relations come for free from your monoidal action coherence relations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, my simple explanation?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;In this way, we can formulate monads functorially as &quot;representations&quot; of the trivial monoidal category.&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One can readily show the two definitions are the same.&lt;/p&gt;&#xA;" OwnerUserId="16" LastEditorUserId="123" LastEditorDisplayName="" LastEditDate="2010-08-19T12:28:32.053" LastActivityDate="2010-08-19T12:28:32.053" CommentCount="8" />
  <row Id="409" PostTypeId="2" ParentId="370" CreationDate="2010-07-21T23:45:18.633" Score="5" ViewCount="" Body="&lt;p&gt;Another book that is more elementary, not requiring any algebraic topology for motivation, and formulating the basics through a question and answer approach is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/052171916X&quot; rel=&quot;nofollow&quot;&gt;Conceptual Mathematics&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An added benefit is that it is written by an expert!&lt;/p&gt;&#xA;" OwnerUserId="16" LastEditorUserId="16" LastEditorDisplayName="" LastEditDate="2010-08-03T15:06:07.790" LastActivityDate="2010-08-03T15:06:07.790" CommunityOwnedDate="2010-08-04T15:59:49.803" CommentCount="5" />
  <row Id="411" PostTypeId="1" AcceptedAnswerId="413" CreationDate="2010-07-22T00:02:34.840" Score="4" ViewCount="321" Body="&lt;p&gt;Say we have a function $\displaystyle f(z)=\sum_{n=0}^\infty a_n z^n$ with radius of convergence R&gt;0. Why is the radius of convergence only R? Can we conclude that there must be a pole, branch cut or discontinuity for some z_0 with |z_0|=R? What does that mean for functions like&lt;br&gt;&#xA;$$f(z)=\begin{cases}&#xA;  0 &amp;amp; \text{for $z=0$} \\&#xA;  e^{-\frac{1}{z^2}} &amp;amp; \text{for $z \neq 0$} \end{cases}$$&lt;br&gt;&#xA;that have a radius of convergence 0?&lt;/p&gt;&#xA;" OwnerUserId="73" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-08-12T13:08:32.587" LastActivityDate="2010-08-12T13:08:32.587" Title="Why do complex functions have a finite radius of convergence?" Tags="&lt;sequences-and-series&gt;&lt;complex-analysis&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="412" PostTypeId="2" ParentId="405" CreationDate="2010-07-22T00:09:50.567" Score="3" ViewCount="" Body="&lt;p&gt;Let $C$ be a category.  Then a monad &lt;em&gt;based at $C$&lt;/em&gt; is a &lt;em&gt;monoid&lt;/em&gt; in the strict monoidal category &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathcal{End}(C)=\mathcal{Hom}_{Cat}(C,C),$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where the natural monoidal product is given by composition of endofunctors, and the monoidal unit is the identity functor.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A monoid in a monoidal category is defined &lt;a href=&quot;http://en.wikipedia.org/wiki/Monoid_object&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you need more explanation, just give me a call.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notation: The category of functors $C\to D$ is also written as $Fun(C,D)$, but this notation is nonstandard.  The standard notations are $\mathcal{Hom}_{Cat}(C,D)$ or simply $Cat(C,D)$.&lt;/p&gt;&#xA;" OwnerUserId="126" LastEditorUserId="126" LastEditorDisplayName="" LastEditDate="2010-07-22T00:17:31.097" LastActivityDate="2010-07-22T00:17:31.097" />
  <row Id="413" PostTypeId="2" ParentId="411" CreationDate="2010-07-22T00:42:49.470" Score="13" ViewCount="" Body="&lt;p&gt;If the radius of convergence is $R$, that means there is a singular point on the circle $|z| = R$. In other words, there is a point $\xi$ on the circle of radius $R$ such that the function cannot be extended via &quot;analytic continuation&quot; in a neighborhood of $\xi$.  This is a straightforward application of compactness of the circle and can be found in books on complex analysis, e.g. Rudin's. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, it does not mean that there is a pole, branch cut, or discontinuity, though those would cause singular values. Indeed, a &quot;pole&quot; on the boundary would only make sense if you can analytically continue the power series to some proper domain containing the disk $D_R(0)$, and this is generally impossible. For instance, the power series $\sum z^{2^j}$ cannot be continued in any way outside the unit disk, because it is unbounded along any ray whose angle is a dyadic fraction.  The unit circle is its natural boundary, though it does not make sense to say that the function has a branch point or pole there.  (More generally, one can show that given any domain in the plane, there is a holomorphic function in that domain which cannot be extended any further, essentially using variations on the same theme.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The function $\sum_j \frac{z^j}{j^2}$, incidentally, is continuous on the closed unit disk, but even though there is a singular point there. So continuity may happen at singular points.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The last function you mention does not have a power series expansion in a neighborhood of zero. In fact, it is not continuous at zero, because it blows up if you approach zero along the imaginary axis.&lt;/p&gt;&#xA;" OwnerUserId="536" LastActivityDate="2010-07-22T00:42:49.470" />
  <row Id="414" PostTypeId="1" AcceptedAnswerId="417" CreationDate="2010-07-22T00:44:57.890" Score="1" ViewCount="227" Body="&lt;p&gt;We define Lie algebra's abstractly as algebras whose multiplication satisfies anti-commutativity and Jacobi's Identity. A particular instance of this is an associative algebra equipped with the commutator bracket [a,b]=ab-ba. However, the notation suggests that this bracket is the one we think about. Additionally, the right adjoint to the functor I just mentioned creates the universal enveloping algebra by quotient-ing the tensor algebra by the tensor version of this bracket; but we could always start with some arbitrary Lie algebra with  some other satisfactory bracket and apply this functor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;why the commutator bracket?&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Is it purely from a historical standpoint(and if so could you explain why)? Or is there a result that says any Lie algebra is essentially one with the commutator bracket(maybe something about the faithfulness of the functor from above)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know of(a colleague told me) a proof that the Jacobi identity is also an artifact of the right adjoint to the universal enveloping algebra. He can show that it is the necessary identity for the universal enveloping algebra to be associative(if someone knows of this in the literature I would also appreciate the link to this!)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope this question is clear, if not, I can revise and try to make it a bit more specific.&lt;/p&gt;&#xA;" OwnerUserId="16" LastActivityDate="2010-07-22T01:03:08.603" Title="Why do we use the commutator bracket for Lie algebra's" Tags="&lt;notation&gt;&lt;lie-algebra&gt;" AnswerCount="1" CommentCount="6" />
  <row Id="415" PostTypeId="1" AcceptedAnswerId="416" CreationDate="2010-07-22T00:47:12.957" Score="6" ViewCount="173" Body="&lt;p&gt;It appears to me that only Triangles, Squares, and Pentagons are able to &quot;tessellate&quot; (is that the proper word in this context?) to become regular 3D convex polytopes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What property of those regular polygons themselves allow them to faces of regular convex polyhedron?  Is it something in their angles?  Their number of sides?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, why are there more Triangle-based Platonic Solids (three) than Square- and Pentagon- based ones? (one each)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, is this the same property that allows certain Platonic Solids to be used as &quot;faces&quot; of regular polychoron (4D polytopes)?&lt;/p&gt;&#xA;" OwnerUserId="51" LastEditorUserId="51" LastEditorDisplayName="" LastEditDate="2010-07-25T07:39:02.780" LastActivityDate="2010-08-05T18:33:17.190" Title="What property of certain regular polygons allows them to be faces of the Platonic Solids?" Tags="&lt;geometry&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="416" PostTypeId="2" ParentId="415" CreationDate="2010-07-22T01:00:59.263" Score="12" ViewCount="" Body="&lt;p&gt;The regular polygons that form the Platonic solids are those for which the measure of the interior angles, say &amp;alpha; for convenience, is such that $3\alpha&amp;lt;2\pi$ (360&amp;deg;) so that three (or more) of the polygons can be assembled around a vertex of the solid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regular (equilateral) triangles have interior angles of measure $\frac{\pi}{3}$ (60&amp;deg;), so they can be assembled 3, 4, or 5 at a vertex ($3\cdot\frac{\pi}{3}&amp;lt;2\pi$, $4\cdot\frac{\pi}{3}&amp;lt;2\pi$, $5\cdot\frac{\pi}{3}&amp;lt;2\pi$), but not 6 ($6\cdot\frac{\pi}{3}=2\pi$--they tesselate the plane).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regular quadrilaterals (squares) have interior angles of measure $\frac{\pi}{2}$ (90&amp;deg;), so they can be assembled 3 at a vertex ($3\cdot\frac{\pi}{2}&amp;lt;2\pi$), but not 4 ($4\cdot\frac{\pi}{2}=2\pi$--they tesselate the plane).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regular pentagons have interior angles of measure $\frac{3\pi}{5}$ (108&amp;deg;), so they can be assembled 3 at a vertex ($3\cdot\frac{3\pi}{5}&amp;lt;2\pi$), but not 4 ($4\cdot\frac{3\pi}{5}&gt;2\pi$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regular hexagons have interior angles of measure $\frac{2\pi}{3}$ (120&amp;deg;), so they cannot be assembled 3 at a vertex ($3\cdot\frac{2\pi}{3}=2\pi$--they tesselate the plane).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any other regular polygon will have larger interior angles, so cannot be assembled into a regular solid.&lt;/p&gt;&#xA;" OwnerUserId="72" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-08-05T18:33:17.190" LastActivityDate="2010-08-05T18:33:17.190" CommentCount="5" />
  <row Id="417" PostTypeId="2" ParentId="414" CreationDate="2010-07-22T01:03:08.603" Score="2" ViewCount="" Body="&lt;p&gt;Well Lie algebras naturally arise from the Lie bracket of vector fields and from taking the Lie algebra of a Lie group.  If we look at a the Lie algebra of a matrix subgroup, then the Lie bracket is the commutator of matrices.&lt;/p&gt;&#xA;" OwnerUserId="9" LastActivityDate="2010-07-22T01:03:08.603" CommentCount="2" />
  <row Id="418" PostTypeId="1" AcceptedAnswerId="420" CreationDate="2010-07-22T01:04:43.687" Score="5" ViewCount="646" Body="&lt;p&gt;Two questions:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Find a bijective function from (0,1) to [0,1]. I haven't find the solution to this since I saw it a few days ago. It strike me as odd--mapping a open set into a closed set.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;S is countable. It's trivial to find a bijective function f:N -&gt; N-S, (when |N| = |N-S|) f(n) = the nth smallest number in N-S. Are they any analogous trivial solutions to f: R-&gt;R-S?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="96" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2010-11-26T15:24:26.753" LastActivityDate="2011-01-17T06:00:31.107" Title="Construct a bijection from R to R-S, where S is countable" Tags="&lt;elementary-set-theory&gt;" AnswerCount="3" CommentCount="10" />
  <row Id="419" PostTypeId="1" AcceptedAnswerId="421" CreationDate="2010-07-22T01:07:25.070" Score="2" ViewCount="241" Body="&lt;p&gt;I just got my statistics test back and I am totally confused about one of the questions!  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A study was done that took a simple random sample of 40 people and measured whether&#xA;  the subjects were right-handed or&#xA;  left-handed, as well as their ages.&#xA;  The study showed that the proportion&#xA;  of left-handed people and the ages had&#xA;  a strong negative correlation. What&#xA;  can we conclude? Explain your answer. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I know that we can't conclude that getting older causes people to become right-handed. Something else might be causing it, not the age. If two things are correlated, we can only conclude association, not causation. So I wrote: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;We can conclude that many people&#xA;  become right-handed as they grow&#xA;  older, but we cannot tell why.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That's exactly what association means, but my teacher marked me wrong! What mistake did I make? Is 40 too small of a sample size to make any conclusions?&lt;/p&gt;&#xA;" OwnerUserId="73" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-07-22T02:27:46.657" LastActivityDate="2010-07-23T01:53:22.163" Title="What can we conclude from correlation?" Tags="&lt;statistics&gt;&lt;homework&gt;&lt;high-school&gt;" AnswerCount="3" CommentCount="16" FavoriteCount="1" />
  <row Id="420" PostTypeId="2" ParentId="418" CreationDate="2010-07-22T01:13:00.127" Score="8" ViewCount="" Body="&lt;p&gt;The proof of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Cantor%E2%80%93Bernstein%E2%80%93Schroeder_theorem&quot; rel=&quot;nofollow&quot;&gt;Schroeder-Bernstein theorem&lt;/a&gt; allows you to get a bijection for 1, since we have an injection (0,1) \to [0,1] and a bijection $f: [0,1] \to [1/4, 3/4] \subset (0,1)$ (say $x \to x/2 +1/4$).  The function's definition will be somewhat messy (basically, it depends on how many times you can lift a point under these to injections already defined, and specifically the parity of the number of times), but it'll do it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For 2, iterate this construction to get a bijection $R \to R - N$.  Then given any countable set $S$, define the map of $R$ that interchanges $N$ and $S$ and leaves every other point fixed.  Then the composition of the first bijection with this second map is your bijection.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Continuity considerations imply that the map can't be continuous: in 1, for instance, we'd otherwise have that (0,1) is compact, which it's not.&lt;/p&gt;&#xA;" OwnerUserId="536" LastActivityDate="2010-07-22T01:13:00.127" CommentCount="4" />
  <row Id="421" PostTypeId="2" ParentId="419" CreationDate="2010-07-22T01:52:19.083" Score="2" ViewCount="" Body="&lt;p&gt;This is wrong: &quot;We can conclude that many people become right-handed as they grow older.&quot; We cannot conclude this at all from the given data. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For one, the study only takes a sample at one point in time, rather than selecting a sample and monitoring their progress through many decades. This is what would be needed for us to even entertain the possibility that aging causes a change in handedness. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other possible causes include that left handed people might have a shorter life expectancy, or perhaps there was a spike in the birth rate of right handed people in the past. There are many other possibilities that have been mentioned in others answers which would also account for the skewed proportions &lt;em&gt;without&lt;/em&gt; requiring people to change handedness with age, which is what you falsely concluded in the test.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, just an observation, but it appears the &quot;study&quot; was conducted under false pretenses. Handedness is a false dichotomy, people can also be ambidextrous.&lt;/p&gt;&#xA;" OwnerUserId="156" LastEditorUserId="156" LastEditorDisplayName="" LastEditDate="2010-07-23T01:53:22.163" LastActivityDate="2010-07-23T01:53:22.163" CommentCount="7" />
  <row Id="422" PostTypeId="2" ParentId="240" CreationDate="2010-07-22T01:57:38.540" Score="2" ViewCount="" Body="&lt;p&gt;In mathematics you can construct a mathematical theory with different sets of axioms. This can be really useful. When mathematicians ignored the parallel line axiom in Euclidian geometry it gave raise to non-Euclidian geometries, which became really important in Einsteinian physics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An axiom of logic is the law of the excluded third which basically says that one statement is either true or false. This means that any theorem that depends completely in this axiom is not valid on mathematical theories that decide to ignore the axiom.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A proof by contradiction is using the axiom directly; if the consequent is false then the antecent is false, then the converse of the consequent is true (because it must be either true or false). If the theorem can be proved in a constructive way, then it does not depend on the Law of Excluded Third and is valid in theories that does not use the law.&lt;/p&gt;&#xA;" OwnerUserId="187" LastActivityDate="2010-07-22T01:57:38.540" CommentCount="1" />
  <row Id="423" PostTypeId="1" AcceptedAnswerId="429" CreationDate="2010-07-22T02:12:22.140" Score="4" ViewCount="331" Body="&lt;p&gt;I've sort of gotten a grasp on the Chain rule with one variable.  If you hike up a mountain at 2 feet an hour, and the temperature decreases at 2 degrees per feet, the temperature would be decreasing for you at 2*2 = 4 degrees per hour.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I'm having a bit more trouble understanding the Chain Rule as applied to multiple variables.  Even the case of 2 dimensions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;z = f(x,y) where x = g(t) and y = h(t).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;dz/dt = dz/dx dx/dt + dz/dy dy/dt&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, this is easy enough to &quot;calculate&quot; (and figure out what goes where).  My teacher taught me a neat tree-based graphical method for figuring out partial derivatives using chain rule.  All-in-all, it was rather hand-wavey.  However, I'm not sure exactly how this works, intuitively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why, intuitively, is the equation above true?  Why &lt;strong&gt;addition&lt;/strong&gt;?  Why not multiplication, like the other chain rule?  Why are some multiplied and some added?&lt;/p&gt;&#xA;" OwnerUserId="51" LastEditorUserId="126" LastEditorDisplayName="" LastEditDate="2010-07-22T07:32:01.610" LastActivityDate="2010-07-22T11:43:11.577" Title="Intuitive reasoning behind the Chain Rule in multiple variables?" Tags="&lt;calculus&gt;&lt;linear-algebra&gt;" AnswerCount="4" FavoriteCount="2" />
  <row Id="424" PostTypeId="2" ParentId="419" CreationDate="2010-07-22T02:13:36.677" Score="0" ViewCount="" Body="&lt;p&gt;We can only conclude that the result is interesting and it deserves more research.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Without further study we can not say if right-handedness causes age (ie. being left handed causes biological alterations that shorten the life span), age causes right-handedness (ie. as people age they become right handed), they are correlated because they are caused by another variable (ie. older people became educated in a different system that discouraged left-handedness), the study had bad luck selecting its sample, the study sample was bad designed, the study was bad designed, etc.&lt;/p&gt;&#xA;" OwnerUserId="187" LastActivityDate="2010-07-22T02:13:36.677" CommentCount="3" />
  <row Id="425" PostTypeId="2" ParentId="423" CreationDate="2010-07-22T02:17:38.100" Score="0" ViewCount="" Body="&lt;p&gt;Changing time is the same as changing x and changing y. If the changes each one caused in z didn't interact, then the total change would be the sum of both changes. If the function is well behaved (differentiatable) then the interaction caused by an infinitesimal change in x with an infinitesimal change in y will be doubly infinitesimal. The proof of the double chain rule just shows this formally.&lt;/p&gt;&#xA;" OwnerUserId="123" LastEditorUserId="123" LastEditorDisplayName="" LastEditDate="2010-07-22T02:24:27.210" LastActivityDate="2010-07-22T02:24:27.210" />
  <row Id="426" PostTypeId="2" ParentId="423" CreationDate="2010-07-22T02:19:45.070" Score="4" ViewCount="" Body="&lt;p&gt;The correct context of the chain rule is that taking the tangent bundle is functorial.  A more down-to-earth answer is provided by working coordinate free using linear algebra.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose $f:X\to Y$ and $g:Y\to Z$ are functions between Banach spaces (these are a generalized version of R^n) such that $f$ is differentiable at $\vec{v}$ and $g$ is differentiable at $f(\vec{v})$ (note that in the general case we must require that their derivatives are toplinear (continuous and linear), since not all linear maps are continuous in the context of infinite dimensional spaces).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then taking the total differential, we see that the chain rule is equivalent to saying that: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$T_\vec{v}(g\circ f)=T_{f(\vec{v})}(g) \circ T_\vec{v}(f).$$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The description you get with coordinates comes from this very much simpler presentation (Where T denotes the total differential) as follows: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To derive the formula with coordinates (say, for example, in three dimensions), we present the total differentials (which are linear transformations) as their Jacobian matrices and test along the column vector ${}^t[x,y,z]$. (Where the leftthand exponent t denotes the matrix transpose).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: When we present linear operators by their matrices, composition of linear transformations becomes matrix multiplication, and evaluation at a vector $\vec{w}$ becomes righthand multiplication by a column matrix.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The main point is that coordinates &lt;em&gt;obscure&lt;/em&gt; what's actually going on here.  The beauty of the coordinate free definition is destroyed by the complicated description of matrix multiplication.&lt;/p&gt;&#xA;" OwnerUserId="126" LastEditorUserId="126" LastEditorDisplayName="" LastEditDate="2010-07-22T07:31:29.990" LastActivityDate="2010-07-22T07:31:29.990" CommentCount="1" />
  <row Id="427" PostTypeId="2" ParentId="240" CreationDate="2010-07-22T02:20:12.463" Score="1" ViewCount="" Body="&lt;p&gt;Proof by contradiction is just as logically valid as any other type of proof. If you are unsure, I think it might help to consider exactly what a proof by contradiction entails.&lt;br&gt;&#xA;Say we have a set of statements $\Gamma$, and that $\Gamma\cup{(\neg\phi)}$ is not consistent. That is, the statement $\neg\phi$ contradicts something in $\Gamma$. (In other words, we supposed $\phi$ was false, and reached a contradiction.) Say that statement was $\psi$. Then $\Gamma\cup{(\neg\phi)}\vdash\psi$ and $\Gamma\cup{(\neg\phi)}\vdash \neg\psi$.&#xA;By the &lt;a href=&quot;http://en.wikipedia.org/wiki/Principle_of_explosion&quot; rel=&quot;nofollow&quot;&gt;principle of explosion&lt;/a&gt;, we conclude that $\Gamma\cup{(\neg\phi)}\vdash\phi$. (We can prove any statement, so we prove $\phi$.&lt;br&gt;&#xA;By &lt;a href=&quot;http://en.wikipedia.org/wiki/Natural_deduction#Proofs_and_type-theory&quot; rel=&quot;nofollow&quot;&gt;deduction&lt;/a&gt;, we know that $\Gamma\vdash(\neg\phi\rarrow\phi)$.&lt;br&gt;&#xA;Most first-order logic systems have an axiom that gives us $((\neg\phi\rarrow\phi)\rarrow\phi$. I hope you can convince yourself that this is true without trouble.&lt;br&gt;&#xA;This yields $\Gamma\vdash\phi$.&lt;br&gt;&#xA;We started with the idea that the negation of the statement $\phi$ was incompatible with your working set of axioms and theorems $\Gamma$, and concluded that therefore $\Gamma$ proves $\phi$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, there is more than one way to prove anything. Other methods can often be more intuitive, more elegant, or may lead to some other useful results. However, that is distinct from &quot;weaker.&quot; Proof by contradiction is perfectly sound.&lt;/p&gt;&#xA;" OwnerUserId="73" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-07-22T02:44:44.980" LastActivityDate="2010-07-22T02:44:44.980" />
  <row Id="428" PostTypeId="2" ParentId="423" CreationDate="2010-07-22T02:21:57.997" Score="1" ViewCount="" Body="&lt;p&gt;Think of it in terms of causality &amp;amp; superposition.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;z = f(x,y)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;If you keep x fixed then dz/dt = df/dx * dx/dt, and if you keep y fixed then dz/dt = df/fy * dy/dt. Superposition says you can just add the two together.&lt;/p&gt;&#xA;" OwnerUserId="120" LastEditorUserId="120" LastEditorDisplayName="" LastEditDate="2010-07-22T11:43:11.577" LastActivityDate="2010-07-22T11:43:11.577" />
  <row Id="429" PostTypeId="2" ParentId="423" CreationDate="2010-07-22T02:26:47.253" Score="6" ViewCount="" Body="&lt;p&gt;The basic reason is that one is simply composing the derivatives just as one composes the functions. Derivatives are linear approximations to functions. When you compose the functions, you compose the linear approximations---not a surprise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm going to try to expand on Harry Gindi's answer, because that was the only way I could grok it, but in somewhat simpler terms.&#xA;The way to think of a derivative in multiple variables is as a linear approximation.   In particular, let $f: R^m \to R^n$ and $q=f(p)$.  Then near $p$, we can write $f$ as $q$ basically something linear plus some &quot;noise&quot; which &quot;doesn't matter&quot; (i.e. is little oh of the distance to $p$).  Call this linear map $L: R^m \to R^n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, suppose $g: R^n \to R^s$ is some map and $r = g(q)$.  We can approximate $g$ near $q$ by $r$ plus some linear map $N$ plus some &quot;garbage&quot; which is, again, small.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For simplicity, I'm going to assume that $p,q,r$ are all zero. This is ok, because one can just move one's origin around a bit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, as before, applying $f$ to a point near zero corresponds loosely to applying the linear transformation $L$. Applying $g$ to a point near zero corresponds loosely to applying $N$.  Hence applying $g \circ f$ corresponds up to some ignorable &quot;garbage&quot; to the map $N \circ L$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means that $N \circ L$ is the linear approximation to $g \circ f$ at zero, in particular this composition is the derivative of $g \circ f$.&lt;/p&gt;&#xA;" OwnerUserId="536" LastActivityDate="2010-07-22T02:26:47.253" />
  <row Id="431" PostTypeId="2" ParentId="419" CreationDate="2010-07-22T02:52:07.877" Score="1" ViewCount="" Body="&lt;p&gt;Here is one example of a plausible explanation that disagrees with your analysis:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cultural expectations for left- and right-handedness have changed over time. Older people may have gone to school at a time where left-handedness was discouraged and students were forced to write with their right hands, training children never to use the left hand instead of the right. Younger participants in the study were in school more recently and learned to write at a time where left-handedness was &lt;em&gt;not&lt;/em&gt; discouraged, creating a positive correlation between left-handedness and youth.&lt;/p&gt;&#xA;" OwnerUserId="45" LastActivityDate="2010-07-22T02:52:07.877" />
  <row Id="432" PostTypeId="1" AcceptedAnswerId="670" CreationDate="2010-07-22T03:03:44.997" Score="6" ViewCount="284" Body="&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Discriminant&quot; rel=&quot;nofollow&quot;&gt;discriminant&lt;/a&gt; &lt;a href=&quot;http://mathurl.com/2dqz7gw&quot; rel=&quot;nofollow&quot;&gt;&lt;img src=&quot;http://mathurl.com/2dqz7gw.png&quot;&gt;&lt;/a&gt; of the cubic polynomial &lt;a href=&quot;http://mathurl.com/24gdf5r&quot; rel=&quot;nofollow&quot;&gt;&lt;img src=&quot;http://mathurl.com/24gdf5r.png&quot;&gt;&lt;/a&gt; &lt;a href=&quot;http://en.wikipedia.org/wiki/Cubic_equation#The_nature_of_the_roots&quot; rel=&quot;nofollow&quot;&gt;indicates not only if there are repeated roots when Δ vanishes&lt;/a&gt;, but also that there are three distinct, real roots if &lt;a href=&quot;http://mathurl.com/2wooxlb&quot; rel=&quot;nofollow&quot;&gt;&lt;img src=&quot;http://mathurl.com/2wooxlb.png&quot;&gt;&lt;/a&gt;, and that there is one real root and two complex roots (complex conjugates) if &lt;a href=&quot;http://mathurl.com/2cuwdfa&quot; rel=&quot;nofollow&quot;&gt;&lt;img src=&quot;http://mathurl.com/2cuwdfa.png&quot;&gt;&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why does &lt;img src=&quot;http://mathurl.com/2cuwdfa.png&quot;&gt; indicate complex roots? I understand that because of the way that the discriminant is defined, it indicates that there is a repeated root if it vanishes, but why does Δ greater than 0 or less than 0 have special meaning, too?&lt;/p&gt;&#xA;" OwnerUserId="184" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-07-26T08:30:47.977" LastActivityDate="2010-07-26T08:30:47.977" Title="Why does the discriminant of a cubic polynomial being less than 0 indicate complex roots?" Tags="&lt;algebra-precalculus&gt;&lt;polynomials&gt;&lt;zeros&gt;&lt;roots&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="433" PostTypeId="2" ParentId="432" CreationDate="2010-07-22T03:11:57.567" Score="6" ViewCount="" Body="&lt;p&gt;The discriminant of any monic polynomial is the product $\prod_{i \neq j} (x_i - x_j)^2$ of the squares of the differences of the roots (in an algebraic closure, e.g. $C$). Cf. the &lt;a href=&quot;http://en.wikipedia.org/wiki/Discriminant#Discriminant_of_a_polynomial&quot; rel=&quot;nofollow&quot;&gt;Wikipedia article&lt;/a&gt; on this. Consequently, if the roots are all real and distinct, this must be positive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(If the polynomial is not monic, the factor $a_0^{2n-2}$ is thrown in, for $a_0$ the leading coefficient and $n$ the degree; this is positive for a real polynomial.)&lt;/p&gt;&#xA;" OwnerUserId="536" LastActivityDate="2010-07-22T03:11:57.567" CommentCount="2" />
  <row Id="434" PostTypeId="1" AcceptedAnswerId="437" CreationDate="2010-07-22T03:22:51.490" Score="7" ViewCount="217" Body="&lt;p&gt;&lt;a href=&quot;http://math.stackexchange.com/questions/250/a-challenge-by-r-p-feynman-give-counter-intuitive-theorems-that-can-be-transl/260#260&quot; rel=&quot;nofollow&quot;&gt;This answer&lt;/a&gt; suggests that there are explicit polynomial equations for which the existence &#xA;(or nonexistence) of integer solutions is unprovable. How can this be?&lt;/p&gt;&#xA;" OwnerUserId="48" LastEditorUserId="38" LastEditorDisplayName="" LastEditDate="2010-07-25T23:39:47.427" LastActivityDate="2010-07-25T23:39:47.427" Title="How can there be explicit polynomial equations for which the existence of integer solutions is unprovable?" Tags="&lt;algebra&gt;&lt;logic&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" />
  <row Id="435" PostTypeId="2" ParentId="250" CreationDate="2010-07-22T03:24:31.227" Score="-2" ViewCount="" Body="&lt;p&gt;Theorem: Any predicate which can be evaluated in polynomial time on a nondeterministic turing machine can be evaluated in polynomial time on a deterministic turing machine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or is asking an open question cheating? :)&lt;/p&gt;&#xA;" OwnerUserId="48" LastEditorUserId="48" LastEditorDisplayName="" LastEditDate="2010-07-22T03:30:16.190" LastActivityDate="2010-07-22T03:30:16.190" CommunityOwnedDate="2010-07-22T03:24:31.227" CommentCount="1" />
  <row Id="437" PostTypeId="2" ParentId="434" CreationDate="2010-07-22T03:32:36.017" Score="8" ViewCount="" Body="&lt;p&gt;Edit: This has been explained better by Joel David Hamkins; cf. &lt;a href=&quot;http://mathoverflow.net/questions/32892/does-anyone-know-a-polynomial-whose-lack-of-roots-cant-be-proved&quot; rel=&quot;nofollow&quot;&gt;this MO thread&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is because given any first-order formula $\phi(x_1, \dots, x_n)$ with $n$ parameters in the integers, there is a polynomial $P(x_1, \dots, x_n, z_1, \dots, z_m)$ which can be proved to have a root in $z_1, \dots, z_m$ in the integers if and only if $\phi(x_1, \dots, x_n)$---this is a version of Matiyasevich's solution to Hilbert's tenth problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, it's possible to construct within any formal system explicit first-order formulas which can't be proved or disproved (for instance, the statement &quot;this formal system is consistent&quot;).  This is  the second incompleteness theorem.  These first-order formulas correspond to polynomials by the first paragraph.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, one can show that you can construct a polynomial $P(z_1, \dots, z_m)$ corresponding to the statement &quot;mathematics* is inconsistent&quot; translated to a formula.  Thus, if mathematics is consistent, there's no mathematical proof that this polynomial doesn't have a root.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;*Mathematics=ZFC here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For more on these lines, see Ebbinghaus-Frum-Thomas's very accessible book on mathematical logic.&lt;/p&gt;&#xA;" OwnerUserId="536" LastEditorUserId="536" LastEditorDisplayName="" LastEditDate="2010-07-22T16:37:55.700" LastActivityDate="2010-07-22T16:37:55.700" CommentCount="10" />
  <row Id="438" PostTypeId="1" AcceptedAnswerId="439" CreationDate="2010-07-22T04:05:54.007" Score="3" ViewCount="340" Body="&lt;p&gt;I know there must be something unmathematical in the following but I don't know where it is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;\sqrt{-1} &amp;amp;= i \\ \\&#xA;\frac1{\sqrt{-1}} &amp;amp;= \frac1i \\ \\&#xA;\frac{\sqrt1}{\sqrt{-1}} &amp;amp;= \frac1i \\ \\&#xA;\sqrt{\frac1{-1}} &amp;amp;= \frac1i \\ \\&#xA;\sqrt{\frac{-1}1} &amp;amp;= \frac1i \\ \\&#xA;\sqrt{-1} &amp;amp;= \frac1i \\ \\&#xA;i &amp;amp;= \frac1i \\ \\&#xA;i^2 &amp;amp;= 1 \\ \\&#xA;-1 &amp;amp;= 1 \quad !!!&#xA;\end{align}&lt;/p&gt;&#xA;" OwnerUserId="187" LastEditorUserId="171" LastEditorDisplayName="" LastEditDate="2010-08-24T18:11:18.780" LastActivityDate="2010-08-24T18:11:18.780" Title="-1 is not 1, so where is the mistake?" Tags="&lt;arithmetic&gt;&lt;complex-numbers&gt;&lt;fake-proofs&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="439" PostTypeId="2" ParentId="438" CreationDate="2010-07-22T04:11:45.640" Score="10" ViewCount="" Body="&lt;p&gt;Between your third and fourth lines, you use sqrt(a)/sqrt(b) = sqrt(a/b).  This is only true when a &gt;= 0 and b &gt; 0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&lt;strong&gt;edit&lt;/em&gt;&lt;/strong&gt;:  As pointed out in the comments, what I meant was that the identity sqrt(a)/sqrt(b) = sqrt(a/b) has domain a &gt;= 0 and b &gt; 0.  Outside that domain, applying the identity is inappropriate, whether or not it &quot;works.&quot;  In general, sqrt(x) where x is a negative real number must first be rewritten as i * sqrt(|x|) before any other algebraic manipulations can be applied (because the identities relating to manipulation of square roots [perhaps exponentiation with non-integer exponents in general] require nonnegative numbers).&lt;/p&gt;&#xA;" OwnerUserId="72" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-07-22T05:04:12.227" LastActivityDate="2010-07-22T05:04:12.227" CommentCount="7" />
  <row Id="440" PostTypeId="1" AcceptedAnswerId="441" CreationDate="2010-07-22T04:22:56.800" Score="8" ViewCount="297" Body="&lt;p&gt;An equivalence relation is defined by three properties: reflexivity, symmetry and transitivity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Doesn't symmetry and transitivity implies reflexivity? Consider the following argument.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For any a and b,&#xA;a R b implies b R a by symmetry. Using transitivity, we have a R a.&lt;/p&gt;&#xA;" OwnerUserId="96" LastActivityDate="2010-07-22T11:28:41.117" Title="Why isn't reflexivity redundant in the definition of equivalence relation?" Tags="&lt;relations&gt;" AnswerCount="3" FavoriteCount="1" />
  <row Id="441" PostTypeId="2" ParentId="440" CreationDate="2010-07-22T04:28:18.623" Score="15" ViewCount="" Body="&lt;p&gt;Actually, without the reflexivity condition,   the empty relation would count as an equivalence relation, which is non-ideal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your argument used the hypothesis that for each $a$, there exists $b$ such that $aRb$ holds. If this is true, then symmetry and transitivity imply reflexivity, but this is not true in general.&lt;/p&gt;&#xA;" OwnerUserId="536" LastActivityDate="2010-07-22T04:28:18.623" />
  <row Id="442" PostTypeId="2" ParentId="440" CreationDate="2010-07-22T04:36:02.390" Score="6" ViewCount="" Body="&lt;p&gt;No.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The missing condition is sometimes called  'seriality' -- for any x there must be a y such that x R y.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you add seriality to the symmetry and transitivity you get a reflexive relation again.&lt;/p&gt;&#xA;" OwnerUserId="106" LastActivityDate="2010-07-22T04:36:02.390" CommentCount="2" />
  <row Id="443" PostTypeId="1" AcceptedAnswerId="445" CreationDate="2010-07-22T05:03:43.867" Score="9" ViewCount="255" Body="&lt;p&gt;1/7 = 0.(142857)...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with the digits in the parentheses repeating.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand that the reason it's a repeating fraction is because 7 and 10 are coprime.  But this...cyclical nature is something that is not observed by any other reciprocal of any natural number that I know of (besides multiples of 7). (if I am wrong, I hope that I may find others through this question)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By &quot;cyclical,&quot; i mean:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&#xA;1/7 = 0.(142857)...&#xA;2/7 = 0.(285714)...&#xA;3/7 = 0.(428571)...&#xA;4/7 = 0.(571428)...&#xA;5/7 = 0.(714285)...&#xA;6/7 = 0.(857142)...&#xA;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Where all of the repeating digits are the same string of digits, but shifted.  Not just a simple &quot;they are all the same digits re-arranged&quot;, but the same digits &lt;strong&gt;in the same order&lt;/strong&gt;, but shifted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or perhaps more strikingly, from the &lt;a href=&quot;http://en.wikipedia.org/wiki/142857_%28number%29&quot; rel=&quot;nofollow&quot;&gt;wikipedia article&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&#xA;1 × 142,857 = 142,857&#xA;2 × 142,857 = 285,714&#xA;3 × 142,857 = 428,571&#xA;4 × 142,857 = 571,428&#xA;5 × 142,857 = 714,285&#xA;6 × 142,857 = 857,142&#xA;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;What is it about the number 7 in relation to the base 10 (and its prime factorization 2*5?) that allows its reciprocal to behave this way?  Is it (and its multiples) unique in having this property?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Cyclic_number&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt; has an article on this subject, and gives a form for deriving them and constructing arbitrary ones, but does little to show the &quot;why&quot;, and finding what numbers have cyclic inverses.&lt;/p&gt;&#xA;" OwnerUserId="51" LastEditorUserId="51" LastEditorDisplayName="" LastEditDate="2010-07-22T07:22:23.383" LastActivityDate="2010-08-05T20:05:24.247" Title="Why is the decimal representation of 1/7 &quot;cyclical&quot;?" Tags="&lt;number-theory&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="2" />
  <row Id="444" PostTypeId="2" ParentId="443" CreationDate="2010-07-22T05:30:54.930" Score="6" ViewCount="" Body="&lt;p&gt;It works with 1/19 = 0.(052631578947368421) too, while n/13 has two cycles: 1/13 = 0.(076923), 2/13 = 0.(153846), 3/13 = 0.(230769), 4/13 = 0.(307692), 5/13 = 0.(384615), and so on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That a cycle must appear when you have a prime number p different from the base in which we work (so in base 10 different from 2 and 5) is clear: if you perform the long division 1/p, sooner or later partial quotients must be repeated, and from that point on the quotients repeat themselves. The length of the cycle must be a divisor of p-1: it may be short (think at 1/11 = 0.(09) ) or have the maximum possible lenght like the cases of 7 and 19.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wikipedia has &lt;a href=&quot;http://en.wikipedia.org/wiki/Cyclic_number&quot; rel=&quot;nofollow&quot;&gt;an article on Cyclic numbers&lt;/a&gt;, and some other example is also &lt;a href=&quot;http://en.wikipedia.org/wiki/Repeating_decimal&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;; &#xA;unfortunately no sufficient rule is given for a number to have its inverse cyclical.&lt;/p&gt;&#xA;" OwnerUserId="89" LastActivityDate="2010-07-22T05:30:54.930" CommentCount="3" />
  <row Id="445" PostTypeId="2" ParentId="443" CreationDate="2010-07-22T05:37:42.593" Score="5" ViewCount="" Body="&lt;p&gt;For a prime p, the length of the repeating block of $\frac{1}{p}$ is the least positive integer k for which $p|(10^k-1)$.  As in mau's answer, $k|(p-1)$, so $k\leq p-1$.  When $k=p-1$, then $\frac{1}{p}$ and its multiples behave as discussed in the question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of the first 100 primes, this is true for 7, 17, 19, 23, 29, 47, 59, 61, 97, 109, 113, 131, 149, 167, 179, 181, 193, 223, 229, 233, 257, 263, 269, 313, 337, 367, 379, 383, 389, 419, 433, 461, 487, 491, 499, 503, 509, 541.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(List generated in Mathematica using &lt;code&gt;Select[Table[Prime[n], {n, 1, 100}], # - 1 == Length[RealDigits[1/#][[1]][[1]]]&amp;amp;]&lt;/code&gt;.)&lt;/p&gt;&#xA;" OwnerUserId="72" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-08-05T20:05:24.247" LastActivityDate="2010-08-05T20:05:24.247" CommentCount="1" />
  <row Id="446" PostTypeId="2" ParentId="418" CreationDate="2010-07-22T06:13:15.787" Score="8" ViewCount="" Body="&lt;p&gt;An explicit bijection (0,1) \to [0,1] for part 1 is given by:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;f(1/2) = 0&lt;/p&gt;&#xA;&#xA;&lt;p&gt;f(1/3) = 1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;f(1/n) = 1/(n-2) for n &gt; 3&lt;/p&gt;&#xA;&#xA;&lt;p&gt;f(x) = x for x not equal to a reciprocal of an integer&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a bijection R to R-S, we can number the elements of S (because S is countable) as s_1, s_2, s_3., .... Choose an infinite sequence t_1, t_2, ... of distinct elements of R, none of which are in S.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then define:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;g(s_i) = t_{2i}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;g(t_i) = t_{2i+1}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;g(x) = x (otherwise)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a bijection from R to R-S.&lt;/p&gt;&#xA;" OwnerUserId="31" LastActivityDate="2010-07-22T06:13:15.787" />
  <row Id="448" PostTypeId="1" AcceptedAnswerId="452" CreationDate="2010-07-22T07:25:11.347" Score="4" ViewCount="163" Body="&lt;p&gt;Let's say, I have 4 yellow and 5 blue balls. How do I calculate in how many different orders I  can place them? And what if I also have 3 red balls?&lt;/p&gt;&#xA;" OwnerUserId="143" LastEditorUserId="143" LastEditorDisplayName="" LastEditDate="2010-07-22T07:53:18.487" LastActivityDate="2010-07-22T09:01:04.033" Title="In how many different ways can I sort balls of two different colors" Tags="&lt;combinatorics&gt;" AnswerCount="3" CommentCount="3" />
  <row Id="449" PostTypeId="2" ParentId="381" CreationDate="2010-07-22T07:25:38.507" Score="12" ViewCount="" Body="&lt;p&gt;Perhaps it's not an entirely practical application, but Fibonacci numbers can be used to &lt;a href=&quot;http://www.catonmat.net/blog/using-fibonacci-numbers-to-convert-from-miles-to-kilometers&quot; rel=&quot;nofollow&quot;&gt;convert from miles to kilometers and vice versa&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Take two consecutive Fibonacci&#xA;  numbers, for example 5 and 8. And&#xA;  you're done converting. No kidding –&#xA;  there are 8 kilometers in 5 miles. To&#xA;  convert back just read the result from&#xA;  the other end - there are 5 miles in 8&#xA;  km!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But why does it work?&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Fibonacci numbers have a property that&#xA;  the ratio of two consecutive numbers&#xA;  tends to the &lt;a href=&quot;http://mathworld.wolfram.com/GoldenRatio.html&quot; rel=&quot;nofollow&quot;&gt;Golden ratio&lt;/a&gt; as&#xA;  numbers get bigger and bigger. The&#xA;  Golden ratio is a number and it&#xA;  happens to be approximately 1.618.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Coincidentally, there are 1.609&#xA;  kilometers in a mile, which is within&#xA;  0.5% of the Golden ratio.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="114" LastActivityDate="2010-07-22T07:25:38.507" CommunityOwnedDate="2010-08-03T18:46:03.740" CommentCount="2" />
  <row Id="450" PostTypeId="1" CreationDate="2010-07-22T07:35:26.143" Score="2" ViewCount="467" Body="&lt;p&gt;I just came back from an intense linear algebra lecture which showed that linear transformations could be represented by transformation matrices; with more generalization, it was later shown that affine transformations (linear + translation) could be represented by matrix multiplication as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This got me to thinking about all those other transformations I've picked up over the past years I've been studying mathematics.  For example, polar transformations -- transforming &lt;code&gt;x&lt;/code&gt; and &lt;code&gt;y&lt;/code&gt; to two new variables &lt;code&gt;r&lt;/code&gt; and &lt;code&gt;theta&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you mapped &lt;code&gt;r&lt;/code&gt; to the &lt;code&gt;x&lt;/code&gt; axis and &lt;code&gt;theta&lt;/code&gt; to the &lt;code&gt;y&lt;/code&gt; axis, you'd basically have a coordinate transformation.  A rather warped one, at that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a way to represent this using a transformation matrix?  I've tried fiddling around with the numbers but everything I've tried to work with has fallen apart quite embarrassingly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More importantly, is there a way to, given a specific non-linear transformation, construct a transformation matrix from it?&lt;/p&gt;&#xA;" OwnerUserId="51" LastActivityDate="2010-09-09T23:00:16.617" Title="Can non-linear transformations be represented as Transformation Matrices?" Tags="&lt;linear-algebra&gt;&lt;transformation&gt;" AnswerCount="5" CommentCount="2" />
  <row Id="451" PostTypeId="2" ParentId="448" CreationDate="2010-07-22T07:40:50.280" Score="2" ViewCount="" Body="&lt;p&gt;The case of two colors is simple: if you have m yellow balls and n blue ones you only need to choose m positions among (m+n) possibilities, that is (m+n)!/(m!&amp;middot;n!). The other balls' positions are automatically set up.&lt;/p&gt;&#xA;" OwnerUserId="89" LastActivityDate="2010-07-22T07:40:50.280" />
  <row Id="452" PostTypeId="2" ParentId="448" CreationDate="2010-07-22T07:48:08.623" Score="7" ViewCount="" Body="&lt;p&gt;This is a standard problem involving the &lt;a href=&quot;http://en.wikipedia.org/wiki/Combination&quot; rel=&quot;nofollow&quot;&gt;combinations of sets&lt;/a&gt;, though perhaps not very obvious intuitively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Firstly consider the number of ways you can rearrange the entire set of balls, counting each ball as indepndent (effectively ignoring colours for now). This is simply &lt;code&gt;(4 + 5)! = 9!&lt;/code&gt;, since the 1st ball can be any of the 9, the 2nd can be any of the remaining 8, and so on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then we calculate how many different ways the yellow balls can be arranged within themselves, since for the purpose of this problem they are considered equivalent. The number of combinations is of course &lt;code&gt;4!&lt;/code&gt;; similarly, for the blue balls the number is &lt;code&gt;5!&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, overall we find:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;total no. of arrangements = number of arrangements of all balls / (number of arrangements of yellow balls * number of arrangements of blue balls)  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Therefore in our case we have:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;total no. of arrangements = 9! / (5! * 4!) = 126&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm sure you can see how this can be easily extended if we also have 3 red balls too. (Hint: the total changes and we have another multiple of identical arrangements to account for.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope that helps...&lt;/p&gt;&#xA;" OwnerUserId="56" LastActivityDate="2010-07-22T07:48:08.623" />
  <row Id="453" PostTypeId="2" ParentId="450" CreationDate="2010-07-22T07:52:24.933" Score="0" ViewCount="" Body="&lt;p&gt;No.  Everything is determined by a choice of basis.  For a more in-depth answer, I would need to explain the first two weeks of linear algebra and draw some commutative diagrams.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you'd like a better explanation, see pages 12-14 of Emil Artin's monograph &lt;em&gt;Geometric Algebra&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="126" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-08-08T20:21:00.297" LastActivityDate="2010-08-08T20:21:00.297" CommentCount="4" />
  <row Id="454" PostTypeId="2" ParentId="356" CreationDate="2010-07-22T07:59:14.977" Score="7" ViewCount="" Body="&lt;h3&gt;The Art of Computer Programming&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;(Donald Knuth)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The legendary book (of multiple volumes, still incomplete) can't go without mention. For learning about algorithms and their complexities, there is no rival. It's written with practicality in mind, though from a largely theoretical perspective.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/The_art_of_computer_programming&quot; rel=&quot;nofollow&quot;&gt;&lt;img src=&quot;http://upload.wikimedia.org/wikipedia/en/6/62/ArtOfComputerProgramming.jpg&quot; alt=&quot;The Art of Computer Programming&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="56" LastActivityDate="2010-07-22T07:59:14.977" CommunityOwnedDate="2010-07-22T07:59:14.977" CommentCount="3" />
  <row Id="455" PostTypeId="2" ParentId="450" CreationDate="2010-07-22T08:00:36.677" Score="1" ViewCount="" Body="&lt;p&gt;You can't represent a non linear transformation with a matrix, however there are some tricks (for want of a better word) available if you use homogenous co-ordinates. For example, 3d translation is a non-linear transformation in a 3x3 3d transformation matrix, but is a linear transformation in 3d homogenous co-ordinates using a 4x4 transformation matrix. The same is true of other things like perspective projections. This is why 4x4 matrices are used in 3d graphics as the homogenous co-ordinate system simplifies things a lot.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To clarify - using homogenous co-ordinates increases the range of transformations representable using matrices from plain linear transformations to affine transformations and some projections, but it doesn't make all non-linear transformations representable using matrices. The non-linear transformation provided as an example is still beyond representation as an affine transformation (Thanks to @Harry for prompting this clarification in the comments)&lt;/p&gt;&#xA;" OwnerUserId="35" LastEditorUserId="35" LastEditorDisplayName="" LastEditDate="2010-07-22T10:39:48.367" LastActivityDate="2010-07-22T10:39:48.367" CommentCount="3" />
  <row Id="456" PostTypeId="2" ParentId="450" CreationDate="2010-07-22T08:19:00.767" Score="3" ViewCount="" Body="&lt;p&gt;As Harry says, you can't (the example of affine transformations can be tweaked to work because they're just linear ones with the origin translated). However, approximating a nonlinear function by a linear one is something we do all the time in calculus through the derivative, and is what we often have to do to make a mathematical model of some real-world phenomenon tractable.&lt;/p&gt;&#xA;" OwnerUserId="76" LastActivityDate="2010-07-22T08:19:00.767" CommentCount="2" />
  <row Id="457" PostTypeId="1" CreationDate="2010-07-22T08:34:29.107" Score="34" ViewCount="1074" Body="&lt;p&gt;I like to use physical demonstrations when teaching mathematics (putting physics in the service of mathematics, for once, instead of the other way around), and it'd be great to get some more ideas to use.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm looking for nontrivial ideas in abstract mathematics that can be demonstrated with some contraption, construction or physical intuition. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, one can restate Euler's proof that $\sum \frac{1}{n^2} = \frac{\pi^2}{6}$ in terms of the flow of an incompressible fluid with sources at the integer points in the plane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or, consider the problem of showing that, for a convex polyhedron whose $i^{th}$ face has area A_i and outward facing normal vector $n_i$, $\sum A_i \cdot n_i = 0$. One can intuitively show this by pretending the polyhedron is filled with gas at uniform pressure. The force the gas exerts on the i_th face is proportional to $A_i \cdot n_i$, with the same proportionality for every face. But the sum of all the forces must be zero; otherwise this polyhedron (considered as a solid) could achieve perpetual motion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For an example showing less basic mathematics, consider &quot;showing&quot; the double cover of SO(3) by SU(2) by needing to rotate your hand 720 degrees to get it back to the same orientation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyone have more demonstrations of this kind?&lt;/p&gt;&#xA;" OwnerUserId="76" LastActivityDate="2010-11-11T16:59:54.103" CommunityOwnedDate="2010-07-22T08:34:29.107" Title="Good Physical Demonstrations of Abstract Mathematics" Tags="&lt;soft-question&gt;&lt;big-list&gt;&lt;physics&gt;&lt;education&gt;&lt;teaching&gt;" AnswerCount="14" CommentCount="3" FavoriteCount="18" />
  <row Id="458" PostTypeId="2" ParentId="381" CreationDate="2010-07-22T08:41:38.497" Score="4" ViewCount="" Body="&lt;p&gt;They appear in &lt;strong&gt;pattern formation&lt;/strong&gt;, for instance in the numbers of petals on a plant, the segements on a pineapple or pinecone, and the structure of nautilus shells. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://library.thinkquest.org/27890/applications5.html&quot; rel=&quot;nofollow&quot;&gt;http://library.thinkquest.org/27890/applications5.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.popmath.org.uk/rpamaths/rpampages/sunflower.html&quot; rel=&quot;nofollow&quot;&gt;http://www.popmath.org.uk/rpamaths/rpampages/sunflower.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.maths.surrey.ac.uk/hosted-sites/R.Knott/Fibonacci/&quot; rel=&quot;nofollow&quot;&gt;http://www.maths.surrey.ac.uk/hosted-sites/R.Knott/Fibonacci/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="166" LastActivityDate="2010-07-22T08:41:38.497" CommunityOwnedDate="2010-08-03T18:46:03.740" CommentCount="1" />
  <row Id="460" PostTypeId="1" AcceptedAnswerId="467" CreationDate="2010-07-22T08:54:37.340" Score="12" ViewCount="285" Body="&lt;p&gt;Are there any interesting and natural examples of semigroups that are not monoids (that is, they don't have an identity element)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To be a bit more precise, I guess I should ask if there any interesting examples of semigroups (X, *) for which there is not a monoid (X, *, e) where e is in X. I don't consider an example like the set of real numbers greater than 10 (considered under addition) to be a sufficiently 'natural' semigroup for my purposes; if the domain can be extended in an obvious way to include an identity element then that's not what I'm after.&lt;/p&gt;&#xA;" OwnerUserId="106" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-07-31T17:49:24.800" LastActivityDate="2010-12-11T01:58:42.147" CommunityOwnedDate="2010-07-22T12:28:57.663" Title="Are there any interesting semigroups that aren't monoids?" Tags="&lt;big-list&gt;&lt;abstract-algebra&gt;&lt;semigroup&gt;&lt;monoid&gt;" AnswerCount="6" CommentCount="7" FavoriteCount="2" />
  <row Id="461" PostTypeId="2" ParentId="448" CreationDate="2010-07-22T09:01:04.033" Score="1" ViewCount="" Body="&lt;p&gt;For some reason I find it easier to think in terms of letters of a word being rearranged, and your problem is equivalent to asking how many permutations there are of the word YYYYBBBBB. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The formula for counting permutations of words with repeated letters (whose reasoning has been described by Noldorin) gives us the correct answer of 9!/(4!5!) = 126.&lt;/p&gt;&#xA;" OwnerUserId="106" LastActivityDate="2010-07-22T09:01:04.033" />
  <row Id="462" PostTypeId="1" AcceptedAnswerId="463" CreationDate="2010-07-22T09:06:01.573" Score="7" ViewCount="266" Body="&lt;p&gt;I'm currently trying to understand the concepts and theory behind some of the common proof verifiers out there, but am not quite sure on the exact nature and construction of the sort of systems/proof calculi they use. Are they essentially based on higher-order logics that use Henkin semantics, or is there something more to it? As I understand, extending Henkin semantics to higher-order logic does not render the formal system any less sound, though I am not too clear on that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Though I'm mainly looking for a general answer with useful examples, here are a few specific questions:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;What exactly is the role of type theory in creating higher-order logics? Same goes with category theory/model theory, which I believe is an alternative.&lt;/li&gt;&#xA;&lt;li&gt;Is extending a) natural deduction, b) sequent calculus, or c) some other formal system the best way to go for creating higher order logics?&lt;/li&gt;&#xA;&lt;li&gt;Where does typed lambda calculus come into proof verification?&lt;/li&gt;&#xA;&lt;li&gt;Are there any other approaches than higher order logic to proof verification?&lt;/li&gt;&#xA;&lt;li&gt;What are the limitations/shortcomings of existing proof verification systems (see below)?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The Wikipedia pages on proof verification programs such as &lt;a href=&quot;http://en.wikipedia.org/wiki/HOL_Light&quot; rel=&quot;nofollow&quot;&gt;HOL Light&lt;/a&gt; &lt;a href=&quot;http://en.wikipedia.org/wiki/Calculus_of_constructions&quot; rel=&quot;nofollow&quot;&gt;Coq&lt;/a&gt;, and &lt;a href=&quot;http://us.metamath.org/&quot; rel=&quot;nofollow&quot;&gt;Metamath&lt;/a&gt; give some idea, but these pages contain limited/unclear information, and there are rather few specific high-level resources elsewhere. There are so many variations on formal logics/systems used in proof theory that I'm not sure quite what the base ideas of these systems are - what is required or optimal and what is open to experimentation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps a good way of answering this, certainly one I would appreciate, would be a brief guide (albeit with some technical detail/specifics) on how one might go about generating a complete proof calculus (proof verification system) from scratch? Any other information in the form of explanations and examples would be great too, however.&lt;/p&gt;&#xA;" OwnerUserId="56" LastEditorUserId="56" LastEditorDisplayName="" LastEditDate="2010-07-26T08:50:09.473" LastActivityDate="2010-07-28T06:12:18.627" Title="How do proof verifiers work?" Tags="&lt;math-software&gt;&lt;logic&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="5" />
  <row Id="463" PostTypeId="2" ParentId="462" CreationDate="2010-07-22T09:29:23.427" Score="5" ViewCount="" Body="&lt;p&gt;I'll answer just part of your question: I think the other parts will become clearer based on this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A proof verifier is essentially a program that takes one argument, a proof representation, and checks that this is properly constructed, and says OK if it is, and either fails silently otherwise, or highlights what is invalid otherwise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In principle, the proof representation could just be a sequence of formulae in a Hilbert system: all logics (at least, first-orderisable logics) can be represented in such a way.  You don't even need to say which rule is specified at each step, since it is decidable whether any formula follows by a rule application from earlier formulae.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In practice, though, the proof representations are more complex.  Metamath is rather close to Hilbert systems, but has a rich set of rules.  Coq and LF use (different) typed lambda calculi with definitions to represent the steps, which are computationally quite expensive to check (IIRC, both are PSPACE hard).  And the proof verifier can do much more: Coq allows ML programs to be extracted from proofs.&lt;/p&gt;&#xA;" OwnerUserId="100" LastEditorUserId="100" LastEditorDisplayName="" LastEditDate="2010-07-22T11:34:40.453" LastActivityDate="2010-07-22T11:34:40.453" CommentCount="4" />
  <row Id="464" PostTypeId="2" ParentId="381" CreationDate="2010-07-22T09:41:32.900" Score="0" ViewCount="" Body="&lt;p&gt;It's not really a mathematical application, but the Fibonacci sequence is closely related to the &lt;a href=&quot;http://en.wikipedia.org/wiki/Golden_ratio#Relationship_to_Fibonacci_sequence&quot; rel=&quot;nofollow&quot;&gt;Golden Ratio&lt;/a&gt;, used in architecture (amongst others).&lt;/p&gt;&#xA;" OwnerUserId="18" LastActivityDate="2010-07-22T09:41:32.900" CommunityOwnedDate="2010-08-03T18:46:03.740" CommentCount="0" />
  <row Id="465" PostTypeId="2" ParentId="83" CreationDate="2010-07-22T10:06:14.170" Score="1" ViewCount="" Body="&lt;p&gt;Seymour Papert's ideas about &quot;maths you can build&quot;.  Look at &lt;a href=&quot;http://scratch.mit.edu/&quot; rel=&quot;nofollow&quot;&gt;Scratch&lt;/a&gt; (much inspired by Papert), and read &lt;a href=&quot;http://en.wikipedia.org/wiki/Mindstorms%3a_Children,_Computers,_and_Powerful_Ideas&quot; rel=&quot;nofollow&quot;&gt;Mindstorms&lt;/a&gt;; note that quite a few of his ideas there are contentious, but there is a lot of maths you can make exciting this way.&lt;/p&gt;&#xA;" OwnerUserId="100" LastActivityDate="2010-07-22T10:06:14.170" CommunityOwnedDate="2010-08-07T22:32:44.343" />
  <row Id="466" PostTypeId="2" ParentId="440" CreationDate="2010-07-22T10:11:04.430" Score="0" ViewCount="" Body="&lt;p&gt;&lt;del&gt;It can be, if you use &lt;a href=&quot;http://en.wikipedia.org/wiki/Euclidean_relation&quot; rel=&quot;nofollow&quot;&gt;Euclideanity&lt;/a&gt; instead of transitivity: &lt;em&gt;a R c&lt;/em&gt; and &lt;em&gt;b R c&lt;/em&gt; implies &lt;em&gt;a R b&lt;/em&gt;.&lt;/del&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an aside, Euclideanity plus seriality characterises the equivalence relations.&lt;/p&gt;&#xA;" OwnerUserId="100" LastEditorUserId="100" LastEditorDisplayName="" LastEditDate="2010-07-22T11:28:41.117" LastActivityDate="2010-07-22T11:28:41.117" CommentCount="6" />
  <row Id="467" PostTypeId="2" ParentId="460" CreationDate="2010-07-22T10:16:54.910" Score="13" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://planetmath.org/encyclopedia/Folded.html&quot; rel=&quot;nofollow&quot;&gt;Convolution&lt;/a&gt; of functions/distributions is useful in a variety of fields, and the identity element, the &lt;a href=&quot;http://en.wikipedia.org/wiki/Dirac_delta_function&quot; rel=&quot;nofollow&quot;&gt;dirac delta&lt;/a&gt;, is not strictly a function.&lt;/p&gt;&#xA;" OwnerUserId="73" LastActivityDate="2010-07-22T10:16:54.910" CommentCount="2" />
  <row Id="468" PostTypeId="2" ParentId="460" CreationDate="2010-07-22T10:29:51.280" Score="0" ViewCount="" Body="&lt;p&gt;Let G be the set of (continuous) functions f: R -&gt; R where f(x) tends to 0 as x tends to infinity: $lim_{x\to \infty}f(x) = 0$. The operator is the usual point-wise multiplication of functions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;G is closed under * since lim(f(x)) = 0 and lim(g(x)) = 0 imply lim(f(x)*g(x)) = 0.  G is a subgroup of {f: R-&gt;R}, so the identity must be the same - the function which is constantly 1. But this identity is not in G.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt; As Harry correctly points out, {f: R-&gt;R} is not a group. Therefore the following correction is needed: consider only functions such that $f(x) \neq 0$ everywhere.&lt;/p&gt;&#xA;" OwnerUserId="26" LastEditorUserId="26" LastEditorDisplayName="" LastEditDate="2010-07-22T11:36:31.260" LastActivityDate="2010-07-22T11:36:31.260" CommentCount="3" />
  <row Id="469" PostTypeId="1" AcceptedAnswerId="478" CreationDate="2010-07-22T10:52:29.193" Score="3" ViewCount="598" Body="&lt;p&gt;What's the meaning of this symbol in mathematical notation? :&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;⊧&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="97" LastEditorUserId="97" LastEditorDisplayName="" LastEditDate="2011-03-30T18:21:29.867" LastActivityDate="2011-03-30T18:21:29.867" ClosedDate="2010-07-26T06:25:07.010" Title="What is the meaning of this symbol (⊧)?" Tags="&lt;logic&gt;&lt;notation&gt;" AnswerCount="4" CommentCount="9" FavoriteCount="1" />
  <row Id="470" PostTypeId="2" ParentId="469" CreationDate="2010-07-22T10:56:04.867" Score="4" ViewCount="" Body="&lt;p&gt;It is a symbol from model theory denoting &lt;a href=&quot;http://en.wikipedia.org/wiki/Entailment&quot; rel=&quot;nofollow&quot;&gt;entailment&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A ⊧ B is read as &quot;A entails B&quot;.&lt;/p&gt;&#xA;" OwnerUserId="126" LastActivityDate="2010-07-22T10:56:04.867" />
  <row Id="472" PostTypeId="2" ParentId="469" CreationDate="2010-07-22T11:07:52.947" Score="3" ViewCount="" Body="&lt;p&gt;It is called a 'double turnstile': &lt;a href=&quot;http://en.wikipedia.org/wiki/Double_turnstile&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Double_turnstile&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="106" LastActivityDate="2010-07-22T11:07:52.947" CommentCount="6" />
  <row Id="473" PostTypeId="2" ParentId="469" CreationDate="2010-07-22T11:22:43.013" Score="2" ViewCount="" Body="&lt;p&gt;$\models$ is also known as the &lt;a href=&quot;http://planetmath.org/encyclopedia/SatisfactionRelation.html&quot; rel=&quot;nofollow&quot;&gt;satisfication relation&lt;/a&gt;. For a structure $\mathcal{M}=(M,I)$ and an $\mathcal{M}$-assignment $\nu$, $(\mathcal{M},\nu)\models \varphi$ means that the formula $\varphi$ is true with the particular assignment $\nu$.&lt;br&gt;&#xA;See &lt;a href=&quot;http://www.trinity.edu/cbrown/topics_in_logic/struct/node2.html&quot; rel=&quot;nofollow&quot;&gt;http://www.trinity.edu/cbrown/topics_in_logic/struct/node2.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="73" LastActivityDate="2010-07-22T11:22:43.013" />
  <row Id="474" PostTypeId="2" ParentId="460" CreationDate="2010-07-22T11:52:19.533" Score="1" ViewCount="" Body="&lt;p&gt;Finite sets of matrices of varying dimensions, where the product A*B={PQ|P in A &amp;amp; Q in B &amp;amp; dim(Q)=codim(P)}, and dim &amp;amp; codim are the dimensions of the source &amp;amp; target spaces of a matrix.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The infinite case has an obvious unit.&lt;/p&gt;&#xA;" OwnerUserId="100" LastActivityDate="2010-07-22T11:52:19.533" />
  <row Id="475" PostTypeId="2" ParentId="457" CreationDate="2010-07-22T11:57:18.243" Score="10" ViewCount="" Body="&lt;p&gt;The book &quot;The Mathematical Mechanic&quot; by Mark Levi is a very good source of such examples, which Levi has been collecting for some time. The first two here are in the book, if I recall correctly.&lt;/p&gt;&#xA;" OwnerUserId="173" LastActivityDate="2010-07-22T11:57:18.243" CommunityOwnedDate="2010-07-22T11:57:18.243" />
  <row Id="476" PostTypeId="1" CreationDate="2010-07-22T12:03:10.743" Score="0" ViewCount="112" Body="&lt;p&gt;Many years ago, I had read a book entitled &quot;Think of a Number&quot; by Malcolm E. Lines, and it was an eminently readable and thought provoking book. In the book, there were topics like Fibonacci numbers (along with the live examples from the nature) and Golden Section. Now I'm looking for a similar book. Can anyone recommend me one?&lt;/p&gt;&#xA;" OwnerUserId="97" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-07-22T20:08:51.293" LastActivityDate="2010-07-22T20:08:51.293" Title="Looking for a book similar to &quot;Think of a Number&quot;" Tags="&lt;soft-question&gt;&lt;big-list&gt;&lt;books&gt;" AnswerCount="4" />
  <row Id="477" PostTypeId="1" AcceptedAnswerId="479" CreationDate="2010-07-22T12:07:38.830" Score="6" ViewCount="389" Body="&lt;p&gt;The set of all ℝ → ℝ continuous functions is &lt;a href=&quot;http://en.wikipedia.org/wiki/Cardinality_of_the_continuum&quot; rel=&quot;nofollow&quot;&gt;&lt;strong&gt;c&lt;/strong&gt;&lt;/a&gt;. How to show that? Is there are bijection between ℝ&lt;sup&gt;n&lt;/sup&gt; and the set of continuous functions?&lt;/p&gt;&#xA;" OwnerUserId="171" LastActivityDate="2010-07-22T12:19:00.337" Title="Cardinality of set of real continuous functions" Tags="&lt;calculus&gt;&lt;set-theory&gt;" AnswerCount="1" FavoriteCount="2" />
  <row Id="478" PostTypeId="2" ParentId="469" CreationDate="2010-07-22T12:11:48.120" Score="10" ViewCount="" Body="&lt;p&gt;Just to enlarge on Harry's answer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your symbol denotes one of two specified notions of implication in formal logic &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\vdash$ -the &lt;strong&gt;turnstile&lt;/strong&gt; symbol denotes &lt;strong&gt;syntactic&lt;/strong&gt; implication (syntactic here means  related to syntax, the structure of a sentence), where the 'algebra' of the logical system in play (for example &lt;a href=&quot;http://en.wikipedia.org/wiki/Sentential_calculus&quot; rel=&quot;nofollow&quot;&gt;sentential calculus&lt;/a&gt;) allows us to 'rearrange and cancel' the stuff we know on the left into the thing we want to prove on the right.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An example might be the classic &quot;all men are mortal $\wedge$ socrates is a man $\vdash$ socrates is mortal&quot; ('$\wedge$' of course here just means 'and'). You can almost imagine cancelling out the 'man bit' on the left to just give the sentence on the right (although the truth may be more complex...).&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;$\models$ -the &lt;strong&gt;double turnstile&lt;/strong&gt;, on the other hand, is not so much about algebra as meaning (formally it denotes &lt;strong&gt;semantic&lt;/strong&gt; implication)- it means that any interpretation of the stuff we know on the left must have the corresponding interpretation of the thing we want to prove on the right true.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An example would be if we had an infinite set of sentences: $\Gamma$:= {&quot;1 is lovely&quot;, &quot;2 is lovely&quot;, ...} in which all numbers appear, and the sentence A= &quot; the natural numbers are precisely {1,2,...}&quot; listing all numbers. Any interpretation would give us B=&quot;all natural numbers are lovely&quot;. So $\Gamma$, A $\models$ B. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Now, the goal of any logician trying to set up a formal system is to have $\Gamma \vdash A \iff \Gamma \models B$, meaning that the 'algebra' must line up with the interpretation, and this is not something we can take as given. Take the second example above- can we be sure that algebraic operations can 'parse' those infinitely many sentences and make the simple sentence on the right (this is to do with a property called &lt;strong&gt;compactness&lt;/strong&gt;)??&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The goal can be split into two distict subgoals:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Soundness:&lt;/strong&gt; $A \vdash B \Rightarrow A \models B$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Completeness:&lt;/strong&gt; $A \models B \Rightarrow A \vdash B$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where the first stops you proving things that aren't true when we interpret them and the second means that everything we know to be true on interpretation, we must be able to prove.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sentential calculus, for example, can be proved complete (and was in Godel's lesser known, but celebrated completeness theorem), but other for other systems Godel's incompleteness theorem, give us a terrible choice between the two.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;In summary:&lt;/strong&gt; The interplay of meaning and axiomatic machine mathematics, captured by the difference between $\models$ and $\vdash$, is a subtle and interesting thing.&lt;/p&gt;&#xA;" OwnerUserId="160" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-09-16T21:42:11.713" LastActivityDate="2010-09-16T21:42:11.713" CommentCount="2" />
  <row Id="479" PostTypeId="2" ParentId="477" CreationDate="2010-07-22T12:19:00.337" Score="12" ViewCount="" Body="&lt;p&gt;The cardinality is at least that of the continuum because every real number corresponds to a constant function.  The cardinality is at most that of the continuum because the set of real continuous functions injects into the sequence space $R^{N}$ by mapping each continuous function to its values on all the rational points. Since the rational points are dense, this determines the function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Cantor%E2%80%93Bernstein%E2%80%93Schroeder_theorem&quot; rel=&quot;nofollow&quot;&gt;Schroeder-Bernstein theorem&lt;/a&gt; now implies the cardinality is precisely that of the continuum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that  then the set of sequences of reals is also of the same cardinality as the same cardinality as the reals.  This is because if we have a sequence of binary representations $.a_1a_2..., .b_1b_2..., .c_1c_2...$, we can splice them together via $.a_1 b_1 a_2 c_1 b_2 a_3...$ so that a sequence of reals can be encoded by one real number.&lt;/p&gt;&#xA;" OwnerUserId="536" LastActivityDate="2010-07-22T12:19:00.337" CommentCount="5" />
  <row Id="480" PostTypeId="2" ParentId="457" CreationDate="2010-07-22T13:05:26.903" Score="6" ViewCount="" Body="&lt;p&gt;There is the hairy ball theorem, which states that no even dimensional sphere admits a nowherevanishing continuous vector field.  In the case of $S^2$, the physical demonstration of this is that one cannot comb the hair of a ball without getting a cowlick.&lt;/p&gt;&#xA;" OwnerUserId="9" LastActivityDate="2010-07-22T13:05:26.903" CommunityOwnedDate="2010-07-22T13:05:26.903" CommentCount="3" />
  <row Id="481" PostTypeId="2" ParentId="476" CreationDate="2010-07-22T13:15:36.210" Score="1" ViewCount="" Body="&lt;p&gt;did you try the books of Eastaway and Wyndham? &lt;em&gt;Why Do Buses Come in Threes?&lt;/em&gt; and &lt;em&gt;How Long Is a Piece of String?&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Blurb for the first one says &quot;An amusing explanation of how maths is relevant to almost everything in life. Citing many examples of the way mathematics can explain common phenomena&quot;; for the second, &quot;This title is for anyone wanting to remind themselves - or discover for the first time - that maths is relevant to almost everything we do. Dating, cooking, travelling by car, gambling and ranking sportsmen all have links with intriguing mathematical problems that are explained in this book&quot;. &lt;/p&gt;&#xA;" OwnerUserId="89" LastEditorUserId="89" LastEditorDisplayName="" LastEditDate="2010-07-22T14:49:47.403" LastActivityDate="2010-07-22T14:49:47.403" />
  <row Id="482" PostTypeId="2" ParentId="438" CreationDate="2010-07-22T13:19:27.403" Score="2" ViewCount="" Body="&lt;p&gt;Isaac's &lt;a href=&quot;http://math.stackexchange.com/questions/438/1-is-not-1-so-where-is-the-mistake/439#439&quot; rel=&quot;nofollow&quot;&gt;answer&lt;/a&gt; is correct, but it can be hard to see if you don't have a strong knowledge of your laws. These problems are generally easy to solve if you examine it line by line and simplify both sides.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;sqrt(-1) = î&#xA;LHS=i, RHS=i&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1/sqrt(-1) = 1/î&#xA;LHS=1/i=-i, RHS=-i&lt;/p&gt;&#xA;&#xA;&lt;p&gt;sqrt(1)/sqrt(-1) = 1/î&#xA;LHS=1/i=-i, RHS=-i&lt;/p&gt;&#xA;&#xA;&lt;p&gt;sqrt(1/-1) = 1/î&#xA;LHS=sqrt(-1)=i, RHS=-i&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can then see that the error must be assuming sqrt(1)/sqrt(-1)=sqrt(1/-1).&lt;/p&gt;&#xA;" OwnerUserId="123" LastActivityDate="2010-07-22T13:19:27.403" CommentCount="2" />
  <row Id="483" PostTypeId="2" ParentId="329" CreationDate="2010-07-22T13:20:23.173" Score="23" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/038797329X&quot; rel=&quot;nofollow&quot;&gt;A Classical Introduction to Modern Number Theory&lt;/a&gt; by Ireland and Rosen  hands down!&lt;/p&gt;&#xA;" OwnerUserId="30" LastEditorUserId="2264" LastEditorDisplayName="" LastEditDate="2011-05-31T15:35:10.310" LastActivityDate="2011-05-31T15:35:10.310" CommunityOwnedDate="2010-07-22T13:20:23.173" CommentCount="1" />
  <row Id="484" PostTypeId="2" ParentId="164" CreationDate="2010-07-22T13:22:16.210" Score="4" ViewCount="" Body="&lt;p&gt;Integration. \int 4pi*r^2 dr = 4/3 pi r^3. &lt;/p&gt;&#xA;" OwnerUserId="30" LastActivityDate="2010-07-22T13:22:16.210" CommentCount="4" />
  <row Id="486" PostTypeId="1" AcceptedAnswerId="495" CreationDate="2010-07-22T13:28:58.457" Score="8" ViewCount="263" Body="&lt;p&gt;When I used to compete in Olympiad Competitions back in high school, a decent number of the easier geometry questions were solvable by what we called a geometry bash. Basically, you'd label every angle in the diagram with the variable then use a limited set of basic geometry operations to find relations between the elements, eliminate equations and then you'd eventually get the result. It seems like the kind of thing you could program a computer to do. So, I'm curious, does there exist any software to do this? I know there is lots of software for solving equations, but is there anything that lets you actually input a geometry problem without manually converting to equations? I'm not looking for anything too advance, even seeing just an attempt would be interesting. If there is anything decent, I think it'd be rather interesting to run the results on various competitions and see how many of the questions it solves.&lt;/p&gt;&#xA;" OwnerUserId="123" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-08-11T06:46:03.067" LastActivityDate="2010-08-11T06:46:03.067" Title="Software for solving geometry questions" Tags="&lt;algebraic-geometry&gt;&lt;logic&gt;&lt;euclidean-geometry&gt;&lt;math-software&gt;&lt;quantifier-elimination&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="487" PostTypeId="2" ParentId="2" CreationDate="2010-07-22T13:29:04.040" Score="0" ViewCount="" Body="&lt;p&gt;A rigorous mathematical treatment follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We say the limit of f(x) as x approaches c is \tau iff for all epsilon in R+ there exists delta in R+ such that |x-c|&amp;lt; \delta implies |f(x)-\tau| &amp;lt; epsilon.&lt;/p&gt;&#xA;" OwnerUserId="30" LastActivityDate="2010-07-22T13:29:04.040" CommentCount="1" />
  <row Id="488" PostTypeId="2" ParentId="75" CreationDate="2010-07-22T13:34:17.930" Score="1" ViewCount="" Body="&lt;p&gt;In the tactile sense, groups and rings do not &quot;feel&quot; very similar. Indeed, groups of grapes feel very plush and juicy, while rings are metallic and hard.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;HAHAHHA Just kidding! xD&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically look. The intro abstract algebra idea is the following (something you might feel is the purpose of the field {nice pun, eh?} if you read as silly a book as Fraleigh): Take a set $S.$ Create an operation and impose some structural conditions. A ring is just more structure on a group (an abelian group - abelian grape hahah!-, in fact) where you introduce a second operation. In this way, a field is the ultimate beast because it has two operations that are invertible.&lt;/p&gt;&#xA;" OwnerUserId="30" LastActivityDate="2010-07-22T13:34:17.930" />
  <row Id="491" PostTypeId="2" ParentId="2" CreationDate="2010-07-22T13:45:06.123" Score="0" ViewCount="" Body="&lt;p&gt;There are two slightly different types of limits (at least, in one real variable, as in single-variable calculus): limits as a variable goes to positive or negative infinity, and limits as a variable goes to some real value.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The limit of f(x) as x goes to infinity (or as x goes to negative infinity) is asking what happens to f(x) at the right (left) end of its graph.  Does f(x) get arbitrarily close to a single real number as x increases (or decreases) without bound--that is, as x gets really big and positive (or really big and negative)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The limit of f(x) as x goes to some real value, say c, is asking what happens to f(x) near, but not at, x = c.  Does f(x) get arbitrarily close to a single real number as x gets arbitrarily close to c?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Arbitrarily close&quot; is used somewhat informally above, but these ideas can be translated into more rigorous definitions (as in other people's answers) by reformulating &quot;arbitrarily close&quot; in terms of inequalities with &amp;epsilon; and/or &amp;delta;.  Also, while a limit is said to not exist if there isn't the single real number as described above, it is somewhat common to talk about the limit being infinity (or negative infinity) if the limit does not exist, but the values of f(x) increase without bound (or decrease without bound).&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-07-22T13:45:06.123" />
  <row Id="493" PostTypeId="2" ParentId="81" CreationDate="2010-07-22T13:46:56.540" Score="1" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://amathew.wordpress.com/&quot; rel=&quot;nofollow&quot;&gt;http://amathew.wordpress.com/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;He climbs Mount Bourbaki with the grace of a leaping gazelle. &lt;/p&gt;&#xA;" OwnerUserId="30" LastEditorUserId="30" LastEditorDisplayName="" LastEditDate="2010-07-22T17:25:02.870" LastActivityDate="2010-07-22T17:25:02.870" CommunityOwnedDate="2010-07-22T13:46:56.540" />
  <row Id="494" PostTypeId="2" ParentId="476" CreationDate="2010-07-22T14:27:22.967" Score="1" ViewCount="" Body="&lt;p&gt;I don't know the book you are speaking of, but for some reason I am reminded of &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0198516231&quot; rel=&quot;nofollow&quot;&gt;1089, and all that&lt;/a&gt; by David Acheson.  This is a charming book about some mathematical curiosities and can be appreciated by the professional and enjoyed by the layperson.&lt;/p&gt;&#xA;" OwnerUserId="121" LastActivityDate="2010-07-22T14:27:22.967" />
  <row Id="495" PostTypeId="2" ParentId="486" CreationDate="2010-07-22T14:30:51.677" Score="7" ViewCount="" Body="&lt;p&gt;This is just a special case of automated theorem proving. A nice thing is some geometry problem can indeed be solved by an algorithm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are theories show such system can be realized algorithmically. &lt;strike&gt;I don't know if anyone have really wrote the program to do it.&lt;/strike&gt; &lt;a href=&quot;http://woody.cs.wichita.edu/&quot; rel=&quot;nofollow&quot;&gt;JGEX&lt;/a&gt; seems to do that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A mechanical geometry proof technique was popularized in China by &lt;a href=&quot;http://woody.cs.wichita.edu/&quot; rel=&quot;nofollow&quot;&gt;Jingzhong Zhang&lt;/a&gt;. He first introduced it as a way for machines to solve geometric problems relating the proportions between areas, lengths or angles. Then some Olympiad people I know start using it to bash that kind of problem. I don't know what the name is in English, but a literal translation of the method is &quot;point removal method&quot;. Although it's not exactly same as what you are talking about, because &quot;input a geometry problem&quot; requires you to provide the construction of the problem from a straight edge and a compass, which is almost like &quot;manually converting to equations&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;the basic idea:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Construct the original problem by compass and straightedge, make a list of every constructed point ordered by the order of construction, let it be L. Record which points are used in the construction of every point.(Only the points used to construct point P are used in the substitution steps 3 and 4)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Translate the theorem into a equivalent form. Usually a/b = 1, where a and b are functions of length and area of certain segments or triangles. Let's call this equation E.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Let P be the last point of L. For every P appears in E, substitute it with another relation using other points from L(P itself is also allowed), usually if we are proving about lengths, we might use area. A list of possible operations are required for this step. It can branches off as a proof tree when the program decide to use different substitutions.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Do another substitution that eliminates the point P. For example, if in the step before, we substitute length to area, then we want to find something involve the length.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Do 3 to 4 over and over until we have 1=1&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;An example: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Angle bisector theorem&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given: $AD$ is the angle bisector of $\angle BAC$ of triangle $\triangle ABC$. Let $XYZ$ be the area of triangle $\triangle XYZ$, and $XY$ be the length of segment $XY$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Prove: $\frac{AB}{AC} = \frac{BD}{DC}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proof:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;First construct $ABC$. Then construct $AD$. The points in the list $L$ are $A,B,C,D$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The equivalent equation to the theorem is&#xA;$\frac{AB}{AC} \frac{DC}{BD} = 1$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$\frac{AB}{AC} \frac{DC}{BD} = \frac{AB}{AC} \frac{ACD}{ABD}$ (substitute length with area)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$=\frac{AB}{AC} \frac{\frac{1}{2} AC\cdot AD \sin \angle CAD}{\frac{1}{2} AB\cdot AD \sin \angle BAD}$, this step successfully remove point $D$ by cancellation.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$=\frac{AB}{AC} \frac{AC}{AB} =1$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;This is only a non-formal explanation of how such automated system would work. I think the following book from Zhang will tell you more about it: Machine proofs in geometry: automated production of readable proofs for geometry theorems. I did not read the book, but the description of it seems like what you are seeking. &lt;a href=&quot;http://woody.cs.wichita.edu/help/pdf/&quot; rel=&quot;nofollow&quot;&gt;A few paper by Zhang and his colleague&lt;/a&gt; can be found in the JGEX's website. &lt;a href=&quot;http://woody.cs.wichita.edu/help/atp.html&quot; rel=&quot;nofollow&quot;&gt;The JGEX documentation on it's automated theorem prover&lt;/a&gt; is also a great resource.&lt;/p&gt;&#xA;" OwnerUserId="96" LastEditorUserId="96" LastEditorDisplayName="" LastEditDate="2010-07-22T14:39:28.130" LastActivityDate="2010-07-22T14:39:28.130" CommentCount="2" />
  <row Id="496" PostTypeId="2" ParentId="476" CreationDate="2010-07-22T14:30:59.960" Score="1" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0805062998&quot; rel=&quot;nofollow&quot;&gt;The Number Devil&lt;/a&gt; may be something like what you're looking for.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Young Robert's dreams have taken a decided turn for the weird. Instead of falling down holes and such, he's visiting a bizarre magical land of number tricks with the number devil as his host. Starting at one and adding zero and all the rest of the numbers, Robert and the number devil use giant furry calculators, piles of coconuts, and endlessly scrolling paper to introduce basic concepts of numeracy, from interesting number sequences to exponents to matrices.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(It's not a watered-down kids' book, even though the description might suggest it.)&lt;/p&gt;&#xA;" OwnerUserId="45" LastActivityDate="2010-07-22T14:30:59.960" />
  <row Id="497" PostTypeId="2" ParentId="81" CreationDate="2010-07-22T14:34:07.113" Score="7" ViewCount="" Body="&lt;p&gt;I find &lt;a href=&quot;http://qchu.wordpress.com/&quot; rel=&quot;nofollow&quot;&gt;Annoying Precision&lt;/a&gt; to be wonderfully readable, and has many many interesting topics. Additionally, &lt;a href=&quot;http://rigtriv.wordpress.com/&quot; rel=&quot;nofollow&quot;&gt;Rigorous Trivialities&lt;/a&gt; is a bit higher level, but has a really useful intro to Algebraic Geometry.&lt;/p&gt;&#xA;" OwnerUserId="16" LastActivityDate="2010-07-22T14:34:07.113" CommunityOwnedDate="2010-07-22T14:34:07.113" CommentCount="2" />
  <row Id="498" PostTypeId="2" ParentId="213" CreationDate="2010-07-22T14:45:31.017" Score="10" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Category theory and algebraic geometry&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I spent a lot of time in undergrad studying things that were kinda nifty, but way to classical to be of any use/interest beyond &quot;fun math&quot;. When I got to grad school, &lt;strong&gt;Category theory&lt;/strong&gt; was assumed and made some of my courses much harder than they should of been. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the words of Ravi Vakil &quot;&lt;strong&gt;Algebraic Geometry&lt;/strong&gt; should be learned slowly over a number of years&quot;. I currently &lt;em&gt;NEED&lt;/em&gt; algebraic geometry, so I don't have this number of years. I wish I would of started that a long time ago. Additionally, both of these topics would of helped me learn the things I was thinking about anyways, in particular &lt;em&gt;commutative algebra&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="16" LastEditorUserId="16" LastEditorDisplayName="" LastEditDate="2010-08-16T18:45:40.760" LastActivityDate="2010-08-16T18:45:40.760" CommunityOwnedDate="2010-07-22T14:45:31.017" CommentCount="4" />
  <row Id="499" PostTypeId="2" ParentId="476" CreationDate="2010-07-22T14:54:21.263" Score="2" ViewCount="" Body="&lt;p&gt;Another more general pop math book covering mathematical curiosities is &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0393059456&quot; rel=&quot;nofollow&quot;&gt;Coincidences, Chaos, and All that Math Jazz&lt;/a&gt;. It is at the level you're talking about. A more focused book on the golden ratio is that by &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0767908163&quot; rel=&quot;nofollow&quot;&gt;Mario Livio&lt;/a&gt;. One nice feature of this second book is that he does a nice job of pointing out places where people believe in coincidences that aren't really there. He stays nicely objective, where many others fail to.&lt;/p&gt;&#xA;" OwnerUserId="16" LastActivityDate="2010-07-22T14:54:21.263" CommentCount="1" />
  <row Id="500" PostTypeId="2" ParentId="213" CreationDate="2010-07-22T15:52:58.950" Score="2" ViewCount="" Body="&lt;p&gt;I did mathematics as an undergrad, and I thought that differential equations were boring and pointless. Type of diff eq -&gt; existence and uniqueness proofs for solutions -&gt; rinse and repeat. Yawn.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But now I find my lack of knowledge of differential equations is hampering my learning some interesting parts of physics that I'd like to know more about...&lt;/p&gt;&#xA;" OwnerUserId="172" LastActivityDate="2010-07-22T15:52:58.950" CommunityOwnedDate="2010-07-22T15:52:58.950" />
  <row Id="501" PostTypeId="1" AcceptedAnswerId="1878" CreationDate="2010-07-22T15:58:45.687" Score="10" ViewCount="530" Body="&lt;p&gt;An extreme form of constructivism is called &lt;em&gt;finitisim&lt;/em&gt;.  In this form, unlike the standard axiom system, infinite sets are not allowed.  There are important mathematicians, such as Kronecker, who supported such a system.  I can see that the natural numbers and rational numbers can easily defined in a finitist system, by easy adaptations of the standard definitions.   But in order to do any significant mathematics, we need to have definitions for the irrational numbers that one is likely to encounter in practice, such as  e  or sqrt(2).  In standard constructions, real numbers are defined as Dedekind cuts, or Cauchy sequences, which are actually sets of infinite cardinality, so they are of no use here. My question is, how would a real number like those be defined in a finitist axiom system.  (Of course we have no hope to construct the entire set of real numbers, since that set is uncountably infinite.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;After doing a little research  I found a constructivist definition in Wikipedia &lt;a href=&quot;http://en.wikipedia.org/wiki/Constructivism_(mathematics)#Example_from_real_analysis&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Constructivism_(mathematics)#Example_from_real_analysis&lt;/a&gt; , but we need a finitist definition of a function for this definition to work. (Because in the standard system, a function over the set of natural numbers is actually an infinite set.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So my question boils down to this:  How can we define a function f over the natural numbers in a finitist axiom system? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Original version of this question, &lt;a href=&quot;http://meta.math.stackexchange.com/questions/172/why-did-you-close-my-question-if-all-sets-were-finite&quot;&gt;which had been closed during private beta&lt;/a&gt;, is as follows:&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;If all sets were finite, how would mahtematics be like?&lt;/strong&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;If we replace the axiom that 'there&#xA;  exists an infinite set' with 'all sets&#xA;  are finite', how would mathematics be&#xA;  like? My guess is that, all the theory&#xA;  that has practical importance would&#xA;  still show up, but everything would be&#xA;  very very unreadable for humans. Is&#xA;  that true?&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;We would have the natural numbers,&#xA;  athough the class of all natural&#xA;  numbers would not be a set. In the&#xA;  same sense, we could have the rational&#xA;  numbers. But could we have the real&#xA;  numbers? Can the standard&#xA;  constructions be adapted to this&#xA;  setting?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="145" LastEditorUserId="145" LastEditorDisplayName="" LastEditDate="2010-08-09T16:19:49.210" LastActivityDate="2010-08-09T16:24:44.743" Title="If all sets were finite, how could the real numbers be defined?" Tags="&lt;logic&gt;&lt;set-theory&gt;&lt;proof-theory&gt;" AnswerCount="6" CommentCount="9" FavoriteCount="2" />
  <row Id="502" PostTypeId="2" ParentId="501" CreationDate="2010-07-22T16:14:27.127" Score="-1" ViewCount="" Body="&lt;p&gt;You wouldn't even have the natural numbers (in totality), let alone the real numbers (or complex numbers) at all. The reals form an &lt;em&gt;uncountably infinite&lt;/em&gt; set and it is the arbitrarily small spacing between them that really gives them their unique properties.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a consequence, you certainly wouldn't have anything like calculus and thus no mathematical analysis whatsoever! Mathematics would not be mathematics in effect. It's almost an inherent contradiction to ask this question. Does that not say enough?&lt;/p&gt;&#xA;" OwnerUserId="56" LastEditorUserId="56" LastEditorDisplayName="" LastEditDate="2010-07-22T16:30:15.240" LastActivityDate="2010-07-22T16:30:15.240" CommentCount="11" />
  <row Id="503" PostTypeId="2" ParentId="501" CreationDate="2010-07-22T16:14:48.077" Score="0" ViewCount="" Body="&lt;p&gt;well: sqrt(2), e and \pi were defined before set theory was even conceived ;-)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think that algebraic numbers could have been defined without resorting to the standard theory of real numbers, just looking at them as roots of polynomials; certain classes of transcendent numbers could also, but I wonder if there is &quot;finitistic&quot; way to define e. I am at loss with this.&lt;/p&gt;&#xA;" OwnerUserId="89" LastEditorUserId="89" LastEditorDisplayName="" LastEditDate="2010-07-22T16:26:06.820" LastActivityDate="2010-07-22T16:26:06.820" CommentCount="6" />
  <row Id="504" PostTypeId="2" ParentId="501" CreationDate="2010-07-22T16:28:14.133" Score="3" ViewCount="" Body="&lt;p&gt;There is a fragment of mathematics that is given by a set of axioms known as the &lt;a href=&quot;http://en.wikipedia.org/wiki/Peano_axioms&quot; rel=&quot;nofollow&quot;&gt;Peano axioms&lt;/a&gt;. Using these rules you can carry out a vast amount of mathematics relating to natural numbers. For example you can prove lots of theorems in number theory using these axioms. The Peano axioms make no reference to sets at all, whether finite or infinite. The only things that exist in this theory are naturals. You can't even form the set of all integers. You can only talk about the naturals themselves. So a vast amount of mathematics would work absolutely fine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even though Peano's axioms are about naturals, you can already use them to talk about finite sets. The idea is that any finite set could be encoded as a finite sequence of symbols which in turn could be represented as naturals using &lt;a href=&quot;http://en.wikipedia.org/wiki/G%C3%B6del_numbering&quot; rel=&quot;nofollow&quot;&gt;Godel numbering&lt;/a&gt;. So questions like &quot;is this set a subset of that one?&quot; could be turned into purely arithmetical statements about Godel numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I'm pretty sure that declaring that there is no infinite set would make little difference to people working within the system defined by Peano's axioms. We'd still have all of the natural numbers to work with, we just wouldn't be able to assemble them into a single entity, the set of all natural numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, there are theorems that make essential use of an infinite set. Like &lt;a href=&quot;http://en.wikipedia.org/wiki/Goodstein%27s_theorem&quot; rel=&quot;nofollow&quot;&gt;Goodstein's theorem&lt;/a&gt;. Without infinite sets (or a substitute of some sort) it would be impossible to prove this result.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the overall result would be, I think, that you could still do lots of mathematics fine. The mathematics you could do wouldn't be all that weird. And you'd simply be depriving yourself of a useful proof technique.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the way, you'd still be able to say many things about real numbers. A real number can be thought of as a Cauchy sequence. A &lt;a href=&quot;http://en.wikipedia.org/wiki/Cauchy_sequence&quot; rel=&quot;nofollow&quot;&gt;Cauchy sequence&lt;/a&gt; is a certain type of sequence of rational numbers. So many statements about real numbers, when unpacked, are really statements about rational, and hence naturals, but in disguise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Update: Uncovering precisely what parts of mathematics you need in order to prove things is a field known as &lt;a href=&quot;http://en.wikipedia.org/wiki/Reverse_mathematics&quot; rel=&quot;nofollow&quot;&gt;reverse mathematics&lt;/a&gt;. Hilbert, and others mathematicians, were interested in trying to prove as much mathematics as possible using finite methods. Although it was ultimately shown that you can't carry out all mathematics using finite methods, it's surprising how much you can. &lt;a href=&quot;http://www.andrew.cmu.edu/user/avigad/Papers/elementary.pdf&quot; rel=&quot;nofollow&quot;&gt;Here&lt;/a&gt;'s a paper that talks about a system called EA which has no infinite sets. Amazingly we can use results from &lt;a href=&quot;http://en.wikipedia.org/wiki/Analytic_number_theory&quot; rel=&quot;nofollow&quot;&gt;analytic number theory&lt;/a&gt; in EA. This is because propositions about analytic functions can be interpreted as statements about natural numbers.&lt;/p&gt;&#xA;" OwnerUserId="80" LastEditorUserId="80" LastEditorDisplayName="" LastEditDate="2010-07-22T16:50:17.980" LastActivityDate="2010-07-22T16:50:17.980" CommentCount="5" />
  <row Id="505" PostTypeId="1" AcceptedAnswerId="507" CreationDate="2010-07-22T16:52:36.447" Score="11" ViewCount="549" Body="&lt;p&gt;Akhil showed that the &lt;a href=&quot;http://math.stackexchange.com/questions/477/cardinality-of-set-of-real-continuous-functions/479#479&quot;&gt;Cardinality of set of real continuous functions&lt;/a&gt; is the same as the continuum, using as a step the observation that continuous functions that agree at rational points must agree everywhere, since the rationals are dense in the reals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This isn't an obvious step, so why is it true?&lt;/p&gt;&#xA;" OwnerUserId="100" LastEditorUserId="100" LastEditorDisplayName="" LastEditDate="2010-08-09T16:01:43.040" LastActivityDate="2011-02-27T21:05:26.510" Title="Can there be two distinct, continuous functions that are equal at all rationals?" Tags="&lt;calculus&gt;&lt;real-analysis&gt;" AnswerCount="3" CommentCount="5" FavoriteCount="2" />
  <row Id="506" PostTypeId="2" ParentId="505" CreationDate="2010-07-22T16:57:33.693" Score="7" ViewCount="" Body="&lt;p&gt;If there were two continuous functions $f(x)$ and $g(x)$ that were equal at all rationals, then (because the rationals are dense) we can show that $\lim_{x \to a} f(x) - g(x) = 0$ for all values of $a$ using a delta-epsilon proof.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the difference of two continuous functions is continuous, we know $\lim_{x \to a} f(x) - g(x) = f(a) - g(a)$ for all $a$, and therefore $f(a) - g(a) = 0$ and $f(x) = g(x)$, proving that $f$ and $g$ must be identical.&lt;/p&gt;&#xA;" OwnerUserId="45" LastEditorUserId="45" LastEditorDisplayName="" LastEditDate="2010-07-22T17:04:22.980" LastActivityDate="2010-07-22T17:04:22.980" CommentCount="2" />
  <row Id="507" PostTypeId="2" ParentId="505" CreationDate="2010-07-22T17:06:50.840" Score="13" ViewCount="" Body="&lt;p&gt;Without resorting to &amp;epsilon;-&amp;delta; arguments: Let f and g be continuous real functions and f(x) = g(x) for all rational x.  For any real number c (in particular, an irrational c), there exists a Cauchy sequence of rational numbers  such that $\lim_{n \to \infty}x_{n}=c$.  Since f and g are continuous, $\lim_{n \to \infty}f({x_{n}})=f({c})$ and $\lim_{n \to \infty}g({x_{n}})=g({c})$.  Since $x_n$ is rational, $f(x_n) = g(x_n)$ for all n, so the two limits must be equal and so f(c) = g(c) for all real c.&lt;/p&gt;&#xA;" OwnerUserId="72" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-08-03T18:01:20.823" LastActivityDate="2010-08-03T18:01:20.823" CommentCount="2" />
  <row Id="508" PostTypeId="2" ParentId="486" CreationDate="2010-07-22T17:24:19.657" Score="7" ViewCount="" Body="&lt;p&gt;You might be interested in Doron Zeilberger's website. He has a page entitled &quot;Plane Geometry: An Elementary Textbook (Circa 2050)&quot; where he envisioned a world in which computers can derive all of plane geometry without human intervention or interference. The accompanying Maple package proves many statements by computer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The page exists at &lt;a href=&quot;http://www.math.rutgers.edu/~zeilberg/PG/gt.html&quot; rel=&quot;nofollow&quot;&gt;http://www.math.rutgers.edu/~zeilberg/PG/gt.html&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="30" LastActivityDate="2010-07-22T17:24:19.657" CommentCount="1" />
  <row Id="509" PostTypeId="2" ParentId="234" CreationDate="2010-07-22T18:06:36.787" Score="4" ViewCount="" Body="&lt;p&gt;This paradox has always interested me.  Something to think about is that there does not exist a uniform probability distribution over the positive real numbers (since they are infinite).  In arriving at your paradox, it seems you are assuming that any real number is equally likely, but this cannot be the case.&lt;/p&gt;&#xA;" OwnerUserId="9" LastActivityDate="2010-07-22T18:06:36.787" />
  <row Id="510" PostTypeId="2" ParentId="234" CreationDate="2010-07-22T18:33:41.163" Score="2" ViewCount="" Body="&lt;p&gt;This puzzle is known as the two envelope paradox. &lt;a href=&quot;http://www.franzdietrich.net/Papers/DietrichList-TwoEnvelopeParadox.pdf&quot; rel=&quot;nofollow&quot;&gt;This paper&lt;/a&gt; contains a nice explanation of the two envelope paradox, and some references to further literature regarding the puzzle.&lt;/p&gt;&#xA;" OwnerUserId="172" LastActivityDate="2010-07-22T18:33:41.163" CommentCount="1" />
  <row Id="511" PostTypeId="2" ParentId="81" CreationDate="2010-07-22T18:37:39.953" Score="0" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://wildaboutmath.com/&quot; rel=&quot;nofollow&quot;&gt;Wild about Math!&lt;/a&gt; and &lt;a href=&quot;http://satmathblog.com/&quot; rel=&quot;nofollow&quot;&gt;SatMathBlog&lt;/a&gt; (more for amateurs)&lt;/p&gt;&#xA;" OwnerUserId="89" LastActivityDate="2010-07-22T18:37:39.953" CommunityOwnedDate="2010-07-22T18:37:39.953" />
  <row Id="512" PostTypeId="2" ParentId="234" CreationDate="2010-07-22T20:01:49.750" Score="2" ViewCount="" Body="&lt;p&gt;Edit: My original answer was slightly misleading. Here is a better explanation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you look at the number on the card, and see that it is x, where x is a real number, you can deduce that your expected return is x if you stay, or 1.25x if you switch. So obviously you should switch, and &lt;strong&gt;there is no paradox the way you've stated the question&lt;/strong&gt;. It's better to switch because there's a fifty-fifty chance of gaining/losing, but if you gain, the difference is greater than if you lose.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The 'paradox' comes from applying that expected return to the situation &lt;strong&gt;without looking at the card first&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If I were to look at the first card, I'd see that it's better to switch.&lt;/li&gt;&#xA;&lt;li&gt;But then if I look at the second card, I'd see it's better to switch.&lt;/li&gt;&#xA;&lt;li&gt;But if it would be better to switch from the second card, I should just take the first card.&lt;/li&gt;&#xA;&lt;li&gt;But then I'd see that I should take the second card...&lt;/li&gt;&#xA;&lt;li&gt;&lt;em&gt;Which card do I pick, if 1.25x is always m y expected return?&lt;/em&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The answer is that you should pick either card, and then switch, because it's not &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;expected return = 1.25x, where x is the number on either card&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The correct expression is&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;expected return = [first card chosen]*1.25&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;So no matter which card you choose, that's your expected return. And if your expected return is always the same, there is no point in switching.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;original answer&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The conclusion you've reached is incorrect - x is a completely random variable, and multiplying it won't help us reach any conclusions. Read the &lt;a href=&quot;http://en.wikipedia.org/wiki/Two_envelopes_problem&quot; rel=&quot;nofollow&quot;&gt;wikipedia article&lt;/a&gt; about the two envelopes problem for more on this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Below is some correct math to demonstrate that switching won't help matters at all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We might have the 'half' envelope H, or we might have the 'double' envelope D. Let E denote the expected value of our earnings if we open the envelope now:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;E = 0.5H + 0.5D&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;D is 2H, so:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;E = 0.5H + 0.5*(2H)&#xA;E = 1.5H&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;If we switch, our expected earnings are still 1.5H because the same math applies, so switching will not help us. QED.&lt;/p&gt;&#xA;" OwnerUserId="181" LastEditorUserId="181" LastEditorDisplayName="" LastEditDate="2010-07-22T23:44:39.753" LastActivityDate="2010-07-22T23:44:39.753" CommentCount="12" />
  <row Id="513" PostTypeId="2" ParentId="1" CreationDate="2010-07-22T20:45:36.930" Score="3" ViewCount="" Body="&lt;p&gt;Infinity is an overloaded term that can mean many things.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One common non-mathematical use of infinity is to refer to everything in the universe. This is &lt;strong&gt;not&lt;/strong&gt; what mathematicians mean when they say infinity. That would be a kin to the set of all sets, which is a paradoxical concept that is not part of mathematical discourse.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mathematicians will use infinity as a way to represent a process that continues indefinitely. This is a kin to saying &quot;take the limit as n goes to infinity&quot;, which is close to saying &quot;continue this process indefinitely.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Infinity is also use infinity to talk about size. All sets are either infinite or finite.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The story doesn't stop there. There is something fundamentally different about sets like the points on a line, where there are no holes, and sets like the integers where there are holes. They are both infinite but one seems denser then the other. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's where whole countable uncountable thing comes in. Infinite sets have a size, but it is not a number in the traditional sense. Its more like &quot;relative size&quot;. Bijections are how we determine size for infinite sets, which are explained well on this page, so I won't repeat the explanation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A more in-depth, but still understandable explanation is given in Computability and Logic by George Boolos.&lt;/p&gt;&#xA;" OwnerUserId="99" LastActivityDate="2010-07-22T20:45:36.930" />
  <row Id="514" PostTypeId="1" AcceptedAnswerId="515" CreationDate="2010-07-22T20:49:28.590" Score="22" ViewCount="826" Body="&lt;p&gt;I just came back from my Number Theory course, and during the lecture there was mention of the Collatz Conjecture.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm sure that everyone here is familiar with it; it describes an operation on a natural number -- n/2 if it is even, 3n+1 if it is odd.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The conjecture states that if this operation is repeated, all numbers will eventually wind up at 1 (or rather, in an infinite loop of 1-4-2-1-4-2-1).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I fired up Python and ran a quick test on this for all numbers up to 5.76 × 10^18 (using the powers of cloud computing and dynamic programming magic).  Which is millions of millions of millions.  And all of them eventually ended up at 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Surely I am close to testing every natural number?  How many natural numbers could there be?  Surely not much more than millions of millions of millions. (I kid)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I explained this to my friend, who told me, &quot;Why would numbers suddenly get different at a certain point?  Wouldn't they all be expected to behave the same?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To which I said, &quot;No, you are wrong!  In fact I am sure there are many conjectures which have been disproved by counterexamples that are extremely large!&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And he said, &quot;It is my conjecture that there are none! (and if any, they are rare)&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please help me, smart math people.  Can you provide a counterexample to his conjecture?  Perhaps, more convincingly, several?  I've only managed to find one! (Polya's)  One, out of the many thousands (I presume) of conjectures.  It's also one that is hard to explain the finer points to the layman.  Are there any more famous or accessible examples?&lt;/p&gt;&#xA;" OwnerUserId="51" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-07-22T22:18:57.237" LastActivityDate="2011-03-31T16:31:18.147" Title="Conjectures that have been disproved with extremely large counterexamples?" Tags="&lt;big-list&gt;&lt;conjectures&gt;&lt;contradiction&gt;" AnswerCount="9" CommentCount="3" FavoriteCount="5" />
  <row Id="515" PostTypeId="2" ParentId="514" CreationDate="2010-07-22T20:59:58.363" Score="15" ViewCount="" Body="&lt;p&gt;The wikipedia article on the Collatz conjecture gives these three examples of conjectures that were disproved with large numbers:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/P%C3%B3lya_conjecture&quot; rel=&quot;nofollow&quot;&gt;Polya conjecture&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Mertens_conjecture&quot; rel=&quot;nofollow&quot;&gt;Mertens conjecture&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Skewes%27_number&quot; rel=&quot;nofollow&quot;&gt;Skewes number&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="9" LastActivityDate="2010-07-22T20:59:58.363" />
  <row Id="516" PostTypeId="2" ParentId="514" CreationDate="2010-07-22T21:03:05.330" Score="11" ViewCount="" Body="&lt;p&gt;The first example which came to my mind is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Skewes%27_number&quot; rel=&quot;nofollow&quot;&gt;Skewes' number&lt;/a&gt;, that is the smallest natural number n for which &amp;pi;(n) &gt; li(n). Wikipedia states that now the limit is near e&lt;sup&gt;727.952&lt;/sup&gt;, but the first estimation was much higher.&lt;/p&gt;&#xA;" OwnerUserId="89" LastActivityDate="2010-07-22T21:03:05.330" />
  <row Id="517" PostTypeId="2" ParentId="118" CreationDate="2010-07-22T21:11:04.467" Score="4" ViewCount="" Body="&lt;p&gt;Calculus is basically a way of calculating rates of changes (similar to slopes, but called derivatives in calculus), and areas, volumes, and surface areas (for starters).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's easy to calculate these kinds of things with algebra and geometry if the shapes you're interested in are simple.  For example, if you have a straight line you can calculate the slope easily.  But if you want to know the slope at an arbitrary point (any random point) on the graph of some function like x-squared or some other polynomial, then you would need to use calculus.  In this case, calculus gives you a way of &quot;zooming in&quot; on the point you're interested in to find the slope EXACTLY at that point.  This is called a derivative.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have a cube or a sphere, you can calculate the volume and surface area easily.  If you have an odd shape, you need to use calculus.  You use calculus to make an infinite number of really small slices of the object you're interested in, determine the sizes of the slices, and then add all those sizes up.  This process is called integration.  It turns out that integration is the reverse of derivation (finding a derivative).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In summary, calculus is a tool that lets you do calculations with complicated curves, shapes, etc., that you would normally not be able to do with just algebra and geometry.&lt;/p&gt;&#xA;" OwnerUserId="195" LastActivityDate="2010-07-22T21:11:04.467" />
  <row Id="518" PostTypeId="1" AcceptedAnswerId="27734" CreationDate="2010-07-22T21:43:15.497" Score="10" ViewCount="500" Body="&lt;p&gt;Let $A$ be a commutative ring. Suppose $P \subset A$ is a minimal prime ideal. Then it is a theorem that $P$ consists of zero-divisors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This can be proved using localization, when $A$ is noetherian: $A_P$ is local artinian, so every element of $PA_P$ is nilpotent.  Hence every element of $P$ is a zero-divisor. (As Matt E has observed, when $A$ is nonnoetherian, one can still use a similar argument: $PA_P$ is the only prime in $A_P$, hence is the radical of $A_P$ by elementary commutative algebra.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can this be proved without using localization?&lt;/p&gt;&#xA;" OwnerUserId="536" LastEditorUserId="536" LastEditorDisplayName="" LastEditDate="2010-07-30T21:44:02.087" LastActivityDate="2011-03-18T04:04:03.567" Title="Why does a minimal prime ideal consist of zerodivisors?" Tags="&lt;commutative-algebra&gt;" AnswerCount="4" CommentCount="4" FavoriteCount="1" />
  <row Id="519" PostTypeId="1" AcceptedAnswerId="597" CreationDate="2010-07-22T21:49:58.990" Score="13" ViewCount="288" Body="&lt;p&gt;The group:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;G = &amp;lt; x, y | x^2 = y^3 = (xy)^7 = 1 &gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is infinite, or so I've been told. How would I go about proving this? (To prove finiteness of a finitely presented group, I could do a coset enumeration, but I don't see how this helps if I want to prove that it's infinite.)&lt;/p&gt;&#xA;" OwnerUserId="31" LastActivityDate="2010-07-30T16:33:34.130" Title="How do you prove that a group specified by a presentation is infinite?" Tags="&lt;group-theory&gt;&lt;presentation&gt;" AnswerCount="4" CommentCount="10" FavoriteCount="6" />
  <row Id="520" PostTypeId="2" ParentId="518" CreationDate="2010-07-22T21:58:37.160" Score="0" ViewCount="" Body="&lt;p&gt;If A has an identity then what about this argument:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take p in P and look at pP.  This is an ideal of A and a subset of P so we must have that pP = P by minimality.  Thus pa = p for some a in P.  a cannot be one for otherwise P = A is not a prime ideal.  Thus p(a-1) = 0 so that p is a zero divisor.&lt;/p&gt;&#xA;" OwnerUserId="9" LastActivityDate="2010-07-22T21:58:37.160" CommentCount="4" />
  <row Id="521" PostTypeId="2" ParentId="11" CreationDate="2010-07-23T00:33:32.400" Score="7" ViewCount="" Body="&lt;p&gt;Okay I burned a lot of reputation points (at least for me) on MathOverflow to gain clarity on how to give some intuition into this problem, so hopefully this answer will be at least be somewhat illuminating. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To gain a deeper understanding of what is going on, first we need to answer the question, &quot;What is a number?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are a lot of ways to define numbers, but in general numbers are thought of as symbols that represent sets. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is easy for things like the natural numbers. So 10 would correspond to the set with ten things -- like a bag of ten stones. Pretty straight forward.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The tricky part is that when we consider ten a subset of the real numbers, we actually redefine it. This is not emphasized even in higher mathematics classes, like real analysis; it just happens when we define the real numbers. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what is 10 when constructed in the real numbers? Well, at least with the Dedekind cut version of the real numbers, &lt;strong&gt;all&lt;/strong&gt; real numbers correspond to a set with an infinite amount of elements. This makes 10 under the hood look drastically different, although in practice it operates exactly the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So let's return to the question: Why is 10 the same as 9.99999? Because the real numbers have this completely surprising quality, where there is no next real number. So when you have two real numbers that are as close together as possible, they are the same. I can't think of any physical object that has this quality, but it's how the real numbers work (makes &quot;real&quot; seem ironic).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With integers (bag of stones version) this is not the same. When you have two integers as close to each other as possible they are still different, and they are distance one apart.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Put another way, 10 bag of stones are not the same as 9.9999999 but 10 the natural number, where natural numbers are a subset of the real numbers is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The bottom line is that the real numbers have these tricky edge cases that are hard to understand intuitively. Don't worry, your intuition is not really failing you. :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I didn't feel confident answering until I got this Terence Tao link: &lt;a href=&quot;http://www.google.com/buzz/114134834346472219368/RarPutThCJv/In-the-foundations-of-mathematics-the-standard&quot; rel=&quot;nofollow&quot;&gt;http://www.google.com/buzz/114134834346472219368/RarPutThCJv/In-the-foundations-of-mathematics-the-standard&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="99" LastEditorUserId="99" LastEditorDisplayName="" LastEditDate="2010-07-23T00:56:33.743" LastActivityDate="2010-07-23T00:56:33.743" CommunityOwnedDate="2010-07-28T06:53:43.987" />
  <row Id="522" PostTypeId="2" ParentId="514" CreationDate="2010-07-23T01:12:25.927" Score="4" ViewCount="" Body="&lt;p&gt;A famous example that is not quite as large as these others is the &lt;a href=&quot;http://mathworld.wolfram.com/ChebyshevBias.html&quot; rel=&quot;nofollow&quot;&gt;prime race&lt;/a&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The conjecture states, roughly: Consider the first n primes, not counting 2 or 3. Divide them into two groups: A contains all of those primes congruent to 1 modulo 3 and B contains those primes congruent to 2 modulo 3. A will never contain more numbers than B. The smallest value of n for which this is false is 23338590792.&lt;/p&gt;&#xA;" OwnerUserId="73" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-07-29T01:15:07.540" LastActivityDate="2010-07-29T01:15:07.540" CommentCount="1" />
  <row Id="523" PostTypeId="1" AcceptedAnswerId="526" CreationDate="2010-07-23T02:09:04.340" Score="0" ViewCount="65" Body="&lt;p&gt;Polygons are, in this question, defined as non-unique if they similar to another (by rotation, reflection, translation, or scaling).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would this answer be any different if similar but non-identical polygons were allowed?  And if only if rotated/translated by rational coefficients?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would this answer be any different if we constrained the length and internal angles of all polygons to rational numbers?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume the number of sides is finite but unbounded, and greater than two.&lt;/p&gt;&#xA;" OwnerUserId="51" LastEditorUserId="51" LastEditorDisplayName="" LastEditDate="2010-07-23T06:41:43.400" LastActivityDate="2010-07-23T06:41:43.400" Title="Is the set of all unique (convex) polygons countable? If so, by what bijection to the natural numbers?" Tags="&lt;geometry&gt;&lt;set-theory&gt;" AnswerCount="1" />
  <row Id="524" PostTypeId="1" AcceptedAnswerId="1206" CreationDate="2010-07-23T02:10:15.867" Score="1" ViewCount="90" Body="&lt;p&gt;I want to find the least squares solution to &lt;strong&gt;Ax&lt;/strong&gt;=&lt;b&gt;b&lt;/b&gt; where &lt;strong&gt;A&lt;/strong&gt; is a highly sparse square matrix.&lt;br&gt;&#xA;I found two methods that look like they might lead me to a solution: &lt;a href=&quot;http://en.wikipedia.org/wiki/QR_decomposition&quot; rel=&quot;nofollow&quot;&gt;QR factorization&lt;/a&gt;, and &lt;a href=&quot;http://en.wikipedia.org/wiki/Singular_value_decomposition#Pseudoinverse&quot; rel=&quot;nofollow&quot;&gt;singular value decomposition&lt;/a&gt;. Unfortunately, I haven't taken linear algebra yet, so I can't really understand most of what those pages are saying. I can calculate both in Matlab though, and it looks like the SVD gave me a smaller squared error. Why did that happen? How can I know which one I should be using in the future?&lt;/p&gt;&#xA;" OwnerUserId="73" LastActivityDate="2010-08-03T06:14:23.543" Title="How can I tell which matrix decomposition to use for OLS?" Tags="&lt;linear-algebra&gt;&lt;math-software&gt;&lt;numerical-methods&gt;" AnswerCount="1" />
  <row Id="525" PostTypeId="2" ParentId="11" CreationDate="2010-07-23T02:10:22.280" Score="-15" ViewCount="" Body="&lt;p&gt;I do not think that 0.999... = 1 unless we're defining the equality operator in some strange way.  Assuming you allow N decimals in the 0.999... term, you can take N as large as you want and it will still be less than 1.  There is no N for which 0.999... is not less than 1.  Sure, if you take the limit as N approaches infinity then you get 1, but &quot;lim 0.999... as N approaches infinity&quot; is not the same as &quot;0.999...&quot; by itself.  By itself, there is always some infinitesimal difference.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the whole point of writing 0.999... is to indicate that it is NOT equal to 1 and to indicate that there is some infinitesimal difference.  Otherwise you would just write 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My opinion is that most people say 0.999... = 1 because that's what they were taught in college and they accepted it without really thinking about it.  All the arguments I've heard or read so far are just contrived ways of making the equality hold, but there is always some subtle flaw that takes a long time to point out, that the proponents will not accept anyway (for reasons I will not get into here).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From a comment I made:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;10x - x != 9.  10x - x would be 8.9999...1.  However infinite the extent of 9s is in x, if we multiply it by 10, the nines are shifted left by one position and a zero inserted at the &quot;last&quot; place, and then when you subtract the other number there is a nine subtracted from a zero at the far right.  Otherwise we'd have to give 0.999.. some unusual properties like automatically increasing the number of nines when it is multiplied.  It would not be just an ordinary number.  Maybe that's the problem.  0.999... might just not be an ordinary type number as some people are using it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The current winning response as of 7/22/10 treats 0.999... as being equivalent to lim (0.999...) as the number of decimals goes to infinity.  But the sequence and the limit of the sequence are not the same thing.&lt;/p&gt;&#xA;" OwnerUserId="195" LastEditorUserId="195" LastEditorDisplayName="" LastEditDate="2010-07-23T02:20:35.140" LastActivityDate="2010-07-23T02:20:35.140" CommunityOwnedDate="2010-07-28T06:53:43.987" CommentCount="4" />
  <row Id="526" PostTypeId="2" ParentId="523" CreationDate="2010-07-23T02:13:47.357" Score="7" ViewCount="" Body="&lt;p&gt;There are uncountably many, because for example one can have rectangles with arbitrary side ratios.  For your second question, if everything is constrained to be rational, there will be countably many, because a polygon is uniquely determined by its ordered collection of sides and angles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is part of a general fact: if $A$ is a countable set, then the collection of ordered $n$-tuples of elements of $A$ for all $n$ is still countable.&lt;/p&gt;&#xA;" OwnerUserId="536" LastActivityDate="2010-07-23T02:13:47.357" CommentCount="6" />
  <row Id="527" PostTypeId="2" ParentId="514" CreationDate="2010-07-23T02:15:52.577" Score="3" ViewCount="" Body="&lt;p&gt;I heard this story from &lt;a href=&quot;http://www.ams.sunysb.edu/~estie/estie.html&quot; rel=&quot;nofollow&quot;&gt;Professor Estie Arkin&lt;/a&gt; at Stony Brook &lt;strike&gt;&lt;em&gt;(sorry, I don't know what conjecture she was talking about)&lt;/em&gt;&lt;/strike&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;For weeks we tried to prove the conjecture (without success) while we left a computer running looking for counter-examples.  One morning we came in to find the computer screen flashing: &lt;em&gt;&quot;Counter-example found&quot;&lt;/em&gt;.  We all thought for sure that there must have been a bug in the algorithm, but sure enough, it was a valid counter-example.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;I tell this story to my students to emphasize that &lt;em&gt;&quot;proof by lack of counter-example&quot;&lt;/em&gt; is not a proof at all!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;[Edit]&lt;/strong&gt; Here was the response from Estie:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;It is mentioned in our paper:&lt;br&gt;&#xA;  &lt;a href=&quot;http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.57.5143&amp;amp;rep=rep1&amp;amp;type=pdf&quot; rel=&quot;nofollow&quot;&gt;Hamiltonian Triangulations for Fast Rendering&lt;/a&gt;&lt;br&gt;&#xA;  &lt;sub&gt;&lt;em&gt;E.M. Arkin, M. Held, J.S.B. Mitchell, S.S. Skiena (1994). Algorithms -- ESA'94, Springer-Verlag, LNCS 855, J. van Leeuwen (ed.), pp. 36-47; Utrecht, The Netherlands, Sep 26-28, 1994.&lt;/em&gt;&lt;/sub&gt; &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Specifically section 4 of the paper, that gives an example of a set of points that does not have a so-called &lt;em&gt;&quot;sequential triangulation&quot;&lt;/em&gt;.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The person who wrote the code I talked about is Martin Held.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="136" LastEditorUserId="136" LastEditorDisplayName="" LastEditDate="2011-03-31T16:31:18.147" LastActivityDate="2011-03-31T16:31:18.147" CommentCount="3" />
  <row Id="528" PostTypeId="2" ParentId="356" CreationDate="2010-07-23T02:45:15.487" Score="0" ViewCount="" Body="&lt;p&gt;I think you should determine what kinds of applications you want to apply CS to and then learn general theory relevant to those applications.  Not all theory is equally applicable everywhere.&lt;/p&gt;&#xA;" OwnerUserId="195" LastActivityDate="2010-07-23T02:45:15.487" CommunityOwnedDate="2010-07-23T02:45:15.487" CommentCount="1" />
  <row Id="529" PostTypeId="1" AcceptedAnswerId="547" CreationDate="2010-07-23T03:11:15.673" Score="14" ViewCount="367" Body="&lt;p&gt;Why are the only (associative) division algebras over the real numbers the real numbers, the complex numbers, and the quaternions?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here a division algebra is an associative algebra where every nonzero number is invertible (like a field, but without assuming commutativity of multiplication).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is an old result proved by Frobenius, but I can't remember how the argument goes.  Anyone have a quick proof?&lt;/p&gt;&#xA;" OwnerUserId="8" LastActivityDate="2010-08-04T01:19:55.590" Title="Why are the only division algebras over the real numbers the real numbers, the complex numbers, and the quaternions?" Tags="&lt;quaternions&gt;&lt;ring-theory&gt;&lt;abstract-algebra&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="3" />
  <row Id="530" PostTypeId="1" CreationDate="2010-07-23T03:17:09.670" Score="10" ViewCount="267" Body="&lt;p&gt;In intro number theory a key lemma is that if a and b are relatively prime integers, then there exist integers x and y such that ax+by=1.  In a more advanced course instead you would use the theorem that the integers are a PID, i.e. that all ideals are principal.  Then the old lemma can be used to prove that &quot;any ideal generated by two elements is actually principal.&quot;  Induction then says that any finitely generated ideal is principal.  But what if all finitely generated ideals are principal but there are some ideals that aren't finitely generated?  Can that happen?&lt;/p&gt;&#xA;" OwnerUserId="8" LastActivityDate="2010-09-08T08:49:41.177" Title="Can you find a domain where ax+by=1 has a solution for all a and b relatively prime, but which is not a PID?" Tags="&lt;number-theory&gt;&lt;abstract-algebra&gt;" AnswerCount="5" CommentCount="4" FavoriteCount="1" />
  <row Id="531" PostTypeId="1" CreationDate="2010-07-23T03:20:13.713" Score="26" ViewCount="746" Body="&lt;p&gt;I know there's something called &quot;ultrafinitism&quot; which is a very radical form of constructivism that I've heard said means people don't believe that really large integers actually exist.  Could someone make this a little bit more precise?  Are there good reasons for taking this point of view?  Can you actually get math done from that perspective?&lt;/p&gt;&#xA;" OwnerUserId="8" LastActivityDate="2011-02-09T09:46:15.163" Title="What is &quot;ultrafinitism&quot; and why do people believe it?" Tags="&lt;soft-question&gt;&lt;philosophy&gt;&lt;constructive-mathematics&gt;" AnswerCount="3" CommentCount="4" FavoriteCount="7" />
  <row Id="532" PostTypeId="2" ParentId="531" CreationDate="2010-07-23T03:30:20.270" Score="16" ViewCount="" Body="&lt;p&gt;The philosophy is explained in Doron Zeilberger's &lt;a href=&quot;http://www.math.rutgers.edu/~zeilberg/mamarim/mamarimPDF/real.pdf&quot; rel=&quot;nofollow&quot;&gt;article&lt;/a&gt;.&#xA; Basically, it's the belief that there is a largest natural number! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've heard a funny story (on Scott Aaronson's blog) about someone who was an ultrafinitist.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-Do you believe in 1?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-Yes, he responded immediately&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-Do you believe in 2?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-Yes, he responded after a brief pause&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-Do you believe in 3? &#xA;-Yes, he responded after a slightly longer pause&#xA;-Do you believe in 4?&#xA;-Yes, after several seconds&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It soon become clear that he would take twice as long to answer the next question as the previous one. (I believe &lt;a href=&quot;http://en.wikipedia.org/wiki/Alexander_Esenin-Volpin&quot; rel=&quot;nofollow&quot;&gt;Alexander Esesin-Volpin&lt;/a&gt; was the person.)&lt;/p&gt;&#xA;" OwnerUserId="536" LastActivityDate="2010-07-23T03:30:20.270" CommentCount="6" />
  <row Id="533" PostTypeId="1" AcceptedAnswerId="561" CreationDate="2010-07-23T03:36:00.833" Score="3" ViewCount="185" Body="&lt;p&gt;I know some references where I can find this, but they seem tedious(Both Hartshorne and Ueno cover this).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am wondering if there is an elegant way to describe these. If this task is too difficult in general, how about just $\mathbb{P}^n$? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="16" LastActivityDate="2010-07-23T11:53:21.343" Title="Classifying Quasi-coherent Sheaves on Projective Schemes" Tags="&lt;algebraic-geometry&gt;&lt;projective-schemes&gt;&lt;quasicoherent-sheaves&gt;&lt;projective-space&gt;" AnswerCount="1" FavoriteCount="0" />
  <row Id="534" PostTypeId="1" AcceptedAnswerId="565" CreationDate="2010-07-23T03:56:46.910" Score="5" ViewCount="82" Body="&lt;p&gt;There is a little triviality that has been referred to as the &lt;a href=&quot;http://www.johndcook.com/blog/2010/01/13/soft-maximum&quot; rel=&quot;nofollow&quot;&gt;&quot;soft maximum&quot;&lt;/a&gt; over on &lt;a href=&quot;http://www.johndcook.com/blog/&quot; rel=&quot;nofollow&quot;&gt;John Cook's Blog&lt;/a&gt; that I find to be fun, at the very least.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The idea is this:  given a list of values, say $x_1,x_2,\ldots,x_n$ , the function &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$g(x_1,x_2,\ldots,x_n) = \log(\exp(x_1) + \exp(x_2) + \cdots + \exp(x_n))$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;returns a value very near the maximum in the list.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This happens because that exponentiation exaggerates the differences between the $x_i$ values.  For the largest $x_i$, $\exp(x_i)$ will be $really$ large.  This largest exponential will significantly outweigh all of the others combined. Taking the logarithm, i.e. undoing the exponentiation,  we essentially recover the largest of the $x_i$'s.  (Of course, if two of the values were very near one another, we aren't guaranteed to get the true maximum, but it won't be far off!) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;About this, John Cook says: &quot;The soft maximum approximates the hard maximum but it also rounds off the corners.&quot;  This couldn't really be said any better.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I recall trying to cleverly construct sequences for proofs in advanced calculus where not-everywhere-differentiable operations would have been great to use if they didn't have that pesky non-differentiable trait.  I can't recall a specific incidence where I was tempted to use $max(x_i)$, but this seems at least plausible that it would have come up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Has anyone used this before or have a scenario off hand where it would be useful?&lt;/p&gt;&#xA;" OwnerUserId="121" LastEditorUserId="121" LastEditorDisplayName="" LastEditDate="2010-07-23T04:58:15.617" LastActivityDate="2010-07-23T12:33:52.603" Title="applications of the &quot;soft maximum&quot;" Tags="&lt;numerical-methods&gt;&lt;analysis&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="535" PostTypeId="1" CreationDate="2010-07-23T03:59:36.487" Score="8" ViewCount="405" Body="&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/pWnq4.png&quot; alt=&quot;complex plot of the zeros&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Diagram and setup from UCSMP &lt;em&gt;Precaluclus and Discrete Mathematics&lt;/em&gt;, 3rd ed.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Above is a partial plot of the zeros of $p_c(x)=4x^4+8x^3-3x^2-9x+c$.  The text stops at showing the diagram and does not discuss the shape of the locus of the zeros or describe the resulting curves.  Are the curves in the locus some specific (named) type of curve?  Is there a simple way to describe the curves (equations)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question need not be limited to the specific polynomial given--a similar sort of locus is generated by the zeros of nearly any quartic polynomial as the constant term is varied.&lt;/p&gt;&#xA;" OwnerUserId="72" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-12-11T06:51:54.967" LastActivityDate="2010-12-11T06:51:54.967" Title="Describe the locus in the complex plane of the zeros of a quartic polynomial as the constant term varies" Tags="&lt;algebra-precalculus&gt;&lt;polynomials&gt;&lt;zeros&gt;" AnswerCount="4" CommentCount="6" />
  <row Id="536" PostTypeId="1" CreationDate="2010-07-23T04:05:29.323" Score="14" ViewCount="782" Body="&lt;p&gt;How does one prove that a simple (steps of length 1 in directions parallel to the axes) symmetric (each possible direction is equally likely) random walk in 1 or 2 dimensions returns to the origin with probability 1?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&lt;strong&gt;edit&lt;/em&gt;&lt;/strong&gt;: note that while returning to the origin is guaranteed (p = 1) in 1 and 2 dimensions, it is not guaranteed in higher dimensions; this means that something in a correct justification for the 1- or 2-d case must fail to extend to 3-d (or fail when the probability for each direction drops from 1/4 to 1/6).&lt;/p&gt;&#xA;" OwnerUserId="72" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-07-23T05:08:37.597" LastActivityDate="2011-05-31T01:28:06.247" Title="Proving that 1- and 2-d simple symmetric random walks return to the origin with probability 1" Tags="&lt;probability-theory&gt;" AnswerCount="7" CommentCount="12" FavoriteCount="4" />
  <row Id="537" PostTypeId="2" ParentId="536" CreationDate="2010-07-23T04:35:13.213" Score="1" ViewCount="" Body="&lt;p&gt;I'll do 1D. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1D walks are building binary strings, 010101, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Say take six steps. Then 111111 is just as likely as 101010. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, how many of the possible sequences have six ones? 1. How many of the possibly sequences have three ones and three zeros? Much more. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That number is called multiplicity, and it grows mighty fast. In the limit its log becomes Shannon entropy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sequences are equally likely, but combinations are not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the limit the combinations with maximum entropy are going dominate all the rest. So the walk is going to have gone an equal number of right and left steps...almost surely.&lt;/p&gt;&#xA;" OwnerUserId="99" LastEditorUserId="99" LastEditorDisplayName="" LastEditDate="2010-07-23T04:54:20.617" LastActivityDate="2010-07-23T04:54:20.617" CommentCount="5" />
  <row Id="538" PostTypeId="2" ParentId="536" CreationDate="2010-07-23T04:51:17.470" Score="1" ViewCount="" Body="&lt;p&gt;I can prove the 1 dimensional case a bit more formally than &lt;a href=&quot;http://math.stackexchange.com/questions/536/proving-that-1-and-2-d-simple-symmetric-random-walks-return-to-the-origin-with-p/537#537&quot; rel=&quot;nofollow&quot;&gt;Jonathan&lt;/a&gt;. First we only look at the absolute values. Let us try to calculate the probability of this never exceeding x. The probability of 2x+1 consecutive moves all being the same is some p&gt;0. If this ever occurs, then the absolute value will exceed x. Consider n groups of 2x+1 moves, the probability that at least one of these is all the same is 1-(1-p)^n, which approaches 1. So the probability of reaching each absolute value x (other than 0) is 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, lets consider the probability of reaching 0 again. Without loss of generality, suppose our first move is +1. We have a 100% chance of reaching a point a distance of 1 from this. There is a 50% that the first such point is 0 and 50% that it is 2. From 2, we have a 100% chance of reaching a point two away from this. 50% chance that this is 0, 50% that this is 4. Repeating, it is easy to see that we have to reach 0 again.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore, after we have reached an absolute value x, the probability of reaching it again is 1. So we reach each absolute value an infinite number of times. Due to symmetry, we can expect to reach each x an infinite number of times.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;: I originally tried to apply it to the 2d case as you can see below. This is incorrect, as even though there will be no maximal absolute value, there is no reason we will necessarily come back down.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Similar reasoning will work in the 2d&#xA;  case. Instead of using symmetry, we&#xA;  simply note that each point with the&#xA;  same Manhattan distance from 0 has an&#xA;  expected value that is a non-zero&#xA;  proportion of the expected value all&#xA;  points a particular Manhattan distance&#xA;  away.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="123" LastEditorUserId="123" LastEditorDisplayName="" LastEditDate="2010-07-23T06:16:39.330" LastActivityDate="2010-07-23T06:16:39.330" CommentCount="6" />
  <row Id="539" PostTypeId="2" ParentId="2" CreationDate="2010-07-23T06:34:30.527" Score="0" ViewCount="" Body="&lt;p&gt;If you were an ant crawling along a curve, at a really small (almost zero) distance away from the point you're taking the limit of, where are you?&lt;/p&gt;&#xA;" OwnerUserId="198" LastActivityDate="2010-07-23T06:34:30.527" />
  <row Id="540" PostTypeId="1" AcceptedAnswerId="832" CreationDate="2010-07-23T06:42:52.100" Score="28" ViewCount="1199" Body="&lt;p&gt;In laymen's terms, as much as possible: What is the Riemann-Zeta function, and why does it come up so often with relation to prime numbers?&lt;/p&gt;&#xA;" OwnerUserId="136" LastEditorUserId="171" LastEditorDisplayName="" LastEditDate="2010-08-14T06:28:36.233" LastActivityDate="2011-02-04T18:00:20.990" Title="What is the Riemann-Zeta function?" Tags="&lt;terminology&gt;&lt;complex-analysis&gt;&lt;prime-numbers&gt;&lt;riemann-zeta&gt;" AnswerCount="5" CommentCount="1" FavoriteCount="13" />
  <row Id="541" PostTypeId="1" AcceptedAnswerId="611" CreationDate="2010-07-23T07:37:42.770" Score="1" ViewCount="281" Body="&lt;p&gt;By this I mean books with math puzzles and problems similar to the ones you would find in mathematical olympiads.&lt;/p&gt;&#xA;" OwnerUserId="191" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-07-29T23:31:18.443" LastActivityDate="2011-05-07T18:43:48.827" CommunityOwnedDate="2010-07-23T16:12:30.563" Title="Which books would you recommend about Recreational Mathematics?" Tags="&lt;soft-question&gt;&lt;big-list&gt;&lt;books&gt;&lt;recreational-mathematics&gt;" AnswerCount="3" CommentCount="5" FavoriteCount="1" />
  <row Id="542" PostTypeId="2" ParentId="540" CreationDate="2010-07-23T07:52:32.097" Score="14" ViewCount="" Body="&lt;p&gt;Giving an explanation in layman's terms is always going to be challenging, given that the Riemann-Zeta function (and related hypothesis) inevitably lies in the domain of abstract mathematics, but I shall do my best.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Riemann_zeta_function&quot; rel=&quot;nofollow&quot;&gt;Riemann-Zeta function&lt;/a&gt; is a &lt;a href=&quot;http://en.wikipedia.org/wiki/Complex_function#Complex_functions&quot; rel=&quot;nofollow&quot;&gt;complex function&lt;/a&gt; that tells us many things about the theory of numbers. Its mystery is increased by the fact it has no &lt;a href=&quot;http://en.wikipedia.org/wiki/Closed-form_expression&quot; rel=&quot;nofollow&quot;&gt;closed form&lt;/a&gt; - i.e. it can't be expressed a single formula that contains other standard (elementary) functions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Although there are many different ways of expressing the Riemann-Zeta function (the Wikipedia article gives several), it can ultimately be derived from the following simple series of real numbers:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://upload.wikimedia.org/math/4/f/e/4fe5e85d36c09cc13a973a39d6387caa.png&quot; alt=&quot;Eq1&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;by extending it into the &lt;a href=&quot;http://en.wikipedia.org/wiki/Complex_plane&quot; rel=&quot;nofollow&quot;&gt;complex plane&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason this strange and esoteric function is so famous and actively discussed in mathematics is due to the &lt;a href=&quot;http://en.wikipedia.org/wiki/Riemann_hypothesis&quot; rel=&quot;nofollow&quot;&gt;Riemann hypothesis&lt;/a&gt; - proposedi n 1859 by the by the great Bernhard Riemann and still unsolved. The Wiki article states the problem in quite simple terms:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The Riemann zeta-function ζ(s) is&#xA;  defined for all complex numbers s ≠ 1.&#xA;  It has zeros at the negative even&#xA;  integers (i.e. at s = −2, −4, −6,&#xA;  ...). These are called the trivial&#xA;  zeros. The Riemann hypothesis is&#xA;  concerned with the non-trivial zeros,&#xA;  and states that:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;The real part of any non-trivial zero of the Riemann zeta function is 1/2.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;  &#xA;  &lt;p&gt;Thus the non-trivial zeros should lie&#xA;  on the critical line, 1/2 + it, where&#xA;  t is a real number and i is the&#xA;  imaginary unit.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Although the conjecture (it is only that at the moment) has many consequences for mathematics (number theory in particular), the primary one, at least the one Riemann originally proposed, is about the &lt;em&gt;distribution of prime numbers&lt;/em&gt;. In other words, it tells us with great precision what the average gaps between primes are as we move to greater and greater numbers. Many of the other implications are rather more esoteric, though perhaps equally important for pure mathematicians.&lt;/p&gt;&#xA;" OwnerUserId="56" LastActivityDate="2010-07-23T07:52:32.097" />
  <row Id="544" PostTypeId="1" AcceptedAnswerId="545" CreationDate="2010-07-23T08:16:24.300" Score="8" ViewCount="413" Body="&lt;p&gt;What is a intuitive explanation of a Markov Chain, and how they work? Please provide at least one practical example.&lt;/p&gt;&#xA;" OwnerUserId="156" LastEditorUserId="156" LastEditorDisplayName="" LastEditDate="2010-07-25T15:12:42.147" LastActivityDate="2010-07-31T05:45:16.580" Title="What is a Markov Chain?" Tags="&lt;terminology&gt;&lt;probability-theory&gt;&lt;intuition&gt;&lt;stochastic-processes&gt;" AnswerCount="5" CommentCount="2" FavoriteCount="3" />
  <row Id="545" PostTypeId="2" ParentId="544" CreationDate="2010-07-23T08:28:36.050" Score="11" ViewCount="" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;A Markov chain is a discrete random&#xA;  process with the property that the&#xA;  next state depends only on the current&#xA;  state (&lt;a href=&quot;http://en.wikipedia.org/wiki/Markov_chain&quot; rel=&quot;nofollow&quot;&gt;wikipedia&lt;/a&gt;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So P(X_n | X_1, X_2, ... X_n-1) = P(X_n | X_n-1). An example could be when you are modelling the weather. You then can take the assumption that the weather of today can be predicted by only using the knowledge of yesterday.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's say we have Rainy and Sunny. When it is rainy on one day the next day is Sunny with probability 0.3. When it is Sunny, the probability for Rain next day is 0.4. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now when it is today Sunny we can predict the weather of the day after tomorrow, by simply calculating the probability for Rain tomorrow, multiplying that with the probablity for Sun after rain plus the probability of Sun tomorrow times the probability of Sun after sun. In total the probability of Sunny of the day after tomorrow is P(R|S)*P(S|R) + P(S|S)*P(S|S) = 0.3*0.4+0.6*0.6 = 0.48&lt;/p&gt;&#xA;" OwnerUserId="143" LastActivityDate="2010-07-23T08:28:36.050" />
  <row Id="546" PostTypeId="2" ParentId="544" CreationDate="2010-07-23T08:30:06.530" Score="5" ViewCount="" Body="&lt;p&gt;In a nutshell, a Markov chain is (the behavior of) a random process which may only find itself in a (not necessarily finite) number of different states. The process moves from a state to another in discrete times (that is, you define a sequence S(t) of states at time t=0,1,2,...), and for which the probability of going from state S to state R depends just from S and R; that is, there is no &quot;memory of the past&quot; and the process is &quot;timeless&quot;. This means that the Markov chain may be modeled as a n*n matrix, where n is the number of possible states. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;An example of a process which may be modeled by a Markov chain is the sequence of faces of a die showing up, if you are allowed to rotate the die wrt an edge. The corresponding matrix is &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;    1   2   3   4   5   6&#xA;   ------------------------ &#xA;1 | 0   1/4 1/4 1/4 1/4 0&#xA;2 | 1/4 0   1/4 1/4 0   1/4&#xA;3 | 1/4 1/4 0   0   1/4 1/4&#xA;4 | 1/4 1/4 0   0   1/4 1/4&#xA;5 | 1/4 0   1/4 1/4 0   1/4&#xA;1 | 0   1/4 1/4 1/4 1/4 0&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;As usual, &lt;a href=&quot;http://en.wikipedia.org/wiki/Markov_chain&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt; and &lt;a href=&quot;http://mathworld.wolfram.com/MarkovChain.html&quot; rel=&quot;nofollow&quot;&gt;MathWorld&lt;/a&gt; are your friends.&lt;/p&gt;&#xA;" OwnerUserId="89" LastActivityDate="2010-07-23T08:30:06.530" />
  <row Id="547" PostTypeId="2" ParentId="529" CreationDate="2010-07-23T08:32:59.613" Score="5" ViewCount="" Body="&lt;p&gt;Essentially one first proves that any real division algebra D is a Clifford algebra (i.e. it's generated by elements of some inner product vector space I subject to relations v&lt;sup&gt;2&lt;/sup&gt;=&amp;lt;v,v&amp;gt;): first one splits D as $\mathbb R\oplus D_0$ where D&lt;sub&gt;0&lt;/sub&gt; is the space of elements with Tr=0 and then one observes that minimal polynomial of a traceless element has the form x&lt;sup&gt;2&lt;/sup&gt;-a=0 (it's quadratic because it's irreducible and the coefficient of x is zero because it is the trace). Now it remains to find out which Clifford algebras are division algebras which is pretty straightforward (well, and it follows from the classification of Clifford algebras).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This proof &lt;a href=&quot;http://en.wikipedia.org/wiki/Frobenius_theorem_(real_division_algebras)&quot; rel=&quot;nofollow&quot;&gt;is written&lt;/a&gt; in Wikipedia.&lt;/p&gt;&#xA;" OwnerUserId="152" LastEditorUserId="152" LastEditorDisplayName="" LastEditDate="2010-07-30T07:33:29.343" LastActivityDate="2010-07-30T07:33:29.343" />
  <row Id="548" PostTypeId="1" CreationDate="2010-07-23T08:51:02.597" Score="8" ViewCount="458" Body="&lt;p&gt;(This was asked due to the comments and downvotes on this &lt;a href=&quot;http://stackoverflow.com/questions/3236489/why-is-0-divided-by-0-an-error/3236541#3236541&quot;&gt;Stackoverflow&lt;/a&gt; answer. I am not that good at maths, so was wondering if I had made any basic mistakes)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ignoring limits, I would like to know if this is a valid explanation for why 0/0 is undefined:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;x = 0/0&lt;br&gt;&#xA;   x * 0 = 0  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Hence There are an infinite number of values for x as anything multiplied by 0 is 0.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;However, it seems to have got comments, with two general themes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once is that you lose the values of x by multiplying by 0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other is that the last line is &lt;code&gt;x * 0 = 0/0 * 0&lt;/code&gt; as it involves a division by 0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any merit to either argument? More to the point, are there any major flaws in my explanation and is there a better way of showing why 0/0 is undefined?&lt;/p&gt;&#xA;" OwnerUserId="108" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-07-26T08:31:38.957" LastActivityDate="2010-08-02T11:38:02.740" Title="Explanation of method for showing that 0 / 0 is undefined" Tags="&lt;algebra&gt;&lt;algebra-precalculus&gt;&lt;zero-divisor&gt;" AnswerCount="7" CommentCount="14" FavoriteCount="2" />
  <row Id="549" PostTypeId="2" ParentId="548" CreationDate="2010-07-23T09:07:55.223" Score="0" ViewCount="" Body="&lt;p&gt;It's undefined because $0/0 = 0$ is against the identity $a/a = 1$. Or at least, that's how I understand it, although there are probably deeper explanations that involve the consistency of algebra. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could also approach it in a simple manner from the direction that what you have is $0^1 * 0^-1$ which would give you $0^0$ which is a very disturbing expression (and could be taken as either 0 or 1 depending on your definition of exponents).&lt;/p&gt;&#xA;" OwnerUserId="35" LastActivityDate="2010-07-23T09:07:55.223" CommentCount="4" />
  <row Id="550" PostTypeId="2" ParentId="392" CreationDate="2010-07-23T09:34:33.213" Score="1" ViewCount="" Body="&lt;p&gt;From first principles,using trig identities and small-angle approximations:&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;sin'(x) = lim(h-&gt;0){(sin(x+h)-sin(x))/h}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;sin(x+h) = sin(x)cos(h)+cos(x)sin(h)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;=&gt; sin'(x) = lim(h-&gt;0){(sin(x)(cos(h)-1) + cos(x)sin(h))/h}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For x small, sin(x)~=x, so lim(h-&gt;0){sin(h)/h}=1, and cos(x)~= 1 -x^2/2, so (h-&gt;0){(cos(h)-1)/h}=0&lt;/p&gt;&#xA;&#xA;&lt;p&gt;=&gt; sin'(x) = cos(x)&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;cos'(x) = lim(h-&gt;0){(cos(x+h)-cos(x))/h}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;cos(x+h) = cos(x)cos(h) - sin(x)sin(h)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;=&gt; cos'(x) = lim(h-&gt;0){(cos(x)(cos(h)-1) - sin(x)sin(h))/h}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;= -sin(x) by the same reasoning above.&lt;/p&gt;&#xA;" OwnerUserId="166" LastActivityDate="2010-07-23T09:34:33.213" CommentCount="1" />
  <row Id="551" PostTypeId="2" ParentId="531" CreationDate="2010-07-23T09:39:52.700" Score="24" ViewCount="" Body="&lt;p&gt;Ultrafinitism is basically resource-bounded  constructivism: proofs have constructive content, and what you get out of these constructions isn't much more than you put in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Looking at the universal and existential quantifier should help clarify things.  Constructively, a universally quantified sentence means that if I am given a parameter, I can construct something that satisfies the quantified predicate.  Ultrafinitistically, the thing you give back won't be much bigger: typically there will be a polynomial bound on the size of what you get back.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For existentially quantified statements, the constructive content is a pair of the value of the parameter, and the construction that satisfies the predicate.  Here the resource is the size of the proof: the size of the parameter and construction will be related to the size of the proof of the existential.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Typically, addition and mutliplication are total functions, but exponentiation is not.  &lt;a href=&quot;http://en.wikipedia.org/wiki/Self-verifying_theories&quot; rel=&quot;nofollow&quot;&gt;Self-verifying theories&lt;/a&gt; are more extreme: addition is total in the strongest of these theories, but multiplication cannot be.  So the resource bound is linear for these theories, not polynomial.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A foundational problem with ultrafinitism is that there aren't nice ultrafinitist logics that support an attractive formulae-as-types correspondence in the way that intuitionistic logic does.  This makes ultrafinitism as less comfy kind of contructivism than intuitionism.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Why do people believe it?&lt;/strong&gt;  For the same kinds of reasons people believe in constructivism: they want mathematical claims to be backed up by something they can regard as concrete.  Just as an intuitionist might be bothered by the idea of cutting a ball into non-measurable pieces and putting them back together into two balls, so too  an ultrafinitist might be concerned about the idea that towers of exponentials are meaningful ways of constructing numbers.  Wittgenstein argued this point in his &quot;Lectures on the Foundations of Mathematics&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Can you actually get math done from that perspective?&lt;/strong&gt; Yes.  If intuitionism is the mathematics of the computable, ultrafinitism is the mathematics of the feasibly computable.  But the difference in ease of working with between ultrafinitism and intuitionism is much bigger than that between intuitionism and classical mathematics.&lt;/p&gt;&#xA;" OwnerUserId="100" LastEditorUserId="100" LastEditorDisplayName="" LastEditDate="2011-02-09T09:46:15.163" LastActivityDate="2011-02-09T09:46:15.163" CommentCount="3" />
  <row Id="552" PostTypeId="2" ParentId="548" CreationDate="2010-07-23T09:42:11.283" Score="3" ViewCount="" Body="&lt;p&gt;I think that ignoring limits is problematic. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there was a limit of the function $f(x,y)=x/y for x,y-&gt;0$ regardless of how the limit is performed, then one would define that value to be $f(0,0)$, even if everything else is strange. Since the limiting value depends on the way the limit is done, choosing a value for $f(0,0)$ is counterproductive as it gives a non-continuous function. Better to have a continuous function over a slightly smaller domain. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This also forces the point that if you do have a limit process that results in the evaluation of $f(0,0)$, you realize early on that you should examine the limit carefully rather than use $\lim g(x) = g(\lim x)$ (which is only true for continuous functions, of course.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the way, this might be too trivial, but I'll give an example of how the limiting value depends on the limit:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\lim_{x\rightarrow0} \frac{\sin(x)}{x} = 1$, (both $\sin(x)$ and $x$ goto 0)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\lim_{x\rightarrow0} \frac{cos x - 1}{x} = 0$ (again, both num and denum goto 0, but num goes &quot;faster&quot;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\lim_{x\rightarrow0} \frac{\sqrt{x}}{x} =+\infty$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS, sorry about the latex notation, hopefully, we could have it working nicely like in MO...&lt;/p&gt;&#xA;" OwnerUserId="165" LastActivityDate="2010-07-23T09:42:11.283" CommentCount="3" />
  <row Id="553" PostTypeId="2" ParentId="548" CreationDate="2010-07-23T09:53:19.190" Score="8" ViewCount="" Body="&lt;p&gt;For &quot;all&quot; x,&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;0/x = 0 ⇒ 0/0 = 0&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;For &quot;all&quot; x,&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;x/x = 1 ⇒ 0/0 = 1&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Moreover, if one could say &lt;code&gt;0/0 = k&lt;/code&gt;, any k, we could then say &lt;code&gt;2 = 3&lt;/code&gt; -- just divide both sides by 0 and get &lt;code&gt;k = k&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since there is no reasonable value &lt;code&gt;0/0&lt;/code&gt; can have, &lt;code&gt;0/0&lt;/code&gt; must be undefined.&lt;/p&gt;&#xA;" OwnerUserId="92" LastEditorUserId="92" LastEditorDisplayName="" LastEditDate="2010-07-30T20:24:53.447" LastActivityDate="2010-07-30T20:24:53.447" CommentCount="3" />
  <row Id="554" PostTypeId="2" ParentId="536" CreationDate="2010-07-23T09:53:30.360" Score="4" ViewCount="" Body="&lt;p&gt;I'll try it for 2D and then we can get 1D as a corollary [excercise!]...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;This is the only proof I know of, there may be a more intuitive (and less messy without tex!) proof out there, but I like this one- it uses generating functions in a really nifty way.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the probability of being at the origin after 2n steps (notice we cannot return in an odd number of steps):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$u_0=1$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;$u_{2n} = p(n,n,0,0)+ p(n-1,n-1,1,1)+...+p(n-k,n-k,k,k)+...+p(0,0,n,n)$ (when $n \neq 0) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here $p(u,d,l,r)$ is the probability of the first 2n steps being u up, d down, l left and r right in any order. Each order has probability $\frac{1}{4^{2n}}$, and there are $\frac{(2n)!}{u!d!l!r!}$ distinct orders, giving $p(n-k,n-k,k,k)=\frac{1}{4^{2n}} \frac{(2n)!}{(n-k)!(n-k)!k!k!}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, since $(2n)!=n! n! \binom{2n}{n}$ we have $p(n-k,n-k,k,k)=\frac{1}{4^{2n}} \binom{2n}{n} \binom{n}{k}^2$ giving&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$u_{2n}= \frac{1}{4^{2n}}  \Sigma_k \binom{2n}{n} \binom{n}{k}^2$ which, by one of those silly binomial results, can be contracted to&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$u_{2n}= \frac{1}{4^{2n}} \binom{2n}{n}^2$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Let us put that in our back pocket for now and consider in stead the probability of &lt;strong&gt;first&lt;/strong&gt; returning after 2n steps $f_{2n}$- this is rather difficult to tackle directly but we can make ourselves a cunning little formula involving it, jazz out some &lt;a href=&quot;http://en.wikipedia.org/wiki/Generating_function&quot; rel=&quot;nofollow&quot;&gt;generating function&lt;/a&gt; fun and seal the deal with some .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The formula in question is:&#xA;$u_{2n}= f_2 u_{2n-2}+ f_4 u_{2n-4} +....+f_{2n-2}u_{2} + f_{2n}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which we shall not prove so much as explain: to return to the origin after 2n steps (LHS) you must either first return after 2 steps and do a 'return to the origin in 2n-2 steps walk' (first term RHS) or first return after 4 steps and do a 'return to the origin in 2n-4 steps walk' (2nd term) or... or first return after 2n-2 steps and do a 'return to the origin in 2 steps walk or first return to the origin after 2n steps.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We shall now tweak our formula just a tiny bit so it has the right symmetry properties for what is to come, we do this by adding $f_0=0$ and $u_0=1$ to give &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$u_{2n}= f_0 u_{2n}+ f_2 u_{2n-2}+ f_4 u_{2n-4} +....+f_{2n-2}u_{2} + f_{2n}u_0$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Which is secretly a statement about generating functions (this is the sweet bit!), see look at the generating functions of u_n, f_n:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$U(x)= \Sigma_m u_{2m} x^{2m}$, $F(x)= \Sigma_m f_{2m} x^{2m}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We see:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$U(x)= 1+ U(x)F(x)$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(where the '1+' is to compensate for the fact that $u_0$ does not appear in the product) Rearranging to:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$F(x)= \frac{U(x)-1}{U(x)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Observe that the probability of return is $\Sigma f_n= F(1)$, which comes out as 1  because $U(1)$ diverges by some tedious stirlings formula bounds that I forget.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; Until Tex comes online, this is pretty unreadable, so &lt;a href=&quot;http://www.dartmouth.edu/~chance/teaching_aids/books_articles/probability_book/Chapter12.pdf&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;'s a link to some lecture notes I found with the same proof (and, fortunately, the same notation!). Enjoy!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit$^2$:&lt;/strong&gt; Hooray, Tex has come online!!! Enjoy.&lt;/p&gt;&#xA;" OwnerUserId="160" LastEditorUserId="160" LastEditorDisplayName="" LastEditDate="2010-08-04T12:42:49.603" LastActivityDate="2010-08-04T12:42:49.603" />
  <row Id="555" PostTypeId="2" ParentId="514" CreationDate="2010-07-23T10:00:35.643" Score="6" ViewCount="" Body="&lt;p&gt;For an old example, Mersenne made the following &lt;a href=&quot;http://en.wikipedia.org/wiki/Mersenne_conjectures&quot; rel=&quot;nofollow&quot;&gt;conjecture&lt;/a&gt; in 1644:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;i&gt;The Mersenne numbers, $M_n=2^n  − 1$, are prime for n = 2, 3, 5, 7, 13, 17, 19, 31, 67, 127 and 257, and no others.&lt;/i&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Euler observed that the Mersenne number at M_61 is prime, so refuting the conjecture.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;M_61 is quite large by the standards of the day: 2 305 843 009 213 693 951.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to Wikipedia, there are 47 known Mersenne primes.&lt;/p&gt;&#xA;" OwnerUserId="100" LastEditorUserId="100" LastEditorDisplayName="" LastEditDate="2010-08-08T07:59:24.667" LastActivityDate="2010-08-08T07:59:24.667" CommentCount="4" />
  <row Id="556" PostTypeId="2" ParentId="540" CreationDate="2010-07-23T10:00:46.830" Score="5" ViewCount="" Body="&lt;p&gt;Here there is another attempt at an explanation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know that the sum of the inverse of the positive numbers, 1 + 1/2 + 1/3 + ..., diverges. Euler shown that the sum of the inverse of the squares,  1/(1&lt;sup&gt;2&lt;/sup&gt;) + 1/(2&lt;sup&gt;2&lt;/sup&gt;) + 1/(3&lt;sup&gt;2&lt;/sup&gt;) + ..., has a finite sum, namely &amp;pi;&lt;sup&gt;2&lt;/sup&gt;/6. Mathematicians love to generalize things, so they thought at the function&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://mathurl.com/35scyna.png&quot; alt=&quot;f(x) = sum for n=1 to infinity of (1/n^x)&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is defined for x &gt; 1. But this was not enough: they decided that the variable could be a complex number and not a real one. There is a standard tecnique (&lt;a href=&quot;http://mathworld.wolfram.com/AnalyticContinuation.html&quot; rel=&quot;nofollow&quot;&gt;Analytic continuation&lt;/a&gt;) which allows us to extend the function to nearly all the complex plane. So we now have a function which formally is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://mathurl.com/5euwuy.png&quot; alt=&quot;zeta(s) = sum for n=1 to infinity of (1/n^s)]&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(the variable being s and not x to show that we are dealing with complex numbers) but is not computed in this way. Just to make an example, &amp;zeta;(0)=1/2, and sum of an infinity of ones is not 1/2 :-) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It may be shown that for s = -2n (n positive integer) &amp;zeta;(s) = 0. But there are infinite other point s'=(x,y) where &amp;zeta;(s') = 0. For all of these points, 0 &amp;lt; x &amp;lt; 1; Riemann's hypothesis says that for all such points x = 1/2. If it were true, we could have the best asymptotic expression to count &amp;pi;(n), that is the number of primes below n.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why does the function pop up when we talk about primes? I don't know, but in the case of integer values Euler proved that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://mathurl.com/34s664v.png&quot; alt=&quot;sum of 1/(n^s) = product over all primes of 1/(1-p^s)&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe this could be a good start.&lt;/p&gt;&#xA;" OwnerUserId="89" LastEditorUserId="89" LastEditorDisplayName="" LastEditDate="2010-07-23T12:07:54.467" LastActivityDate="2010-07-23T12:07:54.467" CommentCount="2" />
  <row Id="558" PostTypeId="2" ParentId="213" CreationDate="2010-07-23T10:22:47.437" Score="4" ViewCount="" Body="&lt;p&gt;Statistics is the topic in which I am still poor and it is still useful to me which I learned so late and that's why I am poor in Statistics.&lt;/p&gt;&#xA;" OwnerUserId="168" LastActivityDate="2010-07-23T10:22:47.437" CommunityOwnedDate="2010-07-23T10:22:47.437" />
  <row Id="559" PostTypeId="2" ParentId="548" CreationDate="2010-07-23T11:16:56.147" Score="0" ViewCount="" Body="&lt;p&gt;It seems as though this is a question of conventions (like the question about why 0^0 is undefined rather than just being 1 &lt;a href=&quot;http://math.stackexchange.com/questions/135/why-is-x0-1-except-when-x-0&quot; rel=&quot;nofollow&quot;&gt;as was asked here&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We could define 0/0 = 0, and I don't think it would break Mathematics. I'm happy to defend my view (or revise it) based on the comments this answer receives.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would not be keen to define it to be anything else, since if we have 0/0 = x then (I'd like the axioms to allow me to say) we have x = 0/0 = (2*0)/0 = 2*(0/0) = 2*x. But it's unclear that we should require (a*b)/c = a*(b/c) to hold even when c = 0.&lt;/p&gt;&#xA;" OwnerUserId="106" LastActivityDate="2010-07-23T11:16:56.147" CommentCount="1" />
  <row Id="560" PostTypeId="2" ParentId="135" CreationDate="2010-07-23T11:41:17.373" Score="1" ViewCount="" Body="&lt;p&gt;If we use the idea of set exponentiation to define exponentiation of cardinals, we have the following natural idea:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;$$A^B:=\{f:B\to A\}$$&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We define the exponential of cardinals as follows: $|A|^|B|:=|A^B|$.  It's easy to check that this agrees with our intuition for exponentiation of natural numbers when $B$ is nonempty.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is only one set representing the cardinal $0$, namely the empty set.  Then we may look at $0^0$ as follows: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;$$0^0=|\emptyset|^{|\emptyset|}=|\emptyset^\emptyset|&lt;br&gt;&#xA;=|\{f:\emptyset\to\emptyset\}|=1$$&lt;/code&gt;&lt;/p&gt;&#xA;" OwnerUserId="126" LastActivityDate="2010-07-23T11:41:17.373" />
  <row Id="561" PostTypeId="2" ParentId="533" CreationDate="2010-07-23T11:53:21.343" Score="2" ViewCount="" Body="&lt;p&gt;Quasi-coherent sheaves on affine schemes (say $Spec(A)$) are obtained by taking an $A$-module $M$ and the associated sheaf (by localizing $M$).  This gives an equivalence of categories between $A$-modules and q-c sheaves on $Spec(A)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $R$ be a graded ring, $R = R_0 + R_1 + \dots$ (direct sum).  Then we can, given a &lt;em&gt;graded&lt;/em&gt; $R$-module $M$, consider its associated sheaf $\tilde{M}$.  The stalk of this at  a homogeneous prime ideal $P$ is defined to be the localization $M_{(P)}$, which is defined as generated by quotients $m/s$ for $s$ homogeneous of the same degree as $m$ and not in $P$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In short, we get sheaves of modules on the affine scheme just as we get the normal  sheaves of rings.  We get sheaves of modules on the projective scheme in the same homogeneous localization way as we get the sheaf of rings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, it's no longer an equivalence of categories. Why? Say you had a graded module $M= M_0 + M_1 + \dots$ (in general, we allow negative gradings as well).  Then it is easy to check that the sheaves associated to $M$ and $M' = M_1 + M_2 + \dots$ are exactly the same.&#xA;Nevertheless, it is possible to get every sheaf on $Proj(R)$ for $R$ a graded ring in this way.  See Proposition II.5.15 in Hartshorne.&lt;/p&gt;&#xA;" OwnerUserId="536" LastActivityDate="2010-07-23T11:53:21.343" CommentCount="5" />
  <row Id="562" PostTypeId="2" ParentId="530" CreationDate="2010-07-23T11:58:10.147" Score="7" ViewCount="" Body="&lt;p&gt;If I'm not mistaken, the integral domain of holomorphic functions on a connected open set $U &#xA;\subset \mathbb{C}$ works. It is a theorem (in Chapter 15 of Rudin's &lt;em&gt;Real and Complex Analysis,&lt;/em&gt; and essentially a corollary of the Weierstrass factorization theorem), that every finitely generated ideal in this domain is principal. This implies that if $a,b$ have no common factor, they generate the unit ideal.  However, for instance, the ideal of holomorphic functions in the unit disk that vanish on all but finitely many of ${1-\frac{1}{n}}$ is nonprincipal.&lt;/p&gt;&#xA;" OwnerUserId="536" LastActivityDate="2010-09-07T12:59:48.080" CommentCount="0" />
  <row Id="563" PostTypeId="2" ParentId="536" CreationDate="2010-07-23T12:01:33.107" Score="8" ViewCount="" Body="&lt;p&gt;See &lt;a href=&quot;http://www.math.cornell.edu/~durrett/PTE/PTE4_Jan2010.pdf&quot; rel=&quot;nofollow&quot;&gt;Durrett, &lt;i&gt;Probability and Examples&lt;/i&gt;&lt;/a&gt; (link goes to online copy of the fourth edition).  On p. 164 Durrett gives a proof that simple random walk is recurrent in two dimensions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First find the probability that simple random walk in &lt;i&gt;one&lt;/i&gt; dimension is at 0 after 2n steps; this is clearly rho_1(2n) = (2n choose n)/2^(2n), since 2n choose n is the number of paths with n right steps and n left steps.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next, the probability that simple random walk in &lt;i&gt;two&lt;/i&gt; dimensions -- call this rho_2(2n) -- is at 0 after 2n steps is the square of the previous probability.  Consider the simple random walk which makes steps to the northeast, northwest, southeast, and southwest with equal probability. The projections of this walk onto the x- and y-axes are independent simple random walks in one dimension.  Roating and rescaling gives the &quot;usual&quot; SRW in two dimensions (with steps north, east, south and west) and doesn't change the probability of being at 0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So rho_2(2n) = ((2n choose n)/2^(2n))^2.  This is asymptotic to 1/(pi*n), and the expected number of returns to the origin is the sum of rho_2(2n) over positive integers n, so the expected number of returns to the origin is infinite. It's not hard to show (and is in fact true, but the proof is unenlightening so I'll leave it to Durrett) that in this case the probability of eventually returning to the origin is 1.&lt;/p&gt;&#xA;" OwnerUserId="173" LastActivityDate="2010-07-23T12:01:33.107" CommentCount="1" />
  <row Id="564" PostTypeId="2" ParentId="544" CreationDate="2010-07-23T12:31:25.573" Score="3" ViewCount="" Body="&lt;p&gt;Markov chains, especially hidden Markov models are huge in computation linguistics. A hidden Markov model is one where we can't directly view the state, but we do have some information about what the state might be. For example, consider breaking down a sentence in parts of speech such as verbs, adjectives, ect. We don't know what the parts of speech are, but we can attempt to deduce them from the word. For example, the word &lt;a href=&quot;http://www.thefreedictionary.com/run&quot; rel=&quot;nofollow&quot;&gt;run&lt;/a&gt; might be used 80% as a verb, 18% of the time as a noun and 2% of the time as an adjective. We also have Markov relations between the parts of speech, so for example an adjective might be followed by a noun 70% of the time and another adjective 30% of the time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can use the &lt;a href=&quot;http://en.wikipedia.org/wiki/Viterbi_algorithm&quot; rel=&quot;nofollow&quot;&gt;Viterbi algorithm&lt;/a&gt; to decide which sequence is most likely to have generated the observed output (the algorithm takes into account both the probability of such a sequence of parts of speech occurring together in a sentence and the relative chance that such parts of speech would be responsible for us observing the given words).&lt;/p&gt;&#xA;" OwnerUserId="123" LastActivityDate="2010-07-23T12:31:25.573" CommentCount="1" />
  <row Id="565" PostTypeId="2" ParentId="534" CreationDate="2010-07-23T12:33:52.603" Score="2" ViewCount="" Body="&lt;p&gt;This is close to being the flipside of the geometric mean, which is the nth root of the product of the numbers, and can be expressed as the exponential of the sum of the logarithms.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another pair of dual mean measures is the regular mean and the harmonic mean (n divided by the sum of the reciprocals).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I say the soft maximum is close to being the flipside of the geometric mean, but it lacks the good property that all of the others have of taking a list of the same value to that value (for definedness, let all values be positive).  Let's call the hyperbolic mean the &quot;soft maximum&quot; of the nth roots of the terms in the list: then this has that good property.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The hyperbolic mean the emphasises large values in a roughly symmetric manner to the way that the geometric mean emphasises small values (which is always smaller than the regular mean), and is, of course, much smaller for long list of large values.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I say, consider it an amplified version of a useful addition to the family of of mean operators.&lt;/p&gt;&#xA;" OwnerUserId="100" LastActivityDate="2010-07-23T12:33:52.603" />
  <row Id="566" PostTypeId="1" CreationDate="2010-07-23T12:44:16.403" Score="19" ViewCount="1008" Body="&lt;p&gt;Okay, so this question was bound to come up sooner or later- the hope was to ask it well before someone asked it badly...&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;We all love a good puzzle&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;To a certain extent, any piece of mathematics is a puzzle in some sense: whether we are classifying the homological intersection forms of four manifolds or calculating the optimum dimensions of a cylinder, it is an element of investigation and inherently puzzlish intrigue that drives us. Indeed most puzzles (cryptic crosswords aside) are somewhat mathematical (the mathematics of sudoku for example is hidden in &lt;a href=&quot;http://en.wikipedia.org/wiki/Latin_squares&quot; rel=&quot;nofollow&quot;&gt;latin squares&lt;/a&gt;). Mathematicians and puzzles get on, it seems, rather well.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;But what is a good puzzle?&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Okay, so in order to make this question worthwhile (and not a ten-page wadeathon through 57 varieties of the men with red and blue hats puzzle), we are going to have to impose some limitations. Not every puzzle-based answer that pops into your head will qualify for answerhood- to do so it must&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;Not be widely known:&lt;/strong&gt; If you have a terribly interesting puzzle that motivates something in cryptography; well done you, but chances are we've seen it. If you saw that &lt;em&gt;hilarious&lt;/em&gt; scene in the film 21, where kevin spacey explains the monty hall paradox badly and want to share, don't do so here. Anyone found posting the liar/truth teller riddle will be immediately disemvowelled.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Be mathematical:&lt;/strong&gt; as much as possible- true: logic &lt;em&gt;is&lt;/em&gt; mathematics, but puzzles beginning 'There is a street where everyone has a different coloured house...' are much of a muchness and tedious as hell. Note: there is a happy medium between this and trig substitutions.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Not be too hard:&lt;/strong&gt; any level is cool but if the answer requires more than two sublemmas, you are misreading your audience&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Actually have an answer:&lt;/strong&gt; crank questions will not be appreciated! You can post the answers/hints in &lt;a href=&quot;http://en.wikipedia.org/wiki/Rot_13&quot; rel=&quot;nofollow&quot;&gt;Rot-13&lt;/a&gt; underneath as comments as on MO if you fancy.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;And should  &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;Ideally include where you found it:&lt;/strong&gt; so we can find more cool stuff like it  &lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Have that indefinable spark that makes a puzzle awesome:&lt;/strong&gt; a situation that seems familiar, requiring unfamiliar thought...&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;For ease of voting- one puzzle per post is bestest.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Some examples to set the ball rolling&lt;/h3&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Simplify $\sqrt{2+\sqrt{3}}$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;From:&lt;/strong&gt; problem solving magazine &lt;strong&gt;Hint:&lt;/strong&gt; GEL N GJB GREZ FBYHGVBA&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Can one make an equilateral triangle with all vertices at integer coordinates?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;From:&lt;/strong&gt; Durham distance maths challenge 2010 &lt;strong&gt;Hint:&lt;/strong&gt; GUVF VF RDHVINYRAG GB GUR ENGVBANY PNFR&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;nxn &lt;a href=&quot;http://en.wikipedia.org/wiki/Magic_squares&quot; rel=&quot;nofollow&quot;&gt;Magic squares&lt;/a&gt; form a vector space over $\mathbb{R}$ prove this, and by way of a linear transformation, derive the dimension of this vector space.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;From:&lt;/strong&gt; Me, I made this up (you can tell, can't you!) &lt;strong&gt;Hint:&lt;/strong&gt;  NCCYL GUR ENAX AHYYVGL GURBERZ&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Happy puzzling!&lt;/p&gt;&#xA;" OwnerUserId="160" LastEditorUserId="97" LastEditorDisplayName="" LastEditDate="2010-10-27T12:34:22.917" LastActivityDate="2011-05-09T22:50:57.603" CommunityOwnedDate="2010-07-23T12:44:16.403" Title="Your favourite maths puzzles" Tags="&lt;soft-question&gt;&lt;big-list&gt;&lt;puzzle&gt;" AnswerCount="18" CommentCount="6" FavoriteCount="19" />
  <row Id="567" PostTypeId="2" ParentId="566" CreationDate="2010-07-23T13:02:28.817" Score="3" ViewCount="" Body="&lt;p&gt;From &lt;em&gt;Mathematical Puzzles&lt;/em&gt; by Peter Winkler:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Divide an hexagon in equilateral triangles, like in the figure. Now fill all the hexagon with the three kinds of diamonds made from two triangles, also shown in the figure. Prove that the number of each kind of diamond is the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i27.tinypic.com/33nkqrr.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="89" LastActivityDate="2010-07-23T13:02:28.817" CommunityOwnedDate="2010-07-23T13:02:28.817" CommentCount="3" />
  <row Id="568" PostTypeId="2" ParentId="566" CreationDate="2010-07-23T13:04:13.867" Score="3" ViewCount="" Body="&lt;p&gt;unknown source:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could the plane be colored with two different colors (say, red and blue) so that there is no equilateral triangle whose vertices are all of the same color?&lt;/p&gt;&#xA;" OwnerUserId="89" LastActivityDate="2010-07-23T13:04:13.867" CommunityOwnedDate="2010-07-23T13:04:13.867" CommentCount="4" />
  <row Id="571" PostTypeId="1" AcceptedAnswerId="578" CreationDate="2010-07-23T13:36:43.717" Score="7" ViewCount="327" Body="&lt;p&gt;You are standing on a cliff at a height &lt;em&gt;h&lt;/em&gt; above the sea. You are capable of throwing a stone with velocity &lt;em&gt;v&lt;/em&gt; at any angle &lt;em&gt;a&lt;/em&gt; between horizontal and vertical. What is the value of &lt;em&gt;a&lt;/em&gt; when the horizontal distance travelled &lt;em&gt;d&lt;/em&gt; is at a maximum?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On level ground, when &lt;em&gt;h&lt;/em&gt; is zero, it's easy to show that &lt;em&gt;a&lt;/em&gt; needs to be midway between horizontal and vertical, and thus π/4 or 45°. As &lt;em&gt;h&lt;/em&gt; increases, however, we can see by heuristic reasoning that &lt;em&gt;a&lt;/em&gt; decreases to zero, because you can put more of the velocity into the horizontal component as the height of the cliff begins to make up for the loss in the vertical component. For small negative values of &lt;em&gt;h&lt;/em&gt; (throwing up onto a platform), &lt;em&gt;a&lt;/em&gt; will actually be greater than 45°.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a fully-solved, closed-form expression for the value of &lt;em&gt;a&lt;/em&gt; when &lt;em&gt;h&lt;/em&gt; is not zero?&lt;/p&gt;&#xA;" OwnerUserId="113" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-07-23T14:50:53.553" LastActivityDate="2010-08-03T09:18:31.527" Title="What is the optimum angle of projection when throwing a stone off a cliff?" Tags="&lt;calculus&gt;&lt;trigonometry&gt;&lt;physics&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="2" />
  <row Id="572" PostTypeId="2" ParentId="541" CreationDate="2010-07-23T13:40:21.850" Score="2" ViewCount="" Body="&lt;p&gt;Art of Problem Solving: &lt;a href=&quot;http://www.artofproblemsolving.com/Store/index.php&quot; rel=&quot;nofollow&quot;&gt;http://www.artofproblemsolving.com/Store/index.php&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anything by Martin Gardner.&lt;/p&gt;&#xA;" OwnerUserId="30" LastActivityDate="2010-07-23T13:40:21.850" CommunityOwnedDate="2010-07-23T16:12:30.563" CommentCount="1" />
  <row Id="573" PostTypeId="1" AcceptedAnswerId="579" CreationDate="2010-07-23T13:40:43.287" Score="5" ViewCount="152" Body="&lt;p&gt;So I know that given a chain complex we can define the d-th cohomology by taking ker{d}/im_{d+1}. But I don't know how this corresponds to the idea of holes in topological spaces (maybe this is homology, I'm a tad confused).&lt;/p&gt;&#xA;" OwnerUserId="30" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-07-24T18:40:46.240" LastActivityDate="2010-07-24T18:40:46.240" Title="varying definitions of cohomology" Tags="&lt;cohomology&gt;&lt;algebraic-topology&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" />
  <row Id="574" PostTypeId="1" AcceptedAnswerId="878" CreationDate="2010-07-23T13:43:25.603" Score="1" ViewCount="260" Body="&lt;p&gt;What does it mean (in a mathematically rigorous way) to claim something is &quot;generic?&quot; How does this coincide with the Zariski topology?&lt;/p&gt;&#xA;" OwnerUserId="30" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-07-24T18:29:49.287" LastActivityDate="2010-07-31T01:07:09.890" Title="genericness and the zariski topology" Tags="&lt;algebraic-geometry&gt;&lt;topology&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="575" PostTypeId="1" AcceptedAnswerId="830" CreationDate="2010-07-23T13:46:38.513" Score="4" ViewCount="116" Body="&lt;p&gt;Given a curve how do you intuitively construct the picture of its projective dual? I know points --&gt; lines, lines--&gt; points but for something like the swallowtail this is not really obvious.&lt;/p&gt;&#xA;" OwnerUserId="30" LastActivityDate="2010-07-28T01:37:45.227" Title="projective duality" Tags="&lt;projective-space&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="576" PostTypeId="2" ParentId="571" CreationDate="2010-07-23T13:52:28.750" Score="0" ViewCount="" Body="&lt;p&gt;I don't have a complete solution, but I attempted to solve this problem using calculus.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;x'=v cos a&#xA;y''= -g and y' (at t=0) = v sin a&#xA;So y'= v sin a -gt&#xA;x0=0, so x=vt cos a&#xA;y0=h, so y=vt sin a - 0.5gt^2+c (integrating with respect to t)&#xA;Subbing in h, y=vt sin a - 0.5gt^2+h&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The ball will hit the ground when y=0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is as far as I got, but it appears that you can find a &lt;a href=&quot;http://math.stackexchange.com/questions/571/what-is-the-optimum-angle-of-projection-when-throwing-a-stone-off-a-cliff/578#578&quot; rel=&quot;nofollow&quot;&gt;closed solution&lt;/a&gt; after all. I originally tried solving the quadratic for t and subbing that into x, but it seems to work much better to do the substitution the other way round. I will leave this solution here in case anyone wants to see how to derive the basic equations for x and y.&lt;/p&gt;&#xA;" OwnerUserId="123" LastEditorUserId="123" LastEditorDisplayName="" LastEditDate="2010-07-23T23:20:49.457" LastActivityDate="2010-07-23T23:20:49.457" CommentCount="4" />
  <row Id="577" PostTypeId="1" AcceptedAnswerId="1760" CreationDate="2010-07-23T14:07:57.573" Score="4" ViewCount="278" Body="&lt;p&gt;According to Wikipedia, Godel's incompleteness theorem states:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;No consistent system of axioms whose&#xA;  theorems can be listed by an&#xA;  &quot;effective procedure&quot; (essentially, a&#xA;  computer program) is capable of&#xA;  proving all facts about the natural&#xA;  numbers.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This obviously includes our current system. So has anyone proposed any additional axioms that seem credible?&lt;/p&gt;&#xA;" OwnerUserId="123" LastActivityDate="2010-08-06T21:00:58.133" Title="Has anyone ever proposed additional axioms?" Tags="&lt;axioms&gt;" AnswerCount="3" CommentCount="4" FavoriteCount="2" />
  <row Id="578" PostTypeId="2" ParentId="571" CreationDate="2010-07-23T14:29:17.023" Score="10" ViewCount="" Body="&lt;p&gt;Assume no friction and uniform gravity &lt;em&gt;g&lt;/em&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you throw a stone at point (0, &lt;em&gt;h&lt;/em&gt;), with velocity (&lt;em&gt;v&lt;/em&gt; cos &lt;em&gt;&amp;theta;&lt;/em&gt;, &lt;em&gt;v&lt;/em&gt; sin &lt;em&gt;&amp;theta;&lt;/em&gt;), then we get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;d &amp;amp;= vt\cos\theta &amp;amp;&amp;amp; (1) \\&#xA;0 &amp;amp;= h + vt\sin\theta - \frac12 gt^2 &amp;amp;&amp;amp; (2)&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only unknown to be solved is &lt;em&gt;t&lt;/em&gt; (total travel time). We could eliminate it by using $t = \frac d{v\cos\theta}$ to get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ 0 = h + d\tan\theta - \frac{gd^2\sec^2\theta}{2v^2}\qquad(3) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then we compute the total derivative with respect to &lt;em&gt;θ&lt;/em&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;0 &amp;amp;= \frac d{d\theta}\left(d\tan\theta\right) - \frac g{2v^2}\frac d{d\theta}\left(d^2\sec^2\theta\right) \\&#xA;&amp;amp;= \ldots&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and then set $\frac{dd}{d\theta}=0$ (because it is maximum) to solve &lt;em&gt;d&lt;/em&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ d = \frac{v^2}{g\tan\theta} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Substitute this back to (3) gives:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;h &amp;amp;= \frac{v^2}g \left( \frac1{2\sin^2\theta} - 1\right) \\&#xA;\Rightarrow \sin\theta &amp;amp;= \left( 2 \left(\frac{gh}{v^2} + 1\right) \right)^{-1/2}&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the closed form of &lt;em&gt;θ&lt;/em&gt; in terms of &lt;em&gt;h&lt;/em&gt;. &lt;/p&gt;&#xA;" OwnerUserId="171" LastEditorUserId="171" LastEditorDisplayName="" LastEditDate="2010-08-03T09:18:31.527" LastActivityDate="2010-08-03T09:18:31.527" CommentCount="2" />
  <row Id="579" PostTypeId="2" ParentId="573" CreationDate="2010-07-23T14:52:06.373" Score="5" ViewCount="" Body="&lt;p&gt;Edited to clear some things up:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Simplicial and singular (co)homology were invented to detect holes in spaces.  To get an intuitive idea of how this works, consider subspaces of the plane.  Here the 2-chains are formal sums of things homeomorphic to the closed disk, and 1-chains are formal sums of things homeomorphic to a line segment.  The operator d takes the boundary of a chain.  For example, the boundary of the closed disk is a circle.  If we take d of the circle we get 0 since a circle has no boundary.  And in general it happens that d^2 = 0, that is boundaries always have no boundaries themselves.  Now suppose we remove the origin from the plane and take a circle around the origin.  This circle is in the kernel of d since it has no boundary.  However, it does not bound any 2-chain in the space (since the origin is removed) and so it is not in the image of the boundary operator on two-dimensions.  Thus the circle represents a non-trivial element in the quotient space ker d / im d.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The way I have defined things makes the above a homology theory simply because the d operator decreases dimension.  Cohomology is the same thing only the operator increases dimension (for example the exterior derivative on differential forms).  Thus algebraically there really is no difference between cohomology and homology since we can just change the grading from i to -i.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From a homology we can get a corresponding cohomology theory by dualizing, that is by looking at maps from the group of chains to the underlying group (e.g. Z or R).  Then d on the cohomology theory becomes the adjoint of the previous boundary operator and thus increases degrees.&lt;/p&gt;&#xA;" OwnerUserId="9" LastEditorUserId="9" LastEditorDisplayName="" LastEditDate="2010-07-23T15:05:02.630" LastActivityDate="2010-07-23T15:05:02.630" />
  <row Id="580" PostTypeId="2" ParentId="566" CreationDate="2010-07-23T15:38:37.710" Score="4" ViewCount="" Body="&lt;p&gt;Most of us know that, being deterministic, computers cannot generate true &lt;a href=&quot;http://en.wikipedia.org/wiki/Pseudorandom_number_generator&quot; rel=&quot;nofollow&quot;&gt;random numbers&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, let's say you have a box which generates truly random binary numbers, but is biased:  it's more likely to generate either a &lt;code&gt;1&lt;/code&gt; or a &lt;code&gt;0&lt;/code&gt;, but you don't know the exact probabilities, or even which is more likely &lt;em&gt;(both probabilities are &gt; 0 and sum to 1, obviously)&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you use this box to create an unbiased random generator of binary numbers?&lt;/p&gt;&#xA;" OwnerUserId="136" LastEditorUserId="114" LastEditorDisplayName="" LastEditDate="2010-09-07T23:29:29.537" LastActivityDate="2010-09-07T23:29:29.537" CommunityOwnedDate="2010-07-23T15:38:37.710" CommentCount="7" />
  <row Id="582" PostTypeId="2" ParentId="566" CreationDate="2010-07-23T15:39:57.557" Score="1" ViewCount="" Body="&lt;p&gt;Assuming you have unlimited time and cash, is there a strategy that's guaranteed to win at roulette?&lt;/p&gt;&#xA;" OwnerUserId="136" LastActivityDate="2010-07-23T15:39:57.557" CommunityOwnedDate="2010-07-23T15:39:57.557" CommentCount="7" />
  <row Id="584" PostTypeId="2" ParentId="566" CreationDate="2010-07-23T15:40:36.243" Score="-2" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Frk n th rd 1&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Y'r n pth n n slnd, cme t  frk n th rd.  Bth pths ld t vllgs f ntvs; th ntr vllg thr lwys tlls th trth r lwys ls &lt;em&gt;(bth villgs cld b trth-tllng r lyng vllgs, r n f ch)&lt;/em&gt;.  Thr r tw ntvs t th frk - thy cld bth b frm th sm vllg, r frm dffrnt vllgs &lt;em&gt;(s bth cld b trth-tllrs, both lrs, r ne f ch)&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;n pth lds t sfty, th thr t dm.  Y'r llwd t sk nly n qstn t ch ntv t fgr t whch pth s whch.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wht d y sk?&lt;/p&gt;&#xA;" OwnerUserId="136" LastEditorUserId="136" LastEditorDisplayName="" LastEditDate="2010-07-23T17:43:23.640" LastActivityDate="2010-07-23T17:43:23.640" CommunityOwnedDate="2010-07-23T15:40:36.243" CommentCount="4" />
  <row Id="585" PostTypeId="2" ParentId="566" CreationDate="2010-07-23T15:40:51.703" Score="0" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Fork in the road 2&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You're once again at a fork in the road, and again, one path leads to safety, the other to doom.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are three natives at the fork.  One is from a village of truth-tellers, one from a village of liars, one from a village of random answerers.  Of course you don't know which is which.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moreover, the natives answer &quot;pish&quot; and &quot;posh&quot; for yes and no, but you don't know which means &quot;yes&quot; and which means &quot;no.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You're allowed to ask only two yes-or-no questions, each question being directed at one native.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What do you ask?&lt;/p&gt;&#xA;" OwnerUserId="136" LastActivityDate="2010-07-23T15:40:51.703" CommunityOwnedDate="2010-07-23T15:40:51.703" CommentCount="1" />
  <row Id="586" PostTypeId="2" ParentId="107" CreationDate="2010-07-23T15:50:56.943" Score="3" ViewCount="" Body="&lt;p&gt;I've always thought this sort of puzzle is a wonderful example of the difference between behaviour as you approach a limit, versus behaviour AT the limit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can make the puzzle more revealing like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Say I have an infinite number of banknotes, each with a unique serial number. I give you bill number 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, you have two options, either you keep that bill and the game ends, or I give you 10 more banknotes, but you have to burn the lowest numbered bill. It seems obvious that option two is much better.&#xA;Now, we repeat this game over and over. You keep getting more and more money.&#xA;But at the limit, every banknote has been burned and you're worse off than you would have been had you just taken the one banknote and left.&lt;/p&gt;&#xA;" OwnerUserId="172" LastActivityDate="2010-07-23T15:50:56.943" />
  <row Id="587" PostTypeId="2" ParentId="577" CreationDate="2010-07-23T15:54:17.940" Score="5" ViewCount="" Body="&lt;p&gt;Adding additional axioms would make more truths provable. But it wouldn't make all truths provable (unless the axiom was inconsistent with the already given ones, in which case all falsehoods would also be provable too).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So adding additional axioms isn't going to help make all the truths provable. I guess you could just add, say, Goldbach's conjecture and the Riemann hypothesis as extra axioms and carry on doing mathematics-PLUS! but why would you want to?&lt;/p&gt;&#xA;" OwnerUserId="172" LastActivityDate="2010-07-23T15:54:17.940" CommentCount="5" />
  <row Id="588" PostTypeId="1" CreationDate="2010-07-23T16:00:43.943" Score="6" ViewCount="392" Body="&lt;p&gt;How do we know if a particular function can be represented as a power series? And once we have come up with a power series representation, how does one figure out its radius of convergence ?&lt;/p&gt;&#xA;" OwnerUserId="105" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-07-28T19:35:02.243" LastActivityDate="2010-07-28T19:35:02.243" Title="What functions can be represented as power series?" Tags="&lt;calculus&gt;&lt;complex-analysis&gt;&lt;sequences-and-series&gt;" AnswerCount="3" />
  <row Id="589" PostTypeId="2" ParentId="588" CreationDate="2010-07-23T16:04:55.110" Score="0" ViewCount="" Body="&lt;p&gt;This is a very general question, as one can create all sorts of power series for different functions. (e.g. Taylor series, Laurent series, Fourier series).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To give the obvious example of Taylor series: a power series representation of a function can be found if the function is infinitely differentiable in the neighbourhood of the given point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With all power series, you will need to find the recursion relation (formula giving a successive term from the current term) and then use the ratio test to solve for the value of the input variable that gives a ratio of convergence of 1.&lt;/p&gt;&#xA;" OwnerUserId="56" LastActivityDate="2010-07-23T16:04:55.110" CommentCount="10" />
  <row Id="590" PostTypeId="2" ParentId="540" CreationDate="2010-07-23T16:13:17.913" Score="3" ViewCount="" Body="&lt;p&gt;The key point is that the Riemann zeta function is a function whose properties encode properties about the prime numbers.  As mentioned by Noldorin, in order to fully understand the Riemann zeta function you need to &quot;analytically continue it to the complex plane&quot; which is a tricky process which takes serious study.  Fortunately for some easier properties of the primes you can just use the definition of the zeta function for real s.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Claim (due to Euler): The fact that $\zeta(s)$ goes to infinity as s-&gt;1 tells you that there are infinitely many primes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sketch of proof: Use the &quot;Euler factorization&quot; mentioned by mau (expand the RHS as a geometric series and then multiply it out using unique factorization into primes):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sum_{n=1}^\infty \frac{1} {n^s} = \prod_{p prime} \frac{1} {1-p^-s}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now take log of both sides to get:&#xA;$\log \zeta(s) = \sum_{p prime} \log \frac{1} {1-p^-s}.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now use the taylor series for \log and send s to one.  You'll get that the left hand side goes to infinity (http://math.stackexchange.com/questions/255/), while the right hand side looks like $\sum 1/p$ + bounded terms.  So there must be infinitely many primes.&lt;/p&gt;&#xA;" OwnerUserId="8" LastEditorUserId="8" LastEditorDisplayName="" LastEditDate="2011-02-04T18:00:20.990" LastActivityDate="2011-02-04T18:00:20.990" CommentCount="2" />
  <row Id="591" PostTypeId="2" ParentId="588" CreationDate="2010-07-23T16:22:15.813" Score="13" ViewCount="" Body="&lt;p&gt;A function can be represented as a power series if and only if it is complex differentiable in an open set. This follows from the general form of Taylor's theorem for complex functions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Being real differentiable--even infinitely many times--is not enough, as the function $e^{-1/x^2}$ on the real line (equal to 0 at 0) is $C^\infty$ yet does not equal its power series expansion since all its derivatives at zero vanish. The reason is that the complexified version of the function is not even continuous at the origin.&lt;/p&gt;&#xA;" OwnerUserId="536" LastActivityDate="2010-07-23T16:22:15.813" CommentCount="2" />
  <row Id="592" PostTypeId="2" ParentId="577" CreationDate="2010-07-23T16:29:37.447" Score="3" ViewCount="" Body="&lt;p&gt;Sure, the &lt;a href=&quot;http://en.wikipedia.org/wiki/Continuum_hypothesis&quot; rel=&quot;nofollow&quot;&gt;continuum hypothesis&lt;/a&gt;.  We know, thanks to the work of Godel and Paul Cohen, that this is a legitimate axiom, in that it doesn't destroy consistency: you can't disprove it (or prove it) using Zermelo-Frank set theory. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;People do actually use the continuum hypothesis in model theory, for instance, where there are several results that depend on it.  For an article about its uses, cf. the &lt;a href=&quot;http://www.tricki.org/article/How_to_use_the_continuum_hypothesis&quot; rel=&quot;nofollow&quot;&gt;Tricki page&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="536" LastEditorUserId="536" LastEditorDisplayName="" LastEditDate="2010-08-06T19:59:50.097" LastActivityDate="2010-08-06T19:59:50.097" CommentCount="3" />
  <row Id="593" PostTypeId="1" AcceptedAnswerId="682" CreationDate="2010-07-23T17:21:44.563" Score="7" ViewCount="118" Body="&lt;p&gt;When I look at the Taylor series for e^x and the volume formula for oriented simplexes, it makes e^x look like it is, at least almost, the sum of simplexes volumes from n to infinity. Does anyone know of a stronger relationship beyond, &quot;they sort of look similar&quot;?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are some links:&lt;br&gt;&#xA;Volume formula&lt;br&gt;&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Simplex#Geometric_properties&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Simplex#Geometric_properties&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Taylor Series&lt;br&gt;&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/E_%28mathematical_constant%29#Complex_numbers&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/E_%28mathematical_constant%29#Complex_numbers&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="99" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-07-30T01:10:28.307" LastActivityDate="2010-07-30T07:28:06.127" Title="Is there a relationship between e and the sum of n-simplexes volumes?" Tags="&lt;geometry&gt;&lt;intuition&gt;&lt;sequences-and-series&gt;" AnswerCount="1" CommentCount="6" />
  <row Id="594" PostTypeId="1" CreationDate="2010-07-23T17:25:13.187" Score="14" ViewCount="807" Body="&lt;p&gt;It is a theorem in elementary number theory that if $p$ is a prime and congruent to 1 mod 4, then it is the sum of two squares. Apparently there is a trick involving arithmetic in the gaussian integers that lets you prove this quickly. Can anyone explain it?&lt;/p&gt;&#xA;" OwnerUserId="536" LastActivityDate="2010-11-22T01:34:17.307" Title="How do you prove that a prime is the sum of two squares iff it is congruent to 1 mod 4?" Tags="&lt;number-theory&gt;&lt;prime-numbers&gt;" AnswerCount="4" CommentCount="4" FavoriteCount="4" />
  <row Id="595" PostTypeId="1" AcceptedAnswerId="596" CreationDate="2010-07-23T17:29:53.710" Score="3" ViewCount="349" Body="&lt;p&gt;I'm learning Linear Algebra using MIT's Open Courseware &lt;a href=&quot;http://ocw.mit.edu/courses/mathematics/18-06-linear-algebra-spring-2005/&quot; rel=&quot;nofollow&quot;&gt;Course 18.06&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Quite often, the professor says &quot;... assuming that the matrix is invertible ...&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Somewhere in the lecture he says that using a determinant on an $n \times n$ matrix is on the order of $O(n!)$ operations, where an operation is a multiplication and a subtraction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Is there a more efficient way?&lt;/strong&gt;  If the aim is to get the inverse, rather than just determine the invertibility, what is the most effecient way to do this?&lt;/p&gt;&#xA;" OwnerUserId="38" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-08-04T00:57:20.783" LastActivityDate="2010-08-04T00:57:20.783" Title="What is the most efficient way to determine if a matrix is invertible?" Tags="&lt;linear-algebra&gt;&lt;matrices&gt;" AnswerCount="3" CommentCount="5" />
  <row Id="596" PostTypeId="2" ParentId="595" CreationDate="2010-07-23T17:38:22.477" Score="8" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Gauss%E2%80%93Jordan_elimination&quot; rel=&quot;nofollow&quot;&gt;Gauss-Jordan elimination&lt;/a&gt; can be used to determine when a matrix is invertible and can be done in polynomial (in fact, cubic) time.  The same method (when you apply the opposite row operation to identity matrix) works to calculate the inverse in polynomial time as wel.&lt;/p&gt;&#xA;" OwnerUserId="536" LastActivityDate="2010-07-23T17:38:22.477" CommentCount="5" />
  <row Id="597" PostTypeId="2" ParentId="519" CreationDate="2010-07-23T17:41:22.817" Score="10" ViewCount="" Body="&lt;p&gt;&amp;lt;x,y|x^2=y^3=1&amp;gt;&amp;cong;PSL_2(Z) and this isomorphism identifies G with $PSL_2/T^7=1$ (where $T:z\mapsto z+1$). Result is the symmetry group of &lt;a href=&quot;http://en.wikipedia.org/wiki/(2,3,7)_triangle_group&quot; rel=&quot;nofollow&quot;&gt;the tiling&lt;/a&gt; of the hyperbolic plane. From this description one can see that G is infinite (e.g. because there are infinitely many triangles in the tiling and G acts on them transitively).&lt;/p&gt;&#xA;" OwnerUserId="152" LastEditorUserId="152" LastEditorDisplayName="" LastEditDate="2010-07-30T07:31:43.497" LastActivityDate="2010-07-30T07:31:43.497" CommentCount="2" />
  <row Id="598" PostTypeId="2" ParentId="594" CreationDate="2010-07-23T17:46:06.803" Score="8" ViewCount="" Body="&lt;p&gt;Let $p$ be a prime congruent to 1 mod 4. Then to write $p = x^2 + y^2$ for $x,y$ integers is the same as writing $p = (x+iy)(x-iy) = N(x+iy)$ for $N$ the norm.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is well-known that the ring of Gaussian integers $\mathbb{Z}[i]$ is a principal ideal domain, even a euclidean domain.   Now I claim that $p$ is not prime in $\mathbb{Z}[i]$. To determine how a prime $p$ of $\mathbb{Z}$ splits in $\mathbb{Z}[i]$ is equivalent to determining how the polynomial $X^2+1$ splits modulo $p$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;First off, $-1$ is a quadratic residue modulo $p$ because $p \equiv 1 \mod 4$.  Consequently, there is $t \in \mathbb{Z}$ with $t^2 \equiv -1 \mod p$, so $X^2+1$ splits modulo $p$, and $p$ does not remain prime in $\mathbb{Z}[i]$.  (Another way of seeing this is to note that&#xA;if $p$ remained prime, then we'd have $p \mid (t+i)(t-i)$, which means that $p \mid t+i$ or $t \mid t-i$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, as a result there is a non-unit $x+iy$ of $\mathbb{Z}[i]$ that properly divides $p$.  This means that the norms properly divide as well.  In particular, $N(x+iy) = x^2+y^2$ properly divides $p^2$, so is $p$ or $1$. It cannot be the latter since otherwise $x+iy$ would be a unit.  So $x^2+y^2 = p$.&lt;/p&gt;&#xA;" OwnerUserId="536" LastActivityDate="2010-07-23T17:46:06.803" CommunityOwnedDate="2010-07-23T17:46:32.027" CommentCount="3" />
  <row Id="599" PostTypeId="2" ParentId="595" CreationDate="2010-07-23T17:49:08.667" Score="6" ViewCount="" Body="&lt;p&gt;A matrix is invertible iff its determinant is non-zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are algorithms which find the determinant in &lt;a href=&quot;http://en.wikipedia.org/wiki/Determinant#Algorithmic_implementation&quot; rel=&quot;nofollow&quot;&gt;slightly worse&lt;/a&gt; than O(n&lt;sup&gt;2&lt;/sup&gt;)&lt;/p&gt;&#xA;" OwnerUserId="136" LastActivityDate="2010-07-23T17:49:08.667" CommentCount="5" />
  <row Id="600" PostTypeId="1" AcceptedAnswerId="825" CreationDate="2010-07-23T18:14:56.057" Score="5" ViewCount="336" Body="&lt;p&gt;Given n distinct objects, there are n! permutations of the objects and n!/n &quot;circular permutations&quot; of the objects (orientation of the circle matters, but there is no starting point, so 1234 and 2341 are the same, but 4321 is different).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given n objects of k types (where the objects within each type are indistinguishable), $r_i$ of the ith type, there are $\frac{n!}{r_1!r_2!\cdots r_k!}$ permutations.  How many circular permutations are there of such a set?&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-07-27T22:02:50.807" Title="Circular permutations with indistinguishable objects" Tags="&lt;combinatorics&gt;" AnswerCount="2" CommentCount="6" FavoriteCount="2" />
  <row Id="601" PostTypeId="1" AcceptedAnswerId="602" CreationDate="2010-07-23T18:21:15.327" Score="5" ViewCount="260" Body="&lt;p&gt;From a standard 52-card deck, how many ways are there to pick a hand of $k$ cards that includes one card from all four suits?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that for any specific $k$, it's possible to break it up into cases based on the partitions of $k$ into four parts. For example, if I want to choose a hand of six cards, I can break it up into two cases based on whether there are (1) three cards from one suit and one card from each of the other three or (2) two cards from each of two suits and one card from each of the other two.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a simpler, more general solution that doesn't require splitting the problem into many different cases?&lt;/p&gt;&#xA;" OwnerUserId="45" LastEditorUserId="100" LastEditorDisplayName="" LastEditDate="2010-07-24T06:08:44.547" LastActivityDate="2010-08-05T19:44:31.013" Title="Counting how many hands of cards use all four suits" Tags="&lt;combinatorics&gt;&lt;card-games&gt;" AnswerCount="1" />
  <row Id="602" PostTypeId="2" ParentId="601" CreationDate="2010-07-23T18:38:18.040" Score="7" ViewCount="" Body="&lt;p&gt;Count the number of hands that do not contain at least one card from every suit and subtract from the total number of k-card hands.  To count the number of hands that do not contain at least one card from every suit, use inclusion-exclusion considering what suits are not in a given hand.  That is, letting $N(\dots)$ mean the number of hands meeting the given criteria, $$\begin{align}&#xA;&amp;amp;N(\mathrm{no\ }\heartsuit)+N(\mathrm{no\ }\spadesuit)+N(\mathrm{no\ }\clubsuit)+N(\mathrm{no\ }\diamondsuit)&#xA;\\&#xA;&amp;amp;\quad\quad-N(\mathrm{no\ }\heartsuit\spadesuit)-N(\mathrm{no\ }\heartsuit\clubsuit)-N(\mathrm{no\ }\heartsuit\diamondsuit)-N(\mathrm{no\ }\spadesuit\clubsuit)-N(\mathrm{no\ }\spadesuit\diamondsuit)-N(\mathrm{no\ }\clubsuit\diamondsuit)&#xA;\\&#xA;&amp;amp;\quad\quad+N(\mathrm{no\ }\heartsuit\spadesuit\clubsuit)+N(\mathrm{no\ }\heartsuit\spadesuit\diamondsuit)+N(\mathrm{no\ }\heartsuit\clubsuit\diamondsuit)+N(\mathrm{no\ }\spadesuit\clubsuit\diamondsuit)&#xA;\\&#xA;&amp;amp;\quad\quad-N(\mathrm{no\ }\heartsuit\spadesuit\clubsuit\diamondsuit)&#xA;\\&#xA;&amp;amp;=4{39 \choose k}-6{26 \choose k}+4{13 \choose k}-{0 \choose k}.&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the number of hands of k cards that include at least one card from every suit is $${52 \choose k}-4{39 \choose k}+6{26 \choose k}-4{13 \choose k}+{0 \choose k}.$$  [Drop terms as appropriate for larger values of k.]&lt;/p&gt;&#xA;" OwnerUserId="72" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-08-05T19:44:31.013" LastActivityDate="2010-08-05T19:44:31.013" CommentCount="2" />
  <row Id="603" PostTypeId="1" AcceptedAnswerId="634" CreationDate="2010-07-23T18:40:32.417" Score="1" ViewCount="198" Body="&lt;p&gt;Congruence transformations (isometries) and similarity transformations (isometries + dilations) should be constructable.  What about other affine transformations?  Other conformal mappings?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&lt;strong&gt;edit&lt;/em&gt;&lt;/strong&gt;: by constructable, I mean given the defining information for the transformation in a geometric way (e.g. a dilation requires a center and a ratio, so the given could be a point and two segments), can you construct the image of a point under the transformation from its preimage?&lt;/p&gt;&#xA;" OwnerUserId="72" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2011-03-21T08:25:12.893" LastActivityDate="2011-03-21T08:25:12.893" Title="What transformations of the plane are geometrically constructable (compass &amp; straight edge)?" Tags="&lt;geometry&gt;&lt;euclidean-geometry&gt;&lt;geometric-construction&gt;&lt;compass-and-straightedge&gt;&lt;transformational-geometry&gt;" AnswerCount="2" FavoriteCount="3" />
  <row Id="604" PostTypeId="1" AcceptedAnswerId="605" CreationDate="2010-07-23T19:15:05.187" Score="7" ViewCount="129" Body="&lt;p&gt;For a long time I've eschewed bulky and inelegant calculators for the use of my trusty trig/log-log slide rule.  For those unfamiliar, here is a simple &lt;a href=&quot;http://www.antiquark.com/sliderule/sim/virtual-slide-rule.html&quot; rel=&quot;nofollow&quot;&gt;slide rule simulator using Javascript&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To demonstrate, find the LL_3 scale, which is on the back of the virtual one.  Let's say we want to solve  &lt;code&gt;3^n&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, you would move the cursor (the red line) over where &lt;code&gt;3&lt;/code&gt; is on the LL_3 scale.  Then, you would &lt;em&gt;slide&lt;/em&gt; the middle slider until the &lt;code&gt;1&lt;/code&gt; on the C scale is lined up to the cursor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And voila, your slide rule is set up to find &lt;code&gt;3^n&lt;/code&gt; for any arbitrary &lt;code&gt;n&lt;/code&gt;!  For example, to find &lt;code&gt;3^2&lt;/code&gt;, move the cursor to &lt;code&gt;2&lt;/code&gt; on the C scale, and your answer is what the cursor is on on the LL_3 scale (&lt;code&gt;9&lt;/code&gt;).  Move your cursor to &lt;code&gt;3&lt;/code&gt; on C, and it should be lined up with &lt;code&gt;27&lt;/code&gt; on LL_3.  To &lt;code&gt;4&lt;/code&gt; on C, it is on &lt;code&gt;81&lt;/code&gt; on LL_3.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can even do this for non-integer exponents (1.3, etc.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can also do this for &lt;strong&gt;exponents less than one&lt;/strong&gt;, by using the LL_2 scale.  For example, to do &lt;code&gt;3^0.5&lt;/code&gt;, you would find &lt;code&gt;5&lt;/code&gt; on the C scale, and look where the cursor is lined up at on the LL_2 scale (which is about &lt;code&gt;1.732&lt;/code&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyways, I was wondering if anyone could explain to me how this all works?  It works, but...why?  What property of logarithms and exponents (and logarithms of logarithms?) allows this to work?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I already understand how the basics of the Slide Rule works (&lt;code&gt;ln(m) + ln(n) = ln(mn)&lt;/code&gt;), with only multiplication, but this exponentiation eludes me.&lt;/p&gt;&#xA;" OwnerUserId="51" LastActivityDate="2010-07-23T19:49:20.130" Title="Why does the log-log scale on my Slide Rule work?" Tags="&lt;intuition&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="2" />
  <row Id="605" PostTypeId="2" ParentId="604" CreationDate="2010-07-23T19:49:20.130" Score="4" ViewCount="" Body="&lt;p&gt;If x = 3&lt;sup&gt;n&lt;/sup&gt;, then log x = n log 3.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The C scale is logarithmic, which means if the reading is p, then the distance is proportional to &amp;nbsp; log p.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, in the LLx scale the distance is proportional to &amp;nbsp; log log p.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, when you align 1 to &quot;3&quot; in LL3, you introduce an &lt;em&gt;offset&lt;/em&gt; of (log log 3). Suppose you get a reading of &lt;em&gt;n&lt;/em&gt; in the C scale, then the corresponding value in LL3 would be:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;log log p = log log 3 + log n&#xA;  (LL3)      (offset)    (C)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;eliminating one level of log gives&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;    log p = log 3 * n&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;eliminating one more level of log gives&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;        p = 3^n&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;LL2 is the same as LL3 except it covers a different range.&lt;/p&gt;&#xA;" OwnerUserId="171" LastActivityDate="2010-07-23T19:49:20.130" CommentCount="2" />
  <row Id="606" PostTypeId="2" ParentId="519" CreationDate="2010-07-23T20:15:16.727" Score="11" ViewCount="" Body="&lt;p&gt;Grigory has already answered your particular question.  However, I wanted to point out that your question &quot;How do you prove that a group specified by a presentation is infinite?&quot; has no good answer &lt;em&gt;in general&lt;/em&gt;.  Indeed, in general the question of whether a group presentation defines &lt;em&gt;the trivial group&lt;/em&gt; is undecidable.&lt;/p&gt;&#xA;" OwnerUserId="8" LastActivityDate="2010-07-23T20:15:16.727" CommentCount="2" />
  <row Id="607" PostTypeId="1" AcceptedAnswerId="608" CreationDate="2010-07-23T20:56:22.497" Score="3" ViewCount="248" Body="&lt;p&gt;Assuming I can play forever, what are my chances of coming out ahead in a coin flipping series?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's say I want &quot;heads&quot;...then if I flip once, and get heads, then I win, because I've reached a point where I have more heads than tails (1-0).  If it was tails, I can flip again.  If I'm lucky, and I get two heads in a row after this, this is another way for me to win (2-1).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously, if I can play forever, my chances are probably pretty decent.  They are at least greater than 50%, since I can get that from the first flip.  After that, though, it starts getting sticky.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've drawn a tree graph to try to get to the point where I could start see the formula hopefully dropping out, but so far it's eluding me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your chances of coming out ahead after 1 flip are 50%.  Fine.  Assuming you don't win, you have to flip at least twice more.  This step gives you 1 chance out of 4.  The next level would be after 5 flips, where you have an addtional 2 chances out of 12, followed by 7 flips, giving you 4 out of 40.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suspect I may be able to work through this given some time, but I'd like to see what other people think...is there an easy way to approach this?  Is this a known problem?&lt;/p&gt;&#xA;" OwnerUserId="175" LastActivityDate="2010-08-07T12:29:15.587" Title="Given enough time, what are the chances I can come out ahead in a coin toss contest?" Tags="&lt;probability&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="608" PostTypeId="2" ParentId="607" CreationDate="2010-07-23T21:08:17.693" Score="16" ViewCount="" Body="&lt;p&gt;100%, for the same reason as the &lt;a href=&quot;http://math.stackexchange.com/questions/536/proving-that-1-and-2-d-simple-symmetric-random-walks-return-to-the-origin-with-p&quot; rel=&quot;nofollow&quot;&gt;1-D walk&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact (again for the same reason), your chances are 100% of eventually reaching X-greater heads than tails (or tails than heads), where X is any non-negative integer.&lt;/p&gt;&#xA;" OwnerUserId="136" LastActivityDate="2010-07-23T21:08:17.693" CommentCount="0" />
  <row Id="610" PostTypeId="2" ParentId="600" CreationDate="2010-07-23T22:57:44.907" Score="2" ViewCount="" Body="&lt;p&gt;This problem is best solved with Pólya's enumeration theorem, which follows from Burnside's lemma.  See the first section of &lt;a href=&quot;http://en.wikipedia.org/wiki/P%C3%B3lya_enumeration_theorem&quot; rel=&quot;nofollow&quot;&gt;this Wikipedia article&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="115" LastActivityDate="2010-07-23T22:57:44.907" CommentCount="1" />
  <row Id="611" PostTypeId="2" ParentId="541" CreationDate="2010-07-23T23:25:35.780" Score="5" ViewCount="" Body="&lt;p&gt;If you're after Olympiad-level books, get &lt;a href=&quot;http://books.google.com/books?id=rVvr2p76cKQC&amp;amp;printsec=frontcover&quot; rel=&quot;nofollow&quot;&gt;The IMO Compendium&lt;/a&gt; which is a collection of problems from the International Math Olympiad, 1959-2004. You can find similar books with national Olympiad problems by going to Amazon and searching for &quot;mathematical Olympiad&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Two books that offer collections of techniques useful for olympiad-level contests are Paul Zeitz's &lt;a href=&quot;http://books.google.com/books?id=Go_iAAAACAAJ&amp;amp;printsec=frontcover&quot; rel=&quot;nofollow&quot;&gt;The Art and Craft of Problem Solving&lt;/a&gt; and Arthur Engel's &lt;a href=&quot;http://books.google.com/books?id=B3EYPeKViAwC&amp;amp;printsec=frontcover&quot; rel=&quot;nofollow&quot;&gt;Problem Solving Strategies&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are lots of other books with similar titles and descriptions. Just follow Google Books's suggestions.&lt;/p&gt;&#xA;" OwnerUserId="115" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2011-05-07T18:43:48.827" LastActivityDate="2011-05-07T18:43:48.827" CommunityOwnedDate="2010-07-23T23:25:35.780" />
  <row Id="612" PostTypeId="1" AcceptedAnswerId="615" CreationDate="2010-07-23T23:38:24.513" Score="4" ViewCount="134" Body="&lt;p&gt;Are there any results from algebraic geometry that have led to an interesting &quot;real world&quot; application? &lt;/p&gt;&#xA;" OwnerUserId="99" LastActivityDate="2010-07-24T00:52:55.863" CommunityOwnedDate="2010-07-24T01:15:56.183" Title="What are some applications outside of mathematics for algebraic geometry?" Tags="&lt;algebraic-geometry&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="1" />
  <row Id="613" PostTypeId="2" ParentId="612" CreationDate="2010-07-23T23:47:06.493" Score="3" ViewCount="" Body="&lt;p&gt;Broadly speaking, algebraic geometry is used a lot in some areas of robotics and mechanical engineering. Real algebraic geometry, for example, is important to the development of CAD systems (think NURBS, computing intersections of primitives, etc.) And AG comes up in robotics when it is important to figure out, say, what motions a robotic arm in a given configuration is capable of, or to construct some kind of linkage that draws a prescribed curve.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Something specific in that vein: Kempe's Universality Theorem gives that any bounded algebraic curve in $\mathbb{R}^2$ is the locus of some linkage. The &quot;locus of a linkage&quot; being the path drawn out by all the vertices of a graph, where the edge lengths are all specified and one or more vertices remains still.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Interestingly, Kempe's orginal proof of the theorem was flawed, and more recent proofs have been more involved. However, &lt;a href=&quot;http://erikdemaine.org/theses/tabbott.pdf&quot; rel=&quot;nofollow&quot;&gt;Timothy Abbott's MIT masters thesis&lt;/a&gt; gives a simpler proof that gives a working linkage for a given curve, and makes for interesting reading concerning the problem in general.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: The NURBS connection is, in part, that can construct a B-spline that approximates a given real algebraic curve, which is crucial in displaying intersection curves, for example. See &lt;a href=&quot;http://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.51.3585&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; for more details (I'm afraid I don't know many on this.)&lt;/p&gt;&#xA;" OwnerUserId="76" LastEditorUserId="76" LastEditorDisplayName="" LastEditDate="2010-07-24T00:52:55.863" LastActivityDate="2010-07-24T00:52:55.863" CommentCount="1" />
  <row Id="614" PostTypeId="1" AcceptedAnswerId="616" CreationDate="2010-07-24T00:06:42.180" Score="1" ViewCount="95" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;A 10-year loan of $500 is repaid with&#xA;  payments at the end of each year. The lender charges interest at an&#xA;  annual effective rate of 10%. &#xA;  Each of the first ten payments is&#xA;  150% of the amount of interest due. &#xA;  Each of the last ten payments is X. &#xA;  Calculate X.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I came across this practice question while studying for my actuary exam. I tried it on my own and got stuck:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\$ $500 will earn $\$ $50 interest each year, so each of the first 10 payments must be $\$ $75.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then after 10 years, a total of 750 has been repaid.&lt;br&gt;&#xA;In 10 years, I can find the accumulated debt by saying &lt;code&gt;PV&lt;/code&gt;=500, &lt;code&gt;I/Y&lt;/code&gt;=10, &lt;code&gt;N&lt;/code&gt;=10, giving me &lt;code&gt;FV&lt;/code&gt;=$\$ $1296.87. So the balance would be $\$ $1296.87-$\$ $750=$\$ $546.97.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I am stuck! How do I find out what the last payments should be? I know I can't just divide $\$ $546.97/10=$\$ $54.697, because the lender is still charging interest while the borrower pays off this remaining debt, so there would still be the interest left over.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This situation isn't mentioned anywhere in my calculator manual! Can one of you give me some explanation about what is going on so that I can do it by hand?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I tried working on it some more, and came up with a really great idea! Since the payments are 1.5x the 10% interest, it's just like paying off 5% of the principal each year! This saves me a lot of time, because I can just set &lt;code&gt;I/Y=-5&lt;/code&gt; and get &lt;code&gt;FV=598.74&lt;/code&gt; on my calculator. I did it the long way by calculating the future value of each interest payment (turns out they were not all $75, because the outstanding principal got smaller), and they were the same. Is this always going to work, or did I just get lucky here?  &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Another update!&lt;br&gt;&#xA;I think I solved it. All I needed to do was to set &lt;code&gt;FV=0&lt;/code&gt;, &lt;code&gt;I/Y=10&lt;/code&gt;, &lt;code&gt;N=10&lt;/code&gt;, &lt;code&gt;PV=598.74&lt;/code&gt;, and then I got &lt;code&gt;PMT=97.44&lt;/code&gt;. I never used the &lt;code&gt;PMT&lt;/code&gt; button before, though, so is there some other way I can check the answer is right?&lt;/p&gt;&#xA;" OwnerUserId="73" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-05-31T20:04:21.637" LastActivityDate="2011-05-31T20:04:21.637" Title="How to determine annual payments on a partially repaid loan?" Tags="&lt;finance&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="615" PostTypeId="2" ParentId="612" CreationDate="2010-07-24T00:08:18.137" Score="2" ViewCount="" Body="&lt;p&gt;The following &lt;a href=&quot;http://www.dms.uaf.edu/~eallman/mega.pdf&quot; rel=&quot;nofollow&quot;&gt;slideshow&lt;/a&gt; gives an explanation of how algebraic geometry can be used in phylogenetics.&#xA;See also &lt;a href=&quot;http://rigtriv.wordpress.com/2008/08/07/phylogenetics-and-algebraic-geometry/&quot; rel=&quot;nofollow&quot;&gt;this post&lt;/a&gt; of Charles Siegel on Rigorous Trivialties. This is not an area I've looked at in much detail at all, but it appears that the idea is to use a graph to model evolutionary processes, and such that the &quot;transition function&quot; for these processes is given by a polynomial map.  In particular, it'd be of interest to look at the potential outcomes, namely the image of the transition function; that corresponds to the image of a polynomial map (which is not necessarily an algebraic variety, but it is a constructible set, so not that badly behaved either).  (In practice, though, it seems that one studies the closure, which is a legitimate algebraic set.)&lt;/p&gt;&#xA;" OwnerUserId="536" LastActivityDate="2010-07-24T00:08:18.137" CommentCount="2" />
  <row Id="616" PostTypeId="2" ParentId="614" CreationDate="2010-07-24T00:24:01.380" Score="2" ViewCount="" Body="&lt;p&gt;This problem is in two stages. For the first stage, notice that you are paying 150% interest, but ending up owing more. This is because you subtracted $\$ $750 from the future value, when in fact each $\$ $75 amount was paid at a time in the past and needs be converted to a future value too. The payment in the nth year has a future value of $75\times(1.1)^{10-n}$. The total future value of the repayment is:&#xA;$$75(1.1^9)+75(1.1^8)+\cdots+75(1.1^0).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that I have assumed that the interest is charged before the repayments are made. This sequence is a &lt;a href=&quot;http://en.wikipedia.org/wiki/Geometric_progression&quot; rel=&quot;nofollow&quot;&gt;geometric progression&lt;/a&gt;. We consider it as a geometric sequence in reverse to make the maths easier. It has first term ($a$) 75, each term 1.1 times the previous ($r$) and 10 terms ($n$). The sum is given by the formula:&#xA;$$\frac{a{r^{n-1}}}{r-1}=\frac{75{1.1^{10-1}}}{0.1}\approx\$1195.31$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After we have solved this first part, then it is just a standard &lt;a href=&quot;http://math.stackexchange.com/questions/618/repayments-of-a-loan-with-compound-interest/685#685&quot;&gt;interest with repayments problem.&lt;/a&gt;. &lt;/p&gt;&#xA;" OwnerUserId="123" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-05-31T20:00:03.727" LastActivityDate="2011-05-31T20:00:03.727" CommentCount="6" />
  <row Id="617" PostTypeId="2" ParentId="607" CreationDate="2010-07-24T00:38:09.387" Score="5" ViewCount="" Body="&lt;p&gt;The question can be answered using &lt;a href=&quot;http://en.wikipedia.org/wiki/Catalan_number&quot; rel=&quot;nofollow&quot;&gt;Catalan numbers&lt;/a&gt;. Let C_n denote the number of sequences of 2n coin tosses in which you are never ahead. Formally, we count sequences in which every prefix has no less &lt;code&gt;T&lt;/code&gt;'s than &lt;code&gt;H&lt;/code&gt;'s. We call this &lt;em&gt;property A&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The number of total sequences of length 2n is $2^{2n}$. We then show that as n→∞, the ratio $C_n / 2^{2n}$ tends to 0. This means that in almost every sequence you will eventually be ahead (the chances of a random sequence having property A tend to 0 as the sequence gets longer).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Indeed,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$C_n = \frac{(2n)!}{(n+1)!n!}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$C_n / 2^{2n} = \frac{(2n)!}{2^{2n}} \cdot \frac{1}{(n+1)!n!}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and it can be shown that this tends to 0 by &lt;a href=&quot;http://en.wikipedia.org/wiki/Stirling%27s_approximation&quot; rel=&quot;nofollow&quot;&gt;Stirling's approximation&lt;/a&gt; (multiply and divide by $(2n/e)^{2n}$).&lt;/p&gt;&#xA;" OwnerUserId="26" LastEditorUserId="26" LastEditorDisplayName="" LastEditDate="2010-08-07T12:29:15.587" LastActivityDate="2010-08-07T12:29:15.587" />
  <row Id="618" PostTypeId="1" AcceptedAnswerId="685" CreationDate="2010-07-24T01:10:30.970" Score="1" ViewCount="92" Body="&lt;p&gt;Suppose I have a loan of M dollars. At the end of each year, I am charged interest at rate R and make a repayment of P. The loan is repaid after n years.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;How long (n) does it take to repay the loan if I am given the other variables?&lt;/li&gt;&#xA;&lt;li&gt;How much are the repayments of P if I am given the other variable?&lt;/li&gt;&#xA;&lt;li&gt;Suppose that the payments were at the start of the year. How would this change the problem?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="123" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-07-24T18:37:49.753" LastActivityDate="2011-05-31T19:38:59.490" Title="Repayments of a loan with compound interest" Tags="&lt;finance&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="1" />
  <row Id="619" PostTypeId="1" AcceptedAnswerId="620" CreationDate="2010-07-24T01:22:34.420" Score="8" ViewCount="224" Body="&lt;p&gt;&lt;em&gt;The Problem:&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Two trains travel on the same track towards each other, each going at a speed of 50 kph. They start out 50km apart. A fly starts at the front of one train and flies at 75 kph to the front of the other; when it gets there, it turns around and flies back towards the first. It continues flying back and forth til the two trains meet and it gets squashed (the least of our worries, perhaps).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How far did the fly travel before it got squashed?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Attempt at a solution:&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can do this by summing the infinite series of the fly's distance for each leg. I get an answer of 37.5 km: but that's so nice! There must be a more intuitive way...is there?&lt;/p&gt;&#xA;" OwnerUserId="76" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2011-01-17T00:58:09.950" LastActivityDate="2011-01-17T00:58:09.950" Title="Problem: Two Trains and a Fly" Tags="&lt;sequences-and-series&gt;&lt;high-school&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="620" PostTypeId="2" ParentId="619" CreationDate="2010-07-24T01:30:46.100" Score="11" ViewCount="" Body="&lt;p&gt;The trains take half an hour to collide, which, at a rate of 75kph, leads to the fly travelling 37.5km.&lt;/p&gt;&#xA;" OwnerUserId="123" LastActivityDate="2010-07-24T01:30:46.100" CommentCount="1" />
  <row Id="621" PostTypeId="2" ParentId="618" CreationDate="2010-07-24T02:20:09.947" Score="2" ViewCount="" Body="&lt;p&gt;Suggest you look into &lt;a href=&quot;http://en.wikipedia.org/wiki/Amortization_calculator&quot; rel=&quot;nofollow&quot;&gt;amortization&lt;/a&gt; calculators.&lt;/p&gt;&#xA;" OwnerUserId="120" LastActivityDate="2010-07-24T02:20:09.947" CommentCount="1" />
  <row Id="622" PostTypeId="1" CreationDate="2010-07-24T02:42:29.813" Score="27" ViewCount="985" Body="&lt;p&gt;Representation theory is a subject I want to like (it can be fun finding the representations of a group), but it's hard for me to see it as a subject that arises naturally or why it is important.  I can think of two mathematical reasons for studying it:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) The character table of a group is packs a lot of information about the group and is concise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) It is practically/computationally nice to have explicit matrices that model a group.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But there are for sure deeper things that I am missing.  I can understand why one would want to study group actions (the axioms for a group beg you to think of elements as operators), but why look at group actions on vector spaces?  Is it because linear algebra is so easy/well-known (when compared to just modules, say)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am also told that representation theory is important in quantum mechanics.  For example, physics should be SO(3) invariant and when we represent this on a Hilbert space of wave-functions, we are led to information about angular momentum.  But this seems to only trivially invoke representation theory since we already start with a subgroup of GL(n) and then extend it to act on wave functions by $\psi(x,t) \mapsto \psi(Ax,t)$ for A in SO(n).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This &lt;a href=&quot;http://en.wikipedia.org/wiki/Particle_physics_and_representation_theory&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Particle_physics_and_representation_theory&lt;/a&gt; wikipedia article claims that if our physical system has G as a symmetry group, then there is a correspondence between particles and representations of G.  I'm not sure if I understand this correspondence since it seems to be saying that if we act an element of G on a state that corresponds to some particle, then this new state also corresponds to the same particle.  So a particle is an orbit of the G action?  Anyone know of good sources that talk about this?&lt;/p&gt;&#xA;" OwnerUserId="9" LastActivityDate="2011-02-23T22:14:44.497" Title="Importance of Representation Theory" Tags="&lt;representation-theory&gt;&lt;physics&gt;" AnswerCount="8" CommentCount="5" FavoriteCount="13" />
  <row Id="623" PostTypeId="1" AcceptedAnswerId="644" CreationDate="2010-07-24T02:54:51.733" Score="8" ViewCount="750" Body="&lt;p&gt;The volume of a cone with height $h$ and radius $r$ is $\frac{1}{3} \pi r^2 h$, which is exactly one third the volume of the smallest cylinder that it fits inside.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This can be proved easily by considering a cone as a &lt;a href=&quot;http://en.wikipedia.org/wiki/Solid_of_revolution&quot; rel=&quot;nofollow&quot;&gt;solid of revolution&lt;/a&gt;, but I would like to know if it can be proved or at least visual demonstrated without using calculus.&lt;/p&gt;&#xA;" OwnerUserId="106" LastActivityDate="2011-03-28T19:05:30.977" Title="Why is the volume of a cone one third of the volume of a cylinder?" Tags="&lt;geometry&gt;" AnswerCount="5" CommentCount="4" FavoriteCount="2" />
  <row Id="624" PostTypeId="1" AcceptedAnswerId="629" CreationDate="2010-07-24T03:00:59.880" Score="7" ViewCount="208" Body="&lt;p&gt;In my AP chem class, I often have to balance chemical equations like the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Al + O&lt;sub&gt;2&lt;/sub&gt; ---&gt; Al&lt;sub&gt;2&lt;/sub&gt;O&lt;sub&gt;3&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The goal is to make both side of the arrow have the same amount of atoms by adding compounds in the equation to each side.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An solution&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4Al + 3O&lt;sub&gt;2&lt;/sub&gt; ---&gt; 2Al&lt;sub&gt;2&lt;/sub&gt;O&lt;sub&gt;3&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the subscripts become really large, or there are lot of atoms involved, trial and error is impossible unless performed by a computer. What if some chemical equation can not be balanced(does such equations exist)? I tried one for a long time only to realize the problem was wrong.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My teacher said trial and error is the only way. Are there other methods?&lt;/p&gt;&#xA;" OwnerUserId="96" LastEditorUserId="96" LastEditorDisplayName="" LastEditDate="2010-07-25T02:30:21.940" LastActivityDate="2010-07-25T03:11:29.843" Title="Balance chemical equations without trial and error?" Tags="&lt;chemistry&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" />
  <row Id="625" PostTypeId="1" CreationDate="2010-07-24T03:01:09.037" Score="8" ViewCount="446" Body="&lt;p&gt;When differentiated with respect to $r$, the derivative of $\pi r^2$ is $2 \pi r$, which is the circumference of a circle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, when the formula for a sphere's volume $\frac{4}{3} \pi r^3$ is differentiated with respect to $r$, we get $4 \pi r^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this just a coincidence, or is there some deep explanation for why we should expect this?&lt;/p&gt;&#xA;" OwnerUserId="106" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-07-24T03:03:29.950" LastActivityDate="2010-07-24T16:59:38.733" Title="Why is the derivative of a circle's area its perimeter (and similarly for spheres)?" Tags="&lt;geometry&gt;&lt;calculus&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="3" />
  <row Id="626" PostTypeId="2" ParentId="622" CreationDate="2010-07-24T03:16:11.290" Score="14" ViewCount="" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The representation theory of finite groups can be used to prove results about finite groups themselves that are otherwise much harder to prove by &quot;elementary&quot; means. For instance, the proof of &lt;a href=&quot;http://en.wikipedia.org/wiki/Burnside%27s_theorem&quot; rel=&quot;nofollow&quot;&gt;Burnside's theorem&lt;/a&gt; (that a group of order $p^a q^b$ is solvable).  A lot of the classification proof of finite simple groups relies on representation theory (or so I'm told, I haven't read the proof...).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Mathematical physics. Lie algebras and Lie groups definitely come up here, but I'm not familiar enough to explain anything. In addition, the classification of complex simple Lie algebras relies on the root space decomposition, which is a significant (and nontrivial) fact about the representation theory of semisimple Lie algebras.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Number theory.  The nonabelian version of L-functions (Artin L-functions) rely on the representations of the Galois group (in the abelian case, these just correspond to sums of 1-dimensional characters).  For instance, the proof that Artin L-functions are meromorphic in the whole plane relies on (I think) &lt;del&gt;Artin&lt;/del&gt; Brauer's theorem (i.e., a corollary of the usual statement) that any irreducible character is an &lt;del&gt;rational&lt;/del&gt; integer combination of induced characters from cyclic subgroups -- this is in Serre's &lt;em&gt;Linear Representations of Finite Groups.&lt;/em&gt; Also, the Langlands program studies  representations of groups $GL_n(\mathbb{A}_K)$ for $\mathbb{A}_K$ the adele ring of a global field. This is a generalization of standard &quot;abelian&quot; class field theory (when $n=1$ and one is determining the character group of the ideles).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Combinatorics. The representation theory of the symmetric group has a lot of connections to combinatorics, because you can parametrize the irreducibles explicitly (via Young diagrams), and this leads to the problem of determining how these Young diagrams interact. For instance, what does the tensor product of two Young diagrams look like when decomposed as a sum of Young diagrams? What is the dimension of the irreducible representation associated to a Young diagram? These problems have a combinatorial flavor.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I should add the disclaimer that I have not formally studied representation theory, and these are likely to be an unrepresentative sample of topics (some of which I have only vaguely heard about).&lt;/p&gt;&#xA;" OwnerUserId="536" LastEditorUserId="536" LastEditorDisplayName="" LastEditDate="2010-08-18T11:34:01.253" LastActivityDate="2010-08-18T11:34:01.253" CommentCount="4" />
  <row Id="627" PostTypeId="2" ParentId="625" CreationDate="2010-07-24T03:23:06.523" Score="1" ViewCount="" Body="&lt;p&gt;Because you use the integral (read: anti-derivative) to find the area under the curve - even a curve in polar coordinates.&lt;/p&gt;&#xA;" OwnerUserId="136" LastEditorUserId="136" LastEditorDisplayName="" LastEditDate="2010-07-24T16:59:38.733" LastActivityDate="2010-07-24T16:59:38.733" CommentCount="4" />
  <row Id="628" PostTypeId="2" ParentId="625" CreationDate="2010-07-24T03:32:36.620" Score="10" ViewCount="" Body="&lt;p&gt;Consider increasing the radius of a circle by an infinitesimally small amount, dx. This increases the area by an &lt;a href=&quot;http://en.wikipedia.org/wiki/Annulus_%28mathematics%29&quot; rel=&quot;nofollow&quot;&gt;annulus&lt;/a&gt; (or ring) with inner radius 2pi*r and outer radius 2pi*(r+dx). At this ring is extremely thin, we can imagine cutting the ring and then flattening it out to form a rectangle with width 2pi*r and height dx (the side of length 2pi*(r+dx) is close enough to 2pi*r that we can ignore that). So the area gain is 2pi*r*dx and to determine the rate of change with respect to x, we divide by dx and so we get 2pi*r. This is of course, just an informative, intuitive explanation and not a formal proof. The same reasoning works with a sphere, we just flatten it out to a rectangular prism instead.&lt;/p&gt;&#xA;" OwnerUserId="123" LastActivityDate="2010-07-24T03:32:36.620" />
  <row Id="629" PostTypeId="2" ParentId="624" CreationDate="2010-07-24T03:34:46.620" Score="8" ViewCount="" Body="&lt;p&gt;Yes; it's possible to write a system of equations that can be solved to find the correct coefficients. Here's an example for the given formula.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We're trying to find coefficients A, B, and C such that&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;A (Al) + B (O_2) -&amp;gt; C (Al_2 O_3)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;In order to do this, we can write an equation for each element based on how many atoms are on each side of the equation.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;for Al:  A = 2C&#xA;for O:  2B = 3C&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This is an uninteresting example, but these will always be linear equations in terms of the coefficients. Note that we have fewer equations than variables. This means that there's more than one way to correctly balance the equation (and there is, because any set of coefficients can be scaled by any factor). We just need to find one integral solution to these equations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To solve, we can arbitrarily set one of the variables to 1 and we'll get a solution with (probably fractional) coefficients. If we add $A=1$, the solution is $(A,B,C) = (1,3/4,1/2)$. To get the smallest solution with integer coefficients, just multiply by the least common multiple of the denominators (4 in this case), giving us $(4,3,2)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the set of equations has no solution where the coefficients are nonzero, then you know that the equation cannot be balanced.&lt;/p&gt;&#xA;" OwnerUserId="45" LastEditorUserId="45" LastEditorDisplayName="" LastEditDate="2010-07-25T03:11:29.843" LastActivityDate="2010-07-25T03:11:29.843" CommentCount="4" />
  <row Id="631" PostTypeId="1" CreationDate="2010-07-24T04:46:02.207" Score="26" ViewCount="848" Body="&lt;p&gt;The 24 game is as follows.  Four numbers are drawn; the player's objective is to make 24 from the four numbers using the four basic arithmetic operations (in any order) and parentheses however one pleases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the following generalization. Given n+1 numbers, determine whether the last one can be obtained from the first n using elementary arithmetical operations as above. This problem admits succint certificates so is in NP.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it NP-complete?&lt;/p&gt;&#xA;" OwnerUserId="536" LastEditorUserId="536" LastEditorDisplayName="" LastEditDate="2010-07-24T13:10:37.653" LastActivityDate="2010-08-04T00:14:22.177" Title="Is the 24 game NP-complete?" Tags="&lt;computer-science&gt;" AnswerCount="3" CommentCount="8" FavoriteCount="12" />
  <row Id="632" PostTypeId="1" AcceptedAnswerId="1245" CreationDate="2010-07-24T05:07:01.957" Score="5" ViewCount="228" Body="&lt;p&gt;I am working on computing phase diagrams for alloys.  These are&#xA;blueprints for a material that show what phase, or combination of&#xA;phases, a material will exist in for a range of concentrations and&#xA;temperatures (see &lt;a&#xA;href=&quot;http://web.cos.gmu.edu/~tstephe3/talks/SIAMMaterialsScience2010.pdf&quot; rel=&quot;nofollow&quot;&gt;this&#xA;pdf presentation&lt;/a&gt;).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The crucial step in drawing the boundaries that separate one phase&#xA;from another on these diagrams involves minimizing a free energy&#xA;function subject to basic physical conservation constraints.  I am&#xA;going to leave out the chemistry/physics and hope that we can move forward&#xA;with the minimization using Lagrange multipliers. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The free energy that is to be minimized is this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\widetilde{G}(x_1, x_2) = f^{(1)}G_{1}(x_1) + f^{(2)}G_{2}(x_2),$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;subject to:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f^{(1)}x_1 + f^{(2)}x_2 = c_1,$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f^{(1)} + f^{(2)} = 1. $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(and also that the $x_{i} &gt; 0$ and $f^{(i)} &gt; 0$, for $i=1,2$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Lagrange formulation is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$L(x_1,x_2,f^{(1)},f^{(2)},\lambda_1, \lambda_2, \lambda_3) =&#xA;f^{(1)}G_{1}(x_1) + f^{(2)}G_{2}(x_2)$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$- \lambda_{1}(f^{(1)}x_1 + f^{(2)}x_2 - c_1)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$- \lambda_{2}(f^{(1)} + f^{(2)} - 1) $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The minimization of $\widetilde{G}$ follows from finding the $x_{i}$'s  that satisfy $\nabla L = 0:$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{\partial L}{\partial x_{1}}   = f^{(1)}G_{1}'(x_1) - \lambda_{1}f^{(1)} = 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{\partial L}{\partial x_2}     = f^{(2)}G_{2}'(x_2) - \lambda_{1}f^{(2)} = 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{\partial L}{\partial f^{(1)}} = G_{1}(x_1) - \lambda_{1}x_{1} - \lambda_2 = 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{\partial L}{\partial f^{(2)}} = G_{2}(x_2) - \lambda_{1}x_{2} - \lambda_2 = 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which yields:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(*) f^{(1)}\left[G_{1}'(x_1) - \lambda_1 \right] = 0$       &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(**) f^{(2)}\left[G_{2}'(x_2) - \lambda_1 \right]= 0 $       &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(***) G_{1}(x_1) - G_{2}(x_2) = \lambda_1 \left[ x_1 - x_2\right]$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because $f^{(1)}$ and $f^{(2)}$ are not to be zero, from (*) and (**) we have that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$G_{1}'(x_1) = G_{2}'(x_2) = \lambda_{1}.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And, a manipulation of equation (***) looks like &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{G_{1}(x_1) -G_{2}(x_2)}{x_1 - x_2} = \lambda_{1}.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, think of $G_{i}$ as an even degree polynomial (which it isn't, but&#xA;it's graph sometimes resembles one) in the plane.  Let the points $x_1$&#xA;and $x_2$ be locations along the x-axis that lie roughly below the&#xA;minima of this curve.  The constraints (*),(*&lt;em&gt;), and (*&lt;/em&gt;*) describe the&#xA;condition that the line drawn between $(x_1,G_{1}(x_1))$ and $(x_2,G_{2}(x_2))$ form a common tangent&#xA;to the &quot;wells&quot; of the curve.  It is these points $x_1$ and $x_2$,&#xA;which represent concentrations of pure components in our alloy, that&#xA;become mapped onto a phase diagram.  It is essentially by repeating this procedure for many&#xA;temperatures that we can trace out the boundaries in the desired phase diagram.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The question is:&lt;/strong&gt;  Looking at this from a purely analytic geometry&#xA;perspective, &lt;strong&gt;how would one derive the &quot;variational&quot; approach to find a common tangent line that we seem to have found using the above Lagrangian?&lt;/strong&gt;  (warning: I don't really know how to&#xA;model things using variational methods.)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;And, secondly:&lt;/strong&gt; I have presented a model of a binary alloy, meaning&#xA;two variables to keep track of representing concentrations.  I have&#xA;been working on ternary alloys, where this free energy $\widetilde{G}$&#xA;is a function of three variables (two independent: $x_1,x_2,x_3$,&#xA;where $x_3 = 1- x_1 - x_2$) and is therefore a surface over a Gibbs&#xA;triangle.  Then $\nabla L = 0$ produces partial derivatives that no&#xA;longer &quot;speak geometry&quot; to me, although the solution is a common tangent&#xA;plane.  (I have attempted to characterize a common tangent plane&#xA;based purely in analytic geometry - completely disregarding the&#xA;Lagrangian - and have come up with several relations between&#xA;directional derivatives... &lt;strong&gt;How might directional derivatives relate&#xA;to the optimality conditions set forth by the Lagrangian?&lt;/strong&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt;  Thank you Greg Graviton for wading through this sub-optimal notation and pointing out several mistakes in the statement of the problem.  (Also, thank you for the &lt;a href=&quot;http://math.stackexchange.com/questions/632/validating-a-mathematical-model-lagrange-formulation-and-geometry/1245#1245&quot;&gt;excellent discussion below&lt;/a&gt;.)&lt;/p&gt;&#xA;" OwnerUserId="121" LastEditorUserId="121" LastEditorDisplayName="" LastEditDate="2010-08-01T18:29:44.957" LastActivityDate="2010-08-01T18:29:44.957" Title="validating a mathematical model (Lagrange formulation and geometry)" Tags="&lt;calculus&gt;&lt;applications&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="633" PostTypeId="2" ParentId="623" CreationDate="2010-07-24T05:55:46.600" Score="1" ViewCount="" Body="&lt;p&gt;You can use Pappus's centroid theorem &lt;a href=&quot;http://math.stackexchange.com/questions/164/why-is-the-volume-of-a-sphere-frac43-pi-r3/174#174&quot; rel=&quot;nofollow&quot;&gt;as in my answer here&lt;/a&gt;, but it does not provide much insight.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If instead of a cylinder and a cone, you consider a cube and a square-based pyramid where the &quot;top&quot; vertex of the pyramid (the one opposite the square base) is shifted to be directly above one vertex of the base, you can fit three such pyramids together to form the complete cube.  (I've seen this as physical toy/puzzle with three pyramidal pieces and a cubic container.)  This may give some insight into the 1/3 &quot;pointy thing rule&quot; (for pointy things with similar, linearly-related cross-sections) that Katie Banks discussed in her comment.&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-07-24T05:55:46.600" />
  <row Id="634" PostTypeId="2" ParentId="603" CreationDate="2010-07-24T06:14:57.360" Score="2" ViewCount="" Body="&lt;p&gt;&lt;em&gt;&lt;strong&gt;edit&lt;/em&gt;&lt;/strong&gt;: The question is much more involved than I'd originally thought.  As implied in the question, I knew that congruence and similarity transformations are constructible.  Immediately below this section is my original answer, which only demonstrates congruence transformations and was intended more to give an idea of what an answer might look like (since, at the time, there was another answer that was not particularly helpful).  In the last section of this answer is my justification that all affine transformations of the plane are constructible.  In re-reading that now, I realize that I'd assumed the ability to construct a point, say P', on a line, say RP, such that RP'/RP is equal to some known ratio.  This is equivalent to being able to construct the dilation of P by the known ratio about center R.  I've added the construction of such a dilation below the congruence transformation section.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;As a partial answer, here are constructions of the basic congruence transformations, assuming basic construction techniques like constructing a line parallel or perpendicular to a given line through a given point and angle-copying:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;reflection&lt;/strong&gt;  Given a point P and a line l, construct the line perpendicular to l through P, and construct the circle centered at the intersection of this new line and l and passing through P.  The image of P under a reflection over the line l is the point of intersection of the circle and the new line (the one not at P).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;translation&lt;/strong&gt;  Given a point P and a vector AB (from A to B), construct the line through A and P, the line through B parallel to line PA, and the line through P parallel to AB.  The image of P under translation by vector AB is the intersection of the two constructed parallels.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;rotation&lt;/strong&gt;  Given a point P, a center of rotation R, and an angle ABC (from A to C), construct the line through P and R, copy angle ABC such that the copy of B coincides with R and the copy of A is on ray RP and let the copy of C be called D, construct the circle with center at R and passing through P.  The image of P under rotation by angle ABC about point R is the intersection of the circle with ray RD.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;dilation&lt;/strong&gt; Given a point P, a center of dilation R, and a ratio AC/AB (where point C lies on ray AB), translate B to B' and C to C' by the translation that takes A to R, construct line B'P, construct the line through C' parallel to B'P, and construct ray RP.  The image of P under a dilation about R by a factor of AC/AB is the intersection of ray RP and the line through C' parallel to B'P.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;All &lt;strong&gt;affine transformations&lt;/strong&gt; are constructible.  Per &lt;a href=&quot;http://mathworld.wolfram.com/AffineTransformation.html&quot; rel=&quot;nofollow&quot;&gt;MathWorld&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/Affine_transformation#Affine_transformation_of_the_plane&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;, an affine transformation of the plane is a transformation of the plane that preserves collinearity and preserves ratios of distances on any given line.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, to show that affine transformations preserve parallelism, suppose that two lines MN and PQ are parallel, and that their images, lines M'N' and P'Q', intersect at X', the image of X.  Since affine transformations preserve collinearity, X must be on MN and on PQ, which is a contradiction, so M'N' and P'Q' cannot intersect.  Thus, affine transformations preserve parallelism.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An affine transformation is determined by a triangle ABC and its image, triangle A'B'C' (per &lt;a href=&quot;http://mathworld.wolfram.com/AffineTransformation.html&quot; rel=&quot;nofollow&quot;&gt;MathWorld&lt;/a&gt;; &lt;a href=&quot;http://en.wikipedia.org/wiki/Affine_transformation#Affine_transformation_of_the_plane&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt; talks about defining an affine transformation by a parallelogram and its image, which is equivalent since affine transformations preserve parallelism).  Given point P and triangles ABC and A'B'C', construct line l_1 through P parallel to AB and line l_2 through P and parallel to AC, call the intersection of l_1 with AC I_1 and call the intersection of l_2 with AB I_2, extend A'B' past B' to a point I'_2 such that AB/AI_2 = A'B'/A'I'_2, extend A'C' past C' to a point I'_1 such that AC/AI_1 = A'C'/A'I'_1, construct line l'_1 through I'_1 parallel to A'B' and line l'_2 through I'_2 parallel to A'C'.  The image of P under the affine transformation mapping ABC onto A'B'C' is the intersection of lines l'_1 and l'_2.&lt;/p&gt;&#xA;" OwnerUserId="72" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-07-26T01:34:29.577" LastActivityDate="2010-07-26T01:34:29.577" CommentCount="4" />
  <row Id="635" PostTypeId="2" ParentId="623" CreationDate="2010-07-24T07:30:06.133" Score="5" ViewCount="" Body="&lt;p&gt;One can cut a cube into 3 pyramids with square bases -- so for such pyramids the volume is indeed 1/3 hS. And then one uses &lt;a href=&quot;http://en.wikipedia.org/wiki/Cavalieris_principle&quot; rel=&quot;nofollow&quot;&gt;Cavalieri's principle&lt;/a&gt; to prove that the volume of any cone is 1/3 hS.&lt;/p&gt;&#xA;" OwnerUserId="152" LastActivityDate="2010-07-24T07:30:06.133" />
  <row Id="636" PostTypeId="1" AcceptedAnswerId="919" CreationDate="2010-07-24T07:36:50.767" Score="9" ViewCount="298" Body="&lt;p&gt;The knight's tour is a sequence of 64 squares on a chess board, where each square is visted once, and each subsequent square can be reached from the previous by a knight's move.  Tours can be cyclic, if the last square is a knight's move away from the first, and acyclic otherwise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are several symmetries among knight's tours.  Both acyclic and cyclic tours have eight reflectional symmetries, and cyclic tours additionally have symmetries arising from starting at any square in the cycle, and from running the sequence backwards.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it known how many knight's tours there are, up to all the symmetries?&lt;/p&gt;&#xA;" OwnerUserId="100" LastActivityDate="2011-01-26T03:11:09.247" Title="How many knight's tours are there?" Tags="&lt;combinatorics&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" />
  <row Id="637" PostTypeId="2" ParentId="623" CreationDate="2010-07-24T08:34:05.543" Score="0" ViewCount="" Body="&lt;p&gt;This is certainly a non-trivial proof if you have to avoid calculus. See the answer over on the &lt;a href=&quot;http://mathforum.org/library/drmath/view/55263.html&quot; rel=&quot;nofollow&quot;&gt;Dr. Math forums&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="56" LastActivityDate="2010-07-24T08:34:05.543" CommentCount="1" />
  <row Id="639" PostTypeId="1" AcceptedAnswerId="1106" CreationDate="2010-07-24T11:03:59.873" Score="20" ViewCount="2550" Body="&lt;p&gt;I just got out from my Math and Logic class with my friend.  During the lecture, a well-known math/logic puzzle was presented:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The King has $1000$ wines, $1$ of which is poisoned.  He needs to identify the poisoned wine as soon as possible, and with the least resources, so he hires the protagonist, a Mathematician.  The king offers you his expendable servants to help you test which wine is poisoned.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The poisoned wine is very potent, so much that one molecule of the wine will cause anyone who drinks it to die.  However, it is slow-acting.  The nature of the slow-acting poison means that there is only time to test one &quot;drink&quot; per servant. (A drink may be a mixture of any number of wines)  (Assume that the King needs to know within an hour, and that any poison in the drink takes an hour to show any symptoms)&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;What is the minimum amount of servants you would need to identify the poisoned wine?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;With enough time and reasoning, one can eventually see that this requires at most &lt;strong&gt;ten&lt;/strong&gt; ($10$) servants (in fact, you could test 24 more wines on top of that 1000 before requiring an eleventh servant).  The proof/procedure is left to the reader.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My friend and I, however, was not content with resting upon this answer.  My friend added the question:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What would be different if there were $2$ wines that were poisoned out of the 1000?  What is the new minimum then?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We eventually generalized the problem to this:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Given $N$ bottles of wine ($N \gt 1$) and, of those, $k$ poisoned wines  ($0 \lt k \lt N$), what is the optimum method to identify the all of the poisoned wines, and how many servants are required ($s(N,k)$)?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;After some mathsing, my friend and I managed to find some (possibly unhelpful) lower and upper bounds:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ log_2 {N \choose k} \le s(N,k) \le N-1 $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is because $log_2 {N \choose k}$ is the minimum number of servants to uniquely identify the $N \choose k$ possible configurations of $k$ poisoned wines in $N$ total wines.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone help us find an optimum strategy?  Besides the trivial one requiring $N-1$ servants.  How about a possible approach to start?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would this problem be any different if you were only required to find a strategy that would for sure find a wine that is &lt;strong&gt;not&lt;/strong&gt; poisoned, instead of identifying all poisoned wines? (other than the slightly trivial solution of $k$ servants)&lt;/p&gt;&#xA;" OwnerUserId="51" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-07-29T23:28:51.460" LastActivityDate="2011-04-02T01:56:02.263" Title="Logic problem: Identifying poisoned wines out of a sample, minimizing test subjects with constraints" Tags="&lt;puzzle&gt;&lt;recreational-mathematics&gt;" AnswerCount="3" CommentCount="16" FavoriteCount="9" />
  <row Id="640" PostTypeId="1" AcceptedAnswerId="820" CreationDate="2010-07-24T11:24:56.920" Score="10" ViewCount="612" Body="&lt;p&gt;When I studied complex analysis, I could never understand how once-differentiable complex functions could be possibly be infinitely differentiable. After all, this doesn't hold for functions from R2 to R2. Can anyone explain what is different about complex numbers?&lt;/p&gt;&#xA;" OwnerUserId="123" LastActivityDate="2011-04-15T23:48:06.473" Title="Why are differentiable complex functions infinitely differentiable?" Tags="&lt;complex-analysis&gt;" AnswerCount="6" CommentCount="5" FavoriteCount="3" />
  <row Id="641" PostTypeId="2" ParentId="640" CreationDate="2010-07-24T11:51:04.973" Score="3" ViewCount="" Body="&lt;p&gt;The proofs I have seen derive this as a corollary of &lt;a href=&quot;http://en.wikipedia.org/wiki/Cauchy%27s_integral_formula&quot; rel=&quot;nofollow&quot;&gt;Cauchy's integral formula&lt;/a&gt;.  Look at the difference quotient as an integral, play around with it, and you get that it converges to what you'd get if you differentiated under the integral sign.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that since harmonic functions also satisfy a similar integral equation, they are also infinitely differentiable in the same way (this also follows since they are real and imaginary parts of holomorphic functions).&lt;/p&gt;&#xA;" OwnerUserId="9" LastEditorUserId="9" LastEditorDisplayName="" LastEditDate="2010-07-24T12:01:33.627" LastActivityDate="2010-07-24T12:01:33.627" CommentCount="3" />
  <row Id="642" PostTypeId="2" ParentId="622" CreationDate="2010-07-24T13:08:06.907" Score="5" ViewCount="" Body="&lt;p&gt;Particles correspond to specific vectors in a representation, not to G-orbits!  The reason has to do with &quot;symmetry breaking.&quot;  The 8 particles in the &lt;a href=&quot;http://en.wikipedia.org/wiki/Eightfold_Way_%28physics%29&quot; rel=&quot;nofollow&quot;&gt;meson octet&lt;/a&gt; correspond to a basis of a certain 8-dimensional representation of the group SU(3) called the &quot;adjoint representation.&quot;  At high enough energies these particles would be indistinguishable.  But at low energies the &quot;SU(3) symmetry has been broken&quot; and the particles become distinguishable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another good physics example that's easier to understand is that the orbital states of electrons in atoms correspond to representations of the group SO(3) of symmetries of space (well, really SU(2) if you want to incorporate spin).  Try reading a standard quantum mechanics textbook for a little bit of this picture and then try thinking about it in terms of representation theory.&lt;/p&gt;&#xA;" OwnerUserId="8" LastActivityDate="2010-07-24T13:08:06.907" CommentCount="2" />
  <row Id="644" PostTypeId="2" ParentId="623" CreationDate="2010-07-24T14:52:32.947" Score="14" ViewCount="" Body="&lt;p&gt;&lt;img src=&quot;http://www.mathematische-basteleien.de/wuerfel16.gif&quot; alt=&quot;alt text&quot;&gt;&lt;br&gt;&#xA;A visual demonstration for the case of a pyramid with a square base. As &lt;a href=&quot;http://math.stackexchange.com/questions/623/why-is-the-volume-of-a-cone-one-third-of-the-volume-of-a-cylinder/635#635&quot; rel=&quot;nofollow&quot;&gt;Grigory states&lt;/a&gt;, &lt;a href=&quot;http://en.wikipedia.org/wiki/Cavalieris_principle&quot; rel=&quot;nofollow&quot;&gt;Cavalieri's principle&lt;/a&gt; can be used to get the formula for the volume of a cone. We just need the base of the square pyramid to have side length r√pi. Such a pyramid has volume 1/3 * h * pi * r&lt;sup&gt;2&lt;/sup&gt;.&lt;br&gt;&#xA;&lt;img src=&quot;http://www.csrjjsmp.com/images/offsite/cone.png&quot; alt=&quot;alt text&quot;&gt;&lt;br&gt;&#xA;Then the area of the base is clearly the same. The cross-sectional area at distance a from the peak is a simple matter of similar triangles: The radius of the cone's cross section will be a/h * r. The side length of the square pyramid's cross section will be a/h *   r√pi.&lt;br&gt;&#xA;Once again, we see that the areas must be equal. So by Cavalieri's principle, the cone and square pyramid must have the same volume: 1/3 * h * pi * r&lt;sup&gt;2&lt;/sup&gt;.&lt;/p&gt;&#xA;" OwnerUserId="73" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-07-24T17:32:17.447" LastActivityDate="2010-07-24T17:32:17.447" />
  <row Id="645" PostTypeId="1" AcceptedAnswerId="647" CreationDate="2010-07-24T15:24:08.620" Score="8" ViewCount="356" Body="&lt;p&gt;Chicken is a famous game where two people drive on a collision course straight towards each other. Whoever swerves is considered a 'chicken' and loses, but if nobody swerves, they will both crash. So the payoff matrix looks something like this:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;                  B swerves               B straight&#xA;A swerves            tie               A loses, B wins&#xA;A straight     B loses, A wins            both lose &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;But I have heard of another situation called the prisoner's dilemma, where two prisoners are each given the choice to testify against the other, or remain silent. The payoff matrix for prisoner's dilemma also looks like  &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;                  B silent               B testify&#xA;A silent            tie               A loses, B wins&#xA;A testify     B loses, A wins            both lose &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I remember hearing that in the prisoner's dilemma, it was always best for both prisoners to testify. But that makes no sense if you try to apply it to chicken: both drivers would crash every time, and in real life, almost always someone ends up swerving. What's the difference between the two situations?&lt;/p&gt;&#xA;" OwnerUserId="73" LastActivityDate="2010-07-24T15:44:51.053" Title="How is prisoner's dilemma different from chicken?" Tags="&lt;game-theory&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="646" PostTypeId="2" ParentId="645" CreationDate="2010-07-24T15:36:48.387" Score="8" ViewCount="" Body="&lt;p&gt;The games aren't just about winning or losing, but also about utility. Here is a more accurate table for chicken:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;                  B swerves                B straight&#xA;A swerves    No gain for either          A loses, B wins&#xA;A straight     B loses, A wins         Both have large loss&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Here is one for prisoners dilemma:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;                    B silent                        B testify&#xA;A silent       Both have small loss         A large loss, B loses nothing&#xA;A testify  B loses nothing, A large loss,        both medium lose &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;In the prisoners dilemma, an individual prisoner will always do better by testifying (look at the table), however, by both testifying they end up in a worse position than if both were silent. In contrast, in chicken, going straight will be better if the other swerves and swerving will be better if the other goes straight. Your tables represent some strategies as being equally good for players when they are not.&lt;/p&gt;&#xA;" OwnerUserId="123" LastEditorUserId="123" LastEditorDisplayName="" LastEditDate="2010-07-24T15:44:51.053" LastActivityDate="2010-07-24T15:44:51.053" />
  <row Id="647" PostTypeId="2" ParentId="645" CreationDate="2010-07-24T15:38:40.977" Score="13" ViewCount="" Body="&lt;p&gt;(See &lt;a href=&quot;http://en.wikipedia.org/wiki/Chicken_%28game%29#Prisoner.27s_dilemma&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Chicken_%28game%29#Prisoner.27s_dilemma&lt;/a&gt;.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The difference is in the payoff. In the &lt;a href=&quot;http://en.wikipedia.org/wiki/Chicken_%28game%29&quot; rel=&quot;nofollow&quot;&gt;&quot;chicken&quot; game&lt;/a&gt;, the payoff matrix is like&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;        Sw       St&#xA;   Sw  0,  0   -1, +1&#xA;   St +1, -1  -10, -10&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;While in the PD game, &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;        Si       Te&#xA;   Si  3, 3     0, 5&#xA;   Te  5, 0     1, 1&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Both games have this structure in the payoff table:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;         C       D&#xA;   C   R, R     S, T&#xA;   D   T, S     P, P&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;But:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;In PD game, the order is T &gt; R &gt; P &gt; S&lt;/li&gt;&#xA;&lt;li&gt;In chicken game, the order is T &gt; R &gt; S &gt; P&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This leads a different Nash equilibria. In the PD game, if A remains silent, B chooses to testify because T &gt; R, while if A testifies, B also testify because S &gt; P. So testifying is the most rational choice of B after considering all possibilities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But in the chicken game, as P &gt; S, if A goes straight, B should swerve. This leads to two Nash equilibria in the pure game: (St, Sw) and (Sw, St). &lt;/p&gt;&#xA;" OwnerUserId="171" LastEditorUserId="171" LastEditorDisplayName="" LastEditDate="2010-07-24T15:44:42.443" LastActivityDate="2010-07-24T15:44:42.443" />
  <row Id="648" PostTypeId="2" ParentId="434" CreationDate="2010-07-24T16:16:57.587" Score="0" ViewCount="" Body="&lt;p&gt;The answer to this question depends on how the problem is defined, but the answer is no, at least without defining the problem in a misleading way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since my first solution was completely off the mark, I have deleted it and posted this new one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider polynomial p. If it has an integer solution, then the solution will eventually be found by random guessing. So if it is impossible to prove the existential status, there must be no solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we know from this link that there is a polynomial, q, that is unsolvable in the integers iff ZFC is consistent. It is well known that ZFC cannot prove its own consistency. So if it ZFC is consistent, then q is unsolvable, but we cannot prove this as then we could prove ZFC. So it seems like it is accurate to say that if mathematics is consistent, we have a polynomial with no integer roots, but we can't prove it. However, if we are assuming maths is consistent, we can use this to prove that the equation is unsolvable (indeed that is what we have done). So, it really isn't accurate an accurate statement at all. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To further clarify, when considering mathematical truth, there are two basic ways of viewing it. The first is where we are assuming that are axioms are true, which necessarily means assuming consistency. If we show any problem is equivalent to consistency, then we consider it to be true.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other is where we are considering a formal set of statements, of which the axioms have been defined to be true and seeing which statements can be derived to be true. From this viewpoint, we don't actually know whether the axioms are consistent or not. In fact, Godel's second incompleteness theorem shows that no &quot;non-trivial&quot; atomic system can prove its own consistency. So showing a problem is equivalent to consistency is actually the same as showing that the problem is unprovable by the atomic system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The confusion comes from assuming ZFC is consistent to eliminate one possibility in a choice, yet not allowing this assumption to be used as an axiom in the proofs.&lt;/p&gt;&#xA;" OwnerUserId="123" LastEditorUserId="123" LastEditorDisplayName="" LastEditDate="2010-07-24T19:39:04.343" LastActivityDate="2010-07-24T19:39:04.343" CommentCount="3" />
  <row Id="649" PostTypeId="2" ParentId="566" CreationDate="2010-07-24T18:13:44.017" Score="5" ViewCount="" Body="&lt;p&gt;A probability problem I love.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take a shuffled deck of cards. Deal off the cards one by one until you reach any Ace. Turn over the next card, and note what it is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The question&lt;/strong&gt;: which card has a higher probability of being turned over, the Ace of Spades or the Two of Hearts?&lt;/p&gt;&#xA;" OwnerUserId="122" LastActivityDate="2010-07-24T18:13:44.017" CommunityOwnedDate="2010-07-24T18:13:44.017" CommentCount="5" />
  <row Id="650" PostTypeId="2" ParentId="123" CreationDate="2010-07-24T19:40:33.433" Score="7" ViewCount="" Body="&lt;p&gt;I don't know if you consider General Relativity &quot;outside acadamia&quot;(and I don't care to argue the point!) but if you do,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the group of symmetries with respect to the Lorentzian Metric can be written as Matrices containing hyperbolic trig functions as elements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note Kenny's comment.&lt;/p&gt;&#xA;" OwnerUserId="16" LastEditorUserId="16" LastEditorDisplayName="" LastEditDate="2010-07-24T20:20:49.873" LastActivityDate="2010-07-24T20:20:49.873" CommentCount="3" />
  <row Id="651" PostTypeId="2" ParentId="640" CreationDate="2010-07-24T20:31:14.640" Score="-1" ViewCount="" Body="&lt;p&gt;It isn't true.  Take any singly but not doubly differentiable function f, with the property that f(x)=f(-x).  Then obtain F from f by rotating f around the origin: F(r e^it)=f(r).  This function is then singly but not doubly differentiable over the whole complex plane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think you are confusing differentiable with holomorphic.&lt;/p&gt;&#xA;" OwnerUserId="100" LastActivityDate="2010-07-24T20:31:14.640" CommentCount="4" />
  <row Id="652" PostTypeId="2" ParentId="370" CreationDate="2010-07-24T20:40:30.187" Score="2" ViewCount="" Body="&lt;p&gt;I've read a fair amount of Sets for Mathematics and found it to be a gentle introduction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0521010608&quot; rel=&quot;nofollow&quot;&gt;http://www.amazon.com/Sets-Mathematics-F-William-Lawvere/dp/0521010608/ref=pd_sim_b_5&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="99" LastActivityDate="2010-07-24T20:40:30.187" CommunityOwnedDate="2010-07-24T20:40:30.187" />
  <row Id="653" PostTypeId="1" CreationDate="2010-07-24T20:59:19.173" Score="4" ViewCount="493" Body="&lt;p&gt;I covered homotopy theory in a recent maths course. However I was never presented with any reasons as to why (or even if) it is useful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any good examples of its use outside academia?&lt;/p&gt;&#xA;" OwnerUserId="126" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-07-24T21:06:29.763" LastActivityDate="2010-11-19T06:01:04.227" Title="Real world uses of homotopy theory" Tags="&lt;soft-question&gt;&lt;big-list&gt;&lt;applications&gt;&lt;homotopy-theory&gt;" AnswerCount="5" CommentCount="2" FavoriteCount="1" />
  <row Id="654" PostTypeId="2" ParentId="653" CreationDate="2010-07-24T21:02:37.217" Score="2" ViewCount="" Body="&lt;p&gt;Robert Ghrist is an amazing applied mathematician who uses a lot of interesting algebraic topology for engineering applications. He uses homology and sheaf theory. I claim this answers your question since homology is a generalization of homotopy theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Relevant link:&#xA;&lt;a href=&quot;http://www.math.uiuc.edu/~ghrist/index_files/research.htm&quot; rel=&quot;nofollow&quot;&gt;http://www.math.uiuc.edu/~ghrist/index_files/research.htm&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="16" LastActivityDate="2010-07-24T21:02:37.217" CommentCount="3" />
  <row Id="655" PostTypeId="1" AcceptedAnswerId="658" CreationDate="2010-07-24T21:30:52.573" Score="3" ViewCount="93" Body="&lt;p&gt;I've learned that multiplying two generating functions f(x) and g(x) gives you the result $\sum_{k=0}^\infty\left(\sum_{j=0}^k a_j\,b_{k-j}\right)x^k$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've used the result, but it was presented in my class without proof and I'm having some trouble tracking one down.  Weak google-foo today, I suppose.  Can anyone give me a pointer to a proof?  If this is a question better answered in book form, that is fine as well.&lt;/p&gt;&#xA;" OwnerUserId="129" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-07-24T22:30:23.300" LastActivityDate="2010-07-25T00:47:31.660" Title="Proof for multiplying generating functions" Tags="&lt;generating-functions&gt;&lt;sequences-and-series&gt;" AnswerCount="2" />
  <row Id="656" PostTypeId="1" AcceptedAnswerId="657" CreationDate="2010-07-24T21:32:04.680" Score="9" ViewCount="464" Body="&lt;p&gt;One thing I have been wondering about lately is how to write an equation that describes a pattern of numbers. What I mean is:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;x   0    1    2 &#xA;y   1    5    9&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;If I have this, I can tell that an equation that describes this would be y=4x+1. In fact, I don't even need the 3rd pair of numbers. It's very easy when the equation is a straight line. But when the equation is a parabola, its not always that easy. For example:   &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;x   0    1    2 &#xA;y   1    2    5  &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I can tell this is y=x&lt;sup&gt;2&lt;/sup&gt;+1, because I recognize the pattern. But I can't always tell just by looking at the numbers what the right equation should be. Is there some way to always know the right equation? I know that if you get the x=0 term you can get the c in y=ax&lt;sup&gt;2&lt;/sup&gt;+bx+c, but that's not enough to let me solve it like I can when the equation is just a line.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, can someone show me how you would do it for  &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;x    0     1     2&#xA;y    5     4     7&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;It's not a homework question, I promise!&lt;/p&gt;&#xA;" OwnerUserId="73" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-07-26T08:29:10.943" LastActivityDate="2010-07-26T08:29:10.943" Title="How can I write an equation that matches any sequence?" Tags="&lt;middle-school&gt;&lt;algebra-precalculus&gt;" AnswerCount="4" CommentCount="3" FavoriteCount="4" />
  <row Id="657" PostTypeId="2" ParentId="656" CreationDate="2010-07-24T21:47:45.883" Score="9" ViewCount="" Body="&lt;p&gt;If you know your relationship is going to be a &lt;strong&gt;polynomial&lt;/strong&gt;, then there are some pretty (conceptually) simple ways you can do this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you know what degree your polynomial is (line, parabola, cubic, etc.) then your job will be much easier.  But if not, then you simply need to look at the number of points you have.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If you are given one point, the best you can do is of degree 0  ( y = k )&lt;/li&gt;&#xA;&lt;li&gt;If you are given two points, the best you can do is of degree 1 ( y = A x + B )&lt;/li&gt;&#xA;&lt;li&gt;If you are given three points, the best you can do is of degree 2 ( y = A x&lt;sup&gt;2&lt;/sup&gt; + B x + C )&lt;/li&gt;&#xA;&lt;li&gt;If you are given four points, the best you can do is of degree 3 ( y = A x&lt;sup&gt;3&lt;/sup&gt; + B x&lt;sup&gt;2&lt;/sup&gt; + C x + D )&lt;/li&gt;&#xA;&lt;li&gt;etc.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;When I say &quot;the best you can do&quot;, what I mean is -- if you have a parabola, but are only given two points, then you really can't identify the parabola.  But you can say that it's a simple line.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's assume you have three points.  The &quot;best you can do&quot; is assume that it is degree 2.  If it is actually of degree one, your answer will magically turn into a line ( your &lt;code&gt;x^2&lt;/code&gt; coefficient will be &lt;code&gt;0&lt;/code&gt; )&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The basic idea of solving relationships/equations is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If you have &lt;code&gt;n&lt;/code&gt; unknowns, you need &lt;code&gt;n&lt;/code&gt; equations/points.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Notice how, in the form of the Parabola ( y = A x&lt;sup&gt;2&lt;/sup&gt; + B x + C ), you have three unknowns?  And also three equations! (points)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's pick three arbitrary points&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&#xA;x   1    2    4&#xA;y   6    7    3&#xA;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;You would set up three equations:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;6 = 1&lt;sup&gt;2&lt;/sup&gt; * A + 1 * B + C&lt;/p&gt;&#xA;&#xA;&lt;p&gt;7 = 2&lt;sup&gt;2&lt;/sup&gt; * A + 2 * B + C&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3 = 4&lt;sup&gt;2&lt;/sup&gt; * A + 4 * B + C&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Three equations, three unknowns.  You should be able to solve this with a combination of most system-of-equation-solving rules.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In our case, we find:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&#xA;A = -1&#xA;B = 4&#xA;C = 3&#xA;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;So our equation is y = -x&lt;sup&gt;2&lt;/sup&gt; + 4 x + 3&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that, if your original three points formed a line, your $A$ would $= 0$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;However, if your equation is &lt;strong&gt;NOT&lt;/strong&gt; a polynomial, then you are left with little more than guess and check, plugging in various coefficients and trials (exponential?  trigonometric?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The beauty of the polynomial approach is that a polynomial of high enough degree will always fit &lt;strong&gt;any&lt;/strong&gt; list of points. (provided that the points form a function)&lt;/p&gt;&#xA;" OwnerUserId="51" LastEditorUserId="51" LastEditorDisplayName="" LastEditDate="2010-07-24T23:46:58.333" LastActivityDate="2010-07-24T23:46:58.333" CommentCount="2" />
  <row Id="658" PostTypeId="2" ParentId="655" CreationDate="2010-07-24T21:50:20.643" Score="2" ViewCount="" Body="&lt;p&gt;It is actually the other way round. A generating function is generally defined to have an addition operation where the components are added and a multiplication operation like that you mentioned. Once we have made these definitions, we observe that polynomials obey the same laws and so that it is convenient to represent generating functions as infinite polynomials rather than just an infinite tuple.&lt;/p&gt;&#xA;" OwnerUserId="123" LastEditorUserId="123" LastEditorDisplayName="" LastEditDate="2010-07-25T00:47:31.660" LastActivityDate="2010-07-25T00:47:31.660" />
  <row Id="659" PostTypeId="1" AcceptedAnswerId="663" CreationDate="2010-07-24T22:11:12.957" Score="0" ViewCount="119" Body="&lt;h3&gt;Motivatation&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;I am presently looking at a structure that I am trying to pin down- my strategy being to pull the thing up into the greatest possible generality (based on the bits I'm sure about) and narrow it down from there. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The situation I have is somewhat similar to the dual space structure in vector spaces, though almost certainly less well-behaved and vector spaces alone will not cut it.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;The Construction&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Consider a vector space $V$ over a field $k$- its dual space $V^*$ appears naturally as the set linear of maps   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$w^* : V \to k$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which, coincidentally, form a vector space in themselves. We can generalise this to arbitrary categories $A$, $B$, $C$ by setting $B=hom(A,C). Then, at least in some sense, $B=A^&lt;em&gt;$. So far so standard, but I want more: a nice property of dual spaces is that an element of $V \otimes V^&lt;/em&gt;$ can be canonically seen as an element of $hom(V,V)$- this is because of the way that $k$ acts on $V$ by multiplication. We can mimic this by letting $C$ be a monoid acting on $A$.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;In summary:&lt;/strong&gt; A category $A$ acted on by a monoid $C$ and a dual A $A^*:=hom(A,C)$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am particularly interested in when $A$ is also a monoid, especially so when $A$ is a space of stochastic matrices.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Questions:&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;So this isn't too unlikely a construction, in fact it's probably forehead-slappingly well known, so:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;What is it called, if anything?&lt;/li&gt;&#xA;&lt;li&gt;In what cases can we have $A=B=C$? Is that necessarily a permutation group for example?&lt;/li&gt;&#xA;&lt;li&gt;Are there any useful canonical examples, besides vector spaces?&lt;/li&gt;&#xA;&lt;li&gt;Better yet, theorems??? Papers???&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;As you can probably tell, I am no category theorist, so any help would be awesome.&lt;/p&gt;&#xA;" OwnerUserId="160" LastActivityDate="2010-08-06T20:21:01.613" Title="Is this the right categorical generalisation of dual space" Tags="&lt;terminology&gt;&lt;reference-request&gt;&lt;category-theory&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="660" PostTypeId="2" ParentId="655" CreationDate="2010-07-24T22:15:09.880" Score="4" ViewCount="" Body="&lt;p&gt;Casebash is correct that this is a definition and not a theorem. But the motivation from 3.48 (Defintion of product of series) of &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/007054235X&quot; rel=&quot;nofollow&quot;&gt;little Rudin&lt;/a&gt; may convince you that this is a good definition:&lt;br&gt;&#xA;$\sum_{n=0}^{\inf} a_n z^n \cdot \sum_{n=0}^{\inf} b_n z^n = (a_0+a_1z+a_2z^2+ \cdots)(b_0+b_1z+b_2z^2+ \cdots)$&lt;br&gt;&#xA;$=a_0b_0+(a_0b_1 + a_1b_0)z + (a_0b_2+a_1b_1+a_2b_0)z^2 + \cdots$&lt;br&gt;&#xA;$=c_0+c_1z+c_2z^2+ \cdots $&lt;br&gt;&#xA;where $c_n=\sum_{k=0}^n a_k b_{n-k}$  &lt;/p&gt;&#xA;" OwnerUserId="73" LastActivityDate="2010-07-24T22:15:09.880" CommentCount="1" />
  <row Id="661" PostTypeId="2" ParentId="656" CreationDate="2010-07-24T22:38:04.883" Score="3" ViewCount="" Body="&lt;p&gt;Given a list of terms of a sequence as you describe, one technique that may be of use (supplementary to Justin's answer) is finite differences.  Calculate the differences between successive terms.  If these first differences are constant, then a linear equation fits the terms you have.  If not, compute the differences of the differences.  If these second differences are constant, then a quadratic equation fits the terms you have.  If not, you can continue to compute differences until you reach a constant difference (in the nth differences means an nth degree polynomial), differences that are a constant multiple of the previous differences (exponential of some sort), or you run out of terms.  In any case, what you find is limited to matching the terms you know, as without some kind of general rule for the sequence, the first unknown term could be anything and completely alter the pattern (and with n known terms, a polynomial of degree n-1 will always perfectly fit).&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-07-24T22:38:04.883" CommentCount="4" />
  <row Id="662" PostTypeId="2" ParentId="656" CreationDate="2010-07-24T22:50:07.723" Score="2" ViewCount="" Body="&lt;p&gt;In the general case of trying to predict some infinite sequence of integers, there is no formula. This is because there is no reason to expect a pattern to continue, since all sequences are possible. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, you can say a certain number is more likely given a set of functions. For instance if you considered all Turing machines, you could say that given n elements of a sequence look at all the Turing machines that predict the current sequence, and then find the most predicted next number. There still isn't a efficient way to compute what the most likely next number is. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ray Solomonoff called this &quot;Universal Probabilistic Induction&quot; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is explained in more depth here:&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Inductive_inference&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Inductive_inference&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="99" LastEditorUserId="99" LastEditorDisplayName="" LastEditDate="2010-07-24T23:19:39.207" LastActivityDate="2010-07-24T23:19:39.207" />
  <row Id="663" PostTypeId="2" ParentId="659" CreationDate="2010-07-24T23:06:21.473" Score="2" ViewCount="" Body="&lt;p&gt;I believe the right generalization of dual spaces is that of a &lt;a href=&quot;http://en.wikipedia.org/wiki/Dual_object&quot; rel=&quot;nofollow&quot;&gt;dual object&lt;/a&gt; in a &lt;a href=&quot;http://en.wikipedia.org/wiki/Tensor_category&quot; rel=&quot;nofollow&quot;&gt;tensor category&lt;/a&gt;, which I will assume symmetric for convenience.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Recall what makes a dual space of a vector space work: We have a map $V \times V^* \to k$ (for $k$ the ground field). The problem is, this isn't a homomorphism in the category of vector spaces; it is rather a bilinear map. So you can think of it as a map $V \otimes V^* \to k$ instead.  This is why you need a tensor structure to think of duals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This isn't enough, though, because we need to know that the pairing is nondegenerate.  One way to express this is that there's a map $k \to V \times V^*$ mapping 1 to the &quot;Casimir element&quot; (which is the sum $\sum e_i \otimes e_i^{\vee}$ where $e_i$ ranges over a basis of $V$ and $e_i^{\vee}$ the dual basis; it is independent of the choice of $e_i$ as a quick computation shows).  The Casimir morphism satisfies the condition that $$V \to (k) \otimes V \to (V \otimes V^&lt;em&gt;) \otimes V \to V \otimes (V^&lt;/em&gt; \otimes V)$ is just the identity.&lt;br&gt;&#xA;Conversely, this is enough to show that the pairing is nondegenerate.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, anyway, how does this make sense in a symmetric tensor category?  Basically, $V$ is the object, $V^*$ the putative dual, and $k$ replaced by the unital object. This definition is entirely arrow-theoretic, and it all goes through as usual.&#xA;It is an exercise to check that the dual is unique.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some examples:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;This coincides with the usual dual in the category of vector spaces&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;This coincides with the dual sheaf if one is working in the category of locally free sheaves on a scheme&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;This corresponds to the dual (contragredient) representation in the (tensor) category of representations of any Hopf algebra (so this includes representations of finite groups and Lie algebras)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Oh, and what happens if you don't have a symmetric tensor category? Then you have to worry about &quot;left&quot; and &quot;right&quot; duals, respectively.  For more about all this, I recommend the &lt;a href=&quot;http://www-math.mit.edu/~etingof/tenscat1.pdf&quot; rel=&quot;nofollow&quot;&gt;notes&lt;/a&gt; of Pavel Etingof on tensor categories.&lt;/p&gt;&#xA;" OwnerUserId="536" LastActivityDate="2010-07-24T23:06:21.473" CommentCount="4" />
  <row Id="664" PostTypeId="1" AcceptedAnswerId="665" CreationDate="2010-07-25T00:19:52.090" Score="8" ViewCount="271" Body="&lt;p&gt;Hi, I just came back from my Introduction to Rotational Kinematics class, and one of the important concepts they described was &lt;em&gt;Rotational Inertia&lt;/em&gt;, or &lt;em&gt;Moment of Inertia&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's basically the equivalent of mass in Netwon's $F = m a$ in linear motion.  The equivalent rotational equation is $\tau = I \alpha$, where $\tau$ is rotational force, $\alpha$ is rotational acceleration, and $I$ is rotational inertia.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a point about an axis, $I$ is $m r^2$, where $r$ is the distance from the point to the axis of rotation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a continuous body, this is an integral -- $I = \int r^2 \,dm$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This really doesn't make any sense to me...you have two independent variables?  I am only used to having one independent variable and one constant.  So I would solve this, using my experience with calculus (which encompasses a read through the Sparks Notes packet) as $ I = m r^2 $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But obviously, this is wrong?  $r$ is not a constant!  How do I deal with it?  Do I need to replace $r$ with an expression that varies with $m$?  But how could $r$ possibly vary with $m$?  Isn't it more likely the other way around?  But how can $m$ vary with $r$?  It's all rather confusing me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could someone help me figure out what to do with all these substitutions for, example, figuring out the Moment of Inertia of a hoop with no thickness and width $w$, with the axis of rotation running through its center orthogonal to its plane?&lt;/p&gt;&#xA;" OwnerUserId="51" LastActivityDate="2010-11-09T08:17:27.780" Title="Finding Moment of Inertia (rotional inertia?) $I$ using integration?" Tags="&lt;calculus&gt;&lt;physics&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="665" PostTypeId="2" ParentId="664" CreationDate="2010-07-25T01:15:03.773" Score="5" ViewCount="" Body="&lt;p&gt;Seeing an expression like $I = \int r^2 dm$ is certainly confusing the first time you see it. When you see an expression like $I = \int x^2 dx$ we effectively iterating over a range on the x axis and adding up the area of an infinite amount of infinitesimally small strips. Note that the area of each strip is approximated as the function value (x^2) times the strip length (dx). When you see an expression like dm, we are iterating over masses instead. We still sum up the function value (r^2), but this time we multiply it by the strip mass (dm). To solve the problem, we usually put m in terms of another variable which we can iterate over more easily.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, consider the moment of inertia of a rod of length L around its center with total mass of L. Each bit of length (dx) has mass (dm) and r=|x|. Solving for $I = \int r^2 dm = \int |x|^2 dx = \int x^2 dx = (x^3)/3+c$. Now, we have definite values for x to sub in (-L/2 and L/2), so we write &lt;code&gt;I=(L/8)/3-(-L/8)/3=L/12&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now lets calculate the moment of inertia of the hoop case you described. We break the hoop up into infinitesimally small rings the same distance from the center. Let the hoop have inner thickness r and outer thickness R. The area is &lt;code&gt;R^2-r^2&lt;/code&gt;. The area density, d, is therefore &lt;code&gt;m/(R^2-r^2)&lt;/code&gt;. A ring at radius k with thickness dk has area &lt;code&gt;2pi k dk&lt;/code&gt;, mass is &lt;code&gt;2pi k dk * m/(R^2-r^2)&lt;/code&gt; and moment of inertia around the central axis &lt;code&gt;2pi k^3 dk * m/(R^2-r^2)&lt;/code&gt;. The integral is &lt;code&gt;pi/2*k^4 *m/(R^2-r^2)+c&lt;/code&gt;. Subbing in the exact values, we get, &lt;code&gt;pi/2*(R^4-r^4) *m/(R^2-r^2)=pi/2*(R^2+r^2)*m&lt;/code&gt;&lt;/p&gt;&#xA;" OwnerUserId="123" LastEditorUserId="123" LastEditorDisplayName="" LastEditDate="2010-11-09T08:17:27.780" LastActivityDate="2010-11-09T08:17:27.780" CommentCount="1" />
  <row Id="667" PostTypeId="1" AcceptedAnswerId="900" CreationDate="2010-07-25T02:14:47.343" Score="8" ViewCount="220" Body="&lt;p&gt;I have heard that one can think of stacks on a site as taking sheaves but instead of the restrictions being equal, we just loosen it to isomorphic, and treat the sheaf conditions with the &quot;obvious&quot; coherence relations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How seriously can one take this analogy? Note that my background is stacks is feeble at best.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope this question isn't too vague. One may choose to respond to this question with &quot;That analogy is stupid because of &lt;strong&gt;__&lt;/strong&gt; gotcha&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance!&lt;/p&gt;&#xA;" OwnerUserId="16" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-07-29T00:57:10.353" LastActivityDate="2010-07-29T00:57:10.353" Title="Stacks are just sheaves up to Isomorphism" Tags="&lt;intuition&gt;&lt;algebraic-geometry&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="2" />
  <row Id="668" PostTypeId="1" AcceptedAnswerId="669" CreationDate="2010-07-25T02:25:11.593" Score="39" ViewCount="1674" Body="&lt;p&gt;In my linear algebra class we just talked about determinants. So far I’ve been understanding the material okay, but now I’m very confused. I get that when the determinant is zero, the matrix doesn’t have an inverse. I can find the determinant of a 2 by 2 matrix by the formula. Our teacher showed us how to compute the determinant of an N by N matrix by breaking it up into the determinants of smaller matrices, and apparently there’s a way by summing over a bunch of permutations. But the notation is really hard for me and I don’t really know what’s going on with them anymore. Can someone help me figure out what a determinant is intuitively, and how all those definitions of it are related?&lt;/p&gt;&#xA;" OwnerUserId="76" LastActivityDate="2011-01-17T09:45:07.343" Title="What's an intuitive way to think about the determinant?" Tags="&lt;linear-algebra&gt;&lt;intuition&gt;" AnswerCount="5" CommentCount="7" FavoriteCount="19" />
  <row Id="669" PostTypeId="2" ParentId="668" CreationDate="2010-07-25T02:25:45.523" Score="29" ViewCount="" Body="&lt;p&gt;Your trouble with determinants is pretty common. They’re a hard thing to teach well, too, for two main reasons that I can see: the formulas you learn for computing them are messy and complicated, and there’s no “natural” way to interpret the value of the determinant, the way it’s easy to interpret the derivatives you do in calculus at first as the slope of the tangent line. It’s hard to believe things like the invertibility condition you’ve stated when it’s not even clear what the numbers mean and where they come from.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rather than show that the many usual definitions are all the same by comparing them to each other, I’m going to state some general properties of the determinant that I claim are enough to specify uniquely what number you should get when you put in a given matrix. Then it’s not too bad to check that all of the definitions for determinant that you’ve seen satisfy those properties I’ll state.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first thing to think about if you want an “abstract” definition of the determinant to unify all those others is that it’s not an array of numbers with bars on the side. What we’re really looking for is a function that takes N vectors (the N columns of the matrix) and returns a number. Let’s assume we’re working with real numbers for now.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remember how those operations you mentioned change the value of the determinant? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) Switching two rows or columns changes the sign. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) Multiplying one row by a constant multiplies the whole determinant by that constant. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(3) The general fact that number two draws from: the determinant is &lt;em&gt;linear in each row&lt;/em&gt;. That is, if you think of it as a function $\det: \mathbb{R}^{n^2} \rightarrow \mathbb{R}$, then $ \det(a \vec{v_1} +b \vec{w_1}, \vec{v_2},...,\vec{v_n}) = a \det(\vec{v_1},\vec{v_2},...,\vec{v_n}) + b \det(\vec{w_1}, \vec{v_2}, ...,\vec{v_n})$, and the corresponding condition in each other slot.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I claim that these facts, together with the fact that the determinant of the identity matrix is one, is enough to define a &lt;em&gt;unique function&lt;/em&gt; that takes in N vectors (each of length N) and returns a real number, the determinant of the matrix given by those vectors. I won’t prove that, but I’ll show you how it helps with some other interpretations of the determinant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, there’s a nice geometric way to think of a determinant. Consider the unit cube in N dimensional space: the set of vectors of length N with coordinates 0 or 1 in each spot. The determinant of the linear transformation (matrix) T is the &lt;em&gt;signed volume of the region gotten by applying T to the unit cube&lt;/em&gt;. (Don’t worry too much if you don’t know what the “signed” part means, for now).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How does that follow from our abstract definition?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, if you apply the identity to the unit cube, you get back the unit cube. And the volume of the unit cube is 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you stretch the cube by a constant factor in one direction only, the new volume is that constant. And if you stack two blocks together aligned on the same direction, their combined volume is the sum of their volumes: this all shows that the signed volume we have is linear in each coordinate when considered as a function of the input vectors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, when you switch two of the vectors that define the unit cube, you flip the orientation. (Again, this is something to come back to later if you don’t know what that means).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So there are ways to think about the determinant that aren’t symbol-pushing. If you’ve studied multivariable calculus, you could think about, with this geometric definition of determinant, why determinants (the Jacobian) pop up when we change coordinates doing integration. Hint: a derivative is a linear approximations of the associated function, and consider a “differential volume element” in your starting coordinate system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It’s not too much work to check that the area of the parallelogram formed by vectors $(a,b)$ and $(c,d)$ is $\det((a,b),(c,d))$, either: you might try that to get a sense for things.&lt;/p&gt;&#xA;" OwnerUserId="76" LastEditorUserId="900" LastEditorDisplayName="" LastEditDate="2010-11-14T18:37:07.847" LastActivityDate="2010-11-14T18:37:07.847" CommunityOwnedDate="2010-07-25T06:05:47.733" CommentCount="3" />
  <row Id="670" PostTypeId="2" ParentId="432" CreationDate="2010-07-25T03:22:36.777" Score="2" ViewCount="" Body="&lt;p&gt;These implications are reached by considering the three, different cases for the roots &lt;a href=&quot;http://mathurl.com/33klydl&quot; rel=&quot;nofollow&quot;&gt;&lt;img src=&quot;http://mathurl.com/33klydl.png&quot;&gt;&lt;/a&gt; of the polynomial: repeated root, all distinct real roots, or two complex roots and one real root.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When one of the roots is repeated, say &lt;a href=&quot;http://mathurl.com/2emr9hn&quot; rel=&quot;nofollow&quot;&gt;&lt;img src=&quot;http://mathurl.com/2emr9hn.png&quot;&gt;&lt;/a&gt; and &lt;a href=&quot;http://mathurl.com/3yf4gya&quot; rel=&quot;nofollow&quot;&gt;&lt;img src=&quot;http://mathurl.com/3yf4gya.png&quot;&gt;&lt;/a&gt;, then it is clear that the discriminant is 0 because the &lt;a href=&quot;http://mathurl.com/2uboj2q&quot; rel=&quot;nofollow&quot;&gt;&lt;img src=&quot;http://mathurl.com/2uboj2q.png&quot;&gt;&lt;/a&gt; term of the product is 0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When one root is a complex number &lt;a href=&quot;http://mathurl.com/3xbav5m&quot; rel=&quot;nofollow&quot;&gt;&lt;img src=&quot;http://mathurl.com/3xbav5m.png&quot;&gt;&lt;/a&gt;, then by the &lt;a href=&quot;http://en.wikipedia.org/wiki/Complex_conjugate_root_theorem&quot; rel=&quot;nofollow&quot;&gt;complex conjugate root theorem&lt;/a&gt;, &lt;a href=&quot;http://mathurl.com/247trlo&quot; rel=&quot;nofollow&quot;&gt;&lt;img src=&quot;http://mathurl.com/247trlo.png&quot;&gt;&lt;/a&gt; is also a root. By the same theorem, the remaining third root must be real. Evaluating the product in the discriminant for this case,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://mathurl.com/2wcefqd&quot; rel=&quot;nofollow&quot;&gt;&lt;img src=&quot;http://mathurl.com/2wcefqd.png&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is less than or equal to 0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, when all roots are real, the product is clearly positive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Putting it all together, Δ:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;less than 0 implies that one root is complex&lt;/li&gt;&#xA;&lt;li&gt;equal to 0 implies that one root is repeated&lt;/li&gt;&#xA;&lt;li&gt;greater than 0 implies that all roots are distinct and real&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="184" LastActivityDate="2010-07-25T03:22:36.777" />
  <row Id="671" PostTypeId="1" AcceptedAnswerId="672" CreationDate="2010-07-25T04:45:55.390" Score="3" ViewCount="264" Body="&lt;p&gt;I just came back from a class on Probability in Game Theory, and was musing over something in my head.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming, for the sake of the question:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Playing cards in their current state have been around for approximately eight centuries&lt;/li&gt;&#xA;&lt;li&gt;A deck of playing cards is shuffled to a random configuration one billion times per day&lt;/li&gt;&#xA;&lt;li&gt;Every shuffle ever is completely (theoretically) random and unaffected by biases caused by human shuffling and the games the cards are used for&lt;/li&gt;&#xA;&lt;li&gt;By &quot;deck of cards&quot;, I refer to a stack of unordered 52 unique cards, with a composition that is identical from deck to deck.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This would, approximately, be on the order of 3 x 10&lt;sup&gt;14&lt;/sup&gt; random shuffles in the history of playing cards.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I were to shuffle a new deck today, completely randomly, what are the probabilistic odds (out of 1) that you create a new unique permutation of the playing cards that has never before been achieved in the history of 3 x 10&lt;sup&gt;14&lt;/sup&gt; similarly random shuffles?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My first thought was to think that it was a simple matter of $\frac{1}{52!} * 3 * 10^{14}$, but then I ran into things like &lt;a href=&quot;http://en.wikipedia.org/wiki/Birthday_Paradox&quot; rel=&quot;nofollow&quot;&gt;Birthday Paradox&lt;/a&gt;.  While it is not analogous (I would have to be asking about the odds that any two shuffled decks in the history of shuffled decks ever matched), it has caused me to question my intuitive notions of Probability.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is wrong in my initial approach, if it is wrong?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the true probability?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And, if the probability is less than 0.5, if we how many more years (centuries?) must we wait, assuming the current rate of one billion shuffles per day, until we reach a state where the probability is 0.5+?   0.9+?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Out of curiosity, it would be neat to know the analogous birthday paradox answer, as well)&lt;/p&gt;&#xA;" OwnerUserId="51" LastEditorUserId="51" LastEditorDisplayName="" LastEditDate="2010-07-25T04:58:07.650" LastActivityDate="2010-07-25T05:12:44.320" Title="When you randomly shuffle a deck of cards, what is the probability that it is a unique permutation never before configured?" Tags="&lt;probability-theory&gt;" AnswerCount="2" />
  <row Id="672" PostTypeId="2" ParentId="671" CreationDate="2010-07-25T05:04:51.993" Score="2" ViewCount="" Body="&lt;p&gt;Your original answer of 3 * 10^14 / 52! is not far from being right. That is in fact the expected number of times any ordering of the cards has occurred.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The probability that any particular ordering of the cards has not occurred, given your initial assumptions, is (1-1/52!)^(3*10^14), and the probability that it has occurred is 1 minus this value.  But for small values of n&amp;epsilon;, (1+&amp;epsilon;)^n is nearly 1+n&amp;epsilon;.  In particular  1-(1-1/52!)^(3*10^14) is nearly 1/52! * 3*10^14.  &lt;/p&gt;&#xA;" OwnerUserId="173" LastActivityDate="2010-07-25T05:04:51.993" CommentCount="3" />
  <row Id="673" PostTypeId="2" ParentId="671" CreationDate="2010-07-25T05:12:44.320" Score="2" ViewCount="" Body="&lt;p&gt;Suppose we shuffle a deck and get a permutation p. For each previous shuffling there is a 1-1/52! chance that p doesn't match it. Each previous shuffling is independent, in that regardless of what p and the other permutations are, the chance of p matching the shuffling is 1-1/52! When probabilities are independent we can simply multiple them to find the chance of all the events happening. In this case, the each event is actually a match &lt;em&gt;not&lt;/em&gt; happening, so the chance of no matches given n previous shuffles is (1-1/52!)^n. We can then complete the calculations as Michael did.&lt;/p&gt;&#xA;" OwnerUserId="123" LastActivityDate="2010-07-25T05:12:44.320" />
  <row Id="674" PostTypeId="1" AcceptedAnswerId="2405" CreationDate="2010-07-25T05:14:24.020" Score="6" ViewCount="508" Body="&lt;p&gt;I have been using Lagrange multipliers in constrained optimization problems, but I don't see &lt;em&gt;how&lt;/em&gt; they actually work to simultaneously satisfy the constraint &lt;em&gt;and&lt;/em&gt; find the lowest possible value of an objective function.&lt;/p&gt;&#xA;" OwnerUserId="121" LastEditorUserId="121" LastEditorDisplayName="" LastEditDate="2010-07-25T06:15:16.997" LastActivityDate="2010-08-14T14:51:28.360" CommunityOwnedDate="2010-07-25T05:14:24.020" Title="How do Lagrange multipliers work to find the lowest value of a function subject to a constraint?" Tags="&lt;calculus&gt;&lt;optimization&gt;" AnswerCount="4" CommentCount="3" FavoriteCount="4" />
  <row Id="675" PostTypeId="2" ParentId="674" CreationDate="2010-07-25T05:18:20.160" Score="5" ViewCount="" Body="&lt;p&gt;This type of problem is generally referred to as &lt;a href=&quot;http://en.wikipedia.org/wiki/Constraint_optimization&quot; rel=&quot;nofollow&quot;&gt;constrained optimization&lt;/a&gt;.  A general technique to solve many of these types of problems is known as the &lt;a href=&quot;http://en.wikipedia.org/wiki/Lagrange_multipliers&quot; rel=&quot;nofollow&quot;&gt;method of Lagrange multipliers&lt;/a&gt;, here is an example of such a problem using Lagrange multipliers and a short justification as to why the technique works.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the parabaloid given by $f(x,y) = x^2 + y^2$.  The global minimum of this surface lies at the origin (at $x=0$, $y=0$).  If we are given the constraint, a requirement on the relationship between $x$ and $y$, that $3x+y=6$, then the origin can no longer be our solution (since $3\cdot 0 + y \cdot 0 \neq 6$).  Yet, there is a lowest point on this function satisfying the given constraint. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What we have so far:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Objective function: $f(x,y) = x^2 + y^2$, &lt;br&gt;&#xA;subject to: $3x+y=6$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From here we can derive the Lagrange formulation of our constrained minimization problem.  This will be a function $L$ of $x$, $y$, and a single Lagrange multiplier $\lambda$ (since we have only a single constraint). &lt;em&gt;It will be this new function that we minimize.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$L(x,y,\lambda) = x^2 + y^2 + \lambda(3x+y-6)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Lagrange formulation incorporates our original function along with our constraint(s).  On the way toward minimizing $L$, we will have to minimize the objective function $x^2 + y^2$, as well as minimize the contribution from the constraint, which is now weighted by a factor of $\lambda$.  If the constraint is met, then the expression $3x+y-6$ will necessarily be zero, and will not contribute anything the value of $L$. This is the &lt;em&gt;trick&lt;/em&gt; of the technique.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Minimizing the Lagrange formulation:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To minimize $L$ we simply find the $x,y$, and $\lambda$ values that make its gradient zero.  (This is exactly analogous to setting the first derivative to zero in calculus.)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\nabla L = 0:$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{\partial L}{\partial x} = 2x + 3 \lambda = 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{\partial L}{\partial y} = 2y + \lambda = 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{\partial L}{\partial \lambda} = 3x + y - 6 = 0$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In our example we have arrived at a system of simultaneous linear equations which can (and should) be solved with matrix algebra.  The solution will be a vector holding values for $x, y$ and $\lambda$.  The lowest value of the objective function, subject to the given constraint, sits at $(x,y,f(x,y))$, and the Lagrange multiplier does not have an immediate physical interpretation.  (The multipliers have meaning when appearing in certain contexts, more info on that &lt;a href=&quot;http://en.wikipedia.org/wiki/Lagrange_multipliers#Interpretation_of_the_Lagrange_multipliers&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.)&lt;/p&gt;&#xA;" OwnerUserId="121" LastEditorUserId="121" LastEditorDisplayName="" LastEditDate="2010-07-25T06:07:09.480" LastActivityDate="2010-07-25T06:07:09.480" CommunityOwnedDate="2010-07-25T05:18:20.160" />
  <row Id="676" PostTypeId="1" AcceptedAnswerId="679" CreationDate="2010-07-25T05:31:54.210" Score="5" ViewCount="382" Body="&lt;p&gt;Randomly break a stick (or a piece of dry spaghetti, etc.) in two places, forming three pieces.  The probability that these three pieces can form a triangle is 1/4 (coordinatize the stick form 0 to 1, call the breaking points x and y, consider the unit square of the coordinate plane, shade the areas that satisfy the triangle inequality &lt;em&gt;edit&lt;/em&gt;: see comments on the question, below, for a better explanation of this).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other day in class&lt;sup&gt;*&lt;/sup&gt;, my professor was demonstrating how to do a Monte Carlo simulation of this problem on a calculator and wrote a program that, for each trial did the following:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Pick a random number x between 0 and 1.  This is the first side length.&lt;/li&gt;&#xA;&lt;li&gt;Pick a random number y between 0 and 1 - x (the remaning part of the stick).  This is the second side length.&lt;/li&gt;&#xA;&lt;li&gt;The third side length is 1 - x - y.&lt;/li&gt;&#xA;&lt;li&gt;Test if the three side lengths satisfy the triangle inequality (in all three permutations).&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;He ran around 1000 trials and was getting 0.19, which he said was probably just random-chance error off 0.25, but every time the program was run, no matter who's calculator we used, the result was around 0.19.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What's wrong with the simulation method?  What is the theoretical answer to the problem actually being simulated?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(&lt;sup&gt;*&lt;/sup&gt; the other day was more than 10 years ago)&lt;/p&gt;&#xA;" OwnerUserId="72" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-07-25T07:03:56.587" LastActivityDate="2010-07-25T07:03:56.587" Title="Probability that a stick randomly broken in two places can form a triangle" Tags="&lt;probability&gt;&lt;monte-carlo&gt;" AnswerCount="1" CommentCount="9" />
  <row Id="677" PostTypeId="1" AcceptedAnswerId="1105" CreationDate="2010-07-25T06:00:35.030" Score="10" ViewCount="173" Body="&lt;p&gt;About two years ago, I started thinking about the following problem: You're given an N and an S, positive integers. You start with an N-gon that has positive integer labels at each vertex, such that the labels sum to S. A &quot;move&quot; is replacing the label at some vertex with the positive difference of the labels at its two adjacent vertices. You win if you can exhibit a sequence of moves such that all of the vertex labels are zero at the end.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a generalization of 2003 USAMO problem B3, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;A positive integer is written at each vertex of a hexagon. A move is to replace a number by the (non-negative) difference between the two numbers at the adjacent vertices. If the starting numbers sum to 2003^2003, show that it is always possible to make a sequence of moves ending with zeros at every vertex.&quot; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The USAMO problem itself can be answered in the affirmative by doing a modified greedy algorithm for most of the moves, showing there's always a move where you can reduce the problem to a smaller sum while changing the parities systematically, then doing casework for the &quot;end game&quot; to show that each configuration of 1s and 0s (or k's and zeros, the cases are of course equivalent) can be gotten down to all zeros.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's easy to check that for n=4 and S odd, there's a winning strategy. I have lots of simulation/numerical evidence and some half baked ideas to suggest that, in general, there's a winning strategy iff N and S have opposite parities. But in two years of coming back to it on and off, I haven't really made progress. It's open, as far as I know, though I knew that more surely two years ago than I do now.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ideas?&lt;/p&gt;&#xA;" OwnerUserId="76" LastActivityDate="2010-07-29T03:30:58.303" Title="Proving that this Game on Polygons Ends" Tags="&lt;open-problem&gt;&lt;contest-math&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="2" />
  <row Id="679" PostTypeId="2" ParentId="676" CreationDate="2010-07-25T06:29:45.050" Score="10" ViewCount="" Body="&lt;p&gt;The three triangle inequalities are&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;x + y &amp;gt; 1-x-y&#xA;x + (1-x-y) &amp;gt; y&#xA;y + (1-x-y) &amp;gt; x&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Your problem is that in picking the smaller number first from a uniform distribution, it's going to end up being bigger than it would if you had just picked two random numbers and taken the smaller one. (You'll end up with an average value of $1/2$ for the smaller instead of $1/3$ like you actually want.) Now when you pick $y$ on $[0, 1-x]$, you're making it smaller than it should be (ending up with average value of $1/4$). To understand this unequal distribution, we can substitute $y (1-x)$ for $y$ in the original inequalities and we'll see the proper distribution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/CARdj.png&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Note that the $y$-axis of the graph doesn't really go from $0$ to $1$; instead the top represents the line $y=1-x$. I'm showing it as a square because that's how the probabilities you were calculating were being generated.) Now the probability you're measuring is the area of the strangely-shaped region on the left, which is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \int_0^{1/2} \frac{1}{2-2x} - \frac{2x-1}{2x-2} \,dx = \ln(2) - 1/2 \approx 0.19314. $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe that's the answer you calculated.&lt;/p&gt;&#xA;" OwnerUserId="45" LastEditorUserId="45" LastEditorDisplayName="" LastEditDate="2010-07-25T06:58:50.747" LastActivityDate="2010-07-25T06:58:50.747" CommentCount="4" />
  <row Id="680" PostTypeId="1" AcceptedAnswerId="681" CreationDate="2010-07-25T06:57:11.093" Score="6" ViewCount="212" Body="&lt;p&gt;Two mathematicians each come into a coffee shop at a random time between 8:00 a.m. and 9:00 a.m. each day. Each orders a cup of coffee then sits at a table, reading a newspaper for 20 minutes before leaving to go to work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On any day, what is the probability that both mathematicians are at the coffee shop at the same time (that is, their arrival times are within 20 minutes of each other)?&lt;/p&gt;&#xA;" OwnerUserId="45" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-07-29T23:29:27.107" LastActivityDate="2010-09-29T04:19:45.697" Title="Probability that two people see each other at the coffee shop" Tags="&lt;probability&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" />
  <row Id="681" PostTypeId="2" ParentId="680" CreationDate="2010-07-25T07:13:22.623" Score="13" ViewCount="" Body="&lt;p&gt;Working in hours and letting 8:00 a.m. be t=0, each mathematician's arrival time is a number between 0 and 1.  The sample space can be represented by the unit square in the coordinate plane with one professor's arrival time as x and the other's as y, where regions with equal areas are equally likely.  We want x - 1/3 &amp;lt; y &amp;lt; x + 1/3 -- that is, the second professor arrives earlier than the first by no more than 1/3 of an hour or later than the first by no more than 1/3 of an hour.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/Z2M3h.png&quot; alt=&quot;plot of the inequalities&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The area of the desired region is 5/9.&lt;/p&gt;&#xA;" OwnerUserId="72" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-09-29T04:19:45.697" LastActivityDate="2010-09-29T04:19:45.697" CommentCount="7" />
  <row Id="682" PostTypeId="2" ParentId="593" CreationDate="2010-07-25T07:48:10.590" Score="4" ViewCount="" Body="&lt;p&gt;The answer is, it's just a fact &amp;ldquo;cone over a simplex is a simplex&amp;rdquo; rewritten in terms of the generating function:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;observe that because n-simplex is a cone over (n-1)-simplex $\frac{\partial}{\partial x}vol(\text{n-simplex w. edge x}) = vol(\text{(n-1)-simplex w. edge x})$; in other words $e(x):=\sum_n vol\text{(n-simplex w. edge x)}$ satisfies an equvation $e'(x)=e(x)$. So $e(x)=Ce^x$ -- and C=1 because e(0)=1.&lt;/p&gt;&#xA;" OwnerUserId="152" LastEditorUserId="152" LastEditorDisplayName="" LastEditDate="2010-07-30T07:28:06.127" LastActivityDate="2010-07-30T07:28:06.127" CommentCount="2" />
  <row Id="683" PostTypeId="1" CreationDate="2010-07-25T07:56:11.163" Score="8" ViewCount="95" Body="&lt;p&gt;I want to randomly generate trees, i.e. undirected acyclic graphs with a single root, making sure that all possible trees with a fixed number of nodes &lt;em&gt;n&lt;/em&gt; are equally likely.&lt;/p&gt;&#xA;" OwnerUserId="191" LastActivityDate="2010-07-25T19:40:11.843" Title="How can I randomly generate trees?" Tags="&lt;probability-theory&gt;&lt;random&gt;" AnswerCount="1" />
  <row Id="684" PostTypeId="2" ParentId="683" CreationDate="2010-07-25T08:24:10.440" Score="4" ViewCount="" Body="&lt;p&gt;Knuth says to look at it as generating all nested parentheses in lexicographic order.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Look here for the details&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www-cs-faculty.stanford.edu/~uno/fasc4a.ps&quot; rel=&quot;nofollow&quot;&gt;http://www-cs-faculty.stanford.edu/~uno/fasc4a.ps&lt;/a&gt;. &lt;/p&gt;&#xA;" OwnerUserId="99" LastEditorUserId="99" LastEditorDisplayName="" LastEditDate="2010-07-25T19:40:11.843" LastActivityDate="2010-07-25T19:40:11.843" CommentCount="6" />
  <row Id="685" PostTypeId="2" ParentId="618" CreationDate="2010-07-25T09:58:14.807" Score="3" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Basic Theory&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The way to solve this problem is to calculate how much each payment reduces your debt after you have been repaying your loan for $n$ years. Let $r=1+R/100$, ie. this converts the interest rate from a percentage to a value you can multiply your debt by to calculate how much you owe after adding one time period's interest.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I make a payment of $P$ at the end of the $k$th year, then we avoid paying interest on this money $n-k$ times and so we reduce our debt by $Pr^{n-k}$. We sum up the future values of all our payments:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sum\limits_{k=1}^n Pr^{n-k}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we reverse this, it is equivalent to:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sum\limits_{k=0}^{n-1} Pr^k$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a &lt;a href=&quot;http://en.wikipedia.org/wiki/Geometric_series&quot; rel=&quot;nofollow&quot;&gt;geometric series&lt;/a&gt;, which can be solved using the formula $\frac{ar^{n-1}}{r-1}$ where $a$ is the first term, $r$ is the factor and $n$ is the number of terms being summed. We then attempt to equate this with the debt owed after $n$ years, which is $Mr^n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We now compare the two equations:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{Pr^{n-1}}{r-1} = Mr^n$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Calculating $n$&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We group the $r^n$ terms:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{P}{r-1} = r^n\frac{M-P}{r-1}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$r^n = \frac{P}{M(r-1)-P}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we just take the $n$th log of the right hand side.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Calculating repayments&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given the principal ($M$) and the interest rate ($r$), what will my payment-per-term ($P$) be over $n$ accruation terms?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$P=\frac{Mr^n(r-1)}{r^{n-1}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Payments made at the start of the year&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case, the future values of our interest payment simply become:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sum\limits_{k=1}^n Pr^k$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We proceed as we did before.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Notes&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We could also solve this problem using present value instead of future value.&lt;/p&gt;&#xA;" OwnerUserId="123" LastEditorUserId="11590" LastEditorDisplayName="" LastEditDate="2011-05-31T19:38:59.490" LastActivityDate="2011-05-31T19:38:59.490" />
  <row Id="686" PostTypeId="1" AcceptedAnswerId="691" CreationDate="2010-07-25T10:54:29.373" Score="4" ViewCount="191" Body="&lt;p&gt;Suppose that I am buying cakes for a party. There are k different types and I intend to buy a total of n cakes. How many different combinations of cakes could I possibly bring to the party?&lt;/p&gt;&#xA;" OwnerUserId="123" LastActivityDate="2010-07-25T15:52:28.867" Title="Combinations of selecting n objects with k different types" Tags="&lt;combinatorics&gt;" AnswerCount="2" CommentCount="8" FavoriteCount="1" />
  <row Id="687" PostTypeId="1" AcceptedAnswerId="1050" CreationDate="2010-07-25T11:08:30.817" Score="5" ViewCount="97" Body="&lt;p&gt;Suppose we model traffic flow between two points with a directed graph. Each route has either a constant travel time or one that linearly increases with traffic. We assume that each driver wishes to minimise their own travel time and we assume that the drivers form a Nash equilibria. Can removing a route ever decrease the average travelling time?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the existence of multiple Nash equilibria makes this question a bit complicated. To clarify, I am looking for a route removal that will guarantee a decrease in the average traveling time regardless of the Nash equilibria that are chosen before and after.&lt;/p&gt;&#xA;" OwnerUserId="123" LastEditorUserId="123" LastEditorDisplayName="" LastEditDate="2010-07-28T21:02:19.803" LastActivityDate="2010-07-28T21:02:19.803" Title="Increasing network throughput by cutting routes" Tags="&lt;graph-theory&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="690" PostTypeId="2" ParentId="686" CreationDate="2010-07-25T14:37:35.910" Score="1" ViewCount="" Body="&lt;p&gt;Let g(n,k) = # combinations of cakes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice that:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;g(n,1) = 1. (all the cakes are the same)&lt;/li&gt;&#xA;&lt;li&gt;g(n,2) = n+1. (e.g. for 5 cakes, the # of cakes of type 1 can be 0, 1, 2, 3, 4, 5)&lt;/li&gt;&#xA;&lt;li&gt;g(1,k) = k.&lt;/li&gt;&#xA;&lt;li&gt;g(2,k) = k*(k-1)/2 + k (the first term is two different cakes; the second term is when both cakes are the same), as long as k &gt; 1. (otherwise g(2,1) = 1)&lt;/li&gt;&#xA;&lt;li&gt;g(3,k) = k * (k-1) * (k-2)/6 + k*(k-1)/2 * 2 + k (the first term is 3 different cakes; the second term is 2 different cakes, with a *2 since there are two choices for which one to duplicate, the third term is when all 3 cakes are the same), as long as k &gt; 2.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;If we think of k as a radix rather than the # of cakes, then this problem is equivalent to expressing the # of distinct n-digit numbers in base k whose digits are in sorted order. (e.g. 1122399 is equivalent to 9921231)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I &lt;em&gt;think&lt;/em&gt; I can express it as a nonrecursive sum:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;g(n,k) = sum from j=1 to max(n,k) of { (k choose j) * h(n,j) } &lt;/p&gt;&#xA;&#xA;&lt;p&gt;where h(n,j) is the # of ways to partition N cakes using j different types. (the term in the sum is when there are j distinct cakes actually chosen.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But that's about as far as I can get... :/&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;edit: looks like it's &lt;a href=&quot;http://en.wikipedia.org/wiki/Combination#Number_of_combinations_with_repetition&quot; rel=&quot;nofollow&quot;&gt;combinations with repetitions&lt;/a&gt; = ((k+n-1) choose n). (same as the wikipedia article with n and k swapped)&lt;/p&gt;&#xA;" OwnerUserId="120" LastActivityDate="2010-07-25T14:37:35.910" />
  <row Id="691" PostTypeId="2" ParentId="686" CreationDate="2010-07-25T15:52:28.867" Score="8" ViewCount="" Body="&lt;p&gt;Using a method that's often called &quot;stars and bars&quot;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We draw $n$ stars in a row to represent the cakes, and $k-1$ bars to divide them up. All of the stars to the left of the first bar are cakes of the first type; stars between the first two bars are of the second type; . . . .&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;**|***||*|&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Here's an example with $n=6$ and $k=5$. We're getting 2 of the first type, 3 of the second type, 0 of the third type, 1 of the fourth type, and 0 of the fifth type.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to solve the problem, we just need to reorder the stars and bars by choosing the $k-1$ spots for the bars out of the $n+k-1$ total spots, so our answer is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \binom{n+k-1}{k-1}. $$&lt;/p&gt;&#xA;" OwnerUserId="45" LastActivityDate="2010-07-25T15:52:28.867" CommentCount="1" />
  <row Id="692" PostTypeId="2" ParentId="154" CreationDate="2010-07-25T15:57:22.297" Score="1" ViewCount="" Body="&lt;p&gt;Complex numbers are the final step in a sequence of increasingly &quot;unreal&quot; extensions to the number system that humans have found it necessary to add over the centuries in order to express significant numerical concepts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first such &quot;unreal number&quot; was zero, back in the mists of time. It seems obvious to us now, but it must have seemed strange at first. How can the number of sheep I have be zero, when I don't actually have any sheep?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Negative numbers are the next most obvious addition to the family of numbers. But what does it mean to have -2 apples? If I have 3 apples and you have 5, it's convenient to be able to say that I have -2 more apples than you. Even so, during the middle ages many mathematicians were very uncomfortable with the idea of negative numbers and tried to arrange their equations so that they didn't occur.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rationals (fractions) seem real enough, since I'm happy to have 2/5 of a pizza. However, this is not the number of pizzas that I have, just a ratio between 0 and 1 pizzas, and so is further removed from the concept of counting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More significant philosophical problems problems arose when irrational numbers were first discovered by the classical Greeks. They were astonished when Euclid (or one of his predecessors) proved that the diagonal of a unit square could not be represented by ratio of two integers. This was such an outrageous idea to them that they called these numbers &quot;irrational&quot;. However, they couldn't easily deny their existence since they have a direct geometric representation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Irrational numbers were first understood as the solutions to algebraic equations such as x^2 = 2 but this still doesn't cover all the possible numbers that we need. For example, the ratio of a circle's diameter to its circumference is π and so has a direct geometric representation. However, in 1882 π was proved to be transcendental, meaning that it can't be defined as the solution to a specific algebraic equation. This begins to seem a lot less &quot;real&quot;, especially when you consider that there are many important transcendental numbers that, unlike π, don't have a geometric interpretation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are of course many algebraic equations that don't have a solution even among the irrational numbers, and in some ways there's no reason why they should. However, when 16th century mathematicians like Gerolamo Cardano began working on solutions to cubic equations they found that the square roots of negative numbers began cropping up very naturally in their procedures, even though the solutions themselves were purely real. This eventually led people to explore the arithmetic of complex numbers and they were surprised to find that it produced a consistent and beautiful theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, complex numbers don't have such an intuitively obvious geometrical meaning as the numbers that came before. They are typically represented graphically as points in the 2D plane, and the rules of addition and multiplication are equivalent to certain operations on lengths and angles, but those operations aren't driven by geometrical necessity in quite the way that squares and circles are. Even so, complex numbers are a perfect representation for various physical phenomena such as the state of particles in quantum mechanics and the behaviour of varying currents in electrical circuits. They are also very useful for reducing the cost of computation in 3D computer graphics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The really special thing about complex numbers, though, is that they are the end of this journey that has been going on for millenia. There is no need to invent further number systems to provide solutions to problems expressed in terms of complex numbers because now every non-contradictory equation, algebraic or transcendental, has a solution within the complex numbers. They are a self-sufficient, consistent system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Physicists have been breaking matter down into smaller and smaller particles over the years. Molecules, atoms, nuclei, protons, quarks. When will the process stop? For several decades it was felt that quarks could be the final, indivisible particle. However, new theoretical frameworks such as string theory are suggesting that there may be more fundamental entities than quarks. Because physics ultimately relies on experimental verification, we can never be sure that there aren't going to be more steps in the sequence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In mathematics, however, we can prove things for all time. Complex numbers are the final step in the sequence, the numbers that we have been reaching for since before the beginning of recorded history. Every other number system is just a subset of the complex numbers, just a part of the true picture. Complex numbers are the real thing.&lt;/p&gt;&#xA;" OwnerUserId="113" LastActivityDate="2010-07-25T15:57:22.297" CommunityOwnedDate="2011-04-15T20:45:29.080" CommentCount="3" />
  <row Id="693" PostTypeId="1" AcceptedAnswerId="694" CreationDate="2010-07-25T16:09:21.277" Score="2" ViewCount="87" Body="&lt;p&gt;I need n cakes for a party. I go to the cake shop and there are k different kinds of cake. For variety, I'd like to get at least one of every cake. How many ways can I do this?&lt;/p&gt;&#xA;" OwnerUserId="45" LastActivityDate="2010-08-05T19:13:10.093" Title="Picking cakes if we need at least one of each type" Tags="&lt;combinatorics&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="694" PostTypeId="2" ParentId="693" CreationDate="2010-07-25T16:18:22.500" Score="6" ViewCount="" Body="&lt;p&gt;Similar to &lt;a href=&quot;http://math.stackexchange.com/questions/686/combinations-of-selecting-n-objects-with-k-different-types/691#691&quot;&gt;the stars and bars technique&lt;/a&gt;, consider the n cakes as a row of n stars *.  Instead of permuting them with k-1 bars | (which allows two bars next to each other, giving 0 of a type, place the k-1 bars (needed to split the n stars into k types) into the n-1 spaces between the stars, allowing at most one bar per space.  The number of ways to do this is ${n-1 \choose k-1}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternately, since you need one of each type, there are only n-k cakes for which you are choosing types.  Using &lt;a href=&quot;http://math.stackexchange.com/questions/686/combinations-of-selecting-n-objects-with-k-different-types/691#691&quot;&gt;the stars and bars technique&lt;/a&gt;, there are ${(n-k)+k-1 \choose k-1} = {n-1 \choose k-1}$ ways to do it.&lt;/p&gt;&#xA;" OwnerUserId="72" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-08-05T19:13:10.093" LastActivityDate="2010-08-05T19:13:10.093" CommentCount="3" />
  <row Id="696" PostTypeId="2" ParentId="656" CreationDate="2010-07-25T20:41:20.947" Score="2" ViewCount="" Body="&lt;p&gt;Of course there are infinite equation (even if you require them to be infinitely differentiable...) that satisfy the given constraints. &#xA;As Isaac and Justin already wrote, you may always find a polynomial of degree at most n-1 (where n is the number of points given) which satisfies the given data; but you cannot be sure that this is the right answer. &#xA;Moreover, if data is not exact but approximate the resulting polynomial may be quite different from the correct function, since it would likely hade huge peaks and falls. In such cases, an approximate method like &lt;a href=&quot;http://en.wikipedia.org/wiki/Least_squares&quot; rel=&quot;nofollow&quot;&gt;least squares&lt;/a&gt; could be more useful.&lt;/p&gt;&#xA;" OwnerUserId="89" LastActivityDate="2010-07-25T20:41:20.947" CommentCount="1" />
  <row Id="697" PostTypeId="1" CreationDate="2010-07-25T22:19:40.330" Score="5" ViewCount="195" Body="&lt;p&gt;In least-squares approximations the normal equations act to project a vector existing in N-dimensional space onto a lower dimensional space, where our problem actually lies, thus providing the &quot;best&quot; solution we can hope for (the orthogonal projection of the N-vector onto our solution space).  The &quot;best&quot; solution is the one that minimizes the &lt;strong&gt;Euclidean distance (two-norm)&lt;/strong&gt; between the N-dimensional vector and our lower dimensional space. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There exist other norms and other spaces besides $\mathbb{R}^d$, what are the analogues of least-squares under a different norm, or in a different space?&lt;/p&gt;&#xA;" OwnerUserId="121" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-19T00:56:38.330" LastActivityDate="2010-11-19T00:56:38.330" CommunityOwnedDate="2010-08-10T18:09:45.803" Title="Are there variations on least-squares approximations?" Tags="&lt;linear-algebra&gt;&lt;big-list&gt;&lt;approximation&gt;&lt;regression&gt;" AnswerCount="3" CommentCount="4" FavoriteCount="1" />
  <row Id="698" PostTypeId="1" AcceptedAnswerId="1472" CreationDate="2010-07-25T22:55:49.980" Score="0" ViewCount="183" Body="&lt;p&gt;Recently I discussed an experiment with a friend. Assume we start a random experiment. At first there is an array with size 100 000, all set to 0. We calculate at each round a random number modulo 2 and select one random position in that array. If the number in the array is 1, nothing is changed and otherwise the pre-computed value is set. The question is: how many distinct hash values would we have added in 1%, 5%, 50%, 95%, 99% of all cases?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example: 4 rounds with array of size 10:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Array                     Position   random number&#xA;[0,...,0]                    5              0&#xA;[0,...,0]                    7              1&#xA;[0,...0,1,0,0,0]             6              1&#xA;[0,..0,.1,1,0,0,0]           6              0&#xA;[0,..0,.1,1,0,0,0]           2              0&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;First we considered this a somehow simple problem. But after thinking about for some hours, searching the web and asking some math students we couldn't find a solution.So do you know a probability distribution for this problem? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remark: Was also posted on &lt;a href=&quot;http://mathoverflow.net/questions/33335/looking-for-a-probability-distribution&quot; rel=&quot;nofollow&quot;&gt;Math Overflow&lt;/a&gt; and got its answer there.&lt;/p&gt;&#xA;" OwnerUserId="203" LastEditorUserId="203" LastEditorDisplayName="" LastEditDate="2010-08-02T11:56:18.977" LastActivityDate="2010-08-03T20:32:23.647" Title="Looking for a probability distribution" Tags="&lt;probability-theory&gt;&lt;probability&gt;" AnswerCount="1" CommentCount="6" />
  <row Id="699" PostTypeId="2" ParentId="697" CreationDate="2010-07-26T00:43:42.073" Score="0" ViewCount="" Body="&lt;p&gt;Sure, there are variations on least-squares approximations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's an engineering answer (not really a pure math answer): On one project I did at my company, we had a thermal model that was approximating the thermal transfer function from measured power to device temperature. The cost of error in the positive direction (underestimating temperature) was worse than the cost of the same error in the negative direction (overestimating temperature) -- so we used a weighting function that was (K * error ^ 2) where K was 1 for negative temperature error and greater than 1 (e.g. 1.5 or 2) for positive temperature error.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We thought about using more complicated mappings (10 degrees underestimate much worse than 1 degree underestimate) but didn't want to go there... I assume this has some analog to a utility function (e.g. expected monetary gain of a system with random outcome has nonlinear mapping to &quot;happiness&quot; or &quot;utility&quot;) where nonlinearity is intentionally introduced.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could do something similar for approximating functions with polynomials: a least-squares fit treats error in a linear way, but there may be places in the function (e.g. at the ends or at the center) where minimizing error is more or less important.&lt;/p&gt;&#xA;" OwnerUserId="120" LastActivityDate="2010-07-26T00:43:42.073" />
  <row Id="700" PostTypeId="2" ParentId="57" CreationDate="2010-07-26T01:05:00.097" Score="1" ViewCount="" Body="&lt;p&gt;The calculus of relations is an algebra of operations over sets of pairs of individuals, where for any relation R, we can express the relation in the usual infix manner: x R y iff (x,y) &amp;isin; R.  This allows all of the properties of relations to be expressed in equational form.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are four fundamental relations we want to define, each of which make non-useless examples of three of your five properties of relations, plus one other useful property, irreflexivity:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Eq = {(x,x) | for each individual x}.  This is the diagonal or equality relation that holds only between something and itself.  It is reflexive, symmetric and transitive, which is to say it is an equivalence relation.&lt;/li&gt;&#xA;&lt;li&gt;All = {(x,y) | for all individuals x,y}.  The whenever relation, it is also an equivalence relation.&lt;/li&gt;&#xA;&lt;li&gt;Neq = {(x,y) | for all individuals x,y for which x &amp;ne; y}: the inequality relation that holds between anything different.  It is symmetric and irreflexive, but not transitive.&lt;/li&gt;&#xA;&lt;li&gt;Never is the empty set.  It is symmetric and transitive, and irreflexive.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Three basic binary operations on relations, and one unary relation:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;R&amp;cap;S = {(x,y) | (x,y) &amp;isin; R and (x,y) &amp;isin; S};&lt;/li&gt;&#xA;&lt;li&gt;R&amp;ndash;S = {(x,y) | (x,y) &amp;isin; R and (x,y) &amp;notin; S};&lt;/li&gt;&#xA;&lt;li&gt;R&amp;sdot;S = {(x,z) | there is some y such that (x,y) &amp;isin; R and (y,z) &amp;isin; S} (composition);&lt;/li&gt;&#xA;&lt;li&gt;tr(R) = {(y,x) | (x,y) &amp;isin; R} (transpose).&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Observe that Never = All&amp;ndash;All, and Neq = All&amp;ndash;Eq.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We say R -&gt; S when S holds whenever R holds.  This is the same as saying either that  R is a subset of S, or that R&amp;cap;S=R, or that R-S=Never.  So Never -&gt; Eq, and Eq -&gt; All.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then we can express your five relations:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;R is reflexive when Eq -&gt; R; dually we have an additional property of a relation, where R is anti-reflexive when R -&gt; Neq;&lt;/li&gt;&#xA;&lt;li&gt;R is symmetric when R -&gt; tr(R);&lt;/li&gt;&#xA;&lt;li&gt;R is anti-symmetric when R &amp;cap; tr(R) -&gt; Eq, or equivalently when R &amp;cap; tr(R) &amp;cap; Neq = 0.&lt;/li&gt;&#xA;&lt;li&gt;R is asymmetric when R &amp;cap; tr(R)=Never: a relation is asymmetric when it is anti-symmetric and anti-reflexive;&lt;/li&gt;&#xA;&lt;li&gt;R is transitive when R&amp;sdot;R -&gt; R.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Also, how can a relation be a- and antisymmetrical at the same time? Don't they cancel each other out?&lt;/em&gt;  &amp;mdash; Look at Eq: it is both symmetric and anti-symmetric, although it is not asymmetric.  In fact, all asymmetric relations are anti-symmetric, but not vice versa: the difference is that asymmetric and anti-symmetric differ in what they assert about the diagonal: anti-symmetric doesn't care about what pairs there might be along the diagonal, whilst asymmetric insists that there are no pairs along the diagonal, which is irreflexivity.&lt;/p&gt;&#xA;" OwnerUserId="100" LastEditorUserId="100" LastEditorDisplayName="" LastEditDate="2010-07-28T21:12:19.983" LastActivityDate="2010-07-28T21:12:19.983" CommentCount="2" />
  <row Id="701" PostTypeId="1" CreationDate="2010-07-26T04:55:18.153" Score="5" ViewCount="223" Body="&lt;p&gt;Hi, I've just came back from my Mathematics of Packing and Shipping lecture, and I've run into a problem I've been trying to figure out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's say I have a rectangle of length $l$ and width $w$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a simple equation that can be used to show me how many circles of radius $r$ can be packed into the rectangle, in &lt;strong&gt;the optimal way&lt;/strong&gt;?  So that no circles overlap. ($r$ is less than both $l$ and $w$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm rather in the dark as to what the optimum method of packing circles together in the least amount of space is, for a given shape.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An equation with a non-integer output is useful to me as long as the truncated (rounded down) value is the true answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I'm not that interested in &lt;strong&gt;how&lt;/strong&gt; the circles would be packed, as I am going to go into business and only want to know how much I can demand from the packers I hire to pack my product)&lt;/p&gt;&#xA;" OwnerUserId="51" LastEditorUserId="51" LastEditorDisplayName="" LastEditDate="2010-07-26T06:22:12.773" LastActivityDate="2010-07-26T12:42:57.450" Title="How many circles of a given radius can be packed into a given rectangular box?" Tags="&lt;geometry&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="2" />
  <row Id="702" PostTypeId="2" ParentId="603" CreationDate="2010-07-26T05:26:53.267" Score="0" ViewCount="" Body="&lt;p&gt;Couldn't all transformation which send each point (x,y) to another point (x',y') which can be computed from the first one by performing only the four operations and extraction of square root?&lt;/p&gt;&#xA;" OwnerUserId="89" LastActivityDate="2010-07-26T05:26:53.267" CommentCount="1" />
  <row Id="704" PostTypeId="1" AcceptedAnswerId="9229" CreationDate="2010-07-26T07:06:24.977" Score="3" ViewCount="138" Body="&lt;p&gt;Is there symbolism (or even a name) for groups of statements in which any fixed-number of them imply all the rest?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, in linear algebra, a basis is sometimes defined as a set of n-dimensional vectors which:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Is linearly independent&lt;/li&gt;&#xA;&lt;li&gt;Spans the vector space&lt;/li&gt;&#xA;&lt;li&gt;Contains exactly n elements&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;However, it is then shown that any TWO of these statements being true implies the third is true.  How would you write this symbolically? (other than $A \wedge B \Rightarrow C, A \wedge C \Rightarrow B, B \wedge C \Rightarrow A$, which does not scale well...)&lt;/p&gt;&#xA;" OwnerUserId="136" LastActivityDate="2010-11-07T03:13:54.083" Title="n out of m theorems (some imply the rest)" Tags="&lt;terminology&gt;&lt;logic&gt;&lt;notation&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="705" PostTypeId="1" CreationDate="2010-07-26T07:13:48.847" Score="2" ViewCount="558" Body="&lt;p&gt;Given&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;A straight line of arbitrary length&lt;/li&gt;&#xA;&lt;li&gt;The ability to construct a straight line in any direction from any starting point with the &quot;unit length&quot;, or the length whose square root of its magnitude yields its own magnitude.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Is there a way to geometrically construct (using only a compass and straightedge) the a line with the length of the square root of the arbitrary-lengthed line?  What is the mathematical basis?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, why can't this be done without the unit line length?&lt;/p&gt;&#xA;" OwnerUserId="51" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-07-26T08:25:37.177" LastActivityDate="2010-07-26T08:52:06.800" Title="Compass-and-straightedge construction of the square root of a given line?" Tags="&lt;geometry&gt;&lt;euclidean-geometry&gt;&lt;geometric-construction&gt;&lt;compass-and-straightedge&gt;" AnswerCount="2" CommentCount="5" />
  <row Id="706" PostTypeId="2" ParentId="705" CreationDate="2010-07-26T07:29:47.623" Score="4" ViewCount="" Body="&lt;p&gt;Without the unit-length segment--that is, without something to compare the first segment to--its length is entirely arbitrary, so can't be valued, so there's no value of which to take the square root.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let the given segment (with length x) be AB and let point C be on ray AB such that BC = 1.  Construct the midpoint M of segment AC, construct the circle with center M passing through A, construct the line perpendicular to AB through B, and let D be one of the intersections of that line with the circle centered at M (call the other intersection E).  BD = sqrt(x).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;AC and DE are chords of the circle intersecting at B, so by the power of a point theorem, AB * BC = DB * BE, so x * 1 = x = DB * BE.  Since DE is perpendicular to AC and AC is a diameter of the circle, AC bisects DE and DB = BE, so x = DB^2 or DB = sqrt(x).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&lt;strong&gt;edit&lt;/em&gt;&lt;/strong&gt;: this is a special case of the more general geometric-mean construction.  Given two lengths AB and BC (arranged as above), the above construction produces the length BD = sqrt(AB * BC), which is the geometric mean of AB and BC.&lt;/p&gt;&#xA;" OwnerUserId="72" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-07-26T07:41:47.890" LastActivityDate="2010-07-26T07:41:47.890" CommentCount="2" />
  <row Id="707" PostTypeId="1" AcceptedAnswerId="789" CreationDate="2010-07-26T07:34:21.030" Score="1" ViewCount="177" Body="&lt;p&gt;Say there are three jars, $j_1, j_2, j_3$ filled with different binary sequences of length two.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The distribution of the binary sequences in each of the jars is given by the $p_i^k(1-p_i)^{n-k}$, where &#xA;$p_i = \frac{i}{m + 1}$ where m is the number of jars, i is the jar index, k is number of 1's and n is the length of the string.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So for three jars we have $p_1 = 0.25, p_2 = 0.5$, and $p_3 = 0.75$ for $j_1, j_2, j_3$ respectively.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are the sequences and their probabilities for $j_1$ with $p_1 = 0.25$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P(00) = 9 / 16&lt;br&gt;&#xA;P(10) = 3 / 16&lt;br&gt;&#xA;P(01) = 3 / 16&lt;br&gt;&#xA;P(11) = 1 / 16&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I tell you that I have selected a binary sequence and the first element is 1 what is the E($p_i$)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, this can be calculated by looking at each of the jars and adding up the probability of candidate sequences times the value of $p_i$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; I wasn't normalizing this conditionally space properly. I'm skipping a step which I'll explain, someone wants.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;E($p_i$) = (4/24 * 1/4) + (8/24 * 1/2) + (12/24 * 3/4) = 14 / 24 = 0.58&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the question is ... what is E($p_i$) when the numbers of jars goes to infinity (or alternatively, when p can take on values between 0 and 1)? Also what happens when the size of the binary strings goes to infinity? Does it have an effect on the outcome? If it does, does the order we take the limits change the answer?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And most importantly what is the general case for when I have $s$ 1's and $r$ 0's?, with a continuous p from 0 to 1 and infinite sequences?&lt;/p&gt;&#xA;" OwnerUserId="99" LastEditorUserId="99" LastEditorDisplayName="" LastEditDate="2010-07-27T19:38:43.117" LastActivityDate="2010-08-03T02:22:09.930" Title="The limiting case of a discrete probability problem" Tags="&lt;probability-theory&gt;&lt;probability&gt;" AnswerCount="1" CommentCount="7" />
  <row Id="708" PostTypeId="2" ParentId="705" CreationDate="2010-07-26T07:40:53.620" Score="3" ViewCount="" Body="&lt;p&gt;If you have a segment AB, place the unit lenght segment on the line where AB lies, starting with A and in the direction opposite to B; let C be the other point of the segment. Now draw a semicircle with diameter AC and the perpendicular to B; this line crosses the semicircle in a point D. BD is then the square root of AB. &#xA;Triangle BCD is a right triangle, like ACD and ABD; all of these are similar, so you find out that AC/AD = AD/AB. But AC=1, so AD = sqrt(AB).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See the drawing below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i26.tinypic.com/2ivgp03.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="89" LastEditorUserId="89" LastEditorDisplayName="" LastEditDate="2010-07-26T08:52:06.800" LastActivityDate="2010-07-26T08:52:06.800" CommentCount="1" />
  <row Id="709" PostTypeId="1" AcceptedAnswerId="6842" CreationDate="2010-07-26T07:49:39.480" Score="4" ViewCount="777" Body="&lt;p&gt;Hi, I'm taking a course on Basic Conic Sections, and one of the ones we are discussing is of a parabola of the form&lt;/p&gt;&#xA;&#xA;&lt;p&gt;y = a x&lt;sup&gt;2&lt;/sup&gt; + b x + c&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My teacher gave me the formula:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x = -\frac{b}{2a}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;as the x coordinate of the vertex.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I asked her why, and she told me not to tell her how to do her job.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My smart friend mumbled something about it involving calculus, but I've always found him a rather odd fellow and I doubt I'd be able to understand a solution involving calculus, because I have no background in it.  If you use something you know from calculus, explain it to someone who has no background in it.  Because I sure don't.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a purely algebraic or geometrical yet &lt;strong&gt;elegant&lt;/strong&gt; derivation for the x coordinate of a parabola of the above form?&lt;/p&gt;&#xA;" OwnerUserId="51" LastEditorUserId="51" LastEditorDisplayName="" LastEditDate="2010-10-16T04:36:44.793" LastActivityDate="2010-10-16T04:36:44.793" Title="Derivation of the formula for the vertex of a Parabola" Tags="&lt;algebra-precalculus&gt;&lt;plane-curves&gt;&lt;conic-sections&gt;" AnswerCount="10" CommentCount="3" FavoriteCount="1" />
  <row Id="710" PostTypeId="2" ParentId="709" CreationDate="2010-07-26T08:02:56.187" Score="13" ViewCount="" Body="&lt;p&gt;By the vertex I assume you mean the minimum/maximum point of the parabola. Indeed, this result can be discovered easily through a bit of calculus, but there is also a simple purely algebraic way, which I will present here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's consider a generic quadratic expression:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$y = ax^2 + bx + c$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We now &lt;a href=&quot;http://en.wikipedia.org/wiki/Completing_the_square&quot; rel=&quot;nofollow&quot;&gt;complete the square&lt;/a&gt; on this formula.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$y = a[x^2 + b/a + c/a]$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$y = a[(x + b/2a)^2 - (b/2a)^2 + c/a]$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The expression $- (b/2a)^2 + c/a$ is a constant (it does not depend on x), so we can replace it with &lt;code&gt;k&lt;/code&gt; for the matter of discussion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$y = a[(x + b/2a)^2 + k]$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, depending on whether a is positive or negative, the parabola given by y will either have a maximum or minimum. Since a and k are fixed, this must occur when $(x + b/2a)^2$ is zero (we know it cannot be less than zero, and it can extend to infinity).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, we know that for $(x + b/2a)^2$ to be zero, $x = -b/2a$. This in turn implies that the function y is at a minimum or a maximum when this is true. Q.E.D.&lt;/p&gt;&#xA;" OwnerUserId="56" LastActivityDate="2010-07-26T08:02:56.187" />
  <row Id="711" PostTypeId="2" ParentId="709" CreationDate="2010-07-26T08:05:03.357" Score="6" ViewCount="" Body="&lt;p&gt;Parabolas of the form you described (y = ...) are symmetric over a vertical line through their vertex.  Let's call that line x = k.  This means that if the graph crosses the x-axis (meaning that $ax^2+bx+c=0$ has real solution(s)), they must be equidistant from x = k, so (k,0) must be the midpoint of the segment with endpoints at the zeros of the quadratic or k is the average of the zeros.  From the quadratic formula, the two zeros of the quadratic are $\frac{-b\pm\sqrt{b^2-4ac}}{2a}$, so their sum is $-\frac{b}{a}$ and their average is $k=-\frac{b}{2a}$.  So, the x-coordinate of the vertex must be $-\frac{b}{2a}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the parabola does not cross the x-axis (no real solutions), there is another parabola with equation $y=ax^2+bx+c'$ for some $c'$ for which the graph is a vertical translation of the graph of the original quadratic, but crosses the x-axis.  Its axis of symmetry is $x=-\frac{b}{2a}$ and because it is a vertical translation of the original, the axis of symmetry of the original is also $x=-\frac{b}{2a}$, so the vertices of both have x-coordinate $-\frac{b}{2a}$.&lt;/p&gt;&#xA;" OwnerUserId="72" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-08-05T20:11:56.420" LastActivityDate="2010-08-05T20:11:56.420" />
  <row Id="712" PostTypeId="1" AcceptedAnswerId="715" CreationDate="2010-07-26T08:11:56.960" Score="10" ViewCount="305" Body="&lt;p&gt;I read &lt;a href=&quot;http://math.stackexchange.com/questions/625/why-is-the-derivative-of-a-circles-area-its-perimeter-and-similarly-for-spheres&quot; rel=&quot;nofollow&quot;&gt;this question&lt;/a&gt; the other day and it got me thinking: the area of a circle is $\pi r^2$, which differentiates to $2 \pi r$, which is just the perimeter of the circle. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Why doesn't the same thing happen for squares? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If we start with the area formula for squares, $l^2$, this differentiates to $2l$ which is sort of right but only &lt;em&gt;half&lt;/em&gt; the perimeter. I asked my calculus teacher and he couldn't tell me why. Can anyone explain???&lt;/p&gt;&#xA;" OwnerUserId="160" LastActivityDate="2010-07-26T09:41:01.373" Title="Why does area differentiate to perimeter for circles and not for squares?" Tags="&lt;geometry&gt;&lt;calculus&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="713" PostTypeId="1" CreationDate="2010-07-26T08:12:43.027" Score="8" ViewCount="484" Body="&lt;p&gt;The notion (rank-2) &quot;tensor&quot; appears in many different parts of physics, e.g. stress tensor, moment of inertia tensor, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know mathematically a tensor can be represented by a 3x3 matrix. But I can't grasp its geometrical picture — unlike scalar (a number) and vector (an arrow with direction and magnitude) which I can easily see what's going on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How to visualize a tensor?&lt;/p&gt;&#xA;" OwnerUserId="171" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-07-28T23:35:58.330" LastActivityDate="2010-08-22T01:44:03.140" Title="How to visualize a rank-2 tensor?" Tags="&lt;geometry&gt;&lt;linear-algebra&gt;&lt;intuition&gt;&lt;tensors&gt;" AnswerCount="4" CommentCount="4" FavoriteCount="1" />
  <row Id="714" PostTypeId="2" ParentId="712" CreationDate="2010-07-26T08:17:48.273" Score="6" ViewCount="" Body="&lt;p&gt;If you use the formula you describe for squares, your measurements are coming from one corner of the square.  Imagine those measurements growing slowly.  The square will grow, but only along the two sides opposite the corner from which you measured, so the derivative of the area formula is only the perimeter on those two sides.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternately, consider measuring the size of a square by the distance d from its center to the midpoint of a side.  This would make the side length 2d, the perimeter 8d and the area 4d^2.  Now, the derivative of the area is the perimeter.  (Also, if you imagine growing the square slowly with this measurement, it grows from the center outward, growing on all four sides.)&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-07-26T08:17:48.273" />
  <row Id="715" PostTypeId="2" ParentId="712" CreationDate="2010-07-26T08:17:57.990" Score="13" ViewCount="" Body="&lt;p&gt;Actually, it is also true for squares (and for regular polygons in general!), the problem you ran into is what the &lt;a href=&quot;http://en.wikipedia.org/wiki/Apothem&quot; rel=&quot;nofollow&quot;&gt;equivalent of &quot;r&quot;&lt;/a&gt; is. The side length of a square is actually more comparable to the circle's diameter.&lt;br&gt;&#xA;Instead, the correct analogue of the circle's radius is the distance from the center of the square to the midpoint of one side, which is only half as long as the square's side.&lt;br&gt;&#xA;&lt;img src=&quot;http://www.csrjjsmp.com/images/offsite/square.png&quot; alt=&quot;alt text&quot;&gt;&lt;br&gt;&#xA;Here, we have A=(2r)&lt;sup&gt;2&lt;/sup&gt;=4r&lt;sup&gt;2&lt;/sup&gt; and P=4*(2r)=8r.&lt;br&gt;&#xA;The perimeter is derivative of the area with respect to r, just as in the case of a circle.&lt;/p&gt;&#xA;" OwnerUserId="73" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-07-26T09:41:01.373" LastActivityDate="2010-07-26T09:41:01.373" CommentCount="0" />
  <row Id="716" PostTypeId="1" AcceptedAnswerId="718" CreationDate="2010-07-26T08:38:16.320" Score="4" ViewCount="455" Body="&lt;p&gt;I know that the harmonic series $$\sum_{k=1}^{\infty}\frac{1}{k}  = \frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \cdots + \frac{1}{n} + \cdots \tag{I}$$ diverges, but what about the alternating harmonic series &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{k}  = \frac{1}{1} - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \frac{1}{6} + \cdots + \frac{(-1)^{n+1}}{n} + \cdots ? \tag{II}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does it converge?  If so, what is its sum?&lt;/p&gt;&#xA;" OwnerUserId="72" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2011-06-05T07:20:58.167" LastActivityDate="2011-06-05T07:20:58.167" Title="Sum of the alternating harmonic series" Tags="&lt;sequences-and-series&gt;" AnswerCount="4" CommentCount="3" FavoriteCount="1" />
  <row Id="717" PostTypeId="2" ParentId="716" CreationDate="2010-07-26T08:50:52.000" Score="6" ViewCount="" Body="&lt;p&gt;it is not absolutely convergent (that is, if you are allowed to reorder terms you may end up with whatever number you fancy).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you consider the associated series formed by summing the terms from 1 to n of the original one, that is you fix the order of summation of the original series, it converges to log(2) (log being the natural logarithm). See &lt;a href=&quot;http://en.wikipedia.org/wiki/Mercator_series&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="89" LastActivityDate="2010-07-26T08:50:52.000" />
  <row Id="718" PostTypeId="2" ParentId="716" CreationDate="2010-07-26T09:14:15.523" Score="7" ViewCount="" Body="&lt;p&gt;Complementary to Mau's answer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Call a series $a_n$ &lt;strong&gt;absolutely convergent&lt;/strong&gt; if $\sum|a_n|$ converges. If $a_n$ converges but is not absolutely convergent we call $a_n$ &lt;strong&gt;conditionally convergent&lt;/strong&gt; &lt;a href=&quot;http://en.wikipedia.org/wiki/Riemann_series_theorem&quot; rel=&quot;nofollow&quot;&gt;The Riemann series theorem&lt;/a&gt; states that any conditionally convergent series can be reordered to converge to any real number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Morally this is because both the positive and negative parts of your series diverge but the divergences cancel each other out, one or other's canceling the other can be staggered by adding on, say,  the negative bits every third term in stead of every other term. This means that in the race for the two divergences to cancel each other out, we give the positive bit something of a head-start and will get a larger positive outcome. Notice how, even in this rearranged version of the series, every term will still come up exactly once.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is also worth noting, on the Wikipedia link Mau provided, that the convergence to $\ln 2$ of your series is at the edge of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Radius_of_convergence&quot; rel=&quot;nofollow&quot;&gt;radius of convergence&lt;/a&gt; for the series expansion of $\ln(1-x)$- this is a fairly typical occurrence: at the boundary of a domain of convergence of a Taylor series, the series is &lt;em&gt;only just&lt;/em&gt; converging- which is why you see this conditional convergence type behavior. &lt;/p&gt;&#xA;" OwnerUserId="160" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2011-06-02T05:00:31.487" LastActivityDate="2011-06-02T05:00:31.487" />
  <row Id="719" PostTypeId="2" ParentId="704" CreationDate="2010-07-26T10:29:09.160" Score="2" ViewCount="" Body="&lt;p&gt;Often the most obvious solution is the best solution. I would write T(s) be 1 if a statement is true and 0 if it is false. I would then call the statements A, B and C and let S={A, B, C}. Then is would write sum T(s) over S &gt;=2 =&gt; all s are true. You can make these statement more symbolic if you really like&lt;/p&gt;&#xA;" OwnerUserId="123" LastEditorUserId="123" LastEditorDisplayName="" LastEditDate="2010-07-26T12:01:45.160" LastActivityDate="2010-07-26T12:01:45.160" CommentCount="2" />
  <row Id="720" PostTypeId="1" AcceptedAnswerId="770" CreationDate="2010-07-26T11:22:17.353" Score="4" ViewCount="200" Body="&lt;p&gt;Given an initial position and a subsequent position, each given by latitude and longitude in the WGS-84 system. How do you determine the heading in degrees clockwise from true north of movement?&lt;/p&gt;&#xA;" OwnerUserId="193" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-07-27T09:03:35.653" LastActivityDate="2010-09-29T03:45:13.390" Title="How to calculate a heading on the earths surface?" Tags="&lt;geometry&gt;&lt;trigonometry&gt;&lt;spherical-geometry&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" />
  <row Id="721" PostTypeId="1" AcceptedAnswerId="722" CreationDate="2010-07-26T11:59:51.513" Score="11" ViewCount="225" Body="&lt;p&gt;Is there an algorithm for working out the best way (i.e. fewest multiplications) of calculating A^n in a structure where multiplication is associative?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, suppose A is a square matrix. Matrix multiplication is associative, and I can compute A^9 with 4 multiplications:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;A^2 = A*A&#xA;A^3 = A^2*A&#xA;A^6 = A^3*A^3&#xA;A^9 = A^6*A^3&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;One method which works is to compute A^(2^i) and use the binary representation of n, but this is not always optimal, e.g. with n=23, we can do it in 6 multiplications:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;A^2 = A*A&#xA;A^3 = A^2*A&#xA;A^5 = A^3*A^2&#xA;A^10 = A^5*A^5&#xA;A^20 = A^10*A^10&#xA;A^23 = A^23*A^3&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;rather than 7:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;A^2 = A*A&#xA;A^4 = A^2*A^2&#xA;A^8 = A^4*A^4&#xA;A^16 = A^8*A^8&#xA;A^20 = A^16*A^4&#xA;A^22 = A^20*A^2&#xA;A^23 = A^22*A&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Is there an algorithm which gives the quickest way?&lt;/p&gt;&#xA;" OwnerUserId="31" LastActivityDate="2010-07-26T21:37:34.503" Title="Algorithm for calculating A^n with as few multiplications as possible" Tags="&lt;algorithms&gt;&lt;abstract-algebra&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="2" />
  <row Id="722" PostTypeId="2" ParentId="721" CreationDate="2010-07-26T12:15:26.850" Score="10" ViewCount="" Body="&lt;p&gt;There seems to be no efficient algorithm for this. Quoting Wikipedia on &lt;a href=&quot;http://en.wikipedia.org/wiki/Addition-chain_exponentiation&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;Addition-chain exponentiation&lt;/em&gt;&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;… the addition-chain method is much more complicated, since the determination of a shortest addition chain seems quite difficult: &lt;strong&gt;no efficient optimal methods are currently known for arbitrary exponents&lt;/strong&gt;, and the related problem of finding a shortest addition chain for a given set of exponents has been proven NP-complete.  Even given a shortest chain, addition-chain exponentiation requires more memory than the binary method, because it must potentially store many previous exponents from the chain simultaneously. In practice, therefore, shortest addition-chain exponentiation is primarily used for small fixed exponents for which a shortest chain can be precomputed and is not too large.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;On the same page, however, it does link to a reference* of some methods better than binary exponentiation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One simple example is to use base-N number instead of base-2. e.g. for A&lt;sup&gt;510&lt;/sup&gt;,&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;with binary:   2 = 1 + 1&#xA;               3 = 2 + 1&#xA;               6 = 3 + 3&#xA;               7 = 6 + 1&#xA;              14 = 7 + 7&#xA;              15 = 14 + 1&#xA;              30 = 15 + 15&#xA;              31 = 30 + 1&#xA;              62 = 31 + 31&#xA;              63 = 62 + 1&#xA;             126 = 63 + 63&#xA;             127 = 126 + 1&#xA;             254 = 127 + 127&#xA;             255 = 254 + 1&#xA;             510 = 255 + 255   (15 multiplications)&#xA;&#xA;with base-8:   2 = 1 + 1&#xA;               3 = 2 + 1&#xA;               4 = 3 + 1&#xA;               5 = 4 + 1&#xA;               6 = 5 + 1&#xA;               7 = 6 + 1&#xA;              14 = 7 + 7&#xA;              28 = 14 + 14&#xA;              56 = 28 + 28&#xA;              63 = 56 + 7&#xA;             126 = 63 + 63&#xA;             252 = 126 + 126&#xA;             504 = 252 + 252&#xA;             510 = 504 + 6      (14 multiplications)&#xA;&#xA;with base-4:   2 = 1 + 1&#xA;               3 = 2 + 1&#xA;               4 = 2 + 2&#xA;               7 = 4 + 3&#xA;              14 = 7 + 7&#xA;              28 = 14 + 14&#xA;              31 = 28 + 3&#xA;              62 = 31 + 31&#xA;             124 = 62 + 62&#xA;             127 = 124 + 3&#xA;             254 = 127 + 127&#xA;             508 = 254 + 254&#xA;             510 = 508 + 2      (13 multiplications)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;(I don't know how to choose the optimal N.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;*: &lt;a href=&quot;http://eref.uqu.edu.sa/files/a_survey_of_fast_exponentiation_methods__64651.pdf&quot; rel=&quot;nofollow&quot;&gt;Daniel M. Gordon, &lt;em&gt;A survey of fast exponentiation methods&lt;/em&gt;, Journal of Algorithms &lt;strong&gt;27&lt;/strong&gt; (1998), pp 129–146&lt;/a&gt;&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="171" LastEditorUserId="171" LastEditorDisplayName="" LastEditDate="2010-07-26T13:17:36.607" LastActivityDate="2010-07-26T13:17:36.607" CommentCount="3" />
  <row Id="723" PostTypeId="2" ParentId="716" CreationDate="2010-07-26T12:28:15.030" Score="2" ViewCount="" Body="&lt;p&gt;Let's  say you have a sequence of nonnegative numbers $a_1 \geq a_2 \geq \dots$ tending to zero. Then it is a &lt;a href=&quot;http://en.wikipedia.org/wiki/Alternating_series_test&quot; rel=&quot;nofollow&quot;&gt;theorem&lt;/a&gt; that the alternating sum $\sum (-1)^i a_i$ converges (not necessarily absolutely, of course).&#xA;This in particular applies to your series.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Incidentally, if you're curious why it converges to $\log(2)$ (which seems somewhat random), it's because of the Taylor series of $\log(1+x)$ while letting $x \to 1$.&lt;/p&gt;&#xA;" OwnerUserId="536" LastEditorUserId="462" LastEditorDisplayName="" LastEditDate="2011-06-02T05:10:47.170" LastActivityDate="2011-06-02T05:10:47.170" CommentCount="1" />
  <row Id="724" PostTypeId="2" ParentId="713" CreationDate="2010-07-26T12:39:20.013" Score="2" ViewCount="" Body="&lt;p&gt;While tensors are generalizations of vectors, I don't think you can really generalize the way you visualize them.  This is because you really want to think of tensors as multi-linear functions and you usually don't think of a vector as a linear function from the dual space to the real numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So even if you can't get as nice a geometric picture of a tensor, you do get a nice grasp on what they are if you view them as multi-linear functions (as opposed to just a collection of numbers) from some copies of your vector space V (and/or its dual V*) into R.  For example, a metric is a type of rank 2 tensor and has a nice geometric meaning-- applying it to two copies of a vector gives the vectors squared length, you can apply to to two vectors to get the angle between them, etc.  The moment of inertia tensor is a 2-tensor I such that I(u,u) is the moment of inertia about the u-axis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are also usually different ways of looking at the same tensor, which may make it easier to get a grasp on it.  For example, a (1,1) tensor is a multilinear map from V x V* into R.  However, this can naturally (i.e. basis independently) be identified with a linear map from V to itself: if T is a (1,1) tensor then contracting it with a vector gives a linear map V* to R, which is just another vector.&lt;/p&gt;&#xA;" OwnerUserId="9" LastActivityDate="2010-07-26T12:39:20.013" />
  <row Id="725" PostTypeId="2" ParentId="701" CreationDate="2010-07-26T12:42:57.450" Score="3" ViewCount="" Body="&lt;p&gt;I had an answer before, but I looked into it a bit more and my answer was incorrect so I removed it. This link may be of interest: &lt;a href=&quot;http://en.wikipedia.org/wiki/Circle_packing_in_a_square&quot; rel=&quot;nofollow&quot;&gt;Circle Packing in a Square (wikipedia)&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It was suggested by KennyTM that there may not be an optimal solution yet to this problem in general. Further digging into this has shown me that this is probably correct. Check out this page: &lt;a href=&quot;http://hydra.nat.uni-magdeburg.de/packing/csq/csq.html&quot; rel=&quot;nofollow&quot;&gt;Circle Packing - Best Known Packings&lt;/a&gt;. As you can see, solutions up to only 30 circles have been found and proven optimal. (Other higher numbers of circles have been proven optimal, but 31 hasn't)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that although problem defined on the wikipedia page and the other link is superficially different than the question asked here, the same fundamental question is being asked, which is &quot;what is the most efficient way to pack circles in a square/rectangle container?&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;...And it seems the answer is &quot;we don't really know&quot; :)&lt;/p&gt;&#xA;" OwnerUserId="181" LastActivityDate="2010-07-26T12:42:57.450" />
  <row Id="726" PostTypeId="1" AcceptedAnswerId="727" CreationDate="2010-07-26T12:51:35.810" Score="8" ViewCount="435" Body="&lt;p&gt;I keep hearing questions about whether something is NP-complete, but they never really mention what it is. Why do people care so much about NP-complete problems?&lt;/p&gt;&#xA;" OwnerUserId="104" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-07-26T13:41:05.957" LastActivityDate="2010-07-30T16:27:13.650" Title="What are NP-complete problems and why are they so important?" Tags="&lt;computer-science&gt;&lt;np-complete&gt;" AnswerCount="3" FavoriteCount="7" />
  <row Id="727" PostTypeId="2" ParentId="726" CreationDate="2010-07-26T12:58:10.257" Score="10" ViewCount="" Body="&lt;p&gt;A problem is in class NP if its solution may be &lt;strong&gt;verified&lt;/strong&gt; in polynomial time, that is if the dimension of the problem is n you may be sure that for large enough n you need less than r&amp;middot;n&lt;sup&gt;k&lt;/sup&gt; operations to verify the solution. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A problem is in class P if its solution may be &lt;strong&gt;found&lt;/strong&gt; in polynomial time, instead. A problem in P is in NP by definition, but the converse may not be the case; probably the most important open question in computer science is whether classes P and NP are the same, that is P=NP. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;NP-complete is a family of NP problems for which you know that if one of them had a polynomial solution then &lt;strong&gt;everyone of them&lt;/strong&gt; has. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(EDITED) For the time being, only known algorithms for NP-&lt;strong&gt;complete&lt;/strong&gt; problems are exponential in number of operations, so they are not practically solvable for n large.&lt;/p&gt;&#xA;" OwnerUserId="89" LastEditorUserId="89" LastEditorDisplayName="" LastEditDate="2010-07-26T17:20:32.843" LastActivityDate="2010-07-26T17:20:32.843" CommentCount="1" />
  <row Id="728" PostTypeId="2" ParentId="726" CreationDate="2010-07-26T13:32:27.070" Score="10" ViewCount="" Body="&lt;p&gt;To expand on Mau's answer, you should care about NP-complete problems because there is an &lt;a href=&quot;http://en.wikipedia.org/wiki/List_of_NP-complete_problems&quot; rel=&quot;nofollow&quot;&gt;entire family&lt;/a&gt; of them that spans a large number of seemingly basic algorithms across a wide range of disciplines. These aren't obscure problems, but extremely important and highly practical questions. For examples, consider the following:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Travelling_salesman_problem&quot; rel=&quot;nofollow&quot;&gt;Travelling salesman problem&lt;/a&gt; - finding the shortest path (on a graph) that allows you to visit every city exactly once.&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Bin_packing_problem&quot; rel=&quot;nofollow&quot;&gt;Bin packing problem&lt;/a&gt; - there are a number of fixed (integer) size bins and objects of varying sizes. Minimise the number of bins required to hold all of the objects&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Knapsack_problem&quot; rel=&quot;nofollow&quot;&gt;Knapsack problem&lt;/a&gt; - given objects of various sizes and values and a knapsack with a fixed integer size, choose the objects that can fit inside with the most value&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Vertex_cover&quot; rel=&quot;nofollow&quot;&gt;Minimal Vertex cover&lt;/a&gt; - finding the smallest set of vertices such that every edge contains at least one chosen vertice&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Clique_problem&quot; rel=&quot;nofollow&quot;&gt;Clique&lt;/a&gt; - finding that largest group of people who all know each other&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Subgraph_isomorphism_problem&quot; rel=&quot;nofollow&quot;&gt;Subgraph isomorphism&lt;/a&gt; - does one graph contain a subgraph isomorphic to another?&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Set_packing&quot; rel=&quot;nofollow&quot;&gt;Set packing&lt;/a&gt; - given a number of sets, what is the maximum number of disjoint sets that can be selected? This is related to set cover, where we are trying to choose sets so that every element is within at least one set&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Subset_sum_problem&quot; rel=&quot;nofollow&quot;&gt;Subset sum&lt;/a&gt; - Given a set of integers, does some subset sum to 0?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Although many of these problems may seem abstract, many more complicated problems can't be efficiently solved with current techniques, as they are equivalent to one of these.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem of NP completeness has received a huge amount of attention. Once you've reduced a problem to NP-complete, you know to give up on an efficient fast algorithm and to start looking at approximations.&lt;/p&gt;&#xA;" OwnerUserId="123" LastEditorUserId="123" LastEditorDisplayName="" LastEditDate="2010-07-26T21:19:05.073" LastActivityDate="2010-07-26T21:19:05.073" CommentCount="4" />
  <row Id="729" PostTypeId="2" ParentId="726" CreationDate="2010-07-26T16:05:34.357" Score="3" ViewCount="" Body="&lt;p&gt;Any problem for which a solution (once found) can be quickly &lt;em&gt;verified&lt;/em&gt; as a solution is said to be &quot;in &lt;strong&gt;NP&lt;/strong&gt;&quot;  (Here, &quot;quickly&quot; means in polynomial-time).  Any problem for which a solution can be &lt;em&gt;found&lt;/em&gt; quickly is said to be &quot;in &lt;strong&gt;P&lt;/strong&gt;.&quot;  &lt;strong&gt;P&lt;/strong&gt; is a subset of &lt;strong&gt;NP&lt;/strong&gt; - that is, any problem for which a solution can be quickly &lt;em&gt;found&lt;/em&gt; can also be quickly &lt;em&gt;verified&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A problem is &lt;strong&gt;NP-complete&lt;/strong&gt; if it is the hardest problem in &lt;strong&gt;NP&lt;/strong&gt;.  Surprisingly, there are many &lt;strong&gt;NP-complete&lt;/strong&gt; problems, which are all equivalent - here, &lt;em&gt;equivalent&lt;/em&gt; means that a quick (polynomial-time) solution to any one of them would give you a quick solution to all the rest.  Also somewhat surprisingly, a quick solution to any &lt;strong&gt;NP-complete&lt;/strong&gt; problem would also give you a quick solution to &lt;em&gt;any&lt;/em&gt; problem in &lt;strong&gt;NP&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So is there a quick (polynomial-time) algorithm to solve &lt;strong&gt;NP-complete&lt;/strong&gt; problems?  That is the &lt;a href=&quot;http://en.wikipedia.org/wiki/P%3DNP&quot; rel=&quot;nofollow&quot;&gt;P=NP problem&lt;/a&gt;, one of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Millennium_Prize_Problems&quot; rel=&quot;nofollow&quot;&gt;greatest unsolved problems&lt;/a&gt; of our time.  However, most sane mathemeticians believe (and hope!) that &lt;strong&gt;P≠NP&lt;/strong&gt;, because proving math-theorems is &lt;strong&gt;NP-Complete&lt;/strong&gt;; so if &lt;strong&gt;P=NP&lt;/strong&gt;, we'd be out of a job! :)&lt;/p&gt;&#xA;" OwnerUserId="136" LastEditorUserId="136" LastEditorDisplayName="" LastEditDate="2010-07-26T21:56:09.773" LastActivityDate="2010-07-26T21:56:09.773" />
  <row Id="730" PostTypeId="1" CreationDate="2010-07-26T17:29:10.320" Score="1" ViewCount="83" Body="&lt;p&gt;In an example I have worked out for my work, I have constructed a category whose objects are graded R-modules(R is a graded ring), and with morphisms the usual morphisms quotient(by quotient I mean simply that this class of morphisms are Isomorphisms, thus creating an eq relation) the following class of morphisms:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Sigma=\left\lbrace f\in Hom_{grR-mod}\left(A,B\right)|ker\left(f\right)_0\neq 0,  cok\left(f\right)_0\neq 0\right\rbrace$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am wondering if this category has a better(more canonical) description, or if I can show it is equivalent to some other interesting category.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="16" LastActivityDate="2010-07-26T17:29:10.320" Title="An elegant description for graded-module morphisms with non-zero zero component." Tags="&lt;modules&gt;&lt;graded-modules&gt;" CommentCount="2" />
  <row Id="731" PostTypeId="2" ParentId="721" CreationDate="2010-07-26T18:11:26.853" Score="1" ViewCount="" Body="&lt;p&gt;Another option would be to use eigendecomposition . It allows you to raise the eigenvalues in the diagonal of the decomposition A = VDV^-1 to a power. It changes the problem from matrix multiplication to the multiplication of the eigenvalues. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once in eigendecomposition form you could perform the same addition-chain exponentiation technique but it would be with scalars, not matrices. Much more efficient, because each matrix multiplication has n^3 multiplies, but with ed you would only have n.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More is explained here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Matrix_decomposition#Eigendecomposition&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Matrix_decomposition#Eigendecomposition&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="99" LastEditorUserId="99" LastEditorDisplayName="" LastEditDate="2010-07-26T21:37:34.503" LastActivityDate="2010-07-26T21:37:34.503" CommentCount="3" />
  <row Id="732" PostTypeId="1" AcceptedAnswerId="733" CreationDate="2010-07-26T20:08:48.103" Score="7" ViewCount="803" Body="&lt;p&gt;Can someone point me to a proof that the set of irrational numbers is uncountable. I know how to show that the set &lt;strong&gt;Q&lt;/strong&gt; is countable but how would you then show that the irrationals are uncountable? &lt;/p&gt;&#xA;" OwnerUserId="180" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2011-01-04T17:43:57.480" LastActivityDate="2011-01-04T17:43:57.480" Title="Proof that the irrational numbers are uncountable" Tags="&lt;elementary-set-theory&gt;&lt;irrational-numbers&gt;" AnswerCount="1" />
  <row Id="733" PostTypeId="2" ParentId="732" CreationDate="2010-07-26T20:12:04.727" Score="22" ViewCount="" Body="&lt;p&gt;Given that the reals are uncountable (can be shown via &lt;a href=&quot;http://en.wikipedia.org/wiki/Cantor_diagonalization&quot; rel=&quot;nofollow&quot;&gt;Cantor diagonalization&lt;/a&gt;) and the rationals are countable, the irrationals are the reals with the rationals removed, which is uncountable.  (Or, since the reals are the union of the rationals and the irrationals, if the irrationals were countable, the reals would be the union of countable sets and would have to be countable, so the irrationals must be uncountable.)&lt;/p&gt;&#xA;" OwnerUserId="72" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-07-26T20:38:10.780" LastActivityDate="2010-07-26T20:38:10.780" />
  <row Id="734" PostTypeId="1" CreationDate="2010-07-26T20:13:26.987" Score="5" ViewCount="267" Body="&lt;p&gt;Are there any real online mathematics (applied math, statistics, ...) degree programs out there?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm full-time employed, thus not having the flexibility of attending an on campus program. I also already have a MSc in Computer Science. My motivation for a math degree is that I like learning and am interested in the subject. I've studied through number of OCW courses on my own, but it would be nice if I could actually be able to have my studying count towards something.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've done my share of Googling for this, but searching for online degrees seems to bring up a lot of institutions that (at least superficially) seem a bit shady (diploma mills?).&lt;/p&gt;&#xA;" OwnerUserId="105" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-07-27T01:14:52.763" LastActivityDate="2010-10-24T16:24:38.453" CommunityOwnedDate="2010-07-26T22:29:23.650" Title="Online Math Degree Programs" Tags="&lt;soft-question&gt;&lt;education&gt;" AnswerCount="3" CommentCount="6" FavoriteCount="4" />
  <row Id="735" PostTypeId="1" AcceptedAnswerId="739" CreationDate="2010-07-26T20:23:17.057" Score="6" ViewCount="324" Body="&lt;p&gt;I was coming back from my Driver's Education class, and something mathsy really stuck out to me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of the essential properties of a car is its current speed.  Or speed at a current time.  For example, at a given point in time in my drive, I could be traveling 40 mph.  But what does that &lt;em&gt;mean&lt;/em&gt;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From my basic algebra classes, I've learned that speed = distance/time.  So if I travel ten miles in half an hour, my average speed would be 20 mph ($\frac{10 mi}{25 h}$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But instantaneous velocity...you aren't measuring average speed for a given amount of time.  You're measuring instantaneous speed over an...instantaneous amount of time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That would be something like (miles) / (time), where time = 0?  Isn't that infinite?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And perhaps, in a difference of time = 0, then I'd be traveling 0 miles.  So would I be said to be going 0 mph at an instantaneous moment in time?  I'd like to be able to tell that to any cops pull me over for &quot;speeding&quot;!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But then if miles = 0 and time = 0, then you have 0/0?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is all rather confusing.  What does it &lt;strong&gt;mean&lt;/strong&gt; to be going 40 mph at a given moment in time, exactly?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've heard this explained using this strange art called &quot;calculus&quot; before, and it's all gone over my head.  Can anyone explain this using terms I (a High School Algebra and Geometry and Driving student) will understand?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I figured that my problem had numbers in it, and therefore has to do with Maths.)&lt;/p&gt;&#xA;" OwnerUserId="51" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-07-27T00:40:48.933" LastActivityDate="2011-05-31T13:11:13.700" Title="What does it mean to be going 40 mph (or 64 kph, etc.) at a given moment?" Tags="&lt;calculus&gt;&lt;intuition&gt;" AnswerCount="7" CommentCount="4" />
  <row Id="737" PostTypeId="2" ParentId="735" CreationDate="2010-07-26T20:36:28.617" Score="7" ViewCount="" Body="&lt;p&gt;If you were to try to measure instantaneous speed as you described, you would in fact have traveled 0 miles in 0 time and 0/0 is undefined  If, however, you look at your average speed over smaller and smaller periods of time around the instant you care about--that is, (distance traveled from $t=t_0$ to $t=t_0+\epsilon$ for various small values of &amp;epsilon;--and these average speeds &quot;converge&quot; (they all get closer to a single value as &amp;epsilon; gets closer to 0), then we say that the instantaneous speed is that single value upon which the average speeds around that point converge.  This is, of course, a somewhat informal explanation; to be more precise requires getting into differential calculus.&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-07-26T20:36:28.617" />
  <row Id="738" PostTypeId="2" ParentId="735" CreationDate="2010-07-26T20:59:48.653" Score="1" ViewCount="" Body="&lt;p&gt;in the real world you need a bit of time to brake or accelerate. This means that in a small amount of time (let's say 1/10 of a second) your speed is more or less constant. So, instead of dividing by 0 seconds, just take a small elapsed time and do the division.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In calculus you will learn how to be more precise, noticing that using a smaller and smaller elapsed time the ratio between distance and time changes less and less; at that point you postulate that &lt;strong&gt;in this scenario only&lt;/strong&gt; you may say that the limit (which is 0/0 and therefore indefinite) is the instantaneous speed. But the odometers don't care for instantaneous speed, so for the moment you shouldn't worry.&lt;/p&gt;&#xA;" OwnerUserId="89" LastActivityDate="2010-07-26T20:59:48.653" CommentCount="2" />
  <row Id="739" PostTypeId="2" ParentId="735" CreationDate="2010-07-26T21:22:38.663" Score="10" ViewCount="" Body="&lt;p&gt;I think there is a very clear meaning in the physical world: If, at some moment, you were going 40 mph, if you were to stop de/accelerating and just hold that velocity, you would cover 40 miles in 1 hour.&lt;/p&gt;&#xA;" OwnerUserId="73" LastActivityDate="2010-07-26T21:22:38.663" />
  <row Id="740" PostTypeId="1" CreationDate="2010-07-26T21:35:48.560" Score="13" ViewCount="741" Body="&lt;p&gt;What are some particularly well-known functions that exhibit pathological behavior at or near at least one value and are particularly useful as examples?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, if f'(a) = b, then f(a) exists, f is continuous at a, f is differentiable at a, but f' need not be continuous at a.  A function for which this is true is f(x) = x^2 sin(1/x) at x=0.&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-08-25T22:10:29.297" CommunityOwnedDate="2010-07-26T21:38:01.230" Title="Useful examples of pathological functions" Tags="&lt;soft-question&gt;&lt;big-list&gt;&lt;calculus&gt;&lt;real-analysis&gt;" AnswerCount="14" CommentCount="2" FavoriteCount="8" />
  <row Id="741" PostTypeId="1" AcceptedAnswerId="743" CreationDate="2010-07-26T21:40:11.743" Score="3" ViewCount="518" Body="&lt;p&gt;I asked this question on Stack Overflow but it was closed as &quot;not programming related&quot;. So I think this is probably the best place for it...&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I read over the wikipedia &lt;a href=&quot;http://en.wikipedia.org/wiki/Linear_programming&quot; rel=&quot;nofollow&quot;&gt;article&lt;/a&gt;, but it seems to be beyond my comprehension. It says it's for optimization, but how is it different than any other method for optimizing things?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An answer that introduces me to linear programming so I can begin diving into some less beginner-accessible material would be most helpful.&lt;/p&gt;&#xA;" OwnerUserId="181" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-07-27T00:39:28.670" LastActivityDate="2010-07-28T04:30:38.603" Title="What is linear programming?" Tags="&lt;linear-algebra&gt;&lt;algorithms&gt;&lt;computer-science&gt;&lt;optimization&gt;&lt;linear-programming&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="2" />
  <row Id="742" PostTypeId="1" AcceptedAnswerId="746" CreationDate="2010-07-26T21:48:35.177" Score="4" ViewCount="246" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;You have sheets of 42-cent stamps and&#xA;  29-cent stamps, but you need at least&#xA;  $3.20 to mail a package.  What is the&#xA;  least amount you can make with the 42-&#xA;  and 29-cent stamps that is sufficient&#xA;  to mail the package?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A contest problem such as this is probably most easily solved by tabulating the possible combinations, using 0 through ceiling(total/greater value) of the greater-value stamp and computing the necessary number of the smaller stamp and the total postage involved.  The particular example above would be solved with a 9-row table, showing the minimum to be $3.23, made with 7 42-cent stamps and 1 29-cent stamp.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a better algorithm for solving this kind of problem?  What if you have more than two values of stamps?&lt;/p&gt;&#xA;" OwnerUserId="72" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-07-27T00:40:02.070" LastActivityDate="2010-08-11T22:01:46.380" Title="Least wasteful use of stamps to achieve a given postage" Tags="&lt;number-theory&gt;&lt;algebra-precalculus&gt;&lt;algorithms&gt;&lt;contest-math&gt;" AnswerCount="5" />
  <row Id="743" PostTypeId="2" ParentId="741" CreationDate="2010-07-26T21:56:42.473" Score="3" ViewCount="" Body="&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Linear_programming#Standard_form&quot; rel=&quot;nofollow&quot;&gt;standard form&lt;/a&gt; (and &lt;a href=&quot;http://en.wikipedia.org/wiki/Linear_programming#Example&quot; rel=&quot;nofollow&quot;&gt;example&lt;/a&gt;) sections pretty well describe what it is.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;How is it different than any other method for optimizing things?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It's, well, just another method.  However, it is somewhat special in that many other optimization algorithms either &lt;a href=&quot;http://en.wikipedia.org/wiki/Linear_programming_relaxation&quot; rel=&quot;nofollow&quot;&gt;use linear programming&lt;/a&gt; as part of their solution, or are in reality a &lt;a href=&quot;http://en.wikipedia.org/wiki/Flow_network&quot; rel=&quot;nofollow&quot;&gt;specialized solution to a linear programming problem&lt;/a&gt;.  &lt;em&gt;In fact&lt;/em&gt;, &lt;a href=&quot;http://en.wikipedia.org/wiki/Linear_programming#Integer_unknowns&quot; rel=&quot;nofollow&quot;&gt;integer linear programming&lt;/a&gt; is &lt;strong&gt;NP-complete&lt;/strong&gt;, meaning that &lt;em&gt;any&lt;/em&gt; problem in &lt;strong&gt;NP&lt;/strong&gt; &lt;a href=&quot;http://math.stackexchange.com/questions/726/what-are-np-complete-problems-and-why-are-they-so-important/729#729&quot; rel=&quot;nofollow&quot;&gt;can be stated as an (integer) linear programming problem&lt;/a&gt;.&lt;br&gt;&#xA;&lt;em&gt;(this also means solving your typical integer linear programming problem is &lt;strong&gt;much&lt;/strong&gt; more difficult than if we didn't restrict ourselves to integers..)&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="136" LastEditorUserId="136" LastEditorDisplayName="" LastEditDate="2010-07-26T22:51:00.397" LastActivityDate="2010-07-26T22:51:00.397" CommentCount="1" />
  <row Id="744" PostTypeId="2" ParentId="735" CreationDate="2010-07-26T22:08:42.607" Score="0" ViewCount="" Body="&lt;p&gt;A state-variable kind of approach: (represents the physical states of a system)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Newton%27s_laws_of_motion&quot; rel=&quot;nofollow&quot;&gt;Newton's first law of motion&lt;/a&gt; says that without external forces, masses will move at constant velocity. Every change in velocity (for a car, whether a change in speed due to braking, or a change in direction due to friction of the tires and the steering mechanism) is due to an external force. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the &quot;state&quot; of the car at any instant in time is its velocity; distance traveled can be derived from velocity:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While you are going 60mi/hr = 26.8m/sec (see &lt;a href=&quot;http://en.wikipedia.org/wiki/Miles_per_hour#Conversions&quot; rel=&quot;nofollow&quot;&gt;conversion factors&lt;/a&gt;) in a particular direction at constant velocity, you can multiply a change in time times velocity and get distance: 26.8m/sec = 26.8 meters in one second. Or 26.8mm in one millisecond. Or 26.8 micrometers in one microsecond.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The process of going from velocity to (directed) distance is called &lt;a href=&quot;http://en.wikipedia.org/wiki/Integral&quot; rel=&quot;nofollow&quot;&gt;integration&lt;/a&gt;, and for those two variables (velocity and directed distance) is just the summing up of distance increments (= velocity * time increments) over infinitesimally short periods of time. For smoothly varying velocities, coarse time increments are a good approximation to reality -- differential equation solvers use this or similar approaches (e.g. &lt;a href=&quot;http://en.wikipedia.org/wiki/Simpson%27s_rule&quot; rel=&quot;nofollow&quot;&gt;Simpson's Rule&lt;/a&gt;) to perform integration.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;But instantaneous velocity...you aren't measuring average speed for a given amount of time&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Well, instantaneous velocity is just the limit of average velocity as the time period becomes infinitely small.&lt;/p&gt;&#xA;" OwnerUserId="120" LastActivityDate="2010-07-26T22:08:42.607" />
  <row Id="745" PostTypeId="2" ParentId="741" CreationDate="2010-07-26T22:09:22.463" Score="2" ViewCount="" Body="&lt;p&gt;BlueRaja's answer is certainly more complete than this one (and gives good references), but here's a rough overview of linear programming.  Suppose that you have a linear function (in high school courses, it's typically a function of two variables) that you want to optimize on a convex &quot;feasible&quot; region bounded by linear equations (again, in high school, the bounds are typically described using linear inequalities in two variables).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because the function to optimize is linear, the set of points for which the function has a particular value, say c, is a line (and all such lines are parallel) and the value of the function anywhere to one side of the line is greater than c and anywhere to the other side of the line is less than c.  So, you can think about moving through this set of constant-value lines to increase/decrease the value of the target function as appropriate to optimize it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the feasible region is bounded by linear equations, as the constant-value line moves through and out of the feasible region, it last touches the feasible region at a vertex (or possibly at all points on an edge connecting two vertices), so the optimal solution must occur at a vertex of the feasible region.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given all that, linear programming comes down to evaluating the target function at all the vertices of the feasible region to find the optimal value.&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-07-26T22:09:22.463" CommentCount="5" />
  <row Id="746" PostTypeId="2" ParentId="742" CreationDate="2010-07-26T22:13:10.403" Score="1" ViewCount="" Body="&lt;p&gt;With two stamps, you can do it in &lt;strike&gt;linear&lt;/strike&gt; pseudo-linear time - &lt;strong&gt;O(totalCost/costOfLargerStamp)&lt;/strong&gt; - by simply enumerating every possibility (there is only one possible count of the smaller stamp for each count of the larger stamp).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, however, solving this is equivalent to solving a general &lt;a href=&quot;http://en.wikipedia.org/wiki/Integer_linear_programming#Integer_unknowns&quot; rel=&quot;nofollow&quot;&gt;integer linear programming problem&lt;/a&gt; written in standard form, which is NP-complete.&lt;/p&gt;&#xA;" OwnerUserId="136" LastEditorUserId="136" LastEditorDisplayName="" LastEditDate="2010-07-26T23:52:29.077" LastActivityDate="2010-07-26T23:52:29.077" CommentCount="1" />
  <row Id="747" PostTypeId="2" ParentId="740" CreationDate="2010-07-26T22:15:40.447" Score="2" ViewCount="" Body="&lt;p&gt;sin(x)/x is useful; it has a singularity at x=0, but if you take the union of y=sin(x)/x with the point (x=0,y=1) then you get sinc(x/&amp;pi;). The &lt;a href=&quot;http://en.wikipedia.org/wiki/Sinc_function&quot; rel=&quot;nofollow&quot;&gt;sinc function&lt;/a&gt; has a lot of applications in signal processing and diffraction.&lt;/p&gt;&#xA;" OwnerUserId="120" LastActivityDate="2010-07-26T22:15:40.447" CommunityOwnedDate="2010-07-26T22:15:40.447" />
  <row Id="748" PostTypeId="2" ParentId="740" CreationDate="2010-07-26T22:27:12.387" Score="4" ViewCount="" Body="&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Dirac_delta_function&quot; rel=&quot;nofollow&quot;&gt;Dirac delta &quot;function.&quot;&lt;/a&gt; It's not a &quot;function,&quot; strictly speaking, but rather a very simple example of a distribution that isn't a function.&lt;/p&gt;&#xA;" OwnerUserId="536" LastActivityDate="2010-07-26T22:27:12.387" CommunityOwnedDate="2010-07-26T22:27:12.387" />
  <row Id="749" PostTypeId="2" ParentId="742" CreationDate="2010-07-26T22:40:45.853" Score="1" ViewCount="" Body="&lt;p&gt;This is a simple variation of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Knapsack_problem&quot; rel=&quot;nofollow&quot;&gt;Knapsack problem&lt;/a&gt;. Which is a NP-Complete problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let n to be the target value, it can be solve using knapsack's dynamic programming solutions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have a finite amount of stamps, then you can add them all up, subtract the target value. Run 0-1 knapsack on the stamps with the resulting number. The stamps outside the knapsack is the solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have a infinite amount of stamps. You can either make them finite: $\lceil \frac{n}{t_i}\rceil$ stamps for stamps with value $t_i$ and run the above process, or you can compute the result by running knapsacks O(log n) times similar to a &lt;a href=&quot;http://en.wikipedia.org/wiki/Binary_search&quot; rel=&quot;nofollow&quot;&gt;binary search&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's no better than a naive &lt;a href=&quot;http://en.wikipedia.org/wiki/Backtracking&quot; rel=&quot;nofollow&quot;&gt;backtracking&lt;/a&gt; solution when n is large. I assume you are not going to pay $100 for stamps. ;)&lt;/p&gt;&#xA;" OwnerUserId="96" LastActivityDate="2010-07-26T22:40:45.853" />
  <row Id="750" PostTypeId="1" AcceptedAnswerId="754" CreationDate="2010-07-26T22:54:07.037" Score="4" ViewCount="130" Body="&lt;p&gt;The absolute value of a 2 by 2 matrix determinant is the area of a corresponding parallelogram with the 2 row vectors as sides.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The absolute value of a 3 by 3 matrix determinant is the volume of a corresponding parallelepiped with the 3 row vectors as sides.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can it be generalized to n-D? The absolute value of a n by n matrix determinant is the volume of a corresponding n-parallelotope?&lt;/p&gt;&#xA;" OwnerUserId="96" LastActivityDate="2010-07-26T23:16:16.113" Title="Determinants and volume of parallelotopes" Tags="&lt;geometry&gt;&lt;linear-algebra&gt;" AnswerCount="3" />
  <row Id="751" PostTypeId="2" ParentId="735" CreationDate="2010-07-26T23:02:12.000" Score="3" ViewCount="" Body="&lt;p&gt;This is similar to other answers. Imagine another car beside yours. That car is covering a distance of 40 miles over the next hour at a constant velocity. At the point at which you keep pace with that car (relative velocity = 0), you are traveling at 40 mph.&lt;/p&gt;&#xA;" OwnerUserId="12" LastActivityDate="2010-07-26T23:02:12.000" />
  <row Id="752" PostTypeId="2" ParentId="750" CreationDate="2010-07-26T23:11:18.070" Score="3" ViewCount="" Body="&lt;p&gt;Yes--see the &lt;a href=&quot;http://en.wikipedia.org/wiki/Determinant&quot; rel=&quot;nofollow&quot;&gt;Wikipedia article&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This follows from the change of variables formula with the Jacobian. However, it is probably more accurate to say that the change-of-variables formula is a (nontrivial) consequence of this.  One argument for this fact, which is given in Rudin's &lt;em&gt;Real and Complex Analysis,&lt;/em&gt; is the fact that both the determinant and the volume function up to sign behave very nicely satisfying certain specific properties (i.e., multilinear, alternating, normalized) hence are the same.&lt;/p&gt;&#xA;" OwnerUserId="536" LastActivityDate="2010-07-26T23:11:18.070" />
  <row Id="753" PostTypeId="2" ParentId="750" CreationDate="2010-07-26T23:14:36.613" Score="1" ViewCount="" Body="&lt;p&gt;Yes, that is because in its most general form the determinate is &lt;strong&gt;defined&lt;/strong&gt; that way, except without the absolute value part. The determinant is a way to compare volumes of subspaces that have the same grade, which is the multi-vector equivalent of the size of matrix.&lt;/p&gt;&#xA;" OwnerUserId="99" LastActivityDate="2010-07-26T23:14:36.613" />
  <row Id="754" PostTypeId="2" ParentId="750" CreationDate="2010-07-26T23:16:16.113" Score="4" ViewCount="" Body="&lt;p&gt;Yes it can. In fact, as Katie Banks noted, a determinant is an &lt;a href=&quot;http://math.stackexchange.com/questions/668/whats-an-intuitive-way-to-think-about-the-determinant/669#669&quot; rel=&quot;nofollow&quot;&gt;intuitive way of thinking about volumes&lt;/a&gt;. To summarise her argument, if we consider the vectors as a matrix, switching two rows, multiplying one by a constant or adding a linear combination will have the same effect on the volume as on the determinate. We can use these operations to transform any n-parallelotope to cube and note that the determinate matches the signed volume here, so it will match it everywhere as well.&lt;/p&gt;&#xA;" OwnerUserId="123" LastActivityDate="2010-07-26T23:16:16.113" />
  <row Id="755" PostTypeId="2" ParentId="742" CreationDate="2010-07-26T23:27:18.700" Score="1" ViewCount="" Body="&lt;p&gt;It is important to understand that NP-completeness is defined in terms of input size, not in terms of either cost to be paid, C, or the number of types of stamps, S, when we are dealing with integers rather than real numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This algorithm can actually be solved in time CS. We take the first stamp value an mark all multiple of it as obtainable (up to the least number greater than the cost). Then, for each subsequent stamp value, we mark as obtainable all numbers that can be obtained by adding a multiple of its value with an obtainable number (up to the least number greater than the cost). We can keep track of the lowest cost postage amount that we have found so far.&lt;/p&gt;&#xA;" OwnerUserId="123" LastActivityDate="2010-07-26T23:27:18.700" />
  <row Id="756" PostTypeId="1" CreationDate="2010-07-27T00:26:20.523" Score="8" ViewCount="410" Body="&lt;p&gt;Here's one that popped into my mind when I was thinking about binary search.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm thinking of an integer between 1 and &lt;em&gt;n&lt;/em&gt;. You have to guess my number. You win as soon as you guess the correct number. If your guess is not correct, I'll give you a hint by saying &quot;too high&quot; or &quot;too low&quot;. What's your best strategy?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is an easy problem if I always tell the truth: by guessing the (rounded) mean of the lower bound and the upper bound, you can find my number in roughly log&lt;sub&gt;2&lt;/sub&gt; &lt;em&gt;n&lt;/em&gt; guesses at most.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But what if I'm allowed to cheat once? What is your best strategy then? To clarify: if you guess my number, you always win instantly. But I'm allowed, at most once, to tell you &quot;too high&quot; when your guess is actually too low, or the opposite. I can also decide not to lie.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a rough upper bound: you can ask each number twice to make sure I'm not cheating, and if I ever give two different answers, just ask a third time and from then on, play the regular game. In this way, you can win with about 2 log&lt;sub&gt;2&lt;/sub&gt; &lt;em&gt;n&lt;/em&gt; guesses at most.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm pretty sure that bound can be improved. Any ideas?&lt;/p&gt;&#xA;" OwnerUserId="115" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-07-27T00:38:44.943" LastActivityDate="2010-11-02T13:11:07.107" Title="Twenty questions against a liar" Tags="&lt;algorithms&gt;&lt;computer-science&gt;&lt;discrete-mathematics&gt;&lt;searching&gt;" AnswerCount="3" CommentCount="5" FavoriteCount="1" />
  <row Id="757" PostTypeId="2" ParentId="740" CreationDate="2010-07-27T00:49:43.627" Score="12" ViewCount="" Body="&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Weierstrass_function&quot; rel=&quot;nofollow&quot;&gt;Weierstrass function&lt;/a&gt; is continuous everywhere and differentiable nowhere.&lt;br&gt;&#xA;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Nowhere_continuous_function&quot; rel=&quot;nofollow&quot;&gt;Dirichlet function&lt;/a&gt; (the indicator function for the rationals) is continuous nowhere.&lt;br&gt;&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Thomae%27s_function&quot; rel=&quot;nofollow&quot;&gt;A modification&lt;/a&gt; of the Dirichlet function is continuous at all irrational values and discontinuous at rational values.&lt;br&gt;&#xA;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Cantor_function&quot; rel=&quot;nofollow&quot;&gt;Devil's Staircase&lt;/a&gt; is uniformly continuous but not absolutely. It increases from 0 to 1, but the derivative is 0 almost everywhere.&lt;/p&gt;&#xA;" OwnerUserId="73" LastEditorUserId="1321" LastEditorDisplayName="" LastEditDate="2010-08-25T22:10:29.297" LastActivityDate="2010-08-25T22:10:29.297" CommunityOwnedDate="2010-07-27T00:49:43.627" CommentCount="4" />
  <row Id="758" PostTypeId="2" ParentId="734" CreationDate="2010-07-27T00:56:57.370" Score="3" ViewCount="" Body="&lt;p&gt;I would recommend against getting a degree from any online-only university. Even if you happen to find one that's not shady, everyone else who hasn't heard of it will assume it is some kind of diploma mill without bothering to do much research. Instead, I think you'd be better off going to a nearby university you're interested in, and ask them if they would be willing to make some kind of special arrangement for you. Many universities allow reduced course loads for students that have families or work full time, and aren't very good about advertising it.&lt;/p&gt;&#xA;" OwnerUserId="73" LastActivityDate="2010-07-27T00:56:57.370" CommunityOwnedDate="2010-07-27T00:56:57.370" />
  <row Id="759" PostTypeId="2" ParentId="740" CreationDate="2010-07-27T00:57:18.650" Score="0" ViewCount="" Body="&lt;p&gt;(1/x)^(1/x). Try graphing it if you dare (including over the negatives)&lt;/p&gt;&#xA;" OwnerUserId="123" LastActivityDate="2010-07-27T00:57:18.650" CommunityOwnedDate="2010-07-27T00:57:18.650" CommentCount="4" />
  <row Id="760" PostTypeId="1" AcceptedAnswerId="762" CreationDate="2010-07-27T01:09:24.273" Score="3" ViewCount="260" Body="&lt;p&gt;I have a 3-D sphere of radius R, centered at the origin.&lt;br&gt;&#xA;(x&lt;sub&gt;1&lt;/sub&gt;,y&lt;sub&gt;1&lt;/sub&gt;,z&lt;sub&gt;1&lt;/sub&gt;) and&#xA;(x&lt;sub&gt;2&lt;/sub&gt;,y&lt;sub&gt;2&lt;/sub&gt;,z&lt;sub&gt;2&lt;/sub&gt;) are two points on the sphere. The Euclidean distance is easy to calculate, but what if I were to restrict myself to traveling on the surface of the sphere?&lt;br&gt;&#xA;Two approaches come to mind: use &lt;a href=&quot;http://en.wikipedia.org/wiki/Arc_length&quot; rel=&quot;nofollow&quot;&gt;arc-length&lt;/a&gt; in some way, or simply use trigonometry: calculate the angle between the two points and get a distance from that.&lt;br&gt;&#xA;Will both/either of these methods work? Which would be easier?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Somewhat related to &lt;a href=&quot;http://math.stackexchange.com/questions/720/how-to-calculate-a-heading-vector-on-the-earths-surface&quot; rel=&quot;nofollow&quot;&gt;this question&lt;/a&gt;. Maybe it will inspire someone to go answer it!&lt;/p&gt;&#xA;" OwnerUserId="73" LastActivityDate="2010-08-11T22:01:26.910" Title="How do I measure distance on a globe?" Tags="&lt;geometry&gt;&lt;calculus&gt;&lt;trigonometry&gt;" AnswerCount="4" CommentCount="2" FavoriteCount="1" />
  <row Id="761" PostTypeId="1" AcceptedAnswerId="763" CreationDate="2010-07-27T01:18:03.910" Score="0" ViewCount="237" Body="&lt;p&gt;Wind resistance -- upwards acceleration, typically varies either linearly or quadratically by the current velocity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a constant downward acceleration due to gravity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can we model the velocity over time of a falling object, subject only to wind resistance and downwards gravity?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't have much experience with differential equations, but I do know that this answer necessarily involves it, so could you possibly explain every step?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you.&lt;/p&gt;&#xA;" OwnerUserId="51" LastEditorUserId="51" LastEditorDisplayName="" LastEditDate="2010-07-27T06:20:24.090" LastActivityDate="2010-07-27T06:20:24.090" Title="Using differential equations to graph velocity over time of a falling object subject to wind resistance" Tags="&lt;calculus&gt;&lt;physics&gt;&lt;differential-equations&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="762" PostTypeId="2" ParentId="760" CreationDate="2010-07-27T01:33:51.347" Score="9" ViewCount="" Body="&lt;p&gt;What you are looking for is what is called the great circle distance: the shortest curve connecting two points on the surface of a sphere is given by traveling along the (there will be exactly one unless the points are polar opposites)  arc of the great circle (that is, the circle of radius R) connecting them.  So you do need to find an arc length but it is easy to do this without invoking calculus if you know about the dot product.  Suppose the two points on the circle are represented by the vectors v and w.  If v*w denotes the dot product of these two vectors, then the angle between them will be arccos(v*w/R^2) (we divide by R^2 since v and w have length R).  Assuming this is in radians, to get the length of the arc connecting them we just multiply this angle by R to get R*arcos(v*w/R^2).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We are quite lucky that there is such a simple formula.  For most manifolds, the curves that minimize distances are not very easy to find since it involves solving a non-linear differential equation (the geodesic equation).  The fact that the sphere is so symmetric helps in this case, and you can maybe convince yourself that an arc of a great circle minimizes distance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You may find this wikipedia article interesting: &lt;a href=&quot;http://en.wikipedia.org/wiki/Geodesic&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Geodesic&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="9" LastEditorUserId="9" LastEditorDisplayName="" LastEditDate="2010-07-27T01:39:57.477" LastActivityDate="2010-07-27T01:39:57.477" />
  <row Id="763" PostTypeId="2" ParentId="761" CreationDate="2010-07-27T01:42:46.160" Score="2" ViewCount="" Body="&lt;p&gt;F = ma  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;--&gt; dv/dt = a = F/m&lt;/p&gt;&#xA;&#xA;&lt;p&gt;F = mg - kv &lt;/p&gt;&#xA;&#xA;&lt;p&gt;--&gt; dv/dt = (g - k/m*v)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a differential equation, solution = v = A + B * exp(-k/m*t)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;dv/dt = - B * k/m * exp(-k/m*t)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;match terms and initial conditions (v = 0 at time t = 0) and you get g - k/m*A - k/m*B*exp(-k/m*t) = -B * k/m*exp(-k/m*t) --&gt; A = g * m / k, B = -g * m / k.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;--&gt; v = mg/k * (1 - exp(-k/m*t))&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's a linear differential equation (dv/dt is a linear function of v); the F = -kv&lt;sup&gt;2&lt;/sup&gt; is a nonlinear differential equation (can't remember off the top of my head how to deal with that one; it may not have a closed form solution). It's a bit difficult to summarize the techniques in general, but any good book on differential equations would cover them.&lt;/p&gt;&#xA;" OwnerUserId="120" LastActivityDate="2010-07-27T01:42:46.160" />
  <row Id="764" PostTypeId="1" AcceptedAnswerId="765" CreationDate="2010-07-27T03:16:25.763" Score="3" ViewCount="274" Body="&lt;p&gt;From Rudin's PMA:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;6.2 Definition&lt;/strong&gt;  Let $\alpha$ be a monotonically increasing function on $[a,b]$. ... Corresponding to each partition $P$ of $[a,b]$, we write&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Delta \alpha_i = \alpha(x_i) - \alpha(x_{i-1})$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;He then goes on to define the Riemann Stieltjes integral of $f$ with respect to $\alpha$, over the interval $[a,b]$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Riemann integral is then pointed out to be a special case of this when $\alpha(x)=x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With $\alpha(x)=x$, I understand $\Delta x = x_i - x_{i-1}$ to represent the directed magnitude of the &quot;base of the approximating rectangle&quot; that we then multiply by the value of $f$ taken somewhere within this interval, thus obtaining the area of an approximating rectangle.  I don't know where to begin to interpret the case where $\alpha(x) \not\equiv x$.  &lt;/p&gt;&#xA;" OwnerUserId="121" LastActivityDate="2010-08-05T17:10:06.220" Title="Riemann vs. Stieltjes Integral" Tags="&lt;real-analysis&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="765" PostTypeId="2" ParentId="764" CreationDate="2010-07-27T04:01:40.240" Score="4" ViewCount="" Body="&lt;p&gt;The idea is that we define a different way of measuring the width of the rectangle. Instead of using the difference of the two coordinates, we take the difference of the &lt;em&gt;images&lt;/em&gt; of those two coordinates under the map $\alpha$.  Basically, the idea is that the $(x_i-x_{i-1})$ is to be interpreted as the width or measure of the interval, a construction which leads more generally to the &lt;a href=&quot;http://en.wikipedia.org/wiki/Lebesgue_measure&quot; rel=&quot;nofollow&quot;&gt;Lebesgue measure&lt;/a&gt; of a subset of the reals.  The $\alpha(x_i)-\alpha(x_{i-1})$ leads to a different measure, starting with a different assignment of &quot;size&quot; to an interval.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that when $\alpha$ is continuously differentiable, then $\alpha(x_i) -\alpha(x_{i-1})$ is equal to $\alpha'(x) (x_i - x_{i-1})$ for $x \in [x_{i-1}, x_i]$ up to higher order terms. So in this case the Stieltjes integral with respect to $\alpha$ is the Riemann integral of the same function times $\alpha'$.&lt;/p&gt;&#xA;" OwnerUserId="536" LastActivityDate="2010-07-27T04:01:40.240" CommentCount="2" />
  <row Id="766" PostTypeId="1" AcceptedAnswerId="767" CreationDate="2010-07-27T06:25:37.183" Score="3" ViewCount="187" Body="&lt;p&gt;RHS is a well known test for determining the &lt;a href=&quot;http://en.wikipedia.org/wiki/Congruence_%28geometry%29#Determining_congruence&quot; rel=&quot;nofollow&quot;&gt;congruency of triangles&lt;/a&gt;. It is easy enough to prove it works, simply use Pythagorus' theorem to reduce to SSS. I thought that it seems strange that this only works for an angle being 90 degrees - or does it? What if I tried changed the given angle to 89 degrees or 91 degrees, would it still be uniquely identified up to congruence?&lt;/p&gt;&#xA;" OwnerUserId="123" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-07-27T06:47:59.760" LastActivityDate="2010-08-05T19:59:38.613" Title="RHS Congruency test - What makes 90 degrees different?" Tags="&lt;geometry&gt;&lt;trigonometry&gt;&lt;euclidean-geometry&gt;&lt;high-school&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="767" PostTypeId="2" ParentId="766" CreationDate="2010-07-27T06:46:59.903" Score="5" ViewCount="" Body="&lt;p&gt;Supposing that we knew two triangles had one angle congruent, a side adjacent to the angle congruent, and the side opposite the angle congruent.  This is sometimes referred to as SSA, which is not a congruence theorem (and I've heard it said that it is &quot;ass-backwards&quot;).  With a little more information, it is possible to determine congruence in some instances.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you'd said about RHS, in that case you can use right-triangle trigonometry to determine that the unknown sides are congruent, then use SSS to establish congruence.  Without the right angle, the technique for determining the length of the third side of the triangle is to use the Law of Sines to determine the measure of the unknown angle opposite the known side, use that to find the measure of the third angle, then use the Law of Cosines to determine the length of the unknown side.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's call one of the triangles ABC with &amp;ang;A and AB and BC being known.  From the Law of Sines, $\frac{\sin A}{BC}=\frac{\sin C}{AB}$ or $\sin C=\frac{AB\cdot\sin A}{BC}$.  There will be two values of C in the range 0&amp;deg; to 180&amp;deg; that satisfy this equation, unless sin &lt;em&gt;C&lt;/em&gt; = 1.  So:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;if sin &lt;em&gt;C&lt;/em&gt; = 1, then C is a right angle, the triangle is uniquely determined, so congruence can be established;&lt;/li&gt;&#xA;&lt;li&gt;if A &amp;ge; 90&amp;deg;, C &amp;lt; 90&amp;deg;, so there is only one solution for C that makes sense in this triangle, the triangle is uniquely determined, and congruence can be established (one might call this SSobtuseA);&lt;/li&gt;&#xA;&lt;li&gt;if A &amp;lt; 90&amp;deg;, but BC &amp;ge; AB, then B &amp;ge; C (in a triangle, the largest/smallest side is opposite the largest/smallest angle), so the only solution for C that makes sense in this triangle is the one with C&amp;nbsp;&amp;lt;&amp;nbsp;90&amp;deg; (if B&amp;nbsp;&amp;ge;&amp;nbsp;C&amp;nbsp;&gt;&amp;nbsp;90&amp;deg;, then A + B + C &gt; 180&amp;deg;), the triangle is uniquely determined, and congruence can be established (one might call this SsA, with the relative sizes of the S/s indicating the relative lengths);&lt;/li&gt;&#xA;&lt;li&gt;otherwise (when A &amp;lt; 90&amp;deg; and BC &amp;lt; AB), there are two possible values for C, both of which lead to triangles, so there are two possible triangles satisfying the given information, and congruence cannot be established.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;So, to your specific question, if the angle were 91&amp;deg; (case 2), congruence would follow; if the angle were 89&amp;deg;, congruence may or may not follow, depending on what you can determine about the other sides.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an aside, RHS is also commonly referred to (at least in the midwestern U.S. in contemporary high school geometry) as hypotenuse-leg or HL.&lt;/p&gt;&#xA;" OwnerUserId="72" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-08-05T19:59:38.613" LastActivityDate="2010-08-05T19:59:38.613" CommentCount="3" />
  <row Id="768" PostTypeId="1" AcceptedAnswerId="773" CreationDate="2010-07-27T08:02:16.660" Score="11" ViewCount="266" Body="&lt;p&gt;Pick's theorem says that given a square grid (that is, all points in the plane with integer coordinates) and a polygon without holes and non selt-intersecting whose vertices are grid points, its area is given by &lt;/p&gt;&#xA;&#xA;&lt;p&gt;i + b/2 - 1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where i is the number of interior lattice points and b is the number of points on its boundary. Theorem and proof may be found on &lt;a href=&quot;http://en.wikipedia.org/wiki/Pick%27s_theorem&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let us suppose that the grid is not square but triangular (or hexagonal). Does a similar theorem hold?&lt;/p&gt;&#xA;" OwnerUserId="89" LastActivityDate="2010-07-30T08:06:23.087" Title="Pick's Theorem on a triangular (or hex) grid" Tags="&lt;geometry&gt;&lt;discrete-geometry&gt;" AnswerCount="3" CommentCount="4" FavoriteCount="3" />
  <row Id="769" PostTypeId="2" ParentId="664" CreationDate="2010-07-27T08:26:39.377" Score="1" ViewCount="" Body="&lt;p&gt;$dm$ takes density fluctuations into account, it's just $dm = \rho(\vec r) d^3r$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;E.g. for a homogeneous cylinder with the rotational axis align parallel to the z-axis it is $\rho = \frac{m}{V}\cdot\theta(R-r)\theta(z^2-a^2)$, where $\theta(x) = \begin{cases} 0 for x&amp;lt;0 and\ 1 for x&gt;0 \end{cases}$ and $r = \sqrt{x^2+y^2}$ is the radial coordinate in &lt;a href=&quot;http://en.wikipedia.org/wiki/Cylindrical_coordinate_system&quot; rel=&quot;nofollow&quot;&gt;cylindrical coordinates&lt;/a&gt;. The result is that your integration over $r$ is from 0 to R and for z from -a to a. $V=\pi R^2 \cdot 2a$ is the volume of the cylinder, $m$ its weight. In cylindrical coordinates, $d^3r = r dr d\phi dz$, so you got&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$I = \int_{r=0}^{R} \int_{z=-a}^a \int_{\phi=0}^{2\pi} r^2 \frac{m}{V} \cdot r d\phi dz dr = 4a\pi \frac{m}{V} \int_{r=0}^R r^3 dr = \frac{am\pi}{V}R^4 = \frac{1}{2}mR^2$.&lt;/p&gt;&#xA;" OwnerUserId="163" LastActivityDate="2010-07-27T08:26:39.377" />
  <row Id="770" PostTypeId="2" ParentId="720" CreationDate="2010-07-27T08:39:51.483" Score="5" ViewCount="" Body="&lt;p&gt;First, some background.  A &quot;line&quot; (the object containing the path of least distance between two points) on a sphere is a great circle; a great circle is the intersection of a sphere with a plane passing through the center of the sphere.  A spherical triangle is the triangle formed by the spherical line segments connecting three non-collinear points on a sphere.  Below is the Spherical Law of Cosines as it appears in UCSMP &lt;em&gt;Functions, Statistics, and Trigonometry&lt;/em&gt;, 3rd ed., copied here because the diagram is good and helps with clarity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/KAOhC.png&quot; alt=&quot;Spherical Law of Cosines (from UCSMP Functions, Statistics, and Trigonometry, 3rd ed.)&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If ABC is a spherical triangle with arcs a, b, and c (meaning the measures of the arcs, not the lengths), then cos c = cos a * cos b + sin a * sin b * cos C.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, to the specific problem at hand.  Let's use the diagram below, also from UCSMP &lt;em&gt;Functions, Statistics, and Trigonometry&lt;/em&gt;, 3rd ed., for reference.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/DRkJ3.png&quot; alt=&quot;globe (from UCSMP Functions, Statistics, and Trigonometry, 3rd ed.)&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let B be the initial point and A be the final point of the movement (and for simplicity, I'll assume they are both in the northern hemisphere, and leave extending the solution to any points as an exercise); N and S are the north and south poles, respectively; C and D are the points on the equator that are on the same line of longitude as A and B, respectively.  Consider spherical triangle ABN.  a = (90&amp;deg; - latitude of point A); b = (90&amp;deg; - latitude of point B). N = positive difference in longitude between points A and B.  Use the Spherical Law of Cosines (cos n = ... form) to determine n, then use the Spherical Law of Cosines (cos b = ... form) to determine B.  If the points are arranged as pictured, the heading of travel from B to A is 360&amp;deg; - B; for other arrangements of the points, the exact method of determining the heading based on the measure of angle B may vary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;(graphics from Lesson 5-10 of UCSMP Functions, Statistics, and Trigonometry, 3rd ed., © 2010 Wright Group/McGraw Hill)&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="72" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-09-29T03:45:13.390" LastActivityDate="2010-09-29T03:45:13.390" CommentCount="1" />
  <row Id="771" PostTypeId="1" AcceptedAnswerId="775" CreationDate="2010-07-27T09:13:21.393" Score="2" ViewCount="464" Body="&lt;p&gt;I am trying to show the following: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://dl.dropbox.com/u/1885087/Expectation.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but I really can't remember what I am supposed to do to get from the LHS to the RHS. I have tried using integration this way&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://dl.dropbox.com/u/1885087/integ.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And then use integration by parts, but even though what I get resembles it, it can't be correct (because $e^{-\gamma W}$ is not the distribution of W).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have also tried using Taylor series expansion, but I think I am way off, and I don't think an approximation here is what I need, because the equality above is exact.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;FYI, this is not homework, I am working through a &lt;a href=&quot;http://www.princeton.edu/~markus/research/papers/liquidity.pdf&quot; rel=&quot;nofollow&quot;&gt;paper&lt;/a&gt; (page 10) and I would really like to know how every step was derived.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone at least point me to the right direction?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;: This expectation on the RHS is very similar to the moment generating function formula (with a negative exponent). If you check &lt;a href=&quot;http://en.wikipedia.org/wiki/Moment-generating_function#Examples&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;, you will see that the moment generating function for the normal distribution is like the LHS (but with a positive sign). So in a way I have my answer, but I still would like to know how to derive it, if there is a way. I know little if anything at all about moment generating functions, so maybe I shouldn't try and derive it but rather just use the result? Does it even make sense to try and derive it?&lt;/p&gt;&#xA;" OwnerUserId="85" LastEditorUserId="85" LastEditorDisplayName="" LastEditDate="2010-08-04T09:53:01.523" LastActivityDate="2010-08-04T09:53:01.523" Title="Expected value of a function of a random variable: help!" Tags="&lt;probability-theory&gt;&lt;taylor-expansion&gt;&lt;probability&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="1" />
  <row Id="772" PostTypeId="2" ParentId="768" CreationDate="2010-07-27T09:22:36.813" Score="2" ViewCount="" Body="&lt;p&gt;This is a very interesting question. I don't have a complete solution yet, but I did get some results. Consider an arbitrary distribution of points. Let P(i,b)=i+b/2-1 where i is the number of internal points and P is the number of boundary points. Let P(A)=P(i_A,b_A), where A is a simple polygon with all its vertices on the points. Wikipedia shows that P(C)=P(A)+P(T) where T is a triangle that shares a single edge with A and C a simple polygon formed by the union of A and T. Since all simple polygons can be triangulated, P(C)=sum P(t) for all t in the triangulation of C.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The proof then has a second part that shows P(t) equals the area for any triangle. So if we want to generalise it for other grids, we have to find a property equal to P(T) for any triangle.&lt;/p&gt;&#xA;" OwnerUserId="123" LastActivityDate="2010-07-27T09:22:36.813" CommentCount="1" />
  <row Id="773" PostTypeId="2" ParentId="768" CreationDate="2010-07-27T09:27:33.757" Score="8" ViewCount="" Body="&lt;p&gt;The short answer is that, no, there can be no formula for polygons with vertices in the hexagonal lattice in terms of just boundary and interior points. This is based on the fact that primitive triangles on this lattice--ones with no lattice points on their boundary (besides the vertices) or in the interior--can have different areas, whereas for the square lattice all primitive triangles have area 1/2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, as Casebash has partly gotten at in his answer, you can approximate things well if you compute what, in the below paper, is called the &quot;boundary characteristic&quot; of the polygon, a number that is somewhat complicated to think to compute, but which gives a decent proxy for how many of each type of primitive triangle the polygon contains.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Kolodziejczyk has been the main one doing work on hexagonal lattice results of this type that I know; he's worth looking up for similar results. Ding Ren is another, and the older work of Grunbaum, etc., still bears on the problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;A Fast Pick-Type Approximation for the Area of H-Polygons,&quot; Ren, Kolodziejczyk, et al., American Mathematical Monthly, 1993.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Available at &lt;a href=&quot;http://www.jstor.org/pss/2323889&quot; rel=&quot;nofollow&quot;&gt;JSTOR&lt;/a&gt; if you have access.&lt;/p&gt;&#xA;" OwnerUserId="76" LastActivityDate="2010-07-27T09:27:33.757" />
  <row Id="774" PostTypeId="2" ParentId="764" CreationDate="2010-07-27T09:33:27.300" Score="2" ViewCount="" Body="&lt;p&gt;To add to Akhil's answer, the Stieltjes integral made a lot more sense to me when I started thinking about it in terms of physics. Think of taking something like the integral you get to determine moment of inertia, where it's mass--a function of position--you're integrating over, and not position itself.&lt;/p&gt;&#xA;" OwnerUserId="76" LastActivityDate="2010-07-27T09:33:27.300" />
  <row Id="775" PostTypeId="2" ParentId="771" CreationDate="2010-07-27T09:51:33.053" Score="3" ViewCount="" Body="&lt;p&gt;If W is randomly chosen with the PDF P(x), then the expectation value should be &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;img src=&quot;http://mathcache.appspot.com/?tex=%5cpng%5c%5bE%5Be%5E%7B-%5Cgamma%20W%7D%5D%3D%5Cint_%7B-%5Cinfty%7D%5E%5Cinfty%20P%28x%29%20e%5E%7B-%5Cgamma%20x%7D%20dx%5c%5d&quot; alt=&quot;$E[e^{-\gamma W}]=\int_{-\infty}^\infty P(x) e^{-\gamma x} dx$&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And I think that equation (E[e&lt;sup&gt;-&amp;gamma;W&lt;/sup&gt;] = e&lt;sup&gt;-&amp;gamma;(E[W] - &amp;frac12;&amp;gamma;Var[W])&lt;/sup&gt;) is correct only when W is a normal distribution.&lt;/p&gt;&#xA;" OwnerUserId="171" LastEditorUserId="171" LastEditorDisplayName="" LastEditDate="2010-07-27T10:22:37.633" LastActivityDate="2010-07-27T10:22:37.633" CommentCount="3" />
  <row Id="776" PostTypeId="1" AcceptedAnswerId="777" CreationDate="2010-07-27T09:52:02.100" Score="3" ViewCount="1161" Body="&lt;p&gt;There is a well known theorem often stated as the angle in a semi-circle being 90 degrees. To be more accurate, any triangle with one of its sides being a diameter and all vertices on the circle has its angle opposite the diameter being 90 degrees. The standard proof uses isosceles triangles and is worth having as an answer, but there is also a much more intuitive proof as well (this proof is more complicated though).&lt;/p&gt;&#xA;" OwnerUserId="123" LastEditorUserId="123" LastEditorDisplayName="" LastEditDate="2010-07-27T20:43:11.277" LastActivityDate="2010-08-06T14:15:55.930" Title="Proof of Angle in a Semi-Circle is 90 degrees" Tags="&lt;geometry&gt;&lt;euclidean-geometry&gt;&lt;circle&gt;" AnswerCount="3" CommentCount="8" />
  <row Id="777" PostTypeId="2" ParentId="776" CreationDate="2010-07-27T10:02:15.607" Score="12" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Nonstandard proof&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the semi-circle with endpoints A and C and center O and the inscribed angle &amp;ang;ABC (B on the semi-circle) together with the rotation image of both about O by 180&amp;deg;.  The image of A is C and vice versa; let B' be the image of B.  The image of a line under a 180&amp;deg; rotation is parallel to the original line, AB is parallel to CB' and BC is parallel to B'A, so ABCB' is a parallelogram.  BO and its image must be parallel, but the image of O is itself, since it is the center of rotation, and if BO and B'O are parallel and contain a point in common, they must lie on the same line, so BB' passes through O.  AC and BB' (the diagonals of ABCB') are both diameters of the circle, so they are congruent.  A parallelogram with congruent diagonals is a rectangle.  Thus, &amp;ang;ABC is a right angle (and has measure 90&amp;deg;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://www.imgftw.net/img/762828246.png&quot; alt=&quot;diagram&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Standard proof&lt;/strong&gt; (or, at least, my guess at it based on the description in the question)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As above, consider the semi-circle with endpoints A and C and center O and the inscribed angle &amp;ang;ABC (B on the semi-circle).  Draw in radius OB.  OA = OB, so &amp;#x25B3;AOB is isosceles and &amp;ang;OAB&amp;cong;&amp;ang;OBA.  OB = OC, so &amp;#x25B3;BOC is isosceles and &amp;ang;OBC&amp;cong;&amp;ang;OCB.  Let &amp;alpha;=m&amp;ang;OAB=m&amp;ang;OBA and &amp;beta;=m&amp;ang;OBC=m&amp;ang;OCB.  In &#xA;&amp;#x25B3;ABC, the measures of the angles are &amp;alpha;, &amp;alpha;+&amp;beta;, and &amp;beta;, so &amp;alpha;+(&amp;alpha;+&amp;beta;)+&amp;beta;=180&amp;deg; or 2(&amp;alpha;+&amp;beta;)=180&amp;deg; or &amp;alpha;+&amp;beta;=90&amp;deg;, so &amp;ang;ABC has measure 90&amp;deg; and is a right angle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://www.imgftw.net/img/319527897.png&quot; alt=&quot;diagram&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;edit:&lt;/em&gt; &lt;strong&gt;Another Nonstandard proof&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Use the labeling as above and apply &lt;a href=&quot;http://en.wikipedia.org/wiki/Stewart%27s_theorem&quot; rel=&quot;nofollow&quot;&gt;Stewart's Theorem&lt;/a&gt; to &amp;#x25B3;ABC: $$(AB)^2(OC) + (BC)^2(AO) = (AC)((BO)^2 + (AO)(OC))$$  Substituting the length r of the radius of the semicircle as appropriate: $$(AB)^2r + (BC)^2r = 2r(r^2 + r^2)=4r^3$$  Dividing both sides by r: $$(AB)^2+(BC)^2=(2r)^2=(AC)^2$$&#xA;So, by the converse of the Pythagorean Theorem, &amp;ang;ABC is a right angle.&lt;/p&gt;&#xA;" OwnerUserId="72" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-08-04T06:22:02.320" LastActivityDate="2010-08-04T06:22:02.320" CommentCount="13" />
  <row Id="778" PostTypeId="2" ParentId="756" CreationDate="2010-07-27T10:20:44.983" Score="7" ViewCount="" Body="&lt;p&gt;Look at &lt;a href=&quot;http://mathoverflow.net/questions/32269&quot; rel=&quot;nofollow&quot;&gt;this answer&lt;/a&gt; on MathOverflow:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Yes, there is a way to guess a number asking 14 questions in worst case. To do it you need a linear code with length 14, dimension 10 and distance at least 3. One such code can be built based on Hamming code (see &lt;a href=&quot;http://en.wikipedia.org/wiki/Hamming_code&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Hamming_code&lt;/a&gt;).&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Here is the strategy.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Let us denote bits of first player's number as &lt;em&gt;a&lt;sub&gt;i&lt;/sub&gt;&lt;/em&gt;, &lt;em&gt;i&lt;/em&gt; ∈ [1..10]. We start with asking values of all those bits. That is we ask the following questions: &quot;is it true that &lt;em&gt;i&lt;/em&gt;-th bit of your number is zero?&quot; Let us denote answers on those questions as &lt;em&gt;b&lt;sub&gt;i&lt;/sub&gt;&lt;/em&gt;, &lt;em&gt;i&lt;/em&gt; ∈ [1..10].&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Now we ask 4 additional questions:&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Is it true that &lt;em&gt;a&lt;sub&gt;1&lt;/sub&gt;&lt;/em&gt; ⊗ &lt;em&gt;a&lt;sub&gt;2&lt;/sub&gt;&lt;/em&gt; ⊗ &lt;em&gt;a&lt;sub&gt;4&lt;/sub&gt;&lt;/em&gt; ⊗ &lt;em&gt;a&lt;sub&gt;5&lt;/sub&gt;&lt;/em&gt; ⊗ &lt;em&gt;a&lt;sub&gt;7&lt;/sub&gt;&lt;/em&gt; ⊗ &lt;em&gt;a&lt;sub&gt;9&lt;/sub&gt;&lt;/em&gt; is equal to zero? (⊗ is sumation modulo 2).&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Is it true that &lt;em&gt;a&lt;sub&gt;1&lt;/sub&gt;&lt;/em&gt; ⊗ &lt;em&gt;a&lt;sub&gt;3&lt;/sub&gt;&lt;/em&gt; ⊗ &lt;em&gt;a&lt;sub&gt;4&lt;/sub&gt;&lt;/em&gt; ⊗ &lt;em&gt;a&lt;sub&gt;6&lt;/sub&gt;&lt;/em&gt; ⊗ &lt;em&gt;a&lt;sub&gt;7&lt;/sub&gt;&lt;/em&gt; ⊗ &lt;em&gt;a&lt;sub&gt;10&lt;/sub&gt;&lt;/em&gt; is equal to zero?&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Is it true that &lt;em&gt;a&lt;sub&gt;2&lt;/sub&gt;&lt;/em&gt; ⊗ &lt;em&gt;a&lt;sub&gt;3&lt;/sub&gt;&lt;/em&gt; ⊗ &lt;em&gt;a&lt;sub&gt;4&lt;/sub&gt;&lt;/em&gt; ⊗ &lt;em&gt;a&lt;sub&gt;8&lt;/sub&gt;&lt;/em&gt; ⊗ &lt;em&gt;a&lt;sub&gt;9&lt;/sub&gt;&lt;/em&gt; ⊗ &lt;em&gt;a&lt;sub&gt;10&lt;/sub&gt;&lt;/em&gt; is equal to zero?&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Is it true that &lt;em&gt;a&lt;sub&gt;5&lt;/sub&gt;&lt;/em&gt; ⊗ &lt;em&gt;a&lt;sub&gt;6&lt;/sub&gt;&lt;/em&gt; ⊗ &lt;em&gt;a&lt;sub&gt;7&lt;/sub&gt;&lt;/em&gt; ⊗ &lt;em&gt;a&lt;sub&gt;8&lt;/sub&gt;&lt;/em&gt; ⊗ &lt;em&gt;a&lt;sub&gt;9&lt;/sub&gt;&lt;/em&gt; ⊗ &lt;em&gt;a&lt;sub&gt;10&lt;/sub&gt;&lt;/em&gt; is equal to zero?&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Let &lt;em&gt;q&lt;sub&gt;1&lt;/sub&gt;&lt;/em&gt;, &lt;em&gt;q&lt;sub&gt;2&lt;/sub&gt;&lt;/em&gt;, &lt;em&gt;q&lt;sub&gt;3&lt;/sub&gt;&lt;/em&gt; and &lt;em&gt;q&lt;sub&gt;4&lt;/sub&gt;&lt;/em&gt; be answers on those additional questions. Now second player calculates &lt;em&gt;t&lt;sub&gt;i&lt;/sub&gt;&lt;/em&gt; (&lt;em&gt;i&lt;/em&gt; ∈ [1..4]) --- answers on those questions based on bits &lt;em&gt;b&lt;sub&gt;j&lt;/sub&gt;&lt;/em&gt; which he previously got from first player.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Now there are 16 ways how bits &lt;em&gt;q&lt;sub&gt;i&lt;/sub&gt;&lt;/em&gt; can differ from &lt;em&gt;t&lt;sub&gt;i&lt;/sub&gt;&lt;/em&gt;. Let &lt;em&gt;d&lt;sub&gt;i&lt;/sub&gt;&lt;/em&gt; = &lt;em&gt;q&lt;sub&gt;i&lt;/sub&gt;&lt;/em&gt; ⊗ &lt;em&gt;t&lt;sub&gt;i&lt;/sub&gt;&lt;/em&gt; (hence &lt;em&gt;d&lt;sub&gt;i&lt;/sub&gt;&lt;/em&gt; = 1 iff &lt;em&gt;q&lt;sub&gt;i&lt;/sub&gt;&lt;/em&gt; ≠ &lt;em&gt;t&lt;sub&gt;i&lt;/sub&gt;&lt;/em&gt; ).&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Let us make table of all possible errors and corresponding values of &lt;em&gt;d&lt;sub&gt;i&lt;/sub&gt;&lt;/em&gt;:&#xA;  position of error -&gt; (&lt;em&gt;d&lt;sub&gt;1&lt;/sub&gt;&lt;/em&gt;, &lt;em&gt;d&lt;sub&gt;2&lt;/sub&gt;&lt;/em&gt;, &lt;em&gt;d&lt;sub&gt;3&lt;/sub&gt;&lt;/em&gt;, &lt;em&gt;d&lt;sub&gt;4&lt;/sub&gt;&lt;/em&gt;)&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;no error -&gt; (0, 0, 0, 0)&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;error in &lt;em&gt;b&lt;sub&gt;1&lt;/sub&gt;&lt;/em&gt; -&gt; (1, 1, 0, 0)&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;error in &lt;em&gt;b&lt;sub&gt;2&lt;/sub&gt;&lt;/em&gt; -&gt; (1, 0, 1, 0)&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;error in &lt;em&gt;b&lt;sub&gt;3&lt;/sub&gt;&lt;/em&gt; -&gt; (0, 1, 1, 0)&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;error in &lt;em&gt;b&lt;sub&gt;4&lt;/sub&gt;&lt;/em&gt; -&gt; (1, 1, 1, 0)&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;error in &lt;em&gt;b&lt;sub&gt;5&lt;/sub&gt;&lt;/em&gt; -&gt; (1, 0, 0, 1)&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;error in &lt;em&gt;b&lt;sub&gt;6&lt;/sub&gt;&lt;/em&gt; -&gt; (0, 1, 0, 1)&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;error in &lt;em&gt;b&lt;sub&gt;7&lt;/sub&gt;&lt;/em&gt; -&gt; (1, 1, 0, 1)&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;error in &lt;em&gt;b&lt;sub&gt;8&lt;/sub&gt;&lt;/em&gt; -&gt; (0, 0, 1, 1)&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;error in &lt;em&gt;b&lt;sub&gt;9&lt;/sub&gt;&lt;/em&gt; -&gt; (1, 0, 1, 1)&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;error in &lt;em&gt;b&lt;sub&gt;10&lt;/sub&gt;&lt;/em&gt; -&gt; (0, 1, 1, 1)&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;error in &lt;em&gt;q&lt;sub&gt;1&lt;/sub&gt;&lt;/em&gt; -&gt; (1, 0, 0, 0)&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;error in &lt;em&gt;q&lt;sub&gt;2&lt;/sub&gt;&lt;/em&gt; -&gt; (0, 1, 0, 0)&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;error in &lt;em&gt;q&lt;sub&gt;3&lt;/sub&gt;&lt;/em&gt; -&gt; (0, 0, 1, 0)&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;error in &lt;em&gt;q&lt;sub&gt;4&lt;/sub&gt;&lt;/em&gt; -&gt; (0, 0, 0, 1)&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;All the values of (&lt;em&gt;d&lt;sub&gt;1&lt;/sub&gt;&lt;/em&gt;, &lt;em&gt;d&lt;sub&gt;2&lt;/sub&gt;&lt;/em&gt;, &lt;em&gt;d&lt;sub&gt;3&lt;/sub&gt;&lt;/em&gt;, &lt;em&gt;d&lt;sub&gt;4&lt;/sub&gt;&lt;/em&gt;) are different. Hence we can find where were an error and hence find all &lt;em&gt;a&lt;sub&gt;i&lt;/sub&gt;&lt;/em&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Answered by &lt;a href=&quot;http://mathoverflow.net/users/7079/falagar&quot; rel=&quot;nofollow&quot;&gt;falagar&lt;/a&gt;.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="92" LastEditorUserId="227" LastEditorDisplayName="" LastEditDate="2010-07-27T21:49:17.880" LastActivityDate="2010-07-27T21:49:17.880" CommunityOwnedDate="2010-07-27T10:20:44.983" CommentCount="2" />
  <row Id="779" PostTypeId="2" ParentId="742" CreationDate="2010-07-27T11:02:18.070" Score="0" ViewCount="" Body="&lt;p&gt;Let &lt;code&gt;t&lt;/code&gt; be your target and &lt;code&gt;a&lt;/code&gt; and &lt;code&gt;b &amp;lt; a&lt;/code&gt; be the values of the stamps you have. Let &lt;code&gt;n = ⌈t/a⌉&lt;/code&gt; be the maximum number of &lt;code&gt;a&lt;/code&gt; stamps it makes sense to use.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For each number &lt;code&gt;i&lt;/code&gt; from &lt;code&gt;n&lt;/code&gt; to &lt;code&gt;0&lt;/code&gt;, consider the cost of the solution with:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;code&gt;i&lt;/code&gt; stamps of value &lt;code&gt;a&lt;/code&gt; and&lt;/li&gt;&#xA;&lt;li&gt;&lt;code&gt;j = ⌈(t - i*a) / b⌉&lt;/code&gt; stamps of value &lt;code&gt;b&lt;/code&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The cost of each solution would obviously be &lt;code&gt;i*a + j*b&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your solution is the pair &lt;code&gt;(i, ⌈(t - i*a) / b⌉ )&lt;/code&gt; that minimizes said cost. (Note there's only one unknown in that pair, &lt;code&gt;i&lt;/code&gt;.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An additional optimization would be to stop immediately as soon as a division without rest is performed, either when determining &lt;code&gt;n&lt;/code&gt; or &lt;code&gt;j&lt;/code&gt;. A division without rest means that we have found a solution that perfectly pays the due amount and cannot thus be improved on. (There may be others like it, but none that is actually better, so there's no need to keep searching).&lt;/p&gt;&#xA;" OwnerUserId="92" LastActivityDate="2010-07-27T11:02:18.070" CommentCount="1" />
  <row Id="780" PostTypeId="2" ParentId="631" CreationDate="2010-07-27T11:21:22.967" Score="3" ViewCount="" Body="&lt;p&gt;This is still WIP. There are a few missing details, still I think it's better than nothing. Feel free to edit in the missing details.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given a problem of &lt;code&gt;SUBSET-SUM&lt;/code&gt;. We have a set of &lt;code&gt;A&lt;/code&gt;={a&lt;sub&gt;1&lt;/sub&gt;,a&lt;sub&gt;2&lt;/sub&gt;,...,a&lt;sub&gt;n&lt;/sub&gt;} numbers, and another number &lt;code&gt;s&lt;/code&gt;. The question we're seeking answer to is, whether or not there's a subset of &lt;code&gt;A&lt;/code&gt; whose sum is &lt;code&gt;s&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm assuming that the 24-game allows you to use rational numbers. Even if it doesn't, I think that it is possible to emulate rational numbers up to denominator of size &lt;code&gt;p&lt;/code&gt; with integers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know that &lt;code&gt;SUBSET-SUM&lt;/code&gt; is NP-complete even for integers only. I think the &lt;code&gt;SUBSET-SUM&lt;/code&gt; problem is &lt;code&gt;NP&lt;/code&gt;-hard even if you allow treating each a&lt;sub&gt;i&lt;/sub&gt; as a negative number. That is even if &lt;code&gt;A&lt;/code&gt; is of the form &lt;code&gt;A&lt;/code&gt;={a&lt;sub&gt;1&lt;/sub&gt;,-a&lt;sub&gt;1&lt;/sub&gt;,a&lt;sub&gt;2&lt;/sub&gt;,-a&lt;sub&gt;2&lt;/sub&gt;,...,a&lt;sub&gt;n&lt;/sub&gt;,-a&lt;sub&gt;n&lt;/sub&gt;}. This is still a wrinkle I need to iron out in this reduction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously, if there's a subset of &lt;code&gt;A&lt;/code&gt; with sum &lt;code&gt;s&lt;/code&gt;, then there's a solution to the &lt;code&gt;24&lt;/code&gt;-problem for how to reach using &lt;code&gt;A&lt;/code&gt; to &lt;code&gt;s&lt;/code&gt;. The solution is only using the &lt;code&gt;+&lt;/code&gt; sign.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is, what happens if there's no solution which only uses the &lt;code&gt;+&lt;/code&gt; sign, but there is a solution which uses other arithmetic operations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let us consider the following problem. Let's take a prime &lt;code&gt;p&lt;/code&gt; which is larger than &lt;code&gt;n&lt;/code&gt;, the total number of elements in &lt;code&gt;A&lt;/code&gt;. Given an oracle which solves the &lt;code&gt;24&lt;/code&gt;-problem, and a &lt;code&gt;SUBSET-SUM&lt;/code&gt; problem of &lt;code&gt;A&lt;/code&gt;={a&lt;sub&gt;1&lt;/sub&gt;,a&lt;sub&gt;2&lt;/sub&gt;,...,a&lt;sub&gt;n&lt;/sub&gt;} and &lt;code&gt;s&lt;/code&gt;. We'll ask the oracle to solve the &lt;code&gt;24&lt;/code&gt;-problem on&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;code&gt;A&lt;/code&gt;={a&lt;sub&gt;1&lt;/sub&gt;+(1/p),a&lt;sub&gt;2&lt;/sub&gt;+(1/p),...,a&lt;sub&gt;n&lt;/sub&gt;+(1/p)}&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;for the following values:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;s&lt;sub&gt;1&lt;/sub&gt;=s+1/p,s&lt;sub&gt;2&lt;/sub&gt;=s+2/p,...,s&lt;sub&gt;n&lt;/sub&gt;=s+n/p.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If the solution includes multiplication, we will have a denominator larger than &lt;code&gt;p&lt;/code&gt; in the end result, and thus we will not be able to reach any s&lt;sub&gt;i&lt;/sub&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given an arithmetric expression that contains a&lt;sub&gt;i&lt;/sub&gt;a&lt;sub&gt;j&lt;/sub&gt;=x+(1/p&lt;sup&gt;2&lt;/sup&gt;), It is impossible that the denominator p&lt;sup&gt;2&lt;/sup&gt; would &quot;cancel&quot; out, since there are at most &lt;code&gt;n&lt;/code&gt; elements in the summation, and thus the numerator would never reach &lt;code&gt;p&lt;/code&gt;, since &lt;code&gt;p&amp;gt;n&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;THIS IS NOT QUIT RIGHT! The expression a&lt;sub&gt;i&lt;/sub&gt;a&lt;sub&gt;j&lt;/sub&gt;-a&lt;sub&gt;k&lt;/sub&gt;a&lt;sub&gt;l&lt;/sub&gt; will be an integer, and therefor our oracle might return answer which includes two multiplications one negative and one positive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What about division? How can we be sure no division will occur. Find another prime &lt;code&gt;q&lt;/code&gt; which is different than p, and larger than the largest a&lt;sub&gt;i&lt;/sub&gt; times &lt;code&gt;n&lt;/code&gt;. Multiply all answers by &lt;code&gt;q&lt;/code&gt;. The set &lt;code&gt;A&lt;/code&gt; will be&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;code&gt;A&lt;/code&gt;={qa&lt;sub&gt;1&lt;/sub&gt;+(1/p),qa&lt;sub&gt;2&lt;/sub&gt;+(1/p),...,qa&lt;sub&gt;n&lt;/sub&gt;+(1/p)}&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We will look for the following values:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;s&lt;sub&gt;1&lt;/sub&gt;=qs+1/p,s&lt;sub&gt;2&lt;/sub&gt;=qs+2/p,...,s&lt;sub&gt;n&lt;/sub&gt;=qs+n/p.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In that case, a&lt;sub&gt;i&lt;/sub&gt;/a&lt;sub&gt;j&lt;/sub&gt; will be smaller than the minimal element in &lt;code&gt;A&lt;/code&gt;, and therefor the end result which will contains a&lt;sub&gt;i&lt;/sub&gt;/a&lt;sub&gt;j&lt;/sub&gt; will never be one of the s&lt;sub&gt;i&lt;/sub&gt; we're looking for.&lt;/p&gt;&#xA;" OwnerUserId="52" LastEditorUserId="52" LastEditorDisplayName="" LastEditDate="2010-07-28T15:25:00.803" LastActivityDate="2010-07-28T15:25:00.803" CommentCount="8" />
  <row Id="781" PostTypeId="1" AcceptedAnswerId="863" CreationDate="2010-07-27T13:14:14.543" Score="4" ViewCount="275" Body="&lt;p&gt;Both &lt;a href=&quot;http://en.wikipedia.org/wiki/Benford%27s_law&quot; rel=&quot;nofollow&quot;&gt;Benford's Law&lt;/a&gt; (if you take a list of values, the distribution of the most significant digit is rougly proportional to the logarithm of the digit) and &lt;a href=&quot;http://en.wikipedia.org/wiki/Zipf%27s_law&quot; rel=&quot;nofollow&quot;&gt;Zipf's Law&lt;/a&gt; (given a corpus of natural language utterances, the frequency of any word is roughly inversely proportional to its rank in the frequency table) are not theorems in a mathematical sense, but they work quite good in the real life. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anyone have an idea why this happens?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(see also &lt;a href=&quot;http://math.stackexchange.com/questions/58&quot; rel=&quot;nofollow&quot;&gt;this question&lt;/a&gt;)&lt;/p&gt;&#xA;" OwnerUserId="89" LastActivityDate="2010-10-13T18:03:48.933" Title="Why does Benford's Law (or Zipf's Law) hold?" Tags="&lt;soft-question&gt;&lt;statistics&gt;&lt;applications&gt;&lt;philosophy&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="782" PostTypeId="2" ParentId="781" CreationDate="2010-07-27T13:16:41.230" Score="4" ViewCount="" Body="&lt;p&gt;For the case of Bendford's Law, of course scale invariance is a necessary condition; it the law must be true either if we measure things in meters or in feet or in furlongs, thus multiplying given data for a constant, the only distribution which allows this is the logarithmic one. But its being necessary does not mean that it is the answer, of course.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Scale invariance is not relevant for Zipf's Law, however, since we have an absolute rank.&lt;/p&gt;&#xA;" OwnerUserId="89" LastActivityDate="2010-07-27T13:16:41.230" CommentCount="2" />
  <row Id="783" PostTypeId="1" AcceptedAnswerId="784" CreationDate="2010-07-27T17:19:25.283" Score="2" ViewCount="170" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a category I have two objects $a$ and $b$ and a morphism $m$ from $a$ to $b$ and one $n$ from $b$ to $a$. Is this always an isomorphism? Why is it emphasized that this has to be true, too: $m \circ n = \mathrm{id}_b$ and $n \circ m = \mathrm{id}_a$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am looking for an example in which the id-part is not true and therefore $m$ and $n$ are not isomorphic.&lt;/p&gt;&#xA;" OwnerUserId="179" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-05-24T16:42:28.010" LastActivityDate="2011-05-24T16:45:21.880" Title="isomorphism and $id$" Tags="&lt;category-theory&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="784" PostTypeId="2" ParentId="783" CreationDate="2010-07-27T17:41:32.347" Score="6" ViewCount="" Body="&lt;p&gt;If you have no restrictions on m and n, then clearly they cannot be isomorphisms in general.  For instance, take any two groups G and H and let m: G --&gt; H, n: H --&gt; G be the zero homomorphisms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even if you say that n and m are monomorphisms, then it is still not true in general that they are isomorphisms.  I believe it is true however if your category is one whose objects are sets with additional structure. See this question: &lt;a href=&quot;http://math.stackexchange.com/questions/309/if-a-is-a-subobject-of-b-and-b-a-subobject-of-a-are-they-isomorphic/327#327&quot; rel=&quot;nofollow&quot;&gt;http://math.stackexchange.com/questions/309/if-a-is-a-subobject-of-b-and-b-a-subobject-of-a-are-they-isomorphic/327#327&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="9" LastActivityDate="2010-07-27T17:41:32.347" CommentCount="4" />
  <row Id="785" PostTypeId="1" AcceptedAnswerId="786" CreationDate="2010-07-27T18:31:44.133" Score="2" ViewCount="823" Body="&lt;p&gt;There is a general formula for solving quadratic equations, namely the Quadratic Formula.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For third degree equations of the form $ax^3+bx^2+cx+d=0$, there is a set of thee equations: one for each root.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Is there a general formula for solving equations of the form&#xA;$ax^4+bx^3+cx^2+dx+e=0$&#xA;?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How about for higher degrees? If not, why not?&lt;/p&gt;&#xA;" OwnerUserId="38" LastActivityDate="2010-08-05T23:00:44.320" Title="Is there a general formula for solving 4th degree equations?" Tags="&lt;algebra-precalculus&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="786" PostTypeId="2" ParentId="785" CreationDate="2010-07-27T18:44:11.520" Score="11" ViewCount="" Body="&lt;p&gt;There is, in fact, a general formula for solving quartic (4th degree polynomial) equations.  As the cubic formula is significantly more complex than the quadratic formula, the quartic formula is significantly more complex than the cubic formula.  &lt;a href=&quot;http://en.wikipedia.org/wiki/Quartic_formula#Solving_a_quartic_equation&quot; rel=&quot;nofollow&quot;&gt;Wikipedia's article on quartic functions&lt;/a&gt; has a lengthy process by which to get the solutions, but does not give an explicit formula.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Beware that in the cubic and quartic formulas, depending on how the formula is expressed, the correctness of the answers likely depends on a particular choice of definition of principal roots for nonreal complex numbers and there are two different ways to define such a principal root.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There cannot be explicit algebraic formulas for the general solutions to higher-degree polynomials, but proving this requires mathematics beyond precalculus (it is typically proved with Galois Theory now, though it was originally proved with other methods).  This fact is known as &lt;a href=&quot;http://en.wikipedia.org/wiki/Abel%E2%80%93Ruffini_theorem&quot; rel=&quot;nofollow&quot;&gt;the Abel-Ruffini theorem&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also of note, Wolfram &lt;a href=&quot;http://library.wolfram.com/examples/quintic/&quot; rel=&quot;nofollow&quot;&gt;sells a poster&lt;/a&gt; that discusses the solvability of polynomial equations, focusing particularly on techniques to solve a quintic (5th degree polynomial) equation.  This poster gives explicit formulas for the solutions to quadratic, cubic, and quartic equations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&lt;strong&gt;edit&lt;/em&gt;&lt;/strong&gt;:  I believe that the formula given below gives the correct solutions for x to $ax^4+bx^3+cx^2+d+e=0$ for all complex a, b, c, d, and e, under the assumption that $w=\sqrt{z}$ is the complex number such that $w^2=z$ and $\arg(w)\in(-\frac{\pi}{2},\frac{\pi}{2}]$ and $w=\sqrt[3]{z}$ is the complex number such that $w^3=z$ and $\arg(w)\in(-\frac{\pi}{3},\frac{\pi}{3}]$ (these are typically how computer algebra systems and calculators define the principal roots).  Some intermediate parameters $p_k$ are defined to keep the formula simple and to help in keeping the choices of roots consistent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let: $$\begin{align*}&#xA;p_1&amp;amp;=2c^3-9bcd+27ad^2+27b^2e-72ace&#xA;\\&#xA;p_2&amp;amp;=p_1+\sqrt{-4(c^2-3bd+12ae)^3+p_1^2}&#xA;\\&#xA;p_3&amp;amp;=\frac{c^2-3bd+12ae}{3a\sqrt[3]{\frac{p_2}{2}}}+\frac{\sqrt[3]{\frac{p_2}{2}}}{3a}&#xA;\end{align*}\quad\quad\quad\quad\begin{align*}&#xA;p_4&amp;amp;=\sqrt{\frac{b^2}{4a^2}-\frac{2c}{3a}+p_3}&#xA;\\&#xA;p_5&amp;amp;=\frac{b^2}{2a^2}-\frac{4c}{3a}-p_3&#xA;\\&#xA;p_6&amp;amp;=\frac{-\frac{b^3}{a^3}+\frac{4bc}{a^2}-\frac{8d}{a}}{4p_4}&#xA;\end{align*}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then: $$\begin{align}&#xA;x&amp;amp;=-\frac{b}{4a}-\frac{p_4}{2}-\frac{\sqrt{p_5-p_6}}{2}&#xA;\\&#xA;\mathrm{or\ }x&amp;amp;=-\frac{b}{4a}-\frac{p_4}{2}+\frac{\sqrt{p_5-p_6}}{2}&#xA;\\&#xA;\mathrm{or\ }x&amp;amp;=-\frac{b}{4a}+\frac{p_4}{2}-\frac{\sqrt{p_5+p_6}}{2}&#xA;\\&#xA;\mathrm{or\ }x&amp;amp;=-\frac{b}{4a}+\frac{p_4}{2}+\frac{\sqrt{p_5+p_6}}{2}&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(These came from having Mathematica explicitly solve the quartic, then seeing what common bits could be pulled from the horrifically-messy formula into parameters to make it readable/useable.)&lt;/p&gt;&#xA;" OwnerUserId="72" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-08-05T23:00:44.320" LastActivityDate="2010-08-05T23:00:44.320" CommentCount="4" />
  <row Id="787" PostTypeId="2" ParentId="785" CreationDate="2010-07-27T18:44:52.423" Score="7" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Quartic_formula#The_general_case.2C_along_Ferrari.27s_lines&quot; rel=&quot;nofollow&quot;&gt;Yes&lt;/a&gt;, there is a quartic formula.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is no such solution by radicals for higher degrees. This is a result of Galois theory, and follows from the fact that the symmetric group $S_5$ is not solvable. It is called &lt;a href=&quot;http://en.wikipedia.org/wiki/Insolubility_of_the_quintic&quot; rel=&quot;nofollow&quot;&gt;Abel's theorem&lt;/a&gt;. In fact, there are specific fifth-degree polynomials whose roots cannot be obtained by using radicals from $\mathbb{Q}$.&lt;/p&gt;&#xA;" OwnerUserId="536" LastActivityDate="2010-07-27T18:44:52.423" />
  <row Id="788" PostTypeId="1" AcceptedAnswerId="792" CreationDate="2010-07-27T19:28:44.280" Score="3" ViewCount="408" Body="&lt;p&gt;The question that follows was inspired by &lt;a href=&quot;http://math.stackexchange.com/questions/785/is-there-a-general-formula-for-solving-4th-degree-equations&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When trying to solve for the roots of a polynomial equation, the quadratic formula is much more simple than the cubic formula and the cubic formula is much more simple than the quartic formula. &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;That the general solutions to various polynomial equations are so complex and difficult to derive seems to suggest a fundamental limitation in the problem solving capabilities of the mathematical machinery. Does this intuition of mine make any sense? What should I make of it?&lt;/li&gt;&#xA;&lt;li&gt;Why is it that with each successive degree in a polynomial equation, the solution becomes so much more complex? Can I gain some intuition about what makes finding the roots so hard?&lt;/li&gt;&#xA;&lt;li&gt;According to the &lt;a href=&quot;http://en.wikipedia.org/wiki/Abel%E2%80%93Ruffini_theorem&quot; rel=&quot;nofollow&quot;&gt;Abel-Ruffini theorem&lt;/a&gt;: &quot;there is no general algebraic solution—that is, solution in radicals— to polynomial equations of degree five or higher.&quot; What is so special about the quintic that makes it the cut-off for finding a general algebraic solution?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="19" LastEditorUserId="19" LastEditorDisplayName="" LastEditDate="2010-07-27T19:37:28.737" LastActivityDate="2011-04-20T06:43:36.493" Title="Why is it so hard to find the roots of polynomial equations?" Tags="&lt;intuition&gt;&lt;polynomials&gt;&lt;roots&gt;" AnswerCount="2" CommentCount="5" FavoriteCount="1" />
  <row Id="789" PostTypeId="2" ParentId="707" CreationDate="2010-07-27T20:21:09.660" Score="1" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;First, sticking to strings of length n=2:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$P(10)=p_i (1-p_i)=\frac{i}{m+1}\left (1-\frac{i}{m+1}  \right )=\frac{i(m+1-i)}{(m+1)^2}$ and $P(11)=p_i^2=\left (\frac{i}{m+1}  \right )^2=\frac{i^2}{(m+1)^2}$; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$E(p_i)=\sum_{i=1}^{m}p_iP(p_i)=\sum_{i=1}^{m}\frac{i}{m+1}\left(\frac{P(10)+P(11)}{\sum_{i=1}^{m}(P(10)+P(11))} \right )$ $=\frac{\sum_{i=1}^{m}\frac{i}{m+1}\left(\frac{i(m+1-i)}{(m+1)^2}+\frac{i^2}{(m+1)^2} \right )}{\sum_{i=1}^{m}\left(\frac{i(m+1-i)}{(m+1)^2}+\frac{i^2}{(m+1)^2} \right )}$ $=\frac{\sum_{i=1}^{m}\frac{i}{m+1}\left(i(m+1-i)+i^2\right )}{\sum_{i=1}^{m}\left(i(m+1-i)+i^2\right )}$ $=\frac{\sum_{i=1}^{m}i^2}{(m+1)\sum_{i=1}^{m}i}$ $=\frac{\left(\frac{m(m+1)(2m+1)}{6} \right )}{(m+1)\frac{m(m+1)}{2}}$ $=\frac{2m+1}{3(m+1)}$; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\lim_{m\to\infty}E(p_i)=\lim_{m\to\infty}\frac{2m+1}{3(m+1)}=\frac{2}{3}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;For general n:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$P(1...)=p_i\sum_{k=0}^{n-1}\left({n-1 \choose k}p_i^k(1-p_i)^{n-1-k} \right )=p_i$ (yes, P(1...) does not depend on n at all!);&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$E(p_i)=\sum_{i=1}^{m}p_iP(p_i)=\sum_{i=1}^{m}p_i\left(\frac{P(1...)}{\sum_{i=1}^{m}P(1...)} \right )=\sum_{i=1}^{m}p_i\left(\frac{p_i}{\sum_{i=1}^{m}p_i} \right )$ $=\frac{\sum_{i=1}^{m}p_i^2}{\sum_{i=1}^{m}p_i}=\frac{\sum_{i=1}^{m}\left(\frac{i}{m+1} \right )^2}{\sum_{i=1}^{m}\frac{i}{m+1}}$ $=\frac{\sum_{i=1}^{m}i^2}{(m+1)\sum_{i=1}^{m}i}=\frac{2m+1}{3(m+1)}$ (no surprise there, given that $P(1...)$ doesn't depend on n);&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\lim_{m\to\infty}E(p_i)=\lim_{m\to\infty}\frac{2m+1}{3(m+1)}=\frac{2}{3}$&lt;/p&gt;&#xA;" OwnerUserId="72" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-08-03T02:22:09.930" LastActivityDate="2010-08-03T02:22:09.930" CommentCount="0" />
  <row Id="790" PostTypeId="1" AcceptedAnswerId="824" CreationDate="2010-07-27T20:26:21.083" Score="1" ViewCount="149" Body="&lt;p&gt;For $n &gt; 1$ an integer, there are well-known formulas for volume of the balls.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the analogous statement in a Banach space/Hilbert Space?&lt;/p&gt;&#xA;" OwnerUserId="218" LastEditorUserId="218" LastEditorDisplayName="" LastEditDate="2010-07-28T13:04:01.453" LastActivityDate="2010-08-05T13:28:22.340" Title="Volume of the sphere in a Banach space" Tags="&lt;real-analysis&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="791" PostTypeId="1" AcceptedAnswerId="913" CreationDate="2010-07-27T20:32:42.910" Score="3" ViewCount="105" Body="&lt;p&gt;What is the volume of intersection of the three cylinders with axes of length $1$ in $x, y, z$ directions starting from the origin, and with radius $1$?&lt;/p&gt;&#xA;" OwnerUserId="218" LastActivityDate="2010-07-28T04:29:55.297" Title="How to compute the volume of this object via integration?" Tags="&lt;calculus&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" />
  <row Id="792" PostTypeId="2" ParentId="788" CreationDate="2010-07-27T20:33:23.530" Score="12" ViewCount="" Body="&lt;p&gt;When you try to solve a degree n equation, there are n roots you have to find (in principle) and none of them is favoured over any of the others, which (in some metaphorical sense) means that you have to break an n-fold symmetry in order to write down the roots.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the symmetry group of the n roots becomes more and more complicated the larger n is.&#xA;For n = 2 it is abelian (and very small!); for n = 3 and 4 it is still solvable (in the technical sense of group theory), which explains the existence of an explicit formula involving radicals (this is due to Galois, and is a part of so-called Galois theory); for n = 5 or more this group is non-solvable (in the technical sense of group theory), and this corresponds to the fact that there is no explicit formula involving radicals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Summary: The complexity of the symmetry group of the n roots leads to a corresponding complexity in explicitly solving the equation.&lt;/p&gt;&#xA;" OwnerUserId="221" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2011-04-20T01:57:08.080" LastActivityDate="2011-04-20T01:57:08.080" />
  <row Id="793" PostTypeId="1" AcceptedAnswerId="813" CreationDate="2010-07-27T20:36:51.667" Score="0" ViewCount="156" Body="&lt;p&gt;How to prove that the exponential function and the logarithm function are the inverses of each other? I want it the following way. We must use the definition as power series, and must verify that all the terms of the composition except the coefficent of $z$ vanish, and that the first degree term is $1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can write down the proof for the coefficient of $x^n$ for arbitrary but fixed $n$ by explicit verification. But how to settle this for all $n$ at one go?&lt;/p&gt;&#xA;" OwnerUserId="218" LastActivityDate="2010-07-27T21:14:16.983" Title="Exponential and log functions compose to identity" Tags="&lt;algebra&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="794" PostTypeId="2" ParentId="462" CreationDate="2010-07-27T20:38:58.120" Score="1" ViewCount="" Body="&lt;p&gt;I don't think the people working in higher-order theorem proving really care about Henkin semantics or models in general, they mostly work with their proof calculi. As long as there are no contradictions or other counterintuitive theorems they are happy. The most important and most difficult theorem they prove is usually that their proof terms terminate, which IIRC can be viewed as a form of soundness.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Henkin semantics is most interesting for people trying to extend their first-order methods to higher-order logic, because it behaves essentially like models of first-order logic. Henkin semantics is somewhat weaker than what you would get with standard set-theoretic semantics, which by Gödels incompleteness theorem can't have a complete proof calculus. I think type theories should lie somewhere in between Henkin and standard semantics.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Where does typed lambda calculus come into proof verification?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To prove some implication &lt;code&gt;P(x) --&amp;gt; Q(x)&lt;/code&gt; with some free variables &lt;code&gt;x&lt;/code&gt; you need to map any proof of &lt;code&gt;P(x)&lt;/code&gt; to a proof of &lt;code&gt;Q(x)&lt;/code&gt;. Syntactically a map (a function) can be represented as a lambda term.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Are there any other approaches than higher order logic to proof verification?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You can also verify proofs in first-order or any other logic, but then you would loose much of the power of the logic. First-order logic is mostly interesting because it is possible to automatically find proofs, if they are not too complicated. The same applies even more to propositional logic.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What are the limitations/shortcomings of existing proof verification systems (see below)?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The more powerful the logic becomes the harder it becomes to construct proofs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the systems are freely available I suggest you play with them, e.g. Isabelle and Coq for a start.&lt;/p&gt;&#xA;" OwnerUserId="210" LastEditorUserId="210" LastEditorDisplayName="" LastEditDate="2010-07-28T06:12:18.627" LastActivityDate="2010-07-28T06:12:18.627" CommentCount="2" />
  <row Id="795" PostTypeId="1" AcceptedAnswerId="867" CreationDate="2010-07-27T20:43:00.637" Score="3" ViewCount="120" Body="&lt;p&gt;There is a  theorem of Riemann to that effect. How to prove it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: This was asked by Kenny in the beta for &quot;calculus&quot;.&lt;/p&gt;&#xA;" OwnerUserId="218" LastActivityDate="2010-07-28T00:23:58.393" Title="How to prove that a conditionally convergent series can be rearranged to sum to any real number?" Tags="&lt;calculus&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="796" PostTypeId="1" CreationDate="2010-07-27T20:44:24.130" Score="3" ViewCount="160" Body="&lt;p&gt;What class of Partial Differential Equations can be solved using the method of separation of variables?&lt;/p&gt;&#xA;" OwnerUserId="218" LastActivityDate="2010-07-28T04:27:18.310" Title="Separation of variables for diff eqns" Tags="&lt;calculus&gt;" AnswerCount="1" />
  <row Id="797" PostTypeId="1" AcceptedAnswerId="817" CreationDate="2010-07-27T20:46:30.707" Score="18" ViewCount="454" Body="&lt;p&gt;There are so many available bases. Why is the strange number 'e' preferred over all else?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course one could integrate 1/x and see this. But is there more to the story?&lt;/p&gt;&#xA;" OwnerUserId="218" LastActivityDate="2010-11-09T08:05:09.417" Title="What's so &quot;natural&quot; about the base of natural logarithms?" Tags="&lt;calculus&gt;" AnswerCount="5" FavoriteCount="3" />
  <row Id="798" PostTypeId="1" CreationDate="2010-07-27T20:48:40.110" Score="0" ViewCount="89" Body="&lt;p&gt;There are quite simple, intuitive and straightforward expressions for evaluating the area or volume of a figure. But why is the expression for the length of a curve so complicated?&lt;/p&gt;&#xA;" OwnerUserId="218" LastActivityDate="2010-07-27T21:15:23.240" Title="Why is the integral expression for the length of a curve more complicated than the expression for area or volume of a figure rotated?" Tags="&lt;calculus&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="799" PostTypeId="1" AcceptedAnswerId="821" CreationDate="2010-07-27T20:51:34.517" Score="2" ViewCount="145" Body="&lt;p&gt;Ok, Ok, I know that in fact the discriminant is defined (up to sign) as a product of differences of the roots of the polynomial.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But why does it then have integral coefficients, if the polynomial you started with had integer coefficients?&lt;/p&gt;&#xA;" OwnerUserId="218" LastEditorUserId="218" LastEditorDisplayName="" LastEditDate="2010-07-28T13:25:57.633" LastActivityDate="2010-07-28T13:25:57.633" Title="Why does the discriminant of an integral polynomial have integral coefficients?" Tags="&lt;algebra&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="800" PostTypeId="2" ParentId="797" CreationDate="2010-07-27T20:57:59.073" Score="19" ViewCount="" Body="&lt;p&gt;Differentiation and integration is precisely why it is considered natural, but not just because $\int 1/x dx=ln x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$e^x$ has the two following nice properties&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ d/dx e^x=e^x $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ \int e^x dx=e^x+c $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we looked at a^x instead, we would get:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$d/dx a^x= d/dx e^(ln a*x)=ln a * a^x$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\int a^x dx= integral e^(ln a*x) dx=a^x/(ln x)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So e is vital to the integration and differentiation of exponentials.&lt;/p&gt;&#xA;" OwnerUserId="123" LastEditorUserId="123" LastEditorDisplayName="" LastEditDate="2010-11-09T08:05:09.417" LastActivityDate="2010-11-09T08:05:09.417" />
  <row Id="801" PostTypeId="2" ParentId="797" CreationDate="2010-07-27T20:58:18.043" Score="2" ViewCount="" Body="&lt;p&gt;The wikipedia page on e, &lt;a href=&quot;http://en.wikipedia.org/wiki/E_%28mathematical_constant%29&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/E_%28mathematical_constant%29&lt;/a&gt;, has many properties of the constant that make it naturally occurring.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the biggest reason it is natural when it comes to exponentiation/logarithms is that it is the only number that satisfies d/dt e^t = e^t, while every other number satisfies d/dt a^t = c a^t, where c is some constant (different than 1).  This makes it &quot;normalized&quot; in a sense.&lt;/p&gt;&#xA;" OwnerUserId="9" LastActivityDate="2010-07-27T20:58:18.043" />
  <row Id="802" PostTypeId="2" ParentId="797" CreationDate="2010-07-27T20:58:19.620" Score="1" ViewCount="" Body="&lt;p&gt;A few months ago when I was learning LaTeX, I wrote a quick page to test its math rendering. I never though it'd help anyone, but here it is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://dl.dropbox.com/u/977994/Stack/Logarithmic%20Growth%20Model.pdf&quot; rel=&quot;nofollow&quot;&gt;Logarithmic Growth Model&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first sentence is key.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The logarithmic growth model is defined by its proportionality of rate of growth to population size.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="227" LastActivityDate="2010-07-27T20:58:19.620" CommentCount="2" />
  <row Id="803" PostTypeId="1" CreationDate="2010-07-27T20:58:47.027" Score="14" ViewCount="928" Body="&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Pythagorean_theorem&quot; rel=&quot;nofollow&quot;&gt;Pythagorean Theorem&lt;/a&gt; is one of the most popular to prove by mathematicians, and there are &lt;a href=&quot;http://www.cut-the-knot.org/pythagoras/&quot; rel=&quot;nofollow&quot;&gt;many proofs available&lt;/a&gt; (including one from &lt;a href=&quot;http://en.wikipedia.org/wiki/James_A._Garfield&quot; rel=&quot;nofollow&quot;&gt;James Garfield&lt;/a&gt;).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What's the most elegant proof?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My favorite is this graphical one:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://upload.wikimedia.org/wikipedia/commons/thumb/1/16/Pythagorean_Proof_%283%29.PNG/800px-Pythagorean_Proof_%283%29.PNG&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to cut-the-knot:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Loomis (pp. 49-50) mentions that the&#xA;  proof &quot;was devised by Maurice Laisnez,&#xA;  a high school boy, in the&#xA;  Junior-Senior High School of South&#xA;  Bend, Ind., and sent to me, May 16,&#xA;  1939, by his class teacher, Wilson&#xA;  Thornton.&quot;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The proof has been published by Rufus&#xA;  Isaac in Mathematics Magazine, Vol. 48&#xA;  (1975), p. 198.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="225" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-07-31T17:47:53.663" LastActivityDate="2010-11-18T18:51:15.880" CommunityOwnedDate="2010-07-27T21:20:09.110" Title="What is the most elegant proof of the Pythagorean theorem?" Tags="&lt;geometry&gt;&lt;big-list&gt;&lt;algebra-precalculus&gt;&lt;euclidean-geometry&gt;" AnswerCount="7" CommentCount="5" FavoriteCount="6" />
  <row Id="804" PostTypeId="1" CreationDate="2010-07-27T21:00:07.103" Score="2" ViewCount="274" Body="&lt;p&gt;I have seen and studied this class equation for a finite group acting on itself by conjugations. The only applications I know are Cauchys' theorm and Sylow's theorem. Are there more?&lt;/p&gt;&#xA;" OwnerUserId="218" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-07-31T17:46:59.970" LastActivityDate="2010-08-04T00:44:09.747" Title="Applications for the class equation from group theory" Tags="&lt;big-list&gt;&lt;abstract-algebra&gt;" AnswerCount="4" FavoriteCount="1" />
  <row Id="805" PostTypeId="2" ParentId="797" CreationDate="2010-07-27T21:00:42.513" Score="9" ViewCount="" Body="&lt;p&gt;The wikipedia article on &lt;a href=&quot;http://bit.ly/FtPFd&quot; rel=&quot;nofollow&quot;&gt;e&lt;/a&gt; tells a bit of the story.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;One example is an account that starts with 1.00 and pays 100% interest per year. If the interest is credited once, at the end of the year, the value is 2.00; but if the interest is computed and added twice in the year, the 1 is multiplied by 1.5 twice, yielding 1.00×1.5² = $2.25. Compounding quarterly yields 1.00×1.25&lt;sup&gt;4&lt;/sup&gt; = 2.4414…, and compounding monthly yields 1.00×(1.0833…)&lt;sup&gt;12&lt;/sup&gt; = 2.613035….&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Bernoulli noticed that this sequence approaches a limit (the force of interest) for more and smaller compounding intervals. Compounding weekly yields 2.692597…, while compounding daily yields 2.714567…, just two cents more. Using n as the number of compounding intervals, with interest of 100%/n in each interval, the limit for large n is the number that came to be known as e; with continuous compounding, the account value will reach 2.7182818…. More generally, an account that starts at $1, and yields (1+R) dollars at simple interest, will yield e&lt;sup&gt;R&lt;/sup&gt; dollars with continuous compounding.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Additionally, it is the base of the exponential function &lt;code&gt;y = k^x&lt;/code&gt;, finding a specific value for &lt;code&gt;k&lt;/code&gt; where &lt;code&gt;d/dx k^x = k^x&lt;/code&gt;.  That is, the rate of change of the exponential curve at any point is equal to the y value of the curve at that point.&lt;/p&gt;&#xA;" OwnerUserId="51" LastActivityDate="2010-07-27T21:00:42.513" />
  <row Id="806" PostTypeId="1" AcceptedAnswerId="933" CreationDate="2010-07-27T21:01:27.537" Score="5" ViewCount="291" Body="&lt;p&gt;There is the notion of class number from algebraic number theory. Why is such a notion defined and what good comes out of it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is nice if it is $1$; we have unique factorization of all ideals; but otherwise?&lt;/p&gt;&#xA;" OwnerUserId="218" LastEditorUserId="218" LastEditorDisplayName="" LastEditDate="2010-07-28T15:26:24.330" LastActivityDate="2010-08-04T00:19:10.063" Title="Applications of class number" Tags="&lt;algebraic-number-theory&gt;" AnswerCount="5" CommentCount="2" FavoriteCount="4" />
  <row Id="807" PostTypeId="2" ParentId="799" CreationDate="2010-07-27T21:02:54.693" Score="5" ViewCount="" Body="&lt;p&gt;The square of the discriminant is &lt;em&gt;symmetric&lt;/em&gt; in the roots of the polynomial; if you permute the roots, it stays the same.  By the &lt;a href=&quot;http://mathworld.wolfram.com/FundamentalTheoremofSymmetricFunctions.html&quot; rel=&quot;nofollow&quot;&gt;fundamental theorem of symmetric functions&lt;/a&gt;, that means it can be expressed as a polynomial, with integer coefficients, in the coefficients of the original polynomial.  (I am using both &quot;coefficients&quot; and &quot;polynomial&quot; in two senses here, so let me know if this doesn't make sense.)  This is a basic observation which will become important if you ever study Galois theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An example will probably make this clearer.  Suppose I have a quadratic polynomial $x^2 + bx + c$ with two roots $r_1, r_2$.  Then $x^2 + bx + c = (x - r_1)(x - r_2)$, so $b = - r_1 - r_2$ and $c = r_1 r_2$.  These are the &lt;em&gt;elementary symmetric functions&lt;/em&gt; in two variables, and the theorem above implies that every polynomial function of $r_1$ and $r_2$ which is invariant under switching the two is actually a polynomial in $b$ and $c$.  For example, the discriminant is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(r_1 - r_2)^2 = r_1^2 - 2r_1 r_2 + r_2^2 = (r_1 + r_2)^2 - 4r_1 r_2 = b^2 - 4c.$$&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-07-27T21:17:44.847" LastActivityDate="2010-07-27T21:17:44.847" CommentCount="3" />
  <row Id="808" PostTypeId="2" ParentId="795" CreationDate="2010-07-27T21:04:59.073" Score="7" ViewCount="" Body="&lt;p&gt;Let the number be r.  The idea is that you first add all the positive elements of the sequence in order, until you get over r, then you add all the negative elements until you get under r, and then all the positive until you get over r, and so forth.&lt;/p&gt;&#xA;" OwnerUserId="9" LastActivityDate="2010-07-27T21:04:59.073" />
  <row Id="809" PostTypeId="2" ParentId="798" CreationDate="2010-07-27T21:05:07.433" Score="0" ViewCount="" Body="&lt;p&gt;Obviously this depends on the shape, as you implied. I would say for a circle the circumference formula is simpler. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also the length of the curve &lt;strong&gt;is a volume&lt;/strong&gt; in some sense, its just 1 dimensional. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Usually in calculus we are calculating the area relative to the the x-axis, and your formula is in terms of x, so that makes things simpler. I imagine that if the curve was parameterized in terms of its arc length finding its length would be quite trivial, and finding the area between the curve and the x-axis would seem very complicated.&lt;/p&gt;&#xA;" OwnerUserId="99" LastEditorUserId="99" LastEditorDisplayName="" LastEditDate="2010-07-27T21:15:23.240" LastActivityDate="2010-07-27T21:15:23.240" />
  <row Id="811" PostTypeId="2" ParentId="803" CreationDate="2010-07-27T21:06:36.277" Score="6" ViewCount="" Body="&lt;p&gt;The proof that uses the fact that shearing a parallelogram parallel to one of its sides preserves area is my favorite.  Here's an animation from &lt;a href=&quot;http://www.arseweb.com/rupe/pythag/proof.html&quot; rel=&quot;nofollow&quot;&gt;this site&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/W9r8D.gif&quot; alt=&quot;animated proof&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="72" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-09-29T04:17:50.523" LastActivityDate="2010-09-29T04:17:50.523" CommunityOwnedDate="2010-09-30T10:29:54.723" CommentCount="1" />
  <row Id="812" PostTypeId="2" ParentId="798" CreationDate="2010-07-27T21:08:31.293" Score="3" ViewCount="" Body="&lt;p&gt;The length of a curve doesn't transform in a nice way under scaling of one of the variables the same way that the area of a 2d figure or the volume of a 3d figure does.  That puts a hard lower bound on how complicated it has to be.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-07-27T21:08:31.293" />
  <row Id="813" PostTypeId="2" ParentId="793" CreationDate="2010-07-27T21:14:16.983" Score="3" ViewCount="" Body="&lt;p&gt;There is a combinatorial proof.  If $A(x) = \sum_{n \ge 0} a_n \frac{x^n}{n!}$ is an exponential generating function for the number of ways $a_n$ to put a certain structure on a set of size $n$, where $a_0 = 0$, then $\exp A(x) = \sum_{n \ge 0} b_n \frac{x^n}{n!}$ has coefficients $b_n$ which count the number of ways to split a set of size $n$ into subsets, each of which is given an $A$-structure.  For example, $A(x) = e^x - 1$ is the structure of &quot;being a nonempty set,&quot; and $\exp A(x)$ counts the number of ways to partition a set into nonempty subsets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now $\log \frac{1}{1 - x} = \sum_{n \ge 1} \frac{x^n}{n}$ can be thought of as the structure of &quot;being a cycle,&quot; which is to say there are $(n-1)!$ ways to arrange $n$ objects into a cycle $s_1 \to s_2 \to ... \to s_n \to s_1$.  So $\exp \log \frac{1}{1 - x}$ counts the number of ways to split $n$ objects up into disjoint cycles, which is the same as the number of permutations of $n$ by cycle decomposition.  So&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\exp \log \frac{1}{1 - x} = \frac{1}{1 - x}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A generalization of this argument lets you compute the &lt;a href=&quot;http://qchu.wordpress.com/2009/06/24/gila-vi-the-cycle-index-polynomials-of-the-symmetric-groups/&quot; rel=&quot;nofollow&quot;&gt;cycle index polynomials of the symmetric groups&lt;/a&gt;.  Note that all of the above manipulations take place in the ring of formal power series over $\mathbb{Z}$.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-07-27T21:14:16.983" CommentCount="1" />
  <row Id="814" PostTypeId="2" ParentId="804" CreationDate="2010-07-27T21:18:22.197" Score="6" ViewCount="" Body="&lt;p&gt;A nice application is Wedderburn's theorem: every finite skewfield is necessarily commutative. Here a skewfield is something which satisfies the same axioms as field, except that multiplication is not required to be commutative; the typical example are quaternions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To see this, let $F$ be a finite skewfield, $Z$ be its center. It is easy to see that $Z$ is a field, hence it must be $\mathbb{F}_q$ for some prime power $q = p^k$. $F$ will then be a vector space of finite dimension $n$ over $\mathbb{F}_q$, hence it will have $q^n$ elements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now write the class equation for the multiplicative group of $F$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$q^n - 1 = q - 1 + \sum_i \frac{q^n - 1}{q^{t_i} - 1}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here $q - 1$ appears as the cardinality of the center, while the sum extends over a set of representatives of the non-trivial conjugacy classes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that for $q^{t_i} - 1$ to divide $q^n - 1$, $t_i$ must divide $n$. Indeed the order of $q$ modulo $q^{t_i} - 1$ is $t_i$, and $q^n = 1 \pmod{q^{t_i} - 1}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let $f_n$ be the $n$-th cyclotomic polynomial. Then $f_n(q)$ divides $q^n - 1$ and it also divides each term in the sum, so $f_n(q)$ divides $q - 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But this is impossible unless $n = 1$ and the sum is empty, in which case $F$ is commutative. Indeed $f_n(q)$ is a product of terms of the form $q - \omega$, where $\omega$ is a root of unity, and this product will have bigger absolute value than $q - 1$ as soon as $n &gt; 1$.&lt;/p&gt;&#xA;" OwnerUserId="233" LastEditorUserId="233" LastEditorDisplayName="" LastEditDate="2010-08-03T12:49:13.113" LastActivityDate="2010-08-03T12:49:13.113" CommentCount="3" />
  <row Id="815" PostTypeId="1" CreationDate="2010-07-27T21:18:45.677" Score="4" ViewCount="513" Body="&lt;p&gt;This question is lifted from &lt;a href=&quot;http://mathoverflow.net/questions/5499/&quot; rel=&quot;nofollow&quot;&gt;Mathoverflow.&lt;/a&gt;. I feel it belongs here too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are mathematicians whose creativity, insight and taste have the power of driving anyone into a world of beautiful ideas, which can inspire the desire, even the need for doing mathematics, or can make one to confront some kind of problems, dedicate his life to a branch of math, or choose an specific research topic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think that this kind of force must not be underestimated; on the contrary, we have the duty to take advantage of it in order to improve the mathematical education of those who may come after us, using the work of those gifted mathematicians (and even their own words) to inspire them as they inspired ourselves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, I'm interested on knowing who (the mathematician), when (in which moment of your career), where (which specific work) and why this person had an impact on your way of looking at math. Like this, we will have an statistic about which mathematicians are more akin to appeal to our students at any moment of their development. Please, keep one mathematician for post, so that votes are really representative.&lt;/p&gt;&#xA;" OwnerUserId="218" LastActivityDate="2011-01-07T14:48:06.590" CommunityOwnedDate="2010-07-27T21:24:12.777" Title="Which mathematicians have influenced you the most?" Tags="&lt;soft-question&gt;&lt;big-list&gt;" AnswerCount="7" CommentCount="2" FavoriteCount="1" />
  <row Id="816" PostTypeId="2" ParentId="740" CreationDate="2010-07-27T21:24:11.607" Score="2" ViewCount="" Body="&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Cauchy_distribution&quot; rel=&quot;nofollow&quot;&gt;Cauchy distribution&lt;/a&gt; is another example.  This distribution pops up frequently in statistical reasoning.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, the ratio of two independent standard normal random variables is a standard Cauchy variable.  We calculate ratios all the time, but often forget to consider that their distribution does not follow a simple normal distribution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is interesting for several reasons:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The mean and variance for the Cauchy distribution are undefined (--&gt; infinity).  If one is trying to estimate these parameters for the ratio of two normal variables, the results may blow up to rather large values that are hard to interpret.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;In such a situation, one would therefore prefer to use more robust estimators of the central tendency (such as the median) and scale (median absolute deviation).  When designing and testing robust estimators (en.wikipedia.org/wiki/Robust_statistics), one way to test them is to try them out on a Cauchy distribution, and make sure that they don't blow up like the usual formulas for mean and variance.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" LastActivityDate="2010-07-27T21:24:11.607" CommunityOwnedDate="2010-07-27T21:24:11.607" CommentCount="1" />
  <row Id="817" PostTypeId="2" ParentId="797" CreationDate="2010-07-27T21:28:11.640" Score="10" ViewCount="" Body="&lt;p&gt;If you know some linear algebra, then here is an abstract reason: $e^x$ is the unique eigenvector of eigenvalue $1$ of the derivative $D$ acting on, say, the space of smooth functions on $\mathbb{R}$.  Why is this important?  The study of solutions of linear differential equations with constant coefficients is equivalent to the study of nullspaces of operators which are polynomials in $D$, e.g. operators of the form $\sum a_k D^k$.  Any such operator automatically commutes with $D$, so this nullspace splits up into eigenspaces of $D$.  That's why solutions to linear differential equations with constant coefficients can be expressed as sums of complex exponentials.  The choice of $e$ makes it particularly easy to see what the eigenvalue is: the eigenvalue of the eigenvector $e^{\lambda x}$ is $\lambda$.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-07-27T21:28:11.640" CommentCount="7" />
  <row Id="818" PostTypeId="2" ParentId="806" CreationDate="2010-07-27T21:33:35.587" Score="6" ViewCount="" Body="&lt;p&gt;We say that a prime p is regular if it does not divide the class number of the p-th cyclotomic field. For regular primes, it is easy to prove Fermat's last theorem, as outlined for instance in &lt;a href=&quot;http://www.jmilne.org/math/CourseNotes/ant.html&quot; rel=&quot;nofollow&quot;&gt;Milne's notes&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically, everything would be easy if the class number was 1, in which case one could use unique factorization. If the class number is prime to p, you use the fact that every ideal whose p-th power is principal must itself be principal. This, together with unique factorization for ideals, turns out to be all that you need in the naif proof.&lt;/p&gt;&#xA;" OwnerUserId="233" LastActivityDate="2010-07-27T21:33:35.587" />
  <row Id="819" PostTypeId="2" ParentId="803" CreationDate="2010-07-27T21:35:14.013" Score="4" ViewCount="" Body="&lt;p&gt;This is not really a proof, in the sense that it depends on certain physical assumptions, but it forces you to think very hard about what those physical assumptions are.  I learned it from Mark Levi's &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0691140200&quot; rel=&quot;nofollow&quot;&gt;The Mathematical Mechanic&lt;/a&gt;, although unfortunately I don't have a diagram.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider a fish tank in the shape of a triangular prism, where the triangle on top has vertices $A, B, C$ ($C$ is the right angle) and the triangle on the bottom has opposite vertices $A', B', C'$.  Let $a, b, c$ denote the lengths of the opposite sides.  Run a rod through the edge $A A'$ on which the fish tank is allowed to pivot, and fill it with water.  By conservation of energy, this system is in mechanical equilibrium.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, I claim (and it is not hard to see) that the total torque about the rod is proportional to $c^2 - b^2 - a^2$, each contribution coming from the pressure of the water in the tank against the corresponding side.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-07-27T21:35:14.013" CommunityOwnedDate="2010-07-27T21:35:14.013" CommentCount="2" />
  <row Id="820" PostTypeId="2" ParentId="640" CreationDate="2010-07-27T21:48:44.120" Score="17" ViewCount="" Body="&lt;p&gt;As Akhil mentions, the keyword is elliptic regularity.  Since I don't know anything about this, let me just say some low-level things and maybe they'll make sense to you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A differentiable function $f : \mathbb{R} \to \mathbb{R}$ can be thought of as a function which behaves locally like a linear function $f(x) = ax + b$.  So, very roughly, it is a collection of tiny vectors which fit together.  These tiny vectors can, however, fit together in a very erratic manner.  That's because since you only have to fit one vector to the two vectors that are its neighbors, there is a lot of room for bad behavior.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A differentiable function $f : \mathbb{C} \to \mathbb{C}$ has to satisfy a much more stringent requirement: locally, it has to behave like a linear function $f(z) = az + b$ where $z, a, b$ are complex, which is a &lt;strong&gt;rotation&lt;/strong&gt; (and scale, and translation).  So, very roughly, it is a collection of tiny rotations which fit together.  Now one rotation has a continuum of neighbors to worry about, and it becomes much harder for erratic behavior to persist.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-07-27T21:48:44.120" CommentCount="1" />
  <row Id="821" PostTypeId="2" ParentId="799" CreationDate="2010-07-27T21:53:18.830" Score="3" ViewCount="" Body="&lt;p&gt;Another definition of discriminant is as the resultant of the polynomial with it's first derivative (up to a scalar), and the resultant of two polynomials vanishes if and only if they have a common root.  So when does a polynomial have a common root with it's derivative? That's when a zero is also a local max or min, precisely a multiple root.&lt;/p&gt;&#xA;" OwnerUserId="176" LastActivityDate="2010-07-27T21:53:18.830" />
  <row Id="822" PostTypeId="2" ParentId="785" CreationDate="2010-07-27T21:58:48.933" Score="1" ViewCount="" Body="&lt;p&gt;Regarding the inability to solve the quintic, this is sort-of true and sort-of false.  No, there is no general solution in terms of $+$, $-$, $\times$ and $\div$, along with $\sqrt[n]{}$.  However, if you allow special theta values (a new operation, not among the standard ones!) then yes, you can actually write down the solutions of arbitrary polynomials this way.  Also, you can do construct lengths equal to the solutions by intersecting lower degree curves (for a quintic, you can do so with a &lt;a href=&quot;http://mathworld.wolfram.com/TridentofDescartes.html&quot; rel=&quot;nofollow&quot;&gt;trident&lt;/a&gt; and a circle.)&lt;/p&gt;&#xA;" OwnerUserId="176" LastActivityDate="2010-07-27T21:58:48.933" CommentCount="1" />
  <row Id="823" PostTypeId="2" ParentId="640" CreationDate="2010-07-27T22:00:11.930" Score="14" ViewCount="" Body="&lt;p&gt;The existence of a complex derivative means that locally a function can only rotate and expand.  That is, in the limit, disks are mapped to disks.  This rigidity is what makes a complex differentiable function infinitely differentiable, and even more, analytic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a complex derivative to exist, it has to exist and have the same value for all ways the &quot;h&quot; term can go to zero in (f(z+h) - f(z))/h.  In particular, h could approach 0 along any radial path, and that's why an analytic function must map disks to disks in the limits.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;By contrast, an infinitely differentiable function of two real variables could map a disk to an ellipse, stretching more in one direction than another.  An analytic function can't do that.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A smooth function of two variables could also flip a disk over, such as f(x, y) = (x, -y).  An analytic function can't do that either.  That's why complex conjugation is not an analytic function.  &lt;/p&gt;&#xA;" OwnerUserId="247" LastEditorUserId="247" LastEditorDisplayName="" LastEditDate="2010-07-28T17:09:07.587" LastActivityDate="2010-07-28T17:09:07.587" />
  <row Id="824" PostTypeId="2" ParentId="790" CreationDate="2010-07-27T22:02:00.617" Score="2" ViewCount="" Body="&lt;p&gt;Maybe this is helpful, a formula for the &lt;a href=&quot;http://www.johndcook.com/blog/2010/07/02/volumes-of-generalized-unit-balls/&quot; rel=&quot;nofollow&quot;&gt;volume of Lp balls in R^n&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="247" LastActivityDate="2010-07-27T22:02:00.617" CommentCount="2" />
  <row Id="825" PostTypeId="2" ParentId="600" CreationDate="2010-07-27T22:02:50.807" Score="4" ViewCount="" Body="&lt;p&gt;I wrote a series of blog posts which explains how to solve questions like this; the relevant one is &lt;a href=&quot;http://qchu.wordpress.com/2009/06/21/gila-v-the-polya-enumeration-theorem-and-applications/&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.  The generating function you want is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{1}{n} \sum_{d | n} (x_1^{n/d} + ... + x_k^{n/d})^d \varphi \left( \frac{n}{d} \right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where the coefficient of $x_1^{r_1} ... x_k^{r_k}$ is the number you want.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-07-27T22:02:50.807" />
  <row Id="826" PostTypeId="2" ParentId="760" CreationDate="2010-07-27T22:04:12.723" Score="3" ViewCount="" Body="&lt;p&gt;Here is an &lt;a href=&quot;http://www.johndcook.com/lat_long_distance.html&quot; rel=&quot;nofollow&quot;&gt;online calculator&lt;/a&gt; for great circle distance and here is a &lt;a href=&quot;http://www.johndcook.com/lat_long_details.html&quot; rel=&quot;nofollow&quot;&gt;detailed explanation&lt;/a&gt; of the calculations behind it.&lt;/p&gt;&#xA;" OwnerUserId="247" LastActivityDate="2010-07-27T22:04:12.723" />
  <row Id="827" PostTypeId="2" ParentId="815" CreationDate="2010-07-27T22:05:59.197" Score="1" ViewCount="" Body="&lt;h2&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Isaac_Newton&quot; rel=&quot;nofollow&quot;&gt;Isaac Newton&lt;/a&gt;&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;I'm not going to give a complete biography, but for those who don't know...&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;He single-handedly invented The Calculus. Of course while at the same time, Leibniz &quot;single-handedly&quot; invented it too. :)&lt;/li&gt;&#xA;&lt;li&gt;He theorized the connection between an object's tendency to fall to the earth and the motion of the heavenly bodies.&lt;/li&gt;&#xA;&lt;li&gt;Studied the laws of motion and developed formulas we commonly use today&lt;/li&gt;&#xA;&lt;li&gt;Investigated light refraction&lt;/li&gt;&#xA;&lt;li&gt;Calculated the speed of sound to less than 1% the experimented value&lt;/li&gt;&#xA;&lt;li&gt;He used the &quot;dot notation&quot; to signify time derivatives, which I prefer far over prime notation. :)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="227" LastActivityDate="2010-07-27T22:05:59.197" CommunityOwnedDate="2010-07-27T22:05:59.197" CommentCount="4" />
  <row Id="828" PostTypeId="2" ParentId="535" CreationDate="2010-07-27T22:06:16.293" Score="1" ViewCount="" Body="&lt;p&gt;Well, you're looking at $p_c(x) = 4x^4 + 8x^3 - 3x^2 - 9x + c=0$, which is just $c=-4x^4-8x^3+3x^2+9x$, so if you graph in the $xc$-plane, you'll get a quartic that opens down with the horizontal cross sections the zeros for given $c$.  In fact, because we can solve for $c$ like this to make it a graph in the affine plane, it is a rational quartic curve, so you might want to start looking there.&lt;/p&gt;&#xA;" OwnerUserId="176" LastActivityDate="2010-07-27T22:06:16.293" CommentCount="2" />
  <row Id="829" PostTypeId="2" ParentId="756" CreationDate="2010-07-27T22:13:54.087" Score="0" ViewCount="" Body="&lt;p&gt;Falagar's answer doesn't seem to be for this question. In this question we only can ask higher/lower, not questions about arbitrary bits. I would think only checking if he lied after a certain number of questions would be best. When playing the game, if he lied at one point, you'd start to get all the answers being either higher or lower (because you're searching in the wrong area) So the point is to maximize the number of times he must change his answer to be able to detect his lie more easily. So I think it would skew your guesses to either the upper or lower end of range. [Sorry for this long post, I can't comment yet and I didn't want to just say people aren't reading the question correctly]&lt;/p&gt;&#xA;" LastActivityDate="2010-07-27T22:13:54.087" />
  <row Id="830" PostTypeId="2" ParentId="575" CreationDate="2010-07-27T22:17:36.657" Score="2" ViewCount="" Body="&lt;p&gt;Answer edited, in response to the comment and a second wind for explaining mathematics:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $F(x_0,x_1,x_2)=0$ be the equation for your curve, and take $(y_0,y_1,y_2)$ to be coordinates on $(\mathbb{P}^2)^*$.  Also, assume that $F$ is irreducible and has no linear factors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then $y_0 x_0+y_1 x_1+y_2 x_2=0$ is the equation of a general line in $\mathbb{P}^2$ (recall, here $y_0,y_1,y_2$ are fixed, and the $x_i$ are the coordinates on the plane) and we look at the open set of $(\mathbb{P}^2)^*$ where $y_2\neq 0$.  On this open set, we can solve the equation of the line for $x_2$, and look at $g(x_0,x_1)=y_2^n F(x_0,x_1,-\frac{1}{y_2}(y_0x_0+y_1x_1))$, a homogeneous polynomial of degree $n$ in $x_0,x_1$ with coefficients homogeneous polynomials in the $y_i$.  This polynomial has zeros the intersections of our curve $C$ with the line $L$ we're looking at.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we want to find points of multiplicity at least two.  So how do we find multiple roots of a polynomial? We take the discriminant! Specifically, we do it for an affinization, and we get a homogeneous polynomial of degree $2n^2-n$ in the $y_i$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All that's left is to factor the polynomial, and kill all the linear factors, just throw them away, the reasons are explained in more computational detail on my &lt;a href=&quot;http://rigtriv.wordpress.com/2008/08/04/dual-curves/&quot; rel=&quot;nofollow&quot;&gt;blogpost&lt;/a&gt;, and there I also do explicit examples, but the method for calculating the equation of the dual curve is as above.&lt;/p&gt;&#xA;" OwnerUserId="176" LastEditorUserId="176" LastEditorDisplayName="" LastEditDate="2010-07-28T01:37:45.227" LastActivityDate="2010-07-28T01:37:45.227" CommentCount="1" />
  <row Id="831" PostTypeId="2" ParentId="815" CreationDate="2010-07-27T22:19:05.907" Score="1" ViewCount="" Body="&lt;h2&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Beno%C3%AEt_Mandelbrot&quot; rel=&quot;nofollow&quot;&gt;Benoît Mandelbrot&lt;/a&gt;&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Although he provided many valuable contributions to the field, I am most in love with his work on Fractals. I find math to be quite beautiful, and the Mandelbrot Set (magnified portion shown below) is a perfect example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://upload.wikimedia.org/wikipedia/commons/3/37/Mandel_zoom_14_satellite_julia_island.jpg&quot; rel=&quot;nofollow&quot;&gt;&lt;img src=&quot;http://upload.wikimedia.org/wikipedia/commons/thumb/3/37/Mandel_zoom_14_satellite_julia_island.jpg/800px-Mandel_zoom_14_satellite_julia_island.jpg&quot; alt=&quot;mandelbrot set zoom 14: julia island&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="59" LastEditorUserId="59" LastEditorDisplayName="" LastEditDate="2011-01-06T16:26:20.040" LastActivityDate="2011-01-06T16:26:20.040" CommunityOwnedDate="2010-07-27T22:19:05.907" CommentCount="5" />
  <row Id="832" PostTypeId="2" ParentId="540" CreationDate="2010-07-27T22:22:26.763" Score="15" ViewCount="" Body="&lt;p&gt;Suppose you want to put a probability distribution on the natural numbers for the purpose of doing number theory.  What properties might you want such a distribution to have?  Well, if you're doing number theory then you want to think of the prime numbers as acting &quot;independently&quot;: knowing that a number is divisible by $p$ should give you no information about whether it's divisible by $q$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That quickly leads you to the following realization: you should choose the exponent of each prime in the prime factorization independently.  So how should you choose these?  It turns out that the probability distribution on the non-negative integers with maximum entropy and a given mean is a geometric distribution, as explained for example by Keith Conrad &lt;a href=&quot;https://docs.google.com/viewer?url=http://www.math.uconn.edu/~kconrad/blurbs/analysis/entropypost.pdf&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.  So let's take the probability that the exponent of $p$ is $k$ to be equal to $(1 - r_p) r_p^k$ for some constant $r_p$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This gives the probability that a positive integer $n = p_1^{e_1} ... p_k^{e_k}$ occurs as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle C \prod_{i=1}^{k} r_p^{e_i}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $C = \prod_p (1 - r_p)$.  So we need to choose $r_p$ such that this product converges.  Now, we'd like the probability that $n$ occurs to be monotonically decreasing as a function of $n$.  It turns out (and this is a nice exercise) that this is true if and only if $r_p = p^{-s}$ for some $s &gt; 1$ (since $C$ has to converge), which gives the probability that $n$ occurs as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{ \frac{1}{n^s} }{ \zeta(s)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\zeta(s)$ is the zeta function.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One way of thinking about this argument is that $\zeta(s)$ is the partition function of a statistical-mechanical system called the &lt;a href=&quot;http://en.wikipedia.org/wiki/Primon_gas&quot; rel=&quot;nofollow&quot;&gt;Riemann gas&lt;/a&gt;.  As $s$ gets closer to $1$, the temperature of this system increases until it would require infinite energy to make $s$ equal to $1$.  But this limit is extremely important to understand: it is the limit in which the probability distribution above gets closer and closer to uniform.  So it's not surprising that you can deduce statistical information about the primes by studying the behavior as $s \to 1$ of this distribution.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Let me mention two other reasons to care about the limit as $s \to 1$ of the above distribution.  First, the basic reason to think of the primes as acting independently is the Chinese Remainder Theorem.  Second, a natural reason to look at a distribution where the probability that a number is divisible by $k$ is $(1 - p^{-1}) p^{-k}$ is that this is precisely the distribution you get on the residues $\bmod p^n$ for $k &amp;lt; n$.  In fact, I believe this can be upgraded to the corresponding statement about Haar measure on the $p$-adic integers.  &lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-08-01T21:26:25.423" LastActivityDate="2010-08-01T21:26:25.423" CommentCount="4" />
  <row Id="833" PostTypeId="2" ParentId="519" CreationDate="2010-07-27T22:30:35.367" Score="0" ViewCount="" Body="&lt;p&gt;Your group is the alternating subgroup of a Coxeter group&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\langle x, y, z | x^2 = y^2 = z^2 = (xy)^2 = (yz)^3 = (zx)^7 = 1 \rangle$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is not on the (well-understood) list of &lt;a href=&quot;http://en.wikipedia.org/wiki/Coxeter_group#Finite_Coxeter_groups&quot; rel=&quot;nofollow&quot;&gt;finite Coxeter groups&lt;/a&gt;, and the alternating subgroup always has index $2$.  (The connection to the hyperbolic plane is that Coxeter groups of rank $3$ always act as symmetries of a tiling of either the sphere, the Euclidean plane, or the hyperbolic plane by triangles.)&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-07-27T22:30:35.367" />
  <row Id="834" PostTypeId="2" ParentId="815" CreationDate="2010-07-27T22:51:14.237" Score="9" ViewCount="" Body="&lt;h1&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Leonhard_Euler&quot; rel=&quot;nofollow&quot;&gt;Leonhard Euler&lt;/a&gt;&lt;/h1&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;He made important discoveries in pretty much &lt;em&gt;every&lt;/em&gt; mathematical field there was at his time.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;He invented graph theory.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;He is responsible for much of the current mathematical notation we use today, including Σ, &lt;em&gt;i&lt;/em&gt;, &lt;em&gt;e&lt;/em&gt;, f(x), π, and sin/cos.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/List_of_topics_named_after_Leonhard_Euler&quot; rel=&quot;nofollow&quot;&gt;&lt;strong&gt;EVERYTHING&lt;/strong&gt; is named after him&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;His &lt;a href=&quot;http://www.math.dartmouth.edu/~euler/tour/tour_08.html&quot; rel=&quot;nofollow&quot;&gt;combined works&lt;/a&gt; fill 80 (!) volumes.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;And last but not least,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://upload.wikimedia.org/math/c/6/6/c669a6c5e0faf3a8ba0befed0f517ae5.png&quot; alt=&quot;(Euler's formula)&quot;&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="136" LastActivityDate="2010-07-27T22:51:14.237" CommunityOwnedDate="2010-07-27T22:51:14.237" />
  <row Id="835" PostTypeId="1" CreationDate="2010-07-27T22:55:17.460" Score="2" ViewCount="386" Body="&lt;p&gt;When I have watched Deal or No Deal (I try not to make a habit of it) I always do little sums in my head to work out if the banker is offering a good deal. Where odds drop below &quot;evens&quot; it's easy to see it's a bad deal, but what would be the correct mathematical way to decide if you're getting a good deal?&lt;/p&gt;&#xA;" OwnerUserId="270" LastActivityDate="2011-05-02T05:21:03.797" Title="Optimal Strategy for Deal or No Deal" Tags="&lt;probability&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="1" />
  <row Id="836" PostTypeId="2" ParentId="806" CreationDate="2010-07-27T23:01:14.113" Score="8" ViewCount="" Body="&lt;p&gt;Well, class field theory states that the class number is the degree of the largest everywhere-unramified abelian extension of a number field (namely, the Hilbert class field).  But class field theory really says a lot more: it says that there's an isomorphism between the Galois group and the ideal class group. And in general, for any abelian extension $L/K$, there's an isomorphism between $G(L/K)$ and a certain &quot;generalized ideal class group&quot; of $K$ where you quotient by ideals that are norms from $L$.  This can be stated somewhat more elegantly using ideles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, you asked about the class number. In class field theory, it's difficult (to my knowledge) to prove directly that the Artin reciprocity map defined from ideals to the Galois group is an isomorphism. You can show that it is surjective and injective, however; to do this, you have to estimate the order of these generalized ideal class groups, which you can do using either analytic (L-function) methods or algebraic (e.g., using the Herbrand quotient) methods. So this is where the size and seemingly &quot;decategorified&quot; properties (like the order) become more important than the groups themselves: the proofs.&lt;/p&gt;&#xA;" OwnerUserId="536" LastEditorUserId="536" LastEditorDisplayName="" LastEditDate="2010-07-28T14:54:32.163" LastActivityDate="2010-07-28T14:54:32.163" CommentCount="6" />
  <row Id="837" PostTypeId="2" ParentId="250" CreationDate="2010-07-27T23:01:37.857" Score="4" ViewCount="" Body="&lt;p&gt;The rational numbers are both a continuum (between any two rationals you can find another rational) and countable (they can be lined up in correspondence with the positive integers).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mathematician missed it for hundreds (thousands?) of years, until Cantor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, the proof of that works both ways, and is equally surprising the other way -- there is a way to order the integers (or any countable set) that makes it into a continuum.&lt;/p&gt;&#xA;" OwnerUserId="271" LastActivityDate="2010-07-27T23:01:37.857" CommunityOwnedDate="2010-07-27T23:01:37.857" CommentCount="4" />
  <row Id="838" PostTypeId="2" ParentId="640" CreationDate="2010-07-27T23:09:13.553" Score="2" ViewCount="" Body="&lt;p&gt;On an intuition level:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Complex numbers are somewhat equivalent to R^2 (homeomorphic to a 2-dimentional real space), on the other hand they are also a 1-dimentional complex space. This gives them an extra structure with consequences like this theorem.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Another view on this is that differentiable complex functions must also satisfy the Cauchy-Reimann equations and this extra hypothesis is what makes them holomorphic&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="272" LastEditorUserId="272" LastEditorDisplayName="" LastEditDate="2010-07-27T23:28:02.027" LastActivityDate="2010-07-27T23:28:02.027" />
  <row Id="839" PostTypeId="2" ParentId="835" CreationDate="2010-07-27T23:12:25.170" Score="0" ViewCount="" Body="&lt;p&gt;In maths, the expected value is the average of how much you'd win if you used the same strategy from the same position a large (approaching infinity) number of times. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We should first note that expected value alone doesn't decide the best option. We also have to take into account risk. Most people (other than gamblers) would prefer a certain dollar rather than a fifty percent chance of two and fifty percent of nothing. To deal with this, we typically define &lt;a href=&quot;http://en.wikipedia.org/wiki/Utility&quot; rel=&quot;nofollow&quot;&gt;utility&lt;/a&gt; instead. Utility varies between individuals and is determined by their risk profile.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the dollar amount are quite large for contestants, it would be logical (in theory) to find someone who can afford to take the risk to insure you. This would allow you to receive the gains from a risker strategy. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For deal or no deal, they encourage you to play by making their offers worse than the expected value at the start of the game. Later, (&lt;a href=&quot;http://en.wikipedia.org/wiki/Deal_or_No_Deal&quot; rel=&quot;nofollow&quot;&gt;according to Wikipedia&lt;/a&gt;) the offers may even exceed the expected value. Without knowing how exactly the offers are calculated (or forming a model), we can't answer this question accurately, but only make general statements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we ignore risk and offers being greater than expected utility, you'd always want to go as far as possible in the game. But, as is, the game is extremely difficult to analyse mathematically.&lt;/p&gt;&#xA;" OwnerUserId="123" LastEditorUserId="123" LastEditorDisplayName="" LastEditDate="2010-07-27T23:20:17.560" LastActivityDate="2010-07-27T23:20:17.560" />
  <row Id="840" PostTypeId="2" ParentId="250" CreationDate="2010-07-27T23:15:09.097" Score="21" ViewCount="" Body="&lt;p&gt;Suppose you have a large collection of books, all of the same size.  &lt;strike&gt;Balance one of them on the edge of a table so that one end of the book is as far from the table as possible.  Balance another book on top of that one, and again try to get as far from the table as possible.&lt;/strike&gt;  Take $n$ of them and try to balance them on top of each other so that the top book is as far as possible away from the edge of the table horizontally.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Theorem:&lt;/strong&gt;  With enough books, you can get arbitrarily far from the table.  If you are &lt;em&gt;really&lt;/em&gt; careful.  This is a consequence of the divergence of the harmonic series.  I think if you haven't heard this one before it's very hard to tell whether it's true or false.&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-08-04T03:15:24.027" LastActivityDate="2010-08-04T03:15:24.027" CommunityOwnedDate="2010-07-27T23:15:09.097" CommentCount="14" />
  <row Id="841" PostTypeId="1" CreationDate="2010-07-27T23:16:48.580" Score="1" ViewCount="198" Body="&lt;p&gt;If I remember rightly there are some integrals of real functions which are easier to compute by using complex analysis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this because of properties of the particular function or because of a lack of a known real analysis technique?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there functions which would require hypercomplex analysis to integrate?&lt;/p&gt;&#xA;" OwnerUserId="276" LastActivityDate="2010-07-27T23:31:45.967" Title="Integral via complex analysis. Integral via hypercomplex analysis" Tags="&lt;complex-analysis&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="842" PostTypeId="1" AcceptedAnswerId="864" CreationDate="2010-07-27T23:19:07.500" Score="11" ViewCount="427" Body="&lt;p&gt;In logic, a semantics is said to be compact iff if every finite subset of a set of sentences has a model, then so to does the entire set. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most logic texts either don't explain the terminology, or allude to the topological property of compactness. I see an analogy as, given a topological space X and a subset of it S, S is compact iff for every open cover of S, there is a finite subcover of S. But, it doesn't seem strong enough to justify the terminology. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there more to the choice of the terminology in logic than this analogy?&lt;/p&gt;&#xA;" OwnerUserId="277" LastActivityDate="2010-08-11T22:03:15.910" Title="Why is compactness in logic called compactness?" Tags="&lt;terminology&gt;&lt;logic&gt;&lt;topology&gt;" AnswerCount="5" CommentCount="1" FavoriteCount="3" />
  <row Id="843" PostTypeId="2" ParentId="841" CreationDate="2010-07-27T23:23:08.977" Score="4" ViewCount="" Body="&lt;p&gt;I'm pretty sure you mean the evaluation of &lt;em&gt;definite&lt;/em&gt; integrals. For instance, $e^{-x^2}$ has no elementary antiderivative, but it's definite integral over the real line can be computed explicitly (it's $\sqrt{\pi}$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You ask what complex analysis has to do with this. Well, the idea is that complex integrals around closed curves (of holomorphic functions) are usually comparatively easy to evaluate: it's because we have the &lt;a href=&quot;http://en.wikipedia.org/wiki/Residue_theorem&quot; rel=&quot;nofollow&quot;&gt;residue theorem&lt;/a&gt; as a tool.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, these real integrals are usually over the whole real line, which doesn't satisfy the hypothesis of a path under the residue theorem. For instance, it's not a finite closed path.  However, we can replace the real number by a large semicircle (to pick one common example) that goes from $-R$ to $R$, then around the upper-half-plane from $R$ to $-R$. This is a closed path and the residue theorem can apply to this. When you let $R$ to $\infty$, the integral around semicircular path often tends to zero (by direct bounding arguments). So what you're left with is the integral along the real line, and it's equal to the residues.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Examples of this will be in any complex analysis textbook, e.g. Ahlfors's.  The Wikipedia article on the residue theorem (link above) also has examples.&lt;/p&gt;&#xA;" OwnerUserId="536" LastActivityDate="2010-07-27T23:23:08.977" />
  <row Id="844" PostTypeId="2" ParentId="842" CreationDate="2010-07-27T23:26:19.687" Score="0" ViewCount="" Body="&lt;p&gt;This isn't a complete answer, partly because I've had discussions with other grad students and we weren't able to work it out satisfactorily, but I've been told that you can actually put a topology on logical statements such that the compactness theorem translates to &quot;The set of true statements is a compact subset of the set of all statements&quot; or something similar.  But as I said, a few of my friends and I weren't able to work out the topology.&lt;/p&gt;&#xA;" OwnerUserId="176" LastActivityDate="2010-07-27T23:26:19.687" />
  <row Id="845" PostTypeId="2" ParentId="842" CreationDate="2010-07-27T23:27:21.497" Score="0" ViewCount="" Body="&lt;p&gt;As far as I know, the link comes from the syntactic theory. You are given a set of symbols of sentence F = {f_i} and you are allowed allowed to form complex statements using them. You can combine the elementary statements with AND, OR, NOT operators and parentheses, in the usual way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you get a set X of composed sentences, like&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(f AND g) OR (NOT h)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or something like that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A syntactic version of the compactness theorem states the following.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Assume that for every finite subset Y of X you can assign truth values to the f_i in such a way that all sentences in Y are true. Than you can do the same for X.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Proof&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Consider the topological space A obtained by taking the product of {0, 1} over the set F. The topology on A is the product topology. By Tychonoff's theorem, A is compact. For every composed statement s, the set of truth values which make s true is a finite intersection of cylinders, hence it is a closed set of A. The hypothesis say that every finite intersection of such closed sets, for s ranging in X, is not empty. Hence the intersection of all such closed set is not empty, which means one can make all statements in X true.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="233" LastEditorUserId="233" LastEditorDisplayName="" LastEditDate="2010-07-27T23:32:26.137" LastActivityDate="2010-07-27T23:32:26.137" CommentCount="1" />
  <row Id="846" PostTypeId="2" ParentId="842" CreationDate="2010-07-27T23:30:16.513" Score="1" ViewCount="" Body="&lt;p&gt;The analogy for the compactness theorem for propositional calculus is as follows. Let $p_i $ be propositional variables; together, they take values in the product space $2^{\mathbb{N}}$.  Suppose we have a collection of statements $S_t$ in these boolean variables such that every finite subset is satisfiable. Then I claim that we can prove that they are all simultaneously satisfiable by using a compactness argument.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $F$ be a finite set. Then the set of all truth assignments (this is a subset of $2^{\mathbb{N}}$) which satisfy $S_t$ for $t \in F$ is a closed set $V_F$ of assignments satisfying the sentences in $F$.  The intersection of any finitely many of the $V_F$ is nonempty, so by the finite intersection property, the intersection of all of them is nonempty (since the product space is compact), whence any truth in this intersection satisfies all the statements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know how this works in predicate logic.&lt;/p&gt;&#xA;" OwnerUserId="536" LastEditorUserId="536" LastEditorDisplayName="" LastEditDate="2010-07-27T23:48:53.003" LastActivityDate="2010-07-27T23:48:53.003" />
  <row Id="847" PostTypeId="1" AcceptedAnswerId="1570" CreationDate="2010-07-27T23:30:47.480" Score="3" ViewCount="442" Body="&lt;p&gt;Is there any mathematical significance to the fact that the law of cosines...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;cos(angle between a and b) = (a^2 + b^2 - c^2) / (2ab)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;... for an impossible triangle yields a cosine &amp;lt; -1 (when c &gt; a+b), or &gt; 1 (when c &amp;lt; |a-b|)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, a = 3, b = 4, c = 8 yields cos(angle ab) = -39/24.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or a = 3, b = 5, c = 1 yields cos(angle ab) = 33/30.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Something to do with hyperbolic geometry/cosines?&lt;/p&gt;&#xA;" OwnerUserId="271" LastActivityDate="2010-08-04T21:40:43.287" Title="Law of cosines with impossible triangles" Tags="&lt;geometry&gt;&lt;trigonometry&gt;" AnswerCount="3" FavoriteCount="1" />
  <row Id="848" PostTypeId="2" ParentId="311" CreationDate="2010-07-27T23:31:34.267" Score="1" ViewCount="" Body="&lt;p&gt;A further sense in which Higher Order Logics with standard or saturated semantics (HOL, hereafter) are less well-behaved than First Order Logic (FOL, hereafter) is a direct consequences of the failure of Completeness (and thus, as explained in other answers, of Compactness). The set of logical truths and the set of correct claims of semantic consequence for these logics are not recursively enumerable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;FOL is Complete, yet not Decidable. So, we determine of an arbitrary sentences and sets of sentences of the language of FOL if those sentences are logical truths or if a set has a given sentence as a consequence. But, since FOL is Complete and proofs are finitely long, we can (in the mathematicians sense of &quot;can&quot;) enumerate the proofs and inspect one by one, checking what sentence the proof shows as a theorem or what sentence the proof derives from what set of assumptions. This gets us a recursive enumeration of the truths and the sentence/set pairs that stand in the consequence relation. (This does not contradict the failure of decidability as we cannot conclude that since we've yet to come across a proof in out enumeration, there isn't one if only we kept looking.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since HOL's are not Complete, this means of showing them recursively enumerable is not available. Indeed, there can be no means; were there such a means, it could be exploited to induce a Complete proof system, and there cannot be such a Complete proof system as the HOL's are not Compact. &lt;/p&gt;&#xA;" OwnerUserId="277" LastActivityDate="2010-07-27T23:31:34.267" CommentCount="2" />
  <row Id="849" PostTypeId="2" ParentId="841" CreationDate="2010-07-27T23:31:45.967" Score="2" ViewCount="" Body="&lt;p&gt;A generic real-valued function carries no additional information other than its values.  But many of the common functions encountered in analysis are just the restriction to $\mathbb{R}$ of holomorphic functions, and these functions carry a lot of additional information, namely, their values in $\mathbb{C}$.  Unlike real-analytic functions, holomorphic functions are extremely rigid: for example, if two holomorphic functions agree on a set of complex numbers with a limit point, they must be equal everywhere.  That means it's possible to deduce information about how a holomorphic function behaves on part of its domain (say, $\mathbb{R}$) from information about how it behaves on other parts of its domain, since the two have to determine each other.  This information is extracted using contour integrals.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-07-27T23:31:45.967" />
  <row Id="851" PostTypeId="2" ParentId="847" CreationDate="2010-07-27T23:38:38.607" Score="2" ViewCount="" Body="&lt;p&gt;One can prove the triangle inequality in any abstract inner product space, such as a Hilbert space; it is a consequence of the Cauchy-Schwarz inequality $\langle a, b \rangle^2 \le ||a||^2 ||b||^2$.  In order for the triangle inequality to fail, the Cauchy-Schwarz inequality has to fail, and in order for Cauchy-Schwarz to fail (which corresponds to the &quot;cosine&quot; no longer being between $1$ and $-1$), some axiom of an abstract inner product space has to be given up.  One choice is to give up positive-definiteness; in other words, lengths of vectors are no longer always non-negative.  This leads to geometries like &lt;a href=&quot;http://en.wikipedia.org/wiki/Minkowski_space&quot; rel=&quot;nofollow&quot;&gt;Minkowski spacetime&lt;/a&gt; which are relevant to relativity.  In Minkowski spacetime, there is a &lt;em&gt;reverse&lt;/em&gt; triangle inequality instead.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit:  I should also mention that the &quot;unit sphere&quot; in Minkowski spacetime is a model for hyperbolic geometry called the &lt;a href=&quot;http://en.wikipedia.org/wiki/Hyperboloid_model&quot; rel=&quot;nofollow&quot;&gt;hyperboloid model&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-07-27T23:45:20.123" LastActivityDate="2010-07-27T23:45:20.123" CommentCount="6" />
  <row Id="852" PostTypeId="2" ParentId="847" CreationDate="2010-07-27T23:41:57.573" Score="0" ViewCount="" Body="&lt;p&gt;For some a, b, and c that form a triangle: increasing the length of c increases the measure of angle C and as m&amp;ang;C approaches 180&amp;deg;, cos C approaches -1; decreasing the length of c increases the measure of angle C and as m&amp;ang;C approaches 0&amp;deg;, cos C approaches 1.  Extending this pattern, it makes sense that if c &gt; a + b, c has gotten bigger past making a triangle with a and b, so cos C &amp;lt; -1, and if c &amp;lt; |a-b|, c has gotten smaller past making a triangle with a and b, so cos C &gt; 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In hyperbolic geometry, the definition of lines (and hence triangles) is different and the sum of the measures of the angles in a triangle is less than 180&amp;deg;.  There is a &lt;a href=&quot;http://en.wikipedia.org/wiki/Hyperbolic_law_of_cosines&quot; rel=&quot;nofollow&quot;&gt;Hyperbolic Law of Cosines&lt;/a&gt;, but it's not quite the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't think there's a sensible way to relate hyperbolic cosine to the Law of Cosines in Euclidean geometry.&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-07-27T23:41:57.573" />
  <row Id="853" PostTypeId="2" ParentId="250" CreationDate="2010-07-27T23:52:55.273" Score="7" ViewCount="" Body="&lt;p&gt;Position-based Cryptography. This is a fun example since it seems very &quot;out of left field&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The setup:&lt;/strong&gt;&#xA;Three servers are positioned in known locations on the globe (their positions can be arbitrary, provided they aren't on top of each other).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A single computer wants to prove its location to the servers. In other words, if the computer is actually located where it claims, then the protocol will accept. However, if the computer is located anywhere else, then the protocol will reject, no matter how the computer cheats (it is even allowed to recruit friends to help it cheat).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All communication is subject to the laws of physics -- information travels at speed c, and quantum mechanics holds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Theorem 1:&lt;/strong&gt; This is impossible if all communication is classical. Cheating is always possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Theorem 2:&lt;/strong&gt; This is possible if quantum communication is possible.&lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2010-07-27T23:52:55.273" CommunityOwnedDate="2010-07-27T23:52:55.273" CommentCount="2" />
  <row Id="854" PostTypeId="2" ParentId="250" CreationDate="2010-07-27T23:56:58.277" Score="1" ViewCount="" Body="&lt;p&gt;Goldbach's Conjecture.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Granted this is open, so it may be cheating a bit. However, this seems like a very hard problem to intuit your way to the &quot;conventional wisdom&quot;. By contrast, &lt;code&gt;P != NP&lt;/code&gt; is &quot;obviously&quot; true. Indeed, Feynman had trouble believing &lt;code&gt;P != NP&lt;/code&gt; was even open.&lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2010-07-27T23:56:58.277" CommunityOwnedDate="2010-07-27T23:56:58.277" CommentCount="3" />
  <row Id="855" PostTypeId="1" AcceptedAnswerId="860" CreationDate="2010-07-27T23:59:09.817" Score="15" ViewCount="1273" Body="&lt;p&gt;I know this is very basic and old hat to many, but I love this question and I am interested in seeing whether there are any proofs beyond the two I already know.&lt;/p&gt;&#xA;" OwnerUserId="12" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-07-28T00:17:44.240" LastActivityDate="2011-05-17T03:21:34.013" Title="For any prime $p &gt; 3$, why is $p^2-1$ always divisible by 24?" Tags="&lt;prime-numbers&gt;&lt;number-theory&gt;" AnswerCount="8" CommentCount="5" FavoriteCount="4" />
  <row Id="856" PostTypeId="2" ParentId="855" CreationDate="2010-07-28T00:02:04.647" Score="16" ViewCount="" Body="&lt;p&gt;P^2-1  = (P+1)(P-1). P must be either 1 or (2 mod 3), so we have a factor of 3 in the product. And P is also either 1 or 3 mod 4. Hence either 2|(P+1) and 4|(P-1) or 2|(P-1) and 4|(P+1). Thus 8*3= 24 divides the product.&lt;/p&gt;&#xA;" OwnerUserId="296" LastActivityDate="2010-07-28T00:02:04.647" />
  <row Id="857" PostTypeId="2" ParentId="855" CreationDate="2010-07-28T00:03:48.777" Score="12" ViewCount="" Body="&lt;p&gt;$p$ must be congruent either to 1,3,5,7 modulo 8.  Then $p^2$ is congruent to $1$ modulo $8$ in either case.  So $8$ divides $p^2-1$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, $p$ is not a multiple of 3, so either $p-1$ or $p+1$ is a multiple of three. So $3$ divides $p^2-1$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Together, it follows that 24 divides $p^2 -1 $.&lt;/p&gt;&#xA;" OwnerUserId="536" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2011-01-08T03:08:28.423" LastActivityDate="2011-01-08T03:08:28.423" CommentCount="1" />
  <row Id="858" PostTypeId="2" ParentId="742" CreationDate="2010-07-28T00:04:11.413" Score="0" ViewCount="" Body="&lt;p&gt;A similar problem is known in mathematics.  It is called the &quot;Postage-Stamp&quot; problem, and usually asks which postal values can be realized and which cannot.  Dynamic programming is a common but not polynomial-time solution for this.  A polynomial time solution using continued fractions in the case of two stamp denominations exists, and more complex algorithms exist for three or more stamp denominations.  Look up &quot;Frobenius Number&quot; for more technical information.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the given problem, I would roughly estimate the Frobenius number, and based on the estimate make an easy guess as to a near-solution and then use a dynamic programming/tabular solution, depending on the amount of time I had to solve the problem.&lt;/p&gt;&#xA;" OwnerUserId="297" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-08-11T22:01:46.380" LastActivityDate="2010-08-11T22:01:46.380" CommentCount="0" />
  <row Id="859" PostTypeId="2" ParentId="81" CreationDate="2010-07-28T00:07:35.650" Score="6" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://sbseminar.wordpress.com/&quot; rel=&quot;nofollow&quot;&gt;Secret Blogging Seminar&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="256" LastActivityDate="2010-07-28T00:07:35.650" CommunityOwnedDate="2010-07-28T00:07:35.650" CommentCount="1" />
  <row Id="860" PostTypeId="2" ParentId="855" CreationDate="2010-07-28T00:08:53.867" Score="45" ViewCount="" Body="&lt;p&gt;The most elementary proof I can think of, without explicitly mentioning any number theory: out of the three consecutive numbers $p – 1$, $p$, $p + 1$, one of them must be divisible by $3$; also, since the neighbours of &lt;em&gt;p&lt;/em&gt; are consecutive even numbers, one of them must be divisible by $2$ and the other by $4$, so their product is divisible by $3 &amp;#183; 2 &amp;#183; 4 = 24$ — and of course, we can throw $p$ out since it's prime, and those factors cannot come from it.&lt;/p&gt;&#xA;" OwnerUserId="205" LastEditorUserId="6622" LastEditorDisplayName="" LastEditDate="2011-05-07T05:02:19.083" LastActivityDate="2011-05-07T05:02:19.083" CommentCount="2" />
  <row Id="861" PostTypeId="1" AcceptedAnswerId="865" CreationDate="2010-07-28T00:13:25.637" Score="3" ViewCount="144" Body="&lt;p&gt;This may be a poorly phrased question - please let me know of it - but what is the correct way to think of the cotangent bundle? It seems odd to think of it as the dual of the tangent bundle (I am finding it odd to reconcile the notions of &quot;maps to the ground field&quot; with this object).&lt;/p&gt;&#xA;" OwnerUserId="298" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-07-28T00:15:36.867" LastActivityDate="2010-07-28T00:53:38.753" Title="Cotangent bundle" Tags="&lt;geometry&gt;&lt;topology&gt;" AnswerCount="3" />
  <row Id="862" PostTypeId="2" ParentId="842" CreationDate="2010-07-28T00:15:14.417" Score="2" ViewCount="" Body="&lt;p&gt;A motto which is related (and sometimes true) is &quot;proofs are finite&quot;.  In most systems of logic under consideration, the statement and a proof of the statement are finite strings of symbols.  One can think of them as &quot;compactly&quot; represented.  How nice then that certain systems will always allow a proof to be found if there is one.  While this does not give a tight connection to topology, it suggests (and you need to work this through on your own to be convinced) that certain infinite abnormalities (like an infinite proof) will not occur.  A similar topological claim is that, for compact sets, an infinite inclusive chain of certain subsets will have nonempty intersection, such a claim easily seen to be false for some non-compact sets, such as the collection { C_n | C_n = {x | x &gt;= n and x is a real number } for n a non-negative integer } .  &lt;/p&gt;&#xA;" OwnerUserId="297" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-08-11T22:03:15.910" LastActivityDate="2010-08-11T22:03:15.910" CommentCount="0" />
  <row Id="863" PostTypeId="2" ParentId="781" CreationDate="2010-07-28T00:15:29.330" Score="1" ViewCount="" Body="&lt;p&gt;As a rough/somewhat-intuitive explanation of why Benford's Law makes sense, consider it with respect to amounts of money.  The amount of time(/effort/work) needed to get from $1000 to $2000 (100% increase) is a lot greater than the amount of time needed to get from $8000 to $9000 (12.5% increase)--increasing money is usually done in proportion to the money one has.  Thought about in the other direction, it should take a fixed amount of time to, say, double one's money, so going from $1000 to $2000 takes as long as from $2000 to $4000 and $4000 to $8000, so the leading digit spends more time at lower values than at higher values.  Because the value growth is exponential, the time spent at each leading digit is roughly logarithmic.&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-07-28T00:15:29.330" />
  <row Id="864" PostTypeId="2" ParentId="842" CreationDate="2010-07-28T00:17:34.070" Score="8" ViewCount="" Body="&lt;p&gt;The Compactness Theorem is equivalent to the compactness of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Stone%27s_representation_theorem_for_Boolean_algebras&quot; rel=&quot;nofollow&quot;&gt;Stone space&lt;/a&gt; of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Lindenbaum%E2%80%93Tarski_algebra&quot; rel=&quot;nofollow&quot;&gt;Lindenbaum&amp;ndash;Tarski algebra&lt;/a&gt; of the first-order language &lt;em&gt;L&lt;/em&gt;. (This is also the space of &lt;a href=&quot;http://en.wikipedia.org/wiki/Type_%28model_theory%29&quot; rel=&quot;nofollow&quot;&gt;0-types&lt;/a&gt; over the empty theory.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A point in the Stone space &lt;em&gt;S&lt;/em&gt;&lt;sub&gt;&lt;em&gt;L&lt;/em&gt;&lt;/sub&gt; is a complete theory &lt;em&gt;T&lt;/em&gt; in the language &lt;em&gt;L&lt;/em&gt;. That is, &lt;em&gt;T&lt;/em&gt; is a set of sentences of &lt;em&gt;L&lt;/em&gt; which is closed under logical deduction and contains exactly one of &amp;sigma; or &amp;not;&amp;sigma; for every sentence &amp;sigma; of the language. The topology on the set of types has for basis the open sets &lt;em&gt;U&lt;/em&gt;(&amp;sigma;) = {&lt;em&gt;T&lt;/em&gt;: &amp;sigma; &amp;isin; &lt;em&gt;T&lt;/em&gt;} for every sentence &amp;sigma; of L. Note that these are all clopen sets since &lt;em&gt;U&lt;/em&gt;(&amp;not;&amp;sigma;) is complementary to &lt;em&gt;U&lt;/em&gt;(&amp;sigma;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To see how the Compactness Theorem implies the compactness of &lt;em&gt;S&lt;/em&gt;&lt;sub&gt;&lt;em&gt;L&lt;/em&gt;&lt;/sub&gt;, suppose the basic open sets &lt;em&gt;U&lt;/em&gt;(&amp;sigma;&lt;sub&gt;i&lt;/sub&gt;), i &amp;isin; I, form a cover of &lt;em&gt;S&lt;/em&gt;&lt;sub&gt;&lt;em&gt;L&lt;/em&gt;&lt;/sub&gt;. This means that every complete theory T contains at least one of the sentences &amp;sigma;&lt;sub&gt;i&lt;/sub&gt;. I claim that this cover has a finite subcover. If not, then the set {&amp;not;&amp;sigma;&lt;sub&gt;i&lt;/sub&gt;: i &amp;isin; I} is finitely consistent. By the Compactness Theorem, the set consistent and hence (by Zorn's Lemma) is contained in a maximally consistent set &lt;em&gt;T&lt;/em&gt;. This theory &lt;em&gt;T&lt;/em&gt; is a point of the Stone space which is not contained in any &lt;em&gt;U&lt;/em&gt;(&amp;sigma;&lt;sub&gt;i&lt;/sub&gt;), which contradicts our hypothesis that the &lt;em&gt;U&lt;/em&gt;(&amp;sigma;&lt;sub&gt;i&lt;/sub&gt;), i &amp;isin; I, form a cover of the space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To see how the compactness of &lt;em&gt;S&lt;/em&gt;&lt;sub&gt;&lt;em&gt;L&lt;/em&gt;&lt;/sub&gt; implies the Compactness Theorem, suppose that {&amp;sigma;&lt;sub&gt;i&lt;/sub&gt;: i &amp;isin; I} is an inconsistent set of sentences in &lt;em&gt;L&lt;/em&gt;. Then &lt;em&gt;U&lt;/em&gt;(&amp;not;&amp;sigma;&lt;sub&gt;i&lt;/sub&gt;), i &amp;isin; I, forms a cover of &lt;em&gt;S&lt;/em&gt;&lt;sub&gt;&lt;em&gt;L&lt;/em&gt;&lt;/sub&gt;. This cover has a finite subcover, which corresponds to a finite inconsistent subset of {&amp;sigma;&lt;sub&gt;i&lt;/sub&gt;: i &amp;isin; I}. Therefore, every inconsistent set has a finite inconsistent subset, which is the contrapositive of the Compactness Theorem.&lt;/p&gt;&#xA;" OwnerUserId="25" LastEditorUserId="25" LastEditorDisplayName="" LastEditDate="2010-07-30T01:03:24.220" LastActivityDate="2010-07-30T01:03:24.220" />
  <row Id="865" PostTypeId="2" ParentId="861" CreationDate="2010-07-28T00:18:21.587" Score="3" ViewCount="" Body="&lt;p&gt;One fruitful way to think about it, if you have physics background, is as phase space.  Your manifold is the configuration space for some system of particles, and the cotangent bundle is then the phases, so the cotangent directions are velocities.  This is helpful also with the symplectic structure on $T^*M$.&lt;/p&gt;&#xA;" OwnerUserId="176" LastActivityDate="2010-07-28T00:18:21.587" CommentCount="4" />
  <row Id="866" PostTypeId="2" ParentId="815" CreationDate="2010-07-28T00:18:44.877" Score="5" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Nicolas_Bourbaki&quot; rel=&quot;nofollow&quot;&gt;&lt;h1&gt;Nicolas Bourbaki&lt;/h1&gt;&lt;/a&gt;&#xA;The story had me from the moment I realized it is true. And as I come to understand more mathematics, I find that I frequently ask myself &quot;What would Bourbaki do?&quot; Also, the &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/3764326506&quot; rel=&quot;nofollow&quot;&gt;individuals&lt;/a&gt; have &lt;a href=&quot;http://en.wikipedia.org/wiki/Jean_Dieudonn%C3%A9&quot; rel=&quot;nofollow&quot;&gt;done&lt;/a&gt; much to expose the &lt;a href=&quot;http://www.fermentmagazine.org/home5.html&quot; rel=&quot;nofollow&quot;&gt;human&lt;/a&gt; side of mathematics.&lt;/p&gt;&#xA;" OwnerUserId="121" LastActivityDate="2010-07-28T00:18:44.877" CommunityOwnedDate="2010-07-28T00:18:44.877" />
  <row Id="867" PostTypeId="2" ParentId="795" CreationDate="2010-07-28T00:23:58.393" Score="1" ViewCount="" Body="&lt;p&gt;More generally, if $\sum_{i=1}^{\infty} c_n$ is a conditionally convergent series of &lt;em&gt;complex&lt;/em&gt; numbers, then the collection of sums of all convergent rearrangements is an affine subspace of the plane $\mathbb{C}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a sketch of why this is true.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose $A$ and $B$ are two rearrangements that converge to the distinct complex numbers $a,b$, respectively. Focus on a partial sum of $A$ that's pretty close to $a$. A sufficiently long partial sum of $B$ will include every term of the former partial sum of $A$, but it will also tend to $b$ as you take more and more terms. It's &quot;clear&quot; then that it's possible to obtain any combination $ta + (1 - t)b$ for $t \in \mathbb{R}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If in addition there is a point $c$ which is the sum of some rearrangement but is not on the line through $a$ and $b$, then the previous result can be reapplied, replacing the numbers $a$ and $b$ with the numbers $c$ and $d$, where $d$ is any point on the line through $a$ and $b$. It follows that in this case the entire plane $\mathbb{C}$ is obtained by rearranging the original series.&lt;/p&gt;&#xA;" OwnerUserId="251" LastActivityDate="2010-07-28T00:23:58.393" />
  <row Id="868" PostTypeId="1" AcceptedAnswerId="904" CreationDate="2010-07-28T00:30:59.580" Score="4" ViewCount="208" Body="&lt;p&gt;I am currently working on a Computer Algebra System and was wondering for suggestions on methods of finding roots/factors of polynomials. I am currently using the Numerical Durand-Kerner method but was wondering if there are any good non-numerical methods (primarily for simplifying fractions etc).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ideally this should work for equations in multiple variables.&lt;/p&gt;&#xA;" OwnerUserId="302" LastEditorUserId="302" LastEditorDisplayName="" LastEditDate="2010-07-28T02:03:51.857" LastActivityDate="2010-07-28T03:44:15.640" Title="What is the best way to factor arbitrary polynomials" Tags="&lt;linear-algebra&gt;&lt;algorithms&gt;&lt;computer-science&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="869" PostTypeId="2" ParentId="861" CreationDate="2010-07-28T00:41:25.627" Score="2" ViewCount="" Body="&lt;p&gt;You might be interested in this MathOverflow post: &lt;a href=&quot;http://mathoverflow.net/questions/17325/why-is-cotangent-more-canonical-than-tangent&quot; rel=&quot;nofollow&quot;&gt;http://mathoverflow.net/questions/17325/why-is-cotangent-more-canonical-than-tangent&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Sorry, I'd leave this as a comment but I just joined this site and don't have enough reputation.)&lt;/p&gt;&#xA;" OwnerUserId="263" LastActivityDate="2010-07-28T00:41:25.627" />
  <row Id="870" PostTypeId="2" ParentId="868" CreationDate="2010-07-28T00:42:24.800" Score="2" ViewCount="" Body="&lt;p&gt;If you're looking to factor exactly, then you'll need to use something that's not one of the fundamental operations of addition, subtraction, multiplication, division and extraction of roots.  The Abel-Ruffini theorem says so for degree five and above.  However, there are numerous other methods to find roots exactly, using more general functions, my favorite being theta functions, as explained in the appendix to Mumford's &quot;Tata Lectures on Theta II&quot;&lt;/p&gt;&#xA;" OwnerUserId="176" LastActivityDate="2010-07-28T00:42:24.800" CommentCount="1" />
  <row Id="871" PostTypeId="2" ParentId="392" CreationDate="2010-07-28T00:45:23.483" Score="6" ViewCount="" Body="&lt;p&gt;If you look carefully and geometrically at the quotient limit that defines sin'(x) in the unit circle, and take the chord and tangent as approximations to the arc (that is the angle; this is the essence of sin(x)/x approaches 1), you will see that limit of the derivative quotient tends exactly to cos(x), that is, it's adjacent/hypotenuse. In other words, it's built into right triangle geometry, like so many phenomena in mathematics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, in that geometry, you'll see lurking the proof for the sin(x+y) formula, which, along with the limit of sin(x)/x, is how the standard proof that sin'(x) = cos(x) goes.  But skipping that algebra and going directly to the geometry is the most straightforward way I know to answer the question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sorry I don't have time or tools to draw the pictures.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suspect this saying the same thing as the physics answer above, but perhaps more directly.  I do think all the answers referring to series expansions miss the point.&lt;/p&gt;&#xA;" OwnerUserId="271" LastEditorUserId="271" LastEditorDisplayName="" LastEditDate="2010-07-28T01:46:01.523" LastActivityDate="2010-07-28T01:46:01.523" CommentCount="2" />
  <row Id="872" PostTypeId="1" AcceptedAnswerId="881" CreationDate="2010-07-28T00:45:46.913" Score="2" ViewCount="217" Body="&lt;p&gt;I understand what the Hodge dual is, but I can't quite wrap my head around the dual space of vector space. They seem very similar, almost the same, but perhaps they are unrelated. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, in R^3, the blade a^b gives you a subspace that's like a plane, and the dual is roughly the normal to the plane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a similarly simple example for the dual space of a vector space, or is there a way to describe the vector space dual in terms of the Hodge dual?&lt;/p&gt;&#xA;" OwnerUserId="99" LastActivityDate="2010-11-22T22:08:07.980" Title="What is the relationship between the Hodge dual of p-vectors and the dual space of an ordinary vector space?" Tags="&lt;linear-algebra&gt;" AnswerCount="2" CommentCount="9" />
  <row Id="873" PostTypeId="2" ParentId="806" CreationDate="2010-07-28T00:45:58.847" Score="6" ViewCount="" Body="&lt;p&gt;Frequently in studying diophantine equations, or related problems, one is forced to look at rings of integers of algebraic number fields, and it may be that the class number of the ring your are forced to deal with is &gt; 1.  In this case, there is nothing you can do to change that fact; you have to live with it.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a response, it is natural to try and develop a theory of the class number and the class group, as a means of finding ways to deal with the failure of unique factorization.  Kummer was the one who invented the notion of class number, and he did in the course of his work on Fermat's Last Theorem; this is what is described in Andrea Ferretti's answer, and is an excellent example of what I am talking about.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another example, quite different in nature, occurs in the proof of Dirichlet's theorem on primes in arithmetic progressions.    This is the theorem that if a and d are coprime natural numbers, then there are infinitely many primes p such that p is congruent to a mod d.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In his proof, he reduces everything to showing that a certain number (the value at 1 of a certain so-called L-function) is non-zero.  He then is able to give a precise formula for this L-function, and shows that it is equal to some non-zero constants times a certain class number.  Since the class number is positive (and so in particular, is non-zero!), he is able to complete his proof.&lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2010-07-28T00:45:58.847" CommentCount="2" />
  <row Id="874" PostTypeId="1" CreationDate="2010-07-28T00:50:36.187" Score="0" ViewCount="111" Body="&lt;p&gt;When drawing an angle line (45 degrees) in a rectangle from a general point &lt;code&gt;p(x,y)&lt;/code&gt; that located on the right or the top line of the rectangle. How can I find the intersection point &lt;code&gt;p2&lt;/code&gt; of this line with the rectangle?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, I want to write the target point, &lt;code&gt;p2&lt;/code&gt;, with my current information: &lt;code&gt;x, y, w, h&lt;/code&gt;. (This variables are described in the picture below).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The point (0,0) is in the top-right corner.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://img690.imageshack.us/img690/9819/rectangleline.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="305" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-07-28T03:32:15.807" LastActivityDate="2010-07-28T04:58:52.313" Title="How to get an equation that output the end point of an angle line in rectangle?" Tags="&lt;geometry&gt;&lt;algebra-precalculus&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="875" PostTypeId="2" ParentId="574" CreationDate="2010-07-28T00:51:57.447" Score="4" ViewCount="" Body="&lt;p&gt;Generic has different meanings in different branches of geometry.  In algebraic geometry,&#xA;it usually means that the property in question holds on a Zariski dense open set.  In other geometric contexts, it could also that the property holds on a dense open set (in whatever is the natural topology under consideration), but can also mean that it holds on a countable intersection of such sets.  (The reason for considering such intersections as being big is&#xA;motivated by the Baire category theorem, which says that in reasonable contexts the complement of such a set will be very &quot;thin&quot;, in some sense.)&lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2010-07-28T00:51:57.447" />
  <row Id="876" PostTypeId="2" ParentId="735" CreationDate="2010-07-28T00:52:00.427" Score="1" ViewCount="" Body="&lt;p&gt;Hmm, apparently the other answerers' algebra classes were a lot more intense than mine...My answer's based on what I would have been comfortable with after basic algebra and geometry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically, if you think graphically, instantaneous velocity is the slope of a line at a single point rather than over an interval. At least that's what you want. So, if you happen to be going at a constant velocity (any straight line on a position graph), you can just use the slope formula. Where calculus comes in is if you're dealing with an inconstant velocity (a curved line).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If that's not making sense right off, just try to realistically graph the movement of a car as it accelerates/decelerates, where the y-axis is position and the x-axis is time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using algebra, you can't take the slope of a curved line. What you can do is take the slope between two points on the curve. The closer these points get to each other, the closer the slope between them will approximate the actual slope of the curve. So, if the slope between t=5.001 min. and t=5.01 min. is 40, then that approximates the actual slope, and instantaneous velocity, at t=5.0055 min.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't think I can get much more specific/accurate without going into (pre-)calculus.&lt;/p&gt;&#xA;" OwnerUserId="220" LastActivityDate="2010-07-28T00:52:00.427" />
  <row Id="877" PostTypeId="2" ParentId="861" CreationDate="2010-07-28T00:53:38.753" Score="1" ViewCount="" Body="&lt;p&gt;I'm not completely sure what you mean by this: &quot;It seems odd to think of it as the dual of the tangent bundle (I am finding it odd to reconcile the notions of &quot;maps to the ground field&quot; with this object),&quot; but maybe the following will help you see why it is natural to consider the dual space of the tangent bundle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given a function f on our manifold, we want to associate something like the gradient of f.  Well, in calculus, what characterized the gradient of a function?  Its the vector field such that when we take its dot product with a vector v at some point p, we get the directional derivative, at p, of f along v.  In a general manifold we don't have a dot product (which is a metric) but we can form a covector field (something which gives an element of the cotangent bundle at any point) such that, when applied to a vector v, we get the directional derivative of f along v.  This covector field is denoted df and is called the exterior derivative of f.&lt;/p&gt;&#xA;" OwnerUserId="9" LastActivityDate="2010-07-28T00:53:38.753" />
  <row Id="878" PostTypeId="2" ParentId="574" CreationDate="2010-07-28T00:54:39.220" Score="5" ViewCount="" Body="&lt;p&gt;The term &lt;em&gt;generic&lt;/em&gt; usually applies to something which happens in an open dense set of some space. The idea is that open dense sets are large subsets of the space. Indeed, they are closed under finite intersections and thus form a base for a &lt;a href=&quot;http://en.wikipedia.org/wiki/Filter_%28mathematics%29#Filter_on_a_set&quot; rel=&quot;nofollow&quot;&gt;filter of subsets&lt;/a&gt; of the space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sometimes, the term is applied more generally to something which happens in a countable intersection of open dense sets (a dense G&lt;sub&gt;&amp;delta;&lt;/sub&gt; set) of some space. Usually, this is in the context of a complete metric space or a locally compact Hausdorff space for which the &lt;a href=&quot;http://en.wikipedia.org/wiki/Baire_category_theorem&quot; rel=&quot;nofollow&quot;&gt;Baire Category Theorem&lt;/a&gt; applies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;em&gt;generic point&lt;/em&gt; is a (sometimes fictitious) point which lies in every open dense set of the space. Irreducible &lt;a href=&quot;http://en.wikipedia.org/wiki/Sober_space&quot; rel=&quot;nofollow&quot;&gt;sober spaces&lt;/a&gt; always have a generic point, it is the unique point whose closure is the whole space. The only Hausdorff space with a generic point is the one-point space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fictitious generic points have a variety of uses. Usually one means a point which lies in all open dense sets which are considered in the current argument, without specifying the open dense sets in question. This is fine because the intersection of finitely many (or even countably many in the case of Baire spaces) open dense sets is guaranteed to be nonempty.&lt;/p&gt;&#xA;" OwnerUserId="25" LastEditorUserId="25" LastEditorDisplayName="" LastEditDate="2010-07-28T01:38:46.330" LastActivityDate="2010-07-28T01:38:46.330" />
  <row Id="879" PostTypeId="2" ParentId="874" CreationDate="2010-07-28T00:56:09.203" Score="0" ViewCount="" Body="&lt;p&gt;Doesn't this depend on the slope of the segment that begins at P = (x,y)? Perhaps assuming some sort of proportionality we can solve this problem. Please provide more information.--&lt;/p&gt;&#xA;" OwnerUserId="298" LastActivityDate="2010-07-28T00:56:09.203" CommentCount="1" />
  <row Id="881" PostTypeId="2" ParentId="872" CreationDate="2010-07-28T01:01:41.547" Score="4" ViewCount="" Body="&lt;p&gt;(Edit:  I have edited this answer several times because my understanding of the situation has been improving.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is always profitable to understand these kind of constructions by understanding exactly what information they depend on.  The Hodge dual depends on a surprising amount of information: you need a vector space $V$ which is equipped with both an inner product and an orientation, which is essentially a choice of which bases of $V$ are &quot;right-handed.&quot;  So let's see what we can say ignoring all this information first.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any abstract vector space $V$ of finite dimension $n$ has exterior powers $\Lambda^2 V, \Lambda^3 V, ... \Lambda^n V$, the last of which is one-dimensional.  The vector spaces $\Lambda^k V$ and $\Lambda^{n-k} V$ always have the same dimension, so we would like to be able to define some sort of &quot;canonical&quot; map between them.  What can we say?  Well, they are always dual: the wedge product defines a natural bilinear map $\Lambda^k V \times \Lambda^{n-k} V \to \Lambda^n V$, and since the latter is one-dimensional this means (once you've proven nondegeneracy) that the two vector spaces are in fact dual.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But duality does not give you a map between them.  When two vector spaces $V, W$ are dual, meaning there is a nondegenerate bilinear map $V \times W \to F$ (where $F$ is the ground field), all you get is an isomorphism $V \simeq W^{\ast}$.  Here you get an isomorphism $\Lambda^k V \simeq \Lambda^{n-k} V^{\ast}$, &lt;strong&gt;once you have specified an isomorphism&lt;/strong&gt; $\Lambda^n V \simeq F$.  This is equivalent to picking out a distinguished vector in $\Lambda^n V$, which there is no way to do in general.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the answer is to introduce extra data.  To identify $\Lambda^{n-k} V^{\ast}$ with $\Lambda^{n-k} V$, we need an inner product.  An inner product gives you &lt;strong&gt;two&lt;/strong&gt; distinguished vectors in $\Lambda^n V$, as follows: take any orthonormal basis $b_1, ... b_n$.  Then wedging together the $b_i$ in any order gets you one of two elements of $\Lambda^n V$, depending on whether the corresponding permutation is even or odd.  But without any extra data, there is no way to identify one of these elements with $1$ and one of these elements with $-1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The extra data that does this is an orientation on $V$, which tells you which bases are &quot;right-handed&quot; and which are &quot;left-handed.&quot;  So an oriented orthonormal basis gives you a distinguished element of $\Lambda^n V$, which gives you a distinguished isomorphism $\Lambda^k V \simeq \Lambda^{n-k} V^{\ast}$, which composed with the isomorphism $\Lambda^{n-k} V^{\ast} \simeq \Lambda^{n-k} V$ is the Hodge dual.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Phew.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is explained in &lt;a href=&quot;http://www-users.math.umd.edu/~toni/hodge.pdf&quot; rel=&quot;nofollow&quot;&gt;these notes&lt;/a&gt; I just found on Google.&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-07-28T01:19:14.083" LastActivityDate="2010-07-28T01:19:14.083" CommentCount="8" />
  <row Id="882" PostTypeId="2" ParentId="835" CreationDate="2010-07-28T01:02:01.250" Score="1" ViewCount="" Body="&lt;p&gt;From what I hear about game-shows in general, if your performance does not make it to air, then you don't get anything.  Hence you cannot just accept the first amount offered (if it turns out to be a better choice) and expect to get it, since it won't make an interesting show.&lt;/p&gt;&#xA;" OwnerUserId="139" LastActivityDate="2010-07-28T01:02:01.250" CommentCount="4" />
  <row Id="886" PostTypeId="1" AcceptedAnswerId="888" CreationDate="2010-07-28T01:20:31.943" Score="3" ViewCount="386" Body="&lt;p&gt;Okay so &lt;a href=&quot;http://math.stackexchange.com/questions/835/optimal-strategy-for-deal-or-no-deal&quot; rel=&quot;nofollow&quot;&gt;this question&lt;/a&gt; reminded me of one my brother asked me a while back about the hit day-time novelty-worn-off-now snoozathon &lt;a href=&quot;http://en.wikipedia.org/wiki/Deal_or_no_deal&quot; rel=&quot;nofollow&quot;&gt;Deal or no deal&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;For the uninitiated:&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;In playing deal or no deal, the player is presented with one of 22 boxes (randomly selected) each containing different sums of money, he then asks in turn for each of the 21 remaining boxes to be opened, occasionally receiving an offer (from a wholly unconvincing 'banker' figure) for the mystery amount in his box.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If he rejects all of the offers along the way, the player is allowed to work his way through several (for some unfathomable reason, emotionally charged) box openings until there remain only two unopened boxes: one of which is his own, the other not. He is then given a choice to stick or switch (take the contents of his own box or the other), something he then agonises pointlessly over for the next 10 minutes.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Monty hall&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;[If you have not seen the monty hall 'paradox' check out this &lt;a href=&quot;http://en.wikipedia.org/wiki/Monty_Hall_problem&quot; rel=&quot;nofollow&quot;&gt;wikipedia link&lt;/a&gt; and prepare to be baffled, then enlightened, then disappointed that the whole thing is so trivial. After which feel free to read on.]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a certain similarity, you will agree, between the situation a deal or no deal player finds himself in having rejected all offers and the dilemma of Monty's contestant in the classic problem: several 'bad choices' have been eliminated and he is left with a choice between a better and worse choice with no way of knowing between them.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;So???&lt;/h3&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; The solution to the monty hall problem is that it is, in fact, better to switch- does the same apply here? Does this depend upon the money in the boxes? Should every player opt for 'switch', cutting the 10 minutes of agonising away???&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="160" LastActivityDate="2011-04-08T18:52:25.027" Title="Deal or no deal: does one switch (to avoid a goat)?/ Should deal or no deal be 10 minutes shorter?" Tags="&lt;probability&gt;&lt;game-theory&gt;" AnswerCount="4" CommentCount="2" />
  <row Id="887" PostTypeId="2" ParentId="22" CreationDate="2010-07-28T01:34:53.483" Score="5" ViewCount="" Body="&lt;p&gt;Here's an explanation in terms of the Hodge dual and the exterior (wedge) product.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let ${e_1, e_2, e_3}$ be the standard orthonormal basis for $\mathbb{R}^3$. Consider the two vectors $a = a_1 e_1 + a_2 e_2 + a_3 e_3$ and $b = b_1 e_1 + b_2 e_2 + b_3 e_3$. From the matrix computation we obtain the familiar formula&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a\times b = (a_2 b_3 - a_3 b_2) e_1 + (a_3 b_1 - a_1 b_3) e_2 + (a_1 b_2 - a_2 b_1) e_3$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But (see note at the bottom)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a \wedge b = (a_1 b_2 - a_2 b_1) e_1 \wedge e_2 + (a_2 b_3 - a_3 b_2) e_2 \wedge e_3 + (a_3 b_1 - a_1 b_3) e_3 \wedge e_1$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where the wedge $\wedge$ represents the exterior product. One can now compute the dual of this latter expression using that the left contraction of $(e_1 \wedge e_2)$ onto $(e_3 \wedge e_2 \wedge e_1)$ is $e_3$ (and similar relations). The result is that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a \times b = (a \wedge b)^*$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;that is, the cross product of $a$ and $b$ is the dual of their exterior product.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Geometrically, this is an incredible picture. The exterior product is the plane element spanned by both $a$ and $b$, and the dual is the vector orthogonal to that plane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is my favorite interpretation of the cross product, but it's only helpful, of course, if you're familiar with exterior algebra and the Hodge dual.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: The wedge product can be found by formally computing &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(a_1 e_1 + a_2 e_2 + a_3 e_3) \wedge (b_1 e_1 + b_2 e_2 + b_3 e_3)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;using the distributivity and anticommutation relations of the exterior product.&lt;/p&gt;&#xA;" OwnerUserId="251" LastEditorUserId="251" LastEditorDisplayName="" LastEditDate="2010-07-28T02:03:15.647" LastActivityDate="2010-07-28T02:03:15.647" CommentCount="1" />
  <row Id="888" PostTypeId="2" ParentId="886" CreationDate="2010-07-28T01:41:01.723" Score="5" ViewCount="" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;I would ceratinly plump for saying they were opened at random&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Then no, the Monty Hall solution doesn't apply.  The whole point is that the door isn't &lt;em&gt;randomly&lt;/em&gt; opened, it's always a goat.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An easy way of seeing this is imagining there are 100 doors, with 99 goats.  If, after you pick a door, the host always opens 98 doors of goats, then switching is very intuitively favorable.  However, if he had just opened 98 doors at random, then most of the time (98 out of 100) he would open the door with the car behind it; and even on the rare occasions he didn't, you still wouldn't be any better off switching than staying.&lt;/p&gt;&#xA;" OwnerUserId="136" LastEditorUserId="136" LastEditorDisplayName="" LastEditDate="2010-12-21T19:51:14.277" LastActivityDate="2010-12-21T19:51:14.277" CommentCount="2" />
  <row Id="889" PostTypeId="2" ParentId="874" CreationDate="2010-07-28T01:43:46.567" Score="1" ViewCount="" Body="&lt;p&gt;If it is just 45 degrees, then the answer is not very difficult. Center a coordinate system at the bottom left hand corner of the rectangle. Hence the coordinates of the (???) point are (q,0) for some q&#xA;&#xA;&lt;p&gt;Note that the because theta is 45 degrees, y = w-q . (Isosceles right triangle). Hence q=w-y, and our point is simply (w-y, 0)...&lt;/p&gt;&#xA;" OwnerUserId="298" LastActivityDate="2010-07-28T01:43:46.567" CommentCount="1" />
  <row Id="890" PostTypeId="2" ParentId="118" CreationDate="2010-07-28T01:50:04.913" Score="0" ViewCount="" Body="&lt;p&gt;Have a look at this explanation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://betterexplained.com/articles/a-gentle-introduction-to-learning-calculus/&quot; rel=&quot;nofollow&quot;&gt;http://betterexplained.com/articles/a-gentle-introduction-to-learning-calculus/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope you love it :)&lt;/p&gt;&#xA;" OwnerUserId="313" LastActivityDate="2010-07-28T01:50:04.913" />
  <row Id="891" PostTypeId="2" ParentId="886" CreationDate="2010-07-28T01:51:46.220" Score="2" ViewCount="" Body="&lt;p&gt;This doesn't have the key features of the Monty Hall problem:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The contestant makes a choice&lt;/li&gt;&#xA;&lt;li&gt;The contestant is given information about other choices they could have made&lt;/li&gt;&#xA;&lt;li&gt;The contestant can change their choice based on this new information&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Without being given extra information, their is no point in changing your choice.&lt;/p&gt;&#xA;" OwnerUserId="123" LastActivityDate="2010-07-28T01:51:46.220" />
  <row Id="892" PostTypeId="1" AcceptedAnswerId="893" CreationDate="2010-07-28T01:55:26.570" Score="0" ViewCount="141" Body="&lt;p&gt;So the projective plane RP^2 is not a vector space.  Is it still isomorphic to its dual?  If not, is there at least an invertible map that takes RP^2 to its dual?&lt;/p&gt;&#xA;" OwnerUserId="314" LastActivityDate="2010-07-28T02:23:49.593" Title="Projective plane and its dual." Tags="&lt;geometry&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="893" PostTypeId="2" ParentId="892" CreationDate="2010-07-28T01:59:42.910" Score="3" ViewCount="" Body="&lt;p&gt;Yes, though the word &quot;dual&quot; is somewhat questionable.  If you mean is $\mathbb{P}(\mathbb{R}^3)$, the standard projective plane, isomorphic to $\mathbb{P}((\mathbb{R}^3)^*)$, the projectivization of the dual, then yes, it follows from the isomorphism of vector spaces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Much more interestingly, the duality allows you to switch points and lines in theorems, such as the &lt;a href=&quot;http://en.wikipedia.org/wiki/Pascal%27s_theorem&quot; rel=&quot;nofollow&quot;&gt;Mystic Hexagon&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/Brianchon%27s_theorem&quot; rel=&quot;nofollow&quot;&gt;Brianchon's Theorem&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="176" LastActivityDate="2010-07-28T01:59:42.910" CommentCount="2" />
  <row Id="894" PostTypeId="1" AcceptedAnswerId="928" CreationDate="2010-07-28T02:06:45.263" Score="3" ViewCount="192" Body="&lt;p&gt;In First Order Logic with Identity (FOL+I), one can express the proposition that there are exactly 3 items that have the property P. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is it not possible to express the proposition that there is a finite number of items that have the property P (in FOL+I)?&lt;/p&gt;&#xA;" OwnerUserId="106" LastActivityDate="2010-07-29T12:37:26.077" Title="Why is the &quot;finitely many&quot; quantifier not definable in First Order Logic?" Tags="&lt;logic&gt;&lt;quantifiers&gt;" AnswerCount="3" CommentCount="0" />
  <row Id="895" PostTypeId="2" ParentId="894" CreationDate="2010-07-28T02:10:44.943" Score="4" ViewCount="" Body="&lt;p&gt;Any property expressible under first-order logic is closed under ultraproducts. The property of finite sets is not, however, closed under ultraproducts.&lt;/p&gt;&#xA;" OwnerUserId="536" LastActivityDate="2010-07-28T02:10:44.943" CommentCount="1" />
  <row Id="896" PostTypeId="2" ParentId="894" CreationDate="2010-07-28T02:11:56.220" Score="0" ViewCount="" Body="&lt;p&gt;Well, it would mean that you have the statement $P_1\vee P_2\vee\ldots$ where $P_i$ stands for &quot;there are $i$ objects with property $P$&quot;, and infinite disjunctions aren't allowed.  As for proving that this isn't equivalent to anything else you can write that IS allowed, I don't know how to do that.&lt;/p&gt;&#xA;" OwnerUserId="176" LastActivityDate="2010-07-28T02:11:56.220" CommentCount="1" />
  <row Id="897" PostTypeId="2" ParentId="835" CreationDate="2010-07-28T02:20:53.437" Score="2" ViewCount="" Body="&lt;p&gt;There are (at least) two factors that mean that simply calculating the average of the remaining options is not enough to describe how someone should play.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Risk_aversion&quot; rel=&quot;nofollow&quot;&gt;Risk aversion&lt;/a&gt; &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Someone's utility is not a predictable function of the amount of money that they win. For instance my utility from winning $\$$5 is more than 100 times my utility from winning 5 cents. However, my utility from winning $\$$100 million is less than 100 times my utility from winning $\$$1 million.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="106" LastEditorUserId="264" LastEditorDisplayName="" LastEditDate="2011-05-02T05:21:03.797" LastActivityDate="2011-05-02T05:21:03.797" CommentCount="1" />
  <row Id="898" PostTypeId="2" ParentId="892" CreationDate="2010-07-28T02:23:49.593" Score="2" ViewCount="" Body="&lt;p&gt;This is a comment more than an answer, mainly in response to Qiaochu's question, but I don't have sufficient rep. to comment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The dual to a projective plane is the set of all lines in the plane, which itself is a&#xA;projective plane (as hinted at in Charles Siegel's answer).  This is an important concept in classical projective geometry.  (Concretely, the equation for a line has the form&#xA;a x + b y + c z = 0, where a,b, and c are some parameters, not all zero, and x,y,z are homogeneous coords. for the points in the proj. plane.  The set of all such lines can thus&#xA;be thought of as the set of all (a,b,c) not all zero; but note that simultaneously multiplying a, b, and c by a non-zero scalar doesn't change the solution set, i.e. doesn't change the line, so the line should really be thought of as corresponding to the homogeneous coordinates (a:b:c); thus the set of all lines is again a projective plane.)&lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2010-07-28T02:23:49.593" />
  <row Id="899" PostTypeId="2" ParentId="154" CreationDate="2010-07-28T02:30:18.093" Score="20" ViewCount="" Body="&lt;p&gt;In my opinion, the most natural way to view complex number is as a class of maps from the plane to itself. Specifically, lets define $(R, \theta)$ to be the map which multiplies every point in the plane by the number $R$, and then rotates it by the angle $\theta$. We may call these maps &quot;dilations with rotations.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Such maps can be added and composed (multiplied) in the obvious way, and its not hard to work out that the sum and product of two such mappings is another dilation with rotation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can also identify the real number $x$ with the map $(x,0)$, i.e. the map which multiplies every point in the plane by $x$.  Then we see that these maps have the magical property that $-1$ has a square root! Namely, if $P$ is the mapping $(1,\pi/2)$ (i.e. rotate every point by angle $\pi/2$), then applying $P$ twice is the same as multiplying every number by $-1$, i.e. $P^2=-1$! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As should be obvious by now, these maps are just complex numbers in disguise. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unsurprisingly, they are &#xA;singularly useful for solving polynomial equations. Indeed, the real number $x'$ is a root of the polynomial equations $a_0 x^n + a_1 x^{n-1} + \cdots + a_n =0$ if and only if the mapping $(x',0)$ satisfies the same equation. So viewing polynomial equations over the set of these mappings loses no solutions, while at the same time giving us additional freedom to do operations such as taking the square roots of negative numbers. &lt;/p&gt;&#xA;" OwnerUserId="317" LastActivityDate="2010-07-28T02:30:18.093" CommunityOwnedDate="2011-04-15T20:45:29.080" CommentCount="3" />
  <row Id="900" PostTypeId="2" ParentId="667" CreationDate="2010-07-28T02:30:26.223" Score="5" ViewCount="" Body="&lt;p&gt;What you are referring to is the &quot;stacks as sheaves of groupoids&quot; point of view. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To illustrate where it comes from, imagine for example that we are talking about the moduli stack of elliptic curves (on the category of schemes).  To give an elliptic curve over a scheme, it is not just enough to specify the elliptic curve over the members of the open cover; we have to explain how we glue the restrictions of the curves on the various opens on their overlaps, and this gluing has to be coherent over triple overlaps.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason for this is that elliptic curves can have non-trivial automorphisms, so that there is no &lt;em&gt;a priori&lt;/em&gt; determined way to make the identifications on the overlaps (because having non-trivial automorphisms is the same as saying that when two curves are isomorphic, they can be isomorphic in more than one way), so it is your job to choose these identifications, and to make sure that you do it in a coherent way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Here elliptic curves can be replaced by any other moduli problem you can think of, of course.)&lt;/p&gt;&#xA;" OwnerUserId="221" LastEditorUserId="221" LastEditorDisplayName="" LastEditDate="2010-07-29T00:13:37.180" LastActivityDate="2010-07-29T00:13:37.180" />
  <row Id="901" PostTypeId="2" ParentId="631" CreationDate="2010-07-28T02:53:48.710" Score="1" ViewCount="" Body="&lt;p&gt;There are a few subtleties that will probably effect the final answer.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Are we required to find the solution, or merely establish existence? By analogy, determining if a number has a prime factorization is trivial, but finding its prime factorization is hard.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Is the runtime being measured in terms of {a_1,...,a_n,s} or {log(a_i),...,log(a_n),log(s)}? By analogy, SUBSET-SUM is in P in the first case, but NP-complete in the second case.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="275" LastEditorUserId="153" LastEditorDisplayName="" LastEditDate="2010-07-28T02:56:07.330" LastActivityDate="2010-07-28T02:56:07.330" CommentCount="6" />
  <row Id="902" PostTypeId="2" ParentId="518" CreationDate="2010-07-28T02:56:36.067" Score="3" ViewCount="" Body="&lt;p&gt;This is a comment on the proof sketch in the question (I don't have sufficient rep. to actually leave comments): the localization A_P is local of dimension zero (its unique maximal ideal is also a minimal prime ideal, and hence is the unique prime ideal of A_P), but need not be Artinian, as far as I can tell.  E.g. if A = C[[x^{1/2},x^{1/3},...]]/(x), then A is local&#xA;with a unique prime ideal (namely the ideal generated by all the x^{1/n}), but is not Artinian (equivalently, not Noetherian), since if (a_i) is any strictly descending sequence of rational numbers in the interval (0,1), then the ideals x^{a_i} A form a strictly descending sequence&#xA;of ideals in A.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Hopefully I'm not blundering here; if I am, someone please let me know!)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Also, I should add that it is still the case that since PA_P is the unique minimal&#xA;prime of A_P, every element of PA_P is nilpotent, and hence every element of P is&#xA;a zero divisor, so my comment is very nitpicky: it is just about the use of the terminology Artinian.)&lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2010-07-28T02:56:36.067" CommentCount="3" />
  <row Id="903" PostTypeId="1" CreationDate="2010-07-28T03:10:02.303" Score="5" ViewCount="542" Body="&lt;p&gt;Which is a better textbook - Dugundji or Munkres? I'm concerned with clarity of exposition and explanation of motivation, etc.&lt;/p&gt;&#xA;" OwnerUserId="298" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-07-28T03:26:01.203" LastActivityDate="2011-05-07T01:54:08.863" CommunityOwnedDate="2010-07-28T03:17:30.603" Title="choosing a topology text" Tags="&lt;soft-question&gt;&lt;books&gt;&lt;topology&gt;" AnswerCount="7" CommentCount="7" FavoriteCount="4" />
  <row Id="904" PostTypeId="2" ParentId="868" CreationDate="2010-07-28T03:38:52.363" Score="5" ViewCount="" Body="&lt;p&gt;If you are interested in the factorization algorithms employed in modern computer algebra systems such as Macsyma, Maple, or Mathematica, then see any of the standard introductions to computer algebra , e.g. Geddes et.al. &quot;Algorithms for Computer Algebra&quot;; Knuth, &quot;TAOCP&quot; v.2; von zur Gathen &quot;Modern Computer Algebra&quot;; Zippel &quot;Effective Polynomial Computation&quot;. See also&#xA;Kaltofen's surveys on polynomial factorization [116,68,56,7] in his publications &lt;a href=&quot;http://www4.ncsu.edu/~kaltofen/bibliography/kaltofen.html&quot; rel=&quot;nofollow&quot;&gt;list&lt;/a&gt;, which contains plenty of theory, history and literature references. Note: Kaltofen's home page appears to be temporarily down so instead see his &lt;a href=&quot;http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.39.7916&amp;amp;rep=rep1&amp;amp;type=pdf&quot; rel=&quot;nofollow&quot;&gt;paper [1]&lt;/a&gt; to get started (see comments)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.39.7916&amp;amp;rep=rep1&amp;amp;type=pdf&quot; rel=&quot;nofollow&quot;&gt;1&lt;/a&gt; Kaltofen, E. Factorization of Polynomials, pp. 95-113 in:&lt;br&gt;&#xA;Computer Algebra, B. Buchberger, R. Loos, G. Collins, editors, Vienna, Austria,  (1982).&lt;br&gt;&#xA;&lt;a href=&quot;http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.39.7916&amp;amp;rep=rep1&amp;amp;type=pdf&quot; rel=&quot;nofollow&quot;&gt;http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.39.7916&amp;amp;rep=rep1&amp;amp;type=pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-07-28T03:44:15.640" LastActivityDate="2010-07-28T03:44:15.640" CommentCount="1" />
  <row Id="905" PostTypeId="2" ParentId="29" CreationDate="2010-07-28T03:39:45.887" Score="3" ViewCount="" Body="&lt;p&gt;Try sage.  It's like python (it is actually written in python, and uses something similar to a python shell), but with a vastly larger set of built-in mathematical operations.&lt;/p&gt;&#xA;" OwnerUserId="326" LastActivityDate="2010-07-28T03:39:45.887" CommunityOwnedDate="2010-07-28T03:39:45.887" />
  <row Id="906" PostTypeId="2" ParentId="29" CreationDate="2010-07-28T03:53:58.273" Score="1" ViewCount="" Body="&lt;p&gt;Some people have suggested Sage or Python. Sage can do the symbolic manipulation that TI calculators can do using Maxima. Python with the right packages can do this as well (SymPy package I believe). Plus these things will be programmable. Sage however is fairly heavy at a few GB if I recall correctly. I'm not sure if there is anyway to interface with Maxima through Python like in Sage without actually having Sage.&lt;/p&gt;&#xA;" OwnerUserId="229" LastEditorUserId="229" LastEditorDisplayName="" LastEditDate="2010-07-28T04:01:11.157" LastActivityDate="2010-07-28T04:01:11.157" CommunityOwnedDate="2010-07-28T03:53:58.273" />
  <row Id="907" PostTypeId="1" AcceptedAnswerId="924" CreationDate="2010-07-28T03:56:11.673" Score="3" ViewCount="569" Body="&lt;p&gt;While I feel quite confident that I've inferred the correct meaning of &quot;In the sequel&quot; from context, I've never heard anyone explicitly tell me, so first off, to remove my niggling doubts: What does this phrase mean?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Someone recently argued to me that &quot;sequel&quot; was actually supposed to refer to a forthcoming second part of a paper, which I found highly unlikely, but I'd just like to make sure. )&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My main questions: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;At what points in the text, and for what kinds of X, is it appropriate to use the phrase &quot;In the sequel, X&quot; in a paper? In a book? Is it ever acceptable to introduce definitions via &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;In the sequel, we introduce the concept of a &quot;blah&quot;, which is a thing satisfying ...&quot; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;at the start of a paper or book &lt;strong&gt;without&lt;/strong&gt; a formal&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Definition. A &quot;blah&quot; is a thing, satsifying ...&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;in the main text of the paper or book? Finally, out of curiosity, I'm wondering how long this phrase has been around, if it's considered out of date or if it's still a popular phrase, and what some good alternatives are.&lt;/p&gt;&#xA;" OwnerUserId="264" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-07-29T01:16:54.680" LastActivityDate="2010-09-08T08:36:29.900" Title="Correct usage of the phrase &quot;In the sequel&quot;? History? Alternatives?" Tags="&lt;soft-question&gt;&lt;terminology&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="2" />
  <row Id="908" PostTypeId="2" ParentId="907" CreationDate="2010-07-28T04:01:41.187" Score="2" ViewCount="" Body="&lt;p&gt;I would never write that (mostly because it really sounds like the definition will be in another paper...). I'm pretty sure I'd write «In what follows, ...».&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2010-07-28T04:01:41.187" CommentCount="2" />
  <row Id="910" PostTypeId="2" ParentId="855" CreationDate="2010-07-28T04:07:58.903" Score="8" ViewCount="" Body="&lt;p&gt;In fact the result holds a bit more generally, namely:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;LEMMA&lt;/strong&gt; $\rm \quad 24\ |\ M^2 - N^2 \;$ if $\rm \; M,N \perp 6, \;$ i.e. coprime to $6.\;$ The proof is easy:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\quad\quad N\perp 2 \;\Rightarrow\; N = \pm 1, \pm 3 \pmod 8 \;\Rightarrow\; N^2 = 1 \pmod 8$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\quad\quad N\perp 3 \;\Rightarrow\; N \;\;= \;\;\;\pm 1\:\quad \pmod 3 \;\Rightarrow\; N^2 = 1 \pmod 3 \;$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So $\rm \quad\; 3, 8\ |\ N^2 - 1 \;\Rightarrow\;  24\ |\ N^2 - 1 \ $ since $\ {\rm lcm}(3,8) = 24.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a special case $\rm\ n = 24\ $ of this much more general result  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;THEOREM&lt;/strong&gt; $\ $  For naturals $\rm\ a,e,n $ with $\rm\ e,n&amp;gt;1 $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad n\ |\ a^e-1$ for all $\rm a\perp n \ \iff\  \phi&amp;#39;(p^k)\:|\:e\ $ for all $\rm\ p^k\:|\:n\:,\ \ p\:$ prime&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with $\rm \;\;\; \phi&amp;#39;(p^k) = \phi(p^k)\ $ for odd primes $\rm p\:,\ $ where $\phi$ is Euler's totient function&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and   $\rm\ \quad \phi&amp;#39;(2^k) = 2^{k-2}\ $ if $\rm  k&amp;gt;2\:,\ $ else $2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The latter exception is due to $\rm \mathbb Z/2^k$ having multiplicative group $\rm C(2) \times C(2^{k-2})$ for $\rm k&amp;gt;2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice that the least such exponent $\rm e$ is given by $\rm \;\lambda(n)\; = \;{\rm lcm}\;\{\phi&amp;#39;(\;{p_i}^{k_i})\}\;$ where $\rm \; n = \prod {p_i}^{k_i}\;$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\lambda(n)$ is called the (universal) &lt;a href=&quot;http://planetmath.org/encyclopedia/Exponent.html&quot; rel=&quot;nofollow&quot;&gt;exponent&lt;/a&gt; of the group $\rm \mathbb Z/n^*,\;$ a.k.a. the &lt;a href=&quot;http://en.wikipedia.org/wiki/Carmichael_function&quot; rel=&quot;nofollow&quot;&gt;Carmichael function&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the case at hand is simply $\rm\ \lambda(24) = lcm(\phi&amp;#39;(2^3),\phi&amp;#39;(3)) = lcm(2,2) = 2\:.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See my &lt;a href=&quot;http://groups.google.com/groups?selm=y8zk55tqcie.fsf%40nestle.csail.mit.edu&quot; rel=&quot;nofollow&quot;&gt;post here&lt;/a&gt; for proofs and further discussion.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2011-01-08T03:36:50.213" LastActivityDate="2011-01-08T03:36:50.213" />
  <row Id="911" PostTypeId="2" ParentId="250" CreationDate="2010-07-28T04:22:44.383" Score="17" ViewCount="" Body="&lt;p&gt;Scott Aaronson once basically did this for a bunch of theorems in computer science &lt;a href=&quot;http://scottaaronson.com/blog/?p=152&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; and &lt;a href=&quot;http://scottaaronson.com/blog/?p=153&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.  I particularly like this one:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Suppose a baby is given some random examples of grammatical and ungrammatical sentences, and based on that, it wants to infer the general rule for whether or not a given sentence is grammatical. If the baby can do this with reasonable accuracy and in a reasonable amount of time, for any “regular grammar” (the very simplest type of grammar studied by Noam Chomsky), then that baby can also break the RSA cryptosystem.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="232" LastEditorUserId="794" LastEditorDisplayName="" LastEditDate="2010-11-26T16:44:21.833" LastActivityDate="2010-11-26T16:44:21.833" CommunityOwnedDate="2010-07-28T04:22:44.383" CommentCount="1" />
  <row Id="912" PostTypeId="2" ParentId="796" CreationDate="2010-07-28T04:27:18.310" Score="2" ViewCount="" Body="&lt;p&gt;There is an extremely beautiful Lie-theoretic approach to separation of variables,&#xA;e.g. see Willard Miller's &lt;a href=&quot;http://www.ima.umn.edu/~miller/separationofvariables.html&quot; rel=&quot;nofollow&quot;&gt;book [1]&lt;/a&gt;. I quote from his introduction:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;This book is concerned with the&#xA;  relationship between symmetries of a &#xA;  linear second-order partial&#xA;  differential equation of mathematical&#xA;  physics,  the coordinate systems in&#xA;  which the equation admits solutions&#xA;  via separation  of variables, and the&#xA;  properties of the special functions&#xA;  that arise in  this manner. It is an&#xA;  introduction intended for anyone with&#xA;  experience in  partial differential&#xA;  equations, special functions, or Lie&#xA;  group theory, such  as group&#xA;  theorists, applied mathematicians,&#xA;  theoretical physicists and  chemists,&#xA;  and electrical engineers. We will&#xA;  exhibit some modem group-theoretic &#xA;  twists in the ancient method of&#xA;  separation of variables that can be &#xA;  used to provide a foundation for much&#xA;  of special function theory. In &#xA;  particular, we will show explicitly&#xA;  that all special functions that arise &#xA;  via separation of variables in the&#xA;  equations of mathematical physics can &#xA;  be studied using group theory. These&#xA;  include the functions of Lam6, Ince, &#xA;  Mathieu, and others, as well as those&#xA;  of hypergeometric type. &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;This is a very critical time in the&#xA;  kistory of group-theoretic methods in &#xA;  special Iunction theory. The basic&#xA;  relations between Lie groups, special &#xA;  functions, and the method of&#xA;  separation of variables have recently&#xA;  been  clarified. One can now construct&#xA;  a group-theoretic machine that, when &#xA;  applied to a given differential&#xA;  equation of mathematical physms,&#xA;  describes  in a rational manner the&#xA;  possible coordinate systems in which&#xA;  the equation  admits solutions via&#xA;  separation of variables and the&#xA;  various expansion  theorems relating&#xA;  the separable (special function)&#xA;  solutions in distinct  coordinate&#xA;  systems. Indeed for themost important&#xA;  linear equations, the  separated&#xA;  solutions are characterized as common&#xA;  eigenfunctions of sets of &#xA;  second-order commuting elements in the&#xA;  umversal enveloping algebra of  the&#xA;  Lie symmetry algebra corresponding to&#xA;  the equation. The problem of &#xA;  expanding one set of separable&#xA;  solutions in terms of another reduces&#xA;  to a  problem in the representation&#xA;  theory of the Lie symmetry algebra.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.ima.umn.edu/~miller/separationofvariables.html&quot; rel=&quot;nofollow&quot;&gt;1&lt;/a&gt; Willard Miller. Symmetry and Separation of Variables.&lt;br&gt;&#xA;Addison-Wesley, Reading, Massachusetts, 1977 (out of print)&lt;br&gt;&#xA;&lt;a href=&quot;http://www.ima.umn.edu/~miller/separationofvariables.html&quot; rel=&quot;nofollow&quot;&gt;http://www.ima.umn.edu/~miller/separationofvariables.html&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gigapedia.com/items:links?id=64401&quot; rel=&quot;nofollow&quot;&gt;http://gigapedia.com/items:links?id=64401&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-07-28T04:27:18.310" />
  <row Id="913" PostTypeId="2" ParentId="791" CreationDate="2010-07-28T04:29:55.297" Score="2" ViewCount="" Body="&lt;p&gt;The answer can be found on pages 27-28 of the following link.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.math.vt.edu/people/plinnell/Vtregional/solutions.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.math.vt.edu/people/plinnell/Vtregional/solutions.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="309" LastActivityDate="2010-07-28T04:29:55.297" CommentCount="0" />
  <row Id="914" PostTypeId="2" ParentId="741" CreationDate="2010-07-28T04:30:38.603" Score="1" ViewCount="" Body="&lt;p&gt;I suggest you &lt;a href=&quot;http://rjlipton.wordpress.com/2010/06/26/stating-pnp-without-turing-machines/&quot; rel=&quot;nofollow&quot; title=&quot;this&quot;&gt;this&lt;/a&gt; article, which talks about Linear Equations, Linear Programming, Integer Programming and P=NP. It's easy to understand and talks about the differences among these things&lt;/p&gt;&#xA;" OwnerUserId="229" LastActivityDate="2010-07-28T04:30:38.603" />
  <row Id="915" PostTypeId="2" ParentId="362" CreationDate="2010-07-28T04:44:48.480" Score="8" ViewCount="" Body="&lt;p&gt;For a nice introduction to the history of ring theory see the following paper&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I. Kleiner. &lt;strong&gt;From numbers to rings: the early history of ring theory&lt;/strong&gt;.&lt;br&gt;&#xA;Elemente der Mathematik 53 (1998) 18-35.&lt;br&gt;&#xA;&lt;a href=&quot;http://retro.seals.ch/cntmng?type=pdf&amp;amp;rid=elemat-001:1998:53::22&amp;amp;subp=hires&quot; rel=&quot;nofollow&quot;&gt;http://retro.seals.ch/cntmng?type=pdf&amp;amp;rid=elemat-001:1998:53::22&amp;amp;subp=hires&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://www.springerlink.com/content/e7p7kxk2y3a71j0d/fulltext.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.springerlink.com/content/e7p7kxk2y3a71j0d/fulltext.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-07-28T04:44:48.480" CommentCount="2" />
  <row Id="916" PostTypeId="2" ParentId="874" CreationDate="2010-07-28T04:58:52.313" Score="0" ViewCount="" Body="&lt;p&gt;Alright, I'm not 100% sure I'm understanding this correctly. You say that p can be located on the right or top line and that p2 can be located on the bottom or left line. Do you mean the rectangle can be rotated? If that's the case, the question should say that p can be on the right or bottom line of the rectangle. Also, are you looking for two separate answers or one that works both when p2 is on the bottom and on the left?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you do mean that the rectangle can be rotated, and want two different answers, it's pretty simple. First I'll deal with when p2 is on the bottom and p is on the right.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since p2 is on the bottom line we know the y-coordinate is h, according to the diagram. We also know that p is (0,y). Because of the 45 degree angle, we know that the distance between p's y-coordinate and the lower right corner is the same as the distance between the lower right corner and p2's x-coordinate, which in this case is p2's x-coordinate. Therefore, the coordinates of p2 are (h-y, h).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If p2 is on the left and p is on the bottom, it's very similar. Since p2 is on the left, it's x-coordinate is h. Because p is on the x-axis, it's (x,0). Because of the 45 degree angle, the distance between the lower left corner and p is the same as the distance between the lower left corner and p2, which this time gives us p2's y-coordinate. Therefore the coordinates of p2 are (h,h-x).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hopefully I understood your intentions correctly. If not, I hope you can use my misunderstandings to further improve your question.&lt;/p&gt;&#xA;" OwnerUserId="220" LastActivityDate="2010-07-28T04:58:52.313" CommentCount="1" />
  <row Id="917" PostTypeId="2" ParentId="250" CreationDate="2010-07-28T05:06:14.937" Score="28" ViewCount="" Body="&lt;p&gt;My favorite would probably be Goodstein's theorem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Start with your favorite number (mine is 37) and express it in hereditary base 2 notation.  That is, write it as a power of 2 with exponents powers of 2, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, 37 = 2^(2^2 + 1) + 2^2 + 1.  This is the first element of the sequence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next, change all the 2's to 3's, and subtract one from what's remaining and express in hereditary base 3 notation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We get 3^(3^3 + 1) + 3^3 + 1 - 1= 3^(3^3 + 1) + 3^3 (which is roughly 2 x 10^13).  This is the second element of the sequence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next, change all 3's to 4's, subtract one, and express in hereditary base 4 notation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We get 4^(4^4 + 1) + 4^4 - 1 = 4^(4^4 + 1) + 3*4^3 + 3*4^2 + 3*4 + 3 (which is roughly 5 x 10^154) .  This is the third element of the sequence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rinse, repeat:  at the nth stage, change all the &quot;n+1&quot; to &quot;n+2&quot;, subtract 1, and reexpress in hereditary base n+2 notation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The theorem is: no matter which number you start with, eventually, your sequence hits 0, despite the fact that it grows VERY quickly at the start.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, if instead of starting with 37, we started with 4, then (according to the wikipedia page), it takes 3*2^(402653211) - 2 steps ( VERY roughly 10^(100,000,000), or a 1 followed by a hundred million 0s).  37 takes vastly longer to drop to 0.&lt;/p&gt;&#xA;" OwnerUserId="331" LastActivityDate="2010-07-28T05:06:14.937" CommunityOwnedDate="2010-07-28T05:06:14.937" CommentCount="2" />
  <row Id="918" PostTypeId="1" AcceptedAnswerId="929" CreationDate="2010-07-28T05:32:29.170" Score="13" ViewCount="1029" Body="&lt;p&gt;Let $S$ be a set of size $n$.  There is an easy way to count the number of subsets with an even number of elements.  Algebraically, it comes from the fact that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle \sum_{k=0}^{n} {n \choose k} = (1 + 1)^n$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;while&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle \sum_{k=0}^{n} (-1)^k {n \choose k} = (1 - 1)^n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It follows that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle \sum_{k=0}^{n/2} {n \choose 2k} = 2^{n-1}$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A direct combinatorial proof is as follows: fix an element $s \in S$.  If a given subset has $s$ in it, add it in; otherwise, take it out.  This defines a bijection between the number of subsets with an even number of elements and the number of subsets with an odd number of elements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The analogous formulas for the subsets with a number of elements divisible by $3$ or $4$ are more complicated, and divide into cases depending on the residue of $n \bmod 6$ and $n \bmod 8$, respectively.  The algebraic derivations of these formulas are as follows (with $\omega$ a primitive third root of unity):  observe that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle \sum_{k=0}^{n} \omega^k {n \choose k} = (1 + \omega)^n = (-\omega^2)^n$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;while&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle \sum_{k=0}^{n} \omega^{2k} {n \choose k} = (1 + \omega^2)^n = (-\omega)^n$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and that $1 + \omega^k + \omega^{2k} = 0$ if $k$ is not divisible by $3$ and equals $3$ otherwise.  (This is a special case of the discrete Fourier transform.)  It follows that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle \sum_{k=0}^{n/3} {n \choose 3k} = \frac{2^n + (-\omega)^n + (-\omega)^{2n}}{3}.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$-\omega$ and $-\omega^2$ are sixth roots of unity, so this formula splits into six cases (or maybe three).  Similar observations about fourth roots of unity show that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle \sum_{k=0}^{n/4} {n \choose 4k} = \frac{2^n + (1+i)^n + (1-i)^n}{4}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $1+i = \sqrt{2} e^{ \frac{\pi i}{4} }$ is a scalar multiple of an eighth root of unity, so this formula splits into eight cases (or maybe four).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt;  Does anyone know a direct combinatorial proof of these identities? &lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="152" LastEditorDisplayName="" LastEditDate="2010-08-03T08:05:40.423" LastActivityDate="2010-08-03T08:05:40.423" Title="How do I count the subsets of a set whose number of elements is divisible by 3? 4?" Tags="&lt;number-theory&gt;&lt;combinatorics&gt;&lt;binomial-coefficients&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="4" />
  <row Id="919" PostTypeId="2" ParentId="636" CreationDate="2010-07-28T05:57:09.560" Score="9" ViewCount="" Body="&lt;p&gt;I was recently surprised to discover that it's actually not known. The number of &lt;em&gt;closed&lt;/em&gt; knight's tours (cyclic) was computed in the 1990s, using &lt;a href=&quot;http://en.wikipedia.org/wiki/Binary_decision_diagram&quot; rel=&quot;nofollow&quot;&gt;Binary decision diagrams&lt;/a&gt;. There are 26,534,728,821,064 closed directed knight's tours, and the number of undirected ones is half that or 13,267,364,410,532. If you count equivalence classes under rotation and reflection, there are slightly more than 1/8th of that: 1,658,420,855,433.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Loebbing and  Wegener (1996) wrote a paper &lt;a href=&quot;http://www.combinatorics.org/Volume_3/Abstracts/v3i1r5.html&quot; rel=&quot;nofollow&quot;&gt;&quot;The Number of Knight's Tours Equals 33,439,123,484,294 — Counting with Binary Decision Diagrams&quot;&lt;/a&gt;; the number in the title in the mistake, as they pointed out in &lt;a href=&quot;http://www.combinatorics.org/Volume_3/Comments/v3i1r5.html&quot; rel=&quot;nofollow&quot;&gt;a comment&lt;/a&gt; to their paper. Brendon McKay independently &lt;a href=&quot;http://www.combinatorics.org/Volume_3/Comments/v3i1r5.01.ps&quot; rel=&quot;nofollow&quot;&gt;computed&lt;/a&gt; the correct number with another method, and the original authors seem to have later &lt;a href=&quot;http://books.google.com/books?id=-DZjVz9E4f8C&amp;amp;pg=PA369&amp;amp;dq=532&quot; rel=&quot;nofollow&quot;&gt;found&lt;/a&gt; the same answer.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finding the exact number of open tours (not cyclic/reentrant) remains &lt;a href=&quot;http://www.ktn.freeuk.com/pa.htm&quot; rel=&quot;nofollow&quot;&gt;open, but it is estimated&lt;/a&gt; to be about 10&lt;sup&gt;15&lt;/sup&gt; or 2×10&lt;sup&gt;16&lt;/sup&gt;.&lt;/p&gt;&#xA;" OwnerUserId="205" LastEditorUserId="205" LastEditorDisplayName="" LastEditDate="2011-01-26T03:11:09.247" LastActivityDate="2011-01-26T03:11:09.247" CommentCount="2" />
  <row Id="920" PostTypeId="2" ParentId="20" CreationDate="2010-07-28T06:04:47.113" Score="4" ViewCount="" Body="&lt;p&gt;I think you were being a little too hard on Isaac.  The truth is that the real numbers are a sophisticated mathematical construction and that any explanation of what they &quot;are&quot; which pretends otherwise is a convenient fiction.  Mathematicians need these kind of sophisticated constructions because they are what is required for rigorous proofs.  Before people explicitly constructed the real numbers and used them to define and prove things about other concepts, it was never totally clear what was true or what was false, and everybody was very confused.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, Cantor proved that the number of points in the plane is the same as the number of  points on a line.  Many people thought that this was impossible before he did it; they had an intuition that you couldn't possibly &quot;fit&quot; the plane into the line.  More generally, people were pretty sure you couldn't fit $\mathbb{R}^n$ into $\mathbb{R}^m$ if $n$ was greater than $m$.  It wasn't until quite a bit later that mathematicians formalized and proved a rigorous mathematical statement which justified this intuition called &lt;a href=&quot;http://en.wikipedia.org/wiki/Invariance_of_domain&quot; rel=&quot;nofollow&quot;&gt;invariance of domain&lt;/a&gt;, which says you can't do this in a &lt;em&gt;continuous&lt;/em&gt; way.  One of the many mathematical constructions you need to even state this theorem is the construction of the real numbers.  (Another is a formal definition of what &quot;continuous&quot; means, but one thing at a time.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, what are the real numbers?  They are a formal way to fill &quot;holes&quot; in the rational numbers, which is necessary for all sorts of things.  The most basic thing they are necessary for is doing geometry.  You probably know that the square root of 2 is irrational.  What this means is that it is impossible to think about the diagonal of a square as being the same kind of object as the sides of a square while only using rational numbers.  But you can rotate a diagonal, and it looks just like the side of a square, only a bit longer.  So you'd like a number system in which you can sensibly talk about any number you can construct geometrically.  You'd also like to be able to talk about rotation!  You can't do that with just rational numbers, either.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So how do you fill in enough holes to do geometry?  &lt;a href=&quot;http://en.wikipedia.org/wiki/Dedekind_cut&quot; rel=&quot;nofollow&quot;&gt;Dedekind&lt;/a&gt; came up with a very clever way to do this.  It starts by observing that a rational number $q$ is completely determined by the set of rational numbers greater than it and the set of rational numbers less than it.  For example, 1/2 is completely determined by the fact that it's always between 1/2 + 1/n and 1/2 - 1/n.  (For the initiated, this is a special case of the Yoneda lemma.)  But there are &quot;numbers,&quot; such as the square root of 2, which aren't rational, and yet have the property that we can always tell what rational numbers are greater than it and what rational numbers are less than it.  For the square root of 2, these are precisely the fractions p/q such that 2q^2 &amp;lt; p^2 and such that 2q^2 &gt; p^2, respectively.  Dedekind's brilliant idea was the following:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;em&gt;Define&lt;/em&gt; a real number to be a partition of the rational numbers!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In Dedekind's construction, the square root of 2 quite literally &lt;em&gt;is&lt;/em&gt; the set of rational numbers that are greater than it, and the set of rational numbers that are less than it. You can define all the usual arithmetic operations on these &quot;numbers,&quot; called Dedekind cuts, and prove all the wonderful theorems you'll find in a standard book on real analysis.  In particular, the property that guarantees that all the holes are filled is called &lt;a href=&quot;http://en.wikipedia.org/wiki/Real_number#Completeness&quot; rel=&quot;nofollow&quot;&gt;completeness&lt;/a&gt;.   &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Figured I might as well add something about the complex numbers.  The story here is beautiful, and if you're really interested you should check out Tristan Needham's &lt;a href=&quot;http://usf.usfca.edu/vca//&quot; rel=&quot;nofollow&quot;&gt;Visual Complex Analysis&lt;/a&gt;.  Some people say that the point of the complex numbers is to let you solve polynomials, but this is really selling them short.  The complex numbers are an inherently &lt;em&gt;geometric&lt;/em&gt; construction, and should be understood as such.  Their geometry and topology just happens to be responsible for the fact that you can solve polynomials with them, but it's also responsible for much more.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a quick sketch.  Now that you've got the real numbers on your hands, you can rigorously talk about plane geometry.  In plane geometry, an important notion is that of &lt;em&gt;similarity&lt;/em&gt;.  Informally, two figures are similar if they have the same shape.  More formally, two figures are similar if you can rotate, translate, and scale one figure so that it matches up with the other.  So similarity is all about a certain collection of transformations of the points in the plane.  It was &lt;a href=&quot;http://en.wikipedia.org/wiki/Erlangen_program&quot; rel=&quot;nofollow&quot;&gt;Klein&lt;/a&gt; who first realized that the important features of different flavors of &quot;geometry&quot; are captured in what kind of transformations are allowed.  So to do geometry the modern way we should focus our attention on these transformations, which form a &lt;a href=&quot;http://en.wikipedia.org/wiki/Group_theory&quot; rel=&quot;nofollow&quot;&gt;group&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To make this easier, let's ignore the translations for now.  We'll pick an origin for our plane, and we'll only allow rotations and scalings about this origin.  Rotations and scalings have the property that they are both linear transformations; this means that if you know what the transformation does to two points $u, v$, you also know what it does to the vector sum $u + v$.  In particular, a linear transformation is determined by what it does to the point $(1, 0)$ and to the point $(0, 1)$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, rotations and scalings satisfy an extra property: they are, in fact, determined by what they do to the point $(1, 0)$.  This is because $(0, 1)$ can be obtained from $(1, 0)$ by a rotation by 90 degrees, and rotations and scalings &lt;em&gt;commute&lt;/em&gt; with each other: if you rotate x degrees then y degrees, that's the same as rotating y degrees then x degrees, which is the same as rotating x+y degrees.  Similarly, if you rotate x degrees then scale by 2, that's the same as scaling by 2, then rotating x degrees.  So if you know what a rotation-and-scaling does to $(1, 0)$, you just rotate that vector by 90 degrees, and you know what it did to $(0, 1)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So to every rotation-and-scaling, we can assign two real numbers: the coordinates of the image of the point $(1, 0)$.  In general, a rotation by $\theta$ angles followed by a scaling by $r$ sends $(1, 0)$ to $(r \cos \theta, r \sin \theta)$.  A different transformation, say a rotation by $\phi$ angles followed by a scaling by $s$, sends $(1, 0)$ to $(s \cos \phi, s \sin \phi)$.  And their &lt;em&gt;composition&lt;/em&gt; sends $(1, 0)$ to $(rs \cos (\theta + \phi), rs \sin (\theta + \phi))$.  In other words, composition of rotations-and-scalings defines a multiplication law on pairs of real numbers.  What is this law, exactly?  Well, by the angle addition formulas, it's&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(a, b) * (c, d) = (ac - bd, ad + bc).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And this is precisely the rule for multiplication in the complex numbers, where $(a, b)$ corresponds to $a + bi$.  You get the rule for addition by observing that not only can you compose two rotations-and-scalings, you can also add their results. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Together, the real numbers and the complex numbers provide a foundation for much of modern mathematics and physics.  For example, the complex numbers turn out (for reasons which are still not well understood) to be fundamental in the description of quantum mechanics.&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-07-28T06:35:01.467" LastActivityDate="2010-07-28T06:35:01.467" />
  <row Id="921" PostTypeId="1" AcceptedAnswerId="926" CreationDate="2010-07-28T06:58:36.537" Score="-1" ViewCount="506" Body="&lt;p&gt;Mathematics is beautiful. Many numbers are interesting such as &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1729 = 1&lt;sup&gt;3&lt;/sup&gt; + 12&lt;sup&gt;3&lt;/sup&gt; = 9&lt;sup&gt;3&lt;/sup&gt; + 10&lt;sup&gt;3&lt;/sup&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please list out the interesting numbers and reasons to make number interesting&lt;/p&gt;&#xA;" OwnerUserId="327" LastEditorUserId="16" LastEditorDisplayName="" LastEditDate="2010-07-28T17:20:37.613" LastActivityDate="2010-07-29T20:38:30.733" CommunityOwnedDate="2010-07-28T06:58:53.930" ClosedDate="2010-07-30T02:30:08.647" Title="Interesting numbers in Maths" Tags="&lt;big-list&gt;&lt;recreational-mathematics&gt;" AnswerCount="4" CommentCount="5" FavoriteCount="1" />
  <row Id="923" PostTypeId="2" ParentId="20" CreationDate="2010-07-28T07:04:41.780" Score="4" ViewCount="" Body="&lt;p&gt;The &lt;em&gt;natural numbers&lt;/em&gt; are defined by the Peano axioms, as in the &lt;a href=&quot;http://math.stackexchange.com/questions/20/can-you-explain-what-a-real-number-is-also-rational-decimal-integer-natural/79#79&quot;&gt;answer of Isaac&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can also view the natural numbers as the cardinalities of finite sets, which implies that zero is a natural number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the other number domains arise because mathematicians want to give values for certain operations which otherwise are only defined partially. Alternatively, they want to give solutions to equations of certain forms, which cannot always be solved in the smaller domain.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;integers&lt;/em&gt; arise from subtraction (solutions of &lt;code&gt;a + x = b&lt;/code&gt;)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;rational numbers&lt;/em&gt; arise from division (solutions of &lt;code&gt;a * x = b&lt;/code&gt;)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;real numbers&lt;/em&gt; arise from taking limits or upper bounds&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;complex numbers&lt;/em&gt; arise from taking roots (solutions of polynomial equations)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;cardinal numbers&lt;/em&gt; arise as the sizes of sets. They can be constructed as isomorphism classes of sets, e.g. the number one would then be the class of all sets containing one element. They are useful for talking about the different sizes of infinite sets, which can get extremely large. Some set theorists compete over who can define the largest cardinal.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;ordinal numbers&lt;/em&gt; arise as the sizes of well-orderings. They can be constructed as isomorphism classes of well-orderings, e.g. the number two would then be the class of all well-orderings containing two elements (with the first smaller than the second in the ordering). Interestingly, the ordinals are themselves well-ordered in a single well-ordering, so they come in a well-defined sequence, and an ordinal number can be viewed as describing its position in this sequence.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="210" LastEditorUserId="210" LastEditorDisplayName="" LastEditDate="2010-07-29T19:29:08.340" LastActivityDate="2010-07-29T19:29:08.340" CommentCount="2" />
  <row Id="924" PostTypeId="2" ParentId="907" CreationDate="2010-07-28T07:12:25.707" Score="12" ViewCount="" Body="&lt;p&gt;Dear Zev,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your interpretation of how &quot;in the sequel&quot; is used in mathematical literature is correct: it means &quot;in what follows&quot;, &quot;in the remainder of the present text&quot;, &quot;from now on&quot;...As you can see, there are many other such expressions, and I don't think that &quot;in the sequel&quot; has any particular nuance of meaning that these other phrasings lack.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On to your question of whether this is acceptable usage in a math paper: acceptable, yes, but not completely recommended.  In more detail:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ths phrase is, as you point out, rather common, so any experienced reader of math will have seen it before.  Also its meaning should be relatively transparent to a literate native speaker of English: this is after all what the dictionary says that &quot;sequel&quot; means.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, there are two drawbacks.  First, as Mariano says, nowadays we hear &quot;sequel&quot; used most commonly for the next movie in a franchise, so its use in a math paper will inevitably make some readers think that you are referring to &lt;strong&gt;[TITLE OF YOUR PAPER] Part II: This Time, It's Personal&lt;/strong&gt;, or whatever.  Second, in my opinion it is somewhat hackneyed language and an idiomatic usage that doesn't add any meaning.  If you read enough math papers you'll find that there are certain linguistic ticks that people pick up from each other (and also sometimes, from writing in other languages), e.g. beginning a sentence with &quot;Remark that...&quot; sounds a little stilted and is probably a semiconscious translation of the french phrase &quot;Remarquer que&quot; (which I would translate as &quot;Notice that&quot;).  Most people agree that good mathematical writing is as stylistically unobstrusive as possible, so using phrases that make people ponder and ask questions about them on websites is, arguably, slightly too distracting.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, it's  no big deal.  If I were a referee or even a copyeditor, I would almost certainly let the author use &quot;in the sequel&quot; if she wants.&lt;/p&gt;&#xA;" OwnerUserId="299" LastActivityDate="2010-07-28T07:12:25.707" CommentCount="3" />
  <row Id="925" PostTypeId="2" ParentId="921" CreationDate="2010-07-28T07:18:03.370" Score="3" ViewCount="" Body="&lt;p&gt;$163$ is interesting because it's the largest &lt;a href=&quot;http://en.wikipedia.org/wiki/Heegner_number&quot; rel=&quot;nofollow&quot;&gt;Heegner number&lt;/a&gt;.  This means that it's the largest positive integer $d$ such that the ring of integers in $\mathbb{Q}(\sqrt{-d})$ has unique factorization.  If you don't know what that means, two other elementary consequences of $163$ being a Heegner number are that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$n^2 + n + 41$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is prime for $n = 0, 1, 2, ... 39$ and that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$e^{\pi \sqrt{163}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is &lt;em&gt;extremely&lt;/em&gt; close to an integer.&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-07-29T20:38:30.733" LastActivityDate="2010-07-29T20:38:30.733" CommunityOwnedDate="2010-07-28T07:18:03.370" />
  <row Id="926" PostTypeId="2" ParentId="921" CreationDate="2010-07-28T07:26:24.550" Score="11" ViewCount="" Body="&lt;p&gt;All positive integers are interesting. Suppose they weren't. Let k be the lowest non-interesting positive integer. That is an interesting property. Therefore contradiction. Generalisation is left as an exercise to the reader.&lt;/p&gt;&#xA;" OwnerUserId="123" LastEditorUserId="123" LastEditorDisplayName="" LastEditDate="2010-07-28T08:22:33.930" LastActivityDate="2010-07-28T08:22:33.930" CommunityOwnedDate="2010-07-28T07:26:24.550" CommentCount="3" />
  <row Id="927" PostTypeId="2" ParentId="921" CreationDate="2010-07-28T07:27:17.107" Score="4" ViewCount="" Body="&lt;p&gt;$196883$ is interesting because it's the smallest dimension of a faithful complex representation of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Monster_group&quot; rel=&quot;nofollow&quot;&gt;Monster group&lt;/a&gt;.  Roughly speaking, a group is a collection of symmetries of something.  A simple group is a group which can't be decomposed further into smaller groups; to borrow a phrase, it's an &quot;atom of symmetry.&quot;  There is a &lt;a href=&quot;http://en.wikipedia.org/wiki/Classification_of_finite_simple_groups&quot; rel=&quot;nofollow&quot;&gt;classification&lt;/a&gt; of these atoms with several infinite families and 26 exceptions, and the Monster group is the largest of these exceptions.  And the simplest way to write it down is using $196883$-by-$196883$ matrices.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$196883$ was crucial to McKay's discovery of &lt;a href=&quot;http://en.wikipedia.org/wiki/Monster_group#Moonshine&quot; rel=&quot;nofollow&quot;&gt;Monstrous moonshine&lt;/a&gt;, which began with the observation that the &lt;a href=&quot;http://en.wikipedia.org/wiki/J-invariant&quot; rel=&quot;nofollow&quot;&gt;j-invariant&lt;/a&gt;, an important function in number theory, has a Taylor series in which $196884$ appears.  The story of this discovery is explained in Mark Ronan's wonderful book &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0192807226&quot; rel=&quot;nofollow&quot;&gt;Symmetry and the Monster&lt;/a&gt;, which I highly recommend.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-07-28T07:27:17.107" CommunityOwnedDate="2010-07-28T07:27:17.107" CommentCount="1" />
  <row Id="928" PostTypeId="2" ParentId="894" CreationDate="2010-07-28T08:01:11.347" Score="6" ViewCount="" Body="&lt;p&gt;We can define formula $P_i$ that says &quot;there are at most $i$ elements satisfying $P$&quot;. Now, if the infinite disjunction of the $P_i$ was definable in FO, it would (by compactness) imply a conjunction of some finite subset of the $P_i$, hence it would imply $P_i$ for some $i$. That is not true, if $P$ can have (say) $i+1$ elements satisfying it.&lt;/p&gt;&#xA;" OwnerUserId="236" LastEditorUserId="236" LastEditorDisplayName="" LastEditDate="2010-07-29T12:37:26.077" LastActivityDate="2010-07-29T12:37:26.077" CommentCount="1" />
  <row Id="929" PostTypeId="2" ParentId="918" CreationDate="2010-07-28T08:25:13.880" Score="7" ViewCount="" Body="&lt;p&gt;Fix two elements s&lt;sub&gt;1&lt;/sub&gt;,s&lt;sub&gt;2&lt;/sub&gt;&amp;isin;S and divide subsets of S into two parts (subsets of S containing only s&lt;sub&gt;2&lt;/sub&gt;)&amp;cup;(subsets of S which contains s&lt;sub&gt;1&lt;/sub&gt; if they contain s&lt;sub&gt;2&lt;/sub&gt;). The second part contains equal number of sets for all reminders mod 3 (because Z/3 acts there adding s&lt;sub&gt;1&lt;/sub&gt;, then s&lt;sub&gt;2&lt;/sub&gt;, then removing both of them) -- namely, 2&lt;sup&gt;n-2&lt;/sup&gt;. And for the first part we have a bijection with subsets &lt;em&gt;(edit: with 2 mod 3 elements)&lt;/em&gt; of a set with (n-2) elements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we get a recurrence relation that gives an answer 2&lt;sup&gt;n-2&lt;/sup&gt;+2&lt;sup&gt;n-4&lt;/sup&gt;+... -- i.e. (2&lt;sup&gt;n&lt;/sup&gt;-1):3 for even and (2&lt;sup&gt;n&lt;/sup&gt;-2):3 for odd n.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Errata.&lt;/strong&gt; For n=0,1,5 mod 6 one should add &quot;+1&quot; to the answer from the previous paragraph (e.g. for n=6 the correct answer is 1+20+1=22 and not 21).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me try to rephrase the solution to make it obvious. For n=2k divide S on k pairs and consider an action of a group Z/3Z on each pair described in a first paragraph. We get an action of (Z/3Z)&lt;sup&gt;k&lt;/sup&gt; on subsets of S, and after removal of it's only fixed point (k-element set consisting of second points from each pair) we get a bijection between subsets which have 0, 1 or 2 elements mod 3. So there are (2&lt;sup&gt;n&lt;/sup&gt;-1):3 sets with i mod 3 elements excluding the fixed point &lt;em&gt;and&lt;/em&gt; to count that point one should add &quot;+1&quot; for i=k mod 3.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And for n=2k+1 there are 2 fixed points&amp;nbsp;&amp;mdash; including or not (2k+1)-th element of S&amp;nbsp;&amp;mdash; with k+1 and k elements respectively.&lt;/p&gt;&#xA;" OwnerUserId="152" LastEditorUserId="152" LastEditorDisplayName="" LastEditDate="2010-07-31T13:46:57.907" LastActivityDate="2010-07-31T13:46:57.907" CommentCount="4" />
  <row Id="930" PostTypeId="2" ParentId="855" CreationDate="2010-07-28T09:09:28.857" Score="10" ViewCount="" Body="&lt;p&gt;This is somewhere between an answer and commentary.  As others have said, the question is equivalent to showing: for any prime $p &amp;gt; 3$, $p^2 \equiv 1 \pmod 3$ and $p^2 \equiv 1 \pmod 8$.  Both of these statements are straightforward to show by just looking at the $\varphi(3) = 2$ reduced residue classes modulo $3$ and the $\varphi(8) = 4$ reduced residue classes modulo $8$.  But what is their significance?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a positive integer $n$, let $U(n) = (\mathbb{Z}/n\mathbb{Z})^{\times}$ be the multiplicative group of units (&quot;reduced residues&quot;) modulo $n$.  Like any abelian group $G$, we have a squaring map&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$[2]: G \rightarrow G$, $g \mapsto g^2$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the image of which is the set of squares in $G$.  So, the question is equivalent to: for $n = 3$ and also $n = 8$, the subgroup of squares in $U(n)$ is the trivial group.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The group $U(3) = \{ \pm 1\}$ has order $2$; since $(-1)^2 = 1$, the fact that the subgroup of squares is equal to $1$ is pretty clear.  But more generally, for any odd prime $p$, the squaring map $[2]$ on $U(p)$ is two-to-one onto its image -- an element of a field has no more than two square roots -- so that precisely half of the elements of $U(p)$ are squares.  It turns out that when $p = 3$, half of $p-1$ is $1$, but of course this is somewhat unusual: it doesn't happen for any other odd prime $p$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The group $U(8) = \{1,3,5,7\}$ has order $4$.  By analogy to the case of $U(p)$, one might expect the squaring map to be two-to-one onto its image so that exactly half of the elements are squares.  But that is not what is happening here: indeed &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$1^2 \equiv 3^2 \equiv 5^2 \equiv 7^2 \equiv 1 \pmod 8$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so the subgroup of squares is again trivial.  What's different?  Since $\mathbb{Z}/8\mathbb{Z}$ is not a field, it is legal for a given element to have more than two square roots, but a more insightful answer comes from the structure of the groups $U(n)$.  For any odd prime $p$, the group $U(p)$ is &lt;em&gt;cyclic&lt;/em&gt; of order $p-1$ (&quot;existence of primitive roots&quot;).  It is easy to see that in any cyclic group of even order, exactly half of the elements are squares.  So $U(8)$ must not be cyclic, so it must be the other abelian group of order $4$, i.e., isomorphic to the Klein $4$-group $C_2 \times C_2$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;More generally, if $p$ is an odd prime number and $a$ is a positive integer, then &#xA;$U(p^a)$ is cyclic of order $p^{a-1}(p-1)$ hence isomorphic to $C_{p^{a-1}} \times C_{p-1}$, whereas for any $a \geq 2$, the group $U(2^a)$ is isomorphic to $C_{2^{a-2}} \times C_2$.  This is one of the first signs in number theory &quot;there is something odd about the prime $2$&quot;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Added&lt;/b&gt;: Note that the above considerations allow us to answer the more general question: &quot;What is the largest positive integer $N$ such that for all primes $p$ with $\operatorname{gcd}(p,N) = 1$, $N$ divides $(p^2-1)$?&quot;  (Answer: $N = 24$.)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Added Later&lt;/b&gt;: I just saw &lt;a href=&quot;http://arxiv.org/PS_cache/arxiv/pdf/1104/1104.5052v1.pdf&quot; rel=&quot;nofollow&quot;&gt;this arxiv preprint&lt;/a&gt; which is entirely devoted to the observation made in the previous paragraph.  I guess the author does not follow this site...&lt;/p&gt;&#xA;" OwnerUserId="299" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2011-05-17T03:21:34.013" LastActivityDate="2011-05-17T03:21:34.013" CommentCount="2" />
  <row Id="931" PostTypeId="2" ParentId="8" CreationDate="2010-07-28T09:21:32.260" Score="6" ViewCount="" Body="&lt;p&gt;Also you can use the &lt;a href=&quot;http://en.wikipedia.org/wiki/Fibonacci_number#Matrix%20form&quot; rel=&quot;nofollow&quot;&gt;matrix equation&lt;/a&gt; for fibonacci numbers:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&#xA;       n &#xA;[ 1 1 ] = [ F_{n+1} F_{n}   ]&#xA;[ 1 0 ] = [ F_{n}   F_{n-1} ]&#xA;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;To calculate $n$-th power of the matrix you can use exponentiation by squaring algorithm:&#xA;_http://en.wikipedia.org/wiki/Exponentiation_by_squaring&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This approach could also be generalized on the case of arbitrary sequence with linear recurrence relation.&lt;/p&gt;&#xA;" OwnerUserId="334" LastActivityDate="2010-07-28T09:21:32.260" CommentCount="1" />
  <row Id="932" PostTypeId="2" ParentId="59" CreationDate="2010-07-28T09:43:00.053" Score="1" ViewCount="" Body="&lt;p&gt;In case you want to implement it on some programming language.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Usually there is a function called like &lt;b&gt;atan2(y, x)&lt;/b&gt; which returns oriented angle between points (x, y) and (1, 0). In that case use could use&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&#xA;atan2(vector product, scalar product).&#xA;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This is usually more stable than using just &lt;b&gt;arccos&lt;/b&gt; or &lt;b&gt;arcsin&lt;/b&gt;.&lt;/p&gt;&#xA;" OwnerUserId="334" LastActivityDate="2010-07-28T09:43:00.053" />
  <row Id="933" PostTypeId="2" ParentId="806" CreationDate="2010-07-28T09:45:24.453" Score="9" ViewCount="" Body="&lt;p&gt;As others have said, often what you want for a particular Diophantine application is that the class number of a certain number field be relatively prime to a certain number.  The famous example of this (as already noted by others) is Kummer's Theorem that for an odd prime $p$, the Fermat equation $x^p + y^p = z^p$ has no integer solutions with $xyz \neq 0$ if the ring of integers of $\mathbb{Q}[e^{\frac{2 \pi i}{p}}]$ has class number prime to $p$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another -- simpler -- nice example is the &lt;strong&gt;Mordell equation&lt;/strong&gt; $y^2 + k = x^3$.  If $k \equiv 1,2 \pmod 4$ and the ring $\mathbb{Z}[\sqrt{-k}]$ has class number prime to $3$, then all of the integer solutions to the Mordell equation can be found.  See Section 4 of&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://math.uga.edu/~pete/4400MordellEquation.pdf&quot; rel=&quot;nofollow&quot;&gt;http://math.uga.edu/~pete/4400MordellEquation.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for an exposition of this which is (I hope) reasonably elementary and accessible to undergraduates. &lt;/p&gt;&#xA;" OwnerUserId="299" LastActivityDate="2010-07-28T09:45:24.453" />
  <row Id="934" PostTypeId="2" ParentId="8" CreationDate="2010-07-28T09:52:46.637" Score="1" ViewCount="" Body="&lt;p&gt;To expand on falagar's answer, my favourite proof of Binet's formula:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;...Which I was going to post a summary of here, but remembered that everything was awful without Tex, so here is a link to &lt;a href=&quot;http://livetoad.org/Courses/Documents/132d/Notes/fibonacci_matrix.pdf&quot; rel=&quot;nofollow&quot;&gt;some notes on it I found on google&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The basic idea is to treat pairs of fibonnacci numbers, adjacent in the sequence, as vectors. Moving on to the next adjacent pair induces a linear transformation not unlike that of the matrix falagar posted. Calculating eigenvalues and eigenvectors can give a complete prediction of where an initial vector will find itself, predicting the whole sequence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's quite a lot of work but I think it's rather illuminating.&lt;/p&gt;&#xA;" OwnerUserId="160" LastActivityDate="2010-07-28T09:52:46.637" />
  <row Id="935" PostTypeId="2" ParentId="594" CreationDate="2010-07-28T10:21:41.717" Score="2" ViewCount="" Body="&lt;p&gt;Here is another proof without complex numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We start with proving that there exists $z \in \mathbb{N}$ such that $z^2 + 1 \equiv 0 \pmod p$. We do this in the same way as Akhil Mathew.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let we have $a^2 + b^2 = pm$. Take $x$ and $y$ such that&lt;br /&gt;&#xA;$x \equiv a \pmod m$ and&lt;br /&gt;&#xA;$y \equiv b \pmod m$ and&lt;br /&gt;&#xA;$x, y \in [-m/2, m/2)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider $u = ax + by$ and $v = ay - bx$. Then $u^2 + v^2 = (a^2 + b^2)(x^2 + y^2)$.&#xA;Moreover, $u$ and $v$ are multiples of $m$.&#xA;Hence $(u/m)^2 + (v/m)^2 = p (x^2 + y^2)/m$. $(x^2 + y^2)/m$ is an integer because of the definition of $x$ and $y$ and that $a^2 + b^2 = pm$.&lt;br /&gt;&#xA;Also $(x^2 + y^2)/m$ is less than $m/2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we change $a$ by $u$ and $b$ by $v$ and continue this process until we get $m=1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice that this is quite efficient way to find representation of $p$ as a sum of two squares - it takes $O(\log p)$ steps to find it provided we have found $z$ such that $z^2 + 1$ is multiple of $p$.&lt;/p&gt;&#xA;" OwnerUserId="334" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-08-09T15:55:17.210" LastActivityDate="2010-08-09T15:55:17.210" CommentCount="1" />
  <row Id="936" PostTypeId="1" AcceptedAnswerId="937" CreationDate="2010-07-28T10:41:41.603" Score="1" ViewCount="219" Body="&lt;p&gt;I am looking for functions and/or constants that when being integrated from minus infinity to infinity produce 1. I think the Dirac delta function is one example but perhaps there are some more? References on useful material is also greatly appreciated.&lt;/p&gt;&#xA;" OwnerUserId="346" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-07-28T14:37:18.693" LastActivityDate="2010-07-28T15:03:54.620" CommunityOwnedDate="2010-08-03T18:47:52.520" Title="Improper Integrals from -oo to oo with solution 1" Tags="&lt;big-list&gt;&lt;calculus&gt;&lt;analysis&gt;" AnswerCount="7" CommentCount="1" />
  <row Id="937" PostTypeId="2" ParentId="936" CreationDate="2010-07-28T10:47:25.527" Score="6" ViewCount="" Body="&lt;p&gt;Any integrable functions that gives a finite nonzero answer can be modified to suit your need.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose ∫ f(x) dx = A, then let g(x) = f(x)/A, automatically we have ∫ g(x) dx = A/A = 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Actually, all continuous &lt;a href=&quot;http://en.wikipedia.org/wiki/List_of_probability_distributions#Continuous_distributions&quot; rel=&quot;nofollow&quot;&gt;probability distribution function&lt;/a&gt; must have this property.)&lt;/p&gt;&#xA;" OwnerUserId="171" LastActivityDate="2010-07-28T10:47:25.527" CommunityOwnedDate="2010-08-03T18:47:52.520" CommentCount="2" />
  <row Id="938" PostTypeId="2" ParentId="653" CreationDate="2010-07-28T10:47:27.460" Score="1" ViewCount="" Body="&lt;p&gt;Homotopy theory is useful in quantum mechanics when (for instance) talking about manifolds of Hamiltonians.  You might have a collection of Hamiltonians that depend continuously on some parameters $p_1, p_2, \ldots, p_n$ such that the matrix representation of the Hamiltonian is periodic in some subset of the parameters $p_{r_1}, p_{r_2}, \ldots, p_{r_m}$.  From this, the fundamental group of the manifold of Hamiltonians can be computed, which has some physical ramifications.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A concrete example of this would be the Hamiltonian that describes the quantum hall effect.  The quantum hall effect is the phenomena that the resistance in a 2-dimensional substrate exposed to a perpendicular electric field at close to zero temperature is quantized.  In condensed matter physics there is a notion called quasimomentum that can be thought of as being related to momentum but is a bit different.  We need something different from the classical definition of momentum because the classical definition depends on translation invariance, and in a crystal there is only discrete translation invariance.  In the quantum hall effect, we have two quasimomenta: $k_x, and k_y$, corresponding to quasimomentum in the $x$ and $y$ directions.  The Hamiltonian is periodic in both of these parameters, leading to a fundamental group of the manifold of Hamiltonians of $\mathbb{Z}^2$, i.e. the fundamental group of the torus $T^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's a lot more to the topic than this.  This paper by Avron, Seiler, and Simon has more details:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://prl.aps.org/abstract/PRL/v51/i1/p51_1&quot; rel=&quot;nofollow&quot;&gt;Homotopy and Quantization in Condensed Matter Physics&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="352" LastActivityDate="2010-07-28T10:47:27.460" CommentCount="4" />
  <row Id="939" PostTypeId="1" AcceptedAnswerId="956" CreationDate="2010-07-28T10:59:44.590" Score="2" ViewCount="278" Body="&lt;p&gt;Let $pi(x)$ be the number of primes not greater than $x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/w/index.php?title=Prime-counting_function&quot; rel=&quot;nofollow&quot;&gt;Wikipedia article&lt;/a&gt; says that $pi(10^{23}) = 1,925,320,391,606,803,968,923$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is how to calculate $pi(x)$ for large $x$ in a reasonable time? What algorithms do exist for that?&lt;/p&gt;&#xA;" OwnerUserId="334" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-09-02T04:33:06.937" LastActivityDate="2010-09-02T04:33:06.937" Title="Counting primes" Tags="&lt;number-theory&gt;&lt;algorithms&gt;&lt;prime-numbers&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="940" PostTypeId="2" ParentId="939" CreationDate="2010-07-28T11:04:33.520" Score="1" ViewCount="" Body="&lt;p&gt;You can use &lt;a href=&quot;http://en.wikipedia.org/wiki/Inclusion%E2%80%93exclusion_principle&quot; rel=&quot;nofollow&quot;&gt;inclusion exclusion principle&lt;/a&gt; to get a boost over the &lt;a href=&quot;http://en.wikipedia.org/wiki/Sieve_of_Eratosthenes&quot; rel=&quot;nofollow&quot;&gt;Eratosthenes sieve&lt;/a&gt;&lt;/p&gt;&#xA;" LastEditorUserId="123" LastEditorDisplayName="" LastEditDate="2010-07-28T11:19:11.453" LastActivityDate="2010-07-28T11:19:11.453" CommentCount="1" />
  <row Id="941" PostTypeId="1" AcceptedAnswerId="945" CreationDate="2010-07-28T11:23:15.767" Score="0" ViewCount="593" Body="&lt;p&gt;This could be a trivial question, but what is exactly the difference of between these two expressions? Am I correct to state the both interchangeably whenever I need to express the approximation of pi? I'm bit confused as &lt;a href=&quot;http://en.wikipedia.org/wiki/Approximation#Mathematics&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; it states pi can be express by ≒ as it's not a rational number, but pi can also be expressed by a series (asymptotic), so it should be ≈ as well.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;π ≈ 3.14..&#xA;π ≒ 3.14..&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="102" LastEditorUserId="102" LastEditorDisplayName="" LastEditDate="2010-07-28T11:48:02.467" LastActivityDate="2010-07-29T05:43:06.793" Title="Approximation symbol: Is π ≈ 3.14.. equal to π ≒ 3.14.. ?" Tags="&lt;definition&gt;&lt;approximation&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" />
  <row Id="942" PostTypeId="1" AcceptedAnswerId="984" CreationDate="2010-07-28T11:41:32.120" Score="5" ViewCount="346" Body="&lt;p&gt;This is a question in Liu's book.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $X$ be a quasi-compact scheme. Show that $X$ contains a closed point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well I'm unable to do this question, so any help would be appreciated. This question also makes me curious to know about the meaning/use of closed points of a scheme in general - by that I mean a scheme which is not an algebraic variety/local scheme over a field, which has a geometric meaning. Thanks!&lt;/p&gt;&#xA;" OwnerUserId="325" LastEditorUserId="16" LastEditorDisplayName="" LastEditDate="2010-07-28T19:47:33.367" LastActivityDate="2011-02-01T11:30:49.857" Title="Meaning of closed points of a scheme" Tags="&lt;intuition&gt;&lt;algebraic-geometry&gt;" AnswerCount="5" FavoriteCount="3" />
  <row Id="944" PostTypeId="1" AcceptedAnswerId="950" CreationDate="2010-07-28T11:49:37.720" Score="3" ViewCount="439" Body="&lt;p&gt;Consider the following assertion from &lt;a href=&quot;http://scottaaronson.com/blog/?p=152#&quot; rel=&quot;nofollow&quot;&gt;Scott Aaronson's blog&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Supposing you do prove the Riemann&#xA;  Hypothesis, it’s possible to convince&#xA;  someone of that fact, without&#xA;  revealing anything other than the fact&#xA;  that you proved it. It’s also possible&#xA;  to write the proof down in such a way&#xA;  that someone else could verify it,&#xA;  with very high confidence, having only&#xA;  seen 10 or 20 bits of the proof.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Can anyone explain where this result comes from?&lt;/p&gt;&#xA;" OwnerUserId="123" LastEditorUserId="536" LastEditorDisplayName="" LastEditDate="2010-07-28T11:59:11.490" LastActivityDate="2010-08-03T03:54:04.270" Title="Proving the Riemann Hypothesis without revealing anything other than you proved it" Tags="&lt;computer-science&gt;&lt;riemann-hypothesis&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="3" />
  <row Id="945" PostTypeId="2" ParentId="941" CreationDate="2010-07-28T11:54:49.487" Score="9" ViewCount="" Body="&lt;p&gt;Any mathematical notation is ok as long as it is common knowledge in your community. For instance, I believe I fully understand the meaning of the $\approx$ symbol. However, I haven't ever seen the second symbol you provided.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To be on the sure side you should provide a definition of any relation symbol you don't consider to be common knowledge. This may happen as a short remark (&quot;..., where $\approx$ denotes ...&quot;) or maybe as a table of the used symbols in the front matter of your work. As with any definition in mathematics, there is no right or wrong in the symbol/notion/etc. you use, only proper or unsound definitions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also: When in doubt, use the symbol that is used more commonly in the standard textbooks of your field. There is no benefit in being avant-garde at notation.&lt;/p&gt;&#xA;" OwnerUserId="347" LastActivityDate="2010-07-28T11:54:49.487" />
  <row Id="946" PostTypeId="1" CreationDate="2010-07-28T11:59:11.490" Score="12" ViewCount="246" Body="&lt;p&gt;Here is another result from &lt;a href=&quot;http://scottaaronson.com/blog/?p=152&quot; rel=&quot;nofollow&quot;&gt;Scott Aaronson's blog&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If every second or so your computer’s&#xA;  memory were wiped completely clean,&#xA;  except for the input data; the clock;&#xA;  a static, unchanging program; and a&#xA;  counter that could only be set to 1,&#xA;  2, 3, 4, or 5, it would still be&#xA;  possible (given enough time) to carry&#xA;  out an arbitrarily long computation —&#xA;  just as if the memory weren’t being&#xA;  wiped clean each second. This is&#xA;  almost certainly not true if the&#xA;  counter could only be set to 1, 2, 3,&#xA;  or 4. The reason 5 is special here is&#xA;  pretty much the same reason it’s&#xA;  special in Galois’ proof of the&#xA;  unsolvability of the quintic equation.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Does anyone have idea of how to show this?&lt;/p&gt;&#xA;" OwnerUserId="123" LastEditorUserId="123" LastEditorDisplayName="" LastEditDate="2010-07-28T21:06:51.040" LastActivityDate="2010-12-06T23:24:29.673" Title="Computation with a memory wiped computer" Tags="&lt;computer-science&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="7" />
  <row Id="947" PostTypeId="2" ParentId="944" CreationDate="2010-07-28T12:03:33.543" Score="1" ViewCount="" Body="&lt;p&gt;The concept behind it is &lt;a href=&quot;http://en.wikipedia.org/wiki/Zero-knowledge_proof&quot; rel=&quot;nofollow&quot;&gt;Zero knowledge proof&lt;/a&gt;. Wikipedia has a good article about it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The similar question was asked in paper &quot;How to Prove a Theorem So No One Else Can Claim It&quot; by M. Blum. Also &lt;a href=&quot;http://euler.nmt.edu/~brian/students/pope.pdf&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; article discusses the question.&lt;/p&gt;&#xA;" OwnerUserId="334" LastActivityDate="2010-07-28T12:03:33.543" />
  <row Id="948" PostTypeId="2" ParentId="942" CreationDate="2010-07-28T12:06:45.673" Score="7" ViewCount="" Body="&lt;p&gt;Zorn's lemma implies there is a minimal nonempty closed set $F \subset X$ with no proper closed subsets (because the closed sets have the finite intersection property in view of quasi-compactness).  It is sufficient to find a closed point in $F$. Now $F$ is in itself a scheme, and it has an open subset $U=\mathrm{Spec}\, A$ for $A$ a ring. This must be all of $F$ by minimality.  A maximal ideal in $A$ gives a closed point in $F$, hence in $X$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are people here who could give a much better answer to your other question, so I'll leave it.&lt;/p&gt;&#xA;" OwnerUserId="536" LastEditorUserId="6725" LastEditorDisplayName="" LastEditDate="2011-02-01T11:30:49.857" LastActivityDate="2011-02-01T11:30:49.857" CommentCount="3" />
  <row Id="949" PostTypeId="1" CreationDate="2010-07-28T12:15:49.507" Score="3" ViewCount="129" Body="&lt;p&gt;Say I have an image, with pixels that can be either 0 or 1. For simplicity, assume it's a 2D image (though I'd be interested in a 3D solution as well). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A pixel has 8 neighbors (if that's too complicated, we can drop to 4-connectedness). Two neighbouring pixels with value 1 are considered to be connected. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I know the probability &lt;code&gt;p&lt;/code&gt; that an individual pixel is 1, and if I can assume that all pixels are independent, how many groups of at least &lt;code&gt;k&lt;/code&gt; connected pixels should I expect to find in an image of size &lt;code&gt;n-by-n&lt;/code&gt;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I really need is a good way of calculating the probability of &lt;code&gt;k&lt;/code&gt; pixels being connected given the individual pixel probabilities. I have started to write down a tree to cover all the possibilities up to &lt;code&gt;k=3&lt;/code&gt;, but even then, it becomes really ugly really fast. Is there a more clever way to go about this?&lt;/p&gt;&#xA;" OwnerUserId="356" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-07-28T13:38:01.037" LastActivityDate="2010-07-28T13:46:55.227" Title="Probability to find connected pixels" Tags="&lt;probability&gt;&lt;graph-theory&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="2" />
  <row Id="950" PostTypeId="2" ParentId="944" CreationDate="2010-07-28T12:26:56.757" Score="6" ViewCount="" Body="&lt;p&gt;The correct answer has already been given by Akhil Mathew in the comments above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The topic belongs to the field of complexity theory in computer science. In complexity theory, there exists an intriguing concept for the problem of deciding whether a given word belongs to a given formal language or not: &lt;a href=&quot;http://en.wikipedia.org/wiki/Interactive_proof_system&quot; rel=&quot;nofollow&quot;&gt;interactive proof systems&lt;/a&gt;. These systems model the interaction between resource-limited verifier (say, you or me) and an almighty prover (say, your much, much smarter older sister). The goal of the interaction is that the prover convinces the verifier from the fact that the given word is or is not an element of the language such that:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;almost surely, the verfier can only be convinced from the true answer and&lt;/li&gt;&#xA;&lt;li&gt;the verifier can only be fooled to believe the opposite within a very small margin of error.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;There are a large number of theoretical results with respect to these interactive systems. These resuts include the follwoing two statements (given informally):&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;everything that can be proven by such an interactive protocoll can also be proven using an interactive protocoll that convinces the verifier (within small margin of error) but it does not reveal &lt;em&gt;any&lt;/em&gt; information about the proof itself. (&lt;a href=&quot;http://en.wikipedia.org/wiki/Zero-knowledge_proof&quot; rel=&quot;nofollow&quot;&gt;Zero-Knowledge-Proof&lt;/a&gt;, &quot;Everything provable is provable in zero-knowledge&quot; by Ben-Or et al, 1988)&lt;/li&gt;&#xA;&lt;li&gt;every proof can be rewritten such that inspection of just a constant number of bits from this proof convinces the verifier within only a small margin of error (&lt;a href=&quot;http://en.wikipedia.org/wiki/PCP_theorem&quot; rel=&quot;nofollow&quot;&gt;PCP-Theorem&lt;/a&gt;, &quot;Proof verification and the hardness of approximation problems&quot; by Arora et al, 1992, and a number of other papers)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Both of these concepts and results are highly non-trivial and beyond the scope of this forum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, in the quote above Scott Aaronson is just using some everyday's problem to illustrate these concepts. To use these results formally, would have to convert the task of &quot;proving the Riemann Hypothesis&quot; to a decision problem of formal languages, as is standard in complexity theory.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;EDIT: There is in fact a small modification in the model between both results stated above which I omitted. First, the interaction between one prover and the verifier can be generalized to multiple provers and a verifier. Next, there is a result that finds that the case of multiple provers can be equivalently reformulated as follows: The provers are removed from the protocol, rather there is a single (possibly very long) string which acts as a written proof for the word problem. Interaction is now looking at an arbitrarily chosen bit of this proof, and the verifier may choose the location of these bits randomly. This is called a &lt;a href=&quot;http://en.wikipedia.org/wiki/Probabilistically_checkable_proof&quot; rel=&quot;nofollow&quot;&gt;Probabilistic Checkable Proof&lt;/a&gt; (hence, PCP). So ,in this scenario, the &quot;written proof&quot; of the Riemann Hypothesis would be interpreted as the proof string.&lt;/p&gt;&#xA;" OwnerUserId="347" LastEditorUserId="347" LastEditorDisplayName="" LastEditDate="2010-07-28T12:52:06.407" LastActivityDate="2010-07-28T12:52:06.407" CommentCount="2" />
  <row Id="951" PostTypeId="1" AcceptedAnswerId="952" CreationDate="2010-07-28T12:28:00.623" Score="5" ViewCount="214" Body="&lt;p&gt;I think that complex analysis is hard because graphs of even basic functions are 4 dimensional. Does anyone have any good visual representations of basic complex functions or know of any tools for generating them?&lt;/p&gt;&#xA;" OwnerUserId="123" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-07-28T15:41:58.513" LastActivityDate="2010-07-28T17:02:13.010" CommunityOwnedDate="2010-07-28T12:28:00.623" Title="Visualising functions from complex numbers to complex numbers" Tags="&lt;big-list&gt;&lt;math-software&gt;&lt;complex-analysis&gt;" AnswerCount="4" CommentCount="1" />
  <row Id="952" PostTypeId="2" ParentId="951" CreationDate="2010-07-28T12:37:01.900" Score="9" ViewCount="" Body="&lt;p&gt;One way that functions from C to C can be represented is to show the image of a grid.  That is, plot the images of the lines x = constant and y = constant under your function, where z = x + yi.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another is what Wikipedia calls &lt;a href=&quot;http://en.wikipedia.org/wiki/Domain_coloring&quot; rel=&quot;nofollow&quot;&gt;domain coloring&lt;/a&gt; (see &lt;a href=&quot;http://www.mai.liu.se/~halun/complex/domain_coloring-unicode.html&quot; rel=&quot;nofollow&quot;&gt;this article by Hans Lundmark&lt;/a&gt; for a more detailed exposition.  The idea here is to color the range of the complex function -- since color space is three-dimensional there are multiple ways to do this -- and then color each point in the domain by the color of the corresponding point in the range.&lt;/p&gt;&#xA;" OwnerUserId="173" LastActivityDate="2010-07-28T12:37:01.900" CommunityOwnedDate="2010-07-28T12:37:01.900" CommentCount="1" />
  <row Id="953" PostTypeId="2" ParentId="936" CreationDate="2010-07-28T12:50:06.147" Score="2" ViewCount="" Body="&lt;p&gt;One good example is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Normal_distribution&quot; rel=&quot;nofollow&quot;&gt;standard Gaussian distribution&lt;/a&gt;, $\phi(x) = \frac{1}{\sqrt{2\pi}} e^{-x^2/2}$.  This is the most straightforward example of a continuous probability distribution function as mentioned by KennyTM above.&lt;/p&gt;&#xA;" OwnerUserId="349" LastActivityDate="2010-07-28T12:50:06.147" CommunityOwnedDate="2010-08-03T18:47:52.520" />
  <row Id="954" PostTypeId="1" CreationDate="2010-07-28T12:51:19.983" Score="0" ViewCount="158" Body="&lt;p&gt;There is a very simple expression for the inverse of Fourier transform. What is the easiest known expression for the inverse Laplace transform?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moreover, what is the easiest way to prove it?&lt;/p&gt;&#xA;" OwnerUserId="218" LastActivityDate="2010-07-29T14:22:08.317" Title="Inverse of Laplace transform" Tags="&lt;analysis&gt;&lt;integral-transforms&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="2" />
  <row Id="955" PostTypeId="1" AcceptedAnswerId="958" CreationDate="2010-07-28T12:52:39.747" Score="0" ViewCount="78" Body="&lt;p&gt;Is there any connection between exact differential equations and forms, or is the similarity in name just an accident?&lt;/p&gt;&#xA;" OwnerUserId="218" LastActivityDate="2010-07-28T12:56:36.820" Title="Exact differential eqns and exact differential forms" Tags="&lt;differential-equations&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="956" PostTypeId="2" ParentId="939" CreationDate="2010-07-28T12:52:46.220" Score="5" ViewCount="" Body="&lt;p&gt;The most efficient prime counting algorithms currently known are all essentially optimizations of the method developed by Meissel in 1870, e.g. see the discussion here &lt;a href=&quot;http://primes.utm.edu/howmany.shtml&quot; rel=&quot;nofollow&quot;&gt;http://primes.utm.edu/howmany.shtml&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-07-28T12:52:46.220" />
  <row Id="957" PostTypeId="1" AcceptedAnswerId="960" CreationDate="2010-07-28T12:54:07.043" Score="5" ViewCount="223" Body="&lt;p&gt;Why are noncommutative nonassociative Hopf algebras are called quantum groups? This seems to be a purely mathematical notion and there is no quantum anywhere in it prima facie.&lt;/p&gt;&#xA;" OwnerUserId="218" LastEditorUserId="218" LastEditorDisplayName="" LastEditDate="2010-07-29T18:34:53.743" LastActivityDate="2010-07-29T18:34:53.743" Title="Why are noncommutative &quot;algebraic&quot; groups called quantum groups?" Tags="&lt;terminology&gt;&lt;noncommutative-geometry&gt;&lt;quantum-groups&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="2" />
  <row Id="958" PostTypeId="2" ParentId="955" CreationDate="2010-07-28T12:56:36.820" Score="1" ViewCount="" Body="&lt;p&gt;The differential equation $I(x,y)dx + J(x,y)dy = 0$ is exact is by definition the same as this differential being an exact form.&lt;/p&gt;&#xA;" OwnerUserId="325" LastActivityDate="2010-07-28T12:56:36.820" />
  <row Id="959" PostTypeId="1" AcceptedAnswerId="962" CreationDate="2010-07-28T13:09:12.457" Score="-2" ViewCount="250" Body="&lt;p&gt;Lets say we have a sphere, cube, and cylinder of same dimensions, that is&#xA;The diameter of the sphere is same as one side of a cube and same as circumference of the cylinder.&#xA;Of these three containers, how can one prove that the sphere has largest capacity?&lt;/p&gt;&#xA;" OwnerUserId="361" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-07-28T13:34:52.137" LastActivityDate="2010-07-29T21:46:44.523" Title="How to prove which amongst a Sphere, Cube and Cylinder of same dimensions has more capacity" Tags="&lt;geometry&gt;&lt;euclidean-geometry&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="960" PostTypeId="2" ParentId="957" CreationDate="2010-07-28T13:18:41.627" Score="2" ViewCount="" Body="&lt;p&gt;One way that Hopf algebras come up is as the algebra of (real or complex) functions on a topological group.  The multiplication is commutative since it is just pointwise multiplication of functions.  However, in non-commutative geometry you want to replace the algebra of functions on a space with a non-commutative algebra, giving a non-commutative Hopf algebra.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This relates to quantum mechanics because there the analog of the classical coordinate functions of position and momentum do not commute.  Therefore we think of the algebra of functions on a quantum &quot;space&quot; as being non-commutative.&lt;/p&gt;&#xA;" OwnerUserId="9" LastEditorUserId="9" LastEditorDisplayName="" LastEditDate="2010-07-29T18:29:02.197" LastActivityDate="2010-07-29T18:29:02.197" CommentCount="9" />
  <row Id="961" PostTypeId="2" ParentId="951" CreationDate="2010-07-28T13:20:53.150" Score="3" ViewCount="" Body="&lt;p&gt;Open two browser windows side-by-side and use &lt;a href=&quot;http://www.wolframalpha.com/&quot; rel=&quot;nofollow&quot;&gt;wolfram alpha&lt;/a&gt;. Recast your function f(z) as f(x+iy) and plot the &lt;a href=&quot;http://bit.ly/bEZgsA&quot; rel=&quot;nofollow&quot;&gt;real part&lt;/a&gt; in one window and the &lt;a href=&quot;http://bit.ly/bfeLyS&quot; rel=&quot;nofollow&quot;&gt;imaginary part&lt;/a&gt; in another. My example links provide plots for z&lt;sup&gt;3&lt;/sup&gt;.&lt;/p&gt;&#xA;" OwnerUserId="167" LastEditorUserId="167" LastEditorDisplayName="" LastEditDate="2010-07-28T13:37:32.940" LastActivityDate="2010-07-28T13:37:32.940" CommunityOwnedDate="2010-07-28T13:20:53.150" />
  <row Id="962" PostTypeId="2" ParentId="959" CreationDate="2010-07-28T13:33:33.073" Score="3" ViewCount="" Body="&lt;p&gt;It is easy to show which has the largest volume (once you've provided the height of the cylinder as per my comment) as long as you accept proofs built on top of the existing verified equations of the volumes of each solid. Then you have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$sphere = \frac{4\pi}{3}(\frac{d}{2})^3 = \frac{\pi d^3}{6}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$cube = d^3$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$cylinder = (\frac{d}{2\pi})^2 \pi h = \frac{d^2}{4\pi^2}h\pi = \frac{hd^2}{4\pi}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The final equation would be:&#xA;$\frac{d^3}{4\pi}$ &#xA;if you make the height of the cylinder equal to it's circumference and therefore equal to the diameter of the sphere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This actually shows a different result to the one you suggested too, namely that the cube has the greatest volume in this setup. This can easily be seen geometrically by envisioning the shapes. The sphere with diameter $d$ will fit inside a cube with side of length $d$. The cylinder with circumference $d$ is the smallest (assuming the height is equal to the circumference) because it is a cylinder with diameter $\frac{d}{2\pi}$ which is a much smaller circle than that described by a cross-section of the sphere taken through a great circle drawn on the surface. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think you may have gotten your assumption from a related but different statement - namely that a sphere has the largest volume given that the &lt;em&gt;surface areas&lt;/em&gt; are equal in the shapes being considered. You can show this in a similar way by simply comparing the equations for volumes again, but calculating them from the surface area equations instead of directly from sides/diameters.&lt;/p&gt;&#xA;" OwnerUserId="35" LastEditorUserId="35" LastEditorDisplayName="" LastEditDate="2010-07-29T21:46:44.523" LastActivityDate="2010-07-29T21:46:44.523" CommentCount="2" />
  <row Id="963" PostTypeId="2" ParentId="949" CreationDate="2010-07-28T13:46:55.227" Score="2" ViewCount="" Body="&lt;p&gt;This looks a bit like percolation theory to me. In the 4-neighbour case, if you look at the dual of the image, the chance that an edge is connected (runs between two pixels of the same colour) is &lt;code&gt;1-2p+2p^2&lt;/code&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't think you can get nice closed-form answer for your question, but maybe a computer can help with some Monte Carlo simulation?&lt;/p&gt;&#xA;" OwnerUserId="360" LastActivityDate="2010-07-28T13:46:55.227" />
  <row Id="964" PostTypeId="2" ParentId="939" CreationDate="2010-07-28T13:50:00.850" Score="1" ViewCount="" Body="&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Sieve_of_Atkin&quot; rel=&quot;nofollow&quot;&gt;Sieve of Atkin&lt;/a&gt; is one of the fastest algorithm used to calculate $pi(x)$. The Wikipedia page says that its complexity is O(N/ log log N).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(edit)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found a &lt;a href=&quot;http://numbers.computation.free.fr/Constants/Primes/Pix/pixproject.html&quot; rel=&quot;nofollow&quot;&gt;distributed computation project&lt;/a&gt; which was able to calculate $\pi(4\times 10^{22})$, maybe it could be useful.&lt;/p&gt;&#xA;" OwnerUserId="354" LastEditorUserId="354" LastEditorDisplayName="" LastEditDate="2010-07-28T16:28:02.443" LastActivityDate="2010-07-28T16:28:02.443" CommentCount="3" />
  <row Id="965" PostTypeId="2" ParentId="936" CreationDate="2010-07-28T14:08:21.867" Score="0" ViewCount="" Body="&lt;p&gt;The function which is 1 on the interval &lt;code&gt;[0;1]&lt;/code&gt;, and 0 elsewhere, is a non-continuous probability distribution function. The function which is 3 on &lt;code&gt;[0;1]&lt;/code&gt; and -1 on (1;3], and so on and on. What kind of answer do you want? What kind of properties do you want your functions to have?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There really are too many functions to list, since multiplying &lt;strong&gt;any&lt;/strong&gt; function by a C^oo function with compact support and then applying Kenny's trick gives you an answer.&lt;/p&gt;&#xA;" OwnerUserId="360" LastActivityDate="2010-07-28T14:08:21.867" CommunityOwnedDate="2010-08-03T18:47:52.520" />
  <row Id="966" PostTypeId="2" ParentId="957" CreationDate="2010-07-28T14:37:05.403" Score="0" ViewCount="" Body="&lt;p&gt;I cannot comment, and this should be a comment...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Observe that the question in your title and the question in the body of your question are quite different! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A non-commutative non-cocommutative Hopf algebra is not the same thing as a non-commutative group, and quantum groups &lt;em&gt;are&lt;/em&gt; usually associative.&lt;/p&gt;&#xA;" OwnerUserId="274" LastEditorUserId="274" LastEditorDisplayName="" LastEditDate="2010-07-28T14:45:05.320" LastActivityDate="2010-07-28T14:45:05.320" CommentCount="9" />
  <row Id="967" PostTypeId="1" CreationDate="2010-07-28T14:42:21.270" Score="10" ViewCount="390" Body="&lt;p&gt;What are some good online/free resources (tutorials, guides, exercises, and the like) for learning Lambda Calculus?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Specifically, I am interested in the following areas:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Untyped lambda calculus&lt;/li&gt;&#xA;&lt;li&gt;Simply-typed lambda calculus&lt;/li&gt;&#xA;&lt;li&gt;Other typed lambda calculi&lt;/li&gt;&#xA;&lt;li&gt;Church's Theory of Types (I'm not sure where this fits in).&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;(As I understand, this should provide a solid basis for the understanding of type theory.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any advice and suggestions would be appreciated.&lt;/p&gt;&#xA;" OwnerUserId="56" LastEditorUserId="100" LastEditorDisplayName="" LastEditDate="2010-07-30T08:19:20.123" LastActivityDate="2011-04-03T14:54:39.093" CommunityOwnedDate="2010-08-07T16:55:42.303" Title="Learning Lambda Calculus" Tags="&lt;logic&gt;&lt;lambda-calculus&gt;&lt;type-theory&gt;" AnswerCount="5" FavoriteCount="8" />
  <row Id="968" PostTypeId="2" ParentId="936" CreationDate="2010-07-28T14:45:22.340" Score="1" ViewCount="" Body="&lt;p&gt;If you take any odd function F differentiable on R and such that F(x)-&gt;l (with l a nonzero real) for x-&gt;infinity, then f(x) = 1/(2l)*F'(x) statisfies your request. For example F(x) = arctan(x)&lt;/p&gt;&#xA;" OwnerUserId="365" LastActivityDate="2010-07-28T14:45:22.340" CommunityOwnedDate="2010-08-03T18:47:52.520" />
  <row Id="969" PostTypeId="2" ParentId="936" CreationDate="2010-07-28T14:49:35.327" Score="2" ViewCount="" Body="&lt;p&gt;This should be a comment but I cannot comment...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Dirac delta function is &lt;em&gt;not&lt;/em&gt; a function!&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2010-07-28T14:49:35.327" CommunityOwnedDate="2010-08-03T18:47:52.520" />
  <row Id="970" PostTypeId="2" ParentId="936" CreationDate="2010-07-28T14:53:46.267" Score="1" ViewCount="" Body="&lt;p&gt;Most practical &lt;a href=&quot;http://en.wikipedia.org/wiki/Wavelet&quot; rel=&quot;nofollow&quot;&gt;mother wavelet&lt;/a&gt; have square norm 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\int_{-\infty}^\infty |\psi(t)|^2 dt = 1$&lt;/p&gt;&#xA;" OwnerUserId="96" LastActivityDate="2010-07-28T14:53:46.267" CommunityOwnedDate="2010-08-03T18:47:52.520" />
  <row Id="971" PostTypeId="2" ParentId="936" CreationDate="2010-07-28T15:03:54.620" Score="3" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Any&lt;/strong&gt; function f(x) which integrates to 1 over &lt;strong&gt;any&lt;/strong&gt; range [a,b] fits this bill, since we can define g(x) = f(x) on [a,b], and 0 everywhere else.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even if you only want continuous functions, restricting ourselves above to f(x) where f(a) = f(b) = 0 still satisfies this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want continuous functions strictly &gt; 0 everywhere, these are known as probability distributions (continuous on [-∞, ∞]).  A large list of such functions can be found &lt;a href=&quot;http://en.wikipedia.org/wiki/List_of_probability_distributions#Supported_on_the_whole_real_line&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.  A few more notable examples are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Normal_distribution&quot; rel=&quot;nofollow&quot;&gt;normal distribution&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Skew_normal_distribution&quot; rel=&quot;nofollow&quot;&gt;skew-normal distribution&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Student%27s_t-distribution&quot; rel=&quot;nofollow&quot;&gt;t-distribution&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Cauchy_distribution&quot; rel=&quot;nofollow&quot;&gt;cauchy distribution&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Fisher-Tippett_distribution&quot; rel=&quot;nofollow&quot;&gt;extreme-value distribution&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="136" LastActivityDate="2010-07-28T15:03:54.620" CommunityOwnedDate="2010-08-03T18:47:52.520" />
  <row Id="972" PostTypeId="2" ParentId="946" CreationDate="2010-07-28T15:16:34.497" Score="3" ViewCount="" Body="&lt;p&gt;As Scott himself states in comments section of &lt;a href=&quot;http://scottaaronson.com/blog/?p=152&quot; rel=&quot;nofollow&quot;&gt;the post in question&lt;/a&gt; (comment #9):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;(4) Width-5 branching programs can compute NC1 (Barrington 1986); corollary pointed out by Ogihara 1994 that width-5 bottleneck Turing machines can compute PSPACE&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, I don't have any ideas how this is proved.&lt;/p&gt;&#xA;" OwnerUserId="347" LastActivityDate="2010-07-28T15:16:34.497" CommunityOwnedDate="2010-07-28T15:16:34.497" CommentCount="1" />
  <row Id="973" PostTypeId="2" ParentId="967" CreationDate="2010-07-28T15:18:53.593" Score="2" ViewCount="" Body="&lt;p&gt;It might be nice to work through &lt;em&gt;Structure and Interpretation of Computer Programs&lt;/em&gt;, which is &lt;a href=&quot;http://mitpress.mit.edu/sicp/&quot; rel=&quot;nofollow&quot;&gt;available online&lt;/a&gt; for free. This book is an introduction to computer science and the programming language Scheme, which is a flavor of the programming language Lisp, which is based on the lambda calculus. Although it is not strictly a book about the lambda calculus, it might be fun or useful to gain some hands-on and &quot;practical&quot; experience with the lambda calculus by reading some of this book and working through some of its exercises.&lt;/p&gt;&#xA;" OwnerUserId="263" LastActivityDate="2010-07-28T15:18:53.593" CommentCount="2" />
  <row Id="974" PostTypeId="2" ParentId="262" CreationDate="2010-07-28T15:28:10.947" Score="1" ViewCount="" Body="&lt;p&gt;I would read a book about Perelman's proof of the Poincaré conjecture (or even the papers themselves). Oh, you mean the book had to be written when I was starting? &lt;/p&gt;&#xA;" OwnerUserId="360" LastActivityDate="2010-07-28T15:28:10.947" CommunityOwnedDate="2010-07-28T15:28:10.947" />
  <row Id="975" PostTypeId="1" CreationDate="2010-07-28T15:46:10.150" Score="1" ViewCount="324" Body="&lt;p&gt;Hi guys, &#xA;In light of the &lt;a href=&quot;http://math.stackexchange.com/questions/81/list-of-interesting-math-blogs&quot;&gt;List of Interesting Math Blogs&lt;/a&gt; I thought it would benefit the community if we have a list of textbooks for specific topics. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Trusted users and/or I will edit this post depending on the answers. Please try to include the textbook name, the field it relates too, the author and if you want, give it a rating. I am not sure about how the rating works but lets say 5 is something a first year mathematics student needs while 1 is for high school and 10 is for advanced levels. If you can think of a better rating system, I am all ears.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I will have the categories setup as following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Calculus&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Linear Algebra:&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="287" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-07-28T15:48:10.693" LastActivityDate="2010-07-28T15:48:10.693" CommunityOwnedDate="2010-07-30T05:57:16.610" ClosedDate="2011-04-21T18:38:15.523" Title="List of interesting and well-explained textbooks." Tags="&lt;soft-question&gt;&lt;big-list&gt;&lt;books&gt;" CommentCount="7" FavoriteCount="1" />
  <row Id="976" PostTypeId="2" ParentId="262" CreationDate="2010-07-28T15:46:35.237" Score="5" ViewCount="" Body="&lt;p&gt;I am not a mathematician but &lt;a href=&quot;http://books.google.com/books?id=R6E0AAAAMAAJ&amp;amp;dq=flatland&amp;amp;printsec=frontcover&amp;amp;source=bn&amp;amp;hl=en&amp;amp;ei=eE9QTLKVF5G5rAex79zVDQ&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=11&amp;amp;ved=0CEQQ6AEwCg#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;Flatland: A Romance of Many Dimensions&lt;/a&gt; blew my mind. I read it when I was a college student in a class on Special Relativity and wish I had read it way earlier.&lt;/p&gt;&#xA;" OwnerUserId="333" LastActivityDate="2010-07-28T15:46:35.237" CommunityOwnedDate="2010-07-28T15:46:35.237" />
  <row Id="977" PostTypeId="2" ParentId="942" CreationDate="2010-07-28T15:54:51.207" Score="4" ViewCount="" Body="&lt;p&gt;Closed points should be thought of as being &quot;actual points&quot;, whereas non-closed points can correspond to all sorts of different things: subvarieties, &quot;fat points&quot;, generic points, etc. You might be interested in reading &lt;a href=&quot;http://www.neverendingbooks.org/index.php/mumfords-treasure-map.html&quot; rel=&quot;nofollow&quot;&gt;this blog post&lt;/a&gt; about Mumford's drawing of $\operatorname{Spec} \mathbb{Z}[x]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One possible way to justify the claim that closed points are the &quot;actual points&quot; is the fact that if we have, for instance, a smooth variety over $\mathbb{C}$, then its &lt;a href=&quot;http://books.google.com/books?id=ZhzXJHUgcRUC&amp;amp;lpg=PA67&amp;amp;ots=aVQoeMkBwc&amp;amp;dq=analytification&amp;amp;pg=PA61#v=onepage&amp;amp;q=analytification&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;analytification&lt;/a&gt; will be a complex manifold. The closed points of the former will then correspond exactly to the points of the latter.&lt;/p&gt;&#xA;" OwnerUserId="263" LastEditorUserId="263" LastEditorDisplayName="" LastEditDate="2010-07-28T19:44:00.380" LastActivityDate="2010-07-28T19:44:00.380" CommentCount="1" />
  <row Id="978" PostTypeId="1" AcceptedAnswerId="981" CreationDate="2010-07-28T16:02:17.240" Score="0" ViewCount="1056" Body="&lt;p&gt;Let A and B be two matrices which can be multiplied.&lt;br&gt;&#xA;Then &lt;strong&gt;$\operatorname{rank}(AB) \leq \operatorname{min}(\operatorname{rank}(A), \operatorname{rank}(B))$&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I proved &lt;strong&gt;$\operatorname{rank}(AB) \leq \operatorname{rank}(B)$&lt;/strong&gt; interpreting AB as a composition of linear maps, observing that $\operatorname{ker}(B) \subseteq \operatorname{ker}(AB)$ and using the kernel-image dimension formula. This also provides, in my opinion, a nice interpretation: if non stable, under subsequent compositions the kernel can only get bigger, and the image can only get smaller, in a sort of &lt;em&gt;loss of information&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How to manage &lt;strong&gt;$\operatorname{rank}(AB) \leq \operatorname{min}(\operatorname{rank}(A), \operatorname{rank}(A))$&lt;/strong&gt;? Is there a nice interpretation like the previous one?&lt;/p&gt;&#xA;" OwnerUserId="365" LastEditorUserId="365" LastEditorDisplayName="" LastEditDate="2010-09-02T10:04:17.533" LastActivityDate="2011-01-16T19:10:07.517" Title="How to prove and interpret rank(AB) &lt;= min(rank(A), rank(B)) ?" Tags="&lt;linear-algebra&gt;" AnswerCount="4" CommentCount="1" FavoriteCount="2" />
  <row Id="979" PostTypeId="2" ParentId="951" CreationDate="2010-07-28T16:02:37.630" Score="1" ViewCount="" Body="&lt;p&gt;For Moebius transformations, check out &lt;a href=&quot;http://www.youtube.com/watch?v=JX3VmDgiFnY&quot; rel=&quot;nofollow&quot;&gt;this nice YouTube video&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="263" LastActivityDate="2010-07-28T16:02:37.630" CommunityOwnedDate="2010-07-28T16:02:37.630" CommentCount="2" />
  <row Id="980" PostTypeId="1" CreationDate="2010-07-28T16:04:45.773" Score="0" ViewCount="868" Body="&lt;p&gt;Especially with relation to topology - rigorous definitions are appreciated, but just as important is the intuition!&lt;/p&gt;&#xA;" OwnerUserId="136" LastEditorUserId="16" LastEditorDisplayName="" LastEditDate="2010-07-28T19:48:10.760" LastActivityDate="2010-07-28T19:57:50.113" Title="What's the difference between open and closed sets?" Tags="&lt;terminology&gt;&lt;set-theory&gt;&lt;intuition&gt;&lt;topology&gt;" AnswerCount="5" CommentCount="8" />
  <row Id="981" PostTypeId="2" ParentId="978" CreationDate="2010-07-28T16:11:23.520" Score="3" ViewCount="" Body="&lt;p&gt;Yes. If you think of A and B as linear maps, then the domain of A is certainly at least as big as the image of B. Thus when we apply A to either of these things, we should get &quot;more stuff&quot; in the former case, as the former is bigger than the latter.&lt;/p&gt;&#xA;" OwnerUserId="263" LastActivityDate="2010-07-28T16:11:23.520" CommentCount="1" />
  <row Id="982" PostTypeId="2" ParentId="978" CreationDate="2010-07-28T16:11:25.330" Score="2" ViewCount="" Body="&lt;p&gt;Prove first that if $f:X\to Y$ and $g:Y\to Z$ are functions between finite sets, then &#xA;$|g(f(X))| \leq \min \{ |f(X)|, |g(Y)| \}.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then use the same idea.&lt;/p&gt;&#xA;" OwnerUserId="274" LastEditorUserId="274" LastEditorDisplayName="" LastEditDate="2011-01-16T19:10:07.517" LastActivityDate="2011-01-16T19:10:07.517" CommentCount="3" />
  <row Id="983" PostTypeId="2" ParentId="980" CreationDate="2010-07-28T16:27:04.723" Score="3" ViewCount="" Body="&lt;p&gt;The rigorous definition of open and closed sets is fundamental to topology: you define a topology by saying what its open sets are.  From this perspective, open and closed sets are axiomatic, like points and lines in geometry.  In any case, closed sets are the complements of open sets and vice versa.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most familiar example of open sets would be open intervals on the real line, intervals of the form {x : a &amp;lt; x &amp;lt; b}. Such sets and their arbitrary unions define the standard topology on the real line.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the paragraph above describes the &lt;em&gt;standard&lt;/em&gt; topology, but not the only one.  You could put a very different topology on the real line.  That's because a topology is determined by what you call open sets and not by the underlying space &lt;em&gt;per se&lt;/em&gt;.  For example, another topology on the real line defines a set to be open if its complement has only a finite number of points. &lt;/p&gt;&#xA;" OwnerUserId="247" LastEditorUserId="247" LastEditorDisplayName="" LastEditDate="2010-07-28T18:22:11.493" LastActivityDate="2010-07-28T18:22:11.493" />
  <row Id="984" PostTypeId="2" ParentId="942" CreationDate="2010-07-28T16:30:13.550" Score="6" ViewCount="" Body="&lt;p&gt;Kevin Lin's answer regarding the meaning of closed points is quite reasonable, especialy in the case when the scheme in question underlies a classical variety.  I want to add some additional remarks and examples for thinking about more general schemes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are some tautological remarks: recall that a point x in a scheme X is called a specialization of y if x lies in the Zariski closure of y (and y is called a generalization of x).  So tautologically, a closed points is one that cannot be specialized any further (just as a generic point cannot be generalized any further).  What does specialization really mean:&#xA;ring theoretically, it means taking the image under a homomorphism; so if p and q are prime&#xA;ideals of a ring A, then q is a specialization of p in Spec A if and only if q contains p, i.e. if A/p surjects onto A/q.  It is perhaps best to think of an example: say A is C[x,y],&#xA;p is the prime ideal gen'd by (x-1) and q is the prime (actually maximal ideal) gen'd by (x-1,y).  Then in A/p, we have &quot;specialized&quot; the value of x to equal 1 (because we have &#xA;declared x-1 = 0) but y is still a free variable.  When we pass to the further quotient A/q, we have specialized both x and y: x is specialized to 1 and y is specialized to 0.   At this point, we can't specialize any more; technically, this is because q is a maximal ideal of A,&#xA;so a closed point of Spec A; intuitively, it is because both x and y have now both been &quot;specialized&quot; to actual numbers, and so we can't specialize any further.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But suppose now we set B = Z[x,y], and take p and q to be the same, i.e. gen'd by x-1 and by&#xA;(x-1,y) respectively.  Then q is &lt;em&gt;not&lt;/em&gt; maximal; there is more capacity for specialization.&#xA;How is this?  Well, x and y are now taking values in Z (rather than the field C) and so we can also reduce both x and y modulo some prime, say 5; this gives a prime ideal r = (x-1,y,5) in B containing q.  Now r &lt;em&gt;is&lt;/em&gt; maximal, and so we are done specializing.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if you have a scheme that is finite type over Z, the closed points will correspond to &#xA;&quot;actual points&quot;, in Kevin's terminology, but defined over finite fields.  The points of the scheme whose coordinates are integers, say, will &lt;em&gt;not&lt;/em&gt; be closed.  One has the choice of thinking them of them as &quot;actual points&quot; which nevertheless can be specialized further by reducing modulo primes, or as subvarieties rather than &quot;actual points&quot;, by identifying them&#xA;with their Zariski closures (for a picture of this, see the drawing of Mumford that Kevin links to).&lt;/p&gt;&#xA;" OwnerUserId="221" LastEditorUserId="221" LastEditorDisplayName="" LastEditDate="2010-07-28T19:56:54.470" LastActivityDate="2010-07-28T19:56:54.470" CommentCount="0" />
  <row Id="985" PostTypeId="2" ParentId="921" CreationDate="2010-07-28T16:37:59.720" Score="1" ViewCount="" Body="&lt;p&gt;I always liked 23 (or more precisely, -23) because Q adjoin the square-root of -23 is the first quadratic imaginary field whose class number is not a power of 2; in fact, it's class group is of order 3. &lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2010-07-28T16:37:59.720" CommunityOwnedDate="2010-07-28T16:37:59.720" />
  <row Id="986" PostTypeId="2" ParentId="980" CreationDate="2010-07-28T16:40:31.667" Score="4" ViewCount="" Body="&lt;p&gt;Intuitively speaking, an open set is a set without a border: every element of the set has, in its neighborhood, other elements of the set. If, starting from a point of the open set, you move away a little, you never exit the set.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A closed set is the complement of an open set (i.e. what stays &quot;outside&quot; from the open set).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that some set exists, that are neither open nor closed.&lt;/p&gt;&#xA;" OwnerUserId="354" LastEditorUserId="354" LastEditorDisplayName="" LastEditDate="2010-07-28T19:47:01.683" LastActivityDate="2010-07-28T19:47:01.683" CommentCount="2" />
  <row Id="987" PostTypeId="2" ParentId="980" CreationDate="2010-07-28T16:43:01.447" Score="2" ViewCount="" Body="&lt;p&gt;A set X is open if for every point p in X, there exists a neighborhood (open ball) N of p such that N is a subset of X. We call the point p of the set X a limit point if every neighborhood of p has another point q which is also in X. The set X is closed if every limit point of X is a point of X. A set can be both open and closed, and such sets are occasionally termed &quot;clopen.&quot; Trivial examples of clopen sets are the empty set (since it has no points, both the above definitions are vacuously true) and the set of all real numbers. You can in turn visualize an open set in R as an open interval on the real line, and a closed set as a closed interval on the real line.&lt;/p&gt;&#xA;" LastActivityDate="2010-07-28T17:10:35.430" CommentCount="3" />
  <row Id="988" PostTypeId="2" ParentId="951" CreationDate="2010-07-28T17:02:13.010" Score="1" ViewCount="" Body="&lt;p&gt;The graphs in the middle of &lt;a href=&quot;http://mathworld.wolfram.com/ConformalMapping.html&quot; rel=&quot;nofollow&quot;&gt;the MathWorld page on Conformal Mapping&lt;/a&gt; show examples of the first method in Michael Lugo's answer as well as something somewhat similar to the second method in that answer.&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-07-28T17:02:13.010" CommunityOwnedDate="2010-07-28T17:02:13.010" />
  <row Id="989" PostTypeId="2" ParentId="250" CreationDate="2010-07-28T17:04:42.087" Score="9" ViewCount="" Body="&lt;p&gt;What is the smallest area of a parking lot in which a car (that is, a segment of) can perform a complete turn (that is, rotate 360 degrees)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(This is obviously the Kakeya Needle Problem. Fairly easy to explain, models an almost reasonable real-life scenario, and has a very surprising answer as you probably know - the lot can have as small an area as you'd like).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Kakeya_set&quot; rel=&quot;nofollow&quot;&gt;Wikipedia entry: Kakeya Set&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="308" LastActivityDate="2010-07-28T17:04:42.087" CommunityOwnedDate="2010-07-28T17:04:42.087" CommentCount="2" />
  <row Id="990" PostTypeId="2" ParentId="980" CreationDate="2010-07-28T17:05:58.433" Score="2" ViewCount="" Body="&lt;p&gt;An open set is a set S for which, given any of its element A, you can find a ball centered in A and whose points are all in S.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A closed set is a set S for which, if you have a sequence of points in S who tend to a limit point B, B is also in S.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Intuitively, a closed set is a set which contains its own boundary, while an open set is a set where you are able not to leave it if you move just a little bit.&lt;/p&gt;&#xA;" OwnerUserId="89" LastActivityDate="2010-07-28T17:05:58.433" CommentCount="1" />
  <row Id="992" PostTypeId="2" ParentId="3" CreationDate="2010-07-28T17:30:41.997" Score="1" ViewCount="" Body="&lt;p&gt;The &lt;a href=&quot;http://www.scienceofbetter.org/podcast/&quot; rel=&quot;nofollow&quot;&gt;Science of Better&lt;/a&gt; is a good applied math podcast.  Features applications of operations research to practical problems.&lt;/p&gt;&#xA;" OwnerUserId="247" LastEditorUserId="247" LastEditorDisplayName="" LastEditDate="2010-07-28T17:39:29.080" LastActivityDate="2010-07-28T17:39:29.080" CommunityOwnedDate="2010-07-28T17:30:41.997" />
  <row Id="993" PostTypeId="2" ParentId="3" CreationDate="2010-07-28T17:38:48.037" Score="1" ViewCount="" Body="&lt;p&gt;Peter Rowlett has a couple mathematical podcasts.  &lt;a href=&quot;http://www.travelsinamathematicalworld.co.uk/&quot; rel=&quot;nofollow&quot;&gt;Travels in a Mathematical World&lt;/a&gt; produced 64 episodes, but recently stopped.  He has a new podcast &lt;a href=&quot;http://www.pulse-project.org/pulsemathsmaths/&quot; rel=&quot;nofollow&quot;&gt;Math / Maths&lt;/a&gt; that he co-hosts with Samuel Hansen.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Samuel Hansen also has a couple other podcasts: &lt;a href=&quot;http://itunes.apple.com/us/podcast/strongly-connected-components/id334982081&quot; rel=&quot;nofollow&quot;&gt;Strongly Connected Components&lt;/a&gt; and &lt;a href=&quot;http://cppodcast.libsyn.com/&quot; rel=&quot;nofollow&quot;&gt;Permutations and Combinations&lt;/a&gt;. Strongly Connected Components is more mathematical and often features interviews.  I've only listened to Permutations and Combinations once or twice.  I believe it's more of a comedy show.&lt;/p&gt;&#xA;" OwnerUserId="247" LastActivityDate="2010-07-28T17:38:48.037" CommunityOwnedDate="2010-07-28T17:38:48.037" />
  <row Id="994" PostTypeId="2" ParentId="903" CreationDate="2010-07-28T17:46:08.790" Score="5" ViewCount="" Body="&lt;p&gt;Try Simmons, Introduction to Topology and Modern Analysis.&lt;/p&gt;&#xA;" OwnerUserId="218" LastActivityDate="2010-07-28T17:46:08.790" CommunityOwnedDate="2010-07-28T17:46:08.790" CommentCount="2" />
  <row Id="995" PostTypeId="2" ParentId="370" CreationDate="2010-07-28T17:47:40.813" Score="1" ViewCount="" Body="&lt;p&gt;First Chapter of Jacobson's Basic Algebra -II.&lt;/p&gt;&#xA;" OwnerUserId="218" LastActivityDate="2010-07-28T17:47:40.813" CommunityOwnedDate="2010-07-28T17:47:40.813" />
  <row Id="996" PostTypeId="1" CreationDate="2010-07-28T17:52:07.437" Score="0" ViewCount="186" Body="&lt;p&gt;I have lifted this from &lt;a href=&quot;http://mathoverflow.net/questions/11669/&quot; rel=&quot;nofollow&quot;&gt;Mathoverflow&lt;/a&gt; since it belongs here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Currently, I'm taking matrix theory, and our textbook is Strang's Linear Algebra. Besides matrix theory, which all engineers must take, there exists linear algebra I and II for math majors. What is the difference, if any, between matrix theory and linear algebra?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&#xA;kolistivra&lt;/p&gt;&#xA;" OwnerUserId="218" LastEditorUserId="536" LastEditorDisplayName="" LastEditDate="2010-07-28T18:50:50.077" LastActivityDate="2010-07-28T18:50:50.077" Title="What is the difference between matrix theory and linear algebra?" Tags="&lt;linear-algebra&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="997" PostTypeId="2" ParentId="329" CreationDate="2010-07-28T17:53:44.890" Score="4" ViewCount="" Body="&lt;p&gt;&lt;em&gt;Problems in Algebraic Number Theory&lt;/em&gt; is written in a style I'd like to see in more textbooks&lt;/p&gt;&#xA;" OwnerUserId="256" LastActivityDate="2010-07-28T17:53:44.890" CommunityOwnedDate="2010-07-28T17:53:44.890" />
  <row Id="998" PostTypeId="1" CreationDate="2010-07-28T18:00:21.333" Score="12" ViewCount="915" Body="&lt;p&gt;Lifted from &lt;a href=&quot;http://mathoverflow.net/questions/2446&quot; rel=&quot;nofollow&quot;&gt;Mathoverflow&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think (almost) everyone agrees that Hartshorne's Algebraic Geometry is still the best.&#xA;Then what might be the 2nd best? It can be a book, preprint, online lecture note, webpage, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One suggestion per answer please. Also, please include an explanation of why you like the book, or what makes it unique or useful.&lt;/p&gt;&#xA;" OwnerUserId="218" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-07-28T19:39:14.497" LastActivityDate="2011-03-01T18:07:15.047" CommunityOwnedDate="2010-07-28T18:00:32.910" Title="Best Algebraic Geometry text book? (other than Hartshorne)" Tags="&lt;big-list&gt;&lt;books&gt;&lt;algebraic-geometry&gt;" AnswerCount="9" CommentCount="1" FavoriteCount="9" />
  <row Id="1002" PostTypeId="1" CreationDate="2010-07-28T18:07:57.030" Score="14" ViewCount="1315" Body="&lt;p&gt;A vague question of Kevin Lin which didn't quite fit at &lt;a href=&quot;http://mathoverflow.net/questions/446&quot; rel=&quot;nofollow&quot;&gt;Mathoverflow&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So ... what is the Fourier transform? What does it do? Why is it useful (both in math and in engineering, physics, etc)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Answers at any level of sophistication are welcome.)&lt;/p&gt;&#xA;" OwnerUserId="218" LastEditorUserId="218" LastEditorDisplayName="" LastEditDate="2010-07-28T20:56:38.927" LastActivityDate="2011-05-08T21:44:41.327" CommunityOwnedDate="2011-04-14T11:45:07.350" Title="Fourier transform for dummies" Tags="&lt;fourier-analysis&gt;" AnswerCount="9" CommentCount="1" FavoriteCount="12" />
  <row Id="1008" PostTypeId="2" ParentId="996" CreationDate="2010-07-28T18:17:43.873" Score="3" ViewCount="" Body="&lt;p&gt;My answer from the MO thread:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A matrix is just a list of numbers, and you're allowed to add and multiply matrices by combining those numbers in a certain way. When you talk about matrices, you're allowed to talk about things like the entry in the 3rd row and 4th column, and so forth. In this setting, matrices are useful for representing things like transition probabilities in a Markov chain, where each entry indicates the probability of transitioning from one state to another. You can do lots of interesting numerical things with matrices, and these interesting numerical things are very important because matrices show up a lot in engineering and the sciences.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In linear algebra, however, you instead talk about linear transformations, which are &lt;strong&gt;not&lt;/strong&gt; (I cannot emphasize this enough) a list of numbers, although sometimes it is convenient to use a particular matrix to write down a linear transformation. The difference between a linear transformation and a matrix is not easy to grasp the first time you see it, and most people would be fine with conflating the two points of view. However, when you're given a linear transformation, you're not allowed to ask for things like the entry in its 3rd row and 4th column because questions like these depend on a choice of basis. Instead, you're only allowed to ask for things that don't depend on the basis, such as the rank, the trace, the determinant, or the set of eigenvalues. This point of view may seem unnecessarily restrictive, but it is fundamental to a deeper understanding of pure mathematics.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-07-28T18:17:43.873" />
  <row Id="1010" PostTypeId="2" ParentId="996" CreationDate="2010-07-28T18:21:49.857" Score="0" ViewCount="" Body="&lt;p&gt;I think of Linear Algebra is a semi-applied Matrix theory. I can't summarize it better then Qiaochu. &lt;/p&gt;&#xA;" OwnerUserId="287" LastActivityDate="2010-07-28T18:21:49.857" />
  <row Id="1012" PostTypeId="1" CreationDate="2010-07-28T18:24:38.307" Score="1" ViewCount="67" Body="&lt;p&gt;An off-topic question posed at &lt;a href=&quot;http://mathoverflow.net/questions/5095/&quot; rel=&quot;nofollow&quot;&gt;Mathoverflow&lt;/a&gt; by Andrew Stacey, but one which fits here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One thing that came out of Terry Tao's recent blog posts on this matter (&lt;a href=&quot;http://terrytao.wordpress.com/2009/10/29/displaying-mathematics-on-the-web/&quot; rel=&quot;nofollow&quot;&gt;first post&lt;/a&gt; and &lt;a href=&quot;http://terrytao.wordpress.com/2009/11/04/displaying-maths-online-ii/&quot; rel=&quot;nofollow&quot;&gt;follow up&lt;/a&gt;) is that it's hard to get an overview of all the different ways of getting one's amazing mathematics onto the web.  I thought it'd be useful to gather together a list of such.  This meant to be a list of ways to do it, not examples of where it's already being done.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Standard community wiki rules: one thing per answer and feel free to edit other's answers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Additional rules: it'd be useful to have a little more than just links.  A brief description, pros and cons (be objective), platforms (does it only work on Linux, sort of thing) - things that might help someone decide which things to examine further.&lt;/p&gt;&#xA;" OwnerUserId="218" LastActivityDate="2010-07-28T18:32:36.403" CommunityOwnedDate="2010-07-28T18:24:38.307" Title="Resources for getting maths on to the web." Tags="&lt;soft-question&gt;" AnswerCount="1" FavoriteCount="2" />
  <row Id="1016" PostTypeId="2" ParentId="1002" CreationDate="2010-07-28T18:25:58.740" Score="3" ViewCount="" Body="&lt;p&gt;I'll give an engineering answer. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have a time series that you think is the result of a additive collection of periodic function, the Fourier transform will help you determine what the dominant frequencies are. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the way guitar tuners work. The perform and FFT on the sound data and pick out the frequency with the greatest power (squares of the real and imaginary parts) and consider that the &quot;note.&quot; This is called the fundamental frequency.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are many other uses, so you might want to add big list as a tag.&lt;/p&gt;&#xA;" OwnerUserId="99" LastActivityDate="2010-07-28T18:25:58.740" CommunityOwnedDate="2011-04-14T11:45:07.350" CommentCount="1" />
  <row Id="1017" PostTypeId="1" AcceptedAnswerId="1262" CreationDate="2010-07-28T18:26:41.203" Score="3" ViewCount="292" Body="&lt;p&gt;Using only precalculus mathematics (including that the area of the triangle with vertices at the origin, (x1,y1), and (x2,y2) is half of the absolute value of the determinant of the 2x2 matrix of the vertices (x1,y1) and (x2,y2), 1/2 * |x1*y2 - x2*y1|) how can one prove that &lt;a href=&quot;http://en.wikipedia.org/wiki/Shoelace_Method&quot; rel=&quot;nofollow&quot;&gt;the shoelace method&lt;/a&gt; works for all non-self-intersecting polygons?&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-07-30T20:44:23.697" Title="Proving the Shoelace Method at the Precalculus Level" Tags="&lt;geometry&gt;&lt;algebra-precalculus&gt;&lt;contest-math&gt;&lt;analytic-geometry&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="1018" PostTypeId="1" CreationDate="2010-07-28T18:26:53.573" Score="3" ViewCount="185" Body="&lt;p&gt;An off-topic question asked at &lt;a href=&quot;http://mathoverflow.net/questions/3036/&quot; rel=&quot;nofollow&quot;&gt;Mathoverflow&lt;/a&gt; by Andrew Stacey; but one which fits here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm interested in hearing of examples of mathematical (or, at a pinch, scientific) websites with serious content where the design of the website actually makes it easy to read and absorb the material. To be absolutely clear, the mathematical content of the website should be on the website itself and not in an electronic article (so meta-sites that make it easy to find material, like MathSciNet or the arXiv, don't count).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: I'm extending this to non-internet material. I want examples where the design of the document/website/whatever actually helped when reading the material.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a little background, I know that LaTeX is meant to help us separate content from context and concentrate on each one in turn, but I often feel when reading an article that the author has concentrated solely on the content and left all of the context to TeX. This is most obvious with websites where there are some really well-designed websites to compare with, but holds as well with articles.&lt;/p&gt;&#xA;" OwnerUserId="218" LastActivityDate="2010-07-28T21:03:46.010" CommunityOwnedDate="2010-07-28T18:26:53.573" Title="Examples of well-displayed mathematics on the internet" Tags="&lt;soft-question&gt;" AnswerCount="2" FavoriteCount="2" />
  <row Id="1019" PostTypeId="2" ParentId="1002" CreationDate="2010-07-28T18:27:08.130" Score="2" ViewCount="" Body="&lt;p&gt;You could think of a Fourier series expanding a function as a sum of sines and cosines analogous to the way a Taylor series expands a function as a sum of powers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or you could think of the Fourier series as a change of variables.  A fundamental skill in engineering and physics is to pick the coordinate system that makes your problem simplest.  Since the derivatives of sines and cosines are more sines and cosines, Fourier series are the right &quot;coordinate system&quot; for many problems involving derivatives.&lt;/p&gt;&#xA;" OwnerUserId="247" LastActivityDate="2010-07-28T18:27:08.130" CommunityOwnedDate="2011-04-14T11:45:07.350" CommentCount="0" />
  <row Id="1022" PostTypeId="2" ParentId="1012" CreationDate="2010-07-28T18:32:36.403" Score="2" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://mathurl.com&quot; rel=&quot;nofollow&quot;&gt;http://mathurl.com&lt;/a&gt; is handy if you want to send someone a quick link to a mathematical expression.&lt;/p&gt;&#xA;" OwnerUserId="247" LastActivityDate="2010-07-28T18:32:36.403" CommunityOwnedDate="2010-07-28T18:32:36.403" />
  <row Id="1024" PostTypeId="2" ParentId="1018" CreationDate="2010-07-28T18:33:22.843" Score="2" ViewCount="" Body="&lt;p&gt;For students in lower level university courses, there are two amazing resources.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;http://tutorial.math.lamar.edu/&quot; rel=&quot;nofollow&quot;&gt;Paul's online math notes&lt;/a&gt;. My linear algebra book was complete garbage. I pretty much used these notes to get through the class. It explains everything very well and does not assume that you already know graduate level mathematics. Presentation wise, its very simple. A plain HTML website, however content is king. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;And ofcourse, Khan Academy. I think everyone knows about Khan. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Last but not least, I would like to add the MIT online courses. I didn't like them that much but they did help me. Maybe its just me but I can not learn anything while staring at a computer screen.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="287" LastActivityDate="2010-07-28T18:33:22.843" CommunityOwnedDate="2010-07-28T18:33:22.843" CommentCount="5" />
  <row Id="1025" PostTypeId="1" AcceptedAnswerId="1027" CreationDate="2010-07-28T18:36:05.147" Score="1" ViewCount="238" Body="&lt;p&gt;Conic sections are a frequent target for dropping when attempting to make room for other topics in advanced algebra and precalculus courses.  A common argument in favor of dropping them is that typical first-year calculus doesn't use conic sections at all.  Do conic sections come up in typical intro-level undergraduate courses?  In typical prelim grad-level courses?  If so, where?&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-07-28T21:32:34.943" Title="Usefulness of Conic Sections" Tags="&lt;algebra-precalculus&gt;&lt;teaching&gt;&lt;high-school&gt;" AnswerCount="4" />
  <row Id="1026" PostTypeId="1" AcceptedAnswerId="1028" CreationDate="2010-07-28T18:39:19.617" Score="3" ViewCount="96" Body="&lt;p&gt;How many ways are there to color faces of a cube with N colors if &#xA;two colorings are the same if it's possible to rotate the cube such that&#xA;one coloring goes to another?&lt;/p&gt;&#xA;" OwnerUserId="382" LastActivityDate="2010-07-28T18:41:56.710" Title="Number of colorings of cube's faces" Tags="&lt;combinatorics&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="1027" PostTypeId="2" ParentId="1025" CreationDate="2010-07-28T18:40:51.893" Score="4" ViewCount="" Body="&lt;p&gt;Conic sections are basic examples in algebraic geometry, since they are (the real forms of) curves of genus zero.  As such, they are also basic examples in number theory, since it is easy to determine the rational points on a conic section, and this is a good warm-up for studying more complicated Diophantine equations.  In fact, curves of genus zero are the &lt;em&gt;only&lt;/em&gt; class of variety for which an algorithm provably exists to determine the rational points!  Even for the next hardest case, elliptic curves, there are no algorithms which provably always work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A great survey of these topics is Bjorn Poonen's &lt;a href=&quot;http://www-math.mit.edu/~poonen/papers/millennial.pdf&quot; rel=&quot;nofollow&quot;&gt;Computing rational points on curves&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit:  There is also Franz Lemmermeyer's &lt;a href=&quot;http://www.fen.bilkent.edu.tr/~franz/publ/conics.pdf&quot; rel=&quot;nofollow&quot;&gt;Conics - a poor man's elliptic curves&lt;/a&gt;, which explains how certain conics can be thought of as &quot;degenerate&quot; elliptic curves.&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-07-28T19:51:03.437" LastActivityDate="2010-07-28T19:51:03.437" />
  <row Id="1028" PostTypeId="2" ParentId="1026" CreationDate="2010-07-28T18:41:56.710" Score="4" ViewCount="" Body="&lt;p&gt;The number of different colorings is equal to $\frac{n^6 + 3n^4 + 12n^3 + 8n^2}{24}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can get this number using &lt;a href=&quot;http://en.wikipedia.org/wiki/Burnside_lemma&quot; rel=&quot;nofollow&quot;&gt;Burnside lemma&lt;/a&gt;.&#xA;The wikipedia article contains solution of your problem as well.&lt;/p&gt;&#xA;" OwnerUserId="334" LastActivityDate="2010-07-28T18:41:56.710" />
  <row Id="1029" PostTypeId="2" ParentId="740" CreationDate="2010-07-28T18:52:19.560" Score="0" ViewCount="" Body="&lt;p&gt;Three examples suffice to show why some modes of convergence don't imply other modes of convergence: pointwise convergence, Lp norm convergence, convergence in measure, etc.  See the counterexamples section &lt;a href=&quot;http://www.johndcook.com/modes_of_convergence.html&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="247" LastActivityDate="2010-07-28T18:52:19.560" CommunityOwnedDate="2010-07-28T18:52:19.560" />
  <row Id="1030" PostTypeId="2" ParentId="740" CreationDate="2010-07-28T18:55:51.020" Score="1" ViewCount="" Body="&lt;p&gt;When I was first learning calculus, the fact that $sin(1/x)$ is continuous on the set $(0,\infty)$ gave me a headache. &lt;/p&gt;&#xA;" OwnerUserId="317" LastActivityDate="2010-07-28T18:55:51.020" CommunityOwnedDate="2010-07-28T18:55:51.020" CommentCount="1" />
  <row Id="1031" PostTypeId="2" ParentId="544" CreationDate="2010-07-28T18:58:18.923" Score="1" ViewCount="" Body="&lt;p&gt;Markov chains are used in Markov Chain Monte Carlo (MCMC).  This computational technique is extremely common in Bayesian statistics.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In Bayesian statistics, you want to compute properties of a posterior distribution.  You'd like to draw independent samples from this distribution, but often this is impractical.  So you construct a Markov chain that has as its limiting distribution the the distribution you want.  So, for example, to get the mean of your posterior distribution you could take the mean of the states of your Markov chain.  (Ergodic theory blesses this process.)&lt;/p&gt;&#xA;" OwnerUserId="247" LastActivityDate="2010-07-28T18:58:18.923" />
  <row Id="1032" PostTypeId="2" ParentId="668" CreationDate="2010-07-28T19:05:34.603" Score="17" ViewCount="" Body="&lt;p&gt;You could think of a determinant as a volume.  Think of the columns of the matrix as vectors at the origin forming the edges of a skewed box.  The determinant gives the volume of that box.  For example, in 2 dimensions, the columns of the matrix are the edges of a rhombus.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can derive the algebraic properties from this geometrical interpretation.  For example, if two of the columns are linearly dependent, you're box is missing a dimension and so it's been flattened to have zero volume.&lt;/p&gt;&#xA;" OwnerUserId="247" LastActivityDate="2010-07-28T19:05:34.603" CommentCount="2" />
  <row Id="1033" PostTypeId="2" ParentId="381" CreationDate="2010-07-28T19:08:36.673" Score="1" ViewCount="" Body="&lt;p&gt;Here's an example of Fibonacci numbers applied to &lt;a href=&quot;http://www.johndcook.com/blog/2008/04/23/fibonacci-numbers-at-work/&quot; rel=&quot;nofollow&quot;&gt;numerical integration&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="247" LastActivityDate="2010-07-28T19:08:36.673" CommunityOwnedDate="2010-08-03T18:46:03.740" />
  <row Id="1034" PostTypeId="2" ParentId="381" CreationDate="2010-07-28T19:10:27.917" Score="1" ViewCount="" Body="&lt;p&gt;Here's a humorous application of Fibonacci numbers to &lt;a href=&quot;http://www.johndcook.com/blog/2009/05/19/golden-ratio-rational-approximation/&quot; rel=&quot;nofollow&quot;&gt;breastfeeding twins&lt;/a&gt;.  Even though the post is somewhat of a joke, it makes a serious point about what are called &quot;almost periodic functions.&quot; &lt;/p&gt;&#xA;" OwnerUserId="247" LastActivityDate="2010-07-28T19:10:27.917" CommunityOwnedDate="2010-08-03T18:46:03.740" />
  <row Id="1035" PostTypeId="2" ParentId="213" CreationDate="2010-07-28T19:14:48.937" Score="6" ViewCount="" Body="&lt;p&gt;I wish I'd understood the importance of inequalities earlier.  I wish I'd carefully gone through the classic book &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0521358809&quot; rel=&quot;nofollow&quot;&gt;Inequalities&lt;/a&gt; by Hardy, Littlewood, and Poyla early on.  Another good book is &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/052154677X&quot; rel=&quot;nofollow&quot;&gt;The Cauchy-Schwarz Masterclass&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can study inequalities as a subject in their own right, often without using advanced math.  But they're critical techniques for advanced math.&lt;/p&gt;&#xA;" OwnerUserId="247" LastActivityDate="2010-07-28T19:14:48.937" CommunityOwnedDate="2010-07-28T19:14:48.937" CommentCount="2" />
  <row Id="1036" PostTypeId="2" ParentId="588" CreationDate="2010-07-28T19:17:44.657" Score="1" ViewCount="" Body="&lt;p&gt;To your question regarding radius of convergence, &lt;a href=&quot;http://en.wikipedia.org/wiki/Radius_of_convergence&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt; gives a good answer.&lt;/p&gt;&#xA;" OwnerUserId="247" LastActivityDate="2010-07-28T19:17:44.657" />
  <row Id="1037" PostTypeId="1" AcceptedAnswerId="1047" CreationDate="2010-07-28T19:19:40.470" Score="5" ViewCount="452" Body="&lt;p&gt;Is the purpose of the derivative notation d/dx strictly for symbolic manipulation purposes?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I remember being confused when I first saw the notation for derivatives - it looks vaguely like there's some division going on and there are some fancy 'd' characters that are added in... I recall thinking that it was a lot of characters to represent an action with respect to one variable. Of course, once you start moving the dx around it makes a little more sense as to why they exist - but is this the only reason?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any history lesson or examples where this notation is helpful or unhelpful is appreciated.&lt;/p&gt;&#xA;" OwnerUserId="82" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-07-28T19:35:35.550" LastActivityDate="2010-12-18T22:36:49.877" Title="Why are derivatives specified as d/dx?" Tags="&lt;reference-request&gt;&lt;notation&gt;&lt;math-history&gt;" AnswerCount="8" FavoriteCount="1" />
  <row Id="1038" PostTypeId="2" ParentId="1037" CreationDate="2010-07-28T19:24:26.903" Score="7" ViewCount="" Body="&lt;p&gt;If you have access to it, the book &lt;em&gt;A History of Mathematical Notations&lt;/em&gt;, by Florian Cajori, has a pretty detailed description of the history of notations for derivatives in its second volume.&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2010-07-28T19:24:26.903" />
  <row Id="1039" PostTypeId="2" ParentId="1037" CreationDate="2010-07-28T19:38:49.083" Score="0" ViewCount="" Body="&lt;p&gt;Adding to Mariano's answer, its worthwhile to look at the modern notation and compare it with the older one. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a link that explains it very well. &#xA;&lt;a href=&quot;http://www.vendian.org/mncharity/dir3/dxdoc/&quot; rel=&quot;nofollow&quot;&gt;http://www.vendian.org/mncharity/dir3/dxdoc/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="287" LastActivityDate="2010-07-28T19:38:49.083" />
  <row Id="1040" PostTypeId="2" ParentId="1002" CreationDate="2010-07-28T19:40:11.203" Score="-2" ViewCount="" Body="&lt;p&gt;A more complicated answer (yet it's going to be imprecise, because I haven't touched this in 15 years...) is the following.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a 3-dimentional space (for example) you can represent a vector &lt;strong&gt;v&lt;/strong&gt; by its end point coordinates, x, y, z, in a very simple way. You choose three vectors which are of unit length and orthogonal with each other (a base), say &lt;strong&gt;i&lt;/strong&gt;, &lt;strong&gt;j&lt;/strong&gt; and &lt;strong&gt;k&lt;/strong&gt;, and calculate the coordinates as such:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;x = &lt;strong&gt;v&lt;/strong&gt; ∙ &lt;strong&gt;i&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;y = &lt;strong&gt;v&lt;/strong&gt; ∙ &lt;strong&gt;j&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;z = &lt;strong&gt;v&lt;/strong&gt; ∙ &lt;strong&gt;k&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In multidimentional space, the equations still hold. In a discrete infinite space, the coordinates and the base vectors become a sequence. The dot product becomes an infinite sum. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a continuous infinite space (like the space of good functions) the coordinates and the bases become functions and the dot product an infinite integral.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the Fourier transform is exactly this kind of operation (based on a set of base functions which are basically a set of sines and cosines). In other words, it is a different representation of the same function in relation to a particular set of base functions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a consequence, for example, functions of time, represented against functions of time and space (in other words integrated over time multiplied by functions of space and time), become functions of space, and so on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope it helps!&lt;/p&gt;&#xA;" OwnerUserId="272" LastActivityDate="2010-07-28T19:40:11.203" CommunityOwnedDate="2011-04-14T11:45:07.350" CommentCount="6" />
  <row Id="1041" PostTypeId="2" ParentId="998" CreationDate="2010-07-28T19:53:44.257" Score="4" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0387977163&quot; rel=&quot;nofollow&quot;&gt;Algebraic Geometry: A First Course&lt;/a&gt; by Joe Harris is a very good book that sits in that region between undergraduate treatments and the prerequisites of Hartshorne. In particular, one does not need to know much commutative algebra to get a lot out of Harris's book. Harris himself recommends reading Hartshorne after his book for the theory of schemes.&lt;/p&gt;&#xA;" OwnerUserId="251" LastActivityDate="2010-07-28T19:53:44.257" CommunityOwnedDate="2010-07-28T19:53:44.257" CommentCount="1" />
  <row Id="1042" PostTypeId="2" ParentId="980" CreationDate="2010-07-28T19:57:50.113" Score="1" ViewCount="" Body="&lt;p&gt;I will not reiterate the very nice definitions found in the other answers, however I think that these &quot;practical&quot; definitions might help you as well on an intuitive level.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Open sets are typically used as domains for functions, as they are more useful for analysing &quot;continuous&quot; properties like differentiability. Also they don't have nasty borders (hence you don't have to deal with functions which are well behaved only on one side of the edge).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Closed sets are useful because, if they are limited, they are compact.&lt;/p&gt;&#xA;" OwnerUserId="272" LastActivityDate="2010-07-28T19:57:50.113" />
  <row Id="1043" PostTypeId="2" ParentId="1025" CreationDate="2010-07-28T19:57:57.367" Score="0" ViewCount="" Body="&lt;p&gt;The study of parabolas (with axis parallel to y-axis) is useful when you have to solve 2nd degree inequations.&lt;/p&gt;&#xA;" OwnerUserId="354" LastActivityDate="2010-07-28T19:57:57.367" CommentCount="1" />
  <row Id="1044" PostTypeId="2" ParentId="1037" CreationDate="2010-07-28T20:04:14.967" Score="1" ViewCount="" Body="&lt;p&gt;This is the Leibniz notation, which is based on the ratio of &quot;infinitesimals&quot;. dy and dx are, respectively, the infinitesimal increment of the dependent variable y and the infinitesimal increment of the variable x. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are other notations: Newton notation, which puts a dot over the variable name, and Cauchy notation, which uses the operator D, as in D(sin(x))=cos(x).&lt;/p&gt;&#xA;" OwnerUserId="354" LastActivityDate="2010-07-28T20:04:14.967" CommentCount="2" />
  <row Id="1045" PostTypeId="1" AcceptedAnswerId="1055" CreationDate="2010-07-28T20:06:54.867" Score="1" ViewCount="91" Body="&lt;p&gt;I have some polygons I would like to map onto the face of a cone.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can see from &lt;a href=&quot;http://www.math.montana.edu/frankw/ccp/multiworld/multipleIVP/cylindrical/body.htm#skip3&quot; rel=&quot;nofollow&quot;&gt;this page&lt;/a&gt; that I can convert the points of the polygon to cylinderical coordinates, which is almost what I want.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do I go about modifying the formulas to work for conical coordinates?&lt;/p&gt;&#xA;" OwnerUserId="386" LastEditorUserId="536" LastEditorDisplayName="" LastEditDate="2010-07-28T20:45:10.763" LastActivityDate="2010-07-28T20:56:53.403" Title="How do I convert from cartesian to conical coordinates?" Tags="&lt;geometry&gt;" AnswerCount="1" />
  <row Id="1046" PostTypeId="2" ParentId="1037" CreationDate="2010-07-28T20:08:43.270" Score="4" ViewCount="" Body="&lt;p&gt;If you're a physics kind of person, then a good reason to like this notation is that it gives the correct units for the derivative: whatever units $f(x)$ is in, the units for $\frac{d}{dx} f(x)$ are obtained by dividing by the units for $x$.  &lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-07-28T20:08:43.270" />
  <row Id="1047" PostTypeId="2" ParentId="1037" CreationDate="2010-07-28T20:10:20.723" Score="11" ViewCount="" Body="&lt;p&gt;Because of their definition:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Start with a function, calculate the difference in value between two points and divide by the size of the interval between the two. You can represent this as such:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;f(x&lt;sub&gt;2&lt;/sub&gt;) - f(x&lt;sub&gt;1&lt;/sub&gt;) / x&lt;sub&gt;2&lt;/sub&gt; - x&lt;sub&gt;1&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or&lt;/p&gt;&#xA;&#xA;&lt;p&gt;∆f(x) / ∆x (where ∆, delta, is the greek capital D and indicates an interval)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;now, take the limit as ∆x goes to zero, and you have the differential. This is indicated by using a lower case d instead of the ∆.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;df(x) / dx&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if this operation is treated as an operator applied to a function, it is usually represented as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;d / dx (applied to f(x) )&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that (typically in physics), you can also use the letter δ to indicate very small intervals and in general you would use the symbol ∂ to represent partial differentials. They are all variations of the letter D.&lt;/p&gt;&#xA;" OwnerUserId="272" LastActivityDate="2010-07-28T20:10:20.723" CommentCount="1" />
  <row Id="1048" PostTypeId="1" AcceptedAnswerId="1052" CreationDate="2010-07-28T20:21:51.977" Score="5" ViewCount="432" Body="&lt;p&gt;In school, we learn that sin is &quot;opposite over hypotenuse&quot; and cos is &quot;adjacent over hypotenuse&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Later on, we learn the power series definitions of sin and cos.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can one prove that these two definitions are equivalent?&lt;/p&gt;&#xA;" OwnerUserId="263" LastActivityDate="2010-08-04T04:31:50.797" Title="Different definitions of trigonometric functions" Tags="&lt;geometry&gt;&lt;calculus&gt;&lt;trigonometry&gt;" AnswerCount="6" CommentCount="2" FavoriteCount="2" />
  <row Id="1049" PostTypeId="2" ParentId="1002" CreationDate="2010-07-28T20:27:01.353" Score="0" ViewCount="" Body="&lt;p&gt;Here's some simple &lt;a href=&quot;http://www.gnu.org/software/octave/&quot; rel=&quot;nofollow&quot;&gt;Matlab&lt;/a&gt; code to play around with if you like.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&#xA;&lt;code&gt;&#xA;% This code will approxmmate the function f using the DFT&#xA;&#xA;clear all &#xA;close all &#xA;&#xA;a=0;b=2*pi; % define interval, i.e. endpoints of domain(f) &#xA;&#xA;N=20; % number of sample points to take from f&#xA;&#xA;% build vector of points in domain(f) to sample from&#xA;for j=1:N+1&#xA;   x(j) = (b-a)*(j-1)/N;&#xA;end&#xA;&#xA;f= cos(x); % approximate cos(x) with resulting Fourier series&#xA;&#xA;&#xA;% build matrix of powers of roots of unity&#xA;for m=1:N+1&#xA;   for n=1:N+1&#xA;      F(m,n) = exp((m-1)*(n-1)*(2*pi*i)/N);&#xA;   end&#xA;end&#xA;&#xA;% solve f = Fc by domng F\f&#xA;c = F\f'; % c is vector of Fourier coefficients&#xA;&#xA;% plot discrete cos(x) using N points &#xA;&#xA;xx=0:0.01:2*pi;&#xA;plot(xx,cos(xx),'g')&#xA;hold on&#xA;&#xA;% build the Fourier series using coefficients from c&#xA;summ=0;&#xA;for k=1:length(c)&#xA;   summ = summ + c(k)*exp(i*(k-1)*x); &#xA;end&#xA;&#xA;% plot the fourier series against the discrete sin function &#xA;plot(x,summ)&#xA;legend('actual','approx.')&#xA;&#xA;&#xA;&lt;/code&gt;&#xA;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;As written you will have the first N=20 terms of the Fourier approximation to the cosine on the interval [a,b]=[0,2*pi].  Not very interesting as is...  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reference: &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0961408804&quot; rel=&quot;nofollow&quot;&gt;Gilbert Strang&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="121" LastActivityDate="2010-07-28T20:27:01.353" CommunityOwnedDate="2011-04-14T11:45:07.350" />
  <row Id="1050" PostTypeId="2" ParentId="687" CreationDate="2010-07-28T20:28:40.367" Score="2" ViewCount="" Body="&lt;p&gt;The form this question is usually asked is whether adding a route can increase the average traveling time, and this is known as &lt;a href=&quot;http://en.wikipedia.org/wiki/Braess_paradox&quot; rel=&quot;nofollow&quot;&gt;Braess's paradox&lt;/a&gt;.  The Wiki article gives an explicit example in which the travel time on some of the routes depends on the traffic.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-07-28T20:28:40.367" />
  <row Id="1051" PostTypeId="2" ParentId="1025" CreationDate="2010-07-28T20:43:13.187" Score="4" ViewCount="" Body="&lt;p&gt;If you're a physics sort of person, conic sections clearly come up when you study how Kepler figured out what the shapes of orbits are, and some of their synthetic properties give useful shortcuts to things like proving &quot;equal area swept out in equal time&quot; that need not involve calculus.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other skills you typically learn while studying conic sections in analytic geometry - polar parametrization of curves, basic facts about various invariants related to triangles and conics, rotations and changing coordinate systems (so as to recognize the equation of a conic in general form as some sort of transformation of a standard done), are all extremely useful in physics. I'd say that plane analytic geometry was the single most useful math tool for me in solving physics problems until I got to fluid dynamics stuff (where that is replaced by complex analysis).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Relatedly, independent of their use in physics, I think they're a great way to show the connections between analytic and synthetic thinking in math, which will come up over and over again for people who go on to study math (coordinate-based versus intrinsic perspectives, respectively).&lt;/p&gt;&#xA;" OwnerUserId="76" LastActivityDate="2010-07-28T20:43:13.187" />
  <row Id="1052" PostTypeId="2" ParentId="1048" CreationDate="2010-07-28T20:43:45.980" Score="5" ViewCount="" Body="&lt;p&gt;Most of the proofs in elementary calculus textbooks use the definition of $\sin x$ via geometry to prove that the derivative of $\sin x$ is $\cos x$ (namely, the fact that $\lim_{x \to 0} \frac{ \sin x}{x} = 1$). Consequently, it follows that $\sin x$ and $\cos x$ are the two linearly independent solutions of $y'' = -y$.  The power series equations are also two linearly independent solutions of this differential equation.  Moreover, $\sin x$ and its derivative coincide with the derivative of the power series for $ \sin x$ at zero (no surprise, it's a Taylor series).  Same for $\cos x$. By uniqueness of solutions to ordinary differential equations, this proves that $\sin x$ and $\cos x$ as defined in school are equal to their power series.  (This is an expansion of Qiaochu's comment.)&lt;/p&gt;&#xA;" OwnerUserId="536" LastActivityDate="2010-07-28T20:43:45.980" />
  <row Id="1053" PostTypeId="2" ParentId="1048" CreationDate="2010-07-28T20:44:50.827" Score="1" ViewCount="" Body="&lt;p&gt;As a rough outline, the circular definitions of sine and cosine (the y- and x-coordinates of the image of (1,0) under a rotation about the origin) lead to being able to differentiate sine and cosine, and once you know how to differentiate them (infinitely), &lt;a href=&quot;http://en.wikipedia.org/wiki/Taylor%27s_theorem&quot; rel=&quot;nofollow&quot;&gt;Taylor's Theorem&lt;/a&gt; justifies that the power series is equal to the function.&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-07-28T20:44:50.827" />
  <row Id="1054" PostTypeId="2" ParentId="1002" CreationDate="2010-07-28T20:54:58.573" Score="8" ViewCount="" Body="&lt;p&gt;Let me partially steal from the accepted answer on MO, and illustrate it with examples I understand:&lt;br&gt;&#xA;&lt;em&gt;The Fourier transform is a different representation that makes convolutions easy.&lt;/em&gt;&lt;br&gt;&#xA;Or, to quote directly from there: &quot;the Fourier transform is a unitary change of basis for functions (or distributions) that diagonalizes all convolution operators.&quot; This often involves expressing an arbitrary function as a superposition of &quot;symmetric&quot; functions of some sort, say functions of the form e&lt;sup&gt;itx&lt;/sup&gt; — in the common signal-processing applications, an arbitrary &quot;signal&quot; is decomposed as a superposition of &quot;waves&quot; (or &quot;frequencies&quot;).&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Example 1: Polynomial multiplication&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;This is the use of the discrete Fourier transform I'm most familiar with. Suppose you want to multiply two polynomials of degree n, given by their coefficients (a&lt;sub&gt;0&lt;/sub&gt;, …, a&lt;sub&gt;n&lt;/sub&gt;) and (b&lt;sub&gt;0&lt;/sub&gt;, …, b&lt;sub&gt;n&lt;/sub&gt;). In their product, the coefficient of x&lt;sup&gt;k&lt;/sup&gt; is c&lt;sub&gt;k&lt;/sub&gt; = &amp;sum;a&lt;sub&gt;i&lt;/sub&gt;b&lt;sub&gt;k-i&lt;/sub&gt;. This is a convolution, and doing it naively would take O(n&lt;sup&gt;2&lt;/sup&gt;) time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instead, suppose we represent the polynomials by their &lt;em&gt;values&lt;/em&gt; at 2n points. Then the value of the product polynomial (the one we want) at any point is simply the product of the values of our original two polynomials. Thus we have reduced convolution to pointwise multiplication. The Fourier transform and its inverse correspond to polynomial evaluation and interpolation respectively, for certain well-chosen points (roots of unity). The Fast Fourier Transform (FFT) is a way of doing both of these in O(n log n) time.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Example 2: Convolution of probability distributions&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Suppose we have two independent (continuous) random variables X and Y, with probability densities f and g respectively. In other words, P(X ≤ x) = ∫&lt;sup&gt;x&lt;/sup&gt;&lt;sub&gt;-∞&lt;/sub&gt; f(t)dt and P(Y ≤ y) = ∫&lt;sup&gt;y&lt;/sup&gt;&lt;sub&gt;-∞&lt;/sub&gt; f(t)dt. We often want the distribution of their sum X+Y, and this is given by a convolution: P(X+Y ≤ z) = ∫f(t)g(z-t)dt. This integration may be hard. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But instead of representing the random variables by their densities, we can also represent them by their &lt;a href=&quot;http://en.wikipedia.org/wiki/Characteristic_function_%28probability_theory%29&quot; rel=&quot;nofollow&quot;&gt;characteristic functions&lt;/a&gt; &amp;phi;&lt;sub&gt;X&lt;/sub&gt;(t) =  E[e&lt;sup&gt;itX&lt;/sup&gt;] and &amp;phi;&lt;sub&gt;Y&lt;/sub&gt;(t) = E[e&lt;sup&gt;itY&lt;/sup&gt;]. Then the characteristic function of X+Y is just:&lt;br&gt;&#xA;&amp;phi;&lt;sub&gt;X+Y&lt;/sub&gt;(t) = E[e&lt;sup&gt;it(X+Y)&lt;/sup&gt;] = &amp;phi;&lt;sub&gt;X&lt;/sub&gt;(t)&amp;phi;&lt;sub&gt;Y&lt;/sub&gt;(t) since they're independent. The characteristic function is the continuous Fourier transform of the density function; it is a change of representation in which convolution becomes pointwise multiplication.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To quote again the answer on MO, many transformations we want to study (translation, differentiation, integration, …) are actually convolutions, so the Fourier transform helps in a wide number of instances. &lt;/p&gt;&#xA;" OwnerUserId="205" LastActivityDate="2010-07-28T20:54:58.573" CommunityOwnedDate="2011-04-14T11:45:07.350" />
  <row Id="1055" PostTypeId="2" ParentId="1045" CreationDate="2010-07-28T20:56:53.403" Score="1" ViewCount="" Body="&lt;p&gt;Presumably you already have the formulas for converting from conical to rectangular coordinates as listed on the &lt;a href=&quot;http://en.wikipedia.org/wiki/Conical_coordinates&quot; rel=&quot;nofollow&quot;&gt;Wikipedia page for conical coordinates&lt;/a&gt;. You'll need to solve for r, μ, and ν in terms of x, y, and z to get your answer. I can't see offhand the easiest way to find a general formula, but if you're trying to find it for particular values of r, μ, and ν, it shouldn't be too hard.&lt;/p&gt;&#xA;" OwnerUserId="45" LastActivityDate="2010-07-28T20:56:53.403" />
  <row Id="1056" PostTypeId="2" ParentId="1048" CreationDate="2010-07-28T20:57:24.090" Score="0" ViewCount="" Body="&lt;p&gt;Call the highschool functions (defined by the &lt;em&gt;right triangle inscribed in a unit circle&lt;/em&gt;, the angle being equal to the &lt;em&gt;length of the arc of the circle&lt;/em&gt;) $\sin_h$ and $\cos_h$, and let $\sin_p$ and $\cos_p$ be the power series definitions. (Note that these functions are continuous and agree at the end points $0$ and $2\pi$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $\sin_p^2(\theta)+\cos_p^2(\theta)=1$ the power series definitions also form a right triangle. Hence $\sin_h = \sin_p \circ \gamma$ and $\cos_h = \cos_p \circ \gamma$ for some parameterization $\gamma$. We know the power series definitions satisfy the arc length criteria so $\gamma$ must be the identity function.&lt;/p&gt;&#xA;" LastActivityDate="2010-08-03T21:45:56.810" CommentCount="2" />
  <row Id="1057" PostTypeId="1" AcceptedAnswerId="1165" CreationDate="2010-07-28T21:01:19.230" Score="1" ViewCount="261" Body="&lt;p&gt;Suppose a finite group has the property that for every $x, y$, it follows that $(xy)^3 = x^3 y^3$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do you prove that it is abelian?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Edit: I recall that the correct exercise needed in addition that the order of the group is not divisible by 3.&lt;/p&gt;&#xA;" OwnerUserId="218" LastEditorUserId="218" LastEditorDisplayName="" LastEditDate="2010-07-29T18:06:58.200" LastActivityDate="2010-08-04T08:17:33.700" Title="Group with an endomorphism that is &quot;almost&quot; abelian is abelian." Tags="&lt;group-theory&gt;" AnswerCount="3" CommentCount="4" />
  <row Id="1058" PostTypeId="2" ParentId="1018" CreationDate="2010-07-28T21:03:46.010" Score="1" ViewCount="" Body="&lt;p&gt;Paul Bourke's &lt;a href=&quot;http://local.wasp.uwa.edu.au/~pbourke/&quot; rel=&quot;nofollow&quot;&gt;website&lt;/a&gt;, in particular his &lt;a href=&quot;http://local.wasp.uwa.edu.au/~pbourke/geometry/&quot; rel=&quot;nofollow&quot;&gt;geometry&lt;/a&gt; section fueled my passion for geometry at a young age, infact I learned the idea of analytic geometry from it. For this reason it holds a special place in my heart. It is a very rich site full of graphics and derivations, very easy to get sucked in and lose a few hours!&lt;/p&gt;&#xA;" LastActivityDate="2010-07-28T21:03:46.010" CommunityOwnedDate="2010-07-28T21:03:46.010" />
  <row Id="1059" PostTypeId="2" ParentId="967" CreationDate="2010-07-28T21:05:58.750" Score="5" ViewCount="" Body="&lt;p&gt;Recommendations:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Barendregt &amp;amp; Barendsen, 1984, &lt;a href=&quot;ftp://ftp.cs.kun.nl/pub/CompMath.Found/lambda.pdf&quot; rel=&quot;nofollow&quot;&gt;Introduction to lambda-calculus&lt;/a&gt;;&lt;/li&gt;&#xA;&lt;li&gt;Girard, Lafont &amp;amp; Taylor, 1987, &lt;a href=&quot;http://www.paultaylor.eu/stable/Proofs+Types.html&quot; rel=&quot;nofollow&quot;&gt;Proofs and Types&lt;/a&gt;;&lt;/li&gt;&#xA;&lt;li&gt;Sørenson &amp;amp; Urzyczyn, 1999, &lt;a href=&quot;http://folli.loria.fr/cds/1999/esslli99/courses/sorensen-urzyczyn.html&quot; rel=&quot;nofollow&quot;&gt;Lectures on the Curry-Howard Isomorphism&lt;/a&gt;.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;All of these are mentioned in &lt;a href=&quot;http://lambda-the-ultimate.org/node/492&quot; rel=&quot;nofollow&quot;&gt;the LtU Getting Started thread&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="100" LastActivityDate="2010-07-28T21:05:58.750" CommentCount="2" />
  <row Id="1061" PostTypeId="2" ParentId="1037" CreationDate="2010-07-28T21:11:54.930" Score="2" ViewCount="" Body="&lt;p&gt;hysterical raisins.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Calculus used to be done with infinitesimals (Archimedes &lt;em&gt;The Method of Mechanical Theorems&lt;/em&gt;, Newtons &lt;em&gt;Fluxions&lt;/em&gt;, ...) but there was some controversy about these &lt;a href=&quot;http://www.maths.tcd.ie/pub/HistMath/People/Berkeley/AnalCont.html&quot; rel=&quot;nofollow&quot;&gt;ghostly quantities&lt;/a&gt; eventually the whole foundation of analysis was rebuilt using limits but the old notations have been kept. So there is (as you noticed) a strange gap between reality (epsilonics) and intuition (infinitesimal quantities) but there are a few more recent redevelopments of the foundation of analysis for example &lt;a href=&quot;http://www.math.wisc.edu/~keisler/calc.html&quot; rel=&quot;nofollow&quot;&gt;Keisler&lt;/a&gt; or &lt;a href=&quot;http://publish.uwo.ca/~jbell/invitation%20to%20SIA.pdf&quot; rel=&quot;nofollow&quot;&gt;Bell&lt;/a&gt;.&lt;/p&gt;&#xA;" LastActivityDate="2010-07-28T21:11:54.930" CommentCount="2" />
  <row Id="1062" PostTypeId="2" ParentId="275" CreationDate="2010-07-28T21:19:25.870" Score="5" ViewCount="" Body="&lt;p&gt;I liked the &lt;a href=&quot;http://press.princeton.edu/titles/8350.html&quot; rel=&quot;nofollow&quot;&gt;Princeton Companion to Mathematics&lt;/a&gt;, though it is a heavy book and not always light reading. I'm a computer scientist with an interest in mathematics.&lt;/p&gt;&#xA;" OwnerUserId="210" LastActivityDate="2010-07-28T21:19:25.870" CommunityOwnedDate="2010-07-28T21:19:25.870" CommentCount="2" />
  <row Id="1063" PostTypeId="2" ParentId="370" CreationDate="2010-07-28T21:23:26.200" Score="2" ViewCount="" Body="&lt;p&gt;The &lt;a href=&quot;http://ncatlab.org/&quot; rel=&quot;nofollow&quot;&gt;nLab&lt;/a&gt; is a great resource for category theory.&lt;/p&gt;&#xA;" OwnerUserId="263" LastActivityDate="2010-07-28T21:23:26.200" CommunityOwnedDate="2010-07-28T21:23:26.200" CommentCount="1" />
  <row Id="1064" PostTypeId="1" CreationDate="2010-07-28T21:30:16.330" Score="4" ViewCount="426" Body="&lt;p&gt;Give an example of a perfect set in $\mathbb R^n$ that does not contain any of the rationals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Or prove that it does not exist).&lt;/p&gt;&#xA;" OwnerUserId="218" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-07-28T22:20:48.400" LastActivityDate="2011-02-04T06:01:21.833" Title="Perfect set without rationals" Tags="&lt;topology&gt;" AnswerCount="5" CommentCount="7" />
  <row Id="1065" PostTypeId="2" ParentId="1025" CreationDate="2010-07-28T21:32:34.943" Score="0" ViewCount="" Body="&lt;p&gt;Conic sections should definitely be retained. If you don't cover conic sections, then what other examples can you cover?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lines? Too simple. General curves? Insufficiently concrete.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Examples are very important for illustrating the general theory and techniques.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, in a multivariable calculus course, typical examples will involve quadric surfaces. Here conic sections will come into play, since hyperplane sections (or &quot;level curves&quot;) of quadric surfaces are conic sections.&lt;/p&gt;&#xA;" OwnerUserId="263" LastActivityDate="2010-07-28T21:32:34.943" />
  <row Id="1066" PostTypeId="1" AcceptedAnswerId="1071" CreationDate="2010-07-28T21:58:08.247" Score="4" ViewCount="191" Body="&lt;p&gt;From &lt;a href=&quot;http://scottaaronson.com/blog/?p=152&quot; rel=&quot;nofollow&quot;&gt;Scott Aaronson's blog&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;There’s a finite (and not&#xA;  unimaginably-large) set of boxes, such&#xA;  that if we knew how to pack those&#xA;  boxes into the trunk of your car, then&#xA;  we’d also know a proof of the Riemann&#xA;  Hypothesis. Indeed, every formal proof&#xA;  of the Riemann Hypothesis with at most&#xA;  (say) a million symbols corresponds to&#xA;  some way of packing the boxes into&#xA;  your trunk, and vice versa.&#xA;  Furthermore, a list of the boxes and&#xA;  their dimensions can be feasibly&#xA;  written down.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;His later commented to explain where he get this from: &quot;3-dimensional bin-packing is NP-complete.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't see how these two are related.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another question inspired by the same article is &lt;a href=&quot;http://math.stackexchange.com/questions/946/computation-with-a-memory-wiped-computer&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="96" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-07-29T00:58:01.943" LastActivityDate="2010-07-29T00:58:01.943" Title="Packing boxes and proof of Riemann Hypothesis" Tags="&lt;computer-science&gt;&lt;proof-theory&gt;" AnswerCount="1" FavoriteCount="2" />
  <row Id="1067" PostTypeId="2" ParentId="1064" CreationDate="2010-07-28T22:01:12.833" Score="15" ViewCount="" Body="&lt;p&gt;An easy example comes from the fact that a number with an infinite &lt;a href=&quot;http://en.wikipedia.org/wiki/Continued_fraction&quot; rel=&quot;nofollow&quot;&gt;continued fraction&lt;/a&gt; expansion is irrational (and conversely). The set of all irrationals with continued fractions consisting only of 1's and 2's in any arrangement is a perfect set of irrational numbers.&lt;/p&gt;&#xA;" OwnerUserId="25" LastActivityDate="2010-07-28T22:01:12.833" CommentCount="2" />
  <row Id="1068" PostTypeId="1" AcceptedAnswerId="2280" CreationDate="2010-07-28T22:19:32.567" Score="8" ViewCount="714" Body="&lt;p&gt;All the integrals I'm familiar with have the form:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$\int f(x)\mathrm{d}x$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And I understand these as the sum of infinite tiny rectangles with an area of: f(x_i) * dx.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it valid to have integrals that do not have a differential, such as dx, or that have the differential elsewhere than as a factor ? Let me give couple of examples on what I'm thinking of:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$\int 1$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If this is valid notation, I'd expect it to sum infinite ones together, thus to go inifinity.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$\int e^{\mathrm{d}x}$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Again, I'd expect this to go to infinity as e^0 = 1, assuming the notation is valid.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$\int (e^{\mathrm{d}x} - 1)$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This I could potentially imagine to have a finite value.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are any such integrals valid ? If so, are there any interesting / enlightening examples of such integrals?&lt;/p&gt;&#xA;" OwnerUserId="105" LastEditorUserId="105" LastEditorDisplayName="" LastEditDate="2010-08-12T15:17:22.003" LastActivityDate="2011-04-25T22:03:36.977" Title="Usage of dx in Integrals" Tags="&lt;calculus&gt;&lt;notation&gt;" AnswerCount="5" CommentCount="2" FavoriteCount="2" />
  <row Id="1069" PostTypeId="1" CreationDate="2010-07-28T22:21:41.067" Score="3" ViewCount="294" Body="&lt;p&gt;I would be glad if someone could explain in intuitive terms what these different derivatives are, and possibly give some practical, understandable examples of how they would produce different results.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To be clear, I would like to understand the geometrical or physical meaning of these operators more than the mathematical or topological subtleties that lead to them!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="272" LastEditorUserId="9" LastEditorDisplayName="" LastEditDate="2010-07-29T00:10:31.523" LastActivityDate="2010-07-29T00:10:31.523" Title="Intuitive explanation of covariant, contravariant and Lie derivatives" Tags="&lt;intuition&gt;&lt;differential-topology&gt;&lt;differential-geometry&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="5" />
  <row Id="1070" PostTypeId="2" ParentId="1068" CreationDate="2010-07-28T22:23:11.300" Score="11" ViewCount="" Body="&lt;p&gt;When you write $\int f(x) dx$, the whole of $\int ... dx$ is an indivisible symbol, just as the $d/dx$ is an indivisible symbol when you write $df/dx$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, there &lt;em&gt;are&lt;/em&gt; reasons why the notation is as it is, but trying to manipulate it like you suggest in $\int e^{dx}$, for example, is simply meaningless. &lt;/p&gt;&#xA;" OwnerUserId="274" LastEditorUserId="274" LastEditorDisplayName="" LastEditDate="2010-08-03T03:38:00.153" LastActivityDate="2010-08-03T03:38:00.153" CommentCount="7" />
  <row Id="1071" PostTypeId="2" ParentId="1066" CreationDate="2010-07-28T22:23:36.590" Score="5" ViewCount="" Body="&lt;p&gt;The question of whether a formal proof of the Riemann Hypothesis exists (with at most a million symbols) is a problem in NP: given such a proof, it can be verified to be correct in polynomial time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bin-packing is NP-complete: this means that every problem in NP can be reduced to bin packing. In particular, the problem mentioned in the previous paragraph can. (This is a reduction that can be made explicit, so once we specify the proof verifier etc., we can carry out the steps of the reduction to get an instance of bin packing. We also need the reduction to be &quot;parsimonious&quot; i.e. solutions correspond one-to-one; I believe it is.)&lt;/p&gt;&#xA;" OwnerUserId="205" LastActivityDate="2010-07-28T22:23:36.590" CommentCount="5" />
  <row Id="1072" PostTypeId="2" ParentId="1068" CreationDate="2010-07-28T22:26:38.787" Score="1" ViewCount="" Body="&lt;p&gt;No, it's not valid. The dx in the integral is a representation of the fact that the integral is obtained as an area, so multiplying the &quot;average&quot; of the function value at each point by an infinitesimal interval.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the manner in which we don't calculate the area does not change, the notation does not change.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are different notations that are used when the integral is over a curve, or over more than variable (thus leading for example to volumes).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The d(variable) notation is also used as a reminder that the integral is against a specific variable and not another, e.g. that int x/y dx differs from int x/y dy.&lt;/p&gt;&#xA;" OwnerUserId="272" LastActivityDate="2010-07-28T22:26:38.787" />
  <row Id="1073" PostTypeId="2" ParentId="381" CreationDate="2010-07-28T22:52:06.230" Score="0" ViewCount="" Body="&lt;p&gt;It is useful when making estimates (for examples for programming tasks) as &lt;/p&gt;&#xA;&#xA;&lt;p&gt;∀ n &gt; 1 ∈ &lt;strong&gt;N&lt;/strong&gt;, F&lt;sub&gt;n+1&lt;/sub&gt; ≠ (F&lt;sub&gt;n+2&lt;/sub&gt; + F&lt;sub&gt;n&lt;/sub&gt;) / 2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and therefore it discourages bad practices like assuming that two equal-sized tasks necessarily take twice as much as one of them (it's false because the larger the scope, the less accurate the estimate).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://scrummethodology.com/scrum-effort-estimation-and-story-points/&quot; rel=&quot;nofollow&quot;&gt;Example&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="272" LastActivityDate="2010-07-28T22:52:06.230" CommunityOwnedDate="2010-08-03T18:46:03.740" />
  <row Id="1074" PostTypeId="2" ParentId="1069" CreationDate="2010-07-28T23:10:38.137" Score="3" ViewCount="" Body="&lt;p&gt;The Lie derivative is a derivative of a vector &lt;em&gt;field&lt;/em&gt; V along another vector &lt;em&gt;field&lt;/em&gt; W.  It is defined at a point p as follows: flow the point p along W for some time t and look at the value of V at this point.  Then push this forward along the flow of W to a vector at p.  Subtract V_p from this, divide by t, and take the limit as t-&gt;0.  So this is a measure of how V changes as it gets pushed around by the flow of W.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The covariant derivative is a derivative of a vector &lt;em&gt;field&lt;/em&gt; V along a vector W.  Unlike the Lie derivative, this does not come for free: we need a connection, which is a way of identifying tangent spaces.  The reason we need this extra data is because if we wanted to take the directional derivative of V along the vector W how we do in Euclidean space, we would be taking something like V_(p+tW) - V_p, which is the difference of vectors living in different tangent spaces.  If we have a metric, then we can impose reasonable conditions that give us a unique connection (the Levi-Civita connection). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have no idea what a contravariant derivative is.  I'd guess it has to do with applying a covariant derivative and lowering indices.&lt;/p&gt;&#xA;" OwnerUserId="9" LastActivityDate="2010-07-28T23:10:38.137" />
  <row Id="1075" PostTypeId="2" ParentId="1068" CreationDate="2010-07-28T23:33:30.033" Score="18" ViewCount="" Body="&lt;p&gt;I think your question here shows that, while you have been using these symbols, you haven't really been given a proper motivation for where they came from.&lt;br&gt;&#xA;Let's go back and consider how we came up with the idea of an integral. In a typical class, you will see a lot of pictures like this:&lt;br&gt;&#xA;&lt;img src=&quot;http://upload.wikimedia.org/wikipedia/commons/thumb/9/91/Integral_approximations.svg/420px-Integral_approximations.svg.png&quot; alt=&quot;alt text&quot;&gt;&lt;br&gt;&#xA;We find the area under the curve by summing up the area of all these little rectangles. If we wanted to write an expression for the area, it would look like:&lt;br&gt;&#xA;&lt;img src=&quot;http://mathurl.com/27gla3n.png&quot; alt=&quot;alt text&quot;&gt;&lt;br&gt;&#xA;The Σ means that we are computing a sum. We are adding the areas of the rectangles, which we have numbered 1 through n, to get the complete area under the curve. The area of each rectangle is given by multiplying the height by the width. The height is given by f(x&lt;sub&gt;i&lt;/sub&gt;) because the base of the rectangle is at 0, and the top of the rectangle is where it meets the function f. The Δx represents the width of each rectangle.&lt;br&gt;&#xA;When we find the integral, we are taking the limit of this sum as the number of rectangles goes to infinity, and each individual rectangle becomes infinitesimally tiny. You can think of the dx as the equivalent of Δx: it represents the infinitesimally small width of each rectangle that we added up to get the area.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you realize this, we can see why integrals only make sense when written ∫f(x)dx. Because we are adding up the areas of rectangles that have height f(x) and width dx. If you try to interpret the expressions you wrote in this way, you will see that they do not really make sense as integrals: you are not summing up rectangles, so you are not finding an area under a curve.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could, of course, define your own notation in which those expressions behave the way you expect them to, but all mathematical notation is driven based on what people find useful, and what people can agree on and easily understand. Your reuse of the integral sign and dx that people are used to seeing in a particular context will probably result in few people adopting your definition.&lt;/p&gt;&#xA;" OwnerUserId="73" LastActivityDate="2010-07-28T23:33:30.033" CommentCount="4" />
  <row Id="1076" PostTypeId="2" ParentId="1048" CreationDate="2010-07-28T23:34:06.787" Score="1" ViewCount="" Body="&lt;p&gt;There is another proof that the derivative of sine is cosine that doesn't use the sandwich theorem mentioned by Qiaochu and Akhil above.  Instead, one can use the definition of arcsine and the standard calculus formula for arc length in terms of an integral to show that arcsine = the integral of (1 - x^2)^(-.5).  It follows that the derivative of arcsine is (1 - x^2)^(-.5), and (by the chain rule) one can use this fact to prove that the derivative of sine is cosine.&#xA;In fact, I'm not sure why this proof is presented less frequently then the one via the sandwich theorem.  The unit circle definition of sine is based on arc length, and in calculus we learn a formula for arc length based on integration.  Why not connect these two concepts for a natural proof that the derivative of sine is cosine?&lt;/p&gt;&#xA;" LastActivityDate="2010-07-28T23:34:06.787" CommentCount="1" />
  <row Id="1077" PostTypeId="2" ParentId="713" CreationDate="2010-07-28T23:42:24.700" Score="0" ViewCount="" Body="&lt;p&gt;One way to &quot;visualize&quot; a 2-tensor (in the presence of an inner product) is as follows: a vector $X$ can be regarded as a function mapping each direction $D$ to a scalar in a linear and homogeneous fashion, namely, the inner product $X\dot D$. In the same vein, a 2-tensor is a function mapping directions to vectors linearly and homogeneously.&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2010-07-28T23:42:24.700" />
  <row Id="1078" PostTypeId="2" ParentId="946" CreationDate="2010-07-28T23:44:41.060" Score="4" ViewCount="" Body="&lt;p&gt;1) Why does a small number of states suffice?&#xA;Regardless of whether the constant is 5 or 500, its still very surprising. Thankfully, it's fairly to prove if you allow the counter to be &lt;code&gt;{1...8}&lt;/code&gt; instead of &lt;code&gt;{1...5}&lt;/code&gt;. [This proof is by Ben-Or and Cleve.]. Start by representing the computation as a circuit, and ignore the whole wiping-clean thing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Define a register machine as follows: It has 3 registers &lt;code&gt;(R_1,R_2,R_3)&lt;/code&gt;, each of which holds a single bit. At each step, the program performs some computation on the registers of the form &lt;code&gt;R_i &amp;lt;-- R_a + x_b R_c&lt;/code&gt; or &lt;code&gt;R_i &amp;lt;-- R_a + x_b R_c + R_d&lt;/code&gt; (where &lt;code&gt;x_1...x_n&lt;/code&gt; is the input).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Initially, set &lt;code&gt;(R_1,R_2,R_3) = (1,0,0)&lt;/code&gt;. The machine should end in the state &lt;code&gt;R_3 + f R_1&lt;/code&gt;. We'll simulate the circuit using a register machine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We now proceed by induction on the depth of the circuit. If the circuit has depth 0, then we just copy the appropriate bit: &lt;code&gt;R_3 &amp;lt;-- R_3 + x_i R_1&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the induction, we have 3 cases, according to whether the final gate is NOT, AND, or OR.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose that the circuit is &lt;code&gt;NOT(f)&lt;/code&gt;. By induction, we can compute &lt;code&gt;f&lt;/code&gt;, yielding the state &lt;code&gt;(R_1,R_2,R_3 + f R_1)&lt;/code&gt;. We can therefore perform the instruction &lt;code&gt;R_3 &amp;lt;-- R_3 + R_1&lt;/code&gt; to get the desired output.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the circuit is &lt;code&gt;AND(f_1,f_2)&lt;/code&gt;, then life is a tad more complicated. By induction, we then execute the following 4 instructions&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;R_2 &amp;lt;-- R_2 + f_1 R_1&lt;/code&gt; &lt;br&gt;&#xA;&lt;code&gt;R_3 &amp;lt;-- R_3 + f_2 R_2&lt;/code&gt; &lt;br&gt;&#xA;&lt;code&gt;R_2 &amp;lt;-- R_2 + f_1 R_1&lt;/code&gt; &lt;br&gt;&#xA;&lt;code&gt;R_3 &amp;lt;-- R_3 + f_2 R_2&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming I haven't made any typos, we are left with the state &lt;code&gt;(R_1,R_2,R_3+f_1f_2R_1)&lt;/code&gt;, as desired. OR works similarly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;QED.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take a moment to process what just happened. It's a slick proof that you have to read 2 or 3 times before it begins to sink in. What we've shown is that we can simulate a circuit by applying a fixed program that stores only 3 bits of information at any time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To convert this into Aaronson's version, we encode the three registers into the counter (that's why we needed the extra 3 spaces). The simple program uses the input and the clock to determine how far we've made it through the computation and then applies the appropriate change to the counter. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) But what's the deal with 5?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To get from 8 states down to 5, you use a similar argument, but are much more careful about exactly how much information needs to be propagated between stages and how it can be encoded. A formal proof requires lots of advanced group theory.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Edit to answer Casebash's questions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Correct. Any computation can be expressed as a circuit composed solely of &quot;NOT&quot;, binary-&quot;AND&quot;, and binary-&quot;OR&quot; gates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) The notation &lt;code&gt;f R_1&lt;/code&gt; means (boolean) multiplication.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) The program for computing &lt;code&gt;f&lt;/code&gt; should take input &lt;code&gt;(R_1,R_2,R_3)&lt;/code&gt; to &lt;code&gt;(R_1,R_2,R_3 + f R_1)&lt;/code&gt;. We insist that the first two registers are unchanged since we use those as temporary storage in the induction. For example, when computing &lt;code&gt;AND&lt;/code&gt;, we compute the first branch and store the result in &lt;code&gt;R_2&lt;/code&gt; while computing the second branch.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4) The single bit of output is the final value of &lt;code&gt;R_3&lt;/code&gt;. Since we started with &lt;code&gt;(1,0,0)&lt;/code&gt;, we end with &lt;code&gt;(1,0,f)&lt;/code&gt;.&lt;/p&gt;&#xA;" OwnerUserId="275" LastEditorUserId="275" LastEditorDisplayName="" LastEditDate="2010-12-06T23:24:29.673" LastActivityDate="2010-12-06T23:24:29.673" CommentCount="3" />
  <row Id="1079" PostTypeId="1" AcceptedAnswerId="1083" CreationDate="2010-07-28T23:44:55.770" Score="11" ViewCount="707" Body="&lt;p&gt;Let f  and g be two periodic functions over R with the following property: If T is a period of f, and S is a period of g, then T/S is irrational. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Conjecture&lt;/strong&gt;: f+g is &lt;em&gt;not&lt;/em&gt; periodic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could you give a proof or a counter example?  It is easier if we assume continuity.  But is it true for arbitrary real valued functions?&lt;/p&gt;&#xA;" OwnerUserId="145" LastEditorUserId="145" LastEditorDisplayName="" LastEditDate="2010-07-29T17:51:22.023" LastActivityDate="2010-08-11T22:05:01.260" Title="Sum of two periodic functions" Tags="&lt;analysis&gt;" AnswerCount="4" CommentCount="3" FavoriteCount="5" />
  <row Id="1080" PostTypeId="2" ParentId="1068" CreationDate="2010-07-28T23:53:16.580" Score="2" ViewCount="" Body="&lt;p&gt;In the context of calculus, $dx$ simply means 'integrate with respect to $x$'. Some books even omit $dx$ entirely because 'of course we're integrating with respect to $x$'. The $dx$-bit does not get a proper meaning before differential forms.&lt;/p&gt;&#xA;" OwnerUserId="401" LastEditorUserId="401" LastEditorDisplayName="" LastEditDate="2011-04-25T22:03:36.977" LastActivityDate="2011-04-25T22:03:36.977" />
  <row Id="1081" PostTypeId="2" ParentId="1079" CreationDate="2010-07-29T00:00:49.343" Score="3" ViewCount="" Body="&lt;p&gt;Pick a basis $B$ of $\mathbb R$ as a $\mathbb Q$ vector space, and split it into two non-empty disjoint parts $B_1$ and $B_2$. Define $\mathbb Q$-linear maps $f,g:\mathbb R\to\mathbb R$ such that $f(x)=x$ and $g(x)=0$ if $x\in B_1$, $f(x)=0$ and $g(x)=x$ if $x\in B_2$. Then $f(x)+g(x)=x$ for all $x\in B$, so that in fact $f+g=\operatorname{id}_{\mathbb R}$, which is not a periodic function. Morever $f$ and $g$ are periodic, and their sets of periods are precisely $B_1$ and $B_2$. Since $B_1\cup B_2$ is linearly independent over $\mathbb Q$, it is easy to see that $x/y\not\in\mathbb Q$ whenever $x\in B_1$ and $y\in B_2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is then an example where the sum is not periodic.&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2010-07-29T00:00:49.343" />
  <row Id="1083" PostTypeId="2" ParentId="1079" CreationDate="2010-07-29T00:32:21.480" Score="8" ViewCount="" Body="&lt;p&gt;Here is a counterexample.  Let $a, b, c \in \mathbb{R}$ be linearly independent over $\mathbb{Q}$.  Let $\text{span}(x, y, z, ...)$ be the $\mathbb{Q}$-vector space in $\mathbb{R}$ spanned by $x, y, z, ...$.  Let $AB = \text{span}(a, b), BC = \text{span}(b, c), AC = \text{span}(a, c)$.  And for a subset $S$ of $\mathbb{R}$, let $\chi_S$ denote the characteristic function of $S$.  Now define&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle f(x) = \chi_{AB} - 2 \chi_{BC}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle g(x) = 3 \chi_{AC} + 2 \chi_{BC}.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then $f$ has period set $\text{span}(b)$, $g$ has period set $\text{span}(c)$, and $f + g$ has period set $\text{span}(a)$.  (I am not sure if the coefficients are necessary; they're just precautions.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are you still interested in the continuous case?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;(Old answer below.  I slightly misunderstood the question when I wrote this.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a simpler example.  I claim that the function $h(x) = \sin x + \sin \pi x$ cannot possibly be periodic.  Why?  Suppose an equation of the form&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sin x + \sin \pi x = \sin (x+T) + \sin \pi (x+T)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;held for all $x$ and some $T &gt; 0$.  Take the second derivative of both sides with respect to $x$ to get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sin x + \pi^2 \sin \pi x = \sin (x+T) + \pi^2 \sin \pi(x+T).$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This implies that $\sin x = \sin (x+T)$ and that $\sin \pi x = \sin \pi(x+T)$, which is impossible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Or is the question whether the sum &lt;em&gt;can&lt;/em&gt; be periodic?)&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-07-31T04:39:10.863" LastActivityDate="2010-07-31T04:39:10.863" CommentCount="3" />
  <row Id="1084" PostTypeId="2" ParentId="668" CreationDate="2010-07-29T00:36:44.637" Score="5" ViewCount="" Body="&lt;p&gt;The top exterior power of an $n$-dimensional vector space $V$ is one-dimensional. Its elements are sometimes called pseudoscalars, and they represent oriented $n$-dimensional volume elements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A linear operator $f$ on $V$ can be extended to a linear map on the exterior algebra according to the rules $f(\alpha) = \alpha$ for $\alpha$ a scalar and $f(A \wedge B) = f(A) \wedge f(B), f(A + B) = f(A) + f(B)$ for $A$ and $B$ blades of arbitrary grade. Trivia: some authors call this extension an &lt;em&gt;outermorphism&lt;/em&gt;. The extended map will be grade-preserving; that is, if $A$ is a homogeneous element of the exterior algebra of grade $m$, then $f(A)$ will also have grade $m$. (This can be verified from the properties of the extended map I just listed.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All this implies that a linear map on the exterior algebra of $V$ once restricted to the top exterior power reduces to multiplication by a constant: the &lt;strong&gt;determinant&lt;/strong&gt; of the original linear transformation. Since pseudoscalars represent oriented volume elements, this means that the determinant is precisely the factor by which the map scales oriented volumes.&lt;/p&gt;&#xA;" OwnerUserId="251" LastEditorUserId="251" LastEditorDisplayName="" LastEditDate="2010-07-29T03:16:51.460" LastActivityDate="2010-07-29T03:16:51.460" CommentCount="1" />
  <row Id="1085" PostTypeId="2" ParentId="998" CreationDate="2010-07-29T00:45:18.067" Score="6" ViewCount="" Body="&lt;p&gt;Before Hartshorne's book there was &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/354063293X&quot; rel=&quot;nofollow&quot;&gt;Mumford's Red Book of Varietie&lt;/a&gt;s.  I think it is a great introductory textbook to modern algebraic geometry (scheme theory).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found that Mumford is quite good at motivating new concepts; in particular I really enjoy his development of nonsingularity and the sheaf of differentials.  I think another great aspect about this book is that it emphasizes how to define things intrinsically (i.e. without reference to a closed or open immersion into affine space) but also explains how to make local arguments (i.e. using immersion into affine space).  A classic example of the above:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(non intrinsic tangent space): Say X is a variety and p is a point of X.  Choose an affine neighborhood so that p corresponds to the origin.  Then this affine neighborhood is spec k[x1, ..., xn]/I for some ideal.  Let I' be all the linear terms of I (i.e. if I = (x,y^2), then I' = (x)).  Then the tangent space at p is spec k[x1,...,xn]/I'.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(intrinsic tangent space): Let m be the maximal ideal of the local ring of the structure sheaf at p, then the tangent space is the dual of the vector space m/m^2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Taking spec of the symmetric algebra of the latter gives you the former.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some drawbacks.  This book doesn't cover nearly as much as Hartshorne's book.  It doesn't have that many exercises.  The notation is slightly different; integral finite type schemes are called pre-varieties and you can remove the `pre' if it's also separated.  Nevertheless I think its a great compliment to reading Hartshorne.&lt;/p&gt;&#xA;" OwnerUserId="403" LastActivityDate="2010-07-29T00:45:18.067" CommunityOwnedDate="2010-07-29T00:45:18.067" />
  <row Id="1086" PostTypeId="2" ParentId="998" CreationDate="2010-07-29T00:55:28.563" Score="0" ViewCount="" Body="&lt;p&gt;Another book I wish I had known about when I was first reading Hartshorne is Miranda's &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0521423538&quot; rel=&quot;nofollow&quot;&gt;Complex Algebraic Curves&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again this book covers much less then Hartshorne and only discusses curves over the complex numbers (and their Jacobians).  But it gives a lot more details and examples of concepts which I found particularly difficult when I first started learning algebraic geometry (sheafs, divisors, cohomology).  It also has a bunch of exercises which I think are often not as challenging as the the exercises in Hartshorne.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It also covers a lot more of the 'classical' theory of curves than Hartshrone does; e.g. &#xA;Weierstrass points. &lt;/p&gt;&#xA;" OwnerUserId="403" LastActivityDate="2010-07-29T00:55:28.563" CommunityOwnedDate="2010-07-29T00:55:28.563" />
  <row Id="1087" PostTypeId="2" ParentId="381" CreationDate="2010-07-29T00:56:04.130" Score="0" ViewCount="" Body="&lt;p&gt;They're a much easier way to evaluate the function&lt;/p&gt;&#xA;&#xA;&lt;p&gt;f(x) = (ø^x -(-ø)^x) / sqrt(5)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;by hand, where ø is the golden ratio and x is an integer :)&lt;/p&gt;&#xA;" OwnerUserId="209" LastActivityDate="2010-07-29T00:56:04.130" CommunityOwnedDate="2010-08-03T18:46:03.740" />
  <row Id="1088" PostTypeId="2" ParentId="998" CreationDate="2010-07-29T01:07:22.173" Score="5" ViewCount="" Body="&lt;p&gt;My last suggestion would be Ravi Vakil's &lt;a href=&quot;http://math.stanford.edu/~vakil/0910-216/&quot; rel=&quot;nofollow&quot;&gt;online notes on the foundations of algebraic geometry&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think these notes might be made into a full on textbook someday.  I haven't looked through all of them but these notes seem to cover as much as Hartshorne does (if not more).  Only rarely do Hartshorne and Vakil define things differently (`projective morphisms' is the only example that comes to mind).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've heard it said that Hartshorne's book is a `baby' version of EGA.  I think Vakil's notes are somewhere between Hartshorne and EGA (probably not the midpoint though).  At least Vakil discusses much more the theory of representable functors, and Noetherian hypothesis are less prevalent in Vakil's notes.  Also Vakil's notes are more complete in that they also include proofs of many of the commutative algebra results that are just stated in Hartshorne.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think Vakil spends a lot more time motivating the material and often the notes are a bit conversational.  Also there are tons of exercises and most of the them are appended with useful qualifiers like (easy but important exercises, unimportant exercise, tedious but useful exercise, etc).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One drawback is that they are very long and they are online notes so there are many typos.  But most of them are grammatical and easy to spot.&lt;/p&gt;&#xA;" OwnerUserId="403" LastActivityDate="2010-07-29T01:07:22.173" CommunityOwnedDate="2010-07-29T01:07:22.173" CommentCount="4" />
  <row Id="1090" PostTypeId="2" ParentId="250" CreationDate="2010-07-29T01:10:24.537" Score="5" ViewCount="" Body="&lt;p&gt;For any five points on the globe, there is an angle in outer space from which you could see at least 4 of the 5 points (assuming the moon or anything isn't in the way).  The proof is pretty simple, too...&lt;/p&gt;&#xA;" OwnerUserId="209" LastActivityDate="2010-07-29T01:10:24.537" CommunityOwnedDate="2010-07-29T01:10:24.537" CommentCount="1" />
  <row Id="1091" PostTypeId="1" AcceptedAnswerId="1095" CreationDate="2010-07-29T01:19:48.370" Score="5" ViewCount="712" Body="&lt;p&gt;The wiki &lt;a href=&quot;http://en.wikipedia.org/wiki/Eigenvalue,_eigenvector_and_eigenspace&quot; rel=&quot;nofollow&quot;&gt;article&lt;/a&gt; on eigenvectors offers the following geometrical interpretation:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Each application of the matrix to an arbitrary vector yields a result which will have rotated towards the eigenvector with the largest eigenvalue.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Qn 1: If there is any other geometrical interpretation particularly in the context of a covariance matrix?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The wiki also discusses the difference between left and right eigenvectors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Qn 2: Do the above geometrical interpretations hold irrespective of whether they are left or right eigenvectors?&lt;/p&gt;&#xA;" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-07-29T01:27:38.477" LastActivityDate="2010-07-30T04:05:04.443" Title="Is there a geometrical interpretation to the notion of eigenvector and eigenvalues?" Tags="&lt;linear-algebra&gt;&lt;intuition&gt;" AnswerCount="4" CommentCount="5" FavoriteCount="2" />
  <row Id="1092" PostTypeId="2" ParentId="1091" CreationDate="2010-07-29T01:29:06.133" Score="0" ViewCount="" Body="&lt;p&gt;Of course! Consider a coordinate transformation of rotation and/or scaling (but not translation):&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;v = Au&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;where v and u are vectors, and A is a transformation matrix. Then the eigenvectors, if they have real components, are the axes which are left unrotated (scaling only) by the transformation. (see &lt;a href=&quot;http://en.wikipedia.org/wiki/Eigenvalue,_eigenvector_and_eigenspace&quot; rel=&quot;nofollow&quot;&gt;wikipedia&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A covariance matrix is a symmetric, positive definite matrix, so it has orthonormal eigenvectors, and these form a tuple of axes; I am fairly sure the eigenvectors form a new basis of linear combinations of the input variables where the basis variables are uncorrelated, but I can't remember how to show this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strike&gt;For example, if w1 = [x;y] is a pair of independent unit-variance zero-mean Gaussian random variables, consider w2 = [u;v] = [1 1; 2 1][x;y] = (x+y,2x+y), so that w1 = [-1 1;2 -1][u;v] = [v-u;2u-v]. Then cov(w2) = [2 3; 3 5]. This has eigenvectors which have sqrt(5) in them, hmmmm...&lt;/strike&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for question 2, I'm not sure.&lt;/p&gt;&#xA;" OwnerUserId="120" LastEditorUserId="120" LastEditorDisplayName="" LastEditDate="2010-07-29T01:55:14.603" LastActivityDate="2010-07-29T01:55:14.603" CommentCount="4" />
  <row Id="1093" PostTypeId="2" ParentId="392" CreationDate="2010-07-29T02:04:40.680" Score="11" ViewCount="" Body="&lt;p&gt;Perhaps the following diagram will provide insight:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://daylateanddollarshort.com/misc/dsindcos.jpg&quot; alt=&quot;(Non)Proof without Words: Derivatives of Sine and Cosine&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The idea is to look at the sine and cosine curves as projections of a helix drawn on a cylinder. If you look at the cylinder itself as a curled planar square of length 2pi, then helix is a curled version of the square's diagonal. A tangent vector along the flat square's diagonal always lies at 45 degrees to the square's sides, say with length-&quot;1&quot; shadows in each direction; after smoothly curling the square into the cylinder, the tangent vector lies at 45 degrees to the cylinder's (z-)axis and the perpendicular (xy-)plane. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Projecting the helix into the zy- and zx-planes gives graphs of sine and cosine. Projecting the helix's tangent vector gives tangent vectors to those graphs. The &quot;dz&quot;s for these projected tangents are always 1 (the &quot;vertical&quot; shadow of the helix's tangent vector). To get at &quot;dy&quot; and &quot;dx&quot; (&quot;v_x&quot; and &quot;v_y&quot; in the diagram) we project &lt;em&gt;down&lt;/em&gt; into the xy-plane where we see a circle, and yet another projected tangent vector.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basic geometry tells us that a tangent to a circle is perpendicular to the radius at the point of tangency. In our circle, the point of tangency --and the radius vector it-- is parameterized as &quot;&amp;lt; cos, sin, 0 &gt;&quot;. The perpendicular tangent line must therefore have a &quot;negative-reciprocal&quot; direction vector: &quot;&amp;lt; -sin, cos, 0 &gt;&quot;, which gives us our &quot;dx&quot; and &quot;dy&quot; for the helix tangent ... and the projected graph tangents as well, so that we may make the following conclusions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The derivative of cosine --by its &lt;em&gt;conceptual definition&lt;/em&gt; as &quot;slope of the tangent line&quot;-- is change-in-x-over-change-in-z = dx/dz = -sin/1 = -sin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Likewise, the derivative of sine is dy/dz = cos/1 = cos.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I like this approach because the conceptual &quot;slope of tangent line&quot; definition of the derivative is used throughout; there are no (obvious) appeals to digressive computational tricks involving trig identities and limits of difference quotients. I also like that the curious negative sign in the derivative of cosine traces back to an elementary property of circle geometry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, this approach doesn't constitute &lt;em&gt;proof&lt;/em&gt; of the formulas. The process of curling the planar square into a cylinder and claiming that the tangent vector behaves as claimed actually &lt;em&gt;assumes&lt;/em&gt; the computational machinery covered by the traditional limit arguments. Nevertheless, on an &lt;em&gt;intuitive&lt;/em&gt; level, I think this argument explains the &quot;why&quot; of the derivatives quite beautifully. Then, knowing what the formulas &lt;em&gt;are&lt;/em&gt; (or &quot;should be&quot;) helps motivate the investigation of the computational tricks needed to provide a rigorous proof.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://daylateanddollarshort.com/math/pdfs/dsindcos.pdf&quot; rel=&quot;nofollow&quot;&gt;Here's a PDF&lt;/a&gt; with a variant of the above discussion (but the same image). &lt;a href=&quot;http://demonstrations.wolfram.com/CalculusFreeDerivativesOfSineAndCosine/&quot; rel=&quot;nofollow&quot;&gt;Here's a Mathematica Demonstration&lt;/a&gt; that animates the various elements, including the square curling into the cylinder.&lt;/p&gt;&#xA;" OwnerUserId="409" LastEditorUserId="409" LastEditorDisplayName="" LastEditDate="2010-07-29T14:25:30.383" LastActivityDate="2010-07-29T14:25:30.383" CommentCount="2" />
  <row Id="1094" PostTypeId="2" ParentId="998" CreationDate="2010-07-29T02:14:13.390" Score="4" ViewCount="" Body="&lt;p&gt;for &lt;em&gt;Undergraduate&lt;/em&gt; algebraic geometry (significantly below the level of Hartshorne), Cox, Little and O'Shea's &lt;em&gt;Ideals, Varieties, and Algorithms&lt;/em&gt; is a pleasant treatment.&lt;/p&gt;&#xA;" OwnerUserId="76" LastActivityDate="2010-07-29T02:14:13.390" CommunityOwnedDate="2010-07-29T02:14:13.390" CommentCount="1" />
  <row Id="1095" PostTypeId="2" ParentId="1091" CreationDate="2010-07-29T02:32:27.370" Score="5" ViewCount="" Body="&lt;p&gt;Here is a partial answer in the case where M is a real symmetric matrix.  This is to ensure, by the real spectral theorem, that M has real eigenvectors with real eigenvalues, so there is a chance for a genuine geometric interpretation which stays in R^n.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;M acts on the unit sphere in R^n in the following way: it sends the unit sphere v^T v = 1 to v^T (M^T M) v = 1.  This modified shape is not generally a sphere, but is generally an ellipsoid.  The axes of this ellipsoid are the eigenvectors of M, and the sizes of each axis are given by the squares of the corresponding eigenvalues.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-07-29T02:32:27.370" CommentCount="3" />
  <row Id="1096" PostTypeId="1" AcceptedAnswerId="1097" CreationDate="2010-07-29T02:38:41.303" Score="7" ViewCount="551" Body="&lt;p&gt;Consider the following:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;1 = 1^2&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;2 + 2 = 2^2&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;3 + 3 + 3 = 3^2&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Therefore,&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;x + x + x + ... (repeated x times) = x^2&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Take the derivative of lhs and rhs and we get:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;1 + 1 + 1 (repeated x times) = 2x&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Which simplifies to:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;x = 2x&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;and hence &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;1 = 2.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Clearly something is wrong but I am unable pinpoint my mistake.&lt;/p&gt;&#xA;" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-08-10T06:25:55.463" LastActivityDate="2010-08-11T04:24:21.853" Title="Where is the flaw in this argument of a proof that 1=2?" Tags="&lt;calculus&gt;&lt;recreational-mathematics&gt;&lt;fake-proofs&gt;" AnswerCount="5" CommentCount="5" />
  <row Id="1097" PostTypeId="2" ParentId="1096" CreationDate="2010-07-29T02:47:45.227" Score="14" ViewCount="" Body="&lt;p&gt;You cannot take the derivative with respect to x of x + x + x + ... (repeated x times) one term at a time because the number of terms depends on x.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even beyond that, if you can express x&lt;sup&gt;2&lt;/sup&gt; as x + x + x + ... (repeated x times), then x must be an integer and if the domain of the expression is the integers, (continuous) differentiation does not make sense and/or the derivatives do not exist.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(&lt;em&gt;edit&lt;/em&gt;: I gave my first reason first because the second reason can be smoothed over by taking &quot;repeated x times&quot; to mean something like $\underset{\lfloor x\rfloor\mathrm{\ addends}}{\underbrace{x+x+\cdots+x}}+(x-\lfloor x\rfloor)\cdot x$.)&lt;/p&gt;&#xA;" OwnerUserId="72" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-08-05T18:40:51.450" LastActivityDate="2010-08-05T18:40:51.450" CommentCount="1" />
  <row Id="1098" PostTypeId="2" ParentId="1096" CreationDate="2010-07-29T02:50:51.570" Score="4" ViewCount="" Body="&lt;p&gt;You cannot differentiate the LHS of your equation&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;x + x + x + ... (repeated x times) = x^2&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is because the LHS is not a continuous function; the number of terms depends on x so the LHS is not well defined when x is not an integer. We can only differentiate continuous functions, so this is not valid.&lt;/p&gt;&#xA;" OwnerUserId="30" LastActivityDate="2010-07-29T02:50:51.570" CommentCount="1" />
  <row Id="1099" PostTypeId="1" AcceptedAnswerId="1110" CreationDate="2010-07-29T02:57:50.193" Score="9" ViewCount="541" Body="&lt;p&gt;Are there any well-studied analogs to the Mandelbrot set using functions other than $f(z)= z^2+c$ in $\mathbb{C}$?&lt;/p&gt;&#xA;" OwnerUserId="72" LastEditorUserId="3936" LastEditorDisplayName="" LastEditDate="2011-05-17T06:33:27.313" LastActivityDate="2011-06-02T22:32:51.343" Title="Mandelbrot-like sets for functions other than $f(z)=z^2+c$?" Tags="&lt;fractals&gt;&lt;dynamical-systems&gt;" AnswerCount="10" CommentCount="1" FavoriteCount="4" />
  <row Id="1100" PostTypeId="2" ParentId="381" CreationDate="2010-07-29T03:05:47.383" Score="4" ViewCount="" Body="&lt;p&gt;How many ways are there to tile a 2 x n grid with 1 x 2 dominos? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This problem can be solved using Fibonacci numbers. Let S&lt;sub&gt;n&lt;/sub&gt; be the number of valid tilings of the 2 x n grid. Each such tiling has either a vertical 1 x 2 domino or two horizontal dominos on the right. Therefore, each tiling of a 2 x (n-1) grid or a 2 x (n-2) grid generates a tiling of the 2 x n grid, and hence we have a recurrence relation S&lt;sub&gt;n&lt;/sub&gt; = S&lt;sub&gt;n-1&lt;/sub&gt; + S&lt;sub&gt;n-2&lt;/sub&gt;. This is precisely the recurrence relation of the Fibonacci numbers. Checking our base cases, we see that there is one way to tile a 1 x 2 grid and two ways to tile a 2 x 2 grid, so S&lt;sub&gt;1&lt;/sub&gt; = 1 and S&lt;sub&gt;2&lt;/sub&gt; = 2. Therefore, the number of tilings is precisely the Fibonacci sequence.&lt;/p&gt;&#xA;" OwnerUserId="30" LastActivityDate="2010-07-29T03:05:47.383" CommunityOwnedDate="2010-08-03T18:46:03.740" />
  <row Id="1101" PostTypeId="2" ParentId="514" CreationDate="2010-07-29T03:11:46.200" Score="8" ViewCount="" Body="&lt;p&gt;Another example: &lt;a href=&quot;http://en.wikipedia.org/wiki/Euler%27s_sum_of_powers_conjecture&quot; rel=&quot;nofollow&quot;&gt;Euler's sum of powers conjecture&lt;/a&gt;, a generalization of Fermat's Last Theorem. It states:&lt;br&gt;&#xA;If the equation &amp;sum;&lt;sub&gt;i=0&lt;/sub&gt;&lt;sup&gt;k&lt;/sup&gt; x&lt;sub&gt;i&lt;/sub&gt;&lt;sup&gt;n&lt;/sup&gt; = z&lt;sup&gt;n&lt;/sup&gt; has a solution in positive integers, then n ≤ k (unless k=1). Fermat's Last Theorem is the k=2 case of this conjecture.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A counterexample for n=5 was found in 1966: it's&lt;br&gt;&#xA;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp; 61917364224 = 27&lt;sup&gt;5&lt;/sup&gt; + 84&lt;sup&gt;5&lt;/sup&gt; + 110&lt;sup&gt;5&lt;/sup&gt; + 133&lt;sup&gt;5&lt;/sup&gt; = 144&lt;sup&gt;5&lt;/sup&gt;.&lt;br&gt;&#xA;The smallest counterexample for n=4 was found in 1988:&lt;br&gt;&#xA;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp; 31858749840007945920321 = 95800&lt;sup&gt;4&lt;/sup&gt; + 217519&lt;sup&gt;4&lt;/sup&gt; + 414560&lt;sup&gt;4&lt;/sup&gt; = 422481&lt;sup&gt;4&lt;/sup&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This example used to be even more useful in the days before FLT was proved, as an answer to the question &quot;Why do we need to prove FLT if it has been verified for thousands of numbers?&quot; :-)&lt;/p&gt;&#xA;" OwnerUserId="205" LastActivityDate="2010-07-29T03:11:46.200" CommentCount="4" />
  <row Id="1102" PostTypeId="2" ParentId="1099" CreationDate="2010-07-29T03:17:46.047" Score="3" ViewCount="" Body="&lt;p&gt;There is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Julia_set&quot; rel=&quot;nofollow&quot;&gt;Julia set&lt;/a&gt;, which can be defined for any complex rational map f(z) = P(z)/Q(z) where P(z) and Q(z) are polynomials. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Julia set for the map f&lt;sub&gt;c&lt;/sub&gt;(z) = z^2 + c is related to the Mandelbrot set in that a point z is in the Mandelbrot set if the Julia set of f&lt;sub&gt;c&lt;/sub&gt;(z) is connected.&lt;/p&gt;&#xA;" OwnerUserId="30" LastActivityDate="2010-07-29T03:17:46.047" CommentCount="1" />
  <row Id="1103" PostTypeId="2" ParentId="1048" CreationDate="2010-07-29T03:25:59.173" Score="1" ViewCount="" Body="&lt;p&gt;If you allow yourself a tiny bit of calculus ( &quot;$\sin x / x \to 1$&quot; as &quot;$x \to 0$&quot; ) and apply some combinatorics, there's a really nice geometric interpretation of the terms of the power series for the functions. Consider this diagram and the polygonal &quot;spiral&quot; that starts at $P_0$ and closes in on the point $P$ (where $|P_0 P| = 1$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://daylateanddollarshort.com/misc/sincosinvolutes.jpg&quot; alt=&quot;The Sine and Cosine Involute Pinwheel&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The horizontal segments $P_{2n} P_{2n+1}$ alternately overshoot and undershoot the length of the cosine segment; the vertical segments $P_{2n+1} P_{2n+2}$ do the same for the sine segment. So, &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$\cos \theta = \sum_{n=0}^{\infty}(-1)^n | P_{2n} P_{2n+1} |$ &lt;br /&gt;&lt;br /&gt;&#xA;  $\sin \theta = \sum_{n=0}^{\infty} (-1)^n | P_{2n+1} P_{2n+2} |$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now, the lengths $|P_{k} P_{k+1}|$ are equal to the lengths of the curves $|I_k|$, which constitute a series of successive &lt;a href=&quot;http://en.wikipedia.org/wiki/Involute&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;involutes&lt;/em&gt;&lt;/a&gt; (with $I_0$ defined to be a segment, and $I_1$ defined to be an arc of the unit circle). Combinatorics and the calculus result I mentioned show that the involute lengths satisfy ...&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$|I_k| = \theta^k / k!$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;... so that the above are, in fact, &lt;em&gt;power series&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Interestingly, the same thing can be done with secant and tangent, using an involute zig-zag:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://daylateanddollarshort.com/misc/sectaninvolutes.jpg&quot; alt=&quot;The Secant and Tangent Involute Zig-Zag&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$\sec \theta = \sum_{n=0}^{\infty} | P_{2n} P_{2n+1} | = \sum_{n=0}^{\infty} | I_{2n} |$&lt;br /&gt;&lt;br /&gt;&#xA;  $\tan \theta = \sum_{n=0}^{\infty} | P_{2n+1} P_{2n+2} | = \sum_{n=0}^{\infty} | I_{2n+1} |$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and the lengths $|I_k|$ turn out to be the appropriate multiples of powers of $\theta^k$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reference to the argument for sine and cosine (attributed to Y. S. Chaikovsky, as reported by Leo Gurin), a complete discussion of the trickier argument for secant and tangent, and then a refinement of the argument for sine and cosine, are in my note &lt;a href=&quot;http://daylateanddollarshort.com/math/pdfs/sectan.pdf&quot; rel=&quot;nofollow&quot;&gt;&quot;Zig-Zag Involutes, Up-Down Permutations, and Secant and Tangent&quot; (PDF)&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;BTW: I have not (yet) cracked the case for cosecant and cotangent.&lt;/p&gt;&#xA;" OwnerUserId="409" LastEditorUserId="409" LastEditorDisplayName="" LastEditDate="2010-08-03T16:23:35.267" LastActivityDate="2010-08-03T16:23:35.267" />
  <row Id="1104" PostTypeId="2" ParentId="1096" CreationDate="2010-07-29T03:28:59.503" Score="4" ViewCount="" Body="&lt;p&gt;Here's my explanation from an old sci.math post:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Zachary Turner  wrote on 26 Jul 2002:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Let D = d/dx = derivative wrt x. Then&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;D[x^2] = D[x  +   x  + ... +   x  (x times)]&#xA;       = D[x] + D[x] + ... + D[x] (x times)&#xA;       =   1  +   1  + ... +   1  (x times)&#xA;       =   x&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;An obvious analogous fallacious argument proves both&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;  D[x f(x)]  =  Df(x) (x  times) = x Df(x)&#xA;&#xA;  D[x f(x)]  =   Dx (f(x) times) = f(x), via  Dx = 1&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;vs.  the correct result: their sum  f(x) + x Df(x)&#xA;as given by the Leibniz product rule (= chain rule for times).&#xA;The error arises from overlooking the dependence upon x in &lt;em&gt;both&lt;/em&gt;&#xA;arguments of the product  x * f(x)  when applying the chain rule.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The source of the error becomes clearer if we consider a&#xA;discrete analog. This will also eliminate any tangential&#xA;concerns on the meaning of &quot;(x times)&quot; for non-integer x.&#xA;Namely, we consider the shift operator  S: n -&gt; n+1  on polynomials  p(n)&#xA;with integer coef's, where  S p(n) = p(n+1). Here's a similar fallacy&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;  S[n^2] =  S[n  +   n  + ... +   n  (n times)]&#xA;         =  S[n] + S[n] + ... + S[n] (n times)&#xA;         =  1+n  + 1+n  + ... + 1+n  (n times)&#xA;         = (1+n)n&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;But correct is  S[n^2] = (n+1)^2. Here the &quot;product rule&quot; is&#xA;simply  S[fg] = S[f] S[g], not  S[f] g  as above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fallacy actually boils down to operator non-commutativity.&#xA;On the space of functions f(x), consider &quot;x&quot; as the linear&#xA;operator of multiplication by  x, so  x: f(x) -&gt; x f(x). Then&#xA;the linear operators  D  and  x  generate an operator algebra&#xA;of polynomials  p(x,D) in NON-commutative indeterminates x,D&#xA;since we have&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;  (Dx)[f] = D[xf] = xD[f] + f = (xD+1)[f], so  Dx = xD+1&#xA;&#xA;  (Sn)[f] = S[nf] = (n+1)S[f], so  Sn = (n+1)S != nS&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This viewpoint reveals the error simply as mistakenly&#xA;assuming commutativity of the operators  x,D  or  n,S.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps something to ponder on boring &lt;em&gt;commutes&lt;/em&gt; !&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-07-29T03:28:59.503" />
  <row Id="1105" PostTypeId="2" ParentId="677" CreationDate="2010-07-29T03:30:58.303" Score="1" ViewCount="" Body="&lt;p&gt;I like your problem. But your opposite parity conjecture doesn't seem to handle the case N=3 and S=2 (even when corrected as in Mau's comment by gcd), with a triangle labeled as follows:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&#xA;        1&#xA;&#xA;     1     0&#xA;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This position appears to be invariant under any move, so you cannot reach all 0s.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More generally, a similar obstacle works for any 3n-gon, with S=2n. If you repeat the pattern 1 1 0 all the way around, it is stable under any move. If $n$ is odd, these will have opposite parities.&lt;/p&gt;&#xA;" OwnerUserId="413" LastActivityDate="2010-07-29T03:30:58.303" CommentCount="1" />
  <row Id="1106" PostTypeId="2" ParentId="639" CreationDate="2010-07-29T03:42:46.310" Score="9" ViewCount="" Body="&lt;p&gt;I asked this question &lt;a href=&quot;http://mathoverflow.net/questions/59939/identifying-poisoned-wines/60312#60312&quot; rel=&quot;nofollow&quot;&gt;on MathOverflow&lt;/a&gt; and got a great answer there.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;For $k = 2$ I can do it with ${\lceil \log_2 N \rceil + 2 \choose 2} - 1$ servants.  In particular for $N = 1000$ I can do it with $65$ servants.  The proof is somewhat long, so I don't want to post it until I've thought about the problem more.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I haven't been able to improve on the above result.  Here's how it works.  Let $n = \lceil \log_2 N \rceil$.  Let me go through the algorithm for $k = 1$ so we're all on the same page.  Number the wines and assign each of them the binary expansion of their number, which consists of $n$ bits.  Find $n$ servants, and have servant $i$ drink all the wines whose $i^{th}$ bit is $1$.  Then the set of servants that die tells you the binary expansion of the poisoned wine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $k = 2$ we need to find $n$ butlers, $n$ maids, and ${n \choose 2}$ cooks.  The cooks will be named $(i, j)$ for some positive integers $1 \le i &amp;lt; j \le n$.  Have butler $i$ drink all the wines whose $i^{th}$ bit is $1$, have maid $i$ drink all the wines whose $i^{th}$ bit is $0$, and have cook $(i, j)$ drink all the wines such that the sum of the $i^{th}$ bit through the $j^{th}$ bit, inclusive, mod 2, is $1$.  This is how the casework breaks down for butlers and maids.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If both butler $i$ and maid $i$ die, then one of the poisoned wines has $i^{th}$ bit $0$ and the other has $i^{th}$ bit $1$.&lt;/li&gt;&#xA;&lt;li&gt;If only butler $i$ dies, then both of the poisoned wines have $i^{th}$ bit $1$.&lt;/li&gt;&#xA;&lt;li&gt;If only maid $i$ dies, then both of the poisoned wines have $i^{th}$ bit $0$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The second two cases are great.  The problem with case 1 is that if it occurs more than once, there's still ambiguity about which wine has which bit.  (The worst scenario is if all the butlers and maids die.)  To fix the issue with case 1, we use the cooks.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $i_1 &amp;lt; ... &amp;lt; i_m$ be the set of bits where case 1 occurs.  We'll say that the poisoned wine whose $(i_1)^{th}$ bit is $1$ is wine A, and the other one is wine B.  Notice that the sum of the $(i_1)^{th}$ through $(i_2)^{th}$ bits of wine A mod 2 is the same as the sum of the $(i_1)^{th}$ through $(i_2)^{th}$ bits of wine B mod 2, and we can determine what this sum is by looking at whether cook $(i_1, i_2)$ died.  The value of this sum determines whether the $(i_2)^{th}$ bit of wine A is 1 or 0 (and the same for wine B).  Similarly, looking at whether cook $(i_j, i_{j+1})$ died tells us the remaining bits of wine A, hence of wine B.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;One last comment for now.  The lower bound is not best possible when $k$ is large compared to $N$; for example, when $k = N-1$ it takes $N-1$ servants.  The reason is that any servant who drinks more than one wine automatically dies, hence gives you no information. &lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2011-04-02T01:56:02.263" LastActivityDate="2011-04-02T01:56:02.263" CommentCount="7" />
  <row Id="1107" PostTypeId="2" ParentId="1099" CreationDate="2010-07-29T04:15:05.790" Score="15" ViewCount="" Body="&lt;p&gt;I have no idea about well-studied, but here's what I've gotten from a bit of playing around (&lt;a href=&quot;http://gist.github.com/497210&quot; rel=&quot;nofollow&quot;&gt;here's the relevant Mathematica code&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;$\boldsymbol{f_c(z)=\cos(z)+c}$ where $\cos(z)$ is defined however Mathematica defines it, using the escape time algorithm with escape radius $10\pi$ and 100 iterations maximum:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/2Ddw7.png&quot; alt=&quot;cosine1&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/pL8nx.png&quot; alt=&quot;cosine2&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;$\boldsymbol{f_c(z)=\sin(z)+c}$ appears to be the same as cosine, but horizontally translated.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;$\boldsymbol{f_c(z)=e^z+c}$, using the escape time algorithm with escape radius 50 and 100 iterations maximum (it's vertically periodic with period $2\pi$):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/1cJU6.png&quot; alt=&quot;exp1&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/dq0cl.png&quot; alt=&quot;exp2&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;edit&lt;/em&gt; (Mathematica code edited, too): $\boldsymbol{f_c(z)=cz(1-z)}$ (since camccann's answer mentions logistic maps) using the escape time algorithm with escape radius 50 and 200 iterations maximum:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/V3csJ.png&quot; alt=&quot;logistic1&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/21gSR.png&quot; alt=&quot;logistic2&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="72" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-09-29T04:33:19.810" LastActivityDate="2010-09-29T04:33:19.810" CommentCount="4" />
  <row Id="1108" PostTypeId="1" AcceptedAnswerId="1114" CreationDate="2010-07-29T04:32:57.897" Score="1" ViewCount="141" Body="&lt;p&gt;I understand the mechanics of the proof of Ostrowski's Theorem, but I'm a little unclear on why one should expect valuations to be related to primes. Is this a special property of number fields and function fields, or do primes of K[x,y] correspond to valuations on K(x,y) in the same way?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm hoping for an answer that can explain what exactly are the algebraic analogs of archimedian valuations, and how to use them - for example, I've heard that the infinite place on K(x) corresponds to the &quot;prime (1/x)&quot; - how does one take a polynomial in K[x] &quot;mod (1/x)&quot; rigorously?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance.&lt;/p&gt;&#xA;" OwnerUserId="264" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-07-29T07:36:36.973" LastActivityDate="2010-07-29T20:22:39.630" Title="Why should one expect valuations to be related to primes? How to treat an infinite place algebraically?" Tags="&lt;number-theory&gt;&lt;algebraic-geometry&gt;" AnswerCount="2" CommentCount="5" FavoriteCount="1" />
  <row Id="1109" PostTypeId="2" ParentId="1064" CreationDate="2010-07-29T04:49:16.473" Score="2" ViewCount="" Body="&lt;p&gt;It can be proven that the Cantor set is perfect. Certainly, this contains infinitely many rationals. How about modifying the construction of the Cantor set by defining:&#xA;$I_1 = [\sqrt{2},\sqrt{2}+1/3] \cup [\sqrt{2}+2/3,\sqrt{2}+1]$, $I_2 = [\sqrt{2},\sqrt{2}+1/9] \cup [\sqrt{2}+2/9,\sqrt{2}+1/3]\cup[\sqrt{2}+2/3,\sqrt{2}+7/9]\cup[\sqrt{2}+8/9,\sqrt{2}+1]$, etc and setting $P = \cap_{i=1}^\infty I_i$? Each of end points of any interval that appears in the construction is a member of $P$ and is irrational. However, is it true that all the members of $P$ must be an end point of a certain interval? I am tempted to think so because we can prove that $P$ does not contain any interval. &lt;/p&gt;&#xA;" LastActivityDate="2010-07-29T04:49:16.473" CommentCount="1" />
  <row Id="1110" PostTypeId="2" ParentId="1099" CreationDate="2010-07-29T05:17:57.723" Score="8" ViewCount="" Body="&lt;p&gt;A variant on the M-set can be defined in straightforward fashion for any iterated function in the complex plane parameterized by a single initial value. For instance, slight modifications produce the &lt;a href=&quot;http://en.wikipedia.org/wiki/Tricorn_%28mathematics%29&quot; rel=&quot;nofollow&quot;&gt;tricorn&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/Burning_Ship_fractal&quot; rel=&quot;nofollow&quot;&gt;burning ship&lt;/a&gt; fractals. However, most such variations tend to be either boring, incoherent, or obviously derived from the Mandelbrot set--nothing particularly novel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some obvious patterns also emerge quickly from many variations: Real exponents alter symmetry, imaginary exponents cause asymmetric twisting, disreputable functions produce misshapen lumps like the burning ship, and so on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, it's more difficult than you might expect to avoid the M-set in the first place: For a well known example, consider &lt;a href=&quot;http://en.wikipedia.org/wiki/Newton%27s_method&quot; rel=&quot;nofollow&quot;&gt;the Newton-Raphson method for approximating roots&lt;/a&gt;, which can be generalized to the complex plane in straightforward fashion. For some polynomial, a point in the complex plane may or may not converge to a particular root after some number of Newton-Raphson iterations. In most cases, plotting the regions of divergence and convergence per root produces a fractal. Modifying the polynomial and the iteration formula produces effects analogous to modifying the constant and iteration function, respectively, of a standard Julia set, and in fact &lt;a href=&quot;http://www.chiark.greenend.org.uk/~sgtatham/newton/&quot; rel=&quot;nofollow&quot;&gt;it turns out that Julia fractals can be considered a special case of Newton-Raphson fractals&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Analogous structure can also be found elsewhere than the complex plane. The period-doubling behavior of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Logistic_map&quot; rel=&quot;nofollow&quot;&gt;logistic map&lt;/a&gt; relates to the behavior of points on the real line of the M-set, and islands of stability for the logistic map correspond to the positions of mini-Mandelbrots. Elaborate Julia-like structures can also be found in the quaternions, although unfortunately visualization of 4-dimensional fractals is somewhat challenging.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My suspicion is that Julia-like structure will arise for &lt;em&gt;any&lt;/em&gt; fractal defined by similar means, e.g., classifying points into sets based on their behavior under repeated iteration of a position-sensitive transformation, but I'm not sure how to define &quot;similar means&quot; precisely enough to formalize that.&lt;/p&gt;&#xA;" OwnerUserId="211" LastActivityDate="2010-07-29T05:17:57.723" CommentCount="2" />
  <row Id="1111" PostTypeId="2" ParentId="1091" CreationDate="2010-07-29T05:25:31.097" Score="3" ViewCount="" Body="&lt;p&gt;If you are interested in covariance matrices, then the eigenvectors of the covariance matrix tell you how to change variables to make your random variables uncorrelated. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Specifically, let $M$ be a covariance matrix of the random variables $X_1,...,X_n$. For simplicity, lets assume that all of these random variables are zero mean. Lets also define $X$ to be the random vector whose $i$&quot;th component is the random variable $X_i$. Let the eigenvectors of $M$ be $v_1, v_2, ..., v_n$. We can assume that these are orthogonal since $M$ is symmetric. Consider the random variables $Y_i$ obtained by taking the dot product of $v_i$ and $X$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then the random variables $Y_1, ..., Y_n$ are uncorrelated!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Indeed:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ E[ Y_i Y_j] = E[ v_i^T X X^T v_j] = v_i^T M v_j = \lambda_j v_i^T v_j = 0 $$&lt;/p&gt;&#xA;" OwnerUserId="317" LastActivityDate="2010-07-29T05:25:31.097" CommentCount="1" />
  <row Id="1112" PostTypeId="2" ParentId="941" CreationDate="2010-07-29T05:43:06.793" Score="1" ViewCount="" Body="&lt;p&gt;While it is certainly true that with the proper definition there is now 'wrong' notation, perhaps it should be mentioned that some notation is more suggestive and/or easier to work with than others, e.g. Arabic numeral vs. Roman numerals, the various symbols for the derivative, and countless others. The actual symbols are arbitrary, but good notation can certainly promote the flow of ideas more easily. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, do I remember correctly that Feynman gave up trying to invent more efficient notation for simple math when he was quite young because nobody could understand what he was doing?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A good notation has a subtlety and suggestiveness which at times make it almost seem like a live teacher.&#xA;    --Bertrand Russell&lt;/p&gt;&#xA;" OwnerUserId="416" LastActivityDate="2010-07-29T05:43:06.793" CommentCount="1" />
  <row Id="1113" PostTypeId="1" AcceptedAnswerId="1387" CreationDate="2010-07-29T06:34:12.370" Score="2" ViewCount="162" Body="&lt;p&gt;Is it true that there are infinitely many nonprime integers $n$ such that $3^{n-1} - 2^{n-1}$ is a multiple of $n$?&lt;/p&gt;&#xA;" OwnerUserId="334" LastActivityDate="2010-08-02T12:18:54.860" Title="Nonprimes with $3^{n-1} \equiv 2^{n-1} \pmod n$" Tags="&lt;number-theory&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="1114" PostTypeId="2" ParentId="1108" CreationDate="2010-07-29T07:03:02.577" Score="2" ViewCount="" Body="&lt;p&gt;I couldn't divine much information on your background (e.g. undergraduate, master's level, PhD student...) from the question, but I recently taught an intermediate level graduate course which had a unit on valuation theory.  Sections 1.6 through 1.8 of&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://math.uga.edu/~pete/8410Chapter1.pdf&quot; rel=&quot;nofollow&quot;&gt;http://math.uga.edu/~pete/8410Chapter1.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;address your questions.  In particular, if your field $K$ is the fraction field of a Dedekind domain $R$, then you can always use each prime ideal $\mathfrak{p}$ of $R$ to define a valuation on $K$, essentially the &quot;order at $\mathfrak{p}$&quot;.  There is also a converse result, Theorem 13: if you have a valuation on $K$ which has the additional property that it is non-negative at every element of the Dedekind domain $R$, then it has to be (up to equivalence) the $\mathfrak{p}$-adic valuation for some $\mathfrak{p}$.   I felt the need to give this additional condition a name, so I called such a valuation &lt;strong&gt;R-regular&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The point is that (as Qiaochu says in his comments), in case $K$ is a number field and $R$ is its ring of integers, every valuation on $K$ is $R$-regular.  However, in the function field setting this is not true and this leads to a discussion of &quot;infinite places&quot;.  Note that I do describe the analogues of Ostrowski's Theorem for finite extensions both of $\mathbb{Q}$ and of $F(t)$ for any field $F$ (in the latter case, one restricts to valuations which are trivial on $F$; when $F$ is finite, this condition is automatic).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would be interested to know whether you find the notes helpful.  If not, I or someone else can probably recommend an alternative reference.    &lt;/p&gt;&#xA;" OwnerUserId="299" LastActivityDate="2010-07-29T07:03:02.577" />
  <row Id="1115" PostTypeId="2" ParentId="1108" CreationDate="2010-07-29T07:06:28.077" Score="1" ViewCount="" Body="&lt;p&gt;Discrete valuations &amp;lt;-&gt; points on a curve&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a nonsingular projective curve over an algebraically closed field, there is a one-one correspondence between the points on it, and the discrete valuations of the function field (i.e. all the meromorphic functions of the curve). The correspondence is point P -&gt; the valuation that sends a function f, to the order of zero/pole of f at P.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maximal ideals &amp;lt;-&gt; points on a curve&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At least for varieties (common zeros of several polynomials) over an algebraically closed field, there is a one-one correspondence between points on it, and the maximal ideal in $k[x_1,\cdots,x_n]$. The correspondence is point $P = (a_1,\cdots,a_n)$ -&gt; the polynomials vanishing at P, which turns out to be $(x_1-a_1,\cdots,x_n-a_n)$. This is something true not only for curves, but for varieties. (Hilbert's Nullstellensatz) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So putting these together, for nonsingular projective curves over an algebraically closed field, you know that there is a one-one correspondence between the maximal ideals (think them as points) and the discrete valuations of the function field. Now the situation here is analogous. You consider a &quot;curve&quot;, whose coordinate ring is $\mathbb{Z}$, with function field $\mathbb{Q}$. The nonarchimedean valuations correspond to discrete valuations in this case. So they should capture order of zeros/poles at some &quot;points&quot;. What are the points? They should correspond to the maximal ideals of $\mathbb{Z}$, which are exactly the primes here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for $K(x)$, look at it as the function field of $K\mathbb{P}^1$. Just like the usual real/complex projective spaces, you should have two pieces here. Let's say $K[x]$ corresponds to the piece where the second coordinate is nonzero. So the corresponding homogeneous coordinates here is like $[x,1]$. We know there is one point missing, which is $[1,0]$. For this, we change our coordinates $[x,1] \to [1,1/x]$, so the piece where the first coordinate is nonzero should be $K[1/x]$. The missing point corresponds to the ideal $(1/x - 0) = (1/x)$, so this is why the infinite place corresponds to (1/x). Of course, a more straight forward interpretation is that for a rational function, you divide both numerator and denominator sufficiently high power of $x$ so that they both become polynomials in 1/x, have nonzero constant term, with an extra term (x to the some power). The infinite place measures this power.&lt;/p&gt;&#xA;" OwnerUserId="325" LastEditorUserId="325" LastEditorDisplayName="" LastEditDate="2010-07-29T20:22:39.630" LastActivityDate="2010-07-29T20:22:39.630" />
  <row Id="1116" PostTypeId="1" AcceptedAnswerId="8447" CreationDate="2010-07-29T07:07:04.470" Score="12" ViewCount="752" Body="&lt;p&gt;How many ways can a rectangle be partitioned by either vertical or horizontal lines into &lt;code&gt;n&lt;/code&gt; sub-rectangles? At first I thought it would be:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;      f(n) = 4f(n-1) - 2f(n-2)  &#xA;where f(0) = 1  &#xA;  and f(1) = 1 &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;but the recurrence relation only counts the cases in which at least one side (either top, bottom, left or right of the original rectangle) is not split into sub-rectangles. There are many other partitions that don't belong to those simple cases like&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://img228.imageshack.us/img228/6572/partitions.png&quot; alt=&quot;sample partitions&quot;&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any other related problem suggestions are welcome. Also it is nice to know how to traverse this partition efficiently.&lt;/p&gt;&#xA;" OwnerUserId="333" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2011-06-03T20:12:13.687" LastActivityDate="2011-06-03T20:12:13.687" Title="Number of ways to partition a rectangle into n sub-rectangles" Tags="&lt;combinatorics&gt;&lt;computer-science&gt;&lt;recurrence-relations&gt;" AnswerCount="4" CommentCount="10" FavoriteCount="8" />
  <row Id="1117" PostTypeId="1" AcceptedAnswerId="1119" CreationDate="2010-07-29T07:13:45.857" Score="5" ViewCount="134" Body="&lt;p&gt;How to find all natural X for that X^2 + (X+1)^2 is a perfect square?&lt;/p&gt;&#xA;" OwnerUserId="382" LastActivityDate="2010-07-29T07:19:22.397" Title="Find all X for that X^2 + (X+1)^2 is a square" Tags="&lt;number-theory&gt;" AnswerCount="1" />
  <row Id="1118" PostTypeId="1" AcceptedAnswerId="1122" CreationDate="2010-07-29T07:15:46.650" Score="10" ViewCount="256" Body="&lt;p&gt;I'd like to characterise the functions that 'have square roots' in the function composition sense. That is, can a given function f be written as f = g.g (where . is function composition)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, the function f(x) = x+10 has a square root g(x) = x+5.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, the function f(x) = 9x has a square root g(x) = 3x.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know if the function f(x) = x^2 + 1 has a square root, but I couldn't think of any.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a way to determine which functions have square roots? To keep things simpler, I'd be happy just to consider functions f: R -&gt; R.&lt;/p&gt;&#xA;" OwnerUserId="106" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-07-29T11:06:06.090" LastActivityDate="2010-07-29T11:06:06.090" Title="Characterising functions f that can be written as f = g.g?" Tags="&lt;algebra&gt;&lt;real-analysis&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="4" />
  <row Id="1119" PostTypeId="2" ParentId="1117" CreationDate="2010-07-29T07:19:22.397" Score="10" ViewCount="" Body="&lt;p&gt;Suppose $x^2 + (x+1)^2 = y^2$. We can rewrite it as $(2x+1)^2 + 1 = 2y^2$ or $(2x+1)^2 - 2y^2 = -1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $z=2x+1$ then we have $z^2 - 2y^2 = -1$. This is &lt;a href=&quot;http://en.wikipedia.org/wiki/Pell_equation&quot; rel=&quot;nofollow&quot;&gt;Pell's equation&lt;/a&gt;.&#xA;Wikipedia article shows how to solve it.&lt;/p&gt;&#xA;" OwnerUserId="334" LastActivityDate="2010-07-29T07:19:22.397" CommentCount="1" />
  <row Id="1120" PostTypeId="1" AcceptedAnswerId="1124" CreationDate="2010-07-29T07:49:22.323" Score="3" ViewCount="154" Body="&lt;p&gt;Are there 2 subsets, say, $A$ and $B$, of the naturals such that&lt;br&gt;&#xA;$$\sum_{n\in A}   f(n)   =  \sum_{n\in B}   f(n)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $f(n)=1/n^2$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $f(n)=1/n$ then there are many counterexamples, which is probably a consequence of the fact that the harmonic series diverges:&#xA;$$\frac23 = \frac12 + \frac16 = \frac14+\frac13+\frac1{12}$$&lt;br&gt;&#xA;And if $f(n)=b^{-n}$ for some base b then it is true because for all $M$, $\sum_{n&gt;M} f(n) &amp;lt; f(M)$. (This is just the base-b representation of a real number.The case b=2 gives a &lt;s&gt;bijection&lt;/s&gt; surjection 2^{\N} \to [0,1]).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we have sort of an in-between case here.&lt;br&gt;&#xA;Also, what if $A$,$B$:&lt;br&gt;&#xA;-are required to be finite sets?&lt;br&gt;&#xA;-are required to be infinite and disjoint?&lt;/p&gt;&#xA;" OwnerUserId="414" LastEditorUserId="414" LastEditorDisplayName="" LastEditDate="2010-07-30T02:15:15.297" LastActivityDate="2010-07-30T02:15:15.297" Title="Summing 1/n^2 over subsets of N." Tags="&lt;sequences-and-series&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="1121" PostTypeId="2" ParentId="370" CreationDate="2010-07-29T08:02:00.330" Score="4" ViewCount="" Body="&lt;p&gt;And when you get bored of reading, let the &lt;a href=&quot;http://www.youtube.com/TheCatsters#p/u/68/xqLgGB7Hv7g&quot; rel=&quot;nofollow&quot;&gt;Catsters&lt;/a&gt; take over. (78 videos on Category theory!)&lt;/p&gt;&#xA;" OwnerUserId="423" LastActivityDate="2010-07-29T08:02:00.330" CommunityOwnedDate="2010-07-29T08:02:00.330" />
  <row Id="1122" PostTypeId="2" ParentId="1118" CreationDate="2010-07-29T08:07:08.430" Score="9" ViewCount="" Body="&lt;p&gt;I showed you the link to the MO question mostly to convince you that this is a hard question.  I will &quot;answer&quot; it in the special case that $f$ is a bijection.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Recall that given a bijection $f : S \to S$, where $S$ is a set, a cycle of $f$ length $n$ is a set of distinct points $x, f(x), ... f^{n-1}(x)$ such that $f^n(x) = x$.  A cycle of infinite length is a set of distinct points $x, f(x), f^2(x), ...$.  It is not hard to see that $S$ is a disjoint union of cycles of $f$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Claim:&lt;/strong&gt;  A bijection $f : S \to S$ has a square root if and only if there are an even number of cycles of $f$ of any given even length.  (For the purposes of this result, infinity is an even number; so there can be an infinite number of cycles, and you need to consider cycles of infinite length.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Proof.&lt;/em&gt;  First we show that any bijection with a square root has this property.  Let $g : S \to S$ be a bijection such that $g(g(x)) = f(x)$.  Then each cycle of $g$ corresponds to either one or two cycles of $f$, as follows.  If the cycle has odd length, it corresponds to one cycle of $f$.  For example, the cycle $1 \to 2 \to 3 \to 1$ of $g$ would correspond to the cycle $1 \to 3 \to 2 \to 1$ of $f$.  If the cycle has even length, it corresponds to two cycles of $f$.  For example, the cycle $1 \to 2 \to 1$ of $g$ would correspond to the pair of cycles $1 \to 1$ and $2 \to 2$, and the cycle $1 \to 2 \to 3 \to ... $ would correspond to the pair of cycles $1 \to 3 \to ... $ and $2 \to 4 \to ...$.  In particular, cycles of $f$ of odd length can come from cycles of $g$ one at a time or two at a time, but cycles of $f$ of even length can only come from cycles of $g$ two at a time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we show the reverse implication.  Given a cycle of $f$ of odd length $2k+1$, consider the corresponding cycle of $f^{k+1}$ of odd length.  Since $f^{2k+2} = f$ when restricted to this cycle, make this a cycle of $g$.  Given a pair of cycles of $f$ of the same even length, just weave them together to get a cycle of $g$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I say &quot;answer&quot; instead of answer because it's not obvious if you can always find the cycle decomposition of some complicated bijection on an infinite set.  In any case, if $f$ isn't assumed to be a bijection this question becomes much harder; the analogue of cycle decomposition is much more difficult to work with.  I suggest you look at some examples where $S$ is finite if you really want to get a grip on this case; best of luck.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-07-29T08:07:08.430" CommentCount="1" />
  <row Id="1123" PostTypeId="2" ParentId="855" CreationDate="2010-07-29T08:07:09.613" Score="3" ViewCount="" Body="&lt;p&gt;Here's a very simplistic proof:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;n^2 = 1 (mod 24) for n=1,5,7 and 11, by checking each case individually.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(n+12)^2 = n^2 + 24n + 144 = n^2 (mod 24).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, n^2 = 1 (mod 24) when n is odd and not divisible by 3, and so n^2-1 is divisible by 24 for these n. You don't need primality of p here!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A slight modification would be to use 1 and 5 as &quot;base cases&quot;, and use the fact that (n+6)^2 = n^2 + 12n + 36 = n^2 + 12(n+3), which is equal to n^2 mod 24 when (n+3) is even, i.e. n is odd.&lt;/p&gt;&#xA;" OwnerUserId="360" LastActivityDate="2010-07-29T08:07:09.613" />
  <row Id="1124" PostTypeId="2" ParentId="1120" CreationDate="2010-07-29T08:38:10.443" Score="5" ViewCount="" Body="&lt;p&gt;Yes to both cases:&#xA;1) $\frac{1}{15^2}+\frac{1}{20^2}=\frac{1}{12^2}$&#xA;2) $\frac{1}{15^2}+\frac{1}{150^2}+\frac{1}{1500^2}+...+\frac{1}{20^2}+\frac{1}{200^2}+\frac{1}{2000^2}+...=\frac{1}{12^2}+\frac{1}{120^2}+\frac{1}{1200^2}...$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for first case - if we have pythagorean triple (a,b,c), such that $a^2+b^2=c^2$, then:&#xA;$\frac{1}{a^2 b^2}=\frac{1}{a^2 c^2}+\frac{1}{b^2 c^2}$&lt;/p&gt;&#xA;" OwnerUserId="425" LastActivityDate="2010-07-29T08:38:10.443" />
  <row Id="1125" PostTypeId="1" AcceptedAnswerId="1161" CreationDate="2010-07-29T08:43:19.217" Score="1" ViewCount="137" Body="&lt;p&gt;I know volume preserving diffeomorphisms of a sphere^2 make a group sdiff(S2). I would to know if it is a Lie group, which I assume if it is that makes interpolation easier (like with rotations). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So that is one question, is it a Lie group?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also is the group path connected? If so, how can I interpolate between two elements in the group?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These are not subjects I know very little about. I apologize if Im phrasing it in some way that sounds ridiculous.  &lt;/p&gt;&#xA;" OwnerUserId="99" LastEditorUserId="99" LastEditorDisplayName="" LastEditDate="2010-07-29T09:15:23.347" LastActivityDate="2010-07-29T18:38:40.070" Title="Interpolating between volume preserving diffeomorphisms of sphere" Tags="&lt;geometry&gt;&lt;abstract-algebra&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="1126" PostTypeId="2" ParentId="1113" CreationDate="2010-07-29T08:43:34.643" Score="0" ViewCount="" Body="&lt;p&gt;as Qiaochu Yuan pointed out, take a &lt;a href=&quot;http://en.wikipedia.org/wiki/Carmichael_number&quot; rel=&quot;nofollow&quot;&gt;Carmichael number&lt;/a&gt; &lt;em&gt;q&lt;/em&gt;; by definition, 3&lt;sup&gt;&lt;em&gt;q&lt;/em&gt;-1&lt;/sup&gt; and 2&lt;sup&gt;&lt;em&gt;q&lt;/em&gt;-1&lt;/sup&gt; are both congruent to 1 mod &lt;em&gt;q&lt;/em&gt;, so their difference is a multiple of &lt;em&gt;q&lt;/em&gt;. Since Carmichael numbers are infinite, you are done.&lt;/p&gt;&#xA;" OwnerUserId="89" LastActivityDate="2010-07-29T08:43:34.643" CommentCount="1" />
  <row Id="1127" PostTypeId="1" AcceptedAnswerId="1128" CreationDate="2010-07-29T08:59:19.303" Score="6" ViewCount="233" Body="&lt;p&gt;I just came up with this problem yesterday.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Problem&lt;/strong&gt;:&#xA;Assume there is an important segment of straight line &lt;code&gt;AB&lt;/code&gt; that needs to be watched at all time. A watchdog can see in one direction in front of itself and must walk at a constant &lt;em&gt;non-zero&lt;/em&gt; speed at all time. (All watchdogs don't need to have the same speed.) When it reaches the end of the segment, it must turn back (at no time) and keep watching the line.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How many watchdogs does it need to guarantee that the line segment is watched at all time? And how (initial positions and speeds of the dogs)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Note&lt;/strong&gt;: &#xA;It's clear that two dogs are not enough. I conjecture that four will suffice and three will not. For example, the below configuration doesn't work from 7.5 second if &lt;code&gt;AB&lt;/code&gt;'s length is 10 meters.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Dog 1 at A               walks to the right with speed 1.0 m/s&#xA;Dog 2 at between A and B walks to the right with speed 1.0 m/s&#xA;Dog 3 at B               walks to the left  with speed 1.0 m/s&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Or it can be illustrated as:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;         A ---------------------------------------- B&#xA;0.0 sec    1 --&amp;gt;               2 --&amp;gt;          &amp;lt;-- 3&#xA;&#xA;2.5 sec              1 --&amp;gt;          &amp;lt;-- 32 --&amp;gt;&#xA;&#xA;5.0 sec                   &amp;lt;-- 31 --&amp;gt;          &amp;lt;-- 2&#xA;&#xA;7.5 sec         &amp;lt;-- 3               &amp;lt;-- 21 --&amp;gt;&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Please provide your solutions, hints, or related problems especially in higher dimensions or looser conditions (watchdogs can walk with acceleration, etc.)&lt;/p&gt;&#xA;" OwnerUserId="333" LastEditorUserId="333" LastEditorDisplayName="" LastEditDate="2010-07-30T04:33:42.937" LastActivityDate="2010-07-30T04:33:42.937" Title="Watchdog Problem" Tags="&lt;discrete-mathematics&gt;&lt;recreational-mathematics&gt;&lt;puzzle&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="1" />
  <row Id="1128" PostTypeId="2" ParentId="1127" CreationDate="2010-07-29T09:29:50.983" Score="8" ViewCount="" Body="&lt;p&gt;Three dogs is enought I think.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let the length of line segment be equal to 1 (with coordinates from 0 to 1).&lt;br &gt;&#xA;First dog: start position = 0, speed = +1/3&lt;br /&gt;&#xA;Second dog: start position = 2/3, speed = +1/3&lt;br /&gt;&#xA;Third dog: start position = 2/3, speed = -1/3&lt;br /&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After 1 second the position becomes similar.&lt;/p&gt;&#xA;" OwnerUserId="334" LastActivityDate="2010-07-29T09:29:50.983" CommentCount="3" />
  <row Id="1130" PostTypeId="1" AcceptedAnswerId="1137" CreationDate="2010-07-29T09:53:03.543" Score="1" ViewCount="59" Body="&lt;p&gt;Suppose we have some function $f(x)$ with local extrema at $x_1, x_2, \dots$, and a second function $g(x)$ which is continuous, strictly increasing and non-zero everywhere over the range of the $x_i$. Will $g(f(x))$ have its local extrema at the same $x_i$ and no others?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If so, are there any obvious loosenings of the constraints on $g$ for which this will remain true?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I'm really thinking of this in the context of signal processing, looking at transformations that preserve the visual structure of an image, but it seems like a general question that must have been trivially proved by someone 250 years ago...)&lt;/p&gt;&#xA;" OwnerUserId="144" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-07-29T11:00:37.177" LastActivityDate="2010-07-29T11:00:37.177" Title="Preserving the extrema of one function after applying another" Tags="&lt;calculus&gt;&lt;signal-processing&gt;" AnswerCount="2" />
  <row Id="1131" PostTypeId="2" ParentId="1130" CreationDate="2010-07-29T10:02:33.837" Score="3" ViewCount="" Body="&lt;p&gt;Since g is continuous and strictly increasing, its inverse $g^{-1}$ is a function and strictly increasing.  Since both are strictly increasing, $a&amp;lt;b\Leftrightarrow g(a)&amp;lt;g(b)\Leftrightarrow g^{-1}(a)&amp;lt;g^{-1}(b)$.  From this, it follows that $x_i$ is a local max (min) of g(f(x)) iff it is a local max (min) of f(x).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If g were continuous and strictly decreasing, it would exchange local maximums and minimums (because $a&amp;lt;b\Leftrightarrow g(a)&amp;gt;g(b)$ and $a&amp;lt;b\Leftrightarrow g^{-1}(a)&amp;gt;g^{-1}(b)$).&lt;/p&gt;&#xA;" OwnerUserId="72" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-07-29T10:16:47.663" LastActivityDate="2010-07-29T10:16:47.663" />
  <row Id="1132" PostTypeId="1" AcceptedAnswerId="13930" CreationDate="2010-07-29T10:12:29.380" Score="6" ViewCount="205" Body="&lt;p&gt;How can I show with a heuristic argument based on a Taylor expansion that for Stratonovich&#xA;stochastic calculus the chain rule takes the form of the classical (Newtonian) one? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Concerning Ito calculus the fact that dX^2 = dt results via a Taylor expansion in Ito's lemma - this fact should stay the same with Stratonovich but it should somehow cancel out in there - I just don't know how...&lt;/p&gt;&#xA;" OwnerUserId="346" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-07-29T11:03:40.633" LastActivityDate="2010-12-11T17:47:37.827" Title="Taylor expansion to show that for Stratonovich stochastic calculus the chain rule takes the form of the classical one" Tags="&lt;calculus&gt;&lt;intuition&gt;&lt;stochastic-processes&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="1133" PostTypeId="1" AcceptedAnswerId="1168" CreationDate="2010-07-29T10:17:31.483" Score="11" ViewCount="587" Body="&lt;p&gt;Bourbaki shows in  a very natural way that every continuous group isomorphism of the additive reals to the positive multiplicative reals is determined by its value at $1$, and in fact, that every such isomorphism is of the form $f_a(x)=a^x$ for $a&gt;0$ and $a\neq 1$.  We get the standard real exponential (where $a=e$) when we notice that for any $f_a$, $(f_a)'=g(a)f_a$ where $g$ is a continuous group isomorphism from the positive multiplicative reals to the additive reals.  By the intermediate value theorem, there exists some positive real $e$ such that $g(e)=1$ (by our earlier classification of continuous group homomorphisms, we notice that $g$ is in fact the natural log).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice that every deduction above follows from a natural question.  We never need to guess anything to proceed.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any natural way like the above to derive the complex exponential?  The only way I've seen it derived is as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Derive the real exponential by some method (inverse function to the natural log, which is the integral of $1/t$ on the interval $[1,x)$, Bourbaki's method, or some other derivation), then show that it is analytic with infinite radius of convergence (where it converges uniformly and absolutely), which means that it is equal to its Taylor series at 0, which means that we can, by a general result of complex analysis, extend it to an entire function on the complex plane.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This derivation doesn't seem natural to me in the same sense as Bourbaki's derivation of the real exponential, since it requires that we notice some analytic properties of the function, instead of relying on its unique algebraic and topological properties.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anyone know of a derivation similar to Bourbaki's for the complex exponential?  &lt;/p&gt;&#xA;" OwnerUserId="126" LastActivityDate="2011-03-26T11:03:57.600" Title="Natural derivation of the complex exponential function?" Tags="&lt;complex-analysis&gt;&lt;real-analysis&gt;&lt;general-topology&gt;" AnswerCount="6" CommentCount="8" FavoriteCount="2" />
  <row Id="1134" PostTypeId="2" ParentId="1125" CreationDate="2010-07-29T10:36:24.243" Score="0" ViewCount="" Body="&lt;p&gt;The term &quot;volume preserving&quot; sounds a bit ambiguous to me: do you mean that your map preserves the total volume or do you mean that its differential at every point preserves volume (i.e. has determinant 1)? The former is weaker than the latter, and gives you more room for interpolation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In any case, there is a famous invariant of continuous maps $S^2\to S^2$ called the &lt;em&gt;degree&lt;/em&gt;. Any two maps with the same degree are homotopic to each other. Being volume preserving (in the former sense) implies that the degree is $1$ (taking orientation into account!), so you can interpolate between any two volume preserving maps. &lt;em&gt;However&lt;/em&gt;, the intermediate maps in this line of reasoning are only continuous, not necessarily diffeomorphisms. I'm confident that with a standard argument &quot;approximate continuous functions by differentiable ones&quot; you can get them to be differentiable, but I don't about &quot;is diffeomorphism&quot; and &quot;is locally volume preserving&quot; parts.&lt;/p&gt;&#xA;" LastActivityDate="2010-07-29T10:36:24.243" CommentCount="1" />
  <row Id="1135" PostTypeId="1" AcceptedAnswerId="1136" CreationDate="2010-07-29T10:47:23.870" Score="5" ViewCount="329" Body="&lt;p&gt;I can never figure out (because the English language is imprecise) which part of &quot;if and only if&quot; means which implication.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(A if and only if B) = (A &amp;lt;=&gt; B), but is the following correct:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(A only if B) = (A =&gt; B)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(A if B) = (A &amp;lt;= B)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The trouble is, one never comes into contact with &quot;A if B&quot; or &quot;A only if B&quot; using those constructions in everyday common speech.&lt;/p&gt;&#xA;" OwnerUserId="126" LastEditorUserId="126" LastEditorDisplayName="" LastEditDate="2010-07-29T10:52:37.360" LastActivityDate="2010-07-29T23:10:35.437" Title="If and only if, which direction is which?" Tags="&lt;terminology&gt;&lt;logic&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" />
  <row Id="1136" PostTypeId="2" ParentId="1135" CreationDate="2010-07-29T10:52:33.460" Score="6" ViewCount="" Body="&lt;p&gt;I wouldn't say that I &lt;em&gt;never&lt;/em&gt; come into contact with those phrasings--they are certainly rare in technical use, but perhaps more common in plain language.  Below is a table of equivalent phrasings of p=&gt;q, from UCSMP &lt;em&gt;Precalculus and Discrete Mathematics&lt;/em&gt;, 3rd ed., &amp;copy; 2010 Wright Group/McGraw Hill (Lesson 1-5).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://www.imgftw.net/img/171340741.png&quot; alt=&quot;table&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-07-29T10:52:33.460" CommentCount="9" />
  <row Id="1137" PostTypeId="2" ParentId="1130" CreationDate="2010-07-29T10:52:44.257" Score="4" ViewCount="" Body="&lt;p&gt;You don't need to assume that $g$ is non-zero, and it could be strictly decreasing as well. Furthermore, the conditions on $g$ only need to hold on the image of $f$ (which doesn't need to the be whole of $\mathbb{R}$, for example).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, if $g$ has a local extremum at $y=f(z)$ and $f$ is strictly increasing around $z$, then you're obviously in trouble, because $g\circ f$ will have a local extremum at $z$. But having no local extrema is equivalent to being strictly monotonous.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only that might relax the conition on $g$ is that it has local extrema exactly where $f$ does and they &quot;cancel each other out&quot; or &quot;amplify each other&quot;.&lt;/p&gt;&#xA;" OwnerUserId="434" LastActivityDate="2010-07-29T10:52:44.257" />
  <row Id="1138" PostTypeId="2" ParentId="1135" CreationDate="2010-07-29T10:53:15.173" Score="6" ViewCount="" Body="&lt;p&gt;This example may be more clear, because apples ⊂ fruits is more obvious:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;This is an apple if it is a fruit&quot; is false.&lt;br&gt;&#xA;&quot;This is an apple only if it is a fruit&quot; is true.&lt;br&gt;&#xA;&quot;This is a fruit if it is an apple&quot; is true.&lt;br&gt;&#xA;&quot;This is a fruit only if it is an apple&quot; is false. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A is an apple =&gt; A is a fruit&lt;/p&gt;&#xA;" OwnerUserId="73" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-07-29T23:10:35.437" LastActivityDate="2010-07-29T23:10:35.437" />
  <row Id="1139" PostTypeId="2" ParentId="740" CreationDate="2010-07-29T11:11:47.873" Score="2" ViewCount="" Body="&lt;p&gt;The function f(x) = x over the rationals and 2x over the irrationals is locally increasing in 0 but it is neither increasing nor decreasing.&lt;/p&gt;&#xA;" OwnerUserId="354" LastActivityDate="2010-07-29T11:11:47.873" CommunityOwnedDate="2010-07-29T11:11:47.873" CommentCount="1" />
  <row Id="1140" PostTypeId="2" ParentId="740" CreationDate="2010-07-29T11:16:22.330" Score="6" ViewCount="" Body="&lt;p&gt;Also &lt;a href=&quot;http://en.wikipedia.org/wiki/Conway_base_13_function&quot; rel=&quot;nofollow&quot;&gt;Conway base 13 function&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This function has the following properties:&lt;br&gt;&#xA;1. On every closed interval $[a, b]$ take every real value.&lt;br&gt;&#xA;2. Is continuous nowhere.  &lt;/p&gt;&#xA;" OwnerUserId="334" LastActivityDate="2010-07-29T11:16:22.330" CommunityOwnedDate="2010-07-29T11:16:22.330" />
  <row Id="1143" PostTypeId="1" CreationDate="2010-07-29T11:58:10.617" Score="4" ViewCount="215" Body="&lt;p&gt;Can you give a definition of the Conway base-13 function better than the one actually present on wikipedia (&lt;a href=&quot;http://www.google.it/url?sa=t&amp;amp;source=web&amp;amp;cd=1&amp;amp;ved=0CBgQFjAA&amp;amp;url=http%3A%2F%2Fen.wikipedia.org%2Fwiki%2FConway_base_13_function&amp;amp;ei=empRTNK-OsqWON2_uPUE&amp;amp;usg=AFQjCNFqIS-UhBV9Miw1QnAZJaxnswI3Yg&amp;amp;sig2=AlFPiLqMkh3gd4VPEHJI-Q&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;), which isn't clear? Maybe with some examples?&lt;/p&gt;&#xA;" OwnerUserId="354" LastActivityDate="2010-07-29T20:58:09.280" Title="A definition of Conway base-13 function" Tags="&lt;calculus&gt;&lt;number-theory&gt;" AnswerCount="3" CommentCount="9" FavoriteCount="2" />
  <row Id="1144" PostTypeId="2" ParentId="740" CreationDate="2010-07-29T12:28:46.330" Score="3" ViewCount="" Body="&lt;p&gt;Have also a look here:&lt;br&gt;&#xA;&lt;a href=&quot;http://mathoverflow.net/questions/22189/what-is-your-favorite-strange-function&quot; rel=&quot;nofollow&quot;&gt;http://mathoverflow.net/questions/22189/what-is-your-favorite-strange-function&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="346" LastActivityDate="2010-07-29T12:28:46.330" CommunityOwnedDate="2010-07-29T12:28:46.330" />
  <row Id="1145" PostTypeId="2" ParentId="514" CreationDate="2010-07-29T12:44:40.587" Score="3" ViewCount="" Body="&lt;p&gt;Further counterexamples can be found here: &lt;a href=&quot;http://mathoverflow.net/questions/15444/the-phenomena-of-eventual-counterexamples&quot; rel=&quot;nofollow&quot;&gt;http://mathoverflow.net/questions/15444/the-phenomena-of-eventual-counterexamples&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="273" LastActivityDate="2010-07-29T12:44:40.587" />
  <row Id="1146" PostTypeId="1" AcceptedAnswerId="1188" CreationDate="2010-07-29T13:12:32.010" Score="6" ViewCount="314" Body="&lt;p&gt;I know that this is meant to explain variance butthe description on Wikpiedia stinks and it is not clear how you can explain variance using this technique&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone explain it in a simple way?&lt;/p&gt;&#xA;" OwnerUserId="437" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-07-29T23:44:52.727" LastActivityDate="2010-07-29T23:44:52.727" Title="Intuitive Way To Understand Principal Component Analysis" Tags="&lt;terminology&gt;&lt;intuition&gt;" AnswerCount="3" CommentCount="5" FavoriteCount="2" />
  <row Id="1147" PostTypeId="1" AcceptedAnswerId="1148" CreationDate="2010-07-29T13:21:02.440" Score="3" ViewCount="152" Body="&lt;p&gt;How to calculate volume of tetrahedron given lengths of all it's edges?&lt;/p&gt;&#xA;" OwnerUserId="334" LastActivityDate="2010-07-29T13:24:25.193" Title="Tetrahedron volume" Tags="&lt;geometry&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="1148" PostTypeId="2" ParentId="1147" CreationDate="2010-07-29T13:24:25.193" Score="5" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://mathworld.wolfram.com/Cayley-MengerDeterminant.html&quot; rel=&quot;nofollow&quot;&gt;Cayley-Menger Determinant&lt;/a&gt; - A generalization of Herons Formula.&lt;/p&gt;&#xA;" LastActivityDate="2010-07-29T13:24:25.193" />
  <row Id="1149" PostTypeId="2" ParentId="1143" CreationDate="2010-07-29T13:36:26.367" Score="4" ViewCount="" Body="&lt;p&gt;You just need to switch back and forth from the lexicographic meaning of the base-13 expansion of the number (think of having ABC instead of .-+) and the loaded meaning you give to the well-formed string as a base-10 number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An example of a number for which Conway base-13 function is 0 is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;0.12-34++1+2-34..11111111111...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where the leftmost . is the threedecimal point (that is, it has a semantic meaning), the three rightmost . mean that the base-13 representation has an infinite number of 1 (that is, they have a metameaning), and the other two . are &quot;digits&quot; of the number (that is, they have a syntactic meaning).&lt;br&gt;&#xA;An example of a number for which Conway base-13 function is not 0 is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;0.12-34++1+2-34.11111111111...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At that number, the function has value -34.11111111111... (in base 10)&lt;/p&gt;&#xA;" OwnerUserId="89" LastActivityDate="2010-07-29T13:36:26.367" />
  <row Id="1150" PostTypeId="2" ParentId="1064" CreationDate="2010-07-29T13:39:56.827" Score="8" ViewCount="" Body="&lt;p&gt;Consider the set of reals x whose binary expansion, if you look only at the even digit places, is some fixed non-eventually-repeating pattern z. This is perfect, since we have branching at the odd digits, but they are all irrational, since z is not eventually repeating.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can draw a picture of this set, and it looks something like the Cantor middle third set, except that you divide into four pieces, and take either first+third or second+fourth, depending on the digits of z.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another solution: Begin with an interval having irrational endpoints, and perform the usual Cantor middle-third construction, except that at stage n, be sure to exclude the n-th rational number (with respect to some fixed enumeration), using a subinterval having irrational endpoints. By systematically excluding all rational numbers, you have the desired perfect set of irrationals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Hi François!)&lt;/p&gt;&#xA;" OwnerUserId="413" LastEditorUserId="413" LastEditorDisplayName="" LastEditDate="2010-07-29T21:09:19.270" LastActivityDate="2010-07-29T21:09:19.270" CommentCount="1" />
  <row Id="1151" PostTypeId="2" ParentId="1133" CreationDate="2010-07-29T13:44:40.870" Score="4" ViewCount="" Body="&lt;p&gt;So, what's unnatural about the complex differential equation...&#xA;f : C -&gt; C satisfying f'(z) = f(z) and f(0)=1 ?&lt;/p&gt;&#xA;" OwnerUserId="442" LastActivityDate="2010-07-29T13:44:40.870" CommentCount="1" />
  <row Id="1152" PostTypeId="2" ParentId="381" CreationDate="2010-07-29T13:55:45.977" Score="9" ViewCount="" Body="&lt;p&gt;Consecutive Fibonacci numbers are the &lt;a href=&quot;http://en.wikipedia.org/wiki/Euclidean_algorithm#Algorithmic%20efficiency&quot; rel=&quot;nofollow&quot;&gt;worst-case&lt;/a&gt; (maximum number of steps) numbers for Euclid's gcd algorithm.&lt;/p&gt;&#xA;" OwnerUserId="334" LastActivityDate="2010-07-29T13:55:45.977" CommunityOwnedDate="2010-08-03T18:46:03.740" />
  <row Id="1153" PostTypeId="2" ParentId="954" CreationDate="2010-07-29T14:22:08.317" Score="1" ViewCount="" Body="&lt;p&gt;The Laplace transform can be simply interpreted as a Wick rotated Fourier transform of a function $f(t)$ which vanishes for $t&amp;lt;0$. Wick rotation means (in this case) changing the frequency $\omega$ of the Fourier transform into an imaginary parameter $s=-i\omega$ of the Laplace transform.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason for the inversion&#xA;    $\mathcal{L}^{-1} {F(s)} = f(t) = \frac{1}{2\pi i}\lim_{T\to\infty}\int\limits_{\gamma-iT}^{\gamma+iT}e^{st}F(s)\,ds, \qquad s=\Re(\gamma)$&#xA;to be a bit more complicated than the inverse Fourier transform (see &lt;a href=&quot;http://en.wikipedia.org/wiki/Inverse_Laplace_transform&quot; rel=&quot;nofollow&quot;&gt;wikipedia&lt;/a&gt;, to quote @Akhil) is this imaginary frequency which, if left purely imaginary, would lead to a non-convergent integral. The formula is still similar to the inverse Fourier transform. In fact I think (but have not verified) you could use it for that purpose to.&lt;/p&gt;&#xA;" OwnerUserId="163" LastActivityDate="2010-07-29T14:22:08.317" />
  <row Id="1154" PostTypeId="2" ParentId="1133" CreationDate="2010-07-29T14:26:04.720" Score="-1" ViewCount="" Body="&lt;p&gt;Let&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&#xA;f(x) = cos(x) + i*sin(x)&#xA;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Then&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&#xA;df/dx = -sin(x) + i*cos(x)&#xA;      = i*f(x)&#xA;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;So that&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&#xA;∫(1/f(x)) df = ∫i dx&#xA;ln(f(x)) = ix  + C&#xA;f(x) = e^(ix + C) = cos(x) + i*sin(x)&#xA;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Since &lt;code&gt;f(0) = 1&lt;/code&gt;, &lt;code&gt;C = 0&lt;/code&gt;, so&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&#xA;e^(ix) = cos(x) + i*sin(x)&#xA;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;(I will update this with LaTeX when that functionality becomes available)&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="136" LastActivityDate="2010-07-29T14:26:04.720" CommentCount="7" />
  <row Id="1155" PostTypeId="2" ParentId="1143" CreationDate="2010-07-29T14:26:35.530" Score="1" ViewCount="" Body="&lt;p&gt;The idea of the Conway base-13 number is to find a function that is not continuous, yet if &lt;code&gt;f(a)&amp;lt;x&amp;lt;f(b)&lt;/code&gt;, then there is some &lt;code&gt;c&lt;/code&gt; between &lt;code&gt;a&lt;/code&gt; and &lt;code&gt;b&lt;/code&gt; with &lt;code&gt;f(c)=x&lt;/code&gt;. This a counterexample to the converse of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Intermediate_value_theorem&quot; rel=&quot;nofollow&quot;&gt;intermediate value theorem&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The function is defined by encoding base-10 values in the tail (the digits left after skipping a finite number). We use &lt;code&gt;+&lt;/code&gt;, &lt;code&gt;-&lt;/code&gt;, &lt;code&gt;.&lt;/code&gt; and the digits to represent an encoded number in the tail and require the encoded number to start with a &lt;code&gt;+&lt;/code&gt; or &lt;code&gt;-&lt;/code&gt;. In base 10, every number ending in an infinite number of &lt;code&gt;9&lt;/code&gt;s can be rewritten to end in an infinite number of &lt;code&gt;0&lt;/code&gt;s instead (ie. 0.999...=1.0). Similarly, we decide we will rewrite each Conway number ending in an infinite number of &lt;code&gt;+&lt;/code&gt;, to ensure that each real number has a unique decimal representation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each number can have up to one base-10 encoded value, which is the result of applying Conway's Base 13 function if it exists. If no such encoding exists for x(ie. &lt;code&gt;+&lt;/code&gt; occurring infinite times in the expansion), then we define f(x)=0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We then show that for each &lt;code&gt;a&lt;/code&gt; and &lt;code&gt;b&lt;/code&gt; that we can find a &lt;code&gt;c&lt;/code&gt; in between with an arbitrary encoded value. We first ensure the number being constructed is between &lt;code&gt;a&lt;/code&gt; and &lt;code&gt;b&lt;/code&gt; by copying enough digits from &lt;code&gt;a&lt;/code&gt; and incrementing a digit that won't matter. This is easier because we have disallowed ending in an infinite number of &lt;code&gt;+&lt;/code&gt;. We then concatenate the digits of the signed base-10 representation of the value we wish the function to have to those digits  we have already fixed.&lt;/p&gt;&#xA;" OwnerUserId="123" LastEditorUserId="123" LastEditorDisplayName="" LastEditDate="2010-07-29T20:58:09.280" LastActivityDate="2010-07-29T20:58:09.280" CommentCount="2" />
  <row Id="1156" PostTypeId="2" ParentId="1143" CreationDate="2010-07-29T14:29:47.020" Score="3" ViewCount="" Body="&lt;p&gt;I understand why the Wikipedia article uses the notation it does, but I find it annoying. Here is a transliteration, with some elaboration.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Expand &lt;em&gt;x&lt;/em&gt;&amp;nbsp;&amp;isin;&amp;nbsp;(0,1) in base 13, using digits {0, 1, ...&amp;nbsp;, 9, &lt;em&gt;d&lt;/em&gt;, &lt;em&gt;m&lt;/em&gt;, &lt;em&gt;p&lt;/em&gt;} --- using the convention &lt;em&gt;d&lt;/em&gt;&amp;nbsp;=&amp;nbsp;10, m&amp;nbsp;=&amp;nbsp;11, &lt;em&gt;p&lt;/em&gt;&amp;nbsp;=&amp;nbsp;12. &lt;strong&gt;N.B.&lt;/strong&gt; for rational numbers whose most reduced expression &lt;em&gt;a/b&lt;/em&gt; is such that &lt;em&gt;b&lt;/em&gt; is a power of 13, there are two such expansions: a terminating expansion, and a non-terminating one ending in repeated &lt;em&gt;p&lt;/em&gt; digits. In such a case, use the terminating expansion.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Let &lt;em&gt;S&lt;/em&gt;&amp;nbsp;&amp;sub;&amp;nbsp;(0,1) be the set of reals whose expansion involves finitely many &lt;em&gt;p&lt;/em&gt;, &lt;em&gt;m&lt;/em&gt;, and &lt;em&gt;d&lt;/em&gt; digits, such that the final &lt;em&gt;d&lt;/em&gt; digit occurs after the final &lt;em&gt;p&lt;/em&gt; digit and the final &lt;em&gt;m&lt;/em&gt; digit. (We may require that there be at least one digit 0--9 between the final &lt;em&gt;p&lt;/em&gt; or &lt;em&gt;m&lt;/em&gt; digit and the final &lt;em&gt;d&lt;/em&gt; digit, but this does not seem to be necessary.) Then, every &lt;em&gt;x&lt;/em&gt;&amp;nbsp;&amp;isin;&amp;nbsp;&lt;em&gt;S&lt;/em&gt; has a base 13 expansion of the form&lt;br&gt;&lt;br&gt;&#xA;0.&lt;em&gt;x&lt;/em&gt;&lt;sub&gt;1&lt;/sub&gt;&lt;em&gt;x&lt;/em&gt;&lt;sub&gt;2&lt;/sub&gt;&amp;nbsp;...&amp;nbsp;&lt;em&gt;x&lt;/em&gt;&lt;sub&gt;n&lt;/sub&gt;&amp;nbsp;[&amp;nbsp;&lt;em&gt;p&lt;/em&gt; or &lt;em&gt;m&lt;/em&gt;&amp;nbsp;]&amp;nbsp;&lt;em&gt;a&lt;/em&gt;&lt;sub&gt;1&lt;/sub&gt;&lt;em&gt;a&lt;/em&gt;&lt;sub&gt;2&lt;/sub&gt;&amp;nbsp;...&amp;nbsp;&lt;em&gt;a&lt;/em&gt;&lt;sub&gt;k&lt;/sub&gt;&amp;nbsp;[&amp;nbsp;&lt;em&gt;d&lt;/em&gt;&amp;nbsp;]&amp;nbsp;&lt;em&gt;b&lt;/em&gt;&lt;sub&gt;1&lt;/sub&gt;&lt;em&gt;b&lt;/em&gt;&lt;sub&gt;2&lt;/sub&gt;&amp;nbsp;...&lt;br&gt;&lt;br&gt;&#xA;for some digits &lt;em&gt;x&lt;/em&gt;&lt;sub&gt;j&lt;/sub&gt;&amp;nbsp;&amp;isin;&amp;nbsp;{0, ...&amp;nbsp;, &lt;em&gt;p&lt;/em&gt;} and where the digits &lt;em&gt;a&lt;/em&gt;&lt;sub&gt;j&lt;/sub&gt; and &lt;em&gt;b&lt;/em&gt;&lt;sub&gt;j&lt;/sub&gt; are limited to {0, ...&amp;nbsp;, 9} for all j. The square brackets above are only intended for emphasis; and in particular the n+1&lt;sup&gt;st&lt;/sup&gt; base-13 digit of &lt;em&gt;x&lt;/em&gt; is the final occurance of either &lt;em&gt;p&lt;/em&gt; or &lt;em&gt;m&lt;/em&gt; in the expansion of &lt;em&gt;x&lt;/em&gt;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;For &lt;em&gt;x&lt;/em&gt;&amp;nbsp;&amp;isin;&amp;nbsp;&lt;em&gt;S&lt;/em&gt;, we define f(&lt;em&gt;x&lt;/em&gt;) by transliterating the string format above. We ignore the digits &lt;em&gt;x&lt;/em&gt;&lt;sub&gt;1&lt;/sub&gt; through &lt;em&gt;x&lt;/em&gt;&lt;sub&gt;n&lt;/sub&gt;&amp;nbsp;, transliterate the &lt;em&gt;p&lt;/em&gt; or &lt;em&gt;m&lt;/em&gt; as a plus-sign or minus-sign, and the &lt;em&gt;d&lt;/em&gt; as a decimal point. This yields a decimal expansion for a real number, either&lt;br&gt;&lt;br&gt; +&lt;em&gt;a&lt;/em&gt;&lt;sub&gt;1&lt;/sub&gt;&lt;em&gt;a&lt;/em&gt;&lt;sub&gt;2&lt;/sub&gt;&amp;nbsp;...&amp;nbsp;&lt;em&gt;a&lt;/em&gt;&lt;sub&gt;k&lt;/sub&gt;&amp;nbsp;.&amp;nbsp;&lt;em&gt;b&lt;/em&gt;&lt;sub&gt;1&lt;/sub&gt;&lt;em&gt;b&lt;/em&gt;&lt;sub&gt;2&lt;/sub&gt;&amp;nbsp;...&lt;br&gt;&lt;br&gt; or&lt;br&gt;&lt;br&gt; &amp;minus;&lt;em&gt;a&lt;/em&gt;&lt;sub&gt;1&lt;/sub&gt;&lt;em&gt;a&lt;/em&gt;&lt;sub&gt;2&lt;/sub&gt;&amp;nbsp;...&amp;nbsp;&lt;em&gt;a&lt;/em&gt;&lt;sub&gt;k&lt;/sub&gt;&amp;nbsp;.&amp;nbsp;&lt;em&gt;b&lt;/em&gt;&lt;sub&gt;1&lt;/sub&gt;&lt;em&gt;b&lt;/em&gt;&lt;sub&gt;2&lt;/sub&gt;&amp;nbsp;...&lt;br&gt;&lt;br&gt;according to whether the n+1&lt;sup&gt;st&lt;/sup&gt; base-13 digit of &lt;em&gt;x&lt;/em&gt; is a &lt;em&gt;p&lt;/em&gt; or an &lt;em&gt;m&lt;/em&gt; respectively. For &lt;em&gt;x&lt;/em&gt;&amp;nbsp;&amp;isin;&amp;nbsp;&lt;em&gt;S&lt;/em&gt;, we set f(&lt;em&gt;x&lt;/em&gt;) to this number; for &lt;em&gt;x&lt;/em&gt;&amp;nbsp;&amp;notin;&amp;nbsp;&lt;em&gt;S&lt;/em&gt;, we set f(&lt;em&gt;x&lt;/em&gt;)&amp;nbsp;=&amp;nbsp;0.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Note: this function is not computable, as there is no way that you can determine in advance whether the base-13 expansion of &lt;em&gt;x&lt;/em&gt;&amp;nbsp;&amp;isin;&amp;nbsp;(0,1) has only finitely many occurances of any of the digits &lt;em&gt;p&lt;/em&gt;, &lt;em&gt;m&lt;/em&gt;, or &lt;em&gt;d&lt;/em&gt;; even if you are provided with a number which is promised to have only finitely many, in general you cannot know when you have found the last one. However, if you are provided with a number &lt;em&gt;x&lt;/em&gt;&amp;nbsp;&amp;isin;&amp;nbsp;(0,1) for which you know the location of the final &lt;em&gt;p&lt;/em&gt;, &lt;em&gt;m&lt;/em&gt;, and &lt;em&gt;d&lt;/em&gt; digits, you can compute f(&lt;em&gt;x&lt;/em&gt;) very straightforwardly.&lt;/p&gt;&#xA;" OwnerUserId="439" LastActivityDate="2010-07-29T14:29:47.020" CommentCount="2" />
  <row Id="1157" PostTypeId="2" ParentId="1057" CreationDate="2010-07-29T14:48:37.347" Score="7" ViewCount="" Body="&lt;p&gt;You don't, as the group is not necessarily abelian! The group of upper triangular 3-by-3 matrices with ones along the diagonal and coefficients in the three-element field &#xA;$\mathbb {Z}/3\mathbb{Z}$ has exponent three, so your equation holds, but it is not abelian.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are lots of examples: the most famous ones are the &lt;a href=&quot;http://en.wikipedia.org/wiki/Burnside%27s_problem#Bounded_Burnside_problem&quot; rel=&quot;nofollow&quot;&gt;Burnside groups&lt;/a&gt; $B(m,3)$: the group I described above is $B(2,3)$.&lt;/p&gt;&#xA;" OwnerUserId="274" LastEditorUserId="274" LastEditorDisplayName="" LastEditDate="2010-07-29T20:54:40.320" LastActivityDate="2010-07-29T20:54:40.320" CommentCount="5" />
  <row Id="1158" PostTypeId="2" ParentId="1133" CreationDate="2010-07-29T15:05:15.577" Score="1" ViewCount="" Body="&lt;p&gt;&lt;em&gt;&lt;strong&gt;Some Assumptions&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I will assume that you are ok with power series being used, just not Taylor's theorem. I will also assume you will allow us to observe a solution to a DE since you used it in your derivation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Defn&lt;/strong&gt; A series of the form $\sum_{n=0}^{\infty}c_n\left(z-z_0\right)^n$ for $c_n,z,z_0\in\mathbb{C}$ is called a &lt;em&gt;power series&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Thm&lt;/strong&gt; There is some $R\in[0,\infty]$ such that the power series above converges absolutely for all $z\in\mathbb{C}$ with $\mid z-z_0\mid &amp;lt; R$ and uniformly in $D\left(z_0,\rho\right)$ for all $\rho &amp;lt; R$. Further, the terms are unbounded for all $z$ with $\mid z-z_0\mid &gt; R$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;pf&lt;/strong&gt; Use the geometric series' convergence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Lemma&lt;/strong&gt; Inside the disk of convergence $\sum_{n=1}^{\infty}nc_n\left(z-z_0\right)^{n-1}$ is the derivative of the power series.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&lt;strong&gt;Construction&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For power functions $y(z)=\sum c_nz^n$ can we find a unique solution to $y'(z)=y(z)$ in $\mathbb{C}$? We can observe that this implies $nc_n=c_{n-1}$. Hence&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Defn&lt;/strong&gt; Let $E\left(z\right):=\sum_{n=0}^{\infty}\frac{1}{n!}z^n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Thm&lt;/strong&gt; 1) $E'=E$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) $E\left(z_1+z_2\right)=E\left(z_1\right)E\left(z_2\right)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3)$E_{\mid_{\mathbb{R}}}$ is strictly increasing and $E(\mathbb{R})=(0,\infty)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4) $x\mapsto E(ix)$ sends $\mathbb{R}$ onto $\mathbb{T}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;5) There is some real $\pi&gt;0$ such that $E\left(\frac{\pi}{2}i\right)=i$ and for all $z_1, z_2\in\mathbb{C}$, then $E\left(z_1\right)=E\left(z_2\right)$ iff $\frac{z_1-z_2}{2\pi i}\in\mathbb{Z}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;6) $E\left(\mathbb{C}\right)=\mathbb{C}\setminus\left\lbrace 0\right\rbrace$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that all of these can be shown purely at the level of power series using no complex analysis. Further the proofs are not hard, if you want more details here let me know.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Corollary&lt;/strong&gt; $E$ is a homomorphism of the additive group $\mathbb{C}$ onto the multiplicative group with $\mathbb{C}\setminus\left\lbrace 0\right\rbrace$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&lt;strong&gt;Application&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you care about loops(which I think you do!) lets observe that $\gamma(t)=e^{it}$ for $t\in[0,2\pi]$, is $\mathbb{T}$ (Note here that $\pi$ is simply the real number we found before, not some existential thing!). We could now push this to get winding numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or you could use this definition of exponential, and the power series definition of inverse to get a (branchless) logarithm. In particular, you can show that the derivative of that fella is $\frac{1}{x}$. So we don't have to &lt;em&gt;define&lt;/em&gt; it that way. :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&lt;strong&gt;Comments&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I agree with you that the definitions of complex exponentials feel contrived and the logarithm is even worse. The branched logarithms is the only part of Palka that I dislike as a complex book. These definitions I can stomach, as they require no Dues ex Machina.&lt;/p&gt;&#xA;" OwnerUserId="16" LastEditorUserId="16" LastEditorDisplayName="" LastEditDate="2010-07-29T15:12:39.570" LastActivityDate="2010-07-29T15:12:39.570" CommentCount="5" />
  <row Id="1159" PostTypeId="2" ParentId="107" CreationDate="2010-07-29T15:21:26.037" Score="7" ViewCount="" Body="&lt;p&gt;[&lt;strong&gt;Edit.&lt;/strong&gt;] I am editing my answer to try to give some more insight, prompted by comments on my original response. I hope to develop a deeper idea into what's going on here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The short answer is that what is important is not the size of the set of balls at any particular time, but rather how the set of balls changes; and ultimately what the limit of those sets are. The key is to determine what that limit is, and then determine how many balls are in that limiting set. The answer is that the limit is the empty set, which has size 0. The rest of this answer is devoted to describing this in some detail.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Part of what I have added to my answer is to point out that although there are multiple ways of measuring convergence &amp;mdash; in terms of various norms on characteristic functions &amp;mdash; only one of these actually defines the limit of the sequence, and in this case the limit is well-defined.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;What matters is not the number of balls, but the set of balls&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;In this problem, we have more than just a number of balls which changes with time. What's different is that each of these balls has a unique &lt;em&gt;identity&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This might not seem like it should matter, but it means that the state of the &quot;system&quot; is not a quantity of balls but a &lt;strong&gt;set&lt;/strong&gt; of balls. That set has a certain size, but the size is a &lt;em&gt;derived&lt;/em&gt; feature of the system; it follows from which particular set of balls is present. So it is important to determine what the limit of the sequence of sets is.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Description of the problem in terms of sets&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Let's consider how the set of balls in the box change with time in the game you present. At step n, you add 2&lt;sup&gt;n&lt;/sup&gt; balls, and remove the n lowest-numbered balls. The &quot;state of the system&quot; is given by the following sets:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;S&lt;/em&gt;(0) = {1}&lt;br&gt;&#xA;&lt;em&gt;S&lt;/em&gt;(1) = {2,3}&lt;br&gt;&#xA;&lt;em&gt;S&lt;/em&gt;(2) = {4,5,6,7}&lt;br&gt;&#xA;&lt;em&gt;S&lt;/em&gt;(3) = {7,8,9,10,11,12,13,14,15}&lt;br&gt;&#xA;&lt;em&gt;S&lt;/em&gt;(4) = {11,12,...,31}&lt;br&gt;&#xA;etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that after step 1, the first ball is removed, never to be added again; so it is not an element of the final set. Similarly, at step 2, the second and third balls are removed, never to be added again; so they aren't elements of the final set. And so on. So... none of the balls are in the final set. So then it must be empty! It doesn't matter that the &lt;em&gt;number&lt;/em&gt; of balls in the sequence are increasing; what matters is that the number of balls which will never again be in the box is also increasing, and in the limit includes &lt;strong&gt;all&lt;/strong&gt; of the balls. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can make this more striking by considering, for each step &lt;em&gt;n&lt;/em&gt;, the set of balls &lt;em&gt;I&lt;/em&gt;(&lt;em&gt;n&lt;/em&gt;) which is in the set &lt;em&gt;S&lt;/em&gt;(&lt;em&gt;t&lt;/em&gt;) for all &lt;em&gt;t&lt;/em&gt;&amp;nbsp;&amp;ge;&amp;nbsp;&lt;em&gt;n&lt;/em&gt;: that is, &lt;em&gt;I&lt;/em&gt;(&lt;em&gt;n&lt;/em&gt;)&amp;nbsp;=&amp;nbsp;&lt;em&gt;S&lt;/em&gt;(&lt;em&gt;n&lt;/em&gt;)&amp;nbsp;&amp;cap;&amp;nbsp;&lt;em&gt;S&lt;/em&gt;(&lt;em&gt;n&lt;/em&gt;+1)&amp;nbsp;&amp;cap;&amp;nbsp;&lt;em&gt;S&lt;/em&gt;(&lt;em&gt;n&lt;/em&gt;+2)&amp;nbsp;&amp;cap;&amp;nbsp;...&amp;nbsp;.&#xA;Because each ball is eventually removed, never to be added again; this means that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I&lt;/em&gt;(0) = &lt;em&gt;I&lt;/em&gt;(1) = &lt;em&gt;I&lt;/em&gt;(2) = ... = &amp;empty;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So while the original description makes it look like, moment-to-moment, the final state of the box should be to hold infinitely many balls, a more &quot;forward looking&quot; approach shows that it's clear that the final state of the box is to be empty.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;An analysis in terms of characteristic functions&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;We can contrast the &quot;intuitive&quot; answer of infinitely many balls in the box, and the more precise answer that there ultimately are no balls in the box, using &lt;a href=&quot;http://en.wikipedia.org/wiki/Characteristic_function&quot; rel=&quot;nofollow&quot;&gt;characteristic functions&lt;/a&gt;: that is, we replace each set &lt;em&gt;S&lt;/em&gt;(&lt;em&gt;n&lt;/em&gt;) by a function &lt;em&gt;f&lt;sub&gt;n&lt;/sub&gt;&lt;/em&gt;&amp;nbsp;:&amp;nbsp;&amp;nbsp;&amp;#8469;&amp;nbsp;&amp;rarr;&amp;nbsp;{0,1} which is 1 for those integers belonging to &lt;em&gt;S&lt;/em&gt;(&lt;em&gt;n&lt;/em&gt;), and 0 otherwise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the various &lt;a href=&quot;http://en.wikipedia.org/wiki/Norm_%28mathematics%29#p-norm&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;p&lt;/em&gt;-norms&lt;/a&gt; on these functions. The cardinality of each set &lt;em&gt;S&lt;/em&gt;(&lt;em&gt;n&lt;/em&gt;) is precisely equal to the 1-norm of the function &lt;em&gt;f&lt;sub&gt;n&lt;/sub&gt;&lt;/em&gt;&amp;nbsp;, which grows without bound. The fact that the 1-norm grows without bound &amp;mdash; and in particular, the &lt;em&gt;distance&lt;/em&gt; between the function &lt;em&gt;f&lt;sub&gt;n&lt;/sub&gt;&lt;/em&gt; and the zero function &lt;strong&gt;0&lt;/strong&gt; in the 1-norm grows without bound &amp;mdash; is essentially the source of most people's intuition about this problem, and exactly the reason why they find it counterintuitive that the final set should be empty.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But just as the size of a set is a derived quantity, the norm of a function &amp;mdash; or of a sequence of functions &amp;mdash; is also a derived quantity; and the norm of the limit of a sequence of functions is not necessarily the limit of the norms. In fact, the functions &lt;em&gt;f&lt;sub&gt;n&lt;/sub&gt;&lt;/em&gt; don't converge &lt;em&gt;to&lt;/em&gt; anything, in any of the &lt;em&gt;p&lt;/em&gt;-norms; it simply diverges to nothing in particular.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But there is at least &lt;strong&gt;one&lt;/strong&gt; notion of convergence which applies to the functions &lt;em&gt;f&lt;sub&gt;n&lt;/sub&gt;&lt;/em&gt;&amp;nbsp;, and that is &lt;a href=&quot;http://en.wikipedia.org/wiki/Pointwise_convergence&quot; rel=&quot;nofollow&quot;&gt;point-wise convergence&lt;/a&gt; &amp;mdash; the form of convergence which is the broadest, in the sense that it applies to the most cases (and with which all other notions of convergence must agree, if they show that a sequence of functions converges at all). We may simply show that for each &lt;em&gt;x&lt;/em&gt;, we have &lt;em&gt;f&lt;sub&gt;n&lt;/sub&gt;&lt;/em&gt;(&lt;em&gt;x&lt;/em&gt;)&amp;nbsp;=&amp;nbsp;0 for sufficiently large &lt;em&gt;n&lt;/em&gt;. It then follows that the sequence &lt;em&gt;f&lt;sub&gt;n&lt;/sub&gt;&lt;/em&gt; converges to &lt;strong&gt;0&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fact that the sequence &lt;em&gt;f&lt;sub&gt;n&lt;/sub&gt;&lt;/em&gt; doesn't converge to &lt;strong&gt;0&lt;/strong&gt; under any of the &lt;em&gt;p&lt;/em&gt;-norms doesn't matter; ultimately what matters is that the sequence converges point-wise, because what we're interested in is the cardinality of the limit itself, which is defined in terms of point-wise convergence. At worst, from a certain aesthetic point of view, one might say that it doesn't converge particularly &lt;em&gt;gracefully&lt;/em&gt; (informally speaking) to &lt;strong&gt;0&lt;/strong&gt;; but it does indeed converge, and that is all that matters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So: using characteristic functions, which is ultimately equivalent to the sets described in the first place, one can show that the sequence of sets does converge, and what they converge to is the empty set. But take comfort that your intuition that they should not reflects a certain awareness of the concept of &lt;em&gt;p&lt;/em&gt;-norms. :-)&lt;/p&gt;&#xA;" OwnerUserId="439" LastEditorUserId="439" LastEditorDisplayName="" LastEditDate="2011-05-25T13:24:02.413" LastActivityDate="2011-05-25T13:24:02.413" CommentCount="10" />
  <row Id="1160" PostTypeId="2" ParentId="740" CreationDate="2010-07-29T15:30:02.437" Score="6" ViewCount="" Body="&lt;p&gt;Here's an example of a strictly increasing function on &amp;#8477; which is continuous exactly at the irrationals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pick your favorite absolutely convergent series &amp;sum;a&lt;sub&gt;n&lt;/sub&gt; in which all the terms are positive (mine is &amp;sum;1/2&lt;sup&gt;n&lt;/sup&gt;) and your favorite enumeration of the rationals: &amp;#8474;={q&lt;sub&gt;1&lt;/sub&gt;,q&lt;sub&gt;2&lt;/sub&gt;,...}. For a real number x, define f(x) to be the sum of all the a&lt;sub&gt;n&lt;/sub&gt; for which q&lt;sub&gt;n&lt;/sub&gt; &amp;le; x.&lt;/p&gt;&#xA;" OwnerUserId="60" LastActivityDate="2010-07-29T15:30:02.437" CommunityOwnedDate="2010-07-29T15:30:02.437" />
  <row Id="1161" PostTypeId="2" ParentId="1125" CreationDate="2010-07-29T15:43:50.367" Score="3" ViewCount="" Body="&lt;p&gt;First, as others have pointed out, the group of volume preserving diffeomorphisms will be infinite dimensional.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the second question, there is a beautiful technique known as Moser's trick which answers it.  Moser's trick, in fancy language, says that if (M,w) and (M,w') are two symplectic structures on the same manifold, and if [w] = [w'] in H^2(M;R) (de Rham cohomology), then there is a family of diffeomorphism f_t:M-&gt;M with f_0 = Id and such that f_1 pulls w' back to w.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a 2-dimensional compact, oriented manifold (like the sphere), we have H^2(M;R) = R (the real numbers), and a symplectic form is nothing but a nonzero element in R (which can be interpreted as the total volume).  Since in this setting, [w] = [w'] iff they both give the same signed volume, it follows from Moser's trick that the group of (signed) volume preserving maps is connected.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we consider unsigned volume, there will be 2 components to the group diffeomorphisms preserving the unsigned volume.  This is because, as others have pointed out, one has the notion of &quot;degree&quot; which shows the map x-&gt; -x is not homotopic to Id, even through just continuous maps (not neccesarily volume preserving).  This shows there are AT LEAST two componenets.  There are at most two components because every volume preserving diffeo can be connected to Id or (x -&gt; -x) by Moser's trick again.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Edit:  I misspoke a little bit.  Moser's trick says that if you have a family w_t of symplectic forms, then there is a family of diffeomorphisms as I described above.  It's not clear to me that what I said (that it's enough to have [w] = [w'] in H^2) is enough to guarantee that there is a family of symplectic forms connecting them.  Further, it seems that Moser's trick only guarantees you have a path of diffeos which starts and ends at a volume preserving diffeo, but may not preserve volume for all time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, in the case of S^2 (or any closed, orientable 2-manifold), I can patch things up. Given w and w', volume forms, with [w] = [w'] (i.e, they have the same volume), then the form w_t = tw + (1-t)w' is a path of symplectic forms which connects them.  For a fixed t, the form w_t is closed since it's a sum of closed forms (or, even easier, because it has top degree), and is nondegenerate because it's a volume form (integration shows the volume given is that of w).  The fact that the volume is constant for each w_t implies that the path of diffeos preserves volume for all time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I wasn't able to immediately convince myself that in general, the convex sum of symplectic forms was nondegenerate, hence my initial hesitation.  In fact, I think that it need not be nondegenerate.)&lt;/p&gt;&#xA;" OwnerUserId="331" LastEditorUserId="331" LastEditorDisplayName="" LastEditDate="2010-07-29T18:38:40.070" LastActivityDate="2010-07-29T18:38:40.070" CommentCount="1" />
  <row Id="1163" PostTypeId="2" ParentId="128" CreationDate="2010-07-29T16:10:13.567" Score="4" ViewCount="" Body="&lt;p&gt;There is a fairly good exposition in &lt;a href=&quot;http://terrytao.wordpress.com/2010/03/28/254b-notes-1-equidistribution-of-polynomial-sequences-in-torii/&quot; rel=&quot;nofollow&quot;&gt;Terry Tao's post&lt;/a&gt;, see Corollaries 4-6. Here is a sketch:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We prove the more general statement: Let $p(n)= \chi n^d + a_{d-1} n^{d-1} + \cdots + a_1 n + a_0$ be any polynomial, with $\chi$ irrational. Then $p(n) \mod 1$ is equidistributed. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Set $e(x) = e^{2 \pi i x}$. By the standard trickery with exponential polynomials, it is enough to show &#xA;$$\sum_{n=0}^{N-1} e(p(n)) = o(N).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Choose a positive integer $h$. With a small error, we can replace the sum by&#xA;$$\sum_{n=0}^{N-1} (1/h) \left( e(p(n)) + e(p(n+1)) + \cdots + e(p(n+h-1)) \right).$$&#xA;By Cauchy-Schwartz, this bounded by&#xA;$$\sqrt{N}/h \left[ \sum_{n=0}^{N-1}  \left( e(p(n)) + e(p(n+1)) + \cdots + e(p(n+h-1)) \right) \overline{ \left( e(p(n)) + e(p(n+1)) + \cdots + e(p(n+h-1)) \right)} \right]^{1/2}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Expanding the inner sum, we get $h^2$ terms of the form $e(p(n) - p(n+k))$. There are $h$ terms where $k=0$; these each sum up to $N$. For the other $h^2-h$ terms, the sum is of the form $\sum_{n=0}^{N-1} e(q(n))$, where $q$ has leading term $\chi d n^{d-1}$. By induction, each of these sums is $o(N)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the quantity in the square root is&#xA;$$hN+o(N)$$&#xA;where the constant in the $o$ depends on $h$ and $\chi$. Putting it all together, we get a bound of&#xA;$$N/\sqrt{h} + o(N).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $h$ was arbitrary, this proves the result.&lt;/p&gt;&#xA;" OwnerUserId="448" LastActivityDate="2010-07-29T16:10:13.567" CommentCount="1" />
  <row Id="1164" PostTypeId="1" CreationDate="2010-07-29T16:18:09.690" Score="1" ViewCount="208" Body="&lt;p&gt;Does combinatorial mean discrete ?&lt;/p&gt;&#xA;" OwnerUserId="450" LastEditorUserId="274" LastEditorDisplayName="" LastEditDate="2010-07-29T16:35:30.860" LastActivityDate="2010-07-29T16:44:44.360" ClosedDate="2010-07-29T21:06:18.763" Title="difference between combinatorics and discrete mathematics" Tags="&lt;soft-question&gt;&lt;combinatorics&gt;&lt;discrete-mathematics&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="1165" PostTypeId="2" ParentId="1057" CreationDate="2010-07-29T16:21:20.203" Score="3" ViewCount="" Body="&lt;p&gt;On the other hand, if the order of your group is not a multiple of 3 then it must be abelian!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can read a proof &lt;a href=&quot;http://groupprops.subwiki.org/wiki/Cube_map_is_endomorphism_iff_abelian_%28if_order_is_not_a_multiple_of_3%29&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="446" LastActivityDate="2010-07-29T16:21:20.203" CommentCount="2" />
  <row Id="1166" PostTypeId="2" ParentId="1133" CreationDate="2010-07-29T16:39:51.603" Score="1" ViewCount="" Body="&lt;p&gt;Also, if you are OK with power series, the Prologue to Walter Rudin's &lt;em&gt;Real and Complex Analysis&lt;/em&gt; seems like exactly what you want.  It's a beautiful development of &lt;em&gt;exp&lt;/em&gt;, as well as &lt;em&gt;sin&lt;/em&gt;, &lt;em&gt;cos&lt;/em&gt;, &lt;em&gt;e&lt;/em&gt;, and even &lt;em&gt;&amp;pi;&lt;/em&gt;, all quite organically.&lt;/p&gt;&#xA;" OwnerUserId="452" LastActivityDate="2010-07-29T16:39:51.603" />
  <row Id="1167" PostTypeId="2" ParentId="1164" CreationDate="2010-07-29T16:44:44.360" Score="5" ViewCount="" Body="&lt;p&gt;Combinatorics is a subset of discrete mathematics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Combinatorics is the study of the ways discrete structures can be arranged/combined. Discrete mathematics is the study of all aspects of discrete structures (thus including combinatorics).&lt;/p&gt;&#xA;" OwnerUserId="393" LastActivityDate="2010-07-29T16:44:44.360" />
  <row Id="1168" PostTypeId="2" ParentId="1133" CreationDate="2010-07-29T17:02:00.233" Score="10" ViewCount="" Body="&lt;p&gt;I think essentially the same characterization holds. The complex exponential is the unique Lie group homomorphism from $\mathbb{C}$ to $\mathbb{C}^*$ such that the (real) derivative at the identity is the identity matrix.&lt;/p&gt;&#xA;" OwnerUserId="455" LastActivityDate="2010-07-29T17:02:00.233" CommentCount="1" />
  <row Id="1169" PostTypeId="2" ParentId="740" CreationDate="2010-07-29T17:05:49.243" Score="2" ViewCount="" Body="&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Cauchy%27s_functional_equation&quot; rel=&quot;nofollow&quot;&gt;Cauchy functionals&lt;/a&gt;, which satisfy the very simple equation &lt;em&gt;f(a+b) = f(a)+f(b)&lt;/em&gt; for all real &lt;em&gt;a,b&lt;/em&gt;.  These are either a line through the origin (the &quot;nice&quot; ones) or really &quot;ugly&quot; functions that are discontinuous and unbounded in every interval.  The latter are possible because the Axiom of Choice implies (actually is equivalent to) that infinite dimensional vector spaces have bases; i.e. the reals over the rationals have a Hamel basis.  A great explanation of all this (including the nice/ugly terminology) is in Horst Herrlich's monograph &lt;em&gt;The Axiom of Choice&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="452" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-08-07T18:28:23.390" LastActivityDate="2010-08-07T18:28:23.390" CommunityOwnedDate="2010-07-29T17:05:49.243" />
  <row Id="1170" PostTypeId="2" ParentId="1146" CreationDate="2010-07-29T18:04:04.790" Score="0" ViewCount="" Body="&lt;p&gt;Have a look here:&lt;br&gt;&#xA;&lt;a href=&quot;http://www.symmys.com/AttilioMeucci/Teaching/Talks/Meucci_ModelingEstimationTechniquesForPortfolioManagement.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.symmys.com/AttilioMeucci/Teaching/Talks/Meucci_ModelingEstimationTechniquesForPortfolioManagement.pdf&lt;/a&gt; esp. p. 22f &lt;br&gt;&lt;/p&gt;&#xA;" OwnerUserId="346" LastEditorUserId="346" LastEditorDisplayName="" LastEditDate="2010-07-29T18:09:19.100" LastActivityDate="2010-07-29T18:09:19.100" />
  <row Id="1171" PostTypeId="2" ParentId="1146" CreationDate="2010-07-29T18:10:06.680" Score="0" ViewCount="" Body="&lt;p&gt;Unfortunately I can also post one hyperlink per answer - so here is another one:&#xA;Also have a look at:&lt;br&gt;&#xA;&lt;a href=&quot;http://papers.ssrn.com/sol3/papers.cfm?abstract_id=1358533&quot; rel=&quot;nofollow&quot;&gt;http://papers.ssrn.com/sol3/papers.cfm?abstract_id=1358533&lt;/a&gt; esp. p. 3f&lt;/p&gt;&#xA;" OwnerUserId="346" LastActivityDate="2010-07-29T18:10:06.680" />
  <row Id="1174" PostTypeId="2" ParentId="1079" CreationDate="2010-07-29T20:14:31.867" Score="2" ViewCount="" Body="&lt;p&gt;If each function has a smallest period, and otherwise fits the conditions, then a proof may be forthcoming by attempting to compute the smallest period of the sum and failing.  However, things become unclear if there is no smallest period, as in the case of the characteristic function of the rationals.  Progress might be made in this case by decomposing such a function as an infinite sum of periodic functions, or at least give more&#xA;counterexamples to study.  (e.g. Write the characteristic function of the rationals as&#xA;an infinite sum of functions of smallest period 1. )&lt;/p&gt;&#xA;" OwnerUserId="297" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-08-11T22:05:01.260" LastActivityDate="2010-08-11T22:05:01.260" CommentCount="4" />
  <row Id="1175" PostTypeId="2" ParentId="1079" CreationDate="2010-07-29T20:22:41.597" Score="2" ViewCount="" Body="&lt;p&gt;as a first start, if f+g is periodic the period cannot be rational wrt to the period of f  and g.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let us suppose that T is the smallest period of f(x), that is &amp;forall;x, f(x+T) = f(x); similarly S is the smallest period of g(x), that is &amp;forall;x, g(x+S) = g(x). If f+g had a period Q, and Q/T = m/n, we have that &amp;forall;x, f(x+nQ)+g(x+nQ) = f(x)+g(x). But f(x+nQ)=f(x+mT)=f(x), thus &amp;forall;x, g(x+nQ)=g(x) and therefore nQ is a period of g, which is impossible since it would mean that T/S is rational.&lt;/p&gt;&#xA;" OwnerUserId="89" LastActivityDate="2010-07-29T20:22:41.597" CommentCount="1" />
  <row Id="1176" PostTypeId="2" ParentId="998" CreationDate="2010-07-29T20:29:17.463" Score="1" ViewCount="" Body="&lt;p&gt;I'm really enjoying &lt;a href=&quot;http://www.mathematik.uni-kl.de/~gathmann/alggeom.php&quot; rel=&quot;nofollow&quot;&gt;Andreas Gathmann's lecture notes&lt;/a&gt;.  They are pretty elementary and surprisingly complete (for lecture notes).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reid also has a really nice text on algebraic geometry («Undergraduate algebraic geometry»).&lt;/p&gt;&#xA;" OwnerUserId="314" LastActivityDate="2010-07-29T20:29:17.463" CommunityOwnedDate="2010-07-29T20:29:17.463" />
  <row Id="1177" PostTypeId="1" AcceptedAnswerId="1205" CreationDate="2010-07-29T20:30:03.983" Score="3" ViewCount="143" Body="&lt;p&gt;What are the differences between a (discrete) cosine transform and a (discrete) Fourier transform? I know the former is used in JPEG encoding while the latter plays a big part is signal and image processing- how related are they?&lt;/p&gt;&#xA;" OwnerUserId="124" LastEditorUserId="1543" LastEditorDisplayName="" LastEditDate="2011-04-17T20:32:35.563" LastActivityDate="2011-04-18T09:52:48.117" Title="Comparing/Contrasting Cosine and Fourier Transforms" Tags="&lt;fourier-analysis&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="1178" PostTypeId="2" ParentId="1127" CreationDate="2010-07-29T20:51:21.433" Score="3" ViewCount="" Body="&lt;p&gt;I'll make the trivial answer: 1 dog at point A, facing point B, walking with a velocity of 0.  Presumably, you should really highlight that the dogs' velocities must be non-zero...this is the kind of side case that math people love to exploit. &lt;/p&gt;&#xA;" OwnerUserId="175" LastActivityDate="2010-07-29T20:51:21.433" CommentCount="1" />
  <row Id="1179" PostTypeId="2" ParentId="123" CreationDate="2010-07-29T21:07:02.397" Score="4" ViewCount="" Body="&lt;p&gt;Velocity addition in (special) relativity is not linear, but becomes linear when expressed in terms of hyperbolic tangent functions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More precisely, if you add two motions in the same direction, such as a man walking at velocity $v_1$ on a train that moves at $v_2$ relative to the ground, the velocity $v$ of the man relative to ground is not $v_1 + v_2$; velocities don't add (otherwise by adding enough of them you could exceed the speed of light).  What does add is the &lt;em&gt;inverse hyperbolic tangent&lt;/em&gt; of the velocities (in speed-of-light units, i.e., $v/c$).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\tanh^{-1}(v/c)=\tanh^{-1}(v_1/c) + \tanh^{-1}(v_2/c)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is one way of deriving special relativity: assume that a velocity addition formula holds, respecting a maximum speed of light and some other assumptions, and show that it has to be the above.&lt;/p&gt;&#xA;" OwnerUserId="467" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-29T11:10:41.143" LastActivityDate="2010-12-29T11:10:41.143" />
  <row Id="1180" PostTypeId="2" ParentId="501" CreationDate="2010-07-29T21:39:50.027" Score="6" ViewCount="" Body="&lt;p&gt;Disclaimer: I am not a finitist --- but as a theoretical computer scientist, I have a certain sympathy for finitism. The following is the result of me openly speculating what an &quot;official&quot; finitist response would be, based on grounds of computability.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The short version is this: &lt;strong&gt;(a)&lt;/strong&gt;&amp;nbsp;It depends on what you mean by a 'number', but there's a reasonable approach which makes it reasonable to talk about finitistic approaches to real numbers; &lt;strong&gt;(b)&lt;/strong&gt;&amp;nbsp;What you can do finitisitically with numbers, real, rational, or otherwise, depends on how you represent those numbers.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;What is a number?&lt;/strong&gt; Is &amp;minus;1 a number? Is sqrt(2) a number? Is &lt;em&gt;i&lt;/em&gt;&amp;nbsp;=&amp;nbsp;sqrt(&amp;minus;1) a number? What about quaternions? --- I'm going to completely ignore this question and suggest a pragmatic, formalist approach: a &quot;number&quot; is an element of a &quot;number system&quot;; and a &quot;number system&quot; is a collection of expressions which you can transform or describe properties of in some given ways (&lt;em&gt;i.e.&lt;/em&gt; certain given arithmetic operations) and test certain properties (&lt;em&gt;e.g.&lt;/em&gt; tests for equality, ordering, &lt;em&gt;etc.&lt;/em&gt;) These expressions don't have to have a meaningful interpretation in terms of quantities or magnitudes as far as I'm concerned; &lt;em&gt;you&lt;/em&gt; get to choose which operations/tests you care about.&lt;br&gt;&lt;br&gt; A finitist would demand that any operation or property be described by an algorithm which provably terminates. That is, it isn't sufficient to prove existence or universality &lt;em&gt;a la&lt;/em&gt; classical logic; existence proofs must be finite constructions --- of a &quot;number&quot;, that is a representation in some &quot;number system&quot; --- and univserality must be shown by a computable test. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Representation of numbers:&lt;/strong&gt; How we represent the numbers matters. A finitist should have no qualms about rational numbers: ratios which ultimately boil down to ordered pairs. Despite this, the decimal expansions of these numbers may be infinitely long: 1/3 = 0.33333... what's going on here?&lt;br&gt;&lt;br&gt; Well, the issue is that we have two representations for the same number, one of which is finite in length (and allows us to perform computations) and another which is not finite in length. However, the decimal expansion can be easily expressed as a function: for all &lt;em&gt;k&lt;/em&gt;, the &lt;em&gt;k&lt;/em&gt;&lt;sup&gt;th&lt;/sup&gt; decimal place after the point is '3'; so you can still characterize it precisely in terms of a finite rule.&lt;br&gt;&lt;br&gt; What's important is that there exists &lt;strong&gt;some&lt;/strong&gt; finite way to express the number. But the way in which we choose to &lt;em&gt;define&lt;/em&gt; the number (as a part of system or numbers, using some way of expressing numbers) will affect what we can do with it...&#xA;there is now a question about what operations we can perform.&lt;br&gt;&lt;br&gt;--- For rationals-as-ratios, we can add/subtract, multiply/divide, and test order/equality. So this representation is a very good one for rationals. &lt;br&gt;&lt;br&gt;--- For rationals-as-decimal-expansions, we can still add/subtract and multiply/divide, by defining a new digit-function which describes how to compute the result from the decimal expansions; these will be messier than the representations as ratios. Order comparisons are still possible for &lt;em&gt;distinct&lt;/em&gt; rationals; but you cannot test equality for arbitrary decimal-expansion representations, because you cannot necessarily verify that all decimal places of the difference |&lt;em&gt;a&lt;/em&gt;&amp;minus;&lt;em&gt;b&lt;/em&gt;| are 0. The best you can do in general is testing &quot;equality up to precision &amp;epsilon;&quot;, wherein you show that |&lt;em&gt;a&lt;/em&gt;&amp;minus;&lt;em&gt;b&lt;/em&gt;|&amp;nbsp;&amp;lt;&amp;nbsp;&amp;epsilon;, for some desired precision &amp;epsilon;. This is a number system which informally we may say has certain amount of &quot;vagueness&quot;; but it is in principle completely specified --- there's nothing wrong with this in principle. It's just a matter of how you wish to define your system of arithmetic.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;What representation of reals?&lt;/strong&gt; Obviously, because there are uncountably many real numbers, you cannot represent all real numbers even if you &lt;em&gt;aren't&lt;/em&gt; a finitist. But we can still express some of them. The same is true if you're a finitist: you just don't have access to as many, and/or you're restricted in what you can do with them, according to what your representation can handle.&lt;br&gt;&lt;br&gt;&#xA;--- Algebraic irrational numbers such as sqrt(2) can be expressed simply like that: &quot;sqrt(2)&quot;. There's nothing wrong with the expressions &quot;sqrt(2)&amp;nbsp;&amp;minus;&amp;nbsp;1&quot; or &quot;[1&amp;nbsp;+&amp;nbsp;sqrt(5)]/2&quot; --- they express quantities perfectly well. You can perform arithmetic operations on them perfectly well; and you can also perform ordering/equality tests by transforming them into a normal form of the type &quot;[sum of integers and roots of integers]/[positive integer]&quot;; if the difference of two quantities is zero, the normal form of the difference will just end up being '0'. For order comparisons, we can compute enough decimal places of each term in the sum to determine whether the result is positive or negative, a process which is guaranteed to terminate.&lt;br&gt;&lt;br&gt;&#xA;--- Numbers such as &amp;pi; and e can be represented by decimal expansions, and computed with in this form, as with the rational numbers. The decimal expansions can be gotten from classical equalities (&lt;em&gt;e.g.&lt;/em&gt; &quot;infinite&quot; series, except computing only &lt;em&gt;partial&lt;/em&gt; sums; a number such as e may be expressed by some finite representation of such an 'exact' formula, together with a computable function which describes how many terms of the series are required to get a correct evaluation of the first &lt;em&gt;k&lt;/em&gt; decimal places.) Of course, what you can do finitistically with these representations is limited in the same way as described above with the rationals; specifically, you cannot always test equality.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="439" LastEditorUserId="439" LastEditorDisplayName="" LastEditDate="2010-07-29T21:54:11.680" LastActivityDate="2010-07-29T21:54:11.680" CommentCount="2" />
  <row Id="1181" PostTypeId="1" AcceptedAnswerId="1182" CreationDate="2010-07-29T22:11:12.637" Score="5" ViewCount="257" Body="&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Burnside%27s_lemma&quot; rel=&quot;nofollow&quot;&gt;Burnside Lemma&lt;/a&gt; looks like it should have an intuitive explanation. Does anyone have one?&lt;/p&gt;&#xA;" OwnerUserId="123" LastEditorUserId="130" LastEditorDisplayName="" LastEditDate="2010-11-18T18:47:12.027" LastActivityDate="2010-11-19T07:22:51.490" Title="Intuitive explanation of the Burnside Lemma" Tags="&lt;group-theory&gt;&lt;intuition&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="2" />
  <row Id="1182" PostTypeId="2" ParentId="1181" CreationDate="2010-07-29T22:11:45.070" Score="2" ViewCount="" Body="&lt;p&gt;As an example, we consider the number of ways of colouring a cube with n colours with uniqueness up to rotation. We call each unique colouring where rotations are not allowed a static colouring and each unique one where they are allowed a dynamic colouring. We define the the set of orbits to be the (disjoint) static colourings that correspond to each dynamic colouring. We will use rotations to mean a rotation that makes the cube occupy the same space, and as being unique if it is a unique function from the cube to the cube. This includes the identity rotation. Intuitively, the lemma says:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proposition 1. #Orbits * #Rotations = sum for each rotation r of #static colourings unchanged by this rotation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We will now consider each orbit O separately. Pick a static colouring c inside O. Suppose two (possibly equal) rotations p, q give the same static colouring, d, when applied on c. Then p^-1*q fixes d. Additionally, suppose r (possibly the identity) fixes d. p^-1*(p*r) will also fix d. So p*r will take c to d. Since p*r is different for each r, and p^-1*q is different for each q, the mapping functions are injective in both directions and there is a bijection between the q and r values.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, for each O, the number of rotations is the sum over each static colorings x in O times the number of rotations producing x. This can be rewritten as the sum over each rotation r of the number of static colourings in O fixed by r (due to the bijection in the previous paragraph). We get proposition 1 by adding over all O.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The general proof is quite similar to this, except that it uses group theory.&lt;/p&gt;&#xA;" OwnerUserId="123" LastActivityDate="2010-07-29T22:11:45.070" CommentCount="1" />
  <row Id="1183" PostTypeId="1" AcceptedAnswerId="1184" CreationDate="2010-07-29T22:12:28.233" Score="8" ViewCount="416" Body="&lt;p&gt;I ran into a problem dividing by imaginary numbers recently. I was trying to simplify:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$2 \over i$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I came up with two methods, which produced different results:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Method 1: ${2 \over i} = {2i \over i^2} = {2i \over -1} = -2i$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Method 2: ${2 \over i} = {2 \over \sqrt{-1}} = {\sqrt{4} \over \sqrt{-1}} = \sqrt{4 \over -1} = \sqrt{-4} = 2i$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know from using the formula from &lt;a href=&quot;http://en.wikipedia.org/wiki/Complex_numbers#Operations&quot; rel=&quot;nofollow&quot;&gt;this Wikipedia article&lt;/a&gt; that method 1 produces the correct result. My question is: &lt;strong&gt;why does method 2 give the incorrect result&lt;/strong&gt;? What is the invalid step?&lt;/p&gt;&#xA;" OwnerUserId="477" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-07-31T17:50:52.350" LastActivityDate="2010-07-31T17:50:52.350" Title="Division by imaginary number" Tags="&lt;algebra-precalculus&gt;" AnswerCount="3" CommentCount="6" FavoriteCount="2" />
  <row Id="1184" PostTypeId="2" ParentId="1183" CreationDate="2010-07-29T22:15:08.030" Score="16" ViewCount="" Body="&lt;p&gt;The incorrect step is saying:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sqrt{4}/\sqrt{-1} = \sqrt{4/-1}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The identity:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sqrt{a}/\sqrt{b} = \sqrt{a/b}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is only justified when $a$ and $b$ are positive.&lt;/p&gt;&#xA;" OwnerUserId="31" LastActivityDate="2010-07-29T22:15:08.030" CommentCount="5" />
  <row Id="1185" PostTypeId="2" ParentId="1183" CreationDate="2010-07-29T22:24:16.773" Score="1" ViewCount="" Body="&lt;p&gt;This is exactly the same issue as in &lt;a href=&quot;http://math.stackexchange.com/questions/438/1-is-not-1-so-where-is-the-mistake/482#482&quot;&gt;this question&lt;/a&gt;. Each non-zero complex number has two numbers that square to give it, with the same magnitude, but with opposite sign. When we define the square root function, we have to decide which of the roots we want. For positive numbers, it is obvious to choose the positive root. For negative number, we choose to have the positive imaginary values, although because of symmetry the choice doesn't mean much anyway.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, to see if the standard multiplication and division laws apply, then we have to consider domain the numbers are in. We already know they apply for non-negative real numbers. It is easy enough to verify that for negative numbers sqrt(a)*sqrt(b)=-sqrt(ab) and sqrt(a)/sqrt(b)=sqrt(a/b). We also see that, if a is positive and b negative, then sqrt(ab)=sqrt(a)sqrt(b), sqrt(a/b)=-sqrt(a)/sqrt(b) and sqrt(b/a)=sqrt(b)/sqrt(a).&lt;/p&gt;&#xA;" OwnerUserId="123" LastEditorUserId="123" LastEditorDisplayName="" LastEditDate="2010-07-29T22:45:08.907" LastActivityDate="2010-07-29T22:45:08.907" CommentCount="2" />
  <row Id="1186" PostTypeId="1" CreationDate="2010-07-29T22:58:13.583" Score="6" ViewCount="387" Body="&lt;p&gt;Anything to speed up or simplify calculations. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A simple example might be to get a multiple of 19, for instance, 38 * 19 = 38 * 20 - 38.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;(This is hard to tag with so few tags in play!)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;mental-calculations tips tricks shortcut cheats time-saver&lt;/p&gt;&#xA;" OwnerUserId="483" LastEditorUserId="483" LastEditorDisplayName="" LastEditDate="2010-07-30T12:32:19.310" LastActivityDate="2010-07-30T13:38:00.860" CommunityOwnedDate="2010-07-30T12:32:19.310" Title="What calculation shortcuts exist to help or speed-up mental (or paper) calculations?" Tags="&lt;big-list&gt;&lt;arithmetic&gt;" AnswerCount="4" CommentCount="3" FavoriteCount="4" />
  <row Id="1187" PostTypeId="1" CreationDate="2010-07-29T22:58:59.543" Score="0" ViewCount="256" Body="&lt;p&gt;I'd like to help a 6 year old who already has a pretty good grasp of 2, 5, and 10 times tables.&lt;/p&gt;&#xA;" OwnerUserId="483" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-07-29T23:00:58.633" LastActivityDate="2010-07-31T00:16:31.383" Title="What's the most effective ways of teaching kids - times tables?" Tags="&lt;soft-question&gt;&lt;teaching&gt;" AnswerCount="2" CommentCount="5" FavoriteCount="1" />
  <row Id="1188" PostTypeId="2" ParentId="1146" CreationDate="2010-07-29T23:41:41.033" Score="3" ViewCount="" Body="&lt;p&gt;Principal component analysis is a useful technique when dealing with large datasets. In some fields, (bioinformatics, internet marketing, etc) we end up collecting data which has many thousands or tens of thousands of dimensions. Manipulating the data in this form is not desirable, because of practical considerations like memory and CPU time. However, we can't just arbitrarily ignore dimensions either. We might lose some of the information we are trying to capture!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Principal component analysis is a common method used to manage this tradeoff. The idea is that we can somehow select the 'most important' directions, and keep those, while throwing away the ones that contribute mostly noise. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, this picture shows a 2D dataset being mapped to one dimension: &#xA; &lt;img src=&quot;http://cmp.felk.cvut.cz/cmp/software/stprtool/examples/pca_example.gif&quot; alt=&quot;alt text&quot;&gt;&lt;br&gt;&#xA;Note that the dimension chosen was not one of the original two: in general, it won't be, because that would mean your variables were uncorrelated to begin with.&lt;br&gt;&#xA;We can also see that the direction of the principal component is the one that maximizes the variance of the projected data. This is what we mean by 'keeping as much information as possible.'&lt;/p&gt;&#xA;" OwnerUserId="73" LastActivityDate="2010-07-29T23:41:41.033" />
  <row Id="1189" PostTypeId="2" ParentId="1186" CreationDate="2010-07-29T23:51:00.967" Score="3" ViewCount="" Body="&lt;p&gt;Art Benjamin is your man! He has many tricks to speed up mental calculation and other fun mathemagical tricks. He also wrote two books on the subject!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a video of him in action: &lt;a href=&quot;http://www.youtube.com/watch?v=M4vqr3_ROIk&quot; rel=&quot;nofollow&quot;&gt;http://www.youtube.com/watch?v=M4vqr3_ROIk&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is his new book: &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0307338401&quot; rel=&quot;nofollow&quot;&gt;http://www.amazon.com/Secrets-Mental-Math-Mathemagicians-Calculation/dp/0307338401&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="16" LastActivityDate="2010-07-29T23:51:00.967" CommunityOwnedDate="2010-08-03T18:44:06.083" />
  <row Id="1190" PostTypeId="2" ParentId="1187" CreationDate="2010-07-30T00:05:16.260" Score="10" ViewCount="" Body="&lt;p&gt;I'm guessing you are looking for a hard data research answer, but unfortunately, the studies I know of are dubious, partly because they rely heavily on &lt;a href=&quot;http://numberwarrior.wordpress.com/2010/04/15/are-learning-styles-dead/&quot; rel=&quot;nofollow&quot;&gt;learning style myths&lt;/a&gt;, partly because some are funded by commercial parties pushing particular systems. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I can give some general advice --&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Give the system for the 9s that the digits add to 9, let your pupil fiddle with the patterns, and they tend to get this one fairly quickly.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Asian countries do fairly well with a &quot;musical&quot; or &quot;chant&quot; method. I've heard of a couple commercial products, including &lt;a href=&quot;http://www.abbeytape.com/ttrap/&quot; rel=&quot;nofollow&quot;&gt;Times Table Rap&lt;/a&gt;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Play with the visual representation using marbles or tokens, especially so the pupil can understand why 3 x 7 is the same as 7 x 3.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The &quot;circle&quot; method of visualizing skip counting (samples of what I mean &lt;a href=&quot;http://overthecrescentmoon.blogspot.com/2009/09/skip-counting.html&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;) can be brought in to help with understanding the &quot;rotating&quot; nature of counting on the times tables.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;There are a fair number of games that help multiplication table prowess, either intentionally or unintentionally. Yahtzee in particular naturally covers the 1-6 portion of the table. &lt;a href=&quot;http://numberwarrior.wordpress.com/2010/03/11/multiplication-table-game/&quot; rel=&quot;nofollow&quot;&gt;I've written a game myself&lt;/a&gt;, although I don't know how well it would work with 6 year olds.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;While I wouldn't necessarily recommend the DS game itself, check out how &lt;a href=&quot;http://bit.ly/9SBY5U&quot; rel=&quot;nofollow&quot;&gt;Professor Kageyama's Maths Training: The Hundred Cell Calculation Method&lt;/a&gt; mixes up the columns and rows of the times table to force students to think about the arrangement. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="295" LastActivityDate="2010-07-30T00:05:16.260" />
  <row Id="1192" PostTypeId="2" ParentId="1186" CreationDate="2010-07-30T00:15:34.210" Score="5" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;To square a number ending in 5:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remove the ending 5. Let the resulting number be n, and compute n(n+1). Append 25 to the end of n(n+1) and that's your answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Example:&lt;/strong&gt;&#xA;85&lt;sup&gt;2&lt;/sup&gt;. Here, we drop the last digit to get 8, compute 8*9 = 72, so 85&lt;sup&gt;2&lt;/sup&gt; = 7225. Similarly, we can compute &#xA;115&lt;sup&gt;2&lt;/sup&gt;. Here, we drop the last digit to get 11, compute 11*12 = 132, so 115&lt;sup&gt;2&lt;/sup&gt; = 13225.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;How does this work?:&lt;/strong&gt;&#xA;Note that&#xA;(10n + 5)&lt;sup&gt;2&lt;/sup&gt; = 100n&lt;sup&gt;2&lt;/sup&gt; + 100n + 25 = 100 * n(n+1) + 25.&lt;/p&gt;&#xA;" OwnerUserId="30" LastActivityDate="2010-07-30T00:15:34.210" CommunityOwnedDate="2010-08-03T18:43:51.840" />
  <row Id="1193" PostTypeId="2" ParentId="566" CreationDate="2010-07-30T00:21:41.217" Score="8" ViewCount="" Body="&lt;p&gt;Here are three of my favorite variations on the hats and prisoners puzzle that I've collected over time:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Fifteen prisoners sit in a line, and hats are placed on their heads. Each hat can be one of two colors: white or black. They can see the colors of the people in front of them but not behind them, and they can’t see their own hat colors. Starting from the back of the line (with the person who can see every hat except his own), each prisoner must try to guess the color of his own hat. If he guesses correctly, he escapes. Otherwise, he is fed to cannibals (because that’s the canonical punishment for failing at hat problems). Each prisoner can hear the guess of each person behind him. By listening for painful screaming and the cheering of cannibals, he can also deduce if each of those guesses was accurate. Of course, this takes place in some magical mathematical universe where people don’t cheat. Assuming that they do not want to be eaten, find the optimal guessing strategy for the prisoners. (The cannibals should eat no more than one prisoner.)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;In the year 3141, Earth’s population has exploded. A countably infinite number of prisoners sit in a line (there exists a back of the line, but the other end extends forever). As in the previous problem, white and black hats are placed on their heads. Due to modern technology, each person can see the hat colors of all infinitely many people in front of them. However, they cannot hear what the people behind them say, and they do not know if those people survive. Assuming that they do not want to be eaten, find the optimal guessing strategy for the prisoners. Assume that there are enough cannibals to eat everyone who fails. (The cannibals should eat no more than finitely many prisoners. Assume the Axiom of Choice.)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;There are seven prisoners, and colored hats will be placed on their heads. The hats have seven possible colors (red, orange, yellow, green, blue, indigo, violet), and may be placed in any arrangement: all the same color, all different colors, or some other arrangement. Each person can see everyone else’s hat color but cannot see his own hat color. They may not communicate after getting their hats, and as in the previous problems, they remain in a magical universe where no one cheats. They must guess their hat colors all at the same time. If at least one person guesses correctly, they are all released. If no one guesses correctly, however, the entire group is fed to cannibals. Assuming that they don’t want to be eaten, find the optimal guessing strategy for the prisoners. (By this point, the cannibals have probably eaten far too much. It would be cruel to force them to eat any more, so spare the cannibals and find a way to guarantee that the seven prisoners survive.)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="30" LastActivityDate="2010-07-30T00:21:41.217" CommunityOwnedDate="2010-07-30T00:21:41.217" CommentCount="0" />
  <row Id="1194" PostTypeId="2" ParentId="187" CreationDate="2010-07-30T01:26:01.123" Score="0" ViewCount="" Body="&lt;p&gt;The Wikipedia article on &lt;a href=&quot;http://en.wikipedia.org/wiki/Honeycomb_(geometry)#Space-filling_polyhedra.5B2.5D&quot; rel=&quot;nofollow&quot;&gt;honeycombs&lt;/a&gt; has several examples of 3d tilings by a single polyhedron which aren't vertex-transitive.  I am not positive if they are vertex-uniform, though, since (again) I don't really understand your definition and can't find a good one online.&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-07-30T01:56:26.340" LastActivityDate="2010-07-30T01:56:26.340" />
  <row Id="1195" PostTypeId="2" ParentId="1187" CreationDate="2010-07-30T01:32:41.653" Score="5" ViewCount="" Body="&lt;p&gt;My personal experience with my 7/8 year old is that Math War works very well for practice. You can play with a normal deck of cards---each player plays two cards, and announces the product, with the higher number winning all the cards. In case of tie, then there is &quot;War&quot;, which means you put two (or more) cards face down and then have aother battle on top, with the winner taking the whole pile. You can also play the game with any set of ordinary flash cards. My son is interested to go for a long time with this...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would also add, however, that although I have heard many people say that one must master the basic math facts before learning more advanced math concepts, I think that this is  nonsense. Go ahead and talk about any kind of mathematics (at the right level) with your child.&lt;/p&gt;&#xA;" OwnerUserId="413" LastEditorUserId="413" LastEditorDisplayName="" LastEditDate="2010-07-30T01:46:30.967" LastActivityDate="2010-07-30T01:46:30.967" />
  <row Id="1196" PostTypeId="1" AcceptedAnswerId="1199" CreationDate="2010-07-30T02:07:43.037" Score="4" ViewCount="1139" Body="&lt;p&gt;I'm trying to freshen up for school in another month, and I'm struggling with the simplest of proofs!&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;h2&gt;Problem:&lt;/h2&gt;&#xA;  &#xA;  &lt;p&gt;For any natural number n , n&lt;sup&gt;3&lt;/sup&gt; + 2n is divisible by 3.&#xA;  &lt;strong&gt;&lt;em&gt;This makes sense&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;h2&gt;Proof:&lt;/h2&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;Basis Step:&lt;/strong&gt; If n = 0, then n&lt;sup&gt;3&lt;/sup&gt; + 2n = 0&lt;sup&gt;3&lt;/sup&gt; +&#xA;  2×0 = 0. So it is divisible by 3.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;Induction:&lt;/strong&gt; Assume that for an arbitrary natural number n,&#xA;  n&lt;sup&gt;3&lt;/sup&gt; + 2n is divisible by 3. &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;Induction Hypothesis:&lt;/strong&gt; To prove this for n+1, first try to express ( n&#xA;  + 1 )&lt;sup&gt;3&lt;/sup&gt; + 2( n + 1 )   in terms of n&lt;sup&gt;3&lt;/sup&gt; + 2n   and use&#xA;  the induction hypothesis. &lt;strong&gt;&lt;em&gt;Got it&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;( n + 1 )&lt;sup&gt;3&lt;/sup&gt; + 2( n + 1 ) = (  n&lt;sup&gt;3&lt;/sup&gt; + 3n&lt;sup&gt;2&lt;/sup&gt; + 3n + 1 ) + ( 2n + 2 )  &lt;strong&gt;&lt;em&gt;Just some simplifying&lt;/em&gt;&lt;/strong&gt;&lt;/li&gt;&#xA;  &lt;li&gt;= ( n&lt;sup&gt;3&lt;/sup&gt; + 2n ) + ( 3n&lt;sup&gt;2&lt;/sup&gt; + 3n + 3 ) &lt;strong&gt;&lt;em&gt;simplifying&#xA;  and regrouping&lt;/em&gt;&lt;/strong&gt;&lt;/li&gt;&#xA;  &lt;li&gt;= ( n&lt;sup&gt;3&lt;/sup&gt; + 2n ) + 3( n&lt;sup&gt;2&lt;/sup&gt; + n + 1 ) &lt;strong&gt;&lt;em&gt;factored out&#xA;  the 3&lt;/em&gt;&lt;/strong&gt;&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;  &#xA;  &lt;p&gt;which is divisible by 3, because (&#xA;  n&lt;sup&gt;3&lt;/sup&gt; + 2n ) is divisible by 3&#xA;  by the induction hypothesis. &lt;strong&gt;&lt;em&gt;What?&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Can someone explain that last part? I don't see how you can claim ( n&lt;sup&gt;3&lt;/sup&gt; + 2n ) + 3( n&lt;sup&gt;2&lt;/sup&gt; + n + 1 ) is divisible by 3. &lt;/p&gt;&#xA;" OwnerUserId="494" LastEditorUserId="1208" LastEditorDisplayName="" LastEditDate="2011-02-28T18:36:39.530" LastActivityDate="2011-03-04T20:33:13.167" Title="Proof that $n^3+2n$ is divisible by 3" Tags="&lt;number-theory&gt;&lt;high-school&gt;" AnswerCount="7" CommentCount="6" FavoriteCount="1" />
  <row Id="1197" PostTypeId="2" ParentId="1196" CreationDate="2010-07-30T02:13:44.880" Score="1" ViewCount="" Body="&lt;p&gt;In a proof by induction, we try to prove that a statement is true for all integers n. To do this, we first check the base case, which is the &quot;Basic Step&quot; above. Then, we have the induction hypothesis, where we assume that the statement is true for an integer k. Using this fact, we prove that the statement is also true for the next integer k+1. This produces a bootstrapping ladder where we use the base case (n=1) to show that the statement is true for n=2 via the inductive hypothesis, and then for n=3, etc, off to infinity; this shows that the statement is true for all integers n.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here, we claimed that ( n&lt;sup&gt;3&lt;/sup&gt; + 2n ) + 3( n&lt;sup&gt;2&lt;/sup&gt; + n + 1 ) is divisible by 3 because this was the inductive hypothesis; we were using this to show that&#xA;( (n+1)&lt;sup&gt;3&lt;/sup&gt; + 2(n+1) ) + 3( (n+1)&lt;sup&gt;2&lt;/sup&gt; + (n+1) + 1 ).&lt;/p&gt;&#xA;" OwnerUserId="30" LastActivityDate="2010-07-30T02:13:44.880" />
  <row Id="1198" PostTypeId="2" ParentId="1196" CreationDate="2010-07-30T02:15:17.760" Score="2" ViewCount="" Body="&lt;p&gt;Why don't you just test the validity of this using modular arithmetic? I.e. take n congruent to 1 mod 3 and we easily get 3=0. If you try n congruent to 2 mod 3, you get 8+4 =12 =0, so you're done. No ugly inductions (although this particular case is not so dirty).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A useful idea when thinking of induction is to think of dominos. If you know something is true for one fixed tile and if you know that it being true for one tile means that it's true for the neighbour on the right, then it's like knocking one over knocks them all over.&lt;/p&gt;&#xA;" OwnerUserId="298" LastEditorUserId="298" LastEditorDisplayName="" LastEditDate="2010-07-30T02:20:51.290" LastActivityDate="2010-07-30T02:20:51.290" CommentCount="4" />
  <row Id="1199" PostTypeId="2" ParentId="1196" CreationDate="2010-07-30T02:26:09.203" Score="7" ViewCount="" Body="&lt;p&gt;In the inductive hypothesis, you assumed that $n^3 + 2n$ was divisible by 3 for some $n$, and now you're proving the same for $n+1$. It's like knocking down dominoes: if you can prove that the first domino falls over (base case) and each domino knocks over the next (inductive step), then that means that all of the dominoes get knocked down eventually.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You know that $(n^3 + 2n) + 3(n^2 + n + 1)$ is divisible by 3 because $n^3 + 2n$ is (because of the inductive hypothesis) and $3(n^2 + n + 1)$ is (because it's 3 times an integer). So, their sum is as well.&lt;/p&gt;&#xA;" OwnerUserId="45" LastEditorUserId="45" LastEditorDisplayName="" LastEditDate="2010-07-30T12:20:44.747" LastActivityDate="2010-07-30T12:20:44.747" CommentCount="1" />
  <row Id="1200" PostTypeId="2" ParentId="1196" CreationDate="2010-07-30T02:34:59.507" Score="3" ViewCount="" Body="&lt;p&gt;Presumably you're only looking for a way to understand the induction problem, but you can note that $n^3+2n = n^3 - 3n^2 + 2n - 3n^2 = (n-2)(n-1)(n) + 3n^2$. Since any three consecutive integers has a multiple of three, we're adding two multiples of three and so get another multiple of 3.&lt;/p&gt;&#xA;" OwnerUserId="45" LastActivityDate="2010-07-30T02:34:59.507" CommentCount="2" />
  <row Id="1201" PostTypeId="2" ParentId="1186" CreationDate="2010-07-30T02:35:19.460" Score="3" ViewCount="" Body="&lt;p&gt;One simple one that you're probably familiar with already: When you multiply a one-digit number, n, by 9, the result has n-1 in the 10s place, and then the ones digit is such that the sum of the digits is 9.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example: 9*6 = 54 because 5 is 6-1, and then 5+4 is 9.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What's really cool is that you can use this trick, plus your fingers, to get the answer instantly. Hold out your hands with all 10 fingers up, then put down the n'th finger (which might be on either hand. Then the number is just (how many fingers there are to the left of the finger you put down)*10 + (how many fingers there are to the right of the finger you put down)&lt;/p&gt;&#xA;" OwnerUserId="496" LastActivityDate="2010-07-30T02:35:19.460" CommunityOwnedDate="2010-08-03T18:45:03.010" />
  <row Id="1202" PostTypeId="1" AcceptedAnswerId="1203" CreationDate="2010-07-30T02:37:40.377" Score="3" ViewCount="101" Body="&lt;p&gt;I was reading &quot;Functional Equations and How to Solve Them&quot; by Small and the following comment pops up without much justification on p. 13: If a(x) is an involution, then f(a(x))=f(x) has as solutions f(x) = T[x,a(x)], where T is an arbitrary symmetric function of u and v.  I was wondering why this was true (it works for examples I've tried, but I am not sure how to prove this -and if there's anything obvious staring at me in the face here).&lt;/p&gt;&#xA;" OwnerUserId="298" LastActivityDate="2010-07-30T02:43:39.330" Title="f(a(x))=f(x) - functional eq" Tags="&lt;functions&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="1203" PostTypeId="2" ParentId="1202" CreationDate="2010-07-30T02:43:39.330" Score="5" ViewCount="" Body="&lt;p&gt;Any function f(x) that is a solution to your functional equation f(a(x)) = f(x) must satisfy the property that it is unchanged when you plug in a(x) instead of x. In addition, f(x) clearly must be a function f(x) = T[x, a(x)] depending on the x and a(x); the question is to see why T must be symmetric.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, T[x, a(x)] = f(x) = f(a(x)) = T[a(x), a(a(x))] = T[a(x), x] since a(x) is an involution. In particular, this means that T must be symmetric in its two variables.&lt;/p&gt;&#xA;" OwnerUserId="30" LastActivityDate="2010-07-30T02:43:39.330" CommentCount="2" />
  <row Id="1204" PostTypeId="1" AcceptedAnswerId="1208" CreationDate="2010-07-30T02:58:21.027" Score="4" ViewCount="208" Body="&lt;p&gt;I know that a superalgebra is a Z/2Z graded algebra and that it behaves nicely. I know very little physics though, so even though I know &lt;em&gt;that&lt;/em&gt; the super- prefix is related to supersymmetry, I don't know what that means; is there a compelling &lt;em&gt;mathematical&lt;/em&gt; reason to consider superalgebras? &lt;/p&gt;&#xA;" OwnerUserId="298" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-07-31T17:46:19.847" LastActivityDate="2010-08-02T22:16:17.183" Title="Why are superalgebras so important?" Tags="&lt;abstract-algebra&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="1205" PostTypeId="2" ParentId="1177" CreationDate="2010-07-30T03:44:04.673" Score="3" ViewCount="" Body="&lt;p&gt;Cosine transforms are nothing more than shortcuts for computing the Fourier transform of a sequence with special symmetry (e.g. if the sequence represents samples from an even function).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To give a concrete example in &lt;em&gt;Mathematica&lt;/em&gt; (&lt;code&gt;$VersionNumber &amp;gt;= 6&lt;/code&gt;), consider the sequence&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;smp = {1., 2., 3., 4., 5., 4., 3. ,2.}&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The sequence has redundancy (e.g. &lt;code&gt;smp[[2]] == smp[[8]]&lt;/code&gt;). A sequence like &lt;code&gt;smp&lt;/code&gt; is termed an &lt;em&gt;even&lt;/em&gt; sequence. The discrete Fourier transform of &lt;code&gt;smp&lt;/code&gt; can be expected to have redundancy as well:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Chop[Fourier[smp]]&#xA;{8.48528137423857, -2.414213562373095, 0, -0.4142135623730949, 0, -0.4142135623730949, 0, -2.414213562373095}&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;and the discrete Fourier transform is itself even. One could hope to have a way to compute the discrete Fourier transform without redundancy, and this is where the (type I) discrete cosine transform comes in:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;FourierDCT[Take[smp, Length[smp]/2 + 1], 1]&#xA;{8.48528137423857, -2.414213562373095, 0., -0.4142135623730949, 0.}&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The other three types of discrete cosine transforms, as well as the four types of discrete sine transforms, are intended to be redundancy-free methods for computing discrete Fourier transforms. To use DCT-I as an example, one can deal with a sequence of length $\frac{N}{2}+1$ instead of a sequence of length $N$, which represents a savings in computational time and effort. (I assume the case of even length here; for odd length, a similar symmetry property can be established.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In any event, I wish to point out two good references on how FFT and the DCTs/DSTs are related: &lt;a href=&quot;http://books.google.com/books?id=ozcvQUqr-GIC&quot; rel=&quot;nofollow&quot;&gt;Van Loan's &lt;em&gt;Computational Frameworks for the Fast Fourier Transform&lt;/em&gt;&lt;/a&gt; and &lt;a href=&quot;http://books.google.com/books?id=coq49_LRURUC&quot; rel=&quot;nofollow&quot;&gt;Briggs/Henson's &lt;em&gt;The DFT: an owner's manual for the discrete Fourier transform&lt;/em&gt;&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="498" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2011-04-18T09:52:48.117" LastActivityDate="2011-04-18T09:52:48.117" />
  <row Id="1206" PostTypeId="2" ParentId="524" CreationDate="2010-07-30T03:53:59.347" Score="2" ViewCount="" Body="&lt;p&gt;It's all dependent on the 2-norm condition number of your matrix (the ratio of the largest singular value to the smallest); as a nice rule of thumb, if the base-10 logarithm of the reciprocal of the condition number is much less than the number of digits your computer uses to store numbers, QR (and maybe even the normal equations) might be sufficient. Otherwise, SVD is a &quot;safer bet&quot;: it always works, but is much slower than the other methods for solving least squares problems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Good references would be the classic &quot;Solving Least Squares Problems&quot; by Lawson and Hanson, and the newer &quot;Numerical Methods for Least Squares Problems&quot; by Björck.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;To add to the answer I gave previously, one way you can proceed for a matrix A whose conditioning you don't know would be as follows:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Compute the QR decomposition of A.&lt;/li&gt;&#xA;&lt;li&gt;Estimate the condition number of R.&lt;/li&gt;&#xA;&lt;li&gt;If R is well conditioned (condition number is &quot;small enough&quot;), stop; else&lt;/li&gt;&#xA;&lt;li&gt;Compute the singular value decomposition of R=U∑V&lt;sup&gt;T&lt;/sup&gt;&lt;/li&gt;&#xA;&lt;li&gt;Multiply Q and U to get the SVD of A.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The advantage of proceeding in this manner is if your matrix is badly conditioned enough to necessitate the use of SVD, the algorithm for computing the SVD has to handle only a triangular matrix instead of the original matrix (which may have more rows than columns as is usual in least-squares applications).&lt;/p&gt;&#xA;" OwnerUserId="498" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-08-03T06:14:23.543" LastActivityDate="2010-08-03T06:14:23.543" CommentCount="2" />
  <row Id="1207" PostTypeId="2" ParentId="1091" CreationDate="2010-07-30T04:05:04.443" Score="1" ViewCount="" Body="&lt;p&gt;Instead of giving an answer, let me point out to you this &lt;a href=&quot;http://www.mathworks.com/moler/eigs.pdf&quot; rel=&quot;nofollow&quot;&gt;chapter&lt;/a&gt; in Cleve Moler's book &quot;Numerical Computing with MATLAB&quot;, there is a nice geometric demonstration in MATLAB on how eigenvalues/eigenvectors (as well as singular values/vectors) of an order-2 square matrix are involved in how a circle is transformed into an ellipse after a linear transformation represented by the matrix.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-07-30T04:05:04.443" />
  <row Id="1208" PostTypeId="2" ParentId="1204" CreationDate="2010-07-30T04:05:14.647" Score="4" ViewCount="" Body="&lt;p&gt;Sure, supermodules over a superalgebra form a nice and nontrivial example of a tensor category. In fact, it is a theorem that Deligne that all tensor categories over $\mathbb{C}$ with suitable growth conditions can be obtained as categories of super-representations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another reason is given by Qiaochu, that supercommutativity arises naturally in a lot of places.&lt;br&gt;&#xA;For instance, the exterior algebra is supercommutative; thus the wedge product on differential forms acts the same way.  Interestingly, the exterior algebra of a direct sum corresponds to taking the super tensor product.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another example: With singular cohomology on a topological space, there is a way to define a cup product, which satisfy the supercommutative law $ab = (-1)^{deg b deg a} ba$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Note that given a $\mathbb{Z}$-graded super-commutative algebra, you can get a $\mathbb{Z}/2$-graded superalgebra as you indicate in your question by taking the sum of the odd parts and the sum of the even parts.)&lt;/p&gt;&#xA;" OwnerUserId="536" LastActivityDate="2010-07-30T04:05:14.647" />
  <row Id="1209" PostTypeId="2" ParentId="1204" CreationDate="2010-07-30T04:34:50.267" Score="3" ViewCount="" Body="&lt;p&gt;I can summarize one really basic reason, which is actually the reason I originally got interested in the definition.  Take a finite-dimensional vector space $V$ of dimension $n$, and let $\left( {n \choose k} \right) = {n+k-1 \choose k}$ denote the number of multisets of size $k$ on a set of size $n$.  (Multisets are like subsets except that more than one copy of a given element is possible.)  Then the symmetric powers of $V$ have dimensions&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle \left( {n \choose 1} \right), \left( {n \choose 2} \right), ... $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;whereas the exterior powers of $V$ have dimensions&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle {n \choose 1}, {n \choose 2}, ....$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now here is a funny identity: it is not hard to see that ${n \choose k} = (-1)^k \left( {-n \choose k} \right)$.  One way we might interpret this identity is that the $k^{th}$ exterior power of a vector space of dimension $n$ is like the $k^{th}$ symmetric power of a vector space of dimension $-n$, whatever that means.  So what could that possibly mean?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer (and I'll let you work this out for yourself, because it's fun) is to work in the category of supervector spaces!  A supervector space $V$ is a direct sum $V_0 \oplus V_1$, and while one notion of dimension is to take $\dim V_0 + \dim V_1$, another (which can be motivated by thinking of $\mathbb{Z}/2\mathbb{Z}$-graded vector spaces as the category of representations of $\mathbb{Z}/2\mathbb{Z}$) is to take $\dim V_0 - \dim V_1$.  So while a purely even vector space has positive dimension, a purely odd vector space has negative dimension.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then: graded-commutativity implies that the symmetric power of a purely odd vector space is the exterior power of the corresponding purely even vector space.  More generally, the $k^{th}$ symmetric power of a vector space of dimension $n$ (for all integers $n$) has dimension $\left( {n \choose k} \right)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(And, of course, the symmetric algebra of a supervector space is naturally a graded-commutative superalgebra.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(The physics connection is that symmetric powers = bosons, exterior powers = fermions, and there is a duality between the two.)&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-07-30T04:34:50.267" CommentCount="3" />
  <row Id="1210" PostTypeId="2" ParentId="631" CreationDate="2010-07-30T05:51:11.413" Score="1" ViewCount="" Body="&lt;p&gt;It's worth considering a few ways of showing that the problem is neither in P, nor NPC. I've marked this answer &quot;community wiki&quot;, so please feel free to add suggestions and flesh out ideas here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Based on my experience playing the 24 game, it seems that most combinations of numbers are solvable. If we could formalize this, we could show that the 24 game is not NPC. Formally, consider the &lt;code&gt;2^n&lt;/code&gt; inputs of length &lt;code&gt;n&lt;/code&gt;. If all but polynomially-many of them solvable, then the language is sparse and cannot be NPC (unless P=NP).&lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2010-07-30T05:51:11.413" CommunityOwnedDate="2010-07-30T05:51:11.413" />
  <row Id="1211" PostTypeId="1" AcceptedAnswerId="1269" CreationDate="2010-07-30T07:46:11.700" Score="4" ViewCount="556" Body="&lt;p&gt;Roots behave strangely over &lt;a href=&quot;http://math.stackexchange.com/questions/1183/division-by-imaginary-number/1185#1185&quot;&gt;complex numbers&lt;/a&gt;. Given this, how do non-integer powers behave over negative numbers? More specifically:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Can we define fractional powers such as (-2)^-1.5?&lt;/li&gt;&#xA;&lt;li&gt;Can we define irrational powers (-2)^pi?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="123" LastEditorUserId="123" LastEditorDisplayName="" LastEditDate="2010-08-21T07:13:37.997" LastActivityDate="2010-08-21T07:13:37.997" Title="Non-integer powers of negative numbers" Tags="&lt;complex-numbers&gt;&lt;powers&gt;" AnswerCount="4" CommentCount="1" FavoriteCount="1" />
  <row Id="1212" PostTypeId="2" ParentId="1211" CreationDate="2010-07-30T08:02:04.923" Score="1" ViewCount="" Body="&lt;p&gt;We can define complex powers as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$z^w := e^{(l(z)*w)}$ for $z,w \in \mathbb{C}$ and a complex logarithm function $l$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;of course you need to make sure there actually exists such a logarithm $l:U\to\mathbb{C}$ function ($U$ needs to be a simply-connected subset of $\mathbb{C}\setminus\{0\}$&lt;/p&gt;&#xA;" OwnerUserId="340" LastActivityDate="2010-07-30T08:02:04.923" />
  <row Id="1213" PostTypeId="2" ParentId="1211" CreationDate="2010-07-30T08:03:21.410" Score="3" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Exponentiation#Powers_of_complex_numbers&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt; says &quot;Complex powers of positive reals are defined via e^x as in section Complex powers of positive real numbers above. These are continuous functions.&#xA;Trying to extend these functions to the general case of noninteger powers of complex numbers that are not positive reals leads to difficulties. Either we define discontinuous functions or multivalued functions.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So let me say a few more words. Essentially what we are looking for is $z^a=e^{a log(t)}$. But as you may or may not know, &lt;a href=&quot;http://en.wikipedia.org/wiki/Complex_logarithm#Problems_with_inverting_the_complex_exponential_function&quot; rel=&quot;nofollow&quot;&gt;complex logarithms&lt;/a&gt; are problematic depending on your viewpoint. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In an &lt;a href=&quot;http://math.stackexchange.com/questions/1133/natural-derivation-of-the-complex-exponential-function/1158#1158&quot;&gt;answer&lt;/a&gt; to Harry's question, I describe a more careful way to define the complex exponential function, and suggest that one should just inverse this to get your logarithm. This way is safe and will yield your logarithm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are less fickle, and are willing to take on the branched logarithms, may I suggest $Log(z)=ln\mid z\mid +iArg(z)$ where Arg is the principle argument. This will lead you to branched exponents. &lt;/p&gt;&#xA;" OwnerUserId="16" LastActivityDate="2010-07-30T08:03:21.410" />
  <row Id="1214" PostTypeId="2" ParentId="768" CreationDate="2010-07-30T08:06:23.087" Score="1" ViewCount="" Body="&lt;p&gt;(For an integer &amp;lambda; and a lattice polygon P) denote by &amp;lambda;P polygon P stretched by &amp;lambda; times. Then number N(&amp;lambda;P) of points inside the polygon &amp;lambda; is a quadratic polynomial in &amp;lambda; with leading coefficient S(P). This is the form of Pick's theorem that holds for any lattice (and obvious analogue works in any dimension&amp;nbsp;&amp;mdash; unlike usual Pick's formula that has no analogue in 3d even for the cubic lattice).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the square lattice it yields ordinary Pick's theorem since for a parallelogram P on the square lattice $N(\lambda P)\approx\lambda^2(i+\frac{b}{2}-1)S(P)$ as &amp;lambda;&amp;rarr;&amp;infin; and general theorem follows from the theorem for parallelograms by additivity.&lt;/p&gt;&#xA;" OwnerUserId="152" LastActivityDate="2010-07-30T08:06:23.087" />
  <row Id="1215" PostTypeId="1" CreationDate="2010-07-30T08:23:48.053" Score="3" ViewCount="323" Body="&lt;p&gt;Can we find f(x) given that 1-f(x) = f(-x) for all real x?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I start by rearranging to: f(-x) + f(x) = 1. I can find an example such as f(x) = abs(x) that works for some values of x, but not all. Is there a method here? Is this possible?&lt;/p&gt;&#xA;" LastActivityDate="2010-07-30T17:13:44.180" Title="Solution to 1-f(x) = f(-x)" Tags="&lt;algebra-precalculus&gt;" AnswerCount="6" CommentCount="5" />
  <row Id="1216" PostTypeId="2" ParentId="1215" CreationDate="2010-07-30T08:32:04.523" Score="11" ViewCount="" Body="&lt;p&gt;f(x)=1/2+(any odd function) (e.g. f(x)=1/2+x or, say, f(x)=1/2+99x^3+7x^5).&lt;/p&gt;&#xA;" OwnerUserId="152" LastActivityDate="2010-07-30T08:32:04.523" CommentCount="2" />
  <row Id="1217" PostTypeId="2" ParentId="1215" CreationDate="2010-07-30T08:32:10.933" Score="2" ViewCount="" Body="&lt;p&gt;Let f(x) = 1 if x&gt;0,   1/2 if x = 0,  and 0 if x&amp;lt;0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If x &gt; 0, then -x &amp;lt; 0 so that f(x)+f(-x)=1+0=1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Likewise with x&amp;lt;0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If x=0, then f(x)+f(-x)=(1/2)+(1/2)=1.&lt;/p&gt;&#xA;" OwnerUserId="264" LastActivityDate="2010-07-30T08:32:10.933" />
  <row Id="1218" PostTypeId="2" ParentId="1215" CreationDate="2010-07-30T08:56:49.720" Score="5" ViewCount="" Body="&lt;p&gt;Clearly, we only have relations between f(x) and f(-x). The relation means that 0 has to have value 1/2. We can divide all non-zero real numbers into disjoint pairs of x and -x and define the function f on each pair separately. For each pair, f(x) can be given any value and then f(-x) has a single valid value. As mentioned by &lt;a href=&quot;http://math.stackexchange.com/questions/1215/solution-to-1-fx-f-x/1216#1216&quot;&gt;Grigory&lt;/a&gt;, the valid functions can be characterised as any odd function plus 1/2.&lt;/p&gt;&#xA;" OwnerUserId="123" LastEditorUserId="123" LastEditorDisplayName="" LastEditDate="2010-07-30T09:47:26.877" LastActivityDate="2010-07-30T09:47:26.877" CommentCount="2" />
  <row Id="1219" PostTypeId="2" ParentId="1215" CreationDate="2010-07-30T08:59:03.443" Score="2" ViewCount="" Body="&lt;p&gt;&lt;em&gt;Usually&lt;/em&gt; simple problems like this ask you to find &lt;strong&gt;a&lt;/strong&gt; function that respects the condition, not &lt;strong&gt;all&lt;/strong&gt; of them. And (again) &lt;em&gt;usually&lt;/em&gt; you start by checking if a simple polynomial function of the first degree could be a solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, if&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;f(x) = a * x + b&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Then&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;f(x) + f(-x) = 1 =&gt; a * x + b + a * (-x) + b = 1 =&gt; 2 * b = 1 =&gt; b = 1/2&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So the condition is satisfied by any function of the type:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;f(x) = a * x + 1/2&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="88" LastEditorUserId="88" LastEditorDisplayName="" LastEditDate="2010-07-30T11:24:47.393" LastActivityDate="2010-07-30T11:24:47.393" />
  <row Id="1220" PostTypeId="1" AcceptedAnswerId="1232" CreationDate="2010-07-30T09:05:52.880" Score="0" ViewCount="145" Body="&lt;p&gt;According to &lt;a href=&quot;http://en.wikipedia.org/wiki/Characteristic_function_%28probability_theory%29&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;, a characteristic function completely determines the properties of a probability distribution. This means it must be unique. However, the definition give is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;Char of X (t)=E[e^itX]&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now e^iz repeats for every 2 pi increase in z. So how can it be unique?&lt;/p&gt;&#xA;" OwnerUserId="123" LastEditorUserId="123" LastEditorDisplayName="" LastEditDate="2010-07-30T23:10:52.143" LastActivityDate="2010-07-30T23:10:52.143" Title="Uniqueness of Characterstic Functions in Probability" Tags="&lt;statistics&gt;" AnswerCount="2" CommentCount="0" />
  <row Id="1221" PostTypeId="2" ParentId="1183" CreationDate="2010-07-30T09:35:56.080" Score="3" ViewCount="" Body="&lt;p&gt;The only soundproof way to be sure to find the right result while dividing two complex numbers&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(a+bi)/(c+di)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is reducing it to a multiplication. The answer is of the form x+yi; therefore &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(c+di)(x+yi) = a+bi&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and you will end up with two linear equations, one for the real coefficient and another for the imaginary one. As Simon and Casebash already wrote, taking a square root leads to problems, since you cannot be sure which value must be chosen.&lt;/p&gt;&#xA;" OwnerUserId="89" LastActivityDate="2010-07-30T09:35:56.080" />
  <row Id="1222" PostTypeId="1" AcceptedAnswerId="1243" CreationDate="2010-07-30T10:10:29.233" Score="4" ViewCount="154" Body="&lt;p&gt;I'm trying to read a document that applies Riemann-Roch left, right and center. I don't know this theorem or the theory it comes from so I need to build up a bit more background before I can tackle this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you please recommend good books or (online) lecture notes which cover &quot;multi-valued functions&quot;, Riemann Surfaces and similar up to (at least) Riemann-Roch? (I also want to pick up a bit about modular forms and the relation between lattices and elliptic curves).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I've done a bit of complex calculus but it was all with analogy to real analysis so I am not sure that it will really give me any head start here. Also apologies for being so vauge with this but I don't know enough about this subject to be any more precise)&lt;/p&gt;&#xA;" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-07-30T10:16:31.023" LastActivityDate="2010-08-14T09:53:42.810" Title="Books/Notes recommendation request: Multivalued functions/Riemann surfaces" Tags="&lt;books&gt;&lt;reference-request&gt;&lt;learning&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="2" />
  <row Id="1223" PostTypeId="2" ParentId="1215" CreationDate="2010-07-30T10:53:45.523" Score="3" ViewCount="" Body="&lt;p&gt;Consider instead the functions g that satisfy the identity -g(x)=g(-x) for all x.  If (x,g(x)) is a point of the function g, then (-x,-g(x)) is also a point (since g(-x)=-g(x)).  Therefore, every function g is symmetric when rotated by 180 degrees about the point (0,0).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do things change for the identity 1-f(x)=f(-x)?  We merely shift the point of symmetry to (0,1/2).  Here the point (x,f(x)) implies the point (-x,1-f(x))&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The function f satisfies the identity 1-f(x)=f(-x) for all real numbers x if and only if it is symmetric when rotated about the point (0,1/2) by 180 degrees.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There's going to be many of these functions; some of which will be polynomials, some of which will not.&lt;/p&gt;&#xA;" OwnerUserId="139" LastActivityDate="2010-07-30T10:53:45.523" />
  <row Id="1224" PostTypeId="2" ParentId="97" CreationDate="2010-07-30T11:08:58.043" Score="6" ViewCount="" Body="&lt;p&gt;I am assuming that you need the error function only for real values. For complex arguments there are other approaches, more complicated than what I will be suggesting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're going the Taylor series route, the best series to use is formula &lt;a href=&quot;http://people.math.sfu.ca/~cbm/aands/page_297.htm&quot; rel=&quot;nofollow&quot;&gt;7.1.6&lt;/a&gt; in Abramowitz and Stegun. It is not as prone to subtractive cancellation as the series derived from integrating the power series for e&lt;sup&gt;-x²&lt;/sup&gt;. This is good only for &quot;small&quot; arguments. For large arguments, you can use either the asymptotic series or the continued fraction representations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Otherwise, may I direct you to &lt;a href=&quot;http://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.90.6481&quot; rel=&quot;nofollow&quot;&gt;these&lt;/a&gt; &lt;a href=&quot;http://homepages.physik.uni-muenchen.de/~Winitzki/erf-approx.pdf&quot; rel=&quot;nofollow&quot;&gt;papers&lt;/a&gt; by S. Winitzki that give nice approximations to the error function.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;(added on 5/4/2011)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wrote about the computation of the (complementary) error function (couched in different notation) in &lt;a href=&quot;http://stats.stackexchange.com/questions/7200/10300#10300&quot;&gt;this answer to a CV question&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="498" LastEditorUserId="45" LastEditorDisplayName="" LastEditDate="2011-05-04T04:59:57.150" LastActivityDate="2011-05-04T04:59:57.150" CommentCount="4" />
  <row Id="1225" PostTypeId="2" ParentId="697" CreationDate="2010-07-30T11:18:20.117" Score="3" ViewCount="" Body="&lt;p&gt;The usual cases treated apart from least squares are the one-norm and infinity-norm (Chebyshev) cases; they crop up in function approximation for instance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Usually both of these are &lt;a href=&quot;http://dx.doi.org/10.1007/BF02162565&quot; rel=&quot;nofollow&quot;&gt;solved via linear programming techniques&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-07-30T11:18:20.117" />
  <row Id="1227" PostTypeId="2" ParentId="220" CreationDate="2010-07-30T12:47:44.910" Score="2" ViewCount="" Body="&lt;p&gt;As I understand it, you are given a sequence $(x_n)_{n\in\mathbb{N}}$ of input values from which you calculate a sequence $f_n$ that is given by the following recurrence relations:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f_0 = 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f_{n+1} = (1-c)f_n + c\cdot x_{n+1}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your &lt;strong&gt;question&lt;/strong&gt; is: given a sine wave $x_n=\sin(\omega n)$, you assert that $f_n$ is also a sine wave and you want to know its amplitude in dependence on the frequency $\omega$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Answer&lt;/strong&gt;: It's easier to calculate the frequency response with exponential functions instead of sine waves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f_n = Ae^{i\omega n}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x_n = e^{i\omega n}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $f_{n+1} = e^{i\omega (n+1)} = e^{i\omega} e^{i\omega n} = e^{i\omega} f_n$, the recurrence relation gives&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$e^{i\omega} A e^{i\omega n} = (1-c)Ae^{i\omega n} + c e^{i\omega} e^{i\omega n}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which implies&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$A(\omega) = \frac{c}{1 - (1-c)e^{-i\omega}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To calculate the response for sine waves, you can represent the sine function as a linear combination of two exponential functions&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x_n = \sin(\omega n) = \frac1{2i}(e^{i\omega n}-e^{-i\omega n})$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and obtain&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f_n = \frac1{2i}(A(\omega)e^{i\omega n}-\bar A(\omega)e^{-i\omega n}) = |A(\omega)| \sin(\omega n + \phi)$ where $A(\omega) =: |A(\omega)| e^{i\phi}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, the complex frequency response $A(\omega)$ encodes both the change in amplitude and the phase shift. I'll leave the exact calculation of $|A(\omega)|$ to you, so that you can get familiar with this method of calculation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Without exponential functions, you'd have to use the addition and subtraction theorems for sine and cosine. Put differently, exponential functions provide a very convenient formulation of the trigonometric identities.&lt;/p&gt;&#xA;" OwnerUserId="434" LastEditorUserId="434" LastEditorDisplayName="" LastEditDate="2010-07-30T15:43:29.187" LastActivityDate="2010-07-30T15:43:29.187" CommentCount="2" />
  <row Id="1228" PostTypeId="2" ParentId="83" CreationDate="2010-07-30T12:48:39.713" Score="2" ViewCount="" Body="&lt;p&gt;Show them large buildings, tallest buildings, pyramids, visualize them about different structure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Explain importance of maths&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Brain remember things easily when comparison is done.&lt;/p&gt;&#xA;" OwnerUserId="327" LastActivityDate="2010-07-30T12:48:39.713" CommunityOwnedDate="2010-08-07T22:32:44.343" />
  <row Id="1229" PostTypeId="2" ParentId="1220" CreationDate="2010-07-30T13:15:03.083" Score="1" ViewCount="" Body="&lt;p&gt;It's just a Fourier transform. E(x) is an integral over the probability distribution. The function is unique; if you focus on the value inside the expectation integral, that's not, but so what?&lt;/p&gt;&#xA;" OwnerUserId="120" LastActivityDate="2010-07-30T13:15:03.083" />
  <row Id="1230" PostTypeId="2" ParentId="1186" CreationDate="2010-07-30T13:38:00.860" Score="4" ViewCount="" Body="&lt;p&gt;When squaring a number, break the calculation into three smaller calculations and add them using the &lt;a href=&quot;http://en.wikipedia.org/wiki/FOIL_method&quot; rel=&quot;nofollow&quot;&gt;FOIL method&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;302^2 = 300^2 + 2*(300*2) + 2^2.&lt;/p&gt;&#xA;" OwnerUserId="19" LastActivityDate="2010-07-30T13:38:00.860" CommunityOwnedDate="2010-07-30T13:38:00.860" CommentCount="1" />
  <row Id="1231" PostTypeId="2" ParentId="220" CreationDate="2010-07-30T13:55:24.143" Score="2" ViewCount="" Body="&lt;p&gt;I don't have enough mojo to comment on Greg's answer.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Greg made a silly calculational mistake: The transfer function $A(\omega)$ should be $c/(1-(1-c)e^{-i\omega})$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;What you want is the modulus of $A(\omega)$. Note that $\sin \omega n$ is precisely the imaginary part of $e^{i\omega n}$. Because the relation between input and output is linear, the response to $\sin\omega n$ will be the imaginary part of $A(\omega)e^{i\omega n}$. That's going to be a sinusoid with some shifting and the amplitude $|A(\omega)|$. &lt;a href=&quot;http://www.wolframalpha.com/input/?i=plot+|.5/%281-.5+exp%28-ix%29%29|&quot; rel=&quot;nofollow&quot;&gt;Here&lt;/a&gt; is a plot for $c=1/2$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;To read more about this sort of things, google &quot;IIR filter&quot; or &quot;infinite impulse response&quot;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="512" LastActivityDate="2010-07-30T13:55:24.143" CommentCount="3" />
  <row Id="1232" PostTypeId="2" ParentId="1220" CreationDate="2010-07-30T15:05:51.883" Score="3" ViewCount="" Body="&lt;p&gt;I thought at first you were asking how the characteristic function can be unique. There's no issue here, because e^ix is a well-defined function: for a given value of x, there is a unique value of e^ix. And so the expectation (an integral) has a unique value as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On second thought, it appears you're asking how it's possible, considering that e^ix is not &lt;em&gt;injective&lt;/em&gt; (i.e., multiple x can have the same value of e^ix), that the original distribution can be completely recovered from the characteristic function. The answer to &lt;em&gt;that&lt;/em&gt; is that you're probably missing the &quot;t&quot; in the expression: the characteristic function is a function of t, and although for a &lt;em&gt;given&lt;/em&gt; value of t (e.g. t=1), the random variables $X$ and $X+2k\pi/t$ &lt;em&gt;would&lt;/em&gt; have the same value of the characteristic function at that point, they would have different values at other t. So the characteristic &lt;em&gt;functions&lt;/em&gt; are different, and yes, the distribution can be completely recovered from the characteristic function.&lt;/p&gt;&#xA;" OwnerUserId="205" LastActivityDate="2010-07-30T15:05:51.883" CommentCount="3" />
  <row Id="1233" PostTypeId="2" ParentId="967" CreationDate="2010-07-30T15:07:00.430" Score="2" ViewCount="" Body="&lt;p&gt;I like &lt;a href=&quot;http://www.cs.kent.ac.uk/people/staff/sjt/TTFP/&quot; rel=&quot;nofollow&quot;&gt;Type Theory and Functional Programming&lt;/a&gt; by Simon Thompson.&lt;/p&gt;&#xA;" OwnerUserId="512" LastActivityDate="2010-07-30T15:07:00.430" />
  <row Id="1234" PostTypeId="1" CreationDate="2010-07-30T16:06:46.553" Score="1" ViewCount="314" Body="&lt;p&gt;This was really inspired by Solitaire, but a few people reacted with ``oh, it's like the towers of Hanoi, isn't it?'' so I'll try to pose the problem in terms of discs here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's start. There are n disks on the real line, one of size 1 at position $x_1$, one of size 2 at position $x_2$, ..., and one of size n at position $x_n$. Your goal is to make a tower with all n discs, consuming as little energy as possible in the process. You are allowed to move a tower whose base is a disk of size k only on top of the disk with size k+1 (which may be the top of another mini-tower). The energy you consume to perform such a move is the distance traveled by the moved mini-tower. For example, the energy consumed by the first move is $|x_k-x_{k+1}|$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, you'd like to write a program that tells you whether the energy you have is enough to perform the task. It just needs to say Y or N. (If the answer is Y, then clearly the list of moves is proof enough that the answer is correct, so the problem is NP. If the answer is N, there's no point in even attempting the task---you are too tired.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What's the fastest such program you can find? Is the problem NP-complete? If it helps, consider a simplifications that restrict $x_k$'s to be rational, integers, integers in a certain range, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's an upper bound: $O(n2^n)$. Represent the initial state by the list $x_1$, $x_2$ ,... ,$x_n$. A move affects the state by deleting a number in this list and the energy consumed by the move is the absolute difference between the deleted number and the number coming after. Clearly there are $2^n$ states with at most $n$ moves each. (See &lt;a href=&quot;http://rgrig.blogspot.com/2010/07/solitaire.html&quot; rel=&quot;nofollow&quot;&gt;my blog post&lt;/a&gt; for an example run of this algorithm if it's not clear. The description there is in terms of cards.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: Some of the comments below refer to older versions of the questions. See the history of edits if they seem confusing.&lt;/p&gt;&#xA;" OwnerUserId="512" LastEditorUserId="512" LastEditorDisplayName="" LastEditDate="2010-07-31T11:47:02.263" LastActivityDate="2010-07-31T12:15:45.357" Title="Is this version of the Hanoi towers problem NP-complete?" Tags="&lt;algorithms&gt;&lt;np-complete&gt;" AnswerCount="1" CommentCount="8" FavoriteCount="2" />
  <row Id="1235" PostTypeId="2" ParentId="83" CreationDate="2010-07-30T16:09:24.400" Score="1" ViewCount="" Body="&lt;p&gt;Check out the book «&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0821804308&quot; rel=&quot;nofollow&quot;&gt;Mathematical circles&lt;/a&gt;».  It's written by some Russian mathematicians who used to run &quot;math circles&quot; for middle and high school students and has a lot of interesting material.&lt;/p&gt;&#xA;" OwnerUserId="314" LastActivityDate="2010-07-30T16:09:24.400" CommunityOwnedDate="2010-08-07T22:32:44.343" />
  <row Id="1236" PostTypeId="1" CreationDate="2010-07-30T16:15:38.950" Score="8" ViewCount="235" Body="&lt;p&gt;How have experts estimated the amount of oil that was shooting out of that pipe in the Gulf? I bet there's some neat math or physics involved here, and some interesting assumptions considering how little concrete data are available.&lt;/p&gt;&#xA;" OwnerUserId="41" LastEditorUserId="9" LastEditorDisplayName="" LastEditDate="2010-07-30T17:17:46.717" LastActivityDate="2010-08-25T18:11:37.750" Title="How do you estimate the flow rate of one fluid into another like the Deep Horizon Oil leak?" Tags="&lt;applications&gt;" AnswerCount="3" CommentCount="5" FavoriteCount="1" />
  <row Id="1237" PostTypeId="1" CreationDate="2010-07-30T16:21:12.173" Score="0" ViewCount="262" Body="&lt;p&gt;A fun question I ask students or interviewees (in engineering) is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;This is not my question, this is an example:&lt;/strong&gt;&lt;br&gt;&#xA;  Using only what you know now, how many&#xA;  cans of soda would you estimate are&#xA;  produced per day (on average) in the&#xA;  United States?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For this question, the result doesn't matter so much as the process you use.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this theme of estimation, what's your favorite question?&lt;/p&gt;&#xA;" OwnerUserId="41" LastEditorUserId="171" LastEditorDisplayName="" LastEditDate="2010-08-16T19:05:37.770" LastActivityDate="2010-08-16T19:05:37.770" CommunityOwnedDate="2010-07-30T16:21:12.173" ClosedDate="2010-07-31T17:35:23.297" Title="What is your favorite estimation exercise?" Tags="&lt;soft-question&gt;" AnswerCount="4" CommentCount="4" />
  <row Id="1238" PostTypeId="2" ParentId="519" CreationDate="2010-07-30T16:24:10.423" Score="1" ViewCount="" Body="&lt;p&gt;Another approach, which can't work in general (see Noah's answer), but will surely work in this case, is to find a normal form for each element of the group, and then see whether there are finitely or infinitely many.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In practice, that means imagining a word in x and y, and then applying the relations as much as possible to simplify it, and then trying to figure out (and prove!) what the possible different &quot;irreducible words&quot; (i.e. words that can no longer be simplified) are.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your case, the first thing one would note is that x can only appear to the 1st power&#xA;(since any higher power can be simplified using x^2 = 1), while y can only appear to the powers +1 or -1 (for the same reason).  Also, we can't have too many expressions of the&#xA;form  xy  or  yx  in a row, because of the third relation.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One can keep going like this.  I didn't, but what I imagine is that one can have &#xA;expressions of the form  x y x y^{-1} x y x y^{-1}   ...   that are arbitrarily long,&#xA;and inequivalent, explaining the infinite order of the group.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It shouldn't be so hard&#xA;to settle the question from this point of view by sitting down with pencil and paper and just playing around with different words to get a feel for what kinds of reductions can&#xA;take place.  (In geometric arguments like Grigory's, one uses a geometric context, such as&#xA;an action on a hyperbolic tiling, as a more conceptual way of understanding the relations and distinguishing inequivalent words.  But in this case I'm sure it won't be hard to see&#xA;everything directly from the presentation.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Added after rereading the question: what I am suggesting is precisely that even when the answer might be infinite, you can still hope to find a coset enumeration, at least in a case&#xA;like this with relatively simple relations.&lt;/p&gt;&#xA;" OwnerUserId="221" LastEditorUserId="221" LastEditorDisplayName="" LastEditDate="2010-07-30T16:33:34.130" LastActivityDate="2010-07-30T16:33:34.130" CommentCount="2" />
  <row Id="1240" PostTypeId="2" ParentId="1237" CreationDate="2010-07-30T16:39:31.703" Score="15" ViewCount="" Body="&lt;p&gt;&quot;How many estimation questions are asked in interviews across the world during a typical 24h period?&quot;&lt;/p&gt;&#xA;" OwnerUserId="512" LastActivityDate="2010-07-30T16:39:31.703" CommunityOwnedDate="2010-07-30T16:39:31.703" CommentCount="2" />
  <row Id="1241" PostTypeId="1" AcceptedAnswerId="1248" CreationDate="2010-07-30T17:09:32.753" Score="4" ViewCount="352" Body="&lt;p&gt;The generalized mean (power mean) with exponent $p$ of $n$ numbers $x_1, x_2, \ldots, x_n$ is defined as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \bar x = (\frac{1}{n} \sum x_i^p)^{1/p}. $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is equivalent to the harmonic mean, arithmetic mean, and root mean square for $p = -1$, $p = 1$, and $p = 2$, respectively. Also its limit at $p = 0$ is equal to the geometric mean.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When should the different means be used? I know harmonic mean is useful when averaging speeds and the plain arithmetic mean is certainly used most often, but I've never seen any uses explained for the geometric mean or root mean square. (Although standard deviation is the root mean square of the deviations from the arithmetic mean for a list of numbers.)&lt;/p&gt;&#xA;" OwnerUserId="45" LastEditorUserId="45" LastEditorDisplayName="" LastEditDate="2010-07-30T21:34:40.073" LastActivityDate="2010-07-30T23:46:52.207" Title="Which average to use? (RMS vs. AM vs. GM vs. HM)" Tags="&lt;average&gt;" AnswerCount="3" CommentCount="6" FavoriteCount="1" />
  <row Id="1242" PostTypeId="2" ParentId="1215" CreationDate="2010-07-30T17:13:44.180" Score="0" ViewCount="" Body="&lt;p&gt;WolframAlpha provides a solution to this (and many other) recurrence equations:&lt;br&gt;&#xA;&lt;a href=&quot;http://www.wolframalpha.com/input/?i=1-f(x)+%3D+f(-x&quot; rel=&quot;nofollow&quot;&gt;http://www.wolframalpha.com/input/?i=1-f(x)+%3D+f(-x&lt;/a&gt;)&lt;/p&gt;&#xA;" OwnerUserId="346" LastActivityDate="2010-07-30T17:13:44.180" CommentCount="1" />
  <row Id="1243" PostTypeId="2" ParentId="1222" CreationDate="2010-07-30T17:14:23.383" Score="1" ViewCount="" Body="&lt;p&gt;The book by Otto Forster on Riemann Surfaces is pretty good. I never finished reading it myself, but it covers things like Riemann-Roch and Abel's theorem from a sheafish viewpoint. In particular, the proof of Riemann-Roch is analogous to the one in Hartshorne; it follows from the Serre duality theorem and an inductive argument. Learning about sheaves is definitely a plus.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, there's a book by Springer, though the level is a bit more elementary.&lt;/p&gt;&#xA;" OwnerUserId="536" LastActivityDate="2010-07-30T17:14:23.383" />
  <row Id="1244" PostTypeId="2" ParentId="1241" CreationDate="2010-07-30T17:24:12.590" Score="2" ViewCount="" Body="&lt;p&gt;I admit I don't really know what type of answer your looking for. So, I'll say something that might very well be entirely irrelevant for your purposes but which I enjoy. At least, it'll provide some context for the power means you asked about.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These generalized power means are basically the discrete (finitary) analogs of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Lp_norm#lp_spaces&quot; rel=&quot;nofollow&quot;&gt;L^p norms&lt;/a&gt;. So, for instance, it's with these norms that you prove (using, say, elementary calculus) the finitary version of &lt;a href=&quot;http://en.wikipedia.org/wiki/Holder%27s_inequality&quot; rel=&quot;nofollow&quot;&gt;Holder's inequality&lt;/a&gt;, which is really important in analysis, because it leads (via a limiting argument) to the more important fact that $L^p$ and $L^q$ spaces (which are continuous analogs of these finitary $l^p$ spaces) are dual for $p,q$ conjugate exponents. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This duality is really important: one example is that if you are trying to prove something about the $L^p$ spaces that is preserved under duality, you just have to restrict yourself to the case $1 \leq p \leq 2$. The theory of &lt;a href=&quot;http://en.wikipedia.org/wiki/Singular_integral&quot; rel=&quot;nofollow&quot;&gt;singular integral operators&lt;/a&gt; provides examples of this: basically, it's easy to prove they are bounded (i.e., reasonably well-behaved) for $p=2$ by Fourier analysis; you prove that they're &quot;weak-bounded&quot; on $L^1$ (in some sense which I won't make precise); then you apply to general results on interpolation to get boundedness in the range $1-2$; finally, this duality operation gives it for $p&gt;2$ as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, &lt;a href=&quot;http://en.wikipedia.org/wiki/Root_mean_square_speed&quot; rel=&quot;nofollow&quot;&gt;root-mean-square speed&lt;/a&gt; is used to define temperature in physics.&lt;/p&gt;&#xA;" OwnerUserId="536" LastActivityDate="2010-07-30T17:24:12.590" />
  <row Id="1245" PostTypeId="2" ParentId="632" CreationDate="2010-07-30T17:26:01.447" Score="4" ViewCount="" Body="&lt;p&gt;Concerning the physical meaning, I take it that $f_1$ and $f_2$ represent the fractions of the two phases in the alloy (this implies $f_1 + f_2 = 1$). I imagine $x_1$ and $x_2$ to correspond to an intensional variable like pressure, whose average $f_1x_1 + f_2x_2 =: \bar x$ is held constant in the experiment. Now, the alloy minimizes the free energy under these constraints.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To &lt;strong&gt;answer&lt;/strong&gt; your first question: there is a geometric reason why the solution is the common tangent to $G_1$ and $G_2$ in the case of two dimensions. Namely, the fractions $f_1$ and $f_2$ are exactly the &lt;a href=&quot;http://en.wikipedia.org/wiki/Barycentric_coordinates_%28mathematics%29&quot; rel=&quot;nofollow&quot;&gt;Barycentric coordinates&lt;/a&gt; of the average $\bar x$ sitting between $x_1$ and $x_2$. In particular, the value of the total energy $\tilde G$ is the height of the line drawn between $(x_1,G_1(x_1))$ and $(x_2,G_2(x_2))$ evaluated at $\bar x$. Here's a sketch:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i30.tinypic.com/2l2fk7.jpg&quot; alt=&quot;Free energy of an alloy of two compounds&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From this picture, it is clear that if this line is &lt;em&gt;not&lt;/em&gt; tangent to both $G_1$ and $G_2$, then you can move it a bit so that the value at $\bar x$ will decrease.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To answer your &lt;strong&gt;second question&lt;/strong&gt;, the geometry readily extends to higher dimensions. For instance, for an alloy of three compounds, one has to consider the triangle enclosed by the three points $(x_1,G_1(x_1))$, $(x_2,G_2(x_2))$ and $(x_3,G_3(x_3))$. The situation is a bit degenerate here, any point inside this triangle whose first coordinate is $\bar x$ represents a valid value of $\tilde G$. Of these, nature will choose the smallest one. Consequently, the lower side of the triangle has to be tangent to two of the individual free energies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i28.tinypic.com/335dwrl.jpg&quot; alt=&quot;alloy of three compounds&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similar reasoning applies when the variable $x$ is not just a number, but, say, a pair of numbers, then we're dealing with a plane tangent to three individual Gibbs functions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While not terribly useful in this case, there is also a very general geometric interpretation of the method of Lagrange multipliers. Namely, consider a goal function $f$ and a holonomic constraint $g$. Then, the Euler-Lagrange-equations give $\nabla f = \lambda \nabla g$ which means that $f$ changes only in directions orthogonal to the surface $g$. But since we're confined to the surface $g$, this must be an extremum. Wikipedia has a &lt;a href=&quot;http://en.wikipedia.org/wiki/File%3aLagrangeMultipliers2D.svg&quot; rel=&quot;nofollow&quot;&gt;picture&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="434" LastEditorUserId="434" LastEditorDisplayName="" LastEditDate="2010-08-01T12:53:29.397" LastActivityDate="2010-08-01T12:53:29.397" CommentCount="4" />
  <row Id="1246" PostTypeId="2" ParentId="595" CreationDate="2010-07-30T17:31:07.103" Score="3" ViewCount="" Body="&lt;p&gt;Computing the determinant and Gaussian elimination are both fine if you are using exact computations, for instance if the entries of your matrix are rational numbers and you are using only rational numbers during the computations. The disadvantage is that the numerator and denominator can get very large indeed. So the number of operations may indeed be O(n&lt;sup&gt;2.376&lt;/sup&gt;) or O(n&lt;sup&gt;3&lt;/sup&gt;), but the cost of every addition and multiplication gets bigger as n grows because the numbers get bigger.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is not an issue if you are using floating point numbers, but then you have the problem that floating point computations are not exact. Some methods are more sensitive to this than others. In particular, checking invertibility by computing the determinant is a bad idea in this setting. Gaussian elimination is better. Even better is to use the singular value decomposition, which will be treated towards the end of the MIT course.&lt;/p&gt;&#xA;" OwnerUserId="520" LastActivityDate="2010-07-30T17:31:07.103" />
  <row Id="1247" PostTypeId="2" ParentId="1237" CreationDate="2010-07-30T17:33:54.413" Score="0" ViewCount="" Body="&lt;p&gt;I like &quot;how many licks does it take to get to the center of a tootsie roll pop&quot;. &lt;/p&gt;&#xA;" OwnerUserId="16" LastActivityDate="2010-07-30T17:33:54.413" CommunityOwnedDate="2010-07-30T17:33:54.413" />
  <row Id="1248" PostTypeId="2" ParentId="1241" CreationDate="2010-07-30T17:52:59.590" Score="2" ViewCount="" Body="&lt;p&gt;One possible answer is for defining unbiased estimators of probability distributions. Often times you want some transformation of the data that gets you closer to, or exactly to, a normal distribution. For example, products of lognormal variables are again lognormal, so the geometric mean is appropriate here (or equivalently, the additive mean on the natural log of the data). Similarly, there are cases where the data are naturally reciprocals or ratios of random variables, and then the harmonic mean can be used to get unbiased estimators. These show up in actuarial applications, for example.&lt;/p&gt;&#xA;" OwnerUserId="452" LastActivityDate="2010-07-30T17:52:59.590" />
  <row Id="1249" PostTypeId="1" CreationDate="2010-07-30T17:55:21.223" Score="9" ViewCount="443" Body="&lt;p&gt;I am in the process of trying to learn algebraic geometry via schemes and am wondering if there are simple motivating examples of why you would want to consider these structures.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think my biggest issue is the following: I understand (and really like) the idea of passing from a space to functions on a space.  In passing from $k^n$ to $R:=k[x_1,\ldots,x_n]$, we may recover the points by looking at the maximal ideas of $R$.  But why consider $\operatorname{Spec} R$ instead of $\operatorname{MaxSpec} R$?  Why is it helpful to have non-closed points that don't have an analog to points in $k^n$?  On a wikipedia article, it mentioned that the Italian school used a (vague) notion of a generic point to prove things.  Is there a (relatively) simple example where we can see the utility of non-closed points?&lt;/p&gt;&#xA;" OwnerUserId="9" LastActivityDate="2011-05-29T16:31:54.350" Title="Motivating Example for Algebraic Geometry/Scheme Theory" Tags="&lt;algebraic-geometry&gt;" AnswerCount="4" FavoriteCount="3" />
  <row Id="1250" PostTypeId="2" ParentId="1249" CreationDate="2010-07-30T18:07:36.283" Score="6" ViewCount="" Body="&lt;p&gt;To start off,  the &lt;a href=&quot;http://sbseminar.wordpress.com/2009/08/06/algebraic-geometry-without-prime-ideals/&quot; rel=&quot;nofollow&quot;&gt;discussion&lt;/a&gt; at Sbseminar has comments from lots of people who actually know algebraic geometry, and if anything I say contradicts something they say, please trust them and not me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One reason is that you lose the functoriality of $Spec$ if you stick to $MaxSpec$: the inverse image of a maximal ideal is not necessarily maximal. Nevertheless, if you stick to schemes of finite type over a field, this is true (it's basically a version of the Nullstellensatz).  In particular, in Serre's FAC paper he defines a &quot;variety&quot; by gluing together regular affine algebraic sets in the sense of classical algebraic geometry.&#xA;But this is less general.  One natural example of a scheme which is not of finite type over a field is simply $Spec \mathbb{Z}$. Then given a scheme $X$ over this (well, admittedly every scheme $X$ is a scheme over $Spec \mathbb{Z}$ in a canonical way), the fibers at the non-closed point of $Spec \mathbb{Z}$ is still interesting and basically amounts to studying polynomial equations over $\mathbb{Q}$ (when $X \to \mathbb{Z}$ is of finite type).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a (simple) example of how generic points can be used, one can prove that a coherent sheaf on a noetherian integral scheme is free on a dense open subset. Why? Because it must be free at the generic point (since the local ring there is a field), and it is a general fact that two coherent sheaves whose stalks are isomorphic are isomorphic in a neighborhood. (This is true actually for sheaves of finite presentation over a ringed space.)&lt;/p&gt;&#xA;" OwnerUserId="536" LastEditorUserId="536" LastEditorDisplayName="" LastEditDate="2011-05-29T16:31:54.350" LastActivityDate="2011-05-29T16:31:54.350" CommentCount="4" />
  <row Id="1252" PostTypeId="2" ParentId="1249" CreationDate="2010-07-30T19:09:42.600" Score="7" ViewCount="" Body="&lt;p&gt;I think I can at least convince you that it's a good idea to work with non-reduced rings, which aren't really captured by their maximal ideals.  The idea is that you get more functors.  Just as $\text{Hom}(-, \mathbb{C})$ is the functor which sends a finitely-generated domain over $\mathbb{C}$ to its set of $\mathbb{C}$-points, it turns out that $\text{Hom}(-, \mathbb{C}[x]/x^2)$ sends a finitely-generated domain over $\mathbb{C}$ to its set of $\mathbb{C}$-points &lt;em&gt;together with a choice of tangent vector.&lt;/em&gt;   In fact this is one way to define the Zariski tangent space.  So, on the scheme side, it's a good idea to study morphisms from $\text{Spec } \mathbb{C}[x]/x^2$ to your scheme, and you can't do this if you identify $\mathbb{C}[x]/x^2$ with its set of $\mathbb{C}$-points.  (You can't even do this if you identify $\mathbb{C}[x]/x^2$ with its prime ideals; you really need the entire structure sheaf.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(A really nice application: an algebraic definition of the Lie algebra of an algebraic group.)&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-07-30T19:09:42.600" />
  <row Id="1253" PostTypeId="1" AcceptedAnswerId="1255" CreationDate="2010-07-30T19:20:49.810" Score="0" ViewCount="101" Body="&lt;p&gt;In my geometry class last year I remember putting down the statement in a column proof &quot;That all isosceles are always and only similar to other isosceles&quot;. I do not remember what I was trying to prove. But, I do remember that I was stressed and that was the only thing I could think of and made a guess thinking I would probably get the proof wrong on my test.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Funny enough though, I didn't get the proof wrong and I was wondering if anyone could show a proof as to why this would be true. I mean I makes sense but, I do not see any way to prove it. Could you please explain how this is true?&lt;/p&gt;&#xA;" LastActivityDate="2010-07-30T19:41:28.683" Title="Are isosceles always and only similar to other isosceles?" Tags="&lt;geometry&gt;" AnswerCount="1" />
  <row Id="1254" PostTypeId="2" ParentId="1249" CreationDate="2010-07-30T19:24:44.747" Score="5" ViewCount="" Body="&lt;p&gt;See my answer &lt;a href=&quot;http://math.stackexchange.com/questions/942/meaning-of-closed-points-of-a-scheme/984#984&quot;&gt;here&lt;/a&gt; for a brief discussion of how points that are closed in one optic (rational solutions to a Diophantine equation, which are closed points on the variety over $\mathbb Q$ attached to the Diophantine equation) become non-closed in another optic (when we clear denominators and think of the Diophantine equation as defining a scheme over $\mathbb Z$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In terms of rings (and connecting to Qiaochu's answer), under the natural map&#xA;$\mathbb Z[x_1,...,x_n] \to \mathbb Q[x_1,...,x_n]$, the preimage of maximal ideals are prime, but not maximal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These examples may give impression that non-closed points are most important in arithmetic situations, but actually that is not the case.  The ring $\mathbb C[t]$ behaves much like $\mathbb Z$, and so one can have the same discussion with $\mathbb Z$ and $\mathbb Q$ replaced by $\mathbb C[t]$ and $\mathbb C(t)$.  Why would one do this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, suppose you have an equation (like $y^2 = x^3 + t$) which you want to study, where you think of $t$ as a parameter.  To study the generic behaviour of this equation, you can think of it as a variety over $\mathbb C(t)$.  But suppose you want to study the geometry for one particular value of $t_0$ of $t$.  Then you need to pass from $\mathbb C(t)$ to $\mathbb C[t]$, so that you can apply the &#xA;homomorphism $\mathbb C[t] \to \mathbb C$ given by $t \mapsto t_0$  (specialization at $t_0$).  This is completely&#xA;analogous to the situation considered in my linked answer, of taking integral solutions to a a Diophantine equation and then reducing them mod $p$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the upshot?  Basically, any serious study of varieties in families (whether arithmetic families, i.e. schemes over $\mathbb Z$, or geometric families, i.e. parameterized families&#xA;of varieties) requires scheme-theoretic techniques and the consideration of non-closed points.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Of course, serious such studies were made by the Italian geometers, by Lefschetz, by Igusa, by Shimura, and by many others before Grothendieck's invention of schemes, but the whole point of schemes is to clarify what came before and to give a precise and workable theory that encompasses all of the contexts considered in the &quot;old days&quot;, and is also more systematic and more powerful than the older techniques.) &lt;/p&gt;&#xA;" OwnerUserId="221" LastEditorUserId="221" LastEditorDisplayName="" LastEditDate="2011-05-29T14:11:45.057" LastActivityDate="2011-05-29T14:11:45.057" />
  <row Id="1255" PostTypeId="2" ParentId="1253" CreationDate="2010-07-30T19:25:36.833" Score="4" ViewCount="" Body="&lt;p&gt;Re-reading your question, I see two possible interpretations of your statement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First (and my original answer), &quot;If &amp;#x25B3;ABC is isosceles and &amp;#x25B3;ABC~&amp;#x25B3;DEF, then &amp;#x25B3;DEF is isosceles.&quot;  Two triangles are similar if and only if the three angles of one are congruent to the three angles of the other.  Since a triangle is isosceles if and only if two of its angles are congruent, if a triangle is similar to an isosceles triangle, then it will also have two congruent angles and must be isosceles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second, &quot;If &amp;#x25B3;ABC and &amp;#x25B3;DEF are isosceles, then they are similar.&quot;  This is not true.  Suppose one triangle has angles with measures 20&amp;deg;, 20&amp;deg;, and 140&amp;deg; and another other triangle has angles with measures 85&amp;deg;, 85&amp;deg;, and 10&amp;deg;.  Both triangles are isosceles (since within each triangle, there is a pair of congruent angles), but the triangles are not similar (because the angles of one are not congruent to the angles of the other).&lt;/p&gt;&#xA;" OwnerUserId="72" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-07-30T19:41:28.683" LastActivityDate="2010-07-30T19:41:28.683" CommentCount="4" />
  <row Id="1256" PostTypeId="2" ParentId="20" CreationDate="2010-07-30T19:42:21.053" Score="0" ViewCount="" Body="&lt;p&gt;There are many types of numbers, though the natural numbers, the integers, the rational, the decimal, the real, and the complex form a nice self-complete expository whole. Hopefully, the following block(s) of text aren't too poorly formatted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) The set of &lt;strong&gt;natural numbers&lt;/strong&gt; consists of numbers with which we count {0,1,2,3,...}. As noted in some of the other answers, some people think that 0 is not a natural number (see &lt;a href=&quot;http://www.marriedtothesea.com/031107/zero.gif&quot; rel=&quot;nofollow&quot;&gt;one of my desktop backgrounds&lt;/a&gt;). Whether or not it is, it's a matter of taste. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is not a matter of taste are the defining properties of the natural numbers. In particular, there is a (binary) operation called +, which takes two numbers a and b and spits out a third number a+b, which is the sum of a and b. It satisfies the usual properties that you would expect from counting: it is commutative (the order of the summands doesn't matter, i.e. a+b=b+a) and associative (the order in which you add summands to each other doesn't matter, i.e. (a+b)+c=a+(b+c), so you can always write a+b+c for a specific number).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is also an order relation &amp;lt; such that for any two different numbers a and b either a&#xA;&#xA;&lt;p&gt;The order plays nice with addition in that a &amp;lt; b implies a+c &amp;lt; b+c. We also have a cancellation property that if a &amp;lt; c, then there exists a number b such that a+b=c.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The order also satisfies one unobvious property called the well-ordering principle and states that if you take any collection of natural numbers, there is a smallest one, i.e. a number that is smaller than all the other numbers in the collection. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The well-ordering principle, together with cancellation, implies that there is a smallest number, and that any two consecutive numbers differ by the same number. From this it follows that either the smallest number is 0 (in the sense that 0 is the number such that 0+a=a+0=a) or that it is 1, where 1 is the common difference between any two consecutive numbers. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Choosing one or the other of the two possibilities defines the natural numbers uniquely as either {0,1,2,...} or {1,2,...}. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is a fun exercise to show that the above axioms are equivalent to the axioms of Peano (note that the axioms of Peano state nothing about addition, so in fact BOTH {0,1,2,...} and {1,2,...} satisfy the axioms; how we define addition with respect to the smallest element is what specifies one of the two sets)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) The set of &lt;strong&gt;integers&lt;/strong&gt;, also known as whole numbers, is {..., -3, -2, -1, 0, 1, 2, 3, ...}.&#xA;It arises as we try to &quot;complete&quot; the arithmetic of the integers in the following sense. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For any two numbers a &amp;lt; b we have a number b-a such that (b-a)+a=b, thus we have subtraction of a smaller number from a bigger number. We do not have such numbers b-a if a&gt;b (i.e. there is no natural number which when you add a to you get b if a&gt;b), but from experience we see that numbers such as ... -3, -2, -1 &#xA;seem to be sensical to add and subtract as we try to solve equations and whatnot. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But while numbers such as b-a when a &amp;lt; b are defined, the same number could be represented in different ways. For example, 2 is both 5-3 and 8-6. So we need to answer the question of when b-a=d-c for a &amp;lt; b, c &amp;lt; d, and by algebra the answer is that they're equal whenever b+c=d+a. Hence, we can define all the integers, positive and negative, by taking them to be pairs of natural numbers (a,b) with the caveat that (a,b)=(c,d) whenever a+c=b+c. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, the integer -2 consists of (among others) the pairs (5,3), (6,8), (0,2). This is all seems like much formal ado about intuitive nothings, but that is only because we can represent any integer as either the pair (a,0), which we write simply as a, which we write as -a, and think of as (a,0) is the integer which when you add a to you get 0, or as (0,a), which we think of as the integer to which which you add 0 you get the natural number a. Hence what we actually do arithmetic with are the pairs {... (3,0), (2,0), (1,0), (0,0), (0,1), (0,2), (0,3),...}, but the fact that we only need to throw in a negative sign to make the arithmetic meaningful is precisely because there exists these special representative pairs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Exercise: define the addition and order on the pairs of natural numbers for the above to work (we lose of course the property that any set of integers has a smallest integers [e.g. the set of negative integers], but the principle still holds for sets of positive integers).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) The &lt;strong&gt;rational numbers (or fractions)&lt;/strong&gt; are obtained by the exact same process of completion as above, except with respect to multiplication. the fraction a/b, where a and b are integers corresponds to the number which when multiplied by b gives you a. Note that a/b=c/d if and only if ad=bc, which mirrors the fact that a-b=c-d if and only if a+b=b+c.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4) The real numbers are best understood as coming from &lt;strong&gt;decimals (or decimal expansions)&lt;/strong&gt;, so what are decimals and where do they come from?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rational numbers as fractions are hard to compare to one another. Is 7/5 bigger or smaller than 4/3? The answer is yes, becayse 7*3&gt;3*5. In general if you have a/b and c/d, we have a/b &amp;lt; c/d, a/b=c=d or a/b&gt;c/d if correspondingly adbc. But that's a tedious process, is there some way of writing them down so that it is clearer who's bigger than who? Better, can the process of writing them down be easier than the process of checking who's bigger than who by cross-multiplication?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer is yes, and we only need consider the usual way in which we write natural numbers.  We can express the number 1729 compactly based on the distributive properties of addition and exponentiation: 1729=1000+700+20+9=10^3+7*10^2+2*10+9, where the digits 1,2,3,4,5,6,7,8,9,10 are the first ten successors of zero, and zero is denoted by 0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we can write rational numbers with denominator 10^n as a.b=a+b/10^n with a and b are natural numbers and b is smaller than 10^n. For example, 1729/100 can be written as 17.29 where 17 and 29 are natural numbers with 29&amp;lt;100.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, for any rational number c/d there exists a rational number with denominator closest 10^n that's closest to and smaller than c/d (this is intuitively obvious and not that hard to formalize). Then we can say that c/d ~ a.b where a+b/10^n is that rational number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, the closest number with denominator 10 to 1/3 is 3/10=.3, with denominator 100 is 33/100=.33, and so on. Then that the infinite string 0.3333... equals 1/3 means that the closest rational number less than or equal to 1/3 with denominator 10^n is 333...3/10^n where you have n 3's.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From this, one can prove that rational numbers correspond have periodic expansions, i.e. that the string a.b has a repeating sequence of digits from some point onward.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(even better, we can compute decimal expansions by the usual process of long division by just adding a decimal point, and we can efficiently compare numbers by looking at the expansion until the first point of difference).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;5) Real numbers come from taking arbitrary infinite sequences of digits as decimals. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From above, we know that decimals actually encode sequences of rational numbers with denominators 1,10, 10^2, 10^3, ...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When should two such sequences be equivalent? Consider the standard question of whether .999...=1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The decimal expansion of 1 is just 1.000... because the sequence of rational numbers with denominators 10, 10^2, 10^3, ... has to be a sequence of rational numbers less than or equal to 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we drop the requirement that the numbers be less than or equal to 1, keep the requirement that the sequence of rational numbers (approximations really) is non-decreasing, and say allow the rational number to be either the closest number less than 1 or 1 itself, then we obtain the other expansion 1=.999...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you had a measuring instrument that was accurate to n decimal places, then upon measuring .999... and 1.000... you would always get .999...9 with n-2 9s.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So it turns out that with finite precision measuring, you can't tell the difference between certain decimal expansions, so we might as way say those two decimal expansions are equal. This is the process of completion by Cauchy sequences put into words.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And those expansions then are the real numbers, and they have addition, multiplication, subtraction and division and ordering just like the rational numbers, except that they're complete in that every sequence of real numbers converges. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;6) Complex numbers&#xA;It turns out that every polynomial with real coefficients of degree &gt;2 has a square root. -1 does not have a a real square root, so we wish to throw a root of -1 in to algebraically complete the real numbers, and thus get the complex numbers. The algebraic construction is best done by explaining concepts such as rings and ideals and quotients and homomorphisms, which is too much machinery.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instead, consider Euclidean geometry. It is a(n advanced) fact that you can coordinize it with real numbers, i.e. you can represent points by pairs of real numbers (a,b), lines by the solutions of the two variable equation ax+by=c, lengths given by square root of ((a-c)^2+(b-d)^2), etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Adding pairs (a,b) and (c,d) gives you (a,b)+(c,d) so addition of pairs corresponds to translations. Another way of interpreting a pair of (a,b) is by interpreting it in polar form (r, theta) where r is the distance of (a,b) form the origin and theta is the angle from the x-axis to (a,b). Then we can think of (a,b)~(r,theta) as a dilation and rotation with center (0,0) and sending (1,0) to (a,b), i.e. a dilation by a factor of r and  a rotation by an angle of theta. But this then defines a multiplication of vectors, which turns out to distribute over addition. Clearly we have inverse rotations/dilations and inverse translations, giving us subtraction and division. Finally, we can identify the real numbers with pairs (r,0) which are only a dilation without any rotation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;These pairs are then our complex numbers and you can deduce all their properties from the above description. Note that the vector (0,1) multiplied by (0,1) gives (-1,0) since it corresponds to a rotation of 90 followed by a rotation of 90, which is a rotation of 180 of (1,0) which equals (-1,0). See Qiaouchu's answer for more details.&lt;/p&gt;&#xA;" OwnerUserId="400" LastActivityDate="2010-07-30T19:42:21.053" CommentCount="1" />
  <row Id="1257" PostTypeId="1" AcceptedAnswerId="1259" CreationDate="2010-07-30T19:59:03.430" Score="8" ViewCount="589" Body="&lt;p&gt;I've solved for it making a computer program, but was wondering there was a mathematical equation that you could use to solve for the nth prime?&lt;/p&gt;&#xA;" LastActivityDate="2010-09-11T17:39:07.933" Title="Is there a known mathematical equation to find the nth prime?" Tags="&lt;prime-numbers&gt;" AnswerCount="5" FavoriteCount="3" />
  <row Id="1258" PostTypeId="2" ParentId="1257" CreationDate="2010-07-30T20:06:31.370" Score="1" ViewCount="" Body="&lt;p&gt;No such formula is known, but there are a few that give impressive results. A famous one is Euler's:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;P(n) = n^2 − n + 41&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Which yields a prime for every natural number lower than 41, though not necessarily the &lt;code&gt;n&lt;/code&gt;th prime.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See more &lt;a href=&quot;http://en.wikipedia.org/wiki/Formula_for_primes&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="456" LastActivityDate="2010-07-30T20:06:31.370" />
  <row Id="1259" PostTypeId="2" ParentId="1257" CreationDate="2010-07-30T20:10:21.033" Score="14" ViewCount="" Body="&lt;p&gt;No, there is no formula that gives the nth prime, except artificial ones you can write that are basically equivalent to &quot;the nth prime&quot;. But if you only want an approximation, the nth prime is roughly around $n \ln n$ (or more precisely, near the number m such that $m/\ln m = n$) by the &lt;a href=&quot;http://en.wikipedia.org/wiki/Prime_number_theorem&quot; rel=&quot;nofollow&quot;&gt;prime number theorem&lt;/a&gt;. In fact, we have the following asymptotic bound on the nth prime $p_n$:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$n \ln n + n(\ln\ln n - 1) &amp;lt; p_n &amp;lt;  n \ln n + n \ln \ln n$ for n≥6  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You can &lt;a href=&quot;http://en.wikipedia.org/wiki/Generating_primes#Prime_sieves&quot; rel=&quot;nofollow&quot;&gt;sieve&lt;/a&gt; within this range if you want the $n$th prime. [Edit: There are better ideas than a sieve, see &lt;a href=&quot;#4314&quot; rel=&quot;nofollow&quot;&gt;the answer by Charles&lt;/a&gt;.]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Entirely unrelated: if you want to see formulae that generate a lot of primes (not the nth prime) up to some extent, like the famous f(n)=n&lt;sup&gt;2&lt;/sup&gt;-n+41, look at the Wikipedia article &lt;a href=&quot;http://en.wikipedia.org/wiki/Formula_for_primes&quot; rel=&quot;nofollow&quot;&gt;formula for primes&lt;/a&gt;, or Mathworld for &lt;a href=&quot;http://mathworld.wolfram.com/PrimeFormulas.html&quot; rel=&quot;nofollow&quot;&gt;Prime Formulas&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="205" LastEditorUserId="205" LastEditorDisplayName="" LastEditDate="2010-09-09T07:30:36.230" LastActivityDate="2010-09-09T07:30:36.230" CommentCount="2" />
  <row Id="1260" PostTypeId="1" AcceptedAnswerId="1284" CreationDate="2010-07-30T20:17:02.063" Score="-1" ViewCount="248" Body="&lt;p&gt;I'm looking for a tool for drawing vectors (without using Microsoft paint, Paint.net etc.) to paste into a Google docs document.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone suggest an easy solution? I need up to 2 column/row vectors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Update:&lt;/em&gt; This is the format I'm after (not arrows, but just the items):  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://upload.wikimedia.org/math/b/6/4/b6459d183a924dad2e5067a15fd45f2c.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="54" LastEditorUserId="171" LastEditorDisplayName="" LastEditDate="2010-08-16T19:05:01.237" LastActivityDate="2010-08-16T19:05:01.237" ClosedDate="2010-08-02T14:27:52.290" Title="Is there a tool for drawing column and row vectors?" Tags="&lt;math-software&gt;" AnswerCount="4" CommentCount="3" />
  <row Id="1261" PostTypeId="2" ParentId="548" CreationDate="2010-07-30T20:19:41.947" Score="1" ViewCount="" Body="&lt;p&gt;EDIT:I botched my original answer, here's what I actually meant:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fractions are equivalence classes of pairs of integers subject to (a,b)=(c,d) iff ad=bc, with addition and multiplication extended from {(a,1)} as a copy of the integers by multiplication defined component-wise in general.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we allow pairs of the form (a,0) for some integer a, we have with the multiplication axioms (a,0)=(b,0) for any a and b, and in particular all of them coincide with (0,1)(a,0)=(0,0) as multiplication is component-wise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But (0,0)=(a,b) for every pair, so our construction gives us the trivial ring. Hence, to get a non-trivial structure, we must disallow pairs of the form (a,0), including (0,0) and that gives us the rationals.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Anyway, the above is the standard reason why 0/0 is undefined. You can actually define it, though to do that you must change some of the fundamental arithmetic properties. For a fun read, check out the following paper on the topic:&#xA;&lt;a href=&quot;http://www2.math.su.se/~jesper/research/wheels/wheels.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www2.math.su.se/~jesper/research/wheels/wheels.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Original answer:&#xA;By definition (or by construction) a/b, where a and b are natural numbers, is defined to be the rational number which when multiplied by b gives you a. From this you have that 0/0=0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Different fractions a/b and c/d are &lt;strong&gt;defined&lt;/strong&gt; to be equal if ad=bc. This allows us to extend the arithmetic of the integers to an arithmetic of fractions. From this you have that 0/0=c/d for every rational number c/d since x0=0 for all x in any system where multiplication distributes over addition (proof: x0=x(0-0)=x0-x0=0).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, if you allow 0/0 among your fractions, and you attempt to extend the arithmetic of the integers, all your fractions must be 0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(note: the reason you cannot allow a/0 among your fractions is when a is not equal to 0 is because if you extend the arithmetic to where you have multiplication distributing over addition, you would have 0x=0 for all x and so the existence of a/0 is only consistent if a=0, which is the case above).&lt;/p&gt;&#xA;" OwnerUserId="400" LastEditorUserId="400" LastEditorDisplayName="" LastEditDate="2010-07-31T17:48:00.530" LastActivityDate="2010-07-31T17:48:00.530" CommentCount="2" />
  <row Id="1262" PostTypeId="2" ParentId="1017" CreationDate="2010-07-30T20:39:10.357" Score="3" ViewCount="" Body="&lt;p&gt;One way is to note that $x_1y_2 - x_2y_1$ is a &lt;em&gt;signed&lt;/em&gt; area, i.e. it may positive or negative. Adding up all the signed areas of the triangles formed by the points $O$, $P_k$ and $P_{k+1}$ will cancel all the superfluous parts, as can be seen from the sketch:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i27.tinypic.com/2hyz7mb.jpg&quot; alt=&quot;Adding signed areas for the sides of a polygon&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same argument also works for trapezoids with the $x$-axis instead of triangles with the origin. (I think this is the most illuminating argument, because the key trick is to give the area a sign depending on orientation.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One could argue that this is not very rigorous, however. A more rigorous proof is to divide the polygon into two smaller polygons (it's not trivial to show that this is possible) and argue that adding the shoelace sums of the two parts gives the shoelace sum of the whole. That's because the two additional terms for the extra side cancel each other. (This cancellation is well-known from line integrals, we are in essence calculating $\frac12 \oint_{polygon} x\,dy - y\,dx$ here.) By induction, you then only have to verify the formula for a triangle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i32.tinypic.com/2d9wm7o.jpg&quot; alt=&quot;Shoelace sums of two polygons&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="434" LastEditorUserId="434" LastEditorDisplayName="" LastEditDate="2010-07-30T20:44:23.697" LastActivityDate="2010-07-30T20:44:23.697" />
  <row Id="1263" PostTypeId="2" ParentId="548" CreationDate="2010-07-30T20:51:10.327" Score="0" ViewCount="" Body="&lt;p&gt;Everyone knows 0/0 is &lt;a href=&quot;http://www.bbc.co.uk/berkshire/content/articles/2006/12/06/divide_zero_feature.shtml&quot; rel=&quot;nofollow&quot;&gt;nullity&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;(yes, this is a joke post, but that was a serious article)&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="136" LastActivityDate="2010-07-30T20:51:10.327" CommentCount="1" />
  <row Id="1264" PostTypeId="2" ParentId="1260" CreationDate="2010-07-30T20:51:48.377" Score="1" ViewCount="" Body="&lt;p&gt;Assuming you want arrows, you can try using Mathematica. In particular, you probably want to use the &lt;a href=&quot;http://reference.wolfram.com/mathematica/ref/Arrow.html&quot; rel=&quot;nofollow&quot;&gt;Arrow&lt;/a&gt; command.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Graphics[{Arrow[{{0, 0}, {1, 0}}], Arrow[{{0, 0}, {1, 2}}]}] &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;gives the two vectors (1, 0) and (1, 2).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming you want 1xn matrices, try LaTeX.&lt;/p&gt;&#xA;" OwnerUserId="30" LastActivityDate="2010-07-30T20:51:48.377" CommentCount="1" />
  <row Id="1265" PostTypeId="2" ParentId="1211" CreationDate="2010-07-30T21:18:07.600" Score="2" ViewCount="" Body="&lt;p&gt;Below is my geometric understanding of why irrational powers of negatives are difficult to define. As such it is probably not rigorous and may be wrong.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The way irrational powers of the real numbers are usually defined is by limits of fractional powers. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For complex numbers, the same is true, except the limiting process is more complicated. We can of course coast as usual on our real numbers result and see that we only need to define irrational powers on the unit circle, since every other points is some positive real multiple of a point on the unit circle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now in general z -&gt; z^n wraps the circle around itself n times. What does z-&gt;z^(1/n) do? Well, it's not clear since each point has n possible points it could have come from, in particular if you partition the circle into n arcs of length 2pi/n, each of those gets mapped to the full circle. Once you choose a starting arc though, z -&gt; z^m maps the starting arc to other arcs in the following way. Partition your starting arc of length 2pi/n into arcs of length 2pi/(nm) and then each of the little arcs gets mapped to a big arc that is 2pim/n away from the previous big arc. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason you choose arcs as opposed to weirdly distributed discrete sets, is because you want exponentiation to be continuous and hence you want the inverse image sets to be as the whole circle is connected.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is no problem with fractional powers of -1, you have n choices for starting arc. But if you want your exponentiation to have some semblance of continuity with respect to the exponent, then you have to be choosing branches (arcs) for z-&gt;z^(1/n) that cohere, i.e. such that for very large n the roots get closer and closer together. This is done by requiring that the nth root of 1 is always 1, which makes all of the arcs into neighborhoods of 1. But this means that all the points with argument between 0 and pi get mapped in the half-arc above 1, and all the points with argument between pi and 2pi get mapped in the half-arc below 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence if you approach -1 from above and from below, the two limits of the nth root will be different, and hence you cannot have continuous exponentiation at -1. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a result (you can do some more visualization if you wish), the irrational powers of -1 cannot even be defined as limits of rational powers.&lt;/p&gt;&#xA;" OwnerUserId="400" LastActivityDate="2010-07-30T21:18:07.600" CommentCount="1" />
  <row Id="1266" PostTypeId="2" ParentId="1257" CreationDate="2010-07-30T21:26:18.713" Score="6" ViewCount="" Body="&lt;p&gt;There are &lt;a href=&quot;http://en.wikipedia.org/wiki/Formula_for_primes&quot; rel=&quot;nofollow&quot;&gt;formulas&lt;/a&gt; on Wikipedia, though they are messy. No polynomial $p(x)$ can output the $n$th prime for all $n$, as is explained in the first section of the article.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is, however, a &lt;a href=&quot;http://primes.utm.edu/glossary/xpage/MatijasevicPoly.html&quot; rel=&quot;nofollow&quot;&gt;polynomial in 26 variables&lt;/a&gt; whose &lt;em&gt;nonnegative&lt;/em&gt; values are precisely the primes. (This is fairly useless as far as computation is concerned.) This comes from the fact that the property of being a prime is decidable, and the &lt;a href=&quot;http://en.wikipedia.org/wiki/Diophantine_set&quot; rel=&quot;nofollow&quot;&gt;theorem of Matiyasevich&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="536" LastActivityDate="2010-07-30T21:26:18.713" CommentCount="2" />
  <row Id="1268" PostTypeId="2" ParentId="756" CreationDate="2010-07-30T21:39:34.520" Score="11" ViewCount="" Body="&lt;p&gt;There is a fairly simple strategy which requires (1&amp;nbsp;+&amp;nbsp;&amp;epsilon;)log(n)&amp;nbsp;+&amp;nbsp;1/&amp;epsilon;&amp;nbsp;+&amp;nbsp;O(1) queries, for any constant &amp;epsilon;&amp;nbsp;&amp;gt;&amp;nbsp;0. I illustrate this strategy below.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;First, I ask you whether or not X, the secret number, is n/2. Without loss of generality, suppose you answer &quot;less&quot;. I learn nothing at first, because you may be lying; but for the moment I give you the benefit of the doubt.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I next ask you whether X is n/4. If you say &quot;less&quot;, I don't know whether or not 0&amp;nbsp;&amp;lt;&amp;nbsp;X&amp;nbsp;&amp;lt;&amp;nbsp;n/4, but I &lt;strong&gt;do&lt;/strong&gt; know that 0&amp;nbsp;&amp;lt;&amp;nbsp;X&amp;nbsp;&amp;lt;&amp;nbsp;n/2, because you can't lie twice. Similarly, if I go about a normal binary search, so long as you continue to answer &quot;less&quot;, I know that your answer to the &lt;em&gt;preceding&lt;/em&gt; query was honest. So we may reduce to the case where you say &quot;more&quot; to my query of whether X is n/4.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If I continue to take you at your word, persue a binary search, and enquire about whether X is 3n/8, you may say either &quot;more&quot; or &quot;less&quot;. If you say &quot;more&quot;, then I don't know whether X&amp;nbsp;&amp;gt;&amp;nbsp;n/2, but I do know that X&amp;nbsp;&amp;gt;&amp;nbsp;n/4, again because you can't lie twice. So again so long as you continue to answer &quot;more&quot; in my normal binary search, I know that your answer to the preceding query was honest.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;More generally:&lt;/strong&gt; if I consistently guess under the hypothesis that you are being honest, in any &quot;monotonic&quot; sequence of responses, I know that all but (possibly) the last of them are honest. So it might seem as though the worst case scenario is where your responses alternate a lot, as would occur for instance if you had chosen something like X&amp;nbsp;=&amp;nbsp;&lt;sub&gt;&amp;lfloor;&lt;/sub&gt;&amp;thinsp;n/3&amp;thinsp;&lt;sub&gt;&amp;rfloor;&lt;/sub&gt;. But in the alternating case too, I can be confident of the honesty of some of your answers:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If you say (non-monotonically) that X is less than 3n/8, I don't know whether X is less than 3n/8 for sure; but I do know that X&amp;nbsp;&amp;lt;&amp;nbsp;n/2, because again you can't have lied about both.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;More generally:&lt;/strong&gt; not only do I know that monotonic subsequences of answers are mostly honest, I know that any time that you answer &quot;more&quot; or &quot;less&quot;, your previous answer of the same kind was also honest. So in fact I should be most suspicious, when I encounter long monotonic subsequences in your answers, that the answer previous to that monotonic subsequence was a lie.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I need is a strategy that will tell me when to revisit an old question, depending on how large n is. Ideally, revisiting old questions would require very little overhead. If I encounter a monotonic sequence of f(n) responses, I revisit the last question before that monotonic sequence started.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;For instance, if your responses to queries about n/4, 3n/8, 7n/16, etc. are all monotonically &quot;more&quot;, I eventually ask about the last time you said &quot;less&quot;, which is for n/2, just in case you lied back then. This allows me to avoid the scenario where you lie about n/2, and keep me guessing at (2&lt;sup&gt;&lt;em&gt;t&lt;/em&gt;&amp;minus;1&lt;/sup&gt;&amp;nbsp;&amp;minus;&amp;nbsp;1)/2&lt;sup&gt;&lt;em&gt;t&lt;/em&gt;&lt;/sup&gt; until I eliminate all possibilities, catch you in your lie, and then redo all of my binary search for queries greater than n/2.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;If I do a double-check like this every time I encounter a monotonic sequence of length r, I will in the worst case query you about (r+1)log(n)/r = (1&amp;nbsp;+&amp;nbsp;1/r)log(n) times. If I catch you in a lie, I will have only &quot;wasted&quot; r queries, and my strategy afterwards can be just a simple binary search without double-checks; so your optimal strategy for maximizing the number of queries I make is actually not to lie at all, or to save your lie for nearly the end of the game to cost me about r additional queries. Here r can be an arbitrarily large integer; thus for any &amp;epsilon;, I can achieve a query rate of (1&amp;nbsp;+&amp;nbsp;&amp;epsilon;)log(n)&amp;nbsp;+&amp;nbsp;1/&amp;epsilon;&amp;nbsp;+&amp;nbsp;O(1) by setting r&amp;nbsp;&amp;gt;&amp;nbsp;1/&amp;epsilon;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Bonus problem #1.&lt;/strong&gt; Without too much extra work, I think you can improve this to a strategy requiring only log(n)&amp;nbsp;+&amp;nbsp;O(log&amp;nbsp;log(n)) queries, but I'm too lazy to work out the details right now.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Bonus problem #2.&lt;/strong&gt; Generalize this strategy to the regime where you are allowed to lie to me some fixed number of times L&amp;nbsp;&gt;&amp;nbsp;0.  &lt;/p&gt;&#xA;" OwnerUserId="439" LastEditorUserId="439" LastEditorDisplayName="" LastEditDate="2010-11-02T13:11:07.107" LastActivityDate="2010-11-02T13:11:07.107" />
  <row Id="1269" PostTypeId="2" ParentId="1211" CreationDate="2010-07-30T21:44:55.380" Score="5" ViewCount="" Body="&lt;p&gt;As other posters have indicated, the problem is that the complex logarithm isn't well-defined on $\mathbb{C}$.  This is related to my comments in &lt;a href=&quot;http://math.stackexchange.com/questions/1183/division-by-imaginary-number&quot;&gt;a recent question&lt;/a&gt; about the square root not being well-defined (since of course $\sqrt{z} = e^{ \frac{\log z}{2} }$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One point of view is that the complex exponential $e^z : \mathbb{C} \to \mathbb{C}$ does not really have domain $\mathbb{C}$.  Due to periodicity it really has domain $\mathbb{C}/2\pi i \mathbb{Z}$.  So one way to define the complex logarithm is not as a function with range $\mathbb{C}$, but as a function with range $\mathbb{C}/2\pi i \mathbb{Z}$.  Thus for example $\log 1 = 0, 2 \pi i, - 2 \pi i, ...$ and so forth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what are we doing when we don't do this?  Well, let us suppose that for the time being we have decided that $\log 1 = 0$.  This is how we get other values of the logarithm: using power series, we can define $\log (1 + z)$ for any $z$ with $|z| &amp;lt; 1$.  We can now pick any number in this circle and take a power series expansion about &lt;em&gt;that&lt;/em&gt; number to get a different power series whose circle of convergence is somewhere else.  And by repeatedly changing the center of our power series, we can compute different values of the logarithm.  This is called &lt;em&gt;analytic continuation&lt;/em&gt;, and typically it proceeds by choosing a (say, smooth) path from $1$ to some other complex number and taking power series around different points in that path.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem you quickly run into is that the value of $\log z$ depends on the choice of path from $1$ to $z$.  For example, the path $z = e^{2 \pi i t}, 0 \le t \le 1$ is a path from $1$ to $1$, and if you analytically continue the logarithm on it you will get $\log 1 = 2 \pi i$.  And that is not what you wanted.  (This is essentially the same as the contour integral of $\frac{1}{z}$ along this contour.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One way around this problem is to arbitrarily choose a ray from the origin and declare that you are not allowed to analytically continue the logarithm through this ray.  This is called choosing a branch cut, and it is not canonical, so I don't like it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is another way to resolve this situation, which is to consider the &lt;a href=&quot;http://en.wikipedia.org/wiki/Riemann_surface&quot; rel=&quot;nofollow&quot;&gt;Riemann surface&lt;/a&gt; $(z, e^z) \subset \mathbb{C}^2$ and to think of the logarithm as the projection to the first coordinate from this surface to $\mathbb{C}$.  So all the difficulties we have encountered above have been due to the fact that we have been trying to pretend that this projection has certain properties that it doesn't have.  A closed path like $z = e^{2\pi i t}$ in which the logarithm starts and ends with different values corresponds to a path on this surface which starts and ends at different &lt;em&gt;points&lt;/em&gt;, so there is no contradiction.  This was Riemann's original motivation for defining Riemann surfaces, and it is this particular Riemann surface that powers things like the residue theorem.&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-07-30T21:55:52.240" LastActivityDate="2010-07-30T21:55:52.240" CommentCount="1" />
  <row Id="1271" PostTypeId="1" AcceptedAnswerId="1272" CreationDate="2010-07-30T21:59:14.623" Score="0" ViewCount="139" Body="&lt;p&gt;In part of my research, the following problem has come up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the system of equations (in complex numbers)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;z^b w^c = 1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;z^d w^e = 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am interested in the solution set when we restrict both z and w to be ath roots of unity, for some positive integer a.  Of course, one immediately sees that (z,w) = (1,1) is a solution.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What are some nice necessary and sufficient conditions on a,b,c,d, and e which guarantee that (z,w) = (1,1) is the ONLY solution?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To give an idea of the flavor of answer I'd be most happy with, one must have gcd(a,b,d) = gcd(c,e) = 1, because if z is any gcd(a,b,d)th root of 1 (which is neccesarily an ath root of 1), then (z,1) is a solution to both equations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It also turns out that z and w must both be gcd(a, be-cd)th roots of 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd love to have an answer like &quot;gcd(a, be-cd) = gcd(a,b,d) = gcd(c,e) = 1 is necessary and sufficient&quot;, with, perhaps, a few more estimates on gcd terms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This problem can also been generalized (and I am interested in that case as well).  Suppose you are given 3 equations&lt;/p&gt;&#xA;&#xA;&lt;p&gt;z^a w^b = 1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;z^c w^d = 1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;z^e w^f = 1, with z and w complex numbers of modulus 1.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What are necessary and sufficient conditions on a,b,c,d,e and f which guarantee that the only simultaneous solution is (z,w) = (1,1)?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The previous problem is a special case of this (which comes from setting b = 0.  Clearly then, z must be an ath root of unity.  It turns that if b is 0, using the fact that gcd(d,f)=1 one can show w must also be an ath root of unity).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And please feel free to retag as appropriate!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you in advance.&lt;/p&gt;&#xA;" OwnerUserId="331" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-07-30T22:06:02.513" LastActivityDate="2010-07-30T22:18:30.490" Title="Solving systems of equations in roots of unity." Tags="&lt;linear-algebra&gt;&lt;number-theory&gt;" AnswerCount="1" />
  <row Id="1272" PostTypeId="2" ParentId="1271" CreationDate="2010-07-30T22:04:14.857" Score="2" ViewCount="" Body="&lt;p&gt;You are computing the nullspace of a $2 \times 2$ matrix over $\mathbb{Z}/a\mathbb{Z}$, so a necessary and sufficient condition is that the matrix in question is invertible, hence $\gcd(a, be-dc) = 1$ should be necessary and sufficient.  In the second case you should compute the &lt;a href=&quot;http://en.wikipedia.org/wiki/Smith_normal_form&quot; rel=&quot;nofollow&quot;&gt;Smith normal form&lt;/a&gt; of your matrix over $\mathbb{Z}$.  I believe a necessary and sufficient condition is that the Smith normal form is [[1 0][0 1][0 0]] (looks like matrices aren't working yet).&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-07-30T22:18:30.490" LastActivityDate="2010-07-30T22:18:30.490" CommentCount="3" />
  <row Id="1273" PostTypeId="2" ParentId="74" CreationDate="2010-07-30T22:04:55.433" Score="2" ViewCount="" Body="&lt;p&gt;Your question: given that X and Z are independent, X is Gaussian (I'll use &quot;normal&quot;), and Y = X+Z, prove that Y is normal iff Z is normal. Right? As you observed, one direction is easy: &lt;em&gt;if&lt;/em&gt; Z is normal, then so is Y=X+Z. So for the other direction, assume that Y is normal. We need to prove that is Z normal too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps there's an even easier way, but it's straightforward to use &lt;a href=&quot;http://en.wikipedia.org/wiki/Characteristic_function_%28probability_theory%29&quot; rel=&quot;nofollow&quot;&gt;characteristic functions&lt;/a&gt;, which completely characterise distributions. Because X and Z are independent,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ \varphi_Y(t) = E[e^{itY}] = E[e^{it(X+Z)}] = E[e^{itX}]E[e^{itZ}]$, and so,&lt;br&gt;&#xA;$ \varphi_Z(t) = E[e^{itZ}] = E[e^{itY}]/E[e^{itX}] $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means that Z has exactly the right characteristic function for a normal variable, and hence it's normal.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;More interestingly and much more generally, there is a theorem of Cramer (e.g. see &lt;a href=&quot;http://books.google.com/books?id=B7Ch-c2G21MC&amp;amp;pg=PA205&amp;amp;dq=cramer+theorem&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;) which says that if X and Z are independent and X+Z is normally distributed, then both X and Z are!&lt;/p&gt;&#xA;" OwnerUserId="205" LastActivityDate="2010-07-30T22:04:55.433" CommentCount="2" />
  <row Id="1275" PostTypeId="2" ParentId="1249" CreationDate="2010-07-30T23:39:46.963" Score="5" ViewCount="" Body="&lt;p&gt;Here's a simple intersection theoretic example.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take the intersection of the line $y=0$ and the parabola $y=x^2$. Classically, the intersection is a point. But note that there is more to the intersection than just the point; there is the fact that the two curves are tangent at that point. Scheme-theoretically, the intersection is $\operatorname{Spec} k[x,y]/(y,y-x^2) \cong \operatorname{Spec} k[x]/(x^2)$. This reflects the tangency. If the intersection were transverse, then the scheme-theoretic intersection would have been just $\operatorname{Spec} k$.&lt;/p&gt;&#xA;" OwnerUserId="263" LastActivityDate="2010-07-30T23:39:46.963" CommentCount="0" />
  <row Id="1276" PostTypeId="2" ParentId="1241" CreationDate="2010-07-30T23:46:52.207" Score="0" ViewCount="" Body="&lt;p&gt;An important special case of the AM-GM inequality is that the product of two (positive) numbers with a constant sum is at a maximum when they are equal. This comes up a lot.&lt;/p&gt;&#xA;" OwnerUserId="123" LastActivityDate="2010-07-30T23:46:52.207" />
  <row Id="1278" PostTypeId="2" ParentId="574" CreationDate="2010-07-31T00:43:33.840" Score="4" ViewCount="" Body="&lt;p&gt;In general*, if something is &quot;generic&quot;, it means it happens or is true &quot;almost all of the time&quot; or &quot;almost everywhere&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In measure theory, for example, when you say &quot;$P(x)$ is true for almost all $x$&quot;, this has the precise meaning that the set of $x$'s for which $P(x)$ does not hold has measure zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One can relate this to the Zariski topology via the fact that Zariski closed subsets of $\mathbb{C}^n$ have Lebesgue measure zero: &lt;a href=&quot;http://mathoverflow.net/questions/25513/zariski-closed-sets-in-cn-are-of-measure-0&quot; rel=&quot;nofollow&quot;&gt;http://mathoverflow.net/questions/25513/zariski-closed-sets-in-cn-are-of-measure-0&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See also these MO posts:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://mathoverflow.net/questions/19688/what-does-generic-mean-in-algebraic-geometry&quot; rel=&quot;nofollow&quot;&gt;http://mathoverflow.net/questions/19688/what-does-generic-mean-in-algebraic-geometry&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://mathoverflow.net/questions/2162/what-are-the-most-important-instances-of-the-yoga-of-generic-points&quot; rel=&quot;nofollow&quot;&gt;http://mathoverflow.net/questions/2162/what-are-the-most-important-instances-of-the-yoga-of-generic-points&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;*or perhaps I should say... generically ;-)&lt;/p&gt;&#xA;" OwnerUserId="263" LastEditorUserId="263" LastEditorDisplayName="" LastEditDate="2010-07-31T01:07:09.890" LastActivityDate="2010-07-31T01:07:09.890" />
  <row Id="1279" PostTypeId="1" AcceptedAnswerId="1282" CreationDate="2010-07-31T01:02:48.323" Score="3" ViewCount="166" Body="&lt;p&gt;I'm looking for the name of a shape that is like a capsule, but where each circle can have different radii. The shape could be described using two circles (two centers and two radii). Something like this (this is a 2D shape, not a lame attempt at 3D):&#xA;&lt;img src=&quot;http://img694.imageshack.us/img694/3212/shape.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="529" LastEditorUserId="529" LastEditorDisplayName="" LastEditDate="2010-07-31T01:38:30.830" LastActivityDate="2010-07-31T06:27:29.753" Title="What is the name for a shape that is like a capsule, but with two different radii?" Tags="&lt;geometry&gt;&lt;terminology&gt;" AnswerCount="2" />
  <row Id="1280" PostTypeId="2" ParentId="405" CreationDate="2010-07-31T01:03:34.703" Score="3" ViewCount="" Body="&lt;p&gt;TheCatsters have short YouTube videos explaining monads, and other parts of category theory: &lt;a href=&quot;http://www.youtube.com/user/TheCatsters&quot; rel=&quot;nofollow&quot;&gt;http://www.youtube.com/user/TheCatsters&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="263" LastActivityDate="2010-07-31T01:03:34.703" />
  <row Id="1282" PostTypeId="2" ParentId="1279" CreationDate="2010-07-31T01:23:26.823" Score="7" ViewCount="" Body="&lt;p&gt;You can just say it is the convex hull of two circles.&lt;/p&gt;&#xA;" OwnerUserId="405" LastActivityDate="2010-07-31T01:23:26.823" CommentCount="5" />
  <row Id="1283" PostTypeId="1" CreationDate="2010-07-31T04:11:00.823" Score="3" ViewCount="470" Body="&lt;p&gt;I am looking for a closed form (ideally expressed as elementary functions) of the function $\exp_n(x) = \sum_{k=0}^n x^k / k!$.  I am already aware of expressing it in terms of the gamma function.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Background / Motivation&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;When counting combinations of objects with generating functions, it is useful to be able to express the partial sum $1 + x + \cdots + x^n$ as $\frac{1-x^{n+1}}{1-x}$. For example, to count the number of ways to pick 5 marbles from a bag of blue, red, and green marbles where we pick at most 3 blue marbles and at most 2 red marbles, we can consider the generating function $f(x) = (1+x+x^2+x^3)(1+x+x^2)(1+x+x^2+\cdots)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;By using the partial sum identity, we can express it as $f(x) = \left(\frac{1-x^4}{1-x}\right)\left(\frac{1-x^3}{1-x}\right)\left(\frac{1}{1-x}\right)$.  Simplify, express as simpler product of series, and find the coefficient of the $x^5$ term.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to be able to do the same for a generating function in the form&#xA;$g(x) = \exp_{n_1}(x)^{p_1} \exp_{n_2}(x)^{p_2} \cdots \exp_{n_j}(x)^{p_j}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The easiest way to extract the coefficient of a given term $x^p / p!$ would be to use a similar closed form expression for $\exp_n(x)$ and a similar technique to $f$.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Attempted Solutions&lt;/h2&gt;&#xA;&#xA;&lt;h3&gt;Differential equation&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Recall that the way to prove the identity $1+x+x^2+\cdots+x^n = \frac{1-x^{n+1}}{1-x}$ is to define $S = 1 + x + x^2 + \cdots + x^n$ and notice that: $S - Sx = 1 - x^{n+1}$. Likewise, notice that $y(x) = \exp_n(x)$ satisfies $y - y' = x^n/n!$.  Via SAGE, the solution is $y(x) = \frac{c+\Gamma(n+1,x)}{n!}e^x$.  Our initial condition $y(0) = 0$ so $c=0$.  By &lt;a href=&quot;http://mathworld.wolfram.com/IncompleteGammaFunction.html&quot; rel=&quot;nofollow&quot;&gt;(2)&lt;/a&gt;, $\Gamma(n+1,x) = n! e^{-x} \exp_n(x)$ so we have $y(x) = \exp_n(x)$.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Recurrence Relation&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Notice that $\exp_n(x) = \exp_{n-1}(x) + x^n/n!$. Using the unilateral Z-Transform and related properties, we find that $\mathcal{Z}[\exp_n(x)] = (z e^{x/z})/(z-1)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, $\exp_n(x) = \mathcal{Z}^{-1}\left[(z e^{x/z})/(z-1)\right] = \frac{1}{2 \pi i} \oint_C z^n e^{x/z}/(z-1)\;dz$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(z^n e^{x/z})/(z-1)$ has two singularities: $z = 1$ and $z = 0$.  The point $z = 1$ is a pole of order one with residue $e^x$.  To find the residue at $z = 0$ consider the product $z^n e^{x/z} (-1/(1-z)) = -z^n \left( \sum_{m=0}^\infty x^m z^{-m} / m! \right) \left( \sum_{j=0}^\infty x^j \right)$.  The coefficient of the $z^{-1}$ term is given when $n - m + j = -1$. The residue of the point $z=0$ is then $-\sum_{m,j} x^m / m! = -\sum_{m=n+1}^\infty x^m / m!$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $C$ by the positively oriented unit circle centered at the origin.  By Cauchy's Residue Theorem, $\frac{1}{2 \pi i} \oint_C z^n e^{x/z}/(z-1)\;dz = \frac{1}{2 \pi i} 2 \pi i \left(e^x - \sum_{m=n+1}^\infty x^m / m!\right) = \exp_n(x)$.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Finite Calculus&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;I've tried to evaluate the sum using finite calculus, but can't seem to make much progress. &lt;/p&gt;&#xA;" OwnerUserId="381" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2011-05-01T13:35:22.443" LastActivityDate="2011-05-01T13:35:22.443" Title="Closed form of a partial sum of the power series of exp(x)" Tags="&lt;combinatorics&gt;&lt;sequences-and-series&gt;" AnswerCount="1" CommentCount="10" FavoriteCount="1" />
  <row Id="1284" PostTypeId="2" ParentId="1260" CreationDate="2010-07-31T04:54:21.547" Score="2" ViewCount="" Body="&lt;p&gt;I'm assuming you want to typeset vectors or matrices in a Google Docs document, not actually draw the arrow. A simple solution is to use the online TeXify tool at &lt;a href=&quot;http://www.texify.com/&quot; rel=&quot;nofollow&quot;&gt;http://www.texify.com/&lt;/a&gt;. You type some LaTeX code, and it gives you some code that you can paste into your Google Docs document.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, give it the following input:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;\begin{pmatrix} x &amp;amp; y &amp;amp; z \\ 1 &amp;amp; 2 &amp;amp; 3 \end{pmatrix}&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;and click the big button. You'll see a 2 x 3 matrix. If you're not familiar with LaTeX notation, just google &quot;pmatrix&quot; and you'll find lots of documentation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A word of caution: what you're pasting into your Google Docs document is actually a link to the Texify site. If it ever goes offline, your equations will disappear as well. &lt;/p&gt;&#xA;" OwnerUserId="115" LastActivityDate="2010-07-31T04:54:21.547" CommentCount="1" />
  <row Id="1285" PostTypeId="1" AcceptedAnswerId="1288" CreationDate="2010-07-31T05:14:01.170" Score="5" ViewCount="133" Body="&lt;p&gt;I found this question a while ago on a SAT practice exam or something, can't quite remember. So given an acute triangle $ABC$ with $P$ a point inside it and $AP$, $BP$, and $CP$ meeting the opposite sides at $D$, $E$, and $F$ respectively:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://imgur.com/tnZew.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can you find the area of triangle $ABC$ given the areas of triangles $x$, $y$, and $z$?&lt;/p&gt;&#xA;" OwnerUserId="215" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-07-31T06:22:37.313" LastActivityDate="2010-07-31T06:40:19.400" Title="Area of a triangle from some of its parts" Tags="&lt;geometry&gt;&lt;contest-math&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="1286" PostTypeId="2" ParentId="544" CreationDate="2010-07-31T05:45:16.580" Score="2" ViewCount="" Body="&lt;p&gt;I had a programming project in college where we generated large amounts of psuedo-English text using Markov chains. The &lt;a href=&quot;http://scf.usc.edu/~csci102b/curr/assgts/PA3/pa3.html&quot; rel=&quot;nofollow&quot;&gt;assignment is here&lt;/a&gt;, although I don't know if that link will be good forever. From that page:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;For example, suppose that [our Markov chains are of length] 2 and the sample file&#xA;  contains&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;I like the big blue dog better than the big elephant with the big blue hat on his tusk. &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;  &#xA;  &lt;p&gt;Here is how the first three words&#xA;  might be chosen:&lt;/p&gt;&#xA;  &#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;&lt;p&gt;A two-word sequence is chosen at random to become the initial prefix.&#xA;  Let's suppose that &quot;the big&quot; is&#xA;  chosen.&lt;/p&gt;&lt;/li&gt;&#xA;  &lt;li&gt;&lt;p&gt;The first word must be chosen based on the probability that it&#xA;  follows the prefix (currently &quot;the&#xA;  big&quot;) in the source. The source&#xA;  contains three occurrences of &quot;the&#xA;  big&quot;. Two times it is followed by&#xA;  &quot;blue&quot;, and once it is followed by&#xA;  &quot;elephant&quot;. Thus, the next word must&#xA;  be chosen so that there is a 2/3&#xA;  chance that &quot;blue&quot; will be chosen, and&#xA;  a 1/3 chance that &quot;elephant&quot; will be&#xA;  chosen. Let's suppose that we choose&#xA;  &quot;blue&quot; this time.&lt;/p&gt;&lt;/li&gt;&#xA;  &lt;li&gt;&lt;p&gt;The next word must be chosen based on the probability that it&#xA;  follows the prefix (currently &quot;big&#xA;  blue&quot;) in the source. The source&#xA;  contains two occurrences of &quot;big&#xA;  blue&quot;. Once it is followed by &quot;dog&quot;,&#xA;  and the other time it is followed by&#xA;  &quot;hat&quot;. Thus, the next word must be&#xA;  chosen so that there is a 50-50&#xA;  probability of choosing &quot;dog&quot; vs.&#xA;  &quot;hat&quot;. Let's suppose that we choose&#xA;  &quot;hat&quot; this time.&lt;/p&gt;&lt;/li&gt;&#xA;  &lt;li&gt;&lt;p&gt;The next word must be chosen based on the probability that it&#xA;  follows the prefix (currently &quot;blue&#xA;  hat&quot;) in the source. The source&#xA;  contains only one occurrence of &quot;blue&#xA;  hat&quot;, and it is followed by &quot;on&quot;.&#xA;  Thus, the next word must be &quot;on&quot; (100%&#xA;  probability).&lt;/p&gt;&lt;/li&gt;&#xA;  &lt;li&gt;&lt;p&gt;Thus, the first three words in the output text would be &quot;blue hat&#xA;  on&quot;.&lt;/p&gt;&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You keep going like that, generating text that is completely nonsensical, but ends up having sort of the same &quot;tone&quot; as the original text. For example, if your sample file is the complete text of Alice In Wonderland (one of the texts we tried it on) then your nonsense comes out kind of whimsical and Carrollian (if that's a word). If your sample file is The Telltale Heart, you get somewhat dark, morbid nonsense.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, while not a rigorous, formal, definition, I hope this helps give you a sense of what a Markov chain is.&lt;/p&gt;&#xA;" OwnerUserId="496" LastActivityDate="2010-07-31T05:45:16.580" CommentCount="1" />
  <row Id="1287" PostTypeId="2" ParentId="1285" CreationDate="2010-07-31T06:04:23.997" Score="4" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Method 1&lt;/strong&gt;: We can perform an &lt;a href=&quot;http://en.wikipedia.org/wiki/Affine_transformation&quot; rel=&quot;nofollow&quot;&gt;affine transformation&lt;/a&gt; to make ADC any triangle we wish, while preserving colinearality and the ratios of areas. In particular, this allows us to assign exact coordinates to A, D and C. We can then use the ratio of x and y to find E and the ratio of z and x+y+z to find P. We can then use these coordinates to find B. Once we know B, we know the ratio of BC to DC and hence the ratio of the area of the whole triangle to the known areas. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Method 2&lt;/strong&gt;: Let Q be the area of the whole triangle&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;AE:EC=x:y=|AEB|:|BEC|&#xA;|APB|=xQ/(x+y)-x&#xA;|BPD|=yQ/(x+y)-y-z&#xA;Now AP:PD=x+y:z&#xA;z(xQ/(x+y)-x)=(x+y)(yQ/(x+y)-y-z)&#xA;Q(zx/(x+y)-y)=xz-xy-y^2-xz-yz=-xy-y^2-xz&#xA;Q(zx-xy-y^2)/(x+y)=(-xy-y^2-xz)&#xA;Q=(-xy-y^2-xz)*(x+y)/(zx-xy-y^2)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="123" LastEditorUserId="123" LastEditorDisplayName="" LastEditDate="2010-07-31T06:39:12.070" LastActivityDate="2010-07-31T06:39:12.070" CommentCount="4" />
  <row Id="1288" PostTypeId="2" ParentId="1285" CreationDate="2010-07-31T06:21:32.323" Score="4" ViewCount="" Body="&lt;p&gt;This was not likely to have been an SAT practice problem, though it is a typical contest problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;AE:EC = x:y (since those two triangles have the same altitude to AC, the ratios of their areas is the ratios of their bases with respect to that altitude) and AP:PD = (x+y):z (same idea as AE:EC).  Knowing these two ratios, apply the technique of &lt;a href=&quot;http://en.wikipedia.org/wiki/Mass_points&quot; rel=&quot;nofollow&quot;&gt;mass points&lt;/a&gt;, putting masses zy at A, zx at C (gives the ratio x:y for AE:EC), and y(x+y) at D (gives (x+y):z for AP:PD).  This results in a mass of y(x+y)-zx at B, so the ratio BD:DC = zx:(y(x+y)-zx).  This must also be the ratio of the areas of &amp;#x25B3;ABD to &amp;#x25B3;ADC (common altitude again), so (area of &amp;#x25B3;ABD):(x+y+z) = zx:(y(x+y)-zx).  Solving from there is a matter of bashing out the algebra.&lt;/p&gt;&#xA;" OwnerUserId="72" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-07-31T06:40:19.400" LastActivityDate="2010-07-31T06:40:19.400" CommentCount="2" />
  <row Id="1289" PostTypeId="2" ParentId="1279" CreationDate="2010-07-31T06:27:29.753" Score="3" ViewCount="" Body="&lt;p&gt;Here is an article at Mathworld on &lt;a href=&quot;http://mathworld.wolfram.com/Circle-CircleTangents.html&quot; rel=&quot;nofollow&quot;&gt;Circle-Circle Tangents&lt;/a&gt;. Perhaps &quot;CirclesWithTangents&quot;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://mathworld.wolfram.com/images/eps-gif/CircleCircleTangentGeneral_1000.gif&quot; alt=&quot;Circle-Circle Tangents&quot; title=&quot;Circle-Circle Tangents&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="208" LastActivityDate="2010-07-31T06:27:29.753" CommentCount="1" />
  <row Id="1290" PostTypeId="2" ParentId="164" CreationDate="2010-07-31T06:38:13.307" Score="2" ViewCount="" Body="&lt;p&gt;To expand on what anosov_diffeomorphism said:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose we agree that the surface area of a sphere is &lt;strong&gt;4 &amp;pi; r&lt;sup&gt;2&lt;/sup&gt;&lt;/strong&gt; (If we do not agree on this, that would be a great question to ask on this site, and then someone can edit this answer to link to it!). Now if I have a sphere of radius &lt;em&gt;r&lt;/em&gt;, and I increase the radius by a tiny amount, &lt;em&gt;dr&lt;/em&gt;, then the new, expanded sphere has a volume that is bigger, by the volume of the thin spherical shell that was just added. If &lt;em&gt;dr&lt;/em&gt; is small enough, then it doesn't much matter whether we look at the inner or outer radius of that shell, and we can just say that the shell's volume is approximately &lt;strong&gt;4 &amp;pi; r&lt;sup&gt;2&lt;/sup&gt; dr&lt;/strong&gt;. So the difference in volume between two spheres of radius $r_1$ and $r_2$ is $\int_{r_1}^{r_2}$ 4 &amp;pi; r&lt;sup&gt;2&lt;/sup&gt; dr, which is 4/3 &amp;pi; r^3&lt;/p&gt;&#xA;" OwnerUserId="496" LastActivityDate="2010-07-31T06:38:13.307" />
  <row Id="1291" PostTypeId="2" ParentId="1237" CreationDate="2010-07-31T06:48:48.637" Score="4" ViewCount="" Body="&lt;p&gt;One that I remember from some book (I think it was Innumeracy by John Allen Paulos) was &quot;How fast does your hair grow, in miles per hour?&quot;&lt;/p&gt;&#xA;" OwnerUserId="496" LastActivityDate="2010-07-31T06:48:48.637" CommunityOwnedDate="2010-07-31T06:48:48.637" CommentCount="1" />
  <row Id="1292" PostTypeId="1" AcceptedAnswerId="1293" CreationDate="2010-07-31T07:09:16.437" Score="14" ViewCount="914" Body="&lt;p&gt;The &quot;&lt;a href=&quot;http://en.wikipedia.org/wiki/Angle_addition_formula#Angle_sum_and_difference_identities&quot; rel=&quot;nofollow&quot;&gt;sum and difference&lt;/a&gt;&quot; formulas often come in handy, but it's not immediately obvious that they would be true.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;\sin(\alpha \pm \beta) &amp;amp;= \sin \alpha \cos \beta \pm \cos \alpha \sin \beta \\&#xA;\cos(\alpha \pm \beta) &amp;amp;= \cos \alpha \cos \beta \mp \sin \alpha \sin \beta&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what I want to know is, &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;How can I prove that these formulas are correct?&lt;/li&gt;&#xA;&lt;li&gt;More importantly, how can I understand these formulas intuitively?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Ideally, I'm looking for answers that make no reference to Calculus, or to &lt;a href=&quot;http://en.wikipedia.org/wiki/Euler%27s_formula&quot; rel=&quot;nofollow&quot;&gt;Euler's formula&lt;/a&gt;, although such answers are still encouraged, for completeness.&lt;/p&gt;&#xA;" OwnerUserId="496" LastEditorUserId="171" LastEditorDisplayName="" LastEditDate="2010-08-13T12:35:28.923" LastActivityDate="2010-08-13T12:35:28.923" Title="How can I understand and prove the &quot;sum and difference formulas&quot; in trigonometry? (cos(a ± b) = ..., etc.)?" Tags="&lt;intuition&gt;&lt;trigonometry&gt;" AnswerCount="6" CommentCount="0" FavoriteCount="11" />
  <row Id="1293" PostTypeId="2" ParentId="1292" CreationDate="2010-07-31T07:19:26.117" Score="10" ViewCount="" Body="&lt;p&gt;The key fact here is that rotation is a linear transformation, e.g. the rotation of $u + v$ is the rotation of $u$ plus the rotation of $v$.  You should draw a diagram that shows this carefully if you don't believe it.  That means a rotation is determined by what it does to $(1, 0)$ and to $(0, 1)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But $(1, 0)$ rotated by $\theta$ degrees counterclockwise is just $(\cos \theta, \sin \theta)$, whereas $(0, 1)$ rotated by $\theta$ degrees counterclockwise is just $(-\sin \theta, \cos \theta)$.  (Again, draw a diagram.)  That means a rotation by $\theta$ is given by a $2 \times 2$ matrix with those entries.  (Matrices don't work here yet.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So take a rotation by $\theta$ and another one by $\theta'$, and multiply the corresponding matrices.  What you get is the sine and cosine angle addition formulas.  (The connection to complex numbers is that one can represent complex numbers as $2 \times 2$ real matrices.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, if you believe that $a \cdot b = |a| |b| \cos \theta$, this implies the cosine angle difference formula when $a$ and $b$ are unit vectors.  Ditto for the cross product and the sine angle difference formula.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-07-31T07:19:26.117" />
  <row Id="1294" PostTypeId="2" ParentId="1292" CreationDate="2010-07-31T07:51:38.223" Score="7" ViewCount="" Body="&lt;p&gt;There are several typical derivations used in high school texts.  Here's one:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://www.imgftw.net/img/400545892.png&quot; alt=&quot;diagram&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take two points on the unit circle, one a rotation of (1,0) by &amp;alpha;, the other a rotation of (1,0) by &amp;beta;.  Their coordinates are as shown in the diagram.  Let c be the length of the segment joining those two points.  By the Law of Cosines (on the blue triangle), $c^2=1^2+1^2-2\cdot1\cdot1\cdot\cos(\alpha-\beta)$.  Using the distance formula, $c=\sqrt{(\cos\alpha-\cos\beta)^2+(\sin\alpha-\sin\beta)^2}$.  Squaring the latter and setting the two equal, $1^2+1^2-2\cdot1\cdot1\cdot\cos(\alpha-\beta)=(\cos\alpha-\cos\beta)^2+(\sin\alpha-\sin\beta)^2$.  Simplifying both sides, $2-2\cos(\alpha-\beta)=\cos^2\alpha-2\cos\alpha\cos\beta+\cos^2\beta+\sin^2\alpha-2\sin\alpha\sin\beta+\sin^2\beta$ $=2-2\cos\alpha\cos\beta-2\sin\alpha\sin\beta$ (using the Pythagorean identity).  Solving for $\cos(\alpha-\beta)$, $\cos(\alpha-\beta)=\cos\alpha\cos\beta+\sin\alpha\sin\beta$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From this identity, the other three can be derived by substituting $\frac{\pi}{2}-\alpha$ for &amp;alpha; (gives sin(&amp;alpha;+&amp;beta;)), then -&amp;beta; for &amp;beta; (gives the remaining two).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As to understanding the formulas intuitively, if you accept that multiplying by a complex number $z_\theta$ for which |z|=1 rotates by &amp;theta;, then you can think about what happens when you multiply $z_\alpha=\cos\alpha+i\sin\alpha$ and $z_\beta=\cos\beta+i\sin\beta$ (by expanding the binomial product), which should result in $\cos(\alpha+\beta)+i\sin(\alpha+\beta)$.&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-07-31T07:51:38.223" CommentCount="6" />
  <row Id="1295" PostTypeId="2" ParentId="1237" CreationDate="2010-07-31T08:16:32.233" Score="0" ViewCount="" Body="&lt;p&gt;What is the mass of the atmosphere?&lt;/p&gt;&#xA;" OwnerUserId="453" LastActivityDate="2010-07-31T08:16:32.233" CommunityOwnedDate="2010-07-31T08:16:32.233" />
  <row Id="1296" PostTypeId="2" ParentId="405" CreationDate="2010-07-31T08:55:15.557" Score="3" ViewCount="" Body="&lt;p&gt;(Sorry for the LaTeX, the symbols &lt;code&gt;\mapsto&lt;/code&gt; and &lt;code&gt;\{&lt;/code&gt; and &lt;code&gt;\}&lt;/code&gt; don't work.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Monads in Haskell and monads in category theory are very much the same: A &lt;em&gt;monad&lt;/em&gt; consists of a functor $T: C \to C$ and two natural transformations $\eta_X : X \to T(X)$ (&lt;code&gt;return&lt;/code&gt; in Haskell) and $\mu_X : T(T(X)) \to T(X)$ (&lt;code&gt;join&lt;/code&gt; in Haskell) subject to the following laws&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\mu_X \circ T(\eta_X) = \mu_X \circ \eta_{T(X)} = 1_{T(X)}$ (left and right unit laws)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\mu_X \circ \mu_{T(X)} = \mu_X \circ T(\mu_X)$ (associativity)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, compared to Haskell, the monad is defined in terms of &lt;code&gt;return&lt;/code&gt;, &lt;code&gt;join&lt;/code&gt; and &lt;code&gt;fmap&lt;/code&gt; instead of &lt;code&gt;return&lt;/code&gt; and &lt;code&gt;(&amp;gt;&amp;gt;=)&lt;/code&gt;. For more details on this, see also the &lt;a href=&quot;http://en.wikibooks.org/wiki/Haskell/Category_theory#Monads&quot; rel=&quot;nofollow&quot;&gt;Haskell wikibook&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Two examples may illuminate this definition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;strong&gt;powerset functor&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\mathcal{P} = X \mapsto \mathcal{P}(X)$ maps a set to the set of its subsets.&lt;/li&gt;&#xA;&lt;li&gt;Functions $f:X \to Y$ are extended point-wise to $\mathcal{P}(f):\mathcal{P}(X) \to \mathcal{P}(Y)$&lt;/li&gt;&#xA;&lt;li&gt;$\eta_X : X \to \mathcal{P}(X)$ is the function $x \mapsto {x}$&lt;/li&gt;&#xA;&lt;li&gt;$\mu_X : \mathcal{P}(\mathcal{P}(X)) \to \mathcal{P}(X)$ flattens the inner layer of subsets: $\mu_X(A) = { b | a \in A, b \in a }$.&lt;/li&gt;&#xA;&lt;li&gt;In Haskell, this is known as the &lt;em&gt;list monad&lt;/em&gt;.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The &lt;strong&gt;closure&lt;/strong&gt; operation on the subsets of a topological space $S$ is a monad, too.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The objects of the category $C$ are the subsets of a given topological space $S$.&lt;/li&gt;&#xA;&lt;li&gt;There is an unique arrow $X \to Y$ between to objects $X$ and $Y$ exactly when $X \subseteq Y$.&lt;/li&gt;&#xA;&lt;li&gt;The monad is given by the functor that maps each object $X$ to its topological closure $\bar X$ and the arrow $X \subseteq Y$ to the arrow $\bar{X}\subseteq \bar{Y}$.&lt;/li&gt;&#xA;&lt;li&gt;Clearly, we have $X \subseteq \bar X$; this is $\eta_X$.&lt;/li&gt;&#xA;&lt;li&gt;Also, we know that $\bar{\bar X} = \bar X$, in particular $\bar{\bar X} \subseteq \bar X$; this is $\mu_X$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="434" LastEditorUserId="434" LastEditorDisplayName="" LastEditDate="2010-07-31T09:01:47.167" LastActivityDate="2010-07-31T09:01:47.167" />
  <row Id="1297" PostTypeId="2" ParentId="1292" CreationDate="2010-07-31T09:24:09.130" Score="8" ViewCount="" Body="&lt;p&gt;You can use the &lt;a href=&quot;http://en.wikipedia.org/wiki/Cosine#Relationship_to_exponential_function_and_complex_numbers&quot; rel=&quot;nofollow&quot;&gt;complex representation,&lt;/a&gt;&lt;br&gt;&#xA;$\cos(x) = \frac{1}{2}(e^{ix} + e^{-ix})$&lt;br&gt;&#xA;$\sin(x) = \frac{1}{2i}(e^{ix} - e^{-ix})$&lt;br&gt;&#xA;and the rules for powers ($a^{x+y}=a^x a^y$)&lt;/p&gt;&#xA;" OwnerUserId="163" LastActivityDate="2010-07-31T09:24:09.130" CommentCount="1" />
  <row Id="1298" PostTypeId="2" ParentId="1292" CreationDate="2010-07-31T11:52:37.290" Score="5" ViewCount="" Body="&lt;p&gt;Though the standard high-school derivations are not the most useful way to remember it in the long run, here's another one which I like because you can &quot;see&quot; it directly without much algebra.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://imgur.com/QnIhq.png&quot; alt=&quot;Angle sum formulae&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let P be the point on the unit circle got by rotating (1,0) by angle α+β. Drop a perpendicular N to the α-rotated line, and R to the x-axis. So from the right triangle ONP, you see &lt;strong&gt;ON = cos &amp;beta;&lt;/strong&gt;. You can see that the angle RPN is &amp;alpha; too: it's the complement of ∠PNQ, and so is ∠QNO = &amp;alpha;. Now,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sin(\alpha + \beta) = \mbox{PR} = \mbox{PQ} + \mbox{QR} = \sin(\beta)\cos(\alpha) + \cos(\beta)\sin(\alpha)$, and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\cos(\alpha + \beta) = \mbox{OR} = \mbox{OM} - \mbox{RM} = \cos(\beta)\cos(\alpha) - \sin(\beta)\sin(\alpha)$.&lt;/p&gt;&#xA;" OwnerUserId="205" LastActivityDate="2010-07-31T11:52:37.290" />
  <row Id="1299" PostTypeId="2" ParentId="1292" CreationDate="2010-07-31T11:56:33.290" Score="4" ViewCount="" Body="&lt;p&gt;I remember that $e^{i\alpha}=\cos\alpha+i\sin\alpha$ and that $i^2=-1$. Both these relations are useful in many other situations and pretty fundamental to understanding complex numbers. Then your equalities are the real and, respectively, the imaginary part of $e^{i(\alpha+\beta)}=e^{i\alpha}e^{i\beta}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is not very different from the other answers, but I actually prefer the algebra perspective. The only place where I think geometrically is in interpreting $e^{i\alpha}=\cos\alpha+i\sin\alpha$ by thinking of the unit circle in the complex plane.&lt;/p&gt;&#xA;" OwnerUserId="512" LastEditorUserId="512" LastEditorDisplayName="" LastEditDate="2010-07-31T20:37:06.617" LastActivityDate="2010-07-31T20:37:06.617" CommentCount="3" />
  <row Id="1300" PostTypeId="2" ParentId="1234" CreationDate="2010-07-31T12:15:45.357" Score="1" ViewCount="" Body="&lt;p&gt;Here's a polynomial time solution given to me by Javi Gomez. Let (i,j) with $i\le j$ represent the situation where disks i, i+1, ..., j are on top of each other in position $x_j$, and let E(i,j) represent the energy needed to obtain that configuration. Clearly E(i,i) is zero for all i. Also, $E(i,j)=\min_k E(i,k)+E(k,j)+|x_k-x_j|$. What we want is E(1,n).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I made this a community wiki since the answer wasn't really found by me. Feel free to edit.)&lt;/p&gt;&#xA;" OwnerUserId="512" LastActivityDate="2010-07-31T12:15:45.357" CommunityOwnedDate="2010-07-31T12:15:45.357" />
  <row Id="1302" PostTypeId="1" AcceptedAnswerId="1305" CreationDate="2010-07-31T13:33:51.307" Score="3" ViewCount="274" Body="&lt;p&gt;Following my &lt;a href=&quot;http://math.stackexchange.com/questions/1279/what-is-the-name-for-a-shape-that-is-like-a-capsule-but-with-two-different-radii&quot;&gt;previous question&lt;/a&gt;, I'm wondering how I can determine if a point is within the convex hull of two circles (a boolean value). There's no problem testing if the point is in either of the two circles, but it can also be &quot;between&quot; them and I don't know how to test for that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Seeing Wolfram MathWorld's article on &lt;a href=&quot;http://mathworld.wolfram.com/Circle-CircleTangents.html&quot; rel=&quot;nofollow&quot;&gt;Circle-Circle Tangeants&lt;/a&gt;, it seems that an inequation that tests if the point is on the internal side the two external circle tangeants would do the trick, but I'm afraid my solving skills are too far away to be able to turn the tangeant equations into a fitting inequality.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm defining the convex hull of two circles using both centers and radii.&lt;/p&gt;&#xA;" OwnerUserId="529" LastActivityDate="2010-08-01T09:56:16.397" Title="How to test if a point is inside the convex hull of two circles?" Tags="&lt;geometry&gt;" AnswerCount="2" />
  <row Id="1303" PostTypeId="2" ParentId="1302" CreationDate="2010-07-31T13:49:21.940" Score="1" ViewCount="" Body="&lt;p&gt;The solution to this problem is indeed to check whether the point is in one of the circles or in the &lt;a href=&quot;http://en.wikipedia.org/wiki/Isosceles_trapezoid&quot; rel=&quot;nofollow&quot;&gt;isosceles trapezoid&lt;/a&gt; determined by the points were circles touch the tangent lines. However, without some care the equations get messy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's start with a circle at (x1', y1') of radius r1', a circle at (x2', y2') with radius r2', and a point (x', y') to test. We may assume that r1' is at least r2'. By shifting followed by rotation followed by scaling this can be transformed into a simpler instance: a circle at (0,0) of radius r1, a circle at (1,0) of radius r2, and a point (x,y) to test. We have: r1=r1'/D, r2=r2'/D, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x=((y'-y'_1)\sin\alpha+(x'-x'_1)\cos\alpha)/D$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$y=((y'-y'_1)\sin\alpha-(x'-x'_1)\cos\alpha)/D$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$D^2=(x'_1-x'_2)^2+(y'_1-y'_2)^2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and $\alpha$ is the angle where (x2'-x1',y2'-y1') lies. (In C, there's a function atan2 that takes the two coordinates of a point and gives the angle. In math, atan doesn't really distinguish between points symmetric wrt (0,0).)&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;If the point is in one of the circles say YES, otherwise continue.&lt;/li&gt;&#xA;&lt;li&gt;If r1=r2 then return (0 &amp;lt; x &amp;lt; 1 and |y| &amp;lt; r1), otherwise continue. (At this point the difference between radii Dr=r1-r2 is strictly between 0 and 1.)&lt;/li&gt;&#xA;&lt;li&gt;If x &amp;lt; r1 Dr, say NO.&lt;/li&gt;&#xA;&lt;li&gt;If x &gt; 1 + r2 Dr, say NO.&lt;/li&gt;&#xA;&lt;li&gt;If $x\Delta r + |y|\sqrt{1-{\Delta r}^2}&gt;r_1$, say NO.&lt;/li&gt;&#xA;&lt;li&gt;Say YES.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;There's only one thing I used, really, namely that certain triangles are similar. Draw the tangent that touches the two circles 'above'. Let's say that it intersects the big circle at A=(x1,y1), the small circle at B=(x2,y2), and the horizontal axis at C=(d,0). Let's also write D=(0,0) and E=(x1,0) and F=(1,0) and G=(x2,0). The similar triangles are ADC, AED, BFC, and BGD. From ADC similar to BFC you find d = r1/Dr. From ADC similar to AED you find x1 = r1 Dr and from BFD similar to BGD you find x2 - 1 = r2 Dr.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To check that a point is on the right side of a line write its equation as ax+by+c=0 and replace equality by inequality. (It's good to remember that the vector $(a,b)^T$ is perpendicular to the line and that the equation can be interpreted as defining all vectors/points whose scalar/dot product with a fixed vector $(a,b)^T$ gives the constant c.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me know if I messed up again :P&lt;/p&gt;&#xA;&#xA;&lt;hr/&gt;&#xA;&#xA;&lt;p&gt;This is the old answer, which is wrong as pointed by the first comment below. It starts with the same transformations (without giving details) but then it uses some line that is clearly &lt;em&gt;not&lt;/em&gt; tangent to the circles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Say the first circle has the center at (0,0) (if not, shift the figure) with radius r_1, and the second at (1,0) (if not, rotate the figure and scale) with radius r_2. If your point is (x,y) and is not in one of the two circles, then x must be between 0 and 1 and |y| must be at most r_1+(r_2-r_1)x.&lt;/p&gt;&#xA;" OwnerUserId="512" LastEditorUserId="512" LastEditorDisplayName="" LastEditDate="2010-08-01T09:56:16.397" LastActivityDate="2010-08-01T09:56:16.397" CommentCount="3" />
  <row Id="1304" PostTypeId="2" ParentId="1260" CreationDate="2010-07-31T14:04:11.890" Score="1" ViewCount="" Body="&lt;p&gt;You can try with &lt;a href=&quot;http://mathurl.com/&quot; rel=&quot;nofollow&quot;&gt;mathurl&lt;/a&gt;, which allows you to write in LaTeX and copy&amp;amp;paste everywhere.&lt;/p&gt;&#xA;" OwnerUserId="354" LastActivityDate="2010-07-31T14:04:11.890" />
  <row Id="1305" PostTypeId="2" ParentId="1302" CreationDate="2010-07-31T14:40:43.003" Score="8" ViewCount="" Body="&lt;p&gt;You can first calculate the &lt;a href=&quot;http://en.wikipedia.org/wiki/Tangent_lines_to_circles&quot; rel=&quot;nofollow&quot;&gt;four points of tangency&lt;/a&gt;, then use the &lt;a href=&quot;http://en.wikipedia.org/wiki/Point_in_polygon&quot; rel=&quot;nofollow&quot;&gt;Point in Polygon&lt;/a&gt; algorithm to determine if your point is inside the quadrilateral (this is assuming you have a programming related problem)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Otherwise, once you have found the four tangent points, you can form four lines and get four simultaneous inequalities.&lt;/p&gt;&#xA;" OwnerUserId="215" LastActivityDate="2010-07-31T14:40:43.003" CommentCount="1" />
  <row Id="1306" PostTypeId="2" ParentId="77" CreationDate="2010-07-31T15:51:53.060" Score="2" ViewCount="" Body="&lt;p&gt;Two more uses that I haven't seen mentioned yet:  if you want to find the area of the parallelogram formed by two vectors (each vector gives a pair of parallel sides), then you would use the magnitude of the cross product of the two vectors.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One use of this is to aid in defining a surface integral.  Let x(u,v) be a parameterization of a surface.  Then at each point, we can find tangent vectors T&lt;sub&gt;u&lt;/sub&gt; =  &amp;part;x/&amp;part;u and T&lt;sub&gt;v&lt;/sub&gt; = &amp;part;x/&amp;part;v.  From the idea of a linear approximation, T&lt;sub&gt;u&lt;/sub&gt; and T&lt;sub&gt;v&lt;/sub&gt; will define a tangent plane at that particular point.  Consider the parallelogram formed with T&lt;sub&gt;u&lt;/sub&gt; and T&lt;sub&gt;v&lt;/sub&gt; as sides.  Informally, we can see that each area element will be |T&lt;sub&gt;u&lt;/sub&gt; x T&lt;sub&gt;v&lt;/sub&gt;| du dv.  Then a function f(u,v) integrated over this surface is &amp;int;&amp;int; f(u,v) |T&lt;sub&gt;u&lt;/sub&gt; x T&lt;sub&gt;v&lt;/sub&gt;| du dv.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the second use, if you wanted to find the volume of the parallelepiped having the three vectors a, b, c as sides, then you would use the magnitude of the scalar triple product |a &amp;sdot; (b x c)|. &lt;/p&gt;&#xA;" OwnerUserId="381" LastActivityDate="2010-07-31T15:51:53.060" />
  <row Id="1307" PostTypeId="1" AcceptedAnswerId="1309" CreationDate="2010-07-31T16:51:50.433" Score="5" ViewCount="122" Body="&lt;p&gt;Let $A$ be a valuation ring of rank two. Then $A$ gives an example of a commutative ring such that $Spec(A)$ is a noetherian topological space, but $A$ is nonnoetherian. (Indeed, otherwise $A$ would be a discrete valuation ring.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a concrete example of such a ring $A$? &lt;/p&gt;&#xA;" OwnerUserId="537" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-07-31T18:17:31.513" LastActivityDate="2010-11-04T00:37:46.830" Title="Valuation rings of rank two" Tags="&lt;abstract-algebra&gt;&lt;commutative-algebra&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" />
  <row Id="1308" PostTypeId="2" ParentId="1307" CreationDate="2010-07-31T17:28:30.750" Score="1" ViewCount="" Body="&lt;p&gt;Take the ring of formal power series (over $\mathbb{C}$, say) with exponents in $\mathbb{Z}^2$ under lex order.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit:  As Robin Chapman mentions, one must be careful about what this means.  The precise construction for any totally ordered abelian group is described &lt;a href=&quot;http://en.wikipedia.org/wiki/Valuation_ring#Construction&quot; rel=&quot;nofollow&quot;&gt;in the Wikipedia article&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-07-31T20:25:21.323" LastActivityDate="2010-07-31T20:25:21.323" />
  <row Id="1309" PostTypeId="2" ParentId="1307" CreationDate="2010-07-31T18:14:22.037" Score="8" ViewCount="" Body="&lt;p&gt;Qiaochu's answer is sound in principle, but in practice one needs to be more careful&#xA;with the definition of the ring. The quotient field $K$ of $A$ consists of&#xA;formal Laurent series of the form&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f=\sum_{r=-r_0}^\infty x^r\sum_{s=-s_0(r)}^\infty a_{r,s}y^s.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here $r_0$ is an integer and for each integer $r$, $s_0(r)$ is an integer&#xA;(depending on $r$).&#xA;So these are the power series where the powers of $x$ are bounded below&#xA;and for each integer $r$ the coefficient of $x^r y^s$ is zero for all&#xA;$s$ below a bound depending on $r$. This complicated-looking condition&#xA;ensures that the product of two elements of $K$ is also an element&#xA;of $K$ (note that one cannot multiply two general Laurent series).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then $A$ will consist of all such series with the additional conditions&#xA;that $r_0=0$ and $s_0(0)=0$. The valuation of an element $f$ is&#xA;the least $(r,s)$ under lexicographic ordering with $a_{r,s}\ne0$.&#xA;Here the ordering is $(r,s)&amp;lt;(r',s')$ if $r &amp;lt; r'$ or $r=r'$ and $s &amp;lt; s'$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A more high-brow interpretation of the condition for memebership of&#xA;$K$ is that the &lt;em&gt;support&lt;/em&gt; of $f$, the setof $(r,s)$ for which $a_{r,s}\ne0$,&#xA;should be well-ordered, that is each subset of the support has a least&#xA;element. (With repsct to this lexicographic ordering of course.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By considering a version of this construction in $n$ variables&#xA;one can construct explicitly a ring with a valuation of rank $n$.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-07-31T18:14:22.037" CommentCount="1" />
  <row Id="1310" PostTypeId="2" ParentId="803" CreationDate="2010-07-31T18:38:40.893" Score="10" ViewCount="" Body="&lt;p&gt;More than dissection proofs, I find the proof using similarity most enlightening and intuitive. See &lt;a href=&quot;http://terrytao.wordpress.com/2007/09/14/pythagoras-theorem/&quot; rel=&quot;nofollow&quot;&gt;post by Terence Tao&lt;/a&gt;: drop a perpendicular from C to the hypotenuse AB.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://www.math.ucla.edu/~tao/pythagoras.jpeg&quot; alt=&quot;Image from Tao&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the figure, the triangles whose are areas are marked x and y are similar to the original triangle (which has area x+y). So accepting that areas of similar right-angled triangles are proportional to the squares of the hypotenuse, x:y:x+y are in ratio a&lt;sup&gt;2&lt;/sup&gt;:b&lt;sup&gt;2&lt;/sup&gt;:c&lt;sup&gt;2&lt;/sup&gt;, which is Pythagoras's theorem.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;See also the linked discussion &lt;a href=&quot;http://math.berkeley.edu/~giventh/papers/eu.pdf&quot; rel=&quot;nofollow&quot;&gt;by Alexander Giventhal&lt;/a&gt; where he remarks that this proof is more general than tiling or dissection proofs, and is even proved by Euclid. If you think of a&lt;sup&gt;2&lt;/sup&gt;+b&lt;sup&gt;2&lt;/sup&gt;=c&lt;sup&gt;2&lt;/sup&gt; as the geometrical result that the sum of areas of squares constructed with sides a and b is the area of a square placed on c, then the Pythagorean theorem is true not just for constructing squares on the sides, but &lt;em&gt;any&lt;/em&gt; similar figures. For instance, as Euclid himself proves, something like the following is true (though it's still just saying a&lt;sup&gt;2&lt;/sup&gt;+b&lt;sup&gt;2&lt;/sup&gt;=c&lt;sup&gt;2&lt;/sup&gt;):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://imgur.com/OpDUA.png&quot; alt=&quot;From Giventhal&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="205" LastActivityDate="2010-07-31T18:38:40.893" CommunityOwnedDate="2010-07-31T18:38:40.893" CommentCount="3" />
  <row Id="1311" PostTypeId="1" AcceptedAnswerId="1312" CreationDate="2010-07-31T19:49:24.733" Score="5" ViewCount="598" Body="&lt;p&gt;I've been told that there are precisely the same number of rationals as there are of integers. The set of rationals is countably infinite, therefore every rational can be associated with a positive integer, therefore there are the same number of rationals as integers. I've ignored sign-related issues, but these are easily handled.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To count the rationals, consider sets of rationals where the denominator and numerator are positive and sum to some constant. If the constant is 2 there's 1/1. If the constant is 3, there's 1/2 and 2/1. If the constant is 4 there's 1/3, 2/2 and 3/1. So far we have counted out 6 rationals, and if we continue long enough, we will eventually count to any specific rational you care to mention.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The trouble is, I find this very hard to accept. I have two reasons. First, this logic seems to assume that infinity is a finite number. You can count to and number any rational, but you cannot number all rationals. You can't even count all positive integers. Infinity is code for &quot;no matter how far you count, you have never counted enough&quot;. If it were possible to count to infinity, it would be possible to count one step less and stop at count infinity-1 which must be different to infinity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second reason is that it's very easy to construct alternative mappings. Between zero and one there are infinitely many rational numbers, between one and two there are infinitely many rational numbers, and so on. To me, this seems a much more reasonable approach, implying that there are infinite rational numbers for every integer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But even then, this is just one of many alternative ways to map between ranges of rationals and ranges of integers. Since you can count the rationals, you can equally count stepping by any amount for each rational. You can use 1..10 for the first rational and 11..20 for the second etc. Or 1..100 and 101..200 etc, or 1..1000 and 1001..2000 etc. You can map finite range of integers of any size to each rational this way and, since there is no finite upper bound to the stepping amount, you could argue there are potentially infinite integers for every single rational.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So... can anyone convince me that there is a single unambiguous correct answer to this question? Are there more rational numbers than integers, or not?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Although I've already accepted an answer, I'll just add some extra context.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My reason for questioning this relates to the Hilbert space-filling curve. I find this interesting because of applications to multi-dimensional indexing data structures in software. However, I found Hilberts claim that the Hilbert curve literally filled a multi-dimensional space hard to accept.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As mentioned in a comment below, a one meter line segment and a two meter line segment can both be seen as sets of points and, but (by the logic in answers below), those two sets are both the same size (cardinality). Yet we would not claim the two line segments are both the same size. The lengths are finite and different. Going beyond this, we most certainly wouldn't claim that the size of &lt;em&gt;any&lt;/em&gt; finite straight line segment is equal to the size of a one-meter-by-one-meter square.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Hilbert curve reasoning makes sense now - the set of points in the curve is equal to the set of points in the space it fills. Previously, I was thinking too much about basic geometry, and couldn't accept the size of a curve as being equal to the size of a space. However, this isn't based on a fallacious counting-to-infinity argument - it's a necessary consequence of an alternative line of reasoning. The two constructs are equal because they both represent the same set of points. The area/volume/etc of the curve follows from that.&lt;/p&gt;&#xA;" OwnerUserId="510" LastEditorUserId="510" LastEditorDisplayName="" LastEditDate="2010-07-31T21:22:29.987" LastActivityDate="2010-07-31T21:22:29.987" Title="Are there more rational numbers than integers?" Tags="&lt;set-theory&gt;&lt;infinity&gt;" AnswerCount="5" CommentCount="4" FavoriteCount="2" />
  <row Id="1312" PostTypeId="2" ParentId="1311" CreationDate="2010-07-31T20:01:41.193" Score="11" ViewCount="" Body="&lt;p&gt;Mathematicians have very precise definitions for terms like &quot;infinite&quot; and &quot;same size&quot;.  The single unambiguous correct answer to this question is that using the standard mathematical definitions, the rationals have the &quot;same size&quot; as the integers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, here are the definitions:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Define &quot;0&quot; = emptyset, &quot;1&quot; = {0}, &quot;2&quot; = {0,1}, &quot;3&quot; = {0,1,2}, etc.  So, the number &quot;n&quot; is really a set with &quot;n&quot; elements in it.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;A set A is called &quot;finite&quot; iff there is some n and a function f:A-&gt;n which is bijective.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;A set A is called &quot;infinite&quot; iff it is not finite.  (Note that this notion says nothing about &quot;counting never stops&quot; or anything like that.)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Two sets A and B are said to have the &quot;same size&quot; if there is a some function f:A-&gt; B which is a bijection.  Note that we do NOT require that ALL functions be bijections, just that there is SOME bijection.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Once one accepts these definitions, one can prove that the rationals and integers have the same size.  One just needs to find a particular bijection between the two sets.  If you don't like the one you mentioned in your post, may I suggest that Calkin-Wilf enumeration of the rationals?  (Simply google search Calkin Wilf counting rationals.  The first .pdf has what I'm talking about).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, these give bijections between the naturals (with out 0) and the rationals, but once you have a bijection like this, it's easy to construct a bijection from the integers to the rationals by composing with a bijection from the naturals to the integers.&lt;/p&gt;&#xA;" OwnerUserId="331" LastActivityDate="2010-07-31T20:01:41.193" CommentCount="6" />
  <row Id="1313" PostTypeId="2" ParentId="1311" CreationDate="2010-07-31T20:04:39.830" Score="0" ViewCount="" Body="&lt;p&gt;You can think about it a different way. Consider the set of real numbers between 0 and 1, and then the set of real numbers between 0 and 2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By intuition, it seems that the set of real numbers between 0 and 2 has double the size of the set between 0 and 1. However, this is not the case, because the two sets have the same &lt;em&gt;cardinality&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the function $f(x) = 2x$. Every real between 0 and 1 is &lt;em&gt;bijected&lt;/em&gt; to a real between 0 and 2. Therefore the sets are of the same size.&lt;/p&gt;&#xA;" OwnerUserId="215" LastActivityDate="2010-07-31T20:04:39.830" CommentCount="1" />
  <row Id="1314" PostTypeId="2" ParentId="803" CreationDate="2010-07-31T20:07:46.967" Score="0" ViewCount="" Body="&lt;p&gt;Not a proof in itself, but the book &lt;a href=&quot;http://www.amazon.com/exec/obidos/ISBN=0873530365/ctksoftwareincA/&quot; rel=&quot;nofollow&quot;&gt;The Pythagorean Proposition&lt;/a&gt; by Loomis has probably the most comprehensive collection of proofs of the Pythagorean theorem.&lt;/p&gt;&#xA;" OwnerUserId="215" LastActivityDate="2010-07-31T20:07:46.967" CommunityOwnedDate="2010-07-31T20:07:46.967" />
  <row Id="1315" PostTypeId="1" AcceptedAnswerId="1320" CreationDate="2010-07-31T20:08:01.177" Score="1" ViewCount="295" Body="&lt;p&gt;Suppose you have a normal distribution with mean=0, and stdev=1.  So the expected value is 0.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now suppose you limit the outcomes, such that no values can be below 0.  So 50% of values now equal 0, and rest of distribution is still normal.  Running 1000000 trials, I come out with an expected value of .4  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is how can I get this expected value through calculation?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="540" LastEditorUserId="540" LastEditorDisplayName="" LastEditDate="2010-08-01T13:53:17.917" LastActivityDate="2010-08-01T13:53:17.917" Title="How do I calculate expected value of partial normal distribution?" Tags="&lt;calculus&gt;&lt;probability-theory&gt;" AnswerCount="1" />
  <row Id="1316" PostTypeId="2" ParentId="1311" CreationDate="2010-07-31T20:09:02.033" Score="1" ViewCount="" Body="&lt;p&gt;The cardinality of the set of rationals is the same as the cardinality of the integers is the same as the cardinality of the natural numbers. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When we count a finite set of elements, we are constructing a one-one map from the set onto a finite initial segment of the natural numbers. If we want to know if two finite sets have the same cardinality (are equi-cardinal) we can either: 1) count both sets and see if we get the same number, or 2) attempt to construct a one-one map from one set onto the other. If we can construct the map aimed at in (2), then the sets are equi-cardinal. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Generalizing that procedure from the finite sets to arbitrary sets, we get that for any two sets, the sets have the same cardinality (are equi-cardinal) if there exists a bijection (a one-one map between the sets that is onto the target rather than merely into). For the finite case, if there is a one-one map that is a bijection, all one-one maps are bijective. That is not the case for infinite sets, which is the root of your second concern.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To address that second concern, consider the map from the negative integers to the positive integers which maps each negative integer to its absolute value. The existence of that map shows that the two sets are equi-cardinal. We can, of course, construct one-one maps from the negative integers to the positive integers that are into rather than on to. (Consider the map that takes each negative integer to its product with -2.) But, the existence of these alternative maps doesn't affect the fact that there is at least one bijection between the sets, and that is all it takes for those sets to be equi-cardinal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for your first concern, I don't see why you think the procedure assumes that &quot;infinity is a finite number&quot;. What it involves is specifying a mapping function from one set to the other that is one-one and onto. That attempt can certainly fail, as Cantor's Diagonalization Argument that the cardinality of a set is always strictly less than the cardinality of its power set shows. (A relevant application of that technique is the well-known proof that the cardinality of the reals is greater than the cardinality of the natural numbers.)&lt;/p&gt;&#xA;" OwnerUserId="277" LastActivityDate="2010-07-31T20:09:02.033" CommentCount="2" />
  <row Id="1317" PostTypeId="2" ParentId="1311" CreationDate="2010-07-31T20:11:34.837" Score="3" ViewCount="" Body="&lt;p&gt;In mathematics a set is called &lt;em&gt;infinite&lt;/em&gt; if it can be put into a 1-1 correspondence with a proper subset of it, and &lt;em&gt;finite&lt;/em&gt; it is not infinite. (I know it seems crazy to have the concept of infinite as primitive and finite as a derivate, but it's simpler to do this, since otherwise you must assume that the integers exist before saying that a set is finite)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for your remarks: &#xA;- with your method (if you don't forget to throw out fractions like 4/6 which is equal to 2/3) you actually &lt;strong&gt;counted&lt;/strong&gt; the rationals, since for each number you have a function which associates it to a natural number. It's true that you cannot count ALL rationals, or all integers; but you cannot either draw a whole straight line, can you?&#xA;- with infinite sets you may build infinite mappings, but you just need a single 1-1 mapping to show that two sets are equal.&lt;/p&gt;&#xA;" OwnerUserId="89" LastActivityDate="2010-07-31T20:11:34.837" CommentCount="3" />
  <row Id="1318" PostTypeId="2" ParentId="329" CreationDate="2010-07-31T20:18:16.647" Score="2" ViewCount="" Body="&lt;p&gt;My favorite is &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0321500318&quot; rel=&quot;nofollow&quot;&gt;Elementary Number Theory&lt;/a&gt; by Rosen, which combines computer programming with number theory, and is accessible at a high school level.&lt;/p&gt;&#xA;" OwnerUserId="215" LastActivityDate="2010-07-31T20:18:16.647" CommunityOwnedDate="2010-07-31T20:18:16.647" />
  <row Id="1319" PostTypeId="2" ParentId="1311" CreationDate="2010-07-31T20:26:04.660" Score="2" ViewCount="" Body="&lt;p&gt;You may not be very satisfied with this answer, but I'll try to explain anyway.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Countability.&lt;/strong&gt; We're not really talking about whether you can &quot;count all of the rationals&quot;, using some finite process. Obviously, if there is an infinite number of elements, you cannot count them in a finite amount of time using any reasonable process. The question is whether there is &lt;em&gt;the same number&lt;/em&gt; of rationals as there are positive integers; this is what it means for a set to be &quot;countable&quot; --- for there to &lt;em&gt;exist&lt;/em&gt; a one-to-one mapping from the positive integers to the set in question. You have described such a mapping, and therefore the rationals are &quot;countable&quot;. (You may disagree with the terminology, but this does not affect whether the concept that it labels is coherent.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Alternative mappings.&lt;/strong&gt; You seem to be dissatisfied with the fact that, unlike the case of a finite set, you can define an injection from the natural numbers to the rationals which is not surjective --- that you can in fact define a more general &lt;em&gt;relation&lt;/em&gt; in which each integer is related to infinitely many rationals, but no two integers are related to the same rational numbers. Well, two can play at that game: you can define a relation in which every rational number is related to infinitely many integers, and no two rationals are related to the same integers! Just define the relation that each positive rational a/b is related to all numbers which are divisible by 2&lt;sup&gt;a&lt;/sup&gt; but not 2&lt;sup&gt;a+1&lt;/sup&gt;, and by 3&lt;sup&gt;b&lt;/sup&gt; but not 3&lt;sup&gt;b+1&lt;/sup&gt;; or more generally respectively 2&lt;sup&gt;ka&lt;/sup&gt; and 3&lt;sup&gt;kb&lt;/sup&gt; for any positive integer k. (There are, as you say, sign issues, but these can be smoothed away.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might complain that the relation I've defined isn't &quot;natural&quot;. Perhaps you have in mind the fact that the integers are a &lt;em&gt;subset&lt;/em&gt; of the rationals --- a subgroup, in fact, taking both of them as additive groups --- and that the factor group &amp;#8474;/&amp;#8484; is infinite. Well, this is definitely interesting, and it's a natural sort of structure to be interested in. But it's more than what the issue of &quot;mere cardinality&quot; is trying to get at: set theory is interested in size regardless of structure, and so we don't restrict to maps which have one or another kind of &quot;naturalness&quot; about them. Of course, if you are interested in mappings which respect some sort of structure, you can build theories of size based on that: this is what is done in measure theory (with measure), linear algebra (with dimension), and indeed group theory (with index). So if you don't like cardinality as set theorists conceive it, you can look at more structured measures of size that you find more interesting!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Immediate predecessors.&lt;/strong&gt; A somewhat unrelated (but still important) complaint that you make is this: &quot;&lt;em&gt;If it were possible to count to infinity, it would be possible to count one step less and stop at count infinity-1 which must be different to infinity.&lt;/em&gt;&quot; The question is: &lt;strong&gt;why&lt;/strong&gt; would you necessarily be able to stop at 'infinity minus one'? This is true for finite collections, but it does not necessarily hold that anything which is true of finite collections is true also for infinite ones. (In fact, obviously, some things necessarily will fail.) --- This is important if you study ordinals, which mirrors the process of counting itself in some ways (labelling things as being &quot;first&quot;, &quot;second&quot;, &quot;third&quot;, and so forth), because of the concept of a limit ordinal: the first &quot;infinitieth&quot; element of a well-ordering doesn't have any immediate predecessors! Again, you are free to say that these are concepts that you are not interested in exploring personally, but this does not mean that they are necessarily incoherent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;To summarize:&lt;/strong&gt; the set theorists measure &quot;the size of a set&quot; using a simple definition which doesn't care about structure, and which may violate your intuitions if you like to take the structure of the integers (and the rational numbers) very seriously, and also want to preserve your intuitions about finite sets. There are two solutions to this: try to stretch your intuition to accomodate the ideas of the set theorists, or study a different branch of math which you find more interesting!&lt;/p&gt;&#xA;" OwnerUserId="439" LastActivityDate="2010-07-31T20:26:04.660" CommentCount="3" />
  <row Id="1320" PostTypeId="2" ParentId="1315" CreationDate="2010-07-31T20:26:47.403" Score="4" ViewCount="" Body="&lt;p&gt;The normal distribution has density function $f(x)=\frac{e^{-\frac{x^2}{2}}}{\sqrt{2\pi}}$; your new distribution has that density function on the positive reals, $P(0)=\frac{1}{2}$, and $P(x)=0$ for the negative reals.  The expected value is $0\cdot\frac{1}{2}+\int_{0}^{\infty}x\cdot f(x)dx=\frac{1}{\sqrt{2\pi}}\approx0.398942$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;edit&lt;/em&gt;: If you were to cut off at $x=c$ (assigning all the probability from below c to c itself) instead of $x=0$, your density function would be $f(x)=\frac{e^{-\frac{x^2}{2}}}{\sqrt{2\pi}}$ for $x&amp;gt;c$, $P(c)=\int_{-\infty}^{c}\frac{e^{-\frac{x^2}{2}}}{\sqrt{2\pi}}dx$, and $P(x)=0$ for $x&amp;lt;c$, so the expected value is $c\cdot P(c) + \int_{c}^{\infty}x\cdot \frac{e^{-\frac{x^2}{2}}}{\sqrt{2\pi}}dx$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;edit 2&lt;/em&gt;: note that the exponent on e in all of the above is $-\frac{x^2}{2}$ (the exponent 2 on the x is, in the current TeX rendering, positioned and sized such as to be somewhat ambiguous)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;edit 3&lt;/em&gt;: my explanation incorrectly mixed probability density functions and literal probabilities--this was solely an issue of terminology and the analytic results still stand, but I have attempted to clarify the language above.&lt;/p&gt;&#xA;" OwnerUserId="72" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-08-01T05:45:33.857" LastActivityDate="2010-08-01T05:45:33.857" CommentCount="4" />
  <row Id="1321" PostTypeId="2" ParentId="460" CreationDate="2010-07-31T20:53:06.657" Score="7" ViewCount="" Body="&lt;p&gt;One source of monoids is given by taking rings with identity, and forgetting about addition.&#xA;So similarly, one source of semigroups that are not monoids is taking rings without identity, and forgetting about addition.  With this in mind, let me explain one basic source of rings without identity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A basic source of rings is given by taking functions satisfying some reasonable condition on a space, e.g. continuous real or complex valued functions on a space, with pointwise addition and multiplication.  Of course, the constant function 1 is continuous, and so this gives a ring with identity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But suppose now that we impose some condition, such as &quot;all functions that are continuous, and which furthermore vanish at some specified point&quot;.  This throws out the constant function 1, and so gives a ring without identity.  Now you could naturally object that this is artificial&#xA;(as per the requirement in the question that there not be an obvious extension to a monoid),&#xA;so let me add more explanation as to why it need not be.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One example of a point to consider is &quot;the point at infinity&quot;, i.e. we could look at all functions which vanish at infinity, i.e. which on the complement of larger and larger compact sets, grow smaller and smaller.  This is a natural condition to impose in many analytic contexts, and so gives a natural example.   (The reason that this kind of growth&#xA;condition is natural in analysis is that, on a non-compact space, e.g. the real line, a random continuous function may not be integrable (just as an example), and imposing some decay at infinity (perhaps of the kind I specified, or perhaps something more quantitive) becomes a way to rescue the situation.)  (Note also that the example that Tomer Vromen gives is exactly of this form.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, note that if your semigroup doesn't have an identity, then you can always formally adjoin one, just by throwing in an extra element e and declaring that ex = x for all x.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One can do a similar thing for rings without identity.  If A is a C-algebra (say) without an identity, then one can form A + C e  (the direct sum), and declare that e acts as a multiplicative identity.  This is a frequently-used technique in the theory of rings-without-identity.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S.  I don't know much literature about semigroups without identity, but for rings without identity, the best literature I know of is in functional analysis books; e.g. Naimark's classic &lt;em&gt;Normed Rings&lt;/em&gt; often treats the case of Banach algebras (and the like) without identity in addition to the case when they do have identity, exactly so as to be able to handle examples such as the ring of continuous functions on a locally compact space that vanish at infinity. &lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2010-07-31T20:53:06.657" CommunityOwnedDate="2010-07-31T20:53:06.657" />
  <row Id="1323" PostTypeId="1" CreationDate="2010-07-31T21:28:34.173" Score="8" ViewCount="188" Body="&lt;p&gt;I was wondering if there was a resource that listed known algebraic groups and their corresponding coordinate rings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: The previous wording was terrible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given an algebraic group $G$, with Borel subgroup $B$ we can form the Flag Variety $G/B$ which is projective. I am hoping for a list of the graded ring $R$ such that $Proj(R)$ corresponds to this Flag Variety.&lt;/p&gt;&#xA;" OwnerUserId="16" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-08-01T17:41:29.873" LastActivityDate="2010-08-01T17:41:29.873" Title="Is there an atlas of Algebraic Groups and corresponding Coordinate rings?" Tags="&lt;reference-request&gt;&lt;algebraic-geometry&gt;&lt;algebraic-groups&gt;" CommentCount="7" FavoriteCount="2" />
  <row Id="1324" PostTypeId="1" AcceptedAnswerId="1325" CreationDate="2010-07-31T21:53:33.887" Score="0" ViewCount="379" Body="&lt;p&gt;What is a primitive polynomial? I was looking into some random number generation algorithms and 'primitive polynomial' came up a sufficient number of times that I decided to look into it in more detail.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm unsure of what a primitive polynomial is, and why it is useful for these random number generators.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd find it particularly helpful if an example of a primitive polynomial could be provided.&lt;/p&gt;&#xA;" OwnerUserId="181" LastActivityDate="2010-07-31T23:40:59.560" Title="What is a primitive polynomial?" Tags="&lt;polynomials&gt;&lt;field-theory&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="1325" PostTypeId="2" ParentId="1324" CreationDate="2010-07-31T22:05:52.727" Score="8" ViewCount="" Body="&lt;p&gt;Consider a finite field $F_p$, then we know that it is cyclic. We call an element primitive if it generates this field. Further, given a field and some polynomial over that field(all the coefficients are in the field), we can form a field extension by any of its roots. This is adjoining on that root and making a field of it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is a simple result of Galois Theory that if we take a field and extend by some root of some polynomial and get a finite extension(one who's degree as a vector space over the original field is finite), that we can find a polynomial $m$ over our ground field such that $m$ vanishes at this root and is minimal(smallest degree, i.e. it divides all other polys which vanish at this root).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we consider a primitive element and its minimal polynomial, that poly is call primitive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;more details on &lt;a href=&quot;http://en.wikipedia.org/wiki/Primitive_polynomial&quot; rel=&quot;nofollow&quot;&gt;wiki&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="16" LastActivityDate="2010-07-31T22:05:52.727" CommentCount="1" />
  <row Id="1326" PostTypeId="1" AcceptedAnswerId="1328" CreationDate="2010-07-31T23:37:24.167" Score="3" ViewCount="158" Body="&lt;p&gt;The title really say's it all, but once again is a curve's curvature invariant under rotation and uniform scaling?&lt;/p&gt;&#xA;" OwnerUserId="99" LastActivityDate="2010-07-31T23:48:16.190" Title="Is a curve's curvature invariant under rotation and uniform scaling?" Tags="&lt;geometry&gt;&lt;differential-geometry&gt;" AnswerCount="1" />
  <row Id="1327" PostTypeId="2" ParentId="1324" CreationDate="2010-07-31T23:40:59.560" Score="7" ViewCount="" Body="&lt;p&gt;BBischof's answer is correct, but unfortunately there's another, quite different possible meaning of the same term: that is a polynomial whose coefficients have no common prime factor (this makes sense over the integers, or other UFDs as well).&lt;/p&gt;&#xA;" OwnerUserId="308" LastActivityDate="2010-07-31T23:40:59.560" CommentCount="3" />
  <row Id="1328" PostTypeId="2" ParentId="1326" CreationDate="2010-07-31T23:43:08.917" Score="8" ViewCount="" Body="&lt;p&gt;A curve's curvature is invariant under rotation. Intuitively, a curve turns just as much no matter how it is oriented. More formally, for a curve $\gamma(s)$ that is parametrized by arc length, the curvature is $\kappa(s) = ||\gamma''(s)||$. Rotation does not change the length of the $\gamma''(s)$ vector, only the direction; therefore, rotation does not affect curvature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A curve's curvature is not invariant under uniform scaling, however. Consider the example of a circle. All circles are the same up to scaling, but they don't all have the same curvature; in general, a circle of radius r has curvature 1/r.&lt;/p&gt;&#xA;" OwnerUserId="30" LastEditorUserId="30" LastEditorDisplayName="" LastEditDate="2010-07-31T23:48:16.190" LastActivityDate="2010-07-31T23:48:16.190" CommentCount="1" />
  <row Id="1329" PostTypeId="1" AcceptedAnswerId="1332" CreationDate="2010-07-31T23:56:42.353" Score="3" ViewCount="122" Body="&lt;p&gt;This is a followup to &lt;a href=&quot;http://math.stackexchange.com/questions/1326/is-a-curves-curvature-invariant-under-rotation-and-uniform-scaling&quot;&gt;this question&lt;/a&gt;, where I learned that curvature is invariant to rotations. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have learned of a version of curvature that is &lt;a href=&quot;http://en.wikipedia.org/wiki/Affine_curvature#Affine_curvature&quot; rel=&quot;nofollow&quot;&gt;invariant under affine transformations&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am wondering if there a is a form of curvature between the two. Invariant under uniform scaling and rotation but not all affine transformations?&lt;/p&gt;&#xA;" OwnerUserId="99" LastEditorUserId="123" LastEditorDisplayName="" LastEditDate="2010-08-01T03:33:21.160" LastActivityDate="2010-08-01T03:33:21.160" Title="What is the form of curvature that is invariant under rotations and uniform scaling" Tags="&lt;geometry&gt;&lt;differential-geometry&gt;" AnswerCount="1" />
  <row Id="1332" PostTypeId="2" ParentId="1329" CreationDate="2010-08-01T02:58:53.723" Score="2" ViewCount="" Body="&lt;p&gt;I don't know if this would suit you, but one thing you can consider (much more naive than&#xA;the notion of affine curvature) is to fix a point P_0 on your curve, and then consider the&#xA;function on the curve given by sending a point P to the quantity&lt;/p&gt;&#xA;&#xA;&lt;p&gt;curvature(P)/curvature(P_0)  .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a kind of relative curvature, where you measure how much everything is curving&#xA;in comparison to the curvature at P_0, and is invariant under scaling and rotation.  &lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2010-08-01T02:58:53.723" CommentCount="5" />
  <row Id="1333" PostTypeId="1" AcceptedAnswerId="1335" CreationDate="2010-08-01T07:45:27.030" Score="10" ViewCount="274" Body="&lt;p&gt;Me and my wife are often not knowing which DVD to watch. If we have two options we have a simple solution, I put one DVD in one hand behind my back and the other DVD in the other hand. She will randomly choose a hand and the DVD I have in that hand will be the one to watch.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This procedure is easy to expand to any power of 2. If we have 4 DVD's I hold 2 in one hand, 2 in the other. When a pair of DVD's is chosen, I split them out to two hands and she choses again.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is, what can we do when we have 3 DVD's. The assumptions we make are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;I am not impartial. If I can influence the result somehow I will try to do that&lt;/li&gt;&#xA;&lt;li&gt;My wife really choses randomly a side every-time, independent of what she chose earlier.&lt;/li&gt;&#xA;&lt;li&gt;I don't have any other place to hide the DVD's, so every DVD is either visible or in one of the two hands.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;As requirement we have that it must be a procedure with a predetermined number of steps. Not more, not less. If this is not possible, a solution that guarantees to finish with an upperbound, this is a good second choice. Off course the DVD we choose must be really random!&lt;/p&gt;&#xA;" OwnerUserId="143" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-08-01T17:56:02.873" LastActivityDate="2010-08-01T21:22:26.807" Title="Is there a possibility to choose fairly from three items when every choice can only have 2 options" Tags="&lt;probability-theory&gt;&lt;algorithms&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="1" />
  <row Id="1334" PostTypeId="2" ParentId="1333" CreationDate="2010-08-01T08:40:07.303" Score="5" ViewCount="" Body="&lt;p&gt;There is no such procedure which has an upperbound on number of steps. Here is proof.&#xA;Let there is such procedure with no more then $N$ steps. On each step you basicly generate random integer between $1$ and $2$. Consider all possible sequences of no more than $N$ generated numbers. Every such sequence has probability of form $\frac{x}{2^N}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider sequences for which the procedure says &quot;1&quot; (the result is equal to $1$). The sum of theirs probabilities is $\frac{x_1}{2_N}$. It also must be equal to $\frac{1}{3}$. But $\frac{x_1}{2_N}$ can't be equal to $\frac{1}{3}$.&lt;/p&gt;&#xA;" OwnerUserId="334" LastActivityDate="2010-08-01T08:40:07.303" />
  <row Id="1335" PostTypeId="2" ParentId="1333" CreationDate="2010-08-01T08:49:35.987" Score="10" ViewCount="" Body="&lt;p&gt;Nice question. Firstly, there's no solution which will always take a fixed number of steps each time: after n independent random trials (choosing a hand), there are 2&lt;sup&gt;n&lt;/sup&gt; possible choices, and 2&lt;sup&gt;n&lt;/sup&gt; is never divisible by 3. (Equivalently: 1/3 does not have a terminating representation in base 2.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But you can come up with solutions that terminate within a fixed number of steps with high probability. Note that you need at least 2 trials, since log&lt;sub&gt;2&lt;/sub&gt;(3) &gt; 1. Here's one simple method:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Hold 2 DVDs in one hand and 1 DVD in other (your wife doesn't know which is which).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If the hand with 2 DVDs is chosen: play them against each other.&lt;br&gt;&#xA;Else: If the 1 DVD is chosen: play it against an empty hand.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Each of the 3 DVDs has equal probability 1/4 of being chosen, and with probability 1/4, you have to repeat from scratch. This solution takes 2 trials with probabilty 3/4, 4 trials with probability (1/4)(3/4), 6 trials with probability (1/4)&lt;sup&gt;2&lt;/sup&gt;(3/4) and so on, so takes an expected number of 8/3 (≈ 2.66) trials. (See &lt;a href=&quot;http://en.wikipedia.org/wiki/Geometric_distribution&quot; rel=&quot;nofollow&quot;&gt;geometric distribution&lt;/a&gt;.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe this is optimal (famous last words?) if you just have to choose out of 3 DVDs once. Probably, if you're going to choose from 3 DVDs many times, you can do better in the long run (amortised) and achieve the lower bound of log&lt;sub&gt;2&lt;/sub&gt; 3 ≈ 1.58 trials on average, by &quot;caching&quot; some random choices from last time. (But will you remember them? :-)) At least you can do something similar when generating random numbers (see  also &lt;a href=&quot;http://stackoverflow.com/questions/137783/given-a-function-which-produces-a-random-integer-in-the-range-1-to-5-write-a-fun&quot;&gt;this mildly related Stack Overflow thread&lt;/a&gt;), but in this case with game-theoretic fairness complications I'm not so sure.&lt;/p&gt;&#xA;" OwnerUserId="205" LastEditorUserId="205" LastEditorDisplayName="" LastEditDate="2010-08-01T08:55:24.973" LastActivityDate="2010-08-01T08:55:24.973" CommentCount="2" />
  <row Id="1336" PostTypeId="2" ParentId="803" CreationDate="2010-08-01T10:03:35.580" Score="6" ViewCount="" Body="&lt;p&gt;My favorite proof of the Pythagorean Theorem is a special case of this picture-proof of the Law of Cosines:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://daylateanddollarshort.com/misc/lawofcos.jpg&quot; alt=&quot;Proof Without Words: The Law of Cosines&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Drop three perpendiculars and let the definition of cosine give the lengths of the sub-divided segments. Then, observe that like-colored rectangles have the same area (computed in slightly different ways) and the result follows immediately.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When C is a right angle, the blue rectangles vanish and we have the Pythagorean Theorem via what amounts to Proof #5 on &lt;a href=&quot;http://www.cut-the-knot.org/pythagoras/&quot; rel=&quot;nofollow&quot;&gt;Cut-the-Knot's Pythagorean Theorem page&lt;/a&gt;. (Note that, as mentioned on CtK, the use of cosine here doesn't amount to an invalid &quot;trigonometric proof&quot;.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The picture works for obtuse C as well. (Proof left as an exercise for the reader.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A final note ... Because the same-colored rectangles have the same area, they're &quot;equidecomposable&quot; (aka &quot;scissors congruent&quot;): it's possible to cut one into a finite number of polygonal pieces that reassemble to make the other. While there's at least one standard procedure for determining how to make the cuts, the resulting pieces aren't necessarily pretty. Some popular dissection proofs of the Pythagorean Theorem --such as Proof #36 on &lt;a href=&quot;http://www.cut-the-knot.org/pythagoras/&quot; rel=&quot;nofollow&quot;&gt;Cut-the-Knot&lt;/a&gt;-- demonstrate a specific, clear pattern for cutting up the figure's three squares, a pattern that applies to all right triangles. I have yet to find a similarly straightforward cutting pattern that would apply to all triangles and show that my same-colored rectangles &quot;obviously&quot; have the same area.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Another exercise for the reader, perhaps? :)&lt;/p&gt;&#xA;" OwnerUserId="409" LastActivityDate="2010-08-01T10:03:35.580" CommunityOwnedDate="2010-08-01T10:03:35.580" CommentCount="0" />
  <row Id="1337" PostTypeId="1" AcceptedAnswerId="1339" CreationDate="2010-08-01T11:22:30.790" Score="2" ViewCount="238" Body="&lt;p&gt;Dot product of two vectors on plane could be defined as product of lengths&#xA;of those vectors and cosine of angle between them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In cartesian coordinates dot product of vectors with coordinates (x1, y1) and&#xA;(x2, y2) is equal to x1*x2 + y1*y2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How to prove it?&lt;/p&gt;&#xA;" OwnerUserId="551" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-08-01T17:29:06.960" LastActivityDate="2010-08-01T17:29:06.960" Title="Dot product in coordinates" Tags="&lt;geometry&gt;&lt;linear-algebra&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" />
  <row Id="1338" PostTypeId="1" AcceptedAnswerId="1345" CreationDate="2010-08-01T11:26:29.567" Score="1" ViewCount="206" Body="&lt;p&gt;Given a list of coordinates of a coplanar plane (&lt;code&gt;pt1&lt;/code&gt;, &lt;code&gt;pt2&lt;/code&gt;, &lt;code&gt;pt3&lt;/code&gt; and so on), how to compute the centroid of the coplanar plane?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One way to do it is to project the plane onto XY and YZ plane, but I don't really favor this approach as you have to check the orientation of the coplanar plane first before doing the projection and computing the centroid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More specifically, I'm looking for a natural extension of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Centroid#Centroid_of_polygon&quot; rel=&quot;nofollow&quot;&gt;2D centroid plane algorithm&lt;/a&gt; in 3D:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://upload.wikimedia.org/math/e/e/1/ee14cbb2b170c4bb435f1d84e78f6d66.png&quot; alt=&quot;alt text&quot;&gt;&#xA;&lt;img src=&quot;http://upload.wikimedia.org/math/a/4/c/a4cee81a1d18e4d067f66d4d40a8a1fe.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://upload.wikimedia.org/math/0/2/a/02aecb75f67f8c7b2fc11fdcbcb6ea80.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any idea?&lt;/p&gt;&#xA;" OwnerUserId="550" LastActivityDate="2010-08-01T15:21:05.333" Title="Compute the Centroid of a 3D Planar Polygon Without Projecting It To Specific Planes" Tags="&lt;geometry&gt;" AnswerCount="1" />
  <row Id="1339" PostTypeId="2" ParentId="1337" CreationDate="2010-08-01T11:39:17.987" Score="7" ViewCount="" Body="&lt;p&gt;I suppose you want to prove that two your definitions of dot product are the same.&#xA;We start with definition of dot product as $(\vec{u}, \vec{v}) = |\vec{u}| |\vec{v}| \cos \theta$. We start with definition of dot product as $(\vec{u}, \vec{v}) = |\vec{u}| |\vec{v}| \cos \theta$ and prove that it also satisfies $(\vec{u}, \vec{v}) = x_1 x_2 + y_1 y_2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At first you can prove that dot product is linear: $(\vec{v_1}, \vec{v_2} + \alpha \vec{v_3}) = (\vec{v_1}, \vec{v_2}) + \alpha (\vec{v_1}, \vec{v_3})$. This is true because $(\vec{v_1}, \vec{v_2})$ is equal to the product of $|\vec{v_1}|$ and projection of $\vec{v_2}$ on $\vec{v_1}$. Projection of sum of vectors is equal to sum of projections. Hence dot product is linear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $\vec{e_1}$ and $\vec{e_2}$ be vectors with coordinates $(1, 0)$ and $(0, 1)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After that if $\vec{v_1} = x_1 \vec{e_1} + y_1 \vec{e_2}$ and $\vec{v_2} = x_2 \vec{e_2} + y_2 \vec{e_2}$ then by linearity of dot product we have $(\vec{v_1}, \vec{v_2}) = x_1 x_2 (\vec{e_1}, \vec{e_1}) + x_1 y_2 (\vec{e_1}, \vec{e_2}) + x_2 y_1 (\vec{e_2}, \vec{e_1}) + x_2 y_2 (\vec{e_2}, \vec{e_2})$.&#xA;Since $(\vec{e_1}, \vec{e_1}) = (\vec{e_2}, \vec{e_2}) = 1$ and $(\vec{e_1}, \vec{e_2}) = (\vec{e_2}, \vec{e_1}) = 0$ we have $(\vec{v_1}, \vec{v_2}) = x_1 x_2 + y_1 y_2$.&lt;/p&gt;&#xA;" OwnerUserId="334" LastEditorUserId="334" LastEditorDisplayName="" LastEditDate="2010-08-01T15:22:23.023" LastActivityDate="2010-08-01T15:22:23.023" CommentCount="9" />
  <row Id="1340" PostTypeId="1" AcceptedAnswerId="1343" CreationDate="2010-08-01T11:42:22.300" Score="0" ViewCount="106" Body="&lt;p&gt;Let's say we have a graph, with a list of edges and vertexes &lt;code&gt;(E,V)&lt;/code&gt;, all the vertexes are connected to at least one edge at one end. There are many ways a complete set of &lt;a href=&quot;http://www.mathreference.com/gph,basis.html&quot; rel=&quot;nofollow&quot;&gt;cycle basis&lt;/a&gt; can be found out from it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the issue is, is it &lt;strong&gt;always&lt;/strong&gt; possible to find a complete set of cycle basis that each edge is shared by at most 2 cycles?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&lt;strong&gt;Edit: There is a &lt;a href=&quot;http://math.stackexchange.com/questions/1340/in-a-graph-is-it-always-possible-to-construct-a-set-of-cycle-basis-with-each-an/1343#1343&quot;&gt;mathematical argument&lt;/a&gt; proving why it is not possible. But admittedly such a highly abstract reasoning is a bit hard for me to grasp. I would appreciate if someone can provide a graphical example of such a graph.&lt;/em&gt;&lt;/strong&gt; &lt;/p&gt;&#xA;" OwnerUserId="550" LastEditorUserId="550" LastEditorDisplayName="" LastEditDate="2010-08-05T05:47:31.717" LastActivityDate="2010-08-05T05:47:31.717" Title="In a graph, is it always possible to construct a set of cycle basis, with each and every edge Is shared by at most 2 cycle bases?" Tags="&lt;graph-theory&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="1341" PostTypeId="2" ParentId="1337" CreationDate="2010-08-01T12:39:02.813" Score="3" ViewCount="" Body="&lt;p&gt;The dot product is invariant under rotations, we may therefore rotate our coordinate system so that v is along the x-axis.  In this case, $v = (|v|, 0)$.  Letting $w = (x,y)$ we have (using the definition of dot product in Cartesian coordinates) $v \cdot w = |v| x$.  But what is $x$?  Well, if you draw the picture and let $\theta$ be the angle between v and w, then we see that $\cos \theta = x/|w||$ so that $x = |w| \cos \theta$.  Thus $v\cdot w = |v||w| \cos \theta$.&lt;/p&gt;&#xA;" OwnerUserId="9" LastActivityDate="2010-08-01T12:39:02.813" CommentCount="2" />
  <row Id="1342" PostTypeId="2" ParentId="1292" CreationDate="2010-08-01T13:25:38.840" Score="8" ViewCount="" Body="&lt;p&gt;Here are my favorite diagrams:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://daylateanddollarshort.com/misc/angaddsub.jpg&quot; alt=&quot;Proof Without Words: Angle Sum and Difference for Sine and Cosine&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As given, the diagrams put certain restrictions on the angles involved: neither angle, nor their sum, can be larger than 90 degrees; and neither angle, nor their difference, can be negative. The diagrams can be adjusted, however, to push beyond these limits.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a bonus mnemonic cheer (which probably isn't as exciting to read as to hear):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Sine, Cosine, &lt;em&gt;Sign&lt;/em&gt;, Cosine, Sine!&lt;br /&gt;&#xA;  Cosine, Cosine, &lt;em&gt;Co-Sign&lt;/em&gt;, Sine, Sine!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The first line encapsulates the sine formulas; the second, cosine. Just drop the angles in (in order $\alpha$, $\beta$, $\alpha$, $\beta$ in each line), and know that &quot;Sign&quot; means to use the &lt;em&gt;same sign&lt;/em&gt; as in the compound argument (&quot;+&quot; for angle sum, &quot;-&quot; for angle difference), while &quot;Co-Sign&quot; means to use the &lt;em&gt;opposite sign&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="409" LastActivityDate="2010-08-01T13:25:38.840" CommentCount="1" />
  <row Id="1343" PostTypeId="2" ParentId="1340" CreationDate="2010-08-01T14:03:03.260" Score="3" ViewCount="" Body="&lt;p&gt;The question was answered in the negative on Math Overflow.  See &lt;a href=&quot;http://mathoverflow.net/questions/30759/in-a-graph-is-it-always-possible-to-construct-a-set-of-cycle-bases-with-each-an&quot; rel=&quot;nofollow&quot;&gt;http://mathoverflow.net/questions/30759/in-a-graph-is-it-always-possible-to-construct-a-set-of-cycle-bases-with-each-an&lt;/a&gt;  &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Edit: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We get a counterexample from any non-planar graph where every edge is part of at least one cycle. Here's a reference: &lt;a href=&quot;http://www.ams.org/journals/proc/1973-037-02/S0002-9939-1973-0313098-X/S0002-9939-1973-0313098-X.pdf&quot; rel=&quot;nofollow&quot;&gt;P. V. O'Neil, Proc. AMS, 37 (2), Feb. 1973, 617-8&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll repeat the argument here.  Take a nonplanar graph with every edge in at least one cycle. If that had a cycle basis like the one you want, then we could generate one for $K_{3,3}$ or $K_{5}$. Suppose we had a basis for $K_{3,3}$. There are 4 cycles in that basis. (A cycle basis has m-n+1 elements, where m is the number of vertices and n is the number of edges.) Also, take the binary sum of the four cycles. The five cycles include every edge exactly twice, so there are a total of 2 $\cdot$ (number of edges) = 18 edges in those five cycles. But each cycle has at least 4 edges, so the five cycles must have at least 20 edges. Contradiction. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let's look at $K_{5}$. A cycle basis has 6 cycles. Also take the binary sum. The 7 cycles include each edge exactly twice, so there are 2 $\cdot$ (number of edges) = 20 edges in the collection. But each cycle has at least three edges, so the 7 cycles have at least 7*3=21 edges. Again, a contradiction. &lt;/p&gt;&#xA;" OwnerUserId="273" LastEditorUserId="273" LastEditorDisplayName="" LastEditDate="2010-08-02T14:51:48.877" LastActivityDate="2010-08-02T14:51:48.877" CommentCount="12" />
  <row Id="1344" PostTypeId="1" AcceptedAnswerId="1373" CreationDate="2010-08-01T14:04:15.177" Score="4" ViewCount="281" Body="&lt;p&gt;Proposition 0.4.5.4 in EGA appears to be a general representability theorem. It reads:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Suppose $F$ is a contravariant functor from the category of locally ringed spaces over $S$ to the category of sets. Suppose given representable sub-functors $F_i$ of $F$, such that the morphisms $F_i \to F$ are representable by open immersions. Suppose furthermore that if $Hom(-, X) \to F$ is a morphism and the functors $F_i \times_F Hom(-,X)$ are representable by $Hom(-,X_i)$, the family $X_i$ forms an open covering of $X$. (That $X_i \to X$ is an open immersion follows from the definitions.)  Finally, suppose that if $U$ ranges over the open subsets of a locally ringed space $X$, the functor $U \to F(U)$ is a sheaf.   Then, $F$ is representable.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I haven't yet been able to grok the proof, but it appears to be some sort of extended gluing construction. This result appears to be used in proving that fibered products exist in the category of schemes. However, it's fairly easy to directly construct fibered products by gluing open affines.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there examples where this result actually makes the life of algebraic geometers easier?  Also, I'd appreciate any links to examples (outside of EGA) where this result is used.&lt;/p&gt;&#xA;" OwnerUserId="536" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-08-01T17:43:27.437" LastActivityDate="2010-09-02T08:01:22.617" Title="Examples/other references for EGA 0.4.5.4" Tags="&lt;reference-request&gt;&lt;algebraic-geometry&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="1345" PostTypeId="2" ParentId="1338" CreationDate="2010-08-01T15:21:05.333" Score="2" ViewCount="" Body="&lt;p&gt;You can take any two orthogonal vectors $\vec{e_1}$ and $\vec{e_2}$ on the plane and use them as a basis. You also need some point $(x_0, y_0, z_0)$ on the plane as origin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given point with coordinates $(x_1, y_1, z_1)$ on your plane you calculate it's coordinates with respect to new basis:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x = (x_1 - x_0) e_{1x} + (y_1 - y_0) e_{1y} + (z_1 - z_0) e_{1z}$&lt;br /&gt;&#xA;$y = (x_1 - x_0) e_{2x} + (y_1 - y_0) e_{2y} + (z_1 - z_0) e_{2z}$&lt;br /&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And after that you can apply your formulae to get $C_x$ and $C_y$. Those coordinates are easyly transformed back into original 3d coordinates:&lt;br /&gt;&#xA;$x = x_0 + e_{1x} C_x + e_{2x} C_y$&lt;br /&gt;&#xA;$y = y_0 + e_{1y} C_x + e_{2y} C_y$&lt;br /&gt;&#xA;$z = z_0 + e_{1z} C_x + e_{2z} C_y$&lt;br /&gt;&lt;/p&gt;&#xA;" OwnerUserId="334" LastActivityDate="2010-08-01T15:21:05.333" CommentCount="5" />
  <row Id="1346" PostTypeId="2" ParentId="164" CreationDate="2010-08-01T18:22:34.037" Score="12" ViewCount="" Body="&lt;p&gt;The volume of a sphere with radius $a$ may be found by evaluating the triple integral&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle V = \int\int\int_{\ S}\ dx\ dy\ dz,$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $S$ is the volume enclosed by the sphere $x^2+y^2+z^2=a^2$. Changing variables to spherical polar coordinates, we obtain &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle V = \int_{0}^{2\pi}d\phi\int_{0}^{\ \pi}d\theta\int_{0}^{\ a}r^2\sin\theta \ dr = \int_{0}^{2\pi}d\phi\int_{0}^{\pi}\sin\theta \ d\theta\int_{0}^{\ a}r^2\ dr = \frac{4\pi a^3}{3},$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;as expected.&lt;/p&gt;&#xA;" OwnerUserId="207" LastEditorUserId="207" LastEditorDisplayName="" LastEditDate="2010-08-01T18:28:21.637" LastActivityDate="2010-08-01T18:28:21.637" />
  <row Id="1347" PostTypeId="1" AcceptedAnswerId="1383" CreationDate="2010-08-01T18:31:43.533" Score="45" ViewCount="633" Body="&lt;p&gt;You've spent your whole life in the hyperbolic plane. It's second nature to you that the area of a triangle depends only on its angles, and it seems absurd to suggest that it could ever be otherwise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But recently a good friend named Euclid has raised doubts about the fifth postulate of Poincaré's &lt;em&gt;Elements&lt;/em&gt;. This postulate is the obvious statement that given a line $L$ and a point $p$ not on $L$ there are at least two lines through $p$ that do not meet $L$. Your friend wonders what it would be like if this assertion were replaced with the following: given a line $L$ and a point $p$ not on $L$, there is exactly one line through $p$ that does not meet $L$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You begin investigating this Euclidean geometry, but you find it utterly impossible to visualize intrinsically. You decide your only hope is to find a model of this geometry within your familiar hyperbolic plane. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What model do you build?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I do not know if there's a satisfying answer to this question, but maybe it's entertaining to try to imagine. For clarity, we Euclidean creatures have built models like the upper-half plane model or the unit-disc model to visualize hyperbolic geometry within a Euclidean domain. I'm wondering what the reverse would be.&lt;/p&gt;&#xA;" OwnerUserId="251" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-08-20T22:48:37.860" LastActivityDate="2010-08-21T03:18:02.510" Title="Hyperbolic critters studying Euclidean geometry" Tags="&lt;geometry&gt;&lt;euclidean-geometry&gt;&lt;hyperbolic-geometry&gt;" AnswerCount="4" CommentCount="1" FavoriteCount="7" />
  <row Id="1348" PostTypeId="1" CreationDate="2010-08-01T18:47:30.570" Score="6" ViewCount="408" Body="&lt;p&gt;I have a Ph.D. in computational and theoretical chemistry with advanced but field-oriented knowledge of mathematics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am fascinated by fractals, but I am unable to understand them from the formal point of view. To my level of understanding, they look like a graphical rendering of an ill-conditioned iterative problem, where small variations of the initial condition lead to huge changes in the final result, but that's just what I got out of it with my current knowledge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How would you explain fractals (such as the Mandelbrot set) to a layperson with basic mathematics knowledge from high school, and how would you instead explain it to someone which has more math training, but not formal. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This question is collateral to a post on the &lt;a href=&quot;http://forthescience.org/blog/2010/07/12/the-mandelbrot-set-in-python/&quot; rel=&quot;nofollow&quot;&gt;Mandelbrot set&lt;/a&gt; I did on my blog some time ago. If you have any comments on what I was doing with my tinkering of the parameters (to get some keywords for further exploration), it's greatly appreciated. I would like to explain it better to my readers, but I am unable to do it. Thanks&lt;/p&gt;&#xA;" OwnerUserId="559" LastActivityDate="2011-05-30T16:42:00.737" Title="How to explain fractals to a layperson and to someone with more math training?" Tags="&lt;fractals&gt;" AnswerCount="4" CommentCount="5" FavoriteCount="1" />
  <row Id="1349" PostTypeId="2" ParentId="1348" CreationDate="2010-08-01T19:09:50.977" Score="4" ViewCount="" Body="&lt;p&gt;There are a lot of examples that are not hard to define rigorously, e.g. the Cantor set. This can be defined rigorously simply by taking a suitable intersection of nested intervals when the middle thirds are successively removed. It is easy to make all sorts of variants, constructing &quot;Cantor-like&quot; sets when you replace &quot;middle third&quot; with &quot;middle fourth&quot; or even with a construction where the ratio of the interval removed changes with the iteration. (This is important because you can then get a set of positive measure.) Leaving aside the remark about measure, all this uses nothing more than elementary mathematics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One way to think of fractals is as of subsets of the phase space of certain dynamical systems. For instance, consider the &lt;a href=&quot;http://en.wikipedia.org/wiki/Smale_horseshoe&quot; rel=&quot;nofollow&quot;&gt;Smale horseshoe&lt;/a&gt;. This is a two-dimensional map that can be visualized geometrically; this is explained in the Wikipedia article. Then the collection of points that stay in the square when you iterate this procedure both forwards and backwards is a fractal set, basically a two-dimensional Cantor set, which is very interesting from the perspective of dynamical systems (it's chaotic---one consequence of this is that knowing a point to arbitrary precision will not tell you what its orbit looks like later on with much precision).  So much of the interesting behavior of dynamical systems is on fractals. You can define a map of the interval into itself such that the points in the interval that stay in the interval no matter how many times you iterate the map is just the ordinary Cantor set, as well.&lt;/p&gt;&#xA;" OwnerUserId="536" LastActivityDate="2010-08-01T19:09:50.977" />
  <row Id="1350" PostTypeId="1" CreationDate="2010-08-01T19:29:28.793" Score="-2" ViewCount="389" Body="&lt;p&gt;Is the set of polynomials of degree n a vector space? If not, are there any similar sets that are vector spaces?&lt;/p&gt;&#xA;" OwnerUserId="560" LastEditorUserId="123" LastEditorDisplayName="" LastEditDate="2010-08-03T07:30:35.280" LastActivityDate="2010-08-03T07:30:35.280" ClosedDate="2010-08-02T08:37:12.670" Title="Are the polynomials of degree n a vector space?" Tags="&lt;linear-algebra&gt;" AnswerCount="1" CommentCount="15" />
  <row Id="1351" PostTypeId="2" ParentId="1348" CreationDate="2010-08-01T19:53:29.330" Score="8" ViewCount="" Body="&lt;p&gt;For a &quot;high-level&quot; explanation, I would say this: Fractals are surprisingly complex patterns that result from the repeated application of relatively simple operations/rules.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of the easiest to visualize examples is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Koch_snowflake&quot; rel=&quot;nofollow&quot;&gt;Koch snowflake&lt;/a&gt;, constructed by adding smaller triangles to each face of the figure at each iteration:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://upload.wikimedia.org/wikipedia/commons/thumb/d/d9/KochFlake.svg/280px-KochFlake.svg.png&quot; alt=&quot;Koch snowflake&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A more real-world example is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Barnsley%27s_fern&quot; rel=&quot;nofollow&quot;&gt;fern leaf&lt;/a&gt;. The DNA in a single plant cell encodes enough information to describe the structure of an entire leaf (and the entire plant, for that matter) &lt;em&gt;without explicitly describing the location of each cell&lt;/em&gt;. Instead, the cells grow according to a set of simple rules that result in the self-similar appearance of the fern, even at smaller and smaller levels:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://upload.wikimedia.org/wikipedia/commons/7/76/Barnsley_fern_plotted_with_VisSim.PNG&quot; alt=&quot;Barnsley fern&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a more complex mathematical explanation that still remains tied to the real world, have a look at the basic &lt;a href=&quot;http://en.wikipedia.org/wiki/Ricker_model&quot; rel=&quot;nofollow&quot;&gt;Ricker model&lt;/a&gt; of population growth and the resulting &lt;a href=&quot;http://www.phaser.com/modules/ecology/ricker/index.html&quot; rel=&quot;nofollow&quot;&gt;bifurcation diagrams&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://www.phaser.com/modules/ecology/ricker/bif.jpg&quot; alt=&quot;bifurcation diagram&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The x-axis on this graph is population growth rate and the y-axis is population density. Although it looks complex, All it takes is a handful of iterations of the basic formula on a hand calculator to see how the results can oscillate between seemingly random population levels.&lt;/p&gt;&#xA;" OwnerUserId="59" LastEditorUserId="59" LastEditorDisplayName="" LastEditDate="2010-08-03T00:47:55.047" LastActivityDate="2010-08-03T00:47:55.047" CommentCount="1" />
  <row Id="1352" PostTypeId="1" CreationDate="2010-08-01T20:11:01.703" Score="9" ViewCount="515" Body="&lt;p&gt;I just learned (thanks to &lt;a href=&quot;http://mathoverflow.net/questions/10014/applications-of-the-chinese-remainder-theorem/10017#10017&quot; rel=&quot;nofollow&quot;&gt;Harry Gindi&lt;/a&gt; and on &lt;a href=&quot;http://www.artofproblemsolving.com/Forum/blog.php?b=10595&quot; rel=&quot;nofollow&quot;&gt;Qiaochu Yuan&lt;/a&gt; MO) that the chinese remainder theorem and lagrange interpolation are really just two instances of the same thing.  Similarly the method of partial fractions can be applied to rationals rather than polynomials. I find that seeing a method applied in different contexts, or just learning a connection that wasn't apparent helps me appreciate a deeper understanding of both.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I ask, can you help me find more examples of this? Especially ones which you personally found inspiring. &lt;/p&gt;&#xA;" LastEditorUserId="16" LastEditorDisplayName="" LastEditDate="2010-08-01T20:35:53.970" LastActivityDate="2010-08-11T22:02:30.823" CommunityOwnedDate="2010-08-02T09:17:13.293" Title="Surprising Generalizations" Tags="&lt;soft-question&gt;&lt;big-list&gt;&lt;intuition&gt;" AnswerCount="8" CommentCount="8" FavoriteCount="5" />
  <row Id="1353" PostTypeId="2" ParentId="1352" CreationDate="2010-08-01T20:22:20.867" Score="1" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Localization&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I learned that you could localize categories(and not just abelian!) I was floored. The general idea that we take a class of morphisms in a category and send them functorially to another category where they are isos is awesome. It is also very important in my work, which is generalizing some ideas of Algebraic Geometry to a more categorical setting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a &lt;a href=&quot;http://ncatlab.org/nlab/show/localization&quot; rel=&quot;nofollow&quot;&gt;link!&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="16" LastActivityDate="2010-08-01T20:22:20.867" CommunityOwnedDate="2010-08-02T15:52:41.847" />
  <row Id="1354" PostTypeId="2" ParentId="1352" CreationDate="2010-08-01T20:26:23.947" Score="6" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Galois Connections&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's be honest, the correspondence between Galois groups and field extension is pretty hott. The first time I saw this I was duly impressed. However, about two years ago, I learned about universal covering spaces. Wow! I swear my understanding of covering spaces doubled when the prof told me that this was a &quot;Galois correspondence for fundamental groups and covering spaces&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again here is a &lt;a href=&quot;http://en.wikipedia.org/wiki/Galois_connection#Algebraic_topology%3a_covering_spaces&quot; rel=&quot;nofollow&quot;&gt;link!&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="16" LastActivityDate="2010-08-01T20:26:23.947" CommunityOwnedDate="2010-08-02T15:52:41.847" CommentCount="3" />
  <row Id="1355" PostTypeId="2" ParentId="1352" CreationDate="2010-08-01T20:27:16.377" Score="5" ViewCount="" Body="&lt;p&gt;Classification of finitely-generated abelian groups and Jordan normal form are two instances of &lt;a href=&quot;http://en.wikipedia.org/wiki/Structure_theorem_for_finitely_generated_modules_over_a_principal_ideal_domain&quot; rel=&quot;nofollow&quot;&gt;the structure theorem for finitely generated modules over a principal ideal domain&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="152" LastActivityDate="2010-08-01T20:27:16.377" CommunityOwnedDate="2010-08-02T15:52:41.847" CommentCount="1" />
  <row Id="1356" PostTypeId="2" ParentId="1352" CreationDate="2010-08-01T21:00:47.437" Score="3" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Model categories&lt;/strong&gt; as a framework for both complexes of R-modules and topological spaces (making precise, for example, analogy between taking projective resolution and replacing a space with weakly homotopy equivalent CW-complex).&lt;/p&gt;&#xA;" OwnerUserId="152" LastActivityDate="2010-08-01T21:00:47.437" CommunityOwnedDate="2010-08-02T15:52:41.847" />
  <row Id="1357" PostTypeId="2" ParentId="1333" CreationDate="2010-08-01T21:22:26.807" Score="3" ViewCount="" Body="&lt;p&gt;In 1976, &lt;a href=&quot;http://www.prismmodelchecker.org/bibitem.php?key=KY76&quot; rel=&quot;nofollow&quot;&gt;Knuth and Yao&lt;/a&gt; proved a result that may help you choose a DVD out of 5, 6, etc.  (I couldn't find a good reference online.) Consider the following problem: You have a fair coin and you must write an algorithm that outputs 1 with probability p1, outputs 2 with probability p2, ..., and outputs n with probability pn. (p1+p2+...+pn=1) Note that &lt;em&gt;any&lt;/em&gt; possible algorithm can be described in terms of certain (possibly infinite) binary trees: A node that has two children means &quot;throw a coin and pick one of the children according to the result&quot;; a leaf is tagged with one of the numbers 1, 2, ..., n and means &quot;output that number&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;An optimal tree that solves the problem has one leaf k on level m if and only if the m-th bit of pk is 1. Otherwise there are zero such leafs.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Optimal&quot; here means not only that the average runtime is as good as possible, but the  stronger guarantee that the probability of running more than m steps is not worse than that of any other algorithm that solves the problem correctly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your case, p1=p2=p3=0.01010101... The digit just before dot corresponds to the root (level 0) and and the subsequent digits to subsequent levels. So a tree that has no leaf on levels 0 and 1, has leafs (1, 2, and 3) on level 2, has no leaf on level 3, ... is optimal. This is exactly the solution given by ShreevatsaR. Such a tree is fairly easy to find once you know what leafs you need to put on each level.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As another example, if p1=...=p6=1/6=0.00101010101..., then you should have leafs (one of each 1, 2, 3, 4, 5, 6) on levels 3, 5, 7, and so on. If you draw a symmetric tree that obeys this you'll find how to choose one DVD out of 6: First grab three with each hand, then do what ShreevatsaR said for three.  You are also guaranteed that it's optimal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now you can mock me and my friends for &lt;a href=&quot;http://www.youtube.com/watch?v=3JBE5ykC1No#t=1m&quot; rel=&quot;nofollow&quot;&gt;using a suboptimal algorithm for allocating hotel rooms&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="512" LastActivityDate="2010-08-01T21:22:26.807" CommentCount="1" />
  <row Id="1358" PostTypeId="2" ParentId="457" CreationDate="2010-08-01T21:26:32.957" Score="4" ViewCount="" Body="&lt;p&gt;Some principles of inversive geometry can be demonstrated using &lt;a href=&quot;http://mathworld.wolfram.com/PeaucellierInversor.html&quot; rel=&quot;nofollow&quot;&gt;Peaucellier's Inversor&lt;/a&gt;, a linkage device designed to transform circular motion into motion in a straight line and vice versa. There are various online demonstrations however, you may wish to build a &quot;real&quot; one for yourself. The book &quot;Mathematical Models&quot; by Cundy and Rollett has an entire chapter devoted to making mechanical models, including the above and related likages. &lt;/p&gt;&#xA;" OwnerUserId="207" LastActivityDate="2010-08-01T21:26:32.957" CommunityOwnedDate="2010-08-01T21:26:32.957" CommentCount="1" />
  <row Id="1359" PostTypeId="2" ParentId="457" CreationDate="2010-08-01T21:46:41.357" Score="3" ViewCount="" Body="&lt;p&gt;I'm not sure how good of an example this is since the original problem already has a natural physical interpretation. Nonetheless this is an example of something that can be proved using Euclidean geometry or mechanics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Starting on one of the edges of a plane polygon $P$ one can set into motion a point-mass which moves along a straight trajectory except on collision with another edge, when it is subjected to an elastic collision response (i.e., its velocity vector is reflected in the edge it has collided with). The trajectory traced out by such a point-mass is its orbit, and an orbit is periodic if the point-mass eventually returns to its starting spot with its starting velocity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of the most basic questions one can ask is whether periodic orbits exist in acute triangles. The answer is yes, and a particularly nice periodic orbit in an acute triangle can be constructed by drawing the three altitudes and connecting their bases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This can be proved using geometry, but it can also be seen in the following physical manner. Assume the triangle's edges are thin wires. Place around each wire a small ring which is free to move along its edge. Now thread a rubber band through the three rings. Tighten this rubber band. This system reaches equilibrium when the rings occupy the positions of the bases of the altitudes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, in the equilibrium state the triangle whose vertices are the rings will be the inscribed triangle of minimal perimeter. There's no such minimum in an obtuse triangle, so the construction doesn't work in that case. As of today it is unknown whether every obtuse triangle admits a periodic orbit, although much partial progress has been achieved.&lt;/p&gt;&#xA;" OwnerUserId="251" LastEditorUserId="251" LastEditorDisplayName="" LastEditDate="2010-11-08T18:32:35.527" LastActivityDate="2010-11-08T18:32:35.527" CommunityOwnedDate="2010-08-01T21:46:41.357" />
  <row Id="1360" PostTypeId="2" ParentId="1350" CreationDate="2010-08-01T21:53:52.680" Score="2" ViewCount="" Body="&lt;p&gt;It is not a vector space because there is no zero element.&lt;/p&gt;&#xA;" OwnerUserId="563" LastActivityDate="2010-08-01T21:53:52.680" CommentCount="0" />
  <row Id="1361" PostTypeId="2" ParentId="1347" CreationDate="2010-08-01T22:01:51.833" Score="7" ViewCount="" Body="&lt;p&gt;Look up &quot;horosphere&quot; (for example, in page 90 of the Princeton Companion to Mathematics).  Wikipedia describes it on its &lt;a href=&quot;http://en.wikipedia.org/wiki/horoball&quot; rel=&quot;nofollow&quot;&gt;Horoball&lt;/a&gt; page. &lt;/p&gt;&#xA;" OwnerUserId="273" LastEditorUserId="273" LastEditorDisplayName="" LastEditDate="2010-08-01T22:15:33.460" LastActivityDate="2010-08-01T22:15:33.460" CommentCount="4" />
  <row Id="1362" PostTypeId="2" ParentId="1348" CreationDate="2010-08-01T22:07:53.733" Score="7" ViewCount="" Body="&lt;p&gt;That's easy.  I would buy them some of this stuff:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://farm2.static.flickr.com/1340/614003477_2d7fd27ca5.jpg&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Then I would tell them to stare at it for awhile.)  It's called Romanesco broccoli, and it grows like this for the same reasons, as explained in e.James' answer, that ferns do.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternately, I would tell them to check out &lt;em&gt;Indra's Pearls&lt;/em&gt; by Mumford, Series, and Wright.&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-08-01T22:22:55.683" LastActivityDate="2010-08-01T22:22:55.683" CommentCount="3" />
  <row Id="1363" PostTypeId="2" ParentId="1352" CreationDate="2010-08-01T22:16:59.357" Score="8" ViewCount="" Body="&lt;p&gt;I agree!  I spend much of my mathematical free time exploring such connections.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a basic one that I constantly ponder.  The rules of matrix multiplication encode two things:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;How to compose a linear transformation $A$ with another linear transformation $B$, with respect to a fixed basis.&lt;/li&gt;&#xA;&lt;li&gt;How to follow an edge of type $A$ on a graph, and then follow an edge of type $B$ (where $A$ and $B$ are just a disjoint partition of the set of edges).&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This means that one can study walks on graphs by studying how a matrix called the &lt;a href=&quot;http://en.wikipedia.org/wiki/Adjacency_matrix&quot; rel=&quot;nofollow&quot;&gt;adjacency matrix&lt;/a&gt; behaves.  This leads into all sorts of &lt;a href=&quot;http://en.wikipedia.org/wiki/Spectral_graph_theory&quot; rel=&quot;nofollow&quot;&gt;beautiful mathematics&lt;/a&gt;; for example, this is the basic tool behind Google's PageRank algorithm, and it also in some sense motivated Heisenberg's &lt;a href=&quot;http://en.wikipedia.org/wiki/Matrix_mechanics&quot; rel=&quot;nofollow&quot;&gt;matrix mechanics&lt;/a&gt; formulation of quantum mechanics.   I often try to recast results in linear algebra in terms of some combinatorial statement about walks on graphs.  &lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-08-01T22:16:59.357" CommunityOwnedDate="2010-08-02T15:52:41.847" CommentCount="4" />
  <row Id="1364" PostTypeId="1" AcceptedAnswerId="1365" CreationDate="2010-08-01T22:33:45.520" Score="1" ViewCount="41" Body="&lt;p&gt;I guarantee there is an easy reference on this, but for some reason I cannot find it. If you can point me to a reference or just write a short proof for me, I would be appreciative.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given a graded ring $R_{\bullet}$ and a localization $R_{\bullet}^{*}$. We also have a graded $R_{\bullet}$-mod, $M_{\bullet}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what I want to know; is &lt;code&gt;$\left(R_{\bullet}^{*}\otimes M_{\bullet}\right)_0=\left(R_{\bullet}^{*}\right)_0\otimes \left(M_{\bullet}\right)_0$&lt;/code&gt;?&lt;/p&gt;&#xA;" OwnerUserId="16" LastEditorUserId="16" LastEditorDisplayName="" LastEditDate="2010-08-01T22:55:55.400" LastActivityDate="2010-08-01T23:13:19.207" Title="The zero component of tensor products of graded mods" Tags="&lt;reference-request&gt;&lt;graded-modules&gt;" AnswerCount="1" CommentCount="7" />
  <row Id="1365" PostTypeId="2" ParentId="1364" CreationDate="2010-08-01T23:13:19.207" Score="1" ViewCount="" Body="&lt;p&gt;For a counterexample, take $R=k[t]$ with its usual grading and $M=R(1)$, the free module of rank one generated in degree $1$. &lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2010-08-01T23:13:19.207" CommentCount="1" />
  <row Id="1366" PostTypeId="2" ParentId="1352" CreationDate="2010-08-01T23:37:04.630" Score="2" ViewCount="" Body="&lt;p&gt;The law of cosines and the equation for the variance of a sum of (possibly correlated) random variables are both consequences of basic inner product space properties. Details &lt;a href=&quot;http://www.johndcook.com/blog/2010/06/17/covariance-and-law-of-cosines/&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="247" LastActivityDate="2010-08-01T23:37:04.630" CommunityOwnedDate="2010-08-02T15:52:41.847" />
  <row Id="1367" PostTypeId="2" ParentId="1352" CreationDate="2010-08-02T00:06:44.387" Score="6" ViewCount="" Body="&lt;p&gt;I loved learning about how differential forms and the exterior derivative generalize 3-d vector calculus (div, grad, curl).  Differential forms are so elegant in comparison, work in arbitrary dimensions, and give rise to beautiful mathematics (e.g. de Rham cohomology, Hodge theory).  And of course, the generalized Stoke's theorem is one of the prettiest equations: $\int_{\partial R} \phi = \int_R d\phi$.&lt;/p&gt;&#xA;" OwnerUserId="9" LastActivityDate="2010-08-02T00:06:44.387" CommunityOwnedDate="2010-08-02T15:52:41.847" />
  <row Id="1368" PostTypeId="1" AcceptedAnswerId="1370" CreationDate="2010-08-02T00:54:21.387" Score="2" ViewCount="733" Body="&lt;p&gt;Struggling with yet another proof:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Prove that, for any positive integer n: (a + b)&lt;sup&gt;n&lt;/sup&gt; &gt;= a&lt;sup&gt;n&lt;/sup&gt; + b&lt;sup&gt;n&lt;/sup&gt; for all a, b &gt; 0:&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I wasted 3 pages of notebook paper on this problem, and I'm getting nowhere slowly. So I need some hints.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;What technique would you use to prove this (e.g. induction, direct, counter example)&lt;/li&gt;&#xA;&lt;li&gt;Are there any tricks to the proof? I've seen some crazy stuff pulled out of nowhere when it comes to proofs...&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="494" LastEditorUserId="536" LastEditorDisplayName="" LastEditDate="2010-08-02T01:00:50.487" LastActivityDate="2010-08-11T22:05:41.697" Title="Prove: (a + b)^n &gt;= a^n + b^n" Tags="&lt;algebra-precalculus&gt;" AnswerCount="6" CommentCount="3" />
  <row Id="1369" PostTypeId="2" ParentId="1368" CreationDate="2010-08-02T01:00:04.590" Score="6" ViewCount="" Body="&lt;p&gt;This follows directly from the &lt;a href=&quot;http://en.wikipedia.org/wiki/Binomial_theorem&quot; rel=&quot;nofollow&quot;&gt;binomial theorem&lt;/a&gt;.&#xA;Alternatively, you can prove it inductively (which is probably more fun): suppose the inequality true for $n-1$. Then $(a+b)^n = (a+b)(a+b)^{n-1} \geq (a+b)(a^{n-1} + b^{n-1})$ by the inductive hypothesis. So $(a+b)^n \geq  a(a^{n-1}+ b^{n-1}) + b(b^{n-1} + a^{n-1})$, and this is at least $a^n + b^n$.&lt;/p&gt;&#xA;" OwnerUserId="536" LastActivityDate="2010-08-02T01:00:04.590" />
  <row Id="1370" PostTypeId="2" ParentId="1368" CreationDate="2010-08-02T01:00:14.090" Score="7" ViewCount="" Body="&lt;p&gt;Hint: Use the binomial theorem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This states that &#xA;$(a + b)^n = \sum_{k = 0}^n {n \choose k} a^{n-k} b^k = a^n + b^n + \sum_{k=1}^{n-1} {n \choose k} a^{n-k} b^k$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, note that every term in the second sum is positive; this is because a, b, and the binomial coefficients are all positive. Therefore, (a+b)&lt;sup&gt;n&lt;/sup&gt; = a&lt;sup&gt;n&lt;/sup&gt; + b&lt;sup&gt;n&lt;/sup&gt; + (sum of positive terms) &gt;= a&lt;sup&gt;n&lt;/sup&gt; + b&lt;sup&gt;n&lt;/sup&gt;.&lt;/p&gt;&#xA;" OwnerUserId="30" LastEditorUserId="153" LastEditorDisplayName="" LastEditDate="2010-08-02T23:55:06.150" LastActivityDate="2010-08-02T23:55:06.150" CommentCount="5" />
  <row Id="1371" PostTypeId="2" ParentId="760" CreationDate="2010-08-02T01:48:26.383" Score="0" ViewCount="" Body="&lt;p&gt;I only wish to add the point made in the &lt;a href=&quot;http://en.wikipedia.org/wiki/Great-circle_distance&quot; rel=&quot;nofollow&quot;&gt;Wikipedia article&lt;/a&gt; for the great circle distance: namely, the classical formula making use of the arccosine can suffer from numerical instability. The formula using the arctangent (which when implemented on a computer should be done using the two-argument arctangent) is numerically robust.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-08-02T01:48:26.383" />
  <row Id="1372" PostTypeId="2" ParentId="1368" CreationDate="2010-08-02T02:10:27.063" Score="6" ViewCount="" Body="&lt;p&gt;It might also be helpful for you to think a little about the geometry of the inequality.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $n=2$, find a way to put an $a \times a$ square and a $b \times b$ square into a $(a+b) \times (a+b)$ square without any &#xA;overlaps. For $n=3$, see if you can fit an $a \times a \times a$ cube and a $b \times b \times b$ cube within an $(a+b) \times (a+b) \times (a+b)$ cube without overlaps. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next, the notion of having more than three dimensions might seem a little weird, but &#xA;think of the box in $n$ dimensions whose sides have length $a+b$. Can you fit two boxes within it, one with side length $a$ and one with side length $b$? &lt;/p&gt;&#xA;" OwnerUserId="317" LastActivityDate="2010-08-02T02:10:27.063" />
  <row Id="1373" PostTypeId="2" ParentId="1344" CreationDate="2010-08-02T04:30:58.563" Score="5" ViewCount="" Body="&lt;p&gt;Dear Akhil,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think one typical situation when this would be used (which you probably already know)&#xA;is when you want to reduce the construction of some object which is supposed to lie over&#xA;a base to the case when the base is affine.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Namely, imagine you want to build a scheme T over a base S, and you cover S by open affines S_i; then if you construct the schemes T_i (pull-back of T over S_i) with appropriate gluing data, you will be done.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, in practice, you won't have T but will instead have the functor it represents on S-schemes, T_i will be the fibre product of that functor and S_i (i.e. just restrict the functor to S_i-schemes), and now if T is a sheaf, and the T_i are representable, you are done.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could imagine applying this to build projective spaces or Grassmanians over T. E.g. suppose you had a locally free sheaf from which you wanted to form the associated projective space bundle.  One could do this directly, with some Proj construction, but one could instead figure out the universal property, and then replace S by an open cover S_i over which the locally free sheaf is actually free, in which case the usual projective space of the appropriate dimension&#xA;(taken over S_i) will obviously represent your functor (provided you figured the functor out correctly!).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I should say that, outside of the context of EGA itself, I don't imagine that people cite this result (or analogous ones) very often.  They are more likely just to write something like &quot;since F is a sheaf, we can reduce our construction to the case when S is affine&quot;.  It is just one of the standard techniques that float around for trying to represent moduli problems.  &lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2010-08-02T04:30:58.563" CommentCount="1" />
  <row Id="1374" PostTypeId="2" ParentId="1368" CreationDate="2010-08-02T06:02:08.590" Score="2" ViewCount="" Body="&lt;p&gt;Let's have a precalculus answer: Consider the function of $a$ depending on the parameter $b$ that is $f_b(a)=(a+b)^n-a^n-b^n$. Its derivative relative to $a$ is $f'_b(a)=n((a+b)^{n-1}-a^{n-1})$ because $b&gt;0$ $f'_b(a,b)$ is nonnegative and $f_b$ is an increasing function of $a$ and you can conclude.&lt;/p&gt;&#xA;" OwnerUserId="441" LastEditorUserId="441" LastEditorDisplayName="" LastEditDate="2010-08-02T10:43:39.783" LastActivityDate="2010-08-02T10:43:39.783" CommentCount="2" />
  <row Id="1375" PostTypeId="2" ParentId="1368" CreationDate="2010-08-02T07:54:15.427" Score="3" ViewCount="" Body="&lt;p&gt;Here is another way to look at the inequality.  Pick the larger of a^n and b^n and divide through by that quantity.  This reduces the problem to showing that (1+r)^n &gt; = 1 + r^n for&#xA;some positive real number r &amp;lt;= 1.  If r &amp;lt; 1, we have r^n &amp;lt; r, so 1 + r^n &amp;lt; 1 + r &amp;lt; (1+r)^n.&#xA;I leave the case r = 1 (and n rational and &gt;= 1)  to others.&lt;/p&gt;&#xA;" OwnerUserId="297" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-08-11T22:05:41.697" LastActivityDate="2010-08-11T22:05:41.697" CommentCount="0" />
  <row Id="1376" PostTypeId="2" ParentId="760" CreationDate="2010-08-02T08:05:28.067" Score="2" ViewCount="" Body="&lt;p&gt;If you have the Euclidean distance d between the two points, and set D = the diameter of the sphere (so D = 2R), then the great circle distance is D*InvSin(d/D), where InvSin is the inverse sine function.  If you know that d/D is usually very small, then you can  approximate the great circle distance using a few terms of the Taylor series for inverse sine.&lt;/p&gt;&#xA;" OwnerUserId="297" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-08-11T22:01:26.910" LastActivityDate="2010-08-11T22:01:26.910" CommentCount="0" />
  <row Id="1377" PostTypeId="2" ParentId="1352" CreationDate="2010-08-02T08:12:36.520" Score="2" ViewCount="" Body="&lt;p&gt;Galois connections are further reaching than one first realizes.  One small application is in model theory, where the relation R between sentences of a theory and models given by&#xA;t R M if sentence t is true in model M, leads to deductively closed theories versus classes of models closed under certain operations, many of which are algebraic in nature.&#xA;They also arise in algebraic geometry and computer science, among other fields.  There&#xA;is a book on Galois connections edited by Marcel Erne; for the strongly inquisitive I recommend checking it out.&lt;/p&gt;&#xA;" OwnerUserId="297" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-08-11T22:02:30.823" LastActivityDate="2010-08-11T22:02:30.823" CommunityOwnedDate="2010-08-02T15:52:41.847" CommentCount="1" />
  <row Id="1378" PostTypeId="1" CreationDate="2010-08-02T08:34:58.713" Score="0" ViewCount="222" Body="&lt;p&gt;While reading one aptitude book, i came across this question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;35% alcohol mixed with 60% alcohol to get a 50% alcohol. In what ratio were they mixed?&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I spent 20 minutes on this question, but I couldn't even figure out what it means! Could anyone explain this to me?&lt;/p&gt;&#xA;" OwnerUserId="569" LastEditorUserId="123" LastEditorDisplayName="" LastEditDate="2010-08-02T09:23:05.357" LastActivityDate="2010-08-04T13:52:32.663" Title="Mixing solutions of specified strength" Tags="&lt;algebra-precalculus&gt;" AnswerCount="3" CommentCount="5" />
  <row Id="1379" PostTypeId="2" ParentId="1378" CreationDate="2010-08-02T08:54:42.007" Score="1" ViewCount="" Body="&lt;p&gt;Call the ratio you want to find r=(volume of 35% alcohol):(volume of 60% alcohol).  Suppose that there are r liters of the 35% alcohol (so there are 0.35 L of alcohol and 0.65 L of water), 1 liter of the 60% alcohol (so the ratio is r), and 1+r liters of the 50% alcohol.  From there, you can construct an equation about the alcohol (or about the water) and solve for r.&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-08-02T08:54:42.007" />
  <row Id="1380" PostTypeId="2" ParentId="1378" CreationDate="2010-08-02T08:55:45.233" Score="-1" ViewCount="" Body="&lt;p&gt;Given &lt;code&gt;a&lt;/code&gt; liters of &lt;code&gt;35%&lt;/code&gt; alcohol solution, and &lt;code&gt;b&lt;/code&gt; liters of &lt;code&gt;60%&lt;/code&gt; alcohol solution, what would be the alcohol ratio after we'll mix this two solutions?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;code&gt;a&lt;/code&gt; liters are actually &lt;code&gt;a*35/100&lt;/code&gt; alcohol, and &lt;code&gt;a*(1-(35/100))&lt;/code&gt; water. Likewise the &lt;code&gt;b&lt;/code&gt; liters contains &lt;code&gt;b*60/100&lt;/code&gt; alcohol and &lt;code&gt;b*(1-(60/100))&lt;/code&gt; water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mixing them together would give us &lt;code&gt;a*35/100+b*60/100&lt;/code&gt; alcohol and &lt;code&gt;a*(1-(35/100))+b*(1-(60/100))&lt;/code&gt; water. The question is for which &lt;code&gt;a,b&lt;/code&gt; we have the ratio&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;((a*35+b*60)/100)/((a*65+b*40)/100) = 50% = 1/2&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;or&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;(a*35+b*60)/(a*65+b*40) = 1/2&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I believe you can take it from here...&lt;/p&gt;&#xA;" OwnerUserId="52" LastEditorUserId="52" LastEditorDisplayName="" LastEditDate="2010-08-04T06:32:22.297" LastActivityDate="2010-08-04T06:32:22.297" CommentCount="3" />
  <row Id="1381" PostTypeId="2" ParentId="1368" CreationDate="2010-08-02T09:21:38.727" Score="0" ViewCount="" Body="&lt;p&gt;Induction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For n=1 it is trivially true&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume true for n=k i.e. (a + b)^k &gt;= a^k + b^k&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider case n=k+1 (a+b)^(k+1)=(a+b)(a+b)^k&gt;=(a+b)(a^k+b^k)=a^(k+1)+b^(k+1)+ab^k+ba^k&gt;=a^(k+1)+b^(k+1) as required&lt;/p&gt;&#xA;" LastActivityDate="2010-08-02T09:21:38.727" />
  <row Id="1382" PostTypeId="1" AcceptedAnswerId="1746" CreationDate="2010-08-02T10:41:04.887" Score="7" ViewCount="359" Body="&lt;p&gt;Some time ago &lt;a href=&quot;http://math.stackexchange.com/users/232/qiaochu-yuan&quot;&gt;Qiaochu Yuan&lt;/a&gt; asked about &lt;a href=&quot;http://math.stackexchange.com/questions/918/how-do-i-count-the-subsets-of-a-set-whose-number-of-elements-is-divisible-by-3-4/&quot;&gt;counting subsets of a set whose number of elements is divisible by 3 (or 4)&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The story becomes even more interesting if one asks about number of subsets of n-element set with r mod 5 elements. Denote this number by, say, P&lt;sub&gt;n&lt;/sub&gt;(r mod 5).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An experiment shows that for small n P&lt;sub&gt;n&lt;/sub&gt;(r mod 5)-P&lt;sub&gt;n&lt;/sub&gt;(r' mod 5) is always a Fibonacci number (recall that for &quot;r mod 3&quot; corresponding difference is always 0 or 1 and for &quot;r mod 2&quot; they are all 0). It's not hard to prove this statement by induction but as always inductive proof explains nothing. &lt;em&gt;Does anybody have a combinatorial proof?&lt;/em&gt; (Or maybe some homological proof&amp;nbsp;&amp;mdash; I've heard one for &quot;r mod 3&quot;-case.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And is there some theorem about P&lt;sub&gt;n&lt;/sub&gt;(r mod l) for arbitrary l (besides that it satisfies some recurrence relation of degree growing with l)?&lt;/p&gt;&#xA;" OwnerUserId="152" LastEditorUserId="152" LastEditorDisplayName="" LastEditDate="2010-08-03T08:05:09.893" LastActivityDate="2010-08-07T07:24:39.447" Title="Counting subsets with r mod 5 elements" Tags="&lt;combinatorics&gt;&lt;binomial-coefficients&gt;" AnswerCount="1" CommentCount="8" FavoriteCount="1" />
  <row Id="1383" PostTypeId="2" ParentId="1347" CreationDate="2010-08-02T11:19:44.533" Score="34" ViewCount="" Body="&lt;p&gt;Here's another version of Doug Chatham's answer, but with details.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you lived in Hyperbolic space, then Euclidean geometry would be natural to you as well.  The reason is that you can take what is called a horosphere (in the half-space model for us, this is just a hyperplane which is parallel to our limiting hyperplane) and this surface actually has a Euclidean geometry on it!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So unlike for us, where the hyperbolic plane cannot be embedded into Euclidean 3-space, the opposite is true: the Euclidean plane can be embedded into hyperbolic 3-space! So this is analogous to our understanding of spherical geometry.  It's no surprise the spherical geometry is slightly different, however, it fits nicely into our Euclidean view of things, because spherical geometry is somewhat contained in three-dimensional geometry because of the embedding.&lt;/p&gt;&#xA;" OwnerUserId="176" LastActivityDate="2010-08-02T11:19:44.533" CommentCount="4" />
  <row Id="1384" PostTypeId="2" ParentId="1260" CreationDate="2010-08-02T11:27:05.767" Score="0" ViewCount="" Body="&lt;p&gt;This was the specific code I needed:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;\begin{bmatrix} &#xA;x_{1} &amp;amp; x_{4} &amp;amp; x_{7} \\&#xA;x_{2} &amp;amp; x_{5} &amp;amp; x_{8} \\&#xA;x_{3} &amp;amp; x_{6} &amp;amp; x_{9} &#xA;\end{bmatrix}&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;$\begin{bmatrix} &#xA;x_{1} &amp;amp; x_{4} &amp;amp; x_{7} \&#xA;x_{2} &amp;amp; x_{5} &amp;amp; x_{8} \&#xA;x_{3} &amp;amp; x_{6} &amp;amp; x_{9} &#xA;\end{bmatrix}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;(Looks like the library the site is using doesn't support bmatrix?)&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Texify's version:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://www.texify.com/img/%5CLARGE%5C%21%5Cbegin%7Bbmatrix%7D%20x_%7B1%7D%20%26%20x_%7B4%7D%20%26%20x_%7B7%7D%20%5C%5Cx_%7B2%7D%20%26%20x_%7B5%7D%20%26%20x_%7B8%7D%20%5C%5Cx_%7B3%7D%20%26%20x_%7B6%7D%20%26%20x_%7B9%7D%20%5Cend%7Bbmatrix%7D.gif&quot; alt=&quot;column vector&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="54" LastActivityDate="2010-08-02T11:27:05.767" />
  <row Id="1385" PostTypeId="2" ParentId="548" CreationDate="2010-08-02T11:38:02.740" Score="3" ViewCount="" Body="&lt;p&gt;The existence of a (multiplicative) inverse for the (additive) zero is &lt;a href=&quot;http://en.wikipedia.org/wiki/Consistency&quot; rel=&quot;nofollow&quot;&gt;inconsistent&lt;/a&gt; with the other &lt;a href=&quot;http://en.wikipedia.org/wiki/Field_%28mathematics%29#Definition_and_illustration&quot; rel=&quot;nofollow&quot;&gt;field axioms&lt;/a&gt;. As you noticed, the crux is that anything multiplied by zero gives zero. Let us then establish this fact first.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;0 is identity for addition: $0+0=0$ &lt;br/&gt;&#xA;multiply by some x: $(0+0)\cdot x=0\cdot x$&lt;br/&gt;&#xA;distributivity: $0\cdot x + 0\cdot x = 0\cdot x$&lt;br/&gt;&#xA;add the additive inverse of $0\cdot x$ to both sides: $(0\cdot x + 0\cdot x) + (-(0\cdot x)) = 0\cdot x + (-(0\cdot x))$&lt;br/&gt;&#xA;associativity of addition: $0\cdot x + (0\cdot x + (-(0\cdot x))) = 0\cdot x + (-(0\cdot x))$&lt;br/&gt;&#xA;definition of &quot;additive inverse&quot;: $0\cdot x + 0 = 0$&lt;br/&gt;&#xA;zero is additive identity: $0\cdot x = 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's assume now that there is a multiplicative inverse of 0, denoted by Z.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(*) $0\times Z=1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the last two relations, 1=0, which contradicts another field axiom (often forgotten), which is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$1\ne0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, you either accept that 0 has no inverse or you change at least one of the field axioms---you can't have both at the same time. In a sense, it is a matter of convention which axioms you choose. In practice, some sets of axioms lead to more useful consequences. For example, if you want 0 to have an inverse and drop the axiom saying that 1 does not equal 0, then the 'arithmetic' you end up doing won't be very interesting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In short, you are right.&lt;/p&gt;&#xA;" OwnerUserId="512" LastActivityDate="2010-08-02T11:38:02.740" />
  <row Id="1386" PostTypeId="2" ParentId="536" CreationDate="2010-08-02T12:06:59.440" Score="0" ViewCount="" Body="&lt;p&gt;If you'd need only the 1d case, you can get this statement from the &lt;a href=&quot;http://en.wikipedia.org/wiki/Law_of_the_iterated_logarithm&quot; rel=&quot;nofollow&quot;&gt;Law of the iterated logarithm&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This does not give much combinatorial insight, and well... requires said law, which is a little more work to proof.&lt;/p&gt;&#xA;" OwnerUserId="505" LastActivityDate="2010-08-02T12:06:59.440" />
  <row Id="1387" PostTypeId="2" ParentId="1113" CreationDate="2010-08-02T12:18:54.860" Score="7" ViewCount="" Body="&lt;p&gt;I think I know how to prove it without using of Carmichael numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take $n = 3^{2^k} - 2^{2^k}$. Then it's not hard to prove by induction on $k$ that $2^k | n - 1$. Hence $3^{2^k} - 2^{2^k} | 3^{n-1} - 2^{n-1}$ (that's true because $a - b | a^r - b^r$ for any natural $a, b$ and $k$).&lt;/p&gt;&#xA;" OwnerUserId="334" LastActivityDate="2010-08-02T12:18:54.860" />
  <row Id="1388" PostTypeId="1" AcceptedAnswerId="1399" CreationDate="2010-08-02T12:42:27.637" Score="7" ViewCount="409" Body="&lt;p&gt;Galois theory tells us that $\frac{z^{11}-1}{z-1} = z^{10} + z^9 + z^8 + z^7 + z^6 + z^5 + z^4 + z^3 + z^2 + z + 1$ can be solved in radicals because its group is solvable. Actually performing the calculation is beyond me, though - here what I have got so far:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let the roots be $\zeta^1,\zeta^2,\ldots,\zeta^{10}$, following Gauss we can split the problem into solving quintics and quadratics by looking at subgroups of the roots. Since 2 is a generator of the group $[2,4,8,5,10,9,7,3,6,1]$ we can partition into the five subgroups of conjugate pairs $[2,9]$,$[4,7]$,$[8,3]$,$[5,6]$,$[10,1]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now put $q_1 = \zeta^2+\zeta^9$,$q_2 = \zeta^4+\zeta^7$,$q_3 = \zeta^8+\zeta^3$,$q_4 = \zeta^5+\zeta^6$,$q_5 = \zeta^{10}+\zeta^1$. So if we can solve the quintic $(q - q_1)(q - q_2)(q - q_3)(q - q_4)(q - q_5) = q^5 + q^4 - 4q^3 - 3q^2 + 3q + 1 = 0$ we would just be left to solve a few quadratic equations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now pari/gp tells me this quintic has the cyclic group C(5):&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;? polgalois(x^5 + x^4 - 4*x^3 - 3*x^2 + 3*x + 1)&#xA;%1 = [5, 1, 1, &quot;C(5) = 5&quot;]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I've worked through examples of solving quadratic and cubic equations based on the galois group but when it comes to this quintic I'm completely stumped so any advice would be tremendously helpful! Thanks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; Thanks to Robin Chapman the problem is reduced significantly. Let $\omega$ be a primitive 5th root of unity (which is easy to express in radicals), it only remains to express $(q_1 + \omega q_2  + \omega^2 q_3  + \omega^3 q_4  + \omega^4 q_5)^5$ in terms of rationals and powers of $\omega$ (and then everything can be substituted back and solved easily). We know this is possible because the term is fixed by the quintics galois group, how to actually perform this evades me but I will try to find a way.&lt;/p&gt;&#xA;" LastActivityDate="2010-08-05T19:34:55.197" Title="How to solve a cyclic quintic in radicals?" Tags="&lt;algebra&gt;&lt;polynomials&gt;" AnswerCount="4" CommentCount="1" FavoriteCount="3" />
  <row Id="1389" PostTypeId="1" AcceptedAnswerId="1391" CreationDate="2010-08-02T13:06:19.443" Score="7" ViewCount="344" Body="&lt;p&gt;Prove that the equation &lt;i&gt;n&lt;sup&gt;a&lt;/sup&gt; + n&lt;sup&gt;b&lt;/sup&gt; = n&lt;sup&gt;c&lt;/sup&gt;&lt;/i&gt;, with &lt;i&gt;a,b,c,n &lt;/i&gt; positive integers, has infinite solutions if &lt;i&gt;n&lt;/i&gt;=2, and no solution if &lt;i&gt;n&lt;/i&gt;&amp;ge;3.&lt;/p&gt;&#xA;" OwnerUserId="89" LastEditorUserId="89" LastEditorDisplayName="" LastEditDate="2010-08-02T13:17:52.630" LastActivityDate="2010-08-02T20:24:33.080" Title="Not quite Fermat's Last Theorem" Tags="&lt;number-theory&gt;&lt;algebra&gt;" AnswerCount="4" CommentCount="2" FavoriteCount="3" />
  <row Id="1390" PostTypeId="2" ParentId="1389" CreationDate="2010-08-02T13:08:19.157" Score="2" ViewCount="" Body="&lt;p&gt;(assume b&amp;ge;a) &lt;code&gt;$n^b&amp;lt;n^a+n^b&amp;lt;n^{b+1}$&lt;/code&gt;.&lt;/p&gt;&#xA;" OwnerUserId="152" LastActivityDate="2010-08-02T13:08:19.157" CommentCount="3" />
  <row Id="1391" PostTypeId="2" ParentId="1389" CreationDate="2010-08-02T13:11:50.193" Score="17" ViewCount="" Body="&lt;p&gt;So this is fermats last theorem upside down? It occurs to me if we have two binary numbers we may add them to get another power of two,&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;   1000000&#xA;   1000000&#xA;+ --------&#xA;  10000000&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;but if we had two numbers in base 3, say&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;  1000000&#xA;  1000000&#xA;+ -------&#xA;  2000000&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;we would not have so much luck.&lt;/p&gt;&#xA;" LastActivityDate="2010-08-02T13:11:50.193" CommentCount="14" />
  <row Id="1392" PostTypeId="2" ParentId="1389" CreationDate="2010-08-02T13:13:21.440" Score="4" ViewCount="" Body="&lt;p&gt;If $n=2$ we can take $a=k, b=k, c=k+1$ for any $k \in \mathbb{N}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $n \ge 3$. We can assume that $a, b, c \ge 0$ because if not we could multiply left and right side by $n^k$ to make them positive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now it's clear that $c \ge a$ and $c \ge b$. Then we have $n^a | n^c$, hence $n^a | n^a + n^b$ and $a \le b$. In the same way $b \le a$. So $a = b$. Hence $2n^a = n^c$ and $n=2$.&lt;/p&gt;&#xA;" OwnerUserId="334" LastActivityDate="2010-08-02T13:13:21.440" CommentCount="1" />
  <row Id="1393" PostTypeId="1" AcceptedAnswerId="1404" CreationDate="2010-08-02T13:14:03.730" Score="1" ViewCount="188" Body="&lt;p&gt;I already asked if there are more rationals than integers here...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://math.stackexchange.com/questions/1311/are-there-more-rational-numbers-than-integers&quot;&gt;http://math.stackexchange.com/questions/1311/are-there-more-rational-numbers-than-integers&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, there is one particular argument that I didn't give before which I still find compelling...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Every integer is also a rational. There exist (many) rationals that are not integers. Therefore there are more rationals than integers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously, in a sense, I am simply choosing one particular bijection, so by the definition of set cardinality this argument is irrelevant. But it's still a compelling argument for &quot;size&quot; because it's based on a trivial/identity bijection.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt; please note that the above paragraph indicates that I know about set cardinality and how it is defined, and accept it as a valid &quot;size&quot; definition, but am asking here about something else.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To put it another way, the set of integers is a proper subset of the set of rationals. It seems strange to claim that the two sets are equal in size when one is a proper subset of the other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there, for example, some alternative named &quot;size&quot; definition consistent with the partial ordering given by the is-a-proper-subset-of operator?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt; clearly it is reasonable to define such a partial order and evaluate it. And while I've use geometric analogies, clearly this is pure set theory - it depends only on the relevant sets sharing members, not on what the sets represent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Helpful answers might include a name (if one exists), perhaps for some abstraction that is consistent with this partial order but defined in cases where the partial order is not. Even an answer like &quot;yes, that's valid, but it isn't named and doesn't lead to any interesting results&quot; may well be correct - but it doesn't make the idea unreasonable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sorry if some of my comments aren't appropriate, but this is pretty frustrating. As I said, it feels like I'm violating some kind of taboo.&lt;/p&gt;&#xA;" OwnerUserId="510" LastEditorUserId="510" LastEditorDisplayName="" LastEditDate="2010-08-02T14:38:15.900" LastActivityDate="2010-08-02T15:20:10.567" Title="Relative sizes of sets of integers and rationals revisited - how do I make sense of this?" Tags="&lt;set-theory&gt;" AnswerCount="7" CommentCount="4" />
  <row Id="1394" PostTypeId="2" ParentId="1393" CreationDate="2010-08-02T13:23:49.073" Score="0" ViewCount="" Body="&lt;p&gt;Actually to elaborate a bit more on:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;It seems strange to claim that the two sets are equal in size&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Let us consider binary representations, every natural number can be written in binary. For example $13 = 1101_2$ but we can define two functions $N(q) = 1+q$, $D(q) = 1/(1+1/q)$ and interpret a binary sequence as a composition of these functions applied to 1, for example $1101_{\mathbb{Q}} = (N \circ N \circ D \circ N) 1 = 8/3$ and by Euclids algorithm this defines every (positive) rational number exactly once.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If all they are, are different interpretations of binary sequences, it would be weird not to claim have equal size!&lt;/p&gt;&#xA;" LastActivityDate="2010-08-02T13:23:49.073" CommentCount="2" />
  <row Id="1395" PostTypeId="2" ParentId="1393" CreationDate="2010-08-02T13:24:46.730" Score="0" ViewCount="" Body="&lt;p&gt;It may seem strange that the set of integers is a proper subset of the set of rationals, but this is exactly the definition of an infinite set.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe it could be easier for you to see that there cannot be a size-based definition when talking about infinite set if you consider this: with two finite sets, in whichever way you choose elements to be removed one at a time from both sets at one, you'll end up with the larger set having still some element while the smaller remained empty. This does not happen with infinite sets; even if you find a way to remove all elements (for example, remove he first at T=0, the second a T=1/2, the third at T=3/4, and so on) you may always choose an ordering in which the &quot;larger&quot; set becomes empty, and the &quot;smaller&quot; has still some element.&lt;/p&gt;&#xA;" OwnerUserId="89" LastActivityDate="2010-08-02T13:24:46.730" CommentCount="6" />
  <row Id="1396" PostTypeId="2" ParentId="1393" CreationDate="2010-08-02T13:32:01.190" Score="1" ViewCount="" Body="&lt;p&gt;I've found &lt;a href=&quot;http://en.wikipedia.org/wiki/Hilbert%27s_paradox_of_the_Grand_Hotel&quot; rel=&quot;nofollow&quot;&gt;Hilbert's Hotel&lt;/a&gt; a useful example to understand (or fail to understand, but on a higher level), how much &quot;infinity&quot; really is, and how much the naive view on things fails when confronted with infinity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It deals with the easier case, comparing the integers with even integers, but maybe it will help. =)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; The wikipedia article linked is not great, but google will surely turn out more useful.&lt;/p&gt;&#xA;" OwnerUserId="505" LastActivityDate="2010-08-02T13:32:01.190" CommentCount="4" />
  <row Id="1397" PostTypeId="2" ParentId="1393" CreationDate="2010-08-02T13:38:42.877" Score="1" ViewCount="" Body="&lt;p&gt;You could consider the &lt;a href=&quot;http://en.wikipedia.org/wiki/Order_relation&quot; rel=&quot;nofollow&quot;&gt;order relation&lt;/a&gt; generated by the operation of inclusion between sets. It is a partial order, and it is, in some way, related to the &quot;size&quot; of the sets.&lt;/p&gt;&#xA;" OwnerUserId="354" LastActivityDate="2010-08-02T13:38:42.877" CommentCount="1" />
  <row Id="1398" PostTypeId="2" ParentId="1393" CreationDate="2010-08-02T13:44:04.250" Score="0" ViewCount="" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Is there, for example, some alternative named &quot;size&quot; definition consistent with the partial ordering given by the is-a-proper-subset-of operator?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I think it is important to keep in mind the context you are looking at something when you want to talk about sizes.  In a topological or geometric context, if $A \subset B$ then we may want to think of $A$ as smaller than $B$.  However, when talking about cardinality of a set, we need to think of the structure of the set as a set (and not part of another set like the real numbers or in its geometric/topological context).  In this way, the only reasonable definition for sets to be the same (isomorphic) is if there is a bijection between them.  Therefore if we want to assign some cardinality to something as a set, it should be the same for all sets that are in bijection with it.&lt;/p&gt;&#xA;" OwnerUserId="9" LastActivityDate="2010-08-02T13:44:04.250" CommentCount="2" />
  <row Id="1399" PostTypeId="2" ParentId="1388" CreationDate="2010-08-02T13:49:38.980" Score="9" ViewCount="" Body="&lt;p&gt;For full details of this and more, the best place to look is the&#xA;following paper (if you have access to it)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;D. S. Dummit, Solving solvable quintics. &lt;em&gt;Math. Comp.&lt;/em&gt; &lt;strong&gt;57&lt;/strong&gt; (1991), 387-401.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The main idea (which extends to any equation with a cyclic Galois group)&#xA;is to consider Lagrange resolvents. Let the equation have roots $x_1,\ldots,x_5$&#xA;with an element $\tau$ of the Galois group permuting them as $x_i\mapsto x_{i+1}$.&#xA;Let $\zeta=\exp(2\pi i/5)$ be the standard fifth root of unity. Then the Lagrange&#xA;resolvents are&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$A_0=x_1+x_2+x_3+x_4+x_5,$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$A_1=x_1+\zeta x_2+\zeta^2 x_3+\zeta^3 x_4+\zeta^4 x_5,$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$A_2=x_1+\zeta^2 x_2+\zeta^4 x_3+\zeta x_4+\zeta^3 x_5,$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$A_3=x_1+\zeta^3 x_2+\zeta x_3+\zeta^4 x_4+\zeta^2 x_5,$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$A_4=x_1+\zeta^4 x_2+\zeta^3 x_3+\zeta^2 x_4+\zeta x_5.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once one has $A_0,\ldots,A_4$ one easily gets $x_1,\ldots,x_5$.&#xA;It's easy to find $A_0$ :-)&#xA;The point is that $\tau$ takes $A_j$ to $\zeta^{-j}A_j$&#xA;and so takes $A_j^5$ to $A_j^5$. Thus $A_j^5$ can be written down in terms&#xA;of rationals (if that's your starting field) and powers of $\zeta$.&#xA;Alas here is where the algebra becomes difficult. The coefficients&#xA;of powers of $\zeta$ in $A_1^5$ are complicated. They can be expressed&#xA;in terms of a root of a &quot;resolvent polynomial&quot; which will have a rational&#xA;root as the equation is cyclic. Once one has done this, you have $A_1$&#xA;as a fifth root of a certain explicit complex number. Then one can&#xA;express the other $A_j$ in terms of $A_1$. The details are not very pleasant,&#xA;but Dummit skilfully navigates through the complexities, and produces&#xA;formulas which are not as complicated as they might be. Alas, I don;t have&#xA;the time nor the energy to provide more details.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-08-02T13:49:38.980" CommentCount="1" />
  <row Id="1400" PostTypeId="1" AcceptedAnswerId="1407" CreationDate="2010-08-02T13:59:04.213" Score="7" ViewCount="259" Body="&lt;p&gt;What is the probability that the convex hull of n+2 random points on n-dimensional sphere contains sphere's center?&lt;/p&gt;&#xA;" OwnerUserId="152" LastEditorUserId="152" LastEditorDisplayName="" LastEditDate="2010-08-02T14:16:35.327" LastActivityDate="2010-08-03T17:08:01.413" Title="Probability that the convex hull of random points contains sphere's center" Tags="&lt;geometry&gt;" AnswerCount="2" CommentCount="6" FavoriteCount="2" />
  <row Id="1401" PostTypeId="1" AcceptedAnswerId="1498" CreationDate="2010-08-02T14:19:57.380" Score="7" ViewCount="454" Body="&lt;p&gt;Why are groups $PSL_3(\mathbb{F}_2)$ and $PSL_2(\mathbb{F}_7)$ isomorphic?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Update.&lt;/em&gt; There is a group-theoretic proof (see answer). But is there any geometric proof? Or some proof using octonions, maybe?&lt;/p&gt;&#xA;" OwnerUserId="152" LastEditorUserId="8575" LastEditorDisplayName="" LastEditDate="2011-05-19T09:00:57.167" LastActivityDate="2011-05-19T09:00:57.167" Title="Why $PSL_3(\mathbb F_2)\cong PSL_2(\mathbb F_7)$?" Tags="&lt;group-theory&gt;&lt;simple-groups&gt;" AnswerCount="7" CommentCount="5" FavoriteCount="3" />
  <row Id="1402" PostTypeId="2" ParentId="1401" CreationDate="2010-08-02T14:37:05.560" Score="8" ViewCount="" Body="&lt;p&gt;Both are simple groups of order 168, and each simple group of order $168$&#xA;is isomorphic to $PSL_2(7)$. An extended exercise, with hints. Prove the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $G$ be a simple group of order $168$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It has $8$ Sylow $7$-subgroups.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It can be identified with a subgroup of $A_8$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Labelling the objects it acts on as $\infty,0,1,\ldots,6$&#xA;one Sylow $7$-subgroup is generated by $g=(0\ 1\ 2\ 3\ 4\ 5\ 6)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The group $G$ is $2$-transitive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The normalizer of $\langle g\rangle$ is generated by $g$ and&#xA;$h=(1\ 2\ 4)(3\ 6\ 5)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The setwise stabilizer $H$ of &lt;code&gt;$\{\infty,0\}$&lt;/code&gt; is generated by $h$&#xA;and another element $k$ which is the product of $(\infty\ 0)$ and&#xA;three other disjoint transpositions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $H$ is cyclic, then the Sylow $2$-subgroup of $G$ would be unique,&#xA;leading to a contradiction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So $H$ is nonabelian and we can take $k=(\infty\ 0)(1\ 6)(2\ 3)(4\ 5)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally $G$ is $PSL_2(7)$.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-08-02T14:37:05.560" CommentCount="2" />
  <row Id="1403" PostTypeId="2" ParentId="1393" CreationDate="2010-08-02T14:47:49.870" Score="1" ViewCount="" Body="&lt;p&gt;Purely set theoretically, cardinality is the right way to think of the &quot;size&quot; of a set. A bijection $f:A\to B$ simply renames each element $x$ in $A$ to $f(x)$, and one reasonably wants the size of a set not to depend on the names given to its elements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are other notions of size if you let your sets have more structure. The &lt;a href=&quot;http://en.wikipedia.org/wiki/Natural_density&quot; rel=&quot;nofollow&quot;&gt;natural density&lt;/a&gt; (if it exists) of a subset $A$ of the natural numbers $\mathbb N$ can be thought as the relative size of $A$ to $\mathbb N$. The natural density of the even numbers is $1/2$, for example, so one might say there are half as many even natural numbers as there are natural numbers altogether. If $A$ and $B$ have natural densities $d(A)$ and $d(B)$, and $A\subseteq B\subseteq \mathbb N$, then $d(A)\leq d(B)$. Not all subsets of $\mathbb N$ have a natural density though, so in particular we can't compare the &quot;sizes&quot; of all sets of naturals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another possibility is considering the (measurable) subsets of a set $X$ equipped with a &lt;a href=&quot;http://en.wikipedia.org/wiki/Measure_%28mathematics%29&quot; rel=&quot;nofollow&quot;&gt;measure&lt;/a&gt; $m$. If $A$ and $B$ are measurable subsets of $X$, and $A\subseteq B$, then $m(A)\leq m(B)$. For example, we can use the Lebesgue measure $m$ on $X=\mathbb R$, which gives measure 1 to the interval $[0,1]$ and measure $1/2$ to the interval $[0,1/2]$. But again, not all subsets of $X$ are measurable, so not all sets can be compared size-wise this way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that in both the approaches above, we can only compare the size of a set relative to some other fixed set ($\mathbb N$ or $X$). Any finite set and the set of rational numbers both have measure 0 with respect to the Lebesgue measure on $\mathbb R$, for example, so we would be forced to admit them to have the same size in this setting.&lt;/p&gt;&#xA;" OwnerUserId="405" LastEditorUserId="405" LastEditorDisplayName="" LastEditDate="2010-08-02T14:56:14.810" LastActivityDate="2010-08-02T14:56:14.810" CommentCount="2" />
  <row Id="1404" PostTypeId="2" ParentId="1393" CreationDate="2010-08-02T14:48:40.757" Score="3" ViewCount="" Body="&lt;p&gt;Of course, there are other notions of size.  In particular, your notion of &quot;a partial order based on inclusion of sets&quot; is a very fruitful concept which has been used frequently.  As a quick example, there is a technique in mathematical logic/set theory called &quot;forcing&quot; which is used to show that certain mathematical statements are unprovable.  Forcing often starts with a partial ordered set where the order is given by inclusion of subsets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In terms of the everyday world interpretation of the word &quot;size&quot;, there are (at least) two problems with the using the partial order given by inclusion of subsets.  The first is, as you said, a &lt;em&gt;partial&lt;/em&gt; order: there are two sets which cannot be compared, i.e., there are 2 sets where you cannot say one is bigger than the other.  The second is that two things will have the same size precisely when the two things are absolutely the same.  There is no notion of different things which happen to be the same size - that can't happen in this partial order.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, lets say we're looking at subsets of the integers.  You pull out your favorite subset: all the odd integers and I pull out mine: all the even integers.  Using the partial order definition of size, these two sets are incomparable.  Mine is neither bigger than, smaller than, or the same size as yours.  To contrast that, using the cardinality notion of size, they have the same size.  This is evidenced by simply taking everything in your set and adding 1 to it to get everything in my set.  For an even more absurd example, consider the set {0} and the set {1}.  One would expect these two sets to have the same notion of &quot;size&quot; (for any notion of &quot;size&quot;!), but using the partial order notion, one cannot compare these two sets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By contrast, cardinality (or, the way I used &quot;size&quot; in the previous link) is defined on ALL sets (assuming the axiom of choice), even those which a priori have no subset relation.  And there are many examples of sets which have the same cardinality, but which are not equal.  (For example, the set of evens and odds, or the sets {0} and {1}).&lt;/p&gt;&#xA;" OwnerUserId="331" LastEditorUserId="331" LastEditorDisplayName="" LastEditDate="2010-08-02T15:20:10.567" LastActivityDate="2010-08-02T15:20:10.567" CommentCount="10" />
  <row Id="1405" PostTypeId="2" ParentId="398" CreationDate="2010-08-02T15:10:37.740" Score="0" ViewCount="" Body="&lt;p&gt;Almost all of this misses the idea that you can't do traditional arithmetic with infinity that way that most of us think of it.  10/2 = 5, five is half of 10, no problem.  If Judy has five apples and Joe has 10 apples, then it is proper to say &quot;Judy has half as many apples as Joe.&quot;  But to say &quot;there are half as many even positive integers as there are positive intergers&quot; is to infer something about infinity/2, or half of infinity, which is not entirely kosher.&lt;/p&gt;&#xA;" OwnerUserId="575" LastActivityDate="2010-08-02T15:10:37.740" />
  <row Id="1406" PostTypeId="2" ParentId="1400" CreationDate="2010-08-02T15:34:25.363" Score="5" ViewCount="" Body="&lt;p&gt;This problem is discussed in J. G. Wendel; A Problem in Geometric Probability, Mathematica Scandinavica 11 (1962) 109-111. Wendel showed, that the probability of $N$ random points lying on the surface of the unit sphere in dimension $n$ all lie on one hemisphere is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$2^{-N+1}\sum_{k=0}^{n-1} {{n-1}\choose k}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've found this &lt;a href=&quot;http://www.math.uni-bielefeld.de/~sillke/PUZZLES/ranpoint.txt&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="505" LastEditorUserId="505" LastEditorDisplayName="" LastEditDate="2010-08-03T06:13:11.830" LastActivityDate="2010-08-03T06:13:11.830" CommentCount="1" />
  <row Id="1407" PostTypeId="2" ParentId="1400" CreationDate="2010-08-02T15:44:39.370" Score="5" ViewCount="" Body="&lt;p&gt;This is one of those old chestnuts that come up again and again.&#xA;To be precise, the probability that the convex hull of $n+2$ points in $S^n$&#xA;(the unit sphere in $\mathbb{R}^{n+1}$) contains the origin is $2^{-n-1}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's a brief argument at &lt;a href=&quot;http://mathworld.wolfram.com/SphereTetrahedronPicking.html&quot; rel=&quot;nofollow&quot;&gt;Wolfram's mathworld&lt;/a&gt; which I don't find&#xA;entirely convincing but which certainly can be patched to form&#xA;a convincing argument. In brief, show that for random points $P_1,\ldots,P_{n+2}$&#xA;on the sphere, then with probability one, exactly one choice of signs&#xA;will put the centre in the convex hull of $\pm P_1,\pm P_2,\ldots,\pm P_{n+1}$&#xA;and $P_{n+2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added&lt;/strong&gt; (3/8/2010)&#xA;Thanks to Grigory for his comment. Changing the notation slightly,&#xA;one can show that under some fairy weak hypotheses, if we choose $m+1$&#xA;points randomly and indepedently in $\mathbb{R}^m$ the probability their convex&#xA;hull contains the origin is $2^{-m}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take a probability distribution on $\mathbb{R}^m$ and choose a sequence&#xA;of points (which we identify with vectors) independently from that distribution.&#xA;Our first condition on this distribution is that $m$ vectors $v_1,\ldots,v_m$&#xA;chosen independently from it are linearly independent with probability one.&#xA;This can fail if say some point occurs with nonzero probability or the&#xA;distribution lies in a hyperplane through the origin. Assume this condition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now a sequence $v_0,v_1,\ldots,v_m$ of random points chosen according to&#xA;our distribution are linearly independent: there are reals $a_i$ not all zero with&#xA;$\sum_i a_i v_i=0$. By our condition, with probability one, the sequence&#xA;$(a_0,\ldots,a_m)$ is unique up to constant multiple, and moreover all the&#xA;$a_i$ are nonzero. So we may assume $a_0=1$ and $a_1,\ldots,a_m$&#xA;are nonzero and uniquely determined. Then the convex hull of the $v_i$&#xA;contains the origin if an only if all the $a_i$ are positive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we introduce another condition: that the distribution is centrally symmetric;&#xA;in detail the probability that a random vector $v$ lies in a set $A$ equals&#xA;the probability that $-v$ lies in $A$. A condition like this is clearly necessary;&#xA;it stops the distribution being supported on a small region far from the origin.&#xA;This condition shows that all the $2^m$ possibilities of signs&#xA;for $a_1,\ldots,a_m$ are equiprobable; since changing the sign of some&#xA;$v_i$ changes the sign of $a_i$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To conclude, if our probability distribution on $\mathbb{R}^m$ satisfies&#xA;these two condition, the probability that the convex hull of $m+1$&#xA;indepdendently chosen points contains the origin is $2^{-m}$.&#xA;These conditions are satisfied by the uniform distribution on a sphere&#xA;with centre at the origin, but also by many others.&lt;/p&gt;&#xA;" OwnerUserId="226" LastEditorUserId="226" LastEditorDisplayName="" LastEditDate="2010-08-03T17:08:01.413" LastActivityDate="2010-08-03T17:08:01.413" CommentCount="2" />
  <row Id="1408" PostTypeId="1" AcceptedAnswerId="1411" CreationDate="2010-08-02T16:20:02.310" Score="6" ViewCount="259" Body="&lt;p&gt;Suppose that $X$ is a topological space with a sheaf of rings $\mathcal{O}_X$. In general, the stalk at a point $p \in X$ is the direct limit of the rings $\mathcal{O}_X(U)$ for all open sets $U$ containing $p$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are two questions on computing stalks - I think both should be true, since a direct limit should be some sort of &quot;limiting process&quot;, but that's far from convincing for me. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Can I compute the stalk of $\mathcal{O}_X$ at a point $p \in X$ by only limiting over basic open sets of $X$ containing $p$?&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Can I compute the stalk of $\mathcal{O}_X$ at a point $p \in X$ by excluding some finite number of &quot;large&quot; open sets around $p$, and then limiting over the remaining open sets around $p$? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="576" LastActivityDate="2010-08-04T17:12:21.593" Title="Computing stalks: do direct limits behave like limits?" Tags="&lt;commutative-algebra&gt;" AnswerCount="3" CommentCount="9" FavoriteCount="4" />
  <row Id="1409" PostTypeId="2" ParentId="1388" CreationDate="2010-08-02T16:43:34.530" Score="1" ViewCount="" Body="&lt;p&gt;I recommend you go and look at this reference. www.passhema.org/proceedings/2007/2007DavisGupta.pdf&lt;/p&gt;&#xA;" OwnerUserId="441" LastActivityDate="2010-08-02T16:43:34.530" CommentCount="1" />
  <row Id="1410" PostTypeId="2" ParentId="398" CreationDate="2010-08-02T16:54:02.130" Score="0" ViewCount="" Body="&lt;p&gt;You can also get it the other way around -- N is (kind of) a subset of E. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Map each integer n into 4n. Then there's a bijection (1-1, onto function) between N and (let's call it) N4, the set of all integers divisible by 4, so N and N4 are, in your intuition, of the same &quot;size&quot;.  But N4 is a proper subset of E !  Of course, you could just as well have mapped N directly to E with bijection f(n) = 2n.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, that shows you how slippery it is to apply concepts of finite size to infinite sets, and why bijection is the way to go.&lt;/p&gt;&#xA;" OwnerUserId="271" LastEditorUserId="271" LastEditorDisplayName="" LastEditDate="2010-08-02T17:06:16.650" LastActivityDate="2010-08-02T17:06:16.650" />
  <row Id="1411" PostTypeId="2" ParentId="1408" CreationDate="2010-08-02T16:55:26.730" Score="4" ViewCount="" Body="&lt;p&gt;Yes. The general statement is the following: limit over a poset is equal to limit over its any &lt;a href=&quot;http://en.wikipedia.org/wiki/Cofinal&quot; rel=&quot;nofollow&quot;&gt;coinitial&lt;/a&gt; subset. Formal proof is easy (hint: construct maps in both directions) and informally it's an analogue of &quot;subsequence has the same limit as a sequence&quot; theorem.&lt;/p&gt;&#xA;" OwnerUserId="152" LastActivityDate="2010-08-02T16:55:26.730" CommentCount="3" />
  <row Id="1412" PostTypeId="2" ParentId="398" CreationDate="2010-08-02T17:01:30.167" Score="0" ViewCount="" Body="&lt;p&gt;The density argument above is not quite to the point, since it involves not only a set but an ordering.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, if you play with orderings, you can get even stranger results, such as the integers and rationals can both be ordered as a continuum (according to the ordering, between any two members in the set there is another member) and a non-continuum (there are two members with no other member in between). Or worse, between any two members is at most a finite set of other members (non-continuum) or an infinite set of other members (continuum).  That's kind of an extreme form of density.&lt;/p&gt;&#xA;" OwnerUserId="271" LastActivityDate="2010-08-02T17:01:30.167" />
  <row Id="1413" PostTypeId="1" AcceptedAnswerId="1414" CreationDate="2010-08-02T17:10:25.987" Score="12" ViewCount="514" Body="&lt;p&gt;This is what I know about why someone might care about fields of positive characteristic:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;they are useful for number theory&lt;/li&gt;&#xA;&lt;li&gt;in algebraic geometry, a theory of &quot;geometry&quot; can be developed over them, and it's fun to see how this geometry works out&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Some people might read this and think, &quot;What more could you need?&quot; But I've never been able to make myself care about number theory, so (1) doesn't help me. (2) is nice for what it is, but I'm hoping there's something more. My understanding of (2) is that this is only geometry in a rather abstract sense and, for instance, there's no generally useful way to directly visually represent these fields or varieties over them the way we can over the reals or complex numbers. (Drawing a curve in R^2 and saying it's the curve over some other field may be helpful for some purposes, but it's not what I'm after here.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there anything else? The ideal (surely impossible) answer for me would be &quot;Yes, such fields are very good models for these common and easy to understand physical systems: A, B, C. Also, we can visualize them and varieties over them quite easily by method D. Finally, here's a bunch of surprising and helpful applications to 500 other areas of mathematics.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;UPDATE: to answer Qiaochu's comment about what I do care about.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's say I care about:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;algebraic &amp;amp; geometric topology&lt;/p&gt;&#xA;&#xA;&lt;p&gt;differential geometry &amp;amp; topology&lt;/p&gt;&#xA;&#xA;&lt;p&gt;applications to physics&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and I certainly care about algebraic geometry over C&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(this is to say I understand the motivations behind these subjects and the general idea, not necessarily that I know them in depth)&lt;/p&gt;&#xA;" OwnerUserId="566" LastEditorUserId="566" LastEditorDisplayName="" LastEditDate="2010-08-02T18:26:49.887" LastActivityDate="2010-08-17T22:50:03.043" Title="Why should I care about fields of positive characteristic?" Tags="&lt;abstract-algebra&gt;&lt;field-theory&gt;" AnswerCount="6" CommentCount="3" FavoriteCount="4" />
  <row Id="1414" PostTypeId="2" ParentId="1413" CreationDate="2010-08-02T17:43:12.907" Score="15" ViewCount="" Body="&lt;p&gt;Let's suppose you care about finite groups, one way or another.  Then you probably care about the classification of finite simple groups.  The bulk of this classification is the groups of Lie type, which were discovered by finding analogues of Lie groups over finite fields.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finite fields are also (as one might guess) very important in computer science.  I'm certainly no expert, but here are some applications I know of:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Cryptographic protocols like &lt;a href=&quot;http://en.wikipedia.org/wiki/Diffie%E2%80%93Hellman_key_exchange&quot; rel=&quot;nofollow&quot;&gt;Diffie-Hellman&lt;/a&gt; have as their basis the simple fact that it is difficult to invert exponentiation in finite fields.&lt;/li&gt;&#xA;&lt;li&gt;The standard way that one factors polynomials over the integers is to apply something like &lt;a href=&quot;http://en.wikipedia.org/wiki/Berlekamp_algorithm&quot; rel=&quot;nofollow&quot;&gt;Berlekamp's algorithm&lt;/a&gt; to factor them over several finite fields first, then combine the factorizations.  &lt;/li&gt;&#xA;&lt;li&gt;The classic theorem that &lt;a href=&quot;http://en.wikipedia.org/wiki/IP_(complexity)#Proof_that_IP_.3D_PSPACE&quot; rel=&quot;nofollow&quot;&gt;IP=PSPACE&lt;/a&gt; requires some work over finite fields.  &lt;/li&gt;&#xA;&lt;li&gt;Elliptic curves over finite fields are used for &lt;a href=&quot;http://en.wikipedia.org/wiki/Elliptic_curve_cryptography&quot; rel=&quot;nofollow&quot;&gt;elliptic curve cryptography&lt;/a&gt;.&lt;/li&gt;&#xA;&lt;li&gt;I have also been told that vector spaces and varieties over finite fields can be used to construct error-correcting codes.  I don't know anything about this, but here is a &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0521771943&quot; rel=&quot;nofollow&quot;&gt;book&lt;/a&gt; on the subject.  For the special case of linear codes this leads to a beautiful analogy between lattice sphere packings and error-correcting codes which is described, for example, by Noam Elkies &lt;a href=&quot;http://www.ams.org/notices/200010/fea-elkies-1.pdf&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Finally, even if you are only interested in varieties over $\mathbb{C}$ (say), if your variety happens to also be nice and defined over $\mathbb{Z}$ then it can be nice and defined over $\mathbb{F}_p$ for all but finitely many $p$ and you can use the &lt;a href=&quot;http://en.wikipedia.org/wiki/Weil_conjectures#Statement_of_the_Weil_conjectures&quot; rel=&quot;nofollow&quot;&gt;Weil conjectures&lt;/a&gt; to compute its Betti numbers by &lt;em&gt;counting&lt;/em&gt;.  This is particularly easy to do for varieties with nice moduli interpretations like flag varieties.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Edit:  You might also be interested in reading Serre's expository article &lt;a href=&quot;http://arxiv.org/abs/0903.0517&quot; rel=&quot;nofollow&quot;&gt;How to use finite fields for problems concerning infinite fields&lt;/a&gt;, as well as Manin's &lt;a href=&quot;http://books.google.de/books?id=ffV07pGT6soC&amp;amp;pg=PA518&amp;amp;lpg=PA518&amp;amp;dq=Manin+%22arithmetical+physics%22&amp;amp;source=bl&amp;amp;ots=Bl0wfHR7iG&amp;amp;sig=kC9sfstSnBdY4eXtZZvWmHBAQgY&amp;amp;hl=de&amp;amp;ei=bA9PS4vOA4PqnAOu38GiCg&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=1&amp;amp;ved=0CAcQ6AEwAA#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;Reflections on arithmetical physics&lt;/a&gt;.  I got the latter link from an excellent &lt;a href=&quot;http://mathoverflow.net/questions/11716/mirror-symmetry-mod-p-physics-mod-p/11739#11739&quot; rel=&quot;nofollow&quot;&gt;answer to an MO question&lt;/a&gt; on mirror symmetry over finite fields.&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-08-02T19:20:56.460" LastActivityDate="2010-08-02T19:20:56.460" CommentCount="3" />
  <row Id="1415" PostTypeId="2" ParentId="1408" CreationDate="2010-08-02T18:09:13.323" Score="6" ViewCount="" Body="&lt;p&gt;Basically, here's a way to think of the stalk that is more &quot;down-to-earth&quot; than direct limits. An element of the stalk $O_x$ is given by a pair $(f, U)$ where $f$ is a section over the open set $U$ and $U$ contains $x$. Two pairs $(f,U), (g,V)$ are considered equivalent if $f=g$ on a neighborhood $W$ of $x$ (contained in $U \cap V$). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;With this definition, it's easy to see that what happens at $x$ doesn't depend on what happens on $F$, where $F$ is any closed set disjoint from $x$. The stalk is a purely local construction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for why this is equivalent to the direct limit: that's a direct corollary of how the construction works in most familiar categories with which one might define a sheaf (sets, groups, rings, etc.)&lt;/p&gt;&#xA;" OwnerUserId="536" LastActivityDate="2010-08-02T18:09:13.323" CommentCount="2" />
  <row Id="1416" PostTypeId="1" AcceptedAnswerId="1424" CreationDate="2010-08-02T18:19:20.480" Score="3" ViewCount="382" Body="&lt;p&gt;I've been looking online (and in teaching journals) for a good introduction to Algebras of Random Variables (on an undergraduate level) and their usage, and have come up short. I know I can find the probability distribution of $h(z)$ where:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$z = x + y$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $x$ and $y$ are from known independent probability distributions (the solution is simply a convolution). Two other operations $z=xy$ and $z=y/x$ can be solved for quite easily as well. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anyone know of any other, more complicated, uses for treating random variables as objects to be manipulated? &lt;/p&gt;&#xA;" OwnerUserId="196" LastActivityDate="2010-08-02T21:13:51.843" Title="Algebra of Random Variables?" Tags="&lt;probability-theory&gt;&lt;probability&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="1417" PostTypeId="2" ParentId="1401" CreationDate="2010-08-02T18:27:42.897" Score="3" ViewCount="" Body="&lt;p&gt;There is a reasonably geometric proof in Section 1.4 of Noam Elkies' &lt;a href=&quot;http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.163.1487&amp;amp;rep=rep1&amp;amp;type=pdf&quot; rel=&quot;nofollow&quot;&gt;The Klein Quartic in number theory&lt;/a&gt;.  A very rough summary is as follows.  The unique simple group $G$ of order $168$ has a $3$-dimensional irreducible representation $V$ which is defined over a number field $k$.  By reducing this representation modulo a prime over $2$ we obtain a $3$-dimensional representation of $G$ over $\mathbb{F}_2$ which identifies $G$ with $\text{GL}_3(\mathbb{F}_2)$ by simplicity (and a counting argument).  By reducing this representation modulo a prime over $7$ we obtain a $3$-dimensional representation of $G$ over $\mathbb{F}_7$ which is, as it turns out, the symmetric square of the defining representation of $\text{PSL}_2(\mathbb{F}_7)$.&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-08-02T18:35:07.403" LastActivityDate="2010-08-02T18:35:07.403" CommentCount="1" />
  <row Id="1418" PostTypeId="2" ParentId="1416" CreationDate="2010-08-02T19:03:01.013" Score="2" ViewCount="" Body="&lt;p&gt;There are nice algebraic relationships for special families.  The sum of normal (Cauchy, Levy) random variables is normal (Cauchy, Levy). The product of log-normal random variables is log-normal.  The sum of gamma random variables is gamma if the distributions have a common scale parameter, etc. See more &lt;a href=&quot;http://www.johndcook.com/distribution_chart.html&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="247" LastActivityDate="2010-08-02T19:03:01.013" CommentCount="1" />
  <row Id="1419" PostTypeId="2" ParentId="1413" CreationDate="2010-08-02T20:10:32.380" Score="8" ViewCount="" Body="&lt;p&gt;If you are interested in algebraic geometry over &lt;strong&gt;C&lt;/strong&gt;, here is another reason. A basic technique in birational geometry is the bend and break, which roughly amounts to taking any curve on a projective variety and deforming it until it becomes a union of curves of lower genus. Eventually this technique can be used to produce rational curves (under appropriate hypothesis).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a major tool, and has been used for instance to study the geometry of Fano varieties, in particular to prove that they are uniruled, or to prove Hartshorne conjecture on the characterization of projective space by the positivity of its tangent bundle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fact is, to make the trick work you have to have a curve whose space of embedded deformations is big enough. This is easy to achieve in positive characteristic: the curve itself may not deform, but if you compose the inclusion with a sufficiently high power of the Frobenius morphism, the map that you obtain will have enough deformations. To prove the result in characteristic 0, a technique of reduction to characteristic p is used.&lt;/p&gt;&#xA;" OwnerUserId="233" LastActivityDate="2010-08-02T20:10:32.380" />
  <row Id="1420" PostTypeId="2" ParentId="1389" CreationDate="2010-08-02T20:24:33.080" Score="2" ViewCount="" Body="&lt;p&gt;Wlog assume $a \le b$. Dividing by $n^a$ yields $1 + n^{b-a} = n^{c-a}$ =&gt; $a=b$ (else $n|1$) =&gt; $ n = 2, c = a+1$.&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-08-02T20:24:33.080" />
  <row Id="1421" PostTypeId="2" ParentId="1413" CreationDate="2010-08-02T20:34:21.840" Score="11" ViewCount="" Body="&lt;p&gt;There is a whole &lt;a href=&quot;http://arxiv.org/abs/0903.0517&quot; rel=&quot;nofollow&quot;&gt;paper&lt;/a&gt; of Serre on this, which I haven't read; I will explain briefly one of my favorite applications.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of the really cool applications of this is to the &lt;a href=&quot;http://en.wikipedia.org/wiki/Ax%E2%80%93Grothendieck_theorem&quot; rel=&quot;nofollow&quot;&gt;Ax-Grothendieck theorem&lt;/a&gt;.  Which is a 100% analytic statement: if $P: \mathbb{C}^n \to \mathbb{C}^n$ is an injective polynomial map, then it's surjective. And yet the proof uses finite fields. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How? Well, first note that the theorem is trivial when $\mathbb{C}$ is replaced by a finite field: an injective map on finite sets is surjective. Using a little algebra (and a simple direct limit argument), it follows that the theorem is true for the algebraically closed field of characteristic $p$ which is the algebraic closure of the integers mod $p$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now apply to a bit of model theory; the theorem of Robinson says that any statement of first-order logic that is true in an alg. closed field of char. $p$ for each $p$ is true in any algebraically closed field of char 0, in particular the complex numbers. The trick is that the Ax-Grothendieck theorem can be phrased as a collection of (infinitely many) statements in first-order logic. And each is true in the algebraic closure of the finite fields. Whence, by Robinson's theorem, it's true in $\mathbb{C}$!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This would be cheating if Robinson's theorem were something obscure. In fact, however, it is a direct application of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Compactness_theorem&quot; rel=&quot;nofollow&quot;&gt;compactness theorem for first-order logic&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For another proof via the Nullstellensatz (but based on finite fields, still), see Terry Tao's &lt;a href=&quot;http://terrytao.wordpress.com/2009/03/07/infinite-fields-finite-fields-and-the-ax-grothendieck-theorem/&quot; rel=&quot;nofollow&quot;&gt;post&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="536" LastActivityDate="2010-08-02T20:34:21.840" />
  <row Id="1422" PostTypeId="1" CreationDate="2010-08-02T20:34:54.600" Score="4" ViewCount="240" Body="&lt;p&gt;Suppose $f \in C(\mathbb{R}^n)$, the space of continuous $\mathbb{R}$-valued functions on $\mathbb{R}^n$.  Are there conditions on $f$ that guarantee it is the pullback of a polynomial under some homeomorphism?  That is, when can I find $\phi:\mathbb{R}^n \to \mathbb{R}^n$ such that $f \circ \phi \in \mathbb{R}[x_1,\ldots, x_n]$?  I have tried playing around with the implicit function theorem but haven't gotten far.  It feels like I may be missing something very obvious.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some related questions:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;A necessary condition in the case of $n = 1$ is that $f$ cannot attain the same value infinitely many times (since a polynomial has only finitely many roots).  Is this sufficient?&lt;/li&gt;&#xA;&lt;li&gt;What if we replace $\mathbb{R}$ by $\mathbb{C}$?  &lt;/li&gt;&#xA;&lt;li&gt;What if we look at smooth functions instead?&lt;/li&gt;&#xA;&lt;li&gt;What about the complex analytic case?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="9" LastActivityDate="2010-08-15T09:34:21.720" Title="Which continuous functions are polynomials?" Tags="&lt;algebraic-geometry&gt;&lt;topology&gt;&lt;differential-topology&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="2" />
  <row Id="1423" PostTypeId="1" CreationDate="2010-08-02T20:58:02.360" Score="8" ViewCount="293" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/List_of_finite_simple_groups&quot; rel=&quot;nofollow&quot;&gt;This Wikipedia article&lt;/a&gt; states that the isomorphism type of a finite simple group is determined by its order, except that:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;L&lt;sub&gt;4&lt;/sub&gt;(2) and L&lt;sub&gt;3&lt;/sub&gt;(4) both have order 20160&lt;/li&gt;&#xA;&lt;li&gt;O&lt;sub&gt;2n+1&lt;/sub&gt;(q) and S&lt;sub&gt;2n&lt;/sub&gt;(q) have the same order for q odd, n &gt; 2&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I think this means that for each integer g, there are 0, 1 or 2 simple groups of order g.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do we need the full strength of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Classification_of_finite_simple_groups&quot; rel=&quot;nofollow&quot;&gt;Classification of Finite Simple Groups&lt;/a&gt; to prove this, or is there a simpler way of proving it?&lt;/p&gt;&#xA;" OwnerUserId="31" LastEditorUserId="8575" LastEditorDisplayName="" LastEditDate="2011-05-19T08:55:22.990" LastActivityDate="2011-05-19T08:55:22.990" Title="Number of finite simple groups of given order is at most 2 - is a classification-free proof possible?" Tags="&lt;group-theory&gt;&lt;simple-groups&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="4" />
  <row Id="1424" PostTypeId="2" ParentId="1416" CreationDate="2010-08-02T21:02:09.167" Score="5" ViewCount="" Body="&lt;p&gt;In the special case that the sample space is the non-negative integers (or a subset thereof), one can think of a probability distribution as a generating function $f(x) = \sum_{n \ge 0} a_n x^n$ where $a_n \ge 0$ and $f(1) = 1$.  Then the sum of random variables corresponds to the product of generating functions, so one can bring generating function techniques (see, for example, &lt;a href=&quot;http://www.math.upenn.edu/~wilf/DownldGF.html&quot; rel=&quot;nofollow&quot;&gt;Wilf&lt;/a&gt;) to bear on such random variables.  For example, it is particularly easy to compute expected values this way: the expected value is $f'(1)$, and the product rule expresses the fact that expected value is additive.  Similarly, the variance is $f''(1) + f'(1) - f'(1)^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't really know a place where these issues are discussed in detail, but one spectacular family of examples is the computation of the expected values and variances of certain statistics on permutations.  For example, suppose we want to compute the expected number of fixed points that a permutation of $n$ elements has.  By Burnside's lemma, the answer is $1$.  But another way to do this computation is to construct the family of polynomials &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle P_n(x) = \frac{1}{n!} \sum_{\pi \in S_n} x^{c_1(\pi)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $c_1(\pi)$ is the number of fixed points.  Then the number we want is $P_n'(1)$.  It turns out we can compute all these numbers at the same time because the bivariate generating function is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle P(x, y) = \sum_{n \ge 0} P_n(x) y^n = \frac{1}{1 - y} \exp \left( xy - y \right).$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then the family of numbers we want is $\frac{\partial}{\partial x} P(x, y)$ evaluated at $x = 1$, which (as it is not hard to verify) is $\frac{1}{1 - y}$.  In fact this is true for the second derivative and all higher derivatives as well; in particular, the variance of the number of fixed points is also $1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What if we want to know the expected number and variance of, say, the total number of cycles?  Now we want to look at the family of polynomials&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle Q_n(x) = \frac{1}{n!} \sum_{\pi \in S_n} x^{c(\pi)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $c(\pi)$ is the total number of cycles.  Then the number we want is $Q_n'(1)$.  Now it turns out that the bivariate generating function is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle Q(x, y) = \sum_{n \ge 0} Q_n(x) y^n = \frac{1}{(1 - y)^x}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(which should be interpreted as $\exp \left( x \log \frac{1}{1 - y} \right)$).  The partial derivative $\frac{\partial}{\partial x} Q(x, y)$ evaluated at $x = 1$ is now &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle \frac{1}{1 - y} \log \frac{1}{1 - y} = \sum_{n \ge 1} H_n y^n$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $H_n$ is the $n^{th}$ harmonic number.  Thus the expected number of cycles of a permutation of $n$ elements is about $\log n$.  (It actually turns out that the expected number of cycles of length $r$ is $\frac{1}{r}$, from which this result immediately follows.)  The second partial derivative evaluated at $x = 1$ is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle \frac{1}{1 - y} \log^2 \frac{1}{1 - y} = \sum_{n \ge 1} G_n y^n$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\displaystyle G_n = \sum_{k=1}^{n-1} \frac{1}{k} H_{n-k}$; I'm not sure of the asymptotic growth of this sequence, though, but whatever it is, the variance of the total number of cycles is $G_n + H_n - H_n^2$.  (In any case $G_n \le H_n^2$, so the variance is less than or equal to $H_n$, and this is probably about right asymptotically.)  One can deduce asymptotics for these kind of sequences using methods such as those in Flajolet and Sedgewick's &lt;a href=&quot;http://algo.inria.fr/flajolet/Publications/books.html&quot; rel=&quot;nofollow&quot;&gt;Analytic Combinatorics&lt;/a&gt;, which is my best guess for more examples of using generating functions in this way.  There are probably examples there related to statistics of trees.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All the generating function identities I used above are a consequence of the exponential formula, one version of which is proven and discussed &lt;a href=&quot;http://qchu.wordpress.com/2009/06/24/gila-vi-the-cycle-index-polynomials-of-the-symmetric-groups/&quot; rel=&quot;nofollow&quot;&gt;in this blog post&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-08-02T21:13:51.843" LastActivityDate="2010-08-02T21:13:51.843" CommentCount="2" />
  <row Id="1425" PostTypeId="2" ParentId="1413" CreationDate="2010-08-02T22:03:53.247" Score="3" ViewCount="" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;You can't avoid finite fields if you do complex algebraic geometry, just as you can't avoid (and it is an advantage to utilize) variables that square to zero.  Some major techniques work by reduction mod $p$ and there aren't always substitutes that accomplish the same directly in characteristic 0. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;A lot of algorithms work by reducing modulo many primes and then patching together the results (Chinese remainder theorem).  For the analysis of what at happens the primes you need to understand finite fields, and in some cases, p-adic fields.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Error-correcting codes have a fundamental significance as constructions of &quot;uniformly distributed&quot; or &quot;well spaced&quot; objects in many dimensions.  This is true irrespective of their additional commercial, cryptographic or computational significance.  And it happens that the comprehensible part of error-correcting code theory is the part related to linear algebra, number theory, and algebraic geometry over finite fields.  To the extent that error-correcting codes are seen to be a model for phenomena in nature or its mathematical idealizations (crystals, packings, coverings, spin glasses, combinatorial phase transitions, etc) then routine use of finite fields is unavoidable, even if you don't care about number theory per se. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="467" LastActivityDate="2010-08-02T22:03:53.247" />
  <row Id="1426" PostTypeId="2" ParentId="1204" CreationDate="2010-08-02T22:16:17.183" Score="1" ViewCount="" Body="&lt;p&gt;If you are willing to build a formalism of supergeometry then some constructions are easier to state in this language, notably differential forms and the De Rham complex.  The first one is just functions from the odd line ($R^{(0 | 1)}$) to your manifold, and the De Rham differential comes from super-diffeomorphisms acting on that line.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, there is a symplectic/orthogonal duality in representation theory that some people (most famously Kontsevich) have advocated as best understood in terms of Lie superalgebras.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Constructions using a formally negative-dimensional object (or calculations that look like they might come from such an object) sometimes can be interpreted in terms of bona fide $Z/2$ graded objects whose superdimension (even dimension minus odd dimension) is the negative dimension in question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;edit: you will get more knowledgeable answers if you post to Mathoverflow, where some of the contributors have written papers on super- or noncommutative geometry.&lt;/p&gt;&#xA;" OwnerUserId="467" LastActivityDate="2010-08-02T22:16:17.183" CommentCount="5" />
  <row Id="1427" PostTypeId="2" ParentId="530" CreationDate="2010-08-02T22:16:47.150" Score="2" ViewCount="" Body="&lt;p&gt;How about this construction:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Define a domain $R_0$ as follows. Take a field $K$, adjoin an indeterminate $x_0$, and localize at $(x_0)$ (that is, adjoin inverses to everything not a multiple of $x_0$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$R_0$ has all its ideals principal and linearly ordered: $(x_0)$ contains $(x_0^2)$ contains $(x_0^3)$...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now given $R_i$, define $R_{i+1}$ inductively: Adjoin an indeterminate $x_{i+1}$, so we have $R_i[x_{i+1}]$. Quotient by $(x_{i+1}^2 - x_i)$. Finally, localize at the prime ideal $(x_{i+1})$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This effectively just gives us one more principal ideal containing all the principal ideals from $R_i : (x_{i+1})$ contains $(x_{i+1}^2)=(x_i)$ contains $(x_i^2)$... &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let $R$ be the union of all the $R_i$, and it's obvious that any finitely generated ideal is principal, but there's a non-fg one generated by all the $x_i$. &lt;/p&gt;&#xA;" OwnerUserId="566" LastEditorUserId="171" LastEditorDisplayName="" LastEditDate="2010-09-07T13:18:30.800" LastActivityDate="2010-09-07T13:18:30.800" CommentCount="1" />
  <row Id="1429" PostTypeId="2" ParentId="1422" CreationDate="2010-08-02T23:03:07.563" Score="1" ViewCount="" Body="&lt;p&gt;Since I can't leave comments I'm writing this here.  I think this question is made difficult by the condition that $\phi$ is just required homeomorphism versus say a diffeomorphism.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case n = 1 you can certainly come up with continuous functions that are not differentiable on a discrete set but can be pulled back to yield a polynomial.  As a baby example consider the function $f$ that is $\sqrt{x}$ on the positive reals and x on the negative reals.  Consider the homeomorphism that is $x^2$ on the positive reals and x on the negative reals, then $f$ pulls back to the polynomial $x$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't think its sufficient that $f$ doesn't attain the same value infinitely many times.  I don't have a counterexample but I think a candidate might be contained in this &lt;a href=&quot;http://projecteuclid.org/DPubS?verb=Display&amp;amp;version=1.0&amp;amp;service=UI&amp;amp;handle=euclid.pja/1200672011&amp;amp;page=record&quot; rel=&quot;nofollow&quot;&gt;article&lt;/a&gt;. The gist is that there are functions everywhere continuous and strictly monotonic but with derivative 0 almost everywhere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think you'd have more luck using the implicit function theorem if you required $\phi$ to be a diffeomorphism.  Also I believe its true that 'most' continuous function from $\mathbb{R} \to \mathbb{R}$ are not very nice (nonwhere differential) so a more tractable question might be the same question but requiring $f$ to be smooth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you replace $\mathbb{R}$ with $\mathbb{C}$ and impose $f$ and $\phi$ both be holomorphic then I think it suffices that $f^{(n)}$ vanish for all sufficiently large $n$ because you can recover $f$ from its taylor series.&lt;/p&gt;&#xA;" OwnerUserId="403" LastActivityDate="2010-08-02T23:03:07.563" CommentCount="1" />
  <row Id="1431" PostTypeId="1" AcceptedAnswerId="1470" CreationDate="2010-08-03T00:13:56.220" Score="12" ViewCount="1300" Body="&lt;p&gt;This is an exercise from &lt;em&gt;Problems from the Book&lt;/em&gt; by Andreescu and Dospinescu.  When it was &lt;a href=&quot;http://www.artofproblemsolving.com/Forum/viewtopic.php?t=290872&quot; rel=&quot;nofollow&quot;&gt;posted on AoPS&lt;/a&gt; a year ago I spent several hours trying to solve it, but to no avail, so I am hoping someone here can enlighten me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Problem:&lt;/strong&gt;  Prove that the function $f : [0, 1) \to \mathbb{R}$ defined by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle f(x) = \log_2 (1 - x) + x + x^2 + x^4 + x^8 + ...$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is bounded.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A preliminary observation is that $f$ satisfies $f(x^2) = f(x) + \log_2 (1 + x) - x$.  I played around with using this functional equation for awhile, but couldn't quite make it work.  &lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="152" LastEditorDisplayName="" LastEditDate="2010-08-04T13:06:33.503" LastActivityDate="2010-08-04T13:06:33.503" Title="Prove that this function is bounded" Tags="&lt;real-analysis&gt;&lt;functional-equations&gt;&lt;asymptotics&gt;" AnswerCount="2" CommentCount="6" FavoriteCount="2" />
  <row Id="1432" PostTypeId="2" ParentId="1401" CreationDate="2010-08-03T00:14:36.033" Score="1" ViewCount="" Body="&lt;p&gt;At the last joint meetings in SF there was a talk about exactly this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.ams.org/amsmtgs/2124_abstracts/1056-z1-379.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.ams.org/amsmtgs/2124_abstracts/1056-z1-379.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I went to the talk but I don't remember the explicit isomorphism; but at the very least a very concrete answer exists.&lt;/p&gt;&#xA;" OwnerUserId="403" LastActivityDate="2010-08-03T00:14:36.033" CommentCount="1" />
  <row Id="1433" PostTypeId="2" ParentId="530" CreationDate="2010-08-03T00:34:29.783" Score="4" ViewCount="" Body="&lt;p&gt;See &lt;a href=&quot;http://en.wikipedia.org/wiki/Bezout_domain&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Bezout_domain&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="589" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2010-09-08T08:49:41.177" LastActivityDate="2010-09-08T08:49:41.177" CommentCount="3" />
  <row Id="1434" PostTypeId="2" ParentId="1423" CreationDate="2010-08-03T01:34:41.807" Score="11" ViewCount="" Body="&lt;p&gt;Although I am not (by any stretch) an expert on finite simple groups, let me flesh out my above comment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the following &lt;strong&gt;QCFSG&lt;/strong&gt; (i.e., &quot;qualitative&quot; CFSG): with only finitely many exceptions, every finite simple group has prime order, is alternating, or is one of the finitely many known infinite families of Lie type.  QCFSG must have been conjectured rather early on, whereas the exact statement of CFSG was much harder to come by, as much of the early work on the classification problem resulted in discovery of new sporadic groups.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I guess that early on someone must have looked at the nonsporadic finite simple groups and noticed that, except for the two exceptions listed above, they have distinct orders. [Assuming this is actually true, that is.  I have no reason to doubt it, but I haven't checked it myself.]  Once you notice that, if you believe QCFSG, then you certainly think that the order of a simple group determines the group &lt;em&gt;up to finitely many exceptions&lt;/em&gt;.  It is very hard for me to imagine how you could prove that the number of exceptions is precisely two without knowing the full CFSG.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I cannot resist conveying a story of Jim Milne, whose moral is that you shouldn't feel too bad when you say something absolutely stupid in public: better mathematicians than you or I have said stupider things.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Finally, a story to keep in mind the next time you ask a totally stupid question at a major lecture. During a Bourbaki seminar on the status of the classification problem for simple finite groups, the speaker mentioned that it was not known whether a simple group (the monster) existed of a certain order. &quot;Could there be more than one simple group of that order?&quot; asked Weil from the audience. &quot;Yes, there could&quot; replied the speaker. &quot;Well, could there be infinitely many?&quot; asked Weil.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For the source, and for some further fun stories, see&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.jmilne.org/math/apocrypha.html&quot; rel=&quot;nofollow&quot;&gt;http://www.jmilne.org/math/apocrypha.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="299" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2010-11-29T15:42:13.990" LastActivityDate="2010-11-29T15:42:13.990" CommentCount="2" />
  <row Id="1435" PostTypeId="2" ParentId="804" CreationDate="2010-08-03T01:35:44.643" Score="0" ViewCount="" Body="&lt;p&gt;Sorry, I don't have enough reputation to comment on Andrea Ferretti's answer, but that proof of Wedderburn's theorem is also given in detail in Herstein's &lt;em&gt;Topics in Algebra&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="589" LastActivityDate="2010-08-03T01:35:44.643" CommentCount="0" />
  <row Id="1436" PostTypeId="2" ParentId="1388" CreationDate="2010-08-03T01:42:09.077" Score="4" ViewCount="" Body="&lt;p&gt;Just for fun: Gap+RadiRoot tells me that if $\zeta_n$ is a primitive $n$th root of unity, and we set&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\omega_1 = \sqrt[5]{\left(\frac{66}{3125}\zeta_{5} + \frac{451}{3125}\zeta_{5}^{2} + \frac{176}{3125}\zeta_{5}^{3} + \frac{286}{3125}\zeta_{5}^{4}\right)}$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\omega_2 = \sqrt[2]{ - \frac{11}{20} + \left(\frac{1}{4}\zeta_{5}^{4}\right)\omega_1 + \left( - \frac{5}{44}\zeta_{5} + \frac{15}{44}\zeta_{5}^{2} + \frac{5}{44}\zeta_{5}^{3} + \frac{5}{44}\zeta_{5}^{4}\right)\omega_1^2 + \left(\frac{25}{121}\zeta_{5} - \frac{75}{242}\zeta_{5}^{2} - \frac{75}{484}\zeta_{5}^{3} + \frac{75}{242}\zeta_{5}^{4}\right)\omega_1^3 + \left( - \frac{375}{2662}\zeta_{5} + \frac{625}{1331}\zeta_{5}^{2} + \frac{625}{2662}\zeta_{5}^{3} + \frac{4375}{5324}\zeta_{5}^{4}\right)\omega_1^4}$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the roots of the polynomial are given by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ - \frac{1}{10} + \frac{1}{2}\omega_1 + \left( - \frac{5}{22}\zeta_{5}^{2} - \frac{5}{11}\zeta_{5}^{3} + \frac{5}{11}\zeta_{5}^{4}\right)\omega_1^2 + \left(\frac{75}{242}\zeta_{5} + \frac{150}{121}\zeta_{5}^{2} + \frac{75}{121}\zeta_{5}^{3} + \frac{125}{121}\zeta_{5}^{4}\right)\omega_1^3 + \left(\frac{1625}{1331}\zeta_{5} + \frac{1000}{1331}\zeta_{5}^{2} + \frac{5125}{2662}\zeta_{5}^{3} + \frac{375}{1331}\zeta_{5}^{4}\right)\omega_1^4-\omega_2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;NB:&lt;/strong&gt; I generated this (in a gap4 session, with RadiRoot properly installed) as follows:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;gap&amp;gt; LoadPackage(&quot;radiroot&quot;);&#xA;---------------------------------------------------------------------------------&#xA;Loading  RadiRoot 2.4 (Roots of a Polynomial as Radicals)&#xA;by Andreas Distler (a.distler@tu-bs.de).&#xA;---------------------------------------------------------------------------------&#xA;true&#xA;gap&amp;gt; g := UnivariatePolynomial( Rationals, [1,1,1,1,1,1,1,1,1,1,1]);&#xA;x_1^10+x_1^9+x_1^8+x_1^7+x_1^6+x_1^5+x_1^4+x_1^3+x_1^2+x_1+1&#xA;gap&amp;gt;  RootsOfPolynomialAsRadicals(g, &quot;latex&quot;);&#xA;&quot;/tmp/tmp.sfoZ6C/Nst.tex&quot;&#xA;gap&amp;gt; &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;That created a $\LaTeX$ file, /tmp/tmp.sfoZ6C/Nst.tex, containing the formulas&lt;/p&gt;&#xA;" OwnerUserId="274" LastEditorUserId="274" LastEditorDisplayName="" LastEditDate="2010-08-04T16:38:07.303" LastActivityDate="2010-08-04T16:38:07.303" />
  <row Id="1437" PostTypeId="1" AcceptedAnswerId="1440" CreationDate="2010-08-03T02:44:53.480" Score="5" ViewCount="440" Body="&lt;p&gt;How to solve this equation?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;x = 10&lt;sup&gt;x/10&lt;/sup&gt;&lt;/p&gt;&#xA;" OwnerUserId="595" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-08-04T16:56:15.487" LastActivityDate="2010-08-05T15:23:43.270" Title="How to solve this exponential equation" Tags="&lt;algebra-precalculus&gt;" AnswerCount="3" CommentCount="10" />
  <row Id="1438" PostTypeId="2" ParentId="1431" CreationDate="2010-08-03T02:47:49.043" Score="8" ViewCount="" Body="&lt;p&gt;Starting from (the natural logarithm of) $(1-x)^{-1} = (1+x)(1+x^2)(1+x^4) \dots$, it becomes clearer where the $\log(2)$ factor comes from.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One has to show that $\Sigma (x^{2^k} - C\log(1 + x^{2^k}))$ is bounded sum of positive terms.  The sum of the first $n$ terms approaches $n - Cn\log(2)$ as $x \to 1-$, so we need $C = 1/\log(2)$ if there is to be boundedness.  &lt;/p&gt;&#xA;" OwnerUserId="467" LastActivityDate="2010-08-03T02:47:49.043" CommentCount="2" />
  <row Id="1440" PostTypeId="2" ParentId="1437" CreationDate="2010-08-03T03:18:33.500" Score="13" ViewCount="" Body="&lt;p&gt;There is an obvious solution $x = 10$.  For $x &gt; 10$ the derivative of the RHS is at least $\log 10 &gt; 1$ so there are no solutions.  For $x \le 0$ there are obviously no solutions.  By the IVT there is a solution in $(0, 10)$, and by convexity this solution is unique.  In fact this solution is in $(1, 2)$.  It can be expressed using the Lambert W-function, but it is really not worth writing down explicitly.  Numerically it is about $1.37$.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-08-03T03:18:33.500" CommentCount="2" />
  <row Id="1441" PostTypeId="2" ParentId="944" CreationDate="2010-08-03T03:54:04.270" Score="4" ViewCount="" Body="&lt;p&gt;The point is that SHORTPROOF is an NP-complete problem: given a sentence of length $n$ in the language of some formal proof system (ZFC, Peano arithmetic, etc), does it have a proof of length at most some fixed polynomial in $n$, such as $(2n)^{100}$?  It's in NP because for a reasonable formal system you can check a given proof fairly quickly.  This problem was considered in Goedel's letter to von Neumann that implicitly stated what we now call the $P \neq NP$ question (the heart of the problem, the universality of concrete NP-complete problems, wasn't known until much later). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any NP-complete problem has a zero-knowledge proof protocol for demonstrating solutions of instances, e.g. that we have a SHORTPROOF of the Riemann hypothesis.  These are &quot;proofs that reveal nothing other than their own validity&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The role of the PCP theorem is to show that the proof protocols (interactive challenge/response games) can be very efficient for any stipulated level of confidence.  The probability that the prover really does have a SHORTPROOF of Riemann, given that we follow the protocol and the prover wins, is at least 99 percent, or whatever specified degree of certainty.&lt;/p&gt;&#xA;" OwnerUserId="467" LastActivityDate="2010-08-03T03:54:04.270" />
  <row Id="1442" PostTypeId="1" AcceptedAnswerId="1655" CreationDate="2010-08-03T04:18:36.063" Score="12" ViewCount="766" Body="&lt;p&gt;The identity&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle (n+1) \text{lcm} \left( {n \choose 0}, {n \choose 1}, ... {n \choose n} \right) = \text{lcm}(1, 2, ... n+1)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is probably not well-known.  The only way I know how to prove it is by using &lt;a href=&quot;http://planetmath.org/encyclopedia/KummersTheorem.html&quot; rel=&quot;nofollow&quot;&gt;Kummer's theorem&lt;/a&gt; that the power of $p$ dividing ${a+b \choose a}$ is the number of carries needed to add $a$ and $b$ in base $p$.  Is there a more direct proof, e.g. by showing that each side divides the other?&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="152" LastEditorDisplayName="" LastEditDate="2010-08-03T08:04:27.773" LastActivityDate="2010-08-11T22:01:07.300" Title="Is there a direct proof of this lcm identity?" Tags="&lt;number-theory&gt;&lt;binomial-coefficients&gt;" AnswerCount="4" FavoriteCount="4" />
  <row Id="1443" PostTypeId="1" AcceptedAnswerId="1444" CreationDate="2010-08-03T04:52:05.810" Score="0" ViewCount="88" Body="&lt;p&gt;In mathematics, geometry is the topic I like. Just out of curiosity I would like to know what computational geometry is about. Who studies this subject and why. What are its applications, or is it confined to a theoretical side of things? How does it relate to computer science?&lt;/p&gt;&#xA;" OwnerUserId="328" LastActivityDate="2010-08-03T05:00:21.960" ClosedDate="2010-08-03T05:18:59.090" Title="What is computational geometry and who should take a course on this subject?" Tags="&lt;geometry&gt;&lt;computer-science&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="1444" PostTypeId="2" ParentId="1443" CreationDate="2010-08-03T05:00:21.960" Score="0" ViewCount="" Body="&lt;p&gt;Computational geometry is a major part of the machinery necessary for creating computer graphics, either in games or movies. There are many uses, but if you are interested in computer graphics, it will be a very helpful class.&lt;/p&gt;&#xA;" OwnerUserId="99" LastActivityDate="2010-08-03T05:00:21.960" />
  <row Id="1445" PostTypeId="2" ParentId="81" CreationDate="2010-08-03T05:30:35.467" Score="10" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://gowers.wordpress.com/&quot; rel=&quot;nofollow&quot;&gt;Timothy Gowers'&lt;/a&gt; blog is excellent.  Like Terence Tao, he is both a Fields medalist and an excellent writer.  Together their blogs were my first real introduction into how professional mathematicians think, and their writing has taught me a lot, both about mathematics and about mathematical writing.  If you are a serious student of mathematics you will find all the blogs you need by scrolling through their blogrolls.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-08-03T05:30:35.467" CommunityOwnedDate="2010-08-03T05:30:35.467" />
  <row Id="1446" PostTypeId="1" CreationDate="2010-08-03T05:41:57.103" Score="2" ViewCount="249" Body="&lt;p&gt;Many people are familiar with some properties of binary relations, such as reflexivity, symmetry and transitivity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are the commonly studied properties of ternary (3-ary) relations? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you could provide a motivating example of why the property is interesting that would also be helpful. &lt;/p&gt;&#xA;" OwnerUserId="106" LastActivityDate="2010-08-04T07:14:22.030" Title="Interesting properties of ternary relations?" Tags="&lt;logic&gt;&lt;relations&gt;" AnswerCount="5" CommentCount="5" FavoriteCount="2" />
  <row Id="1447" PostTypeId="2" ParentId="1446" CreationDate="2010-08-03T07:49:47.823" Score="2" ViewCount="" Body="&lt;p&gt;One interesting kind of ternary relation is the &quot;betweenness&quot; relation characterised by the Axioms of Order in Hilbert's Foundations of Geometry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I expect ternary relations are typically studied less because identifying an interesting one requires much more involved definitions than is normally the case for binary relations...&lt;/p&gt;&#xA;" OwnerUserId="172" LastActivityDate="2010-08-03T07:49:47.823" />
  <row Id="1448" PostTypeId="2" ParentId="1446" CreationDate="2010-08-03T08:38:36.867" Score="2" ViewCount="" Body="&lt;p&gt;One interesting example is &quot;being &lt;a href=&quot;http://en.wikipedia.org/wiki/Steiner_system&quot; rel=&quot;nofollow&quot;&gt;Steiner triple system&lt;/a&gt;&quot; (and this is has a connection with Qiaochu Yuan's comment: any Steiner triple system defines commutative quasigroup).&lt;/p&gt;&#xA;" OwnerUserId="152" LastActivityDate="2010-08-03T08:38:36.867" CommunityOwnedDate="2010-08-10T17:16:40.313" />
  <row Id="1449" PostTypeId="2" ParentId="1442" CreationDate="2010-08-03T08:47:17.063" Score="0" ViewCount="" Body="&lt;p&gt;It will need something clever.  For n + 1 = 6, you need 6 times the lcm of 1 5 and 10 in&#xA;order to get enough powers of 2 ( and 3 ).  Is there a characterization of those n + 1 where the&#xA;entire factor of n + 1 is needed, and not the lcm, for the left hand side?&lt;/p&gt;&#xA;" OwnerUserId="297" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-08-11T22:01:07.300" LastActivityDate="2010-08-11T22:01:07.300" CommentCount="0" />
  <row Id="1450" PostTypeId="2" ParentId="1401" CreationDate="2010-08-03T08:56:30.503" Score="2" ViewCount="" Body="&lt;p&gt;(Idea of a proof from V.Dotsenko's paper in &lt;a href=&quot;http://www.mccme.ru/free-books/matprosd.html&quot; rel=&quot;nofollow&quot;&gt;http://www.mccme.ru/free-books/matprosd.html&lt;/a&gt; (in Russian))&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the set of 4-tuples of points on $\mathbb{P}^1(\mathbb{F}_7)$ with cross-ratio equal to 3. This set contains exactly 28 elements. Identifying 4-tuple with its complement (recall that there are exactly 8 points on $\mathbb{P}^1(\mathbb{F}_7)$) one gets 14-element set — exactly the number of points + the number of lines on $\mathbb{P}^2(\mathbb{F}_2)$. Define two tuples to be incident if they intersect. Claim: 1) result is indeed $\mathbb{P}^2(\mathbb{F}_2)$; 2) induced homomorphism $PSL_2(\mathbb{F}_7)\to PSL_3(F_2)$ is an isomorphism.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Hope to update the answer with more details when I understand them. Anyway, the first step already looks somewhat mysterious — does anybody have any comments?)&lt;/p&gt;&#xA;" OwnerUserId="152" LastActivityDate="2010-08-03T08:56:30.503" />
  <row Id="1451" PostTypeId="2" ParentId="1442" CreationDate="2010-08-03T09:12:48.513" Score="-1" ViewCount="" Body="&lt;p&gt;Couldn't it be proved by induction? For n=1 or 2 it's clear. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's suppose the result is true for n-1; now, ${ n \choose k } = {n\over k} { n-1 \choose k }$ and ${ n \choose n } = 1$, so &lt;/p&gt;&#xA;&#xA;&lt;p&gt;lcm $({ n \choose 0, }, { n \choose 1 }, \ldots, { n \choose n }$ = n &amp;middot; lcm $({ n-1 \choose 0, }, { n-1 \choose 1 }, \ldots, { n-1 \choose n-1 }$ = lcm (1,2,...n) for induction hypothesis. n+1 does not divide neither $({ n \choose 0, }, { n \choose 1 }, \ldots, { n \choose n }$ nor {1, 2, ... n}, so the thesis is true also for n.&lt;/p&gt;&#xA;" OwnerUserId="89" LastActivityDate="2010-08-03T09:12:48.513" CommentCount="2" />
  <row Id="1452" PostTypeId="2" ParentId="1437" CreationDate="2010-08-03T09:18:29.967" Score="8" ViewCount="" Body="&lt;p&gt;You can study and graph the two functions &lt;em&gt;y&lt;/em&gt; = &lt;em&gt;x&lt;/em&gt; and &lt;em&gt;y&lt;/em&gt; = 10&lt;sup&gt;&lt;em&gt;x&lt;/em&gt;/10&lt;/sup&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://img594.imageshack.us/img594/3081/function.png&quot; alt=&quot;Graph of y=x and y=10^(x/10)&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From which you can see that there are only two solutions.&lt;/p&gt;&#xA;" OwnerUserId="354" LastActivityDate="2010-08-03T09:18:29.967" CommentCount="2" />
  <row Id="1453" PostTypeId="1" AcceptedAnswerId="1606" CreationDate="2010-08-03T09:21:02.880" Score="2" ViewCount="131" Body="&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/PCP_theorem&quot; rel=&quot;nofollow&quot;&gt;PCP theorem&lt;/a&gt; states that:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Every decision problem in &lt;a href=&quot;http://en.wikipedia.org/wiki/NP_%28complexity%29&quot; rel=&quot;nofollow&quot;&gt;NP&lt;/a&gt; has&#xA;  &lt;a href=&quot;http://en.wikipedia.org/wiki/Probabilistically_checkable_proof&quot; rel=&quot;nofollow&quot;&gt;probabilistically checkable proofs&lt;/a&gt; of&#xA;  constant &lt;a href=&quot;http://en.wikipedia.org/wiki/Query_complexity&quot; rel=&quot;nofollow&quot;&gt;query complexity&lt;/a&gt; and&#xA;  logarithmic randomness complexity.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Can anyone give an intuitive explanation of how this can be done?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Links&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;This is a follow up from &lt;a href=&quot;http://math.stackexchange.com/questions/944/proving-the-riemann-hypothesis-without-revealing-anything-other-than-you-proved-i&quot;&gt;this&#xA;question&lt;/a&gt;.&lt;/li&gt;&#xA;&lt;li&gt;There is a proof &lt;a href=&quot;http://eccc.uni-trier.de/report/2005/046/&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="123" LastActivityDate="2010-08-05T12:55:01.437" Title="Intutive explanation of the PCP Theorem" Tags="&lt;proof-theory&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="1454" PostTypeId="1" AcceptedAnswerId="1544" CreationDate="2010-08-03T11:01:03.730" Score="1" ViewCount="190" Body="&lt;p&gt;I was reading &lt;a href=&quot;http://people.ischool.berkeley.edu/~hal/Papers/2006/position.pdf&quot; rel=&quot;nofollow&quot;&gt;this paper&lt;/a&gt; on position auctions for web ads. Basically, there are N slots each with an expected number of clicks (in a particular time period) $x$. Each agent makes a bid $B_i$ of how much they are willing to pay per click. The bids are put in decreasing order and agent who makes the $i^\text{th}$ highest bid receives receives the slot with the $i^\text{th}$ highest click through rate for a price $P_i$ equal to $B_{i+1}$ (except for the last agent who pays nothing).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To obtain the least price to win the $i^\text{th}$ slot, we note that we have to beat the $i^\text{th}$ agent's &lt;em&gt;bid&lt;/em&gt; if we are moving up, but that we only have to beat the $i^\text{th}$ agents &lt;em&gt;price&lt;/em&gt; if we are moving down. We easily obtain the following equations for Nash Equilibria:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$(v_s-p_s)x_s\ge(v_s-p_t)x_t$ for $t&gt;s$&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$(v_s-p_s)x_s\ge(v_s-P_{t-1})x_t$ for $t &amp;lt; s$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The paper then defines the symmetric Nash equilibrium to be a set of prices with:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(v_s-p_s)x_s\ge(v_s-p_t)x_t$ for all $t$ and $s$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically, instead of having the second part of the previous conditions, the first part of the previous equations is valid anywhere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is the symmetric Nash equilibrium defined more generally? In particular, is it the same as the &lt;a href=&quot;http://en.wikipedia.org/wiki/Symmetric_equilibrium&quot; rel=&quot;nofollow&quot;&gt;symmetric equilibrium in this Wikipedia article&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="123" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-08-03T20:24:34.067" LastActivityDate="2010-08-04T22:33:52.443" Title="Symmetric nash equilibrium" Tags="&lt;game-theory&gt;&lt;nash-equilibrium&gt;" AnswerCount="1" CommentCount="7" />
  <row Id="1456" PostTypeId="2" ParentId="1446" CreationDate="2010-08-03T12:27:15.567" Score="0" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Pythagorean_triple&quot; rel=&quot;nofollow&quot;&gt;Pythagorean triples&lt;/a&gt; induce a ternary relation that has many interesting properties.&lt;/p&gt;&#xA;" OwnerUserId="589" LastActivityDate="2010-08-03T12:27:15.567" CommentCount="4" />
  <row Id="1457" PostTypeId="2" ParentId="457" CreationDate="2010-08-03T12:30:56.930" Score="2" ViewCount="" Body="&lt;p&gt;How about probability &amp;amp; statistics?  Not exactly physics, but lots of applications which can be demonstrated with empirical data.  Any example where &quot;taking an average&quot; seems reasonable is amenable to finding a distribution.  Many examples: frequencies of arrival (traffic, say) as Poisson or negative binomial; arrival times as geometric; insurance claims as lognormal or gamma (or other more complex skewed distributions, but no need to get that complicated); percentiles as beta; human physical characteristics as normal.  Depending upon your course, you could even take empirical data and try fitting distributions using various techniques, which employ calculus, numerical methods, power series (e.g. moments), etc.&lt;/p&gt;&#xA;" OwnerUserId="452" LastActivityDate="2010-08-03T12:30:56.930" CommunityOwnedDate="2010-08-03T12:30:56.930" />
  <row Id="1458" PostTypeId="2" ParentId="457" CreationDate="2010-08-03T12:58:50.547" Score="12" ViewCount="" Body="&lt;p&gt;I cannot resist mentioning the waiter's trick as a physical demonstration of the fact that $SO(3)$ is not simply connected. For those who don't know it, it is the following: you can hold a dish on your hand and perform two turns (one over the elbow, one below) in the same direction and come back in the original position. I guess one can find it on youtube if it is not clear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To see why the two things are related, I borrow the following explanation by Harald Hanche-Olsen on MathOverflow:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Draw a curve through your body from a stationary point, like your foot, up the leg and torso and out the arm, ending at the dish. Each point along the curve traces out a curve in SO(3), thus defining a homotopy. After you have completed the trick and ended back in the original position, you now have a homotopy from the double rotation of the dish with a constant curve at the identity of SO(3). You can't stop at the halfway point, lock the dish and hand in place, now at the original position, and untwist your arm: This reflects the fact that the single loop in SO(3) is not null homotopic.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="233" LastActivityDate="2010-08-03T12:58:50.547" CommunityOwnedDate="2010-08-03T12:58:50.547" CommentCount="1" />
  <row Id="1459" PostTypeId="1" AcceptedAnswerId="1491" CreationDate="2010-08-03T13:04:36.257" Score="1" ViewCount="286" Body="&lt;p&gt;Exercise 10 in Geroch's Mathematical Physics asks whether direct products distribute over direct sums in arbitrary categories. (They do in the category of sets, which is what motivates the question). That is, (using AxB to mean the direct product and A+B to be the direct sum), whether there is an isomorphism between Ax(B+C) and (AxB)+(AxC).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An earlier question asked you to show that associativity of products holds. (AxB)xC is isomorphic to Ax(BxC). This is true, even when not all objects have products (i.e. it is true that &lt;em&gt;if&lt;/em&gt; both Ax(BxC) and (AxB)xC exist, then they are isomorphic...). So I take this question to not require that all objects have products...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't believe there is the relevant isomorphism, but I'm not sure what I've done proves it. So my question is: Is the following argument a good way of approaching category theory questions?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I wrote out a A,B,C,AxB,AxC,B+C,Ax(B+C),(AxB)+(AxC) and drew in all the arrows I know exist from the definitions of products and sums (e.g. AxB guarantees that there is an arrow from AxB to A and one to B...). Then I observed that this diagram has no arrows that go from Ax(B+C) to (AxB)+(AxC) or the other way. That is, even allowing for composition of arrows, there need not be a morphism between the two above mentioned objects. So, I said, this means that this diagram is a diagram of a category where Ax(B+C) is not isomorphic to (AxB)+(AxC), since if there are no morphisms between them, there can't be any isomorphisms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this a good way of constructing counterexamples in category theory? Is there any rigorous discussion of &quot;diagrams&quot; used this way? (I've heard it mentioned, but I don't know where to look).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a question about whether the strategy I am using is a good one, not really about the actual truth value of the statement in question.&lt;/p&gt;&#xA;" OwnerUserId="172" LastEditorUserId="172" LastEditorDisplayName="" LastEditDate="2010-08-04T16:41:47.360" LastActivityDate="2010-08-04T16:41:47.360" Title="Constructing a counterexample in category theory" Tags="&lt;category-theory&gt;&lt;universal-algebra&gt;&lt;proof-strategy&gt;" AnswerCount="5" CommentCount="5" />
  <row Id="1460" PostTypeId="2" ParentId="1459" CreationDate="2010-08-03T13:45:35.753" Score="0" ViewCount="" Body="&lt;p&gt;I am not an expert on this topic, so forgive me if this is obvious.  I see what you have done, but isn't this a rather &quot;vacuous&quot; category?  Products and coproducts satisfy universal properties in that we can factor through them with respect to other morphisms.  For example, if there are arrows from B and C each to A, then suddenly you have a new arrow from B+C to A which is not in your diagram&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now here's where I lose the thread: if you assume products and coproducts exist for all objects A, B, and C, must not some other arrows exist between A, B, and C?  If not, then you may be OK, but I just don't know if you can assume that.  The definitions of (co)products reference such other arrows, so if there are none, then aren't all the (co)products vacuous?&lt;/p&gt;&#xA;" OwnerUserId="452" LastActivityDate="2010-08-03T13:45:35.753" CommentCount="1" />
  <row Id="1461" PostTypeId="1" AcceptedAnswerId="1469" CreationDate="2010-08-03T14:13:02.740" Score="1" ViewCount="207" Body="&lt;p&gt;Let $O(f(n))$ be the minimum time complexity for output a real sequence ${a_i}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The input of the algorithm is a integer $n$. It output the finite sequence $a_1, a_2, \ldots, a_n$. (Clearly $f(n)\geq n$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What can we say about the time complexity of $\max(a_1, a_2, \ldots, a_n)$? Is there always an algorithm to solve this problem in time complexity less than $O(f(n))$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, the algorithm that output the sequence of natural numbers is a $O(n)$ algorithm. Finding the maximum value between 1 and n is $O(1)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we have some more information, like Kolmogorov complexity of the sequence, how will it change the time bounds? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems that if a sequence can be described easily, we can find a way to find the maximum value easier than naively compute the sequence and compare.&lt;/p&gt;&#xA;" OwnerUserId="96" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-08-03T19:24:20.967" LastActivityDate="2010-08-03T19:24:20.967" Title="Time complexity to find the extremal value of a function" Tags="&lt;computer-science&gt;&lt;algorithms&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="1463" PostTypeId="2" ParentId="1459" CreationDate="2010-08-03T16:25:47.593" Score="1" ViewCount="" Body="&lt;p&gt;The dual to the statement that $A\times(B+C)$ is isomorphic to&#xA;$(A\times B)+(A\times C)$ is that $A+(B\times C)$ is isomorphic&#xA;to $(A+B)\times(A+C)$. Now if this statement fails in some&#xA;category $\mathcal{C}$ then the original fails in the opposite&#xA;category $\mathcal{C}^{\mathrm{op}}$.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-08-03T16:25:47.593" CommentCount="2" />
  <row Id="1464" PostTypeId="2" ParentId="1347" CreationDate="2010-08-03T16:59:22.630" Score="4" ViewCount="" Body="&lt;p&gt;An alternative to the horosphere model ...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In &lt;a href=&quot;http://www.jstor.org/pss/2323257&quot; rel=&quot;nofollow&quot;&gt;&quot;A Euclidean Model for Euclidean Geometry&quot;&lt;/a&gt;, Adolf Madur discusses a Disk model of the Euclidean plane. (Madur says that David Gans has priority for discussing this model, so I'll call it the &quot;Gans Disk&quot;.) The &quot;lines&quot; consist of diameters of the Disk, and half-ellipses that have a diameter as a major axis; the measure of the angle between two &quot;lines&quot; is defined as the traditional measure of the angle between their respective major axes. With an appropriate metric (which I have forgotten, and which is &lt;em&gt;just&lt;/em&gt; missing in the document preview linked), we get all of the Euclidean plane crammed into the Disk.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Overlaying the Gans Disk on the Poincare Disk provides another way for Hyperbolians to study Euclidean geometry. They just have to agree to treat these half-ellipse paths (which I don't think are &lt;em&gt;ellipses&lt;/em&gt; to them) as &quot;lines&quot;, and to alter their concept of angle measure and length accordingly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This model might be considerably harder for Hyperbolians to wrap their minds around than the horosphere model, though.&lt;/p&gt;&#xA;" OwnerUserId="409" LastActivityDate="2010-08-03T16:59:22.630" />
  <row Id="1465" PostTypeId="1" AcceptedAnswerId="1468" CreationDate="2010-08-03T17:04:24.330" Score="2" ViewCount="128" Body="&lt;p&gt;...and encoding it as a probability distribution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose we have a sequence of non-negative integers that is periodic with period N:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$A_{1},A_{2},...,A_{N},A_{1}...$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each $A_{k}$ takes on a value no greater than some constant B:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$0 \leq A_{k} \leq B$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We then take this sequence and do a simple convolution, for some constant $L &gt; 0$ and $1 \leq n \leq N$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$S_{L}(n) = A_{n} + A_{n+1} +...+ A_{n+L-1}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From $S_{L}(n)$ we then form a probability distribution $P(n)$ which gives the frequency of each of its values. Let $e_{j}(k) = 1$ if $j = k$ and $0$ otherwise. Then:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$P(n) = (e_{n}(S_{L}(1)) + e_{n}(S_{L}(2)) +...+ e_{n}(S_{L}(N))) / N$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I would like to find out is the extent to which this process can be reversed. I have two data points:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) I know (pretty much) everything about the probability distribution $P(n)$: the distribution itself, its mean, range, variance, skewness, kurtosis, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) I can tell you the frequency of values of $A_{k}$ in one period, so that if the sequence is 1,0,2,3,1,0, I can tell you there are two 0's, two 1's, one 2, and one 3.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To what extent am I able to reconstruct the sequence $A_{k}$ from these two data points?&lt;/p&gt;&#xA;" OwnerUserId="592" LastActivityDate="2010-08-03T18:10:59.330" Title="Recreating an Integer Sequence After Convolution..." Tags="&lt;number-theory&gt;&lt;combinatorics&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="1466" PostTypeId="2" ParentId="1461" CreationDate="2010-08-03T17:15:43.063" Score="1" ViewCount="" Body="&lt;p&gt;Not exactly what you're looking for, but probably related: global optimization of one-variable real functions is NP-complete.&lt;/p&gt;&#xA;" OwnerUserId="589" LastActivityDate="2010-08-03T17:15:43.063" />
  <row Id="1467" PostTypeId="1" AcceptedAnswerId="1471" CreationDate="2010-08-03T18:07:41.393" Score="2" ViewCount="487" Body="&lt;p&gt;Let &lt;em&gt;C&lt;/em&gt; = {0, 1, 2, …, &amp;alefsym;&lt;sub&gt;0&lt;/sub&gt;, &amp;alefsym;&lt;sub&gt;1&lt;/sub&gt;, &amp;alefsym;&lt;sub&gt;2&lt;/sub&gt;, …}. What is |&lt;em&gt;C&lt;/em&gt;|? Or is it even well-defined?&lt;/p&gt;&#xA;" OwnerUserId="171" LastActivityDate="2010-08-04T15:45:45.260" Title="Cardinality of all cardinalities" Tags="&lt;set-theory&gt;" AnswerCount="5" CommentCount="5" FavoriteCount="3" />
  <row Id="1468" PostTypeId="2" ParentId="1465" CreationDate="2010-08-03T18:10:59.330" Score="2" ViewCount="" Body="&lt;p&gt;No, you cannot recover the sequence $A_k$. As a trivial example, note that any cyclic permutation of $(A_1, A_2, \dots, A_N)$ would result in the same distribution (and frequencies). But since this is a periodic sequence, you probably don't care about distinguishing between cyclic permutations of the same sequence, so here's another example.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider L=1. Then your probability distribution is just equivalent to the frequencies, so &lt;em&gt;any&lt;/em&gt; permutation of the $A_k$s would give the same distribution. If L=1 is too degenerate, here's another example with L=2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Say $L=2$, and $N=10$. Then, both sequences $(1, 1, 0, 0, 0, 1, 1, 0, 0, 0)$ and $(1, 1, 0, 0, 0, 0, 1, 1, 0, 0)$ would have the same distribution of sums $S_L$: 2 twice, 1 four times, and 0 four times. You can easily extend this example to any $L$.&lt;/p&gt;&#xA;" OwnerUserId="205" LastActivityDate="2010-08-03T18:10:59.330" CommentCount="1" />
  <row Id="1469" PostTypeId="2" ParentId="1461" CreationDate="2010-08-03T18:46:51.930" Score="2" ViewCount="" Body="&lt;p&gt;Let $a_n = $ number of 1's in the first $n$ terms of another given 0-1 sequence, $b_n$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, $a_n$ could equal the number of odd perfect numbers between 1 and (2n-1), or the number of Fermat primes $2^{2^k} + 1$ with $k \leq n$, or the number of the first $n$ LISP programs that terminate within 4000*(program-length)^4 steps.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a_n$ is non-decreasing, but to find $a_n$ (which is the maximum of the first $n$ terms) is as difficult as calculating all the previous $a_i$.&lt;/p&gt;&#xA;" OwnerUserId="467" LastActivityDate="2010-08-03T18:46:51.930" />
  <row Id="1470" PostTypeId="2" ParentId="1431" CreationDate="2010-08-03T19:02:05.137" Score="14" ViewCount="" Body="&lt;p&gt;OK, a second trick is needed (but it actually finishes the problem).  It is nice and simple enough that it's probably what the authors intended by a &quot;Book&quot; solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $f(x) = x \log(2) - \log(1+x)$.  We want to show that $S(x) = f(x) + f(x^2) + f(x^4) + \dots$ is bounded.  Because $f(0)=f(1)=0$ and $f$ is differentiable, we can find a constant $A$ such that $|f(x)| \leq Ax(1-x) = Ax - Ax^2$.  The sum of this bound over the powers $x^{2^k}$ is telescopic. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice that the role of $\log(2)$ was to ensure that $f(1)=0$.   &lt;/p&gt;&#xA;" OwnerUserId="467" LastActivityDate="2010-08-03T19:02:05.137" CommentCount="3" />
  <row Id="1471" PostTypeId="2" ParentId="1467" CreationDate="2010-08-03T20:01:12.813" Score="10" ViewCount="" Body="&lt;p&gt;C is not a set - it is infact a proper class. If C were a set, then |C| would be defined. It then follows that |C| would be the largest cardinality, since there is a total order between all the cardinalities, and $|C| &gt; \kappa$ for every cardinality $\kappa$ (every cardinality is equivalent to the set of all smaller cardinalities). But $2^{|C|} &gt; |C|$ and so there cannot be a largest cardinal.&lt;/p&gt;&#xA;" OwnerUserId="26" LastActivityDate="2010-08-03T20:01:12.813" CommentCount="6" />
  <row Id="1472" PostTypeId="2" ParentId="698" CreationDate="2010-08-03T20:05:13.227" Score="2" ViewCount="" Body="&lt;p&gt;As answered by &lt;a href=&quot;http://mathoverflow.net/users/6579/t&quot; rel=&quot;nofollow&quot;&gt;T.&lt;/a&gt; on &lt;a href=&quot;http://mathoverflow.net/questions/33335/looking-for-a-probability-distribution/33342#33342&quot; rel=&quot;nofollow&quot;&gt;MathOverflow.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;This is equivalent to (among other&#xA;  names) the Coupon Collector problem.&#xA;  Your are asking about the distribution&#xA;  of the number of coupons collected&#xA;  after t steps, when the total number&#xA;  of possible coupons is n. &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Coupon_collector%27s_problem&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Coupon_collector%27s_problem&lt;/a&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;ADDED: this and related distributions&#xA;  are also studied under other names&#xA;  such as Birthday Problem, random&#xA;  mappings, and random hashing.&#xA;  Kolchin-Sevastyanov-Chistyakov &lt;em&gt;Random&#xA;  Allocations&lt;/em&gt;, Knuth &lt;em&gt;The Art Of Computer&#xA;  Programming, vol. 2&lt;/em&gt;, and Flajolet &amp;amp;&#xA;  Sedgwick &lt;a href=&quot;http://algo.inria.fr/flajolet/Publications/book.pdf&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;Analytic Combinatorics&lt;/em&gt;&lt;/a&gt; all&#xA;  discuss these problems and may contain&#xA;  the precise asymptotics of the&#xA;  distribution you are looking for.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;III.10 in Flajolet and Sedgewick gives&#xA;  the Poisson answer $1−\exp(−t/n)$ when&#xA;  the ratio is held constant, but other&#xA;  asymptotic regimes are also of&#xA;  interest especially in hashing&#xA;  problems. Birthday problem is when&#xA;  $t=O(n^{1/2})$ and one gets statistics&#xA;  of the number of collisions. For t=n^k&#xA;  with 1/2&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="73" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-08-03T20:32:23.647" LastActivityDate="2010-08-03T20:32:23.647" CommunityOwnedDate="2010-08-03T20:05:13.227" />
  <row Id="1473" PostTypeId="2" ParentId="1467" CreationDate="2010-08-03T21:09:51.713" Score="1" ViewCount="" Body="&lt;p&gt;It is not possible to answer the question as such. The best I can do is to redirect you to the wikipedia article on &lt;a href=&quot;http://en.wikipedia.org/wiki/Grothendieck_universe&quot; rel=&quot;nofollow&quot;&gt;Grothendieck universes&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a more detailed explanation, someone more knowledgeable in logic has to come forth.&lt;/p&gt;&#xA;" OwnerUserId="218" LastActivityDate="2010-08-03T21:09:51.713" />
  <row Id="1474" PostTypeId="2" ParentId="1467" CreationDate="2010-08-03T21:33:36.150" Score="3" ViewCount="" Body="&lt;p&gt;The most common way to define the cardinal number $|X|$ of a set $X$ is as the least ordinal which is in bijection with $X$. Then $C$ is an unbounded class of ordinals, and any such is necessarily a proper class. Since $C$ is not a set, it does not have a cardinality.&lt;/p&gt;&#xA;" OwnerUserId="405" LastActivityDate="2010-08-03T21:33:36.150" />
  <row Id="1475" PostTypeId="1" CreationDate="2010-08-03T22:09:46.883" Score="42" ViewCount="1272" Body="&lt;p&gt;&lt;a href=&quot;http://math.gmu.edu/~eobrien/Venn4.html&quot; rel=&quot;nofollow&quot;&gt;This page&lt;/a&gt; gives a few examples of Venn diagrams for 4 sets. Some examples:&lt;br&gt;&#xA;&lt;img src=&quot;http://math.gmu.edu/~eobrien/image/Vennflower.gif&quot; alt=&quot;alt text&quot;&gt; &#xA;&lt;img src=&quot;http://math.gmu.edu/~eobrien/image/Venn4r.gif&quot; alt=&quot;alt text&quot;&gt;&lt;br&gt;&#xA;Thinking about it for a little, it is impossible to partition the plane into the 16 segments required for a complete 4-set Venn diagram using only circles as we could do for &amp;lt;4 sets. Yet it is doable with ellipses or rectangles, so we don't require non-convex shapes as &lt;a href=&quot;http://en.wikipedia.org/wiki/Venn_diagram#Edwards.27_Venn_diagrams&quot; rel=&quot;nofollow&quot;&gt;Edwards&lt;/a&gt; uses.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what properties of a shape determine its suitability for n-set Venn diagrams? Specifically, why are circles not good enough for the case n=4?&lt;/p&gt;&#xA;" OwnerUserId="73" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-08-03T22:15:57.890" LastActivityDate="2011-05-10T02:22:01.187" Title="Why can a venn diagram for 4+ sets not be constructed using circles?" Tags="&lt;geometry&gt;&lt;combinatorics&gt;&lt;logic&gt;" AnswerCount="2" CommentCount="7" FavoriteCount="8" />
  <row Id="1476" PostTypeId="2" ParentId="1467" CreationDate="2010-08-03T23:20:37.197" Score="6" ViewCount="" Body="&lt;p&gt;This is &quot;Fact 20&quot; on page 10 of &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://math.uga.edu/~pete/settheorypart1.pdf&quot; rel=&quot;nofollow&quot;&gt;http://math.uga.edu/~pete/settheorypart1.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These are notes on infinite sets from the most &quot;naive&quot; perspective (e.g., one of the facts is that every infinite set has a countable subset, so experts will see that some weak form of the Axiom of Choice is being assumed without comment.  But this is consistent with the way sets are used in mainstream mathematics).  It is meant to be accessible to undergraduates.  In particular, ordinals are not mentioned, although there are some further documents -- replace &quot;1&quot; in the link above with &quot;2&quot;, &quot;3&quot; or &quot;4&quot; -- which describe such things a bit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I don't see why it is necessary or helpful to speak of ordinals (or universes!) to answer this question.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added&lt;/strong&gt;: to be clear, I wish to recast the question in the following way:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;There is no set $C$ such that for every set $X$, there exists $Y \in C$ and a bijection from $X$ to $Y$.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is easy to prove via Cantor's diagonalization and it sidesteps the &quot;reification problem for cardinalities&quot;, i.e., we do not need to say what a cardinality of a set &lt;em&gt;is&lt;/em&gt;, only to know when two sets have the same cardinality.  I believe this is appropriate for a general mathematical audience.&lt;/p&gt;&#xA;" OwnerUserId="299" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2010-08-04T10:46:05.350" LastActivityDate="2010-08-04T10:46:05.350" />
  <row Id="1477" PostTypeId="2" ParentId="804" CreationDate="2010-08-03T23:56:37.217" Score="2" ViewCount="" Body="&lt;p&gt;The class equation implies a nontrivial finite $p$-group has a nontrivial center, or more generally that a nontrivial normal subgroup of a nontrivial finite $p$-group $G$ contains a nontrivial element of the center of $G$.  (Note: there are infinite groups where every element has $p$-power order and the center is trivial, so the finiteness assumption on the group is important.) This has standard further consequences for finite $p$-groups, although they are no longer direct consequences of the class equation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The class equation itself is a special case of a more general result: the orbit-stabilizer formula for group actions.  &lt;em&gt;That&lt;/em&gt; has lots of uses.&lt;/p&gt;&#xA;" OwnerUserId="619" LastActivityDate="2010-08-03T23:56:37.217" />
  <row Id="1479" PostTypeId="2" ParentId="806" CreationDate="2010-08-04T00:19:10.063" Score="7" ViewCount="" Body="&lt;p&gt;The class group of a number field $K$ can be used to &lt;em&gt;parametrize&lt;/em&gt; other objects. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) If $[L:K] = n$, the possible $O_K$-module structure of $O_L$ is described by the ideal classes of $K$, although it is still an open question in general to show for each $n &gt; 1$ and each ideal class of $K$ that there's an extension $L/K$ with degree $n$ such that $O_L$ as an $O_K$-module corr. to that ideal class.  (This is known for small $n$, but not for general $n$.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) The orbits of the action of $\text{SL}_2(O_K)$ on ${\mathbf P}^1(K)$ are in bijection with ideals classes in $K$.  For instance, the action is transitive iff $K$ has class number 1. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) When $O$ is a quadratic order with discriminant $d$, the (narrow) class group of $O$ describes the primitive quadratic forms of discriminant $d$ up to proper equivalence.  Here we need a slightly more general concept than the usual ideal class group (unless $O = O_K$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4) Weierstrass equations for an elliptic curve over $K$ up to a standard change of variables are related to ideal classes in $K$ (see Silverman's first book on ell. curves, Chap. VIII).&lt;/p&gt;&#xA;" OwnerUserId="619" LastActivityDate="2010-08-04T00:19:10.063" />
  <row Id="1480" PostTypeId="2" ParentId="370" CreationDate="2010-08-04T00:36:28.577" Score="3" ViewCount="" Body="&lt;p&gt;I'm also a fan of Tom Leinster's lecture notes, available on his webpage &lt;a href=&quot;http://www.maths.gla.ac.uk/~tl/msci/&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;. In difficulty level, I would say these are harder than Conceptual Mathematics but easier than Categories and Sheaves, and at a similar level as Categories for the Working Mathematician.&lt;/p&gt;&#xA;" LastActivityDate="2010-08-04T00:36:28.577" CommunityOwnedDate="2010-08-04T00:36:28.577" />
  <row Id="1481" PostTypeId="2" ParentId="804" CreationDate="2010-08-04T00:44:09.747" Score="0" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Burnside&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Burnside&lt;/a&gt;'s_lemma&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Burnside's lemma can be easily deduced from the class equation and is useful in combinatorics&lt;/p&gt;&#xA;" LastActivityDate="2010-08-04T00:44:09.747" />
  <row Id="1482" PostTypeId="1" CreationDate="2010-08-04T01:00:29.497" Score="0" ViewCount="250" Body="&lt;p&gt;I want to solve $0&amp;lt;1−an/(mb^2)e^{−r(T−t)}&amp;lt;1$, where $r, a, b, T, t&gt;0$. The solution is that either $an\leq mb^2$ or $mb^2\leq an\leq mb^2e^{rT}$ and $t&amp;lt; T − (\ln(an) − \ln(mb^2 ))/r$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My thoughts are that the first part $0&amp;lt;1−\frac{an}{mb^2}e^{−r(T−t)}$ gives me $an \leq mb^2$ because $e^{−r(T−t)}&gt;0$, so I have the first part. The second part  $1−\frac{an}{mb^2}e^{−r(T−t)}&amp;lt;1$ does not give me useful information since $\frac{an}{mb^2}e^{−r(T−t)}&gt;0$ always. How do I get the other half of the solution ( $mb^2\leq an\leq mb^2e^{rT}$ and $t&amp;lt; T − (\ln(an) − \ln(mb^2 ))/r$)? I also realise that the problem I have to solve reduces to solving $xy&amp;lt;1$ where both $x,y&gt;0$.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Merged from: &lt;a href=&quot;http://math.stackexchange.com/questions/1543/tricky-inequality&quot;&gt;http://math.stackexchange.com/questions/1543/tricky-inequality&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do I go about solving $0&amp;lt;1−\frac{an}{mb^2}e^{−r(T−t)}&amp;lt;1$, where a,b,T&gt;t&gt;0? I have been stuck here for some time now.&lt;/p&gt;&#xA;" OwnerUserId="626" LastEditorUserId="626" LastEditorDisplayName="" LastEditDate="2010-08-05T00:57:28.763" LastActivityDate="2010-08-05T10:41:55.200" Title="problem with inequality" Tags="&lt;algebra-precalculus&gt;" AnswerCount="1" CommentCount="8" FavoriteCount="0" />
  <row Id="1483" PostTypeId="2" ParentId="529" CreationDate="2010-08-04T01:19:55.590" Score="2" ViewCount="" Body="&lt;p&gt;There is a simple proof of Frobenius's theorem in Lam's book on noncommutative rings, pp. 208--209.  He attributes the argument to Palais. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One should consider this theorem to be two theorems: (1) C is the only C-central division algebra and (2) R and H are the only R-central division algebras. The reason there are so few choices is that C is alg. closed and R is nearly so. Division algebras with center equal to a particular field can be created using cyclic Galois extensions* and since Q has such extensions of arb. high degree there are Q-central division alg. of arb. high dimension.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;*There are further technical conditions to be satisfied on the cyclic extension in order for the construction of a division algebra to work, e.g., a finite field has a cyclic extension of each degree but there are no central div. alg. of dim. &gt; 1 over a finite field.  The relevant technical conditions are satisfied when the base field is the rational numbers. &lt;/p&gt;&#xA;" OwnerUserId="619" LastActivityDate="2010-08-04T01:19:55.590" />
  <row Id="1484" PostTypeId="2" ParentId="301" CreationDate="2010-08-04T01:23:26.147" Score="3" ViewCount="" Body="&lt;p&gt;The general theorem giving the relationship between the arithmetical hierarchy and computability is known as Post's theorem [1]. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In part, Post's theorem states that a set of natural numbers is $\Sigma^0_1$ if and only if it is recursively enumerable. If the set is also $\Pi^0_1$ then its complement is recursively enumerable too. So a $\Delta^0_1$ set of natural numbers must be computable. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A more general consequence of Post's theorem is that a set of natural numbers is $\Delta^0_{n+1}$ if and only if the set is computable from the $n$th Turing jump of the empty set, $\emptyset^{(n)}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1: &lt;a href=&quot;http://en.wikipedia.org/wiki/Post%27s_theorem&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Post%27s_theorem&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="630" LastActivityDate="2010-08-04T01:23:26.147" />
  <row Id="1485" PostTypeId="2" ParentId="1283" CreationDate="2010-08-04T01:31:28.547" Score="1" ViewCount="" Body="&lt;p&gt;I'm not sure you'll like this, but in terms of the incomplete $\Gamma$ function, one can get a closed form as&#xA;$$\frac{e^{x}\Gamma(n+1,x)}{\Gamma(n+1)}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Incomplete_gamma_function&quot; rel=&quot;nofollow&quot;&gt;incomplete $\Gamma$&lt;/a&gt; function is defined as&#xA;$$\Gamma(s,x) = \int_x^{\infty} t^{s-1}e^{-t}dt$$. &lt;/p&gt;&#xA;" OwnerUserId="620" LastActivityDate="2010-08-04T01:31:28.547" CommentCount="4" />
  <row Id="1486" PostTypeId="2" ParentId="198" CreationDate="2010-08-04T01:45:39.907" Score="4" ViewCount="" Body="&lt;p&gt;Proof theorists have obtained several &quot;relative consistency&quot; proofs between classical and constructive theories. These show that if certain theories of classical mathematics are inconsistent, then corresponding theories of constructive mathematics are also inconsistent. These relative consistency results are proved constructively. They show that the consistency problem does not simply disappear if we switch to constructive mathematics. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of the more famous relative consistency techniques uses a &quot;double negation translation&quot;. This method assigns each formula $\phi$ of a system a corresponding formula $\phi^N$ (the &quot;translation&quot; of $\phi$). The exact definition of the translation varies from author to author, depending on the system at hand. But the name is somewhat accurate: the definition of $\phi^N$ involves adding additional negation symbols to $\phi$ in the right places. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In 1933, Gödel proved there is a translation $N$ of formulas of Peano arithmetic so that whenever a formula $\phi$ is provable in Peano arithmetic, the corresponding formula $\phi^N$ is provable in the constructive system of Heyting arithmetic. Moreover, if $\phi$ is of the form $A \land \lnot A$ then Gödel's translation assigns it the formula $\phi^N = A^N \land \lnot A^N$, which is still contradictory. This means that if Peano arithmetic is inconsistent, so is its constructive counterpart Heyting arithmetic. Gödel's proof is constructive, like you would hope.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if we were only worried about consistency, there would be no advantage to working in Heyting arithmetic instead of Peano arithmetic. But people who work in constructive mathematics do not do it only for the sake of consistency. Constructive proofs carry more information than classical proofs, so constructive provability is interesting even to classical mathematicians. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can read a little more about Gödel's result at &lt;a href=&quot;http://en.wikipedia.org/wiki/G%C3%B6del%E2%80%93Gentzen_negative_translation&quot; rel=&quot;nofollow&quot;&gt;this wikipedia article&lt;/a&gt;. &lt;/p&gt;&#xA;" OwnerUserId="630" LastActivityDate="2010-08-04T01:45:39.907" />
  <row Id="1487" PostTypeId="1" AcceptedAnswerId="1520" CreationDate="2010-08-04T01:59:21.300" Score="1" ViewCount="151" Body="&lt;p&gt;I am trying to understand this problem and yes this is from my assignment and I should be doing it myself, but I have been staring at it for 2 hours and not getting anywhere, so decided to post it here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let G be a connected cubic simple graph that contains 2 edge-disjoint spanning trees show that |G| = 4.&lt;/p&gt;&#xA;" OwnerUserId="634" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-08-04T02:10:08.813" LastActivityDate="2011-05-17T10:43:46.880" Title="Connected simple cubic graph" Tags="&lt;graph-theory&gt;&lt;homework&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="1488" PostTypeId="2" ParentId="1487" CreationDate="2010-08-04T02:20:30.137" Score="5" ViewCount="" Body="&lt;p&gt;Express the number of edges in a connected cubic simple graph in terms of the number of vertices. Then do the same for a graph that contains 2 edge-disjoint spanning trees. Solve this equation for the number of vertices.&lt;/p&gt;&#xA;" OwnerUserId="635" LastActivityDate="2010-08-04T02:20:30.137" CommentCount="1" />
  <row Id="1489" PostTypeId="2" ParentId="1048" CreationDate="2010-08-04T04:21:00.310" Score="0" ViewCount="" Body="&lt;p&gt;To do this rigorously you have essentially no choice as to the logic of the proof.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first definition constructs (sin,cos) as a pair of functions on the unit circle X^2 + Y^2 = 1.  The functions are just Y and X respectively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second definition constructs another pair (Sin,Cos), of functions on the real line.  The functions are specific power series.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To show that these pairs are &quot;equal&quot; has a unique meaning: to construct some local identification of the two spaces (a locally invertible parametrization of the circle by the line, and vice versa, i.e., a covering map) such that under this identification (sin,cos) corresponds to (Sin,Cos).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This identification is itself unique: the rotationally invariant angle measure in the plane (Haar measure on SO(2,R) in its conventional normalization to have total length 2*Pi).  If angle measure is already available by other means, such as length measurement on the circle, one gets a second construction of (sin,cos) through their inverse functions (using integrals) and one then has to check that arcsin^{-1} = Sin where the function on the left is an integral and the right is a power series.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Otherwise we have only the map in one direction, (Cos(t),Sin(t)) from the line to the circle using power series, and one has to check that  Cos^2(t) + Sin^2(t) = 1  and the local invertibility (everywhere nonzero velocity vector) of the map.  This would &lt;em&gt;define&lt;/em&gt; angle measure on the circle as &quot;t&quot;, ie., the inverse of the power-series parametrization of the circle by the line. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Edited to remove Tex, click on the edit-history to see it with the TeX.) &lt;/p&gt;&#xA;" OwnerUserId="467" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-08-04T04:31:50.797" LastActivityDate="2010-08-04T04:31:50.797" />
  <row Id="1490" PostTypeId="1" AcceptedAnswerId="1494" CreationDate="2010-08-04T04:29:44.467" Score="2" ViewCount="157" Body="&lt;p&gt;Is there a collective term for the XY, YZ and ZX planes in 3D co-ordinate geometry? I was thinking &quot;principal planes&quot; but I'm not sure where I heard that.&lt;/p&gt;&#xA;" OwnerUserId="90" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-08-04T04:33:12.560" LastActivityDate="2010-08-04T05:29:36.147" Title="Collective Term for XY, YZ and ZX planes" Tags="&lt;geometry&gt;&lt;terminology&gt;&lt;euclidean-geometry&gt;&lt;analytic-geometry&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" />
  <row Id="1491" PostTypeId="2" ParentId="1459" CreationDate="2010-08-04T04:32:56.350" Score="1" ViewCount="" Body="&lt;p&gt;What you have done is a sensible start. You have tried proving the result and you suspect it isn't true. In this situation you want to look for a counter example (and just one is enough).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My suggestion is the category of non-commutative rings. You still have direct sum. However the product is the free product. (For commutative rings it is the tensor product.)&lt;/p&gt;&#xA;" OwnerUserId="638" LastActivityDate="2010-08-04T04:32:56.350" CommentCount="2" />
  <row Id="1492" PostTypeId="2" ParentId="187" CreationDate="2010-08-04T04:59:26.443" Score="0" ViewCount="" Body="&lt;p&gt;Transitive action on the vertices is usually the definition of &quot;looking the same from each vertex&quot;.   So maybe you have two different groups in mind:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The automorphism group of the combinatorics of the tiling; the abstract structure of vertices, faces, edges, etc.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The group of Euclidean motions that leave the tiling invariant. Motions meaning isometries of space, where one should also specify whether orientation-reversal is allowed or not.  Any instance of this is also an automorphism of the tiling combinatorics.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;If tilings that are vertex-transitive under group #1 are &quot;vertex-uniform&quot; and those that are vertex-transitive in the stricter sense of group #2 are &quot;vertex transitive&quot;, then it is easy to give examples where some of the polyhedra are deformations of others (so not isometrically transitive in sense #2).  For example, a one-dimensional periodic tiling with several different sizes of interval will have all vertices equivalent combinatorially but a finite number of distinct vertex types under geometric equivalence.&#xA;This is maybe too simple to be what Grunbaum had in mind, can you quote the book?&lt;/p&gt;&#xA;" OwnerUserId="467" LastActivityDate="2010-08-04T04:59:26.443" />
  <row Id="1493" PostTypeId="2" ParentId="1401" CreationDate="2010-08-04T05:01:44.113" Score="1" ViewCount="" Body="&lt;p&gt;Can't leave comments yet, but the details of there being only one simple group of order 168, and why PSL(2,7) and PSL(3,2) are order 168 and simple, are spelled out on pages 141-147 in Smith and Tabachnikova's &quot;Topics in Group Theory&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Steve&lt;/p&gt;&#xA;" OwnerUserId="641" LastActivityDate="2010-08-04T05:01:44.113" CommentCount="1" />
  <row Id="1494" PostTypeId="2" ParentId="1490" CreationDate="2010-08-04T05:29:36.147" Score="10" ViewCount="" Body="&lt;p&gt;&quot;Principal planes&quot; (and principal axes, principal projections) are terms used in engineering, architecture and technical drawing when using orthogonal projections (usually several of them) to represent a 3-dimensional image.  Names of other fields where the same words could arise: descriptive geometry, perspective drawing, computer graphics, axonometry, surveying, geodesy, artistic perspective.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In mathematics, &quot;coordinate planes&quot; is completely standard, as can be seen from the first comment and its multiple upvotes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Apart from being non-standard, use of &quot;principal&quot; to refer to the coordinate axes or planes is imprecise, and it conflicts with a web of other (useful because mutually consistent) terms such as principal axes of an ellipsoid or of a stress, principal component analysis in statistics, principal diagonal of a matrix.  One wants to keep the latter terms as they are because they all are expressing the same circle of concepts related to diagonalization of quadratic forms.  So &quot;coordinate planes&quot; it is.&lt;/p&gt;&#xA;" OwnerUserId="467" LastActivityDate="2010-08-04T05:29:36.147" />
  <row Id="1495" PostTypeId="1" AcceptedAnswerId="1524" CreationDate="2010-08-04T05:48:12.653" Score="0" ViewCount="199" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am very weak in solving problem related to probability and permutation and combination. Can you please guide me the good content by which I can understand it better.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance.&lt;/p&gt;&#xA;" OwnerUserId="644" LastActivityDate="2010-08-04T12:42:58.247" ClosedDate="2010-08-04T12:50:35.423" Title="Learn Probability and Permutation and combination" Tags="&lt;probability&gt;&lt;learning&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="1496" PostTypeId="2" ParentId="250" CreationDate="2010-08-04T05:51:50.697" Score="0" ViewCount="" Body="&lt;p&gt;Every set can be well ordered, I once sat on the bar of my favourite place and ran into this girl I haven't seen for years. I explained a bit about AC, Zorn's Lemma and the Well-Ordering principle and that they are all equivalent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(In another time, my friend told me that if every set can be well-ordered I should tell him the successor of 0 in the real numbers, I answered that 1 is. He then argued that he means the real numbers, and not the natural numbers. I told him that my well-order puts the natural numbers with the usual ordering first, then the rationals and then the irrationals. But I can shift a finite number of positions if he wants me to.)&lt;/p&gt;&#xA;" OwnerUserId="622" LastActivityDate="2010-08-04T05:51:50.697" CommunityOwnedDate="2010-08-04T05:51:50.697" />
  <row Id="1497" PostTypeId="2" ParentId="1446" CreationDate="2010-08-04T06:19:17.207" Score="0" ViewCount="" Body="&lt;p&gt;One class of example arises in Lie theory. Take $L$ a simple Lie algebra. Then there is a particular $\mathfrak{sl}(2)\subset L$, name take $E$ to be a highest root, $F$ a lowest root, and $H=[E,F]$. Then decompose $L$ as a representation of this subalgebra. You get $L=L_0\oplus L_1\otimes T \oplus \mathfrak{sl}(2)$. Then $L_1$ is a ternary system. This satisfies a (complicated) identity. You can reconstruct $L$ from the ternary system and you need this identity for the Jacobi identity.&lt;/p&gt;&#xA;" OwnerUserId="638" LastActivityDate="2010-08-04T06:19:17.207" />
  <row Id="1498" PostTypeId="2" ParentId="1401" CreationDate="2010-08-04T06:26:47.567" Score="3" ViewCount="" Body="&lt;p&gt;The group $G=\operatorname{PSL}_2(7)$ acts on $X=P^1(\mathbb{F}_7)$. Fix $p\in X$, and consider the action of the stabilizer subgroup $G_p=\{g\in G:g\cdot p=p\}$ on the set $\binom{X\setminus\{p\}}{3}$ of $3$-element subsets of $X\setminus\{p\}$. It has three orbits, of sizes $7$, $7$ and $21$; this can be checked by considering cross-ratios. Pick one of the small ones: one can check that it is a triple Steiner system $S(2,3,7)$, so it is isomorphic as a design, to $P^2(\mathbb{F}_2)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Playing a bit with this construction can be used to realize the isomorphism explicitely.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Later.&lt;/strong&gt; An observation, which helps explainwhy this works, is that if $o\in\binom{X}{4}$ is one of the $G$-orbits of size $14$, then the automorphism group of $o$ (that is, the set of permutations of $X$ which map elements of $o$ to elements of $o$) is $\operatorname{GL}_3(\mathbb{F}_2)\rtimes\mathbb{F}_2^3$, the group of affine maps of $\mathbb{F}_2^3$ (here  &lt;code&gt;\rtimes&lt;/code&gt; is supposed to mean &lt;em&gt;crossed product&lt;/em&gt;). Fixing an element $p\in X$, as I did above, amounts to picking a 'zero' in $\mathbb{F}_2^3$ viewed as an affine space, that is, looking at it as a vector space. (This puts a structure of affine $3$-space over $\mathbb{F}_2$ on $X$, and if we start with the other $14$-element orbit $o'\in\binom{X}{4}$ we get &lt;em&gt;another&lt;/em&gt; structure of affine $3$-space over $\mathbb{F}_2$; $\operatorname{PGL}_2(7)$ is precisely the set of permutations of $X$ which preserves both affine structures)&lt;/p&gt;&#xA;" OwnerUserId="274" LastEditorUserId="274" LastEditorDisplayName="" LastEditDate="2010-08-04T07:29:19.783" LastActivityDate="2010-08-04T07:29:19.783" CommentCount="3" />
  <row Id="1499" PostTypeId="1" AcceptedAnswerId="1517" CreationDate="2010-08-04T06:30:18.277" Score="0" ViewCount="174" Body="&lt;p&gt;If $(\mathbb{R},+)$ is a group and $H$ is a proper subgroup of $\mathbb{R}$ then prove that $H$ is of measure zero.&lt;/p&gt;&#xA;" LastActivityDate="2010-08-04T13:24:13.907" Title="Proper Measurable subgroups of R" Tags="&lt;real-analysis&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="1501" PostTypeId="2" ParentId="1499" CreationDate="2010-08-04T06:57:33.497" Score="6" ViewCount="" Body="&lt;p&gt;Your title talks about &quot;proper measurable subgroups of R&quot;, but the body of your post doesn't require that H be measurable.  The following outline shows that if H is a proper subgroup of R and is measurable, then it must be measure 0.  I'm not sure if every subgroup must be measurable or not....&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a rough outline of the proof:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lemma 1:  If H is a subset of R and has positive measure, then H-H = {a-b| a,b, in H} contains an interval around 0.&#xA;Proof:  See &lt;a href=&quot;http://unapologetic.wordpress.com/2010/04/23/lebesgue-measurable-sets/&quot; rel=&quot;nofollow&quot;&gt;http://unapologetic.wordpress.com/2010/04/23/lebesgue-measurable-sets/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For Lemma 2 and 3, I'll leave the proofs to you (but I can add details if you need them).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lemma 2:  If H is a subgroup of G, then H-H=H.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lemma 3:  If H is a subgroup of the real numbers R and contains an interval around 0, then H = R.&lt;/p&gt;&#xA;" OwnerUserId="331" LastActivityDate="2010-08-04T06:57:33.497" />
  <row Id="1502" PostTypeId="1" AcceptedAnswerId="1505" CreationDate="2010-08-04T07:00:33.320" Score="5" ViewCount="331" Body="&lt;p&gt;What is a set? I know that results such as &lt;a href=&quot;http://en.wikipedia.org/wiki/Russell%27s_paradox&quot; rel=&quot;nofollow&quot;&gt;Russell's paradox&lt;/a&gt; mean that the definition isn't as straight forward as one might expect.&lt;/p&gt;&#xA;" OwnerUserId="123" LastActivityDate="2010-08-04T15:58:12.053" Title="Definition of a set" Tags="&lt;set-theory&gt;&lt;definition&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="0" />
  <row Id="1503" PostTypeId="2" ParentId="1446" CreationDate="2010-08-04T07:14:22.030" Score="1" ViewCount="" Body="&lt;p&gt;So far nobody is actually giving properties, but just examples.  I'll continue that theme.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When Gauss defined composition of quadratic forms, on the level of quadratic forms what he defined was not really a law of composition but a ternary relation (three quadratic forms $Q_1$, $Q_2$, and $Q_3$ are &quot;in composition&quot; if $Q_1(x,y)Q_2(x',y') = Q_3(B,B')$ where $B$ and $B'$ are linear in $xx', xy', yx', yy'$). At the level of proper equivalence classes of quadratic forms this becomes a group law.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could say any group law is defined by a ternary relation $ghk = 1$ on the group. &#xA;This fits the geometric description and addition of points on an elliptic curve or Bhargava's interpretation of Gauss's composition.&lt;/p&gt;&#xA;" OwnerUserId="619" LastActivityDate="2010-08-04T07:14:22.030" />
  <row Id="1505" PostTypeId="2" ParentId="1502" CreationDate="2010-08-04T07:26:07.573" Score="7" ViewCount="" Body="&lt;p&gt;The intuitive notion of a set is of a collection whose identity is entirely determined by its members. Russell's paradox resulted, in effect, from taking any condition at all to determine a collection of members and thus a set. (Russell formulated his paradox against &lt;a href=&quot;http://plato.stanford.edu/entries/frege-logic/#S2&quot; rel=&quot;nofollow&quot;&gt;Gottlob Frege's Basic Law V&lt;/a&gt; which has, as a consequence, that every condition determines a set.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a way, there is no &quot;one notion of a set.&quot; In response to Russell's Paradox (and others, such as the &lt;a href=&quot;http://mathworld.wolfram.com/Burali-FortiParadox.html&quot; rel=&quot;nofollow&quot;&gt;Burali-Forti Paradox&lt;/a&gt;), various axiomatizations of set theory have been developed to try to capture and render precise the notion of a set. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Results due to Goedel and Cohen show that widely accepted principles governing sets as captured by the &lt;a href=&quot;http://mathworld.wolfram.com/Zermelo-FraenkelSetTheory.html&quot; rel=&quot;nofollow&quot;&gt;Zermelo–Fraenkel axiomatization&lt;/a&gt; fail to decide some interesting and contentious &quot;higher&quot; claims about sets (the &lt;a href=&quot;http://mathworld.wolfram.com/AxiomofChoice.html&quot; rel=&quot;nofollow&quot;&gt;Axiom of Choice&lt;/a&gt; and the &lt;a href=&quot;http://mathworld.wolfram.com/ContinuumHypothesis.html&quot; rel=&quot;nofollow&quot;&gt;Continuum Hypothesis&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, in a real way, there is not consensus in the mathematical community about what exactly a set is nor about what principles give a full description of sets. (In practice, most mathematicians use Zermelo–Fraenkel set theory and make appeal to the Axiom of Choice, though many prefer to avoid the later when possible. Some refuse to accept it outright.)&lt;/p&gt;&#xA;" OwnerUserId="277" LastEditorUserId="277" LastEditorDisplayName="" LastEditDate="2010-08-04T15:58:12.053" LastActivityDate="2010-08-04T15:58:12.053" />
  <row Id="1506" PostTypeId="2" ParentId="370" CreationDate="2010-08-04T07:39:54.490" Score="0" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Category_theory#External_links&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt; has some nice free texts linked at the bottom. There's an online version of &lt;a href=&quot;http://katmat.math.uni-bremen.de/acc/acc.pdf&quot; rel=&quot;nofollow&quot;&gt;Abstract and Concrete Categories&lt;/a&gt;, for example.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Steve Awodey has some &lt;a href=&quot;http://www.math.uchicago.edu/~may/VIGRE/VIGRE2009/Awodey.pdf&quot; rel=&quot;nofollow&quot;&gt;lecture notes&lt;/a&gt; available online too. (Awodey's newish book is expensive, but probably rather good)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Patrick Schultz's answer, and BBischoff's comment on an earlier answer also have good links to freely available resources.&lt;/p&gt;&#xA;" OwnerUserId="172" LastActivityDate="2010-08-04T07:39:54.490" CommunityOwnedDate="2010-08-04T07:39:54.490" />
  <row Id="1507" PostTypeId="2" ParentId="1057" CreationDate="2010-08-04T08:17:33.700" Score="2" ViewCount="" Body="&lt;p&gt;I wrote a short proof &lt;a href=&quot;http://www.artofproblemsolving.com/Forum/viewtopic.php?f=61&amp;amp;t=346228&quot; rel=&quot;nofollow&quot; title=&quot;here&quot;&gt;here&lt;/a&gt; .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Steve&lt;/p&gt;&#xA;" OwnerUserId="641" LastActivityDate="2010-08-04T08:17:33.700" />
  <row Id="1508" PostTypeId="2" ParentId="1475" CreationDate="2010-08-04T08:19:49.513" Score="28" ViewCount="" Body="&lt;p&gt;The short answer, from a &lt;a href=&quot;http://www.ams.org/notices/200611/ea-wagon.pdf&quot; rel=&quot;nofollow&quot;&gt;paper&lt;/a&gt; by Frank Ruskey, Carla D. Savage, and Stan Wagon is as follows:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;... it is impossible to draw a Venn diagram with circles that will represent all the possible intersections of four (or more) sets. This is a simple consequence of the fact that circles can finitely intersect in at most two points and &lt;a href=&quot;http://en.wikipedia.org/wiki/Planar_graph#Euler.27s_formula&quot; rel=&quot;nofollow&quot;&gt;Euler’s relation&lt;/a&gt; F − E + V = 2 for the number of faces, edges, and vertices in a plane graph.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The same paper goes on in quite some detail about the process of creating Venn diagrams for higher values of &lt;em&gt;n&lt;/em&gt;, especially for simple diagrams with rotational symmetry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a simple summary, the best answer I could find was on &lt;a href=&quot;http://wiki.answers.com/Q/How_do_you_solve_a_four_circle_venn_diagram&quot; rel=&quot;nofollow&quot;&gt;WikiAnswers&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Two circles intersect in at most two points, and each intersection creates one new region. (Going clockwise around the circle, the curve from each intersection to the next divides an existing region into two.)&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Since the fourth circle intersects the first three in at most 6 places, it creates at most 6 new regions; that's 14 total, but you need 2^4 = 16 regions to represent all possible relationships between four sets.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;But you can create a Venn diagram for four sets with four ellipses, because two ellipses can intersect in more than two points.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Both of these sources indicate that the critical property of a shape that would make it suitable or unsuitable for higher-order Venn diagrams is the number of possible intersections (and therefore, sub-regions) that can be made using two of the same shape.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To illustrate further, consider some of the complex shapes used for &lt;em&gt;n&lt;/em&gt;=5, &lt;em&gt;n&lt;/em&gt;=7 and &lt;em&gt;n&lt;/em&gt;=11 (from &lt;a href=&quot;http://mathworld.wolfram.com/VennDiagram.html&quot; rel=&quot;nofollow&quot;&gt;Wolfram Mathworld&lt;/a&gt;):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://mathworld.wolfram.com/images/gifs/VennDiagrams.jpg&quot; alt=&quot;Venn diagrams for n=5, 7 and 11&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The structure of these shapes is chosen such that they can intersect with each-other in as many different ways as required to produce the number of unique regions required for a given &lt;em&gt;n&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See also: &lt;a href=&quot;http://www.brynmawr.edu/math/people/anmyers/PAPERS/Venn.pdf&quot; rel=&quot;nofollow&quot;&gt;Are Venn Diagrams Limited to Three or Fewer Sets?&lt;/a&gt;  &lt;/p&gt;&#xA;" OwnerUserId="59" LastEditorUserId="59" LastEditorDisplayName="" LastEditDate="2010-08-04T09:19:53.013" LastActivityDate="2010-08-04T09:19:53.013" CommentCount="1" />
  <row Id="1509" PostTypeId="2" ParentId="370" CreationDate="2010-08-04T08:23:26.307" Score="4" ViewCount="" Body="&lt;p&gt;Awodey's new book, while pricey, is a really pleasant read and a good tour of Category Theory from a logician's perspective all the way up to topos theory, with a more up to date view on categories than Mac Lane.&lt;/p&gt;&#xA;" OwnerUserId="666" LastActivityDate="2010-08-04T08:23:26.307" CommunityOwnedDate="2010-08-04T08:23:26.307" CommentCount="2" />
  <row Id="1510" PostTypeId="2" ParentId="370" CreationDate="2010-08-04T08:28:13.943" Score="0" ViewCount="" Body="&lt;p&gt;Barr and Wells, in addition to &lt;em&gt;Toposes, Triples and Theories&lt;/em&gt;, have written &lt;em&gt;Category Theory for the Computing Sciences&lt;/em&gt;, a comprehensive tome which goes through most of the interesting aspects of category theory with a constant explicit drive to relate everything to computer science whenever possible.&lt;/p&gt;&#xA;" OwnerUserId="666" LastActivityDate="2010-08-04T08:28:13.943" CommunityOwnedDate="2010-08-04T08:28:13.943" />
  <row Id="1512" PostTypeId="1" AcceptedAnswerId="1556" CreationDate="2010-08-04T09:40:40.957" Score="9" ViewCount="311" Body="&lt;p&gt;This is Problem 1.7 from &lt;a href=&quot;http://www.colby.edu/personal/fqgouvea/deform.dvi&quot; rel=&quot;nofollow&quot;&gt;Gouvea's lecture notes on deformations of Galois representations&lt;/a&gt;. In particular, he asks you to show that it has many subgroups of finite index which are not closed. So here's what I've got so far, which may be wrong.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can write the compositum as F = &lt;b&gt;Q&lt;/b&gt;[&amp;radic;–1, &amp;radic;2, &amp;radic;3, &amp;radic;5, &amp;radic;7 ...] (can I?) and then the Galois group G = Gal(F/&lt;b&gt;Q&lt;/b&gt;)  is isomorphic to a direct product &amp;Pi;&lt;sub&gt;p&lt;/sub&gt; (&lt;b&gt;Z&lt;/b&gt;/2&lt;b&gt;Z&lt;/b&gt;) where the product is taken over all primes p, as well as p=-1, and the pth component is generated by the conjugation &amp;sigma;&lt;sub&gt;p&lt;/sub&gt; defined by &amp;radic;p -&gt; –&amp;radic;p.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An example of a subgroup which isn't closed would be the subgroup H consisting of &lt;i&gt;finite&lt;/i&gt; products of conjugations, since for example, H contains the sequence &amp;sigma;&lt;sub&gt;2&lt;/sub&gt;, &amp;sigma;&lt;sub&gt;2&lt;/sub&gt;&amp;sigma;&lt;sub&gt;3&lt;/sub&gt;, &amp;sigma;&lt;sub&gt;2&lt;/sub&gt;&amp;sigma;&lt;sub&gt;3&lt;/sub&gt;&amp;sigma;&lt;sub&gt;5&lt;/sub&gt;, &amp;sigma;&lt;sub&gt;2&lt;/sub&gt;&amp;sigma;&lt;sub&gt;3&lt;/sub&gt;&amp;sigma;&lt;sub&gt;5&lt;/sub&gt;&amp;sigma;&lt;sub&gt;7&lt;/sub&gt;... which converges to the automorphism &quot;conjugate everything&quot;, and this automorphism is not contained in H.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However this subgroup is nowhere near being finite index--it has the cardinality of the natural numbers, whereas G has the cardinality of the reals. The only finite index subgroups I can think of take are of the form Gal(F/K) where K is a finite extension of &lt;b&gt;Q&lt;/b&gt;, but of course these are by definition all closed. So I guess I've stuffed up somewhere, and I'd be really grateful for any help?! In know this may seem a bit &quot;homework questiony&quot; but it's not, it's just something that's really bugging me!&lt;/p&gt;&#xA;" OwnerUserId="668" LastActivityDate="2010-08-05T22:02:31.707" Title="How to describe the Galois group of the compositum of all quadratic extensions of Q?" Tags="&lt;number-theory&gt;&lt;group-theory&gt;&lt;galois-theory&gt;" AnswerCount="4" CommentCount="3" FavoriteCount="1" />
  <row Id="1513" PostTypeId="2" ParentId="1502" CreationDate="2010-08-04T11:30:07.073" Score="5" ViewCount="" Body="&lt;p&gt;In very naive set theory (say in the late 19th century), a set was taken to be an arbitrary collection of objects. The difficulty is in telling which things that seem like they should be collections actually are well-defined collections.  The paradoxes show that it's unclear whether this concept of set is coherent, although it &lt;em&gt;is&lt;/em&gt; the natural-language meaning of the word &quot;set&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because the concept seems poorly defined, almost all contemporary set theory deals with a more restrictive notion: pure, well-founded sets. These are the sets that can be constructed starting with the empty set and taking powersets and subsets. The only elements of these sets are other sets. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;These sets are defined in stages. At the first stage, you only have the empty set. At every larger stage, you add the powerset of every set that has already been constructed. There is one stage for every ordinal number, and the collection of all sets available after stage $\alpha$ is named $V_\alpha$. Symbolically, we have $V_0 = \emptyset$ and, in general, &#xA;$$&#xA;V_\alpha = \bigcup_{\beta &amp;lt; \alpha} P(V_\beta)&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For each ordinal $\alpha$, $V_\alpha$ is a set. The union $V = \bigcup_\alpha V_\alpha$ is a proper class. The sequence $( V_\alpha )$, indexed by ordinals, is known as the &lt;em&gt;cumulative hierarchy&lt;/em&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &quot;sets&quot; that mathematicians study and that are formalized in Zermelo-Fraenkel set theory are exactly the sets in $V$.  Moreover, the formalization of mathematics into set theory does not require any other sets than those in $V$ (which is why essentially all mathematics can be formalized into ZFC). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, for all practical mathematical purposes (outside of set theory), the answer to &quot;what is a set&quot; is &quot;a set is an element of $V$&quot;. &lt;/p&gt;&#xA;" OwnerUserId="630" LastActivityDate="2010-08-04T11:30:07.073" CommentCount="2" />
  <row Id="1514" PostTypeId="1" AcceptedAnswerId="1516" CreationDate="2010-08-04T11:38:47.437" Score="3" ViewCount="284" Body="&lt;p&gt;Given the sum of first &lt;code&gt;N&lt;/code&gt; elements of sequence $A$: $S_{n} = n^{2} + 3n + 4$.&lt;br&gt;&#xA;Compute $A_{1} + A_{3} + \ldots + A_{21}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know this problem can be tackled by carefully calculating each value of the sequence. But I wonder what are the better ways to solve it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance!&lt;/p&gt;&#xA;" OwnerUserId="696" LastEditorUserId="696" LastEditorDisplayName="" LastEditDate="2010-08-04T11:43:53.730" LastActivityDate="2010-08-11T22:00:44.973" Title="A high school competition-level problem concerning sum and sequence" Tags="&lt;sequences-and-series&gt;&lt;high-school&gt;&lt;contest-math&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="1" />
  <row Id="1515" PostTypeId="2" ParentId="162" CreationDate="2010-08-04T11:44:44.743" Score="1" ViewCount="" Body="&lt;p&gt;how about a set that contains everything with the exception of himself o_O&lt;/p&gt;&#xA;" OwnerUserId="699" LastActivityDate="2010-08-04T11:44:44.743" />
  <row Id="1516" PostTypeId="2" ParentId="1514" CreationDate="2010-08-04T11:45:10.577" Score="5" ViewCount="" Body="&lt;p&gt;$A_n=S_n-S_{n-1}=2n+2$. Now one has to find a sum of an arithmetic progression $4+8+\dots+44$.&lt;/p&gt;&#xA;" OwnerUserId="152" LastActivityDate="2010-08-04T11:45:10.577" CommentCount="0" />
  <row Id="1517" PostTypeId="2" ParentId="1499" CreationDate="2010-08-04T11:46:58.903" Score="2" ViewCount="" Body="&lt;p&gt;As was noted by Jason DeVito if H is measurable then measure of H is 0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the other hand, if we suppose, that the axiom of choice holds, there is a possibility, that H is not measurable. The proof is quite simple. $\mathbb{R}$ is a vector space over $\mathbb{Q}$. Therefore there is some basis. Suppose e is one of it's elements and H is a subspace in $\mathbb{R}$, generated by others. Then H is a subgroup of ($\mathbb{R}$, +).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lemma: H is not measurable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose H is measurable. Then as was noted above, it's measure m(H)=0. Then every set of the form $H+q e = {h+qe, h\in H}$, where $q\in Q$, has measure 0 (because it is just a shift of H). But $\mathbb{R}$ is equal to union of countable many sets with measure 0: $\mathbb{R} = \cup_{q\in\mathbb{Q}}(H+qe)$. Therefore $m(\mathbb{R})=0$. We have come to a contradiction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also there is simple proof of the fact, that if H is measurable, then H is of measure 0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lemma: If H is measurable proper subgroup of $\mathbb{R}$, then m(H)=0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If H={0} then proposition of the lemma is obvious. Otherwise we can find positive element z in H. Suppose, $H_0 = H\cap [0,z)$. If $m(H_0)=0$ then $m(H)=0$. Otherwise $m(H_0)=\delta&gt;0$. Let's take integer N, such that $\delta N &gt; z+1$ (we will see later, why).&#xA;Note that if x is not in H, then x/n (for every positive integer n) and -x are also not in H. Therefore, using the fact that H is proper, we can find positive x&amp;lt;1, such that $x\notin H$. Suppose y=x/N!. Then for every positive integer n, number ny obeys the following properties:&lt;br&gt;&#xA;1. 1&gt;ny&gt;0.&lt;br&gt;&#xA;2. ny is not in H.&lt;br&gt;&#xA;Then sets H_0, H_0+y, ..., H_0+(N-1)y do not intersect. Therefore we can write&#xA;$1+z = m( [0, 1+z) ) \leq m(\cup_{n=0}^{N-1} (H_0 + n y) = N \delta.$ Here we have a contradiction with definition of N.&lt;/p&gt;&#xA;" OwnerUserId="695" LastEditorUserId="695" LastEditorDisplayName="" LastEditDate="2010-08-04T13:24:13.907" LastActivityDate="2010-08-04T13:24:13.907" CommentCount="1" />
  <row Id="1518" PostTypeId="2" ParentId="1514" CreationDate="2010-08-04T11:57:40.667" Score="-1" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Arithmetic_progression&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Arithmetic_progression&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Sn = n/2(A1+An)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="699" LastEditorUserId="699" LastEditorDisplayName="" LastEditDate="2010-08-04T12:14:12.310" LastActivityDate="2010-08-04T12:14:12.310" />
  <row Id="1519" PostTypeId="1" CreationDate="2010-08-04T11:58:10.697" Score="7" ViewCount="240" Body="&lt;h3&gt;For the uninitiated&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Morse_theory&quot; rel=&quot;nofollow&quot;&gt;Morse theory&lt;/a&gt;, as many other early alebraic-topology widgets, leads to a picture of smooth manifolds as being built up from 'cells', copies of $\mathbb{D}^n$ for varying $n$, 'glued' to each other by the usual topological tools; giving rise to (in some sense) a more natural picture of &lt;a href=&quot;http://en.wikipedia.org/wiki/Homology_%28mathematics%29&quot; rel=&quot;nofollow&quot;&gt;homology&lt;/a&gt; as 'coming from' &lt;a href=&quot;http://en.wikipedia.org/wiki/Cellular_homology&quot; rel=&quot;nofollow&quot;&gt;cellular homology&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Example&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;As an example, consider the torus $\mathbb{T}^2$: we begin with empty space, attach a 0-cell ($\mathbb{D}^0=$ a point), attach a 1-cell ($\mathbb{D}^1=$ a line) to your point (both ends of the line are attached to the point, creating a circle), attach another 1-cell (in the same way, to the same point, creating a sort of figure 8).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The hardest bit to visualise is next: attaching a 2-cell ($\mathbb{D}^2=$ a disk, which we will think of as its homeomorphism equivalent, a square). Begin by twisting your figure 8 so that one circle is in the xy-plane, the other in the xz-plane, now attach the top and bottom of your square (coloured red in picture below) to the xy circle (creating a 'curling round' tube) and the left and right edges (coloured blue below) of your square (now a tube) either side of the xz circle, completing the torus.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://inperc.com/wiki/images/thumb/8/8c/Torus.JPG/150px-Torus.JPG&quot; alt=&quot;Torus cell decomposition&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Problem&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;The above takes some thinking, but a little reading around shows that this is fairly easy to see. What makes it so easy is that the cells we are attaching are of &lt;strong&gt;adjacent dimensions&lt;/strong&gt;, that is; we may easily identify the boundary of one with the entirety of another. Where it gets harder to visualise is when the dimensions of the cells we are attaching to one another differ by &gt;1- the canonical example of this is the complex projective plane $\mathbb{CP}^2$, a 4-manifold built by attaching a disk to a point (making a sphere) and then attaching a 4-ball to that sphere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The latter attaching map (wherein points are identified with their images), I know, may be thought of as the &lt;a href=&quot;http://en.wikipedia.org/wiki/Hopf_map&quot; rel=&quot;nofollow&quot;&gt;Hopf fibration&lt;/a&gt; $\partial \mathbb{D}^4=S^3 \to \mathbb{CP}^1=S^2 $, but I have no way of visualising this, particularly with regard to the interior of the 4-disk. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;How does a 4 cell wrap around a 2 cell without producing a singularity of some kind? Is this analagous in some sense to &lt;a href=&quot;http://en.wikipedia.org/wiki/Dehn_surgery&quot; rel=&quot;nofollow&quot;&gt;Dehn surgery&lt;/a&gt; in which one uses a thickening? Is there a right way to think about this or can it only really be thought of 'intellectually'?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="160" LastEditorUserId="160" LastEditorDisplayName="" LastEditDate="2010-08-04T12:23:49.190" LastActivityDate="2010-09-27T16:30:14.270" Title="Visualising CP^2- a problem of attaching cells with a dimension gap &gt;1" Tags="&lt;intuition&gt;&lt;topology&gt;&lt;algebraic-topology&gt;" AnswerCount="5" CommentCount="3" FavoriteCount="4" />
  <row Id="1520" PostTypeId="2" ParentId="1487" CreationDate="2010-08-04T12:03:10.313" Score="3" ViewCount="" Body="&lt;p&gt;Let $G$ be a simple cubic graph, with two edge-disjoint spanning trees, that has $n$ vertices and $m$ edges. Since $G$ is regular of degree 3, we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$3n = 2m.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now any spanning tree has $n-1$ edges, and $G$ has two disjoint spanning trees. So we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$m \ge 2n - 2.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Combining the first equation and the second inequality, we get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$n \le 4$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now there are no cubic simple graphs on 1, 2 or 3 vertices, so the result follows. (It is debatable whether there is one on 0 vertices.)&lt;/p&gt;&#xA;" OwnerUserId="615" LastEditorUserId="615" LastEditorDisplayName="" LastEditDate="2011-05-17T10:43:46.880" LastActivityDate="2011-05-17T10:43:46.880" CommentCount="2" />
  <row Id="1521" PostTypeId="2" ParentId="1519" CreationDate="2010-08-04T12:20:56.650" Score="6" ViewCount="" Body="&lt;p&gt;1) Near a point of $S^2$ the picture looks like $\mathbb{C}$ sitting inside, naturally $\mathbb{C}^2$ (because $\mathbb{C}P^2\cong(\mathbb{C}^3\setminus{0})/\mathbb{C}^{\times}$ and $S^2=\mathbb{C}P^1$ is the image of a hyperplane in $\mathbb{C}^3$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Think first about $S^2$: one attaches a disk to a point, contracting the circle on the disk's boundary. Now, to make $\mathbb{C}P^2$ one takes 4-disk and do pretty much the same but not for one circle but for all, well, fibers of the Hopf fibration at once — or, in other words, for all points of $S^2$ at once. So, since there were no singularities after gluing $S^2$, there will be no singularities here either.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Not sure if it's an answer, but only hope it helps.)&lt;/p&gt;&#xA;" OwnerUserId="152" LastActivityDate="2010-08-04T12:20:56.650" CommentCount="1" />
  <row Id="1522" PostTypeId="2" ParentId="1459" CreationDate="2010-08-04T12:36:17.137" Score="0" ViewCount="" Body="&lt;p&gt;The category of of all groups is also a counterexample. Letting $A,B,C:=\mathbb{Z}$ , the LHS is $\mathbb{Z}\times (\mathbb{Z} * \mathbb{Z})$ and the RHS is $(\mathbb{Z}\times \mathbb{Z})*(\mathbb{Z}\times \mathbb{Z})$, which are not isomorphic, as their Abelianizations are $\mathbb{Z}^3$, resp. $\mathbb{Z}^4$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another example is the category of Vector spaces. If $A,B,C$ are just one dimensional vector spaces the dimension of the LHS and the RHS don't agree (Note $\times\neq \otimes$).&lt;/p&gt;&#xA;" OwnerUserId="707" LastEditorUserId="707" LastEditorDisplayName="" LastEditDate="2010-08-04T12:41:44.363" LastActivityDate="2010-08-04T12:41:44.363" />
  <row Id="1523" PostTypeId="2" ParentId="1459" CreationDate="2010-08-04T12:40:08.627" Score="0" ViewCount="" Body="&lt;p&gt;Note that in your example (as far as I understand it, you have a category with only 8 objects) your category does not have products and coproducts (you have them only for some special pairs of objects (for example, you don't have product $(A\times B)\times(A\times C)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if your exercise was about categories with products and coproducts, then your example is not an answer on it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The simplest way to construct a counterexample is to check some known categories. It will be quite useful for studying category theory, because you will check, what do abstract notions mean in concrete examples. In this case this works, you can check for example category of vector spaces over fixed field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another method is to use some method to construct new category by those you know (for example, as was it was shown by Robin Chapman).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another method is just create some simple category --- it is the method you have tried.&lt;/p&gt;&#xA;" OwnerUserId="695" LastActivityDate="2010-08-04T12:40:08.627" />
  <row Id="1524" PostTypeId="2" ParentId="1495" CreationDate="2010-08-04T12:42:58.247" Score="1" ViewCount="" Body="&lt;p&gt;You learn to swim by swimming, learn to talk by talking and &lt;strong&gt;you learn combinatorics by solving problems.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can get problems at &lt;a href=&quot;http://www.artofproblemsolving.com/&quot; rel=&quot;nofollow&quot;&gt;&lt;strong&gt;Art of Problem Solving&lt;/strong&gt;&lt;/a&gt;, try to solve as many &quot;different&quot; problems as you can.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you need help you know &lt;a href=&quot;http://math.stackexchange.com/&quot;&gt;&lt;strong&gt;where to get it&lt;/strong&gt;&lt;/a&gt; ;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;May the force be with you! &lt;/p&gt;&#xA;" OwnerUserId="705" LastActivityDate="2010-08-04T12:42:58.247" />
  <row Id="1525" PostTypeId="2" ParentId="967" CreationDate="2010-08-04T13:06:13.490" Score="9" ViewCount="" Body="&lt;p&gt;&lt;img src=&quot;http://worrydream.com/AlligatorEggs/pieces_1.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://worrydream.com/AlligatorEggs/&quot; rel=&quot;nofollow&quot;&gt;&lt;strong&gt;Alligator Eggs&lt;/strong&gt;&lt;/a&gt; is a cool way to learn lambda calculus.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also learning functional programming languages like Scheme, Haskell etc. will be added fun.&lt;/p&gt;&#xA;" OwnerUserId="705" LastActivityDate="2010-08-04T13:06:13.490" CommentCount="2" />
  <row Id="1526" PostTypeId="1" AcceptedAnswerId="1536" CreationDate="2010-08-04T13:33:43.277" Score="3" ViewCount="647" Body="&lt;p&gt;I have to divide 2860 by 3186. The question gives only 2 minutes and that division is only half part of question. Now I can't possibly make that division in or less than 2 minutes by applying traditional methods, which I can't apply on that division anyways.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So anyone can perform below division using faster technique?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2860/3186&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for reading, hoping to get some answers. :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a multiple choice question, with answers 6/7, 7/8, 8/9 and 9/10.&lt;/p&gt;&#xA;" OwnerUserId="712" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-08-04T17:51:29.800" LastActivityDate="2011-02-08T00:59:58.920" Title="Need faster division technique for 4 digit numbers" Tags="&lt;number-theory&gt;&lt;arithmetic&gt;&lt;numerical-methods&gt;" AnswerCount="8" CommentCount="7" FavoriteCount="2" />
  <row Id="1527" PostTypeId="2" ParentId="1378" CreationDate="2010-08-04T13:44:23.413" Score="0" ViewCount="" Body="&lt;p&gt;Mixture1 : 60%&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mixture2 : 35%&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let us say you take mixture using a pot of 100 units.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;also you take Mixture1 'a' times and Mixture2 'b' times using the pot.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rightarrow$&#xA;$\frac{(a  60 + b  35)}{(a 100 + b  100)} = \frac{50}{100}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;solving this you get &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{a}{b} = \frac{3}{2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-&gt; 300 units of Mixture1 and 200 units of Mixture2 will give you 500 units of mixture with alcohol concentration of 50% because 3 x 60 + 2 x 35 = 250.  &lt;/p&gt;&#xA;" OwnerUserId="705" LastEditorUserId="705" LastEditorDisplayName="" LastEditDate="2010-08-04T13:52:32.663" LastActivityDate="2010-08-04T13:52:32.663" />
  <row Id="1528" PostTypeId="2" ParentId="1526" CreationDate="2010-08-04T13:44:29.670" Score="0" ViewCount="" Body="&lt;p&gt;You could maybe use &lt;a href=&quot;http://en.wikipedia.org/wiki/Euclidean_algorithm&quot; rel=&quot;nofollow&quot;&gt;Euclid's Algorithm&lt;/a&gt; to find the GCD of these two numbers, and use this to reduce the fraction. With luck, the result will be easier to manage.&lt;/p&gt;&#xA;" OwnerUserId="505" LastActivityDate="2010-08-04T13:44:29.670" CommentCount="2" />
  <row Id="1529" PostTypeId="2" ParentId="1526" CreationDate="2010-08-04T14:05:00.470" Score="2" ViewCount="" Body="&lt;p&gt;If this problem appears on a standardized test you may want to take into account some broader considerations...  How many of these digits are even significant (relative to other quantities you will be computing with in &quot;second half&quot; of the problem)? What further operations will you be performing with this quantity - i.e. will you be significantly amplifying your error with nonlinear operations?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;At this point you may find that you can simply approximate the quantity as $\frac{2.9}{3.2}$, which can quickly be converted to the decimal 0.906, or so.  Seeing that the original number is approximately 0.898, you should be good to go.&lt;/p&gt;&#xA;" OwnerUserId="121" LastActivityDate="2010-08-04T14:05:00.470" CommentCount="1" />
  <row Id="1530" PostTypeId="2" ParentId="1514" CreationDate="2010-08-04T14:25:03.850" Score="0" ViewCount="" Body="&lt;p&gt;Determining a nice form for A(n) is a good start.  This is S(n) - S(n-1) = 2n + 2 .&#xA;Now the desired sum is sum {k= 1 to 10} A(2k-1) = sum {k=1 to 10} 4k = 4 * 10 * 11 / 2 = 220 .&lt;/p&gt;&#xA;" OwnerUserId="297" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-08-11T22:00:44.973" LastActivityDate="2010-08-11T22:00:44.973" CommentCount="0" />
  <row Id="1531" PostTypeId="2" ParentId="1526" CreationDate="2010-08-04T14:32:05.737" Score="3" ViewCount="" Body="&lt;p&gt;If you are given options like those in your comment (&lt;strong&gt;ie. where the numerators/denominators are of comparable size&lt;/strong&gt;) you can use the fact that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{2860}{3186}=\frac{a}{b} \iff 2860b= 3186a$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To see which $\frac{a}{b}$ is best, calculate $2860b- 3186a$ for each starting from one of the two midsize fractions and repeating for larger or smaller fractions depending on the sign of the result (if your answer is positive a larger fraction would approximate better, if negative: a smaller one). The choice of a and b giving the smallest answer is the correct one.&lt;/p&gt;&#xA;" OwnerUserId="160" LastEditorUserId="160" LastEditorDisplayName="" LastEditDate="2010-08-04T17:33:29.313" LastActivityDate="2010-08-04T17:33:29.313" CommentCount="3" />
  <row Id="1532" PostTypeId="2" ParentId="201" CreationDate="2010-08-04T14:35:50.003" Score="1" ViewCount="" Body="&lt;p&gt;I found this proof on this &lt;a href=&quot;http://www.matrix67.com/blog/archives/2700&quot; rel=&quot;nofollow&quot;&gt;Chinese website&lt;/a&gt;. Here is the translation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a proof with information theory (the proof can be done without information theory, but this is just cool). $\log(n)$ always means $\log_2(n)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Uniformly pick a integer N between 1 and n. We have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$H(N) = \log(n)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $H(x)$ is the entropy function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$N$ can be represented by $m$ integers $X_1,\ldots,X_m$ uniquely as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$N = p_i^{X_i}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $m$ is the amount of primes no larger than $n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$H(N) = H(X_1, X_2, \ldots, X_m)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$H(N) \leq H(X_1)+H(X_2)+\ldots+H(X_m)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $X_i \leq \log(N)$ ($2^{\log_2(N)} = N$), we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$H(N) \leq m H(log(N))$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{\log(n)}{\log(\log(n)+1)} \leq m$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The lhs can increase indefinitely. This even give a bound for the amount of primes smaller than $n$&lt;/p&gt;&#xA;" OwnerUserId="96" LastActivityDate="2010-08-04T14:35:50.003" CommentCount="1" />
  <row Id="1533" PostTypeId="1" CreationDate="2010-08-04T14:41:49.183" Score="6" ViewCount="1829" Body="&lt;p&gt;As a programmer, mathematics is important basic knowledge to study some topics, especially Algorithms. Many websites, and my fellows suggest me to study Discrete Mathematics before going to Algorithms, so I want to know which Discrete Mathematics book is suitable for my needs?&lt;/p&gt;&#xA;" OwnerUserId="725" LastEditorUserId="123" LastEditorDisplayName="" LastEditDate="2010-08-14T07:10:56.690" LastActivityDate="2010-08-14T07:19:25.220" CommunityOwnedDate="2010-08-04T20:33:55.820" Title="What is the best book for studying discrete mathematics?" Tags="&lt;big-list&gt;&lt;books&gt;&lt;discrete-mathematics&gt;" AnswerCount="5" CommentCount="0" FavoriteCount="10" />
  <row Id="1534" PostTypeId="2" ParentId="262" CreationDate="2010-08-04T14:44:22.797" Score="4" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Men_of_Mathematics&quot; rel=&quot;nofollow&quot;&gt;Men of Mathematics by E T Bell&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://img.fkcdn.com/img/185/9780671628185.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="705" LastActivityDate="2010-08-04T14:44:22.797" CommunityOwnedDate="2010-08-04T14:44:22.797" CommentCount="2" />
  <row Id="1535" PostTypeId="2" ParentId="1533" CreationDate="2010-08-04T14:46:13.760" Score="18" ViewCount="" Body="&lt;p&gt;By Donald Knuth himself! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://ecx.images-amazon.com/images/I/518GBVWEBYL._SL500_AA300_.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="705" LastActivityDate="2010-08-04T14:46:13.760" CommunityOwnedDate="2010-08-04T20:33:55.820" CommentCount="9" />
  <row Id="1536" PostTypeId="2" ParentId="1526" CreationDate="2010-08-04T14:46:56.800" Score="8" ViewCount="" Body="&lt;p&gt;if an approximate answer is enough, you may just throw out digits at the right of the two numbers; so 2860/3186 is more or less 28/31 (it's better to leave two digits, especially because 2860 is much more than 2000 while 3186 is not so much more than 3000)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A second kind of approximation consists in adding or subtracting from both sides two numbers whose ratio is more or less what you expect the answer is. If you start from 2860/3186, which is near 1/1, you may subtract 186 from both sides, ending up with 2674/3000 ~ .891; if you are quick in additions and subtractions you may first add 14 to both sides, obtaining 2874/3200, then subtracting 180 and 200 (whose ratio is .9) obtaining 2694/3000 ~ .896.&lt;/p&gt;&#xA;" OwnerUserId="89" LastActivityDate="2010-08-04T14:46:56.800" CommentCount="1" />
  <row Id="1537" PostTypeId="1" AcceptedAnswerId="1539" CreationDate="2010-08-04T15:01:54.853" Score="24" ViewCount="835" Body="&lt;p&gt;There are lots (an infinitude) of smooth functions that coincide with f(n)=n! on the integers.  Is there a simple reason why Euler's Gamma function $\Gamma (z) = \int_0^\infty t^{z-1} e^t dt$ is &quot;best&quot;?  In particular, I'm looking for reasons that I can explain to first-year calculus students.&lt;/p&gt;&#xA;" OwnerUserId="726" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-08-28T09:52:18.710" LastActivityDate="2010-08-28T09:52:18.710" Title="Why is Euler's Gamma function the &quot;best&quot; extension of the factorial function to the reals?" Tags="&lt;real-analysis&gt;&lt;education&gt;&lt;gamma-function&gt;&lt;special-functions&gt;" AnswerCount="7" CommentCount="3" FavoriteCount="10" />
  <row Id="1539" PostTypeId="2" ParentId="1537" CreationDate="2010-08-04T15:41:57.133" Score="32" ViewCount="" Body="&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Euler%25E2%2580%2593Mascheroni_constant&quot; rel=&quot;nofollow&quot;&gt;Bohr–Mollerup theorem&lt;/a&gt; shows that the gamma function is the &lt;em&gt;only&lt;/em&gt; function that satisfies the properties &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$f(1)=1$;&lt;/li&gt;&#xA;&lt;li&gt;$f(x+1)=xf(x)$ for every $x\geq 0$;&lt;/li&gt;&#xA;&lt;li&gt;$\log f$ is a convex function. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The condition of log-convexity is particularly important when one wants to prove various inequalities for the gamma function. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;By the way, the gamma function is &lt;em&gt;not the only meromorphic function satisfying&lt;/em&gt;&#xA;$$f(z+1)=z f(z),\qquad f(1)=1,$$&#xA;&lt;em&gt;with no zeroes and no poles other than the points&lt;/em&gt; $z=-n$, $n=0,1,2\dots$. There is a whole family of such functions, which, in general, have the form&#xA;$$f(z)=\exp{(-g(z))}\frac{1}{z\prod\limits_{n=1}^{\infty} \left(1+\frac{z}{m}\right)e^{-z/m}},$$&#xA;where $g(z)$ is an entire function such that&#xA;$$g(z+1)-g(z)=\gamma+2k\pi i,\quad k\in\mathbb Z, $$&#xA;($\gamma$ is &lt;a href=&quot;http://en.wikipedia.org/wiki/Euler%25E2%2580%2593Mascheroni_constant&quot; rel=&quot;nofollow&quot;&gt;Euler's constant&lt;/a&gt;). The gamma function corresponds to the simplest choice&#xA;$g(z)=\gamma z$.&lt;/p&gt;&#xA;" OwnerUserId="723" LastEditorUserId="723" LastEditorDisplayName="" LastEditDate="2010-08-04T22:08:58.947" LastActivityDate="2010-08-04T22:08:58.947" CommentCount="2" />
  <row Id="1541" PostTypeId="2" ParentId="1537" CreationDate="2010-08-04T15:42:02.950" Score="0" ViewCount="" Body="&lt;p&gt;Doesn't one choose the Gamma Function because it satisfies the functional equation $\Gamma(x) = (x-1)*\Gamma(x-1)$ for all positive $x$ and therefore looks a lot like the factorial?&lt;/p&gt;&#xA;" OwnerUserId="505" LastActivityDate="2010-08-04T15:42:02.950" CommentCount="2" />
  <row Id="1542" PostTypeId="2" ParentId="1467" CreationDate="2010-08-04T15:45:45.260" Score="0" ViewCount="" Body="&lt;p&gt;I may be wrong but it seems that C is not a set of all sets or anything similar. It has IMHO a trivial bijection into $\mathbb{Z}$ and therefore into $\mathbb{N}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I haven't met any paradox to forbid $|C| \in C$ ($\{1\}$ is such set).&lt;/p&gt;&#xA;" OwnerUserId="728" LastActivityDate="2010-08-04T15:45:45.260" CommentCount="4" />
  <row Id="1543" PostTypeId="1" CreationDate="2010-08-04T15:55:08.090" Score="0" ViewCount="191" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Possible Duplicate:&lt;/strong&gt;&lt;br&gt;&#xA;  &lt;a href=&quot;http://math.stackexchange.com/questions/1482/problem-with-inequality&quot;&gt;problem with inequality&lt;/a&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;p&gt;How do I go about solving $0&amp;lt;1−\frac{an}{mb^2}e^{−r(T−t)}&amp;lt;1$, where a,b,T&gt;t&gt;0? I have been stuck here for some time now.&lt;/p&gt;&#xA;" OwnerUserId="626" LastEditorUserId="626" LastEditorDisplayName="" LastEditDate="2010-08-04T16:38:08.100" LastActivityDate="2010-08-04T17:54:26.630" ClosedDate="2010-08-04T20:37:43.003" Title="tricky inequality" Tags="&lt;algebra-precalculus&gt;" AnswerCount="0" CommentCount="0" FavoriteCount="0" />
  <row Id="1544" PostTypeId="2" ParentId="1454" CreationDate="2010-08-04T16:08:31.747" Score="1" ViewCount="" Body="&lt;p&gt;The notion &quot;symmetric equilibrium&quot; (the one from Wikipedia article) is not applicable here, because the game is not symmetric (different players have different &quot;profits per click&quot;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've a look at the paper and I think, that &quot;symmetric Nash equilibrium&quot; in your case is nothing but technically convenient case of Nash equilibrium (the latter is not unique in your case).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also I've noted a probable misprint in the proof of the &quot;Fact 1&quot;.&#xA;It should be $(v_s-p_s)x_s \geq (v_s-p_{S+1})x_{S+1}$ instead of $(v_s-p_s)x_s \geq (v_{S+1}-p_{S+1})x_{S+1}$.&lt;/p&gt;&#xA;" OwnerUserId="695" LastEditorUserId="695" LastEditorDisplayName="" LastEditDate="2010-08-04T22:33:52.443" LastActivityDate="2010-08-04T22:33:52.443" />
  <row Id="1545" PostTypeId="2" ParentId="1537" CreationDate="2010-08-04T16:11:34.100" Score="18" ViewCount="" Body="&lt;p&gt;Actually there are other (less-frequently) used extensions to the factorial, with different properties from the gamma function which may be desirable in some contexts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://www.luschny.de/math/factorial/hadamard/f1.gif&quot; alt=&quot;Euler's Gamma Function&quot;&gt;&lt;br&gt;&#xA;&lt;em&gt;Euler's Gamma Function&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://www.luschny.de/math/factorial/hadamard/hadgamma1.gif&quot; alt=&quot;Hadamard's Gamma function&quot;&gt;&lt;br&gt;&#xA;&lt;em&gt;Hadamard's Gamma function&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://www.luschny.de/math/factorial/hadamard/LuschnyFactorial5.gif&quot; alt=&quot;Luschny's factorial function&quot;&gt;&lt;br&gt;&#xA;&lt;em&gt;Luschny's factorial function&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;See &lt;a href=&quot;http://www.luschny.de/math/factorial/hadamard/HadamardsGammaFunction.html&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; for more information.&lt;/p&gt;&#xA;" OwnerUserId="136" LastActivityDate="2010-08-04T16:11:34.100" />
  <row Id="1546" PostTypeId="2" ParentId="1519" CreationDate="2010-08-04T16:25:23.167" Score="1" ViewCount="" Body="&lt;p&gt;Concerning the Hopf fibration, the film &quot;&lt;a href=&quot;http://www.dimensions-math.org/Dim_E.htm&quot; rel=&quot;nofollow&quot;&gt;Dimensions&lt;/a&gt;&quot; does a superb job of visualizing it (chapters 7-8 in the &lt;a href=&quot;http://www.dimensions-math.org/Dim_tour_E.htm&quot; rel=&quot;nofollow&quot;&gt;table of contents&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Concerning the visualization of the glueing, maybe it helps to visualize other, similar constructions to get at least some intuition. For instance, you could try to get a picture of how the torus looks like after every &quot;meridian&quot; (blue circle in your picture) has been squashed to a point.&lt;/p&gt;&#xA;" OwnerUserId="434" LastActivityDate="2010-08-04T16:25:23.167" CommentCount="0" />
  <row Id="1547" PostTypeId="2" ParentId="198" CreationDate="2010-08-04T16:40:50.263" Score="0" ViewCount="" Body="&lt;p&gt;Here is one elementary result you can't prove without the Axiom of Choice (that I used to think was not part of constructive mathematics).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Everybody knows how to prove that, if two sets $X$ and $Y$ are non-empty, then their cartesian product $X \times Y$ is non-empty too: since $X$ is non-empty, you can choose one element from it: $x \in X$. For the same reason, you can choose also one $y \in Y$. So you have an element $(x,y) \in X\times Y$ and therefore it is non-empty. Right?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a finite family of non-empty sets $X_1, \dots , X_n$ this procedure still works: you choose one element in each set starting from the first: $x_1 \in X_1$. And then keep on doing the same for the rest: choose $x_2$ in the second set $X_2$, $x_3$ in the third one $X_3$,... And so on, till you reach $x_n \in X_n$. You have thus produced an element of the cartesian product: $(x_1, \dots , x_n) \in X_1 \times \dots \times X_n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even if you had an infinite &lt;em&gt;countable&lt;/em&gt; family of non-empty sets $X_1, X_2, \dots , X_n, \dots$, you could produce an element of their product $\prod_{n=1}^{\infty} X_n$ this way: choose successively elements $x_1 \in X_1$, $x_2 \in X_2$,..., $x_n \in X_n$... And you would obtain your element $(x_n)$ of your infinite cartesian product. Hence $\prod_{n=1}^{\infty} X_n$ is non-empty too if all the $X_n$, $n=1, 2, \dots $ are non-empty.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, try to do the same with an &lt;em&gt;arbitrary&lt;/em&gt; family of sets $X_i$, with the indexes $i$ belonging to an &lt;em&gt;arbitrary&lt;/em&gt; indexing set $I$. That is, set $I$ can be infinite and &lt;em&gt;not&lt;/em&gt;  countable. This kind of cartesian products do exist in &quot;nature&quot;. For instance: $I$ could be the set of real numbers $\mathbb{R}$. Take $X_i = \mathbb{R}$ for all $i \in \mathbb{R}$ too. This cartesian product $\prod_{i\in \mathbb{R}} X_i = \mathbb{R}^\mathbb{R}$ is the same as the set of all (not necessarily continuous) functions from $\mathbb{R}$ to $\mathbb{R}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, you can prove that $\mathbb{R}^\mathbb{R}$ is non-empty by showing an element of it (for instance, $f(x) = x$ for all $x\in \mathbb{R}$), but how can you prove, in general, that the product of an arbitrary family of non-empty sets $\prod_{i\in I}X_i$ is non-empty too?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you try to imitate what you have just done in the finite or countable cases, you'll find asking yourself: &lt;em&gt;where&lt;/em&gt; do I start? Which is the &lt;em&gt;first&lt;/em&gt;  $\ i \in I$? Assuming there is a &lt;em&gt;first&lt;/em&gt; $i$, which is the &lt;em&gt;next&lt;/em&gt; one?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer is that, for an infinite, non-countable, set of indexes $I$ there is in general no such a thing as a &lt;em&gt;first&lt;/em&gt; index $i$, nor a &lt;em&gt;next&lt;/em&gt; one. So you can't use the previous algorithm in order to produce an element of the product $\prod_{i\in I} X_i$...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Except you have something that allows you to &lt;em&gt;choose&lt;/em&gt;, one particular element of each set $x_i \in X_i$ -no matter how, even with no precise constructive algorithm as in the finite or countable cases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you were able to do that for no matter which set of indexes $I$, then you would have your element $(x_i)_{i\in I}$ and you would have proved that, if every $X_i \neq \emptyset$, so is their cartesian product $\prod_{i\in I} X_i$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What do &lt;em&gt;allow&lt;/em&gt; you for an infinite, non-countable, set of indexes $I$ to &lt;em&gt;choose&lt;/em&gt; those $x_i \in X_i$ for every $I$? Well, this is exactly what the &lt;em&gt;Axiom of Choice&lt;/em&gt; says.&lt;/p&gt;&#xA;" OwnerUserId="664" LastEditorUserId="664" LastEditorDisplayName="" LastEditDate="2010-08-17T01:27:45.070" LastActivityDate="2010-08-17T01:27:45.070" CommentCount="5" />
  <row Id="1548" PostTypeId="1" AcceptedAnswerId="1625" CreationDate="2010-08-04T16:53:25.383" Score="9" ViewCount="242" Body="&lt;p&gt;Here is a classic theorem of Herstein:&#xA;$G$ is a finite group, $M$ a maximal subgroup, which is abelian.  Then $G$ is solvable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The proof is pretty easy, but it uses character theory (specifically, Frobenius' theorem on Frobenius groups).  Is there a character-theory-free proof?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To get things going, note that we can reduce to the case where:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i) $M$ is core-free and a Hall subgroup of $G$;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ii) $Z(G)=1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Steve&lt;/p&gt;&#xA;" OwnerUserId="641" LastActivityDate="2010-08-05T14:19:01.960" Title="G a finite group, M a maximal subgroup; M abelian implies G solvable?" Tags="&lt;group-theory&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="3" />
  <row Id="1549" PostTypeId="2" ParentId="1537" CreationDate="2010-08-04T16:55:16.410" Score="10" ViewCount="" Body="&lt;p&gt;For whatever reason, Nature (by which I mean integrals) seems to prefer the Gamma function as the &quot;correct&quot; substitute for the factorial in various integrals, which seems to come more or less from its integral definition.  For example, for non-negative integers $a, b$, it's not hard to show (and there's a really nice probabilistic argument) that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle \int_{0}^1 t^a (1 - t)^b \, dt = \frac{a! b!}{(a+b+1)!}.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For (non-negative?) real values of $a$ and $b$ the correct generalization is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle \int_0^1 t^a (1 - t)^b \, dt = \frac{\Gamma(a-1) \Gamma(b-1)}{\Gamma(a+b)}.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And, of course, integrals are important, so the Gamma function must also be important.  For example, the Gamma function appears in the general formula for the &lt;a href=&quot;http://en.wikipedia.org/wiki/N-sphere#Volume_and_surface_area&quot; rel=&quot;nofollow&quot;&gt;volume of an n-sphere&lt;/a&gt;.  But the result that, for me, really forces us to take the Gamma function seriously is its appearance in the &lt;a href=&quot;http://en.wikipedia.org/wiki/Riemann_zeta_function#The_functional_equation&quot; rel=&quot;nofollow&quot;&gt;functional equation for the Riemann zeta function&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-08-04T16:55:16.410" CommentCount="3" />
  <row Id="1550" PostTypeId="2" ParentId="262" CreationDate="2010-08-04T17:06:00.823" Score="17" ViewCount="" Body="&lt;p&gt;Gödel, Escher, Bach: An Eternal Golden Braid  by Douglas Hofstadter:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://upload.wikimedia.org/wikipedia/en/thumb/f/f1/GEBcover.jpg/200px-GEBcover.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="2263" LastActivityDate="2010-08-04T17:06:00.823" CommunityOwnedDate="2010-08-04T17:06:00.823" />
  <row Id="1551" PostTypeId="2" ParentId="1408" CreationDate="2010-08-04T17:12:21.593" Score="3" ViewCount="" Body="&lt;p&gt;I think there is a missing word in Akhil Mathew's answer: and it's &quot;filtered&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can do that because stalks are &lt;em&gt;filtered&lt;/em&gt; colimits (aka &quot;direct limits&quot;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For filtered colimits, $\varinjlim_i X_i$, you can take representatives of elements $x \in \varinjlim_i X_i$ for some $i$ belonging to the set of indexes $I$ (in our case, the open sets $U$). That is, you can find some $i$ and $x_i \in X_i$ that goes to your $x$ through the universal arrow $X_i \longrightarrow \varinjlim_i X_i$. For instance, every element of the stalk $O_{X,p}$ can be represented by a section $f \in O_X(U)$ for some open set $U$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But this is not true for other kinds of colimits.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, take the push-out of two arrows $f: A \longrightarrow B$ and $g: A \longrightarrow C$ in the category of, say, abelian groups. Elements of this push out $B \oplus_A C$ are classes of pairs $(b,c) \in B\oplus C$, where you quotient out elements of the form $(f(a), 0) - (0, g(a))$, for all $a\in A$. That is, $(f(a),0) = (0,g(a))$ in $B\oplus_A C$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Elements of $B\oplus_A C$ cannot be represented, in general, by elements coming from just $B$ or $C$, which are of the form $(b,0)$ or $(0,c)$, respectively: so, for a general $(b,c) \in B\oplus_A C$ there is no $b \in B$, nor $c\in C$, that represents it.&lt;/p&gt;&#xA;" OwnerUserId="664" LastActivityDate="2010-08-04T17:12:21.593" CommentCount="1" />
  <row Id="1552" PostTypeId="2" ParentId="1512" CreationDate="2010-08-04T17:48:00.320" Score="2" ViewCount="" Body="&lt;p&gt;The finite index subgroups that you say that you can think of are precisely&#xA;the closed (equivalently open) finite index subgroups of $Gal(F/\mathbb{Q})$ &#xA;with respect to the Krull topology on the group. These are really the only relevant&#xA;ones in terms of Galois theory. But the group has non-closed subgroups too,&#xA;but their existence requires some non-constructive principle such as Zorn's&#xA;lemma. These subgroups won't be open or closed in the Krull topology.&lt;/p&gt;&#xA;" OwnerUserId="226" LastEditorUserId="226" LastEditorDisplayName="" LastEditDate="2010-08-04T18:46:40.320" LastActivityDate="2010-08-04T18:46:40.320" />
  <row Id="1553" PostTypeId="2" ParentId="1482" CreationDate="2010-08-04T17:54:26.630" Score="4" ViewCount="" Body="&lt;p&gt;There are many unnecessary variables. Let&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\alpha = \frac{an}{mb^2}.\qquad(1)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then the inequality becomes&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ 0 &amp;lt; 1 - \alpha e^{-r (T - t)} &amp;lt; 1 $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first obvious step is perform “1 &amp;minus;” on every parts,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ 1 &gt; \alpha e^{-r (T-t)} &gt; 0 $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the exponential function's range is positive and &amp;lt; 1 (since &lt;em&gt;r&lt;/em&gt; &gt; 0 and &lt;em&gt;T&lt;/em&gt; &gt; &lt;em&gt;t&lt;/em&gt; &gt; 0), we can ensure &lt;em&gt;&amp;alpha;&lt;/em&gt; is positive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If 0 &amp;lt; &lt;em&gt;&amp;alpha;&lt;/em&gt; ≤ 1, then every &lt;em&gt;t&lt;/em&gt; will satisfy the inequality (the first solution).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So assume &lt;em&gt;&amp;alpha;&lt;/em&gt; &gt; 1. Now it's pretty obvious on how to solve &lt;em&gt;t&lt;/em&gt; in terms of &lt;em&gt;r&lt;/em&gt;, &lt;em&gt;T&lt;/em&gt; and &lt;em&gt;&amp;alpha;&lt;/em&gt;. Substitute (1) again to get back &lt;em&gt;a&lt;/em&gt;, &lt;em&gt;n&lt;/em&gt;, &lt;em&gt;m&lt;/em&gt; and &lt;em&gt;b&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Things you may consider:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;em&gt;e&lt;sup&gt;x&lt;/sup&gt;&lt;/em&gt; is strictly increasing.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="171" LastEditorUserId="171" LastEditorDisplayName="" LastEditDate="2010-08-05T10:41:55.200" LastActivityDate="2010-08-05T10:41:55.200" CommentCount="6" />
  <row Id="1554" PostTypeId="2" ParentId="1537" CreationDate="2010-08-04T17:54:55.770" Score="2" ViewCount="" Body="&lt;p&gt;Comment on Jens: You can take any function on $[0,1)$ and extend to $\mathbb{R}$ using this functional equation. You have three examples above. These can obviously be made continuous. More surprising is that there are several ways of getting a convex function. However if you insist on log-convex there is a unique function.&lt;/p&gt;&#xA;" OwnerUserId="638" LastActivityDate="2010-08-04T17:54:55.770" CommentCount="2" />
  <row Id="1555" PostTypeId="1" AcceptedAnswerId="1560" CreationDate="2010-08-04T18:14:41.537" Score="5" ViewCount="333" Body="&lt;p&gt;I'm trying to understand better the following addition formula: $$\int_0^a \frac{\mathrm{d}x}{\sqrt{1-x^2}} + \int_0^b \frac{\mathrm{d}x}{\sqrt{1-x^2}} = \int_0^{a\sqrt{1-b^2}+b\sqrt{1-a^2}} \frac{\mathrm{d}x}{\sqrt{1-x^2}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The term $a\sqrt{1-b^2}+b\sqrt{1-a^2}$ can be derived from trigonometry (since $\sin(t) = \sqrt{1 - \cos^2(t)}$) but I have not been able to find any way to derive this formula without trigonometry, how could it be done?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;edit:&lt;/strong&gt; fixed a mistake in my formula.&lt;/p&gt;&#xA;" LastActivityDate="2010-08-06T20:34:09.730" Title="Finding an addition formula without trigonometry" Tags="&lt;calculus&gt;" AnswerCount="2" CommentCount="9" FavoriteCount="1" />
  <row Id="1556" PostTypeId="2" ParentId="1512" CreationDate="2010-08-04T18:14:50.697" Score="7" ViewCount="" Body="&lt;p&gt;Let's forget the Galois group structure: it is enough to consider the infinite product $G = \mathbb{Z}/(2)^\mathbb{N}$ with its profinite topology and exhibit non-closed subgroups of finite index.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An element of $G$ is naturally identified with a subset of $\mathbb{N}$: the string $(a_i)$ corresponds to the subset $A$ such that $n \in A$ if and only if $a_n = 1$. With this interpretation, the operation on $G$ just becomes the symmetric difference of sets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now consider the filter $F'$ of all cofinite subsets of $\mathbb{N}$, and let $F$ be any ultrafilter containing it. I show that we can associate to $F$ a subgroup of index $2$. The $F$ will be non closed since $F$ is not a principal ultrafilter (indeed any non-principal ultrafilter will do).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the characteristic function of $F$: $$f = \chi_F \colon G = \mathcal{P}(\mathbb{N}) \to \mathbb{Z}/(2)$$. I claim that this is a homomorphism, and its kernel is the desired subgroup.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is easy to verify directly. Let $A, B \in G$ and assume $f(A) = f(B) = 0$. This means that neither $A$ nor $B$ belong to $F$, so their complementary sets do. It follows that $A^c \cap B^c \in F$, so that $A \cup B \notin F$. A fortiori the symmetric difference $A \Delta B \notin F$, whic means $f(A \Delta B) = 0$. The other cases are similar.&lt;/p&gt;&#xA;" OwnerUserId="233" LastActivityDate="2010-08-04T18:14:50.697" />
  <row Id="1557" PostTypeId="2" ParentId="1526" CreationDate="2010-08-04T18:36:03.180" Score="10" ViewCount="" Body="&lt;p&gt;Another very specific trick, based on the multiple choices: since all the choices are of the form $\displaystyle \frac{n-1}{n} = 1 - \frac{1}{n}$, you're trying to find $n$ for which $1/n$ is closest to 1&amp;minus;(your fraction). So you'd consider (denominator&amp;minus;numerator)/denominator = (3186&amp;minus;2860)/3186 which is (around 320)/3186, clearly closest to 1/10.&lt;/p&gt;&#xA;" OwnerUserId="205" LastActivityDate="2010-08-04T18:36:03.180" />
  <row Id="1560" PostTypeId="2" ParentId="1555" CreationDate="2010-08-04T19:40:04.787" Score="3" ViewCount="" Body="&lt;p&gt;Replace the first integral by the same thing from $-a$ to $0$, and consider the points W,X,Y,Z on the unit circle above $-a,0,b$ and $c = a\sqrt{1-b^2} + b \sqrt{1-a^2}$.  Draw the family of lines parallel to XY (and WZ). This family sets up a map from the circle to itself; through each point, draw a parallel and take the other intersection of that line with the circle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your formula says that this map [edit: or rather the map it induces on the $x$-coordinates of points on the circle] is a change of variables converting the integral on $[-a,0]$ to the same integral on $[b,c]$.  Whatever differentiation you perform in the process of proving this, will be the verification that $dx/y$ is a rotation-invariant differential on the circle $x^2 + y^2 = 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[The induced map on x-coordinates is: $x \to$ point on semicircle above $x \to$ corresponding point on line parallel to XY $\to x$-coordinate of the second point.  Here were are just identifying $[-1,1]$ with the semicircle above it.]&lt;/p&gt;&#xA;" OwnerUserId="467" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-08-06T20:34:09.730" LastActivityDate="2010-08-06T20:34:09.730" CommentCount="4" />
  <row Id="1562" PostTypeId="2" ParentId="262" CreationDate="2010-08-04T20:12:40.433" Score="0" ViewCount="" Body="&lt;p&gt;I'll recommend two, which are similar in that they take fairly elementary mathematical problems and give very thorough and careful &quot;talking out loud&quot; illustrations of how a proper mathematician would go about thinking them through - what's really going on, what's a good example, what's a definitive counterexample, how to generalise, how to realise you've reached a dead end, and so on.  &quot;Proofs and refutations&quot; by Imre Lakatos (just one, geometrical, problem, in glorious detail).  &quot;Mathematics and plausible reasoning Vol 1&quot; by G. Polya (a little more advanced, and much more satisfying, than &quot;How to solve it&quot;).&lt;/p&gt;&#xA;" LastActivityDate="2010-08-04T20:12:40.433" CommunityOwnedDate="2010-08-04T20:12:40.433" />
  <row Id="1563" PostTypeId="1" CreationDate="2010-08-04T21:01:22.910" Score="5" ViewCount="447" Body="&lt;p&gt;I just started learning model theory on my own, and I was wondering if there are any interesting examples of two structures of a language L which are not isomorphic, but are elementarily equivalent (this means that any L-sentence satisfied by one of them is satisfied by the second).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am using the notaion of David Marker's book &quot;Model theory: an introduction&quot;.&lt;/p&gt;&#xA;" LastActivityDate="2010-09-14T05:40:03.343" Title="Example of non-isomorphic structures which are elementarily equivalent" Tags="&lt;logic&gt;&lt;model-theory&gt;" AnswerCount="5" FavoriteCount="1" />
  <row Id="1564" PostTypeId="2" ParentId="1563" CreationDate="2010-08-04T21:13:07.820" Score="0" ViewCount="" Body="&lt;p&gt;Take a look at the wiki article on &lt;a href=&quot;http://en.wikipedia.org/wiki/Real_closed_field&quot; rel=&quot;nofollow&quot;&gt;Real closed fields&lt;/a&gt;. Briefly, they are fields that are first-order equivalent to the field of the real numbers. An example is the field of real numbers that are roots of polynomials with rational coefficients. That's pretty much all I know about them, though.&lt;/p&gt;&#xA;" OwnerUserId="400" LastActivityDate="2010-08-04T21:13:07.820" />
  <row Id="1565" PostTypeId="2" ParentId="1563" CreationDate="2010-08-04T21:13:08.320" Score="2" ViewCount="" Body="&lt;p&gt;There are a real numbers worth of countable models of Peano arithmetic, all elementarily equivalent to the usual model, but all pairwise nonisomorphic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Every nonstandard example (nonstandard means not isomorphic to the usual model) is characterized by the fact that it contains &quot;infinitely large&quot; numbers.  That is, each model M has a canonical copy of N = usual naturals inside.  However, each nonstandard model M has an element p (in fact, countably many such elements) which the model things is bigger than everything in its copy of N.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They are created by standard compactness arguments paired with the downward Lowenheim-Skolem theorem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let PA denote the axioms for Peano Arithmetic and set T = Th(N), the theory of N (i.e., the set of all first order statements which are true in the usual interpretation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, add a constant c to the language and let phi_n be the statement c &gt; n, or more appropriately, c &gt; SSSS.....S(0), where S is the successor function of PA and there are n S's in the expression.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, consider the set of axioms PA union T union phi_n for every n.  By a standard compactness argument, this set of axioms is consistent, and so by the Godel completeness theorem, there is a model M of all the axioms simultaneously.  By downward Lowenheim-Skolem, we can assume M is countable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, ask yourself, what can c be?  Well, since M satisfies phi_n for every n, we must have c &gt; n for each &quot;usual&quot; natural number in M, i.e., c is infinite!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What's really cool is that by playing around with different kinds of phi_n, one can find models that have elements which are divisible by NO &quot;standard&quot; prime number.  One can also find models that have elements which are simultaneously divisible by EVERY &quot;standard&quot; prime number!&lt;/p&gt;&#xA;" OwnerUserId="331" LastActivityDate="2010-08-04T21:13:08.320" />
  <row Id="1566" PostTypeId="2" ParentId="262" CreationDate="2010-08-04T21:20:39.347" Score="2" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0471608394&quot; rel=&quot;nofollow&quot;&gt;Geometric Algebra&lt;/a&gt; by Emil Artin. Though not for the beginner, it can do wonders for an intermediate undergraduate in terms of expanding their horizons and helping them appreciate the beauty and interconnectedness of mathematics. It did for me and I think convinced me that I'm a geometer at heart.&lt;/p&gt;&#xA;" OwnerUserId="400" LastActivityDate="2010-08-04T21:20:39.347" CommunityOwnedDate="2010-08-04T21:20:39.347" CommentCount="1" />
  <row Id="1567" PostTypeId="2" ParentId="1563" CreationDate="2010-08-04T21:23:33.447" Score="6" ViewCount="" Body="&lt;p&gt;Laws, yes&lt;sup&gt;[1]&lt;/sup&gt;: if there weren't such examples, there wouldn't be a subject called model theory and therefore Marker's book would not exist.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are, however, not very many explicit examples which can be given, with proof, right after the definition of elementary equivalence, a pedagogical problem that I encounered recently when I taught a short summer course on model theory.  When I first gave the definition, all I was able to come up with was the following argument: for any language $L$, the class of $L$-structures is a proper class [any nonempty set can be made into the &quot;universe&quot;, or underlying set of, an $L$-structure] whereas since there are at most &#xA;&lt;code&gt;$2^{\max|L|,\aleph_0}$&lt;/code&gt; different theories in the language $L$, the $L$-structures up to elementary equivalence must form a set.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if you just want examples without proof, sure, here goes:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) In the empty language, two sets are elementarily equivalent iff they are both infinite or both finite of the same cardinality.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Any two dense linear orders without endpoints are elementarily equivalent.  [The same holds for DLOs &lt;em&gt;with&lt;/em&gt; endpoints, but the two classes of structures are not elementarily equivalent to each other.]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) Any two algebraically closed fields of the same characteristic are elementarily equivalent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4) Any two real-closed fields are elementarily equivalent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;s&gt;5) Any two infinite models of the theory of finite fields are elementarily equivalent.&lt;/s&gt; (Nope! See Dave Marker's answer.)   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In each case, such structures exist of every infinite cardinality, so the class of isomorphism classes of such structures is a proper class, not a set. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;[1]: &quot;M-O-O-N, that spells model theory!&quot;&lt;/p&gt;&#xA;" OwnerUserId="299" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2010-08-06T09:24:27.740" LastActivityDate="2010-08-06T09:24:27.740" CommentCount="1" />
  <row Id="1568" PostTypeId="1" AcceptedAnswerId="1574" CreationDate="2010-08-04T21:23:48.410" Score="2" ViewCount="195" Body="&lt;p&gt;As a follow up to &lt;a href=&quot;http://math.stackexchange.com/questions/1416/algebra-of-random-variables&quot;&gt;algebra-of-random-variables&lt;/a&gt;, is it possible to compute:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$z = \frac{dx}{dy}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $x$ and $y$ are drawn from a known, independent probability distribution $x \in f(x)$, $y \in g(y)$? Geometrically the problem makes sense, one has a function $h(x,y) = f(x)g(y)$ that is the joint probability distribution and we are essentially asking for the gradient along one direction. I can get this to the expression ($z \in q(z)$):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$q(z) = \int f(x) g( \int{[z dx]} ) ( \frac{d}{dz} \int{[z dx]} ) dx$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I'm not quite sure what to make of this, or even how to compute it given specific PDF's.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;: For simplicity, assume all the PDF's considered are continuous and smooth.&lt;/p&gt;&#xA;" OwnerUserId="196" LastEditorUserId="196" LastEditorDisplayName="" LastEditDate="2010-08-04T22:00:03.460" LastActivityDate="2010-08-04T22:19:26.333" Title="Derivatives of random variables" Tags="&lt;probability-theory&gt;&lt;probability&gt;" AnswerCount="2" />
  <row Id="1570" PostTypeId="2" ParentId="847" CreationDate="2010-08-04T21:40:43.287" Score="4" ViewCount="" Body="&lt;p&gt;This is not a directly a matter of hyperbolic geometry but of &lt;em&gt;complex Euclidean geometry&lt;/em&gt;.  The construction of &quot;impossible&quot; triangles is the same as the construction of square roots of negative numbers, when considering the coordinates the vertices of those triangles must have.  If you calculate the coordinates of a triangle with sides 1,3,5 or 3,4,8 you get complex numbers.  In ordinary real-coordinate Euclidean geometry this means there is no such triangle.  If complex coordinates are permitted, the triangle exists, but not all its points are visible in drawings that only represent the real points.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In plane analytic geometry where the Cartesian coordinates are allowed to be complex, concepts of point,line, circle, squared distance, dot-product, and (with suitable definitions) angle and cosine can be interpreted using the same formulas.  This semantics extends the visible (real-coordinate) Euclidean geometry to one where any two circles intersect, but possibly at points with complex coordinates.  We &quot;see&quot; only the subset of points with real coordinates, but the construction that builds a triangle with given distances between the sides continues to work smoothly, and some formulations of the law of Cosines will continue to hold.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are certainly relations of this picture to hyperbolic geometry.  One is that $cos(z)=cosh(iz)$ so you can see the hyperbolic cosine and cosine as the same once complex coordinates are permitted.  Another is that the Pythagorean metric on the complex plane, considered as a 4-dimensional real space, is of the form $x^2 + y^2 - w^2 - u^2$, so that the locus of complex points at distance $0$ from the origin contains copies of the hyperboloid model of hyperbolic geometry.  But there is no embedding of the hyperbolic plane as a linear subspace of the complex Euclidean plane, so we don't get from this an easier way of thinking about hyperbolic geometry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To help visualize what is going on it is illuminating to calculate the coordinates of a triangle with sides 3,4,8 or other impossible case, and the dot-products of the vectors involved.&lt;/p&gt;&#xA;" OwnerUserId="467" LastActivityDate="2010-08-04T21:40:43.287" CommentCount="2" />
  <row Id="1571" PostTypeId="2" ParentId="1568" CreationDate="2010-08-04T21:46:00.977" Score="1" ViewCount="" Body="&lt;p&gt;You can take a derivative of a function $x$ with respect to one of its variables and the function has to be smooth w.r.t. this parameter.   If $x$ is being drawn from a set of smooth functions then it is certainly possible to consider the derivative of a random variable $x$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I'm really not sure what $\frac{dx}{dy}$ would even mean in your case.&lt;/p&gt;&#xA;" OwnerUserId="780" LastActivityDate="2010-08-04T21:46:00.977" />
  <row Id="1572" PostTypeId="2" ParentId="29" CreationDate="2010-08-04T22:05:50.883" Score="3" ViewCount="" Body="&lt;p&gt;SpeQ is really nice. If you close it's left/right areas it's just like Notepad, but a Notepad which can compute stuff. You can change any line displayed, press Enter at it's end and the line will be recomputed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just like in Notepad, you can select any part of what's displayed and delete it. You can also paste non-mathematical text into it. It will also remember what's displayed inside it the next time you start it. It has an option to clear it's content on Exit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A lot of people recommended Python. It's is my main programming language and I love it, but SpeQ it's MUCH more powerful and easy to use than Python as a calculator.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://dl.dropbox.com/u/190212/public/speq.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.speqmath.com/index.php?id=1&quot; rel=&quot;nofollow&quot;&gt;http://www.speqmath.com/index.php?id=1&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="775" LastEditorUserId="775" LastEditorDisplayName="" LastEditDate="2010-08-04T22:13:24.340" LastActivityDate="2010-08-04T22:13:24.340" CommunityOwnedDate="2010-08-04T22:05:50.883" />
  <row Id="1573" PostTypeId="2" ParentId="262" CreationDate="2010-08-04T22:08:04.640" Score="2" ViewCount="" Body="&lt;p&gt;When I was in my fourth year of high school I got a copy of &lt;em&gt;What is Mathematics?&lt;/em&gt; by Courant and Robbins. That book showed to me that Mathematics is far more than a &quot;boring tool&quot; to do Physics and opened up new worlds. I would recommend it to any bright high school kid with an interest in math and sciences.&lt;/p&gt;&#xA;" OwnerUserId="688" LastActivityDate="2010-08-04T22:08:04.640" CommunityOwnedDate="2010-08-04T22:08:04.640" />
  <row Id="1574" PostTypeId="2" ParentId="1568" CreationDate="2010-08-04T22:08:56.793" Score="4" ViewCount="" Body="&lt;p&gt;Radon-Nikodym derivative of the measures represented by variables $x$ and $y$ is one answer to your question, and is somewhat similar to the reasoning you sketched about a product density $h(x,y)=f(x)g(y)$.   For &quot;change in $x$ per change in $y$&quot; you perhaps want the correlation coefficient (multiplied by standard deviation of $x$). &lt;/p&gt;&#xA;" OwnerUserId="467" LastActivityDate="2010-08-04T22:08:56.793" CommentCount="0" />
  <row Id="1576" PostTypeId="1" AcceptedAnswerId="1602" CreationDate="2010-08-04T23:05:09.973" Score="2" ViewCount="191" Body="&lt;p&gt;I have an application that wants controllable random functions from $\mathbb{Z}^2$ and $\mathbb{Z}^3$ to $2^{32}$ , where by controllable I basically mean seedable by some parameters (say, on the order of 3 to 5 32-bit integers) such that the same seeds will always produce the same functions.  The most obvious way of doing this (for the two-dimensional case, say) would seem to be computing the value at some point $(x,y)$ by using $x$, $y$, and the seed parameters as seeds for something like an LFSR generator or a Mersenne Twister, then running the RNG for some fixed number of steps and taking the resultant value as the value of the function at that point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is, how can I be certain that this procedure won't keep too much correlation between adjacent 'seed points', and is there either a straightforward analysis or even just some general guideline for how many iterations would be necessary to eliminate that correlation?  My first back-of-the-envelope guess would be that each iteration roughly doubles the decorrelation between given seed values, so that 32 iterations would be necessary to achieve the requisite decorrelation over a range of $2^{32}$ values (and in practice I'd probably double it to 64 iterations), but that's strictly a guess and any proper analysis would be welcome!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edited for clarification:&lt;/strong&gt; To further outline the issue, I may be sampling this random function $f$ (for some given seed parameters) at arbitrary values, and need those samples to be identical between passes; so for instance, if a first application computes $f(0, 0)$, $f(437, 61)$, $f(-23, 129)$, and then $f(5,3)$, and a second (potentially concurrent) application computes $f(1,0)$ and then $f(5,3)$, both passes need to find the same value of $f$ at $(5,3)$.  I may also be sampling $f$ at arbitrary points, so I'd like the evaluation to take constant time (and in particular, evaluating $f(x,y)$ shouldn't take time linear in $x+y$).&lt;/p&gt;&#xA;" OwnerUserId="785" LastEditorUserId="785" LastEditorDisplayName="" LastEditDate="2010-08-05T18:02:49.817" LastActivityDate="2010-08-05T18:02:49.817" Title="'(Pseudo)-random functions' by seeding of PRNGs?" Tags="&lt;number-theory&gt;&lt;algorithms&gt;&lt;computer-science&gt;&lt;numerical-methods&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="1577" PostTypeId="1" AcceptedAnswerId="1578" CreationDate="2010-08-05T00:36:13.250" Score="0" ViewCount="71" Body="&lt;p&gt;Say we have a family {$A_\alpha$} of disjoint Baire spaces. Also suppose that each $A_\alpha$ is disjoint from the closure of the union of the other sets. Show that $\bigcup_{\alpha} A_\alpha$ is a Baire space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think we can prove this by transfinite induction. Suppose the property holds for all $\beta$ &amp;lt; $\alpha$, show it holds for $\alpha$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $\alpha$ is a limit ordinal: we want to show that for every sequence of open dense subsets of $\bigcup_{\beta&amp;lt;\alpha} A_\alpha$ their intersection is open dense. This sequence of open dense sets in the union is defined as the union of all open dense sets in each of the sets since they all are Baire spaces and by induction hypothesis the union up to $\beta$ is a Baire space. Suppose this is not the case. Then there exists an open set $U$ such that the intersection of these open dense sets misses $U$. There is some ordinal $\beta$ such that $U \in A_\beta$, but this is a contradiction since we supposed that the union up to $\beta$ was a Baire space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now how do I handle the successor case? it seems trickier.&lt;/p&gt;&#xA;" OwnerUserId="786" LastActivityDate="2010-08-05T00:47:57.033" Title="disjoint union of Baire spaces which is a Baire space" Tags="&lt;topology&gt;" AnswerCount="1" />
  <row Id="1578" PostTypeId="2" ParentId="1577" CreationDate="2010-08-05T00:47:57.033" Score="3" ViewCount="" Body="&lt;p&gt;I'm pretty sure it's simpler and doesn't require transfinite induction or anything nearly as difficult. In this case the $A_\alpha$ are open sets (as the complement of the closure of the union of the others). So suppose $U_1, \dots,$ are open dense sets in this union. Then each intersection $U_i \cap A_\alpha$ is open and dense in $A_\alpha$. The countable collection has intersection which is dense in each $A_\alpha$ by the Baire property. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means that the countable intersection, call it $F$, is dense in the union $\bigcup A_\alpha$. (Any point in the union belongs one of the $A_\alpha$, and there is a point of $F \cap A_\alpha$ nearby, hence a point of $F$ nearby.)&lt;/p&gt;&#xA;" OwnerUserId="536" LastActivityDate="2010-08-05T00:47:57.033" />
  <row Id="1579" PostTypeId="1" AcceptedAnswerId="1581" CreationDate="2010-08-05T00:49:39.170" Score="4" ViewCount="383" Body="&lt;p&gt;This question was motivated by &lt;a href=&quot;http://math.stackexchange.com/questions/1442/is-there-a-direct-proof-of-this-lcm-identity&quot;&gt;pondering this lcm identity&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider that gcd(1,6,15) = 1, but lcm(1,6,15)=30, but 1*6*15 = 90.  (2,6,15) shows a similar phenomenon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what is the correct identity for n-ary gcd/lcm?&lt;/p&gt;&#xA;" OwnerUserId="620" LastActivityDate="2010-08-18T02:11:55.883" Title="n-ary version of gcd(a,b)lcm(a,b) = ab" Tags="&lt;number-theory&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="1580" PostTypeId="2" ParentId="29" CreationDate="2010-08-05T00:53:40.923" Score="2" ViewCount="" Body="&lt;p&gt;If I am at school (and in my lab) I will just have Maple running for symbolic stuff. It's pretty good at it. Other then that, for simple calculations I use windows calculator (its a hotkey on my keyboard) and for more advanced stuff, I'll use WolframAlpha.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, my plan is to learn mathematica. &lt;/p&gt;&#xA;" OwnerUserId="287" LastActivityDate="2010-08-05T00:53:40.923" CommunityOwnedDate="2010-08-05T00:53:40.923" />
  <row Id="1581" PostTypeId="2" ParentId="1579" CreationDate="2010-08-05T01:06:02.443" Score="7" ViewCount="" Body="&lt;p&gt;For $n \ge 2$ there isn't just one identity; there are actually $n+1$ identities, half of which are dual to the other half, and two of which are the trivial identity $a_1 ... a_n = a_1 ... a_n$.  The first two are as follows.  Let $P = a_1 ... a_n$.  Then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle a_1 ... a_n = \text{gcd}(a_1, a_2, ... a_n) \text{lcm} \left( \frac{P}{a_1}, \frac{P}{a_2}, ... \right).$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle a_1 ... a_n = \text{gcd}(a_1 a_2, a_1 a_3, ..., a_{n-1} a_n) \text{lcm} \left( \frac{P}{a_1 a_2}, \frac{P}{a_1 a_3}, ..., \frac{P}{a_{n-1} a_n} \right).$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other identities are similar.  This is because &quot;pointwise&quot; (e.g. at a particular prime $p$) the two-variable identity is just the identity $\text{min}(a, b) + \text{max}(a, b) = a + b$ for non-negative integers $a, b$, and the above identities are the natural generalization of this one. &lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-08-05T01:11:48.877" LastActivityDate="2010-08-05T01:11:48.877" CommentCount="2" />
  <row Id="1582" PostTypeId="2" ParentId="1533" CreationDate="2010-08-05T01:56:21.643" Score="4" ViewCount="" Body="&lt;p&gt;Discrete Math knowledge is needed to become adept in proving the correctness and deriving the complexity of algorithms and data structures. You will be taught those in Algo/DS books, but you can only get the mathematical proficiency by practicing just discrete math.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Knuth book is very good for that. But IMHO, you will only need it if you for doing advanced proofs in DS/Algorithms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a beginner, it would be great to go over &quot;Grimaldi&quot; &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0201199122&quot; rel=&quot;nofollow&quot;&gt;http://www.amazon.com/Discrete-Combinatorial-Mathematics-Applied-Introduction/dp/0201199122&lt;/a&gt;&#xA;and then quickly move to Algorithms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Otherwise, you will continue going deep in Discrete Math and never get to Algorithms/DS.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remember, Discrete Math does not teach you how to design algorithms or Data structures. That will come only by practicing Algorithm problems @ topcoder, acm icpc , spoj etc and reading books on Algos/DS or courses on those.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My 2 cents.&lt;/p&gt;&#xA;" OwnerUserId="791" LastActivityDate="2010-08-05T01:56:21.643" CommunityOwnedDate="2010-08-05T01:56:21.643" CommentCount="1" />
  <row Id="1583" PostTypeId="2" ParentId="29" CreationDate="2010-08-05T02:02:54.517" Score="7" ViewCount="" Body="&lt;p&gt;A few people have mentioned Sage, but I think deserves a bit more of a plug. Firstly someone mentioned that Sage is quite large to install. This is true, the binary downolad comes in at around 1 gig, however you can use it for free online, without having to install a thing! Head on over to  &lt;a href=&quot;http://www.sagenb.org&quot; rel=&quot;nofollow&quot;&gt; sagenb.org &lt;/a&gt;, and sign up for a free account, which will allow you to create and save your own worksheets. I think this is really nice--it feels like it's your &quot;gmail&quot; but for maths, so you can just log in from anywhere and &quot;check your maths&quot; if you want :-D&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a very popular option--I believe that in the last 9  months around 30,000 people have created accounts! It's perfectly suited for high school students and undergraduates, since no installation is necessary, and hence it can be used straight away in any computer lab with internet access. Here's a screenshot of what the notebook looks like. (I just googled &quot;sage notebook screenshot.&quot;) It's been designed to look and work like the Mathematica notebook system, and it's surprisingly slick and easy to use.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://sidk.info/wp-content/uploads/2007/11/sage_screenshot2.jpeg&quot; alt=&quot;Sage screenshot&quot;&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There really is a lot to recommend it. I used the online notebook for around 6 months before I installed a copy on my own computer, and this was because I became interested in developing Sage. Although Sage is based on Python (so in particular any Python syntax will work), you really don't need to know Python to get started using Sage--I didn't when I started. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sage is quite powerful too. Aside from using Maxima for symbolic integration, I belive Sage uses Pari for highly optimised polynomial arithmatic, and I have used Sage for some pretty intense linear algebra computations (over the integers as well as over finite fields), although for this kind of thing you'll probably not want to rely on the online server which can be a little slow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And let's not forget that you also get the moral boost of being part of something really positive--an movement to create an open source alternative to costly and &quot;closed&quot; alternatives like Mathematica, Maple and Matlab.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Disclaimer: I just came back from the Sage Days 23 sage developers workshop in Leiden, and I'm a bit pumped!)&lt;/p&gt;&#xA;" OwnerUserId="668" LastActivityDate="2010-08-05T02:02:54.517" CommunityOwnedDate="2010-08-05T02:02:54.517" />
  <row Id="1584" PostTypeId="2" ParentId="1563" CreationDate="2010-08-05T02:19:19.103" Score="4" ViewCount="" Body="&lt;p&gt;Take any complete theory $T$ that has two models of of different cardinalities. Then the models are elementary equivalent (they both model $T$) but they cannot be isomorphic because isomorphisms preserve cardinality.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the Lowenheim-Skolem theorem, every theory with infinite models has models of different cardinalities, so really any complete theory with infinite models will work for $T$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Personal note: Whenever I teach students about cardinality, I always point out this sort of application. Because cardinality is preserved by bijections, showing two objects (groups, rings, models, etc) have different cardinalities is an easy way to show they are not isomorphic.  I find this a very compelling reason to care about cardinality outside of set theory. &lt;/p&gt;&#xA;" OwnerUserId="630" LastEditorUserId="630" LastEditorDisplayName="" LastEditDate="2010-08-05T11:18:54.957" LastActivityDate="2010-08-05T11:18:54.957" CommentCount="6" />
  <row Id="1585" PostTypeId="2" ParentId="1533" CreationDate="2010-08-05T04:10:42.853" Score="2" ViewCount="" Body="&lt;p&gt;Theres many different areas to discrete math, and many good books. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;theres Graph Theory by Diestel, which has a free pdf version available at &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://diestel-graph-theory.com/&quot; rel=&quot;nofollow&quot;&gt;diestel-graph-theory.com&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;theres generatingfunctionology by wilf, free pdf version at &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.math.upenn.edu/~wilf/DownldGF.html&quot; rel=&quot;nofollow&quot;&gt;math.upenn.edu/~wilf/DownldGF.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other books that are good include Enumerative combinatorics 1 and 2 by Richard P Stanley (a book which is sufficiently dense that having at least 1 analysis and algebra course each will help).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;that being said, for more introductory expositions in terms of expected mathematical maturity, I'd suggest googling around and looking at various lecture notes of the &quot;intro to combinatorics&quot; or &quot;mathematics for computer scientists&quot; sorts. I found that MIT OCW's &quot;mathematics for Computer Scientists&quot; notes were quite nice when I looked at them several years ago. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://ocw.mit.edu/courses/electrical-engineering-and-computer-science/6-042j-mathematics-for-computer-science-fall-2005/readings/&quot; rel=&quot;nofollow&quot;&gt;ocw.mit.edu/courses/electrical-engineering-and-computer-science...&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;has a link to the lecture notes. There are some really funny asides in it. One of my favorites &quot;... anyone who says that is wrong, and you should make fun of them until they cry&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, If you want to dig even deeper into discrete math/ combinatorics, the value of building up a wee bit of mathematical basics in other areas of math. Complex Analysis, real analysis (at the level of at least baby rudin, and perhaps even up to functional analysis), maybe some probability up to its measure theory formulation level, and at least a smidge of abstract algebra.  Then you can do stuff like look at the combinatorics of random processes (great for analyzing randomized algorithms) and look at cool problems like percolation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;theres probably other things I should suggest, but the point is discrete math is accessible without that much of a background, but is also rewards you for enriching that mathematics background with some amazingly beautiful stuff thats 1) awesome and fun 2) useful.&lt;/p&gt;&#xA;" OwnerUserId="798" LastEditorUserId="121" LastEditorDisplayName="" LastEditDate="2010-08-05T04:56:48.560" LastActivityDate="2010-08-05T04:56:48.560" CommunityOwnedDate="2010-08-05T04:10:42.853" />
  <row Id="1586" PostTypeId="2" ParentId="1512" CreationDate="2010-08-05T04:52:56.257" Score="5" ViewCount="" Body="&lt;p&gt;One way to think of the situation is that the countable product of $\mathbf Z/2\mathbf Z$s that you are considering&#xA;is a vector space $V$ over $\mathbf F_2$ that is uncountable, and thus has an (even more) uncountable dual space.  If $f:V \to \mathbf F_2$ is a non-zero element of the dual space, its kernel is an index 2 subgroup of $V$.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the open subgroups that you know about correspond to (i.e. are the kernels of) the projections onto the various factors&#xA;in the product, and these are countable in number.  So there is a massive host of other index 2 subgroups, corresponding to all the other non-zero elements $f$ of the dual space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As others have noted, you won't be able to write these down explicitly though.&lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2010-08-05T04:52:56.257" CommentCount="3" />
  <row Id="1587" PostTypeId="2" ParentId="903" CreationDate="2010-08-05T05:41:21.980" Score="1" ViewCount="" Body="&lt;p&gt;As BBischof says: Willard is nice. I also learned a lot from Mendelson, a little dover book. I think that you should use whatever gets you through the subject &quot;quickest&quot;. Point Set topology is now just basic language that is very necessary, but it seems there is not much research in the field. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My dad did not go into math because of a course out of munkres, he had no intuition for the topic the way it was taught. I dont see this as a criticism of Munkres because I think one of the major goals of any early course is to untrain the student. There are spaces that are pathological that you have to deal with. You have to be rigorous. I think that is one of the main points of such a course, that and preparing the student for what comes next. Which is why mariano asks what your goals are.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One thing about willard though is that you should make sure you do the exercises, they contain key facts!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: I felt it pertinent to add that I did go through a course from munkres (by the time I did it was pointless for me, I had already been using the relevant stuff for my field daily) and it was quite well done I thought. It is the classic, I don't think you can go wrong with it.&lt;/p&gt;&#xA;" OwnerUserId="627" LastEditorUserId="627" LastEditorDisplayName="" LastEditDate="2010-08-05T15:07:36.513" LastActivityDate="2010-08-05T15:07:36.513" CommunityOwnedDate="2010-08-05T05:41:21.980" CommentCount="4" />
  <row Id="1588" PostTypeId="2" ParentId="903" CreationDate="2010-08-05T05:57:26.957" Score="5" ViewCount="" Body="&lt;p&gt;Good and Free &lt;br /&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://uob-community.ballarat.edu.au/~smorris/topology.htm&quot; rel=&quot;nofollow&quot;&gt;&lt;strong&gt;Topology Without Tears by Sidney A. Morris&lt;/strong&gt;&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="705" LastActivityDate="2010-08-05T05:57:26.957" CommunityOwnedDate="2010-08-05T05:57:26.957" CommentCount="3" />
  <row Id="1589" PostTypeId="1" AcceptedAnswerId="1591" CreationDate="2010-08-05T06:00:14.783" Score="7" ViewCount="318" Body="&lt;p&gt;I asked this question a while back on MO : &lt;a href=&quot;http://mathoverflow.net/questions/32689/how-should-a-homotopy-theorist-think-about-sheaf-cohomology&quot; rel=&quot;nofollow&quot;&gt;http://mathoverflow.net/questions/32689/how-should-a-homotopy-theorist-think-about-sheaf-cohomology&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One thing that really helped in learning the Serre SS was doing particular computations (like $H^*(CP^{\infty})$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am curious, as a sort of followup if anyone can suggest:&#xA;1. a reference where small computations are carried out? or&#xA;2. a specific computation to do with a small enough sheaf an some simple topological space that would be able to give one a feel for sheaf cohomology. So this space that we are working over need not be a scheme, in fact it would probably be best if it were not a scheme since i dont understand them quite yet. And are there tricks of the trade to computing these things? or do people just hammer away ate injective resolutions? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;in short, please suggest a space and a sheaf on it that i should work on computing the sheaf cohomology of.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;thanks for your time,&#xA;Sean&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS I of course welcome any other suggestions for understanding how to compute sheaf cohomology.&lt;/p&gt;&#xA;" OwnerUserId="627" LastEditorUserId="627" LastEditorDisplayName="" LastEditDate="2010-08-26T16:05:08.073" LastActivityDate="2010-08-26T16:05:08.073" Title="toy Sheaf Cohomology Computation" Tags="&lt;algebraic-geometry&gt;&lt;algebraic-topology&gt;&lt;cohomology&gt;" AnswerCount="3" CommentCount="6" FavoriteCount="1" />
  <row Id="1590" PostTypeId="2" ParentId="1519" CreationDate="2010-08-05T06:08:32.050" Score="0" ViewCount="" Body="&lt;p&gt;I think you understand attaching cells that have dimension gap better than you think! Consider any $S^n$ for $n&gt;&gt;0$, we can think of this as being the attachment of $D^n$ to a point which is 0-dimensional.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, I would recommend that you think of this attaching cells thing a bit more carefully. if you understand the hopf fibration &lt;em&gt;then you do&lt;/em&gt; understand the attaching of the cell. The hopf fibration is &lt;em&gt;not&lt;/em&gt; easy to picture (as far as I know). I would recommend thinking about the map in terms of what happens when you work with coordinates. I will be back later to add more... I promise.&lt;/p&gt;&#xA;" OwnerUserId="627" LastActivityDate="2010-08-05T06:08:32.050" />
  <row Id="1591" PostTypeId="2" ParentId="1589" CreationDate="2010-08-05T06:39:00.947" Score="3" ViewCount="" Body="&lt;p&gt;Any de Rham cohomology (or Dolbeault cohomology) computation is a computation in sheaf cohomology. Actually --- any computation in singular cohomology is a computation in sheaf cohomology!! ;-) We're just taking different resolutions of the appropriate constant sheaf.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;IIRC, there are some good Cech cohomology computations and examples in Bott-Tu. Also, have you read section 3.H of Hatcher's algebraic topology book, on &quot;local coefficients&quot;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a simple example from algebraic geometry, compute the cohomology of the structure sheaf of $\mathbb{A}^2$ minus a point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I seem to recall an exercise or an example in Hartshorne in which the genus of a degree $d$ curve in $\mathbb{P}^2$ is computed using Cech cohomology.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The section in Hartshorne on the cohomology of $\mathbb{P}^n$ uses Cech cohomology, and I remember finding it pretty instructive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Eisenbud's commutative algebra book probably has lots of good examples. However, he might not use words like &quot;sheaf cohomology&quot; (since I don't think the book introduces sheaves), although that's what it is.&lt;/p&gt;&#xA;" OwnerUserId="263" LastEditorUserId="263" LastEditorDisplayName="" LastEditDate="2010-08-05T06:48:53.283" LastActivityDate="2010-08-05T06:48:53.283" />
  <row Id="1592" PostTypeId="2" ParentId="653" CreationDate="2010-08-05T06:43:35.837" Score="1" ViewCount="" Body="&lt;p&gt;Here is an answer to your meta question since i can not yet post there:&#xA;I disagree that such questions should be closed without thought. Some people may genuinely be interested in real world applications and some may come up and be particularly helpful in ways I might not have conceived of. Perhaps the &quot;questioner&quot; meant something broader like what comes out of homtopy theory, then certainly the Ghrist work would be relevant (despite the fact the that homoglogy is definitely not a generalization of homotopy, not evenof $\pi_1$).The work that Carlsson is doing is sort of a real world application, although not specifically of any model categorical stuff (... maybe it is, i dunno).&lt;/p&gt;&#xA;" OwnerUserId="627" LastActivityDate="2010-08-05T06:43:35.837" />
  <row Id="1593" PostTypeId="2" ParentId="1576" CreationDate="2010-08-05T07:10:07.140" Score="0" ViewCount="" Body="&lt;p&gt;This does not strictly answer your question, but might be too long for a comment. =)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe you should rethink your approach. Most randon number generators are designed to be seeded once, and than return a sequence of &quot;random&quot; numbers, as close to independent and uniformly distributed as possible. I don't think that there are many results about the correlation of the 64th sequence element for different seeds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, (at least in the implementations I know) the period of the algorithm is a lot longer than the number of possible seeds. If you have to reuse seeds in your case, you could introduce correlations that are avoidable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instead of precomputing the entire grid, you could get a random value just-in-time and store it for future reference, saving memory if you don't access every node.&lt;/p&gt;&#xA;" OwnerUserId="505" LastActivityDate="2010-08-05T07:10:07.140" CommentCount="1" />
  <row Id="1594" PostTypeId="2" ParentId="709" CreationDate="2010-08-05T07:33:05.207" Score="1" ViewCount="" Body="&lt;p&gt;&lt;img src=&quot;http://hotmath.com/images/gt/lessons/genericalg1/parabs.gif&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$y = ax^2 + bx + c$&lt;br /&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for parabola written in &lt;strong&gt;Vertex Form&lt;/strong&gt; vertex is (h,k).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$y = m(x-h)^2 + k = \underline{m}x^2 + \underline{(-2mh)}x+ \underline{mh^2 +k}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rightarrow$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a = m$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$b = -2mh = -2ah \rightarrow h = -b/2a$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$c = mh^2 + k = ah^2 + k \rightarrow k = c - b^2/4a$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://hotmath.com/hotmath_help/topics/vertex-of-a-parabola.html&quot; rel=&quot;nofollow&quot;&gt;&lt;strong&gt;source&lt;/strong&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="705" LastActivityDate="2010-08-05T07:33:05.207" />
  <row Id="1595" PostTypeId="2" ParentId="1512" CreationDate="2010-08-05T07:35:10.837" Score="1" ViewCount="" Body="&lt;p&gt;Here is an easy point which is implicit in the other answers, but which is part of the question (as I read it).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Put $(p_1,p_2,p_3,p_4,\dots):=(-1,2,3,5,\dots)$, and define the fields $K_0,K_1,\dots$ by $K_0:=\mathbb Q$ and $K_n:=K_{n-1}(\sqrt{p_n})$ for $n\ge 1$. [It's clear that any quadratic field is contained into some $K_n$.] To make sure that the Galois group in question is the indicated one, we need to check that $\sqrt{p_n}$ is &lt;strong&gt;not&lt;/strong&gt; in $K_{n-1}$. To do that we prove the following apparently stronger claim.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $n$ be a positive integer and $a$ the product of the $p_k^{e(k)}$, where $k$ runs over the integers $\ge n$ and $e(k)$ is an integer equal to $0$ for almost all $k$. Assume $\sqrt a\in K_{n-1}$. Then all the $e(k)$ are even.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is easily proved by induction. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT.&lt;/strong&gt; Things can be described in this way: The obvious morphism from $\mathbb Q^{\times}/\mathbb Q^{\times 2}$ into our Galois group induces an isomorphism between the profinite completions. [It's nice to obtain a Galois group by applying a simple construction to the base field.]&lt;/p&gt;&#xA;" OwnerUserId="660" LastEditorUserId="660" LastEditorDisplayName="" LastEditDate="2010-08-05T09:27:07.193" LastActivityDate="2010-08-05T09:27:07.193" />
  <row Id="1596" PostTypeId="1" CreationDate="2010-08-05T08:56:51.800" Score="0" ViewCount="394" Body="&lt;p&gt;Can someone explain to me in simple terms what a non-linear transformation is in maths?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know some single-variable calculus, but I read it has to do with multi-variable calculus, which I'm not familiar with.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If someone could explain it in simple words, that would be helpful.&lt;/p&gt;&#xA;" OwnerUserId="807" LastActivityDate="2010-08-05T11:41:51.863" Title="Non-Linear Transformation" Tags="&lt;calculus&gt;&lt;transformation&gt;" AnswerCount="2" CommentCount="5" FavoriteCount="2" />
  <row Id="1597" PostTypeId="2" ParentId="1537" CreationDate="2010-08-05T09:45:56.930" Score="1" ViewCount="" Body="&lt;p&gt;Wielandt's theorem says that the gamma-function is the only function $f$ that satisfies the properties:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$f(1)=1$&lt;/li&gt;&#xA;&lt;li&gt;$f(z+1)=zf(z)$ for all $z&gt;0$&lt;/li&gt;&#xA;&lt;li&gt;$f(z)$ is analytic for $\operatorname{Re}z&gt;0$&lt;/li&gt;&#xA;&lt;li&gt;$f(z)$ is bounded for $1\leq \operatorname{Re}z\leq 2$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;(See also the related MathOverflow thread &lt;a href=&quot;http://mathoverflow.net/questions/23229/importance-of-log-convexity-of-the-gamma-function&quot; rel=&quot;nofollow&quot;&gt;Importance of Log Convexity of the Gamma Function&lt;/a&gt;, where I learned about the above theorem.)&lt;/p&gt;&#xA;" OwnerUserId="405" LastActivityDate="2010-08-05T09:45:56.930" />
  <row Id="1598" PostTypeId="2" ParentId="903" CreationDate="2010-08-05T10:29:55.607" Score="1" ViewCount="" Body="&lt;p&gt;Sutherland's &lt;em&gt;Metric and topological spaces&lt;/em&gt; is nice. I know a guy who swears by Kelley's &lt;em&gt;General topology&lt;/em&gt; but I've not read it myself.&lt;/p&gt;&#xA;" OwnerUserId="172" LastActivityDate="2010-08-05T10:29:55.607" CommunityOwnedDate="2010-08-05T10:29:55.607" />
  <row Id="1599" PostTypeId="2" ParentId="1596" CreationDate="2010-08-05T10:33:57.537" Score="2" ViewCount="" Body="&lt;p&gt;Let $V_1, V_2$ be two &lt;a href=&quot;http://en.wikipedia.org/wiki/Vector_space&quot; rel=&quot;nofollow&quot;&gt;vector spaces&lt;/a&gt; over the &lt;a href=&quot;http://en.wikipedia.org/wiki/Field_%28mathematics%29&quot; rel=&quot;nofollow&quot;&gt;field&lt;/a&gt; $F$. A transformation $T: V_1 \to V_2$ is linear if for every $x, y \in V_1$ and every $\alpha \in F$ it is true that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(*) $T(x + \alpha y) = T(x) + \alpha T(y)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;T is not a linear transformation if there are some $x, y, \alpha$ such that (*) is not true.&lt;/p&gt;&#xA;" OwnerUserId="26" LastActivityDate="2010-08-05T10:33:57.537" />
  <row Id="1600" PostTypeId="2" ParentId="1537" CreationDate="2010-08-05T10:44:36.710" Score="3" ViewCount="" Body="&lt;p&gt;This is a comment posted as an answer for lack of reputation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Following Qiaochu Yuan, the gamma function shows up in the functional equation of the zeta function as the factor in the Euler product corresponding to the &quot;prime at infinity&quot;, and it occurs there as the Mellin transform of some gaussian function. (Gaussian functions occur in turn as eigenvectors of the Fourier transform.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is at least as old as Tate's thesis, and a possible reference is Weil's &lt;strong&gt;Basic Number Theory&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT.&lt;/strong&gt; Artin was one of the first people to popularize the log-convexity property of the gamma function (see his book on the function in question), and also perhaps the first mathematician to fully understand this Euler-factor-at-infinity aspect of the same function (he was Tate's thesis advisor). I thought his name had to be mentioned in a discussion about the gamma function.&lt;/p&gt;&#xA;" OwnerUserId="660" LastEditorUserId="660" LastEditorDisplayName="" LastEditDate="2010-08-05T11:13:14.637" LastActivityDate="2010-08-05T11:13:14.637" CommentCount="1" />
  <row Id="1601" PostTypeId="1" AcceptedAnswerId="1702" CreationDate="2010-08-05T11:14:05.370" Score="2" ViewCount="464" Body="&lt;p&gt;In a recent discussion, I came across the idea of proving lower bound for number of comparisons required to find the largest element in an array. The bound is $n - 1$. This is so because, the set of comparisons performed by every such algorithm looks like a tournament tree, which always has $n - 1$ internal nodes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The obvious question is then, What should be the lower bound for finding 2nd largest elelment from an array.&lt;/p&gt;&#xA;" OwnerUserId="813" LastEditorUserId="123" LastEditorDisplayName="" LastEditDate="2010-08-05T23:10:33.557" LastActivityDate="2010-08-06T08:47:12.300" Title="lower bound for finding second largest element" Tags="&lt;algorithms&gt;&lt;computer-science&gt;&lt;searching&gt;" AnswerCount="3" CommentCount="8" FavoriteCount="1" />
  <row Id="1602" PostTypeId="2" ParentId="1576" CreationDate="2010-08-05T11:36:06.083" Score="1" ViewCount="" Body="&lt;p&gt;Note that when an &lt;a href=&quot;http://www.newwaveinstruments.com/resources/articles/m_sequence_linear_feedback_shift_register_lfsr.htm&quot; rel=&quot;nofollow&quot;&gt;LFSR&lt;/a&gt; is used to generate a sequence of +1 / -1 values, the correlation between that sequence and any time-shifted version of that sequence is near zero: (sum from i = 0 to N-1 of a[i]a[i+k]) is either N if k is a multiple of N, or -1 otherwise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The initial state of an LFSR has a one-to-one mapping to the time shift -- this is analogous to a discrete logarithm problem in Galois fields; given a time shift k, it's easy to find the initial state of an LFSR (O(log(k), I think, as it's just computing exponentiation in the appropriate Galois field), but given the initial state of an LFSR, it's hard to find the time shift k in anything shorter than O(k). So your PRNG could be an LFSR with very long period e.g. 64 bits or greater, and the &quot;seed&quot; could be a time shift used to derive a different initial state.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as 2- or 3- or k-dimensional mapping, interleave the bits of the coordinates, and use a hash function to map those bits to an integer for use in deriving LFSR seeds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure what kind of correlation you're looking to avoid between functions. (e.g. correlation in a statistical sense, or cryptographic independence e.g. one PRNG cannot be used to predict future output of another).&lt;/p&gt;&#xA;" OwnerUserId="120" LastActivityDate="2010-08-05T11:36:06.083" CommentCount="4" />
  <row Id="1603" PostTypeId="2" ParentId="1596" CreationDate="2010-08-05T11:41:51.863" Score="0" ViewCount="" Body="&lt;p&gt;In addition to the definition of linear map that Tomer remind you, here are two examples.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, $f(x,y) = x^2y$ is &lt;em&gt;not&lt;/em&gt; a linear map $f: \mathbb{R}^2 \longrightarrow \mathbb{R}$ because&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;f(2x,2y) = 4x^22y \neq 2x^2y = 2f(x,y) \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More generally, the linear maps $f: \mathbb{R}^m \longrightarrow \mathbb{R}^n$ are &lt;em&gt;necesseraly&lt;/em&gt; of the form &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;f(x_1, \dots , x_m) = (a_1^1 x_1 + \dots + a_1^m x_m , \dots , a_n^1 x_1 + \dots + a_n^m x_m)&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with $a^i_j$ constant coefficients.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, two more examples:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$f(x,y) = x + 2y$ is a linear map.&lt;/li&gt;&#xA;&lt;li&gt;$f(x,y,z) = 3x + 1$ is a non-linear map&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="664" LastActivityDate="2010-08-05T11:41:51.863" />
  <row Id="1604" PostTypeId="2" ParentId="1589" CreationDate="2010-08-05T11:42:04.530" Score="0" ViewCount="" Body="&lt;p&gt;Rotman does some very elementary explicit computations of Cech cohomology in his book Homological Algebra.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I remember correctly he does these computations using resolutions, spectral sequences, and by just starting with some sequence. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a complete beginner to this material, I was able to understand his treatment and compute some specific examples on my own.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope this helps :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S. I just looked at the first edition, and it seems to be different slightly. For your information I used the second edition.&lt;/p&gt;&#xA;" OwnerUserId="16" LastActivityDate="2010-08-05T11:42:04.530" CommentCount="3" />
  <row Id="1605" PostTypeId="2" ParentId="198" CreationDate="2010-08-05T11:42:51.033" Score="2" ViewCount="" Body="&lt;p&gt;&quot;Taking the principle of excluded middle from the mathematician would be the same, say, as proscribing the telescope to the astronomer or to the boxer the use of his fists. To prohibit existence statements and the principle of excluded middle is tantamount to relinquishing the science of mathematics altogether.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-David Hilbert&lt;/p&gt;&#xA;" OwnerUserId="126" LastActivityDate="2010-08-05T11:42:51.033" CommunityOwnedDate="2010-08-05T11:42:51.033" />
  <row Id="1606" PostTypeId="2" ParentId="1453" CreationDate="2010-08-05T11:57:21.947" Score="3" ViewCount="" Body="&lt;p&gt;A detailed explanation can be found in many places. I'll to provide an intuitive one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By &lt;a href=&quot;http://en.wikipedia.org/wiki/Cook%E2%80%93Levin_theorem&quot; rel=&quot;nofollow&quot;&gt;Cook-Levin theorem&lt;/a&gt;, the &lt;a href=&quot;http://en.wikipedia.org/wiki/Boolean_satisfiability_problem&quot; rel=&quot;nofollow&quot;&gt;Boolean satisfiability problem&lt;/a&gt; is &lt;a href=&quot;http://en.wikipedia.org/wiki/Np_complete&quot; rel=&quot;nofollow&quot;&gt;NP-complete&lt;/a&gt; - ie. every decision problem in NP can be reduced to it. We will ask of our prover to supply the input and output of every gate in the circuit and consider this as the proof that $x\in L$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we were to stop there, then the prover could cheat by changing a single bit in his proof, which would require us to check a large (non-constant) number of bits in it. So we must ask something more from the prover. This something more will be the encoding of his proof using some (special) error-correcting code. Intuitively, this &quot;smudges&quot; the false bits onto a large number of bits in the code-word. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The verifier receives a word from the prover, there are 2 ways in which the prover might try and cheat, this word might not be a code-word in the chosen code, or it might be the encoding of something which isn't a proof. We'll examine the (slightly different) separation of cases of the word either being far from every codeword (a large number of bits must be changed to reach a codeword) or that it is close to codeword which isn't an encoding of a proof. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To be able to detect these we demand that our code has the following 2 properties:&#xA;1) locally checkable - we can, by reading a constant number of bits, detect w.h.p if a word is far from any codeword.&#xA;2) locally decodable - we can, by reading a constant number of bits, decode w.h.p a bit from the encoded word.&#xA;(finding such codes is hard, hadamard has this properties but the code-word's size is exponential, RM codes have these properties as well (to a lesser degree) and they are the ones generally used.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the verifier checks if the word is close to being a code-word (using 1), and if the test succeeds, picks a random gate and decodes it's outputs and output (using 2). this is sufficient to achieve constant query complexity and logarithmic randomness.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It should be mentioned that  (&lt;a href=&quot;http://www.wisdom.weizmann.ac.il/~dinuri/mypapers/combpcp.pdf&quot; rel=&quot;nofollow&quot;&gt;Dinur&lt;/a&gt;) has a combinatorial proof of PCP which is quite different from what i've discribed, but her version is (for me, at least) less intuitive.  &lt;/p&gt;&#xA;" OwnerUserId="816" LastEditorUserId="123" LastEditorDisplayName="" LastEditDate="2010-08-05T12:55:01.437" LastActivityDate="2010-08-05T12:55:01.437" CommentCount="3" />
  <row Id="1607" PostTypeId="2" ParentId="1601" CreationDate="2010-08-05T11:57:32.260" Score="2" ViewCount="" Body="&lt;p&gt;An observation: it will always require at least n-1. Draw a graph. If we test two nodes, then we draw an arrow from the smaller to the larger. We know a is greater than b if and only if there is a path from a to b along the arrows. We will always need at least n-1 arrows to connect the graph and hence n-1 comparisons to find a lower bound. It is easy enough to see that this isn't the minimal upper bound though - we will only know the second highest element if our arrows form a line (ie. we can go from a &quot;start&quot; node to an &quot;end&quot; node by passing through each node and each arrow).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another fact - there are algorithms for finding the &lt;a href=&quot;http://en.wikipedia.org/wiki/Selection_algorithm&quot; rel=&quot;nofollow&quot;&gt;kth largest number&lt;/a&gt; in guaranteed linear time. That could give you some information, but there could always be a special algorithm for finding the second largest element.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't have a solution for the problem yet. But one thing we can try is breaking the problem into two groups and combining them. Suppose we have two groups, A and B, where we know both the largest and second largest elements, but no knowledge of any relationships between the groups. Then it takes two arrows to find the second highest element. First we compare the highest elements in A and B. Without loss of generality assume A has the highest element. Then we would compare B with the highest element in A. Showing that the groups can't be combined with a single comparison is trivial. This gives us a recursion relation for the time taken to solve by combining groups which have been solved separately: $f(n)=min f(k)+f(n-k)+2$ where $ 0 &amp;lt; k &amp;lt; n $.&lt;/p&gt;&#xA;" OwnerUserId="123" LastEditorUserId="123" LastEditorDisplayName="" LastEditDate="2010-08-05T23:08:25.010" LastActivityDate="2010-08-05T23:08:25.010" CommentCount="3" />
  <row Id="1608" PostTypeId="1" AcceptedAnswerId="1616" CreationDate="2010-08-05T11:58:30.107" Score="1" ViewCount="102" Body="&lt;p&gt;How can I calculate the length of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Cathetus&quot; rel=&quot;nofollow&quot;&gt;cathetus&lt;/a&gt; in a triangle if I know the length of the hypotenuse and the ratio between the two catheti?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The hypotenuse is 5cm long, and the ratio between the catheti is 4:3 - how long are the catheti or either cathetus?&lt;/p&gt;&#xA;" OwnerUserId="470" LastEditorUserId="123" LastEditorDisplayName="" LastEditDate="2010-08-05T12:39:05.777" LastActivityDate="2010-08-05T12:45:53.860" Title="Get length of cathetus/i from length of hypothenuse and ratio between catheti?" Tags="&lt;geometry&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="1609" PostTypeId="1" AcceptedAnswerId="1635" CreationDate="2010-08-05T12:05:17.973" Score="9" ViewCount="381" Body="&lt;p&gt;when trying to explain AG-codes to computer scientists, the major points of contention i am faced with are the concepts of divisors, Riemann-Roch space and the genus of a function field. are there any intuitive explanations for these concepts? or at least ones that are less dependent on knowledge of algebraic-geometry/topology? &lt;/p&gt;&#xA;" OwnerUserId="816" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-08-05T19:47:57.917" LastActivityDate="2010-08-09T20:47:20.013" Title="intuitive explantions for the concepts of divisor and genus" Tags="&lt;intuition&gt;&lt;algebraic-geometry&gt;&lt;algebraic-curves&gt;" AnswerCount="4" FavoriteCount="8" />
  <row Id="1610" PostTypeId="2" ParentId="1608" CreationDate="2010-08-05T12:07:38.157" Score="1" ViewCount="" Body="&lt;p&gt;You can call one cathetus 4*x*, the other 3*x*, and apply Pythagora's theorem: (4*x*)&lt;sup&gt;2&lt;/sup&gt;+(3*x*)&lt;sup&gt;2&lt;/sup&gt;=5&lt;sup&gt;2&lt;/sup&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You will obtain 25*x*&lt;sup&gt;2&lt;/sup&gt;=25, which yields &lt;em&gt;x&lt;/em&gt;=1. So one cathetus is 4cm, and the other 3cm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Remember that 3,4,5 is a Pythagorean triple)&lt;/p&gt;&#xA;" OwnerUserId="354" LastActivityDate="2010-08-05T12:07:38.157" CommentCount="2" />
  <row Id="1611" PostTypeId="2" ParentId="1609" CreationDate="2010-08-05T12:11:59.867" Score="2" ViewCount="" Body="&lt;p&gt;Well, the concepts of genus and divisors come from geometry/topology intuition originally, so those explanations are generally going to be the fastest.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, we can give very rough versions without needing to say &quot;genus is the number of holes in the surface that the curve looks like.&quot;  You can think of genus as a measure of how complicated the function field is.  Naturally $k(x)$ is going to be the simplest possible one, and it happens to be genus zero.  Genus one function fields look like $k(x,y)$ but where $y^2$ is a cubic in $x$.  Now, this isn't very precise, but it's a rough thing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for divisors, the topology/geometry of them is &quot;linear combinations of points, modulo zero-$\infty$ loci of functions,&quot; but to handle it purely algebraically, you'll want to change the word &quot;point&quot; to &quot;discrete valuations&quot;, so these are integer linear combinations of the various ways to describe the order of an element of the function field, modulo some set of trivial ones.&lt;/p&gt;&#xA;" OwnerUserId="176" LastActivityDate="2010-08-05T12:11:59.867" />
  <row Id="1612" PostTypeId="2" ParentId="1609" CreationDate="2010-08-05T12:28:46.967" Score="3" ViewCount="" Body="&lt;p&gt;Charles has already explained the notion of genus (which comes from topology--there's also a purely algebraic notion of genus as a first cohomology group, but I at least would find it less intuitive). So I'll talk about divisors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First of all, the analytic notion of a nonsingular projective curve is a compact Riemann surface. &quot;Nonsingular&quot; translates to &quot;complex manifold,&quot; &quot;curve&quot; translates to &quot;dimension 1,&quot; and &quot;projective&quot; translates to &quot;compact.&quot; So I will talk about Riemann surfaces.*&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have a Riemann surface, it's locally the same as the complex plane. And in the complex plane, you have a way to measure the order of the zero of a holomorphic function. More generally, you can measure (in $\mathbb{Z}$) the order of a zero of a meromorphic function (which is negative if it has a pole, zero if it is nonvanishing and analytic, positive if it has a zero).  So to any meromorphic function (of course, we don't restrict to holomorphic ones---they're all constant) on a compact Riemann surface, we can assign a finite set of points with multiplicities consisting of the places where the order is nonzero (i.e., the zeros and poles).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A divisor is more general. It's just a formal sum of points on the Riemann surfaces with multiplicities.  A divisor may come from a function as above; then it's called principal. In general, however, it does not. Yet given a divisor $D$, we can associate a vector space of meromorphic functions such that $div(f)+ D$ is a divisor with nonnegative coefficients. This is denoted $L(D)$ and its dimension by $l(D)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Riemann-Roch_Theorem&quot; rel=&quot;nofollow&quot;&gt;Riemann-Roch theorem&lt;/a&gt; is then a statement about the dimensions $l(D)$. One of its consequences, in particular, is that if $|D|$ (the sum of the multiplicities) is large, then you can always find a (nonzero) element of $L(D)$. This is intuitively easy to understand.  If $D$ is large, then you are allowing lots of leeway for $f$, potentially having poles at many places. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;*I'm being informal here, but there is a general &quot;GAGA&quot; theory about these equivalences that goes into much more detail, which I know next to nothing about.&lt;/p&gt;&#xA;" OwnerUserId="536" LastActivityDate="2010-08-05T12:28:46.967" />
  <row Id="1613" PostTypeId="2" ParentId="1589" CreationDate="2010-08-05T12:33:54.223" Score="1" ViewCount="" Body="&lt;p&gt;This is rather scheme-y, but there's  a really nice paper by &lt;a href=&quot;http://projecteuclid.org/DPubS?service=UI&amp;amp;version=1.0&amp;amp;verb=Display&amp;amp;handle=euclid.rmjm/1250128841&quot; rel=&quot;nofollow&quot;&gt;Kempf&lt;/a&gt; (hopefully you have institutional access :() that gives a very basic and elementary proof that the higher cohomology of a quasi-coherent sheaf on an affine scheme is trivial. The first part of the paper uses nothing more than the basic properties  (e.g. long exact sequence) of cohomology, and might be fun. I thought it was fun, anyway; it's also nice because it shows that Hartshorne is unnecessarily restrictive in sticking to noetherian affine schemes in chapter III (even if one wants to avoid anything fancy).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;OK, update: here is the proof &lt;a href=&quot;https://amathew.wordpress.com/2010/08/05/the-paper-of-kempf/&quot; rel=&quot;nofollow&quot;&gt;explained&lt;/a&gt; (admittedly by a beginner :)).&lt;/p&gt;&#xA;" OwnerUserId="536" LastEditorUserId="536" LastEditorDisplayName="" LastEditDate="2010-08-06T02:51:39.390" LastActivityDate="2010-08-06T02:51:39.390" CommentCount="2" />
  <row Id="1614" PostTypeId="1" AcceptedAnswerId="1622" CreationDate="2010-08-05T12:36:01.677" Score="1" ViewCount="157" Body="&lt;p&gt;&lt;strong&gt;Stromquist's Theorem&lt;/strong&gt;:  If the simple closed curve J is &quot;nice enough&quot; then it has an inscribed square.&lt;br&gt;&#xA;&quot;Nice enough&quot; includes polygons.&lt;br&gt;&#xA;Read more about it here:  www.webpages.uidaho.edu/~markn/squares&lt;br&gt;&#xA;An &quot;inscribed square&quot; means that the &lt;em&gt;corners&lt;/em&gt; of a square overlap with the curve.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to suggest a counter-example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The curve connected by the points:&#xA;(.2,0)&#xA;    (1,0)&#xA;    (1,1)&#xA;    (0,1)&#xA;    (0, .2)&#xA;    (-.2, -.2)&#xA;    (.2,0)&lt;br&gt;&#xA;Link to plot: &lt;a href=&quot;http://www.freeimagehosting.net/uploads/5b289e6824.png&quot; rel=&quot;nofollow&quot;&gt;http://www.freeimagehosting.net/uploads/5b289e6824.png&lt;/a&gt;&lt;br&gt;&#xA;Can this curve be incribed by a square?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(An older version of this question had another example: a triangle on top of a square (without their mutual side.) )&lt;/p&gt;&#xA;" OwnerUserId="821" LastEditorUserId="821" LastEditorDisplayName="" LastEditDate="2010-08-05T13:49:20.537" LastActivityDate="2010-08-05T14:01:55.283" Title="Is this a counter example to Stromquist's Theorem?" Tags="&lt;geometry&gt;" AnswerCount="2" />
  <row Id="1615" PostTypeId="2" ParentId="1614" CreationDate="2010-08-05T12:42:49.153" Score="8" ViewCount="" Body="&lt;p&gt;What about the square on the bottom- with it and the triangle's mutual side returned?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; Btw. Don't be disheartened that your counterexample didn't come through- looking for counterexamples to proved theorems will never give a genuine counterexample (unless ZFC turns out to be inconsistent!!;)), but it can provide valuable insight into what the theorem means. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here you can see from experience that 'most of the square' may be contained within the curve- this is something that may not be obvious looking at the examples of Stormquist in action in books on the topic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Trying to find counterexamples, IMHO, is a great way to develop as a mathematician- just do not be afraid to be wrong ;)&lt;/p&gt;&#xA;" OwnerUserId="160" LastEditorUserId="160" LastEditorDisplayName="" LastEditDate="2010-08-05T12:54:36.057" LastActivityDate="2010-08-05T12:54:36.057" CommentCount="1" />
  <row Id="1616" PostTypeId="2" ParentId="1608" CreationDate="2010-08-05T12:45:53.860" Score="1" ViewCount="" Body="&lt;p&gt;Let the ratio be 1:r and the hypotenuse be h. Then the sides are then x and rx for some x. By the &lt;a href=&quot;http://en.wikipedia.org/wiki/Pythagorean_theorem&quot; rel=&quot;nofollow&quot;&gt;Pythagorean Theorem&lt;/a&gt; we get $x^2+r^2*x^2=h^2$. So $x=\sqrt{{h^2}/(1+r^2)}$. We can then calculate rx, the other side.&lt;/p&gt;&#xA;" OwnerUserId="123" LastActivityDate="2010-08-05T12:45:53.860" />
  <row Id="1617" PostTypeId="1" CreationDate="2010-08-05T13:09:38.787" Score="0" ViewCount="383" Body="&lt;p&gt;Pardon me for not knowing LaTeX code.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A 3x3 matrix A:&lt;br&gt;&#xA;[  2  -1   1  ]&lt;br&gt;&#xA;| -2   3  -2  |&lt;br&gt;&#xA;[ -1   1   0  ]&lt;p&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;...produces this characteristic equation:&lt;br&gt;&#xA;&amp;lambda;&lt;sup&gt;2&lt;/sup&gt; - 4&amp;lambda; + 3 = 0&lt;p&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;...these eigenvalues:&lt;br&gt;&#xA;&amp;lambda;&lt;sub&gt;1&lt;/sub&gt; = 1&lt;br&gt;&#xA;&amp;lambda;&lt;sub&gt;2&lt;/sub&gt; = $\dfrac{-1 + \sqrt{13}}{2}$&lt;br&gt;&#xA;&amp;lambda;&lt;sub&gt;3&lt;/sub&gt; = $\dfrac{-1 - \sqrt{13}}{2}$&lt;p&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;...and these eigenvectors:&lt;br&gt;&#xA;@ &amp;lambda;&lt;sub&gt;1&lt;/sub&gt;, &lt;b&gt;x&lt;/b&gt; = s&amp;lt;1, 1, 0&gt; + t&amp;lt;-1, 0, 1&gt; &lt;b&gt;or&lt;/b&gt; span{&amp;lt;1, 1, 0&gt;, &amp;lt;-1, 0, 1&gt;}&lt;br&gt;&#xA;&lt;b&gt;x&lt;sub&gt;1&lt;/sub&gt;&lt;/b&gt; = &amp;lt;1, 1, 0&gt;&lt;br&gt;&#xA;&lt;b&gt;x&lt;sub&gt;2&lt;/sub&gt;&lt;/b&gt; = &amp;lt;-1, 0, 1&gt;&lt;br&gt;&#xA;@ &amp;lambda;&lt;sub&gt;3&lt;/sub&gt;, &lt;b&gt;x&lt;sub&gt;3&lt;/sub&gt;&lt;/b&gt; = &amp;lt;0, 0, 0&gt;&lt;br&gt;&#xA;@ &amp;lambda;&lt;sub&gt;4&lt;/sub&gt;, &lt;b&gt;x&lt;sub&gt;4&lt;/sub&gt;&lt;/b&gt; = &amp;lt;0, 0, 0&gt;&lt;p&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After normalizing the already orthogonal &lt;b&gt;x&lt;sub&gt;1&lt;/sub&gt;&lt;/b&gt;, I get span(, &amp;lt; -($\sqrt{2}$/2), 0, $\sqrt{2}$/2&gt;}.&lt;p&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Setting up my P matrix for diagonalization, it appears to be a 3x4 matrix that is singular:&lt;br&gt;&#xA;$[\left(&#xA;\begin{array}{ccc}&#xA;\sqrt{2}/2 &amp;amp; - \sqrt{2}/2 &amp;amp; 0 &amp;amp; 0\&#xA;\sqrt{2}/2 &amp;amp; 0 &amp;amp; 0 &amp;amp; 0\&#xA;0 &amp;amp; \sqrt{2}/2 &amp;amp; 0 &amp;amp; 0\end{array}&#xA;\right)]$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Questions: Why do I have four vectors, and why are two of them zero vectors? Symmetric matrices are &lt;b&gt;always&lt;/b&gt; diagonalizable, where there's a P such that P&lt;sup&gt;-1&lt;/sup&gt; = P&lt;sup&gt;T&lt;/sup&gt;.&lt;/p&gt;&#xA;" LastActivityDate="2010-08-05T14:02:41.820" Title="Symmetric matrices and orthogonal diagonalization." Tags="&lt;linear-algebra&gt;" AnswerCount="2" CommentCount="6" />
  <row Id="1618" PostTypeId="1" AcceptedAnswerId="1628" CreationDate="2010-08-05T13:15:27.857" Score="8" ViewCount="393" Body="&lt;p&gt;Let $X,Y$ be $\sigma$-finite measure spaces, and let $L^p(X) \otimes L^p(Y)$ be the algebraic tensor product.  The product has a natural map into $L^p(X \times Y)$ which takes $\sum a_{ij} f_i \otimes g_j$ to the function $F(x,y) = \sum a_{ij} f_i(x) g_j(y)$.  A moment's thought shows that this map is well-defined.  Is it also injective?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems that this should be true, but I can't see how to prove it.  Intuitively, one needs to show that if $\sum a_{ij} f_i(x) g_j(y) = 0$ a.e., then one should be able to cancel all the terms in the sum using bilinearity.  It is not quite clear how to do this without knowing anything about the terms.&lt;/p&gt;&#xA;" OwnerUserId="822" LastEditorUserId="822" LastEditorDisplayName="" LastEditDate="2010-08-05T13:41:13.250" LastActivityDate="2010-08-10T21:06:56.040" Title="Is the natural map $L^p(X) \otimes L^p(Y) \to L^p(X \times Y)$ injective?" Tags="&lt;linear-algebra&gt;&lt;real-analysis&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="2" />
  <row Id="1619" PostTypeId="2" ParentId="1617" CreationDate="2010-08-05T13:16:13.050" Score="2" ViewCount="" Body="&lt;p&gt;Well, first off, a vector isn't orthogonal, a set of vectors is.  Remember that for two vectors to be orthogonal, it means that they are at right angles to each other, and orthonormal means that plus they're unit vectors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if I had to say where I think your first error is, you took a 3 x 3 matrix and got a quadratic equation somehow, but you should have a cubic.  And also, as for your eigenvectors, where did $\lambda_4$ come from? What is it? The symbol just appears from nowhere.  Perhaps you have typos and it's $\lambda_2,\lambda_3$ rather than 3 and 4, and each of them must have a nonzero eigenvector, because they are eigenvalues of multiplicity 1 (though with the equation error, they might not be eigenvalues), and so you would have to be incorrect about having a two dimensional eigenspace in the first place.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, both of the eigenvectors for 1 check out, which means that you've incorrectly calculated the eigenvalues.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(instead of just an answer, I put in all the thinking I did to get there, because I thought it might help clarify how to check your work in the future)&lt;/p&gt;&#xA;" OwnerUserId="176" LastActivityDate="2010-08-05T13:16:13.050" CommentCount="6" />
  <row Id="1620" PostTypeId="2" ParentId="790" CreationDate="2010-08-05T13:28:22.340" Score="3" ViewCount="" Body="&lt;p&gt;In an infinite-dimensional normed space with a translation-invariant measure, the measure of a ball must be either $0$ or $\infty$.  This fact is sometimes summarized as &quot;&lt;a href=&quot;http://en.wikipedia.org/wiki/There_is_no_infinite-dimensional_Lebesgue_measure&quot; rel=&quot;nofollow&quot;&gt;there is no infinite-dimensional Lebesgue measure&lt;/a&gt;.&quot;  So unless you have some other notion of &quot;volume&quot; in mind, only the finite-dimensional case (i.e. $\mathbb{R}^n$ with some other norm) has any content.  And with regard to Hilbert spaces, the only finite-dimensional Hilbert spaces are $\mathbb{R}^n$ with the Euclidean inner product, so of course we know about this.&lt;/p&gt;&#xA;" OwnerUserId="822" LastActivityDate="2010-08-05T13:28:22.340" />
  <row Id="1622" PostTypeId="2" ParentId="1614" CreationDate="2010-08-05T14:01:55.283" Score="4" ViewCount="" Body="&lt;p&gt;Regarding your edit: (0.2, 0) — (1, 0.2) — (0.8, 1) — (0, 0.8) (and many others)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.imgftw.net/img/326639277.png&quot; rel=&quot;nofollow&quot;&gt;&lt;img src=&quot;http://www.imgftw.net/img/326639277.png&quot; alt=&quot;&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="171" LastActivityDate="2010-08-05T14:01:55.283" />
  <row Id="1623" PostTypeId="2" ParentId="1526" CreationDate="2010-08-05T14:01:56.563" Score="0" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Continued_fraction&quot; rel=&quot;nofollow&quot;&gt;Continued fractions&lt;/a&gt; give the best approximations to a number using smaller terms in the fraction. This may take a bit of practice to calculate at speed, but for the number in question there are only 8 terms:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;[0;1,8,1,3,2,2,7],&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;with convergents:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;0, 1, 8/9, 9/10, 35/39, 79/88, 193/215, 1430/1593&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One can clearly see that the option 9/10 is the best one from the choices on offer. Note the convergents get progressively more accurate and oscillate between being over-estimates and under-estimates. The final term is, of course, the original fraction itself in its simplest form.&lt;/p&gt;&#xA;" OwnerUserId="166" LastActivityDate="2010-08-05T14:01:56.563" />
  <row Id="1624" PostTypeId="2" ParentId="1617" CreationDate="2010-08-05T14:02:41.820" Score="0" ViewCount="" Body="&lt;p&gt;Your characteristic polynomial is incorrect.&#xA;Go to wolframalpha.com and enter: {{2, -1, 1},{-2, 3, 2}, {-1, 1, 0}}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;www.wolframalpha.com/input/?i={{2,+-1,+1},{-2,+3,+2},+{-1,+1,+0}}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You will see that the characteristic polynomial is of 3rd degree.&#xA;It must be of 3rd degree because your matrix is 3x3.&#xA;Here is what you'll get:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;p(x)=-x^3+5 x^2-3 x-1&lt;br&gt;&#xA;lambda_1 = 4.23607&lt;br&gt;&#xA;lambda_2 = 1&lt;br&gt;&#xA;lambda_3 = -0.236068  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And you will also get 3 eigenvectors. (Never 4 for a 3x3 matrix).&lt;/p&gt;&#xA;" OwnerUserId="821" LastActivityDate="2010-08-05T14:02:41.820" CommentCount="1" />
  <row Id="1625" PostTypeId="2" ParentId="1548" CreationDate="2010-08-05T14:19:01.960" Score="3" ViewCount="" Body="&lt;p&gt;You can apply &lt;a href=&quot;http://groupprops.subwiki.org/wiki/Burnside%27s_normal_p-complement_theorem&quot; rel=&quot;nofollow&quot;&gt;Burnside's normal $p$-complement theorem&lt;/a&gt; to get a normal complement $N$ of $M$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then take an element $m$ of $M$ with prime order.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Case 1: The centralizer $C_N(m)$ of $m$ in $N$ is nontrivial.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As $M$ centralizes $m$, it acts on the fixed points of $m$ (in the action by conjugation on $N$), i.e., $C_N(m)$ is an $M$-invariant subgroup of $N$. If $C_N(m) = N$, then $m \in Z(G)$ contradicts ii). Otherwise $M &amp;lt; C_N(m)M &amp;lt; G$ contradicting the maximality of $M$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Case 2: $N$ has the fixed point free automorphism $m$ of prime order.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By Thompson's thesis $N$ is nilpotent, hence $G$ solvable.&lt;/p&gt;&#xA;" OwnerUserId="669" LastActivityDate="2010-08-05T14:19:01.960" />
  <row Id="1626" PostTypeId="1" AcceptedAnswerId="1634" CreationDate="2010-08-05T14:24:36.597" Score="4" ViewCount="569" Body="&lt;p&gt;How can I calculate $27^{41}\ \mathrm{mod}\ 77$ as simple as possible?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I already know that $27^{60}\ \mathrm{mod}\ 77 = 1$ because of &lt;a href=&quot;http://en.wikipedia.org/wiki/Euler%27s_theorem&quot; rel=&quot;nofollow&quot;&gt;Euler’s theorem&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ a^{\phi(n)}\ \mathrm{mod}\ n = 1 $$&#xA;and&#xA;$$ \phi(77) = \phi(7 \cdot 11) = (7-1) \cdot (11-1) = 60 $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also know from using modular exponentiation that $27^{10} \mathrm{mod}\ 77 = 1$ and thus&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ 27^{41}\ \mathrm{mod}\ 77 = 27^{10} \cdot 27^{10} \cdot 27^{10} \cdot 27^{10} \cdot 27^{1}\ \mathrm{mod}\ 77 = 1 \cdot 1 \cdot 1 \cdot 1 \cdot 27 = 27 $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But can I derive the result of $27^{41}\ \mathrm{mod}\ 77$ using $27^{60}\ \mathrm{mod}\ 77 = 1$ somehow?&lt;/p&gt;&#xA;" OwnerUserId="659" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-08-05T15:30:14.937" LastActivityDate="2011-02-28T01:35:01.600" Title="Modular exponentiation using Euler’s theorem" Tags="&lt;number-theory&gt;&lt;modular-arithmetic&gt;" AnswerCount="4" CommentCount="2" FavoriteCount="1" />
  <row Id="1627" PostTypeId="1" AcceptedAnswerId="1636" CreationDate="2010-08-05T14:52:14.937" Score="1" ViewCount="69" Body="&lt;p&gt;I try to learn the theory of category from &lt;a href=&quot;http://katmat.math.uni-bremen.de/acc/acc.pdf&quot; rel=&quot;nofollow&quot;&gt;The Joy of Cats&lt;/a&gt;. I got stacked with the first exercise (3A a).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we have a simple graph with one vertex and 2 nodes all we know is that in category there is 2 objects and one homomorphism (beside identities). Hence:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\[C = (\{\{x, y\}, \{1, 2\}\}, \{f, id\}, id, \cdot), f = \{(x, 1), (y, 2)\}\]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and &lt;/p&gt;&#xA;&#xA;&lt;p&gt;\[C = (\{\{x\}, \{1\}, \{f, id\}, id, \cdot), f = \{(x, 1)\}\]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Should have the same graph but are they isomorphic?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Note about notation:&lt;/strong&gt; I treat category as quadruple $(O, H, id, \cdot)$ where $O$ is class of objects (here is set), $H$ is class of homomorphisms (in example set). $id$ is identity and $\cdot$ is composition of homomorphisms.&lt;/p&gt;&#xA;" OwnerUserId="728" LastEditorUserId="728" LastEditorDisplayName="" LastEditDate="2010-08-05T16:02:29.750" LastActivityDate="2010-08-05T16:39:26.027" Title="Thin categories and graphs (isomorphism of categories)" Tags="&lt;category-theory&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="1628" PostTypeId="2" ParentId="1618" CreationDate="2010-08-05T14:52:25.310" Score="2" ViewCount="" Body="&lt;p&gt;The previous version of this answer was incorrect. It has generously been accepted by Nate Eldredge on the ground that &quot;it definitely contains the right idea&quot;. This previous version has better remain forgotten forever. The reader should consult &lt;a href=&quot;http://math.stackexchange.com/questions/1618/is-the-natural-map-lpx-otimes-lpy-to-lpx-times-y-injective/1678#1678&quot;&gt;Nate's answer&lt;/a&gt;, which is correct and complete. I'll just try here to spell out the underlying linear algebra lemma (which is trivial, but which I failed to see). &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;UNDERLYING LEMMA.&lt;/strong&gt; Let $U,V,W$ vector spaces over some field $K$, and $\alpha:U\otimes V\to W$ a linear map. Assume that for each nonzero vector $u_0$ in $U$ there are linear maps $\beta:W\to V$ and $\gamma:U\to K$ such that $\gamma(u_0)\not=0$ and &#xA;  $$\beta(\alpha(u\otimes v))=\gamma(u)v\ \forall\ u,v.$$ Then $\alpha$ is injective. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The proof can be very easily extracted from &lt;a href=&quot;http://math.stackexchange.com/questions/1618/is-the-natural-map-lpx-otimes-lpy-to-lpx-times-y-injective/1678#1678&quot;&gt;Nate's answer&lt;/a&gt;. Here is the most standard application: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;APPLICATION.&lt;/strong&gt; Let $U,V,W$ be respectively the spaces of $K$-valued functions on the sets $X,Y,X\times Y$, and $\beta,\gamma$ are given by $(\beta(w))(y)=w(x_0,y)$, $\gamma(u)=u(x_0)$ where $x_0$ is chosen so that $u_0(x_0)\not=0$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT.&lt;/strong&gt; Here is a mild generalization.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $X$ be a measure space, let $\mathcal V(X)$ be the vector space of all measurable complex valued functions on $X$, let $\mathcal N$ be the subspace of functions vanishing almost everywhere, and let $V(X)$ be the quotient. [Is there a standard notation for these spaces?]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $Y$ be another measure space. The bilinear map from $\mathcal V(X)\times\mathcal V(Y)$ to $\mathcal V(X\times Y)$ sending $(f,g)$ to $(x,y)\mapsto f(x)g(y)$ induces a linear map $\Phi$ from $V(X)\otimes V(Y)$ to $V(X\times Y)$.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;THEOREM.&lt;/strong&gt; Assume $X$ and $Y$ are $\sigma$-finite. Then $\Phi$ is injective.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This slightly more general than the statement proved in &lt;a href=&quot;http://math.stackexchange.com/questions/1618/is-the-natural-map-lpx-otimes-lpy-to-lpx-times-y-injective/1678#1678&quot;&gt;Nate's answer&lt;/a&gt; because no integrability assumptions are made on the functions.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;LEMMA.&lt;/strong&gt; Let $f_i,\dots,f_n$ be in $\mathcal V(X)$, and $S\subset X^n$ the set of $x\in X^n$ such that $\det(f_i(x_j))\not=0$. Then the $f_i$ are linearly independent in $V(X)$ if and only if the measure of $S$ is positive.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Denote by $|A|$ the measure of $A$, and by $S_g$ the support of the function $g$, that is the set of points where it is nonzero. Put $d(x):=\det(f_i(x_j))$ for $x\in X^n$. The above set $S$ is now $S_d$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To prove the if part of the lemma, assume the $f_i$ are linearly dependent in $V(X)$ and check $|S_d|=0$ as follows. Let $\lambda$ be a nonzero vector of $\mathbb C^n$ such that $g:=\sum \lambda_i f_i$ satisfies $|S_g|=0$. Denote again by $\lambda$ the 1 by $n$ matrix formed by the $\lambda_i$, and by $A(x)$ the matrix $(f_i(x_j))$. For $x$ in $(S_g^c)^n$, where the superscript c mean &quot;complement&quot;, we have $\lambda A(x)=0$. Multiplying on the right by the adjugate of $A(x)$ we get $d(x)=0$. This shows that $S_d$ is contained in the measure zero subset $((S_g^c)^n)^c$ of $X^n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To prove the only if part of the lemma, assume $|S_d|=0$ and check that the $f_i$ are linearly dependent in $V(X)$ as follows. Expand $d(x)$ as $$d(x)=\sum\ d_i(x')\ f_i(x_n)$$ with $x':=(x_1,\dots,x_{n-1})$. Arguing by induction, we can assume $|S_{d_i}|&gt;0$ for all $i$. Let $U$ be the union of the $S_{d_i}$. Denoting by $d(x',\bullet)$ the function $x_n\mapsto d(x)$, we have&#xA;$$S_d=\bigcup_{x'\in X^{n-1}}\ {x'}\times S_{d(x',\bullet)}.$$ Fubini yields&#xA;$$0=|S_d|=\int_{X^{n-1}}\ |S_{d(x',\bullet)}|\ dx'.$$&#xA;If the $f_i$ were linearly independent in $V(X)$, we would have $|S_{d(x',\bullet)}|&gt;0$ for all $x'$ in $U$, a contradiction. The lemma is proved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's prove the theorem. In the next lines, $i,j,k$ will lie between 1 and $m$, whereas $p,q,r$ will lie between 1 and $n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $f_1,\dots,f_m$ be in $\mathcal V(X)$; let $g_1,\dots,g_n$ be in $\mathcal V(Y)$; and put $F_{ip}(x,y):=f_i(x)g_p(y)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The theorem is equivalent to the statement that the linear independence of the $f_i$ and $g_p$ in $V(X)$ and $V(Y)$ implies that of the $F_{ip}$ in $V(X\times Y)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $S\subset X^m$ and $T\subset Y^n$ be defined by the conditions $\det(f_i(x_j))\not=0$ and $\det(g_p(y_q))\not=0$. Embedd $U:=S^n\times T^m$ into $(X\times Y)^{mn}$ in the obvious way, and denote the $(i,p)$-th coordinate of $z\in(X\times Y)^{mn}$ by $z_{ip}=(x_{pi},y_{ip})$. Then $z$ is in $U$ iff $x_p\in S$ and $y_i\in T$ for all $(i,p)$, that is iff&#xA;$$\det\Big(f_j(x_{pk})\Big)_{jk}\not=0\not&#xA;=\det\Big(g_q(y_{ir})\Big)_{qr}$$&#xA;for all $(i,p)$. We claim that the validity of this double nonequality for all $(i,p)$ implies the nonequality&#xA;$$\det\Big(f_i(x_{qj})\ g_p(y_{jq})\Big)_{(i,p)(j,q)}\not=0.$$&#xA;In view of the lemma, the theorem is equivalent to the claim. But the claim doesn't depend on the measures defined on $X$ and $Y$. As the theorem holds when $X$ and $Y$ are equipped with the counting measure (see the application of the &quot;underlying lemma&quot; at the beginning of the post), we are done.&lt;/p&gt;&#xA;" OwnerUserId="660" LastEditorUserId="660" LastEditorDisplayName="" LastEditDate="2010-08-10T08:41:44.417" LastActivityDate="2010-08-10T08:41:44.417" CommunityOwnedDate="2010-08-10T08:43:03.587" />
  <row Id="1629" PostTypeId="2" ParentId="653" CreationDate="2010-08-05T15:11:55.163" Score="3" ViewCount="" Body="&lt;p&gt;Homotopy groups and homotopy classes of maps are used in physics to study topological defects.  Roughly speaking, $\pi_k(X)$ classifies codimension $k+1$ defects in textures modeled on $X$ (maps from $R^d$ to $X$ continuous except at the loci of defects).  Other homotopy classes of maps can be used to study linked defects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See the following beautiful review paper by N.D. Mermin for an introduction: &lt;a href=&quot;http://rmp.aps.org/abstract/RMP/v51/i3/p591_1&quot; rel=&quot;nofollow&quot;&gt;http://rmp.aps.org/abstract/RMP/v51/i3/p591_1&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="826" LastEditorUserId="826" LastEditorDisplayName="" LastEditDate="2010-08-06T19:47:28.357" LastActivityDate="2010-08-06T19:47:28.357" />
  <row Id="1630" PostTypeId="2" ParentId="1626" CreationDate="2010-08-05T15:13:57.283" Score="2" ViewCount="" Body="&lt;p&gt;You can use &lt;a href=&quot;http://en.wikipedia.org/wiki/Exponentiation_by_squaring&quot; rel=&quot;nofollow&quot;&gt;exponentiation by squaring&lt;/a&gt; (all operations are modulo 77):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$27^{41} = 27^{32+8+1} = 27 \cdot 27^8 \cdot (27^8)^4 = (*)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\big[ 27^8 = ((27^2)^2)^2 = (36^2)^2 = 64^2 = 15 \big]$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(*) = 27 \cdot 15 \cdot 15^4 = 27 \cdot 15 \cdot (15^2)^2 = 27 \cdot 15 \cdot 71^2 = 27 \cdot 15 \cdot 36 = 27$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This uses only 7 multiplications instead of 41.&lt;/p&gt;&#xA;" OwnerUserId="26" LastActivityDate="2010-08-05T15:13:57.283" />
  <row Id="1631" PostTypeId="2" ParentId="709" CreationDate="2010-08-05T15:23:07.250" Score="2" ViewCount="" Body="&lt;p&gt;The &lt;em&gt;vertex&lt;/em&gt; of a parabola is the point on the axis of symmetry that intersects it (&lt;a href=&quot;http://en.wikipedia.org/wiki/Parabola&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The point $(x,y)=\left( 0,0\right) $ is the vertex of the parabola given by $y=ax^{2}$.&#xA;By making the change of coordinates&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$X=x+\dfrac{b}{2a},Y=y+\dfrac{b^{2}-4ac}{4a}$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;equation $y=ax^{2}+bx+c$ is transformed into $Y=aX^{2}$, the vertex of which is the point $(X,Y)=(0,0)$, i.e. $(x,y)=\left(-\dfrac{b}{2a},-\dfrac{b^{2}-4ac}{4a}\right)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(This change of coordinates is a &lt;em&gt;translation&lt;/em&gt; of both axes $x$, $y$, respectively, by $-\dfrac{b}{2a},-\dfrac{b^{2}-4ac}{4a}$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, the $x$-coordinate of the vertex of the parabola is $-\dfrac{b}{2a}$.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit.&lt;/strong&gt; &lt;em&gt;Example&lt;/em&gt;. Plot of $y=2x^2-x+4$; blue axes: $X=x-1/4,Y=y-31/8$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://problemasteoremas.files.wordpress.com/2010/08/parabola.jpg?w=490&amp;amp;h=392&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="752" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2010-08-29T10:30:47.447" LastActivityDate="2010-08-29T10:30:47.447" />
  <row Id="1632" PostTypeId="1" CreationDate="2010-08-05T15:23:30.370" Score="1" ViewCount="62" Body="&lt;p&gt;A discrete one-dimensional model of optical imaging looks like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$I(r) = \sum_i e_i P(r - r_i)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here, the $e_i$ are point light sources at locations $r_i$ in the object and $P$ is a point spread function that blurs each point. We can assume that $P$ is even, non-negative and has a finite extent, ie $P(x) = 0$ for $|x| &gt; p$. The $e_i$ are all positive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A more complex imaging process instead produces an image like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$I(r)^2 = \sum_{i,j} e_i e_j P(r - r_i) P(r - r_j) \cos (r_i - r_j)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(The $^2$ is a consequence of the reconstruction method. In practice we normally take the square root, but per &lt;a href=&quot;http://math.stackexchange.com/questions/1130/preserving-the-extrema-of-one-function-after-applying-another&quot;&gt;my earlier question here&lt;/a&gt;, I think the salient structural features of the resulting image should be unchanged by this transformation.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Numerical simulations suggest that the latter method allows better resolution of the points when $p \sim \pi$, and is not materially different when $p \ll \pi$. (We ignore the case where $p \gg \pi$ as not physically tenable.) Intuitively, this is because the &quot;trough&quot; in the $\cos (r_i - r_j)$ term reduces the interaction between points at some intermediate separations. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd like to be able to be more analytical about this. We can readily extend the above to a continuous model with an object function $O$ in place of the $e_i$. The ordinary image becomes a convolution integral&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$I(r) = \int O(s) \: P(r - s) \; \mathrm{d}s$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for which there are standard analytical approaches available. However, the more complex reconstruction looks something like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$I(r)^2 = \int \int  O(s) \; O(t) \; P(r-s) \; P(r-t) \; \cos (s - t) \mathrm{d}s \mathrm{d}t$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This looks fairly intractable to me, and I'm not really sure where to begin with it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, my first question is: is there any point in trying to look at the continuous model, or should I just concentrate on the discrete? On the one hand, every single practical use case for the technique will actually be using discrete measurements, so in that sense the discrete model is more realistic. On the other, it may be that there are things one could demonstrate using the continuous model that are not available otherwise. (But even if that's the case, I'm probably not capable of doing so without at least being nudged in the right direction!)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And I guess my second question is: continuous or not, does anyone have any other suggestions of useful ways to analyse this model?&lt;/p&gt;&#xA;" OwnerUserId="144" LastActivityDate="2010-08-05T15:23:30.370" Title="Analysing an optics model in discrete and continuous forms" Tags="&lt;analysis&gt;&lt;signal-processing&gt;&lt;physics&gt;" />
  <row Id="1633" PostTypeId="2" ParentId="1437" CreationDate="2010-08-05T15:23:43.270" Score="0" ViewCount="" Body="&lt;p&gt;Open a graphic calculator such as fooplot.com&lt;br&gt;&#xA;Enter a function:&lt;br&gt;&#xA;y(x) = 10^(x/10)&lt;br&gt;&#xA;and another function:&lt;br&gt;&#xA;y(x) = x&lt;br&gt;&#xA;Click the intersection button and then the intersections on the image.&lt;/p&gt;&#xA;" OwnerUserId="821" LastActivityDate="2010-08-05T15:23:43.270" CommentCount="1" />
  <row Id="1634" PostTypeId="2" ParentId="1626" CreationDate="2010-08-05T15:26:21.820" Score="6" ViewCount="" Body="&lt;p&gt;As suggested in the comment above, you can use the Chinese Remainder Theorem, by using Euler's theorem / Fermat's theorem on each of the primes separately. You know that $27^{10} \equiv 1 \mod 11$, and you can also see that modulo 7, $27 \equiv -1 \mod 7$, so $27^{10} \equiv (-1)^{10} \equiv 1\mod 7$ as well. So $27^{10} \equiv 1 \mod 77$, and $27^{41} = 27^{40+1} \equiv 27 \mod 77$. (We've effectively found the order of 27 as 10, but a more mechanical procedure may be to use that $27^{41} \equiv 27 \equiv 5 \mod 11$ and $27^{6} \equiv 1 \mod 7$ to see that $27^{41} = 27^{42-1} \equiv 27^{-1} \equiv -1 \mod 7$, and put 5 mod 11 and -1 mod 7 together to get 27.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is if you're doing it by hand, for this case. In general, algorithmically, you would just use repeated squaring to exponentiate numbers. You don't gain much by using Euler's theorem, since finding the order of a number mod $n$ is as hard as factoring $n$.&lt;/p&gt;&#xA;" OwnerUserId="205" LastEditorUserId="205" LastEditorDisplayName="" LastEditDate="2010-08-05T15:32:29.663" LastActivityDate="2010-08-05T15:32:29.663" CommentCount="3" />
  <row Id="1635" PostTypeId="2" ParentId="1609" CreationDate="2010-08-05T15:42:56.583" Score="14" ViewCount="" Body="&lt;p&gt;If you want to avoid algebraic geometry and topology, then you will probably be forced to use algebra.  This is okay, because the things you ask about have algebraic interpretations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First of all, it probably helps to write the function field as $k(x,y),$ where $x$ and $y$ are related by some equation $f(x,y) = 0$.  One problem is going to be that $f$ can't be chosen to be smooth in general (because not every curve can be embedded as a smooth plane curve), but it's probably best to ignore this; if it becomes important, then probably your colleagues will be delving deeper into the theory anyway, and so the whole level of explanation can be ramped up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now to explain divisors, you can have them imagine intersecting $f(x,y) = 0$ with some other curve $g(x,y) = 0$; the intersections will be a bunch of points, possibly with multiplicity.&#xA;This is a divisor.   So divisors are just natural ways of encoding how curves intersect one another.  (Again, I am ignoring here the issue of points at infinity; you will have to decide whether that is appropriate, or whether you need a higher level of precision in your explanations.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the genus, it's a little more complicated to explain algebraically, but possible; here goes:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose that the equation $f(x,y) = 0$ has degree $d$, so your function field corresponds to a degree $d$ curve in the plane.  Now let $V_n$ be the vector space of all polynomials of degree $\leq n$ in $x$ and $y$.  Let's suppose that $n \geq d$ (and in general, we should&#xA;think that $n$ is large).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A simple computation shows that $V_n$ has dimension $(n+2)(n+1)/2$.  Inside $V_n$, we have&#xA;a subspace consisting of all the multiples of $f$.  This subspace is obtained by taking&#xA;the elements of $V_{n-d}$ (i.e. polynomials of degree at most $n-d$) and multiplying them by $f$, i.e. it is the subspace $f V_{n-d}$ of $V_n$, and so has dimension $(n-d+2)(n-d+1)/2$.  If we look at the quotient $V_n/fV_{n-d},$ this then has dimension &#xA;$n d + 1 - (d-1)(d-2)/2$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the meaning of this quotient?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $g \in V_n$ represents a non-zero element of $V_n/fV_{n-d}$, then it is a degree $n$&#xA;equation that is &lt;em&gt;not&lt;/em&gt; divisible by $f$, so it &lt;em&gt;doesn't&lt;/em&gt; vanish identically on $f(x,y) = 0$, so by Bezout's theorem, the intersection of $g(x,y) = 0$ and $f(x,y) = 0$ is $n d$ &#xA;points, i.e. a divisor of degree $n d$.  So we see that non-zero elements of &#xA;$V_n/f V_{n-d}$ correspond to those divisors of degree $n d$ that are obtained by &#xA;intersecting $f(x,y) = 0$ with a degree $n$ curve.  (In fact, we should think about non-zero elements up to scaling, because if we multiply $g$ by a non-zero scalar,&#xA;the curve $g(x,y) = 0$ doesn't change.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus among all the degree $n d$ divisors on $f(x,y) = 0$, those that come by intersecting with a degree $n$ curve form a space of dimension $n d - (d-1)(d-2)/2.$  (Here I have&#xA;subtracted 1, because rescaling accounts for 1 of the dimensions in the above formula.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now what is the dimension of the space of all divisors of degree $n d$ (with non-negative coefficients,&#xA;which are the only kind that can possibly arise as intersections; divisors with non-negative coefficients are called effective)?  Well, we just have to choose $n d$ points and add them together.  We are choosing the points from a curve, which is one-dimensional, so that means there is an $n d$-dimensional space of effective divisors of degree $n d$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus we see that, in the $n d$-dimensional space of all effective divisors of degree $n d$,&#xA;those that arise by intersecting with a degree $n$ curve are a subspace of dimension&#xA;$n d -(d-1)(d-2)/2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The quantity $(d-1)(d-2)/2$ is precisely the genus.&#xA;So what we see is that the bigger the genus is, the harder it is for a degree $n d$&#xA;divisor on $f(x,y) = 0$ to be obtained by intersecting with another curve.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, if the degree $d$ is $1$ or $2$, then the genus vanishes, and &#xA;every degree $n d$ divisor comes from intersecting with a degree $n$ curve.&lt;br&gt;&#xA;E.g. on a conic (i.e. when $d = 2$) any two points come from intersecting with a line&#xA;(the line that passes through those two points), any four points come from intersecting &#xA;with a conic, and so on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, if the degree $d = 3$, then not all triples of points on $f(x,y) = 0$&#xA;come from intersecting with a line: in fact if you give yourself two points, then&#xA;they determine a line, which in turn determines the 3rd point of intersection.&#xA;Similarly (going up to the case $n =3$), a general set of 9 points on the cubic doesn't come from&#xA;intersecting with another cubic; instead, if you give yourself 8 points on the cubic&#xA;curve, you can find another cubic passing through these 8 points, and its 9th point of intersection is uniquely determined by the given 8 points.  (Since 9 = 8 + 1, this &#xA;is a particular manifestation of the fact that our cubic curve has genus 1.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you can see, I've lapsed into a more geometrical way of thinking (using notions such as dimension), but I don't think one can avoid this completely: the concept of genus arose historically from essentially the kind of computations that I've just been making, and at some point you have to think about spaces of divisors and their dimensions if you want to understand it.  Still, I hope that this gives you an avenue to explaining the concept which is more algebraic, and so more accessible to your colleagues.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One more technical remark: The space $V_{n}/f V_{n-d}$ is an example of a Riemann--Roch space, and the formula for its dimension ($n d + 1 - $ the genus) is a special case of the Riemann-Roch formula.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[Technical remark added in response to a question of T. in the comments below; feel free to ignore it if it is at too high a level:]  Note that the above discussion works even if $f$ is allowed to be singular (and irreducible, say, so that&#xA;we can't have polynomials $g$ which vanish on one component of $f(x,y) = 0$ without vanishing &#xA;on the whole curve).  The reason is that $(d-1)(d-2)/2$ is always the &lt;em&gt;arithmetic genus&lt;/em&gt; of a plane curve of degree $d$, whether or not the curve is smooth, and it is the arithmetic genus that intervenes in the Riemann--Roch formula.  (I believe that this is the origin of the adjective &lt;em&gt;arithmetic&lt;/em&gt; in arithmetic genus: it is this version of the genus for singular curves that comes up when you make Riemann--Roch-type calculations of the dimensions of various spaces of divisors.)&lt;/p&gt;&#xA;" OwnerUserId="221" LastEditorUserId="221" LastEditorDisplayName="" LastEditDate="2010-08-09T03:26:02.570" LastActivityDate="2010-08-09T03:26:02.570" CommentCount="6" />
  <row Id="1636" PostTypeId="2" ParentId="1627" CreationDate="2010-08-05T15:53:59.867" Score="1" ViewCount="" Body="&lt;p&gt;It is a bit hard to understand your notation. If understand correctly, the isomorphism is realized by the functor $F$ which sends $\{x, y\}$ to $\{ x \}$ and $\{ 1, 2 \}$ to $\{ 1 \}$ (on objects) and send the unique non trivial morphism $f$ to $f$.&lt;/p&gt;&#xA;" OwnerUserId="233" LastEditorUserId="233" LastEditorDisplayName="" LastEditDate="2010-08-05T16:39:26.027" LastActivityDate="2010-08-05T16:39:26.027" CommentCount="3" />
  <row Id="1637" PostTypeId="2" ParentId="764" CreationDate="2010-08-05T17:10:06.220" Score="0" ViewCount="" Body="&lt;p&gt;In Stieltjes integral you assign different importance to different parts of the set you are integrating over.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The usefulness of this would become clearer when you know the theory of Lebesgue measure, which generalizes this even further. For example there is the Dirac measure which when used for integrating functions cares only about the value of a function at a point(typically the origin). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Understanding it in the Stieljes form is not any harder; moreover it will pave the way for measure theory.&lt;/p&gt;&#xA;" OwnerUserId="218" LastActivityDate="2010-08-05T17:10:06.220" />
  <row Id="1638" PostTypeId="2" ParentId="998" CreationDate="2010-08-05T17:11:06.403" Score="3" ViewCount="" Body="&lt;p&gt;Nobody has mentioned the book of Shafarevich; so I mention myself.&lt;/p&gt;&#xA;" OwnerUserId="218" LastActivityDate="2010-08-05T17:11:06.403" CommunityOwnedDate="2010-08-05T17:11:06.403" />
  <row Id="1639" PostTypeId="2" ParentId="262" CreationDate="2010-08-05T17:13:27.977" Score="2" ViewCount="" Body="&lt;p&gt;Anybody who wants to be a serious mathematician better read W. Rudin's &quot;Principles of mathematical Analysis&quot;. It gives a rigorous foundation to the basic notions analysis and introduces the reader to the world of rigor, after the sloppy days of calculus courses. One must learn the notion of rigor properly if one wants to be a mathematician. More than anything else, it is an exercise in the rectitude of thought. No other book is so universally used that would teach this notion, than Rudin. &lt;/p&gt;&#xA;" OwnerUserId="218" LastActivityDate="2010-08-05T17:13:27.977" CommunityOwnedDate="2010-08-05T17:13:27.977" />
  <row Id="1640" PostTypeId="2" ParentId="1413" CreationDate="2010-08-05T17:15:31.403" Score="1" ViewCount="" Body="&lt;p&gt;I do not want to be mean or rude here; but my personal belief is that if you do not appreciate number theory, you are missing the most important topic in mathematics. It is literally the heart of mathematics. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you really need a concrete example, I can give one from the point of view of algebra. To have an example of a purely non-separable polynomial, you go to char p.&lt;/p&gt;&#xA;" OwnerUserId="218" LastActivityDate="2010-08-05T17:15:31.403" CommentCount="3" />
  <row Id="1641" PostTypeId="2" ParentId="457" CreationDate="2010-08-05T17:16:54.430" Score="2" ViewCount="" Body="&lt;p&gt;The Fourier transform of the Dirac delta function is a constant function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore whatever frequency your radio is turned to during a storm, or your TV, you will receive a noise.&lt;/p&gt;&#xA;" OwnerUserId="218" LastActivityDate="2010-08-05T17:16:54.430" CommunityOwnedDate="2010-08-05T17:16:54.430" />
  <row Id="1642" PostTypeId="2" ParentId="450" CreationDate="2010-08-05T17:53:53.310" Score="1" ViewCount="" Body="&lt;p&gt;As others have already mentioned, the Jacobian determinant transforms one coordinate system to another by relating infinitesimal areas (or volumes) from one system to another. Consider going from Cartesian to Polar coordinates:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;J = \det\frac{\partial(x,y)}{\partial(r,\theta)}&#xA;=\begin{vmatrix}&#xA; \frac{\partial x}{\partial r} &amp;amp; \frac{\partial x}{\partial \theta} \\&#xA; \frac{\partial y}{\partial r} &amp;amp; \frac{\partial y}{\partial \theta} \\&#xA;\end{vmatrix}&#xA;$$&#xA;$$&#xA;=\begin{vmatrix}&#xA; \cos\theta &amp;amp; -r\sin\theta \\&#xA; \sin\theta &amp;amp; r\cos\theta \\&#xA;\end{vmatrix}&#xA;=r\;\cos^2\theta + r\;\sin^2\theta = r&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is useful because:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathrm{d}A = J\;\mathrm{d}r\mathrm{d}\theta = r\mathrm{d}r\mathrm{d}\theta$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\iint_\mathbf{R} f(r,\theta)\mathrm{d}A = \int_a^b \int_0^{r(\theta)} f(r,\theta) r\mathrm{d}r\mathrm{d}\theta$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which tells you that if you have a function $f(r, \theta)$ you can compute the integral as long as you remember to add a factor of $r$. The common transformations have all been worked out and can be found &lt;a href=&quot;http://en.wikipedia.org/wiki/List_of_canonical_coordinate_transformations&quot; rel=&quot;nofollow&quot;&gt;here on Wikipedia&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="196" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-09-09T23:00:16.617" LastActivityDate="2010-09-09T23:00:16.617" CommentCount="3" />
  <row Id="1643" PostTypeId="2" ParentId="903" CreationDate="2010-08-05T17:58:54.490" Score="1" ViewCount="" Body="&lt;p&gt;I would like to recommend &lt;em&gt;&lt;a href=&quot;http://www.pearson.ch/HigherEducation/MathematicsStatistics/AdvancedMathematics/1471/9780131848696/Introduction-to-Topology-Pure-and.aspx&quot; rel=&quot;nofollow&quot;&gt;Topology: Pure and Applied&lt;/a&gt;&lt;/em&gt;, by Colin Adams and Robert Franzosa.&#xA;Anyone familiar with Colin Adams's &lt;em&gt;The Knot Book&lt;/em&gt; will expect this to be equally accessible.&#xA;And they will not be disappointed.  In terms of your criteria&amp;mdash;clarity and motivation&amp;mdash;I&#xA;find it unsurpassed.&lt;/p&gt;&#xA;" OwnerUserId="237" LastActivityDate="2010-08-05T17:58:54.490" CommunityOwnedDate="2010-08-05T17:58:54.490" />
  <row Id="1644" PostTypeId="2" ParentId="1626" CreationDate="2010-08-05T18:17:21.653" Score="6" ViewCount="" Body="&lt;p&gt;If you are serious about &quot;as simple as possible&quot; then observe that $27^{41} = 3^{123}$ and use  &lt;a href=&quot;http://en.wikipedia.org/wiki/Carmichael_function&quot; rel=&quot;nofollow&quot;&gt;Carmichael's theorem&lt;/a&gt; (a strengthening of Euler's theorem which actually gives a tight bound) to deduce that $3^{30} \equiv 1 \bmod 77$ and hence $3^{123} \equiv 3^3 \equiv 27 \bmod 77$.  But I do not think this is the right question to ask; you should really be asking &quot;as general as possible,&quot; and then the answer you have accepted is most appropriate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Of course it suffices to use Euler's theorem for the above computation, but few people seem to learn Carmichael's theorem and I always like to point it out when I can.)&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-08-05T18:17:21.653" CommentCount="4" />
  <row Id="1645" PostTypeId="2" ParentId="262" CreationDate="2010-08-05T18:18:27.687" Score="1" ViewCount="" Body="&lt;p&gt;I've been rereading &lt;i&gt;Littlewood's Miscellany&lt;/i&gt; recently. It's a very readable collection of the writings of J. E. Littlewood, carefully edited by Béla Bollobás. Any budding mathematician will draw much inspiration from it. I like &lt;i&gt;A Mathematician's Apology&lt;/i&gt;, but if I was forced into choosing only one book, it would be &lt;i&gt;Littlewood's Miscellany&lt;/i&gt;.&lt;/p&gt;&#xA;" OwnerUserId="823" LastActivityDate="2010-08-05T18:18:27.687" CommunityOwnedDate="2010-08-05T18:18:27.687" />
  <row Id="1646" PostTypeId="2" ParentId="450" CreationDate="2010-08-05T18:33:14.460" Score="0" ViewCount="" Body="&lt;p&gt;You can represent some non-linear transforms (like translation) of a n dimensional vector with a n+1 dimensional matrix. However, converting the vector to its n+1 dimensional homogeneous version and back is not a linear transformation and also not representable as a matrix.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More is explained here: &#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Transformation_matrix&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Transformation_matrix&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="99" LastActivityDate="2010-08-05T18:33:14.460" />
  <row Id="1647" PostTypeId="1" CreationDate="2010-08-05T18:57:39.233" Score="3" ViewCount="72" Body="&lt;p&gt;I am writing a Gaussian blur filter in graphics shader code.  I want to make the blur parameterized by radius from the users perspective.  The best method I can figure to do this is to pick a suitable stopping point for y, say .001, and solve for the variance to plug into the normal distribution function that will achieve that value of y.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately I cannot for the life of me solve this equation for v...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x = 10$ (blur radius)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$.001 = \frac{1}{2 \pi v^2}e^{-\frac{x^2}{2v^2}}$$.&lt;/p&gt;&#xA;" OwnerUserId="836" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-09-04T19:41:48.183" LastActivityDate="2010-09-05T23:27:59.197" Title="How do I calculate a specific variation for a known value of the normal distribution function?" Tags="&lt;algorithms&gt;&lt;computer-science&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="1648" PostTypeId="1" CreationDate="2010-08-05T19:05:28.907" Score="4" ViewCount="625" Body="&lt;p&gt;i've sampled a real world process, network ping times. The &quot;round-trip-time&quot; is measured in milliseconds. Results are plotted in a histogram:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i36.tinypic.com/2qtfp6f.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ping times have a minimum value, but a long upper tail.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i want to know what statistical distribution this is, and how to estimate its parameters. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even though the distribution is not a normal distribution, i can still show what i am trying to achieve.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The normal distribution uses the function:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://upload.wikimedia.org/math/9/e/1/9e1e4a3af93c9680ba75669a0b69fbf6.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with the two parameters&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;μ (mean)&lt;/li&gt;&#xA;&lt;li&gt;σ&lt;sup&gt;2 &lt;/sup&gt; (variance)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;h2&gt;Parameter estimation&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The formulas for estimating the two parameters are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://upload.wikimedia.org/math/3/2/4/324081bc08a66fbeba64f43a3dbd5e1a.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Applying these formulas against the data i have in Excel, i get:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;μ = 10.9558 (mean)&lt;/li&gt;&#xA;&lt;li&gt;σ&lt;sup&gt;2 &lt;/sup&gt; = 67.4578 (variance)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;With these parameters i can plot the &quot;&lt;em&gt;normal&lt;/em&gt;&quot; distribution over top my sampled data:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i34.tinypic.com/2mqu3wl.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously it's not a normal distribution. A normal distribution has an infinite top and bottom tail, and is symmetrical. This distribution is not symmetrical.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;What principles would i apply, what flowchart, would i apply to determine what kind of distribution this is?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And cutting to the chase, what is the formula for that distribution, and what are the formulas to estimate its parameters?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;i want to get the distribution so i can get the &quot;average&quot; value, as well as the &quot;spread&quot;:&#xA;&lt;img src=&quot;http://i38.tinypic.com/2i27sp4.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i am actually plotting the histrogram in software, and i want to overlay the theoretical distribution:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i38.tinypic.com/nqpnck.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Tags&lt;/strong&gt;: sampling, statistics, parameter-estimation, normal-distribution&lt;/p&gt;&#xA;" OwnerUserId="835" LastEditorUserId="536" LastEditorDisplayName="" LastEditDate="2010-08-06T19:48:35.717" LastActivityDate="2010-08-10T02:40:44.470" Title="How do I figure out what kind of distribution this is?" Tags="&lt;discrete-mathematics&gt;&lt;probability&gt;&lt;average&gt;&lt;gamma-function&gt;" AnswerCount="4" CommentCount="7" FavoriteCount="1" />
  <row Id="1649" PostTypeId="1" AcceptedAnswerId="1657" CreationDate="2010-08-05T19:06:08.530" Score="0" ViewCount="282" Body="&lt;p&gt;Are there any examples of solving for the global maximum of a non-differentiable function where you:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Construct a series of differentiable functions that approach the non-differentiable function in the limit&lt;/li&gt;&#xA;&lt;li&gt;Show the maximum of each differentiable function converges to some value, which is thus your answer.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;For all I know, the procedure above is fatally flawed (or there are trivial examples, I would be most interested in non-trivial examples) in some way, if that is the case let me know.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am specifically interested in examples involving absolute values.&lt;/p&gt;&#xA;" OwnerUserId="99" LastEditorUserId="76" LastEditorDisplayName="" LastEditDate="2010-08-05T22:05:53.933" LastActivityDate="2010-08-06T11:46:42.377" Title="Finding the Extrema Non-differentiable Functions." Tags="&lt;calculus&gt;&lt;sequences-and-series&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="1650" PostTypeId="2" ParentId="1648" CreationDate="2010-08-05T19:13:22.353" Score="1" ViewCount="" Body="&lt;p&gt;The go-to distribution for things like wait times is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Exponential_distribution&quot; rel=&quot;nofollow&quot;&gt;Exponential&lt;/a&gt;. Yours doesn't look exactly the same because of the tiny lower tail, but I would be inclined to attribute that to noise/measurement error. (The assumption of independence of events is almost certainly wrong for ping times, but it's probably still your best choice.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, you would probably be better off asking this sort of question on the &lt;a href=&quot;http://stats.stackexchange.com&quot;&gt;stats site&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; As pointed out by &lt;a href=&quot;http://stats.stackexchange.com/questions/1315/how-do-i-figure-out-what-kind-of-distribution-this-is/1316#1316&quot;&gt;Srikant Vadali&lt;/a&gt;, the &lt;a href=&quot;http://en.wikipedia.org/wiki/Gamma_distribution&quot; rel=&quot;nofollow&quot;&gt;Gamma distribution&lt;/a&gt; is more general and can account for a non-negligible short tail, so may be a more appropriate choice. It's easier to estimate the parameter for the exponential, though.&lt;/p&gt;&#xA;" OwnerUserId="144" LastEditorUserId="144" LastEditorDisplayName="" LastEditDate="2010-08-05T22:04:55.963" LastActivityDate="2010-08-05T22:04:55.963" CommentCount="3" />
  <row Id="1651" PostTypeId="1" CreationDate="2010-08-05T19:18:33.297" Score="1" ViewCount="184" Body="&lt;p&gt;I am seeking a listing of the distinct &lt;a href=&quot;http://mathworld.wolfram.com/HamiltonianCycle.html&quot; rel=&quot;nofollow&quot;&gt;Hamiltonian cycles&lt;/a&gt; following the edges of the icosahedron and the dodecahedron.  By &lt;em&gt;distinct&lt;/em&gt; I mean they are not congruent by some symmetry&#xA;of the icosahedron or dodecahedron (respectively). So they do not make the same sequence of angular turns. For example (as Gerhard corrected me in the comments), there is just one distinct Hamiltonian cycle on the cube.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hamiltonian cycles of the Platonic solids are all over the web, but I am not finding a definitive list of the number and a description of each.  Thanks to anyone who can point me in the right direction!&lt;/p&gt;&#xA;" OwnerUserId="237" LastEditorUserId="237" LastEditorDisplayName="" LastEditDate="2010-08-05T22:38:20.607" LastActivityDate="2010-08-05T22:38:20.607" Title="Distinct Hamiltonian cycles of the icosahedron and dodecahedron" Tags="&lt;geometry&gt;&lt;reference-request&gt;&lt;euclidean-geometry&gt;&lt;graph-theory&gt;&lt;discrete-geometry&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="1652" PostTypeId="2" ParentId="1647" CreationDate="2010-08-05T19:20:20.717" Score="0" ViewCount="" Body="&lt;p&gt;I don't think the radius has to involve the variance. You could just use the radius to scale the x-axis. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you could sample the distribution from -1 to 1, always. If the radius is 4, would sample 9 regions. For 10 you sample 21 times between -1 and 1. The variance can stay the same. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might want to try this question here too &lt;a href=&quot;http://gamedev.stackexchange.com/&quot;&gt;http://gamedev.stackexchange.com/&lt;/a&gt;. &lt;/p&gt;&#xA;" OwnerUserId="99" LastActivityDate="2010-08-05T19:20:20.717" />
  <row Id="1653" PostTypeId="2" ParentId="1651" CreationDate="2010-08-05T19:28:05.820" Score="2" ViewCount="" Body="&lt;p&gt;There is a listing of a hamilton cycle count of 2560 at this website: &lt;a href=&quot;http://mathworld.wolfram.com/IcosahedralGraph.html&quot; rel=&quot;nofollow&quot;&gt;http://mathworld.wolfram.com/IcosahedralGraph.html&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="771" LastActivityDate="2010-08-05T19:28:05.820" CommentCount="1" />
  <row Id="1654" PostTypeId="2" ParentId="1388" CreationDate="2010-08-05T19:34:55.197" Score="2" ViewCount="" Body="&lt;p&gt;Solving $x^{11} = 1$ in radicals was first accomplished by Vandermonde, whose method is described on p.24-26 of Edwards' book on Galois theory:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://books.google.com/books?id=0bH6SUHSvloC&amp;amp;pg=PA24&quot; rel=&quot;nofollow&quot;&gt;http://books.google.com/books?id=0bH6SUHSvloC&amp;amp;pg=PA24&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="467" LastActivityDate="2010-08-05T19:34:55.197" />
  <row Id="1655" PostTypeId="2" ParentId="1442" CreationDate="2010-08-05T19:44:33.743" Score="8" ViewCount="" Body="&lt;p&gt;Consider &lt;em&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Leibniz_harmonic_triangle&quot; rel=&quot;nofollow&quot;&gt;Leibnitz harmonic triangle&lt;/a&gt;&lt;/em&gt; — a table that is like &amp;laquo;Pascal triangle reversed&amp;raquo;: on it's sides lie numbers $\frac{1}{n}$ and each number is the sum of two beneath it (see the &lt;a href=&quot;http://upload.wikimedia.org/math/5/8/1/581098eb5e9213bf6c66e932ed218e08.png&quot; rel=&quot;nofollow&quot;&gt;picture&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One can easily proove by induction that m-th number in n-th row of Leibnitz triangle is $\frac{1}{(n+1)\binom{n}{m}}$. So LHS of our identity is just lcd of fractions in n-th row of the triangle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But it's not hard to see that any such number is an integer linear combination of fractions on triangle's sides (i.e. $1/1,1/2,\dots,1/n$) — and vice versa. So LHS is equal to $lcd(1/1,\dots,1/n)$ — and that is exactly RHS.&lt;/p&gt;&#xA;" OwnerUserId="152" LastActivityDate="2010-08-05T19:44:33.743" CommentCount="2" />
  <row Id="1656" PostTypeId="1" AcceptedAnswerId="1671" CreationDate="2010-08-05T19:51:01.563" Score="1" ViewCount="54" Body="&lt;p&gt;Say P ~ Q (P and Q are «projectively equivalent») iff there is a projective transformation f such that f(P) = Q.  Then ~ is an equivalence relation.  I read that the space of inscribed n-gons modulo projective equivalence has dimension n-3.  Why is this?  Also, are there any related results?&lt;/p&gt;&#xA;" OwnerUserId="314" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-08-05T23:08:49.410" LastActivityDate="2010-08-05T23:08:49.410" Title="Space of inscribed n-gons modulo projective transformations." Tags="&lt;geometry&gt;&lt;projective-geometry&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="1657" PostTypeId="2" ParentId="1649" CreationDate="2010-08-05T19:53:53.507" Score="1" ViewCount="" Body="&lt;p&gt;A simple example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $F_n(x) = \sqrt{x^2+2^{-n}}$. It is not hard to show that $F_n(x) \to \sqrt{x^2} = |x|$.&#xA;Every $F_n$ is differentiable and has a local minimum at 0, and indeed so does |x|.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me know if this is what you're looking for.&lt;/p&gt;&#xA;" OwnerUserId="26" LastActivityDate="2010-08-05T19:53:53.507" CommentCount="5" />
  <row Id="1658" PostTypeId="1" AcceptedAnswerId="1661" CreationDate="2010-08-05T20:07:03.340" Score="5" ViewCount="90" Body="&lt;p&gt;Whenever I think about how a function behaves, I always try to identify a general pattern of behavior with some common numbers (somewhere between 5 and 100 maybe) and then I try to see if anything interesting happens around 1, 0 and into negative numbers if applicable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If that all works out, I essentially assume that I know that the function is going to behave similarly for very large numbers as it does for those relatively small numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there notable (famous, clever or common) functions where very large numbers would cause them to behave significantly differently than would initially be thought if I followed my regular experimental pattern? If so, are there any warning signs I should be aware of?&lt;/p&gt;&#xA;" OwnerUserId="82" LastActivityDate="2010-08-05T20:56:26.957" Title="Function behavior with very large variables" Tags="&lt;functions&gt;" AnswerCount="4" CommentCount="4" FavoriteCount="1" />
  <row Id="1659" PostTypeId="1" AcceptedAnswerId="1663" CreationDate="2010-08-05T20:09:41.823" Score="2" ViewCount="158" Body="&lt;p&gt;Suppose I have a point $x \in \mathbb{R}^n$ on an &lt;a href=&quot;http://en.wikipedia.org/wiki/N-sphere&quot; rel=&quot;nofollow&quot;&gt;n-sphere&lt;/a&gt;. Suppose I divide the n-sphere into 4 sections (I think this makes sense in $n$ dimensions), how do I know which section $x$ lies on? &lt;/p&gt;&#xA;" OwnerUserId="825" LastEditorUserId="825" LastEditorDisplayName="" LastEditDate="2010-08-05T22:50:30.577" LastActivityDate="2010-08-05T22:50:30.577" Title="Locating the quadrant containing a point on an n-sphere" Tags="&lt;geometry&gt;" AnswerCount="1" CommentCount="9" />
  <row Id="1660" PostTypeId="1" AcceptedAnswerId="1719" CreationDate="2010-08-05T20:18:00.743" Score="5" ViewCount="216" Body="&lt;p&gt;Denote by $D(a_1,\dots,a_n)$ the product $\prod_{j&gt;i}(a_j-a_i)$. Assuming that $a_i$ are integers s.t. $a_1\le a_2\le\dots\le a_n$, proove that $D(a_1,...,a_n)/D(1,...,n)$ is the number of &lt;em&gt;Gelfand-Zetlin triangles&lt;/em&gt; (that is, &lt;a href=&quot;http://imrn.oxfordjournals.org/content/vol2010/issue13/images/medium/rnp223ueq3.gif&quot; rel=&quot;nofollow&quot;&gt;triangles&lt;/a&gt; consisting of $\frac{n(n+1)}2$ integers, s.t. each number is greater it's lower-left neighbor but not greater than lower-right neighbor) with the base $a_i$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, for n=3 one needs to prove that number of b&lt;sub&gt;1&lt;/sub&gt;, b&lt;sub&gt;2&lt;/sub&gt;, b' s.t. $a_1\le b_1&amp;lt;a_2\le b_2&amp;lt;a_3$, $b_1\le b'&amp;lt;b_2$ is exactly $\frac{(a_3-a_2)(a_3-a_1)(a_2-a_1)}{3}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As one can guess from the name &amp;ldquo;Gelfand-Zetlin&amp;rdquo;, this fact is well-known in representation theory (namely, in LHS we count dimension of a $gl_n$-representation by Weyl formula, and in RHS we count elements in Gelfand-Zetlin basis of the same representation). But maybe someone can come with more or less direct proof? (Some kind of bijective proof, maybe.)&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h3&gt;Informal probabilistic argument&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;For simplicity, consider the case $n=3$: D(a_1,a_2,a_3) counts the number of triangles s.t. $a_1\le b_1&amp;lt;a_2\le b_2&amp;lt;a_3$, $a_1\le b'&amp;lt;a_3$ — and we're interested only in G-Z triangles, i.e. in triangles s.t. $b_1\le b'&amp;lt; b_2$. Now, mathematical expectation of the length of the interval $(b_1,b_2)$ is exactly one half of the length of the interval from which $b'$ is chosen. So one may &lt;em&gt;expect&lt;/em&gt; that the probability that random triangle is G-Z is $1/2$ — and the answer is indeed $D(a_1,a_2,a_3)/D(1,2,3)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(The main problem with this computation is that we're multiplying probabilities for events that are clearly not independent. And although for n=3 it's not hard to transform this heuristic argument into a formal proof, even for n=4 I failed to do such thing.)&lt;/p&gt;&#xA;" OwnerUserId="152" LastEditorUserId="152" LastEditorDisplayName="" LastEditDate="2010-08-06T07:54:36.650" LastActivityDate="2010-08-19T19:02:35.283" Title="Direct proof of Gelfand-Zetlin identity" Tags="&lt;combinatorics&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="1" />
  <row Id="1661" PostTypeId="2" ParentId="1658" CreationDate="2010-08-05T20:19:02.367" Score="4" ViewCount="" Body="&lt;p&gt;The &lt;a href=&quot;http://www.geatbx.com/docu/fcnindex-01.html#P160_7291&quot; rel=&quot;nofollow&quot;&gt;Griewank function&lt;/a&gt;, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ f(\mathbf x) = \frac1{4000}\sum_{i=1}^n x_i^2 - \prod_{i=1}^n \cos\left(\frac{x_i}{\sqrt i}\right) + 1 $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is one of the objective functions used in testing optimization algorithms, looks completely different in large scale (dominated by &lt;strong&gt;x&lt;/strong&gt;&lt;sup&gt;2&lt;/sup&gt;) and small scale (dominated by cos &lt;em&gt;x&lt;/em&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://www.geatbx.com/docu/fcnindex-msh_f8_500-19.gif&quot; alt=&quot;&quot;&gt;&#xA;&lt;img src=&quot;http://www.geatbx.com/docu/fcnindex-msh_f8_50-20.gif&quot; alt=&quot;&quot;&gt;&#xA;&lt;img src=&quot;http://www.geatbx.com/docu/fcnindex-msh_f8_8-21.gif&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="171" LastActivityDate="2010-08-05T20:19:02.367" />
  <row Id="1662" PostTypeId="2" ParentId="1658" CreationDate="2010-08-05T20:19:13.113" Score="2" ViewCount="" Body="&lt;p&gt;Many rational functions $f(x)=\frac{n(x)}{d(x)}=q(x)+\frac{r(x)}{d(x)}$ (where deg(r) &amp;lt; deg(d)) behave very differently in the general vicinity of the zeros of d(x) than for large (positive or negative) values x.  Near the zeros of d(x), the values of $\frac{r(x)}{d(x)}$ dominate the values of q(x) (that is, f(x) behaves like $\frac{r(x)}{d(x)}$), whereas for large (positive or negative) values of x, the values of q(x) dominate the values of $\frac{r(x)}{d(x)}$ (that is, f(x) behaves like q(x)).&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-08-05T20:19:13.113" />
  <row Id="1663" PostTypeId="2" ParentId="1659" CreationDate="2010-08-05T20:30:29.270" Score="3" ViewCount="" Body="&lt;p&gt;This is just a rephrasing of ShreevastasR's answer; no credit to me.  It does make sense to divide an $n$-sphere into quadrants, as you explain in $\mathbb{R}^3$: partition by two coordinate planes.&#xA;But then deciding which quadrant is, as ShreevastasR says, simply looking at the signs of the coordinates of $x$.  If $x_1$ and $x_2$ are both positive, you are in the first, $++$, quadrant;&#xA;if $x_1$ is negative and $x_2$ positive, you are in the second, $-+$, quadrant. And so on.  If instead you partition the sphere into $2^n$ orthants, then you consider all the signs of the coordinates of $x$.&lt;/p&gt;&#xA;" OwnerUserId="237" LastEditorUserId="237" LastEditorDisplayName="" LastEditDate="2010-08-05T22:40:30.963" LastActivityDate="2010-08-05T22:40:30.963" CommentCount="1" />
  <row Id="1664" PostTypeId="2" ParentId="1658" CreationDate="2010-08-05T20:33:45.130" Score="1" ViewCount="" Body="&lt;p&gt;The &lt;a href=&quot;http://mathworld.wolfram.com/ChebyshevBias.html&quot; rel=&quot;nofollow&quot;&gt;Chebyshev bias&lt;/a&gt; behavior is properly understood only by examining up to very very large numbers.&lt;/p&gt;&#xA;" OwnerUserId="218" LastActivityDate="2010-08-05T20:33:45.130" />
  <row Id="1665" PostTypeId="1" AcceptedAnswerId="1669" CreationDate="2010-08-05T20:39:29.283" Score="2" ViewCount="264" Body="&lt;p&gt;Let $p$ be a prime. If $\frac{p-1}{4}$ and $\frac{p+1}{2}$ are also primes then prove that $p=13$.&lt;/p&gt;&#xA;" LastActivityDate="2010-08-06T15:40:28.367" Title="Prime satisfying a given condition" Tags="&lt;number-theory&gt;" AnswerCount="3" CommentCount="3" />
  <row Id="1666" PostTypeId="2" ParentId="1658" CreationDate="2010-08-05T20:44:45.497" Score="2" ViewCount="" Body="&lt;p&gt;The values of a function $f(x)$ for relatively small values of its argument $x$ is &lt;em&gt;typically&lt;/em&gt; a very bad predictor of asymptotic behavior of $f(x)$ for large $x$. This is true even when $f(x)$ is an &lt;a href=&quot;http://en.wikipedia.org/wiki/Analytic_function&quot; rel=&quot;nofollow&quot;&gt;analytic function&lt;/a&gt; which is uniquely determined by its values on any small interval $x\in[-\epsilon,\epsilon]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Have a look at this excerpt from &lt;a href=&quot;http://books.google.co.uk/books?id=pntQAAAAMAAJ&amp;amp;q=concrete+mathematics+Ronald+L.+Graham,+Donald+Ervin+Knuth,+Oren+Patashnik&amp;amp;dq=concrete+mathematics+Ronald+L.+Graham,+Donald+Ervin+Knuth,+Oren+Patashnik&amp;amp;hl=en&amp;amp;ei=pSBbTJ7nHcj64Aapq-38AQ&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=1&amp;amp;ved=0CC4Q6AEwAA&quot; rel=&quot;nofollow&quot;&gt;&quot;Concrete Mathematics&quot;&lt;/a&gt; for a (not the worst possible) example of how deceptive the &quot;small argument values&quot;  intuition could be.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;It helps to cultivate an expansive attitude when we're doing asymptotic analysis: We should think big, when imagining a variable that approaches infinity. For example, the hierarchy says that $\log n\prec n^{0.0001}$; this might seem wrong if we limit our horizons to teeny-tiny numbers like one googol, $n = 10^{100}$. For in that case, $\log n = 100$, while $n^{0.0001}$ is only $10^{0.01}\approx 1.0233$. But if we go up to a googolplex, $n = 10^{10^{100}}$, then $\log n = 10^{100}$ pales in comparison with $n^{0.0001} = 10^{10^{96}}$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="723" LastEditorUserId="723" LastEditorDisplayName="" LastEditDate="2010-08-05T20:56:26.957" LastActivityDate="2010-08-05T20:56:26.957" />
  <row Id="1667" PostTypeId="2" ParentId="1649" CreationDate="2010-08-05T20:56:58.657" Score="0" ViewCount="" Body="&lt;p&gt;Here's a stab at why the problem is hard for highly non-differentiable but continuous functions. So say $f$ is nowhere differentiable on $[a,b]$. I claim that there are either infinitely or no local extrema of $f$. (By local extrema, I mean extrema that occur in the interior of the interval $[a,b]$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Indeed, suppose there were finitely many, say $c_1 &amp;lt; c_2&amp;lt;\dots f(c_2)$. Then we can take the global maximum of $f$ on $[c_1, c_2]$, which occurs at an &lt;em&gt;interior&lt;/em&gt; point; it is thus a local extremum of $f$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $f$ is a monotone function, then it is a theorem of Lebesgue that it is differentiable a.e. In particular, the above reasoning shows that the existence of finitely many local extrema implies that $f$ is differentiable a.e.&lt;/p&gt;&#xA;" OwnerUserId="536" LastEditorUserId="536" LastEditorDisplayName="" LastEditDate="2010-08-06T11:46:42.377" LastActivityDate="2010-08-06T11:46:42.377" CommentCount="4" />
  <row Id="1668" PostTypeId="2" ParentId="1649" CreationDate="2010-08-05T21:21:07.437" Score="2" ViewCount="" Body="&lt;p&gt;The approach you outlined is commonly used in practice.  If your original problem has some nice properties, such as convexity, the approach will work well.  For example, the &lt;a href=&quot;http://www.johndcook.com/blog/2010/01/13/soft-maximum/&quot; rel=&quot;nofollow&quot;&gt;soft maximum&lt;/a&gt; is a common way to construct a series of smooth, convex approximations to the maximum function. &lt;/p&gt;&#xA;" OwnerUserId="247" LastActivityDate="2010-08-05T21:21:07.437" CommentCount="1" />
  <row Id="1669" PostTypeId="2" ParentId="1665" CreationDate="2010-08-05T21:22:35.970" Score="3" ViewCount="" Body="&lt;p&gt;nice one.&#xA;For (p-1)/4 to be odd, p must be of the form 8k+5, so (p-1)/4 is of the form 2k+1 and (p+1)/2 of the form 4k+3.&lt;br&gt;&#xA;If k is = 0,1,2 (mod 3) then the three numbers are respectively congruent to&lt;br&gt;&#xA;k=0(mod 3): 2,1,0 (mod 3)&lt;br&gt;&#xA;k=1(mod 3): 1,0,1 (mod 3)&lt;br&gt;&#xA;k=2(mod 3): 0,2,2 (mod 3)&lt;br&gt;&#xA;This means that the only way all three of them are prime numbers is that one of them is 3. &#xA;For k=0 we have 5,1,3 which is ruled out; for k=1 we have 13,3,7 which satisfies hypotheses; for k&gt;1 all numbers are &gt; 3.&#xA;The only other case to be checked is (p-1)/4 = 2; in this case however p = 9, so this could not be a solution.&lt;/p&gt;&#xA;" OwnerUserId="89" LastActivityDate="2010-08-05T21:22:35.970" />
  <row Id="1670" PostTypeId="2" ParentId="1660" CreationDate="2010-08-05T22:03:21.940" Score="0" ViewCount="" Body="&lt;p&gt;Just to make sure: you know that what you are asking for is actually Weyl's dimension formula for the highest weight representation in the case of SL(n,C) resp. SU(n)? I suppose there is no direct combinatorial proof that does not make use of this fact.&lt;/p&gt;&#xA;" LastActivityDate="2010-08-05T22:03:21.940" CommentCount="2" />
  <row Id="1671" PostTypeId="2" ParentId="1656" CreationDate="2010-08-05T22:54:57.777" Score="2" ViewCount="" Body="&lt;p&gt;Of the projective transformations fixing a conic, there is a unique one sending any given ordered triple of points to any other given ordered triple of points.  So you are free to determine the location of the first 3 vertices of the n-gon, modulo projective equivalence, but any two placements of the remaining (n-3) points are projectively inequivalent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, the space of n-gons up to projective equivalence can be thought of as the space of (n-3) points on a projective line (or conic, it is isomorphic) punctured at three given points.  This has dimension (n-3).&lt;/p&gt;&#xA;" OwnerUserId="467" LastActivityDate="2010-08-05T22:54:57.777" />
  <row Id="1672" PostTypeId="2" ParentId="1601" CreationDate="2010-08-05T22:56:14.577" Score="3" ViewCount="" Body="&lt;p&gt;The (tight) lower bound is $n + \lceil \lg n \rceil - 2$ (where $\lg n$ means $\log_2 n$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll prove tightness first: that this can be achieved (apparently the idea is due to Lewis Carroll!). First find the maximum using a &quot;tennis tournament&quot; structure: first compare the $n$ elements in pairs, then compare the $n/2$ &quot;winners&quot; in pairs, and so on. Since every element except the winner loses exactly once, this takes $n-1$ comparisons. But now note that the second largest element must be one which lost to the winner, as it couldn't have been defeated by any other element. So you need to find the maximum among all the $\lg n$ elements that were defeated by the winner, and finding this maximum can be done in $\lg n - 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can prove this is a lower bound as well. Firstly, you need to find the maximum, since  cannot be sure some element is the second maximum without knowing which element is the maximum. Further, for each element that lost to the maximum ($m$ of them, say), this comparison was useless in determining whether it was or not the second maximum. In other words, all elements other than the maximum must lose at least once, &lt;em&gt;and&lt;/em&gt; all but one ($m-1$) of the elements that directly lost to the maximum must lose at least once more, no matter in which order the comparisons were made. So we need at least $n-1 + m-1$ comparisons. (To restate: $n-2$ of them must be found less than the second-largest element — comparisons with the largest element do not help here — plus $m$ of them must lose directly to the maximum by definition.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;An adversary can make sure you always have at least $\lg n$ elements that lost to the winner (proved &lt;a href=&quot;http://www.cse.unsw.edu.au/~cs2011/lect/2711_Adversary.pdf&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; or &lt;a href=&quot;http://euler.slu.edu/~goldwasser/class/loyola/comp363/2003_Spring/handouts/selectionproofs.pdf&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;), so this proves the $n + \lg n  - 2$ lower bound. The adversarial argument for the lower bound: the adversary keeps for each element a list of elements known to be less than or equal to it. Whenever a query's result is already &quot;known&quot;, the adversary gives that answer, else it declares the one which has a larger list the winner (breaking ties arbitrarily). The maximum is determined when the size of its list grows to n, and with each comparison the size can only double (can prove this…), so the number of comparisons was at least $\lg n$.&lt;/p&gt;&#xA;" OwnerUserId="205" LastEditorUserId="205" LastEditorDisplayName="" LastEditDate="2010-08-06T08:47:12.300" LastActivityDate="2010-08-06T08:47:12.300" CommentCount="5" />
  <row Id="1673" PostTypeId="2" ParentId="1648" CreationDate="2010-08-05T23:17:39.257" Score="1" ViewCount="" Body="&lt;p&gt;I'd vote for a &lt;a href=&quot;http://en.wikipedia.org/wiki/Poisson_distribution&quot; rel=&quot;nofollow&quot;&gt;Poisson distribution&lt;/a&gt; with a constant offset. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A hand-wavy reasoning might be that the round trip time is due to a constant offset being the best-case round-trip time assuming no delays due to router queues (= wave propagation velocity over physical distance, + minimum processing time), with &quot;rare events&quot; (see wikipedia page) corresponding to queueing delays in one or more routers that make up the network path(s).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as parameter estimation goes, I'm not familiar with how to do it for samples taken from a (suspected) Poisson distribution, but I'm sure you could find something on the Internet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;aha, here we go: &lt;a href=&quot;http://en.wikipedia.org/wiki/Poisson_distribution#Parameter_estimation&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Poisson_distribution#Parameter_estimation&lt;/a&gt; -- you could use this after subtracting off the minimum of a large number of samples.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;drat, stupid me, I glossed over the fact that Poisson = discrete probability distribution. &lt;/p&gt;&#xA;" OwnerUserId="120" LastActivityDate="2010-08-05T23:17:39.257" CommentCount="2" />
  <row Id="1674" PostTypeId="1" AcceptedAnswerId="1679" CreationDate="2010-08-05T23:30:46.287" Score="1" ViewCount="108" Body="&lt;p&gt;If X and Y are normed spaces I know that the space of a bounded linear functions from X to Y , B(X,Y),  is complete if Y is complete.  Is there an example of a normed spaces X,Y s.t. B(X,Y) is not complete?  &lt;/p&gt;&#xA;" OwnerUserId="832" LastActivityDate="2010-08-06T00:19:56.443" Title="Bounded Linear Operators" Tags="&lt;real-analysis&gt;" AnswerCount="1" />
  <row Id="1675" PostTypeId="1" AcceptedAnswerId="1690" CreationDate="2010-08-05T23:34:27.810" Score="2" ViewCount="94" Body="&lt;p&gt;In DotA, there is a character called &quot;&lt;a href=&quot;http://www.dotastrategy.com/hero-71-MogulKhanAxe.html&quot; rel=&quot;nofollow&quot;&gt;Axe&lt;/a&gt;&quot;. Every time he is attacked, he has a chance to spin (17%) his blade and deal damage based on what level the skill is, 100 / 125 / 150 / 175 damage for levels 1, 2, 3, and 4. When the spin activates, it triggers the cooldown of 0.7 / 0.65 / 0.6 / 0.55 seconds, so that attacking Axe does not generate a chance to spin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was trying to calculate an average damage per second that this skill generates, given that most of the time you will find Axe taking around 5 attacks per second (average 1.667 attacks per second from creeps, 3 creeps per camp), so I figured that would mean the probability of him spinning in any second is 1 - 0.83&lt;sup&gt;5&lt;/sup&gt;, so given the damage from earlier, we should expect him to deal 60.61 / 75.76 / 90.91 / 106.07 damage per second?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I got this calculation, but then I realized that I have to factor in the cooldown somewhere, but I have no idea where to start. &lt;/p&gt;&#xA;" OwnerUserId="850" LastEditorUserId="850" LastEditorDisplayName="" LastEditDate="2010-08-06T03:44:01.317" LastActivityDate="2010-08-06T03:44:01.317" Title="How can I calculate this expected rate?" Tags="&lt;statistics&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="1676" PostTypeId="1" CreationDate="2010-08-05T23:38:10.800" Score="9" ViewCount="337" Body="&lt;p&gt;Somewhat related to &lt;a href=&quot;http://math.stackexchange.com/questions/1249/motivating-example-for-algebraic-geometry-scheme-theory&quot;&gt;this question&lt;/a&gt;, but almost infinitely more basic.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;A Confession&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;I am, should classification prove essential, a differential geometer and a topologist by inclination and by training: as an undergraduate I shunned any ring that wasn't $\mathbb{Z}_n$ or a ring of differential operators and held close the differentiable and the non-singular. It did not seem to matter then that these exotic 'schemes' and their exciting projective morphisms were beyond me, and to a certain extent it does not seem to matter now; but increasingly my old uni friends, fellow MOers (and, hey, even math.stack exchangers) are talking about nothing else but schemes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In recent months (after a &lt;a href=&quot;http://mathoverflow.net/questions/27971/why-is-there-no-cayleys-theorem-for-rings&quot; rel=&quot;nofollow&quot;&gt;frighteningly eye-opening MO question&lt;/a&gt;) I have found myself becoming more amenable to rings, and am less daunted by my paucity of understanding than previously. In spite of this, though, I remain &lt;strong&gt;entirely&lt;/strong&gt; in the dark about schemes.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Where I Sit&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;I am not a complete novice. I completed an undergraduate course in algebraic geometry: illuminating, interesting, but all classical beyond belief. I have read and re-read the &lt;a href=&quot;http://en.wikipedia.org/wiki/Scheme_%28mathematics%29&quot; rel=&quot;nofollow&quot;&gt;wikipedia page&lt;/a&gt; on schemes several times- taking in all of the neccessary components: &lt;a href=&quot;http://en.wikipedia.org/wiki/Spectrum_of_a_ring&quot; rel=&quot;nofollow&quot;&gt;the spectrum of a ring&lt;/a&gt;, &lt;a href=&quot;http://en.wikipedia.org/wiki/Locally_ringed_space&quot; rel=&quot;nofollow&quot;&gt;a locally ringed space&lt;/a&gt; et al, but have no idea how these fit together to make the objects I fiddled with over a semester two years ago.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have made numerous guesses about generalised nulstellensatze and structure sheaves, but to explain any would probably be to complicate matters further unneccessarily. I am aware there are probably brilliant texts that do exactly what I am asking for, but I am not currently affiliated to a university and my current library would require ordering in, which for the sort of toe-dipping excercise I intend here would be overkill. So I ask:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Can anyone provide me with a canonical example of a scheme, pointing along the way the topology and the spectra associated to each open set. Perhaps deeper, if it pleases: what I am looking for is a sort of 'scheme jargon safari'.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am aware this is silly, and perhaps asking for a verbatim quotation of page 2 of any decent algebraic geometry text, but I would be ever so grateful. Can anyone help? &lt;/p&gt;&#xA;" OwnerUserId="160" LastEditorUserId="160" LastEditorDisplayName="" LastEditDate="2010-08-05T23:50:37.597" LastActivityDate="2010-12-06T18:26:48.483" Title="An example of a scheme in the language of schemes" Tags="&lt;intuition&gt;&lt;algebraic-geometry&gt;" AnswerCount="5" CommentCount="3" FavoriteCount="5" />
  <row Id="1678" PostTypeId="2" ParentId="1618" CreationDate="2010-08-06T00:14:48.557" Score="1" ViewCount="" Body="&lt;p&gt;EDIT: Here is a cleaned-up and corrected version of this answer, based on Pierre-Yves' suggestion (thanks!).  His answer above contains a much more complete version.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $\sum_{i=1}^n a_{i} f_i \otimes g_i$ is not the zero element of $L^p(X) \otimes L^p(Y)$, we may assume without loss of generality that the $f_i$ are linearly independent.  We can also assume that $a_1 \ne 0$ and $g_1 \ne 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose that the corresponding function $F(x,y) = \sum_{i=1}^n a_{i} f_i(x) g_i(y) = 0$ a.e.  Since $g_1 \ne 0$, there is a measurable $B \subset Y$ of positive finite measure such that $\int_B g_1 \ne 0$ (the integral is finite by H&amp;ouml;lder).    Then by Fubini's theorem, for a.e. $x$ we have&#xA;$$&#xA;0 = \int_{B} F(x,y)dy = \sum_{i=1}^n a_i \left(\int_{B} g_i\right) f_i(x).&#xA;$$&#xA;This contradicts the assumed linear independence of the $f_i$.&lt;/p&gt;&#xA;" OwnerUserId="822" LastEditorUserId="822" LastEditorDisplayName="" LastEditDate="2010-08-10T19:56:55.087" LastActivityDate="2010-08-10T19:56:55.087" CommentCount="1" />
  <row Id="1679" PostTypeId="2" ParentId="1674" CreationDate="2010-08-06T00:19:56.443" Score="2" ViewCount="" Body="&lt;p&gt;Let $X = \mathbb{R}$ with the Euclidean norm and let $Y$ be a normed space which is not complete.  You should find that $B(X, Y) \simeq Y$.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-08-06T00:19:56.443" CommentCount="1" />
  <row Id="1680" PostTypeId="2" ParentId="1676" CreationDate="2010-08-06T00:24:09.603" Score="10" ViewCount="" Body="&lt;p&gt;Dear Tom,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am going to consider everything here over the field $\mathbb C$.  You can replace $\mathbb C$ by any algebraically closed field (or even any field) with essentially no changes, but working over $\mathbb C$ is the natural starting point, and has the advantage that one can connect with the kind of geometry/topololgy with which you are more familiar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given a classical variety $V$, you can consider all the closed subvarieties.  These satisfy the axioms of a topology, called the Zariski topology.   For definiteness, let's say that our&#xA;variety is an affine variety, so it is being cut out by polynomials in $\mathbb C^n$, for some $n$.  The usual topology on $\mathbb C^n$ induces a topology on $V$, which has many more open sets than the Zariski topology (unless $V$ is $0$-dimensional).  The point is that to be closed in the Zariski topology, you have to really be the zero locus of some polynomial, i.e. another variety, so it is hard to be Zariski closed, and hence similarly hard to be Zariski open.    (Just to be absolutely clear, let's look at an example: the real line is closed in $\mathbb C$, but is not Zariski closed; there is no polynomial in one variable over $\mathbb C$ that vanishes precisely on the points of the real line; indeed,&#xA;such a polynomial either vanishes at only finitely many points, or else is identically zero, &#xA;and so vanishes everywhere.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You also have the notion of rational function on the variety (just think of the restriction of a ratio of polynomials in $n$-variables to $V$, such that the denominator does not vanish&#xA;identically on $V$); a rational function is called &lt;em&gt;regular&lt;/em&gt; at a point $P$ of the variety if it has no singularity at that point.  Being a singularity is a Zariski closed condition (singularities occur where the denominator of the rational function, which is a polynomial, vanish), so being regular at a point is a Zariski open condition.  If we fix a Zariski open&#xA;set in advance, we can look at the ring of all rational functions that are regular on that&#xA;open set.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;These form a sheaf on $V$ (with its Zariski topology).  It is much &quot;smaller&quot; than the sheaves you are used to, like smooth or continuous functions.  Not only are there many&#xA;fewer open sets to think about (just the Zariski open ones), but on a given open set,&#xA;there will be incredibly more continuous or smooth functions than regular functions, just because being the ratio of polynomials is a very restrictive condition on a function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we look at the global sections of this sheaf, i.e. the functions that are regular on the whole of $V$, we in particular get a ring which is called the affine ring of $V$.  If I just hand you this ring (as a $\mathbb C$-algebra), it turns out that you can recover $V$, namely&#xA;$V$ is the maximal spectrum of this ring (i.e. point of $V$ are in natural bijection with&#xA;maximal ideals of $V$).  The map one way is easy: given a point, we can look at all regular functions on $V$ that vanish at the point; this gives a maximal ideal in the ring of all&#xA;regular functions.  That this is a bijection is harder, and is essentially equivalent to the Nullstellensatz.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To see the role of the entire spectrum of the ring (i.e. the prime ideals as well as the maximal ideals) one has to say and think about more, but this is probably enough for now. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To learn more, you should google &quot;affine ring of a variety&quot; or similar expressions, and you should find troves of information, at a great range of levels.  Once you understand this &#xA;basic connection, it makes sense to look at schemes in more detail.&lt;/p&gt;&#xA;" OwnerUserId="221" LastEditorUserId="221" LastEditorDisplayName="" LastEditDate="2010-12-06T16:38:23.850" LastActivityDate="2010-12-06T16:38:23.850" CommentCount="1" />
  <row Id="1681" PostTypeId="2" ParentId="1676" CreationDate="2010-08-06T00:24:12.847" Score="5" ViewCount="" Body="&lt;p&gt;$\text{Spec } \mathbb{C}[x]$ is probably as basic as it gets.  As a set, this is the collection of maximal ideals $(x - a), a \in \mathbb{C}$ together with the prime ideal $(0)$.  As a topological space, this is $\mathbb{C}$ in the cofinite topology (the closed sets are the finite ones) together with a point $(0)$ whose closure is the entire space (and which is in every open set except the empty set).  The local ring at $(x - a)$ is the subring of $\mathbb{C}(x)$ of rational functions which are defined at $a$, and the local ring at $(0)$ is all of $\mathbb{C}(x)$; more generally, the section of the structure sheaf over a Zariski-open set $U$ is the subring of $\mathbb{C}(x)$ of rational functions which are defined at every $a \in U$.  &lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-08-06T00:24:12.847" CommentCount="1" />
  <row Id="1682" PostTypeId="2" ParentId="1676" CreationDate="2010-08-06T00:33:40.130" Score="9" ViewCount="" Body="&lt;p&gt;You should read David Eisenbud and Joe Harris's &lt;em&gt;The Geometry of Schemes&lt;/em&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Really.&lt;/em&gt; :)&lt;/p&gt;&#xA;" OwnerUserId="274" LastEditorUserId="274" LastEditorDisplayName="" LastEditDate="2010-08-06T01:08:58.673" LastActivityDate="2010-08-06T01:08:58.673" CommentCount="3" />
  <row Id="1683" PostTypeId="2" ParentId="709" CreationDate="2010-08-06T00:53:35.193" Score="2" ViewCount="" Body="&lt;p&gt;Echoing Americo's and Isaac's answers, but without appealing to the quadratic equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First treat the special case f(x)=ax²+d (i.e. no &quot;x&quot; term); by remembering that the line that divides the parabola symmetrically passes through the vertex, and is in fact the y-axis (x=0), we see that 0 is the abscissa of the vertex.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To treat the more general f(x)=ax²+bx+c, we try a substitution of the form x=x'-h (geometrically, this corresponds to translating your coordinate system horizontally by h units). Expanding this, you will get something like a(x')²+(2ah+b)x'+(constant term). To go back to the special case we first treated, we find the value of h such that the coefficient of x' is 0. Thus you obtain the expression for the vertex's x-coordinate.&lt;/p&gt;&#xA;" OwnerUserId="498" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-08-06T01:28:54.390" LastActivityDate="2010-08-06T01:28:54.390" />
  <row Id="1684" PostTypeId="2" ParentId="1676" CreationDate="2010-08-06T01:01:55.487" Score="3" ViewCount="" Body="&lt;p&gt;Well, Qiaochu lists the  affine line, so I'm going to do the projective line. (Everything here is over the complex numbers.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The projective line $P^1$, or equivalently the Riemann sphere.  This is the space of all lines through the origin in $\mathbb{C}^2$.  Recall that the Riemann sphere can be obtained by gluing two copies of the complex plane. Namely, if $S^2 = \mathbb{C} \cup { \infty }$, then one copy is just the subset $\mathbb{C}$. The other copy is $\mathbb{C}^* \cup {\infty}$, which is identified with the complex numbers by taking reciprocals. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is obviously a Riemann surface.  In fact, the maps defining this &quot;gluing&quot; can be checked to be algebraic (they're just reciprocals), so it is in fact a scheme, and a toy example which isn't affine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What's a regular function over an open subset of $S^2$? Well, one defines a holomorphic function over an open subset of $S^2$ by saying that the pull-back to each chart is holomorphic. Now here we say that the pull-back to each chart is regular in the sense of its being a rational function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Interestingly, there are only constant functions which are regular on all of $P^1$. Indeed, such a function (considered as a map $P^1 \to \mathbb{C}$) would be a holomorphic map from a compact Riemann surface, hence constant. Here's an algebraic argument. Consider the open set $\mathbb{C}^*$. On this (affine) open set, it can be checked that the only regular functions are polynomials in $z $ and $1/z$ (or otherwise the denominator would blow up).  However, if the function is regular everywhere, then $1/z$ can't occur (that would blow up at the origin) and $z$ can't occur (that isn't defined at ${\infty}$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The projective line is important because it is compact in the complex topology. The algebraic version of this is that it is &lt;em&gt;proper&lt;/em&gt; over $\mathbb{C}$. In particular, any map out of it into a complex variety is a &lt;em&gt;emph&lt;/em&gt; closed map. However, with affine varieties, you don't have this closed property anymore. For instance, the hyperbola $xy=1$ in the affine plane (clearly a closed set in the Zariski topology) projects to the non-open complement of the origin of $A^1$ via $(x,y) \to x$. This doesn't happen for the projective line.  (&quot;Proper&quot; is an algebraic analog of &quot;compact,&quot; just as &quot;separated&quot; is an analog of &quot;Hausdorff.&quot;)&lt;/p&gt;&#xA;" OwnerUserId="536" LastActivityDate="2010-08-06T01:01:55.487" CommentCount="1" />
  <row Id="1685" PostTypeId="1" AcceptedAnswerId="1697" CreationDate="2010-08-06T01:04:40.570" Score="4" ViewCount="189" Body="&lt;p&gt;Are there books or article that develop (or sketch the main points) of Euclidean geometry without fudging the hard parts such as angle measure, but might at times use coordinates, calculus or other means so as to maintain rigor or avoid the detail involved in Hilbert-type axiomatizations?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am aware of Hilbert's foundations and the book by Moise.   I was wondering if there is anything more modern that tries to stay (mostly) in the tradition of synthetic geometry.   &lt;/p&gt;&#xA;" OwnerUserId="467" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-08-06T01:22:47.757" LastActivityDate="2010-08-06T06:59:41.937" Title="Rigorous synthetic geometry without Hilbert axiomatics" Tags="&lt;geometry&gt;&lt;euclidean-geometry&gt;&lt;trigonometry&gt;" AnswerCount="1" />
  <row Id="1686" PostTypeId="2" ParentId="1563" CreationDate="2010-08-06T01:46:44.667" Score="10" ViewCount="" Body="&lt;p&gt;First, I'm glad you are reading my book! :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me make a couple of comments on Pete's answer--this is my first time here and I don't see how to leave comments.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Any two dense linear orders without endpoints are elementarily equivalent.&#xA;In particular $(Q,&amp;lt;)$ and $(R,&amp;lt;)$ are elementarily equivalent.  So there is no first order way of expressing the completeness of the reals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) Any two algebraically closed fields of the same characteristic are elementarily equivalent.  So the algebraic numbers&#xA;is elementarily equivalent to the complex numbers.  This means you can prove first order things about the algebraic numbers using complex analysis or the complex numbers using Galois Theory or countability.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;4) Similarly the reals field is elementarily equivalent to the real algebraic numbers or to the field of real Puiseux series.&#xA;One can for example use the Puiseux series to prove asymptotic properties of semialgebraic functions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, Pete's comment 5) about infinite models of the theory of finite fields being elementarily equivalent isn't quite right.  This is only true if the relative algebraic closure of the prime fields are isomorphic.&#xA;For example, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;a) take an  ultrapower of finite fields $F_p$ where the ultrafilter containing {2,4,8,...} then the resulting model has &#xA;characteristic 2, while if the ultrafilter contains the set of primes, then the ultraproduct has characteristic 0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;b) if the ultrapower contains the set of primes congruent to 1 mod 4, in the ultraproduct -1 is a square, while if&#xA;the ultraproduct contains the set of primes congruent to 3 mod 4 then in the ultraproduct -1 is not a square..&lt;/p&gt;&#xA;" OwnerUserId="851" LastEditorUserId="851" LastEditorDisplayName="" LastEditDate="2010-08-08T13:44:06.557" LastActivityDate="2010-08-08T13:44:06.557" CommentCount="5" />
  <row Id="1687" PostTypeId="1" AcceptedAnswerId="1689" CreationDate="2010-08-06T01:52:18.913" Score="2" ViewCount="65" Body="&lt;p&gt;I want to try to emulate what this application can do:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://img530.imageshack.us/img530/5756/bezier.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given the red round dots (from the mouse) it is able to solve for the bezier handles given that the tension of the curve is set to 0.6. How could I solve for the bezier handles? Is there some way I could do this with the slope or perpendicular of the previous and next curve?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;&#xA;&lt;p&gt;*I should also add that each point is equadistant meaning point i + 1 is the distance to point i or point i + 2.&lt;/p&gt;&#xA;" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-08-26T01:32:00.180" LastActivityDate="2010-08-26T01:32:00.180" Title="Generating bezier handles based on constraints" Tags="&lt;geometry&gt;&lt;algebra-precalculus&gt;&lt;bezier-curve&gt;" AnswerCount="1" />
  <row Id="1688" PostTypeId="2" ParentId="240" CreationDate="2010-08-06T01:54:45.447" Score="11" ViewCount="" Body="&lt;p&gt;Over at &lt;a href=&quot;http://mathoverflow.net/questions/12342/reductio-ad-absurdum-or-the-contrapositive/12400#12400&quot; rel=&quot;nofollow&quot;&gt;this MathOverflow question&lt;/a&gt;, I posted the following answer to a similar question (and there are several other interesting answers there):&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;em&gt;With good reason&lt;/em&gt;, we mathematicians prefer a direct proof of an implication over a proof by contradiction, when such a proof is available. (all else being equal)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;What is the reason? The reason is the &lt;em&gt;fecundity&lt;/em&gt; of the proof, meaning our ability to use the proof to make further mathematical conclusions. When we prove an implication (p implies q) directly, we assume p, and then make some intermediary conclusions r&lt;sub&gt;1&lt;/sub&gt;, r&lt;sub&gt;2&lt;/sub&gt;, before finally deducing q. Thus, our proof not only establishes that p implies q, but also, that p implies r&lt;sub&gt;1&lt;/sub&gt; and r&lt;sub&gt;2&lt;/sub&gt; and so on. Our proof has provided us with additional knowledge about the context of p, about what else must hold in any mathematical world where p holds. So we come to a fuller understanding of what is going on in the p worlds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, when we prove the contrapositive (&amp;not;q implies &amp;not;p) directly, we assume &amp;not;q, make intermediary conclusions  r&lt;sub&gt;1&lt;/sub&gt;, r&lt;sub&gt;2&lt;/sub&gt;, and then finally conclude &amp;not;p. Thus, we have also established not only that &amp;not;q implies &amp;not;p, but also, that it implies r&lt;sub&gt;1&lt;/sub&gt; and r&lt;sub&gt;2&lt;/sub&gt; and so on. Thus, the proof tells us about what else must be true in worlds where q fails. Equivalently, since these additional implications can be stated as (&amp;not;r&lt;sub&gt;1&lt;/sub&gt; implies q), we learn about many different hypotheses that all imply q. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;These kind of conclusions can increase the value of the proof, since we learn not only that (p implies q), but also we learn an entire context about what it is like in a mathematial situation where p holds (or where q fails, or about diverse situations leading to q). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;With reductio, in contrast, a proof of (p implies q) by contradiction seems to carry little of this extra value. We assume p and &amp;not;q, and argue  r&lt;sub&gt;1&lt;/sub&gt;, r&lt;sub&gt;2&lt;/sub&gt;, and so on, before arriving at a contradiction. The statements r&lt;sub&gt;1&lt;/sub&gt; and r&lt;sub&gt;2&lt;/sub&gt; are all deduced under the contradictory hypothesis that p and &amp;not;q, which ultimately does not hold in any mathematical situation. The proof has provided extra knowledge about a nonexistant, contradictory land. (Useless!) So these intermediary statements do not seem to provide us with any greater knowledge about the p worlds or the q worlds, beyond the brute statement that (p implies q) alone.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe that this is the reason that sometimes, when a mathematician completes a proof by contradiction, things can still seem unsettled beyond the brute implication, with less context and knowledge about what is going on than would be the case with a direct proof.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For an example of a proof where we are led to false expectations in a proof by contradiction, consider Euclid's proof that there are infinitely many primes. In a common proof by contradiction, one assumes that p&lt;sub&gt;1&lt;/sub&gt;, ..., p&lt;sub&gt;n&lt;/sub&gt; are &lt;em&gt;all&lt;/em&gt; the primes. It follows that since none of them divide the product-plus-one p&lt;sub&gt;1&lt;/sub&gt;...p&lt;sub&gt;n&lt;/sub&gt;+1, that this product-plus-one is also prime. This contradicts that the list was exhaustive. Now, many beginner's falsely expect after this argument that whenever p&lt;sub&gt;1&lt;/sub&gt;, ..., p&lt;sub&gt;n&lt;/sub&gt; are prime, then the product-plus-one is also prime. But of course, this isn't true, and this would be a misplaced instance of attempting to extract greater information from the proof, misplaced becasue this is a proof by contradiction, and that conclusion relied on the assumption that p&lt;sub&gt;1&lt;/sub&gt;, ..., p&lt;sub&gt;n&lt;/sub&gt; were &lt;em&gt;all&lt;/em&gt; the primes. If one organizes the proof, however, as a direct argument showing that whenever p&lt;sub&gt;1&lt;/sub&gt;, ..., p&lt;sub&gt;n&lt;/sub&gt; are prime, then there is yet another prime not on the list, then one is lead to the true conclusion, that p&lt;sub&gt;1&lt;/sub&gt;...p&lt;sub&gt;n&lt;/sub&gt;+1 has merely a prime divisor not on the original list.&lt;/p&gt;&#xA;" OwnerUserId="413" LastActivityDate="2010-08-06T01:54:45.447" CommentCount="2" />
  <row Id="1689" PostTypeId="2" ParentId="1687" CreationDate="2010-08-06T02:00:53.407" Score="2" ViewCount="" Body="&lt;p&gt;From the graphic, it appears that the handles at a given point $p_k$ are parallel to the line through $p_{k-1}$ and $p_{k+1}$ and that the lengths of the handles at every point are equal.  I suspect that the length of the handles is determined by the &quot;tension&quot; that you mention.&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-08-06T02:00:53.407" CommentCount="1" />
  <row Id="1690" PostTypeId="2" ParentId="1675" CreationDate="2010-08-06T02:56:12.237" Score="1" ViewCount="" Body="&lt;p&gt;Let r be the chance an attack caused axe to spin and d be the damage it does. Since we know the number of attacks on Axe per second, we just need to find the expected spin damage per attack, call this x. If someone attacks Axe once, we expect the retaliation damage to be rd. However, each attack blocks a certain number of attacks depending on cooldown, call this number b. This reduces our expected value by brx.  So we have:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;x=rd-brx&#xA;x(1+br)=rd&#xA;x=rd/(1+br)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Note that this assumes the combat goes for an infinite amount of attacks. When the combat is shorter, the expected damage will be higher. Given the speed that combat happens in Dota, this effect will often be significant&lt;/p&gt;&#xA;" OwnerUserId="123" LastActivityDate="2010-08-06T02:56:12.237" />
  <row Id="1691" PostTypeId="1" AcceptedAnswerId="1695" CreationDate="2010-08-06T04:36:39.063" Score="2" ViewCount="160" Body="&lt;p&gt;If we have a curve $C:\{ P(x,y) = 0 \}$ and define $\omega=\frac{\mathrm{d}x}{y}$ then is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_0^A \omega + \int_0^B \omega = \int_0^{A \oplus B} \omega$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(with $\oplus$ being addition on a group on the curve) a theorem?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm quite sure this is what's happening when C is an elliptic curve, but I have failed to make this work out when C is defined by $P(x,y) = x^2 + y^2 - 1$ (unit circle) and the group law is defined by firing a ray through $(3/5,4/5)$ (chosen arbitrarily) parallel to $AB$ and taking it's intersection with the circle as $A \oplus B$.&lt;/p&gt;&#xA;" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-08-06T06:19:12.360" LastActivityDate="2010-08-06T12:17:16.223" Title="Trying to piece together an integral addition theorem" Tags="&lt;calculus&gt;&lt;algebraic-geometry&gt;&lt;conic-sections&gt;&lt;elliptic-curves&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="1692" PostTypeId="2" ParentId="709" CreationDate="2010-08-06T04:51:20.560" Score="3" ViewCount="" Body="&lt;p&gt;Here is the simplest way I know, using a bit of algebra.  Starting with Isaac's observation that the vertex lies on the axis of symmetry, the line x = k.  Question is, what's k? That's your answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;if f(x) = ax&lt;sup&gt;2&lt;/sup&gt; + bx + c&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then the symmetry around x = k means that, for any h &lt;/p&gt;&#xA;&#xA;&lt;p&gt;f(k+h) = f(k-h)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From that, just multiply it all out and some simple additive cancellation gives&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-2akh - bh = 2akh + bh&lt;/p&gt;&#xA;&#xA;&lt;p&gt;h cancels out, and a bit more algebra gives&lt;/p&gt;&#xA;&#xA;&lt;p&gt;k = -$\frac{b}{2a}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To really complete this, however, we should show that x = k = -$\frac{b}{2a}$ is a maximum or minimum point. Again, with more algebra:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;f(k+h) - f(k) boils down to ah&lt;sup&gt;2&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So when a &gt; 0, then f(k+h) &gt; f(k) for any h != 0, establishing that x = k is a minimum point&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And likewise when a &amp;lt; 0, x = k is a maximum. This also shows the symmetry around x = k, and how the coefficient a determines whether the curve is concave upward or downward.&lt;/p&gt;&#xA;" OwnerUserId="271" LastEditorUserId="271" LastEditorDisplayName="" LastEditDate="2010-08-06T12:20:35.610" LastActivityDate="2010-08-06T12:20:35.610" />
  <row Id="1694" PostTypeId="1" CreationDate="2010-08-06T05:56:39.303" Score="8" ViewCount="841" Body="&lt;p&gt;Wikipedia &lt;a href=&quot;http://en.wikipedia.org/wiki/Logarithm&quot; rel=&quot;nofollow&quot;&gt;sez&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The natural logarithm of x is often written &quot;ln(x)&quot;, instead of log_e(x) especially in disciplines where it isn't written &quot;log(x)&quot;. However, some mathematicians disapprove of this notation. In his 1985 autobiography, Paul Halmos criticized what he considered the &quot;childish ln notation,&quot; which he said no mathematician had ever used. In fact, the notation was invented by a mathematician, Irving Stringham, professor of mathematics at University of California, Berkeley, in 1893.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Apparently the notation &quot;ln&quot; first appears in Stringham's book &lt;em&gt;Uniplanar algebra: being part I of a propædeutic to the higher mathematical analysis&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But this doesn't explain why &quot;ln&quot; has become so pervasive. I'm pretty sure that most high schools in the US at least still use the notation &quot;ln&quot; today, since all of the calculus students I come into contact with at Berkeley seem to universally use &quot;ln&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How did this happen?&lt;/p&gt;&#xA;" OwnerUserId="263" LastActivityDate="2010-08-07T21:47:12.533" Title="How did the notation &quot;ln&quot; for &quot;log base e&quot; become so pervasive?" Tags="&lt;notation&gt;&lt;math-history&gt;" AnswerCount="7" CommentCount="7" />
  <row Id="1695" PostTypeId="2" ParentId="1691" CreationDate="2010-08-06T06:01:55.567" Score="2" ViewCount="" Body="&lt;p&gt;$0$, the lower limit for the integrals, should be whatever point on the curve is $0$ for the group law.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then $dx/y$ should be rotation-invariant for any circle. It will not be true for a rotated ellipse; there is an invariant differential but it is not $dx/y$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For an elliptic curve, $dx/y$ being the invariant differential relies on the curve being written as $y^2=f(x)$ with $\deg(f)=3$. If you move the curve the invariant differential will be a different one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For higher genus the curve does not have an addition law.&lt;/p&gt;&#xA;" OwnerUserId="467" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-08-06T06:31:16.967" LastActivityDate="2010-08-06T06:31:16.967" CommentCount="1" />
  <row Id="1697" PostTypeId="2" ParentId="1685" CreationDate="2010-08-06T06:59:41.937" Score="2" ViewCount="" Body="&lt;p&gt;You might look at Hartshorne's&#xA;&lt;a href=&quot;http://books.google.co.uk/books?id=EJCSL9S6la0C&amp;amp;lpg=PP1&amp;amp;dq=hartshorne&amp;amp;pg=PP1#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;Geometry: Euclid and Beyond&lt;/em&gt;&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-08-06T06:59:41.937" CommentCount="0" />
  <row Id="1698" PostTypeId="2" ParentId="1694" CreationDate="2010-08-06T07:06:06.650" Score="1" ViewCount="" Body="&lt;p&gt;One simple possible reason&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;It is very difficult to come up with a&#xA;  notation that is concise, correct and&#xA;  understandable by at least two people.&#xA;  - P A M Dirac / Richard Feynmann&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This one is concise, correct and is understandable at least by you and me! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Generally, notations used in popular textbooks spread like virus. Books become popular because of good notation and notation becomes widespread because book is popular.  &lt;/p&gt;&#xA;" OwnerUserId="705" LastEditorUserId="705" LastEditorDisplayName="" LastEditDate="2010-08-06T07:14:23.187" LastActivityDate="2010-08-06T07:14:23.187" CommentCount="1" />
  <row Id="1699" PostTypeId="2" ParentId="67" CreationDate="2010-08-06T07:16:04.540" Score="4" ViewCount="" Body="&lt;p&gt;Elliptic curve cryptography (ECC), or at least some version of it, essentially follows the same idea as RSA.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Modulo some details, here's how RSA works. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a &quot;public key&quot; $N$ which is a product of two large prime numbers $p$ and $q$. Now pick two numbers $e$ and $d$ such that $ed = 1$ mod $(p-1)(q-1)$. The number $e$ is a &quot;public key&quot;. The number $d$ is a &quot;secret key&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $A$ be our secret message (a number). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then the encrypted message is $B := A^e$ mod $N$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is a fact that $A$ is equal to $B^d$ mod $N$. (It follows by &lt;a href=&quot;http://en.wikipedia.org/wiki/Fermat%27s_little_theorem#Generalizations&quot; rel=&quot;nofollow&quot;&gt;a generalization of Fermat's little theorem&lt;/a&gt;.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus if we know $d$ and $B$, we can decrypt the message (that is, we can recover $A$ from $B$) by computing $B^d$ mod $N$. The point is that, even if an adversary knows $B$, $N$, and $e$, presumably he also needs $d$ to recover the secret message $A$. Moreover, to find $d$ he presumably needs to know $p$ and $q$, and finding $p$ and $q$ is presumably difficult.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;RSA works by exploiting the properties of $\mathbb{Z}/N\mathbb{Z}$ (or actually its group of units). ECC is analogous, except we use an elliptic curve (over a finite field) instead of $\mathbb{Z}/N\mathbb{Z}$. An elliptic curve is, in fancy terms, an example of an &quot;abelian variety&quot;. In less fancy terms, you can think of an elliptic curve as simply a set of &quot;things&quot; with a funny multiplication law $\ast$: that is, given two &quot;things&quot; $a$ and $b$, we can &quot;multiply&quot; them together to get a new &quot;thing&quot; $a \ast b$. Now in ECC, our secret message $A$ is such a &quot;thing&quot;, and the way we encrypt it is by multiplying $A$ with itself $e$ times (using the funny multiplication) to get a new &quot;thing&quot; $B$. The way we decrypt $B$ and recover $A$ is by multiplying $B$ with itself $d$ times (for some appropriate $d$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I hope I am getting this right. I am working mainly from memory. My memory of this stuff might be inaccurate, or flat out wrong...)&lt;/p&gt;&#xA;" OwnerUserId="263" LastEditorUserId="263" LastEditorDisplayName="" LastEditDate="2010-08-06T07:34:18.187" LastActivityDate="2010-08-06T07:34:18.187" CommentCount="1" />
  <row Id="1700" PostTypeId="2" ParentId="1694" CreationDate="2010-08-06T07:39:09.190" Score="7" ViewCount="" Body="&lt;p&gt;It is probably at least in part an instance of consumer lock-in, the phenomenon where the advantages of the majority of the market using the same or compatible products outweigh sufficiently minor differences of one version of the product versus another.  (When I learned this term, the canonical example was VHS versus Beta.  I am just about old enough to remember that when VCR terminology first came out, vendors would carry both.  Already when the first video stores opened up there was more product available in VHS, so a lot of stores would have just one rack with all the Beta products.  And eventually of course Beta died.  However those who remember and care seem to largely agree that Beta was the superior technology.  Apologies for giving such an old-fogey example.  Can someone suggest something more current?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, when it comes to electronic calculators, having everyone agree what's going to happen when you press a certain button is a good thing.  (In fact, another lock-in phenomenon is that when I was in high school in the early 90's, most students had Texas Instruments calculators of one kind or other.  Among the real geeks it was known that the &quot;cadillac of calculators&quot; was actually the Hewlett-Packard, which used &lt;a href=&quot;http://en.wikipedia.org/wiki/Reverse_Polish_notation&quot; rel=&quot;nofollow&quot;&gt;reverse polish notation&lt;/a&gt;.  Serious CS people appreciate RPN, but the problem is that if you're a high school kid and pick up such a calculator for the first time, it's very hard to figure out what's going on.  I haven't seen an HP calculator for many years.)  The notation $\ln$ is simple and unambiguous: you don't have to like it (and I don't, especially), but you know what it means, and it's easier to fit on a small calculator key than $\log_e$.  I think if you're first learning about logarithms, then base ten is probably the simplest (to have any clue what $e$ is other than &quot;about $2.71828...$&quot; requires some calculus, and is in my experience one of the more subtle concepts of first year calculus), so it's reasonable to have that be the standard base for the logarithm for a general audience.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, I'm sure everyone here knows this but I wish my calculus students had a better appreciation of it: exactly what base you take for the logarithm only matters up to a multiplicative constant anyway, since $\log_c x = \frac{\log_a x}{\log_a c}$.  So it's no big deal either way.&lt;/p&gt;&#xA;" OwnerUserId="299" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2010-08-06T08:51:54.617" LastActivityDate="2010-08-06T08:51:54.617" CommentCount="6" />
  <row Id="1701" PostTypeId="1" AcceptedAnswerId="1708" CreationDate="2010-08-06T07:43:00.133" Score="6" ViewCount="385" Body="&lt;p&gt;I came across the statement, that Yoneda-lemma is a generalization of Cayley`s theorem which states, that every group is isomorphic to a group of permutations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How exactly does generalizes Yoneda-lemma Cayley`s theorem? Can Cayley's theorem be deduced from Yoneda lemma, is it a generalization of a special case of Yoneda, or is this rather a philosiophical statement?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To me it seems that Yoneda embedding is more canonical than Cayleys theorem, because in the latter you have to choose whether the group acts from the left or from the right on itself. But maybe this is an optical illusion.&lt;/p&gt;&#xA;" OwnerUserId="667" LastActivityDate="2010-08-06T12:37:14.343" Title="Yoneda-Lemma as generalization of Cayley`s theorem?" Tags="&lt;category-theory&gt;&lt;group-theory&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="6" />
  <row Id="1702" PostTypeId="2" ParentId="1601" CreationDate="2010-08-06T07:53:39.567" Score="2" ViewCount="" Body="&lt;p&gt;The definitive answer for any $k$ (not just the second largest) is discussed in &lt;a href=&quot;http://compgeom.cs.uiuc.edu/~jeffe/teaching/497/02-selection.pdf&quot; rel=&quot;nofollow&quot;&gt;these notes&lt;/a&gt;. In short, if $V(n,k)$ is the number of comparisons needed to determine the kth largest element in a set of size $n$, then &#xA;$$ V(n,k) \ge n + R(n,k) - 2\sqrt{R(n,k)} $$&#xA;where&#xA;$$ R(n,k) = \log \binom{n}{k} - \log (n-k+1) + 3$$&lt;/p&gt;&#xA;" OwnerUserId="865" LastActivityDate="2010-08-06T07:53:39.567" CommentCount="1" />
  <row Id="1703" PostTypeId="2" ParentId="1694" CreationDate="2010-08-06T08:01:17.010" Score="0" ViewCount="" Body="&lt;p&gt;The &lt;code&gt;log&lt;/code&gt; abbreviation was more understandable then the &lt;code&gt;lg&lt;/code&gt; abbreviation, and also was more natural compared to the &lt;code&gt;sin&lt;/code&gt;, &lt;code&gt;cos&lt;/code&gt;, &lt;code&gt;tan&lt;/code&gt; abbreviations. Compare the &lt;code&gt;tg&lt;/code&gt; notation versus &lt;code&gt;tan&lt;/code&gt;. They are both in use, however &lt;code&gt;tan&lt;/code&gt; is more common. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of &lt;code&gt;log_a(x)&lt;/code&gt; we have two important special cases: &lt;code&gt;log_10&lt;/code&gt; and &lt;code&gt;log_e&lt;/code&gt;. &lt;code&gt;Log_10&lt;/code&gt; was so common, especially in high school, that it became abbreviated to &lt;code&gt;log&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;Log_e&lt;/code&gt; is/was more difficult to understand at high school level  - I can remember that. The notation &lt;code&gt;ln&lt;/code&gt; to distinguish this special case is helpful especially for beginners. Two different set of rules for two different scenarios (&lt;code&gt;log=log_10&lt;/code&gt; and &lt;code&gt;ln=log_e&lt;/code&gt;). Later on the can &lt;em&gt;perhaps&lt;/em&gt; understand the general case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To add to the confusion: my teachers at the university used the notation &lt;code&gt;log&lt;/code&gt; to mean &lt;code&gt;ln&lt;/code&gt;!! From there point of view any extra notation should be avoided and &lt;code&gt;log_10&lt;/code&gt; had no particular importance to them. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the &lt;code&gt;ln&lt;/code&gt; notation turns out to be useful from a didactical perspective.&lt;/p&gt;&#xA;" OwnerUserId="706" LastEditorUserId="706" LastEditorDisplayName="" LastEditDate="2010-08-06T08:06:38.560" LastActivityDate="2010-08-06T08:06:38.560" />
  <row Id="1704" PostTypeId="1" AcceptedAnswerId="1707" CreationDate="2010-08-06T08:19:05.400" Score="3" ViewCount="109" Body="&lt;p&gt;In group cohomology theory, we know that the cohomological dimension cd(G) for a profinite group is a fundamental numerical invariant. We say G is of virtual cohomological dimension n (denote it by vcd(G)) if there exists an open subgroup H such that cd(H)=n. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems that vcd(G) is not always a fixed integer and my question is: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any criterion when vcd(G) become fixed ,or say, when cd(G)=cd(H) for all open subgroup H of G ? If there are any related references please let me know, thx !&lt;/p&gt;&#xA;" OwnerUserId="866" LastActivityDate="2010-08-06T09:09:30.197" Title="When does the virtual cohomological dimension become a numerical invariant?" Tags="&lt;cohomology&gt;" AnswerCount="1" />
  <row Id="1705" PostTypeId="2" ParentId="1694" CreationDate="2010-08-06T08:27:29.157" Score="7" ViewCount="" Body="&lt;p&gt;Cajori, in his &lt;em&gt;History of mathematical notations&lt;/em&gt;, Vol. II, as far as I can see, mentions the notation «$\operatorname{ln}$» once when he is talking about notations for logarithm. He refers to [Irving Stringham, &lt;em&gt;Uniplanar Algebra&lt;/em&gt;, (San Francisco 1893), p. xiii] as someone who used that notation; the way this guy is mentioned makes me doubt he was the first or alone in this, though (I'd love to know what 'uniplanar algebra' is (was?)!)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The mention of «$\operatorname{ln}$» is quite minor, and I would guess that at the time Cajori was writing (the volume was completed in August 1925) the notation was essentially not used, for otherwise he would have been more interested in it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;PS:&lt;/strong&gt; I think I got the link from someone here, or on MO... &lt;a href=&quot;http://spikedmath.com/043.html&quot; rel=&quot;nofollow&quot;&gt;http://spikedmath.com/043.html&lt;/a&gt; is pretty relevant!&lt;/p&gt;&#xA;" OwnerUserId="274" LastEditorUserId="274" LastEditorDisplayName="" LastEditDate="2010-08-06T08:33:04.000" LastActivityDate="2010-08-06T08:33:04.000" CommentCount="1" />
  <row Id="1706" PostTypeId="1" AcceptedAnswerId="1731" CreationDate="2010-08-06T09:03:33.143" Score="5" ViewCount="616" Body="&lt;p&gt;I've been trying to find answer to this question for some time but in every document I've found so far it's taken for granted that reader know what $\mathbf ℝ^+$ is.&lt;/p&gt;&#xA;" OwnerUserId="348" LastEditorUserId="630" LastEditorDisplayName="" LastEditDate="2010-11-03T01:05:00.710" LastActivityDate="2010-11-03T01:05:00.710" Title="Does set $\mathbb{R}^+$ include zero?" Tags="&lt;notation&gt;&lt;number-systems&gt;" AnswerCount="5" CommentCount="13" FavoriteCount="1" />
  <row Id="1707" PostTypeId="2" ParentId="1704" CreationDate="2010-08-06T09:09:30.197" Score="3" ViewCount="" Body="&lt;p&gt;Although it is not obvious from the definition, in fact -- assuming that there exists at least one open subgroup of finite cohomological dimension, otherwise there is nothing to say -- the virtual cohomological dimension is always a fixed integer.  In other words, if there exists a single open subgroup $H_0$ of $G$ such that $\operatorname{cd}(H_0) &amp;lt; \infty$, then for all open subgroups $H$ of $G$ with $\operatorname{cd}(H) &amp;lt; \infty$, we have &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\operatorname{cd}(H) = \operatorname{cd}(H_0)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Moreover, this also holds for the $p$-cohomological dimension at any prime $p$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe this result was first proved by Serre.  In any case, it follows easily from Proposition I.14 in Serre's &lt;em&gt;Galois Cohomology&lt;/em&gt;.  In fact, a slightly stronger result is given as Proposition I.14$'$: again assuming that there exists at least one open subgroup $H_0$ of finite cohomological dimension, an open subgroup $H$ has infinite cohomological dimension if $H$ has nontrivial elements of finite order and cohomological dimension equal to $\operatorname{cd}(H_0)$ otherwise.  &lt;/p&gt;&#xA;" OwnerUserId="299" LastActivityDate="2010-08-06T09:09:30.197" CommentCount="2" />
  <row Id="1708" PostTypeId="2" ParentId="1701" CreationDate="2010-08-06T10:38:14.537" Score="8" ViewCount="" Body="&lt;p&gt;Just as Cayley's theorem states that every group is a subgroup of a symmetric group, Yoneda's lemma states that every (locally small) category $C$ embeds into a category of functors defined on $C$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Specifically, Yoneda's lemma states that if $F:C\to Set$ is an arbitrary set-valued functor, then $F(A) = Nat(\hom(A,-), F)$, so that natural transformations from a hom-set functor are in bijective correspondence with the elements in the functor image.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For this to be a straight generalization, we may consider a group $G$ as a category $C_G$ (actually a groupoid) with a single object $\ast$, and each group element a morphism. Then, there is only a single object, so a set-valued functor is the same thing as a set $S$ with a map from $G$ to $Bij(S,S)$, the set of bijections from $S$ to $S$. We can see this by unrolling the definition of set-valued functor: $\ast$ goes to $S$, and each element of $G$ goes to some map $S\to S$; all of which maps have to be bijections since otherwise the group properties suffer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose now that we have some such $G$-set $S$, Yoneda's lemma tells us that its elements are bijective with natural transformations from $G$ to $S$; so what is a natural transformation here? Our two functors are $S$, that takes $\ast$ to $S$, and $\hom(\ast,-)$ that takes $\ast$ to $G$; both as sets. A natural transformation of these is a set-valued map from one image to the other, such that the 'obvious' square of induced maps from morphisms in $C_G$ commutes - thus, $Nat(\hom(\ast,-),S)$ is the collection of $G$-set maps from $G$ as a left $G$-representation to $S$ itself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of the things Yoneda brings is a bijection $Nat(\hom(a,-),\hom(b,-)) = \hom(b,a)$: the Yoneda embedding. Applied to the group situation, this tells us that $\hom_G(G,G)=G$. Certainly, left multiplication by an element is a group endomorphism; and what this tells us is that these are all there is. This bijection is exactly what is used in the proof of Cayley's theorem on the wikipedia page.&lt;/p&gt;&#xA;" OwnerUserId="666" LastEditorUserId="666" LastEditorDisplayName="" LastEditDate="2010-08-06T11:27:56.573" LastActivityDate="2010-08-06T11:27:56.573" CommentCount="1" />
  <row Id="1709" PostTypeId="2" ParentId="1648" CreationDate="2010-08-06T10:45:04.263" Score="2" ViewCount="" Body="&lt;p&gt;Sounds an awful lot like the conditions you would expect for &lt;a href=&quot;http://en.wikipedia.org/wiki/Erlang_distribution&quot; rel=&quot;nofollow&quot;&gt;an Erlang distribution&lt;/a&gt; to me- it also looks a lot like one... &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Erlang distributions model the times between occurrences in poisson processes and are frequently used as parts of models of internet traffic. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My interpretation is this: as a site sending back a signal, one processes and sends stuff for a given user in an approximately poisson process (the approximate 'limit' of bernoilli trials p-&gt; stuff for user 1-p -&gt; stuff for a different user) and the time spent waiting for one to occur is therefore distributed Erlangwise, with a shift to the right (to account for the user's sending of the signal). This gives the shape you have above :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; This should be Erlang-2 if that is not already clear, since receiving and sending are two poisson occurrences from the same distribution depending (as indicated above) on traffic [That is: occurrence 1- server has free bit to process receiving, occurrence 2- computer has free bit to process sending]&lt;/p&gt;&#xA;" OwnerUserId="160" LastEditorUserId="160" LastEditorDisplayName="" LastEditDate="2010-08-06T12:21:45.543" LastActivityDate="2010-08-06T12:21:45.543" CommentCount="2" />
  <row Id="1710" PostTypeId="1" AcceptedAnswerId="1712" CreationDate="2010-08-06T11:35:44.143" Score="5" ViewCount="246" Body="&lt;p&gt;I've been experimenting with the summation of polynomials. My line of attack is to treat the subject the way I would for calculus, but not using limits.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By way of a very simple example, suppose I wish to add the all numbers between 10 and 20 inclusive, and find a polynomial which I can plug the numbers into to get my answer.  I suspect its some form of polynomial with degree 2. So I do a integer 'differentiation':&lt;/p&gt;&#xA;&#xA;&lt;p&gt;diff(x^2) = x^2-(x-1)^2 = 2x-1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can see from this that I nearly have my answer, so assuming an inverse 'integration' operation and re-arranging:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1/2)diff(x^2+int(1)) = x&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, I know that the 'indefinite integral' of 1 is just x, from 'differentiating' x-(x-1) = 1. So ultimately:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1/2)(x^2+x) = int(x)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So to get my answer I take the 'definite' integral: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;int(x):10,20 = (1/2)(20^2+20)-(1/2)(9^2+9) = 165 (the lower bound needs decreasing by one)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is, is there a general way I can 'integrate' any polynomial, in this way?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please excuse my lack of rigour and odd notation.&lt;/p&gt;&#xA;" OwnerUserId="869" LastActivityDate="2010-08-06T16:17:25.780" Title="Indefinite Summation of Polynomials" Tags="&lt;calculus&gt;" AnswerCount="4" CommentCount="1" />
  <row Id="1711" PostTypeId="2" ParentId="540" CreationDate="2010-08-06T11:45:39.807" Score="3" ViewCount="" Body="&lt;p&gt;The above answers give excellent explanations about why the zeta function has close connections to number theory, but I thought I'd mention something about why the Riemann Hypothesis should matter so much.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By taking the logarithm and then differentiating the zeta function, one gets the formula&#xA;$$\frac{\zeta'(s)}{\zeta(s)}=\sum_{n=1}^\infty\frac{\Lambda(n)}{n^s}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for $\Re(s)&gt;1$, where $\Lambda(n)$ is the von Mangoldt function which takes the value $\log p$ at powers of primes $p$, and is 0 everywhere else. Think of it as a weighted way of counting the primes (the prime number theorem tells us that $\log p$ is the natural weight to choose).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Much of analytic number theory proceeds by choosing a weight of the set we wish to consider (often the primes), and then encoding this weighting in a so-called Dirichlet series (an infinite sum of the form above). We can then use analysis to study this series and get lots of useful information.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case, then, the function we need to study to get information about the primes is $\frac{\zeta'(s)}{\zeta(s)}$, which we can study using complex analysis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In complex analysis, a good slogan is 'the only things that matter are zeros and poles' (effectively points where the function shoots off to infinity). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence to understand the prime numbers, we just need to understand the zeros and poles of $\frac{\zeta'(s)}{\zeta(s)}$ - we know about the simple pole at $s=1$, we know there aren't any other zeros where it counts, and we also know that the only other poles are at zeros of $\zeta(s)$ (roughly because dividing by zero causes infinity). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, if we knew where these zeros are (i.e. the Riemann hypothesis) we can work with $\frac{\zeta'(s)}{\zeta(s)}$ in all kinds of clever ways to get good results on the prime numbers.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;More specifically, in the usual contour proof of the prime number theorem, knowing that there aren't any other zeros in $\Re(s)&gt;1/2$ would allow us to shift the contour further to the left, reducing the error term in the result to (roughly) $O(\sqrt{x})$.&lt;/p&gt;&#xA;" OwnerUserId="870" LastActivityDate="2010-08-06T11:45:39.807" />
  <row Id="1712" PostTypeId="2" ParentId="1710" CreationDate="2010-08-06T11:58:31.347" Score="6" ViewCount="" Body="&lt;p&gt;Your &quot;diff&quot; is actually called &lt;a href=&quot;http://en.wikipedia.org/wiki/Finite_difference&quot; rel=&quot;nofollow&quot;&gt;(backward) finite difference&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;\nabla_1 [ P ](x) &amp;amp;= P(x) - P(x-1) \\&#xA;&amp;amp;= P(x-1+1) - P(x-1) \\&#xA;&amp;amp;= \Delta_1[ P ](x-1)&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The inverse the forward finite difference is called &lt;a href=&quot;http://en.wikipedia.org/wiki/Indefinite_sum&quot; rel=&quot;nofollow&quot;&gt;indefinite sum&lt;/a&gt;. Extracted from Wikipedia, the useful formulae for polynomials is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;\Delta^{-1}_1 x^n &amp;amp;= \frac{B_{n+1}(x)}{n+1} + C \\&#xA;\Delta^{-1}_1 af(x) &amp;amp;= a \Delta^{-1}_1 f(x)&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(B&lt;sub&gt;n+1&lt;/sub&gt;(x) is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Bernoulli_polynomials&quot; rel=&quot;nofollow&quot;&gt;Bernoulli polynomial&lt;/a&gt;.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &amp;Delta;&lt;sup&gt;-1&lt;/sup&gt; can be converted back to your &quot;int&quot; by substituting $x \mapsto x + 1$.&lt;/p&gt;&#xA;" OwnerUserId="171" LastEditorUserId="171" LastEditorDisplayName="" LastEditDate="2010-08-06T12:06:41.143" LastActivityDate="2010-08-06T12:06:41.143" CommentCount="1" />
  <row Id="1713" PostTypeId="2" ParentId="1710" CreationDate="2010-08-06T12:01:14.540" Score="5" ViewCount="" Body="&lt;p&gt;You seem to be reaching for the&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Finite_difference&quot; rel=&quot;nofollow&quot;&gt;calculus of finite differences&lt;/a&gt;, once a well-known topic but rather&#xA;unfashionable these days. The answer to your question is yes: given a polynomial&#xA;$f(x)$ there is a polynomial $g(x)$ (of degree one greater than $f$) such that&#xA;$$f(x)=g(x)-g(x-1).$$&#xA;This polynomial $g$ (like the integral of $f$) is unique save for&#xA;its constant term.&#xA;Once one has $g$ then of course&#xA;$$f(a)+f(a+1)+\cdots+f(b)=g(b)-g(a-1).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When $f(x)=x^n$ is a monomial, the coefficients&#xA;of $g$ involve the endlessly fascinating &lt;a href=&quot;http://en.wikipedia.org/wiki/Bernoulli_number&quot; rel=&quot;nofollow&quot;&gt;Bernoulli numbers&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="226" LastEditorUserId="226" LastEditorDisplayName="" LastEditDate="2010-08-06T12:12:04.237" LastActivityDate="2010-08-06T12:12:04.237" />
  <row Id="1714" PostTypeId="2" ParentId="1710" CreationDate="2010-08-06T12:09:46.660" Score="2" ViewCount="" Body="&lt;p&gt;As other answers have noted, you are about to discover the &lt;a href=&quot;http://en.wikipedia.org/wiki/Finite_difference&quot; rel=&quot;nofollow&quot;&gt;calculus of finite differences&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For practical calculations, here a most useful fact: the rule&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{d}{dx} x^n = n x^{n-1}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;corresponds to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Delta_1 x(x-1)(x-2)\cdots(x-(n-1)) = n x(x-1)(x-2)\cdots(x-(n-2))$&lt;/p&gt;&#xA;" OwnerUserId="434" LastActivityDate="2010-08-06T12:09:46.660" CommentCount="2" />
  <row Id="1715" PostTypeId="2" ParentId="1691" CreationDate="2010-08-06T12:17:16.223" Score="1" ViewCount="" Body="&lt;p&gt;muad,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You're right! This is what's happening on elliptic curves.  And, if you look at higher genus, as T.. said, there is no group law on the curve.  But something silly like there not being a group law on curves of high genus wasn't enough to stop the 19th century mathematicians, they discovered that it's the group law on an abelian variety related to the curve that matters, and the curve and the abelian variety happen to be the same for cubics!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can find all this worked out in any good book that talks about elliptic integrals or abelian integrals, and I'm personally a big fan of the exposition in Stillwell's &quot;Mathematics and its History&quot;&lt;/p&gt;&#xA;" OwnerUserId="176" LastActivityDate="2010-08-06T12:17:16.223" CommentCount="2" />
  <row Id="1716" PostTypeId="2" ParentId="709" CreationDate="2010-08-06T12:25:40.793" Score="1" ViewCount="" Body="&lt;p&gt;I like Americo's answer using translation of y = ax^2. But to finish it, we should at least note that ax^2 has a maximum or minimum at &amp;lt;0,0&gt;, which is quite obvious, and that translation preserves order and therefore maxima and minima, which is pretty easy.&lt;/p&gt;&#xA;" OwnerUserId="271" LastActivityDate="2010-08-06T12:25:40.793" />
  <row Id="1717" PostTypeId="2" ParentId="1701" CreationDate="2010-08-06T12:37:14.343" Score="3" ViewCount="" Body="&lt;p&gt;I can also add that both Yoneda and Cayley are results which follow from the general philosophy of investigating algebraic structures by letting them act on themselves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) If you let a group $G$ act on itself, you realize it as a subgroup of $\mathfrak{S}_n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) If you let a ring with unit act on itself, you realize it as a subring of $E$, where $E$ is the underlying additive group.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) Similarly if you let a $k$-algebra act on itself, you realize it as a subalgebra of $\mathcal{M}_n(k)$. In particular this gives the classical realization of $\mathbb{C}$ as a matrix algebra over $\mathbb{R}$ and of the quaternions as a matrix algebra over $\mathbb{C}$ or $\mathbb{R}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4) You can let a Lie algebra act on itself, but unfortunately this action need not be faithful (Lie algebra don't have units...). So you only obtain the easy first step of Ado's theorem about embedding Lie algebras into mtraix algebras.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;5) If you let a category $\mathcal{C}$ act on itself, you obtain an embedding into $Fun(\mathcal{C}^{op}, Set)$, which is the content of Yoneda's lemma.&lt;/p&gt;&#xA;" OwnerUserId="233" LastActivityDate="2010-08-06T12:37:14.343" CommentCount="3" />
  <row Id="1718" PostTypeId="2" ParentId="1648" CreationDate="2010-08-06T12:42:56.610" Score="1" ViewCount="" Body="&lt;p&gt;From the &lt;a href=&quot;http://stats.stackexchange.com/questions/1315&quot;&gt;comments on stats.stackexchange&lt;/a&gt;, it seems like you may not care too much about the distribution, but just a pretty curve to overlay on your graph. In which case, some kind of spline is your best bet. Use some kind of curves with asymptotes at y=0 for your upper- and lower-most segments, and whatever fits best in between.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you do actually care about the underlying distribution:&lt;br&gt;&#xA;The first step would be to use whatever outside knowledge you have to characterize the distribution. For example:&lt;br&gt;&#xA;Network ping is a sum of independent wait times (the individual nodes in the network). This would suggest a &lt;a href=&quot;http://en.wikipedia.org/wiki/Gamma_distribution&quot; rel=&quot;nofollow&quot;&gt;Gamma&lt;/a&gt;/Erlang distribution if each of these steps is identical, and a &lt;a href=&quot;http://www.math.bme.hu/~balazs/sumexp.pdf&quot; rel=&quot;nofollow&quot;&gt;more complex distribution&lt;/a&gt; if they are not.&lt;br&gt;&#xA;Ping is a measure of time until the computer at the other end responds to your request, the likelihood of which is proportional to the time elapsed. This would suggest a &lt;a href=&quot;http://en.wikipedia.org/wiki/Weibull_distribution&quot; rel=&quot;nofollow&quot;&gt;Weibull&lt;/a&gt; distribution.&lt;br&gt;&#xA;Ping time is the accumulation of a large number of factors that all have a multiplicative effect on the result. Then a &lt;a href=&quot;http://en.wikipedia.org/wiki/Log-normal_distribution&quot; rel=&quot;nofollow&quot;&gt;log-normal&lt;/a&gt; distribution would be best. &#xA;I don't know enough about networking to say anything about the accuracy of any of the above models, and it's also perfectly likely that ping time follows some other model which I haven't thought of. I just wanted to demonstrate the idea: that you should think about what factors contribute to the thing you are trying to model, and how they interact.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And, of course, the distribution does not necessarily have to be a known one! In which case the above won't get you very far! In this case you might want to come up with your own empirical distribution, for which a variety of methods exist. The most common are to take your measurements as the distribution (as long as you have a sufficiently large number) or to take each of those data points and treat it as the center of some uniform/normal/other distribution, and sum everything with appropriate scaling.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After you know the type of distribution, you may also be able to use domain knowledge to estimate some of its parameters. For example, you might guess at the number of exponentials being summed based on the shape of the network. You can also use your measured mean and variance to form estimates of the distribution parameters. For example, if you thought that your distribution was a Gamma(3,θ), then you could use your measured variance to estimate θ=4.74182454 based on our known formula for variance of a Gamma Distribution. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you have your guess at a distribution, you will want to test its goodness of fit. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For this, the standard method would be to apply the one-sample &lt;a href=&quot;http://en.wikipedia.org/wiki/Kolmogorov-Smirnov_test&quot; rel=&quot;nofollow&quot;&gt;Kolmogorov-Smirnov test&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other potentially applicable tests are the &lt;a href=&quot;http://en.wikipedia.org/wiki/Cramer-von-Mises_test&quot; rel=&quot;nofollow&quot;&gt;Cramer-von-Mises&lt;/a&gt;, &lt;a href=&quot;http://en.wikipedia.org/wiki/Anderson-Darling_test&quot; rel=&quot;nofollow&quot;&gt;Anderson-Darling&lt;/a&gt;, or &lt;a href=&quot;http://en.wikipedia.org/wiki/Pearson%27s_chi-square_test&quot; rel=&quot;nofollow&quot;&gt;chi-square&lt;/a&gt; tests.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is incomplete, I will add more later.&lt;/p&gt;&#xA;" OwnerUserId="73" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-08-10T02:40:44.470" LastActivityDate="2010-08-10T02:40:44.470" />
  <row Id="1719" PostTypeId="2" ParentId="1660" CreationDate="2010-08-06T12:46:36.323" Score="3" ViewCount="" Body="&lt;p&gt;As Qiaochu points out the formula given by Gjergji Zaimi is certainly relevant.&#xA;It realizes $D(a_1,\ldots,a_n)/D(1,\ldots,n)$ as the&#xA;determinant of the $n$-by-$n$ matrix $M$ with $(i,j)$-entry&#xA;$${a_i\choose j-1}.$$&#xA;By row operations this equals the determinant of the $(n-1)$-by-$(n-1)$  matrix $N$&#xA;with $(i,j)$-entry&#xA;$${a_{i+1}\choose j}-{a_i\choose j}.$$&#xA;The $i$-th row of $n$ is the sum of vectors $v_{a_i},v_{a_i+1},\ldots,v_{a_{i+1}-1}$&#xA;where the $j$-th entry of $v_c$ is&#xA;$${c\choose j-1}.$$&#xA;By the multilinearity of the determinant as a function of its rows, $N$&#xA;is the sum the determinants with rows $v_{b_1},\ldots,v_{b_{n-1}}$&#xA;where $a_1\le b_i &amp;lt; a_2\le b_2 &amp;lt; a_3\le\cdots$. These tuples&#xA;$(b_1,\ldots,b_{n-1})$ are precisely the admissible penultimate rows&#xA;in GZ triangles with bottom row $(a_1,\ldots,a_n)$. Hence both sides of the&#xA;sought equality satisfy the same recurrence.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-08-06T12:46:36.323" CommentCount="1" />
  <row Id="1720" PostTypeId="2" ParentId="1526" CreationDate="2010-08-06T12:46:39.427" Score="0" ViewCount="" Body="&lt;p&gt;Well 3186-2860 = 326.  That is very nearly a tenth of 3186, but 3186/9 = 354. But 326 is closer to 318.6 than to 354, so I'd go for 9/10 instead of 8/9.&lt;/p&gt;&#xA;" OwnerUserId="869" LastActivityDate="2010-08-06T12:46:39.427" />
  <row Id="1721" PostTypeId="2" ParentId="903" CreationDate="2010-08-06T12:53:00.723" Score="2" ViewCount="" Body="&lt;p&gt;I think Munkres' is a great book, though you can safely skip some sophisticated counter-examples in it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Janick's &quot;Topology&quot; is more intuitive, less technical or just less narrowly focused on the specific subject of Topology -so it links to other parts of mathematics. But I've never dared to base my lectures on it, precisely because elementary Topology can be explained on its own, out of the blue, without assuming any previous knowledge at all, whereas mixing it with Differential Geometry, for instance, can be more confusing than useful for undergraduate students.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nevertheless, Janick's book has some masterpieces; for instance, the proof of Urysohn's lemma: the first time I read it I realised that there are people in this world who &lt;em&gt;actually&lt;/em&gt; understand maths, and others, well... we just repeat, more or less succesfully, what we have learnt (or so we hope). The same proof in Munkres' simply doesn't compare.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for Dugundji, it's a great book, but nowadays it seems to me much too difficult for undergraduate students.&lt;/p&gt;&#xA;" OwnerUserId="664" LastActivityDate="2010-08-06T12:53:00.723" CommunityOwnedDate="2010-08-06T12:53:00.723" />
  <row Id="1722" PostTypeId="2" ParentId="776" CreationDate="2010-08-06T13:00:43.033" Score="3" ViewCount="" Body="&lt;p&gt;In fact it is a corollary of the theorem stating that the angle subtended at the centre of the circle (here 180 degrees) is double the angle subtended at the centre. If this is not satisfactory use the contrapositive instead: if angle ABC is not 90 degrees, the angle AOB cannot be 180 degrees, i.e. A,O,B are not collinear.&lt;/p&gt;&#xA;" OwnerUserId="871" LastActivityDate="2010-08-06T13:00:43.033" />
  <row Id="1723" PostTypeId="2" ParentId="1665" CreationDate="2010-08-06T13:29:49.513" Score="3" ViewCount="" Body="&lt;p&gt;The cases $p=2,3$ can be trivially checked. I'll assume $p\ge 5$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note $\left( \displaystyle\frac{p-1}{4} \right) \left( \displaystyle\frac{p+1}{2}\right)=\displaystyle\frac{p^2-1}{8t}=a$ (say). Then $a$ has only two prime divisors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now it is well known that $24|p^2-1$ for $p&gt;3$. Let $p^2-1=24t$. Then $a=3t$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus $3$ is a prime divisor of $a$, implying $3$ equals one of $\displaystyle\frac{p-1}{4},\frac{p+1}{2}$. Direct substitution shows that $p=13$ is the only solution.&lt;/p&gt;&#xA;" OwnerUserId="871" LastActivityDate="2010-08-06T13:29:49.513" />
  <row Id="1724" PostTypeId="2" ParentId="1555" CreationDate="2010-08-06T13:51:27.147" Score="0" ViewCount="" Body="&lt;p&gt;I will try to summarize here what I've found so far on this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the form being integrated is $\frac{\mathrm{d}x}{\sqrt{1 - x^2}} =: \omega$ let $y = \sqrt{1 - x^2}$, we can also write this as $x^2 + y^2 - 1 = 0$ which makes it more obvious that this is a unit circle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $N$ be the point $(1,0)$ and we can define a group structure on the curve as follows. To add $A$ and $B$, fire a ray from $N$ parallel to $AB$ and pick the point it intersects the curve as $A \oplus B$. Symbolically, this means $A \oplus B = N - k(B - A)$ for some nonzero $k$, since $A \oplus B$ lies on the circle we can expand it into the equation of the curve to solve for $k$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xA;(- k (x_b - x_a))^2 + (1 - k (y_b - y_a))^2 - 1 &amp;amp;= 0 \\&#xA;k ((x_b - x_a)^2 + (y_b - y_a)^2) - 2 (y_b - y_a) &amp;amp;= 0 \\&#xA;\frac{2 (y_b - y_a)}{(x_b - x_a)^2 + (y_b - y_a)^2} &amp;amp;= k&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;hence $x_{a \oplus b} = - 2 \frac{(x_b - x_a)(y_b - y_a)}{(x_b - x_a)^2 + (y_b - y_a)^2} = x_a \sqrt{1-x_b^2} + x_b \sqrt{1-x_a^2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I think by &lt;a href=&quot;http://math.stackexchange.com/questions/1691/trying-to-piece-together-an-integral-addition-theorem&quot;&gt;the theorem&lt;/a&gt;, if &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\omega$ is invariant along the curve&lt;/li&gt;&#xA;&lt;li&gt;$\oplus$ is the group law on points of the curve&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;we can conclude now that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_0^a \omega + \int_0^b \omega = \int_0^{a \oplus b} \omega$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Certainly, the previous equation holds - but for this post to count as a proof the theorem is needed. Furthermore I believe this same idea works for any conic section and also for cubic curves and perhaps not any others curves.&lt;/p&gt;&#xA;" LastActivityDate="2010-08-06T13:51:27.147" CommentCount="1" />
  <row Id="1725" PostTypeId="2" ParentId="776" CreationDate="2010-08-06T14:15:55.930" Score="3" ViewCount="" Body="&lt;p&gt;Really short vector proof:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Center the circle at the origin, and scale to have radius 1. Let the vertex of the right triangle be at vector $v$, and let the diameter be the segment from the vector $w$ to $-w$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then&#xA;$(v-w) \cdot (v-(-w)) = (v-w) \cdot (v+w) = (v \cdot v) - (w \cdot w) = 1 - 1 = 0$, so the angle formed by $vw$ and $v(-w)$ is a right angle.&lt;/p&gt;&#xA;" OwnerUserId="831" LastActivityDate="2010-08-06T14:15:55.930" CommentCount="1" />
  <row Id="1726" PostTypeId="2" ParentId="1626" CreationDate="2010-08-06T14:44:42.393" Score="1" ViewCount="" Body="&lt;p&gt;By little Fermat: $\; 6,10|120 \Rightarrow 3^{120} \equiv 1 \pmod{7, 11} \Rightarrow 3^{123} \equiv 3^3 \pmod{77}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See also these &lt;strong&gt;Fermat-Euler-Carmichael&lt;/strong&gt; generalizations of little Fermat-Euler from my &lt;a href=&quot;http://groups.google.com/groups?selm=y8zk55tqcie.fsf%40nestle.csail.mit.edu&quot; rel=&quot;nofollow&quot;&gt;sci.math post on Apr 10 2009&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;THEOREM 1&lt;/strong&gt;.  For natural numbers $ a,e,n $ with $ e,n&amp;gt;1 $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\;\; n|a^e-a$ for all $a \iff n$ is squarefree and prime $ p|n \Rightarrow p-1|e-1 $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;REMARK&lt;/strong&gt; The special case  $e = n$ is &lt;a href=&quot;http://en.wikipedia.org/wiki/Carmichael_number#Korselt.27s_criterion&quot; rel=&quot;nofollow&quot;&gt;Korselt's criterion&lt;/a&gt; for Carmichael numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;THEOREM 2&lt;/strong&gt;.  For natural numbers $ a,e,n $ with $ e,n&amp;gt;1 $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\;\; n|a^e-1$ for all $a$ coprime to $n \iff  p$ prime, $\; p^k||n \Rightarrow E(p^k)|e $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with $\;\;\; E(p^k) = \phi(p^k) $ for odd primes $p$, or $p=2,\; k \le 2 $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and   $\quad E(2^k) = 2^{k-2} $ for $ k&amp;gt;2 $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The latter exception is due to $\mathbb Z/2^k$ having multiplicative group $C(2) \times C(2^{k-2})$ for $k&amp;gt;2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice that the least such exponent $e$ is given by $\ \lambda(n)\ = {\rm lcm}\{E(\;{p\_i}^{k\_i})\}\;$ where $\; n = \prod {p\_i}^{k\_i}\;$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\lambda(n)$ is called the (universal) &lt;a href=&quot;http://planetmath.org/encyclopedia/Exponent.html&quot; rel=&quot;nofollow&quot;&gt;exponent&lt;/a&gt; of the group $\mathbb Z/n^*,\;$ a.k.a. the &lt;a href=&quot;http://en.wikipedia.org/wiki/Carmichael_function&quot; rel=&quot;nofollow&quot;&gt;Carmichael function&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See my post &lt;a href=&quot;http://groups.google.com/groups?selm=y8zk55tqcie.fsf%40nestle.csail.mit.edu&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; for proofs and further discussion.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2011-02-28T01:35:01.600" LastActivityDate="2011-02-28T01:35:01.600" />
  <row Id="1727" PostTypeId="1" AcceptedAnswerId="1729" CreationDate="2010-08-06T14:58:42.553" Score="4" ViewCount="242" Body="&lt;p&gt;Related to &lt;a href=&quot;http://math.stackexchange.com/questions/1659/locating-the-quadrant-containing-a-point-on-an-n-sphere&quot;&gt;this question&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that I'm using the &lt;a href=&quot;http://mathworld.wolfram.com/Hypersphere.html&quot; rel=&quot;nofollow&quot;&gt;geometer definition&lt;/a&gt; of an $n$-sphere of radius $r$, i.e.$ \{ x \in \mathbb{R}^n : \|x\|_2 = r  \} $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose I have an $n$-sphere centered at $\bf 0$ in $\mathbb{R}^n$ with radius $r$ which has been divided into $2^k$ orthants by $k$ axis-aligned hyperplanes (note, $k \le n$) in $\mathbb{R}^{n-1}$ passing through $\bf 0$. For e.g., if $k=1$, we have $2$ $n$-hemispheres.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's the &lt;strong&gt;question&lt;/strong&gt;: how do I find the center of mass of such an orthant? Or (since I'm still working on it), how would you find the center of mass of an $n$-hemisphere?&lt;/p&gt;&#xA;" OwnerUserId="825" LastEditorUserId="825" LastEditorDisplayName="" LastEditDate="2010-08-06T16:27:32.287" LastActivityDate="2010-08-06T19:21:31.747" Title="Center of mass of an $n$-hemisphere" Tags="&lt;geometry&gt;&lt;physics&gt;" AnswerCount="1" CommentCount="7" />
  <row Id="1728" PostTypeId="2" ParentId="1526" CreationDate="2010-08-06T15:01:58.817" Score="1" ViewCount="" Body="&lt;p&gt;This isn't so much a math answer as a test-taking answer: &lt;strong&gt;You don't have to compute the fraction, you just have to determine which is the right answer.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;2860/3186&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;This is a multiple choice question, with answers 6/7, 7/8, 8/9 and 9/10.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you reduce the fraction to get a/b, then b must be a divisor of 3186. This allows you to immediately eliminate some choices. It can't be 9/10 because 10 doesn't divide 3186. You can quickly check that 7 and 8 don't divide 3186, but 9 does, so the only one of the choices that &lt;em&gt;has a shot&lt;/em&gt; at being the correct answer is 8/9.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Incidentally, &lt;strong&gt;none of those answer is correct&lt;/strong&gt;; it's not 8/9 either. The furthest you can reduce the fraction is 1430/1593, so you must have an error in your question. Either the fraction is wrong or you're supposed to find the &lt;em&gt;best approximation&lt;/em&gt; rather than the actual value.&lt;/p&gt;&#xA;" OwnerUserId="60" LastActivityDate="2010-08-06T15:01:58.817" />
  <row Id="1729" PostTypeId="2" ParentId="1727" CreationDate="2010-08-06T15:14:15.180" Score="2" ViewCount="" Body="&lt;p&gt;One finds the centre of mass for these bodies just as one&#xA;does for any body, by integration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The $x_i$-coordinate of the centre of mass of a region $A$ in $\mathbb{R}^n$&#xA;is&#xA;$$\frac{\int_A x_i\ dx_1\cdots dx_n}{\int_A\ dx_1\cdots dx_n}.$$&#xA;Here the region might as well be that between the planes $x_n=a$ and $x_n=b$&#xA;in the sphere of  radius $r$ centred at the origin. Using the symmetry&#xA;of $A$ this ratio of integrals equals&#xA;$$\frac{\gamma_{n-1}\int_a^b x_n(r^2-x_n^2)^{(n-1)/2} dx_n}&#xA;{\gamma_{n-1}\int_a^b (r^2-x_n^2)^{(n-1)/2} dx_n}$$&#xA;(in the $x_n$ direction) where $\gamma_{n-1}$ is the volume of the $(n-1)$-dimensional&#xA;unit ball (and obligingly cancels). These integrals can be attacked by&#xA;trig substitutions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edited&lt;/strong&gt;&#xA;It's now clear that my original interpretation of your question&#xA;was wrong. However it's still not clear whether your centre of mass&#xA;is for a solid orthant or its curved surface. In any case if your orthant&#xA;is defined by the conditions $x_1,\ldots,x_k\ge0$ then its centre&#xA;of mass has the form $(a,\ldots,a,0\ldots,0)$ where $a$ depends on $r$&#xA;and $n$ but &lt;strong&gt;not&lt;/strong&gt; on $k$. One sees this from the symmetry of the problem.&#xA;Thus the problem reduces to the hemispheric case. In the solid case the&#xA;answer is&#xA;$$a=\frac{\int_0^r x(r^2-x^2)^{(n-1)/2} dx}&#xA;{\int_0^r (r^2-x^2)^{(n-1)/2} dx}$$&#xA;while in the &quot;shell&quot; case it is&#xA;$$a=\frac{\int_0^r x(r^2-x^2)^{(n-3)/2} dx}&#xA;{\int_0^r (r^2-x^2)^{(n-3)/2} dx}$$&#xA;(if I've done my sums right).&lt;/p&gt;&#xA;" OwnerUserId="226" LastEditorUserId="226" LastEditorDisplayName="" LastEditDate="2010-08-06T19:21:31.747" LastActivityDate="2010-08-06T19:21:31.747" CommentCount="3" />
  <row Id="1730" PostTypeId="1" CreationDate="2010-08-06T15:30:44.160" Score="11" ViewCount="798" Body="&lt;p&gt;I'm teaching a College Algebra class in the upcoming semester, and only a small portion of the students will be moving on to further mathematics. The class is built around functions, so I need to start with the definition of one, yet many &quot;official&quot; definitions I have found too convoluted (or poorly written) for general use.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's one of the better &quot;light&quot; definitions I've found:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A function is a relationship which&#xA;  assigns to each input (or domain)&#xA;  value, a unique output (or range)&#xA;  value.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This sounds simple enough on the surface, but putting myself &quot;in the head&quot; of a student makes me pause. It's almost &lt;em&gt;too&lt;/em&gt; compact with potentially ambiguous words for the student (relationship? assigns? unique?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's my personal best attempt, in 3 parts. Each part of the definition would include a discussion and examples before moving to the next part.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A relation is a set of links between&#xA;  two sets.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Each link of a relation has an input&#xA;  (in the starting set) and an output&#xA;  (in the ending set).&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;A function is a relation where every&#xA;  input has one and only one possible&#xA;  output.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm somewhat happier here: starting with a relation gives some natural examples and makes it easier to impart the special importance of a function (which is &quot;better behaved&quot; than a relation in practical circusmtances).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I'm also still uneasy (&quot;links&quot;? A set between sets?) and I was wanting to see if anyone had a better solution.&lt;/p&gt;&#xA;" OwnerUserId="295" LastActivityDate="2010-12-22T07:13:24.917" Title="How do you define functions for non-mathematicians?" Tags="&lt;algebra-precalculus&gt;&lt;teaching&gt;&lt;definition&gt;&lt;functions&gt;" AnswerCount="9" FavoriteCount="4" />
  <row Id="1731" PostTypeId="2" ParentId="1706" CreationDate="2010-08-06T15:33:47.170" Score="10" ViewCount="" Body="&lt;p&gt;It depends on the choice of the person using the notation: sometimes it does, sometimes it doesn't. It is just a variant of the situation with $\mathbb N$, which half the world (the mistaken half!) considers to include zero.&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2010-08-06T15:33:47.170" />
  <row Id="1732" PostTypeId="2" ParentId="1730" CreationDate="2010-08-06T15:40:20.550" Score="8" ViewCount="" Body="&lt;p&gt;The way you've restated the definition is fairly common in contemporary high school books in the U.S. (perhaps changing &quot;links between two sets&quot; to &quot;ordered pairs&quot;).  What I've seen a lot of in middle school and earlier algebra settings is the idea of a &quot;function machine.&quot;  The function machine graphic below is from &lt;a href=&quot;http://etc.usf.edu/clipart/41800/41849/function_41849.htm&quot; rel=&quot;nofollow&quot;&gt;FCIT (&amp;copy;2009)&lt;/a&gt;, but &lt;a href=&quot;http://www.google.com/images?q=function+machine&quot; rel=&quot;nofollow&quot;&gt;a google image search for function machine&lt;/a&gt; will show you many different ways the concept can be visualized.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/BgduF.gif&quot; alt=&quot;function machine&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While this probably pushes the idea that a function has a formula, I'd claim that &quot;rule&quot; could be as general as a specific listing of which inputs map to which outputs, as in your definition.  To me, the prevalence of this machine metaphor in middle school contexts suggests that it works well for students who do not necessarily yet have a sense of symbolic algebra.  I've seen function machines used as low as 3rd grade.&lt;/p&gt;&#xA;" OwnerUserId="72" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-09-29T04:16:05.283" LastActivityDate="2010-09-29T04:16:05.283" CommentCount="4" />
  <row Id="1733" PostTypeId="2" ParentId="1665" CreationDate="2010-08-06T15:40:28.367" Score="5" ViewCount="" Body="&lt;p&gt;The primes have product (p^3 - p)/8 which is divisible by 3. So one prime = 3. &#xA;The rest is trivial.&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-08-06T15:40:28.367" />
  <row Id="1734" PostTypeId="2" ParentId="1694" CreationDate="2010-08-06T15:43:58.553" Score="1" ViewCount="" Body="&lt;p&gt;I suggest that everybody from now on uses ln, lg, and lb respectively for the natural, decimal, and binary logarithmic functions, reserving log for when the base is displayed explicitly or is some fixed arbitrary number if the scale doesn't matter. I do (with explanation if necessary at the first instance) . What could be simpler?&lt;/p&gt;&#xA;" OwnerUserId="875" LastActivityDate="2010-08-06T15:43:58.553" CommentCount="5" />
  <row Id="1735" PostTypeId="1" CreationDate="2010-08-06T15:44:16.930" Score="2" ViewCount="134" Body="&lt;p&gt;Let $f$ be a function defined on an open interval $(a,b)$ which has continuous derivatives of order $1,2, \dots , n-1$, and there is a point $c \in (a,b)$ such that $f(c), f^{{1}}(c), \dots , f^{(n-1)}(c)$ are all $0$.  However $f^{n}(c)$ is not $0$ and we can assume that $f^{n}(x) &gt; 0$ for $x \in (a,b)$, however &lt;strong&gt;no assumptions about the continuity of&lt;/strong&gt; $f^{(n)}(x)$ &lt;strong&gt;are made&lt;/strong&gt;. &#xA;In this case is it true that $\lim_{x\to c}\frac{f(x)}{(x-c)^n} = \frac{f^{(n)}(c)}{n!}$?&lt;/p&gt;&#xA;" OwnerUserId="404" LastActivityDate="2010-08-06T15:58:28.373" Title="A particular case of L'Hospital" Tags="&lt;calculus&gt;&lt;analysis&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="1736" PostTypeId="2" ParentId="1735" CreationDate="2010-08-06T15:53:02.417" Score="2" ViewCount="" Body="&lt;p&gt;L'Hospital's rule does not depend on the continuity of the derivative.  Under your conditions, $\lim_{x\to c}f^{(k)}(x)=0$ for $k\le n-1$, so L'Hospital's rule can be applied to give:&#xA;$$\lim_{x\to c}\frac{f(x)}{(x-c)^n}&#xA;=\lim_{x\to c}\frac{f'(x)}{n(x-c)^{n-1}}&#xA;=\cdots&#xA;=\lim_{x\to c}\frac{f^{(n-1)}(x)}{n!(x-c)}&#xA;=\lim_{x\to c}\frac{f^{(n)}(x)}{n!}.&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, to assert that $\lim_{x\to c}\frac{f^{(n)}(x)}{n!}=\frac{f^{(n)}(c)}{n!}$ requires that $f^{(n)}$ be continuous at c. (n! is independent of x, so $\lim_{x\to c}\frac{f^{(n)}(x)}{n!}=\frac{f^{(n)}(c)}{n!}$ and $\lim_{x\to c}f^{(n)}(x)=f^{(n)}(c)$ are equivalent and the latter is the definition of continuity for $f^{(n)}$ at c.)&lt;/p&gt;&#xA;" OwnerUserId="72" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-08-06T15:58:28.373" LastActivityDate="2010-08-06T15:58:28.373" CommentCount="0" />
  <row Id="1737" PostTypeId="2" ParentId="1706" CreationDate="2010-08-06T16:11:30.543" Score="3" ViewCount="" Body="&lt;p&gt;As a rule of thumb most mathematicians of the anglo saxon school consider that positive numbers (be it $\mathbb{N}$ or $\mathbb{R}^{+}$) do not include while the latin (French, Italian) and russian schools make a difference between positive and strictly positive and between negative and strictly negative. This means by the way that $0$ is the intersection of positive and negative numbers. One needs to know upfront the convention.&lt;/p&gt;&#xA;" OwnerUserId="441" LastActivityDate="2010-08-06T16:11:30.543" CommentCount="6" />
  <row Id="1738" PostTypeId="2" ParentId="1730" CreationDate="2010-08-06T16:14:41.257" Score="3" ViewCount="" Body="&lt;p&gt;Isaac's answer is almost exactly the first definition that I give. But what comes after is similar to what you are describing, I discuss each ambiguous term in the definition at great length, replacing the word with synonyms. Hopefully this lets the students take in the slight abstraction of the definition. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;After I do the defining and explaining of words I do about 9 examples. Three from each of three classes:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;little point-set diagrams, i.e. ovals with points inside them and lines going between them.&#xA;algebraically with f(x)=&#xA;graphs&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Two from each class of example are functions and I point out all the parts of the definition and what they correspond to. The third example in each class is a non-example, i.e. not a function. I point out where the issue is, thus showing that one point has two arrows coming out of it, one number can be plugged in to get two, and the vertical line test respectively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I find that this is very successful. &lt;/p&gt;&#xA;" OwnerUserId="16" LastActivityDate="2010-08-06T16:14:41.257" />
  <row Id="1739" PostTypeId="2" ParentId="1710" CreationDate="2010-08-06T16:17:25.780" Score="4" ViewCount="" Body="&lt;p&gt;For any particular polynomial there is an easier way to do indefinite summation than using the Bernoulli numbers, going off of Greg Graviton's answer.  Here we'll use the forward difference $\Delta f(x) = f(x+1) - f(x)$.  Then &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle \Delta {x \choose n} = {x \choose n-1}.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This implies that we can perform a &quot;Taylor expansion&quot; on any polynomial to write it in the form $f(x) = \sum a_n {x \choose n}$ by evaluating the finite differences $\Delta^n f(0)$ at zero.  For any particular polynomial $f$ it is easy to write these finite differences down by constructing a table.  In general, the formula is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle a_n = \Delta^n f(0) = \sum_{k=0}^{n} (-1)^{n-k} {n \choose k} f(k)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;as one can readily prove by writing $\Delta = S - I$ where $S$ is the shift operator $S f(x) = f(x+1)$ and $I$ is the identity operator $I f(x) = f(x)$.  Then the indefinite sum of $f$ is just $\sum a_n {x \choose n+1}$.  This is the easiest way I know how to do such computations by hand, and it also leads to a fairly easy method for polynomial interpolation given the values of a polynomial at consecutive integers.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-08-06T16:17:25.780" CommentCount="3" />
  <row Id="1740" PostTypeId="2" ParentId="1706" CreationDate="2010-08-06T17:08:32.327" Score="11" ViewCount="" Body="&lt;p&gt;You will often find $\mathbf R^+$ for the positive reals, and $\mathbf R^+_0$ for the positive reals and the zero.&lt;/p&gt;&#xA;" OwnerUserId="354" LastActivityDate="2010-08-06T17:08:32.327" />
  <row Id="1741" PostTypeId="1" CreationDate="2010-08-06T18:09:59.140" Score="-1" ViewCount="232" Body="&lt;p&gt;Classes are larger than sets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What's larger than classes ?&lt;/p&gt;&#xA;" OwnerUserId="882" LastActivityDate="2010-08-08T20:42:53.020" Title="Sets. Classes. ... ?" Tags="&lt;set-theory&gt;" AnswerCount="4" CommentCount="5" />
  <row Id="1742" PostTypeId="2" ParentId="514" CreationDate="2010-08-06T18:16:48.003" Score="5" ViewCount="" Body="&lt;p&gt;Another class of examples arise from diophantine equations with huge minimal solutions. Thus the conjecture that such an equation is unsolvable in integers has only huge counterexamples. Well-known examples arise from Pell equations, e.g. the smallest solution to the classic &lt;a href=&quot;http://en.wikipedia.org/wiki/Archimedes%27_cattle_problem&quot; rel=&quot;nofollow&quot;&gt;Archimedes Cattle problem&lt;/a&gt; has &lt;strong&gt;206545 decimal digits&lt;/strong&gt;, namely 77602714 ... 55081800.&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-08-06T18:16:48.003" />
  <row Id="1743" PostTypeId="2" ParentId="1741" CreationDate="2010-08-06T18:20:53.190" Score="0" ViewCount="" Body="&lt;p&gt;A class is a collection of things. Some classes are sets (in some sense, only &quot;smaller&quot; classes can be sets). Since &lt;em&gt;any&lt;/em&gt; collection is a class, there can't be anything larger.&lt;/p&gt;&#xA;" OwnerUserId="26" LastActivityDate="2010-08-06T18:20:53.190" CommentCount="1" />
  <row Id="1744" PostTypeId="2" ParentId="1730" CreationDate="2010-08-06T18:36:32.957" Score="24" ViewCount="" Body="&lt;p&gt;For fun, I like to liven-up the &quot;black box&quot;/machine view of a function by putting a monkey into the box. (I got pretty good at chalkboard-sketching a monkey that looked a little bit like &lt;a href=&quot;http://pbskids.org/curiousgeorge/&quot; rel=&quot;nofollow&quot;&gt;Curious George&lt;/a&gt;, but with a tail.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Give the Function Monkey an input and he'll cheerfully give you an output. The Function Monkey is smart enough to read and follow rules, and make computations, but he's not qualified to make decisions: his rules &lt;em&gt;must&lt;/em&gt; provide for exactly one output for a given input. &lt;em&gt;(Never let a Monkey choose!)&lt;/em&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can continue the metaphor by discussing the monkey's &quot;domain&quot; as the inputs he understands (what he can control); giving him an input outside his domain just confuses and frightens him ... or, depending upon the nature of the audience, kills him. &lt;em&gt;(What? You gave the Reciprocal Monkey a Zero? You killed the Function Monkey!)&lt;/em&gt; Of course, it's probably more appropriate to say that the Function Monkey simply &lt;em&gt;ignores&lt;/em&gt; such inputs, but students seem to like the drama. (As warnings go, &lt;em&gt;&quot;Don't kill the Function Monkey!&quot;&lt;/em&gt; gets more attention than &lt;em&gt;&quot;Don't bore the Function Monkey!&quot;&lt;/em&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Function Monkey comes in handy later when you start graphing functions: imagine that the x-axis is covered with coconuts (one coconut per &quot;x&quot; value). The Function Monkey strolls along the axis, picks up a &quot;x&quot; coconut, computes the associated &quot;y&quot; value (because that's what he &lt;em&gt;does&lt;/em&gt;), and then throws the coconut up (or down) the appropriate height above (or below) the axis, where it magically &lt;em&gt;sticks&lt;/em&gt; (or &lt;em&gt;hovers&lt;/em&gt; or whatever). So, if you ever want to plot a function, just &quot;Be a Function Monkey and throw some coconuts around&quot;. (Warning: Students may insist that that's not a &lt;em&gt;coconut&lt;/em&gt; the Monkey is throwing.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Further on, you can make the case that we're &lt;em&gt;smarter&lt;/em&gt; than monkeys (at least, we should strive to be): We don't always have to mindlessly plot points to know what the graph of an equation looks like; we can sometimes anticipate the outcome by studying the equation. This motivates manipulating an equation to tease out clues about the shape of its graph, explaining, for instance, our interest in the slope-intercept form of a line equation (and the almost-never-taught intercept-intercept form, which I personally like a lot), the special forms of conic section equations (which aren't all &lt;em&gt;functions&lt;/em&gt;, of course), and all that stuff related to translations and scaling.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Parametric equations can be presented as a way to let the Function Monkey plot elaborate curves ... both in the plane and in space (and beyond).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All in all, I find that the Function Monkey can make the course material more engaging &lt;em&gt;without dumbing it down&lt;/em&gt;; he provides a fun way to interpret the definitions and behaviors of functions, not a way to avoid them. Now, is the Function Monkey too cutesy for a College Algebra class? My high school students loved him, even at the Calculus level. One former student told me that he would often invoke the Function Monkey when tutoring his college peers. If it's clear to the students that the instructor isn't trying to patronize them, the Function Monkey may prove quite helpful.&lt;/p&gt;&#xA;" OwnerUserId="409" LastActivityDate="2010-08-06T18:36:32.957" CommentCount="3" />
  <row Id="1745" PostTypeId="1" AcceptedAnswerId="1757" CreationDate="2010-08-06T19:02:27.273" Score="3" ViewCount="141" Body="&lt;p&gt;I needed (for my research) to solve a Diophantine equation, in particular,&#xA;$$ 2 a + 3 b + 4 c + 5 d = 12 .$$&#xA;And I could easily solve it&#xA;(for example, on solution is $a=2, b=1, c=0, d=1$).&#xA;But this made me wonder if such equations, with their coefficients increasing sequences of&#xA;natural numbers, are a special case of Diophantine equations that are &lt;em&gt;always&lt;/em&gt; explicitly solvable, &#xA;despite the negative solution to Hilbert's 10th problem.&lt;/p&gt;&#xA;" OwnerUserId="237" LastEditorUserId="1778" LastEditorDisplayName="" LastEditDate="2010-09-21T03:30:34.870" LastActivityDate="2010-09-21T03:30:34.870" Title="Diophantine equations whose coefficients are increasing sequences of integers" Tags="&lt;number-theory&gt;&lt;diophantine-equations&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="1746" PostTypeId="2" ParentId="1382" CreationDate="2010-08-06T19:09:37.260" Score="1" ViewCount="" Body="&lt;p&gt;(Sketch of a bijective solution.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Recall that binomial coefficients count number of walks with steps (+1,+1) and (+1,-1) from the origin to different points (e.g. the number of walks to the point (2n,0) is $\binom{2n}{n}$). Consider the following involution on the set of all such walks: if the path intersects with the line y=l-1 or the line y=-1, reflect its part starting from the first intersection point (w.r.t. corresponding line).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This involution almost gives a bijection between P&lt;sub&gt;n&lt;/sub&gt;(r mod l) and P&lt;sub&gt;n&lt;/sub&gt;(-r-2 mod l) (and moving the strip we can get other correspondences of this kind). But it has some fixed points — namely, paths that lie inside the strip 0&amp;le;y&amp;le;l-2 (aka walks on the path graph of length l-1 mentioned by &lt;a href=&quot;http://math.stackexchange.com/users/232/qiaochu-yuan&quot;&gt;Qiaochu Yuan&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now to answer original question one only needs to recall that numbers of such paths for l=5 are exactly Fibonacci numbers.&lt;/p&gt;&#xA;" OwnerUserId="152" LastEditorUserId="152" LastEditorDisplayName="" LastEditDate="2010-08-07T07:24:39.447" LastActivityDate="2010-08-07T07:24:39.447" CommentCount="2" />
  <row Id="1747" PostTypeId="1" CreationDate="2010-08-06T19:12:23.137" Score="33" ViewCount="2227" Body="&lt;p&gt;Not a math student, so forgive me if the question seems trivial or if I pose it &quot;wrong&quot;. Here goes...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Say I'm flipping a coin a &lt;em&gt;n&lt;/em&gt; times. I am not sure if it's a &quot;fair&quot; coin, meaning I am not sure if it will come up heads and tails each with a propability of exactly 0.5. Now, if after &lt;em&gt;n&lt;/em&gt; throws it has come up heads exactly as many times as it has come up tails, then obviously there's nothing to indicate that the coin is not fair. But my intuition tells me that it would be improbable even for a completely fair coin to come up with heads and tails an exact even number of times given a large amount of tosses. My question is this: How &quot;off&quot; should the result be for it to be probable that the coin is not fair? IOW, how many more tosses should come up heads rather than tails in a series of &lt;em&gt;n&lt;/em&gt; throws before I should assume the coin is weighted?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Someone mentioned Pearson's chi-square test but then for some reason deleted their answer. Can someone confirm if that is indeed the right place to look for the answer?&lt;/p&gt;&#xA;" OwnerUserId="886" LastEditorUserId="886" LastEditorDisplayName="" LastEditDate="2010-08-07T09:07:19.153" LastActivityDate="2011-02-17T20:08:54.877" Title="How to determine if coin comes up heads more often than tails?" Tags="&lt;probability&gt;" AnswerCount="9" CommentCount="11" FavoriteCount="3" />
  <row Id="1748" PostTypeId="1" CreationDate="2010-08-06T19:40:32.847" Score="7" ViewCount="700" Body="&lt;p&gt;What are the best algebraic geometry textbooks for undergraduate students?&lt;/p&gt;&#xA;" LastActivityDate="2011-03-01T18:41:26.537" CommunityOwnedDate="2010-08-06T19:48:54.623" Title="(undergraduate) Algebraic Geometry Textbook Recomendations" Tags="&lt;algebraic-geometry&gt;" AnswerCount="8" CommentCount="1" FavoriteCount="9" />
  <row Id="1749" PostTypeId="2" ParentId="1741" CreationDate="2010-08-06T19:41:44.047" Score="2" ViewCount="" Body="&lt;p&gt;Classes are defined differently in systems, but essentially--&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sets are defined from bottom-up (adding elements starting from nothing) while classes are defined from top-down (starting with the universal class &lt;em&gt;V&lt;/em&gt; and then deciding from there which classes are sets). So it doesn't make sense to talk about things larger than the universal class, and if notions of comparing sizes are possible, you're simply dealing with classes that are also sets. (&lt;em&gt;V&lt;/em&gt; itself is not a set.)&lt;/p&gt;&#xA;" OwnerUserId="295" LastActivityDate="2010-08-06T19:41:44.047" CommentCount="1" />
  <row Id="1750" PostTypeId="2" ParentId="1748" CreationDate="2010-08-06T19:49:21.067" Score="7" ViewCount="" Body="&lt;p&gt;See the MathOverflow question about exactly this: &lt;a href=&quot;http://mathoverflow.net/questions/2446/best-algebraic-geometry-text-book-other-than-hartshorne&quot; rel=&quot;nofollow&quot;&gt;Best Algebraic Geometry text book?&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Personally, from what little I've seen, &lt;a href=&quot;http://www.cs.amherst.edu/~dac/iva.html&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;Ideals, Varieties and Algorithms&lt;/em&gt;&lt;/a&gt; by Cox, Litle and O'Shea is a very nice book for undergraduates. It requires little background, and has a computational perspective that's especially good for those with concrete or computational interests. I don't know how good it is as a preparation for further &quot;serious&quot; study in algebraic geometry, but it's certainly engaging.&lt;/p&gt;&#xA;" OwnerUserId="205" LastEditorUserId="205" LastEditorDisplayName="" LastEditDate="2010-08-06T20:09:47.640" LastActivityDate="2010-08-06T20:09:47.640" CommunityOwnedDate="2010-08-06T19:49:21.067" CommentCount="4" />
  <row Id="1752" PostTypeId="1" AcceptedAnswerId="1754" CreationDate="2010-08-06T19:57:01.113" Score="0" ViewCount="109" Body="&lt;p&gt;It seems like an extremely easy question, but I can't solve it. All I know is that it is a $\frac{1}{2^n}$ chance that all tosses will result in heads and $\sum_{k=0}^{n} f(k, n)$ is equal to 1 given the function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even though school has just started for me, this is not a homework question.&lt;/p&gt;&#xA;" OwnerUserId="227" LastActivityDate="2010-08-06T20:03:04.660" Title="What is the probability that exactly $i$ heads will result from $n$ coin tosses?" Tags="&lt;probability&gt;&lt;statistics&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="1754" PostTypeId="2" ParentId="1752" CreationDate="2010-08-06T20:03:04.660" Score="4" ViewCount="" Body="&lt;p&gt;This probability is equal to $f(i, n) = \binom{n}{i}2^{-n}$, where $\binom{n}{i} = \frac{n!}{i! (n-i)!}$ is &lt;a href=&quot;http://en.wikipedia.org/wiki/Binomial_coefficient&quot; rel=&quot;nofollow&quot;&gt;binomial coefficient&lt;/a&gt;. This is true because there are $\binom{n}{i}$ ways to choose which coins will be heads up and each such combinaton of $i$ heads up coins has the same probability $2^{-n}$.&lt;/p&gt;&#xA;" OwnerUserId="334" LastActivityDate="2010-08-06T20:03:04.660" CommentCount="1" />
  <row Id="1755" PostTypeId="2" ParentId="1747" CreationDate="2010-08-06T20:12:48.480" Score="8" ViewCount="" Body="&lt;p&gt;Given your prefatory comment, I'm going to avoid talking about the normal curve and the associated variables and use as much straight probability as possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's do a side problem first. If on a A-D multiple choice test you guess randomly, what's the probability you get 8 out of 10 questions right?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each problem you have a 25% (.25) chance of getting right and a 75% (.75) chance of getting wrong.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You want to first choose which eight problems you get right. That can be done in &lt;a href=&quot;http://www.wolframalpha.com/input/?i=10+choose+8&quot; rel=&quot;nofollow&quot;&gt;10 choose 8&lt;/a&gt; ways.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You want .25 to happen eight times [$(.25)^8$] and .75 to happen twice [$(.75)^2$]. This needs to be multiplied by the possible number of ways to arrange the eight correct problems, hence your odds of getting 8 out of 10 right is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;${10 \choose{8}}(.25)^8(.75)^2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ok, so let's say you throw a coin 3000 times. What's the probability that it comes up heads only 300 times? By the same logic as the above problem that would be&lt;/p&gt;&#xA;&#xA;&lt;p&gt;${3000 \choose{300}}(.5)^{300}(.5)^{2700}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or a rather unlikely 6.92379... x 10^-482.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given throwing the coin n times, the probability it comes up heads x times is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;${n \choose{x}}(.5)^n$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or if you want to ask the probability it comes up heads x times or less&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sum_{i=0}^{x}{{n \choose{i}}(.5)^n}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so all you have to do is decide now how unlikely are you willing to accept?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(This was a &lt;a href=&quot;http://en.wikipedia.org/wiki/Binomial_probability&quot; rel=&quot;nofollow&quot;&gt;Binomial Probability&lt;/a&gt; if you want to read more and all the fancier methods involving an integral under the normal curve and whatnot start with this concept.)&lt;/p&gt;&#xA;" OwnerUserId="295" LastEditorUserId="295" LastEditorDisplayName="" LastEditDate="2010-08-06T20:18:58.093" LastActivityDate="2010-08-06T20:18:58.093" CommentCount="1" />
  <row Id="1756" PostTypeId="2" ParentId="659" CreationDate="2010-08-06T20:21:01.613" Score="0" ViewCount="" Body="&lt;p&gt;You can discuss duals in a monoidal category (which may not be symmetric).&#xA;This has been mentioned by Akhil.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $V$ and $W$ be objects (in your monoidal category) with $K$ the identity for tensor product. Then you require morphisms $K\rightarrow V\otimes W$ and $W\otimes V\rightarrow K$&#xA;which satisfy the zig-zag identities (so-called because this becomes clear if you draw string diagrams). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Formally the zig-zag identities are &#xA;$$V=K\otimes V\rightarrow V\otimes W\otimes V\rightarrow V\otimes K=V$$&#xA;is the identity map and&#xA;$$W=W\otimes K\rightarrow W\otimes V\otimes W\rightarrow K\otimes W=k$$&#xA;is the identity map.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is equivalent to $Hom(W\otimes X,Y)=Hom(X,V\otimes Y)$ and $Hom(X\otimes V,Y)=Hom(V,Y\otimes W)$ (both natural in $X$ and in $Y$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then you say $V$ is left/right dual to $W$ and $W$ is right/left dual to $V$&#xA;(I can never remember which). Then we can define $V$ to be dual to $W$ if it is left dual and right dual.&lt;/p&gt;&#xA;" OwnerUserId="638" LastActivityDate="2010-08-06T20:21:01.613" />
  <row Id="1757" PostTypeId="2" ParentId="1745" CreationDate="2010-08-06T20:38:46.367" Score="8" ViewCount="" Body="&lt;p&gt;Linear Diophantine equations are always &lt;strike&gt;solvable&lt;/strike&gt; decidable (in linear time).  If the coefficients are $a_1, a_2, ... a_n$ then the numbers that can appear on the RHS are precisely the multiples of $\text{gcd}(a_1, ... a_n)$, and one can find solutions using the extended Euclidean algorithm.&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-08-06T21:15:48.970" LastActivityDate="2010-08-06T21:15:48.970" CommentCount="3" />
  <row Id="1758" PostTypeId="1" AcceptedAnswerId="1792" CreationDate="2010-08-06T20:40:58.017" Score="5" ViewCount="346" Body="&lt;p&gt;Can you divide a disk into 7 pieces of equal area, with 3 line segments? (You can surely divide it into 7 pieces, but could those have equal areas?)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(This question was left unanswered at another forum. I can see with some visual arguments that the answer should be &lt;em&gt;no&lt;/em&gt;, but I couldn't find a nice way to write it down.)&lt;/p&gt;&#xA;" OwnerUserId="145" LastEditorUserId="145" LastEditorDisplayName="" LastEditDate="2010-08-06T21:01:39.757" LastActivityDate="2010-08-09T21:23:41.117" Title="Dividing a disk into 7 equal pieces with 3 line segments" Tags="&lt;euclidean-geometry&gt;" AnswerCount="4" FavoriteCount="1" />
  <row Id="1759" PostTypeId="2" ParentId="1748" CreationDate="2010-08-06T20:51:34.690" Score="2" ViewCount="" Body="&lt;p&gt;Everyone is linking to that MO question, so allow me to link to another MO &lt;a href=&quot;http://mathoverflow.net/questions/14371/looking-for-an-introductory-textbook-on-algebraic-geometry-for-an-undergraduate-l&quot; rel=&quot;nofollow&quot;&gt;question&lt;/a&gt; which is more relevant to this specific request.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, I should mention that it partially depends on what you are looking for. Should this be for self study with little help from professors? Then your best bet would be to follow Ravi Vakil's &lt;a href=&quot;http://math216.wordpress.com/&quot; rel=&quot;nofollow&quot;&gt;course next semester&lt;/a&gt; or use his old notes &lt;a href=&quot;http://math.stanford.edu/~vakil/0708-216/&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you intend on having a professor over see the self study then I think that Ueno's books: &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0821808621&quot; rel=&quot;nofollow&quot;&gt;Alg Geom 1&lt;/a&gt; &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0821813579&quot; rel=&quot;nofollow&quot;&gt;and 2&lt;/a&gt;, which I think are great for self study.(I should mention that his lower level &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0821811444&quot; rel=&quot;nofollow&quot;&gt;book&lt;/a&gt; is not nearly as good and I would NOT recommend it.) Another strong candidate for led self study would be &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/144193099X&quot; rel=&quot;nofollow&quot;&gt;Harris' nice intro&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are going to offer a course or take a course then I think &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/1441928073&quot; rel=&quot;nofollow&quot;&gt;everybody's buddy&lt;/a&gt; would be nice augmented by &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0387986375&quot; rel=&quot;nofollow&quot;&gt;this masterpiece&lt;/a&gt; would be a sweet course, but likely too challenging for undergrad.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think that covers the basics. :D&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Two more comments, &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0471050598&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; book is frustrating and I do not think it is appropriate to look at for undergrads trying to learn algebraic geometry, especially if they are moving to modern scheme theory. And another book with a very different feeling, but a really fun way to start thinking about algebraic geometry is Katz' &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0821836870&quot; rel=&quot;nofollow&quot;&gt;Enumerative Geometry book&lt;/a&gt;. It was written especially for undergrads, and covers some really neat material that can give some good motivation to think about algebraic geometry and topology.&lt;/p&gt;&#xA;" OwnerUserId="16" LastActivityDate="2010-08-06T20:51:34.690" CommunityOwnedDate="2010-08-06T20:51:34.690" CommentCount="14" />
  <row Id="1760" PostTypeId="2" ParentId="577" CreationDate="2010-08-06T21:00:58.133" Score="9" ViewCount="" Body="&lt;p&gt;Set theory is completely full of new axioms, some of them expressing fundamental set principles, and many of them having consequences even in natural number arithmetic that are not provable without them (via consistency strength statements).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the past fifty years of research in set theory, a major lesson that has been learned is that many many fundamental questions about set theory are simply independent of the usual ZFC axioms. This includes most all of the questions about infinite cardinal arithmetic, but also subtle questons about infinite combinatorics.  For example, Is the exponential function (size of power set function) injective on infinite cardinalities? Independent of ZFC. Do Souslin trees exist?  Independent of ZFC. Is the size of the smallest dominating family of functions $f:\omega\to\omega$ necessarily $\aleph_1$?  Independent of ZFC. Can Lebesgue measure be more than countably additive, when CH fails?  Independent of ZFC. &#xA;There are hundreds of examples. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The response to this phenomenon was naturally to investigate the various hypotheses that go beyond ZFC. So we now know a great deal about what happens when CH holds, or when it fails, or when CH fails but the dominating number is low, and so on. The effect of this is to treat these hypotheses as semi-temporary axioms within a domain of research, which may not be the same usage of axiom that you inquired about, but the effect is the same. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The powerful method of forcing, invented by Cohen in order to prove the consistency of $ZFC+\neg CH$, allows us to show that many other theories are also consistent. This method has given rise to a number of forcing axioms, such as Martin's Axiom, the Proper Forcing Axiom, Martin's Maximum and many others. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A much larger and very interesting class of axioms are provided by the large cardinal hierarchy, involving such notions as inaccessible cardinals, measurable cardinals, Woodin cardinals and so on. A curious feature of these strong axioms of infinity is that they imply certain highly regular features to occur among the projective sets of reals. Thus, although the axioms themselves seem to have nothing to do with sets of reals, they imply very nice properties for the sets of reals that we can easily define. Many set theorists take this as some kind of positive evidence for their truth. Meanwhile, the large cardinal hypotheses are intensely studied as a fundamental research effort to understand the nature of mathematical infinity. But you cannot do this without assuming that these cardinals exist, since this is provably not provable in ZFC, if consistent, and so these large cardinal hypotheses constitute new axioms. &lt;/p&gt;&#xA;" OwnerUserId="413" LastActivityDate="2010-08-06T21:00:58.133" CommentCount="1" />
  <row Id="1761" PostTypeId="2" ParentId="1730" CreationDate="2010-08-06T21:06:19.310" Score="1" ViewCount="" Body="&lt;p&gt;I start with the notion of an expression. An &lt;em&gt;expression&lt;/em&gt; is a grammatically meaningful combination of variables and constants. I don't need to tell &lt;em&gt;this&lt;/em&gt; audience what I mean by that. An &lt;em&gt;equation&lt;/em&gt; relate two expressions. To solve an equation for a variable one manipulates the equation according to rules until one variable is written unambiguously in terms of the others. This may not be always be done. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A &lt;em&gt;function&lt;/em&gt; is an equation in two variables in which one variable (y) can be solved as an unambiguous expression in the other variable (x). Thus y can be written as an expression in x. &#xA;Then I let the students know (a) this definition is not quite good enough for mathematicians, (b) it will work pretty well for all of the applications that we have in mind. Throughout the process examples are given.&lt;/p&gt;&#xA;" OwnerUserId="722" LastActivityDate="2010-08-06T21:06:19.310" />
  <row Id="1762" PostTypeId="2" ParentId="1748" CreationDate="2010-08-06T21:12:05.710" Score="1" ViewCount="" Body="&lt;p&gt;My favorite book at undegraduate level is Harris' Algebraic Geometry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For something more advanced, with a strongly geometric point of view, I advise Griffiths and Harris Principles of algebraic geometry. It is far more demanding, though.&lt;/p&gt;&#xA;" OwnerUserId="233" LastActivityDate="2010-08-06T21:12:05.710" CommunityOwnedDate="2010-08-06T21:12:05.710" />
  <row Id="1763" PostTypeId="2" ParentId="1741" CreationDate="2010-08-06T21:17:17.673" Score="3" ViewCount="" Body="&lt;p&gt;From one possible point of view, conglomerate come next.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take a model of set theory $\mathcal{U}$ and assume you have an inaccesible cardinal; let $S \in \mathcal{U}$ be a set of that cardinality. Then $S$ itself is another model of set theory. So you can (re)-define:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Sets as elements of $S$. That is, $A \in \mathcal{U}$ is a set if and only if $A \in S$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Classes as subsets of $S$. That is, $A \in \mathcal{U}$ is a class if and only if $A \subset S$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) Conglomerates as elements of $\mathcal{U}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this way sets and classes will satisfy the usual axioms, and you have conglomerates as even bigger collections.&lt;/p&gt;&#xA;" OwnerUserId="233" LastActivityDate="2010-08-06T21:17:17.673" CommentCount="3" />
  <row Id="1764" PostTypeId="2" ParentId="1748" CreationDate="2010-08-06T21:21:05.740" Score="7" ViewCount="" Body="&lt;p&gt;Cox, Little, and O'Shea is an excellent book.  I'd also like to recommend Miles Reid's &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0521356628&quot; rel=&quot;nofollow&quot;&gt;Undergraduate Algebraic Geometry&lt;/a&gt;, which unlike Hartshorne or many other standard references was written specifically with undergraduates in mind.  You should also check out his book on commutative algebra; I found both very enlightening.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-08-06T21:21:05.740" CommunityOwnedDate="2010-08-06T21:21:05.740" />
  <row Id="1765" PostTypeId="2" ParentId="1758" CreationDate="2010-08-06T21:31:15.733" Score="1" ViewCount="" Body="&lt;p&gt;Each line segment is a chord of the circle, with 3 of the 7 regions in the segment of the circle determined by the chord (the segment is the region between the chord and the minor arc between its endpoints).  The area of a segment with arc measure &amp;alpha; radians of a circle with radius r is $\frac{r^2}{2}(\alpha-\sin\alpha)$&amp;mdash;for simplicity, let $r=\sqrt{2}$ so the segment area is $\alpha-\sin\alpha$ and the area of the circle is $2\pi$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If each of the 7 regions has equal area, then the area of the segment determined by each chord must be $\frac{3}{7}$ of the area of the circle, so for each chord, $\alpha-\sin\alpha=\frac{6\pi}{7}\Rightarrow\alpha\approx2.91624$.  With 3 such chords, there is no way to get the central triangular region to have area more than about $\frac{1}{10}$ of the circle, so no way to get 7 regions of equal area.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Below left is just about the largest the central triangular region can be; below right is a more symmetric configuration, which actually doesn't make the central triangle much smaller.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://www.imgftw.net/img/225458569.png&quot; alt=&quot;largest central triangle&quot;&gt; &lt;img src=&quot;http://www.imgftw.net/img/193455867.png&quot; alt=&quot;somewhat symmetric configuration&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&lt;strong&gt;edit&lt;/em&gt;&lt;/strong&gt;:  When I said &quot;empirically&quot; in a comment below, that was a bit imprecise: given an angle of measure 2.91624 radians, I constructed (compass and straightedge) a circle with one chord meeting the given criteria.  I then constructed a few dozen possible locations for the second chord and for each of those, a few hundred locations for the third chord, measuring the area of the central triangle, when it existed.  There are 4 distinct configurations where it exists, corresponding to the 4 regions in the contour plot in my other answer.  The area function is continuous on these regions, so a well-distributed large sample of values across each of the regions gives an accurate depiction of the behavior of the function.  When a triangle has some aspects free and some aspects constrained, the maximum area is typically at an extreme value of the free aspects or at a configuration of maximal symmetry.  The multitude of constructions indicated local extrema for the configuration above left (narrowest but tallest) and for a configuration where the side of the triangle on the other chord is nearly the whole chord (shortest but widest).  The configuration shown above left had the largest area (this is verified computationally in my other answer).  Neither of these extreme configurations, however, are anywhere near 7 equal-area regions&amp;mdash;in each case, at least one region is reduced to the degenerate case of a point.  Above right is the most symmetrical configuration, which is intuitively the most likely to generate 7 equal-area regions.  This corresponds to the middle of the lighter-colored region in the middle of the contour plot in my other answer and is at or near a local minimum of the area function.&lt;/p&gt;&#xA;" OwnerUserId="72" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-08-07T17:52:15.070" LastActivityDate="2010-08-07T17:52:15.070" CommentCount="6" />
  <row Id="1766" PostTypeId="1" AcceptedAnswerId="1783" CreationDate="2010-08-06T22:28:06.517" Score="10" ViewCount="623" Body="&lt;p&gt;Suppose $a_{n}&gt;0$ and the following series converges&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sum_{n=1}^{\infty} a_{n}^{3}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does this imply that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sum_{n=1}^{\infty} \frac{a_{n}}{n}$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;converges?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was able to prove that the second series also converges by using the limit comparision test.   Is there another way to show the second series converges (e.g. root or ratio test)?&lt;/p&gt;&#xA;" OwnerUserId="832" LastEditorUserId="832" LastEditorDisplayName="" LastEditDate="2010-08-07T02:35:37.280" LastActivityDate="2010-08-08T22:07:07.723" Title="If $\sum_{n=1}^{\infty} a_{n}^{3}$ converges does $\sum_{n=1}^{\infty} \frac{a_{n}}{n}$ converge?" Tags="&lt;real-analysis&gt;" AnswerCount="2" CommentCount="5" FavoriteCount="2" />
  <row Id="1767" PostTypeId="1" AcceptedAnswerId="1770" CreationDate="2010-08-06T22:59:41.843" Score="5" ViewCount="186" Body="&lt;p&gt;Hi All,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This one has been bugging me for a while - I have a number of definitions for a symmetric group, none of which I understand. Clearly, $S_n$ is a group, which I follow. Now, the wikipedia definition (most readily to hand) is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The symmetric group on a set X is the&#xA;  group whose underlying set is the&#xA;  collection of all bijections from X to&#xA;  X and whose group operation is that of&#xA;  function composition.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Of that, I can see (clearly) we have an underlying set which combined with an operation follows set rules as per the definition of a group. I understand a bijection from X to X implies that $\forall x \in X$, $x \mapsto x$. So, just as in geometric symmetry a set is symmetrical if it becomes itself again after some map.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All well and good, but how do composite functions relate to this? Surely taking the bijective requirement into consideration the only operation permitted is one that effectively leaves the set unchanged?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, can anyone please either explain how composite functions fit in my definition or (equally fine) provide an example of a function that works so I can try to fit the idea to the definitions I've come across.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many thanks.&lt;/p&gt;&#xA;" OwnerUserId="892" LastActivityDate="2010-08-07T16:22:59.393" Title="What is a symmetric group?" Tags="&lt;group-theory&gt;" AnswerCount="3" />
  <row Id="1768" PostTypeId="1" AcceptedAnswerId="1769" CreationDate="2010-08-06T23:07:31.847" Score="5" ViewCount="163" Body="&lt;p&gt;&lt;a href=&quot;http://math.stackexchange.com/questions/1766/convergent-series&quot;&gt;This question&lt;/a&gt; just reminded me of a conundrum I posed myself in my first year of university. I never did get a satisfactory answer...&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Let $a_n$ be a &lt;a href=&quot;http://www.answers.com/topic/null-sequence&quot; rel=&quot;nofollow&quot;&gt;null sequence&lt;/a&gt;. Does it follow that $\sum \frac{a_n}{n}$ converges?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Any ideas?&lt;/p&gt;&#xA;" OwnerUserId="160" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-08-07T01:06:31.277" LastActivityDate="2010-08-07T01:06:31.277" Title="More convergent series..." Tags="&lt;sequences-and-series&gt;&lt;analysis&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="1769" PostTypeId="2" ParentId="1768" CreationDate="2010-08-06T23:20:13.870" Score="12" ViewCount="" Body="&lt;p&gt;If by null sequence you mean a sequence that converges to 0, then no. Try $a_n=1/\log n.$ By integral comparison, the series diverges:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_2^\infty\dfrac1{n\log n}\geq\int_2^\infty\dfrac{dx}{x\log x}=\int_{\log 2}^\infty\dfrac{du}u=\infty,$$&#xA;where I've used the change of variables $u=\log x$.&lt;/p&gt;&#xA;" OwnerUserId="405" LastEditorUserId="405" LastEditorDisplayName="" LastEditDate="2010-08-07T00:03:50.947" LastActivityDate="2010-08-07T00:03:50.947" CommentCount="3" />
  <row Id="1770" PostTypeId="2" ParentId="1767" CreationDate="2010-08-06T23:25:07.577" Score="6" ViewCount="" Body="&lt;p&gt;I think you misunderstood the definition/use of &lt;em&gt;bijection&lt;/em&gt;. I'll use an example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $X = \{1, 2, 3\}$, the appropriate symmetric group is $S_3$, and it is the set of all possible permutations of X:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$f_1: 1\to 1, 2\to 2, 3\to 3$&lt;/li&gt;&#xA;&lt;li&gt;$f_2: 1\to 1, 2\to 3, 3\to 2$&lt;/li&gt;&#xA;&lt;li&gt;$f_3: 1\to 3, 2\to 2, 3\to 1$&lt;/li&gt;&#xA;&lt;li&gt;$f_4: 1\to 2, 2\to 1, 3\to 3$&lt;/li&gt;&#xA;&lt;li&gt;$f_5: 1\to 2, 2\to 3, 3\to 1$&lt;/li&gt;&#xA;&lt;li&gt;$f_6: 1\to 3, 2\to 1, 3\to 2$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Generally, bijections from a set X to itself are &lt;em&gt;permutations&lt;/em&gt;, and intuitively they are &quot;re-orderings&quot; of the members of the set. It then follows that for a set of size n there are exactly n! permutations, and thus $|S_n| = n!$.&lt;/p&gt;&#xA;" OwnerUserId="26" LastActivityDate="2010-08-06T23:25:07.577" CommentCount="2" />
  <row Id="1771" PostTypeId="2" ParentId="1767" CreationDate="2010-08-06T23:27:06.823" Score="1" ViewCount="" Body="&lt;p&gt;Each member of $S_n$ is a bijection and the composition of two bijections is another bijection. Therefore $S_n$ is closed under composition.&lt;/p&gt;&#xA;" OwnerUserId="207" LastActivityDate="2010-08-06T23:27:06.823" />
  <row Id="1772" PostTypeId="2" ParentId="518" CreationDate="2010-08-06T23:33:46.173" Score="0" ViewCount="" Body="&lt;p&gt;In the early days of commutative algebra people did not use the language of localization, so probably such proof exists. In particular, I have been told Kaplansky's early book (around 1950?) might be such reference, but I don't have it right now.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In any case, &lt;a href=&quot;http://books.google.com/books?id=Fm_yPgZBucMC&amp;amp;lpg=PA79&amp;amp;ots=7Ap1zxS3Ig&amp;amp;dq=localization%20without%20admitting%20it&amp;amp;pg=PA79#v=onepage&amp;amp;q=localization%20without%20admitting%20it&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;Exercise 2.3&lt;/a&gt; of Eisenbud tells you how to cheat and localize without admitting it. I might be wrong here, but what you want to prove is about some multiplicative properties of elements in $R$, so perhaps any proof you can find is just localization in disguise. &lt;/p&gt;&#xA;" OwnerUserId="223" LastActivityDate="2010-08-06T23:33:46.173" />
  <row Id="1774" PostTypeId="1" CreationDate="2010-08-07T00:52:37.047" Score="4" ViewCount="693" Body="&lt;p&gt;I don't know whether the books metioned in &lt;a href=&quot;http://math.stackexchange.com/questions/329/best-ever-book-on-number-theory&quot;&gt;http://math.stackexchange.com/questions/329/best-ever-book-on-number-theory&lt;/a&gt; are beyond undergraduate/high-school-olympiad level.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please recommend your favourite.&lt;/p&gt;&#xA;" OwnerUserId="696" LastActivityDate="2010-08-17T21:44:34.547" CommunityOwnedDate="2010-08-07T00:55:26.913" Title="Undergraduate/High-School-Olympiad Level Introductory Number Theory Books For Self-Learning" Tags="&lt;number-theory&gt;&lt;books&gt;" AnswerCount="4" CommentCount="6" FavoriteCount="5" />
  <row Id="1775" PostTypeId="2" ParentId="362" CreationDate="2010-08-07T00:57:13.837" Score="0" ViewCount="" Body="&lt;p&gt;There's also the books &lt;a href=&quot;http://www.springer.com/birkhauser/historyofscience/book/978-0-8176-4684-4&quot; rel=&quot;nofollow&quot;&gt;A History of Abstract Algebra&lt;/a&gt; and &lt;a href=&quot;http://www.ams.org/bookstore-getitem/item=hmath-32&quot; rel=&quot;nofollow&quot;&gt;Episodes in the History of Modern Algebra (1800-1950)&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="589" LastActivityDate="2010-08-07T00:57:13.837" />
  <row Id="1776" PostTypeId="2" ParentId="740" CreationDate="2010-08-07T02:31:06.337" Score="4" ViewCount="" Body="&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Bump_function1.1.&quot; rel=&quot;nofollow&quot;&gt;bump function&lt;/a&gt; $\psi\colon \mathbb{R} \to \mathbb{R}$ given by $\psi(x) = e^{-1/(1-x^2)}$ for $|x|\leq 1$ and $\psi(x) = 0$ for $|x| &gt; 1$ is not exactly pathological per se, but it's very useful for (at least) two things:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;It's an example of a function that is smooth (all derivatives exist) but not analytic (not equal to its Taylor series in a neighbourhood of every point) -- look at the points $x=\pm 1$.&lt;/li&gt;&#xA;&lt;li&gt;You can use it to build partitions of unity on manifolds, which is important when you want to build a global object (like a Riemannian metric) out of local data (defined in each chart).&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="876" LastActivityDate="2010-08-07T02:31:06.337" CommunityOwnedDate="2010-08-07T02:31:06.337" CommentCount="2" />
  <row Id="1777" PostTypeId="2" ParentId="1747" CreationDate="2010-08-07T02:31:23.823" Score="-1" ViewCount="" Body="&lt;p&gt;There is a very simple probability formula called the Binomial Distribution.&#xA;The Binomial Distribution works as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Two start off with, I need a game with only two possible outcomes: win or lose. This is where the &quot;bi&quot; (meaning two) of the &quot;binomial distribution&quot; comes from. Next, each possible outcome needs to have an associated probability. In this case, both the heads and tails have a probability of 50%, which we write as $0.5$. The binomial distribution will tell us the probability of getting $k$ heads in $n$ trials. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is the formula:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$B(k,n,p) = \frac{n!}{k!(n-k)!}  {p^{k}}  {{(1-p)}^{n-k}}$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remember, k is the number of heads, n is the number of trials, and p the probability of getting a heads. The probability of getting a tails is thus 1-p, which is also 0.5.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose we roll the dice 100 times and get only 35 heads.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$B(35,100,0.5) = \frac{100!}{35!(100-35)!}  {0.5^{35}}  {{(1-0.5)}^{100-35}}$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is equal to 0.000863855665741652, which is a 0.0863855666% probability (out of 100). That's more than enough to show that this dice is probably loaded. You can now use this formula with any number of heads in any number of flips to calculate the probability. It's up to you to decide how low a percentage you're willing to take, but I'd generally advise below two standard deviations from the mean: below 4.54%.&lt;/p&gt;&#xA;" OwnerUserId="647" LastActivityDate="2010-08-07T02:31:23.823" CommentCount="1" />
  <row Id="1778" PostTypeId="1" CreationDate="2010-08-07T02:49:12.603" Score="6" ViewCount="425" Body="&lt;p&gt;Motivation: A friend asked me this question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Problem: Suppose you start off with a dollar. You flip a fair coin, if it lands on heads you win 50 cents otherwise you lose 50 cents. If after n flips you have a nonzero amount of money, you win. Whats the probability you win? What about the limiting case as n tends to infinity?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;edit: In this game you are not allowed to have negative money. Thanks, Jonathan Fischoff, the linked helped greatly.&lt;/p&gt;&#xA;" OwnerUserId="899" LastEditorUserId="899" LastEditorDisplayName="" LastEditDate="2010-08-07T09:10:11.407" LastActivityDate="2010-09-05T10:17:19.317" Title="A probability game" Tags="&lt;probability&gt;" AnswerCount="4" CommentCount="3" FavoriteCount="3" />
  <row Id="1779" PostTypeId="2" ParentId="1774" CreationDate="2010-08-07T03:27:42.397" Score="4" ViewCount="" Body="&lt;p&gt;There are several elementary number theory books which you could use and which do not assume a level of knowledge beyond high school math. I would strongly recommend Underwood Dudley's Elementary Number Theory and Harold Stark's An Introduction to Number Theory. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;They're both beautifully written and cover most of the things that are usually covered in any introductory number theory course (at a basic level of course). Some of the other books that have already been suggested are excelent. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Particularly Katie Banks' suggestion of Ireland and Rosen, although this book makes some use of the language of groups, rings and fields, so it may be more advanced for a high school student.&lt;/p&gt;&#xA;" OwnerUserId="900" LastActivityDate="2010-08-07T03:27:42.397" CommunityOwnedDate="2010-08-07T03:27:42.397" CommentCount="1" />
  <row Id="1781" PostTypeId="1" AcceptedAnswerId="1784" CreationDate="2010-08-07T04:11:28.180" Score="6" ViewCount="235" Body="&lt;p&gt;This is the equation I'm having trouble with:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$G \frac{M m}{r^2} = m \frac{d^2 r}{dt^2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's the non-vector form of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Newton%27s_law_of_universal_gravitation&quot; rel=&quot;nofollow&quot;&gt;universal law of gravitation&lt;/a&gt; on the left and &lt;a href=&quot;http://en.wikipedia.org/wiki/Newton%27s_laws_of_motion&quot; rel=&quot;nofollow&quot;&gt;Newton's second law of motion&lt;/a&gt; on the right. I assume that upon correctly modeling and solving this, I will have a function of time that outputs the distance from a spherical mass in space (ie. distance from the Earth given an initial condition of $r(0) = 10,000 \mathrm{km}$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, WolframAlpha gives a &lt;a href=&quot;http://tinyurl.com/35vuvoe&quot; rel=&quot;nofollow&quot;&gt;hell of an answer&lt;/a&gt;, which leads me to believe that I'm modeling this equation completely wrong. Can anyone shed some light on this problem?&lt;/p&gt;&#xA;" OwnerUserId="227" LastActivityDate="2010-08-07T15:55:51.887" Title="Can this gravitational field differential equation be solved, or does it not show what I intended?" Tags="&lt;physics&gt;&lt;differential-equations&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="2" />
  <row Id="1782" PostTypeId="2" ParentId="1747" CreationDate="2010-08-07T04:44:33.790" Score="4" ViewCount="" Body="&lt;p&gt;This question is one of statistics (esp. statistical inference), not probability per se.  Keywords: &quot;binomial sampling&quot;; &quot;confidence interval for a proportion&quot;.  Asking this at &lt;a href=&quot;http://stats.stackexchange.com&quot;&gt;http://stats.stackexchange.com&lt;/a&gt; will get more complete answers. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The related probability fact is: if a coin has probability $p$ of coming up heads and is tossed $n$ times, then the observed number of heads will on average be $np$, but we expect the observed number to fluctuate (if the experiment with $n$ tosses is repeated many times) around the average by an amount on the order of $\sqrt{np(1-p)}$.  Keywords: normal distribution, bell curve, Central Limit Theorem, binomial distribution, convergence of binomial distribution to normal (Gaussian) distribution.&lt;/p&gt;&#xA;" OwnerUserId="467" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-08-07T06:41:51.903" LastActivityDate="2010-08-07T06:41:51.903" CommentCount="3" />
  <row Id="1783" PostTypeId="2" ParentId="1766" CreationDate="2010-08-07T05:10:35.953" Score="6" ViewCount="" Body="&lt;p&gt;$3|a_n|/n \leq (|a_n|^3 + (2/n^{3/2})$ by a well known inequality.  Sum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This also shows the range of values that the second infinite sum can assume, given the value of the first, and assuming all terms are non-negative.&lt;/p&gt;&#xA;" OwnerUserId="467" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-08-07T08:09:22.480" LastActivityDate="2010-08-07T08:09:22.480" />
  <row Id="1784" PostTypeId="2" ParentId="1781" CreationDate="2010-08-07T06:21:08.480" Score="10" ViewCount="" Body="&lt;p&gt;Yes you're modeling it wrong. Gravity is an attractive force, so if the mass is at r = 0, and r is nonnegative, then the gravity force should point towards zero, i.e. negative.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ - \frac{GMm}{r^2} = m\ddot r$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But now Alpha even refuses to solve it. That's fine, because Mathematica can't solve many things. The usual approach is to notice that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \frac{d^2r}{dt^2} = \frac{dv}{dt} = \frac{dv}{dr}\frac{dr}{dt} = v \frac{dv}{dr} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;to convert that 2nd-order ODE to a 1st-order ODE:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ -GM \frac{dr}{r^2} = v \,dv $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This gives the solution v(r).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \frac{GM}r + \text{constant} = \frac{v^2}2 \qquad (*) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But recall $v = dr/dt$. So we have another 1st-order ODE to solve, which gives r(t). The solution should look like that “hell of answer” because of the square root.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&amp;nbsp;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: If you rearrange the terms you should see that Equation (*) is just conservation of energy. &lt;/p&gt;&#xA;" OwnerUserId="171" LastEditorUserId="171" LastEditorDisplayName="" LastEditDate="2010-08-07T15:55:51.887" LastActivityDate="2010-08-07T15:55:51.887" />
  <row Id="1786" PostTypeId="2" ParentId="1778" CreationDate="2010-08-07T07:21:51.247" Score="3" ViewCount="" Body="&lt;p&gt;This type problem on random walk is usually solved with the Reflection Principle, with the walk visualized as a lattice path.  Strangely, I can't find an online reference to the solution but it is given in Feller's book on probability theory, volume 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here, measuring the money in units of 0.5 dollars,  the walk, drawn in the $(x,y)$ plane, starts at (2,0), moves by  $(+1,\pm 1)$ at each step, and the question is what is the probability that over $n$ moves the walk is always $\geq 1$.&lt;/p&gt;&#xA;" OwnerUserId="467" LastActivityDate="2010-08-07T07:21:51.247" CommentCount="1" />
  <row Id="1788" PostTypeId="2" ParentId="1747" CreationDate="2010-08-07T08:20:59.570" Score="1" ViewCount="" Body="&lt;p&gt;In typical math problems involving coins we know the probability ahead of time. If you know the probability of heads and tails with absolute certainty the sequence of heads and tails you get from flipping the coin tells you &lt;strong&gt;nothing&lt;/strong&gt; about the what the probability should be (because you already know it). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand if you have absolutely no idea what the probability is, and all you know is that the there are two sides you will end up with something called the &quot;Rule of Succession&quot; or $\frac{Heads + 1}{Total + 2}$. &lt;a href=&quot;http://en.wikipedia.org/wiki/Rule_of_succession&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Rule_of_succession&lt;/a&gt; for assigning the probability.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your case it sounds like you are in between the two. You have some information that leads you to believe that the coin is balanced but you are not positive, and the thus the frequencies from flipping will affect your probability assignment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is why this question is tricky. You have to take everything you know about the coin and encode mathematically.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One why to think about doing that is assume that you have already flipped the coin x number of times. So in the case of knowing the probability for certain, it is as though we have flipped the coin an infinite number of times. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So maybe you would start by assuming that your prior knowledge is equivalent to having flipped the coin 10,000 times with half heads and half tails. That way the frequency will still will affect it, but not so much initially.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't of any mathematically texts that have attempted at solving anything but the two extreme cases (complete information, zero information). A physicist, Jaynes attempted to broach this problem here &lt;a href=&quot;http://www-biba.inrialpes.fr/Jaynes/cc18i.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www-biba.inrialpes.fr/Jaynes/cc18i.pdf&lt;/a&gt; (I can only partially follow his reasoning), but I don't know if there is an accepted mathematically rigorous way to solve this type of problem.&lt;/p&gt;&#xA;" OwnerUserId="99" LastActivityDate="2010-08-07T08:20:59.570" CommentCount="6" />
  <row Id="1789" PostTypeId="1" AcceptedAnswerId="1790" CreationDate="2010-08-07T08:41:44.397" Score="2" ViewCount="367" Body="&lt;p&gt;I just finished reading the Wikipedia article on the &lt;a href=&quot;http://en.wikipedia.org/wiki/Cauchy%27s_condensation_test&quot; rel=&quot;nofollow&quot;&gt;Cauchy condensation test&lt;/a&gt;. I understand the trapezoidal view, but apparently &quot;the 'condensation' of terms is analogous to a substitution of an exponential function&quot;. Can anyone explain what is meant by this?&lt;/p&gt;&#xA;" OwnerUserId="123" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-08-07T18:02:50.093" LastActivityDate="2010-08-07T18:02:50.093" Title="Cauchy condensation test - Exponential function explanation" Tags="&lt;calculus&gt;&lt;analysis&gt;&lt;convergence&gt;" AnswerCount="2" />
  <row Id="1790" PostTypeId="2" ParentId="1789" CreationDate="2010-08-07T08:46:27.163" Score="4" ViewCount="" Body="&lt;p&gt;Explanation is quite straightforward. Consider integral $\int f(x)dx$. After change of variable $x=2^t$ it becomes $const\cdot\int 2^t f(2^t)dt$ — compare with $\sum 2^n f(2^n)$ from Cauchy condensation test.&lt;/p&gt;&#xA;" OwnerUserId="152" LastActivityDate="2010-08-07T08:46:27.163" />
  <row Id="1792" PostTypeId="2" ParentId="1758" CreationDate="2010-08-07T08:57:11.433" Score="4" ViewCount="" Body="&lt;p&gt;Let's focus on two of the chords. These have to split the circle in&#xA;the ratio $4:3$ so the minor arc subtended by each is equal. Call&#xA;one $AB$ and regard it as &quot;fixed&quot; and the other $CD$ which we think&#xA;of as moving. As I can't draw diagrams, look at Isaac's left diagram.&#xA;Let the two chords start in the position given by Isaac's two chords&#xA;meeting on the circumference. Let $AB$ be the one on the left and $CD$ be&#xA;the one on the right with $B=C$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now rotate $CD$ clockwise. Let $P$ be the point of intersection.&#xA;Then the angle $APD$ increases, the point $P$ moves monotonely from $B$ towards&#xA;$A$ on the line $AB$, and $C$ moves monotonely from $B$ towards $A$ on the minor&#xA;arc. Thus the area of the &quot;sector&quot; (bounded by two&#xA;lines and an arc of the circle) $PBC$ steadily increases, since at a later&#xA;time it will contain the sector at an earlier time. There is a unique&#xA;time at which the &quot;sector&quot; has area $1/7$ of the circle. Let $\theta_0$&#xA;be the angle $APD$ at this time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the sought configuration, all three angles of the central triangle have&#xA;area $\pi/3$ and so it exists if and only if $\theta_0=\pi/3$. So&#xA;deciding whether the configuration exists reduces to a single calculation:&#xA;what is the area of &quot;sector&quot; $PBC$ when angle $APD$ equals $\pi/3$;&#xA;this is a calculation that I'm not going to do :-)&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-08-07T08:57:11.433" CommentCount="8" />
  <row Id="1793" PostTypeId="2" ParentId="1442" CreationDate="2010-08-07T09:10:43.950" Score="3" ViewCount="" Body="&lt;p&gt;More generally, for $0 \leq k \leq n$, there is an identity&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(n+1) {\rm lcm} ({n \choose 0}, {n \choose 1}, \dots {n \choose k}) = {\rm lcm} (n+1,n,n-1, \dots n+1-k)$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is simply the fact that any integer linear combination of $f(x), \Delta f(x), \Delta^2 f(x), \dots \Delta^k f(x)$ is an integer linear combination of $f(x), f(x-1), f(x-2), \dots f(x-k)$ where $\Delta$ is the difference operator, $f(x) = 1/x$, and $x = (n+1)$.&lt;/p&gt;&#xA;" OwnerUserId="467" LastActivityDate="2010-08-07T09:10:43.950" CommentCount="3" />
  <row Id="1794" PostTypeId="2" ParentId="1789" CreationDate="2010-08-07T09:37:57.243" Score="3" ViewCount="" Body="&lt;p&gt;&quot;Condensation&quot; because the series $a_n$ is compressed to a subseries $a_{2^n}$ carrying all the same information about convergence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The proof of the condensation test is easy, as Grigory M's answer indicates and whenever it works so does a comparison of sum with integral, which is a more fundamental idea.  So you might naturally wonder why this seemingly trivial convergence test deserves its own name, other than the fame of its supposed inventor.  The reason it is retained in textbooks is that it is often a convenient criterion to use for series when the terms or the partial sums have some dependence on $\log (n)$.  The proof of divergence of the harmonic series is an example.&lt;/p&gt;&#xA;" OwnerUserId="467" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-08-07T09:47:49.810" LastActivityDate="2010-08-07T09:47:49.810" CommentCount="3" />
  <row Id="1795" PostTypeId="1" AcceptedAnswerId="1796" CreationDate="2010-08-07T10:24:47.263" Score="1" ViewCount="108" Body="&lt;p&gt;Can anyone explain the syntax from the following Wikipedia article on the &lt;a href=&quot;http://en.wikipedia.org/wiki/Convergence_of_random_variables#Definition&quot; rel=&quot;nofollow&quot;&gt;convergence of random variables&lt;/a&gt;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://upload.wikimedia.org/math/1/b/d/1bd88f4785429841e1d24c2996cbb6e0.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;What is the squiggly line?&lt;/li&gt;&#xA;&lt;li&gt;What are $d$, $D$, $L$, $L_X$?&lt;/li&gt;&#xA;&lt;li&gt;Why does it say $X_n$ implies X? They don't seem to be statements&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="123" LastEditorUserId="123" LastEditorDisplayName="" LastEditDate="2010-08-07T10:49:47.747" LastActivityDate="2010-08-07T11:43:46.200" Title="Syntax for the convergence of random variables" Tags="&lt;probability-theory&gt;&lt;notation&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="1796" PostTypeId="2" ParentId="1795" CreationDate="2010-08-07T11:25:25.680" Score="4" ViewCount="" Body="&lt;p&gt;All of these expressions are just notations for &quot;X_n converges to $X$ in distribution&quot;.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The squiggly line is really just a notation (a rather strange one) for convergence in distribution.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The symbols $d$ and $\mathcal D$ stand for &lt;em&gt;distribution&lt;/em&gt;, $\mathcal L$ and $\mathcal L_X$ stand for &lt;em&gt;law&lt;/em&gt;. (One also speaks of &quot;convergence in law&quot; instead of &quot;convergence in distribution&quot;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;This is also just a notation for convergence in distribution (also a rather strange one); the arrow does not symbolize an implication of statements.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="367" LastEditorUserId="367" LastEditorDisplayName="" LastEditDate="2010-08-07T11:43:46.200" LastActivityDate="2010-08-07T11:43:46.200" CommentCount="1" />
  <row Id="1797" PostTypeId="2" ParentId="1781" CreationDate="2010-08-07T11:34:49.583" Score="2" ViewCount="" Body="&lt;p&gt;I think you are doing it all wrong. Your Newtonian gravitational force equation looks wrong to me in spherical coordinates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's &lt;a href=&quot;http://en.wikiversity.org/wiki/Advanced_Classical_Mechanics/Central_Forces&quot; rel=&quot;nofollow&quot;&gt;a different approach&lt;/a&gt; using Lagrangian and Hamiltonian approach. The solution is &#xA;&lt;img src=&quot;http://upload.wikimedia.org/math/0/4/6/04604f267d01b2296779593b47ca1d71.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where &lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{aligned}&#xA;\V(r)\ &amp;amp;= -\frac1r\&#xA;\end{aligned}&lt;/p&gt;&#xA;" OwnerUserId="550" LastActivityDate="2010-08-07T11:34:49.583" />
  <row Id="1798" PostTypeId="2" ParentId="1694" CreationDate="2010-08-07T15:06:07.213" Score="0" ViewCount="" Body="&lt;p&gt;Well, we call $e^x$ natural exponentation and therefore &lt;code&gt;\log_e(x)&lt;/code&gt; natural logarithm. This one is abbreviated as &lt;code&gt;\ln&lt;/code&gt; (for logarithmus naturalis).&lt;/p&gt;&#xA;" OwnerUserId="917" LastActivityDate="2010-08-07T15:06:07.213" />
  <row Id="1799" PostTypeId="1" AcceptedAnswerId="1803" CreationDate="2010-08-07T15:44:42.270" Score="2" ViewCount="245" Body="&lt;p&gt;For a given $k&gt;0$ constant, assuming that $x,y&gt;0$. Also this equation has a particular name, or some mathematician associated with it?&lt;/p&gt;&#xA;" OwnerUserId="918" LastActivityDate="2010-08-07T20:05:20.733" Title="How to solve the equation $x^y=y^x=k$?" Tags="&lt;calculus&gt;&lt;real-analysis&gt;" AnswerCount="1" />
  <row Id="1800" PostTypeId="2" ParentId="1767" CreationDate="2010-08-07T16:22:59.393" Score="1" ViewCount="" Body="&lt;p&gt;The original question claims that the author doesn't yet understand the definitions of the symmetric group.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case, I think it's best to start with (what I think) is the most accessible definition.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$S_n$, the symmetric group on the $n$ numbers $1,\ldots,n$ is the set of all possible ways to permute the set of numbers $1,\ldots,n$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For example, one possible permutation (element of $S_n$) is &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$g = (1,2,\ldots,n) \Rightarrow (2,3,\ldots,n,1)$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This set of permutations forms a group: the multiplication is simply performing one permutation, and then the next. So &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$g\cdot g = g^2: (1,2,\ldots,n) \Rightarrow (3,4,\ldots,n,1,2)$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I think you should get comfortable with this definition first. Once you think about it for a while, you'll realize that each permutation can be represented as a function on the set ${1,\ldots,n}$. For example, the permutation $g$ above would be the function that sends $1 \mapsto 2$, $2 \mapsto 3$, and so on (in general, $n \mapsto n + 1$, where $n$ maps back to $1$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you think a little further, you'll realize that any function you get from a permutation in such a fashion must be bijective (that is, a permutation can't, say, send all the numbers $1,\ldots,n$ to $1$, since that's clearly not a permutation).&lt;/p&gt;&#xA;" OwnerUserId="576" LastActivityDate="2010-08-07T16:22:59.393" />
  <row Id="1801" PostTypeId="2" ParentId="740" CreationDate="2010-08-07T16:28:12.663" Score="2" ViewCount="" Body="&lt;p&gt;I like $f(x,y) = \frac{xy}{x^{2}+y{2}}$ for $(x,y) \ne (0,0)$ and $f(x,y)=0$ for $(x,y)=(0,0)$.  Here $f$ has partial derivatives at (0,0) but it is not differentiable at $(0,0)$&lt;/p&gt;&#xA;" OwnerUserId="832" LastEditorUserId="832" LastEditorDisplayName="" LastEditDate="2010-08-07T16:34:16.977" LastActivityDate="2010-08-07T16:34:16.977" CommunityOwnedDate="2010-08-07T16:28:12.663" CommentCount="1" />
  <row Id="1802" PostTypeId="1" AcceptedAnswerId="1805" CreationDate="2010-08-07T16:53:06.537" Score="0" ViewCount="163" Body="&lt;p&gt;Let $A$, $X$, and $Y$ be arbitrary topological spaces. Let $F:X\times A\rightarrow Y$ be a continuous function. Let $P$ be the map from $X\times A$ to $Y\times A$ taking $(x,a)$ to $(F(x,a),a)$. Does it follow from continuity of $F$ that $P$ is continuous?&lt;/p&gt;&#xA;" OwnerUserId="921" LastActivityDate="2010-08-07T17:01:14.323" Title="Is this function continuous?" Tags="&lt;general-topology&gt;" AnswerCount="1" />
  <row Id="1803" PostTypeId="2" ParentId="1799" CreationDate="2010-08-07T16:56:17.637" Score="7" ViewCount="" Body="&lt;p&gt;WLOG suppose that $x \ge y$ and let $x = y^n$ for some $n \ge 1$.  Then (ignoring $k$) the equation becomes $y^{ny} = y^{y^n}$, or $ny = y^n$, or $y^{n-1} = n$, which gives &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle y = n^{ \frac{1}{n-1} }, x = n^{ \frac{n}{n-1} }$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now it only remains to find the values of $n$ such that $x^y = k$, which you should just do numerically.  (I guess one must also account for the solutions where $x = y$; the graph $x^y = y^x$ has two components which intersect at $(e, e)$, which one gets from the above by taking the limit as $n \to 1$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am almost positive this equation does not have a name because, in all honesty, it is not interesting.  &lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-08-07T17:34:18.657" LastActivityDate="2010-08-07T17:34:18.657" CommentCount="2" />
  <row Id="1804" PostTypeId="2" ParentId="1758" CreationDate="2010-08-07T16:59:37.510" Score="0" ViewCount="" Body="&lt;p&gt;Each line segment is a chord of the circle, with 3 of the 7 regions in the segment of the circle determined by the chord (the segment is the region between the chord and the minor arc between its endpoints).  The area of a segment with arc measure $\theta$ radians of a circle with radius r is $\frac{r^2}{2}(\theta-\sin\theta)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If each of the 7 regions has equal area, then the area of the segment determined by each chord must be $\frac{3}{7}$ of the area of the circle, so for each chord, $\frac{r^2}{2}(\theta-\sin\theta)=\frac{3\pi r^2}{7}\Rightarrow\theta\approx 2.91624$.  Going forward, I'll use $\omega$ to refer to this value.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let's work with the unit circle centered at the origin and place one chord so that its minor arc would be swept out by the rotation images of (1,0) about the origin by angles in the interval $[0,\omega]$.  Let the other two chords be placed such that their minor arcs would be swept out by the rotation images of (1,0) about the origin by angles in the intervals $[\alpha,\alpha+\omega]$ and $[\beta,\beta+\omega]$ where $0&amp;lt;\alpha&amp;lt;\beta&amp;lt;2\pi$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In any arrangement of chords that yields 7 regions, there will be a central triangular region.  The lines containing the chords have equations $y=\frac{\sin\omega}{\cos\omega-1}(x-1)$, $y-\sin\alpha=\frac{\sin\alpha-\sin(\alpha+\omega)}{\cos\alpha-\cos(\alpha+\omega)}(x-\cos\alpha)$, and $y-\sin\beta=\frac{\sin\beta-\sin(\beta+\omega)}{\cos\beta-\cos(\beta+\omega)}(x-\cos\beta)$.  The vertices of the central triangular region are at the pairwise points of intersection of those three chords:&#xA;$\left(\cos\frac{\omega}{2}\cos\frac{\alpha+\omega}{2}\sec\frac{\alpha}{2},&#xA;\cos\frac{\omega}{2}\sin\frac{\alpha+\omega}{2}\sec\frac{\alpha}{2}\right)$, &#xA;$\left(\cos\frac{\omega}{2}\cos\frac{\beta+\omega}{2}\sec\frac{\beta}{2},&#xA;\cos\frac{\omega}{2}\sin\frac{\beta+\omega}{2}\sec\frac{\beta}{2}\right)$, &#xA;and $\left(\cos\frac{\omega}{2}\cos\frac{\alpha+\beta+\omega}{2}\sec\frac{\alpha-\beta}{2},\frac{\csc(\alpha-\beta)}{2}(-\cos\alpha+\cos\beta-\cos(\alpha+\omega)+\cos(\alpha-\beta))\right)$.  The area of this triangle is $\frac{1}{2}\left|\left(\cos\left(\alpha-\frac{\beta}{2}\right)-\cos\frac{\beta}{2}\right)\cos^2\frac{\omega}{2}\sec\frac{\alpha}{2}\sec\frac{\alpha-\beta}{2}\tan\frac{\beta}{2}\right|$ (restricted to only values of $\alpha$ and $\beta$ for which the three points of intersection are within the circle).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A contour plot of the area is shown below (white areas do not satisfy the constraints described above; darker colors represent smaller values of the area).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://www.imgftw.net/img/455062883.png&quot; alt=&quot;contour plot&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The greatest possible area for the triangle is approximately 0.226346 when $\alpha\approx 2.91624$ and $\beta\approx 3.36694$ (numerical approximation from Mathematica), which is well short of one seventh the area of the circle, $\frac{\pi}{7}\approx 0.448799$.  Therefore, it is not possible to divide the circle into 7 regions of equal area using 3 chords.&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-08-07T16:59:37.510" />
  <row Id="1805" PostTypeId="2" ParentId="1802" CreationDate="2010-08-07T17:01:14.323" Score="7" ViewCount="" Body="&lt;p&gt;Yes.  This follows from the fact that a function $U \to V \times W$ is continuous if and only if its component functions $U \to V, U \to W$ are, and from the fact that the projection maps $V \times W \to V$ and $V \times W \to W$ are continuous.  Both of these facts in turn follow  from the universal property of the product topology.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-08-07T17:01:14.323" CommentCount="1" />
  <row Id="1806" PostTypeId="2" ParentId="1116" CreationDate="2010-08-07T17:46:55.937" Score="1" ViewCount="" Body="&lt;p&gt;I don't have enough rep to comment yet, so I'll make this CW since it's not a complete answer, just a remark.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would it overcome some of the difficulty by using some sort of graph representation of the subdivided rectangle? Each rectangle is a vertex and an edge represents sharing a side, i.e.&#xA;the first rectangle above would be represented by the edges ${(1,2),(1,3),(2,4),(3,4)}$ and the second by ${(1,2),(1,3),(1,4),(2,4),(2,5),(3,4),(3,5)}$, numbering the rectangles from top-left to bottom-right. The tricky part here would then be to implement the operations for subdividing by a vertical or horizontal line segment. Starting instead with lines instead of segments could be promising.&lt;/p&gt;&#xA;" OwnerUserId="854" LastActivityDate="2010-08-07T17:46:55.937" CommunityOwnedDate="2010-08-07T17:46:55.937" CommentCount="1" />
  <row Id="1807" PostTypeId="2" ParentId="1778" CreationDate="2010-08-07T18:05:43.473" Score="0" ViewCount="" Body="&lt;p&gt;As n goes to infinity, you run out of money with probability 1.  This is known as &lt;a href=&quot;http://en.wikipedia.org/wiki/Gambler%27s_ruin&quot; rel=&quot;nofollow&quot;&gt;gambler's ruin&lt;/a&gt; or the drunk walk--if you are winning/losing money or walking side to side such that you win/lose/go left/go right with probability 1/2 (a 1-d simple symmetric &lt;a href=&quot;http://en.wikipedia.org/wiki/Random_walk&quot; rel=&quot;nofollow&quot;&gt;random walk&lt;/a&gt;), and there is a boundary like going broke or ending up in the gutter, as the number of games/steps goes to infinity, the probability of hitting that boundary goes to 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know how to prove it, but per Joshua Zucker's comment on &lt;a href=&quot;http://www.research.att.com/~njas/sequences/A001405&quot; rel=&quot;nofollow&quot;&gt;A001405 in the OEIS&lt;/a&gt;, and agreeing with T..'s answer, the probability of ending up with a positive amount of money after n flips if you started with no money would be $\frac{1}{2^n}{n\choose \left\lfloor\frac{n}{2}\right\rfloor}$.  Because you start with one dollar instead of no money, this is offset so that the probability of having money after n flips is $\frac{1}{2^{n-1}}{n-1\choose \left\lfloor\frac{n-1}{2}\right\rfloor}$ for n&amp;ge;1.&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-08-07T18:05:43.473" />
  <row Id="1808" PostTypeId="1" CreationDate="2010-08-07T18:47:03.710" Score="7" ViewCount="517" Body="&lt;p&gt;I'm mainly a CS major, but I also want to learn more advanced  mathematics. I see lot of cross over between CS and applied math, same can't be said about pure math.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Definition of 'advanced': Something a pure math major learns during and after late undergraduate years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are the fields that combines advanced mathematical topics with computer science? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Theoretical computer science and computational group theory are good examples.&lt;/p&gt;&#xA;" OwnerUserId="96" LastActivityDate="2010-08-09T08:23:58.353" CommunityOwnedDate="2010-08-07T18:47:03.710" Title="Fields that require both CS and pure math" Tags="&lt;soft-question&gt;&lt;big-list&gt;" AnswerCount="10" CommentCount="2" FavoriteCount="5" />
  <row Id="1809" PostTypeId="2" ParentId="1808" CreationDate="2010-08-07T19:13:39.587" Score="1" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Number Theory&lt;/strong&gt;: many large numbers were tested by using computers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Linear programming&lt;/strong&gt;: many algorithms have been developed.&lt;/p&gt;&#xA;" OwnerUserId="369" LastActivityDate="2010-08-07T19:13:39.587" CommunityOwnedDate="2010-08-07T19:13:39.587" />
  <row Id="1810" PostTypeId="2" ParentId="1808" CreationDate="2010-08-07T19:22:17.247" Score="4" ViewCount="" Body="&lt;p&gt;Computational algebraic geometry is the first example that comes to mind. It's used in phylogenetics (http://siam.org/pdf/news/1146.pdf can also look up the book Algebraic Statistics for Computational Biology). It also shows up in robotics and control theory, but I know a lot less about that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Polytopes (see Coxeter - Regular Polytopes or Grunbaum - Convex Polytopes) also appear in a lot of applications. The only specific one that comes to mind at the moment is the problem of figuring out whether an object (like a robot) can traverse a given maze. The algorithms use the convex hull of the robot to test whether or not it could traverse the maze. I'm sure there are a lot of people more knowledgeable than me on the subject.&lt;/p&gt;&#xA;" OwnerUserId="249" LastActivityDate="2010-08-07T19:22:17.247" CommunityOwnedDate="2010-08-07T19:22:17.247" />
  <row Id="1812" PostTypeId="2" ParentId="1808" CreationDate="2010-08-07T20:30:21.323" Score="2" ViewCount="" Body="&lt;p&gt;Robotics.  I don't know much about this field but from a theoretical point of view a lot of lie group theory is used (to describe joint motion and position).  In particular the lie theory of rigid motions i.e. of the Euclidean group $SE(3)$ is relevant.  I would imagine CS would come in more from the computational side when larger calculations are involved.  Also its robots which need to be built so plenty of mechanical and electrical engineering going on here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is mostly based on a quick glance at &lt;a href=&quot;http://www.springer.com/computer/ai/book/978-0-387-20874-9&quot; rel=&quot;nofollow&quot;&gt;http://www.springer.com/computer/ai/book/978-0-387-20874-9&lt;/a&gt;.  Which is written for an undergraduate math or engineering major.&lt;/p&gt;&#xA;" OwnerUserId="403" LastActivityDate="2010-08-07T20:30:21.323" CommunityOwnedDate="2010-08-07T20:30:21.323" />
  <row Id="1813" PostTypeId="2" ParentId="1808" CreationDate="2010-08-07T21:40:36.610" Score="0" ViewCount="" Body="&lt;p&gt;Today is an open question to find an efficient algorithm/data structure that can be used to index/search information in a high dimensional space (ie d &gt; 20).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've seen approaches using computational geometry, space filling curves, principal component analysis, etc. &lt;/p&gt;&#xA;" OwnerUserId="918" LastActivityDate="2010-08-07T21:40:36.610" CommunityOwnedDate="2010-08-07T21:40:36.610" />
  <row Id="1814" PostTypeId="1" CreationDate="2010-08-07T21:44:25.857" Score="2" ViewCount="138" Body="&lt;p&gt;Are there any books or online notes that cover the basics of lie groups using nonstandard analysis? Another thing I would like is a to see these things set in category theory (along the lines of &lt;em&gt;Algebra: Chapter 0&lt;/em&gt; except for differential geometry?)&lt;/p&gt;&#xA;" LastActivityDate="2010-08-08T06:13:16.587" CommunityOwnedDate="2010-08-08T13:00:02.230" Title="Books on Lie Groups via nonstandard analysis?" Tags="&lt;differential-geometry&gt;&lt;lie-algebra&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" />
  <row Id="1815" PostTypeId="2" ParentId="1694" CreationDate="2010-08-07T21:47:12.533" Score="2" ViewCount="" Body="&lt;p&gt;I don't have enough rep. on this site to post a comment on Gerard's answer, so I'll write this as an answer. Those who say &quot;tan&quot; is more common than tg should keep in mind where they live. In Russian the standard abbreviation is tg, not tan. They write sin and cos, but the trig functions tan, cot, and csc have alternate abbreviations. &#xA;See the first line at &lt;a href=&quot;http://ru.wikipedia.org/wiki&quot; rel=&quot;nofollow&quot;&gt;http://ru.wikipedia.org/wiki&lt;/a&gt;/Тригонометрические_функции, for instance, where the trigonometric functions are named and then abbreviated in parentheses. (You may need to cut and paste, rather than click, on that link since the cyrillic part of the link is not working properly.) &lt;/p&gt;&#xA;" OwnerUserId="619" LastActivityDate="2010-08-07T21:47:12.533" CommentCount="1" />
  <row Id="1816" PostTypeId="2" ParentId="83" CreationDate="2010-08-07T21:49:50.613" Score="2" ViewCount="" Body="&lt;p&gt;My biggest hobby throughout junior high was video game development. I taught myself how to program in C/C++ and first wrote pong. This introduced me to vectors and I saw first-hand how convenient they could be for geometric computations. I then gradually made more and more complex programs. By high school I was mainly interested in computer graphics, which led to my learning linear algebra.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Having lots of programming experience is in itself extremely beneficial. In today's computer age, you can never know enough about programming.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So this is obviously a quite indirect approach and it's not exactly a teaching technique, but I would &lt;em&gt;highly&lt;/em&gt; recommend any young child at least look into game development. It's very intellectually stimulating, it will expose them to a variety of fields that they might otherwise have to wait years and years to be exposed to, it will condition them to think logically and rigorously from a young age, and above all it's lots of fun to see computer science and mathematics work together to produce a video game--which, by the way, you can then play!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These days there are simply incredible resources for teaching oneself the techniques of game development. Anyone interested should visit &lt;a href=&quot;http://www.gamedev.net&quot; rel=&quot;nofollow&quot;&gt;http://www.gamedev.net&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="251" LastActivityDate="2010-08-07T21:49:50.613" CommunityOwnedDate="2010-08-07T22:32:44.343" />
  <row Id="1817" PostTypeId="1" AcceptedAnswerId="1825" CreationDate="2010-08-07T23:49:06.073" Score="2" ViewCount="181" Body="&lt;p&gt;Let $X$ be a compact metric space.  Given a sequence $x_n \in X$ and an ultrafilter $\mathcal{U}$ on $\mathbb{N}$, we can define the &quot;Banach limit&quot; of ${x_n}$ with respect to $\mathcal{U}$. This limit is the unique element of $X$ such that every neighborhood of it contains a $\mathcal{U}$-large collection of the sequence ${x_n}$.  However, this process is not necessarily translation-invariant. When $X$ is a closed interval in the reals, then one can define a Banach limit by using an ultrafilter and taking the ultrafilter-limit of the Cesaro means.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a way to define a translation-invariant &quot;limit&quot; with reasonable properties for sequences taking values in an arbitrary compact metric space?&lt;/p&gt;&#xA;" OwnerUserId="536" LastActivityDate="2011-03-15T16:11:57.970" Title="Banach limits in metric spaces" Tags="&lt;real-analysis&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="0" />
  <row Id="1818" PostTypeId="2" ParentId="1116" CreationDate="2010-08-08T00:25:55.657" Score="1" ViewCount="" Body="&lt;p&gt;Here is a non-answer which may prove interesting and tractable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given a configuration of rectangles that properly tile a surrounding rectangle, one can&#xA;represent the tiling in a number of ways.  Two that come to mind I will call C and LL.&#xA;C represents each rectangle by giving its dimensions (x-span and y-span) and the coordinates for the center of the rectangle, and LL does the same, except it specifies the coordinate for the lower left vertex (instead of the center) of each rectangle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With plenty of rectangles in a tiling, there is some redundancy in the LL form of a tiling.  For example, it should be possible to deduce the dimensions of the rectangle in the extreme lower left corner of the tiling by noting stored coordinates of the rectangle above and the rectangle to the right of this corner rectangle.  No such boundaries exist for the upper rightmost rectangle in a tiling, but perhaps an &quot;end point&quot;&#xA;could be added to indicate the maximum x and maximum y values in the tiled rectangle.  Suppose we add such an end point, and let LL' be a representation like LL that contains the lower left coordinates and the end point, and does NOT contain the dimension information.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Statement 1: LL' is an equivalent representation to LL.  That is, given a representation in LL form, one can construct a representation in LL' form and use it to recreate the tiling; conversely, one can build an LL form from an LL' form to represent the same tiling.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Statement 1 appears to be true; it is clear how to make an LL' form from an LL form, while for the converse, it should be possible to check all the lower left points to see which would help determine the dimensions for a given rectangle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is statement 1 true?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now consider C and make C' from C by again forgetting the dimensions, except that you can&#xA;provide two endpoints representing the upper right and lower left corners of the tiled rectangle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Statement 2: C and C' are equivalent in the sense used above for LL and LL' .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is clear how to get a C' form from a C form.  It is NOT clear that it is possible to get a C form from a C' form where you have no dimension information at all, except for the&#xA;entire rectangle as implied by the two endpoints.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is Statement 2 true?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Statements 1 and 2 have a bearing on the problem above: it may be possible to enumerate geometrically inequivalent tilings with n rectangles by looking at normalized forms of either C, C', LL, or LL'.  Also, there may be advantages to switching between the forms of the tiling in determining equivalence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are interesting variations that can occur.  For example, C'' might have the center of the tiled rectangle instead of the two endpoints, or even give just the dimensions of&#xA;the tiled rectangle instead of the coordinates for its upper right and lower left corners.  What information is lost by such representations?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Meta-question:  Is there a geometric property at work here that might explain why LL' uses less information but is still equivalent to C (assuming Statement 1 is true), while C' does not (assuming Statement 2 is false)?  Can something be said in general about what properties of a form for tiling are needed in order to represent the tiling both minimally and faithfully?&lt;/p&gt;&#xA;" OwnerUserId="297" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-08-11T22:02:54.053" LastActivityDate="2010-08-11T22:02:54.053" CommentCount="0" />
  <row Id="1819" PostTypeId="1" CreationDate="2010-08-08T00:32:12.913" Score="2" ViewCount="280" Body="&lt;p&gt;What is the difference between convergence in distribution, convergence in probability and almost sure convergence?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;References&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Convergence_of_random_variables&quot; rel=&quot;nofollow&quot;&gt;Convergence of random variables&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="123" LastActivityDate="2010-09-11T07:20:19.580" Title="Convergence of random variables" Tags="&lt;probability-theory&gt;&lt;real-analysis&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="1" />
  <row Id="1820" PostTypeId="2" ParentId="1819" CreationDate="2010-08-08T02:01:56.513" Score="4" ViewCount="" Body="&lt;p&gt;Convergence in distribution means pointwise convergence of the CDFs.  It's not a form of convergence of the random variables per se. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Convergence in probability says that for every epsilon &gt; 0, the probability of the sequence of random variables being more than epsilon from the limiting random variable goes to zero. Convergence in probability is also called convergence in measure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Convergence almost everywhere = almost sure convergence = pointwise convergence of the random variables except possibly on a set of measure zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a &lt;a href=&quot;http://www.johndcook.com/modes_of_convergence.html&quot; rel=&quot;nofollow&quot;&gt;diagram&lt;/a&gt; showing which modes of convergence imply which other modes.&lt;/p&gt;&#xA;" OwnerUserId="247" LastActivityDate="2010-08-08T02:01:56.513" />
  <row Id="1821" PostTypeId="2" ParentId="1814" CreationDate="2010-08-08T02:26:21.643" Score="2" ViewCount="" Body="&lt;p&gt;I never looked at that part of the book in any detail at all, but Abraham Robinson's &lt;em&gt;Nonstandard Analysis&lt;/em&gt; includes some material on Lie groups via nonstandard analysis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A little bit of googling found a &lt;a href=&quot;http://arxiv.org/abs/funct-an/9205003&quot; rel=&quot;nofollow&quot;&gt;paper&lt;/a&gt; on the arXiv that might be relevant.&lt;/p&gt;&#xA;" OwnerUserId="536" LastActivityDate="2010-08-08T02:26:21.643" />
  <row Id="1822" PostTypeId="1" AcceptedAnswerId="1823" CreationDate="2010-08-08T02:34:27.177" Score="6" ViewCount="1133" Body="&lt;p&gt;I have just started auditing Linear Algebra, and the first thing we learned was how to solve a system of linear equations by reducing it to row echelon form (or reduced row echelon form). I know how to solve a system of linear equations using determinants, and so the using row echelon form seems to be very inefficient and an easy way to make mistakes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The lecturer seemed to indicate that this will be useful in the future for something else, but a friend who has a PhD in mathematics said it is totally useless.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this true? Is this technique totally useless or is there some &quot;higher mathematics&quot; field/ technique where row echelon form will be useful?&lt;/p&gt;&#xA;" OwnerUserId="85" LastActivityDate="2010-08-08T10:27:53.353" Title="Is reducing a matrix to row echelon form useful at all?" Tags="&lt;linear-algebra&gt;" AnswerCount="5" />
  <row Id="1823" PostTypeId="2" ParentId="1822" CreationDate="2010-08-08T02:56:55.597" Score="14" ViewCount="" Body="&lt;p&gt;Go beyond 2 x 2 and row echelon form is &lt;em&gt;the&lt;/em&gt; way to solve systems of linear equations.&#xA;This stuff is not done by hand, but by a computer. Row echelon form is much, much faster than &#xA;using determinants (Cramer's formula).  Ask someone who works in numerical analysis about this.  Your friend with the Ph.D. is clearly unaware of the needs of applied math. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In pure math, Cramer's formula is very important for some proofs (and I personally prefer it for conceptual reasons over the algorithmic row echelon business, but then I don't work in numerical analysis). In applied math problems that give rise to systems of linear equations, you typically face equations in hundreds of variables.  You are not going to solve that with a determinant.  Moreover, if a computer needed to find a determinant of a 100 x 100 matrix it would &lt;em&gt;not&lt;/em&gt; use the sum of 100! terms as in the Laplace expansion formula for determinants, but instead apply faster algorithms (the QR algorithm).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your profile says you are studying economics.  At some point when you may need to deal with some system of linear equations in economics and have the computer solve it for you, the method of solution could be treated as a black box in the same way most people drive  without knowing how a car really works. But the reality is that if it were not for the algorithm of reducing to row echelon form, the computer couldn't solve systems of hundreds of equations rapidly. &lt;/p&gt;&#xA;" OwnerUserId="619" LastEditorUserId="619" LastEditorDisplayName="" LastEditDate="2010-08-08T03:06:06.027" LastActivityDate="2010-08-08T03:06:06.027" CommentCount="4" />
  <row Id="1824" PostTypeId="2" ParentId="1822" CreationDate="2010-08-08T03:10:50.993" Score="10" ViewCount="" Body="&lt;p&gt;Gaussian elimination on an $n \times n$ matrix takes takes on the order of $n^3$ operations, where an operation is an addition, subtraction, multiplication, or a division. By contrast, computing determinants using the cofactor expansion takes $n!$ operations, which is much, much longer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, consider a $100 \times 100$ matrix. Then $100^3$ is a million, and a million operations can be done by a typical laptop in under a second; but $100!$ is approximately $1$ followed by $158$ zeros, which would take even a supercomputer more time than the age of the universe. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, I believe the opposite of your claim is true: if you want to compute determinants, the best way is to to reduce to echelon form. &lt;/p&gt;&#xA;" OwnerUserId="317" LastEditorUserId="317" LastEditorDisplayName="" LastEditDate="2010-08-08T06:21:36.890" LastActivityDate="2010-08-08T06:21:36.890" CommentCount="6" />
  <row Id="1825" PostTypeId="2" ParentId="1817" CreationDate="2010-08-08T03:45:38.183" Score="4" ViewCount="" Body="&lt;p&gt;Let $X = \lbrace 0,1\rbrace$ be the compact metric space with exactly two points.  Then the space of sequences is just $\Sigma_2^+ = \lbrace 0,1 \rbrace^\mathbb{N}$, and an ultrafilter $\mathcal{U}$ defines a Banach limit in a fairly straightforward way:  for $i=0,1$, write $A_i = \lbrace n \in \mathbb{N} \mid x_n = i \rbrace$, and $\lim_\mathcal{U} x_n = i$ if and only if $A_i \in \mathcal{U}$.  This limit is translation invariant if and only if the ultrafilter itself is, so your question reduces to the existence of a translation invariant ultrafilter on $\mathbb{N}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe that no such ultrafilter exists, and hence the answer to your question is &quot;no&quot;; however, I'm not particularly knowledgeable about ultrafilters, so I welcome correction or confirmation on this last point.&lt;/p&gt;&#xA;" OwnerUserId="876" LastActivityDate="2010-08-08T03:45:38.183" CommentCount="1" />
  <row Id="1826" PostTypeId="2" ParentId="1747" CreationDate="2010-08-08T05:07:50.303" Score="1" ViewCount="" Body="&lt;p&gt;Others have provided fine answers for the standard method of calculating the probability of getting &lt;em&gt;k&lt;/em&gt; or more heads in &lt;em&gt;n&lt;/em&gt; tosses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I think you are approaching the question from the wrong direction. Few people have tossed a single coin enough times. Pearson did it around 1900: He tossed a coin 24,000 times and came up with 12,012 heads. Count Buffon tossed one 4040 times and got 2048 heads and Kerrich did it 10,000 times in a German POW camp and got 5067 heads. (Moore et al. &quot;Introduction to the Practice of Statistics&quot;, 6th ed, p.240).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to make sure a given coin is fair, you should use physical measurements and analyze the surface and composition of the coin to make sure its weight is uniformly distributed, it's not bent etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If someone offers you a bet where you pay 10 dollars if the coin comes up heads, and he pays 10,000 dollars if it comes up tails, you can be assured that the coin is not fair ;-)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Otherwise, we assume a generic coin picked without any special procedure is fair. Without that assumption, life as we know it would not be possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, &lt;a href=&quot;http://www.state.nj.us/casinos/actreg/reg/docs_chapter46/c46s01sec15to19b.pdf&quot; rel=&quot;nofollow&quot;&gt;casinos are not required to roll their dice thousands of times&lt;/a&gt; to ascertain their fairness. In fact, doing so may wear out the corners or edges. Rather, there are specific manufacturing process requirements that need to be met.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, no one wastes time trying to ensure that the coin used to determine who gets the ball at the start of a football game is fair. &lt;a href=&quot;http://www.nfl.com/rulebook/cointoss&quot; rel=&quot;nofollow&quot;&gt;NFL's rules on the coin flip are simple&lt;/a&gt; and they do not involve ascertaining the fairness of the coin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Probability will tell you that if 1,000 people each toss their &lt;em&gt;fair&lt;/em&gt; coins 30 times, most of the percentages will be very close to 50%. Determining whether an individual coin is fair is not a task for Statistics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See also &lt;a href=&quot;http://www-stat.stanford.edu/~cgates/PERSI/papers/dyn_coin_07.pdf&quot; rel=&quot;nofollow&quot;&gt;Dynamical Bias in Coin Tossing&lt;/a&gt; and &lt;a href=&quot;http://www.dartmouth.edu/~chance/chance_news/recent_news/chance_news_11.02.html#item2&quot; rel=&quot;nofollow&quot;&gt;Chance News 11.02&lt;/a&gt; referenced in that article.&lt;/p&gt;&#xA;" OwnerUserId="930" LastEditorUserId="930" LastEditorDisplayName="" LastEditDate="2010-08-08T15:18:34.913" LastActivityDate="2010-08-08T15:18:34.913" CommentCount="7" />
  <row Id="1827" PostTypeId="2" ParentId="1822" CreationDate="2010-08-08T05:53:27.510" Score="5" ViewCount="" Body="&lt;p&gt;Aside from its incredible utility in practical computations, echelon form is also a special case of a general construction called &lt;a href=&quot;http://en.wikipedia.org/wiki/Bruhat_decomposition&quot; rel=&quot;nofollow&quot;&gt;Bruhat decomposition&lt;/a&gt; which is important in the theory of Lie groups and algebraic groups.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-08-08T05:53:27.510" CommentCount="1" />
  <row Id="1828" PostTypeId="2" ParentId="1814" CreationDate="2010-08-08T06:13:16.587" Score="2" ViewCount="" Body="&lt;p&gt;You can find references to such literature in Isaac Goldbring's &lt;a href=&quot;http://www.math.uiuc.edu/~igoldbr2/researchstatement.pdf&quot; rel=&quot;nofollow&quot;&gt;statement of research&lt;/a&gt;. In particular see the work of Robinson [13], Bate [1], Singer [14] and Hirschfeld [6] referenced there.&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-08-08T06:13:16.587" />
  <row Id="1829" PostTypeId="1" CreationDate="2010-08-08T08:02:29.217" Score="2" ViewCount="217" Body="&lt;p&gt;I am working a bit on a collection of Linear Algebra examples, &#xA;as well as some examples on induction. This is what is taught freshman year at our university.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I intend to release this to the public, either by selling printed copies or releasing it online. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since I do not have experience using such material myself, there are some questions I would like some opinions on:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;How much theory should I include? Is references to course litterature enough?&lt;/li&gt;&#xA;&lt;li&gt;Is there a format preference? Small text, so that the collection is more enviromental-friendly, or with big marginals for notes?&lt;/li&gt;&#xA;&lt;li&gt;Best way to deal with misprints?&lt;/li&gt;&#xA;&lt;li&gt;Should induction and Linear algeba be separate pieces? &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Please share your experience if you have done something similar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT:&#xA;An answer I seek is something along the lines of:&#xA;&quot;I am a &quot;something&quot; stident, and I prefer &quot;something&quot;, and would like to see more of &quot;something&quot;.&lt;/p&gt;&#xA;" OwnerUserId="934" LastEditorUserId="934" LastEditorDisplayName="" LastEditDate="2010-08-15T20:33:19.260" LastActivityDate="2010-08-20T21:06:15.857" CommunityOwnedDate="2010-08-08T12:30:22.677" Title="Tips for writing math solutions for others" Tags="&lt;linear-algebra&gt;&lt;examples&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="1830" PostTypeId="2" ParentId="1808" CreationDate="2010-08-08T08:11:36.853" Score="0" ViewCount="" Body="&lt;p&gt;I am working with the Schroedinger equation,&#xA;which suprisinly leads to graph theory (via cell decompositions).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To get a good grip on these graphs, I did a lot of programming,&#xA;since doing it by hand takes way too much time. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I generated about 100 examples, and from there, I could easily see what needed to be proved,&#xA;something which was quite hard prior to that discovery.&lt;/p&gt;&#xA;" OwnerUserId="934" LastActivityDate="2010-08-08T08:11:36.853" CommunityOwnedDate="2010-08-08T08:11:36.853" CommentCount="2" />
  <row Id="1831" PostTypeId="2" ParentId="1822" CreationDate="2010-08-08T08:12:02.610" Score="3" ViewCount="" Body="&lt;ol&gt;&#xA;&lt;li&gt;I think the only &lt;em&gt;real&lt;/em&gt; utility of Cramer's rule -besides resolving 2x2 systems by hand- is that, for a general system of linear equations $AX = B$ with $\det A \neq 0$, it shows blatantly &lt;a href=&quot;http://en.wikipedia.org/wiki/Cramer%27s_rule&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Cramer%27s_rule&lt;/a&gt; the continuous dependence of the solution with respect both $A$ and $B$.&lt;/li&gt;&#xA;&lt;li&gt;But, for the same reason, Cramer's rule is very dangerous because that $\det A$ appearing in the denominator: using it with a computer for small values of $\det A$ makes errors grow quickly.&lt;/li&gt;&#xA;&lt;li&gt;Nevertheless, I must confess that I feel somewhat unconfortable telling these kind of things to my students, since they could (and sometimes do) easely argue: (a) Are we going to &lt;em&gt;actually&lt;/em&gt; solve some, say, 5x5 system of linear equations &lt;em&gt;by hand&lt;/em&gt;? (b) Who cares what the computer really does in order to solve the system? -The only thing I need is the solution.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;For these embarrassing objections, I have the following problem: try to solve some &lt;em&gt;simultaneuous&lt;/em&gt; systems of linear equations. That is to say, for instance, two systems $AX = B_1$ and $AX = B_2$, with the same $A$. Of course, you can apply Cramer's rule two times, but the reducing row echelon algorithm allows you to solve it with one go, since the operations for reducing the system depend &lt;em&gt;only&lt;/em&gt; on $A$. In manual computations, you can take advantage of this putting both matrices $B_1$ and $B_2$ together like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;(A \vert B_1 B_2)&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, this can be exploited further for any number of simultaneous systems $AX = B_1, \dots , AX=B_n$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;(A \vert B_1 B_2 \dots B_n) \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, inverting a matrix is a problem of solving a simultaneous system of linear equations:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;AX = I  \quad \Longleftrightarrow \quad (A \vert e_1 \dots e_n)  \ ,&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $I$ is the identity matrix and $e_1, \dots , e_n$ is the standard basis of $\mathbb{R}^n$.&lt;/p&gt;&#xA;" OwnerUserId="664" LastEditorUserId="664" LastEditorDisplayName="" LastEditDate="2010-08-08T08:44:56.617" LastActivityDate="2010-08-08T08:44:56.617" CommentCount="2" />
  <row Id="1832" PostTypeId="1" AcceptedAnswerId="1843" CreationDate="2010-08-08T09:55:43.383" Score="2" ViewCount="141" Body="&lt;p&gt;Let $m,n \geq 2$ be integers. Find all functions $f:[0,\infty) \to \mathbb{R}$ continuous at at least one point in $[0,\infty)$ and such that $$f\biggl(\frac{1}{n} \sum\limits_{i=1}^{n} x_{i}^{m} \biggr)=\frac{1}{n} \sum\limits_{i=1}^{n}(f(x_i))^{m} \quad \text{for} \ x_{i} \geq 0, i=1,2,...,n$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not getting any idea to proceed.&lt;/p&gt;&#xA;" LastEditorUserId="536" LastEditorDisplayName="" LastEditDate="2010-08-08T13:37:43.530" LastActivityDate="2010-08-08T14:04:07.957" Title="Functions satisfying a given condition" Tags="&lt;real-analysis&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="1833" PostTypeId="2" ParentId="1808" CreationDate="2010-08-08T09:58:14.643" Score="2" ViewCount="" Body="&lt;p&gt;Cryptography is a field which requires lot of Number theory. Formal languages and automata theory requires some Graph theory, and these automata things can be connected to computer science as well.&lt;/p&gt;&#xA;" LastActivityDate="2010-08-08T09:58:14.643" CommunityOwnedDate="2010-08-08T09:58:14.643" />
  <row Id="1836" PostTypeId="2" ParentId="1822" CreationDate="2010-08-08T10:27:53.353" Score="4" ViewCount="" Body="&lt;p&gt;alex already covered much of what I had wanted to say, so allow me to share something that had been written by Forman Acton in his (excellent!) &quot;Numerical Methods that Work&quot; on the matter:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;...the fact that this sequence of operations takes (much more) labor on the part of the computer as a direct solution of the problem by (Gaussian elimination) has never penetrated (the student's) consciousness. The glibness of the formal mathematical notation has obscured the realities of the arithmetic process. The cure is simple enough--the student should be asked to solve a 4-by-4 system, by hand, both ways. But this sort of realism is currently unfashionable--arithmetic labor being deemed ony suitable for machines.&quot;&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-08-08T10:27:53.353" CommentCount="1" />
  <row Id="1838" PostTypeId="1" AcceptedAnswerId="1858" CreationDate="2010-08-08T12:15:59.717" Score="5" ViewCount="267" Body="&lt;p&gt;I understand what the problem with Gimbal Lock is, such that at the North Pole, all directions are south, there's no concept of east and west. But what I don't understand is why this is such an issue for navigation systems? Surely if you find you're in Gimbal Lock, you can simply move a small amount in any direction, and then all directions are right again?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why does this cause such a problem for navigation?&lt;/p&gt;&#xA;" OwnerUserId="32" LastEditorUserId="32" LastEditorDisplayName="" LastEditDate="2010-08-08T12:25:55.697" LastActivityDate="2010-08-10T02:52:13.350" Title="Why is Gimbal Lock an issue?" Tags="&lt;geometry&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="2" />
  <row Id="1839" PostTypeId="1" AcceptedAnswerId="1841" CreationDate="2010-08-08T12:33:18.440" Score="5" ViewCount="183" Body="&lt;p&gt;Let $a_n\geq 0$ be a sequence of non-negative numbers. Consider the following two statements:&#xA;$$&#xA;\text{(I)}\qquad\qquad \lim_{n\to\infty} \frac{1}{n^2}\sum_{i=1}^n a_i =0,&#xA;$$&#xA;$$&#xA;\text{(II)}\qquad\qquad\qquad \sum_{n=1}^\infty \frac{a_n}{n^2}&amp;lt;\infty.&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Questions:&lt;/strong&gt; Does (I) imply (II)? Does (II) imply (I)? Otherwise, please provide counterexamples.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Motivation: Both statements occur in the context of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Law_of_large_numbers&quot; rel=&quot;nofollow&quot;&gt;law of large numbers&lt;/a&gt; for non-identically distributed random variables. With $a_n=\mathrm{Var}(X_n)$, one can conclude the weak LLN if the $X_n$ are pairwise uncorrelated and condition (I) holds. The strong LLN can be concluded if the $X_n$ are stochastically independent and condition (II) holds. Therefore, one might expect that (II) implies (I).&lt;/p&gt;&#xA;" OwnerUserId="367" LastEditorUserId="367" LastEditorDisplayName="" LastEditDate="2010-08-08T16:27:41.513" LastActivityDate="2010-08-08T16:27:41.513" Title="Comparison of two convergence conditions for sequences of non-negative numbers" Tags="&lt;sequences-and-series&gt;" AnswerCount="1" />
  <row Id="1840" PostTypeId="1" CreationDate="2010-08-08T12:52:58.083" Score="0" ViewCount="116" Body="&lt;p&gt;Given the following:  &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\sum_{i=1}^{n} a_i = 1, a_i \geq 0 \forall i$&lt;/li&gt;&#xA;&lt;li&gt;$\sum_{i=1}^{n} b_i = 1, b_i \geq 0 \forall i$&lt;/li&gt;&#xA;&lt;li&gt;$\sum_{i=1}^{n} |a_i - b_i| \leq e$ where $e \ll 1$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;What is the upper bound on:  &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\sum_{i,j=1}^{n} |a_i\cdot a_j - b_i\cdot b_j|$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I am almost sure it is bounded by $2e$ but am not able to prove it.&lt;/p&gt;&#xA;" OwnerUserId="938" LastEditorUserId="171" LastEditorDisplayName="" LastEditDate="2010-08-08T13:56:21.143" LastActivityDate="2010-08-08T17:36:59.260" Title="a proof on the upperbound of this expression:" Tags="&lt;sequences-and-series&gt;" AnswerCount="1" CommentCount="8" />
  <row Id="1841" PostTypeId="2" ParentId="1839" CreationDate="2010-08-08T12:57:02.083" Score="9" ViewCount="" Body="&lt;p&gt;II implies I:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(we deal with the partial sums here)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One applies the Cauchy-Schwarz inequality to get:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle \left( \sum_{i=1}^n \frac{a_i}{i^2} \right)\left(\sum_{i=1}^n i^2\right) \ge (\sum_{i=1}^n a_i)^2 $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can see that $ \sum_{i=1}^n i^2 $ is of the magnitude $n^3$ as $n \rightarrow \infty$. So when we divide both sides by $n^4$, LHS will converge to 0 as $n \rightarrow 0$, which implies that $\displaystyle  \frac{\sum a_n}{n^2}$ converges to 0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I does NOT imply II:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;e.g. take $a_1,a_2$ to be anything, $\displaystyle a_n = \frac{n}{\log n}$ for $n \ge 3$. This does not satisfy II (well known). But it does satisfy I, because asymptotically, &#xA;$\displaystyle \frac{1}{n^2} \sum_{i=1}^n a_i \sim \frac{1}{n^2} \int_3^n \frac{x}{\log x} dx$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Logarithmic_integral_function&quot; rel=&quot;nofollow&quot;&gt;logarithmic integral&lt;/a&gt; has a well known approximation&#xA;$\displaystyle \int_3^n \frac{1}{\log x}dx = O\left(\frac{n}{\log n}\right)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So $\displaystyle \frac{1}{n^2} \int_3^n \frac{x}{\log x} dx \leq \frac{n}{n^2} \int_3^n \frac{1}{\log x} dx = O\left(\frac{1}{\log n}\right) \rightarrow 0$ as $n \rightarrow \infty$.&lt;/p&gt;&#xA;" OwnerUserId="325" LastEditorUserId="325" LastEditorDisplayName="" LastEditDate="2010-08-08T13:04:50.720" LastActivityDate="2010-08-08T13:04:50.720" CommentCount="1" />
  <row Id="1842" PostTypeId="2" ParentId="1808" CreationDate="2010-08-08T13:24:37.680" Score="4" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Semantics of programming languages&lt;/strong&gt;: the modern version is a mixture of type theory and category theory.  Also, there is a lot of subtle issues in logic (linear logic in particular) that crops up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Design of programming languages&lt;/strong&gt;: Things like &lt;em&gt;linear logic&lt;/em&gt; have influenced the addition of a number of features to languages (or reasonable restrictions).  &lt;em&gt;monads&lt;/em&gt; in Haskell come straight out of category theory.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Analysis of programming languages&lt;/strong&gt;: static analysis is nowadays all based on &lt;em&gt;abstract interpretation&lt;/em&gt;, which is fundamentally an application of lattices and of Galois connections.&lt;/p&gt;&#xA;" OwnerUserId="620" LastActivityDate="2010-08-08T13:24:37.680" CommunityOwnedDate="2010-08-08T13:24:37.680" CommentCount="1" />
  <row Id="1843" PostTypeId="2" ParentId="1832" CreationDate="2010-08-08T13:28:49.353" Score="1" ViewCount="" Body="&lt;p&gt;Let $x_1=x_2=\ldots=x_n=x$, then $f(x^m)=f(x)^m$, and $f(0)=f(0)^m$, so $f(0)$ is either 0,1 or -1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let $y_i=x_i^m$, then we have $f(\frac{1}{n}\sum_{i=1}^n y_i)=\frac{1}{n}\sum_{i=1}^n f(y_i)$ for any $y_i\in \mathbb{R}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $f(0)=0$, then $y_1=y$, $y_2=\ldots=y_n=0$ implies that $f(\frac{1}{n} y)=\frac{1}{n}f(y)$.&#xA;This implies $\frac{1}{n}f(\sum_{i=1}^n y_i)=f(\frac{1}{n}\sum_{i=1}^n y_i)=\frac{1}{n}\sum_{i=1}^n f(y_i)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Letting $y_3=\ldots=y_n=0$, we get that $f(y_1+y_2)=f(y_1)+f(y_2)$, which is &lt;a href=&quot;http://en.wikipedia.org/wiki/Cauchy%27s_functional_equation&quot; rel=&quot;nofollow&quot;&gt;Cauchy's Functional Equation&lt;/a&gt;. Since $f$ is continuous at one point, it's linear, so $f(x)=ax$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We get $a=a^m$, so the solutions in this case are $f(x)=0$, $f(x)=x$ and $f(x)=-x$ (if $m$ is odd).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $f(0)\neq 0$, WLOG let $f(0)=1$ (or else replace $f$ by $-f$ if $m$ is odd).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Analogously $f(\frac{1}{n}y)=\frac{f(y)+n-1}{n}$, and $\frac{f(\sum_{i=1}^n y_i)+n-1}{n}=f(\frac{1}{n}\sum_{i=1}^n y_i)=\frac{1}{n}\sum_{i=1}^n f(y_i)$, so that $f(\sum_{i=1}^n y_i)+n-1=\sum_{i=1}^n f(y_i)\Rightarrow g(\sum_{i=1}^n y_i)=\sum_{i=1}^n g(y_i)$ where $g(y)=f(y)-1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From this we have that $g$ satisfies Cauchy's functional equation, so that $f(y)=ay+1$. But $ay^m+1=f(y^m)=f(y)^m=(ay+1)^m$, which implies $a=0$ by comparing coefficients. So in this case the solutions are $f(x)=1$ and $f(x)=-1$ (if $m$ is odd).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We conclude that the solutions are $f(x)=0$, $f(x)=x$, $f(x)=1$ and if $m$ is odd $f(x)=-x$ and $f(x)=-1$.&lt;/p&gt;&#xA;" OwnerUserId="446" LastEditorUserId="446" LastEditorDisplayName="" LastEditDate="2010-08-08T14:04:07.957" LastActivityDate="2010-08-08T14:04:07.957" CommentCount="3" />
  <row Id="1844" PostTypeId="1" AcceptedAnswerId="1849" CreationDate="2010-08-08T14:52:41.617" Score="2" ViewCount="634" Body="&lt;p&gt;On a 2D plane, line X is at x radians angle, an incoming light travels at y radians angle, how to calculate the angle of the outgoing light reflected off line X? How to do this in a way to cover all cases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: I was trying to figure out &lt;a href=&quot;http://projecteuler.net/index.php?section=forum&amp;amp;id=144&quot; rel=&quot;nofollow&quot;&gt;Project Euler problem 144&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="369" LastEditorUserId="369" LastEditorDisplayName="" LastEditDate="2010-08-09T03:35:27.663" LastActivityDate="2010-08-09T03:35:27.663" Title="How to calculate reflected light angle?" Tags="&lt;geometry&gt;&lt;trigonometry&gt;&lt;project-euler&gt;" AnswerCount="4" CommentCount="4" FavoriteCount="1" />
  <row Id="1845" PostTypeId="2" ParentId="1844" CreationDate="2010-08-08T15:47:01.057" Score="1" ViewCount="" Body="&lt;p&gt;Let the line has equation $ax + by + c = 0$, point where light comes on the line has coordinates $(x_0, y_0)$ and direction of the light is $(v_x, v_y)$. Suppose also that line and light direction are normed: $a^2 + b^2 = 1$ and $v_x^2 + v_y^2 = 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Vector $(-b, a)$ is a vector parallel to the line. Let $d = -bv_x + av_y$ --- projection of light's direction to the line.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then new direction of the light is $((-b)\cdot(2d) - v_x, (a)\cdot(2d) - v_y)$.&lt;/p&gt;&#xA;" OwnerUserId="334" LastActivityDate="2010-08-08T15:47:01.057" CommentCount="1" />
  <row Id="1846" PostTypeId="2" ParentId="709" CreationDate="2010-08-08T15:51:33.067" Score="1" ViewCount="" Body="&lt;p&gt;There are nice answers already but I would like to add another way to look at this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The quadratic equation $a(x-r_1)(x-r_2) = ax^2 - a(r_1 + r_2)x + r_1 r_2 = 0$ defines a parabola, $r_1, r_2$ are the roots (intersections with the x-axis).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The middle point is between the two roots, $\frac{r_1 + r_2}{2} = -\frac{- a(r_1 + r_2)}{2a}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This leads to an interesting question about what happens when the parabola does not have real roots.&lt;/p&gt;&#xA;" LastActivityDate="2010-08-08T15:51:33.067" />
  <row Id="1848" PostTypeId="2" ParentId="1838" CreationDate="2010-08-08T16:08:00.047" Score="0" ViewCount="" Body="&lt;p&gt;You seem to have a very idealised idea of what is likely to happen. In the vicinity of a pole the magnetic field lines are nearly vertical. Any compass that needs gimbals is not going to be reliable over a wide area.&lt;/p&gt;&#xA;" OwnerUserId="638" LastActivityDate="2010-08-08T16:08:00.047" CommentCount="2" />
  <row Id="1849" PostTypeId="2" ParentId="1844" CreationDate="2010-08-08T16:14:33.067" Score="3" ViewCount="" Body="&lt;p&gt;Since you've stated all three angles in similar terms, and want a formula that works in all cases, lets use the angle with the x axis, in 360 degree terms, which accomplishes both purposes and is good for computation.  So here's the picture, with z the angle you are seeking, and a the angle of reflection...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;p&gt;&lt;img src=&quot;http://djlewis.smugmug.com/photos/962600635_sUzmn-O.png&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, using the two triangles with the x axis as base and the fact that an exterior angle of a triangle is the sum of the other interior angles, you get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;z = x + a&lt;/p&gt;&#xA;&#xA;&lt;p&gt;y = $\pi$ - 2a + z&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And solving those for z in terms of x and y gives&lt;/p&gt;&#xA;&#xA;&lt;p&gt;z = $\pi$ + 2x - y&lt;/p&gt;&#xA;&#xA;&lt;p&gt;OK, this is actually far from a general analysis -- angles x, y and z could occur in any order on the x axis, but we may assume that y is to the right of x without losing generality, so there are only two other cases. Also angle x could &gt; $\frac{\pi}{2}$, but since the lines are undirected, we may assume that x &amp;lt; $\pi$. Finally, angle x could be 0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hmmm... thinking about this some more.  When z falls to the right of y, the same formula follows because the roles of z and y are interchanged. But when z falls to the left of x, it's because the line of the reflected light intersects the x axis &quot;backwards&quot;.  And then the geometry yields z = 2x - y, or the prior formula if you take the angle of the reflected light as the supplement of z.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we really need vectors for the light rays, not undirected lines, and/or the original problem is not altogether well-formed, that is the notion &quot;angle&quot; of a light ray needs to be defined as the direction of its vector. If you do that, then the angle labeled z in my diagram is not the correct vector directional angle. It should be $\pi$ + z, so the true formula, in vector direction is z = 2x - y, and that works for all cases. (Haven't checked the degenerate cases, but no reason to expect them to fail).&lt;/p&gt;&#xA;" OwnerUserId="271" LastEditorUserId="271" LastEditorDisplayName="" LastEditDate="2010-08-08T22:10:26.597" LastActivityDate="2010-08-08T22:10:26.597" CommentCount="3" />
  <row Id="1850" PostTypeId="2" ParentId="514" CreationDate="2010-08-08T16:21:28.287" Score="1" ViewCount="" Body="&lt;p&gt;I don't know if I would consider this accessible or 'large', but the counterexample of Adyan to the famous General Burnside Problem (http://en.wikipedia.org/wiki/Burnside%27s_problem) in group theory requires an odd exponent greater than or equal to 665. The &quot;shorter&quot; counterexample (proof) due to Olshanskii requires an exponent greater than 10^10. The reason for the large number in the latter proof is essentially due to 'large scale' consequences of Gauss-Bonnet theorem for certain planar graphs expressing relations in groups. It may be that a finer analysis can show that a counterexample can occur at exponent as low as 5, but this is still not known. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is probably essentially different than what you are asking, since we aren't forced to consider 665 because the cases 1-664 are known to be true. I thought it may be fun to point out, here, though! &lt;/p&gt;&#xA;" OwnerUserId="941" LastActivityDate="2010-08-08T16:21:28.287" />
  <row Id="1851" PostTypeId="2" ParentId="709" CreationDate="2010-08-08T16:47:32.927" Score="4" ViewCount="" Body="&lt;p&gt;An alternative approach, which isn't necessarily the &lt;em&gt;best&lt;/em&gt; in and of itself, but stick with me ...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instead of using the fact that the vertex is the point where a &lt;em&gt;vertical&lt;/em&gt; line cuts the parabola exactly in half, we'll use the fact that it's also the point where a &lt;em&gt;horizontal&lt;/em&gt; line &lt;em&gt;just&lt;/em&gt; brushes the curve, meeting it in a single point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider how the horizontal line $y=d$ crosses the parabola as you adjust $d$. If $d$ places the line above or below the vertex, there are either 2 or 0 points of intersection (which is which depends upon whether the parabola points up or down); when $d$ places the line at the same height as the vertex, the parabola seems to sit comfortably on the line. (The line is &lt;em&gt;tangent&lt;/em&gt; to the parabola the point they have in common, in the same way that lines can be tangent to circles.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can find the point of tangency (the vertex), then, by using algebra to determine when the intersection of our parabola and line equations have a single solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We begin with the standard find-the-intersection strategy, setting our formulas for $y$ equal:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$d = a x^2 + b x + c$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When does this equation have a &lt;em&gt;single&lt;/em&gt; solution for $x$? Let's see. Rearranging and invoking the Quadratic Formula, we have:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$a x^2 + b x + c - d = 0$&lt;br /&gt;&lt;br /&gt;&#xA;  $x = \frac{1}{2a}\left( - b \pm \sqrt{ b^2 - 4 a (c-d) } \right)$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now, when the stuff under the radical &lt;em&gt;matters&lt;/em&gt;, then we have either 2 or 0 values of $x$: two values if the stuff is positive (the square root gives a quantity to add and subtract from $-b$); no roots if the stuff is negative (the square root gives an imaginary number we can't use here). We'll have our sought-after single value of $x$ when $d$ is whatever-it-has-to-be (its exact value doesn't matter) to make the stuff under the radical becomes zero, making the radical term vanish; when that happens, our value of $x$ falls out as&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$x = -\frac{b}{2a}$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Just like with the other approaches.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With &lt;em&gt;this&lt;/em&gt; approach, however, we can consider a more sophisticated situation. It may be somewhat advanced for your class, but --if you can make sense of my rambling-- you can use it to blow your teacher's mind. :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Before we begin: I wouldn't necessarily recommend this exploration for every student just beginning to learn about parabola equations. However, I think having it &quot;in the archive&quot;, as it were, can be worthwhile for people who come across the question to see that the answer can lead to some interesting places.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We ask ourselves this question:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What if we &lt;em&gt;tilt the line&lt;/em&gt;?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That is: What can we say about the $x$-coordinate of the point of tangency where the parabola is touched by a line with slope $m$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let the slanted line's equation be $y = m x + d$, where $d$ is adjustable. We can make the same observations as in the horizontal case: if $d$ places the line above or below the point of tangency, the line will meet the parabola in either 2 or 0 places; for a &lt;em&gt;special&lt;/em&gt; value of $d$, however, the line meets the parabola at the single point, the point of tangency.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As before, we set our formulas for $y$ equal to each other ...&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$m x + d = a x^2 + b x + c$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;... then rearrange and invoke the Quadratic Formula ...&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$a x^2 + ( b - m ) x + ( c - d ) = 0$&lt;br /&gt;&lt;br /&gt;&#xA;  $x = \frac{1}{2a}\left( -(b-m) \pm \sqrt{stuff} \right) = \frac{1}{2a}\left(m-b\pm \sqrt{stuff}\right)$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And as before, we don't really care about the complicated mess that &quot;$stuff$&quot; might be, since we're only interested in the situation where it goes away anyhow.[*] Once the radical term vanishes and the dust settles, we have this:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;For the parabola described by the equation $y = a x^2 + b x + c$, the $x$-coordinate of the point where a line of slope $m$ lies tangent to the curve is given by &lt;br /&gt;&lt;br /&gt;&#xA;  $x = \frac{1}{2a}\left( m - b \right)$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here's the kicker that your smart friend[**] can appreciate: That statement &lt;em&gt;really is&lt;/em&gt; a fact that most people don't see until Calculus. Except there, the issue tends to be phrased a little differently: It's not about finding the $x$-coordinate of the point-of-tangency based on a given slope, but about finding the slope of the line based on the $x$-coordinate for the point-of-tangency. So, we re-phrase the above equation to yield $m$:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;For the parabola described by the equation $y = a x^2 + b x + c$, the slope $m$ of the line tangent to the curve at a point with $x$-coordinate $x$ is given by&lt;br /&gt;&lt;br /&gt;&#xA;  $m = 2 a x + b$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You'll be way ahead of the game if you can see a neat way to get directly from the parabola equation to the slope formula. I'll give you a hint by showing how the formula fits in with a couple of facts you already know about the slopes of lines, and show you one step further, with a few &lt;em&gt;unnecessary&lt;/em&gt; zeros thrown in:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$y = a$ &amp;emsp; slope = $0$.&lt;br /&gt;&#xA;  $y = a x + b$ &amp;emsp; slope = $a + 0$.&lt;br /&gt;&#xA;  $y = a x^2 + b x + c$ &amp;emsp; slope = $2 a x + b + 0$&lt;br /&gt;&#xA;  $y = a x^3 + b x^2 + c x + d$ &amp;emsp; slope = $3 a x^2 + 2 b x + c + 0$&lt;br /&gt;&#xA;  $y = a x^4 + b x^3 + c x^2 + d x + e$ &amp;emsp; slope = &lt;em&gt;you tell me&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Detect a pattern?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[*] Actually, you should care enough to convince yourself that you can find a value of $d$ that &lt;em&gt;makes&lt;/em&gt; &quot;$stuff$&quot; go away, but for now you can trust me that such a $d$ exists.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[**] Keep in mind that &quot;smart&quot; doesn't always mean &quot;having the right answers&quot;; just as often, it means &quot;asking the right questions&quot;. The fact that you chose not to accept a formula blindly and asked your teacher &quot;why&quot; --and then steadfastly continued your quest for an answer when she wouldn't provide one-- suggests to me that your smart friend has a smart friend, too.&lt;/p&gt;&#xA;" OwnerUserId="409" LastActivityDate="2010-08-08T16:47:32.927" CommentCount="1" />
  <row Id="1852" PostTypeId="1" CreationDate="2010-08-08T16:48:13.563" Score="2" ViewCount="168" Body="&lt;p&gt;Imagine a process with two variables &lt;em&gt;min&lt;/em&gt; and &lt;em&gt;max&lt;/em&gt;, and two counters &lt;em&gt;hi&lt;/em&gt; and &lt;em&gt;lo&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We initialize &lt;em&gt;min&lt;/em&gt; and &lt;em&gt;max&lt;/em&gt; by selecting two random numbers (assume a uniform (0,1) distribution for convenience), and sorting them.  We also set the &lt;em&gt;hi&lt;/em&gt; and &lt;em&gt;lo&lt;/em&gt; counters to 0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At each step we select a new random number &lt;em&gt;r&lt;/em&gt; from our distribution and do the following&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;if (r &amp;lt; min) {&#xA;  lo++;&#xA;} else if (r &amp;gt; max) {&#xA;  hi++;&#xA;} else if (lo &amp;gt; hi) {&#xA;  max = r;&#xA;  hi++;&#xA;} else if (hi &amp;gt; lo) {&#xA;  min = r;&#xA;  lo++;&#xA;} else {&#xA;  if (rand(0,1) &amp;gt; 0.5 ){&#xA;    max = r;&#xA;    hi++;&#xA;  } else {&#xA;    min = r;&#xA;    lo++;&#xA;  }&#xA;}&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;If the random number is outside our range, we increment the appropriate counter, and if it lies inside our range we increment the lower counter and adjust our range appropriately.  The question is, after &lt;em&gt;n&lt;/em&gt; iterations, where do we expect min and max to end up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It can be seen that the difference between min and max (assuming a uniform(0, 1) distribution) is distributed as the minimum of &lt;em&gt;n + 2&lt;/em&gt; uniform (0, 1) random variables.  However, it should be possible to say considerably more about the location of &lt;em&gt;min&lt;/em&gt; and &lt;em&gt;max&lt;/em&gt; over time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;  This algorithm is only vaguely related to median finding.  The process given here just takes a sequence of unifomly distributed random numbers and attempts to guess the median by remembering only two values.  This, of course is not an effective method, but similar algorithms are used (remembering more than two variables).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The process that I want to analyze has two values and two counters, and takes a sequence of random numbers, if the random number is larger than both, then increase the &lt;em&gt;hi&lt;/em&gt; counter, if the number is smaller than both then increase the &lt;em&gt;lo&lt;/em&gt; counter, and if the number lies between the two, either replace &lt;em&gt;max&lt;/em&gt; with the current number and increase &lt;em&gt;hi&lt;/em&gt; or else replace &lt;em&gt;min&lt;/em&gt; with the current number and increase &lt;em&gt;lo&lt;/em&gt;, whichever makes the &lt;em&gt;hi&lt;/em&gt; or &lt;em&gt;lo&lt;/em&gt; counters closer (in cases where either choice could be made, flip a coin).  &lt;em&gt;hi&lt;/em&gt; and &lt;em&gt;lo&lt;/em&gt; are counting the numbers that we have seen that are larger than &lt;em&gt;max&lt;/em&gt; or smaller that &lt;em&gt;min&lt;/em&gt; respectively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I want to know, is how to figure out the distribution of &lt;em&gt;min&lt;/em&gt; and &lt;em&gt;max&lt;/em&gt; as the number of iterations becomes large, also of interest is the distribution of &lt;em&gt;hi - lo&lt;/em&gt;.  I can find the distribution of &lt;em&gt;max - min&lt;/em&gt; using elementary order statistics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If &lt;em&gt;min = max&lt;/em&gt;, then &lt;em&gt;hi - lo&lt;/em&gt; is an ordinary one dimensional random walk, and is easy to analyze.  When they are different the walk is subtly biased towards 0.  Similarly the values of &lt;em&gt;min&lt;/em&gt; and &lt;em&gt;max&lt;/em&gt; are biased towards 0.5, I want to know how to find out by how much they are biased.&lt;/p&gt;&#xA;" OwnerUserId="943" LastEditorUserId="943" LastEditorDisplayName="" LastEditDate="2010-08-08T21:20:01.863" LastActivityDate="2010-08-08T21:20:01.863" Title="Analysis of a biased random walk related to median finding" Tags="&lt;algorithms&gt;&lt;probability&gt;&lt;stochastic-processes&gt;" CommentCount="0" FavoriteCount="0" />
  <row Id="1854" PostTypeId="2" ParentId="1844" CreationDate="2010-08-08T17:12:42.310" Score="1" ViewCount="" Body="&lt;p&gt;&lt;img src=&quot;http://www.imgftw.net/img/412079240.png&quot; alt=&quot;diagram&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that regardless of what line the angles are measured from, there is a parallel line through the point of reflection on the mirror for which the angle measures will be the same.  My diagram is not fully general as drawn (in terms of the angles), but the argument below is fully general, provided that all angle measures are directed&amp;mdash;that is, counterclockwise angles are positive and clockwise angles are negative.  x is the angle from a reference line to (one half of) the mirror, y is the angle from that same reference line to the incoming light ray.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The angle from the measured half of the mirror to the incoming light ray is $y-x$ (note that if the diagram is as I'd drawn, except that x were to the other half of the mirror, $x&gt;y$ so $y-x&amp;lt;0$, which makes sense since $y-x$ would be going counterclockwise from the other half of the mirror to the incoming light ray).  Because the magnitude of the angles of incidence and reflection are the same, but the angle of reflection is the opposite direction from the other half of the mirror, the angle from the other half of the mirror to the outgoing light ray is $-(y-x)=x-y$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This makes the directed angle from the measured half of the mirror to the outgoing light ray $\pi+(x-y)$ (start at the measured half of the mirror, rotate $\pi$ counterclockwise, then rotate $x-y$, which is directed back clockwise).  From the angle reference line to the outgoing light ray is then $x+\pi+x-y=\pi+2x-y$ (as in David Lewis's answer).&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-08-08T17:12:42.310" />
  <row Id="1855" PostTypeId="2" ParentId="1840" CreationDate="2010-08-08T17:36:59.260" Score="3" ViewCount="" Body="&lt;p&gt;$\sum_{i,j=1}^{n} |a_i\cdot a_j - b_i\cdot b_j|$ = $\sum_{i,j=1}^{n} |a_i\cdot a_j - b_i.a_j + b_i.a_j - b_i\cdot b_j|$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$= \sum_{i,j=1}^{n} |(a_i - b_i).a_j + (a_j - b_j).b_i|$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\leq \sum_{i,j=1}^{n} [|(a_i - b_i).a_j| + |(a_j - b_j).b_i|]$ ........ (using the triangle inequality ,i.e., $|x+y| \leq |x|+|y|$)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$= (\sum_{i,j=1}^{n} |(a_i - b_i).a_j|) + (\sum_{i,j=1}^{n} |(a_j - b_j).b_i|)$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\leq (\sum_{i}^{n} |(a_i - b_i).1|) + (\sum_{i}^{n} |e.b_i|)$ .... (By solving the inner summation over j and using the given (in)equations).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\leq e + e$&lt;br&gt;&#xA;$= 2.e$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence Proved.&lt;/p&gt;&#xA;" OwnerUserId="938" LastActivityDate="2010-08-08T17:36:59.260" />
  <row Id="1856" PostTypeId="2" ParentId="1808" CreationDate="2010-08-08T19:55:32.963" Score="1" ViewCount="" Body="&lt;p&gt;Computer graphics is a prime example of a field that needs sophisticated CS and math.&lt;/p&gt;&#xA;" OwnerUserId="589" LastActivityDate="2010-08-08T19:55:32.963" CommunityOwnedDate="2010-08-08T19:55:32.963" />
  <row Id="1857" PostTypeId="2" ParentId="1778" CreationDate="2010-08-08T20:00:06.867" Score="0" ViewCount="" Body="&lt;p&gt;As Issac pointed out, this is known as the Gamler's ruin problem. I recently wrote a couple of blog posts (&lt;a href=&quot;http://randomdeterminism.wordpress.com/2010/07/09/gamblers-ruin-2/&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; and the post linked from there) explaining how to calculate the ruin probability. I'll repeat part of one of the proofs here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Problem Formulation&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A gambler enters a casino with $n$ dollars in cash and starts playing a game where he wins with probability  $p$ and looses with probability $q = 1-p$  The gampler plays the game repeatedly, betting $1$ dollar in each round. He leaves the gave it his total fortune reaches $N$ or he runs out of money (he is &lt;em&gt;ruined&lt;/em&gt;), whichever happens first. What is the probability that the gambler is ruined.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A gambler's ruin can be modeled as a one-dimensional random walk in which we are interested in the hitting probability of the absorbing states. Calculating these probabilities is fairly straightforward. Let $P_N(n)$ denote the probability that the gambler's fortune reaches $N$ dollars before he is ruined on the condition that his current fortune is $n$. Then,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$P_N(n) = p P_N(n+1) + q P_N(n-1)$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;which can be rewritten as&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$\displaystyle [P_N(n+1) - P_N(n)] = \left(\frac q p \right)[ P_N(n) - P_N(n-1)]$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Since $P_N(0) = 0$, we have that&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$\displaystyle P_N(2) - P_N(1) = \left(\frac qp \right) P_N(1)$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and similarly&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$\displaystyle P_N(3) - P_N(2) = \left(\frac qp \right) [P_N(2) - P_N(1)] = \left( \frac qp \right)^2 P_N(1)$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Continuing this way, we get that&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$\displaystyle P_N(n) - P_N(n-1) = \left( \frac qp \right)^{n-1} P_N(1) $.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and therefore, by adding the first $n$ such terms, we get&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$\displaystyle P_N(n) = \sum_{k=0}^{n-1} \left( \frac qp \right)^k P_N(1)$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Moreover, we know that&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$\displaystyle P_N(N) = \sum_{k=0}^{N-1} \left( \frac qp \right)^k P_N(1) = 1$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Thus,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$\displaystyle P_N(1) = \frac 1{\sum_{k=0}^{N-1} \left( \frac qp \right)^k} = \frac { 1 - (q/p)}{\strut 1 - (q/p)^N}, \quad  p \neq q $&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$P_N(1) = \frac 1N, \quad p = q $.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Combining with the previous expression for $P_N(n)$ we get,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$\displaystyle P_N(n) = \begin{cases} \frac{ 1 - (q/p)^n} {\strut 1 - (q/p)^N}, &amp;amp; p \neq 1/2 \ \frac{n}{N}, &amp;amp; p = 1/2 \end{cases}$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For ease of representation let $\lambda = q/p$. Then, the probability of &lt;em&gt;winning&lt;/em&gt; are&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$\displaystyle P_N(n) = \frac{ 1 - \lambda^n} {\strut 1 - \lambda^N}, \quad \lambda \neq 1 $&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$P_N(n) = \frac{n}{N}, \quad \lambda = 1 $.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="763" LastActivityDate="2010-08-08T20:00:06.867" CommentCount="2" />
  <row Id="1858" PostTypeId="2" ParentId="1838" CreationDate="2010-08-08T20:20:18.283" Score="5" ViewCount="" Body="&lt;p&gt;I don't imagine that this is a practical issue for navigation any longer, given the advent of GPS technology. However, it is of practical concern in 3-d animation and robotics. To get back to your navigation example, suppose that I have a mechanical gyro mounted in an airplane flying over the North Pole. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the gyro is only mounted on three gimbals, one of the gimbals will freeze because moving smoothly to the proper orientation would require at least one of the gimbals to flip 180 degrees instantaneously. The gimbal lock problem can be countered by adding a redundant degree of freedom in the form of an extra gimbal, an extra joint in a robotic arm, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you pointed out, it's the singularity at the poles of the representation that's the problem. Having a redundant degree of freedom helps because you can have enough information at the pole to move the correct direction. In 3-d graphics, if an axis-angle representation (Euler axis and angle) or quaternions are used instead of a triple-axis representation (Euler angles), then weird rotation artifacts due to gimbal lock are eliminated (performing a YouTube search for &quot;gimbal lock&quot; yields several visual demonstrations of the problem).&lt;/p&gt;&#xA;" OwnerUserId="950" LastActivityDate="2010-08-08T20:20:18.283" />
  <row Id="1859" PostTypeId="2" ParentId="1741" CreationDate="2010-08-08T20:42:53.020" Score="1" ViewCount="" Body="&lt;p&gt;Sets are first-order objects in a system like ZFC, although there are systems like second-order arithmetic where the first-order objects are non-set individuals and sets of individuals are the second-order objects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bernays-Gödel set theory (BG set theory), which I see both WP and Wolfram prefer to call by the unwieldy term &lt;a href=&quot;http://mathworld.wolfram.com/vonNeumann-Bernays-GoedelSetTheory.html&quot; rel=&quot;nofollow&quot;&gt;Von Neumann–Bernays–Gödel set theory&lt;/a&gt;, adds second-order quantifiers over classes.  The result is conservative over ZFC (i.e., its first-order theorems are the same), but allows much shorter proofs of theorems.  You can formalise talk of classes in first-order set theory by using &lt;em&gt;class terms&lt;/em&gt;, where you treat one-place predicates as defining a set by extension, but this is weaker than using second-order quantifiers, and an important issue in second-order set theories is what classes a particular set theory is committed to.  This question can be addressed axiomatically in third-order set theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've seen the usage &quot;second-order set&quot; and &quot;third-order set&quot; to talk about classes and their third-order generalisation.  &lt;/p&gt;&#xA;" OwnerUserId="100" LastActivityDate="2010-08-08T20:42:53.020" />
  <row Id="1860" PostTypeId="1" CreationDate="2010-08-08T20:43:43.737" Score="3" ViewCount="344" Body="&lt;p&gt;Prove (or disprove) the following statement:&#xA;For any positive integers $x,y,t$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle\sum_{i=1}^{t(y+1)-1} \frac{1}{t(xy+x-1)-x+i}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is an increasing function of $t$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My attempts:&#xA;The statement appears to be true numerically.&#xA;Tried some obvious bounds to compare the sums for consecutive values of $t$ but didn't find one that was strong enough to prove the statement. &lt;/p&gt;&#xA;" LastActivityDate="2010-08-12T17:36:50.857" Title="Comparing sums of reciprocals" Tags="&lt;number-theory&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="1861" PostTypeId="2" ParentId="1766" CreationDate="2010-08-08T20:53:10.920" Score="3" ViewCount="" Body="&lt;p&gt;Here's a more conceptual answer that doesn't resort to inequalities ad hoc.  Assume $a_n &gt; 0$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem can be rephrased, by writing $(a_n)^3 = 1/(n^{3/2 + p})$ (with $p$ a function of $n &gt; 1$, and ignoring $a_1$), as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;If $\Sigma 1/(n^{3/2 + p})$ converges then $\Sigma 1/(n^{3/2 + (p/3)})$ converges&lt;/em&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The transformation moves the series toward the (convergent) one with $p=0$.  This preserves convergence, because the series can be split into the terms with $p \leq 0$ and $p&gt;0$.  For the first set, convergence is improved, and for the second set, the sum is dominated by $\Sigma 1/n^{3/2}$. &lt;/p&gt;&#xA;" OwnerUserId="467" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-08-08T22:07:07.723" LastActivityDate="2010-08-08T22:07:07.723" />
  <row Id="1862" PostTypeId="1" AcceptedAnswerId="1863" CreationDate="2010-08-08T20:53:50.873" Score="3" ViewCount="165" Body="&lt;p&gt;I've been reading about tropical geometry and many papers reference polyhedral fans. I feel like I have a decent intuitive picture of what they are from reading articles but I still haven't been able to guess the general definition. All the ones I've encountered have been systems of linear inequalities, so that is my best guess at a general definition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any comments on where they appeared first historically or links/books to general resources on learning about them would be appreciated. Also, I'm curious to know what other areas of math these show up in?&lt;/p&gt;&#xA;" OwnerUserId="249" LastActivityDate="2010-08-21T14:39:53.640" Title="What is the rigorous definition of polyhedral fan? What are some good resources to learn about them? What context do they arise naturally in?" Tags="&lt;geometry&gt;" AnswerCount="1" />
  <row Id="1863" PostTypeId="2" ParentId="1862" CreationDate="2010-08-08T21:12:44.340" Score="5" ViewCount="" Body="&lt;p&gt;A polyhedral cone is a subset of a real vector space which is the intersection of finitely many closed half spaces. (The defining planes of these half spaces must pass through $0$.) A fan is a finite set $F$ of polyhedral cones, all living in the same vector space, such that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) if $\sigma$ is a cone in $F$, and $\tau$ is a face of $\sigma$, then $\tau$ is in $F$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) if $\sigma$ and $\sigma'$ are in $F$, then $\sigma \cap \sigma'$ is a face of both $\sigma$ and of $\sigma'$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This &lt;a href=&quot;http://sbseminar.wordpress.com/2009/02/18/toric-varieties-and-fans/&quot; rel=&quot;nofollow&quot;&gt;blog post&lt;/a&gt; of mine might help you visualize these definitions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most mathematicians I know learned fans from Fulton's &lt;em&gt;Toric Varieties&lt;/em&gt;. This would involve learning a lot of algebraic geometry on top of your combinatorics, although it is algebraic geometry that is very relevant to tropical geometry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a pure combinatorics reference, have you tried Chapter 2 of &lt;a href=&quot;http://www.math.ucdavis.edu/~deloera/BOOK/final.pdf&quot; rel=&quot;nofollow&quot;&gt;De Loera, Rambau and Santos&lt;/a&gt;? They focus on polyhedral complexes, which is the more general setup where you don't require that the half spaces pass through $0$, but they talk about fans as well. I haven't had a chance to look at it yet but, based on my knowledge of the authors, I expect it is very good.&lt;/p&gt;&#xA;" OwnerUserId="448" LastEditorUserId="448" LastEditorDisplayName="" LastEditDate="2010-08-21T14:39:53.640" LastActivityDate="2010-08-21T14:39:53.640" CommentCount="3" />
  <row Id="1864" PostTypeId="2" ParentId="1116" CreationDate="2010-08-08T21:14:51.033" Score="1" ViewCount="" Body="&lt;p&gt;This problem is tantalizing. It may be a &quot;toy&quot;-version of some substantial ongoing research in topology, so it's valuable and interesting.  It has immediate analogs in areas of classical topology, knot theory, universal algebra, and topological field theory.  By toy-version, I don't mean it's necessarily easy to solve, but that it can be used to present some advanced ideas in a non-technical way.  Forgive me because what follows is not an answer to your question, but a rambling list of related problems and possible applications. I only hope it inspires someone here to investigate the connections in algebraic topology. (Thus, community wiki.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are asking to count equivalence classes of parametrized rectangle framings (where a parametrized framing is a framing with precise line positions).  In some sense, this type of enumeration problem comes up often in algebraic topology when one wants to understand a cellular decomposition of a complicated space.  Take for instance, the space of n-planes in R^d.  These spaces are called Grassmannians and their topology can be analyzed by breaking them down into subsets of various dimensions.  These subsets are called Schubert cells.  Counting these Schubert cells and understanding how they fit together was an important advancement.  See the wikipedia entries on &lt;strong&gt;Grassmannian&lt;/strong&gt; and &lt;strong&gt;Enumerative Geometry&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your parametrized rectangle framing spaces (denoted perhaps Fi^{n} where i indexes the equivalence classes of framings having n sub-rectangles) are analogous to the cells of the Grassmannian, but the resulting stratified space is some universal space of parametrized rectangular framings which includes framings with an arbitrarily large but finite number of sub-rectangles.  That is, each of your framing spaces F (the equivalence classes you want to count)  represents a &quot;cell&quot; of some fixed dimension which degenerates into lower degree framings at the boundaries.  By carefully studying the simple shape of each cell and how they degenerate into other cells, we can create a universal space $F^{\infty}$ = $\bigcup$ $F^{n}_{i}$ (&lt;em&gt;modulo the boundary case equivalence relations&lt;/em&gt;).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sometimes one can approach the topology of this universal space by other means, and that global knowledge can be used to count the number of cells needed in each stratum.  Finally, as you point out, the spaces have certain symmetries which may pass up to the full space in an interesting way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This stratification trick comes up all the time  If you want to understand some large space, find a way to break it down into strata which fit together in an analyzable way.  Or, find a related space which has a natural stratification, analyze this stratified space, and make inferences about the original space.  A great, non-trivial use of this in the 90's was when Vassiliev realized that knots could be studied by analyzing the space of non-knots which admitted a very clear stratification (essentially stratify the non-knots by the number of points where the circle map is non-injective)  Vassiliev found clear topological structure in this space and this allowed him to make claims about the structure of the set of knots.  This led others like Kontsevich and Bar Natan to provide real computational tools using tricks for counting and integrating over cells.  For instance, Mathworld has two good introductory articles on &lt;strong&gt;Vassiliev Invariants&lt;/strong&gt; and &lt;strong&gt;Kontsevich Integrals&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thirdly, your rectangle framing space is evocative of the little-disks operad which encodes an algebraic structure.  See the wikipedia page:&lt;br&gt;&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Operad_theory#.22Little_something.22_operads&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Operad_theory#.22Little_something.22_operads&lt;/a&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whenever we have a set of topological spaces such that elements of the spaces can be geometrically combined to get higher degree spaces, it hints that the geometry can be studied using techniques from algebra.  If you imagine putting variables in each of your frame's sub-rectangles, you've sort of described an algebraic operation.  Imagine that every one of your parametrized n-framings defined a map from n inputs to 1 output, an &quot;n-ary operation.&quot;  Further assume that the geometry forces some consistency condition that demands that when you place the result of a p-ary operation into one of the sub-rectangles in a q-ary frame, you get just what you'd get from the corresponding (p+q-1)-ary frame operation.  This means your algebra must satisfy certain standard relations (maybe up to homotopy) and then these operations may descend to operations on an appropriately defined cohomology theory.  That is, the topology of your universal framing space may index various operations which satisfy certain relations exactly.  See for instance the ncatlab.org entries on &lt;strong&gt;operad&lt;/strong&gt;, &lt;strong&gt;L-infinity-algebra&lt;/strong&gt;, and &lt;strong&gt;A-infinity-algebra&lt;/strong&gt;.&lt;/p&gt;&#xA;" OwnerUserId="948" LastActivityDate="2010-08-08T21:14:51.033" CommunityOwnedDate="2010-08-08T21:16:11.643" />
  <row Id="1865" PostTypeId="2" ParentId="307" CreationDate="2010-08-08T21:27:37.610" Score="3" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\rm\ (p-1)!\ mod\ p\:$ is the product of all elts of $\rm\: {\mathbb F}_p^*$. The map $\rm n \mapsto n^{-1}$ is a permutation of $\rm\:{\mathbb F}_p^*\:$ of order 2 so it decomposes into cycles of length 1 or 2, which partition the product. The 2-cycles $\rm (n, n^{-1})$ have product 1 so they can be deleted from the product, leaving only the product of 1-cycles $\rm (n)$. They satisfy $\rm\: n^{-1} = n \Rightarrow n^2 = 1 \Rightarrow n = \pm 1\:$ via $\rm{\mathbb F}_p$ a field. So the product reduces to $\:-1*1 = -1$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This generalizes: if a finite abelian group has a unique elt of order 2 then it is equal to the product of all the elts; otherwise the product is 1, e.g. see this &lt;a href=&quot;http://groups.google.com/group/sci.math/browse_frm/thread/784a5eb7087387eb&quot; rel=&quot;nofollow&quot;&gt;thread&lt;/a&gt; for hints.    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice how we've exploited the existence of a symmetry - here an involution that induces a natural pairing of elts. Frequently involution and reflection symmetries lie at the heart of elegant proofs, e.g. see the elegant proof by Liouville, &lt;a href=&quot;http://eprints.maths.ox.ac.uk/677/1/invariant.pdf&quot; rel=&quot;nofollow&quot;&gt;Heath-Brown&lt;/a&gt; and &lt;a href=&quot;http://mathoverflow.net/questions/31113&quot; rel=&quot;nofollow&quot;&gt;Zagier&lt;/a&gt;, that every prime $\equiv 1 \pmod 4$ is a sum of 2 squares, or the little-known beautiful &lt;a href=&quot;http://mathoverflow.net/questions/33726&quot; rel=&quot;nofollow&quot;&gt;reflective&lt;/a&gt; generation of the ternary tree of primitive Pythagorean triples due to Aubry.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-10-14T22:30:00.247" LastActivityDate="2010-10-14T22:30:00.247" />
  <row Id="1866" PostTypeId="2" ParentId="307" CreationDate="2010-08-08T22:02:25.603" Score="1" ViewCount="" Body="&lt;p&gt;Here are a couple possible proofs of Wilson's theorem for $p&gt;2$ ($p=2$ is easily checked):&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;We have that $x^{p-1}-1$ has roots $1,2,\ldots,p-1$ over $\mathbb{Z}/p\mathbb{Z}$ (by &lt;a href=&quot;http://en.wikipedia.org/wiki/Fermat%27s_little_theorem&quot; rel=&quot;nofollow&quot;&gt;Fermat's Little Theorem&lt;/a&gt;). But as $\mathbb{Z}/p\mathbb{Z}$ is a field, we have unique factorization of polynomials, so that $x^{p-1}-1=(x-1)(x-2)\ldots(x-(p-1))$. Comparing constant terms wields Wilson's theorem.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Let $g$ be a &lt;a href=&quot;http://en.wikipedia.org/wiki/Primitive_root_modulo_n&quot; rel=&quot;nofollow&quot;&gt;primitive root&lt;/a&gt; modulo $p$. Then $(p-1)!\equiv g\times g^2\times \ldots \times g^{p-1}=g^{p\frac{p-1}{2}}\equiv g^{\frac{p-1}{2}}\bmod{p}$ by Fermat's Little Theorem, and $g^{\frac{p-1}{2}}\equiv -1 \bmod{p}$ because if $(g^{\frac{p-1}{2}})^2=g^{p-1}\equiv 1 \bmod{p}$ and $g^{\frac{p-1}{2}}\not \equiv 1 \bmod{p}$ by the definition of primitive root.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="446" LastActivityDate="2010-08-08T22:02:25.603" />
  <row Id="1867" PostTypeId="2" ParentId="1844" CreationDate="2010-08-08T23:08:50.270" Score="0" ViewCount="" Body="&lt;p&gt;Ok, here's another way to look at it, using vector directions and kinda intuitive, but I think it is pretty close to a real proof. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Start with the mirror line X horizontal, that is, its angle x = 0. The it's clear that the vector angle, z, of the reflected light is the negative of the vector angle, y, of the light itself: z = -y.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://djlewis.smugmug.com/photos/962899093_ZPQ96-O.png&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now rotate line X by d degrees around the point of reflection, either way, leaving the original light ray (angle y) fixed and letting the reflected ray (angle z) rotate around the point of reflection to maintain the angle of reflection equal to the angle of incidence. Assuming counterclockwise rotation, d &gt; 0, this &quot;pushes&quot; the reflected line by 2d, one d for a direct push to maintain the angle fo reflection, and another d because the angle of incidence also increases by d, so the reflected light must rotate that much more to keep the two angles of the light equal. Likewise when d &amp;lt; 0 for clockwise rotation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we are increasing (or decreasing, for counterclockwise rotation) angle x by d, but angle z (vector angle) by 2d. Hence...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;z = 2d - y = 2x - y&lt;/p&gt;&#xA;" OwnerUserId="271" LastActivityDate="2010-08-08T23:08:50.270" />
  <row Id="1869" PostTypeId="2" ParentId="1196" CreationDate="2010-08-09T00:07:09.457" Score="1" ViewCount="" Body="&lt;p&gt;Alternatively, $f:{\mathbb N} \rightarrow {\mathbb Z}/3{\mathbb Z} \;$ is constant since $f(n+1) = f(n)$, so $f(n) = f(0) = 0$. This reduces the induction to a trivial lemma: $f(n)$ is constant if $f(n+1) = f(n)$ for all $n\in \mathbb N$, a trivial telescopy.&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-08-09T00:07:09.457" />
  <row Id="1870" PostTypeId="1" AcceptedAnswerId="2781" CreationDate="2010-08-09T00:29:13.923" Score="6" ViewCount="503" Body="&lt;p&gt;What is the the most &lt;em&gt;educational&lt;/em&gt; software for high school and college math?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Not the one that just gives you the answer, but has &lt;strike&gt;any of&lt;/strike&gt; the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Edit: Software mentioned in answer added in brackets&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;provides step-by-step solutions [&lt;em&gt;WolframAlpha&lt;/em&gt;];&lt;/li&gt;&#xA;&lt;li&gt;can verify your step-by-step solution [&lt;em&gt;empty&lt;/em&gt;];&lt;/li&gt;&#xA;&lt;li&gt;can give you hints [&lt;em&gt;empty&lt;/em&gt;];&lt;/li&gt;&#xA;&lt;li&gt;can generate problems similar to given one [&lt;em&gt;empty&lt;/em&gt;];&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;or something along those lines.&lt;/li&gt;&#xA;&lt;/p&gt;&#xA;" OwnerUserId="956" LastEditorUserId="956" LastEditorDisplayName="" LastEditDate="2010-08-09T17:12:26.533" LastActivityDate="2011-01-17T00:51:39.790" CommunityOwnedDate="2010-08-09T00:57:28.670" Title="Educational Math Software" Tags="&lt;math-software&gt;&lt;education&gt;" AnswerCount="7" CommentCount="1" FavoriteCount="7" />
  <row Id="1871" PostTypeId="2" ParentId="1870" CreationDate="2010-08-09T00:44:04.687" Score="8" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://www.wolframalpha.com/&quot; rel=&quot;nofollow&quot;&gt;Wolfram Alpha&lt;/a&gt;. Try, for instance &quot;solve x^2-x-1=0&quot; and click &quot;show steps&quot;.&lt;/p&gt;&#xA;" OwnerUserId="405" LastActivityDate="2010-08-09T00:44:04.687" CommunityOwnedDate="2010-08-09T02:57:41.033" />
  <row Id="1872" PostTypeId="1" CreationDate="2010-08-09T01:52:39.957" Score="4" ViewCount="227" Body="&lt;p&gt;Both matrix multiplication and quaternion multiplication are non-commutative; hence the use of terms like &quot;premultiplication&quot; and &quot;postmultiplication&quot;. After encountering the concept of &quot;quaternion matrices&quot;, I am a bit puzzled as to how one may multiply two of these things, since there are at least four ways to do this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some searching has netted &lt;a href=&quot;http://en.cnki.com.cn/Article_en/CJFDTOTAL-LXXB198402006.htm&quot; rel=&quot;nofollow&quot;&gt;this paper&lt;/a&gt;, but not having any access to it, I have no way towards enlightenment except to ask this question here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there are indeed these four ways to multiply quaternion matrices, how does one figure out which one to use in a situation, and what shorthand might be used to talk about a particular version of a multiplication?&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-08-12T04:11:07.923" Title="On multiplying quaternion matrices" Tags="&lt;linear-algebra&gt;&lt;quaternions&gt;&lt;matrices&gt;" AnswerCount="3" CommentCount="10" FavoriteCount="1" />
  <row Id="1873" PostTypeId="1" CreationDate="2010-08-09T03:46:53.487" Score="0" ViewCount="205" Body="&lt;p&gt;I need to find eigenvalues/eigenvectors of different kinds of nxn matrices. For example as listed below. What is the typical process? Should I always go by the route of finding eigenvalues by finding roots of characteristic polynomial and then getting eigenvectors by solving (A-lambda*I)*x = 0?&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;     [,1] [,2] [,3]&#xA;[1,]    2    0    0&#xA;[2,]    1    2    0&#xA;[3,]    0    1    2&#xA;&#xA;      [,1] [,2] [,3] [,4]&#xA; [1,]    4    1    1    1&#xA; [2,]    1    4    1    1&#xA; [3,]    1    1    4    1&#xA; [4,]    1    1    1    4&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;These are just examples. Typically I want to find eigenvectors of nxn matrices. If you can show me the process of finding solution of one of these matrices, that would be helpful.&lt;/p&gt;&#xA;" OwnerUserId="957" LastEditorUserId="171" LastEditorDisplayName="" LastEditDate="2010-08-09T07:38:35.497" LastActivityDate="2010-08-09T11:10:52.327" Title="process of finding eigenvalues and eigenvectors" Tags="&lt;linear-algebra&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="1874" PostTypeId="1" AcceptedAnswerId="1875" CreationDate="2010-08-09T03:55:53.467" Score="10" ViewCount="225" Body="&lt;p&gt;I recently noticed a few things in some recent questions on MO:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) the fundamental group of $S^2$ minus, say, 4 points, is $\langle a,b,c,d\ |\ abcd=1\rangle$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) The fundamental group of a torus minus a point is $\langle a,b,c\ |\ [a,b]c=1\rangle$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was just wondering if you have a manifold $M$, and you know a presentation of its fundamental group, can you quickly get a presentation of the fundamental group of $M$ minus $n$ points?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, this could merely be a coincidence.  These are both surfaces, and both their fundamental groups only have one relation.  But perhaps there is more to it than that?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any and all helpful comments appreciated,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Steve&lt;/p&gt;&#xA;" OwnerUserId="641" LastActivityDate="2010-08-09T04:21:41.677" Title="Presentation of the fundamental group of a manifold minus some points" Tags="&lt;algebraic-topology&gt;" AnswerCount="1" FavoriteCount="3" />
  <row Id="1875" PostTypeId="2" ParentId="1874" CreationDate="2010-08-09T04:21:41.677" Score="10" ViewCount="" Body="&lt;p&gt;The basic tool here is Van Kampen's theorem.  Let $M$ be the manifold, and let $M' = M\setminus {x}.$  Let $D$ be a small ball around $x$.&#xA;Then $M = M' \cup D,$ and $M \cap D = D\setminus {x}$ is homotopic to $S^{n-1}$ (here $n$ &#xA;is the dimension of $M$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So Van Kampen's theorem says that $\pi_1(M) = \pi_1(M')*_{\pi_1(S^{n-1})} \pi_1(D)$.&#xA;If $n &gt; 2,$ then $S^{n-1}$ is simply connected, and of course $\pi_1(D)$ is simply connected,&#xA;and so this reduces to $\pi_1(M) = \pi_1(M')$; in other words, deleting points doesn't change&#xA;$\pi_1$ in dimensions $&gt; 2$.  (You can probably convince yourself of this directly: imagine you have some loop contracting in a 3-dimensional space; then if you remove a point, you can always just perturb the contraction slightly so that it misses the point.  Of course,&#xA;this is not the case in two dimensions.)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $n = 2$, then $\pi_1(S^1) = \mathbb Z$ is infinite cyclic, while $\pi_1(D)$ is trivial&#xA;again.  So we see that $\pi_1(M)$ is obtained from $\pi_1(M')$ by killing a loop.   One can be more precise, of course (and I will restrict myself to the orientable case, so as to make life easier):  if you have a genus $g$ surface, with $r&gt;0$ punctures, and also&#xA;$g + r &gt; 1$, then $\pi_1$ is a free group on $2 g + r - 1$ generators. (Every puncture after the first adds another independent loop, namely the loop around that puncture.) If $g \geq 1,$ and $r = 0$, then $\pi_1$ is obtained via Van Kampen as above: you&#xA;begin with a free group on $2g$ generators for $M'$, and then you kill off the loop&#xA;around the puncture when you fill it in (so you get the standard presentation for the&#xA;$\pi_1$ of a compact orientable surface of genus $g \geq 1$).  (Conversely, going from&#xA;the compact surface $M$ to the once-punctured surface $M'$ does not add any generators&#xA;to $\pi_1$, but gets rid of a relation.)&#xA;If $g = 0$ and $r \leq 1,$ then you have either a disk ($r = 1$) or a sphere ($r = 0$)&#xA;and so $\pi_1$ is trivial.&lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2010-08-09T04:21:41.677" CommentCount="1" />
  <row Id="1876" PostTypeId="1" AcceptedAnswerId="1911" CreationDate="2010-08-09T05:11:09.750" Score="1" ViewCount="156" Body="&lt;p&gt;I am not sure if this question is appropriate here but would appreciate any help. Let $n&gt;2$ and $x_1,\cdots,x_n$ be real numbers. What is the infimum of:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$A = \sum_{i=1}^n \frac{1}{(1+x_i)^2} $$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;subject to the constraint $\prod_{i=1}^nx_i=1$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similar question for: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$B = \sum_{i=1}^n \frac{1}{(1-x_i)^2} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance!&lt;/p&gt;&#xA;" OwnerUserId="223" LastEditorUserId="223" LastEditorDisplayName="" LastEditDate="2010-08-09T06:19:44.423" LastActivityDate="2010-08-09T14:28:37.840" Title="Infimum of some quantities" Tags="&lt;calculus&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="1877" PostTypeId="1" AcceptedAnswerId="1883" CreationDate="2010-08-09T05:37:43.387" Score="3" ViewCount="121" Body="&lt;h2&gt;Background&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;It is a standard and important fact in basic calculus/real analysis that a continuous function on a compact metric space is in fact uniformly continuous.  That is, suppose (X,d) is a compact metric space and f: X &amp;rarr; &amp;#8477; is such that for every x&amp;isin;X and &amp;epsilon;&gt;0 there exists &amp;delta;&gt;0 such that d(x,y)&amp;lt;&amp;delta; implies |f(x)-f(y)|&amp;lt;&amp;epsilon;.  Then in fact, such a &amp;delta; can be chosen independently of x.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Question&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Does a similar statement hold regarding &lt;em&gt;semi&lt;/em&gt;-continuous functions?  For concreteness, let's consider upper semi-continuous functions, so suppose (X,d) is compact and f: X &amp;rarr; &amp;#8477; has the property that for every x&amp;isin;X and &amp;epsilon;&gt;0 there exists &amp;delta;&gt;0 such that d(x,y)&amp;lt;&amp;delta; implies f(y) &amp;lt; f(x)+&amp;epsilon;.  (Note the asymmetry of x and y in this definition.)  Then is it true that &amp;delta;=&amp;delta;(&amp;epsilon;) can be chosen independently of x?&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Reformulation&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Given $\delta, \epsilon &gt; 0$, consider the set&#xA;$$&#xA;X_\delta^\epsilon := \lbrace x\in X \mid f(y) &amp;lt; f(x) + \epsilon \text{ for every } y\in B(x,\delta) \rbrace.&#xA;$$&#xA;Then $f$ is upper semi-continuous if and only if $\bigcup_{\delta&gt;0} X_\delta^\epsilon = X$ for every $\epsilon &gt; 0$, and $f$ is &lt;em&gt;uniformly&lt;/em&gt; upper semi-continuous if and only if this union stabilises -- that is, if for every $\epsilon &gt; 0$ there exists $\delta&gt;0$ such that $X_\delta^\epsilon = X$.&lt;/p&gt;&#xA;" OwnerUserId="876" LastEditorUserId="876" LastEditorDisplayName="" LastEditDate="2010-08-09T15:11:17.877" LastActivityDate="2010-08-09T15:11:17.877" Title="Uniform semi-continuity" Tags="&lt;real-analysis&gt;" AnswerCount="2" />
  <row Id="1878" PostTypeId="2" ParentId="501" CreationDate="2010-08-09T05:59:45.900" Score="8" ViewCount="" Body="&lt;p&gt;Set theory with all sets finite has been studied, is a familiar theory in disguise, and is enough for most/all concrete real analysis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Specifically, Zermelo-Fraenkel set theory with the Axiom of Infinity replaced by its negation (informally, &quot;there is no infinite set&quot;) is equivalent to first-order Peano Arithmetic.  Call this system &lt;em&gt;finite ZF&lt;/em&gt;, the theory of hereditarily finite sets.  Then under the Goedel arithmetic encoding of finite sets, Peano Arithmetic can prove all the theorems of Finite ZF, and under any of the standard constructions of integers from finite sets, Finite ZF proves all the theorems of Peano Arithmetic.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The implication is that theorems unprovable in PA involve intrinsically infinitary reasoning.  Notably, finite ZF was used as an equivalent of PA in the Paris-Harrington paper &quot;A Mathematical Incompleteness in Peano Arithmetic&quot; which proved that their modification of the finite Ramsey theorem can't be proved in PA.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Real numbers and infinite sequences are not directly objects of the finite ZF universe, but there is a clear sense in which real (and complex, and functional) analysis can be performed in finite ZF or in PA.  One can make statements about $\pi$ or any other explicitly defined real number, as theorems about a specific sequence of rational approximations ($\forall n P(n)$) and these can be formulated and proved using a theory of finite sets.  PA can perform very complicated induction proofs, i.e., transfinite induction below $\epsilon_0$.   In practice this means any concrete real number calculation in ordinary mathematics.  For the example of the prime number theorem, using complex analysis and the Riemann zeta function, see Gaisi Takeuti's &lt;em&gt;Two Applications of Logic to Mathematics&lt;/em&gt;.    More discussion of this in a MO thread and my posting there:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://mathoverflow.net/questions/31846/is-the-riemann-hypothesis-equivalent-to-a-pi-1-sentence&quot; rel=&quot;nofollow&quot;&gt;http://mathoverflow.net/questions/31846/is-the-riemann-hypothesis-equivalent-to-a-pi-1-sentence&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://mathoverflow.net/questions/31846/is-the-riemann-hypothesis-equivalent-to-a-pi-1-sentence/31942#31942&quot; rel=&quot;nofollow&quot;&gt;http://mathoverflow.net/questions/31846/is-the-riemann-hypothesis-equivalent-to-a-pi-1-sentence/31942#31942&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proof theory in general and reverse mathematics in particular contain analyses of the logical strength of various theorems in mathematics (when made suitably concrete as statements about sequences of integers), and from this point of view PA, and its avatar finite set theory, are very powerful systems.  &lt;/p&gt;&#xA;" OwnerUserId="467" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-08-09T06:22:45.603" LastActivityDate="2010-08-09T06:22:45.603" CommentCount="0" />
  <row Id="1879" PostTypeId="2" ParentId="1876" CreationDate="2010-08-09T06:12:51.163" Score="0" ViewCount="" Body="&lt;p&gt;As it is, this question doesn't quite make sense.  The &lt;a href=&quot;http://en.wikipedia.org/wiki/Infimum&quot; rel=&quot;nofollow&quot;&gt;infimum&lt;/a&gt; of a set X of real numbers is its greatest lower bound.  Currently you don't actually have a set of numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If your after inf(X) of the set $$X:=\{\sum_{i=1}^n \frac{1}{(1+x_i)^2}:x_1 \cdot x_2 \cdots x_n=1\},$$ then the two obvious tries are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;When $x_1=x_2=\cdots=x_n=1$ which identifies that the element $n/2^2 \in X$.&lt;/li&gt;&#xA;&lt;li&gt;When $x_1 \rightarrow 0$ from the positive direction and $x_2=x_3=\cdots=x_n=(1/x_1)^{n-1} \rightarrow \infty$ (if n is fixed).  Hence inf(X) is at most 1.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The first case is better when $n \in \{1,2,3\}$.  This suggests to me that finding inf(X) is not going to be straightforward.&lt;/p&gt;&#xA;" OwnerUserId="139" LastEditorUserId="139" LastEditorDisplayName="" LastEditDate="2010-08-09T09:04:19.363" LastActivityDate="2010-08-09T09:04:19.363" CommentCount="6" />
  <row Id="1880" PostTypeId="1" AcceptedAnswerId="1882" CreationDate="2010-08-09T06:21:19.367" Score="2" ViewCount="897" Body="&lt;p&gt;This seems counterintuitive, but 22/7 is closer to pi than 3.14=314/100 which has a significantly greater denominator.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Why is 22/7 a better approximation for pi than 3.14?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This has important implications:  e.g. Should &quot;pi day&quot; by the 14th of March or 22nd of July?&lt;/p&gt;&#xA;" OwnerUserId="139" LastActivityDate="2010-10-04T16:11:35.890" Title="Why is 22/7 a better approximation for pi than 3.14?" Tags="&lt;approximation&gt;&lt;pi&gt;" AnswerCount="4" CommentCount="11" FavoriteCount="3" />
  <row Id="1881" PostTypeId="2" ParentId="1880" CreationDate="2010-08-09T06:39:44.057" Score="10" ViewCount="" Body="&lt;p&gt;Well, just measure |pi - 22/7| and |pi-3.14| ...&lt;/p&gt;&#xA;" OwnerUserId="934" LastActivityDate="2010-08-09T06:39:44.057" CommentCount="2" />
  <row Id="1882" PostTypeId="2" ParentId="1880" CreationDate="2010-08-09T06:42:08.903" Score="12" ViewCount="" Body="&lt;p&gt;It has to do with the continued fraction expansion of $\pi$.  Suppose $[a_1, a_2, \ldots]$ is the continued fraction of an irrational number $\alpha$ -- that is, if $a_n$ is the (essentially unique) sequence of natural numbers such that if we define partial convergents by $x_1 = a_1$, $x_2 = a_1 + 1/a_2$, $x_3 = a_1 + 1/(a_2 + 1/a_3)$, $x_4 = a_1 + 1/(a_2 + 1/(a_3 + 1/a_4))$, and so on, then $\alpha = \lim_{n\to\infty} x_n$.  Then the partial convergents $x_n$ are rational numbers that approximate $\alpha$ better than anything that is not a partial convergent, in the following sense:  a rational number $\frac pq$ satisfies the inequality $|\alpha - \frac pq| &amp;lt; \frac 1{2q^2}$ if and only if $\frac pq$ is one of the convergents $x_n$.  (One could, of course, come up with different notions of what constitutes a &quot;good&quot; approximation.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The continued fraction expansion of $\pi$ is $[3,7,15,1,292,1,1,\dots]$, so the first few convergents are $3$, $\frac{22}{7}$, $\frac{333}{106}$, $\frac{355}{113}$, etc.  Thus $\frac{22}{7}$ is a better approximation than $\frac{314}{100}$ (in the above sense) because it appears in the list of partial convergents, while $\frac{314}{100}$ does not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Incidentally, the approximation $x_n$ is best when the coefficient $a_{n+1}$ is quite large, so the size of $a_5 = 292$ means that $x_4 = \frac{355}{113}$ is a particularly good approximation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the risk of self-promotion, I wrote a brief exposition of all this in a bit more detail -- you can find it on my website if you're interested, at &lt;a href=&quot;http://www.math.psu.edu/climenha/contfrac.html&quot; rel=&quot;nofollow&quot;&gt;http://www.math.psu.edu/climenha/contfrac.html&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="876" LastEditorUserId="876" LastEditorDisplayName="" LastEditDate="2010-08-21T03:59:26.697" LastActivityDate="2010-08-21T03:59:26.697" CommentCount="17" />
  <row Id="1883" PostTypeId="2" ParentId="1877" CreationDate="2010-08-09T07:15:14.900" Score="2" ViewCount="" Body="&lt;p&gt;f(x)=0 (x&amp;le;0), f(x)=-1/x (x&amp;gt;0) is upper semi-continuous on [-1;1] — but not uniformly.&lt;/p&gt;&#xA;" OwnerUserId="152" LastActivityDate="2010-08-09T07:15:14.900" CommentCount="2" />
  <row Id="1884" PostTypeId="1" AcceptedAnswerId="1885" CreationDate="2010-08-09T07:27:08.177" Score="0" ViewCount="202" Body="&lt;p&gt;How would you find eigenvalues/eigenvectors of a nxn matrix where each diagonal entry is scalar 'd' and all other entries are '1' ? I am looking for a decomposition but cannot find anything for this. For example:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;&amp;gt; a&#xA;[,1] [,2] [,3] [,4]&#xA;[1,]    2    1    1    1&#xA;[2,]    1    2    1    1&#xA;[3,]    1    1    2    1&#xA;[4,]    1    1    1    2&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="957" LastActivityDate="2010-08-09T07:45:18.133" Title="find eigenvectors/eigenvalues of this matrix" Tags="&lt;linear-algebra&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="1885" PostTypeId="2" ParentId="1884" CreationDate="2010-08-09T07:45:18.133" Score="5" ViewCount="" Body="&lt;p&gt;The matrix is $(d-1)I + J$ where $I$ is the identity matrix and $J$ is the all-ones matrix, so once you have the eigenvectors and eigenvalues of $J$ the eigenvectors of $(d-1)I + J$ are the same and the eigenvalues are each $d-1$ greater.  (Convince yourself that this works.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But $J$ has rank $1$, so it has eigenvalue $0$ with multiplicity $n-1$.  The last eigenvalue is $n$, and it's quite easy to write down all the eigenvectors.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-08-09T07:45:18.133" CommentCount="1" />
  <row Id="1886" PostTypeId="2" ParentId="594" CreationDate="2010-08-09T07:50:15.627" Score="3" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Taken From I.N. Herstein&lt;/strong&gt; There are 2 results which i am going to use.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Let $p$ be a prime integer and suppose that for some integer $c$ relatively prime to $p$ we can find integers $x$ and $y$ such that $x^{2}+y^{2}=cp$. Then $p$ can be written as the sum of 2 squares.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If $p$ is a prime of the form $4n+1$, then we can solve the congruence $x^{2} \equiv \ -1 \  (mod)  \ p$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Now the main result. If $p$ is a prime of the form $4n+1$ then $p$ is the sum of 2 squares.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proof. By 2 there exists and $x$ such that $x^{2} \equiv -1 \text{mod} \ p$. So $x$ can be chose such that $0 \leq x \leq (p-1)$. We can restrict the size of $p$ even further, namely to satisfy $|x| \leq \frac{p}{2}$. For if $x &gt; p/2$ then, $y=p-x$ satisfies $y^{2} \equiv -1 \text{mod} \ p$ but $|y| \leq p/2$. Thus we may assume that we have an integer $x$ such that $|x| \leq p/2$ and $x^{2}+1$ is a multiple of $p$ say $cp$. Now $cp=x^{2}+1 \leq p^{2}/4 +1 &amp;lt; p^{2}$, hence $c &amp;lt; p$ and hence $(c,p)=1$. Invoking (1) we have $p=a^{2}+b^{2}$.&lt;/p&gt;&#xA;" LastActivityDate="2010-08-15T18:34:46.983" CommentCount="2" />
  <row Id="1887" PostTypeId="2" ParentId="1808" CreationDate="2010-08-09T07:56:10.043" Score="2" ViewCount="" Body="&lt;p&gt;Much of data mining requires an understanding of high dimensional geometry. Lots of machine learning also requires an understandind of advanced probability theory.&lt;/p&gt;&#xA;" OwnerUserId="865" LastActivityDate="2010-08-09T07:56:10.043" CommunityOwnedDate="2010-08-09T07:56:10.043" CommentCount="1" />
  <row Id="1889" PostTypeId="2" ParentId="1808" CreationDate="2010-08-09T08:23:58.353" Score="0" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Computable_analysis&quot; rel=&quot;nofollow&quot;&gt;Computable analysis&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="100" LastActivityDate="2010-08-09T08:23:58.353" CommunityOwnedDate="2010-08-09T08:23:58.353" />
  <row Id="1892" PostTypeId="1" AcceptedAnswerId="1895" CreationDate="2010-08-09T10:47:27.760" Score="6" ViewCount="927" Body="&lt;p&gt;Although whether $$ P = NP $$ is important from theoretical computer science point of view, but I fail to see any practical implication of it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose that we can prove all questions that can be verified in polynomial time have polynomial time solutions, it won't help us in finding the actual solutions. Conversely, if we can prove that $$ P \ne NP,$$ then this doesn't mean that our current NP-hard problems have no polynomial time solutions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From practical point of view ( practical as in the sense that we can immediately use the solution to real world scenario), it shouldn't bother me whether P vs NP is proved or disproved any more than whether my &lt;strong&gt;&lt;em&gt;current problem&lt;/em&gt;&lt;/strong&gt; has a polynomial time solution. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Am I right?&lt;/p&gt;&#xA;" OwnerUserId="550" LastEditorUserId="171" LastEditorDisplayName="" LastEditDate="2010-08-09T11:32:17.017" LastActivityDate="2010-08-10T01:04:04.810" Title="The Practical Implication of P vs NP Problem" Tags="&lt;computer-science&gt;&lt;np-complete&gt;" AnswerCount="6" CommentCount="4" FavoriteCount="3" />
  <row Id="1893" PostTypeId="2" ParentId="1873" CreationDate="2010-08-09T10:48:31.377" Score="1" ViewCount="" Body="&lt;p&gt;Just a little warning about the first example: let's call that matrix $J$. It is a (non-diagonal) Jordan matrix, so it is not diagonalizable. That is, there is no base of eigenvectors for this matrix, although, as J. Mangaldan observed, 2 is the only eigenvalue and the last vector of the standard $\mathbb{R}^3$ basis is an eigenvector, since, as you can see, $Je_3 = 2e_3$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can convince yourself that $J$ is not diagonalizable computing its characteristic polynomial: $Q(t) = (2-t)^3$ and observing that $2$ has multiplicity $3$, but the eigenspace $\ker (J-2I)$ has dimension $1$.&lt;/p&gt;&#xA;" OwnerUserId="664" LastEditorUserId="664" LastEditorDisplayName="" LastEditDate="2010-08-09T11:10:52.327" LastActivityDate="2010-08-09T11:10:52.327" CommentCount="1" />
  <row Id="1894" PostTypeId="2" ParentId="1892" CreationDate="2010-08-09T11:11:54.320" Score="2" ViewCount="" Body="&lt;p&gt;Actually $P \ne NP$ &lt;em&gt;does&lt;/em&gt; mean that our current NP-hard problems have no polynomials time solutions. NP-complete problems are the hardest problems in NP and NP-hard problems are at least as hard as this. So if $P \ne NP$, then all these NP-hard problems must be harder than P.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whether the proof helps us find solutions will of course depend on the proof. If $P \ne NP$, then we know not to waste time looking for polynomial solutions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $P=NP$, then the real practical benefits would of course come from the solution, rather than the proof. That is fine - there is no reason why all theoretical computer science needs to be &lt;em&gt;directly&lt;/em&gt; practical.&lt;/p&gt;&#xA;" OwnerUserId="123" LastEditorUserId="123" LastEditorDisplayName="" LastEditDate="2010-08-09T21:45:46.140" LastActivityDate="2010-08-09T21:45:46.140" CommentCount="2" />
  <row Id="1895" PostTypeId="2" ParentId="1892" CreationDate="2010-08-09T11:11:57.830" Score="8" ViewCount="" Body="&lt;p&gt;Many of the problems we know to be in NP or NP-complete are problems that we actually want to solve, problems that arise, say, in circuit design or in other industrial design applications. Furthermore, since the diverse NP-complete problems are all polynomial time related to one another, if we should ever learn a feasible means of solving any of them, we would have feasible means for all of them. The result of this would be extraordinary, something like a second industrial revolution. It would be as though we suddenly had a huge permanent increase in computational power, allowing us to solve an enormous array of practical problems heretofore out of our computational reach. The P vs. NP question is important in part because of this tantalizing possibility. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it were proved that P = NP and the proof provided a specific polynomial time algorithm for an NP-complete problem, then because of the existing reduction proofs, we could immediately produce polynomial time algorithms for all our other favorite NP problems. Of course, a proof may be indirect, and not provide a specific polynomial time algorithm, but you can be sure that if we have a proof of P=NP, then enormous resources will be put into extracting from the proof a speciffic algorithm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conversely, if someone were to prove $P \neq NP$, then it would mean that there could be no polynomial time solution for any NP complete problem. (In particular, the last sentence of your second paragraph is not correct.) &lt;/p&gt;&#xA;" OwnerUserId="413" LastActivityDate="2010-08-09T11:11:57.830" CommentCount="7" />
  <row Id="1896" PostTypeId="1" AcceptedAnswerId="1920" CreationDate="2010-08-09T11:22:02.447" Score="0" ViewCount="92" Body="&lt;p&gt;If $k_1,k_2,...,k_n$ are principal curvatures of a hypersurface in $\mathbb{R}^{n+1}$ then one can apparently locally parametrize the manifold as $(x_1,x_2,...,x_n,y)$ such that,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$y = \frac{1}{2}(k_1 x_1^2 +...+k_n x_n^2) + O(\vert x \vert ^3) $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to know how this is proven. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I guess the coordinates are obtained by &quot;integrating&quot; the eigen-vectors of the shape operator whose eigen-values are the principal curvatures anyway. I suppose one has to be able to find out curves passing through a point which have as velocity at that point the principal directions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I have no experience of doing such a calculation on arbitrary manifolds and hence I can't see how to do this. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Further I would like to understand how for dimension 2 the principal curvatures (as defined above) become the minimum and the maximum of geodesic curvatures and also minimum and maximum of curvatures of curves obtained by intersecting the manifold with planes orthogonal to the tangent space. (in elementary surface theory the principal curvatures get defined as the maximum and the minimum of the above quantities)&lt;/p&gt;&#xA;" OwnerUserId="965" LastActivityDate="2010-10-04T04:19:18.463" Title="Local equation for the manifold from its principal curvatures" Tags="&lt;differential-geometry&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="1897" PostTypeId="1" CreationDate="2010-08-09T11:46:26.987" Score="0" ViewCount="97" Body="&lt;p&gt;Suppose we have a functional question, such as the &lt;a href=&quot;http://math.stackexchange.com/questions/1890/real-analysis-problem&quot;&gt;one here&lt;/a&gt;. Often a key step in these problems is to try to equate or chain results together (&lt;a href=&quot;http://math.stackexchange.com/questions/1890/real-analysis-problem&quot;&gt;see my hint&lt;/a&gt;). From looking at these questions, it seems that there is often only a finite number of ways that we can attempt to join them together. Can this fact be examined more formally?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: I am not asking you to turn my answer from that question into something algebraic. It is that way purposely as I was only trying to give a hint, not a complete solution. What I am asking, is whether there are any results that show that the functions can only be &quot;combined&quot; or &quot;equated&quot; in a certain number of ways on some reasonably general class of problems.&lt;/p&gt;&#xA;" OwnerUserId="123" LastActivityDate="2010-08-09T12:44:41.940" Title="Solving functional questions" Tags="&lt;real-analysis&gt;&lt;functions&gt;" AnswerCount="1" />
  <row Id="1898" PostTypeId="2" ParentId="1892" CreationDate="2010-08-09T12:40:34.537" Score="1" ViewCount="" Body="&lt;p&gt;Currently, if a manager asks their software engineering team to look at implementing some utility, and the team says that requirements are NP hard, that's a reason that the project requirements need to be changed before work on implementation can begin.  That's because no-one knows how to give feasible solutions to such problems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The plurality of complexity theorists furthermore believe P =/= NP, so that means that there's a widespread belief among experts that feasible solutions to these problems will never be found.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If someone shows P=NP, then if the team says the requirements are NP-complete, then the manager and the team will start to move from talk of theory to possible realisations and their efficiency.  &lt;/p&gt;&#xA;" OwnerUserId="100" LastActivityDate="2010-08-09T12:40:34.537" />
  <row Id="1899" PostTypeId="2" ParentId="1897" CreationDate="2010-08-09T12:44:41.940" Score="1" ViewCount="" Body="&lt;p&gt;One example is that there's an easy way to get uniqueness of functional equations in certain cases. Consider the &quot;cohomological&quot; equation $f(Tx) - f(x) = g(x)$ (which is important in dynamical systems, e.g. it comes up when you are trying to find an invariant $n$-form). The goal, of course, is to prove that there exists $f$ given $g$. If $T: X \to X$ is topologically transitive, then any two solutions differ  by a constant. This is because if $x_0$ has a dense orbit, then the value of $f(x_0)$ determines $f(Tx_0), f(T^2x_0)$, and so on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More generally, there is the same uniqueness for the &quot;twisted&quot; cohomological equation $f(Tx) - \alpha f(x) = g(x)$, where $\alpha$ is a nonzero real number. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(This is not, I think, relevant to the question, but I'll mention that one can get existence of a Holder solution in the untwisted case for transitive Anosov diffeomorphisms of compact manifolds if the sum of $g$ over every finite orbit is zero, though it requires a clever trick. This is a result of Livsic.)&lt;/p&gt;&#xA;" OwnerUserId="536" LastActivityDate="2010-08-09T12:44:41.940" CommentCount="1" />
  <row Id="1901" PostTypeId="1" CreationDate="2010-08-09T13:22:20.880" Score="4" ViewCount="412" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have heard that some issues in group theory prevent classifying all manifolds upto homotopy using the fundamental group invariant. Does anyone know what are those issues?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks,&#xA;K.&lt;/p&gt;&#xA;" LastActivityDate="2010-11-29T20:51:56.813" Title="Why is fundamental group insufficient to classify manifolds?" Tags="&lt;group-theory&gt;&lt;topology&gt;" AnswerCount="5" CommentCount="3" />
  <row Id="1902" PostTypeId="2" ParentId="1872" CreationDate="2010-08-09T13:24:25.917" Score="0" ViewCount="" Body="&lt;p&gt;This gives quite an intuitive idea about what is going on:&lt;br&gt;&#xA;&lt;a href=&quot;http://plus.maths.org/content/curious-quaternions&quot; rel=&quot;nofollow&quot;&gt;http://plus.maths.org/content/curious-quaternions&lt;/a&gt;&lt;br&gt;&lt;/p&gt;&#xA;" OwnerUserId="346" LastActivityDate="2010-08-09T13:24:25.917" CommentCount="1" />
  <row Id="1905" PostTypeId="2" ParentId="1872" CreationDate="2010-08-09T13:27:30.013" Score="0" ViewCount="" Body="&lt;p&gt;...unfortunately I am only allowed to post one link per answer - so here is the sequel&#xA;&lt;a href=&quot;http://plus.maths.org/content/ubiquitous-octonions&quot; rel=&quot;nofollow&quot;&gt;http://plus.maths.org/content/ubiquitous-octonions&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="346" LastActivityDate="2010-08-09T13:27:30.013" CommentCount="1" />
  <row Id="1907" PostTypeId="2" ParentId="1901" CreationDate="2010-08-09T13:31:35.700" Score="2" ViewCount="" Body="&lt;p&gt;Well, there are no reasons to expect that manifolds with equal &amp;pi;&lt;sub&gt;1&lt;/sub&gt; are homotopy equivalent, and in general, they indeed aren't. Even for closed manifolds of fixed dimension: say, both CP&lt;sup&gt;2&lt;/sup&gt; and S&lt;sup&gt;4&lt;/sup&gt; have trivial fundamental group.&lt;/p&gt;&#xA;" OwnerUserId="152" LastEditorUserId="152" LastEditorDisplayName="" LastEditDate="2010-08-10T16:53:10.693" LastActivityDate="2010-08-10T16:53:10.693" />
  <row Id="1909" PostTypeId="1" AcceptedAnswerId="1915" CreationDate="2010-08-09T13:44:35.387" Score="4" ViewCount="188" Body="&lt;p&gt;Let $(\mathbb{R},+)$ be the group of Real Numbers under addition. Let $H$ be a proper subgroup of $\mathbb{R}$. Prove that either $H$ is dense in $\mathbb{R}$ or there is an $a \in \mathbb{R}$ such that $H={ na : n=0, \pm{1},\pm{2},...}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not able to proceed. &lt;/p&gt;&#xA;" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-08-10T06:26:38.223" LastActivityDate="2010-08-10T06:26:38.223" Title="Characterizing Dense Subgroups of the Reals" Tags="&lt;abstract-algebra&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="1910" PostTypeId="1" AcceptedAnswerId="1912" CreationDate="2010-08-09T13:46:07.163" Score="1" ViewCount="173" Body="&lt;p&gt;School starts in 1.5 weeks and I'm taking a course in Symbolic Logic. Here is a quote from the syllabus:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;You need to keep a notebook with all your homework in it. This notebook can take one of two forms. 1. A looseleaf (or some kind of removable/replaceable page notebook) or 2. A folder (aka subdirectory) on your computer. The latter is only recommended for true computer geeks who like ﬁguring out how to work with symbols (e.g., ∃, ∀, ⊃, ⊢, →, ↔). I will periodically collect homework.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I would like to try and do homework on my computer, if only to see if I can do it, even though in the long run I'll probably get frustrated and just use paper.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What do you recommend for this type of document? I assume it'll be a mix of words and equations with the above symbols. Is there a good special program for this, or instructions for doing this in MS Office 2010 or OpenOffice? Is this a good application for L&lt;sup&gt;A&lt;/sup&gt;T&lt;sub&gt;E&lt;/sub&gt;X, and if so, how does someone with no L&lt;sup&gt;A&lt;/sup&gt;T&lt;sub&gt;E&lt;/sub&gt;X knowledge (me) learn enough to do my homework? Is it even worth the trouble or should I just stick with deadtree?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My primary OS is Windows 7, if it makes a difference.&lt;/p&gt;&#xA;" OwnerUserId="137" LastActivityDate="2010-08-12T17:58:59.880" Title="How to do homework for a class in Symbolic Logic on my computer?" Tags="&lt;logic&gt;&lt;math-software&gt;" AnswerCount="3" CommentCount="3" />
  <row Id="1911" PostTypeId="2" ParentId="1876" CreationDate="2010-08-09T13:46:38.487" Score="3" ViewCount="" Body="&lt;p&gt;I assume you intend for all the $x_i$ to be positive. The infimum is $\min(n/4, 1)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a good example of how to use convexity to minimize an expression.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Make the change of variables $x_i = e^{t_i}$, so $\sum t_i=0$. Define $f(t) = 1/(1+e^t)^2$. We want to minimize $\sum f(t_i)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, $f(t)$ is convex for $t&gt; - \log 2$ and concave for $t &amp;lt; - \log 2$. This is an easy calculation with the double derivative of $f$. (You might find this &lt;a href=&quot;http://www.wolframalpha.com/input/?i=Plot%5B1%2F%281%2Be%5Et%29%5E2%2C%20%7Bt%2C-Infinity%2C%20Infinity%7D%5D&quot; rel=&quot;nofollow&quot;&gt;graph&lt;/a&gt; helpful.) So, if all of the $t_i$ lie in $[- \log 2, \infty)$, then &lt;a href=&quot;http://en.wikipedia.org/wiki/Jensen%27s_inequality&quot; rel=&quot;nofollow&quot;&gt;Jensen's inequality&lt;/a&gt; shows that the minimum occurs when all the $t_i$ are $0$. In that case, we get $n/4$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We claim that we can restrict to the case that at most one of the $t_i$ lies in $(- \infty, - \log 2)$. Proof: Consider any solution $(t_i)$, with $t_1$, $t_2$, ..., $t_k &amp;lt; - \log 2$. Replace these values by $k-1$ copies of $- \log 2$ and $t_1+t_2+\cdots+t_k + (k-1) \log 2$. Since $f$ is concave in this range, we have decreased the sum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, we may assume that there is one $t_i$, call it $u$, in $(- \infty, - \log 2)$, and all the others are in $[- \log 2, \infty)$. For a fixed value of $u$, by convexity, we will obtain the minimum when all the other values are equal to each other, say $v$. So $u+(n-1)v=0$. Our sum is&#xA;$$g(v) :=(1+e^{-(n-1)v})^{-2} + (n-1) (1+e^v)^{-2}.$$&#xA;We want to minimize this quantity, subject to the condition that $v \geq -\log 2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As $v \to \infty$, the function $g$ goes to $1$. We'd like to know whether there is any point in $[\log 2, \infty)$ where $g$ dips below $1$. Such a point must be a local minimum. Here are some plots for $n$ &lt;a href=&quot;http://www.wolframalpha.com/input/?i=Plot%5B%281%2Be%5E%28-%282%29v%29%29%5E%28-2%29%20%2B%20%282%29%20%281%2Be%5Ev%29%5E%28-2%29%2C%20%7Bt%2C-0.5%2C%200.5%7D%5D&quot; rel=&quot;nofollow&quot;&gt;equal to 3&lt;/a&gt; and &lt;a href=&quot;http://www.wolframalpha.com/input/?i=Plot%5B%281%2Be%5E%28-%289%29v%29%29%5E%28-2%29%20%2B%20%289%29%20%281%2Be%5Ev%29%5E%28-2%29%2C%20%7Bt%2C-0.5%2C%200.5%7D%5D&quot; rel=&quot;nofollow&quot;&gt;equal to 10&lt;/a&gt;. Numerically, it looks like there is a local minimum at $0$, where $g$ is $n/4$ and a local maximum at some positive value. If we can confirm this, the infimum is $\min(1, n/4)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's put the cherry on this sundae and prove that there is no other local minimum. The derivative $g'$ is&#xA;$$\frac{-2(n-1)e^{-(n-1)v}}{(1+e^{-(n-1)v})^3} + \frac{(n-1) e^v}{(1+e^v)^3}$$&#xA;Set this equal to zero and clear out denominators to get&#xA;$$(1+e^v)^3 e^{2 (n-1)v} = (1+e^{(n-1)v})^3 e^v.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Set $w=e^v$. For $n \geq 4$, this is a polynomial in $w$ with two sign changes so, by &lt;a href=&quot;http://en.wikipedia.org/wiki/Descartes%27_rule_of_signs&quot; rel=&quot;nofollow&quot;&gt;Descartes' rule of signs&lt;/a&gt;, it has at most two positive roots. There also turn out to be two real roots for $n=2$ and $3$, by computer check.  These are the local minimum we have already noticed at $v=0$ ($w=1$), and the local maximum nearby. So there are no other local minima, and we are done.&lt;/p&gt;&#xA;" OwnerUserId="448" LastEditorUserId="171" LastEditorDisplayName="" LastEditDate="2010-08-09T14:28:37.840" LastActivityDate="2010-08-09T14:28:37.840" CommentCount="3" />
  <row Id="1912" PostTypeId="2" ParentId="1910" CreationDate="2010-08-09T14:02:10.557" Score="2" ViewCount="" Body="&lt;p&gt;For &quot;a mix of words and equations with symbols&quot;  I'd clearly recommend using latex. &lt;a href=&quot;http://www.ctan.org/tex-archive/info/lshort/english/lshort.pdf&quot; rel=&quot;nofollow&quot;&gt;This&lt;/a&gt; is a good introduction.&lt;/p&gt;&#xA;" OwnerUserId="367" LastActivityDate="2010-08-09T14:02:10.557" />
  <row Id="1913" PostTypeId="1" CreationDate="2010-08-09T14:07:06.677" Score="3" ViewCount="77" Body="&lt;p&gt;Let $X $ be a set, $|X|=n$ and $G$ be a group with a 2-transitive action on $X$.&#xA;what can be said about the size of $G$?&lt;/p&gt;&#xA;" OwnerUserId="816" LastEditorUserId="816" LastEditorDisplayName="" LastEditDate="2010-08-09T14:20:05.083" LastActivityDate="2010-08-09T15:08:58.757" Title="minimal size of 2-transitive groups" Tags="&lt;combinatorics&gt;&lt;group-theory&gt;" AnswerCount="2" />
  <row Id="1914" PostTypeId="2" ParentId="1913" CreationDate="2010-08-09T14:12:53.167" Score="3" ViewCount="" Body="&lt;p&gt;When $n$ is a prime (or indeed a prime power), there are two-transitive&#xA;subgroups of Sym$(X)$ (where $|X|=n$) of order $n(n-1)$. In any case $n(n-1)$&#xA;is a lower bound for the order of $G$, since this is the number of ordered pairs&#xA;$(x,x')$ of distinct elements of $X$. Alas I can't see where your $n^2+n$ comes from.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-08-09T14:12:53.167" CommentCount="1" />
  <row Id="1915" PostTypeId="2" ParentId="1909" CreationDate="2010-08-09T14:14:28.950" Score="9" ViewCount="" Body="&lt;p&gt;If there is a smallest positive element, then we are done, since any positive element must be an integer multiple of it, or otherwise we could use a euclidean-type-algorithm to get a positive element with smaller value. (I.e., suppose $a$ is the smallest positive element, and $b$ a   positive element which is not an integer multiple of $a$---keep subtracting copies of $a$ until you get something that is strictly between $0 $ and $a$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So assume there is a sequence $a_n$ contained in the group that consists of positive numbers tending to zero. Then the group contains each ${\mathbb{Z} a_n}$. This means that for each $n$, any number in $\mathbb{R}$ is within $|a_n|$ of an element of the group. Since the $|a_n|$ can be small, we find that the group is dense.&lt;/p&gt;&#xA;" OwnerUserId="536" LastActivityDate="2010-08-09T14:14:28.950" CommentCount="2" />
  <row Id="1916" PostTypeId="2" ParentId="1901" CreationDate="2010-08-09T14:34:00.183" Score="6" ViewCount="" Body="&lt;p&gt;I think that what you are thinking of is the insolvability of the so-called word problem in group theory.  What this means is  (among other things) that it is not possible to write an algorithm (i.e. a computer program) which can tell whether two groups, each presented by finitely many generators and relations, are or aren't isomorphic.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are given a manifold (presented say by gluing together various open balls), you can get&#xA;such a presentation of its fundamental group, but (at least when the dimension is 4 or higher --- and I think that 4 is the right bound here) in fact you can get an arbitrary presentation in this way, and so the insolvability of the word problem means that, in practice, you don't have any way to figure out what the fundamental group of your manifold is (even whether or not it is trivial!)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strike&gt;&#xA;As a consequence, you can't find an algorithm to determine whether two manifolds (of dimension 4 or higher) are homotopic (let along homeomorphic or diffeomorphic), since homotopic manifolds will have isomorphic&#xA;$\pi_1$s, and so we could use such an algorithm to solve the word problem, by encoding two finitely presented groups as the $\pi_1$ of some manifolds, and then applying our hypothetical manifold algorithm. &#xA;&lt;/strike&gt;&#xA;Following Pete Clark's comment below, I'm no longer sure about the correctness of the preceding statement.  What is true is that there is no general algorithm to determine the fundamental group of a manifold (in dimension $\geq 4$), or even to determine if a manifold &#xA;is simply connected.  (Whether this actually prohibits a classification is not clear to me; with luck, someone with more expertise will weigh in.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For this reason, a lot (although not all) investigations of higher dimensional manifolds restrict attention to simply connected manifolds.  Then there is no obstruction to classification coming from issues of the uncomputatibility of $\pi_1$.  Of course,&#xA;as others have noted in their answers, there can be many non-diffeomorphic, non-homeomorphic manifolds, and non-homotopic manifolds of a given dimension, so the problem with $\pi_1$ is far from the only problem when it comes to classifying manifolds.  But it is certainly one problem, and you are correct that (at least from the point of view I'm explaining here) it is a problem of group theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Added: See Damiano's answer to this question for a precise explanation of the relationship between the word problem for groups and the classification problem for manifolds.&lt;/p&gt;&#xA;" OwnerUserId="221" LastEditorUserId="221" LastEditorDisplayName="" LastEditDate="2010-08-10T20:35:28.583" LastActivityDate="2010-08-10T20:35:28.583" CommentCount="3" />
  <row Id="1917" PostTypeId="2" ParentId="1901" CreationDate="2010-08-09T14:34:42.367" Score="3" ViewCount="" Body="&lt;p&gt;For a better pair of examples take the once punctured torus and a pair of pants. Both have fundamental groups free of rank 2. They are both homotopy equivalent to a figure 8, but as manifolds they are not homeomorphic since one has 3 boundary components and the other has one boundary component. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are simply connected closed manifolds of dimension 4 and higher that are not homeomorphic. In even dimensions they can be distinguished by their middle dimensional homology intersection forms.&lt;/p&gt;&#xA;" OwnerUserId="722" LastActivityDate="2010-08-09T14:34:42.367" />
  <row Id="1918" PostTypeId="1" CreationDate="2010-08-09T14:42:46.870" Score="3" ViewCount="381" Body="&lt;p&gt;What are some interesting mathematical paradoxes?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I have in mind are things like the Banach-Tarski paradox, Paradox of Zeno of Elea, Russel's paradox, etc..&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: As an additional restriction, let us focus on paradoxes that are not already in the list at:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://secure.wikimedia.org/wikipedia/en/wiki/Category:Mathematics_paradoxes&quot; rel=&quot;nofollow&quot;&gt;https://secure.wikimedia.org/wikipedia/en/wiki/Category:Mathematics_paradoxes&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="977" LastEditorUserId="977" LastEditorDisplayName="" LastEditDate="2010-08-10T01:05:13.373" LastActivityDate="2011-02-25T20:18:11.753" CommunityOwnedDate="2010-08-09T14:44:53.277" Title="Mathematical paradoxes?" Tags="&lt;soft-question&gt;&lt;big-list&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="1" />
  <row Id="1919" PostTypeId="1" CreationDate="2010-08-09T14:54:03.613" Score="12" ViewCount="388" Body="&lt;p&gt;The simple group of order 60 can be generated by the permutations (1,2)(3,4) and (1,3,5), but all you need to do is square the first one and it becomes the identity.  Can't we find a version of the simple group where the elements of small order can be ignored?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a group H, define Ω&lt;sub&gt;n&lt;/sub&gt;(H) to be the subgroup generated by elements of order less than n.  For instance, if n=3 and H=SL(2,5) is the perfect group of order 120, then Ω&lt;sub&gt;3&lt;/sub&gt;(SL(2,5)) has order 2, and H/Ω&lt;sub&gt;n&lt;/sub&gt;(H) is the simple group of order 60.  If n=4 and H=SL(2,5)⋅3&lt;sup&gt;4&lt;/sup&gt; is the perfect group of order (60)⋅(162) whose 3-core is not complemented, then Ω&lt;sub&gt;n&lt;/sub&gt;(H) has order 162 and H/Ω&lt;sub&gt;n&lt;/sub&gt;(H) is again the simple group of order 60.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My first question is if there are smaller examples for n=4, since the jump 1, 2, 162 seems a bit drastic for n=2, 3, 4.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Is there a group H of order less than (60)⋅(162) such that H/Ω&lt;sub&gt;4&lt;/sub&gt;(H) is the simple group of order 60?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Probably, for each positive integer n, there is a finite group H such that H/Ω&lt;sub&gt;n&lt;/sub&gt;(H) is the simple group of order 60.  I am interested in whether such H can be chosen to be &quot;small&quot; somehow.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Is there a sequence of finite groups H&lt;sub&gt;n&lt;/sub&gt; and a constant C such that H&lt;sub&gt;n&lt;/sub&gt;/Ω&lt;sub&gt;n&lt;/sub&gt;(H&lt;sub&gt;n&lt;/sub&gt;) is the simple group of order 60 and such that |H&lt;sub&gt;n&lt;/sub&gt;| ≤ C⋅n?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I would also be fine with some references to where such a problem is discussed.  It would be nice if there was some sort of analogue to the Schur multiplier describing the largest non-silly kernel, and a clear definition of what a silly kernel is (I think it is too much to ask for a non-silly kernel to be contained in the Frattini subgroup, and I think it might be unreasonable to ask for the maximum amongst minimal kernels).&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;In case it helps, here are some reduced cases that I know can be handled:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A simpler example: if instead of the simple group of order 60, we concentrate on the simple group of order 2, then we can choose H&lt;sub&gt;n&lt;/sub&gt; to be the cyclic group of order 2&lt;sup&gt;1+lg(n−1)&lt;/sup&gt; when n≥2, and the order of H&lt;sub&gt;n&lt;/sub&gt; is bounded above and below by multiples of n.  We can create much larger H&lt;sub&gt;n&lt;/sub&gt; for n≥3 by taking the direct product of our small H&lt;sub&gt;n&lt;/sub&gt; with an elementary abelian 2-group of large order, but then Ω&lt;sub&gt;n&lt;/sub&gt;(H&lt;sub&gt;n&lt;/sub&gt;×2&lt;sup&gt;n&lt;/sup&gt;) =  Ω&lt;sub&gt;n&lt;/sub&gt;(H&lt;sub&gt;n&lt;/sub&gt;)×2&lt;sup&gt;n&lt;/sup&gt; has just become silly since the entire elementary abelian 2-group part, 2&lt;sup&gt;n&lt;/sup&gt;, is unrelated and uses a lot of extra generators, that is, it is not contained within the Frattini subgroup.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A moderate example: if instead of the simple group of order 60, we take the non-abelian group of order 6, then I can find a natural choice of H&lt;sub&gt;n&lt;/sub&gt; with |H&lt;sub&gt;n&lt;/sub&gt;| ≤ C⋅n, but I am not sure if there are other reasonable choices.  My choice of H&lt;sub&gt;n&lt;/sub&gt; has Φ(H&lt;sub&gt;n&lt;/sub&gt;)=1, which suggests to me that Frattini extensions may not be the right idea.&lt;/p&gt;&#xA;" OwnerUserId="583" LastEditorUserId="8575" LastEditorDisplayName="" LastEditDate="2011-05-19T08:26:20.600" LastActivityDate="2011-05-19T08:26:20.600" Title="Ignoring elements of small order in the simple group of order 60" Tags="&lt;group-theory&gt;&lt;simple-groups&gt;" CommentCount="1" FavoriteCount="1" />
  <row Id="1920" PostTypeId="2" ParentId="1896" CreationDate="2010-08-09T15:01:08.527" Score="1" ViewCount="" Body="&lt;p&gt;This won't be a complete derivation, but I will post it here on the chance that it is useful.&#xA;If no-one else comes along with a more complete explanation, I can try to expand on it at some later point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $P$ be a point on your hypersurace in $\mathbb R^{n+1}$, and let $\Pi$ be the tangent plane to your hypersurface at $P$, interpreted as an actual plane in $\mathbb R^n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Change coordinates so that $P$ is the origin, and the tangent plane is given by $y = 0$.&#xA;(Here $y$ is the $n+1$st coordinate of $\mathbb R^{n+1}$.)  Now choose the first $n$-coordinates&#xA;$x_1, \ldots, x_n$ so that they are orthogonal and point in the direction of the various principal curvatures, with principal curvatures $k_1,\ldots,k_n$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The claim, then, is that your hypersurface is given locally at $P$ by the equation&#xA;$y = (1/2)(k_1^2 x_1^2 + \cdots + k_n x_n^2) + O(x^3),$ or in other words, that the quadric&#xA;$y = (1/2)(k_1^2 x_1^2 + \cdots + k_n x_n^2)$ has contact order 2 with the hypersurface at the point $P$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This will follow more-or-less from the definition of the principal curvatures as eigenvalues of the second fundamental form: they encode the shape of your manifold at the point $P$ up to second order, and you can check that quadric we are writing down has the same principal curvatures at the origin, and so has the same shape as your manifold at the point $P$.  Thus they coincide up to second order.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you haven't actually checked this, that would be a useful exercise: take the quadric&#xA;$y = (1/2)(k_1x_1^2 + \cdots + k_n x_n^2),$ and compute its second fundamental form at the origin; you should find that it is diagonal, with eigenvalues precisely $k_1,\ldots,k_n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for your questions about the various different interpretations of the principal curvatures when $n = 2$, these are discussed extensively in one of the volumes of Spivak&#xA;(I think the second volume; in any case, if you look in the contents you will find the&#xA;place where he first starts to discuss curvature, and he talks about principal curvatures&#xA;of surfaces in a lot of detail.)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a brief hint: a 2x2 symmetrix matrix $A$ can be thought of as defining a conic section centered at the origin (via the equation $x^T A x = 1$), and the two eigendirections&#xA;are the major and minor axes of the conic.  (This is just linear algebra and basic analytic geometry.)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you now apply this statement in the context of the second fundamental form for a surface in space, you will be able to derive that the two principal curvatures at a point correspond to maximizing and minizing the curvature of curves passing through that point.&lt;/p&gt;&#xA;" OwnerUserId="221" LastEditorUserId="221" LastEditorDisplayName="" LastEditDate="2010-10-04T04:19:18.463" LastActivityDate="2010-10-04T04:19:18.463" />
  <row Id="1921" PostTypeId="2" ParentId="1913" CreationDate="2010-08-09T15:02:50.927" Score="4" ViewCount="" Body="&lt;p&gt;As a complement to Robin Chapman's answer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since G is 2-transitive, its order is divisible by n*(n-1), not just bounded below by, and so it would be interesting to bound |G|/(n*(n-1)).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If n is not a prime power, then it is quite possible for the lower bound to be huge.  The smallest reasonable non-prime, n=6, has its smallest 2-transitive group with order 60 = 6*5*2.  The next, n=10, has its smallest 2-transitive group with order 10*9*4.  For most n, the smallest multiple is (n-2)!/2, that is, the alternating group on n points is the smallest 2-transitive group.  This happens already at n=22, 33, 34, 35, and asymptotically takes over.  Cameron–Neumann–Teague showed this in their 1982 paper &lt;a href=&quot;http://www.ams.org/mathscinet-getitem?mr=661693&quot; rel=&quot;nofollow&quot;&gt;MR661693&lt;/a&gt;, and I believe it is covered in Dixon–Mortimer's textbook.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So on the one hand the lower bound for a 2-transitive group on n points is n*(n-1) for prime powers n, but for most n the lower bound is n!/2.&lt;/p&gt;&#xA;" OwnerUserId="583" LastEditorUserId="583" LastEditorDisplayName="" LastEditDate="2010-08-09T15:08:58.757" LastActivityDate="2010-08-09T15:08:58.757" />
  <row Id="1922" PostTypeId="2" ParentId="1877" CreationDate="2010-08-09T15:08:12.190" Score="2" ViewCount="" Body="&lt;p&gt;A little further thought reveals the following:  &lt;em&gt;uniform semi-continuity implies uniform continuity&lt;/em&gt;.  Thus the answer to my question is a resounding &quot;no&quot;, since any function that is upper semi-continuous but not continuous &lt;em&gt;cannot&lt;/em&gt; be uniformly upper semi-continuous.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof&lt;/strong&gt;.  Let $f$ be uniformly upper semi-continuous.  Then for every $\epsilon&gt;0$ there exists $\delta&gt;0$ such that for every $x\in X$, we have $f(y) &amp;lt; f(x) + \epsilon$ whenever $y\in B(x,\delta)$.  However, since this statement holds for every $x$, it also holds with $x$ and $y$ reversed; in the language of the original post, both $x$ and $y$ are contained in the set $X_\delta^\epsilon = X$.  Since $y$ is in this set and $x\in B(y,\delta)$, we also have $f(x) &amp;lt; f(y) + \epsilon$, and thus $|f(x) - f(y)| &amp;lt; \epsilon$.  But this is just the definition of uniform continuity.&lt;/p&gt;&#xA;" OwnerUserId="876" LastActivityDate="2010-08-09T15:08:12.190" CommentCount="1" />
  <row Id="1923" PostTypeId="1" CreationDate="2010-08-09T15:21:11.143" Score="2" ViewCount="76" Body="&lt;p&gt;Noncommutative algebraic geometry is a developing field. Things have not yet got the final form as in commutative geometry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But one might wonder whether things are any better in the case of skew-fields, ie division rings, ie possibly noncommutative rings in which each nonzero element has a multiplicative inverse. Algebraic geometry is much simpler in the case of fields. So are things better in the case of skew fields? For instance, is there any particularly nice geometry over the quaternions, like there is one over the complex numbers?&lt;/p&gt;&#xA;" OwnerUserId="977" LastActivityDate="2010-08-09T15:21:11.143" Title="Noncommutative algebraic geometry in the case of skew fields" Tags="&lt;algebraic-geometry&gt;&lt;noncommutative-geometry&gt;" CommentCount="3" FavoriteCount="3" />
  <row Id="1924" PostTypeId="2" ParentId="1918" CreationDate="2010-08-09T15:43:59.210" Score="4" ViewCount="" Body="&lt;p&gt;One example not in the above list is &lt;a href=&quot;http://groups.google.com/groups?selm=WGD.95Dec11023450%40martigny.ai.mit.edu&quot; rel=&quot;nofollow&quot;&gt;Goodstein's theorem&lt;/a&gt;, a highly nonintuitive concrete number theoretic theorem which is unprovable in Peano arithmetic (or, similarly, the Hercules vs. Hydra game). They essentially encode induction up to the ordinal $\epsilon_0 = \omega^{\omega^{\omega^{\cdot^{\cdot^{\cdot}}}}}$ - something that is not at all intuitive to those who are not familiar with such ordinals - especially their Cantor normal form.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2011-02-25T20:18:11.753" LastActivityDate="2011-02-25T20:18:11.753" CommunityOwnedDate="2010-08-09T15:43:59.210" CommentCount="6" />
  <row Id="1925" PostTypeId="2" ParentId="594" CreationDate="2010-08-09T15:46:07.027" Score="5" ViewCount="" Body="&lt;p&gt;Perhaps my favorite argument (other than any arguably &quot;correct&quot; arguments, such as the one Akhil has given, or arguments starting from the fact that $x^2 + y^2$ is the unique binary quadratic form of discriminant $-4$ up to equivalence) uses continued fractions.  Suppose that $u^2 \equiv -1 \pmod{p}$, and consider the continued fraction expansion of the rational number $u/p$.  Let $r/s$ be the last convergent to $u/p$ with the property that $s &amp;lt; \sqrt{p}$.  Then, setting $x=s$ and $y = rp-us$, one has $x^2 + y^2 = p$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's the argument: let $r'/s'$ be the convergent following $r/s$.  Then the basic theory of continued fractions gives the estimate $|r/s - u/p| &amp;lt; 1/ss'$, and the right-hand side is less than $1/s\sqrt{p}$ by hypothesis.  Clearing denominators gives $y &amp;lt; \sqrt{p}$, so that $0 &amp;lt; x^2 + y^2 &amp;lt; 2p$.  On the other hand $x^2 + y^2$ is checked to be divisible by $p$ (by choice of $u$), so must be equal to $p$.&lt;/p&gt;&#xA;" OwnerUserId="631" LastEditorUserId="631" LastEditorDisplayName="" LastEditDate="2010-08-09T16:13:29.633" LastActivityDate="2010-08-09T16:13:29.633" CommentCount="3" />
  <row Id="1926" PostTypeId="1" AcceptedAnswerId="1943" CreationDate="2010-08-09T15:47:19.053" Score="2" ViewCount="231" Body="&lt;p&gt;I know a little bit of the theory of compact Riemann surfaces, wherein there is a very nice divisor -- line bundle correspondence. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But when I take up the book of Hartshorne, the notion of &lt;a href=&quot;http://planetmath.org/encyclopedia/CartierDivisor.html&quot; rel=&quot;nofollow&quot;&gt;Cartier divisor&lt;/a&gt; there is very confusing. It is certainly not a direct sum of points; perhaps it is better to understand it in terms of line bundles. But Cartier divisors do not seem to be quite the same thing as line bundles. The definition is hard to figure out. Can someone clear the misunderstanding for me and explain to me how best to understand Cartier divisors?&lt;/p&gt;&#xA;" OwnerUserId="977" LastEditorUserId="977" LastEditorDisplayName="" LastEditDate="2010-08-09T15:54:12.560" LastActivityDate="2010-08-09T19:00:55.773" Title="Divisor -- line bundle correspondence in algebraic geometry" Tags="&lt;algebraic-geometry&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="2" />
  <row Id="1927" PostTypeId="1" AcceptedAnswerId="1930" CreationDate="2010-08-09T16:18:20.740" Score="0" ViewCount="108" Body="&lt;p&gt;Let's take field $(K, +,\times,0,1)$. Let's take set $S(K) \subset K$ such that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\[ \forall_{e \in K}\exists_{a \in S(K)}\exists_{p \in K} e = a \times p \times p \]&#xA;\[ \lnot \exists_{a, b \in S(K)}\exists_{p \in K} (a \times p \times p = b \land a \neq b) \]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example $S(\mathbb{R}) = \{-1, 1\}$, $S(\mathbb{C}) = \{1\}$, $S(\mathbb{Z}_3) = \{1, 2\}$, $S(\mathbb{Z}_5) = \{1, 2\}$. Of course sets are not unique like $S(\mathbb{R}) = \{-e,\pi\}$ is also possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Sorry if I misuse terms - I haven't found english translation - but I found it useful to generalize the signature of quadratic polynomials).&lt;/p&gt;&#xA;" OwnerUserId="728" LastActivityDate="2010-08-09T16:24:36.990" Title="What's the name of this set" Tags="&lt;polynomials&gt;" AnswerCount="1" />
  <row Id="1928" PostTypeId="2" ParentId="1910" CreationDate="2010-08-09T16:21:34.123" Score="0" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://www.youtube.com/watch?v=atIhEDQSI1U&quot; rel=&quot;nofollow&quot;&gt;GNU Emacs + AUCTeX&lt;/a&gt; works fine for me.&lt;/p&gt;&#xA;" OwnerUserId="705" LastActivityDate="2010-08-09T16:21:34.123" />
  <row Id="1930" PostTypeId="2" ParentId="1927" CreationDate="2010-08-09T16:24:36.990" Score="2" ViewCount="" Body="&lt;p&gt;If I am deciphering your notation correctly, I would call $S$ &quot;a set of coset representatives for $K^{\times}/(K^{\times})^2$.&quot;&lt;/p&gt;&#xA;" OwnerUserId="631" LastActivityDate="2010-08-09T16:24:36.990" CommentCount="1" />
  <row Id="1931" PostTypeId="2" ParentId="501" CreationDate="2010-08-09T16:24:44.743" Score="2" ViewCount="" Body="&lt;p&gt;Finitism still allows you to use infinitary definitions of real numbers, because a finitist is content with finite &lt;em&gt;proofs&lt;/em&gt; even if the concepts mentioned by those proofs would seem to require infinite sets. For example, a finitist would still recognize that &quot;ZFC proves that every bounded nonempty set of reals has a least upper bound&quot; even if the finitist does not accept that infinite sets exist. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proofs in various infinitary systems are of interest to finitists because of conservation results. In this setting, a conservation result would show that if a sentence about the natural numbers of a certain form is provable in some infinitary system, the sentence is actually provable in a finitistic system.  For example, there are finitistic proofs that if any $\Pi^0_2$ sentence about the natural numbers is provable in the infinitary system $\text{WKL}_0$ of second order arithmetic, that sentence is also provable in the finitistic system $\text{PRA}$ of primitive-recursive arithmetic. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many consistency results are proven finitistically. For example, there is a finitistic proof that if ZF set theory without the axiom of choice is consistent, then ZFC set theory with the axiom of choice is also consistent. This proof studies infinitary systems of set theory, but the objects actually handled are finite formal proofs rather than infinite sets.&lt;/p&gt;&#xA;" OwnerUserId="630" LastActivityDate="2010-08-09T16:24:44.743" CommentCount="0" />
  <row Id="1932" PostTypeId="1" CreationDate="2010-08-09T16:46:05.870" Score="3" ViewCount="146" Body="&lt;p&gt;In &lt;a href=&quot;http://meta.math.stackexchange.com/questions/172/why-did-you-close-my-question-if-all-sets-were-finite/178#178&quot;&gt;the comment thread of an answer&lt;/a&gt;, I said:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The computable numbers are based on the intuitionistic continuum, and are not finitary. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To which T.. replied:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Computable numbers are not based on the intuitionistic continuum.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This disagreement contains, I think, a good example of a philosophical question: are the computable reals within the scope of finitistic mathematics?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;References&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Bendegem, 2010, &lt;a href=&quot;http://plato.stanford.edu/entries/geometry-finitism/&quot; rel=&quot;nofollow&quot;&gt;Finitism in Geometry&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;Edalat, 2009, &lt;a href=&quot;http://www.doc.ic.ac.uk/~ae/papers/longo-f.pdf&quot; rel=&quot;nofollow&quot;&gt;A computable approach to measure and integration theory&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;Zach, 2001, &lt;a href=&quot;http://www.ucalgary.ca/rzach/papers/hilbert.html&quot; rel=&quot;nofollow&quot;&gt;Hilbert's Finitism&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;Zach, 2003, &lt;a href=&quot;http://plato.stanford.edu/entries/hilbert-program&quot; rel=&quot;nofollow&quot;&gt;Hilbert's Program&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="100" LastActivityDate="2010-10-13T18:03:36.857" Title="Are the computable reals finitary?" Tags="&lt;soft-question&gt;&lt;constructive-mathematics&gt;&lt;philosophy&gt;&lt;real-number-computation&gt;&lt;finitism&gt;" AnswerCount="2" />
  <row Id="1933" PostTypeId="1" AcceptedAnswerId="1945" CreationDate="2010-08-09T16:57:54.247" Score="3" ViewCount="141" Body="&lt;p&gt;I have seen a lot of places where the group cohomology when a group acts on a module, is extensively used. But beyond seeing the definition and some claims of partial results, I havent seen any uses of the case when we replace action on a module by action on noncommutative group. It seems hard to believe that much can be made out of it as H_1 is just a set, not even a group. Can somebody explain some uses of bringing up and studying this notion?&lt;/p&gt;&#xA;" OwnerUserId="977" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-08-09T23:02:01.287" LastActivityDate="2010-08-09T23:02:01.287" Title="Use of noncommutative group cohomology" Tags="&lt;abstract-algebra&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="1934" PostTypeId="2" ParentId="29" CreationDate="2010-08-09T17:14:56.107" Score="2" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Microsoft_Math&quot; rel=&quot;nofollow&quot;&gt;Microsoft Math&lt;/a&gt; looks like TI and shows the history of items&lt;/p&gt;&#xA;" OwnerUserId="956" LastActivityDate="2010-08-09T17:14:56.107" CommunityOwnedDate="2010-08-09T17:14:56.107" />
  <row Id="1935" PostTypeId="1" CreationDate="2010-08-09T17:22:03.380" Score="4" ViewCount="351" Body="&lt;p&gt;Given a polynomial over a field, what are the methods to see it is irreducible? Only two comes to my mind now. First is Eisenstein criterion. Another is that if a polynomial is irreducible mod p then it is irreducible. Are there any others?&lt;/p&gt;&#xA;" OwnerUserId="977" LastEditorUserId="2614" LastEditorDisplayName="" LastEditDate="2011-05-04T08:10:01.077" LastActivityDate="2011-05-04T08:10:01.077" Title="Methods to see if a polynomial is irreducible" Tags="&lt;abstract-algebra&gt;&lt;polynomials&gt;" AnswerCount="5" CommentCount="1" FavoriteCount="5" />
  <row Id="1936" PostTypeId="2" ParentId="1918" CreationDate="2010-08-09T17:24:02.893" Score="3" ViewCount="" Body="&lt;p&gt;A nice paradox (in the sense of going against the common opinion) which is not in that list is &lt;a href=&quot;http://en.wikipedia.org/wiki/Arrow%27s_impossibility_theorem&quot; rel=&quot;nofollow&quot;&gt;Arrow's theorem&lt;/a&gt;. More or less, it states the following. Let be given a set of people who vote on some issue, and have a finite number of alternatives (at least 3). Each person orders the alternatives according to her preferences; the outcome of the vote is an order on the set of alternatives which is supposed to reflect the common consensus.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More formally, a preference is a total order on the set of alternatives, and a voting system is a function which associates to each $n$-uple of preferences another preference. It turns out that the only voting system which satisfies some innocent-looking hypothesis is the projection on some factor, that is, the dictatorship of one of the people.&lt;/p&gt;&#xA;" OwnerUserId="233" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-08-09T18:22:05.807" LastActivityDate="2010-08-09T18:22:05.807" CommunityOwnedDate="2010-08-09T17:24:02.893" />
  <row Id="1937" PostTypeId="2" ParentId="1892" CreationDate="2010-08-09T17:25:02.970" Score="1" ViewCount="" Body="&lt;p&gt;Not directly related to the question, but definitely relevant. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Three days ago &lt;a href=&quot;http://www.hpl.hp.com/personal/Vinay_Deolalikar/Papers/pnp_preliminary.pdf&quot; rel=&quot;nofollow&quot;&gt;proof&lt;/a&gt; for P != NP is published. Community thinks it looks serious.&lt;/p&gt;&#xA;" OwnerUserId="956" LastEditorUserId="956" LastEditorDisplayName="" LastEditDate="2010-08-09T23:11:20.820" LastActivityDate="2010-08-09T23:11:20.820" CommentCount="5" />
  <row Id="1938" PostTypeId="2" ParentId="29" CreationDate="2010-08-09T17:30:41.760" Score="1" ViewCount="" Body="&lt;p&gt;Here are three links you may find useful:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://wims.unice.fr/wims/wims.cgi?lang=en&amp;amp;+session=YV6390EF2C.1&amp;amp;+module=home&quot; rel=&quot;nofollow&quot;&gt;http://wims.unice.fr/wims/wims.cgi?lang=en&amp;amp;+session=YV6390EF2C.1&amp;amp;+module=home&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.numbertheory.org/php/php.html&quot; rel=&quot;nofollow&quot;&gt;http://www.numbertheory.org/php/php.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.alpertron.com.ar/NUMBERT.HTM&quot; rel=&quot;nofollow&quot;&gt;http://www.alpertron.com.ar/NUMBERT.HTM&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="660" LastActivityDate="2010-08-09T17:30:41.760" CommunityOwnedDate="2010-08-09T17:30:41.760" />
  <row Id="1939" PostTypeId="2" ParentId="1935" CreationDate="2010-08-09T17:32:50.803" Score="5" ViewCount="" Body="&lt;p&gt;To better understand the Eisenstein and related irreducibility tests you should learn about Newton polygons. It's the &lt;strong&gt;master theorem&lt;/strong&gt; behind all these related results. A good place to start is Filaseta's notes - see the links below. Note: you may need to write to Filaseta to get access to his interesting book [1] on this topic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[1] &lt;a href=&quot;http://www.math.sc.edu/~filaseta/gradcourses/Math788F/latexbook/&quot; rel=&quot;nofollow&quot;&gt;http://www.math.sc.edu/~filaseta/gradcourses/Math788F/latexbook/&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;[2] &lt;a href=&quot;http://www.math.sc.edu/~filaseta/gradcourses/Math788F/NewtonPolygonsTalk.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.math.sc.edu/~filaseta/gradcourses/Math788F/NewtonPolygonsTalk.pdf&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;[3] Newton Polygon Applet &#xA;&lt;a href=&quot;http://www.math.sc.edu/~filaseta/newton/newton.html&quot; rel=&quot;nofollow&quot;&gt;http://www.math.sc.edu/~filaseta/newton/newton.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[4] Abhyankar, Shreeram S.&lt;br&gt;&#xA;Historical ramblings in algebraic geometry and related algebra.&lt;br&gt;&#xA;Amer. Math. Monthly 83 (1976), no. 6, 409-448.&lt;br&gt;&#xA;&lt;a href=&quot;http://links.jstor.org/sici?sici=0002-9890(197606/07)83:6%3C409:HRIAG&quot; rel=&quot;nofollow&quot;&gt;http://links.jstor.org/sici?sici=0002-9890(197606/07)83:6%3C409:HRIAG&lt;/a&gt;... &lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-08-09T17:32:50.803" CommentCount="3" />
  <row Id="1940" PostTypeId="2" ParentId="29" CreationDate="2010-08-09T17:35:54.930" Score="2" ViewCount="" Body="&lt;p&gt;For those of us on a Mac, there's &lt;a href=&quot;http://www.acqualia.com/soulver/&quot; rel=&quot;nofollow&quot;&gt;Soulver&lt;/a&gt;, which is more like a notepad with calculating in mind. Lots of interesting features like conversions and instant stock prices, but no graphing if you're reliant on that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're interested in having a history of disconnected equations over the course of a day, you could just open a new document, use it for the day and then discard it when you quit. However, it's more designed to create a set of calculations which you can save as a document and then return to later. You can link calculations together as well, letting you change a single value and recomputing all your answers instantly.&lt;/p&gt;&#xA;" OwnerUserId="82" LastActivityDate="2010-08-09T17:35:54.930" CommunityOwnedDate="2010-08-09T17:35:54.930" />
  <row Id="1941" PostTypeId="2" ParentId="1932" CreationDate="2010-08-09T17:53:26.567" Score="3" ViewCount="" Body="&lt;p&gt;A real number $r$ is defined to be computable if there is an algorithm that, given any natural number $n$, will produce a rational number within distance $1/n$ of $r$. Thus each individual computable number is defined using a finite amount of information (the algorithm). On the other hand, there is a well-known intuitive sense in which an arbitrary real number can contain an infinite amount of information. In this sense, computable real numbers are finitary in a way that arbitrary real numbers are not. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The definition of a computable real number can be stated in either classical mathematics or in intuitionistic mathematics. Nothing in the concept of a computable real number ties you to one logic or the other. &lt;/p&gt;&#xA;" OwnerUserId="630" LastActivityDate="2010-08-09T17:53:26.567" CommentCount="2" />
  <row Id="1942" PostTypeId="1" CreationDate="2010-08-09T17:58:32.297" Score="2" ViewCount="232" Body="&lt;p&gt;We have a pool of items.  Each item has a set of characteristics.  Each characteristic is a string, integer, or floating-point number.  Each item is also of a particular type.  Types are hierarchical.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We also have a set of bins.  Each bin has a requirement, which is a boolean expression comprised of atomic expressions regarding item characteristics.  These atomic expressions can be string, integer, or floating-point comparisons and type comparisons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We wish to put the items in the bins such that each item placed in a bin satisfies the requirement of that bin.  Each bin also must contain a specified number of items.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, we might have the following type hierarchy:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;shape &#xA;&lt;ul&gt;&#xA;&lt;li&gt;triangle &#xA;&lt;ul&gt;&#xA;&lt;li&gt;equilateral triangle&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;rectangle &#xA;&lt;ul&gt;&#xA;&lt;li&gt;square&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;We could have the following items:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;         &lt;li&gt;LargeRedTriangle&lt;/li&gt;&#xA;    &lt;ul&gt;&#xA;         &lt;li&gt;type = triangle&lt;/li&gt;&#xA;         &lt;li&gt; color = &quot;red&quot;&lt;/li&gt;&#xA;         &lt;li&gt; area = 1000&lt;/li&gt;&#xA;    &lt;/ul&gt;&#xA; &lt;li&gt;SmallShape&lt;/li&gt;&#xA;    &lt;ul&gt;&#xA; &lt;li&gt;type = shape&lt;/li&gt;&#xA; &lt;li&gt;area = 5&lt;/li&gt;&#xA;    &lt;/ul&gt;&#xA;&lt;li&gt;LargeBlueSquare&#xA;    &lt;ul&gt;&#xA; &lt;li&gt;type = square&lt;/li&gt;&#xA; &lt;li&gt;color = &quot;blue&quot;&lt;/li&gt;&#xA; &lt;li&gt;area = 2000&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Here is an example bin:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;LargeRectangles&#xA;&lt;ul&gt;&#xA;&lt;li&gt;(type == rectangle) and (area &gt; 750)&lt;/li&gt;&#xA;&lt;li&gt;item count = 5&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;We could put LargeBlueSquare in the LargeRectangles bin.  We want exactly 5 items in this bin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is another example bin:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;NonBlueTriangles&#xA;&lt;ul&gt;&#xA;&lt;li&gt;(type == triangle) and (not (color == &quot;blue&quot;))&lt;/li&gt;&#xA;&lt;li&gt;item count = 10&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;LargeRedTriangle can go in the NonBlueTriangles bin.  We want exactly 10 items in this bin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, we have a set of items and a set of bins.  The question is to find out how many ways there are to put the items in the bins while satisfying the constraints (i.e., the requirements for items in bins and the number of items in each bin).  (We don't need to enumerate all the ways to do this; we only need to know how many different ways of doing this there are.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there an efficient way of determining this number?  Or do we just have to 'brute-force' it and try every possible combination?&lt;/p&gt;&#xA;" OwnerUserId="984" LastEditorUserId="984" LastEditorDisplayName="" LastEditDate="2010-08-09T18:23:25.147" LastActivityDate="2010-09-16T21:58:48.003" Title="Counting placing items in bins according to boolean restrictions" Tags="&lt;combinatorics&gt;" AnswerCount="3" />
  <row Id="1943" PostTypeId="2" ParentId="1926" CreationDate="2010-08-09T18:10:04.580" Score="7" ViewCount="" Body="&lt;p&gt;When discussing divisors, a helpful distinction to make at the beginning is effective divisors vs. all divisors.  Normally effective divisors have a more geometric description; all divisors can then be obtained from the effective ones by allowing some minus signs to come into the picture.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An irreducible effective Weil divisor on a variety $X$ is the same thing as an irreducible codimension one subvariety, which in turn is the same thing as a height one point $\eta$ of $X$.  (We get $\eta$ as the generic point of the irred. codim'n one subvariety, and we recover the subvariety as the closure of $\eta$.)&lt;br&gt;&#xA; An effective Weil divisor is a non-negative integral linear combination of irreducible ones, so you can think of it as a non-negative integral linear combination of height one points $\eta$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Typically, one restricts to normal varieties, so that all the local rings at height one points  are DVRs.    Then, given any pure codimension one subscheme $Z$ of $X$, you can attach a Weil&#xA;divisor to $Z$, in the following way:&#xA;because the local rings at height one points are DVRs, if $Z$ is any codimension one subscheme of $X$, cut out by an ideal sheaf $\mathcal I_Z$,  and $\eta$ is a height one point, then the stalk $\mathcal I_{Z,\eta}$ is&#xA;an ideal in the DVR $\mathcal O_{X,\eta}$, thus is just some power of the maximal ideal&#xA;$\mathfrak m_{\eta}$ (using the DVR property), say $\mathcal I_{Z,\eta} = \mathfrak m_{\eta}^{m_{Z,\eta}},$ and so the multiplicity $m$ of $Z$ at $\eta$ is well-defined.&lt;br&gt;&#xA;Thus the effective Weil divisor $$div(Z) := \sum_{\eta \text{ of height one}} m_{Z,\eta}\cdot  \eta$$&#xA;is well-defined.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that this recipe is only goes one way: starting with the Weil divisor, we can't recover $Z$, because the Weil divisor does not remember all the scheme structure (i.e. the whole&#xA;structure sheaf, or equivalently, the whole ideal sheaf) of $Z$, but only its behaviour at its generic points (which amounts to the same thing as remembering the irreducible components and their multiplicities). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;An effective Cartier divisor is actually a more directly geometric objects, namely, &#xA;it is a &lt;em&gt;locally principal&lt;/em&gt; pure codimension one subscheme, that is,&#xA;a subscheme, each component of which is codimension one, and which, locally around each point, is the zero locus of a section of the structure sheaf.   Now in order to cut out&#xA;a pure codimension one subscheme as its zero locus, a section of the structure&#xA;sheaf should be regular (in the commutative algebra sense), i.e. a non-zero divisor.&#xA;Also, two regular sections will cut out the same zero locus if their ratio is a unit&#xA;in the structure sheaf.  So if we let $\mathcal O_X^{reg}$ denote the subsheaf of&#xA;$\mathcal O_X$ whose sections are regular elements (i.e. non-zero divisors in each stalk),&#xA;then the equation of a Cartier divisor is a well-defined global section of the quotient&#xA;sheaf&#xA;$\mathcal O_X^{reg}/\mathcal O_X^{\times}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now suppose that we are on a smooth variety.  Then any irreducible codimension one subvariety&#xA;is in fact locally principal, and so given a Weil divisor &#xA;$$D = \sum_{\eta \text{ of height one}} m_{\eta} \cdot\eta,$$&#xA;we can actually find attach a canonical Cartier divisor to it, in the following way:&#xA;in a n.h. of some point $x$, let $f_{\eta}$ be a local equation for the Zariski closure&#xA;of $\eta$; then if $Z(D)$ is cut out locally by $\prod_{\eta} f_{\eta}^{m_{\eta}} = 0,$&#xA;then $Z(D)$ is locally principal by construction, and, again by construction,&#xA;$div(Z(D)) = D.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in the smooth setting,&#xA;we see that $Z \mapsto div(Z)$ and $D \mapsto Z(D)$ establish a bijection between&#xA;effective Cartier divisors and effective Weil divisors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, on a singular variety, it can happen that an irreducible codimension one subvariety need not be locally principal in the neighbourhood of a singular point (e.g. a generating line on the cone $x^2 +y^2 + z^2 = 0$&#xA;in $\mathbb A^3$ is not locally&#xA;principal in any neighbourhood of the cone point).  Thus there can be Weil divisors that&#xA;are not of the form $div(Z)$ for any Cartier divisor $Z$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To go from effective Weil divisor to all Weil divisors, you just allow negative coefficients.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To go from effective Cartier divisors to all Cartier divisors, you have to allow yourself &#xA;to invert the functions $f$ that cut out the effective Cartier divisors, or equivalently, to go from the sheaf of monoids $\mathcal O_X^{reg}/\mathcal O_X^{\times}$ to the associated sheaf of groups, which is $\mathcal K_X^{reg}/\mathcal O_X^{\times}.$&#xA;(Here, it helps to remember that $\mathcal K_X$ is obtained from $\mathcal O_X$&#xA;by inverting non-zero divisors.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, for the connection with line bundles: if $\mathcal L$ is a line bundle,&#xA;and $s$ is a regular section (i.e. a section whose zero locus is pure codimension one,&#xA;or equivalently, a section which, when we choose a local isomorphism $\mathcal L_{| U}&#xA;\cong \mathcal O_U$, is not a zero divisor), then the zero locus $Z(s)$ of $s$&#xA;is an effective Cartier divisor, essentially by definition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we have a map $(\mathcal L,s) \mapsto Z(s)$ which sends line bundles with regular sections to effective Cartier divisors.  This is in fact an isomorphism of monoids&#xA;(where on the left we consider pairs $(\mathcal L,s)$ up to isomorphism of pairs):&#xA;given an effective Cartier divisor $D$, we can define $\mathcal O(D)$ to be the&#xA;subsheaf of $\mathcal K_X$ consisting (locally) of sections $f$ such that the locus of&#xA;poles of $f$ (a well-defined Cartier divisor) is contained (as a subscheme)in the Cartier&#xA;divisor $D$ (perhaps less intuitively, but more concretely: if $D$ is locally cut out&#xA;by the equation $g = 0$, then $\mathcal O(D)$ consists (locally) of sections $f$&#xA;of $\mathcal K_X$ such that $fg$ is in fact a section of $\mathcal O_X$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The constant function $1$ certainly lies in $\mathcal O(D)$, and (thought of as a section&#xA;of $\mathcal O(D)$ -- not as a function!) its zero locus is exactly $D$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus $D \mapsto (\mathcal O(D), 1)$ is an inverse to the above map $(\mathcal L,s) \mapsto&#xA;Z(s)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, if we choose two different regular sections of the same line bundle,&#xA;the corresponding Cartier divisors are linearly equivalent.  Thus we are led to the&#xA;isomorphism &quot;line bundles up to isomorphism = Cartier divisors up to linear equivalence&quot;.&#xA;But, just to emphasize, to understand this it is best to restrict first to line bundles&#xA;which admit a regular section, and then think of the corresponding Cartier divisor as being&#xA;the zero locus of that section.   This brings out the geometric nature of the Cartier divisor quite clearly.&lt;/p&gt;&#xA;" OwnerUserId="221" LastEditorUserId="221" LastEditorDisplayName="" LastEditDate="2010-08-09T19:00:55.773" LastActivityDate="2010-08-09T19:00:55.773" CommentCount="1" />
  <row Id="1944" PostTypeId="2" ParentId="1935" CreationDate="2010-08-09T18:23:40.420" Score="5" ViewCount="" Body="&lt;p&gt;One method for polynomials over $\mathbb{Z}$ is to use complex analysis to say something about the location of the roots.  Often Rouche's theorem is useful; this is how Perron's criterion is proven, which says that a monic polynomial $x^n + a_{n-1} x^{n-1} + ... + a_0$ with integer coefficients is irreducible if $|a_{n-1}| &gt; 1 + |a_{n-2}| + ... + |a_0|$ and $a_0 \neq 0$.  A basic observation is that knowing a polynomial is reducible places constraints on where its roots can be; for example, if a monic polynomial with prime constant coefficient $p$ is reducible, one of its irreducible factors has constant term $\pm p$ and the rest have constant term $\pm 1$.  It follows that the polynomial has at least one root inside the unit circle and at least one root outside.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An important thing to keep in mind here is that there exist irreducible polynomials over $\mathbb{Z}$ which are reducible modulo every prime.  For example, $x^4 + 16$ is such a polynomial.  So the modular technique is not enough in general.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-08-09T18:23:40.420" CommentCount="1" />
  <row Id="1945" PostTypeId="2" ParentId="1933" CreationDate="2010-08-09T18:29:26.133" Score="8" ViewCount="" Body="&lt;p&gt;Often elements of $H^1$ are used to classify objects of a certain kind.  E.g. if $k$ is a field and $k^s$ its separable closure, and $G_k = Gal(k^s/k)$, then if $X$ is a variety&#xA;over $k$, and $Aut(X)$ is the group of automorphisms of $X_{/k^s}$,&#xA;then the set $H^1(G_k,Aut(X))$ classifies varieties over $k$ that become isomorphic to&#xA;$X$ when both are regarded as varieties over $k^s$.   The base point of $H^1$ corresponds &#xA;to the original variety $X$ itself. (Such varieties are called &lt;em&gt;twists&lt;/em&gt; of $X$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is an important construction, which (along with variants) is used all the time in&#xA;arithmetic geometry, including (perhaps especially) the theory of algebraic groups.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As one concrete example, let me mention&#xA;a famous example, namely the result that $H^1(G_k, GL_n(k^s)) = 1$.  (A generalized form of Hilbert's Thm. 90.)   If we now consider the short exact sequence&#xA;$1 \to (k^s)^{\times} \to GL_n(k^s) \to PGL_n(k^s) \to 1,$&#xA;we then obtain an injection $H^1(G_k, PGL_n(k^s))\hookrightarrow H^2(G_k,(k^s)^{\times}).$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now $PGL_n(k^s)$ acts as automorphisms of $\mathbb P^{n-1}$, so we see that the twists of&#xA;projective spaces (such twists are called Brauer--Severi varieties) are classified by certain elements of the Brauer group.  For example, smooth conics in the plane (which are twists&#xA;of $\mathbb P^1$, because over $k^s$ they acquire rational points, and so are isomorphic&#xA;to $\mathbb P^1$) correspond to quaternion algebras.  (To see this concretely --- or rather,&#xA;to go the other way ---&#xA;if $D$ is a quaternion algebra, then $D^{tr = 0}$, the subspace of elements of reduced trace zero, is 3-dimensional, and the reduced norm gives a quadratic form on this space.&#xA;Passing to the associated two-dimensional projective space, we get a conic in a projective&#xA;plane.   This is more concrete than the cohomological story, but the cohomological story has the advantage of being very general, and relies just on a general fact, namely Hilbert's Thm. 90, rather than specific knowledge of the geometric situation.  This is one typical advantage of cohomological arguments and constructions, when you can find them/make them.)&lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2010-08-09T18:29:26.133" />
  <row Id="1946" PostTypeId="1" AcceptedAnswerId="2049" CreationDate="2010-08-09T18:29:59.503" Score="0" ViewCount="262" Body="&lt;p&gt;Was reading &quot;Taylor's Expansion&quot; when i came across this interesting problem, interesting for me. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose that $f^{n+2}$ is continuous on $[0,x]$. Prove that there is a $\theta \in (0,1)$ such that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;f(x) &amp;amp;= f(0) + \frac{f'(0)}{1!}x + \cdots + \frac{f^{n-1}(0)}{(n-1)!}x^{n-1} \\&#xA;&amp;amp; \quad + \frac{f^{n}\left(\frac{x}{n+1}\right)}{n!}x^{n} + \frac{n}{2(n+1)}f^{n+2}(\theta x)x^{n+2}{(n+2)!}&#xA;\end{align}&lt;/p&gt;&#xA;" LastEditorUserId="76" LastEditorDisplayName="" LastEditDate="2010-08-10T20:41:20.487" LastActivityDate="2010-08-10T20:41:20.487" Title="Existence of a $\theta$ - Taylor Expansion Problem" Tags="&lt;real-analysis&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="1947" PostTypeId="1" AcceptedAnswerId="1950" CreationDate="2010-08-09T18:36:38.833" Score="0" ViewCount="84" Body="&lt;p&gt;Before starting algebraic geometry, I got some understanding of Compact Riemann Surfaces which is more or less rigorous; and then I attended a couple of lectures on analytic geometry. I did not quite fully grasp the analytic geometry rigorously; but it stuck in my mind and this previous experiences helped me greatly when learning algebraic geometry. However I am not able to imagine what would be the precise analogue of Cartan's Theorem A for algebraic varieties over a field. There is the wikipedia article:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Cartan&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Cartan&lt;/a&gt;'s_theorems_A_and_B&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which tells me the precise analogue of Cartan Theorem B in the book of Hartshorne. And indeed it is really analogous. But it does not mention Theorem A. Note that here I am restricting the situation to varieties out of fear of any pathologies schemes might have.&lt;/p&gt;&#xA;" OwnerUserId="977" LastEditorUserId="977" LastEditorDisplayName="" LastEditDate="2010-08-09T18:43:21.677" LastActivityDate="2010-08-09T19:00:17.023" Title="Analogue of Cartan theorem A in algebraic geometry" Tags="&lt;algebraic-geometry&gt;" AnswerCount="1" />
  <row Id="1948" PostTypeId="1" AcceptedAnswerId="1997" CreationDate="2010-08-09T18:55:11.073" Score="2" ViewCount="325" Body="&lt;p&gt;These are two famous identities of &lt;strong&gt;Ramanujan&lt;/strong&gt;.  Where can I find the proofs of them:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;$ \displaystyle \frac{1}{\pi} = \frac{2\sqrt{2}}{9801} \sum\limits_{k=0}^{\infty} \frac{(4k!)(1103 + 26390k)}{(k!)^{2} (396)^{4k}}$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$\displaystyle \int\limits_{0}^{\infty} \frac{1 + x^{2}/(b+1)^{2}}{1+x^{2}/a^{2}} \times \frac{1+ x^{2}/(b+2)^{2}}{1 + x^{2}/(a+1)^{2}} \times  \cdots dx= \frac{\sqrt{\pi}}{2} \times \frac{\Gamma(a+1) \Gamma(b+\frac{1}{2}) \Gamma(b-a+\frac{1}{2}}{\Gamma(a)\Gamma(b+\frac{1}{2} \Gamma(b-a+1)}$ for $0 &amp;lt; a &amp;lt; b+\frac{1}{2}$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" LastActivityDate="2010-08-10T00:57:12.550" Title="Where can I find the proofs of the following" Tags="&lt;reference-request&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="1949" PostTypeId="1" CreationDate="2010-08-09T18:56:26.907" Score="2" ViewCount="210" Body="&lt;p&gt;I would like to know how to explicitly prove that Riemann Curvature,Ricci Curvature, Sectional Curvature and Scalar Curvature are left invariant under an isometry. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can't see this explained in most books I have looked at. &#xA;They atmost explain preservation of the connection. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I guess doing an explicit proof for the sectional curvature should be enough (and easiest?) since all the rest can be written in terms of it. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Given Akhil's reply I think I should try to understand the connection invariance proof better and here goes my partial attempt. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $\nabla$ be the connection on the manifold $(M,g)$ and $\nabla '$ be the Riemann connection on the manifold $(M',g')$ and between these two let $\phi$ be the isometry. Then one wants to show two things,&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$D\phi [\nabla _ X Y] = \nabla ' _{D\phi[X]} D\phi [Y]$&lt;/li&gt;&#xA;&lt;li&gt;$R(X,Y)Z = R'(D\phi [X],D\phi [Y]) D\phi [Z]$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Where $R$ and $R'$ are the Riemann connection on $(M,g)$ and $(M',g')$ respectively. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One defines the map $\nabla ''$ on M which maps two vector fields on M to another vector field using $\nabla '' _X Y = D\phi ^{-1} (\nabla' _{D\phi[X]} D\phi [Y]$. By the uniqueness of the Riemann connection the proof is complete if one can show that this $\nabla ''$ satisfies all the conditions of being a Riemann connection on M. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am getting stuck after a few steps while trying to show the Lebnitz property of $\nabla ''$. Let $f$ be some smooth function on M and then one would like to show that,&#xA;$\nabla '' _X fY = X(f)Y + f\nabla '' _X Y$ which is equivalent to showing that,&#xA;$D\phi ^{-1} (\nabla' _{D\phi[X]} D\phi [fY]) = X(f) + f D\phi ^{-1} (\nabla' _{D\phi[X]} D\phi [Y])$ knowing that $\nabla '$ satisfies the Leibnitz property on $M'$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some how I am not being able to unwrap the above to prove this.&#xA;I can get the second term of the equation but not the first one. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proving symmetry of $\nabla ''$ is easy but again proving metric compatibility is stuck for me. If $X,Y,Z$ are 3 vector fields on M then one would want to show that, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$Xg(Y,Z) = g(\nabla ''_X Y,Z) + g(Y, \nabla '' _X Z)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is equivalent to showing that, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$Xg(Y,Z) = g(D\phi ^{-1} (\nabla' _{D\phi[X]} D\phi [Y]),Z) + g(Y,D\phi ^{-1} (\nabla' _{D\phi[X]} D\phi [Z]) )$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;knowing that $\nabla'$ satisfies metric compatibility equation on $M'$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It would be helpful if someone can help me fill in the steps. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then one is left with proving the curvature endomorphism equation. &lt;/p&gt;&#xA;" OwnerUserId="965" LastEditorUserId="965" LastEditorDisplayName="" LastEditDate="2010-08-12T16:04:32.893" LastActivityDate="2010-08-16T17:16:07.940" Title="Explicitly proving invariance of curvatures under isometry" Tags="&lt;differential-geometry&gt;" AnswerCount="1" CommentCount="12" FavoriteCount="3" />
  <row Id="1950" PostTypeId="2" ParentId="1947" CreationDate="2010-08-09T19:00:17.023" Score="3" ViewCount="" Body="&lt;p&gt;Over an affine scheme any quasi-coherent sheaf is generated by its global sections, so I guess Theorem A is always true.&lt;/p&gt;&#xA;" OwnerUserId="223" LastActivityDate="2010-08-09T19:00:17.023" CommentCount="2" />
  <row Id="1951" PostTypeId="2" ParentId="1942" CreationDate="2010-08-09T19:02:48.920" Score="0" ViewCount="" Body="&lt;p&gt;If you only have one bin, it's actually quite easy.  Let N be how many items qualify to go into the bin, and K be the number of items the bin is required to have.  Your answer would be nCr(N, K).  The nCr (a.k.a. &lt;a href=&quot;http://en.wikipedia.org/wiki/Binomial_coefficient&quot; rel=&quot;nofollow&quot;&gt;binomial coefficient&lt;/a&gt;) function is defined as:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;               n!&#xA;nCr n k = ------------&#xA;           k! * (n-k)!&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;(&lt;code&gt;n!&lt;/code&gt; means n &lt;a href=&quot;http://en.wikipedia.org/wiki/Factorial&quot; rel=&quot;nofollow&quot;&gt;factorial&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for multiple bins, I doubt that knowledge of the type hierarchy or the predicates in detail will be needed to make it efficient (but I may be wrong).  The approach I'd take is to, for each item, list all the bins that item qualifies for.  Then, just find out how many combinations of filled-up bins you get.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sorry I don't have the answer to having multiple bins (yet), but I hope this gets you started.  &lt;/p&gt;&#xA;" OwnerUserId="985" LastEditorUserId="985" LastEditorDisplayName="" LastEditDate="2010-08-09T19:08:20.437" LastActivityDate="2010-08-09T19:08:20.437" />
  <row Id="1952" PostTypeId="1" CreationDate="2010-08-09T19:28:49.687" Score="1" ViewCount="331" Body="&lt;p&gt;The correspondence between algebraic and analytic geometry is thought-provoking. The GAGA makes this precise to some extent. But there is more of this analogy. As I checked an article on &lt;a href=&quot;http://eom.springer.de/a/a012430.htm&quot; rel=&quot;nofollow&quot;&gt;analytic space&lt;/a&gt;, most of the notions there seems to have an algebraic analogue. I had been wondering why people studied analytic geometry separately to the extent it has been. It would be nice to know about some theorems in analytic geometry that have no analogue in algebraic geometry, and this would justify the study of this subject separately. Here I mean theorems that do need to use the full power of analytic geometry; not things like Hodge decomposition for which Kahler is sufficient.&lt;/p&gt;&#xA;" OwnerUserId="977" LastActivityDate="2010-08-11T15:07:38.980" Title="Non-algebraic theorems in analytic geometry" Tags="&lt;algebraic-geometry&gt;" AnswerCount="2" CommentCount="7" FavoriteCount="1" />
  <row Id="1953" PostTypeId="1" CreationDate="2010-08-09T19:40:43.120" Score="0" ViewCount="270" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am trying to figure out what the correct notation for a multiplicative algebra should be.  I've seen a bunch of weird and conflicting ways for writing these things in the literature, but they are all either too specialized or bring in extra baggage.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To be more specific, given a ring R and an R-module, M, with a finite basis B, we can identify the elements of M with functions of the form $f, g : B \to R$.  Then we construct the &lt;em&gt;multiplicative algebra&lt;/em&gt; on M according to the following rule:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ (f \, g)(x) = f(x) g(x) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is simple:  What is the standard notation for this construction?  I've seen a bunch of weird forms.  For example, in the case where R is the field of complex numbers, we can write something like [; C^k(B) ;] for this algebra.  Or if we are dealing with matrices, we can talk about Hadamard products.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But can we write something simpler?  I would really just like to do something along the lines of say; define R(B) to be this multiplicative algebra and be done with it.  The reason for this is that if we are working with something like Pontryagin duality or representation theory, the set B may be a group/monoid/whatever, and there could be multiple algebras we defined over it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To illustrate why this is an issue, let $B = \mathbb R$ be the group of reals under addition, then we can define both a convolution algebra, $R[ \mathbb R ]$ and a multiplicative algebra $ R( \mathbb R )$ which are identified with the same R-module, but have totally different structures.  (Though they are related by Pontryagin duality in the case where $R = \mathbb C$.)&lt;/p&gt;&#xA;" OwnerUserId="986" LastEditorUserId="986" LastEditorDisplayName="" LastEditDate="2010-08-09T23:51:54.267" LastActivityDate="2011-05-22T17:00:33.073" Title="Notation for a multiplicative algebra" Tags="&lt;abstract-algebra&gt;" AnswerCount="1" CommentCount="7" FavoriteCount="1" />
  <row Id="1955" PostTypeId="1" CreationDate="2010-08-09T19:59:49.377" Score="1" ViewCount="164" Body="&lt;p&gt;I have seen at many places the notions that Lie Algebras are infinitesimal objects and they look really close at a point. But I never understood this. They are abstract algebraic objects different from rings in that they are equipped with a weird sort of product and a weird Jacobi identity. Any hints on how to make this connection to infinitesimals?&lt;/p&gt;&#xA;" OwnerUserId="987" LastActivityDate="2010-08-10T04:47:54.727" Title="Lie algebras and infinitesimals" Tags="&lt;lie-algebra&gt;" AnswerCount="3" CommentCount="3" />
  <row Id="1956" PostTypeId="1" AcceptedAnswerId="1959" CreationDate="2010-08-09T20:06:44.197" Score="13" ViewCount="692" Body="&lt;p&gt;The following integral,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \int_0^1 \frac{x^4(1-x)^4}{x^2 + 1} \mathrm{d}x = \frac{22}{7} - \pi $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is clearly positive, which proves that $\pi &amp;lt; 22/7$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a similar integral which proves $\pi &gt; 333/106$?&lt;/p&gt;&#xA;" LastActivityDate="2010-08-09T21:07:19.760" Title="Is there an integral that proves $\pi &gt; 333/106$?" Tags="&lt;calculus&gt;&lt;approximation&gt;&lt;pi&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="6" />
  <row Id="1957" PostTypeId="2" ParentId="1955" CreationDate="2010-08-09T20:11:16.883" Score="1" ViewCount="" Body="&lt;p&gt;Lie algebras are infinitesimal in the sense that they naturally arise (over $\mathbb{R}$) as tangent space at the identity of a Lie group, which is also the same as the left-invariant vector fields (infinitesimal flows) on the Lie group.&lt;/p&gt;&#xA;" OwnerUserId="176" LastActivityDate="2010-08-09T20:11:16.883" CommentCount="2" />
  <row Id="1958" PostTypeId="2" ParentId="1609" CreationDate="2010-08-09T20:47:20.013" Score="0" ViewCount="" Body="&lt;p&gt;If your CS friends are like me, they might still find the answers above a little overwhelming. So you could start as follows: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, show them the &lt;a href=&quot;http://en.wikipedia.org/wiki/Divisor&quot; rel=&quot;nofollow&quot;&gt;Wiki's divisor&lt;/a&gt; page (it always works!). Then explain to them that by the &lt;a href=&quot;http://en.wikipedia.org/wiki/Fundamental_theorem_of_arithmetic&quot; rel=&quot;nofollow&quot;&gt;fundamental theorem of arithmetic&lt;/a&gt;, any divisor is just a bunch of prime numbers with multiplicities. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next, tell them that completely similar to $\mathbb Z$, the FTA works over $\mathbb C[x]$ (which you can think of as a line). Except now that each &quot;divisor&quot; (polynomial) can be thought of as a bunch of points (roots of the polynomial) on that line, with multiplicities. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But why stop as a straight line? One can do the same thing for a (reasonably nice) curve on the plane, and a natural way to get a bunch of points is to intersect with another curve. By the way, a bunch of points &quot;divide&quot; the curve, further justifying the terminology! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Back to $\mathbb C[x]$, you can point out that the total numbers of points (with mult.) is the degree of your polynomial $p(x)$, or the dimension of the vector space $\mathbb C[x]/(p(x))$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;At this point, if they still follow you, show them Matt E's splendid answer (-: &lt;/p&gt;&#xA;" OwnerUserId="223" LastActivityDate="2010-08-09T20:47:20.013" />
  <row Id="1959" PostTypeId="2" ParentId="1956" CreationDate="2010-08-09T20:58:48.610" Score="16" ViewCount="" Body="&lt;p&gt;I think this integral would do the job&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \int\limits_{0}^{1} \frac{x^{5}(1-x)^{6}(197+462x^{2})}{530(1+x^{2})}= \pi -\frac{333}{106}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also you can refer to S.K. Lucas Integral proofs that $335/113 &gt; \pi$ Gazette Aust.math.soc  &lt;strong&gt;32&lt;/strong&gt; (2005), 263-266.&lt;/p&gt;&#xA;" OwnerUserId="9413" LastEditorUserId="9413" LastEditorDisplayName="" LastActivityDate="2010-08-09T20:58:48.610" CommentCount="3" />
  <row Id="1960" PostTypeId="2" ParentId="1956" CreationDate="2010-08-09T21:07:19.760" Score="8" ViewCount="" Body="&lt;p&gt;Although this is not exactly an answer to the question, it seems sufficiently related to mention: there are some direct generalizations, given on the &lt;a href=&quot;http://en.wikipedia.org/wiki/Proof_that_22/7_exceeds_%CF%80&quot; rel=&quot;nofollow&quot;&gt;Wikipedia page about this integral&lt;/a&gt;. For instance, $$0 &amp;lt; \frac14\int_0^1\frac{x^8(1-x)^8}{1+x^2}\ dx=\pi -\frac{47171}{15015}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, &#xA;$$\frac1{2^{2n-1}}\int_0^1 x^{4n}(1-x)^{4n}\ dx&#xA;&amp;lt;\frac1{2^{2n-2}}\int_0^1\frac{x^{4n}(1-x)^{4n}}{1+x^2}\ dx&#xA;&amp;lt;\frac1{2^{2n-2}}\int_0^1 x^{4n}(1-x)^{4n}\ dx$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which for n=1 (the integral in the question) gives slightly better bounds than just $\pi &amp;lt; 22/7$:&#xA;$$ \frac{1}{1260} &amp;lt; \frac{22}{7} - \pi &amp;lt; \frac{1}{630}$$&lt;/p&gt;&#xA;" OwnerUserId="205" LastActivityDate="2010-08-09T21:07:19.760" CommentCount="2" />
  <row Id="1961" PostTypeId="2" ParentId="1933" CreationDate="2010-08-09T21:17:32.873" Score="3" ViewCount="" Body="&lt;p&gt;Matt E's response is the canonical first answer.  As someone (else) who works with nonabelian group (in particular, Galois) cohomology frequently, let me give a second answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Certain interesting maps between &lt;em&gt;commutative&lt;/em&gt; cohomology groups are defined via a non-commutative intermediary.  The justification for this is that, though one does not in general have anything like a &quot;long exact sequence&quot; in non-commutative cohomology, if one has an extension of $\mathfrak{g}$-modules&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$1 \rightarrow Z \rightarrow G \rightarrow A \rightarrow 1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $A$ is commutative and $Z$ is &lt;strong&gt;central&lt;/strong&gt;, then one gets a connecting map in cohomology &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Delta: H^1(\mathfrak{g},A) \rightarrow H^2(\mathfrak{g},Z).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Often elements in an $H^2$ may be viewed as &quot;obstructions&quot; to something desirable happening at the $H^1$-level.  In particular, this is the case for the &lt;strong&gt;period-index obstruction map&lt;/strong&gt; in the Galois cohomology of abelian varieties.  See for instance &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://math.uga.edu/~pete/wc1.pdf&quot; rel=&quot;nofollow&quot;&gt;http://math.uga.edu/~pete/wc1.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and also publications [12], [14], [16] on&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.math.uga.edu/~pete/papers.html&quot; rel=&quot;nofollow&quot;&gt;http://www.math.uga.edu/~pete/papers.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="299" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2010-08-09T21:33:50.297" LastActivityDate="2010-08-09T21:33:50.297" />
  <row Id="1962" PostTypeId="2" ParentId="1758" CreationDate="2010-08-09T21:23:41.117" Score="0" ViewCount="" Body="&lt;p&gt;(One more attempt at a simple and complete solution.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each chord must divide the circle into the ratio 4:3, so each chord must correspond to an arc of measure approximately 2.91624.  Consider two such chords, one in a fixed location, and the other moving through all possible configurations in which it intersects the first chord.  For simplicity, let the area of the circle be 7, so that the fixed chord divides the circle into regions with area 3 and 4.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://www.imgftw.net/img/114911250.gif&quot; alt=&quot;animation of moving chord&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Note that if the chord were to continue rotating counterclockwise, what is currently the &quot;bottom&quot; of the chord would intersect the fixed chord, but any such configuration is the reflection image over the perpendicular bisector of the fixed chord of a configuration in the animation shown.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the chord rotates counterclockwise, the point of intersection of the two chords and the endpoint of the moving chord on the upper half of the circle both move solely leftward, so the area of the blue region is monotonically increasing and the area of the red region is monotonically decreasing as the chord rotates counterclockwise.  The areas of the colored regions vary between 0 and 3, and are continuous functions of the rotation of the chord.  There are 2 positions of the rotating chord for which one of the two colored regions has area 1:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://www.imgftw.net/img/284449919.png&quot; alt=&quot;1:2:1:3 configuration&quot;&gt;&#xA;&lt;img src=&quot;http://www.imgftw.net/img/843189976.png&quot; alt=&quot;2:1:2:2 configuration&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In each configuration, we can determine the areas of the regions by symmetry (exchange the moving chord and the fixed chord).  In the left diagram, the areas are (counterclockwise, starting with the blue region) 1:2:1:3.  In the right diagram, the areas are 2:1:2:2.  Since the final chord must divide 3 of the existing 4 regions in order to have 7 regions, the configuration with areas 1:2:1:3 (above left) won't work, since dividing either region with area 1 won't give us 7 regions with area 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, working with the configuration with areas 2:1:2:2 (above right), add a third chord, rotating through all possible positions that divide the lower right region.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://www.imgftw.net/img/936642181.gif&quot; alt=&quot;adding the third chord&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As before, the area of the purple region is monotone and continuous as the chord rotates and is between 0 and 2, so there is a unique configuration where the area of the purple region is 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://www.imgftw.net/img/948982901.png&quot; alt=&quot;purple region has area 1&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Considering the position of the purple chord relative to the original fixed chord, the small wedge region at the right cannot have area 1, and this can be verified computationally.  (It is also the case that the area of the triangle in the center cannot have are 1 in this configuration.)  So, it is not possible to divide the circle into 7 regions of equal area using 3 chords.&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-08-09T21:23:41.117" CommentCount="3" />
  <row Id="1965" PostTypeId="2" ParentId="1955" CreationDate="2010-08-09T21:47:59.633" Score="3" ViewCount="" Body="&lt;p&gt;This is all carefully explained in Chapter 8 of Fulton and Harris.  The key fact is that if $G$ is a connected Lie group, then $G$ is generated by the elements in any neighborhood of the identity.  This implies that a morphism out of $G$ is determined by what it does to elements arbitrarily close to the identity (and it is a general principle in category theory that an object is determined by the morphisms out of it).  Since we have a tangent space at the identity, we can say even more: it turns out that a morphism $f : G \to H$ is determined by its differential $df : T_e G \to T_e H$, where $T_e$ is the tangent space at the identity, otherwise known as the Lie algebra.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This differential is just a linear map between finite-dimensional vector spaces, so it's much easier to handle than the original map $f$.  The problem is then to characterize which linear maps can occur.  As a necessary condition, $f$ must preserve the Lie bracket on $T_e G$, and if $G$ is simply connected this is both necessary and sufficient.  So essentially the whole point of the wacky definition of a Lie algebra is to make this theorem true.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means, roughly speaking, that Lie algebras capture the local, or infinitesimal, structure of a Lie group.  The Lie algebra of a Lie group can't capture the global topological structure, but being able to separate out the easy part and the hard part of understanding a Lie group is very valuable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(The connection to Charles' answer is that a tangent vector at the identity determines, by translation, a left-invariant vector field on $G$.  You can think of such a vector field as specifying, at each element of $G$, a direction in which something can flow.)&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-08-09T21:47:59.633" CommentCount="1" />
  <row Id="1966" PostTypeId="1" AcceptedAnswerId="1968" CreationDate="2010-08-09T21:49:35.953" Score="0" ViewCount="174" Body="&lt;p&gt;Can every compact metric space be realized as the continuous image of a cantor set?&lt;/p&gt;&#xA;" LastActivityDate="2010-08-09T21:54:56.707" Title="Cantor Set and Compact Metric Spaces" Tags="&lt;real-analysis&gt;" AnswerCount="2" />
  <row Id="1967" PostTypeId="2" ParentId="1955" CreationDate="2010-08-09T21:51:18.123" Score="4" ViewCount="" Body="&lt;p&gt;The direct infinitesimal analogue of a given Lie group is not the Lie algebra with its bracket operation and Jacobi identity, but the Lie algebra (thought of as just the space of tangent vectors at the identity, without the added structure of a bracket operation) with addition of elements being the group operation.  Addition is commutative but the group generally is not. To capture the noncommutativity you need to squeeze additional information from the group down to the infinitesimal level of the Lie algebra (by taking second-order information in the series expansion of group elements near the identity; the tangent space is first-order).&#xA;The Lie multiplication $[x,y]$ is the second-order infinitesimal analogue of the commutator $x y x^{-1} y^{-1}$, and the Jacobi identity is an analogue of an identity for the commutator.  Historically, the Jacobi identity for algebras (that is, for Lie algebras whose bracket is $XY - YX$ in an associative algebra) must have come first, and is used mainly in algebras, but you can think of it as coming from the group.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Locally, second-order information is enough: the Lie algebra determines the structure of the group, up to some questions of a different, &quot;global&quot; nature (topology) about connectivity and covering spaces. &lt;/p&gt;&#xA;" OwnerUserId="467" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-08-10T04:47:54.727" LastActivityDate="2010-08-10T04:47:54.727" />
  <row Id="1968" PostTypeId="2" ParentId="1966" CreationDate="2010-08-09T21:54:41.607" Score="4" ViewCount="" Body="&lt;p&gt;Yes (assuming it's nonempty, of course). Moreover if you google &quot;continuous image of the Cantor set&quot;, the first hit takes you to &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Cantor_set&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Cantor_set&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where you can read that this theorem is true and a reference is given to Willard's &lt;em&gt;General Topology&lt;/em&gt;.  (The article does not say this and perhaps it should, but it is specifically Theorem 30.7 on p. 217 of the Dover edition.)  &lt;/p&gt;&#xA;" OwnerUserId="299" LastActivityDate="2010-08-09T21:54:41.607" CommentCount="1" />
  <row Id="1969" PostTypeId="1" AcceptedAnswerId="1976" CreationDate="2010-08-09T21:54:55.473" Score="3" ViewCount="640" Body="&lt;p&gt;If i were to ask this: Which quote by a &quot;Mathematician&quot; do you like, then what would be your answer!&lt;/p&gt;&#xA;" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-08-09T23:01:30.400" LastActivityDate="2010-08-10T11:20:54.510" CommunityOwnedDate="2010-08-09T21:56:50.373" ClosedDate="2010-08-10T14:53:14.440" Title="Best Quotes by a Mathematician" Tags="&lt;soft-question&gt;&lt;big-list&gt;" AnswerCount="12" CommentCount="8" FavoriteCount="3" />
  <row Id="1970" PostTypeId="2" ParentId="1966" CreationDate="2010-08-09T21:54:56.707" Score="5" ViewCount="" Body="&lt;p&gt;Yes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://mathoverflow.net/questions/5357/theorems-for-nothing-and-the-proofs-for-free/5388#5388&quot; rel=&quot;nofollow&quot;&gt;http://mathoverflow.net/questions/5357/theorems-for-nothing-and-the-proofs-for-free/5388#5388&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And the comment of Harald-Hanche Olsen.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Surprising, yes, but once you know about it, it seems easy enough to cook up a proof. Just write the set as a union of two closed subsets, decide to map the left half of the Cantor set onto one and the right half to the other, then do the same to each of these two sets, and so on. In the limit you have the map you want, provided you have arranged for the diameters of the parts to go to zero.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="218" LastActivityDate="2010-08-09T21:54:56.707" />
  <row Id="1971" PostTypeId="2" ParentId="1969" CreationDate="2010-08-09T21:56:26.007" Score="2" ViewCount="" Body="&lt;p&gt;For me this is one which i like a lot:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Perfect numbers like perfect men are very rare.&quot; Rene Descartes&lt;/p&gt;&#xA;" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-08-09T21:57:03.183" LastActivityDate="2010-08-09T21:57:03.183" CommunityOwnedDate="2010-08-09T21:57:03.183" />
  <row Id="1972" PostTypeId="2" ParentId="1969" CreationDate="2010-08-09T22:06:01.067" Score="4" ViewCount="" Body="&lt;blockquote&gt;Mathematics is one of the few disciplines that teaches us about the power of thought as distinct from the power of authority.&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This appears on my university's &lt;a href=&quot;http://math.gmu.edu/&quot; rel=&quot;nofollow&quot;&gt;department website&lt;/a&gt;, many of us attribute it to our late chair Klaus Fischer.  &lt;/p&gt;&#xA;" OwnerUserId="121" LastActivityDate="2010-08-09T22:06:01.067" CommunityOwnedDate="2010-08-09T22:06:01.067" CommentCount="1" />
  <row Id="1973" PostTypeId="2" ParentId="1932" CreationDate="2010-08-09T22:23:55.967" Score="2" ViewCount="" Body="&lt;p&gt;The following are unrelated concepts:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Finitism&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The &quot;intuitionistic continuum&quot; (does that mean Weyl's book? the real numbers as defined in Bishop's book on constructive analysis?)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Computable real numbers&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;For example, you can read the definition of computable real numbers under a classical or an &quot;intuitionistic&quot; interpretation of the words, and they will differ as to whether &quot;1 if Goldbach conjecture is true and 2 if it's false&quot; is a computable real number.  But the theory of computable reals will make sense in both environments.&lt;/p&gt;&#xA;" OwnerUserId="467" LastActivityDate="2010-08-09T22:23:55.967" />
  <row Id="1974" PostTypeId="2" ParentId="1969" CreationDate="2010-08-09T22:23:58.870" Score="6" ViewCount="" Body="&lt;p&gt;I'd probably go with this nice one from Richard Askey:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If things are nice there is probably a&#xA;  good reason why they are nice: and if&#xA;  you do not know at least one reason&#xA;  for this good fortune, then you still&#xA;  have work to do.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-08-09T22:23:58.870" CommunityOwnedDate="2010-08-09T22:23:58.870" />
  <row Id="1975" PostTypeId="2" ParentId="1946" CreationDate="2010-08-09T22:39:04.460" Score="3" ViewCount="" Body="&lt;p&gt;Something is wrong with this formula. It leads to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$e = 1 + 1 + \frac{e^{\frac{1}{3}}}{2} + e^\theta 8 &gt; e$.&lt;/p&gt;&#xA;" OwnerUserId="370" LastActivityDate="2010-08-09T22:39:04.460" CommentCount="3" />
  <row Id="1976" PostTypeId="2" ParentId="1969" CreationDate="2010-08-09T22:39:30.577" Score="7" ViewCount="" Body="&lt;p&gt;&quot;If only I had the theorems! Then I should find the proofs easily enough.&quot; - Riemann&lt;/p&gt;&#xA;" OwnerUserId="956" LastActivityDate="2010-08-09T22:39:30.577" CommunityOwnedDate="2010-08-09T22:39:30.577" CommentCount="1" />
  <row Id="1977" PostTypeId="2" ParentId="29" CreationDate="2010-08-09T22:48:00.983" Score="1" ViewCount="" Body="&lt;p&gt;I tend to use &lt;em&gt;Mathematica&lt;/em&gt; heavily for its symbolic capabilities, but I have a soft spot for &lt;a href=&quot;http://www.gnu.org/software/octave/&quot; rel=&quot;nofollow&quot;&gt;Octave&lt;/a&gt;, an open-source MATLAB clone. For simple number crunching, it works nicely.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-08-09T22:48:00.983" CommunityOwnedDate="2010-08-09T22:48:00.983" />
  <row Id="1978" PostTypeId="2" ParentId="1935" CreationDate="2010-08-09T22:52:09.540" Score="3" ViewCount="" Body="&lt;p&gt;Below is another method for irreducibility testing - excerpted from one of my old sci.math posts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In 1918 Stackel published the following simple observation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;THEOREM&lt;/strong&gt;  If $ p(x) $ is a composite integer coefficient polynomial&lt;/p&gt;&#xA;&#xA;&lt;p&gt;then  $ p(n) $ is  composite for all $|n| &gt; B $, for some bound $B$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;in fact $ p(n) $ has at most $ 2d $ prime values, where  $ d = {\rm deg}(p)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The simple proof can be found online in &lt;a href=&quot;http://web.math.fsu.edu/~aluffi/archive/paper134.ps&quot; rel=&quot;nofollow&quot;&gt;Mott &amp;amp; Rose [3]&lt;/a&gt;, p. 8.&#xA;I highly recommend this delightful and stimulating 27 page paper&#xA;which discusses prime-producing polynomials and related topics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Contrapositively, $ p(x) $ is prime (irreducible) if it assumes a prime value&#xA;for large enough $ |x| $. Conversely Bouniakowski conjectured (1857)&#xA;that prime $ p(x) $ assume infinitely many prime values (except in trivial&#xA;cases where the values of $p$ have an obvious common divisor, e.g. $ 2 | x(x+1)+2$ ).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an example, Polya-Szego popularized A. Cohn's irreduciblity test, which&#xA;states that $ p(x) \in {\mathbb Z}[x]$ is prime if  $ p(b) $&#xA;yields a prime in radix $b$ representation (so necessarily $0 \le p_i &amp;lt; b$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example $f(x) = x^4 + 6 x^2 + 1 \pmod p$ factors for all primes $p$,&#xA;yet  $f(x)$ is prime since  $f(8) = 10601$ octal $= 4481$ is prime.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: Cohn's test fails if, in radix $b$, negative digits are allowed, e.g.&#xA;$f(x) = x^3 - 9 x^2 + x-9 = (x-9)(x^2 + 1)$ but $f(10) = 101$ is prime.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For further discussion see my prior &lt;a href=&quot;http://groups.google.com/groups?selm=y8zvg4m9yhm.fsf%40nestle.ai.mit.edu&quot; rel=&quot;nofollow&quot;&gt;post [1]&lt;/a&gt;, along with Murty's online &lt;a href=&quot;http://www.mast.queensu.ca/~murty/polya4.dvi&quot; rel=&quot;nofollow&quot;&gt;paper [2]&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[1] Dubuque, sci.math 2002-11-12, On prime producing polynomials&lt;br&gt;&#xA;&lt;a href=&quot;http://groups.google.com/groups?selm=y8zvg4m9yhm.fsf%40nestle.ai.mit.edu&quot; rel=&quot;nofollow&quot;&gt;http://groups.google.com/groups?selm=y8zvg4m9yhm.fsf%40nestle.ai.mit.edu&lt;/a&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;[2] Murty, Ram.  Prime numbers and irreducible polynomials.&lt;br&gt;&#xA;Amer. Math. Monthly, Vol. 109 (2002), no. 5, 452-458.&lt;br&gt;&#xA;&lt;a href=&quot;http://www.mast.queensu.ca/~murty/polya4.dvi&quot; rel=&quot;nofollow&quot;&gt;http://www.mast.queensu.ca/~murty/polya4.dvi&lt;/a&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;[3] Mott, Joe L.; Rose, Kermit.&#xA;Prime producing cubic polynomials&lt;br&gt;&#xA;Ideal theoretic methods in commutative algebra, 281-317.&lt;br&gt;&#xA;Lecture Notes in Pure and Appl. Math., 220, Dekker, New York, 2001.&lt;br&gt;&#xA;&lt;a href=&quot;http://web.math.fsu.edu/~aluffi/archive/paper134.ps&quot; rel=&quot;nofollow&quot;&gt;http://web.math.fsu.edu/~aluffi/archive/paper134.ps&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-08-09T22:52:09.540" />
  <row Id="1979" PostTypeId="2" ParentId="1918" CreationDate="2010-08-09T22:58:18.237" Score="2" ViewCount="" Body="&lt;p&gt;One of the consequences of &lt;a href=&quot;http://en.wikipedia.org/wiki/G%C3%B6del%27s_incompleteness_theorems&quot; rel=&quot;nofollow&quot;&gt;Goedel's incompleteness theorem&lt;/a&gt; is that if $T$ is a finitely axiomatizable theory of arithmetic, then &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$T$ proves that $T$ is consistent&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;if and only if &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$T$ is inconsistent!&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The reason is that an inconsistent theory proves anything, and a consistent theory never proves its own consistency.&lt;/p&gt;&#xA;" OwnerUserId="413" LastEditorUserId="413" LastEditorDisplayName="" LastEditDate="2010-08-09T23:03:19.977" LastActivityDate="2010-08-09T23:03:19.977" CommunityOwnedDate="2010-08-09T22:58:18.237" />
  <row Id="1980" PostTypeId="2" ParentId="1942" CreationDate="2010-08-09T23:22:24.327" Score="0" ViewCount="" Body="&lt;p&gt;This is an example of a Constraint Satisfaction Problem.  If you have more objects&#xA;than bins, you can sometimes treat the bins individually (as a poster did with the case of one bin) and multiply the counts together, or use the individual counts for each bin in some similar fashion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another way is to do some form of divide and conquer; for example, suppose, among all the other constraints and attributes that are&#xA;present, all the objects belong to exactly one of five different classes.  Then an attempt at counting can begin with enumerating all the possibilities for partially filling bins with class 1 items.  Each possibility leads to ways to enumerating the wayss to fill the remaining space fully or partially with class 2 items, and so on.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another way to process things is to bundle certain items together: fries, a burger, and a drink can (setting certain things like nutrition aside) be called a meal, and the constraints may be that enumeration is simplified if an attempt is made to pack the bins with bundled groups of items.  However, you will have a hard time getting nice lower and upper bounds that are close if you have any reasonable complexity in your constraints.&#xA;If your constraints are such that there are very few ways to pack some of the&#xA;bins, then a depth first search approach which packs those bins first may give you a&#xA;good handle on approximate counts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Besides Constraint Satisfaction Problem, you might try Polya Enumeration, reduction to one of a number of basic counting problems, Inclusion-Exclusion formula, and topics in enumerative combinatorics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Good Luck.&lt;/p&gt;&#xA;" OwnerUserId="297" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-08-11T22:00:11.167" LastActivityDate="2010-08-11T22:00:11.167" CommentCount="0" />
  <row Id="1981" PostTypeId="2" ParentId="1892" CreationDate="2010-08-09T23:27:08.967" Score="4" ViewCount="" Body="&lt;p&gt;If $P = NP$, computational revolution (once a specific algorithm is identified for an NP-hard problem, with explicit asymptotic runtime bounds).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $P &amp;lt; NP$ &lt;em&gt;and one can prove it&lt;/em&gt;, secure (classical) cryptography provably exists, and a huge missing piece in our understanding of computation is filled in.  The first already has significant implications for daily life, and developing the second would have much larger implications.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You should also understand that after 40 years of research, today P=NP carries a host of related ideas like: easy-to-hard phase transition in combinatorial problems; quantifiable boundaries between easy and hard approximate versions of specific NP-complete problems (so getting within 7/8 of the optimal solution is easy but anything closer is NP-complete); counting and randomly sampling combinatorial objects are the same problem; zero-knowledge proofs &quot;that reveal nothing but their own validity&quot; (unforgeable ID cards).  It's a very rich universe of ideas and it doesn't run out of questions once you know the answer to P=NP.&lt;/p&gt;&#xA;" OwnerUserId="467" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-08-09T23:41:38.657" LastActivityDate="2010-08-09T23:41:38.657" CommentCount="2" />
  <row Id="1982" PostTypeId="2" ParentId="1969" CreationDate="2010-08-10T00:09:56.260" Score="5" ViewCount="" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;em&gt;Cubum autem in duos cubos, aut quadratoquadratum in duos quadratoquadratos, et generaliter nullam in infinitum ultra quadratum potestatem in duos eiusdem nominis fas est dividere cuius rei demonstrationem mirabilem sane detexi. Hanc marginis exiguitas non caperet.&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Pierre de Fermat&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;em&gt;I have discovered a truly marvelous proof that it is impossible to separate a cube into two cubes, or a fourth power into two fourth powers, or in general, any power higher than the second into two like powers. This margin is too narrow to contain it.&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Ref: &lt;a href=&quot;http://en.wikipedia.org/wiki/Fermat%27s_Last_Theorem&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Fermat%27s_Last_Theorem&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="918" LastActivityDate="2010-08-10T00:09:56.260" CommunityOwnedDate="2010-08-10T00:09:56.260" CommentCount="1" />
  <row Id="1983" PostTypeId="1" CreationDate="2010-08-10T00:17:32.873" Score="3" ViewCount="118" Body="&lt;p&gt;The &lt;a href=&quot;http://eom.springer.de/p/p072630.htm&quot; rel=&quot;nofollow&quot;&gt;Phragmen-Lindelöf theorem&lt;/a&gt; gives a consequence of the Riemann hypothesis, viz, the &lt;a href=&quot;http://en.wikipedia.org/wiki/Lindel%C3%B6f_hypothesis&quot; rel=&quot;nofollow&quot;&gt;Lindelöf hypothesis&lt;/a&gt;. As such this is weaker than Riemann hypothesis; but it is still considered that even a proof of this weaker result will be a breakthrough.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Question:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What is the strongest known result yet on the Lindelöf hypothesis?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="977" LastActivityDate="2010-08-10T07:54:41.083" Title="Latest known result on Lindelöf hypothesis" Tags="&lt;number-theory&gt;&lt;complex-analysis&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="1984" PostTypeId="1" CreationDate="2010-08-10T00:21:00.137" Score="7" ViewCount="860" Body="&lt;p&gt;Hello. I am A Neophyte Amateur Mathematician. I have beed reading a lot about Journals and the topic of peer-review in Mathematics Journals. Does one have to have Professional crediendtials or have a Doctorate in order to Publish in Peer-review Mathematics Jourals or just the desire to competenyly solve Mathematical problems?  &lt;/p&gt;&#xA;" OwnerUserId="993" LastEditorUserId="766" LastEditorDisplayName="" LastEditDate="2011-01-09T19:19:21.650" LastActivityDate="2011-01-09T19:19:21.650" Title="Does a Person Need a Mathematics Degree in order to Publish in a Mathematics jounal?" Tags="&lt;soft-question&gt;" AnswerCount="9" CommentCount="2" FavoriteCount="8" />
  <row Id="1985" PostTypeId="1" CreationDate="2010-08-10T00:24:04.763" Score="13" ViewCount="893" Body="&lt;p&gt;What software do you use to &lt;strong&gt;accurately draw geometry diagrams&lt;/strong&gt;?&lt;/p&gt;&#xA;" OwnerUserId="215" LastActivityDate="2011-01-18T07:25:55.923" CommunityOwnedDate="2010-08-10T00:24:04.763" Title="Software for drawing geometry diagrams" Tags="&lt;geometry&gt;&lt;big-list&gt;&lt;math-software&gt;" AnswerCount="15" FavoriteCount="10" />
  <row Id="1986" PostTypeId="2" ParentId="1984" CreationDate="2010-08-10T00:31:50.300" Score="25" ViewCount="" Body="&lt;p&gt;No, no sort of degree or specific credentials are required.  It is far from unheard of that someone publishes one or more papers while still an undergraduate.  There are even a few instances of important work being published by high school students.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are an uncredentialled mathematician you should take extra pains to make sure that your work passes superficial tests for looking serious and competent.  Off the top of my head, this would include the following:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Make sure that your grammar and spelling is virtually flawless (in particular, you have some spelling mistakes in your post, which is probably not such a big deal for this site, but would create a bad impression in a journal submission).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Use some form of TeX to format your document.  (This is freely available online and just takes a little practice to become reasonably proficient.  Not to do so is a big warning sign of outsider status.)&lt;sup&gt;[1]&lt;/sup&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Make sure that your work contains a bibliography which gives due acknowledgment to previous work in the field.  Wherever possible, you should try to cite primary sources (i.e., journal papers) or standard texts, and you should try to avoid citing things like wikipedia and popular books.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;[1] I am aware that there are a few leading mathematicians who do not use TeX.  The geometer Cliff Taubes is a popular example.  However, Professor Taubes began his mathematical career before TeX became widespread and, in any event, does not need any help in getting his work taken seriously.  I am not saying that someone who does not follow the above precepts is not a serious mathematician, just that they look less like one at a first impression.  &lt;/p&gt;&#xA;" OwnerUserId="299" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2011-01-09T08:33:54.067" LastActivityDate="2011-01-09T08:33:54.067" CommentCount="2" />
  <row Id="1987" PostTypeId="2" ParentId="1985" CreationDate="2010-08-10T00:36:20.663" Score="2" ViewCount="" Body="&lt;p&gt;One of the best is &lt;a href=&quot;http://poincare.matf.bg.ac.rs/~janicic//gclc/&quot; rel=&quot;nofollow&quot;&gt;WinGCLC&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Graph quality is really exceptional, it's easy to use, can be easily exported to SVG/EPS/BMP/LaTeX, supports command line for scripting, supports animation...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is really useful is that once you define all the points/lines/intersections/circles, you can easily change the location of one of the starting objects and the whole diagram will be accurately redrawn.&lt;/p&gt;&#xA;" OwnerUserId="956" LastActivityDate="2010-08-10T00:36:20.663" CommunityOwnedDate="2010-08-10T00:36:20.663" />
  <row Id="1988" PostTypeId="2" ParentId="1984" CreationDate="2010-08-10T00:38:35.027" Score="7" ViewCount="" Body="&lt;p&gt;Pete Clark has given an excellent answer. I just want to add two more things:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Try to discuss with trustworthy professional mathematicians before publication who will be sympathetic and honorable.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;First publish in a website like the arXiv. It is free-for-all, and you will get to know in short order if your paper is really interesting.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="977" LastActivityDate="2010-08-10T00:38:35.027" />
  <row Id="1989" PostTypeId="1" CreationDate="2010-08-10T00:38:45.760" Score="1" ViewCount="256" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Morera%27s_theorem&quot; rel=&quot;nofollow&quot;&gt;Morera's theorem&lt;/a&gt; is an underappreciated theorem in complex analysis. I have been struck by the simplicity of its proof and some clever applications of it and I had been interested in finding out more of such. Please contribute examples. One example is the Weierstrass theorem that if a sequence of holomorphic functions converge absolutely and uniformly on every compact subset in a domain, then the limit is also holomorphic. And there are numerous applications of this latter fact.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, please come ahead and contribute clever and slick applications of Morera's theorem that will impress people!&lt;/p&gt;&#xA;" OwnerUserId="977" LastActivityDate="2010-08-10T12:55:38.543" Title="Striking applications of Morera's theorem" Tags="&lt;complex-analysis&gt;" AnswerCount="2" />
  <row Id="1990" PostTypeId="2" ParentId="1969" CreationDate="2010-08-10T00:39:57.677" Score="2" ViewCount="" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;em&gt;There is no branch of mathematics, however abstract, which may not some&#xA;  day be applied to phenomena of the&#xA;  real world.&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www-groups.dcs.st-and.ac.uk/~history/Quotations/Lobachevsky.html&quot; rel=&quot;nofollow&quot;&gt;Nikolai Ivanovich Lobachevsky&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="752" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2010-08-10T00:49:33.940" LastActivityDate="2010-08-10T00:49:33.940" CommunityOwnedDate="2010-08-10T00:39:57.677" />
  <row Id="1991" PostTypeId="2" ParentId="1985" CreationDate="2010-08-10T00:42:42.320" Score="5" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://asymptote.sourceforge.net/&quot; rel=&quot;nofollow&quot;&gt;Asymptote&lt;/a&gt; is my tool of choice for virtually any mathematics diagram, geometry included.  &lt;/p&gt;&#xA;" OwnerUserId="876" LastActivityDate="2010-08-10T00:42:42.320" CommunityOwnedDate="2010-08-10T00:42:42.320" />
  <row Id="1992" PostTypeId="2" ParentId="1969" CreationDate="2010-08-10T00:43:54.017" Score="7" ViewCount="" Body="&lt;p&gt;&quot;It's like if you want to be a good pianist, &#xA;you have to do a lot of scales and a lot of practice, &#xA;and a lot of that is kind of boring, it's work.&#xA;But you need to do that before you can really be very expressive and really play beautiful music.&#xA;You have to go through that phase of practice and drill.&quot;&#xA;- Terry Tao&lt;/p&gt;&#xA;" OwnerUserId="249" LastActivityDate="2010-08-10T00:43:54.017" CommunityOwnedDate="2010-08-10T00:43:54.017" />
  <row Id="1993" PostTypeId="2" ParentId="1985" CreationDate="2010-08-10T00:46:58.173" Score="7" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://www.dynamicgeometry.com/&quot; rel=&quot;nofollow&quot;&gt;Geometer's Sketchpad&lt;/a&gt;&amp;mdash; Sketchpad 5 exports/copy-pastes diagrams much better than 4 did.&lt;/p&gt;&#xA;" OwnerUserId="72" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-08-10T01:55:27.060" LastActivityDate="2010-08-10T01:55:27.060" CommunityOwnedDate="2010-08-10T00:46:58.173" />
  <row Id="1994" PostTypeId="2" ParentId="1984" CreationDate="2010-08-10T00:46:58.783" Score="15" ViewCount="" Body="&lt;p&gt;There is usually no technical requirement for any particular credentials, and it seems unlikely to me that editors and referees will check for this. The usual publication process is that you submit a paper to a journal editor, who then forwards it one or more competent referees, who make a recommendations on the suitability for publication. Often, the referees make suggestions for improving the paper, whether or not it is found acceptable for publication.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nevertheless, an amateur mathematician will find him or herself at a disadvantage in several respects of this process. First, for someone with less experience in the professional mathematical community, it may be more difficult to recognize which accomplishments actually merit publication, and editors sometimes receive articles submitted in good faith from amateurs, which are trivial or seriously wrong in some respect. Second, even when the result is correct and worthwhile, if the presentation of the paper deviates from the accepted norms, it may be found wanting. For example, it is more difficult for an amateur to know which topics need more careful explaining in a paper and which do not, and inappropriate decisions in such cases can hurt the reception of the paper. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I see (as I write this) that others have now given some concrete advice. The most important advice I can give is to take the suggestions of the editors and referees seriously. If someone claimes that part of your paper is unclear or wrong, then you should try to understand exactly why they thought so; doing so will inevitably lead you to a better paper.&lt;/p&gt;&#xA;" OwnerUserId="413" LastEditorUserId="413" LastEditorDisplayName="" LastEditDate="2010-08-10T00:52:13.483" LastActivityDate="2010-08-10T00:52:13.483" CommentCount="2" />
  <row Id="1995" PostTypeId="2" ParentId="1985" CreationDate="2010-08-10T00:48:03.897" Score="2" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://www.adobe.com/products/illustrator/&quot; rel=&quot;nofollow&quot;&gt;Adobe Illustrator&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-08-10T00:48:03.897" CommunityOwnedDate="2010-08-10T00:48:03.897" />
  <row Id="1996" PostTypeId="2" ParentId="1985" CreationDate="2010-08-10T00:54:08.483" Score="1" ViewCount="" Body="&lt;p&gt;If you are a mathematician and wants to use such pictures for your notes or presentations using TeX files, use the pstricks package enhanced for use of easy by the package &lt;a href=&quot;http://latexdraw.sourceforge.net/&quot; rel=&quot;nofollow&quot;&gt;LaTeXdraw&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then there is a very useful linux program called &lt;a href=&quot;http://www.google.com/url?sa=t&amp;amp;source=web&amp;amp;cd=1&amp;amp;ved=0CBwQFjAA&amp;amp;url=http%3A%2F%2Fwww.xfig.org%2F&amp;amp;ei=JqNgTLDDO5WHOK-gwb0J&amp;amp;usg=AFQjCNHK2HkYQJAR_5YssWZqz9QguYesng&quot; rel=&quot;nofollow&quot;&gt;xfig&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="977" LastActivityDate="2010-08-10T00:54:08.483" CommunityOwnedDate="2010-08-10T00:54:08.483" />
  <row Id="1997" PostTypeId="2" ParentId="1948" CreationDate="2010-08-10T00:57:12.550" Score="1" ViewCount="" Body="&lt;p&gt;Since you seem to be in Chennai, walk into any mathematics library and pick up the collected works of Ramanujan. That is the best. Or look into the notes edited by Bruce Berndt.&lt;/p&gt;&#xA;" OwnerUserId="977" LastActivityDate="2010-08-10T00:57:12.550" CommentCount="0" />
  <row Id="1998" PostTypeId="2" ParentId="1985" CreationDate="2010-08-10T01:00:54.523" Score="0" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://latexdraw.sourceforge.net/&quot; rel=&quot;nofollow&quot;&gt;LaTeXDraw&lt;/a&gt; is a graphical front end for &lt;a href=&quot;http://en.wikipedia.org/wiki/PSTricks&quot; rel=&quot;nofollow&quot;&gt;PSTricks&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="207" LastActivityDate="2010-08-10T01:00:54.523" CommunityOwnedDate="2010-08-10T01:00:54.523" CommentCount="0" />
  <row Id="1999" PostTypeId="2" ParentId="1984" CreationDate="2010-08-10T01:01:45.330" Score="4" ViewCount="" Body="&lt;p&gt;I've never published a paper, so I'm the wrong person to answer your question; this answer is really more of a long comment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A few years ago, I had worked on something independently that I thought I'd try to get published. Looking back, it wasn't very interesting--basically an attempt to generalize Riemann integration to an abstract setting---but I tried submitting it to the Proceedings of the AMS. It was rejected, but I got about two pages of helpful and interesting comments from a referee (suggesting that I try a non-researchish journal; I never actually had the time to do that though) as well as kind words of advice from the person sending the email. So it was a good experience in the long run. I also got practice in writing concise mathematics and with LaTeX (though I was fairly fluent by that point).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, the point is that even if it gets rejected (which it won't necessarily), you'll probably learn something about mathematical life from the experience. And, besides, you can put it on the arXiv if someone &quot;endorses&quot; you.&lt;/p&gt;&#xA;" OwnerUserId="536" LastActivityDate="2010-08-10T01:01:45.330" />
  <row Id="2000" PostTypeId="2" ParentId="1892" CreationDate="2010-08-10T01:04:04.810" Score="0" ViewCount="" Body="&lt;p&gt;There is an interesting heuristic to suggest that P is actually not NP. It is that, roughly, the task of finding out a proof of a statement is an NP task, but that of verifying it is a P task. From our actual experience that verifying a proof is far easier than finding one, we can intuitively expect P != NP to hold true.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The practical application is that a result which would agree with our intuition would be a great thing, and psychologically satisfying moreover.&lt;/p&gt;&#xA;" OwnerUserId="977" LastActivityDate="2010-08-10T01:04:04.810" CommentCount="1" />
  <row Id="2001" PostTypeId="1" CreationDate="2010-08-10T01:10:21.193" Score="2" ViewCount="221" Body="&lt;p&gt;What is a simple example, without getting into the mess of triangulated categories, of an additive category that is not abelian?&lt;/p&gt;&#xA;" LastActivityDate="2010-08-10T20:42:55.163" Title="Additive category that is not abelian" Tags="&lt;category-theory&gt;" AnswerCount="4" FavoriteCount="3" />
  <row Id="2002" PostTypeId="2" ParentId="1985" CreationDate="2010-08-10T01:13:06.020" Score="7" ViewCount="" Body="&lt;p&gt;For geometry I've always used &lt;a href=&quot;http://www.geogebra.org/cms/&quot; rel=&quot;nofollow&quot;&gt;Geogebra&lt;/a&gt;, and I think it's pretty good.&lt;/p&gt;&#xA;" OwnerUserId="446" LastActivityDate="2010-08-10T01:13:06.020" CommunityOwnedDate="2010-08-10T01:13:06.020" />
  <row Id="2003" PostTypeId="2" ParentId="2001" CreationDate="2010-08-10T01:14:56.330" Score="3" ViewCount="" Body="&lt;p&gt;The category of finitely generated modules over a non-Noetherian ring.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The category of filtered modules over a ring is an example given in Gelfand-Manin.&lt;/p&gt;&#xA;" OwnerUserId="977" LastEditorUserId="977" LastEditorDisplayName="" LastEditDate="2010-08-10T10:54:52.823" LastActivityDate="2010-08-10T10:54:52.823" CommentCount="6" />
  <row Id="2004" PostTypeId="2" ParentId="1984" CreationDate="2010-08-10T01:16:06.310" Score="8" ViewCount="" Body="&lt;p&gt;Implicit in the previous answers: &lt;em&gt;do your best to pick an appropriate journal&lt;/em&gt;. Even professionals don't always know whether a particular journal will publish their paper (even if the results are perfectly correct). There are too many factors to count that can affect the editors' decision. For an amateur, the choice can be even more opaque. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Graduate students often rely on advisors to guide them in picking the right journals for their first papers. If you have academic contacts, they may be willing to give you advice about possible journals. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There has been a heavy emphasis on undergraduate mathematics research recently, at least in the United States. Several journals have sprung up to publish papers written by students. If this applies to you, search for &quot;undergraduate mathematics journal&quot; on Google to find several possibilities. &lt;/p&gt;&#xA;" OwnerUserId="630" LastActivityDate="2010-08-10T01:16:06.310" />
  <row Id="2005" PostTypeId="2" ParentId="1969" CreationDate="2010-08-10T01:23:00.960" Score="2" ViewCount="" Body="&lt;p&gt;In translation from German:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;We must know, we will know.&quot; - David Hilbert&lt;/p&gt;&#xA;" OwnerUserId="176" LastActivityDate="2010-08-10T01:23:00.960" CommunityOwnedDate="2010-08-10T01:23:00.960" />
  <row Id="2006" PostTypeId="2" ParentId="1969" CreationDate="2010-08-10T01:24:34.887" Score="3" ViewCount="" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;The only advice we offer to new researchers is read! read! read! many papers with complete proofs.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;-- Iwaniec &amp;amp; Kowalski.&lt;/p&gt;&#xA;" OwnerUserId="977" LastActivityDate="2010-08-10T01:24:34.887" CommunityOwnedDate="2010-08-10T01:24:34.887" />
  <row Id="2007" PostTypeId="2" ParentId="1984" CreationDate="2010-08-10T01:44:55.617" Score="0" ViewCount="" Body="&lt;p&gt;(A bit too long for comments, so...)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I may: /. recently discussed something of this sort: &lt;a href=&quot;http://ask.slashdot.org/story/10/06/24/1739247&quot; rel=&quot;nofollow&quot;&gt;http://ask.slashdot.org/story/10/06/24/1739247&lt;/a&gt; . Apparently, most journals don't reject submissions just because the submitter has no degree to speak of. Good luck! Make sure however that you have already done a thorough review of the related literature to make sure you are not merely duplicating anybody's results.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-08-10T01:44:55.617" />
  <row Id="2008" PostTypeId="2" ParentId="1985" CreationDate="2010-08-10T01:48:31.973" Score="0" ViewCount="" Body="&lt;p&gt;I've used &lt;a href=&quot;http://www.satimage.fr/software/en/downloads/index.html&quot; rel=&quot;nofollow&quot;&gt;Smile by Satimage&lt;/a&gt; on my Mac. (Not SmileLab, just the free Smile. And not to be confused with PDF utility company &lt;a href=&quot;http://www.smileonmymac.com/&quot; rel=&quot;nofollow&quot;&gt;&quot;Smile on My Mac&quot;&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Smile uses AppleScript for drawing. Some &lt;a href=&quot;http://www.satimage.fr/software/en/smile/graphics/graphics_samples.html&quot; rel=&quot;nofollow&quot;&gt;example files&lt;/a&gt; show an amazing facility for geometric drawings: named points, marked angles, primitives for things like barycenter and circumcenter, TeX-styled labels, etc. A fully-labeled illustration of the &lt;a href=&quot;http://www.satimage.fr/software/en/smile/graphics/graphics_sample_euler.html&quot; rel=&quot;nofollow&quot;&gt;Euler Circle&lt;/a&gt; can be created with a few lines of script. Outputs to PDF, JPG, etc. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I always have lots of trouble navigating the online documentation, so I know I'm not getting the full benefit of the program. Nevertheless, the ability to describe a drawing precisely with a script is what I need, and it's what the app delivers.&lt;/p&gt;&#xA;" OwnerUserId="409" LastEditorUserId="409" LastEditorDisplayName="" LastEditDate="2010-08-10T02:40:09.650" LastActivityDate="2010-08-10T02:40:09.650" CommunityOwnedDate="2010-08-10T01:48:31.973" />
  <row Id="2009" PostTypeId="2" ParentId="1935" CreationDate="2010-08-10T02:01:21.663" Score="1" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/3540407146&quot; rel=&quot;nofollow&quot;&gt;Polynomials&lt;/a&gt; by Prasolov covers amoung others:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Eisenstein's criterion&lt;/li&gt;&#xA;&lt;li&gt;Duma's criterion&lt;/li&gt;&#xA;&lt;li&gt;Irreducibility of polynomials attaining small values&lt;/li&gt;&#xA;&lt;li&gt;Hilbert's criterion&lt;/li&gt;&#xA;&lt;li&gt;Irreducibility of trinomials and fournomials&lt;/li&gt;&#xA;&lt;li&gt;A few algorithms for factorization&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="956" LastActivityDate="2010-08-10T02:01:21.663" />
  <row Id="2010" PostTypeId="1" AcceptedAnswerId="2017" CreationDate="2010-08-10T02:34:40.443" Score="5" ViewCount="238" Body="&lt;p&gt;I'm optimizing a family of objectives over x1,...,xn with a single constraint that x's add up to 1, and when using method of Lagrange multipliers, the single multiplier always ends up 0 for members of this family...does this tell us anything interesting about these objectives?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: the answer was -- it means the constraint it inactive, ie, removing the constraint doesn't change the answer of optimization problem&lt;/p&gt;&#xA;" OwnerUserId="998" LastEditorUserId="998" LastEditorDisplayName="" LastEditDate="2010-08-11T20:00:41.400" LastActivityDate="2010-08-11T20:00:41.400" Title="Meaning of Lagrange multiplier always being 0?" Tags="&lt;calculus&gt;&lt;optimization&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" />
  <row Id="2011" PostTypeId="2" ParentId="1952" CreationDate="2010-08-10T02:42:47.647" Score="2" ViewCount="" Body="&lt;p&gt;I'm not sure that the main motivation for studying analytic geometry is that there are theorems true there that are not true in algebraic geometry; if anything, the reverse is true, since any complex variety is also a complex analytic space.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But there are analytic varieties that are not algebraic, so analytic geometry is a broader subject than complex algebraic geometry, which has its own appeal.  Thinking analytically also&#xA;gives a source of techniques that are not available by purely algebraic methods (and maybe this is what you mean by theorems that are true in analytic geometry but not algebraic geometry).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, let me also mention that &#xA;very interesting phenomena can occur when you study the interaction between the algebraic and analytic worlds, for example of the following kind:  the moduli space of analytic K3 surfaces is a connected 20 dimensional space.  The moduli space of algebraic K3 surfaces is a closed subspace, which is the union of countably many 19 dimensional connected components.   This shows that some problems, like the study of moduli spaces, can become&#xA;simpler in the analytic regime than they are if you restrict to the algebro-geometric setting.&lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2010-08-10T02:42:47.647" CommentCount="5" />
  <row Id="2012" PostTypeId="2" ParentId="1838" CreationDate="2010-08-10T02:52:13.350" Score="2" ViewCount="" Body="&lt;p&gt;One problem that I have come across is when roller coasters are being designed. If you have the pitch at +/-90 degrees (pointing straight up/down) then using normal Euler-based orientation you can't easily specify an angle of banking as you have no reference to 'up'. To solve this Quaternions are often used.&lt;/p&gt;&#xA;" OwnerUserId="999" LastActivityDate="2010-08-10T02:52:13.350" />
  <row Id="2013" PostTypeId="1" AcceptedAnswerId="2019" CreationDate="2010-08-10T03:03:40.500" Score="1" ViewCount="351" Body="&lt;p&gt;How are operations such as the sum, the product, the quotient, exponentiation, etc. of random variables solved or approached?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://math.stackexchange.com/questions/1416/algebra-of-random-variables&quot;&gt;This question&lt;/a&gt; addresses a similar problem but starts one step further: knowing how to find the distribution of a function of random variables. I'm asking for the previous step.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, my question is how, in general, all the operations over random variables are handled? &lt;/p&gt;&#xA;" OwnerUserId="1000" LastEditorUserId="1000" LastEditorDisplayName="" LastEditDate="2010-11-16T15:16:35.000" LastActivityDate="2010-11-16T15:16:35.000" Title="How to work with operations/algebra of random variables?" Tags="&lt;probability-theory&gt;&lt;probability&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="2" />
  <row Id="2015" PostTypeId="2" ParentId="1984" CreationDate="2010-08-10T03:10:10.187" Score="4" ViewCount="" Body="&lt;p&gt;No professional credentials or degree is required. However, lacking such will subject your paper to &lt;em&gt;much&lt;/em&gt; higher scrutiny so it is essential to adequately prepare for such. Since many uncredentialed submissions are by cranks it is important to differentiate your paper from such submissions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, be sure to choose an appropriate journal. Browse the recent publications of various journals to see if they accept papers on your subject at the level of your work. Seek advice from professors at local universities or in online math forums. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second, make it clear that you are aware of prior work by properly referencing such. Consult mathscinet and Zentralblatt as well as web searches (e.g. Google Scholar) to find related prior work. You may be able to access mathscinet at a local university library.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Crank submissions usually fail on both accounts. Namely, they often submit to journals whose prestige far exceeds their results and they usually fail to adequately cite prior work. So if you succeed on both these points you're off to a good start differentiating your work from a crank submission.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Best of luck!&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-08-10T03:10:10.187" />
  <row Id="2017" PostTypeId="2" ParentId="2010" CreationDate="2010-08-10T03:18:49.253" Score="3" ViewCount="" Body="&lt;p&gt;One of two things that I can think of are happening here.  The global minimum (or max, whatever your goal is here) of your function may actually lie within the simplex defined by $\sum_{i=1}^{4} q_{i} = 1$.  In this case your algorithm is finding the correct solution and the constraint is trivially satisfied.  (Consider minimizing $x^2+y^2$ constrained to the line $x = 0$, so the Lagrange formulation is $L(x,y,\nu) = x^2 + y^2 + \nu x$.  Solving $\nabla L=0$ demands that $\nu=0$. )&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively, your algorithm could be sampling only points that already satisfy the constraint - so you are only checking those points (intentionally or otherwise).  Then the algorithm is hopefully finding a local minimum (or max) along this constraint and your multiplier will always be zero.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope this helps you identify where a problem may be in your model - or clears up a conceptual issue and allows you to move forward with the solution you have.  As far as interpreting the meaning of Lagrange multipliers, &lt;a href=&quot;http://en.wikipedia.org/wiki/Lagrange_multipliers#Interpretation_of_the_Lagrange_multipliers&quot; rel=&quot;nofollow&quot;&gt;look here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="121" LastEditorUserId="121" LastEditorDisplayName="" LastEditDate="2010-08-10T03:37:22.337" LastActivityDate="2010-08-10T03:37:22.337" CommentCount="1" />
  <row Id="2018" PostTypeId="2" ParentId="1952" CreationDate="2010-08-10T03:56:33.960" Score="1" ViewCount="" Body="&lt;p&gt;How about Yau's proof of the Calabi conjecture?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Keywords: Geometric analysis; geometric PDE.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As Matt E says, we have many (beautiful, interesting, and useful) tools in the analytic category which don't exist in the algebraic category; my favorite example is Dolbeault resolutions. There are also various PDE methods that can be used. Conversely, analytic geometry also gives PDE-ists interesting examples to study.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many things that are now studied in the algebraic category were first studied or discovered in the analytic category. Two examples off the top of my head: Hodge theory, Kodaira-Spencer deformation theory. At least when these theories first came out, they had no analogues in the algebraic category.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A somewhat more &quot;modern&quot; example is the Bogomolov-Tian-Todorov theorem: deformations of Calabi-Yau manifolds are unobstructed. Until &lt;a href=&quot;http://arxiv.org/abs/0902.0732&quot; rel=&quot;nofollow&quot;&gt;very recently&lt;/a&gt;, I think this had no algebraic proof.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Working in the analytic category also allows us to use the theory of (real) Lie groups and Lie algebras. See for example &lt;a href=&quot;http://mathoverflow.net/questions/16207/why-can-the-dolbeault-operators-be-realised-as-lie-algebra-actions&quot; rel=&quot;nofollow&quot;&gt;http://mathoverflow.net/questions/16207/why-can-the-dolbeault-operators-be-realised-as-lie-algebra-actions&lt;/a&gt; and &lt;a href=&quot;http://mathoverflow.net/questions/16880/dolbeault-operators-for-cp1-as-mathfraksu2-actions&quot; rel=&quot;nofollow&quot;&gt;http://mathoverflow.net/questions/16880/dolbeault-operators-for-cp1-as-mathfraksu2-actions&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, the analytic theory is of independent interest in mathematical physics and in string theory. See for instance the famous paper &quot;Kodaira-Spencer Theory of Gravity and Exact Results for Quantum String Amplitudes&quot; by Bershadsky-Cecotti-Ooguri-Vafa.&lt;/p&gt;&#xA;" OwnerUserId="263" LastEditorUserId="263" LastEditorDisplayName="" LastEditDate="2010-08-10T20:36:18.940" LastActivityDate="2010-08-10T20:36:18.940" CommentCount="1" />
  <row Id="2019" PostTypeId="2" ParentId="2013" CreationDate="2010-08-10T05:47:41.463" Score="1" ViewCount="" Body="&lt;p&gt;In abstract probability based on measure theory, you are asking about the proof that any function from your list is measurable, such as $f(X,Y)=X+Y$ or $XY$ or $|X|^Y$.  You can see proofs of this for specific $f$ in any textbook that discusses Lebesgue measure theory.&#xA;Any continuous multivariable function is measurable, for example.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In practice $f$ is some concrete function and you need only that for any reasonable subset $S$ of the range of $f$ (an open set, a compact set, countable or &quot;tame&quot; unions and intersections of these), $f^{-1}(S)$ is reasonable in a similar sense and therefore &quot;obviously measurable&quot;.   Explicitly proving the measurability of a specific function would be unusual.&lt;/p&gt;&#xA;" OwnerUserId="467" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-08-10T20:08:10.503" LastActivityDate="2010-08-10T20:08:10.503" />
  <row Id="2021" PostTypeId="2" ParentId="1985" CreationDate="2010-08-10T06:28:53.213" Score="0" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://www.livegeometry.com/&quot; rel=&quot;nofollow&quot;&gt;&lt;strong&gt;Live Geometry&lt;/strong&gt;&lt;/a&gt; for Windows users.&lt;/p&gt;&#xA;" OwnerUserId="705" LastActivityDate="2010-08-10T06:28:53.213" CommunityOwnedDate="2010-08-10T06:28:53.213" />
  <row Id="2022" PostTypeId="2" ParentId="1985" CreationDate="2010-08-10T06:32:27.540" Score="1" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://www.wolfram.com/products/mathematica/index.html&quot; rel=&quot;nofollow&quot;&gt;Mathematica&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-08-10T06:32:27.540" CommunityOwnedDate="2010-08-10T06:32:27.540" />
  <row Id="2025" PostTypeId="2" ParentId="1983" CreationDate="2010-08-10T07:54:41.083" Score="3" ViewCount="" Body="&lt;p&gt;The best result currently known along these lines, due to Hermann Weyl, is that&#xA; $$|\zeta(\frac{1}{2}+it)|=\mathcal{O}(\tau^{16+\epsilon})$$ for any $\epsilon &gt; 0$ as $\tau \to \infty$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can refer more about this on: &lt;a href=&quot;http://www.openquestions.com/oq-ma014.htm&quot; rel=&quot;nofollow&quot;&gt;http://www.openquestions.com/oq-ma014.htm&lt;/a&gt;&lt;/p&gt;&#xA;" LastActivityDate="2010-08-10T07:54:41.083" CommentCount="3" />
  <row Id="2026" PostTypeId="2" ParentId="1989" CreationDate="2010-08-10T08:02:41.627" Score="0" ViewCount="" Body="&lt;p&gt;For example, suppose that $f_{1},f_{2},\cdots,f_{n}$ is a sequence of holomorphic functions converging uniformly to a function $f$ on an open disc. By cauchy's theorem we know that $$\oint\limits_{C} f(z) \ dz = 0 $$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;for every $n$, along any closed curve $C$ in the disc. Then the uniform convergence implies that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\oint\limits_{C} f(z) \ dz = \lim_{n \to \infty} \oint\limits_{C} f_{n}(z) \ dz =0$$&lt;/p&gt;&#xA;" LastActivityDate="2010-08-10T08:02:41.627" CommentCount="2" />
  <row Id="2027" PostTypeId="2" ParentId="1236" CreationDate="2010-08-10T08:31:31.807" Score="-1" ViewCount="" Body="&lt;p&gt;I think this may be of some help: &lt;a href=&quot;http://iopscience.iop.org/1873-7005/15/3/A03/pdf/1873-7005_15_3_A03.pdf&quot; rel=&quot;nofollow&quot;&gt;http://iopscience.iop.org/1873-7005/15/3/A03/pdf/1873-7005_15_3_A03.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" LastActivityDate="2010-08-10T08:31:31.807" CommentCount="1" />
  <row Id="2028" PostTypeId="1" AcceptedAnswerId="2039" CreationDate="2010-08-10T08:36:44.617" Score="2" ViewCount="168" Body="&lt;p&gt;I recently read a very good inequality concerning the no of primes $\pi(x)$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\pi(n)&gt;\frac{1}{6}\frac{n}{\log n}\mathrm{\ for\ }n\ge 2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are any other such elementary inequalities concerning the primes?&lt;/p&gt;&#xA;" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-08-10T16:39:53.177" LastActivityDate="2010-08-10T16:39:53.177" ClosedDate="2010-08-10T16:49:37.720" Title="Inequalities Concerning Primes" Tags="&lt;number-theory&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="2029" PostTypeId="1" AcceptedAnswerId="2030" CreationDate="2010-08-10T09:08:10.110" Score="4" ViewCount="332" Body="&lt;p&gt;A prime number not equal to 2 and 5 can't have last digit equal to 2,4,5,6 and 8.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it true that this is the only restriction on last digits of prime numbers?&#xA;I mean if its true that for any sequence of digits with last digit not equal to 2,4,5,6 and 8&#xA;there exists a prime number with given sequence of digits?&lt;/p&gt;&#xA;" OwnerUserId="1006" LastActivityDate="2010-08-11T06:40:59.707" Title="Last digits of primes" Tags="&lt;prime-numbers&gt;" AnswerCount="5" />
  <row Id="2030" PostTypeId="2" ParentId="2029" CreationDate="2010-08-10T09:14:33.193" Score="13" ViewCount="" Body="&lt;p&gt;I think that you forgot about zero. It's clear that prime number can't have zero as last digit. And if you add zero to your list then your statement is true.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $x_1, x_2, \cdots x_n$ be your sequence. Take $x = x_1 10^{n-1} + \cdots + x_n$ ---&#xA;number formed by this sequence. Then $x$ and $10^n$ are coprime since last digit of $x$ is odd and is not divisible by 5.&#xA;Hence by &lt;a href=&quot;http://en.wikipedia.org/wiki/Dirichlet_theorem&quot; rel=&quot;nofollow&quot;&gt;Dirichlet theorem&lt;/a&gt; there exists&#xA;infinitely many primes $p$ with $p \equiv x \pmod {10^n}$. Those primes have required last digits.&lt;/p&gt;&#xA;" OwnerUserId="334" LastActivityDate="2010-08-10T09:14:33.193" CommentCount="1" />
  <row Id="2031" PostTypeId="2" ParentId="2029" CreationDate="2010-08-10T09:16:00.590" Score="4" ViewCount="" Body="&lt;p&gt;Don't forget that the last digit could also be $0$, but there are no such prime numbers (since anything ending in $0$ is divisible by $10$).   Otherwise, yes, this is true:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have a string of final digits $a_n \cdots a_1$ with $a_1 \neq 0,2,4,5,6,8$, then the number $k = 10^{n-1} a_n + \ldots + a_1$ is not divisible by $2$ or $5$ so is prime to $10^n$.  By a &lt;a href=&quot;http://en.wikipedia.org/wiki/Dirichlet%27s_theorem_on_primes_in_arithmetic_progressions&quot; rel=&quot;nofollow&quot;&gt;celebrated theorem of Dirichlet&lt;/a&gt;, the arithmetic progression $k, k + 10^n, k + 2 \cdot 10^n,\ldots$ then contains infinitely many prime numbers.  Any one of these is &#xA;congruent to $k$ modulo $10^n$, which is equivalent to saying that the sequence of the last $n$ digits is $a_n \cdots a_1$.  &lt;/p&gt;&#xA;" OwnerUserId="299" LastActivityDate="2010-08-10T09:16:00.590" />
  <row Id="2032" PostTypeId="2" ParentId="1984" CreationDate="2010-08-10T09:21:59.440" Score="2" ViewCount="" Body="&lt;p&gt;There have already been a lot of good advices. Let me stress the importance of chosing the journal right. There are too difficulties here:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The journal is interested in your subject.&lt;/li&gt;&#xA;&lt;li&gt;Your paper fits the quality standards of the journal.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;As for (1) there is an easy algorithm: you sure cite other works in you bibliography, right? Look at the journals in which they were published. If an specific journal does appear more often than others, there is a good chance that your subject interests that journal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second issue is more difficult to grasp for beginners without the advise of a senior colleague. It takes some time to develop the sense of how &lt;em&gt;good&lt;/em&gt; is your paper and, consulting the JCR list of mathematical journals, find the appropriate journal according to its quality. Of course, you could try another easy &quot;algorithm&quot; for this: start with the highest quality journal in your field and, as they reject your paper, go down the list. There is nothing wrong in doing this, but some experience could help you in not trying Annals of Maths as your first option.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And a final advice: I'm sure that almost &lt;em&gt;every&lt;/em&gt; professional mathematician has had at least one of his/her papers rejected (in my case, I think that 30% of my papers were rejected by the first journal where I've sent them).&lt;/p&gt;&#xA;" OwnerUserId="664" LastActivityDate="2010-08-10T09:21:59.440" />
  <row Id="2033" PostTypeId="2" ParentId="1969" CreationDate="2010-08-10T09:35:55.887" Score="2" ViewCount="" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;A mathematician is a device for&#xA;  turning coffee into theorems.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Paul Erdős (or Alfréd Rényi)&lt;/p&gt;&#xA;" OwnerUserId="960" LastActivityDate="2010-08-10T09:35:55.887" CommunityOwnedDate="2010-08-10T09:35:55.887" />
  <row Id="2034" PostTypeId="2" ParentId="1969" CreationDate="2010-08-10T10:42:08.673" Score="6" ViewCount="" Body="&lt;p&gt;I like Newton's:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If I have seen a little further it is by standing on the shoulders of Giants&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If only because of its lesser known interpretation. The above quote is from a letter by Newton to Robert Hooke, a man born with a severe stoop who criticized Newton for having stolen his theory of optics. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Newton's response, below the waterline, can be seen as saying &quot;I didn't steal optics from you, shorty!&quot; Hiding such snideness in a concession to modesty, even if terribly mean and apocryphal as an interpretation, is pretty stunning...&lt;/p&gt;&#xA;" OwnerUserId="160" LastActivityDate="2010-08-10T10:42:08.673" CommunityOwnedDate="2010-08-10T10:42:08.673" />
  <row Id="2035" PostTypeId="2" ParentId="1985" CreationDate="2010-08-10T10:53:04.290" Score="1" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://www.inkscape.org/&quot; rel=&quot;nofollow&quot;&gt;Inkscape&lt;/a&gt; is a popular open source drawing package.  One feature that makes it particularly nice for drawing mathematical diagrams is that you can easily &lt;a href=&quot;http://www.johndcook.com/blog/2009/12/22/including-latex-in-inkscape/&quot; rel=&quot;nofollow&quot;&gt;insert LaTeX output&lt;/a&gt; in your drawings.&lt;/p&gt;&#xA;" OwnerUserId="247" LastActivityDate="2010-08-10T10:53:04.290" CommunityOwnedDate="2010-08-10T10:53:04.290" />
  <row Id="2036" PostTypeId="2" ParentId="1969" CreationDate="2010-08-10T11:20:54.510" Score="4" ViewCount="" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;em&gt;Young man, in mathematics you don't understand things. You just get used to them.&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;John von Neumann&lt;/p&gt;&#xA;" OwnerUserId="1011" LastActivityDate="2010-08-10T11:20:54.510" CommunityOwnedDate="2010-08-10T11:20:54.510" />
  <row Id="2037" PostTypeId="1" AcceptedAnswerId="2069" CreationDate="2010-08-10T11:22:22.450" Score="6" ViewCount="454" Body="&lt;p&gt;What is the probability of getting &lt;a href=&quot;http://en.wikipedia.org/wiki/Yahtzee#Lower_section&quot; rel=&quot;nofollow&quot;&gt;yahtzee&lt;/a&gt; using N dices with X sides in Y throws in a single round?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which side of the dice one gets yahtzee with doesn't matter (e.g. it doesn't matter if it is ones or twos etc.). I also assume perfect strategy is used, that is, after each throw one saves the number it is the most of (and if equal, just pick one at random).&lt;/p&gt;&#xA;" OwnerUserId="893" LastEditorUserId="893" LastEditorDisplayName="" LastEditDate="2010-08-11T08:53:17.867" LastActivityDate="2010-10-11T14:03:30.083" Title="What is the probability of getting yahtzee?" Tags="&lt;probability&gt;&lt;game-theory&gt;" AnswerCount="3" CommentCount="6" />
  <row Id="2038" PostTypeId="2" ParentId="2001" CreationDate="2010-08-10T11:26:13.287" Score="3" ViewCount="" Body="&lt;p&gt;There are at least two kinds of (interesting) examples.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I. When we can get an abelian category, but have to add more (co)kernels: 1) the category of free modules over a ring; 2) the category of projective modules over a ring; 3) the category of vector bundles on a topological space (if fact, 3 is a particular case of 2).&#xA;(From 1 or 2 one gets abelian category of all modules over the ring, from 3 — abelian category of sheafs of vector spaces on X.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;II. When we already have (co)kernels (&quot;category is pre-abelian&quot;) but not all mono-/epimorphisms are normal. As explained in another answer, an example is the category of filtered modules.&lt;/p&gt;&#xA;" OwnerUserId="152" LastEditorUserId="152" LastEditorDisplayName="" LastEditDate="2010-08-10T16:51:14.300" LastActivityDate="2010-08-10T16:51:14.300" CommunityOwnedDate="2010-08-11T11:29:36.717" />
  <row Id="2039" PostTypeId="2" ParentId="2028" CreationDate="2010-08-10T11:42:22.387" Score="0" ViewCount="" Body="&lt;p&gt;What is your question? This is well-known stuff. It appears for instance in section 4.5 of Apostol's &lt;em&gt;Introduction to Analytic Number Theory&lt;/em&gt;, which also contains an upper bound. You could at least say where you got yours from. (This should have been a comment, not an answer, but I don't have enough reputation for adding comments.)&lt;/p&gt;&#xA;" OwnerUserId="589" LastActivityDate="2010-08-10T11:42:22.387" CommentCount="1" />
  <row Id="2040" PostTypeId="1" AcceptedAnswerId="2041" CreationDate="2010-08-10T11:47:45.387" Score="3" ViewCount="146" Body="&lt;p&gt;We should have $\mathbb{Q} = \lim_\rightarrow \frac 1n \mathbb{Z}$ but a few things are confusing me. Since the index category is a set, we should get the coproduct: $\bigsqcup \frac 1n \mathbb{Z}$. But in this coproduct we have $1/2 \neq 2/4$ which clearly is wrong. This make me suspect that the index category is not an index set. What we probably should get is a union (and not a coproduct) in the category of rings (in which $1/2 = 2/4$). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; Help me understand what the index category is (eg, what are its maps) and am I correct when believing we should get an ordinary union? &lt;/p&gt;&#xA;" LastActivityDate="2010-08-10T12:57:11.487" Title="Colimit of $\frac{1}{n} \mathbb{Z}$" Tags="&lt;category-theory&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="2041" PostTypeId="2" ParentId="2040" CreationDate="2010-08-10T11:52:28.670" Score="7" ViewCount="" Body="&lt;p&gt;Index category is the category of positive integers with a morphism from n to m iff n|m. The limit is not coproduct but union of $\frac{1}{n}\mathbb{Z}$ with natural embedings (which is the same as union of $\frac{1}{n}\mathbb{Z}$ inside $\mathbb{Q}$ — hence the equality).&lt;/p&gt;&#xA;" OwnerUserId="152" LastEditorUserId="152" LastEditorDisplayName="" LastEditDate="2010-08-10T12:57:11.487" LastActivityDate="2010-08-10T12:57:11.487" />
  <row Id="2042" PostTypeId="2" ParentId="1985" CreationDate="2010-08-10T12:25:19.743" Score="0" ViewCount="" Body="&lt;p&gt;I use the programme CaRMetal. Here are a few screenshots:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://i34.tinypic.com/2vwcits.jpg&quot; rel=&quot;nofollow&quot;&gt;http://i34.tinypic.com/2vwcits.jpg&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://i35.tinypic.com/oa42ts.jpg&quot; rel=&quot;nofollow&quot;&gt;http://i35.tinypic.com/oa42ts.jpg&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://i35.tinypic.com/2lxfin8.jpg&quot; rel=&quot;nofollow&quot;&gt;http://i35.tinypic.com/2lxfin8.jpg&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://i35.tinypic.com/rble9l.jpg&quot; rel=&quot;nofollow&quot;&gt;http://i35.tinypic.com/rble9l.jpg&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I guess most geometric constructions programmes are good, although I wouldn't recommend Kig.&lt;/p&gt;&#xA;" OwnerUserId="871" LastActivityDate="2010-08-10T12:25:19.743" CommunityOwnedDate="2010-08-10T12:25:19.743" />
  <row Id="2043" PostTypeId="1" AcceptedAnswerId="2047" CreationDate="2010-08-10T12:36:57.627" Score="5" ViewCount="251" Body="&lt;p&gt;It's easy for scalars, $(\exp(a(x)))' = a' e^a$. But can anything be said about matrices? Do $A(x)$ and $A'(x)$ commute such that $(\exp(A(x)))' = A' e^A = e^A A'$ or is this only a special case?&lt;/p&gt;&#xA;" OwnerUserId="163" LastActivityDate="2010-08-11T13:50:17.213" Title="Is there a general formula for the derivative of $\exp(A(x))$ when $A(x)$ is a matrix?" Tags="&lt;calculus&gt;&lt;matrices&gt;" AnswerCount="2" CommentCount="6" />
  <row Id="2044" PostTypeId="2" ParentId="2029" CreationDate="2010-08-10T12:39:56.780" Score="0" ViewCount="" Body="&lt;p&gt;When you write down numbers in base $10 = 2\cdot 5$, the last digit only allows to determine divisibility by the base's factors 2 and 5, indicated by $0,2,4,5,6,8$. Would you pick another base, e.g. $30=2\cdot 3\cdot 5$ you could read the divisibility by 3 in addition.&lt;/p&gt;&#xA;" OwnerUserId="163" LastActivityDate="2010-08-10T12:39:56.780" />
  <row Id="2045" PostTypeId="1" AcceptedAnswerId="2053" CreationDate="2010-08-10T12:53:04.397" Score="3" ViewCount="395" Body="&lt;p&gt;I'm working on a problem from a past exam and I'm stuck, so I'm asking for help.&#xA;Here it is:  $A = \frac12 \left[\begin{array}{rrrr}&#xA;1 &amp;amp;  1 &amp;amp;  1 &amp;amp;  1 \\&#xA;1 &amp;amp;  1 &amp;amp; -1 &amp;amp; -1 \\&#xA;1 &amp;amp; -1 &amp;amp;  1 &amp;amp; -1 \\&#xA;1 &amp;amp; -1 &amp;amp; -1 &amp;amp;  1&#xA;\end{array}\right]$ find  $\mathbf A^{-1}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My problem isn't the inverse matrix itself. We just get the determinant, see if it's zero or not, get the adjoint matrix and divide it by determinant.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My problem is space. As you can see, it's a 4x4 matrix meaning that I'd have to do 4x4 3x3 determinants to get the adjoint matrix plus 2 3x3 determinants to get determinant of the matrix. Now we get one A3 piece of paper for 6 problems. The problems are printed on one side and the other side is blank. This and the fact that inverse matrix is $A = \frac12 \left[\begin{array}{rrrr}&#xA;1 &amp;amp;  1 &amp;amp;  1 &amp;amp;  1 \\&#xA;1 &amp;amp;  1 &amp;amp; -1 &amp;amp; -1 \\&#xA;1 &amp;amp; -1 &amp;amp;  1 &amp;amp; -1 \\&#xA;1 &amp;amp; -1 &amp;amp; -1 &amp;amp;  1&#xA;\end{array}\right]$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;led me to believe that there's some catch that I do not see. Any ideas what could it be?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;STRIKE&gt;Also if someone could edit these matrices from MATLAB format into something that this site will parse would be great!&lt;/STRIKE&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;STRIKE&gt;&lt;strong&gt;EDIT&lt;/strong&gt; Unfortunately it seem that TeX code for matrices doesn't work here. Here's the matrix in MATLAB form, if anyone wants it A=(1/2)*[1,1,1,1;1,1,-1,-1;1,-1,1,-1;1,-1,-1,1];&lt;/STRIKE&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT 2&lt;/strong&gt; Answer by Jack Schmidt contains code for matrices. &lt;/p&gt;&#xA;" OwnerUserId="348" LastEditorUserId="348" LastEditorDisplayName="" LastEditDate="2010-08-10T13:56:05.043" LastActivityDate="2010-08-11T00:42:38.457" Title="How to get inverse of this matrix using least amount of space?" Tags="&lt;matrices&gt;" AnswerCount="3" CommentCount="5" />
  <row Id="2046" PostTypeId="2" ParentId="1989" CreationDate="2010-08-10T12:55:38.543" Score="1" ViewCount="" Body="&lt;p&gt;Ok, please see this link, i hope this helps: &lt;a href=&quot;http://www.jstor.org/page/termsConfirm.jsp?redirectUri=/stable/pdfplus/2309592.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.jstor.org/page/termsConfirm.jsp?redirectUri=/stable/pdfplus/2309592.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" LastActivityDate="2010-08-10T12:55:38.543" CommentCount="2" />
  <row Id="2047" PostTypeId="2" ParentId="2043" CreationDate="2010-08-10T13:09:40.663" Score="2" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://math.stackexchange.com/users/121/tom-stephens&quot;&gt;Tom Stephens&lt;/a&gt; linked to &lt;a href=&quot;http://en.wikipedia.org/wiki/Matrix_exponential#The_exponential_map&quot; rel=&quot;nofollow&quot;&gt;the exponential map&lt;/a&gt;, which states that&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$ \frac{d}{dt}e^{X(t)} = \int\limits_0^1 e^{\alpha X(t)} \frac{dX(t)}{dt} e^{(1-\alpha) X(t)} d\alpha $&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If $X(t)$ and $\frac{d}{dt}X(t)$ commute, the latter also commutes with $\exp(X(t))$ and &lt;em&gt;then&lt;/em&gt; it simplifies into $ \frac{d}{dt}e^{X(t)} = \frac{d X(t)}{dt} e^{X(t)}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A counter-example is&#xA;$$X(t) = \begin{pmatrix} \cos(t) &amp;amp; \sin(t) \\ \sin(t) &amp;amp; -\cos(t) \end{pmatrix}$$&#xA;at $t=0$, i.e. $X(0) = \sigma_3, \dot X(0) = \sigma_1$ ($\sigma_i$ are the non-commuting &lt;a href=&quot;http://en.wikipedia.org/wiki/Pauli_matrices&quot; rel=&quot;nofollow&quot;&gt;Pauli Matrices&lt;/a&gt;)&lt;/p&gt;&#xA;" OwnerUserId="163" LastEditorUserId="163" LastEditorDisplayName="" LastEditDate="2010-08-10T14:59:38.213" LastActivityDate="2010-08-10T14:59:38.213" />
  <row Id="2048" PostTypeId="1" CreationDate="2010-08-10T13:13:35.100" Score="2" ViewCount="145" Body="&lt;p&gt;Hi everyone,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure a 'power equation' is the right name for the equation I'd like to know more about (and, specifically, about its solutions), but I don't know the 'proper' way to name it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The simplest 'power polynomial' is : $P(x) = x^x $. The simplest 'power equation is: (1) $x^x = c$ for some $c \in \mathbb{R} $. What are the &lt;em&gt;exact&lt;/em&gt; solutions to (1) of x (in $\mathbb{C}$) with regards to c, apart from the 'obvious' solutions (x=0, c=0), (x=1, c=1) and (x=2, c=4) and all the other solutions of the form $ x^x = n^n$ for $n \in \mathbb{N}$ ?    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We could extend the power polynomial: $P_{2}(x) = x^{{ax}^{bx}} + x^{cx} $. What are the &lt;em&gt;exact&lt;/em&gt; solutions of $P_{2}(x) = d$ for $a,b,c,d \in \mathbb{R}, x\in \mathbb{C}$ ? Or perhaps I should ask what the exact form of the solutions is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We could generalize the power polynomial even further to $P_{3}(x)$ and $P_{n}(x)$, but I don't know how to write down the latter, general polynomial. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Max&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(P.S. I References are always welcome. II If you think this question belongs to MO, please tell me).&lt;/p&gt;&#xA;" OwnerUserId="1014" LastActivityDate="2010-08-10T13:21:19.947" Title="Solution(s) to 'power equations'" Tags="&lt;algebra&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="1" />
  <row Id="2049" PostTypeId="2" ParentId="1946" CreationDate="2010-08-10T13:15:33.817" Score="2" ViewCount="" Body="&lt;p&gt;@ Julio Cesar: I don't know about the problem, but this is why i think that the result could be correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Theorem:&lt;/strong&gt; Let $f$ and $g$ be continuous on $[a,b]$, and let $g$ have constant sign on $[a,b]$. Then there is a $c \in (a,b)$ such that $$\int\limits_{a}^{b} f(x)g(x) \ dx = f(c) \int\limits_{a}^{b} g(x) \ dx$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please refer &quot;E.I. Poffald, Amer.Math.Monthly 97 (1990), 205-213&quot; for a proof of this result.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using this result, by Taylor's formula with integral remainder we have $$f^{n}\Bigl(\frac{x}{n+1}\Bigr)=f^{n}(0) + f^{n+1}(0)\frac{x}{n+1} + \int\limits_{0}^{\frac{x}{n+1}} f^{n+2}(t) \Bigl(\frac{x}{n+1}-t\Bigr) \ dt$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence $$f(0)+ \frac{f'(0)}{1!}x + \cdots + \frac{f^{n-1}(0)}{(n-1)!}x^{n-1} + \frac{f^{n}(\frac{x}{n+1})}{n!}x^{n} = \text{ Our sum with the Integral Remainder}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then one can define $g$ in this manner.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$g(t)= \frac{(x-t)^{n+1}}{n+1} - x^{n}\Bigl(\frac{x}{n+1}-t\Bigr) \quad ; t \in [0,x]$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have $g'(t)&gt;0$ for $t \in (0,x)$ and $g(0)=0$. Thus $g$ is positive on the open interval $(0,x)$. Then using our theroem we have $$\int\limits_{0}^{\frac{x}{n+1}} f^{n+2}(t)(g(t))=f^{n+2}(c) \int\limits_{0}^{\frac{x}{n+1}} g(t) \ dt$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is where i am stuck.&lt;/p&gt;&#xA;" LastActivityDate="2010-08-10T13:15:33.817" />
  <row Id="2050" PostTypeId="2" ParentId="2048" CreationDate="2010-08-10T13:21:19.947" Score="3" ViewCount="" Body="&lt;p&gt;Maybe you should read about the &lt;a href=&quot;http://en.wikipedia.org/wiki/Lambert_W_function&quot; rel=&quot;nofollow&quot;&gt;Lambert W-function&lt;/a&gt; which gives the solutions to expressions like $z=x^x$. However, I am not sure what to do in case of such a &quot;power tower&quot; like $P_2$.&lt;/p&gt;&#xA;" OwnerUserId="347" LastActivityDate="2010-08-10T13:21:19.947" />
  <row Id="2051" PostTypeId="1" AcceptedAnswerId="2063" CreationDate="2010-08-10T13:24:08.287" Score="0" ViewCount="258" Body="&lt;p&gt;Give an example of sequence of Measurable functions defined on some measurable subset $E$ of $\mathbb{R}$  such that $f_{n} \to f$ pointwise almost everywhere on $E$ but $$\int\limits_{E} f \ dm \not\leq \lim_{n} \inf \int\limits_{E} f_{n} \ dm$$&lt;/p&gt;&#xA;" LastActivityDate="2010-08-11T19:51:08.943" Title="Fatou's Lemma Counterexample" Tags="&lt;measure-theory&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="2052" PostTypeId="2" ParentId="2051" CreationDate="2010-08-10T13:39:18.907" Score="0" ViewCount="" Body="&lt;p&gt;For $E=[0,1]$ set: $f_n(x)=n- n^2x$ for $x\in [0,\frac1n]$ $f_n(x)=0$ else.&lt;/p&gt;&#xA;" OwnerUserId="160" LastActivityDate="2010-08-10T13:39:18.907" CommentCount="3" />
  <row Id="2053" PostTypeId="2" ParentId="2045" CreationDate="2010-08-10T13:44:13.857" Score="2" ViewCount="" Body="&lt;p&gt;I guess A is &lt;a href=&quot;http://en.wikipedia.org/wiki/Orthogonal_matrix&quot; rel=&quot;nofollow&quot;&gt;orthogonal&lt;/a&gt; and symmetric, so that tells you A&lt;sup&gt;−1&lt;/sup&gt; = A&lt;sup&gt;T&lt;/sup&gt; = A, but uh, that's not a very common situation to my mind.  Maybe someone else has a better &quot;test-taking strategy&quot; explanation, but me personally, I would just row reduce or whatever method you use in general.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An orthogonal matrix is defined to be a matrix whose transpose is its inverse.  However, for us the better (almost) definition is a matrix whose rows (or columns) are orthogonal, as in, perpendicular.  So (1,1,1,1) is orthogonal to (1,-1,1,-1) since their dot product is (1)(1)+(1)(-1)+(1)(1)+(1)(-1) = 1 - 1 + 1 - 1 is zero.  You also should check that the length of the vector in each row is 1, $\sqrt{(1/2)^2 + (1/2)^2 + (1/2)^2 + (1/2)^2} = 1$ so good, but even if not, that part is easily fixed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sometimes you can tell just by looking that a matrix is orthogonal.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;As far as parsing goes:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is the matrix:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;$A = \frac12 \begin{pmatrix}&#xA;1 &amp;amp;  1 &amp;amp;  1 &amp;amp;  1 \\\\&#xA;1 &amp;amp;  1 &amp;amp; -1 &amp;amp; -1 \\\\&#xA;1 &amp;amp; -1 &amp;amp;  1 &amp;amp; -1 \\\\&#xA;1 &amp;amp; -1 &amp;amp; -1 &amp;amp;  1&#xA;\end{pmatrix}$&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;$A = \frac12 \begin{pmatrix}1&amp;amp;1&amp;amp;1&amp;amp;1\\1&amp;amp;1&amp;amp;-1&amp;amp;-1\\1&amp;amp;-1&amp;amp;1&amp;amp;-1\\1&amp;amp;-1&amp;amp;-1&amp;amp;1\end{pmatrix}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is using the array environment:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;$A = \frac12 \left(\begin{array}{rrrr}&#xA;1 &amp;amp;  1 &amp;amp;  1 &amp;amp;  1 \\\\&#xA;1 &amp;amp;  1 &amp;amp; -1 &amp;amp; -1 \\\\&#xA;1 &amp;amp; -1 &amp;amp;  1 &amp;amp; -1 \\\\&#xA;1 &amp;amp; -1 &amp;amp; -1 &amp;amp;  1&#xA;\end{array}\right)$&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;$A = \frac12 \left(\begin{array}{rrrr}1&amp;amp;1&amp;amp;1&amp;amp;1\\1&amp;amp;1&amp;amp;-1&amp;amp;-1\\1&amp;amp;-1&amp;amp;1&amp;amp;-1\\1&amp;amp;-1&amp;amp;-1&amp;amp;1\end{array}\right)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The backslashes get eaten by the markdown software, so you just double them.&lt;/p&gt;&#xA;" OwnerUserId="583" LastEditorUserId="583" LastEditorDisplayName="" LastEditDate="2010-08-10T14:00:12.193" LastActivityDate="2010-08-10T14:00:12.193" CommentCount="6" />
  <row Id="2054" PostTypeId="2" ParentId="2029" CreationDate="2010-08-10T13:52:45.317" Score="2" ViewCount="" Body="&lt;p&gt;Yes, including 0. Generally  $\pmod m$ the number of elts coprime to $m$ is counted by the &lt;a href=&quot;http://en.wikipedia.org/wiki/Euler%27s_totient_function&quot; rel=&quot;nofollow&quot;&gt;Euler totient function&lt;/a&gt; $\phi(m)$. These elts are precisely the units, i.e. invertible elts $\pmod m$. Here $\phi(10) = 4$ with units $\{1,3,7,9\}$ Generally each such unit may be represented by a prime - this is the famous &lt;a href=&quot;http://en.wikipedia.org/wiki/Dirichlet%27s_theorem_on_arithmetic_progressions&quot; rel=&quot;nofollow&quot;&gt;Dirichlet theorem&lt;/a&gt; on arithmetic progressions. These units form a group and the study of such unit groups plays a fundamental role in number theory and algebra.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-08-10T14:05:29.407" LastActivityDate="2010-08-10T14:05:29.407" />
  <row Id="2055" PostTypeId="1" CreationDate="2010-08-10T14:06:03.147" Score="0" ViewCount="250" Body="&lt;p&gt;I'd like to write down proofs. I use first-order logic and natural deduction. Formulas tend to be long, too long for LaTeX. Writing formulas with the computer is also a slow process. Handwriting is a lot faster and more natural, but I can't trust paper. Time, water, dirt, fire are its enemies. I could write down something and then immediately scan it, but I'm an eco-guy and I'm worried about using a whole page for only two lines. How do you write math? Thanks.&lt;/p&gt;&#xA;" LastActivityDate="2010-08-10T16:14:13.213" CommunityOwnedDate="2010-08-10T14:49:09.253" Title="How to write down proofs?" Tags="&lt;soft-question&gt;" AnswerCount="6" CommentCount="5" FavoriteCount="2" />
  <row Id="2056" PostTypeId="2" ParentId="2055" CreationDate="2010-08-10T14:09:10.473" Score="1" ViewCount="" Body="&lt;p&gt;On a blackboard.&lt;/p&gt;&#xA;" OwnerUserId="160" LastActivityDate="2010-08-10T14:09:10.473" />
  <row Id="2057" PostTypeId="2" ParentId="2055" CreationDate="2010-08-10T14:22:24.327" Score="2" ViewCount="" Body="&lt;p&gt;There are a number of possible approaches. One can become quite proficient even with a text-based editor if it is user extensible - e.g. see the (La)TeX extensions for Emacs. These are easily user customizable if you have basic programming knowledge. Alternatively see the WYSIWYG LaTeX front-ends, e.g. Lyx. Another possibility is to employ a graphics / digitizing tablet (e.g. Wacom) for input, but this may require writing your own software to obtain optimal productivity.&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-08-10T14:22:24.327" CommentCount="2" />
  <row Id="2058" PostTypeId="2" ParentId="2055" CreationDate="2010-08-10T14:38:30.917" Score="0" ViewCount="" Body="&lt;p&gt;If you are a user of Windows 7 and have a graphics tablet , you could use it in combination with hand-writing recognition using the Math Input Panel.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It will take some time to train your computer to correctly recognize your handwriting, but if you are only going to use it for formulas, then it shouldn't take more that few days. You can also use it with mouse, but it's less natural then with a real pen and tablet. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another factor that can impact your experience is the choice of the tablet. Good ones are mainly from Wacom and are expensive (about 50€ here for cheapest models!) but come with good pens and good drivers. They are also powered using EM induction, so the mass distribution in the pen is more like in real pens. (BTW, I really don't want to advertise Wacom, but I'm not aware of any other manufacturer whose tablets have good drivers and whose pens are small and light enough.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also used a tablet made by Trust and had problems with drivers and weight distribution. Pen is powered by 2 AAA batteries which are in top part of the pen making it very heavy. Also pen is larger that a real pen so it feels clumsy.&lt;/p&gt;&#xA;" OwnerUserId="348" LastActivityDate="2010-08-10T14:38:30.917" CommunityOwnedDate="2010-08-10T15:01:02.720" />
  <row Id="2059" PostTypeId="2" ParentId="2055" CreationDate="2010-08-10T14:58:40.807" Score="1" ViewCount="" Body="&lt;p&gt;If you plan to enter the scientific community you should begin to learn LaTeX at precisely this moment.  I recommend text-based editors such as emacs or vim for the sole reason that they are available on &lt;em&gt;every&lt;/em&gt; unix-based machine, whether accessed remotely or on your own personal shiny macbook. &lt;/p&gt;&#xA;" OwnerUserId="121" LastActivityDate="2010-08-10T14:58:40.807" CommunityOwnedDate="2010-08-10T14:58:40.807" />
  <row Id="2060" PostTypeId="2" ParentId="2043" CreationDate="2010-08-10T15:19:39.843" Score="1" ViewCount="" Body="&lt;p&gt;The question comes down to computing what's called the &quot;derivative of&quot;, the &quot;differential of&quot;, or the &quot;tangent map to&quot;, the exponential map from $M_n(\mathbb R)$ into itself at a given matrix $A$ (not necessarily the zero matrix). There is a classical formula for this. Here is the first reference I found: pages 1 and 2 of&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.math.columbia.edu/~woit/notes4.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.math.columbia.edu/~woit/notes4.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;by Peter Woit. Here it is (with Peter Woit's notation) &#xA;$$\exp_*(X)\ Y=\exp(X)\ \frac{1-e^{-ad(X)}}{ad(X)}\ Y.$$ &#xA;Here is a reference for the Chain Rule:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Chain_rule#The_fundamental_chain_rule&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Chain_rule#The_fundamental_chain_rule&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It reads, in Peter Woit's notation and under appropriate assumptions,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(f\circ g)_*(x)=f_*(g(x))\circ g_*(x).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[Thank you to &lt;a href=&quot;http://math.stackexchange.com/users/171/kennytm&quot;&gt;KennyTM&lt;/a&gt; for having edited this formula.]  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT 1.&lt;/strong&gt; Here are the two formulas written in another notation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\exp'(X)=\exp(X)\ \frac{1-e^{-ad(X)}}{ad(X)}\quad,$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(f\circ g)'(x)=f'(g(x))\circ g'(x).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT 2.&lt;/strong&gt; Here is another reference. This is a post by &lt;a href=&quot;http://math.stackexchange.com/users/536/akhil-mathew&quot;&gt;Akhil Mathew&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://deltaepsilons.wordpress.com/2009/11/07/helgasons-formula-ii/&quot; rel=&quot;nofollow&quot;&gt;http://deltaepsilons.wordpress.com/2009/11/07/helgasons-formula-ii/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="660" LastEditorUserId="660" LastEditorDisplayName="" LastEditDate="2010-08-11T13:50:17.213" LastActivityDate="2010-08-11T13:50:17.213" CommentCount="4" />
  <row Id="2062" PostTypeId="1" AcceptedAnswerId="2407" CreationDate="2010-08-10T15:45:55.957" Score="2" ViewCount="247" Body="&lt;p&gt;We know that if $ \displaystyle d(n)= \sum\limits_{d \mid n} 1$, then we have &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \sum\limits_{n \leq x} d(n)= x\log{x} + (2C-1)x + \mathcal{O}(\sqrt{x})$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have referred Apostol's &quot;Analytic Number theory&quot; and i understood the first half of the proof where the error term is $\mathcal{O}(x)$, but please tell me as to how to improve the error term to $\sqrt{x}$.&lt;/p&gt;&#xA;" LastEditorUserId="171" LastEditorDisplayName="" LastEditDate="2010-10-27T18:05:27.470" LastActivityDate="2010-10-27T18:05:27.470" Title="Dirichlet's Divisor Problem" Tags="&lt;number-theory&gt;&lt;analytic-number-theory&gt;" AnswerCount="3" />
  <row Id="2063" PostTypeId="2" ParentId="2051" CreationDate="2010-08-10T15:49:54.207" Score="1" ViewCount="" Body="&lt;p&gt;Since Fatou’s Lemma &lt;em&gt;holds&lt;/em&gt; for non-negative measurable functions, I suppose you are looking for an example involving not necessarily non-negative functions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take, for instance, $f_n(x)=-1_{n\leq x\leq n+1}$ (just $-1$ times the indicator function on the interval $[n,n+1]$). Then $f=\lim\limits_{n\to\infty}\inf f_n=0$. Hence,&#xA;$$&#xA;0=\int f dm \not\leq \lim_{n\to\infty}\inf\int f_n dm = -1.&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="367" LastEditorUserId="367" LastEditorDisplayName="" LastEditDate="2010-08-10T15:58:30.197" LastActivityDate="2010-08-10T15:58:30.197" CommentCount="2" />
  <row Id="2064" PostTypeId="2" ParentId="1985" CreationDate="2010-08-10T15:59:10.417" Score="2" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://tug.org/metapost.html&quot; rel=&quot;nofollow&quot;&gt;Metapost&lt;/a&gt;. There are a few packages to simplify drawing geometry drawings.&lt;/p&gt;&#xA;" OwnerUserId="763" LastActivityDate="2010-08-10T15:59:10.417" CommunityOwnedDate="2010-08-10T15:59:10.417" />
  <row Id="2065" PostTypeId="2" ParentId="2037" CreationDate="2010-08-10T16:04:41.960" Score="2" ViewCount="" Body="&lt;p&gt;According to &lt;a href=&quot;http://en.wikipedia.org/wiki/Yahtzee#Probabilities&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt; &quot;the probability&#xA;of a Yahtzee for any three-roll turn is about 0.04603 (or $\frac{347897}{7558272}$), or roughly 1 in 22 attempts.&quot;&lt;/p&gt;&#xA;" OwnerUserId="367" LastEditorUserId="367" LastEditorDisplayName="" LastEditDate="2010-08-10T16:15:30.687" LastActivityDate="2010-08-10T16:15:30.687" />
  <row Id="2066" PostTypeId="2" ParentId="2055" CreationDate="2010-08-10T16:07:04.687" Score="-1" ViewCount="" Body="&lt;p&gt;On book margins.&lt;/p&gt;&#xA;" OwnerUserId="664" LastActivityDate="2010-08-10T16:07:04.687" CommunityOwnedDate="2010-08-10T16:07:04.687" CommentCount="1" />
  <row Id="2067" PostTypeId="2" ParentId="2055" CreationDate="2010-08-10T16:14:13.213" Score="1" ViewCount="" Body="&lt;p&gt;Judging from your question, I presume you want to write down &lt;em&gt;formal&lt;/em&gt; proofs in natural deduction. If so, a proof assistant (check Wikipedia for examples) will help you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you don't care that your proofs are formally checked, you can use $\LaTeX$ with the &lt;a href=&quot;http://math.ucsd.edu/~sbuss/ResearchWeb/bussproofs/&quot; rel=&quot;nofollow&quot;&gt;bussproofs package&lt;/a&gt;.&lt;/p&gt;&#xA;" LastActivityDate="2010-08-10T16:14:13.213" CommunityOwnedDate="2010-08-10T16:14:13.213" />
  <row Id="2068" PostTypeId="2" ParentId="2062" CreationDate="2010-08-10T16:20:26.110" Score="3" ViewCount="" Body="&lt;p&gt;I'd post this as a comment, but I don't have the reputation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There usual route is via the Hyperbola Method: &lt;a href=&quot;http://myyn.org/m/article/dirichlet-hyperbola-method/&quot; rel=&quot;nofollow&quot;&gt;http://myyn.org/m/article/dirichlet-hyperbola-method/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="256" LastActivityDate="2010-08-10T16:20:26.110" />
  <row Id="2069" PostTypeId="2" ParentId="2037" CreationDate="2010-08-10T16:25:04.423" Score="4" ViewCount="" Body="&lt;p&gt;What follows is a solution to a smaller problem; I don't have a good way to determine the probabilities of exactly k of some kind being the most of any kind on the first roll, but the work below could be used from that point to get an answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $f(N,X,Y)$ be the probability of getting all 1s with N dice, each with X sides, in at most Y rolls, where after each roll, 1s are retained and only non-1 dice are re-rolled.  Working 1 roll at a time, the first roll could have exactly $k=0,\dots,N$ 1s.  The probability of exactly k 1s in a single roll is ${N\choose k}\left(\frac{1}{X}\right)^k\left(\frac{x-1}{x}\right)^{N-k}$ and having k 1s in the first roll, we then need $N-k$ 1s in the remaining $Y-1$ rolls.  So, &#xA;$$f(N,X,Y)=\sum_{k=0}^{N}{N\choose k}\left(\frac{1}{X}\right)^k\left(\frac{x-1}{x}\right)^{N-k}f(N-k,X,Y-1).$$&#xA;Also, the probability of all 1s on 0 dice is 1 (0 of the 0 dice are guaranteed to be 1), so $f(0,X,Y)=1$, and the probability of all 1s on $N\ge 0$ dice in 0 rolls is 0 (can't get any 1s without rolling some dice), so $f(N,X,0)=0$ for $N\ge 0$.  This is a complete definition of $f(N,X,Y)$, though it does not lend itself to easy computation.  (However, software like Mathematica may be able to compute from this definition.  In Mathematica: &lt;code&gt;f[n_, x_, y_] := If[n == 0, 1, If[y == 0, 0, &#xA;       Sum[&#xA;    Binomial[n, k]*(1/x)^k*(1 - 1/x)^(n - k)*f[n - k, x, y - 1], &#xA;         {k, 0, n}]]]&lt;/code&gt;.)&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-08-10T16:25:04.423" CommentCount="1" />
  <row Id="2070" PostTypeId="2" ParentId="2001" CreationDate="2010-08-10T16:25:11.100" Score="2" ViewCount="" Body="&lt;p&gt;Another nice example is the category of finite-dimensional vector bundles over a fixed base space (with bundle maps over the identity as morphisms).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the base space is not too simple (the interval suffices), then this category is not (pre-)Abelian because there are, in general, neither kernels nor cokernels. Intuitively speaking, the obvious candidate for the kernel of a bundle map need not form a vector bundle because its dimension need not be locally constant.&lt;/p&gt;&#xA;" OwnerUserId="367" LastEditorUserId="367" LastEditorDisplayName="" LastEditDate="2010-08-10T18:07:57.147" LastActivityDate="2010-08-10T18:07:57.147" CommentCount="3" />
  <row Id="2071" PostTypeId="1" CreationDate="2010-08-10T16:52:26.263" Score="6" ViewCount="195" Body="&lt;p&gt;Recently I found &lt;a href=&quot;http://www.reddit.com/r/programming/comments/cy4m6/and_e_appears_from_nowhere_quick_numeric/&quot; rel=&quot;nofollow&quot;&gt;this post&lt;/a&gt; on Reddit. It describes the following algorithm to find e: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Here is an example of e turning up&#xA;  unexpectedly. Select a random number&#xA;  between 0 and 1. Now select another&#xA;  and add it to the first. Keep doing&#xA;  this, piling on random numbers. How&#xA;  many random numbers, on average, do&#xA;  you need to make the total greater&#xA;  than 1? The answer is e.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This means that you need on average ~2.7 random real numbers to make the sum greater than 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, a random number between 0 and 1 would on average be equal to 0.5. So intuitively I would think that, on average, only 2 random numbers would be required to have a sum &gt; 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So where did I go wrong in my thinking? &lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Update&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;I just figured it out myself: You need at least two numbers to have a sum &gt; 1, but often you'll need three, sometimes you'll need four, sometimes five, etc... So it only natural that the average required numbers is above 2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for the replies!&lt;/p&gt;&#xA;" OwnerUserId="1025" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2011-05-02T06:33:36.627" LastActivityDate="2011-05-02T06:33:36.627" Title="Explanation of numeric experiment that approximates e?" Tags="&lt;probability&gt;&lt;stochastic-processes&gt;" AnswerCount="5" CommentCount="4" FavoriteCount="2" />
  <row Id="2072" PostTypeId="2" ParentId="2071" CreationDate="2010-08-10T17:01:39.853" Score="4" ViewCount="" Body="&lt;p&gt;It is never possible to get greater than 1 on the first try, so the average must be strictly greater than 2.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-08-10T17:01:39.853" />
  <row Id="2073" PostTypeId="2" ParentId="2071" CreationDate="2010-08-10T17:16:22.543" Score="0" ViewCount="" Body="&lt;p&gt;&quot;How many..., on average, ...&quot; means the expected number of random numbers.  That is, the sum of each possible number of random numbers multiplied by its probability.  As Qiaochu Yuan's answer says, it is impossible to be greater than 1 on the first try, so the expected number is&#xA;$$2\cdot P(\mathrm{it\ takes\ exactly\ 2\ numbers})+3\cdot P(\mathrm{it\ takes\ exactly\ 3\ numbers})$$ $$+\;4\cdot P(\mathrm{it\ takes\ exactly\ 4\ numbers})+\cdots$$&#xA;Since each of those probabilities is nonzero, but the sum of those probabilities is 1, the expected number must be strictly greater than 2.&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-08-10T17:16:22.543" />
  <row Id="2074" PostTypeId="2" ParentId="1942" CreationDate="2010-08-10T17:28:36.283" Score="1" ViewCount="" Body="&lt;p&gt;See &lt;a href=&quot;http://stackoverflow.com/questions/3442770/counting-placing-items-in-bins-according-to-boolean-restrictions&quot;&gt;Moron's answer&lt;/a&gt; to this question on StackOverflow.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If I understood correctly your problem boils down to the problem of finding the number of &lt;a href=&quot;http://en.wikipedia.org/wiki/Matching_%28graph_theory%29&quot; rel=&quot;nofollow&quot;&gt;Perfect Matchings&lt;/a&gt; in a bipartite graph.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Items form the left set. Bins form the right. If a bin needs k items, you make k copies of that bin.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Now you form an edge between an item and a bin if the item is a possible candidate for the bin.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Now you need to find number of perfect matchings in this graph.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Unfortunately, this is a hard problem. Calculating the number of perfect matchings in a graph is equivalent to finding the &lt;a href=&quot;http://en.wikipedia.org/wiki/Permanent&quot; rel=&quot;nofollow&quot;&gt;Permanent&lt;/a&gt; of an associated matrix, which is &lt;a href=&quot;http://en.wikipedia.org/wiki/Sharp-P&quot; rel=&quot;nofollow&quot;&gt;#P-Complete&lt;/a&gt;. See: &lt;a href=&quot;http://en.wikipedia.org/wiki/Computation_of_the_permanent_of_a_matrix&quot; rel=&quot;nofollow&quot;&gt;Computation of the Permanent&lt;/a&gt;. &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Your best bet might be randomized/approximation algorithms.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="984" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-09-16T21:58:48.003" LastActivityDate="2010-09-16T21:58:48.003" CommentCount="1" />
  <row Id="2075" PostTypeId="2" ParentId="674" CreationDate="2010-08-10T17:42:12.703" Score="3" ViewCount="" Body="&lt;p&gt;You probably figured it out already, but for posterity -- this &lt;a href=&quot;http://www-diglib.stanford.edu/~klein/lagrange-multipliers.pdf&quot; rel=&quot;nofollow&quot;&gt;document&lt;/a&gt; gives intuition which I didn't see in Boyd book, esp. Figure 3 on page 3.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To solve equality constrained problem you look for points where gradient of objective is in the same direction as gradient of the constraint function. Why those points? Well, suppose we have a feasible point that's a maximum of objective and the gradients of objective/constraint functions are not collinear. Then you can take an infinitesimal step in the direction orthogonal to the gradient of constraint function to increase the objective. And because you are moving in the direction orthogonal to gradient of the constraint function, your constraint is still satisfied. Hence you get a feasible point with larger objective value, contradiction.&lt;/p&gt;&#xA;" OwnerUserId="998" LastActivityDate="2010-08-10T17:42:12.703" CommunityOwnedDate="2010-08-10T17:42:12.703" />
  <row Id="2076" PostTypeId="2" ParentId="2071" CreationDate="2010-08-10T17:45:59.780" Score="-1" ViewCount="" Body="&lt;p&gt;Sorry, can't explain it either. It's obvious you need at least two random numbers to go greater than 1. And if the random numbers are on average .5 you'll need (edit) on average (/edit) maximum three numbers. So we end up in the range [2..3], but that's as close as I can get to e.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;While the method sure looks curious, it must be about the least efficient way ever to calculate e digits. A quick-and-dirty test program I wrote only gave 4 correct significant digits after a billion iterations. Or maybe Delphi's PRNG isn't perfect.&lt;/p&gt;&#xA;" OwnerUserId="923" LastEditorUserId="923" LastEditorDisplayName="" LastEditDate="2010-08-11T05:52:37.793" LastActivityDate="2010-08-11T05:52:37.793" CommentCount="3" />
  <row Id="2077" PostTypeId="2" ParentId="5" CreationDate="2010-08-10T17:49:48.213" Score="3" ViewCount="" Body="&lt;p&gt;If $\sqrt 2$ were rational, we could write it as a fraction $a/b$ in&#xA;&lt;strong&gt;lowest terms&lt;/strong&gt;. Then&#xA;$$a^2 = 2 b^2.$$&#xA;Look at the last digit of $a^2$. It has to be $0$, $1$, $4$, $5$, $6$ or $9$.&#xA;Now look at the last digit of $2b^2$. It has to be $0$, $2$ or $8$.&#xA;As $a^2$ and $2b^2$ are the same number, its last digit must be $0$.&#xA;But that's only possible if $a$ ends in $0$ and $b$ ends in $0$ or $5$.&#xA;Either way both $a$ and $b$ are multiples of $5$ contradicting $a/b$&#xA;being in lowest terms.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-08-10T17:49:48.213" CommentCount="2" />
  <row Id="2078" PostTypeId="2" ParentId="697" CreationDate="2010-08-10T17:51:50.707" Score="2" ViewCount="" Body="&lt;p&gt;For regression, least sum of squares criterion tried to fit your function through the mean of the data. In other words, with enough degrees of freedom, value of your fitted function for x will be the average of all observed values at x. Least sum of absolute values instead produces a function that'll go through the median of observed values. Some &lt;a href=&quot;http://hunch.net/?p=269&quot; rel=&quot;nofollow&quot;&gt;discussion&lt;/a&gt; on this&lt;/p&gt;&#xA;" OwnerUserId="998" LastActivityDate="2010-08-10T17:51:50.707" />
  <row Id="2079" PostTypeId="1" AcceptedAnswerId="2081" CreationDate="2010-08-10T18:05:11.567" Score="2" ViewCount="349" Body="&lt;p&gt;Is there a isomorphism between the additive group of real algebraic numbers and the multiplicative group of positive real algebraic numbers, which is order preserving.&lt;/p&gt;&#xA;" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-08-18T04:55:45.747" LastActivityDate="2010-09-02T16:16:44.620" Title="Isomorphism between algebraic Numbers." Tags="&lt;abstract-algebra&gt;" AnswerCount="2" />
  <row Id="2080" PostTypeId="2" ParentId="1901" CreationDate="2010-08-10T18:08:08.363" Score="7" ViewCount="" Body="&lt;p&gt;This would have belonged as a comment to MattE answer, but I do not have the reputation to leave comments. The reason it is impossible to algorithmically classify topological manifolds (up to homotopy) is, as MattE very clearly explained, that the word problem for groups can be embedded in the question of classifying manifolds. Indeed, suppose that you want to decide if a finitely presented group $G$ is trivial. First, there are several constructions that yield a compact 4-manifold whose fundamental group is isomorphic to $G$; choose your favorite construction and let $M$ be the resulting manifold. Observe that the intersection pairing in homology is computable, so we may just as well assume that we know what this is for our manifold $M$. It turns out that the homotopy type of a simply connected manifold is completely determined by the intersection form on the second integral homology group (and that any intersection form can be realised). So if on the side we construct the simply connected manifold $N$ with intersection form coinciding with the one of $M$, then deciding whether the homotopy types of $M$ and $N$ coincide will decide if the fundamental group of $M$ is trivial or not, i.e. if $G$ is the trivial group or not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are some issues on how you compute the intersection pairing and how you find the manifold $N$ given the intersection form, and there are tricks to simplify all these questions, but I will not go into this, in the hope that what I wrote will suffice! Of course, if you want more details, feel free to ask, and I will try to answer!&lt;/p&gt;&#xA;" OwnerUserId="1028" LastActivityDate="2010-08-10T18:08:08.363" CommentCount="2" />
  <row Id="2081" PostTypeId="2" ParentId="2079" CreationDate="2010-08-10T18:24:44.647" Score="19" ViewCount="" Body="&lt;p&gt;Any such isomorphism would have to extend to a continuous morphism from the additive group of the real numbers to the multiplicative group of the positive real numbers because the order-preserving condition means convergent sequences get sent to convergent sequences.  Any such morphism is $a^x$ for some positive real $a$, and this is impossible e.g. by the &lt;a href=&quot;http://en.wikipedia.org/wiki/Gelfond%E2%80%93Schneider_theorem&quot; rel=&quot;nofollow&quot;&gt;Gelfond-Schneider theorem&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-08-10T18:24:44.647" CommentCount="2" />
  <row Id="2082" PostTypeId="2" ParentId="2071" CreationDate="2010-08-10T18:30:37.350" Score="1" ViewCount="" Body="&lt;p&gt;I believe the answer is related to this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$e = \sum_{n = 0}^\infty \frac{1}{n!} = \frac{1}{0!} + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \frac{1}{4!} + \cdots$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In that, by choosing random numbers, you're approximating a sampling from that series.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For your amusement, here's a little Python program that does a brute force calculation:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;#!/usr/bin/env python&#xA;import random&#xA;iterations = 100000000&#xA;count = 0.0&#xA;for i in range(iterations):&#xA;    sum = 0&#xA;    while sum &amp;lt; 1:&#xA;            sum += random.random()&#xA;            count += 1&#xA;print count / iterations&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="208" LastActivityDate="2010-08-10T18:30:37.350" CommentCount="4" />
  <row Id="2083" PostTypeId="2" ParentId="2045" CreationDate="2010-08-10T18:59:50.740" Score="1" ViewCount="" Body="&lt;p&gt;Have you tried reducing the matrix to row echelon form? Maybe my answer to this question:&#xA; &lt;a href=&quot;http://math.stackexchange.com/questions/1822/is-reducing-a-matrix-to-row-echelon-form-useful-at-all/1831#1831&quot;&gt;http://math.stackexchange.com/questions/1822/is-reducing-a-matrix-to-row-echelon-form-useful-at-all/1831#1831&lt;/a&gt; could be helpful.&lt;/p&gt;&#xA;" OwnerUserId="664" LastActivityDate="2010-08-10T18:59:50.740" CommentCount="6" />
  <row Id="2084" PostTypeId="2" ParentId="2071" CreationDate="2010-08-10T19:21:35.857" Score="4" ViewCount="" Body="&lt;p&gt;The other answers should have resolved why the number is &gt;2, but for the exact reason why it is &lt;em&gt;e&lt;/em&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For having N random numbers whose sum is &amp;lt;1, the probability is exactly the same as the volume of a unit &lt;a href=&quot;http://en.wikipedia.org/wiki/Simplex&quot; rel=&quot;nofollow&quot;&gt;standard N-simplex&lt;/a&gt;, i.e. 1/N!.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, the probability that it takes exactly N random numbers for a sum of ≥1, the probability is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \frac1{(N-1)!} - \frac1{N!} = \frac1{N(N-2)!} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore the expected number is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \sum_{N=1}^\infty \frac N{N(N-2)!} = e $$&lt;/p&gt;&#xA;" OwnerUserId="171" LastEditorUserId="171" LastEditorDisplayName="" LastEditDate="2010-08-10T19:35:54.823" LastActivityDate="2010-08-10T19:35:54.823" CommentCount="3" />
  <row Id="2085" PostTypeId="1" CreationDate="2010-08-10T20:22:17.710" Score="0" ViewCount="439" Body="&lt;p&gt;Please explain Ito's Lemma in layman's terms if possible and how this relates to Taylor series&lt;/p&gt;&#xA;" OwnerUserId="1032" LastActivityDate="2010-08-10T20:22:17.710" ClosedDate="2010-08-11T00:34:47.973" Title="What is Ito's Lemma?" Tags="&lt;stochastic-processes&gt;" AnswerCount="0" CommentCount="6" />
  <row Id="2086" PostTypeId="2" ParentId="1984" CreationDate="2010-08-10T20:33:38.477" Score="3" ViewCount="" Body="&lt;p&gt;No, but reputation counts for a lot and not having an academic track record means that you are missing many of the easy credibility markers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Interestingly, when I was working towards my PhD back in the mid 1990s my supervisor made an interesting observation that some of &lt;em&gt;his&lt;/em&gt; best work in fact had been the hardest to get published.  He felt that really innovative, ground breaking ideas were often the most difficult to get accepted and were subjected to far greater scrutiny than normal.&lt;/p&gt;&#xA;" OwnerUserId="1032" LastActivityDate="2010-08-10T20:33:38.477" />
  <row Id="2087" PostTypeId="2" ParentId="2001" CreationDate="2010-08-10T20:42:55.163" Score="3" ViewCount="" Body="&lt;p&gt;In infinite dimensions, all hell breaks loose. For example, neither the category of Banach spaces nor the category of Hilbert spaces, although additive, are abelian.&lt;/p&gt;&#xA;" OwnerUserId="1033" LastActivityDate="2010-08-10T20:42:55.163" />
  <row Id="2093" PostTypeId="1" CreationDate="2010-08-10T21:32:53.563" Score="5" ViewCount="241" Body="&lt;p&gt;How can I determine the size of the largest collection of k-element subsets of an n-element set such that each pair of subsets has at most m elements in common?&lt;/p&gt;&#xA;" OwnerUserId="984" LastEditorUserId="984" LastEditorDisplayName="" LastEditDate="2010-08-10T21:57:37.837" LastActivityDate="2010-10-12T07:19:33.643" Title="Choosing subsets of a set with a specified amount of maximum overlap between them" Tags="&lt;combinatorics&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="1" />
  <row Id="2094" PostTypeId="2" ParentId="2013" CreationDate="2010-08-10T21:35:57.213" Score="1" ViewCount="" Body="&lt;p&gt;See Wikipedia &lt;a href=&quot;http://en.wikipedia.org/wiki/Random_variable#Functions_of_random_variables&quot; rel=&quot;nofollow&quot;&gt;Functions of random variables&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="247" LastActivityDate="2010-08-10T21:35:57.213" CommentCount="1" />
  <row Id="2095" PostTypeId="2" ParentId="206" CreationDate="2010-08-10T21:44:15.657" Score="0" ViewCount="" Body="&lt;p&gt;In practical application of mathematical analysis, the most useful &quot;well-behavedness&quot; condition on a function would be &lt;a href=&quot;http://en.wikipedia.org/wiki/Lipschitz_continuity&quot; rel=&quot;nofollow&quot;&gt;Lipshitz continuity&lt;/a&gt;. This ensures that the variations of a function are not too wild. The most important consequence is that differential equations will have a unique solution. All sorts of modelling of natural and physical situations uses differential equations, and it is valuable to know abstractly that they would have solutions under some tame conditions. Given such a theorem, one can concentrate exclusively on finding the solutions.&lt;/p&gt;&#xA;" OwnerUserId="977" LastActivityDate="2010-08-10T21:44:15.657" />
  <row Id="2096" PostTypeId="2" ParentId="227" CreationDate="2010-08-10T21:50:30.897" Score="0" ViewCount="" Body="&lt;p&gt;You should certainly know ordinary differential equations, partial differential equations, probability theory and statistics. &lt;/p&gt;&#xA;" OwnerUserId="977" LastActivityDate="2010-08-10T21:50:30.897" CommunityOwnedDate="2010-08-12T14:58:43.903" />
  <row Id="2098" PostTypeId="2" ParentId="1096" CreationDate="2010-08-10T22:53:09.750" Score="0" ViewCount="" Body="&lt;p&gt;Lets define what is &lt;code&gt;x+x+x+... x times for x - real&lt;/code&gt;. Natural definition is &lt;code&gt;x+x+x.. := x*x&lt;/code&gt; (note - just the same as Isaac has wrote in his edit).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose we want left our initial definition as is. We don't know what is &lt;code&gt;x+x+.. repeat x times for x - real&lt;/code&gt; (and note we don't have rule how to obtain derivative from such func). So lets use definition of derivative. &lt;code&gt;f(x):=x+x+x.. repeat x times&lt;/code&gt;, &lt;code&gt;Df(x)=(f(x+h)-f(x))/h, h-&amp;gt;0&lt;/code&gt;. &lt;code&gt;Df(x)=((x+h+x+h+x+h.. repeat x+h times) - (x+x+x.. repeat x times))/h, h-&amp;gt;0&lt;/code&gt;. Suppose &lt;code&gt;x+x+... repeat a+b times := (x+x+.. repeat a times) + (x+x+.. repeat b times)&lt;/code&gt; we have &lt;code&gt;Df(x)=((x+h+x+h+x+h.. repeat x times) - (x+x+x.. repeat x times) + (x+h+x+h+x+h.. repeat h times))/h, h-&amp;gt;0&lt;/code&gt;, &lt;code&gt;Df(x)=((h+h+h.. repeat x times) + (x+h+x+h+x+h.. repeat h times))/h, h-&amp;gt;0&lt;/code&gt;, or &lt;code&gt;Df(x)=(1+1+1.. repeat x times) + (x+h+x+h+x+h.. repeat 1 times), h-&amp;gt;0&lt;/code&gt; and at last &lt;code&gt;Df(x)=x + x+h, h-&amp;gt;0 = 2x&lt;/code&gt;&lt;/p&gt;&#xA;" OwnerUserId="1035" LastActivityDate="2010-08-10T22:53:09.750" />
  <row Id="2099" PostTypeId="1" CreationDate="2010-08-10T23:08:39.053" Score="4" ViewCount="204" Body="&lt;p&gt;From:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Moduli_space&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Moduli_space&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;we see that moduli of curves is a very algebro-geometric topic. It is easy to understand its relevance and importance in algebraic geometry. But the mind boggles when we try to imagine how on earth such a topic from pure and abstruse mathematics is relevant in physics. I will be thankful if somebody can give some explanation.&lt;/p&gt;&#xA;" LastActivityDate="2010-08-11T14:15:58.043" Title="How is moduli of curves relevant in physics?" Tags="&lt;algebraic-geometry&gt;&lt;physics&gt;" AnswerCount="2" FavoriteCount="2" />
  <row Id="2100" PostTypeId="1" AcceptedAnswerId="2105" CreationDate="2010-08-10T23:31:38.860" Score="1" ViewCount="415" Body="&lt;p&gt;I am trying to brush up my graph theory skills. I have not done any in over 4 years and i am rusty...If someone could help me out with this simple proof i would appreciate it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Prove that for any graph $G$ of order at least 2, the degree sequence has at least one pair of repeated entries.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the degree sequence if a list of the degrees of each vertex (usually written in descending order).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that the sum of the degrees of the vertices of a graph is equal to $2|E|$ and that the number of vertices of odd degree is even.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If someone could help me out and point me in the right direction i would appreciate it.&lt;/p&gt;&#xA;" OwnerUserId="43" LastActivityDate="2010-08-11T22:04:10.543" Title="Degree Sequence of a Graph" Tags="&lt;graph-theory&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="2101" PostTypeId="2" ParentId="2099" CreationDate="2010-08-10T23:39:27.287" Score="6" ViewCount="" Body="&lt;p&gt;In string theory, a particle is actually a string.  In closed string theory, it is a closed string, i.e. a loop.  As time progresses, this loop will trace out a surface (its &lt;em&gt;worldsheet&lt;/em&gt;).  If you follow the particle from its creation to its destruction, this surface will be closed.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For certain computations of physical quantities attached to the worldsheet (and perhaps at this point someone with more expertise can add details) one has to choose a conformal structure on the worldsheet; but to get an answer independent of the choice, one then integrates over all possible conformal structures; or sometimes one can show that the quantity computed is independent of the conformal structure (perhaps because it is computed by a horizontal section of a bundle with connection over the space of conformal structures). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, we note that the space of all conformal structures on a closed topological surface of genus $g$ is the same as the moduli space of algebraic curves of genus $g$.  &lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2010-08-10T23:39:27.287" />
  <row Id="2102" PostTypeId="2" ParentId="2100" CreationDate="2010-08-10T23:50:24.413" Score="2" ViewCount="" Body="&lt;p&gt;If there were no repeats, what would the degree sequence look like?  Why could it not look like that?&lt;/p&gt;&#xA;" OwnerUserId="297" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-08-11T22:04:10.543" LastActivityDate="2010-08-11T22:04:10.543" CommentCount="4" />
  <row Id="2103" PostTypeId="2" ParentId="2045" CreationDate="2010-08-11T00:11:33.450" Score="2" ViewCount="" Body="&lt;p&gt;If you have a $n\times n$ matrix $A$ , the inverse matrix $A^{-1}$ can be&#xA;computed by using a &lt;em&gt;compact&lt;/em&gt; method, that consists of&#xA;the following steps:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Augment your matrix with the identity matrix $I$: $\left( A|I\right) $&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Use &lt;a href=&quot;http://en.wikipedia.org/wiki/Gaussian_elimination&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;Gaussian elimination&lt;/em&gt;&lt;/a&gt; to get an upper triangular matrix on the left&#xA;and a matrix $J$ on the right: $\left( U|J\right) $, where $J=\left(&#xA;J_{1}|J_{2}|\ldots |J_{n}\right) $. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;To get the $n$ vector columns of the inverse matrix: $A^{-1}=\left(&#xA;x_{1}|x_{2}|\ldots |x_{n}\right) $, solve for $x_{k}$, with $k=1,2,\ldots ,n$, $n$ systems of equations $U\cdot x_{k}=J_{k}$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Note 1: in your problem $n=4$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note 2: there is a variant of this method called &quot;Gaussian elimination with&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Gaussian_elimination&quot; rel=&quot;nofollow&quot;&gt;partial pivoting&lt;/a&gt;&quot; that gives a better accuracy, when the entries are not&#xA;rational numbers, which is not the present situation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Until now I have not yet computed matrix $A^{-1}$ in this case. Instead I have used&#xA;The Scientific Notebook (included in the Scientific Work Place) to find that your matrix satisfies $A=A^{-1}$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$A=\begin{pmatrix}1/2&amp;amp;1/2&amp;amp;1/2&amp;amp;1/2 \\ 1/2&amp;amp;1/2&amp;amp;-1/2&amp;amp;-1/2\\1/2&amp;amp;-1/2&amp;amp;1/2&amp;amp;-1/2\\1/2&amp;amp;-1/2&amp;amp;-1/2&amp;amp;1/2\end{pmatrix}=A^{-1}$&lt;/p&gt;&#xA;" OwnerUserId="752" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2010-08-11T00:42:38.457" LastActivityDate="2010-08-11T00:42:38.457" CommentCount="4" />
  <row Id="2104" PostTypeId="2" ParentId="153" CreationDate="2010-08-11T00:51:11.700" Score="14" ViewCount="" Body="&lt;p&gt;As someone who uses both Matlab and Mathematica regularly, I'd like to give some strengths of Matlab over Mathematica (since Mehper C. Palavuzlar gave such a good overview of Mathematica's strengths over Matlab)&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;You can make Matlab scripts fully stand-alone. Yes, it will be a 500 MB executable for a program that prints &quot;Hello World&quot;, but it can be run by someone who doesn't have Matlab. This also makes it possible to run your calculations in parallel on a server farm like Amazon's EC2 (is it even possible to use EC2 for Mathematica's computations?)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Matlab has a much larger and more active community. There are more Matlab blogs out there to browse for tips and you can get answers to questions on Matlab Central within 10 minutes sometimes. Mathematica has a moderated newsgroup which is high quality, but slow -- because of moderation it typically takes one day for your question to be seen by other people&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Bigger community means more ready-to-use code out there. For instance, I started using Matlab for some visual recognition tasks, and I could get pretty much all the pieces for segmentation, low level feature extraction, adaptive thresholding etc from the web, whereas in Mathematica I would have to implement them from scratch. It's common for researchers to publish Matlab code they used for their computations on their homepages, whereas I'm yet to see anyone (at least in vision/statistics/optimization) publishing Mathematica code to go with along with their papers&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Matlab's language is easier to learn (and to debug!). Mathematica offers an &quot;rewriting system&quot; paradigm unfamiliar to many, and even after years of programming in Mathematica I occasionally run into unexpected behaviors due to some intricacy of the evaluation sequence&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Overall, I'd recommend to go with Mathematica if you want to play around by yourself, and Matlab if you want to share your code or use code of others.&lt;/p&gt;&#xA;" OwnerUserId="998" LastEditorUserId="998" LastEditorDisplayName="" LastEditDate="2010-08-11T21:01:40.567" LastActivityDate="2010-08-11T21:01:40.567" CommentCount="1" />
  <row Id="2105" PostTypeId="2" ParentId="2100" CreationDate="2010-08-11T01:43:59.523" Score="4" ViewCount="" Body="&lt;p&gt;Yes, it just came to me after i posted the comment to the previous answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A graph with at least 2 vertices and no edges the 0 degree is repeated.&#xA;A graph with 2 vertices and 1 edge the degree 1 is repeated.&#xA;Each vertex has at least one entry in the degree sequence, so there are a total of $N$ entries. But each degree can have a maximum of degree = $N-1$. Therefore according to the pigeonhole principle at least one of the degrees has to repeat.&lt;/p&gt;&#xA;" OwnerUserId="43" LastEditorUserId="43" LastEditorDisplayName="" LastEditDate="2010-08-11T01:57:23.813" LastActivityDate="2010-08-11T01:57:23.813" CommentCount="2" />
  <row Id="2107" PostTypeId="2" ParentId="2099" CreationDate="2010-08-11T03:27:04.563" Score="4" ViewCount="" Body="&lt;p&gt;In quantum field theory many quantities are calculated as (formal) Feynman path integrals, that is, integrals &quot;over the space of all paths&quot;.  To make sense of this in dimensions higher than 2, one uses a perturbation expansion (in powers of Planck's constant) of the integral which leads to a sequence of finite dimensional integrals described by Feynman diagrams: finite graphs with (among other decorations) some &quot;input&quot; and &quot;output&quot; vertices --- think of a process where some particles collide and some are emitted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In string theory instead of a sum over paths of a point there is a sum over trajectories of a closed loop (a string), i.e., integration over some space of surfaces.  The analogue of the finite-dimensional perturbative Feynman diagrams are surfaces connecting some oriented loops. Loops with positive orientation relative to the surface are the &quot;inputs&quot; and the others are the &quot;outputs&quot;.  The surface can also have empty input or output.  Integration over all surfaces of this type is still an infinite-dimensional problem.  To reduce to a finite-dimensional integral, one integrates over the moduli space of conformal structures on each topological type of bordered surface.  This is finite-dimensional and there are standard measures on it (Weil-Petersson).  For problems with conformal invariance (&quot;conformal field theory&quot;) this recipe was given by Polyakov and is basic for string theory.   As far as I understand it, the worldsheet is always governed by a conformal field theory, so the Polyakov recipe is &quot;the&quot; method for defining the observables in perturbative string theory.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, for open strings one might want the boundaries to be open circles, or punctures, thus cutting out closed discs or points from the surface.   So the general moduli space is that of  conformal structures on surfaces of a fixed finite genus with a given finite type of boundary and one can at least define a measure on this space.  Rigorous calculations can be extremely complicated as seen in the arxiv papers by D'Hoker and Phong where they went through a huge tour-de-force to construct the perturbative superstring measure through 2 and 3 loops.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As Matt explained, the relation to algebraic geometry is that (at least for compact surfaces without boundary) the moduli space of conformal structures is the same as that of algebraic structures; in each conformal class there is an algebraic curve of that genus.  I am less sure of this in the non-compact case of punctured surfaces, and for bordered Riemann surfaces where the boundary contains loops, the moduli space is of geometric/symplectic/complex-analytic nature and I don't know if it has an algebraic analogue.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(edit: I think that there is a basic difference between QFT and string theory in this analogy, because in string theory only the perturbative measure is known at present, while in QFT path-space integrals are thought of as an underlying non-perturbative theory and one can make sense of that to a large extent even if Feynman's &quot;measure&quot; on path space doesn't exist as an integration measure in the ordinary mathematical sense.   In string theory the non-perturbative definition of the theory is presently unknown.  There are objects in string theory that are considered non-perturbative and thus hints of the underlying theory, but even conjecturally or heuristically there is not a consensus as to what perturbative string theory is a perturbation of.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(added: &#xA;According to this paper, for bosonic string theory in Polyakov's prescription, the moduli spaces in the perturbative integrals are those of smooth closed surfaces, without punctures or boundary loops.  This is interesting because conformal field theory does use the more complicated surfaces, but I don't know the details of how CFT is used in string theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://projecteuclid.org/DPubS?service=UI&amp;amp;version=1.0&amp;amp;verb=Display&amp;amp;handle=euclid.cmp/1104116138&quot; rel=&quot;nofollow&quot;&gt;http://projecteuclid.org/DPubS?service=UI&amp;amp;version=1.0&amp;amp;verb=Display&amp;amp;handle=euclid.cmp/1104116138&lt;/a&gt;   )&lt;/p&gt;&#xA;" OwnerUserId="467" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-08-11T14:15:58.043" LastActivityDate="2010-08-11T14:15:58.043" />
  <row Id="2108" PostTypeId="2" ParentId="1096" CreationDate="2010-08-11T04:24:21.853" Score="7" ViewCount="" Body="&lt;p&gt;I think the discrete/continuous issue is sort of a red herring.  To me, the problem is forgetting to use the chain rule!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To un-discretize, think of the function $F(u,v) = uv$, which we could think of as $u + \dots + u$, $v$ times.  Then $x^2 = F(x,x)$.  Differentiating both sides gives $2x = F_u(x,x) + F_v(x,x)$, which is perfectly true.  In the fallacious example, the problem is essentially that the $F_v$ term has been omitted.  In some sense, one has forgotten to differentiate the operation &quot;$x$ times&quot; with respect to $x$!  Of course, the notation makes this easier to do.&lt;/p&gt;&#xA;" OwnerUserId="822" LastActivityDate="2010-08-11T04:24:21.853" CommentCount="1" />
  <row Id="2110" PostTypeId="1" CreationDate="2010-08-11T06:29:51.057" Score="2" ViewCount="150" Body="&lt;p&gt;Here is a time sample: $Q = \{(t_i, x_i) | 0 \leq x_i \leq x_{i+1}, 1 \leq i \leq n\}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and rules:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) $T_1 \leq t_{i+1} - t_i &amp;lt; T_2$ where  $T_1, T_2 &gt; 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) $x_i$ comes with error:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x_i = \lfloor x(t_i) \rfloor$ where $\lfloor x \rfloor$ - whole part of $x$, $x(t_i)$ - voluntary unknown law of object motion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How we can find (approximately but sufficiently accurate) speed of the object at the given time?&lt;/p&gt;&#xA;" OwnerUserId="1041" LastEditorUserId="1041" LastEditorDisplayName="" LastEditDate="2010-11-21T21:00:22.003" LastActivityDate="2010-11-21T21:00:22.003" Title="Determine speed of the object at the current time by the non-uniform time sample" Tags="&lt;sequences-and-series&gt;&lt;statistics&gt;&lt;numerical-methods&gt;&lt;approximation&gt;" AnswerCount="3" CommentCount="5" FavoriteCount="1" />
  <row Id="2111" PostTypeId="2" ParentId="2029" CreationDate="2010-08-11T06:40:59.707" Score="0" ViewCount="" Body="&lt;p&gt;There is infinite may constrains on digits of the prime number. Each one may be deduced as follows ( from Euclid method): suppose You have number $S =  \sum 10^i *a_i$ which is divisible by $p$. Then You have $S= ( \sum 10^j * b_j )*p = \sum 10^j b_j*p $ . You may perform multiplication and look for additional constrains. So as You have criteria for divisibility by 2, 4, or 5, which gives You constraints on last digits, You may construct similar criteria for divisibility by 13 etc.  as well. It is possible, that some of them gives You constrains of the form:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for S&gt;p last digit has to be in a set M, where M is proper subset of {1,3,7}&lt;/p&gt;&#xA;" OwnerUserId="1043" LastActivityDate="2010-08-11T06:40:59.707" />
  <row Id="2112" PostTypeId="1" AcceptedAnswerId="2150" CreationDate="2010-08-11T06:55:31.287" Score="3" ViewCount="676" Body="&lt;p&gt;The reason I ask this is because difficult mathematical proofs are just not plain self-evident. You would need a few years of intensive study before you can get to the point of understanding the topics and the proofs. The problem is that for a difficult math theorem, maybe only a handful of mathematicians have studied and agreed on the validity of the proof. There is a chance, given that the topic is highly esoteric, that there are bugs in the proof that render it invalid. We just don't know. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This would give us a conundrum; we thought that mathematics is the most secure branch of knowledge, but it isn't; because a lot of the proofs are convoluted and esoteric to the point that only a handful a people who can understand them. So we can never be 100% sure that a proof is always valid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Am I right?&lt;/p&gt;&#xA;" OwnerUserId="550" LastActivityDate="2010-11-05T11:26:06.570" CommunityOwnedDate="2010-08-12T13:15:22.263" Title="Is It True that We Can Never Be Sure of Validity of a Mathematical Proof?" Tags="&lt;meta-math&gt;" AnswerCount="6" CommentCount="7" FavoriteCount="2" />
  <row Id="2113" PostTypeId="2" ParentId="2112" CreationDate="2010-08-11T06:59:51.727" Score="0" ViewCount="" Body="&lt;p&gt;You'll be sure when you'll understand it.&lt;/p&gt;&#xA;" OwnerUserId="480" LastActivityDate="2010-08-11T06:59:51.727" CommunityOwnedDate="2010-08-12T13:15:22.263" CommentCount="3" />
  <row Id="2114" PostTypeId="2" ParentId="2093" CreationDate="2010-08-11T07:16:03.263" Score="0" ViewCount="" Body="&lt;p&gt;Hi Please refer this link: &lt;a href=&quot;http://gilkalai.wordpress.com/2008/10/06/extremal-combinatorics-iv-shifting/&quot; rel=&quot;nofollow&quot;&gt;http://gilkalai.wordpress.com/2008/10/06/extremal-combinatorics-iv-shifting/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can also look here as well to get more idea for this type of problems: &lt;a href=&quot;http://mathoverflow.net/questions/25067/given-n-k-element-subsets-of-n-is-there-a-small-subset-a-of-n-which-intersects-t&quot; rel=&quot;nofollow&quot;&gt;http://mathoverflow.net/questions/25067/given-n-k-element-subsets-of-n-is-there-a-small-subset-a-of-n-which-intersects-t&lt;/a&gt;&lt;/p&gt;&#xA;" LastActivityDate="2010-08-11T07:16:03.263" />
  <row Id="2115" PostTypeId="1" AcceptedAnswerId="30460" CreationDate="2010-08-11T07:32:20.277" Score="12" ViewCount="772" Body="&lt;p&gt;We know that there exist real functions which are continuous at each irrational and dis-&#xA;continuous at each rational number. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But does there exist a function $f: \mathbb{R} \to \mathbb{R}$ that is differentiable at every irrational and discontinuous at every rational?&lt;/p&gt;&#xA;" OwnerUserId="9413" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2011-04-02T05:07:08.280" LastActivityDate="2011-04-02T05:07:08.280" Title="Discontinuous at rationals and differentiable at irrationals?" Tags="&lt;real-analysis&gt;" AnswerCount="2" CommentCount="5" FavoriteCount="6" />
  <row Id="2116" PostTypeId="1" AcceptedAnswerId="2177" CreationDate="2010-08-11T07:41:57.820" Score="0" ViewCount="221" Body="&lt;p&gt;For any natural number $n &gt; 1$, define $E(n)$,to be the highest exponent to&#xA;which a prime divides it. For instance, $E(12)=E(36)=2$. Show that $$\lim_{N \to \infty} \frac{1}{N} \sum\limits_{n=2}^{N} E(n)$$ exists and find its value &lt;/p&gt;&#xA;" LastEditorUserId="171" LastEditorDisplayName="" LastEditDate="2010-08-11T15:13:30.920" LastActivityDate="2010-08-11T22:02:04.303" Title="Mean of highest exponent in prime factorization of all integers ≥ 2" Tags="&lt;number-theory&gt;" AnswerCount="1" CommentCount="6" />
  <row Id="2120" PostTypeId="1" AcceptedAnswerId="2122" CreationDate="2010-08-11T08:30:49.793" Score="8" ViewCount="3091" Body="&lt;p&gt;We know that $\displaystyle\zeta(2)=\sum\limits_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6}$ and it converges.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now does there exists a bijective map $f:\mathbb{N} \to \mathbb{N}$ such that the sum $$\sum\limits_{n=1}^{\infty} \frac{f(n)}{n^2}$$ converges.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If our $s=2$ was not fixed, then can we have a function such that $\displaystyle \zeta(s)=\sum\limits_{n=1}^{\infty} \frac{f(n)}{n^s}$ converges&lt;/p&gt;&#xA;" OwnerUserId="9413" LastEditorUserId="226" LastEditorDisplayName="" LastEditDate="2010-10-07T18:44:22.227" LastActivityDate="2010-11-09T18:25:33.667" Title="Does there exist a bijective $f:\mathbb{N} \to \mathbb{N}$ such that $\sum f(n)/n^2$ converges?" Tags="&lt;number-theory&gt;&lt;convergence&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="3" />
  <row Id="2121" PostTypeId="1" AcceptedAnswerId="2153" CreationDate="2010-08-11T08:41:14.963" Score="13" ViewCount="479" Body="&lt;p&gt;Does there exists a finite group $G$ and a normal subgroup $H$ of $G$ such that $|Aut(H)|&gt;|Aut(G)|$&lt;/p&gt;&#xA;" OwnerUserId="9413" LastEditorUserId="9413" LastEditorDisplayName="" LastActivityDate="2011-03-27T18:42:24.940" Title="Existence of a normal subgroup with $|Aut(H)|&gt;|Aut(G)|$" Tags="&lt;group-theory&gt;" AnswerCount="6" FavoriteCount="3" />
  <row Id="2122" PostTypeId="2" ParentId="2120" CreationDate="2010-08-11T08:41:34.213" Score="32" ViewCount="" Body="&lt;p&gt;For $s&gt;2$ you can take $f(n)=n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $s=2$ if you have $m &amp;lt; n$ and $a=f(m) &gt; b=f(n)$ then&#xA;$$\frac{a}{m^2}+\frac{b}{n^2}&gt;\frac{b}{m^2}+\frac{a}{n^2}$$&#xA;(proved either naively or as a case of the &quot;rearrangement inequality&quot;)&#xA;so the sum $\sum f(n)/n^2$ can be reduced by swapping the values $f(m)$ and $f(n)$.&#xA;Hence&#xA;$$\sum\frac{f(n)}{n^2}\ge\sum\frac{n}{n^2}$$&#xA;which is divergent.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-08-11T08:41:34.213" />
  <row Id="2123" PostTypeId="2" ParentId="2121" CreationDate="2010-08-11T08:48:44.993" Score="6" ViewCount="" Body="&lt;p&gt;An elementary Abelian group can have a large automorphism group.&#xA;Let $H$ be a largish elementary Abelian group, and $G$&#xA;be an extension of $H$, say be a semidirect product with&#xA;a cyclic group acting faithfully on $H$. Then it is possible&#xA;for Aut$(G)$ to be smaller than Aut$(H)$.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-08-11T08:48:44.993" CommentCount="4" />
  <row Id="2124" PostTypeId="2" ParentId="2120" CreationDate="2010-08-11T08:50:44.927" Score="9" ViewCount="" Body="&lt;p&gt;For $s=2$ the answer is negative. This series doesn't converge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To prove this we can use &lt;a href=&quot;http://en.wikipedia.org/wiki/Abel_transformation&quot; rel=&quot;nofollow&quot;&gt;Abel transformation&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\sum_{n=1}^{n=N} \frac{f(n)}{n^2} = \sum_{n=1}^{n=N} (\sum_{k=1}^{k=n} f(k)) (\frac{1}{n^2} - \frac{1}{(n + 1)^2}) + (\sum_{n=1}^{N} \frac{1}{n^2})\frac{1}{(N+1)^2}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $f$ is bijection, $\sum_{k=1}^{k=n} f(k) \ge \frac{n^2}{2}$ hence the first sum is greater than $\sum_{n=1}^{N}\frac{c}{n}$ for some $c &gt; 0$.&lt;/p&gt;&#xA;" OwnerUserId="334" LastActivityDate="2010-08-11T08:50:44.927" />
  <row Id="2125" PostTypeId="1" AcceptedAnswerId="2127" CreationDate="2010-08-11T09:05:06.373" Score="6" ViewCount="201" Body="&lt;p&gt;&lt;strong&gt;Problem&lt;/strong&gt;: Let $n$ be an integer and $p$ a prime dividing $5(n^2-n+\frac{3}{2})^2-\frac{1}{4}$. Prove that $p \equiv 1 \pmod{10}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The polynomial can be re-written as $(\sqrt{5}(n^2-n+\frac{3}{2})-\frac{1}{2})(\sqrt{5}(n^2-n+\frac{3}{2})+\frac{1}{2})$. If this vanishes mod $p$ then $5$ is a quadratic residue mod $p$, which shows that $p \equiv \pm 1 \pmod{5}$ (the primes 2 and 5 are easily ruled out). It feels like the problem should be solvable by understanding the splitting of primes in the splitting field of this polynomial, but I can't find an appropriate &quot;reciprocity law&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The things I'm not sure about are:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;How does one rule out the primes congruent to -1 mod 5?&lt;/li&gt;&#xA;&lt;li&gt;Under what circumstances is it the case that the set {rational primes that split in the ring of integers of some number field} is the union of arithmetic progressions? This a kind of generalized reciprocity law but I don't know in what generality they are known to hold.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="308" LastActivityDate="2010-08-11T16:52:12.510" Title="Primes dividing the values of integer polynomials" Tags="&lt;number-theory&gt;&lt;algebraic-number-theory&gt;" AnswerCount="2" FavoriteCount="2" />
  <row Id="2126" PostTypeId="2" ParentId="2121" CreationDate="2010-08-11T09:14:31.163" Score="7" ViewCount="" Body="&lt;p&gt;The quasidihedral group of order 16, QD16, has automorphism group of order 16, but it has Q8 (the quaternion group) as a normal subgroup, and | Aut(Q8) | = 24.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Steve&lt;/p&gt;&#xA;" OwnerUserId="641" LastActivityDate="2010-08-11T09:14:31.163" />
  <row Id="2127" PostTypeId="2" ParentId="2125" CreationDate="2010-08-11T09:16:17.280" Score="5" ViewCount="" Body="&lt;p&gt;The only way that this situation could possibly arise&#xA;is if the splitting field of $f(x)=5(x^2-x+3/2)^2-1/4$&#xA;is the cyclotomic field generated by the fifth roots of unity.&#xA;To show this, it helps to get your hands dirty and actually&#xA;solve the equation $f(x)=0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, number fields are determined up to isomorphism&#xA;by their splitting laws. The splitting law will be determined by a&#xA;congruence condition iff the extension is an Abelian extension of&#xA;$\mathbb{Q}$ (by class field theory).&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-08-11T09:16:17.280" CommentCount="5" />
  <row Id="2129" PostTypeId="2" ParentId="392" CreationDate="2010-08-11T10:36:44.690" Score="0" ViewCount="" Body="&lt;p&gt;In the spirit of the question, this answer addresses the remark: &quot;Or, even more amazingly, the right triangle definition alone&quot;. In essence the same &lt;a href=&quot;http://math.stackexchange.com/questions/392/intuitive-understanding-of-the-derivatives-of-sinx-and-cosx/871#871&quot;&gt;answer as David Lewis'&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Geometrically &lt;code&gt;d(sinθ)/dθ&lt;/code&gt; can be derived in a right triangle by enlarging the right triangle from &lt;code&gt;θ-&amp;gt;θ+dθ&lt;/code&gt; while keeping &lt;code&gt;a:=adj&lt;/code&gt; and the right angle fixed. In first order &lt;code&gt;d(sinθ)=(o+do)/(h+dh)-o/h≈do/h&lt;/code&gt;, where &lt;code&gt;o:=opp&lt;/code&gt;, &lt;code&gt;h:=hyp&lt;/code&gt;.  The small part of the circle with radius &lt;code&gt;h&lt;/code&gt; that defines &lt;code&gt;dθ&lt;/code&gt; is, again in first order, equal to the opposite side of a triangle with the perpendicular projection of h on h+dh, so that &#xA;&lt;code&gt;dθ=do┴/h&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we see that the derivative of &lt;code&gt;sinθ&lt;/code&gt; equals the proportion between &lt;code&gt;do&lt;/code&gt; and &lt;code&gt;do┴&lt;/code&gt;.&#xA;One can immediately see that this proportion equals &lt;code&gt;sin(π/2-θ)=cos(θ)&lt;/code&gt; in the small triangle in the upper right corner.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So &quot;the proportion between &lt;code&gt;do&lt;/code&gt; and &lt;code&gt;do┴&lt;/code&gt; equals the proportion of the two adjoining sides of the angle &lt;code&gt;θ&lt;/code&gt;&quot; would be the intuitive, geometrical meaning of &lt;code&gt;sin'(θ)=cos(θ)&lt;/code&gt;.&lt;/p&gt;&#xA;" OwnerUserId="706" LastActivityDate="2010-08-11T10:36:44.690" />
  <row Id="2130" PostTypeId="1" CreationDate="2010-08-11T11:01:04.907" Score="2" ViewCount="263" Body="&lt;p&gt;Sometimes one encounter the requirement that the objects of a category needs to be a set. What if it was not, could you provide examples of what could go wrong? (One example per answer).&lt;/p&gt;&#xA;" LastActivityDate="2010-11-20T02:51:11.153" Title="What's so special with small categories?" Tags="&lt;big-list&gt;&lt;set-theory&gt;&lt;category-theory&gt;" AnswerCount="5" CommentCount="8" FavoriteCount="2" />
  <row Id="2131" PostTypeId="1" CreationDate="2010-08-11T11:06:30.357" Score="3" ViewCount="186" Body="&lt;p&gt;I have a data set &lt;code&gt;2 3 3 4 4 4 4&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to find the number of unique numbers of 3 digit numbers that can be formed using this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was thinking of doing &lt;code&gt;7 P 3 /(4!*2!)&lt;/code&gt;, but this doesn't seem right.&lt;/p&gt;&#xA;" LastEditorUserId="705" LastEditorDisplayName="" LastEditDate="2010-08-11T13:11:36.807" LastActivityDate="2010-08-11T15:07:35.173" Title="Permutations with duplicates" Tags="&lt;combinatorics&gt;" AnswerCount="4" CommentCount="1" />
  <row Id="2132" PostTypeId="2" ParentId="2112" CreationDate="2010-08-11T11:07:33.227" Score="4" ViewCount="" Body="&lt;p&gt;I guess the point that mathematicians are trying to make is that a theorem is &lt;em&gt;always&lt;/em&gt; valid.  However, there may be human flaws in the process, which end up labelling non-theorems as theorems.  A scientific model, on the other hand, could be flawed in some (possibly untestable) instances.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In practice, I'm 100% convinced, for example, that Euclid's proof of an infinite number of primes is correct.  But for more complicated proofs, errors are reduced by publishing (and in particular, peer review).  Some theorems (particularly important ones) have multiple published proofs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS. Computer proofs are just as invalid on the philosophical level since humans make computers, etc.&lt;/p&gt;&#xA;" OwnerUserId="139" LastActivityDate="2010-08-11T11:07:33.227" CommunityOwnedDate="2010-08-12T13:15:22.263" CommentCount="2" />
  <row Id="2133" PostTypeId="2" ParentId="2131" CreationDate="2010-08-11T11:22:30.067" Score="1" ViewCount="" Body="&lt;p&gt;If you start off with the 7P3 ways of choosing three distinct ordered elements from the multiset {2,3,3,4,4,4,4} then you will need to compensate for the multiply-counted 3-digit numbers in different ways depending on the structure of the number itself.  Consequently, you're likely to be stuck using an almost brute-force attack on the problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, you could generate all the unordered 3-digit numbers:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;233, 234, 244, 334, 344, 444&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Then consider one-by-one how many ordered 3-digit numbers have those particular digits.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;3!/2! + 3! + 3!/2! + 3!/2! + 3!/2! + 3!/3!&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Once you're done, you could check your result using &lt;a href=&quot;http://www.gap-system.org/&quot; rel=&quot;nofollow&quot;&gt;GAP&lt;/a&gt; by:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Arrangements([2,3,3,4,4,4,4],3);&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="139" LastActivityDate="2010-08-11T11:22:30.067" />
  <row Id="2134" PostTypeId="2" ParentId="2131" CreationDate="2010-08-11T11:27:33.710" Score="3" ViewCount="" Body="&lt;p&gt;If you have &lt;code&gt;2&lt;/code&gt; as the first digit, you can have &lt;code&gt;3&lt;/code&gt; and &lt;code&gt;4&lt;/code&gt; as the second and third digits, so this gives &lt;code&gt;233 234 243 244&lt;/code&gt;, &lt;strong&gt;four&lt;/strong&gt; numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have &lt;code&gt;3&lt;/code&gt; as the first digit, you can have &lt;code&gt;2&lt;/code&gt; and &lt;code&gt;3&lt;/code&gt; and &lt;code&gt;4&lt;/code&gt; as second and third digits (but &lt;code&gt;2&lt;/code&gt; only once), giving &lt;code&gt;323 332 324 342 334 343 344&lt;/code&gt;, &lt;strong&gt;seven&lt;/strong&gt; numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have &lt;code&gt;4&lt;/code&gt; as the first digit, you can have &lt;code&gt;2&lt;/code&gt;, &lt;code&gt;3&lt;/code&gt; and &lt;code&gt;4&lt;/code&gt; as second and third digits, giving &lt;code&gt;424 442 423 432 433 434 443 444&lt;/code&gt;, &lt;strong&gt;eight&lt;/strong&gt; numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4 + 7 + 8 = 19 unique 3 digit numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rather brute force, but it works pretty easily for this problem. &lt;/p&gt;&#xA;" OwnerUserId="456" LastActivityDate="2010-08-11T11:27:33.710" />
  <row Id="2135" PostTypeId="2" ParentId="2093" CreationDate="2010-08-11T11:59:40.373" Score="0" ViewCount="" Body="&lt;p&gt;In some instances, such a collection of sets would be a t-design (a generalised &lt;a href=&quot;http://en.wikipedia.org/wiki/Block_design&quot; rel=&quot;nofollow&quot;&gt;block design&lt;/a&gt;).&lt;/p&gt;&#xA;" OwnerUserId="139" LastActivityDate="2010-08-11T11:59:40.373" />
  <row Id="2136" PostTypeId="1" AcceptedAnswerId="2142" CreationDate="2010-08-11T12:07:35.583" Score="3" ViewCount="121" Body="&lt;p&gt;Is possible to construct infinitely differentiable functions that interpolate through arbitrary points, the way polynomial splines do? If so, do they have a name and is there an algorithm for constructing the curves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suppose a infinite degree polynomial spline might qualify but I was hoping there might be something more like a bump function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I definitely am a novice when it comes to this stuff, so please do not hesitate to correct me if I misusing any terminology.&lt;/p&gt;&#xA;" OwnerUserId="99" LastActivityDate="2010-08-11T14:26:11.637" Title="Are there smooth analogs to polynomial splines" Tags="&lt;numerical-methods&gt;" AnswerCount="3" CommentCount="5" FavoriteCount="1" />
  <row Id="2137" PostTypeId="2" ParentId="2131" CreationDate="2010-08-11T12:21:52.233" Score="1" ViewCount="" Body="&lt;p&gt;Alternatively, start from all 3-digit permutations of {2,3,4}, with repeats. That's $3^3=27$. Then remove numbers that have too many 2's or 3's.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take out numbers with exactly 2 2's: 2 choices for the remaining digit and 3 ways to permute the 3 digits is 6.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take out numbers with exactly 3 2's, of which there's only 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take out numbers with 3 3's, of which there's only 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then what's left is 27-6-1-1 = 19&lt;/p&gt;&#xA;" OwnerUserId="541" LastEditorUserId="541" LastEditorDisplayName="" LastEditDate="2010-08-11T14:58:17.253" LastActivityDate="2010-08-11T14:58:17.253" />
  <row Id="2138" PostTypeId="2" ParentId="2112" CreationDate="2010-08-11T12:36:29.550" Score="5" ViewCount="" Body="&lt;p&gt;I think there are two issues floating around here. Very complicated informal proofs (i.e. what most mathematicians write, using ordinary languages and verbal reasoning) may indeed contain hard to spot errors. However, a correct proof could, in theory, be translated into a formal language of set theory, say ZFC. Such a proof would be enormous, and might look a lot like machine language (being a combination of symbolic first order logic and the membership symbol of ZFC set theory).  But every step of reasoning in such a proof is trivial to verify.  Thus, we can in theory check such proofs with computers (or by hand) and be convinced they are correct with very high certainty.  Of course, most proofs are not done this way, but there are projects (e.g. Metamath, Mizer, etc.) to formalize large parts of mathematics.  I think this is what you are getting at with your question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second issue, however, is the more complex question having to do with Godel incompleteness.  In brief, to construct a formal proof, we need a foundation, which is typically ZFC set theory.  But by incompleteness, we cannot know that ZFC is consistent &lt;em&gt;inside of ZFC&lt;/em&gt;!  Most set theorist strongly believe ZFC is consistent, however, as no one has ever exhibited an inconsistency.  So mathematicians tacitly assume that ZFC is consistent, and from that assumption write proofs.  In set theory itself, you often see proofs that read &quot;Assuming ZFC is consistent, X is true.&quot;  But by contraposition, this yields the tantalizing possibility that we might find, via other means, that X is &lt;em&gt;false&lt;/em&gt;, and thus that ZFC is &lt;em&gt;inconsistent&lt;/em&gt;!  This would be an absolutely explosive event in mathematics, of course.  I don't think this is what your question was getting at, but I thought it would be useful to know that there is indeed a &quot;fundamental&quot; incompleteness issue floating around here.&lt;/p&gt;&#xA;" OwnerUserId="452" LastActivityDate="2010-08-11T12:36:29.550" CommunityOwnedDate="2010-08-12T13:15:22.263" CommentCount="1" />
  <row Id="2139" PostTypeId="2" ParentId="2136" CreationDate="2010-08-11T12:38:54.977" Score="3" ViewCount="" Body="&lt;p&gt;Yes - if you have a finite list of points on the plane that pass the vertical line test, you can make a smooth (infinitely differentiable) function that goes through all of them. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This can actually be done somewhat explicitly, using the formulas at [1]. Define:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;f(x) = &#xA;\begin{cases}&#xA;0 &amp;amp; x \leq 0 \\&#xA;e^{-1/x}&amp;amp; x &gt; 0 \\&#xA;\end{cases}.&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It can be proved that this is a smooth function. Then define&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;g(x) = \frac{f(x)}{f(x)+f(1-x)}.&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The function $g(x)$ is smooth, defined for all real $x$, and has the property that $g(x) = 0$ for all $x \leq 0$ while $g(x) = 1$ for all $x \geq 1$. By adding together shifted and scaled copies of $g(x)$, you can make a smooth function that goes through any finite list of points that pass the vertical line test. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1: &lt;a href=&quot;http://en.wikipedia.org/wiki/Non-analytic_smooth_function&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Non-analytic_smooth_function&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="630" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-08-11T14:26:11.637" LastActivityDate="2010-08-11T14:26:11.637" CommentCount="7" />
  <row Id="2140" PostTypeId="2" ParentId="2112" CreationDate="2010-08-11T12:40:42.320" Score="3" ViewCount="" Body="&lt;p&gt;I have been convinced for years of the following claim: &lt;em&gt;Every compact topological space is sequentially compact.&lt;/em&gt; (The converse is trivially false.) I'm positive I have even been taught this in my topology course as an undergraduate, and I can provide at least two wrong, but very subtly so, &quot;proofs&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So this is one thing simple enough for me to understand 100% what is going on, still I did not spot the mistake. By the way, I'm not alone in this respect: a long conversation about the issue &lt;a href=&quot;http://www.math.ntnu.no/~stacey/Vanilla/nForum/comments.php?DiscussionID=1217&quot; rel=&quot;nofollow&quot;&gt;has been going&lt;/a&gt; on at the NForum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, before this episode I was much more confident in the absolute certainty of mathematical proofs.&lt;/p&gt;&#xA;" OwnerUserId="233" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2010-11-05T09:00:55.780" LastActivityDate="2010-11-05T09:00:55.780" CommunityOwnedDate="2010-08-12T13:15:22.263" CommentCount="3" />
  <row Id="2142" PostTypeId="2" ParentId="2136" CreationDate="2010-08-11T12:52:34.993" Score="3" ViewCount="" Body="&lt;p&gt;For &lt;em&gt;n&lt;/em&gt; data points, there is a unique polynomial of degree &lt;em&gt;n&lt;/em&gt;−1 that passes through the points; one way to get it is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Lagrange_polynomial&quot; rel=&quot;nofollow&quot;&gt;Lagrange polynomial&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another class of interpolating functions is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Thin_plate_spline&quot; rel=&quot;nofollow&quot;&gt;thin plate spline&lt;/a&gt;, which is typically used for 2D image warping, but can easily be applied to 1D interpolation as well, which is a strictly simpler problem. The Wikipedia article on &lt;a href=&quot;http://en.wikipedia.org/wiki/Polyharmonic_spline&quot; rel=&quot;nofollow&quot;&gt;polyharmonic splines&lt;/a&gt;, of which the thin plate spline is a special case, has examples of other infinitely differentiable splines.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe that in principle, you can choose any basis function whose translations are linearly independent, and find an interpolating function as a sum of translations and scalings of the basis. The above are essentially examples of particular basis functions people have used because of their other nice properties.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Edit:&lt;/em&gt; Sorry, thin plate splines and other examples in the article, apart from the Gaussian, are infinitely differentiable &lt;em&gt;almost&lt;/em&gt; everywhere. However, if you really want infinite differentiability everywhere, you can use any infinitely differentiable function as basis instead, such as the Gaussian, or $x^2/\sqrt{x^2+1}$.&lt;/p&gt;&#xA;" OwnerUserId="856" LastEditorUserId="856" LastEditorDisplayName="" LastEditDate="2010-08-11T13:59:12.107" LastActivityDate="2010-08-11T13:59:12.107" />
  <row Id="2145" PostTypeId="1" AcceptedAnswerId="2148" CreationDate="2010-08-11T13:42:30.323" Score="-1" ViewCount="170" Body="&lt;p&gt;if a&amp;lt; b which of the following are always correct&lt;br&gt;&#xA;1) a^3&amp;lt; a^2b&lt;br&gt;&#xA;the answer to this question would be true because there is theorm that &quot;if a&amp;lt; b and we add the same postive number to both sides of the inequalities , this would not effect the inequality&quot;&lt;br&gt;&#xA;Now the second problem is interesting&lt;br&gt;&#xA;2) a^2 &amp;lt; ab&lt;br&gt;&#xA;Now there is also another theorm that says &quot;if a &amp;lt; b and we multiply the same positive number with both sides this would not effect the inequality&quot;&lt;br&gt;&#xA;now as it is specified on the top of the question that a&amp;lt; b , now in this question we are multiplying a on both sides which makes it a^2 &amp;lt; ab so this should be correct. but in the answers at the end of the book the question number of this question is not given. please help.&lt;/p&gt;&#xA;" OwnerUserId="1050" LastActivityDate="2010-08-11T15:31:28.653" Title="i am confused with this question whoes answer would be either yes or no" Tags="&lt;calculus&gt;" AnswerCount="4" CommentCount="1" />
  <row Id="2146" PostTypeId="2" ParentId="2145" CreationDate="2010-08-11T13:50:24.920" Score="0" ViewCount="" Body="&lt;p&gt;1) a^3&amp;lt; a^2b&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is always true&#xA;because even if 'a' is negative &lt;/p&gt;&#xA;&#xA;&lt;p&gt;a^3 will always be less than a^2b (positive) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In 2, if you have for example&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a = -10&lt;/p&gt;&#xA;&#xA;&lt;p&gt;b = 2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;then&#xA;a &amp;lt; b = true&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a^2 = 20&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ab = -20&lt;/p&gt;&#xA;&#xA;&lt;p&gt;then a^2 &amp;lt; ab becomes false&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so 1 is always true, 2 is not.&lt;/p&gt;&#xA;" LastActivityDate="2010-08-11T13:50:24.920" CommentCount="1" />
  <row Id="2148" PostTypeId="2" ParentId="2145" CreationDate="2010-08-11T13:57:41.923" Score="2" ViewCount="" Body="&lt;p&gt;I think the easiest way to see this is to bring everything to the left side.  For inequality 1 you get:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a^3 - a^2 b &amp;lt; 0&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now factor out a^2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a^2( a - b) &amp;lt; 0&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a true statement - provided a != 0 - since a^2 is positive and a-b is negative.  Recall that a &amp;lt; b.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For inequality 2 you get:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a (a - b) &amp;lt; 0&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While (a - b) is negative we can't say whether or not a is positive or negative.  The inequality is true only if a is positive and is false if a is negative.&lt;/p&gt;&#xA;" OwnerUserId="1054" LastActivityDate="2010-08-11T13:57:41.923" CommentCount="1" />
  <row Id="2149" PostTypeId="2" ParentId="2145" CreationDate="2010-08-11T14:06:42.090" Score="0" ViewCount="" Body="&lt;p&gt;Both are false; try for instance for a = 0 &amp;lt; 1 = b.&lt;/p&gt;&#xA;" OwnerUserId="1056" LastActivityDate="2010-08-11T14:06:42.090" />
  <row Id="2150" PostTypeId="2" ParentId="2112" CreationDate="2010-08-11T14:12:45.460" Score="13" ViewCount="" Body="&lt;p&gt;Most philosophers have agreed for centuries that we can be &lt;em&gt;completely sure&lt;/em&gt; of very little, if anything.  Descartes famously argued &lt;a href=&quot;http://en.wikipedia.org/wiki/Cogito_ergo_sum&quot; rel=&quot;nofollow&quot;&gt;cogito ergo sum&lt;/a&gt;, i.e., that the action of contemplating the possibility of his nonexistence confirms his existence (a nice early example of a &quot;self-defeating proof by contradiction&quot;).  This particular argument is no longer thought to be unassailable: see the linked article. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus the idea that you cannot be &lt;em&gt;completely sure&lt;/em&gt; that a mathematical proof is correct is true, but in the same way that you cannot be completely sure of virtually anything: i.e., in a not very interesting way and one which has nothing to do with mathematical knowledge in particular.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You seem to be arguing that a special feature of mathematical verifications is that they can be long and complicated.  I disagree that this is particular to mathematics.  Suppose I wanted to create a list of all days between September 1, 2010 and September 1, 2020 in which the temperature in my condo never exceeds 80 degrees Fahrenheit.  What a pain in the butt it would be to be even reasonably certain about this: possibly I would want someone to remain in my condo at all times, probably I would want to buy some scientific equipment (I don't trust my thermostat that much, especially to distinguish between 79.5 and 80 degrees); certainly the task will take ten years to complete.  This is a long, difficult, esoteric verification that has nothing to do with mathematics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As I have tried to point out, nothing that you say argues against &quot;we thought that mathematics is the most secure branch of knowledge&quot;.  You also don't see who &quot;we&quot; are or why we thought that, but as it happens &lt;em&gt;I&lt;/em&gt; do think that mathematics is, relatively speaking, an especially secure branch of knowledge.  If I spent ten years thinking about the proof of the Poincare Conjecture, then at the end I would be much more secure in my opinion that it was correct [or, possibly, incorrect!] than I would be at the end of my condo temperature experiment.  For PC I would have a coherent, well-understood argument that I could turn around, adapt to other situations, explore consequences of, etc.  I could burn all the original documents I read during those 10 years and not be much worse off.  For my condo temperature experiment I would just have reams of data and greater or lesser amounts of experimental doubt.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Complicated things are indeed complicated, and I think we must be honest that there is always the possibility (and very often, the reality) of human error.  Coming back to Descartes, he believed that he had discovered a &lt;em&gt;method&lt;/em&gt; by which it was possible to reason perfectly without error, and he applied his method to develop the new field of analytic geometry.  This was fantastic, pioneering work.  But entirely without error?  Of course not; as is well documented, he made some mistakes.  So do almost all of us.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I would claim that a relatively simple mathematical proof has one of the highest degrees of certainty that we can muster.  In another response, someone mentioned the infinitude of prime numbers and indeed there was a previous question on this site with the title &quot;Is it possible that there are only finitely many primes?&quot;  When we answer &lt;strong&gt;No!&lt;/strong&gt; we are more sure of this than a researcher in any other field I can think of could be.  &lt;/p&gt;&#xA;" OwnerUserId="299" LastActivityDate="2010-08-11T14:12:45.460" CommunityOwnedDate="2010-08-12T13:15:22.263" />
  <row Id="2151" PostTypeId="2" ParentId="2136" CreationDate="2010-08-11T14:22:40.427" Score="2" ViewCount="" Body="&lt;p&gt;It's a bit disingenuous to say polynomial splines are infinitely differentiable; a spline with nth order pieces can only have at most C^{n-1} continuity. That being said, a different basis set (e.g. the Fourier basis or rational functions) would in principle be infinitely differentiable, but might be a very &quot;wiggly&quot; way to &quot;connect the dots&quot; so to speak. Polynomial interpolation, whatever method you use, will also suffer from this &quot;wiggliness&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why would you need your approximant to be infinitely differentiable anyway? Unless you're going to be doing something special I don't see it being a necessity.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-08-11T14:22:40.427" CommentCount="3" />
  <row Id="2152" PostTypeId="2" ParentId="227" CreationDate="2010-08-11T14:35:20.960" Score="1" ViewCount="" Body="&lt;p&gt;You have a choice of a course called &quot;Group Theory and Linear Algebra&quot;?  You want linear algebra but not group theory if you have limited time.  If you have an interest in understanding stochastic models in a serious way then you want to understand stochastic differential equations, which requires serious doses of measure-theoretic probability (not lower-level probability using no measure theory).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Have you looked at Greg Mankiw's blog?  He has a section of links called Advice for Students and one is about math courses.  The link is &lt;a href=&quot;http://gregmankiw.blogspot.com/2006/05/which-math-courses.html&quot; rel=&quot;nofollow&quot;&gt;http://gregmankiw.blogspot.com/2006/05/which-math-courses.html&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="619" LastActivityDate="2010-08-11T14:35:20.960" CommunityOwnedDate="2010-08-12T14:58:43.903" />
  <row Id="2153" PostTypeId="2" ParentId="2121" CreationDate="2010-08-11T14:39:12.480" Score="12" ViewCount="" Body="&lt;p&gt;A version of Robin Chapman's answer that might be easier to verify:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The additive group H of a field of order 2&lt;sup&gt;n&lt;/sup&gt; has two nice types of automorphism: multiplication by a scalar, and the Frobenius automorphism.  The semi-direct product G is called AΓL(1,2&lt;sup&gt;n&lt;/sup&gt;).  It has order (2&lt;sup&gt;n&lt;/sup&gt;)⋅(2&lt;sup&gt;n&lt;/sup&gt;−1)⋅(n) ≤ 2&lt;sup&gt;3n&lt;/sup&gt;, and for n ≥ 2 it is its own automorphism group.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It has the additive group H of the field as a normal subgroup.  However, as just an additive group, H has automorphism group all n×n matrices over the field with 2 elements, which has size (2&lt;sup&gt;n&lt;/sup&gt;−1)⋅(2&lt;sup&gt;n&lt;/sup&gt;−2)⋅⋅⋅(2&lt;sup&gt;n&lt;/sup&gt;−2&lt;sup&gt;n−1&lt;/sup&gt;) ≥ 2&lt;sup&gt;(n−1)&lt;sup&gt;2&lt;/sup&gt;&lt;/sup&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For large n, |Aut(G)| = (2&lt;sup&gt;n&lt;/sup&gt;)⋅(2&lt;sup&gt;n&lt;/sup&gt;−1)⋅(n) ≤ 2&lt;sup&gt;3n&lt;/sup&gt; ≤ 2&lt;sup&gt;(n−1)&lt;sup&gt;2&lt;/sup&gt;&lt;/sup&gt; ≤ (2&lt;sup&gt;n&lt;/sup&gt;−1)⋅(2&lt;sup&gt;n&lt;/sup&gt;−2)⋅⋅⋅(2&lt;sup&gt;n&lt;/sup&gt;−2&lt;sup&gt;n−1&lt;/sup&gt;) = |Aut(H)|&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For n = 4, G = AΓL(1,16), Aut(G) = G has order 960, and Aut(H) = GL(4,2) has order 20160.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For n = 10, G = AΓL(1,1024), Aut(G) = G has order 10475520, and Aut(H) = GL(10,2) has order 366440137299948128422802227200.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, Aut(H) can be enormously bigger than Aut(G).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is reasonably important in finite group theory:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;G is a very rigid group with lots of structure.  Because it contains all the &quot;automorphisms&quot; of the field, the group itself determines the field.  An automorphism of the group will have to be an automorphism of the field, and we've already listed them all.  Sometimes this expressed by saying the group G determines the geometry of the affine line on which it acts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;H is a very floppy group to which you can do nearly anything.  Without the maps encoding scalar multiplication, H no longer remembers the field that defined it.  It is just a vector space, and so instead of the (very few) field automorphisms, you are now allowed to use any vector space automorphism.  H has lost its structure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think H is the canonical example of a horribly structureless group.  Groups like GL and AΓL are pretty standard examples of groups with very clear structure.  The symmetric group is very similar.  Except for a few early cases, a symmetric group is its own automorphism group because it already contains within itself the set of points on which it is acting.&lt;/p&gt;&#xA;" OwnerUserId="583" LastEditorUserId="583" LastEditorDisplayName="" LastEditDate="2010-08-11T15:07:07.497" LastActivityDate="2010-08-11T15:07:07.497" />
  <row Id="2154" PostTypeId="2" ParentId="2125" CreationDate="2010-08-11T14:43:38.303" Score="2" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; Your polynomial $p(n)$ splits over ${\mathbb Q}(w), w = \zeta_5$, namely  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ 125 \; p(x) = 125 \; (5 x^4-10 x^3+20 x^2-15 x+11) $  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\quad\quad\quad\quad\quad\; =  \;\; (5 x+3 w^3-4 w^2-w-3) (5 x+4 w^3+3 w^2+7 w+1)$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\quad\quad\quad\quad\quad\quad\; * \; (5 x-3 w^3+4 w^2+w-2)  (5 x-4 w^3-3 w^2-7 w-6) $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regarding the other questions in the query and the comments: there has been much research on various ways of characterizing number fields by splitting behavior, norm sets, etc - going all the way back back to Kronecker. Searching on the terms &quot;Kronecker equivalent&quot; or &quot;arithmetically equivalent&quot; will find pertinent literature. E.g. below is one enlightening review&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.ams.org/mathscinet-getitem?mr=485790&quot; rel=&quot;nofollow&quot;&gt;MR0485790 (58 #5595)&lt;/a&gt; 12A65 (12A75)&lt;br&gt;&#xA;Gauthier, François&lt;br&gt;&#xA;Ensembles de Kronecker et représentation des nombres premiers par une forme quadratique binaire.&lt;br&gt;&#xA;Bull. Sci. Math. (2) 102 (1978), no. 2, 129--143.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;L. Kronecker [Berlin Monatsber. 1880, 155--162; Jbuch 12, 65] first tried to characterize algebraic number fields by the decomposition behavior of primes. Recently, the Kronecker classes of algebraic number fields have been studied by W. Jehne [J. Number Theory 9 (1977), no. 2, 279--320; MR0447184 (56 #5499)] and others. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This article deals with the following types of questions:&lt;br&gt;&#xA;(a) When does the set of primes having a given splitting type in an algebraic number field contain (up to a finite set) an arithmetic progression?&lt;br&gt;&#xA;(b) When is this set a union of arithmetic progressions? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $K$ is an algebraic number field, let $\text{spl}(K)$ denote the set of rational primes which split completely in $K$ and let $\text{spl}^1(K)$ denote the set of rational primes which have at least one linear factor in $K$. Moreover, if $K/Q$ is a Galois extension with Galosis group $G$, let ${\text Art}_{K/Q}$ denote the Artin map which assigns a conjugacy class of $G$ to almost all rational primes $p$. If $C$ is a conjugacy class of $G$ then $\text{Art}_{K/Q}^{-1}(C)$ is the set of primes having Artin symbol $C$. Finally a set $S$ of rational primes is said to contain an arithmetic progression or to be the union of arithmetic progressions if the set of primes in the arithmetic progression(s) differs from $S$ by at most a finite set. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $G'$ denote the commutator subgroup of the Galois group $G$. Two results proved in the article are:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Theorem A.&lt;/strong&gt; The following statements are equivalent:&lt;br&gt;&#xA;(a) $|C|=|G'|$;&lt;br&gt;&#xA;(b) $\text{Art}_{K/Q}^{-1}(C)$ is the union of arithmetic progressions;&lt;br&gt;&#xA;(c) $\text{Art}_{K/Q}^{-1}(C)$ contains an arithmetic progression.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Theorem B.&lt;/strong&gt; The following statements are equivalent:&lt;br&gt;&#xA;(a) $K/Q$ is abelian;&lt;br&gt;&#xA;(b) $\text{spl}(K)$ contains an arithmetic progression;&lt;br&gt;&#xA;(c) $\text{spl}(K)$ is the union of arithmetic progressions;&lt;br&gt;&#xA;(d) there exist a modulus $m$ and a subgroup ${r_1,\cdots,r_t}$ of the multiplicative group modulo $m$ such that $\text{spl}(K)$ is the union of the arithmetic progressions $mx+r_i\ (i=1,\cdots,t)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When $K/Q$ is a non-Galois extension it is well known that $\text{spl}(K)=\text{spl}(\overline K)$ where $\overline K$ denotes the normal closure of $K$. It follows from Theorem B that $\text{spl}(K)$ cannot contain an arithmetic progression. However, the author gives two conditions, one necessary and the other sufficient, for $\text{spl}^1(K)$ to be the union of arithmetic progressions when $K/Q$ is non-Galois. As a final application of his result the author gives a necessary and sufficient condition for the set of primes represented by a quadratic form to be the union of arithmetic progressions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The proofs use class field theory, properties of the Artin map and the Čebotarev density theorem. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reviewed by Charles J. Parry&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-08-11T16:52:12.510" LastActivityDate="2010-08-11T16:52:12.510" CommentCount="1" />
  <row Id="2155" PostTypeId="2" ParentId="2145" CreationDate="2010-08-11T15:05:12.900" Score="0" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; If $b-a &gt; 0 $ then $c(b-a) &gt; 0 \iff c &gt; 0$. Now specialize $c = a, \; a^2$.&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-08-11T15:05:12.900" />
  <row Id="2156" PostTypeId="2" ParentId="2131" CreationDate="2010-08-11T15:07:35.173" Score="0" ViewCount="" Body="&lt;p&gt;It seems to me that a general formula for this simple but irregular question would be an overkill. Perhaps it is more efficient to simply count the possibilities:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;for each of the possible 3 locations of 2 there are 4 such numbers, for a total of 12;&lt;/li&gt;&#xA;&lt;li&gt;there are a total of 8 different 3-digit numbers which have digits 3 or 4; however exactly one of them, namely 333, is not permissible; thus there are 7 different allowable 3-digit numbers, which have 3 and 4 as their only digits;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;We see that all together there are 12+7 = 19 different solutions (3-digit numbers, which satisfy the stated assumptions).&lt;/p&gt;&#xA;" OwnerUserId="1056" LastActivityDate="2010-08-11T15:07:35.173" />
  <row Id="2158" PostTypeId="1" AcceptedAnswerId="2171" CreationDate="2010-08-11T15:25:13.767" Score="10" ViewCount="608" Body="&lt;p&gt;I have a partition of a positive integer (p). How can I prove that the factorial of p can always be divided by the product of the factorials of the parts?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a quick example 9!/(2!3!4!) = 1260 (no remainder), where 9=2+3+4.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can &lt;em&gt;nearly&lt;/em&gt; see it by looking at factors, but I can't see a way to guarantee it.&lt;/p&gt;&#xA;" OwnerUserId="869" LastActivityDate="2010-10-02T07:09:57.843" Title="Division of Factorials" Tags="&lt;number-theory&gt;" AnswerCount="8" CommentCount="3" FavoriteCount="3" />
  <row Id="2159" PostTypeId="1" AcceptedAnswerId="2163" CreationDate="2010-08-11T15:25:58.790" Score="5" ViewCount="142" Body="&lt;p&gt;Now, I may have only slept two hours last night and would currently struggle to discern a 'proof' by induction of FLT from a piece of genuine mathematics, but that doesn't stop mathematics from bugging me. At present I am puzzled by &lt;a href=&quot;http://mathoverflow.net/questions/34982/are-there-space-filling-curves-for-the-hilbert-cube&quot; rel=&quot;nofollow&quot;&gt;something I saw on MO this morning&lt;/a&gt;...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The linked question concerns the hilbert cube $[0,1]^\mathbb{N}$ (an infinite product of intervals) and the exitence of space filling curves thereof- that is: continuous images of the unit circle that are surjections on the hilbert cube. &lt;a href=&quot;http://mathoverflow.net/questions/34982/are-there-space-filling-curves-for-the-hilbert-cube/34983#34983&quot; rel=&quot;nofollow&quot;&gt;The accepted answer&lt;/a&gt;, together with another answer (which actually constructs such a map) and various comments, seems to allude toward an answer in the affirmative. However, the linked theorem (the 'Hahn–Mazurkiewicz theorem') which states:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A nonempty Hausdorff topological space is a continuous image of the unit interval if and only if it is a compact, connected, locally connected second-countable space.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;seems in direct contradiction to this since (and I may be mistaken for reasons explained above):&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The hilbert cube is a subset of a normed space and hence a metric space&lt;/li&gt;&#xA;&lt;li&gt;The sequence (1,0,0...), (0,1,0...), (0,0,1,...) has no convergent subsequence&lt;/li&gt;&#xA;&lt;li&gt;So the hilbert cube is not sequentially compact, therefore non-compact (the two are equivalent in metric spaces)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Which seems at odds with the only if portion of the theorem's statement. Maybe this is wikipedia taking me for a ride. Maybe I am just hallucinating a portion of this argument. Either way, this is annoying me. Thanks in advance for clearing this up...&lt;/p&gt;&#xA;" OwnerUserId="160" LastActivityDate="2010-08-11T15:42:51.023" Title="A curious compactness confusion: space filling curves in the hilbert cube that contradict a bona fide theorem?!?!?" Tags="&lt;general-topology&gt;" AnswerCount="2" CommentCount="5" FavoriteCount="0" />
  <row Id="2161" PostTypeId="2" ParentId="2158" CreationDate="2010-08-11T15:34:25.657" Score="0" ViewCount="" Body="&lt;p&gt;Hi--&#xA; Please look at this link:&#xA;mathdl.maa.org/images/upload.../00029890.di021346.02p0064l.pdf&lt;/p&gt;&#xA;" LastActivityDate="2010-08-11T15:34:25.657" CommentCount="2" />
  <row Id="2162" PostTypeId="2" ParentId="2158" CreationDate="2010-08-11T15:35:45.000" Score="13" ViewCount="" Body="&lt;p&gt;These quotients are integers since they solve counting problems.&#xA;For instance, how many nine-letter words are there with 2 As, 3 Bs and 4 Cs?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the full story see the &lt;a href=&quot;http://en.wikipedia.org/wiki/Multinomial_theorem&quot; rel=&quot;nofollow&quot;&gt;multinomial theorem&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-08-11T15:35:45.000" CommentCount="2" />
  <row Id="2163" PostTypeId="2" ParentId="2159" CreationDate="2010-08-11T15:41:26.763" Score="6" ViewCount="" Body="&lt;p&gt;$(1,0,0,\ldots), (0,1,0,\ldots)$ converges to $(0,0,0,\ldots)$, so your example doesn't contradict the compactness of the Hilbert cube.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$[0,1]^{\mathbb N}$ is homeomorphic to $[0,1]\times[0,1/2]\times[0,1/3]\times\cdots$ with the $\ell_2$-norm. So the norm of $(a_1,a_2,a_3,\ldots)$ is $\sum\left(\dfrac{a_n}{n}\right)^2$, and therefore your sequence converges to $(0,0,0,\ldots)$ since the norm of $(0,\ldots,0,1,0,\ldots)$ (with the 1 in the $n$th position) is $1/n^2$, which tends to 0.&lt;/p&gt;&#xA;" OwnerUserId="405" LastActivityDate="2010-08-11T15:41:26.763" />
  <row Id="2164" PostTypeId="2" ParentId="2159" CreationDate="2010-08-11T15:42:51.023" Score="6" ViewCount="" Body="&lt;p&gt;The sequence you give converges to $0$ in the product topology.  For instance, convergence in the product topology is equivalent to pointwise (or coordinatewise) convergence, and your sequence converges coordinatewise to $0$.  See e.g.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Pointwise_convergence&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Pointwise_convergence&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(&lt;b&gt;Added&lt;/b&gt;: Samuel's answer proceeds differently from mine -- via an explicit metric on the &quot;Hilbert cube&quot; -- and is also correct.)  &lt;/p&gt;&#xA;" OwnerUserId="299" LastActivityDate="2010-08-11T15:42:51.023" />
  <row Id="2165" PostTypeId="2" ParentId="2158" CreationDate="2010-08-11T15:46:39.837" Score="7" ViewCount="" Body="&lt;p&gt;If you believe (:-) in the two-part Newton case, then the rest is easily obtained by induction. For instance (to motivate you to write a full proof):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$9!/(2!3!4!) = (9!/(5!\cdot 4!)) \cdot (5!/(2!\cdot 3!))$$&lt;/p&gt;&#xA;" OwnerUserId="1056" LastEditorUserId="152" LastEditorDisplayName="" LastEditDate="2010-08-11T17:02:51.137" LastActivityDate="2010-08-11T17:02:51.137" CommentCount="4" />
  <row Id="2166" PostTypeId="1" AcceptedAnswerId="2173" CreationDate="2010-08-11T15:59:34.917" Score="1" ViewCount="220" Body="&lt;p&gt;&lt;strong&gt;Statement&lt;/strong&gt;  For every $n &gt; 1$ there is always at least one prime $p$ such that $n &amp;lt; p &amp;lt; 2n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am curious to know that if I replace that $2n$ by $2n-\epsilon$, ($\epsilon&gt;0$) then what is the  $\inf (\epsilon)$ so that the inequality still holds, meaning there is always a prime between $n$ and $2n-\epsilon$&lt;/p&gt;&#xA;" LastActivityDate="2010-08-13T20:39:56.267" Title="Bertrand's Postulate" Tags="&lt;number-theory&gt;" AnswerCount="3" />
  <row Id="2167" PostTypeId="2" ParentId="2130" CreationDate="2010-08-11T16:05:58.640" Score="1" ViewCount="" Body="&lt;p&gt;In the theory of abelian categories, the problem with non-small categories (or at least categories without some smallness assumption) is that the collection of inclusions might not be a set.  This can prevent you from taking constructions you might want, such as taking the union over all submodules with a given property.  This happens, for instance, when trying to construct essential extensions.  This is usually fixed by requiring that the collection of inclusions into an object forms a set, rather than requiring smallness.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suspect that even outside the realm of abelian categories, non-smallness wreaks havoc with categorical constructions which involve a 'for all' such as equalizers, fibered products and more general limits.  However, I don't know of any specific examples of this.&lt;/p&gt;&#xA;" OwnerUserId="1062" LastActivityDate="2010-08-11T16:05:58.640" />
  <row Id="2168" PostTypeId="2" ParentId="2166" CreationDate="2010-08-11T16:10:23.003" Score="1" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Bertrand%27s_postulate&quot; rel=&quot;nofollow&quot;&gt;Bertrand's postulate&lt;/a&gt; is &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;if n &gt; 3 is an integer, then there always exists at least one prime number p with n &amp;lt; p &amp;lt; 2n − 2.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Thus &amp;epsilon; &amp;lt; 2 for n &gt; 3. What if n ≤ 3?&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;For n = 3, 3 &amp;lt; 5 &amp;lt; 6 - &amp;epsilon; &amp;rArr; &amp;epsilon; &amp;lt; 1&lt;/li&gt;&#xA;&lt;li&gt;For n = 2, 2 &amp;lt; 3 &amp;lt; 4 - &amp;epsilon; &amp;rArr; &amp;epsilon; &amp;lt; 1&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Hence we have 0 &amp;lt; &amp;epsilon; &amp;lt;  1, if &amp;epsilon; is a constant.&lt;/p&gt;&#xA;" OwnerUserId="171" LastEditorUserId="171" LastEditorDisplayName="" LastEditDate="2010-08-11T17:39:21.873" LastActivityDate="2010-08-11T17:39:21.873" CommentCount="3" />
  <row Id="2169" PostTypeId="2" ParentId="392" CreationDate="2010-08-11T16:16:16.417" Score="6" ViewCount="" Body="&lt;p&gt;I agree with David (+1), this is the pertinent graph, and it works for me:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://www.animations.physics.unsw.edu.au/jw/graphics/calculus8%28circle%29.gif&quot; alt=&quot;Sin and cosine on the unit circle&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Explanation &lt;a href=&quot;http://www.animations.physics.unsw.edu.au/jw/calculus.htm#Trigonometric&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="312" LastActivityDate="2010-08-11T16:16:16.417" />
  <row Id="2171" PostTypeId="2" ParentId="2158" CreationDate="2010-08-11T17:08:55.947" Score="3" ViewCount="" Body="&lt;p&gt;The key observation is that the product of $n$ consecutive integers is divisible by $n!$. This can be proved by induction.&lt;/p&gt;&#xA;" OwnerUserId="589" LastActivityDate="2010-08-11T17:08:55.947" CommentCount="2" />
  <row Id="2172" PostTypeId="2" ParentId="2158" CreationDate="2010-08-11T17:27:28.460" Score="4" ViewCount="" Body="&lt;p&gt;Besides the obvious combinatorial interpretations, one also has the has the following reduction from multinomial coefficient to products of binomial coefficients. Namely for $n = i+j+k+\cdots + m $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \frac{n!}{i!j!k!\cdots m!} = \binom{n}{i} \frac{(n-i)!}{j!k!\cdots m!} = \binom{n}{i}\binom{n-i}{j} \frac{(n-i-j)!}{k!\cdots m!} = \;\cdots$$&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-08-11T17:27:28.460" CommentCount="1" />
  <row Id="2173" PostTypeId="2" ParentId="2166" CreationDate="2010-08-11T17:29:31.483" Score="1" ViewCount="" Body="&lt;p&gt;Three related points are worthy of mention, showing that epsilon can be close to n.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a result of Finsler that approximates how many primes lie between n and 2n, which is of order o(n/log(n)) as is to be expected by the Prime Number Theorem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Literature on prime gaps will tell you the exponent delta such that there is (for sufficiently large n) at least one prime in the interval (n , n + n^delta).  I think delta is less than 11/20.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Observed data suggests that n^delta can be replaced by something much smaller: for n between something like 3 and 10^14 , some function like 2(log(n))^2 works.&lt;/p&gt;&#xA;" OwnerUserId="297" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-08-11T22:03:45.550" LastActivityDate="2010-08-11T22:03:45.550" CommentCount="0" />
  <row Id="2174" PostTypeId="1" AcceptedAnswerId="2179" CreationDate="2010-08-11T18:17:45.760" Score="2" ViewCount="132" Body="&lt;p&gt;As a teenager I was given this problem which took me a few years to solve.  I'd like to know if this hae ever been published.  When I presented my solution I was told that it was similar to one of several he had seen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For an &lt;code&gt;n&lt;/code&gt; dimensional space, develop a formula that evaluates the maximum number of &lt;code&gt;n&lt;/code&gt; dimensional regions when divided by &lt;code&gt;k&lt;/code&gt; &lt;code&gt;n-1&lt;/code&gt; dimensional (hyper)planes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example:  A line is partitioned by points: 1 point, 2 line segments. 10 points, 11 line segments, and so one.  &lt;/p&gt;&#xA;" OwnerUserId="1067" LastEditorUserId="1067" LastEditorDisplayName="" LastEditDate="2010-08-11T23:17:25.603" LastActivityDate="2010-08-11T23:17:25.603" Title="Are there publications or references describing N dimensional space divided by N-1 dimensional (hyper)planes?" Tags="&lt;geometry&gt;&lt;algebra-precalculus&gt;&lt;reference-request&gt;" AnswerCount="3" CommentCount="7" />
  <row Id="2175" PostTypeId="2" ParentId="2013" CreationDate="2010-08-11T18:23:29.867" Score="1" ViewCount="" Body="&lt;p&gt;Here is a (smaller) concrete answer for the sum, ie. what is the PDF of $h(z)$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ z = x + y $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;if $x$ and $y$ are drawn from independent probability distributions $f(x)$ and $g(y)$. It is very useful to realize that if $x,y$ are independent, then the function $p(x,y)=f(x)g(y)$ is the joint PDF, that is when you have two independent events the chance of both of them occurring is simply their product. Now think of the 2D plane where the point $(x,y)$ is the value $f(x)g(y)$. Integrating some area in this plane is equal to the probability of the event occouring over the area (just like in 1D).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To solve for $z=x+y$ consider all the points where $z$ is constant - this defines a line with slope $-1$ and a y-intercept of $z$. We need to get an area for this, so we can integrate the area defined by a line that is very close to it. That is, we need to find the are between the two lines:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ y_1 = -x + z $$&#xA;$$ y_2 = -x + z + dx $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The integral is then:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ h(z) = \int_{x=-\infty}^{\infty} \int_{y=-x+z}^{y=-x+z+dx} f(x)g(y) dy dx $$&#xA;$$ h(z) = \int_{x=-\infty}^{\infty} f(x) \int_{y=-x+z}^{y=-x+z+dx} g(y) dy dx $$&#xA;$$ h(z) = \int_{x=-\infty}^{\infty} f(x)g(z-x) dx $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a lot swept under the rug here and for that I'd refer you to the &lt;a href=&quot;http://www.dartmouth.edu/~chance/teaching_aids/books_articles/probability_book/Chapter7.pdf&quot; rel=&quot;nofollow&quot;&gt;reference given by Kaestur Hakarl&lt;/a&gt; in the comments. Also, if you're physics minded (like myself) I got introduced to the subject in Arfken &amp;amp; Weber's Mathematical Methods for Physicists vol 6, pg 1126. Here they introduce the change in area through the Jacobian, which helps you generalize to other functions such as the product and ratio.&lt;/p&gt;&#xA;" OwnerUserId="196" LastActivityDate="2010-08-11T18:23:29.867" CommentCount="0" />
  <row Id="2176" PostTypeId="2" ParentId="2174" CreationDate="2010-08-11T18:27:47.330" Score="2" ViewCount="" Body="&lt;p&gt;Though not discussed in the full generality you describe, the problem is discussed at length for n = 1, 2, and 3 in &lt;em&gt;Mathematics for High School Teachers: An Advanced Persepective&lt;/em&gt; by Usiskin, Peressini, Machisotto, and Stanley (section 5.1.4; &amp;copy; 2003; published by Pearson Education).  Specifically, the text uses that problem as an example for discussing how induction can be applied to prove the formulae using the geometry of the problem in the inductive step (when you add the next (n-1)-dimensional boundary object, how many additional regions are created?).&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-08-11T18:27:47.330" CommentCount="3" />
  <row Id="2177" PostTypeId="2" ParentId="2116" CreationDate="2010-08-11T18:28:41.760" Score="1" ViewCount="" Body="&lt;p&gt;Here are some suggestions to approximate the limit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider F(n) to be the greater of 1 and the highest power of 2 that appears in the prime factorization of n.  Show what the limit of the average over n of F(n) is as n gets large.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let G_p(n) be defined like F(n), except replace 2 by an arbitrary prime p.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that F(n) &amp;lt;= E(n), and that E(n) = max(G_p(n) over all primes p), so that if limit&#xA;of the average over n of E(n) exists, you can bound it using a sum of limits involving G_p.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For more accurate estimates, consider inclusion-exclusion.&lt;/p&gt;&#xA;" OwnerUserId="297" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-08-11T22:02:04.303" LastActivityDate="2010-08-11T22:02:04.303" CommentCount="1" />
  <row Id="2178" PostTypeId="1" AcceptedAnswerId="2226" CreationDate="2010-08-11T18:32:51.797" Score="0" ViewCount="250" Body="&lt;p&gt;Let $G$ be a graph such that $\delta(G) \geq k$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a) Prove that $G$ has a path of length at least $k$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;solution: We know that &#xA;$\delta(G) = min\lbrace deg(v) | v \in V(G) \rbrace$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;if $\delta(G) = k$ then there exists some $v \in V(G)$ such that $deg(v) = k$. This means all other vertices $u \in V(G)$ have $deg(u) \geq k$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now i know this must be part of the proof. How would i prove that there exists a path of at least k?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;b)if $k \geq 2$, prove that $G$ has a cycle of length at least $k+1$.&lt;/p&gt;&#xA;" OwnerUserId="43" LastEditorUserId="43" LastEditorDisplayName="" LastEditDate="2010-08-12T11:45:19.260" LastActivityDate="2010-08-12T17:10:49.260" Title="Graph Path Length Problem" Tags="&lt;graph-theory&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="2179" PostTypeId="2" ParentId="2174" CreationDate="2010-08-11T18:43:50.527" Score="4" ViewCount="" Body="&lt;p&gt;Section 3 of &lt;a href=&quot;http://www.macalester.edu/~bressoud/pub/Art_of_Counting/Art_of_Counting.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.macalester.edu/~bressoud/pub/Art_of_Counting/Art_of_Counting.pdf&lt;/a&gt; discusses the problem for $n=3$ and poses the problem of generalizing to higher dimensions.  This article states that the first published solution was by Jakob Steiner in 1826.  Other references are cited.&lt;/p&gt;&#xA;" OwnerUserId="273" LastActivityDate="2010-08-11T18:43:50.527" />
  <row Id="2180" PostTypeId="1" AcceptedAnswerId="2185" CreationDate="2010-08-11T18:50:05.977" Score="0" ViewCount="133" Body="&lt;p&gt;&lt;code&gt;Ax = b&lt;/code&gt;.  I need a way to analyze a square matrix &lt;code&gt;A&lt;/code&gt; to see if its solution vector &lt;code&gt;x&lt;/code&gt; will always be positive when &lt;code&gt;b&lt;/code&gt; is positive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This question arises from solving the radiosity equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://www.siggraph.org/education/materials/HyperGraph/radiosity/images/slide11.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm interested to know when A is &lt;em&gt;incorrect&lt;/em&gt;, which would be when &lt;code&gt;x&lt;/code&gt; has negative values even though &lt;code&gt;b&lt;/code&gt; is positive.&lt;/p&gt;&#xA;" OwnerUserId="1069" LastActivityDate="2010-08-11T20:17:21.750" Title="How can I characterize the type of solution vector that comes out of a matrix?" Tags="&lt;linear-algebra&gt;&lt;matrices&gt;" AnswerCount="1" />
  <row Id="2181" PostTypeId="2" ParentId="2178" CreationDate="2010-08-11T19:20:10.953" Score="1" ViewCount="" Body="&lt;p&gt;You can construct path and cycle inductively. Start at an arbitrary vertex $v_0$, and move along an arbitrary edge to $v_1$. Since vertex $v_1$ has degree k or greater, it has at least $k-1$ neighbours other than $v_0$. Move along an edge to one of these neighbours $v_2$. Since vertex $v_2$ has degree $k$ or greater, it has at least $k-2$ neighbours other than $v_0$ and $v_1$. Move along an edge to one of these neighbours $v_3$. And so on... until you reach vertex $v_k$. At this stage, you already have a path of length $k$. There may then be no unvisited vertices left to move to, in which case v_0 must be one of them, and you can return to it for a cycle of length k+1. Otherwise, just keep on moving to new vertices. If when you get to vertex v_n you find that it is connected to one of the vertices $v_j$ for $0 \leq j \leq (n-k)$ then you can go back to that vertex for a cycle of length k+1 or greater. If not, then v_n will have a neighbour that you have not yet visited, and the journey can continue. Assuming that G has only finitely many vertices, you must eventually complete a cycle.&lt;/p&gt;&#xA;" LastActivityDate="2010-08-11T19:20:10.953" CommentCount="1" />
  <row Id="2182" PostTypeId="1" AcceptedAnswerId="2184" CreationDate="2010-08-11T19:45:00.923" Score="0" ViewCount="157" Body="&lt;p&gt;Suppose $G$ is a locally compact abelian group, with Haar Measure $\mu$, then is $\mu(E)=\mu(E^{-1})$? for all subsets $E$ of $G$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have seen that this is true for all Borel subsets of $G$, but i am in need of a proof and also an example as to why it doesn't hold for other subsets.&lt;/p&gt;&#xA;" LastActivityDate="2010-08-12T02:02:56.197" Title="Locally Compact Group with Haar Measure" Tags="&lt;fourier-analysis&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="2183" PostTypeId="2" ParentId="2174" CreationDate="2010-08-11T19:49:27.857" Score="1" ViewCount="" Body="&lt;p&gt;I presume you are asking for the number of regions into which $n$-space&#xA;is divided into by $k$ hyperplanes &quot;in general position&quot;. &#xA;See the notes of Richard Stanley on &lt;a href=&quot;http://math.mit.edu/~rstan/arrangements/arr.html&quot; rel=&quot;nofollow&quot;&gt;hyperplane arrangements&lt;/a&gt;.&#xA;The answer to your problem is Proposition 2.4, and there are lots&#xA;more goodies too!&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-08-11T19:49:27.857" CommentCount="1" />
  <row Id="2184" PostTypeId="2" ParentId="2182" CreationDate="2010-08-11T19:53:20.593" Score="2" ViewCount="" Body="&lt;p&gt;Not necessarily. This condition is equivalent to each left-invariant Haar measure&#xA;being right-invariant. A group having this property is called &lt;em&gt;unimodular&lt;/em&gt;.&#xA;Since the concept has a name, then there must be groups where it fails.&#xA;See the Wikipedia article on Haar measure for an example.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-08-11T19:53:20.593" CommentCount="3" />
  <row Id="2185" PostTypeId="2" ParentId="2180" CreationDate="2010-08-11T19:53:52.773" Score="4" ViewCount="" Body="&lt;p&gt;If all entries of $A^{-1}$ are positive numbers, then $A$ has the property you desire.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(&lt;strong&gt;Edit:&lt;/strong&gt; This condition is both &lt;em&gt;sufficient&lt;/em&gt; and &lt;em&gt;necessary&lt;/em&gt;. If one entry $A^{-1}_{ij}$ is negative, then the choice $b=e_j$ will make the $i$-th component of $x=A^{-1}b$ equal to $A^{-1}_{ij}$, but this is negative.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The following argument shows that your particular example matrix does have this property: your matrix $A$ has the form&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$A = 1 - R F$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $R=\mathrm{diag}(\rho_1,\rho_2,\dots,\rho_n)$ is the diagonal matrix of reflectivities and $F$ the matrix of form factors. The physical interpretation of $R$ and $F$ makes sure that the norm of $RF$ will be smaller than 1, so that $A^{-1}$ can be expressed as a geometric series&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$A^{-1} = (1- RF)^{-1} = 1 + RF + (RF)^2 + (RF)^3 + \dots$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, since all entries of $RF$ are positive numbers, this formula implies that the entries of $A^{-1}$ are positive numbers as well.&lt;/p&gt;&#xA;" OwnerUserId="434" LastEditorUserId="434" LastEditorDisplayName="" LastEditDate="2010-08-11T20:17:21.750" LastActivityDate="2010-08-11T20:17:21.750" />
  <row Id="2186" PostTypeId="2" ParentId="2158" CreationDate="2010-08-11T20:02:36.403" Score="10" ViewCount="" Body="&lt;p&gt;The &quot;high-level&quot; way to see this is to recall that whenever a finite group $G$ has a subgroup $H$, we know that $|H|$ divides $|G|$.  Then note that $S_n$ clearly contains $S_{n_1} \times ... \times S_{n_k}$ as a subgroup for any partition $n_1 + ... + n_k = n$.  (This is actually the same as the combinatorial interpretation in Robin Chapman's answer, since what we are counting is the number of cosets $G/H$, and these cosets are precisely what the multinomial theorem is counting.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This basic lemma is surprisingly useful.  For example, it is not hard to use it to show that $m! (n!)^m$ divides $(mn)!$.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-08-11T20:02:36.403" CommentCount="1" />
  <row Id="2187" PostTypeId="1" AcceptedAnswerId="2201" CreationDate="2010-08-11T20:13:42.307" Score="0" ViewCount="203" Body="&lt;p&gt;Supose we have a set S of natural numbers. We define: &#xA;f(x) = x / 2 if x is even and f(x) = (x - 1) / 2 if x is odd&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let a[0] = a[1] = 1 and a[n] = a[n - 1] + a[n - 2] if n does not belong to S and a[n] = f(a[n - 1] + a[n - 2]) otherwise, n &gt; 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to calculate the remainder of a[n] modulus a prime p. The problem is that the numbers become quite large so I can not store then. I must work with the reminders.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My fist idea was to work with the modular inverse of 2 and e kept the remainder mod p at each step and instead of diving by 2 I multiply by 2^(-1). The problem is that is does not work if S has more than one element.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the solution is to keep the values mod 2^y*p where y is the numbers of elements in S (#S). And in the end take do remainder of a[n] mod p. It seems to work but I don't understand why...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm sorry if I was not clear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you.&lt;/p&gt;&#xA;" OwnerUserId="1072" LastActivityDate="2010-08-11T21:59:04.537" Title="Calculate the remainder when there are division" Tags="&lt;number-theory&gt;&lt;algorithms&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="2189" PostTypeId="1" AcceptedAnswerId="2195" CreationDate="2010-08-11T20:22:06.607" Score="2" ViewCount="280" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;... octonions are the crazy old uncle nobody lets out of the attic: they are nonassociative.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Comes from a a quote by John Baez. Clearly, the sucessor to quaterions from the Caley-Dickson process is a numerical beast, but has anybody found any real-world uses for them? For example, quaterions have a nice connection to computer graphics through the connection to SO(4), and that alone makes them worth studying. What can be done with a nonassociative algebra like the octonions?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: simply mentioning that they&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;have applications in fields such as string theory, special relativity, and quantum logic.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;is not what I'm looking for (I can read wikipedia too). A specific example, especially one that is geared to someone who is &lt;strong&gt;not&lt;/strong&gt; a mathematician by trade would be nice!&lt;/p&gt;&#xA;" OwnerUserId="196" LastEditorUserId="1028" LastEditorDisplayName="" LastEditDate="2010-08-23T15:17:39.730" LastActivityDate="2010-08-23T15:17:39.730" CommunityOwnedDate="2010-08-11T20:55:10.957" Title="What are some real-world uses of Octonions?" Tags="&lt;abstract-algebra&gt;&lt;octonions&gt;&lt;division-algebra&gt;" AnswerCount="3" FavoriteCount="3" />
  <row Id="2190" PostTypeId="1" AcceptedAnswerId="2197" CreationDate="2010-08-11T20:25:53.260" Score="0" ViewCount="206" Body="&lt;p&gt;&lt;em&gt;This problem is taken from &lt;a href=&quot;http://imc2009.elte.hu/Problems/imc2009-day2.pdf&quot; rel=&quot;nofollow&quot;&gt;International Mathematics Competition for University Students 2009&lt;/a&gt; (IMC 2009), Day 2, Problem 5.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Let $\mathbb{M}$ be the vector space of $m \times p$ real matrices. For a vector subspace $S \subset \mathbb{M}$, denote $\delta(S)$ the dimesion of the vector space generated by all columns of all matrices in $S$. Say that a vector space $T \subset \mathbb{M}$ is a covering space, if $$ \bigcup\limits_{A \in T, A \neq 0} ker(A) = \mathbb{R}^{p}$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Such a $T$ is minimal if it does not contain a proper vector subspace $S \subset T$ which is also a covering matrix space. Let $T$ be a minimal covering matrix space and let $n= \text{dim}(T)$. Prove that $$\delta(T) \leq { n \choose 2 } $$&lt;/p&gt;&#xA;" LastEditorUserId="16" LastEditorDisplayName="" LastEditDate="2010-08-15T22:22:50.653" LastActivityDate="2010-08-15T22:22:50.653" Title="Dimension of Vector space" Tags="&lt;linear-algebra&gt;&lt;contest-math&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="2191" PostTypeId="2" ParentId="2189" CreationDate="2010-08-11T20:29:33.297" Score="0" ViewCount="" Body="&lt;p&gt;Hi--&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please refer this link: &lt;a href=&quot;http://www.citeulike.org/user/Scis0000002/article/6831691&quot; rel=&quot;nofollow&quot;&gt;http://www.citeulike.org/user/Scis0000002/article/6831691&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are lots of applications, and one of them is this.&lt;/p&gt;&#xA;" LastActivityDate="2010-08-11T20:29:33.297" CommunityOwnedDate="2010-08-11T20:55:10.957" />
  <row Id="2192" PostTypeId="2" ParentId="2158" CreationDate="2010-08-11T20:39:58.810" Score="10" ViewCount="" Body="&lt;p&gt;Below is a sketch of a little-known purely arithmetical proof that binomial coefficients are integral. I purposely constructed the proof so that it would be comprehensible to an educated layperson. The proof gives an algorithm to rewrite a binomial coefficient as a product of fractions whose denominators are coprime to any given prime. The method of proof is  best comprehended by applying the algorithm to a specific example.&lt;br&gt;&#xA;[Note: It may prove helpful to first read &lt;a href=&quot;http://math.stackexchange.com/questions/5829&quot;&gt;this simpler example&lt;/a&gt; before proceeding to the exposition below].&lt;/p&gt;&#xA;&#xA;&lt;p&gt;E.g. consider $\ \ \binom{39}{17}\: =\: \frac{39!}{22! \; 17!}\: =\: \frac{23 \cdot 24 \cdots\; 39}{1 \cdot 2 \cdots\; 17}\:.\ $ When this fraction is reduced to lowest terms $\rm\:a/b\:,\ $ no prime $\rm\ p &gt; 17\ $ can divide its denominator $\rm\: b\:,\: $ since $\rm\ b\:|\:17\:!\ \:$ Hence, to show that $\rm\ a/b\ $ is an integer, it suffices to show that no prime $\rm\ p \le 17\ $ divides its denominator $\rm\: b\:$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;E.g. we show that $2$ doesn't divide $\rm b$. The highest power of $2$ in the &#xA;denominator terms is $16 &amp;lt; 17$. Now align the numerator and denominator terms $\rm (mod\ 16)$&#xA;by shifting the 1st numerator term so it lies above its value $\rm (mod\ 16)$,&#xA;viz. $23 \equiv 7 \pmod{16}$ so right-shift the numerator terms until $23\:$ lies above $7$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{}{1}\frac{}{2}\frac{}{3}\frac{}{4}\frac{}{5}\frac{}{6}\frac{23}{7}\frac{24}{8}\frac{25}{9}\frac{26}{10}\frac{27}{11}\frac{28}{12}\frac{29}{13}\frac{30}{14}\frac{31}{15}\frac{32}{16}\frac{33}{17}\frac{34}{}\frac{35}{}\frac{36}{}\frac{37}{}\frac{38}{}\frac{39}{}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I claim that $2$ doesn't divide the reduced denominator of each aligned fraction. Indeed $\ 24/8 = 3$, $\: 26/10 = 13/5 $,  $\:\ 28/12 = 7/3 $,  $\:\ 30/14 = 15/7 $,  $\:\ 32/16 = 2$.&#xA;This holds because these fractions $\rm\; c/d \;$  satisfy  $\rm c \equiv d\ (mod\ 16)\:$&#xA;i.e.  $\rm c = d + 16\: n \;$  so  $\rm 2|d \Rightarrow 2|c$, $\rm\; 4|d \Rightarrow 4|c$, $\:\cdots $, $\rm\; 16|d \Rightarrow 16|c$,&#xA;i.e. any power of $2\:$ below $16$ that divides $\rm d$ must also divide $\rm c$, so it cancels out upon reduction. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore to prove that $2$ doesn't divide the reduced denominator of $\binom{39}{17}$ &#xA;it suffices to prove the same for the &quot;leftover&quot; fraction composed &#xA;of the above non-aligned terms $(34 \cdots 39)/(1 \cdots 6) = \binom{39}{6}$.&#xA;Being an $\rm\binom{n}{k}$ with smaller $\rm k = 6 &amp;lt; 17\:,\:$ this follows by induction. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the same proof works for any prime $\rm p$, we conclude that no prime divides the &#xA;reduced denominator of $\binom{39}{17}$, therefore it is an integer.$\quad$ &lt;strong&gt;QED&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Informally, the reason that this works is because the denominator sequence starts at $1$, which is coprime to every prime $\rm p$. This ensures that it is the &quot;greediest&quot; possible contiguous sequence of integers, in the sense that its product contains the least power of $\rm\:p\:$ compared to other contiguous sequences of equal length.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The algorithm extends to multinomials by using the simple reduction of multinomials to products of binomials mentioned in my prior post here.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-10-02T07:09:57.843" LastActivityDate="2010-10-02T07:09:57.843" CommentCount="1" />
  <row Id="2193" PostTypeId="1" AcceptedAnswerId="2196" CreationDate="2010-08-11T20:42:22.503" Score="0" ViewCount="171" Body="&lt;p&gt;Given a finite group $G$ and its subgroups $H,K$ such that $$G \times H \cong G \times K$$ does it imply that $H=K$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clearly, one can see that this doesn't work out for all subgroups. Is there any condition by which this can remain true.&lt;/p&gt;&#xA;" LastActivityDate="2010-08-11T23:41:29.340" Title="Cancellation of Direct Products" Tags="&lt;group-theory&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="1" />
  <row Id="2194" PostTypeId="1" AcceptedAnswerId="2216" CreationDate="2010-08-11T20:44:57.440" Score="14" ViewCount="276" Body="&lt;p&gt;The Hilbert Cube $H$ is defined to be $[0,1]^{\mathbb{N}}$, i.e., a countable product of unit intervals, topologized with the product topology.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, I've read that the Hilbert Cube is homogeneous.  That is, given two points $p, q\in H$, there is a homeomorphism $f:H\rightarrow H$ with $f(p)=q$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What's confusing to me is that it seems like there seems to be a stratification of points.  That is, there are&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Points contained in $(0,1)^{\mathbb{N}}$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Points which have precisely n coordinate a 0 or 1 for n a fixed natural number.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Point which have countably many coordinates equaling 0 or 1 and countably many not and&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Points which have n many coordinates NOT equal to 0 or 1.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Now, for fixed $p$ and $q$ both in class 1 or 3 (or fix an n and use class 2 or 4), it's clear to me that there is a homeomorphism taking $p$ to $q$, simply by swapping around factors and using the fact that $(0,1)$ is clearly homogeneous.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But what are the homeomorphisms which mix the classes?  In particular, what homemorphism takes (0,0,0,....) to (1/2, 1/2,1/2,....)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Said another way, for any natural number $n&gt;1$, $[0,1]^n$ is NOT homogeneous, precisely because of these boundary points.  What allows you to deal with the boundary points in the infinite product case?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As always, feel free to retag, and thanks in advance!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;  In the off chance that someone stumbles across this question, I just wanted to provide a rough idea of the answer, as garnered from the link Pete provided in his answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If one has a point of the form $(1,p)$ in $[0,1] \times [0,1]$, then there is a self homeomorphism of $[0,1]\times[0,1]$ taking $(1,p)$ to $(q,1)$ with $q\neq 0, 1$.  For example, one can use a &quot;square rotation&quot;.  From here, the idea is simple:  given a point in $H$ of the form $(1, p_2, p_3, p_4,...)$, apply the square rotation on the first two factors to get a new point of the form $(q_1, 1, p_2, p_3,...)$.  Now, apply the square rotation on the second two factors to get a new point of the form $(q_1, q_2, 1, p_3,...)$.  The point is that after k iterations, the first k coordinates are all in the interior.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now one proves a techinical lemma that states that the infinite composition of these homeomorphisms is a well defined homeomorphism.  The infinite composition maps the point $(1, p_2, ....)$ to a point of the form $(q_1, q_2, ....)$ which lies on the &quot;interior&quot; of $H$.  Finally, using the fact that $(0,1)$ is clearly homogeneous, one can easily map $(q_1, q_2, ...)$ to $(1/2,1/2,...)$.&lt;/p&gt;&#xA;" OwnerUserId="331" LastEditorUserId="331" LastEditorDisplayName="" LastEditDate="2010-08-12T06:17:56.357" LastActivityDate="2011-03-07T17:39:48.530" Title="Why is the Hilbert Cube homogeneous?" Tags="&lt;topology&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="2" />
  <row Id="2195" PostTypeId="2" ParentId="2189" CreationDate="2010-08-11T20:50:27.850" Score="4" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://math.ucr.edu/home/baez/octonions/octonions.html&quot; rel=&quot;nofollow&quot;&gt;John Baez&lt;/a&gt; has a long online article about uses of the octonions, at least some of which is concerned with their relationship to physics.  You might also want to read his papers with Huerta, &lt;a href=&quot;http://arxiv.org/abs/0909.0551&quot; rel=&quot;nofollow&quot;&gt;Division Algebras and Supersymmetry I&lt;/a&gt; and &lt;a href=&quot;http://arxiv.org/abs/1003.3436&quot; rel=&quot;nofollow&quot;&gt;Division Algebras and Supersymmetry II&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't think you'll be able to find an easy application to explain to a layman, since the octonions are naturally connected to geometry in higher dimensions than most people can be bothered to care about.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-08-11T20:50:27.850" CommunityOwnedDate="2010-08-11T20:55:10.957" />
  <row Id="2196" PostTypeId="2" ParentId="2193" CreationDate="2010-08-11T21:03:38.630" Score="3" ViewCount="" Body="&lt;p&gt;Yes.  This is an easy consequence of the Krull-Schmidt theorem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://planetmath.org/encyclopedia/KrullRemakSchmidtTheorem.html&quot; rel=&quot;nofollow&quot;&gt;http://planetmath.org/encyclopedia/KrullRemakSchmidtTheorem.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Edit&lt;/b&gt;: To be clear, what I am claiming is that if one has finite groups $H,K,G$ such that $G \times H \cong G \times K$, then $H \cong K$.  As Steve D points out below, the hypothesis and conclusion of the OP's literal question are a bit different than this.  But &lt;em&gt;because&lt;/em&gt; of the Krull-Schmidt theorem, the OP's literal question becomes: let $H$ and $K$ be isomorphic subgroups of a finite group $G$.  When do we have $H = K$?  It is quite clear that the answer is &quot;not always&quot;, and I find implausible that there would be a clean necessary and sufficient condition for this.  But let's see what transpires...&lt;/p&gt;&#xA;" OwnerUserId="299" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2010-08-11T22:39:00.277" LastActivityDate="2010-08-11T22:39:00.277" CommentCount="4" />
  <row Id="2197" PostTypeId="2" ParentId="2190" CreationDate="2010-08-11T21:06:40.073" Score="4" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://www.imc-math.org.uk/imc2009/imc2009-day2-solutions.pdf&quot; rel=&quot;nofollow&quot;&gt;Problem 5&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1028" LastActivityDate="2010-08-11T21:06:40.073" CommentCount="1" />
  <row Id="2198" PostTypeId="2" ParentId="2121" CreationDate="2010-08-11T21:12:25.850" Score="1" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://www.imc-math.org.uk/imc2008/day1_solutions.pdf&quot; rel=&quot;nofollow&quot;&gt;Problem 5&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1028" LastActivityDate="2010-08-11T21:12:25.850" CommentCount="1" />
  <row Id="2199" PostTypeId="2" ParentId="2193" CreationDate="2010-08-11T21:13:01.870" Score="2" ViewCount="" Body="&lt;p&gt;Here's an entry point into the literature, from the introduction to Lam's &lt;a href=&quot;http://math.berkeley.edu/~lam/ohio.ps&quot; rel=&quot;nofollow&quot;&gt;paper [1]&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the study of any algebraic system in which there is a notion of a direct sum, the  theme of cancellation arises very naturally: if $A \oplus B \cong A\oplus C$ in the given system, can we conclude that $B \cong  C$? (For an early treatment of this problem, see the work of Jonsson  and Tarski [JT] in 1947.) The answer is, perhaps not surprisingly, sometimes &quot;yes&quot; and  sometimes &quot;no&quot;: it all depends on the algebraic system, and it depends heavily on the  choice of A as well.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Starting with a simple example, we all know that, by the Fundamental Theorem of  Abelian Groups, the category of finitely generated abelian groups satisfies cancellation.  But a little more is true, which solved what would have been the &quot;Third Test Problem&quot;  for §6 in Kaplansky's book [Ka 1] (see the Notes in [Ka_1:§20]): if A is a f.g. (finitely  generated) abelian group, then for any abelian groups $B$ and $C$, $A\oplus B \cong  A\oplus C$ still implies $B \cong C$. Thus, f.g. abelian groups A remain &quot;cancellable&quot; (with respect  to direct sums) in the category of all abelian groups. This takes a proof, which was  first given, independently, by P. M.Cohn [Co] and E. A. Walker [W]. And yet, there exist  many torsionfree abelian groups of rank 1 (that is, nonzero subgroups of the rational  numbers Q ) that are not cancellable in the category of torsionfree abelian groups of  finite rank, according to B. Jonsson [Jo].  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;[1] T.Y. Lam. A Crash Course on Stable Range, Cancellation, Substitution, and Exchange&lt;br&gt;&#xA;University of California, Berkeley, Ca 94720&lt;br&gt;&#xA;&lt;a href=&quot;http://math.berkeley.edu/~lam/ohio.ps&quot; rel=&quot;nofollow&quot;&gt;http://math.berkeley.edu/~lam/ohio.ps&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-08-11T22:03:17.360" LastActivityDate="2010-08-11T22:03:17.360" CommentCount="1" />
  <row Id="2200" PostTypeId="2" ParentId="2194" CreationDate="2010-08-11T21:18:54.387" Score="4" ViewCount="" Body="&lt;p&gt;This appears to be a difficult result.  It was first proven in 1931 by O. H. Keller (&lt;a href=&quot;http://www.springerlink.com/content/r8nqu8v469110152/&quot; rel=&quot;nofollow&quot;&gt;reference&lt;/a&gt;), but the paper is in German.  Keller actually proves a stronger result: any infinite-dimensional compact convex set in $\ell^2$ is homeomorphic to the Hilbert cube.  (I guess there is an obvious choice of such a set which is clearly homogeneous, but I can't think of one at the moment.)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is another proof of the theorem in &lt;a href=&quot;http://iopscience.iop.org/1468-4810/42/3/A08/pdf/1468-4810_42_3_A08.pdf&quot; rel=&quot;nofollow&quot;&gt;this paper&lt;/a&gt; of Ageev, which is in English.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-08-11T21:18:54.387" CommentCount="10" />
  <row Id="2201" PostTypeId="2" ParentId="2187" CreationDate="2010-08-11T21:47:44.570" Score="1" ViewCount="" Body="&lt;p&gt;This sounds similar to computing the last nonzero base-10 digit of factorial(n).  The fact that you are, according to some scheme, dropping the last bit of what would otherwise be a Fibonacci sequence expressed in binary makes me think there is no hope for finding any periodicity in the values mod p, which is what happens in the Fibonacci sequence itself.  If you drop the last bit only finitely many times, then continue after that, your sequence modulo p will become periodic.&lt;/p&gt;&#xA;" OwnerUserId="297" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-08-11T21:59:04.537" LastActivityDate="2010-08-11T21:59:04.537" CommentCount="2" />
  <row Id="2202" PostTypeId="2" ParentId="2193" CreationDate="2010-08-11T23:41:29.340" Score="1" ViewCount="" Body="&lt;p&gt;Concerning Steve D's interpretation of the question, here is a partial answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the following property (P) of a group: for any two subgroups $H$ and $K$ of $G$, if $H \cong K$, then $H = K$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Claim: For a finite group $G$, the following are equivalent:&lt;br&gt;&#xA;(i) $G$ has property (P).&lt;br&gt;&#xA;(ii) $G$ is cyclic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cyclic groups are characterized among finite groups by having at most one subgroup of any given order, so certainly (ii) $\implies$ (i).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conversely, assume $G$ has property (P).  Then it is a &lt;strong&gt;Dedekind group&lt;/strong&gt;: all of its subgroups are normal (for otherwise it has two subgroups which are conjugate -- hence isomorphic -- but unequal).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Case 1: $G$ is abelian, of exponent $N$.  Then $G$ is isomorphic to $C_N \times G'$ for some subgroup $G'$ (this is a step in the classification of finite abelian groups; it also follows easily from the theorem).  If $G'$ is not cyclic, then $G'$ is not trivial hence contains an element of order $p | N$ and thus over all $G$ contains at least two subgroups of order $p$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Case 2: $G$ is a nonabelian Dedekind group, i.e., a &lt;a href=&quot;http://planetmath.org/encyclopedia/QuasiHamiltonianGroup.html&quot; rel=&quot;nofollow&quot;&gt;Hamiltonian group&lt;/a&gt;.  As Dedekind showed, a Hamiltonian group (finite or otherwise!) must contain a subgroup isomorphic to the quaternion group $Q_8$.  But $Q_8$ contains three cyclic subgroups of order $4$, namely those generated by the elements $i$, $j$ and $k$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are some noncyclic infinite abelian groups with property (P), e.g. $\mathbb{Q}/\mathbb{Z}$, but probably one can classify them as well: I haven't thought much about it.  &lt;/p&gt;&#xA;" OwnerUserId="299" LastActivityDate="2010-08-11T23:41:29.340" CommentCount="4" />
  <row Id="2204" PostTypeId="1" AcceptedAnswerId="2207" CreationDate="2010-08-12T00:16:43.700" Score="12" ViewCount="344" Body="&lt;p&gt;There are $N$ bugs in a plane. All bugs are moving at the same constant (nonzero) speed, but no two bugs are moving in the same direction (velocity vectors are of the same speed, but no two are parallel). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Prove that at some point in time $N$ bugs will form &lt;strong&gt;convex&lt;/strong&gt; polygon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: Can you loosen up any of the conditions so that the statement still holds?&lt;/p&gt;&#xA;" OwnerUserId="956" LastEditorUserId="76" LastEditorDisplayName="" LastEditDate="2010-08-17T05:50:32.830" LastActivityDate="2010-08-22T08:00:11.617" Title="Bugs walking in a plane" Tags="&lt;geometry&gt;&lt;intuition&gt;" AnswerCount="4" CommentCount="7" FavoriteCount="3" />
  <row Id="2205" PostTypeId="2" ParentId="2204" CreationDate="2010-08-12T00:24:59.923" Score="9" ViewCount="" Body="&lt;p&gt;The counterexample I thought I had here doesn't work.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a proof.  Since none of the bugs are moving in the same direction, any pair of lines determined by the velocity vectors intersect.  Let $C$ denote the convex hull of these intersection points.  Since after waiting a sufficiently long period of time the bugs will be arbitrarily far away from $C$, if we &quot;zoom out&quot; far enough $C$ will become arbitrarily small with respect to the convex hull of the location of the bugs.  It follows that we can assume that $C$ is arbitrarily small to begin with.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We now claim that the bugs eventually form a convex polygon in which the angle at each vertex is strictly less than $\pi$.  To do this it suffices to examine a configuration of three bugs $a, b, c$ in consecutive counterclockwise order.  Pick a coordinate system in which the centroid of $C$ is the origin and $b$ travels in the positive $y$-direction (hence $a$ travels to the right and $c$ travels to the left).  Then it is easy to see that regardless of where $a, b, c$ initially begin along their routes, $b$ will eventually have $y$-coordinate greater than either $a$ or $c$, so angle $abc$ will eventually be strictly less than $\pi$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It follows that by waiting sufficiently long the bugs will always form a convex polygon.  In fact, the bugs are approximating the convex polygon whose vertices are the unit velocity vectors of the bugs.&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-08-12T22:10:51.063" LastActivityDate="2010-08-12T22:10:51.063" CommentCount="10" />
  <row Id="2206" PostTypeId="2" ParentId="2204" CreationDate="2010-08-12T00:47:41.033" Score="1" ViewCount="" Body="&lt;p&gt;Suppose the bugs all started at the origin, then they would form a convex polygon at all times after $t = 0$. It should be noted we have some wiggle room - we may move any bug a small distance (proportional to time) and the polygon will remain convex.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since our bugs lie initially within a circle, let them walk far enough that they are within the wiggle room of the convex polygon that would have been formed had they started at the origin. Hence the bugs form a convex polygon.&lt;/p&gt;&#xA;" LastActivityDate="2010-08-12T00:47:41.033" CommentCount="0" />
  <row Id="2207" PostTypeId="2" ParentId="2204" CreationDate="2010-08-12T01:02:43.157" Score="8" ViewCount="" Body="&lt;p&gt;Assuming no bugs get squashed in the process:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At $\lim_{t\to\infty}$ when $t$ is time, the bugs' beginning points shrink to $\frac{P}{t} = 0$ as observed when zoomed out. This means the final position of each bug lies on $\sqrt{r^2 + (vt)^2}$, or on the edge of a huge circle. Thus, you can see that all the bugs make a convex polygon (or $N$-gon) since a circle can be thought of as a regular (and convex) $\infty$-gon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course this isn't the most vigorous proof in the world, but it's written in the style &lt;a href=&quot;http://www.smbc-comics.com/index.php?db=comics&amp;amp;id=1914&quot; rel=&quot;nofollow&quot;&gt;that most humans can understand&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="227" LastEditorUserId="227" LastEditorDisplayName="" LastEditDate="2010-08-12T01:35:35.513" LastActivityDate="2010-08-12T01:35:35.513" CommentCount="4" />
  <row Id="2209" PostTypeId="1" AcceptedAnswerId="2213" CreationDate="2010-08-12T01:30:02.050" Score="11" ViewCount="234" Body="&lt;p&gt;There are 100 ropes in a bag. In each step, two rope ends are picked at random, tied together and put back into a bag. The process is repeated until there are no free ends.&lt;br&gt;&#xA;What is the expected number of loops at the end of the process?&lt;/p&gt;&#xA;" OwnerUserId="956" LastActivityDate="2010-08-12T02:51:06.847" Title="Expected number of loops" Tags="&lt;probability&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" />
  <row Id="2210" PostTypeId="1" AcceptedAnswerId="2214" CreationDate="2010-08-12T01:49:36.043" Score="4" ViewCount="194" Body="&lt;p&gt;I was wondering if anyone had any experience with an undergraduate course that emphasized the building of mathematical theories or if they'd ever heard of this being done? How did the class work (did the professor list axioms and definitions each week and see what the students were able to prove, did you discuss a specific mathematical theory and the various problems that arose during its creation?) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does this sort of thing work well for undergraduates? I'd be very interested in taking a class like this because I've never really had any experience with that aspect of mathematics. So if there are any classes that are commonly taught in this sort of style, I'd be interested to know that as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="249" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-08-12T01:54:25.923" LastActivityDate="2010-08-12T02:26:19.027" CommunityOwnedDate="2010-08-12T01:54:25.923" Title="undergraduate courses emphasizing theory building?" Tags="&lt;soft-question&gt;&lt;education&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" />
  <row Id="2211" PostTypeId="1" AcceptedAnswerId="2215" CreationDate="2010-08-12T02:00:21.523" Score="1" ViewCount="162" Body="&lt;p&gt;Some points on (integer, rectangular) grid in a plane are colored white, some black and some are not colored.&lt;br&gt;&#xA;In each step, one vertical or horizontal line can be selected, and all colored points on that line would toggle their color.&lt;br&gt;&#xA;Prove there is a finite sequence of steps after which on every vertical or horizontal line number of black and number of white points differs by 1 at most.&lt;/p&gt;&#xA;" OwnerUserId="956" LastActivityDate="2010-08-12T04:14:10.027" Title="Black and White points on a grid" Tags="&lt;discrete-geometry&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="2212" PostTypeId="2" ParentId="2182" CreationDate="2010-08-12T02:02:56.197" Score="1" ViewCount="" Body="&lt;p&gt;Of course, a measure is in general only defined on some $\sigma$-algebra of sets, and for Haar measure that's the Borel sets.  As Pete mentioned you can complete the measure if you like, but that still won't yield all sets.  So for an arbitrary set $E$, $\mu(E)$ may not even be defined.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can see this is necessary with an example you already know: let $G = S^1$ thought of as $[0,1]$ mod endpoints, and let $E$ be the usual example of a non-measurable set (the Vitali set).  There's no value for $\mu(E)$ that will let $\mu$ be translation invariant, finite, and nontrivial, which are required for a Haar measure, and so there's no &quot;Haar measure&quot; defined for all subsets.  To be able to have a Haar measure, we need to define it only on &quot;reasonable&quot;, i.e. Borel, sets.&lt;/p&gt;&#xA;" OwnerUserId="822" LastActivityDate="2010-08-12T02:02:56.197" />
  <row Id="2213" PostTypeId="2" ParentId="2209" CreationDate="2010-08-12T02:24:00.063" Score="10" ViewCount="" Body="&lt;p&gt;Suppose you start with $n$ ropes. You pick two free ends and tie them together:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;If you happened to pick two ends of the same rope, you've added one additional loop (which you can set aside, since you'll never pick it now on), and have $n-1$ ropes&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If you happened to pick ends of different ropes, you've added no loop, and effectively replaced the two ropes with a longer rope, so you have $n-1$ ropes in this case too.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Of the $\binom{2n}{2}$ ways of choosing two ends, $n$ of them result in the first case, so the first case has probability $\frac{n}{2n(2n-1)/2} = 1/(2n-1)$. So the expected number of loops you &lt;strong&gt;add in the first step&lt;/strong&gt;, when you start with $n$ ropes, is $$\left(\frac{1}{2n-1}\right)1 + \left(1-\frac{1}{2n-1}\right)0 = \frac{1}{2n-1}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After this, you start over with $n-1$ ropes. Since what happens in the first step and later are independent (and expectation is linear anyway), the expected number of loops is $$ \frac{1}{2n-1} + \frac{1}{2n-3} + \dots + \frac{1}{3} + 1 = H_{2n} - \frac{H_n}{2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, for $n=100$, the answer is roughly $3.28$, which come to think of it seems surprisingly small for the number of loops!&lt;/p&gt;&#xA;" OwnerUserId="205" LastEditorUserId="205" LastEditorDisplayName="" LastEditDate="2010-08-12T02:51:06.847" LastActivityDate="2010-08-12T02:51:06.847" CommentCount="6" />
  <row Id="2214" PostTypeId="2" ParentId="2210" CreationDate="2010-08-12T02:26:19.027" Score="5" ViewCount="" Body="&lt;p&gt;I have benefited from more than one professor who taught more or less according to the &lt;a href=&quot;http://en.wikipedia.org/wiki/Moore_method&quot; rel=&quot;nofollow&quot;&gt;Moore Method&lt;/a&gt;.  These professors happened to come out of SUNY Binghamton in the early 1980's.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The courses required lots of input from the students.  One professor in particular would ask the class to make a conjecture about some construction he would put on the board and then name the conjecture after the student - we were then personally responsible for success or failure of our ideas.  After the first few humiliations, this became wonderfully empowering and fun.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The biggest impact I have noticed from taking these courses has been on my independent studying in other classes.  I notice that I am able to replicate the types of attacks on my own reasoning that my professor used to - hence leading me closer toward the boundaries of my own understanding.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't really know how to respond to your idea of theory building... it seems to me that mathematics is not created upward from luckily chosen axioms, but in a highly nonlinear and back-and-forth fashion.  The only example I know of to illustrate this clearly is the development of topology, starting in the late 1800's with Cantor and more or less culminating in the early 1900's with Hausdorff (excellent discussions with references appear in &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0800202090&quot; rel=&quot;nofollow&quot;&gt;Engelking, General Topology&lt;/a&gt;).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I agree that it would be a great benefit to you to find courses which at least approach what you have in mind.  Good luck to you. &lt;/p&gt;&#xA;" OwnerUserId="121" LastActivityDate="2010-08-12T02:26:19.027" CommunityOwnedDate="2010-08-12T02:26:19.027" CommentCount="1" />
  <row Id="2215" PostTypeId="2" ParentId="2211" CreationDate="2010-08-12T03:21:46.837" Score="5" ViewCount="" Body="&lt;p&gt;What sequence of horizontal or vertical steps would lead&lt;/p&gt;&#xA;&#xA;&lt;p&gt;B W&lt;/p&gt;&#xA;&#xA;&lt;p&gt;W W&lt;/p&gt;&#xA;&#xA;&lt;p&gt;to have the number of black and white points almost even?&lt;/p&gt;&#xA;" OwnerUserId="557" LastEditorUserId="557" LastEditorDisplayName="" LastEditDate="2010-08-12T04:14:10.027" LastActivityDate="2010-08-12T04:14:10.027" CommentCount="1" />
  <row Id="2216" PostTypeId="2" ParentId="2194" CreationDate="2010-08-12T03:49:35.893" Score="3" ViewCount="" Body="&lt;p&gt;When I first saw the question, I intended to post the following link as an answer, but then I saw that Qiaochu had come first with an essentially equivalent answer.  I still don't see anything lacking in his answer (I was the first to upvote it), but if someone is looking for a different reference, here is one:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.narcis.info/publication/RecordID/oai:cwi.nl:7603&quot; rel=&quot;nofollow&quot;&gt;http://www.narcis.info/publication/RecordID/oai:cwi.nl:7603&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="299" LastActivityDate="2010-08-12T03:49:35.893" CommentCount="1" />
  <row Id="2218" PostTypeId="2" ParentId="1872" CreationDate="2010-08-12T04:11:07.923" Score="2" ViewCount="" Body="&lt;p&gt;I guess I should expand my comment into an answer.  Given two matrices $a_{ij}$ and $b_{ij}$ with entries in any (associative) ring $R$, the natural definition of the product has entries&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle c_{ij} = \sum_k a_{ik} b_{kj}.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This multiplication is associative, and it also agrees with the multiplication one obtains from any finite-dimensional matrix representation of $R$ by replacing each entry by the corresponding matrix.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I do not see any particular reason to consider a different notion of multiplication.  Changing the order of some of the multiplications seems nonsensical to me, and multiplying in the opposite order gives you essentially the same multiplication.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This definition does &lt;em&gt;not&lt;/em&gt; agree with the definition in my first comment; multiplication by one of the above matrices does not define an $R$-module homomorphism when $R$ is noncommutative.  &lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-08-12T04:11:07.923" CommentCount="8" />
  <row Id="2221" PostTypeId="1" CreationDate="2010-08-12T05:30:45.557" Score="6" ViewCount="304" Body="&lt;p&gt;Suppose that $c_k$ is an decreasing sequence of non-negative real numbers, such that $c_0=1$&#xA;and $c_{k}\leq \frac{1}{2}(c_{k-1}+c_{k+1})$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it true that the generated function of $c_k$ admit an integral representation as below&#xA;$$&#xA;\sum_{k=0}^{\infty}c_kz^k=\int_{\partial\mathbb D} \frac{1}{1-\zeta z}d\mu(\zeta),&#xA;$$&#xA;where $\mu$ is a Borel Probability Measure in $\partial\mathbb D$ ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Motivation:&lt;/b&gt; This question is related a possible slight different solution for the question &#xA;asked in &lt;a href=&quot;http://math.stackexchange.com/questions/2188/complex-analysis-question&quot;&gt;http://math.stackexchange.com/questions/2188/complex-analysis-question&lt;/a&gt; whose the answer, as pointed out by damiano, can be found at the IMC website.&lt;/p&gt;&#xA;" OwnerUserId="633" LastEditorUserId="633" LastEditorDisplayName="" LastEditDate="2010-08-12T11:56:39.613" LastActivityDate="2011-05-27T13:03:06.517" Title="Convex sequences and Integral representation for the generating function" Tags="&lt;complex-analysis&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="2222" PostTypeId="2" ParentId="154" CreationDate="2010-08-12T07:14:29.090" Score="4" ViewCount="" Body="&lt;p&gt;As phrased, your question invites a philosophical answer but you said that's not what you want. One way to approach this question is to ask if complex numbers correspond to anything in the real world. What can you count or measure with them?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as I can tell, complex numbers are most directly useful for measuring things that rotate or oscillate. They're used by electrical engineers because voltages and currents can oscillate, and could be used for measuring springs or pendulums, or for anything that behaves like a wave. There are not that many situations in day-to-day life where you'd use complex numbers, but they're used extensively in physics because waves and oscillations show up everywhere.&lt;/p&gt;&#xA;" OwnerUserId="1086" LastActivityDate="2010-08-12T07:14:29.090" CommunityOwnedDate="2011-04-15T20:45:29.080" />
  <row Id="2223" PostTypeId="2" ParentId="535" CreationDate="2010-08-12T09:39:01.460" Score="0" ViewCount="" Body="&lt;p&gt;One way to look at your problem is that you are trying to find the graph of an equation p(x,c)=0. In other words, you are trying to solve an implicit function equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general the solutions to implicit polynomial equations in two variables are called Algebraic Varieties, and can be very complicated (http://en.wikipedia.org/wiki/Algebraic_variety). So you can't really say necessarily what kind of curve you will get. However, maybe for this specific polynomial it is a simple curve.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What can be said is that if you know p(x, c) = 0 for some given (x,c), then for all points d near enough to c one can find a root x(d) such that p(x(d), d) = 0. This is called the implicit function theorem, and another result of the theorem is that x(d) is even differentiable with respect to d.&lt;/p&gt;&#xA;" OwnerUserId="1089" LastActivityDate="2010-08-12T09:39:01.460" />
  <row Id="2224" PostTypeId="2" ParentId="1236" CreationDate="2010-08-12T09:44:03.210" Score="0" ViewCount="" Body="&lt;p&gt;The following page should be of some help:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Diffusion&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Diffusion&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1089" LastActivityDate="2010-08-12T09:44:03.210" CommentCount="1" />
  <row Id="2225" PostTypeId="1" AcceptedAnswerId="2228" CreationDate="2010-08-12T10:36:39.460" Score="1" ViewCount="374" Body="&lt;p&gt;If $f: \mathbb{R} \to \mathbb{R}$ is a continuous function and satisfies $f(x)=f(2x+1)$, then its not to hard to show that $f$ is a constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is suppose $f$ is continuous and it satisfies $f(x)=f(2x+1)$, then can the domain of $f$ be restricted so that $f$ doesn't remain a constant. If yes, then give an example of such a function.&lt;/p&gt;&#xA;" LastEditorUserId="8297" LastEditorDisplayName="" LastEditDate="2011-05-15T08:55:23.660" LastActivityDate="2011-05-15T08:55:23.660" Title="Function Satisfying $f(x)=f(2x+1)$" Tags="&lt;real-analysis&gt;&lt;functional-equations&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="2226" PostTypeId="2" ParentId="2178" CreationDate="2010-08-12T10:42:13.323" Score="1" ViewCount="" Body="&lt;p&gt;b) consider a path $v_0\dots v_k$ of length $k$ (existing for part a) and for Chandru's construction) such that $v_i \neq v_j$ $\forall i\neq j$.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If $v_0$ and $v_k$ are adjacent, you have a cycle of lenght $k+1$.&lt;/li&gt;&#xA;&lt;li&gt;Otherwise, since $deg(v_k)\geq k$, $v_k$ has at least one adjacent vertex $v_l$ with $l&gt;k$ and you can extend the path with this new vertex. If $v_l$ is adjacent with $v_0$ or $v_1$ you're done, otherwise you can extend the path again with a vertex that doesn't belong to the path yet. Since $G$ has only a finite number of vertices, at some point you will find a vertex $v_n$ whose adjacent vertices are all part of the path. $deg(v_n)\geq k$, so $v_n$ is adjacent to $v_i$ for some $i&amp;lt; n-k$ and you can finally close the path with a cycle of lenght $n-i \geq k+1$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1065" LastEditorUserId="1065" LastEditorDisplayName="" LastEditDate="2010-08-12T12:58:45.760" LastActivityDate="2010-08-12T12:58:45.760" CommentCount="1" />
  <row Id="2227" PostTypeId="1" AcceptedAnswerId="2348" CreationDate="2010-08-12T10:46:33.783" Score="0" ViewCount="355" Body="&lt;p&gt;Can one think of an example of a continuous $2\pi$ periodic function whose Fourier series fails to converge on $\mathbb{R}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I referred this in the wikipedia page but no avail: &lt;strong&gt;It might be interesting to note that Jean-Pierre Kahane and Yitzhak Katznelson proved that for any given set E of measure zero, there exists a continuous function ƒ such that the Fourier series of ƒ fails to converge on any point of E.&lt;/strong&gt;&lt;/p&gt;&#xA;" LastActivityDate="2010-08-13T14:16:42.610" Title="Example of a function whose Fourier Series fails to converge at One point" Tags="&lt;fourier-analysis&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="2228" PostTypeId="2" ParentId="2225" CreationDate="2010-08-12T10:56:07.593" Score="5" ViewCount="" Body="&lt;p&gt;Let $f$ have value $1$ on $[0,\infty)$ and value $0$ on $(-\infty,-1]$. This function is not constant (although it is locally constant), and satisfies $f(x)=f(2x+1)$ whenever $x$ is in its domain. &lt;/p&gt;&#xA;" OwnerUserId="413" LastActivityDate="2010-08-12T10:56:07.593" CommentCount="4" />
  <row Id="2229" PostTypeId="1" AcceptedAnswerId="2246" CreationDate="2010-08-12T10:56:17.673" Score="5" ViewCount="201" Body="&lt;p&gt;Using the duality between locally compact Hausdorff spaces and commutative $C^*$-algebras one can write down a vocabulary list translating topological notions regarding a locally compact Hausdorff space $X$ into algebraic notions ragarding its ring of functions $C_0(X)$ (see Wegge-Olsen's book, for instance). For example, we have the following correspondences:&#xA;\begin{align*}&#xA;\text{open subset of $X$}\quad &amp;amp;\longleftrightarrow\quad\text{ideal in $C_0(X)$}\newline&#xA;\text{dense open subset of $X$}\quad &amp;amp;\longleftrightarrow\quad\text{essential ideal in $C_0(X)$}\newline&#xA;\text{closed subset of $X$}\quad &amp;amp;\longleftrightarrow\quad\text{quotient of $C_0(X)$}\newline&#xA;\text{locally closed subset of $X$}\quad &amp;amp;\longleftrightarrow\quad\text{subquotient of $C_0(X)$}\newline&#xA;\text{???}\quad &amp;amp;\longleftrightarrow\quad\text{$C^*$-subalgebra in $C_0(X)$}&#xA;\end{align*}&#xA;By ideal I always mean a two-sided closed (and hence self-adjoint) ideal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, I can't quite see &lt;strong&gt;how to reconvert a $C^*$-subalgebra in $C_0(X)$ into something topological involving only the space $X$.&lt;/strong&gt; Can you come up with something handy?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Example:&lt;/strong&gt; A simple example of a subalgebra of a commutative $C^*$-algebra not being an ideal is&#xA;$$&#xA;\mathbb C\cdot(1,1)\subset \mathbb C\oplus\mathbb C.&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;(Alternatively, we could think about this question within the duality of affine algebraic varieties and finitely generated commutative reduced algebras or even within the duality between affine schemes and commutative rings.)&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Edit: Since I was not completely satisfied by the response I got here, I reposted this question on &lt;a href=&quot;http://mathoverflow.net/questions/35507/what-is-the-commutative-analogue-of-a-c-subalgebra&quot; rel=&quot;nofollow&quot;&gt;MO&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="367" LastEditorUserId="367" LastEditorDisplayName="" LastEditDate="2010-08-15T13:17:56.627" LastActivityDate="2010-08-15T13:17:56.627" Title="What is the commutative analogue of a $C^*$-subalgebra?" Tags="&lt;functional-analysis&gt;&lt;noncommutative-geometry&gt;&lt;c-star-algebras&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" />
  <row Id="2231" PostTypeId="2" ParentId="2130" CreationDate="2010-08-12T11:29:04.123" Score="1" ViewCount="" Body="&lt;p&gt;If a category is not small, its presheaf category is not locally small and this can lead to all sorts of problems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Not despair though, as there are several work-arounds. The best of all may simply be to pretend the problem does not exist and proceed naively; 99% percent of the time this works like a charm as one can always clean up the act through the use of more or less contorted devices / axioms.&lt;/p&gt;&#xA;" OwnerUserId="1033" LastActivityDate="2010-08-12T11:29:04.123" />
  <row Id="2232" PostTypeId="2" ParentId="2227" CreationDate="2010-08-12T11:55:20.037" Score="7" ViewCount="" Body="&lt;p&gt;No. Any continuous function is in $L^2$. The &lt;a href=&quot;http://en.wikipedia.org/wiki/Carleson%27s_theorem&quot; rel=&quot;nofollow&quot;&gt;Carleson-Hunt theorem&lt;/a&gt; states that an $L^2$ function's Fourier series converges almost everywhere to the function.&lt;/p&gt;&#xA;" OwnerUserId="536" LastActivityDate="2010-08-12T11:55:20.037" CommentCount="5" />
  <row Id="2234" PostTypeId="1" AcceptedAnswerId="2236" CreationDate="2010-08-12T12:31:33.753" Score="2" ViewCount="1241" Body="&lt;p&gt;some guy asked for formula for sum of harmonic progression.&#xA;So i have done some calculation and give him some approx formula.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;integration of 1/x from 1 to n = 1/2(y1+y2)+1/2(y2+y3)..........1/2(yn-1+yn)               where yi is ith term of HP&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ln(n) = (y1)/2+y2+y3..............(yn)/2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;sum of HP = ln(n)+(first term+last&#xA;  term)/2&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;e.g. 1+1/2+..............+1/10 = 2.8525                  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;actual result = 2.9289&lt;/p&gt;&#xA;&#xA;&lt;p&gt;my question is how to correct this formula&lt;/p&gt;&#xA;" OwnerUserId="1093" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-08-13T03:25:28.067" LastActivityDate="2010-10-27T21:17:37.567" Title="sum of harmonic progression?" Tags="&lt;sequences-and-series&gt;" AnswerCount="4" />
  <row Id="2235" PostTypeId="2" ParentId="2225" CreationDate="2010-08-12T12:35:48.427" Score="4" ViewCount="" Body="&lt;p&gt;As in &lt;a href=&quot;http://mathoverflow.net/questions/31990/continuous-functions-remaining-constant&quot; rel=&quot;nofollow&quot;&gt;the previous proof&lt;/a&gt; of $f$ being constant on $\mathbb{R}$, define $g(x) = f(x-1)$, so that $g(x) = g(2x)$; the domains of $f$ and $g$ are just shifted versions of each other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Certainly, if the domain of $g$ is small enough, say $[2,3]$, then $g$ can be any continuous function, because the domain contains no $x$ and $2x$ at the same time. A more interesting question is: how large can we make the domain so that $g$ will still not be constant? The answer to this is suggested by &lt;a href=&quot;http://math.stackexchange.com/questions/2225/function-satisfying-fxf2x1/2228#2228&quot;&gt;JDH's answer&lt;/a&gt;: if we remove only the single point 0, making the domain $\mathbb{R} \setminus \{0\}$, it is disconnected into two components which can independently have constant values.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How big can a domain be on which $g$ is not even locally constant? Remove an arbitrarily small interval around $0$. Take any non-constant continuous function $h$ which is periodic with unit period, and let $g(x) = h(\log_2 x)$. Then $g(x) = g(2x)$ for all $x$, and is continuous everywhere.&lt;/p&gt;&#xA;" OwnerUserId="856" LastActivityDate="2010-08-12T12:35:48.427" />
  <row Id="2236" PostTypeId="2" ParentId="2234" CreationDate="2010-08-12T12:35:49.753" Score="8" ViewCount="" Body="&lt;p&gt;The partial sums of the harmonic series are called &quot;harmonic numbers.&quot; The difference between the nth harmonic number and ln(n) tends to a limit as n increases, and that limit is called &lt;a href=&quot;http://en.wikipedia.org/wiki/Euler%27s_constant&quot; rel=&quot;nofollow&quot;&gt;Euler's constant&lt;/a&gt; or gamma.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's a great book about all this called &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0691141339&quot; rel=&quot;nofollow&quot;&gt;Gamma: Exploring Euler's Constant&lt;/a&gt;. &lt;/p&gt;&#xA;" OwnerUserId="247" LastEditorUserId="247" LastEditorDisplayName="" LastEditDate="2010-08-13T00:41:43.427" LastActivityDate="2010-08-13T00:41:43.427" CommentCount="2" />
  <row Id="2237" PostTypeId="1" AcceptedAnswerId="2259" CreationDate="2010-08-12T13:19:17.043" Score="3" ViewCount="181" Body="&lt;p&gt;Hi there,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am trying to find the answer to the following question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I am playing poker and am dealt two cards neither of which are spades, what are the chances of two or more of the flop cards (the flop is the first three community cards) being spades.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I answered this question in the following way:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As I have no spades, there are still 13 spades unseen and a total of 50 cards remaining unseen (52 cards in the pack less my two cards which I've been dealt.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the probability of the first flop card being a spade is p(A) and the second p(B) and the third p(C) then I need to find:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;p((A&amp;cap;B)&amp;cup;(A&amp;cap;C)&amp;cup;(B&amp;cap;C))&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so given the formula:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;p(X&amp;cup;Y&amp;cup;Z) = p(X) + p(Y) + p(Z) - p(X&amp;cap;Y) - p(X&amp;cap;Z) - p(Y&amp;cap;Z) + p(X&amp;cap;Y&amp;cap;Z)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I calculated:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;p(2 or more spades on flop) = 3(13/50 x 12/49) - 3((13/50 x 12/49) x 11/48) + (13/50 x 12/49 x 11/48)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which gives &lt;strong&gt;16.18%&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I wanted to do next was solve the question using combinations. Through a process of trial and error I got the same answer, but am a bit confused as to why it works this way. My method was:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;13C2 * 48 = 3744&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3744 - (3 * 13C3) + 13C3 = 3172&lt;/p&gt;&#xA;&#xA;&lt;p&gt;total number possible 3 card combinations on flop = 50C3 = 19600&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3172 / 19600 = &lt;strong&gt;16.18%&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I get the total number of 2 card combination of spades by 13C2 = 78. Then there are 48 other cards which can form combinations with each of these 2 card combination so 13C2 * 48.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I'm losing my grasp of whats happening. Why does 13C2 = p(A) + p(B) + p(C)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All I have done to follow is find the number of combos for p(A&amp;cap;B), p(A&amp;cap;C), p(B&amp;cap;C) and p(A&amp;cap;B&amp;cap;C) and then crunched the numbers as in the calculation shown above. But even though I (think I've) got the right answer, I've come to it in a try-it-and-see way and can't understand for instance why 13C2 * 48 doesn't give the right answer straight off. Surely that gives the number of 3 card combinations on the flop with 2 or more spades which is what I'm looking for, doesn't it? But I don't get the same answer as with the original calculation if I use that figure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sorry for the waffle. I hope someone can see where my thinking is going wrong and can help explain the combinations part a bit better for me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many thanks in advance.&lt;/p&gt;&#xA;" OwnerUserId="1094" LastEditorUserId="1094" LastEditorDisplayName="" LastEditDate="2010-08-12T16:29:32.857" LastActivityDate="2010-08-12T17:05:38.147" Title="I've got the same answer to my question via probability and combinations, but I don't know why!" Tags="&lt;combinatorics&gt;&lt;probability&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="2238" PostTypeId="1" AcceptedAnswerId="2244" CreationDate="2010-08-12T13:20:59.503" Score="2" ViewCount="90" Body="&lt;p&gt;There are $N&gt;1$ points in a plane. Consider the set of all distances between pairs of the points. Let $n$ be the number of elements of this set.&#xA;We know that $$n \leq {N \choose 2}.$$&#xA;What is the lower bound estimate on $n$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the sake of clearity, if points are $A_1, A_2, \dots, A_N$, and $S = \{d(A_i, A_j)| 1\leq i &amp;lt; j \leq N \}$, then $n = |S|$.&lt;/p&gt;&#xA;" OwnerUserId="956" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-08-13T06:53:59.167" LastActivityDate="2010-08-13T06:53:59.167" Title="Lower bound for a set of distances between pairs of points in a plane" Tags="&lt;discrete-geometry&gt;&lt;extremal-combinatorics&gt;" AnswerCount="1" CommentCount="6" />
  <row Id="2239" PostTypeId="1" AcceptedAnswerId="2247" CreationDate="2010-08-12T13:23:30.853" Score="4" ViewCount="292" Body="&lt;p&gt;Despite appearances, this is not a question on computational aspects of number theory. The background is as follows. I once asked a number theorist about what he considered to be the most important unsolved problems in arithmetic geometry. He told me about a few, but along with some well-known problems he told me the following one also:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;How to determine conceptually when a number is squarefree or not?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When I protested that this sounded like a computational question, he told me that no, this is not so, and demonstrated that this has a rather nice solution for the ring of polynomials over a field, which is in many senses analogous to the ring of integers. Take the polynomial, take its derivative and compute the gcd using the Euclidean algorithm. But for the ring of integers there is nothing analogous to the derivative, and he wanted a solution of the problem by constructing a good notion of a differential in this case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Question: What are the known investigations along this line? What well-known topics in arithmetic geometry are related to this? And what would be some other interesting consequences of a successful development of such a method?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any other comments that might enlighten me further would be received with gratitude.&lt;/p&gt;&#xA;" OwnerUserId="1119" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-08-12T14:07:17.677" LastActivityDate="2011-04-21T16:42:37.367" Title="Importance of determining whether a number is squarefree, using geometry." Tags="&lt;arithmetic-geometry&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="1" />
  <row Id="2240" PostTypeId="1" AcceptedAnswerId="2241" CreationDate="2010-08-12T13:30:08.577" Score="2" ViewCount="215" Body="&lt;p&gt;We know that $(\mathbb{Q},+,\times)$ is a subfield of $(\mathbb{R},+,\times)$. It is easy to see that the automorphism of $\mathbb{Q}$ is only the identity. For a quick proof, lets go through the main steps:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$f(1)=1 \Longrightarrow f(n)=n$ for all $n \in \mathbb{N}$.&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$f(-1)=-1$ which says that $f(x)=x$ for all $x \in \mathbb{Z}$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$\displaystyle f \Bigl(\frac{p}{q}\Bigr) = \frac{p}{q}$, where $q \neq 0$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;One, then uses the continuity of $f$ and the denseness of $\mathbb{Q}$ to prove that the Automorphism of $\mathbb{R}$ is also trivial.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My Question&lt;/strong&gt; Given a subfield $K$ of $\mathbb{C}$ is it and an automorphism of $K$ can it be extended to the whole of $\mathbb{C}$.&lt;/p&gt;&#xA;" LastActivityDate="2010-08-12T19:51:29.600" Title="Extending Automorphism of a Field." Tags="&lt;field-theory&gt;" AnswerCount="3" />
  <row Id="2241" PostTypeId="2" ParentId="2240" CreationDate="2010-08-12T13:35:57.453" Score="5" ViewCount="" Body="&lt;p&gt;Firstly, you need a little more to show that the only automorphism of $\mathbb R$ is the identity. You have to prove that any such automorphism is continuous. For a proof, let $x$ be a positive number; so it is the square of some number; so it is taken to another positive number under the automorphism. Since positive numbers go to positive numbers, the automorphism is order-preserving, and hence the identity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Secondly, this result is not true if you replace $\mathbb R$ by $\mathbb C$. The automorphism group of $\mathbb C$ over $\mathbb Q$ is uncountable. This is because $\mathbb C$ can be constructed by attaching uncountably many algebraicaly-independent transcendentals to $\mathbb Q$ and then taking the algebraic closure. Any automorphism of $\mathbb Q$ can be extended to an automorphism of $\mathbb C$, by first extending to the algebraic closure, and then to $\mathbb C$. The same argument would work for any algebraic number field. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;More general subfields are obtained as follows: You attach some number of transcendentals to $\mathbb Q$, and then take an algebraic extension of it. An automorphism of this field can be similarly extended to the whole of $\mathbb C$ as above.&lt;/p&gt;&#xA;" OwnerUserId="1119" LastEditorUserId="1119" LastEditorDisplayName="" LastEditDate="2010-08-12T13:45:01.363" LastActivityDate="2010-08-12T13:45:01.363" CommentCount="2" />
  <row Id="2243" PostTypeId="2" ParentId="2239" CreationDate="2010-08-12T13:49:05.317" Score="5" ViewCount="" Body="&lt;p&gt;The problem of finding some sort of analogue of differentiation of polynomials, in the context of integers, is a celebrated and difficult one.&#xA;I don't have anything useful to say about it, except to remark that the standard context in which this problem arises is that of the ABC conjecture: namely, the polynomial analogue of the ABC conjecture is a theorem, and one of the tools in its proof involves differentiating polynomials.  This has led people to try to construct some sort of analogous tool over the integers, with the goal of proving the ABC conjecture.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a short article of Faltings (called &quot;Does there exist an arithmetic Kodaira--Spencer class&quot;) in which he writes down some reasonable axioms such a notion of differentiation would satisfy (in scheme theoretic language over Spec $\mathbb Z$) and shows that it is impossible to satisfy them.  Nevertheless, one might (and many do!) hope for some more subtle structure, more complicated than the direct analogue that Faltings rules out, but still having the same basic nature as a derivative.  &lt;/p&gt;&#xA;" OwnerUserId="221" LastEditorUserId="221" LastEditorDisplayName="" LastEditDate="2011-04-21T16:42:37.367" LastActivityDate="2011-04-21T16:42:37.367" CommentCount="5" />
  <row Id="2244" PostTypeId="2" ParentId="2238" CreationDate="2010-08-12T13:54:34.057" Score="5" ViewCount="" Body="&lt;p&gt;You might want to read &lt;a href=&quot;http://www.math.ucsd.edu/~fan/mypaps/fanpap/67distances.pdf&quot; rel=&quot;nofollow&quot;&gt;this article&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The abstract of this article says:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A classical problem in combinatorial geometry is that of determining&#xA;the minimum number $f(n)$ of different distances determined by $n$ points in&#xA;the Euclidean plane. In 1952, L. Moser proved that $f(n) &gt; \frac{n^{2/3}}{2\cdot3^{1/3}} - 1$ and this has remained for 30 years as the best lower bound known for f(n). It is shown that $f(n) &gt; cn^{5/7}$ for some fixed constant c.&lt;/p&gt;&#xA;" OwnerUserId="334" LastActivityDate="2010-08-12T13:54:34.057" />
  <row Id="2246" PostTypeId="2" ParentId="2229" CreationDate="2010-08-12T14:08:25.817" Score="5" ViewCount="" Body="&lt;p&gt;Roughly, the answer will be that closed C*-subalgebras will correspond to quotient spaces&#xA;(via pull-back of functions).  In your example, the quotient map is one which identifies&#xA;the two points into a single point.   I haven't thought through, though, whether this is a completely correct statement as it stands, or whether one has to add additional caveats.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;[Added to answer a question in comments:]  The idea is that if $X$ surjects onto $Y$&#xA;then we get an injection $C_0(Y) \to C_0(X)$, and conversely.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;[Additional discussion added after more thought:] Let me say something about the analogous situation in algebraic geometry, where I am more comfortable with the technical issues:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Affine algebraic sets over $\mathbb C$ correspond to finite type reduced $\mathbb C$-algebras.&#xA;Giving an inclusion $A \hookrightarrow B$ of finite type reduced $\mathbb C$-algebras&#xA;corresponds to giving a map $X \to Y$ of algebraic sets which is dominant, i.e. the image is dense.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now in your set-up: if $X \to Y$ is a map of locally compact Hausdorff spaces with dense image, then again the map $C_0(X) \to C_0(Y)$ will be injective; so I might have been too hasty when I asserted that we get a surjective map.  On the other hand, perhaps the&#xA;image of $C_0(X) \to C_0(Y)$ will not be closed in this level of generality; it's a while since I've thought carefully about these sorts of things, so I don't think I can say more&#xA;right now with any certainty.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, I'm not so used to working in the case of rings without unit, so my suggestion has more chance to be correct in the case when the spaces are compact.&#xA;So perhaps it would be easiest to think about the case when $X$ and $Y$ are compact first; note then a map with dense image will automatically be surjective, and so this case might be simpler to understand for this reason too.  (In fact, thinking about your example of an ideal that you mention in comments, it might be easier to pass to one-point compactifications --- and thus add a unit --- before proceeding.  Because indeed I think&#xA;that in the ideal case, what will happen is that we will get a map from the 1 point compactification of the space to the one point compactification of the open set which crushes the complement of the open set down to the point at infinity, exactly as you suggest in your comment.)&lt;/p&gt;&#xA;" OwnerUserId="221" LastEditorUserId="221" LastEditorDisplayName="" LastEditDate="2010-08-12T14:36:26.687" LastActivityDate="2010-08-12T14:36:26.687" CommentCount="3" />
  <row Id="2247" PostTypeId="2" ParentId="2239" CreationDate="2010-08-12T14:11:09.117" Score="3" ViewCount="" Body="&lt;p&gt;The standard example is the proof, by differentiation, of Fermat's last theorem for polynomials.  More generally, the proof of the ABC conjecture for polynomials (Mason's theorem) by differentiation.   In light of the analogies between algebraic number theory and algebraic geometry this suggests the hope that some kind of arithmetic differentiation, if it exists, could be a missing structure leading to breakthroughs such as a proof of the ABC conjecture or a simple proof of Fermat.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As somebody once said, &quot;define PDE's over a number field and you'll be a rich man&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as I know the furthest-developed approach to this problem at present is in a series of works by Alexander Buium on arithmetic differential operators: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.math.unm.edu/~buium/prebook.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.math.unm.edu/~buium/prebook.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="467" LastActivityDate="2010-08-12T14:11:09.117" CommentCount="2" />
  <row Id="2248" PostTypeId="1" AcceptedAnswerId="2301" CreationDate="2010-08-12T14:36:27.107" Score="3" ViewCount="355" Body="&lt;p&gt;Bernhard Elsner, alias MathOMan, posted this exercise in plane Geometry, &lt;a href=&quot;http://www.mathoman.com/en/index.php/1529-theorem-about-a-circle-three-chords-and-a-midpoint&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;Theorem about a circle, three chords and a midpoint&lt;/em&gt;&lt;/a&gt; on January 29th, 2010.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Let $\mathcal{C}$ be a circle, $A,B$ two distinct points on $\mathcal{C}$ and $M$ be the midpoint of the chord $[AB]$. Take two other chords,$[PQ]$ and $[SR]$, that pass through $M$. Let $C$ (resp. $D$) be the intersection of $[AB]$ with $[PS]$ (resp. $[RQ]$).&#xA;Prove that $M$ is the midpoint of the chord $[CD]$.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://www.latroika.com/mathoman/pix/exercice-cercle-cordes-milieu.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To prove it I've written the following (failed) argument, in the &lt;a href=&quot;http://www.mathoman.com/de/index.php/1529-verschiedene-sehnen-in-einem-kreis&quot; rel=&quot;nofollow&quot;&gt;German version&lt;/a&gt; of this post (translation of mine):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://calcauxprobteor.files.wordpress.com/2010/04/cordas.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The figure is symmetric with respect to $M$: $\overline{AM}=\overline{MB}$, $\overline{PM}=\overline{MU}$, $\overline{RM}=\overline{MW}$, $\overline{QM}=\overline{MV}$, $\overline{QR}=\overline{VW}$, $\overline{SM}=\overline{MT}$. From $\dfrac{\overline{SC}}{\overline{DT}}=\dfrac{\overline{CM}}{\overline{MD}}=1$ follows that $\overline{CM}=\overline{MD}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is an extract of the author's reply (translation of mine): &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;It is not clear that  $\overline{QM}=\overline{MV}$. Is the point $V$ on the line $(QM)$ defined by this equality or is $V$ defined as the intersection point of the lines $(QM)$ and $(WC)$? Why are both definitions to give the same point?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(...)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $C^{\prime }$ be the intersection of $(SP)$ and $(VW)$, and $D^{\prime }$&#xA;the intersection of $(TU)$ and $(QR)$. (...)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One still has to show that $C^{\prime }=C$ and $D^{\prime }=D$.&quot; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have agreed with these objections.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Until now no proof has been posted. The author considers that the &quot;proof is not quite simple&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Q. What is the theorem this exercise refers to? Or how does one prove it?&lt;/p&gt;&#xA;" OwnerUserId="752" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2010-08-12T23:47:43.147" LastActivityDate="2010-08-12T23:47:43.147" Title="How to prove the midpoint of a chord is also the midpoint of the line segment defined by the points of intersection of other two chords with it?" Tags="&lt;geometry&gt;&lt;euclidean-geometry&gt;" AnswerCount="1" CommentCount="10" FavoriteCount="0" />
  <row Id="2249" PostTypeId="2" ParentId="2239" CreationDate="2010-08-12T15:06:11.030" Score="10" ViewCount="" Body="&lt;p&gt;No feasible (polynomial time) algorithm is currently known for &#xA;recognizing squarefree integers or for computing the squarefree &#xA;part of an integer. In fact it may be the case that this problem &#xA;is no easier than the general problem of integer factorization. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This problem is important because one of the main tasks &#xA;of computational algebraic number theory reduces to it (in &#xA;deterministic polynomial time). Namely the problem of computing &#xA;the ring of integers of an algebraic number field depends upon &#xA;the square-free decomposition of the polynomial discriminant &#xA;when computing an integral basis, e.g. [2] S.7.3 p.429 or [1] &#xA;This is due to Chistov [0]. See also Problems 7,8, p.9 in [3], &#xA;which lists 36 open problems in number theoretic complexity. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason that such problems are simpler in function fields versus number fields&#xA;is due to the availability of derivatives. This opens up a powerful&#xA;toolbox that is not available in the number field case. For example&#xA;once derivatives are available so are Wronskians - which provide powerful&#xA;measures of dependence in transcendence theory and diophantine approximation.&#xA;A simple yet stunning example is the elementary proof of the polynomial case of Mason's ABC theorem, which yields as a very special case FLT for polynomials, cf.&#xA;my recent &lt;a href=&quot;http://mathoverflow.net/questions/35147&quot; rel=&quot;nofollow&quot;&gt;MO post&lt;/a&gt; and my old &lt;a href=&quot;http://google.com/groups?selm=WGD.96Jul17041312@berne.ai.mit.edu&quot; rel=&quot;nofollow&quot;&gt;sci.math post [4]&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[0] A. L. Chistov. The complexity of constructing the ring of integers&lt;br&gt;&#xA;of a global field. Dokl. Akad. Nauk. SSSR, 306:1063--1067, 1989.&lt;br&gt;&#xA;English Translation: Soviet Math. Dokl., 39:597--600, 1989. 90g:11170&lt;br&gt;&#xA;&lt;a href=&quot;http://citeseerx.ist.psu.edu/showciting?cid=854849&quot; rel=&quot;nofollow&quot;&gt;http://citeseerx.ist.psu.edu/showciting?cid=854849&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[1] Lenstra, H. W., Jr.  Algorithms in algebraic number theory.&lt;br&gt;&#xA;Bull. Amer. Math. Soc. (N.S.) 26 (1992), no. 2, 211--244.  93g:11131&lt;br&gt;&#xA;&lt;a href=&quot;http://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.105.8382&quot; rel=&quot;nofollow&quot;&gt;http://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.105.8382&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[2] Pohst, M.; Zassenhaus, H.  Algorithmic algebraic number theory.&lt;br&gt;&#xA;Cambridge University Press, Cambridge, 1997.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;[3] Adleman, Leonard M.; McCurley, Kevin S.&lt;br&gt;&#xA;Open problems in number-theoretic complexity. II.&lt;br&gt;&#xA;Algorithmic number theory (Ithaca, NY, 1994), 291--322,&lt;br&gt;&#xA;Lecture Notes in Comput. Sci., 877, Springer, Berlin, 1994. 95m:11142&lt;br&gt;&#xA;&lt;a href=&quot;http://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.48.4877&quot; rel=&quot;nofollow&quot;&gt;http://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.48.4877&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[4] sci.math.research, 1996/07/17&lt;br&gt;&#xA; poly FLT, abc theorem, Wronskian formalism [was: Entire solutions of f^2+g^2=1]&lt;br&gt;&#xA;&lt;a href=&quot;http://groups.google.com/group/sci.math/msg/4a53c1e94f1705ed&quot; rel=&quot;nofollow&quot;&gt;http://groups.google.com/group/sci.math/msg/4a53c1e94f1705ed&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://google.com/groups?selm=WGD.96Jul17041312@berne.ai.mit.edu&quot; rel=&quot;nofollow&quot;&gt;http://google.com/groups?selm=WGD.96Jul17041312@berne.ai.mit.edu&lt;/a&gt;  &lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-08-12T16:48:22.523" LastActivityDate="2010-08-12T16:48:22.523" />
  <row Id="2250" PostTypeId="1" AcceptedAnswerId="2380" CreationDate="2010-08-12T15:24:39.323" Score="0" ViewCount="204" Body="&lt;p&gt;Given the following:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;an $(n \times z)$ matrix $A = {(a_1,a_2, ... ,a_n)}^{T}$ where $z \geq n$ and every $a_i$ is a $z$-dimensional row vector.  &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$a_i = [a_{i1}  a_{i2} ...  a_{iz}]$ where $a_{ij} \geq 0,  \forall j$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$\sum_{i=1}^{z}a_{ri} = 1, \forall r \in ${$1,2,...,n$}.  &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$\sum_{i=1}^{z}|a_{pi} - a_{qi}| \leq \epsilon, \forall p,q$ where $\epsilon &amp;lt;&amp;lt; 1$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Find a provable upper bound on:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\sum_{i,j=1}^{z}|(1/n)*\sum_{k=1}^{n}[a_{ki}.(a_{f(k)j} - a_{g(k)j})]|$  &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;where f and g are permutations over the set {$1,2,...,n$} such that $f(i) \neq g(i) \forall i$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am expecting the bound to be $\epsilon^2$ but I have no idea how to prove it.&lt;/p&gt;&#xA;" OwnerUserId="938" LastEditorUserId="938" LastEditorDisplayName="" LastEditDate="2010-08-14T08:49:27.763" LastActivityDate="2010-08-14T08:49:27.763" Title="a provable upper bound on the summation" Tags="&lt;sequences-and-series&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="2251" PostTypeId="1" AcceptedAnswerId="2253" CreationDate="2010-08-12T15:51:04.227" Score="5" ViewCount="216" Body="&lt;p&gt;Is it guaranteed that there will be some p s.t. p|2^n-1 but not p|2^m-1 for any m&#xA;&#xA;&lt;p&gt;In other words, does each 2^x-1 introduce a &lt;em&gt;new&lt;/em&gt; prime factor?&lt;/p&gt;&#xA;" OwnerUserId="796" LastActivityDate="2010-08-12T16:35:58.173" Title="a prime factor of 2^n-1 which is not a factor of 2^m-1 for m &lt; n" Tags="&lt;prime-numbers&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="2252" PostTypeId="2" ParentId="2251" CreationDate="2010-08-12T16:12:37.367" Score="1" ViewCount="" Body="&lt;p&gt;No. $2^6-1 = 3^2 \cdot 7$. But we have that $3|2^2-1$ and $7|2^3-1$&lt;/p&gt;&#xA;" OwnerUserId="256" LastActivityDate="2010-08-12T16:12:37.367" />
  <row Id="2253" PostTypeId="2" ParentId="2251" CreationDate="2010-08-12T16:16:30.310" Score="5" ViewCount="" Body="&lt;p&gt;Yes, it's true (except for $2^6-1=7\times 9$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is known as Bang's theorem, and is a corollary of &lt;a href=&quot;http://en.wikipedia.org/wiki/Zsigmondy%27s_theorem&quot; rel=&quot;nofollow&quot;&gt;Zsigmondy's Theorem&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can find a proof &lt;a href=&quot;http://www.ams.org/journals/proc/1997-125-07/S0002-9939-97-03981-6/S0002-9939-97-03981-6.pdf&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; (Theorem 3).&lt;/p&gt;&#xA;" OwnerUserId="446" LastActivityDate="2010-08-12T16:16:30.310" />
  <row Id="2254" PostTypeId="1" AcceptedAnswerId="2257" CreationDate="2010-08-12T16:20:04.363" Score="2" ViewCount="217" Body="&lt;p&gt;Let $H,K$ be proper subgroups of a group $G$ having a complete set $S$ of representatives of left cosets in common, that is,&#xA;$$&#xA;G = \bigsqcup_{s \in S} sH = \bigsqcup_{s \in S} sK&#xA;$$&#xA;It seems in general one cannot expect any serious relation on $H,K.$ But I am afraid of overlooking some general result here. Any information on the subject will be warmly accepted. Regards, Olod&lt;/p&gt;&#xA;" OwnerUserId="1100" LastEditorUserId="1100" LastEditorDisplayName="" LastEditDate="2010-08-12T18:25:36.237" LastActivityDate="2010-08-12T18:25:36.237" Title="If two subgroups have a complete set of left coset representatives in common, then ..." Tags="&lt;group-theory&gt;" AnswerCount="2" CommentCount="6" />
  <row Id="2255" PostTypeId="2" ParentId="2251" CreationDate="2010-08-12T16:29:45.867" Score="2" ViewCount="" Body="&lt;p&gt;This result is due to Zsigmondy (1892), with special cases (b=1) &#xA;(re)discovered by Bang (1886), Birkhoff &amp;amp; Vandiver (1904) ... see &#xA;Ribenboim, The New Book of Prime Number Records, p. 43, 67-68, 338, 437. &#xA;Vandiver, a &lt;a href=&quot;http://www-groups.dcs.st-and.ac.uk/~history/Mathematicians/Vandiver.html&quot; rel=&quot;nofollow&quot;&gt;prolific researcher&lt;/a&gt; on FLT,&#xA;applied it to Fermat's Last Theorem (first case) and related diophantine equations, &#xA;e.g. see Ribenboim's book 13 Lectures on FLT, pages 52,&lt;em&gt;161&lt;/em&gt;,206,234,236. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Such results have many applications: a MathSciNet search on &#xA;Anywhere=(Zsigmondy or (Birkhoff and Vandiver))  will find &#xA;over 35 related Math Reviews. Schinzel (1974, MR 93k:11107) &#xA;extended the theorem to arbitrary algebraic number fields $K$: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If $A$ and $B$ are algebraic integers in $K$, $(A,B)=1$, and $A/B$ is of degree $d$ &#xA;  and not a root of unity, then there exists  $n_0 = n_0(d)$ such that &#xA;  for all  $n &gt; n_0,  A^n - B^n $ has a prime ideal factor $P$&lt;br&gt;&#xA;  not dividing   $A^m - B^m $ for all $ m &amp;lt; n$. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Later (1993, same MR) &quot;he generalized this to show that every &#xA;algebraic number which is not a root of unity satisfies only a &#xA;finite number of independent generalized cyclotomic equations &#xA;considered by the reviewer [in Structural properties of &#xA;polylogarithms, Chapter 11, see p. 236, Amer. Math. Soc., &#xA;Providence, RI, 1991; see MR 93b:11158]&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are also &lt;a href=&quot;http://arxiv.org/abs/math/0409540&quot; rel=&quot;nofollow&quot;&gt;elliptic&lt;/a&gt; and &lt;a href=&quot;http://arxiv.org/abs/1002.4829&quot; rel=&quot;nofollow&quot;&gt;polynomial&lt;/a&gt; generalizations.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-08-12T16:35:58.173" LastActivityDate="2010-08-12T16:35:58.173" />
  <row Id="2256" PostTypeId="2" ParentId="2254" CreationDate="2010-08-12T16:41:39.040" Score="2" ViewCount="" Body="&lt;p&gt;If $S$ is a complete set of (left) coset representatives for $H$, then for every $x\in G$ there exists $s\in S$ such that $xH=sH$, and moreover, if $s_1,s_2\in S$ are such that $s_1H=s_2H$, then $s_1=s_2$. That is: there is one and exactly one representative from each coset of $H$ in $G$ in the set $S$. As such, your condition is trivially satisfied under the assumption that $S$ is a complete set of coset representatives for both $H$ and $K$, since for $s_1,s_2\in S$, you have&#xA;$$s_1H=s_2H \Longleftrightarrow s_1=s_2\Longleftrightarrow s_1K=s_2K.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So either you meant something else, or you are just asking for condition under which two subgroups $H$ and $K$ can have the same complete set of coset representatives. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Jack Schmidt already gave an example with $H$ and $K$ not conjugate, and asked if there is an example in which $H$ and $K$ are not isomorphic. I think this does it: take $G=\mathbb{Z}_2\times \mathbb{Z}_2\times \mathbb{Z}_4$, the product of two cyclic groups of order two and one of order four. Take $H={0}\times{0}\times\mathbb{Z}_4$, and $K={0}\times\mathbb{Z}_2\times\langle 2\rangle$ (so $H$ is cyclic of order $4$, and $K$ is the Klein $4$-group). Let $S={(0,0,1), (1,0,1), (1,1,0), (0,1,0)}$. If I did not make some silly mistake, then this is a complete set of coset representatives for both $H$ and $K$. &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-08-12T17:00:55.610" LastActivityDate="2010-08-12T17:00:55.610" CommentCount="6" />
  <row Id="2257" PostTypeId="2" ParentId="2254" CreationDate="2010-08-12T16:50:50.353" Score="3" ViewCount="" Body="&lt;p&gt;|S| = [G:H] = [G:K], so at least their indexes are equal. I don't think much else is true, because of the following example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If G = 2 × 2 is the Klein four group, then H = 2 × 1 and K = 1 × 2 are two subgroups with a common set of coset representatives: S = { (0,0), (1,1) }.  H and K are not conjugate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general if G is a semi-direct product H⋉N, then H has S=N as a set of coset representatives.  It is very possible for N to have more than one &quot;complement&quot; H, that is, another subgroup K such that G=K⋉N.  In nice situations like G nonabelian of order 6, all complements are conjugate, but in general they need not be as the dihedral 2-groups (including the Klein four group) show.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It would be nice to have an example where H and K are not even isomorphic, but semi-direct products won't do that.  I don't believe H, K need to be complemented to have a common transversal, but I guess that is another reasonable guess to rule out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; Well, the dihedral group G of order 8 has a cyclic normal subgroup H and Klein four normal subgroup K, the union of which is not all of G.  Since a set of coset representatives S has only 2 elements, we just need to take the identity and an element neither in H nor K.  In particular, H need not be isomorphic to K.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also the dihedral group of order 16 has a similar pair (H cyclic order 4, K a four-group), and so neither H nor K need be complemented.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One can also use Abelian examples as Arturo points out, and one can extend the D8 example to S4 as Steve points out.  A fun problem.&lt;/p&gt;&#xA;" OwnerUserId="583" LastEditorUserId="583" LastEditorDisplayName="" LastEditDate="2010-08-12T17:13:55.200" LastActivityDate="2010-08-12T17:13:55.200" CommentCount="4" />
  <row Id="2259" PostTypeId="2" ParentId="2237" CreationDate="2010-08-12T17:05:38.147" Score="1" ViewCount="" Body="&lt;p&gt;OK, I think I've got the answer. I tripped myself up initially because I didn't really understand what I was doing when I was multplying ${13\choose2}$ by 48. I'd reasoned that if I had Ace, King of spades then I could times by 48 because there were 48 other cards to make 48 combinations with AK spades in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But what I now realise is that multiplying the whole of ${13\choose2}$ by 48 causes an overlap in counting. For instance if I have A7 of spades then one of the other 48 cards is the King of spades. And in fact I need to apply the Inclusion/Exclusion rules just the same as if I'd been using probability to come to the right answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So ${13\choose2}\times48$ does equal p(A) + p(B) + p(C) and I need to use the inclusion exclusion rules to take away any combinations that have been counted twice. SO:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;p(A&amp;cap;B) = p(A&amp;cap;C) = p(B&amp;cap;C) = ${13\choose3}$ = 286&lt;/p&gt;&#xA;&#xA;&lt;p&gt;p(A&amp;cap;B&amp;cap;C) = ${13\choose3}$ = 286&#xA;and as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;p(A&amp;cup;B&amp;cup;C) = p(A) + p(B) + p(C) - p(A&amp;cap;B) - p(A&amp;cap;C)- p(B&amp;cap;C) + p(A&amp;cap;B&amp;cap;C)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;= (${13\choose2}\times48$) - (3 x ${13\choose3}$) + ${13\choose3}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;= 3744 - 858 + 286&lt;/p&gt;&#xA;&#xA;&lt;p&gt;= 3172&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3172 / 19,600 (total number of possible flops) = 0.1618 = 16.18%&lt;/p&gt;&#xA;" OwnerUserId="1094" LastActivityDate="2010-08-12T17:05:38.147" />
  <row Id="2260" PostTypeId="1" AcceptedAnswerId="2263" CreationDate="2010-08-12T17:15:42.560" Score="5" ViewCount="2087" Body="&lt;p&gt;Apparently 1+2+3+4+...+n = 1/2*n*(n+1).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How? What's the proof? Or maybe it is self apparent just looking at the above? Does this problem have a name and maybe a presence on the net? And what is this math called so I can add the right tags.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thx if you can help.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: &#xA;This problem is know as &quot;The sum of the first n positive integers&quot;. Do a search. &lt;/p&gt;&#xA;" OwnerUserId="1103" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-08-12T22:00:24.160" LastActivityDate="2011-02-27T14:01:12.243" Title="Proof for formula for sum of sequence 1+2+3+...+n?" Tags="&lt;algebra-precalculus&gt;&lt;sequences-and-series&gt;" AnswerCount="8" CommentCount="1" FavoriteCount="3" />
  <row Id="2261" PostTypeId="2" ParentId="2260" CreationDate="2010-08-12T17:18:31.213" Score="4" ViewCount="" Body="&lt;p&gt;Once you have a formula like this, you can prove it by induction.&#xA;But that begs the question as to how you get such a formula.&#xA;In this case you might ask: (a) what's the &quot;average&quot; term?&#xA;and (b) how many terms are there?&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-08-12T17:18:31.213" />
  <row Id="2262" PostTypeId="2" ParentId="2260" CreationDate="2010-08-12T17:19:45.953" Score="2" ViewCount="" Body="&lt;p&gt;The most general case of this is called an &lt;em&gt;arithmetic progression&lt;/em&gt; or &lt;em&gt;(finite) arithmetic series&lt;/em&gt;. There are many, many, many proofs. An easy one: write all the summands in a row; write them again just below, but from right to left now (so $1$ is under $n$, $2$ is under $n-1$, etc). Add them up, and figure out how it relates to the quantity you are looking for.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-08-12T18:07:02.437" LastActivityDate="2010-08-12T18:07:02.437" CommentCount="1" />
  <row Id="2263" PostTypeId="2" ParentId="2260" CreationDate="2010-08-12T17:23:29.567" Score="12" ViewCount="" Body="&lt;p&gt;Let S = 1 + 2 + ... + (n-1) + n.&#xA;Write it backwards: S = n + (n-1) + ... + 2 + 1.&#xA;Add the two equations, term by term; each term is n+1, so&#xA;2S = (n+1) + (n+1) + ... + (n+1) = n(n+1).&#xA;Divide by 2: S = n(n+1)/2&lt;/p&gt;&#xA;" OwnerUserId="1105" LastActivityDate="2010-08-12T17:23:29.567" CommentCount="3" />
  <row Id="2264" PostTypeId="2" ParentId="566" CreationDate="2010-08-12T17:31:18.823" Score="7" ViewCount="" Body="&lt;p&gt;The Blue-Eyed Islander problem is one of my favorites.  You can read about it &lt;a href=&quot;http://terrytao.wordpress.com/2008/02/05/the-blue-eyed-islanders-puzzle/&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; on Terry Tao's website, along with some discussion.  I'll copy the problem here as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is an island upon which a tribe resides. The tribe consists of 1000 people, with various eye colours. Yet, their religion forbids them to know their own eye color, or even to discuss the topic; thus, each resident can (and does) see the eye colors of all other residents, but has no way of discovering his or her own (there are no reflective surfaces). If a tribesperson does discover his or her own eye color, then their religion compels them to commit ritual suicide at noon the following day in the village square for all to witness. All the tribespeople are highly logical and devout, and they all know that each other is also highly logical and devout (and they all know that they all know that each other is highly logical and devout, and so forth).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[For the purposes of this logic puzzle, &quot;highly logical&quot; means that any conclusion that can logically deduced from the information and observations available to an islander, will automatically be known to that islander.]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of the 1000 islanders, it turns out that 100 of them have blue eyes and 900 of them have brown eyes, although the islanders are not initially aware of these statistics (each of them can of course only see 999 of the 1000 tribespeople).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One day, a blue-eyed foreigner visits to the island and wins the complete trust of the tribe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One evening, he addresses the entire tribe to thank them for their hospitality.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, not knowing the customs, the foreigner makes the mistake of mentioning eye color in his address, remarking “how unusual it is to see another blue-eyed person like myself in this region of the world”.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What effect, if anything, does this faux pas have on the tribe?&lt;/p&gt;&#xA;" OwnerUserId="1101" LastActivityDate="2010-08-12T17:31:18.823" CommunityOwnedDate="2010-08-12T17:31:18.823" CommentCount="1" />
  <row Id="2265" PostTypeId="2" ParentId="1860" CreationDate="2010-08-12T17:36:50.857" Score="2" ViewCount="" Body="&lt;p&gt;You should be able to use the fact that the $n^{th}$ &lt;a href=&quot;http://en.wikipedia.org/wiki/Harmonic_number&quot; rel=&quot;nofollow&quot;&gt;Harmonic Number&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$H_n = \ln n + \gamma + \frac{1}{2n} - O(\frac{1}{n^2})$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your sum is a difference of two such numbers and so is approximately of the form $\ln\frac{at+b}{ct+d}$ where $a &gt; c$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sorry, haven't done the complete math, but this approach looks promising.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-08-12T17:36:50.857" CommentCount="4" />
  <row Id="2266" PostTypeId="2" ParentId="2240" CreationDate="2010-08-12T17:52:19.690" Score="0" ViewCount="" Body="&lt;p&gt;First of all thanks a lot to George S. Secondly i recently found a paper entitled, which discusses something on Automorphisms of Complex Numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is the link for that paper: mathdl.maa.org/images/upload_library/22/Ford/PaulBYale.pdf&lt;/p&gt;&#xA;" LastActivityDate="2010-08-12T17:52:19.690" />
  <row Id="2267" PostTypeId="2" ParentId="2260" CreationDate="2010-08-12T17:56:23.783" Score="3" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; Pair each summand $k$ with it's &quot;reflection&quot; $n+1-k$. This is simply a discrete analog of the method of computing the area of the triangle under the diagonal of a square by reflecting a subtriangle through the midpoint of the diagonal to form an $n$ by $n/2$ rectangle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Like the analogous proof of &lt;a href=&quot;http://math.stackexchange.com/questions/1865&quot;&gt;Wilson's theorem&lt;/a&gt;, the method exploits the existence of a nontrivial symmetry. In Wilson's theorem we exploit the symmetry $n \mapsto n^{-1}$ which exists due to the fact that ${\mathbb F}_p^*$ forms a group. Here we exploit a reflection through a line - a symmetry that exists due to the &lt;em&gt;linear&lt;/em&gt; nature of the problem (which doesn't work for nonlinear sums, e.g. $\sum n^2$).  Symmetries often lead to elegant proofs. One should always look for innate symmetries when first pondering problems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Generally there are (Galois) theories and algorithms for &lt;strong&gt;summation in closed form&lt;/strong&gt;, in analogy to the differential case (Ritt, Kolchin, Risch et al.). A &lt;em&gt;very&lt;/em&gt; nice motivated introduction can be found in the introductory chapter of Carsten Schneider's thesis &lt;a href=&quot;http://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.67.9442&quot; rel=&quot;nofollow&quot;&gt;Symbolic Summation in Difference Fields&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-08-12T19:32:30.147" LastActivityDate="2010-08-12T19:32:30.147" CommentCount="1" />
  <row Id="2268" PostTypeId="2" ParentId="1910" CreationDate="2010-08-12T17:58:59.880" Score="1" ViewCount="" Body="&lt;p&gt;I would concur with the suggestion that you should use LaTeX. I'm bothering with a separate answer to &lt;em&gt;strongly discourage&lt;/em&gt; the use of MS Word with symbol fonts for logical notation. Before I learned LaTeX, I'd tried to write up logic in word processors (both Word and, later, OpenOffice). It was deeply frustrating when my documents got quite messed up when I moved from one computer to another and found myself working without the symbol fonts I had at my disposal when I wrote the documents. (Yes, there are ways around that problem, but it nevertheless seems worth the warning.)&lt;/p&gt;&#xA;" OwnerUserId="277" LastActivityDate="2010-08-12T17:58:59.880" CommentCount="1" />
  <row Id="2269" PostTypeId="1" AcceptedAnswerId="2271" CreationDate="2010-08-12T18:10:12.557" Score="3" ViewCount="291" Body="&lt;p&gt;I was just reading a paper of Ramanujan entitled &quot; On question 330 of Professor Sanjana&quot; when i got stuck up with a Proposition which i am unable to answer. The proposition is if &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ \displaystyle f(p) = \frac{\pi}{2^{p+1}} \frac{\Gamma(p+1)}{ \Bigl[ \Gamma(\frac{P+2}{2}) \Bigr]^{2}}$, then $$\log{f(p)}= \log(\frac{\pi}{2}) - p\log{2}  + \frac{p^2}{2} \Bigl(1-\frac{1}{2}\Bigr)S_{2} - \frac{p^3}{3} \Bigl(1-\frac{1}{2^2}\Bigr)S_{3} + \cdots$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\displaystyle S_{n} = \frac{1}{1^n} + \frac{1}{2^n} + \cdots \text{ad inf.}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could anyone explain me as to how can i deduce this.&lt;/p&gt;&#xA;" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-08-27T02:53:37.263" LastActivityDate="2010-08-27T02:53:37.263" Title="Question Relating Gamma Function to Riemann Zeta function evaluated at integers" Tags="&lt;gamma-function&gt;&lt;riemann-zeta&gt;&lt;special-functions&gt;&lt;classical-analysis&gt;&lt;ramanujan&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="2" />
  <row Id="2270" PostTypeId="1" AcceptedAnswerId="2275" CreationDate="2010-08-12T18:19:22.863" Score="10" ViewCount="962" Body="&lt;p&gt;Does $S_k= \sum_{n=1}^{\infty}\sin(n^k)/n$ converge for all $k&gt;0$? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Motivation&lt;/strong&gt;: I recently learned that $S_1$ &lt;a href=&quot;http://en.wikipedia.org/wiki/Dirichlet%27s_test&quot; rel=&quot;nofollow&quot;&gt;converges&lt;/a&gt;. I think $S_2$ converges by the integral test. Was the question known in general?   &lt;/p&gt;&#xA;" OwnerUserId="223" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-08-19T22:00:09.730" LastActivityDate="2010-11-12T13:24:02.667" Title="Convergence of $\sum_{n=1}^{\infty}\sin(n^k)/n$" Tags="&lt;calculus&gt;&lt;sequences-and-series&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="5" />
  <row Id="2271" PostTypeId="2" ParentId="2269" CreationDate="2010-08-12T18:26:59.620" Score="4" ViewCount="" Body="&lt;p&gt;I presume this is a Maclaurin series. I don't know how it comes&#xA;about, but my instinct would be to expand out the infinite product&#xA;for $f(p)$ (which becomes an infinite sum for $\log f(p)$) in powers&#xA;of $p$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added&lt;/strong&gt; (13/8/2010)&#xA;As a hint: recall that&#xA;$$\Gamma(p+1)=p\Gamma(p)&#xA;=e^{-\gamma p}\prod_{k=1}^\infty\left(\frac{n}{n+p}\right)e^{p/n}$$&#xA;by the infinite product for the gamma function. Applying&#xA;this to $\Gamma(p+1)/\Gamma(p/2+1)^2$ one gets some nice&#xA;cancellation. Then take logs and apply the series for $\log(1+x)$.&#xA;You get a double sum; change the order of summation and hope that&#xA;the inner sum comes to something like a zeta series.&lt;/p&gt;&#xA;" OwnerUserId="226" LastEditorUserId="226" LastEditorDisplayName="" LastEditDate="2010-08-13T17:13:44.230" LastActivityDate="2010-08-13T17:13:44.230" CommentCount="6" />
  <row Id="2273" PostTypeId="1" CreationDate="2010-08-12T19:26:07.727" Score="2" ViewCount="132" Body="&lt;p&gt;Sometimes when I am reading a mathematical text I will be unable to decipher the steps necessary to translate one mathematical statement into the next. Usually I have to ask for help, but I was wondering if there is a software package that would let me enter two mathematical statements and the it would find the steps necessary to get from the start to the finish. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I realize that this not tractable in general, however, it would seem with the right heuristics it could still be possible many times.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a response both to my own experience and this question:&#xA;&lt;a href=&quot;http://math.stackexchange.com/questions/2269/question-relating-gamma-function&quot;&gt;http://math.stackexchange.com/questions/2269/question-relating-gamma-function&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="99" LastActivityDate="2010-10-05T17:25:21.780" Title="Is are there software package that can help one find a path between two mathematical statements?" Tags="&lt;math-software&gt;&lt;learning&gt;" AnswerCount="4" CommentCount="1" FavoriteCount="1" />
  <row Id="2274" PostTypeId="2" ParentId="2273" CreationDate="2010-08-12T19:35:37.237" Score="0" ViewCount="" Body="&lt;p&gt;It is probably not exactly on things You are looking for, but prover9-mace4 tries to find prof in FOL for given statements starting from given axioms ( prover9), or opposite, tries to find counterexample (mace4). I can imagine that for simple statements, like in group theory, it gives what You are looking for. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Prover9&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Prover9&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1043" LastActivityDate="2010-08-12T19:35:37.237" CommentCount="1" />
  <row Id="2275" PostTypeId="2" ParentId="2270" CreationDate="2010-08-12T19:41:11.530" Score="21" ViewCount="" Body="&lt;p&gt;This is a replacement for my previous answer. The sum converges, and this fact needs even more math than I believed before.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Begin by using summation by parts. This gives &#xA;$$\sum_{n=1}^N \left(\sum_{m=1}^N \sin(m^k) \right) \left( \frac{1}{n}-\frac{1}{n+1}\right) + \frac{1}{N+1} \left(\sum_{m=1}^N \sin(m^k) \right).$$ &#xA;Write $S_n:= \left(\sum_{m=1}^n \sin(m^k) \right)$. So this is&#xA;$$\sum_{n=1}^N S_n/(n(n+1)) + s_N/(N+1).$$&#xD;&#xA;The second term goes to zero by Weyl's &lt;a href=&quot;http://terrytao.wordpress.com/2010/03/28/254b-notes-1-equidistribution-of-polynomial-sequences-in-torii/&quot; rel=&quot;nofollow&quot;&gt;polynomial equidistribution theorem&lt;/a&gt;. So your question is equivalent to the question of whether $\sum s_n/(n(n+1))$ converges. We may as well clean this up a little: Since $|S_n| \leq n$, we know that $\sum S_n \left( 1/n(n+1) - 1/n^2 \right)$ converges. So the question is whether&#xA;$$\sum \frac{S_n}{n^2}$$&#xD;&#xA;converges.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I will show that $S_n$ is small enough that $\sum S_n/n^2$ converges absolutely.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The way I want to prove this is to use &lt;a href=&quot;http://en.wikipedia.org/wiki/Weyl%27s_inequality&quot; rel=&quot;nofollow&quot;&gt;Weyl's inequality&lt;/a&gt;. Let $p_i/q_i$ be an infinite sequence of rational numbers such that $|1/(2 \pi) - p_i/q_i| &amp;lt; 1/q_i^2$. Such a sequence exists by a standard lemma. Weyl inequality gives that&#xA;$$S_N = O\left(N^{1+\epsilon} (q_i^{-1} + N^{-1} + q_i N^{-k})^{1/2^{k-1}} \right)$$&#xD;&#xA;for any $\epsilon&amp;gt;0$.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Thanks to George Lowther for pointing out the next step: According to &lt;a href=&quot;http://mathworld.wolfram.com/IrrationalityMeasure.html&quot; rel=&quot;nofollow&quot;&gt;Salikhov&lt;/a&gt;, for $q$ sufficiently large, we have&#xA;$$|\pi - p/q| &amp;gt; 1/q^{7.6304+\epsilon}.$$&#xD;&#xA;Since $x \mapsto 1/(2x)$ is Lipschitz near $\pi$, and since $p/q$ near $\pi$ implies that $p$ and $q$ are nearly proportional, we also have the lower bound $|1/(2 \pi) - p/q|&amp;gt; 1/q^{7.6304+\epsilon}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $p_i/q_i$ be the convergents of the continued fraction of $1/(2 \pi)$. By a standard result, $|1/(2 \pi) - p_i/q_i| \leq 1/(q_i q_{i+1})$. Thus, $q_{i+1} \leq q_i^{6.6304 + \epsilon}$ for $i$ sufficiently large. Thus, the intervals $[q_i, q_i^{7}]$ contain all sufficiently large integers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For any large enough $N$, choose $q_i$ such that $N^{k-1} \in [q_i, q_i^7]$. Then Weyl's inequality gives the bound&#xA;$$S_N = O \left( N^{1+\epsilon} (N^{-1/(7(k-1))} + N^{-1} + N^{-1} \right)^{1/2^{k-1}}$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So $S_N = O(N^{1-1/(7(k-1)2^{k-1}) + \epsilon})$, which is enough to make sure the sum converges.&lt;/p&gt;&#xA;" OwnerUserId="448" LastEditorUserId="448" LastEditorDisplayName="" LastEditDate="2010-11-12T13:24:02.667" LastActivityDate="2010-11-12T13:24:02.667" CommentCount="11" />
  <row Id="2276" PostTypeId="2" ParentId="2240" CreationDate="2010-08-12T19:51:29.600" Score="1" ViewCount="" Body="&lt;p&gt;This and related questions about automorphism groups of algebraically closed fields (a topic I find interesting and have spent some time thinking about) are discussed in Section 9.1 of &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://math.uga.edu/~pete/FieldTheory.pdf&quot; rel=&quot;nofollow&quot;&gt;http://math.uga.edu/~pete/FieldTheory.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Specifically, Theorem 77 answers the OP's question affirmatively.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(These notes are still very rough.  In particular there is not yet a bibliography.  When this gets remedied, a citation to Paul Yale's paper will be in order: it was definitely something I read when writing these notes.)  &lt;/p&gt;&#xA;" OwnerUserId="299" LastActivityDate="2010-08-12T19:51:29.600" CommentCount="1" />
  <row Id="2277" PostTypeId="2" ParentId="566" CreationDate="2010-08-12T20:05:42.640" Score="3" ViewCount="" Body="&lt;p&gt;The odd town puzzle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have a town with $m$ clubs formed by $n$ citizens of the town.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The clubs are so formed that&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Each club has an odd number of members.&lt;/li&gt;&#xA;&lt;li&gt;Any two clubs have an even number of common members. (Could be zero too).&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Show that $m \le n$.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="555" LastEditorDisplayName="" LastEditDate="2010-09-06T19:12:27.333" LastActivityDate="2010-09-06T19:12:27.333" CommunityOwnedDate="2010-08-12T20:05:42.640" CommentCount="1" />
  <row Id="2278" PostTypeId="1" AcceptedAnswerId="2302" CreationDate="2010-08-12T20:06:37.630" Score="4" ViewCount="152" Body="&lt;p&gt;Let us be given a $n\times n$ matrix containing only zeros and ones.Now, the goal is to remove some 'ones' from the matrix (i.e. replace them with zeros) so that in each row and each column there is at most one 'one'. Of course, it is preferable that we leave as much 'ones' as it is possible (maximum is obviously $n$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, let us define the following algorithm: if there is a 'one' which is all alone in a column or a row, leave it there, and replace all 'ones' in its column and row with zeros. We repeat the process as long as there are single 'ones' in rows or columns. If there are no single 'ones' left, we take the first row or column (first is counted from up (for rows) and from left (for columns)) containing two 'ones' (if there are no pairs, we go for the first triplet and so on). In that pair (or triplet, or ...) we leave the first 'one' intact (first is counted from the left for elements of row, from the top for elements of a column) and change all other 'ones' in the row and column of that first 'one' into zeros.&#xA;Now, we check are there now any single 'ones' as defined above - if yes, we repeat that part of the algorithm, if no, we look for the first pair, triplet, etc. The algorithm ends when all 'ones' are singles both in their column and row.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does this algorithm always give the maximum covering, i.e. does it always terminate with maximal number of 'ones' in the final matrix? In other words, can there be a matrix with a possible exclusion of some ones such that it ends up with $n$ ones using some other exclusion scheme, while ending up with less than $n$ 'ones' using the algorithm described above?&lt;/p&gt;&#xA;" OwnerUserId="553" LastEditorUserId="171" LastEditorDisplayName="" LastEditDate="2010-08-12T20:20:08.753" LastActivityDate="2010-08-13T08:00:47.697" Title="Can this algorithm on removing 1's from a (0,1)-matrix fail?" Tags="&lt;combinatorics&gt;&lt;algorithms&gt;" AnswerCount="3" />
  <row Id="2279" PostTypeId="1" AcceptedAnswerId="2282" CreationDate="2010-08-12T20:09:58.917" Score="0" ViewCount="118" Body="&lt;p&gt;Prove that a number in the sequence $2,3,4,...,n \ (n&gt;2$, is relatively prime to all other numbers if and only if it is a prime that exceeds $\displaystyle\frac{n}{2}$. Does such a prime always exist?&lt;/p&gt;&#xA;" LastActivityDate="2010-08-12T20:18:51.050" Title="Proving that a Number is Relatively Prime in a sequence." Tags="&lt;number-theory&gt;&lt;prime-numbers&gt;" AnswerCount="1" />
  <row Id="2280" PostTypeId="2" ParentId="1068" CreationDate="2010-08-12T20:17:43.890" Score="2" ViewCount="" Body="&lt;p&gt;First of all &lt;a href=&quot;http://www.math.wisc.edu/~keisler/calc.html&quot; rel=&quot;nofollow&quot;&gt;Non-standard Analysis&lt;/a&gt; makes nonsense like $\frac{\mathrm{d}y}{\mathrm{d}x}$ into a meaningful fraction rather than just a &quot;symbol&quot; (whatever that means).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first and simplest example of differentials being used beyond the $\int f(x)\mathrm{d}x$ format is would be multidimensional integration:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \int_V \mathrm{d}x \mathrm{d}y \mathrm{d}z = \int \left(\int \left(\int \mathrm{d}x \right) \mathrm{d}y \right) \mathrm{d}z $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which gives the volume of the solid $V$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The next example is Greens Theorem from vector calculus,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \int A(x,y) \mathrm{d}x - B(x,y) \mathrm{d}y = \int (\partial_1 B -  \partial_2 A) \mathrm{d}x \mathrm{d}y $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is also possible to forget integration completely and just use &lt;a href=&quot;http://www.math.oregonstate.edu/bridge/papers/CMJdifferentials.pdf&quot; rel=&quot;nofollow&quot;&gt;equations with differentials in them&lt;/a&gt; to solve calculus problems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you can see the standard format is not even close to the whole picture, If you want to integrate terms like $\int (e^{\mathrm{d}x} - 1)$ &lt;em&gt;please do&lt;/em&gt;! There is absolutely nothing to stop you figuring out from scratch how to solve this sort of integral and making the theory rigorous.&lt;/p&gt;&#xA;" LastActivityDate="2010-08-12T20:17:43.890" CommentCount="7" />
  <row Id="2281" PostTypeId="2" ParentId="2278" CreationDate="2010-08-12T20:18:00.207" Score="5" ViewCount="" Body="&lt;p&gt;I'm not sure about whether this works, but I'd be doubtful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another way to express this problem is finding maximum&#xA;partial matchings in a bipartite graph. (Just interpret&#xA;the matrix an adjacency matrix for the graph).&#xA;By a maximum partial matching I mean a maximal-sized&#xA;set of edges no two having a vertex in common.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason for my scepticism is that if your algorithm worked,&#xA;then it would give an algorithm for finding the&#xA;solution in the situation of &lt;a href=&quot;http://en.wikipedia.org/wiki/Hall%27s_marriage_theorem&quot; rel=&quot;nofollow&quot;&gt;Hall's marriage theorem&lt;/a&gt;.&#xA;Your algorithm is &quot;monotone&quot; in the sense that it adds&#xA;edges but never removes them, but all the algorithms I have&#xA;seen for Hall's theorem, one needs to remove edges in certain&#xA;circumstances in the process.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the problem is hand, an algorithm comes from any that solves&#xA;the maximum flow problem in a network. Attach a source to each&#xA;vertex on one side of the bipartition and a sink to each vertex&#xA;on the other side. Give all edges unit capacity and find&#xA;a maximum flow.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-08-12T20:18:00.207" />
  <row Id="2282" PostTypeId="2" ParentId="2279" CreationDate="2010-08-12T20:18:51.050" Score="1" ViewCount="" Body="&lt;p&gt;If it's not a prime, all its divisors are there.&lt;br&gt;&#xA;If it's not greater than n/2, than 2p is not greater than n, and it's in the list.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Every prime greater than n/2 trivially satisfies the condition.&#xA;It always exists by Bertrand's postulate ($n/2 &amp;lt; p &amp;lt; 2\cdot n/2$).&lt;/p&gt;&#xA;" OwnerUserId="956" LastActivityDate="2010-08-12T20:18:51.050" />
  <row Id="2283" PostTypeId="2" ParentId="2273" CreationDate="2010-08-12T20:21:50.293" Score="0" ViewCount="" Body="&lt;p&gt;There's only a handful of math software that deals with human solving/thinking process. Most apply some kind of normalization to more abstract concepts or more convenient data structures, run an algorithm on that, and then try to simplify the answer and present it in more human form.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you were to get more into this, I suspect that you could use any powerful CAS to normalize both expressions and get the same canonical form, and from there you could trace back to both expressions.&lt;br&gt;&#xA;This of course would be very painful process.&lt;/p&gt;&#xA;" OwnerUserId="956" LastActivityDate="2010-08-12T20:21:50.293" />
  <row Id="2284" PostTypeId="1" AcceptedAnswerId="2285" CreationDate="2010-08-12T20:41:22.647" Score="4" ViewCount="401" Body="&lt;p&gt;Everyone knows that $\pi$ is an irrational number, and one can refer to this page &lt;a href=&quot;http://planetmath.org/encyclopedia/PiAndPi2AreIrrational.html&quot; rel=&quot;nofollow&quot;&gt;http://planetmath.org/encyclopedia/PiAndPi2AreIrrational.html&lt;/a&gt; for the proof that $\pi^{2}$ is also irrational.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What about the Highers powers of $\pi$, meaning is $\pi^{n}$ irrational for all $n \in \mathbb{N}$ or does there exists a $m \in \mathbb{N}$ when $\pi^{m}$ is rational.&lt;/p&gt;&#xA;" LastActivityDate="2011-03-27T17:29:26.160" Title="Irrationality of powers of $\pi$" Tags="&lt;number-theory&gt;&lt;real-analysis&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="2285" PostTypeId="2" ParentId="2284" CreationDate="2010-08-12T20:50:36.010" Score="16" ViewCount="" Body="&lt;p&gt;What Robin hinted at:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $\pi^{n}$ was rational, then $\pi$ would not be transcendental, as it would be the root of $ax^{n}-b = 0$ for some integers $a,b$.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-08-12T20:50:36.010" CommentCount="6" />
  <row Id="2286" PostTypeId="2" ParentId="2260" CreationDate="2010-08-12T20:51:39.690" Score="0" ViewCount="" Body="&lt;p&gt;You know, this method of writing backwards and finding the sum of an arithmetic series was invented by Karl Friedrich Gauss in a matter of minutes when he was a schoolboy. He is one of the greatest mathematicians that ever lived!&lt;/p&gt;&#xA;" LastActivityDate="2010-08-12T20:51:39.690" CommentCount="4" />
  <row Id="2287" PostTypeId="2" ParentId="2273" CreationDate="2010-08-12T20:53:20.950" Score="2" ViewCount="" Body="&lt;p&gt;The likely outcome of applying software to the reading of ordinary proofs would be to discover that most proofs contain steps that are not formally correct.  For example, in classical geometry many steps in the arguments are &quot;generically true&quot; but false for special degenerate configurations.  Sometimes these configurations, such as a triangle with all three vertices the same point, are implicitly excluded from the cases considered in the theorem, or can be handled in the proof using additional steps that might be considered trivial and are not articulated in the written argument.  In such cases the theorem would be correct in substance but some steps could not be filled in by a computerized proof assistant, because the steps are wrong.    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Less often, there are more difficult cases, where the non-generic configurations and components can't be ignored, or the intuitive idea of what is generic behavior is incorrect.  In this case a theorem may be correct, but a proof that doesn't deal with the special cases errs not only in the details but in its overall argument.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A similar thing happens in computational linguistics.  When trying to process natural language with formal grammars, it turns out that simple sentences are multiply ambiguous, with a large number of unintended but correct interpretations.  People constantly use additional knowledge to disambiguate the language they hear, but software does not yet have that capability. &lt;/p&gt;&#xA;" OwnerUserId="467" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-08-12T20:58:25.960" LastActivityDate="2010-08-12T20:58:25.960" />
  <row Id="2288" PostTypeId="2" ParentId="2260" CreationDate="2010-08-12T21:19:17.387" Score="10" ViewCount="" Body="&lt;p&gt;My favourite proof is the one given &lt;a href=&quot;http://mathoverflow.net/questions/8846/proofs-without-words/8847#8847&quot; rel=&quot;nofollow&quot;&gt;here on MathOverflow&lt;/a&gt;.  I'm copying the picture here for easy reference, but full credit goes to &lt;a href=&quot;http://mathoverflow.net/users/1409/mariano-suarez-alvarez&quot; rel=&quot;nofollow&quot;&gt;Mariano Su&amp;aacute;rez-Alvarez&lt;/a&gt; for this answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://img31.imageshack.us/img31/919/triangle1y.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Takes a little bit of looking at it to see what's going on, but it's nice once you get it.  Observe that if there are n rows of yellow discs, then:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;there are a total of 1 + 2 + ... + n yellow discs;  &lt;/li&gt;&#xA;&lt;li&gt;every yellow disc corresponds to a unique pair of blue discs, and vice versa;&lt;/li&gt;&#xA;&lt;li&gt;there are ${n+1 \choose 2} = \frac 12 n(n+1)$ such pairs.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="876" LastActivityDate="2010-08-12T21:19:17.387" CommunityOwnedDate="2010-08-12T21:19:17.387" CommentCount="2" />
  <row Id="2290" PostTypeId="2" ParentId="2260" CreationDate="2010-08-12T21:31:15.130" Score="3" ViewCount="" Body="&lt;p&gt;Another &quot;picture proof&quot; I just thought of... but without a picture, since I can't draw:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose you want to add up all the integers from 1 to n. Then draw n rows on a board, put 1 unit in the first row, 2 in the second, and so on. If you draw a right triangle of height and length n to try and contain this shape, you will cut off half of each unit on the diagonal. So let's add all of these up; you have n^2/2 units inside the triangle, and you have n/2 unites cut off on the diagonal (there are n squares on the diagonal, and half of each one is cut off). Adding these gives n^2/2 + n/2 = n(n+1)/2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you draw it out it makes more sense, and I think it's geometrically a bit more straightforward than the other picture proof. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Side note: I have no idea how to write math on this site... I'll go consult the meta, I'm sure there's something there about it.)&lt;/p&gt;&#xA;" OwnerUserId="423" LastActivityDate="2010-08-12T21:31:15.130" CommentCount="2" />
  <row Id="2291" PostTypeId="1" CreationDate="2010-08-12T21:34:56.583" Score="1" ViewCount="400" Body="&lt;p&gt;How is the power series expansion of an analytic function at a point constructed, without using Cauchy's theorem(or formula)?&lt;/p&gt;&#xA;" LastActivityDate="2010-08-12T22:21:25.127" Title="Power series expansion without Cauchy theorem" Tags="&lt;complex-analysis&gt;" AnswerCount="1" />
  <row Id="2292" PostTypeId="1" AcceptedAnswerId="2298" CreationDate="2010-08-12T21:48:13.887" Score="0" ViewCount="392" Body="&lt;p&gt;&lt;em&gt;This problem is taken from Problem 2.4.31 (page 84) from &lt;a href=&quot;http://books.google.com/books?id=OBK7D1qcY0oC&amp;amp;pg=PA84&amp;amp;lpg=PA84&amp;amp;dq=%22Give+an+example+of+a+bounded+function%22+%22which+is+not+Riemann+Integrable,+but+is+a+derivative+of+some+function%22&amp;amp;source=bl&amp;amp;ots=JgTTEAPY3o&amp;amp;sig=4qcimfqsmlE-5EWKVuEJFlapn7A&amp;amp;hl=en&amp;amp;ei=1UNoTM_3I4OYuAOMvKD-Aw&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=2&amp;amp;ved=0CBYQ6AEwAQ#v=onepage&amp;amp;q=%22Give%20an%20example%20of%20a%20bounded%20function%22%20%22which%20is%20not%20Riemann%20Integrable%2C%20but%20is%20a%20derivative%20of%20some%20function%22&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;Problems in Mathematical Analysis: Integration&lt;/a&gt; by W. J. Kaczor, Wiesława J. Kaczor and Maria T. Nowak.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Give an example of a bounded function $f:[0,1] \to \mathbb{R}$ which is not Riemann Integrable, but is a derivative of some function $g$ on $[0,1]$.&lt;/p&gt;&#xA;" LastEditorUserId="171" LastEditorDisplayName="" LastEditDate="2010-08-15T19:46:48.653" LastActivityDate="2010-08-15T19:46:48.653" Title="Bounded Function Which is Not Riemann Integrable." Tags="&lt;measure-theory&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="2293" PostTypeId="1" CreationDate="2010-08-12T21:48:21.733" Score="5" ViewCount="481" Body="&lt;p&gt;What are some nice applications of algebraic topology that can be presented to beginning students? To give examples of what I have in mind: Brouwer's fixed point theorem, Borsuk-Ulam theorem, Hairy Ball Theorem, any subgroup of a free group is free.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The deeper the methods used, the better. All the above can be proved with just the fundamental group. More involved applications would be nice.&lt;/p&gt;&#xA;" OwnerUserId="1119" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-08-13T00:19:08.030" LastActivityDate="2010-12-18T07:52:59.350" CommunityOwnedDate="2010-08-12T21:55:25.927" Title="Applications of algebraic topology" Tags="&lt;big-list&gt;&lt;applications&gt;&lt;algebraic-topology&gt;" AnswerCount="11" CommentCount="0" FavoriteCount="5" />
  <row Id="2294" PostTypeId="2" ParentId="2293" CreationDate="2010-08-12T21:50:57.630" Score="4" ViewCount="" Body="&lt;p&gt;The Jordan-Brouwer separation theorem is a nice application of singular homology.&lt;/p&gt;&#xA;" OwnerUserId="536" LastActivityDate="2010-08-12T21:50:57.630" CommunityOwnedDate="2010-08-12T21:50:57.630" />
  <row Id="2295" PostTypeId="2" ParentId="2293" CreationDate="2010-08-12T21:52:33.523" Score="1" ViewCount="" Body="&lt;p&gt;Hi--&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is one interesting application of Algebraic Topology in Molecular Biology: www.springerlink.com/index/h55gv40w17306n57.pdf&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a computer related application please see this link: &lt;a href=&quot;http://www.lix.polytechnique.fr/~sanjeevi/atmcs/&quot; rel=&quot;nofollow&quot;&gt;http://www.lix.polytechnique.fr/~sanjeevi/atmcs/&lt;/a&gt;&lt;/p&gt;&#xA;" LastActivityDate="2010-08-12T21:52:33.523" CommunityOwnedDate="2010-08-12T22:28:12.873" />
  <row Id="2296" PostTypeId="2" ParentId="2291" CreationDate="2010-08-12T21:53:37.893" Score="2" ViewCount="" Body="&lt;p&gt;Below are a couple of possible approaches. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;MR0123687 (23 #A1010)  30.20&lt;br&gt;&#xA;Porcelli, P.; Connell, E. H.&lt;br&gt;&#xA;&lt;strong&gt;A proof of the power series expansion without Cauchy's formula.&lt;/strong&gt;&lt;br&gt;&#xA;Bull. Amer. Math. Soc. 67 1961 177--181.&lt;br&gt;&#xA;&lt;a href=&quot;http://projecteuclid.org/euclid.bams/1183524076&quot; rel=&quot;nofollow&quot;&gt;http://projecteuclid.org/euclid.bams/1183524076&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Starting with the basic result of topological analysis that a differentiable function of a complex variable generates an open mapping, the authors succeed in establishing in remarkably simple fashion, independent of any integration theory, the validity of the power-series expansion of such a function. A series of lemmas leads up to the theorem that differentiability of a function $f(z)$ on $0&amp;lt;|z-z_0| &amp;lt; r $ plus continuity at $z_0$ implies differentiability also at $z_0$. Using this, along with existence of higher-ordered derivatives, it is then shown that if $f(z)$ is continuous and $|f(z)|\leq 1$ for $|z|\leq 1$ and differentiable for $|z|&amp;lt;1$, then $|f^{(n)}(0)|\leq n!2^n$, for all $n$, and the Taylor series for $f$ converges to $f$ for $|z|&amp;lt;1/2$.&lt;br&gt;&#xA;Reviewed by G. T. Whyburn&lt;/p&gt;&#xA;&#xA;&lt;p&gt;MR0993637 (90m:30002a)   30A99 (26A39 30B10 42A20)&lt;br&gt;&#xA;Shisha, Oved(1-RI)&lt;br&gt;&#xA;&lt;strong&gt;Proof of power series and Laurent expansions of complex differentiable functions without use of Cauchy's integral formula or Cauchy's integral theorem.&lt;/strong&gt;&lt;br&gt;&#xA;J. Approx. Theory 57 (1989), no. 2, 117--135.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;MR1006337 (90m:30002b)  30A99 (26A39 30B10 42A20)&lt;br&gt;&#xA;Shisha, Oved&lt;br&gt;&#xA;&lt;strong&gt;Erratum: ``Proof of power series and Laurent expansions of complex differentiable functions without use of Cauchy's integral formula or Cauchy's integral theorem''.&lt;/strong&gt;&lt;br&gt;&#xA;J. Approx. Theory 58 (1989), no. 2, 246.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The author shows how to establish the basic results of complex function theory in the context of real variable Fourier analysis. His basic tool is a generalization of the classical Riemann integral, which he proposes should become the standard integral of the working analyst in place of currently used integrals (including the Lebesgue integral), all of which it contains as a special case. A topological development, eliminating all use of integrals, is due to \n E. Connell\en, \n R. L. Plunket\en, \n P. Porcelli\en, \n A. H. Read\en and \n G. T. Whyburn\en [Whyburn, Topological analysis, revised edition, Princeton Univ. Press, Princeton, NJ, 1964; MR0165476 (29 #2758)]. The present development generalizes the work of \n P. R. Beesack\en [Canad. Math. Bull. 15 (1972), 473--480; MR0310199 (46 #9301)], who placed various conditions on the derivative to achieve his results. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The relevant property of the new integral is the fact that if $f$ is an arbitrary differentiable function on an interval $[a, b]$, then $\int^b_a f'(x)\,dx=f(b)-f(a)$. For any function $f$ on $[a,b]$, the new integral is defined as the number $I$ such that for every $\varepsilon &gt;0$, there exists a positive function $\delta_\varepsilon(x)$ on $[a,b]$, such that for every partition $a=x_0 &amp;lt; x_1 &amp;lt; \cdots &amp;lt; x_n=b$ and sequence $s_1,\cdots, s_n$, $x_{k-1} \le s_k\le x_k$, for $k=1,\cdots,n$, such that $x_k-x_{k-1}&amp;lt;\delta_{\epsilon}(k)$ for $k=1,\cdots,n$, one has $|I-\sum ^n_{k=1}f(s_k)(x_k-x_{k-1})|&amp;lt;\varepsilon$. Let $f$ be a complex differentiable function on an annulus $0\le R' &amp;lt; |z| &amp;lt; R''&amp;lt;\infty$. Then by Dini's test, since $f$ is everywhere differentiable, for each $r$, $R' &amp;lt; r &amp;lt; R''$, the function $f(re^{i\phi})$, $0\le \phi\le 2\pi$, can be expanded in a Fourier series $f(re^{i\phi}) =\sum^{+\infty}_{n=-\infty}c_k(r)e^{i\phi}$. The objective of the paper is achieved if it can be shown that the coefficients $a_k(r)$ are independent of $r$. The basic lemma used states that $$ \iint_{\begin{array}{} R_1\le r\le R \\ 0\le \varphi\le 2\pi \end{array}} f'(re^{i\varphi})drd\varphi =\int_0^{2\pi}e^{-i\varphi}[f(Re^{i\varphi})-F(R_1e^{-i\varphi})]d\varphi,$$ $$ \widehat{\iint}_{\begin{array}{} R_1\le r\le R \\ 0\le \varphi\le 2\pi \end{array}} f'(re^{i\varphi})drd\varphi =\int_{R_1}^{R}r^{-1}\int_0^{2\pi}e^{-i\varphi}f(re^{i\varphi})d\varphi dr,$$ where the integrals on the left are two-dimensional generalized Riemann integrals. The exposition is quite detailed and fully self-contained, including the development of all relevant characteristics of the generalized Riemann integral.&lt;br&gt;&#xA;\edref {Two minor errors are corrected in the erratum.}&lt;br&gt;&#xA;Reviewed by Kenneth O. Leland&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-08-12T22:21:25.127" LastActivityDate="2010-08-12T22:21:25.127" />
  <row Id="2297" PostTypeId="2" ParentId="2293" CreationDate="2010-08-12T21:58:09.763" Score="5" ViewCount="" Body="&lt;p&gt;Brouwer's fixed point theorem can be illustrated using the game &lt;a href=&quot;http://en.wikipedia.org/wiki/Hex_(board_game)#Theory_and_Proofs&quot; rel=&quot;nofollow&quot;&gt;Hex&lt;/a&gt;.  In fact, the determinacy of Hex is equivalent to the 2-dimensional version.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-08-12T21:58:09.763" CommunityOwnedDate="2010-08-12T21:58:09.763" />
  <row Id="2298" PostTypeId="2" ParentId="2292" CreationDate="2010-08-12T22:01:12.430" Score="3" ViewCount="" Body="&lt;p&gt;I gave an answer to this question on Math Overflow some months ago:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://mathoverflow.net/questions/6711/integrability-of-derivatives/6716#6716&quot; rel=&quot;nofollow&quot;&gt;http://mathoverflow.net/questions/6711/integrability-of-derivatives/6716#6716&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See, in particular, the following paper:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.math.uga.edu/~pete/Goffman77.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.math.uga.edu/~pete/Goffman77.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="299" LastActivityDate="2010-08-12T22:01:12.430" />
  <row Id="2299" PostTypeId="2" ParentId="2293" CreationDate="2010-08-12T22:02:26.017" Score="2" ViewCount="" Body="&lt;p&gt;You can prove the Grushko decomposition theorem, as well as the Kurosh subgroup theorem.  See for example the last two sections of Massey's &quot;Algebraic Topology: an Introduction&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Steve&lt;/p&gt;&#xA;" OwnerUserId="641" LastActivityDate="2010-08-12T22:02:26.017" CommunityOwnedDate="2010-08-12T22:02:26.017" />
  <row Id="2300" PostTypeId="1" AcceptedAnswerId="2307" CreationDate="2010-08-12T22:18:51.603" Score="0" ViewCount="264" Body="&lt;p&gt;It is said that there is a proof of fundamental theorem of algebra using Lie Theory. I have seen this claim at various places. But I could never find such a proof. Can anybody help me out?&lt;/p&gt;&#xA;" OwnerUserId="1119" LastEditorUserId="1119" LastEditorDisplayName="" LastEditDate="2010-08-14T09:41:05.400" LastActivityDate="2010-08-14T09:41:05.400" Title="Fundamental theorem of algebra using Lie Theory" Tags="&lt;complex-variables&gt;&lt;lie-groups&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="1" />
  <row Id="2301" PostTypeId="2" ParentId="2248" CreationDate="2010-08-12T22:24:32.410" Score="3" ViewCount="" Body="&lt;p&gt;The theorem is known as the &quot;&lt;a href=&quot;http://www.cut-the-knot.org/pythagoras/Butterfly.shtml&quot; rel=&quot;nofollow&quot;&gt;Butterfly Theorem&lt;/a&gt;&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One fairly simple proof is the following (from &lt;em&gt;Selected Problems and Theorems of Elementary Mathematics&lt;/em&gt; by D. O. Shklyarsky, N. N. Chentsov and I. M. Yaglom).&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Let O be the center of the given circle. Since OM ⊥ CD, in order to show that CM = MD, we have to prove that ∠COM = ∠DOM.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Drop perpendiculars OK and ON from O onto PS and QR, respectively. Obviously, K is the midpoint of PS and N is the midpoint of QR. Further,&#xA;    ∠PSR = ∠PQR and ∠QPS = ∠QRS,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;as angles subtending equal arcs. Triangles SPM and QRM are therefore similar, and SP/SM = QR/QM, or SK/SM = QN/QM. In other words, in triangles SKM and QNM two pairs of sides are proportional. Also, the angles between the corresponding sides are equal. We infer that the triangles SKM and QNM are similar. Hence, ∠SKM = ∠QNM.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, have a look at the quadrilaterals OKCM and ONDM. Both have a pair of opposite straight angles, which implies that both are inscribable in a circle. In OKCM, ∠SKM = ∠COM. In ONDM, ∠QNM = ∠DOM. From which we get what we've been looking for: ∠COM = ∠DOM.&lt;/p&gt;&#xA;" OwnerUserId="446" LastActivityDate="2010-08-12T22:24:32.410" CommentCount="3" />
  <row Id="2302" PostTypeId="2" ParentId="2278" CreationDate="2010-08-12T22:31:14.480" Score="5" ViewCount="" Body="&lt;p&gt;Yes. Consider a large square matrix with two large square blocks filled with ones and one empty row and one empty column.  Delete one one from each of the blocks off the diagonal, and place a corresponding one in the empty row and column.  You now have one row with two elements, the rest having many elements, and one column having only two elements, the rest having many.  Your procedure will deal with the sparesely populated row and column first.  Permute the rows and columns so that you procedure links the two choices with different blocks.  Now the remainder of each block has one more row than it has columns or vice versa, as a perfect matching cannot exist.  It is easy to see however that a perfect matching existed before your algorithm matched the sparse row and column.&lt;/p&gt;&#xA;" OwnerUserId="943" LastActivityDate="2010-08-12T22:31:14.480" CommentCount="0" />
  <row Id="2303" PostTypeId="2" ParentId="2278" CreationDate="2010-08-12T22:35:05.880" Score="3" ViewCount="" Body="&lt;p&gt;Diagram to give an example for deinst's answer. I tried putting this in a comment, but it stuffed up the formatting. The Xs marks ones that give a complete match &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;X01100001&#xA;1X1100000&#xA;11X100000&#xA;111X00000&#xA;00001011X&#xA;0000X1110&#xA;000011X10&#xA;0000111X0&#xA;01000X000&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Now, suppose we take the two squares marked by Y's&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;*0**0000Y&#xA;111100000&#xA;111100000&#xA;111100000&#xA;00001011*&#xA;00001*110&#xA;00001*110&#xA;00001*110&#xA;0*000Y000&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;We have two 3 by 4 squares left, so we can take 6 more items. This give a total of 8 instead of 9. We can enforce those two blocks to be taken by using a permutation.&lt;/p&gt;&#xA;" OwnerUserId="123" LastEditorUserId="123" LastEditorDisplayName="" LastEditDate="2010-08-13T08:00:47.697" LastActivityDate="2010-08-13T08:00:47.697" CommentCount="0" />
  <row Id="2304" PostTypeId="1" AcceptedAnswerId="3445" CreationDate="2010-08-12T22:39:10.180" Score="3" ViewCount="220" Body="&lt;p&gt;Suppose $M_1$ and $M_2$ are invariant subspaces of the unilateral shift U such that $M_1$ subset $M_2$ and $M_1$ is of codimension strictly larger than $1$ in $M_2$. Show that there exists $M$ invariant under $U$ satisfying $M_1 \subset M \subset M_2$ where the inclusions are strict. All subspaces are closed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This problem is from the Springer GTM: &quot;An introduction to operators on the Hardy-Hilbert space&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: Perhaps I can take $M := U M_2$? Maybe I should give that some more thought.&lt;/p&gt;&#xA;" OwnerUserId="1120" LastEditorUserId="1120" LastEditorDisplayName="" LastEditDate="2010-08-18T23:50:07.387" LastActivityDate="2010-08-27T08:03:53.797" Title="Invariant Subspaces (Hardy Space)" Tags="&lt;complex-analysis&gt;&lt;functional-analysis&gt;" AnswerCount="1" />
  <row Id="2305" PostTypeId="1" AcceptedAnswerId="2337" CreationDate="2010-08-12T22:40:24.547" Score="6" ViewCount="235" Body="&lt;p&gt;I need an algorithm to produce all strings with the following property. Here capital letter refer to strings, and small letter refer to characters. $XY$ means the concatenation of string $X$ and $Y$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $\Sigma = \{a_0, a_1,\ldots,a_n,a_0^{-1},a_1^{-1},\ldots,a_n^{-1}\}$ be the set of usable characters. Every string is made up of these symbols.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Out put any set $S_n$ with the following property achieves the goal.($n\geq 2$)&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;If $W\in S_n$, then any cyclic shift of $W$ is not in $S_n$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If $W\in S_n$, then $|W| = n$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If $W\in S_n$, then $W \neq Xa_ia_i^{-1}Y$, $W \neq Xa_i^{-1}a_iY$, $W \neq a_iXa_i^{-1}$ and $W \neq a_i^{-1}Xa_i$ for any string $X$ and $Y$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If $W\not \in S_n$, $S_n \cup \{W\}$ will violate at least one of the above 3 properties. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Clearly any algorithm one can come up with is an exponential algorithm. but I'm still searching for a fast algorithm because this have some practical uses. At least for $\Sigma=\{a_0,a_1,a_0^{-1},a_1^{-1}\}$ and $n&amp;lt;25$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The naive approach for my practical application requires $O(4^n)$ time. It generate all strings of length n. When ever a new string is generated, the program create all cyclic permutations of the string and check if it have been generated before though a hash table. If not, add to the list of the result strings. Total amount of operation are $O(n4^n)$, and that's assuming perfect hashing. 12 is the limit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there better approaches? clearly a lot of useless strings were generate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: The practical usage is to find the maximum of minimum self intersection of a curve on a torus with a hole. Every curve can be characterized by a string described above. Therefore I have to generate every string and feed it to a program that calculate the minimum self intersection.&lt;/p&gt;&#xA;" OwnerUserId="96" LastEditorUserId="96" LastEditorDisplayName="" LastEditDate="2010-08-14T00:08:33.477" LastActivityDate="2010-08-22T12:49:47.297" Title="A fast way to generate cyclic strings with few restrictions?" Tags="&lt;algorithms&gt;" AnswerCount="2" CommentCount="5" FavoriteCount="3" />
  <row Id="2306" PostTypeId="2" ParentId="2284" CreationDate="2010-08-12T22:40:31.647" Score="0" ViewCount="" Body="&lt;p&gt;To nitpick Moron's answer - for any polynomial over the rational numbers, $\pi$ won't be a root of the said polynomial. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, any linear combination of positive powers of $\pi$ with rational coefficients is non-zero.&#xA;In particular, $\pi^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(My problem with Moron's answer is that for example the unit roots of order 3 won't be the root of any polynomial $ax^n + b$ for rational $a,b$.)&lt;/p&gt;&#xA;" OwnerUserId="622" LastActivityDate="2010-08-12T22:40:31.647" CommentCount="3" />
  <row Id="2307" PostTypeId="2" ParentId="2300" CreationDate="2010-08-12T22:40:42.393" Score="2" ViewCount="" Body="&lt;p&gt;The references can be found in the comments here: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://mathoverflow.net/questions/34699/approaches-to-riemann-hypothesis-using-methods-outside-number-theory/34718#34718&quot; rel=&quot;nofollow&quot;&gt;http://mathoverflow.net/questions/34699/approaches-to-riemann-hypothesis-using-methods-outside-number-theory/34718#34718&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;For those who don't know it, the (well, a) Lie-theoretic proof of Fund. Thm. of Algbera due to Witt is on p. 245 of the book &quot;Numbers&quot; by Ebbinghaus et al. – KConrad Aug 6 at 4:44&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Witt's Lie-theoretic proof of Fund. Thm. of Algbera seems to be Witt (Ernst), Über einen Satz von Ostrowski, Arch. Math. (Basel) 3, (1952). 334. – Chandan Singh Dalawat Aug 6 at 6:11&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="223" LastActivityDate="2010-08-12T22:40:42.393" CommentCount="6" />
  <row Id="2308" PostTypeId="2" ParentId="2300" CreationDate="2010-08-12T22:41:57.773" Score="0" ViewCount="" Body="&lt;p&gt;Muad: Are you talking about this www.fc.up.pt/mp/jcsantos/PDF/artigos/FTA.pdf&lt;/p&gt;&#xA;" LastActivityDate="2010-08-12T22:41:57.773" CommentCount="3" />
  <row Id="2310" PostTypeId="2" ParentId="2260" CreationDate="2010-08-12T22:57:36.210" Score="3" ViewCount="" Body="&lt;p&gt;How many ways are there to choose a $2$-element subset out of an $n$-element set?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the one hand, you can choose the first element of the set in $n$ ways, then the second element of the set in $n-1$ ways, then divide by $2$ because it doesn't matter which you choose first and which you choose second.  This gives $\frac{n(n-1)}{2}$ ways.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, suppose the $n$ elements are $1, 2, 3, ... n$, and suppose the larger of the two elements you choose is $j$.  Then for every $j$ between $2$ and $n$ there are $j-1$ possible choices of the smaller of the two elements, which can be any of $1, 2, ... j-1$.  This gives $1 + 2 + ... + (n-1)$ ways.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the two expressions above count the same thing, they must be equal.  This is known as the principle of double counting, and it is one of a combinatorialist's favorite weapons.  A generalization of this argument allows one to deduce the sum of the first $n$ squares, cubes, fourth powers...&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-08-12T22:57:36.210" />
  <row Id="2312" PostTypeId="2" ParentId="1533" CreationDate="2010-08-12T23:50:23.967" Score="2" ViewCount="" Body="&lt;p&gt;A very good textbook for discrete mathematics at an undergraduate level is the &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0073229725&quot; rel=&quot;nofollow&quot;&gt;Kenneth Rosen book titled Discrete Mathematics and Its Applications&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The book provides solutions to half of the problems. You can also buy the &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0073107794&quot; rel=&quot;nofollow&quot;&gt;Student's Solutions Guide&lt;/a&gt;. I don't own it, but I would suspect that it either provides the answers to the other half of the questions or provides a step-by-step guide to solving the problems (the book only provides final answers with minimal explanations of those answers).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's used for the two-quarter sequence in Discrete Mathematics that is taken by computer science and software engineering majors, as well as a number of mathematics programs at my university. I kept this book around even after I took the course, and I'm currently using it to brush up on my discrete math skills for my Certified Software Development Associate exam.&lt;/p&gt;&#xA;" OwnerUserId="91" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2010-08-13T07:49:11.017" LastActivityDate="2010-08-13T07:49:11.017" CommunityOwnedDate="2010-08-12T23:50:23.967" />
  <row Id="2314" PostTypeId="2" ParentId="2293" CreationDate="2010-08-13T00:16:50.343" Score="6" ViewCount="" Body="&lt;p&gt;One can prove the fundamental theorem of algebra using the fundamental group of the circle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, as a generalization of the Hairy Ball Theorem, one may compute, for all spheres, the maximum number of linearly independent vector fields that may live on that sphere. (the hairy ball theorem says that this number is 0 for even dimensional spheres, and at least 1 for odd-dimensional spheres). For dimensions less than 15 (I think), one can compute this number using only cohomology operations. The general result was proven by Adams, but I'm not sure how he did it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a side note: I don't think you can prove the general versions of Borsuk-Ulam, Brouwer, etc. with just the fundamental group... You need either higher homotopy groups or higher homology groups. &lt;/p&gt;&#xA;" OwnerUserId="423" LastActivityDate="2010-08-13T00:16:50.343" CommunityOwnedDate="2010-08-13T00:16:50.343" CommentCount="1" />
  <row Id="2315" PostTypeId="1" AcceptedAnswerId="2316" CreationDate="2010-08-13T00:24:38.533" Score="3" ViewCount="900" Body="&lt;p&gt;I am auditing a Linear Algebra class, and today we were taught about the rank of a matrix. The definition was given from the row point of view: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;The rank of a matrix A is the number&#xA;  of non-zero rows in the reduced&#xA;  row-echelon form of A&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The lecturer then explained that if the matrix $A$ has size $m * n$, then $rank(A) \leq m$ and $rank(A) \leq n$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The way I had been taught about rank was that it was the smallest of &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;the number of rows bringing new information&lt;/li&gt;&#xA;&lt;li&gt;the number of columns bringing new information. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I don't see how that would change if we transposed the matrix, so I said in the lecture:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;then the rank of a matrix is the same of its transpose, right?&quot; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And the lecturer said: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;oh, not so fast! Hang on, I have to think about it&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the class has about 100 students and the lecturer was just substituting for the &quot;normal&quot; lecturer, he was probably a bit nervous, so he just went on with the lecture.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have tested &quot;my theory&quot; with one matrix and it works, but even if I tried with 100 matrices and it worked, I wouldn't have proven that it always works because there might be a case where it doesn't.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So my question is first whether I am right, that is, whether the rank of a matrix is the same as the rank of its transpose, and second, if that is true, how can I prove it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks  :)&lt;/p&gt;&#xA;" OwnerUserId="85" LastEditorUserId="85" LastEditorDisplayName="" LastEditDate="2010-08-13T00:42:24.177" LastActivityDate="2010-08-13T00:42:24.177" Title="Is the rank of a matrix the same of its transpose? If yes, how can I prove it?" Tags="&lt;linear-algebra&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="2316" PostTypeId="2" ParentId="2315" CreationDate="2010-08-13T00:31:40.423" Score="7" ViewCount="" Body="&lt;p&gt;The answer is yes.  This statement often goes under the name &quot;row rank equals column rank&quot;.  Knowing that, it is easy to to search the internet for proofs, e.g.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://homepage.mac.com/ehgoins/ma265/lecture_28.pdf&quot; rel=&quot;nofollow&quot;&gt;http://homepage.mac.com/ehgoins/ma265/lecture_28.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also any reputable linear algebra text should prove this: it is indeed a rather important result.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, since you said that you had only a substitute lecturer, I won't castigate him, but this would be a distressing lacuna of knowledge for someone who is a regular linear algebra lecturer.  &lt;/p&gt;&#xA;" OwnerUserId="299" LastActivityDate="2010-08-13T00:31:40.423" CommentCount="1" />
  <row Id="2317" PostTypeId="2" ParentId="2315" CreationDate="2010-08-13T00:33:11.120" Score="1" ViewCount="" Body="&lt;p&gt;Yes, it is a fact. This is true over any commutative field. See for instance the first chapter of Emil Artin, Galois Theory for a very elementary argument.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you need to phrase that argument in more conceptual terms, consider the matrices as linear transformations. If A is the matrix, then let $A^t$ be the transpose, and then $A^tA$ and $A$ have the same domain, and use the fact that they have the same null space, and use the dimension theorem rank + nullity = dimension of the space.&lt;/p&gt;&#xA;" OwnerUserId="1119" LastActivityDate="2010-08-13T00:33:11.120" CommentCount="5" />
  <row Id="2318" PostTypeId="2" ParentId="2269" CreationDate="2010-08-13T00:34:31.103" Score="1" ViewCount="" Body="&lt;p&gt;Take a look at equation 34 on &lt;a href=&quot;http://mathworld.wolfram.com/GammaFunction.html&quot; rel=&quot;nofollow&quot;&gt;http://mathworld.wolfram.com/GammaFunction.html&lt;/a&gt; After taking the log of both sides, the zeta function appears in essentially the same form as in your equation. They reference a 1968 paper by Wrench for more information. I haven't worked through the details myself, but the similarities are striking, and it looks like a promising approach.&lt;/p&gt;&#xA;" OwnerUserId="1123" LastActivityDate="2010-08-13T00:34:31.103" />
  <row Id="2319" PostTypeId="2" ParentId="2293" CreationDate="2010-08-13T00:55:15.570" Score="7" ViewCount="" Body="&lt;p&gt;The only finite-dimensional division algebras over $\mathbb{R}$ have dimensions 1, 2, 4, and 8.  Note that we do not require that there be an identity element. For weaker versions of this result, see pages 173 and 222 in Hatcher's book &lt;em&gt;Algebraic Topology&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: An earlier version of this answer stated, mistakenly, that the only finite-dimensional division algebras are $\mathbb{R}$, $\mathbb{C}$, $\mathbb{H}$, and $\mathbb{O}$.  For this to be true, one needs the additional hypothesis that there is an identity element.&lt;/p&gt;&#xA;" OwnerUserId="455" LastEditorUserId="455" LastEditorDisplayName="" LastEditDate="2010-08-13T13:58:34.090" LastActivityDate="2010-08-13T13:58:34.090" CommunityOwnedDate="2010-08-13T00:55:15.570" CommentCount="1" />
  <row Id="2320" PostTypeId="2" ParentId="2293" CreationDate="2010-08-13T01:18:36.973" Score="4" ViewCount="" Body="&lt;p&gt;This is my favorite.  One can show that for any continuous map from $S^{1}$ to $R^{3}$ there is a direction along which the map has at least 4 extrema (in particular, at least 2 global minima and 2 global maxima.)  More colloquially, one can show that every potato chip can be placed on a table so its edge touches the table in at least two points and its edge simultaneously has two points of maximum height. &lt;/p&gt;&#xA;" OwnerUserId="948" LastEditorUserId="948" LastEditorDisplayName="" LastEditDate="2010-08-13T02:51:31.960" LastActivityDate="2010-08-13T02:51:31.960" CommunityOwnedDate="2010-08-13T01:18:36.973" CommentCount="1" />
  <row Id="2321" PostTypeId="1" AcceptedAnswerId="2327" CreationDate="2010-08-13T02:27:14.560" Score="3" ViewCount="312" Body="&lt;p&gt;Let $p_N/q_N$ be the $N^\text{th}$ convergent of the continued fraction for some irrational number $\alpha$. It turns out that for any other approximation $p/q$ (with $q \le q_N$) which isn't a convergent $|\alpha q - p| &gt; |\alpha q_{N-1} - p_{N-1}|$. I'm wondering if there are any nice proofs for this result?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;In my book this is proved by picking $x,y$ that solves&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{pmatrix} p_N &amp;amp; p_{N-1} \\ q_N &amp;amp; q_{N-1} \end{pmatrix}&#xA;\begin{pmatrix} x \\ y \end{pmatrix} =&#xA;\begin{pmatrix} p \\ q \end{pmatrix}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;since $x$ and $y$ have opposite sign, as well as $\alpha q_N - p_N$ and $\alpha q_{N-1} - p_{N-1}$ have opposite sign we can conclude that $|\alpha q - p| = |x (\alpha q_N - p_N) + y (\alpha q_{N-1} - p_{N-1})| = |x| |\alpha q_N - p_N| + |y| |\alpha q_{N-1} - p_{N-1}|$ which proves the theorem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am looking for different proofs than this one.&lt;/p&gt;&#xA;" LastActivityDate="2010-08-23T13:16:01.510" Title="A nicer proof of Lagrange's 'best approximations' law?" Tags="&lt;continued-fractions&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="2322" PostTypeId="2" ParentId="457" CreationDate="2010-08-13T03:04:55.733" Score="2" ViewCount="" Body="&lt;p&gt;Katie Banks wrote:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, one can restate Euler's proof that $\sum \frac{1}{n^2} = \frac{\pi^2}{6}$ in terms of the flow of an incompressible fluid with sources at the integer points in the plane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could somebody please elaborate or provide a reference. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks. &lt;/p&gt;&#xA;" OwnerUserId="1127" LastActivityDate="2010-08-13T03:21:37.753" CommunityOwnedDate="2010-08-13T03:04:55.733" CommentCount="2" />
  <row Id="2323" PostTypeId="2" ParentId="2093" CreationDate="2010-08-13T04:53:40.637" Score="1" ViewCount="" Body="&lt;p&gt;I think this problem is still open, but the following might be useful:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Ray-Chaudhuri-Wilson Theorem&lt;/strong&gt;: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $L$ be a set of $m$ integers and $F$ be an $L$-intersecting $k$-uniform family of subsets of a set of $n$ elements, where $m \le k$, then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$|F| \le {n \choose m}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\bullet$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$k$-uniform family is a set of subsets, each subset being of size k.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An $L$-intersecting family is such that the intersection size of any two distinct sets in the family is in $L$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The following result of Frankl gives us a lower bound&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Frankl's Result:&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For every $k \ge m \ge 1$ and $n \ge 2k^{2}$ there exists a $k$-uniform family $F$ of size $&gt; (\frac{n}{2k})^{m}$ on $n$ points such that $|A \cap B| \le m-1$ for any two distinct sets $A,B \in F$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\bullet$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For an algorithm for constructing such sets (based on Frankl's result) refer: &lt;a href=&quot;http://stackoverflow.com/questions/2955318/creating-combinations-that-have-no-more-one-intersecting-element/2955527#2955527&quot;&gt;http://stackoverflow.com/questions/2955318/creating-combinations-that-have-no-more-one-intersecting-element/2955527#2955527&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-08-13T06:00:40.287" LastActivityDate="2010-08-13T06:00:40.287" />
  <row Id="2324" PostTypeId="1" CreationDate="2010-08-13T05:32:28.170" Score="2" ViewCount="126" Body="&lt;p&gt;What might be some potential methods to encode a 100-point signal (curve) for input to a Artificial Neural Network?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example: we have a large number of 100-pt 'curves' ranging from flat-line to approximately a half-sine wave with a wide range of max amplitude.  The curves have some noise and occasional anomalous 'bumps'.  These training set curves can be broken down into known positives or negatives for 'truthing'.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I wanted to use an Artificial Neural Network to give a result for untrained data curves, what are potential method to encode the input data sets?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would transforming each of the 100 points in a curve to a [0.0000 .... 1.0000] range be useful (and thus we'd have 100 inputs to the ANN?  Or another method might produce better results?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;NOTE: I understand that FFT and Power Spectrum analysis might be a completely different alternative here as well, but am focussing right now on a possible neural network method.&lt;/p&gt;&#xA;" OwnerUserId="1131" LastActivityDate="2010-12-14T18:44:25.953" Title="Encoding a Discrete Signal to an Artificial Neural Network?" Tags="&lt;signal-processing&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="2325" PostTypeId="1" CreationDate="2010-08-13T05:55:57.990" Score="2" ViewCount="87" Body="&lt;p&gt;I have recently been thnking some about algebraic groups and reading parts of Humphreys book on them, and I was wondering if there is a general process to showing they are simply connected. In particular I was wondering over other fields than C but if the answer only works in C I will settle for that. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One idea I had was that may e using borel-weil-bott one could make a slick arguement for when the fundamental group is trivial. I would just like to get away from the ad-hoc thinking process I am using.  &lt;/p&gt;&#xA;" OwnerUserId="16" LastActivityDate="2010-08-13T09:00:56.663" Title="Checking if Algebraic Groups are simply connected." Tags="&lt;algebraic-geometry&gt;&lt;algebraic-groups&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="2326" PostTypeId="2" ParentId="457" CreationDate="2010-08-13T06:27:59.577" Score="8" ViewCount="" Body="&lt;p&gt;From wikipedia article about &lt;a href=&quot;http://en.wikipedia.org/wiki/Theorema_Egregium&quot; rel=&quot;nofollow&quot;&gt;Gauss Theorema Egregium&lt;/a&gt;: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;An application of the Theorema Egregium is seen in a common pizza-eating strategy: A slice of pizza can be seen as a surface with constant Gaussian curvature 0. Gently bending a slice must then roughly maintain this curvature (assuming the bend is roughly a local isometry). If one bends a slice horizontally along a radius, non-zero principal curvatures are created along the bend, dictating that the other principal curvature at these points must be zero. This creates rigidity in the direction perpendicular to the fold, an attribute desirable when eating pizza (since it prevents the pizza toppings from falling off).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1043" LastActivityDate="2010-08-13T06:27:59.577" CommunityOwnedDate="2010-08-13T06:27:59.577" />
  <row Id="2327" PostTypeId="2" ParentId="2321" CreationDate="2010-08-13T06:32:09.227" Score="3" ViewCount="" Body="&lt;p&gt;Of course there is a nicer proof! In fact, it's almost obvious if one thinks about geometric interpretation of continued fraction: consider the line $y=\alpha x$; then the best approximation (i.e. approximation that minimizes $|\alpha q-p|=q|\alpha-\frac{p}{q}|$) is the point of integer lattice nearest to this line; finally observe that convergents with even/odd numbers of the continued fraction give coordinates of [verices of] the convex hull of [points of] lattice lying over/under the line.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;One can also (as &lt;a href=&quot;http://math.stackexchange.com/users/498/j-mangaldan&quot;&gt;J. Mangaldan&lt;/a&gt; suggests) take a look at &lt;em&gt;Lorentzen, Waadeland. Continued Fractions with Applications&lt;/em&gt;, I.2.1 (esp. figure 1 and the text near it; there are no words &quot;convex hull&quot; there but implicitly everything is explained, more or less).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Upd. One more reference is Davenport's &quot;The Higher Arithmetics&quot; (section IV.12).&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Finally, an illustration (from Arnold's book).&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Bold line $y=\alpha x$ (on the picture $\alpha=\frac{10}7$) is approximated by vectors $e_i$ corresponding to convergents (namely, $e_i$ ends at $(p_i,q_i)$);&lt;/li&gt;&#xA;&lt;li&gt;each vector (starting from $e_3$) is a linear combination of two preciding ones: $e_{i+2}=e_i+a_{i-1}e_{i+1}$, where $a$ is the maximal integer s.t. the new vector still doesn't intersect the line;&lt;/li&gt;&#xA;&lt;li&gt;this is exactly the algorithm for representing $\alpha$ by a continued fraction: $\alpha=[a_0,a_1,\dots]$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/coRwt.png&quot; alt=&quot;geometry of continued fractions&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="152" LastEditorUserId="152" LastEditorDisplayName="" LastEditDate="2010-08-23T13:16:01.510" LastActivityDate="2010-08-23T13:16:01.510" CommentCount="9" />
  <row Id="2329" PostTypeId="2" ParentId="155" CreationDate="2010-08-13T07:34:38.773" Score="9" ViewCount="" Body="&lt;p&gt;Have you ever heard of Galois theory? It is a theory that studies the solutions of equations over fields.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As it turns out, there is a special type of Galois theory called Differential Galois theory, which studies fields with a differential operator on them:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Differential_galois_theory&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Differential_galois_theory&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using this theory, one can prove that functions like sin(x)/x and x^x don't have an indefinite integral.&lt;/p&gt;&#xA;" OwnerUserId="1089" LastActivityDate="2010-08-13T07:34:38.773" CommentCount="2" />
  <row Id="2330" PostTypeId="1" CreationDate="2010-08-13T07:58:30.463" Score="1" ViewCount="163" Body="&lt;p&gt;I need to find the minimum of a function f(t) = int g(t,x) dx over [0,1]. What I did in mathematica is as follows:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;f[t_] = NIntegrate[g[t,x],{x,-1,1}]&#xA;FindMinimum[f[t],{t,t0}]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;However mathematica halts at the first try, because NIntegrate does not work with the symbolic t. Although Plot[f[t],{t,0,1}] works perferctly, FindMinimum stops at the initial point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any way to get around it? Thanks!&lt;/p&gt;&#xA;" OwnerUserId="1132" LastActivityDate="2010-08-13T18:49:53.220" Title="find minimum of a function defined by integration in Mathematica" Tags="&lt;math-software&gt;&lt;mathematica&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="2331" PostTypeId="2" ParentId="2325" CreationDate="2010-08-13T09:00:56.663" Score="1" ViewCount="" Body="&lt;p&gt;Please see &lt;a href=&quot;http://mathoverflow.net/questions/19262/simply-connectedness-of-algebraic-group&quot; rel=&quot;nofollow&quot;&gt;http://mathoverflow.net/questions/19262/simply-connectedness-of-algebraic-group&lt;/a&gt;&lt;/p&gt;&#xA;" LastActivityDate="2010-08-13T09:00:56.663" CommentCount="1" />
  <row Id="2334" PostTypeId="2" ParentId="155" CreationDate="2010-08-13T10:37:36.773" Score="6" ViewCount="" Body="&lt;p&gt;A key result in this area is a &lt;a href=&quot;http://en.wikipedia.org/wiki/Liouville%27s_theorem_%28differential_algebra%29#Basic_theorem&quot; rel=&quot;nofollow&quot;&gt;theorem by Liouville&lt;/a&gt;. You'll find easy expositions of this result in the references below:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Antoine Chambert-Loir. A Field Guide to Algebra, last chapter, Springer, 2005.&lt;/li&gt;&#xA;&lt;li&gt;M Rosenlicht. Integration in finite terms, Amer.    Math. Monthly,&#xA;    (1972), pp 963-972.&lt;/li&gt;&#xA;&lt;li&gt;Toni Kasper. Integration in ﬁnite&#xA;terms: the Liouville theory. Math.&#xA;Mag.,  53(4):195–201, 1980.&lt;/li&gt;&#xA;&lt;li&gt;D.    H. Potts. Elementary&#xA;integrals. Amer.    Math. Monthly,&#xA;63:545–554, 1956.&lt;/li&gt;&#xA;&lt;li&gt;A. D. Fitt and G. T. Q. Hoare. The closed-form integration of arbitrary functions,&#xA;The Mathematical Gazette, Vol. 77, No. 479 (Jul., 1993), pp. 227-236&lt;/li&gt;&#xA;&lt;li&gt;Elena Anne Marchisotto and Gholam-Ali Zakeri. An invitation to integration in finite terms, The College Mathematics Journal, Vol. 25, No. 4 (Sep., 1994), pp. 295-308&lt;/li&gt;&#xA;&lt;li&gt;D. G. Mead. Integration, The American Mathematical Monthly, Vol. 68, No. 2 (Feb., 1961), pp. 152-156 &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="589" LastActivityDate="2010-08-13T10:37:36.773" CommentCount="1" />
  <row Id="2336" PostTypeId="2" ParentId="2130" CreationDate="2010-08-13T11:53:49.693" Score="0" ViewCount="" Body="&lt;p&gt;As a rule, &quot;small&quot; things live most naturally in the &quot;thing of things.&quot; In this case, the category Cat of small categories is locally small, but the category CAT of all categories is not. As most categorical constructions (adjunctions, universals, ...) assume hom(A,B) is a set, CAT can be problematic to reason about.&lt;/p&gt;&#xA;" OwnerUserId="1136" LastActivityDate="2010-08-13T11:53:49.693" />
  <row Id="2337" PostTypeId="2" ParentId="2305" CreationDate="2010-08-13T12:37:22.273" Score="3" ViewCount="" Body="&lt;p&gt;Making explicit what is implicit in Qiaochu Yuan's comment, and demonstrating that &lt;a href=&quot;http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.97.4244&amp;amp;rep=rep1&amp;amp;type=pdf&quot; rel=&quot;nofollow&quot;&gt;someone else's work&lt;/a&gt; has failed to evade my eyes. (It is a neat article, read it.)  I present this adaptation of Duval's algorithm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assign an order to your symbols, say $a_1, a_2, a_1^{-1}, a_2^{-1}$ let first_symbol and _last_symbol be the first and last symbols in the set.  Let next be a function that gives the next symbol in sequence. The function conflict checks to see if the two symbols are inverses of each other.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;w[1] &amp;lt;- first_symbol&#xA;i &amp;lt;- 1&#xA;repeat&#xA;  for j = 1 to n–i&#xA;    do w[i+j] &amp;lt;- w[j]&#xA;  if i = n and not conflict(w[1], w[n])&#xA;    then output w[1] ... w[n]&#xA;  i &amp;lt;- n&#xA;  while i &amp;gt; 0 and w[i] = last_symbol&#xA;    do i &amp;lt;- i–1&#xA;  if i &amp;gt; o  &#xA;     then w[i] &amp;lt;- next(w[i])&#xA;  if i &amp;gt; 1 and conflict(w[i-1], w[i]) &#xA;     then w[i] &amp;lt;- next(w[i])&#xA;until i = 0&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This is just Duval's algorithm for generating a list of the lexicographically minimal cyclic shifts with extra checks to step over the cases where a conflict should occur. I have neither bothered to work out either a formal proof that this works, or implemented it in actual code.  Caveat Emptor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt; As expected, I missed a corner case.  The following python code appears to work. It takes the length of the cycle and a list of integers (I use integers for the group)&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;def cycles(n,l):&#xA;    w = range(n+1)&#xA;    m = len(l) - 1&#xA;    w[1] = 0&#xA;    i = 1&#xA;    while i &amp;gt; 0:&#xA;        for j in range(n-i):&#xA;            w[j + i + 1] = w[j + 1]&#xA;        if i == n and l[w[1]] + l[w[n]] != 0:&#xA;            print [l[w[i]] for i in xrange(1,n+1)]&#xA;        i = n&#xA;        while i &amp;gt; 0 and w[i] == m:&#xA;            i = i - 1&#xA;        while i &amp;gt; 0:&#xA;            if i &amp;gt; 0:&#xA;                w[i] = w[i] + 1&#xA;            if i &amp;gt; 1 and l[w[i-1]] + l[w[i]] == 0:&#xA;                w[i] = w[i] + 1&#xA;            if w[i] &amp;lt;= m:&#xA;                break&#xA;            i = i - 1&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;to get the length four cycles for {-2, -1, 1, 2} call&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;cycles(4, [-2, -1, 1, 2])&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;resulting in&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;[-2, -2, -2, -1]&#xA;[-2, -2, -2, 1]&#xA;[-2, -2, -1, -1]&#xA;[-2, -2, 1, 1]&#xA;[-2, -1, -2, 1]&#xA;[-2, -1, -1, -1]&#xA;[-2, -1, 2, -1]&#xA;[-2, -1, 2, 1]&#xA;[-2, 1, 1, 1]&#xA;[-2, 1, 2, -1]&#xA;[-2, 1, 2, 1]&#xA;[-1, -1, -1, 2]&#xA;[-1, -1, 2, 2]&#xA;[-1, 2, 1, 2]&#xA;[-1, 2, 2, 2]&#xA;[1, 1, 1, 2]&#xA;[1, 1, 2, 2]&#xA;[1, 2, 2, 2]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Ahem&lt;/strong&gt; Didn't I say&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;def cycles(n,l):&#xA;    w = range(n+1)&#xA;    m = len(l) - 1&#xA;    w[1] = 0&#xA;    i = 1&#xA;    while i &amp;gt; 0:&#xA;        for j in range(n-i):&#xA;            w[j + i + 1] = w[j + 1]&#xA;        if (i == n) and ((l[w[1]] + l[w[n]]) != 0):&#xA;            print [l[w[i]] for i in xrange(1,n+1)]&#xA;        i = n&#xA;        while i &amp;gt; 0 and w[i] == m:&#xA;            i = i - 1&#xA;        while i &amp;gt; 0:&#xA;            if i &amp;gt; 0:&#xA;                w[i] = w[i] + 1&#xA;            if (i &amp;gt; 1) and ((l[w[i-1]] + l[w[i]]) == 0):&#xA;                w[i] = w[i] + 1&#xA;            if w[i] &amp;lt;= m:&#xA;                break&#xA;            i = i - 1&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;That's what I should have said if I took my own advice.  Sorry.&lt;/p&gt;&#xA;" OwnerUserId="943" LastEditorUserId="943" LastEditorDisplayName="" LastEditDate="2010-08-15T01:44:37.713" LastActivityDate="2010-08-15T01:44:37.713" CommentCount="5" />
  <row Id="2338" PostTypeId="1" AcceptedAnswerId="2342" CreationDate="2010-08-13T12:45:26.733" Score="7" ViewCount="202" Body="&lt;p&gt;Is it true that sin(n&lt;sup&gt;k&lt;/sup&gt;) ≠ (sin n)&lt;sup&gt;k&lt;/sup&gt; for any positive integers n and integers k ≠ 1?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What if n (&gt; 0), k (≠ 1) are rational?&lt;/p&gt;&#xA;" OwnerUserId="171" LastActivityDate="2010-08-13T13:12:44.917" Title="Is sin(n^k) ≠ (sin n)^k in general?" Tags="&lt;number-theory&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="2" />
  <row Id="2339" PostTypeId="1" AcceptedAnswerId="2343" CreationDate="2010-08-13T12:54:33.847" Score="4" ViewCount="345" Body="&lt;p&gt;How can you prove that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{k=1}^{n-1}\tan^{2}\frac{k \pi}{2n} = \frac{(n-1)(2n-1)}{3}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for every integer $n\geq 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS: no, it's not a homework... :-)&lt;/p&gt;&#xA;" OwnerUserId="1030" LastEditorUserId="171" LastEditorDisplayName="" LastEditDate="2010-08-13T13:27:13.773" LastActivityDate="2010-08-13T13:34:58.093" Title="Prove that $\sum_{k=1}^{n-1}\tan^{2}\frac{k \pi}{2n} = \frac{(n-1)(2n-1)}{3}$" Tags="&lt;trigonometry&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="2340" PostTypeId="2" ParentId="2293" CreationDate="2010-08-13T13:04:14.997" Score="2" ViewCount="" Body="&lt;p&gt;Beginning students might appreciate applications outside mathematics. Here are a couple of current (undergraduate) projects: &lt;a href=&quot;http://inperc.com/wiki/index.php?title=The_topology_of_data&quot; rel=&quot;nofollow&quot; title=&quot;The topology of data&quot;&gt;The topology of data&lt;/a&gt; and  3D image analysis. The tool used is called &quot;persistent homology&quot;.&lt;/p&gt;&#xA;" OwnerUserId="1138" LastEditorUserId="1138" LastEditorDisplayName="" LastEditDate="2010-08-13T16:20:49.523" LastActivityDate="2010-08-13T16:20:49.523" CommunityOwnedDate="2010-08-13T13:04:14.997" />
  <row Id="2341" PostTypeId="1" CreationDate="2010-08-13T13:11:30.677" Score="0" ViewCount="296" Body="&lt;p&gt;Hey guys,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm new at learning boolean algebra and understand the basic laws. But now I'm trying to proof:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;x'z+ xyz + xy'z = z&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Can someone help me with this please ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance.&lt;/p&gt;&#xA;" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-08-13T13:37:17.447" LastActivityDate="2010-08-14T03:38:08.367" Title="boolean algebra proof" Tags="&lt;logic&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="2342" PostTypeId="2" ParentId="2338" CreationDate="2010-08-13T13:12:44.917" Score="11" ViewCount="" Body="&lt;p&gt;If $\sin(n^k)=\sin(n)^k$, then $\frac{e^{in^k}-e^{-in^k}}{2i}=(\frac{e^{in}-e^{-in}}{2i})^k$, so that $e^i$ would be algebraic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But that directly contradicts the &lt;a href=&quot;http://en.wikipedia.org/wiki/Lindemann%E2%80%93Weierstrass_theorem&quot; rel=&quot;nofollow&quot;&gt;Lindemann-Weierstrass theorem&lt;/a&gt; because $i$ is algebraic.&lt;/p&gt;&#xA;" OwnerUserId="446" LastActivityDate="2010-08-13T13:12:44.917" CommentCount="1" />
  <row Id="2343" PostTypeId="2" ParentId="2339" CreationDate="2010-08-13T13:21:50.230" Score="8" ViewCount="" Body="&lt;p&gt;By well know formula we have&#xA;$$&#xA;(\cos{\frac{k\pi}{2n}} + i\sin{\frac{k\pi}{2n}})^{2n}=(-1)^{k}&#xA;$$&#xA;Hence by &lt;a href=&quot;http://en.wikipedia.org/wiki/Binomial_formula&quot; rel=&quot;nofollow&quot;&gt;Binomial theorem&lt;/a&gt; we have ($[x]$ is not an integer part, brackets are added for clarity)&#xA;$$&#xA;\sum_{t=0}^{2n}\binom{2n}{t}[\cos{\frac{k\pi}{2n}}]^t \cdot [i\sin{\frac{k\pi}{2n}}]^{2n-t}=(-1)^{k}&#xA;$$&#xA;Now we consider only imaginary part of this:&#xA;$$&#xA;\sum_{r=0}^{n-1}\binom{2n}{2r+1}[\cos{\frac{k\pi}{2n}}]^{2r+1} \cdot [i\sin{\frac{k\pi}{2n}}]^{2n-2r-1}=0&#xA;$$&#xA;Divide it by $[\cos{\frac{k\pi}{2n}}]^{2n}$:&#xA;$$&#xA;\sum_{r=0}^{n-1}\binom{2n}{2r+1}[i\tan{\frac{k\pi}{2n}}]^{2n-2r-1}=0&#xA;$$&#xA;Now multiply by $i\tan{\frac{k\pi}{2n}}$:&#xA;$$&#xA;\sum_{r=0}^{n-1}\binom{2n}{2r+1}[i\tan{\frac{k\pi}{2n}}]^{2n-2r}=0&#xA;$$&#xA;So $[\tan{\frac{k\pi}{2n}}]^2$ are roots of the following polynomial:&#xA;$$&#xA;\sum_{r=0}^{n-1}\binom{2n}{2r+1}[i\cdot x]^{n-r}=0&#xA;$$&#xA;Hence by &lt;a href=&quot;http://en.wikipedia.org/wiki/Vieta%27s_formulas&quot; rel=&quot;nofollow&quot;&gt;Vieta's_formulas&lt;/a&gt; sum of it roots is equal to&#xA;$$&#xA;\frac{\binom{2n}{3}}{\binom{2n}{1}} =\frac{(2n-1)(n-1)}{3}&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="334" LastActivityDate="2010-08-13T13:21:50.230" CommentCount="1" />
  <row Id="2344" PostTypeId="2" ParentId="2339" CreationDate="2010-08-13T13:24:16.607" Score="0" ViewCount="" Body="&lt;p&gt;I am doing a similar thing for $\cot$ i hope you can reciprocate it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have $$(\cot(\theta)-i)^{n} = \frac{ \cos{n\theta} - \sin{n \theta}}{\sin^{n}(\theta)}$$&#xA;Equating the real and imaginary parts on both sides we have $$ \frac{\sin(n\theta)}{\sin{\theta}} = \sum\limits_{s} {n \choose 2s+1}  (-1)^{s} \cot^{n-2s-1}(\theta)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now take $2n+1$ instead of $n$ we have, $$\sin{(2n+1)\theta} = \sin^{2n+1}(\theta)P_{n}\cot^{2}(\theta)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for $\displaystyle 0 &amp;lt; \theta &amp;lt; \frac{\pi}{2}$ and where $P_{n}$ is the polynomial given by $${ 2n+1 \choose 1}T^{n} - {2n+1 \choose 3}T^{n-1} + \cdots$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Noting that the zeros of $P_{n}$ are precisely $\displaystyle \frac{r\pi}{2n+1},\ n =1,2,..$, we have the first identity from the sum of the roots formula. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the inequality $\sin{x} &amp;lt; x &amp;lt; \tan{x}$ we have $$\cot^{2m}(x) &amp;lt; \frac{1}{x^{2m}} &amp;lt; (1 + \cot^{2}(x))^{m}$$ we have $$ \sum\limits_{r=1}^{n} \cot^{2m} \frac{r\pi}{2n+1}  &amp;lt; \frac{(2n+1)^2m}{\pi^{2m}} \sum\limits_{r=1}^{n} \frac{1}{r^{2m}} &amp;lt; \sum\limits_{r=1}^{n} \Bigl( 1 + \cot^{2} \frac{r\pi}{2n+1} \Bigr)^{m}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, $$\sum\limits_{r=1}^{n} \Bigl(1+\cot^{2} \frac{r\pi}{2n+1} \Bigr)^{m} =  \sum\limits_{r=1}^{n} \cot^{2m} \frac{r\pi}{2n+1} + \mathcal{O}(n^{2m-1})$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words to find $c_{2m}$ where $ \sum\limits_{r=1}^{n} \cot^{2m} \frac{r\pi}{2n+1} = c_{2m}n^{2m} + \mathcal{O}(n^{2m-1})$ it suffices to look at the sum, $$ \cot^{2m} \frac{\pi}{2n+1} + \cot^{2m} \frac{2\pi}{2n+1} + \cdots + cot^{2m} \frac{n\pi}{2n+1}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is the sum $s_{m}$ of $m-th$ powers of the roots of the polynomial $P_{n}$. Then you can complete it using newtons formula for the sum of the roots.&lt;/p&gt;&#xA;" LastActivityDate="2010-08-13T13:34:58.093" CommentCount="0" />
  <row Id="2345" PostTypeId="2" ParentId="2341" CreationDate="2010-08-13T13:29:00.290" Score="1" ViewCount="" Body="&lt;p&gt;Assuming you mean what is more commonly written as $(\lnot x\land z)\lor(x\land y\land z)\lor(x\land\lnot y\land z)=z$, then it is equivalent, as Tom Stephens points out, to proving that $\lnot x\lor(x\land(y\lor\lnot y))$ is valid (that is, evaluates to true for all x and y). But that's clearly the same as $\lnot x \lor x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I used&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;distributivity: $x\lor(y\land z)=(x\lor y)\land(x\lor z)$ and also $x\land(y\lor z)=(x\land y)\lor(x\land z)$&lt;/li&gt;&#xA;&lt;li&gt;excluded middle: $x\lor\lnot x$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="512" LastActivityDate="2010-08-13T13:29:00.290" CommentCount="3" />
  <row Id="2346" PostTypeId="1" AcceptedAnswerId="2388" CreationDate="2010-08-13T13:33:55.397" Score="8" ViewCount="421" Body="&lt;p&gt;How does one set out to find all continuous functions $f:\mathbb{R} \to \mathbb{R}$ which satisfy $f^{k}(x)=f(x^k)$ , where $k \in \mathbb{N}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Motivation : &lt;a href=&quot;http://math.stackexchange.com/questions/2338/is-sinnk-sin-nk-in-general&quot;&gt;http://math.stackexchange.com/questions/2338/is-sinnk-sin-nk-in-general&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="9413" LastEditorUserId="9413" LastEditorDisplayName="" LastActivityDate="2010-09-02T08:01:05.733" Title="Continuous function Satisfying $f^{k}(x)=f(x^k)$" Tags="&lt;real-analysis&gt;" AnswerCount="3" CommentCount="9" FavoriteCount="3" />
  <row Id="2347" PostTypeId="2" ParentId="2346" CreationDate="2010-08-13T13:47:42.883" Score="4" ViewCount="" Body="&lt;p&gt;For a continuous $f:[0,1] \rightarrow (0, \infty)$ it is easy to show that $f(x)=1$ for $\forall x \in [0,1]$.&lt;/p&gt;&#xA;" OwnerUserId="832" LastActivityDate="2010-08-13T13:47:42.883" CommentCount="3" />
  <row Id="2348" PostTypeId="2" ParentId="2227" CreationDate="2010-08-13T14:15:46.170" Score="0" ViewCount="" Body="&lt;p&gt;Hi, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;All, please see this example. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $G_{n}$ denote the grouping of this $2n$ numbers, $$\frac{1}{2n-1},\frac{1}{2n-3},...,\frac{1}{3},1,-1,-\frac{1}{3},\cdots,-\frac{1}{2n-1}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We take a strictly increasing sequence of positive integers ${\lambda_n}$ and consider the groups $G_{\lambda_1},G_{\lambda_2},\cdots,$. We multiply each number of the group $G_{\lambda_n}$ by $n^{-2}$ and obtain the sequence $$\frac{1}{1^{2}(2\lambda_{1}-1)}, \cdots,-\frac{1}{1^{2}(2\lambda_{1}-1)}, \frac{1}{2^{2}(2\lambda_{2}-1)},...,-\frac{1}{2^2(2\lambda_{2}-1)},....,$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;say $\alpha_{1},\alpha_{2},\cdots$. Our aim is to show that $$\sum\limits_{n=1}^{\infty} \alpha_{n} \cos{nx}$$&#xA;is the fourier series of a continuous function. We group the terms in the following way $$\sum\limits_{n=1}^{2\lambda_{1}} \alpha_{n} \cos{nx} + \sum\limits_{n=2\lambda_{1}+1}^{2\lambda_{1}+2\lambda_{2}} \alpha_{n}\cos{nx} + \sum\limits_{n=2\lambda_{1}+2\lambda_{2}+1}^{2\lambda_{1}+2\lambda_{2}+2\lambda_{3}} \alpha_{n} \cos{nx}\cdots$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The last series can be written as $$\sum\limits_{n=1}^{2\lambda_{1}} \alpha_{n} \cos{nx} + \sum\limits_{n=2}^{\infty} \frac{\phi(\lambda_{n},2\lambda_{1}+2\lambda_{2} + \cdots + 2\lambda_{n-1},x)}{n^2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $$\phi(n,r,x)= \frac{\cos{(r+1)x}}{2n-1} + \frac{\cos{(r+2)x}}{2n-3} + \cdots + \frac{\cos{(r+n)x}}{1} - \frac{\cos{(r+n+1)x}}{1} - \cdots - \frac{\cos{(r+2n)x}}{2n-1}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now one can show that there is a constant $M$ (independent of $n,r$ and $x$) such that $|\phi(n,r,x)|\leq M$. From this it follows that the grouped series $$\sum\limits_{n=1}^{2\lambda_{1}} \alpha_{n} \cos{nx} + \sum\limits_{n=2}^{\infty} \frac{\phi(\lambda_{n},2\lambda_{1}+2\lambda_{2} + \cdots + 2\lambda_{n-1},x)}{n^2}$$&#xA;converges absolutely on $\mathbb{R}$, say to $f(x)$, and $f$ is continuous on $\mathbb{R}$. It is also easy to check that $$f(x) \sim \sum\limits_{n=1}^{\infty} \alpha_{n} \cos{nx}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We shall finally show that ${\lambda_n}$ can be chose so that the above series diverges at zero, that is $S_{n} = \alpha_{1} + \alpha_{2} +  \cdots + \alpha_{n}$ diverges to infinity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $$S_{2\lambda_{1}+2\lambda_{2}+ \cdots + 2 \lambda_{n-1} + \lambda_{n}} = \frac{1}{n^2} \Bigl( \frac{1}{2\lambda_{n}-1} + \cdots + \frac{1}{3} + 1 \Bigr)$$ behaves as $\frac{\ln{\lambda_{n}}}{{2n^{2}}}$ as $n \to \infty$, it is enough to take $\lambda_{n}=n^{n^2}$. Then the fourier series does not converge to $f$ at $x=2k\pi, \ k\in \mathbb{Z}$.&lt;/p&gt;&#xA;" LastActivityDate="2010-08-13T14:15:46.170" />
  <row Id="2349" PostTypeId="1" AcceptedAnswerId="2350" CreationDate="2010-08-13T14:33:08.500" Score="3" ViewCount="503" Body="&lt;p&gt;I'm doing exercise on discrete mathematics and I'm stuck with question:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;if $f$ is an invertible function from Y to Z and $g$ is an invertible function from X to Y&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Show that composition of $(f \circ\ g)$ is given by $(f \circ\ g) ^{-1} = g^{-1} \circ\ f^{-1}$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I've no idea how to prove this, please help me by give me some reference or hint to its solution.&lt;/p&gt;&#xA;" OwnerUserId="725" LastEditorUserId="205" LastEditorDisplayName="" LastEditDate="2010-08-14T04:23:46.953" LastActivityDate="2010-08-16T13:05:17.240" Title="How to prove $(f \circ\ g) ^{-1} = g^{-1} \circ\ f^{-1}$ ?" Tags="&lt;algebra-precalculus&gt;&lt;functions&gt;&lt;homework&gt;" AnswerCount="5" CommentCount="6" />
  <row Id="2350" PostTypeId="2" ParentId="2349" CreationDate="2010-08-13T14:37:19.533" Score="6" ViewCount="" Body="&lt;p&gt;Use the definition of an inverse and associativity of composition to show that the right hand side is the inverse of $(f \circ g)$.&lt;/p&gt;&#xA;" OwnerUserId="423" LastActivityDate="2010-08-13T14:37:19.533" CommentCount="2" />
  <row Id="2351" PostTypeId="2" ParentId="2349" CreationDate="2010-08-13T14:38:14.617" Score="1" ViewCount="" Body="&lt;p&gt;This is straightforward. Take x in the domain of f. It goes to f(x) = y. And g takes y to z = g(y). Therefore $g^{-1}$ takes z to y and $f^{-1}$ takes y to x. Both sides of your equation takes $z$ to $x$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please try to think more before asking. This was not hard, was it? :) &lt;/p&gt;&#xA;" OwnerUserId="1119" LastEditorUserId="1119" LastEditorDisplayName="" LastEditDate="2010-08-13T14:48:23.940" LastActivityDate="2010-08-13T14:48:23.940" CommentCount="3" />
  <row Id="2352" PostTypeId="1" CreationDate="2010-08-13T14:54:26.093" Score="4" ViewCount="246" Body="&lt;p&gt;This is a strengthening of Chandru's question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://math.stackexchange.com/questions/2227/&quot;&gt;http://math.stackexchange.com/questions/2227/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a nice and concrete example of a Fourier series that fails to converge on some &quot;big&quot; set of measure zero, for instance on the Cantor ternary set?&lt;/p&gt;&#xA;" OwnerUserId="1119" LastEditorUserId="1119" LastEditorDisplayName="" LastEditDate="2010-08-14T09:39:49.957" LastActivityDate="2010-10-13T11:01:25.923" Title="A Fourier series failing to converge on the Cantor Set" Tags="&lt;real-analysis&gt;&lt;fourier-series&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" />
  <row Id="2353" PostTypeId="2" ParentId="2130" CreationDate="2010-08-13T15:00:30.383" Score="1" ViewCount="" Body="&lt;p&gt;I will give one example of a statement that will go wrong if you are not careful with set theoretical foundations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $\mathcal C$ be a category. Then the Yoneda lemma gives an embedding $\mathcal C \hookrightarrow Funct(\mathcal{C}, Sets)$ into the category of all functors from $\mathcal C$ to the category $Sets$ of suitably &quot;small&quot; sets. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let $CAT$ be the category of all &quot;small&quot; categories. Again by the naturality of the Yoneda Construction, you get an embedding $CAT \hookrightarrow Funct(CAT, Sets)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are not careful, various set-theoretical and logical problems might arise with the above statement. Here to really make sense of things without any problems anywhere, you have to fix a Grothendieck Universe and work within it, as far as I understand things.&lt;/p&gt;&#xA;" OwnerUserId="1119" LastActivityDate="2010-08-13T15:00:30.383" />
  <row Id="2354" PostTypeId="2" ParentId="2352" CreationDate="2010-08-13T15:04:30.670" Score="2" ViewCount="" Body="&lt;p&gt;Hi--&#xA; I don't have an exact answer for this but i think this might help: &lt;a href=&quot;http://www.springerlink.com/content/p5t566u325582613/fulltext.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.springerlink.com/content/p5t566u325582613/fulltext.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;One more link: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this paper the authors show that for any set of zero measure there exists a continuous function on the circle whose Fourier series diverges on that set. In French.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Sergei Vladimirovich Konyagin, On divergence of trigonometrique Fourier series everywhere, C. R. Acad. Sci. Paris 329 (1999), 693-697&lt;/em&gt;.&lt;/p&gt;&#xA;" LastActivityDate="2010-08-13T15:04:30.670" CommentCount="3" />
  <row Id="2355" PostTypeId="1" AcceptedAnswerId="2531" CreationDate="2010-08-13T15:14:33.010" Score="22" ViewCount="766" Body="&lt;p&gt;The number of elements of order 2 in a group is fairly restricted: 0, odd, or infinity.  All such possibilities occur already in the trivial group and in dihedral groups.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The number of elements of order 3 in a group can be shown to be similarly restricted: 0, 2 mod 6, or infinity.  However, something strange happens: not all possibilities can be realized.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even worse, there is a fairly small number that I cannot decide whether it is or is not the number of elements of order 3 in a finite group:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Is there a group with exactly 92 elements of order 3?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;More boldly, I would like to know (but feel free to answer only the first question):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Exactly which numbers occur as the number of elements of order 3 in a group?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Background&lt;/strong&gt;: Such questions were studied a bit by Sylow and more heavily by Frobenius.  The theorem that the number of elements of order p is equal to −1 mod p is contained in one of Frobenius's 1903 papers.  Since elements of order 3 come in pairs, this doubles to give 2 mod 6 for p=3.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, Frobenius's results were improved some 30 years later by P. Hall who showed that if the Sylow p-subgroups are not cyclic, then the number of elements of order p is −1 mod p&lt;sup&gt;2&lt;/sup&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the Sylows are cyclic of order p&lt;sup&gt;n&lt;/sup&gt;, then the number of subgroups of order p is congruent to 1 mod p&lt;sup&gt;n&lt;/sup&gt; by the standard counting method.  If the Sylow itself is order p, then the subgroup generated by the elements of order p acts faithfully and transitively on the Sylow subgroups, so for small enough numbers, the subgroup can just be looked up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In all cases, we can assume the group is finite since the subgroup generated by the elements of a fixed order is finite (assuming there are only finitely many elements of that fixed order).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Easier example&lt;/strong&gt;: For instance there is no group with exactly 68 elements of order 3, since such a group would have cyclic Sylow 3-subgroups by Hall, order 3 Sylows by the counting, but then would have 34 Sylow 3-subgroups, and so (the subgroup generated by the elements of order 3 would) be a primitive group of degree 34.  One checks the list of primitive groups of degree 34 (that is, A&lt;sub&gt;34&lt;/sub&gt; and S&lt;sub&gt;34&lt;/sub&gt;, both with ginormous Sylow 3-subgroups) to see no such group exists.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One could also try 140, but the action need not be primitive so the table lookup is harder.  Such a group has Sylows of order 3, but is not solvable, so is somewhat restricted.&lt;/p&gt;&#xA;" OwnerUserId="583" LastActivityDate="2010-08-17T04:17:46.750" Title="Is there a group with exactly 92 elements of order 3?" Tags="&lt;group-theory&gt;" AnswerCount="2" CommentCount="7" FavoriteCount="3" />
  <row Id="2356" PostTypeId="1" AcceptedAnswerId="2357" CreationDate="2010-08-13T15:16:50.157" Score="11" ViewCount="657" Body="&lt;p&gt;When flipping a coin to make important decisions in life you can flip once to choose between 2 possible outcomes. (Heads I eat cake, Tails I eat chocolate!)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can also flip twice to choose between 4 outcomes. (Heads-Heads, Tails-Tails, Heads-Tails, Tails-Heads)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you use a coin to choose evenly between three possible choices? If so how?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Ignoring slight abnormalities in weight)&lt;/p&gt;&#xA;" OwnerUserId="1099" LastEditorUserId="121" LastEditorDisplayName="" LastEditDate="2010-08-14T17:59:44.437" LastActivityDate="2010-08-25T20:07:17.217" Title="Is it possible to 'split' coin flipping 3 ways?" Tags="&lt;probability&gt;&lt;fair-division&gt;" AnswerCount="5" CommentCount="3" />
  <row Id="2357" PostTypeId="2" ParentId="2356" CreationDate="2010-08-13T15:21:12.017" Score="15" ViewCount="" Body="&lt;p&gt;If you throw your coin $n$ times you have $2^n$ outcomes, the probability of each of which is $\frac{1}{2^n}$. The larger $n$ is, the better you can divide $2^n$ into three approximately equal parts:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just define $a_n=[2^n/3]$ and $b_n=[2\cdot 2^n/3]$, where $[\cdot]$ denotes rounding off (or on). Since $\frac{a_n}{2^n}\to\frac{1}{3}$ and $\frac{b_n}{2^n}\to\frac{2}{3}$ as $n\to\infty$, each of the three outcomes&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;the number of Heads is between $0$ and $a_n$&quot;,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;the number of Heads is between $a_n$ and $b_n$&quot;, and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;the number of Heads is between $b_n$ and $2^n$&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;has approximately the probability $\frac{1}{3}$.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Alternatively&lt;/em&gt;, you could apply your procedure to get four outcomes with the same probability (Heads-Heads, Tails-Tails, Heads-Tails, Tails-Heads) to your problem in the following way:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Associate the three outcomes Heads-Heads, Tails-Tails, Heads-Tails with your three possible choices. In the case that Tails-Heads occurs, just repeat the experiment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sooner or later you will find an outcome different from Tails-Heads.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Indeed, by symmetry, the probability for first Heads-Heads, first Tails-Tails, or first Heads-Tails is $\frac{1}{3}$, respectively.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;(&lt;em&gt;Alternatively&lt;/em&gt;, you could of course throw a die and select your first choice if the outcome is 1 or 2, select your second choice if the outcome is 3 or 4, and select your third choice if the outcome is 5 or 6.)&lt;/p&gt;&#xA;" OwnerUserId="367" LastEditorUserId="367" LastEditorDisplayName="" LastEditDate="2010-08-13T22:19:18.823" LastActivityDate="2010-08-13T22:19:18.823" CommentCount="4" />
  <row Id="2358" PostTypeId="2" ParentId="1533" CreationDate="2010-08-13T15:28:08.370" Score="0" ViewCount="" Body="&lt;p&gt;Mathematical Thinking: Problem-Solving and Proofs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;John P. D'Angelo, Douglas B. West.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Available at &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0130144126&quot; rel=&quot;nofollow&quot;&gt;Amazon&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is supposed to be an introduction to mathematical proofs. As such it is not restricted to discrete mathematics. But it does a very good job for discrete mathematics. You would also see some proof in real analysis; but you can focus only on the discrete part ignoring this.&lt;/p&gt;&#xA;" OwnerUserId="1119" LastActivityDate="2010-08-13T15:28:08.370" CommunityOwnedDate="2010-08-13T15:28:08.370" />
  <row Id="2360" PostTypeId="1" AcceptedAnswerId="2362" CreationDate="2010-08-13T15:40:04.273" Score="3" ViewCount="537" Body="&lt;p&gt;I'm looking for the equation of a family of functions that roughly resembles the sketch below (with apologies for the crudeness of said sketch):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://img202.imageshack.us/img202/6907/functionsketch.png&quot; /&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Properties I'm looking for:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\lim_{x\to-\infty}f(x)=y_1$ (i.e. approaches asymptote y=y&lt;sub&gt;1&lt;/sub&gt;)&lt;/li&gt;&#xA;&lt;li&gt;$\lim_{x\to+\infty}f(x)=y_2$ (i.e. approaches asymptote y=y&lt;sub&gt;2&lt;/sub&gt;)&lt;/li&gt;&#xA;&lt;li&gt;$|f'(x)|$ is at a maximum at &lt;em&gt;x&lt;/em&gt;=0&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I remember something involving e&lt;sup&gt;&lt;em&gt;x&lt;/em&gt;&lt;/sup&gt;/(??), but I can't remember exactly and can't seem to come up with the formula. I want to use it to weight tags in a tag cloud (i.e. the more recently a tag was used (smaller &lt;em&gt;x&lt;/em&gt;), the more weight (larger &lt;em&gt;y&lt;/em&gt;) the tag gets). The whole graph would be shifted up and to the right of what is shown in the sketch, but I know how to do that part once I have a formula.&lt;/p&gt;&#xA;" OwnerUserId="477" LastEditorUserId="477" LastEditorDisplayName="" LastEditDate="2010-08-13T15:45:40.687" LastActivityDate="2010-08-13T17:03:56.273" Title="Family of functions with two horizontal asymptotes" Tags="&lt;functions&gt;&lt;asymptotics&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="2361" PostTypeId="2" ParentId="2356" CreationDate="2010-08-13T15:43:15.317" Score="3" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt; Oh dear, Rasmus has extended his answer and rendered this one obsolete!  In fact, stop reading this right now and go see what he has done.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I interpret what you are asking as trying to find a way to decide without introducing bias (as would be introduced by counting tails-heads as the same as heads-tails).  &lt;a href=&quot;http://math.stackexchange.com/questions/2356/is-it-possible-to-split-coin-flipping-3-ways/2357#2357&quot;&gt;Rasmus's suggestion&lt;/a&gt; of repeating the experiment for a certain configuration seems the best choice.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I drew a little tree and tried to group outcomes into three &quot;equally-likely&quot; sets and then realized this is impossible because $3$ does not divide $2^n$ for any $n$.  (The quantity $2^n$ being the number of unique outcomes after $n$ &quot;flips&quot;.)&lt;/p&gt;&#xA;" OwnerUserId="121" LastActivityDate="2010-08-13T15:43:15.317" />
  <row Id="2362" PostTypeId="2" ParentId="2360" CreationDate="2010-08-13T15:47:35.490" Score="6" ViewCount="" Body="&lt;p&gt;I think what you're looking for is a form of &lt;a href=&quot;http://en.wikipedia.org/wiki/Logistic_function&quot; rel=&quot;nofollow&quot;&gt;logistic function&lt;/a&gt;, such as $$f(x)=\frac{2}{1+e^x}-1.$$&#xA;&lt;img src=&quot;http://i.imgur.com/gyurj.png&quot; alt=&quot;graph&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;edit&lt;/em&gt;: For your specific criteria with $y_1$ and $y_2$: $$f(x)=\frac{y_1-y_2}{1+e^x}+y_2.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;edit 2&lt;/em&gt;: For comparison of my answer to the other two answers:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/WHMcb.png&quot; alt=&quot;comparison graph&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;dashed/black: $\frac{2}{1+e^x}-1$; blue: $-\frac{2}{\pi}\arctan x$; red: $-\frac{x}{\sqrt{x^2+1}}$&lt;/p&gt;&#xA;" OwnerUserId="72" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-08-13T17:03:56.273" LastActivityDate="2010-08-13T17:03:56.273" />
  <row Id="2363" PostTypeId="2" ParentId="2356" CreationDate="2010-08-13T15:50:05.550" Score="1" ViewCount="" Body="&lt;p&gt;If you are at a crossroads in life requiring you to choose one of three options, then coin tossing is not the correct thing to do.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rather, pick up a dice and roll it. Go for first option if the diece turns up 1 or 2. Go for second option  if the dice turns up 3 or 4. Go for the third option if the dice turns up 5 or 6.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: Oops, I saw that Rasmus already gave this option. Anyway I strongly suggest that you prefer this one to the other two methods given by him.&lt;/p&gt;&#xA;" OwnerUserId="1119" LastActivityDate="2010-08-13T15:50:05.550" CommentCount="5" />
  <row Id="2364" PostTypeId="2" ParentId="2360" CreationDate="2010-08-13T15:54:38.450" Score="5" ViewCount="" Body="&lt;p&gt;Just take $- tan^{-1} x$. Vary it appropriately for additional constraints.&lt;/p&gt;&#xA;" OwnerUserId="1119" LastActivityDate="2010-08-13T15:54:38.450" CommentCount="1" />
  <row Id="2366" PostTypeId="2" ParentId="2360" CreationDate="2010-08-13T16:36:17.043" Score="4" ViewCount="" Body="&lt;p&gt;Another famous family of functions that behave as you describe is those of form $y=\dfrac{x}{\sqrt{x^2+1}}$. (This function is actually the sine of the arctan function &lt;a href=&quot;http://math.stackexchange.com/questions/2360/family-of-functions-with-two-horizontal-asymptotes/2364#2364&quot;&gt;George suggested&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Graph of $y=-\dfrac{x}{\sqrt{x^2+1}}$:&lt;br&gt;&#xA;&lt;img src=&quot;http://i.imgur.com/JBNSo.gif&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a general y&lt;sub&gt;1&lt;/sub&gt; and y&lt;sub&gt;2&lt;/sub&gt;, the formula would be $y=-\dfrac{y_1-y_2}{2}*\dfrac{x}{\sqrt{x^2+1}}+\dfrac{y_1+y_2}{2}$&lt;/p&gt;&#xA;" OwnerUserId="73" LastActivityDate="2010-08-13T16:36:17.043" />
  <row Id="2367" PostTypeId="1" AcceptedAnswerId="2410" CreationDate="2010-08-13T16:44:35.100" Score="2" ViewCount="263" Body="&lt;p&gt;How does one prove that ring of Hamilton Quaternions with coefficients coming from the field $\mathbb{Z}/p\mathbb{Z}$ is not a divison ring.&lt;/p&gt;&#xA;" LastActivityDate="2010-08-13T20:58:40.043" Title="A question on Hamilton Quaternions." Tags="&lt;abstract-algebra&gt;&lt;ring-theory&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="2368" PostTypeId="2" ParentId="2367" CreationDate="2010-08-13T16:53:58.287" Score="2" ViewCount="" Body="&lt;p&gt;The sledgehammer is to invoke &lt;a href=&quot;http://en.wikipedia.org/wiki/Wedderburn%27s_little_theorem&quot; rel=&quot;nofollow&quot;&gt;Wedderburn's theorem&lt;/a&gt;&#xA;to the effect that every finite division ring is commutative.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a slightly lighter sledgehammer invoke &lt;a href=&quot;http://en.wikipedia.org/wiki/Four_square_theorem&quot; rel=&quot;nofollow&quot;&gt;Legendre's four&#xA;square theorem&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="226" LastEditorUserId="226" LastEditorDisplayName="" LastEditDate="2010-08-13T17:15:27.457" LastActivityDate="2010-08-13T17:15:27.457" CommentCount="3" />
  <row Id="2369" PostTypeId="1" AcceptedAnswerId="2371" CreationDate="2010-08-13T16:57:34.710" Score="0" ViewCount="106" Body="&lt;p&gt;I heard that one can construct field based on $p^n$ elements where $p$ is prime. I tried with $p = n = 2$. It seemed easy as there are 2 groups which $\{0,1,2,3\}$ can form and 1 group formed by $\{1,2,3\}$. However each time I get into contradiction. Is there anything I missed or it is not possible to form $p^n$ element group?&lt;/p&gt;&#xA;" OwnerUserId="728" LastActivityDate="2010-08-13T17:08:52.320" Title="Construction of $p^n$ field" Tags="&lt;set-theory&gt;&lt;prime-numbers&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="2370" PostTypeId="2" ParentId="2369" CreationDate="2010-08-13T17:02:01.893" Score="0" ViewCount="" Body="&lt;p&gt;Did you refer the wikipedia article for constructing fields. Please refer the section of explicitly constructing finite fields. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Finite_field&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Finite_field&lt;/a&gt;. Or if you want to be more advance you can refer this also: www.emis.de/journals/IJOPCM/Vol/09/IJOPCM(vol.2.3.1.S.9).pdf&lt;/p&gt;&#xA;" LastActivityDate="2010-08-13T17:02:01.893" />
  <row Id="2371" PostTypeId="2" ParentId="2369" CreationDate="2010-08-13T17:04:42.777" Score="3" ViewCount="" Body="&lt;p&gt;There are finite fields of order $p^n$ for all prime powers $p^n$. They&#xA;are often called &lt;a href=&quot;http://en.wikipedia.org/wiki/Galois_field&quot; rel=&quot;nofollow&quot;&gt;Galois fields&lt;/a&gt;. But their additive groups&#xA;are not cyclic (when $n\ne 1$). So for order $4$ one can have an&#xA;addition table&#xA;$$\begin{array}{cccc}&#xA;0&amp;amp;1&amp;amp;a&amp;amp;b\\&#xA;1&amp;amp;0&amp;amp;b&amp;amp;a\\&#xA;a&amp;amp;b&amp;amp;0&amp;amp;1\\&#xA;b&amp;amp;a&amp;amp;1&amp;amp;0&#xA;\end{array}$$&#xA;but not an addition table&#xA;$$\begin{array}{cccc}&#xA;0&amp;amp;1&amp;amp;2&amp;amp;3\\&#xA;1&amp;amp;2&amp;amp;3&amp;amp;0\\&#xA;2&amp;amp;3&amp;amp;0&amp;amp;1\\&#xA;3&amp;amp;0&amp;amp;1&amp;amp;2&#xA;\end{array}.$$&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-08-13T17:04:42.777" />
  <row Id="2372" PostTypeId="1" AcceptedAnswerId="2500" CreationDate="2010-08-13T17:07:54.007" Score="6" ViewCount="590" Body="&lt;p&gt;How can I find the number of &lt;em&gt;k&lt;/em&gt;-permutations of &lt;em&gt;n&lt;/em&gt; objects, where there are &lt;em&gt;x&lt;/em&gt; types of objects, and r&lt;sub&gt;1&lt;/sub&gt;, r&lt;sub&gt;2&lt;/sub&gt;, r&lt;sub&gt;3&lt;/sub&gt; ... r&lt;sub&gt;x&lt;/sub&gt; give the number of each type of object?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I have 20 letters from the alphabet. There are some duplicates - 4 of them are &lt;em&gt;a&lt;/em&gt;, 5 of them are &lt;em&gt;b&lt;/em&gt;, 8 of them are &lt;em&gt;c&lt;/em&gt;, and 3 are &lt;em&gt;d&lt;/em&gt;. How many unique 15-letter permutations can I make?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In the example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;n = 20&lt;br&gt;&#xA;k = 15&lt;br&gt;&#xA;x = 4&lt;br&gt;&#xA;r&lt;sub&gt;1&lt;/sub&gt; = 4 | r&lt;sub&gt;2&lt;/sub&gt; = 5 | r&lt;sub&gt;3&lt;/sub&gt; = 8 | r&lt;sub&gt;4&lt;/sub&gt; = 3&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some background...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was originally going to solve this problem in order to solve a simpler problem, but I managed to find a simpler solution to that problem instead. Now I'm still looking for the solution to this more general problem out of interest.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've done some more work on this problem but haven't really come up with anything useful. Intuition tells me that as Douglas suggests below there will probably not be an easy solution. However, I haven't been able to prove that for sure - does anyone else have any ideas?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've now re-asked this question (&lt;a href=&quot;http://mathoverflow.net/questions/37211/permutations-with-identical-objects&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;) on MO.&lt;/p&gt;&#xA;" OwnerUserId="181" LastEditorUserId="181" LastEditorDisplayName="" LastEditDate="2010-08-31T03:20:40.913" LastActivityDate="2010-08-31T06:32:02.473" Title="Permutations with identical objects" Tags="&lt;combinatorics&gt;" AnswerCount="2" CommentCount="5" FavoriteCount="2" />
  <row Id="2373" PostTypeId="2" ParentId="2369" CreationDate="2010-08-13T17:08:52.320" Score="2" ViewCount="" Body="&lt;p&gt;First, any field of order $p^n$ will have characteristic $p$, so the underlying &lt;em&gt;additive&lt;/em&gt; structure of the group is necessarily $(\mathbb{Z}_p)^n$; for order $2^2$, it follows that the additive structure has to be the Klein $4$-group (it cannot be cyclic of order $4$). Second, the multiplicative group of nonzero elements of any finite field is necessarily &lt;em&gt;cyclic&lt;/em&gt;, so the multiplicative structure will always be isomorphic to $\mathbb{Z}_{p^n-1}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That said, trying to fit together the two structures abstractly is likely to prove very hard. Instead, the easiest way to construct such a field is as an extension of degree $n$ of the field $\mathbb{F}_p$ of $p$ elements. This is achieved by finding a monic irreducible polynomial of degree $n$ with coefficients in $\mathbb{F}_p$, call it $f(x)$, and considering the quotient $\mathbb{F}_p[x]/(f(x))$, where $\mathbb{F}_p[x]$ is the ring of polynomials with coefficients in $\mathbb{F}_p$, and $(f(x))$ is the ideal of all multiples of $f(x)$. Arithmetic of this field is easily computed using polynomial multiplication and long division (inverses can be found using the Euclidean algorithm). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It may not be immediately obvious that there is always a monic irreducible polynomial of degree $n$ over $\mathbb{F}_p$ for all primes $p$ and all positive integers $n$; but you can deduce that this is the case by showing that there in fact is a (unique up to isomorphism) field of order $p^n$ (consider the splitting field of $x^{p^n}-x$ over $\mathbb{F}_p$), and then showing that if $\alpha$ is a generator for the multiplicative group of this field, then the field is given by $\mathbb{F}_p(\alpha)$. General theory of field extensions then tells you the monic irreducible of $\alpha$ will do.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-08-13T17:08:52.320" />
  <row Id="2374" PostTypeId="1" CreationDate="2010-08-13T17:33:04.730" Score="4" ViewCount="182" Body="&lt;p&gt;I am learning the axioms of Zermelo-Fraenkel (ZF) set theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One axiom schema basically says that given any set S and any formula phi(x), there is a set T consisting of all those elements x of S such that phi(x). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I find this axiom schema unsatisfying because it only guarantees that subsets of S &lt;em&gt;definable by a formula&lt;/em&gt; are really sets.  It's like saying that a function from X to Y only exists if you can write down a formula for it rather than just allowing arbitrary single-valued subsets of X cross Y.  Is there a way in the language of ZF to say &quot;given a set S, if T is a subset of S then T is a set?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Related is the power set axiom.  Given a set S, there is a set P(S) consisting of exactly &lt;em&gt;all&lt;/em&gt; the subsets of S.  But in ZF the objects of the theory are sets (no urelements).  Everything is a set.  So, the elements of P(S) are all sets.  Doesn't this mean that &lt;em&gt;every&lt;/em&gt; subset of a given set S is a set?  If so, why the need for the subset axioms?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can anticipate some difficulty.  I want to say that &quot;if x is a member of P(S) then x is a set,&quot; but I cannot express the predicate &quot;is a set&quot; in the language of set theory, and strangely enough there is no need to since everything is a set!  I attempted:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;(for all S)(for all x)[x subset S --&gt; (there exists y)(x = y)]&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where &quot;x subset S&quot; is an abbreviation for &quot;(for all z)[z in x --&gt; z in S].&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But this attempt is silly because when I say &quot;for all x,&quot; x is automatically a set and there is no need to say it is a set.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm confused.  &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Are the subset axioms necessary?&lt;/li&gt;&#xA;&lt;li&gt;Is there a way to remove reference to a defining formula so that &lt;em&gt;every&lt;/em&gt; subset of a given set is a set?&lt;/li&gt;&#xA;&lt;li&gt;If the answer to 1 is &quot;yes&quot; and the answer to 2 is &quot;no&quot; then why is set theory so weird as to allow only definable subsets on the one hand and yet allow for a set of all subsets on the other?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" LastActivityDate="2010-08-13T19:15:25.333" Title="Is the Subset Axiom Schema in ZF necessary?" Tags="&lt;set-theory&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="1" />
  <row Id="2375" PostTypeId="1" AcceptedAnswerId="34744" CreationDate="2010-08-13T17:47:57.870" Score="1" ViewCount="85" Body="&lt;p&gt;Let U be a unitary transformation from Hilbert space H to Hilbert space K.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do you call a *-homomorphism f from B(H) to B(K), defined by $f(a) = U a U^{-1}$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm interested both in a symbol, which can be used in formulas, and a name for it, which can be used in texts or in speech.&lt;/p&gt;&#xA;" OwnerUserId="695" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-08-18T04:53:28.340" LastActivityDate="2011-04-23T17:59:15.553" Title="Notation of homeomorphism from B(H) to B(K), corresponding to unitary transformation of Hilbert spaces." Tags="&lt;abstract-algebra&gt;&lt;terminology&gt;&lt;notation&gt;&lt;functional-analysis&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="2376" PostTypeId="2" ParentId="2346" CreationDate="2010-08-13T17:48:06.590" Score="2" ViewCount="" Body="&lt;p&gt;Define $g$ on $[2,2^{k}]$ by $(g(2))^{k}=g(2^k)$.  Then using the functional equation you can extend to an $f$ on all numbers $\geq 1$. We can then use the similar trick for $(0,1)$ and for the negative numbers as well.&lt;/p&gt;&#xA;" LastActivityDate="2010-08-13T17:48:06.590" />
  <row Id="2377" PostTypeId="2" ParentId="2375" CreationDate="2010-08-13T17:52:01.777" Score="1" ViewCount="" Body="&lt;p&gt;If $H=K$ and $U$ is a unitary, then $f$ is called an &lt;a href=&quot;http://en.wikipedia.org/wiki/Inner_automorphism#Ring_case&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;inner automorphism&lt;/em&gt;&lt;/a&gt; of $\mathbb B(H)$,&#xA;because it is given by conjugation with a unitary element of your algebra $\mathbb B(H)$ (these are the &quot;obviously existing automorphism&quot;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some notation (for the group case) is proposed &lt;a href=&quot;http://en.wikipedia.org/wiki/Inner_automorphism#Notation&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;. At least for rings this notation isn't standard, however.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The above naming doesn't nicely generalise to your more general situation and I don't think there's a common name for isomorphisms given by conjugation with a unitary transformation.&lt;/p&gt;&#xA;" OwnerUserId="367" LastEditorUserId="367" LastEditorDisplayName="" LastEditDate="2010-08-15T14:23:40.163" LastActivityDate="2010-08-15T14:23:40.163" CommentCount="2" />
  <row Id="2378" PostTypeId="1" AcceptedAnswerId="2379" CreationDate="2010-08-13T18:03:12.893" Score="0" ViewCount="131" Body="&lt;p&gt;How did the definition of Algebraic objects like group, ring and field come up? When groups were first introduced, were they given the 4 axioms as we give now. And what made Mathematicians to think of something like this.&lt;/p&gt;&#xA;" LastActivityDate="2010-08-21T05:44:06.157" CommunityOwnedDate="2010-08-13T18:42:59.240" Title="Definition of an Algebraic Objects" Tags="&lt;soft-question&gt;&lt;math-history&gt;" AnswerCount="4" CommentCount="2" />
  <row Id="2379" PostTypeId="2" ParentId="2378" CreationDate="2010-08-13T18:18:34.250" Score="3" ViewCount="" Body="&lt;p&gt;I'm certainly no historian, so let me give my impression of the history in a CW answer and others can improve on this answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Groups were understood as permutations of objects long before they were understood as abstract groups, just as manifolds were understood as nice subsets of $\mathbb{R}^n$ long before they were understood as abstract manifolds.  Most notably, Galois worked with permutations of the roots of a polynomial in order to understand the solvability of polynomials by radicals.  Similarly, number theorists like Kummer worked with concrete examples of number rings and fields.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some of the people responsible for the spread of the abstract point of view are (in no particular order) Hilbert, Noether, Artin, van der Waerden, (Dedekind?), .... together they more or less determined the abstract algebra curriculum as we now understand it.  Their reasons for doing everything axiomatically include that the proofs are shorter and more powerful this way; in particular, algebraic geometry needed rigorous tools and commutative algebra provided them (and Zariski used them).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One big payoff of the abstract point of view is that people now recognized fundamental groups and homology groups in topology as groups.  Before the abstract study of groups these objects, while defined in some sense, weren't recognized as groups because they didn't permute anything.  &lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-08-13T18:18:34.250" CommunityOwnedDate="2010-08-13T18:18:34.250" CommentCount="2" />
  <row Id="2380" PostTypeId="2" ParentId="2250" CreationDate="2010-08-13T18:38:49.407" Score="2" ViewCount="" Body="&lt;p&gt;The sum in question is at most ε&lt;sup&gt;2&lt;/sup&gt;.  (We do not need the condition that the row sum equals 1 or the condition f(i)≠g(i) to obtain this.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof&lt;/strong&gt;.  Since&#xA;$$\sum_{k=1}^n(a_{f(k)j}-a_{g(k)j})=\sum_{k=1}^na_{f(k)j}-\sum_{k=1}^na_{g(k)j}=0,$$&#xA;we have&#xA;$$\left|\sum_{k=1}^na_{ki}(a_{f(k)j}-a_{g(k)j})\right|&#xA;=\left|\sum_{k=1}^n(a_{ki}-a_{1i})(a_{f(k)j}-a_{g(k)j})\right|$$&#xA;$$\le\sum_{k=1}^n|a_{ki}-a_{1i}||a_{f(k)j}-a_{g(k)j}|.$$&#xA;Therefore, the sum in question is at most&#xA;$$\frac1n\sum_{i,j=1}^z\sum_{k=1}^n|a_{ki}-a_{1i}||a_{f(k)j}-a_{g(k)j}|&#xA;=\frac1n\sum_{k=1}^n\left(\sum_{i=1}^z|a_{ki}-a_{1i}|\right)\left(\sum_{j=1}^z|a_{f(k)j}-a_{g(k)j}|\right)$$&#xA;$$\le\frac1n\sum_{k=1}^n\epsilon^2=\epsilon^2.$$&lt;/p&gt;&#xA;" OwnerUserId="926" LastEditorUserId="926" LastEditorDisplayName="" LastEditDate="2010-08-13T19:09:27.280" LastActivityDate="2010-08-13T19:09:27.280" CommentCount="1" />
  <row Id="2381" PostTypeId="2" ParentId="2330" CreationDate="2010-08-13T18:49:53.220" Score="1" ViewCount="" Body="&lt;p&gt;&lt;em&gt;OP's &lt;a href=&quot;http://stackoverflow.com/questions/3474961/find-minimum-of-a-function-defined-by-integration-in-mathematica&quot;&gt;equivalent question on Stack Overflow&lt;/a&gt; was already answered. This is &lt;a href=&quot;http://stackoverflow.com/questions/3474961/find-minimum-of-a-function-defined-by-integration-in-mathematica/3475061#3475061&quot;&gt;the accepted answer by Andrew Moylan&lt;/a&gt;.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Try this:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;In[58]:= g[t_, x_] := t^3 - t + x^2&#xA;&#xA;In[59]:= f[t_?NumericQ] := NIntegrate[g[t, x], {x, -1, 1}]&#xA;&#xA;In[60]:= FindMinimum[f[t], {t, 1}]&#xA;&#xA;Out[60]= {-0.103134, {t -&amp;gt; 0.57735}}&#xA;&#xA;In[61]:= Plot[f[t], {t, 0, 1}]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Two relevant changes I made to your code:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Define f with &lt;code&gt;:=&lt;/code&gt; instead of with &lt;code&gt;=&lt;/code&gt;. This effectively gives a definition for f &quot;later&quot;, when the user of f has supplied the values of the arguments. See &lt;a href=&quot;http://reference.wolfram.com/mathematica/ref/SetDelayed.html&quot; rel=&quot;nofollow&quot;&gt;SetDelayed&lt;/a&gt;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Define f with &lt;code&gt;t_?NumericQ&lt;/code&gt; instead of &lt;code&gt;t_&lt;/code&gt;. This says, t can be anything numeric (Pi, 7, 0, etc). But not anything non-numeric (t, x, &quot;foo&quot;, etc).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="171" LastActivityDate="2010-08-13T18:49:53.220" CommunityOwnedDate="2010-08-13T18:49:53.220" />
  <row Id="2382" PostTypeId="2" ParentId="2374" CreationDate="2010-08-13T18:51:15.450" Score="4" ViewCount="" Body="&lt;p&gt;The subtle point you are speaking about is exactly the difference between first order logic and second order logic. In first order logic, we quantify only over objects in the universe. In second order logic, we can quantify also over subcollections of the universe. ZFC is a first order theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the ZFC subset axiom is &lt;em&gt;not&lt;/em&gt; saying that every subset of a set is describable in the way you mention. Rather, it is making the much milder claim that if we can describe a subset, then it is a set. And this seems totally unobjectionable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You evidently want to consider a stronger version of the axiom, which says somehow that all the subcollections that exist anywhere are also sets. This is a second order assertion, since it is quantifying over subsets of the universe. This second order subset axiom, however, is not expressible in first order logic. The reason is that any model with an infinite set, satisfying the second-order subset axiom would have to uncountable, since it would truly contain all the subsets of its sets. Thus, there could be no countable models of your theory, which would violate the Lowenheim Skolem theorem if the axiom were expressible in first order logic. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nevertheless, it is natural to look into the possibilities of second order ZFC, and this is studied in set theory. It turns out that the models of your second order ZFC (using second order Replacement also) are exactly the same as the universes known as $H_\kappa$ for an inaccessible cardinal $\kappa$, one of the large cardinal notions. These models are also known as the Grothendieck universes, and are used throughout  category theory as a convenient way of formalizing the large/small distinction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It might be a good exercise for a beginner in this area to prove that if $\kappa$ is a (strongly) inaccessible cardinal, then the collection $H_\kappa$ of sets whose transitive closure has size less than $\kappa$ is a model of second order ZFC. When $\kappa$ is inaccessible, then this universe $H_\kappa$ is the same as $V_\kappa$ in the Levy cumulative hierarchy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, let me remark that some might find it to be somewhat incoherent to propose a second-order theory as the main axiomatization of set theory. The purpose of the axiomatization, after all, is to set forth certain minimal bedrock principles that the set concept should obey, but if we make these principles second-order, then we would seem to need a prior concept of set to interpret them. That is, a second order logic only makes sense in the context of a larger set-theoretic background. So it wouldn't seem to work foundationally to provide such an axiomatization as the principal axiomatization of set theory.&lt;/p&gt;&#xA;" OwnerUserId="413" LastEditorUserId="413" LastEditorDisplayName="" LastEditDate="2010-08-13T19:15:25.333" LastActivityDate="2010-08-13T19:15:25.333" CommentCount="2" />
  <row Id="2383" PostTypeId="2" ParentId="2374" CreationDate="2010-08-13T18:52:50.937" Score="1" ViewCount="" Body="&lt;p&gt;So, maybe I should write this as a response. So, let's take the notions in order. First, I will reserve the word &quot;set&quot; for objects in the theory, and use &quot;collection&quot; for the more intuitive notion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that &quot;$A$ is a subset of $B$&quot; is an abbreviation for two statements: (i) $A$ is a set; and (ii) for every $x$, if $x\in A$ then $x\in B$. So &quot;if $A$ is a subset of $B$ then $A$ is a set&quot; is a tautology of the form $P\wedge Q\Rightarrow P$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As user80 points out, you are misinterpreting the Subset Axiom schema slightly; it does not state that &lt;em&gt;only&lt;/em&gt; those subcollections which you express as ${a\in A: \phi(a)}$ are sets (it is not an &quot;if and only if&quot; statement). It 'merely' states that all those are sets, leaving open that perhaps there may be &quot;other&quot; subcollections that may also be sets. So in your analogy, it would be like saying that if you can write down a formula for a function from $X$ to $Y$ then the result will &lt;em&gt;certainly&lt;/em&gt; be a function, but it does not tell you that you get a function &lt;em&gt;only&lt;/em&gt; if you can write down a formula: just that writing down a formula is &lt;em&gt;a&lt;/em&gt; way of getting a function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It it &quot;necessary&quot;? Well, it is necessary if you wish to be able to construct new sets by specifying certain elements from something that you already know to be a set. Otherwise, you'll have no warrant for saying that such objects are sets, only that they are collections. The Axiom of Separation/Subset is about one way to take a set, pare it down, and get a set.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As to the axiom of the power set, you are again misinterpreting it. It's not that the axiom of the power set says that subsets of S are sets; that's already in the notion of &quot;subset&quot;. Rather, the Power Set Axiom tells you that you have a set whose elements &lt;em&gt;are&lt;/em&gt; the subsets of your given set $S$. It guarantees the existence of a set which is &quot;one level up&quot; from $S$, in the sense that its elements are subsets of $S$. From the Axiom of Pairs and other issues it is not hard to show that if you have a finite collection of things that you know are sets, then there is a set whose elements are exactly those in your original collection (that is, it allows you to pull together those things and put them in a set). But what if you have an infinite number of things that you know are subsets of $A$? In order to make sure you have a set that contains all of them as elements, you need something beyond knowing that they are sets; that's where the Axiom of the Power Set comes in: it gives you an overset (which is a set) in which all these objects will be elements. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How does Power Set interact with Separation/Subset? Anything you obtain using Separation will necessarily be an element of the Power Set. It does not say that those are the only things in the power set, just that those are &lt;em&gt;guaranteed&lt;/em&gt; to be in the Power Set.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-08-13T18:52:50.937" CommentCount="3" />
  <row Id="2384" PostTypeId="1" CreationDate="2010-08-13T18:58:45.980" Score="9" ViewCount="327" Body="&lt;p&gt;I am tutoring several talented students, middle school level and early high school level, in mathematics. I am always looking for new sources from which to draw questions. Can anyone recommend books, web-sites, etc. with a interesting questions?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know of the ArtofProblemSolving.com. I am currently using the books &quot;Challenging Problems in Algebra&quot; and &quot;Challenging Problems in Geometry&quot; by Posamentier and Salkind.&lt;/p&gt;&#xA;" OwnerUserId="1151" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-08-23T01:33:43.623" LastActivityDate="2010-11-16T09:32:51.477" CommunityOwnedDate="2010-08-13T19:38:28.320" Title="Sources of problems for teaching/tutoring young mathematicians." Tags="&lt;education&gt;&lt;teaching&gt;&lt;high-school&gt;" AnswerCount="9" FavoriteCount="8" />
  <row Id="2385" PostTypeId="2" ParentId="2384" CreationDate="2010-08-13T19:00:54.273" Score="1" ViewCount="" Body="&lt;p&gt;Try Yakov Perelman, &quot;Mathematics can be Fun&quot;.&lt;/p&gt;&#xA;" OwnerUserId="1119" LastActivityDate="2010-08-13T19:00:54.273" CommunityOwnedDate="2010-08-13T19:38:28.320" CommentCount="1" />
  <row Id="2386" PostTypeId="2" ParentId="2384" CreationDate="2010-08-13T19:04:25.593" Score="0" ViewCount="" Body="&lt;p&gt;There is this &quot;Math Help Forum&quot; which discusses about high school level and undergraduate level mathematics. You can find lots and lots of things. Here is one problem website : &lt;a href=&quot;http://amc.maa.org/e-exams/e8-usamo/usamo.shtml&quot; rel=&quot;nofollow&quot;&gt;http://amc.maa.org/e-exams/e8-usamo/usamo.shtml&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Calculus by Thomas finney is a Classic book for teaching Calculus.&lt;/p&gt;&#xA;" LastActivityDate="2010-08-13T19:04:25.593" CommunityOwnedDate="2010-08-13T19:38:28.320" CommentCount="1" />
  <row Id="2388" PostTypeId="2" ParentId="2346" CreationDate="2010-08-13T19:14:21.933" Score="3" ViewCount="" Body="&lt;p&gt;Case $k=1$ is trivial. So, suppose $k&gt;1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose k is odd. For even k one should correct this solution a bit.&#xA;First note, that $a^k=a$ is equivalent to $a\in{-1,0,1}$. We have this equation for a=f(-1), a=f(0), a=f(1).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that for $x\in (-1,1)$ sequence $x,x^k,(x^k)^k,\dots$ goes to 0. Denote this sequence by $b_n$: $b_n = x^{(k^n)}$. From continuity of f we know, that $f(b_n)\rightarrow f(0)$ and $(f(x))^{(k^n)}\rightarrow f(0)$. Therefore for $x\in(-1,1)$ we know, that $f(x) \in (-1,1)$ or f is constant on this interval and equal to 1 or -1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now suppose x is in $(0,\infty)$. By the same reason the sequence $(f(x))^{(k^{-n})}$ goes to f(1). It is true iff  (f(x)=0 and f(1)=0) or (f(x) is in $(-\infty,0)$ and f(1)=-1) or (f(x) is in $(0,\infty)$ and f(1)=1). The same is for x in $(-\infty,0)$ and f(-1).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we have the following cases&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;f(0)=1, then f(x)=1 for $x \in&#xA;[-1,1]$. Other values of f can be&#xA;uniquely determined by values on&#xA;$[-2^k,-2]\cup[2,2^k]$. This values can be chosen arbitrary to form continuous function&lt;br&gt;&#xA;to $(0,\infty)$ such that&#xA;$f(-2^k)=(f(-2))^k$ and&#xA;$f(2^k)=(f(2))^k$.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Now I will just list all other cases&#xA;    without going to details, which are&#xA;    similar to shown in the first case.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;f(0)=-1&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;f(0)=0,f(-1)=1, f(1)=1&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;f(0)=0,f(-1)=1, f(1)=-1&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;f(0)=0,f(-1)=1, f(1)=0&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;f(0)=0,f(-1)=-1, f(1)=1&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;f(0)=0,f(-1)=-1, f(1)=-1&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;f(0)=0,f(-1)=-1, f(1)=0&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;f(0)=0,f(-1)=1, f(1)=1&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;f(0)=0,f(-1)=1, f(1)=-1&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;f(0)=0,f(-1)=1, f(1)=0&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="695" LastActivityDate="2010-09-02T08:01:05.733" />
  <row Id="2389" PostTypeId="1" CreationDate="2010-08-13T19:28:36.330" Score="1" ViewCount="100" Body="&lt;p&gt;How do you calculate the following summation? $\sum_{i=k}^n i$&lt;/p&gt;&#xA;" OwnerUserId="1153" LastActivityDate="2010-08-14T19:08:20.490" Title="Calculate integer summation when lower bound is a variable" Tags="&lt;sequences-and-series&gt;&lt;algorithms&gt;&lt;computer-science&gt;" AnswerCount="3" />
  <row Id="2390" PostTypeId="2" ParentId="2389" CreationDate="2010-08-13T19:31:10.833" Score="4" ViewCount="" Body="&lt;p&gt;This is a special case for the formula for the sum of n terms in an arithmetic progression:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Arithmetic_progression&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Arithmetic_progression&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1119" LastActivityDate="2010-08-13T19:31:10.833" CommentCount="1" />
  <row Id="2391" PostTypeId="2" ParentId="2389" CreationDate="2010-08-13T19:38:36.243" Score="1" ViewCount="" Body="&lt;p&gt;Okay. I was able to find it here, thanks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Summation#Some_summations_of_polynomial_expressions&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Summation#Some_summations_of_polynomial_expressions&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1153" LastActivityDate="2010-08-13T19:38:36.243" />
  <row Id="2392" PostTypeId="1" AcceptedAnswerId="2394" CreationDate="2010-08-13T19:41:56.033" Score="1" ViewCount="125" Body="&lt;p&gt;Familiar Exercise: Suppose $G$ is a finite group and $T$ is an automorphism of $G$ which sends more than three quarters of elements of $G$ onto their inverses, then prove that $G$ is abelian.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The group of Quarternions is an example of a group which sends exactly $\displaystyle \frac{3}{4}$ elements onto their inverses. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a finite group with an automorphism $T$ which sends exactly $\displaystyle\frac{4}{5}$ of elements of $G$ onto their inverses?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly can we find groups with Automorphism $T$ which sends exactly $\displaystyle \frac{n}{n+1}$ of elements of $G$ onto their inverses.&lt;/p&gt;&#xA;" LastActivityDate="2010-08-13T20:20:53.380" Title="Finding groups with given Automorphism" Tags="&lt;group-theory&gt;&lt;abstract-algebra&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="2393" PostTypeId="2" ParentId="2384" CreationDate="2010-08-13T19:50:43.110" Score="2" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://www.mathcounts.org/&quot; rel=&quot;nofollow&quot;&gt;Mathcounts&lt;/a&gt; is aimed at talented middle school students. If you are affiliated with a school, you can register as a coach with them, and they can provide you with many helpful resources. If not, they still have problem sets available for sale from their catalog, as well as coaching handbooks, which include solutions and teaching advice.&lt;/p&gt;&#xA;" OwnerUserId="73" LastActivityDate="2010-08-13T19:50:43.110" CommunityOwnedDate="2010-08-13T19:50:43.110" CommentCount="1" />
  <row Id="2394" PostTypeId="2" ParentId="2392" CreationDate="2010-08-13T19:55:32.613" Score="4" ViewCount="" Body="&lt;p&gt;Prove that&#xA;if $G$ is abelian and $\phi$ is an automorphism of $G$&#xA;then $\{g\in G : \phi(g)=g^{-1} \}$&#xA;is a subgroup of $G$. &lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-08-13T19:55:32.613" CommentCount="2" />
  <row Id="2395" PostTypeId="1" CreationDate="2010-08-13T19:57:38.023" Score="4" ViewCount="803" Body="&lt;p&gt;I've already taken my calculus sequence and I'm interested in brushing up and staying sharp on the basics. So far, my calculus background is limited to single-variable calculus, which I applied in my physics sequence and engineering statistics. I learned a little bit of vector math in physics as well, but I'm not strong in it. My mathematical strengths lie in the discrete mathematics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm particularly interested in studying for my IEEE Certified Software Development Associate exam - 10% of the exam is mathematics (calculus, differential equations, and statistics). Calculus is one of the mathematics knowledge areas that a software engineer should be competent in, so I'm looking for not only solidifying the courses that I've already taken, but moving forward with slightly more advanced topics, at an undergraduate mathematics level.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm currently using the textbook from my calculus sequence, but I'm interested to know what books other people used to learn calculus and if the book is potentially worth checking out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What books are available to help me?&lt;/p&gt;&#xA;" OwnerUserId="91" LastEditorUserId="91" LastEditorDisplayName="" LastEditDate="2010-08-13T22:40:17.210" LastActivityDate="2010-09-11T00:32:10.590" CommunityOwnedDate="2010-08-13T19:57:45.683" Title="What are the recommended textbooks for introductory calculus?" Tags="&lt;calculus&gt;&lt;books&gt;" AnswerCount="7" CommentCount="6" FavoriteCount="3" />
  <row Id="2396" PostTypeId="2" ParentId="2395" CreationDate="2010-08-13T19:59:50.700" Score="4" ViewCount="" Body="&lt;p&gt;Thomas &amp;amp; Finney, Calculus and Analytic Geometry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Spivak, Calculus.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Apostol, Calculus.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I recommend Apostol the most.&lt;/p&gt;&#xA;" OwnerUserId="1119" LastActivityDate="2010-08-13T19:59:50.700" CommunityOwnedDate="2010-08-13T19:59:50.700" CommentCount="4" />
  <row Id="2397" PostTypeId="2" ParentId="2395" CreationDate="2010-08-13T20:03:22.160" Score="1" ViewCount="" Body="&lt;p&gt;My university uses &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0495011665&quot; rel=&quot;nofollow&quot;&gt;Calculus: Early Transcendentals by James Stewart&lt;/a&gt; in the standard calculus 3 quarter sequence, the prolonged 4 quarter calculus sequence, as well as in multivariable calculus. This book covers topics such as functions and models, limits and derivatives, differentiation and applications of differentiation, integrals, techniques of integration, vectors, vector functions, partial derivatives, multiple integrals, vector calculus, and second-order differential equations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also used a different James Stewart book in high school during pre-calculus and calculus. I don't remember the name of that book - it might have even been an earlier edition of this book.&lt;/p&gt;&#xA;" OwnerUserId="91" LastActivityDate="2010-08-13T20:03:22.160" CommunityOwnedDate="2010-08-13T20:03:22.160" CommentCount="2" />
  <row Id="2399" PostTypeId="2" ParentId="2384" CreationDate="2010-08-13T20:06:48.377" Score="0" ViewCount="" Body="&lt;p&gt;Check out the &lt;a href=&quot;http://blossoms.mit.edu/&quot; rel=&quot;nofollow&quot;&gt;Blossoms&lt;/a&gt; videos.  Although aimed more at a classroom setting there may be interesting snippets to grab.&lt;/p&gt;&#xA;" OwnerUserId="943" LastActivityDate="2010-08-13T20:06:48.377" CommunityOwnedDate="2010-08-13T20:06:48.377" />
  <row Id="2400" PostTypeId="1" AcceptedAnswerId="2408" CreationDate="2010-08-13T20:11:16.260" Score="7" ViewCount="265" Body="&lt;p&gt;Having picked up a rudimentary understanding of tensors from reading mechanics papers and Wikipedia, I tend to think of rank-2 tensors simply as square matrices (along with appropriate transformation rules). Certainly, if the distinction between vectors and dual vectors is ignored, a rank 2 tensor $T$ seems to be simply a multilinear map $V \times V \rightarrow \mathbb{R}$, and (I think) any such map can be represented by a matrix $\mathbf{A}$ using the mapping $(\mathbf{v},\mathbf{w}) \mapsto \mathbf{v}^T\mathbf{Aw}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is this: Is this a reasonable way of thinking about things, at least as long as you're working in $\mathbb{R}^n$? Are there any obvious problems or subtle misunderstandings that this naive approach can cause? Does it break down when you deal with something other than $\mathbb{R}^n$? In short, is it &quot;morally wrong&quot;?&lt;/p&gt;&#xA;" OwnerUserId="856" LastActivityDate="2010-08-13T20:54:14.643" Title="Is it misleading to think of rank-2 tensors as matrices?" Tags="&lt;linear-algebra&gt;&lt;intuition&gt;&lt;tensors&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="2" />
  <row Id="2401" PostTypeId="2" ParentId="2395" CreationDate="2010-08-13T20:14:02.383" Score="1" ViewCount="" Body="&lt;p&gt;Elementary Analysis : The theory of calculus by Kenneth Ross is a beautiful written book. Link is here : &lt;a href=&quot;http://www.springer.com/mathematics/analysis/book/978-0-387-90459-7&quot; rel=&quot;nofollow&quot;&gt;http://www.springer.com/mathematics/analysis/book/978-0-387-90459-7&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Diferential and Integral Calculus by R.Courant is again a very good book which one can learn.&#xA;&lt;a href=&quot;http://books.google.co.in/books?id=eyC1nk9-YjkC&amp;amp;printsec=frontcover&amp;amp;dq=Calculus+R.Courant&amp;amp;source=bl&amp;amp;ots=v0rLcDvVW_&amp;amp;sig=VPB6ZBXfMm6yj0vLG4YalVkpV7I&amp;amp;hl=en&amp;amp;ei=_6ZlTLXAN47IvQOhtZHyDA&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=2&amp;amp;ved=0CCEQ6AEwAQ#v=onepage&amp;amp;q=Calculus%20R.Courant&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;http://books.google.co.in/books?id=eyC1nk9-YjkC&amp;amp;printsec=frontcover&amp;amp;dq=Calculus+R.Courant&amp;amp;source=bl&amp;amp;ots=v0rLcDvVW_&amp;amp;sig=VPB6ZBXfMm6yj0vLG4YalVkpV7I&amp;amp;hl=en&amp;amp;ei=_6ZlTLXAN47IvQOhtZHyDA&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=2&amp;amp;ved=0CCEQ6AEwAQ#v=onepage&amp;amp;q=Calculus%20R.Courant&amp;amp;f=false&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Advanced Calculus by J.M.H Ohmsted is also a very good book. Although title says advanced its not really advanced.&lt;/p&gt;&#xA;" LastActivityDate="2010-08-13T20:14:02.383" CommunityOwnedDate="2010-08-13T20:14:02.383" />
  <row Id="2402" PostTypeId="2" ParentId="2400" CreationDate="2010-08-13T20:22:32.910" Score="10" ViewCount="" Body="&lt;p&gt;It's not misleading as long as you change your notion of equivalence.  When a matrix represents a linear transformation $V \to V$, the correct notion of equivalence is similarity: $M \simeq B^{-1} MB$ where $B$ is invertible.  When a matrix represents a bilinear form $V \times V \to \mathbb{R}$, the correct notion of equivalence is &lt;em&gt;congruence&lt;/em&gt;: $M \simeq B^TMB$ where $B$ is invertible.  As long as you keep this distinction in mind, you're fine.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-08-13T20:22:32.910" CommentCount="1" />
  <row Id="2403" PostTypeId="1" AcceptedAnswerId="2412" CreationDate="2010-08-13T20:24:29.923" Score="5" ViewCount="280" Body="&lt;p&gt;The statements of Poincare duality for manifolds and Serre Duality for coherent sheaves on algebraic varieties or analytic spaces look tantalizingly similar. I have heard tangential statements from some people that there is indeed some connection between the two. But I was never able to figure it out on myself. For instance for a naive attempt on a smooth complex manifold, the dimensions don't match. Can somebody help me out? &lt;/p&gt;&#xA;" OwnerUserId="1119" LastEditorUserId="1119" LastEditorDisplayName="" LastEditDate="2010-08-14T09:40:34.137" LastActivityDate="2010-08-14T09:40:34.137" Title="Precise connection between Poincare Duality and Serre Duality." Tags="&lt;algebraic-geometry&gt;&lt;differential-topology&gt;&lt;coherent-sheaves&gt;&lt;duality-theorems&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="2" />
  <row Id="2404" PostTypeId="2" ParentId="2395" CreationDate="2010-08-13T20:28:58.213" Score="1" ViewCount="" Body="&lt;p&gt;Even though I took a course out of it more than 10 years ago, I still refer back to Spivak's Calculus all the time.&lt;/p&gt;&#xA;" OwnerUserId="1151" LastActivityDate="2010-08-13T20:28:58.213" CommunityOwnedDate="2010-08-13T20:28:58.213" CommentCount="4" />
  <row Id="2405" PostTypeId="2" ParentId="674" CreationDate="2010-08-13T20:39:55.737" Score="3" ViewCount="" Body="&lt;p&gt;This elaborates on Mr. Bulatov's answer (note: Mr. Bulatov's link is broken, but I infer it is supposed to link to Dan Klein's notes &lt;a href=&quot;http://www.cs.berkeley.edu/~klein/papers/lagrange-multipliers.pdf&quot; rel=&quot;nofollow&quot;&gt;Lagrange Multipliers without Permanent Scarring&lt;/a&gt;, which are indeed excellent).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, for the sake of simplicity let's work with functions in 2 variables, i.e. we want to find critical points of f(x,y) where the constraint is g(x,y)=c. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take any point (a,b) in the xy-plane such that g(a,b)=c. Suppose that there was a (unit) vector $\vec v$ tangent to g at (a,b) whose dot product with the gradient vector of f at (a,b) was non-zero. We can take $\vec v$ such that the dot product with $\vec v$ is positive since $\nabla f(a,b)\cdot \vec v=-(\nabla f(a,b)\cdot -\vec v)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now what this means is that the the partial derivative of f with respect to $\vec v\quad$ is positive and that points along $\vec v\quad$ close enough to (a,b) have higher f-value. I (and Mr. Bulatov) claim that this means that there are points on g(x,y)=c close to (a,b) with higher f-value, meaning that f is not a maximum. Mr. Bulatov phrases the argument well in terms of infinitesimals, but I feel that a little bit more rigor is illuminating.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider a sequence of points (a&lt;sub&gt;i&lt;/sub&gt;,b&lt;sub&gt;i&lt;/sub&gt;) on g(x,y)=c that converge to (a,b). Each point (a&lt;sub&gt;i&lt;/sub&gt;,b&lt;sub&gt;i&lt;/sub&gt;) comes with a unit vector $\vec u_i\quad$ that corresponds to the direction (is a scalar multiple of) (a,b)-(a&lt;sub&gt;i&lt;/sub&gt;,b&lt;sub&gt;i&lt;/sub&gt;). Because $\vec v\quad$ is a tangent vector, we can choose the (a&lt;sub&gt;i&lt;/sub&gt;,b&lt;sub&gt;i&lt;/sub&gt;) such that the $\vec u_i$ converge to $\vec v$ . But because f is differentiable in more than one variable, its partial derivatives are continuous, and so we have that the partial derivatives $\nabla f(a,b)\cdot \vec u_i$ converge to $\nabla f(a,b) \cdot \vec v$, and so for all sufficiently large i the partial derivative in the u&lt;sub&gt;i&lt;/sub&gt; direction is positive and so $f(a_i,b_i)&gt;f(a,b)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, if f(a,b) was a maximum (or a minimum) of f on g(x,y)=c, gradient vector of f at (a,b) would be have dot product zero with every vector in the line tangent to g at (a,b). But that means that the gradient vector of f is orthogonal to the tangent line of g, which means that the gradient of f must be parallel to (i.e. a scalar multiple of) the gradient of g.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From here we obtain the equations $\nabla f(x,y)=\lambda\nabla g(x,y)$ and $g(x,y)=c$, which are satisfied if and only if $(x,y)$ is a minimum of $f(x,y)+\lambda (g(x,y)-c)$ for positive $\lambda$ and a maximum for negative $\lambda$.&lt;/p&gt;&#xA;" OwnerUserId="400" LastActivityDate="2010-08-13T20:39:55.737" CommunityOwnedDate="2010-08-13T20:39:55.737" />
  <row Id="2406" PostTypeId="2" ParentId="2166" CreationDate="2010-08-13T20:39:56.267" Score="0" ViewCount="" Body="&lt;p&gt;The answer depends if you want an answer that is true &quot;for all &lt;em&gt;n&lt;/em&gt;&quot; or an answer that is true &quot;for all sufficiently large &lt;em&gt;n&lt;/em&gt;.&quot; For instance, there is not always a prime in an interval of the form (n, 3n/2). Take n=7, for instance. There is always a prime in such an interval &quot;for sufficiently large &lt;em&gt;n&lt;/em&gt;,&quot; however.&lt;/p&gt;&#xA;" OwnerUserId="1151" LastActivityDate="2010-08-13T20:39:56.267" />
  <row Id="2407" PostTypeId="2" ParentId="2062" CreationDate="2010-08-13T20:51:30.950" Score="6" ViewCount="" Body="&lt;p&gt;You can use inclusion/exclusion:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{n\leq x} d(n) = \sum_{mn\le x} 1 = \sum_{m\le \sqrt{x}} \ \sum_{n\le x/m} 1 + \sum_{n\le \sqrt{x}} \ \sum_{m\le x/n} 1 - \sum_{m\le\sqrt{x}} 1 \sum_{n\le\sqrt{x}} 1.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the first two double sums are the same (with the roles of $m$ and $n$ interchanged). Hence&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \sum_{n\leq x} d(n) = 2 \sum_{n\le \sqrt{x}} \Big\lfloor\frac{x}{n}\Big\rfloor - \Big( \big\lfloor\sqrt{x}\big\rfloor \Big)^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\lfloor x \rfloor$ denotes the largest integer less than or equal to $x$. Now use the fact that $\lfloor x \rfloor=x+O(1)$ to finish the proof. You will have to use the identity&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \sum_{n\leq x} \frac{1}{n} = \log x +\gamma + O\Big(\frac{1}{x}\Big)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\gamma$ is Euler's constant.&lt;/p&gt;&#xA;" OwnerUserId="1151" LastEditorUserId="1151" LastEditorDisplayName="" LastEditDate="2010-09-30T19:53:52.943" LastActivityDate="2010-09-30T19:53:52.943" />
  <row Id="2408" PostTypeId="2" ParentId="2400" CreationDate="2010-08-13T20:54:14.643" Score="4" ViewCount="" Body="&lt;p&gt;You're absolutely right.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe someone will find useful a couple of remarks telling the same story in coordinate-free way:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;what happens here is indeed identification of space with it's dual: so a bilinear map $T\colon V\times V\to\mathbb{R}$ is rewritten as $V\times V^*\to\mathbb{R}$ — which is exactly the same thing as a linear operator $A\colon V\to V$;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;an identification of $V$ and $V^*$ is exactly the same thing as a scalar product on $V$, and using this scalar product one can write $T(v,w)=(v,Aw)$;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;so orthogonal change of basis preserves this identification — in terms of Qiaochu Yuan's answer one can see this from the fact that for orthogonal matrix $B^T=B^{-1}$ (moral of the story: if you have a canonical scalar product, there is no difference between $T$ and $A$ whatsoever; and if you don't have one — see Qiaochu Yuan's answer.)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="152" LastActivityDate="2010-08-13T20:54:14.643" />
  <row Id="2410" PostTypeId="2" ParentId="2367" CreationDate="2010-08-13T20:58:40.043" Score="3" ViewCount="" Body="&lt;p&gt;This answer builds on damiano's comment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because of Wedderburn's theorem, we know that there are no finite-dimensional non-commutative division algebras over $\mathbb{F}_p$.  In particular, every quaternion algebra over $\mathbb{F}_p$ is split, i.e., isomorphic to the algebra $M_2(\mathbb{F}_p)$ of $2 \times 2$ matrices over $\mathbb{F}_p$.  Can we see this directly by arguments particular to quaternion algebras?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yes.  For instance, a quaternion algebra is split iff its associated norm form&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$N(x) = x \overline{x} = q(x_1,x_2,x_3,x_4)$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;is isotropic -- i.e., there exists $(x_1,\ldots,x_4) \neq (0,\ldots,0)$ such that $q(x_1,\ldots,x_4) = 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of the Hamiltonian quaternion algebra,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$N(x) = x_1^2 + x_2^2 + x_3^2 + x_4^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As Robin points out, one way to see that this form is isotropic over $\mathbb{F}_p$ is to &#xA;realize that by Lagrange's Four Squares Theorem, $p$ is a sum of four integral squares, and then reduce modulo $p$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is overkill.  The fact that any quadratic form in at least three variables over a finite field is isotropic is a special case of the (easy to prove) Chevalley-Warning theorem: see e.g.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://math.uga.edu/~pete/4400ChevalleyWarning.pdf&quot; rel=&quot;nofollow&quot;&gt;http://math.uga.edu/~pete/4400ChevalleyWarning.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even this is more than is necessary: it follows from a simple counting argument that any nondegenerate quadratic form in at least two variables over $\mathbb{F}_p$ is &lt;em&gt;universal&lt;/em&gt;, i.e., represents every nonzero element of the field.  For the details, see e.g. p. 5 of &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://math.uga.edu/~pete/quadraticforms2.pdf&quot; rel=&quot;nofollow&quot;&gt;http://math.uga.edu/~pete/quadraticforms2.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From this we get that every nondegenerate quadratic form in at least three variables over a finite field is isotropic, and in particular the norm form of any quaternion algebra over a finite field is split.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact the universality of binary quadratic forms over finite fields is the key to this result.  Assuming the characteristic is not $2$, the quaternion algebra may be represented as $\langle a, b \rangle$.  For a quaternion algebra $Q = \langle a, b \rangle$ over any field $F$ (of characteristic not $2$), a necessary and sufficient criterion for $Q$ to be split is that $b$ is a norm in the extension $F(\sqrt{a})/F$.  Thus the universality of the norm form of $\mathbb{F}_p(\sqrt{a})/\mathbb{F}_p$ implies that all quaternion algebras over $\mathbb{F}_p$ are split.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For that matter, from the perspective of Galois cohomology, Wedderburn's theorem is not especially deep or difficult to prove.  Using standard properties of cohomology of cyclic groups, it comes down to showing that for any finite degree extension $\mathbb{F}_{q^n}/\mathbb{F}_q$ of finite fields, the norm map is surjective, and this may also be proved by an elementary counting argument (or simply by a straightforward calculation).   &lt;/p&gt;&#xA;" OwnerUserId="299" LastActivityDate="2010-08-13T20:58:40.043" />
  <row Id="2411" PostTypeId="2" ParentId="2062" CreationDate="2010-08-13T21:04:49.873" Score="1" ViewCount="" Body="&lt;p&gt;@Jim: You can represent the floor function with \lfloor and \rfloor (e.g. $\lfloor x \rfloor$). Would post this as a comment, but I don't have the reputation.&lt;/p&gt;&#xA;" OwnerUserId="1016" LastActivityDate="2010-08-13T21:04:49.873" />
  <row Id="2412" PostTypeId="2" ParentId="2403" CreationDate="2010-08-13T21:13:46.427" Score="5" ViewCount="" Body="&lt;p&gt;As far as I know, to make a precise connection, one has to invoke Hodge theory.  Suppose&#xA;that $X$ is a compact smooth projective variety of dimension $d$.  Then Poincare duality pairs $H^n(X,\mathbb C)$ with $H^{2d-n}(X,\mathbb C),$ for any $n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the Hodge decomposition gives&#xA;$$H^n(X,\mathbb C) = \oplus_{p+q = n}&#xA;H^q(X,\Omega^p)$$&#xA;and &#xA;$$H^{2d-n}(X,\mathbb C) = \oplus_{p'+q' = 2 d - n} H^{q'}(X,\Omega^{p'})&#xA;= \oplus_{p + q = n}H^{d-q}(X,\Omega^{d - p}).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now Serre duality gives a duality between&#xA;$H^q(X,\Omega^p)$ and $H^{d - q}(X,\Omega^{d-p}),$&#xA;and the compatibility statement is that Poincare duality between $H^n$ and $H^{2 d - n}$&#xA;is induced by the direct sum of the pairings on the various summands in the Hodge decomposition given by Serre duality.  (Perhaps up to signs and powers of $2 \pi i$,&#xA;which I'm not brave enough to work out right now.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Added: A good case to think about for a newcomer to Hodge theory is the case when $X$ is&#xA;a compact Riemann surface (or equivalently, an algebraic curve).  If the genus of $X$ is $g$,&#xA;then $H^1(X,\mathbb C)$ is $2g$-dimensional, and is endowed with a symplectic pairing via&#xA;Poincare duality.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hodge theory breaks $H^1(X,\mathbb C)$ up into the sum of two $g$-dimensional subspaces, namely $H^0(X,\Omega^1)$ and $H^1(X,\mathcal O)$.  These are isotropic under Poincare duality (i.e. the Poincare duality pairing vanishes when restricted to either of them),&#xA;but the become dual to one another under Poincare duality, and that pairing agrees with&#xA;the Serre duality pairing (up to a factor of $2\pi i$, perhaps).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The easiest part of this to understand is the inclusion $H^0(X,\Omega^1) \subset&#xA;H^1(X,\mathbb C)$: a holomorphic differential gives a cohomology class just via de Rham&#xA;theory (i.e. we integrate the holomorphic one form over 1-cycles); note that holomorphic&#xA;1-forms are automatically exact, because if you apply the exterior derivative, you get a holomorphic 2-form, which must vanish (because $X$ is a curve, i.e. of complex dimension &#xA;one).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To see why $H^0(X,\Omega^1)$ is isotropic under Poincare duality, note that in the de Rham&#xA;picture, the Poincare duality pairing corresponds to wedging forms. But wedging two holomorphic 1-forms again&#xA;gives a holomorphic 2-form, which must vanish (as we already noted).&lt;/p&gt;&#xA;" OwnerUserId="221" LastEditorUserId="221" LastEditorDisplayName="" LastEditDate="2010-08-14T00:18:48.353" LastActivityDate="2010-08-14T00:18:48.353" CommentCount="5" />
  <row Id="2415" PostTypeId="2" ParentId="2378" CreationDate="2010-08-13T21:43:04.133" Score="3" ViewCount="" Body="&lt;p&gt;Let me address the question: &quot;what made Mathematicians ... think of something like this&quot;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer is: Galois, in studying the problem of factorization of polynomials, realized that&#xA;reasoning about the symmetries of the roots could be more powerful a tool than studying explicit formulas (which, loosely speaking, had been the basic method in the theory of equations up to that time).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As this structural/conceptual point of view began to reveal its power in solving difficult concrete problems, more and more mathematicians began to think in this way.  The structural concepts were then isolated from their concrete settings, and this is how they are taught today.  But the motivation was, and for many remains, the applications of these abstract notions to concrete problems.  (A standard but helpful example is the pivotal role that group theory, Galois theory, cohomology, and ring theory played in the proof of Fermat's Last Theorem.)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore, nowadays we train ourselves and our students to recognize any hint of structure in a problem, and to exploit such structure to the hilt.  So these concepts have become basic tools in the problem-solving toolkits of contemporary mathematicians.&lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2010-08-13T21:43:04.133" CommunityOwnedDate="2010-08-13T21:43:04.133" />
  <row Id="2416" PostTypeId="2" ParentId="2378" CreationDate="2010-08-13T21:52:07.157" Score="3" ViewCount="" Body="&lt;p&gt;See the books &lt;a href=&quot;http://www.springer.com/birkhauser/historyofscience/book/978-0-8176-4684-4&quot; rel=&quot;nofollow&quot;&gt;A History of Abstract Algebra&lt;/a&gt; and &lt;a href=&quot;http://www.ams.org/bookstore-getitem/item=hmath-32&quot; rel=&quot;nofollow&quot;&gt;Episodes in the History of Modern Algebra (1800-1950)&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="589" LastActivityDate="2010-08-13T21:52:07.157" CommunityOwnedDate="2010-08-13T21:52:07.157" />
  <row Id="2417" PostTypeId="2" ParentId="2349" CreationDate="2010-08-13T21:56:35.393" Score="6" ViewCount="" Body="&lt;p&gt;You put your socks first and then your shoes but you take off your shoes before taking off your socks.&lt;/p&gt;&#xA;" OwnerUserId="589" LastActivityDate="2010-08-13T21:56:35.393" CommentCount="3" />
  <row Id="2418" PostTypeId="1" CreationDate="2010-08-13T22:37:03.223" Score="16" ViewCount="877" Body="&lt;p&gt;I had seen this problem a long time back and wasn't able to solve it. For some reason I was reminded of it and thought it might be interesting to the visitors here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Apparently, this problem is from a mathematics magazine of some university in the United States (sorry, no idea about either).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the problem is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose $S \subset \mathbb{Z}$ (set of integers) such that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) $|S| = 15$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) $\forall s \in S, \exists a,b \in S$ such that $s = a+b$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Show that for every such $S$, there is a non-empty subset $T$ of $S$ such that the sum of elements of $T$ is zero and $|T| \leq 7$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update&lt;/strong&gt; (Sep 13)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is an approach which seems promising and others might be able to take it ahead perhaps.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you look at the set as a vector $s$, then there is a matrix $A$ with the main diagonal being all $1$, each row containing exactly one $1$ and one $-1$ (or a single $2$) in the non-diagonal position such that $As = 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem becomes equivalent to proving that for any such matrix $A$ the row space of $A$ contains a vector with all zeroes except for a $1$ and $-1$ or a vector with all zeroes except $\leq 7$ ones.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This implies that the numbers in the set $S$ themselves don't matter and we can perhaps replace them with elements from a different field (like say reals, or complex numbers).&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-09-13T17:56:43.500" LastActivityDate="2010-09-13T17:56:43.500" Title="A zero sum subset of a sum-full set." Tags="&lt;combinatorics&gt;" AnswerCount="2" CommentCount="11" FavoriteCount="7" />
  <row Id="2419" PostTypeId="1" CreationDate="2010-08-13T22:40:19.287" Score="4" ViewCount="219" Body="&lt;p&gt;A friend of mine is subediting a book. The author of the book died in 1999. The author, at some point, uses the word &quot;trillion&quot; which is, unfortunately, an ambiguous word in the UK: when I was at school it used to mean $10^{18}$ but nowadays it means $10^{12}$. My friend is faced with the following paragraph:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;'According to Asimov, the amino acids of the proteins behave in a much freer way than our words do: they can be rearranged in any manner and always retain some meaning. A simple protein is made up of eight amino acids, which can be classified putting the numbers one to eight in a series, changing the order of sequence by one digit each time. Out of the same number of &quot;words&quot; we can construct a little over 40,000 organised &quot;biological phrases&quot; from the same genetic code, each one with its own meaning, which is the mission of every protein. But if the chains become longer, as in the case of more complicated molecules such as insulin, which consists of 30 amino acids, the tally rises to a staggering eight million trillion possibilities.'&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The editor doesn't like &quot;eight million trillion&quot; and wants to replace it with &quot;800000..000&quot;. The question, of course, is &quot;how many zeros&quot;? More precisely, the question is: does someone with a better understanding of chemistry than me able to reconstruct Asimov's calculation and see whether the answer is approximately $8\times 10^{24}$ or $8\times 10^{18}$?&lt;/p&gt;&#xA;" OwnerUserId="1162" LastEditorUserId="171" LastEditorDisplayName="" LastEditDate="2010-08-14T08:17:16.233" LastActivityDate="2010-08-14T08:17:16.233" Title="Asimov quote about &quot;eight million trillion&quot; arrangements of amino acids" Tags="&lt;combinatorics&gt;&lt;chemistry&gt;" AnswerCount="2" CommentCount="6" />
  <row Id="2420" PostTypeId="2" ParentId="713" CreationDate="2010-08-13T23:06:53.523" Score="0" ViewCount="" Body="&lt;p&gt;A common approach for displaying symmetric tensors for scientific data visualization is the use of what are called &quot;tensor glyphs&quot; (see for example &quot;&lt;a href=&quot;http://www.cs.utah.edu/~gk/papers/vissym04/&quot; rel=&quot;nofollow&quot;&gt;Superquadric Tensor Glyphs&lt;/a&gt;&quot; by Kindlmann, 2004). The basic idea helps for visualizing tensors intuitively, too. It really only makes sense for &lt;em&gt;symmetric&lt;/em&gt; tensors, but then, many tensors that arise in physics are symmetric.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Since my familiarity is only with linear algebra and not tensor analysis, I'm going to equate &quot;tensor&quot; and &quot;matrix&quot; in the following. According to the answers to &lt;a href=&quot;http://math.stackexchange.com/questions/2400/is-it-misleading-to-think-of-rank-2-tensors-as-matrices&quot;&gt;this question&lt;/a&gt;, this is not a horribly wrong thing to do.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose the tensor $T$ is symmetric and positive definite. Treat it as a matrix, and transform the unit sphere $S = \{x : \|x\| = 1\}$ with it to get an ellipsoid $TS = \{x : \|T^{-1}x\| = 1\}$. The ellipsoid's axes are parallel to $T$'s eigenvectors, with lengths proportional to the corresponding eigenvalues. This has some other nice properties:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The identity matrix is represented by a unit sphere.&lt;/li&gt;&#xA;&lt;li&gt;Scalar multiplication $\alpha T$ uniformly scales the ellipsoid by $\alpha$.&lt;/li&gt;&#xA;&lt;li&gt;A tensor with one large and two small eigenvalues looks like a line segment pointing along the large eigenvector.&lt;/li&gt;&#xA;&lt;li&gt;A tensor with two large and one small eigenvalues looks like a disk normal to the small eigenvector.&lt;/li&gt;&#xA;&lt;li&gt;The mapping between symmetric positive definite matrices/tensors and ellipsoids centered on the origin is bijective.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;If your tensors are symmetric but &lt;em&gt;not&lt;/em&gt; positive definite, you can still visualize them in terms of deformations of the unit sphere, corresponding to $I + \epsilon T$. Now a negative eigenvalue corresponds to shrinking, and a positive one to stretching.&lt;/p&gt;&#xA;" OwnerUserId="856" LastActivityDate="2010-08-13T23:06:53.523" CommentCount="1" />
  <row Id="2421" PostTypeId="2" ParentId="2419" CreationDate="2010-08-13T23:11:13.547" Score="2" ViewCount="" Body="&lt;p&gt;What I think the author intends in the first part is that some piece of genetic code gives you $8$ distinct amino acids and you want to count the number of ways to rearrange them to get a protein, which is where the $8! \approx 40000$ number comes from.  Since there are $20$ amino acids, the best you can do for $30$ amino acids is to use each of them once and ten of them twice, giving an answer of $\frac{30!}{2^{10}} \approx 2.59 \times 10^{29}$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am unable to find the actual list of amino acids in insulin (which is ridiculous to me; why isn't this information on a wiki somewhere?), but the above calculation suggests that the $10^{24}$ number is closer.  If anyone wants to help me out, the actual polypeptide chain in question is the B chain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit:  According to &lt;a href=&quot;http://www.greatcom.org/resources/reasons_skeptics/ch_15/default.htm&quot; rel=&quot;nofollow&quot;&gt;this citation&lt;/a&gt;, Asimov's actual estimate is $8 \times 10^{27}$.  So it looks like the author misquoted him.&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-08-13T23:48:10.583" LastActivityDate="2010-08-13T23:48:10.583" CommentCount="4" />
  <row Id="2423" PostTypeId="2" ParentId="2356" CreationDate="2010-08-13T23:24:17.020" Score="2" ViewCount="" Body="&lt;p&gt;Here's another method: consider H to be 0 and T to be 1. Consider the results that you get from your coins as successive bits (after the binary point) in a binary number.  Assign numbers in the interval [0, 1/3) to the first choice; [1/3, 2/3) to the second choice; [2/3, 1) to the third choice.  Flip coins until you know, no matter what happens on the remaining flips, in which of these intervals the resulting real number in [0, 1) will lie.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More concretely, this means that if you flip two coins and observe HH, make the first choice; no matter what happens the result will be between .0000... = 0 and .001111... = 1/4.  Similarly if you observe TT, make the third choice; the result will be between 3/4 and 1.  If you observe HT, you can't commit to a choice yet; the result will be between 1/4 and 1/2, which overlaps two of the intervals. Something similar is true for TH.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In either case flip a third time.  HTT corresponds to the interval [3/8, 1/2) which lies entirely in [1/3, 2/3), so HTT corresponds to the second choice; similarly for THH.  If you see HTH you're still undecided between the first and second choices; THT is still undecided between the second and third.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Continuing in this way, HTHH corresponds to the first choice, HTHT and THTH are still undecided, THTT corresponds to the third choice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, for the cut-points 1/3 and 2/3, you flip coins until you get either two heads or two tails in a row. If you first get two heads in a row, this corresponds to the first choice if the initial flip was a head, the second if the initial flip was a tail. If you first get two tails in a row, this corresponds to the second choice if the initial flip was a head, the third choice if the initial flip was a tail.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might wonder how many flips this takes, on average, to give a result. With probability 1/2 you get a result on the second flip; with probability 1/4, on the third flip; with probability 1/8, on the fourth flip, and in general with probability $1/2^{n-1}$ for $n \ge 2$.  The sum $\sum_{n \ge 2} n/2^{n-1}$ has value 3, so on average this method takes three flips.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a bit slower on average than the method Rasmus suggested, which takes on average 8/3 flips.  However, the same method works even if the three choices are not equally likely! (It won't have such a clean way of being expressed as &quot;flip until you get two in a row&quot;, though.)&lt;/p&gt;&#xA;" OwnerUserId="173" LastActivityDate="2010-08-13T23:24:17.020" />
  <row Id="2424" PostTypeId="2" ParentId="2419" CreationDate="2010-08-13T23:43:46.443" Score="2" ViewCount="" Body="&lt;p&gt;The second sentence of that paragraph is meaningless, I think. The number of eight-amino-acid polypeptides (those are chains of amino acids) is 20^8, which is much larger than 40,000.  And insulin isn't really big enough to be a good example of a typical protein; typical proteins have hundreds of amino acids.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That being said, Qiaochu's interpretation seems correct.  I was able to track down &lt;a href=&quot;http://www.indulgedfurries.com/petdiabetes/insulin_sequence.htm&quot; rel=&quot;nofollow&quot;&gt;the structure of human insulin&lt;/a&gt; -- note that the image is for bovine insulin, and human insulin differs by substituting Thr for Ala in the final position.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the B-chain of human insulin has the sequence&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Phe Val Asn Gln His Leu Cys Gly Ser His Leu Val Glu Ala Leu Tyr Leu Val Cys Gly Glu Arg Gly Phe Phe Tyr Thr Pro Lys Ala&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which contains &lt;/p&gt;&#xA;&#xA;&lt;p&gt;4 Leu&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3 each Phe, Val, Gly,  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2 each His, Cys, Glu, Ala, Tyr, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1 each Asn, Gln, Ser, Arg, Thr, Pro , Lys&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and the number of ways that all of these can be rearranged is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \frac{30!}{4! 3!^3 2!^5} \approx 1.6 \times 10^{27} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is not all that close to either of the proposed replacements.  Then again, it's possible that whoever did the calculation was working from the sequence of some insulin other than human, or just made a mistake.&lt;/p&gt;&#xA;" OwnerUserId="173" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-08-13T23:53:45.237" LastActivityDate="2010-08-13T23:53:45.237" CommentCount="4" />
  <row Id="2425" PostTypeId="1" CreationDate="2010-08-14T00:16:55.830" Score="4" ViewCount="307" Body="&lt;p&gt;Experimentally, this limit appears to converge to 1/e, but I can't figure out how to solve it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again, it's the limit as x goes to infinity of (x^(x+1) / (x+1)^x) - ((x-1)^x / x^(x-1))&lt;/p&gt;&#xA;" OwnerUserId="1164" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-08-14T00:25:02.350" LastActivityDate="2010-08-14T14:38:01.353" Title="What is the limit as x goes to infinity of x^(x+1)/(x+1)^x - (x-1)^x/x^(x-1)" Tags="&lt;calculus&gt;" AnswerCount="3" CommentCount="3" />
  <row Id="2426" PostTypeId="2" ParentId="2349" CreationDate="2010-08-14T00:42:41.660" Score="4" ViewCount="" Body="&lt;p&gt;$$\begin{align}&#xA;&amp;amp; \text{id} \\&#xA;=&amp;amp; f \circ f^\circ \\&#xA;=&amp;amp; f  \circ \text{id} \circ f^\circ \\&#xA;=&amp;amp; f \circ (g \circ g^\circ) \circ f^\circ \\&#xA;=&amp;amp; f \circ g \circ g^\circ \circ f^\circ \\&#xA;=&amp;amp; (f \circ g) \circ (g^\circ \circ f^\circ)&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore $(f \circ g)^\circ = g^\circ \circ f^\circ$.&lt;/p&gt;&#xA;" LastActivityDate="2010-08-14T00:42:41.660" />
  <row Id="2427" PostTypeId="1" AcceptedAnswerId="2498" CreationDate="2010-08-14T00:45:44.867" Score="17" ViewCount="471" Body="&lt;p&gt;Is there any overarching reason why, after excluding the infinite classes of finite simple groups&#xA;(cyclic, alternating, Lie-type),&#xA;what remains---the &lt;em&gt;sporadic&lt;/em&gt;, exceptional finite simple groups, is in fact a finite list (just 26)?  In some sense, the prime numbers can be viewed as &quot;sporadic,&quot; but there is an infinite supply.  Is there some principle that indicates that there must be only a finite number of these&#xA;exceptional groups, and the &quot;only issue&quot; (to minimize a huge, multi-year community effort) was to identify them?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I ask in relative ignorance of modern group theory, and apologize in advance for the naiveness of my question.&lt;/p&gt;&#xA;" OwnerUserId="237" LastEditorUserId="8575" LastEditorDisplayName="" LastEditDate="2011-05-19T08:51:55.657" LastActivityDate="2011-05-19T08:51:55.657" Title="Why are there only a finite number of sporadic simple groups?" Tags="&lt;group-theory&gt;&lt;simple-groups&gt;" AnswerCount="3" CommentCount="4" FavoriteCount="3" />
  <row Id="2428" PostTypeId="2" ParentId="2341" CreationDate="2010-08-14T00:48:42.210" Score="3" ViewCount="" Body="&lt;p&gt;Hi User. Are you still stuck? I'll use your terminology, which seems to be used in electrical engineering.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) x'z + xyz + xy'z&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) x'z + xz(y+y')       &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since y+y' = 1, and xz1 = xz then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) x'z + xz&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4) (x'+x)z&lt;/p&gt;&#xA;&#xA;&lt;p&gt;5) z&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope that helps.&lt;/p&gt;&#xA;" OwnerUserId="592" LastEditorUserId="592" LastEditorDisplayName="" LastEditDate="2010-08-14T03:38:08.367" LastActivityDate="2010-08-14T03:38:08.367" />
  <row Id="2429" PostTypeId="1" AcceptedAnswerId="2431" CreationDate="2010-08-14T01:08:02.280" Score="0" ViewCount="355" Body="&lt;p&gt;How should I rotate a point (x,y) to location (a,b) on the co-ordinate by any angle?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/LQiiA.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1167" LastActivityDate="2010-08-14T07:05:03.973" Title="Rotate a point in circle about an angle" Tags="&lt;trigonometry&gt;" AnswerCount="3" />
  <row Id="2430" PostTypeId="2" ParentId="2429" CreationDate="2010-08-14T01:11:47.250" Score="1" ViewCount="" Body="&lt;p&gt;See:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Rotation_matrix#Two_dimensions&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Rotation_matrix#Two_dimensions&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here since the rotation is clockwise rather than counter clockwise, you would have to substitute $- \theta$ in place of $\theta$.&lt;/p&gt;&#xA;" OwnerUserId="1119" LastActivityDate="2010-08-14T01:11:47.250" CommentCount="3" />
  <row Id="2431" PostTypeId="2" ParentId="2429" CreationDate="2010-08-14T01:13:38.663" Score="2" ViewCount="" Body="&lt;p&gt;The rotation transformation of a point, treated as a vector, would be (you seem to be going clockwise so I'll swing with that):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\bigl(\begin{smallmatrix}&#xA;x'\\&#xA;y'&#xA;\end{smallmatrix}\bigr)=\bigl(\begin{smallmatrix}&#xA;\cos(\theta)&amp;amp;\sin(\theta)\\ &#xA;-\sin(\theta)&amp;amp;\cos(\theta)&#xA;\end{smallmatrix}\bigr)\cdot \bigl(\begin{smallmatrix}&#xA;x\\ &#xA;y&#xA;\end{smallmatrix}\bigr)$&lt;/p&gt;&#xA;" OwnerUserId="498" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-08-14T06:24:27.070" LastActivityDate="2010-08-14T06:24:27.070" CommentCount="1" />
  <row Id="2432" PostTypeId="2" ParentId="2425" CreationDate="2010-08-14T01:19:12.943" Score="3" ViewCount="" Body="&lt;p&gt;Making the change of variables $\; z = 1/x \;$ and collecting together like exponents yields&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\quad\quad\quad\quad \displaystyle &#xA;\lim_{z\to 0^+} {\frac{(1+z)^{-1/z} - (1-z)^{1/z}}{z}}\; = \;\lim_{z\to 0^+} {\frac{f(-z)-f(z)}{z}} $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Applying the well-known $\rm exp$ and $\rm log$ taylor series to $f(z) =  (1-z)^{1/z}\; = \; e^{{\rm log}(1-z)/z}$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f(z) = e^{-1-\frac{z}{2}+\;\cdots} = e^{-1} (1 + (-\frac{z}{2} +\;\cdots) + (-\frac{z}{2} +\cdots)^2 + \;\cdots) =  \frac{1}{e} - \frac{z}{2e} + O(z^2)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore  $\quad\quad\quad\quad\quad\quad\quad\quad\displaystyle&#xA;\frac{f(-z)-f(z)}{z}\; = \;\frac{1}{e} + O(z)\quad\quad\quad$  as confirmed by Macsyma:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i33.tinypic.com/bi1m6h.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-08-14T14:38:01.353" LastActivityDate="2010-08-14T14:38:01.353" CommentCount="1" />
  <row Id="2433" PostTypeId="2" ParentId="2341" CreationDate="2010-08-14T01:19:50.180" Score="2" ViewCount="" Body="&lt;p&gt;If you think of the Boolean operations as complement, intersection and union of sets, which is equivalent since every Boolean algebra is isomorphic to a field of sets, then the equation is obvious. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Namely, on the right side we have the set $z$. On the left side, we have the elements of $z$ separated into three categories: those that are not in $x$, those that are in $x$ and also in $y$, and those that are in $x$ but not in $y$. Since this accounts for all possible elements of $z$, we may deduce the equality.&lt;/p&gt;&#xA;" OwnerUserId="413" LastActivityDate="2010-08-14T01:19:50.180" />
  <row Id="2434" PostTypeId="2" ParentId="2425" CreationDate="2010-08-14T01:49:29.687" Score="1" ViewCount="" Body="&lt;p&gt;&lt;b&gt;HINT&lt;/b&gt; You can write the difference as $x \left( y^{-x} - z^x \right)=x\left(1-(yz)^x\right)/y^z$, where $y$ and $z$ are rational functions of $x$ such that the limiting value of $y^x$ (and $z^x$) is well-known, finite, and non-zero. Now focus on finding the limit of $x\left(1-(yz)^x\right)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Restricting $x$ to integers and invoking the Binomial Theorem helps, but that only shows what the limit --&lt;em&gt;if&lt;/em&gt; it exists-- would have to be, leaving open the possibility that the limit might not exist when considering non-integer $x$.)&lt;/p&gt;&#xA;" OwnerUserId="409" LastActivityDate="2010-08-14T01:49:29.687" CommentCount="1" />
  <row Id="2435" PostTypeId="1" AcceptedAnswerId="2674" CreationDate="2010-08-14T04:11:46.293" Score="2" ViewCount="234" Body="&lt;p&gt;I have a function that (more or less) is supposed to select a small number m of random number from the range [1,n] (for some n &gt;&gt; m) and I need to test that it work. Is there an easy to implement test that will give good confidence that it is working correctly? (For reference the full on chi squared test is not simple enough.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;m in [5,500], n is small enough I can use normal int/float types for most math.&lt;/li&gt;&#xA;&lt;li&gt;I can get as many set as I'm willing to wait for (&gt;100s per second)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The reason I'm looking for something simple is that the code I'm checking isn't that complicated. If the test code is more complex than the code under test, then the potential for the test being wrong becomes a real problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit 2:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I took another look at Chi-squared and it's not as bad as I remembered... as long as I have a fixed number of buckets and a fixed significance threshold. (Or a canned CDF, and I don't think I do.)&lt;/p&gt;&#xA;" OwnerUserId="1171" LastEditorUserId="1171" LastEditorDisplayName="" LastEditDate="2010-08-18T18:59:13.183" LastActivityDate="2010-08-20T15:11:14.547" Title="Is there a simple test for uniform distributions?" Tags="&lt;statistics&gt;" AnswerCount="3" CommentCount="5" />
  <row Id="2436" PostTypeId="2" ParentId="2425" CreationDate="2010-08-14T06:47:28.473" Score="4" ViewCount="" Body="&lt;p&gt;A fairly mechanical approach is to write the limit as&#xA;$$\lim_{x\to\infty}(f(x)-f(x-1))$$&#xA;where&#xA;$$f(x)=\frac{x^{x+1}}{(x+1)^x}.$$&#xA;Then&#xA;$$\log f(x)=(x+1)\log x - x\log(x+1)&#xA;= \log x - x \log(1+1/x)$$&#xA;and so&#xA;$$\log f(x)= \log x - 1 + 1/(2x) + O(x^{-2})$$&#xA;as $x\to\infty$ (using the Maclaurin series for $\log(1+t)$).&#xA;Therefore&#xA;$$f(x) = (x/e)(1+1/(2x)+O(x^{-2}))=x/e+1/(2e)+O(x^{-1})$$&#xA;and so&#xA;$$f(x-1) =(x-1)/e+1/(2e) +O((x-1)^{-1}).$$&#xA;Subtracting,&#xA;$$f(x)-f(x-1)= 1/e+O(x^{-1})).$$&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-08-14T06:47:28.473" />
  <row Id="2437" PostTypeId="2" ParentId="2429" CreationDate="2010-08-14T06:50:23.117" Score="1" ViewCount="" Body="&lt;p&gt;Assuming that you ask where will &lt;em&gt;x&lt;/em&gt; end up after being rotated clockwise for an angle $\theta$ around the origin, then the answer is as given by George S. and J. Mangaldan. I'd like to give a brief explanation of &lt;em&gt;why&lt;/em&gt; that is the answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;The key is that rotation is a linear transformation.&lt;/em&gt; If we write &lt;em&gt;Ax&lt;/em&gt; for the point &lt;em&gt;x&lt;/em&gt; rotated, then then $A(ax+bx) = a(Ax) + b(Ax)$, where &lt;em&gt;a&lt;/em&gt; and &lt;em&gt;b&lt;/em&gt; are some real numbers. Here &lt;em&gt;ax&lt;/em&gt; is a point obtained by stretching both coordinates, and lies on the line that contains x and the origin. It's a scaling operation. Also, $x+y$ is obtained by adding the coordinates, which you can imagine as the diagonal of a parallelogram whose edges are &lt;em&gt;x&lt;/em&gt; and &lt;em&gt;y&lt;/em&gt;. So the equation $A(ax+bx) = a(Ax) + b(Ax)$ says that if you scale a figure with two points, then `add' the points by taking the parallelogram diagonal, and then rotate you get the same result you get if you start by rotating, and then you scale and add.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://imgur.com/c5TbD.png&quot; alt=&quot;&quot; title=&quot;Hosted by imgur.com&quot; /&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, assuming that equation is OK when &lt;em&gt;A&lt;/em&gt; represents a rotation, note that every vector $x=(a,b)$ can be written as $a(1,0)+b(0,1)$, so its rotation will be $a(A(1,0))+b(A(0,1))$. In other words, all you need to know is where do the points (1,0) and (0,1) end up after being rotated (namely, you need the points A(1,0) and A(0,1), each having two coordinates). So four numbers are enough to figure out where &lt;em&gt;all&lt;/em&gt; points end up after being rotated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Linear transformations are &lt;em&gt;the&lt;/em&gt; reason matrices are important.&lt;/p&gt;&#xA;" OwnerUserId="512" LastEditorUserId="512" LastEditorDisplayName="" LastEditDate="2010-08-14T07:05:03.973" LastActivityDate="2010-08-14T07:05:03.973" CommentCount="1" />
  <row Id="2439" PostTypeId="1" CreationDate="2010-08-14T07:20:02.330" Score="8" ViewCount="215" Body="&lt;p&gt;Here's a question.  Have there been any theoretical results showing that if P != NP, there must exist some problems in NP which are not NP-complete and which are not in P either?  Just curious because I've never seen this question addressed.&lt;/p&gt;&#xA;" LastEditorUserId="334" LastEditorDisplayName="" LastEditDate="2010-08-14T07:36:53.197" LastActivityDate="2010-08-14T07:36:53.197" Title="Assuming P != NP, do we know whether there are problems which are in NP, not in P and are not NP complete ?" Tags="&lt;computer-science&gt;&lt;np-complete&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" />
  <row Id="2441" PostTypeId="2" ParentId="2439" CreationDate="2010-08-14T07:29:19.173" Score="9" ViewCount="" Body="&lt;p&gt;Yes, this is known as Ladner's theorem, proved by Richard Ladner in 1975. The class of such problems are called &lt;a href=&quot;http://en.wikipedia.org/wiki/NP-Intermediate&quot; rel=&quot;nofollow&quot;&gt;NP-intermediate&lt;/a&gt;. &lt;a href=&quot;http://books.google.com/books?id=47GwSJfb1mQC&amp;amp;pg=PA38&quot; rel=&quot;nofollow&quot;&gt;Here's one proof&lt;/a&gt; that does &quot;cut and paste&quot; between some problem in P and some NP-complete problem, &lt;a href=&quot;http://blog.computationalcomplexity.org/2003/03/foundations-of-complexity-lesson-16.html&quot; rel=&quot;nofollow&quot;&gt;here are two&lt;/a&gt;, and there are &lt;a href=&quot;http://www.google.com/search?q=ladner%27s+theorem&quot; rel=&quot;nofollow&quot;&gt;others&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are also problems that are already &lt;em&gt;believed&lt;/em&gt; to be neither NP-complete nor in P, like integer factorisation and graph isomorphism, as well as &lt;a href=&quot;http://mathoverflow.net/questions/31821/problems-known-to-be-in-both-np-and-conp-but-not-known-to-be-in-p&quot; rel=&quot;nofollow&quot;&gt;problems known to be in NP ∩ coNP&lt;/a&gt; but not known to be in P.&lt;/p&gt;&#xA;" OwnerUserId="205" LastActivityDate="2010-08-14T07:29:19.173" CommentCount="2" />
  <row Id="2443" PostTypeId="2" ParentId="1870" CreationDate="2010-08-14T07:53:39.613" Score="1" ViewCount="" Body="&lt;p&gt;There is &lt;a href=&quot;http://www.microsoft.com/learningspace/&quot; rel=&quot;nofollow&quot;&gt;Microsoft Math&lt;/a&gt;. I haven't actually used it, but I know that it can provide step by step solutions. &lt;a href=&quot;http://www.squarecirclez.com/blog/microsoft-math-30-review/713&quot; rel=&quot;nofollow&quot;&gt;Here&lt;/a&gt;'s a nice review. Also it's $20.&lt;/p&gt;&#xA;" OwnerUserId="348" LastActivityDate="2010-08-14T07:53:39.613" CommunityOwnedDate="2010-08-14T07:53:39.613" />
  <row Id="2444" PostTypeId="1" AcceptedAnswerId="2456" CreationDate="2010-08-14T08:28:27.470" Score="0" ViewCount="212" Body="&lt;p&gt;I'd like to prove that $\forall x\left(x\not\in x\right)$ in the context of Morse-Kelley set theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's call $A=\left\{y:y\not\in y\right\}$. I can easily prove that $A\not\in A$. In fact, if you suppose that $A\in A$, it follows that $A\not\in A$ by definition of class comprehension. Since this is a contradiction, then the hypothesis is wrong, that is, $A\not\in A$. But I cannot do the same for an arbitrary class $k$, since I know nothing about $k$ and I cannot apply the definition of class comprehension.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any ideas? Thanks.&lt;/p&gt;&#xA;" OwnerUserId="1175" LastActivityDate="2010-08-15T21:21:09.563" Title="How to prove that nothing is a member of itself?" Tags="&lt;set-theory&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="0" />
  <row Id="2445" PostTypeId="2" ParentId="2395" CreationDate="2010-08-14T08:31:56.090" Score="0" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://college.cengage.com/mathematics/ostebee/calculus_graphical/2e/students/&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;Calculus from Graphical, Numerical, and&#xA;Symbolic Points of View&lt;/em&gt; by Ostebee and Zorn&lt;/a&gt; is my favorite book from which to teach (er, actually, I have a slight preference for the 1st edition over the 2nd, but they're both good).  It's aimed a little above average (i.e. harder than the Harvard Consortium book [Hughes-Hallett, Gleason, et al]), but it's much more of a modern reform-calculus than Spivak.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, I think the conversational tone of the book makes it more readable and the variety of problems (especially the fact that it has challenging problems rather than just repetitive exercised) makes it a good book from which to study.&lt;/p&gt;&#xA;" OwnerUserId="72" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-08-14T08:37:59.163" LastActivityDate="2010-08-14T08:37:59.163" CommunityOwnedDate="2010-08-14T08:31:56.090" />
  <row Id="2446" PostTypeId="2" ParentId="2395" CreationDate="2010-08-14T08:32:23.250" Score="0" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://bcs.wiley.com/he-bcs/Books?action=index&amp;amp;itemId=0471448761&amp;amp;itemTypeId=BKS&amp;amp;bcsId=1404&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;Calculus: Single Variable, Course Advantage Edition&lt;/em&gt; by Hughes-Hallett, Gleason, et al&lt;/a&gt;, aka the &quot;Harvard Consortium&quot; book, is perhaps the standard &quot;reform calculus&quot; book.&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-08-14T08:32:23.250" CommunityOwnedDate="2010-08-14T08:32:23.250" CommentCount="2" />
  <row Id="2447" PostTypeId="2" ParentId="2418" CreationDate="2010-08-14T08:35:13.163" Score="7" ViewCount="" Body="&lt;p&gt;A weaker statement, where we allow elements in $T$ to be repeated, can be proved as below: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since we can look at the set $\{-s | s\in S \}$ we may assume there are at most $7$ positive numbers in $S$. Let each positive number be a vertex, from each vertex $s$ we draw an arrow to any vertex $a$ such that $s=a+b$. Since if $s&gt;0$, one of $a,b$ must be positive, there is at least one arrow from any vertex. So there must be a cycle $s_1,\cdots,s_n=s_1$ with $n\leq 8$. We can let $T$ consists of $s_i-s_{i+1}$, $1\leq i\leq n-1$.   &lt;/p&gt;&#xA;" OwnerUserId="223" LastEditorUserId="274" LastEditorDisplayName="" LastEditDate="2010-09-09T10:06:33.777" LastActivityDate="2010-09-09T10:06:33.777" />
  <row Id="2448" PostTypeId="1" AcceptedAnswerId="2449" CreationDate="2010-08-14T08:45:14.577" Score="1" ViewCount="333" Body="&lt;p&gt;The Sum of 2 sets of Measure zero might well be very large for example, the sum of $x$-axis and $y$-axis, is nothing but the whole plane. Similarly one can ask this question about Cantor sets:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $C$ is the cantor set, then what is the measure of $C+C$?&lt;/p&gt;&#xA;" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2010-12-11T09:05:12.607" LastActivityDate="2010-12-11T09:05:12.607" Title="Measure of the Cantor set plus the Cantor set" Tags="&lt;real-analysis&gt;&lt;measure-theory&gt;" AnswerCount="2" />
  <row Id="2449" PostTypeId="2" ParentId="2448" CreationDate="2010-08-14T08:52:47.383" Score="10" ViewCount="" Body="&lt;p&gt;If you are asking the case where $C$ is the Cantor ternary set, then you can show that $C+ C$ is actually $[0,2]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For more general cantor sets, you can find a description in the paper: On the topological structure of the arithmetic sum of two Cantor sets, P Mendes and F Oliveira, available at &lt;a href=&quot;http://iopscience.iop.org/0951-7715/7/2/002&quot; rel=&quot;nofollow&quot;&gt;http://iopscience.iop.org/0951-7715/7/2/002&lt;/a&gt; .&lt;/p&gt;&#xA;" OwnerUserId="1119" LastEditorUserId="1119" LastEditorDisplayName="" LastEditDate="2010-08-14T09:20:40.417" LastActivityDate="2010-08-14T09:20:40.417" CommentCount="2" />
  <row Id="2450" PostTypeId="1" AcceptedAnswerId="2457" CreationDate="2010-08-14T09:14:31.940" Score="3" ViewCount="513" Body="&lt;p&gt;The tradition way for two children to divide a piece of cake fairly between them is &quot;you cut, I choose&quot;, What process accomplishes this is intuitively clear. Is there a similar process which works for $3$ children.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose that three children $(1),(2)$ and $(3)$ have associated measures (ways of evaluating the pieces of cake) $\alpha_{1},\alpha_{2}$ and $\alpha_{3}$. Assume for the simplicity of the notation that the sum of total pieces of the cake is equal to $1$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Question : Does the cake have three subsets $C_{1},C_{2}$ and $C_{3}$  that are pairwise disjoint and their union is equal to $C$, and that have moreover the property that $$\alpha(C_{1}) \geq \frac{1}{3}, \quad \alpha(C_{2} )\geq \frac{1}{3} , \quad \alpha(C_{3}) \geq \frac{1}{3}?$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Solution:&lt;/strong&gt; Let one of the three children say $(1)$, divide the cake into what he regards three equal pieces $C_{1},C_{2}$ and $C_{3}$: in other words $\alpha(C_{1})= \alpha(C_2)=\alpha(C_{3})=\frac{1}{3}$. Let each of $(2)$ and $(3)$ claim dibs on one of those three pieces, the one he prefers the most, the one he would be quite satisfied to have. It is possible that $(2)$ and $(3)$ will claim different pieces, and it is possible that they will claim the same one, but no matter, atleast one of the three pieces, $C_{1},C_{2},C_{3}$ will be left unclaimed. Let $(1)$ remove that piece (one of those pieces) and eat it. So far everyone is satisfied. Child (1) got what he believes to be exactly a third, and in the judgement of the children $(2)$ and $(3)$ the best piece (the claimed piece) is still available. The problem reduces therefore, to dividing whats fairly left, - and that's the classical problem of dividing a piece of cake between two claimants. Let that problem be solved by &quot;you cut, i choose&quot; and everybody is happy.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Solution&lt;/strong&gt; The Proposed solution is wrong; whats wrong with it?&lt;/p&gt;&#xA;" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-08-14T17:55:31.793" LastActivityDate="2010-08-15T00:16:12.787" Title="Cutting the Cake Problem." Tags="&lt;set-theory&gt;&lt;fair-division&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="2" />
  <row Id="2451" PostTypeId="2" ParentId="1222" CreationDate="2010-08-14T09:53:42.810" Score="0" ViewCount="" Body="&lt;p&gt;You might like Raghavan Narasimhan's &quot;Compact Riemann Surfaces&quot;. It does things like the construction of the Jacobian. It seems to suit your interests best.&lt;/p&gt;&#xA;" OwnerUserId="1119" LastActivityDate="2010-08-14T09:53:42.810" CommentCount="1" />
  <row Id="2452" PostTypeId="2" ParentId="1037" CreationDate="2010-08-14T10:02:40.683" Score="0" ViewCount="" Body="&lt;p&gt;It is possible to attach a precise meaning to &quot;dx&quot; as an infinitesimal using nonstandard analysis. There using model theory it is possible to rigorously make intuitive statements in calculus that we use all the time. This is a theory created by Abraham Robinson.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a brief informal discussion, see the appendix(by Stewart) of &quot;What is mathematics&quot; by Courant, Robbins and Stewart. Or see Robinson's book &quot;Nonstandard Analysis&quot;.&lt;/p&gt;&#xA;" OwnerUserId="1119" LastEditorUserId="1119" LastEditorDisplayName="" LastEditDate="2010-08-14T10:13:58.117" LastActivityDate="2010-08-14T10:13:58.117" />
  <row Id="2453" PostTypeId="2" ParentId="1" CreationDate="2010-08-14T10:40:09.497" Score="4" ViewCount="" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;How there can be different kinds of infinities?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is very simple to see. This is because of:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Claim: A given set $X$ and its power set $P(X)$ can never be in bijection. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proof: By contradiction. Let $f$ be any function from $X$ to $P(X)$. It suffices to prove $f$ cannot be surjective. That means that some member of $P(X)$ i.e., some subset of $S$, is not in the image of $f$.  Consider the set:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$T={ x\in X: x\not\in f(x) }.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For every $x$ in $X$ , either $x$ is in $T$ or not. If $x$ is in $T$, then by definition of $T$, $x$ is not in $f(x)$, so $T$ is not equal to $f(x)$. On the other hand,  if $s$ is not in $T$, then by definition of $T$, $x$ is in $f(x)$, so again $T$ is not equal to $f(x)$. Q.E.D.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus take any infinite set you like. Then take its power set, its power set, and so on. You get an infinite sequence of sets of increasing cardinality(Here I am skipping a little; but a use of the Schroeder-Bernstein theorem will fix things).&lt;/p&gt;&#xA;" OwnerUserId="1119" LastActivityDate="2010-08-14T10:40:09.497" />
  <row Id="2454" PostTypeId="2" ParentId="154" CreationDate="2010-08-14T10:54:17.793" Score="1" ViewCount="" Body="&lt;p&gt;Even natural numbers are abstract notions. Through abstract constructions the complex numbers can be demonstrated. So asking whether complex numbers is something like asking whether the natural numbers exist, whether the real numbers exist, or not. They are all things in the abstract domain of thought.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to reconcile it with your physical intuition, try the following. I suppose that your mental picture of the real numbers is by associating a line to it. Similarly, you can associate a plane to the set of complex numbers. Points are complex numbers. The real numbers will be the x-axis and imaginary numbers would be the y-axis. Addition is straightforward to visualize. For visualizing multiplication, write a complex number in the form $re^{i\theta}$. Then multiplication is a combination of stretching by a magnitude of $r$ and rotation by an angle of $\theta$. I hope this helps you in mental visualization.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In electrical networks, complex numbers are very good for analysis of alternating current networks. If some current has an imaginary component, it means that the current is leading or lagging the voltage by some &quot;phase difference&quot;, or &quot;angle&quot;. A resistance having imaginary components mean that it has capacitance or inductance. Only the &quot;real&quot; parts contribute to energy loss. But for imepdance calculations you have to consider capacitance and inductance too. A similar analysis can be made for every kind of control systems. So complex number are all very much there in nature and all sorts of engineering problems. I cannot stress their importance enough.&lt;/p&gt;&#xA;" OwnerUserId="1119" LastActivityDate="2010-08-14T10:54:17.793" CommunityOwnedDate="2011-04-15T20:45:29.080" CommentCount="1" />
  <row Id="2455" PostTypeId="1" AcceptedAnswerId="2618" CreationDate="2010-08-14T11:22:11.197" Score="5" ViewCount="366" Body="&lt;p&gt;I was quite impressed by a &lt;a href=&quot;http://mathoverflow.net/questions/8846/proofs-without-words/8847#8847&quot; rel=&quot;nofollow&quot;&gt;beautiful proof&lt;/a&gt;&lt;sup&gt;1&lt;/sup&gt;  of the formula for n'th triangular number — it's kind of a bijective proof built on the top of geometric summation idea.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question&lt;/strong&gt;: does this argument generalize to higher dimensions? i.e. can someone prove the formula for n-th &amp;laquo;k-dimensional simplex number&amp;raquo; this way?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/km6YV.png&quot; alt=&quot;the proof&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, the statement for k=3 is that n-th &lt;a href=&quot;http://en.wikipedia.org/wiki/Tetrahedral_number&quot; rel=&quot;nofollow&quot;&gt;pyramidal number&lt;/a&gt; is $\binom{n+2}{3}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(One obvious fallback would be observing that any ball in the pyramid is characterized by its coordinates&lt;sup&gt;2&lt;/sup&gt;, i.e. balls are counted by a &lt;a href=&quot;http://en.wikipedia.org/wiki/Multiset#Multiset_coefficients&quot; rel=&quot;nofollow&quot;&gt;negative binomial coefficient&lt;/a&gt; — which is exactly the desired result. That's a good proof, but needless to say, that's not the proof I'm looking for.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;1&lt;/sup&gt; original post by &lt;a href=&quot;http://math.stackexchange.com/users/274/mariano-suarez-alvarez&quot;&gt;Mariano Suárez-Alvarez&lt;/a&gt;, &lt;a href=&quot;http://math.stackexchange.com/questions/2260/proof-for-formula-for-sum-of-sequence-123-n/2288#2288&quot;&gt;cited&lt;/a&gt; by &lt;a href=&quot;http://math.stackexchange.com/users/876/vaughn-climenhaga&quot;&gt;Vaughn Climenhaga&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;2&lt;/sup&gt; k-simplex is the subset $x_0+\dots+x_k=1$, $x_i\ge0$ in $\mathbb{R}^{k+1}$, and n-th k-simplex number is the number of non-negative integer solutions of $x_0+\dots+x_k=n$&lt;/p&gt;&#xA;" OwnerUserId="152" LastEditorUserId="152" LastEditorDisplayName="" LastEditDate="2010-08-15T12:09:28.520" LastActivityDate="2010-08-16T19:40:37.027" Title="Geometric Proof of the Formula for Simplex Numbers" Tags="&lt;combinatorics&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="0" />
  <row Id="2456" PostTypeId="2" ParentId="2444" CreationDate="2010-08-14T11:38:27.783" Score="4" ViewCount="" Body="&lt;p&gt;This can be proved using the foundation axiom. Suppose there exists a class $A$ such that $A \in A$, then $A$ is a set and we can consider the set $\left\{A\right\}$. The foundation axiom says that $\exists B \in \left\{A\right\}$ such that $B \cap \left\{A\right\} = \emptyset$. Since $B \in \left\{A\right\}$, $B$ must be $A$. It follows that $A \in A=B$, thus $A \in B \cap \left\{A\right\} \neq \emptyset$. And you have a contraddiction.&lt;/p&gt;&#xA;" OwnerUserId="1065" LastEditorUserId="1065" LastEditorDisplayName="" LastEditDate="2010-08-15T21:21:09.563" LastActivityDate="2010-08-15T21:21:09.563" CommentCount="4" />
  <row Id="2457" PostTypeId="2" ParentId="2450" CreationDate="2010-08-14T12:02:33.697" Score="3" ViewCount="" Body="&lt;p&gt;After giving to (1) one of unclaimed pieces, less than 2/3 of cake can be left (in measure of (2) or (3)), so one of them will be unsatisfied.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; The right solution is the following.&lt;br&gt;&#xA;1. Let (1) divide cake into three equal pieces.&lt;br&gt;&#xA;2. Ask (2) and (3) &quot;What is the best one?&quot;. If their answers are different, we are done: each one of (2) and (3) gets what he (or she) wants, and (1) takes unclaimed piece. Otherwise go to step 3.&lt;br&gt;&#xA;3. Ask (2) and (3) &quot;Is there any other piece, you would satisfied with?&quot;. If at least one of the answers is &quot;Yes&quot; we are done similar to step 2. Otherwise go to step 4.&lt;br&gt;&#xA;4. (1) takes any of two unclaimed pieces.&lt;br&gt;&#xA;5. Thanks to the answer &quot;No&quot; at the step 3 we know, that at least 2/3 of the cake left in both measures (the one of (2) and the one of (3)). Therefore we can apply &quot;you cut, I choose&quot; method to the leaving 2 pieces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; This problem can be solved with arbitrary number of children using the algorithm for finding maximum matching in a bipartite graph. Suppose, number of children is n. Cutting the cake algorithm is the following.&lt;br&gt;&#xA;1. Let (1) divide cake into three equal pieces.&lt;br&gt;&#xA;2. Ask others the following: &quot;Please, list all the pieces, you will be satisfied with.&quot;&lt;br&gt;&#xA;3. Consider bipartite graph with 2n-1 vertices: n pieces of cake and all children except (1). Children is connected with a cake iff it will be satisfied with it.&#xA;4. Find maximum matching in this graph. Lets call vertex &quot;free&quot; iff it is not matched in this matching with another one.&lt;br&gt;&#xA;5. If there are no free children, give pieces of the cake according to the matching (and free piece of cake is for (1)). In this case we are done. Otherwise go to step 6.&lt;br&gt;&#xA;6. In the algorithm of maximum matching we are able to find its certificate, i.e. pair of sets (U,V) with U containing all free children and some non-free, and V containing some pieces of cake, such that 1) amount of elements in V is equal to amount of non-free elements in U; 2) every vertex in U has edges only to vertexes form V 3) there are no edges from chosen matching, connecting vertex from V to some vertex outside U. In our case it means that every piece of cake outside V is less than 1/n in all measures of children from U.&lt;br&gt;&#xA;7. Give pieces of cake to children outside U (except (1) ) according to chosen matching.&lt;br&gt;&#xA;8. Give any free piece of cake outside V to (1).&lt;br&gt;&#xA;9. Note, that according to 6, all pieces of cake given at 7 and 8 are unimportant for elements of U. Therefore there is enough cake left for them. Divide it according to this algorithm between elements of U (note, that (1) is not in U, and, therefore size of U is less than n).&lt;/p&gt;&#xA;" OwnerUserId="695" LastEditorUserId="695" LastEditorDisplayName="" LastEditDate="2010-08-14T20:52:36.297" LastActivityDate="2010-08-14T20:52:36.297" CommentCount="0" />
  <row Id="2458" PostTypeId="2" ParentId="2450" CreationDate="2010-08-14T12:21:25.233" Score="0" ViewCount="" Body="&lt;p&gt;Two things:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;You are assuming that (1) cut the cake into three pieces of equal value to him, and then took one piece. Now there are two pieces left for (2) and (3) to choose between, and you claimed that this problem could be solved by the original &quot;you cut, I choose&quot; method. But, since the cake is already cut, they cannot use that method.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;(1) cut the cake into three pieces of equal value to himself and then took a piece for himself, but this piece may have been more valuable to (2) or (3) than any of the other pieces, so they would not be satisfied.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I have an idea of how to do it, but its not complete:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let (1) take three pieces of string to mark off where he intends to cut the cake into three slices of what he believes are equally valued. Then there are two cases for people (2) and (3). If (2) and (3)'s optimal slices are different, then they should just take those slices and (1) can take the last one- everyone is satisfied.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If both (2) and (3) want the same slice, then remove the string and start again with (2) setting the string on the cake this time instead.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By doing this eventually you reduce the problem to the case where when any person cuts the cake into equal slices, the other two people will want the same piece.&lt;/p&gt;&#xA;" OwnerUserId="1089" LastActivityDate="2010-08-14T12:21:25.233" />
  <row Id="2459" PostTypeId="1" AcceptedAnswerId="2460" CreationDate="2010-08-14T12:57:43.883" Score="1" ViewCount="59" Body="&lt;p&gt;Hello, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;perhaps a somewhat silly question, but is there a generally accepted formal notation for denoting the number of columns a matrix has? Something I could use in the description of an algorithm like: &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;if horizontaldim(V) &amp;gt; x then&#xA;end if&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;or &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;if size(V,2) &amp;gt; x then &#xA;end if&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;or should I just use a description like&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;if number of columns in V &amp;gt; x then&#xA;end if&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;?&lt;/p&gt;&#xA;" OwnerUserId="358" LastActivityDate="2010-08-14T13:16:04.470" Title="formal notation for number of rows/columns in a matrix (e.g to use in pseudocode/algorithm environment in latex)" Tags="&lt;notation&gt;&lt;matrices&gt;" AnswerCount="1" />
  <row Id="2460" PostTypeId="2" ParentId="2459" CreationDate="2010-08-14T13:16:04.470" Score="3" ViewCount="" Body="&lt;p&gt;None that I know of, but I've seen numerical linear algebra books (e.g. Golub and Van Loan) just say something like $V\in\mathbb{R}^{m\times n}$ for a matrix V with m rows and n columns, and then use m and n in the following algorithm description.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;MATLAB notation, which some other people use, just says rows(V) and columns(V).&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-08-14T13:16:04.470" CommentCount="1" />
  <row Id="2462" PostTypeId="1" AcceptedAnswerId="2466" CreationDate="2010-08-14T13:22:01.570" Score="0" ViewCount="122" Body="&lt;p&gt;Is it true that for $n \in \mathbb{N}$ we can have $4n = x^{2} + y^{2}$ or $4n = x^{2} - y^{2}$ for $x,y \in \mathbb{N} \cup (0)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was just working out and this came out to be true from $n=1$ to $n=20$. After that i didn't try, but i would like to see as what will be a counter example.&lt;/p&gt;&#xA;" LastActivityDate="2010-08-14T14:13:38.433" Title="Multiples of 4 as sum or difference of 2 squares" Tags="&lt;number-theory&gt;&lt;abstract-algebra&gt;&lt;algebraic-number-theory&gt;" AnswerCount="2" />
  <row Id="2463" PostTypeId="2" ParentId="2450" CreationDate="2010-08-14T13:44:16.883" Score="0" ViewCount="" Body="&lt;p&gt;Fiktor is correct. &#xA;In &lt;a href=&quot;http://en.wikipedia.org/wiki/Proportional_%28fair_division%29&quot; rel=&quot;nofollow&quot;&gt;proportional division&lt;/a&gt;, you want to achieve the result, that each of the n player will end up with at lest 1/n cake in her own valuation. The procedure above does not achieve this. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Counter example: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Total valuation of cake&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Value P1 P2 P3&#xA;       3  5  8&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;So each player should end up with at least&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Value P1 P2 P3&#xA;       1  5/3  8/3&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Player 1 makes first cut, valuation after first cut:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Value P1 P2 P3&#xA;C1    1   3  1&#xA;C2    1   2  4&#xA;C3    1   1  3&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;P3 always chooses C2, P2 chooses C1.  So P1 is chooses C3. &#xA;So it remains:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;   Value P2 P3&#xA;    C1    3   1&#xA;    C2    2   4&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Say P3 now makes the second cut, so that C1 and C2 have same valuation:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;   Value P2 P3&#xA;    C1    &amp;gt;3   5/2&#xA;    C2    &amp;lt;2   5/2&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;so P2 chooses C1. Now P3 is left with a piece that she values at 5/2 which is less than the proportional 8/3. &lt;/p&gt;&#xA;" OwnerUserId="358" LastEditorUserId="358" LastEditorDisplayName="" LastEditDate="2010-08-15T00:16:12.787" LastActivityDate="2010-08-15T00:16:12.787" CommentCount="1" />
  <row Id="2465" PostTypeId="2" ParentId="2462" CreationDate="2010-08-14T13:52:14.960" Score="2" ViewCount="" Body="&lt;p&gt;It is true because&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ (n+1)^2 - (n-1)^2 = 4n $$&lt;/p&gt;&#xA;" OwnerUserId="171" LastActivityDate="2010-08-14T13:52:14.960" />
  <row Id="2466" PostTypeId="2" ParentId="2462" CreationDate="2010-08-14T13:55:09.913" Score="1" ViewCount="" Body="&lt;p&gt;Here is a hint: the form $x^2-y^2$ factors as $(x+y)(x-y)$.  Therefore, if you want to represent an integer $N$ as $x^2-y^2$, you can attempt to do so by choosing a factorization $N = ab$ and solving the linear system&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x+y = a$&lt;br&gt;&#xA;$x-y = b$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This system has the unique rational solution $x = \frac{a+b}{2}$, $y = \frac{a-b}{2}$.  This gives an integral solution iff $a$ and $b$ have the same parity.  Use this to show:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A positive integer $N$ is of the form $x^2 - y^2$ for $x,y \in \mathbb{Z}$ (possibly $0$) iff $N$ is odd or $N$ is divisible by $4$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, $4n$ is always of the form $x^2-y^2$.  You want a little more: that $x$ and $y$ are both nonzero.  Clearly $x$ cannot be zero, so you need to analyze the case $y = 0$ and show that whenever all possible solutions to $n = x^2 - y^2$ have $y = 0$, then there are nonzero $X$ and $Y$ such that $4n = X^2 + Y^2$.  This is not so hard...&lt;/p&gt;&#xA;" OwnerUserId="299" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2010-08-14T14:13:38.433" LastActivityDate="2010-08-14T14:13:38.433" CommentCount="5" />
  <row Id="2468" PostTypeId="1" CreationDate="2010-08-14T14:09:32.343" Score="6" ViewCount="221" Body="&lt;p&gt;I discovered while teaching Calc 2 that if you apply L'Hopital's rule to $\frac{x}{\sqrt{x^2+1}}$  you get $\frac{\sqrt{x^2+1}}{x}$, and if you apply L'Hopital &lt;em&gt;again&lt;/em&gt; you get $\frac{x}{\sqrt{x^2+1}}$ back.  In other words the L'Hopital &lt;em&gt;operator&lt;/em&gt; has a cycle of order two. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT (Thanks KennyTM): &quot;I suppose the L'Hopital operator should be defined on equivalence classes of pairs $(f(x),g(x)$ of differentiable functions with the fractional equivalence:  $(f(x),g(x))\equiv(h(x),k(x))$ if and only if $fk=gh$.&quot;  &lt;em&gt;This does not work&lt;/em&gt;.  But it doesn't really take pairs of functions to pairs of function, either.  So the first problem is to find out &lt;em&gt;how&lt;/em&gt; it is an operator.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Has anyone ever studied this operator?  Wikipedia tells me nothing. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yes, I know it is easier to find the limit by dividing through by $x$, but some students want to apply L'Hopital to &lt;em&gt;everything&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="318" LastEditorUserId="318" LastEditorDisplayName="" LastEditDate="2010-08-14T15:27:54.437" LastActivityDate="2010-08-14T17:18:30.607" Title="Has L'Hopital's Rule been studied as an operator?" Tags="&lt;calculus&gt;&lt;real-analysis&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="1" />
  <row Id="2469" PostTypeId="1" AcceptedAnswerId="2491" CreationDate="2010-08-14T14:46:48.510" Score="1" ViewCount="99" Body="&lt;p&gt;Let $G$ be a group. A normal subgroup $N$ of $G$ is called &lt;i&gt;ultracharacteristic&lt;/i&gt; if for every normal subgroup $U$ of $G$ the condition&#xA;$$&#xA;G/U \cong G/N&#xA;$$&#xA;implies that $U \ge N.$ What are instances of natural conditions on $G/N$ that make $N$ ultracharacteristic? For example, suppose $G/N$ has only inner automorphisms; is then $N$ ultracharacteristic? &lt;/p&gt;&#xA;" OwnerUserId="1100" LastActivityDate="2010-08-15T00:10:38.180" Title="Sufficient conditions for ultracharactericity" Tags="&lt;group-theory&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="2470" PostTypeId="2" ParentId="674" CreationDate="2010-08-14T14:51:28.360" Score="1" ViewCount="" Body="&lt;p&gt;&lt;em&gt;Two variables and one constraint&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we have an objective function of two real variables $f(x,y)$ subject to the constraint $g(x,y)=0$ and want to find the minima or maximum, provided that there is any, then the three following conditions must be satisfied:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f_{x}(x,y)=0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f_{y}(x,y)=0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$g(x,y)=0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\lambda g(x,y)=0$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\lambda g_{x}(x,y)=0$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\lambda g_{y}(x,y)=0$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\lambda $ is the Lagrange multiplier. Thus&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f_{x}(x,y)-\lambda g_{x}(x,y)=0$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f_{y}(x,y)-\lambda g_{y}(x,y)=0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for some $(x,y,\lambda )=(x^{\ast },y^{\ast },\lambda ^{\ast })$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we set $F(x,y,\lambda )=f(x,y)-\lambda g(x,y)$. &lt;em&gt;Observe that&lt;/em&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(This is the key point of my explanation)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$F_{x}(x,y,\lambda )=f_{x}(x,y)-\lambda g_{x}(x,y)=0$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$F_{y}(x,y,\lambda )=f_{y}(x,y)-\lambda g_{y}(x,y)=0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$F_{\lambda }(x,y,\lambda )=-g(x,y)=0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence we must find the solution $(x^{\ast },y^{\ast },\lambda ^{\ast })$ of&#xA;the following system of 2+1 equations:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$F_{x}(x,y,\lambda )=0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$F_{y}(x,y,\lambda )=0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$F_{\lambda }(x,y,\lambda )=0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Maximum or Minimum?&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To establish whether the function $f(x,y)$ has a maximum or a minimum one must define additional conditions that take into account the 2nd derivatives of $f$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$n$ &lt;em&gt;variables and&lt;/em&gt; $m$ &lt;em&gt;constraints&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a function $f$ of $n$ real variables, subject to $m$ constraints $g_{k}$, with $1\leq k\leq m$, we have $m$ Lagrange multipliers and must solve a system of $n+m$ equations. &lt;/p&gt;&#xA;" OwnerUserId="752" LastActivityDate="2010-08-14T14:51:28.360" CommunityOwnedDate="2010-08-14T14:51:28.360" />
  <row Id="2474" PostTypeId="1" CreationDate="2010-08-14T17:17:30.843" Score="3" ViewCount="569" Body="&lt;p&gt;In May's &quot;A Concise Course in Algebraic Topology&quot; I am supposed to calculate the fundamental group of the double torus. Can this be done using van Kampen's theorem and the fact that for (based) spaces $X, Y$: $\pi_1(X\times Y) = \pi_1(X)\times\pi_1(Y)$? Or do I need other theorems to prove this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe that this should equal $\pi_1(T)\times\pi_1(T)\times\pi_1(S^1)$ where $T$ is the torus minus a closed disc on the surface, but I do not know how to calculate $\pi_1(T)$.&lt;/p&gt;&#xA;" LastActivityDate="2010-08-18T19:53:56.677" Title="Fundamental group of the double torus" Tags="&lt;algebraic-topology&gt;&lt;homotopy-theory&gt;" AnswerCount="4" />
  <row Id="2475" PostTypeId="2" ParentId="2468" CreationDate="2010-08-14T17:18:30.607" Score="2" ViewCount="" Body="&lt;p&gt;I dunno if its been studied as a differential operator, and I kind of doubt it, but I think you could define it in a way similar to the way that you suggested.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You consider two pairs (f,g) and (h,k) equivalent if $f/g = h/k$ or rather $fk = gh$. Then the operator L acts on equivalence classes by the following operation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ L(f,g) = (Df, Dg) $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But for this operator to be well defined we must enforce that $L(h,k)$ is equivalent to $L(f,g)$. This means that we must have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$Df/Dg = Dh/Dk$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is not always true. Suppose that $(f,g) = (x, x^2)$ which is equivalent to $(h,k) = (1,x)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$Df/Dg = \frac{1}{2x} \neq 0 = Dh/Dk$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the L'Hospital operator is well defined on the equivalence classes of pairs $(f,g)$ with $g$ non-constant with the equivalence relation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(f,g)$ equiv $(h,k) \Leftrightarrow fk = hg $ and $ Df Dk = Dh Dg $.&lt;/p&gt;&#xA;" OwnerUserId="1089" LastActivityDate="2010-08-14T17:18:30.607" />
  <row Id="2479" PostTypeId="2" ParentId="2474" CreationDate="2010-08-14T17:40:02.647" Score="0" ViewCount="" Body="&lt;p&gt;Hi: Please see this link. This has a nice explanation as well as some more information : &lt;a href=&quot;http://math.ucr.edu/home/baez/algebraic_topology/Math205B_Mar02.pdf&quot; rel=&quot;nofollow&quot;&gt;http://math.ucr.edu/home/baez/algebraic_topology/Math205B_Mar02.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" LastActivityDate="2010-08-14T17:40:02.647" CommentCount="4" />
  <row Id="2481" PostTypeId="2" ParentId="107" CreationDate="2010-08-14T18:04:46.580" Score="1" ViewCount="" Body="&lt;p&gt;As other people has said, we model the situation by making a function $F(t)$ that returns the set of balls in the box at time $t$. Within this model, we have $F(\text{11:00}) = \emptyset$. That much is unambiguous. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But it's important to keep in mind that this is only a mathematical model. In the real world, there are many reasons to doubt the model will be accurate. For example, how can you move balls arbitrarily quickly, and how can the box have unlimited volume?.  There's no reason to think that, just because the limit is well-defined within our simplistic model, it has to be well-defined in the physical world. &lt;/p&gt;&#xA;" OwnerUserId="630" LastActivityDate="2010-08-14T18:04:46.580" />
  <row Id="2482" PostTypeId="2" ParentId="2474" CreationDate="2010-08-14T18:09:25.603" Score="9" ViewCount="" Body="&lt;p&gt;By van Kampen's theorem, what you get is actually&#xA;$$\pi_1(T)\ast_{\pi_i(S^1)}\pi_1(T)$$&#xA;which is an amalgamated product (a pushout in the category of groups).&#xA;Roughly speaking if you have two groups $G_1$ and $G_2$ and&#xA;embeddings $i_1$ and $i_2$ of a group $H$ in both then $G_1\ast_H\ast G_2$&#xA;is the group freely generated by the elements of $G_1$ and $G_2$&#xA;but identifying elements $i_1(h)$ and $i_2(h)$ for $h\in H$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now $\pi_1(T)$ can be computed using the fact that $T$ deformation retracts&#xA;to a bouquet of two circles. (Think about the standard torus; fix a point&#xA;and look at the circles through it going round the torus in the two natural&#xA;ways.)&lt;/p&gt;&#xA;" OwnerUserId="226" LastEditorUserId="226" LastEditorDisplayName="" LastEditDate="2010-08-14T18:16:03.217" LastActivityDate="2010-08-14T18:16:03.217" CommentCount="2" />
  <row Id="2483" PostTypeId="1" AcceptedAnswerId="16800" CreationDate="2010-08-14T18:58:24.240" Score="34" ViewCount="1293" Body="&lt;p&gt;This is a pet idea of mine which I thought I'd share.  Fix a prime $q$ congruent to $1 \bmod 4$ and define a sequence $F_n$ by $F_0 = 0, F_1 = 1$, and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle F_{n+2} = F_{n+1} + \frac{q-1}{4} F_n.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then $F_n = \frac{\alpha^n - \beta^n}{\alpha - \beta}$ where $\alpha, \beta$ are the two roots of $f(x) = x^2 - x - \frac{q-1}{4}$.  When $q = 5$ we recover the ordinary Fibonacci numbers.  The discriminant of $f(x)$ is $q$, so it splits $\bmod p$ if and only if $q$ is a quadratic residue $\bmod p$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $\left( \frac{q}{p} \right) = -1$, then the Frobenius morphism $x \mapsto x^p$ swaps $\alpha$ and $\beta$ (working over $\mathbb{F}_p$), hence $F_p \equiv -1 \bmod p$.  And if $\left( \frac{q}{p} \right) = 1$, then the Frobenius morphism fixes $\alpha$ and $\beta$, hence $F_p \equiv 1 \bmod p$.  In other words,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle F_p \equiv \left( \frac{q}{p} \right) \bmod p.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Quadratic reciprocity in this case is equivalent to the statement that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle F_p \equiv \left( \frac{p}{q} \right) \bmod p.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt;  Does anyone have any ideas about how to prove this directly, thereby proving quadratic reciprocity in the case that $q \equiv 1 \bmod 4$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My pet approach is to think of $F_p$ as counting the number of ways to tile a row of length $p-1$ by tiles of size $1$ and $2$, where there is one type of tile of size $1$ and $\frac{q-1}{4}$ types of tiles of size $2$.  The problem is that I don't see, say, an obvious action of the cyclic group $\mathbb{Z}/p\mathbb{Z}$ on this set.  Any ideas?&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2011-01-08T16:59:32.680" Title="Quadratic reciprocity via generalized Fibonacci numbers?" Tags="&lt;number-theory&gt;&lt;combinatorics&gt;" AnswerCount="1" CommentCount="14" FavoriteCount="14" />
  <row Id="2484" PostTypeId="2" ParentId="2389" CreationDate="2010-08-14T19:08:20.490" Score="3" ViewCount="" Body="&lt;p&gt;$\sum_{i=k}^n i = \sum_{i=0}^n i - \sum_{i=0}^{k-1} i = \frac{1}{2} \left[ n(n+1)  - (k-1)k \right]$&lt;/p&gt;&#xA;" OwnerUserId="1089" LastActivityDate="2010-08-14T19:08:20.490" />
  <row Id="2485" PostTypeId="1" AcceptedAnswerId="2762" CreationDate="2010-08-14T21:32:46.747" Score="1" ViewCount="335" Body="&lt;p&gt;A measurable rectangle in the cartesian product $X \times Y$ of 2 measurable spaces in $\mathbb{R}^{2}$ is of the form $A \times B$, where $A$ and $B$ are measurable subsets of $X$ and $Y$ respectively. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question&lt;/strong&gt; Can a measurable function on $\mathbb{R}^{2}$ be essentially unbounded on every measurable rectangle of postive measure?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What i did was to recognize a measurable function as limit of sequence of bounded measurable functions, but don't really know how to proceed from there.&lt;/p&gt;&#xA;" LastActivityDate="2010-08-19T16:10:32.930" Title="Measurable Functions behaviour on $\mathbb{R}^{2}$." Tags="&lt;analysis&gt;&lt;measure-theory&gt;" AnswerCount="3" CommentCount="4" />
  <row Id="2486" PostTypeId="2" ParentId="107" CreationDate="2010-08-14T22:03:53.230" Score="2" ViewCount="" Body="&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;As others have pointed out, there is no mathematical contradiction here. We have a sequence of sets $S_i$ with increasing cardinality whose intersection is the empty set; this is perfectly consistent with the axioms of set theory. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Nevertheless, there is a serious philosophical problem here. I understand that J. Thompson argued in his paper &lt;/p&gt;&#xA;&#xA;&lt;p&gt;‘Tasks and Super-Tasks’, &lt;em&gt;Analysis&lt;/em&gt;, XV, pp. 1–13, 1954-55,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;that there is indeed a &lt;em&gt;logical&lt;/em&gt; contradiction in this scenario. He concluded that such supertasks (i.e. tasks which involve doing an infinite number of actions in finite time) are not merely &lt;em&gt;physically impossible&lt;/em&gt; but also &lt;em&gt;logically impossible&lt;/em&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In lieu of precise definitions of these terms, consider, for comparison, traveling faster than the speed of light. If the theory of relativity is correct, it may very well be &lt;em&gt;physically impossible&lt;/em&gt; but its not &lt;em&gt;logically impossible&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, consider a cake that you eat at dinner &lt;em&gt;and&lt;/em&gt; that you also don't eat at dinner. This may violate the particular laws of the universe, but its also &lt;em&gt;logically impossible&lt;/em&gt;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Other philosophers have disputed Thompson's argument, for example: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;‘Tasks, Super-Tasks, and Modern Eleatics’, P. Benacerraf, &lt;em&gt;Journal of Philosophy&lt;/em&gt;, LIX, pp. 765–784, 1962.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand that Benacerraf argued that there is no logical connection between what is inside the box at 11 pm and what is inside the box at any time before 11 pm. Our universe tends to be a &quot;continuous&quot; one, but that is not logically required. Thus Benacerraf concludes there is no obvious reason why supertasks are logically impossible. I believe &#xA;Benacerraf's refutation is accepted by most of today's philosophers.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I have not read the original papers. My knowledge of this comes from the book &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0521896320&quot; rel=&quot;nofollow&quot;&gt;Paradoxes&lt;/a&gt; by R.M. Sainsbury, which was assigned reading in my freshman philosophy class. A good source for further reading is the &lt;a href=&quot;http://plato.stanford.edu/entries/spacetime-supertasks/&quot; rel=&quot;nofollow&quot;&gt;Supertask&lt;/a&gt; page of the Stanford Encyclopedia of Philosophy.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="317" LastEditorUserId="317" LastEditorDisplayName="" LastEditDate="2010-08-14T22:18:37.447" LastActivityDate="2010-08-14T22:18:37.447" />
  <row Id="2487" PostTypeId="1" AcceptedAnswerId="2495" CreationDate="2010-08-14T22:35:00.130" Score="8" ViewCount="495" Body="&lt;p&gt;There is a famous literature on $\displaystyle e = \sum\limits_{n=0}^{\infty} \frac{1}{n!}$. We know that $e$ is irrational as well as transcendental. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Question is: For each  $x_{n}$, sequence of $\pm{1}'s$, let $$f(x_{n}) = \sum\limits_{n=0}^{\infty} \frac{x_{n}}{n!}$$ for which sequences $(x_n)$ is $f(x_{n})$ rational?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also can we write $e$ as $\displaystyle e = \sum\limits_{n=1}^{\infty} \frac{x_{n}}{n}$ where $x_{n}$ is a sequence as above.&lt;/p&gt;&#xA;" LastActivityDate="2010-12-18T20:56:52.337" Title="Rationality of series $\sum \frac{1}{n!}$" Tags="&lt;real-analysis&gt;&lt;analysis&gt;" AnswerCount="3" FavoriteCount="1" />
  <row Id="2488" PostTypeId="2" ParentId="2487" CreationDate="2010-08-14T22:51:09.097" Score="5" ViewCount="" Body="&lt;p&gt;&lt;b&gt;Suggestion for the first question&lt;/b&gt; [I have not checked all the details myself]: Have you tried to modify the proof of the irrationality of $e$ to deal with your similar-looking series?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Specifically, referring to the standard proof given in &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Proof_that_e_is_irrational&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Proof_that_e_is_irrational&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which steps require modification if $\sum_{n=0}^{\infty} \frac{1}{n!}$ is replaced by &#xA;$\sum_{n=0}^{\infty} \frac{x_n}{n!}$ for $x_n \in \{ \pm 1 \}$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Hint for the second question&lt;/b&gt;: are you familiar with the proof of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Riemann_rearrangement_theorem&quot; rel=&quot;nofollow&quot;&gt;Riemann Rearrangement Theorem&lt;/a&gt;?  Can you adapt it here?&lt;/p&gt;&#xA;" OwnerUserId="299" LastEditorUserId="171" LastEditorDisplayName="" LastEditDate="2010-08-15T06:23:42.993" LastActivityDate="2010-08-15T06:23:42.993" CommentCount="2" />
  <row Id="2489" PostTypeId="1" AcceptedAnswerId="2490" CreationDate="2010-08-14T23:29:46.930" Score="1" ViewCount="213" Body="&lt;p&gt;When writing software, there are often situations where I need a parameter to be a floating point number x in the range 0 &amp;lt;= x &amp;lt;= 1. I don't know of a name for that category, but I think there must be one because it's such a useful categories. Perhaps there's a name in probability theory?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(If they don't have a name, I hereby declare them to be &quot;wombat numbers&quot;.)&lt;/p&gt;&#xA;" LastActivityDate="2010-08-14T23:37:23.640" Title="Is there a name for numbers between zero and one, inclusive?" Tags="&lt;terminology&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="2490" PostTypeId="2" ParentId="2489" CreationDate="2010-08-14T23:37:23.640" Score="4" ViewCount="" Body="&lt;p&gt;Sometimes I've heard it called the &quot;unit interval&quot; as it &quot;probability measures are functions from a boolean algebra to the unit interval&quot;&lt;/p&gt;&#xA;" OwnerUserId="172" LastActivityDate="2010-08-14T23:37:23.640" CommentCount="4" />
  <row Id="2491" PostTypeId="2" ParentId="2469" CreationDate="2010-08-15T00:10:38.180" Score="5" ViewCount="" Body="&lt;p&gt;If $K$ is any group, then $N = K \times 1$ is a normal subgroup of $G = K \times K$, with quotient $G/N$ isomorphic to $K$. But $N$ cannot be ultracharacteristic unless $K$ is trivial, since $G/(K \times 1)$ and $G/(1 \times K)$ are isomorphic. So the only condition that references $G/N$ alone and that guarantees $N$ to be ultracharacteristic is &quot;$G/N$ is trivial&quot;.&lt;/p&gt;&#xA;" OwnerUserId="631" LastActivityDate="2010-08-15T00:10:38.180" CommentCount="1" />
  <row Id="2492" PostTypeId="1" CreationDate="2010-08-15T00:14:14.180" Score="1" ViewCount="121" Body="&lt;p&gt;This question concerns  an implementation of the topmoumoute natural gradient (tonga) algorithm as described in page 5 in the paper Le Roux et al 2007 &lt;a href=&quot;http://research.microsoft.com/pubs/64644/tonga.pdf&quot; rel=&quot;nofollow&quot;&gt;http://research.microsoft.com/pubs/64644/tonga.pdf&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand that the basic idea is to augment stochastic gradient ascent with the covariance of the stochastic gradient estimates. Basically, the natural gradient approach multiplies a stochastic gradient with the inverse of the covariance of the gradient estimates in order to weight each component of the gradient by the variance of this component. We prefer moving into directions that show less variance during the stochastic gradient estimates: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ng \propto C^{-1} g$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since updating and inverting the covariance in an online optimisation setting is costly, the authors describe a ``cheap'' approximate update algorithm as described on page 5 as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$C_t = \gamma \hat{C}_{t-1} + g_tg_t^T$ where  $C_{t-1}$ is the low rank approximation at time step t-1. Writing  $\hat{C}_{t} = X_tX_t^T$ with $X_t =\sqrt{\gamma} X_{t-1}\ \ g_t]$ they use an iterative update rule for the  gram matrix $X_t^T X_T = G_t = &#xA;\begin{pmatrix}\gamma G_{t-1}&amp;amp;\sqrt{\gamma} X^T_{t-1}g_t\\ \sqrt{\gamma} g^T_t X_{t-1}&amp;amp;g_t^Tg_t\end{pmatrix}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They then state ``To keep a low-rank estimate of  $\hat{C}_{t} = X_tX_t^T$, we can compute its eigendecomposition and keep only the first k eigenvectors. This can be made low cost using its relation to that of the Gram matrix&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$G_t= X_t^T X_T$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$G_t = VDV^T$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$C_t = (X_tVD^{-\frac12})D(X_tVD^{-\frac12})^T$''&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because it's cheaper than updating and decomposing G at every step, they then suggest that you should update X for several steps using &#xA;$C_{t+b} = X_{t+b}X_{t+b}^T$ with $X_{t+b} = \left[\gamma U_t, \ \gamma^{\frac{b-1}{2}g_{t+1}},...\ \  \gamma^{-\frac12}g_{t+b-1}, \ \gamma^{\frac{t+b}{2}g_{t+b}}\right]$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can see why you can get $C_t$ from $G_t$ using the eigendecomposition. But I'm unsure about their update rule for X. The authors don't explain where U is coming from. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My first assumptions (by notation) was, that this is the first k eigenvectors of $C_t$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But this makes little sense. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My second assumption was that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$U = (X_tVD^{-\frac12})D^{\frac12}$ because then $UU^T \approx XX^T = C$, but again I'm not sure if this is reasonable, because I don't think that then&#xA;$[\gamma U, g]$ is a good approximation for $[\gamma X_{t-1}, g]$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if anyone has any advice as to what U should be, that would be much appreciated. I'm not getting good results using this method and want to make sure I have implemented it as intended. &lt;/p&gt;&#xA;" OwnerUserId="358" LastEditorUserId="358" LastEditorDisplayName="" LastEditDate="2010-08-15T00:21:54.913" LastActivityDate="2010-12-14T00:44:25.977" Title="tonga: low rank approximation of the natural gradient, question regarding Le Roux et al. 2007" Tags="&lt;linear-algebra&gt;&lt;algorithms&gt;&lt;optimization&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="2493" PostTypeId="2" ParentId="1819" CreationDate="2010-08-15T01:48:56.800" Score="1" ViewCount="" Body="&lt;p&gt;Consider a series of random variables $X_i$ to $X$. Let $F(x)$ be the cumulative distribution function of $X$ an $F_{n}$ be the cumulative distribution function of $X_i$. Convergence in distribution occurs when $F(x)$ is the pointwise limit of $F_{n}$, ie. $\lim_{n \to \infty} F_{n}(x)=F(x)$, whenever F is continuous at x.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Convergence in probability is defined as follows $\lim_{n \to \infty } Pr(|X_n-X| \ge e)=0$. It is stronger than convergence in distribution - I don't know how to prove this, but I can give the following explanation. For convergence in distribution, we only looked at the cumulative functions, so it didn't matter whether the variables were dependent or not. For convergence in probability, $X$ needs to either be a constant or $X_n$ must be dependent on X.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Almost sure convergence is defined as follows $Pr(\lim_{n \to \infty} X_n=X)=1$. On this &lt;a href=&quot;http://mathoverflow.net/questions/35211/reducing-limits-to-a-canoncial-form&quot; rel=&quot;nofollow&quot;&gt;MathOverflow thread&lt;/a&gt; I asked if there was a way to reduce limits to some kind of canonical form. I obtained the following results where $dif_x=|X_n-X|$:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Convergence in probability: $\forall e, d, n&gt;N_1(e,d): dif_x \ge e \text{ with } p &amp;lt; d $&lt;/li&gt;&#xA;&lt;li&gt;Almost sure convergence: ($\forall e, n&gt;N_2(e): dif_x &amp;lt; e) \text{ with } p=1$&lt;/li&gt;&#xA;&lt;li&gt;Alternative form: ($\forall e, n&gt;N_2(e): dif_x \ge e) \text{ with } p=0$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;We notice that for convergence in probability, we can reduce $p$ below any arbitrary positive value by choosing values of $n$ high enough, while in the second formula we can reduce it to 0. Additionally, the probability in the first formula looks at individual probabilities, while the second looks at probabilities over all possible values of e and n. From these forms it is clear that Almost Sure Convergence implies Convergence in probability. If convergence in probability is not occurring, then we can find some e, d, with with $p \ge d \forall n$. This alone will cause the $p$ in almost sure convergence to not be 0.&lt;/p&gt;&#xA;" OwnerUserId="123" LastEditorUserId="123" LastEditorDisplayName="" LastEditDate="2010-09-11T07:20:19.580" LastActivityDate="2010-09-11T07:20:19.580" />
  <row Id="2494" PostTypeId="2" ParentId="107" CreationDate="2010-08-15T02:25:27.037" Score="4" ViewCount="" Body="&lt;p&gt;No balls.  If $B(t)$ is the number of balls in the box at time $t$, then as you say, $B(t)$ is increasing, and $\lim_{t \to 11}B(t) = +\infty$.  But since every ball has been removed, $B(11)=0$.  All this is fine.  The &quot;paradox&quot; as I see it is that at the outset you expect the function $B$ to be continuous in some sense, but &lt;em&gt;a priori&lt;/em&gt; there's no reason why it should be, and indeed it is not.&lt;/p&gt;&#xA;" OwnerUserId="822" LastActivityDate="2010-08-15T02:25:27.037" />
  <row Id="2495" PostTypeId="2" ParentId="2487" CreationDate="2010-08-15T03:22:34.900" Score="14" ViewCount="" Body="&lt;p&gt;The numbers you describe are in fact all irrational.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given a fixed rational number $p/q$, any other rational number $a/b$ that approximates $p/q$ (but is not equal to $p/q$) can only do so to order $1/b$.  This can be seen as follows:&#xA;$$ p/q - a/b = (pb - qa)/qb,$$&#xD;&#xA;which is at least $(qb)^{-1}$ if it is not zero. So rationals cannot be approximated too well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now consider the series $$\sum x_n/n!$$ and the partial sums $S_j$.  The partial sums $S_k$ have denominator $k!$. Their error from the infinite sum is at most $\sum_{n&amp;gt;k} 1/n!$ which (from standard upper bounds via geometric series) little oh of $1/k!$ (as in the standard proof for $e$).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So basically, the problem is that the series converges &quot;too rapidly&quot; for the sum to be rational. A rational number cannot be approximated too well (well in the denominators) by other rationals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that there are similar Diophantine approximation results for algebraic numbers, cf. &lt;a href=&quot;http://en.wikipedia.org/wiki/Thue%E2%80%93Siegel%E2%80%93Roth_theorem&quot; rel=&quot;nofollow&quot;&gt;Roth's theorem&lt;/a&gt;. Unfortunately that's not applicable here, though, since the approximations $S_k$ are little oh of the denominator, not little oh of a power of the denominator.&lt;/p&gt;&#xA;" OwnerUserId="536" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-12-18T20:56:52.337" LastActivityDate="2010-12-18T20:56:52.337" CommentCount="7" />
  <row Id="2496" PostTypeId="1" AcceptedAnswerId="2507" CreationDate="2010-08-15T03:23:21.240" Score="1" ViewCount="291" Body="&lt;p&gt;A Banach Space is a vector space which is normed and is complete with respect to its norm (every cauchy sequence converges). Examples of Banach spaces include $\mathbb{R}^n$ and the space of bounded continuous functions with the norm $\Vert f(x) \Vert = \sup \lbrace \vert f(x) \vert \rbrace$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The set of all Bounded Linear Operators $\Omega$ on the space of all $C^\infty$ functions over $\mathbb{R}$ forms a Banach space:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;For two operators $A, B \in \Omega$ define $(A+B)(f) = Af + Bf$, and for a constant $c \in \mathbb{R}$ define $(cA)(f) = c Af$. This makes $\Omega$ into a vector space.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Define a norm by $ \Vert A \Vert_\Omega = \sup \lbrace \Vert Af \Vert : f \in C^\infty(\mathbb{R}), \Vert f \Vert \leq 1 \rbrace $, where $\Vert f \Vert$ for $f \in C^\infty(\mathbb{R})$ is some norm on that space.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Now my question is: Does the norm chosen for $C^\infty(\mathbb{R})$ affect the structure of $\Omega$ (i.e. its topology or whatever), and is there a natural choice for such a norm in this situation?&lt;/p&gt;&#xA;" OwnerUserId="1089" LastActivityDate="2010-08-15T14:38:52.453" Title="How to define a Norm on a Banach space of Operators" Tags="&lt;real-analysis&gt;&lt;complex-analysis&gt;&lt;functional-analysis&gt;" AnswerCount="2" CommentCount="8" />
  <row Id="2497" PostTypeId="1" AcceptedAnswerId="2499" CreationDate="2010-08-15T04:36:35.660" Score="0" ViewCount="366" Body="&lt;p&gt;Probably a proof (if any exist) that calls upon Knuth's up-arrow &lt;a href=&quot;http://en.wikipedia.org/wiki/Knuth%27s_up-arrow_notation&quot; rel=&quot;nofollow&quot;&gt;notation&lt;/a&gt; or &lt;a href=&quot;http://en.wikipedia.org/wiki/Busy_beaver&quot; rel=&quot;nofollow&quot;&gt;Busy Beaver&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="19" LastActivityDate="2010-08-15T22:44:06.537" CommunityOwnedDate="2010-08-15T14:44:00.880" Title="What is the biggest number ever used in a mathematical proof?" Tags="&lt;big-numbers&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" />
  <row Id="2498" PostTypeId="2" ParentId="2427" CreationDate="2010-08-15T06:00:01.373" Score="11" ViewCount="" Body="&lt;p&gt;Gerhard Michler has worked on a research program to show fairly convincingly that the possibility of infinitely many sporadic groups (with a uniform construction, but highly non-uniform properties) was quite real.  Roughly speaking the second round of sporadic groups was discovered looking for special configurations of centralizers of involutions, and he shows how this search can be continued, how it constructs almost all of the sporadic simple groups in a uniform fashion, and how it does not obviously stop there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is discussed in some detail in his books &lt;a href=&quot;http://www.ams.org/mathscinet-getitem?mr=2266036&quot; rel=&quot;nofollow&quot;&gt;MR2266036&lt;/a&gt; and &lt;a href=&quot;http://www.ams.org/mathscinet-getitem?mr=2583258&quot; rel=&quot;nofollow&quot;&gt;MR2583258&lt;/a&gt;, the Theory of Finite Simple Groups, volumes I and II.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, at least according to him, it should not be taken for granted that there are only finitely many sporadic groups, as there is a fairly reasonable procedure for possibly producing an infinite collection of basically unrelated finite simple groups (at least, nowhere near as related as groups of a fixed Lie type and rank).&lt;/p&gt;&#xA;" OwnerUserId="583" LastActivityDate="2010-08-15T06:00:01.373" CommentCount="1" />
  <row Id="2499" PostTypeId="2" ParentId="2497" CreationDate="2010-08-15T06:04:22.780" Score="4" ViewCount="" Body="&lt;p&gt;The mathematician Harvey Friedman observed a special finite form of Kurskal's Tree Theorem. Regarding this form, Friedman discusses the existence of a rapidly growing function he calls $TREE(n)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The $TREE$ sequence begins $TREE(1)=1$ and $TREE(2)=3$, but $TREE(3)$ is a number so extremely large that its weak lower bound is $(A(...A(1)...))$, where the number of A's is $A(187196)$, and $A()$ is a version of Ackermann's function: $A(x) = 2↑↑...↑x$ with $x-1 ↑s$ (Knuth up-arrows). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whereas Graham's Number is $A^{64}(4)$, the above mentioned lower bound is $A^{A(187196)}$. As you can imagine, the $TREE$ function keeps on growing rather quickly. For a discussion on the hierarchy of fast growing functions see here:&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Fast-growing_hierarchy&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Fast-growing_hierarchy&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are other examples of numbers greater than Graham's Number, as can be seen here: &lt;a href=&quot;http://en.wikipedia.org/wiki/Goodstein_function#Sequence_length_as_a_function_of_the_starting_value&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Goodstein_function#Sequence_length_as_a_function_of_the_starting_value&lt;/a&gt;, although I'm not sure if this number is larger than Friedman's $TREE(3)$   &lt;/p&gt;&#xA;" OwnerUserId="647" LastEditorUserId="647" LastEditorDisplayName="" LastEditDate="2010-08-15T22:44:06.537" LastActivityDate="2010-08-15T22:44:06.537" />
  <row Id="2500" PostTypeId="2" ParentId="2372" CreationDate="2010-08-15T06:14:50.973" Score="4" ViewCount="" Body="&lt;p&gt;There's probably not going to be an easy way to do this...  Consider two different examples of 15-letter &quot;permutations&quot;.  Then the number of permutations with that multiset of digits depends on the proportions of the chosen digits.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you really want to do this, you can sum $k!/(s(1)! s(2)! \cdots s(x)!)$ (called the &lt;a href=&quot;http://en.wikipedia.org/wiki/Multinomial_theorem&quot; rel=&quot;nofollow&quot;&gt;mulitnomial coefficient&lt;/a&gt;) over all partitions of $s(1)+s(2)+ \cdots +s(x)=k$ [in this partition s(i) is allowed to be zero and order is important] such that $s(i) \leq r(i)$ for all i.  The part s(i) says you have s(i) copies of the i-th letter.  The number of permutations with s(i) i-th letters is given as above, by the &lt;a href=&quot;http://en.wikipedia.org/wiki/Group_action&quot; rel=&quot;nofollow&quot;&gt;Orbit Stabiliser Theorem&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Although, this is only one step better than the caveman's counting formula:  sum_P 1 where P is the set of permutations you want to count.  I.e. just count them one-by-one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: While I'm making a few touch-ups, here's GAP code that implements the above formula.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;NrPermIdent:=function(k,T)&#xA;  local PSet,x;&#xA;  x:=Size(T);&#xA;  PSet:=Filtered(OrderedPartitions(k+x,x)-1,p-&amp;gt;ForAll([1..x],i-&amp;gt;p[i]&amp;lt;=T[i]));&#xA;  return Sum(PSet,p-&amp;gt;Factorial(k)/Product(p,i-&amp;gt;Factorial(i)));&#xA;end;;&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;where T is a list of bounds and k is the number of terms in the partition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;gap&amp;gt; NrPermIdent(15,[4,5,8,3]);&#xA;187957770&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;As another indication that finding a simple formula for these numbers is not going to be easy, observe that NrPermIdent(n,[n,k]) is equal to $\sum_{0 \leq i \leq k} {n \choose k}$ (which is considered a difficult sum to find -- see: &lt;a href=&quot;http://mathoverflow.net/questions/17202/&quot; rel=&quot;nofollow&quot;&gt;http://mathoverflow.net/questions/17202/&lt;/a&gt;).  I remember reading somewhere (most likely in &lt;a href=&quot;http://www.math.upenn.edu/~wilf/AeqB.html&quot; rel=&quot;nofollow&quot;&gt;A=B&lt;/a&gt;) that you can prove there is no &quot;closed-form&quot; solution for this.&lt;/p&gt;&#xA;" OwnerUserId="139" LastEditorUserId="139" LastEditorDisplayName="" LastEditDate="2010-08-31T06:32:02.473" LastActivityDate="2010-08-31T06:32:02.473" CommentCount="0" />
  <row Id="2502" PostTypeId="1" AcceptedAnswerId="2516" CreationDate="2010-08-15T08:58:00.283" Score="3" ViewCount="221" Body="&lt;p&gt;I've heard large cardinals talked about, and I (think I) understand a little about how you define them, but I don't understand why you would bother.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What's the simplest proof or whatever that requires the use of large cardinals? Is there some branch of mathematics that makes particularly heavy use of them?&lt;/p&gt;&#xA;" OwnerUserId="172" LastActivityDate="2010-08-15T16:07:52.910" Title="What are large cardinals for?" Tags="&lt;set-theory&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="2503" PostTypeId="2" ParentId="1422" CreationDate="2010-08-15T09:34:21.720" Score="0" ViewCount="" Body="&lt;p&gt;Hi--&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Eric, i had asked a similar question, in MO, which includes the condition of Differentiablility. Please check this one out. It may be of some use. &lt;a href=&quot;http://mathoverflow.net/questions/34059/if-f-is-infinitely-differentiable-then-f-coincides-with-a-polynomial&quot; rel=&quot;nofollow&quot;&gt;http://mathoverflow.net/questions/34059/if-f-is-infinitely-differentiable-then-f-coincides-with-a-polynomial&lt;/a&gt;&lt;/p&gt;&#xA;" LastActivityDate="2010-08-15T09:34:21.720" />
  <row Id="2504" PostTypeId="1" AcceptedAnswerId="2512" CreationDate="2010-08-15T11:03:08.890" Score="2" ViewCount="224" Body="&lt;p&gt;I just started reading some Schapira notes on &lt;a href=&quot;http://people.math.jussieu.fr/~schapira/lectnotes/AlTo.pdf&quot; rel=&quot;nofollow&quot;&gt;Algebra and Topology&lt;/a&gt;. Statement 1.7 is the following: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hom (Z, Ker(f, g)) ≃ Ker(Hom (Z, X) ⇉ Hom (Z, Y ))&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ker(f, g) = {x ∈ X; f(x) = g(x)}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried verifying this myself by drawing a diagram, but failed. Can anyone explain why this is true?&lt;/p&gt;&#xA;" OwnerUserId="123" LastEditorUserId="123" LastEditorDisplayName="" LastEditDate="2010-08-16T08:58:15.393" LastActivityDate="2010-08-16T09:04:56.390" Title="Prove Hom (Z, Ker(f, g)) ≃ Ker(Hom (Z, X) ⇉ Hom (Z, Y )) for sets" Tags="&lt;abstract-algebra&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="1" />
  <row Id="2507" PostTypeId="2" ParentId="2496" CreationDate="2010-08-15T11:51:49.750" Score="1" ViewCount="" Body="&lt;p&gt;First off there is no norm that can be put on $C^\infty(\mathbb{R})$, or even $C^\infty([0,1])$, that will make it into a Banach Space. I mention the last one because there are lots of obvious norms, sup norm, $L^p$ norms etc. But it will not be complete with respect to any norm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's fairly simple to see this, for the $L^2$ norm. Differentiation would then be an operator defined on the whole Banach space that has an adjoint. In fact, it is self-adjoint (this is by integration by parts). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now it is a theorem that any operator on a Banach space with an adjoint is automatically bounded (this is a consequence of Total Boundedness Principle). But of course differentiation is unbounded. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for your main question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you put some norm on $C^\infty(\mathbb{R})$ then yes it would greatly affect the set of bounded operators (even as a set, not just as a banach space). For example putting different $L^p$-norms would give you different duals spaces. While if you say restricted to bounded smooth functions and used the sup-norm, the dual would be measures on $\mathbb{R}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I guess the take-away message is that the definition of bounded operators involves the norm on the domain. So if you change that norm you can greatly change the bounded operators. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the way most of this stuff can be found in pretty much any Functional Analysis textbook.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I like Analysis NOW by Gert Pederson.&lt;/p&gt;&#xA;" OwnerUserId="1193" LastActivityDate="2010-08-15T11:51:49.750" CommentCount="2" />
  <row Id="2508" PostTypeId="1" AcceptedAnswerId="2509" CreationDate="2010-08-15T12:23:51.247" Score="2" ViewCount="258" Body="&lt;p&gt;Hay dear!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am solving the following inequality, please look at it and tell me am i write or not because this is an example in howard anton's book, i solved it on my own as given below but in the book it is solved differently, please look at my solution and tell me am i right or not and if not then explain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/bOwwL.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1050" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-08-15T20:14:52.360" LastActivityDate="2010-08-15T20:14:52.360" Title="Solving a quadratic inequality" Tags="&lt;algebra-precalculus&gt;" AnswerCount="4" CommentCount="5" />
  <row Id="2509" PostTypeId="2" ParentId="2508" CreationDate="2010-08-15T12:35:09.083" Score="4" ViewCount="" Body="&lt;p&gt;For ab to be positive either&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;a and b are both positive&lt;/li&gt;&#xA;&lt;li&gt;a and b are both negative&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Here, a=x-5 and b=x+2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They are both positive if x&gt;5. They are both negative if x&amp;lt;-2. Either of these will solve the problem&lt;/p&gt;&#xA;" OwnerUserId="123" LastEditorUserId="123" LastEditorDisplayName="" LastEditDate="2010-08-15T12:47:44.873" LastActivityDate="2010-08-15T12:47:44.873" CommentCount="3" />
  <row Id="2510" PostTypeId="2" ParentId="2508" CreationDate="2010-08-15T12:35:54.060" Score="0" ViewCount="" Body="&lt;p&gt;also have to look at the case where both are negative.&lt;/p&gt;&#xA;" LastActivityDate="2010-08-15T12:35:54.060" />
  <row Id="2512" PostTypeId="2" ParentId="2504" CreationDate="2010-08-15T13:09:29.157" Score="2" ViewCount="" Body="&lt;p&gt;Maybe something is missing in the original statement and is: whose kernel is that on the right hand side?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First of all, let's recall that $f$ and $g$ are maps from $X$ to $Y$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then that kernel on the right is obviously&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\ker (f_{*}, g_{*}) \subset \mathrm{Hom}(Z,X) \ ,&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;f_{*}, g_{*} : \mathrm{Hom}(Z,X) \longrightarrow \mathrm{Hom}(Z,Y)&#xA;$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;are the maps defined by $\varphi \mapsto f\circ \varphi$ and $\varphi \mapsto g\circ \varphi$, respectively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That said, the bijection you ask for simply sends every map &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\varphi : Z \longrightarrow \ker (f,g) \subset X&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;to $i\circ \varphi$, where $i: \ker (f,g) \hookrightarrow X$ is the inclusion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the other direction:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\psi \in \ker (f_{*}, g_{*}) \ \Longleftrightarrow \ f\circ \psi = g\circ \psi \ \Longleftrightarrow \ f(\psi (z)) = g(\psi (z))  \ \text{for all} \ z \in Z&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which means:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\psi (z) \in \ker (f,g)   \ \text{for all} \ z \in Z \ \Longleftrightarrow \ \mathrm{im} (\psi ) \subset \ker (f, g) \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, $\psi : Z \longrightarrow X$ factorises through the inclusion $i: \ker (f,g) \hookrightarrow X$. That is, since $i$ is injective, there exists a &lt;em&gt;unique&lt;/em&gt; $\varphi : Z \longrightarrow \ker (f,g)$ such that $\psi = i \circ \varphi$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, the map $LHS \longrightarrow RHS$ sends $\varphi $ to $i \circ \varphi$ and the map $RHS \longrightarrow LHS$ sends $\psi$ to $\varphi$ such that $\psi = i \circ \varphi$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both compositions are the identity by definition:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\psi \mapsto \varphi \mapsto i \circ \varphi = \psi&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\varphi \mapsto i \circ \varphi \mapsto \varphi \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conclusion: I've never seen &quot;two&quot; sets that were so much the same.   :-)&lt;/p&gt;&#xA;" OwnerUserId="664" LastEditorUserId="664" LastEditorDisplayName="" LastEditDate="2010-08-16T01:32:45.270" LastActivityDate="2010-08-16T01:32:45.270" CommentCount="2" />
  <row Id="2513" PostTypeId="2" ParentId="2504" CreationDate="2010-08-15T13:13:59.240" Score="0" ViewCount="" Body="&lt;p&gt;I think Hom (Z, X) ⇉ Hom (Z, Y) means some pair from this domain, rather than all possible pairs. If I am wrong, then this solution will need to be changed. If it is a single, pair, then it is pretty clear that it will be f◦, g◦&lt;/p&gt;&#xA;&#xA;&lt;p&gt;LHS=$K_1$ is all k:Z→X'=Ker(f,g)$\in$X and f(x)=g(x) for x$\in$X'&#xA;RHS=$K_2$ is all k:Z→X with f◦(k)=g◦(k) or f◦(k)(z)=g◦(k)(z) for z$\in$Z&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider $k \in K_1$. As f(x)=g(x),f◦(k)(z)=g◦(k)(z) for all z. So $k \in K_2$&#xA;Now consider $k \in K_2$. f◦(k)=g◦(k). If k(z) not in Ker(f,g) for some z, then f◦(k)(z)$\ne$g◦(k)(z), so contradiction. So $k \in K_1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, there is a bijection between $k \in K_1$ and $k \in K_2$&lt;/p&gt;&#xA;" OwnerUserId="123" LastEditorUserId="123" LastEditorDisplayName="" LastEditDate="2010-08-16T09:04:56.390" LastActivityDate="2010-08-16T09:04:56.390" CommentCount="1" />
  <row Id="2514" PostTypeId="1" AcceptedAnswerId="2523" CreationDate="2010-08-15T13:23:59.043" Score="2" ViewCount="573" Body="&lt;p&gt;I'm currently studying Polynomial Rings, but I can't figure out why they are Rings, not Fields. In the definition of a Field, a Set builds a Commutative Group with Addition and Multiplication. This implies an inverse multiple for every Element in the Set. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The book doesn't elaborate on this, however. I don't understand why a Polynomial Ring couldn't have an inverse multiplicative for every element (at least in the Whole numbers, and it's already given that it has a neutral element). Could somebody please explain why this can't be so?&lt;/p&gt;&#xA;" OwnerUserId="44" LastActivityDate="2010-08-17T04:13:15.363" Title="Why can't the Polynomial Ring be a Field?" Tags="&lt;ring-theory&gt;" AnswerCount="4" CommentCount="6" FavoriteCount="2" />
  <row Id="2516" PostTypeId="2" ParentId="2502" CreationDate="2010-08-15T13:58:31.957" Score="4" ViewCount="" Body="&lt;p&gt;I will leave the explaining of large cardinals to someone more knowleadgeable and explain one place where they are useful in tidying up things: category theory. In category theory, you are constantly facede with proper classes (the category of all sets, of all groups, etc.). To make things worse, you want to form functor categories but due to the sizes of the classes involved there is no way to do that in ZFC (and other set theories like NBG or MK soon hit a wall of their own). The Grothendieck axiom of universes is a large cardinal axiom (and I am told that it is rather mild compared to the large cardinals that set theorists routinely consider) that allows us to tidy up things for this and other constructions without paying attention to set sizes. Truth be told, this is mainly a convenience, as if one payed proper attention, and at the cost of circumlocutions (and vast tribulations) one could avoid them. But really, why go to all this trouble to settle what is a minor technicality not germane to the problem at hand, when you have this labor-savibng device at hand?&lt;/p&gt;&#xA;" OwnerUserId="1033" LastActivityDate="2010-08-15T13:58:31.957" />
  <row Id="2517" PostTypeId="2" ParentId="2514" CreationDate="2010-08-15T14:34:19.203" Score="6" ViewCount="" Body="&lt;p&gt;Because by definition, the only polynomial that can have a negative degree is 0, which is defined to have a degree of -$\infty$. Non-zero constants have degree 0.  You then have the degree equation: &lt;em&gt;deg (fg) = deg f + deg g&lt;/em&gt;  for any polynomials &lt;em&gt;f,g&lt;/em&gt;.  By inspection, any polynomial of degree &lt;em&gt;n&lt;/em&gt; $\geq$ 1 would need as an inverse a polynomial of degree -&lt;em&gt;n&lt;/em&gt;, which does not exist (i.e. what Agusti Roig said!) The set you want does exist, however: it is called the &lt;em&gt;field of rational functions&lt;/em&gt;, and is precisely the set of ratios of polynomials.  It is constructed the same way that the field of rational numbers is from the ring of integers.&lt;/p&gt;&#xA;" OwnerUserId="452" LastActivityDate="2010-08-15T14:34:19.203" />
  <row Id="2518" PostTypeId="2" ParentId="2514" CreationDate="2010-08-15T14:37:23.080" Score="1" ViewCount="" Body="&lt;p&gt;Consider $\mathbb{C}[x]$ the ring of polynomials with coefficients from $\mathbb{C}$. This is an example of polynomial ring which is not a field, because $x$ has no multiplicativ inverse.&lt;/p&gt;&#xA;" LastActivityDate="2010-08-15T14:37:23.080" />
  <row Id="2519" PostTypeId="2" ParentId="2496" CreationDate="2010-08-15T14:38:52.453" Score="5" ViewCount="" Body="&lt;p&gt;I'm not aware of any natural choices for norms on $C^\infty(\mathbb{R})$ at all.  (As mentioned in comments, things like the sup or $L^p$ norms are infinite for some $C^\infty(\mathbb{R})$ functions.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are &lt;em&gt;unnatural&lt;/em&gt; choices, contrary to Owen Sizemore's assertion.  For example, it is known that all infinite-dimensional separable Fr&amp;eacute;chet spaces (such as $C^\infty(\mathbb{R})$ in its usual topology) have Hamel dimension $2^{\aleph_0}$, hence are isomorphic as vector spaces.  (I believe this is due to Mazur, but couldn't find the exact reference.) So if $(X, ||\cdot||_X)$ is your favorite infinite-dimensional separable normed space, there is a linear isomorphism $T : C^\infty(\mathbb{R}) \to X$ (of course it is in general horribly discontinuous).  For $f \in C^\infty(\mathbb{R})$, set $||f|| := ||Tf||_X$.  Voil&amp;agrave;, a norm on $C^\infty(\mathbb{R})$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, under this norm $C^\infty(\mathbb{R})$ is isometrically isomorphic to $X$, so for all intents and purposes you are really working with $X$.  You can even make $C^\infty(\mathbb{R})$ into a Banach space or even a Hilbert space by choosing $X$ accordingly.  And the space of bounded operators on $(C^\infty(\mathbb{R}), ||\cdot||)$ is isometrically isomorphic to the space of those on $X$. Thus for any infinite-dimensional separable normed space $X$ you can realize the space of bounded operators $L(X)$ in this way, so the choice of norm on $C^\infty(\mathbb{R})$ certainly does affect the structure of $L(C^\infty(\mathbb{R}))$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This illustrates in some sense that the norm contains almost all of the structure of a normed space; if you drop it, you are left with just the vector space structure, which determines almost nothing.  &quot;Choosing a norm&quot; on a space is not really a sensible thing to do, since in effect it is really just choosing a normed space which could be almost totally unrelated to the original space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, it should be pointed out that the space of bounded linear operators on an &lt;em&gt;incomplete&lt;/em&gt; normed space is not a Banach space.  Proof: Let $X$ be an incomplete normed space, let $\lbrace x_n\rbrace$ be a Cauchy sequence in $X$ with no limit, and let $f \in X^*$ be a nonzero bounded linear functional on $X$.  Define operators $T_n$ by $T_n x = f(x) x_n$.  Then $\lbrace T_n \rbrace$ is Cauchy in operator norm, but does not converge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The correct theorem is that for normed spaces $X,Y$, the space $L(X,Y)$ of bounded linear operators from $X$ to $Y$ is a Banach space iff $Y$ is a Banach space.&lt;/p&gt;&#xA;" OwnerUserId="822" LastActivityDate="2010-08-15T14:38:52.453" />
  <row Id="2520" PostTypeId="2" ParentId="2514" CreationDate="2010-08-15T14:52:27.693" Score="0" ViewCount="" Body="&lt;p&gt;The units of $D[x]$ are exactly the units of $D$, when $D$ is a domain.&lt;/p&gt;&#xA;" OwnerUserId="589" LastEditorUserId="589" LastEditorDisplayName="" LastEditDate="2010-08-15T20:27:53.500" LastActivityDate="2010-08-15T20:27:53.500" CommentCount="2" />
  <row Id="2521" PostTypeId="2" ParentId="2508" CreationDate="2010-08-15T15:25:54.520" Score="2" ViewCount="" Body="&lt;p&gt;Casebash's answer is very good.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a second answer. You can apply the following&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Theorem: If the roots $x_{1},x_{2}$ of $f(x)=ax^{2}+bx+c$ are real and $x_{1}\neq x_{2}$ (with $x_{1} &amp;lt; x_{2}$), then, the signal of $f(x)$ is:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;opposite to the signal of $a$ for $x\in \left[ x_{1},x_{2}\right] $,&lt;/li&gt;&#xA;&lt;li&gt;the same of $a$ for $x\in \left] -\infty ,x_{1}\right[ \vee x\in \left]&#xA;x_{2},-\infty \right[ $.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Since in your case $a=1&gt;0$, $x_{1}=-2&amp;lt;5=x_{2}$, you have $x^{2}-3x-10&gt;0$ for &#xA;$x\in \left] -\infty ,-2\right[ \vee x\in \left] 5,\infty \right[ $.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Addendum:&lt;/em&gt; A possible proof of this theorem is to use the explanation of Casebash, taking into consideration that $ax^{2}+bx+c=a(x-x_1)(x-x_2)$&lt;/p&gt;&#xA;" OwnerUserId="752" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2010-08-15T17:05:16.697" LastActivityDate="2010-08-15T17:05:16.697" />
  <row Id="2522" PostTypeId="2" ParentId="2502" CreationDate="2010-08-15T16:07:52.910" Score="2" ViewCount="" Body="&lt;p&gt;G. Rodrigues's specific answer gets at the general issue: large cardinals are used to examine how much more one can proof in ZFC set theory.  The first time I discovered large cardinals (in Jech's 2000 book &lt;em&gt;Set Theory&lt;/em&gt;), I was amazed.  A large cardinal is just a &quot;very big&quot; set, after all, but I did not realize that the existence of such a set changed the nature of what was mathematically provable.  For example, there is, according to Jech, the event that started it all: Ulam's work on the &lt;em&gt;problem of measure&lt;/em&gt;.  It is well-known that Lebesque measure over the reals is not defined for all sets, but it turns out to be undecidable in ZFC alone if &lt;em&gt;any&lt;/em&gt; non-trivial measure on the reals exists at all.  In order to get such a measure, one must assume the existence of a large cardinal, which is now called a &lt;em&gt;measurable cardinal&lt;/em&gt;.  So I think of large cardinals as things that change the very nature of the mathematical &quot;plumbing&quot;.  Deep stuff.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On a more practical level, I think it was Dudley who said that large cardinals can be useful for seeing why a proof is failing: if a proof is not working, seeing if it fails at a large cardinal can provide insight.&lt;/p&gt;&#xA;" OwnerUserId="452" LastActivityDate="2010-08-15T16:07:52.910" />
  <row Id="2523" PostTypeId="2" ParentId="2514" CreationDate="2010-08-15T16:10:41.893" Score="26" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\;\;\quad\rm  x \; f(x) = 1 \;$ in $\rm R[x] \;\Longrightarrow \; 0 = 1 \;$ in $\rm R \;\;$  by evaluating at $\; x = 0 $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;NOTE&lt;/strong&gt; This has an instructive &lt;strong&gt;universal&lt;/strong&gt; interpretation: if $\rm x$ is a unit in $\rm R[x]$ then so too is every $\rm R$-algebra element $\rm r\:$, as follows by evaluating $\;\rm x \; f(x) = 1 \;$ at $\rm\; x = r \;$. Therefore to present a counterexample we need only exhibit a nonunit in some $\rm R$-algebra. A natural choice is the nonunit $\;\rm 0\in R\;$.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-08-17T04:13:15.363" LastActivityDate="2010-08-17T04:13:15.363" CommentCount="2" />
  <row Id="2524" PostTypeId="2" ParentId="2504" CreationDate="2010-08-15T16:14:14.103" Score="4" ViewCount="" Body="&lt;p&gt;This is a special case of the theorem that covariant hom-functors&#xA;(here $\mathrm{Hom}(Z,\underline{\ })$) preserve limits.&#xA;This is Theorem 1 in section V.4 of Mac Lane's&#xA;&lt;em&gt;Categories for the Working Mathematician&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-08-15T16:14:14.103" CommentCount="5" />
  <row Id="2525" PostTypeId="2" ParentId="2508" CreationDate="2010-08-15T16:41:06.920" Score="2" ViewCount="" Body="&lt;p&gt;If you graph the function $y=x^2-3x-10$, you can see that the solution is $x&amp;lt;-2$ or $x&gt;5$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/UIwCt.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="354" LastActivityDate="2010-08-15T16:41:06.920" />
  <row Id="2527" PostTypeId="1" CreationDate="2010-08-15T18:25:46.097" Score="2" ViewCount="90" Body="&lt;p&gt;For a while I've been looking for some kind of software that will let me create and cross-reference definitions. The way I imagine this working would be, for instance: I create a file named &quot;Hopf algebra&quot;, I type in &quot;A Hopf algebra over a field k is a k-bialgebra such that...&quot; and follow with the definition of a Hopf algebra, and things like &quot;bialgebra&quot; and &quot;field&quot; would be links to those definitions. Essentially something like a DIY maths encyclopedia. Bonus things would be allowing me to put things into categories (e.g. ring theory vs. group theory) and colour-coding, and some manner of LaTeX support would be pretty essential.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I cannot be the only one who's had this idea, but so far all my Google searches have been turning up empty - I figure there's some key word I'm missing. Has anyone heard of something like this before?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: What I'm looking for is a sort of better-organised and better-crossref'd software version of the giant stack of index cards I use right now, not a wiki or anything communal like that.&lt;/p&gt;&#xA;" OwnerUserId="1203" LastEditorUserId="1203" LastEditorDisplayName="" LastEditDate="2010-08-15T19:38:44.367" LastActivityDate="2010-08-16T08:59:16.920" Title="Software for creating a definition library" Tags="&lt;math-software&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="2528" PostTypeId="2" ParentId="2527" CreationDate="2010-08-15T18:49:56.067" Score="2" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://groupprops.subwiki.org/wiki/Main_Page&quot; rel=&quot;nofollow&quot;&gt;http://groupprops.subwiki.org/wiki/Main_Page&lt;/a&gt; is something like this for group theory.  There is also the &lt;a href=&quot;http://ncatlab.org/nlab/show/HomePage&quot; rel=&quot;nofollow&quot;&gt;nLab&lt;/a&gt;.  Both of them run on wiki software; do you have something else in mind?&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-08-15T18:59:15.063" LastActivityDate="2010-08-15T18:59:15.063" CommentCount="3" />
  <row Id="2529" PostTypeId="1" AcceptedAnswerId="2546" CreationDate="2010-08-15T19:00:18.010" Score="1" ViewCount="144" Body="&lt;p&gt;This problem is taken from I.N. Herstein&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Problem&lt;/strong&gt; If $G$ is a group and $a \in G$ if of finite order and has only a finite number of conjugates in $G$, prove that these conjugates of $a$ generate a finite normal subgroup of $G$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to see a solution for this and also i would like to know whether the conjugates always generate a normal subgroup. &lt;/p&gt;&#xA;" LastActivityDate="2010-08-15T23:34:23.133" Title="Conjugates Generating a normal subgroup." Tags="&lt;group-theory&gt;&lt;abstract-algebra&gt;" AnswerCount="1" CommentCount="8" />
  <row Id="2530" PostTypeId="1" AcceptedAnswerId="2556" CreationDate="2010-08-15T19:14:02.973" Score="2" ViewCount="553" Body="&lt;p&gt;Although i am very much new to &quot;Analytic Number Theory&quot;, there are some non mathematical questions which puzzle me. First of all, why was &lt;strong&gt;G.H.Hardy&lt;/strong&gt; so much keen to have an elementary proof of the &lt;strong&gt;Prime Number Theorem&lt;/strong&gt;. He also stated that producing such a proof, will change the complexion of Mathematics, but nothing like that has happened. What was on Hardy's mind? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The elementary proof although has some intricate tricks involved, i am curious to know whether the methodology can be applied for attacking more complex problems. I have seen that the analytic proof has a continuation and is not over and discuss some more interesting properties regarding the $\zeta$ function. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also saw this thread &lt;em&gt;http://mathoverflow.net/questions/16735/is-a-non-analytic-proof-of-dirichlets-theorem-on-primes-known-or-possible&lt;/em&gt;. Are there any such theorem's which piques peoples curiosity in getting an elementary proof.(While writing this FLT comes to my mind!!).&lt;/p&gt;&#xA;" LastActivityDate="2010-08-16T06:35:38.190" Title="Elementary proof of the Prime Number Theorem - Need?" Tags="&lt;soft-question&gt;&lt;intuition&gt;&lt;applications&gt;" AnswerCount="1" CommentCount="10" FavoriteCount="2" />
  <row Id="2531" PostTypeId="2" ParentId="2355" CreationDate="2010-08-15T19:34:06.407" Score="4" ViewCount="" Body="&lt;p&gt;Let p,p' = 2 mod 3 be prime.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose that G is a group with the following properties:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(i)  The 3-Sylow subgroup of G is cyclic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(ii) The number of elements of order 3 in G is 2*p*p'&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Claim: There exists a group which _in_addition_ satisfies either:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(iii)(a); G is simple; OR (iii)(b): 3 exactly divides |G|, and G surjects onto Z/3Z.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remark: Suppose that G is a group with cyclic 3-Sylow subgroups. Since all&#xA;3-Sylow subgroups are conjugate, this implies that every subgroup of&#xA;order 3 in G is also conjugate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proof: We may assume that G is not simple, and hence admits a proper normal&#xA;subgroup H.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose that H is a normal subgroup of G. If 3 divides |H|, then&#xA;H contains a subgroup of G of order 3. All such subgroups are&#xA;conjugate in G, and since H is normal, they all lie in H. Thus&#xA;G and H have the same number of elements of order 3.&#xA;Moreover, the 3-Sylow of H is clearly cyclic, and so we may&#xA;replace G by H.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose that (|H|,3) = 1. Then G/H still has a cyclic 3-Sylow subgroup,&#xA;and hence every element of order 3 in G/H is conjugate. This implies that&#xA;every element of order 3 in G/H lifts to an element of order 3 in G. Thus&#xA;G/H has at most&#xA;2*p*p' elements of order three. Yet the number of elements&#xA;of order 3 is (by some theorem of Frobenius?)&#xA;divisible by 2 and congruent to -1 mod 3. Thus the&#xA;number of element of G/H of order 3 is either 2 or 2*p*p'. In&#xA;the latter case, we replace G by G/H. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We may now assume that G/H has exactly 2 elements of order 3.&#xA;Clearly G/H has a unique subgroup of order 3, which must be normal.&#xA;Thus G/H and G have a quotient of order |G/H|/3, and thus G contains&#xA;a normal subgroup F of order 3|H|. As above, we may replace G by F.&#xA;Note, however, that 3 exactly divides |F|, and &#xA;F surjects onto Z/3Z, and thus, by Schur-Zassenhaus&#xA;(overkill in this case, of course), F is a semi-direct product.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;My understanding of Jack's argument:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose that p' = 2, so G has 4p elements of order 3.&#xA;We still assume that the 3-Sylow of G is cyclic.&#xA;G acts by conjugation on the 2p subgroups of order p, giving&#xA;a map from G--&gt;S_2p. Let Q be one of these subgroups,&#xA;and let M be the normalizer of Q. Let P be a 3-Sylow containing Q.&#xA;Certainly [G:M] = 2p, by the orbit-stabilizer formula.&#xA; If X is a subgroup of G containing M,&#xA;then [G:X] = 1,2,p or 2p. Let N be the normalizer of P. Clearly&#xA;M contains N. Thus P is contained in M, and thus P is a 3-Sylow of M.&#xA;Similarly, P is also a 3-Sylow subgroup of X. The Sylow theorems applied&#xA;to M and X thus imply that [X:N] and [M:N] are both 1 mod 3, and&#xA;thus [X:M] is 1 mod 3. Hence, since [X:M] = 1,2,p or 2p, and, since&#xA;p = 2 mod 3, either X = M or X = G. Thus M is a maximal subgroup&#xA;of G, and hence the action of G is primitive. Now we suppose:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assumption (*): The only primitive subgroups of S_2p are A_2p and S_2p.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, we deduce that some quotient of G is either A_2p or S_2p.&#xA;If G is simple, we deduce that G = A_2p, which doesn't have cyclic&#xA;3-Sylows if 2p &gt; 5, and if G is a semi-direct product of Z/3Z with&#xA;a  group of order co-prime to 3, then G cannot surject onto A_2p or&#xA;S_2p if p &gt; 2. It seems to follow that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If p = 5,8 mod 9, and Assumption (*) holds, then G cannot have 4p elements&#xA;of order 3. (The congruence conditions on p ensure that the 3-Sylow of G is&#xA;cyclic).&lt;/p&gt;&#xA;" LastActivityDate="2010-08-17T04:17:46.750" CommunityOwnedDate="2010-08-15T23:46:45.433" CommentCount="10" />
  <row Id="2532" PostTypeId="1" CreationDate="2010-08-15T19:42:56.243" Score="-3" ViewCount="113" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Possible Duplicate:&lt;/strong&gt;&lt;br&gt;&#xA;  &lt;a href=&quot;http://math.stackexchange.com/questions/2508/solving-a-quadratic-inequality&quot;&gt;Solving a quadratic inequality&lt;/a&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;p&gt;Suppose we have (x-5)(x+2) &gt; 0 &lt;br&gt;&#xA;can we solve it like this &lt;br&gt;&#xA;(x-5) &gt; 0 and (x+2) &gt;0 &lt;br&gt;&#xA;x &gt; 5     and  x &gt; -2&lt;br&gt;&#xA;please help.&lt;/p&gt;&#xA;" OwnerUserId="1050" LastActivityDate="2010-08-15T19:59:45.830" ClosedDate="2010-08-15T20:09:19.847" Title="solving inequalities" Tags="&lt;algebra-precalculus&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="2534" PostTypeId="2" ParentId="2532" CreationDate="2010-08-15T19:45:41.293" Score="1" ViewCount="" Body="&lt;p&gt;If $ab&gt;0$ does it always imply that $a&gt;0$ or $b&gt;0$. You can also have both $a$ and $b$ to be less than $0$. So in your question it can also be $(x-5)&amp;lt;0$ along with $(x+2)&amp;lt;0$.&lt;/p&gt;&#xA;" LastActivityDate="2010-08-15T19:59:45.830" CommentCount="5" />
  <row Id="2535" PostTypeId="2" ParentId="1829" CreationDate="2010-08-15T20:51:41.570" Score="1" ViewCount="" Body="&lt;p&gt;On previous web sites, I've seen many people asking for more examples on Singular Value Decomposition (SVD).  Strang's LA book does this, but some folk want more.  Another thing that some undergraduates do, but then forget, but then need to do again in other courses is QR factorization (Gram-Schmidt); so, you should have multiple examples.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Those are 2 things that are pretty straight-forward for most, but examples (after the theory) would be good.&lt;/p&gt;&#xA;" LastActivityDate="2010-08-15T20:51:41.570" CommunityOwnedDate="2010-08-15T20:51:41.570" />
  <row Id="2536" PostTypeId="1" AcceptedAnswerId="2541" CreationDate="2010-08-15T20:54:45.790" Score="7" ViewCount="463" Body="&lt;p&gt;Please see the problem &lt;strong&gt;5&lt;/strong&gt; of the given link: &lt;a href=&quot;http://www.artofproblemsolving.com/Forum/resources.php?c=2&amp;amp;cid=59&amp;amp;year=2005&amp;amp;sid=722231ab4ec5ce280584eb8f24f07656&quot; rel=&quot;nofollow&quot;&gt;http://www.artofproblemsolving.com/Forum/resources.php?c=2&amp;amp;cid=59&amp;amp;year=2005&amp;amp;sid=722231ab4ec5ce280584eb8f24f07656&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It asks us to prove that $$\sum\limits_{n=1}^{\infty} \frac{1}{n^n} = \int\limits_{0}^{1} x^{x} \ dx $$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Natural question which one gets by seeing this type of question is, find all continuous functions which satisfy $$\int\limits_{-\infty}^{\infty} f(x) \ dx = \sum\limits_{n= -\infty}^{\infty} f(n)$$&lt;/p&gt;&#xA;" LastEditorUserId="171" LastEditorDisplayName="" LastEditDate="2010-08-16T07:38:54.367" LastActivityDate="2010-08-16T08:13:16.903" Title="Finding $f$ such that $ \int f = \sum f$" Tags="&lt;real-analysis&gt;&lt;sequences-and-series&gt;&lt;complex-analysis&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="2" />
  <row Id="2537" PostTypeId="1" AcceptedAnswerId="2545" CreationDate="2010-08-15T21:06:02.550" Score="5" ViewCount="152" Body="&lt;p&gt;In a &lt;a href=&quot;http://pauli.uni-muenster.de/~munsteg/arnold.html&quot; rel=&quot;nofollow&quot;&gt;talk by V.I. Arnold&lt;/a&gt;, this is said:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;When I was a first-year student at the Faculty of Mechanics and Mathematics of the Moscow State University, the lectures on calculus were read by the set-theoretic topologist L.A. Tumarkin, who conscientiously retold the old classical calculus course of French type in the Goursat version. He told us that integrals of rational functions along an algebraic curve can be taken if the corresponding Riemann surface is a sphere and, generally speaking, cannot be taken if its genus is higher, and that for the sphericity it is enough to have a sufficiently large number of double points on the curve of a given degree (which forces the curve to be unicursal: it is possible to draw its real points on the projective plane with one stroke of a pen).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I would like to understand the mathematical part of this. What do I need to know to see why this makes sense? Where can I get enough of the background to understand it fully?&lt;/p&gt;&#xA;" LastActivityDate="2010-08-16T02:39:15.080" Title="Toward &quot;integrals of rational functions along an algebraic curve&quot;" Tags="&lt;calculus&gt;&lt;reference-request&gt;&lt;riemann-surface&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="3" />
  <row Id="2538" PostTypeId="2" ParentId="2324" CreationDate="2010-08-15T21:06:29.757" Score="1" ViewCount="" Body="&lt;p&gt;You might get &lt;em&gt;some&lt;/em&gt; results with the method you mentioned with 100 inputs to the ANN. It might be good enough for your application, but if it's not, you'll need to apply some pre-processing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What kind of pre-processing will do the trick is largely dependent on the nature of your signal and what you're trying to get. Try thinking about what should not be important (what characteristics of the signal) to end result and try to filter that out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, if DC component is not important, remove it from all signals.&#xA;If amplitude is not important, scale everything to [0,1] independently.&#xA;If you have some high frequency noise, filter the signal with low-pass.&#xA;More generally, think what characteristics of the spectra of the signal might be important.&#xA;One of the tricks is to look at auto-correlation of the signal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With or without pre-processing, coarser quantization or coding with Gray's code might additionally help.&lt;/p&gt;&#xA;" OwnerUserId="956" LastActivityDate="2010-08-15T21:06:29.757" />
  <row Id="2539" PostTypeId="2" ParentId="2527" CreationDate="2010-08-15T21:19:00.933" Score="2" ViewCount="" Body="&lt;p&gt;You could try using Wikipedia with &lt;a href=&quot;http://www.google.com/sidewiki/intl/en/index.html&quot; rel=&quot;nofollow&quot;&gt;Google SideWiki&lt;/a&gt; or any other software that lets you annotate a site.&lt;/p&gt;&#xA;" OwnerUserId="123" LastEditorUserId="123" LastEditorDisplayName="" LastEditDate="2010-08-16T08:59:16.920" LastActivityDate="2010-08-16T08:59:16.920" CommentCount="1" />
  <row Id="2541" PostTypeId="2" ParentId="2536" CreationDate="2010-08-15T21:49:52.980" Score="1" ViewCount="" Body="&lt;p&gt;Pick any sequence $f_k, -\infty &amp;lt; k &amp;lt; \infty$ such that $\sum_{n=-\infty}^{\infty} f_n$ exists, and any other two sequences $g_k, h_k, -\infty &amp;lt; k &amp;lt; \infty$ such that $\sum_{n=-\infty}^{\infty} g_n = 0$ and $h_k\rightarrow 0, k\rightarrow \pm \infty$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Put $f(x) = f_k, x = k$ and construct $f(x)$ on non-integer points however you want to satisfy $f(m)=f_m$, $f(m+1)=f_{m+1}$ and $\int_{m}^{m+1} f(x) dx = f_m+g_m$ and $|\int_{m}^{y} f(x) - (y-m)(f_m-g_m) dx| &amp;lt; h_m, y\in (m,m+1)$ for all $m \in \mathbb{Z}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is just as hard as finding all continuous functions on $g$ on $[0,1]$ whose integral is $1$ and $g(0)=0$, $g(1)=1$ and $|\int_0^y g(x)-y| dx &amp;lt; \varepsilon, y\in (0,1)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Motivation: When we pick $f_k$, we have fixed values for $f(x)$ on all integers, and also RHS of equality. To satisfy LHS = RHS, we have freedom to construct $f(x)$ however we want on non-integer points between fixed values on integer points since $f(x)$ should be continious.&lt;br&gt;&#xA;When we consider partial sums $\sum_{n=0}^{m} f_n$ and $\int_0^m f(x)dx$, these two can differ but they should have the same limits. This is controlled by sequence $g_n$.&lt;br&gt;&#xA;The rest is just technicality to ensure that $\int_{-\infty}^{\infty}f(x)dx$ is indeed defined. Since we choose practically values for $\int_0^m f(x)dx$ we need to ensure that $\int_0^y f(x)dx$ do not 'wiggle' too much between integer points when $m\rightarrow \infty$. This is what we use $h_n$ for.&lt;br&gt;&#xA;All given conditions should be sufficient and necessary.&lt;/p&gt;&#xA;" OwnerUserId="956" LastEditorUserId="956" LastEditorDisplayName="" LastEditDate="2010-08-16T08:13:16.903" LastActivityDate="2010-08-16T08:13:16.903" CommentCount="0" />
  <row Id="2542" PostTypeId="1" AcceptedAnswerId="2564" CreationDate="2010-08-15T22:17:01.810" Score="10" ViewCount="306" Body="&lt;p&gt;Friend told me this one, I'm completely stuck but also completely fascinated:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are 100 boxes with apples, oranges and bananas (mixed). &lt;strike&gt;How to&lt;/strike&gt; Prove that you can pick 51 boxes and to get at least half of all apples, at least half of all oranges and at least half of all bananas?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: You can take a look in the boxes.&lt;/p&gt;&#xA;" OwnerUserId="956" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-08-27T22:44:44.250" LastActivityDate="2010-08-27T22:44:44.250" Title="100 boxes of fruits - pick 51 and get at least half of each type?" Tags="&lt;combinatorics&gt;&lt;contest-math&gt;" AnswerCount="1" CommentCount="15" FavoriteCount="3" />
  <row Id="2543" PostTypeId="1" AcceptedAnswerId="2551" CreationDate="2010-08-15T22:18:19.423" Score="2" ViewCount="517" Body="&lt;p&gt;I have a symmetric matrix A. How do I compute a matrix B such that B^tB=A where B^t is the transpose of B. I cannot figure out if this is at all related to the square root of A. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've gone through wikimedia links of square root of a matrix.&lt;/p&gt;&#xA;" OwnerUserId="957" LastActivityDate="2010-08-16T02:20:27.840" Title="square root of symmetric matrix and transposition" Tags="&lt;linear-algebra&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" />
  <row Id="2544" PostTypeId="2" ParentId="2543" CreationDate="2010-08-15T22:22:33.383" Score="1" ViewCount="" Body="&lt;p&gt;What you apparently want here is the Cholesky decomposition, which factors a matrix A into $BB^T$ where $B$ is a triangular matrix. However, this only works if your matrix is positive definite.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-08-15T22:22:33.383" CommentCount="11" />
  <row Id="2545" PostTypeId="2" ParentId="2537" CreationDate="2010-08-15T22:42:31.813" Score="8" ViewCount="" Body="&lt;p&gt;What Arnold is describing (in a slightly oblique manner) is the theory of algebraic curves.&#xA;The idea is that if one wants to integrate some rational expression $R(x,y)dx + S(x,y) dy$&#xA;over the curve $f(x,y) = 0$, then the question of whether one can find an antiderivative&#xA;in terms of elementary functions has a positive or negative answer depending on whether&#xA;the geometric genus of the curve $f(x,y) = 0$ is zero or positive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One direction is not so hard to see directly: if $f(x,y) = 0$ has geometric genus zero,&#xA;this means that we can trace out this curve in terms of a single parameter, i.e.&#xA;we can find parametric expressions $x = x(t)$ and $y = y(t)$ so that $f(x(t),y(t)) = 0$.&#xA;Then if we rewrite the integral in terms of the variable $t$, basic integral calculus (the substitution rule) lets us rewrite the integrand as a rational function of $t$, and we can&#xA;always integrate a rational function in terms of elementary functions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is less obvious is that if $f(x,y) = 0$ has positive geometric genus, then it is &lt;em&gt;not&lt;/em&gt; possible to find such a parameterization of the curve (this is a non-trivial statement),&#xA;and it is &lt;em&gt;not&lt;/em&gt; possible to find an elementary antiderivative (this is related to the previous statement, but is another non-trivial deduction).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first example is the curve $y^2 = (1-x^2)(1-kx^2)$ (here $k$ is some constant, neither 0 nor 1),&#xA;with the integral being $\int dx/y = \int dx/\sqrt{(1-x^2)(1 - kx^2)}$.  This is what is called an elliptic integral, and (for more or less 150 years, beginning with the invention&#xA;of calculus) people tried to find an elementary expression for it, until finally Abel and Jacobi showed that this wasn't possible, because this curve has geometric genus one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you don't know any algebraic geometry, then a good place to start is Miles Reid's &quot;Undergaduate algebraic geometry&quot;.  The theorem you need is the one which says that&#xA;there is no rational map from a genus zero curve to a positive genus zero curve,&#xA;which I'm pretty sure is proved in that book, at least for genus one curves.&#xA;(It is not so difficult to pass from this theorem in the case of smooth curves to the &#xA;case of singular curves, but you will have more difficulty finding a treatment of the&#xA;singular curve case, which is what Arnold is talking about when he mentions double points.)&#xA;Depending on the level at which you're beginning, you might want to consult one of the algebraic geometry road-map questions on Mathoverflow; there is one asking for an undergraduate road-map for learning algebraic geometry, and a second asking for a graduate road-map.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you do already know some algebraic geometry, then what you want is a historical source&#xA;that relates the geometry you know to its historical origins.  Dieudonne wrote a history of algebraic geometry, which must surely discuss this.  There will be many other historical&#xA;sources too.  For the best guide to the historical literature, you might want to ask on Mathoverflow, where there will probably be more people reading who are familiar with historical treatments of the theory.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I should say that if you are beginning from a position of knowing no algebraic geometry, then it will take some time and effort to learn what is needed to fully understand what Arnold is discussing, especially from a standard textbook (which will likely not proceed in a straight line to where you want to go, but rather develop a more general theory, which will then be specialized to the situation you are interested in).&lt;br&gt;&#xA;So even if that is your situation, I recommend that you also do some historical reading,&#xA;to help get a better feeling for exactly what parts of an algebraic geometry text-book&#xA;you will need to read to satisfy your interest in Arnold's statement.&lt;/p&gt;&#xA;" OwnerUserId="221" LastEditorUserId="221" LastEditorDisplayName="" LastEditDate="2010-08-16T02:39:15.080" LastActivityDate="2010-08-16T02:39:15.080" CommentCount="2" />
  <row Id="2546" PostTypeId="2" ParentId="2529" CreationDate="2010-08-15T23:34:23.133" Score="2" ViewCount="" Body="&lt;p&gt;This called Dicman's lemma and is 14.5.7 on page 425 (1st ed) or &lt;a href=&quot;http://books.google.com/books?id=lqyCjUFY6WAC&amp;amp;lpg=PA442&amp;amp;pg=PA442&quot; rel=&quot;nofollow&quot;&gt;page 442 (2nd ed)&lt;/a&gt; of Robinson's Course in the Theory of Groups.&lt;/p&gt;&#xA;" OwnerUserId="583" LastActivityDate="2010-08-15T23:34:23.133" />
  <row Id="2547" PostTypeId="2" ParentId="2492" CreationDate="2010-08-15T23:37:51.053" Score="1" ViewCount="" Body="&lt;p&gt;So it turns out the first assumption was actually correct:  U is indeed the first k eigenvectors of C, that we calculate from G by means of the eigendecompposition $(X_tVD^{-\frac12})$.  Thanks to J. Mangaldan for his comments and especially the pointers to Sherman-Morrison-Woodbury. &lt;/p&gt;&#xA;" OwnerUserId="358" LastActivityDate="2010-08-15T23:37:51.053" />
  <row Id="2548" PostTypeId="2" ParentId="2355" CreationDate="2010-08-16T00:15:12.397" Score="3" ViewCount="" Body="&lt;p&gt;&lt;em&gt;Here is my attempt, fixed by Septimus Harding:&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If H is a group with 92 elements of order 3, then let G be the subgroup generated by those elements.  G has finite order by Dicman's lemma, and so G has a Sylow 3-subgroup P.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P must be cyclic, lest Hall's result imply 46≡4 mod 9.  Since P is cyclic, 46≡1 mod |P|, so |P| divides 45, so |P| is 1, 3, or 9.  1 is impossible since there would be 0 elements of order 3.  3 is impossible, since then G has 46 Sylow 3-subgroups but no finite group has 46 Sylow p-subgroups.  So P is cyclic of order 9.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now G acts by conjugation on the 46 subgroups of order 3, and since the Sylow 3-subgroups are cyclic and (all of) their order 3-subgroups are conjugate, G acts transitively on the 46 subgroups.  Let Q = Ω(P) ≤ P be one of the subgroups of order 3, and let M = N&lt;sub&gt;G&lt;/sub&gt;(Q) be its normalizer.  Then N = N&lt;sub&gt;G&lt;/sub&gt;(P) ≤ M and [G:M] = 46.  If M ≤ X ≤ G, then both [M:N] and [X:N] are ≡1 mod 3 by Sylow, but [X:N] = [X:M][M:N], so [X:M] ≡ 1 mod 3.  However [X:M] divides 46, so [X:M] = 1 or 46, so X = M or G, and so M is a maximal subgroup of G.  So G acts primitively on 46 points (the cosets of N, aka, the subgroups of order 3), so G/Core(G,M) is a primitive group of degree 46, that is, G/Core(G,M) is the alternating or the symmetric group on 46 points.  Since the Sylow 3-subgroups of the alternating and symmetric groups (on 6 or more points) are not cyclic, we have a contradiction, and no such G exists.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;This works to prove that: if n=pq for primes p,q ≡ {5,8} mod 9, and if the only primitive groups of degree pq have non-cyclic Sylow 3-subgroups (like Alt(n) and Sym(n)), then there is no group with exactly 2n elements of order 3, even though 2n ≡ 2 mod 6.  I believe primitive groups of degree pq are classified, so this could probably be made completely explicit.  140 elements is not immediately handled, since 70 = 2*5*7, but I suspect the same idea will work.&lt;/p&gt;&#xA;" OwnerUserId="583" LastEditorUserId="583" LastEditorDisplayName="" LastEditDate="2010-08-16T05:43:21.307" LastActivityDate="2010-08-16T05:43:21.307" CommunityOwnedDate="2010-08-16T00:15:12.397" />
  <row Id="2549" PostTypeId="1" AcceptedAnswerId="2557" CreationDate="2010-08-16T00:39:05.017" Score="6" ViewCount="242" Body="&lt;p&gt;I recently learned that on Riemannian manifolds, one can define the &lt;em&gt;sectional curvature&lt;/em&gt; (&lt;a href=&quot;http://en.wikipedia.org/wiki/Sectional_curvature&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Sectional_curvature&lt;/a&gt;) of a (2-dimensional) plane section.  I was wondering if a similar concept exists for higher dimensional &quot;space sections.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is what got me thinking about this: For 2-dimensional manifolds (surfaces), the sectional curvature is equal to $\kappa_1\kappa_2$, where $\kappa_1$ and $\kappa_2$ are the principal curvatures.  Is there a name for the quantity $\kappa_1\kappa_2\kappa_3$ for 3-manifolds, etc., and does it carry similar geometric significance?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Edit: Typesetting fixed)&lt;/p&gt;&#xA;" OwnerUserId="640" LastEditorUserId="640" LastEditorDisplayName="" LastEditDate="2010-08-30T04:50:54.123" LastActivityDate="2010-08-30T04:50:54.123" Title="Are there higher-dimensional analogues of sectional curvature?" Tags="&lt;reference-request&gt;&lt;differential-geometry&gt;" AnswerCount="2" CommentCount="7" FavoriteCount="2" />
  <row Id="2550" PostTypeId="1" AcceptedAnswerId="3152" CreationDate="2010-08-16T01:10:52.327" Score="6" ViewCount="371" Body="&lt;p&gt;Meta: Since there seems be a dearth of questions here (I am used to stackoverflow!), I think we should all try and contribute some questions...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Part a) of the following problem appeared in one of the Putnam Exams (sorry, don't know which year exactly).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to solve Part a) don't read Part b).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have a painting device, which given the co-ordinates of a points in the 2D plane, will colour all points on that plane black, which are at an irrational distance from the given point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Initially you start out with the 2D plane being white. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;a) You want to colour the whole plane black. What is the minimum number of points you need to feed to the painting device?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;b) Show that it is sufficient to feed $(0,0), (1,0), (\sqrt{2},0)$. &lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-08-16T01:16:01.427" LastActivityDate="2010-08-23T20:00:21.917" Title="Irrational painting device" Tags="&lt;puzzle&gt;&lt;putnam&gt;" AnswerCount="3" CommentCount="7" />
  <row Id="2551" PostTypeId="2" ParentId="2543" CreationDate="2010-08-16T02:20:27.840" Score="4" ViewCount="" Body="&lt;p&gt;As J. Mangaldan says, you need your matrix $A$ to be positive definite. Since $A$, being symmetric, is always diagonalizable, this is the same as saying that it has non-negative eigenvalues. If this is the case, you can adapt alex's comment almost literally for the real case: as we've said, $A$ is diagonalizable, but, also, there exists an orthonormal base of eigenvectors of $A$. That is, there is an invertible matrix $S$ and a diagonal matrix $D$ such that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;D = SAS^t , \quad \text{with} \quad SS^t = I \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;D = \mathrm{diag} (\lambda_1, \lambda_2, \dots , \lambda_n) \ ,&#xA;$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;is a diagonal matrix and has only non-negative eigenvalues $\lambda_i$, you can take its square root&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\sqrt{D} = \mathrm{diag} (\sqrt{\lambda_1}, \sqrt{\lambda_2}, \dots , \sqrt{\lambda_n} ) \ ,&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and then, on one hand, you have:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\left( S^t \sqrt{D} S   \right)^2 = \left( S^t \sqrt{D} S\right) \left(S^t \sqrt{D} S \right) = S^t \left( \sqrt{D}\right)^2 S = S^t D S = A \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, $S^t \sqrt{D} S$ is a symmetric matrix too:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\left( S^t \sqrt{D} S   \right)^t = S^t (\sqrt{D})^t S^{tt} = S^t \sqrt{D^t} S = S^t \sqrt{D} S  \ ,&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so you have your $B = S^t \sqrt{D} S$ such that $B^t B = A$.&lt;/p&gt;&#xA;" OwnerUserId="664" LastActivityDate="2010-08-16T02:20:27.840" />
  <row Id="2552" PostTypeId="1" CreationDate="2010-08-16T02:43:17.580" Score="1" ViewCount="189" Body="&lt;p&gt;i'm reading &lt;em&gt;3D math primer for graphics and game development&lt;/em&gt; By Fletcher Dunn and Ian Parberry, on page &lt;a href=&quot;http://books.google.com.hk/books?id=wCfWkc_E3GkC&amp;amp;lpg=PP1&amp;amp;ots=zIZTXKhbGJ&amp;amp;dq=3D%20Math%20Primer%20for%20Graphics%20and%20Game%20Development&amp;amp;hl=en&amp;amp;pg=PA170#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;170&lt;/a&gt;, the logarithm of quaternion is defined as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;\log \mathbf q &amp;amp;= \log \left( \begin{bmatrix} \cos \alpha &amp;amp; \mathbf n \sin \alpha \end{bmatrix} \right) \\&#xA;&amp;amp;\equiv \begin{bmatrix} 0 &amp;amp; \alpha \mathbf n \end{bmatrix}&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i don't see how does $\log \left( \begin{bmatrix} \cos \alpha &amp;amp; \mathbf n \sin \alpha \end{bmatrix} \right)$ equals to $\begin{bmatrix} 0 &amp;amp; \alpha \mathbf n \end{bmatrix}$, can anyone help me out?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;thanks.&lt;/p&gt;&#xA;" LastEditorUserId="171" LastEditorDisplayName="" LastEditDate="2010-08-16T08:00:43.533" LastActivityDate="2010-08-16T08:34:16.463" Title="the logarithm of quaternion" Tags="&lt;quaternions&gt;" AnswerCount="2" />
  <row Id="2553" PostTypeId="2" ParentId="2112" CreationDate="2010-08-16T03:58:23.020" Score="3" ViewCount="" Body="&lt;p&gt;If I understand you correctly, you aren't bringing up the issues raised by Gödel. You are raising the point that since it's always possible a mistake has been made, we can't be sure of the validity of a mathematical proof. There is always the possibility that even our computational examination of a proof is forgetting some facet that renders it untrue. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One must understand, however, that this is not a flaw in mathematics but an inherent element of reality. Maths does remain the most secure branch of knowledge, except that the way we practically do math may not be as secure as previously imagined. A fair point. &lt;/p&gt;&#xA;" OwnerUserId="647" LastActivityDate="2010-08-16T03:58:23.020" CommunityOwnedDate="2010-08-16T03:58:23.020" />
  <row Id="2554" PostTypeId="2" ParentId="2552" CreationDate="2010-08-16T04:17:46.630" Score="3" ViewCount="" Body="&lt;p&gt;I can't see the page in Google books, but what you apparently have there is the logarithm of a &lt;strong&gt;unit&lt;/strong&gt; quaternion q, which has scalar part $\cos(\theta)$ and vector part $\sin(\theta)\vec{n}$ where $\vec{n}$ is a unit vector.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the logarithm of an arbitrary quaternion $q=(s\;\;v)$ is defined as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ln q=(\ln\,|q|\;\;\frac{v}{\|v\|}\arccos\frac{s}{|q|})$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $|q|$ is the norm of the quaternion and $\|v\|$ is the norm of the vector part, applying that formula to a unit quaternion yields a scalar part of 0 (the logarithm of the norm of a unit quaternion is zero), and you should now be able to derive the formula for the vector part.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-08-16T04:17:46.630" />
  <row Id="2555" PostTypeId="2" ParentId="2536" CreationDate="2010-08-16T04:18:23.793" Score="0" ViewCount="" Body="&lt;p&gt;I found a similar question, answered by Ramanujan. The paper is entitled &quot;Some Integrals in Ramaujan's lost notebook&quot; by Bruce C Brendt. Here is the link: &lt;a href=&quot;http://www.jstor.org/stable/4097261?cookieSet=1&quot; rel=&quot;nofollow&quot;&gt;http://www.jstor.org/stable/4097261?cookieSet=1&lt;/a&gt;&lt;/p&gt;&#xA;" LastActivityDate="2010-08-16T04:18:23.793" />
  <row Id="2556" PostTypeId="2" ParentId="2530" CreationDate="2010-08-16T04:25:23.533" Score="4" ViewCount="" Body="&lt;p&gt;As KCd explains in a comment, the proof of the PNT in Hardy's time seemed to be intimately connected to the complex analytic theory of the $\zeta$-function.  In fact, it was known to&#xA;be equivalent to the statement that $\zeta(s)$ has no zeroes on the half-plane $\Re s \geq  1$.&#xA;Although this equivalence may seem strange to someone unfamiliar with the subject, it is in&#xA;fact more or less straightforward.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, the equivalence of PNT and the zero-freeness of $\zeta(s)$ in the region&#xA;$\Re s \geq 1$ does not lie as deep as the fact that these results are actually true.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The possibility that one could then prove PNT in a direct way, avoiding complex analysis, seemed unlikely to Hardy, since then one would also be giving a proof, avoiding complex&#xA;analysis, of the fact that the complex analytic function $\zeta(s)$ has no zero in the region $\Re s \geq 1$, which would be a peculiar state of affairs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What added to&#xA;the air of mystery surrounding the idea of an elementary proof was the possibility of accessing the Riemann hypothesis this way.  After all, if one could prove in an elementary way that&#xA;$\zeta(s)$ was zero free when $\Re s \geq 1$, perhaps the insights gained that way&#xA;might lead to a proof that $\zeta(s)$ is zero free in the region $\Re s &gt; 1/2$ (i.e.&#xA;the Riemann hypothesis), a statement which had resisted (and continues to resist)&#xA;attack from the complex analytic perspective,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, when the elementary proof of PNT was finally found, it didn't have the ramifications that Hardy anticipated (as KCd pointed out in his comment).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a modern perspective on the elementary proof, and a comparison with the complex analytic proof, I strongly recommend Terry Tao's &lt;a href=&quot;http://www.math.ucla.edu/~tao/preprints/Expository/prime.dvi&quot; rel=&quot;nofollow&quot;&gt;exposition of the prime number theorem&lt;/a&gt;.  In this exposition,&#xA;Tao is not really concerned with elementary vs. complex analytic techniques as such, but rather with&#xA;explaining what the mathematical content of the two arguments is, in a way which makes&#xA;it easy to compare them.  Studying Tao's article should help you develop a deeper sense of the mathematical content of the two arguments, and their similarities and differences.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As Tao's note explains, both arguments involve a kind of &quot;harmonic analysis&quot; of the primes,&#xA;but the elementary proof works primarily in &quot;physical space&quot; (i.e. one works directly with the prime counting function and its relatives), while the complex analytic proof works much more in &quot;Fourier space&quot; (i.e. one works much more with the Fourier transforms of the prime counting function and its relatives).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My understanding (derived in part from Tao's note) is that Bombieri's sieve is (at least in part) an outgrowth of the elementary proof, and it is in sieve methods that one can look to find modern versions of the type of arguments that appear in the elementary proof.  (As one example, see Kevin Ford's paper &lt;a href=&quot;http://www.ams.org/journals/tran/2005-357-04/S0002-9947-04-03579-2/home.html&quot; rel=&quot;nofollow&quot;&gt;On Bombieri's asymptotic sieve&lt;/a&gt;, which in its first two pages includes a&#xA;discussion of the relationship between certain sieving problems and the elementary proof.)&#xA; But I should note that modern analytic number theorists don't pursue sieve methods out of some desire for having &quot;elementary&quot; proofs.  Rather, some results can be proved by $\zeta$- or $L$-function methods, and others by sieving methods; each has their strengths and weaknesses.  They can be combined, or played off, one against the other. (The &#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Bombieri%E2%80%93Vinogradov_theorem&quot; rel=&quot;nofollow&quot;&gt;Bombieri--Vinogradov theorem&lt;/a&gt; is an example&#xA;of a result proved by sieve methods which, as far as I understand, is stronger than what&#xA;could be proved by current $L$-function methods; indeed, it an averaged form of the&#xA;Generalized Riemann Hypothesis.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To see how this mixing of methods is possible, I again recommend Tao's note.  Looking at it should give you a sense of how, in modern number theory, the methods of the two proofs of PNT (elementary and complex analytic) are not living in different, unrelated worlds, but are just two different, but related, methods for approaching the &quot;harmonic analysis of the primes&quot;.  &lt;/p&gt;&#xA;" OwnerUserId="221" LastEditorUserId="221" LastEditorDisplayName="" LastEditDate="2010-08-16T06:35:38.190" LastActivityDate="2010-08-16T06:35:38.190" CommentCount="3" />
  <row Id="2557" PostTypeId="2" ParentId="2549" CreationDate="2010-08-16T06:31:47.137" Score="2" ViewCount="" Body="&lt;p&gt;Sure, there's all kinds of generalizations.  One way to think of sectional curvature is you take a 2-dimensional subspace of the tangent space, exponentiate a small neighbourhood of the origin, and take the Gauss curvature at the point of tangency. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given a $k$-dimensional subspace of the tangent space, exponentiate a small neighbourhood of $0$, and take the scalar curvature at the point of tangency.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another analogy you could build on is to use the comparison definition of sectional curvature -- measuring the infinitesimal deviation of the length of a circle (exponentiated from a 2-dimensional tangent space) of a given radius from that of a Euclidean circle.  You could do the same for content of spheres of arbitrary dimension in subspaces of the tangent space, etc. I imagine these give very related notions of curvature although I've never worked out the details. &lt;/p&gt;&#xA;" OwnerUserId="642" LastActivityDate="2010-08-16T06:31:47.137" CommentCount="2" />
  <row Id="2558" PostTypeId="1" AcceptedAnswerId="2560" CreationDate="2010-08-16T06:32:03.643" Score="8" ViewCount="1007" Body="&lt;p&gt;You're probably thinking, &quot;Why?&quot;  Please let me explain...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is (&lt;em&gt;very&lt;/em&gt;) well-known that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \forall (a,b,c,x) \in \mathbb{C}^* \times \mathbb{C}^3: ax^2 + bx + c = 0 \Leftrightarrow x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}. $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For some bizarre reason, I decided to try to solve $ bx + c = 0 $ using this formula by introducing a term $ \alpha x^2 $ and removing it in the limit $ \alpha \to 0 $.  Doing so with L'Hopital's rule, I find these solutions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \displaystyle x_1 = \lim_{\alpha \to 0} {\frac{-b + \sqrt{b^2 - 4c \alpha}}{2 \alpha}} = \lim_{\alpha \to 0} {\frac{-c}{\sqrt{b^2 - 4c \alpha}}} = \frac{-c}{b}, $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \displaystyle x_2 = \lim_{\alpha \to 0} {\frac{-b - \sqrt{b^2 - 4c \alpha}}{2 \alpha}} = \infty. $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first was to be expected, but I still haven't been able to explain the second cleanly (that is, in a way other than &quot;since $ -c/b $ is gone, it couldn't be a true number&quot;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In addition, carrying out the analogous process one degree lower yields a root at either zero or infinity, depending on the constant.  The latter possibility (which occurs when $ c \neq 0 $) corresponds to the unsolvable case, while the former (in which $ c = 0 $) corresponds to the trivially satisfied one, so a root at zero here appears to have a vastly different meaning from $ x_1 = 0 $ above, where $ x_1 $ gives the location of the unique, genuine root of $ bx + 0 = 0 $ provided $ b \neq 0 $.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;why a solution at zero can have either of the two meanings just described, and&lt;/li&gt;&#xA;&lt;li&gt;whether the phantom root $ x_2 = \infty $ (obtained by treating the first-degree polynomial $ bx + c $ as a degenerate case of the second-degree one) has a meaningful interpretation.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Thank you all in advance, and sorry if my typesetting doesn't render nicely (this is my first experience).&lt;/p&gt;&#xA;" OwnerUserId="336" LastActivityDate="2010-08-16T07:46:52.787" Title="On applying the quadratic formula to a first-degree equation" Tags="&lt;polynomials&gt;&lt;roots&gt;&lt;infinity&gt;&lt;limit&gt;" AnswerCount="3" />
  <row Id="2559" PostTypeId="2" ParentId="2558" CreationDate="2010-08-16T06:38:51.787" Score="10" ViewCount="" Body="&lt;p&gt;Just a note on your attempt to solve a degenerate quadratic: remember that the quadratic formula can be derived in two ways: solving ax²+bx+c for x, or solving a+b/x+c/x² for 1/x and then reciprocating the result. Viewing it in this manner, one equation's &quot;infinite root&quot; is the reversed equation's 0 root.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-08-16T06:38:51.787" CommentCount="2" />
  <row Id="2560" PostTypeId="2" ParentId="2558" CreationDate="2010-08-16T07:30:12.310" Score="10" ViewCount="" Body="&lt;p&gt;I think the answer is to work projectively.  Rather than consider the solutions to $ax^2 + bx + c = 0$ in $\mathbb{C}$ one should think of the solutions to $aX^2 + bXY + cY^2 = 0$ in $\mathbb{P}^1(\mathbb{C})$.  Then the $a = 0$ case is easy to explain; the corresponding equation $bXY + cY^2 = 0$ has one root $(c : -b)$ which is expected and another $(1 : 0)$ which is the point at infinity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This seems reasonable to me because the degeneration at $a = 0$ is something like a failure of Bezout's theorem, which is repaired precisely by working projectively.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-08-16T07:30:12.310" CommentCount="7" />
  <row Id="2561" PostTypeId="1" AcceptedAnswerId="2590" CreationDate="2010-08-16T07:40:51.880" Score="4" ViewCount="318" Body="&lt;p&gt;Good example of a polynomial which produces a finite number of primes is: $$x^{2}+x+41$$ which produces primes for every integer $ 0 \leq x \leq 39$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a paper H.Stark proves the following result: $X_{n}$(is the ring of &quot;algebraic integers&quot; in $Q(\sqrt{-n})$ is a prinicpal ideal domain for positive $n$ if and only if &#xA;$n = 1,2,3,7,11,19,43,67,163. $  For a reference one can see: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Harold Stark, A complete determination of the complex quadratic fields of class-number one, Michigan Math J., 14 (1967) 1-27&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider in general the polynomial $x^{2}+x + K= (x+ \alpha)(x+ \bar{\alpha})$  which we can factorize where $\alpha$ is given by $$ \alpha = \frac{{1} + \sqrt{1-4K}}{2}, \quad \bar{\alpha} = \frac{1 - \sqrt{1-4K}}{2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One can get some relationships between polynomials which produce prime in the field $Q(\sqrt{-n})$. Question is if a polynomial produces a prime, then will $X_{n}$ as defined above be a PID?&lt;/p&gt;&#xA;" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2011-02-22T04:08:32.693" LastActivityDate="2011-02-22T04:14:01.470" Title="Euler's remarkable prime-producing polynomial and quadratic UFDs" Tags="&lt;abstract-algebra&gt;&lt;algebraic-number-theory&gt;" AnswerCount="1" FavoriteCount="4" />
  <row Id="2563" PostTypeId="2" ParentId="2558" CreationDate="2010-08-16T07:46:52.787" Score="9" ViewCount="" Body="&lt;p&gt;Informally, the &quot;quadratic&quot; polynomial with a=0 has a second zero at the compactification point at infinity.  Graphically (working in the reals):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/gOMTn.gif&quot; alt=&quot;animated graph&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, as a goes through zero, the &quot;quadratic&quot; goes through the linear special case, where the second zero goes through infinity, crossing between the positive and negative ends of the real axis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe that this parallels the more technical explanation given by Qiaochu Yuan.&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-08-16T07:46:52.787" CommentCount="3" />
  <row Id="2564" PostTypeId="2" ParentId="2542" CreationDate="2010-08-16T07:50:50.187" Score="6" ViewCount="" Body="&lt;p&gt;This is apparently a (hard) problem from the Russian Math Olympiad which no one in the exam solved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See here for a list of questions in that exam: &lt;a href=&quot;http://www.artofproblemsolving.com/Forum/viewtopic.php?f=125&amp;amp;t=32171&quot; rel=&quot;nofollow&quot;&gt;http://www.artofproblemsolving.com/Forum/viewtopic.php?f=125&amp;amp;t=32171&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A solution for this problem is here: &lt;a href=&quot;http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1367869#p1367869&quot; rel=&quot;nofollow&quot;&gt;http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1367869#p1367869&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A hint that was given (by Fedor Petrov):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If we have $2k$ boxes, we may&#xA;  partition them into two groups of $k$&#xA;  boxes in such a way that number of&#xA;  apples in both groups differ by at&#xA;  most the maximal number of apples in a&#xA;  single box, and the same for oranges.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-08-16T08:02:11.877" LastActivityDate="2010-08-16T08:02:11.877" CommentCount="2" />
  <row Id="2565" PostTypeId="2" ParentId="2552" CreationDate="2010-08-16T08:34:16.463" Score="1" ViewCount="" Body="&lt;p&gt;Just recall $\exp(\alpha i) = \cos \alpha + i \sin \alpha$ for complex numbers, the quaternion (remember a quaternion is just 3 complex numbers which all have the same real part) version is by direct analogue and take logarithm of both sides.&lt;/p&gt;&#xA;" LastActivityDate="2010-08-16T08:34:16.463" />
  <row Id="2567" PostTypeId="1" CreationDate="2010-08-16T09:36:59.887" Score="1" ViewCount="239" Body="&lt;p&gt;Hi, I would like to know if the Lebesgue spaces $L_p$ with $ 0 &amp;lt; p &amp;lt; 1 $ are separable or not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that this is true for $1 \leq p &amp;lt; + \infty$, but I do not find any references for the &lt;/p&gt;&#xA;&#xA;&lt;p&gt;case  $ 0 &amp;lt; p &amp;lt; 1 $.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you &lt;/p&gt;&#xA;" OwnerUserId="1222" LastEditorUserId="171" LastEditorDisplayName="" LastEditDate="2010-08-16T11:07:59.933" LastActivityDate="2010-08-31T04:42:53.600" Title="Separability of $ L_p $ spaces" Tags="&lt;topology&gt;&lt;functional-analysis&gt;" AnswerCount="2" CommentCount="0" />
  <row Id="2568" PostTypeId="2" ParentId="2550" CreationDate="2010-08-16T09:53:01.387" Score="0" ViewCount="" Body="&lt;p&gt;I decided to give this problem a go, but I couldn't get a complete solution. Lets look at the distances squared:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$x_1^2+y_1^2$&lt;/li&gt;&#xA;&lt;li&gt;$(x_1-1)^2+y_1^2$&lt;/li&gt;&#xA;&lt;li&gt;$(x_1 - \sqrt{2})^2+y_1^2$&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;These must all be rational. The difference between any two rational numbers must be rational, so $-2x_1+1$ is rational and so is $x_1$. Subtracting 2. and 3., we get: $(-2 \sqrt(2)-2)x_1+3$, implying that $x_1$ must be 0 if it is rational.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This then gives us:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$y_1^2, y_1^2+1, y_1^2+2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose $y_1^2=a^2/b^2$ where a and b are integers. Clearly, $y_1^2+1, y_1^2+2$ are rational if and only if $a^2+b^2$ and $a^2+2*b^2$ are perfect squares. Using modular arithmetic it is pretty clear that b needs to be divisible by 2 (squares are only 0 or 1 mod 4), implying $b^2$ is divisible by 4.&lt;/p&gt;&#xA;" OwnerUserId="123" LastEditorUserId="123" LastEditorDisplayName="" LastEditDate="2010-08-16T21:33:44.997" LastActivityDate="2010-08-16T21:33:44.997" CommentCount="4" />
  <row Id="2569" PostTypeId="2" ParentId="2550" CreationDate="2010-08-16T10:19:06.770" Score="3" ViewCount="" Body="&lt;p&gt;Suppose that there exists a point A that is left uncolored after feeding in the three points (0,0), (1,0), and $(\sqrt{2},0)$.  The distances from point A to each of these three points (call these b, c, and d, respectively) must be rational numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/ryXfw.png&quot; alt=&quot;diagram&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Apply &lt;a href=&quot;http://en.wikipedia.org/wiki/Stewart%27s_theorem&quot; rel=&quot;nofollow&quot;&gt;Stewart's Theorem&lt;/a&gt;: $$b^2(\sqrt{2}-1)+c^2\cdot 1=\sqrt{2}(d^2+(\sqrt{2}-1)\cdot 1).$$ So $$\sqrt{2}(b^2-d^2+1)=b^2-c^2+2$$ or $$\sqrt{2}=\frac{b^2-c^2+2}{b^2-d^2+1}.$$  Since b, c, and d are rational numbers, the right side is rational, but the left side is irrational, so there is no such point A.  Thus, to color the entire plane black, it is sufficient to feed in the three points (0,0), (1,0), and $(\sqrt{2},0)$.&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-08-16T10:19:06.770" CommentCount="1" />
  <row Id="2570" PostTypeId="2" ParentId="2567" CreationDate="2010-08-16T11:13:27.227" Score="0" ViewCount="" Body="&lt;p&gt;Please refer this article. It talks about $L_{p}$ spaces for $0 &amp;lt; p \leq 1$. Link: &#xA;&lt;a href=&quot;http://www.jstor.org/stable/2041603?seq=2&quot; rel=&quot;nofollow&quot;&gt;http://www.jstor.org/stable/2041603?seq=2&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Look at the step functions, the ones that take rational values  and whose steps have rational endpoints there should be only countably many of those. And then you can perhaps apply the same argument, you use to prove it for $L_{p}$ spaces for $1 &amp;lt; p &amp;lt; \infty$.&lt;/p&gt;&#xA;" LastActivityDate="2010-08-16T12:09:50.020" CommentCount="2" />
  <row Id="2571" PostTypeId="1" AcceptedAnswerId="2572" CreationDate="2010-08-16T11:16:39.657" Score="2" ViewCount="285" Body="&lt;p&gt;Square of an irrational number can be a rational number e.g. $\sqrt{2}$ is irrational but its square is 2 which is rational.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But is there a irrational number square root of which is a rational number?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it safe to assume, in general, that $n^{th}$-root of irrational will always give irrational numbers?&lt;/p&gt;&#xA;" OwnerUserId="705" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2011-01-03T19:33:56.897" LastActivityDate="2011-01-03T19:33:56.897" Title="About powers of irrational numbers" Tags="&lt;number-theory&gt;&lt;irrational-numbers&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="2572" PostTypeId="2" ParentId="2571" CreationDate="2010-08-16T11:20:43.627" Score="13" ViewCount="" Body="&lt;p&gt;Obviously, if p is rational, then p&lt;sup&gt;2&lt;/sup&gt; must also be rational (trivial to prove).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ p \in \mathbb Q \Rightarrow p^2 \in \mathbb Q. $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take the &lt;a href=&quot;http://en.wikipedia.org/wiki/Contraposition&quot; rel=&quot;nofollow&quot;&gt;contraposition&lt;/a&gt;, we see that if x is irrational, then &amp;radic;x must also be irrational.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ p^2 \notin \mathbb Q \Rightarrow p \notin \mathbb Q. $$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;By negative power I assume you mean (1/n)-th power (it is obvious that $(\sqrt2)^{-2} = \frac12\in\mathbb Q$). It is true by the statement above — just replace &lt;sup&gt;2&lt;/sup&gt; by &lt;sup&gt;n&lt;/sup&gt;.&lt;/p&gt;&#xA;" OwnerUserId="171" LastActivityDate="2010-08-16T11:20:43.627" CommentCount="4" />
  <row Id="2574" PostTypeId="1" AcceptedAnswerId="2577" CreationDate="2010-08-16T12:09:56.557" Score="2" ViewCount="317" Body="&lt;p&gt;In a related &lt;a href=&quot;http://math.stackexchange.com/questions/2571/about-powers-of-irrational-numbers&quot;&gt;question&lt;/a&gt; we discussed raising numbers to powers.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am interested if anybody knows any results for raising numbers to irrational powers.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, we can easily show that there exists an irrational number raised to an irrational power such that the result is a rational number. Observe ${\sqrt 2 ^ {\sqrt 2}}$. Since we do not know if ${\sqrt 2 ^ {\sqrt 2}}$ is rational or not, there are two cases.  &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;${\sqrt 2 ^ {\sqrt 2}}$ is rational, and we are finished.  &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;${\sqrt 2 ^ {\sqrt 2}}$ is irrational, but if we raise it by ${\sqrt 2}$ again, we can see that&#xA;$$\left ( \sqrt 2 ^ \sqrt 2 \right ) ^ \sqrt 2 = \sqrt 2 ^ {\sqrt 2 \cdot \sqrt 2} = \sqrt 2 ^ 2 = 2.$$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Either way, we have shown that there exists an irrational number raised to an irrational power such that the result is rational.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can more be said about raising irrational numbers to irrational powers?  &lt;/p&gt;&#xA;" OwnerUserId="557" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2011-01-04T06:05:47.477" LastActivityDate="2011-01-04T06:05:47.477" Title="Real Numbers to Irrational Powers" Tags="&lt;number-theory&gt;&lt;irrational-numbers&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="1" />
  <row Id="2575" PostTypeId="1" CreationDate="2010-08-16T12:17:17.150" Score="2" ViewCount="139" Body="&lt;p&gt;I have statistics of 100 questions which can be answered either &quot;yes&quot; or &quot;no&quot;:&lt;br&gt;&#xA;1) 63.3 - 36.7 (63.3% respondents answered &quot;yes&quot; and 36.7% answered &quot;no&quot;)&lt;br&gt;&#xA;2) 30.1 - 69.9&lt;br&gt;&#xA;...&lt;br&gt;&#xA;100) 88.0 - 12.0&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That 100 answers would be our sample.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then I ask the same 100 questions to 101st respondent and get new set of answers:&lt;br&gt;&#xA;1) yes&lt;br&gt;&#xA;2) yes&lt;br&gt;&#xA;...&lt;br&gt;&#xA;100) no&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I need to do is some how calculate &quot;correlation&quot; value between this respondent and overall sample. Any method will be OK, I just need to get some number. Therefore there may be different &quot;right&quot; answers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance.&lt;/p&gt;&#xA;" OwnerUserId="1137" LastActivityDate="2011-05-21T15:59:57.380" Title="How to calculate correlation between poll statistics and one's answers?" Tags="&lt;statistics&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="2577" PostTypeId="2" ParentId="2574" CreationDate="2010-08-16T12:24:37.353" Score="2" ViewCount="" Body="&lt;p&gt;Some examples:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Eulers Identity $e^{i \pi} + 1 = 0$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Gelfond-Schneider Constant: $2^\sqrt{2} = 2.66514414269022518865\cdots$ is trancendental by Gelfond-Schneider.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$i^i = 0.2078795763507619085469556198\cdots$ is also trancendental &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Ramanujan constant: $e^{\pi \sqrt{163}} = 62537412640768743.999999999999250072597\cdots$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$e^\gamma$ where $\gamma$ is the Euler–Mascheroni constant. (okay nobody has proved it's irrational yet, but surely is)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" LastActivityDate="2010-08-16T12:31:45.267" CommunityOwnedDate="2010-08-16T12:48:38.833" CommentCount="2" />
  <row Id="2580" PostTypeId="2" ParentId="2574" CreationDate="2010-08-16T12:44:21.417" Score="0" ViewCount="" Body="&lt;p&gt;Forgive my lack of knowledge of how to correctly implement equations in TeX! I always particularly liked that... &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$e^{\pi} - \pi = 19.9990999791894757672664\cdots$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think it would also be possible to prove that result is transcendental as well. The two cases being $e^{\pi}$ is transcendental or $e^{\pi}$ isn't. The difference of a non-transcendental number and a transcendental number is transcendental, and the only time the difference of two transcendental numbers wouldn't be transcendental is when you could extract the second from the first - That is to say, you could find some non-transcendental $x$  that would satisfy $e^{\pi} = x + \pi$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, my formal proof skills are not what they once were. I'm sure this is trivial, and really the result isn't even that close to 20. But I always liked it :)&lt;/p&gt;&#xA;" OwnerUserId="1200" LastEditorUserId="1200" LastEditorDisplayName="" LastEditDate="2010-08-16T12:58:36.757" LastActivityDate="2010-08-16T12:58:36.757" CommentCount="3" />
  <row Id="2581" PostTypeId="1" CreationDate="2010-08-16T12:46:51.133" Score="42" ViewCount="1876" Body="&lt;p&gt;This was asked on sci.math ages ago, and never got a satisfactory answer (AFAICR)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given a number of sticks of integral length &gt;= $n$ whose lengths add to $n(n+1)/2$&#xA;Can these be divided into sticks of lengths $1,2,3, \ldots ,n$.  &lt;/p&gt;&#xA;" OwnerUserId="943" LastActivityDate="2010-10-02T21:44:44.500" Title="Cutting sticks puzzle" Tags="&lt;combinatorics&gt;&lt;puzzle&gt;" AnswerCount="3" CommentCount="8" FavoriteCount="16" />
  <row Id="2583" PostTypeId="2" ParentId="213" CreationDate="2010-08-16T12:53:14.677" Score="9" ViewCount="" Body="&lt;p&gt;Graph Theory is a fantastic field. First, the fact that abstract concepts can be readily visualized makes it engaging for new students. And second, I believe it provides solid foundations into mathematical thinking like proofs, and engages the student to explore other related fields.&lt;/p&gt;&#xA;" OwnerUserId="1010" LastActivityDate="2010-08-16T12:53:14.677" CommunityOwnedDate="2010-08-16T12:53:14.677" />
  <row Id="2584" PostTypeId="1" AcceptedAnswerId="2854" CreationDate="2010-08-16T13:02:04.063" Score="2" ViewCount="379" Body="&lt;p&gt;I've got a fixed solution space defined by a minimum and maximum float which has no divisible slice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You then have 0..N &lt;a href=&quot;http://en.wikipedia.org/wiki/Normal_distribution&quot; rel=&quot;nofollow&quot;&gt;Normal distribution&lt;/a&gt; probability sets which I need to combine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the end I neeed&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Method to define probability set with a numeric range (not sliceable) and 0..N Gaussian functions&#xA;A function which can generate a random number in the range as defined by the calculated probabilities.&#xA;Also I know it is possible that some combinations will generate a zero solution space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I'm thinking the way to do it is take the normalised probability density functions and multiply them to get a new space then normalising the result. I just can't seem to break it down into algorithmic form.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any ideas?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Extra Clarification&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Let's examine the following height rules.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;STATUS Height [ 1.2 : 2.4 ] &#xA;  MODIFIER Gender[ Male ] {Height [ 1.9 ~ 0.4 ] }&#xA;  MODIFIER Nation[ Red ] { Height [ 1.7 ~ 0.2 ] }&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Now assuming that we have a man from the Red nation we need to resolve these. &#xA;  We know that the original status declaration defines the entire possibility space.&#xA;  We cannot leave that space.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/xDRMJ.png&quot; alt=&quot;alt text&quot;&gt;&#xA;&lt;img src=&quot;http://i.imgur.com/1ndjv.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now basically I need to find a way of combining these to get a new probability.&#xA;What I meant by slice was because its a floating point number I can't break it into a set of elements of calculate the new probability for each element. I can't slice it up into pieces.&lt;/p&gt;&#xA;" OwnerUserId="1194" LastEditorUserId="1194" LastEditorDisplayName="" LastEditDate="2010-08-17T12:54:36.897" LastActivityDate="2010-08-26T10:07:28.947" Title="Combine n Normal distribution Probability Sets in a limited float range" Tags="&lt;probability&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="2" />
  <row Id="2585" PostTypeId="2" ParentId="2349" CreationDate="2010-08-16T13:05:17.240" Score="1" ViewCount="" Body="&lt;p&gt;Heres a hint: The jacket is put on after the shirt, but is taken off before&#xA;it.&lt;/p&gt;&#xA;" LastActivityDate="2010-08-16T13:05:17.240" />
  <row Id="2586" PostTypeId="1" AcceptedAnswerId="2589" CreationDate="2010-08-16T13:11:38.907" Score="2" ViewCount="209" Body="&lt;p&gt;Hi, I was going to ask this on SO but I think its more math than programming:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given the sidelength, number of vertices and vertex angle in the polygon, how can I calculate the radius of its circumscribed circle. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The polygon may have any number of sides greater than or equal to 3.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The wikipedia entry only discusses circumscribed circle of a triangle...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;edit:&lt;/strong&gt; &#xA;Also, the polygon is centered around the point (0,0). So I guess I'm asking what is the distance from the origin to any of its points..&lt;/p&gt;&#xA;" OwnerUserId="1238" LastEditorUserId="1238" LastEditorDisplayName="" LastEditDate="2010-08-16T13:17:11.907" LastActivityDate="2010-08-16T13:35:37.463" Title="Radius of the circumscribed circle of a regular polygon." Tags="&lt;geometry&gt;" AnswerCount="2" CommentCount="5" />
  <row Id="2587" PostTypeId="2" ParentId="2586" CreationDate="2010-08-16T13:22:02.907" Score="1" ViewCount="" Body="&lt;p&gt;Each corner of a polygon must be at equal distance from the center of the circumscribing circle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, find equation for perpendicular bisectors of any to sides of the polygon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Intersection of perpendicular bisectors will give you the center of the circle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then distance between center and any corner of the polygon is the radius of the circle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT(response to edited question):&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If length of a side is $L$ and number of vertices is $N$ then&#xA;Suppose points A and B defines a side and C is the center.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then angle ACB is $360/N$ &#xA;Suppose D is midpoint of a edge AB then angle DCA = angle DCB = 180/ N.&#xA;This implies $sin(180/N) = L/(2R) $ where R is radius of the circle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, $ R = \frac{L}{2sin(180/N)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Pseudocode:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;def get_circumradius(L,N):&#xA;    return L/(2 * math.sin(180/N))&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="705" LastEditorUserId="705" LastEditorDisplayName="" LastEditDate="2010-08-16T13:35:37.463" LastActivityDate="2010-08-16T13:35:37.463" CommentCount="1" />
  <row Id="2589" PostTypeId="2" ParentId="2586" CreationDate="2010-08-16T13:30:18.350" Score="5" ViewCount="" Body="&lt;p&gt;Let $s$ be sidelength and $\alpha$ be vertex angle. Then using simple trigonometry you can find that&#xA;$$&#xA;\sin{\frac{\alpha}{2}} = \frac{s/2}{r}&#xA;$$&#xA;Hence&#xA;$$&#xA;r = \frac{s}{2 \sin{\frac{\alpha}{2}}}&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="334" LastActivityDate="2010-08-16T13:30:18.350" CommentCount="1" />
  <row Id="2590" PostTypeId="2" ParentId="2561" CreationDate="2010-08-16T14:04:08.643" Score="10" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;THEOREM&lt;/strong&gt; $\ $ The polynomial $\rm\ f(x)\ =\ (x-\alpha)\:(x-\alpha&amp;#39;)\ =\ x^2 + x + k\ $ assumes only prime values for $\rm\ 0\ \le\ x\ \le\ k-2 \ \iff\ \mathbb Z[\alpha]\ $ is a PID.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\ (\Rightarrow)\ $ Show all primes $\rm\ p \le \sqrt{n},\; n = 1-4k\ $ satisfy $\rm\ (n/p) = -1\ $ so no primes split/ramify.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For proofs, see e.g. Cohn, Advanced Number Theory, pp. 155-156, or Ribenboim, My numbers, my friends, 5.7 p.108.  Note: both proofs employ the bound $\rm\ p &amp;lt; \sqrt{n}\ $ without explicitly mentiioning that this is a consequence of the Minkowski bound - presumably assuming that is obvious to the reader based upon earlier results. Thus you'll need to read the prior sections on the Minkowski bound. Compare Stewart and Tall, Algebraic number theory and FLT, 3ed, Theorem 10.4 p.176 where the use of the Minkowski bound is mentioned explicitly. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively see the self-contained &lt;a href=&quot;http://www.jstor.org/stable/2689515&quot; rel=&quot;nofollow&quot;&gt;paper [1]&lt;/a&gt; which proceeds a bit more simply, employing Dirichlet approximation to obtain a generalization of the Euclidean algorithm  (the Dedekind-Rabinowitsch-Hasse criterion for a PID). If memory serves correct, this is close to the approach originally employed by Rabinowitsch when he first published this theorem in 1913.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[1] Daniel Fendel, &lt;a href=&quot;http://www.jstor.org/stable/2689515&quot; rel=&quot;nofollow&quot;&gt;Prime-Producing Polynomials and Principal Ideal Domains,&lt;/a&gt;&lt;br&gt;&#xA;Mathematics Magazine, Vol. 58, 4, 1985, 204-210&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2011-02-22T04:14:01.470" LastActivityDate="2011-02-22T04:14:01.470" CommentCount="1" />
  <row Id="2591" PostTypeId="1" CreationDate="2010-08-16T15:11:02.367" Score="7" ViewCount="357" Body="&lt;p&gt;Quoting from &lt;a href=&quot;http://math.stackexchange.com/questions/907/correct-usage-of-the-phrase-in-the-sequel-history-alternatives/924#924&quot;&gt;this excellent answer&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If you read enough math papers you'll find that there are certain linguistic ticks that people pick up from each other&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So here's a question (primarily for you native English speakers out there, I guess):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Are there any particularly annoying language mistakes that you often see in mathematical writing (presumably by foreigners who are copying the mistake from other foreigners)?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A list of common errors to avoid might be useful to us non-native speakers. As an example of what I have in mind, I can mention a phrase that sounds wrong to my ears, but which I have come across countless times: &quot;This allows to prove...&quot;.&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2010-08-16T17:03:53.687" CommunityOwnedDate="2010-08-16T16:49:13.317" ClosedDate="2010-08-16T22:32:16.803" Title="Common English language mistakes in mathematical writing" Tags="&lt;soft-question&gt;&lt;big-list&gt;&lt;terminology&gt;" AnswerCount="1" CommentCount="8" FavoriteCount="2" />
  <row Id="2592" PostTypeId="1" AcceptedAnswerId="2750" CreationDate="2010-08-16T15:25:42.560" Score="4" ViewCount="214" Body="&lt;p&gt;A long time ago I was curious about the closed-form solutions to the equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ \frac{d^{n}y}{dx^n} y = 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $n$ an odd number, try $y = A x^k$. Then $y^{(n)} = A k(k-1)...(k-n)x^{k-n}$. This gives the formula&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$A^2 k(k-1)...(k-n) x^{2k - n - 1} = 1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which can only be true if $k = \frac{n+1}{2}$ and $n$ is odd ($k$ cannot be an integer for the formula to work, check this yourself). Furthermore one has to have that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$A = (k(k-1)...(k-n))^{-1/2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is real if $n$ is odd.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus there are closed-form solutions to my problem for $n$ odd, and my question is if anyone can find a closed-form solution for $n=2$ or in general if $n$ is even.&lt;/p&gt;&#xA;" OwnerUserId="1089" LastActivityDate="2010-08-19T03:31:32.473" Title="Solutions to the equation $y^{(n)} y = 1$ for even $n$." Tags="&lt;real-analysis&gt;&lt;differential-equations&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="2593" PostTypeId="1" AcceptedAnswerId="2597" CreationDate="2010-08-16T15:26:42.000" Score="3" ViewCount="156" Body="&lt;p&gt;Given a function $f$ defined on the set of all natural numbers $\mathbb{N}$ with three conditions:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;If $m,n$ relatively prime, then $f(mn) = f(m)f(n)$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$f$ strictly increasing.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$f(2) = 2$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Find a 4th condition such that the result will be that $f(n)$ must equal $n$ for any natural number $n$. (Of course all the conditions together are needed. Your 4th condition should not make any of these three conditions redundant.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is posted in my university website:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[1]: http:// mathstat.uohyd.ernet.in/noticeboard/generaldetails.php?id=12&lt;/p&gt;&#xA;" LastActivityDate="2010-08-16T17:33:19.360" Title="Finding Extra Condition for a function to satisfy $f(n)=n$." Tags="&lt;number-theory&gt;&lt;analysis&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="2595" PostTypeId="1" CreationDate="2010-08-16T16:38:36.623" Score="6" ViewCount="323" Body="&lt;p&gt;Hello,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a Image processing application in which I have a Matrix equation as below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A*R=I&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A = 3x3 matrix (Constants)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;R = 3x1 matrix (Column Vector) Lets call this as actual output.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I = 3x1 matrix . Lets call I as Ideal output&lt;/p&gt;&#xA;&#xA;&lt;p&gt;* denotes matrix multiplication.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know the values of matrix I, and R. I have to find what matrix A, if post multiplied by R would give me matrix I&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I set his situation up in matrix algebra and solve it compute A?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any pointers would be helpful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank You.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-AD.&lt;/p&gt;&#xA;" OwnerUserId="1249" LastActivityDate="2010-08-22T20:24:00.713" Title="Need to compute values of all the entries of a 3x3 matrix?" Tags="&lt;matrices&gt;" AnswerCount="5" CommentCount="2" />
  <row Id="2596" PostTypeId="2" ParentId="2595" CreationDate="2010-08-16T16:50:21.757" Score="1" ViewCount="" Body="&lt;p&gt;As Rasmus said, if you have only one (I,R) pair, then there are multiple possibilities for A.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are interested in any one such A, you can use the following.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $R = (x,y,z)$ and assume that $x \neq 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $I = (a,b,c)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can choose the matrix to be&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;[a/x 0 0]&#xA;[b/x 0 0]&#xA;[c/x 0 0]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-08-16T16:50:21.757" CommentCount="6" />
  <row Id="2597" PostTypeId="2" ParentId="2593" CreationDate="2010-08-16T16:51:44.330" Score="3" ViewCount="" Body="&lt;p&gt;In fact, it's a theorem of Erdös that every non-decreasing multiplicative function $f:\mathbb{N}\to\mathbb{R}$ is $f(n)=n^c$ for some constant $c$. This implies that those three conditions are already enough to prove that $f(n)=n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can read a proof by Pierre Pbornsztein &lt;a href=&quot;http://www.artofproblemsolving.com/Forum/viewtopic.php?p=179524#p179524&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="446" LastEditorUserId="205" LastEditorDisplayName="" LastEditDate="2010-08-16T17:33:19.360" LastActivityDate="2010-08-16T17:33:19.360" CommentCount="7" />
  <row Id="2598" PostTypeId="2" ParentId="2591" CreationDate="2010-08-16T17:03:53.687" Score="3" ViewCount="" Body="&lt;p&gt;One that confused me greatly when I first found it was &quot;We ignore if X is true.&quot;  I read it as &quot;We &lt;em&gt;disregard&lt;/em&gt; whether X is true,&quot; which made no sense in context.  Eventually I realized the authors meant &quot;We &lt;em&gt;do not know&lt;/em&gt; if X is true.&quot;  I suppose this corresponds to the fact that &quot;to ignore&quot; and &quot;to be ignorant&quot; have different meanings in English despite having the same (Latin?) root.  &lt;/p&gt;&#xA;" OwnerUserId="822" LastActivityDate="2010-08-16T17:03:53.687" CommunityOwnedDate="2010-08-16T17:03:53.687" />
  <row Id="2600" PostTypeId="2" ParentId="1949" CreationDate="2010-08-16T17:16:07.940" Score="1" ViewCount="" Body="&lt;p&gt;Take a look at the last big displayed equation under &quot;formal definition&quot; here.  It shows you Gauss's explicit form for a Levi-Civita connection in terms of the metric.  Since you know how the metric transforms under an isometry, and how a Lie bracket transforms under a diffeomorphism, working out how the connection transforms under an isometry amounts to putting those ingredients together. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Levi-Civita_connection&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Levi-Civita_connection&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="642" LastActivityDate="2010-08-16T17:16:07.940" />
  <row Id="2601" PostTypeId="1" CreationDate="2010-08-16T17:35:02.150" Score="9" ViewCount="669" Body="&lt;p&gt;Frequently, we introduce a new concept with a formal definition, then immediately say &quot;Intuitively, what this means is...&quot; What are the absolute best metaphors you've seen (for concepts of any level)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, I think the comparison of modular arithmetic to arithmetic with times, on a clock face, is fantastic, because it corresponds so perfectly and because clocks are so widely known.&lt;/p&gt;&#xA;" OwnerUserId="629" LastActivityDate="2011-01-27T23:02:21.027" CommunityOwnedDate="2010-08-16T17:35:02.150" Title="Best intuitive metaphors for math concepts (of any level)" Tags="&lt;soft-question&gt;&lt;big-list&gt;&lt;intuition&gt;&lt;education&gt;" AnswerCount="6" CommentCount="10" FavoriteCount="6" />
  <row Id="2602" PostTypeId="1" AcceptedAnswerId="2624" CreationDate="2010-08-16T17:40:59.877" Score="3" ViewCount="320" Body="&lt;p&gt;I've taken an undergraduate course in ring and group theroy, but haven't studied number theory formally.   I've noticed many important results in number theory have been generalized in group/ring theory (e.g. Lagrange Thm generalizing Fermats Little Thereom).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is are there any results in elementary number theory that have not been generalized using elementary group and ring theory?&lt;/p&gt;&#xA;" OwnerUserId="832" LastActivityDate="2010-08-17T00:05:51.077" Title="Elementary Number theory results that are not generalized by ring or group theory" Tags="&lt;soft-question&gt;&lt;abstract-algebra&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="2" />
  <row Id="2603" PostTypeId="2" ParentId="2595" CreationDate="2010-08-16T17:53:01.507" Score="3" ViewCount="" Body="&lt;p&gt;Actually if you have 24 sets of &lt;strong&gt;I&lt;/strong&gt; and &lt;strong&gt;R&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \mathsf A \cdot \mathbf R_i = \mathbf I_i \qquad i = 1, 2, \dotsc, 24 $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;since there are only 9 variables for each element of A but 24&amp;times;3 = 72 linear equations, the problem is &lt;a href=&quot;http://en.wikipedia.org/wiki/Overdetermined_system&quot; rel=&quot;nofollow&quot;&gt;overdetermined&lt;/a&gt; and there may be no solutions. However, if you accept some errors &lt;strong&gt;&amp;epsilon;&lt;/strong&gt;,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \mathsf A \cdot \mathbf R_i + \boldsymbol \epsilon_i = \mathbf I_i \qquad i = 1, 2, \dotsc, 24 $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is possible to find such an A, as this introduces 72 more variables (the problem becomes underdetermined). The best thing is of course to minimize the total error. We first write A as a column of row vectors:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \mathsf A = \begin{bmatrix} \mathbf A_1 \\ \mathbf A_2 \\ \mathbf A_3 \end{bmatrix} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;then we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \mathbf R_i \cdot \mathbf A_j + \epsilon_{ij} = I_{ij} \qquad i = 1, \dotsc, 24; j = 1, 2, 3 $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;this is exactly like a &lt;a href=&quot;http://en.wikipedia.org/wiki/Linear_regression_model&quot; rel=&quot;nofollow&quot;&gt;linear regression problem&lt;/a&gt; (if we minimize the sum of squares of errors)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \mathsf X \cdot \boldsymbol \beta_j + \boldsymbol \epsilon_j = \mathbf y_j $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;\mathsf X &amp;amp;= \begin{bmatrix}&#xA;  R_{1,1} &amp;amp; R_{1,2} &amp;amp; R_{1,3} \\&#xA;  \vdots &amp;amp; \vdots &amp;amp; \vdots \\&#xA;  R_{24,1} &amp;amp; R_{24,2} &amp;amp; R_{24,3}&#xA;\end{bmatrix} \\&#xA;\boldsymbol\beta_j &amp;amp;= \begin{bmatrix}&#xA;  A_{j,1} \\&#xA;  A_{j,2} \\&#xA;  A_{j,3}&#xA;\end{bmatrix} \\&#xA;\mathbf y_j &amp;amp;= \begin{bmatrix}&#xA;  I_{1,j} \\&#xA;  \vdots \\&#xA;  I_{24,j}&#xA;\end{bmatrix}&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With these we can use any software supporting regression (or by hand) to find the best approximation of &lt;strong&gt;A&lt;/strong&gt;&lt;sub&gt;j&lt;/sub&gt;, and thus the whole matrix A.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(This can be easily generalized to $N \gg 24$ pixels.)&lt;/p&gt;&#xA;" OwnerUserId="171" LastEditorUserId="171" LastEditorDisplayName="" LastEditDate="2010-08-16T18:46:54.267" LastActivityDate="2010-08-16T18:46:54.267" CommentCount="2" />
  <row Id="2604" PostTypeId="2" ParentId="2595" CreationDate="2010-08-16T17:53:02.143" Score="1" ViewCount="" Body="&lt;p&gt;Let $a_{ij}$ be the number in $i$-th row and $j$-th column in the matrix $A$. Then each equation of the form $AR=I$ is basicly three linear equations with unknowns $a_{ij}$. If there are 24 sets then you have $24 \cdot 3$ linear equations in $9$ unknowns. This system might have no solution in general. But, you can use &lt;a href=&quot;http://en.wikipedia.org/wiki/Linear_least_squares&quot; rel=&quot;nofollow&quot;&gt;Linear least squares&lt;/a&gt; to find a solution that fits best all those equations.&lt;/p&gt;&#xA;" OwnerUserId="334" LastActivityDate="2010-08-16T17:53:02.143" />
  <row Id="2605" PostTypeId="1" CreationDate="2010-08-16T18:05:44.990" Score="1" ViewCount="94" Body="&lt;p&gt;When showing that two functors $F:A\rightarrow B$ and $G:B\rightarrow A$ are adjoint, one defines a natural bijection $Mor(X,G(Y)) \rightarrow Mor(F(X),Y)$. What if one do not require the bijection to be natural, what issues would arise? &lt;/p&gt;&#xA;" OwnerUserId="1253" LastActivityDate="2010-08-16T18:05:44.990" Title="Adjoint functors requiring a natural bijection." Tags="&lt;category-theory&gt;" CommentCount="3" />
  <row Id="2606" PostTypeId="1" AcceptedAnswerId="2609" CreationDate="2010-08-16T18:29:08.860" Score="7" ViewCount="197" Body="&lt;p&gt;I have three questions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does the groupification of a semigroup always exist? I believe this should be yes because for every $x$ in the semigroup one could just define an element $x'$ that should work as its inverse. But what would then happen to the product $x'y$ for $x,y$ elements of the semigroup? It feels like we get choices (or maybe not) here that messes things up. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When defining the groupification, $G$, of a semigroup $S$ one require it to come with a morphism (of semigroups) $S \rightarrow G$ such that any other morphism (of semigroups) from $S$ to another group $G'$ factorizes through the previous map. Exactly which type of objects can be groupified? I guess one cannot groupify a topological space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a broad question but is there some sense of -ification? In the example one could replace &quot;group&quot; by &quot;topological space&quot; and talk about topologyfication. Now, no such word seem to exist so I guess one could not &quot;topologyfy&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can (i think) consider the groupification functor from the category of semigroups to the category of groups and it should be adjoint to the forgetful functor from the category of groups to the category of semigroups. This would suggest that we need some sense of a forgetful functor in the first place to talk about a -ification. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Apologies for this bad question, sometimes asking the right question is just as hard as answering it. &lt;/p&gt;&#xA;" OwnerUserId="1253" LastActivityDate="2011-01-16T10:35:48.130" Title="Is there some universal sense of -ification (eg, groupification) in category theory" Tags="&lt;category-theory&gt;" AnswerCount="4" CommentCount="4" FavoriteCount="3" />
  <row Id="2607" PostTypeId="2" ParentId="2595" CreationDate="2010-08-16T18:31:47.980" Score="1" ViewCount="" Body="&lt;p&gt;FIRST CASE. If you had only &lt;em&gt;one&lt;/em&gt; set of values for $I$ and $R$, as you said in your answer to Rasmus, then you'd have a system of linear equations with infinitely many solutions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For psychological reasons, let's write it this way:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;XA = B \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here $A$ is your $R$, $B$ your $I$ and $X$ your $A$. If we transpose we get a simultaneous system of linear equations&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;A^t X^t = B^t \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $A^t = (a_1 \ a_2 \ a_3)$ and $B^t = (b_1 \ b_2 \ b_3)$, and $(x_i \ y_i \ z_i)$ is the i-th colum $X^t$, we have the linear equations&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;a_1 x_i + a_2 y_i + a_3 z_i = b_i&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for $i = 1, 2, 3$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which, assuming $a_1 \neq 0$, you can solve like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;x_i = \frac{b_i}{a_1} - \frac{a_2}{a_1} y_i - \frac{a_3}{a_1} z_i \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, give $y_i$ and $z_i$ the values you want and you have a solution for your problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Maybe I could develop a little bit more my answer, in order to &lt;em&gt;really&lt;/em&gt; include your problem, which is one of overdeterminated simultaneous systems of linear equations, as KennyTM says, since you don't have a prefered pair of values $(R,I)$, do you? In order to handle all the $24$ pairs of values $(R,I)$ you have, maybe you should take into account this SECOND CASE.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;SECOND CASE. I'm sorry, but I'm changing slightly the notation again. In the end, I'll write the solution with yours.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;AX = B&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;be a (simultaneous) system of linear equations such as yours, with $A$ a $24\times 3$ matrix ($24$ rows, $3$ columns), $X$ a $3\times 3$ matrix and $B$ a $24 \times 3$ matrix.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;*&lt;em&gt;Hypothesis: Let's assume that our matrix $A$ (that is, your $R$) has rang $ 3$. *&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(If this is not the case, the problem is more involved.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's write the first system this way:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;x_1a_1 + y_1a_2 + z_1a_3 = b_1  \ .  \qquad [1]&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here, $a_i , i = 1,2,3$ are the columns of $A$, $X_1 = \begin{pmatrix} x_1 &amp;amp; y_1 &amp;amp; z_1 \end{pmatrix}^t$ the first column of $X$ and also $b_1$ is the first column of $B$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we see in one go a geometrical interpretation of our system of equations: system [1] has a solution $\begin{pmatrix} x_1 &amp;amp; y_1 &amp;amp; z_1 \end{pmatrix}$ if and only if the vector $b_1$ belongs to the linear span generated by the columns of $A$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;AX_1 = b_1 \quad \text{has a solution} \qquad \Longleftrightarrow \qquad b_1 \in [a_1, a_2, a_3]&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What can we do if this is &lt;em&gt;not&lt;/em&gt; the case? -To look for the &lt;em&gt;nearest&lt;/em&gt; vector in $[a_1, a_2, a_3]$ to $b_1$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This &lt;em&gt;nearest&lt;/em&gt; vector is, of course, the &lt;em&gt;orthogonal projection&lt;/em&gt; of $b_1$ onto the subspace $[a_1, a_2, a_3] $.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to Wikipedia, &lt;a href=&quot;http://en.wikipedia.org/wiki/Orthogonal_projection&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Orthogonal_projection&lt;/a&gt; , the matrix of this orthogonal projection in the standard basis of $\mathbb{R}^{24}$ is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;P_A = A (A^tA)^{-1}A^t \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the best $X = \begin{pmatrix} X_1 &amp;amp; X_2 &amp;amp; X_3\end{pmatrix}$ for you is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;X = A (A^tA)^{-1}A^tB&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $B = \begin{pmatrix} b_1 &amp;amp; b_2 &amp;amp; b_3 \end{pmatrix}$. Now, let's go back to your notation: $A = X^t$, $R = A^t$ and $I = B^t$. So&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;A =  \left( R^t (R^{tt}R^t)^{-1}R^{tt}I^t \right)^t = I R^t ((RR^t)^{-1})^t R&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $R$ and $I$ are now the matrices with &lt;em&gt;all&lt;/em&gt; of your $R$'s and $I$'s as columns.&lt;/p&gt;&#xA;" OwnerUserId="664" LastEditorUserId="664" LastEditorDisplayName="" LastEditDate="2010-08-17T09:16:23.623" LastActivityDate="2010-08-17T09:16:23.623" />
  <row Id="2608" PostTypeId="2" ParentId="213" CreationDate="2010-08-16T18:41:03.000" Score="5" ViewCount="" Body="&lt;p&gt;I wish I'd learned logic much, much earlier. Obviously young students couldn't handle much depth, but at least a basic introduction to a few concepts would be nice. Just understanding the concept of axioms and deductive rules would put all of math into some perspective. When I finally understood that math was constructed with formal definitions and proofs (or, for example, that there was more than one useful way to axiomatize), I felt I'd been kept in the dark my whole life, doing something (math) that I had absolutely no understanding of.&lt;/p&gt;&#xA;" OwnerUserId="629" LastActivityDate="2010-08-16T18:41:03.000" CommunityOwnedDate="2010-08-16T18:41:03.000" />
  <row Id="2609" PostTypeId="2" ParentId="2606" CreationDate="2010-08-16T18:46:26.903" Score="6" ViewCount="" Body="&lt;p&gt;I take the liberty of answering only the parts of your question, to which I think I can give a precise answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1: Yes, it is called the &lt;a href=&quot;http://en.wikipedia.org/wiki/Grothendieck_group&quot; rel=&quot;nofollow&quot;&gt;Grothendieck group&lt;/a&gt; $G(N)$ of the given semigroup $S$. This construction is functorial.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2: There is always a canonical semigroup homomorphism $S\to G(S)$, but this need not be injective in general. For example, the Grothendieck group corresponding to $\mathbb N\cup\{\infty\}$ with the obvious addition $(n+\infty=\infty)$ is trivial!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3: Concerning -ification in general, at the moment I have nothing to add to BBischof's great comments above.&lt;/p&gt;&#xA;" OwnerUserId="367" LastEditorUserId="367" LastEditorDisplayName="" LastEditDate="2011-01-16T10:35:48.130" LastActivityDate="2011-01-16T10:35:48.130" />
  <row Id="2610" PostTypeId="2" ParentId="2602" CreationDate="2010-08-16T18:47:01.927" Score="2" ViewCount="" Body="&lt;p&gt;I contend that every result that could reasonably be called &quot;elementary&quot; number theory also holds for $\mathbb{F}_q[x]$, the ring of polynomials in one variable over a finite field.  Quadratic reciprocity fits the bill here, for example.  This is the subject of &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0387953353&quot; rel=&quot;nofollow&quot;&gt;at least one book&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-08-16T18:47:01.927" CommentCount="4" />
  <row Id="2611" PostTypeId="1" CreationDate="2010-08-16T18:50:43.073" Score="1" ViewCount="298" Body="&lt;p&gt;If given the cost to play, and the average win.  Can I calculate the edge? (probability of winning)&lt;/p&gt;&#xA;" OwnerUserId="540" LastActivityDate="2010-12-05T10:28:01.937" Title="How do I calculate gambling edge using average win?" Tags="&lt;probability&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="2612" PostTypeId="1" AcceptedAnswerId="2617" CreationDate="2010-08-16T19:08:16.283" Score="2" ViewCount="179" Body="&lt;p&gt;I've recently been interested in Monte Carlo simulations and their uses, unfortunately most of the examples I find are difficult to understand for a beginner. What are some simple examples of using Monte Carlo simulations used to solve math problems? Bonus points for showing a simple example and using it to understand a more complex one.&lt;/p&gt;&#xA;" OwnerUserId="82" LastEditorUserId="121" LastEditorDisplayName="" LastEditDate="2010-08-16T19:33:23.800" LastActivityDate="2010-08-16T22:31:18.863" Title="What are some uses for Monte Carlo simulations in mathematics?" Tags="&lt;big-list&gt;&lt;monte-carlo&gt;" AnswerCount="4" CommentCount="3" FavoriteCount="1" />
  <row Id="2614" PostTypeId="2" ParentId="2606" CreationDate="2010-08-16T19:13:26.227" Score="4" ViewCount="" Body="&lt;p&gt;To add to Rasmus's answer to 1. One can't avoid elements like&#xA;$x^{-1}y$ in the Grothendieck group. Consider say the natural numbers.&#xA;It's Grothendieck group is the integers. As it's additive we&#xA;write say four plus the inverse of 6 as $4-6$. Things aren't too&#xA;bad in the commutative case: (in additive notation) all elements&#xA;of $G(S)$ have the form $a-b$ with $a$, $b\in S$. Moreover&#xA;$a-b=c-d$ iff there is $s\in S$ with $a+d+s=b+c+s$&#xA;(and we can forget about $s$ if $S$ is cancellative).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But things are nastier in the noncommutative world. Returning&#xA;to multiplicative notation, you now get elements in $G(s)$ like $ab^{-1}cd^{-1}e$&#xA;which perhaps don't simplify to shorter ones. And it's not so&#xA;easy to give a criterion for when two elements are the same&#xA;(or equivalently when one element equals the identity).&#xA;&lt;em&gt;C'est la vie!&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-08-16T19:13:26.227" />
  <row Id="2615" PostTypeId="2" ParentId="2606" CreationDate="2010-08-16T19:20:30.440" Score="7" ViewCount="" Body="&lt;p&gt;As BBischof says, the standard notion of -ification is to take the left adjoint of a forgetful functor.  This includes the following as special cases:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The groupification of a monoid or semigroup,&lt;/li&gt;&#xA;&lt;li&gt;The free group on a set, the free vector space on a set, etc.&lt;/li&gt;&#xA;&lt;li&gt;The abelianization of a group,&lt;/li&gt;&#xA;&lt;li&gt;The group ring of a group (the forgetful functor here sends a ring to its group of units),&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;and many other examples.  I do not think one can reasonably talk about universal -ification without a specific choice of forgetful functor; if no good choice exists, you won't get a good notion of -ification, and on the other hand there may be more than one choice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Notably, I remember reading that in at least one example the natural construction is to take the &lt;em&gt;right&lt;/em&gt; adjoint, but I don't remember what this example was.)&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-08-16T19:20:30.440" CommentCount="2" />
  <row Id="2616" PostTypeId="2" ParentId="2612" CreationDate="2010-08-16T19:34:37.587" Score="0" ViewCount="" Body="&lt;p&gt;EDIT: a previous comment posted the same answer, I just noticed&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could use montecarlo method to approximate pi.&#xA;You basically define a square of side length 2 and inscribe a circle of radius 1 in it. Let the center of the circle be at the origin.&#xA;Use a random number generator to pick an x-coordinate between 0 and 1 and a y-coordinate between 0 and 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;number of pts in the circle / number of pts in square ~ pi/4&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the simplest example of a monte carlo code I know and I think it is probably the standard example.&lt;/p&gt;&#xA;" OwnerUserId="249" LastEditorUserId="249" LastEditorDisplayName="" LastEditDate="2010-08-16T19:42:11.813" LastActivityDate="2010-08-16T19:42:11.813" />
  <row Id="2617" PostTypeId="2" ParentId="2612" CreationDate="2010-08-16T19:34:46.120" Score="7" ViewCount="" Body="&lt;p&gt;Monte Carlo methods are very useful in numerically evaluating high-dimensional integrals. With traditional integration methods, the number of integrand evaluations required to maintain accuracy grows quickly as dimension increases.  With Monte Carlo integration, the number of integrand evaluations needed is &lt;em&gt;independent&lt;/em&gt; of dimension.  For many high-dimensional integrals, Monte Carlo methods are the only practical choice.&lt;/p&gt;&#xA;" OwnerUserId="247" LastActivityDate="2010-08-16T19:34:46.120" CommentCount="3" />
  <row Id="2618" PostTypeId="2" ParentId="2455" CreationDate="2010-08-16T19:40:37.027" Score="2" ViewCount="" Body="&lt;p&gt;Place the balls so that the corners of the pyramid have coordinates (1,1,1), (n,1,1), (1,n,1), (1,1,n). Then for the ball at (x,y,z) we see that (x,x+y,x+y+z) is an ordered triple of distinct numbers between 1 and n+2 inclusive. There are $\binom{n+2}{3}$ such triples, and they are clearly in one-to-one correspondence with the balls. Geometrically, this corresponds to drawing planes x=constant, x+y=constant, and x+y+z=constant through a given ball, and seeing where these planes intersect the x axis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(To compare with the case $k=2$, &quot;straighten&quot; the picture above into a right triangle. Then the diagonal lines become lines x=constant and x+y=constant, and the bottom row of balls (the blue ones) can be identified with the points where these lines intersect the x axis.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This also works in higher dimensions, giving $\binom{n+k-1}{k}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whether this proof really is any different from the one you suggested in the question can of course be debated...&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2010-08-16T19:40:37.027" CommentCount="0" />
  <row Id="2619" PostTypeId="1" CreationDate="2010-08-16T19:58:20.230" Score="3" ViewCount="197" Body="&lt;p&gt;Couple of days ago, i actually asked this question : &lt;a href=&quot;http://math.stackexchange.com/questions/2121/existence-of-a-normal-subgroup-with-authautg&quot;&gt;http://math.stackexchange.com/questions/2121/existence-of-a-normal-subgroup-with-authautg&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was thinking about the converse of this statement. Suppose $G$ is a finite group, with subgroups $H$ and satisfies $|Aut(G)|=|Aut(H)|$, then can anything be said about, $H$ or $G$ in terms of their structure. Will $H$ be normal or ....? Or one can even ask, that find all finite groups $G$ such that $|Aut(H)|=|Aut(G)|$ for every subgroup $H$ of $G$! I am not interested that much in this question as there seems one can have lot of Groups of this type, but i am more curious to know the behaviour of $H$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, anyone interested can very well read this article in MO: &lt;a href=&quot;http://mathoverflow.net/questions/1075/order-of-an-automorphism-of-a-finite-group&quot; rel=&quot;nofollow&quot;&gt;http://mathoverflow.net/questions/1075/order-of-an-automorphism-of-a-finite-group&lt;/a&gt;&lt;/p&gt;&#xA;" LastActivityDate="2010-08-16T19:58:20.230" Title="Solving $|Aut(H)| = |Aut(G)|$" Tags="&lt;abstract-algebra&gt;&lt;group-theory&gt;" CommentCount="7" />
  <row Id="2620" PostTypeId="2" ParentId="2606" CreationDate="2010-08-16T20:12:45.263" Score="4" ViewCount="" Body="&lt;p&gt;The name of the really &lt;em&gt;really&lt;/em&gt; general construction (which is indeed in many cases the left adjoint of the forgetful functor) is called the &lt;strong&gt;completion&lt;/strong&gt; of an object. The basic idea is to find the 'most natural' or 'smallest' object having the original as a subobject, which means we can extend the idea to, say, the closure of a subset of a topological space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://ncatlab.org/nlab/show/completion&quot; rel=&quot;nofollow&quot;&gt;Here's the n-lab page if you're interested...&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="160" LastActivityDate="2010-08-16T20:12:45.263" CommentCount="1" />
  <row Id="2621" PostTypeId="1" CreationDate="2010-08-16T20:48:51.997" Score="3" ViewCount="164" Body="&lt;p&gt;Let $A$ be a commutative Banach algebra with unit. &#xA;It is well known that if the Gelfand transform $\hat{x}$ of $x\in A$ is non-zero, then $x$ is invertible in $A$ (the so called Wiener Lemma in the case when $A$ is the Banach algebra of absolutely convergent Fourier series).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a converse of the above, let $B$ be a Banach space contained in $A$ and suppose $B$ is closed under inversion - i.e.: If $x\in B$ and $x^{-1}\in A$ then $x^{-1}\in B$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) Prove that $B$ is a Banach algebra.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) Must $A$ and $B$ have the same norm? If not are the norms similar?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(3) Do $A$ and $B$ have the same maximal ideal space?&lt;/p&gt;&#xA;" OwnerUserId="1154" LastActivityDate="2010-08-17T07:55:26.113" Title="Converse of Wiener's lemma" Tags="&lt;functional-analysis&gt;&lt;fourier-series&gt;" AnswerCount="1" CommentCount="6" />
  <row Id="2622" PostTypeId="2" ParentId="2612" CreationDate="2010-08-16T21:09:11.103" Score="2" ViewCount="" Body="&lt;p&gt;Monte Carlo methods are used extensively in &lt;a href=&quot;http://en.wikipedia.org/wiki/Monte_Carlo_methods_in_finance&quot; rel=&quot;nofollow&quot;&gt;financial mathematics&lt;/a&gt; for the pricing of complex or &quot;exotic&quot; financial derivatives.  With equity options, for example, the value of the stocks in the option contract is simulated using a stochastic process and parameters that can be observed or derived from other financial instruments.  This process is computationally very intensive resulting in many investment banks having server farms of tens of thousands of machines dedicated to the pricing and risk management of derivatives.&lt;/p&gt;&#xA;" OwnerUserId="1032" LastActivityDate="2010-08-16T21:09:11.103" />
  <row Id="2624" PostTypeId="2" ParentId="2602" CreationDate="2010-08-16T22:17:26.140" Score="5" ViewCount="" Body="&lt;p&gt;Problems for integers whose generalizations in algebraic number theory have not been developed (or at least considered) are rare.  The only examples I know of are: &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Ancient problems without a strong relation to modern algebraic techniques.  This includes the existence of odd perfect numbers, and compositeness of Fermat numbers $2^{2^k}+1$.   Keep in mind, though, that most of the very old problems have been understood or finished off by modern techniques, such as the congruent number problem and Fermat's Last Theorem being reduced to facts about elliptic curves, or the probabilistic approach to the distribution of special types of primes (twin, Mersenne, k-term progressions, etc), and these better-understood problems are more amenable to generalization beyond the integers.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Questions of combinatorial nature, such as covering congruences or arithmetic progressions (van der Waerden theorems).  Here the problems can be generalized to other rings, but most interest has focused on the integer case.  &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Uses of number theory in logic and computer science, such as Goedel coding, complexity of Presburger arithmetic, or other problems where iterated exponentiation appears.   Algebraic number theory is primarily about problems described by polynomial equations, and the reduction of exponential Diophantine equations to polynomial ones (as in the solution of Hilbert's 10th problem) is not direct enough to allow the use of the standard algebraic tools.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;For everything else, where standard algebraic techniques or analytic ones (zeta functions, Diophantine approximation, sieves, transcendence theory, complex analysis, etc) are applicable to a problem over the integers, there has usually been an effort to search for analogues of those methods in algebraic number theory, p-adic number theory, geometry over finite fields, and complex geometry.   This applies to all the major techniques and theories developed since 1800.  So to find examples one needs to move away from &quot;central core&quot; number theory and look at questions that are not, so far, primarily studied by the methods of algebraic number theory and algebraic geometry. &lt;/p&gt;&#xA;" OwnerUserId="467" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-08-16T23:18:11.877" LastActivityDate="2010-08-16T23:18:11.877" />
  <row Id="2625" PostTypeId="2" ParentId="2612" CreationDate="2010-08-16T22:31:18.863" Score="1" ViewCount="" Body="&lt;p&gt;Tangential to John D. Cook's reply, and also somewhat related: Monte Carlo also finds application in the solution of (partial, stochastic) differential equations, of which cubature (nobody ever uses MC in the one-dimensional case practically ;) ) is but a specialized case. As already mentioned by John, the pain dealt by the &quot;curse of dimensionality&quot; stings less with Monte Carlo.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-08-16T22:31:18.863" />
  <row Id="2626" PostTypeId="1" AcceptedAnswerId="2644" CreationDate="2010-08-16T22:39:14.420" Score="5" ViewCount="425" Body="&lt;p&gt;I'd like to create a basic model of the forces acting on a sail (wind sail, like a tail ship)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A couple of things I was thinking about: &#xA;1) can create a very simple model where wind is 'one' force acting on the a uniform body.&#xA;2) model wind as vectors. This is where I am a little confused on how to start. &#xA;3) adding in multiple sails. &#xA;4) I know that this would be a differential equation but after that I can't really see how it would be modeled. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any pointers? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not looking for someone to actually do the modelling for me, just a place to start. Like maybe some Wikipedia articles, etc&lt;/p&gt;&#xA;&#xA;&lt;p&gt;thanks&lt;/p&gt;&#xA;" OwnerUserId="608" LastActivityDate="2010-08-18T21:45:12.413" Title="Modelling forces acting on a sail" Tags="&lt;differential-equations&gt;" AnswerCount="5" CommentCount="10" FavoriteCount="1" />
  <row Id="2627" PostTypeId="1" CreationDate="2010-08-16T22:54:18.983" Score="2" ViewCount="184" Body="&lt;p&gt;Let $X$ denote the blowup of $\mathbb P^2$, $E$ the exceptional divisor, and $H$ the pullback of the hyperplane class.  How can I compute $H^0(X,mE+nH)$, $H^1(X,mE+nH)$, and $H^2(X,mE+nH)$ for $m,n \in \mathbb Z$?  If I'm working analytically, how can I think of these geometrically (e.g. &quot;an element of $H^1(X,2E)$ is equivalent to a $1$-form on $\mathbb P^2$ such that...&quot;)?&lt;/p&gt;&#xA;" LastActivityDate="2010-08-17T16:11:58.447" Title="Cohomology of line bundles on the blowup of $\mathbb P^2$" Tags="&lt;algebraic-geometry&gt;&lt;cohomology&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="1" />
  <row Id="2628" PostTypeId="1" AcceptedAnswerId="2699" CreationDate="2010-08-16T23:26:57.880" Score="3" ViewCount="412" Body="&lt;p&gt;By stressing my manipulative powers and a bit of help from &lt;em&gt;Mathematica&lt;/em&gt; I was able to show that the triple integral&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac1{8\pi}\int_0^{2\pi}\int_0^\pi\int_0^\pi\,(2\cos\theta\cos\varphi-\sin\theta\sin\varphi\cos\psi)^{2k}\sin\theta\sin\varphi \,\mathrm{d}\theta\mathrm{d}\varphi\mathrm{d}\psi $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where k is a positive integer is equivalent to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$${\left(\frac{2^k}{2k+1}\right)^2}\sum_{j=0}^k\frac{\binom{2j}{j}}{\binom{2k}{k}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;by making use of the binomial expansion and then using a hypergeometric identity that degenerates to a ratio of gamma functions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For some reason I think there was no need for &lt;em&gt;Mathematica&lt;/em&gt; to invoke the Gauss hypergeometric function ${}_2 F_1\left(a,b;c;z\right)$ in performing the triple integration, so I'm wondering if there might be a way to show that this identity holds, using nothing more complicated than a gamma function.&lt;/p&gt;&#xA;" OwnerUserId="498" LastEditorUserId="8246" LastEditorDisplayName="" LastEditDate="2011-05-26T09:47:51.773" LastActivityDate="2011-05-26T09:47:51.773" Title="An elementary way of simplifying a trigonometric triple integral?" Tags="&lt;gamma-function&gt;&lt;closed-form&gt;&lt;integral&gt;" AnswerCount="2" CommentCount="5" FavoriteCount="2" />
  <row Id="2629" PostTypeId="2" ParentId="2626" CreationDate="2010-08-17T00:15:17.253" Score="1" ViewCount="" Body="&lt;p&gt;What about $F = \sum_i W \cdot S_i$, where $W$ is the wind velocity vector and $S$ is a list of all the sail unit vectors?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't see this at all being a differential equation, but if you further specify your idea of this model, I can help you a little more.&lt;/p&gt;&#xA;" OwnerUserId="227" LastActivityDate="2010-08-17T00:15:17.253" CommentCount="2" />
  <row Id="2630" PostTypeId="1" AcceptedAnswerId="2631" CreationDate="2010-08-17T01:13:38.287" Score="2" ViewCount="151" Body="&lt;p&gt;Suppose A is some algebraic structure, and x and y are two elements of the underlying set.  Is there any more concise way of stating, &quot;there exists an automorphism in A which maps x to y&quot;?.  It seems like the obvious thing to say is &quot;x and y are symmetric in A&quot;, but I've never come across this phrasing or any other in the literature, which is amazing to me since it's such an enormously useful concept.&lt;/p&gt;&#xA;" OwnerUserId="1262" LastActivityDate="2010-08-17T16:50:47.770" Title="Algebra terminology question" Tags="&lt;terminology&gt;&lt;universal-algebra&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="2" />
  <row Id="2631" PostTypeId="2" ParentId="2630" CreationDate="2010-08-17T02:28:05.643" Score="2" ViewCount="" Body="&lt;p&gt;Some authors write that $x$ and $y$ are conjugate under $G$, or $G$-conjugate, e.g. from Bourbaki, Algebra:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Definition 5.&lt;/strong&gt; Let $G$ be a group, $E$ a $G$-set and $x \in G$. An element $y \in E$ is&#xA;  &lt;em&gt;conjugate&lt;/em&gt; to $x$ under the operation of $G$ if there exists an element $\alpha \in G$ such &#xA;  that $y = \alpha x$. The set of conjugate elements of $x$ is called the &lt;em&gt;orbit&lt;/em&gt; of $x$ in $E$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-08-17T16:50:47.770" LastActivityDate="2010-08-17T16:50:47.770" />
  <row Id="2633" PostTypeId="1" AcceptedAnswerId="2640" CreationDate="2010-08-17T04:58:58.507" Score="0" ViewCount="326" Body="&lt;p&gt;Let $(0,\infty)$ be equipped with the Lebesgue measure, and let $1 &amp;lt; p &amp;lt; \infty$. For each $ f \in L_{p}(\lambda)$ let $$T(f)(x) = x^{-1} \int f \chi_{(0,x)} \ d\lambda \quad \text{for} \ x &gt;0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then prove that $T$ is a bounded linear operator from $L_{p}(0,\infty)$ into itself.&lt;/p&gt;&#xA;" LastActivityDate="2010-08-17T17:24:17.560" Title="Proving that $T$ is a bounded linear operator" Tags="&lt;real-analysis&gt;&lt;functional-analysis&gt;&lt;measure-theory&gt;" AnswerCount="4" CommentCount="2" />
  <row Id="2634" PostTypeId="1" CreationDate="2010-08-17T05:09:19.613" Score="1" ViewCount="156" Body="&lt;p&gt;A Liouville number is a number which can be approximated very closely be a sequence of rational numbers (here is the rigorous definition I am working off of: &lt;a href=&quot;http://en.wikipedia.org/wiki/Liouville_number&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Liouville_number&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm looking for an example of a Liouville number which cannot be approximated by a sequence of rational numbers with a denominators which are all a constant c multiplied by powers of some number a.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, the Louiville constant (0.110001000000000000000001...) can be approximated by the sequence 1/10, 11/10^2, 110001/10^6, etc, which is not what I am looking for because in each case the denominator is a power of 10. In this case, we would say that c=1, a=10, and the denominator is always of the form c*a^n for some positive n.&lt;/p&gt;&#xA;" OwnerUserId="796" LastActivityDate="2010-08-17T05:09:19.613" Title="Certain Liouville Numbers" Tags="&lt;number-theory&gt;" CommentCount="2" FavoriteCount="1" />
  <row Id="2638" PostTypeId="2" ParentId="2633" CreationDate="2010-08-17T07:13:51.533" Score="0" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt; This is just an approach, it doesn't actually work. I don't know if it's possible to make it correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $q = \frac{1}{1 - \frac{1}{p}}$ so $\frac{1}{p} + \frac{1}{q} = 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By &lt;a href=&quot;http://en.wikipedia.org/wiki/Holder_inequality&quot; rel=&quot;nofollow&quot;&gt;Hölder's inequality&lt;/a&gt; we have&#xA;$$&#xA;\int_{0}^{x} |f(\lambda)| d\lambda \le (\int_{0}^{x} |f(\lambda)|^{p} d\lambda)^{\frac{1}{p}} (\int_{0}^{x} |1|^{q} d\lambda)^{\frac{1}{q}} \le ||f||_{p} x^{\frac{1}{q}}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;|\frac{1}{x} \int_{0}^{x} f(\lambda) d\lambda|^p \le |\frac{1}{x} \int_{0}^{x} |f(\lambda)| d\lambda|^p \le ||f||^{p}_{p} x^{\frac{1}{q}-1}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now norm of $T(f)$ has the following form:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;||T(f)||_{p} = (\int _{0}^{\infty} |T(f)(x)|^{p}dx)^{1/p} = (\int _{0}^{\infty}|\frac{1}{x} \int _{0}^{x} f(\lambda) d\lambda|^pdx)^{1/p}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $||f||^{p} = c$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;||T(f)||_{p} \le (\int _{0}^{\infty} c^{p} x^{\frac{1}{q}-1} dx)^{1/p} = ||f||_p (\int _{0}^{\infty} x^{-1/p})^{p}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $(\int _{0}^{\infty} x^{-1/p})^{p}$ is some constant we have $||T(f)|| _{p} \le ||f|| _{p} const$, hence this operator is bounded.&lt;/p&gt;&#xA;" OwnerUserId="334" LastEditorUserId="334" LastEditorDisplayName="" LastEditDate="2010-08-17T08:14:26.457" LastActivityDate="2010-08-17T08:14:26.457" CommentCount="3" />
  <row Id="2639" PostTypeId="2" ParentId="2633" CreationDate="2010-08-17T07:32:38.337" Score="0" ViewCount="" Body="&lt;p&gt;Following falagar's solution one can also get the value of $ \displaystyle ||T|| = \frac{p}{p-1}$&lt;/p&gt;&#xA;" LastActivityDate="2010-08-17T07:32:38.337" />
  <row Id="2640" PostTypeId="2" ParentId="2633" CreationDate="2010-08-17T07:33:09.847" Score="3" ViewCount="" Body="&lt;p&gt;One (direct) way to see this is that $T$ is dominated by (a slightly modfied version of) the &lt;a href=&quot;http://en.wikipedia.org/wiki/Hardy-Littlewood_maximal_operator&quot; rel=&quot;nofollow&quot;&gt;Hardy-Littlewood maximal operator&lt;/a&gt;, and it is a theorem that this is a bounded operator on $L^p$ for $p$ greater than 1. Or, directly, one can see that $T$ is bounded on $L^\infty$ and weak-bounded on $L^1$ (because if $f$ belongs to $ L^1$ then $Tf(x) &amp;lt;= ||f||_1/x$), so the &lt;a href=&quot;http://en.wikipedia.org/wiki/Marcinkiewicz_interpolation_theorem&quot; rel=&quot;nofollow&quot;&gt;Marcinkiewicz interpolation theorem&lt;/a&gt; applies. This is really much too indirect, though, but I'd like to point out that this is a special case of something more general. A direct approach is in page 242 of &lt;a href=&quot;http://books.google.com/books?id=u5fkpAJ5Rc0C&amp;amp;pg=PA239&amp;amp;lpg=PA239&amp;amp;dq=hardy%27s+inequality+proof&amp;amp;source=bl&amp;amp;ots=MyVsFH7dqx&amp;amp;sig=JCc4As_ifrCAFHMf8Xq-MMBqufk&amp;amp;hl=en&amp;amp;ei=STlqTOiRDIGBlAfsxoC5Ag&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=3&amp;amp;ved=0CCQQ6AEwAjhG#v=onepage&amp;amp;q=hardy%27s%20inequality%20proof&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;Inequalities&lt;/a&gt; (the page is available).&lt;/p&gt;&#xA;" OwnerUserId="536" LastEditorUserId="536" LastEditorDisplayName="" LastEditDate="2010-08-17T08:04:15.810" LastActivityDate="2010-08-17T08:04:15.810" />
  <row Id="2641" PostTypeId="2" ParentId="2621" CreationDate="2010-08-17T07:55:26.113" Score="1" ViewCount="" Body="&lt;p&gt;For now, I shall assume that $B$ is a closed subspace of $A$ (and is treated simply as a topological vector space for the moment).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) Let $x \in B$; we need to prove that $x^2 \in B$. From this elementary algebra will imply that $B$ is closed under multiplication. Now, when $t$ is close to zero, we have $1-tx$ invertible in $A$, hence in $B$; so $1+tx + t^2x^2 + \dots$ lies in $B$ for $t$ close to zero. Taking the second derivative at $0$ (recall that $B$ is a &lt;em&gt;closed&lt;/em&gt; subspace) shows that $x^2 \in B$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't fully understand what you are asking in (2) yet, so:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(3) No. Take $A$ to be any Banach algebra not equal to the complex numbers and $B$ the subalgebra spanned by the identity.&lt;/p&gt;&#xA;" OwnerUserId="536" LastActivityDate="2010-08-17T07:55:26.113" CommentCount="4" />
  <row Id="2643" PostTypeId="2" ParentId="2626" CreationDate="2010-08-17T09:55:56.080" Score="2" ViewCount="" Body="&lt;p&gt;The post &lt;em&gt;&lt;a href=&quot;http://terrytao.wordpress.com/2009/03/23/sailing-into-the-wind-or-faster-than-the-wind/&quot; rel=&quot;nofollow&quot;&gt;The Sailing into the wind, or faster than the wind&lt;/a&gt;&lt;/em&gt;  by Professor Terence Tao and the comments to it may be of interest to you. (See my comment above and cbrulak's reply)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To explain &quot;the power of the wind to sail in a direction against that of the wind or to sail with a speed faster than the wind itself&quot; the author describes the one-, two- and three-dimensional sailing models.&lt;/p&gt;&#xA;" OwnerUserId="752" LastActivityDate="2010-08-17T09:55:56.080" />
  <row Id="2644" PostTypeId="2" ParentId="2626" CreationDate="2010-08-17T10:21:24.860" Score="5" ViewCount="" Body="&lt;p&gt;Assuming no edge effects/turbulence, we can calculate the force of wind on a single sail using the &lt;a href=&quot;http://en.wikipedia.org/wiki/Drag_equation&quot; rel=&quot;nofollow&quot;&gt;drag equation&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\vec{F_D} = \tfrac12 \rho \vec{u}^2 C_D A$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where the drag coefficient $C_D$ is going to be fairly high, for a concave sail.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, however, the wind may not be hitting the sail orthogonally, but rather at some angle. We can generalise the drag equation as such:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\vec{F_{sail}} = \tfrac12 \rho \vec{u}^2 C_D (\hat{\vec{u}} \cdot \vec{S})$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $S% is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Vector_area&quot; rel=&quot;nofollow&quot;&gt;vector area&lt;/a&gt; of the sail. We take the dot product of this vector with the unit velocity vector to resolve the wind force in the direction of the sail. This is of course an approximation, as the concave shape of the sail will come into play, but probably a good enough one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That pretty much explains the basic situation for a single sail. Now, for multiple sails, you would of course simply combine the forces on the individual sails, which will all have their own values of $C_D$ and $S$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\vec{F_{sails,total}} = \sum \vec{F_{sail}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suggest you familiarise yourself with the mechanics and specific equations used here, and use that as a basic model to start.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For some general information on the physics of sailing, you may find &lt;a href=&quot;http://www.animations.physics.unsw.edu.au/jw/sailing.html&quot; rel=&quot;nofollow&quot;&gt;this page&lt;/a&gt; helpful.&lt;/p&gt;&#xA;" OwnerUserId="56" LastActivityDate="2010-08-17T10:21:24.860" CommentCount="2" />
  <row Id="2645" PostTypeId="1" AcceptedAnswerId="2646" CreationDate="2010-08-17T10:48:58.560" Score="10" ViewCount="230" Body="&lt;p&gt;Suppose there are two correlated random variable and having very small correlation coefficient (order of 10&lt;sup&gt;-1&lt;/sup&gt;). Is it valid to approximate it as  independent random variables? &lt;/p&gt;&#xA;" OwnerUserId="1273" LastEditorUserId="171" LastEditorDisplayName="" LastEditDate="2010-08-17T12:14:59.223" LastActivityDate="2010-08-18T08:38:58.927" Title="probability and statistics: Does having little correlation imply independence?" Tags="&lt;probability-theory&gt;&lt;statistics&gt;&lt;probability&gt;" AnswerCount="4" CommentCount="1" FavoriteCount="2" />
  <row Id="2646" PostTypeId="2" ParentId="2645" CreationDate="2010-08-17T10:52:51.410" Score="5" ViewCount="" Body="&lt;p&gt;Independence of random variables implies that they are uncorrelated, but the reverse is not true. Hence no, such an approximation is not valid (given that information alone).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A clear description is given on &lt;a href=&quot;http://en.wikipedia.org/wiki/Uncorrelated&quot; rel=&quot;nofollow&quot;&gt;this Wikipedia page&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If X and Y are independent, then they&#xA;  are uncorrelated. However, not all&#xA;  uncorrelated variables are&#xA;  independent. For example, if X is a&#xA;  continuous random variable uniformly&#xA;  distributed on [−1, 1] and Y = X²,&#xA;  then X and Y are uncorrelated even&#xA;  though X determines Y and a particular&#xA;  value of Y can be produced by only one&#xA;  or two values of X.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="56" LastEditorUserId="56" LastEditorDisplayName="" LastEditDate="2010-08-17T13:54:46.760" LastActivityDate="2010-08-17T13:54:46.760" CommentCount="3" />
  <row Id="2647" PostTypeId="1" AcceptedAnswerId="2649" CreationDate="2010-08-17T11:57:00.337" Score="6" ViewCount="341" Body="&lt;p&gt;i) Does there exists a nonempty subset of integers $S$, such that for all $a \in S$, there exists $b,c \in S$ such that $a = b + c$, where $a, b, c$ are &lt;em&gt;distinct&lt;/em&gt; integers. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edited to add:&#xA;ii) Can an $S$ with these properties be &lt;em&gt;finite&lt;/em&gt;, and also have the property that $a \in S$ implies $-a \notin S$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;This question is inspired by &lt;a href=&quot;http://math.stackexchange.com/questions/2418/a-zero-sum-subset-of-a-sum-full-set&quot;&gt;another&lt;/a&gt; question, which has not been fully answered yet.&lt;/em&gt; &lt;/p&gt;&#xA;" OwnerUserId="145" LastEditorUserId="145" LastEditorDisplayName="" LastEditDate="2010-08-17T12:36:30.647" LastActivityDate="2010-08-19T12:18:20.387" Title="Sets of integers in which every element is the sum of two others" Tags="&lt;algebra&gt;" AnswerCount="4" CommentCount="1" />
  <row Id="2648" PostTypeId="2" ParentId="2647" CreationDate="2010-08-17T12:02:50.043" Score="1" ViewCount="" Body="&lt;p&gt;The only possibilities I can think of at the moment are $ S = n \mathbb{Z} $, so for now I guess the answer is &lt;em&gt;technically&lt;/em&gt; yes.  I'll try to think of others and possibly characterize all such sets.  Great question!&lt;/p&gt;&#xA;" OwnerUserId="336" LastActivityDate="2010-08-17T12:02:50.043" CommentCount="1" />
  <row Id="2649" PostTypeId="2" ParentId="2647" CreationDate="2010-08-17T12:07:02.000" Score="10" ViewCount="" Body="&lt;p&gt;Yes, you can take the set of integers between $-n$ and $n$ inclusive for any $n \ge 3$.&#xA;Here $0 = 1 + (-1)$, $1 = 2 + (-1)$ and $x = (x - 1) + 1$ for $x \ge 2$. Similar formulae for negative values.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt; Answer for new version of the question&#xA;Set found by hand:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\{ -22, -20, -18, -16, -14, -12, -10, -2, 1, 3, 7, 8, 15, 23 \}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt; Answer with 8 elements. If you add -12 to this set you will get set with odd number of elements.&#xA;$$&#xA;\{ -10, -8, -6, -2, 1, 3, 4, 5 \}&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="334" LastEditorUserId="334" LastEditorDisplayName="" LastEditDate="2010-08-17T14:40:41.767" LastActivityDate="2010-08-17T14:40:41.767" CommentCount="3" />
  <row Id="2650" PostTypeId="1" AcceptedAnswerId="2660" CreationDate="2010-08-17T13:25:05.303" Score="2" ViewCount="354" Body="&lt;p&gt;In the towers of Hanoi game, how do we know that we have the optimal algorithm for solving it? I thought about this and it seemed like any deviation from the standard strategies would be putting you back a step, but I had no idea how to demonstrate this rigorously.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All I know is that the proof involves the Lucas correspondence between the Hanoi graph and the odd coefficients in Pascal's triangle. How is Pascal's triangle turned into a graph? I assume the coefficient are the vertices, but I don't see how you form the edges?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was also wondering how to generalize the strategy to n discs and k rods because it seems like this correspondence argument wouldn't really work in the general case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically, I am wondering how the odd coefficients of the Pascal triangle are turned into a graph and whether or not there is a similar method to find and prove optimal strategy when we increase the number of rods.&lt;/p&gt;&#xA;" OwnerUserId="249" LastEditorUserId="249" LastEditorDisplayName="" LastEditDate="2010-08-17T15:16:56.017" LastActivityDate="2010-08-17T16:37:10.110" Title="How to prove the optimal Towers of Hanoi strategy?" Tags="&lt;computer-science&gt;&lt;graph-theory&gt;&lt;puzzle&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="2651" PostTypeId="2" ParentId="2645" CreationDate="2010-08-17T13:55:05.090" Score="2" ViewCount="" Body="&lt;p&gt;If X and Y are Jointly Gaussian and X and Y are uncorrelated then X and Y are independent&lt;/p&gt;&#xA;" OwnerUserId="832" LastActivityDate="2010-08-17T13:55:05.090" CommentCount="0" />
  <row Id="2652" PostTypeId="2" ParentId="2647" CreationDate="2010-08-17T13:57:07.097" Score="2" ViewCount="" Body="&lt;p&gt;I posted an answer to this question previously, giving a finite set. That answer was incorrect: this is a different answer, which contains &lt;em&gt;lower bounds&lt;/em&gt; on finite sets of the sort you ask for.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A set S such that (&amp;forall;a&amp;nbsp;&amp;isin;&amp;nbsp;S)(&amp;exist;b,c&amp;nbsp;&amp;isin;&amp;nbsp;S):(b&amp;ne;c&amp;nbsp;&amp;nbsp;&lt;em&gt;and&lt;/em&gt;&amp;nbsp;&amp;nbsp;a&amp;nbsp;=&amp;nbsp;b&amp;nbsp;+&amp;nbsp;c) we will call &lt;strong&gt;self-supporting&lt;/strong&gt;: a self-supporting set S which is disjoint from &amp;minus;S will be &lt;strong&gt;strongly&lt;/strong&gt; self-supporting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proposition 1.&lt;/strong&gt; A self-supporting set must contain at least three positive elements, and three negative elements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[&lt;em&gt;Proof&lt;/em&gt;]&#xA;A self-supporting set S is non-empty; because the property of self-support is preserved by negation of the elements, without loss of generality we may consider sets containing positive elements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The largest positive element of S  can only be formed as a sum of two other positive elements; thus it has at least three positive elements 0&amp;lt;c&amp;lt;b&amp;lt;a such that a&amp;nbsp;=&amp;nbsp;b&amp;nbsp;+&amp;nbsp;c. The smallest positive element cannot be formed as a sum of smaller positive elements, so S contains negative elements. By similar arguments, it must contain at least three negative elements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the set {&amp;minus;3,&amp;nbsp;&amp;minus;2,&amp;nbsp;&amp;minus;1,&amp;nbsp;1,&amp;nbsp;2,&amp;nbsp;3} is self-supporting, this bound is optimal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proposition 2.&lt;/strong&gt; A strongly self-supporting set must contain at least four positive elements, and four negative elements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[&lt;em&gt;Proof&lt;/em&gt;] Suppose that S is self-supporting, but only has three positive elements 0&amp;lt;c&amp;lt;b&amp;lt;a. Then a&amp;nbsp;=&amp;nbsp;b&amp;nbsp;+&amp;nbsp;c. We cannot form b as a sum of c with some other positive element of S; it then follows that b can only be formed as a sum of a and &amp;minus;c, which implies that S is not &lt;em&gt;strongly&lt;/em&gt; self-supporting. Thus, a strongly self-supporting set contains at least four positive elements; and similarly for negative elements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The smaller set in flagar's answer, {&amp;minus;10,&amp;nbsp;&amp;minus;8,&amp;nbsp;−6,&amp;nbsp;−2,&amp;nbsp;1,&amp;nbsp;3,&amp;nbsp;4,&amp;nbsp;5}, is thus a minimal-size strongly self-supporting set --- and may also be the set having elements of the smallest absolute value.&lt;/p&gt;&#xA;" OwnerUserId="439" LastEditorUserId="439" LastEditorDisplayName="" LastEditDate="2010-08-19T12:14:05.437" LastActivityDate="2010-08-19T12:14:05.437" CommentCount="1" />
  <row Id="2653" PostTypeId="2" ParentId="201" CreationDate="2010-08-17T14:14:06.620" Score="8" ViewCount="" Body="&lt;p&gt;Here is a sweet concise reformulation of Euclid's proof that deserves to be better known.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;THEOREM&lt;/strong&gt; $\rm\ \ N \: (N+1) \;$ has a larger set of prime factors than does $\:\rm N &gt; 0\:$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof&lt;/strong&gt; $\ $ Since $\rm N+1 &gt; 1\:$ it has a prime factor $\rm P$. But $\rm P$ cannot divide $\rm N$ since $\rm N$ is coprime to $\rm N+1\ $ (viz. if $\rm\: P$ divides both $\rm N+1$ and $\rm N$ then $\rm P$ divides their difference $\rm N+1 - N = 1\:,\: $ contradiction). Thus the prime factors of $\rm\: N(N+1)$ include all those of $\rm N$ plus at least one prime $\rm P$ not dividing $\rm N$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;COROLLARY&lt;/strong&gt; $\ \ $ There are infinitely many primes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof&lt;/strong&gt; $\ $ Iterating $\rm\ N\to N\: (N+1)\ $ produces integers with an unbounded number of prime factors.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-09-25T22:58:45.210" LastActivityDate="2010-09-25T22:58:45.210" CommentCount="5" />
  <row Id="2654" PostTypeId="2" ParentId="2650" CreationDate="2010-08-17T15:58:50.313" Score="4" ViewCount="" Body="&lt;p&gt;In general, with n rods and n disks, the problem is equivalent to finding a Hamiltonian path on an n-hypercube. There is no proven optimal strategy known for 4 rods. You could also read the &lt;a href=&quot;http://mathworld.wolfram.com/TowerofHanoi.html&quot; rel=&quot;nofollow&quot;&gt;MathWorld article&lt;/a&gt; which has references for quite a few papers on the subject.&lt;/p&gt;&#xA;" OwnerUserId="1016" LastActivityDate="2010-08-17T15:58:50.313" CommentCount="1" />
  <row Id="2655" PostTypeId="1" AcceptedAnswerId="2657" CreationDate="2010-08-17T16:01:46.350" Score="4" ViewCount="350" Body="&lt;p&gt;2 is the first prime number.&#xA;3 is the second.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I give a prime number such as 1151024046313875220631 is there any software/website which can give the position of the prime number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know there are resources to find Nth prime. But I am having a hard time finding the reverse.&lt;/p&gt;&#xA;" OwnerUserId="1279" LastActivityDate="2010-09-08T20:28:19.610" Title="Is it possible to find the position of a prime number online?" Tags="&lt;prime-numbers&gt;" AnswerCount="6" CommentCount="4" />
  <row Id="2656" PostTypeId="2" ParentId="2627" CreationDate="2010-08-17T16:11:58.447" Score="2" ViewCount="" Body="&lt;p&gt;I assume you're only blowing up at one point, so here's at least a nice geometric description for $H^0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $p$ is the point blown up to $E$, then $dH-E$ is the system of plane curves of degree $d$ passing through $p$, $dH-2E$ are those that have a double point at $p$, etc.  This works with any number of blownup points, and a good exercise is using this interpretation to find all 27 lines on a cubic surface (which is the blowup at 6 points)&lt;/p&gt;&#xA;" OwnerUserId="176" LastActivityDate="2010-08-17T16:11:58.447" />
  <row Id="2657" PostTypeId="2" ParentId="2655" CreationDate="2010-08-17T16:12:55.640" Score="2" ViewCount="" Body="&lt;p&gt;If you have access to Mathematica, PrimePi[x] will give you the number of primes less than x. Combined with PrimeQ, which verifies that x is indeed prime, will give you which prime number x is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: I have no idea how long Mathematica would take (or if it could in fact compute it) for a number that high.&lt;/p&gt;&#xA;" OwnerUserId="1016" LastActivityDate="2010-08-17T16:12:55.640" CommentCount="3" />
  <row Id="2658" PostTypeId="2" ParentId="154" CreationDate="2010-08-17T16:19:03.323" Score="18" ViewCount="" Body="&lt;p&gt;One need only consult the history of algebra to find many informal&#xA;discussions on the existence and consistence of complex numbers. &#xA;Any &lt;em&gt;informal&lt;/em&gt; attempt to justify the existence of $\mathbb C$ &#xA;will face  the same obstacles that existed in earlier times. Namely, the lack of any rigorous (set-theoretic) foundation makes it difficult to be precise - both syntactically and semantically. Nowadays the set-theoretic foundation of algebraic structures &#xA;is so subconscious that is is easy to overlook just how much power it provides &#xA;for such purposes. But this oversight is easily remedied. One need only consult&#xA;some of the older literature where even leading mathematicians struggled&#xA;immensely to rigorously define complex numbers. For example, see the&#xA;quote below by Cauchy and Hankel's scathing critique - which is guaranteed &#xA;to make your jaw drop! (Below is an excerpt from my &lt;a href=&quot;http://groups.google.com/groups?selm=y8zhbxkjhil.fsf@nestle.csail.mit.edu&quot; rel=&quot;nofollow&quot;&gt;post&lt;/a&gt; on the notion &#xA;of formal polynomial rings and their quotients).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A major accomplishment of the set-theoretical definition of &#xA;algebraic structures was to eliminate imprecise syntax and semantics. &#xA;By eliminating the syntactic polynomial term $\rm\ a+b\cdot x+c\cdot x^2\ \ $ &#xA;in favor of its set-theoretic semantic reduction &#xA;$\rm\:(a,b,c,0,0,\ldots)\:$, there can no longer be any doubt about the precise denotation of the symbols $\rm\: x,\; +,\;\cdot\:,$ or about their equality, since, by set theory definition, tuples are equal iff their components are equal. Similarly for complex numbers  $\rm\:a + b\cdot i\:$ &#xA;vs. their set-theoretic pair reduction $\rm\:(a,b)\:$ discovered by Hamilton. &#xA;Before Hamilton gave this semantic reduction of $\mathbb C$ to pairs &#xA;of reals, prior syntactic constructions (e.g. by Cauchy) as &#xA;formal expressions or terms $\rm\:a+b\cdot i\:$ were subject to heavy criticism regarding &#xA;the precise denotation of their constituent symbols, e.g. &#xA;precisely what is the meaning of the symbols $\rm\;i, +, = \;$ ? &#xA;In modern language, Cauchy's construction of $\mathbb C$ is simply the &#xA;the quotient ring $\rm\:\mathbb R[x]/(x^2+1)\:,\ $ which he described essentially &#xA;as real polynomial expressions modulo $\rm\:x^2+1\:$. However, in Cauchy's time &#xA;mathematics lacked the necessary (set-theory) foundations to &#xA;rigorously define the syntactic expressions comprising the&#xA;polynomial ring term algebra $\rm\mathbb R[x]$, and its quotient ring of&#xA;congruence classes $\rm\:(mod\ x^2+1)\:$. The best that Cauchy could &#xA;do was to attempt to describe the constructions in terms of &#xA;imprecise &lt;em&gt;natural&lt;/em&gt; (human) language, e.g, in 1821 Cauchy wrote: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In analysis, we call a symbolic expression any combination of &#xA;  symbols or algebraic signs which means nothing by itself but &#xA;  which one attributes a value different from the one it should &#xA;  naturally be [...] Similarly, we call symbolic equations those &#xA;  that, taken literally and interpreted according to conventions &#xA;  generally established, are inaccurate or have no meaning, but &#xA;  from which can be deduced accurate results, by changing and &#xA;  altering, according to fixed rules, the equations or symbols &#xA;  within [...] Among the symbolic expressions and equations &#xA;  whose theory is of considerable importance in analysis, one &#xA;  distinguishes especially those that have been called imaginary. $\quad$ -- Cauchy, &lt;a href=&quot;http://books.google.com/books?id=M0or-HGe7D0C&amp;amp;pg=PA117&quot; rel=&quot;nofollow&quot;&gt;Cours d'analyse,1821, S.7.1&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;While nowadays, using set theory, we can rigorously interpret such &quot;symbolic expressions&quot; &#xA;as terms of formal languages or term algebras, it was far too &#xA;imprecise in Cauchy's time to have any hope of making sense &#xA;to his colleagues, e.g. Hankel replied scathingly: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If one were to give a critique of this reasoning, we can not &#xA;  actually see where to start. There must be something &quot;which &#xA;  means nothing,&quot; or &quot;which is assigned a different value than &#xA;  it should naturally be&quot; something that has &quot;no sense&quot; or is &#xA;  &quot;incorrect&quot;, coupled with another similar kind, producing &#xA;  something real. There must be &quot;algebraic signs&quot; - are these &#xA;  signs for quantities or what? as a sign must designate something &#xA;  - combined with each other in a way that has &quot;a meaning.&quot; I do &#xA;  not think I'm exaggerating in calling this an unintelligible &#xA;  play on words, ill-becoming of mathematics, which is proud &#xA;  and rightly proud of the clarity and evidence of its concepts. $\quad$-- Hankel&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Thus it comes as no surprise that Hamilton's elimination &#xA;of such &quot;meaningless&quot; symbols - in favor of pairs of reals - &#xA;served as a major step forward in placing such numbers on a &#xA;foundation more amenable to his contemporaries. &#xA;Although there was not yet any theory of sets in which to &#xA;rigorously axiomatize the notion of pairs, they were far easier &#xA;to accept naively - esp. given the already known closely &#xA;associated geometric interpretation of complex numbers. &#xA;Hamilton introduced pairs as 'couples' in &lt;a href=&quot;http://www.maths.tcd.ie/pub/HistMath/People/Hamilton/PureTime/PureTime.pdf&quot; rel=&quot;nofollow&quot;&gt;1837 [1]&lt;/a&gt;: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;p. 6: The author acknowledges with pleasure that he agrees with &#xA;  M. Cauchy, in considering every (so-called) Imaginary Equation &#xA;  as a symbolic representation of two separate Real Equations: &#xA;  but he differs from that excellent mathematician in his method &#xA;  generally, and especially in not introducing the sign  sqrt(-1) &#xA;  until he has provided for it, by his  Theory of Couples, &#xA;  a possible and real meaning, as a symbol of the couple (0,1) &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;p. 111:  But because Mr. Graves employed, in his reasoning, the &#xA;  usual principles respecting about Imaginary Quantities, and &#xA;  was content to prove the symbolical necessity without showing &#xA;  the interpretation, or inner meaning, of his formulae, the &#xA;  present Theory of Couples is published to make manifest that &#xA;  hidden meaning: and to show, by this remarkable instance, that &#xA;  expressions which seem according to common views to be merely &#xA;  symbolical, and quite incapable of being interpreted, may pass &#xA;  into the world of thoughts, and acquire reality and significance, &#xA;  if Algebra be viewed as not a mere Art or Language, but as the &#xA;  Science of Pure Time.  $\quad$ -- Hamilton, 1837&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Not until the much later development of set-theory was it explicitly realized &#xA;that ordered pairs and n-tuples serve a fundamental foundational role.&#xA;Indeed, as Akihiro Kanamori wrote on p. 289 (17) of &#xA;his very interesting &lt;a href=&quot;http://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.95.9839&quot; rel=&quot;nofollow&quot;&gt;paper [2]&lt;/a&gt; on the history of set theory: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In 1897 Peano explicitly formulated the ordered pair using &#xA;  $\rm\:(x, y)\:$ and moreover raised the two main points about the &#xA;  ordered pair: First, equation 18 of his Definitions stated &#xA;  the instrumental property which is all that is required of &#xA;  the ordered pair: &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$$\rm (x,y) = (a,b) \ \ \iff \ \ x = a \ \ and\ \ y = b $$&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Second, he broached the possibility of reducibility, writing: &#xA;  &quot;The idea of a pair is fundamental, i.e., we do not know how &#xA;  to express it using the preceding symbols.&quot; &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Once set-theory was fully developed one had the raw materials &#xA;(syntax and semantics) to provide rigorous constructions of &#xA;algebraic structures and precise languages for term algebras. The polynomial ring $\rm\:R[x]\:$ is nowadays just a special case of much more general constructions of free algebras. Such equationally axiomatized algebras and their genesis via so-called 'universal mapping properties'  are topics discussed at length in any course on Universal Algebra - &#xA;e.g. see &lt;a href=&quot;http://math.berkeley.edu/~gbergman/245/&quot; rel=&quot;nofollow&quot;&gt;Bergman [3]&lt;/a&gt; for a particularly lucid presentation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;[1] William Rowan Hamilton. Theory of conjugate functions, or algebraic couples; with a preliminary and elementary essay on algebra as the science of pure time&lt;br&gt;&#xA;Trans. Royal Irish Academy, v.17, part 1 (1837), pp. 293-422.)&lt;br&gt;&#xA;&lt;a href=&quot;http://www.maths.tcd.ie/pub/HistMath/People/Hamilton/PureTime/PureTime.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.maths.tcd.ie/pub/HistMath/People/Hamilton/PureTime/PureTime.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[2] Akihiro Kanamori. The Empty Set, the Singleton, and the Ordered Pair&lt;br&gt;&#xA;The Bulletin of Symbolic Logic, Vol. 9, No. 3. (Sep., 2003), pp. 273-298.&lt;br&gt;&#xA;&lt;a href=&quot;http://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.95.9839&quot; rel=&quot;nofollow&quot;&gt;http://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.95.9839&lt;/a&gt;&lt;br&gt;&#xA;PS  &lt;a href=&quot;http://www.math.ucla.edu/~asl/bsl/0903/0903-001.ps&quot; rel=&quot;nofollow&quot;&gt;http://www.math.ucla.edu/~asl/bsl/0903/0903-001.ps&lt;/a&gt;&lt;br&gt;&#xA;PDF &lt;a href=&quot;http://ifile.it/b20c48j&quot; rel=&quot;nofollow&quot;&gt;http://ifile.it/b20c48j&lt;/a&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;[3] George M. Bergman. An Invitation to General Algebra and Universal Constructions.&lt;br&gt;&#xA;PS  &lt;a href=&quot;http://math.berkeley.edu/~gbergman/245/&quot; rel=&quot;nofollow&quot;&gt;http://math.berkeley.edu/~gbergman/245/&lt;/a&gt;&lt;br&gt;&#xA;PDF &lt;a href=&quot;http://ifile.it/yquj5w1&quot; rel=&quot;nofollow&quot;&gt;http://ifile.it/yquj5w1&lt;/a&gt;  &lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2011-01-26T20:48:25.920" LastActivityDate="2011-01-26T20:48:25.920" CommunityOwnedDate="2011-04-15T20:45:29.080" CommentCount="3" />
  <row Id="2659" PostTypeId="2" ParentId="2655" CreationDate="2010-08-17T16:22:14.383" Score="2" ViewCount="" Body="&lt;p&gt;What you are looking for is the &quot;Prime Counting Function&quot;.  The closest thing to it you will find online is Wolfram Alpha:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.wolframalpha.com/examples/PrimeNumbers.html&quot; rel=&quot;nofollow&quot;&gt;http://www.wolframalpha.com/examples/PrimeNumbers.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1280" LastActivityDate="2010-08-17T16:22:14.383" />
  <row Id="2660" PostTypeId="2" ParentId="2650" CreationDate="2010-08-17T16:37:10.110" Score="7" ViewCount="" Body="&lt;p&gt;I will address your first question, but not the one for larger number of rods; as far as I know, it's still generally wide open what the optimal strategy might be even for $4$ rods and a smallish number of disks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To show the optimal strategy for $n$ disks in $3$ rods is the &quot;usual&quot; one, you can do it by induction (which yields a recursive solution). I'm sure there are other ways of proving it, perhaps with Lucas numbers as you suggest.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clearly, the optimal strategy with $n=1$ is to simply move the disk directly. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume you already have the optimal strategy for moving $k$ disks. To move $k+1$ disks, you need to move the largest disk from the initial rod to the terminal rod, but that is the only time it needs to move (it cannot help you with the other disks, since it must lie at the bottom at any given time, so any other moves only require further moves in the end); to move the bottom ($k+1$)st disk from the initial rod $I$ to the terminal rod $T$, you must first move the top $k$ disks out of the way; this requires moving the $k$ disks from the initial rod $I$ to the auxiliary rod $A$, and the optimal way of doing this is the optimal strategy you know for $k$ disks. Then you move the $k+1$st disk, and then you want to move the remaining $k$ disks from the auxiliary rod to the terminal one in as few moves as possible (the optimal way). So the optimal strategy for $k+1$ disks is to move the top $k$ using the optimal strategy for $k$ from $I$ to $A$, then move the largest disk from $I$ to $T$, then move the top $k$ disks using the optimal strategy for $k$ from $A$ to $T$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;By keeping track of the actual number of moves needed at each step, you can give the number. For $n=1$, the number is $1=2^1-1$. Assume that moving $k$ disks requires $2^k-1$ moves in the optimal strategy. The optimal strategy for $k+1$ described above takes $(2^k-1) + 1 + (2^k-1) = 2^k+2^k - 1 = 2^{k+1}-1$ steps; thus, the optimal solution for $n$ disks and $3$ rods requires $2^n-1$ moves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(This does not generalize easily to more than $3$ rods for presumably obvious reasons).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A bit more interesting is trying to prove that the non-recursive solution gives an optimal solution; this solution only requires you to remember the last disk you moved at any given time (the recursive solution is more memory intensive, of course). Number the rods $0$, $1$, and $2$. We have three rules:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Never move the same disk twice in succession.&lt;/li&gt;&#xA;&lt;li&gt;You can only move a disk from the top of one rod to the top of another rod.&lt;/li&gt;&#xA;&lt;li&gt;Moving a disk from rod $i$ to rod $j$ is only valid if $i\neq j$, and either rod $j$ is empty or the top disk in rod $j$ is larger than the top disk in rod $i$.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;With these rules, the non-recursive algorithm has two simple steps:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If you are moving the disk from rod $i$, and the two other rods are valid destinations, then move it to rod $i+1\mod 3$. Otherwise, move it to the only valid destination.&lt;/li&gt;&#xA;&lt;li&gt;If no move is possible, stop. Otherwise, continue.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This process will solve the puzzle with $3$ rods in the minimum number of moves.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-08-17T16:37:10.110" CommentCount="4" />
  <row Id="2661" PostTypeId="2" ParentId="2626" CreationDate="2010-08-17T16:57:49.430" Score="1" ViewCount="" Body="&lt;p&gt;Since you are asking for just a place to start:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Try Yakov Perelman, &quot;Physics can be fun&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The book has two volumes. It has an article somewhere, the precise location of which I am unable to locate as I do not have the book handy, on how to explain the nautical method of &quot;tacking&quot;, ie sailing against the wind. It explains the forces on the sails using diagrams, etc.. It has also many other articles on nautical themes. I think this book is the best place to start for an amateur.&lt;/p&gt;&#xA;" OwnerUserId="1119" LastActivityDate="2010-08-17T16:57:49.430" />
  <row Id="2662" PostTypeId="2" ParentId="2633" CreationDate="2010-08-17T17:24:17.560" Score="1" ViewCount="" Body="&lt;p&gt;It's a classic exercise, see for example Rudin's Real and Complex analysis (3rd ed.), exercise 14, chap.3, p. 72. There are detailed step-by-step hints, you should be fine.&lt;/p&gt;&#xA;" OwnerUserId="1197" LastActivityDate="2010-08-17T17:24:17.560" CommentCount="1" />
  <row Id="2663" PostTypeId="1" AcceptedAnswerId="2667" CreationDate="2010-08-17T18:13:51.290" Score="2" ViewCount="311" Body="&lt;p&gt;I was wondering if the following is true:&#xA;In a topological space with partial order, the inf and sup for a closed subset are achievable inside the subset so that they become minimum and maximum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it still true if I replace &quot;a topological space with partial order&quot; with &quot;a topological space with total order&quot;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a topological space, what other kinds of condition can make inf and sup of a subset achievable in itself?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks and regards!&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;More question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In an &quot;ordered&quot; (not sure what kinds of order is proper here) topological space, are inf and sup of a subset accumulation points of the subset?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;More questions again:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In Euclidean space, are inf and sup for a closed subset inside the subset? Are they accumulation points of the subset? What if in metric space? Thanks!&lt;/p&gt;&#xA;" OwnerUserId="1281" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2010-11-25T17:06:04.900" LastActivityDate="2010-11-25T17:06:04.900" Title="when inf and sup of a subset achievable in itself?" Tags="&lt;topology&gt;&lt;general-topology&gt;&lt;order-theory&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="2664" PostTypeId="1" AcceptedAnswerId="2665" CreationDate="2010-08-17T18:21:29.653" Score="2" ViewCount="254" Body="&lt;p&gt;In the article about &lt;a href=&quot;http://en.wikipedia.org/wiki/Metric_space&quot; rel=&quot;nofollow&quot;&gt;metric space&lt;/a&gt; on wikipedia:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If X is a complete subset of the metric space M, then X is closed in M. Indeed, a space is complete iff it is closed in any containing metric space.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Do the second sentences say that: &#xA;in a metric space, a subset is complete if and only if the subset is closed in the metric space?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If not, what is the relation between completeness and closedness for a subset in a metric space?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks and regards!&lt;/p&gt;&#xA;" OwnerUserId="1281" LastActivityDate="2010-08-17T18:27:18.033" Title="completeness and closedness for a subset in a metric space" Tags="&lt;topology&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="2665" PostTypeId="2" ParentId="2664" CreationDate="2010-08-17T18:24:18.617" Score="3" ViewCount="" Body="&lt;p&gt;Notice: A subset is complete iff it is closed in &lt;em&gt;any&lt;/em&gt; metric space in which it is a subspace.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A closed subspace of a complete metric space is complete. On the other hand, a closed subspace of a metric space need not be complete. Take the space $(0,1)$ which is closed in itself. But it is not complete. So closedness is not enough for completeness.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However a compact metric space is always complete. Compactness is the right property.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For this and more, try out G. F. Simmons, &quot;Introduction to Topology and Modern Analysis&quot;.&lt;/p&gt;&#xA;" OwnerUserId="1119" LastActivityDate="2010-08-17T18:24:18.617" CommentCount="2" />
  <row Id="2666" PostTypeId="2" ParentId="2664" CreationDate="2010-08-17T18:27:18.033" Score="3" ViewCount="" Body="&lt;p&gt;No; if that were the case, then $\mathbb{Q}$ would be complete, since it is closed in itself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What the second sentence says is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Let $X$ be a metric space. Then $X$ is complete if and only if for every metric space $M$, if $X$ is contained in $M$ then $X$ is closed in $M$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In particular, if you have a fixed complete $X$ and a fixed metric space $M$ that contains $X$, then you can conclude that $X$ is closed in $M$ (the first sentence).&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-08-17T18:27:18.033" CommentCount="1" />
  <row Id="2667" PostTypeId="2" ParentId="2663" CreationDate="2010-08-17T18:36:52.023" Score="4" ViewCount="" Body="&lt;p&gt;1 and 2:  No.  Take an interval $(\alpha, \beta)$ in $\mathbb{Q}$ where $\alpha, \beta$ are irrational.  You can verify that such an interval is closed and that it contains neither its infimum nor its supremum.  The property you want, for totally ordered sets, is called &lt;a href=&quot;http://en.wikipedia.org/wiki/Total_order#Completeness&quot; rel=&quot;nofollow&quot;&gt;completeness&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3:  I will assume you mean a totally ordered set with its &lt;a href=&quot;http://en.wikipedia.org/wiki/Order_topology&quot; rel=&quot;nofollow&quot;&gt;order topology&lt;/a&gt;.  In that case, this is not always true if the infimum or supremum are contained in the subset but true otherwise; this is just a matter of working through the definitions.&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-08-18T00:30:40.383" LastActivityDate="2010-08-18T00:30:40.383" CommentCount="12" />
  <row Id="2668" PostTypeId="2" ParentId="11" CreationDate="2010-08-17T18:58:40.717" Score="-2" ViewCount="" Body="&lt;p&gt;This should have been a comment; but I post as an answer anyway. I just want to remark that in spite of all the criticisms, all the statements in the most popular proof in the list above are true. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is, in the real number system, it is indeed true that:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;1/3 = 0.333333...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It is again true that:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;0.33333.... * 3 = 0.99999....&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Etc..&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would still stand by the superiority of Noah Snyder's answer, however. That's the one addressing the subtleties.&lt;/p&gt;&#xA;" OwnerUserId="1119" LastActivityDate="2010-08-17T18:58:40.717" CommunityOwnedDate="2010-08-17T18:58:40.717" />
  <row Id="2669" PostTypeId="2" ParentId="2435" CreationDate="2010-08-17T20:27:29.120" Score="-1" ViewCount="" Body="&lt;p&gt;I am not at all sure that this would work, but it seems reasonable... (if it doesn't work, I'd like to know why)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can very simply use Monte Carlo integration to approximate pi by setting up a known area and inscribing a circle in it, and sampling the entire area with a &lt;em&gt;uniformly distributed sequence*&lt;/em&gt;, and considering the ratio of &quot;hits&quot; to the total number of &quot;shots&quot;.  (I elaborated on this slightly in &lt;a href=&quot;http://math.stackexchange.com/questions/297/simple-numerical-methods-for-calculating-the-digits-of-pi/372#372&quot;&gt;this answer&lt;/a&gt; to another question.)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;* I think the uniform distribution of points is necessary in order to guarantee that there is no bias toward any particular area of the sampling space.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If this is indeed the salient feature of uniformly distributed sequences in this context, then what would happen if we sampled with a non-uniformly distributed sequence?  I presume that we would not approximate the digits of pi very well... If this turns out to be true, then you may throw your sequence at a Monte Carlo problem with a known solution and see how it does.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a quick Matlab script approximating pi using a 2-d square.  In principle, this could be done in any dimension.  If you just have a sequence of points (1-d) you could try to approximate the length of a given segment in an interval of known length.  It also seems to me that the more points you have the better idea you will get to how far your points are from being uniformly distributed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Let me know if something didn't make sense here and I will try to clarify.)&lt;/p&gt;&#xA;" OwnerUserId="121" LastActivityDate="2010-08-17T20:27:29.120" CommentCount="3" />
  <row Id="2670" PostTypeId="1" AcceptedAnswerId="2678" CreationDate="2010-08-17T20:29:40.317" Score="9" ViewCount="118" Body="&lt;p&gt;Let $X$ and $Y$ be Banach spaces. A bounded operator $T\colon X\to Y$ is called Fredholm iff&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;The dimension of $\ker(T)$ is finite,&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The codimension of the image $\mathrm{im}(T)$ is finite,&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The image $\mathrm{im}(T)$ is closed in $Y$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; &lt;em&gt;Is the third condition redundant?&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some lecture notes I'm working through claim that the third condition follows from the second one together with the &lt;a href=&quot;http://en.wikipedia.org/wiki/Open_mapping_theorem_%28functional_analysis%29&quot; rel=&quot;nofollow&quot;&gt;open mapping theorem&lt;/a&gt;. I've checked some books on functional-analysis without finding a proof of this.&lt;/p&gt;&#xA;" OwnerUserId="367" LastEditorUserId="367" LastEditorDisplayName="" LastEditDate="2010-08-17T21:44:21.037" LastActivityDate="2010-08-17T22:41:23.163" Title="Is the closedness of the image of a Fredholm operator implied by the finiteness of the codimension of its image?" Tags="&lt;functional-analysis&gt;" AnswerCount="2" />
  <row Id="2671" PostTypeId="2" ParentId="2655" CreationDate="2010-08-17T21:01:43.203" Score="4" ViewCount="" Body="&lt;p&gt;You can use the function prime_pi in Sage (http://sagemath.org), which is also available for free online at &lt;a href=&quot;http://sagenb.org&quot; rel=&quot;nofollow&quot;&gt;http://sagenb.org&lt;/a&gt;.   For example, &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;   sage: prime_pi(2011)&#xA;   305&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Like Mathematica, Sage's prime_pi function is too slow to solve your problem above.   It's also somewhat slower than Mathematica's still.  &lt;/p&gt;&#xA;" OwnerUserId="1286" LastActivityDate="2010-08-17T21:01:43.203" />
  <row Id="2672" PostTypeId="2" ParentId="1774" CreationDate="2010-08-17T21:08:43.110" Score="3" ViewCount="" Body="&lt;p&gt;My favorite elementary number theory book is the one I published with Springer: &lt;a href=&quot;http://wstein.org/ent/&quot; rel=&quot;nofollow&quot;&gt;http://wstein.org/ent/&lt;/a&gt;   This isn't a completely shameless plug, because I was just allowed to release the PDF version legally for free (available at the above URL), which the original poster might appreciate. &lt;/p&gt;&#xA;" OwnerUserId="1286" LastActivityDate="2010-08-17T21:08:43.110" CommunityOwnedDate="2010-08-17T21:08:43.110" />
  <row Id="2673" PostTypeId="2" ParentId="1774" CreationDate="2010-08-17T21:24:49.773" Score="0" ViewCount="" Body="&lt;p&gt;There is a list at &lt;a href=&quot;http://www.imomath.com/index.php?options=mbb|knjige&amp;amp;p=0&quot; rel=&quot;nofollow&quot;&gt;http://www.imomath.com/index.php?options=mbb|knjige&amp;amp;p=0&lt;/a&gt; . There's also the book   &lt;a href=&quot;http://www.worldscibooks.com/mathematics/7242.html&quot; rel=&quot;nofollow&quot;&gt;Problems of number theory in mathematical competitions&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="589" LastActivityDate="2010-08-17T21:24:49.773" CommunityOwnedDate="2010-08-17T21:24:49.773" />
  <row Id="2674" PostTypeId="2" ParentId="2435" CreationDate="2010-08-17T21:39:49.947" Score="3" ViewCount="" Body="&lt;p&gt;You could use a chi square test.  Basically, this tests whether the number of draws that fall into various intervals is consistent with a uniform random distribution.   See Knuth's &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0201896842&quot; rel=&quot;nofollow&quot;&gt;TAOCP volume 2, Seminumerical Algorithms&lt;/a&gt; for an explanation of the chi-square test.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Update: If the chi square test is too complicated, you could plot a histogram of your output and eyeball it to see if it looks uniform.  But if you'd like to quantify whether it's even enough, that's exactly what a chi square test is.  The test will also tell you whether the data are &lt;em&gt;too&lt;/em&gt; evenly spread out, i.e. not random enough.&lt;/p&gt;&#xA;" OwnerUserId="247" LastEditorUserId="247" LastEditorDisplayName="" LastEditDate="2010-08-18T12:24:23.793" LastActivityDate="2010-08-18T12:24:23.793" CommentCount="6" />
  <row Id="2675" PostTypeId="2" ParentId="2645" CreationDate="2010-08-17T21:43:24.767" Score="2" ViewCount="" Body="&lt;p&gt;It depends on what else you know about the relationship between the variables.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the correlation coefficient is the full extent of your information, then the approximation is unsafe, as Noldorin points out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If, on the other hand, you have good external evidence that the coefficient adequately captures the level of a small linear relationship (eg, a slight dependence on some third quantity that is not germane to your analysis), then it may well be valid to approximate them as independent for some purposes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;RVs about which you know nothing are useful abstractions -- and this is, after all, the maths site -- but real world data often exist in less of vacuum. If you're analysing in the context of a model, that may help you to work out what approximations you can get away with.&lt;/p&gt;&#xA;" OwnerUserId="144" LastActivityDate="2010-08-17T21:43:24.767" CommentCount="1" />
  <row Id="2676" PostTypeId="2" ParentId="1774" CreationDate="2010-08-17T21:44:34.547" Score="1" ViewCount="" Body="&lt;p&gt;Davenport's The Higher Arithmetic was my first number theory book. I think its very accessible to a high school student or beginning undergraduate student. It's quite short and very quickly readable.&#xA;If you find this treatment too informal, Niven and Zuckerman's an Introduction to the Theory of Numbers is a standard text that I think is a very well written undergraduate text, but this has already been mentioned.&lt;/p&gt;&#xA;" OwnerUserId="249" LastActivityDate="2010-08-17T21:44:34.547" CommunityOwnedDate="2010-08-17T21:44:34.547" />
  <row Id="2677" PostTypeId="2" ParentId="2670" CreationDate="2010-08-17T21:54:05.153" Score="1" ViewCount="" Body="&lt;p&gt;The third condition is redundant.  The proof I've seen is the following.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Throughout $X,Y$ are Banach spaces and $E$ is a closed subspace of $X$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lemma: if $F \subset X$ is finite dimensional then $E+F$ is closed and the image of $E$ in $X/F$ is closed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;proof: reduce to the case $\dim F = 1$, i.e. $F = \mathbb{C} x_0$ with $x_0 \not \in E$ (otherwise $E+F$ is trivially closed).  We have to show every convergent subsequence converges to a point in $E + F$.  Let $z_n = y_n + c_n x_0$ be such a sequence with limit $z_0$.  Let $\delta = dist(x_0, E) &gt; 0$.  Now $(z_n)$ is Cauchy so $|z_n - z_m| \to 0$ for $n,m$ sufficiently large.  That is, $|(c_n - c_m)x_0 - (y_n -y_m)| \ge |c_n -c_m|\delta \to 0$.  It follows that $(c_n)$ is Cauchy so converges to $c_0$.  So $y_n$ converges to $z_0 - c_0x_0 \in E$ hence $z_0 \in E + F$ as desired.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So $E+F$ is a Banach space hence so is $E+F/F = Im(E) \in X/F$ which says exactly that its closed in $X/F$. QED&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we can show $TE$ is closed in $Y$.  Let $y_0 + TX, ..., y_n + TX$ be a basis for $Y/TX$ and $Y_0 = span(y_i)$.  Consider $T' \colon X \oplus Y_0 \to Y$ given by $T'(x\oplus y) = Tx + y$.  This is surjective so $T'' \colon X \oplus Y_0/(\ker T \oplus 0) \to Y$ is invertible.  By the lemma $E$ is closed in $X \oplus Y_0/(\ker T \oplus 0)$ and maps to $TE$ under the invertible map $T''$. QED.  &lt;/p&gt;&#xA;" OwnerUserId="403" LastActivityDate="2010-08-17T21:54:05.153" />
  <row Id="2678" PostTypeId="2" ParentId="2670" CreationDate="2010-08-17T22:06:44.210" Score="6" ViewCount="" Body="&lt;p&gt;Assume that $T$ is injective, because if it is not we know that $\textrm{ker}(T)$ is closed so we can replace $T$ by the induced map from $X/\textrm{ker}(T)$ (which is a Banach space).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now define $T' := X \oplus C \to Y$ by $T'(x, c) = T(x) + c$ where $C$ is a closed complement for the range of $T$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This $T'$ is clearly bounded, linear and an isomorphism. So by the open mapping theorem $T'$ is open. Note that $\textrm{im}(T) = T'(X \oplus {0})$ which is closed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Offtopic: (How do you guys use LaTeX in the posts? I can't find it in the FAQ)&lt;/p&gt;&#xA;" OwnerUserId="1120" LastEditorUserId="1120" LastEditorDisplayName="" LastEditDate="2010-08-17T22:41:23.163" LastActivityDate="2010-08-17T22:41:23.163" CommentCount="1" />
  <row Id="2679" PostTypeId="2" ParentId="1413" CreationDate="2010-08-17T22:50:03.043" Score="2" ViewCount="" Body="&lt;p&gt;The construction of the real field from the rationals using Cauchy sequences can be mimicked to construct other (characteristic zero) complete fields not isometric to the reals. Namely, instead of starting out with the metric defined by the usual (archimedean) absolute value, one can consider the $p$-adic metric ($p$ a fixed prime) and proceed in the same lines. A theorem of Ostrowski says that up to equivalence these are in fact the only ways to complete $\Bbb Q$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Being complete, the fields ${\Bbb Q}_p$ thus obtained can be used to develop an analytic theory which is similar to the classical theory &quot;over $\Bbb R$&quot; but has some subtle differences.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A source of difference is that the fields ${\Bbb Q}_p$ have--unlikely $\Bbb R$--a natural subring, the radius $1$ sphere centered in $0$, a.k.a. the $p$-adic integers $\Bbb Z_p$, which is a local ring with residue field isomorphic to the field with $p$ elements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a clue to the fact that &quot;$p$-adic analysis&quot; is deeply intertwined with the theory of finite fields.&lt;/p&gt;&#xA;" OwnerUserId="688" LastActivityDate="2010-08-17T22:50:03.043" />
  <row Id="2680" PostTypeId="1" AcceptedAnswerId="2686" CreationDate="2010-08-18T00:23:22.067" Score="2" ViewCount="116" Body="&lt;p&gt;I need to create a simple formula for determining the popularity of an item based on votes and age.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is my current formula, which needs some work:&lt;br/&gt;&#xA;30 / (days between post date and now) * (vote count) = weighted vote&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whenever a vost is cast for an item it checks if its weighted vote is &gt; 300. If an item has a weighted vote more than 300 then it is promoted to the front page of my website.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is that this formula makes it very hard for older items to be promoted.&lt;br/&gt;&#xA;30 / 1 day * 10 votes = 300 (promoted)&lt;br/&gt;&#xA;30 / 5 days * 15 votes = 90 (not promoted)&lt;br/&gt;&#xA;30 / 30 days * 30 votes = 30 (not promoted)&lt;br/&gt;&#xA;30 / 80 days * 40 votes = 15 (not promoted)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I alter the formula to make it relatively easier for older items to be promoted (IE. make the above four weighted values fairly close together)?&lt;/p&gt;&#xA;" OwnerUserId="1288" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-08-18T04:52:12.647" LastActivityDate="2010-08-18T16:36:45.980" Title="Popularity formula (using votes and age)" Tags="&lt;algebra-precalculus&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="2681" PostTypeId="1" CreationDate="2010-08-18T01:38:33.857" Score="2" ViewCount="106" Body="&lt;p&gt;Let $G$ be an infinite simple group. Suppose there is a subset $X$ of $G$ with&#xA;$$&#xA;X^g = gXg^{-1}=X\qquad (g \in G)&#xA;$$&#xA;which is closed under taking inverses and which even generates $G$ in finitely many steps:&#xA;$$&#xA;X^m = G&#xA;$$&#xA;for a suitable $m \ge 2.$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We somehow suspect that $X$ might be equal to $G.$ &lt;i&gt;If so&lt;/i&gt;, we have to establish (in the end!) that $X$ is actually closed under the multiplication; this would make $X$ a nonidentity normal subgroup of $G,$ and hence $X=G.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But suppose this direct approach meets serious difficulties. What are then possible &lt;i&gt;&lt;b&gt;indirect&lt;/b&gt;&lt;/i&gt; approaches here? Anyone who met this situation in their own research, or elsewhere is kindly requested to share their ideas. &lt;/p&gt;&#xA;" OwnerUserId="1100" LastEditorUserId="8575" LastEditorDisplayName="" LastEditDate="2011-05-19T09:13:51.543" LastActivityDate="2011-05-19T09:13:51.543" Title="Inverse-closed subsets generating simple groups in finitely many steps" Tags="&lt;group-theory&gt;&lt;simple-groups&gt;" CommentCount="7" FavoriteCount="2" />
  <row Id="2682" PostTypeId="1" AcceptedAnswerId="2685" CreationDate="2010-08-18T01:56:19.353" Score="1" ViewCount="180" Body="&lt;p&gt;I have several points on a graph. The graph is of a continuous curve that flows through points (0, inf), (11, 5000), and (3, 200000). Can I find the equation for that line?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Let's try this:&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm looking to find a function that given an input (x) becomes closer to infinity as x grows closer to 0, and closer to 0 the closer x gets to infinity. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The issue with &lt;code&gt;y = log(x)&lt;/code&gt; is the scale is to small (#1) and the curve it too pronounced (#2).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For constraints I would say &lt;code&gt;x &amp;gt;= 0 &amp;amp;&amp;amp; y &amp;gt;= 0&lt;/code&gt; at all times.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The purpose of this (for those who &lt;em&gt;must&lt;/em&gt; know or are simply curious) is as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a program that loops through an array of values a certain number of times. The number of times &lt;em&gt;should&lt;/em&gt; be dependent on the number of values in the array.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, an array with 5000 values should be processed 11 times. An array with 250k values should be processed 3 times.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;int y = // Function goes here, given x;&#xA;for(int i = 0; i &amp;lt; y; i++)&#xA;{&#xA;    // Process&#xA;}&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;I've been playing with a graph utility&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{0.3}{\frac{log(x)}{24}} = y$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gives me approximately what I'm looking for, though doesn't give a proper values as $x \to 0$ (or &gt; 4 actually). However &lt;em&gt;there has to be a &quot;proper&quot; way to do this&lt;/em&gt;. Rather then punch numbers in until it &quot;looks&quot; right.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Looks like the &quot;right&quot; way gives the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$y=\frac{C}{x^n}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solving for $C$ and $n$ gives $C ~= 19.3896$ and $n ~= 0.3522$ using $(5, 11), (200, 3)$ as points for simplicity.&lt;/p&gt;&#xA;" OwnerUserId="1289" LastEditorUserId="1289" LastEditorDisplayName="" LastEditDate="2010-08-18T14:59:03.957" LastActivityDate="2010-08-18T14:59:03.957" Title="Fitting a curve to a decreasing data set undefined at zero" Tags="&lt;algebra-precalculus&gt;" AnswerCount="2" CommentCount="22" />
  <row Id="2683" PostTypeId="2" ParentId="1579" CreationDate="2010-08-18T02:11:55.883" Score="6" ViewCount="" Body="&lt;p&gt;Such generalizations occur frequently as problems. Below are a couple of examples - which include references to prior literature, e.g. to an inclusion-exclusion max/min proof in Polya &amp;amp; Szego's problem book. They're problem E2229 from the Monthly 77 1970 p. 402; problem 1344 from Math. Mag. 64 1990 p. 134.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i33.tinypic.com/rlgdb9.jpg&quot; alt=&quot;alt text&quot;&gt;&#xA;&lt;img src=&quot;http://i38.tinypic.com/2i7mkvc.jpg&quot; alt=&quot;alt text&quot;&gt;&#xA;&lt;img src=&quot;http://i37.tinypic.com/5ck5ms.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-08-18T02:11:55.883" />
  <row Id="2685" PostTypeId="2" ParentId="2682" CreationDate="2010-08-18T03:44:50.280" Score="2" ViewCount="" Body="&lt;p&gt;If y = 4525365/x^2.83916, that is,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$y = \frac{4525365}{x^2.83916}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;then when x = 3, y = 200000.02 and when x = 11, y = 5000.04 and when x = ∞, y = 0, and when x = 0, y = ∞.  It also satisfies y ≥ 0 whenever x &gt; 0 and x ≥ 0 whenever y &gt; 0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is based on the idea that the longer the array the less times you can afford to scan the array.  The amount of work might be equal to x*y, so you want to keep the amount of work constant, x*y = C, or y = C/x.  That does not quite fit your points, so I decided to use x^n*y = C instead.  I solved for n using:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3^n*200000 = 11^n*5000, so 200000/5000 = 11^n/3^n, so 40 = (11/3)^n, so n = log(40)/log(11/3) ≈ 2.83916.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then C = 3^2.83916*200000 ≈ 4525365.&lt;/p&gt;&#xA;" OwnerUserId="583" LastActivityDate="2010-08-18T03:44:50.280" CommentCount="5" />
  <row Id="2686" PostTypeId="2" ParentId="2680" CreationDate="2010-08-18T05:05:00.257" Score="2" ViewCount="" Body="&lt;p&gt;One possibility is to take the cube root of the number of days:&#xA;$$\begin{align}&#xA;\frac{3}{\sqrt[3]{1\text{ day}}}\cdot 10\text{ votes}&amp;amp;=300&#xA;\\&#xA;\frac{3}{\sqrt[3]{5\text{ days}}}\cdot 15\text{ votes}&amp;amp;\approx 263.2&#xA;\\&#xA;\frac{3}{\sqrt[3]{30\text{ days}}}\cdot 30\text{ votes}&amp;amp;\approx 289.6&#xA;\\&#xA;\frac{3}{\sqrt[3]{80\text{ days}}}\cdot 40\text{ votes}&amp;amp;\approx 278.5&#xA;\\&#xA;\end{align}$$&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-08-18T05:05:00.257" />
  <row Id="2687" PostTypeId="1" AcceptedAnswerId="2689" CreationDate="2010-08-18T05:08:37.223" Score="5" ViewCount="189" Body="&lt;p&gt;I saw a statement on &lt;a href=&quot;http://en.wikipedia.org/wiki/Metric_space#Continuous_maps&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Metric_space#Continuous_maps&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;f is continuous if and only if it is continuous on every compact subset of M1.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It is stated for the case that f is a mapping between two metric spaces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was wondering if the statement is true for mappings between general topological spaces?&#xA;Why?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks and regards!&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;More questions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What makes the statement true for a mapping between two metric spaces? Can you give a sketch for the proof?&lt;/p&gt;&#xA;" OwnerUserId="1281" LastEditorUserId="1281" LastEditorDisplayName="" LastEditDate="2010-08-18T06:52:51.053" LastActivityDate="2010-12-11T17:25:19.837" Title="Generalize continuity from compact subsets to the whole domain space" Tags="&lt;topology&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="2" />
  <row Id="2688" PostTypeId="2" ParentId="622" CreationDate="2010-08-18T05:25:06.697" Score="3" ViewCount="" Body="&lt;p&gt;Representation theory is one of the keys to the Fourier transform on groups (and co-set spaces etc.). For this matter, I recommend this book:&#xA;&lt;a href=&quot;http://books.google.com/books?id=yanslHJYvAIC&amp;amp;pg=PA415&amp;amp;lpg=PA415&amp;amp;dq=sugiura+representaion+theory&amp;amp;source=bl&amp;amp;ots=j8sf2rsfnD&amp;amp;sig=Z4wjbz316q-yfVh5WO0DDQcl6Es&amp;amp;hl=en&amp;amp;ei=mWxrTP2eApCRjAfMuITiAQ&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=1&amp;amp;ved=0CBQQ6AEwAA#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;Unitary representations and harmonic analysis: an introduction&#xA; By Mitsuo Sugiura&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you take a look at the &lt;a href=&quot;http://en.wikipedia.org/wiki/Representation_theory&quot; rel=&quot;nofollow&quot;&gt;Wikipedia page&lt;/a&gt; about representation theory, you will find several applications.&lt;/p&gt;&#xA;" OwnerUserId="1154" LastActivityDate="2010-08-18T05:25:06.697" />
  <row Id="2689" PostTypeId="2" ParentId="2687" CreationDate="2010-08-18T05:45:10.800" Score="9" ViewCount="" Body="&lt;p&gt;The statement is false in general.  Let $X$ be an uncountable set with the &lt;a href=&quot;http://en.wikipedia.org/wiki/Cocountable_topology&quot; rel=&quot;nofollow&quot;&gt;cocountable topology&lt;/a&gt;.  Then all the countable subsets of $X$ have the discrete topology and are therefore closed but not compact, so the only compact subsets of $X$ are the finite subsets, which also carry the discrete topology.  It follows that every function $X \to Y$, where $Y$ is any topological space, is continuous on compact subsets of $X$.  (But, for example, if $Y$ has the same underlying set with the discrete topology, the &quot;identity&quot; function $X \to Y$ is not continuous.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit:  In response to your second question, the property you are looking at is equivalent to being &lt;a href=&quot;http://en.wikipedia.org/wiki/Compactly_generated_space&quot; rel=&quot;nofollow&quot;&gt;compactly generated&lt;/a&gt;.  A topological space $X$ is compactly generated if it satisfies the following condition: a subset of $X$ is open if and only if its intersection with every compact subset is open.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So why are metric spaces compactly generated?  Recall that a space is &lt;a href=&quot;http://en.wikipedia.org/wiki/First-countable_space&quot; rel=&quot;nofollow&quot;&gt;first-countable&lt;/a&gt; if every point has a countable neighborhood basis.  Metric spaces are first-countable because at any point the sequence of open balls of radius $\frac{1}{n}, n \in \mathbb{N}$ forms a countable neighborhood basis.  And first-countable spaces are always compactly generated; see the proof &lt;a href=&quot;http://topospaces.subwiki.org/wiki/First-countable_implies_compactly_generated&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-08-18T07:04:40.013" LastActivityDate="2010-08-18T07:04:40.013" CommentCount="1" />
  <row Id="2690" PostTypeId="2" ParentId="622" CreationDate="2010-08-18T06:12:16.750" Score="2" ViewCount="" Body="&lt;p&gt;Certainly, group actions used to model dynamical systems, for instance, Markov Chains, find extremely handy the results from representation theory.&#xA;For example, suppose you have a deck of cards, and every time you choose two cards at random and transpose them. How long you have to wait for the deck to be randomized? The (exact) answer to this question lies exclusively in the area of representation theory! See for example, &lt;a href=&quot;http://www.cambridge.org/catalogue/catalogue.asp?isbn=9780521883368&quot; rel=&quot;nofollow&quot;&gt;http://www.cambridge.org/catalogue/catalogue.asp?isbn=9780521883368&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="1295" LastActivityDate="2010-08-18T06:12:16.750" />
  <row Id="2691" PostTypeId="2" ParentId="2687" CreationDate="2010-08-18T06:55:27.180" Score="4" ViewCount="" Body="&lt;p&gt;Let $X$ be a topological space.&#xA;Then $X$ has the property that a function $f:X\to Y$&#xA;(where $Y$ is an abritrary topological space) is continuous if&#xA;its restriction to each compact subspace is continuous if and only if&#xA;$X$ is &lt;a href=&quot;http://en.wikipedia.org/wiki/Compactly_generated_space&quot; rel=&quot;nofollow&quot;&gt;compactly generated&lt;/a&gt; topological space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The space $X$ is compactly generated if a subset $A$ whose intersection&#xA;with any compact subspace $K$ of $X$ is open in $K$, is open in $X$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most familiar topological spaces are compactly generated&#xA;(Euclidean spaces, manifolds, CW-complexes etc.) but not all;&#xA;see Qiaochu's answer for an example.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-08-18T06:55:27.180" />
  <row Id="2692" PostTypeId="2" ParentId="622" CreationDate="2010-08-18T07:02:16.650" Score="21" ViewCount="" Body="&lt;p&gt;One comment about your sentence &quot;this seems to only trivially invoke representation theory&quot;.&#xA;It might be surprising, but such obvious representations are actually the source of much&#xA;interesting mathematics, and a lot of effort of representation theorists is devoted to studying them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More precisely: the general set up here is that we have a group (in your example $SO(n)$) acting&#xA;on a space $X$ (in your example $\mathbb R^n$), and we look at the space of functions on $X$&#xA;(let me write it $\mathcal F(X)$; in a careful treatment, one would have to think about whether&#xA;we wanted continuous, smooth, $L^2$, or some other kind of functions, but I will suppress that&#xA;kind of technical consideration here).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, as you observe, there is a natural representation of $G$ on $\mathcal F(X)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are right that from a certain point of view this seems trivial, because the representation is obvious.  Unlike when one first learns rep'n theory of finite groups,&#xA;where one devotes a lot of effort to constructing reps., in this context, the rep. stares you in&#xA;the face.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So how can this be interesting?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, the representation $\mathcal F(X)$ will almost never be irreducible, so an important&#xA;question becomes: how does this representation decompose into irreducible representations?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suddenly we have a quite non-trivial representation theoretic problem:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, we have to work out the list of irreducible representations of $G$ (which is much&#xA;like what one does in a first course on rep'n theory of finite groups).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second, we have to figure out how $\mathcal F(X)$ decomposes, which involves representation theory (among other things, you have to develop methods for investigating this sort of question), and also often a lot of analysis (because typically $\mathcal F(X)$ will be&#xA;infinite dimensional, and may be a Hilbert space, or have some other similar sort of topological vector space structure which should be incorporated into the picture). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't think I should say too much more here, but I will just give some illustrative examples:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(a) If $ X = G = S^1$ (the circle group, say thought of as $\mathbb R/\mathbb Z$)&#xA;acting on itself by addition, then the solution to the problem of decomposing&#xA;$\mathcal F(S^1)$ is the theory of Fourier series.  (Note that a function on $S^1$ is the&#xA;same as a periodic function on $\mathbb R$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(b) If $ X = G = \mathbb R$, with $G$ acting on itself by addition, then&#xA;the solution to the above question (how does $\mathcal F(\mathbb R)$ decompose under the&#xA;action of $\mathbb R$) is the theory of the Fourier transform.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(c) If $ X = S^2$ and $G = SO(3)$ acting on $X$ via rotations, then decomposing&#xA;$\mathcal F(S^2)$ into irreducible representations gives the theory of spherical harmonics.&#xA;(This is an important example in quantum mechanics; it comes up for example in the theory&#xA;of the hydrogen atom, when one has a spherical symmetry because the electron orbits the&#xA;nucleus, which one thinks of as the centre of the sphere.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(d) If $ X = SL_2(\mathbb R)/SL_2(\mathbb Z)$ (this is the quotient of a Lie group&#xA;by a discrete subgroup, so is naturally a manifold, in this case of dimension 3),&#xA;with $G = SL_2(\mathbb R)$ acting by left multiplication, then the problem of decomposing&#xA;$\mathcal F(X)$ leads to the theory of modular forms and Maass forms, and is the first&#xA;example in the more general theory of automorphic forms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Added: Looking over the other answers, I see that this is an elaboration on  AD.'s answer.&lt;/p&gt;&#xA;" OwnerUserId="221" LastEditorUserId="221" LastEditorDisplayName="" LastEditDate="2010-08-18T08:33:03.437" LastActivityDate="2010-08-18T08:33:03.437" CommentCount="0" />
  <row Id="2693" PostTypeId="2" ParentId="2655" CreationDate="2010-08-18T07:03:36.457" Score="2" ViewCount="" Body="&lt;p&gt;If an approximation suffices, you could use the &lt;a href=&quot;http://en.wikipedia.org/wiki/Prime_number_theorem#The_prime-counting_function_in_terms_of_the_logarithmic_integral&quot; rel=&quot;nofollow&quot;&gt;offset logarithmic integral function&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="505" LastActivityDate="2010-08-18T07:03:36.457" />
  <row Id="2694" PostTypeId="1" AcceptedAnswerId="10608" CreationDate="2010-08-18T07:47:43.093" Score="22" ViewCount="1168" Body="&lt;p&gt;I have been fascinated by this problem since I first heard about it in high school.  From the Wikipedia article &lt;a href=&quot;http://en.wikipedia.org/wiki/Collatz_problem&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Collatz_problem&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Take any natural number n. If n is even, divide it by 2 to get n / 2, if n is odd multiply&#xA;  it by 3 and add 1 to obtain 3n + 1. Repeat the process indefinitely. The conjecture is&#xA;  that no matter what number you start with, you will always eventually reach 1. [...]&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Paul Erdős said about the Collatz conjecture: &quot;Mathematics is not yet ready for such&#xA;  problems.&quot; He offered $500 USD for its solution.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My question is: how important do you consider the answer to this question, and why?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And would you speculate on what might have possessed Paul Erdős to make such an offer?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Is there any reason to think that a proof of the Collatz Conjecture would be&#xA;complex (like FLT) rather than simple (like PRIMES is in P)?  And can this&#xA;characterization of FLT vs. PRIMES is in P be made more specific than a bit-length comparison?&lt;/p&gt;&#xA;" OwnerUserId="1284" LastEditorUserId="1284" LastEditorDisplayName="" LastEditDate="2010-09-01T02:59:39.917" LastActivityDate="2011-04-27T08:17:32.963" CommunityOwnedDate="2010-08-18T14:51:51.837" Title="What is the importance of the Collatz conjecture?" Tags="&lt;number-theory&gt;&lt;soft-question&gt;" AnswerCount="9" CommentCount="6" FavoriteCount="5" />
  <row Id="2695" PostTypeId="2" ParentId="2694" CreationDate="2010-08-18T07:53:21.800" Score="-2" ViewCount="" Body="&lt;p&gt;I would say that it is a very interesting question, so simple and yet so hard to prove.&#xA;That is what all important questions should be like, but compared to &quot;Why are we here?&quot;, &#xA;we might actually get an answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the same time, one can spend a great deal of time trying to prove it,&#xA;maybe Erdös spent too much time on it, so he just wanted a proof of it so that he could move on.&lt;/p&gt;&#xA;" OwnerUserId="934" LastActivityDate="2010-08-18T07:53:21.800" CommunityOwnedDate="2010-08-18T14:51:51.837" />
  <row Id="2696" PostTypeId="2" ParentId="2645" CreationDate="2010-08-18T08:38:58.927" Score="3" ViewCount="" Body="&lt;p&gt;The correlation coefficient only measures &lt;strong&gt;linear&lt;/strong&gt; dependence of two random variables. So if they depend on each other in a non-linear way, the correlation coefficient will not catch it.&lt;/p&gt;&#xA;" OwnerUserId="505" LastActivityDate="2010-08-18T08:38:58.927" CommentCount="1" />
  <row Id="2698" PostTypeId="1" AcceptedAnswerId="2714" CreationDate="2010-08-18T09:19:22.427" Score="3" ViewCount="84" Body="&lt;p&gt;Suppose, we have f1(n) = o(f2(n)).. (small o)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, we have to compare f1(f2(n)) and f2(f1(n)) asymptotically..&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What can we say about it? Which one will be greater &amp;amp; why?&lt;/p&gt;&#xA;" OwnerUserId="933" LastActivityDate="2010-08-18T18:10:31.223" Title="Comparing composite functions asymptotically" Tags="&lt;algorithms&gt;" AnswerCount="2" />
  <row Id="2699" PostTypeId="2" ParentId="2628" CreationDate="2010-08-18T09:36:34.793" Score="6" ViewCount="" Body="&lt;p&gt;Okay ... here we go ...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To save typing, I'm going to use $A$, $B$, and $C$ for $\theta$, $\phi$, and $\psi$. So, our integral starts as this ...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\int\int\int \left(2 \cos A \cos B - \sin A \sin B \cos C\right)^{2k} \sin A \sin B&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;... and expands to this ...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int\int\int \sum_{j=0}^{2k} {2k \choose j} \; 2^j (-1)^{2k-j} \cos^j A \; \sin^{2k-j+1} A \; \cos^j B \; \sin^{2k-j+1} B \; \cos^{2k-j} C$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The integral over the sum will be a sum of integrals over the terms; and, since the $A$, $B$, and $C$ factors are independent, each integrated term is a product of integrals. Moreover, as mentioned in my comment on the original question, we're integrating $\cos^{2k-j}C$ over the full period of cosine, so terms with odd powers of $\cos C$ --those corresponding to odd values of $j$-- vanish. So, we take $j=2h$ and for convenience define $m := k-h$, then we write ...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{h=0}^{k} 2^{2h} {2k \choose 2h} \; \left( \int \cos^{2h} A \; \sin^{2m+1} A \right) \left( \int \cos^{2h} B \; \sin^{2m+1} B \right) \left( \int \cos^{2m} C \right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first two integrals are identical, so we're really looking at the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{h=0}^{k} 2^{2h} {2k \choose 2h} \; \left( \int \cos^{2h} A \; \sin^{2m+1} A \right)^2 \left( \int \cos^{2m} C \right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the pieces ...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the reduction formula for the integral of powers of cosine, we have &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_{0}^{2\pi} \cos^{2m}C \; dC = \frac{(2m-1)(2m-3)\cdots(1)}{(2m)(2m-2)\cdots(2)}\int_{0}^{2\pi} dC = \frac{2\pi(2m-1)!!}{2^{m} m!} = \frac{2\pi(2m)!}{2^{2m} m!^2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $x!!$ is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Double_factorial#Double_factorial&quot; rel=&quot;nofollow&quot;&gt;double factorial&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the other factor ...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_{0}^{\pi} \cos^{2h} A \; \sin^{2m+1} A \; dA = \int_{0}^{\pi} \cos^{2h} A \; \left( 1 - \cos^{2}A\right)^m \; \sin A \; dA = \int_{-1}^{1} u^{2h} \left( 1 - u^2 \right)^m \; du$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$= \int_{-1}^{1} u^{2h} \sum_{n=0}^{m} {m \choose n} (-1)^n u^{2n} \; du = \sum_{n=0}^{m} (-1)^n {m \choose n} \int_{-1}^{1} u^{2n+2h} \; du = \sum_{n=0}^{m} {m \choose n} \frac{2 (-1)^n}{2n+2h+1} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mathematica evaluates the sum into a form we manipulate into something we can use: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\Gamma(h+\frac{1}{2}) m!}{\Gamma(m+h+\frac{3}{2})}  = \frac{\sqrt{\pi} (2h-1)!!}{2^h}\frac{2^{k+1}m!}{\sqrt{\pi}(2k+1)!!}= \frac{(2h)!}{2^{2h}h!}\frac{2^{k+1}m!}{(2k+1)}\frac{2^{k}k!}{(2k)!}= \frac{2^{2m+1}m!}{2k+1}\frac{(2h)!}{h!}\frac{k!}{(2k)!}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This leads to a simplification bonanza in the formula for the integral ...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{h=0}^{k} \;\; 2^{2h} \;\;\;\; \frac{(2k)!}{(2h)!(2m)!} \;\;\;\;  \frac{2^{4m+2}m!^2}{(2k+1)^2}\frac{(2h)!^2}{h!^2}\frac{k!^2}{(2k)!^2} \;\;\;\; \frac{2\pi(2m)!}{2^{2m} m!^2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$= 8 \pi \sum_{h=0}^{k} \frac{2^{2(m+h)}}{(2k+1)^2}\frac{(2h)!}{h!^2}\frac{k!^2}{(2k)!}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$= 8 \pi  \frac{2^{2k}}{(2k+1)^2} \frac{k!^2}{(2k)!} \sum_{h=0}^{k}\frac{(2h)!}{h!^2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$= 8 \pi  \left(\frac{2^{k}}{2k+1}\right)^2 {2k \choose k}^{-1} \;\; \sum_{h=0}^{k}{2h \choose h}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;... as expected (well, &quot;expected&quot; from your having previously forced it out of Mathematica).&lt;/p&gt;&#xA;" OwnerUserId="409" LastActivityDate="2010-08-18T09:36:34.793" CommentCount="2" />
  <row Id="2700" PostTypeId="2" ParentId="2698" CreationDate="2010-08-18T09:36:40.517" Score="0" ViewCount="" Body="&lt;p&gt;What happens if $f(x)=x^t$ and $g(x)=e^x$&#xA;where $t$ is a positive real?&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-08-18T09:36:40.517" />
  <row Id="2701" PostTypeId="2" ParentId="2628" CreationDate="2010-08-18T10:40:48.200" Score="0" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Here is a hint&lt;/strong&gt; to an other approach: Search for the key words &quot;convolution formula&quot; &quot;Legendre polynomial&quot; &quot;addition formula&quot; and you will be en-lighted... &lt;/p&gt;&#xA;&#xA;&lt;p&gt;--- A little explanation and some spherical harmonics ---&lt;/p&gt;&#xA;&#xA;&lt;p&gt;@J. Mangaldan You are totally right in particular when I was wrong in my suggestion! :/&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I thought about was convolution on the double coset space $X=K\backslash G/K$ of $G= SO(3)$ where $K=SO(2)$. The space $X$ may be identified with $[-1,1]$ and the convolution (as induced from the translation operator on $G$) is then given by&#xA;$$f*g(x)= \frac{1}{4\pi}\int_{-1}^1\int_{0}^{2\pi}f(y)g(xy+\sqrt{1-x^2}\sqrt{1-y^2}\cos\theta) d\theta dy.$$&#xA;An interesting fact is that $f*g=g*f$ even though $G$ is not Abelian.&#xA;Choosing $f(x)=1$ and $g(x)=x^{2k}$ we get&#xA;$$\frac{1}{8\pi}\int_{-1}^1\int_{-1}^1\int_{0}^{2\pi}(xy+\sqrt{1-x^2}\sqrt{1-y^2}\cos\theta)^{2k} d\theta dydx=$$&#xA;$$\int_{-1}^1f*g(x)dx/2 = \int_{-1}^1 g*f(x)dx/2=\int_{-1}^1x^{2k}dx/2= 1/(2k+1).$$&#xA;In our case the integrand is (after changing variables) &#xA;$$(2xy + \sqrt{1-x²}\sqrt{1-y²}\cos\theta)^{2k}=$$&#xA;$$=\sum_{j=0}^{2k}{2k \choose j}(xy)^j(xy+\sqrt{1-x²}\sqrt(1-y²)\cos\theta)^{2k-j}$$&#xA;Using this we get an other convolution - but I can not see how this can be used directly. :(&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Legendre polynomials might be good since they are the characters on the space $X$.&lt;/p&gt;&#xA;" OwnerUserId="1154" LastEditorUserId="1154" LastEditorDisplayName="" LastEditDate="2010-08-21T08:29:46.533" LastActivityDate="2010-08-21T08:29:46.533" CommentCount="3" />
  <row Id="2702" PostTypeId="2" ParentId="2682" CreationDate="2010-08-18T10:57:09.520" Score="1" ViewCount="" Body="&lt;p&gt;As everybody told you, there are infinitely many solutions for your problem. One particularly simple kind of them are functions like:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;y = \frac{a}{x} + bx \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can obtain which constants $a, b$ fit with your pair of data $(11, 5.000)$ and $(3, 200.000)$ imposing them:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align*}&#xA;5.000      &amp;amp;=    \frac{1}{11}a + 11b&lt;br&gt;&#xA;\\ 200.000    &amp;amp;=    \frac{1}{3}a + 3b&#xA;\end{align*}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You're getting a linear system of equations in which $a,b$ are the unknowns. Since $\frac{3}{11} - \frac{11}{3} \neq 0$, you have a unique solution.&lt;/p&gt;&#xA;" OwnerUserId="664" LastActivityDate="2010-08-18T10:57:09.520" />
  <row Id="2703" PostTypeId="1" CreationDate="2010-08-18T10:58:26.740" Score="1" ViewCount="1332" Body="&lt;p&gt;I have an examples book with a limit exercise that I can't understand.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The limit in question is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f(x,y)=\frac{x}{x+y}$$&#xA;with $x\ne-y$;&#xA;$$\lim_{(x,y)\to(0,0)} f(x,y)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And then to solve it, it goes:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\lim_{(x,y)\to(0,0)} f(x,y)&#xA;= \lim_{x\to0} f(x,mx) =\lim_{x\to 0}\frac{x}{x+mx}=\frac{1}{1+m}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you help me understand that?&#xA;Thanks,&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;UPDATE&lt;/strong&gt;: Ok, just to make sure that I got it right. I have a very similar test exercise with 4 different options.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The following limit $$\lim_{(x,y)\to(0,0)}\frac{-x^3+3xy^2}{x^2+y^2}$$ equals:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A. 0&lt;/p&gt;&#xA;&#xA;&lt;p&gt;B. -infinite&lt;/p&gt;&#xA;&#xA;&lt;p&gt;C. Doesn't exist&lt;/p&gt;&#xA;&#xA;&lt;p&gt;D. +infinite&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My doubt is: if I consider it normally I'd say that it doesn't exist, but if I solve it using the same approach (i.e. y=mx) then the limit equals 0.&#xA;&lt;strong&gt;Which one is the right answer?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="1298" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-08-19T21:54:07.533" LastActivityDate="2010-08-19T21:54:07.533" Title="0/0 limit question" Tags="&lt;calculus&gt;&lt;limit&gt;" AnswerCount="6" CommentCount="2" FavoriteCount="2" />
  <row Id="2704" PostTypeId="2" ParentId="2703" CreationDate="2010-08-18T11:09:53.087" Score="8" ViewCount="" Body="&lt;p&gt;The point is that this limit doesn't exist.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For it to exist, then if you take &lt;strong&gt;any&lt;/strong&gt; sequence of points $(x_n,y_n)$&#xA;converging to $(0,0)$ for which $f(x_n,y_n)$ is defined, then&#xA;$$\lim_{n\to\infty}f(x_n,y_n)$$&#xA;must exist and also be &lt;strong&gt;independent&lt;/strong&gt; of the sequence $(x_n,y_n)$.&#xA;The book is saying in essence that if you have a sequence of points of the form&#xA;$(x_n,m x_n)$ (so lying on the line with gradient $m$ though the origin)&#xA;and converging to the origin, then&#xA;$$\lim_{n\to\infty}f(x_n,y_n)=\frac1{1+m}\qquad\qquad\qquad(*)$$&#xA;which is &lt;strong&gt;not&lt;/strong&gt; independent of the sequence of points $(x_n,y_n)$,&#xA;as changing $m$ will change the limit $(*)$.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-08-18T11:09:53.087" CommentCount="7" />
  <row Id="2705" PostTypeId="2" ParentId="2474" CreationDate="2010-08-18T11:39:32.540" Score="0" ViewCount="" Body="&lt;p&gt;This is a response to Robin Chapman's answer. (For some reason I am not able to ask this directly under his question.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why do we get &lt;em&gt;that&lt;/em&gt; formula from van Kampen? The double torus is the union of the two open subsets that are homeomorphic to $T$ and whose intersection is $S^1$. So by van Kampen this should equal the colimit of $\pi_1(W)$ with $W \in {T,T,S^1}$. I thought the colimit in the category of groups is just the direct sum, hence the result should be $\pi_1(T) \oplus \pi_1(T) \oplus \pi_1(S^1)$.&lt;/p&gt;&#xA;" OwnerUserId="1253" LastActivityDate="2010-08-18T11:39:32.540" CommentCount="4" />
  <row Id="2706" PostTypeId="2" ParentId="2703" CreationDate="2010-08-18T12:08:30.610" Score="3" ViewCount="" Body="&lt;p&gt;I will elaborate on Robin's answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;There is no&lt;/strong&gt; unique limit as (&lt;em&gt;x,y&lt;/em&gt;) goes to (0,0) --- &lt;em&gt;i.e.&lt;/em&gt; the limit does not exist. For instance, if we approach (0,0) on the &lt;em&gt;x&lt;/em&gt; axis, we obtain the limit&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$\lim\limits_{(x,y) \to (0,0)} \dfrac{x}{x+y} = \lim\limits_{x \to 0} \dfrac{x}{x} = 1,$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and if we approach (0,0) on the &lt;em&gt;y&lt;/em&gt; axis, we obtain the limit&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$\lim\limits_{(x,y) \to (0,0)} \dfrac{x}{x+y} = \lim\limits_{y \to 0} \dfrac{0}{y} = 0.$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;However,&lt;/strong&gt; we can still consider limits along different curves. The two above limits, along the &lt;em&gt;x&lt;/em&gt; and &lt;em&gt;y&lt;/em&gt; axes, are two examples: the limit you describe in your question is a similar limit, on the locus of the equation &lt;em&gt;y&lt;/em&gt;&amp;nbsp;=&amp;nbsp;&lt;em&gt;mx&lt;/em&gt;. &lt;em&gt;This is something different&lt;/em&gt; than whether there is &quot;a limit&quot; of f(&lt;em&gt;x,y&lt;/em&gt;) as we approach (0,0) --- in this case, because there are different limits depending on how we approach the origin, we may consider the question of what limit one has as one approaches the origin on a particular curve.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case, a &quot;curve&quot; is a function  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$\big(x(t),y(t)\big) = c(t)$,&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;for some function $c: \mathbb R \to \mathbb R^2$. &quot;Approaching&quot; (0,0) along the curve c(&lt;em&gt;t&lt;/em&gt;) entails specifying some domain for c in which it does not cross itself (for the example of the line &lt;em&gt;y&lt;/em&gt;&amp;nbsp;=&amp;nbsp;&lt;em&gt;mx&lt;/em&gt; one could take c(&lt;em&gt;t&lt;/em&gt;)&amp;nbsp;=&amp;nbsp;(&lt;em&gt;t&lt;/em&gt;,&amp;nbsp;&lt;em&gt;mt&lt;/em&gt;), in which case the domain can be the real numbers) and for which c(&lt;em&gt;t&lt;/em&gt;)&amp;nbsp;=&amp;nbsp;(0,0) for some (unique) value &lt;em&gt;t&lt;/em&gt;&amp;nbsp;=&amp;nbsp;T. Then, evaluating the limit of f(&lt;em&gt;x,y&lt;/em&gt;) along the curve (&lt;em&gt;x,y&lt;/em&gt;)&amp;nbsp;=&amp;nbsp;c(&lt;em&gt;t&lt;/em&gt;) means just taking the limit of the &lt;em&gt;composite&lt;/em&gt; function f(c(&lt;em&gt;t&lt;/em&gt;)) as &lt;em&gt;t&lt;/em&gt; approaches T.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of approaching (0,0) on the curve c(&lt;em&gt;t&lt;/em&gt;)&amp;nbsp;=&amp;nbsp;(&lt;em&gt;t&lt;/em&gt;,&amp;nbsp;&lt;em&gt;mt&lt;/em&gt;), we just take T&amp;nbsp;=&amp;nbsp;0, and evaluate the limit of f(&lt;em&gt;t&lt;/em&gt;,&amp;nbsp;&lt;em&gt;mt&lt;/em&gt;) as &lt;em&gt;t&lt;/em&gt; approaches 0. Up to a change of variables, this is just what you have done above.&lt;/p&gt;&#xA;&#xA;&#xA;" OwnerUserId="439" LastActivityDate="2010-08-18T12:08:30.610" CommentCount="3" />
  <row Id="2707" PostTypeId="2" ParentId="2694" CreationDate="2010-08-18T12:25:38.487" Score="5" ViewCount="" Body="&lt;p&gt;Paul Erdös offered cash prizes for the solution to problems according to his assessment of their difficulty and importance. I believe his judgement of the difficulty of this problem has been shown to be correct by the fact that it remains open.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think this problem is very important in the sense that a large proportion of the people reading this response will have had a go at it, at one time or another. Thus its solution would be of interest to many.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another reason for its importance is that, like Fermat's Last Theorem (Wiles's Theorem), it's easy to state and understand and thus has the potential to attract young people towards mathematics. I too learned about it in high school and could not resist its allure.&lt;/p&gt;&#xA;" OwnerUserId="1301" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2010-08-18T22:07:05.553" LastActivityDate="2010-08-18T22:07:05.553" CommunityOwnedDate="2010-08-18T14:51:51.837" />
  <row Id="2708" PostTypeId="2" ParentId="2703" CreationDate="2010-08-18T12:36:34.347" Score="0" ViewCount="" Body="&lt;p&gt;De l'Hôpital explains the last step.&lt;br&gt;&#xA;If both numerator and denominator approach zero, you can apply de l'Hôpital, provided the derivative of the functions in numerator and denominator exist for x=0, which is the case.&lt;/p&gt;&#xA;" OwnerUserId="923" LastActivityDate="2010-08-18T12:36:34.347" CommentCount="4" />
  <row Id="2709" PostTypeId="2" ParentId="2655" CreationDate="2010-08-18T12:50:38.967" Score="0" ViewCount="" Body="&lt;p&gt;For primes smaller than your example you can use Wolfram|Alpha, as Adam S pointed out. Wolfram|Alpha has the prime Pi function:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.wolframalpha.com/input/?i=pi(55252335667&quot; rel=&quot;nofollow&quot;&gt;http://www.wolframalpha.com/input/?i=pi(55252335667&lt;/a&gt;)&lt;/p&gt;&#xA;" OwnerUserId="1302" LastActivityDate="2010-08-18T12:50:38.967" />
  <row Id="2710" PostTypeId="1" AcceptedAnswerId="2730" CreationDate="2010-08-18T12:55:12.593" Score="33" ViewCount="600" Body="&lt;p&gt;The &lt;em&gt;Mandelbrot set&lt;/em&gt; is the set of points of the complex plane whos orbits do not diverge. An point $c$'s &lt;em&gt;orbit&lt;/em&gt; is defined as the sequence $z_0 = c$, $z_{n+1} = z_n^2 + c$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The shape of this set is well known, &lt;strong&gt;why is it that if you zoom into parts of the filaments you will find slightly deformed copies of the original shape&lt;/strong&gt;, for example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/7HMU1.jpg&quot; alt=&quot;Detail of the Mandelbrot set&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I measured some points on the Mandelbrot, and the corresponding points from one of these smaller Mu-molecules. Comparing the orbit sequences it was possible to find points on each sequence which were very close - but this experiment did not really help me to understand anything new.&lt;/p&gt;&#xA;" LastActivityDate="2010-08-20T20:08:03.883" Title="Why does the Mandelbrot set contain (slightly deformed) copies of itself?" Tags="&lt;fractals&gt;" AnswerCount="2" CommentCount="5" FavoriteCount="3" />
  <row Id="2711" PostTypeId="2" ParentId="2703" CreationDate="2010-08-18T13:06:31.493" Score="5" ViewCount="" Body="&lt;p&gt;For your second question: is&#xA;$$\lim_{(x,y)\to(0,0)}\frac{-x^3+3xy^2}{x^2+y^2}=0?$$&#xA;imagine a sequence of points $(x_n,y_n)$&#xA;converging to the origin, and think whether $f(x_n,y_n)\to0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would be tempted to write these points in polar form:&#xA;$x_n=r_n\cos t_n$ and $y_n=r_n\sin t_n$ as then the denominator&#xA;becomes $r_n^2$ (nice) and moreover $r_n\to0$ (why?).&lt;/p&gt;&#xA;" OwnerUserId="226" LastEditorUserId="226" LastEditorDisplayName="" LastEditDate="2010-08-18T13:13:25.613" LastActivityDate="2010-08-18T13:13:25.613" CommentCount="14" />
  <row Id="2712" PostTypeId="2" ParentId="2694" CreationDate="2010-08-18T13:13:11.853" Score="14" ViewCount="" Body="&lt;p&gt;So many mathematicians, and famous ones among them, have tried various ways to attack this problem, and it is still as elusive as it was when first posed. So the importance of the problem is that genuinely new mathematical ideas will have to be created to solve it, and such ideas may be helpful in other domains where &quot;truly important&quot; problems are at stake. Note that Erdös himself has said something along the lines that &quot;we don't have the mathematics yet to solve this problem&quot;.&lt;/p&gt;&#xA;" OwnerUserId="1303" LastActivityDate="2010-08-18T13:13:11.853" CommunityOwnedDate="2010-08-18T14:51:51.837" CommentCount="4" />
  <row Id="2713" PostTypeId="2" ParentId="2703" CreationDate="2010-08-18T13:26:33.867" Score="2" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; The limit becomes that of $x/f(x)$ along the path $y = f(x)-x$ where $f(x)\to 0$. But this limit can be whatever you desire, by choosing $f$ weaker, comparable, or stronger than $x$ at $0$. Explicitly: choosing $f$ to be one of $\; x^{1/2},\; x/c,\; x^2$ yields $\; x/f(x)\to 0,\; c,\; \infty\;$ respectively, as $x\to0^+$.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-08-19T21:48:58.767" LastActivityDate="2010-08-19T21:48:58.767" CommentCount="3" />
  <row Id="2714" PostTypeId="2" ParentId="2698" CreationDate="2010-08-18T13:54:56.763" Score="2" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $f_1 = \log x,\; f_2 = x^n \;$ for $\; n = 1/2,\; 1,\; 2\;$ shows that the answer is &quot;nothing&quot;&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-08-18T13:54:56.763" />
  <row Id="2715" PostTypeId="1" AcceptedAnswerId="2717" CreationDate="2010-08-18T14:14:08.527" Score="4" ViewCount="250" Body="&lt;p&gt;This was asked to me by a friend. I tried this problem for a long time, but i couldn't solve it. Seems interesting though. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Prove that it is impossible to construct two different seven digit numbers, one of which is divisible by the other, out of the digits 1,2,3,4,5,6,7 (All seven digits must be in each number)&lt;/p&gt;&#xA;" LastActivityDate="2010-08-18T18:56:42.970" Title="Constructing a seven digit number such that one divides the other." Tags="&lt;number-theory&gt;&lt;combinatorics&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="2717" PostTypeId="2" ParentId="2715" CreationDate="2010-08-18T14:28:19.350" Score="10" ViewCount="" Body="&lt;p&gt;Say the numbers are x and y and y divides x.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the remainder upon dividing both numbers by 9. Both leave a remainder 1 (as sum of digits leaves a remainder 1).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So x = ky where 2 &amp;lt;= k &amp;lt;= 8 is not possible.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-08-18T14:28:19.350" CommentCount="2" />
  <row Id="2719" PostTypeId="1" AcceptedAnswerId="2724" CreationDate="2010-08-18T15:11:54.973" Score="14" ViewCount="332" Body="&lt;p&gt;consdier a cubic polynomial, p(x)=k(x-a)(x-b)(x-c) where k is some constant and a,b,c its 3 roots (not necessarily distinct, not necessarily real). It is very simple to show that if you average two roots of a cubic polynomial and compute the tangent line at their average that it will intersect the cubic polynomial at the remaining root. I tried to generalize this result to other odd degree polynomials and it did not seem to work well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is so special about cubic polynomials that allows this to work that isn't true about higher degree odd polynomials?&lt;/p&gt;&#xA;" OwnerUserId="249" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-08-18T19:54:11.293" LastActivityDate="2010-08-18T19:54:11.293" Title="Averaging 2 roots of a cubic polynomial" Tags="&lt;calculus&gt;&lt;algebra-precalculus&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="3" />
  <row Id="2720" PostTypeId="2" ParentId="2715" CreationDate="2010-08-18T15:16:38.297" Score="2" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt;  $\; ax = b,\; (a,m)=1 \;\Rightarrow\; x = b/a \;\;{\rm mod}\; m \;$ is the &lt;strong&gt;unique&lt;/strong&gt; solution such that $\; \; 0\le x &amp;lt; m. \;$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now put $\; m = 9, \;$ i.e. cast out nines. This is a ubiquitous technique for recreational problems of this ilk.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;CHALLENGE&lt;/strong&gt; Find a natural generalization to arbitrary length integers.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-08-18T15:46:03.100" LastActivityDate="2010-08-18T15:46:03.100" />
  <row Id="2722" PostTypeId="2" ParentId="855" CreationDate="2010-08-18T17:27:14.237" Score="2" ViewCount="" Body="&lt;p&gt;let prime number p=2k+1,&#xA;p^2-1=4k(k+1), thus: 8|p^2-1&#xA;by theorem, p^2=1(mod 3), thus 3|p^2-1&#xA;thus, 24|p^2-1&lt;/p&gt;&#xA;" OwnerUserId="1059" LastEditorUserId="1059" LastEditorDisplayName="" LastEditDate="2010-08-20T18:09:26.853" LastActivityDate="2010-08-20T18:09:26.853" CommentCount="2" />
  <row Id="2723" PostTypeId="1" AcceptedAnswerId="4569" CreationDate="2010-08-18T17:45:56.700" Score="1" ViewCount="268" Body="&lt;p&gt;I have a non-mathematician friend who is interested in re-learning algebra. I am more than happy to help, but I am in no position to judge what is a &lt;strong&gt;good&lt;/strong&gt; introductory text --- only to identify when a text is a very &lt;strong&gt;bad&lt;/strong&gt; one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My friend is interested in starting from &quot;basics&quot; --- he's comfortable with order-of-operations, minus-times-minus-is-plus, and positive integer exponents. He's shaky with negative exponents, logarithms, and roots.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any recommendations for good, freely-available resources?&lt;/p&gt;&#xA;" OwnerUserId="439" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-08-18T20:02:44.677" LastActivityDate="2010-09-27T02:52:58.960" CommunityOwnedDate="2010-08-18T17:51:23.057" Title="Free introductory resources for learning algebra?" Tags="&lt;algebra-precalculus&gt;&lt;education&gt;&lt;learning&gt;" AnswerCount="3" CommentCount="5" FavoriteCount="2" />
  <row Id="2724" PostTypeId="2" ParentId="2719" CreationDate="2010-08-18T17:52:48.137" Score="9" ViewCount="" Body="&lt;p&gt;This result &lt;em&gt;does&lt;/em&gt; hold for higher-degree polynomials; you just didn't generalize it correctly.  The general statement is as follows.  Let $p(x) = (x - a) q(x)$ where $q(x)$ is &lt;em&gt;any&lt;/em&gt; differentiable function, and let $r$ be such that $q'(r) = 0$.  Then the tangent line at $(r, p(r))$ intersects the $x$-axis at $(a, 0)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a fairly simple computation.  Since $p'(x) = (x - a) q'(x) + q(x)$, it follows that $p'(r) = q(r)$, so the tangent line at $(r, p(r))$ has slope $q(r)$ and hence it intersects the $x$-axis at $(r - \frac{p(r)}{q(r)}, 0) = (a, 0)$.  In particular, the above result holds for $q(x)$ a polynomial of any degree greater than or equal to $2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Intuitively, when $q'(r) = 0$, the linear approximation to $p(x)$ at $r$ is $(x - a) q(r)$, so of course it has to hit the $x$-axis at $(a, 0)$.  In fact, this way of thinking about it tells you that the $r$ such that $q'(r) = 0$ are the only $r$ for which this occurs.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the special case that $q(x)$ is quadratic, $r$ happens to be equal to the average of the roots of $q$.  For higher-degree polynomials this is no longer the case.  Instead, all you know is that the roots of $q'(x)$ interlace between the roots of $q(x)$.&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-08-18T18:24:01.443" LastActivityDate="2010-08-18T18:24:01.443" />
  <row Id="2725" PostTypeId="2" ParentId="2723" CreationDate="2010-08-18T18:01:34.673" Score="0" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://www.curriki.org/xwiki/bin/view/Main/WebHome&quot; rel=&quot;nofollow&quot;&gt;Curriki&lt;/a&gt; is supposed to be a good resource for this kind of stuff, but I haven't tried it out myself.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-08-18T18:01:34.673" CommunityOwnedDate="2010-08-18T18:01:34.673" />
  <row Id="2727" PostTypeId="2" ParentId="2723" CreationDate="2010-08-18T18:28:06.880" Score="0" ViewCount="" Body="&lt;p&gt;There's a pretty good video course on YouTube EDU, &lt;a href=&quot;http://www.youtube.com/user/UMKC#grid/user/DE28CF08BD313B2A&quot; rel=&quot;nofollow&quot;&gt;&quot;College Algebra&quot;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1137" LastActivityDate="2010-08-18T18:28:06.880" CommunityOwnedDate="2010-08-18T18:28:06.880" />
  <row Id="2728" PostTypeId="2" ParentId="2715" CreationDate="2010-08-18T18:56:42.970" Score="1" ViewCount="" Body="&lt;p&gt;7654321/1234567 = 6. something &lt;/p&gt;&#xA;&#xA;&lt;p&gt;so x = ky where 2&amp;lt;=k&amp;lt;=6&lt;/p&gt;&#xA;&#xA;&lt;p&gt;sum of the digits are not divisible by 3 so no 3 and 6.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;now x = ky where k = 2,4 or 5 so not possible.&lt;/p&gt;&#xA;" OwnerUserId="1093" LastActivityDate="2010-08-18T18:56:42.970" />
  <row Id="2729" PostTypeId="2" ParentId="2485" CreationDate="2010-08-18T19:28:46.130" Score="0" ViewCount="" Body="&lt;p&gt;Maybe the following works:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $R_n$ be a countable basis of rectangles (for example rational coordinates) for $\mathbb{R}^2$. Define $f_0 = 0$ and given $f_n$ define $f_{n + 1}$ as $f_n$ outside $R_{n + 1}$ and $n + 1$ inside $R_{n + 1}$. $(f_n)$ is an increasing sequence of measurable functions so $f = \lim f_n$ is measurable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now take any open rectangle $R$ and given $M &gt; 0$ take $n &gt; M$ such that $R_n \subset R$. Now $\mu(|f| &gt; M) \geq \mu(|f_n| &gt; M) \geq \textrm{area}(R_n) &gt; 0$. So it is essentially unbounded.&lt;/p&gt;&#xA;" OwnerUserId="1120" LastActivityDate="2010-08-18T19:28:46.130" />
  <row Id="2730" PostTypeId="2" ParentId="2710" CreationDate="2010-08-18T19:41:30.467" Score="18" ViewCount="" Body="&lt;p&gt;I really like this question!  I can't yet upvote, so I'll offer an answer instead.  This is only a partial answer, as I don't fully understand this material myself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose that $f$ is a quadratic polynomial.  Suppose that there is an integer $n$ and a domain $U \subset \mathbb{C}$ so that the $n$-th iterate, $f^n$, restricted to $U$ is a &quot;quadratic-like map.&quot;  Then we'll call $f$ &lt;em&gt;renormalizable&lt;/em&gt;.   (See Chapter 7 of McMullen's book &quot;Complex dynamics and renormalization&quot; for more precise definitions.) Now renormalization preserves the property of having a connected Julia set.  Also the parameter space of &quot;quadratic-like maps&quot; is basically a copy of $\mathbb{C}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, fix a quadratic polynomial $f$ and suppose that it renormalizes.  Then, in the generic situation, all $g$ close to $f$ also renormalize using the same $n$ and almost the same $U$.  This gives a map from a small region about $f$ to the space of quadratic-like maps.  This gives a partial map from the small region to the Mandelbrot set and so explains the &quot;local&quot; self-similarity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To sum up: all of the quadratic polynomials in a baby Mandelbrot set renormalize and all renormalize in essentially the same way.  (I believe that there are issues as you approach the place where the baby is attached to the parent.)  Thus renormalization explains why the baby Mandelbrot set appears.&lt;/p&gt;&#xA;" OwnerUserId="1307" LastEditorUserId="1307" LastEditorDisplayName="" LastEditDate="2010-08-19T22:30:51.903" LastActivityDate="2010-08-19T22:30:51.903" />
  <row Id="2731" PostTypeId="1" CreationDate="2010-08-18T19:45:10.507" Score="3" ViewCount="250" Body="&lt;p&gt;You know that a couple has two children. You go to the couple's house and one of their children, a young boy, opens the door. What is the probability that the couple's other child is a girl?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you list all possibilities for the sexes of two children, BB, BG, GB, GG, you see that 2 of the 3 pairs that have B (for boy) in them also have a girl, so the answer one could argue is 2/3.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, one could argue that the answer is 1/2, since the probability that any one child is a girl is 1/2, and intuitively (?) should be independent of the gender of its siblings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some background to possibly justify posting it here: the question was asked at an interview for an actuarial/insurance type position, and the interviewer was the answer was 2/3, whereas my friend who was being interviewed (and has a masters in math) thought the answer was 1/2, even after the interviewer explained his logic. My friend felt that the interviewer wasn't taking into account the fact that it is not equally likely that a boy will open the door in the BB versus the BG combination, and one has to take into account that fact. I have no idea which is the correct answer, both sound somewhat convincing to me (I have a Ph.D. in math, but I won't mention from where in an effort to avoid embarrassing my degree granting institution!). Anyways, any help would be appreciated and I apologize if this is too simple a question for this forum.&lt;/p&gt;&#xA;" LastActivityDate="2010-08-22T01:47:49.467" Title="very simple conditional probability question" Tags="&lt;probability&gt;" AnswerCount="6" CommentCount="2" />
  <row Id="2732" PostTypeId="2" ParentId="2474" CreationDate="2010-08-18T19:53:56.677" Score="0" ViewCount="" Body="&lt;p&gt;As a futher hint: the fundamental group of the torus $\mathbb{T}$ is generated by $a, b$ and has the relation $abAB$, where capitals denote inverses.  If you remove an open disk then you get a once-holed torus $T$.  Now the fundamental group is free (why?) and the boundary is the homotopic to the element $abAB$ (why?).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you can take another copy of the torus, say $\mathbb{S}$, with fundamental group generated by $c, d$ and having relation $cdCD$.  Again remove a disk to get a once holed torus $S$. Now carefully follow the answer already given, gluing $T$ and $S$, and so on. &lt;/p&gt;&#xA;" OwnerUserId="1307" LastActivityDate="2010-08-18T19:53:56.677" />
  <row Id="2733" PostTypeId="2" ParentId="2731" CreationDate="2010-08-18T20:07:25.803" Score="6" ViewCount="" Body="&lt;p&gt;This was also posted on MathOverflow, where the answer of 1/2 was confirmed several ways.  My favorite is this: There are really &lt;em&gt;three&lt;/em&gt; independent &quot;choices&quot; are being made: the sex of the first (say, older) child, the sex of the younger child, and whether the older or younger child opens the door.  So there are &lt;em&gt;eight&lt;/em&gt; equally likely outcomes, and by considering them it's easy to get 1/2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to mention a related game, which I believe I first saw in Martin Gardner's &lt;em&gt;Aha! Gotcha&lt;/em&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have three cards; one is black on both sides, one is red on both sides, and one is black on one side and red on the other.  I will let you select a card at random and look at one side, and I'll bet you even money that the other side is the same color.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This sounds fair.  Suppose you see black; now the card is either the black-red (which would win the game for you) or black-black (which wins for me), and at the outset, you were equally likely to have picked either one.  But in fact I win the game 2/3 of the time.  As with the puzzle, considering &lt;em&gt;all&lt;/em&gt; possible outcomes, there are 6 total &lt;em&gt;sides&lt;/em&gt; of cards.  You are equally likely to be looking at any of the 3 black sides, but only one of them (the black side of the black-red card) wins the game for you.  Another argument is to note that if you pick the black-black card, you are twice as likely to see a black side as if you pick the black-red card, so I am twice as likely to win.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I recommend you find this interviewer and offer to play this game with him.  You could even sweeten the odds a little to make it sound more attractive.  With high probability, you can clean him out.  Don't forget to kick back a &quot;finder's fee&quot; to your friend, or at least a nice dinner (as consolation for the job).&lt;/p&gt;&#xA;" OwnerUserId="822" LastEditorUserId="822" LastEditorDisplayName="" LastEditDate="2010-08-18T20:31:54.450" LastActivityDate="2010-08-18T20:31:54.450" CommentCount="3" />
  <row Id="2734" PostTypeId="2" ParentId="2731" CreationDate="2010-08-18T20:13:58.283" Score="3" ViewCount="" Body="&lt;p&gt;Let A be the event &quot;Family has at least one girl&quot;.   Let B be the event &quot;Boy opens door&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pr(A and B)=1/4 *1/2 + 1/4*1/2 = 1/4&lt;/p&gt;&#xA;&#xA;&lt;p&gt;AND &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pr(B)=1*1/4 + 1/4*1/2 +1/4*1/2 = 1/2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So Pr(A|B) = 1/4 / ( 1/2) = 1/2.&lt;/p&gt;&#xA;" OwnerUserId="832" LastActivityDate="2010-08-18T20:13:58.283" />
  <row Id="2735" PostTypeId="1" AcceptedAnswerId="2748" CreationDate="2010-08-18T20:17:36.730" Score="3" ViewCount="371" Body="&lt;p&gt;Say you have a very dense matrix that is 30000x30000 elements. The very dense matrix comes from the radiosity equation, which I discussed &lt;a href=&quot;http://math.stackexchange.com/questions/2180/how-can-i-characterize-the-type-of-solution-vector-that-comes-out-of-a-matrix&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Say you have Ax = B.  You have B, and A is 30000x30000 elements.  Solving:  You need to find x given B.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solving a matrix like this in C code actually introduces an interesting problem:  you cannot actually store it all in memory at once.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, I want to know about ways to solve this problem &quot;in pieces&quot; -- I want to mathematically decompose the 30000x30000 matrix into smaller sub-matrices (perhaps in chunks of 30000x20?), and somehow solve that way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What algorithms exist to break down / solve a matrix &quot;in pieces&quot; or steps, that can be solved regardless of memory restrictions?  It can be an iterative technique.&lt;/p&gt;&#xA;" OwnerUserId="1069" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-08-20T14:51:31.840" LastActivityDate="2011-05-12T12:05:01.077" Title="Solving very large matrices in &quot;pieces&quot;" Tags="&lt;linear-algebra&gt;&lt;algorithms&gt;&lt;matrices&gt;" AnswerCount="4" CommentCount="11" />
  <row Id="2736" PostTypeId="2" ParentId="2731" CreationDate="2010-08-18T20:18:02.937" Score="3" ViewCount="" Body="&lt;p&gt;(Pasting my mathoverflow answer...)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We will assume all the obvious implicit assumptions (eg. random child being boy of girl is 50/50, boys and girls open the door uniformly, etc.).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you had a slightly different question, i.e. if you asked the couple if they have at least one boy, and the answer is yes, then the chance of the other one being a girl is 2/3. Intuitively, the probability is not 1/2 because in this case the answer depends on &lt;em&gt;both&lt;/em&gt; the children, i.e. it is a function of both of them considered together.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if you asked the couple to pick a child at random, then she/he bears no information about the other child, and consequently his/her gender does not give you any information about the sibling.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your case is the second case, where the child opening the door is selected at random, and she happens to be female. This does not bear any information regarding the other child.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So answer is 1/2 and &lt;strong&gt;your friend is correct&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mathematically,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$P(Other\ is\ B|G\ opens\ door) = P(BG|G\ opens\ door) =$&#xA;$\frac{P(BG\ and\ G\ opens)}{P(GG\ and\ G\ opens\ door) + P(BG\ and\ G\ opens\ door)} = \frac{1/2*1/2}{1/4+1/2*1/2} = 1/2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This can be graphically understood as below -&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://imgur.com/AcS9S.png&quot; /&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;As a digression, a twist in the question can be brought about - if you take probabilities for a boy and girl to be different for opening the door.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Eg. suppose boy opens with probability $p$, girl with $q=1-p$, in a family with BG.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then $P(Other\ is\ B|G\ opened\ door) = P(BG)/P(G\ opened\ door) = $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{P(BG\ and\ G\ opens)}{P(GG\ and\ G\ opens\ door) + P(BG\ and\ G\ opens\ door)} = \frac{1/2*q&#xA;}{1/4+1/2*q} = \frac{2q}{2q+1}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(This means $q = 0$ implies $P(BG|G\ opens)=0$. That makes sense, since girls don't open the door if there is a boy, so definitely the other one is girl too.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Table added.&lt;/p&gt;&#xA;" OwnerUserId="1310" LastEditorUserId="1310" LastEditorDisplayName="" LastEditDate="2010-08-18T21:52:37.230" LastActivityDate="2010-08-18T21:52:37.230" />
  <row Id="2737" PostTypeId="2" ParentId="2694" CreationDate="2010-08-18T20:25:47.257" Score="12" ViewCount="" Body="&lt;p&gt;I do not think this is a conceptually important problem.  It is an example of a down-to-earth problem that can be checked numerically up to a large value and it has resisted a solution for many years.  Not all such problems are automatically important (e.g., nonexistence of odd perfect numbers).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An analogy with the significance of Fermat's last theorem is apt. Before the link was made between FLT and deep conjectures of elliptic curves, there was no over-arching significance to knowing whether or not FLT was true. (I think the link between FLT and the abc-conjecture was made at around the same time.) Yes, work on FLT was responsible for useful developments in algebraic number theory, but all the same it was not clear for a long time that settling the problem, or rather finding a counterexample, would have any other repercussions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If tomorrow someone showed that the Collatz conjecture were a consequence of the abc-conjecture or some other recognizably important unsolved problem, then I would change my mind about its importance (because, as with the link between the modularity conjecture and FLT, a counterexample to Collatz would then have real implications elsewhere in mathematics).  But as long as it stays isolated, having no implications to other problems, I don't think on its own it is a mathematically profound question. The same goes for odd perfect numbers: unless someone shows the existence of an odd perfect number has effects elsewhere that we do not expect (like a counterexample to FLT having a very unexpected implication for elliptic curves), I don't think the mainstream would consider odd perfect numbers to be important either. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the pedagogical side, however, I will definitely grant that this is a nice problem to show students unfamiliar with advanced mathematics that there really are unsolved math problems. People don't necessarily realize this, e.g., they may think that everything can be solved by computers or something. &lt;/p&gt;&#xA;" OwnerUserId="619" LastActivityDate="2010-08-18T20:25:47.257" CommunityOwnedDate="2010-08-18T20:25:47.257" CommentCount="2" />
  <row Id="2738" PostTypeId="2" ParentId="2626" CreationDate="2010-08-18T20:57:35.183" Score="1" ViewCount="" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;When your analysis is on a simple basis where the sail can be modeled as a plane, the force will be perpendicular to the plane.  If you need a 3-D model of sail, it's more complex.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If you take what might be called the 20th century model in sailboat design, the force on the sail is resolved into a vector perpendicular to the approaching wind (lift) and a force parallel to the wind (drag). Likewise the forces on the hull are resolve to lift perpendicular to the direction of travel, and drag parallel to the direction of travel. You can find diagrams like this going back to Manfred Curry. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I don't know about square riggers, but for the normal sloop rig, the jib and main interact too closely to consider them as independent. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;One math professor who wrote about sailing was c. stanley ogilvy. I wouldn't rush out to buy his books, but if you can find them in a library, you might find them interesting. The most widely available heavy duty treatments are by Marchaj (http://en.wikipedia.org/wiki/Czes%C5%82aw_Marchaj).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="1312" LastEditorUserId="1312" LastEditorDisplayName="" LastEditDate="2010-08-18T21:45:12.413" LastActivityDate="2010-08-18T21:45:12.413" />
  <row Id="2739" PostTypeId="1" AcceptedAnswerId="2743" CreationDate="2010-08-18T21:00:44.567" Score="3" ViewCount="500" Body="&lt;p&gt;Problem 63 of the 2001 St. Petersburg Mathematical Olympiad, Second Round, 11th grade. Are there three different numbers x, y, z in [0,pi/2] such that the numbers sin x, sin y, sin z, cos x, cos y, cos z can be divided into three pairs with equal sums?&lt;/p&gt;&#xA;" OwnerUserId="1313" LastEditorUserId="1313" LastEditorDisplayName="" LastEditDate="2010-08-19T18:58:25.337" LastActivityDate="2010-08-19T18:58:25.337" Title="sin x, sin y, sin z, cos x, cos y, cos z in [0,pi/2]" Tags="&lt;trigonometry&gt;" AnswerCount="1" CommentCount="0" />
  <row Id="2743" PostTypeId="2" ParentId="2739" CreationDate="2010-08-18T22:10:58.343" Score="4" ViewCount="" Body="&lt;p&gt;Without loss of generality each of $x$, $y$, $z$ are $\le\pi/4$, since if for example $x\gt\pi/4$, then we can replace $x$ with $\pi/2-x$ which will not change the set $\{\cos(x),\sin(x)\}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So now $0\lt x \lt y \lt z \le \pi/4$, and we have the following ordering:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sin(x)\lt\sin(y)\lt\sin(z)\lt\cos(z)\lt\cos(y)\lt\cos(x)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence the only possible pairings must be:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$A=\sin(x)+\cos(x)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$B=\sin(y)+\cos(y)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$C=\sin(z)+\cos(z)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With $A=B=C$ &#xA;(any other pairing will make one side heavier, term by term).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now note $f(t)=\sin(t)+\cos(t)$ is a concave function since $\frac{d^2f(t)}{dt^2}=-\sin(t)-\cos(t)&amp;lt;0$ for $t \in (0,\pi/4)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So it follows that $B\gt \frac{A+C}{2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So they can never be equal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: Fixed glitch spotted by Tom&lt;/p&gt;&#xA;" OwnerUserId="1310" LastEditorUserId="1310" LastEditorDisplayName="" LastEditDate="2010-08-19T16:44:30.277" LastActivityDate="2010-08-19T16:44:30.277" CommentCount="13" />
  <row Id="2744" PostTypeId="1" AcceptedAnswerId="2749" CreationDate="2010-08-18T22:24:17.720" Score="2" ViewCount="190" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Possible Duplicate:&lt;/strong&gt;&lt;br&gt;&#xA;  &lt;a href=&quot;http://math.stackexchange.com/questions/2561/behaviour-of-polynomials-in-a-pid&quot;&gt;Behaviour of Polynomials in a PID!&lt;/a&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;p&gt;Prove: if $p$ is a prime, and if $n^2+n+p$ is prime for $0\leq n \leq \sqrt{p/3}$, then it is also prime for $0 \leq n \leq p-2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This appeared on reddit recently, but no proof was posted. With $p=41$, it is Euler's famous prime-generating polynomial.&lt;/p&gt;&#xA;" OwnerUserId="1320" LastActivityDate="2010-08-18T22:57:56.430" ClosedDate="2010-08-19T17:06:06.377" Title="When is $n^2+n+p$ prime?" Tags="&lt;number-theory&gt;" AnswerCount="1" CommentCount="0" />
  <row Id="2745" PostTypeId="2" ParentId="2592" CreationDate="2010-08-18T22:24:44.833" Score="1" ViewCount="" Body="&lt;p&gt;Well I tried for $\frac{d^2y}{dx^2}y=1$, and I have partial solution. I will use this reply to trace the work since it supports LaTeX, I will try to complete it later (or someone else can take over).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose $x=f(y)$, so $y=f^{-1}(x)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then $dx=f'(y)dy$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or $\frac{dy}{dx}=\frac{1}{f'(y)}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So $\frac{d^2y}{dx^2}=\frac{d}{dy}\left(\frac{1}{f'(y)}\right)\frac{dy}{dx}=-\frac{f''(y)}{(f'(y))^2}\frac{dy}{dx}=-\frac{f''(y)}{(f'(y))^3}=-\frac{g'(y)}{(g(y))^3}$ where $g(y)=f'(y)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solving $-\frac{g'(y)}{(g(y))^3}=\frac{1}{y}$, we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$g(y)=f'(y)=\frac{1}{\sqrt{\log(y^2)+c}}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we need to solve for $f(y)$, and invert the function to get $y=f^{-1}(x)$.&lt;/p&gt;&#xA;" OwnerUserId="1310" LastActivityDate="2010-08-18T22:24:44.833" />
  <row Id="2746" PostTypeId="1" AcceptedAnswerId="2747" CreationDate="2010-08-18T22:49:43.523" Score="18" ViewCount="878" Body="&lt;p&gt;If $n&gt;1$ is an integer, then $\sum_{k=1}^n\frac 1k$ is not an integer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you know &lt;a href=&quot;http://en.wikipedia.org/wiki/Bertrand%27s_postulate&quot; rel=&quot;nofollow&quot;&gt;Bertrand's Postulate&lt;/a&gt;, then you know there must be a prime $p$ between $n/2$ and $n$, so $\frac 1p$ appears in the sum, but $\frac{1}{2p}$ does not. Aside from $\frac 1p$, every other term $\frac 1k$ has $k$ divisible only by primes smaller than $p$. We can combine all those terms to get $\sum_{k=1}^n\frac 1k = \frac 1p + \frac ab$, where $b$ is not divisible by $p$. If this were an integer, then (multiplying by $b$) $\frac bp +a$ would also be an integer, which it isn't since $b$ isn't divisible by $p$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anybody know an elementary proof of this which doesn't rely on Bertrand's Postulate? For a while, I was convinced I'd seen one, but now I'm starting to suspect whatever argument I saw was wrong.&lt;/p&gt;&#xA;" OwnerUserId="60" LastActivityDate="2011-02-28T06:04:58.010" Title="Is there an elementary proof that $∑_{k=1}^n 1/k$ is never an integer?" Tags="&lt;number-theory&gt;&lt;sequences-and-series&gt;" AnswerCount="4" CommentCount="3" FavoriteCount="5" />
  <row Id="2747" PostTypeId="2" ParentId="2746" CreationDate="2010-08-18T22:53:47.060" Score="15" ViewCount="" Body="&lt;p&gt;An elementary proof uses the following fact:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $2^s$ is the highest power of $2$ in the set $S = \{1,2,...,n\}$, then $2^s$ is not a divisor of any other integer in $S$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To use that, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;consider the highest power of $2$ which divides $n!$. Say that is $t$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the number can be rewritten as &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle \frac{\sum_{k=1}^{n}{\frac{n!}{k}}}{n!}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The highest power of $2$ which divides the denominator is $t$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the highest power of $2$ that divides $\displaystyle \frac{n!}{k}$ is at least $t-s$. If $k \neq 2^{s}$, then this is atleast $t-s+1$ as the highest power of $2$ that divides $k$ is atmost $s-1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In case $k=2^s$, the highest power of $2$ that divides $\displaystyle \frac{n!}{k}$ is exactly $t-s$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus the highest power of $2$ that divides the numerator is atmost $t-s$. If $s &gt; 0$ (which is true if $n &gt; 1$), we are done. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact the above proof shows that the number is of the form $\frac{odd}{even}$.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-09-07T15:16:57.343" LastActivityDate="2010-09-07T15:16:57.343" CommentCount="2" />
  <row Id="2748" PostTypeId="2" ParentId="2735" CreationDate="2010-08-18T22:55:00.147" Score="7" ViewCount="" Body="&lt;p&gt;The first question that should be asked of any person presenting you with a large-ish matrix: &quot;is the matrix dense or sparse?&quot; In the latter case, one definitely does not need to allocate space for the zero entries, and should thus look into special techniques  to storing them(which as I understand nowadays rely on a liberal dose of graph theory in the general case, though band matrices are still handled by storing their bands in an appropriate format).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if even after that you have the perfect excuse for having a large dense matrix (which I still believe is quite unlikely), there is a way to invert and take the determinant of a large matrix via partitioning.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Say we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\textbf{A}=\begin{pmatrix}\textbf{E}&amp;amp;\textbf{F}\\ \textbf{G}&amp;amp;\textbf{H}\end{pmatrix}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\textbf{E}$ and $\textbf{H}$ are square matrices with dimensions $m\times m$ and $n\times n$ respectively, and $\textbf{F}$ and $\textbf{G}$ are dimensioned appropriately (so the dimension of $\textbf{A}$ is $(m+n)\times(m+n)$). The inverse can then be computed as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\textbf{A}^{-1}=\begin{pmatrix}\textbf{E}^{-1}+\left(\textbf{E}^{-1}\textbf{F}\right)(\textbf{H}-\textbf{G}\textbf{E}^{-1}\textbf{F})^{-1}\left(\textbf{G}\textbf{E}^{-1}\right)&amp;amp;-\left(\textbf{E}^{-1}\textbf{F}\right)(\textbf{H}-\textbf{G}\textbf{E}^{-1}\textbf{F})^{-1}\\-(\textbf{H}-\textbf{G}\textbf{E}^{-1}\textbf{F})^{-1}\left(\textbf{G}\textbf{E}^{-1}\right)&amp;amp;(\textbf{H}-\textbf{G}\textbf{E}^{-1}\textbf{F})^{-1}\end{pmatrix}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where the parentheses emphasize common factors that you might profit from computing only once.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the determinant, it is given by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\det\;\textbf{A}=\det\left(\textbf{H}-\textbf{G}\textbf{E}^{-1}\textbf{F}\right)\;\det\;\textbf{E}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not entirely sure what seemed to be unsatisfactory with this answer, so here's a bit more of exposition: as always, the &quot;inverse&quot; here is merely notation! One would most certainly first perform LU decomposition on $\textbf{E}$ and $\textbf{H}$ first. One would also partition the right-hand side $\textbf{b}$ accordingly:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\textbf{b}=\begin{pmatrix}\textbf{c}\\ \textbf{d}\end{pmatrix}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so $\textbf{c}$ is a length-m vector and $\textbf{d}$ is a length-n vector.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Formally multiplying the partitioned inverse with the partitioned right-hand side gives&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\begin{pmatrix}\textbf{E}^{-1}+\left(\textbf{E}^{-1}\textbf{F}\right)(\textbf{H}-\textbf{G}\textbf{E}^{-1}\textbf{F})^{-1}\left(\textbf{G}\textbf{E}^{-1}\right)&amp;amp;-\left(\textbf{E}^{-1}\textbf{F}\right)(\textbf{H}-\textbf{G}\textbf{E}^{-1}\textbf{F})^{-1}\\-(\textbf{H}-\textbf{G}\textbf{E}^{-1}\textbf{F})^{-1}\left(\textbf{G}\textbf{E}^{-1}\right)&amp;amp;(\textbf{H}-\textbf{G}\textbf{E}^{-1}\textbf{F})^{-1}\end{pmatrix}\cdot\begin{pmatrix}\textbf{c}\\ \textbf{d}\end{pmatrix}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which when expanded and simplified is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\begin{pmatrix}\textbf{E}^{-1}\textbf{c}+\left(\textbf{E}^{-1}\textbf{F}\right)(\textbf{H}-\textbf{G}\textbf{E}^{-1}\textbf{F})^{-1}\left(\textbf{G}\textbf{E}^{-1}\textbf{c}-\textbf{d}\right)\\-(\textbf{H}-\textbf{G}\textbf{E}^{-1}\textbf{F})^{-1}\left(\textbf{G}\textbf{E}^{-1}\textbf{c}-\textbf{d}\right)\end{pmatrix}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At this point you should be able to figure out how you would use an available decomposition of $\textbf{E}$ or $\textbf{H}$, and which common subexpressions can be just computed once.&lt;/p&gt;&#xA;" OwnerUserId="498" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-08-20T13:01:28.767" LastActivityDate="2010-08-20T13:01:28.767" CommentCount="7" />
  <row Id="2749" PostTypeId="2" ParentId="2744" CreationDate="2010-08-18T22:57:56.430" Score="1" ViewCount="" Body="&lt;p&gt;This follows by employing in &lt;a href=&quot;http://math.stackexchange.com/questions/2590&quot;&gt;Rabinowitsch's&lt;/a&gt; proof a Gauss bound, e.g. see Theorem 9.1 &lt;a href=&quot;http://www.fen.bilkent.edu.tr/~franz/ant-st.pdf#page=18&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-08-18T22:57:56.430" CommentCount="1" />
  <row Id="2750" PostTypeId="2" ParentId="2592" CreationDate="2010-08-18T23:03:53.050" Score="5" ViewCount="" Body="&lt;p&gt;Let $y=y(x)$ and write primes for derivatives w.r.t $x$. In the equation $$y''y=1$$ the independent variable $x$ does not appear, so we can do a (somewhat magical) change of variables: in the new equation, the &lt;em&gt;independent&lt;/em&gt; variable will be $y$ and the dependent variable will be $p=y'$. If we use dots to denote derivatives with respect to $y$, a little computation using the chain rule shows that $p''=p\dot p$, so the equation becomes $$\dot ppy=1.$$ This can be rewritten as $$p\,\mathrm dp=\frac{\mathrm dy}{y},$$ which has its variables separated. It can be integrated at once, to get the relation $$\frac{p^2}{2}=\log y+c$$ with $c$ a constant. Now we recall what $p$ was, that is $y'$, and this becomes a new (first order!) ODE on $y(x)$. We have «simplified» things... (We could have obtained the same reduction od order by multiplying the original equation by $y'$, dividing it by $y$ and integrating: the trick with the $p$ substitution is, though, a general method)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same approach will similarly simplify the general case.&lt;/p&gt;&#xA;" OwnerUserId="274" LastEditorUserId="274" LastEditorDisplayName="" LastEditDate="2010-08-18T23:11:46.453" LastActivityDate="2010-08-18T23:11:46.453" />
  <row Id="2752" PostTypeId="2" ParentId="2746" CreationDate="2010-08-18T23:28:43.727" Score="13" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\ $ Since there is a unique denominator $\rm\:\color{red} {2^K}\:$ having maximal power of $2$, upon multiplying all terms through by $\rm\:2^{K-1}$ one deduces the contradiction that $\rm\ 1/2\ =\ c/d \;$ with $\rm\: d \:$ odd. For example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\quad\quad\quad\quad\quad\quad \color{green}{m}\ =\ \:  1\: +\: \frac{1}{2} + \frac{1}{3} +\: \color{red}{\frac{1}{4}} \;+\; \frac{1}{5} + \frac{1}{6} + \frac{1}{7} $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\quad\quad\Rightarrow\quad\;\; \color{green}{2m}\ =\ 2 +\; 1 + \frac{2}{3} +\; \color{red}{\frac{1}{2}} \;+\; \frac{2}{5} + \frac{1}{3} + \frac{2}{7} $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\quad\quad\Rightarrow\quad -\color{red}{\frac{1}{2}}\  =\ \:2 +\: 1 + \frac{2}{3} - \color{green}{2m}  + \frac{2}{5} + \frac{1}{3} + \frac{2}{7}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The prior sum has all odd denominators so reduces to a fraction with odd denominator $\rm\:d\ |\ 3\cdot 5\cdot 7$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Note&lt;/strong&gt; $\ $ I purposely avoided any use of valuation theory because Anton requested an &quot;elementary&quot; solution. The above proof can easily be made comprehensible to a high-school student.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2011-02-28T06:04:58.010" LastActivityDate="2011-02-28T06:04:58.010" />
  <row Id="2755" PostTypeId="1" AcceptedAnswerId="2761" CreationDate="2010-08-19T00:08:12.373" Score="127" ViewCount="26004" Body="&lt;p&gt;My nephew was folding laundry, and turning the occasional shirt right-side-out. I showed him a  &quot;trick&quot; where I turned it right-side-out by pulling the whole thing through a sleeve instead of the bottom or collar of the shirt. He thought it was really cool (kids are easily amused, and so am I).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So he learned that you can turn a shirt or pants right-side-out by pulling the material through any hole, not just certain ones. I told him that even if there was a rip in the shirt, you could use that to turn it inside-out or right-side-out, and he was fascinated by this and asked &quot;why?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't really know the answer to this. Why is this the case? What if the sleeves of a long-sleeve shirt were sewn together at the cuff, creating a continuous tube from one sleeve to the other? Would you still be able to turn it right-side-out? Why? What properties must a garment have so that it can be turned inside-out and right-side-out?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sorry if this is a lame question, but I've always wondered. I wouldn't even know what to google for, so that is why I am asking here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;If you know the answer to this, could you please put it into layman's terms?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update:&lt;/strong&gt; Wow, I really appreciate all the participation. This is a really pleasant community and I have learned a lot here. It seems that the answer is that you need at least one puncture in the garment through which to push or pull the fabric. It appears that you can have certain handles, although it's not usually practical with clothing due to necessary stretching.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Accepted (a while ago actually -- sorry for not updating sooner) Dan's answer because among the answers that I understand, it is the highest ranked by this community.&lt;/p&gt;&#xA;" OwnerUserId="1319" LastEditorUserId="1319" LastEditorDisplayName="" LastEditDate="2010-09-03T17:02:47.693" LastActivityDate="2010-12-06T12:33:56.443" Title="Why can you turn clothing right-side-out?" Tags="&lt;topology&gt;&lt;algebraic-topology&gt;&lt;general-topology&gt;" AnswerCount="6" CommentCount="12" FavoriteCount="59" />
  <row Id="2756" PostTypeId="2" ParentId="2746" CreationDate="2010-08-19T00:08:20.847" Score="8" ViewCount="" Body="&lt;p&gt;What the heck -- I'll leave my comment as an answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See the Example on p. 13 of &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://math.uga.edu/~pete/4400intro.pdf&quot; rel=&quot;nofollow&quot;&gt;http://math.uga.edu/~pete/4400intro.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is discussed, together with (as a footnote) the strange phenomenon that this is often solved by an appeal to Bertrand's Postulate.  The discussion in the above text is intended to be &quot;didactic&quot; in that a few details are left to the reader, and I recommend it as a good exercise to flesh them out.  &lt;/p&gt;&#xA;" OwnerUserId="299" LastActivityDate="2010-08-19T00:08:20.847" CommentCount="6" />
  <row Id="2757" PostTypeId="2" ParentId="2601" CreationDate="2010-08-19T00:09:10.703" Score="5" ViewCount="" Body="&lt;p&gt;The number line. For whatever elementary school grade it is, it illustrates why you need zero, and why you need negative numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The venerable cake, or probably pizza nowadays, to illustrate fractions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Riemann Sphere. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The coin flip as a prototypical random event.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Markov chain as a frog hopping from lily pad to lily pad.&lt;/p&gt;&#xA;" OwnerUserId="1312" LastEditorUserId="1312" LastEditorDisplayName="" LastEditDate="2010-08-19T18:24:14.407" LastActivityDate="2010-08-19T18:24:14.407" CommunityOwnedDate="2010-08-19T00:09:10.703" CommentCount="2" />
  <row Id="2758" PostTypeId="2" ParentId="2755" CreationDate="2010-08-19T00:29:24.437" Score="6" ViewCount="" Body="&lt;p&gt;A shirt with no hole, or all holes sealed is homeomorphic to a sphere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A shirt with holes is homeomorphic to a sphere with punctures.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now you intuitively see that a rubber sphere with punctures can be turned inside out by continuous motion, whereas a perfect sphere cannot.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To understand why, let's concentrate on 2-dimentional plane, taking a circle instead of a sphere. It can be proven that any continuous deformation that inverts the circle will make it cross itself. However if the circle has 'holes', or gaps, then it topologically is equivalent to one or more line segments, which can all be inverted in place without touching each other (appropriate scaling may be necessary).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similar feature holds for a sphere too. Hence you cannot turn a shirt inside out if all the holes are sealed, but you can do it even if you have one hole (and provided that the shirt is flexible enough to go through that hole if the hole is small).&lt;/p&gt;&#xA;" OwnerUserId="1310" LastActivityDate="2010-08-19T00:29:24.437" CommentCount="4" />
  <row Id="2759" PostTypeId="2" ParentId="2601" CreationDate="2010-08-19T00:41:46.580" Score="0" ViewCount="" Body="&lt;p&gt;I can remember the following at school level -&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Cartesian coordinates. Pythagoras theorem. (Metaphor - distance from the wall or from the edge of the floor.)&lt;/li&gt;&#xA;&lt;li&gt;Variables used in basic algebra. (Metaphor - common nouns, eg. &quot;an &lt;em&gt;animal&lt;/em&gt; is green and has two legs...&quot;)&lt;/li&gt;&#xA;&lt;li&gt;Irrationals and the completeness axiom. (Metaphor - can zoom in the real line without limit.)&lt;/li&gt;&#xA;&lt;li&gt;Complex number and complex plane. (&quot;Add just solution to $x^2+1=0$ to the real line, suddenly every polynomial is solvable!&quot;)&lt;/li&gt;&#xA;&lt;li&gt;Sets, relations. (Venn diagrams are helpful for sets.)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="1310" LastActivityDate="2010-08-19T00:41:46.580" CommunityOwnedDate="2010-08-19T00:41:46.580" CommentCount="1" />
  <row Id="2760" PostTypeId="2" ParentId="2755" CreationDate="2010-08-19T00:42:01.497" Score="10" ViewCount="" Body="&lt;p&gt;I can at least answer the question about the rip, and point you to some terms to look up.  What you are looking at is not geometry as it is usually understood but &lt;a href=&quot;http://en.wikipedia.org/wiki/Topology&quot; rel=&quot;nofollow&quot;&gt;topology&lt;/a&gt; (particularly of &lt;a href=&quot;http://en.wikipedia.org/wiki/Surface&quot; rel=&quot;nofollow&quot;&gt;surfaces with boundary&lt;/a&gt;), which informally is the study of properties of things that are invariant under deformation.  It is often said that a topologist can't tell a coffee mug from a doughnut (it is possible to deform one into the other) and this is the attitude you should take towards questions of this kind.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From this perspective, a rip is the same thing as the bottom of the shirt is the same thing as a sleeve: you can deform each of them into the other, so topologically they're the same thing.  (Imagine making the rip bigger, or squashing a long sleeve into a short sleeve.)  They are all just examples of what topologists call punctures.    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know the answer to your general question, but one property a surface must have to be able to be turned inside out is that it must be &lt;a href=&quot;http://en.wikipedia.org/wiki/Orientability&quot; rel=&quot;nofollow&quot;&gt;orientable&lt;/a&gt;; in other words, it must have an inside and an outside to begin with!  Surfaces such as the &lt;a href=&quot;http://en.wikipedia.org/wiki/M%C3%B6bius_strip&quot; rel=&quot;nofollow&quot;&gt;Mobius strip&lt;/a&gt; famously don't have this property, so it doesn't even make sense to ask about whether it's possible to turn such things inside-out or not.  (And some non-orientable surfaces can't even be realized in three dimensions...)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Orientable surfaces happen to have a particularly simple description: they are all topologically just spheres with some &lt;a href=&quot;http://content.answcdn.com/main/content/img/McGrawHill/Encyclopedia/images/CE701100FG0010.gif&quot; rel=&quot;nofollow&quot;&gt;handles sewn on&lt;/a&gt; and with some punctures.  To turn them inside out, at least one puncture is necessary, and if you can turn one handle inside-out then you can presumably turn all of them inside-out.  If what I said in the comments is wrong and you can't turn handles inside-out, then a necessary and sufficient condition is that there is at least one puncture and no handles.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-08-19T00:42:01.497" CommentCount="10" />
  <row Id="2761" PostTypeId="2" ParentId="2755" CreationDate="2010-08-19T00:42:56.970" Score="41" ViewCount="" Body="&lt;p&gt;Everything I'm wearing is a topological sphere with holes (t-shirts have 4, pants 3, shoes and socks 1) in which cases any hole works.  Instead of a long-sleeve shirt with the arms sewn together, consider a pair of pants with legs sewn together to form a topological torus with a hole (so if you were to wear them your feet would be touching and it would be impossible to put your shoes on).  This pair of pants has two parameters which are roughly constant, the circumference of the leg and the total length of the two legs.  When it is turned inside out through the waist, these parameters swap roles, so you will have a tube about the length of a pant leg with an opening on each end, the same as if you had turned one leg inside-out and pushed the other leg through it prior to sewing.  I think this would be possible to do with real toroidal clothing as long as it is thin enough, because the process doesn't require any stretching.&lt;/p&gt;&#xA;" OwnerUserId="1284" LastEditorUserId="1284" LastEditorDisplayName="" LastEditDate="2010-08-19T03:38:12.677" LastActivityDate="2010-08-19T03:38:12.677" CommentCount="16" />
  <row Id="2762" PostTypeId="2" ParentId="2485" CreationDate="2010-08-19T01:06:22.040" Score="1" ViewCount="" Body="&lt;p&gt;The answer to this is yes! Let $g\colon\mathbb{R}\to\mathbb{R}$ be essentially unbounded on every nontrivial open interval and set $f(x,y)=g(x+y)$. Then, f is essentially unbounded on every measurable rectangle $A\times B$ of positive Lebesgue measure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To construct g you can take $g_0(x)=1_{\{1&gt;x&gt;0\}}x^{-1/2}$ which is integrable but not square integrable on any neighborhood of 0. Then, letting $q_1,q_2,\ldots$ be an enumeration of the rationals, $g(x)=\sum_n2^{-n}g_0(x-q_n)$ is integrable (so, is finite almost everywhere) but not square integrable on any open interval, so is not essentially bounded. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To see that this construction of f works, pick any two sets $A,B\subseteq\mathbb{R}$ with positive Lebesque measure. By intersecting with a bounded interval, we can restrict to sets with finite measure. I'll write $I_A(x)$ for the indicator function of a set A and $(I_A*I_B)(x)=\int I_A(x-y)I_B(y)\,dy$ for the convolution, which is a continuous function. For any $K&gt;0$ let $C=\{x\in\mathbb{R}\colon\vert g(x)\vert\ge K\}$, which has positive measure in any nontrivial open interval. Then,&#xA;$$&#xA;\begin{align}&#xA;\mu((A\times B)\cap \{x\in\mathbb{R}^2\colon \vert f(x)\vert\ge K\})&amp;amp;=\int\int I_A(x)I_B(y)I_C(x+y)\,dx\,dy\\&#xA;&amp;amp;=\int (I_A*I_B)(x)I_C(x)\,dx.&#xA;\end{align}&#xA;$$&#xA;The convolution $I_A*I_B$ is a nonnegative function with positive integral $\int (I_A*I_B)(x)\,dx=\mu(A)\mu(B)&gt;0$, so it must be strictly positive at some point. By continuity, it is bounded below by a positive number on some open interval, so the integral displayed above is positive. Therefore, f is essentially unbounded on $A\times B$.&lt;/p&gt;&#xA;" OwnerUserId="1321" LastActivityDate="2010-08-19T01:06:22.040" CommentCount="2" />
  <row Id="2763" PostTypeId="1" AcceptedAnswerId="2764" CreationDate="2010-08-19T01:12:52.383" Score="5" ViewCount="197" Body="&lt;p&gt;Standard deck has 52 cards, 26 Red and 26 Black. A run is a maximum contiguous block of cards, which has the same color.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Eg.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;(R,B,R,B,...,R,B) has 52 runs.&lt;/li&gt;&#xA;&lt;li&gt;(R,R,R,...,R,B,B,B,...,B) has 2 runs.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;What is the expected number of runs in a shuffled deck of cards?&lt;/p&gt;&#xA;" OwnerUserId="1310" LastActivityDate="2010-08-19T04:02:37.850" Title="What is the expected number of runs of same color in a standard deck of cards?" Tags="&lt;probability-theory&gt;&lt;statistics&gt;&lt;probability&gt;" AnswerCount="1" />
  <row Id="2764" PostTypeId="2" ParentId="2763" CreationDate="2010-08-19T01:32:50.857" Score="8" ViewCount="" Body="&lt;p&gt;The expected number of runs is 27.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $X_n$ be the color of the n'th card. For n&amp;lt;52, the n'th card is the end of a run if and only if $X_n\not=X_{n+1}$ and the last card in the pack is always the end of a run. So, the total number of runs is&#xA;$$&#xA;N=\sum_{n=1}^{51}1_{\{X_n\not=X_{n+1}\}}+1.&#xA;$$&#xA;The expected number of runs is&#xA;$$&#xA;\mathbb{E}[N]=\sum_{n=1}^{51}\mathbb{P}(X_n\not=X_{n+1})+1.&#xA;$$&#xA;Whatever colour the n'th card is, there are 51 remaining cards in the deck of which 26 of them are a different colour from $X_n$. So, $\mathbb{P}(X_n\not=X_{n+1})=26/51$, giving&#xA;$$&#xA;\mathbb{E}[N]=51 (26/51) + 1=27.&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="1321" LastActivityDate="2010-08-19T01:32:50.857" CommentCount="2" />
  <row Id="2769" PostTypeId="2" ParentId="2746" CreationDate="2010-08-19T03:10:34.177" Score="8" ViewCount="" Body="&lt;p&gt;I never heard of the Bertrand postulate approach before. Anton, the argument for the n-th harmonic sum not being an integer when n &gt; 1 goes back to Taeisinger in 1915. In fact, the n-th harmonic sum tends to infinity 2-adically. This naturally raises the question of the p-adic behavior of harmonic sums for odd primes p, which quickly leads into unsolved problems. I wrote a discussion of that at &lt;a href=&quot;http://www.math.uconn.edu/~kconrad/blurbs/ugradnumthy/padicharmonicsum.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.math.uconn.edu/~kconrad/blurbs/ugradnumthy/padicharmonicsum.pdf&lt;/a&gt;. &lt;/p&gt;&#xA;" OwnerUserId="619" LastActivityDate="2010-08-19T03:10:34.177" CommentCount="1" />
  <row Id="2771" PostTypeId="2" ParentId="2755" CreationDate="2010-08-19T04:21:20.650" Score="64" ViewCount="" Body="&lt;p&gt;First, a warning.  I suspect this response is likely not going to be immediately comprehensibe.  There is a formal set-up for your question, there are tools available to understand what's going on.  They're not particularly light tools, but they exist and they're worthy of being mentioned.  Before I write down the main theorem, let me set-up some terminology.  The tools belong to a subject called &lt;em&gt;manifold theory&lt;/em&gt; and &lt;em&gt;algebraic topology&lt;/em&gt;.  The names of the tools I'm going to use are called things like: the isotopy extension theorem, fibre-bundles, fibrations and homotopy-groups.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have a surface $\Sigma$, it's your shirt or whatever else you're interested in, some surface in 3-dimensional space.  Surfaces have automorphism groups, let me call it $Aut(\Sigma)$.  These are, say, all the self-homeomorphisms or diffeomorphisms of the surface.   And surfaces can sit in space.  A way of putting a surface in space is called an embedding.  Let's call all the embeddings of the surface $Emb(\Sigma, \mathbb R^3)$.  $Emb(\Sigma, \mathbb R^3)$ is a set, but in the subject of topology these sets have a natural topology as well.  We think of them as a space where &quot;nearby&quot; embeddings are almost the same, except for maybe a little wiggle here or there.  The topology on the set of embeddings is called the compact-open topology (see Wikipedia, for details on most of these definitions). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Okay, so now there's some formal nonsense.  Look at the quotient space $Emb(\Sigma, \mathbb R^3)/Aut(\Sigma)$.  You can think of this as all ways $\Sigma$ can sit in space, but without any labelling -- the surface has no parametrization.  So it's the space of all subspaces of $\mathbb R^3$ that just happen to be homeomorphic to your surface. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Richard Palais has a really nice theorem that puts this all into a pleasant context.  The preamble is we need to think of everything as living in the world of smooth manifolds -- smooth embeddings, $Aut(\Sigma)$ is the diffeomorphism group of the surface, etc. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are two locally-trivial fibre bundles (or something more easy to prove -- Serre fibrations), this is the &quot;global&quot; isotopy-extension theorem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$Diff(\mathbb R^3, \Sigma) \to Diff(\mathbb R^3) \to Emb(\Sigma, \mathbb R^3)/Aut(\Sigma)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$Diff(\mathbb R^3 fix \Sigma) \to Diff(\mathbb R^3, \Sigma) \to Aut(\Sigma)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;here $Diff(\mathbb R^3)$ indicates diffeomorphisms of $\mathbb R^3$ that are the identity outside of a sufficiently large ball, say.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the Palais theorem, once you apply the homotopy long exact sequence of a fibration is giving you a language that allows you to translate between automorphisms of your surface, and motions of the surface in space. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's a theorem of Jean Cerf's that $Diff(\mathbb R^3)$ is connected.  So a little diagram chase says that you that an automorphism of a surface can be realized as by a motion of that surface in 3-space if and only if that automorphism of the surface extends to an automorphism of 3-space.  For closed surfaces, the Jordan-Brouwer separation theorem gives you an obstruction to turning your surface inside-out.  But for non-closed surfaces you're out of tools.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To figure out if you can realize an automorphism as a motion, you literally have to try to extend it &quot;by hands&quot;.  This is a very general phenomena -- you have one manifold sitting in another, but rarely does an automorphism of the submanifold extend to the ambient manifold.  You see this phenomena happening in various other branches of mathematics as well -- an automorphism of a subgroup does not always extend to the ambient group, etc. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you try your luck and try to build the extension yourself.  In some vague sense that's a formal analogy between the visceral mystery of turning the surface inside-out and a kind of formalized mathematical problem, but of a fundamentally analogous feel. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We're looking for automorphisms that reverse orientation. For an arbitrary surface with boundary in 3-space, it's not clear if you &lt;em&gt;can&lt;/em&gt; turn the surface inside out.  This is because the surface might be knotted. Unknotted surfaces are examples like your t-shirt.  Let's try to cook up something that &lt;em&gt;can't&lt;/em&gt; be turned inside-out. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The automorphism group of a 3-times punctured sphere has 12 path-components (12 elements up to isotopy).  There are 6 elements that preserve orientation, and 6 that reverse.  In particular the orientation-reversing automorphisms reverse the orientation of all the boundary circles.  So if you could come up with a knotted pair-of-pants (3-times punctured surface) so that its boundary circles did not admit a symmetry that reversed the orientations of all three circles simultaneously, you'd be done. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe this doesn't seem like a reduction to you, but it is. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, there are things called non-invertible knots: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Invertible_knot&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Invertible_knot&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So how do we cook up a knotted pair-of-pants from that? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's the idea.  The non-invertible knot in the link above is sometimes called $8_{17}$.  Here is another picture of it:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://katlas.org/wiki/8_17&quot; rel=&quot;nofollow&quot;&gt;http://katlas.org/wiki/8_17&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a variant on that. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://rybu.org/photo/mathish/webstuff/ribbonsurf.png&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Interpret this image as a ribbon of paper that has three boundary circles.  One boundary circle is unknotted.  One is $8_{17}$.  The other is some other knot. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It turns out that other knot isn't trivial, nor is it $8_{17}$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So why can't this knotted pair of pants be turned inside-out?  Well, the three knots are distinct, and $8_{17}$ can't be reversed. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason why I know the other knot isn't $8_{17}$?  It's a hyperbolic knot and it has a different ($4.40083...$) hyperbolic volume than $8_{17}$ ($10.9859...$). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;FYI: in some sense this is one of the simplest surfaces with non-trivial boundary that can't be turned inside-out.  All discs can be turned inside-out.  Similarly, all annuli (regardless of how they're knotted) can be turned inside-out. So for genus zero surfaces, 3 boundary components is the least you can have if you're looking for a surface that can't be turned inside-out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;edited to correct for Jason's comment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;comment added later: I suggest if you purchase a garment of this form you return it to the manufacturer. &lt;/p&gt;&#xA;" OwnerUserId="642" LastEditorUserId="642" LastEditorDisplayName="" LastEditDate="2010-08-19T07:50:27.730" LastActivityDate="2010-08-19T07:50:27.730" CommentCount="13" />
  <row Id="2772" PostTypeId="1" CreationDate="2010-08-19T04:25:04.937" Score="5" ViewCount="380" Body="&lt;p&gt;What exactly does blow up mean, when people say, for example, that a solution (to a pde (say)) blows up. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks. &lt;/p&gt;&#xA;" OwnerUserId="1127" LastEditorUserId="152" LastEditorDisplayName="" LastEditDate="2010-08-19T08:47:09.170" LastActivityDate="2010-08-20T02:09:13.277" Title="Blow up of a solution" Tags="&lt;soft-question&gt;&lt;terminology&gt;&lt;differential-equations&gt;" AnswerCount="6" CommentCount="5" FavoriteCount="1" />
  <row Id="2773" PostTypeId="2" ParentId="2703" CreationDate="2010-08-19T04:38:13.893" Score="1" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;EDITED&lt;/strong&gt;. The version I originally posted was wrong.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the second problem, use the fact that squares are non-negative to show that&#xA;$$|f(x,y)| \le 3|x|$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From which it follows (by using the sandwich theorem) that,&#xA;$$\lim_{(x,y) \to (0,0)} f(x,y)=0$$.&lt;/p&gt;&#xA;" OwnerUserId="1195" LastEditorUserId="1195" LastEditorDisplayName="" LastEditDate="2010-08-19T05:50:34.263" LastActivityDate="2010-08-19T05:50:34.263" />
  <row Id="2774" PostTypeId="2" ParentId="2772" CreationDate="2010-08-19T05:11:47.980" Score="-2" ViewCount="" Body="&lt;p&gt;It means that it costs more to solve the problem, relative to a smaller or more typical one.  In a practical context it has a negative connotation, but in a theoretical context it may be used in a constructive way.  For example maybe I can transform one problem into another which is known to be in P, with a polynomial &quot;blow up&quot;, thereby proving that the first problem is also in P.  On the other hand the transformation might incur a super-polynomial &quot;blow up&quot; meaning I failed to prove that.&lt;/p&gt;&#xA;" OwnerUserId="1284" LastEditorUserId="1284" LastEditorDisplayName="" LastEditDate="2010-08-19T05:43:32.467" LastActivityDate="2010-08-19T05:43:32.467" CommentCount="1" />
  <row Id="2775" PostTypeId="2" ParentId="2772" CreationDate="2010-08-19T05:21:03.893" Score="1" ViewCount="" Body="&lt;p&gt;In the context I usually hear it used, it means that a quantity is undefined or goes to infinity in the limit under consideration. For example, a function which has a singularity at that point blows up at that point (or &quot;as &lt;em&gt;x&lt;/em&gt; goes to&quot; that point).  Essentially, &quot;blows up&quot; indicates that something is being divided by zero. But I never considered it a technical term, just a bit of mathematical street lingo ;-) I do not know if this is &lt;em&gt;exactly&lt;/em&gt; what is meant when it's used in the context of PDE analysis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By way of example, I would say the integral&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\int_{r_0}^\infty \frac{1}{r^2}\mathrm{d}r$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;blows up when $r_0 = 0$. Or that the function $\frac{1}{r^2}$ blows up at $r = 0$.&lt;/p&gt;&#xA;" OwnerUserId="1190" LastEditorUserId="1190" LastEditorDisplayName="" LastEditDate="2010-08-20T02:09:13.277" LastActivityDate="2010-08-20T02:09:13.277" CommentCount="9" />
  <row Id="2776" PostTypeId="2" ParentId="2772" CreationDate="2010-08-19T05:28:37.000" Score="1" ViewCount="" Body="&lt;p&gt;It denotes a singularity or discontinuity - as opposed to decaying or stable solutions.&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-08-19T05:28:37.000" />
  <row Id="2777" PostTypeId="1" CreationDate="2010-08-19T06:41:00.527" Score="2" ViewCount="315" Body="&lt;p&gt;Let $\theta_{kl}$ be an angle such that $\cos\theta_{kl}=\frac{1}{2}(\cos(\frac{2\pi k}{n})+\cos(\frac{2\pi l}{n}))$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given that definition, if I introduce a new variable $t$ is the following a correct?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\cos(t\theta_{kl})\approx\frac{1}{2}(\cos(\frac{2\pi kt}{n})+\cos(\frac{2\pi lt}{n}))$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update:&lt;/strong&gt; I'm actually interested in the asymptotics of $\theta_{kl}$. By a second order approximation $\theta_{kl}^2=O(\frac{k^2+l^2}{n^2})$. Is it correct? If it is, then the above holds, right? But $\theta_{kl}$ needs to be small.&lt;/p&gt;&#xA;" OwnerUserId="1330" LastEditorUserId="1330" LastEditorDisplayName="" LastEditDate="2010-08-19T07:30:48.557" LastActivityDate="2011-01-16T14:44:58.133" Title="Approximating a cosine" Tags="&lt;trigonometry&gt;&lt;approximation&gt;" AnswerCount="3" CommentCount="0" />
  <row Id="2779" PostTypeId="2" ParentId="2777" CreationDate="2010-08-19T07:01:54.663" Score="0" ViewCount="" Body="&lt;p&gt;If you expand the cosines in both sides of the equation as series, you get something like $1-\frac{(t\theta_{kl})^2}{2}+\frac{(t\theta_{kl})^4}{24}+\dots\approx 1-\frac{(2\pi kt/n)^2+(2\pi lt/n)^2}{4}+\dots$ ; one does not really resemble the other much...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Seeing your update, have a look at this Wolfram Alpha &lt;a href=&quot;http://tinyurl.com/35la8ov&quot; rel=&quot;nofollow&quot;&gt;result&lt;/a&gt;, you can see that the leading term is of the form $\sqrt{k^2+l^2}\frac{t}{n}$.&lt;/p&gt;&#xA;" OwnerUserId="498" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-08-19T07:39:27.040" LastActivityDate="2010-08-19T07:39:27.040" CommentCount="1" />
  <row Id="2780" PostTypeId="1" AcceptedAnswerId="2782" CreationDate="2010-08-19T07:24:15.563" Score="2" ViewCount="208" Body="&lt;p&gt;When I was studying topology I remember being able to demonstrate that the set of topological surfaces with any number of punctures (including the projective plane, Klein bottle, Moebius strip, double torus, etc.), together with their associated inverses, was &lt;em&gt;not&lt;/em&gt; a group with respect to the operations of connection and disconnection.  I could show with a sequence of drawings that it is possible to connect two Klein bottles, deform the surface, and disconnect a torus from a single remaining Klein bottle, i.e. K+K = K+T, so inverses are not well defined even though the operator is otherwise associative on that set, has an identity (the sphere), and is also Abelian.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I never made much progress in my understanding beyond this point so I am interested in any answer that can simply explain what is going on here (am I misinterpreting something else as a topological property?), else provide a reference (I'm not even sure what branch of topology addresses this).  One slightly more specific question I have is: what is an example of a collection of topologies which &lt;em&gt;does&lt;/em&gt; form a group under connection and is this ever a useful property to have?&lt;/p&gt;&#xA;" OwnerUserId="1284" LastEditorUserId="1284" LastEditorDisplayName="" LastEditDate="2010-08-19T07:55:28.203" LastActivityDate="2010-08-19T08:01:23.057" Title="Basic questions about the algebra of surfaces" Tags="&lt;soft-question&gt;&lt;topology&gt;&lt;algebraic-topology&gt;" AnswerCount="2" CommentCount="6" FavoriteCount="1" />
  <row Id="2781" PostTypeId="2" ParentId="1870" CreationDate="2010-08-19T07:33:12.373" Score="4" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://www.maplesoft.com/&quot; rel=&quot;nofollow&quot;&gt;Maple&lt;/a&gt; can give step-by-step solutions as well as give hints on what rule you could apply next.&lt;/p&gt;&#xA;" OwnerUserId="505" LastActivityDate="2010-08-19T07:33:12.373" CommunityOwnedDate="2010-08-19T07:33:12.373" />
  <row Id="2782" PostTypeId="2" ParentId="2780" CreationDate="2010-08-19T07:46:32.920" Score="3" ViewCount="" Body="&lt;p&gt;Right, surfaces under the connect-sum operation form a commutative monoid, but it's not a free commutative monoid.  It's the quotient of the free commutative monoid on two generators (projective space, torus) by the relation you mentioned (in these generators you'd express the klein bottle as the connect-sum of two projective spaces).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a monoid where the only invertible element is the sphere. The proof is that the &quot;genus&quot; is additive under connect sum, and is only zero for the sphere.  Here the 'genus' is the sum of &amp;lt; the number of projective-space summands &gt; + &amp;lt; twice the number of torus summands &gt; (for this to make sense you have to express your surface as a connect-sum of only projective spaces and tori. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not quite sure what your question is. In the sense that there's no question mark in your 1st paragraph. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think there's a terminology problem.  &quot;Collection of topologies&quot; literally means &quot;a collection of subsets of various sets satisfying some rules&quot;.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you replace surfaces with various other objects, the connect-sum operation can be a group operation.  For example, if your objects are smooth embeddings of $S^3$ in $S^6$, taken up to isotopy, then the connect-sum operation turns this collection into an infinite-cyclic group. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Strangely enough, embeddings of $S^j$ in $S^n$, taken up to isotopy and with operation given by connect sum form a group whenever $n &gt; j+2$.  Sometimes it's a group for silly reasons -- there is only one such embedding up to isotopy provided $2n-3j-3&gt;0$.  But when $2n-3j-3 \leq 0$ generally there are many. &lt;/p&gt;&#xA;" OwnerUserId="642" LastActivityDate="2010-08-19T07:46:32.920" />
  <row Id="2783" PostTypeId="2" ParentId="2777" CreationDate="2010-08-19T07:57:27.480" Score="0" ViewCount="" Body="&lt;p&gt;My reasoning is the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By definition $\cos\theta_{kl}=\frac{1}{2}\cos\frac{2\pi k}{n}+\frac{1}{2}\cos\frac{2\pi l}{n}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given that $\theta_{kl}$ is small, we take the Taylor expansion on both sides up to the second term and obtain&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$1-\theta_{kl}^2\approx\frac{1}{2}(1-\frac{4\pi^2k^2}{n^2}+1-\frac{4\pi^2l}{n^2})=1-\frac{2\pi^2k^2+2\pi^2l^2}{n^2}$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which implies&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\theta_{kl}=O(\frac{k+l}{n})$.&lt;/p&gt;&#xA;" OwnerUserId="1330" LastActivityDate="2010-08-19T07:57:27.480" />
  <row Id="2784" PostTypeId="2" ParentId="2780" CreationDate="2010-08-19T08:01:23.057" Score="3" ViewCount="" Body="&lt;p&gt;Connected sum usually gives a structure of a commutative monoid but rarely of a group. Absence of inverse elements is explained (e.g. for manifolds or for knots) by Mazur swindle (&lt;a href=&quot;http://en.wikipedia.org/wiki/Eilenberg%E2%80%93Mazur_swindle&quot; rel=&quot;nofollow&quot;&gt;wiki&lt;/a&gt;, &lt;a href=&quot;http://terrytao.wordpress.com/2009/10/05/mazurs-swindle/&quot; rel=&quot;nofollow&quot;&gt;Terence Tao&lt;/a&gt;).&lt;/p&gt;&#xA;" OwnerUserId="152" LastActivityDate="2010-08-19T08:01:23.057" />
  <row Id="2785" PostTypeId="2" ParentId="2772" CreationDate="2010-08-19T08:50:39.287" Score="1" ViewCount="" Body="&lt;p&gt;It means that as time goes to infinity, the response-- be it displacement, velocity, heat, field or whatever-- will also go to infinity. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is physically impossible and is usually due to the poor numerical/approximated methods use to solve the PDE. &lt;/p&gt;&#xA;" OwnerUserId="550" LastActivityDate="2010-08-19T08:50:39.287" CommentCount="2" />
  <row Id="2786" PostTypeId="1" CreationDate="2010-08-19T09:36:07.700" Score="4" ViewCount="462" Body="&lt;p&gt;Hi,&#xA;I'm starting to read Baby Rudin (Principles of mathematical analysis) now and I wonder whether you know of any companions to it. Another supplementary book would do too. I tried Silvia's notes, but I found them a bit too &quot;logical&quot; so to say. Are they good? What else do you recommend?&lt;/p&gt;&#xA;" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-08-19T10:57:24.307" LastActivityDate="2010-08-28T16:21:03.560" CommunityOwnedDate="2010-08-19T14:14:10.333" Title="Companions to Rudin?" Tags="&lt;soft-question&gt;&lt;analysis&gt;" AnswerCount="5" CommentCount="2" FavoriteCount="7" />
  <row Id="2787" PostTypeId="2" ParentId="2786" CreationDate="2010-08-19T10:22:30.667" Score="4" ViewCount="" Body="&lt;p&gt;Hi--&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Introduction to real analysis by Bartle and Sherbert&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Methods of Real Analysis by R.R. Goldberg&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) Mathematical Analysis by Tom Apostol&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4) Real and Abstract Analysis by Karl Stromberg.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;5) A radical approach to real analysis by David M Bressoud by MAA.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first book is a very good book for a beginner. The next two are classics. (4) is also very good in case you want to read something advanced. The last one keeps entertaining you with some interesting examples as well as some interesting history of Real Analysis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Happy Reading!!&lt;/p&gt;&#xA;" LastActivityDate="2010-08-19T10:22:30.667" CommunityOwnedDate="2010-08-19T14:14:10.333" />
  <row Id="2788" PostTypeId="2" ParentId="622" CreationDate="2010-08-19T10:24:54.877" Score="0" ViewCount="" Body="&lt;p&gt;The idea of considering the representations of a group $G$ as a group of linear operators to ultimately gain informations on the group $G$ itself is an important istance of the general process of &quot;linearization&quot; that is employed systematically in many areas of mathematics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In down-to-earth (and possibly vague) terms, I would say that the only general class of problems that we are able to resolve fully are those which are linear, meaning that involve only vector spaces--best if finitely dimensional :-) --and linear maps between them. When we deal with objects that are intrinsically non-linear, a widespread strategy is to try to attach to them some linear &quot;gadgets&quot; that in some way preserve, in a different form, some of the structure of the original object.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Two basic examples that should be well-known to every math student:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The construction of the tensor product reduces multilinear algebra to linear algebra.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The construction of the homotopy and (co)homology groups of a topological space is an attempt to &quot;understand topology via groups and linear spaces&quot;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The linear representations of a group $G$ respond to the same principle: they are a &quot;linear&quot; set of data attached to $G$ which hopefully can serve to characterize $G$.  &lt;/p&gt;&#xA;" OwnerUserId="688" LastActivityDate="2010-08-19T10:24:54.877" />
  <row Id="2789" PostTypeId="2" ParentId="2786" CreationDate="2010-08-19T10:49:36.800" Score="2" ViewCount="" Body="&lt;p&gt;There is a set of notes and additional exercises due to George Bergman.  See his web page...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://math.berkeley.edu/~gbergman/ug.hndts/&quot; rel=&quot;nofollow&quot;&gt;http://math.berkeley.edu/~gbergman/ug.hndts/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="442" LastActivityDate="2010-08-19T10:49:36.800" CommunityOwnedDate="2010-08-19T14:14:10.333" CommentCount="3" />
  <row Id="2790" PostTypeId="2" ParentId="2647" CreationDate="2010-08-19T11:19:58.280" Score="1" ViewCount="" Body="&lt;p&gt;A further answer, generalizing flagar's example: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For any N&amp;nbsp;&amp;ge;&amp;nbsp;10, there exists a set of the sort you ask for which contains at most eleven elements (with fewer than eleven in the case that N&amp;nbsp;&amp;lt;&amp;nbsp;13), and with all elements having absolute value at most N:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;{&amp;minus;N,&amp;nbsp;&amp;minus;N+2,&amp;nbsp;&amp;minus;N+4,&amp;nbsp;&amp;minus;2,&amp;nbsp;1,&amp;nbsp;3,&amp;nbsp;4,&amp;nbsp;5,&amp;nbsp;N&amp;minus;7,&amp;nbsp;N&amp;minus;6,&amp;nbsp;N&amp;minus;5}.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The shorter example presented by flagar corresponds to the case N&amp;nbsp;=&amp;nbsp;10.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The key to this construction was to think in terms of small sequences of &lt;em&gt;almost mutually-supporting&lt;/em&gt; subsequences. For instance, the sequence &amp;minus;N, &amp;minus;N+2, &amp;minus;N+4 consist of integers which are sums of themselves, plus one of &amp;minus;2 or 4. We can include 4 in the set by including it in the sequence 3, 4, 5: we can support this by including 1 in the set (where 1 itself is supported as the sum of 3 and &amp;minus;2). We then only need a way of supporting &amp;minus;2: we do this with the other subsequence which includes N&amp;minus;6.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each time we have a short sequence, we have a collection of integers which is easy to &quot;support&quot; using only a small number of other integers. So one strategy to constructing such self-supporting sets of integers is to think in terms of one or more short sequences of integers, with each sequence working to support another.&lt;/p&gt;&#xA;" OwnerUserId="439" LastEditorUserId="439" LastEditorDisplayName="" LastEditDate="2010-08-19T12:18:20.387" LastActivityDate="2010-08-19T12:18:20.387" />
  <row Id="2791" PostTypeId="1" AcceptedAnswerId="2792" CreationDate="2010-08-19T11:31:55.080" Score="4" ViewCount="298" Body="&lt;p&gt;In section 4 of the article by Afred van der Poorten's &lt;a href=&quot;http://www.ega-math.narod.ru/Apery1.htm&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;A Proof That Euler Missed ...&lt;/em&gt;&lt;/a&gt;  the following inequality is used:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\nu_{p}\displaystyle\binom{n}{m}\leq\left\lfloor\dfrac{\ln n}{\ln p}\right\rfloor-\nu_{p}(m)\qquad(\ast)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(&lt;em&gt;In the original denoted&lt;/em&gt; $\text{ord}$ $_{p}(\cdot)$ &lt;em&gt;instead of&lt;/em&gt; $\nu_p(\cdot)$). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\nu_{p}(k)$ is the $p$-adic valuation of $k\in\mathbb{Q}$, i. e. the exponent of the prime $p$ in the prime factorization of $k$. &#xA;I know some properties of the foor function and that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\nu_{p}(a/b)=\nu_{p}(a)-\nu_{p}(b),$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\nu_{p}(a\cdot b)=\nu_{p}(a)\cdot \nu_{p}(b)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\nu_{p}(n!)=\displaystyle\sum_{i\geq 1}\left\lfloor \dfrac{n}{p^{i}}\right\rfloor $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but I didn't convince myself on the correct argument I should use to prove $(\ast )$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question&lt;/strong&gt;: How can this inequality be proved?&lt;/p&gt;&#xA;" OwnerUserId="752" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2010-10-05T21:32:16.860" LastActivityDate="2010-10-05T21:32:16.860" Title="Binomial coefficients: how to prove an inequality on the $p$-adic valuation?" Tags="&lt;number-theory&gt;&lt;binomial-coefficients&gt;" AnswerCount="2" FavoriteCount="0" />
  <row Id="2792" PostTypeId="2" ParentId="2791" CreationDate="2010-08-19T11:58:18.307" Score="4" ViewCount="" Body="&lt;p&gt;Since $\nu_p(n!)=\sum_i\lfloor n/p^i\rfloor$, $\nu_p\binom nm=\sum_i(\lfloor n/p^i\rfloor-\lfloor m/p^i\rfloor-\lfloor (n-m)/p^i\rfloor)$. Observe that each summand is $\le 1$, and for $i&gt;\left\lfloor\frac{\ln n}{\ln p}\right\rfloor$ it's clearly $0$. That gives $v_{p}\binom{n}{m}\le \left\lfloor\frac{\ln n}{\ln p}\right\rfloor$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally observe that for $i\le\nu_p(m)$ $\lfloor m/p^i\rfloor=m/p^i$, so ($\lfloor m/p^i\rfloor+\lfloor x/p^i\rfloor=\lfloor(m+x)/p^i\rfloor$ and) the corresponding summand is also $0$. That gives the inequality in question.&lt;/p&gt;&#xA;" OwnerUserId="152" LastActivityDate="2010-08-19T11:58:18.307" CommentCount="1" />
  <row Id="2793" PostTypeId="2" ParentId="622" CreationDate="2010-08-19T12:57:43.290" Score="8" ViewCount="" Body="&lt;p&gt;Groups very, very rarely appear abstractly in (mathematical) &lt;em&gt;nature&lt;/em&gt;: when we encounter them, it is almost always in the form of a representation (a linear representation, a permutation representation, a non-linear representation in the form of automorphisms of an algebra, of a variety, of a manifold, of a group (!)).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is thus only natural that the study of representation theory be important! I would go further, turn your question backwards, and claim that group theory is important because it allows us to study the representations of groups.&lt;/p&gt;&#xA;" OwnerUserId="274" LastEditorUserId="274" LastEditorDisplayName="" LastEditDate="2010-08-19T14:28:27.367" LastActivityDate="2010-08-19T14:28:27.367" />
  <row Id="2796" PostTypeId="1" AcceptedAnswerId="2797" CreationDate="2010-08-19T13:11:15.227" Score="2" ViewCount="159" Body="&lt;p&gt;I'm starting a chapter on Functions and they had the steps shown to reach the p-q equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ x_{1,2} = -\frac{p}{2} \pm\sqrt{\left(\frac{p}{2}\right)^2 - q}$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I wanted to do the same with the Quadratic Equation. I'm using the base linear equation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ax+by+c = 0.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The solution I have so far is as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x^2 + \frac{b}{a}x + \frac{c}{a}= 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x^2 + \frac{b}{a}x = -\frac{c}{a}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x^2 + \frac{b}{a}x + \left(\frac{b}{2a}\right)^2 = -\frac{c}{a} + \left(\frac{b}{2a}\right)^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left(x + \frac{b}{2a}\right)^2 = \left(\frac{b}{2a}\right)^2 - \frac{c}{a}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left(x + \frac{b}{2a}\right) = \pm\sqrt{\left(\frac{b}{2a}\right)^2 - \frac{c}{a}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x = -\frac{b}{2a} \pm\sqrt{\left(\frac{b}{2a}\right)^2 - \frac{c}{a}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My problem comes from trying to solve the insides of the square root:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sqrt{(\frac{b}{2a})^2 - \frac{c}{a}} = \sqrt{\frac{b^2}{4a2} - \frac{c}{a}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$= \sqrt{\frac{b^2}{4a^2} - \frac{c}{a} \left(\frac{4a}{4a}\right)} = \sqrt{\frac{b^2 - 4ac}{4a^2}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$= \sqrt{\frac{b^2 - 4ac}{\left(2a\right)^2}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x_{1,2} = \frac{-\left(\frac{b}{2a}\right) \pm\sqrt{b^2 -4ac}}{2a}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but there is still the problem of the -(b/2a) outside of the sqrt. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What am I doing wrong? Also, Tex is awesome; is there a better way to do the 1,2 subscripts than _1,_2?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Answer:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instead of &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x_{1,2} = \frac{-\left(\frac{b}{2a}\right) \pm\sqrt{b^2 -4ac}}{2a}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The solution goes&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{-b}{2a}\pm \sqrt{\frac{b^{2}-4ac}{4a^{2}}}=\frac{-b}{2a}\pm \frac{\sqrt{b^{2}-4ac}}{2a}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$= \frac{-b \pm\sqrt{b^2 - 4ac}}{2a}$$&lt;/p&gt;&#xA;" OwnerUserId="44" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-08-23T13:48:17.433" LastActivityDate="2010-08-23T13:48:17.433" Title="Help in getting the Quadratic Equation." Tags="&lt;polynomials&gt;" AnswerCount="4" CommentCount="3" />
  <row Id="2797" PostTypeId="2" ParentId="2796" CreationDate="2010-08-19T13:29:18.430" Score="3" ViewCount="" Body="&lt;p&gt;Sorry I don't know how to do tex on websites, but I'm trying to learn.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You just made a small mistake on the final step. In the second to last step, we actually have our full equation as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{-b}{2a}\pm \sqrt{\frac{b^{2}-4ac}{4a^{2}}}=\frac{-b}{2a}\pm \frac{\sqrt{b^{2}-4ac}}{2a}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we can collect the common factor of 1/2a and get:&#xA;$\frac{-b\pm \sqrt{b^{2}-4ac}}{2a}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does that make sense?&lt;/p&gt;&#xA;" OwnerUserId="249" LastEditorUserId="249" LastEditorDisplayName="" LastEditDate="2010-08-19T14:13:47.840" LastActivityDate="2010-08-19T14:13:47.840" CommentCount="4" />
  <row Id="2798" PostTypeId="2" ParentId="2786" CreationDate="2010-08-19T13:41:38.353" Score="3" ViewCount="" Body="&lt;p&gt;I find Terrence Tao's notes to be a great companion, but he deviates from the order Rudin's presented in.&#xA;&lt;a href=&quot;http://www.math.ucla.edu/~tao/resource/general/131ah.1.03w/&quot; rel=&quot;nofollow&quot;&gt;Course 1&lt;/a&gt;&#xA;&lt;a href=&quot;http://www.math.ucla.edu/~tao/resource/general/131bh.1.03s/&quot; rel=&quot;nofollow&quot;&gt;course 2&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="249" LastActivityDate="2010-08-19T13:41:38.353" CommunityOwnedDate="2010-08-19T14:14:10.333" />
  <row Id="2799" PostTypeId="2" ParentId="2796" CreationDate="2010-08-19T13:47:45.610" Score="0" ViewCount="" Body="&lt;p&gt;You factored 2a out of the square root and put it in the denominator without factoring it out of -(b/2a).&lt;/p&gt;&#xA;" OwnerUserId="796" LastActivityDate="2010-08-19T13:47:45.610" />
  <row Id="2800" PostTypeId="2" ParentId="2796" CreationDate="2010-08-19T13:57:07.787" Score="0" ViewCount="" Body="&lt;p&gt;The typical TeX way to do the 1,2 subscripts would be x_{1,2}, which yields $x_{1,2}$ as desired.  In other words, you use curly braces, like so: &quot;_{&lt;em&gt;subscripts go here&lt;/em&gt;}&quot;.&lt;/p&gt;&#xA;" OwnerUserId="506" LastActivityDate="2010-08-19T13:57:07.787" CommentCount="1" />
  <row Id="2801" PostTypeId="2" ParentId="2796" CreationDate="2010-08-19T14:06:19.403" Score="1" ViewCount="" Body="&lt;p&gt;Because I don't have edit privileges (yet?), I can't edit WWright's answer (which is correct) in order to TeX-ify it (perhaps someone with editing privileges will?), but just for the sake of making WWright's answer easier to read, here it is in TeX:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$-\frac{b}{2a} \pm \sqrt{\frac{b^2 - 4ac}{(2a)^2}} = -\frac{b}{2a} \pm \frac{\sqrt{b^2 - 4ac}}{2a}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As these two terms have a common denominator, we can add them:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$-\frac{b}{2a} \pm \frac{\sqrt{b^2 - 4ac}}{2a} = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$&lt;/p&gt;&#xA;" OwnerUserId="506" LastActivityDate="2010-08-19T14:06:19.403" CommentCount="2" />
  <row Id="2802" PostTypeId="1" AcceptedAnswerId="2836" CreationDate="2010-08-19T14:43:54.850" Score="0" ViewCount="165" Body="&lt;p&gt;In 1D case, R1 as a typical case of 1D Euclidean space is totally ordered. I was wondering if any 1D Euclidean space E1 is ordered as well? What is it&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For higher dimensional case, are both Rn and En ordered? What are the orders?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks and regards!&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Update:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I meant for an order for an Euclidean space:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;that can induce or be compatible with the topology of the Euclidean space;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or that is defined in their definition, for example, the order of R1 is the one defined in one of R1's definition as an ordered field with supreme property;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or that is used by default or most commonly. &lt;/p&gt;&#xA;" OwnerUserId="1281" LastEditorUserId="1281" LastEditorDisplayName="" LastEditDate="2010-08-19T17:40:40.437" LastActivityDate="2010-08-19T23:00:21.523" Title="Order on Euclidean space" Tags="&lt;set-theory&gt;&lt;topology&gt;" AnswerCount="2" CommentCount="6" />
  <row Id="2803" PostTypeId="2" ParentId="622" CreationDate="2010-08-19T14:44:41.260" Score="2" ViewCount="" Body="&lt;p&gt;Representation theory plays a big role in the group-theoretic approach to special functions.&#xA;For example, Willard Miller showed that the powerful Infeld-Hull factorization / ladder method - widely exploited by physicists - is equivalent to the representation theory of four local Lie groups. This lie-theoretic approach served to powerfully unify and &quot;explain&quot; all prior similar attempts to provide a unfied theory of such classes of special functions, e.g. Truesdell's influential book &lt;a href=&quot;http://books.google.com/books?id=p1X4KKyU6kEC&quot; rel=&quot;nofollow&quot;&gt;An Essay Toward a Unified Theory of Special Functions&lt;/a&gt;. Below is the first paragraph of the introduction to Willard Miller's classic monograph &lt;a href=&quot;http://www.ima.umn.edu/~miller/lietheoryspecialfunctions.html&quot; rel=&quot;nofollow&quot;&gt;Lie theory and special functions&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;This monograph is the result of an attempt to understand the role &#xA;  played by special function theory in the formalism of mathematical &#xA;  physics. It demonstrates explicitly that special functions which arise in &#xA;  the study of mathematical models of physical phenomena and the &#xA;  identities which these functions obey are in many cases dictated by &#xA;  symmetry groups admitted by the models. In particular it will be shown &#xA;  that the factorization method, a powerful tool for computing eigenvalues &#xA;  and recurrence relations for solutions of second order ordinary differential &#xA;  equations (Infeld and Hull), is equivalent to the representation &#xA;  theory of four local Lie groups. A detailed study of these four groups and &#xA;  their Lie algebras leads to a unified treatment of a significant proportion &#xA;  of special function theory, especially that part of the theory which is &#xA;  most useful in mathematical physics. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;See also Miller's sequel &lt;a href=&quot;http://www.ima.umn.edu/~miller/separationofvariables.html&quot; rel=&quot;nofollow&quot;&gt;Symmetry and Separation of Variables&lt;/a&gt;. Again I quote from the preface:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;This book is concerned with the relationship between symmetries of a &#xA;  linear second-order partial differential equation of mathematical physics, &#xA;  the coordinate systems in which the equation admits solutions via separation &#xA;  of variables, and the properties of the special functions that arise in &#xA;  this manner. It is an introduction intended for anyone with experience in &#xA;  partial differential equations, special functions, or Lie group theory, such &#xA;  as group theorists, applied mathematicians, theoretical physicists and &#xA;  chemists, and electrical engineers. We will exhibit some modern group-theoretic &#xA;  twists in the ancient method of separation of variables that can be &#xA;  used to provide a foundation for much of special function theory. In &#xA;  particular, we will show explicitly that all special functions that arise &#xA;  via separation of variables in the equations of mathematical physics can &#xA;  be studied using group theory. These include the functions of Lame, Ince, &#xA;  Mathieu, and others, as well as those of hypergeometric type. &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;This is a very critical time in the history of group-theoretic methods in &#xA;  special function theory. The basic relations between Lie groups, special &#xA;  functions, and the method of separation of variables have recently been &#xA;  clarified. One can now construct a group-theoretic machine that, when &#xA;  applied to a given differential equation of mathematical physics, describes &#xA;  in a rational manner the possible coordinate systems in which the equation &#xA;  admits solutions via separation of variables and the various expansion &#xA;  theorems relating the separable (special function) solutions in distinct &#xA;  coordinate systems. Indeed for the most important linear equations, the &#xA;  separated solutions are characterized as common eigenfunctions of sets of &#xA;  second-order commuting elements in the universal enveloping algebra of &#xA;  the Lie symmetry algebra corresponding to the equation. The problem of &#xA;  expanding one set of separable solutions in terms of another reduces to a &#xA;  problem in the representation theory of the Lie symmetry algebra. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;See Koornwinder's &lt;a href=&quot;http://projecteuclid.org/DPubS/Repository/1.0/Disseminate?view=body&amp;amp;id=pdf_1&amp;amp;handle=euclid.bams/1183544920&quot; rel=&quot;nofollow&quot;&gt;review&lt;/a&gt; of this book for a very nice concise introduction to the group-theoretic approach to separation of variables.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2011-02-23T22:14:44.497" LastActivityDate="2011-02-23T22:14:44.497" />
  <row Id="2805" PostTypeId="2" ParentId="2735" CreationDate="2010-08-19T15:51:50.460" Score="2" ViewCount="" Body="&lt;p&gt;This is too long for a comment, so I'll throw in &lt;a href=&quot;http://books.google.com/books?id=YHXU4W3Ez2MC&amp;amp;pg=PA253&quot; rel=&quot;nofollow&quot;&gt;this choice piece&lt;/a&gt; by Forman Acton in his &lt;em&gt;Numerical Methods That Work&lt;/em&gt; on the subject of dealing with large matrices:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Whenever a person eagerly inquires if my computer can solve a set of 300&#xA;  equations in 300 unknowns, I must quickly suppress the temptation to&#xA;  retort, &quot;Yes, but why bother?&quot; There are, indeed, legitimate sets of&#xA;  equations that large. They arise from replacing a partial differential&#xA;  equation on a set of grid points, and the person who knows enough to tackle&#xA;  this type of problem also usually knows what kind of computer he needs.&#xA;  The odds are all too high that our inquiring friend is suffering from a&#xA;  quite different problem: he probably has collected a set of experimental&#xA;  data and is now attempting to fit a 300-parameter model to it - by Least&#xA;  Squares! The sooner this guy can be eased out of your office, the sooner&#xA;  you will be able to get back to useful work - but these chaps are persistent.&#xA;  They have fitted three-parameter models on desk machines with no&#xA;  serious difficulties and now the electronic computer permits them more&#xA;  grandiose visions. They leap from the known to the unknown with a terrifying&#xA;  innocence and the perennial self-confidence that every parameter&#xA;  is totally justified. It does no good to point out that several parameters&#xA;  are nearly certain to be competing to &quot;explain&quot; the same variations in the&#xA;  data and hence the equation system will be nearly indeterminate. It does&#xA;  no good to point out that all large least-squares matrices are striving&#xA;  mightily to be proper subsets of the Hilbert matrix-which is virtually&#xA;  indeterminate and uninvertible—and so even if all 300 parameters were&#xA;  beautifully independent, the fitting equations would still be violently&#xA;  unstable. All of this, I repeat, does no good—and you end up by getting&#xA;  angry and throwing the guy out of your office...&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;...The computer center's director must prevent the looting of valuable&#xA;  computer time by these would-be fitters of many parameters. The task&#xA;  is not a pleasant one, but the legitimate computer users have rights,&#xA;  too. The alternative commits everybody to a miserable two weeks of sloshing&#xA;  around in great quantities of &quot;Results&quot; that are manifestly&#xA;  impossible, with no visible way of finding the trouble. The trouble, of&#xA;  course, arises from looking for a good answer to a poorly posed problem, but&#xA;  a computer director seldom knows enough about the subject matter to win&#xA;  any of those arguments with the problem's proposer, and the impasse&#xA;  finally has to be broken by violence—which therefore might as well&#xA;  be used in the very beginning.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="498" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2011-05-12T12:05:01.077" LastActivityDate="2011-05-12T12:05:01.077" CommunityOwnedDate="2010-08-19T15:51:50.460" CommentCount="2" />
  <row Id="2808" PostTypeId="2" ParentId="2735" CreationDate="2010-08-19T16:02:01.610" Score="1" ViewCount="" Body="&lt;p&gt;If all you are interested in is solving $A x = B$, I would suggest looking at iterative and/or block-iterative methods.  Noting your other question asking about positivity of solutions, I would suggest that you look at &lt;a href=&quot;http://faculty.uml.edu/cbyrne/aclafctext.pdf&quot; rel=&quot;nofollow&quot;&gt;Applied and Computational&#xA;Linear Algebra: A First Course&lt;/a&gt; by Charlie Byrne.  The MART algorithm described therein is used for finding positive $x$ solutions to $A x = B$m, and addresses many similar problems from tomography and signal processing.  If you like it get his 'Applied Iterative Methods'.&lt;/p&gt;&#xA;" OwnerUserId="943" LastActivityDate="2010-08-19T16:02:01.610" CommentCount="4" />
  <row Id="2809" PostTypeId="2" ParentId="2485" CreationDate="2010-08-19T16:10:32.930" Score="0" ViewCount="" Body="&lt;p&gt;Please refer Paul Halmos's book entitled &quot;Problems for Mathematicians Young and Old&quot;. Its somewhere located in the Measure theory Section. Although, i couldn't comprehend the solution fully, i would like to see a better example.&lt;/p&gt;&#xA;" LastActivityDate="2010-08-19T16:10:32.930" CommentCount="1" />
  <row Id="2811" PostTypeId="2" ParentId="2802" CreationDate="2010-08-19T16:44:56.127" Score="0" ViewCount="" Body="&lt;p&gt;If you are meaning a locally Euclidean whose topology is the one given by a linear order then $S^1$ is a counter-example.&lt;/p&gt;&#xA;" OwnerUserId="370" LastActivityDate="2010-08-19T16:44:56.127" />
  <row Id="2814" PostTypeId="2" ParentId="2786" CreationDate="2010-08-19T17:00:40.223" Score="2" ViewCount="" Body="&lt;p&gt;The book &lt;a href=&quot;http://community.middlebury.edu/~abbott/UA/UA.html&quot; rel=&quot;nofollow&quot;&gt;Understanding Analysis&lt;/a&gt; by Stephen Abbott is very good. So is &lt;a href=&quot;http://www.ams.org/bookstore-getitem/item=gsm-62&quot; rel=&quot;nofollow&quot;&gt;A Companion to Analysis&lt;/a&gt; by T. W. Körner.&lt;/p&gt;&#xA;" OwnerUserId="589" LastActivityDate="2010-08-19T17:00:40.223" CommunityOwnedDate="2010-08-19T17:00:40.223" />
  <row Id="2815" PostTypeId="1" AcceptedAnswerId="2826" CreationDate="2010-08-19T17:06:53.283" Score="15" ViewCount="401" Body="&lt;p&gt;This is a final exam question in my algorithms class:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$k$ is a taxicab number if $k = a^3+b^3=c^3+d^3$, $a,b,c,d$ are distinct positive integers. Find all taxicab number $k$ such that $a,b,c,d &amp;lt; n$ in $O(n)$ time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know if the problem had a typo or not, because $O(n^3)$ seems more reasonable. The best I can come up with is $O(n^2 \log n)$, and that's the best anyone I know can come up with. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The $O(n^2 \log n)$ algorithm: &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Try all possible $a^3+b^3=k$ pairs, for each $k$, store $(k,1)$ into a binary tree(indexed by $k$) if $(k,i)$ doesn't exist, if $(k,i)$ exists, replace $(k,i)$ with $(k,i+1)$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Transverse the binary tree, output all $(k,i)$ where $i\geq 2$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Are there any faster methods? This should be the best possible method without using any number theoretical result because the program might output $O(n^2)$ taxicab numbers. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is $O(n)$ even possible? One have to prove there are only $O(n)$ taxicab numbers lesser than $2n^3$ in order to prove there exist a $O(n)$ algorithm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;: The professor admit it was a typo, it should have been $O(n^3)$. I'm happy he made the typo, since the answer Tomer Vromen suggested is amazing.&lt;/p&gt;&#xA;" OwnerUserId="96" LastEditorUserId="96" LastEditorDisplayName="" LastEditDate="2011-01-13T21:31:16.170" LastActivityDate="2011-01-13T21:31:16.170" Title="Find taxicab numbers in $O(n)$ time" Tags="&lt;number-theory&gt;&lt;algorithms&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="5" />
  <row Id="2816" PostTypeId="1" AcceptedAnswerId="2818" CreationDate="2010-08-19T17:21:15.697" Score="1" ViewCount="70" Body="&lt;p&gt;In a densely and totally ordered set, induce a order topology from the order.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are the dense in terms of the order and the dense in terms of the order topology equivalent?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks and regards!&lt;/p&gt;&#xA;" OwnerUserId="1281" LastActivityDate="2010-08-19T18:20:59.140" Title="dense in terms of order and in terms of the order topology" Tags="&lt;set-theory&gt;&lt;topology&gt;" AnswerCount="1" />
  <row Id="2817" PostTypeId="2" ParentId="2755" CreationDate="2010-08-19T17:38:30.020" Score="13" ViewCount="" Body="&lt;p&gt;The wikipedia page for the &lt;a href=&quot;http://en.wikipedia.org/wiki/Torus&quot; rel=&quot;nofollow&quot;&gt;torus&lt;/a&gt; has a very nice &lt;a href=&quot;http://en.wikipedia.org/wiki/File%3aInside-out_torus_%28animated,_small%29.gif&quot; rel=&quot;nofollow&quot;&gt;animation&lt;/a&gt; of a punctured torus turning inside out.  In just the same way, if you stitch together the two cuffs of a pair of pants together you can turn the result inside out.  You have to attach the cuffs in the simplest way, without knotting, to get a torus with a hole, not a Klein bottle with a hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As Ryan says, it is not possible to turn every piece of clothing inside out,  However such &quot;bad&quot; pieces of clothing have to be knotted in some fashion.  Here is an example, perhaps a bit simpler than Ryan's, that avoids hyperbolic geometry :).  Take a pair of pants, tie the legs in an overhand knot, and stitch together the cuffs to get a knotted punctured torus.  The knot in this case is a &lt;a href=&quot;http://Trefoil&quot; rel=&quot;nofollow&quot;&gt;trefoil&lt;/a&gt; knot.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll call this piece of clothing $S$.  The proof I know that $S$ cannot be turned inside out uses some ideas from low-dimensional topology. Notice that the curve of stitches &lt;a href=&quot;http://en.wikipedia.org/wiki/Incompressible_surface&quot; rel=&quot;nofollow&quot;&gt;compresses&lt;/a&gt; (bounds a disk $D$) in the inner direction.  If you could turn $S$ inside out then that motion of $S$ would also give a motion of the disk $D$.  Thus the motion sends the curve of stitches to some curve on $S$ (still not parallel to the boundary!) that compresses in the &lt;em&gt;outer direction&lt;/em&gt;.  But since the trefoil is not the &lt;a href=&quot;http://en.wikipedia.org/wiki/Unknot&quot; rel=&quot;nofollow&quot;&gt;unknot&lt;/a&gt; there is no such curve.&lt;/p&gt;&#xA;" OwnerUserId="1307" LastEditorUserId="1307" LastEditorDisplayName="" LastEditDate="2010-08-19T19:42:34.100" LastActivityDate="2010-08-19T19:42:34.100" CommentCount="5" />
  <row Id="2818" PostTypeId="2" ParentId="2816" CreationDate="2010-08-19T17:44:26.470" Score="3" ViewCount="" Body="&lt;p&gt;Yes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let X be dense in terms of order.  Let U be an open subset.  We need to show X intersects U.  Since U is open, it contains a subset of the form (a,b) with a &amp;lt; b, since these sets form a basis for the topology (by definition of order topology).  Since X is dense in terms of order, there is an x in X such that a &amp;lt; x &amp;lt; b.  Then x is in X and (a,b), and hence x is in X and U.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conversely, suppose X is dense in terms of topology, meaning it intersects every open set.  Choose two elements a and b and suppose a &amp;lt; b.  Then (a,b) is an open set.  Since X is dense in terms of topology, there is an x in (a,b), and thus this x will satisfy a &amp;lt; x &amp;lt; b.  Hence, X is dense in terms of order.&lt;/p&gt;&#xA;" OwnerUserId="331" LastEditorUserId="331" LastEditorDisplayName="" LastEditDate="2010-08-19T18:20:59.140" LastActivityDate="2010-08-19T18:20:59.140" CommentCount="2" />
  <row Id="2819" PostTypeId="2" ParentId="1880" CreationDate="2010-08-19T18:05:08.953" Score="2" ViewCount="" Body="&lt;p&gt;It only seems odd to you because you are used to representing numbers in base 10. What if you used base 7?&lt;/p&gt;&#xA;" OwnerUserId="1312" LastActivityDate="2010-08-19T18:05:08.953" CommentCount="3" />
  <row Id="2820" PostTypeId="2" ParentId="162" CreationDate="2010-08-19T18:19:21.250" Score="0" ViewCount="" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;Why is &quot;the set of all sets&quot; a&#xA;  paradox? It seems like it would be&#xA;  fine, to me.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The lesson of this and other set theory paradoxes is that when you define a set in a proof, you are obliged to demonstrate that the set exists and is well-defined. To me, a paradox of Russell's Paradox is that he fell into this trap since he pretty much wore himself out trying to generate a rigorous foundation for Mathematics.&lt;/p&gt;&#xA;" OwnerUserId="1312" LastActivityDate="2010-08-19T18:19:21.250" />
  <row Id="2821" PostTypeId="2" ParentId="2791" CreationDate="2010-08-19T18:25:57.900" Score="3" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Theorem (Kummer):&lt;/strong&gt;  $\nu_p {n \choose m}$ is the number of carries it takes to add $m$ and $n-m$ in base $p$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now note that $\lfloor \frac{\ln n}{\ln p} \rfloor$ is the maximum possible number of carries and the last $\nu_p(m)$ digits of $m$ cannot be associated to any carries.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Kummer's theorem itself is not hard to prove; it more or less follows from the last identity you list.&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-08-20T21:30:40.267" LastActivityDate="2010-08-20T21:30:40.267" CommentCount="1" />
  <row Id="2822" PostTypeId="2" ParentId="2815" CreationDate="2010-08-19T18:42:21.907" Score="2" ViewCount="" Body="&lt;p&gt;Apparently this is a solved problem and every rational solution to $x^3 + y^3 = z^3 + w^3$ is proportional to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x = 1 − (p − 3q)(p^2 + 3q^2)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$y = −1 + (p + 3q)(p^2 + 3q^2)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$z = (p + 3q) − (p^2 + 3q^2)^2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$w = −(p − 3q) + (p^2 + 3q^2)^2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See: &lt;a href=&quot;http://129.81.170.14/~erowland/papers/koyama.pdf&quot; rel=&quot;nofollow&quot;&gt;http://129.81.170.14/~erowland/papers/koyama.pdf&lt;/a&gt;, Page 2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also see: &lt;a href=&quot;http://sites.google.com/site/tpiezas/010&quot; rel=&quot;nofollow&quot;&gt;http://sites.google.com/site/tpiezas/010&lt;/a&gt; (search for J. Binet).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So it seems like an O(n^2) algorithm might be possible. Perhaps using this more cleverly can give us an O(n) time algorithm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope that helps. Please do update us with the solution given by your professor.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-08-19T18:42:21.907" />
  <row Id="2823" PostTypeId="2" ParentId="1660" CreationDate="2010-08-19T19:02:35.283" Score="0" ViewCount="" Body="&lt;h3&gt;Schur polynomials proof&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Schur_polynomial&quot; rel=&quot;nofollow&quot;&gt;Observe&lt;/a&gt; that $D(a_1,\ldots,a_n)/D(1,\ldots,n)=s_\lambda(1,\ldots,1)$ (for $a_i=\lambda_i+i$). But by Giambelli (aka Jacobi-Trudi) &lt;a href=&quot;http://en.wikipedia.org/wiki/Giambelli_formula&quot; rel=&quot;nofollow&quot;&gt;formula&lt;/a&gt; $s_\lambda=\det h_{\lambda_i+j-1}$, so $s_\lambda(1,\ldots,1)=\det\binom{a_i}{j-1}$. Now by &lt;a href=&quot;http://qchu.wordpress.com/2009/11/17/the-lindstrom-gessel-viennot-lemma/&quot; rel=&quot;nofollow&quot;&gt;Lindstrom-Gessel-Viennot lemma&lt;/a&gt; last determinant counts non-intersecting lattice paths from the set $0,1,\ldots,n$ on Y-axis to the set $a_1,\ldots,a_n$ on X-axis. Finally, observe that any such path is characterized by x-coordinates of &quot;steps down&quot; — which are subject to conditions from the definition of GZ triangle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;/* Well, it's not that direct and it's more or less the proof from Robin Chapman's answer. Nevertheless, it explains something, I hope. */&lt;/p&gt;&#xA;" OwnerUserId="152" LastActivityDate="2010-08-19T19:02:35.283" />
  <row Id="2824" PostTypeId="2" ParentId="2786" CreationDate="2010-08-19T19:39:15.230" Score="3" ViewCount="" Body="&lt;p&gt;Gelbaum and Olmsted, Counterexamples in Analysis. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first real analysis/advanced calculus class is full of theorems with multiple conditions, and it can be difficult to tell which ones are necessary for what parts of the theorem. This book will provide examples for why the theorems are as they are and not otherwise.&lt;/p&gt;&#xA;" OwnerUserId="76" LastActivityDate="2010-08-19T19:39:15.230" CommunityOwnedDate="2010-08-19T19:39:15.230" />
  <row Id="2825" PostTypeId="2" ParentId="162" CreationDate="2010-08-19T19:57:08.697" Score="4" ViewCount="" Body="&lt;p&gt;Russel's paradox arises if you consider the set $U=\left\{x:x\not\in x\right\}$. Ask yourself if $U\in U$. If you suppose so, then by the definition of unrestricted set comprehension $U\not\in U$. You have a contradiction, so it must be the opposite of what you supposed, that is, $U\not\in U$. But this is the same as saying $U$ belongs to the complement of itself, that is, $U\in U$. You now have another contradiction, but this is far worse, since you have no hypotheses. The whole theory is logically inconsistent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In set theory there are two ways for getting rid of the Russel's paradox: either you disallow the set of all sets and other similar sets (see for example the Zermelo-Fraenkel set theory), or you allow them, but you also restrict the way they are used (see for example the Morse-Kelley set theory).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the first case, set comprehension says if you have a set $A$ you can have $\left\{x\in A:\phi\left(x\right)\right\}$ (notice: writing $\left\{x:\phi\left(x\right)\right\}$ is just wrong in this case, because you have to have an initial set). If you now define $U=\left\{x\in A:x\not\in x\right\}$ and you repeat the same passages as before, it only follows that $U\not\in A$. There's no contradiction and the theory is consistent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the second case, you consider classes, not just sets. Sets are classes that belong to some other class, while proper classes are classes that belong to no class. Set comprehension, in this case, says you can have $\left\{x:\phi\left(x\right)\right\}$, but all its members are sets by definition. If try to reproduce Russel's paradox, you get that $U\not\in U$. If you then suppose that $U$ is a set, then you have a contradiction, so $U$ must be a proper class. This is all you get. No contradictions. The theory is consistent.&lt;/p&gt;&#xA;" OwnerUserId="1175" LastActivityDate="2010-08-19T19:57:08.697" />
  <row Id="2826" PostTypeId="2" ParentId="2815" CreationDate="2010-08-19T20:06:14.433" Score="10" ViewCount="" Body="&lt;p&gt;I don't know about O(n), but I can do it in O(n^2). The main idea is to use &lt;a href=&quot;http://eli.thegreenplace.net/2008/08/23/initializing-an-array-in-constant-time/&quot; rel=&quot;nofollow&quot;&gt;initialization of an array in O(1)&lt;/a&gt; (this is the best reference that I've found, which is weird since this seems like a very important concept). Then you iterate through all the possible (a,b) pairs and do the same as step 1 in your proposed algorithm. Since a^3+b^3 &amp;lt;= 2n^3, the array needs to be of size 2n^3, but it's still initialized in O(1). Accessing an array element is O(1) like in a regular array.&lt;/p&gt;&#xA;" OwnerUserId="26" LastEditorUserId="26" LastEditorDisplayName="" LastEditDate="2010-08-19T20:20:06.613" LastActivityDate="2010-08-19T20:20:06.613" CommentCount="6" />
  <row Id="2827" PostTypeId="1" AcceptedAnswerId="2828" CreationDate="2010-08-19T20:13:39.360" Score="9" ViewCount="634" Body="&lt;p&gt;What is a good PDE book suitable for self study?   I'm looking for a book that doesn't require much prerequisite knowledge beyond undergraduate-level analysis.  My goal is to understand basic solutions techniques as well as some basic theory.&lt;/p&gt;&#xA;" OwnerUserId="832" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-08-19T20:51:50.203" LastActivityDate="2010-08-20T02:40:13.453" CommunityOwnedDate="2010-08-19T20:51:04.793" Title="Good 1st PDE book for self study" Tags="&lt;books&gt;&lt;differential-equations&gt;" AnswerCount="7" CommentCount="1" FavoriteCount="5" />
  <row Id="2828" PostTypeId="2" ParentId="2827" CreationDate="2010-08-19T20:20:44.007" Score="2" ViewCount="" Body="&lt;p&gt;The book by &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0470054565&quot; rel=&quot;nofollow&quot;&gt;Strauss&lt;/a&gt; is pretty good for a first course. For a second one the book by &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0821849743&quot; rel=&quot;nofollow&quot;&gt;Evans&lt;/a&gt; is nice but it requires some knowledge of measure theory and functional analysis.&lt;/p&gt;&#xA;" OwnerUserId="1120" LastEditorUserId="1120" LastEditorDisplayName="" LastEditDate="2010-08-19T22:31:16.443" LastActivityDate="2010-08-19T22:31:16.443" CommunityOwnedDate="2010-08-19T20:51:04.793" CommentCount="3" />
  <row Id="2829" PostTypeId="2" ParentId="2827" CreationDate="2010-08-19T20:26:48.157" Score="1" ViewCount="" Body="&lt;p&gt;I like Karl E. Gustafson's &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0486612716&quot; rel=&quot;nofollow&quot;&gt;Introduction to Partial Differential Equations and Hilbert Space Methods&lt;/a&gt;.  It was certainly readable after an advanced calc sequence.  You will find a few short and worthwhile conversational paragraphs throughout the book.  He also uses the technique of revisiting interesting concepts from different perspectives throughout the book.  And it's a Dover paperback, so it's cheap.&lt;/p&gt;&#xA;" OwnerUserId="121" LastActivityDate="2010-08-19T20:26:48.157" CommunityOwnedDate="2010-08-19T20:26:48.157" />
  <row Id="2830" PostTypeId="1" AcceptedAnswerId="2850" CreationDate="2010-08-19T20:27:10.637" Score="2" ViewCount="83" Body="&lt;p&gt;Suppose p and q are discrete distributions over k outcomes, q is given, for which p does the following hold?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$p_1 \log q_1 + \ldots + p_k \log q_k \ge q_1 \log q_1 + \ldots + q_k \log q_k$&lt;/p&gt;&#xA;" OwnerUserId="998" LastActivityDate="2010-08-20T15:15:16.400" Title="When does entropy bound cross-entropy?" Tags="&lt;probability-theory&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="2831" PostTypeId="2" ParentId="2827" CreationDate="2010-08-19T20:33:53.087" Score="1" ViewCount="" Body="&lt;p&gt;Logan's &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0387209530&quot; rel=&quot;nofollow&quot;&gt;Applied Partial Differential Equations&lt;/a&gt; might be suitable for you if you want a (relatively) quick overview of the subject, since it's not very long (about 200 pages). It's aimed at undergraduates in math, engineering and the sciences.&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2010-08-19T20:33:53.087" CommunityOwnedDate="2010-08-19T20:51:04.793" />
  <row Id="2832" PostTypeId="2" ParentId="2827" CreationDate="2010-08-19T20:58:53.103" Score="0" ViewCount="" Body="&lt;p&gt;I would recommend: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fritz John, &lt;a href=&quot;http://books.google.com/books?id=cBib_bsGGLYC&amp;amp;printsec=frontcover&amp;amp;dq=Fritz+john&amp;amp;hl=en&amp;amp;ei=Y5ptTP-MG6aHOPyC8LAL&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=1&amp;amp;ved=0CCUQ6AEwAA#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;Partial Differential Equations&lt;/a&gt; (Applied Mathematical Sciences) ISBN: 0387906096. It is a classical Springer book that contains what you ask for. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Google Books might be a good start before you make your final decision. &lt;/p&gt;&#xA;" OwnerUserId="1154" LastActivityDate="2010-08-19T20:58:53.103" CommunityOwnedDate="2010-08-19T20:58:53.103" />
  <row Id="2833" PostTypeId="2" ParentId="2827" CreationDate="2010-08-19T21:26:46.720" Score="0" ViewCount="" Body="&lt;p&gt;I haven't read a lot of PDE material, but I enjoyed Taylor's &lt;a href=&quot;http://books.google.com/books?id=uphqFYqYm0oC&amp;amp;printsec=frontcover&amp;amp;dq=taylor+partial+differential+equations&amp;amp;hl=en&amp;amp;ei=YaFtTNPMBYH7lwe_gZmvDQ&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=1&amp;amp;ved=0CCgQ6AEwAA#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;book&lt;/a&gt;. It's quite well-written, and also contains introductory material (like Lie derivatives), since it does things on manifolds.&lt;/p&gt;&#xA;" OwnerUserId="536" LastActivityDate="2010-08-19T21:26:46.720" CommunityOwnedDate="2010-08-19T21:26:46.720" />
  <row Id="2834" PostTypeId="2" ParentId="2772" CreationDate="2010-08-19T22:17:58.437" Score="0" ViewCount="" Body="&lt;p&gt;&quot;Blows up&quot; means it goes to infinity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1/x &quot;blows up&quot; at x = 0.&lt;/p&gt;&#xA;" OwnerUserId="1069" LastActivityDate="2010-08-19T22:17:58.437" />
  <row Id="2835" PostTypeId="2" ParentId="2772" CreationDate="2010-08-19T22:34:10.013" Score="7" ViewCount="" Body="&lt;p&gt;The meaning is, of course, context-dependent...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the context of differential equations, that a solution to an equation with a &quot;time&quot; variable blows up usually means that the maximal domain for which it is defined is finite, so that at the endpoint of that interval something `bad' happens: either the solution goes to infinity, or it stops being smooth (in a way that makes the differential equation stop having sense, maybe), or something. This is an important phenomenon, one which causes trouble. A couple of examples:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Perelman's solution of the Poicaré conjecture—in a very vague sense—consists of a way to `work around' the fact that certain solutions of a (very complicated non-linear) PDE blow up; &lt;/li&gt;&#xA;&lt;li&gt;the third `Millenium' Clay &lt;a href=&quot;http://www.claymath.org/millennium/Navier-Stokes_Equations/&quot; rel=&quot;nofollow&quot;&gt;problem&lt;/a&gt; is (very roughly) the question «do the solutions of the Navier-Stokes equation blow up?».&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Consider, as a very simple example, the equation $$\frac{\mathrm dx}{\mathrm dt}=x^2.$$ This equation makes sense and satisfies the conditions for existence and uniqueness of solutions on all of the $(t,x)$-plane, but if you solve it (which is easy to do explicitely, as it has the two variabls separated) you'll see that all of its solutions have a maximal interval which is a half-line (which is bounded on the left or on the right, depending on the initial condition) and that at the finite end of that interval the solutions become unbounded. We thus say that all solutions of our equation blow up in finite time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are also equations which have &lt;em&gt;some&lt;/em&gt; solutions which blow up and some which live forever. One example is $$\frac{\mathrm dx}{\mathrm dt}=\begin{cases}x^2&amp;amp;\text{if $x\geq0$}\\0&amp;amp;\text{if $x\leq0$}\end{cases}$$ and you'll surely find lots of fun in trying to concoct examples where even more interesting phenomena occur.&lt;/p&gt;&#xA;" OwnerUserId="274" LastEditorUserId="274" LastEditorDisplayName="" LastEditDate="2010-08-19T22:44:33.370" LastActivityDate="2010-08-19T22:44:33.370" />
  <row Id="2836" PostTypeId="2" ParentId="2802" CreationDate="2010-08-19T23:00:21.523" Score="6" ViewCount="" Body="&lt;p&gt;I think the OP is asking whether, for $n &gt; 1$, the topology on $\mathbb{R}^n$ is an &lt;a href=&quot;http://en.wikipedia.org/wiki/Order_topology&quot; rel=&quot;nofollow&quot;&gt;order topology&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer is &lt;strong&gt;no&lt;/strong&gt;.  Here is one simple way to see this: recall that a point $x$ in a topological space $X$ is a &lt;a href=&quot;http://en.wikipedia.org/wiki/Cut-point&quot; rel=&quot;nofollow&quot;&gt;cut-point&lt;/a&gt; if $X \setminus x$ is disconnected.  For $n &gt; 1$, $\mathbb{R}^n$ has no cut-points.  However, for any order topology, all but at most two points are cut-points, namely the points corresponding to the largest and smallest elements of the order, if any.  [To be clear, these points &lt;em&gt;may&lt;/em&gt; still be cut points, as for instance with $\mathbb{Q} \cap [0,1]$.] Otherwise $(-\infty,x)$, $(x,\infty)$ gives a separation of $X$.  Since $\mathbb{R}^n$ has more than two points (!), for $n &gt; 1$ the Euclidean topology is not an order topology.  &lt;/p&gt;&#xA;" OwnerUserId="299" LastActivityDate="2010-08-19T23:00:21.523" CommentCount="2" />
  <row Id="2837" PostTypeId="1" AcceptedAnswerId="5246" CreationDate="2010-08-19T23:28:10.120" Score="5" ViewCount="1261" Body="&lt;p&gt;Given a line segment, denoted by it's 2 endpoints (X1, Y1) and (X2, Y2), and a circle, denoted by it's center point (Xc, Yc) and a radius R, how can I tell if the line segment is a tangent of or runs through this circle? I don't need to be able to discern between tangent or running through a circle, I just need to be able to discern between the line segment making contact with the circle in any way and no contact. If the line segment enters but does not exit the circle (if the circle contains an endpoint), that meets my specs for it making contact.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In short, I need a function to find if any point of a line segment lies in or on a given circle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My application is that I'm using the circle as a proximity around a point. I'm basically testing if one point is within R distance of any point in the line segment. And it must be a line segment, not a line.&lt;/p&gt;&#xA;" OwnerUserId="757" LastEditorUserId="757" LastEditorDisplayName="" LastEditDate="2010-08-19T23:54:44.083" LastActivityDate="2010-09-28T17:01:14.923" Title="How to tell if a line segment intersects with a circle?" Tags="&lt;geometry&gt;" AnswerCount="8" CommentCount="8" />
  <row Id="2838" PostTypeId="2" ParentId="2837" CreationDate="2010-08-19T23:32:48.927" Score="8" ViewCount="" Body="&lt;p&gt;Find the intersections of the line containing the segment and the circle. This amounts to solving a quadratic equation. If there are no intersections (&lt;em&gt;i.e.&lt;/em&gt; the solutions of the corresponding equation are non-real), then your segment does not intersect the circle. Now, if there &lt;em&gt;are&lt;/em&gt; intersections, see whether they are inside the segment or not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To implement this, let $P$ and $Q$ be the endpoints of your segment and $C$ and $r$ be the center and the radius of your circle. Then every point of the line through $P$ and $Q$ is given by the formula $$t P + (1-t) Q$$ for exactly one value of $t\in\mathbb R$, and the points in the segment are precisely those for which the corresponding $t$ is in the interval $[0,1]$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the point corresponding to $t\in\mathbb R$ is on the circle if and only if $$\langle t P + (1-t) Q - C,t P + (1-t) Q - C\rangle = r^2,$$ where $\langle\mathord\cdot,\mathord\cdot\rangle$ is the usual inner (dot) product of vectors. If you solve this equation—this is easy, as it is a quadratic equation for $t$—, you find the $t$'s corresponding to the points of intersection of the circle and the line, and if at least one of those $t$'s belong to the interval $[0,1]$, then the &lt;em&gt;segment&lt;/em&gt; intersects the circle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Later:&lt;/strong&gt; I tried to have &lt;em&gt;Mathematica&lt;/em&gt; do the computation for me. Assuming $C=(0,0)$ and $r=1$, as we may up to translation and rescaling, and letting $P=(p1,p2)$, $Q=(q1,q2)$, the following computes the $t$'s:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;P = {p1, p2};&#xA;Q = {q1, q2};&#xA;ts = t /. Solve[Norm[t P + (1 - t) Q]^2 == 1, t];&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This ends up with the variable ts having the value&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left\{\frac{-p_1 q_1-p_2 q_2-\sqrt{p_1^2&#xA;   \left(-q_2^2\right)-2 p_1 q_1+2 p_2 p_1 q_1&#xA;   q_2-p_2^2 q_1^2-2 p_2&#xA;   q_2+p_1^2+p_2^2+q_1^2+q_2^2}+q_1^2+q_2^2}{-2 p_1&#xA;   q_1-2 p_2 q_2+p_1^2+p_2^2+q_1^2+q_2^2},\frac{-p_1&#xA;   q_1-p_2 q_2+\sqrt{p_1^2 \left(-q_2^2\right)-2 p_1&#xA;   q_1+2 p_2 p_1 q_1 q_2-p_2^2 q_1^2-2 p_2&#xA;   q_2+p_1^2+p_2^2+q_1^2+q_2^2}+q_1^2+q_2^2}{-2 p_1&#xA;   q_1-2 p_2 q_2+p_1^2+p_2^2+q_1^2+q_2^2}\right\}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(This was a huge formula, which will likely not fit on your browser window...)&#xA;If the expression inside the square roots is negative, there are no real $t$'s, so the line (hence, &lt;em&gt;a fortiori&lt;/em&gt; the segment) and the circle are disjoint. If not, now we need to see if at least one of the roots in in $[0,1]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I now tell &lt;em&gt;Mma&lt;/em&gt; to tell me when then first of the roots is in $[0,1]$, by telling&#xA;her to compute&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Reduce[0 &amp;lt;= ts[[1]] &amp;lt;= 1, Reals]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;it works for a while, and comes up with a huge answer, presumably equivalent to the original&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;0 &amp;lt;= ts[[1]] &amp;lt;= 1&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Can anyone make sense of the huge answer? (if one asks instead for the more meaningful&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Reduce[0 &amp;lt;= ts[[1]] &amp;lt;= 1 || 0 &amp;lt;= ts[[2]] &amp;lt;= 1, t, Reals]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;the same thing happens)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;PS:&lt;/strong&gt; Please notice that I have made absolutely no attempt at being fast or particularly smart with this idea: it is just the straightforward was to set up the problem and solve it.&lt;/p&gt;&#xA;" OwnerUserId="274" LastEditorUserId="274" LastEditorDisplayName="" LastEditDate="2010-08-20T14:57:34.153" LastActivityDate="2010-08-20T14:57:34.153" CommentCount="9" />
  <row Id="2839" PostTypeId="2" ParentId="2837" CreationDate="2010-08-20T00:35:20.227" Score="5" ViewCount="" Body="&lt;p&gt;Maybe this is what you are getting at...  Given a line through points $(x_1,y_1)$ and $(x_2,y_2)$, and given a single point $p$ &lt;em&gt;not necessarily on the line&lt;/em&gt;, we want to find out: &lt;em&gt;If we draw a circle of radius $r$ around $p$, does that circle intersect the line?&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If this is what you want then you are looking for the &lt;a href=&quot;http://mathworld.wolfram.com/Point-LineDistance2-Dimensional.html&quot; rel=&quot;nofollow&quot;&gt;perpendicular distance from the point to the line&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The derivation on MathWorld may not be want you want to look at right now considering your stated familiarity with so many symbols - so let's just get right down to it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given that your line passes through the two points $(x_1,y_1)$ and $(x_2,y_2)$, a point $(a,b)$ not necessarily on the line, and some radius $r$ around the point $(a,b)$, in order to see if your point $(a,b)$ is within $r$ from the line, you need to check&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;if&lt;/code&gt; $r \geq \frac{|(x_1 - x_2)(a-x_1) + (y_2 - y_1)(b-y_1)|}{\sqrt{(x_1-x_2)^2 + (y_2-y_1)^2}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;then&lt;/code&gt; $(a,b)$ is within $r$ of the line (taken as an orthogonal distance).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt; I don't think this will work if you are working with just a line segment... This solution assumes your line is extended infinitely in both directions. &lt;strong&gt;(update: there are scenarios where this would fail)&lt;/strong&gt; I think the case of the segment itself being further away than $r$ can be taken care of in cases before you arrive at the &quot;if&quot; statement above.  It should be something along the lines of &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;if&lt;/code&gt; $r \leq \sqrt{(x_1 - a)^2 + (y_1 - b)^2}$ &lt;code&gt;or&lt;/code&gt; $r \leq \sqrt{(x_2-a)^2 + (y_2 - b)^2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;then&lt;/code&gt; the segment is too far away to even consider checking the longer inequality above. &lt;strong&gt;(again, the previous if/then is insufficient... still thinking)&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Final Edit:&lt;/strong&gt; This approach has been adapted and completed far more satisfactorily by &lt;a href=&quot;http://math.stackexchange.com/questions/2837/how-to-tell-if-a-line-segment-intersects-with-a-circle/2844#2844&quot;&gt;Isaac&lt;/a&gt;.  I will offer a &lt;a href=&quot;http://math.stackexchange.com/questions/2837/how-to-tell-if-a-line-segment-intersects-with-a-circle/2862#2862&quot;&gt;second approach&lt;/a&gt; that may be exactly what &lt;a href=&quot;http://math.stackexchange.com/questions/2837/how-to-tell-if-a-line-segment-intersects-with-a-circle/2838#2838&quot;&gt;Mariano&lt;/a&gt; did, but I'll try to keep it short sweet and simple.&lt;/p&gt;&#xA;" OwnerUserId="121" LastEditorUserId="121" LastEditorDisplayName="" LastEditDate="2010-08-20T16:24:25.783" LastActivityDate="2010-08-20T16:24:25.783" CommentCount="4" />
  <row Id="2840" PostTypeId="2" ParentId="2827" CreationDate="2010-08-20T01:34:57.837" Score="0" ViewCount="" Body="&lt;p&gt;I think you cannot get anything better than &lt;a href=&quot;http://www.ams.org/bookstore-getitem/item=gsm-19-r&quot; rel=&quot;nofollow&quot;&gt;Evans' book&lt;/a&gt;. Its size may be a little scaring, but it is the most clear and well written book on the subject I ever met.&lt;/p&gt;&#xA;" OwnerUserId="233" LastActivityDate="2010-08-20T01:34:57.837" CommunityOwnedDate="2010-08-20T01:34:57.837" />
  <row Id="2841" PostTypeId="2" ParentId="2827" CreationDate="2010-08-20T02:40:13.453" Score="1" ViewCount="" Body="&lt;p&gt;Partial Differential Equations for Scientists and Engineers by Farlow. It's Dover, so it's cheap. And it's a great first intro - very applied. If you want to follow on with a more rigorous one, you can't beat Evans (Springer - ISBN13: 978-0821207729)&lt;/p&gt;&#xA;" LastActivityDate="2010-08-20T02:40:13.453" CommunityOwnedDate="2010-08-20T02:40:13.453" CommentCount="1" />
  <row Id="2843" PostTypeId="1" AcceptedAnswerId="2863" CreationDate="2010-08-20T06:51:52.177" Score="2" ViewCount="506" Body="&lt;p&gt;I am interested in image analysis and am looking for an approachable tutorial to the Wiener filter. At some point I am interested in implementing such a filter but I would like to have a deeper understanding of the algorithms I'll be writing. Does anyone have recommendations for papers, web sites, etc. that present good coverage on the topic?&lt;/p&gt;&#xA;" OwnerUserId="124" LastEditorUserId="171" LastEditorDisplayName="" LastEditDate="2010-08-20T12:13:22.767" LastActivityDate="2010-08-20T16:32:07.287" Title="Wiener filter: A good tutorial" Tags="&lt;signal-processing&gt;&lt;fourier-analysis&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" />
  <row Id="2844" PostTypeId="2" ParentId="2837" CreationDate="2010-08-20T07:08:49.000" Score="2" ViewCount="" Body="&lt;p&gt;If I understand your edit correctly, you have a point $C=(x_c,y_c)$ and the line segment with endpoints $P=(x_1,y_1)$ and $Q=(x_2,y_2)$ and you want to know whether or not $C$ is within a distance $r$ of any point on the line segment $\overline{PQ}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Working from &lt;a href=&quot;http://math.stackexchange.com/questions/2837/how-to-tell-if-a-line-segment-intersects-with-a-circle/2839#2839&quot;&gt;Tom Stephens's answer&lt;/a&gt;, let's look at the whole line $\overleftrightarrow{PQ}$ containing the line segment $\overline{PQ}$.  The distance between $C$ and $\overleftrightarrow{PQ}$ is $d=\frac{|(x_1 - x_2)(x_c-x_1) + (y_2 - y_1)(y_c-y_1)|}{\sqrt{(x_1-x_2)^2 + (y_2-y_1)^2}}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) If $d&gt;r$, then $C$ is too far from $\overleftrightarrow{PQ}$ to be within $r$ of any point on $\overline{PQ}$, finished.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $d\le r$, then $C$ is close enough to $\overleftrightarrow{PQ}$, but may not be close enough to $\overline{PQ}$.  $d$ is the perpendicular distance from $C$ to $\overleftrightarrow{PQ}$.  Consider two cases&amp;mdash;first, that $C$ and $\overline{PQ}$ are arranged such that the perpendicular distance is to a point on $\overline{PQ}$; second, that the perpendicular distance is to a point not on $\overline{PQ}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) In the first case, $\angle PQC$ and $\angle QPC$ each have measure &amp;le;90&amp;deg;, so the cosine of each angle must be nonnegative so $\frac{(x_1-x_2)^2-(x_1-x_c)^2+(x_2-x_c)^2+(y_1-y_2)^2-(y_1-y_c)^2+(y_2-y_c)^2}{2\sqrt{((x_1-x_2)^2+(y_1-y_2)^2)((x_2-x_c)^2+(y_2-y_c)^2)}}\ge 0$ and $\frac{(x_1-x_2)^2+(x_1-x_c)^2-(x_2-x_c)^2+(y_1-y_2)^2+(y_1-y_c)^2-(y_2-y_c)^2}{2\sqrt{((x_1-x_2)^2+(y_1-y_2)^2)((x_1-x_c)^2+(y_1-y_c)^2)}}\ge 0$ (from the Law of Cosines).  [ &lt;em&gt;edit 2&lt;/em&gt;: After a more careful reading of &lt;a href=&quot;http://math.stackexchange.com/questions/2837/how-to-tell-if-a-line-segment-intersects-with-a-circle/2848#2848&quot;&gt;damiano's answer&lt;/a&gt;, it occurred to me that these are much more hideously complicated than necessary.  The denominators do not affect the sign at all and the numerators are somewhat simpler when expanded, so the two inequalities can be replaced by: $x_1x_2-x_1x_c+x_2x_c-x_2^2+y_1y_2-y_1y_c+y_2y_c-y_2^2\ge0$ and $x_1x_2+x_1x_c-x_2x_c-x_1^2+y_1y_2+y_1y_c-y_2y_c-y_1^2\ge0$.  These are more clearly equivalent to damiano's vector-dot-product criteria. ]  If both of these inequalities hold, then $d\le r$ is the distance from $C$ to $\overline{PQ}$, so $C$ is within $r$ of $\overline{PQ}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(3) If those two inequalities from the Law of Cosines do not both hold, then $d$ is not the distance from $C$ to any point on $\overline{PQ}$, so the shortest distance from $C$ to any point on $\overline{PQ}$ is the distance to the closer of $P$ and $Q$: $d'=\min(\sqrt{(x_1-x_c)^2+(y_1-y_c)^2},\sqrt{(x_2-x_c)^2+(y_2-y_c)^2})$.  If $d'\le r$, then $C$ is within $r$ of $\overline{PQ}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&lt;strong&gt;edit&lt;/strong&gt; (adding the numbers above, also)&lt;/em&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given &lt;a href=&quot;http://math.stackexchange.com/questions/2837/how-to-tell-if-a-line-segment-intersects-with-a-circle/5246#5246&quot;&gt;Corey Ogburn's answer&lt;/a&gt;, I thought it might be helpful to add a diagram and discuss what parts of my original answer address which regions.  The numbers below discuss the correspondingly-numbered part of my original answer, above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/rxaPF.png&quot; alt=&quot;diagram&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) If $d&gt;r$, then the point $C$ is in the beige region.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) $d\le r$, so if $\angle PQC$ and $\angle QPC$ each have measure &amp;le;90&amp;deg;, then $C$ is in the purple region.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(3) $C$ is not in the beige or purple regions, so we only need to see whether it's in the blue or red regions (based on whether $C$ is close enough to the endpoints of the segment).&lt;/p&gt;&#xA;" OwnerUserId="72" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-09-23T15:12:20.280" LastActivityDate="2010-09-23T15:12:20.280" CommentCount="0" />
  <row Id="2845" PostTypeId="1" AcceptedAnswerId="2853" CreationDate="2010-08-20T09:18:11.370" Score="3" ViewCount="146" Body="&lt;p&gt;Suppose I have a matrix $S$ having a one-dimensional nullspace $\lbrace e \rbrace$ such that $S + ee^\top$ is a positive definite symmetric matrix.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let $b \in Range(S)$ and suppose I solve the equation $(S + ee^\top)x = b$ is there anyway I can derive the solution $x'$ of the equation $Sx' = b$?   I was trying a Sherman Morrison Woodbury type formula, but this fails since the denominator is 0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any help would be appreciated.&lt;/p&gt;&#xA;" OwnerUserId="780" LastActivityDate="2010-08-20T18:19:34.023" Title="Solving a linear equation given the solution of another." Tags="&lt;linear-algebra&gt;&lt;matrices&gt;" AnswerCount="1" CommentCount="0" />
  <row Id="2847" PostTypeId="2" ParentId="2815" CreationDate="2010-08-20T09:26:17.627" Score="1" ViewCount="" Body="&lt;p&gt;For $O(n^2)$ (randomized) time, you can also use a hashtable of size $\Theta(n^2)$. Looking up will be constant time because the number of taxicab numbers is $O(n^2)$.&lt;/p&gt;&#xA;" OwnerUserId="512" LastActivityDate="2010-08-20T09:26:17.627" />
  <row Id="2848" PostTypeId="2" ParentId="2837" CreationDate="2010-08-20T09:38:51.833" Score="2" ViewCount="" Body="&lt;p&gt;If I understood what you want correctly, you need to find conditions on a line segment so that it lies entirely outside a circle; a possible reason for confusion is that mathematicians tend to call &quot;circle&quot; just the one-dimensional boundary of the &quot;disk&quot; and they call &quot;disk&quot; its interior.  In this terminology, I think that you want the segment to lie completely outside of the closed disk; this is the question that I will an answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, a geometric description of the answer.  Let $P$ and $Q$ be the endpoints of the segment, let $C$ be the center of the circle and let $r$ be the radius of the circle $C$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) We rule out the possibility that at least one among $P$ and $Q$ is contained inside the closed disk.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) If in the triangle $PQC$ one of the two angles on the segment $PQ$ is not acute, then moving along the segment $PQ$ and away from the non-acute angle the distance from the center $C$ increases, so that there is no intersection in this case because of 1).  Otherwise, the two angles on the segment $PQ$ are acute, so that the minimum distance between the center $C$ and the points on the segment $PQ$ is achieved inside the segment.  In this case (i.e. when the two angles on $PQ$ are acute), we make sure that the distance between the center $C$ and the line joining $P$ and $Q$ is larger than $r$.  In view of 1) and the previous discussion this implies that there is no contact between the segment and the disk.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Translation into formulas:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) $dist(P,C)&gt;r$ and $dist(Q,C)&gt;r$;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) if the inequalities $PC \cdot PQ &gt; 0$ and $QC \cdot QP &gt; 0$ hold, then also the inequality $||PC \times PQ|| &gt; r ||PQ||$ must hold (in the formulas, we denote by $\cdot$ the scalar product, by $\times$ the cross product, and by $||-||$ the length of a vector).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Explicit equations, assuming that $C$ is the origin (which you can always achieve by translations):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) $x_P^2+y_P^2 &gt; r^2$ and $x_Q^2+y_Q^2 &gt; r^2$;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) if both $x_P^2 + y_P^2 \geq x_P x_Q + y_P y_Q$ and $x_Q^2 + y_Q^2 \geq x_P x_Q + y_P y_Q$ hold, then check that also $(x_Py_Q-x_Qy_P)^2 &gt; r^2 ((x_P-x_Q)^2 + (y_P-y_Q)^2)$ holds.&lt;/p&gt;&#xA;" OwnerUserId="1028" LastEditorUserId="1028" LastEditorDisplayName="" LastEditDate="2010-08-20T10:23:47.617" LastActivityDate="2010-08-20T10:23:47.617" CommentCount="1" />
  <row Id="2850" PostTypeId="2" ParentId="2830" CreationDate="2010-08-20T10:04:50.640" Score="2" ViewCount="" Body="&lt;p&gt;p such that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\prod_i q_i^{p_i-q_i} \ge 1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I arrive at this point with the following calculations:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First of all I rewrite the equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_i p_i \log q_i \ge \sum_i q_i \log q_i$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_i (p_i \log q_i - q_i \log q_i) = \sum_i (\log q_i^{p_i-q_i}) = \log \prod_i q_i^{p_i-q_i} \ge 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most simple deduction about p is that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$p_i \ge q_i , \forall i$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$q_i \ge 1$$&lt;/p&gt;&#xA;" OwnerUserId="422" LastEditorUserId="171" LastEditorDisplayName="" LastEditDate="2010-08-20T15:15:16.400" LastActivityDate="2010-08-20T15:15:16.400" CommentCount="1" />
  <row Id="2851" PostTypeId="1" AcceptedAnswerId="2875" CreationDate="2010-08-20T10:06:11.717" Score="2" ViewCount="114" Body="&lt;p&gt;I am crossposting this question from Stack Overflow since it is more math then programming related.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In Reference to &lt;a href=&quot;http://stackoverflow.com/questions/2500499/howto-project-a-planar-polygon-on-a-plane-in-3d-space&quot;&gt;my question&lt;/a&gt; about projecting a planar Polygon to a plane, i came to another problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to project a base contour which is aligned to the x-y-plane along a projection vector from from a point P1 to a point P2. This already works, but now i also need to change the alignment of my base contour using the position of P1  and two vectors defining the new alignement base's x and y should be transformed to.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to use a transformation matrix M for this, which premultiplied to my projecton matrix p should give me the transformation matrix from a vertex v in the base contour to a vertex v' on the projection plane such as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;v' = (P x M) * v&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, my question is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How to create a transformation matrix which transforms a vertex v relative to a coordinate system with O = (0,0,0), xdir = (1,0,0), and ydir = (0,1,0) to a vertex relative to a coordinate system with O'=(ox, oy, oz) and xdir and ydir as two arbitrary perpendicular unit vectors?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Clarifications&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;I use 4x4 matrices&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1357" LastActivityDate="2010-08-20T23:42:27.787" Title="Constructing an Extrusion between two non-parallel planes" Tags="&lt;matrices&gt;&lt;transformation&gt;" AnswerCount="1" />
  <row Id="2853" PostTypeId="2" ParentId="2845" CreationDate="2010-08-20T11:14:27.970" Score="5" ViewCount="" Body="&lt;p&gt;Suppose $x$ is the solution to $(S+ee^T)x = b$. Since $b\in Range(S)$, we may write $b = Sz$ for some z. Now compute:&#xA;$$&#xA;e^T(S+ee^T)x = e^Tb = e^TSz&#xA;$$&#xA;Since S is symmetric, and e is in its nullspace, we have $e^TS = 0$. So the above equation simplifies to $e^Tx = 0$. But this implies&#xA;$$&#xA;(S+ee^T)x = Sx&#xA;$$&#xA;So x is a solution to the equation $Sx=b$ as well. As noted above, the solution to $Sx=b$ is not unique; $x + \lambda e$ is also a solution for any real $\lambda$.&lt;/p&gt;&#xA;" OwnerUserId="1356" LastEditorUserId="1356" LastEditorDisplayName="" LastEditDate="2010-08-20T18:19:34.023" LastActivityDate="2010-08-20T18:19:34.023" CommentCount="6" />
  <row Id="2854" PostTypeId="2" ParentId="2584" CreationDate="2010-08-20T13:35:46.543" Score="1" ViewCount="" Body="&lt;p&gt;First let's make sure I understood your question correctly:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;You have a probability function that is expressed as a sum/product of N parametrized one-dimensional gaussians, each with different mean and standard deviation.&lt;/li&gt;&#xA;&lt;li&gt;You want to generate stochastic variables according to this distribution.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Is this correct?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If this is the case, I reccommend you use a variation of &lt;a href=&quot;http://en.wikipedia.org/wiki/Rejection_sampling&quot; rel=&quot;nofollow&quot;&gt;rejection sampling&lt;/a&gt;. The recipe is quite straightforward, but you might have to iterate a bit before you get an answer out of it. This is the basic outline.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;You generate a uniformly distributed random number in your desired interval, x.&lt;/li&gt;&#xA;&lt;li&gt;You calculate the value of your probability distribution, p(x)&lt;/li&gt;&#xA;&lt;li&gt;You generate another uniformly distributed random number between 0 and 1, q&lt;/li&gt;&#xA;&lt;li&gt;If q &amp;lt; p(x), return x&lt;/li&gt;&#xA;&lt;li&gt;If not, start from step 1.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;No matter how large the temptation, do not re-use q for different iterations.&lt;/p&gt;&#xA;" OwnerUserId="2684" LastActivityDate="2010-08-20T13:35:46.543" CommentCount="5" />
  <row Id="2855" PostTypeId="1" AcceptedAnswerId="2870" CreationDate="2010-08-20T15:03:59.053" Score="1" ViewCount="248" Body="&lt;p&gt;Could someone verify this proof?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given the following definition of divisibility $(\forall{a,b \in \mathbb{N}})[a | b \equiv (\exists{c \in \mathbb{N}})(ac=b)]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Statement: $(\forall{a,b,c \in \mathbb{N}})[(a|b) \rightarrow (ac|bc)] $ .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proof:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $a,b \in \mathbb{N}$ be arbitrary.  Assuming that $a|b$, we can conclude that $\exists{d \in \mathbb{N}}$ such that $ad = b$.  By Algebra we can deduce the following sequence of statements.  If $(ad)=b$, then $(ad)c=bc$.  Therefore $(ac)d=bc$.  Hence by the definition of divisibility we know that $ac|bc$.&lt;/p&gt;&#xA;" OwnerUserId="1360" LastActivityDate="2010-08-20T22:28:45.517" Title="Divisibility Proof" Tags="&lt;number-theory&gt;" AnswerCount="3" CommentCount="8" FavoriteCount="0" />
  <row Id="2856" PostTypeId="1" AcceptedAnswerId="2860" CreationDate="2010-08-20T15:10:28.713" Score="2" ViewCount="115" Body="&lt;p&gt;I'm am trying to define a data structure to represent road networks.  The immediately obvious structure is that of a graph - a set of nodes and edges that connect pairs of nodes.  The nodes would represent things like intersections, and the edges would represent lanes.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the basic concept of a graph is insufficient to describe what I want.  I also want to be able to describe how certain lanes are &quot;reachable&quot; or &quot;adjacent&quot; to other lanes.  (Imagine parallel lanes with no median between them.  You can just switch lanes.)  This seems to imply a need to have some sort of meta-edge that connects pairs of edges.  Is there any mathematical structure that sounds like this?  Is this reducible to the standard concept of a graph? Thanks.&lt;/p&gt;&#xA;" OwnerUserId="1358" LastActivityDate="2010-08-20T17:44:11.297" Title="Looking for a particular generalization of the concept of a mathematical graph" Tags="&lt;graph-theory&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="1" />
  <row Id="2857" PostTypeId="2" ParentId="2435" CreationDate="2010-08-20T15:11:14.547" Score="0" ViewCount="" Body="&lt;p&gt;You can use the Kolmogorov Smirnov test too. It is a non parametric test, and will work on many distributions - including Uniform. The advantage it has over other tests is that it looks at the whole distribution. Also it is fairly easy to use.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The idea is to compare the CDFs of the $H_0$ distribution with the observed empirical distribution, and find the maximum difference. This difference does not depend on the distribution (hence it is non-parametric), and depends only on $n$, the number of observations you are drawing. Then you look it up against a table to see how different it is from expected range of the difference.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Empirical CDF of your observations $\{x_0,x_1,...,x_n\}$ is $F_n(x)=\frac{1}{n}\sum_1^nI_{x_i\lt x}$. Then your test statistic will be $D_n=\sup|F(x)-F_n(x)|$. Assuming you sort your $x_n$'s in ascending order, and assuming your numbers come from Uniform[0,1] (wlg since you can scale them appropriately) it will be $D_n=\max_i(max(|x_i-\frac{i}{n}|,|x_i-\frac{i-1}{n}))$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you compute that, it's only a matter of looking up table or using the right software. I recommend &lt;a href=&quot;http://www.ciphersbyritter.com/JAVASCRP/NORMCHIK.HTM&quot; rel=&quot;nofollow&quot;&gt;this site&lt;/a&gt; which has an excellent tool to compute KS along with many other tests. For KS, you need to select &quot;Dn&quot;, put in the value you got, press the button, and assume interpret the result as it is truly uniform if P (computed in the form) &amp;lt; $\alpha$.&lt;/p&gt;&#xA;" OwnerUserId="1310" LastActivityDate="2010-08-20T15:11:14.547" CommentCount="2" />
  <row Id="2858" PostTypeId="2" ParentId="2856" CreationDate="2010-08-20T15:33:39.653" Score="0" ViewCount="" Body="&lt;p&gt;May be you can use a colored graph with the convention that if there are more than edges from point A to point B, they are switchable inbetween.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Edit (after comment that A and B may not be same):&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me know if I am wrong, but geographically close points may be quite far apart topologically (i.e. when they have many intermediate closely situated points). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the problem boils down to given a standard graph, for each edge how to best store which edges are 'reachable'.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reachability seems to be a reflexive and commutative relation (i.e. a is reachable from b implies b is reachable from a). If reachability is also a transitive relation (i.e. if i can reach b from a and c from b then i can reach c from a), then you can still use the coloring approach. All you need to do is to divide the edges into equivalence classes, and assign separate colors for each equivalence class.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, if it is not transitive, then the only option I see is for each edge to store a list of edges that are reachable from it.&lt;/p&gt;&#xA;" OwnerUserId="1310" LastEditorUserId="1310" LastEditorDisplayName="" LastEditDate="2010-08-20T16:52:58.803" LastActivityDate="2010-08-20T16:52:58.803" CommentCount="5" />
  <row Id="2860" PostTypeId="2" ParentId="2856" CreationDate="2010-08-20T15:39:48.353" Score="3" ViewCount="" Body="&lt;p&gt;You could make the lanes themselves nodes as well&lt;/p&gt;&#xA;" OwnerUserId="136" LastActivityDate="2010-08-20T15:39:48.353" CommentCount="2" />
  <row Id="2861" PostTypeId="2" ParentId="2855" CreationDate="2010-08-20T16:11:54.817" Score="0" ViewCount="" Body="&lt;p&gt;Yes, your proof is correct. Since it employs only the associative and commutativite axioms for multiplication it remains true in any commutative semigroup. The proof employs only the inference rules of first-order equational logic, namely that equality is an equivalence relation (reflexive, symmetric, transitive) and the rules of &lt;strong&gt;substitution&lt;/strong&gt;: $p(x) = q(x) \rightarrow p(a) = q(a)$ and &lt;strong&gt;replacement&lt;/strong&gt;: $a = b \rightarrow p(a) = p(b)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By a famous theorem of Birkhoff, for any algebraic structure that can be defined by axioms - like comutativity, associativity, etc - that are all universally quantified &lt;strong&gt;equations&lt;/strong&gt;, said inference rules are &lt;em&gt;complete&lt;/em&gt;, i.e. an equation is provable from the axioms using said inference rules iff it is true in every structure that satisfies the axioms. However, finding such a purely equational proof can be highly difficult. For example, consider the famous Jacobson theorem that rings satisfying the axiom $x^n = x \;$ are commutative $xy = yx$.  it's notoriously difficult for $n&gt;2$ to find equational proofs of this theorem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Returning to the specific problem: generally scaling a polynomial $f(x)$ by a cancelable element $c$ does not change its solution set since $c f(d) = 0 \iff f(d) = 0$. In particular the  solvability of $f(x) = 0$ is not altered - which is what you are concerned with in the definition of divisibility, i.e. the solvability of $ax = b$. E.g. consider the less trivial case of quadratic $f(x)$. Note that the formula $g(a,b,c)$ for the solutions of a quadratic $ax^2+bx+c$ remains unaltered by nonzero coefficient scalings $g(ad,bd,cd) = g(a,b,c)$. As for solvability, scaling the polynomial by $d$ multiplies the discriminant $b^2-4ac$ by $d^2$ so it does not affect the solvability criterion - that the discriminant be a perfect square. But we don't need to know these explicit formulas to deduce this. Rather, we can deduce it at a conceptually higher-level, simply from the cancelability of $c$, i.e the invertibility of  scaling transformations $x\to cx$.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-08-20T18:05:38.343" LastActivityDate="2010-08-20T18:05:38.343" />
  <row Id="2862" PostTypeId="2" ParentId="2837" CreationDate="2010-08-20T16:21:56.037" Score="1" ViewCount="" Body="&lt;p&gt;Well, what the heck... there has been so much virtual ink spilled at this point that what's another gallon, really?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(&lt;strong&gt;update&lt;/strong&gt;: This is precisely the solution &lt;a href=&quot;http://math.stackexchange.com/users/274/mariano-suarez-alvarez&quot;&gt;Mariano proposed&lt;/a&gt; long before I did, it just happens that I couched this in a different notation.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The principle behind &lt;em&gt;this&lt;/em&gt; approach is that we can find out algebraically whether the set of coordinates defining the perimeter of a circle with center $(h,k)$ and radius $r$ contains any point of the line segment connecting the points $(x_1,y_1)$ and $(x_2,y_2)$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, writing the set of points $(x,y)$ lying along the line segment as&#xA;$$tx_1 + (1-t)x_2 = x$$&#xA;$$ty_1 + (1-t)y_2 = y$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $t$ ranges between $0$ and $1$ inclusive, and writing the equation for the circle centered at $(h,k)$ with radius $r$ as&#xA;$$(x-h)^2 + (y-k)^2 = r^2,$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;then we can simply substitute our expressions for $x$ and $y$ into the equation for the circle, and then figure out if there at least one such $t \in [0,1] \subset \mathbb{R}$.  If there is, then the perimeter of the circle and the line segment share at least one point.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since we agreed that we have seen enough virtual ink today, I will simply write down what that substitution looks like and then the final computation that needs to be done.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The substitution&lt;/strong&gt;: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left(tx_1 + (1-t)x_2 - h \right)^2 + \left(ty_1 + (1-t)y_2 - k \right)^2 = r^2.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(This is a quadratic equation in $t$, note that the values of $x_1,x_2,y_1,y_2,h,k,$ and $r$ are known.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The computation&lt;/strong&gt;: &#xA;$$At^2 + Bt + C = 0$$&#xA;with &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$A = x_{1}^{2} - 2x_{1}x_{2} + x_{2}^{2},$&lt;br&gt;&#xA;$B = 2x_{1}x_{2} - 2x_{2}^{2} - 2hx_{2} + 2hx_{2},$ and &lt;br&gt;&#xA;$C = x_{2}^{2} - 2hx_{2} + h^{2} - r^{2}.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solve for $t$ using the quadratic formula and then test whether at least one $t$ is a real number between $0$ and $1$, inclusive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;PS:&lt;/strong&gt; If I have stomped on another's solution that has already been posted it is because I have not done due diligence and verified all of the solutions given previously.  Please let me know in the comments.&lt;/p&gt;&#xA;" OwnerUserId="121" LastEditorUserId="121" LastEditorDisplayName="" LastEditDate="2010-08-20T16:46:51.480" LastActivityDate="2010-08-20T16:46:51.480" CommentCount="2" />
  <row Id="2863" PostTypeId="2" ParentId="2843" CreationDate="2010-08-20T16:32:07.287" Score="1" ViewCount="" Body="&lt;p&gt;Detection Estimation and Modulation Theroy Volume 1 by Harry Van Trees is a good start if you have a basic understanding of Probability and Random Variables.  Any statistical signal processing book should also cover Wiener Filtering&lt;/p&gt;&#xA;" OwnerUserId="832" LastActivityDate="2010-08-20T16:32:07.287" />
  <row Id="2864" PostTypeId="1" AcceptedAnswerId="2867" CreationDate="2010-08-20T16:33:21.137" Score="7" ViewCount="546" Body="&lt;p&gt;A &lt;a href=&quot;http://en.wikipedia.org/wiki/Hyperbolic_geometry&quot; rel=&quot;nofollow&quot;&gt;hyperbolic geometry&lt;/a&gt; is a non-Euclidian geometry with constant negative curvature. It has the property that given a line and a point, many lines can be drawn containing the point that never meet the given line. The picture below (Circle Limit by M.C.Escher) is a conformal map of hyperbolic geometry to Euclidian plane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/rklhN.png&quot; /&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do you show that in a space with hyperbolic geometry, area of a triangle with angles $\alpha$, $\beta$, and $\gamma$ is $\Delta\propto\pi-\alpha-\beta-\gamma$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A similar result exists for sphere: $\Delta=R^2(\alpha+\beta+\gamma-\pi)$. I could prove this, but I suspect my proof cannot be modified for hyperbolic geometry. My proof for the sphere is fairly trivial, it makes use of symmetry and a Venn diagram.&lt;/p&gt;&#xA;" OwnerUserId="1310" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-08-20T22:48:14.833" LastActivityDate="2010-08-20T23:32:20.923" Title="Area of a triangle $\propto\pi-\alpha-\beta-\gamma$" Tags="&lt;geometry&gt;&lt;hyperbolic-geometry&gt;" AnswerCount="4" CommentCount="2" />
  <row Id="2865" PostTypeId="1" AcceptedAnswerId="5110" CreationDate="2010-08-20T16:33:23.493" Score="7" ViewCount="336" Body="&lt;p&gt;If $(G_i,f_{ij})$ is a direct system of topological groups, is it always the case that the  topological\group-theoretical direct limit $G:=\varinjlim_iG_i$ is a topological group? (The topology on $G$ is the final topology with respect to the canonical homomorphisms $\psi_i:G_i\rightarrow G$). It is immediate from the group structure on $G$ and the definition of its topology that the inversion map is continuous. Moreover, I'm pretty certain that if the transition maps $f_{ij}$ are open, then the $\psi_i$ are open, and this gives continuity of multiplication. It's not at all clear to me that this condition is necessary though. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have always sort of assumed that this was true, but it's never really been an issue because I never start with a system of topological groups and take the direct limit; I always have a topological group and express it as a direct limit (e.g. the idele group of a global field, or the Cartier dual of a finite free $\mathbb{Z}_p$-module).&lt;/p&gt;&#xA;" OwnerUserId="628" LastActivityDate="2010-09-21T01:16:32.377" Title="Is a direct limit of topological groups always a topological group?" Tags="&lt;general-topology&gt;" AnswerCount="1" FavoriteCount="2" />
  <row Id="2866" PostTypeId="2" ParentId="2837" CreationDate="2010-08-20T16:37:41.633" Score="1" ViewCount="" Body="&lt;p&gt;There are too many answers already, but since no one mentioned this, perhaps this might still be useful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can consider using &lt;a href=&quot;http://mathforum.org/dr.math/faq/formulas/faq.polar.html&quot; rel=&quot;nofollow&quot;&gt;Polar Coordinates&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Translate so that the center of the circle is the origin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The equation of a line in polar coordinates is given by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$r = p \sec (\theta - \omega)$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;See the above web page for what $\omega$ is. You can compute $p$ and $\theta$ by using the two endpoints of the segment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If R is the radius of the circle, you need to find all $\theta$ in $[0, 2\pi]$ such that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$R = p \sec (\theta - \omega)$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now all you need to check is if this will allow the point to fall within the line segment.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-08-20T17:03:14.483" LastActivityDate="2010-08-20T17:03:14.483" />
  <row Id="2867" PostTypeId="2" ParentId="2864" CreationDate="2010-08-20T16:39:48.820" Score="4" ViewCount="" Body="&lt;p&gt;I direct you to chapter 1 of &lt;em&gt;Fuchsian Groups&lt;/em&gt; by Svetlana Katok. It has the following proof of the theorem (at least where I learned it) with illustrations:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let us define hyperbolic area to be: $$ \mu(A)=\int_{A}\frac{dxdy}{y^2}. $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It can be verified that for all $T \in$ $\text{PSL}(2,\mathbb{R})$ that $\mu(A)=\mu(T(A)).$ (This fact will be useful to us later).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So let us consider a triangle in this hyperbolic plane $H$. We will use the upper half plane model.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Case 1:&lt;/strong&gt; One vertex of the triangle belongs to $\mathbb{R}\cup {\infty}.$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then the angle at that vertex is 0. We can use transformations (recall $T$ from above?) from $\text{PSL}(2,\mathbb{R})$ to change two sides of $A$ into vertical geodesics; hence the base of the triangle is a Euclidean semicircle orthogonal to  $\mathbb{R}.$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then we can see that $$\mu(A)=\int_{A}\frac{dxdy}{y^2}=\int_{a}^{b}dx\int_{\sqrt{1-x^2}}^{\infty}\frac{dy}{y^2}=\int_{a}^{b}\frac{dx}{\sqrt{1-x^2}}.$$ When we substitute $x=cos(\theta),$ this integral becomes: $$\mu(A)=\int_{\pi - \alpha}^{\beta}\frac{-\sin(\theta)d\theta}{\sin\theta}=\pi-\alpha-\beta.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Case 2:&lt;/strong&gt;  $A$ has no vertices in $\mathbb{R}\cup{\infty}.$ &#xA;Let the triangle have vertices $A, B,$ and $C.$ Then let the geodesic connecting $A$ and $B$ intersect the real axis at $D$. Then $\mu(A)=\mu(ACD)-\mu(BCD)$ and both are triangles as in the previous case. (You can work out the rest of the details accordingly -  draw a picture)&lt;/p&gt;&#xA;" OwnerUserId="298" LastEditorUserId="298" LastEditorDisplayName="" LastEditDate="2010-08-20T17:03:35.217" LastActivityDate="2010-08-20T17:03:35.217" />
  <row Id="2868" PostTypeId="2" ParentId="2864" CreationDate="2010-08-20T17:00:31.773" Score="5" ViewCount="" Body="&lt;ol&gt;&#xA;&lt;li&gt;Observe that &lt;em&gt;defect&lt;/em&gt; of a (hyperbolic or spherical) polygon (i.e. the difference between the sum of its angles and the sum of angles of a Euclidean polygon with the same number of vertices) is an additive congruence-invariant function.&lt;/li&gt;&#xA;&lt;li&gt;Up to proportionality, there exists only one additive congruence-invariant function (for Euclidean plane it is sometimes called &lt;a href=&quot;http://en.wikipedia.org/wiki/Bolyai%E2%80%93Gerwien_theorem&quot; rel=&quot;nofollow&quot;&gt;Bolyai–Gerwien theorem&lt;/a&gt;; one may say that it just means that area is well-defined).&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Hence the result.&lt;/p&gt;&#xA;" OwnerUserId="152" LastActivityDate="2010-08-20T17:00:31.773" CommentCount="2" />
  <row Id="2869" PostTypeId="1" AcceptedAnswerId="2935" CreationDate="2010-08-20T17:02:46.573" Score="3" ViewCount="122" Body="&lt;p&gt;The process below yields a distribution of &quot;response times&quot; (RT), and I'd like to know if there is an analytic solution to obtain the density function of this distribution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An RT is recorded at the earliest time ($t_1$) at which a function, $f(t)$, reaches a threshold.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The threshold varies across iterations of the process according to a random uniform distribution with bounds specified by the parameters $P_1$ and $P_2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f(t)$ begins at $t=0$ and has a form whose derivative is defined by the function $g(t)$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$g(t) = P_3\left (\frac{P_4}{P_4+P_5}\right )\left (1-\exp\left [-(P_4+P_5)\int_{0}^{t}h(s)ds\right ]\right )$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $h(t)$ is defined as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$h(t) = P_6\Gamma(t,P_8,n)[1 - \Gamma(t-P_7,P_9,n)]$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and the gamma function defined as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Gamma(t,\beta,n) = 1- \left ( e^{-\beta t} \right )\left ( \sum_{n-1}^{j=0} \frac{(\beta t)^{j}}{j!}  \right )$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, there are in total 10 parameters in the process: $P_1$ through $P_9$, and $n$. I could of course perform a Euler simulation to generate a distribution of RT values for given a particular set of 10 parameter values, but it would be nice to avoid this because I'm hoping to use a Nelder–Mead search algorithm to find the set of 10 parameter values that maximizes the likelihood of a set of observed RT data points, and it really slows down the search if I have to compute estimates of the density function for every candidate set of parameter values.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for any help you're able to provide!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pursuant to comments, here is a further description of the process:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The process is an adaptation of the model presented by &lt;a href=&quot;http://dx.doi.org/10.1037/a0015156&quot; rel=&quot;nofollow&quot;&gt;Smith &amp;amp; Racliff&lt;/a&gt; (&lt;a href=&quot;http://star.psy.ohio-state.edu/coglab/People/roger/pdf/psychrev09b.pdf&quot; rel=&quot;nofollow&quot;&gt;PDF&lt;/a&gt;) which models perception, attention and decision in the context of a speeded detection task. Function $h(t)$ models perception: accumulation of information about a brief visual stimulus. Function $g(t)$ models the transfer of this information to memory via attention, and function $h(t)$ is a linear ballistic decision process in the vein of &lt;a href=&quot;http://dx.doi.org/10.1016/j.cogpsych.2007.12.002&quot; rel=&quot;nofollow&quot;&gt;Brown &amp;amp; Heathcote&lt;/a&gt; (&lt;a href=&quot;http://psychology.stanford.edu/~jlm/pdfs/BrownHeathcote08.pdf&quot; rel=&quot;nofollow&quot;&gt;PDF&lt;/a&gt;). I thought that supplementing an LBA as the decision mechanism in place of Smith &amp;amp; Ratcliff's diffusion process might simplify the derivation of the final density function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$P_1$ and $P_2$ are the upper and lower bounds of the random uniform threshold.&#xA;$P_3$ is the asymptote of $g(t)$.&#xA;$P_4$ is the intensity of the stimulus.&#xA;$P_5$ is the intensity of the background.&#xA;$P_6$ is the amplitude of the stimulus.&#xA;$P_7$ is the duration of the stimulus.&#xA;$P_8$ is the rate at which the stimulus comes on.&#xA;$P_9$ is the rate at which the stimulus turns off.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Smith &amp;amp; Ratcliff (2009) note that $n$ might be reasonably fixed to a value of 3, so possibly that can further simplify things. Additionally, while Smith &amp;amp; Ratcliff sought to fit their model to different stimulus intensities, I don't need this particular feature, so I'm not sure if I need parameters $P_4$ and $P_5$ (though I'm not sure how to properly remove them either).&lt;/p&gt;&#xA;" OwnerUserId="1361" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-08-30T23:05:38.743" LastActivityDate="2010-08-30T23:05:38.743" Title="Is there an analytic solution for the density function of this complex random variable?" Tags="&lt;functional-analysis&gt;&lt;gamma-function&gt;&lt;functions&gt;&lt;probability&gt;&lt;stochastic-processes&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="0" />
  <row Id="2870" PostTypeId="2" ParentId="2855" CreationDate="2010-08-20T17:05:57.410" Score="0" ViewCount="" Body="&lt;p&gt;The proof is valid.&lt;/p&gt;&#xA;" OwnerUserId="1310" LastActivityDate="2010-08-20T17:05:57.410" CommentCount="2" />
  <row Id="2871" PostTypeId="2" ParentId="2864" CreationDate="2010-08-20T17:13:55.900" Score="4" ViewCount="" Body="&lt;p&gt;There is also a &lt;strong&gt;differential-geometric proof&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a surface with constant Gaussian curvature $K$ &lt;a href=&quot;http://en.wikipedia.org/wiki/Gauss%E2%80%93Bonnet_theorem&quot; rel=&quot;nofollow&quot;&gt;Gauss-Bonnet formula&lt;/a&gt; yields $\text{area}\cdot K+\text{sum of exterior angles}=2\pi$ aka $\text{area}=\text{defect}/K$. (This gives not only proportionality but also the coefficient.)&lt;/p&gt;&#xA;" OwnerUserId="152" LastActivityDate="2010-08-20T17:13:55.900" />
  <row Id="2872" PostTypeId="1" AcceptedAnswerId="2873" CreationDate="2010-08-20T17:24:57.020" Score="1" ViewCount="109" Body="&lt;p&gt;How can I use the chain rule $\frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx}$ to differentiate an expression like $\frac{d}{dx}(y^3)$ ?   The solution in the book is $\frac{d}{dx}(y^3) = 3y^2 \frac{dy}{dx}$.  I'm not sure how they got it though.&lt;/p&gt;&#xA;" OwnerUserId="1360" LastEditorUserId="1360" LastEditorDisplayName="" LastEditDate="2010-08-20T17:43:03.153" LastActivityDate="2010-08-20T18:33:34.493" Title="How do you take the derivative of a variable with respect to a different variable:" Tags="&lt;calculus&gt;" AnswerCount="2" CommentCount="8" />
  <row Id="2873" PostTypeId="2" ParentId="2872" CreationDate="2010-08-20T17:28:02.883" Score="5" ViewCount="" Body="&lt;p&gt;In the problem, you want to differentiate the expression $\color{green}y^{\color{blue}3}$ with the variable $\color{red}x$. But there is no information on what $\color{green}y$ is. So we must assume $\color{green}y$ is an arbitrary function, i.e. $\color{green}y := \color{green}y(\color{red}x)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the crucial point is to note that &lt;strong&gt;the &lt;sup&gt;3&lt;/sup&gt; itself is also a function&lt;/strong&gt;, i.e. we can define $\color{blue}z(\color{green}y) = \color{green}y^{\color{blue}3}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this way, we could apply the chain rule &lt;sub&gt;&lt;sup&gt;(some names changed to be consistent with the problem)&lt;/sup&gt;&lt;/sub&gt;,&#xA;$$ &#xA;\frac{d \color{blue} z}{d \color{red} x} = \frac{d \color{blue} z}{d \color{green} y} \cdot \frac{d \color{green} y}{d \color{red} x}&#xA;$$&#xA;i.e. explictly,&#xA;$$&#xA;\frac{d \Bigl( \color{blue} z\bigl(\color{green}y(\color{red}x)\bigr) \Bigr)}{d \color{red} x}&#xA;= \frac{d \Bigl( \color{blue} z \bigl(\color{green}y\color{silver}{(x)}\bigr) \Bigr) }{d\bigl( \color{green}y\color{silver}{(x)}\bigr)}&#xA;\cdot \frac{d \bigl( \color{green} y (\color{red}x)\bigr) }{d \color{red} x}&#xA;$$&#xA;In $d\color{blue}z/d\color{green}y$, since no other variables (like $\color{red}x$) is directly involved, we could treat $\color{green}y$ as a variable instead of an implicit function. In $d\color{green}y/d\color{red}x$, as the function $\color{green}y(\color{red}x)$ is arbitrary, the differentiated form is the best representation already. Therefore, the expression above simplifies to&#xA;$$&#xA;\frac{d \left( \color{green}y^{\color{blue}3} \right) }{d \color{red} x}&#xA;= \underbrace{  \color{teal}{ 3 \color{green}y^2 }  }_{d(\color{green}y^{\color{blue}3})/d\color{green}y} \cdot &#xA;\frac{d \color{green} y}{d \color{red} x}&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="171" LastEditorUserId="171" LastEditorDisplayName="" LastEditDate="2010-08-20T18:32:45.230" LastActivityDate="2010-08-20T18:32:45.230" CommentCount="2" />
  <row Id="2874" PostTypeId="2" ParentId="2856" CreationDate="2010-08-20T17:33:48.177" Score="-1" ViewCount="" Body="&lt;p&gt;It's not clear whether nodes for intersections are necessary.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You need a data structure that can accomodate (state transitions resulting from) the operations &lt;em&gt;continue&lt;/em&gt; and &lt;em&gt;switch lanes&lt;/em&gt;.   The minimal set of states needed is: those pieces of road where, if you continue driving along them, the set of &quot;switch&quot; options is constant. This is the smallest set of states such that any switch operation is between different states. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could impose additional nodes thought of as STATIONARY states of motion or FIXED-LOCATION geographic data, to express starting points, destinations, intersections, stopovers or geographic locations, but only the lanes on which a traveller moves are logically necessary.  At a road intersection, one can switch from lane A on one road to lane B on another road, which is expressible as a single state transition, without going through an &quot;inside the intersection&quot; state.  Having an intersection state may be troublesome, as passing through the intersection within different lane-to-lane transition paths will equate situations inside the intersection that are logically different (they support different switch options). &lt;/p&gt;&#xA;" OwnerUserId="467" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-08-20T17:44:11.297" LastActivityDate="2010-08-20T17:44:11.297" />
  <row Id="2875" PostTypeId="2" ParentId="2851" CreationDate="2010-08-20T17:50:59.723" Score="2" ViewCount="" Body="&lt;p&gt;You have $O\rightarrow O'$, $e_1\rightarrow e_1'$, $e_2\rightarrow e_2'$. Also using cross product to find the orthogonal, we have $e_1 \times e_2 =e_3\rightarrow e_1'\times e_2'$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Explicitly -&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$e_1=(1,0,0,1)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$e_2=(0,1,0,1)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$e_3=(0,0,1,1)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$O=(0,0,0,1)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$e_1'=(xdir_x,xdir_y,xdir_z,1)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$e_2'=(ydir_x,ydir_y,ydir_z,1)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$e_3'=e_1'\times e_2'$ [here $\times$ indicates cross product - and you need to append 1 to make it a 4-vector]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$O'=(o_x,o_y,o_z,1)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here $xdir_x$, $xdir_y$ etc. are xdir and ydir's coordinates in the first system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you need to solve an equation in 4x4 matrices for getting the right transformation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\textbf{R}=\begin{pmatrix}e_1\\e_2\\e_3\\O\end{pmatrix}$, &#xA;$\textbf{S}=\begin{pmatrix}e_1'\\e_2'\\e_3'\\O'\end{pmatrix}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To find the required transformation $\textbf{T}$, we have $\textbf{R}\textbf{T}=\textbf{S}$ and so $\textbf{T}=\textbf{R}^{-1}\textbf{S}$.&lt;/p&gt;&#xA;" OwnerUserId="1310" LastEditorUserId="1310" LastEditorDisplayName="" LastEditDate="2010-08-20T23:42:27.787" LastActivityDate="2010-08-20T23:42:27.787" CommentCount="3" />
  <row Id="2877" PostTypeId="2" ParentId="2872" CreationDate="2010-08-20T18:21:23.147" Score="1" ViewCount="" Body="&lt;p&gt;I think there is a bit of a mix up over notation - specifically, the problem is using $y$ in two different contexts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The definition of the chain rule that you need (and what you have correctly stated):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{dy}{dx} = \frac{dy}{du} \frac{du}{dx}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, for a composite function $y(u(x)) = [u(x)]^3$ (here, $y$ is the 'cubing' function), in order to take the derivative of $y$ with respect to $x$ we first take the derivative of $y$ w.r.t. $u$, then since $u$ is a function of $x$ (and should be written as $u(x)$ at this point) we use the chain rule as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{dy}{du} = 3u^{2}$, and $\frac{du}{dx} = u'(x)$ since we don't have the expression of the function $u(x)$ itself. So it looks like&#xA;$$\frac{d}{dx}([u(x)]^3) = 3[u(x)]^{2}u'(x).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This looks different because I have chosen to relabel the function $y^3$ as $u^{3}$ in order for the stated chain rule to 'look right.'  Alternatively, we could state the chain rule as $\frac{dh}{dx} = \frac{dh}{dy}\frac{dy}{dx}$ where $h = h(y(x)) = [y(x)]^3$.  Then $\frac{dh}{dy} = 3y^2$ and $\frac{dy}{dx}$ is just $y'$ again because we don't have a specific expression for that function.  Here we would have $\frac{dh}{dx} = 3y^2y'$.&lt;/p&gt;&#xA;" OwnerUserId="121" LastEditorUserId="121" LastEditorDisplayName="" LastEditDate="2010-08-20T18:33:34.493" LastActivityDate="2010-08-20T18:33:34.493" CommentCount="4" />
  <row Id="2878" PostTypeId="2" ParentId="1880" CreationDate="2010-08-20T19:03:42.390" Score="7" ViewCount="" Body="&lt;p&gt;Just for fun...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a proof that $\displaystyle \frac{22}{7}$ is a better approximation than $\displaystyle 3.14$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First we consider the amazing and well known integral formula for $\displaystyle \frac{22}{7} -\pi$ (for instance see this page: &lt;a href=&quot;http://en.wikipedia.org/wiki/Proof_that_22/7_exceeds_%CF%80&quot; rel=&quot;nofollow&quot;&gt;Proof that 22/7 exceeds pi&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_{0}^{1}\frac{x^{4}(1-x)^{4}}{1+x^2}dx = \frac{22}{7} -\pi$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We will to show that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$0 &amp;lt; \frac{22}{7} -\pi &amp;lt; \pi - 3.14$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That $\displaystyle 0 &amp;lt; \frac{22}{7} - \pi$ follows trivially from the above integral.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We will now show that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_{0}^{1}\frac{x^{4}(1-x)^{4}}{1+x^2}dx &amp;lt; \frac{1}{700}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We split this up as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_{0}^{1}\frac{x^{4}(1-x)^{4}}{1+x^2}dx = \int_{0}^{\frac{1}{2}}\frac{x^{4}(1-x)^{4}}{1+x^2} + \int_{\frac{1}{2}}^{1}\frac{x^{4}(1-x)^{4}}{1+x^2}dx$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first integral can be upper-bounded by replacing $\displaystyle x$ in the denominator with $\displaystyle 0$ and the second integral can be upper-bounded by replacing $\displaystyle x$ in the denominator with $\displaystyle \frac{1}{2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus we have that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_{0}^{1}\frac{x^{4}(1-x)^{4}}{1+x^2}dx &amp;lt; \int_{0}^{\frac{1}{2}}x^{4}(1-x)^{4}dx + \int_{\frac{1}{2}}^{1} \frac{4x^{4}(1-x)^{4}}{5}dx $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now $$\int_{0}^{\frac{1}{2}}x^{4}(1-x)^{4}dx = \int_{\frac{1}{2}}^{1}x^{4}(1-x)^{4}dx$$ as $\displaystyle x^4(1-x)^4$ is symmetric about $\displaystyle x = \frac{1}{2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is also known that $$\int_{0}^{1}x^{4}(1-x)^{4}dx = \frac{1}{630}$$ (see the above page again)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus we have that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_{0}^{1}\frac{x^{4}(1-x)^{4}}{1+x^2}dx &amp;lt; \frac{1}{2*630} + \frac{4}{5*2*630} = \frac{1}{700}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus we have that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{22}{7} - \pi &amp;lt; \frac{1}{700}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i.e&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$2\pi &gt; 2(\frac{22}{7} - \frac{1}{700})$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$2\pi &gt; \frac{22}{7} + \frac{22}{7} -  \frac{2}{700}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$2\pi &gt; \frac{22}{7} + \frac{2200}{700} -  \frac{2}{700}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$2\pi &gt; \frac{22}{7} + \frac{2198}{700}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$2\pi &gt; \frac{22}{7} + \frac{314}{100}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus we have that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$0 &amp;lt; \frac{22}{7} - \pi &amp;lt; \pi - \frac{314}{100}$$&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-10-04T16:11:35.890" LastActivityDate="2010-10-04T16:11:35.890" />
  <row Id="2879" PostTypeId="1" CreationDate="2010-08-20T20:05:33.780" Score="7" ViewCount="238" Body="&lt;p&gt;Take any natural number $n$, and factor it as $n=2^{e_1} 3^{e_2} 5^{e_3} ... p^{e_i}$, &#xA;where $i$ is the $i$-th prime.&#xA;Now map $n$ to the point $n \mapsto (e_1,e_2,\ldots,e_i,0,\ldots)$, where $i$ is the last prime in the factorization&#xA;of $n$.&#xA;For example, &#xA;$$n=123456789&#xA;\mapsto&#xA;(0,2,\ldots,1,\ldots,1,\ldots)$$ &#xA;because $123456789=2^0 3^2 \cdots 3607^1 \cdots 3803^1$.&#xA;So every number in $\mathbb{N}$ is mapped to a point in an arbitrarily high dimensional space.&#xA;This mapping has the property that addition of the vectors corresponds to multiplication of&#xA;the numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My (extremely vague!) question is: Does this viewpoint helps gain any insights into the &#xA;structure/properties of natural numbers?&#xA;Do line/planes/curves in this space mark out numerically interesting regions?&#xA;Perhaps allowing real or complex numbers?&#xA;The numbers in $\mathbb{N}$ fill this infinite-dimensional space very sparsely.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is (very!) far from my research expertise, so any comments/references/links would be appreciated.&lt;/p&gt;&#xA;" OwnerUserId="237" LastEditorUserId="237" LastEditorDisplayName="" LastEditDate="2010-08-22T13:59:34.660" LastActivityDate="2010-08-22T13:59:34.660" Title="Mapping natural numbers into prime-exponents space" Tags="&lt;number-theory&gt;&lt;prime-numbers&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="2880" PostTypeId="2" ParentId="2710" CreationDate="2010-08-20T20:08:03.883" Score="0" ViewCount="" Body="&lt;p&gt;Incurring the risk of sounding too simplistic, isn't that one of the fundamental properties of a fractal [&lt;a href=&quot;http://en.wikipedia.org/wiki/Fractal&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;]:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Quasi-self-similarity.&lt;/strong&gt; This is a looser form of self-similarity; the fractal appears approximately (but not exactly) identical at different scales. Quasi-self-similar fractals contain small copies of the entire fractal in distorted and degenerate forms. Fractals defined by &lt;a href=&quot;http://en.wikipedia.org/wiki/Recurrence_relation&quot; rel=&quot;nofollow&quot;&gt;recurrence relations&lt;/a&gt; are usually quasi-self-similar but not exactly self-similar. The &lt;a href=&quot;http://en.wikipedia.org/wiki/Mandelbrot_set&quot; rel=&quot;nofollow&quot;&gt;Mandelbrot set&lt;/a&gt; is quasi-self-similar, as the satellites are approximations of the entire set, but not exact copies.&lt;/p&gt;&#xA;" OwnerUserId="1366" LastActivityDate="2010-08-20T20:08:03.883" CommentCount="1" />
  <row Id="2881" PostTypeId="2" ParentId="2879" CreationDate="2010-08-20T20:29:12.000" Score="5" ViewCount="" Body="&lt;p&gt;This is a really broad question, so I'll just point you to two closely related notions.  (About which, I should admit, I know very little.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the function field setting, the objects you're mapping into go by the name of &lt;a href=&quot;http://en.wikipedia.org/wiki/Divisor_(algebraic_geometry)#Weil_divisor&quot; rel=&quot;nofollow&quot;&gt;divisors&lt;/a&gt;.  Divisors are an important tool in algebraic geometry and the theory of Riemann surfaces, e.g. this is how people talk about &lt;a href=&quot;http://en.wikipedia.org/wiki/Riemann%E2%80%93Roch_theorem&quot; rel=&quot;nofollow&quot;&gt;Riemann-Roch&lt;/a&gt;.  Back in the number field setting one can still define divisors and even ask for an arithmetic analogue of Riemann-Roch, which is described in the relevant chapter of Neukirch (I don't have it on hand at the moment).  This type of stuff is the domain of &lt;a href=&quot;http://en.wikipedia.org/wiki/Arakelov_theory&quot; rel=&quot;nofollow&quot;&gt;Arakelov theory&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternately, instead of just taking the exponents of the prime factorization you could embed the integers into their &lt;a href=&quot;http://en.wikipedia.org/wiki/Adele_ring&quot; rel=&quot;nofollow&quot;&gt;adele ring&lt;/a&gt;; this is, for example, the modern way to do class field theory.  &lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-08-20T20:29:12.000" CommentCount="1" />
  <row Id="2882" PostTypeId="1" CreationDate="2010-08-20T20:57:03.113" Score="3" ViewCount="172" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Possible Duplicate:&lt;/strong&gt;&lt;br&gt;&#xA;  &lt;a href=&quot;http://math.stackexchange.com/questions/600/circular-permutations-with-indistinguishable-objects&quot;&gt;Circular permutations with indistinguishable objects&lt;/a&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;p&gt;We can compute the total number of k-ary necklaces/bracelets of length n using the Euler totient function:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Necklace_%28combinatorics%29&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Necklace_%28combinatorics%29&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Apparently there is a formula for the number of k-ary necklaces/bracelets when the number of beads for each color is fixed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, the number of necklaces/bracelets of length 5, made up of three colors (red, green, blue), with 2 red beads, 2 green beads, and 1 blue bead. etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone provide the formula or a reference?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="592" LastActivityDate="2010-08-20T20:57:03.113" ClosedDate="2010-08-20T21:51:05.340" Title="Number of Necklace/Bracelets With Fixed Number of Beads" Tags="&lt;combinatorics&gt;&lt;reference-request&gt;" CommentCount="2" />
  <row Id="2883" PostTypeId="1" AcceptedAnswerId="2885" CreationDate="2010-08-20T21:01:36.990" Score="1" ViewCount="324" Body="&lt;p&gt;This is a problem from &quot;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0131848682&quot; rel=&quot;nofollow&quot;&gt;Introduction to Mathematics - Algebra and Number Systems&lt;/a&gt;&quot; (specifically, exercise set 2 #9), which is one of my math texts. Please note that this isn't homework, but I would still appreciate hints rather than a complete answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem reads as follows:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If 3p&lt;sup&gt;2&lt;/sup&gt; = q&lt;sup&gt;2&lt;/sup&gt;, where $p,q \in \mathbb{Z}$, show that 3 is a common divisor of p and q.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am able to show that 3 divides q, simply by rearranging for p&lt;sup&gt;2&lt;/sup&gt; and showing that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$p^2 \in \mathbb{Z} \Rightarrow q^2/3 \in \mathbb{Z} \Rightarrow 3|q$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I'm not sure how to show that 3 divides p.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moron left a comment below in which I was prompted to apply the solution to this question as a proof of $\sqrt{3}$'s irrationality. Here's what I came up with...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strike&gt;[incorrect solution...]&lt;/strike&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;...is this correct?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Edit:&lt;/b&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The correct solution is provided in the comments below by Bill Dubuque.&lt;/p&gt;&#xA;" OwnerUserId="181" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-06T18:43:03.210" LastActivityDate="2011-01-06T18:43:03.210" Title="Show that $3p^2=q^2$ implies $3|p$ and $3|q$" Tags="&lt;number-theory&gt;&lt;proof-strategy&gt;" AnswerCount="4" CommentCount="5" />
  <row Id="2884" PostTypeId="2" ParentId="1829" CreationDate="2010-08-20T21:06:15.857" Score="8" ViewCount="" Body="&lt;p&gt;I am an undergrad student, and I just finished taking Application of Linear Algebra. Here is what I want from a linear algebra text:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EXAMPLES WHICH ACTUALLY DEMONSTRATE THE POINT.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An example of what I don't ever want to see: the first example in my textbook was to show how to do matrix multiplication. Simple, right? No. They used a square symmetric matrix multiplied by itself to demonstrate this, and didn't walk through the individual steps. It just gave some matrix A and said look! A.A = B. See? And of course we didn't.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So please! Check that your examples can only be understood one way, and that this way is the way you intend.&lt;/p&gt;&#xA;" LastActivityDate="2010-08-20T21:06:15.857" CommunityOwnedDate="2010-08-20T21:06:15.857" CommentCount="1" />
  <row Id="2885" PostTypeId="2" ParentId="2883" CreationDate="2010-08-20T21:14:31.180" Score="2" ViewCount="" Body="&lt;p&gt;Write $q$ as $3r$ and see what happens.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-08-20T21:14:31.180" CommentCount="3" />
  <row Id="2886" PostTypeId="2" ParentId="2879" CreationDate="2010-08-20T21:20:14.063" Score="6" ViewCount="" Body="&lt;p&gt;In algebraic number theory such rings of &lt;strong&gt;valuation vectors&lt;/strong&gt; or &lt;strong&gt;adèles&lt;/strong&gt; or &lt;strong&gt;repartitions&lt;/strong&gt; play a key role in globally structuring local information for the global study of number fields. In particular, the topological properties of such adèle rings imply key results in algebraic number theory such as the structure of unit groups, finiteness of the class number, various results at the foundation of the Riemann-Roch theorem, etc. For example, see Tate's below review of a paper by Iwasawa (esp. the penultimate paragraph).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;MR0053970&lt;/strong&gt; (14,849a) 10.0X&lt;br&gt;&#xA;Iwasawa, Kenkichi. &lt;strong&gt;On the rings of valuation vectors&lt;/strong&gt;.&#xA;Ann. of Math. (2) 57, (1953). 331--356.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $K$ be a finite algebraic number field, or an algebraic function field of one variable over a finite constant field. The ring $R$ of valuation vectors over $K$ has the following properties: (1) $R$ is a semi-simple commutative ring with unit element 1; (2) $R$ is locally compact, but neither compact nor discrete; (3) $R$ has a subfield $K$ containing 1, such that $K$ is discrete in $R$ and the residue class space $R/K$ is compact.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the first half of his paper the author proves conversely that any topological ring having these properties is the ring of valuation vectors over a field $K$ of the type described above. The main tool used is the notion of the norm $N(\sigma,G)$ of an automorphism $\sigma$ of a locally compact group $G$, which is the factor by which the Haar measure in $G$ is stretched by $\sigma$. It is multiplicative in $\sigma$; $N(\sigma,G)=N(\sigma,H)N(\sigma,G/H)$ if $\sigma H\subset H$; and $N(\sigma,G)=1$ if $G$ is compact or discrete. If $F$ is a nondiscrete locally compact field, then the function $N(\alpha,F)$ for $\alpha\in F^\ast$ gives the &quot;normed&quot; valuation of $F$. The intersection of all closed maximal ideals $M$ of $R$ is 0, and $N(x,R)=\prod_MN(x,R/M)$ for regular $x\in R$. In particular, if $\xi\in K^\ast$, then from (3), $N(\xi,R)=1$, so that the valuations $N(\xi,R/M)$ of $K$ satisfy the ``product formula.'' From this it follows by more or less familiar methods that $K$ is an arithmetic field and $R$ its valuation vector ring. The next section contains a proof that valuation vector rings do enjoy properties (1), (2), and (3).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, the author shows that the two central theorems of algebraic number theory can be proved directly from the topological properties of $R$. The first of these is the self-duality of $R$ with respect to $K$, on which the analog of the Riemann-Roch theorem for number fields rests. If $\chi$ is any non-trivial additive character of $R$ vanishing on $K$, then the map $x\rightarrow\chi(xy)$ is an isomorphism of $R$ onto its character group, and $\chi(x\xi)=1$ for all $\xi\in K$ if and only if $x\in K$. The essential step in the proof is to show that the kernel of the map $x\rightarrow\chi(xy)$ is a compact ideal of $R$, and is therefore 0. The second theorem states the compactness of the group of idèle classes of norm 1, which is equivalent to the finiteness of class number and unit theorem. The idèle group $J$ is the multiplicative group of regular elements of $R$. It is a locally compact group in the topology for which the convergence of $a_i$ in $J$ means the convergence of $a_i$ and $a_i{}^{-1}$ in $R$. The map $a\rightarrow N(a,R)$ is a homomorphism of $J$ with kernel $J_1$, and the compactness of $J_1/K^\ast$ is proved by a Minkowski-type argument using Haar measure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Throughout the paper, the case of function fields over non-finite constant fields is treated in parallel, using linear compactness instead of compactness.&lt;br&gt;&#xA;Reviewed by J. T. Tate&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-08-20T21:20:14.063" CommentCount="1" />
  <row Id="2887" PostTypeId="1" CreationDate="2010-08-20T21:43:59.903" Score="3" ViewCount="197" Body="&lt;p&gt;I solved the following problem from by book, but the answer of this problem at the end of book is x &amp;lt; = 3 Please tell me how can i get this answer.&#xA;&lt;br&gt;&#xA;&lt;img src=&quot;http://i.imgur.com/n5TTa.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1050" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-08-20T22:30:56.840" LastActivityDate="2010-08-20T23:55:42.040" Title="Question about solving absolute values" Tags="&lt;algebra-precalculus&gt;" AnswerCount="3" CommentCount="3" />
  <row Id="2888" PostTypeId="2" ParentId="2883" CreationDate="2010-08-20T21:46:12.053" Score="1" ViewCount="" Body="&lt;p&gt;Moron's answer certainly covers your question, but as someone who's not your instructor I'd like to see a few more details in your 'proof' of the first half - can you be more specific about how $q^2/3 \in \mathbb{Z} \Rightarrow 3|q$?  While that's easy, it's not necessarily trivial, and you've elided some details there...&lt;/p&gt;&#xA;" OwnerUserId="785" LastActivityDate="2010-08-20T21:46:12.053" CommentCount="4" />
  <row Id="2890" PostTypeId="2" ParentId="2883" CreationDate="2010-08-20T21:50:16.997" Score="1" ViewCount="" Body="&lt;p&gt;Think about how many times each prime factor must appear on each side of the equation, if you were to break p and q into their prime factorizations. The left side has a 3 in it, how many must the right side have, at least?&lt;/p&gt;&#xA;" OwnerUserId="76" LastActivityDate="2010-08-20T21:50:16.997" CommentCount="1" />
  <row Id="2891" PostTypeId="2" ParentId="2887" CreationDate="2010-08-20T21:55:17.990" Score="1" ViewCount="" Body="&lt;p&gt;By definition of absolute value:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$|x| = x$ if $x &gt; 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$|x| = -x$ otherwise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are given $|x-3| = 3 - x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now given a real $x$, either $x&gt;3$ or $x \le 3$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(The reason for splitting it this way is that we have $|x-3|$ and in order to get rid of the || we need to decide whether $x-3 &gt; 0$ or not)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we split into two cases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;Case 1)&lt;/code&gt;  $x &gt; 3$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then we have that $x-3 &gt; 0$ and so by definition of absolute value, $|x-3| = x-3$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore you equation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$|x-3| = 3 - x$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is same as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x-3 = 3 -x$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is same as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$2x = 6$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is same as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x = 3$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since we assumed $x &gt; 3$, there is no solution to your equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;Case 2)&lt;/code&gt; $x \le 3$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then we have that $x - 3 \le 0$ and so by definition of absolute value&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$|x-3| = -(x-3) = 3-x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore your equation is same as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$3-x = 3-x$ which is true for any $x$ (but keep in mind that we are only considering $x \le 3$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence any $x \le 3$ satisfies this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Combine the two solutions for both the cases and you get $x \le 3$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The way you solved it, you get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x = 3$ or $x &amp;lt; 3$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you combine the two, you can say $x \le 3$.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-08-20T22:22:45.540" LastActivityDate="2010-08-20T22:22:45.540" CommentCount="4" />
  <row Id="2893" PostTypeId="2" ParentId="2887" CreationDate="2010-08-20T22:26:49.327" Score="7" ViewCount="" Body="&lt;p&gt;I'll answer by editing your solution slightly:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Depending on the sign of $x-3$:&#xA;$$\begin{align}&#xA;x-3=3-x&amp;amp;\text{ and }x-3\ge 0&amp;amp;\quad\text{ or }\quad&amp;amp;&amp;amp;-(x-3)=3-x\text{ and }&amp;amp;x-3&amp;lt;0&#xA;\\&#xA;x-3-3+x=0&amp;amp;\text{ and }x\ge 3&amp;amp;\quad\text{ or }\quad&amp;amp;&amp;amp; x-3=x-3\text{ and }&amp;amp;x&amp;lt;3&#xA;\\&#xA;2x-6=0&amp;amp;\text{ and }x\ge 3&amp;amp;\quad\text{ or }\quad&amp;amp;&amp;amp; x-3-x+3=0\text{ and }&amp;amp;x&amp;lt;3&#xA;\\&#xA;2x=6&amp;amp;\text{ and }x\ge 3&amp;amp;\quad\text{ or }\quad&amp;amp;&amp;amp; 0=0\text{ and }&amp;amp;x&amp;lt;3&#xA;\\&#xA;x=3&amp;amp;\text{ and }x\ge 3&amp;amp;\quad\text{ or }\quad&amp;amp;&amp;amp; \text{(true) and }&amp;amp;x&amp;lt;3&#xA;\\&#xA;&amp;amp;x=3&amp;amp;\quad\text{ or }\quad&amp;amp;&amp;amp; x&amp;lt;3&amp;amp;&#xA;\\&#xA;&amp;amp;&amp;amp;x\le 3&amp;amp;&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;edit&lt;/em&gt;: As a further explanation of the problem as a whole, consider the graph below, where $|x-3|$ is shown in blue and $3-x$ is shown in red.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/YQdIF.png&quot; alt=&quot;graph&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The graphs coincide for $x\le 3$ and the blue graph is higher for $x&gt;3$, so the original equation is true for $x\le 3$.&lt;/p&gt;&#xA;" OwnerUserId="72" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-08-20T22:45:56.460" LastActivityDate="2010-08-20T22:45:56.460" />
  <row Id="2894" PostTypeId="2" ParentId="2855" CreationDate="2010-08-20T22:28:45.517" Score="0" ViewCount="" Body="&lt;p&gt;We know that $\displaystyle\frac{b}{a}$ is an integer. Therefore $\displaystyle\frac{bc}{ac}=\frac{b}{a}\in \mathbb{Z}$ i.e. $ac|bc$. Otherwise $c=0$ and then the result is trivial.&lt;/p&gt;&#xA;" OwnerUserId="871" LastActivityDate="2010-08-20T22:28:45.517" />
  <row Id="2897" PostTypeId="2" ParentId="2887" CreationDate="2010-08-20T23:06:05.897" Score="3" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; It's obvious by a shift: put $\; z = x-3 \;$ in $\; |z| = -z \iff z \le 0 \; $ Making this substitution yields $|x-3| = 3-x \iff x-3 \le 0 \iff x \le 3$&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-08-20T23:55:42.040" LastActivityDate="2010-08-20T23:55:42.040" CommentCount="4" />
  <row Id="2898" PostTypeId="2" ParentId="2864" CreationDate="2010-08-20T23:32:20.923" Score="1" ViewCount="" Body="&lt;p&gt;This is really a comment which complements earlier responses:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For spheres the &quot;formula&quot; for the area of spherical triangle is known as Girard's Theorem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Girard%27s_theorem#Girard.27s_theorem&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Girard%27s_theorem#Girard.27s_theorem&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1369" LastActivityDate="2010-08-20T23:32:20.923" CommentCount="1" />
  <row Id="2899" PostTypeId="1" AcceptedAnswerId="2905" CreationDate="2010-08-20T23:40:42.890" Score="7" ViewCount="623" Body="&lt;p&gt;Recently I stumbled upon someone who said he wanted to understand why&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\arctan x = \int\dfrac{dx}{1+x^2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At first I was confused. This is an easy result in any integral calculus course. But then he explained that although he understood the proof, he wanted to understand it &quot;intuitively&quot;. He wanted to see why it was in terms of arclength and addition and subtraction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is: Is there an &quot;intuitive&quot; way to explain this identity?&lt;/p&gt;&#xA;" OwnerUserId="565" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-08-21T00:04:55.253" LastActivityDate="2011-01-05T10:45:32.977" Title="Can this standard calculus result be explained &quot;intuitively&quot;" Tags="&lt;calculus&gt;&lt;intuition&gt;&lt;teaching&gt;" AnswerCount="7" CommentCount="4" FavoriteCount="5" />
  <row Id="2901" PostTypeId="1" AcceptedAnswerId="2912" CreationDate="2010-08-21T00:00:30.363" Score="4" ViewCount="223" Body="&lt;p&gt;Let $f(x)$ be a continuous function from $\mathbb{R}\rightarrow\mathbb{R}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's denote $k$-times repeated application of the function, $f(f(f(...f(x)...)))$ as $f^{(k)}(x)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's call any $x$ a &lt;strong&gt;periodic point&lt;/strong&gt; with period $n$ if $f^{(n)}(x)=x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it true that if a point with period 3 exists, then points with all possible periods exist?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words is it true that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\exists x:f^{(3)}(x)=x\Rightarrow \forall n&gt;0 \exists y:f^{(n)}(y)=y$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and if so, why?&lt;/p&gt;&#xA;" OwnerUserId="1310" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-08-21T01:01:03.197" LastActivityDate="2010-08-21T20:03:24.383" Title="Period of 3 implies chaos" Tags="&lt;real-analysis&gt;&lt;dynamical-systems&gt;" AnswerCount="4" FavoriteCount="1" />
  <row Id="2902" PostTypeId="2" ParentId="2901" CreationDate="2010-08-21T00:05:07.673" Score="8" ViewCount="" Body="&lt;p&gt;I believe you want &lt;a href=&quot;http://mathworld.wolfram.com/SharkovskysTheorem.html&quot; rel=&quot;nofollow&quot;&gt;Sharkovsky's Theorem&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A reasonable explanation is in:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Kaplan, Harvey. &quot;A cartoon-assisted proof of Sarkowskiĭ's theorem.&quot;&#xA;Amer. J. Phys. 55 (1987), no. 11, 1023–1032.&#xA;MR&lt;a href=&quot;http://www.ams.org/mathscinet-getitem?mr=917121&quot; rel=&quot;nofollow&quot;&gt;917121&lt;/a&gt;&#xA;DOI:&lt;a href=&quot;http://dx.doi.org/10.1119/1.14928&quot; rel=&quot;nofollow&quot;&gt;10.1119/1.14928&quot;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="583" LastActivityDate="2010-08-21T00:05:07.673" CommentCount="2" />
  <row Id="2903" PostTypeId="1" AcceptedAnswerId="2904" CreationDate="2010-08-21T00:15:57.890" Score="9" ViewCount="210" Body="&lt;p&gt;I came across this 'paradox' -&#xA;$$1=e^{2\pi i}\Rightarrow 1=(e^{2\pi i})^{2\pi i}=e^{2\pi i \cdot 2\pi i}=e^{-4\pi^2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I realized the fallacy lies in the fact that in general $(x^y)^z\ne x^{yz}$. Why doesn't it work with complex numbers even though it is valid in real case? Is it related to the fact that logarithm of complex number is not unique?&lt;/p&gt;&#xA;" OwnerUserId="1310" LastEditorUserId="152" LastEditorDisplayName="" LastEditDate="2010-08-21T00:32:15.953" LastActivityDate="2010-08-21T20:09:52.950" Title="Why can't I combine complex powers" Tags="&lt;complex-analysis&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="4" />
  <row Id="2904" PostTypeId="2" ParentId="2903" CreationDate="2010-08-21T00:28:07.273" Score="12" ViewCount="" Body="&lt;p&gt;Even without any complex numbers: $-1=(-1)^{2\cdot\frac12}\ne((-1)^2)^{\frac12}=1^{\frac12}=1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But you're right, the problem is that raising to a (non-integer) power is essentially a &lt;em&gt;multivalued&lt;/em&gt; function.&lt;/p&gt;&#xA;" OwnerUserId="152" LastActivityDate="2010-08-21T00:28:07.273" CommunityOwnedDate="2010-08-21T00:28:07.273" CommentCount="2" />
  <row Id="2905" PostTypeId="2" ParentId="2899" CreationDate="2010-08-21T00:48:21.647" Score="17" ViewCount="" Body="&lt;p&gt;The geometric picture is as follows.  Let $O = (0, 0), A = (1, 0), X = (1, x)$.  Then $\arctan x = \angle AOX$.  We want to understand why $\angle AOY \approx \arctan x + \frac{h}{1 + x^2}$ where $Y = (1, x + h)$ and $h$ is small; equivalently, we want to understand why $\angle XOY \approx \frac{h}{1 + x^2}$.  Since this angle is small, we equivalently want to understand why $\sin \angle XOY \approx \frac{h}{1 + x^2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now $\triangle XOY$ evidently has area $\frac{h}{2}$.  On the other hand, it has area $\frac{1}{2} |OX| |OY| \sin \angle XOY$ where $|OX| = \sqrt{1 + x^2}$ and $|OY| \approx |OX|$.  The result follows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(The derivative follows, anyway.  The integral follows by dividing up $AX$ into little pieces and drawing a bunch of lines to $O$, then summing up all of the contributions.)&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-08-21T00:48:21.647" CommentCount="1" />
  <row Id="2906" PostTypeId="2" ParentId="2901" CreationDate="2010-08-21T01:02:48.233" Score="5" ViewCount="" Body="&lt;p&gt;See the Monthly paper of that exact title &lt;a href=&quot;http://pb.math.univ.gda.pl/chaos/pdf/li-yorke.pdf&quot; rel=&quot;nofollow&quot;&gt;Period three implies chaos&lt;/a&gt; - which made the JSTOR &lt;a href=&quot;http://www.maa.org/news/062607jstor.html&quot; rel=&quot;nofollow&quot;&gt;All-Stars&lt;/a&gt; list as the third most frequently accessed Monthly article. See also &lt;a href=&quot;http://www.jstor.org/stable/4145161&quot; rel=&quot;nofollow&quot;&gt;A simple proof of Sharkovsky's theorem&lt;/a&gt; and this &lt;a href=&quot;http://links.jstor.org/sici?sici=0002-9890%28199711%29104%3A9%3C846%3AROST%3E2.0.CO%3B2-G&quot; rel=&quot;nofollow&quot;&gt;paper&lt;/a&gt; for some history on Sharkovsky's theorem.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-08-21T01:11:11.897" LastActivityDate="2010-08-21T01:11:11.897" />
  <row Id="2908" PostTypeId="2" ParentId="2899" CreationDate="2010-08-21T02:55:32.420" Score="1" ViewCount="" Body="&lt;p&gt;The real answer I think involves complex numbers. Factor $1+x^2 = (i-x)(-i-x)$, decompose in partial fractions, get an expression involving logarithms that ends up being $\arctan x$ as described in &lt;a href=&quot;http://en.wikipedia.org/wiki/Inverse_trigonometric_functions#Logarithmic_forms&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Inverse_trigonometric_functions#Logarithmic_forms&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="589" LastActivityDate="2010-08-21T02:55:32.420" CommentCount="3" />
  <row Id="2910" PostTypeId="2" ParentId="1347" CreationDate="2010-08-21T03:18:02.510" Score="1" ViewCount="" Body="&lt;p&gt;In a sense there is one model for Euclidean geometry.  However, the geometry of the sphere can be studied on spheres with different radii and, thus, different curvature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For critters who grew up on a hyperbolic plane there is also a parameter that measures the curvature of their world.  Some nice visuals about this and technical details can be found here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.math.cornell.edu/~dwh/papers/crochet/crochet.html&quot; rel=&quot;nofollow&quot;&gt;http://www.math.cornell.edu/~dwh/papers/crochet/crochet.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1369" LastActivityDate="2010-08-21T03:18:02.510" />
  <row Id="2911" PostTypeId="2" ParentId="2378" CreationDate="2010-08-21T05:44:06.157" Score="0" ViewCount="" Body="&lt;p&gt;With respect to the introduction of fields, Erich Reck in his &lt;a href=&quot;http://plato.stanford.edu/entries/dedekind-foundations/#FunGroLat&quot; rel=&quot;nofollow&quot;&gt;&quot;Dedekind's Contributions to the Foundations of Mathematics&quot;&lt;/a&gt;, cites Dedekind as being the first to introduce this concept into mathematics, in the context of his work in algebraic number theory. Reck's article goes into some detail about Dedekind's other contributions to mathematics that seem like they may be of interest, for example, his work on lattices.&lt;/p&gt;&#xA;" OwnerUserId="1179" LastActivityDate="2010-08-21T05:44:06.157" CommunityOwnedDate="2010-08-21T05:44:06.157" />
  <row Id="2912" PostTypeId="2" ParentId="2901" CreationDate="2010-08-21T06:06:04.483" Score="2" ViewCount="" Body="&lt;p&gt;I found &lt;a href=&quot;http://books.google.com/books?id=48YHnbHGZAgC&amp;amp;dq=chaos+an+introduction+to+dynamical+systems&amp;amp;printsec=frontcover&amp;amp;source=bn&amp;amp;hl=en&amp;amp;ei=wVtvTLaRE4nAsAPVt52uCw&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=4&amp;amp;sqi=2&amp;amp;ved=0CDAQ6AEwAw#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;Chaos: An Introduction to Dynamical Systems&lt;/a&gt; by Kathleen T. Alligood, Tim D. Sauer, and James A. Yorke, to be a sensible text.  A small section of the text contains a brief explanation of Sharkovsky's Theorem.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In short, Sharkovsky uses this unique ordering of the natural numbers in such a way as to ensure that for each natural number n, the existence of a period-n point implies the existence of periods containing all the periods of orders higher than n. In my opinion, the text provides a reasonable simplification of this ordering and gently walks one through a decipherable step by step proof.&lt;/p&gt;&#xA;" OwnerUserId="219" LastActivityDate="2010-08-21T06:06:04.483" />
  <row Id="2913" PostTypeId="2" ParentId="815" CreationDate="2010-08-21T06:17:42.040" Score="4" ViewCount="" Body="&lt;h2&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/David_Hilbert&quot; rel=&quot;nofollow&quot;&gt;David Hilbert&lt;/a&gt;&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Hilbert worked in many areas of mathematics (both pure and applied) and his work on the &quot;Hilbert Program&quot; contributed significantly to the development of modern logic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I find him particularly inspiring because he serves a reminder that creativity and imagination are important qualities for mathematicians to possess-according to one story, a mathematics student decided to instead become a novelist, to which Hilbert is reported to have replied &quot;He did not have enough imagination for mathematics, but he had enough for novels&quot; (see Constance Reid's book&quot; Hilbert&quot;).&lt;/p&gt;&#xA;" OwnerUserId="1179" LastActivityDate="2010-08-21T06:17:42.040" CommunityOwnedDate="2010-08-21T06:17:42.040" />
  <row Id="2914" PostTypeId="1" AcceptedAnswerId="2920" CreationDate="2010-08-21T06:43:40.787" Score="4" ViewCount="229" Body="&lt;p&gt;Regarding &lt;a href=&quot;http://math.stackexchange.com/questions/2903/why-cant-i-combine-complex-powers&quot;&gt;this thread&lt;/a&gt;,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;it is not possible to combine complex powers in the usual way:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ (x^y)^z = x^{yz} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There was mention of multi-valued functions, is there some theory that makes this all work out correctly?&lt;/p&gt;&#xA;" LastActivityDate="2010-08-21T14:15:16.903" Title="How to combine complex powers?" Tags="&lt;complex-analysis&gt;&lt;complex-numbers&gt;" AnswerCount="4" CommentCount="1" FavoriteCount="2" />
  <row Id="2916" PostTypeId="2" ParentId="2899" CreationDate="2010-08-21T07:01:11.173" Score="2" ViewCount="" Body="&lt;p&gt;Since $D \tan(x) = 1 + \tan(x)^2$ and $\tan(\arctan(x)) = x$ we see that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xA;\tan(\arctan(x)) &amp;amp;= x \\&#xA;(1 + \tan(\arctan(x))^2 \cdot D \arctan(x) &amp;amp;= 1 \\&#xA;D \arctan(x) &amp;amp;= \frac{1}{(1 + x)^2} \\&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So by the fundamental theorem of calculus we recover the integral.&lt;/p&gt;&#xA;" LastActivityDate="2010-08-21T07:01:11.173" CommentCount="1" />
  <row Id="2917" PostTypeId="1" AcceptedAnswerId="2919" CreationDate="2010-08-21T07:12:07.843" Score="6" ViewCount="227" Body="&lt;p&gt;For two matrices $\textbf{S}$ and $\textbf{T}$, a proof of $\det(\textbf{ST})=\det(\textbf{S})\det(\textbf{T})$ is given below in the &lt;a href=&quot;http://en.wikipedia.org/wiki/Penrose_graphical_notation&quot; rel=&quot;nofollow&quot;&gt;diagrammatic tensor notation&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://imgur.com/75Hzw.jpg&quot; /&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here $\det$ denotes the determinant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why can the antisymmetrizing bar be inserted in the middle because &quot;there is already antisymmetry in the index lines&quot;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For an introduction to the notation, you can refer to Figures 12.17 and 12.18 below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://imgur.com/UIw4m.png&quot; /&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://imgur.com/a3lZi.png&quot; /&gt;&lt;/p&gt;&#xA;" OwnerUserId="1310" LastActivityDate="2010-08-21T08:17:14.950" Title="Proof of det(ST)=det(S)det(T) in Penrose graphical notation" Tags="&lt;linear-algebra&gt;&lt;matrices&gt;&lt;tensors&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="2918" PostTypeId="2" ParentId="2914" CreationDate="2010-08-21T07:13:40.883" Score="6" ViewCount="" Body="&lt;p&gt;For complex numbers the general way of defining $x^y$ (with $x$ nonzero)&#xA;is as $\exp(ly)$ where $l$ is one of the logarithms of $x$. But $x$ does&#xA;not have a unique logarithm, it has many. If $l$ is one logarithm of $x$&#xA;then the general one is $l+2n\pi i$ with $n$ an integer. Hence the general&#xA;value of $x^y$ is&#xA;$$\exp(ly+2\pi iny)=\exp(ly)\exp(2\pi i ny).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Before proceeding further note that if $y$ is an integer then $\exp(2\pi i ny)=1$&#xA;so that $x^n$ is uniquely determined. More generally if $y$ is a rational&#xA;with denominator $d$ then $\exp(2\pi i ny)$ (and so $x^y$) takes $d$ different values.&#xA;But if $y$ is irrational, then $x^y$ takes infinitely many variables.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the problem in hand $x^y$ takes the values $\exp(ly)\exp(2\pi i ny)$&#xA;as above, and $x^{yz}$ the values&#xA;$$\exp(lyz)\exp(2\pi inyz).$$&#xA;However in general each value of $x^y$ gives rise to infinitely many values&#xA;of $(x^y)^z$, namely&#xA;$$\exp((ly+2\pi iny)z+2\pi i mz)$$&#xA;so $(x^y)^z$ has a &quot;doubly-infinite&quot; family of values. In general, there&#xA;will be values of $(x^y)^z$ which are not values of $x^{yz}$&#xA;so the &quot;paradox&quot; in the &lt;a href=&quot;http://math.stackexchange.com/questions/2903/&quot;&gt;previous posting&lt;/a&gt; should be expected.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-08-21T07:13:40.883" />
  <row Id="2919" PostTypeId="2" ParentId="2917" CreationDate="2010-08-21T07:43:17.920" Score="2" ViewCount="" Body="&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/KS4YK.png&quot; alt=&quot;3!&quot;&gt;&lt;/p&gt;&#xA;" LastActivityDate="2010-08-21T08:17:14.950" CommentCount="3" />
  <row Id="2920" PostTypeId="2" ParentId="2914" CreationDate="2010-08-21T08:01:09.897" Score="3" ViewCount="" Body="&lt;p&gt;Since people keep hearing about Riemann surfaces in the three threads thus far about complex exponentiation and the logarithm being multivalued, but are probably wondering about the multivalued part, here's a visualization courtesy of Michael Trott:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/tSh7x.gif&quot; alt=&quot;logarithm branch cut&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a plot of the logarithm's imaginary part over the complex plane. Note the branch cut running through the negative real axis. Analytic continuation consists of joining together copies of this cut surface above and below (corresponding to adding or subtracting integer multiples of $2\pi i$) and matching up the cuts of each piece to maintain continuity, thus:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/goruf.gif&quot; alt=&quot;Riemann surface of logarithm&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(the right hand side is the analytically continued imaginary part)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(from &lt;a href=&quot;http://functions.wolfram.com/ElementaryFunctions/Log/visualizations/&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;)&lt;/p&gt;&#xA;" OwnerUserId="498" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-08-21T08:14:57.857" LastActivityDate="2010-08-21T08:14:57.857" CommunityOwnedDate="2010-08-21T08:01:09.897" />
  <row Id="2921" PostTypeId="2" ParentId="2914" CreationDate="2010-08-21T08:05:35.317" Score="1" ViewCount="" Body="&lt;p&gt;I guess the problem in general arises because $a^b$ can have multiple values. We have&#xA;$$\begin{eqnarray}&#xA;\log((x^y)^z) &amp;amp; = &amp;amp; z(\log(x^y)+2\pi im)=z(y(\log(x)+2 \pi in)+2\pi im)\\&#xA;&amp;amp; = &amp;amp; zy\log(x)+2\pi i(ny + m)z\\&#xA;&amp;amp; = &amp;amp; \log(x^{yz}e^{2\pi i(ny + m)z})&#xA;\end{eqnarray}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems a &lt;strong&gt;correction factor&lt;/strong&gt; of $e^{2\pi i(ny + m)z}$ is necessary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So all that can be said is that $\exists m,n\in\mathbb{Z}:(x^y)^z=x^{yz}e^{2\pi i(ny + m)z}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, we can say that the correction term is 1 if $(ny+m)z$ is always an integer, which happens for example if we know $y,z\in\mathbb{Z}$.&lt;/p&gt;&#xA;" OwnerUserId="1310" LastActivityDate="2010-08-21T08:05:35.317" />
  <row Id="2922" PostTypeId="2" ParentId="2189" CreationDate="2010-08-21T08:38:23.397" Score="1" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Freudenthal_magic_square&quot; rel=&quot;nofollow&quot;&gt;The Freudenthal–Tits magic square&lt;/a&gt;&lt;/p&gt;&#xA;" LastActivityDate="2010-08-21T08:38:23.397" CommunityOwnedDate="2010-08-21T08:38:23.397" />
  <row Id="2923" PostTypeId="1" AcceptedAnswerId="2948" CreationDate="2010-08-21T09:59:47.107" Score="1" ViewCount="398" Body="&lt;p&gt;There are at least 2 places in Wikipedia saying that X_n converges to X in mean &lt;strong&gt;in and only if&lt;/strong&gt; X_n converges to X &lt;strong&gt;in probability&lt;/strong&gt; and X_n is uniformly integrable. See the following link for example:&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Convergence_of_random_variables#Properties_4&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Convergence_of_random_variables#Properties_4&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I found a reference in Billingsley's book saying that we only need &lt;strong&gt;convergence in distribution&lt;/strong&gt; along with uniform integrability to imply the convergence in mean. Is Wikipedia wrong, or am I missing something here?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="1380" LastEditorUserId="1380" LastEditorDisplayName="" LastEditDate="2010-08-21T10:12:00.060" LastActivityDate="2010-11-28T19:38:16.860" Title="Do convergence in distribution along with uniform integrability imply convergence in mean?" Tags="&lt;probability-theory&gt;&lt;measure-theory&gt;" AnswerCount="4" FavoriteCount="1" />
  <row Id="2924" PostTypeId="1" AcceptedAnswerId="2931" CreationDate="2010-08-21T10:02:35.623" Score="2" ViewCount="122" Body="&lt;p&gt;I am looking for an element $(a_n)_{n\in\mathbb Z}\in\ell^1(\mathbb Z)$ with the property&#xA;$$&#xA;\lVert(a_n)\rVert_{\ell^1(\mathbb Z)} &gt; \lVert\sum_{n\in\mathbb Z}a_n z^n\rVert_\infty,&#xA;$$&#xA;where the norm on the right hand side denotes the sup-norm on $\mathcal C(\mathbb T)$ ($\mathbb T$ is the 1-Torus).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Motivation: I want to prove that the Gelfand transformation on $\ell^1(\mathbb Z)$ is not isometric.&lt;/p&gt;&#xA;" OwnerUserId="367" LastEditorUserId="367" LastEditorDisplayName="" LastEditDate="2010-08-21T12:20:54.107" LastActivityDate="2010-08-21T13:28:03.473" Title="The Gelfand transformation on $\ell^1(\mathbb Z)$ is not isometric. Do you have an example?" Tags="&lt;functional-analysis&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="2925" PostTypeId="2" ParentId="2899" CreationDate="2010-08-21T10:06:39.503" Score="4" ViewCount="" Body="&lt;p&gt;Put $\arctan x=:\alpha$, let $n$ be large, choose points $t_k:=\tan(k\alpha/n)$  $\ (0\leq k\leq n)$ and put  $\tau_k:=\sqrt{t_k t_{k-1}}$. Then&#xA;$$\sum_{k=1}^n  {t_k - t_{k-1} \over 1 + \tau_k^2}=\sum_{k=1}^n  {t_k - t_{k-1} \over 1 + t_k t_{k-1}}= n \tan(\alpha/n).$$&#xA;Here the left side is a Riemann sum for the integral $$\int_0^{\tan\alpha}{dt \over 1+t^2}=\int_0^x{dt \over 1+t^2},$$ and the right side has limit $\alpha=\arctan x$ when $n\to\infty$.&lt;/p&gt;&#xA;" OwnerUserId="1303" LastActivityDate="2010-08-21T10:06:39.503" CommentCount="3" />
  <row Id="2926" PostTypeId="2" ParentId="2575" CreationDate="2010-08-21T10:09:15.117" Score="1" ViewCount="" Body="&lt;p&gt;[In short, any of the two quantities will satisfy your need- $d$ (easy to calculate), or a monotonic function of $d$ $P(D\lt d)$ (needs bootstrapping to calculate). Both share the property that smaller they are, bigger the chance that new point comes from surveyed data. are described below. For a detailed discussion, read on.]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem can essentially be reformulated as a statistical hypothesis testing problem, and we need to test if the new observation comes from the surveyed population.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I will outline the construction of one such test. For this I will assume the questions are independent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$H_0$: New observation comes from the population. Assuming independence, the population can be characterized by a vector of probabilities $(p_1,p_2,...,p_n)$ for the $n$ questions. These $p_i$'s correspond to your calculated % of &quot;yes&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$H_1$: New observation does not come from the population.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose the new observation is $B=(B_1,...,B_n)$, where each $B_i$ is 1 if &quot;yes&quot; and 0 if &quot;no&quot; was entered.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A reasonable test statistic would be $D=\sum (B_i-p_i)^2$, which looks at the total squared deviation from the expected answers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can then judge if you want to label the new sample as coming from your population by evaluating $P(D\lt d)$ - if this probability is &amp;lt;5%, you can say with 95% confidence that it comes from the sample.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Computing the exact distribution of $D$ under $H_0$ will be challenging, to determine $P(D\lt d)$ you can use bootstrapping - i.e. simulating observations under $H_0$ a million times and checking what proportion of times your simulated $D$ is less than observed $d$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you need a 0 to 1 valued 'correlation' like statistic, you can look at $1-P(D\lt d)$.&lt;/p&gt;&#xA;" OwnerUserId="1310" LastActivityDate="2010-08-21T10:09:15.117" />
  <row Id="2927" PostTypeId="2" ParentId="2923" CreationDate="2010-08-21T10:17:06.047" Score="0" ViewCount="" Body="&lt;p&gt;Convergence of mean (or any other feature of the distribution for that matter) does not imply weak convergence of the random variables.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take the example of an infinite sequence of iid random variables of any non degenerate distribution and see what is going wrong.&lt;/p&gt;&#xA;" OwnerUserId="1310" LastActivityDate="2010-08-21T10:17:06.047" CommentCount="2" />
  <row Id="2929" PostTypeId="2" ParentId="2923" CreationDate="2010-08-21T11:12:06.877" Score="2" ViewCount="" Body="&lt;p&gt;Slight misunderstanding in KalEl's answer: convergence &lt;i&gt;in&lt;/i&gt; mean - which is convergence in the $L^1$ norm associated with the probability measure, and as such is a feature of the sequence of random variables itself rather than only the sequence of distributions - isn't the same thing as convergence &lt;i&gt;of&lt;/i&gt; mean.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, no, convergence in distribution as such isn't enough, even given uniform integrability - for instance, take two identically distributed nonzero random variables with a.s. disjoint support and alternate between them for your sequence, say (take $[0,1]^2$ with the usual Lebesgue measure for your probability space and use the indicator functions of the top and bottom half, for example), if KalEl's suggestion of an iid sequence doesn't immediately help you.&lt;/p&gt;&#xA;" OwnerUserId="1381" LastActivityDate="2010-08-21T11:12:06.877" CommentCount="1" />
  <row Id="2930" PostTypeId="2" ParentId="535" CreationDate="2010-08-21T12:03:20.307" Score="2" ViewCount="" Body="&lt;p&gt;I presume that you are restricting $c$ to be real, so your question is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Which complex numbers $z$ can be roots of equations&#xA;  $$4z^4+8x^3 -3z^2 -9z+c=0$$&#xA;  as $c$ varies through the reals?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is the same question as&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;For which complex numbers is&#xA;  $$g(z)=4z^4+8x^3 -3z^2 -9z$$&#xA;  real?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Certainly if $z$ is real then $g(z)$ is real, so we are really interested&#xA;in non-real solutions for $z$.&#xA;Now a complex number $w$ is real iff $w-\overline{w} =0$ where $\overline{w}$&#xA;is the complex conjugate of $w$. So we want to solve the equation&#xA;$$g(z)-\overline{g(z)}=0,$$&#xA;equivalently&#xA;$$4(z^4-\overline{z}^4)+6(z^3-\overline{z}^3)&#xA;-3(z^2-\overline{z}^2)-9(z-\overline{z})=0.\qquad\qquad(*)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we set $z=x+yi$ with $y$ real we can simplify this using&#xA;$$z-\overline{z}=2yi,$$&#xA;$$z^2-\overline{z}^2=4xyi,$$&#xA;$$z^3-\overline{z}^3=2(3x^2-y^2)yi$$&#xA;and&#xA;$$z^4-\overline{z}^4=2(4x^3-4xy^2)yi.$$&#xA;Thus $(*)$ becomes, on cancelling $2i$,&#xA;$$y[16(x^3-xy^2)+16(3x^2-y^2)-6x-9]=0.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The factor of $y$ in the above equation&#xA;accounts for the $x$-axis (so all real&#xA;values of $z$) and so the other solutions lie on the curve defined by&#xA;the equation&#xA;$$16(x^3-xy^2)+16(3x^2-y^2)-6x-9=0.$$&#xA;I'm not going to analyse this equation further, but it looks&#xA;a fairly typical cubic, so it probably defines an elliptic curve.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-08-21T12:03:20.307" />
  <row Id="2931" PostTypeId="2" ParentId="2924" CreationDate="2010-08-21T12:25:33.990" Score="4" ViewCount="" Body="&lt;p&gt;Take for instance $(1, -1, -1)$. The $\ell^1$-norm is three. However, there is no real $x$ with $|1 - e^{ix} - e^{2ix}| $ equalling 3. Indeed, this would imply (by the equality case of the triangle inequality) that $-1 = e^{ix} = e^{2ix}$, which is impossible.&lt;/p&gt;&#xA;" OwnerUserId="536" LastActivityDate="2010-08-21T12:25:33.990" CommentCount="1" />
  <row Id="2932" PostTypeId="2" ParentId="2924" CreationDate="2010-08-21T12:31:00.187" Score="1" ViewCount="" Body="&lt;p&gt;You corrected the misprint good! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose $f=(a_n)\in\ell^1$, the Gelfand transform $\hat{f}$ of $f$ is given by&#xA;$$\hat{f}(t)=\sum a_ne^{int}$$&#xA;hence $$\sup_t |\hat{f}(t)|=\sup_t\left|\sum a_ne^{int}\right|\leq \sum |a_n|=\|f\|_{\ell^1}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can also take a closer look at the Dirichlet kernel or the Féjer kernel in order to find examples where you have strict inequality. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, given any commutative Banach algebra we have the spectral radius formula&#xA;$$&#xA;\sup|\hat{f}|=\inf_{n\geq1} \|f^n\|^{1/n}$$&#xA;and by the Banach algebra inequality we have &#xA;$$\|f^n\|^{1/n}\leq\|f\|$$&lt;/p&gt;&#xA;" OwnerUserId="1154" LastEditorUserId="1154" LastEditorDisplayName="" LastEditDate="2010-08-21T13:28:03.473" LastActivityDate="2010-08-21T13:28:03.473" CommentCount="1" />
  <row Id="2933" PostTypeId="2" ParentId="535" CreationDate="2010-08-21T12:54:49.657" Score="2" ViewCount="" Body="&lt;p&gt;We need to find all $z:\exists c\in \mathbb{R}:p_c(z)=0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some manipulation gives&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\begin{array}&#xA;&amp;amp;&amp;amp;p_c(z)=0\\&#xA;\Leftrightarrow&amp;amp;p_0(z)+c=0\\&#xA;\Leftrightarrow&amp;amp;Re(p_0)+i Im(p_0)+c=0\\&#xA;\Leftrightarrow&amp;amp;Re(p_0)+c=0\text{ and } Im(p_0)=0\\&#xA;\end{array}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now $Im(p_0(z))=0$ is a fixed set of points on the complex plane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$Re(p_0)+c$ defines a surface on on $\mathbb{C}$ which varies with $c$. The surfaces that we get varying $c$ are just the translation of the surface $z=Re(p_0)$, along $z$ axis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Our required set of points are those that lie in intersection of these the sets&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\{z\in \mathbb{C}:Im(p_0(z))=0\}\cap\{z\in \mathbb{C}:\exists c:Re(p_0(z))+c=0\}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$=\{z\in \mathbb{C}:Im(p_0(z))=0\}\cap\mathbb{C}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$=\{z\in \mathbb{C}:Im(p_0(z))=0\}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Example 1:&lt;/strong&gt; As an example let's take $p_c(z)=z^2+c$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Plot of $Im(p_0(z))=2 x y = 0$ is just the two axes, as given below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://imgur.com/ksZ9g.png&quot; /&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the required locus.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A plot of $Re(p_0(x))$ is also given below. This doesn't help in finding the locus though, and this is attached only to see why the second term in the intersection above is entire $\mathbb{C}$. Evidently, the points which intersect with the complex plane as we move the surface up and down is the whole complex plane. This will hold true for any polynomial as polynomials are holomorphic functions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://imgur.com/Iuyxg.png&quot; /&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Example 2:&lt;/strong&gt; Let's take your example $p_c(z)=z^4+8z^3−3z^2−9z+c$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\begin{array}\&#xA; Im(p_0(z))&amp;amp;=&amp;amp;Im(p_0(x+iy))\\&#xA;&amp;amp;=&amp;amp;16 x^3 y+24 x^2 y-16 x y^3-6 x y-8 y^3-9 y\\&#xA;&amp;amp;=&amp;amp;-y \left(-16 x^3-24 x^2+16 x y^2+6 x+8 y^2+9\right)\end{array}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Plot of $p_0(z)=0$ (which gives your locus) is shown below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://imgur.com/8Bmly.png&quot; /&gt;&lt;/p&gt;&#xA;" OwnerUserId="1310" LastEditorUserId="1310" LastEditorDisplayName="" LastEditDate="2010-08-21T13:00:58.130" LastActivityDate="2010-08-21T13:00:58.130" />
  <row Id="2934" PostTypeId="2" ParentId="2914" CreationDate="2010-08-21T13:45:53.047" Score="5" ViewCount="" Body="&lt;p&gt;The original question asked about a &lt;em&gt;theory&lt;/em&gt; that would make this all work out correctly.  The previous answers are all correct, but they did not point out the underlying, unifying factor which makes all these computations much simpler: the &lt;a href=&quot;http://www.google.ca/url?sa=t&amp;amp;source=web&amp;amp;cd=3&amp;amp;ved=0CCMQFjAC&amp;amp;url=http%3A%2F%2Fciteseerx.ist.psu.edu%2Fviewdoc%2Fdownload%3Fdoi%3D10.1.1.49.3350%26rep%3Drep1%26type%3Dpdf&amp;amp;ei=zNVvTP_9HMmRnwex0JSGCA&amp;amp;usg=AFQjCNGOK8yueIBVHnvGITOxqnxkh-2s_w&amp;amp;sig2=_Sm-9lIGcJ-evsY3BxlY0Q&quot; rel=&quot;nofollow&quot;&gt;Unwinding number&lt;/a&gt;.  The &lt;em&gt;unwinding number&lt;/em&gt; $\mathcal{K}(z)$ is defined to be&#xA;$$ \ln(\exp(z)) = z + 2\pi i\mathcal{K}(z).$$&#xA;It is an integer which tells you how many times you need to 'unwind' $z$ before you get into the range where $\ln\circ \exp$ is the identity.  This then allows a lot of identities to be 'fixed' with relative ease.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also find the paper &lt;a href=&quot;http://www.apmaths.uwo.ca/~djeffrey/Offprints/AISC2000.pdf&quot; rel=&quot;nofollow&quot;&gt;Reasoning about the Elementary Functions of Complex Analysis&lt;/a&gt; by Corless, Davenport, Jeffrey, Litt and Watt quite instructive in this domain.&lt;/p&gt;&#xA;" OwnerUserId="620" LastEditorUserId="620" LastEditorDisplayName="" LastEditDate="2010-08-21T14:15:16.903" LastActivityDate="2010-08-21T14:15:16.903" CommentCount="3" />
  <row Id="2935" PostTypeId="2" ParentId="2869" CreationDate="2010-08-21T14:13:28.297" Score="1" ViewCount="" Body="&lt;p&gt;I can't find a closed-form, but I can get partway.  Note that your $\Gamma$ can be expressed as $1-\Gamma(n,\beta t)/\Gamma(n)$ using the incomplete gamma function; it can also be expressed (in more complicated form) using the WhittakerM function (aka 1F1).  Thus so can $h(t)$.  What this really means is that $h(t)$ is holonomic and it particular satisfies a 4th order linear ordinary differential equation with polynomial coefficients.  The product-integral in $g$'s definition bumps the order by one, and then integration (to get $f$) by another order, so $f(t)$ satisfies an order $6$ LODE.  With all those parameters, it is hopeless to solve.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have a few relations - like if $P_8$ and $P_9$ are related - then you might have a chance.  Basically, unless you can knock down the complexity of what $h$ actually is, it is rather unlikely that the whole process will 'collapse' in the presence of that many free parameters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming that $n=3$ makes things much simpler.  Then $h(t)$ is a simple sum of exponentials.  It is however integrable in closed-form (but is a medium-sized mess, so I won't post it here).  But that's where things are likely to stop, since exponentials-of-exponentials do not tend to integrate in closed-form very often.  Numerically, one ought to be able to generate some very good approximations for evaluating $f(t)$ though.  I guess it depends on what you are eventually trying to do with your equations.&lt;/p&gt;&#xA;" OwnerUserId="620" LastEditorUserId="620" LastEditorDisplayName="" LastEditDate="2010-08-30T19:30:28.917" LastActivityDate="2010-08-30T19:30:28.917" />
  <row Id="2936" PostTypeId="1" CreationDate="2010-08-21T15:07:43.070" Score="2" ViewCount="179" Body="&lt;p&gt;I can understand how the Greek alphabet came to be prominent in mathematics as the Greeks had a huge influence in the math of today. Certain letters came to have certain implications about their meaning (i.e. θ is almost always an angle, never a function).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But why did x and y come to prominence? They seem like 2 arbitrary letters for input and output, and I can't think why we began to use them instead of a and b. Why did they become the de facto standard for Cartesian coordinates?&lt;/p&gt;&#xA;" OwnerUserId="104" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-08-21T17:19:00.267" LastActivityDate="2010-08-21T21:49:32.057" Title="Why are x and y such common variables in todays equations? How did their use originate?" Tags="&lt;notation&gt;&lt;math-history&gt;" AnswerCount="3" CommentCount="4" FavoriteCount="1" />
  <row Id="2937" PostTypeId="1" AcceptedAnswerId="2940" CreationDate="2010-08-21T15:15:33.200" Score="2" ViewCount="93" Body="&lt;p&gt;In a proposal, assume $p$ is the condition and $q$ the conclusion, so the proposal is $p \to q$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $\neg$ be the negation on either condition or conclusion.&lt;br&gt;&#xA;what is the name for the operation from $p\to q$ to $\neg p\to\neg q$?&#xA;And what is the name for the operation from $p \to q$ to $p\to\neg q$? And from $p\to q$ to $\neg q\to\neg p$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What topics in logic cover these stuffs?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to prove $p\to q$, is it equivalent to disprove $p\to\neg q$? Is this the so-called &quot;proof by contradiction&quot;? And is it equiv to prove $\neg q\to\neg p$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks and regards!&lt;/p&gt;&#xA;" OwnerUserId="1281" LastEditorUserId="1281" LastEditorDisplayName="" LastEditDate="2010-08-21T15:33:59.387" LastActivityDate="2011-04-06T17:44:07.603" Title="logic operations on proposals" Tags="&lt;logic&gt;" AnswerCount="2" />
  <row Id="2938" PostTypeId="2" ParentId="2936" CreationDate="2010-08-21T15:20:21.217" Score="1" ViewCount="" Body="&lt;p&gt;Classically &quot;x&quot; was always used in the sciences for denoting an unknown quantity, e.g. the &quot;X-rays&quot; of Röntgen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cajori has a nice discussion in &quot;A History of Mathematical Notations&quot;. In brief, Descartes's convention was to use letters from the earlier half (a,b,c...) for known quantities and from the latter half (x,y,z...) for unknowns. This was in the 1600s if I remember correctly.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-08-21T15:20:21.217" />
  <row Id="2939" PostTypeId="2" ParentId="2937" CreationDate="2010-08-21T15:24:36.503" Score="2" ViewCount="" Body="&lt;p&gt;Briefly, $p\rightarrow q$ can be read as &quot;if $p$, then $q$&quot;; &quot;if not p then not q&quot; would be the &quot;inverse&quot;, and &quot;if not q then not p&quot; is the &quot;contrapositive&quot;, which has the same truth value as the original proposal.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-08-21T15:24:36.503" CommentCount="4" />
  <row Id="2940" PostTypeId="2" ParentId="2937" CreationDate="2010-08-21T15:26:58.467" Score="5" ViewCount="" Body="&lt;p&gt;You can find a table of these &lt;a href=&quot;http://en.wikipedia.org/wiki/Contraposition#Comparisons&quot; rel=&quot;nofollow&quot;&gt;on Wikipedia&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;\begin{array}{c|l} \hline&#xD;&#xA;p \to q &amp;amp; \text{Implication (Conditional)} \\\\ &#xD;&#xA;\neg p \to \neg q &amp;amp; \text{Inverse} \\\\ &#xD;&#xA;q \to p &amp;amp; \text{Converse} \\\\ &#xD;&#xA;\neg q \to \neg p &amp;amp; \text{Contrapositive} \\\\ &#xD;&#xA;\phantom\neg p \to \neg q &amp;amp; \text{Contradiction} \\\\ \hline&#xD;&#xA;\end{array}&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Recall that $p \to q$ is equivalent to $\neg p \vee q$, Therefore (Implication &amp;hArr; Contrapositive):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;(p \to q) &amp;amp;\equiv (\neg p \vee q) \\&#xA;&amp;amp;\equiv (\neg\neg q \vee \neg p) \\&#xA;&amp;amp;\equiv (\neg q \to \neg p)&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, if we can disprove $p\to\neg q$, then we are sure both $p$ and $q$ are true, which is stronger than proving $p\to q$ that allows $p$ to be false. So they are not &quot;equivalent&quot;.&lt;/p&gt;&#xA;" OwnerUserId="171" LastEditorUserId="6622" LastEditorDisplayName="" LastEditDate="2011-04-06T17:44:07.603" LastActivityDate="2011-04-06T17:44:07.603" CommentCount="0" />
  <row Id="2941" PostTypeId="2" ParentId="2883" CreationDate="2010-08-21T15:35:22.520" Score="2" ViewCount="" Body="&lt;p&gt;Below is a &lt;strong&gt;conceptual&lt;/strong&gt; proof of the irrationality of square-roots. It shows that this result follows immediately from &lt;strong&gt;unique fractionization&lt;/strong&gt; -- the uniqueness of the denominator of any reduced fraction -- i.e. the least denominator divides every denominator. This in turn follows from the key fact that the set of all possible denominators of a fraction is closed under subtraction so comprises an &lt;strong&gt;ideal&lt;/strong&gt; of $\mathbb Z$, necessarily principal, since $\mathbb Z$ is a $\rm PID$. But we can easily eliminate this highbrow language to obtain the following conceptual high-school level proof:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;THEOREM&lt;/strong&gt; $\ $ Let $\;\rm n\in\mathbb N.\;$ Then $\;\rm r = \sqrt{n}\;$ is integral if rational.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof&lt;/strong&gt; $\ $ Consider the set $\rm D$ of all the possible denominators $\rm d$ for $\rm r, \;$ i.e. $\;\rm D = \{ d\in\mathbb Z \;:\: dr \in \mathbb Z \;\}$.&#xA;Notice $\rm D$ is closed under subtraction: $\rm d,e \in D \Rightarrow dr,er\in\mathbb Z \Rightarrow (d-e)\:r = dr - er \in\mathbb Z.\;$&#xA;Further $\rm d\in D \Rightarrow dr\in D$ since $\rm (dr)r = dn\in\mathbb Z \;$ via $\;\rm r^2 = n\in\mathbb Z.\;$ Therefore, invoking the Lemma below, &#xA;with $\rm d $ the least positive element in $\rm D,$ we infer that $\;\rm d|dr \;$ in $\mathbb Z,\;$ i.e. $\rm\ r = (dr)/d \in\mathbb Z\:.\quad\quad$ &lt;strong&gt;QED&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;LEMMA&lt;/strong&gt; $\ $ Suppose $\;\rm D\subset\mathbb Z \;$ is closed under subtraction and that $\rm D$ contains a nonzero element.&lt;br&gt;&#xA;Then $\rm D \:$ has a positive element and the least positive element of $\rm D$ divides every element of $\rm D\:$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof&lt;/strong&gt; $\rm\ \ 0 \ne d\in D \Rightarrow d-d = 0\in D\Rightarrow 0-d = -d\in D\:.\ $ Hence $\rm D$ contains a positive element. Let $\rm d$ be the least positive element in $\rm D$. Since $\rm d|n \iff d\;|-n, \;$ if $\rm c\in D$ is not divisible by $\rm d$ then we &#xA;may assume that $\rm c$ is positive, and the least such element. But $\rm c-d$ is a positive element of $\rm D$ not divisible by $\rm d$ &#xA;and smaller than $\rm c$, contra leastness of $\rm c$. So $\rm d$ divides every element of $\rm D\:.\ $ &lt;strong&gt;QED&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The proof exploits the fact that the denominator ideal $\rm D$ has the special property that is is closed under multiplication by $\rm\: r\:.\ $ The fundamental role that this property plays becomes clearer when one learns about Dedekind's notion of a &lt;strong&gt;conductor ideal&lt;/strong&gt;. Employing such yields a trivial one-line proof of the generalization that a Dedekind domain is integrally closed since conductor ideals are invertible so cancellable. This viewpoint serves to generalize and unify all of the ad-hoc proofs of this class of results - esp. those proofs that proceed essentially by descent on denominators. This conductor-based &lt;strong&gt;structural&lt;/strong&gt; viewpoint is not as well known as it should be - e.g. even some famous number theorists have overlooked this. See my &lt;a href=&quot;http://mathoverflow.net/questions/30313&quot; rel=&quot;nofollow&quot;&gt;post here&lt;/a&gt; for further details.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2011-01-06T07:36:21.130" LastActivityDate="2011-01-06T07:36:21.130" />
  <row Id="2942" PostTypeId="2" ParentId="2936" CreationDate="2010-08-21T15:57:44.300" Score="0" ViewCount="" Body="&lt;p&gt;I read somewhere this convention was started by &lt;a href=&quot;http://en.wikipedia.org/wiki/Ren%C3%A9_Descartes&quot; rel=&quot;nofollow&quot;&gt;Rene Descartes&lt;/a&gt;. While conceptualizing the &lt;a href=&quot;http://en.wikipedia.org/wiki/Cartesian_coordinate_system&quot; rel=&quot;nofollow&quot;&gt;coordinate system&lt;/a&gt;, he used $x$ and $y$ to denote the axes. It took root from there on and has been used ever since.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am sorry I can't remember the source now, but I will cite if I do remember later.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/qwfnZ.jpg&quot; alt=&quot;alt Rene Descartes&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1310" LastActivityDate="2010-08-21T15:57:44.300" />
  <row Id="2943" PostTypeId="1" AcceptedAnswerId="3084" CreationDate="2010-08-21T16:44:15.903" Score="7" ViewCount="213" Body="&lt;p&gt;Exercise 3B.4 on &lt;a href=&quot;http://books.google.com/books?id=pCLhYaMUg8IC&amp;amp;pg=PA84&amp;amp;lpg=PA84&quot; rel=&quot;nofollow&quot;&gt;page 84&lt;/a&gt; of Isaacs's Finite Group Theory asks one to prove a precursor of the Sylow theorems for π-subgroups, and I only know how to do it using the Sylow theorems, so I have missed the point.  I actually needed a similar (false) statement in my research, so I thought it would be wise to get the point.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Suppose G is a finite group, and N is a normal subgroup whose index and order are coprime.  Let U be a subgroup of G whose order is coprime to the order of N.  If either N or U is solvable, then show that U is contained in some subgroup H whose order is the index of N in G.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Available tools:&lt;/strong&gt; We already have Schur–Zassenhaus, and so we know that G has a subgroup whose order is the index of N.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, if G has a normal Hall π-subgroup, then every π′-subgroup is contained in some Hall π′-subgroup.  However Hall subgroups are the next section, and containment and π-separability are even in the section after that, so appealing to Hall's generalization of Sylow's theorem will miss the point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Attempt:&lt;/strong&gt; Let M be a maximal subgroup of G containing U.  If N is not contained in M, then M∩N is a normal subgroup of M whose order is coprime to its index, and so we can find H ≤ M with U ≤ H and |H| = [M:M∩N]=[G:N], and we are done.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If N is contained in every maximal subgroup of G containing U, then, …. It'd be nice if N was in the Frattini, but that's not true.  Perhaps at this point it would be wise to use solvability?&lt;/p&gt;&#xA;" OwnerUserId="583" LastActivityDate="2010-08-23T03:22:03.437" Title="Schur–Zassenhaus exercise: coprime subgroup contained in a complement" Tags="&lt;group-theory&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="2944" PostTypeId="1" AcceptedAnswerId="2947" CreationDate="2010-08-21T17:11:29.450" Score="2" ViewCount="174" Body="&lt;p&gt;Let $f$ be a diffeomorphism, say from $\mathbb R^n$ to $\mathbb R^n$ , such as the transition map between two coordinate charts on a differentiable manifold.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A differential $n$-form (or rather its coefficient function which is obtained by using the canonical one-chart atlas on $\mathbb R^n$) then transforms essentially by multiplication with $\mathrm{det}(Df)$, while integrals transform essentially by multiplication of the integrand with $\lvert\mathrm{det}(Df)\rvert$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(This is the reason for the necessity to choose an orientation in order to define the integral of a top form on a differentiable manifold.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Question: What is an intuitive or conceptional reason for these different transformation behaviours of forms and integrands?&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="367" LastEditorUserId="367" LastEditorDisplayName="" LastEditDate="2010-08-22T08:54:54.207" LastActivityDate="2010-08-22T15:54:49.553" Title="Why do differential forms and integrands have different transformation behaviours under diffeomorphisms?" Tags="&lt;analysis&gt;" AnswerCount="2" CommentCount="5" FavoriteCount="1" />
  <row Id="2945" PostTypeId="1" AcceptedAnswerId="2946" CreationDate="2010-08-21T17:37:07.533" Score="3" ViewCount="149" Body="&lt;p&gt;Hi, I am reading p.22 of Dan Bump's Automorphihic forms and representations. A cusp of the congruence group acting on the upper half plane is defined to be an orbit of the action of the congruence subgroup on $\mathbb{Q} \mathbb{P}^1$. It also says that intuitively, the cusps are the places where a fundamental domain of the congruence subgroup touching the boundary of the upper half plane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Question: Why does the definition mean what the intuition says?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="325" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-22T02:11:34.137" LastActivityDate="2010-11-22T02:11:34.137" Title="Definition of cusp of a congruence group" Tags="&lt;number-theory&gt;&lt;modular-forms&gt;" AnswerCount="1" />
  <row Id="2946" PostTypeId="2" ParentId="2945" CreationDate="2010-08-21T17:41:41.080" Score="5" ViewCount="" Body="&lt;p&gt;You should identify the upper half plane with a subspace of $\mathbb{C}$ with a subspace of the Riemann sphere.  In this identification $\mathbb{P}^1(\mathbb{R})$ is a great circle separating the upper half plane and lower half plane, and $\mathbb{P}^1(\mathbb{Q})$ is the orbit of $\infty$ under $\text{PSL}_2(\mathbb{Z})$.  This orbit breaks up into a union of orbits under any congruence subgroup and these are the points at which a fundamental domain can touch the boundary because &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;under the action of $\text{PSL}_2(\mathbb{Z})$ the fundamental domain touches the boundary only at $\infty$, and&lt;/li&gt;&#xA;&lt;li&gt;a fundamental domain for any congruence subgroup is a union of translates of fundamental domains for $\text{PSL}_2(\mathbb{Z})$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-08-21T17:41:41.080" CommentCount="1" />
  <row Id="2947" PostTypeId="2" ParentId="2944" CreationDate="2010-08-21T18:09:10.060" Score="2" ViewCount="" Body="&lt;p&gt;Transforming by a diffeomorphism $f$ shouldn't make an (unsigned) integral negative.  For instance, say that the integral of $f$ over a subset $A$ of $\mathbb{R}^n$ is just the usual Lebesgue integral, no worries about differential forms of orientation.  So shifting between $f$ and $f \circ \alpha$ for $\alpha$ some map cannot possibly change the sign. This is why the absolute value signs appear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One can consider a &quot;volume element:&quot; this is something that transforms via $|Det f|$ for a diffeomorphism $f$.  Any differential form leads to a volume element (take its absolute value).  But the opposite is not always true. One can still define (unsigned) integration with respect to a volume element.  However, note that the cancellations in signed integration are useful. These cancellations imply that the integral of an exact form on a compact oriented manifold is zero, for instance. So restricting to volume elements loses interesting and useful information.&lt;/p&gt;&#xA;" OwnerUserId="536" LastEditorUserId="536" LastEditorDisplayName="" LastEditDate="2010-08-22T15:54:49.553" LastActivityDate="2010-08-22T15:54:49.553" CommentCount="2" />
  <row Id="2948" PostTypeId="2" ParentId="2923" CreationDate="2010-08-21T18:16:02.743" Score="2" ViewCount="" Body="&lt;p&gt;The Wikipedia statement is right, and the statement which you attribute to Billingsley is wrong, in some sense inherently so.  Convergence in distribution is a notion of convergence of &lt;em&gt;measures&lt;/em&gt;. A sequence of random variables can converge in distribution even if they live on totally different probability spaces, and their joint distribution never enters.  In contrast, convergence in mean (i.e. $L^1$) only makes sense for a sequence of random variables defined on the &lt;em&gt;same&lt;/em&gt; probability space (which is to say, you have to know all their joint distributions).  So it doesn't make sense for the former to imply the latter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even if your random variables happen to be defined on the same measure space, it fails in general, as Jacob Katz pointed out.  For a very explicit counterexample, let $\lbrace X_i \rbrace_{i=1}^\infty$ be iid Bernoulli.  Trivially $X_i$ converges in distribution, and they are uniformly integrable because they are uniformly bounded.  But $E|X_i - X_j| = 1/2$ for all $i \ne j$, so the $X_i$ are not Cauchy in mean.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's possible that Billingsley is in error, but I think it's more likely that you have missed some context.  Can you post the relevant text?&lt;/p&gt;&#xA;" OwnerUserId="822" LastActivityDate="2010-08-21T18:16:02.743" CommentCount="1" />
  <row Id="2949" PostTypeId="1" CreationDate="2010-08-21T18:51:34.310" Score="37" ViewCount="3327" Body="&lt;p&gt;A large part of my fascination in mathematics is because of some very surprising results that I have seen there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I remember one I found very hard to swallow when I first encountered it, was what is known as the &lt;a href=&quot;http://en.wikipedia.org/wiki/Banach%E2%80%93Tarski_paradox&quot; rel=&quot;nofollow&quot;&gt;Banach Tarski Paradox&lt;/a&gt;. It states that you can separate a ball $x^2+y^2+z^2 \le 1$ into finitely many disjoint parts, rotate and translate them and rejoin (by taking &lt;em&gt;disjoint&lt;/em&gt; union), and you end up with exactly two complete balls of the same radius!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I ask you which are your most surprising moments in maths?&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Chances are you will have more than one. May I request post multiple answers in that case, so the voting system will bring the ones most people think as surprising up. Thanks!&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1310" LastEditorUserId="5116" LastEditorDisplayName="anon" LastEditDate="2011-02-14T02:17:34.833" LastActivityDate="2011-05-21T15:06:20.890" CommunityOwnedDate="2010-08-21T19:06:41.580" Title="Which one result in maths has surprised you the most?" Tags="&lt;big-list&gt;" AnswerCount="88" CommentCount="8" FavoriteCount="20" />
  <row Id="2950" PostTypeId="2" ParentId="2949" CreationDate="2010-08-21T19:02:49.637" Score="46" ViewCount="" Body="&lt;p&gt;If a function of a complex variable is once differentiable, it's infinitely differentiable.&lt;/p&gt;&#xA;" OwnerUserId="247" LastActivityDate="2010-08-21T19:02:49.637" CommunityOwnedDate="2010-08-21T21:22:23.977" CommentCount="4" />
  <row Id="2951" PostTypeId="2" ParentId="2949" CreationDate="2010-08-21T19:05:30.913" Score="6" ViewCount="" Body="&lt;p&gt;There exists a non-reflexive Banach space that is isomorphic to its dual.&lt;/p&gt;&#xA;" OwnerUserId="247" LastActivityDate="2010-08-21T19:05:30.913" CommunityOwnedDate="2010-08-21T21:22:23.977" CommentCount="4" />
  <row Id="2952" PostTypeId="1" CreationDate="2010-08-21T19:45:15.200" Score="5" ViewCount="127" Body="&lt;p&gt;Let $\omega:\mathbb{Z}\to (0,\infty)$ and let $1\leq p&amp;lt;\infty$. Consider the space  $\ell^p(\mathbb{Z},\omega)$ of complex valued sequences $f=(a_n)_{n \in \mathbb{Z}}$ such that &#xA;$$\|f\|=\|f\|_{\ell^p(\mathbb{Z},\omega)}:=\left(\sum_{n\in\mathbb{Z}}|a_n|^p\omega(n)^p\right)^{1/p}&amp;lt;\infty.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next, given two complex sequences $f=(a_n)_{n \in \mathbb{Z}}$ and $g=(b_n)_{n \in \mathbb{Z}}$ their formal convolution is defined by $f*g=(c_n)_{n\in\mathbb{Z}}$ where $c_n=\sum_{k\in\mathbb{Z}}a_kb_{n-k}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is to find necessary and sufficient conditions on $\omega$ such that $\ell^p(\mathbb{Z},\omega)$ is a Banach algebra. &#xA;In other words, if $f,g\in\ell^p(\mathbb{Z},\omega)$ then $f*g\in \ell^p(\mathbb{Z},\omega)$ and $\|f*g\|\leq\|f\|\cdot\|g\|$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $p=1$ the condition is $\omega(n+k)\leq\omega(n)\omega(k)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;--- Reformulated the problem so that $f\in\ell^p(\omega)$ is the same as $f\omega\in\ell^p$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe this is an open problem, there are however sufficient conditions:  $\omega^{-p'}*\omega^{-p'}\leq \omega^{-p'}$ where $1/p̈́'+1/p=1$ (the history of the condition is hard to tell but it is given as Lemma 8.11 in Acta Mathematica Volume 174, Number 1, 1-84, &quot;Completeness of translates in weighted spaces on the half-line&quot; by Alexander Borichev and Håkan Hedenmalm). The proof is based on Hölder's inequality:&#xA;$$\|f*g\|^p =\sum_n \left|\sum_k a_kb_{n-k}\right|^p\omega(n)^p\leq $$&#xA;$$\leq\sum_n \left(\sum_k |a_k|^p|b_{n-k}|^p\omega(n-k)^p\omega(k)^p\right)\left(\sum_k\frac{1}{\omega(n-k)^{p'}\omega(k)^{p'}}\right)^{\frac{p}{p'}}\omega(n)^p\leq $$&#xA;$$\qquad\leq\|f\|^p\|g\|^p$$ &lt;/p&gt;&#xA;" OwnerUserId="1154" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-08-24T00:34:45.487" LastActivityDate="2010-08-25T16:41:11.567" Title="When is the weighted space $\ell^p(\mathbb{Z},\omega)$ a Banach algebra ($p&gt;1$)?" Tags="&lt;sequences-and-series&gt;&lt;functional-analysis&gt;&lt;fourier-series&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="3" />
  <row Id="2953" PostTypeId="2" ParentId="2949" CreationDate="2010-08-21T19:51:39.807" Score="25" ViewCount="" Body="&lt;p&gt;Maybe this is too obvious, but the fact that the Rationals are countable blew my mind.&lt;/p&gt;&#xA;" OwnerUserId="16" LastActivityDate="2010-08-21T19:51:39.807" CommunityOwnedDate="2010-08-21T19:51:39.807" CommentCount="5" />
  <row Id="2954" PostTypeId="2" ParentId="2949" CreationDate="2010-08-21T19:52:53.657" Score="4" ViewCount="" Body="&lt;p&gt;I would not rate this example as surprising, but it did provoke in me a little epiphany when I finally understood it. There is a theorem of category theory that characterizes adjunctions as a pair of functors and a pair of natural transformations satisfying a bunch of equations. Now in some sense, this is a pure formality (the proof is easy), but on the other hand, an adjunction encodes a parameterized universal property, with some implicit quantifiers (over potentially proper classes) floating around. Now think of all the adjunctions you have come across that encode huge amounts of information. The characterization theorem says that this is the same as a pair of 2-cells in a 2-category satisfying a pair of equations. Look, Ma, no quantifiers, no isomorphisms, no nothing. Just a bunch of equations in a 2-category. The single most important concept of category theory and what do we end up with? a pair of equations...&lt;/p&gt;&#xA;" OwnerUserId="1033" LastActivityDate="2010-08-21T19:52:53.657" CommunityOwnedDate="2010-08-21T19:52:53.657" />
  <row Id="2955" PostTypeId="2" ParentId="2944" CreationDate="2010-08-21T19:53:33.110" Score="1" ViewCount="" Body="&lt;p&gt;A manifold is not necessary orientable. But the &lt;strong&gt;orientation bundle&lt;/strong&gt; always exists. You can construe it as a real line bundle. Then the &lt;strong&gt;density bundle&lt;/strong&gt; is obtained by tensoring the bundle of top forms with the orientation bundle. By definition, a nonvanishing section of the orientation bundle is an &lt;strong&gt;orientation&lt;/strong&gt;, and a section of the density bundle is a &lt;strong&gt;density&lt;/strong&gt;. Compactly supported densities can be integrated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Locally, orientations &quot;transform&quot; by the sign of the determinant (of the jacobian matrix), top forms by the determinant, and densities by the absolute value of the determinant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It suffices to understand the case of the real line.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT.&lt;/strong&gt; I've just realized that, in your question, you mention manifolds only in a parenthesis, and concentrate on $\mathbb R^n$. Open subsets of $\mathbb R^n$ are of course always orientable. But the key point, again, is to understand what happens on the real line.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the real line, it's natural to write a 1-form as $f\ dx$, and a density as $f\ |dx|$ (say with $f$ continuous). If you have a compact interval $I$, then you can integrate $f\ |dx|$ over $I$, and $f\ dx$ from one bound to the other. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT 2.&lt;/strong&gt; You can write the standard orientation of $\mathbb R$ as $|dx|/dx$, and define the integral of the form $f\ dx$ over the interval $I$ equipped with the orientation $\pm|dx|/dx$ as the integral over the nonoriented interval $I$ of the density $\pm f\ |dx|$ equal to the orientation times the form.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you only need to define the integral of a density over a nonoriented (compact) interval. [See for instance de Rham's book on differentiable manifolds.] &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT 3.&lt;/strong&gt; In the same spirit, to define the $L^2$ space of a manifold, you consider &lt;strong&gt;half-densities&lt;/strong&gt;. More precisely, you complete the pre-hilbert space of compactly supported half-densities. The half-densities &quot;transform&quot; by the square root of the determinant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The general principle is this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $M$ is an $n$-manifold, then each action of $G:=GL_n(\mathbb R)$ on a manifold $F$ induces a bundle over $M$ with fiber $F$, called the &lt;strong&gt;associated bundle&lt;/strong&gt;. If the action is an $r$-dimensional $\mathbb R$-linear representation, then the associated bundle is a rank $r$ vector bundle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most basic example is the tangent bundle, associated to the natural representation on $\mathbb R^n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $\chi$ be the characters of $G$ given by the inverse of the determinant. The bundles of $n$-forms, orientations, densities, and half-densities, are the line bundles respectively associated to the following characters of $G$: $\chi$, the sign of $\chi$, its absolute value, and the square root of its absolute value.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the definition of the associated bundle, see&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Associated_bundle#Fiber_bundle_associated_to_a_principal_bundle&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Associated_bundle#Fiber_bundle_associated_to_a_principal_bundle&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In our case, the principal bundle is the &lt;strong&gt;frame bundle&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Frame_bundle&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Frame_bundle&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="660" LastEditorUserId="660" LastEditorDisplayName="" LastEditDate="2010-08-22T04:58:49.807" LastActivityDate="2010-08-22T04:58:49.807" />
  <row Id="2956" PostTypeId="2" ParentId="2221" CreationDate="2010-08-21T20:01:41.377" Score="0" ViewCount="" Body="&lt;p&gt;I think this is not possible. We need a stronger condition on the $c_k$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume such a measure exists, then $L:C(S^1) \to \mathbb{C}$ given by&#xA;$$L(f) = \int_{S^1} f \, d\mu$$&#xA;is a linear functional so we can extend it to $L^2(S^1)$. To see it is continuous note that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$|L(f)| \leq \|f\|_1 \leq \|f\|_2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also write this extension as $L$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By Riesz representation theorem we know that $L(f) = \langle f, x \rangle$ for some $x \in L^2(S^1)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So $L(\sum \zeta^n z^n) = \sum b_n z^n$ where $b_n = \langle \zeta^n, x \rangle$ and we know that this is equal to $\sum c_n z^n$. So that $c_n = \langle \zeta^n, x \rangle$.&#xA;But we also know that $\sum |c_n|^2 &amp;lt; \infty$. This is a stronger condition than convexity as George pointed out below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope it is correct this time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem looks a lot like the Hausdorff moment problem and the standard work on moment problems is probably the book by Akhiezer: The classical moment problem and some related questions in analysis. &#xA;Hafner Publishing Co., New York 1965.&lt;/p&gt;&#xA;" OwnerUserId="1120" LastEditorUserId="1120" LastEditorDisplayName="" LastEditDate="2010-08-24T21:46:14.487" LastActivityDate="2010-08-24T21:46:14.487" CommentCount="8" />
  <row Id="2957" PostTypeId="2" ParentId="2901" CreationDate="2010-08-21T20:03:24.383" Score="0" ViewCount="" Body="&lt;p&gt;This blog entry has a nice intro to the theorem, although no proof.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://divisbyzero.com/2008/12/18/sharkovskys-theorem/&quot; rel=&quot;nofollow&quot;&gt;http://divisbyzero.com/2008/12/18/sharkovskys-theorem/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1310" LastActivityDate="2010-08-21T20:03:24.383" />
  <row Id="2958" PostTypeId="2" ParentId="2903" CreationDate="2010-08-21T20:09:52.950" Score="4" ViewCount="" Body="&lt;p&gt;This is related to a note by Euler, maybe he was the first to realize that $i^i$ is real. Actually, &#xA;$$i^i = (e^{i\pi/2})^i = e^{-\pi/2}$$&#xA;on the other hand &#xA;$$i^i = (e^{i(\pi/2+2\pi n})^i = e^{-\pi/2 -2\pi n},\ n\in\mathbb{Z}.$$&#xA;So maybe it is better to say that $i^i$ is a subset of the $\mathbb{R}$ and that certain equality signs are to be understood as congruences.&lt;/p&gt;&#xA;" OwnerUserId="1154" LastActivityDate="2010-08-21T20:09:52.950" />
  <row Id="2959" PostTypeId="2" ParentId="2731" CreationDate="2010-08-21T20:10:04.777" Score="1" ViewCount="" Body="&lt;p&gt;The answer is 1/2 since whether first or second child opens the door is independent of if he is a boy or a girl.&lt;/p&gt;&#xA;" OwnerUserId="1310" LastActivityDate="2010-08-21T20:10:04.777" />
  <row Id="2960" PostTypeId="2" ParentId="83" CreationDate="2010-08-21T20:14:41.240" Score="1" ViewCount="" Body="&lt;p&gt;For people this age and older I know of no better way to introduce someone to mathematics than to have them look at Martin Gardner's books.  A sample of his books are listed here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.york.cuny.edu/~malk/biblio/martin-gardner-biblio.html&quot; rel=&quot;nofollow&quot;&gt;http://www.york.cuny.edu/~malk/biblio/martin-gardner-biblio.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1369" LastActivityDate="2010-08-21T20:14:41.240" CommunityOwnedDate="2010-08-21T20:14:41.240" />
  <row Id="2961" PostTypeId="2" ParentId="2731" CreationDate="2010-08-21T20:28:02.213" Score="0" ViewCount="" Body="&lt;p&gt;Here's a simple way to think about it...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know one of the children is a boy (he opened the door). Therefore there are 3 possible sets of children (numbered for later reference), which you can trivially derrive from the Boy-Girl probability tree:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;B&lt;sub&gt;1&lt;/sub&gt;B&lt;sub&gt;2&lt;/sub&gt;&lt;/li&gt;&#xA;&lt;li&gt;B&lt;sub&gt;3&lt;/sub&gt;G&lt;sub&gt;4&lt;/sub&gt;&lt;/li&gt;&#xA;&lt;li&gt;G&lt;sub&gt;5&lt;/sub&gt;B&lt;sub&gt;6&lt;/sub&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So there are 4 possibilities for the second child:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;B&lt;sub&gt;1&lt;/sub&gt; (B&lt;sub&gt;2&lt;/sub&gt; opened the door)&lt;/li&gt;&#xA;&lt;li&gt;B&lt;sub&gt;2&lt;/sub&gt; (B&lt;sub&gt;1&lt;/sub&gt; opened the door)&lt;/li&gt;&#xA;&lt;li&gt;G&lt;sub&gt;4&lt;/sub&gt; (B&lt;sub&gt;3&lt;/sub&gt; opened the door)&lt;/li&gt;&#xA;&lt;li&gt;G&lt;sub&gt;5&lt;/sub&gt; (B&lt;sub&gt;4&lt;/sub&gt; opened the door)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Then the  chance is obviously 1/2 that the other child is a girl.&lt;/p&gt;&#xA;" OwnerUserId="181" LastEditorUserId="181" LastEditorDisplayName="" LastEditDate="2010-08-21T20:33:39.830" LastActivityDate="2010-08-21T20:33:39.830" />
  <row Id="2962" PostTypeId="2" ParentId="2949" CreationDate="2010-08-21T21:03:36.197" Score="5" ViewCount="" Body="&lt;p&gt;As an undergraduate, the fact that |P(x)| &gt; |X|.  I recall being surprised at how both how short and easy this was to prove, and that it implied there were infinitely many &quot;sizes&quot; of infinity.   &lt;em&gt;(The standard diagonalization of decimals proof only showed there were two sizes and took more time.)&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="1390" LastActivityDate="2010-08-21T21:03:36.197" CommunityOwnedDate="2010-08-21T21:03:36.197" CommentCount="4" />
  <row Id="2963" PostTypeId="2" ParentId="2949" CreationDate="2010-08-21T21:18:59.750" Score="26" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Cauchy%27s_integral_formula&quot; rel=&quot;nofollow&quot;&gt;Cauchy's Integral Formula&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fact that the values of an analytic function on the edge of disk (or a simple closed curve) are enough to determine all the values within the curve was very surprising to me.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-08-21T21:18:59.750" CommunityOwnedDate="2010-08-21T21:18:59.750" CommentCount="3" />
  <row Id="2965" PostTypeId="2" ParentId="2936" CreationDate="2010-08-21T21:49:32.057" Score="3" ViewCount="" Body="&lt;p&gt;This question has been asked previously on MathOverflow, and answered (by Mariano Suárez-Alvarez). You can follow &lt;a href=&quot;http://mathoverflow.net/questions/30307/explanation-why-x-y-z-are-always-variables/30414#30414&quot; rel=&quot;nofollow&quot;&gt;this link&lt;/a&gt;, and I quote his response below.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;You'll find details on this point (and precise references) in Cajori's History of mathematical notations, ¶340. He credits Descartes in his La Géometrie for the introduction of x, y and z  (and more generally, usefully and interestingly, for the use of the first letters of the alphabet for known quantities and the last letters for the unknown quantities) He notes that Descartes used the notation considerably earlier: the book was published in 1637, yet in 1629 he was already using x as an unknown (although in the same place y is a known quantity...); also, he used the notation in manuscripts dated earlier than the book by years.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;It is very, very interesting to read through the description Cajori makes of the many, many other alternatives to the notation of quantities, and as one proceeds along the almost 1000 pages of the two volume book, one can very much appreciate how precious are the notations we so much take for granted!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1321" LastActivityDate="2010-08-21T21:49:32.057" />
  <row Id="2966" PostTypeId="1" AcceptedAnswerId="2975" CreationDate="2010-08-21T21:56:15.267" Score="2" ViewCount="247" Body="&lt;p&gt;Zero is finite. If n is finite, then n+1 is finite. Hence, by induction, all numbers are finite. What is the catch?&lt;/p&gt;&#xA;" OwnerUserId="1385" LastEditorUserId="1385" LastEditorDisplayName="" LastEditDate="2010-08-21T22:46:28.147" LastActivityDate="2010-08-21T23:34:09.600" Title="Induction versus Natural Numbers" Tags="&lt;infinity&gt;&lt;axioms&gt;" AnswerCount="3" CommentCount="3" />
  <row Id="2967" PostTypeId="2" ParentId="2966" CreationDate="2010-08-21T22:02:11.440" Score="2" ViewCount="" Body="&lt;p&gt;You might call each &lt;em&gt;natural number&lt;/em&gt; finite (that's what induction applies to).  Wait until you hear about &lt;a href=&quot;http://en.wikipedia.org/wiki/Inaccessible_cardinal&quot; rel=&quot;nofollow&quot;&gt;these&lt;/a&gt;, though.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT:  As the comment to the question points out, the natural numbers are collectively infinite in number, if that's what you're referring to.&lt;/p&gt;&#xA;" OwnerUserId="336" LastActivityDate="2010-08-21T22:02:11.440" CommentCount="2" />
  <row Id="2968" PostTypeId="1" AcceptedAnswerId="3272" CreationDate="2010-08-21T22:09:20.263" Score="4" ViewCount="434" Body="&lt;p&gt;Each of n persons draws a card from a shuffled deck of k cards numbered from 1 to k.  There are at least as many cards as persons.  The winner is the person who is holding the largest card.  If everyone is honest, how can they mutually determine the identity of the winner, without any person learning additional information beyond that which can necessarily be inferred from n, k, the value of that person's own card, and the identity of the winner?  In the general situation with k &gt;&gt; n for any person this information includes the values of all cards except for the one held by that person, and also the relative value of any two non-winning persons' cards.  If this is not possible, in what ways can this information be minimized?&lt;/p&gt;&#xA;" OwnerUserId="1284" LastEditorUserId="1284" LastEditorDisplayName="" LastEditDate="2010-09-06T18:40:04.197" LastActivityDate="2010-09-06T18:40:04.197" Title="Mental card game" Tags="&lt;puzzle&gt;&lt;cryptography&gt;" AnswerCount="4" CommentCount="7" FavoriteCount="3" />
  <row Id="2969" PostTypeId="1" CreationDate="2010-08-21T22:09:55.783" Score="15" ViewCount="610" Body="&lt;p&gt;I have this wooden puzzle composed of 25 y-shaped &lt;a href=&quot;http://en.wikipedia.org/wiki/Pentomino&quot; rel=&quot;nofollow&quot;&gt;pentominoes&lt;/a&gt;, where the objective is to assemble them into a 5x5x5 cube. After spending quite a few hours unsuccessfully trying to solve the puzzle, I finally gave up and wrote a program to try all possible combinations using backtracking. Analyzing the results revealed that for every solution found, the computer made - on average - 50 million placements and removals of pieces. This is obviously beyond my capabilities as a human, even if I can see a few steps ahead that a partial solution leads to a &quot;dead end&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, my question is this: given that the puzzle is so symmetrical, is there some way to significantly prune the search tree (maybe even making it possible for me to solve the puzzle on my own)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/XrLj1.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(sorry for the poor quality)&lt;/p&gt;&#xA;" OwnerUserId="26" LastActivityDate="2011-03-14T13:39:18.073" Title="Tiling pentominoes into a 5x5x5 cube" Tags="&lt;puzzle&gt;&lt;tiling&gt;" AnswerCount="5" CommentCount="10" FavoriteCount="4" />
  <row Id="2970" PostTypeId="2" ParentId="713" CreationDate="2010-08-21T22:11:02.430" Score="1" ViewCount="" Body="&lt;p&gt;The other solutions posted already do a good job of explaining tensors as dyads, linear transformations, or using the ellipsoid interpretation. I thought I'd give an engineering perspective.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In engineering applications, physical quantities (mass, velocity, force, etc.) are either &quot;scalars&quot;, &quot;vectors&quot;, or &quot;tensors&quot;. What determines which one, is how the quantity changes under a coordinate transformation. In other words: If you rotate your world view, what happens to the quantity in question? Suppose $U$ is a rotation (read: orthogonal) matrix that rotates vectors from the initial frame $F_1$ to another frame $F_2$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Scalars, such as mass, do not change at all. An object's mass does not depend on the orientation of your frame, so $m_1$ = $m_2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Vectors, such as force or velocity, do change. If $v_1$ is expressed in your starting frame, then $v_2 = U^Tv_1$. Rotating the frame amounts to rotating the vector. All vector quantities tranform in exactly this way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Tensors, such as stress or inertia, change as well. If $J_1$ is expressed in your starting frame, then $J_2 = U^TJ_1U$. Any quantity that transforms in this manner is a tensor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Angular momentum (the product of the moment of inertia and the angular velocity) is a vector, because:&#xA;$$&#xA;J_2 w_2 = U^TJ_1U U^Tw_1 = U^T(J_1 w_1)&#xA;$$&#xA;Kinetic energy is a scalar, which is intuitive, but you can verify it using the transformations above. For an object in translational and rotational motion,&#xA;$$&#xA;E = \frac{1}{2} m_1 v_1^T v_1 + \frac{1}{2} w_1^T J_1 w_1 = \frac{1}{2} m_2 v_2^T v_2 + \frac{1}{2} w_2^T J_2 w_2&#xA;$$&#xA;And so on.&lt;/p&gt;&#xA;" OwnerUserId="1356" LastEditorUserId="1356" LastEditorDisplayName="" LastEditDate="2010-08-22T01:44:03.140" LastActivityDate="2010-08-22T01:44:03.140" />
  <row Id="2971" PostTypeId="1" AcceptedAnswerId="2972" CreationDate="2010-08-21T22:33:24.153" Score="4" ViewCount="111" Body="&lt;p&gt;Let the &lt;em&gt;class number&lt;/em&gt; $h(d)$ denote the number of distinct binary quadratic forms with discriminant $d &amp;lt; 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a better algorithm for $h$ than brute force?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To be precise, by brute force I meant to generate enough forms to completely cover the space and then reducing them down to see how many equivalence classes there are.&lt;/p&gt;&#xA;" LastActivityDate="2010-08-22T02:21:26.950" Title="Algorithms to compute the class number" Tags="&lt;number-theory&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="2" />
  <row Id="2972" PostTypeId="2" ParentId="2971" CreationDate="2010-08-21T22:45:10.550" Score="6" ViewCount="" Body="&lt;p&gt;Yes, there are algorithms that are much better than brute force. For example, see section 5.4 in Henri Cohen's book &quot;A course in computational algebraic number theory&quot; for Shanks's baby-step giant-step algorithm $O(|D|^{1/4+\epsilon})$ - which is practical for negative discriminants $D$ up to 25 digits or more and, further, McCurley's sub-exponential algorithm (including Atkin's variant) which is $O(L(|D|)^\alpha)$ for  $\alpha = \sqrt 2 \;$ or perhaps even $\alpha = \sqrt{9/8},$ where $\; L(x) = e^{\sqrt {\ln x \ln\ln x}}$. This can handle $D$ up to 50 digits or more (nowadays, with various improvements, probably around 80 digits or more - the prior numbers are quoted from the 1993 edition of Cohen's book - currently the bible for computational algebraic number theory).&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2010-08-22T02:21:26.950" LastActivityDate="2010-08-22T02:21:26.950" CommentCount="2" />
  <row Id="2973" PostTypeId="2" ParentId="2966" CreationDate="2010-08-21T22:58:52.217" Score="1" ViewCount="" Body="&lt;p&gt;The Axiom of Infinity may be relevant to your understanding, because it concerns the existence of an infinite set, which is also an infinite &lt;em&gt;cardinal&lt;/em&gt; number.  Without it all sets are finite and equivalent to natural numbers, for the reason that you state (induction is not powerful enough).  Starting with Cantor's Theorem you can build even larger cardinals by taking the power set of the infinite set asserted to exist by the Axiom of Infinity.  There is also a technique called transfinite induction which is used to prove propositions about ordinals and cardinals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Transfinite_induction&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Transfinite_induction&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1284" LastEditorUserId="1284" LastEditorDisplayName="" LastEditDate="2010-08-21T23:10:47.983" LastActivityDate="2010-08-21T23:10:47.983" />
  <row Id="2974" PostTypeId="2" ParentId="2949" CreationDate="2010-08-21T23:18:03.930" Score="16" ViewCount="" Body="&lt;p&gt;PRIMES is in P.  This was surprising to me both because I knew it as an open problem before it was proved, and because the proof is simple enough that I can follow the outline and understand some of the details.  The proof of FLT was not as surprising to me because comprehending it seems to require a lot of background that I don't have.&lt;/p&gt;&#xA;" OwnerUserId="1284" LastEditorUserId="1284" LastEditorDisplayName="" LastEditDate="2010-08-22T00:36:51.370" LastActivityDate="2010-08-22T00:36:51.370" CommunityOwnedDate="2010-08-21T23:18:03.930" />
  <row Id="2975" PostTypeId="2" ParentId="2966" CreationDate="2010-08-21T23:34:09.600" Score="7" ViewCount="" Body="&lt;p&gt;You have to be careful with your wording in this type of question. There is more than one type of number. Usually, what is being referred to is clear from the context, but here it is not. Natural numbers (0,1,2,3,...,etc) are all finite, and standard induction only applies to these.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are other types of numbers, such as &lt;a href=&quot;http://en.wikipedia.org/wiki/Cardinal_number&quot; rel=&quot;nofollow&quot;&gt;cardinal numbers&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/Ordinal_number&quot; rel=&quot;nofollow&quot;&gt;ordinal numbers&lt;/a&gt; which can be infinite. The principle of induction does not apply to these. There is an extension of induction, called &lt;a href=&quot;http://en.wikipedia.org/wiki/Transfinite_induction&quot; rel=&quot;nofollow&quot;&gt;transfinite induction&lt;/a&gt;. On top of showing that the desired property holds for n+1 whenever it holds for n, transfinite induction also involves proving that the property holds in a limiting sense. The property of being finite does not survive this limit, so there is no problem with having infinite ordinal numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Usually when someone says &lt;em&gt;number&lt;/em&gt; without being clear about what they are referring to, they mean natural numbers, integers, real numbers or, possibly, complex numbers. In that case, there are no infinite numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is not the kind of thing which should cause problems in a serious mathematical discussion, but I've seen this ambiguity create confusion many times before on internet forums.&lt;/p&gt;&#xA;" OwnerUserId="1321" LastActivityDate="2010-08-21T23:34:09.600" />
  <row Id="2976" PostTypeId="2" ParentId="2968" CreationDate="2010-08-21T23:45:07.570" Score="7" ViewCount="" Body="&lt;p&gt;If any person pulls the k value card, they immediately identify themselves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Should this not happen, after a second the individual with the k-1 card identifies himself. Similarly, the k-t person identifies himself after t seconds if no one has been identified previously. On this strategy, it would take at most k-n seconds for someone to identify himself as the highest card holder, with no other information being passed besides everyone aware the value of highest card is k-n and who drew it.&lt;/p&gt;&#xA;" OwnerUserId="557" LastActivityDate="2010-08-21T23:45:07.570" CommentCount="8" />
  <row Id="2977" PostTypeId="2" ParentId="2949" CreationDate="2010-08-21T23:46:58.300" Score="11" ViewCount="" Body="&lt;p&gt;Oh, I've been surprised a lot of times, but a particularly memorable one for me was learning the &lt;a href=&quot;http://mathworld.wolfram.com/MaximumModulusPrinciple.html&quot; rel=&quot;nofollow&quot;&gt;maximum modulus principle&lt;/a&gt; of complex analysis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the numerics front, I still find it amazing that the humble trapezoidal rule is the best one to use for &lt;a href=&quot;http://www.jstor.org/pss/2695765&quot; rel=&quot;nofollow&quot;&gt;integrating periodic functions over a period&lt;/a&gt;, better than Simpson's rule or the other fancier quadrature methods. This can be seen by appealing to Euler-Maclaurin.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-08-21T23:46:58.300" CommunityOwnedDate="2010-08-21T23:46:58.300" CommentCount="7" />
  <row Id="2978" PostTypeId="2" ParentId="2949" CreationDate="2010-08-22T00:41:30.583" Score="3" ViewCount="" Body="&lt;p&gt;The connection between syntax and model theory.  For example, you can tell that you can't define &quot;field&quot; (the algebraic structure) by equations because the category of fields doesn't have products.  In other words, a &lt;em&gt;property of the models&lt;/em&gt; controls the &lt;em&gt;logical connectives you must use to say what it is&lt;/em&gt;.  There are many results like this. &lt;/p&gt;&#xA;" OwnerUserId="318" LastActivityDate="2010-08-22T00:41:30.583" CommunityOwnedDate="2010-08-22T00:41:30.583" />
  <row Id="2979" PostTypeId="2" ParentId="2949" CreationDate="2010-08-22T00:45:50.527" Score="30" ViewCount="" Body="&lt;p&gt;The fact that you can &lt;a href=&quot;http://new.math.uiuc.edu/optiverse/&quot; rel=&quot;nofollow&quot;&gt;turn a sphere inside out&lt;/a&gt; differentiably.&lt;/p&gt;&#xA;" OwnerUserId="318" LastActivityDate="2010-08-22T08:12:08.317" CommunityOwnedDate="2010-08-22T00:45:50.527" CommentCount="1" />
  <row Id="2980" PostTypeId="1" AcceptedAnswerId="2984" CreationDate="2010-08-22T00:45:51.510" Score="2" ViewCount="144" Body="&lt;p&gt;I'm working on a game which relies on &quot;flinging&quot; an object. That is, click and hold on the object, and then drag and release it, and it continues on the path you were dragging it. Of course, the most well-known example of flinging is with iPhone and Android devices, where you can quickly scroll down a list by quickly swiping your finger upward, giving the illusion of &quot;flinging&quot; the list.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm tracking mouse positions (x,y) and timestamps. But I'm drawing a blank as to how I can take a list of positions and times and get out of it a velocity or curve that an object should follow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are my options? Right now I am looking only for a straight-line fling action, but if it's easy to implement some sort of curve that better fits the fling, that would be good information that I might be able to integrate into the design of the game.&lt;/p&gt;&#xA;" OwnerUserId="137" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-10-27T17:35:23.797" LastActivityDate="2010-10-27T17:35:23.797" Title="Math behind a &quot;fling&quot;? (i.e. on a mobile touch device)" Tags="&lt;approximation&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" />
  <row Id="2981" PostTypeId="1" CreationDate="2010-08-22T01:02:34.860" Score="13" ViewCount="409" Body="&lt;p&gt;The elementary real-valued functions are not closed under integration.  (Elementary function has a precise definition --  see Risch algorithm in Wikipedia).  This means there are elementary functions whose integrals are not elementary.  So we can construct a larger class of functions by adjoining all the integrals of elementary functions.  You can repeat this process indefinitely.  If I understand things correctly, the set of functions that is the countable closure of this process is closed under integration.  Does any &lt;em&gt;finite&lt;/em&gt; iteration of the process achieve closure under integration?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My guess is no.  Has anyone thought about this?&lt;/p&gt;&#xA;" OwnerUserId="318" LastActivityDate="2010-11-14T20:12:40.690" Title="Closing up the elementary functions under integration." Tags="&lt;calculus&gt;&lt;real-analysis&gt;" AnswerCount="4" CommentCount="2" FavoriteCount="3" />
  <row Id="2982" PostTypeId="2" ParentId="2980" CreationDate="2010-08-22T01:03:15.547" Score="0" ViewCount="" Body="&lt;p&gt;Why not compute a cubic spline from the 5 most recent samples equidistant in time?  This will give you a smooth curve and you can use the parametric derivative of the spline as a velocity.  &lt;a href=&quot;http://mathworld.wolfram.com/CubicSpline.html&quot; rel=&quot;nofollow&quot;&gt;http://mathworld.wolfram.com/CubicSpline.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1284" LastActivityDate="2010-08-22T01:03:15.547" CommentCount="2" />
  <row Id="2983" PostTypeId="2" ParentId="2981" CreationDate="2010-08-22T01:12:48.163" Score="1" ViewCount="" Body="&lt;p&gt;I'm not entirely sure it can be closed; apparently more and more complicated integrals of compositions of elementary functions require further generalizations of the &quot;hypergeometric function&quot; to several variables; for instance, the integral of $\sin(\sin(x))$ requires a &lt;a href=&quot;http://mathworld.wolfram.com/KampedeFerietFunction.html&quot; rel=&quot;nofollow&quot;&gt;Kampé de Fériet&lt;/a&gt; hypergeometric function for its closed form, and the elliptic integrals (integrals containing square roots of cubic or quartic polynomials) require the &lt;a href=&quot;http://mathworld.wolfram.com/LauricellaFunctions.html&quot; rel=&quot;nofollow&quot;&gt;Lauricella&lt;/a&gt; or &lt;a href=&quot;http://mathworld.wolfram.com/AppellHypergeometricFunction.html&quot; rel=&quot;nofollow&quot;&gt;Appell&lt;/a&gt; hypergeometrics. As to whether there is &quot;the one hypergeometric to rule them all&quot;, I think that's an open question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To clarify the little discussion Qiaochu and I had in the comments for other readers: the reason you should be interested in the hypergeometrics is that virtually all the elementary functions, and combinations thereof (sum, products, compositions, etc.) are expressible as hypergeometrics. As I have shown in the example I gave in the comments, integrating a hypergeometric function with p+q+1 arguments can require a hypergeometric function with p+q+3 arguments. If you imagine this procedure being applied by multiplying a hypergeometric function with each term of a power series, you get &lt;a href=&quot;http://functions.wolfram.com/HypergeometricFunctions/MeijerG/06/01/01/01/0002/&quot; rel=&quot;nofollow&quot;&gt;even more complicated hypergeometric functions&lt;/a&gt; (in that link, you have the multivariate generalization of the hypergeometric function known as Meijer's G function. Integrating a Meijer function can give you a &lt;a href=&quot;http://functions.wolfram.com/HypergeometricFunctions/MeijerG/21/01/01/0001/&quot; rel=&quot;nofollow&quot;&gt;Meijer function with even more arguments&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the question one should be asking, I think, is that if there exists &quot;the one hypergeometric to rule them all&quot;; or to (mis)use another colloquialism: &quot;the mother of all hypergeometrics&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT×2&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I guess all I have been trying to say, after all this elaboration: unless one would be satisfied with an &quot;it's turtles all the way down&quot; sort of answer, then I think there's no way to satisfactorily resolve this question.&lt;/p&gt;&#xA;" OwnerUserId="498" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-08-22T05:08:12.987" LastActivityDate="2010-08-22T05:08:12.987" CommentCount="3" />
  <row Id="2984" PostTypeId="2" ParentId="2980" CreationDate="2010-08-22T01:30:21.920" Score="4" ViewCount="" Body="&lt;p&gt;In the physical world, once an object is released from all external forces, it will travel in a straight line. UI design strongly suggests that interfaces work better when the user is using previous knowledge of motor control, eye movement, physics, etc. Unless you have a legitimate reason to curve after the object is released, I would personally go with linear only.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In that case, take the last few coordinates and timestamps, and average the velocity components together. Then run a quadratic or exponential function of time on each component until the change of position is negligibly small. The latter functions relates to throwing an object upward and watching it being slowed by gravity, and the former relates to shoving an object on the ground and seeing it come to a stop by friction. Since the exponential function seems like the best for the job, here's an example.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Find the last few velocities for the $x$ and $y$ components, and average them:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$v = \frac{\Delta x}{\Delta t}; y = \cdots$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x = \frac{\sum v_x}{n}; y = \cdots$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Set time from release for $t$, $\beta$ as some friction constant, $v_x$ as the average velocity for each component, and $x_0$ as the initial release point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x = v_x (1 - e^{-\beta t})$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The purpose of the 1 is to align the function to start at zero when time is zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I hope this helps. I'm interested in what you're doing, so you'll have to show me when you're done. ;)&lt;/p&gt;&#xA;" OwnerUserId="227" LastActivityDate="2010-08-22T01:30:21.920" CommentCount="4" />
  <row Id="2985" PostTypeId="2" ParentId="2731" CreationDate="2010-08-22T01:47:49.467" Score="2" ViewCount="" Body="&lt;p&gt;1/2. Imagine there are four houses in a row, each with two children. One has two boys; one has two girls; one has an older boy and a younger girl; one has a younger boy and an older girl. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, you go up to one of these houses at random and a child (chosen at random, say by coin flip) comes to the door. In four of the possible scenarios, this child is a boy. In two of &lt;em&gt;those&lt;/em&gt; scenarios, the &lt;em&gt;other&lt;/em&gt; child is also a boy. The answer is then two divided by four, or one-half.&lt;/p&gt;&#xA;" OwnerUserId="173" LastActivityDate="2010-08-22T01:47:49.467" CommentCount="2" />
  <row Id="2986" PostTypeId="2" ParentId="2949" CreationDate="2010-08-22T02:09:02.170" Score="16" ViewCount="" Body="&lt;p&gt;Euler's Polyhedral Formula: $\text{vertices} + \text{faces} - \text{edges} = 2$ for convex (more generally, sphere-like) polyhedra.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Euler discovered this about 1750 though the Greeks might well have discovered this fact. The first proof, however, was given by Legendre, using spherical geometry.&lt;/p&gt;&#xA;" OwnerUserId="1369" LastEditorUserId="1369" LastEditorDisplayName="" LastEditDate="2010-08-22T11:24:30.980" LastActivityDate="2010-08-22T11:24:30.980" CommunityOwnedDate="2010-08-22T02:09:02.170" CommentCount="1" />
  <row Id="2991" PostTypeId="1" AcceptedAnswerId="2995" CreationDate="2010-08-22T05:47:14.873" Score="2" ViewCount="330" Body="&lt;p&gt;Hi this is my first time posting on here... so please bear with me :P&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was just wondering how I can solve something like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;25x ≡ 3 (mod 109)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If someone can give a break down on how to do it would be appreciated (I'm a slow learner...)!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is proof that I've attempted:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Using definition of modulus we can rewrite 25x ≡ 3 (mod 109) as 25x = 3 + 109y (for some integer y). We can rearrange that to 25x - 109y = 3.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;We use Extended Euclidean Algorithm (not sure about this part, I keep messing things up), so this is where I'm stuck at.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-08-22T05:53:33.413" LastActivityDate="2011-03-07T00:11:35.763" Title="Not understanding Simple Modulus Congruency" Tags="&lt;number-theory&gt;&lt;discrete-mathematics&gt;&lt;modular-arithmetic&gt;" AnswerCount="4" CommentCount="0" FavoriteCount="2" />
  <row Id="2992" PostTypeId="2" ParentId="2755" CreationDate="2010-08-22T06:12:15.813" Score="50" ViewCount="" Body="&lt;p&gt;I'm going to try to give a lighter-flavoured version of my previous answer.  I'd rather not edit the previous one anymore so here goes another response.  I want to make clear, this response is to &lt;em&gt;you&lt;/em&gt;, not your 10-year-old nephew.  How you translate this response to any person depends more on you and that person than anything else. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take a look at the &lt;a href=&quot;http://en.wikipedia.org/wiki/Diffeomorphism&quot; rel=&quot;nofollow&quot;&gt;Wikipedia page for diffeomorphism&lt;/a&gt;. In particular,the lead image &lt;img src=&quot;http://upload.wikimedia.org/wikipedia/commons/thumb/5/51/Diffeomorphism_of_a_square.svg/220px-Diffeomorphism_of_a_square.svg.png&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I look at that image I see the standard Cartesian coordinate grid, but deformed a little. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://upload.wikimedia.org/wikipedia/commons/c/c9/2D_Cartesian_Coordinates.PNG&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's a &quot;big theorem&quot; in a subject called &lt;a href=&quot;http://en.wikipedia.org/wiki/Differentiable_manifold&quot; rel=&quot;nofollow&quot;&gt;Manifold Theory&lt;/a&gt; and it's name is the &quot;Isotopy Extension Theorem&quot;.  Moreover, it has a lot to do with these kinds of pictures. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The isotopy extension theorem is roughly this construction: say you have some rubber, and it's sitting in a medium of liquid epoxy that's near-set.  Moreover, imagine the epoxy to be multi-coloured.  So when you move the rubber bit around in the epoxy, the epoxy will &quot;track&quot; the rubber object.  If your epoxy had a happy-face coloured into it originally, after you move the rubber, you'll see a deformed happy-face.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://minimatefactory.com/blog/wp-content/uploads/2008/11/paint.jpg&quot;&gt;&#xA;&lt;img src=&quot;http://www.deloitte.com/assets/Dcom-UnitedStates/Local%20Assets/Images/Full%20Size%20Images/us_risk_paint%20swirl_200x200.jpg&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you get images that look a lot like mixed paint.  Stir various blotches of paint, and the paint gets distorted.  The more you stir, the more it mixes and it gets harder and harder to see the original image.  The important thing is that the mixed paint is something of a &quot;record&quot; of how you moved your rubber object.  And if your motion of the rubber object returns it to its initial position, there is a function&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ f : X \to X $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $X$ is all positions outside your rubber object.  Given $x \in X$ you can ask where the particle of paint at position $x$ went after the mixing, and call that position $F(x)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;All my talk about fibre bundles and homotopy-groups in the previous response was a &quot;high level&quot; encoding of the above idea.  An intermediate step in the formalization of this idea is the solution of an ordinary differential equation, and that differential equation is essentially the &quot;paint-mixing idea&quot; above, in case you want to look at this subject in more detail later. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what does this mean?  A motion of an object from an initial position &lt;em&gt;back to the initial position&lt;/em&gt; gives you an idea of how to &quot;mix paint&quot; outside the object.  Or said another way, it gives you an &lt;a href=&quot;http://en.wikipedia.org/wiki/Automorphism&quot; rel=&quot;nofollow&quot;&gt;Automorphism&lt;/a&gt; of the complement, in our case that's a 1-1, continuous bijective function between 3-dimensional space remove the garment and itself. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You may find it odd but mathematicians have been studying &quot;paint mixing&quot; in all kinds of mathematical objects, including &quot;the space outside of garments&quot; and far more bizarre objects for well over 100 years.  This is the subject of &lt;a href=&quot;http://en.wikipedia.org/wiki/Dynamical_system&quot; rel=&quot;nofollow&quot;&gt;dynamical systems&lt;/a&gt;.  &quot;Garment complements&quot; are a very special case, as these are subsets of 3-dimensional euclidean space and so they're 3-manifolds.  Over the past 40 years our understanding of 3-manifolds has changed and seriously altered our understanding of things.  To give you a sense for what this understanding is, let's start with the basics.  3-manifolds are things that on small scales look just like &quot;standard&quot; 3-dimensional Euclidean space.   So 3-manifolds are an instance of &quot;the flat earth problem&quot;.  Think about the idea that maybe the earth is like a flat sheet of paper that goes on forever.  Some people (apparently) believed this at some point. And superficially, as an idea, it's got some things going for it.  The evidence that the earth isn't flat requires some build-up. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://upload.wikimedia.org/wikipedia/commons/thumb/9/97/Triangles_%28spherical_geometry%29.jpg/300px-Triangles_%28spherical_geometry%29.jpg&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyhow, so 3-manifolds are the next step.  Maybe all &lt;em&gt;space&lt;/em&gt; isn't flat in some sense.  That's a tricky concept to make sense of as space isn't &quot;in&quot; anything -- basically by definition whatever space is in we'd call space, no?  Strangely, it's not this simple.  A guy named Gauss &lt;a href=&quot;http://en.wikipedia.org/wiki/Theorema_Egregium&quot; rel=&quot;nofollow&quot;&gt;discovered that there is a way to make sense of space being non-flat&lt;/a&gt; &lt;em&gt;without&lt;/em&gt; space sitting in something larger. Meaning curvature is a relative thing, not something judged by some exterior absolute standard.  This idea was a revelation and spawned the idea of an &lt;em&gt;abstract manifold&lt;/em&gt;.  To summarize the notion, here is a little thought experiment. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine a rocket with a rope attached to its tail, the other end of the rope fixed to the earth.  The rocket takes off and goes &lt;em&gt;straight&lt;/em&gt; away from the earth.  Years later, the rocket returns from some other direction, and we grab both loose ends of the rope and pull.  We pull and pull, and soon the rope is tight.  And the rope doesn't move, it's taught. &lt;em&gt;as if&lt;/em&gt; it was stuck on something.  But the rope isn't touching anything except your hands.  Of course you can't see all the rope at one time as the rope is tracing out the (very long) path of the rocket. But if you climb along the rope, after years you can verify: it's finite length, it's not touching anything except where it's pinned-down on the earth.  And it can't be pulled in.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is what a topologist might call a &lt;em&gt;hole in the universe&lt;/em&gt;.  We have abstract conceptions of these types of objects (&quot;holes in the universe&quot;) but by their nature they're not terribly easy to visualize -- not impossible either, but it takes practice and some training.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the 1970's by the work of many mathematicians we started to achieve an understanding of what we expected 3-manifolds to be like.  In particular we had procedures to construct them all, and a rough idea of how many varieties of them there should be.  The conjectural description of them was called the &lt;a href=&quot;http://en.wikipedia.org/wiki/Geometrization_conjecture&quot; rel=&quot;nofollow&quot;&gt;geometrization conjecture&lt;/a&gt;. It was a revelation in its day, since it implied that many of our traditional notions of geometry from studying surfaces in 3-dimensional space translate to the description of all 3-dimensional manifolds.   The geometriztion conjecture was recently &lt;a href=&quot;http://en.wikipedia.org/wiki/Solution_of_the_Poincar%C3%A9_conjecture&quot; rel=&quot;nofollow&quot;&gt;proven&lt;/a&gt; in 2002. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The upshot of this theory is that in some sense 3-dimensional manifolds &quot;crystalize&quot; and break apart in certain standard ways.  This forces any kind of dynamics on a 3-manifold (like &quot;paint mixing outside of a garment&quot;) to respect that crystalization. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So how do I find a garment you can't turn inside-out?  I manufacture one so that its exterior crystalizes in a way I understand.  In particular I find a complement that won't allow for this kind of turning inside-out.  The fact that these things exist is rather delicate and takes work to see.  So it's not particularly easy to explain the proof.  But that's the essential idea. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: To say a tad more, there is a certain way in which this &quot;crystalization&quot; can be extremely beautiful.  One of the simplest types of crystalizations happens when you're dealing with a finite-volume &lt;a href=&quot;http://en.wikipedia.org/wiki/Hyperbolic_3-manifold&quot; rel=&quot;nofollow&quot;&gt;hyperbolic manifold&lt;/a&gt;.  This happens more often than you might imagine -- and it's the key idea working in the example in my previous response.  The decomposition in this case is very special as there's something called the &quot;Epstein-Penner decomposition&quot; which gives a canonical way to cut the complement into convex polytopes.  Things like tetrahedra, octahedra, icosahedra, etc, very standard objects.  So understanding the dynamics of &quot;garments&quot; frequently gets turned into (ie the problem &quot;reduces to&quot;) the understanding of the geometry of convex polytopes -- the kind of things Euclid was very comfortable with.  In particular there's software called &quot;SnapPea&quot; which allows for rather easy computations of these things. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://www.math.utk.edu/~morwen/dirichlet/gif/2vol3_dirichlet.gif&quot;&gt;&#xA;&lt;img src=&quot;http://www.math.utk.edu/~morwen/dirichlet/gif/v2678_animated.gif&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Images taken from &lt;a href=&quot;http://www.math.utk.edu/~morwen/&quot; rel=&quot;nofollow&quot;&gt;Morwen Thistlethwaite's&lt;/a&gt; webpage.  These are images of the closely-related notion of a &quot;Dirichlet domain&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is an image of the Dirichlet domain for the complement of $8_{17}$, the key idea in the construction of my previous post.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://rybu.org/photo/mathish/chalk/MyScreenshot.png&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Technically, this in the Poincare model for hyperbolic space, which gives it the jagged/curvy appearance. &lt;/p&gt;&#xA;" OwnerUserId="642" LastEditorUserId="642" LastEditorDisplayName="" LastEditDate="2010-08-22T07:55:53.637" LastActivityDate="2010-08-22T07:55:53.637" CommentCount="3" />
  <row Id="2993" PostTypeId="2" ParentId="2991" CreationDate="2010-08-22T06:22:15.963" Score="3" ViewCount="" Body="&lt;p&gt;You need to just 'divide' by 25 and get the solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$25x=3(mod\ 109)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Rightarrow 25^{-1}25x=25^{-1}3 (mod\ 109)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Rightarrow x=25^{-1}3 (mod\ 109)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now $25^{-1}=48$, since $25*48=1200=1(mod\ 109)$. So we have -&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x=48*3=35(mod\ 109)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Refer to &lt;a href=&quot;http://en.wikipedia.org/wiki/Modular_multiplicative_inverse&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Modular_multiplicative_inverse&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1310" LastActivityDate="2010-08-22T06:22:15.963" CommentCount="5" />
  <row Id="2994" PostTypeId="2" ParentId="2981" CreationDate="2010-08-22T06:51:34.907" Score="0" ViewCount="" Body="&lt;p&gt;Alternate between applying the integral to each element and closing the set with respect to the elementary operations.  Let's call the smallest number of such compound iterations necessary to construct a function its &quot;rank&quot;.  So exp has rank 0, but erf, erf(exp), and erf + exp all have rank 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The integral of a function with rank R may have rank R or R+1.  The derivative of a function with rank R may have rank R or R-1, and the rank of the derivative of a function with rank 0 is always 0.  If the derivative of a function with rank R has rank R-1, let's call that function &quot;primitive&quot; to distinguish between the two phases of the iteration.  A non-primitive function can by definition be written in terms of primitive functions of the same rank (and possibly other functions of smaller rank).  For example, erf is primitive, but erf(erf) is not, although they are of the same rank 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose R is the largest rank.  Then the integral f of any rank-R function f' is a non-primitive rank-R function.  It is possible to express f in terms of functions that have a rank smaller than R or are primitive and have rank equal to R, and at least one function of the latter type is required.  The sum of any two primitive functions remains primitive, so in the general case we have such a function appearing in a non-additive expression.  For the same reason we can find a case where it is not the root of the expression tree.  Now we can write f as a composition of a product:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;f = g(h * j)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where j is a primitive function of rank R and g and h are some other functions.&#xA;The derivatives of f are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;f' = (h * j)' * g'(h * j)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;f'' = (h * j)'' * g'(h * j) + (h * j)' * g''(h * j)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;f'' contains the same terms that define f' and is therefore also of rank R, so f' is not primitive.  But there must be at least one such primitive function f', so the hypothesis of a largest rank is contradicted.&lt;/p&gt;&#xA;" OwnerUserId="1284" LastEditorUserId="1284" LastEditorDisplayName="" LastEditDate="2010-08-22T09:50:54.503" LastActivityDate="2010-08-22T09:50:54.503" CommentCount="3" />
  <row Id="2995" PostTypeId="2" ParentId="2991" CreationDate="2010-08-22T07:52:27.233" Score="2" ViewCount="" Body="&lt;p&gt;The extended euclidean algorithm is used to find x and y such that ax + by = gcd of a and b.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In our case a = 109 and b = 25.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we start as follows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Find remainder and quotient when we divide 109 by 25 and write the remainder on the left hand side.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;9 = 109 - 25*4.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we get two new numbers 25 and 9. Write the remainder on the left hand side again.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;7 = 25 - 9*2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we have two new numbers, 9 and 7.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the extended algorithm, we use the formula for 9 in the first step &lt;/p&gt;&#xA;&#xA;&lt;p&gt;7 = 25 - (109 - 25*4)*2 = 25*9 - 109*2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2 = 9 - 7*1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;= (109-25*4) - (25*9 - 109*2) = 109*3 - 25*13&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now write&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1 = 7 - 3*2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i.e.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1 = (25*9 - 109*2) - 3*(109*3 - 25*13) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;i.e.&#xA;1 = 25*48 - 109*11&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus 25x - 109y = 1 for x = 48 and y = 11.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So 25x - 109y = 3 for x = 48*3 = 144 and y = 11*3 = 33.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore 144*25 = 3 (mod 109).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you need a number &amp;lt;= 109,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;144 = 109 + 35.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we have (109+35)*25 = 3 (mod 109).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which implies 35*25 = 3 (mod 109).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus x = 35 is a solution to your equation, which we found using the extended euclidean algorithm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope that helps.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-08-22T07:52:27.233" CommentCount="4" />
  <row Id="2996" PostTypeId="2" ParentId="2204" CreationDate="2010-08-22T08:00:11.617" Score="1" ViewCount="" Body="&lt;p&gt;Ha, that's funny, I read &quot;plane&quot; as &quot;airplane&quot;. :-)&lt;br&gt;&#xA;(That's what happens if you talk about bugs in a math problem!)&lt;/p&gt;&#xA;" OwnerUserId="1383" LastActivityDate="2010-08-22T08:00:11.617" CommentCount="1" />
  <row Id="2997" PostTypeId="2" ParentId="370" CreationDate="2010-08-22T08:07:14.007" Score="2" ViewCount="" Body="&lt;p&gt;The first few chapters of Goldblatt's &lt;a href=&quot;http://books.google.com/books?id=VUrePAAACAAJ&quot; rel=&quot;nofollow&quot;&gt;Topoi: the categorial analysis of logic&lt;/a&gt; provide another fairly elementary introduction to the basics of category theory.&lt;/p&gt;&#xA;" OwnerUserId="1399" LastActivityDate="2010-08-22T08:07:14.007" CommunityOwnedDate="2010-08-22T08:07:14.007" />
  <row Id="2998" PostTypeId="2" ParentId="2949" CreationDate="2010-08-22T08:08:28.683" Score="6" ViewCount="" Body="&lt;p&gt;I recall vividly the moment I learnt of &lt;a href=&quot;http://en.wikipedia.org/wiki/Thomae%27s_function&quot; rel=&quot;nofollow&quot;&gt;Thomae's function&lt;/a&gt;, which is continuous at all irrational numbers and discontinuous at all rational numbers.&lt;/p&gt;&#xA;" OwnerUserId="1301" LastActivityDate="2010-08-22T08:08:28.683" CommunityOwnedDate="2010-08-22T08:08:28.683" CommentCount="1" />
  <row Id="2999" PostTypeId="2" ParentId="2595" CreationDate="2010-08-22T08:14:50.683" Score="2" ViewCount="" Body="&lt;p&gt;Given some $A$, you can measure how well it performs on the ensemble of $R_1,...,R_{24}$ and $I_1,...,I_{24}$ by computing the error. For example:&#xA;$$&#xA;J = \sum_{i=1}^{24} ||A R_i - I_i ||^2&#xA;$$&#xA;Where we have used the standard Euclidean norm. The goal is to find the $A$ that makes $J$ as small as possible. We can express this more compactly by stacking the vectors $R_i$ and $I_i$ into larger 3x24 matrices. So if we let: $R = [R_1 R_2 \dots R_{24}]$ and similarly for $I$, we can rewrite our error as:&#xA;$$&#xA;J = ||AR - I||_F^2&#xA;$$&#xA;where we are now using the &lt;a href=&quot;http://en.wikipedia.org/wiki/Matrix_norm&quot; rel=&quot;nofollow&quot;&gt;Frobenius norm&lt;/a&gt;. This optimization problem can be solved in many different ways. For example, you can &lt;a href=&quot;http://en.wikipedia.org/wiki/Matrix_calculus&quot; rel=&quot;nofollow&quot;&gt;differentiate&lt;/a&gt; J with respect to A and set the result equal to zero. This yields the optimality condition:&#xA;$$&#xA;ARR^T = IR^T&#xA;$$&#xA;So any A that satisfies the above equation will be optimal in the sense that it will minimize J. If R is a full-rank matrix (likely), then the optimal A is given by:&#xA;$$&#xA;A = IR^T(RR^T)^{-1}&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="1356" LastActivityDate="2010-08-22T08:14:50.683" CommentCount="1" />
  <row Id="3000" PostTypeId="2" ParentId="998" CreationDate="2010-08-22T09:36:03.033" Score="2" ViewCount="" Body="&lt;p&gt;The &lt;a href=&quot;http://books.google.com/books?id=8zQYxx7PjqsC&quot; rel=&quot;nofollow&quot;&gt;Invitation to algebraic geometry&lt;/a&gt; by Smith et al. is also very readable. &lt;/p&gt;&#xA;" OwnerUserId="1399" LastActivityDate="2010-08-22T09:36:03.033" CommunityOwnedDate="2010-08-22T09:36:03.033" />
  <row Id="3001" PostTypeId="1" AcceptedAnswerId="3004" CreationDate="2010-08-22T09:45:04.413" Score="3" ViewCount="160" Body="&lt;p&gt;In my course of linear algebra I studied that if V is a finite dimensional vector space on the complex field, then every endomorphism of V has an eigenvector. The proof is simple: taken a polynomial f that is null on the endomorphism A (it exists because of the finite dimension of V), exploiting the algebraic closure of C we are there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I gave the following geometrical interpretation: such a space can't be &lt;em&gt;fully twisted&lt;/em&gt; by any of its linear operators, we always have at least an invariant subspace of dimenson 1. This fact still appears a bit magical to me. Is there some deep or different reason for this in other parts of mathematics such as geometry or algebra? Are there other interesting geometrical consequences? Does it remain true if the space is infinite dimensional?&lt;/p&gt;&#xA;" OwnerUserId="365" LastActivityDate="2010-08-22T11:02:31.687" Title="Is there something deep in the fact that an endomorphism of a finite dimensional complex vector space has an eigenvector?" Tags="&lt;linear-algebra&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="3002" PostTypeId="2" ParentId="2949" CreationDate="2010-08-22T09:54:49.523" Score="5" ViewCount="" Body="&lt;p&gt;When I started studying elliptic curves and modular forms I was really amazed by the fact that for a normalized eigenform the Fourier coefficients are the Hecke eigenvalues.&lt;/p&gt;&#xA;" OwnerUserId="688" LastActivityDate="2010-08-22T09:54:49.523" CommunityOwnedDate="2010-08-22T09:54:49.523" />
  <row Id="3003" PostTypeId="2" ParentId="3001" CreationDate="2010-08-22T10:03:36.643" Score="3" ViewCount="" Body="&lt;h3&gt;Topological argument&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Suppose $A$ has no kernel. Then it induces an automorphism of the projective space $\mathbb P(V)$. Now the fact that A has an eigenvector follows from the (purely topological) &lt;a href=&quot;http://en.wikipedia.org/wiki/Lefschetz_fixed_point_theorem&quot; rel=&quot;nofollow&quot;&gt;Lefschetz formula&lt;/a&gt; (because $\mathbb P(V)$ has non-zero Euler characteristic and any linear transformation lies in the connected component of the identity).&lt;/p&gt;&#xA;" OwnerUserId="152" LastEditorUserId="152" LastEditorDisplayName="" LastEditDate="2010-08-22T10:40:43.977" LastActivityDate="2010-08-22T10:40:43.977" CommentCount="7" />
  <row Id="3004" PostTypeId="2" ParentId="3001" CreationDate="2010-08-22T10:18:31.477" Score="3" ViewCount="" Body="&lt;p&gt;As to the depth and significance of the existence of eigenvectors for finite-dimensional linear operators over $\mathbb{C}$, I think you have already identified the core algebraic reason: it is that the field $\mathbb{C}$ is algebraically closed.  I think it is fair to call this a &quot;deep fact&quot; -- the task of giving a rigorous proof was the topic of Gauss's thesis work, and in fact by modern standards Gauss's (first) proof is still not completely rigorous.  (Some feel this way, anyway -- there is not universal agreement here.) Whole books have been written on various proofs of this result.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course the depth here lies in the fact that the definition of the complex field is topological / analytic, ultimately relying on the completeness of $\mathbb{R}$.  If you start with a field $K$, the following is not very deep:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proposition: For a field $K$, the following are equivalent:&lt;br&gt;&#xA;(i) Every linear endomorphism of a finite dimensional $K$-vector space has an eigenvector.&lt;br&gt;&#xA;(ii) $K$ is algebraically closed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To see (i) &amp;rArr; (ii), use the fact that every monic polynomial of degree $n$ is the chracteristic and minimal polynomial of its &lt;a href=&quot;http://en.wikipedia.org/wiki/Companion_matrix&quot; rel=&quot;nofollow&quot;&gt;companion matrix&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To answer your second question: no, there are linear operators on infinite dimensional $\mathbb{C}$-vector spaces without eigenvalues, even &lt;em&gt;bounded&lt;/em&gt; linear operators.  A fundamental example is the Hilbert space $L^2([0,1])$ of square integrable $\mathbb{C}$-valued functions on the unit interval.  Then the &lt;strong&gt;multiplication operator&lt;/strong&gt; $M: f \mapsto xf$ has norm $1$ but is easily seen to have no eigenvectors, since the equation $xf = \lambda f$ forces $f = 0$ almost everywhere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In functional analysis, there is a suitable generalization of the notion of the set of eigen*values* of a linear operator on certain infinite dimensional spaces, namely the &lt;a href=&quot;http://en.wikipedia.org/wiki/Spectrum_%28functional_analysis%29&quot; rel=&quot;nofollow&quot;&gt;spectrum&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Addendum&lt;/b&gt;: inspired by damiano's comment, here is what is in some sense the simplest possible example of a linear operator on an infinite dimensional space without an eigenvector: Let $K$ be any field, and let $V$ be a vector space of countably infinite dimension with basis ${e_n}_{n=1}^{\infty}$.  Then consider the &lt;strong&gt;shift operator&lt;/strong&gt; $T$ &#xA;on $V$, i.e., the unique linear operator such that for all $n \in \mathbb Z^+$, $T(e_n) = e_{n+1}$.  $T$ is very simple and easy to visualize: it just happens never to &quot;cycle back&quot; on itself.  Indeed, if $v = \sum_{n=1}^{\infty} a_n e_n \in V$ (with $a_n = 0$ &#xA;for all but finitely many $n$), then assuming that $Tv = \lambda v$ gives &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sum_{n=1}^{\infty} \lambda a_n e_n = \sum_{n=2}^{\infty} a_{n-1} e_n$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular $\lambda a_1 = 0$.  It is clear that the kernel of $T$ is $0$, so if $v \neq 0$, $\lambda \neq 0$ and thus $a_1 = 0$ and then the above equation implies that $a_n = 0$ for all $n$, i.e., $v = 0$. (If $V = K[x]$ is the space of polynomials with $K$-coefficients, then with respect to the natural basis $e_n = x^{n-1}$, multiplication by $x$ &lt;i&gt;is&lt;/i&gt; the shift operator $T$.)  &lt;/p&gt;&#xA;" OwnerUserId="299" LastEditorUserId="152" LastEditorDisplayName="" LastEditDate="2010-08-22T11:02:31.687" LastActivityDate="2010-08-22T11:02:31.687" CommentCount="2" />
  <row Id="3005" PostTypeId="2" ParentId="2949" CreationDate="2010-08-22T10:44:26.500" Score="5" ViewCount="" Body="&lt;p&gt;The Thom-Pontrjagin theorem: $\Omega_n^{SO} \cong \pi_n(MSO)$. The group of equivalence classes of n-manifolds with respect to oriented cobordism is isomorphic to the n-th homotopy group of the Thom spectrum MSO. This can be generalized to include different cobordisms (unoriented, ...) and different Thom spectra. See for example &lt;a href=&quot;http://www.math.umass.edu/~weston/oldpapers/cobord.pdf&quot; rel=&quot;nofollow&quot; title=&quot;the minor thesis&quot;&gt;the minor thesis&lt;/a&gt; by T. Weston.&lt;/p&gt;&#xA;" LastActivityDate="2010-08-22T10:44:26.500" CommunityOwnedDate="2010-08-22T10:44:26.500" CommentCount="0" />
  <row Id="3006" PostTypeId="1" AcceptedAnswerId="3010" CreationDate="2010-08-22T10:54:24.820" Score="11" ViewCount="861" Body="&lt;p&gt;As the title implies, It is seems that $e^x$ is the only function whoes derivative is the same as itself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;thanks.&lt;/p&gt;&#xA;" OwnerUserId="1382" LastEditorUserId="171" LastEditorDisplayName="" LastEditDate="2010-08-22T11:00:55.777" LastActivityDate="2010-11-22T19:27:45.473" Title="Could you explain why $de^x/dx = e^x$ &quot;intuitively&quot;?" Tags="&lt;calculus&gt;&lt;intuition&gt;" AnswerCount="12" CommentCount="8" FavoriteCount="7" />
  <row Id="3007" PostTypeId="2" ParentId="3006" CreationDate="2010-08-22T11:10:25.387" Score="0" ViewCount="" Body="&lt;p&gt;$d/dx (e^{-x} \cdot y)=-e^{-x} \cdot y+e^{-x} \cdot dy/dx=e^{-x}(dy/dx-y)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if $dy/dx=y$, then $d/dx (e^{-x} \cdot y)=0$, ie $e^{-x} \cdot y=c$ or $y=c \cdot e^x$&lt;/p&gt;&#xA;" OwnerUserId="123" LastEditorUserId="123" LastEditorDisplayName="" LastEditDate="2010-08-22T20:59:36.120" LastActivityDate="2010-08-22T20:59:36.120" CommentCount="4" />
  <row Id="3008" PostTypeId="1" AcceptedAnswerId="3156" CreationDate="2010-08-22T11:11:38.677" Score="3" ViewCount="406" Body="&lt;p&gt;When I was first learning about the axiom of choice, a very helpful and fun exercise was to try to find a rule that maps every nonempty subset of $\mathbb R$ to one of its members.  No matter how hard you try, it was impossible to find one.  (If you try the same for the nonempty subsets of $\mathbb N$, it is pretty easy, just take the smallest member of each set.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My first question is, has it been proved in some formal way that such a choice function cannot be completely described with a finite number of symbols? (I think this is equivalent to  the undecidability of the existence of a choice function on  $\mathcal P (\mathbb R) \setminus \{ \emptyset \}$ in ZF.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another question is, in ZF with the negation of AC, is it provable that there exists no choice function on $\mathcal P (\mathbb R) \setminus \{ \emptyset \}$ ?  (My guess is negative.)&lt;/p&gt;&#xA;" OwnerUserId="145" LastEditorUserId="145" LastEditorDisplayName="" LastEditDate="2010-08-23T12:30:14.347" LastActivityDate="2010-08-23T23:00:25.553" Title="Choice function on $\mathcal P (\mathbb R) \setminus \{ \emptyset\}$" Tags="&lt;set-theory&gt;" AnswerCount="2" CommentCount="6" />
  <row Id="3009" PostTypeId="2" ParentId="3006" CreationDate="2010-08-22T11:25:12.227" Score="4" ViewCount="" Body="&lt;p&gt;Doesn't quite explain $e^x$, but looking at the series of $2^n$, i.e. $\{1,2,4,8,16,32,...,2^k,2^{k+1},...\}$ might help.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You see there if you take the first order difference you get back the same series, as $2^{k+1}-2^k=2^k$.&lt;/p&gt;&#xA;" OwnerUserId="1310" LastActivityDate="2010-08-22T11:25:12.227" CommentCount="1" />
  <row Id="3010" PostTypeId="2" ParentId="3006" CreationDate="2010-08-22T11:25:58.480" Score="11" ViewCount="" Body="&lt;p&gt;Well, think of &lt;a href=&quot;http://en.wikipedia.org/wiki/Exponential_growth&quot; rel=&quot;nofollow&quot;&gt;exponential growth&lt;/a&gt; (like e.g. bacteria grow):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know, the more bacteria exist in a colony, the faster the colony will grow. More precisely: The growth speed of the colony $B$ is proportional to it's size ... Double size, double speed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{dB}{dt} \sim B$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore we know the growth is exponential, since bacteria clone themselves in fixed amounts of time, i.e.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$B \sim 2^{k\cdot t}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Putting it together, we can deduce that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{d}{dt}2^{kt} = c \cdot 2^{kt}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or with $a = 2^k$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{d}{dt}a^t = c \cdot a^t$&lt;/p&gt;&#xA;&#xA;&lt;hr/&gt;&#xA;&#xA;&lt;p&gt;Now, how do we get the $e$? We just ask: &lt;em&gt;What base $a$ do we have to take such that $c = 1$, i.e. $\dot{B} = B$?&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We simply &lt;strong&gt;call&lt;/strong&gt; that base $e$. Having such an $e$ is quite useful. We could use its special derivation traits we found above to define all exponential functions to the base $e$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a^x = e^{x \cdot \ln a}$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This shows that the factor $c$ we encountered in the above equations equals $\ln a = \log_e a$ and therefore, we can easily derive all kinds of exponential terms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After all, $e$ turns out to be $lim_{x \to \infty} \left(1 + \frac{1}{x}\right)^x \approx 2.718\ldots$&lt;/p&gt;&#xA;" OwnerUserId="917" LastEditorUserId="917" LastEditorDisplayName="" LastEditDate="2010-08-22T11:44:00.183" LastActivityDate="2010-08-22T11:44:00.183" />
  <row Id="3011" PostTypeId="2" ParentId="3006" CreationDate="2010-08-22T11:38:42.360" Score="4" ViewCount="" Body="&lt;p&gt;I'll act on Casebash's proposal later, but for now... care for a movie?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/nLLL7.gif&quot; alt=&quot;tangent line&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(granted, I cheated and spaced the two plots of $\exp(x)$ to clearly show the moving tangent line... but you guys should be able to understand this).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(thanks to Stan Wagon!)&lt;/p&gt;&#xA;" OwnerUserId="498" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-08-22T11:51:36.440" LastActivityDate="2010-08-22T11:51:36.440" />
  <row Id="3012" PostTypeId="2" ParentId="3006" CreationDate="2010-08-22T11:41:27.830" Score="0" ViewCount="" Body="&lt;p&gt;There are two components to the original question:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Why do exponential functions generally (the class $x \mapsto ab^x$ for $b&gt;0$) have the property of being eigenfunctions of the derivative operator? That is: why do these functions f have the property $\frac{\mathrm d}{\mathrm dx} f(x) = \lambda f(x)$?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Why are the functions $x \mapsto a \mathrm e^x$ in particular +1 eigenfunctions --- i.e. why is the scalar $\lambda$ equal to 1 for these functions in particular? (To put it another way: why is this the case when the base is equal to the particular constant e&amp;nbsp;&amp;asymp;&amp;nbsp;2.71828?)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I can give an intuitive answer to the first of these questions, by combining two observations.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Consider the slope at a paticular point, &lt;em&gt;e.g.&lt;/em&gt; at x&amp;nbsp;=&amp;nbsp;0. Multiplying the function f(x) by some scalar now only amplifies the function by that factor, but also amplifies the slope by that same factor. (This is just a restatement of the product rule, for constant scalar factors.)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Exponential functions f(x) are self-similar: multiplying them by a positive scalar is equivalent to translating them to the left, or to the right, depending on the scalar by which you multiply: more precisely, if you multiply $\mathrm e^x$ by &lt;em&gt;c&lt;/em&gt;, why you get is $\mathrm e^{[x + \ln(c)]}$, or the same function translated  ln(&lt;em&gt;c&lt;/em&gt;) to the left. More generally, for $f(x) = ab^x$, we have &lt;em&gt;c&lt;/em&gt;f(x) = f(x + log&lt;sub&gt;&lt;em&gt;b&lt;/em&gt;&lt;/sub&gt;(&lt;em&gt;c&lt;/em&gt;)); that is, multiplying f(x) by &lt;em&gt;c&lt;/em&gt; is equivalent to a shift to the left by log&lt;sub&gt;&lt;em&gt;b&lt;/em&gt;&lt;/sub&gt;(&lt;em&gt;c&lt;/em&gt;).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Now, combine these two facts. The slope of f(x) at x&amp;nbsp;=&amp;nbsp;log&lt;sub&gt;&lt;em&gt;b&lt;/em&gt;&lt;/sub&gt;(&lt;em&gt;c&lt;/em&gt;) is &lt;em&gt;c&lt;/em&gt; times the slope of f(x) at x&amp;nbsp;=&amp;nbsp;0, because of the self-similarity of the exponential function, and because the slope of &lt;em&gt;c&lt;/em&gt;f(x) at x&amp;nbsp;=&amp;nbsp;0 is &lt;em&gt;c&lt;/em&gt; times the slope of f(x) at x&amp;nbsp;=&amp;nbsp;0. Thus the slope of f(x) is a function which is proportional to f(x), QED. &lt;/p&gt;&#xA;" OwnerUserId="439" LastEditorUserId="439" LastEditorDisplayName="" LastEditDate="2010-08-22T21:12:22.567" LastActivityDate="2010-08-22T21:12:22.567" CommentCount="3" />
  <row Id="3014" PostTypeId="2" ParentId="3006" CreationDate="2010-08-22T11:43:44.817" Score="7" ViewCount="" Body="&lt;p&gt;As Mangaldan says,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!} = 1 + \frac{x}{1!}  + \frac{x^2}{2!} + \cdots + \frac{x^n}{n!} + \cdots &#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you accept (it's true, but it could take a while to explain why) that this series, like polynomials, can be derived &lt;em&gt;term by term&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;p(x) = a_0 + a_1 x + a_2 x^2 + \cdots + a_n x^n \  \Longrightarrow \ p'(x) = a_1 + 2a_2x + \cdots + na_n x^{n-1}  \ ,&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\frac{d}{dx} e^x = \frac{1}{1!} + \frac{2x}{2!} + \cdots + \frac{nx^{n-1}}{n!} + \cdots = \sum_{n=0}^{\infty} \frac{x^n}{n!} = e^x \ .&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="664" LastActivityDate="2010-08-22T11:43:44.817" CommentCount="9" />
  <row Id="3015" PostTypeId="2" ParentId="3006" CreationDate="2010-08-22T11:45:06.467" Score="3" ViewCount="" Body="&lt;p&gt;Suppose $\frac{d}{dx}f(x)=f(x)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then for small $h$, $f(x+h)=f(x)+hf(x)=f(x)(1+h)$. If we do this for a lot of small intervals of length $h$, we see $f(x+a)=(1+h)^{a/h}f(x)$. (Does this ring a bell already?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Setting $x=0$ in the above, and fixing $f(0)=1$, we then have $f(1)=(1+h)^{1/h}$, which in limit as $h\rightarrow 0$ goes to $e$. And continuing $f(x)=(1+h)^{x/h}$, which goes to $e^x$.&lt;/p&gt;&#xA;" OwnerUserId="1310" LastActivityDate="2010-08-22T11:45:06.467" CommentCount="2" />
  <row Id="3016" PostTypeId="2" ParentId="2969" CreationDate="2010-08-22T12:03:50.773" Score="2" ViewCount="" Body="&lt;p&gt;First, a few thoughts about this specific box packing problem. It might be possible to pack this box by packing the &quot;y&quot; polycube into 25 cells in 5 identical layers. This corresponds to collapsing this 3-D shape into a polyomino in the plane, and then using it to pack a 5x5 rectangle. However, this polyomino can not tile a 5x5 rectangle. (The terminology for polyominoes is not totally standard but some use the order of a polyomino to mean the smallest number of copies of this shape to tile a rectangle with congruent copies of itself.) The order of this polyomino is 10. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second, there is lots of material about polyominoes and to a lesser extent box packings in the books:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;S. Golomb, Polyominoes: Puzzles, Patterns, Problems, and Packings (Revised Edition), Princeton U. Press, 1994.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;G. Martin Polyominoes, MAA, 1991.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Third, there are a variety of papers related to polyominoes and boxes at Michael Reid's very nice site with links to others:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.math.ucf.edu/~reid/Polyomino/&quot; rel=&quot;nofollow&quot;&gt;http://www.math.ucf.edu/~reid/Polyomino/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.math.ucf.edu/~reid/Research/index.html&quot; rel=&quot;nofollow&quot;&gt;http://www.math.ucf.edu/~reid/Research/index.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fourth, there are &quot;complexity&quot; results which govern the general problem of given a collection of tiles (either in 2-D or 3-D) when can one use them to tile some particular shape (say a rectangle or rectangular box). Here is a paper related to this of Erik Demaine:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://erikdemaine.org/papers/Jigsaw_GC/&quot; rel=&quot;nofollow&quot;&gt;http://erikdemaine.org/papers/Jigsaw_GC/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also see:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://archive.ite.journal.informs.org/Vol5No3/Chlond/index.php&quot; rel=&quot;nofollow&quot;&gt;http://archive.ite.journal.informs.org/Vol5No3/Chlond/index.php&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1369" LastEditorUserId="1369" LastEditorDisplayName="" LastEditDate="2010-08-22T20:18:29.367" LastActivityDate="2010-08-22T20:18:29.367" CommentCount="1" />
  <row Id="3017" PostTypeId="2" ParentId="2949" CreationDate="2010-08-22T12:14:15.260" Score="11" ViewCount="" Body="&lt;p&gt;One of the most surprising results in numerical math is &lt;a href=&quot;http://en.wikipedia.org/wiki/Wilkinson%27s_polynomial&quot; rel=&quot;nofollow&quot;&gt;Wilkinson's polynomial&lt;/a&gt;. Wilkinson gave an example in which a very tiny change to one coefficient of a polynomial can have a drastic impact on the location of the zeros.  The change in the location of the roots is seven orders of magnitude larger than the change in the coefficient.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(This is an exact result.  The impact of the coefficient change is not due to numerical precision.  The point of the example, however, is that changes such as the perturbation of the coefficient are inevitable in numerical computing.)&lt;/p&gt;&#xA;" OwnerUserId="247" LastActivityDate="2010-08-22T12:14:15.260" CommunityOwnedDate="2010-08-22T12:14:15.260" CommentCount="2" />
  <row Id="3018" PostTypeId="2" ParentId="2949" CreationDate="2010-08-22T12:33:30.527" Score="1" ViewCount="" Body="&lt;p&gt;Another example from the numerics front: it's surprising that despite the theoretical fact that Gaussian elimination can be unstable (even with pivoting!), &lt;a href=&quot;http://dx.doi.org/10.1137/0610012&quot; rel=&quot;nofollow&quot;&gt;examples&lt;/a&gt; that trigger this instability are in fact &lt;a href=&quot;http://dx.doi.org/10.1137/0611023&quot; rel=&quot;nofollow&quot;&gt;very rare in practice&lt;/a&gt;, and can be handled by a &lt;a href=&quot;http://dx.doi.org/10.1023/A:1022314201484&quot; rel=&quot;nofollow&quot;&gt;simple fix&lt;/a&gt; if they do arise.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-08-22T12:33:30.527" CommunityOwnedDate="2010-08-22T12:33:30.527" CommentCount="1" />
  <row Id="3019" PostTypeId="2" ParentId="2305" CreationDate="2010-08-22T12:49:47.297" Score="1" ViewCount="" Body="&lt;p&gt;First of all, you might be interested in the work of Chas and Phillips: &quot;Self-intersection of curves on the punctured torus&quot;.  I've only skimmed their paper, but they seem to be doing something closely related to what you want.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second I want to guess, for some reason, that the average time to compute self-intersection number is a lot slower than the average time to generate a word.  (Is that the case?  Could you tell me how you are computing minimal self-intersection numbers?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If so, I guess that you want to generate as few strings as possible.  I'll use $a, A, b, B$ as the generating set for $\pi_1 = \pi_1(T)$.  Looking at Lyndon words is essentially the same as applying inner automorphisms (conjugation, ie cyclic rotation) to your words.   You might also try replacing a word $w$ by its inverse $W$.  If some rotation of $W$ beats $w$ [sic], then you can throw $w$ away. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are also other &quot;geometric automorphisms&quot; (elements of the mapping class group) &#xA;of $\pi_1$ which are very useful eg rotation of $T$ by one-quarter: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$a \mapsto b \mapsto A \mapsto B \mapsto a.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are also two nice reflections: either fix $b, B$ and swap $a$ with $A$, or the other way around.  Composing these gives the hyperelliptic which swaps $a$ with $A$ and swaps $b$ with $B$.  (I use python's swapcase function for this -- very simple!) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If any of these operations (or any compositions of these, eg the reverse of a word) produces a word $w'$ that is lexicographically before $w$, then you can throw $w$ away.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please let me know if this is helpful -- I'm interested in this kind of problem. &lt;/p&gt;&#xA;" OwnerUserId="1307" LastActivityDate="2010-08-22T12:49:47.297" />
  <row Id="3020" PostTypeId="1" AcceptedAnswerId="3079" CreationDate="2010-08-22T13:51:38.763" Score="3" ViewCount="197" Body="&lt;p&gt;I am interested in the Gelfand transformation&#xA;$$&#xA;\Phi\colon\ell^1(\mathbb Z)\to\mathcal C(\mathbb T),\quad a\mapsto\sum_{n\in\mathbb Z}a_n z^n.&#xA;$$&#xA;This is an injective homomorphism of Banach algebras. It is neither isometric nor surjective. However, its image---the Wiener algebra $W$ consisting of all continuous functions on $\mathbb T$ whose Fourier series is absolutely convergent---is a subalgebra of $\mathcal C(\mathbb T)$ which is dense in the subspace topology.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Question: Can we prove of disprove that $\Phi$ has a continuous inverse on its image $W$?&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words: Is $\Phi\colon\ell^1(\mathbb Z)\to W$ an isomorphism of topological algebras?&#xA;(Here $W$ carries the topology induced by the sup-norm from $\mathcal C(\mathbb T)$.&lt;/p&gt;&#xA;" OwnerUserId="367" LastEditorUserId="367" LastEditorDisplayName="" LastEditDate="2010-08-24T08:43:39.657" LastActivityDate="2010-08-24T08:43:39.657" Title="Does the Gelfand transformation on $\ell^1(\mathbb Z)$ possess a continuous inverse on its image?" Tags="&lt;functional-analysis&gt;" AnswerCount="2" CommentCount="6" FavoriteCount="1" />
  <row Id="3021" PostTypeId="2" ParentId="2869" CreationDate="2010-08-22T14:11:01.767" Score="0" ViewCount="" Body="&lt;p&gt;I can't hope to do better than Jacques's answer, so I'll just offer some little numerical advice on computing your function to be fitted (but you probably already know this): make sure that whatever system you're doing your computations in has support for both the upper incomplete gamma function $\Gamma(a,x)$ and the lower incomplete gamma function $\gamma(a,x)$. With these two, we can represent your $h(t)$ as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$h(t)=\frac1{(n-1)!^2}P_6 \gamma(n,P_8 t)\Gamma(n,P_9(t-P_7))$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason for doing things this way is that we mitigate the possibility of any subtractive cancellation happening when your downhill simplex optimization wanders too far out. Which brings me to my second piece of advice: you would do well to find good initial values for your $P_i$ parameters. Nelder-Mead isn't really known for being fast (at least there's no need to assume continuity!), and it also tends to wander around your parameter space unless suitably restricted.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-08-22T14:11:01.767" CommentCount="2" />
  <row Id="3022" PostTypeId="2" ParentId="2991" CreationDate="2010-08-22T14:30:45.563" Score="1" ViewCount="" Body="&lt;p&gt;I meant this as a comment to the discussion after Student's answer but it seems that I don't have the option (reputation too low?) so I'll post it as an answer. Sorry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to compute quickly the inverse of 25 mod 109, note that $25=5^2$. Thus $25^{-1}=t^2$ where $t=5^{-1}$ mod 109. On the other hand, computing the inverse of 5 modulo any number $N$ ending with 9 (or 4) is immediate since it is just $(N+1)/5$.&#xA;Thus $25^{-1}=((109+1)/5)^2=22^2=48$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moral: when performing actual computations always look for easy tricks that allow shortcuts. &lt;/p&gt;&#xA;" OwnerUserId="688" LastActivityDate="2010-08-22T14:30:45.563" />
  <row Id="3023" PostTypeId="2" ParentId="2549" CreationDate="2010-08-22T14:34:36.897" Score="2" ViewCount="" Body="&lt;p&gt;In general the term &quot;sectional curvature&quot; is used in the n-dimensional setting. Basically one computes what is known in theory of surfaces as &quot;Gauss Curvature&quot; for the surface gotten by exponentiating every two plane inside the tangent space of the n-manifold. Then one can show that the following things to build intuition,&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;If two Riemann curvatures are giving the same sectional curvature for every two plane then the Riemann curvatures are equal as endomorphisms. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Ricci curvature along a direction is the sum of the sectional curvatures of all possible two planes spanned by that direction and one more from a basis gotten by extending the given direction. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Scalar curvature at a point is the sum of the sectional curvatures of all possible two planes spanned by a chosen basis. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;A theorem of Schur says that sectional curvature if constant on all 2-planes at a point is constant at all point of the manifold. Sectional curvature is the strongest notion of curvature and will made constant under very strong conditions like maximal symmetry or locally geodesic reflecting isometry or transitive action of the isometry group on orthonormal frames. &lt;/p&gt;&#xA;" OwnerUserId="965" LastActivityDate="2010-08-22T14:34:36.897" CommentCount="0" />
  <row Id="3024" PostTypeId="1" AcceptedAnswerId="3494" CreationDate="2010-08-22T14:45:07.203" Score="3" ViewCount="133" Body="&lt;p&gt;I keep coming across calculations like this,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider a metric on an $n+2$ dimensional manifold given as,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ds^2 = 2dudr + 2L(u,r)du^2 -r^2d\Omega_n^2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then apparently once can write down the Ricci and Einstein and other tensors as a function of n. Like for the above the Einstein tensor apparently has the following non-zero components,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$G_{01} = \frac{n}{r}L_r+\frac{n(n-1)(2L-1)}{2r^2}$ &#xA;$G_{22} = (n-1)[(2L-1)(\frac{2-n}{2})-2rL_r]-r^2L_{rr}$&#xA;$G_{00} = -\frac{nL_u}{r} + \frac{2nLL_r}{r} + \frac{n(n-1)L(2L-1)}{r^2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and $G^2_2 = G^3_3 = ... = G^{n+1}_{n+1}$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(where the subscripts of L denote partial derivatives with respect to those variables)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a fixed given n I can imagine doing the calculation either by hand or some software but I would like to know who these expressions are derived for a general n. &lt;/p&gt;&#xA;" OwnerUserId="965" LastActivityDate="2010-08-28T00:26:33.527" Title="How to compute curvature tensors for general n-dimensions?" Tags="&lt;differential-geometry&gt;" AnswerCount="2" FavoriteCount="2" />
  <row Id="3025" PostTypeId="2" ParentId="3006" CreationDate="2010-08-22T14:55:47.370" Score="1" ViewCount="" Body="&lt;p&gt;Let us define a function $\exp$ as the solution to $D \exp(x) = \exp(x)$ with initial value $\exp(0) = 1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clearly $D \exp \circ f = D f \cdot (\exp \circ f)$ but also notice that&#xA;$$ \begin{align}&#xA;D (\exp(a(x)) \exp(b(x))) &amp;amp;= D a(x) \exp(a(x)) \exp(b(x)) + D b(x) \exp(a(x)) \exp(b(x)) \\&#xA;&amp;amp;= (D (a + b) (x)) \cdot \exp(a(x)) \exp(b(x))&#xA;\end{align} $$&#xA;so we see that $\exp(a)\exp(b)=\exp(a+b)$ which lets us write $\exp(x) = e^x$ for some as yet undetermined constant $e$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Geometrically we can see that the gradient of the tangent line of the curve $y = e^x$ must be between the finite differences, i.e. $\frac{e^x - e^{x-h}}{h} &amp;lt; e^x &amp;lt; \frac{e^{x+h} - e^x}{h}$. With some numerical exploration we see that $e$ is between $2$ and $3$. Specifically, the finite differences for $e=2$ and $h=0.9$ are too small and the finite differences for $e=3$ and $h=0.1$ are too big. To narrow it down more let $e=2.7$ and $h = 0.01$ to see that $2.7 &amp;lt; e &amp;lt; 3$.&lt;/p&gt;&#xA;" LastActivityDate="2010-08-22T15:02:02.880" />
  <row Id="3026" PostTypeId="1" CreationDate="2010-08-22T14:58:39.593" Score="1" ViewCount="191" Body="&lt;p&gt;Hello!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not very fluent in mathematical proofs. High School has, sadly, not taught me any kind of proof-theory. That's why I would like your help with my proof of &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$m = S_m (mod  3).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $S_m$ is the digit sum of m and $\wedge$ is a logical-AND.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm going to use the very little that I know:  Mathematical induction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Basis:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$A(1): S_m = 1 \wedge m = 1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$m = S_m (mod 3) \Rightarrow 1 = 1 (mod 3)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Inductive Step:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$A(n)\colon m = n \wedge S_m = \sum\limits_{k=0}^K m_k = (m_0 + m_1 + ... + m_k)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$m = S_m (mod 3)$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$A(n+1)\colon m = n+1 \wedge S_m = \sum\limits_{k=0}^K m_k = (m_0 + m_1 + ... + m_k + m_k+1)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$m = S_m (mod 3)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm having trouble at the step of A(n+1) because I don't know how to actually prove that it's valid? Please note that $m_0, ..., m_k$ refers to the summation of the digits of the number m. I just didn't find any good documentation that showed how to depict that in mathematical terms. Feel free to correct!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, how do curly braces work? Documentation says \{, \} should do the trick, but they are invisible to me. &lt;/p&gt;&#xA;" OwnerUserId="44" LastEditorUserId="44" LastEditorDisplayName="" LastEditDate="2010-08-23T08:48:14.290" LastActivityDate="2010-08-23T08:48:14.290" Title="Prove that $m = S_m (\mod 3)$?" Tags="&lt;induction&gt;" AnswerCount="4" CommentCount="8" />
  <row Id="3027" PostTypeId="2" ParentId="3006" CreationDate="2010-08-22T15:04:05.933" Score="5" ViewCount="" Body="&lt;p&gt;One conceptual way to understand the standard definition $e^x = \lim_{n\to\infty}(1+x/n)^n$ is to view it as arising from applying &lt;a href=&quot;http://en.wikipedia.org/wiki/Euler_method&quot; rel=&quot;nofollow&quot;&gt;Euler's approximation method&lt;/a&gt; to $y' = y$. The same method also works arbitrary higher-order constant coefficiant ODEs by converting them to linear system form and employing matrix exponentials. This is described quite nicely in Arnold's beautiful textbook &quot;Ordinary differential equations&quot;. Here's an excerpt:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/gK98d.jpg&quot; alt=&quot;alt text&quot;&gt; &#xA;&lt;img src=&quot;http://i.imgur.com/PV2vn.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-08-22T15:04:05.933" CommentCount="3" />
  <row Id="3029" PostTypeId="2" ParentId="3026" CreationDate="2010-08-22T15:08:48.340" Score="1" ViewCount="" Body="&lt;p&gt;I'm having a little trouble understanding your formatting/notation, but I am guessing that what you want to show is that $m = S_m$ (mod 3) where $S_m$ is the sum of the digits of the number $m$.  In the inductive, step, it looks like you are saying that $S_{n+1} = S_n + 1$.  This is not always true (e.g. it is true if $n = 23$ but not true if $n = 59$).  So you should consider the two cases where the last digit is not a 9 and where the last digit is a 9.  You'll easily see you get the same thing mod 3 and this will complete the proof.&lt;/p&gt;&#xA;" OwnerUserId="9" LastActivityDate="2010-08-22T15:08:48.340" />
  <row Id="3030" PostTypeId="2" ParentId="3026" CreationDate="2010-08-22T15:16:56.663" Score="3" ViewCount="" Body="&lt;p&gt;The best way to get the result is that to remember that writing $m=a_n...a_1a_0$ in base 10 is actually a shorthand for&#xA;$$&#xA;m=a_0+a_1\cdot10+...+a_n\cdot 10^n.&#xA;$$&#xA;Now it's enough to observe that $10^n\equiv1\bmod 3$ for all $n$ and thus&#xA;$$&#xA;m\equiv a_0+a_1+...+a_n \bmod 3.&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="688" LastActivityDate="2010-08-22T15:16:56.663" />
  <row Id="3031" PostTypeId="1" AcceptedAnswerId="3109" CreationDate="2010-08-22T15:22:18.313" Score="16" ViewCount="519" Body="&lt;p&gt;Groups can be defined abstractly as sets with a binary operation satisfying certain identities, or concretely as a collection of permutations of a set.  Cayley's theorem ensures that these two definitions are equivalent: any abstract group acts as a collection of permutations of its underlying set, and this action is faithful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, rings can be defined abstractly as sets with a pair of binary operations satisfying certain identities, or concretely as a collection of endomorphisms of an abelian group.  There is a &quot;Cayley's theorem&quot; here as well: any abstract ring acts as a collection of endomorphisms of its underlying abelian group, and this action is faithful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The situation for Lie algebras seems much less clear to me.  The adjoint representation is not generally faithful, and Ado's theorem comes with qualifications and doesn't have the simplicity of the two theorems above.  For me, the problem is that I don't have a good sense of what the concrete definition of a Lie algebra is supposed to be.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suspect that a good concrete definition of a Lie algebra is as a space of derivations on some algebra closed under commutator.  In that case, is it correct to say that a Lie algebra acts faithfully as derivations on its universal enveloping algebra?  Is this a good analogue of Cayley's theorem?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Motivation: in the books on Lie algebras I have read, the authors verify that Lie algebras which occur in nature satisfy alternativity and the Jacobi identity, but I have never seen any simple justification that these axioms are &quot;enough&quot; in the same way that Cayley's theorem tells you that the axioms for a group or a ring are &quot;enough.&quot;  There is just Ado's theorem which, again, comes with qualifications and is hard.)&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-08-23T12:05:30.773" Title="&quot;Cayley's theorem&quot; for Lie algebras?" Tags="&lt;abstract-algebra&gt;&lt;lie-algebra&gt;" AnswerCount="4" CommentCount="3" FavoriteCount="10" />
  <row Id="3032" PostTypeId="2" ParentId="3026" CreationDate="2010-08-22T15:22:49.937" Score="3" ViewCount="" Body="&lt;p&gt;The reason this works is as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$10 = 9 + 1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;modulo 3 this says that $10 = 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Numbers like $3157$ really mean $3 \cdot 10^3 + 1 \cdot 10^2 + 5 \cdot 10 + 7$, so applying the previous congruence we get $3 \cdot 10^3 + 1 \cdot 10^2 + 5 \cdot 10 + 7 = 3 + 1 + 5 + 7$ modulo 3.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Now that we have a strong understanding of why the theorem is true let head toward a detailed proof.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let us use the principle that all numbers can be &lt;em&gt;written down&lt;/em&gt; in base 10, for example the sequence of digits $(3,6,8,4)$ represents the number $4863$ (and a technicality, the empty sequence represents $0$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First let us define the interpretation from sequences to numbers as a recursive function:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \begin{align}&#xA;\text{interp}() &amp;amp;= 0 \\&#xA;\text{interp}(ds,d) &amp;amp;= d + 10 \cdot \text{interp}(ds)&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \begin{align}&#xA;\text{digitalroot}() &amp;amp;= 0 \\&#xA;\text{digitalroot}(ds,d) &amp;amp;= d + \text{digitalroot}(ds)&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can now prove the theorem (that the interpretation of a number is equal to its digital root) by induction on the length of the number:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Base case $\text{interp}() = 0 = \text{digitalroot}()$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Inductive case, assume that $\text{interp}(ds) = \text{digitalroot}(ds)$, we would like to now prove that $d + 10 \cdot \text{interp}(ds) = d + \text{digitalroot}(ds)$, cancelling d and reducing the 10 to a 1 we simply have the hypothesis, which is known to be true. That completes the proof.&lt;/p&gt;&#xA;" LastActivityDate="2010-08-22T15:22:49.937" />
  <row Id="3033" PostTypeId="2" ParentId="3026" CreationDate="2010-08-22T15:26:45.863" Score="1" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt;: It suffices to work $\rm\pmod 9. \;$ Put $\rm\; S_N := $ the sum of the decimal digits of $\rm N$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice $\rm\;N = 10\: n + r \;\Rightarrow\; S_N = S_n + r \;\Rightarrow\; N - S_N  = n - S_n + 9 \: n$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence $\rm\; N - S_N \equiv n - S_n \pmod 9$ -- the inductive step.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;SIMPLER&lt;/strong&gt;: For  $\rm N = a_k 10^k + \;\cdots + 10\; a_1 + a_0 \;$ put $\rm\; P(N) = a_k X^k + \;\cdots+ a_1 \; x + a_0$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;FACTOR THEOREM&lt;/strong&gt; $\rm\;\Rightarrow\; X-1|P(X)-P(1) \;\Rightarrow\; 9|P(10)-P(1) \;\:$ i.e. $\rm\; 9|N - S_N$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So &lt;strong&gt;casting nines&lt;/strong&gt; follows from the Factor Theorem purely by the &lt;strong&gt;polynomial form&lt;/strong&gt; of radix notation.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yet another example of the &lt;strong&gt;power of polynomials!&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-08-22T15:59:39.470" LastActivityDate="2010-08-22T15:59:39.470" />
  <row Id="3034" PostTypeId="1" AcceptedAnswerId="3050" CreationDate="2010-08-22T15:34:31.163" Score="2" ViewCount="72" Body="&lt;p&gt;One can see &lt;a href=&quot;http://books.google.com/books?id=rIyIlQwVCgkC&amp;amp;pg=PA92&amp;amp;lpg=PA92&amp;amp;dq=TBL+spacetimes&amp;amp;source=bl&amp;amp;ots=uu2rVqvCtn&amp;amp;sig=JqwYqPgtBEXK_SIwEOjZFxhNKzY&amp;amp;hl=en&amp;amp;ei=50FxTPz-Mo_mvQPI8NxB&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=3&amp;amp;ved=0CB8Q6AEwAg#v=onepage&amp;amp;q=TBL%20spacetimes&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;this reference&lt;/a&gt; for TBL space-times.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to know how the explicit expression for the function called $G$ in equations $3.108,3.108,3.110$ in the above reference is obtained. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also it would be nice to see some further references about TBL space-times. &lt;/p&gt;&#xA;" OwnerUserId="965" LastActivityDate="2010-08-22T19:58:38.613" Title="Tolman-Bondi-Lemaitre space times" Tags="&lt;differential-geometry&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" />
  <row Id="3035" PostTypeId="2" ParentId="3031" CreationDate="2010-08-22T15:45:16.143" Score="3" ViewCount="" Body="&lt;p&gt;A finite-dimensional real Lie algebra always arises as the Lie algebra of&#xA;a Lie group. Of course, the proof of this does use Ado's theorem, as well&#xA;as some Lie group theory (a subalgebra of a Lie algebra of a Lie group $G$&#xA;is also a Lie algebra of a Lie group, but not necessarily the Lie&#xA;algebra of a closed subgroup of $G$). The upshot is that over $\mathbb{R}$&#xA;the definition of (finite-dimensional) Lie algebra suffices to deal&#xA;with all Lie groups.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-08-22T15:45:16.143" CommentCount="2" />
  <row Id="3036" PostTypeId="2" ParentId="3031" CreationDate="2010-08-22T16:49:37.383" Score="1" ViewCount="" Body="&lt;p&gt;Classically, the &quot;concrete&quot; definition of what a Lie algebra is is simply a subspace of $\mathrm{End}(V)$, for some vector space $V$, which is closed under commutators. It is a theorem, then, that such a thing is the same thing as a pair $(\text{vector space},\text{antisymmetric bracket})$ satisfying the Jacobi identity. This theorem is quite important when one develops the theory that way, because it means that one can describe abstractly Lie algebras through &lt;em&gt;equations&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(This should be compared with the situation for &lt;a href=&quot;http://en.wikipedia.org/wiki/Jordan_algebra&quot; rel=&quot;nofollow&quot;&gt;Jordan algebras&lt;/a&gt;, as explained in Jacobson's amazing book on the subject: &lt;a href=&quot;http://en.wikipedia.org/wiki/Jordan_algebra#Special_Jordan_algebras&quot; rel=&quot;nofollow&quot;&gt;special Jordan algebras&lt;/a&gt;, those subspaces of $\mathrm{End}(V)$'s closed under the product $A\bullet B\dot=\tfrac12(AB+BA)$, are not characterizable through equations only; this is a consequence of the fact that the class of special Jordan algebras is not closed under quotients)&lt;/p&gt;&#xA;" OwnerUserId="274" LastEditorUserId="274" LastEditorDisplayName="" LastEditDate="2010-08-23T04:57:37.360" LastActivityDate="2010-08-23T04:57:37.360" CommentCount="4" />
  <row Id="3037" PostTypeId="2" ParentId="3008" CreationDate="2010-08-22T17:09:14.917" Score="2" ViewCount="" Body="&lt;p&gt;In Kunen's book on set theory, he leads one through an exercise where one creates a model of ZF (assuming ZF is consistent) in which the power set of the natural numbers is not well orderable.  Since one can prove that the power set of the naturals is in bijective correspondence with the reals, in this model, the real numbers cannot be well orderable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I don't have Kunen at home with me now, and it's been a year or two since I even glanced at the relevant section.  If I remember, I'll update the reference next time I have a copy of the book nearby).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, there can (I think, but I'm not positive) be models where choice fails and the reals &lt;em&gt;are&lt;/em&gt; well orderable.  Note that choice refers to ALL sets, so one simply needs to rig things to fail at REALLY huge cardinal sizes but have choice work at lower levels.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, there are models where choice functions are definable: Godel's universe $L$.  For example, in $L$, EVERY set is definable.  That is, given a set $X$, there is a first order formula (with parameters) $\phi$ such that $x\in X$ iff $\phi(x)$.  Now, a function from $X$ to $Y$ (including a choice function), is nothing but a special subset of $X\times Y$, and hence will be definable in $L$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I learned about this last paragraph from assorted posting by Joel David Hamkins over at MO.  For example, see &lt;a href=&quot;http://mathoverflow.net/questions/23393/set-theory-and-vl&quot; rel=&quot;nofollow&quot;&gt;http://mathoverflow.net/questions/23393/set-theory-and-vl&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;edit&lt;/em&gt; I realize that this doesn't seem to address your question directly.  However, it's true that for any set $X$, the powerset of $X$ has a choice function iff $X$ can be well ordered.  See Levy's book &quot;Basic Set Theory&quot; on page 160 for details.&lt;/p&gt;&#xA;" OwnerUserId="331" LastEditorUserId="331" LastEditorDisplayName="" LastEditDate="2010-08-22T17:54:24.283" LastActivityDate="2010-08-22T17:54:24.283" />
  <row Id="3038" PostTypeId="1" AcceptedAnswerId="3080" CreationDate="2010-08-22T17:24:14.270" Score="2" ViewCount="344" Body="&lt;p&gt;Suppose we define a topology on the real numbers as that obtained by defining open sets to be those such that the complement is finite. How does one show that this topology cannot be obtained from a distance function?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks  in advance. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;p.s: This is not a homework question. I am trying to self study point set topology. &lt;/p&gt;&#xA;" OwnerUserId="1127" LastActivityDate="2010-08-23T03:49:41.713" Title="A topology which cannot be obtained by a distance function" Tags="&lt;topology&gt;" AnswerCount="3" CommentCount="4" />
  <row Id="3039" PostTypeId="2" ParentId="3038" CreationDate="2010-08-22T17:25:26.670" Score="5" ViewCount="" Body="&lt;p&gt;Try showing that metric topologies are Hausdorff and that this topology isn't.&lt;/p&gt;&#xA;" OwnerUserId="331" LastActivityDate="2010-08-22T17:25:26.670" CommentCount="3" />
  <row Id="3040" PostTypeId="2" ParentId="2571" CreationDate="2010-08-22T17:25:51.473" Score="1" ViewCount="" Body="&lt;p&gt;This is true precisely because the rationals $\mathbb Q$ comprise a multiplicative &lt;strong&gt;subsemigroup&lt;/strong&gt; of the reals $\mathbb R,$&lt;br&gt;&#xA;i.e. the subset of rationals is closed under the multiplication operation of $\mathbb R$. Your statement arises by taking the contrapositive of this statement - which transfers it into an equivalent statement in the complement set $\mathbb R \backslash \mathbb Q$ of irrational reals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus $\rm\quad\quad\quad r_1,\ldots,r_n \in \mathbb Q \;\Rightarrow\; r_1 \cdots r_n \in \mathbb Q$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Contra+ $\rm\quad\; r_1 r_2\cdots r_n \not\in \mathbb Q \;\Rightarrow\; r_1\not\in \mathbb Q \;\:$ or $\rm\;\cdots\;$ or $\rm\;r_n\not\in\mathbb Q$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your case $\rm\;\;\; r^n\not\in \mathbb Q \;\Rightarrow\; r\not\in \mathbb Q \;$ is the special constant case $\rm r_i = r$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously the same is true if we replace $\rm\mathbb Q\subset \mathbb R$ by any subsemigroup chain $\rm G\subset H$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The contrapositive form is important in algebra since it characterizes &lt;strong&gt;prime&lt;/strong&gt; ideals in semigroups, rings, etc.&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-08-22T17:25:51.473" />
  <row Id="3042" PostTypeId="2" ParentId="3038" CreationDate="2010-08-22T18:46:07.260" Score="3" ViewCount="" Body="&lt;p&gt;Your topology is called the &lt;em&gt;finite complement topology&lt;/em&gt; of the real line: &lt;a href=&quot;http://en.wikipedia.org/wiki/Finite_complement_topology&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Finite_complement_topology&lt;/a&gt; . To show that it's not Hausdorff, try to intersect &lt;em&gt;any&lt;/em&gt; two open sets -you don't even need the points of the Hausdorff definition-, remembering the rule&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;(X \backslash A ) \cap (X \backslash B) = X \backslash (A \cup B)  \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Having any Topology standard book at hand, like Munkres, might help your self study. For instance, I'm quite shure that the proof that every metric space is Hausdorff can be found there.&lt;/p&gt;&#xA;" OwnerUserId="664" LastEditorUserId="664" LastEditorDisplayName="" LastEditDate="2010-08-23T03:49:41.713" LastActivityDate="2010-08-23T03:49:41.713" />
  <row Id="3043" PostTypeId="2" ParentId="3006" CreationDate="2010-08-22T18:54:27.917" Score="6" ViewCount="" Body="&lt;p&gt;(Looks like Niel and &quot;J&quot; beat me to the punch while I was composing this, but I'll post my answer anyway since I like my diagram. :) ) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can't get too much more intuitive than the Law of Exponents ($a^x a^c = a^{x+c}$) right?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To reduce symbolic (and mental) clutter for a while, let's write $k$ for $a^c$, to get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$k \; a^x = a^{x+c}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The point is that &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Multiplying the value of $a^x$ by &lt;em&gt;something&lt;/em&gt; yields the same result as adding &lt;em&gt;some other thing&lt;/em&gt; to the exponent.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Consider what this bit of algebra tells us about the geometry of the graph of the function $y=a^x$ (shown in red in the figure below). Recall some fundamental notions:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;$y=k a^x$ is the result of &lt;em&gt;vertically stretching&lt;/em&gt; (scaling) the original graph by a factor of $k$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$y = a^{x+c}$ is the result of &lt;em&gt;horizontally sliding&lt;/em&gt; (translating) the original graph by a (signed) distance of $-c$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The Law of Exponents tells us that (provided $k$ and $c$ are related appropriately), &lt;strong&gt;these transformed graphs are identical!&lt;/strong&gt; In the figure, the blue graph represents &lt;strong&gt;both&lt;/strong&gt; results.[*] Importantly, what points move where aren't the same under both actions; scaling moves the red point $P$ vertically onto the blue point $Q$; translating moves the red point $R$ horizontally onto $Q$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/SHrcE.jpg&quot; alt=&quot;Diagram of Exponential Functions&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, as suggested by the diagram, imagine a tangent vector poking out of each point of the original graph, and consider what the transformations to do those vectors:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Vertically stretching scales (only) the &quot;rise&quot; of the vector by factor $k$; that is, a vector with slope $m$ is pulled to achieve a slope of $m k$.&lt;/li&gt;&#xA;&lt;li&gt;Horizontally translating &lt;em&gt;has no effect&lt;/em&gt; on the vector's slope.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;What can we conclude here? Why, something pretty remarkable:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;For any point $P$ on the original graph, the point $R$ --located $c$ units to the right[**] of $P$-- is such that the slope of the tangent vector at $R$ is $k$-times the slope of the tangent vector at $P$ (where $k$ and $c$ are related appropriately).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Let me take this opportunity to retire $k$, since &lt;em&gt;it&lt;/em&gt; is beginning to become clutter; I'll just make the appropriate relationship explicit. Also, I'm going to retire the point name $R$, opting to describe the point instead. I'll summarize things this way:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;On the graph $y=a^x$, if the slope of tangent vector at any $P$ is $m$, then the slope of the tangent vector at the point $c$ units to the right of $P$ is $m\;a^c$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Notice that there's &lt;strong&gt;nothing special&lt;/strong&gt; about the players in this game. $P$ is &lt;em&gt;any&lt;/em&gt; point on the graph, $c$ is any (horizontal) distance you care to choose; heck, even the exact value of $a$ is up for grabs. Let's use this to our advantage.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose we take $P$ to be the point where the (original, red) graph crosses the $y$-axis; that is, we take $P$ to have $x$-coordinate $0$ (and $y$-coordinate $1$, but this doesn't really matter). Then, &quot;the point $c$ units to the right of $P$&quot; can be described more simply as &quot;the point with $x$-coordinate $c$&quot;, and we have&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;On the graph $y = a^x$, if the slope of the tangent vector at the point with $x$-coordinate $0$ is $m$, then the slope of the tangent vector at the point with $x$-coordinate $c$ is $m\;a^c$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In Calculusian prose:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If $f(x) = a^x$, then $f^{\prime}(c) = a^c f^{\prime}(0)$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Observe that we don't even need &quot;$c$&quot; in the above formula, since it's just taking the place of some $x$-coordinate. We can simply write:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If $f(x) = a^x$, then $f^{\prime}(x) = a^x f^{\prime}(0)$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;From here, it's pretty much a matter of definition to get to the final answer to your specific question. After all, the above holds for any (non-negative) value of $a$. Clearly, some values of $a$ correspond to graphs that cross the $y$-axis very steeply; some values correspond to graphs that cross the $y$-axis very shallowly; it's not un-reasonable to believe that there's a &lt;em&gt;convenient&lt;/em&gt; value would cause the graph to cross the $y$-axis &lt;em&gt;juuuuuuuuust right&lt;/em&gt; ... with a slope of $1$. Of course (as I mix my folklore), to name a demon is to control him, so we'll simply assign a symbol to this &quot;just right&quot; value of $a$.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Let &quot;e&quot; be the number such that the tangent vector to $y=e^x$ at $x=0$ has slope $1$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Assuming that such a number really does exist, you don't even have to know its exact value to conclude&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If $f(x) = e^x$, then $f^{\prime}(x) = e^x$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You can then turn your attention to figuring out why $e$ happens to have the value $2.718...$ . Other answers in this thread provide insights on how those arguments proceed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also left as an exercise is to determine why, in the formula for the derivative of $a^x$, the &quot;$f^{\prime}(0)$&quot; factor is in fact &quot;$\log a$&quot; (the natural logarithm of $a$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[*] As Niel mentions, this illustrates that the graph of an exponential function is &quot;self-similar&quot;. I think it's important to add &quot;uni-directionally&quot; to the description, in order to distinguish it from (conventional) similarity transformations where scaling occurs &quot;omni-directionally&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;[**] &quot;to the left&quot; works, too, with appropriate changes to the discussion.&lt;/p&gt;&#xA;" OwnerUserId="409" LastActivityDate="2010-08-22T18:54:27.917" CommentCount="1" />
  <row Id="3044" PostTypeId="2" ParentId="2384" CreationDate="2010-08-22T19:01:07.340" Score="0" ViewCount="" Body="&lt;p&gt;I'm working on www.BuzzMath.com, could fit your needs.&lt;/p&gt;&#xA;" LastActivityDate="2010-08-22T19:01:07.340" CommunityOwnedDate="2010-08-22T19:01:07.340" />
  <row Id="3047" PostTypeId="1" AcceptedAnswerId="3056" CreationDate="2010-08-22T19:18:38.943" Score="5" ViewCount="225" Body="&lt;p&gt;This is from Feller's &lt;em&gt;Introduction to Probability Theory and Its Applications&lt;/em&gt;. In the context of Bernoulli trials, we define:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$b(k;n,p) = \binom{n}{k}p^kq^{n-k},$$&#xA;$$P\{S_n \ge r\} = \sum_{v=0}^{\infty}b(r+v;n,p).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The latter being the probability of having at least $r$ successes. Now, supposing $r \gt np$ and knowing that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{b(k; n,p)}{b(k-1;n,p)}=\frac{(n-k+1)p}{kq}=1+\frac{(n+1)p-k}{kq},$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;show that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$P\{S_n \ge r\} \le b(r;n,p)\frac{rq}{r-np}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to Feller, it follows from the obvious fact that the terms of the series decrease faster than the terms of a geometric series with ratio $1-\frac{r-np}{rq}$. However, it's not obvious for me and I don't see how the upper bound follows.&lt;/p&gt;&#xA;" OwnerUserId="519" LastActivityDate="2010-08-22T23:11:32.360" Title="Binomial distribution and upper bound" Tags="&lt;probability&gt;" AnswerCount="1" />
  <row Id="3048" PostTypeId="1" AcceptedAnswerId="3052" CreationDate="2010-08-22T19:52:57.813" Score="4" ViewCount="289" Body="&lt;p&gt;What is the proper definition of a Countable Set?&lt;/p&gt;&#xA;" OwnerUserId="1414" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2010-11-25T17:00:41.440" LastActivityDate="2010-11-25T17:19:43.410" Title="Definition of a countable set" Tags="&lt;terminology&gt;&lt;discrete-mathematics&gt;&lt;elementary-set-theory&gt;" AnswerCount="3" CommentCount="5" FavoriteCount="0" />
  <row Id="3049" PostTypeId="1" AcceptedAnswerId="3057" CreationDate="2010-08-22T19:54:55.140" Score="2" ViewCount="134" Body="&lt;p&gt;I'm quite new at measure theory and all I know I read by myself, so bear with me if this question is too trivial. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given the definition of asymptotic, can we say that for $x \rightarrow \infty$ two almost everywhere equal functions are asymptotic? I really think that this is not the case, but I haven't been able to find a counterexample. Maybe, if in general this does not hold, would it be right for real functions?&lt;/p&gt;&#xA;" OwnerUserId="1415" LastActivityDate="2010-08-22T20:28:52.937" Title="Are almost everywhere equal functions asymptotic?" Tags="&lt;analysis&gt;&lt;measure-theory&gt;" AnswerCount="1" />
  <row Id="3050" PostTypeId="2" ParentId="3034" CreationDate="2010-08-22T19:58:38.613" Score="3" ViewCount="" Body="&lt;p&gt;It actually just comes from integration. Equation 3.106 from that book is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \dot R^2 = \left( - \frac{\partial R}{\partial t} \right)^2 = \frac FR + f $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;rearranging the terms gives (note that $\dot R &amp;lt; 0$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ dt = - \frac{dR}{\sqrt{\frac FR + f}} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now perform the substitution $z = fR/F$, giving&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ dt = -\frac F{f^{3/2}} \frac{dz}{\sqrt{1+\frac1z}}. $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Integrating gives&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ t - t_0 = \frac F{f^{3/2}} \left( \sinh^{-1}\sqrt z - \sqrt{z(z+1)}\right) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the rest is just algebra.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am no expert on general relativity, but the common term of that TBL spacetime seems start with &quot;&lt;a href=&quot;http://en.wikipedia.org/wiki/Lemaitre%E2%80%93Tolman_metric&quot; rel=&quot;nofollow&quot;&gt;Lemaitre-Tolman&lt;/a&gt;&quot;. There is a review article &lt;a href=&quot;http://arxiv.org/abs/0709.2044&quot; rel=&quot;nofollow&quot;&gt;on arXiv&lt;/a&gt;&lt;sup&gt;[1]&lt;/sup&gt; which might help.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;[1]: Kari Enqvist (2008). &lt;em&gt;Lemaitre–Tolman–Bondi model and accelerating expansion&lt;/em&gt;. General Relativity and Gravitation &lt;strong&gt;40&lt;/strong&gt;, 2–3, pp 451–466. DOI: 10.1007/s10714-007-0553-9.&lt;/sup&gt;&lt;/p&gt;&#xA;" OwnerUserId="171" LastActivityDate="2010-08-22T19:58:38.613" />
  <row Id="3051" PostTypeId="2" ParentId="3048" CreationDate="2010-08-22T20:00:55.863" Score="6" ViewCount="" Body="&lt;p&gt;A non-empty set $X$ is countable if and only if there exists a surjective function $f$ from $\mathbb{N}$ onto $X$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Countable_set&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Countable_set&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="622" LastActivityDate="2010-08-22T20:00:55.863" CommentCount="5" />
  <row Id="3052" PostTypeId="2" ParentId="3048" CreationDate="2010-08-22T20:04:28.443" Score="9" ViewCount="" Body="&lt;p&gt;A plain English definition from Kenneth Rosen's &quot;Discrete Mathematics and its Applications&quot;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A set that is either finite or has the same cardinality as the set of positive integers is called countable.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="91" LastActivityDate="2010-08-22T20:04:28.443" CommentCount="2" />
  <row Id="3053" PostTypeId="1" AcceptedAnswerId="3058" CreationDate="2010-08-22T20:16:05.537" Score="4" ViewCount="259" Body="&lt;p&gt;Let C&lt;sup&gt;0&lt;/sup&gt; be the set of all of all (edit: bounded) continuous functions from R-&gt;R with the sup norm.  Then a &quot;path&quot; in C&lt;sup&gt;0&lt;/sup&gt; is a continuous function f:[0,1]-&gt;C&lt;sup&gt;0&lt;/sup&gt;.  The question boils down to if there is path (or continuous deformation) which transforms any one bounded continuous function into another.&lt;/p&gt;&#xA;" OwnerUserId="1417" LastEditorUserId="1417" LastEditorDisplayName="" LastEditDate="2010-08-22T20:30:09.640" LastActivityDate="2010-08-23T08:32:23.463" Title="Is the set of continous functions from R-&gt;R path connected?" Tags="&lt;analysis&gt;" AnswerCount="3" CommentCount="4" FavoriteCount="2" />
  <row Id="3054" PostTypeId="2" ParentId="2981" CreationDate="2010-08-22T20:19:10.803" Score="1" ViewCount="" Body="&lt;p&gt;In a differential Galois field, integration is the antiderivative operation and not the infinite summation, but I will still call it integration.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we start out with simple elementary functions, say rational functions with integer coefficients, then there will be many functions that do not have integrals. If the original set of functions is finite, then only a finite number of functions need to be added.  The process can then be repeated at the next level.  The only catch is that at every stage one has to deal with constants (think of the roots of the denominators). Most functions force the addition of several integrals.  The number of functions added grows geometrically with the levels, so I would expect there is no easy way to define a countable closure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there initial sets for which the process stops? Yes, polynomials for example.  Are there initial sets for which it goes on forever?  Yes. &lt;/p&gt;&#xA;" OwnerUserId="1413" LastActivityDate="2010-08-22T20:19:10.803" CommentCount="4" />
  <row Id="3055" PostTypeId="2" ParentId="3006" CreationDate="2010-08-22T20:23:37.783" Score="0" ViewCount="" Body="&lt;p&gt;Here's my 2 cents, just look at the def of derivative&lt;/p&gt;&#xA;&#xA;&lt;p&gt;df/dx = lim f(x+d)-f(x)/d&lt;/p&gt;&#xA;&#xA;&lt;p&gt;now the defining property of e&lt;sup&gt;x&lt;/sup&gt; is that for very small x, e&lt;sup&gt;x&lt;/sup&gt; ~ 1 + x&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and since e&lt;sup&gt;x + d&lt;/sup&gt; = e&lt;sup&gt;x&lt;/sup&gt;e&lt;sup&gt;d&lt;/sup&gt; ~ e&lt;sup&gt;x&lt;/sup&gt;(1+d), it's clear that by plugging everything into the definition of derivative you will get the required result.&lt;/p&gt;&#xA;" OwnerUserId="1417" LastActivityDate="2010-08-22T20:23:37.783" CommentCount="1" />
  <row Id="3056" PostTypeId="2" ParentId="3047" CreationDate="2010-08-22T20:27:11.830" Score="4" ViewCount="" Body="&lt;p&gt;First notice that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $k &gt; r$ we have that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{(n-k+1)p}{kq} &amp;lt;\frac{(n-k+1)p}{rq}$ (as $n \ge k-1$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now $\frac{(n-k+1)p}{rq} \leq \frac{(n-(r+1)+1)p}{rq} = \frac{(n-r)p}{rq}$ as the numerator is the largest when $k = r+1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now $1 - \frac{(n-r)p}{rq} = \frac{rq - np + rp}{rq} = \frac{r -np}{rq}$ (as $p+q=1$). &lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-08-22T23:11:32.360" LastActivityDate="2010-08-22T23:11:32.360" CommentCount="2" />
  <row Id="3057" PostTypeId="2" ParentId="3049" CreationDate="2010-08-22T20:28:52.937" Score="5" ViewCount="" Body="&lt;p&gt;The short answer is no; I assume that you are working with the real numbers and the usual measure with the usual definition of asymptotic. The counterexample goes like this: any countable set has measure 0, and so choose some sequence x&lt;sub&gt;i&lt;/sub&gt; such that x&lt;sub&gt;i&lt;/sub&gt; goes to infinity (say, the positive integers). Then take any function f and define a new function g by g(x)=f(x) if x is not one of the x&lt;sub&gt;i&lt;/sub&gt; and g(x&lt;sub&gt;i&lt;/sub&gt;)=2f(x&lt;sub&gt;i&lt;/sub&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The two functions are equal almost everywhere since they differ only on the sequence x&lt;sub&gt;i&lt;/sub&gt;, which as a sequence is countable and thus have measure 0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then f(x)/g(x) equals 1 when x is not one of the x&lt;sub&gt;i&lt;/sub&gt; and equals 2 on the x&lt;sub&gt;i&lt;/sub&gt;, which means that the value as you go to infinity doesn't converge, i.e. it always oscillates between 1 and 2 as you go farther and farther out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even more awesomely, you can do the same trick not with the sequence x&lt;sub&gt;i&lt;/sub&gt;, but say, with the rational numbers, which are also countable and also have a measure of 0. Then the function given by h(x)=f(x)/g(x) is going to be continuous nowhere in the extended real line (i.e. no limit of h(x) as x goes to any number, or to positive or negative infinity will exist).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The long answer, which I lack time to think through and write up right now, would discuss why countable sets have measures 0 and the interplay between Borel measures, topology, and the equivalence classes of functions given by a.e. equality.&lt;/p&gt;&#xA;" OwnerUserId="400" LastActivityDate="2010-08-22T20:28:52.937" CommentCount="1" />
  <row Id="3058" PostTypeId="2" ParentId="3053" CreationDate="2010-08-22T20:39:20.857" Score="6" ViewCount="" Body="&lt;p&gt;It suffices to find a path to the 0 function from any other continuous function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $f:\mathbb{R}\rightarrow\mathbb{R}$ be any continuous function and define $F(x,t) = tf(x)$.  Assuming we can prove this is, in fact, continuous, note that $F(x,1) = f(x)$ and $F(x,0)$ is the $0$ function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, why is it continuous?  Let $\epsilon &gt; 0$.  Let $K = sup|f(x)|$  Choose $\delta &gt; 0 $ such that $\delta K &amp;lt; \epsilon$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, we have $|t_1 - t_2| &amp;lt; \delta$ implies $|t_1f(x) - t_2 f(x)| = |(t_1 - t_2)f(x)| \leq \delta K = \epsilon$, so that this map is continuous (in fact, uniformly continuous).&lt;/p&gt;&#xA;" OwnerUserId="331" LastActivityDate="2010-08-22T20:39:20.857" />
  <row Id="3059" PostTypeId="1" AcceptedAnswerId="3078" CreationDate="2010-08-22T20:51:17.727" Score="3" ViewCount="120" Body="&lt;p&gt;It is often said that differential forms (sections of an exterior power of the cotangent bundle) are the things that you can integrate. But unless I'm being thoroughly dense differential forms are not the only things that you can integrate, c.f. the arclength form (on a 2d manifold) $ds=\sqrt{dx^2+dy^2}$, the unsigned 1-d forms $|f(x,y)dx+g(x,y)dy|$, or the unsigned area forms $|h(x,y)dx\wedge dy|$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Where do the arclength form $ds=\sqrt{dx^2+dy^2}$, the unsigned 1-d forms |f(x,y)dx+g(x,y)dy|, and the unsigned area forms $|h(x,y)dx\wedge dy|$ live relative to the differentials $dx$ and $dy$, which I understand to live in the cotangent bundle of some 2-dimensional manifold?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="400" LastActivityDate="2010-08-22T22:45:55.973" Title="What is the theory of non-linear forms (as contrasted to the theory of differential forms)?" Tags="&lt;differential-geometry&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="2" />
  <row Id="3060" PostTypeId="1" AcceptedAnswerId="3076" CreationDate="2010-08-22T21:02:01.293" Score="6" ViewCount="235" Body="&lt;p&gt;How can we prove that it is posible to arrange numbers 1,2,3,4.........n in a row so that the average of any two of these numbers never appears between them?&lt;/p&gt;&#xA;" OwnerUserId="1414" LastActivityDate="2010-08-23T19:00:44.020" Title="Arrangement of Numbers" Tags="&lt;discrete-mathematics&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="3061" PostTypeId="1" CreationDate="2010-08-22T21:26:09.130" Score="5" ViewCount="233" Body="&lt;p&gt;I haven't studied properly the theory of infinities yet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $A_0$ denote the set of natural numbers. Let $A_{i+1}$ denote the set whose elements are all the subsets of $A_i$ for $i=0,...,n,...$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand well that the cardinality of $A_{i+1}$ is always greater than the cardinality of $A_i$ for all $i \in \mathbb{N}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which is the simplest argument which proves that there exists a set whose cardinality is greater than $A_i$ for all $i \in \mathbb{N}$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" LastEditorUserId="10557" LastEditorDisplayName="" LastEditDate="2011-05-15T06:02:28.700" LastActivityDate="2011-05-15T10:39:01.787" Title="large sets. set theory." Tags="&lt;set-theory&gt;" AnswerCount="2" />
  <row Id="3062" PostTypeId="2" ParentId="3061" CreationDate="2010-08-22T21:30:34.853" Score="7" ViewCount="" Body="&lt;p&gt;Let $X = \bigcup_{i=0}^\infty A_i$  Then $A_i\subseteq  X$ for all i so $|A_i|\leq |X|$ for all i.  Now, consider the powerset of $X$.  Then we have $|A_i|\leq |X|&amp;lt;|P(X)|$.&lt;/p&gt;&#xA;" OwnerUserId="331" LastActivityDate="2010-08-22T21:30:34.853" CommentCount="3" />
  <row Id="3065" PostTypeId="2" ParentId="3061" CreationDate="2010-08-22T21:36:58.887" Score="5" ViewCount="" Body="&lt;p&gt;Your question is closely linked to Beth numbers, their definition is:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$\beth_0 = \aleph_0 =$ the cardinality of the non-negative integers&lt;/li&gt;&#xA;&lt;li&gt;For a successor ordinal $\alpha +1$ put $\beth_{\alpha+1} =$ the cardinality of the power-set of $\beth_{\alpha}$&lt;/li&gt;&#xA;&lt;li&gt;For a limit ordinal $\delta$ put $\beth_\delta = \bigcup_{\alpha\lt\delta} \beth_\alpha$&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;What you're asking about is $\beth_\omega$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Further reading: Wikipedia page on &lt;a href=&quot;http://en.wikipedia.org/wiki/Beth_number&quot; rel=&quot;nofollow&quot;&gt;Beth number&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="622" LastEditorUserId="5363" LastEditorDisplayName="" LastEditDate="2011-05-15T10:39:01.787" LastActivityDate="2011-05-15T10:39:01.787" CommentCount="2" />
  <row Id="3069" PostTypeId="1" CreationDate="2010-08-22T21:53:17.017" Score="5" ViewCount="277" Body="&lt;p&gt;I would like to know whether the number of infinity sets of different cardinalities is uncountable or countable. Is the proof simple?&#xA;Thanks.&lt;/p&gt;&#xA;" LastActivityDate="2010-08-23T12:24:06.480" Title="number of infinite sets with different cardinalities" Tags="&lt;set-theory&gt;" AnswerCount="4" CommentCount="1" FavoriteCount="2" />
  <row Id="3070" PostTypeId="2" ParentId="3069" CreationDate="2010-08-22T21:59:23.127" Score="2" ViewCount="" Body="&lt;p&gt;If by the number of cardinalities you mean the set of all cardinalities - then there is none.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The class of all cardinal numbers is a proper class, as it contains a copy of the class of ordinal numbers - which is not a set.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So to answer you question, it's not countable. It's so big it's not even a set.&lt;/p&gt;&#xA;" OwnerUserId="622" LastActivityDate="2010-08-22T21:59:23.127" />
  <row Id="3071" PostTypeId="2" ParentId="3053" CreationDate="2010-08-22T22:00:02.610" Score="3" ViewCount="" Body="&lt;p&gt;Any real vector space $V$ with a &quot;reasonable&quot; topology is path connected. Given a vector $v_1\in V$ move along the line spanned by $v_1$ to $0$ and from there move to any other vector $v_2$ moving along le line spanned by it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reasonable means that for any $w\in V$ every open $U\ni w$ must contain an open segment around $v$ of the line spanned by $w$, which is certainly true for the case under consideration (topology under the sup norm).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I guess that $L^2({\Bbb R})$ is not reasonable, so maybe my terminology could be improved :-)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;*&lt;em&gt;Edit : *&lt;/em&gt; Of course $L^2({\Bbb R})$ IS reasonable!&#xA;$$&#xA;(\int_{\Bbb R}(f-tf)^2dx)^{1/2}=|1-t|(\int_{\Bbb R}f^2dx)^{1/2}&#xA;$$&#xA;gets as close as needed to 0 as $t\to1$.&lt;/p&gt;&#xA;" OwnerUserId="688" LastEditorUserId="688" LastEditorDisplayName="" LastEditDate="2010-08-22T23:45:40.043" LastActivityDate="2010-08-22T23:45:40.043" CommentCount="2" />
  <row Id="3072" PostTypeId="1" AcceptedAnswerId="3116" CreationDate="2010-08-22T22:00:52.627" Score="2" ViewCount="250" Body="&lt;p&gt;I'm making a game which you can see here, if you are on Windows or Linux: &lt;a href=&quot;http://insertnamehere.org/birdsofprey/&quot; rel=&quot;nofollow&quot;&gt;http://insertnamehere.org/birdsofprey/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you click and hold your mouse on a bird, you can see I'm just swinging the bird back and forth in pendulum motion. I would like to, instead, implement a more realistic motion, where the movement of your mouse affects the swinging of the bird like a pendulum with a moving pivot.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found &lt;a href=&quot;http://www.maplesoft.com/applications/view.aspx?SID=4888&amp;amp;view=html&quot; rel=&quot;nofollow&quot;&gt;a document&lt;/a&gt; on this topic but the equations rely on knowing the pivot's acceleration (X'' and Y''), which I do not; I am only repeatedly translating the bird graphic to the current mouse position.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have the bird's angle (-180 to 180 degrees), angular velocity and acceleration. I will need to alter these three variables each time the mouse is moved, so I will also have the last (x,y) and the new mouse (x,y).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this enough to make a good simulation of a pendulum with moving pivot?&lt;/p&gt;&#xA;" OwnerUserId="137" LastActivityDate="2010-08-23T14:49:51.430" Title="Pendulum with moving pivot" Tags="&lt;physics&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="3073" PostTypeId="2" ParentId="3069" CreationDate="2010-08-22T22:00:53.657" Score="4" ViewCount="" Body="&lt;p&gt;Technically, neither one is true: the class of cardinalities is too large to form a set.  (However, having to choose between &quot;countable&quot; and &quot;uncountable&quot;, certainly &quot;uncountable&quot; seems better: there is at least &lt;em&gt;an&lt;/em&gt; uncountable set of cardinalities of sets.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think this is equivalent to a previous question on this site, but anyway: please see Fact 20 of &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://math.uga.edu/~pete/settheorypart1.pdf&quot; rel=&quot;nofollow&quot;&gt;http://math.uga.edu/~pete/settheorypart1.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and let me know if you have any further questions.  &lt;/p&gt;&#xA;" OwnerUserId="299" LastActivityDate="2010-08-22T22:00:53.657" />
  <row Id="3074" PostTypeId="2" ParentId="3072" CreationDate="2010-08-22T22:07:54.817" Score="2" ViewCount="" Body="&lt;p&gt;The mouse velocity is [(current mouse position) - (previous mouse position)]/timestep (presumably you're using a constant timestep of 10ms or something?) and then the acceleration is similarly [(current mouse velocity) - (previous mouse velocity)]/timestep. (At least, approximately.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you have the acceleration, I guess you can use the formulas you already have.&lt;/p&gt;&#xA;" OwnerUserId="496" LastActivityDate="2010-08-22T22:07:54.817" />
  <row Id="3075" PostTypeId="2" ParentId="2949" CreationDate="2010-08-22T22:08:17.687" Score="21" ViewCount="" Body="&lt;p&gt;Brouwer's fixed point theorem, which has several non intuitive consequences in the real-world such as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fact that if you lay a piece of paper on your desk and trace around its outline, then crumble/wad the paper up and put it back inside the lines that there will always be a point on the paper exactly above where it started relative to the desk&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And, no matter how you stir your coffee there will always be some point in the liquid that ends in the same place that it was before mixing.&lt;/p&gt;&#xA;" OwnerUserId="1352" LastActivityDate="2010-08-22T22:08:17.687" CommunityOwnedDate="2010-08-22T22:08:17.687" CommentCount="7" />
  <row Id="3076" PostTypeId="2" ParentId="3060" CreationDate="2010-08-22T22:10:48.200" Score="7" ViewCount="" Body="&lt;p&gt;Assume the inductive hypothesis that it is true for lists smaller than n.  To reorder 1...n this way as well, split it into evens and odds, and apply the function floor((x+1)/2) to these sets to create two half-problems of the same type which by the hypothesis can be solved.  Now unapply the transformation to each half-solution using the function 2x to recover the evens and the function 2x-1 to recover the odds.  These functions are linear so the condition is still satisfied.  Now concatenate these two lists together to form the solution. This concatenation always works because the average of an even and an odd is not an integer.&lt;/p&gt;&#xA;" OwnerUserId="1284" LastEditorUserId="1284" LastEditorDisplayName="" LastEditDate="2010-08-23T19:00:44.020" LastActivityDate="2010-08-23T19:00:44.020" CommentCount="3" />
  <row Id="3077" PostTypeId="2" ParentId="2384" CreationDate="2010-08-22T22:22:54.347" Score="3" ViewCount="" Body="&lt;p&gt;If you haven't been using &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0471789011&quot; rel=&quot;nofollow&quot;&gt;&quot;The Art and Craft of Problem Solving&quot; by Zeitz&lt;/a&gt;, you should. Like, right at this moment. :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For neat problems... you might want to look into &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0883853256&quot; rel=&quot;nofollow&quot;&gt;&quot;Which Way Did The Bicycle Go?&quot; by Konhauser, Wagon and Velleman&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="498" LastEditorUserId="2264" LastEditorDisplayName="" LastEditDate="2010-11-16T09:32:51.477" LastActivityDate="2010-11-16T09:32:51.477" CommunityOwnedDate="2010-08-22T22:22:54.347" />
  <row Id="3078" PostTypeId="2" ParentId="3059" CreationDate="2010-08-22T22:45:55.973" Score="6" ViewCount="" Body="&lt;p&gt;The answer to &quot;what kinds of things can you integrate&quot; depends on the context. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;b&gt;&lt;em&gt;Measurable functions&lt;/em&gt;&lt;/b&gt; are things you can integrate over measure spaces, which includes in particular measurable subsets of R^n.  &lt;/li&gt;&#xA;&lt;li&gt;&lt;b&gt;&lt;em&gt;Differential forms&lt;/em&gt;&lt;/b&gt; are things you can integrate over oriented smooth manifolds -- the key thing about them is that their integrals are invariant under smooth, orientation-preserving changes of coordinates.&lt;/li&gt;&#xA;&lt;li&gt;&lt;b&gt;&lt;em&gt;Densities&lt;/em&gt;&lt;/b&gt; are things that can be integrated in a coordinate-independent way on any smooth manifold, regardless of whether it has an orientation or not.  &lt;/li&gt;&#xA;&lt;li&gt;Coming full circle, every Riemannian manifold (i.e., smooth manifold endowed with a Riemannian metric) has a naturally-defined density dV, so in that context you can integrate measurable functions again: the integral of the function f is defined to be the integral of the density f dV.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;All three of the expressions you asked about are examples of densities.  For details, see my book &lt;em&gt;Introduction to Smooth Manifolds&lt;/em&gt;, pp. 375-382.&lt;/p&gt;&#xA;" OwnerUserId="1421" LastActivityDate="2010-08-22T22:45:55.973" />
  <row Id="3079" PostTypeId="2" ParentId="3020" CreationDate="2010-08-23T00:02:57.457" Score="2" ViewCount="" Body="&lt;p&gt;No, because if there were a continuous inverse, then $\Phi$ would be bounded below, and from this it would follow that $W$ is complete.  But it isn't, because there are continuous functions that have Fourier series that are uniformly but not absolutely convergent, and hence Cauchy sequences (the partial sums of the Fourier series) in $W$ with no limit in $W$.  There are even examples of my claim from the last sentence in the disk algebra, i.e. functions in $C(\mathbb{T})$ whose negatively indexed Fourier coefficients vanish, as was first shown by Hardy and as I discovered when trying to answer a MathOverflow question &lt;a href=&quot;http://mathoverflow.net/questions/12968/mode-of-convergence-of-a-power-series&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-08-23T00:02:57.457" CommentCount="3" />
  <row Id="3080" PostTypeId="2" ParentId="3038" CreationDate="2010-08-23T00:07:55.107" Score="3" ViewCount="" Body="&lt;p&gt;As mentioned above, the topologies you want are those which are not metrizalbe. There are several metrication theorems that are given in terms of separation properties of the spaces. &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;A space is first countable if every point has a neighborhood basis&lt;/li&gt;&#xA;&lt;li&gt;A space is second countable if it has a countable basis&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;All metric spaces are first countable, so if your space is not first countable then it is not metrizable. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Probably the most commonly known theorem is Urysohn's metrization theorem: every second countable, regular, Hausdorff space is metrizable. Hausdorff just means that points can be separated by open sets and regular means that (nonempty) closed sets and points can be separated by open sets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A more powerful condition is that of normality. A space is normal if disjoint closed sets can be separated by open sets. It turns out that all metrizable space are normal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At first glance, these non-metrizable spaces just seem like pathological examples, but some are very important. For example, the Zariski topology is not metrizable, but is essential to the field of algebraic geometry.&lt;/p&gt;&#xA;" OwnerUserId="1423" LastActivityDate="2010-08-23T00:07:55.107" />
  <row Id="3081" PostTypeId="2" ParentId="83" CreationDate="2010-08-23T00:47:53.277" Score="1" ViewCount="" Body="&lt;p&gt;Personal experiences as an elementary school math resource teacher and a 1st grade teacher have led me to believe that mathematics learning should be engaging and fun at any level (including the graduate level), but clearly it is of the utmost importance to effectively engage students at an early age, so as to hopefully instill in them a lifelong appreciation for the subject.  Most children are quite active in nature, so providing hands-on activities that keep their minds and bodies actively involved is an important task. It seems that geometry is like the fail-proof mathematical concept that can pique the interest of even the toughest audiences.  Probably because geometry is naturally hands-on and therefore easily applicable to real-life situations.  Children can see and touch geometric concepts.  It may seem trivial to adults, but numbers and number systems can in reality be quite abstract concepts to wrap some little brains around.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another way to engage children is to give them a sense of ownership in the mathematics learning process, providing them with a keener awareness of mathematical applications, and thereby providing them with a more personal encounter with learning mathematics.  Children feel empowered when they are permitted to study things that are of interest to them.  Mathematical connections can be made to just about anything that children find curious in our present day society.  Be it the arts, dance, music, video games, computers, ipods, facebook, twitter, sports, or games, organic &quot;aha&quot; or &quot;teachable&quot; moments can easily be generated by harnessing children's intuitve curiosities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have found literature (specifically picture books) a useful tool for harnessing some of these curiosities in children of all ages. Additionally, actively listening to a narrative can engage the mind and promote purposeful thinking, learning, and even critical analysis.    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, another reason that I enjoy using literature to teach math is that literature tends to provide a portal for asking and generating thought provoking and meaningful questions.  One can kind of let the narrative guide the instruction.  If the kids aren't interested in it, there is no point in belaboring the issue, so move on until you find a piece of literature that does interest them.  Some books that I would recommend (based on student ratings) are &lt;a href=&quot;http://books.google.com/books?id=5BfjuWQBV6IC&amp;amp;printsec=frontcover&amp;amp;dq=the+very+greedy+triangle&amp;amp;source=bl&amp;amp;ots=VRZPvz19ql&amp;amp;sig=0-Yz0lhnPBJhwXQ_JF7Tp_w9dxU&amp;amp;hl=en&amp;amp;ei=irxxTMr-JYT6sAP96oTkDA&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=7&amp;amp;ved=0CD0Q6AEwBg#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;The Very Greedy Triangle&lt;/a&gt;, &lt;a href=&quot;http://books.google.com/books?id=MzOvangyz-EC&amp;amp;dq=the+number+devil&amp;amp;printsec=frontcover&amp;amp;source=bn&amp;amp;hl=en&amp;amp;ei=87xxTNuWN5OksQO2lfm_BA&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=11&amp;amp;sqi=2&amp;amp;ved=0CEIQ6AEwCg#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;The Number Devil&lt;/a&gt;, &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0698117530&quot; rel=&quot;nofollow&quot;&gt;Anno's Mysterious Multiplying Jar&lt;/a&gt;, &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0590300121&quot; rel=&quot;nofollow&quot;&gt;Amanda Bean's Amazing Dream&lt;/a&gt;, &lt;a href=&quot;http://books.google.com/books?id=SkU7BGazcusC&amp;amp;printsec=frontcover&amp;amp;dq=grapes+of+math&amp;amp;source=bl&amp;amp;ots=NL1kZAvPEg&amp;amp;sig=9P4JGvSSLVie2dHxKpJAKRl07Go&amp;amp;hl=en&amp;amp;ei=Q75xTNHVJIHWtQP8voTNCw&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=2&amp;amp;sqi=2&amp;amp;ved=0CCYQ6AEwAQ#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;The Grapes of Math&lt;/a&gt;, &lt;a href=&quot;http://books.google.com/books?id=Tncckz5fkEoC&amp;amp;dq=how+much+is+a+million&amp;amp;printsec=frontcover&amp;amp;source=bn&amp;amp;hl=en&amp;amp;ei=gb5xTN6aHIa4sAP_5YzFCw&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=4&amp;amp;sqi=2&amp;amp;ved=0CCoQ6AEwAw#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;How Much Is a Million&lt;/a&gt;, &lt;a href=&quot;http://www.curriculumconnection.net/sircumdragonpi.htm&quot; rel=&quot;nofollow&quot;&gt;Sir Cumference and the Dragon of Pi&lt;/a&gt;, &lt;a href=&quot;http://www.gregtang.com/mathstore.asp&quot; rel=&quot;nofollow&quot;&gt;The Best of Times&lt;/a&gt;, &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/059093998X&quot; rel=&quot;nofollow&quot;&gt;One Grain of Rice&lt;/a&gt;, &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0590944592&quot; rel=&quot;nofollow&quot;&gt;Spaghetti and Meatballs For All&lt;/a&gt;, and we can't forget how Lewis Carroll's &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0451527747&quot; rel=&quot;nofollow&quot;&gt;Alice in Wonderland&lt;/a&gt; is densely saturated with mathematical concepts...There are oodles more!  However, authors that I tend to come back to on a consistent basis are &lt;a href=&quot;http://www.gregtang.com/&quot; rel=&quot;nofollow&quot;&gt;Greg Tang&lt;/a&gt;, &lt;a href=&quot;http://www.jsworldwide.com/&quot; rel=&quot;nofollow&quot;&gt;Jon Scieszka&lt;/a&gt;, &lt;a href=&quot;http://www2.scholastic.com/browse/collection.jsp?id=492&quot; rel=&quot;nofollow&quot;&gt;Marilyn Burns&lt;/a&gt;, &lt;a href=&quot;http://cneuschwander.com/&quot; rel=&quot;nofollow&quot;&gt;Cindy Nueschwander&lt;/a&gt;, and &lt;a href=&quot;http://www.davidschwartz.com/&quot; rel=&quot;nofollow&quot;&gt;David M. Schwartz&lt;/a&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the end of the day, just try to find ways to make mathematics learning engaging and thought provoking but also meaningful and fun!  And by all means, one doesn't always have to use a mathematics picture book to teach math concepts.  Once I read, &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0590558498&quot; rel=&quot;nofollow&quot;&gt;It's Pumpkin Time&lt;/a&gt;, a book about the life cycle of pumpkins to my students and ended up creating measurement activities that corresponded with the book rather than science actitivies.  Just get creative! Children love that!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More &lt;a href=&quot;http://www.childrenspicturebooks.info/articles/picture_books_for_math.htm&quot; rel=&quot;nofollow&quot;&gt;resources&lt;/a&gt; if you're interested.&lt;/p&gt;&#xA;" OwnerUserId="219" LastEditorUserId="555" LastEditorDisplayName="" LastEditDate="2010-08-24T06:38:23.577" LastActivityDate="2010-08-24T06:38:23.577" CommunityOwnedDate="2010-08-23T00:47:53.277" CommentCount="1" />
  <row Id="3082" PostTypeId="2" ParentId="2384" CreationDate="2010-08-23T01:35:04.603" Score="0" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://www.keypress.com/x5488.xml&quot; rel=&quot;nofollow&quot;&gt;Problem Solving Strategies: Crossing the River with Dogs and Other Mathematical Adventures&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-08-23T01:35:04.603" CommunityOwnedDate="2010-08-23T01:35:04.603" />
  <row Id="3083" PostTypeId="2" ParentId="2384" CreationDate="2010-08-23T01:36:11.713" Score="0" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://arml.com/books.php&quot; rel=&quot;nofollow&quot;&gt;Books of old ARML contests&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-08-23T01:36:11.713" CommunityOwnedDate="2010-08-23T01:36:11.713" />
  <row Id="3084" PostTypeId="2" ParentId="2943" CreationDate="2010-08-23T03:22:03.437" Score="3" ViewCount="" Body="&lt;p&gt;Let $K$ be a complement to $N$ in $G$, which exists by Schur-Zassenhaus.  Then $|K|=|G|/|N|$.  Consider the subgroup $UN$.  Since $K(UN)=G$, $|K\cap UN|=|K|\cdot |UN|/|G|=|K|\cdot|U|\cdot|N|/|G|=|U|$, since $K$ is a complement to $N$.  Thus, $K\cap UN$ is a complement to $N$ in $UN$.  Since either $U$ or $N$ is solvable, all complements to $N$ in $UN$ are conjugate, so $K\cap UN=U^g$.  Thus, $K^{g^{-1}}$ contains $U$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Steve&lt;/p&gt;&#xA;" OwnerUserId="641" LastActivityDate="2010-08-23T03:22:03.437" CommentCount="4" />
  <row Id="3085" PostTypeId="1" CreationDate="2010-08-23T04:44:29.543" Score="5" ViewCount="1092" Body="&lt;p&gt;Is there a comprehensive resource listing binomial identities? I am more interested in combinatorial proofs of such identities, but even a list without proofs will do.&lt;/p&gt;&#xA;" OwnerUserId="813" LastActivityDate="2010-10-07T19:39:11.610" Title="A comprehensive list of binomial identities?" Tags="&lt;combinatorics&gt;&lt;discrete-mathematics&gt;&lt;binomial-coefficients&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="1" />
  <row Id="3086" PostTypeId="2" ParentId="3020" CreationDate="2010-08-23T04:53:18.473" Score="0" ViewCount="" Body="&lt;p&gt;No,no,no..&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Wiener algebra not normed by the sup-norm. &#xA;As I see it, it might be very confusing for other readers if we just guess here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Common notations are: &#xA;$W$ or $A(\mathbb{T})$ for the algebra of absolute convergent Fourier series normed by &#xA;$$\|f\|_{W}= \sum_{n=-\infty}^\infty |a_n| \quad(&amp;lt;\infty)$$ &#xA;where $f(t)=\sum_{n=-\infty}^\infty a_ne^{int}$ ($0\leq t\leq2\pi$).&#xA;That is $W=\ell¹(\mathbb{Z})$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$W^+$ or $A^+(\mathbb{T})$ for the algebra of absolute convergent Taylor series normed by &#xA;$$\|f\|_{W^+}= \sum_{n=0}^\infty |a_n| \quad(&amp;lt;\infty)$$ &#xA;where $f(z)=\sum_{n=0}^\infty a_nz^{n}$ ($|z|\leq1$). That is $W^+=\ell¹(\mathbb{N})$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some references.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;http://books.google.com/books?id=ZZk4AAAAIAAJ&amp;amp;pg=PA167&amp;amp;dq=Wiener+algebra+absolute+convergent+fourier&amp;amp;hl=en&amp;amp;ei=gfpxTL3LOovOswavkpS6Bg&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=5&amp;amp;ved=0CEAQ6AEwBA#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;http://books.google.com/books?id=ZZk4AAAAIAAJ&amp;amp;pg=PA167&amp;amp;dq=Wiener+algebra+absolute+convergent+fourier&amp;amp;hl=en&amp;amp;ei=gfpxTL3LOovOswavkpS6Bg&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=5&amp;amp;ved=0CEAQ6AEwBA#v=onepage&amp;amp;q&amp;amp;f=false&lt;/a&gt;&#xA;(page 169)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Wiener_algebra&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Wiener_algebra&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;http://mathworld.wolfram.com/WienerAlgebra.html&quot; rel=&quot;nofollow&quot;&gt;http://mathworld.wolfram.com/WienerAlgebra.html&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="1154" LastEditorUserId="1154" LastEditorDisplayName="" LastEditDate="2010-08-23T05:05:58.843" LastActivityDate="2010-08-23T05:05:58.843" CommentCount="3" />
  <row Id="3087" PostTypeId="1" CreationDate="2010-08-23T05:09:35.357" Score="6" ViewCount="529" Body="&lt;p&gt;&quot;&lt;em&gt;Life is open book.&lt;/em&gt;&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With the advent of widely accessible, inexpensive (or even free) computational tools and Computer Algebra Systems (TI-89, Wolfram|Alpha, etc.), much of what traditionally comprises a high school math curriculum can now easily be done by almost everyone.  Factoring polynomials, solving inequalities, graphing linear equations, differentiation and integration -- these are the types of skills high school math students spend most of their time learning, and yet all of it can be done for free by anyone with a web browser.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What does this mean for the high school math curriculum?  On the one hand, we could leave it more-or-less the same, insisting that today's student learn what we learned decades ago, while banning or carefully regulating the use of these new tools.  On the other hand, we could embrace the tools and the opportunities they create to spend more math class time on different topics and skills, perhaps focusing more on analytic and synthetic problem solving and less on mechanical symbolic manipulation -- but at the risk of students never learning some basic foundations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So how about it?  Binomial coefficients?  The angle-addition formulas for trig functions?  The conditions under which a function has an inverse?  Basic computer programming?  Keeping in mind that the vast majority of high school students do &lt;em&gt;not&lt;/em&gt; go on to become professional mathematicians, what should the high school math curriculum consist of?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Btw, I post this question (inspired by &lt;a href=&quot;http://www.theodoregray.com/BrainRot/&quot; rel=&quot;nofollow&quot;&gt;this discussion&lt;/a&gt;) here because this is a community of thoughtful mathematicians.  I recognize this discussion may belong in a different forum, but I don't what/where that forum is.  Any suggestions are welcome.&lt;/p&gt;&#xA;" OwnerUserId="506" LastActivityDate="2010-10-31T17:30:22.860" CommunityOwnedDate="2010-08-23T11:37:46.497" Title="What should the high school math curriculum consist of?" Tags="&lt;big-list&gt;&lt;education&gt;&lt;teaching&gt;&lt;high-school&gt;" AnswerCount="6" CommentCount="5" FavoriteCount="2" />
  <row Id="3088" PostTypeId="2" ParentId="3053" CreationDate="2010-08-23T05:28:43.003" Score="2" ViewCount="" Body="&lt;p&gt;I think it's worth to develop a little bit more Akhil's interesting remark.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;1.&lt;/strong&gt; If instead of ${\cal C}^0(\mathbb{R}, \mathbb{R})$, you take ${\cal C}^0([a,b], \mathbb{R})$, where $[a,b] \subset \mathbb{R}$ is a compact interval, then something interesting happens: you can give ${\cal C}^0([a,b], \mathbb{R})$ the &lt;em&gt;compact-open topology&lt;/em&gt;, &lt;a href=&quot;http://en.wikipedia.org/wiki/Compact-open_topology&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Compact-open_topology&lt;/a&gt; , and this topology coincides with the one induced by the sup norm (also called the topology of &lt;em&gt;uniform convergence&lt;/em&gt;, &lt;a href=&quot;http://en.wikipedia.org/wiki/Uniform_convergence&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Uniform_convergence&lt;/a&gt; ). More generally, the same is true for ${\cal C}^0 (X , Y)$, where $X$ is any compact topological space and $Y$ any metric space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;2.&lt;/strong&gt; Now, for psychological reasons, let's use the notation $Y^X$ instead of ${\cal C}^0 (X , Y) $, and let $X$, $Y$ be any pair of topological spaces. Put the compact-open topology on $Y^X$ and let $Z$ be a third topological space. To every map $F: X\times Z \longrightarrow Y$ you can associate the map $\Phi (F) : Z \longrightarrow Y^X$ defined by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;(\Phi (F) (z)) (x)= F(x,z) \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reciprocally, to every map $\gamma :  Z \longrightarrow Y^X$ you can associate the map $\Psi (\gamma ): X \times Z \longrightarrow Y$ defined by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\Psi (\gamma) (x,z) = (\gamma (z)) (x) \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can check easily that $\Phi$ and $\Psi$ are inverses one each other. Moreover, if $F$ is continuous, so is $\Phi (F)$. If $X$ is a locally compact Hausdorff space, the same is true for $\Psi$: $\gamma$ continuous implies $\Psi (\gamma )$ continuous. (See Munkres' &quot;Topology&quot;, theorem 46.11.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, when $X$ is a locally compact Hausdorff space, you have bijections, inverses one each other:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\Phi : Y^{X \times Z} \longrightarrow (Y^X)^Z  \qquad \text{and} \qquad \Psi: (Y^X)^Z \longrightarrow Y^{X \times Z}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, take $Z = I$, the unit interval. This bijection tells us that, when $X$ is a locally compact Hausdorff space, &lt;em&gt;paths&lt;/em&gt; $\gamma : I \longrightarrow Y^X$ and &lt;em&gt;homotopies&lt;/em&gt; $F : X \times I \longrightarrow Y$ are the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;3.&lt;/strong&gt; An easy topological exercise: any map $f: X \longrightarrow Y$ is homotopic to a constant map and all the constant maps are homotopic, either if&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$X$ is contractile and $Y$ is path-connected, or&lt;/li&gt;&#xA;&lt;li&gt;$Y$ is contractile.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;So, coming back to your situation, we could have said that ${\cal C}^0 (\mathbb{R}, \mathbb{R})$, with the compact-open topology, is always path-connected because $X = \mathbb{R}$ is locally compact, Hausdorff and contractile, and $Y= \mathbb{R}$ is path-connected (or because $X=\mathbb{R}$ is locally compact and Hausdorff and $Y= \mathbb{R}$ is contractile).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Notice that Jason has used the second possibility: he has constructed a &lt;em&gt;homotopy&lt;/em&gt; from the constant function $0$ to $f$. We could use the first one: $F(x,t) = f(xt)$ is a homotopy from the constant function $f(0)$ to $f$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, Jason's solution is more straightforward and doesn't need all this elementary machinery of point-set topology. But now we can say that also other function spaces, with the compact-open topology, are path-connected: every time you take as $X$ a locally compact Hausdorff space and any of the two previous possibilities, ${\cal C}^0(X,Y)$ is path-connected. For instance,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;{\cal C}^0(\mathbb{R}, Y) ,\ {\cal C}^0([a,b],Y) , \ {\cal C}^0(\mathbb{R}^n,Y) ,\ {\cal C}^0(D^2,Y)  \dots&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for any path-connected space $Y$, are path-connected. Also&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;{\cal C}^0(X,\mathbb{R}) , \ {\cal C}^0(X, [a,b]) , \ {\cal C}^0(X,\mathbb{R}^n) ,\ {\cal C}^0(X,D^2)  \dots&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for any locally compact Hausdoff space $X$, are path-connected.&lt;/p&gt;&#xA;" OwnerUserId="664" LastEditorUserId="664" LastEditorDisplayName="" LastEditDate="2010-08-23T08:32:23.463" LastActivityDate="2010-08-23T08:32:23.463" />
  <row Id="3089" PostTypeId="2" ParentId="2949" CreationDate="2010-08-23T06:45:49.170" Score="5" ViewCount="" Body="&lt;p&gt;Gold's theorem provides pretty convincing mathematical evidence supporting the universal grammar hypothesis in linguistics. This hypothesis is two-fold: (1) children are not presented logically with enough information to actually learn their native language; (2) hence there exists a &lt;em&gt;universal grammar&lt;/em&gt; which is encoded somehow in the human brain and which facilitates the logical gap between the positive data given to the child and the data necessary to determine the language's grammar. While the universal grammar hypothesis isn't universally accepted, it has been one of the most important ideas in linguistics so far.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gold's theorem shows that certain classes of languages are logically not learnable. Of course, it operates in a purely formal setting. I'll provide up this setting now following the definitions and notations of &lt;a href=&quot;http://www.math.harvard.edu/theses/senior/carroll/gabriel_carroll.pdf&quot; rel=&quot;nofollow&quot;&gt;Gabriel Carroll&lt;/a&gt;, pg. 41.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Start with a finite alphabet $\Sigma$ and let $\Sigma^*$ designate the set of finite sequences of elements of $\Sigma$. A language $L$ is a subset of $\Sigma^*$. A text of $L$ is an infinite string $w_1, w_2, \dots$ of elements of $L$ such that every element of $L$ occurs at least once. A learner for a class $\mathcal{L}$ of languages is a function $\Lambda : (\Sigma^*)^* \rightarrow \mathcal{L}$ that intuitively takes a sequence of strings of $\Sigma$ and guesses the language in $\mathcal{L}$ in which all these strings are grammatically correct. The learner $\Lambda$ learns the language $L \in \mathcal{L}$ if for every text $w_1,w_2,\dots$ of $L$ there exists a natural number $N$ such that $\Lambda(w_1,w_2,\dots,w_n) = L$ for $n \geq N$. The learner $\Lambda$ learns the class $\mathcal{L}$ if it learns each language in $\mathcal{L}$, and the class $\mathcal{L}$ is learnable if there exists a learner which learns it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is Gold's theorem, first proved by Gold in his seminal &lt;a href=&quot;http://www3.isrl.illinois.edu/~junwang4/langev/localcopy/pdf/gold67limit.pdf&quot; rel=&quot;nofollow&quot;&gt;paper&lt;/a&gt; (but my wording is taken from Carroll):&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If the class $\mathcal{L}$ contains all finite languages and at least one infinite language, then $\mathcal{L}$ is not learnable.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In particular, any finite language is regular. Hence the class of regular languages is unlearnable, and it follows at once that every class of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Chomsky_hierarchy&quot; rel=&quot;nofollow&quot;&gt;Chomsky Hierarchy&lt;/a&gt; is unlearnable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The proof of Gold's theorem is, as Carroll shows, not very hard, although certainly not intuitive, and it can be reduced to a corollary of the following characterization of learnable classes of languages (Carroll, Lemma 9):&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;A countable class $\mathcal{L}$ of nonempty languages is learnable if and only if, for each $L \in \mathcal{L}$, there is a finite ''telltale'' subset $T \subseteq L$ such that $L$ is minimal in $\{L' \in \mathcal{L} : T \subseteq L'\}$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="251" LastEditorUserId="251" LastEditorDisplayName="" LastEditDate="2010-08-23T16:51:24.807" LastActivityDate="2010-08-23T16:51:24.807" CommunityOwnedDate="2010-08-23T06:45:49.170" />
  <row Id="3090" PostTypeId="2" ParentId="370" CreationDate="2010-08-23T06:51:29.907" Score="0" ViewCount="" Body="&lt;p&gt;(This is really a comment to &lt;a href=&quot;http://math.stackexchange.com/questions/370/good-book-lecture-notes-about-category-theory/2997#2997&quot;&gt;mathphysicist's answer&lt;/a&gt;, but I don't have enough rep to post comments yet.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Goldblatt's book (which is very beautifully written, by the way) is available online in its entirety &lt;a href=&quot;http://dlxs2.library.cornell.edu/m/math/browse/author/g.php&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2010-08-23T06:51:29.907" CommunityOwnedDate="2010-08-23T06:51:29.907" />
  <row Id="3091" PostTypeId="2" ParentId="2949" CreationDate="2010-08-23T07:20:57.630" Score="2" ViewCount="" Body="&lt;p&gt;I got really struck by duality, when my professor lectured about it the first time. I think that even though the algebraic concept is easy to understand, to think that there exists a space such that all inclusions are switched always had a special place in my mind.&lt;/p&gt;&#xA;" OwnerUserId="1415" LastActivityDate="2010-08-23T07:20:57.630" CommunityOwnedDate="2010-08-23T07:20:57.630" CommentCount="4" />
  <row Id="3092" PostTypeId="2" ParentId="3069" CreationDate="2010-08-23T07:40:15.533" Score="1" ViewCount="" Body="&lt;p&gt;Contrary to the responses thus far (and with no disrespect intended to their authors), I think we can say quite unequivocally that there are &lt;strong&gt;&lt;em&gt;strictly more&lt;/em&gt;&lt;/strong&gt; than countably-many sets with distinct infinite cardinalities, and that (ipso facto) there are uncountably many such sets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answers before mine fixate on the idea of cardinality itself as the concept of quantity, and correctly point out that that there isn't a &quot;set of infinite cardinalities&quot;. (There is a &lt;em&gt;class&lt;/em&gt; of infinite cardinalities, if you like that sort of thing, but perhaps not everyone likes talking about proper classes.) Because there cannot be a set of infinite cardinalities, the reasoning goes, there cannot be a cardinality (a well-defined sense of &quot;the number of&quot;) of infinite cardinalities. However, the original question isn't about &quot;cardinality&quot; &lt;em&gt;per se&lt;/em&gt;: it's about a less formal notion of &quot;the number of something&quot;, which we conventionally define by cardinality in the case of sets. As we are not talking about &quot;the set of X&quot;, however, perhaps we shouldn't be using this definition of number. What then?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, the original question isn't asking us &lt;em&gt;precisely&lt;/em&gt; how many infinite cardinalities there are: just whether this number is &lt;strong&gt;countable&lt;/strong&gt; or &lt;strong&gt;uncountable&lt;/strong&gt; --- a dichotomy of ordering; whether it is less-than-or-equal-to the &quot;number&quot; of natural numbers (&lt;em&gt;i.e.&lt;/em&gt; whether it can be placed in bijective correspondance with a subset of the natural numbers), or not. This is, as far as I'm aware, the definition of the terms &quot;countable&quot; and &quot;uncountable&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the number of infinite cardinalities cannot be put into one-to-one correspondence with any subset of the natural numbers, it follows that the number of infinities is uncountable. The fact that they cannot be put into a set is irrelevant, in my opinion: at worst, one only needs to identify a single uncountable set U of infinities, note that there is no surjection onto it from the natural numbers, and note incidentally that there are more infinities outside of the set U anyway. Any cardinal &amp;alefsym;&lt;sub&gt;&lt;em&gt;C&lt;/em&gt;&lt;/sub&gt; (identified in the usual way with an ordinal), for &lt;em&gt;C&lt;/em&gt; itself an uncountable ordinal, will suffice for U.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, there are uncountably many different infinities, QED.&lt;/p&gt;&#xA;" OwnerUserId="439" LastEditorUserId="439" LastEditorDisplayName="" LastEditDate="2010-08-23T07:49:51.910" LastActivityDate="2010-08-23T07:49:51.910" CommentCount="3" />
  <row Id="3093" PostTypeId="1" CreationDate="2010-08-23T07:47:44.270" Score="6" ViewCount="332" Body="&lt;p&gt;The Banach--Tarski theorem applies only in the case of three or more dimensions. In 3D, there are five regular solids, two of them being not at all obvious, and the 4D case is also interesting; but the higher-dimensional cases each yield just three &quot;solids&quot;, only one of which isn't obvious. In dynamics, particles tend to coalesce in one or two dimensions, while in four or more dimensions they tend to disperse drearily; only in 3D do they move freely but with significant local interaction. And the Poincare conjecture proved to be much harder in the 3D case than in the others.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the question has three parts: What other examples are there of 3D richness? Are there any underlying reasons for it? And are there fields where richness begins at a higher, but still small, number of dimensions? I would be particularly interested to learn of any theorems that hold in just a particular number (not 0, 1, or 2) of dimensions. &lt;/p&gt;&#xA;" OwnerUserId="875" LastEditorUserId="367" LastEditorDisplayName="" LastEditDate="2010-08-23T10:10:47.050" LastActivityDate="2010-08-23T23:23:27.653" CommunityOwnedDate="2010-08-23T18:37:37.637" Title="Why is the 3D case so rich?" Tags="&lt;big-list&gt;" AnswerCount="8" CommentCount="4" FavoriteCount="3" />
  <row Id="3094" PostTypeId="2" ParentId="3024" CreationDate="2010-08-23T08:02:52.710" Score="1" ViewCount="" Body="&lt;p&gt;The full solution to this problem (with a slight difference in notation) is given in appendix A: of the following  &lt;a href=&quot;http://arxiv.org/PS_cache/gr-qc/pdf/9910/9910109v2.pdf&quot; rel=&quot;nofollow&quot;&gt;article:&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1430" LastActivityDate="2010-08-23T08:02:52.710" />
  <row Id="3095" PostTypeId="2" ParentId="3087" CreationDate="2010-08-23T08:15:36.647" Score="2" ViewCount="" Body="&lt;p&gt;I would include more probability, such as how to conduct a simple probabilistic experiment using statistical tests. Once you understand that, you can do a lot by simply looking up various tests. Without understanding probability, people are easily manipulated using statistics or make poor choices.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would include definitely include more use of technology, including Wolfram Alpha and basic mathematical programming. These tools allow people to achieve much more.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some basic mathematical philosophy would be good as well. Game theory can be very thought provoking and is quite simple as well. We could look at other areas even though we wouldn't be able to examine them rigorously. I think students should understand that maths is based on axioms, that not all maths problems can be solved (assuming ZFC is consistent) and that many problems can't be solved efficiently (P vs NP).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would focus more on exploring various mathematical puzzles or unexpected results: &lt;a href=&quot;http://math.stackexchange.com/questions/234/card-doubling-paradox&quot;&gt;Card doubling paradox&lt;/a&gt;, &lt;a href=&quot;http://www.relisoft.com/science/hats.html&quot; rel=&quot;nofollow&quot;&gt;3 hats puzzle&lt;/a&gt;, &lt;a href=&quot;http://en.wikipedia.org/wiki/Monty_Hall_problem&quot; rel=&quot;nofollow&quot;&gt;Monty hall problem&lt;/a&gt; and other results like this.&lt;/p&gt;&#xA;" OwnerUserId="123" LastEditorUserId="123" LastEditorDisplayName="" LastEditDate="2010-08-23T22:43:15.997" LastActivityDate="2010-08-23T22:43:15.997" CommunityOwnedDate="2010-08-23T18:46:03.097" CommentCount="3" />
  <row Id="3096" PostTypeId="2" ParentId="3093" CreationDate="2010-08-23T08:32:07.923" Score="6" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Exotic_R4&quot; rel=&quot;nofollow&quot;&gt;Exotic $\mathbb R^4$&lt;/a&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are infinitely many non-diffemorphic smooth structures on the topological space $\mathbb R^n$ if and only if $n=4$. (Otherwise there is only one diffeomorphism class.)&lt;/p&gt;&#xA;" OwnerUserId="367" LastActivityDate="2010-08-23T08:32:07.923" CommunityOwnedDate="2010-08-23T18:37:37.637" CommentCount="1" />
  <row Id="3097" PostTypeId="1" AcceptedAnswerId="3100" CreationDate="2010-08-23T09:18:23.443" Score="5" ViewCount="253" Body="&lt;p&gt;I am aware that on a finite set the number of equivalence relations is the n-th &lt;a href=&quot;http://en.wikipedia.org/wiki/Bell_number&quot; rel=&quot;nofollow&quot;&gt;Bell's Number&lt;/a&gt;. On the other hand, the only reference I could find on the web for infinite sets was this: &lt;a href=&quot;http://www.informaworld.com/smpp/content~db=all~content=a750850336&quot; rel=&quot;nofollow&quot;&gt;On counting equivalence relations&lt;/a&gt; by D.J. Baylis, in which he proves that the set of equivalence relations on $\mathbb{N}$ is uncountable. I am interested in any sort of theorem that gives some rule about counting equivalence relations on an infinite set -something like $&amp;#x2136;_0 = \aleph_0$ and &amp;#x2136;&lt;sub&gt;n&lt;/sub&gt; = 2&lt;sup&gt;&amp;#x2136;&lt;sub&gt;n-1&lt;/sub&gt;&lt;/sup&gt; - if there are any.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: Mathjax does not seem to be working on the second formula with beth, feel free to edit and correct it.&lt;/p&gt;&#xA;" OwnerUserId="1415" LastEditorUserId="1415" LastEditorDisplayName="" LastEditDate="2010-08-23T10:47:25.220" LastActivityDate="2010-08-23T11:33:51.977" Title="Counting equivalence relations" Tags="&lt;set-theory&gt;" AnswerCount="3" />
  <row Id="3098" PostTypeId="2" ParentId="3097" CreationDate="2010-08-23T10:06:25.573" Score="0" ViewCount="" Body="&lt;p&gt;Let $X$ be an infinite set. Any function $f : X \to X$ determines an equivalence relation on $X$:&#xA;$$&#xA;x_1 \equiv x_2 \iff f(x_1) = f(x_2).&#xA;$$&#xA;It follows that there are at least $2^{|X|}$ equivalence relations on $X$ (see the comments below). On the other hand, any equivalence relation defines a function from $X$ to $X,$ take a complete set $S$ of representatives, send every element of a given class to its representative in $S.$ Thus there are no more than $|X|^{|X|}$ equivalence relations on $X.$ Therefore there are exactly&#xA;$|X|^{|X|}=2^{|X|}$ equivalence relations on $X.$ &lt;/p&gt;&#xA;" OwnerUserId="1100" LastEditorUserId="1100" LastEditorDisplayName="" LastEditDate="2010-08-23T11:33:51.977" LastActivityDate="2010-08-23T11:33:51.977" CommentCount="7" />
  <row Id="3099" PostTypeId="2" ParentId="3097" CreationDate="2010-08-23T10:16:57.570" Score="0" ViewCount="" Body="&lt;p&gt;I will assume that the axiom of choice holds: this may not be necessary, but I am not sure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Start by observing that equivalence relations on a set $S$ are the &quot;same&quot; as partitions of $S$ (recall that a partition of $S$ is a set $B$ of subsets of $S$ with the properties that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) $\cup_{b \in B}b = S$, and &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) for all $b,b'\in B$ we have $b \cap b'=\emptyset$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $S$ be an infinite set, let $B(S)$ denote the set of partitions of $S$ and $P(S)$ the power set of $S$ (namely, the set of all subsets of $S$). I am going to show that $|B(S)| = |P(S)|$; note that $|P(S)| = 2^{|S|}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First of all, use the axiom of choice to fix, for every non-empty subset $A$ of $S$, an element $c(A) \in A$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To any function $f \colon S \to S$ we associate a partition $B_f$ of $S$ by letting $B_f = \{ f^{-1}(s) | s \in f(S) \}$; in words, $B_f$ is the partition whose elements are the non-empty fibers of the function $f$.  Any partition $B$ is obtained by this construction: we define a function $f_B \colon S \to S$ by letting $f_B(s) = c(b_s)$, where $b_s$ is the part of $B$ containing $s$. Clearly, the partition $B_{f_B}$ coincides with the partition $B$. We deduce that there is a surjection from the set of all functions $S \to S$ to the set of partitions; thus we conclude that $|B(S)| \leq |S^S| = |P(S)| = |2^S|$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the other inequality, let $P'(S)$ denote the set of unordered pairs of subsets of $S$ of the form $\{ A , S \setminus A \}$, where $A \subset S$ is a proper non-empty subset of $S$. Each unordered pair in $P'(S)$ determines uniquely a partition of $S$ with two parts, so that there is an injection $P'(S) \to B(S)$. We deduce that $|B(S)| \geq |P'(S)|$. To conclude it suffices to show that $|P'(S)| \geq |P(S)|$. This is easy, since there is a natural surjection $P(S) \setminus \{ S , \{\emptyset\} \} \to P'(S)$ whose fibers consist of exactly two elements, and $P(S)$ is an infinite set.&lt;/p&gt;&#xA;" OwnerUserId="1028" LastActivityDate="2010-08-23T10:16:57.570" />
  <row Id="3100" PostTypeId="2" ParentId="3097" CreationDate="2010-08-23T10:17:50.270" Score="3" ViewCount="" Body="&lt;p&gt;Firstly, to address Beth numbers (as you noted, the math markup doesn't work with them. I'll use words instead)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Beth numbers are cardinal numbers defined as followed:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Beth null = Aleph null&lt;/li&gt;&#xA;&lt;li&gt;Beth alpha+1 = Cardinality of the power set of Beth alpha&lt;/li&gt;&#xA;&lt;li&gt;Beth delta, for a limit ordinal delta = the union of all smaller Beth numbers.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Now, given an infinite set $X$ and assuming the axiom of choice, we have that $|X\times X|=|X|$, therefore $|P(X\times X)| = |P(X)|$.&#xA;Define $R(X)$ as the set of all equivalence relations on $X$, then clearly $R(X) \subset P(X\times X)$ and therefore $|R(X)| \le |P(X)|$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand there are $2^{|X|}$ partitions of $X$ into two sets (an easy exercise in cardinal arithmetic), so for every partition as such $\{X', X\backslash X'\}$ we can define an equivalence relation with two equivalence classes, namely $X'$ and its complement in $X$. Therefore $|R(X)| \ge |P(X)|$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All in all we have that the number of equivalence relations on $X$ is $2^{|X|}$.&lt;/p&gt;&#xA;" OwnerUserId="622" LastEditorUserId="622" LastEditorDisplayName="" LastEditDate="2010-08-23T10:29:54.373" LastActivityDate="2010-08-23T10:29:54.373" CommentCount="3" />
  <row Id="3101" PostTypeId="1" AcceptedAnswerId="3549" CreationDate="2010-08-23T10:20:26.163" Score="5" ViewCount="441" Body="&lt;p&gt;Hi! I'm studying for math exam and one of the questions that often appears is related to derivative of a product of two functions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The theorem says that $(f(x)g(x))'=f'(x)g(x)+f(x)g'(x)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The proof goes like this:&#xA;$f(x+h)g(x+h)-f(x)g(x)=(f(x+h)-f(x))g(x)+(g(x+h)-g(x))f(x)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After that we divide the equation by h and let h approach 0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now what I don't understand is how they got the right side of the proof.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Same problem with quotient:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Theorem says:&#xA;$\left(\frac{f(x)} {g(x)}\right)'=\frac{f'(x)g(x)-f(x)g'(x)} {g^2(x)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The above comes from&#xA;$\frac {f(x+h)} {g(x+h)} - \frac {f(x)} {g(x)} = \frac{(f(x+h)-f(x))g(x)-(g(x+6h)-g(x))f(x)} {g(x+h)g(x)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can see from where $f(x+h)g(x)$ comes, but I can't see from where $f(x)g(x)$ came.&lt;/p&gt;&#xA;" OwnerUserId="348" LastActivityDate="2010-08-28T14:38:22.457" Title="Derivative of a product and derivative of quotient of functions theorem: I don't understand its proof." Tags="&lt;calculus&gt;" AnswerCount="9" CommentCount="4" FavoriteCount="2" />
  <row Id="3103" PostTypeId="2" ParentId="3101" CreationDate="2010-08-23T10:45:10.060" Score="2" ViewCount="" Body="&lt;p&gt;I'll let someone else take care of proving the product formula; the quotient formula follows from applying the product formula and the chain rule.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, remember that you can express $\frac{f(x)}{g(x)}$ as $f(x)\cdot\left(\frac1{g(x)}\right)$. You can now apply the product formula as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{\mathrm{d}}{\mathrm{d}x}\frac{f(x)}{g(x)}=\left(\frac{\mathrm{d}}{\mathrm{d}x} f(x)\right)\left(\frac1{g(x)}\right)+f(x)\left(\frac{\mathrm{d}}{\mathrm{d}x}\frac1{g(x)}\right)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The left addend is easily simplified:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{\mathrm{d}}{\mathrm{d}x}\frac{f(x)}{g(x)}=\frac{f^{\prime}(x)}{g(x)}+f(x)\left(\frac{\mathrm{d}}{\mathrm{d}x}\frac1{g(x)}\right)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the right addend requires the chain rule:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{\mathrm{d}}{\mathrm{d}x}\frac{f(x)}{g(x)}=\frac{f^{\prime}(x)}{g(x)}+f(x)\left(-\frac1{g(x)^2}\right)\left(\frac{\mathrm{d}}{\mathrm{d}x}g(x)\right)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and the formula follows from there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you will be doing the proof using limits, just apply the steps you took in proving the product rule, with the terms being $f(x)$ and $\frac1{g(x)}$&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-08-23T10:45:10.060" />
  <row Id="3104" PostTypeId="2" ParentId="3093" CreationDate="2010-08-23T10:57:03.550" Score="1" ViewCount="" Body="&lt;p&gt;I think it's to do something with that in dimension 4 and above, you can have 2 commuting rotations about different axes (e.g. rotating around the xy-plane or zw-plane), so that things can be 'broken up' into separate pieces. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Related is the fact that you need 3 parameters to describe a rotation in 3-space, but 6 in 4-space.&lt;/p&gt;&#xA;" OwnerUserId="360" LastActivityDate="2010-08-23T10:57:03.550" CommunityOwnedDate="2010-08-23T18:37:37.637" CommentCount="2" />
  <row Id="3105" PostTypeId="2" ParentId="3101" CreationDate="2010-08-23T11:09:04.903" Score="7" ViewCount="" Body="&lt;p&gt;There is a more comprehensible (I hope) proof coming from thinking of derivative as a linear approximation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the definition of derivative&lt;sup&gt;1&lt;/sup&gt;, $f(x)=f(x_0)+f'(x_0)(x-x_0)+o(x-x_0)$ and $g(x)=g(x_0)+g'(x_0)(x-x_0)+o(x-x_0)$. Multiplying the equalities we get&#xA;\begin{multline}&#xA;f(x)g(x)=f(x_0)g(x_0)+(f'(x_0)g(x_0)+f(x_0)g'(x_0))(x-x_0)+\\&#xA;+f'(x_0)g'(x_0)(x-x_0)^2+o(x-x_0)&#xA;\end{multline}&#xA;and since $f'(x_0)g'(x_0)(x-x_0)^2=o(x-x_0)$, we can rewrite it as \[(fg)(x)=(fg)(x_0)+(f'g+fg')(x_0)\cdot(x-x_0)+o(x-x_o)\] — which exactly means that (fg)'=f'g+fg'.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;1&lt;/sup&gt; $\lim_{x\to x_0}g(x)=a$ iff $g(x)=a+o(1)$, hence $\lim_{x\to x_0}\frac{f(x)-f(x_0)}{x-x_0}=f'(x_0)$ iff $\frac{f(x)-f(x_0)}{x-x_0}=f'(x_0)+o(1)$ which can be rewritten as $f(x)-f(x_0)=f'(x_0)(x-x_0)+o(x-x_0)$ (see e.g. &lt;a href=&quot;http://en.wikipedia.org/wiki/Little-o_notation&quot; rel=&quot;nofollow&quot;&gt;wikipedia&lt;/a&gt; if &quot;o(1)&quot; notation is unfamiliar to you).&lt;/p&gt;&#xA;" OwnerUserId="152" LastEditorUserId="123" LastEditorDisplayName="" LastEditDate="2010-08-23T22:48:30.890" LastActivityDate="2010-08-23T22:48:30.890" CommentCount="5" />
  <row Id="3106" PostTypeId="2" ParentId="3031" CreationDate="2010-08-23T11:28:53.687" Score="3" ViewCount="" Body="&lt;p&gt;Due to the Peter-Weyl theorem every finite dimensional Lie group is isomorphic to a subgroup of the&#xA;orthogonal group O(m) for some m. Therefore its Lie algebra is a subalgebra of the &#xA;Lie algebra of the orthogonal group which is the same as the Lie algebra of the spin group Spin(m).&#xA;Now, the Lie algebra of the spin group is the bivector subalgebra of the clifford algebra Cl(m).&#xA;Thus every finite dimensional Lie algebra is isomorphic to a Lie subalgebra of the bivector subalgebra of a clifford algebra.&#xA;This point of view seems analogous to the Cayley's theorem for the following reasons:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;There exists a projective representation of the symmetric group in the Clifford algebra:&#xA;(a,b) --&gt; e_a - e_b. Thus the Clifford algebra may be seen as a kind of quantization of the symmetric group.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;This construction generalizes at least to the case of affine Kac-Moody algebras which have similar&#xA;realizations in the infinite dimensional Clifford algebra.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="1430" LastActivityDate="2010-08-23T11:28:53.687" />
  <row Id="3107" PostTypeId="2" ParentId="3101" CreationDate="2010-08-23T11:49:38.350" Score="1" ViewCount="" Body="&lt;p&gt;(Adding and subtracting the same amount, like $f(x)g(x)$, is very common in mathematics. Think of what you do when you solve a linear equation - add+subtract multiply+divide.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a geometrical proof of the product rule. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Denote by $T_f$ the tangent of $f(x)$ at $x_0$. It is given by $T_f(x)=f'(x_0)(x-x_0) + f(x_0)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Denote by $T_g$ the tangent of $g(x)$ at $x_0$. It is given by $T_g(x)=g'(x_0)(x-x_0) + g(x_0)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The slope of $f(x)g(x)$ at $x_0$ is $(fg)'(x_0)$ - which we want to find. That slope is the same as the slope of the product of the tangents at $x_0$: that is $(fg)'(x_0)=(T_fT_g)'(x_0)$. Multiplying through and differentiate we get&#xA;$$(T_fT_g)'(x) = ((f'(x_0)(x-x_0) + f(x_0))\cdot(g'(x_0)(x-x_0) + g(x_0)))'=$$&#xA;$$=2(x-x_0)f'(x_0)g'(x_0)+ f'(x_0)g(x_0) + f(x_0)g'(x_0)$$&#xA;so &#xA;$$(fg)'(x_0)=(T_fT_g)'(x_0) =f'(x_0)g(x_0) + f(x_0)g'(x_0).$$&lt;/p&gt;&#xA;" OwnerUserId="1154" LastEditorUserId="1154" LastEditorDisplayName="" LastEditDate="2010-08-23T14:18:22.180" LastActivityDate="2010-08-23T14:18:22.180" />
  <row Id="3108" PostTypeId="2" ParentId="3093" CreationDate="2010-08-23T11:59:55.523" Score="6" ViewCount="" Body="&lt;p&gt;This is a well-known phenomenon among topologists, and although I'm not an expert, I'll give one standard answer: 3- and 4-dimensional topology are very different from topology in 5 or more dimensions because &lt;a href=&quot;http://en.wikipedia.org/wiki/Surgery_theory&quot; rel=&quot;nofollow&quot;&gt;surgery theory&lt;/a&gt; works in 5 or more dimensions.  In 3 and 4 dimensions one does not have enough &quot;wiggle room&quot; for surgery theory to be effective and this is responsible for some anomalous behavior.  This leads to some remarkable phenomena connecting 3-manifold topology to other branches of mathematics, some of which are listed in the &lt;a href=&quot;http://en.wikipedia.org/wiki/3-manifold&quot; rel=&quot;nofollow&quot;&gt;Wikipedia article&lt;/a&gt;, and it makes the 4-dimensional case rather special as well.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-08-23T11:59:55.523" CommunityOwnedDate="2010-08-23T18:37:37.637" />
  <row Id="3109" PostTypeId="2" ParentId="3031" CreationDate="2010-08-23T12:05:30.773" Score="9" ViewCount="" Body="&lt;p&gt;Here is a way of avoiding Ado's theorem, at the expense of using the Poincare-Birkhoff-Witt Theorem. The PBW theorem has no finite dimensionality or characteristic hypotheses, so you may like this better. Note, however, that I will realize finite dimensional Lie algebras as endomrophisms of infinite dimensional vector spaces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's define a concrete Lie algebra to be a vector space $V$, and a vector subspace $\mathfrak{h}$ of $\mathrm{End}(V)$ closed under commutator. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Theorem:&lt;/strong&gt; Every Lie algebra is isomorphic to a concrete Lie algebra.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof:&lt;/strong&gt; Let $\mathfrak{g}$ be a Lie algebra and $U$ its universal enveloping algebra. The Lie algebra $\mathfrak{g}$ acts on $U$ by left multiplication, so this gives a map $\mathfrak{g} \to U$ taking bracket to commutator. We must prove this map is injective. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Choose a basis ${ v_i }$ for $\mathfrak{g}$. Suppose that left multiplication by $\sum a_i v_i$ is $0$. Then $\left( \sum a_i v_i \right) \cdot 1 = \sum a_i v_i$ would be zero in $U$. But, by the PBW theorem, the $v_i$ are linearly independent in $U$, a contradiction. QED&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as I know, this special case of PBW is as hard as the whole theorem. &lt;/p&gt;&#xA;" OwnerUserId="448" LastActivityDate="2010-08-23T12:05:30.773" CommentCount="1" />
  <row Id="3110" PostTypeId="2" ParentId="3069" CreationDate="2010-08-23T12:24:06.480" Score="-1" ViewCount="" Body="&lt;p&gt;For any set X, the set P(X) of all subsets of X has a bigger cardinality than X itself (for X is finite this is easy, for X infinite you need a clever argument from Cantor, obtaineble in any set theory text). You can do this again, obtaining:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;card(X) &amp;lt; card(P(X)) &amp;lt; card(P(P(X))) &amp;lt; ...&lt;/p&gt;&#xA;" LastActivityDate="2010-08-23T12:24:06.480" CommentCount="1" />
  <row Id="3111" PostTypeId="1" CreationDate="2010-08-23T12:24:28.010" Score="-2" ViewCount="345" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Possible Duplicate:&lt;/strong&gt;&lt;br&gt;&#xA;  &lt;a href=&quot;http://math.stackexchange.com/questions/2887/question-about-solving-absolute-values&quot;&gt;Question about solving absolute values&lt;/a&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;p&gt;I know the method of solving the equation like this $|2x+1|=|3x+9|$ but the problem is if the same equation would be like this $| 2x+1 | = x | 3x+9 |$, how can I solve this?&lt;/p&gt;&#xA;" OwnerUserId="1050" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-08-23T13:10:22.613" LastActivityDate="2010-08-23T14:25:39.810" ClosedDate="2010-08-23T14:34:36.357" Title="Solving equations with multiple absolute values" Tags="&lt;algebra-precalculus&gt;" AnswerCount="2" CommentCount="8" />
  <row Id="3112" PostTypeId="2" ParentId="3111" CreationDate="2010-08-23T12:37:49.043" Score="2" ViewCount="" Body="&lt;p&gt;To solve $|f(x)|=g(x)|h(x)|$ in general, find the sets $P=\{x:f(x)g(x)\ge 0\}$ and $N=\{x:f(x)g(x)\lt 0\}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then solve the two equations&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$f(x)=g(x)h(x)$ - but discard solutions that are not in $P$.&lt;/li&gt;&#xA;&lt;li&gt;$f(x)=-g(x)h(x)$ - but discard solutions that are not in $N$.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;You will be left with the solutions of $|f(x)|=g(x)|h(x)|$.&lt;/p&gt;&#xA;" OwnerUserId="1310" LastActivityDate="2010-08-23T12:37:49.043" />
  <row Id="3113" PostTypeId="2" ParentId="3093" CreationDate="2010-08-23T12:39:36.560" Score="1" ViewCount="" Body="&lt;p&gt;The following is just a toy example and of course not as deep as the examples already mentioned:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One-dimensional space is special in the sense that $\mathbb R^1\setminus\{0\}$ is not connected---it has two connected components. In all other dimension $n\neq 1$, the space $\mathbb R^n\setminus\{0\}$ is connected.&lt;/p&gt;&#xA;" OwnerUserId="367" LastActivityDate="2010-08-23T12:39:36.560" CommunityOwnedDate="2010-08-23T18:37:37.637" CommentCount="4" />
  <row Id="3114" PostTypeId="2" ParentId="3093" CreationDate="2010-08-23T12:51:14.180" Score="1" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Hurwitz%27s_theorem_%28normed_division_algebras%29&quot; rel=&quot;nofollow&quot;&gt;Hurwitz's theorem&lt;/a&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only normed division algebras over the real numbers appear in dimensions 1, 2, 4 and 8. &lt;/p&gt;&#xA;" OwnerUserId="367" LastActivityDate="2010-08-23T12:51:14.180" CommunityOwnedDate="2010-08-23T18:37:37.637" />
  <row Id="3115" PostTypeId="2" ParentId="3093" CreationDate="2010-08-23T13:07:57.140" Score="2" ViewCount="" Body="&lt;p&gt;One example of $n \geq 2$-D richness is simply the fact that $n$-by-$n$ matrices don't commute.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is trivial, but it means for example that every nonabelian finite group has an irreducible representation of degree $&gt;1$.  &lt;/p&gt;&#xA;" OwnerUserId="536" LastEditorUserId="536" LastEditorDisplayName="" LastEditDate="2010-08-23T23:23:27.653" LastActivityDate="2010-08-23T23:23:27.653" CommunityOwnedDate="2010-08-23T18:37:37.637" CommentCount="2" />
  <row Id="3116" PostTypeId="2" ParentId="3072" CreationDate="2010-08-23T13:37:21.200" Score="1" ViewCount="" Body="&lt;p&gt;The actual solution may not be what you want, since it will come with perks like the birds swinging almost a full $-180^\circ$ to $180^\circ$ everytime when pulled hard and then immediately held still. Also you will have to solve a differential equation in real time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;May I offer an alternative instead which will make it look real while being easy to implement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since you have a simple oscillation, I assume you simply vary the angle as $\alpha(t)=r\cos(2\pi f t)+3\pi/2$ for some range $r$ and frequency $f$ in your code.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To add a touch of reality, you can vary the 'down' angle based on the horizontal velocity of the mouse. If the velocity is $v$, then calculate $\beta(v)=3\pi/2-\tan^{-1}(Av)$. This will give some pseudo-direction for the adjusted direction of pull.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also decrease the range of the oscillation as the velocity increases: $r(v)=1/(1+(Bv)^2)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally use the following function to get the angle: $\alpha(t)=r(v)\cos(2\pi f t)+\beta(v)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You will have to choose fractions $A$ and $B$ till you hit upon the right motion that pleases you. $A=B=0$ will give your original motion back. Larger $A$ will make the change in angle more sensitive to speed, and larger $B$ will dampen the oscillations more at a given velocity.&lt;/p&gt;&#xA;" OwnerUserId="1310" LastActivityDate="2010-08-23T13:37:21.200" />
  <row Id="3117" PostTypeId="2" ParentId="3111" CreationDate="2010-08-23T13:38:13.210" Score="2" ViewCount="" Body="&lt;p&gt;Both equations you mention can be solved in the same way.  As others have said in the comments on your question here and the comments and answers on &lt;a href=&quot;http://math.stackexchange.com/questions/2887/question-about-solving-absolute-values&quot;&gt;this question&lt;/a&gt;, $|ax-b|=\begin{cases}&#xA;ax-b &amp;amp; \text{ if } x\ge \frac{b}{a} \\ &#xA;b-ax &amp;amp; \text{ if } x&amp;lt; \frac{b}{a} &#xA;\end{cases}$, where I'll call $\frac{b}{a}$ a &quot;turning point&quot; value.  So, for any equation an absolute value, there is at least one turning point value of the variable for each absolute value.  In both of your examples, the turning point values are $-\frac{1}{2}$ and $-3$.  The $n$ (2 in your examples) turning point values divide the number line into $n+1$ (3 in your examples) regions where, within each region, each absolute value can be replaced by either the expression inside the absolute value or its opposite.  Each resulting equation can then be solved, restricting solutions to the corresponding region on the number line.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's consider a different example: $x|4x-1|=|3x-2|$.  The turning points are $\frac{1}{4}$ and $\frac{2}{3}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{matrix}&#xA;\leftarrow  &amp;amp; \frac{1}{4} &amp;amp; \text{---} &amp;amp; \frac{2}{3} &amp;amp; \rightarrow&#xA;\\&#xA;x(1-4x)=(2-3x) &amp;amp; | &amp;amp; x(4x-1)=(2-3x) &amp;amp; | &amp;amp; x(4x-1)=(3x-2)&#xA;\\&#xA;x-4x^2=2-3x &amp;amp; | &amp;amp; 4x^2-x=2-3x &amp;amp; | &amp;amp; 4x^2-x=3x-2&#xA;\\&#xA;0=4x^2-4x+2 &amp;amp; | &amp;amp; 4x^2+2x-2=0 &amp;amp; | &amp;amp; 4x^2-4x+2=0&#xA;\\&#xA;\text{no }x\in\mathbb{R} &amp;amp; | &amp;amp; x=-1\text{ or }x=\frac{1}{2} &amp;amp; | &amp;amp; \text{no }x\in\mathbb{R}&#xA;\end{matrix}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(&lt;em&gt;edit&lt;/em&gt;: Note that between any two adjacent regions, the only change in the equations at the tops of the regions is the one term with turning point at the boundary between the two regions.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the solutions for each region-specific equation apply only to that region, so the solutions -1 and $\frac{1}{2}$ in the middle region only apply to the interval $[\frac{1}{4},\frac{2}{3}]$, so -1 is not a solution (it's not in the interval) but $x=\frac{1}{2}$ is a solution (and it's the only solution to $x|4x-1|=|3x-2|$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regardless of the algebraic method of solution, graphing both sides of the equation and looking for points of intersection is a good way to check your solution.&lt;/p&gt;&#xA;" OwnerUserId="72" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-08-23T14:25:39.810" LastActivityDate="2010-08-23T14:25:39.810" />
  <row Id="3118" PostTypeId="1" AcceptedAnswerId="3121" CreationDate="2010-08-23T13:44:55.477" Score="5" ViewCount="650" Body="&lt;p&gt;Can anyone help me out here? Can't seem to find the right rules of divisibility to show this.&lt;/p&gt;&#xA;" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2010-08-23T14:00:09.563" LastActivityDate="2010-12-03T07:14:13.777" Title="If a | m and (a + 1) | m, prove a(a + 1) | m." Tags="&lt;number-theory&gt;" AnswerCount="4" CommentCount="3" FavoriteCount="2" />
  <row Id="3119" PostTypeId="1" AcceptedAnswerId="3302" CreationDate="2010-08-23T13:49:57.680" Score="5" ViewCount="243" Body="&lt;p&gt;In a recent discussion, someone told me tuples in the modern meaning (in particular,  tuples are heterogeneous: that is, different elements of a tuple can belong to different sets/have different types) first appeared in Codd's &lt;a href=&quot;http://en.wikipedia.org/wiki/Tuple_relational_calculus&quot; rel=&quot;nofollow&quot;&gt;tuple calculus&lt;/a&gt;. I was surprised it would be so late, but &lt;a href=&quot;http://www.google.ru/search?tbs=bks:1,bkv%3aa,cdr:1,cd_min:03.01.0001,cd_max:31.12.1969&amp;amp;tbo=p&amp;amp;q=tuple&amp;amp;num=10&quot; rel=&quot;nofollow&quot;&gt;searching Google Books&lt;/a&gt; before 1970, I can't see any clearly heterogeneous examples, and quite a few clearly homogeneous ones (&quot;tuple of ones and zeros&quot;, &quot;tuple of natural numbers&quot;, etc.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Сan anybody confirm that Codd introduced heterogeneous tuples or point out an earlier appearance?&lt;/p&gt;&#xA;" OwnerUserId="1437" LastActivityDate="2010-08-25T19:10:08.443" Title="When did the term &quot;tuple&quot; get its current meaning?" Tags="&lt;terminology&gt;&lt;math-history&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" />
  <row Id="3120" PostTypeId="1" AcceptedAnswerId="3127" CreationDate="2010-08-23T13:51:32.327" Score="3" ViewCount="508" Body="&lt;p&gt;Given:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A = (0,0)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;B = (0,-10)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;AB = AC&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the angle between AB and AC, &lt;strong&gt;how are the coordinates at C calculated?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="1439" LastEditorUserId="1439" LastEditorDisplayName="" LastEditDate="2010-08-23T14:36:08.180" LastActivityDate="2010-08-23T23:02:04.647" Title="Find the coordinates in an isosceles triangle" Tags="&lt;geometry&gt;&lt;trigonometry&gt;&lt;euclidean-geometry&gt;" AnswerCount="3" CommentCount="7" />
  <row Id="3121" PostTypeId="2" ParentId="3118" CreationDate="2010-08-23T13:54:44.253" Score="7" ViewCount="" Body="&lt;p&gt;It is not surprising that you are finding this difficult, because it goes beyond basic divisibility rules -- it rather requires something which is essentially equivalent to the uniqueness of prime factorization.  [&lt;b&gt;Edit&lt;/b&gt;: Actually this is comment is incorrect -- as Robin Chapman's answer shows, it &lt;em&gt;is&lt;/em&gt; possible to prove this using just divisibility rules.  In particular it is true in any integral domain.]  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I assume $a$ and $m$ are positive integers.  The first observation is that $a$ and $a+1$ are relatively prime: i.e., there is no integer $d &gt; 1$ -- or equivalently, no prime number-- which divides both $a$ and $a+1$, for then $d$ would have to divide $(a+1) - a = 1$, so $d = 1$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the key step: since $a$ divides $m$, we may write $m = aM$ for some positive integer $M$.  So $a+1$ divides $aM$ and is relatively prime to $a$.  &lt;strong&gt;I claim&lt;/strong&gt; that this implies $a+1$ divides $M$.  Assuming this, we have $M = (a+1)N$, say, so altogether &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$m = aM = a(a+1)N$, so $a(a+1)$ divides $m$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The claim is a special case of:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Generalized) Euclid's Lemma: Let $a,b,c$ be positive integers.  Suppose $a$ divides $bc$ and $a$ is relatively prime to $b$.  Then $a$ divides $c$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A formal proof of this requires some work!  See for instance&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Euclid%27s_lemma&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Euclid's_lemma&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, proving this is essentialy as hard as proving the &lt;a href=&quot;http://en.wikipedia.org/wiki/Fundamental_theorem_of_arithmetic&quot; rel=&quot;nofollow&quot;&gt;fundamental theorem of arithmetic&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="299" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2010-08-23T18:06:24.263" LastActivityDate="2010-08-23T18:06:24.263" CommentCount="7" />
  <row Id="3122" PostTypeId="1" AcceptedAnswerId="3124" CreationDate="2010-08-23T14:06:32.077" Score="2" ViewCount="286" Body="&lt;p&gt;Some days ago as i had asked as to how to test the Riemann Integrability of the function. Now i was recently given this question about proving that the given function is Riemann Integrable. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/BqmUO.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can i show that this function is Riemann Integrable or not in the interval $[0,1]$. I tried using partitions, but it didn't work. I here don't want to use the Riemann Lebesgue lemma as i want to understand the methodology behind selecting the partitions.&lt;/p&gt;&#xA;" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-08-24T00:24:52.060" LastActivityDate="2010-09-12T03:10:37.317" Title="Showing $\frac{1}{x}-\left[\frac{1}{x}\right]$ is Riemann Integrable" Tags="&lt;real-analysis&gt;&lt;analysis&gt;" AnswerCount="3" CommentCount="7" FavoriteCount="1" />
  <row Id="3123" PostTypeId="2" ParentId="3122" CreationDate="2010-08-23T14:19:35.577" Score="0" ViewCount="" Body="&lt;p&gt;To show that the function in Riemann-integrable you have to show that it is bounded and has at most a countable set of discontinuities in the interval you are integrating. But this is the case of an improper integral, since $1/x$ is not defined in $x = 0$, you have to follow the criterions for improper integrals, which is the existence of the limit $\lim_{x\rightarrow 0} \int_x^1 \frac{1}{x} dx$ and its finiteness. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;del&gt;Assuming that by $[\frac{1}{x}]$ you mean the integer part of x (the closest integer less then or equal to x), we can always assume that $[\frac{1}{x}]$ is zero, because if x less than or greater than zero, you always have a number less than one, the integer part of which is always zero. So your integral is the same as $\int_0^1 \frac{1}{x} dx - \int_0^1 [\frac{1}{x}] dx = \int_0^1 \frac{1}{x} dx$, which does not converge on that interval.&lt;/del&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: promt comments showed that I'm way over my head...Sorry for the wrong answer.&lt;/p&gt;&#xA;" OwnerUserId="1415" LastEditorUserId="1415" LastEditorDisplayName="" LastEditDate="2010-08-23T14:29:54.090" LastActivityDate="2010-08-23T14:29:54.090" CommentCount="2" />
  <row Id="3124" PostTypeId="2" ParentId="3122" CreationDate="2010-08-23T14:28:05.997" Score="4" ViewCount="" Body="&lt;p&gt;The function is Riemann-integrable because it is bounded (it takes values in $[0,1]$) and has countably many discontinuities, namely, the points of the form $\frac{1}{n}$ and $0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This uses &lt;a href=&quot;http://en.wikipedia.org/wiki/Riemann_integral#Properties&quot; rel=&quot;nofollow&quot;&gt;Lebesgue's criterion for Riemann integrability&lt;/a&gt; which you probably meant with &lt;a href=&quot;http://en.wikipedia.org/wiki/Riemann%E2%80%93Lebesgue_lemma&quot; rel=&quot;nofollow&quot;&gt;Riemann-Lebesgue-Lemma&lt;/a&gt; and hence unfortunately didn't want to use.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for doing it by hand with partitions, try Akhil Mathew's hint above.&lt;/p&gt;&#xA;" OwnerUserId="367" LastEditorUserId="367" LastEditorDisplayName="" LastEditDate="2010-08-23T19:42:43.340" LastActivityDate="2010-08-23T19:42:43.340" />
  <row Id="3125" PostTypeId="2" ParentId="3118" CreationDate="2010-08-23T14:28:49.740" Score="3" ViewCount="" Body="&lt;p&gt;An alternate route: We show that, if $ax+by=1$ and $m$ is divisible by $a$ and $b$ then $m$ is divisible by $ab$. (Then apply this with $b=a+1$, $x=-1$ and $y=1$.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof:&lt;/strong&gt; Let $m=ak=bl$. Then $ab(xl+ky)=(ax+by)m=m$. QED&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The point here is that the hypothesis $\exists_{x,y}: ax+by=1$ is often easier to use than $GCD(a,b)=1$. The equivalence between these two is basically equivalent to unique factorization, and you can often dodge unique factorization by figuring out which of these two you really need.&lt;/p&gt;&#xA;" OwnerUserId="448" LastActivityDate="2010-08-23T14:28:49.740" CommentCount="9" />
  <row Id="3126" PostTypeId="2" ParentId="3072" CreationDate="2010-08-23T14:32:15.317" Score="1" ViewCount="" Body="&lt;p&gt;In case you really need to simulate accurate physics, you can use the solution given below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/l8GKb.png&quot; alt=&quot;alt diagram&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the picture above, vector $A = g-a$ where $g$ is acceleration due to gravity, and $a$ is the acceleration of mouse which you need to compute in real time. You will have to translate $g=9.8m/s^2$ into $pixel/s^2$ and store as a constant. For $A$, you need to calculate $A_x$ and $A_y$ based on the last two frames.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We need to find $\alpha(t)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have $\beta=tan^{-1}\frac{A_x}{A_y}$ and $A_1=|A|\cos(\alpha-\beta)$. Angular acceleration will be $-A_1/h$ ($h$ will vary from bird to bird).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You will need to solve the differential equation $\frac{d^2\alpha}{dt^2}=A_1/h$ in real time to make the bird swing realistically.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The obvious way to solve the differential equation numerically is given below -&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Store angular velocity $\omega=\frac{d}{dt}\alpha$. At each frame of the animation, assuming $\Delta t$ time has passed since last frame, do&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$\omega\leftarrow\omega+\frac{A_1}{h}\Delta t$&lt;/li&gt;&#xA;&lt;li&gt;$\alpha\leftarrow\alpha+\omega\Delta t$&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="1310" LastEditorUserId="1310" LastEditorDisplayName="" LastEditDate="2010-08-23T14:49:51.430" LastActivityDate="2010-08-23T14:49:51.430" />
  <row Id="3127" PostTypeId="2" ParentId="3120" CreationDate="2010-08-23T14:36:52.330" Score="5" ViewCount="" Body="&lt;p&gt;&lt;em&gt;&lt;strong&gt;edit&lt;/strong&gt; (to match revised question)&lt;/em&gt;: Given your revised question, there is still the issue of C being on either side of the y-axis, but you have specified that AB=AC and that you are given $\mathrm{m}\angle BAC$ (the angle between AB and AC), so as in my original answer (below), the directed (trigonometric) measure of the angle from the positive x-axis to AC is $\mathrm{m}\angle BAC-90^{\circ}$ and AC=AB=10, so C has coordinates $(10\cos(\mathrm{m}\angle BAC-90^{\circ}),10\sin(\mathrm{m}\angle BAC-90^{\circ}))$.  (This matches up to one of the answers in Moron's solution; the other corresponds to the other side of the y-axis.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&lt;strong&gt;original answer&lt;/strong&gt; (when it was not specified that AB=AC and when the given angle was C)&lt;/em&gt;:  As suggested in the comments, there are several cases.  First, C could be on either side of the y-axis; let's assume that C has positive x-coordinate (leaving the case where it has negative x-coordinate for you to solve).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second, ABC could be isosceles with AB=AC, AB=BC, or AC=BC.  In the first case, $\angle B\cong \angle C$ (which cannot happen unless C is acute) and $\mathrm{m}\angle BAC=180^{\circ}-2\mathrm{m}\angle C$, so the directed (trigonometric) measure of the angle from the positive x-axis to AC is $90^{\circ}-2\mathrm{m}\angle C$ and AC=AB=10, so C has coordinates $(10\cos(90^{\circ}-2\mathrm{m}\angle C),10\sin(90^{\circ}-2\mathrm{m}\angle C))$.  The second case is similar to the first (so it's left for you to solve).  In the third case, C is equidistant from A and B, so C must lie on the perpendicular bisector of AB (as in J. Mangaldan's comment), and by symmetry this perpendicular bisector of AB also bisects $\angle ACB$; from there, you can use right triangle trigonometry to determine the coordinates of C (left for you to solve).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The cases where AB=AC (blue), AB=BC (red), and AC=BC (green) (lighter versions on the left side of the y-axis) are shown below for measures of angle C between 0 and 180&amp;deg;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/JFKWo.gif&quot; alt=&quot;animated diagram&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="72" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-08-23T16:55:07.517" LastActivityDate="2010-08-23T16:55:07.517" CommentCount="10" />
  <row Id="3128" PostTypeId="2" ParentId="3120" CreationDate="2010-08-23T14:58:45.250" Score="3" ViewCount="" Body="&lt;p&gt;Use &lt;a href=&quot;http://en.wikipedia.org/wiki/Polar_coordinate_system&quot; rel=&quot;nofollow&quot;&gt;Polar Co-ordinates&lt;/a&gt;. A point with polar co-ordinates $(r,\theta)$ is the same point in $x,y$ co-ordinates (or as also called, rectangular co-ordinates) as $(r\cos \theta, r\sin \theta)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case, point $C$ lies at a distance $10$ from $A$ which is the origin, so $r = 10$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If given angle CAB is $\alpha$, the the polar angle is either $\frac{3pi}{2}-\alpha$  or $\frac{3pi}{2}+\alpha$, i.e in polar co-ordinates $C$ is either $(10,\frac{3pi}{2}-\alpha)$ or $(10,\frac{3pi}{2}+\alpha)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(It might help to draw a figure).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now convert back to $x,y$ co-ordinates.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-08-23T15:04:42.740" LastActivityDate="2010-08-23T15:04:42.740" CommentCount="7" />
  <row Id="3129" PostTypeId="2" ParentId="2949" CreationDate="2010-08-23T15:07:44.840" Score="22" ViewCount="" Body="&lt;p&gt;The infinite-dimensional sphere is contractible.&lt;/p&gt;&#xA;" OwnerUserId="455" LastActivityDate="2010-08-23T15:07:44.840" CommunityOwnedDate="2010-08-23T15:07:44.840" CommentCount="3" />
  <row Id="3130" PostTypeId="2" ParentId="2949" CreationDate="2010-08-23T15:11:12.760" Score="5" ViewCount="" Body="&lt;p&gt;There exists $f\colon \mathbb{N}\times\mathbb{N}\to\mathbb{N}$ which is bijective.&lt;/p&gt;&#xA;" OwnerUserId="622" LastActivityDate="2010-08-23T15:11:12.760" CommunityOwnedDate="2010-08-23T15:11:12.760" CommentCount="1" />
  <row Id="3131" PostTypeId="2" ParentId="3120" CreationDate="2010-08-23T15:37:13.820" Score="1" ViewCount="" Body="&lt;p&gt;Let $a,b$ and $c$ be the side lengths and $A,B$ and $C$ the angles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a^{2}=x^{2}+\left( y+10\right) ^{2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$b^{2}=x^{2}+y^{2}=10^{2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$b=c=10$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the (Neper) theorem of tangents (corollary of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Law_of_tangents&quot; rel=&quot;nofollow&quot;&gt;Law of tangents&lt;/a&gt;):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\tan \frac{A-B}{2}=\frac{a-b}{a+b}\cot \frac{C}{2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{A+B}{2}=\frac{\pi }{2}-\frac{C}{2}\quad C&amp;lt;\pi $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and by the theorem of sinus&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$c\sin A=a\sin C\iff \left( x^{2}+\left( y+10\right) ^{2}\right) \sin&#xA;C=10\sin A$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Compiling, we get:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{A-B}{2}=\arctan (\frac{\sqrt{x^{2}+\left( y+10\right) ^{2}}-10}{\sqrt{%&#xA;x^{2}+\left( y+10\right) ^{2}}+10}\cot \frac{C}{2})$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{A+B}{2}=\frac{\pi }{2}-\frac{C}{2}\quad C&amp;lt;\pi $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(x^{2}+(\sqrt{100-x^{2}}+10)^{2})\sin C=10\sin A$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$y^{2}=10^{2}-x^{2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have to solve this system of four equations and four unknowns $x,y,A,B$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: I started this approach before the question has been updated.&lt;/p&gt;&#xA;" OwnerUserId="752" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2010-08-23T23:02:04.647" LastActivityDate="2010-08-23T23:02:04.647" CommentCount="7" />
  <row Id="3132" PostTypeId="2" ParentId="1706" CreationDate="2010-08-23T15:37:14.147" Score="3" ViewCount="" Body="&lt;p&gt;I write, e.g., $\mathbb R_{&gt;0}$, $\mathbb R_{\geq0}$, $\mathbb N_{&gt;0}$.&lt;/p&gt;&#xA;" OwnerUserId="1303" LastActivityDate="2010-08-23T15:37:14.147" />
  <row Id="3133" PostTypeId="2" ParentId="3093" CreationDate="2010-08-23T15:40:33.723" Score="6" ViewCount="" Body="&lt;p&gt;There is a very specific reason why one needs 3 dimensions or more for the Banach Tarski paradox.  In dimension 3 or higher one can make rotations in independent directions,&#xA;and so the group $SO(3)$ of rotations of space contains a copy of $F_2$, the free group on&#xA;two generators.  This fact is what underlies the Banach Tarski paradox.  (The group&#xA; $F_2$ is &lt;em&gt;not amenable&lt;/em&gt;.)&lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2010-08-23T15:40:33.723" CommunityOwnedDate="2010-08-23T18:37:37.637" />
  <row Id="3134" PostTypeId="2" ParentId="1706" CreationDate="2010-08-23T15:53:58.173" Score="2" ViewCount="" Body="&lt;p&gt;I met  (in IBDP programme, UK and Poland) the followin notation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\[\mathbb{R}^{+} = \{ x | x \in \mathbb{R} \land x &gt; 0 \} \]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\[\mathbb{R}^{+} \cup {0} = \{ x | x \in \mathbb{R} \land x \geq 0 \} \]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With the explanation that $\mathbb{R}^{+}$ denotes set of positive reals and $0$ is neither positive nor negative.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\mathbb{N}$ is possibly a sligthly different case and it usually differ from branch of mathematics to branch of mathematics. I belive that is &lt;em&gt;usually&lt;/em&gt; includes $0$ but I belive theory of numbers is easier without it. It can be easilly extended in such was to have $\mathbb{N}^+ = \mathbb{Z}^+$ denoting positive integers/naturals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course as noted before it is mainly a question of notation.  &lt;/p&gt;&#xA;" OwnerUserId="728" LastActivityDate="2010-08-23T15:53:58.173" />
  <row Id="3135" PostTypeId="1" AcceptedAnswerId="3151" CreationDate="2010-08-23T16:11:10.090" Score="3" ViewCount="149" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;At 5am, a pedestrian starts walking from A to B, a distance of 30 km. At 7 am a bicyclist whose speed is 2 times the pedestrian's speed, started riding from A to B, too. After some time, the bicyclist met and passed the pedestrian. Two hours after the meeting &lt;strong&gt;the pedestrian&lt;/strong&gt; reached his destination (B). (The bicyclist reached B before the pedestrian reached it.)&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;What is the speed of the pedestrian?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So, first I found the distance that the pedestrian went until the bike rider reached him like that:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Marked x = the pedestrian's speed.&lt;/li&gt;&#xA;&lt;li&gt;Marked y = the distance that the pedestrian went until the bike rider reached him.&lt;/li&gt;&#xA;&lt;li&gt;Time until the bike rider reached the pedestrian = y/x.&lt;/li&gt;&#xA;&lt;li&gt;&lt;code&gt;x(y/x + 2) = 30&lt;/code&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;code&gt;y = 30 - 2x&lt;/code&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So the distance that the pedestrian went until the bike rider reached him was &lt;code&gt;30 - 2x&lt;/code&gt;. But how I continue from here?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="1445" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-08-23T17:07:22.643" LastActivityDate="2010-08-23T19:09:57.760" Title="Traffic Homework Help" Tags="&lt;algebra-precalculus&gt;&lt;homework&gt;" AnswerCount="2" CommentCount="5" />
  <row Id="3136" PostTypeId="2" ParentId="3093" CreationDate="2010-08-23T16:16:02.517" Score="-2" ViewCount="" Body="&lt;p&gt;I'm going to be a bit contrary: is this really true? Or is it just that we started studying the 3D case before we could study anything higher-dimensional, and that our brains are obviously optimized for thinking in three dimensions?&lt;/p&gt;&#xA;" OwnerUserId="173" LastActivityDate="2010-08-23T16:16:02.517" CommunityOwnedDate="2010-08-23T18:37:37.637" CommentCount="1" />
  <row Id="3138" PostTypeId="1" CreationDate="2010-08-23T16:19:31.030" Score="2" ViewCount="574" Body="&lt;p&gt;I am interested in continuous functions which are not smooth, i.e. which have only finitely many derivatives (for now just taking the case f:R-&gt;R).  As far as I can tell, there are only 2 features or properties of such a function which can make it non-smooth.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The derivative of some order diverges to infinity at one or more points.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The derivative of some order oscillates like sin(1/x) does at x = 0 at one or more points.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Q1: Are there any other ways for a continuous function to be non-smooth?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Q2: What are some example's of non-smooth functions which satisfy a differential equation?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In response to Qiaochu comment:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A simple &quot;description of a limit failing to exist which is better than &quot;the limit fails to exist.&quot; is to specify, for example, if the limit of a sequence of natural numbers diverges to infinity or not. for example:  1,2,3,... and 1,2,1,2,1,...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These two examples are similar to my 1. and 2., i.e. the derivative diverges to infinity or it oscillates around without settling down.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't see the problem with trying to classify the ways continuous functions cannot be smooth by looking at their graphs.  One example of a non smooth function is sin(1/x), it has a certain property at x=0, another non-smooth function is |x| and it has another property at x=0.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand you feel the question is not well defined because &quot;being like sin(1/x) at x=0&quot; is an intuitive concept.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ok, so what are examples of non-smooth continuous functions without &quot;kinks&quot; like |x| has at 0?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are two that I can think of, sin(1/x) and the Weierstrass function.  But if you think about the self similarity of the graph of the Weirstrauss function, then it's clear that the derivative of this function osscilates rapidly at every point like sin(1/x) does at x=0.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regardless of whether or not it's well-defined to say the Weierstrass function is like sin(1/x) in some sense, I am just curious to know of some other examples of non-smooth continuous functions without kinks in their graph, can someone please help?&lt;/p&gt;&#xA;" OwnerUserId="1417" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-08-24T19:08:24.283" LastActivityDate="2010-09-27T22:53:45.540" Title="Ways for a continuous function to not be smooth" Tags="&lt;real-analysis&gt;&lt;analysis&gt;" AnswerCount="4" CommentCount="5" FavoriteCount="2" />
  <row Id="3139" PostTypeId="2" ParentId="3138" CreationDate="2010-08-23T16:28:23.623" Score="1" ViewCount="" Body="&lt;p&gt;For Q1)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Depends on your definition of oscillate I suppose.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f(x) = |x|$ does not have a derivative at $0$ and would not fit your criteria for 1 or 2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In any case, your question seems to be equivalent to the question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;In what ways can the limit of a function fail to exist at a point&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As to the comment of derivative of |x| being a step function and so the second derivative of |x| at 0 is infinity is nonsense, IMO. The first derivative itself does not exist, so talking about the second makes no sense. Perhaps you need to clarify exactly what you mean.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-08-23T16:51:29.970" LastActivityDate="2010-08-23T16:51:29.970" CommentCount="5" />
  <row Id="3140" PostTypeId="2" ParentId="3138" CreationDate="2010-08-23T16:38:54.610" Score="0" ViewCount="" Body="&lt;p&gt;Re Q2: For smooth ODEs (i.e. something like y′=f(y) for f smooth, possibly with y taking values in Rn but depending on one parameter), existence of a solution implies regularity (i.e. smoothness) of it. This isn't particularly easy to prove, though; I know a proof is in Lang's &lt;em&gt;Differential and Riemannian Manifolds,&lt;/em&gt; and if I recall correctly the idea is to prove inductively that $y'$ is $C^1$ (or at least Lipschitz) by some sort of Gronwall inequality argument, and then get that $y'$ satisfies its own differential equation, and repeat inductively. The details are very hazy in my mind.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is part of the general theory of ordinary differential equations, though, so there are probably plenty of sources. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For PDEs, automatic regularity is not necessarily the case, though it is for elliptic PDEs with smooth coefficients. More generally, regularity holds for certain &lt;a href=&quot;http://en.wikipedia.org/wiki/Hypoelliptic_operator&quot; rel=&quot;nofollow&quot;&gt;hypoelliptic&lt;/a&gt; operators (e.g. the heat operator). &lt;/p&gt;&#xA;" OwnerUserId="536" LastActivityDate="2010-08-23T16:38:54.610" />
  <row Id="3141" PostTypeId="2" ParentId="3138" CreationDate="2010-08-23T16:40:43.780" Score="0" ViewCount="" Body="&lt;p&gt;To answer your second question, I offer two classical examples:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1: The cycloid&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\left(y^{\prime}\right)^2=\frac{2a-y}{y}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2: The tractrix&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$y^{\prime}=-\frac{\sqrt{a^2-x^2}}{x}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;($a$ is a parameter in both examples)&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-08-23T16:40:43.780" />
  <row Id="3143" PostTypeId="2" ParentId="3118" CreationDate="2010-08-23T17:14:25.550" Score="20" ViewCount="" Body="&lt;p&gt;The other answers put this in a general context, but in this example one&#xA;can be absolutely explicit. If $a\mid m$ and $(a+1)\mid m$ then there are&#xA;integers $r$ and $s$ such that&#xA;$$m=ar=(a+1)s.$$&#xA;Then&#xA;$$a(a+1)(r-s)=(a+1)[ar]-a[(a+1)s]=(a+1)m-am=m.$$&#xA;As $r-s$ is an integer, then $a(a+1)\mid m$.&lt;/p&gt;&#xA;" OwnerUserId="226" LastEditorUserId="205" LastEditorDisplayName="" LastEditDate="2010-08-23T18:00:19.000" LastActivityDate="2010-08-23T18:00:19.000" CommentCount="1" />
  <row Id="3144" PostTypeId="2" ParentId="3135" CreationDate="2010-08-23T17:18:58.087" Score="1" ViewCount="" Body="&lt;p&gt;In problems like this (where there are multiple moving objects that start at different times), I usually start by assigning a variable to the quantity that the question asks for: $$\text{let }x=\text{the speed of the pedestrian}$$&#xA;then assigning a variable to elapsed time from a specific instant: $$\text{let }t=\text{time (in hours) since 5am}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With these variables, I then try to write expressions for the positions of each moving object in terms of $x$ and $t$, and set the position expressions equal corresponding to when the objects are in the same place.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If this is not enough to get you unstuck, please comment on this answer and let me know what you've done so far using these ideas.&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-08-23T17:18:58.087" />
  <row Id="3145" PostTypeId="1" AcceptedAnswerId="3146" CreationDate="2010-08-23T17:51:01.950" Score="3" ViewCount="170" Body="&lt;p&gt;If I want to show a topological subspace is closed in an ambient space, does it suffice to know what happens on an open cover of the ambient space? More specifically,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Let $X$ is a topological space with a given open cover ${ U_i }$. Suppose that $Z \subset X$ is a set such that $Z \cap U_i$ is closed in $U_i$ for all $i$. Does it follow that $Z$ is closed in X?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is clearly true if there are finitely many ${ U_i }$. At first thought, it seems unlikely to be true in the infinite case, but I'm having trouble coming up with a suitable counter-example.&lt;/p&gt;&#xA;" OwnerUserId="576" LastActivityDate="2010-09-06T11:02:47.323" Title="Is whether a set is closed or not a local property?" Tags="&lt;topology&gt;" AnswerCount="2" FavoriteCount="3" />
  <row Id="3146" PostTypeId="2" ParentId="3145" CreationDate="2010-08-23T18:03:03.350" Score="6" ViewCount="" Body="&lt;p&gt;Yes, it's true: being closed is &lt;del&gt;a local property&lt;/del&gt; local on the base. Indeed, suppose $x \notin Z$. Then $x$ is in a neighborhood $U_i$ which is one of the sets in the open cover. Now $x \in U_i - Z \cap U_i$, which is an open set, so there is a neighborhood of $x$, contained in $U_i$, which does not intersect $Z \cap U_i$. This means that this neighborhood of $x$ does not intersect $Z$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is not true for closed covers (i.e. with the $U_i$ closed) in general, but it is true when they form a locally finite cover.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Incidentally, if each point $x \in Z$ has a neighborhood $U$ such that $Z \cap U$ is closed in $U$, then $Z$ is called locally closed: this means it is the intersection of a closed subset and an open subset, but is not itself necessarily closed. &lt;/p&gt;&#xA;" OwnerUserId="536" LastEditorUserId="536" LastEditorDisplayName="" LastEditDate="2010-08-23T19:47:31.740" LastActivityDate="2010-08-23T19:47:31.740" CommentCount="5" />
  <row Id="3149" PostTypeId="2" ParentId="3118" CreationDate="2010-08-23T18:51:37.610" Score="7" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;PROOF&lt;/strong&gt; $\;\rm\quad\displaystyle \frac{m}{a},\; \frac{m}{a+1}\in\mathbb{Z} \;\;\Rightarrow\;\; \frac{m}{a} - \frac{m}{a+1} \; = \;\frac{m}{a(a+1)} \in \mathbb Z\quad\;\;$ &lt;strong&gt;QED&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;NOTE&lt;/strong&gt; $\ $ More generally, for $\:\rm n = bc \:-\; ad \;$ a linear combination of $\rm a, b$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;we have $\rm\quad\quad\quad\displaystyle \frac{m}{a},\; \frac{m}{b}\in\mathbb{Z} \;\;\Rightarrow\;\; \frac{m}{a}\frac{bc}{b} - \frac{ad}{a}\frac{m}{b} = \frac{mn}{ab} \in \mathbb Z$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By Bezout's identity, $\rm\ n = gcd(a,b)\ $  is the least positive linear combination, so the above yields&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\quad\quad\quad\quad\quad\quad\quad\ &#xD;&#xA; a,b|m \;\Rightarrow\; ab\:|\:m\;gcd(a,b) \;\Rightarrow\; \mathfrak{m}_{a,b} |\:m\;\;\;$  for $\;\;\rm \mathfrak{m}_{a,b} := \frac{ab}{gcd(a,b)}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i.e. $\:$ every common multiple $\rm m$ of $\rm a,b$ is a multiple of $\;\rm \mathfrak{m}_{a,b}. \;$ But $\rm\:\mathfrak{m}_{a,b}\:$ is also a common multiple, i.e. $\rm\ a,b|\mathfrak{m}_{a,b} \;$ viz. $\rm \frac{\mathfrak{m}_{a,b}}{a} = \;\frac{a}{a}\frac{b}{gcd(a,b)}\in\mathbb Z \;\Rightarrow\; a|\mathfrak{m}_{a,b} \;$. By symmetry, also $\rm\ b|\mathfrak{m}_{a,b} \;.\ $ Therefore $\rm \mathfrak{m}_{a,b} = lcm(a,b)$ since it's a common multiple of $\rm a,b$ that's divisibility-least, i.e. that divides every common multiple. This is the general definition of LCM in an arbitrary domain (ring without zero-divisors), namely we make the following &lt;strong&gt;universal&lt;/strong&gt; dual definitions of LCM and GCD:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;DEFINITION&lt;/strong&gt; of &lt;strong&gt;LCM&lt;/strong&gt; $\quad$ If $\quad\rm a,b|c \;\iff\; [a,b]\:|\:c \quad\;$ then $\quad\rm [a,b] \;$ is an &lt;strong&gt;LCM&lt;/strong&gt; of $\;\rm a,b$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;DEFINITION&lt;/strong&gt; of &lt;strong&gt;GCD&lt;/strong&gt; $\quad$ If $\quad\rm c|a,b \;\iff\; c\:|\:(a,b) \quad$ then $\quad\rm (a,b) \;$ is an &lt;strong&gt;GCD&lt;/strong&gt; of $\;\;\rm a,b$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: that $\;\rm a,b\:|\:[a,b] \;$ follows by putting $\;\rm c = [a,b] \;$ in the definition. Dually $\;\rm (a,b)|\:a,b \;$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Such $\iff$ definitions provide slick unified proofs of both arrow directions, e.g. the fundamental&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;THEOREM&lt;/strong&gt;  $\rm\;\; (a,b) = ab/[a,b] \;\;$  if $\;\rm\ [a,b] \;$  exists.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof&lt;/strong&gt;:  $\rm\quad\quad d|\:a,b \;\iff\; a,b\:|\:ab/d \;\iff\; [a,b]\:|\:ab/d \;\iff\; d\:|\:ab/[a,b] \quad\;\;$ &lt;strong&gt;QED&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The conciseness of this proof arises by exploiting to the hilt the $\iff$ definition of LCM and GCD.  Compare this slick proof to many obfuscated proofs found in elementary number theory textbooks!  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the theorem, GCDs exist if LCMs exist. But common multiples clearly comprise an ideal, being closed under subtraction and multiplication by any ring element. Hence in a PID the generator of an ideal of common multiples is clearly an LCM. In Euclidean domains this can be proved directly by a simple descent, e.g. in $\:\mathbb Z \;$ we have the following high-school level proof of the existence of LCMs (and, hence, of GCDs), after noting the set $\rm M$ of common multiples of $\rm a,b$ is closed under subtraction and contains $\:\rm ab \ne 0\:$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;LEMMA&lt;/strong&gt; $\ $ Suppose $\;\rm M\subset\mathbb Z \;$ is closed under subtraction and that $\rm M$ contains a nonzero element.&lt;br&gt;&#xA;Then $\rm M \:$ has a positive element and the least such positive element of $\;\rm M$ divides every element.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof&lt;/strong&gt; $\rm\ \ 0 \ne m\in M \Rightarrow m-m = 0\in M\Rightarrow 0-m = -m\in M, \;$ so $\rm M$ has a positive element.&lt;br&gt;&#xA;Let $\rm m$ be the least positive element in $\rm M$. Since $\rm m|n \iff m\;|-n, \;$ if some $\rm n\in M$ isn't divisible by $\rm m$ then we &#xA;may assume this $\rm n$ is positive, and the least such element. But $\rm\: n-m \:$ is a positive element of $\rm M$ not divisible by $\rm m$ &#xA;and smaller than $\rm n$, contra leastness of $\rm n\:$. Hence $\rm m$ divides every element of $\rm M.\;\;$ &lt;strong&gt;QED&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-12-03T07:14:13.777" LastActivityDate="2010-12-03T07:14:13.777" />
  <row Id="3151" PostTypeId="2" ParentId="3135" CreationDate="2010-08-23T19:09:57.760" Score="1" ViewCount="" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;At 5am, a pedestrian starts walking from A to B, a distance of 30 km.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Let speed of pedestrian be $v$. At time $t$, pedestrian is at distance $vt$. (Where $t$ denotes time since pedestrian started.)&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;At 7 am a bicyclist whose speed is 2 times the pedestrian's speed, started riding from A to B, too.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Speed of cyclist is $2v$. At time $t$, cyclist is at distance $(2v)(t-2)$. (Why?)&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;After some time, the bicyclist met and passed the pedestrian.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Let they meet at time $t_1$. Then $vt_1=2v(t_1-2)$. (Why? Notice that $v$ cancels out, hence solve for $t_1$ from here.)&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Two hours after the meeting the pedestrian reached his destination.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The pedestrian covers 30km in $t_1+2$ hours, with $t_1$ known. Hence deduce his speed.&lt;/p&gt;&#xA;" OwnerUserId="1310" LastActivityDate="2010-08-23T19:09:57.760" CommentCount="3" />
  <row Id="3152" PostTypeId="2" ParentId="2550" CreationDate="2010-08-23T20:00:21.917" Score="0" ViewCount="" Body="&lt;p&gt;Just so I don't forget, here is the solution I was thinking of, for part b).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $(x,y)$ be a point which is at a rational distance from $(0,0), (1,0)$ and $(\sqrt{2},0)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then we have that $x^2 + y^2$, $(x-1)^2 + y^2$ and $(x-\sqrt{2})^2 + y^2$ are all squares of rational numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The difference of the first and second implies $x$ is rational.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The difference of the second and third implies that $x = 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus $y$ is rational.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus we have that for some integers $a,b,p,q$, $a \neq 0$, $b \neq 0$ that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) $a^2 + b^2 = p^2$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) $a^2 + 2b^2 = q^2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Square (1) and multiply (2) by $a^2$ and subtract.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$b^4 = p^4 - (aq)^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is well known that the equation $x^4 + y^2 = z^4$ has no non-trivial solutions (and in fact can be used to show that Fermat's Last theorem is true for $n=4$). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;See here for instance: &lt;a href=&quot;http://planetmath.org/encyclopedia/X4Y4z2HasNoSolutionsInPositiveIntegers.html&quot; rel=&quot;nofollow&quot;&gt;http://planetmath.org/encyclopedia/X4Y4z2HasNoSolutionsInPositiveIntegers.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-08-23T20:00:21.917" />
  <row Id="3153" PostTypeId="1" AcceptedAnswerId="3154" CreationDate="2010-08-23T22:00:32.883" Score="9" ViewCount="283" Body="&lt;p&gt;An absolute value expression such as $|ax-b|$ can be rewritten in two cases as $|ax-b|=\begin{cases}&#xA;ax-b &amp;amp; \text{ if } x\ge \frac{b}{a} \\ &#xA;b-ax &amp;amp; \text{ if } x&amp;lt; \frac{b}{a} &#xA;\end{cases}$, so an equation with $n$ separate absolute value expressions can be split up into $2^n$ cases, but is there a better way?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, with $|2x-5|+|x-1|+|4x+3|=13$, is there a better way to handle all the possible combinations of $x\ge\frac{5}{2}$ versus $x&amp;lt;\frac{5}{2}$, $x\ge 1$ versus $x&amp;lt; 1$, and $x\ge-\frac{3}{4}$ versus $x&amp;lt;-\frac{3}{4}$?&lt;/p&gt;&#xA;" OwnerUserId="72" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2011-03-31T17:10:07.593" LastActivityDate="2011-03-31T17:10:07.593" Title="What is the best way to solve an equation involving multiple absolute values?" Tags="&lt;algebra-precalculus&gt;&lt;faq&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="3154" PostTypeId="2" ParentId="3153" CreationDate="2010-08-23T22:04:09.137" Score="13" ViewCount="" Body="&lt;p&gt;For an equation with $n$ absolute values, the $n$ places where each absolute value splits into 2 cases divide the number line into $n+1$ regions where, within each region, each absolute value can be replaced by either the expression inside the absolute value or its opposite. Each resulting equation can then be solved, restricting solutions to the corresponding region on the number line.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the given example,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\small{\begin{matrix}&#xA;\leftarrow  &amp;amp; -\frac{3}{4} &amp;amp; \text{---}   &amp;amp; 1 &amp;amp; \text{---} &amp;amp; \frac{5}{2} &amp;amp; \rightarrow&#xA;\\&#xA;\begin{matrix}-(2x-5)-(x-1)\\-(4x+3)=13\end{matrix}&#xA;&amp;amp; \begin{matrix}|\\|\end{matrix} &amp;amp;&#xA;\begin{matrix}-(2x-5)-(x-1)\\+(4x+3)=13\end{matrix} &#xA;&amp;amp; \begin{matrix}|\\|\end{matrix} &amp;amp;&#xA;\begin{matrix}-(2x-5)+(x-1)\\+(4x+3)=13\end{matrix} &#xA;&amp;amp; \begin{matrix}|\\|\end{matrix} &amp;amp;&#xA;\begin{matrix}(2x-5)+(x-1)\\+(4x+3)=13\end{matrix} &#xA;\\&#xA;-7x+3=13&#xA;&amp;amp; | &amp;amp;&#xA;x+9=13&#xA;&amp;amp; | &amp;amp;&#xA;3x+7=13&#xA;&amp;amp; | &amp;amp;&#xA;7x-3=13&#xA;\\&#xA;x=-\frac{10}{7}&#xA;&amp;amp; | &amp;amp;&#xA;x=4\notin[-\frac{3}{4},1]&#xA;&amp;amp; | &amp;amp;&#xA;x=2&#xA;&amp;amp; | &amp;amp;&#xA;x=\frac{16}{7}\notin[\frac{5}{2},\infty)&#xA;\end{matrix}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the solutions are $x=-\frac{10}{7}$ and $x=2$ (the values that were solutions to an equation for a particular region and were within that region).&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-08-23T22:04:09.137" CommentCount="3" />
  <row Id="3155" PostTypeId="1" AcceptedAnswerId="3157" CreationDate="2010-08-23T22:42:04.720" Score="13" ViewCount="612" Body="&lt;p&gt;This was part of a &lt;a href=&quot;http://books.google.com/books?id=9LtfZr1snG0C&amp;amp;lpg=PP1&amp;amp;pg=PA110#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;homework problem&lt;/a&gt; from J.B. Conway's complex analysis text which I was assigned long ago but didn't get.  A few years later I was a TA for a course where the problem was assigned.  I still didn't know how to solve it, nor did any of my students.  If you can give me a simple answer, my gratefulness will far outweigh my embarrassment.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Define $f(z)=z\sin(\frac{1}{z})$ on the punctured plane.  What is the image under $f$ of a (small) punctured disk at the origin?  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Remarks:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The corresponding problem for $\sin(\frac{1}{z})$ is straightforward to solve using the definition of $\sin$ in terms of exponentials, the fact that $\exp$ is onto the punctured plane, and periodicity.  The answer in this case is the whole plane.&lt;/li&gt;&#xA;&lt;li&gt;The corresponding problem for $z\cos(\frac{1}{z})$ is straightforward to solve using Picard's theorem, because the oddness of the function implies that 0 is the only possible excluded point, and it is easy to see that 0 is in the image by taking reciprocals of large zeros of $\cos$.  Thus the answer in this case is also the whole plane. (The same argument could be applied to the previous remark, but there no big theorems are needed.  Perhaps Picard's theorem isn't needed here either.)&lt;/li&gt;&#xA;&lt;li&gt;Another part of the problem I couldn't solve generalizes this, taking $z^n\sin(\frac{1}{z})$ when $n$ is a positive integer.  When $n$ is even, there is no problem, because the argument from the previous remark applies.  But I don't know how to solve it when $n$ is odd.  &lt;/li&gt;&#xA;&lt;li&gt;By Picard's theorem, there is at most one point missing. I would be surprised if there is a point missing, but I have no argument to back up this expectation.&lt;/li&gt;&#xA;&lt;li&gt;Although an elementary argument would be nice, I do not care whether the argument is &quot;appropriate&quot; for what is covered up to that point in the text.  This is pure curiosity.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2011-01-11T09:37:14.837" Title="What is the image near the essential singularity of z sin(1/z)?" Tags="&lt;complex-analysis&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" />
  <row Id="3156" PostTypeId="2" ParentId="3008" CreationDate="2010-08-23T23:00:25.553" Score="2" ViewCount="" Body="&lt;p&gt;If there is a definable wellordering of $\mathbb{R}$, then every real number is &lt;a href=&quot;http://en.wikipedia.org/wiki/Ordinal_definable_set&quot; rel=&quot;nofollow&quot;&gt;ordinal definable&lt;/a&gt;. Conversely, if every real number is ordinal definable, then there is a definable wellordering of $\mathbb{R}$. Thus the assumption that there is a definable wellordering of $\mathbb{R}$ is equivalent to $\mathbb{R} \subseteq \mathrm{OD}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This assumption is completely independent of AC in the sense that all four combinations AC + $\mathbb{R} \subseteq \mathrm{OD}$, AC + $\mathbb{R} \not\subseteq \mathrm{OD}$, &amp;not;AC + $\mathbb{R} \subseteq \mathrm{OD}$, &amp;not;AC + $\mathbb{R} \not\subseteq \mathrm{OD}$ are relatively consistent with ZF! The details of this are somewhat involved, see Thomas J. Jech's &lt;em&gt;The Axiom of Choice&lt;/em&gt; for an introduction to the various methods to prove independence from ZF.&lt;/p&gt;&#xA;" OwnerUserId="25" LastActivityDate="2010-08-23T23:00:25.553" CommentCount="2" />
  <row Id="3157" PostTypeId="2" ParentId="3155" CreationDate="2010-08-23T23:24:32.597" Score="10" ViewCount="" Body="&lt;p&gt;It can (apparently) be shown that the equation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sin z = Az$ has infinitely many roots for any complex number $A$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(For instance see this: &lt;a href=&quot;http://books.google.com/books?id=DYCOCBCBwoIC&amp;amp;pg=PA270&quot; rel=&quot;nofollow&quot;&gt;http://books.google.com/books?id=DYCOCBCBwoIC&amp;amp;pg=PA270&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given any $\delta &gt; 0$ there must be such a root with $|z| &gt; \frac{1}{\delta}$, otherwise we can show that $\sin z - Az$ must be identically zero on some bounded disk.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I suppose this shows that there is no value missed by $z \sin\frac{1}{z}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Am a bit rusty, please pardon if incorrect).&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-08-24T00:54:52.997" LastActivityDate="2010-08-24T00:54:52.997" CommentCount="3" />
  <row Id="3158" PostTypeId="2" ParentId="3153" CreationDate="2010-08-24T00:32:54.837" Score="5" ViewCount="" Body="&lt;p&gt;This is merely a very special case of the powerful &lt;strong&gt;CAD (cylindrical algebraic decomposition)&lt;/strong&gt; algorithm for quantifier elimination in real-closed fields, e.g. see Jirstrand's &lt;a href=&quot;http://webber.physik.uni-freiburg.de/~hon/vorlss02/Literatur/ljung/publications/1995/1807.ps.Z&quot; rel=&quot;nofollow&quot;&gt;paper [1]&lt;/a&gt; for a nice introduction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[1] M. Jirstrand. &lt;strong&gt;Cylindrical algebraic decomposition - an introduction&lt;/strong&gt;. 1995&lt;br&gt;&#xA;Technical report S-58183, Automatic Control group, Department of Electrical Engineering&lt;br&gt;&#xA;Linkoping University, Linkoping, Sweden.&lt;br&gt;&#xA;Freely available &lt;a href=&quot;http://webber.physik.uni-freiburg.de/~hon/vorlss02/Literatur/ljung/publications/1995/1807.ps.Z&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; or &lt;a href=&quot;http://lib.org.by/get/_Papers/Computer_algebra/Solving%20equations/Jirstrand%20M.%20Introduction%20to%20cylindrical%20algebraic%20decomposition%20%281995%29%2840s%29.ps.gz&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-08-24T01:26:00.093" LastActivityDate="2010-08-24T01:26:00.093" CommentCount="5" />
  <row Id="3161" PostTypeId="2" ParentId="3085" CreationDate="2010-08-24T01:05:41.900" Score="4" ViewCount="" Body="&lt;p&gt;If you do possess Concrete Mathematics, check out &lt;code&gt;5.1 Basic Identities&lt;/code&gt; and &lt;code&gt;5.2 Basic Practice&lt;/code&gt;. The list can be found at P174 &lt;code&gt;Table 174 The top ten binomial coefficient identities&lt;/code&gt;.&lt;/p&gt;&#xA;" OwnerUserId="696" LastActivityDate="2010-08-24T01:05:41.900" CommentCount="1" />
  <row Id="3162" PostTypeId="2" ParentId="3155" CreationDate="2010-08-24T01:29:45.390" Score="5" ViewCount="" Body="&lt;p&gt;The image of $f(z)=z^n\sin(\frac1z)$ (for n odd) on any punctured disc about the origin is the whole of $\mathbb{C}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The following is (I think) a simplified proof along the lines of that given in Moron's reference. I'll use proof by contradiction, so suppose that there is an $a$ such that $f(z^{-1})=z^{-n}\sin(z)\not=a$ for all large $z$. By symmetry, $f(z^{-1})=g(z^2)$ for an analytic function $g$ with, for $n&gt;1$, a pole at the origin of order $(n-1)/2$. As $g(z)-a$ can have only finitely many zeros, we can decompose it as&#xA;$$&#xA;g(z)=a+z^{(1-n)/2}h(z)\prod_{i=1}^m(z-b_i)&#xA;$$&#xA;where $b_i$ are the zeros of $g(z)-a$.&#xA;Writing out $f(z^{-1})=\frac{1}{2i}z^{-n}(e^{iz}-e^{-iz})$ shows that $f(z^{-1})$ is of order $1$. So $g$ and, hence, $h$ are of order 1/2. Any entire and nonzero function of order less than 1 is constant, so $h$ is constant and we see that $f$ is a rational function, which is false.&lt;/p&gt;&#xA;" OwnerUserId="1321" LastActivityDate="2010-08-24T01:29:45.390" CommentCount="1" />
  <row Id="3163" PostTypeId="1" AcceptedAnswerId="3167" CreationDate="2010-08-24T01:32:41.687" Score="1" ViewCount="176" Body="&lt;p&gt;Let $p &gt;5$ be a prime number. Prove that every algebraic integer of the $p$th cyclotomic field can be represented as a sum of (finitely many) distinct units of the ring of algebraic integers of the field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reference: &lt;a href=&quot;http://www.artofproblemsolving.com/Forum/resources.php?c=2&amp;amp;cid=152&amp;amp;year=1977&amp;amp;sid=151602f87027a7ce87d3aa9421a666e9&quot; rel=&quot;nofollow&quot;&gt;http://www.artofproblemsolving.com/Forum/resources.php?c=2&amp;amp;cid=152&amp;amp;year=1977&amp;amp;sid=151602f87027a7ce87d3aa9421a666e9&lt;/a&gt;          Question No: 4&lt;/p&gt;&#xA;" LastActivityDate="2010-08-24T02:22:44.350" Title="Algebraic Integers in a Cyclotomic Field." Tags="&lt;number-theory&gt;&lt;algebraic-number-theory&gt;" AnswerCount="1" CommentCount="7" />
  <row Id="3164" PostTypeId="2" ParentId="2110" CreationDate="2010-08-24T02:00:21.773" Score="1" ViewCount="" Body="&lt;p&gt;If I understand the question correctly, we are given real numbers $0 \leq x_1 \leq \dots \leq x_n$, and real numbers $0 &amp;lt; t_1 &amp;lt; \dots  &amp;lt; t_n$. We also have some unknown differentiable function $x(t)$ that satisfies:&#xA;$$&#xA;x_i \leq x(t_i) &amp;lt; x_i+1  \quad for\quad i=1,2,\dots ,n&#xA;$$&#xA;Our goal is to estimate $\frac{d}{dt}x(t)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose we are interested in the derivative at some point $t_0$. It is clear that the constraints above do not constrain the possible values of the derivative at $t_0$. Indeed, given any set of $x_i$ and $t_i$, one could find an admissible function $x(t)$ for which the derivative at $t_0$ is &lt;em&gt;anything&lt;/em&gt; we want it to be. So unless we know something more about the &quot;law of motion&quot; governing x, there isn't a reasonable way to estimate the velocity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the absence of more information, I'll just assume that x is a polynomial. This will enforce some sort of smoothness. Finding the lowest-degree polynomial that satisfies all the constraints can be cast as a linear program. To see how, suppose that our polynomial is of degree k. Namely, $x(t) = a_0 + a_1 t + \dots + a_{k}t^k$. Then, consider the LP:&#xA;\begin{align*}&#xA;minimize &amp;amp;\quad z \\&#xA;\text{subject to:}&amp;amp;\quad 0 \leq x(t_i) \leq z\quad i=1,\dots,n&#xA;\end{align*}&#xA;Start with $n=0$, and keep increasing until you find an optimal $z$ that is less than 1. Now that you have your polynomial approximation for $x$, you can easily evaluate its derivative anywhere.&lt;/p&gt;&#xA;" OwnerUserId="1356" LastActivityDate="2010-08-24T02:00:21.773" CommentCount="7" />
  <row Id="3165" PostTypeId="2" ParentId="1730" CreationDate="2010-08-24T02:13:09.173" Score="3" ViewCount="" Body="&lt;p&gt;I would describe it as a &lt;strong&gt;verb&lt;/strong&gt;.  People are familiar with nouns and verbs so you get to shovel in a bunch of formal understanding for free that way.  Take a &lt;strong&gt;noun&lt;/strong&gt;, a real number or a vector or, even better, a food item, a TV show, or a person &amp;ndash; and think of something that operates on it.  Then the verb &lt;strong&gt;transforms&lt;/strong&gt; it and it's different.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Really the best thing to do is give them a positive, &lt;strong&gt;creative&lt;/strong&gt;, self-expressive assignment &amp;ndash; like an &lt;strong&gt;essay&lt;/strong&gt;.  Offer extra credit to people who find examples of functions in the real world ongoing throughout the semester.  If they bring in examples of two- or three-place operators then you get to explain why they're &lt;em&gt;right&lt;/em&gt; in front of the class and they've just brought up the lecture point for you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Look at &lt;strong&gt;the covers of magazines in the grocery store&lt;/strong&gt; and you will get the material for examples you should be using.  Take weight loss.  You could graph the number of calories taken in versus weight gain/loss.  Give them a lot of examples of functions and then the definition at the end &amp;ndash; &lt;em&gt;then&lt;/em&gt; have them look throughout the semester for more examples.  This should be in concert with or as part of a weekly one-page (or one-paragraph) essay that they submit to you with an example of something from class that related to something from life.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Here's another trick I've used in teaching about $(x-1)^2$ versus $x^2 - 1$, which will surely come up as well.  &lt;strong&gt;Decompose&lt;/strong&gt; it into two functions.  $x \mapsto x-1$ &lt;em&gt;relabels the abcissa&lt;/em&gt; and you've prepared the board by using a dotted or &lt;strong&gt;no ordina&lt;/strong&gt;.  Then $\cdot \mapsto \cdot^2$ allows you to plot the parabola &amp;ndash; and it's conveniently shifted to the correct place when you put the original abcissa beneath.&lt;/p&gt;&#xA;" OwnerUserId="1457" LastEditorUserId="1457" LastEditorDisplayName="" LastEditDate="2010-12-22T07:13:24.917" LastActivityDate="2010-12-22T07:13:24.917" CommentCount="3" />
  <row Id="3167" PostTypeId="2" ParentId="3163" CreationDate="2010-08-24T02:22:44.350" Score="1" ViewCount="" Body="&lt;p&gt;Miklos Schweitzer is a &lt;em&gt;very hard&lt;/em&gt; contest.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, solution for this (and other problems) can be found in the book:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Contests in Higher Mathematics, published by Springer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Google books has it:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://books.google.com/books?id=2wwXImJ2HocC&quot; rel=&quot;nofollow&quot;&gt;http://books.google.com/books?id=2wwXImJ2HocC&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And this particular problem's solution appears here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://books.google.com/books?id=2wwXImJ2HocC&amp;amp;pg=PA88&quot; rel=&quot;nofollow&quot;&gt;http://books.google.com/books?id=2wwXImJ2HocC&amp;amp;pg=PA88&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-08-24T02:22:44.350" CommentCount="4" />
  <row Id="3168" PostTypeId="2" ParentId="2110" CreationDate="2010-08-24T02:24:39.633" Score="0" ViewCount="" Body="&lt;p&gt;Since you mention that your $x_i$ is error-contaminated; the sanest solution (in the absence of other pertinent information that could be used to simplify your problem, e.g. a conjectured functional form for $x(t)$) would be probably a smoothing spline; this is a similar but probably more amenable proposal to Laurent's, with the advantage that any bad nonpolynomial behavior your data might exhibit can be confined to only a few intervals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know what computing environment you're using, but MATLAB at least has smoothing splines in its Spline Toolbox (check the documentation for syntax and proper use); and there are FORTRAN programs developed by Carl de Boor (in fact, the MATLAB toolbox was based on these old FORTRAN programs) @ &lt;a href=&quot;http://www.netlib.org/&quot; rel=&quot;nofollow&quot;&gt;NetLib&lt;/a&gt; ; &lt;a href=&quot;http://www.netlib.org/pppack/index.html&quot; rel=&quot;nofollow&quot;&gt;pppack&lt;/a&gt; is the particular library you want.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-08-24T02:24:39.633" />
  <row Id="3170" PostTypeId="1" AcceptedAnswerId="3180" CreationDate="2010-08-24T02:41:43.313" Score="7" ViewCount="156" Body="&lt;p&gt;Serre's criterion for affineness (Hartshorne III.3.7) states that:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Let $X$ be a noetherian scheme. Suppose $H^1(X, \mathcal{F})= 0$ for every quasi-coherent sheaf on $X$. Then $X$ is affine.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There is a more general statement (EGA II.5.2.1) that reaches the same conclusion with the hypothesis that $X$ is a separated quasi-compact scheme or one where the topological space is noetherian.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't, however, understand why we need these hypotheses.  The idea of the proof is to find a set of $g_i \in \Gamma(X, O_X)$ generating the unit ideal and such that the $X_{g_i}$ are affine.  This can be done by showing the $X_{g_i}$ form a basis at each closed point using the cohomology statement (an argument which I'm pretty sure works without &lt;em&gt;any&lt;/em&gt; conditions on the scheme). Then, you take open affine sets of the form $X_f$ containing each closed point and take their union; this is an open set whose complement is closed and must contain no closed point, hence is empty. This part of the argument relies on the fact that every closed subset contains a closed point, a fact which is true under noetherian hypotheses since a scheme is a $T_0$-space. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But isn't it true that every quasi-compact $T_0$  space has a closed point? A minimal closed set must be reduced to a point. So I'm unclear why &quot;quasi-compactness&quot; alone is not enough. (It &lt;em&gt;shouldn't&lt;/em&gt; be enough. Grothendieck is not one to mince hypotheses!)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Questions:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Where does the standard argument break down for schemes which are only quasi-compact? &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Is this true if $X$ is only quasi-separated and quasi-compact? (Are there counterexamples otherwise?)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="536" LastActivityDate="2010-08-24T06:00:11.900" Title="Why do we need noetherianness (or something like it) for Serre's criterion for affineness?" Tags="&lt;algebraic-geometry&gt;&lt;schemes&gt;" AnswerCount="1" FavoriteCount="2" />
  <row Id="3171" PostTypeId="2" ParentId="154" CreationDate="2010-08-24T02:47:06.750" Score="7" ViewCount="" Body="&lt;p&gt;Here's a way to visualize a vector &quot;pointing&quot; in the $i$ direction.  If electrical power is traveling 100% efficiently from power generator to your home, then it's pointing in the direction $\langle 1 \rangle$ (or flip-flopping $\langle 1 \rangle$ and $\langle -1 \rangle$ for A/C).  If the power points in the direction of $\langle \sqrt{-1} \rangle$ then the wire will heat up but transmit no useful electricity.&lt;/p&gt;&#xA;" OwnerUserId="1457" LastEditorUserId="1457" LastEditorDisplayName="" LastEditDate="2011-03-18T11:06:07.173" LastActivityDate="2011-03-18T11:06:07.173" CommunityOwnedDate="2011-04-15T20:45:29.080" CommentCount="2" />
  <row Id="3173" PostTypeId="1" CreationDate="2010-08-24T03:34:03.660" Score="0" ViewCount="350" Body="&lt;p&gt;By &quot;isomorphism&quot; I mean any structure-preserving map with a structure-preserving inverse.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Please accept my advance apology if this question is out of bounds. I sense that it's borderline, but I'm hoping it'll be considered in better taste than the typical &quot;What is your favorite X?&quot; question. I think a collection of great isomorphisms would be interesting because of what isomorphisms uniquely have the power to do: reveal deep and astonishing connections between seemingly unrelated fields of study; open a channel through which techniques and ideas can pass between disciplines; collapse two or many problems into one.)&lt;/p&gt;&#xA;" OwnerUserId="1458" LastEditorUserId="1458" LastEditorDisplayName="" LastEditDate="2010-08-24T06:10:08.730" LastActivityDate="2010-08-24T15:16:21.577" CommunityOwnedDate="2010-08-24T03:34:03.660" ClosedDate="2010-08-24T12:01:34.183" Title="What is your favorite isomorphism?" Tags="&lt;big-list&gt;&lt;math-history&gt;" AnswerCount="4" CommentCount="2" />
  <row Id="3174" PostTypeId="2" ParentId="3173" CreationDate="2010-08-24T03:42:19.613" Score="0" ViewCount="" Body="&lt;p&gt;I like the idea that an $n$ dimensional vector space over the field $\mathbb{C}$ is isomorphic to $\mathbb{C}^n$&lt;/p&gt;&#xA;" OwnerUserId="832" LastEditorUserId="832" LastEditorDisplayName="" LastEditDate="2010-08-24T15:16:21.577" LastActivityDate="2010-08-24T15:16:21.577" CommunityOwnedDate="2010-08-24T03:42:19.613" CommentCount="2" />
  <row Id="3177" PostTypeId="1" AcceptedAnswerId="3179" CreationDate="2010-08-24T04:18:05.863" Score="5" ViewCount="180" Body="&lt;p&gt;I was having a look at &lt;a href=&quot;http://stackoverflow.com/questions/2792443/finding-the-centroid-of-a-polygon&quot;&gt;this question on SO&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From what I know, the centroid is the center of mass of an object. so, by definition its position is given by a simple mean of the positions of all the points in the object.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a polygon, it only has mass at the vertices. So, the centroid should be given by the arithmetic mean of the coordinates of the vertices.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But &lt;a href=&quot;http://en.wikipedia.org/wiki/Centroid#Centroid_of_polygon&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt; says centroid is given by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/MX2km.png&quot; alt=&quot;alt text&quot;&gt;&lt;img src=&quot;http://i.imgur.com/S5n2P.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where A is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/u5kDg.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Why doesn't a simple arithmetic mean work?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="324" LastActivityDate="2010-11-11T00:13:28.900" Title="Why doesn't a simple mean give the position of a centroid in a polygon?" Tags="&lt;geometry&gt;" AnswerCount="2" />
  <row Id="3178" PostTypeId="2" ParentId="3101" CreationDate="2010-08-24T04:28:41.467" Score="4" ViewCount="" Body="&lt;p&gt;The &lt;strong&gt;derivative product rule&lt;/strong&gt; is merely a special case of the &lt;strong&gt;congruence product rule&lt;/strong&gt; for rings, i.e.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;PRODUCT RULE&lt;/strong&gt; $\rm\quad\:\:\:\: a\equiv a_1,\; b\equiv b_1 \;\Rightarrow\;\; ab \equiv a_1 b_1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof&lt;/strong&gt;: $\rm\quad\; ab-a_1 b_1 \equiv a(b-b_1)+(a-a_1)b_1 \equiv 0 \quad$ &lt;strong&gt;QED&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus $\rm\quad\quad\quad\;\;\: f(x+t) \equiv \: f(x)\: + \: f'(x) \; t \;\pmod {t^2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and  $\rm\quad\quad\quad\quad g(x+t) \equiv g(x) + g'(x) \; t \;\pmod {t^2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\quad\quad\;\Rightarrow\quad\: f(x+t)g(x+t) \:\equiv\: f(x)g(x) + (f'(x)g(x) + f(x)g'(x)) \; t \:\;\pmod {t^2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\displaystyle\quad\quad\quad\:\Rightarrow\quad \frac{f(x+t)g(x+t)\: - \:f(x)g(x)}{t} \equiv\: f'(x)g(x) + f(x)g'(x) \quad\:\pmod t$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact this is how one universally defines derivatives in formal polynomial rings $\rm R[x]$, c.f. my post &lt;a href=&quot;http://mathoverflow.net/questions/32343&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;. This yields a &lt;strong&gt;purely algebraic&lt;/strong&gt; approach to polynomial derivatives - devoid of limits or other topological notions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The ring $\rm R[t]/t^2 \;$ is known as the algebra of dual numbers over the ring $\rm R$. &#xA;This ring and its higher order analogs $\;\rm R[t]/t^n \;$  prove quite useful when studying (higher) derivations algebraically since they provide convenient &lt;strong&gt;algebraic models of tangent / jet spaces&lt;/strong&gt;. E.g. as above, they permit easy transfer of properties of homomorphisms to derivations -- see for example section 8.15 in Jacobson, Basic Algebra II. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dual numbers have been applied in many contexts, e.g. &lt;a href=&quot;http://people.maths.ox.ac.uk/szendroi/defth.pdf&quot; rel=&quot;nofollow&quot;&gt;deformation theory [2]&lt;/a&gt;, &#xA;&lt;a href=&quot;http://bt.pa.msu.edu/cgi-bin/display.pl?name=dahap&quot; rel=&quot;nofollow&quot;&gt;numerical analysis [3]&lt;/a&gt; (along with &lt;a href=&quot;http://bt.pa.msu.edu/pub/papers&quot; rel=&quot;nofollow&quot;&gt;Levi-Civita fields&lt;/a&gt;), where &#xA;they're viewed simply as truncated Taylor power series, and also in &#xA;Synthetic Differential Geometry &lt;a href=&quot;http://www.jstor.org/stable/20116534&quot; rel=&quot;nofollow&quot;&gt;(SDG) [1]&lt;/a&gt;, another rigorization of &#xA;inifinitesimals based on work of Lawvere and Kock. Note that SDG employs these&#xA;nilpotent infinitesimals, unlike Abraham Robinson's nonstandard &#xA;analysis, which employs invertible infinitesimals (hence contains infinite elements). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;[1] Bell, J. L. Infinitesimals. Synthese 75 (1988) #3, 285--315.&lt;br&gt;&#xA;&lt;a href=&quot;http://www.jstor.org/stable/20116534&quot; rel=&quot;nofollow&quot;&gt;http://www.jstor.org/stable/20116534&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[2] Szendroi, B.  The unbearable lightness of deformation theory, &#xA;a tutorial introduction.&lt;br&gt;&#xA;&lt;a href=&quot;http://people.maths.ox.ac.uk/szendroi/defth.pdf&quot; rel=&quot;nofollow&quot;&gt;http://people.maths.ox.ac.uk/szendroi/defth.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[3] M. Berz, Differential Algebraic Techniques,&lt;br&gt;&#xA;in &quot;Handbook of Accelerator &#xA;Physics and Engineering, M. Tigner, A.Chao (Eds.)&quot; (World Scientific, 1998)&lt;br&gt;&#xA;&lt;a href=&quot;http://bt.pa.msu.edu/cgi-bin/display.pl?name=dahape&quot; rel=&quot;nofollow&quot;&gt;http://bt.pa.msu.edu/cgi-bin/display.pl?name=dahape&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://bt.pa.msu.edu/NA/&quot; rel=&quot;nofollow&quot;&gt;http://bt.pa.msu.edu/NA/&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://bt.pa.msu.edu/pub/papers/&quot; rel=&quot;nofollow&quot;&gt;http://bt.pa.msu.edu/pub/papers/&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-08-24T14:34:48.027" LastActivityDate="2010-08-24T14:34:48.027" CommentCount="8" />
  <row Id="3179" PostTypeId="2" ParentId="3177" CreationDate="2010-08-24T04:33:04.377" Score="16" ViewCount="" Body="&lt;p&gt;The centroid of a polygon is indeed its center of mass -- but the mass of a polygon is uniformly distributed over its surface, &lt;em&gt;not&lt;/em&gt; only at the vertices. You're right that if the mass were split evenly among the vertices only, the centroid would be the arithmetic mean of the coordinates of the vertices.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It just so happens that both definitions are equivalent (mass evenly distributed over the surface vs mass at the vertices only) for simple shapes like triangles and rectangles.&lt;/p&gt;&#xA;" OwnerUserId="1356" LastActivityDate="2010-08-24T04:33:04.377" CommentCount="2" />
  <row Id="3180" PostTypeId="2" ParentId="3170" CreationDate="2010-08-24T05:36:18.693" Score="5" ViewCount="" Body="&lt;p&gt;The short answer is that quasi-compactness is enough (for the statement you asked about): see lemma 3.1 in &lt;a href=&quot;http://www.math.columbia.edu/algebraic_geometry/stacks-git/coherent.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.math.columbia.edu/algebraic_geometry/stacks-git/coherent.pdf&lt;/a&gt; (but its essentially the argument in Hartshorne)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;this business of extra hypothesis comes up because Hartshorne and Gronthendieck are proving an iff statement; that is, these hypothesis are needed to prove that if $X$ is affine then $H^1(X,F) = 0$ for every quasicoherent sheaf.  In the case of Hartshorne you need notherian hypothesis to prove &lt;/p&gt;&#xA;&#xA;&lt;p&gt;lemma II.3.3 If $I$ is an injective $A$-module and $f \in A$ then $I \to I_f$ is surjective.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;with this lemma you can then show if $I$ is injective then $\tilde I$ is flasque an so you can use them to calculate cohomology: $F$ is quasicoherent $\Rightarrow F = \widetilde{M}$.  If $M \to I_0 \to I_1 \to ...$ (1) is an injective resolution then $\widetilde{M} \to \widetilde{I_0} \to \widetilde{I_1} \to ...$ is flasque resolution and applying global sections just recovers (1) so all the higher cohomology vanishes.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not familiar with Gronthendieck's argument and I don't know if you can replace separated with quasi-separated.&lt;/p&gt;&#xA;" OwnerUserId="403" LastEditorUserId="403" LastEditorDisplayName="" LastEditDate="2010-08-24T06:00:11.900" LastActivityDate="2010-08-24T06:00:11.900" CommentCount="1" />
  <row Id="3181" PostTypeId="2" ParentId="3173" CreationDate="2010-08-24T05:41:25.393" Score="1" ViewCount="" Body="&lt;p&gt;The natural isomorphism between a finite-dimensional space and its double-dual.&lt;/p&gt;&#xA;" OwnerUserId="589" LastActivityDate="2010-08-24T05:41:25.393" CommunityOwnedDate="2010-08-24T05:41:25.393" />
  <row Id="3182" PostTypeId="2" ParentId="3177" CreationDate="2010-08-24T05:47:07.537" Score="3" ViewCount="" Body="&lt;p&gt;Laurent is right, a polygon is not just the vertices, but the whole region. The arithmetic mean of the vertices would give the centroid if the (equal) masses were concentrated at the vertices. While that answers your question, perhaps for the future, the below might be useful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In case of a mass distributed over a region (not just a polygon), &lt;a href=&quot;http://en.wikipedia.org/wiki/Green%27s_theorem&quot; rel=&quot;nofollow&quot;&gt;Green's Theorem&lt;/a&gt; might be helpful in calculating the area and the centroid:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/5ZxGv.png&quot; alt=&quot;Green's Theorem&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your case,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $\displaystyle D$ is the region of the polygon, then the x-coordinate of the center of mass is given by the area of the polygon times&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle \iint_{D} x dxdy$ which by Green's theorem is same as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle \oint_{C} \frac{x^2}{2} dy$, where the line integral is taken over the perimeter of the polygon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For centroid we choose $\displaystyle M(x,y) = \frac{x^2}{2}$ and $\displaystyle L(x,y) = 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The area of the polygon is given by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle \iint_{D} 1 dxdy$ and can be written as a line integral.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For area we choose $\displaystyle M(x,y) = \frac{x}{2}$ and $\displaystyle L(x,y) = \frac{-y}{2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For more details with the computation details worked out, see the following webpage: &lt;a href=&quot;http://www.worldserver.com/turk/computergraphics/Moments.pdf&quot; rel=&quot;nofollow&quot;&gt;Computing the moments of a 2D polygon&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-11-11T00:13:28.900" LastActivityDate="2010-11-11T00:13:28.900" CommentCount="2" />
  <row Id="3183" PostTypeId="1" AcceptedAnswerId="3221" CreationDate="2010-08-24T07:34:57.887" Score="4" ViewCount="169" Body="&lt;p&gt;Here algebraic group means affine algebraic group in both instances.  Also I'm mainly interested in groups over $\mathbb{C}$.  In fact I'm taking $\pi_1(G)$ to mean the fundamental group of $G_{an}$, the analytification.  So I guess my question only applies to the base field being either $\mathbb{C}$ or $\mathbb{R}$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case these groups are also Lie groups with Lie algebras.  If $\mathfrak{g}$ is a semisimple Lie algebra then there is a connection with the weight and root lattices: there is a 1-1 correspondence between connected Lie groups with Lie algebra $G$ and lattices $\Lambda$ with $\Lambda_W \supset \Lambda \supset \Lambda_R$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The group corresponding to $\Lambda = \Lambda_R$ is always algebraic because it is the adjoint group.  A slightly more general question is then, for $\Lambda_W \supset \Lambda \supset \Lambda_R$ is the corresponding group $G_\Lambda$ affine algebraic?  &lt;/p&gt;&#xA;" OwnerUserId="403" LastActivityDate="2010-08-24T18:20:13.960" Title="Is the universal cover of an algebraic group an algebraic group?" Tags="&lt;algebraic-geometry&gt;&lt;algebraic-groups&gt;&lt;lie-groups&gt;" AnswerCount="2" CommentCount="0" />
  <row Id="3185" PostTypeId="1" AcceptedAnswerId="3188" CreationDate="2010-08-24T08:51:19.890" Score="6" ViewCount="245" Body="&lt;p&gt;&lt;a href=&quot;http://math.stackexchange.com/questions/3163/algebraic-integers-in-a-cyclotomic-field&quot;&gt;This question&lt;/a&gt; made me realize I had a misconception about the cyclotomic integers: I thought the units were exactly the roots of unity. There are only finitely many units but infinitely many integers so the question is impossible to solve unless there are more units. So what are the units of cyclotomic integers?&lt;/p&gt;&#xA;" LastEditorUserId="8575" LastEditorDisplayName="" LastEditDate="2011-05-16T09:11:02.373" LastActivityDate="2011-05-16T09:11:02.373" Title="What are the units of cyclotomic integers?" Tags="&lt;number-theory&gt;&lt;abstract-algebra&gt;&lt;algebraic-number-theory&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="3" />
  <row Id="3186" PostTypeId="1" AcceptedAnswerId="3187" CreationDate="2010-08-24T09:02:21.577" Score="1" ViewCount="175" Body="&lt;p&gt;From &lt;a href=&quot;http://en.wikipedia.org/wiki/Einstein_notation&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;First, we can use Einstein notation in&#xA;  linear algebra to distinguish easily&#xA;  between vectors and covectors: upper&#xA;  indices are used to label components &#xA;  (coordinates) of vectors, while lower&#xA;  indices are used to label components&#xA;  of covectors. However, vectors&#xA;  themselves (not their components) have&#xA;  lower indices, and covectors have&#xA;  upper indices.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am trying to read the Wikipedia article, but I am constantly getting confused between what represents a vector/covector and what represents a component of one of these. How can I tell?&lt;/p&gt;&#xA;" OwnerUserId="123" LastEditorUserId="171" LastEditorDisplayName="" LastEditDate="2010-08-24T09:15:34.567" LastActivityDate="2010-08-24T09:15:34.567" Title="Einstein notation - difference between vectors and scalars" Tags="&lt;linear-algebra&gt;&lt;notation&gt;&lt;tensors&gt;" AnswerCount="1" />
  <row Id="3187" PostTypeId="2" ParentId="3186" CreationDate="2010-08-24T09:11:54.077" Score="4" ViewCount="" Body="&lt;p&gt;A vector component is always written with 1 upper index $a^i$, while a covector component is written with 1 lower index $a_i$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In Einstein notation, if the same index variable appear in both upper and lower positions, an implicit summation is applied, i.e.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ a_i b^i = a_1 b^1 + a_2 b^2 + \dotsb \qquad (*) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, a vector is constructed from its component as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \mathbf a = a^1 \mathbf{\hat e}_1 + a^2 \mathbf{\hat e}_2 + \dotsb $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\mathbf{\hat e}_i$ are the basis vectors. But this takes the form like (*), so if we make basis vectors to take lower indices, we will get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \mathbf a = a^i \mathbf{\hat e}_i $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is likely what Wikipedia means.&lt;/p&gt;&#xA;" OwnerUserId="171" LastActivityDate="2010-08-24T09:11:54.077" CommentCount="2" />
  <row Id="3188" PostTypeId="2" ParentId="3185" CreationDate="2010-08-24T09:38:54.397" Score="6" ViewCount="" Body="&lt;p&gt;I don't think there's a straightforward way to describe all the units. If $\zeta$ is a root of unity in the field, then $\frac{\zeta^k-1}{\zeta-1}$ is a unit whenever $k$ is relatively prime to the order of $\zeta$, and therefore products of such elements are units as well. This already gives you infinitely many units, and though they don't necessarily cover all units they do come close (this is a finite-index subgroup of the group of units). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Look up Kummer's Lemma, for instance &lt;a href=&quot;http://www.math.uconn.edu/~kconrad/blurbs/gradnumthy/kummer.pdf&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;, and &quot;&lt;a href=&quot;http://en.wikipedia.org/wiki/Cyclotomic_unit&quot; rel=&quot;nofollow&quot;&gt;Cyclotomic unit&lt;/a&gt;&quot;.&lt;/p&gt;&#xA;" OwnerUserId="308" LastEditorUserId="308" LastEditorDisplayName="" LastEditDate="2010-08-24T18:38:36.347" LastActivityDate="2010-08-24T18:38:36.347" CommentCount="2" />
  <row Id="3190" PostTypeId="2" ParentId="3185" CreationDate="2010-08-24T09:59:29.370" Score="3" ViewCount="" Body="&lt;p&gt;Maybe this &quot;answer&quot; will help you to clarify the concepts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1:By definition, units in a given number field $K$ are the &quot;norm one&quot; integers. That is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$O_K^*$={$ u\in O_K|\ |N_{K|\mathbb{Q}}(u)|=1 $}.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2:For the structure of the group of units, one has the famous &lt;a href=&quot;http://en.wikipedia.org/wiki/Dirichlet%27s_unit_theorem&quot; rel=&quot;nofollow&quot;&gt;Dirichlet's unit theorem&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3:However, even in the case of cyclotomic fields, there is no explicit formula to produce the whole group of units. But we can describe its subgroup: &lt;a href=&quot;http://planetmath.org/encyclopedia/IndexOfTheGroupOfCyclotomicUnitsInTheFullUnitGroup.html&quot; rel=&quot;nofollow&quot;&gt;&quot;cyclotomic units&quot;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="866" LastActivityDate="2010-08-24T09:59:29.370" />
  <row Id="3191" PostTypeId="2" ParentId="3173" CreationDate="2010-08-24T11:29:05.967" Score="2" ViewCount="" Body="&lt;p&gt;Isomorphism between the &lt;a href=&quot;http://en.wikipedia.org/wiki/File%3aRiemann_surface_log.jpg&quot; rel=&quot;nofollow&quot;&gt;infinite spiral given by&lt;/a&gt; Riemann surface of $\log(z)$ and the punctured sphere.&lt;/p&gt;&#xA;" OwnerUserId="1310" LastActivityDate="2010-08-24T11:29:05.967" CommunityOwnedDate="2010-08-24T11:29:05.967" />
  <row Id="3192" PostTypeId="1" AcceptedAnswerId="3202" CreationDate="2010-08-24T11:37:08.740" Score="3" ViewCount="159" Body="&lt;p&gt;I am aware of gauge transformations and covariant derivatives as understood in Quantum Field Theory and I am also familiar with deRham derivative for vector valued differential forms.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I thinking of the gauge field A of the gauge group G as a Lie(G) valued 1-form on the manifold. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I can't see why under a gauge transformation on A by an element $g\in G$ amounts to the following change, $A \mapsto A^g = gAg^{-1} -dgg^{-1}$ (if G is thought of as a matrix Lie Group) or in general $A_g = Ad(g)A + g^* \omega$  (where $\omega$ is the left invariant Maurer-Cartan form on G and I guess $g^*$ is pull-back of $\omega$ along left translation map by $g$). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Curvature is defined as $F = dA + \frac{1}{2}[A,A]$ and using this one wants to now see why does $F \mapsto F_g = gFg^{-1}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Firstly is the expression for $A_g$ a definition or is there a derivation for that?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I try proving this (assuming matrix Lie groups) I am getting stuck in multiple places like what is $dA_g$ ? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would be happy if someone can explain the explicit calculations and/or give a reference where such things are explained. Usual books which explain differential forms or connections on principal bundles don't seem to help with such calculations. &lt;/p&gt;&#xA;" OwnerUserId="965" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2010-08-24T17:23:54.667" LastActivityDate="2010-08-25T08:36:17.133" Title="Gauge transformations in differential forms." Tags="&lt;differential-geometry&gt;" AnswerCount="1" FavoriteCount="0" />
  <row Id="3193" PostTypeId="2" ParentId="3173" CreationDate="2010-08-24T11:47:05.577" Score="0" ViewCount="" Body="&lt;p&gt;To follow previous answers, I would say the isomorphism between a finite-dimensional vector space and its dual, even though it isn't natural (&lt;a href=&quot;http://en.wikipedia.org/wiki/Dual_space#CITEREFMacLaneBirkhoff1999&quot; rel=&quot;nofollow&quot;&gt;MacLane &amp;amp; Birkhoff 1999&lt;/a&gt;, §VI.4, from Wikipedia).&lt;/p&gt;&#xA;" OwnerUserId="1415" LastActivityDate="2010-08-24T11:47:05.577" CommunityOwnedDate="2010-08-24T11:47:05.577" />
  <row Id="3194" PostTypeId="1" CreationDate="2010-08-24T12:08:06.877" Score="1" ViewCount="151" Body="&lt;p&gt;Can anyone kindly give some reference on taking trace of vector valued differential forms?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Like if $A$ and$B$ are two vector valued forms then I want to understand how/why this equation is true?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$dTr(A\wedge B) = Tr(dA\wedge B) - Tr(A\wedge dB)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One particular case in which I am interested in will be when $A$ is a Lie Algebra valued one-form on some 3-manifold. Then I would like to know what is the precise meaning/definition of $Tr(A)$ or $Tr(A\wedge dA)$ or $Tr(A\wedge A \wedge A)$? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In how general a situation is a trace of a vector valued differential form defined? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It would be great if someone can give a local coordinate expression for such traces. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any references to learn this would be of great help. &lt;/p&gt;&#xA;" OwnerUserId="965" LastEditorUserId="965" LastEditorDisplayName="" LastEditDate="2010-08-24T19:57:58.287" LastActivityDate="2010-08-24T19:57:58.287" Title="Taking trace of vector valued differential forms" Tags="&lt;differential-geometry&gt;" AnswerCount="1" CommentCount="6" FavoriteCount="1" />
  <row Id="3195" PostTypeId="2" ParentId="3194" CreationDate="2010-08-24T12:23:05.983" Score="2" ViewCount="" Body="&lt;p&gt;I do not know what you mean by vector-valued form, exactly. But your equation follows from two facts:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$d$ satisfies the (graded) Leinbiz equation $$d(A\wedge B) = dA\wedge B + (-1)^{\text{degree of $A$}}A\wedge dB,$$&lt;/li&gt;&#xA;&lt;li&gt;and $\mathrm{Tr}$ is linear and commutes with $d$, so that $$d\,\mathrm{Tr}(A)=\mathrm{Tr}\,dA.$$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="274" LastActivityDate="2010-08-24T12:23:05.983" CommentCount="5" />
  <row Id="3196" PostTypeId="2" ParentId="3183" CreationDate="2010-08-24T12:39:45.100" Score="4" ViewCount="" Body="&lt;p&gt;For groups over $\mathbb{R}$ the answer is no.  The group of real points of an affine algebraic group has a faithful finite-dimensional linear representation over $\mathbb{R}$, but it is well-known that the universal cover of $\text{SL}_2(\mathbb{R})$ has no faithful finite-dimensional linear representations (unfortunately I don't know a reference for this).&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-08-24T16:30:05.880" LastActivityDate="2010-08-24T16:30:05.880" CommentCount="4" />
  <row Id="3197" PostTypeId="1" CreationDate="2010-08-24T12:42:00.503" Score="5" ViewCount="139" Body="&lt;p&gt;If G is a simply connected Lie Group then why is every G-bundle over an orientable 3-manifold trivial? (Why is orientability important?) &lt;/p&gt;&#xA;" OwnerUserId="965" LastActivityDate="2010-08-24T13:40:16.203" Title="Principal bundles on 3-manifolds" Tags="&lt;differential-geometry&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="3198" PostTypeId="1" CreationDate="2010-08-24T13:39:23.017" Score="6" ViewCount="691" Body="&lt;p&gt;Is there a proof that the ratio of a circle's diameter and the circumference is the same for all circles, that doesn't involve some kind of limiting process, e.g. a direct geometrical proof?&lt;/p&gt;&#xA;" OwnerUserId="1470" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-08-24T19:33:09.483" LastActivityDate="2010-08-24T20:15:12.113" Title="Proof that Pi is constant (the same for all circles), without using limits" Tags="&lt;geometry&gt;&lt;pi&gt;" AnswerCount="3" CommentCount="16" FavoriteCount="1" />
  <row Id="3199" PostTypeId="2" ParentId="3197" CreationDate="2010-08-24T13:40:16.203" Score="1" ViewCount="" Body="&lt;p&gt;Are you sure that orientability is necessary? The result is proved in lemma 4.1.1.  &lt;a href=&quot;http://www.math.uni-bonn.de/people/himpel/himpel_cstheory.pdf&quot; rel=&quot;nofollow&quot; title=&quot;here&quot;&gt;here&lt;/a&gt; and I do not see where orientability is used.&lt;/p&gt;&#xA;" LastActivityDate="2010-08-24T13:40:16.203" />
  <row Id="3200" PostTypeId="2" ParentId="3198" CreationDate="2010-08-24T13:52:18.240" Score="2" ViewCount="" Body="&lt;p&gt;Intuitively, all circles are similar and therefore doubling the diameter also doubles the circumference.  The same applies to ratios other than 2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To make this rigorous, we have to consider what we mean by “the length of the circumference.”  The usual rigorous definition uses integration and therefore relies on the notion of limits.  I guess that any rigorous definition of the length of a curve ultimately requires the notion of limits.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: Rephrased a little to make the connection between the two paragraphs clearer.&lt;/p&gt;&#xA;" OwnerUserId="926" LastEditorUserId="926" LastEditorDisplayName="" LastEditDate="2010-08-24T14:07:52.803" LastActivityDate="2010-08-24T14:07:52.803" CommentCount="10" />
  <row Id="3201" PostTypeId="2" ParentId="3198" CreationDate="2010-08-24T15:22:00.390" Score="1" ViewCount="" Body="&lt;p&gt;It seems -as far as I can understand what he was doing- that even Euclid used some sort of limiting process (the principle of exhaustion): &lt;a href=&quot;http://aleph0.clarku.edu/~djoyce/java/elements/bookXII/propXII2.html&quot; rel=&quot;nofollow&quot;&gt;http://aleph0.clarku.edu/~djoyce/java/elements/bookXII/propXII2.html&lt;/a&gt; . What Euclid is proving here is the following: let $d_1$, $d_2$ be the diameters of two circles and $A_1$, $A_2$ their areas. Then &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\frac{A_1} {A_2} = \frac{d_1^2}{d_2^2} \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which is the same as saying that the proportion between the area of a circle and the square of its radius is constant: if $r_1$ and $r_2$ are the radii of our circles&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\frac{A_1}{d_1^2} = \frac{A_2}{d_2^2} \quad \Longleftrightarrow \quad \frac{A_1}{r_1^2} = \frac{A_2}{r_2^2}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is to say, &quot;$\pi$ is constant&quot;.&lt;/p&gt;&#xA;" OwnerUserId="664" LastActivityDate="2010-08-24T15:22:00.390" />
  <row Id="3202" PostTypeId="2" ParentId="3192" CreationDate="2010-08-24T15:24:29.700" Score="1" ViewCount="" Body="&lt;p&gt;The method is to use the Leibniz rule in the differentiation and and change the sign whenever the exterior derivative moves past an odd form. In addition the following identity must be used $ dgg^{-1} + g dg^{-1} = 0$, and remember that the commutator is is between odd forms thus it is with a plus sign.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are the intermediate results:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{1}{2}[A_g, A_g] = \frac{1}{2}g[A, A] g^{-1} -[dgg^{-1}, g A g^{-1}] + dgg^{-1}\wedge dgg^{-1}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$dA_g = g dA g^{-1} + [dgg^{-1}, g A g^{-1}] - dgg^{-1}\wedge dgg^{-1}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are the required details:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ d(gAg^{-1})$&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Application of the Leibniz rule (Please observe the minus sign in the last term)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;$ d(gAg^{-1}) = dg \wedge A g^{-1} + g dA g^{-1} - g A \wedge dg^{-1}$&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Using the identities $g g^{-1} = 1$ in the first term and $ dg^{-1} = - g^{-1}dg g^{-1} $&#xA;in the last term&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;$ = dg g^{-1} g \wedge A g^{-1} + g dA g^{-1} +g A g^{-1} \wedge dg g^{-1} $ &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Collection of the first and last term into a commutator:&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;$  = g dA g^{-1} +[ dg g^{-1}, g A g^{-1} ]$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ d(dg g^{-1})$ &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Application of the Leibniz rule (Please observe the minus sign in the last term)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;$ d(dg g^{-1}) = ddg g^{-1} - dg \wedge dg^{-1}$ &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Using the identities $dd = 0$ and again $ dg^{-1} = - g^{-1}dg g^{-1} $, we obtain:&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;$ d(dg g^{-1}) = + dg g^{-1}\wedge dg g^{-1}$ &lt;/p&gt;&#xA;" OwnerUserId="1430" LastEditorUserId="1430" LastEditorDisplayName="" LastEditDate="2010-08-25T08:36:17.133" LastActivityDate="2010-08-25T08:36:17.133" CommentCount="5" />
  <row Id="3203" PostTypeId="1" AcceptedAnswerId="3206" CreationDate="2010-08-24T15:38:00.727" Score="4" ViewCount="336" Body="&lt;p&gt;Does a limit that has the value of infinite exist or not?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've recently come across certain sources that say that if the value of a limit is infinite, then that limit does not exist. This contradicts what my calculus teachers and lecturers taught me however, that a limit doesn't exist if the right hand limit and left hand limit differ. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So which one is it? &lt;/p&gt;&#xA;" OwnerUserId="456" LastEditorUserId="171" LastEditorDisplayName="" LastEditDate="2010-08-24T15:43:52.223" LastActivityDate="2010-08-25T02:53:40.783" Title="Infinite limits" Tags="&lt;calculus&gt;&lt;definition&gt;&lt;limit&gt;" AnswerCount="6" CommentCount="1" FavoriteCount="1" />
  <row Id="3204" PostTypeId="2" ParentId="3203" CreationDate="2010-08-24T15:43:19.213" Score="1" ViewCount="" Body="&lt;p&gt;The language use in my calculus class was that this is an &lt;em&gt;improper limit.&lt;/em&gt; So that strictly speaking there is no limit in the proper sense.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The criterion with right hand limits and left hand limits works at points really belonging to the real line (and then first of all both one-sided limits have to exist for themselves).&lt;/p&gt;&#xA;" OwnerUserId="367" LastActivityDate="2010-08-24T15:43:19.213" />
  <row Id="3205" PostTypeId="1" AcceptedAnswerId="3207" CreationDate="2010-08-24T15:52:03.503" Score="1" ViewCount="65" Body="&lt;p&gt;We have a set of items $I = \{i_1, i_2, ..., i_n\}$.  Each of these items has what we call a &lt;em&gt;p&lt;/em&gt; value, which is some real number.  We want to choose a subset of $I$, call it $I'$, of size $m$ (for some $m$ with $1 \le m \le n$) such that the average of the &lt;em&gt;p&lt;/em&gt; values of the items in $I'$ falls within some specified range, $[p_l, p_u]$.  (For example, we might require an average &lt;em&gt;p&lt;/em&gt; value between 0.70 and 0.74.)  Moreover, we want to do this efficiently.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We hope to do this in O(n) time, but any polynomial time algorithm is good enough.  We certainly do not want to just try every possible subset of I of size $m$ and then check whether it satisfies the average p-value constraint.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, we will be doing this repeatedly and we want the subsets chosen to be a uniformly random distribution over all the possible such subsets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a way of doing this?&lt;/p&gt;&#xA;" OwnerUserId="984" LastEditorUserId="171" LastEditorDisplayName="" LastEditDate="2010-08-24T16:07:01.457" LastActivityDate="2010-08-24T16:10:24.967" Title="Choosing subsets of a set such that the subsets satisfy a global constraint" Tags="&lt;combinatorics&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="3206" PostTypeId="2" ParentId="3203" CreationDate="2010-08-24T15:57:43.333" Score="6" ViewCount="" Body="&lt;p&gt;Obviously it depends on the definition of &quot;exists&quot;. Some authors explicitly work over the extended real line with $\pm\infty$ adjoined, so that such infinite limits do explicitly &quot;exist&quot; as first-class values. But there is no consensus. One needs to pay attention to the author's definitions and conventions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps it is worth mention - even though this case is rather trivial - that adjoining points at infinity is a special case of various constructions that attempt to simplify matters by some type of existential closure. Below I append an excerpt from an old &lt;a href=&quot;http://google.com/groups?selm=y8zohi4mrfs.fsf_-_@martigny.ai.mit.edu&quot; rel=&quot;nofollow&quot;&gt;1996 post&lt;/a&gt; where I give a little further discussion:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This thread originated in a query as to whether infinity or 1/0 &#xA;could be admitted as a &quot;value&quot;, and soon drifted into discussion &#xA;of the Riemann sphere and other topological manifestations of &#xA;infinity via compactification. Below I point out a couple of &#xA;marvelous references on these topics; further I would like to &#xA;bring to your attention a much wider perspective on such topics, &#xA;namely that of existential closure as studied in model theory. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a beautiful exposition of points at infinity, projective &#xA;closure, compactifications, modifications, etc. in [FM][1] Chapter 7, &#xA;Points at Infinity, by H. Behnke and H. Grauert. This is volume III &#xA;in the excellent &quot;Fundamentals of Mathematics&quot; series,&#xA;which deserves to be on the bookshelf of every budding mathematician. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A much deeper appreciation of the methodology behind these constructions &#xA;can be had by studying them from a model-theoretic perspective, in &#xA;particular from the standpoint of existential closure and model &#xA;completion. Kenneth Manders has written a series of thought &#xA;provoking papers [2],[3] from this perspective.&#xA;I close with an excerpt from the introduction to [2]:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;The systematic adjunction of roots, or solutions to other simple &#xA;  conditions, as in formation of the complex numbers by adjoining &#xA;  imaginaries, or in adjunction of points &quot;at infinity&quot; in traditional &#xA;  geometry, may be analysed as &lt;em&gt;existential closure&lt;/em&gt; and &lt;em&gt;model &#xA;  completion&lt;/em&gt;. 'Existential closure' refers to a class of processes &#xA;  which attempt to round off a domain and simplify its theory by &#xA;  adjoining elements -- more properly, it refers to the formal &#xA;  relationship that obtains in such a process. 'Model completion' is &#xA;  one of the terms employed when this process is successful. The &#xA;  formation of the complex numbers, and the move from affine to &#xA;  projective geometry, are successes of this kind. Thus, the theory of &#xA;  existential closure gives a theoretical basis of Hilbert's &quot;method &#xA;  of ideal elements.&quot; I now sketch the theory of existential closure, &#xA;  to bring out when, how, and in what sense existential closure gives &#xA;  conceptual simplification.&quot; &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;[FM] Fundamentals of mathematics. Vol. III. Analysis.&lt;br&gt;&#xA;Edited by H. Behnke, F. Bachmann, K. Fladt and W. Suss.&lt;br&gt;&#xA;Translated from the second German edition by S. H. Gould.&lt;br&gt;&#xA;Reprint of the 1974 edition. MIT Press,&lt;br&gt;&#xA;Cambridge, Mass.-London, 1983. xiii+541 pp. ISBN: 0-262-52095-8 00A05 &lt;/p&gt;&#xA;&#xA;&lt;p&gt;[2] Manders, Kenneth&lt;br&gt;&#xA;Domain extension and the philosophy of mathematics.&lt;br&gt;&#xA;J. Philos. 86 (1989), no. 10, 553--562.&lt;br&gt;&#xA;&lt;a href=&quot;http://www.jstor.org/stable/2026666&quot; rel=&quot;nofollow&quot;&gt;http://www.jstor.org/stable/2026666&lt;/a&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;[3] Manders, Kenneth L.&lt;br&gt;&#xA;Logic and conceptual relationships in mathematics.&lt;br&gt;&#xA;Logic colloquium '85 (Orsay, 1985), 193--211,&lt;br&gt;&#xA;Stud. Logic Found. Math., 122,&lt;br&gt;&#xA;North-Holland, Amsterdam-New York, 1987.&lt;br&gt;&#xA;&lt;a href=&quot;http://dx.doi.org/10.1016/S0049-237X(09)70554-3&quot; rel=&quot;nofollow&quot;&gt;http://dx.doi.org/10.1016/S0049-237X(09)70554-3&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-08-24T17:36:14.030" LastActivityDate="2010-08-24T17:36:14.030" />
  <row Id="3207" PostTypeId="2" ParentId="3205" CreationDate="2010-08-24T16:03:59.697" Score="1" ViewCount="" Body="&lt;p&gt;Assuming $m$ is an input, your problem is NP-Complete.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Subset sum can be reduced to it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given $A = \{x_{1}, ..., x_{n}\}$ for which we need to find a subset of sum $S$, we can create $n$ inputs to your algorithm, with input $A$, $m$ and $p_{L} = p_{U} = \frac{S}{m}$, for $1 \leq m \leq n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $m$ is fixed then the brute force algorithm (which you don't want) is $O(n^m)$ and so has a polynomial time algorithm.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-08-24T16:03:59.697" />
  <row Id="3208" PostTypeId="2" ParentId="3205" CreationDate="2010-08-24T16:10:24.967" Score="0" ViewCount="" Body="&lt;p&gt;One solution I was considering was to choose items for I' successively such that the average &lt;em&gt;p&lt;/em&gt; value of the items selected so far always falls within some upper and lower bound.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously, we could just choose p_l and p_u as our lower and upper bounds, respectively, and this would satisfy the average &lt;em&gt;p&lt;/em&gt; value constraint.  But this would also be too restrictive.  It could leave out a large number of possible I' where small &lt;em&gt;p&lt;/em&gt; values might balance out high &lt;em&gt;p&lt;/em&gt; values.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I was considering letting the upper and lower bounds start out far apart when the first items are selected and then let them narrow down to p_l and p_u when the final mth item is selected.  So we would have a cone of ranges that would become narrower as more items are picked.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More precisely, we would define two functions &lt;em&gt;u&lt;/em&gt; and &lt;em&gt;l&lt;/em&gt; such that the average of the first i items picked (call this p_i) would satisfy &lt;em&gt;l&lt;/em&gt;(i) &amp;lt;= p_i &amp;lt;= &lt;em&gt;u&lt;/em&gt;(i) for 0 &amp;lt;= i &amp;lt;= m.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not really sure how to define these two functions, though.  I know that &lt;em&gt;l&lt;/em&gt;(m) = p_l and &lt;em&gt;u&lt;/em&gt;(m) = p_u, but I don't know what &lt;em&gt;l&lt;/em&gt;(0) or &lt;em&gt;u&lt;/em&gt;(0) should be.  Also, should the functions be linear?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Comments or suggestions about this possible solution are welcome.&lt;/p&gt;&#xA;" OwnerUserId="984" LastActivityDate="2010-08-24T16:10:24.967" />
  <row Id="3209" PostTypeId="2" ParentId="3203" CreationDate="2010-08-24T16:10:57.900" Score="2" ViewCount="" Body="&lt;p&gt;It is important to consider the context in which you are taking the limit.  When taking the limit of a sequence $\{ x_{n}\}$, we must consider what set the elements of that sequence are coming from.  For example, if $\{ x_{n}\} \subset \mathbb{R}$, then for the limit $L$ to exist (in $\mathbb{R}$) we must have that $L \in \mathbb{R}$.  If $\{ x_{n}\} \to \infty$ as $n \to \infty$, then the limit doesn't exist (in $\mathbb{R}$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, there is another notion of limit called $\text{lim sup}$.  According to &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0070856133&quot; rel=&quot;nofollow&quot;&gt;Baby Rudin&lt;/a&gt; (slightly modified):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;&lt;strong&gt;3.16 Definition&lt;/strong&gt; Let $\{s_{n}\}$ be a sequence of real numbers. Let $E$ be the set of numbers $x$ (in the extended real number system [which includes $\pm \infty$]) such that $s_{n_{k}} \to x$ for some subsequence $\{s_{n_k}\}$.  This set $E$ contains all subsequential limits* plus possibly the numbers $+ \infty$, $-\infty$. &#xA;&lt;br&gt;&lt;br&gt;&#xA;Now define $$\begin{align}\text{lim sup} \;\; s_{n} &amp;= \sup E, \;\;\;\;\text{and} \\\ &#xA;\text{lim inf} \;\; s_{n} &amp;= \inf E.\end{align}$$&#xA;* A subsequential limit is just the limit of the subsequence, if that subsequence converges.&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;With this definition we can discuss infinite limits (superior or inferior). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt; Removed my erroneous claim that finite $\text{lim sup}$'s and $\text{lim inf}$'s collapse to the regular old $\lim$.  (Counterexample: $\text{lim sup}_{x \to \infty} \sin(x) = 1$ but $\lim _{x \to \infty}\sin(x) \neq 1$.&lt;/p&gt;&#xA;" OwnerUserId="121" LastEditorUserId="121" LastEditorDisplayName="" LastEditDate="2010-08-24T16:28:29.933" LastActivityDate="2010-08-24T16:28:29.933" />
  <row Id="3210" PostTypeId="1" CreationDate="2010-08-24T16:12:13.577" Score="0" ViewCount="225" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Possible Duplicate:&lt;/strong&gt;&lt;br&gt;&#xA;  &lt;a href=&quot;http://math.stackexchange.com/questions/438/1-is-not-1-so-where-is-the-mistake&quot;&gt;-1 is not 1, so where is the mistake?&lt;/a&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;p&gt;I'm trying to understand the exact point of failure in the following reasoning:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$1 = \sqrt{1} = \sqrt{(-1)(-1)} = \sqrt{\sqrt{-1}^2\sqrt{-1}^2} = \sqrt{(\sqrt{-1}\sqrt{-1})^2} = \sqrt{-1}\sqrt{-1} = \sqrt{-1}^2 = -1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've been previously told that the problem is due to square root not being a function in C; which I found totally unhelpful. Could someone please explain the problem here in simpler terms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you all for your comments in trying to help me understand this. I finally do. Following is the explanation on my problems in understanding this, in case it'll be of any help to anyone else.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My problem was really due to using an incorrect definition of i: $i = \sqrt{-1}$. While the correct definition would be: $i^2 = -1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My incorrect definition led me reasoning, such as (which superficially seemed to give expected results. I see now that this is incorrect, too):&#xA;$\sqrt{-9} = \sqrt{9 * (-1)} = \sqrt{\sqrt{9}^2 \sqrt{-1}^2} = \sqrt{(\sqrt{9} \sqrt{-1})^2} = \sqrt{9} \sqrt{-1} = 3i$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instead, had I used the correct definition of i:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;${(xi)}^2 = x^2i^2 = -x^2 = -9$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x^2 = 9$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x = +- 3$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, analyzing the equations in the original problem, I can see at least the following two errors:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) In the third =, I'm relying on $-1 = {\sqrt{-1}}^2$, while I should be relying on: $-1 = (+-\sqrt{-1})^2$ which would of course give two different branches. Hmm.. on the second reading, this isn't really a problem, as even with the two separate branches, both of them will lead to the result in the next step.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) In the fifth =, I'm relying on $\sqrt{i^4} = i^2$, which would be correct, if i was a non-negative number in R. But as i is the imaginary unit and in C: $\sqrt{i^4} = \sqrt{i} = +-(1/\sqrt{2})(1 + i)$.&lt;/p&gt;&#xA;" OwnerUserId="105" LastEditorUserId="105" LastEditorDisplayName="" LastEditDate="2010-08-24T20:22:18.370" LastActivityDate="2010-08-24T20:22:18.370" ClosedDate="2010-08-24T16:53:39.307" Title="Simple Complex Number Problem: 1 = -1" Tags="&lt;complex-numbers&gt;" AnswerCount="1" CommentCount="8" />
  <row Id="3211" PostTypeId="2" ParentId="3210" CreationDate="2010-08-24T16:18:59.033" Score="1" ViewCount="" Body="&lt;p&gt;You need to pay attention to branches of multivalued functions, e.g. see the Wikipedia explanation &lt;a href=&quot;http://en.wikipedia.org/wiki/Imaginary_unit&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;. Similar less-trivial questions often arise when symbolic mathematical sotfware systems exhibit bugs due to failure to stay on principal branches, e.g. see this &lt;a href=&quot;http://groups.google.com/group/sci.math.symbolic/browse_frm/thread/9c0201d4dda46458&quot; rel=&quot;nofollow&quot;&gt;thread&lt;/a&gt; where John McKay asks what your favorite system returns for $(-1)^{5/9} - (-1)^{2/9} - (-1)^{8/9}$. You may find such discussions instructive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the reader who may be interested in algorithms see for example&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thomas Breuer. sam@math.rwth-aachen.de&lt;br&gt;&#xA;&lt;strong&gt;Integral Bases for Subfields of Cyclotomic Fields.&lt;/strong&gt;&#xA;AAECC 8, 1997, 279-289&#xA;&lt;a href=&quot;http://www.springerlink.com/content/qnuu0knap4fl3kjt/fulltext.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.springerlink.com/content/qnuu0knap4fl3kjt/fulltext.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Abstract&lt;/strong&gt;. Integral bases of cyclotomic fields are constructed that allow to&#xA;determine easily the smallest cyclotomic field in which a given sum of roots of&#xA;unity lies.  For subfields of cyclotomic fields integral bases are constructed&#xA;that consist of orbit sums of Galois groups on roots of unity. These bases are&#xA;closely related to the bases of the enveloping cyclotomic fields mentioned above. &#xA;In both situations bases over the rationals and over cyclotomic fields are treated.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-08-24T17:04:51.540" LastActivityDate="2010-08-24T17:04:51.540" />
  <row Id="3212" PostTypeId="2" ParentId="3203" CreationDate="2010-08-24T16:32:14.977" Score="0" ViewCount="" Body="&lt;p&gt;It depends on what you are working with. For example, one definition of the derivative of a function that the limit: $\lim_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}$ exists and it is finite. On the other hand, if you're talking about the limit of a sequence ${ x_n }$, then you may want to distinguish between cases when the limit is infinite, or it doesn't exist, which means, like you said, that taken two subsequences you get different values for the limit -in this case you say that the sequence oscillates-. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Definitions, of course, solve a lot of problems, you define in $\mathbb{R}$ $\lim_{x \rightarrow \infty} f(x) = \infty$ as follows: $\forall N &gt;0, \; \exists M &gt;0: \; x &gt; M \Rightarrow f(x) &gt; N$, then you don't really have to call in the extended number line.&lt;/p&gt;&#xA;" OwnerUserId="1415" LastEditorUserId="1415" LastEditorDisplayName="" LastEditDate="2010-08-24T17:12:54.333" LastActivityDate="2010-08-24T17:12:54.333" CommentCount="2" />
  <row Id="3214" PostTypeId="2" ParentId="3101" CreationDate="2010-08-24T17:04:53.583" Score="5" ViewCount="" Body="&lt;p&gt;An equivalent way to state the product rule is $\frac{(fg)'}{fg} = \frac{f'}{f} + \frac{g'}{g}$.  I prefer this statement because it is more intuitive: it says precisely that the &lt;em&gt;relative&lt;/em&gt; instantaneous change in $fg$ is the sum of the relative instantaneous change in $f$ and the relative instantaneous change in $g$.  In other words, this is just an expression of the approximation $(1 + a)(1 + b) \approx 1 + a + b$ when $a, b$ are both small (&quot;multiplication near the identity is addition&quot;).  In other other words, this is an expression of the familiar fact that if you increase something by 5%, then by 5% again, the total increase is just a little more than 10%.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for turning this into a formal proof, divide both sides of the expression you're confused about by $fg$.  (This is not strictly valid if $f$ or $g$ is equal to zero, but I'm going for clarity here.)&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-08-24T17:04:53.583" CommentCount="1" />
  <row Id="3215" PostTypeId="1" AcceptedAnswerId="3220" CreationDate="2010-08-24T17:46:48.180" Score="11" ViewCount="499" Body="&lt;p&gt;Consider the sequence defined as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x_1 = 1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x_{n+1} = \sin x_n$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think I was able to show that the sequence $\sqrt{n} x_{n}$ converges to $\sqrt{3}$ by a tedious elementary method which I wasn't too happy about.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I think I did this by showing that $\sqrt{\frac{3}{n+1}} &amp;lt; x_{n} &amp;lt; \sqrt{\frac{3}{n}}$, don't remember exactly)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This looks like it should be a standard problem. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anyone know a simple (and preferably elementary) proof for the fact that the sequence $\sqrt{n}x_{n}$ converges to $\sqrt{3}$?&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-08-24T17:54:50.097" LastActivityDate="2010-08-25T16:33:37.837" Title="Convergence of $\sqrt{n}x_{n}$ where $x_{n+1} = \sin(x_{n})$" Tags="&lt;sequences-and-series&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="4" />
  <row Id="3216" PostTypeId="2" ParentId="3198" CreationDate="2010-08-24T17:58:10.050" Score="3" ViewCount="" Body="&lt;p&gt;Limits are not involved in the problem of proving that $\pi(C)$ is independent of the circle $C$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In geometrical definitions of $\pi$, to a circle $C$ is associated a sequence of finite polygonal objects and thus a sequence of numbers (or lengths, or areas, or ratios of those) $\pi_k(C)$.  This sequence is thought of as a set of approximations converging to $\pi$, but that doesn't concern us here; what is important is that the sequence is &lt;em&gt;independent of the circle C&lt;/em&gt;.  Any further aspects of the sequence such as its limit or the rate of convergence will also be the same for any two circles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(edit: an example of a &quot;geometrical definition&quot; of a sequence of approximants $\pi_k(C)$ is: perimeter of a regular $k$-sided polygon inscribed in circle C, divided by the diameter of C.  Also, the use of words like &lt;em&gt;limit&lt;/em&gt; and &lt;em&gt;approximation&lt;/em&gt; above does not reflect any assumption that the sequences have limits or that an environment involving limits has been set up. We are demonstrating that if $\pi(C)$ is defined using some construction on the sequence, then whether that construction involves limits or not, it must produce the same answer for any two circles.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The proof that $\pi_k(C_1) = \pi_k(C_2)$  of course would just apply the similarity of polygons and the behavior of length and area with respect to changes of scale.  This argument does not assume a limit-based theory of length and area, because the theory of length and area for polygons in Euclidean geometry only requires dissections and rigid motions (&quot;cut-and-paste equivalence&quot; or &lt;em&gt;equidecomposability&lt;/em&gt;).  Any polygonal arc or region can be standardized to an interval or square by a finite number of (area and length preserving) cut-and-paste dissections.  Numerical calculations involving the $\pi_k$, such as ratios of particular lengths or areas, can be understood either as applying to equidecomposability classes of polygons, or to the standardizations.  In both interpretations, due to the similitude, the results will be the same for $C_1$ and $C_2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(You might think that this is proving a different conclusion, that the equidecomposability version of $\pi$ for the two circles is equal, and not the numerical equality of $\pi$ within a theory that has real numbers as lengths and areas for arbitrary curved figures.  However, any real number-based theory, including elementary calculus, Jordan measure, and Lebesgue measure, is set up with a minimum requirement of compatibility with the geometric operations of dissection and rigid motion, so once equidecomposability is known, numerical equality will also follow.)   &lt;/p&gt;&#xA;" OwnerUserId="467" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-08-24T20:15:12.113" LastActivityDate="2010-08-24T20:15:12.113" CommentCount="15" />
  <row Id="3217" PostTypeId="1" AcceptedAnswerId="3224" CreationDate="2010-08-24T17:59:50.733" Score="1" ViewCount="193" Body="&lt;p&gt;$$ \frac{4^n-1}{n-1} &gt; 4 &amp;times; 10^6 $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is an easy way to solve it by hand? I need to find the minimum integral value of &lt;code&gt;n&lt;/code&gt; to satisfy this equation.&lt;/p&gt;&#xA;" OwnerUserId="324" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-08-28T17:28:11.747" LastActivityDate="2010-08-28T17:28:11.747" Title="How to solve this equation?" Tags="&lt;calculus&gt;&lt;algebra-precalculus&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="3218" PostTypeId="2" ParentId="3101" CreationDate="2010-08-24T18:05:51.363" Score="3" ViewCount="" Body="&lt;p&gt;This answer is really very similar to the others, but I hope that&#xA;the emphasis is sufficiently different to be worth posting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fix a number $x$. The &lt;strong&gt;definition&lt;/strong&gt; of the derivative of $f$ is that&#xA;$$f'(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}.$$&#xA;Let's rewrite this as&#xA;$$f'(x)=\lim_{h\to 0}F(h)$$&#xA;where we define&#xA;$$F(h)=\frac{f(x+h)-f(x)}{h}$$&#xA;for $h\ne0$. We can rewrite this definition as&#xA;$$f(x+h)=f(x)+hF(h).$$&#xA;For the same reason,&#xA;$$g(x+h)=g(x)+hG(h)$$&#xA;where&#xA;$$G(h)=\frac{g(x+h)-g(x)}{h}$$&#xA;and also&#xA;$$g'(x)=\lim_{h\to 0}G(h).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We now need to investigate the derivative of $f(x)g(x)$. Now by definition&#xA;this is&#xA;$$\lim_{h\to0}\frac{f(x+h)g(x+h)-f(x)g(x)}{h}.$$&#xA;Manipulating this quotient gives&#xA;$$\frac{f(x+h)g(x+h)-f(x)g(x)}{h}=\frac{(f(x)+hF(h))(g(x)+hG(h))-f(x)g(x)}{h}$$&#xA;$$=\frac{hf(x)G(h)+hF(h)g(x)+h^2F(h)G(h)}{h}=f(x)G(h)+F(h)g(x)+hF(h)G(h).$$&#xA;Now this yields to the &quot;algebra of limits&quot;: using $\lim_{h\to 0}F(h)=f'(x)$&#xA;and $\lim_{h\to 0}G(h)=g'(x)$ we get&#xA;$$\lim_{h\to0}\frac{f(x+h)g(x+h)-f(x)g(x)}{h}=f(x)g'(x)+f'(x)g(x)+0f'(x)g'(x)$$&#xA;which simplifies to $f(x)g'(x)+f'(x)g(x)$ which is exactly what you want.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Actually this argument is just the same as your original but expressed&#xA;in a more pedestrian way, without any cunning sleights of hand.&#xA;As  follow-up a good exercise is to obtain the quotient rule&#xA;in a similar fashion.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-08-24T18:05:51.363" />
  <row Id="3219" PostTypeId="2" ParentId="3101" CreationDate="2010-08-24T18:07:47.240" Score="2" ViewCount="" Body="&lt;p&gt;Here is an answer based on Carathéodory's formulation of differentiability. Let $a\in A\subset\mathbb R$ and $f:A\to\mathbb R$. Assume there a sequence in $A\setminus\{a\}$ converging to $a$. Then there is at most one function $\varphi:A\to\mathbb R$ which is continuous at $a$ and satisfies $$f(x)=f(a)+(x-a)\ \varphi(x)$$ for all $x$ in $A$. If such a $\varphi$ exists, we say that $f$ is &lt;strong&gt;differentiable at&lt;/strong&gt; $a$, and we put $f'(a):=\varphi(a)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's easy to see that this definition is equivalent to the usual one, and, using it, it's easy to verify that the product of two functions differentiable at $a$ is differentiable at $a$ and satisfies the famous formula for the derivative of a product. (Same thing for the Chain Rule.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;See &quot;The Derivative à la Carathéodory&quot;, Stephen Kuhn, &lt;em&gt;The American Mathematical Monthly&lt;/em&gt;, Vol. 98, No. 1 (Jan., 1991), pp. 40-44: &lt;a href=&quot;http://www.jstor.org/pss/2324035&quot; rel=&quot;nofollow&quot;&gt;http://www.jstor.org/pss/2324035&lt;/a&gt; or &lt;a href=&quot;http://www.mediafire.com/?xyl23spk3ui0cga&quot; rel=&quot;nofollow&quot;&gt;http://www.mediafire.com/?xyl23spk3ui0cga&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="660" LastEditorUserId="660" LastEditorDisplayName="" LastEditDate="2010-08-26T06:48:52.990" LastActivityDate="2010-08-26T06:48:52.990" CommentCount="2" />
  <row Id="3220" PostTypeId="2" ParentId="3215" CreationDate="2010-08-24T18:09:24.653" Score="16" ViewCount="" Body="&lt;p&gt;Before getting into the details, let me say: The ideas I'm talking about, including this exact exapmle, can be found in chapter 8 of &lt;i&gt;Asymptotic Methods in Analaysis&lt;/i&gt; (second edition), by N. G. de Brujin. This is a really superb book, and I recommend it to anyone who wants to learn how to approximate quantities in &quot;calculus-like&quot; settings. (If you want to do approximation in combinatorial settings, I recommend Chapter 9 of &lt;i&gt;Concrete Mathematics&lt;/i&gt;.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, this isn't just about $\sin$. Let $f$ be a function $0 \leq f(u) \leq u$ on an interval $[0,c]$ then the sequence $x_n:=f(f(f(\cdots f(c)\cdots)$ approaches $0$. If $f(u)=u-a u^{k+1} + O(u^{k+2})$ (with $a&gt;0$) then $x_n \approx \alpha n^{-1/k}$ and you can prove that by the same methods here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Having said that, the answer to your question. On $[0,1]$, we have&#xA;$$\sin x=x-x^3/6+O(x^5).$$&#xA;Setting $y_n=1/x_n^2$, we have&#xA;$$1/x_{n+1}^2 = x_n^{-2} \left(1-x_n^2/6+O(x_n^4) \right)^{-2} =  1/x_n^2 + 1/3 + O(x_n^2)$$&#xA;so&#xA;$$y_{n+1} = y_n + 1/3 + O(y_n^{-1}).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We see that&#xA;$$y_n = \frac{n}{3} + O\left( \sum_{k=1}^n y_k^{-1} \right)$$&#xA;and&#xA;$$\frac{1}{n}y_n = \frac{1}{3} + \frac{1}{n} O\left( \sum_{k=1}^n y_k^{-1} \right)$$&#xA;Since we already know that $x_n \to 0$, we know that $y_n^{-1} \to 0$, so the average goes to zero and we get $\lim_{n \to \infty} y_n/n=1/3$. Transforming back to $\sqrt{n} x_n$ now follows by the continuity of $1/\sqrt{t}$.  &lt;/p&gt;&#xA;" OwnerUserId="448" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-08-25T16:33:37.837" LastActivityDate="2010-08-25T16:33:37.837" CommentCount="2" />
  <row Id="3221" PostTypeId="2" ParentId="3183" CreationDate="2010-08-24T18:12:44.720" Score="4" ViewCount="" Body="&lt;p&gt;Over $\mathbf{C}$ I believe the answer is yes (the universal cover is algebraic), although I'm not really an expert. Here's the story as I understand it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A connected[*] semisimple linear algebraic group over a field $k$ is called simply connected if it admits no nontrivial isogeny from another connected group. (An isogeny is a surjective, flat homomorphism of algebraic $k$-groups with finite kernel.) Now just like in the Lie group story, nice algebraic groups are classified by combinatorial data. Precisely, a reductive $k$-group $G$ together with a split maximal torus $T$, is classified by a &quot;root datum&quot;, which is roughly the roots and coroots of $G$ with respect to $T$, plus the lattices of characters and cocharacters of $T$. (&quot;Roughly&quot; because when $k$ is not algebraically closed, you need to keep track of the action of the Galois group of $k$ on the lattices, too.) Correspondingly, isogenies between such groups match bijectively with appropriate &quot;morphisms&quot; between the respective root data. These theorems are a bit involved, but can certainly be found in Borel's book on linear algebraic groups, for the case you care about. (Algebraically closed field of characteristic zero, namely $\mathbf{C}$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A consequence of this theory is that for any connected semisimple linear algebraic $\mathbf{C}$-group, there exists an &quot;algebraic universal cover&quot;, i.e. an isogeny $\tilde{G}\to G$ from a simply connected $\mathbf{C}$-group $\tilde{G}$. (In fact, we don't need to be working over $\mathbf{C}$ for this to be true.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now here's the crux. I claim that if $G$ is a simply connected $\mathbf{C}$-group, then the closed points $G(\mathbf{C})$ are simply connected in the classical topology. The hypothesis that we are over $\mathbf{C}$ is crucial: $\mathrm{Sp}(2n)$ is a simply connected $\mathbf{R}$-group, but the universal cover of $\mathrm{Sp}(2n)(\mathbf{R})$ is the non-algebraic metaplectic group. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me sketch the proof of the claim, which is suprisingly hard. (This was explained to me by Brian Conrad; any errors I introduce are, of course, my own.) First, it's a fact, although not a tautology, that since $G$ is connected, $G(\mathbf{C})$ is connected in the classical topology. So that's a good start. Next, by classical Lie theory, the complex Lie group $G(\mathbf{C})$ is homotopy equivalent to a maximal compact real Lie-subgroup $K$. Since $K$ is a compact manifold, this implies that $H^1(G(\mathbf{C}),\mathbf{Z})$ is finitely generated. But this group is the abelianization of $\pi_1(G(\mathbf{C}))$ (Hurewicz); and the fundamental group is already abelian because this is true for all topological groups. So $\pi_1(G(\mathbf{C}))$ is finitely generated (and abelian). In particular, it has a finite quotient. So if $G(\mathbf{C})$ were not (classically) simply connected, there would be a &lt;em&gt;finite&lt;/em&gt; covering map of complex Lie groups $G'\to G(\mathbf{C})$. Now a hard theorem of Grauert (relating $\pi_1(G(\mathbf{C}))$ to the &quot;\'etale fundamental group&quot; of $G$, which classifies the algebraic analogue of finite covering maps) implies that $G'$, as well as its analytic group structure, can uniquely be given an algebraic structure. In other words, there is an algebraic $\mathbf{C}$-group $G_0'$ with $G'=G_0'(\mathbf{C})$, and an isogeny $G_0'\to G$, such that the induced map&#xA;on $\mathbf{C}$-points is $G'\to G(\mathbf{C})$. In particular $G_0'\to G$ is a nontrivial isogeny, since on $\mathbf{C}$-points it is a nontrivial finite convering map. And this contradicts the (algebraic) simple-connectedness of $G$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whew! So in summary, for connected semisimple linear algebraic $\mathbf{C}$-groups $G$, the universal cover of $G(\mathbf{C})$ is precisely $\tilde G(\mathbf{C})$ where $\tilde G$ is the simply connected form of $G$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[*] Warning: In this answer, &quot;connected&quot; means Zariski-connected. For algebraic groups over $\mathbf{R}$, this is VERY different from the $\mathbf{R}$-points being connected in the classical topology!&lt;/p&gt;&#xA;" OwnerUserId="1265" LastEditorUserId="1265" LastEditorDisplayName="" LastEditDate="2010-08-24T18:20:13.960" LastActivityDate="2010-08-24T18:20:13.960" />
  <row Id="3222" PostTypeId="1" AcceptedAnswerId="3229" CreationDate="2010-08-24T18:38:34.100" Score="0" ViewCount="144" Body="&lt;p&gt;Suppose $f:[0,1]-&gt;R$ is defined as f(x) = 1 if x is irrational and f(x)=0 if x is rational.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know from Real Analysis that $\int_{0}^{1} f =1$, but intuitively it seems that the &quot;area under curve&quot; is 0.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there some intuitive explaination as to why $\int_{0}^{1} f =1$  (without using measure theory)?&lt;/p&gt;&#xA;" OwnerUserId="832" LastEditorUserId="832" LastEditorDisplayName="" LastEditDate="2010-08-24T18:55:33.827" LastActivityDate="2010-08-24T19:25:52.243" Title="Integral of 1 minus Dirichlet's function" Tags="&lt;real-analysis&gt;" AnswerCount="2" CommentCount="6" />
  <row Id="3223" PostTypeId="2" ParentId="3217" CreationDate="2010-08-24T18:40:13.867" Score="1" ViewCount="" Body="&lt;p&gt;If n has to be an integer, you can find the solution by brute force. Once you have the solution, you can show that it is correct and unique by showing that the function f(n) = (4^n - 1)/(n-1) is increasing for n &gt;= 2. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If n is allowed to be real, you can use your integer guess as the starting point for Newton's method and find the solution numerically to as much precision as you'd like.  Since f(n) is increasing, you can even find a good interval estimate.  if f(a) &amp;lt; 4*10^6 and f(b) &gt; 4*10^6, you know the solution is in the interval (a, b).&lt;/p&gt;&#xA;" OwnerUserId="247" LastActivityDate="2010-08-24T18:40:13.867" />
  <row Id="3224" PostTypeId="2" ParentId="3217" CreationDate="2010-08-24T18:41:16.707" Score="1" ViewCount="" Body="&lt;p&gt;Instead of solving $\frac{4^n-1}{n-1} &gt; 4&amp;times;10^6$ try solving&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{4^n}{n-1} &gt; 4&amp;times;10^6 = 4^4 &amp;times;5^6$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i.e&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{4^{n-4}}{n-1} &gt; 5^6$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clearly $n&gt; 10$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Try a binary search method.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$n = 20$ is too large.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$n=15$ too large.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$n=12$ is too small.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$n=13$ seems to give the answer.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-08-24T18:41:16.707" />
  <row Id="3225" PostTypeId="2" ParentId="3217" CreationDate="2010-08-24T18:49:31.323" Score="1" ViewCount="" Body="&lt;p&gt;Wolfram Alpha will solve it for you if you write: solve &quot;your inequality&quot;.&#xA;But as you want to solve it by hand you can rearrange it to:&#xA;$$ n &gt; (4+3\log_2{5}) + \frac{1}{2} \log_2(n-1) - \log\left(1-\frac{1}{2^{2n}}\right)/2\log(2) $$&#xA;Now the first term in brackets is about 10.96578, use that value for n in the next log term to get 1.65849... This shows you that n&gt;12.6, and so you get n=13. Note that the last term is too small to worry about.&lt;/p&gt;&#xA;" OwnerUserId="1301" LastActivityDate="2010-08-24T18:49:31.323" CommentCount="1" />
  <row Id="3226" PostTypeId="2" ParentId="3222" CreationDate="2010-08-24T18:53:05.153" Score="1" ViewCount="" Body="&lt;p&gt;The rationals are a set of Lebesgue measure zero, so they don't impact the value of integrals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's how you might see that.  Start with an epsilon &gt; 0.  You can count the rationals, so put an interval of length epsilon/2 over the first rational.  Put an interval of length epsilon/4 over the second rational, etc.  You've created a sequence of intervals of total length epsilon that cover the rationals, so they must have measure less than epsilon.  But epsilon was arbitrary, so they must have measure zero.&lt;/p&gt;&#xA;" OwnerUserId="247" LastActivityDate="2010-08-24T18:53:05.153" CommentCount="4" />
  <row Id="3228" PostTypeId="2" ParentId="3203" CreationDate="2010-08-24T18:58:59.387" Score="1" ViewCount="" Body="&lt;p&gt;There is no ambiguity about infinite limits. For instance, when we &lt;em&gt;define&lt;/em&gt; that a sequence or real numbers $(x_n)$ &quot;tends to infinity&quot;, $(x_n) \longrightarrow +\infty$, we just mean that, for each real number $N$ there is some $n_0 \in \mathbb{N}$ such that, for all $n\geq n_0$, we have $x_n &gt; N$. That is, the sequence $(x_n)$ grows indifinitely. There is no implication here that $+\infty$ &quot;exists&quot;, or doesn't, nor any problem with the fact that, certainly, $+\infty$ is not a real number. In fact, there is &lt;em&gt;no&lt;/em&gt; $\infty$ on the right hand side of the definition, but just real numbers. And symbols on the left hand side of a definition mean just you want them to mean according with what you put on the right hand side.&lt;/p&gt;&#xA;" OwnerUserId="664" LastEditorUserId="664" LastEditorDisplayName="" LastEditDate="2010-08-25T02:37:49.383" LastActivityDate="2010-08-25T02:37:49.383" CommentCount="1" />
  <row Id="3229" PostTypeId="2" ParentId="3222" CreationDate="2010-08-24T19:25:52.243" Score="3" ViewCount="" Body="&lt;p&gt;Some of this answer is implicit in many people's comments, but let's make it explicit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a question. It seems to you that the area under the curve --- of Dirichelt's function --- is zero. But Dirichelt's function is not a &quot;curve&quot;, precisely because it is discontinuous. What then do you mean?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's consider that idea of &quot;under the curve&quot;, for some possible failures of intuition.&#xA;If you take the phrase &quot;under the curve&quot; very seriously, you'll probably be thinking of something like the &lt;a href=&quot;http://en.wikipedia.org/wiki/Darboux_integral&quot; rel=&quot;nofollow&quot;&gt;lower Darboux sum&lt;/a&gt;. This, indeed, is zero for any interval, because there is no space for any rectangles of finite width underneath &quot;the curve&quot;. The problem is that the Darboux integral is not well-defined, because the &lt;em&gt;upper&lt;/em&gt; Darboux sum for any interval is the length of the interval --- because the shortest rectangles of finite width which envelop the function are the rectangles of height 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, Dirichlet's function is not integrable by techniques such as the Darboux integral, because the upper and lower Darboux sums do not coincide. Additional techniques are required --- such as measure theory --- to define the integral at all.&lt;/p&gt;&#xA;" OwnerUserId="439" LastActivityDate="2010-08-24T19:25:52.243" />
  <row Id="3230" PostTypeId="2" ParentId="2991" CreationDate="2010-08-24T19:38:35.290" Score="1" ViewCount="" Body="&lt;p&gt;Here's an alternative method that is due to Gauss. Scale the congruence so to reduce the leading coefficient. Hence we seek a multiple of $\:25\:$ that is smaller $\rm(mod\ 109)\:.\ $ Clearly $4\:$ works, since $\; 4\cdot25\equiv 100 \equiv -9 \;$ has smaller absolute value than $25$. Scaling by $4$ yields $\rm-9\ x \equiv 12.\;$ Similarly, scaling this by $12$ yields $\rm\ x \equiv 144 \equiv 35$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This always works if the modulus is prime, i.e. it will terminate with leading coefficient $1$ (versus $0$, else the leading coefficient would properly divide the prime $\rm\:p\:$). It's a special case of the Euclidean algorithm that computes inverses mod $\:\rm p\:$ prime. This is the way that Gauss proved that irreducible integers are prime (i.e. that $\rm p|ab\Rightarrow p|a$ or $\rm p|b$) - hence unique factorization; it's essentially Gauss: Disq. Arith. Art.5, 1801. It iterates  $\rm (a,p) \to (p \;mod\; a, p)\;$ i.e. $\rm a\to a&amp;#39; \to a&amp;#39;&amp;#39; \to \cdots,\; n&amp;#39; = p \;mod\; n \;$ instead of $\rm (a,p) \to (p \;mod\; a,\: a)$ as in the Euclidean algorithm. It generates a descending chain of multiples of $\rm\ a\ (mod\ p)\:$. See my &lt;a href=&quot;http://groups.google.com/groups?selm=y8zu1hn1v14.fsf@nestle.ai.mit.edu&quot; rel=&quot;nofollow&quot;&gt;sci.math post on 2002\12\9&lt;/a&gt; for further discussion.  &lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2011-03-07T00:11:35.763" LastActivityDate="2011-03-07T00:11:35.763" />
  <row Id="3231" PostTypeId="1" AcceptedAnswerId="3234" CreationDate="2010-08-24T19:49:51.553" Score="3" ViewCount="574" Body="&lt;p&gt;How can I convert a hexadecimal number, for example &lt;code&gt;0x1A03&lt;/code&gt; to its octal value?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that one way is to convert it to decimal and then convert it to octal&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;0x1A03 = 6659 = 0o15003&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Is there a simple way to do it without the middle step (conversion to decimal or conversion to binary)?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Why do we tend to convert it to Base10 every time?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="324" LastActivityDate="2010-08-25T19:17:48.303" Title="How to convert a hexadecimal number to an octal number?" Tags="&lt;arithmetic&gt;" AnswerCount="5" CommentCount="1" FavoriteCount="1" />
  <row Id="3232" PostTypeId="1" AcceptedAnswerId="3237" CreationDate="2010-08-24T19:51:11.240" Score="1" ViewCount="67" Body="&lt;p&gt;I am having some difficulty following the concept of shrinking a set of node as given in the paper &lt;a href=&quot;http://www.zib.de/Publications/Reports/ZR-06-22.pdf&quot; rel=&quot;nofollow&quot;&gt;On the Bottleneck Shortest Path Problem&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It says:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Shrinking a set of nodes can be done&#xA;  in linear time. More precisely, given&#xA;  a set S ⊆ V of nodes of an&#xA;  (undirected) graph G = (V,E), one can&#xA;  construct in linear time another graph&#xA;  with nodes (V \S)$\cup${vnew} (where vnew&#xA;  represents the shrunken set S), where&#xA;  v,w ∈ V \ S are adjacent if and only&#xA;  if v and w are adjacent in G (in this&#xA;  case,the edge keeps its weight), and&#xA;  vnew and w ∈ V \ S are adjacent if and&#xA;  only if there is some v ∈ S such that&#xA;  v and w are adjacent in G (in which&#xA;  case the edge receives the biggest&#xA;  weight of any edge connecting S and w&#xA;  in G).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am particularly confused about the last sentence of how vnew and w are adjacent.  Actually I am pretty much confused about it all and any clarity would be appreciated.&lt;/p&gt;&#xA;" OwnerUserId="1453" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2010-11-27T14:03:07.433" LastActivityDate="2010-11-27T14:03:07.433" Title="Deciphering this explanation for 'shrinking a set of nodes'?" Tags="&lt;graph-theory&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="3233" PostTypeId="1" AcceptedAnswerId="3242" CreationDate="2010-08-24T19:54:08.357" Score="2" ViewCount="247" Body="&lt;p&gt;I would like help with the following problem (chapter 9, #4) from Lee's Smooth Manifolds [its not homework, I'm reading it and I got stuck on this one]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If a Lie group $G$ acts smoothly and freely on a smooth manifold $M$ and the orbit space $M/G$ has a smooth manifold structure such that the quotient map $\pi: M\to M/G$ is a smooth submersion, then $G$ acts properly. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Its kind of a converse to the standard theorem about quotienting a manifold by a group action.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any hints/help?&lt;/p&gt;&#xA;" OwnerUserId="379" LastActivityDate="2010-08-24T21:05:40.653" Title="Converse To Quotient Manifold Theorem [Exercise in Lee Smooth Manifolds]" Tags="&lt;differential-geometry&gt;&lt;differential-topology&gt;" AnswerCount="1" FavoriteCount="2" />
  <row Id="3234" PostTypeId="2" ParentId="3231" CreationDate="2010-08-24T19:58:01.593" Score="9" ViewCount="" Body="&lt;p&gt;A simpler way is to go through binary (base 2) instead of base 10.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;0x1A03 = 0001 1010 0000 0011&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now group the bits in bunches of 3 starting from the right&lt;/p&gt;&#xA;&#xA;&lt;p&gt;0 001 101 000 000 011&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This gives&lt;/p&gt;&#xA;&#xA;&lt;p&gt;0 1 5 0 0 3&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which is your octal representation.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-08-24T19:58:01.593" />
  <row Id="3235" PostTypeId="2" ParentId="3231" CreationDate="2010-08-24T20:00:24.187" Score="4" ViewCount="" Body="&lt;p&gt;When you're converting among bases $2^n$, you can often do so more quickly with a comprehensive conversion table.  For example, between binary and octal, each block of 3 binary digits will convert to one octal digit:&#xA;$$\begin{matrix}&#xA;000_2=0_8 &amp;amp; 001_2=1_8 &amp;amp; 010_2=2_8 &amp;amp; 011_2=3_8&#xA;\\ &#xA;100_2=4_8 &amp;amp; 101_2=5_8 &amp;amp; 110_2=6_8 &amp;amp; 111_2=7_8&#xA;\end{matrix}$$&#xA;Similarly, when converting between binary and hexadecimal, each block of 4 binary digits converts to a single hexadecimal digit.  Because of these two facts, each block of 12 (least common multiple of 3 and 4) binary digits is 4 octal digits and 3 hexadecimal digits, so each block of 3 hexadecimal digits can be converted to a block of 4 octal digits.  The table for doing hexadecimal&amp;lt;-&gt;octal directly is quite large, though, so it's usually simpler to convert to binary as an intermediate form.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is also possible to do the conversion directly by performing division in octal or hexadecimal, though this can be tricky to get used to.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think we tend to use decimal as an intermediate form because we are most familiar and comfortable with base 10 (since we generally have 10 fingers and use it most often).&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-08-24T20:00:24.187" />
  <row Id="3236" PostTypeId="2" ParentId="3231" CreationDate="2010-08-24T20:02:15.063" Score="4" ViewCount="" Body="&lt;p&gt;I think the easiest way is to go through binary.  A hex digit corresponds to 4 bits, an octal digit to 3.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;0x1A03 in binary is 0001 1010 0000 0011 (grouped into 4-bit nibbles).  If I regroup into 3-bit groups (from the right), I have 001 101 000 000 011.  That's octal 0o15003.  &lt;/p&gt;&#xA;" OwnerUserId="822" LastActivityDate="2010-08-24T20:02:15.063" />
  <row Id="3237" PostTypeId="2" ParentId="3232" CreationDate="2010-08-24T20:08:08.513" Score="1" ViewCount="" Body="&lt;p&gt;Draw some graph with vertex set $V$. Mark off a subset of those vertices and call it $S$. You now have two types of vertices: those in $S$ and those outside of $S$, and you have three types of edges: those connecting vertices of $S$, those connecting vertices outside of $S$, and those on the &quot;boundary&quot;, connecting vertices of $S$ to vertices outside of $S$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the &quot;shrinking&quot; operation, we're removing all the vertices of $S$ and replacing them with a single new vertex which you're calling vnew. Visually, think of squishing all the vertices of $S$ together. So in the new graph all the vertices outside of $S$ stay put and all the vertices in $S$ merge into this new guy vnew.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What happens to the edges?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The edges that used to connect vertices of $S$ among themselves are just gone.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The edges that used to connect two vertices outside of $S$ stay put. Both their ends are still there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, the boundary edges: each one used to connect some vertex $s$ in $S$ to some vertex $v$ outside of $S$. Quite naturally, such an edge is now replaced by a new one connecting our vnew to $v$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, if $v$ (outside of $S$) was connected to one or more vertices in $S$, it is now connected to vnew. If there are weights on the edges, the weight of this new edge between $v$ and vnew is taken to be the max of all the weights of edges connecting $v$ to anyone in $S$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(If $v$ is not connected to anything in $S$, it stays unconnected to vnew).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does this make sense now?&lt;/p&gt;&#xA;" OwnerUserId="308" LastActivityDate="2010-08-24T20:08:08.513" />
  <row Id="3238" PostTypeId="2" ParentId="3232" CreationDate="2010-08-24T20:17:12.933" Score="1" ViewCount="" Body="&lt;p&gt;Consider the following example,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Say you had a set of companies and for each pair of companies, you record the maximum number of emails sent by one to the other (i.e For example, when considering exchanges between companies $A$ and $B$ you pick maximum of number of emails from $A$ to $B$, and number of emails from $B$ to $A$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can model this as a graph, with the companies as nodes, join two companies by an edge if they had an email exchange. Mark the edge with the maximum number of email sent by one to the other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now suppose some companies decided to merge and wanted to form a new company called $vnew$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How would the graph model change?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For two companies which were not involved in the merger, the edge between them remains the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose companies $A,B,C$ were part of the merger, while $D$ was not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you now consider the email exchange between $D$ and $vnew$, you need to consider the previous exchange between $D$ and $A$, $D$ and  $B$ and $D$ and $C$. Pick the maximum among those three, and create an edge between $D$ and the new company $vnew$ recording that exchange.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope that helps.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-08-24T20:17:12.933" />
  <row Id="3241" PostTypeId="2" ParentId="3231" CreationDate="2010-08-24T20:40:39.253" Score="3" ViewCount="" Body="&lt;p&gt;There is a trick for hex to/from octal because they relate to binary but lets ignore that and consider the general idea (which works for any bases).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since we want to convert to octal lets start using octal as our number system - as if were always were. So we count like this 1,2,3,4,5,6,7,10,11,12,13,14,15,16,17,20,21,22,...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And our multiplication table is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{array}{|c|c|c|c|c|c|c|} &#xA;  \hline &#xA;     \times &amp;amp; \mathbf 1 &amp;amp; \mathbf 2 &amp;amp; \mathbf 3 &amp;amp; \mathbf 4 &amp;amp; \mathbf 5 &amp;amp; \mathbf 6 &amp;amp; \mathbf 7\\&#xA;  \hline &#xA;     \mathbf 1 &amp;amp; 1 &amp;amp; 2 &amp;amp; 3 &amp;amp; 4 &amp;amp; 5 &amp;amp; 6 &amp;amp; 7 \\&#xA;     \mathbf 2 &amp;amp; 2 &amp;amp; 4 &amp;amp; 6 &amp;amp; 10 &amp;amp; 12 &amp;amp; 14 &amp;amp; 16 \\&#xA;     \mathbf 3 &amp;amp; 3 &amp;amp; 6 &amp;amp; 11 &amp;amp; 14 &amp;amp; 17 &amp;amp; 22 &amp;amp; 25 \\&#xA;     \mathbf 4 &amp;amp; 4 &amp;amp; 10 &amp;amp; 14 &amp;amp; 20 &amp;amp; 24 &amp;amp; 30 &amp;amp; 34 \\&#xA;     \mathbf 5 &amp;amp; 5 &amp;amp; 12 &amp;amp; 17 &amp;amp; 24 &amp;amp; 31 &amp;amp; 36 &amp;amp; 43 \\&#xA;     \mathbf 6 &amp;amp; 6 &amp;amp; 14 &amp;amp; 22 &amp;amp; 30 &amp;amp; 36 &amp;amp; 44 &amp;amp; 52\\&#xA;     \mathbf 7 &amp;amp; 7 &amp;amp; 16 &amp;amp; 25 &amp;amp; 34 &amp;amp; 43 &amp;amp; 52 &amp;amp; 61 \\&#xA;  \hline &#xA;\end{array} &#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the hexadecimal digits are $0\text x1 = 1$, $0\text x2 = 2$, $0\text x3 = 3$, $0\text x4 = 4$, $0\text x5 = 5$, $0\text x6 = 6$, $0\text x7 = 7$, $0\text x8 = 10$, $0\text x9 = 11$, $0\text {xA} = 12$, $0\text xB = 13$, $0\text xC = 14$, $0\text xD = 15$, $0\text xE = 16$, $0\text xF = 17$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, we want to re-interpret the number &lt;code&gt;0x1A03&lt;/code&gt; which is equal to $ 20^3 \cdot 1 + 20^2 \cdot 12 + 20 \cdot 0 + 3$. Now we just &quot;do the math&quot;, multiply everything out and add it up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{align}&#xA;&amp;amp; 20^3 \cdot 1 + 20^2 \cdot 12 + 20 \cdot 0 + 3 \\&#xA;=&amp;amp; 10000 + 400 \cdot 12 + 3 \\&#xA;=&amp;amp; 10000 + 5000 + 3 \\&#xA;=&amp;amp; 10000 + 5000 + 3 \\&#xA;=&amp;amp; 15003&#xA;\end{align}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Now the fact that we have all these zeros suggests there would be a quicker method for these particular bases (normally we would not get so many zeros in our calculations).&lt;/p&gt;&#xA;" LastActivityDate="2010-08-24T21:10:39.473" CommentCount="5" />
  <row Id="3242" PostTypeId="2" ParentId="3233" CreationDate="2010-08-24T21:05:40.653" Score="2" ViewCount="" Body="&lt;p&gt;I'm pretty sure this is true. Here's my suggested approach. There's a step I don't know how to do, so hopefully someone can fill in the gap, or you can figure it out as an exercise. (And please post a comment if you do, as I'm curious!)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, show that $M$ is a principal fiber bundle with structure group $G$. That is, locally on $M/G$, $M$ looks like $(M/G)\times G$. To do this, use that $M\to  M/G$ is a submersion to find neighborhoods $U$ on $M/G$ and $V$ on $M$ where $M\to M/G$ looks like the projection onto several coordinates. The coordinates you &quot;forget&quot; in this projection give a chart on some neighborhood of $G$. (This is why the fibers of submersions are smooth manifolds; in this case, the fiber is $G$ since the action is free.)  Now use translation by the group law to &quot;spread $V$ out&quot; so that the entire preimage of $U$ in $M$ is diffeomorphic to $U\times G$, with the restriction of $M\to M/G$ to this open being the projection onto the first factor. This ensures that $M$ is a principal $G$-bundle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This reduces the problem to showing that $G$ acts properly and freely on any principal fiber bundle with structure group $G$ (freeness being clear). To show that $G\times M\to M\times M$ is proper, it suffices to show that it is closed and that the preimage of a point in compact. The latter condition is obvious, by working locally in a trivializing neighborhood of that point. I don't immediately see how to get the closedness of the map, but I've convinced myself it's true in some examples.&lt;/p&gt;&#xA;" OwnerUserId="1265" LastActivityDate="2010-08-24T21:05:40.653" CommentCount="1" />
  <row Id="3243" PostTypeId="1" CreationDate="2010-08-24T21:20:15.723" Score="4" ViewCount="313" Body="&lt;p&gt;I'm trying to find out for which natural numbers n does 9^n + 1 have all of its prime factors less than 40.  If I can provide a positive answer to my title question, then I will have a proof that only n = 1 causes 9^n + 1 to have all of its prime factors less than 40.  Thank you for any help.&lt;/p&gt;&#xA;" LastActivityDate="2010-11-11T00:22:17.157" Title="Does 9^(2^n) + 1 always have a prime factor larger than 40?" Tags="&lt;number-theory&gt;" AnswerCount="2" />
  <row Id="3244" PostTypeId="2" ParentId="3243" CreationDate="2010-08-24T21:44:27.093" Score="10" ViewCount="" Body="&lt;p&gt;Let $\displaystyle a_n:=9^{2^n}+1$; reducing $\displaystyle a_n$ modulo the primes in the interval $\displaystyle [2,40]$, it is easy to find that the only times we get zero is when either the prime is 2 and $\displaystyle n$ is arbitrary, or when the prime is 17 and $\displaystyle n$ is 2.  Thus, for $\displaystyle n&gt;2$, the number $\displaystyle a_n$ has prime factors larger than 2 (&lt;strong&gt;and hence larger than 40&lt;/strong&gt;) provided it is not a power of 2 itself. Since $\displaystyle a_n \equiv 2 \pmod{4}$, we see that $\displaystyle a_n$ is never a power of two. Finally, $\displaystyle a_1=82=2 \cdot 41$ and $\displaystyle a_2=6562=2 \cdot 17 \cdot 193$, and we conclude that $\displaystyle a_n$ has a prime factor larger than 40 for $\displaystyle n \geq 1$.&lt;/p&gt;&#xA;" OwnerUserId="1028" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-11-11T00:19:12.070" LastActivityDate="2010-11-11T00:19:12.070" CommentCount="0" />
  <row Id="3245" PostTypeId="1" AcceptedAnswerId="3247" CreationDate="2010-08-24T21:59:44.630" Score="4" ViewCount="236" Body="&lt;p&gt;The $\Gamma (x)$ function has just one minimum for $x&gt;0$ . This result uses some&#xA;properties of the gamma function:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\Gamma ^{\prime \prime }(x)&gt;0$ and $\Gamma (x)&gt;0$ for all $x&gt;0$&lt;/li&gt;&#xA;&lt;li&gt;$\Gamma (1)=\Gamma (2)=1$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Observing the following graph (created in SWP) of $y=\Gamma (x)$ this minimum is near $x=3/2$, but likely $\min \Gamma (x)\neq \Gamma \left( 3/2\right) =\dfrac{1}{2}\Gamma \left( 1/2\right) =\dfrac{1}{2}\sqrt{\pi }$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/hNj21.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think that it is not possible to find analytically the exact value of $x_{\min }$, even by converting to an adequate problem in the interval $]0,1]$ and using the functional equation $\Gamma (x+1)=x\Gamma (x)$ and the reflection formula&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Gamma (p)\Gamma (p-1)=\dfrac{\pi }{\sin px}\qquad $( $0\lt p\lt 1$)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;a) Which is the best way to find $\min_{[1,2]}\Gamma (x)$ and does $x_{\min }$ lay in $[1,3/2]$ or in $[3/2,2]$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;b) Is there some useful series expansion of $\Gamma (x)$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;c) Which numeric method do you suggest?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt;  Due to the shape of $\Gamma (x)$ I thought on the one-dimensional &lt;a href=&quot;http://ntu-cg.ntu.edu.sg/ysong/conference/MemeticGradientSearch.pdf&quot; rel=&quot;nofollow&quot;&gt;Davies-Swann-Campey&lt;/a&gt; method of direct search for unconstrained optimization, which approximates a function near a minimum by successive approximating quadratic polynomials.&lt;/p&gt;&#xA;" OwnerUserId="752" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2010-08-29T10:20:35.633" LastActivityDate="2010-08-29T10:20:35.633" Title="Minimum of the Gamma Function $\Gamma (x)$ for $x&gt;0$. How to find $x_{\min}$?" Tags="&lt;calculus&gt;&lt;approximation&gt;&lt;gamma-function&gt;&lt;numerical-methods&gt;&lt;special-functions&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="0" />
  <row Id="3246" PostTypeId="2" ParentId="3245" CreationDate="2010-08-24T22:08:24.590" Score="3" ViewCount="" Body="&lt;p&gt;According to &lt;a href=&quot;http://mathworld.wolfram.com/GammaFunction.html&quot; rel=&quot;nofollow&quot;&gt;MathWorld&lt;/a&gt; the minimum of the Gamma function for positive $x$ is 1.46163...; in particular I guess this is enough to deduce that it is smaller than $3/2$. You can follow the links along to find some references where this is proved.&lt;/p&gt;&#xA;" OwnerUserId="1028" LastActivityDate="2010-08-24T22:08:24.590" CommentCount="1" />
  <row Id="3247" PostTypeId="2" ParentId="3245" CreationDate="2010-08-24T22:09:24.120" Score="4" ViewCount="" Body="&lt;p&gt;There indeed is no closed-form for the gamma function's minimum; what you can do instead, however, is to find the positive root of the digamma function (the logarithmic derivative of the gamma function), which should be available in your computing environment.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-08-24T22:09:24.120" CommentCount="6" />
  <row Id="3248" PostTypeId="1" CreationDate="2010-08-24T22:43:14.593" Score="1" ViewCount="77" Body="&lt;p&gt;I have been thinking about some conjectures of mine regarding non-uniformly convergent sequences of functions.  I'm happy to discuss my work so far, if anyone wants to know, but for brevity let me just say that I am particularly interested in finding a reference which discusses some theory of non-uniformly convergent sequences of functions, specifically for functions from R&lt;sup&gt;2&lt;/sup&gt;-&gt;R.  Does anyone know of a good place to look for this?.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have been trying to find a reference today for this subject and could not, is that because nobody thinks it's very interesting?  I tried google, wiki, googlebooks, ect... but there are SO many ref's for uniform convergence that come up, it's hard to find what I'm looking for.&lt;/p&gt;&#xA;" OwnerUserId="1417" LastActivityDate="2010-08-24T22:43:14.593" Title="Reference for theory of non-uniformly convergent sequences of functions" Tags="&lt;real-analysis&gt;" CommentCount="6" FavoriteCount="1" />
  <row Id="3249" PostTypeId="1" AcceptedAnswerId="3258" CreationDate="2010-08-24T23:04:20.163" Score="1" ViewCount="355" Body="&lt;p&gt;In a &lt;a href=&quot;http://math.stackexchange.com/questions/3138/ways-for-a-continuous-function-to-not-be-smooth&quot;&gt;recent post&lt;/a&gt; I asked what are the ways a continuous function cannot be smooth.  Apparently my question was not well posed, but since I am still thinking about these types of functions I will try one more time to ask my question in a proper way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I define a &quot;kink&quot; in the graph of a function as an abrupt, discontinuous change in the first derivative.  For example, the function f(x) = |x| has what I call a kink at x=0.  I apologize if this is not totally rigorous, but I think the idea of a &quot;kink&quot; is clear enough that someone could make it rigorous with little effort.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Besides sin(1/x) and the Weirstrauss function, what are some examples of non-smooth continuous functions without kinks, either in the graph of the function or the graph of any of it's derivatives?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(i.e. I am not looking for any function for which the graph of one of the derivatives has a kink)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In case anyone is interested, I edited the linked post yesterday to respond to Qiaochu's comment.&lt;/p&gt;&#xA;" OwnerUserId="1417" LastActivityDate="2010-08-25T01:52:29.063" Title="What are some examples of non smooth continuous functions without &quot;kinks&quot;?" Tags="&lt;analysis&gt;" AnswerCount="3" CommentCount="4" FavoriteCount="1" />
  <row Id="3250" PostTypeId="1" CreationDate="2010-08-24T23:12:11.743" Score="6" ViewCount="586" Body="&lt;p&gt;I am interested in how today's professional mathematicians view the Pythagorean theorem, in terms of how the theorem fits within the axiomatic framework of mathematics.  I often come across textbooks that define length by the Pythagorean theorem, so that the theorem is in essence a definition or axiom.  In more modern mathematics such as linear algebra, is the Pythagorean theorem generally just used as the &lt;em&gt;definition&lt;/em&gt; of length?  Is it more conventional today to treat the Pythagorean theorem as a definition (or axiom) rather than a theorem?  Are there any modern proofs of the Pythagorean theorem that don't rely on Euclidean geometry (like a proof that utilizes linear algebra/the dot product, etc.)?&lt;/p&gt;&#xA;" OwnerUserId="1477" LastActivityDate="2010-08-28T17:10:05.323" Title="Where does the Pythagorean theorem &quot;fit&quot; within modern mathematics?" Tags="&lt;linear-algebra&gt;&lt;euclidean-geometry&gt;" AnswerCount="5" CommentCount="2" />
  <row Id="3251" PostTypeId="2" ParentId="3243" CreationDate="2010-08-24T23:14:52.873" Score="12" ViewCount="" Body="&lt;p&gt;Here is a proof with very few computations involved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is well known that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $\displaystyle x^2 + 1 = 0 (\mod p)$ for some prime $\displaystyle p$, then $\displaystyle p = 1 (\mod 4)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps a little less well known than the above is that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $\displaystyle x^4 + 1 = 0 (\mod p)$ for some prime $\displaystyle p$, then $\displaystyle p = 1 (\mod 8)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $\displaystyle 9^{2^n}+1 = (3^{2^{n-1}})^4 + 1$ we must have that the only possible prime  &amp;lt; 40 which is a candidate is $\displaystyle 17$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus we need to check the residues only for $\displaystyle 17$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now since $\displaystyle 3$ is a primitive root of $\displaystyle 17$, we have that $\displaystyle 3^{8} + 1 = 0 (\mod 17)$. Since $\displaystyle 3^{2^3}  = -1 ( \mod 17)$, we have that $\displaystyle 3^{2^n} + 1 = 2 (\mod 17)$ for all $\displaystyle n &gt; 3$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus we only need to check $\displaystyle 3^{2^3} +1 = 9^{2^2} + 1 = 2&amp;times;17&amp;times;193$, which has a factor $\displaystyle &gt; 40$ and we are done.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Interestingly, for any number $N$, we can show that at most a &lt;em&gt;finite&lt;/em&gt; number of numbers of the form $\displaystyle 9^{2^n}+1$ have all their prime factors  &amp;lt; $\displaystyle N$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is because for $\displaystyle m$ &amp;lt; $\displaystyle n$, $\displaystyle 9^{2^m}-1$ divides $\displaystyle 9^{2^n}-1$ which implies that $\displaystyle (9^{2^n} + 1, 9^{2^m} + 1) = 2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i.e. any two numbers in the sequence have a gcd of $\displaystyle 2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus the number of numbers with all prime factors &amp;lt; $\displaystyle N$ is finite.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Another proof of the same fact is that if $\displaystyle 9^{2^m}+1 = 0 (\mod p)$ then $\displaystyle 9^{2^n}+1 = 2 (\mod p)$ for $n &gt; m$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, this is a different proof that the number of primes is infinite!&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-11-11T00:22:17.157" LastActivityDate="2010-11-11T00:22:17.157" CommentCount="4" />
  <row Id="3252" PostTypeId="2" ParentId="3250" CreationDate="2010-08-24T23:27:27.327" Score="4" ViewCount="" Body="&lt;p&gt;The Pythagorean theorem follows &lt;em&gt;directly&lt;/em&gt; from the basic properties of the dot product: if $x \cdot v = 0$, then $(x + v) \cdot (x + v) = x \cdot x + v \cdot v$.  There is not much to say about this proof from a modern point of view except that it does not depend on the finite-dimensionality of the underlying space, e.g. in &lt;a href=&quot;http://en.wikipedia.org/wiki/Hilbert_space&quot; rel=&quot;nofollow&quot;&gt;Hilbert spaces&lt;/a&gt; it even generalizes to &lt;a href=&quot;http://en.wikipedia.org/wiki/Parsevals_identity&quot; rel=&quot;nofollow&quot;&gt;Parseval's identity&lt;/a&gt;.  (Of course one then has to prove that dot products induce norms, but this is the standard Cauchy-Schwarz argument.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to interpret your title question more broadly.  My impression is that mathematicians do not really think about the Pythagorean theorem directly (unless, perhaps, they are functional analysts!).  Rather, following Klein's &lt;a href=&quot;http://en.wikipedia.org/wiki/Erlangen_program&quot; rel=&quot;nofollow&quot;&gt;Erlangen philosophy&lt;/a&gt;, they think about the &lt;a href=&quot;http://en.wikipedia.org/wiki/Group_(mathematics)&quot; rel=&quot;nofollow&quot;&gt;group of symmetries&lt;/a&gt; that the use of the Pythagorean theorem (really, dot products) implies.  In the plane, this is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Euclidean_group&quot; rel=&quot;nofollow&quot;&gt;Euclidean group&lt;/a&gt; $\text{E}(2)$, in which sits the &lt;a href=&quot;http://en.wikipedia.org/wiki/Orthogonal_group&quot; rel=&quot;nofollow&quot;&gt;orthogonal group&lt;/a&gt; $\text{O}(2)$ of linear transformations preserving the dot product (rotations and reflections).  The geometry determined by the Euclidean group is one where it makes sense to talk about lengths and angles and all the familiar tools of Euclidean geometry.  There are other &lt;a href=&quot;http://en.wikipedia.org/wiki/Lie_group&quot; rel=&quot;nofollow&quot;&gt;Lie groups&lt;/a&gt; which determine other types of geometries (the &quot;non-Euclidean&quot; ones).&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-08-24T23:27:27.327" CommentCount="12" />
  <row Id="3253" PostTypeId="2" ParentId="3250" CreationDate="2010-08-24T23:29:45.503" Score="-1" ViewCount="" Body="&lt;p&gt;It all has to do with non-euclidean geometry.  The basic idea is that the distance function, or to say the same thing as a professional mathematician, the &quot;metric&quot;, is super important in geometry.  You can have any kind of distance function you like, as long as it has three key properties (look them up).  This is a pretty big subject, and it all goes back to the independence of Eulcid's 5th axiom on parallel lines.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The distance between two points depends on the curvature of the underlying space, that's the key idea.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's not possible to prove that the distance between two points is given by the Pythagorean theorem; because the actual distance between two actual points is the domain of physics, not axiomatic mathematics.  The best we can do in mathematics is define what we mean by distance (with a distance function), and since it's a definition, there's nothing to prove. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The story goes that Gauss realized Euclid's 5th postulate was independent because there are many geometries where it fails, e.g. on a circle or hyperboloid.  But Gauss didn't want to just tell the world he had solved the oldest problem in math, so what he did was when Riemann came to him looking for a job as a professor, he told Reimann to list 3 possible topics to give a talk on as a sort of interview for the job.  So Riemann wrote two topics down, but he couldn't think of a good third topic, and he put down &quot;foundations of geometry&quot; even though he hadn't thought very hard about it, just because he had some nice idea's which he had not fully formulated.  And since Gauss already been thinking about that, he picked the foundations of geometry; and the rest is math history.  Riemann went home and came up with a theory significantly beyond what Gauss had already known.  &lt;/p&gt;&#xA;" OwnerUserId="1417" LastEditorUserId="1417" LastEditorDisplayName="" LastEditDate="2010-08-24T23:44:58.990" LastActivityDate="2010-08-24T23:44:58.990" CommentCount="7" />
  <row Id="3254" PostTypeId="1" CreationDate="2010-08-24T23:34:56.947" Score="4" ViewCount="867" Body="&lt;p&gt;The fact that rotation about an angle is a linear transformation is both important (for example, this is used to prove the sine/cosine angle addition formulas; see &lt;a href=&quot;http://math.stackexchange.com/questions/1292/how-can-i-understand-and-prove-the-sum-and-difference-formulas-in-trigonometry&quot;&gt;http://math.stackexchange.com/questions/1292/how-can-i-understand-and-prove-the-sum-and-difference-formulas-in-trigonometry&lt;/a&gt;) and somewhat intuitive geometrically.  However, even if this fact seems fairly obvious (at least from a diagram), how does one turn the picture proof into a formal proof?  On a related note, it seems likely that many formal proofs using a diagram will end up relying on Euclidean geometry (using angle/side congruence properties), but isn't one of the points of linear algebra to avoid using Euclidean geometry explicitly?&lt;/p&gt;&#xA;" OwnerUserId="1477" LastActivityDate="2010-08-25T23:55:01.290" Title="How do you formally prove that rotation is a linear transformation?" Tags="&lt;linear-algebra&gt;" AnswerCount="7" CommentCount="6" FavoriteCount="4" />
  <row Id="3255" PostTypeId="2" ParentId="3254" CreationDate="2010-08-24T23:42:17.133" Score="1" ViewCount="" Body="&lt;p&gt;First how do you non-geometrically define rotation? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps in the 2D case, you can define it as multiplication by a complex number (which is in fact true!)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you do that, use the definition of a Linear Transformation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Show that the map T satisfies:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) $T(a + b) = T(a) + T(b)$ for any vectors $a,b$ (or complex numbers if you chose to go that route) &lt;br/&gt;&#xA;2) $T(\alpha b) = \alpha T(b)$ for any scalar $\alpha$ and any vectors $a,b$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that, you can define complex numbers as vectors in the abstract without having to rely on any Euclidean Axioms.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-08-25T05:47:09.090" LastActivityDate="2010-08-25T05:47:09.090" CommentCount="2" />
  <row Id="3256" PostTypeId="2" ParentId="3249" CreationDate="2010-08-24T23:50:37.747" Score="2" ViewCount="" Body="&lt;p&gt;One rigorous definition of a kink that I can think of is that the left and right derivatives exist and do not agree.  In that case, plenty of functions &quot;don't have kinks&quot; at points where they are continuous but not differentiable.  $\sin \frac{1}{x}$ and the Weierstrass function &quot;don't have kinks&quot; in the sense that neither the left nor the right derivative exist (at $0$ for the first function and everywhere for the second), and one can come up with arbitrarily badly behaved examples (so again, I do not see the point of this question) as explained for example in damiano's comment to your previous post.  If you do not understand this comment, I can elaborate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please don't take this the wrong way, but the fact that you ask this question and put the Weierstrass function in as an example indicates to me that you don't understand the construction of the Weierstrass function.  The point of the construction of the Weierstrass function is that although a uniformly convergent sequence of continuous functions is continuous, a uniformly convergent sequence of differentiable functions need not be differentiable.  One can imitate this construction for an extremely wide class of functions (e.g. by replacing the sines with another periodic function) and get any kind of examples you want.  So when you say &quot;the Weierstrass function,&quot; do you literally mean &quot;the Weierstrass function,&quot; or do you mean &quot;any function constructed sort of like the Weierstrass function&quot;?  If the latter, you should be more precise about this.&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-08-24T23:56:01.853" LastActivityDate="2010-08-24T23:56:01.853" CommentCount="1" />
  <row Id="3257" PostTypeId="2" ParentId="3254" CreationDate="2010-08-24T23:59:01.077" Score="0" ViewCount="" Body="&lt;p&gt;You can prove this by looking at the appropriate diagram and using trig.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, what is interesting is that it is not too difficult to show that &lt;em&gt;any&lt;/em&gt; isometry of $\mathbb R^n$ (with its usual inner product) is necessarily linear.  So if you accept that rotations are isometries (which to me at least is much more self-evident than them being linear) then you get linearity immediately.&lt;/p&gt;&#xA;" OwnerUserId="9" LastActivityDate="2010-08-24T23:59:01.077" CommentCount="1" />
  <row Id="3258" PostTypeId="2" ParentId="3249" CreationDate="2010-08-25T00:05:34.170" Score="2" ViewCount="" Body="&lt;p&gt;This has already been said before, but let me iterate this.  The value of the derivative of a function at a point is evaluated by computing a limit, if it exists.  The limit of a sequence may not exist if at least one of the following happens:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;the absolute value of some of the terms grows without bounds;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;you can extract two subsequences which both converge, but they converge to different values.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;It is important to keep in mind that both behaviors may happen for the same limit, and that for 2 you might be able to extract countably many subsequences each converging to a different value (only &quot;countably many&quot; as I am talking now of a limit of a &lt;em&gt;sequence&lt;/em&gt;). &lt;strong&gt;EDIT: in fact, you can have uncountably many different limits (see Jason's comment below).&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The situation for limits of functions is not simpler than the above, since you can &quot;embed&quot; any sequence into the limit defining the derivative of a continuous function at a point.  One of the &lt;em&gt;good&lt;/em&gt; consequences of the fact that you are dealing with a limit of a &lt;em&gt;continuous&lt;/em&gt; function is that if situation 2 occurs and the two subsequences you found have limits a&lt;code&gt;&amp;lt;&lt;/code&gt;b, then also every intermediate value $c \in [a,b]$ is the limit of a subsequence.  An elaboration of this is that the derivative of a once differentiable function satisfies the intermediate value property (&lt;a href=&quot;http://en.wikipedia.org/wiki/Darboux%27s_theorem_(analysis)&quot; rel=&quot;nofollow&quot;&gt;Darboux's Theorem&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, if you analyze the behavior of the values of the limit that defines a derivative at a point you may find that, as you zoom in, the values you obtain &lt;/p&gt;&#xA;&#xA;&lt;p&gt;a. also zoom in on a single real number - the derivative exists;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;b. are always contained in some finite interval, no matter how close to the actual limiting point you get - e.g. the function $x^2 \sin(1/x)$ near 0;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;c. are unbounded (possibly with alternating signs) - e.g. the function $\sqrt{x} \sin (1/x)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note also that further pathologies include the fact that the behaviors as you approach the limit from below or from above may well be different.  Even crazier, the &quot;overall&quot; interval you find in b, or an infinite interval in the case c, may contain inside it also countably many of other intervals nested more or less arbitrarily.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To conclude, derivatives of continuous functions are so immensely complicated that there is very little you can say about them!&lt;/p&gt;&#xA;" OwnerUserId="1028" LastEditorUserId="1028" LastEditorDisplayName="" LastEditDate="2010-08-25T01:52:29.063" LastActivityDate="2010-08-25T01:52:29.063" CommentCount="3" />
  <row Id="3259" PostTypeId="2" ParentId="3254" CreationDate="2010-08-25T00:17:27.097" Score="1" ViewCount="" Body="&lt;p&gt;If you want to be formal &lt;em&gt;and&lt;/em&gt; avoid Euclidean geometry, it seems to me that the right way to go is the other way around: first define certain linear transformations (the ones which preserve the dot product and have determinant $1$) and then prove that they have the properties you expect them to have.  More precisely, the group $\text{SO}(2)$ is abelian with Lie algebra $\mathbb{R}$, and the parameterization $\theta \mapsto \left[ \begin{array}{cc} \cos \theta &amp;amp; \sin \theta \\ - \sin \theta &amp;amp; \cos \theta \end{array} \right]$ is the exponential map from the Lie algebra to the group.  The methods for doing this should be covered in any book on Lie theory.&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-08-25T00:23:32.170" LastActivityDate="2010-08-25T00:23:32.170" CommentCount="9" />
  <row Id="3260" PostTypeId="2" ParentId="3249" CreationDate="2010-08-25T00:20:49.260" Score="2" ViewCount="" Body="&lt;p&gt;A real valued function is said to have a discontinuity of the first kind at a point if the left and right-hand limits at that point exist but are unequal. There are, however, only countably many such points.&#xA;These points, applied to the derivative, are kind of like &quot;kinks.&quot; A discontinuity of the second kind is not a &quot;kink&quot; since one of the  sided limits doesn't exist.  (I am attempting, possibly incorrectly, to make precise your terminology.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The derivative of $x^2 \sin(1/x^2)$ at zero is zero, so the function is differentiable everywhere, yet the derivative is unbounded in any neighborhood of the origin, which is thus a discontinuity of the second kind. More generally, &lt;a href=&quot;http://en.wikipedia.org/wiki/Volterra%27s_function&quot; rel=&quot;nofollow&quot;&gt;Volterra's function&lt;/a&gt; gives a differentiable function with a derivative with an uncountable number (even positive measure) of discontinuities of the second kind.&lt;/p&gt;&#xA;" OwnerUserId="536" LastEditorUserId="536" LastEditorDisplayName="" LastEditDate="2010-08-25T00:31:45.913" LastActivityDate="2010-08-25T00:31:45.913" />
  <row Id="3261" PostTypeId="2" ParentId="3254" CreationDate="2010-08-25T00:33:31.567" Score="2" ViewCount="" Body="&lt;p&gt;This depends on your definitions of &quot;rotation&quot; and &quot;angle&quot;, but, if you are willing to accept the sum and difference formulas of your link &lt;a href=&quot;http://math.stackexchange.com/questions/1292/how-can-i-understand-and-prove-the-sum-and-difference-formulas-in-trigonometry&quot;&gt;http://math.stackexchange.com/questions/1292/how-can-i-understand-and-prove-the-sum-and-difference-formulas-in-trigonometry&lt;/a&gt; , then you may follow the reasoning in Quiochu Yuan's answer there the other way around: given a vector $(x,y) \in \mathbb{R}^2$, let $\alpha$ be the angle determined by $(x,y)$ and the $x$-axis and $r= \sqrt{x^2 + y^2}$ its length. Then, of course,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{eqnarray*}&#xA;x &amp;amp;=&amp;amp; r \cos \alpha \\&#xA;y &amp;amp;=&amp;amp; r \sin \alpha&#xA;\end{eqnarray*}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you rotate $(x,y)$ an angle $\theta$, you'll obtain the vector $(x',y')$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{eqnarray*}&#xA;x' &amp;amp;=&amp;amp; r \cos (\alpha + \theta ) \\&#xA;y' &amp;amp;=&amp;amp; r \sin (\alpha + \theta )  \ .&#xA;\end{eqnarray*}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now you apply those sum and difference formulas and get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{eqnarray*}&#xA;x' &amp;amp;=&amp;amp; r \cos\alpha\cos\theta - r \sin\alpha\sin\theta &amp;amp;=&amp;amp; \cos\theta \cdot x - \sin\theta \cdot y\\&#xA;y' &amp;amp;=&amp;amp; r \sin\alpha\cos\theta + r \cos\alpha\sin\theta &amp;amp;=&amp;amp; \sin\theta \cdot x + \cos\theta\cdot y \ .&#xA;\end{eqnarray*}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which is the same as saying that $(x',y')$ is obtained from $(x,y)$ multiplying with the matrix&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{pmatrix}&#xA;\cos\theta &amp;amp; -\sin\theta   \\&#xA;\sin\theta  &amp;amp; \cos\theta&#xA;\end{pmatrix}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the rotation $(x,y) \mapsto (x',y')$ is the same as multiplication by a matrix, it is a linear transformation.&lt;/p&gt;&#xA;" OwnerUserId="664" LastEditorUserId="664" LastEditorDisplayName="" LastEditDate="2010-08-25T00:57:22.327" LastActivityDate="2010-08-25T00:57:22.327" />
  <row Id="3262" PostTypeId="2" ParentId="3254" CreationDate="2010-08-25T00:56:10.967" Score="6" ViewCount="" Body="&lt;p&gt;The problem comes in defining rotation so that this whole thing doesn't wind up circular, doing so fully would undoubtedly tie us all in knots, but one can prove the linearity of rotations by observing a few choice properties:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Rotations preserve angles&lt;/li&gt;&#xA;&lt;li&gt;They fix the origin&lt;/li&gt;&#xA;&lt;li&gt;And they preserve lengths&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Now just draw a parallelogram $OACB$ (spanned by vectors $\vec{OB}, \vec{OA}$- with $\vec{OC}=\vec{OB}+\vec{OA}$), rotate and chase the angles and lengths to observe that $R(\vec{OB})+R(\vec{OA})= R(\vec{OC})= R(\vec{OB}+ \vec{OA})$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's sort of Euclidianish but it's the only decent way to play it (especially if we want higher dimensions) without the 'lucky guess' approach championed by Qiaochu.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Addendum:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you feel pushed, you can turn the angle and length chase into norms and inner products but that would be a tragic waste of an isomorphism- here we are dealing with a geometric object in the language of vectors and to treat it non-geometrically would put us at an unnecessary disadvantage. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A good deal of higher mathematics depends on finding 'the best place' to solve a given problem, but the best place need not be the most abstract one: translating something about PDE into hilbert space theory may help find a solution, but it may also rob us of our geometric techniques and intuition. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A great example is something called the h-cobordism theorem, which can (and does- Milnor's take is a gem) take up a small book, using formal techniques (morse theory) to prove it, but can be done in half the time with some geometric (handlebody) jazz.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Vector spaces are another string to our bow, a new place to do math, and at times a more rigorous place- but they are not (at least from my perspective) the only place. &lt;/p&gt;&#xA;" OwnerUserId="160" LastEditorUserId="160" LastEditorDisplayName="" LastEditDate="2010-08-25T01:18:09.377" LastActivityDate="2010-08-25T01:18:09.377" CommentCount="2" />
  <row Id="3263" PostTypeId="2" ParentId="3203" CreationDate="2010-08-25T02:53:40.783" Score="0" ViewCount="" Body="&lt;p&gt;While I'll defer to other's answers (e.g. &lt;a href=&quot;http://math.stackexchange.com/questions/3203/infinite-limits/3206#3206&quot;&gt;Bill Dubuque's&lt;/a&gt;) that in the most general sense &quot;it depends&quot; is the correct answer, it may be worth noting that in the context of AP Calculus and across numerous common AP Calculus texts (at both the AB and BC level), a limit exists if and only if it has a real value, so saying that $\underset{x\to c}\lim f(c)=\infty$ or $\underset{x\to c}\lim f(c)=-\infty$ is saying that the limit as $x$ goes to $c$ of $f(x)$ does not exist and simultaneously giving more information about how the limit does not exist.&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-08-25T02:53:40.783" />
  <row Id="3264" PostTypeId="1" AcceptedAnswerId="3268" CreationDate="2010-08-25T04:04:14.540" Score="6" ViewCount="300" Body="&lt;p&gt;Somewhat prompted by the discussions of &lt;a href=&quot;http://math.stackexchange.com/users/232&quot;&gt;Qiaochu Yuan&lt;/a&gt; and &lt;a href=&quot;http://math.stackexchange.com/users/1102&quot;&gt;Moron&lt;/a&gt; in this &lt;a href=&quot;http://math.stackexchange.com/questions/3254&quot;&gt;question&lt;/a&gt;, I realized that my understanding of linear/affine transformations thus far had been built on a convoluted series of circular arguments. I will now be asking a question in order to patch the gaps in my knowledge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Due to my innate tendency to view things geometrically, I had always taken for granted the fact that rotations, reflections, and translations, among other affine transformations, are isometric (That is to say, no matter how you move, spin, or reflect an object, all the lengths and angles do not change at all).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To put it another way, I considered these transformations being isometries as postulates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question, then, is if there is a rigorous (so probably non-geometric) way of justifying that these transformations are isometries.&lt;/p&gt;&#xA;" OwnerUserId="498" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-08-25T23:39:34.410" LastActivityDate="2010-09-01T22:42:01.117" Title="On isometric affine transformations" Tags="&lt;geometry&gt;&lt;linear-algebra&gt;" AnswerCount="5" CommentCount="1" FavoriteCount="2" />
  <row Id="3265" PostTypeId="2" ParentId="3254" CreationDate="2010-08-25T04:43:38.243" Score="2" ViewCount="" Body="&lt;p&gt;Maybe my first answer was not enough &quot;formal&quot;, since I didn't say explicitely what I was assuming a &quot;rotation&quot; is -nor &quot;angle&quot;. (Exercise: (1) find in which places I was using those unsaid definitions of &quot;rotation&quot; and &quot;angle&quot; and (2) which definitions are they.   :-)   )&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, it was a sort of &quot;high-school proof&quot;. The problem about the definition of &quot;rotation&quot;, as some answers have already pointed out, or implied, is that, if you &lt;em&gt;really&lt;/em&gt; want to be formal, you end saying that a rotation is some kind of linear transformation &lt;em&gt;by definition&lt;/em&gt;. Full stop.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nevertheless, I think I can produce an &quot;undergraduate&quot; proof of this fact, avoiding circular arguments. It has the merit that is valid in any dimension. For which I need the following&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Convention.&lt;/strong&gt; Whatever &quot;rotation&quot; means,&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;It is a map of some vector space $V$,&lt;/li&gt;&#xA;&lt;li&gt;Which has a way of measuring &quot;lengths&quot; and &quot;angles&quot; of its vectors, and&lt;/li&gt;&#xA;&lt;li&gt;&quot;Rotations&quot; preserve those &quot;lengths&quot; and &quot;angles&quot;.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Now, a fancy way to have &quot;lengths&quot; and &quot;angles&quot; in a vector space is to have a dot product in it. So let's assume that $V$ is an euclidian vector space; for instance, a real vector space such as $\mathbb{R}^n$ with the standard dot product.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, in such a $V$, &quot;length&quot; means &quot;norm&quot;, which is $\| v \| = +\sqrt{v\cdot v}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have to be more careful with &quot;angles&quot; because the standard definition already involves rotations. To avoid a circular argument, we define the (non-oriented) angle determined by two vectors $v, w$ as the &lt;em&gt;unique&lt;/em&gt; real number $\theta = \widehat{vw} \in [0, \pi ]$ such that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\cos \theta = \frac{v\cdot w}{\|v\| \|w\|} \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice that this definition makes sense since, because of Schwartz inequality &lt;a href=&quot;http://en.wikipedia.org/wiki/Schwartz_Inequality&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Schwartz_Inequality&lt;/a&gt; , we always have $-1 \leq \frac{v\cdot w}{\|v\| \|w\|} \leq 1$ and $\cos : [0, \pi] \longrightarrow [-1,1]$ is bijective.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, &quot;rotation&quot; is some kind of map $f: V \longrightarrow V$ which preserves norms and angles. Since the dot product can be expressed in terms of norms and (cosines of) angles, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;v\cdot w = \|v\| \|w\| \cos\widehat{vw}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;rotations preserve dot products:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;f(v) \cdot f(w) = v\cdot w \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, let's show that a map that preserves the dot product is necessarily linear:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{align}&#xA;\| f(v+w) - f(v) -f(w) \|^2  &amp;amp;= \left( f(v+w) -f(v) -f(w) \right) \cdot \left( f(v+w) -f(v) -f(w)  \right)   \\&#xA;          &amp;amp;=  \left( f(v+w)\cdot f(v+w) - f(v+w) \cdot f(v) - f(v+w)\cdot f(w) \right) \\&#xA;          &amp;amp;{} \qquad + \left( -f(v)\cdot f(v+w) + f(v) \cdot f(v) + f(v)\cdot f(w) \right) \\&#xA;          &amp;amp;{} \qquad + \left( -f(w)\cdot f(v+w) + f(w)\cdot f(v) + f(w)\cdot f(w) \right) \\&#xA;          &amp;amp;= \left( (v+w)\cdot (v+w) - (v+w) \cdot v - (v+w)\cdot w \right) \\&#xA;          &amp;amp;{} \qquad + \left( -v\cdot (v+w) + v \cdot v + v\cdot w \right) \\&#xA;          &amp;amp;{} \qquad + \left( -w\cdot (v+w) + w\cdot v + w\cdot w \right) \\&#xA;          &amp;amp;= \| v+w - v -w \|^2 = 0  \ .&#xA;\end{align}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;f(v+w) = f(v) + f(w) \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A similar computation shows that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\|f(\lambda v ) - \lambda f(v) \|^2 = \|\lambda v - \lambda v\|^2 = 0 \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;f(\lambda v) = \lambda f(v) \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, our rotation $f$ is a linear transformation.&lt;/p&gt;&#xA;" OwnerUserId="664" LastEditorUserId="664" LastEditorDisplayName="" LastEditDate="2010-08-25T23:55:01.290" LastActivityDate="2010-08-25T23:55:01.290" CommentCount="2" />
  <row Id="3266" PostTypeId="2" ParentId="3264" CreationDate="2010-08-25T04:45:15.577" Score="3" ViewCount="" Body="&lt;p&gt;As in my comment on the question to which you refer, exactly how you define things is critical.  Suppose we consider transformations of the plane where the plane is defined to be $\mathbb{C}$ and a transformation of the plane is a 1-1 function $f:\mathbb{C}\to\mathbb{C}$.  An isometry is a transformation that preserves distances, so a transformation $f$ is an isometry if and only if $|f(a)-f(b)|=|a-b|$ for all $a,b\in\mathbb{C}$.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;em&gt;Theorem:&lt;/em&gt; Any function of the form $f(z)=cz+d$ (direct isometry) or $f(z)=c\bar{z}+d$ (opposite isometry) where $c,d\in\mathbb{C}$ and $|c|=1$ is an isometry.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;em&gt;proof:&lt;/em&gt; (direct isometry) $|f(a)-f(b)|$ $=|ca+d-(cb+d)|$ $=|ca-cb|$ $=|c(a-b)|$ $=|c|\cdot|a-b|$ $=1\cdot|a-b|$ $=|a-b|$.  (opposite isometry) $|f(a)-f(b)|$ $=|c\bar{a}+d-(c\bar{b}+d)|$ $=|c\bar{a}-c\bar{b}|$ $=|c(\bar{a}-\bar{b})|$ $=|c|\cdot|\overline{a-b}|$ $=1\cdot|a-b|$ $=|a-b|$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Also note that if $f$ and $g$ are isometries, so is $f\circ g$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For our purposes and for simplicity, define a translation by $a+bi$ to be the function $T_{a+bi}(z)=z+a+bi$, a rotation about 0 by $\theta$ to be the function $R_\theta(z)=(\cos\theta+i\sin\theta)z$, and reflection over the real axis to be $r_\mathbb{R}(z)=\bar{z}$.  (It should be relatively easy to show that these definitions align with any geometric definitions.)  Each of these is clearly an isometry, based on the above theorem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a rotation about some point $p$ rather than about 0, apply $T_p\circ R_\theta\circ T_{-p}$.  For a reflection over a line parallel to the real axis and passing through $bi$, apply $T_{bi}\circ r_\mathbb{R}\circ T_{-bi}$.  For a reflection over a line not parallel to the real axis, intersecting the real axis at $a$ at an angle $\theta$, apply $T_a\circ R_\theta\circ r_\mathbb{R}\circ R_{-\theta}\circ T_{-a}$.  Each of these is also an isometry because it is the composition of isometries.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, any rotation, reflection, or translation of the plane is an isometry.&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-08-25T04:45:15.577" CommentCount="3" />
  <row Id="3267" PostTypeId="2" ParentId="2968" CreationDate="2010-08-25T05:09:51.157" Score="4" ViewCount="" Body="&lt;p&gt;You should be able to use a &lt;a href=&quot;http://en.wikipedia.org/wiki/Homomorphic_encryption&quot; rel=&quot;nofollow&quot;&gt;Fully Homomorphic Encryption Scheme&lt;/a&gt; which was recently proved to be possible by Craig Gentry from IBM (and was an important open problem in cryptography till 2009!)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using a fully homomorphic scheme, any circuit can be evaluated homomorphically, effectively allowing computation on encrypted input without having to decrypt it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To use it in your case, each can encrypt their card value and do a homomorphic comparison (for instance see this: &lt;a href=&quot;http://portal.acm.org/citation.cfm?id=1356047&quot; rel=&quot;nofollow&quot;&gt;http://portal.acm.org/citation.cfm?id=1356047&lt;/a&gt;, caveat: I have only read the abstract). We can find out the person with the greatest value (or sort them in order) without having to divulge any non-encrypted information.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, this is still theoretical, it might be a while before this becomes practical.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See the wiki page for more details.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-08-25T13:13:23.530" LastActivityDate="2010-08-25T13:13:23.530" />
  <row Id="3268" PostTypeId="2" ParentId="3264" CreationDate="2010-08-25T05:13:29.397" Score="6" ViewCount="" Body="&lt;p&gt;The first point I'd make is that translations are &lt;em&gt;not&lt;/em&gt; linear transformations because every linear transformation must take 0 to 0, and the only translation that does this is the identity map.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, here's a general way to see exactly which linear transformations are isometries (including rotations and reflections).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One way is to start with the Euclidean inner product.  In the &quot;standard basis&quot;, it looks like $\langle v,w\rangle = \sum_i v_i w_i$.  Once we have this, the length of a vector $x$ is defined to be $\sqrt{\langle x,x\rangle}$.  The notation for this is $|x|$.  The angle between two vectors $x$ and $y$ is defined by $cos\theta = \frac{\langle x,y\rangle}{|x||y|}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We'll say a function &quot;preserves the inner product&quot; if $\langle x , y\rangle = \langle f(x), f(y)\rangle$ for all $x$ and $y$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Claim 1.  A linear function preserves the inner product iff the linear function preserves all lengths.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The proof one way is trivial.  Assuming the linear function preserves all lengths, notice that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$|x|^2 + 2\langle x,y\rangle + |y|^2 = \langle x+y,x+y\rangle = \langle f(x) + f(y), f(x) + f(y)\rangle $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$= |f(x)|^2 + 2\langle f(x),f(y)\rangle + |f(y)|^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, using the fact that $|x|^2 = |f(x)|^2$, we learn that $\langle x,y\rangle = \langle f(x), f(y)\rangle$.  (Note that, as an added bonus, if a linear transformation preserves length, it preserves the inner product and hence it automatically preserves angles as well.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The next thing worth noting is that you can move a linear transformation around in an inner product.  That is, if $A$ is any matrix, then there is a unique matrix $B$ such that $\langle Ax, y\rangle = \langle x,By\rangle$ for all $x$ and $y$.  In fact, in an orthonormal basis, $B$ is simply given as the transpose of $A$ - that is, $B = A^t$.  The proof is simple:  let $e_i$ be an orthonormal basis.  Then $A_{ij} = \langle Ae_i, e_j\rangle = \langle e_i, Be_j\rangle = B_{ji}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, we come to exactly how to recognize isometries.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Claim 2:  the matrix $A$ is an isometry iff $A^t A = Id$.  It turns out, this will automatically imply $AA^t = Id$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's the proof.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume $A$ is an isometry.  Then $\langle x,y\rangle = \langle Ax, Ay\rangle = \langle x, A^t A y\rangle$.  Thus, we have $\langle x,y\rangle = \langle x, A^tA y\rangle$ for all $x$ and $y$.  If we set $y = x$, then we get $|x|^2 = |x||A^tA x|cos\theta$.  But since $A$ is an isometry, so is $A^t$ and hence $|A^t A x| = |x|$.  Thus, we get $|x|^2 = |x|^2 cos\theta$ which tells us $\theta = 0$, i.e, that $A^tA x = x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conversely, if $A^tA = Id$, then $\langle x,y\rangle = \langle x, A^t A y\rangle = \langle Ax, Ay\rangle$, showing that $A$ is an isometry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, one just verifies that every rotation matrix and reflection matrix satisfies $A^tA = Id$.  &lt;/p&gt;&#xA;" OwnerUserId="331" LastActivityDate="2010-08-25T05:13:29.397" CommentCount="4" />
  <row Id="3269" PostTypeId="2" ParentId="2899" CreationDate="2010-08-25T06:24:28.423" Score="0" ViewCount="" Body="&lt;p&gt;This was actually &lt;a href=&quot;http://blog.hiremebecauseimsmart.com/post/977613775/arctan#disqus_thread&quot; rel=&quot;nofollow&quot; title=&quot;my question&quot;&gt;my question&lt;/a&gt; originally.  To put the demand for intuitiveness in starker terms:  &lt;strong&gt;what does the arc of a circle have to do with the area of a square and the operation of reciprocation?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If someone can show me a geometric figure with area $1 \over 1 + x^2$ that might be a good start.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or, they can explain how this sequence&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Times a number by itself;&lt;/li&gt;&#xA;&lt;li&gt;Add one to it;&lt;/li&gt;&#xA;&lt;li&gt;Flip that result&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;leads to &quot;a little bit of&quot; circular arc.&lt;/p&gt;&#xA;" OwnerUserId="1457" LastActivityDate="2010-08-25T06:24:28.423" CommentCount="8" />
  <row Id="3270" PostTypeId="1" AcceptedAnswerId="3288" CreationDate="2010-08-25T06:33:37.280" Score="8" ViewCount="332" Body="&lt;p&gt;I understand the technical and logical distinction between &quot;if&quot; and &quot;only if&quot; and &quot;if and only if&quot;.  But I have always been troubled by the phrase &quot;only if&quot; even though I am able to parse and interpret it.  Also in my posts on this and other sites I have frequently had to make edits to migrate the term &quot;only&quot;, sometimes across multiple structural boundaries of a sentence, which is empirical evidence to myself that I don't intuitively know the meaning of the word. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any simple rule that I can use to determine whether or not it is appropriate to use this word in a particular context in order to achieve more clarity?  In mathematical discourse, what are some other common lexical contexts, meaningful or not, in which appears the word &quot;only&quot;?  Why do I often write &quot;only&quot; in the wrong place?&lt;/p&gt;&#xA;" OwnerUserId="1284" LastEditorUserId="1284" LastEditorDisplayName="" LastEditDate="2010-08-25T08:52:28.513" LastActivityDate="2010-08-25T16:33:03.593" Title="What does &quot;only&quot; mean?" Tags="&lt;soft-question&gt;&lt;terminology&gt;" AnswerCount="2" CommentCount="11" FavoriteCount="2" />
  <row Id="3271" PostTypeId="1" AcceptedAnswerId="3274" CreationDate="2010-08-25T06:54:34.950" Score="3" ViewCount="498" Body="&lt;p&gt;How do I evaluate this function for given $s$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\zeta(s) = \sum_{n=1}^\infty \frac{1}{n^s} = \frac{1}{1^s} + \frac{1}{2^s} + \frac{1}{3^s} + \cdots. $&lt;/p&gt;&#xA;" OwnerUserId="705" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-08-27T01:20:37.343" LastActivityDate="2010-12-03T13:53:07.570" Title="How to evaluate Riemann Zeta function" Tags="&lt;complex-analysis&gt;&lt;numerical-methods&gt;&lt;riemann-zeta&gt;&lt;special-functions&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="5" />
  <row Id="3272" PostTypeId="2" ParentId="2968" CreationDate="2010-08-25T07:10:43.157" Score="5" ViewCount="" Body="&lt;p&gt;If I understand your question correctly, your problem can be seen as a multi-party extension of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Yao%27s_Millionaires%27_Problem&quot; rel=&quot;nofollow&quot;&gt;Millionaires' problem&lt;/a&gt;. In the Millionaires' problem problem, two millionaires want to know which of them is richer without revealing their wealth. There is an secure two-party protocol (due to &lt;a href=&quot;http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.103.7844&amp;amp;rep=rep1&amp;amp;type=pdf&quot; rel=&quot;nofollow&quot;&gt;Andy Yao, 1982&lt;/a&gt;), and there may be even more sophisticated algorithms around these days. Maybe a simple extension of this protocol will solve your problem (every peer talks to everybody else to lean that the other guy has a higher card). However, I don't know if this simple extension is still secure.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;EDIT: After looking over Yao's paper I realize that he is also already addressing the multi-party problem. So you should find your solution in his paper, or follow some of the references given in &lt;a href=&quot;http://en.wikipedia.org/wiki/Secure_multi-party_computation&quot; rel=&quot;nofollow&quot;&gt;this article at Wikipedia&lt;/a&gt;. Of course, the recent homomorphic encryption scheme will also solve your problem, as has been pointed out by user Moron before. However, since the homomorphic encryption scheme is very general, it might be a little bit too complicated for your problem at hand if you are looking for a practical scheme.&lt;/p&gt;&#xA;" OwnerUserId="347" LastEditorUserId="347" LastEditorDisplayName="" LastEditDate="2010-08-25T09:52:37.130" LastActivityDate="2010-08-25T09:52:37.130" CommentCount="3" />
  <row Id="3273" PostTypeId="1" CreationDate="2010-08-25T07:32:51.077" Score="6" ViewCount="206" Body="&lt;p&gt;Hi all,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Weierstrass' approximation theorem for continuous functions on a compact space by using polynomials is well-known. As far as I know, there are some variants of this theorem, e.g. Stone-Weierstrass that does not refer only to polynomials as approximator functions. Where could I find these Weierstrass-like approximation theorems? On-line references are OK, but one might also point to some books.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance,&#xA;Lucian&lt;/p&gt;&#xA;" OwnerUserId="1268" LastEditorUserId="1268" LastEditorDisplayName="" LastEditDate="2010-08-25T11:27:30.100" LastActivityDate="2010-12-24T05:29:11.530" Title="Approximation theorems" Tags="&lt;approximation&gt;&lt;approximation-theory&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="2" />
  <row Id="3274" PostTypeId="2" ParentId="3271" CreationDate="2010-08-25T08:48:24.983" Score="6" ViewCount="" Body="&lt;p&gt;Assuming you're talking about numerical evaluation, along the critical strip $0&amp;lt;\Re s&amp;lt;1$ and &quot;large&quot; $\Im s$, (which is the region of interest for many) the &lt;a href=&quot;http://www.ams.org/journals/tran/1988-309-02/S0002-9947-1988-0961614-2/S0002-9947-1988-0961614-2.pdf&quot; rel=&quot;nofollow&quot;&gt;Riemann-Siegel formula&lt;/a&gt; is standard; off the strip, what you can manage is a polyalgorithm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $\Re s\leq0$, one can use the reflection formula for $\zeta$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\zeta(1-s)=\frac{2}{(2\pi)^s}\cos\left(\frac{s\pi}{2}\right)\Gamma(s)\zeta(s)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so we can consider evaluation for $\Re s&amp;gt;0$ in what follows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that if $|s|$ is &quot;large enough&quot; (how large is &quot;large&quot; depends on the computing environment you're in), one can simply use the defining series, since its terms quickly diminish in magnitude. That leaves the problem of what to do for small to medium-sized $s$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The key is to use the &lt;a href=&quot;http://mathworld.wolfram.com/DirichletEtaFunction.html&quot; rel=&quot;nofollow&quot;&gt;Dirichlet $\eta$ function&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\eta(s):=-\sum_{j=1}^\infty \frac{(-1)^j}{j^s}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is related to $\zeta$ by the following identity:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\eta(s)=\left(1-2^{1-s}\right)\zeta(s)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason for our interest in $\eta$ is that although this sum is slowly convergent, it is an alternating series, and a number of algorithms exist for quickly finding the sum of an alternating series numerically.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One method is the &lt;a href=&quot;http://math.stackexchange.com/questions/6625&quot;&gt;Levin transformation&lt;/a&gt;; another one, which is one of the simplest methods for numerically summing alternating series (and my personal favorite) is the Cohen-Rodriguez Villegas-Zagier algorithm. The algorithm is a bit too long to describe here, so I will just have to point you to the &lt;a href=&quot;http://projecteuclid.org/euclid.em/1046889587&quot; rel=&quot;nofollow&quot;&gt;original paper&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is in fact identical to the approach taken by Borwein in this &lt;a href=&quot;http://www.cecm.sfu.ca/personal/pborwein/PAPERS/P155.pdf&quot; rel=&quot;nofollow&quot;&gt;paper&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="498" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-03T13:53:07.570" LastActivityDate="2010-12-03T13:53:07.570" CommentCount="7" />
  <row Id="3275" PostTypeId="1" AcceptedAnswerId="3282" CreationDate="2010-08-25T10:00:34.200" Score="3" ViewCount="171" Body="&lt;p&gt;I'm reading Shelah's book &quot;Proper and Improper Forcing&quot; (the first two chapters were recommended for learning the basics of forcing)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given a quasi-order $P$ we say that $\mathcal{I}$ is a &lt;em&gt;dense&lt;/em&gt; subset of $P$  if&#xA;$$(\forall p \in P) (\exists q \in \mathcal{I}) (p\le q)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We say that it is open if for any $p,q \in P$ we have that $p \in \mathcal{I} \wedge p\le q$ then $q \in \mathcal{I}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$G$ is called &lt;em&gt;directed&lt;/em&gt; if every two elements in $G$ has an upper bound in $G$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$G$ is called &lt;em&gt;downward closed&lt;/em&gt; if for every $p \in G$ and $q\in P$ if $q \le p$ then $q \in G$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A subset $G$ of $P$ is called &lt;em&gt;generic&lt;/em&gt; over $V$ if it is directed, downward closed and for any dense and open subset of $P$ that is in $V$, the intersection with $G$ is non-empty.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now they give proof that if $P$ has no trivial branches then a generic set cannot exist in the universe, saying that if $G \in V$ then $P\backslash G \in V$ and it is dense and open, how come? I can't figure that out.&lt;/p&gt;&#xA;" OwnerUserId="622" LastEditorUserId="622" LastEditorDisplayName="" LastEditDate="2010-08-25T15:41:10.587" LastActivityDate="2010-08-25T15:46:11.657" Title="Generic sets in ZFC" Tags="&lt;set-theory&gt;" AnswerCount="1" CommentCount="6" />
  <row Id="3276" PostTypeId="2" ParentId="2899" CreationDate="2010-08-25T10:07:38.140" Score="3" ViewCount="" Body="&lt;p&gt;Stereographic projection offers a geometric interpretation of this formula. In fact the factor&#xA;$(1+x^2)^{-1}$ is the metric of the circle in the stereographic projection coordinate. The construction is as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take a circle of radius 1/2, call the angle of (the radius vector of) a general &#xA;point on the circle with respect to the radius to the south pole $2\theta$.&#xA;Connect the point with the north pole and call the distance of the point of intersection of the extension&#xA;of this chord with the tangent at the south pole $x$. One can parameterize the points on the circle&#xA;by the coordinate $x$ which is called the stereographic projection coordinate. It is easy to see that $ x = tan(\theta)$.&#xA;The length of an arc of the circle between $\theta_1$ and$\theta_2$ (having the stereographic projection coordinates $x_1$ and $x_2$) is given by:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$s = 2 \int_{\theta_1}^{\theta_2} d\theta = 2\int_{x_1}^{x_2} \frac{dx}{1+x^2}$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This construction has further nice properties&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The group of plane rotations $SO(2)$ acts on $x$ by a Mobius transformation.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;This construction generalizes to spheres in higher dimensions.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="1430" LastActivityDate="2010-08-25T10:07:38.140" />
  <row Id="3277" PostTypeId="2" ParentId="1730" CreationDate="2010-08-25T11:00:25.667" Score="5" ViewCount="" Body="&lt;p&gt;I just wanted to add a few cents to this post to say that Isaac's statement &quot;&lt;em&gt;I've seen function machines used as low as 3rd grade&lt;/em&gt;&quot; is quite true.  I teach kindergarten and 1st grade, and I use function machines with my students, mostly when introducing the idea of complements in relation to addition and subtraction.  It's a standard part of the &quot;patterns and algebra&quot; portion of the Everyday Mathematics curriculum.  Functions are honestly NOT a challenging concept for my students to grasp when presented in this manner, hence, I am absolutely confident that your College Algebra students will be just fine! :)&lt;/p&gt;&#xA;" OwnerUserId="219" LastActivityDate="2010-08-25T11:00:25.667" />
  <row Id="3278" PostTypeId="1" AcceptedAnswerId="3310" CreationDate="2010-08-25T11:44:49.830" Score="3" ViewCount="301" Body="&lt;p&gt;In high school, we were often given questions of the form: &quot;What is the probability of A given B?&quot; For example:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What is the probability that two&#xA;  people were born on the same day given&#xA;  that one was born on a Tuesday?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Intuitively, most people expect that knowing someone was born on a Tuesday doesn't give you any information, as it is no different from being born on a Monday or Wednesday. However, the answer given is that we are looking at pairs of people with at least one born on Tuesday (13 out of 49) and only one of these pairs has both people born on the same day, so the answer is 1 in 13. If by &quot;given one was born on Tuesday&quot;, you mean that we the birthdays of the two people are uniformly distributed between all possible pairs where at least one was born on a Tuesday&quot;, then this analysis is correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the question stated just says that we are &quot;given&quot; this fact. It doesn't explain how we came to know this fact. Lets suppose someone blurted this fact out randomly. Generally it wouldn't be because they looked at groups of people and discarded them until they found a pair where at least one was born on a Tuesday. Instead, I think it would be better to model it as someone randomly getting a pair of people, then making a true statement about them. For simplicity, we will assume they statements are of the form: &quot;One of these people was born on a (INSERT DAY)&quot;. We will assume that if they were born on different days, each is equally likely. In this model, the statement that someone was born on a Tuesday actually makes no difference to the chance they were born on the same day.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems that part of the problem comes from giving the word &quot;given&quot; (often represented by the symbol |) a formal definition. People expect it to have the same meaning as the English word given. Anyway:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Is there a word I can use instead of &quot;given&quot; to be clearer?&lt;/li&gt;&#xA;&lt;li&gt;Do mathematician's object to the problem being stated this way, or is this considered to be clear? If they do, why is the symbol &quot;|&quot; often pronounced as &quot;given&quot;?&lt;/li&gt;&#xA;&lt;li&gt;Have people who attended high school in other countries had this difference made clear to them?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="123" LastEditorUserId="123" LastEditorDisplayName="" LastEditDate="2010-08-25T21:53:00.590" LastActivityDate="2010-08-26T11:48:14.930" Title="Probability of A given B" Tags="&lt;probability&gt;" AnswerCount="3" CommentCount="6" FavoriteCount="1" />
  <row Id="3279" PostTypeId="2" ParentId="3273" CreationDate="2010-08-25T13:40:38.083" Score="3" ViewCount="" Body="&lt;p&gt;For approximation with Polynomials, a Weierstrass like theorem is the &lt;a href=&quot;http://mathworld.wolfram.com/MuentzsTheorem.html&quot; rel=&quot;nofollow&quot;&gt;Muntz's Theorem&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moving away from polynomials, we have the classic &lt;a href=&quot;http://en.wikipedia.org/wiki/Fourier_series&quot; rel=&quot;nofollow&quot;&gt;Fourier Series&lt;/a&gt;. The &lt;a href=&quot;http://mathworld.wolfram.com/GeneralizedFourierSeries.html&quot; rel=&quot;nofollow&quot;&gt;Generalization&lt;/a&gt; of Fourier series gives rise to many approximation schemes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sorry, I wasn't able to find a single page...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope that helps.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-08-25T13:40:38.083" CommentCount="3" />
  <row Id="3280" PostTypeId="2" ParentId="3273" CreationDate="2010-08-25T13:47:12.893" Score="2" ViewCount="" Body="&lt;p&gt;Functions belonging to reproducing kernel Hilbert spaces can be approximated by weighted discrete sums of the reproducing kernels evaluated at discrete points of the dual variable. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;See the following two articles: &lt;a href=&quot;http://math.bu.edu/people/mkon/M510-1-05.pdf&quot; rel=&quot;nofollow&quot;&gt;article-1&lt;/a&gt; &lt;a href=&quot;http://www.nt.ntnu.no/users/skoge/prost/proceedings/ifac2002/data/content/00753/753.pdf&quot; rel=&quot;nofollow&quot;&gt;article-2&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1430" LastActivityDate="2010-08-25T13:47:12.893" />
  <row Id="3281" PostTypeId="2" ParentId="3270" CreationDate="2010-08-25T14:40:51.300" Score="2" ViewCount="" Body="&lt;p&gt;I think analogies in plain English are the way to internalize this... so here's one:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given that you want to wear socks with your shoes, put your shoes on &lt;em&gt;only if&lt;/em&gt; you have already put your socks on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The idea is that &lt;em&gt;there is no other way&lt;/em&gt; to arrive at the state of having your socks and shoes on (aside from the ridiculous possibility of placing your socks over your shoes).&lt;/p&gt;&#xA;" OwnerUserId="121" LastActivityDate="2010-08-25T14:40:51.300" CommunityOwnedDate="2010-08-25T14:40:51.300" CommentCount="0" />
  <row Id="3282" PostTypeId="2" ParentId="3275" CreationDate="2010-08-25T15:35:09.833" Score="2" ViewCount="" Body="&lt;p&gt;I'm going to use the def of &quot;directed&quot; that I found in Shelah's &lt;a href=&quot;http://projecteuclid.org/DPubS?service=UI&amp;amp;version=1.0&amp;amp;verb=Display&amp;amp;handle=euclid.pl/1235419814&quot; rel=&quot;nofollow&quot;&gt;book&lt;/a&gt;&lt;sup&gt;1&lt;/sup&gt;:  A set $G$ is directed if it is downward closed (if $p$ in G and if $q\leq p$ then $q\in G$) &lt;strong&gt;and&lt;/strong&gt; every two memebers of $G$ have a common upperbound in $G$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To see that $P-G$ is open is not too hard:  Let $p\in P-G$ and suppose $q\in G$ with $p\leq q$.  If $q$ were in $G$, then since $G$ is downwardly closed, we'd have $p\in G$, but that contradicts $p\in P-G$.  Thus, $q\notin G$ so $q\in P-G$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, why is $P-G$ dense?  Well, let $p\in P$.  We want to find a $q\in P-G$ with $p\leq q$.  Since there are no trivial branches, we know that above $p$ and are two points $r$ and $s$ with no common upperbound.  Since $G$ is directed, $r$ and $s$ cannot both be in $G$.  Hence, say, $r\notin G$.  Then $r\in P-G$, so we're done.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;1&lt;/sup&gt;I'm using chapter 1 and finding the definition on page 3 of the .pdf.&lt;/p&gt;&#xA;" OwnerUserId="331" LastEditorUserId="152" LastEditorDisplayName="" LastEditDate="2010-08-25T15:46:11.657" LastActivityDate="2010-08-25T15:46:11.657" CommentCount="4" />
  <row Id="3283" PostTypeId="1" AcceptedAnswerId="3286" CreationDate="2010-08-25T16:00:33.537" Score="11" ViewCount="657" Body="&lt;p&gt;I asked myself this question when I was a young boy playing around with the calculator.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Today, I think I know the answer, but I'm not sure whether I'd be able to explain it to a child or layman playing around with a calculator.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence I'm interested in answers suitable for a person that, say, knows what the square root of a square number is, but doesn't know about sequences, functions, convergence and the like.&lt;/p&gt;&#xA;" OwnerUserId="367" LastActivityDate="2010-08-26T00:00:03.893" Title="&quot;Why do I always get 1 when I keep hitting the square root button on my calculator?&quot;" Tags="&lt;calculus&gt;&lt;convergence&gt;&lt;calculator&gt;" AnswerCount="7" CommentCount="8" />
  <row Id="3284" PostTypeId="2" ParentId="3283" CreationDate="2010-08-25T16:11:39.617" Score="13" ViewCount="" Body="&lt;p&gt;Make a picture of the usual spiral converging to the fixed point, in the style of &lt;img src=&quot;http://i.imgur.com/WyeV5.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where does this little obsession aparent on this site with 'layman' explanations? :P&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;NB.&lt;/strong&gt; The picture comes from the Wikipedia page on &lt;a href=&quot;http://en.wikipedia.org/wiki/Fixed_point_%28mathematics%29&quot; rel=&quot;nofollow&quot;&gt;fixed points&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Later.&lt;/strong&gt; To make the picture it is better to use the cosine (so you do not even have to &lt;em&gt;make&lt;/em&gt; the picture because that is what's in the wikipedia page) mainly because the iteration is seen more clearly: for the square root, the sequence converges boringly and fast to be interesting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;PS.&lt;/strong&gt; You can draw this kind of pictures, assuming you have acces to &lt;em&gt;Mathematica&lt;/em&gt;, with the following code:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;f[t_] := Cos[t];&#xA;g = Plot[{f[t], t}, {t, -\[Pi]/2, \[Pi]/2}, &#xA;  AspectRatio -&amp;gt; 1, PlotStyle -&amp;gt; {Thick}, Epilog -&amp;gt;&#xA;   Module[{pts = NestList[f, .1, 10]},&#xA;    {&#xA;     Thin, PointSize[0.015],  &#xA;     Line@ Flatten[Map[{#, {#[[2]], #[[2]]}} &amp;amp;, Partition[pts, 2, 1], 1], 1],&#xA;     Red, Point[Partition[pts, 2, 1]]&#xA;     }&#xA;   ]&#xA;]    &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="274" LastEditorUserId="274" LastEditorDisplayName="" LastEditDate="2010-08-25T16:42:41.767" LastActivityDate="2010-08-25T16:42:41.767" CommentCount="5" />
  <row Id="3286" PostTypeId="2" ParentId="3283" CreationDate="2010-08-25T16:22:10.600" Score="10" ViewCount="" Body="&lt;p&gt;Here is how I had justified it to myself when I was a kid (I was convinced :-))&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $x &gt; 1$ then $\sqrt x &gt; 1$ and $x &gt; \sqrt x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we keep reducing the number while still being $&gt; 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, we cannot end up at a number $&gt;1$ as then taking the square root would reduce it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we do end up at $1$, we stay there. Since the calculator has a limited precision, we end up at $1$, and pretty quickly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, the assumption here is that we do end up somewhere :-) &#xA;(Which seemed justified by the fact that the first non-zero digit after the decimal point always seemed to reduce)&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-08-25T16:27:34.983" LastActivityDate="2010-08-25T16:27:34.983" CommentCount="4" />
  <row Id="3287" PostTypeId="2" ParentId="3283" CreationDate="2010-08-25T16:25:27.060" Score="7" ViewCount="" Body="&lt;p&gt;If they know about logarithms, try this:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;taking repeated square roots of a positive real number is the same as repeatedly dividing its logarithm by 2,&lt;/li&gt;&#xA;&lt;li&gt;repeatedly dividing something by 2 gets you to 0, and&lt;/li&gt;&#xA;&lt;li&gt;the exponential of 0 is 1.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-08-25T16:25:27.060" CommentCount="4" />
  <row Id="3288" PostTypeId="2" ParentId="3270" CreationDate="2010-08-25T16:33:03.593" Score="12" ViewCount="" Body="&lt;p&gt;The terms &quot;only&quot; and &quot;only if&quot; mean different things.  The first is English and the second is mathematical.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) &quot;only&quot; in English means &quot;just this and not others&quot;, where &quot;others&quot; could be an object or an action, depending on what &quot;only&quot; is connected with.  For example, &quot;We only proved that $p$ is prime&quot; means you showed $p$ is prime but not anything further, which I think is synonymous with &quot;We proved only that $p$ is prime&quot;; it is just a stylistic judgment as to which of those you use and I prefer the second version at the moment. However, &quot;We proved that only $p$ is prime&quot; means you proved $p$ is prime while other numbers in the argument (that may possibly have been prime) are not prime.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The classical example, which is not meant as a personal remark, is to insert &quot;only&quot; in front of each word of the sentence &quot;I love you&quot;. Each version has a completely different meaning.  This is discussed under the topic &quot;Modifiers&quot; on the page&#xA;&lt;a href=&quot;http://www.kaptest.com/GMAT/Learn-and-Discuss/Community/blogs/tag/verbal/&quot; rel=&quot;nofollow&quot;&gt;http://www.kaptest.com/GMAT/Learn-and-Discuss/Community/blogs/tag/verbal/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) In math, &quot;only if&quot; has the same meaning as &quot;implies&quot;.  (Logically, it's the direction that is not &quot;if&quot;, which is how I first was able to remember its meaning.) In terms of Tom's example, &quot;shoes are on ==&gt; socks are on&quot; has the same meaning as &quot;your shoes are on only if your socks are on.&quot;&lt;/p&gt;&#xA;" OwnerUserId="619" LastActivityDate="2010-08-25T16:33:03.593" />
  <row Id="3289" PostTypeId="2" ParentId="2952" CreationDate="2010-08-25T16:41:11.567" Score="2" ViewCount="" Body="&lt;p&gt;This is supposed to be a comment, but I don't seem to be able to write one :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm quite sure it is an open problem : it was a few years ago when I last checked up on it. It's a very difficult one. There are indeed sufficient conditions, but necessary and sufficient ones seem very hard to find.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I recommend you take a look at the recent work of Kuznetsova, Yu. N. on the subject.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Malik&lt;/p&gt;&#xA;" OwnerUserId="1197" LastActivityDate="2010-08-25T16:41:11.567" CommentCount="2" />
  <row Id="3290" PostTypeId="2" ParentId="3254" CreationDate="2010-08-25T16:50:54.517" Score="2" ViewCount="" Body="&lt;p&gt;As an exercise in trigonometry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/OWu6S.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The transformation of rectangular co-ordinates $(x,y)$ of $P$ into the&#xA;rectangular co-ordinates $(x',y')$ of $P$ by a &lt;em&gt;rotation of the axes&lt;/em&gt; is given by:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x=x^{\prime }\cos \phi -y^{\prime }\sin \phi $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$y=x^{\prime }\sin \phi +y^{\prime }\cos \phi $.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This transformation is invertible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: This way of proving the transformation of rectangular co-ordinates does not &quot;avoid using Euclidean geometry explicitly&quot; (in this case trigonometry), as writen in the question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However the same method can be used to derive the transformation of rectangular into spherical co-ordinates, which is a non-linear transformation.&lt;/p&gt;&#xA;" OwnerUserId="752" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2010-08-25T20:42:21.720" LastActivityDate="2010-08-25T20:42:21.720" />
  <row Id="3291" PostTypeId="1" AcceptedAnswerId="3292" CreationDate="2010-08-25T16:52:54.200" Score="3" ViewCount="107" Body="&lt;p&gt;When calculating some probabilities, I got sums of the form&#xA;$$\sum_{j=0}^c {j+a+b \choose a} p^j,$$ for integers $a, b, c &gt; 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does someone know closed forms for these values?&lt;/p&gt;&#xA;" OwnerUserId="669" LastEditorUserId="152" LastEditorDisplayName="" LastEditDate="2010-08-25T17:47:20.837" LastActivityDate="2010-08-26T04:20:07.293" Title="Estimates involving sums with binomials" Tags="&lt;sequences-and-series&gt;&lt;binomial-coefficients&gt;" AnswerCount="2" CommentCount="9" FavoriteCount="1" />
  <row Id="3292" PostTypeId="2" ParentId="3291" CreationDate="2010-08-25T16:59:16.430" Score="3" ViewCount="" Body="&lt;p&gt;I don't have time right now to work out the exact answer, but the methods and identities in the middle third of &lt;a href=&quot;http://www.artofproblemsolving.com/Forum/blog.php?u=14052&amp;amp;b=10590&quot; rel=&quot;nofollow&quot;&gt;this blog post&lt;/a&gt; will give a closed form for fixed $a$.  Use the geometric series formula to find $\sum_{j=0}^c p^{a+b+j}$, differentiate $a$ times with respect to $p$, then divide by $a! p^b$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If a closed form for fixed $a$ isn't good enough, you should probably be more precise about which of your parameters are large and which are small.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit:  Still don't have time to give a complete answer, but here's a fun trick.  Instead of computing the answer for fixed $a$ we can write down a generating function&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle P_{b,c}(x) = \sum_{a=0}^{\infty} x^a \sum_{j=0}^c {a+b+j \choose a} p^j$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;then exchange the order of summation, giving&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle \begin{align}&#xA;P_{b,c}(x) &amp;amp;= \sum_{j=0}^c p^j \sum_{a=0}^{\infty} {a+b+j \choose a} x^a \\&#xA; &amp;amp;= \sum_{j=0}^c p^j \frac{1}{(1 - x)^{b+j+1}} \\&#xA; &amp;amp;= \frac{1}{(1 - x)^{b+1}} \sum_{j=0}^c \left( \frac{p}{1-x} \right)^j \\&#xA; &amp;amp;= \frac{1}{(1 - x)^{b+1}} \frac{1 - \left( \frac{p}{1-x} \right)^{c+1}}{1 - \frac{p}{1-x}} \\&#xA; &amp;amp;= \frac{(1-x)^{c+1} - p^{c+1}}{(1 - x)^{b+c+1}(1 - p - x)}. \end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then the coefficient of $x^a$ of this rational function is the number you want.  Not sure how useful that is for you, but you might be able to extract a useful asymptotic from it.  The identity in the second line is the binomial theorem for negative exponents.&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-08-26T04:20:07.293" LastActivityDate="2010-08-26T04:20:07.293" CommentCount="2" />
  <row Id="3293" PostTypeId="2" ParentId="3291" CreationDate="2010-08-25T17:00:57.120" Score="0" ViewCount="" Body="&lt;p&gt;If you don't need an exact closed form or just care about the asymptotic growth &lt;a href=&quot;http://en.wikipedia.org/wiki/Stirling%27s_approximation&quot; rel=&quot;nofollow&quot;&gt;Stirling's Approximation&lt;/a&gt; might be useful. I've found it useful in the past for generating good asymptotic bounds on similar functions.&lt;/p&gt;&#xA;" OwnerUserId="1487" LastActivityDate="2010-08-25T17:00:57.120" CommentCount="1" />
  <row Id="3294" PostTypeId="1" AcceptedAnswerId="3983" CreationDate="2010-08-25T17:45:07.927" Score="4" ViewCount="367" Body="&lt;p&gt;I subdivide a cubic bezier curve at a given &lt;em&gt;t&lt;/em&gt; value using de Casteljau’s algorithm, which yields two cubic bezier curves. Afterwards I “scale” the second curve (proportionally).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I’d like to reconnect or approximate the two curves to/with a single curve in a third step. Is that possible?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/ujFOM.png&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This illustrates what I’m intending to do.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I guess reversing de Casteljau’s algorithm won’t work because I don’t have one of the intermediate points.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there are multiple approaches, I’d favor a simpler (faster to compute) strategy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update:&lt;/strong&gt;&#xA;Maybe this figure makes it more clear; it shows all the points I have:&#xA;&lt;img src=&quot;http://i.imgur.com/0hh6f.png&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The original cubic bezier curve is defined by the points $ p_{0}, p_{1}, p_{2}, p_{3} $. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is divided at a given $ t $ (timing) value using De Casteljau’s algorithm, which yields the points $ q_{1}, r_{2}, i_{1}, q_{2}, r_{1}, k $ where $ k $ is the division point. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The two subcurves are defined by the control points $ p_{0}, q_{1}, q_{2}, k $ and $ k, r_{1}, r_{2}, p_{3} $, respectively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The scaled second subcurve is defined by the points $ k, {r}' _{1}, {r}' _{2}, {r}' _{3} $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Scaling is applied as follows: $ {p}' = k + (p - k) \cdot factor $ for $ r_{1}, r_{2}, p_{3} $&lt;/p&gt;&#xA;" OwnerUserId="1488" LastEditorUserId="1488" LastEditorDisplayName="" LastEditDate="2010-09-01T13:11:56.463" LastActivityDate="2010-09-03T22:36:41.800" Title="How to approximate/connect two continuous cubic bezier curves with/to a single one?" Tags="&lt;approximation&gt;&lt;bezier-curve&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" />
  <row Id="3295" PostTypeId="2" ParentId="3283" CreationDate="2010-08-25T18:03:26.050" Score="4" ViewCount="" Body="&lt;p&gt;Dualizing reduces it to: repeatedly squaring $1+\epsilon$ goes to $\infty$. But that's clear by the Binomial theorem.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Namely: given any positive $\;\epsilon, \;$ we can force $\;\;\; 2^n\epsilon &gt; x \;\;$ by choosing $\: n \:$ large enough.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore we have that $\;\;\; (1+\epsilon)^{2^n} \ge 1+2^n \epsilon &gt; x \ge 1\quad$ by the Binomial Theorem. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, taking $\;2^n\:$'th roots: $\;\;\; 1+\epsilon \;\ge\; x^{1/2^n} \;\ge\; 1. \;\;$ Hence $\; x^{1/2^n}\to 1\;$ as $\;n\to\infty$&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-08-25T20:36:45.307" LastActivityDate="2010-08-25T20:36:45.307" CommentCount="2" />
  <row Id="3296" PostTypeId="2" ParentId="1236" CreationDate="2010-08-25T18:11:37.750" Score="1" ViewCount="" Body="&lt;p&gt;One interesting fact told to me by my father, a chem engineer, is that if you have a high pressure gas leaking into a low pressure gas through a small hole, there is a upper limit to the rate of flow. That is, no matter how high the pressure gets on the high pressure side, the rate of flow does not surpass some finite limit. I suppose this relates to the speed of sound, but I don't know. I also don't know if there a corresponding phenomenon with liquids.&lt;/p&gt;&#xA;" OwnerUserId="1312" LastActivityDate="2010-08-25T18:11:37.750" CommentCount="1" />
  <row Id="3297" PostTypeId="2" ParentId="3283" CreationDate="2010-08-25T18:22:12.597" Score="0" ViewCount="" Body="&lt;p&gt;I think a better question is to analyze how the roots of a number X approach 1, depending on X.  I.e. x&lt;sup&gt;a&lt;/sup&gt; ~ 1 + a(lnX).  Not only does this introduce the log, it gives a natural derivation of e which is easy to remember.  I never understood what was &quot;natural&quot; about the natural log until I was an undergraduate, and I think the real value of this sequence is the way it leads so memorably and naturally to the derivation of e. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's easy to see why you arrive at one from repeatedly taking square roots (as moron answered), since the sequence is strictly decreasing and bounded below by 1.  I think a motivated person of almost any age can understand that, as long as it's explained slowly and clearly.&lt;/p&gt;&#xA;" OwnerUserId="1417" LastActivityDate="2010-08-25T18:22:12.597" />
  <row Id="3299" PostTypeId="2" ParentId="3264" CreationDate="2010-08-25T18:57:33.837" Score="2" ViewCount="" Body="&lt;p&gt;Ok, this one is pretty easy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Check out the theory of orthogonal matricies on wiki.  The basic idea is that the matrix corresponding to the isometric linear transformation must be orthogonal.  The proof is really pretty, but basically the idea is to show that an inner product of two vectors&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(X,Y) is equal to (AX,AY) for a matrix A, if and only if the transpose of A is the same as it's inverse.  i.e. A is orthogonal.  The next step is to show that the rotation matrix is orthogonal, and that's a rigorous proof.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This idea is easily generalized to non-euclidean metrics.  Although it might be at too high a level, I absolutely love matrix lie groups by brian C hall.  The first chapter is on &lt;a href=&quot;http://books.google.com/books?id=0SQRcucskaIC&amp;amp;printsec=frontcover&amp;amp;dq=lie+groups+brian+hall&amp;amp;hl=en&amp;amp;ei=Kmd1TK2GDoH98AbApuWnBw&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=1&amp;amp;ved=0CDEQ6AEwAA#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;google books&lt;/a&gt;, and contains some simple proofs.  This book only gets super hard in chapter 3;lie groups/algebras is one of my favorite subjects and is certainly where this line of questioning will lead you.&lt;/p&gt;&#xA;" OwnerUserId="1417" LastActivityDate="2010-08-25T18:57:33.837" />
  <row Id="3300" PostTypeId="2" ParentId="3119" CreationDate="2010-08-25T19:05:46.920" Score="1" ViewCount="" Body="&lt;p&gt;I'm not sure having really understood your question, but it seems to me that mathematicians make no difference between heterogeneous and homogeneous tuples. Consider $\left(5,b,f,8,3\right)$: one can say it is an heterogeneous tuple because the set $\left\{5,b,f,8,3\right\}$ can be partitioned into $\left\{b,f\right\}$ and $\left\{5,8,3\right\}$, letters and digits respectively. But are letters and digits of different type? What do you mean by type? In mathematics, as far as I know, there is only one type: the set type. Everything is a set. In this example letters and digits should be both defined as (particular) sets. So every tuple is homogeneous by default. Also, consider $\left(2,1,9,7\right)$: there are only digits this time, so one can say it is an homogeneous tuple. But what if I split the set of all digits into those less then $5$ (the &quot;low&quot; digit type) and those equal to or greater than $5$ (the &quot;high&quot; digit type)? It becomes heterogeneous. I would say the context is of great importance here.&lt;/p&gt;&#xA;" OwnerUserId="1175" LastActivityDate="2010-08-25T19:05:46.920" CommentCount="1" />
  <row Id="3301" PostTypeId="2" ParentId="3250" CreationDate="2010-08-25T19:08:45.050" Score="7" ViewCount="" Body="&lt;p&gt;Length, in Euclidean geometry, is a relation, not a number. We say that two segments are of the same length if they are congruent to each other, and congruence is one of the undefined notions of Euclidean geometry. The Pythagorean theorem comes about as soon as you decide to map lengths to positive real numbers, and it does so because of how length and angles are related by the congruence and similarity axioms (e.g. how the Pythagorean theorem is proved).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lengths, in vector spaces, can be anything that satisfies the axioms for a &lt;a href=&quot;http://en.wikipedia.org/wiki/Normed_vector_space&quot; rel=&quot;nofollow&quot;&gt;norm&lt;/a&gt;. But to have a Pythagorean theorem, you need a notion of perpendicularity of vectors, that is, &lt;a href=&quot;http://en.wikipedia.org/wiki/Inner_product&quot; rel=&quot;nofollow&quot;&gt;an inner product&lt;/a&gt;, which exists only if the norm satisfies &lt;a href=&quot;http://en.wikipedia.org/wiki/Parallelogram_law&quot; rel=&quot;nofollow&quot;&gt;the parallelogram law&lt;/a&gt;. For example, &lt;a href=&quot;http://en.wikipedia.org/wiki/Taxicab_metric&quot; rel=&quot;nofollow&quot;&gt;the taxicab norm&lt;/a&gt;, which is given by |(a,b)|=|a|+|b| does not come from an inner product and there is no Pythagorean theorem for that notion of length.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The natural question is why does the Pythagorean theorem in Euclidean geometry correspond to the Pythagorean theorem in a finite dimensional vector space over the reals with an inner product? (there is a theorem that states that in a finite-dimensional vector space there is only one inner product up to isomorphism, so we really can talk about THE Pythagorean theorem).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer is that the finite dimensional vector space over the reals with an inner product encodes the congruence and similarity axioms of Euclidean geometry. The basic exposition of (most of) this can be found in the first half of Chapter 2 of Emil Artin's wonderful book Geometric Algebra. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take a geometry of points and lines for which the following statements are true:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Any two points determine a line.&lt;/li&gt;&#xA;&lt;li&gt;Parallel postulate: for any point p not on line l&lt;sub&gt;1&lt;/sub&gt; there exists a unique line l&lt;sub&gt;2&lt;/sub&gt; that passes through p but doesn't intersect l&lt;sub&gt;1&lt;/sub&gt;.&lt;/li&gt;&#xA;&lt;li&gt;Non-triviality: there exist 3 non-collinear points.&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Desargues%27_theorem&quot; rel=&quot;nofollow&quot;&gt;Desargues' Theorem&lt;/a&gt; (the wiki article is horrible)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The first three statements allow you to define translations (transformations of the plane that send lines to parallel lines and have no fixed points), while the fourth allows you to a construct a field over which the space of translations is a 2-dimensional vector space, and such that if you associate translations OP and OQ with (0,1) and (1,0), then any point R has a translation OR which can be written as (a,b) with a and b in the field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this way, any affine Desarguesian plane (thing that satisfies statements 1,2,3 and 4) can be identified with a 2-dimensional vector space over some field (and conversely any 2-dimensional vector space over some field corresponds to an affine Desarguesian plane). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alright. Now, Euclidean geometry satisfies Desargues' Theorem, so which field does it correspond to? Well, it turns out that the fact that the geometry is ordered (i.e. we have a notion of an ordering of points on a line) means that the field has to be ordered and thus is a subfield of the real numbers (this is where we need something like the continuity or Hilbert's completeness axiom which roughly state that things that should intersect do intersect, and imply that the base field is all of the real numbers). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then the whole inner product shebang turns out to just be a codification of the various congruence axioms. The norm is given by choosing some vector to make a unit vector (segment) and then considering to which vector in the same direction other vectors (segments) are congruent to. The inner product is a matter of encoding the notion of angle between vectors as an inner product, i.e. as a function linear in both vectors, which essentially takes the notions of similarity, congruences and unit circles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It turns out that you can do all of this in the reverse direction, and so the Euclidean plane really is a 2-d vector space over the reals with an inner product.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Addendum: a comment on symmetry groups that is too long to leave as a comment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Every norm on a vector space has its own unit blob, that is, the set of vectors with norm less than one. Geometrically, blobs that correspond to a norm satisfy the following properties: they are convex, they are absorbing (every vector is a multiple of a vector in the blob), and they don't contain any lines through the origin. It is a theorem that any such blob corresponds to a norm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As we know, the transformations of a vector space are linear transformations, which means that the symmetry group of an object in your vector space is going to consist of the linear transformations that bijectively send an object to itself. In other words, the group of symmetries of an object is the set of invertible linear transformations that fix that object.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the blob of the Euclidean norm is the unit disk, and the unit circle is the boundary of the disk. The unit circle is the set of vectors with Euclidean norm 1, and in general, the boundary of the unit blob for a norm will be the set of vectors with norm 1. The group of symmetries of the boundary of the blob then will be the set of invertible linear transformations that send vectors with norm 1 to vectors with norm 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A slight caveat is that you don't want to consider just any old invertible linear transformations because when you add a norm to a vector space, you are in effect defining a topology, that is, a way to define continuity. As a result you want to work with &lt;em&gt;continuous&lt;/em&gt; linear transformations. BUT! In a finite dimensional vector space, no matter what the norm, we have that &lt;em&gt;every&lt;/em&gt; linear transformation is continuous.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What this means is that for any norm on, say the 2d vector space over the reals, the symmetry groups of the boundary of the unit blob are all invertible linear transformations of the same space. Hence, we can directly compare these groups to each other since they live in the same space (of invertible linear transformations of R&lt;sup&gt;2&lt;/sup&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example: the boundary of the unit blob for the Euclidean norm is the unit circle, i.e. all points such that x&lt;sup&gt;2&lt;/sup&gt;+y&lt;sup&gt;2&lt;/sup&gt;=1. The boundary of the unit blob of the Taxicab norm is the unit diamond, i.e. all points such that |x|+|y|=1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now if you think about it, any linear transformation that sends the unit diamond to itself will also send the unit circle to itself. Hence, the symmetry group for the Euclidean norm contains the symmetry group for the Taxicab norm, and in fact it contains it properly since, for example, rotation by 45 degrees fixes the circle, but doesn't fix the diamond.&lt;/p&gt;&#xA;" OwnerUserId="400" LastEditorUserId="400" LastEditorDisplayName="" LastEditDate="2010-08-26T21:19:15.677" LastActivityDate="2010-08-26T21:19:15.677" CommentCount="6" />
  <row Id="3302" PostTypeId="2" ParentId="3119" CreationDate="2010-08-25T19:10:08.443" Score="3" ViewCount="" Body="&lt;p&gt;Tuples appear in essentially all formal treatments of set theory, and those go back to way before the 70s!&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2010-08-25T19:10:08.443" CommentCount="2" />
  <row Id="3303" PostTypeId="2" ParentId="3231" CreationDate="2010-08-25T19:17:48.303" Score="1" ViewCount="" Body="&lt;p&gt;There's a fast procedure involving no intermediate representation.  You need only four lookup tables totaling 544 entries.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Earlier answers have established that the conversion can be done in blocks of 3 (working right to left).  Consider the rightmost block in your example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;0xA03 = 1010 0000 0011 B&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You need to break this binary string into four groups of three bits, which I will number 1, 2, 3, 4 from the right to the left:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1: 011 B&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2: 000 B&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3: 000 B&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4: 101 B&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first depends only on the rightmost hex digit 0x3.  The second depends only on the two rightmost digits 0x03 (it gets its rightmost bit from the 0x3 and its first two bits from 0x0).  The third depends only on the second and third digits 0xA0.  The fourth depends only on the third digit 0xA.  Whence, you only need to perform the following conversions, each of which can be stored in its own static lookup table:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1: 0x3 --&gt; 03 [16 entries cover all possibilities]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2: (0x0, 0x3) --&gt; 00 [16 * 16 entries]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3: (0xA, 0x0) --&gt; 00 [16 * 16 entries]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4: 0xA --&gt; 5 [16 entries].&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now you repeat with the next block, padding with zeros on the left as needed.  Continuing the example, the next block is 0x001:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1: 0x1 --&gt; 01&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2: (0x0, 0x1) --&gt; 00.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can stop here because all the original input has been consumed.  The output, working backwards, is 0015003.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This directly answers both parts of the original question: it provides a simple conversion without the middle step and it avoids base 10 (which is rarely used for computer conversion anyway: usually the job is done with bit shifting and masking, essentially a binary operation).  In case this procedure still looks too much like the other proposed solutions, please note that it performs absolutely no arithmetic (apart from decrementing pointers to input and output as it proceeds): it takes a string representation (hex) as its input, uses the characters to index its tables, and outputs an octal string.&lt;/p&gt;&#xA;" OwnerUserId="1489" LastActivityDate="2010-08-25T19:17:48.303" CommentCount="2" />
  <row Id="3304" PostTypeId="2" ParentId="3283" CreationDate="2010-08-25T19:33:45.380" Score="4" ViewCount="" Body="&lt;p&gt;A nice series of solutions has already been given.  It's hard to improve on any of them.  But because this is an elementary question, it deserves an elementary answer.  (Although &quot;elementary&quot; is subjective, being able to punch buttons on a calculator requires little mathematical sophistication, so we should attempt to minimize the formality and maximize the intuition in the explanation.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you begin with a value x on the calculator with x &gt; 1, note that its square root lies between 1 and the midpoint of the interval [1, x].&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Proofs: The midpoint is (x+1)/2.  Because ((x+1)/2)^2 - x = ((x-1)/2)^2 &gt;= 0, the midpoint is no greater than the root of x.  Non-algebraic version: draw graphs of the relations x = y^2 and y = (1+x)/2 on the same plot and note that the latter never falls below the former.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It follows immediately (by induction if you want to be formal) that the iterated root, although never falling below 1, decreases faster than the iterated average with 1, which obviously (on geometric or arithmetic grounds) converges to 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The case of 0 &amp;lt; x &amp;lt; 1 is reduced to the case 1/x &gt; 1 because Sqrt(1/x) = 1/Sqrt(x).&lt;/p&gt;&#xA;" OwnerUserId="1489" LastActivityDate="2010-08-25T19:33:45.380" />
  <row Id="3305" PostTypeId="1" AcceptedAnswerId="3307" CreationDate="2010-08-25T20:05:05.883" Score="4" ViewCount="103" Body="&lt;p&gt;A function $f$ is superadditive if $f(x) + f(y) \le f(x+y)$. The question is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does a real number $a$ exists such that for all &#xA;real numbers with $x, y\ \ge \ a $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \Gamma(x) + \Gamma(y) \le \Gamma(x+y) \quad ?$$&lt;/p&gt;&#xA;" LastActivityDate="2010-08-25T20:32:05.253" Title="Is the Gamma function superadditive?" Tags="&lt;real-analysis&gt;&lt;gamma-function&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="3306" PostTypeId="2" ParentId="2356" CreationDate="2010-08-25T20:07:17.217" Score="4" ViewCount="" Body="&lt;p&gt;A simple (practical, low-computation) approach to choosing among three options with equal probability exploits the fact that in a run of independent flips of an unbiased coin, the chance of encountering THT before TTH occurs is 1/3.  So:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Flip a coin repeatedly, keeping track of the last three outcomes.  (Save time, if you like, by assuming the first flip was T and proceeding from there.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Stop whenever the last three are THT or TTH.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the last three were THT, select option 1.  Otherwise flip the coin one more time, choosing option 2 upon seeing T and option 3 otherwise.&lt;/p&gt;&#xA;" OwnerUserId="1489" LastActivityDate="2010-08-25T20:07:17.217" />
  <row Id="3307" PostTypeId="2" ParentId="3305" CreationDate="2010-08-25T20:16:34.340" Score="5" ViewCount="" Body="&lt;p&gt;$a = 2$ will do, because (letting $x \ge y$ wlg),&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Gamma(x+y) \ge \Gamma(x+2) = (x+1)x \Gamma(x) \ge 6 \Gamma(x) \ge \Gamma(x) + \Gamma(x) \ge \Gamma(x) + \Gamma(y)$. &lt;/p&gt;&#xA;" OwnerUserId="1489" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-08-25T20:32:05.253" LastActivityDate="2010-08-25T20:32:05.253" CommentCount="8" />
  <row Id="3308" PostTypeId="2" ParentId="3271" CreationDate="2010-08-25T20:32:52.090" Score="2" ViewCount="" Body="&lt;p&gt;You should look at the work of Wadim Zudilin. In particular you should look at &#xA;&lt;a href=&quot;http://wain.mi.ras.ru/PS/zeta5-11%24.pdf&quot; rel=&quot;nofollow&quot;&gt;&quot;One of the numbers ζ(5), ζ(7), ζ(9), ζ(11) is irrational&quot; (Turpion link, pdf 91k, gzip ps 80k) in Russian Math. Surveys  56:4 (2001), 774--776;&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="1154" LastActivityDate="2010-08-25T20:32:52.090" CommentCount="4" />
  <row Id="3309" PostTypeId="2" ParentId="3264" CreationDate="2010-08-25T20:35:02.703" Score="4" ViewCount="" Body="&lt;p&gt;You implicitly define rotations, translations, and reflections as certain kinds of affine transformations.  Thus it suffices to check algebraically that they preserve distances.  (There is no need to check that they preserve angles: that follows immediately by polarization.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's not terribly deep.  More insight is afforded through the usual geometric demonstration that reflections and translations are composed of a finite number of reflections.  That a reflection is an isometry is not only geometrically obvious but also easy to show algebraically.&lt;/p&gt;&#xA;" OwnerUserId="1489" LastActivityDate="2010-08-25T20:35:02.703" />
  <row Id="3310" PostTypeId="2" ParentId="3278" CreationDate="2010-08-25T20:58:55.200" Score="3" ViewCount="" Body="&lt;p&gt;It is very true that statements and problems on conditional probability are often presented in an ambiguous way. The problem isn't specifically with the term &quot;given&quot;, I believe, but rather with the fact that the presentation does not make it clear what the sample space is and what are the distributions (typically some things are assumed, without comment, to be uniform).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, specifically to answer your questions:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;No, not that I'm aware of. If the variables are clearly defined, the use of &quot;given&quot; to indicate conditional probability is common and perfectly fine.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If I understand the question correctly (there seems to be a word missing), then most mathematicians do not regard such phrasings as you have mentioned to be sufficiently clear. See &lt;a href=&quot;http://blog.tanyakhovanova.com//?p=234&quot; rel=&quot;nofollow&quot;&gt;Peter Winkler's comments&lt;/a&gt; on exactly this kind of problems.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Not in the country I went to high school in :-)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="308" LastActivityDate="2010-08-25T20:58:55.200" CommentCount="1" />
  <row Id="3311" PostTypeId="1" AcceptedAnswerId="3313" CreationDate="2010-08-25T21:12:06.687" Score="1" ViewCount="109" Body="&lt;p&gt;This is a follow-up to the MathStackexchange question 3305 and MathOverflow question 23229.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Bohr-Mollerup theorem states that the Gamma function is the unique function that satisfies: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(1)\ f(x+1) = xf(x), \ (2)\ f(1) = 1, \ (3)\ \ln \circ f \ $ is convex.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume a function $f\colon \mathbb{R^{+}} \to \mathbb{R}$  that satisfies: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(1)\ f(x+1) = xf(x), \ (2)\ f(1) = 1, \ (3)\ $ f is superadditive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a (natural) condition which makes this function unique?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit 1: The answer is 'no' as Moron explains. And if we assume $f$ superadditive only for $x,y \ge a$ for some real $a$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit 2: I accept Moron's answer because it the correct answer to my question. Intended was the question in the sense of my first edit (and my first question). I am also curious to see an answer if condition (3) reads: (3'') $\ \ln \circ f \ $ is superadditive. Thanks to whuber for suggesting this.&lt;/p&gt;&#xA;" LastActivityDate="2010-08-26T21:05:24.057" Title="Importance of Superadditivity of the Gamma Function" Tags="&lt;real-analysis&gt;&lt;gamma-function&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="3312" PostTypeId="2" ParentId="3278" CreationDate="2010-08-25T21:13:24.437" Score="1" ViewCount="" Body="&lt;p&gt;I'm hesitant to use an answer to respond, instead of a comment, because the answer seems to merely be a matter of semantics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's look at a similar question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;What are the odds that two people are of the same gender?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And its counterpart:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;What are the odds that two people of the same gender, if we knew that one of them is a boy?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The questions are inherently different.  The answer to the first can be seen by charting all possibilities within the limitations (there are no limitations), and it can be seen that it's 2/4, or 50%&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second can be seen by charting all possibilities and eliminating the ones that aren't allowed (two girls), and we'll see that the answer is 1/3, or about 33%.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;For a somewhat intuitive answer of &quot;how&quot; we know something, we can use a statistical model.  Let's say we wanted to find the odds of the second question.  We could conduct a survey, asking a perfect world: (where both genders are equally likely)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Anybody who has two children, with one being a boy, answer if they have two boys or not.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This survey would give you the 33% that our chart-all-the-possibilities method would lead us to expect.  There are no ambiguities because we see the answer very clearly real and evident.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But this is a maths site, and I brought up statistics, so I'm probably going to be killed in my sleep.&lt;/p&gt;&#xA;" OwnerUserId="51" LastActivityDate="2010-08-25T21:13:24.437" CommentCount="1" />
  <row Id="3313" PostTypeId="2" ParentId="3311" CreationDate="2010-08-25T21:30:03.417" Score="4" ViewCount="" Body="&lt;p&gt;There is no such function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f(2) = f(1+1) = 1&amp;times;f(1) = 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By superadditivity $f(1+1) \ge f(1) + f(1)$ i.e. $ 1 = f(2) \ge 2f(1) = 2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Not possible.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-08-25T21:30:03.417" />
  <row Id="3314" PostTypeId="1" CreationDate="2010-08-25T23:49:05.113" Score="2" ViewCount="193" Body="&lt;p&gt;Given &lt;code&gt;e1..e5&lt;/code&gt; as the exponents of the base units, are there any well known methods for producing a simplified form?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an example &lt;code&gt;m^2*kg*s^-1&lt;/code&gt; simplifies to &lt;code&gt;W*s^2&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a set of derived units as Name/exponent-vector pairs as input and it can be assumed all exponents are rational (and in fact most being integers and almost all not having a denominator more than about 3-4).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know this can be reduced to a linear algebra problem but it is VERY under defined and in fact has several possible solutions.&lt;/p&gt;&#xA;" OwnerUserId="1171" LastEditorUserId="1171" LastEditorDisplayName="" LastEditDate="2010-08-26T00:02:22.610" LastActivityDate="2011-01-23T01:46:53.560" Title="How to determine the units of a value." Tags="&lt;linear-algebra&gt;" AnswerCount="1" CommentCount="7" />
  <row Id="3315" PostTypeId="1" CreationDate="2010-08-25T23:50:35.547" Score="11" ViewCount="1102" Body="&lt;p&gt;If i is the square root of -1, is the square root of i imaginary? Is it used or considered often in mathematics? How is it notated?&lt;/p&gt;&#xA;" OwnerUserId="104" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-08-26T00:24:42.643" LastActivityDate="2010-08-31T13:10:23.783" Title="what is the square root of i?" Tags="&lt;complex-numbers&gt;" AnswerCount="8" CommentCount="4" FavoriteCount="1" />
  <row Id="3316" PostTypeId="2" ParentId="3315" CreationDate="2010-08-25T23:56:00.383" Score="10" ViewCount="" Body="&lt;p&gt;$i^\frac1{2}=\left(e^{\pi i/2}\right)^\frac1{2}=e^{\pi i/4}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$e^{\pi i/4}=\cos\left(\frac{\pi}{4}\right)+i \sin\left(\frac{\pi}{4}\right)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or simplified,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{1+i}{\sqrt{2}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is of course the &quot;principal value&quot;; the other value (thanks Matt E!) is the &quot;negative square root&quot;, $-\frac{1+i}{\sqrt{2}}$ or in exponential form, $-e^{\pi i/4}=e^{-3\pi i/4}$&lt;/p&gt;&#xA;" OwnerUserId="498" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-08-26T01:36:29.243" LastActivityDate="2010-08-26T01:36:29.243" CommentCount="5" />
  <row Id="3318" PostTypeId="2" ParentId="3283" CreationDate="2010-08-26T00:00:03.893" Score="0" ViewCount="" Body="&lt;pre&gt;&lt;code&gt;if(x == 0): sqrt(x) == 0&#xA;if(0 &amp;lt; x &amp;lt; 1): x &amp;lt; sqrt(x) &amp;lt; 1&#xA;if(x == 1): sqrt(x) == 1&#xA;if(1 &amp;lt; x ): 1 &amp;lt; sqrt(x) &amp;lt; x&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Except for the cases &lt;code&gt;x == 0&lt;/code&gt; and &lt;code&gt;x == 1&lt;/code&gt;, &lt;code&gt;x != sqrt(x)&lt;/code&gt; so it must move, and it must move towards 1, so it will approach 1 to any accuracy you choose.&lt;/p&gt;&#xA;" OwnerUserId="1171" LastActivityDate="2010-08-26T00:00:03.893" CommentCount="2" />
  <row Id="3319" PostTypeId="1" CreationDate="2010-08-26T00:00:28.557" Score="5" ViewCount="446" Body="&lt;p&gt;I've heard a lot about this number e. Why is it so important? How does it fit into the 'bigger picture' of mathematics? How is it calculated and used?&lt;/p&gt;&#xA;" OwnerUserId="104" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-08-26T00:03:03.863" LastActivityDate="2010-08-28T23:31:39.980" Title="Why is the number e so important in mathematics?" Tags="&lt;calculus&gt;" AnswerCount="7" CommentCount="1" />
  <row Id="3320" PostTypeId="1" CreationDate="2010-08-26T00:05:25.460" Score="7" ViewCount="375" Body="&lt;p&gt;Pi appears a LOT in trigonometry, but only because of its 'circle-significance'. Does pi ever matter in things not concerned with circles? Is its only claim to fame the fact that its irrational and an important ratio?&lt;/p&gt;&#xA;" OwnerUserId="104" LastActivityDate="2011-01-13T22:01:03.543" Title="Does the number pi have any significance besides being the ratio of a circle's diameter to its circumference?" Tags="&lt;pi&gt;" AnswerCount="4" CommentCount="7" FavoriteCount="2" />
  <row Id="3321" PostTypeId="2" ParentId="3320" CreationDate="2010-08-26T00:13:19.570" Score="17" ViewCount="" Body="&lt;p&gt;It is difficult to know if a circle is not lurking somewhere, whenever there is $\pi$, but the values of the Riemann zeta function at the positive even integers have a lot to do with powers of $\pi$: see &lt;a href=&quot;http://en.wikipedia.org/wiki/Riemann_zeta_function&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; for the values.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, you can prove that the probability that two &quot;randomly chosen&quot; positive integers are coprime is $\frac{1}{\zeta(2)} = \frac{6}{\pi^2}$.&lt;/p&gt;&#xA;" OwnerUserId="1028" LastActivityDate="2010-08-26T00:13:19.570" CommentCount="1" />
  <row Id="3322" PostTypeId="2" ParentId="3319" CreationDate="2010-08-26T00:19:33.767" Score="2" ViewCount="" Body="&lt;p&gt;Quite simply, because of the identity&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{d}{dx} (e^x) = e^x$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is in fact the primary way of defining the constant &lt;em&gt;e&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The consequences of this simple fact within mathematics are myriad, and can only really be properly appreciated by studying enough calculus and other related fields.&lt;/p&gt;&#xA;" OwnerUserId="56" LastActivityDate="2010-08-26T00:19:33.767" />
  <row Id="3323" PostTypeId="2" ParentId="3320" CreationDate="2010-08-26T00:22:19.617" Score="10" ViewCount="" Body="&lt;p&gt;$\pi$ appears in &lt;a href=&quot;http://en.wikipedia.org/wiki/Stirlings_approximation&quot; rel=&quot;nofollow&quot;&gt;Stirling's approximation&lt;/a&gt;, which is not obviously related to circles.  This means that $\pi$ appears in asymptotics related to binomial coefficients, such as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle {2n \choose n} \approx \frac{4^n}{\sqrt{\pi n}}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, the probability of flipping exactly $n$ heads and $n$ tails after flipping a coin $2n$ times is about $\frac{1}{\sqrt{\pi n}}$.  This asymptotic also suggests that on average you should flip between $n + \sqrt{\pi n}$ and $n - \sqrt{\pi n}$ heads. &lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-08-26T00:22:19.617" CommentCount="3" />
  <row Id="3324" PostTypeId="2" ParentId="3314" CreationDate="2010-08-26T00:28:42.330" Score="1" ViewCount="" Body="&lt;p&gt;In general the problem of &quot;simplifying&quot; units (mathematical expressions in general) has no well defined solution - it is not even a well-defined process, in most cases. For example, there are multiple ways of reducing your given example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;m^2 kg s^-1 =&amp;gt; N m s&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;m^2 kg s^-1 =&amp;gt; J s&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;m^2 kg s^-1 =&amp;gt; W s^2&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The best you can really do is recursively enumerate over all such possibilities and use some heuristic to choose the &quot;best&quot; (or give the option of using any of them).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Note: intuitively I would guess that these units are representing angular momentum, in which case &lt;code&gt;N m s&lt;/code&gt; probably makes most sense, though it's quite subjective.)&lt;/p&gt;&#xA;" OwnerUserId="56" LastActivityDate="2010-08-26T00:28:42.330" CommentCount="1" />
  <row Id="3325" PostTypeId="2" ParentId="3319" CreationDate="2010-08-26T00:36:12.893" Score="3" ViewCount="" Body="&lt;p&gt;One of the first exposures students have to $e$ is in the formula for simple interest compounded continuously: $$A = Pe^{rt}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This formula can be easily derived from the formula for simple interest compounded discretely: $$A = P(1 + \frac{r}{n})^{nt}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The derivation depends on the identity $$e = \lim_{x \to \infty} (1 + \frac{1}{x})^x$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So although this may not be the &lt;em&gt;most&lt;/em&gt; important application, it is one way in which $e$ is used and calculated.&lt;/p&gt;&#xA;" OwnerUserId="506" LastActivityDate="2010-08-26T00:36:12.893" />
  <row Id="3326" PostTypeId="2" ParentId="3319" CreationDate="2010-08-26T00:45:37.513" Score="5" ViewCount="" Body="&lt;p&gt;One could easily write entire textbooks about this subject, and I think the only way to properly appreciate the answer is just to learn a lot of mathematics.  But I will mention one perspective which suggests that $e$ ought to be fundamental to mathematics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the perspective of &lt;a href=&quot;http://ncatlab.org/nlab/show/groupoid+cardinality&quot; rel=&quot;nofollow&quot;&gt;groupoid cardinality&lt;/a&gt;.  The definition of a groupoid is slightly technical, but for the purposes of this discussion a groupoid is a collection of groups.  In many counting problems involving groups it turns out to be natural not to count all of the objects involved, but to divide each object by the number of symmetries it has.  For example, if you draw five dots in a line on a paper and then fold the paper in half through the middle dot, two pairs of the dots have been identified and one dot has one half identified with the other half; in other words, the folding symmetry has cut the middle dot in half, so one might say that we now have $\frac{5}{2} = 2 + \frac{1}{2}$ dots.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the basic idea behind groupoid cardinality: the cardinality of a groupoid is the sum $\sum_x \frac{1}{|\text{Aut}(x)|}$, where the sum runs over all the isomorphism classes of objects of the groupoid (here, over all the groups in the collection, where $|\text{Aut}(x)|$ is the size of the corresponding group).  All groups here are finite.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now here is the fundamental statement.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;The groupoid cardinality of the groupoid of finite sets (e.g. the collection of symmetric groups $S_1, S_2, ...$) is $e = \sum_{n \ge 0} \frac{1}{n!}$.&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In other words, $e$ somehow embodies a fundamental property of finite sets.  It is possible to think about the whole theory of exponential generating functions in this way, in particular to think about $e^x$ and its derivative property this way.  Part of this story is described, I believe, in John Baez's &lt;a href=&quot;http://math.ucr.edu/home/baez/groupoidification/&quot; rel=&quot;nofollow&quot;&gt;the Tale of Groupoidification&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of the more concrete ways to think about the relationship between the symmetric groups and $e$ is via &lt;a href=&quot;http://qchu.wordpress.com/2009/06/24/gila-vi-the-cycle-index-polynomials-of-the-symmetric-groups/&quot; rel=&quot;nofollow&quot;&gt;this generating function&lt;/a&gt;.  The identity described in that blog post, which sometimes goes by the name of the &quot;exponential formula,&quot; explains, among many &lt;em&gt;many&lt;/em&gt; other things, why the probability that a permutation of a large set has no fixed points is about $\frac{1}{e}$.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-08-26T00:45:37.513" CommentCount="4" />
  <row Id="3327" PostTypeId="2" ParentId="3264" CreationDate="2010-08-26T00:53:28.987" Score="2" ViewCount="" Body="&lt;p&gt;Let's consider $\mathbb{R}^3$ as a vector space on $\mathbb{R}$, endowed with the standard scalar product $\langle,\rangle$ (also know as $\mathbb{E}^3$, the Euclidean space). We define a function $f: \mathbb{R}^3 \to \mathbb{R}^3$ to be an &lt;em&gt;isometry&lt;/em&gt; if $||f(u)-f(w)|| = ||u-w||$ for all $v,w \in \mathbb{R}^3$, i. e. if $f$ &lt;em&gt;preserve distances&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Transaltions are isometries&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We define a &lt;em&gt;translation by $z$&lt;/em&gt; the map $\tau_{z}: \mathbb{R}^3 \to \mathbb{R}^3$ such that $\tau_{z}(u) = u+z$ for all $u \in \mathbb{R}^3$. Then we have $$||\tau_{z}(u)-\tau_{z}(w)|| = ||(u+z) - (w+z)|| = ||u-w||$$ for all $u,w \in \mathbb{R}^3$ so we are done.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we consider reflections and rotations. They are both &lt;em&gt;real unitary linear maps&lt;/em&gt; (also known as &lt;em&gt;orthogonal&lt;/em&gt;, or members of $O(3)$). The following are equivalent propieties of these transformations of the space:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\langle L(u),L(w)\rangle = \langle u,w\rangle$ for all $L \in O(3)$ $u,w \in \mathbb{R}^3$ or they preserve scalar product&lt;/li&gt;&#xA;&lt;li&gt;$||L(u)|| = ||u||$ for all $L \in O(3)$ $u \in \mathbb{R}^3$ or they preserve lenghts&lt;/li&gt;&#xA;&lt;li&gt;$||L(u)|| = ||u||$ for all $L \in O(3)$ and $u \in \mathbb{R}^3$ with $||u|| = 1$ or the unit sphere is $L$-invariant&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Now, we'll see that all real unitary transformation of the space are isometries, proving then that reflections and rotations are of this type we have finished.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Real unitary transformations are isometries&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $L \in O(3)$ then $$||L(u)-L(w)|| = \sqrt{\langle L(u)-L(w),L(u)-L(w)\rangle} = \sqrt{\langle L(u-w),L(u-w)\rangle}$$ by definition of norm and using the linearity of L, and $$\sqrt{\langle L(u-w),L(u-w)\rangle} = \sqrt{\langle u-w,u-w\rangle} = ||u-w||$$ by using one of the three proprieties above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Reflections are unitary transformations (and thus isometries)&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $S\subseteq\mathbb{R}^3$ is a subspace, the we define reflection in respect to $S$ the linear map $\rho_{S}: S \oplus S^{\bot} \to S \oplus S^{\bot}$ such that if $u = u_{S}+u_{S^{\bot}}$ then $\rho_{S}(u) = u_{S}-u_{S^{\bot}}$. So we have $$||u_{S}±u_{S^{\bot}}|| = \sqrt{||u_{S}||^2 + ||u_{S^{\bot}}||^2}$$ and then $||\rho_{S}(u)|| = ||u||$ for all $u \in \mathbb{R}^3$, that is one of the three equivalent proprieties above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Rotations are unitary transformations (and thus isometries)&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a little weird. As you can read &lt;a href=&quot;http://math.stackexchange.com/questions/3254/how-do-you-formally-prove-that-rotation-is-a-linear-transformation/3259#3259&quot; title=&quot;here&quot;&gt;here&lt;/a&gt;, perhaps the best way to &lt;em&gt;define&lt;/em&gt; rotations is to consider them as paricular members of $O(3)$, so there is nothing to prove.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: what has been said remains true for an arbitrary space $V$ of dimension $n$ on $\mathbb{R}$ endowed with a positive product $\langle ,\rangle$ because of the isomorphism with $\mathbb{E}^n$, this is the power of linear algebra!&lt;/p&gt;&#xA;" OwnerUserId="365" LastEditorUserId="365" LastEditorDisplayName="" LastEditDate="2010-09-01T22:42:01.117" LastActivityDate="2010-09-01T22:42:01.117" />
  <row Id="3328" PostTypeId="2" ParentId="3319" CreationDate="2010-08-26T01:36:41.880" Score="0" ViewCount="" Body="&lt;p&gt;The first thing to note is that e&lt;sup&gt;x&lt;/sup&gt; is defined by the well known infinite series.  But that's not actually the crucially important thing from my point of view.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The real significance comes from the exponential mapping in the the theory of continuous symmetries (Lie Groups).  The exponential mapping is a function from the &quot;infinitesimal&quot; elements of the group (Lie Algebra) to the elements of the group itself.  This is not so easy to explain unfortunately, but you may be interested to wiki the principle axis theorem for some analogies from physics that are easy to visualize intuitively (the principle axis are the lie algebra).  Clearly, the theory of continuous symmetries is of fundamental significance, and the appearance of the infinite series for e at the heart of this theory is for me quite profound.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore, for very general Lie Groups, we have the amazing identity&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A = e&lt;sup&gt;t&lt;sub&gt;1&lt;/sub&gt;L&lt;sub&gt;1&lt;/sub&gt;&lt;/sup&gt;e&lt;sup&gt;t&lt;sub&gt;2&lt;/sub&gt;L&lt;sub&gt;2&lt;/sub&gt;...&lt;/sup&gt;e&lt;sup&gt;t&lt;sub&gt;N&lt;/sub&gt;L&lt;sub&gt;N&lt;/sub&gt;&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For an element of the Lie Group A with a lie algebra L&lt;sub&gt;i&lt;/sub&gt; of size N, and some real numbers t&lt;sub&gt;i&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We see the connection to the real number defined by e by representing a continuous group as a group of matrices (under matrix multiplication), and noticing that the exponential mapping is the infinite series for e&lt;sup&gt;X&lt;/sup&gt;, where X is now a matrix.&lt;/p&gt;&#xA;" OwnerUserId="1417" LastEditorUserId="1417" LastEditorDisplayName="" LastEditDate="2010-08-26T01:42:17.923" LastActivityDate="2010-08-26T01:42:17.923" CommentCount="1" />
  <row Id="3329" PostTypeId="1" CreationDate="2010-08-26T02:23:20.127" Score="4" ViewCount="335" Body="&lt;p&gt;Back in the day, I was absolutely enthralled by the study of plane curves and their properties (I have &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0521055857&quot; rel=&quot;nofollow&quot;&gt;Lockwood&lt;/a&gt; and &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0486442764&quot; rel=&quot;nofollow&quot;&gt;Zwikker&lt;/a&gt; to thank). I learned early on that for the purposes of generating plots on a computer (and for that matter deducing equations of &quot;derived curves&quot; and determining other special properties), one should try to find a representation in parametric equations for your plane curve.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As I recall, in dealing with algebraic curves represented by an implicit Cartesian equation, I knew of only three tricks to derive parametric equations from an implicit equation (listed in decreasing order of effectiveness; I note that I did all these investigations even before I knew computer algebra systems existed):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1: Convert to polar coordinates to express in the form $r=r(\theta)$; the parametric equations are then &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\begin{align*}x&amp;amp;=r(\theta)\cos(\theta)\\y&amp;amp;=r(\theta)\sin(\theta)\end{align*}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2: The $y=mx$ &quot;trick&quot; (I never did get to learn the formal name for this technique); to use the implicit equation for the &lt;a href=&quot;http://mathworld.wolfram.com/FoliumofDescartes.html&quot; rel=&quot;nofollow&quot;&gt;folium of Descartes&lt;/a&gt; as an example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x^3+y^3=3xy$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x^3+(mx)^3=3x(mx)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and then by solving for x and using the relation $y=mx$ again,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\begin{align*}x&amp;amp;=\frac{3m}{1+m^3}\\y&amp;amp;=\frac{3m^2}{1+m^3}\end{align*}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I remember this worked especially well for curves whose (only?) singular points are at the origin, but not very well for other curves; can anybody explain why?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3: Randomly replacing x or y with any of the six trigonometric functions (maybe multiplied by a convenient constant), and hope that I can easily solve for the other variable. For instance, I managed to derive the parametric equation for the &lt;a href=&quot;http://mathworld.wolfram.com/Bicorn.html&quot; rel=&quot;nofollow&quot;&gt;bicorn&lt;/a&gt; and the &lt;a href=&quot;http://mathworld.wolfram.com/DuerersConchoid.html&quot; rel=&quot;nofollow&quot;&gt;Dürer conchoid&lt;/a&gt; in this way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Probably the only other thing I learned way after I had moved on to other things was that elliptic curves can for instance be represented as parametric equations involving the Weierstrass ℘ function or the elliptic exponential, but this is apparently limited to elliptic curves only.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now for my question: did I miss any other useful (general?) methods for turning an implicit Cartesian equation for an algebraic curve into parametric equations?&lt;/p&gt;&#xA;" OwnerUserId="498" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-08-28T13:11:38.130" LastActivityDate="2010-08-28T13:11:38.130" Title="Parametrizing implicit algebraic curves" Tags="&lt;geometry&gt;&lt;plane-curves&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" />
  <row Id="3331" PostTypeId="2" ParentId="3250" CreationDate="2010-08-26T03:46:43.757" Score="1" ViewCount="" Body="&lt;p&gt;Based on the previous answers, I have come to the following (possible incomplete or incorrect) conclusion: It seems that the Pythagorean theorem can be dealt with in one of two ways, either it is taken as the definition of length/distance, or it is proved by invoking the technical definitions of an inner product space.  However, proving Pythagoras via the definitions of an inner product space really just shifts the question to how one motivates these definitions (namely, why does one define the length of a vector to be the square root of its dot product with itself?).  Thus, one can say that the Pythagorean theorem is generally taken to be the &lt;em&gt;definition&lt;/em&gt; of length or distance, and the only thing left to do is &lt;em&gt;motivate&lt;/em&gt; this definition; that is, show why the Euclidean norm is such a natural system for determining length.  Qiaochu has alluded to the fact that the Euclidean norm is natural because of its large symmetry group, but I'd like someone to elaborate on this a bit if possible...&lt;/p&gt;&#xA;" LastActivityDate="2010-08-26T03:46:43.757" CommentCount="6" />
  <row Id="3332" PostTypeId="2" ParentId="3315" CreationDate="2010-08-26T04:07:29.057" Score="16" ViewCount="" Body="&lt;p&gt;Let $z=(a+bi)$ be a complex number which is a square root of $i$, that is &#xA;$$i=z^2=(a^2-b^2)+2abi.$$ &#xA;Equating real and imaginary parts we have,&#xA;$$a^2-b^2=0, 2ab=1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The two real solutions to this pair of equations are&#xA;$a={1 \over \sqrt{2}},b={1 \over \sqrt{2}}$ and $a=-{1 \over \sqrt{2}},b=-{1 \over \sqrt{2}}$. The two square roots of $i$ therefore are &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$$\pm {1 \over \sqrt{2}} (1+i)$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1195" LastActivityDate="2010-08-26T04:07:29.057" />
  <row Id="3333" PostTypeId="1" AcceptedAnswerId="3334" CreationDate="2010-08-26T04:09:09.350" Score="3" ViewCount="113" Body="&lt;p&gt;Ramsey number R(3,6)=18. How to construct a graph of 17 nodes which does not contain neither a clique of order 3 or an independent set of order 6. could you show me the tactics or the adjacency matrix?&lt;/p&gt;&#xA;" OwnerUserId="1498" LastActivityDate="2010-08-26T10:02:54.803" Title="how to construct 17-vertexed graph for Ramsey number R(3,6)=18" Tags="&lt;graph-theory&gt;" AnswerCount="2" />
  <row Id="3334" PostTypeId="2" ParentId="3333" CreationDate="2010-08-26T04:18:09.307" Score="3" ViewCount="" Body="&lt;p&gt;Look at &lt;a href=&quot;http://cs.anu.edu.au/~bdm/data/ramsey.html&quot; rel=&quot;nofollow&quot;&gt;http://cs.anu.edu.au/~bdm/data/ramsey.html&lt;/a&gt; for the R(3,6)'s with &lt;em&gt;less&lt;/em&gt; than 18 vertices. The ones with 17 are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/Mw2xq.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Apartently, an example is constructed by hand in&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;J. G. Kalbfleish, &lt;em&gt;Chromatic graphs and Ramsey's theorem&lt;/em&gt;, Ph. D. thesis, University of Waterloo, January 1966.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="274" LastEditorUserId="274" LastEditorDisplayName="" LastEditDate="2010-08-26T06:15:05.003" LastActivityDate="2010-08-26T06:15:05.003" CommentCount="1" />
  <row Id="3335" PostTypeId="1" AcceptedAnswerId="3496" CreationDate="2010-08-26T04:20:02.753" Score="2" ViewCount="234" Body="&lt;p&gt;I love R.Courant and F.John's &lt;em&gt;Introduction to Calculus and Analysis&lt;/em&gt; because of its wide coverage, precise description and friendly written style.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there any classic textbooks like it on ODE/PDE?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;thanks.&lt;/p&gt;&#xA;" OwnerUserId="1382" LastEditorUserId="121" LastEditorDisplayName="" LastEditDate="2010-08-26T04:24:26.587" LastActivityDate="2010-08-28T00:46:55.493" CommunityOwnedDate="2010-08-26T04:27:38.983" Title="Could you recommend some classic textbooks on ordinary/partial differential equation?" Tags="&lt;reference-request&gt;&lt;differential-equations&gt;" AnswerCount="5" CommentCount="0" FavoriteCount="3" />
  <row Id="3336" PostTypeId="1" AcceptedAnswerId="3344" CreationDate="2010-08-26T04:27:18.300" Score="1" ViewCount="326" Body="&lt;p&gt;Sometimes, there are many ways to give defintion to a mathmatical principle, for example the natural base logarithm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How about sine and cosine?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;thanks.&lt;/p&gt;&#xA;" OwnerUserId="1382" LastActivityDate="2010-08-26T20:49:06.333" Title="How many ways to define sine and cosine?" Tags="&lt;analysis&gt;&lt;trigonometry&gt;" AnswerCount="5" CommentCount="4" />
  <row Id="3337" PostTypeId="2" ParentId="3335" CreationDate="2010-08-26T04:36:10.410" Score="0" ViewCount="" Body="&lt;p&gt;I am a big fan of Bentley and Cooke's &lt;a href=&quot;http://www.abebooks.com/servlet/BookDetailsPL?bi=1174162092&amp;amp;searchurl=an%3Dbentley%2Band%2Bcooke%26sts%3Dt%26x%3D82%26y%3D18&quot; rel=&quot;nofollow&quot;&gt;Linear Algebra with Differential Equations&lt;/a&gt;. It was published in the early 70's, it is black and white (i.e. not flashy), and has excellent references throughout the text (such as a footnote pointing to &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/082840044X&quot; rel=&quot;nofollow&quot;&gt;Kamke&lt;/a&gt;).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore, the topics of differential equations and linear algebra are introduced side by side at the introductory level.  I think this is an ideal text for the beginning engineering student that holds him or herself to learning a bit more mathematics than their peers.&lt;/p&gt;&#xA;" OwnerUserId="121" LastActivityDate="2010-08-26T04:36:10.410" CommunityOwnedDate="2010-08-26T04:36:10.410" CommentCount="1" />
  <row Id="3338" PostTypeId="2" ParentId="3335" CreationDate="2010-08-26T04:46:49.703" Score="1" ViewCount="" Body="&lt;p&gt;Good books on ODE would be:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Ordinary Differential Equations by Morris Tenenbaum and Harry Pollard&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Differential Equations by George F. Simmons&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) Ordinary Differential Equations by E.Coddington&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For PDE, I request you to take a look at this link: &lt;a href=&quot;http://www.physicsforums.com/showthread.php?t=175382&quot; rel=&quot;nofollow&quot;&gt;http://www.physicsforums.com/showthread.php?t=175382&lt;/a&gt;&lt;/p&gt;&#xA;" LastActivityDate="2010-08-26T04:46:49.703" CommunityOwnedDate="2010-08-26T04:46:49.703" />
  <row Id="3339" PostTypeId="2" ParentId="3336" CreationDate="2010-08-26T04:52:09.907" Score="4" ViewCount="" Body="&lt;p&gt;I like to define them by the differential equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$y''=-y$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and then, choosing the initial conditions we get these two functions. This also give rise nice definition for $\pi$ being the fundamental period for the solutions of these equations.&lt;/p&gt;&#xA;" OwnerUserId="633" LastActivityDate="2010-08-26T04:52:09.907" CommentCount="1" />
  <row Id="3340" PostTypeId="1" AcceptedAnswerId="3341" CreationDate="2010-08-26T05:07:18.977" Score="1" ViewCount="116" Body="&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/86j1p.jpg&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is why do we define $f(\xi+0)$ as $\lim_{\varepsilon \rightarrow 0} f(\xi + \varepsilon^2)$ and $f(\xi-0)$ as $\lim_{\varepsilon \rightarrow 0} f(\xi - \varepsilon^2)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;thanks.&lt;/p&gt;&#xA;" OwnerUserId="1382" LastEditorUserId="226" LastEditorDisplayName="" LastEditDate="2010-08-26T06:19:26.777" LastActivityDate="2010-08-26T06:19:26.777" Title="Why we assign $1/2[f(\epsilon+0) + f(\epsilon-0)]$ as the value at the discontinuity point?" Tags="&lt;calculus&gt;&lt;fourier-series&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="3341" PostTypeId="2" ParentId="3340" CreationDate="2010-08-26T05:13:50.477" Score="7" ViewCount="" Body="&lt;p&gt;It is not that we &lt;em&gt;assign&lt;/em&gt; that value: Fourier series simply converge to that independently of our desires!&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2010-08-26T05:13:50.477" CommentCount="1" />
  <row Id="3342" PostTypeId="2" ParentId="3122" CreationDate="2010-08-26T05:24:08.700" Score="0" ViewCount="" Body="&lt;p&gt;Let $\epsilon &gt; 0$ be given. There is a positive integer $n_{0}$ such that $1/n &amp;lt; \epsilon/2$ for $n &gt; n_{0}$. Choose a partition $P$ determined by $$ 0 = x_{0} &amp;lt; x_{1}= \frac{1}{n_{0}+1} &amp;lt; x_{2} &amp;lt; \cdots &amp;lt; x_{n'_{0}}= \frac{1}{n_0}$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ &amp;lt; x_{n'_{0}+1} &amp;lt; \cdots &amp;lt; x_{n_{1}}'= \frac{1}{n_{0}-1} &amp;lt; \cdots &amp;lt; x_{n_{n_{0}-1}'}=1$$ such that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \displaystyle x_{i} - x_{i-1} &amp;lt; \frac{\epsilon}{4n_{0}}$$   Then $$U(P,f) - L(P,f) = \frac{1}{n_{0}+1} + \sum\limits_{i=2}^{n_{0}'} (M_{i}-m_{i})(x_{i}=x_{i-1})$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ + \sum\limits_{k=0}^{n_{0}-2} \sum\limits_{i=n_{k}'+1}^{n_{k+1}'}(M_{i}-m_{i})(x_{i}-x_{i-1})$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ &amp;lt; \frac{\epsilon}{2} + 2n_{0} \frac{\epsilon}{4n_{0}} = \epsilon$$&lt;/p&gt;&#xA;" LastActivityDate="2010-09-12T03:10:37.317" />
  <row Id="3343" PostTypeId="2" ParentId="3319" CreationDate="2010-08-26T06:30:38.517" Score="10" ViewCount="" Body="&lt;p&gt;The number $e$ is &lt;strong&gt;not&lt;/strong&gt; very important in mathematics!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the contrary it's the &lt;em&gt;exponential function&lt;/em&gt;&#xA;$$\exp(x)=\sum_{n=0}^\infty\frac{x^n}{n!}$$&#xA;that's important. The number $e$ is just $\exp(1)$ - just one&#xA;value of the exponential function. Compare how often one&#xA;sees the exponential function as opposed to $e$ &lt;strong&gt;by itself&lt;/strong&gt;.&#xA;In the four previous answers to this question, three focus on the&#xA;exponential function, and only one on $e$ itself.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-08-26T06:30:38.517" />
  <row Id="3344" PostTypeId="2" ParentId="3336" CreationDate="2010-08-26T06:38:53.573" Score="3" ViewCount="" Body="&lt;p&gt;(This is by no means a comprehensive list.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Right-triangle definition&lt;/strong&gt;: the sine (cosine) of an acute angle is the ratio of the lengths of the leg opposite (adjacent to) the given angle to the hypotenuse of the triangle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Bizarre Geometric Definition&lt;/strong&gt;: the cosine of an angle in a triangle is $\frac{a^2+b^2-c^2}{2ab}$, where $a$ and $b$ are the lengths of the sides adjacent to the angle and $c$ is the length of the side opposite the angle; the sine of an acute angle is the cosine of its complement; the sine of an obtuse angle is the cosine of its supplement's complement. (&lt;em&gt;edit: this one might be even worse than I'd originally thought as a definition, so perhaps just ignore it.&lt;/em&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Circular definition (1)&lt;/strong&gt;: the sine (cosine) of a magnitude of rotation is the vertical (horizontal) coordinate of the image of the point (1,0) under a rotation of the given magnitude centered at the origin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Circular definition (2)&lt;/strong&gt;: the sine (cosine) of a directed angle with vertex at the origin and initial ray on the positive x-axis is the y-coordinate (x-coordinate) of the point of intersection of the terminal ray of the angle with the unit circle centered at the origin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Power series definition&lt;/strong&gt;:&#xA;$$\begin{align}&#xA;\sin x&amp;amp;=\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)!}x^{2n+1}=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\cdots&#xA;\\&#xA;\cos x&amp;amp;=\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n)!}x^{2n}=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\cdots&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Exponential definition&lt;/strong&gt;:&#xA;$$\sin x=\frac{e^{ix}-e^{-ix}}{2i};\quad\quad\cos x=\frac{e^{ix}+e^{-ix}}{2}$$&lt;/p&gt;&#xA;" OwnerUserId="72" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-08-26T06:47:49.147" LastActivityDate="2010-08-26T06:47:49.147" CommentCount="2" />
  <row Id="3345" PostTypeId="2" ParentId="3336" CreationDate="2010-08-26T07:19:19.593" Score="0" ViewCount="" Body="&lt;p&gt;I like good old definition using right angled triangle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/Sxit7.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$sin\theta = b/c$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$cos\theta = a/c$&lt;/p&gt;&#xA;" OwnerUserId="705" LastActivityDate="2010-08-26T07:19:19.593" CommentCount="1" />
  <row Id="3346" PostTypeId="2" ParentId="3315" CreationDate="2010-08-26T08:09:07.367" Score="5" ViewCount="" Body="&lt;p&gt;More generally, if you want to compute all the $n$-th roots of a complex number $z_0$, that is, all the complex numbers $z$ such that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;z^n = z_0 \ ,  \qquad \qquad \qquad \qquad [1]&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;you should write this equation in exponential form: $z = re^{i\theta}, \ , z_0 = r_0 e^{i\theta_0}$. Then [1] becomes&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\left( r e^{i \theta}\right)^n = r_0 e^{i\theta}   \qquad \Longleftrightarrow \qquad r^n e^{in\theta} = r_0 e^{i\theta_0}  \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if you have two complex numbers in polar coordinates which are equal, their moduluses must be equal clearly:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;r^n = r_0 \qquad \Longrightarrow \qquad r = +\sqrt[n]{r_0}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;since $r, r_0 \geq 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the arguments, we cannot simply conclude that $n\theta = \theta_0$, but just that they differ in an integer multiple of $2\pi$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;n\theta = \theta_0 + 2k\pi   \qquad \Longleftrightarrow \qquad  \theta = \frac{\theta_0 + 2k \pi}{n} \quad \text{for} \quad k = 0, \pm 1 , \pm 2, \dots&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It would seem that we have an infinite number of $n$-th roots, but we have enough with $k = 0, 1, \dots , n-1$, since for instance for $k=0$ and $k=n$ we obtain the same complex numbers. Thus, finally&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\sqrt[n]{r_0 e^{i\theta_0}} = +\sqrt[n]{r_0} e^{i \frac{0_0 + 2k\pi}{n}} \ , \quad k = 0, 1, \dots , n-1 &#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;are all the complex $n$-th roots of $z_0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Examples&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;(1)&lt;/strong&gt; For $n=2$, we obtain that every complex number has exactly two square roots:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{align}&#xA;\sqrt{z_0} &amp;amp;= +\sqrt{r_0}e^{i\frac{\theta_0 + 2k\pi}{2}} \ , k = 0,1  \\&#xA;           &amp;amp;= +\sqrt{r_0}e^{i\frac{\theta_0}{2}} \quad \text{and} \quad +\sqrt{r_0}e^{i\left(\frac{\theta_0}{2} + \pi \right)}  \ .&#xA;\end{align}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, since $i = e^{i\frac{\pi}{2}}$, we obtain&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\sqrt{i} = &#xA;\begin{cases}&#xA;e^{i\frac{\pi}{4}} = \cos\frac{\pi}{4} +i \sin\frac{\pi}{4} = \frac{\sqrt{2}}{2} + i \frac{\sqrt{2}}{2}    \\&#xA;e^{i(\frac{\pi}{4} + \pi)} = \cos\frac{5\pi}{4} +i \sin\frac{5\pi}{4} = -\frac{\sqrt{2}}{2} - i \frac{\sqrt{2}}{2}  \ .&#xA;\end{cases}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, if $z_0 = -1 = e^{i\pi}$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\sqrt{-1} = e^{i \frac{\pi}{2}} = i \quad \text{and} \quad e^{i\left( \frac{\pi}{2} + \pi\right)} = e^{i\frac{3\pi}{2}} = -i \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;(2)&lt;/strong&gt; For $z_0 = 1 = e^{i \cdot 0}$ and any $n$, we obtain the $n$-th &lt;em&gt;roots of unity&lt;/em&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\sqrt[n]{1} = e^{i\frac{2k\pi}{n}} \ , \quad k= 0, 1, \dots , n-1 \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, if $n= 2$, we get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\sqrt{1}= e^{i \cdot 0} = 1 \quad \text{and} \quad  e^{i\pi}= -1&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and for $n= 4$, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\sqrt[4]{1} = e^{i\frac{2k\pi}{4}} \ , \quad k = 0, 1, 2, 3 \ ,&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;that is,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\sqrt[4]{1} = 1, i, -1 , -i \ .&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="664" LastEditorUserId="664" LastEditorDisplayName="" LastEditDate="2010-08-26T13:04:44.737" LastActivityDate="2010-08-26T13:04:44.737" CommentCount="3" />
  <row Id="3347" PostTypeId="2" ParentId="3329" CreationDate="2010-08-26T09:00:33.243" Score="3" ViewCount="" Body="&lt;p&gt;This is not intended as a full answer, but as a way to try to decide if it is feasible to be able to parameterize a plane curve in an easy way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The two examples you give, namely the folium and the bicorn, are examples of &quot;irreducible curves of geometric genus zero&quot;. For the purpose of this discussion you can take &quot;irreducible curve&quot; to mean the locus of zeros of an irreducible polynomial in two variables. Geometric genus zero is harder to define, though see &lt;a href=&quot;http://math.stackexchange.com/questions/1609/intuitive-explantions-for-the-concepts-of-divisor-and-genus&quot;&gt;this&lt;/a&gt; question and the various answers, especially the answer of Matt E. In the linked answer, the &quot;genus&quot; that is defined is the &quot;arithmetic genus&quot;, whereas we are interested in the &quot;geometric genus&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The two genera are not unrelated, and the discrepancy between them is concentrated at the singular points of the curve. A singular point of a plane curve $C$ given as the vanishing set of a polynomial $f(x,y)$ is a point $p$ in the plane that lies on the curve (i.e. $f(p)=0$) and where both partial derivatives of $f$ vanish (i.e. $df/dx(p)=df/dy(p)=0$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, while the arithmetic genus of an (irreducible) plane curve $C$ of degree $d$ is simply $p_a:=(d-1)(d-2)/2$, the geometric genus of $C$ is an integer less than or equal to $p_a$. I will not dwell too much on how to compute the geometric genus, for the moment I will simply say that every singular point decreases the arithmetic genus by at least one (counting also the &quot;esoteric&quot; singular points at infinity, as you need to explain the bicorn). There are algorithmic way to decide the drop in genus that every singular point determines and thus the computation of the geometric genus can be taken as &quot;easy&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Over the complex numbers.&lt;/strong&gt; It is a fact that, whenever the geometric genus of an irreducible curve $C$ is zero then you can find a parameterization of $C$ by &lt;em&gt;rational functions of a single variable $t$&lt;/em&gt;, provided that you are thinking about the complex solutions of your curve. This parameterization can be found algorithmically, and while there are various tricks to use, the substitution $y=mx$ is at the core: depending on context it might be called &quot;projection&quot; or &quot;blow up&quot;. The other trick is that you can &quot;reembed&quot; your curve by considering the vector space whose coordinates are all the monomials of some given degree $n$ and take the image of the curve under the map that sends a point in the plane in the $N$-tuple of all the evaluations of the monomials you have chosen. Instead of being more explicit, let me give an example: if the curve is the $x$-axis in the plane, and you choose the monomials $1,x,y,x^2,xy,y^2$ to reembed, then we image of the $x$-axis (i.e. the locus where $y=0$) is the set of points with coordinates $(1,x,0,x^2,0,0)$ in a 6-dimensional space. Forgetting the useless coordinates (this is an instance of projection) gives us a plane curve parameterized by $(x,x^2)$, namely the parabola $x_1^2=x_2$. In this example we started with a parameterization and ended with a parameterization. Being more refined you can also start with an implicit equation and end with (lots of) implicit equations in more variables.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you have projections and reembeddings you can play them against one another: embed your curve in large enough space. Project away from singular points until you get back to the plane; reembed and reproject. There is a way of making sure that &quot;singular points get resolved&quot; by doing this, so that eventually you get to a curve having no singular points. Once you are there, further projections will do the trick of getting you to a line.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Over the real numbers.&lt;/strong&gt; Of course things are somewhat trickier. Certainly two possibilities are the folium, where the rational parameterization you had over the complex numbers &quot;just works&quot; over the reals, but there is the slightly more elaborate one of the bicorn, where you have a rational parameterization in terms of trigonometric functions. Again, by modification of the tricks discussed above, you can get every irreducible curve of genus zero to become a plane conic. Once in this form, you will find a rational parameterization if and only if you can find a point, and deciding if there is a point reduces to the computation of whether a matrix constructed out of the coefficients of the equation has eigenvalues of different signs or not. The distinction between rational parameterization in terms of a variable or in terms of trigonometric functions has to do with personal taste.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So far, this has dealt with curves of genus zero and I will not say much about higher genus. If the genus is one, there is, as you mentioned, the possibility of using the Weierstrass $\wp$. Here things become more complicated: the function used to parameterize everything is less familiar, you will probably need some transcendental numbers to get this parameterization to work, and the topological description of the space of solutions gets more in the way. In the genus zero case, the only thing that mattered was: is it empty, is it not empty. In the genus one case there is &quot;one more&quot; possibility. Of course, higher genera become harder and harder, and I will stop here, given that this is already &lt;em&gt;way&lt;/em&gt; too much for this!&lt;/p&gt;&#xA;" OwnerUserId="1028" LastActivityDate="2010-08-26T09:00:33.243" CommentCount="7" />
  <row Id="3348" PostTypeId="1" AcceptedAnswerId="3360" CreationDate="2010-08-26T09:11:02.423" Score="2" ViewCount="192" Body="&lt;p&gt;Construct a string by using only n characters, the string should do not contain any consecutive repeated pattern, in other words, the string must have not any substring matching /(.+)\1/, what is the length l(n) of the longest string S(n)?&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;n    S(n)    l(n)&#xA;1    a       1&#xA;2    aba     3&#xA;3    ?       ?&#xA;4    ?       ?&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Edit:&lt;br&gt;&#xA;&lt;code&gt;abcabacbabcabacabcac&lt;/code&gt; has none consecutive  repeated pattern,so l(3)&gt;=20&lt;/p&gt;&#xA;" OwnerUserId="1498" LastEditorUserId="171" LastEditorDisplayName="" LastEditDate="2010-08-26T12:36:55.710" LastActivityDate="2010-08-26T13:14:38.157" Title="The longest string of none consecutive repeated pattern" Tags="&lt;combinatorics&gt;&lt;counting&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="1" />
  <row Id="3349" PostTypeId="2" ParentId="3348" CreationDate="2010-08-26T09:41:32.183" Score="0" ViewCount="" Body="&lt;p&gt;To find a maximal string for n letters, add a new letter in every other place of a maximal string for n-1 letters.  It is still the case that no consecutive identical substrings exist.  No more letters can be added because two of them would have to appear consecutively, so this is the best that is possible.  The length of the string is 2^n-1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a&#xA;bab&#xA;cbcacbc&#xA;dcdbdcdadcdbdcd&lt;/p&gt;&#xA;" OwnerUserId="1284" LastActivityDate="2010-08-26T09:41:32.183" CommentCount="4" />
  <row Id="3350" PostTypeId="2" ParentId="3333" CreationDate="2010-08-26T10:02:54.803" Score="0" ViewCount="" Body="&lt;p&gt;Here is an explicit example that is very concrete, but a near miss: a graph $G$  with 16 vertices containing no triangle and no independent set with six elements. I think that it is not possible to extend the graph $G$ by a single vertex preserving this property, but I would be interested if I could be proved wrong! For those who are interested and know about these things, the graph $G$ is the graph of configurations of the exceptional curves on a del Pezzo surface of degree four. None of this will be needed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Construction of&lt;/strong&gt; $G$. Start with a circle $C$ in the plane, five distinct points $p_1,\ldots,p_5$ on the circle $C$ and the 10 lines $\{\ell_{ij}\}_{1 \leq i &amp;lt; j \leq 5}$ through the five points. The graph has as vertices the circle $C$, the points $\{p_i\}$ and the lines $\{\ell_{ij}\}$, for a total of $1+5+10=16$ vertices. The edges are as follows:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;the vertex $C$ is joined to each point $p_1,\ldots,p_5$, but to no line;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;the vertex $p_i$ is joined to each line through itself and to the circle, but nothing else (for instance, there is no edge between different points and no edge between $p_1$ and $\ell_{23}$);&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;the vertex $\ell_{ij}$ is joined to the points $p_i$ and $p_j$, as well as all the lines $\ell_{kl}$ for $i,j\neq k,l$;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;where, of course, $i$ and $j$ range in $\{1,\ldots,5\}$, and $i$&lt;code&gt;&amp;lt;&lt;/code&gt;$j$ when both appear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can now check that there are no triangles and no independent sets with six elements in $G$. The checks are easier once you realize that there is a large group of symmetries acting on $G$ (the group is larger than the obvious permutations of the indices: it has order 1920).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, this graph has 16 vertices! As I mentioned above, I think that any extension of this graph will have either a triangle or an independent set of size six, but I am not sure: anyone can confirm this? One way to go about this would be to see if this graph appears as a subgraph in one of the examples that Mariano gives in his answer. My argument was not this one, and I would prefer a more conceptual argument.&lt;/p&gt;&#xA;" OwnerUserId="1028" LastActivityDate="2010-08-26T10:02:54.803" />
  <row Id="3351" PostTypeId="2" ParentId="2584" CreationDate="2010-08-26T10:07:28.947" Score="0" ViewCount="" Body="&lt;p&gt;First up, you should accept Ranieri's answer, which is obviously correct. The question of &quot;slicing&quot; seems like a temporary misunderstanding that shouldn't stand in your way. Rejection sampling works with continuous distributions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second, it seems you are attempting to randomly generate individuals from a population with some number of subgroups, each of which has an expected distribution of some continuous traits. Since the traits are distributed normally, $\forall{x}: P(x) &gt; 0$, so it is notionally possible to have outlier individuals with &lt;em&gt;any&lt;/em&gt; trait combination. However, it is very likely that arbitrary distributional constraints will not be satisfiable at the population level.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, imagine that you have a population divided into two genders and two colours, and specified to be 50/50 male/female and 50/50 red/blue. Then require that some trait, say foolishness, is distributed $N(1,0.1)$ for each of the subgroups men, red and blue, but $N(0.5, 0.1)$ for women. Is this possible? I'm pretty sure not. And that's without even trying to clip values into a fixed range.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, I suspect that you might need a somewhat better model for your probability distribution. My first inclination would be to weight the different criteria in some fashion to prioritise the distribution of some traits over others. However, I'd have to think that through properly, which I don't have time to do just now. Maybe later...&lt;/p&gt;&#xA;" OwnerUserId="144" LastActivityDate="2010-08-26T10:07:28.947" CommentCount="1" />
  <row Id="3352" PostTypeId="1" CreationDate="2010-08-26T10:22:08.570" Score="2" ViewCount="714" Body="&lt;p&gt;Let $f: D \to \mathbb{R}\ $ be a function on non-singular, convex domain $D \subseteq \mathbb{R}^d$ and let us assume the second-order derivatives of $f$ exist. It is well known that $f$ is convex if and only if its Hessian $\nabla^2 f(x)$ is positive semi-definite for all $x \in D$. It is also known that if $\nabla^2 f(x)$ is positive definite for all $x \in D$, we may conclude that $f$ is strictly convex (for a reference, see &lt;a href=&quot;http://www.stanford.edu/~boyd/cvxbook/&quot; rel=&quot;nofollow&quot;&gt;Boyd and Vandenberghe, 2004&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, if $f$ is strictly convex, we still merely know that $\nabla^2 f(x)$ is positive semi-definite for all $x \in D$. That is, there may be $x \in D$ such that $y^T \nabla^2 f(x) y = 0\ $ for some $y\not=0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an example, consider $f(x)=x^4$. In this case, $f$ is strictly convex, but $f''(x)=12x^2$ and, hence, $yf''(x)y=0$ for $x=0$ and $yf''(x)y&gt;0$ for all $x\not=0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yet, these points that ruin the complete positive definiteness seem to be very sparsely distributed within $D$. So, my question is as follows:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If $f$ is strictly convex, how can we characterize the set of points $X$ for which&#xA;  $\nabla^2f(x)$ is not positive definite for $x \in X$, and &#xA;  $\nabla^2f(x)$ is positive definite for $x \in D \setminus X$?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That is, has such a set $X$ been investigated before, what properties are known, and where can I learn more about it? Any reference is welcome.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, my guess is that on can state the following:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Conjecture 1:&lt;/strong&gt; The set $X$ is merely a discrete subset of $D$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;EDIT: Since Conjecture 1 has obviously been disproved by George Lowther below, allow me to restate my guess as the following (less bold) statement:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Conjecture 2:&lt;/strong&gt; The set $X$ is does not contain a non-empty open ball.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;or, even more cautious:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Conjecture 3:&lt;/strong&gt; The set $D \setminus X$ does contain a non-empty open ball.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="347" LastEditorUserId="347" LastEditorDisplayName="" LastEditDate="2010-08-27T11:15:34.763" LastActivityDate="2010-08-27T11:15:34.763" Title="Positive definite Hessians from strictly convex functions" Tags="&lt;real-analysis&gt;&lt;convex-analysis&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="3" />
  <row Id="3353" PostTypeId="1" AcceptedAnswerId="3365" CreationDate="2010-08-26T10:40:47.077" Score="2" ViewCount="326" Body="&lt;p&gt;In the Portuguese book &lt;em&gt;Análise Matemática&lt;/em&gt; (Mathematical Analysis) by C. Sarrico, it is proved there exist&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle\lim_{n\rightarrow +\infty}f_{n}(x)=\lim_{n\rightarrow +\infty }e^{\log f_{n}(x)},\qquad(1)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f_{n}(x)=\dfrac{n!n^{x}}{x(x+1)(x+2)\cdots (x+n)}.\qquad(2)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And it is stated there that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle\lim_{n\rightarrow +\infty }f_{n}(x)=\Gamma (x)=\int_{0}^{+\infty}t^{x-1}e^{-t}dt\qquad x&gt;0.\qquad(3)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The author writes that a proof can be found in &quot;Principles of Mathematical Analysis&quot; by W. Rudin. Since I don't have it, I ask the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Question: How is the sketch of such a proof or other proofs of the same result?&lt;/p&gt;&#xA;" OwnerUserId="752" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2010-10-04T21:48:09.397" LastActivityDate="2010-10-04T21:48:09.397" Title="Two ways of defining the gamma function $\Gamma (x)$. How to show they are equivalent?" Tags="&lt;real-analysis&gt;&lt;sequences-and-series&gt;&lt;special-functions&gt;&lt;gamma-function&gt;" AnswerCount="2" FavoriteCount="0" />
  <row Id="3354" PostTypeId="2" ParentId="3336" CreationDate="2010-08-26T11:01:49.430" Score="1" ViewCount="" Body="&lt;p&gt;Once you define sine, you can define cosine to be the sine of the complementary angle. i.e. cos(theta) = sin(pi/2 - theta). So you could reduce your question to asking for various ways of defining sine.&lt;/p&gt;&#xA;" OwnerUserId="247" LastActivityDate="2010-08-26T11:01:49.430" />
  <row Id="3355" PostTypeId="1" CreationDate="2010-08-26T11:10:46.623" Score="1" ViewCount="143" Body="&lt;p&gt;Determine the Euler characteristic of the surface&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ M=\left\{(x,y,z); \sqrt{x^2+y^2}=1+z^{2n}, 0&amp;lt; z&amp;lt; 1\right\} $$&lt;/p&gt;&#xA;" LastEditorUserId="171" LastEditorDisplayName="" LastEditDate="2010-08-26T11:52:44.073" LastActivityDate="2011-05-13T21:38:17.360" Title="Euler characteristic of a compact surface" Tags="&lt;differential-geometry&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="3356" PostTypeId="1" AcceptedAnswerId="3358" CreationDate="2010-08-26T11:27:15.630" Score="5" ViewCount="331" Body="&lt;p&gt;I'm struggling with this former Putnam Exam problem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose $f$ and $g$ are nonconstant, differentiable, real-valued functions on $R$.  Furthermore, suppose that for each pair of real numbers $x$ and $y$, $f(x + y) = f(x)f(y) - g(x)g(y)$ and $g(x + y) = f(x)g(y) + g(x)f(y)$.  If $f'(0) = 0$, prove that $(f(x))^2 + (g(x))^2 = 1$ for all $x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Right.  So obviously, $f(x) = \cos x$ and $g(x) = \sin x$ satisfy the conditions and also the conclusion of the problem.  But are these the unique such functions, and if so, how to prove it?  And if not, then how to prove the conclusion otherwise?&lt;/p&gt;&#xA;" OwnerUserId="506" LastActivityDate="2010-08-26T18:21:07.040" Title="Uniqueness of $\sin x$ and $\cos x$? (Putnam Exam problem)" Tags="&lt;calculus&gt;&lt;putnam&gt;" AnswerCount="3" CommentCount="9" FavoriteCount="1" />
  <row Id="3357" PostTypeId="2" ParentId="3278" CreationDate="2010-08-26T11:48:14.930" Score="1" ViewCount="" Body="&lt;p&gt;I don't think there is any ambiguity in the use of the word “given” here, but I do think that one needs to learn the way it is used in conditional probability.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If people are not familiar with how terms are used then of course there could be misunderstanding. There are some interesting comments about ambiguity in such questions in &lt;a href=&quot;http://en.wikipedia.org/wiki/Boy_or_Girl_paradox&quot; rel=&quot;nofollow&quot;&gt;this wikipedia article.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(This sentence from the article made me smile: Many people, including professors of mathematics, argued strongly for both sides with a great deal of confidence, sometimes showing disdain for those who took the opposing view.) &lt;/p&gt;&#xA;" OwnerUserId="1301" LastActivityDate="2010-08-26T11:48:14.930" />
  <row Id="3358" PostTypeId="2" ParentId="3356" CreationDate="2010-08-26T12:05:20.903" Score="6" ViewCount="" Body="&lt;p&gt;Let $u(x) = f(x)^2 + g(x)^2$. It should be easy to show that $u(x+y) = u(x)u(y)$, thus $u$ must be an exponential function $u(x) = e^{Bx}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, as $f(2x) = f(x)^2 - g(x)^2$, we should get $0 = g(0) g'(0)$. Comparing this with $u'(0)$ we should get $B = 0$, thus $f(x)^2 + g(x)^2 = 1$.&lt;/p&gt;&#xA;" OwnerUserId="171" LastActivityDate="2010-08-26T12:05:20.903" CommentCount="3" />
  <row Id="3359" PostTypeId="2" ParentId="3356" CreationDate="2010-08-26T12:23:48.873" Score="3" ViewCount="" Body="&lt;p&gt;Since KennyTM posted a very elegant proof I will post an uncreative one that requires no thought to produce.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First write out all the obvious identities that come from the information told:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;$f(x + y) = f(x)f(y) - g(x)g(y)$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$f(x) = f(x)f(0) - g(x)g(0)$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$f(0) = f(0)f(0) - g(0)g(0)$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$g(x + y) = f(x)g(y) + g(x)f(y)$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$g(x) = f(x)g(0) + g(x)f(0)$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$g(0) = f(0)g(0) + g(0)f(0)$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$Df(x + c) = Df(x)f(c) - Dg(x)g(c)$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$Dg(x + c) = Df(x)g(c) + Dg(x)f(c)$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$Df(0) = 0 = Df(0)f(0) - Dg(0)g(0) = -Dg(0)g(0)$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$Dg(0) = Df(0)g(0) + Dg(0)f(0) = Dg(0)f(0)$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The $Df(0)$ equation tells us that $Dg(0) = 0 \lor g(0) = 0$ Since we don't want $Dg(0) = 0$ lets discharge that automatically now by proving it contradictory: It would imply that $Dg(0 + c) = Df(0)g(c) + Dg(0)f(c) = 0$ so $g(x)$ would be a constant but that is not allowed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have seen that $g(0) = 0$ and now we update the identifies with this information, and that tells us is that $f(0) = 1$. We cannot determine the value of $Dg(0)$ so just call it some constant $k$. Now it is easy to recognize that we have a simple ODE problem that could be solved symbolically by a calculator or by taking eigenvalues/vectors of the matrix that defines the system. Giving the solutions $\sin(kx)$ &amp;amp; $\cos(kx)$.&lt;/p&gt;&#xA;" LastActivityDate="2010-08-26T14:05:57.623" />
  <row Id="3360" PostTypeId="2" ParentId="3348" CreationDate="2010-08-26T12:28:11.080" Score="3" ViewCount="" Body="&lt;p&gt;A theorem of Thue's (not Thue's theorem!) states that for $n = 3$ (and therefore, also for larger $n$) an infinite sequence of characters (chosen out of $n$) without repeated sub-sequences (a.k.a square-free) exists. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: here's a slightly related reference: J.D. Currie's paper &lt;a href=&quot;http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.11.965&amp;amp;rep=rep1&amp;amp;type=pdf&quot; rel=&quot;nofollow&quot;&gt;'There are ternary circular square-free words of length&#xA;n for n  18.'&lt;/a&gt;, which at least has a reference to Thue's paper from 1906. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit 2: Another reference, with explicit construction of sequences of length 3, 9, 27 and so on: &lt;a href=&quot;http://arxiv.org/abs/0712.0139&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/0712.0139&lt;/a&gt;. &#xA;That paper gives the following sequence of length 27: 123213231213123132312132123&lt;/p&gt;&#xA;" OwnerUserId="360" LastEditorUserId="360" LastEditorDisplayName="" LastEditDate="2010-08-26T12:47:05.130" LastActivityDate="2010-08-26T12:47:05.130" />
  <row Id="3361" PostTypeId="2" ParentId="3348" CreationDate="2010-08-26T13:14:38.157" Score="1" ViewCount="" Body="&lt;p&gt;Let me apologize, for it's not an answer, but merely an extended comment. I like the problem very much. It sounds very classical, and most likely the answer must be known. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Being reformulated, the problem is as follows: if $X$ is an $n$-letter alphabet, we have a rewriting rule&#xA;$$&#xA;w_1 s s w_2=w_1 s^2 w_2 \to w_1 w_2&#xA;$$&#xA;where $s$ is nonempty and at least one of $w_1,w_2$ must be nonempty. (In effect, any word that satisfies the condition set by a-boy must be square-free so to speak.) One can also a semigroup with $n$ generators whose relations are determined by the rewriting rule above. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem as set asks whether all words over $X$ are reduced to words of certain length $l(n).$ One wonders, however, whether $n=2$ is an exceptional case, and actually there are arbitrary large reduced words. For instance, the following word of length 68 in letters $0,1,2$ is reduced:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;12012101201020120212012101202101210201202120121012010201202120121012&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a 4-letter alphabet it is easy to find words of quite large length (I've found by means of computing such words of lenght $\ge 1000.$)&lt;/p&gt;&#xA;" OwnerUserId="1100" LastActivityDate="2010-08-26T13:14:38.157" CommunityOwnedDate="2010-08-26T13:14:38.157" CommentCount="1" />
  <row Id="3362" PostTypeId="1" AcceptedAnswerId="3407" CreationDate="2010-08-26T13:23:08.630" Score="3" ViewCount="120" Body="&lt;p&gt;Let $f : [0, 1] \to \mathbb{R}$ be a 'well behaved' function and define $c_n$ by the Riemann sums&#xA;$$&#xA;c_n := \frac{1}{n} \sum_{k = 1}^n f(\frac{k - \theta}{n})&#xA;$$&#xA;for some given $\theta \in [0, 1]$. Then obviously $c_n$ will converge to the integral of $f$. Under which conditions on $f$ will this convergence be monotone? In particular, will the convergence be monotone if $f$ is twice differentiable with $f &gt; 0$, $f' &amp;lt; 0$ and $f'' &gt; 0$?&lt;/p&gt;&#xA;" OwnerUserId="960" LastEditorUserId="960" LastEditorDisplayName="" LastEditDate="2010-08-26T18:15:34.910" LastActivityDate="2010-08-26T19:39:44.927" Title="When do Riemann sums converge monotonically?" Tags="&lt;real-analysis&gt;&lt;convergence&gt;&lt;integral&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="3363" PostTypeId="1" AcceptedAnswerId="3402" CreationDate="2010-08-26T13:39:19.927" Score="7" ViewCount="101" Body="&lt;p&gt;Walking on a lattice.  The number of various paths from (0,0) to (m,n)  using north and east steps is binomial coefficient&lt;/p&gt;&#xA;&#xA;&lt;p&gt;C(m+n,m)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;if he needs to go back (0,0) using south and west steps, and doesn't pass by the passed points. Then what is the number of various loops walking from (0,0) to (m,n) then returning to (0,0)? Any algebraic expression for this?&#xA;&lt;img src=&quot;http://i.imgur.com/OS8Zx.png&quot; alt=&quot;alt text&quot;&gt;&#xA;btw:i asked this question before, but had not get an answer yet. Maybe I can get a good answer at here.&lt;/p&gt;&#xA;" OwnerUserId="1498" LastEditorUserId="1498" LastEditorDisplayName="" LastEditDate="2010-08-26T14:08:08.060" LastActivityDate="2010-08-26T18:22:44.583" Title="the number of loops on lattice?" Tags="&lt;combinatorics&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="3364" PostTypeId="1" CreationDate="2010-08-26T13:41:22.033" Score="2" ViewCount="351" Body="&lt;p&gt;A continuous function does not always map open sets to open sets, but a continuous function will map compact sets to compact sets.  One could make list of such preservations of topological &#xA;properties by a continuous function $f$:&#xA;$$ f( \mathrm{open} ) \neq \mathrm{open} \;,$$&#xA;$$ f( \mathrm{closed} ) \neq \mathrm{closed} \;,$$&#xA;$$ f( \mathrm{compact} ) = \mathrm{compact} \;,$$&#xA;$$ f( \mathrm{convergent \; sequence} ) = \mathrm{convergent \; sequence} \;.$$&#xA;Could you please help in extending this list?&#xA;(And correct the above if I've erred!)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Edit&lt;/b&gt;.  Thanks for the several comments and answers extending my list.&#xA;I was hoping that I could see some common theme among the properties preserved by a continuous&#xA;mapping, separating those that are not preserved.  But I don't see such a pattern.&#xA;If anyone does, I'd appreciate a remark.  Thanks!&lt;/p&gt;&#xA;" OwnerUserId="237" LastEditorUserId="237" LastEditorDisplayName="" LastEditDate="2010-08-27T11:16:14.497" LastActivityDate="2010-09-28T07:03:30.743" Title="Topological properties preserved by continuous maps" Tags="&lt;topology&gt;" AnswerCount="3" CommentCount="4" FavoriteCount="1" />
  <row Id="3365" PostTypeId="2" ParentId="3353" CreationDate="2010-08-26T13:53:23.343" Score="3" ViewCount="" Body="&lt;p&gt;I don't have a copy of Rudin, but can give you a proof of this. There's various ways of playing about with the integral form to get what you want, although I'm not sure what the cleanest method is.&#xA;One way is to use the fact that $1_{\{t\le n\}}(1-t/n)^n\to e^{-t}$ to write $\Gamma(x)$ as&#xA;$$&#xA;\begin{align}&#xA;\Gamma(x)&amp;amp;=\lim_{n\to\infty}\int_0^nt^{x-1}(1-t/n)^{n}\,dt\\&#xA;&amp;amp;=\lim_{n\to\infty}n^x\int_0^1s^{x-1}(1-s)^n\,ds.&#xA;\end{align}&#xA;$$&#xA;To prove that the first integral commutes with the limit, you could use the dominated convergence theorem. The second integral is just using the substitution $t=ns$.&#xA;It needs to be shown that this integral is equal to your function $f_n(x)$. In fact, I recognize both the function $f_n$ and the integral as &lt;a href=&quot;http://en.wikipedia.org/wiki/Beta_function&quot; rel=&quot;nofollow&quot;&gt;beta functions&lt;/a&gt;.&#xA;$$&#xA;\int_0^1s^{x-1}(1-s)^n\,ds=B(x,n+1)&#xA;$$&#xA;To find the explicit form for this, you can either use $B(x,y)=\Gamma(x)\Gamma(y)/\Gamma(x+y)$ (a proof is given on the &lt;a href=&quot;http://en.wikipedia.org/wiki/Beta_function&quot; rel=&quot;nofollow&quot;&gt;Wikipedia page&lt;/a&gt;) or repeatedly apply the identity $B(x,n+1)=B(x+1,n)n/x$ (which follows from integration by parts) and $B(x,1)=1/x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: You can also apply a very similar argument to the one above using the limit $(1+t/n)^{-n}\to e^{-t}$, which I did in my first version of this answer. Using $(1-t/n)^n\to e^{-t}$ seems cleaner though, so I edited the answer accordingly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An alternative method is to write&#xA;$$&#xA;\begin{align}&#xA;\Gamma(x) &amp;amp;= \frac{\Gamma(x+n+1)}{\Gamma(n+1)}\frac{\Gamma(n+1)\Gamma(x)}{\Gamma(x+n+1)}\\&#xA;&amp;amp;= \frac{\Gamma(x+n+1)}{\Gamma(n+1)}\frac{n!}{x(x+1)\cdots(x+n)}.&#xA;\end{align}&#xA;$$&#xA;Then, the limit you need would follow as long as it can be shown that $\Gamma(x+n+1)/\Gamma(n+1)$ approaches $n^x$ asymptotically as $n\to\infty$.&#xA;$$&#xA;\Gamma(n+x+1)=\int_0^\infty t^x t^ne^{-t}\,dt&#xA;$$&#xA;Then, the required expression depends on showing that, in the limit $n\to\infty$, to leading order, the integral only contributes for values of $t/n$ close to 1 (you can see that $t^ne^{-t}$ has its maximum at $t=n$).&lt;/p&gt;&#xA;" OwnerUserId="1321" LastEditorUserId="1321" LastEditorDisplayName="" LastEditDate="2010-08-26T18:19:13.920" LastActivityDate="2010-08-26T18:19:13.920" CommentCount="1" />
  <row Id="3366" PostTypeId="1" AcceptedAnswerId="3371" CreationDate="2010-08-26T13:53:26.640" Score="1" ViewCount="113" Body="&lt;p&gt;I was given this question (there were prior questions defining the equivelence relation ~)&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;let E be the set of equivelence classes of ~&#xA; define f from E to Reals&#xA;    f:={([(x,y)],r)|x=yr,\: x,y,r in Reals}&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I'm trying to understand what it means; is the following correct?&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;f:[(x,y)] mapsto r&#xA;&#xA;E=(Reals X {Reals/0})/~ &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;ie the Quitient set of (Reals X {Reals/0}) by ~&#xA;((x,y) was ealier defined as an element of (Reals X {Reals/0}) ))&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Am I understanding this right?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My function takes in a equivlence class, [(x,y)]&#xA;and reaturns a real value r, furfilling the condition that x=yr?&lt;/p&gt;&#xA;" OwnerUserId="1505" LastActivityDate="2010-08-26T14:40:07.670" Title="Am I understanding this question right? (function defined with set notation)" Tags="&lt;functions&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="3367" PostTypeId="2" ParentId="3353" CreationDate="2010-08-26T13:55:04.663" Score="1" ViewCount="" Body="&lt;p&gt;First, it should be noted that the integral representation is valid only for x with positive real part.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Having gotten that disclaimer away, one sketch of a proof goes as follows: the limit representation can be shown to be equivalent to the Weierstrass infinite product:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac1{\Gamma(z)}=ze^{\gamma z}\prod_{j=1}^\infty{\left(1+\frac{z}{j}\right)\exp\left(-\frac{z}{j}\right)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;($\gamma$ here is the Euler-Mascheroni constant; the representation $\lim_{k\to\infty}{\sum_{j=1}^k{\frac1{j}}-\ln(k)}$ is useful here).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You now show that the infinite product/limit is the same as the integral $n^z \int_0^1 (1-u)^n u^{z-1} \mathrm{d}u$, and then find a substitution for which &lt;a href=&quot;http://en.wikipedia.org/wiki/Watson%27s_lemma&quot; rel=&quot;nofollow&quot;&gt;Watson's lemma&lt;/a&gt; can be applied.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-08-26T13:55:04.663" CommentCount="4" />
  <row Id="3368" PostTypeId="1" AcceptedAnswerId="3369" CreationDate="2010-08-26T14:10:47.007" Score="2" ViewCount="198" Body="&lt;p&gt;Hi there,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a programming problem I'm trying to solve where, in order to be efficient, I have to find the probability of some events occuring. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lets say I have 3 transactions. At each transaction there is a 5% chance that the transaction will be declined. I want to know what are the chances of 0, 1, 2 and 3 of the transactions being declined.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have being trying to visualise this with a venn diagram and trying to work out the areas of each event, but unfortunately when I add up the separate events I keep on getting more than or less than 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The event that 0 of the transactions will be declined is simply:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;p(0 transactions declined) = .95 * .95* .95 = 6859/8000&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The event that all 3 of the transactions will be declined should be:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;p(3 transactions declined) = .05 * .05 * .05 = 1/8000&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I start running into trouble:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;p(2 transactions only declined) = (3 x (.05 * .05)) -3/8000 (because the A&amp;cap;B&amp;cap;C has been counted three times and we only want to know about when there are 2 transactions, and NOT when there are 2 or 3) = 57/8000&lt;/p&gt;&#xA;&#xA;&lt;p&gt;p(1 transaction only declined) = Well p(A&amp;cap;(A'&amp;cup;B'&amp;cup;C')) x 3 would give us the answer. I make this out to be 3 x (0.05 -(2 x 20/8000) - 1/8000) = 1077/8000&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I add all those events up, then I get 7994/8000. I'm not sure what the right way to go about this is. I'd like to work it out for myself, but I really need a point in the right direction. I don't just want to make this work for three transactions, but for as many as I want so I need a general formula.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many thanks&lt;/p&gt;&#xA;" OwnerUserId="1094" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-08-26T16:21:44.043" LastActivityDate="2010-08-26T16:21:44.043" Title="Probability of a certain number of independent events occurring" Tags="&lt;probability&gt;" AnswerCount="3" />
  <row Id="3369" PostTypeId="2" ParentId="3368" CreationDate="2010-08-26T14:26:47.563" Score="3" ViewCount="" Body="&lt;p&gt;You need to look up the binomial distribution. One online resource is Grinstead and Snell's &lt;a href=&quot;http://www.dartmouth.edu/~chance/teaching_aids/books_articles/probability_book/book.html&quot; rel=&quot;nofollow&quot;&gt;book&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1195" LastActivityDate="2010-08-26T14:26:47.563" CommentCount="1" />
  <row Id="3370" PostTypeId="2" ParentId="3368" CreationDate="2010-08-26T14:28:37.073" Score="3" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Binomial_distribution&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Binomial_distribution&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="822" LastActivityDate="2010-08-26T14:28:37.073" CommentCount="1" />
  <row Id="3371" PostTypeId="2" ParentId="3366" CreationDate="2010-08-26T14:36:25.543" Score="2" ViewCount="" Body="&lt;p&gt;Yes; the function is defined to take equivalence classes as inputs, and return reals as values. If you put in the class of $(x,y)$, which is $[(x,y)] = \{(a,b)\in \mathbb{R}\times\mathbb{R}^{\star}\,|\, (x,y)\sim (a,b)\}$, it returns a real number $r$. Which real number? The real number $r$ such that $x=yr$. (I'm using $\mathbb{R}^{\star}$ to denote the nonzero real numbers) The function takes equivalence classes as inputs, and has real numbers as outputs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You would need to make sure that this function is &lt;em&gt;well defined&lt;/em&gt;, though; that is, the definition of the function depends on the name you give the equivalence class; but the class $[(x,y)]$ has many different &quot;names&quot;; it can also be called $[(a,b)]$ for some other real numbers $a$ and $b$. So you need to make sure that if $[(x,y)]=[(a,b)]$ (that is, if $(x,y)\sim (a,b)$) then the real number $r$ such that $x=yr$ is the same as the real number $s$ such that $a=bs$. Otherwise, what you describe would not be a function.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-08-26T14:36:25.543" CommentCount="1" />
  <row Id="3372" PostTypeId="2" ParentId="3364" CreationDate="2010-08-26T14:37:01.227" Score="3" ViewCount="" Body="&lt;ol&gt;&#xA;&lt;li&gt;Connectedness and path connectedness&lt;/li&gt;&#xA;&lt;li&gt;If $f$ is a local homeomorphism, then $f$ is an open and closed map.&lt;/li&gt;&#xA;&lt;li&gt;If $f$ is onto and the domain is normal (can separate closed sets) or the map has compact fibers, then the image will be Hausdorff if the domain is.&lt;/li&gt;&#xA;&lt;li&gt;Second-countability is preserved under open maps.&lt;/li&gt;&#xA;&lt;li&gt;The image of a simply connected space need not be simply connected. &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;That's off the top of my head.. there are many more. Also, this should probably be community wiki.&lt;/p&gt;&#xA;" OwnerUserId="423" LastActivityDate="2010-08-26T14:37:01.227" />
  <row Id="3373" PostTypeId="2" ParentId="3368" CreationDate="2010-08-26T14:37:42.957" Score="3" ViewCount="" Body="&lt;p&gt;Your calculation for p(1), which I don't completely follow, is somewhat off. Another, more straightforward way of calculating this would be &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$p(\textrm{1 transaction declined}) = {3 \choose 1} \times .05 \times .95 \times .95 = \frac{1083}{8000}$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that this exactly accounts for the $\frac{6}{8000}$ you are missing.&lt;/p&gt;&#xA;" OwnerUserId="1507" LastActivityDate="2010-08-26T14:37:42.957" CommentCount="1" />
  <row Id="3375" PostTypeId="2" ParentId="3366" CreationDate="2010-08-26T14:40:07.670" Score="2" ViewCount="" Body="&lt;p&gt;That looks right to me.  Recall that a function $f : X \to Y$ is formally a subset of $X \times Y$, i.e. a set of ordered pairs $(x,y), x \in X, y \in Y$, with the property that if $(x,y) \in f$ and $(x,z) \in f$ then $y=z$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here $X$ is the set $E$ of equivalence classes, and $Y = \mathbb{R}$.  You are told that $f$ is the set of ordered pairs $([(x,y)],r)$ for which $x = yr$.  One has to check that this is well-defined with respect to the equivalence relation ~: if $x=yr$ and $(x,y)~(x',y')$, then $x'=y'r$.  Then $f$ makes sense as a set.  To check that it is a function, one should check that if $([(x,y)], r) \in f$ and $([(x,y)], r') \in f$ (i.e. $x=yr$ and $x=yr'$) then $r=r'$ (which should follow since $y \ne 0$ by assumption).&lt;/p&gt;&#xA;" OwnerUserId="822" LastActivityDate="2010-08-26T14:40:07.670" />
  <row Id="3376" PostTypeId="1" AcceptedAnswerId="3377" CreationDate="2010-08-26T14:45:56.637" Score="8" ViewCount="310" Body="&lt;p&gt;Suppose $f$ and $g$ are are two polynomials with complex coefficents (i.e $f,g \in \mathbb{C}[x]$).&#xA;Let $m$ be the order of $f$ and let $n$ be the order of $g$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there some general conditions where  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$fg= \alpha x^{n+m}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for some non-zero $\alpha \in \mathbb{C}$ &lt;/p&gt;&#xA;" OwnerUserId="832" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2011-03-04T22:15:59.647" LastActivityDate="2011-03-04T22:15:59.647" Title="When does the product of two polynomials = $x^{k}$ ?" Tags="&lt;abstract-algebra&gt;&lt;polynomials&gt;" AnswerCount="4" CommentCount="3" FavoriteCount="1" />
  <row Id="3377" PostTypeId="2" ParentId="3376" CreationDate="2010-08-26T14:52:12.307" Score="14" ViewCount="" Body="&lt;p&gt;Polynomials over $\mathbb{C}$ (in fact, over any field) are a Unique Factorization Domain (see &lt;a href=&quot;http://en.wikipedia.org/wiki/Unique_factorization_domain&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Unique_factorization_domain&lt;/a&gt;); since $x$ is an irreducible, the only way for that to happen is for $f=ax^m$ and $g=bx^n$, with $ab=\alpha$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(If you don't want to bring in the sledgehammer of unique factorization, you can just do it explicitly: look at the lowest nonzero term in $f$ and the lowest nonzero term in $g$; their product will be the lowest nonzero term in $fg$, hence must be of degree $m+n$. Since the degree of the lowest nonzero term of $f$ is at most $m$ and the one of $g$ is at most $n$, you have that they must be exactly of degree $m$ and $n$, respectively, and you get the result)&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-08-26T15:56:52.853" LastActivityDate="2010-08-26T15:56:52.853" CommentCount="1" />
  <row Id="3378" PostTypeId="2" ParentId="3376" CreationDate="2010-08-26T14:53:17.450" Score="4" ViewCount="" Body="&lt;p&gt;The answer just occured to me.   The roots of $f$ and $g$ must be be at 0. &lt;/p&gt;&#xA;" OwnerUserId="832" LastActivityDate="2010-08-26T14:53:17.450" CommentCount="0" />
  <row Id="3379" PostTypeId="2" ParentId="3335" CreationDate="2010-08-26T14:55:32.783" Score="0" ViewCount="" Body="&lt;p&gt;I am not sure if it would be considered a classic or not, but I absolutely love &lt;a href=&quot;http://books.google.com/books?id=MSbMZCWX2OQC&amp;amp;printsec=frontcover&amp;amp;dq=differential+equations+with+boundary-value+problems&amp;amp;source=bl&amp;amp;ots=-uLwOGzUzB&amp;amp;sig=s5kdyM3Sq0txRfdt9VeIc-XmiKI&amp;amp;hl=en&amp;amp;ei=I352TIvBNoK8lQe12IjsCw&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=6&amp;amp;sqi=2&amp;amp;ved=0CEMQ6AEwBQ#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;Differential Equations with Boundary-Value Problems&lt;/a&gt; by Dennis G. Zill and Michael R. Cullen.  I feel as though the definitions are written and explained in such a way, that if one was not a mathematician, they could still grasp a basic understanding of many differential equations concepts.  I am not familiar with many other elementary differential equations textbooks, but liked that this text goes into some details regarding &lt;a href=&quot;http://planetmath.org/encyclopedia/FourierSineAndCosineSeries.html&quot; rel=&quot;nofollow&quot;&gt;Fourier series&lt;/a&gt;, &lt;a href=&quot;http://planetmath.org/encyclopedia/ApproximatingFourierIntegralsWithDiscreteFourierTransforms.html&quot; rel=&quot;nofollow&quot;&gt;integrals&lt;/a&gt;, and &lt;a href=&quot;http://planetmath.org/encyclopedia/FourierTransform.html&quot; rel=&quot;nofollow&quot;&gt;transforms&lt;/a&gt;, Legendre series, Bessel functions, and the &lt;a href=&quot;http://planetmath.org/encyclopedia/EigenvalueProblem.html&quot; rel=&quot;nofollow&quot;&gt;Sturm-Louiville problem&lt;/a&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I still use this text as a reference for some of my graduate level ODE issues, as it is more comprehensible than most graduate level Diff-EQ texts. I also use it quite frequently when tutoring students in undergraduate level differential equations, as it contains a myriad of detailed examples and worked problems as well as a variety of real-world applications.&lt;/p&gt;&#xA;" OwnerUserId="219" LastActivityDate="2010-08-26T14:55:32.783" CommunityOwnedDate="2010-08-26T14:55:32.783" />
  <row Id="3380" PostTypeId="2" ParentId="3376" CreationDate="2010-08-26T14:57:11.063" Score="6" ViewCount="" Body="&lt;p&gt;Yes, the intuitively evident ones: all other terms in f and g must vanish.  To see this, note that the product of the constant terms of f and g equals the constant term of fg, which is zero, whence at least one of these polynomials is multiple of x.  Without any loss of generality assume it is f.  Then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;fg = x (f/x) g,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;implying (f/x) g is a multiple of x^(n+m-1).  By induction this reduces us to the case n+m=0, which is trivial (because f and g then have no other terms). QED.&lt;/p&gt;&#xA;" OwnerUserId="1489" LastActivityDate="2010-08-26T14:57:11.063" />
  <row Id="3381" PostTypeId="2" ParentId="3376" CreationDate="2010-08-26T15:00:54.843" Score="8" ViewCount="" Body="&lt;p&gt;It's true over any domain $\rm D$ since then $x$ is prime (via $\rm D[x]/x = D$ a domain), and products of primes always factor uniquely in domains. It fails over non-domains, e.g. $\rm x = (2x+3)(3x+2) \in \mathbb Z/6[x].$&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-08-28T23:33:11.693" LastActivityDate="2010-08-28T23:33:11.693" />
  <row Id="3382" PostTypeId="1" AcceptedAnswerId="3385" CreationDate="2010-08-26T15:06:22.447" Score="6" ViewCount="289" Body="&lt;p&gt;A couple of years ago I found the following continued fraction for $\frac{1}{e-2}$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{1}{e-2} = 1+\frac{1}{2 + \frac{2}{3 + \frac{3}{4 + \frac{4}{5 + \frac{5}{6 + \frac{6}{7 + \frac{7}{\cdots}}}}}}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;from fooling around with the well-known continued fraction for $\phi$. Can anyone here help me figure out why this equality holds? &lt;/p&gt;&#xA;" OwnerUserId="1507" LastEditorUserId="1507" LastEditorDisplayName="" LastEditDate="2010-08-26T15:23:00.980" LastActivityDate="2010-10-04T15:43:12.023" Title="Continued fraction for $\frac{1}{e-2}$" Tags="&lt;continued-fractions&gt;" AnswerCount="2" CommentCount="17" FavoriteCount="1" />
  <row Id="3383" PostTypeId="2" ParentId="3335" CreationDate="2010-08-26T15:21:54.960" Score="0" ViewCount="" Body="&lt;p&gt;Though I'm not sure if it exactly meets your criteria, my favorite PDE book is &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0470054565&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;Partial Differential Equations: An Introduction&lt;/em&gt; by Walter A. Strauss&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-08-26T15:21:54.960" CommunityOwnedDate="2010-08-26T15:21:54.960" />
  <row Id="3384" PostTypeId="2" ParentId="3315" CreationDate="2010-08-26T15:50:37.520" Score="1" ViewCount="" Body="&lt;p&gt;If you understand Argand diagrams (the representation of complex numbers in the complex plane) and can envision the unit circle in it, you can easily do this in your head:&lt;br&gt;&#xA;-1 is 1 rotated over $\pi$ radians. The square root of a number on the unit circle is the number rotated over half the angle, so $i$, or $\sqrt{-1}$ is 1 rotated over $\pi/2$ radians. To find $\sqrt{i}$ you just half the angle again: $\pi/4$ radians. The corresponding real and imaginary parts are $\cos\frac{\pi}{4}$ and $\sin\frac{\pi}{4}$ resp.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;edit&lt;/strong&gt;&lt;br&gt;&#xA;LaTexified (thanks Agosti)&lt;/p&gt;&#xA;" OwnerUserId="923" LastEditorUserId="923" LastEditorDisplayName="" LastEditDate="2010-08-28T05:38:14.267" LastActivityDate="2010-08-28T05:38:14.267" CommentCount="8" />
  <row Id="3385" PostTypeId="2" ParentId="3382" CreationDate="2010-08-26T15:52:19.177" Score="8" ViewCount="" Body="&lt;p&gt;Euler proved in &quot;&lt;em&gt;De Transformatione Serium in Fractiones Continuas&lt;/em&gt;&quot; &lt;strong&gt;Reference: The Euler Archive, Index number E593&lt;/strong&gt; (On the&#xA;Transformation of Infinite Series to Continued Fractions) [Theorem VI, §40 to §42] that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$s=\dfrac{1}{1+\dfrac{2}{2+\dfrac{3}{3+\cdots }}}=\dfrac{1}{e-1}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here his an explanation on how he proceeded. He stated that if&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\dfrac{a}{a+\dfrac{b}{b+\dfrac{c}{c+\cdots }}}=s,$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$a+\dfrac{a}{a+\dfrac{b}{b+\dfrac{c}{c+\cdots }}}=\dfrac{s}{1-s}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since, in this case, we have $a=1,b=2,c=3,\ldots $ it follows&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$1+\dfrac{1}{1+\dfrac{2}{2+\dfrac{3}{3+\cdots }}}=\dfrac{1}{e-2}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: Euler proves first how to transform an alternating series of a particular type into a continued fraction and then uses the expansion&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$e^{-1}=1-\dfrac{1}{1}+\dfrac{1}{1\cdot 2}-\dfrac{1}{1\cdot 2\cdot 3}+\ldots &#xA; .$$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;REFERENCES&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The Euler Archive, Index number E593&lt;/strong&gt;, &lt;a href=&quot;http://www.math.dartmouth.edu/~euler/&quot; rel=&quot;nofollow&quot;&gt;http://www.math.dartmouth.edu/~euler/&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Translation of Leonhard Euler's paper by Daniel W. File&lt;/strong&gt;, The Ohio State University. &lt;/p&gt;&#xA;" OwnerUserId="752" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2010-10-04T15:43:12.023" LastActivityDate="2010-10-04T15:43:12.023" CommentCount="2" />
  <row Id="3386" PostTypeId="1" AcceptedAnswerId="3394" CreationDate="2010-08-26T16:10:28.410" Score="4" ViewCount="171" Body="&lt;p&gt;Hello guys,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've got a question in my task sheet. The question is as follows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(43*93*47*97)/3007=X&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Find the exact value of X. I've tried a lot, but couldn't find easier way to do it without calculator, which of course, is not allowed in exam. There are no options, they're just asking the value of X.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would love if someone could help to give &lt;em&gt;Method&lt;/em&gt; to solve the problem. As I said, I know how to solve above problem with the help of calculators and I've already found the factorization with help of calc, but no luck in manual mode. :(&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance. :)&lt;/p&gt;&#xA;" OwnerUserId="712" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-08-26T16:25:25.350" LastActivityDate="2010-08-26T20:11:45.057" Title="Division with 4 digit number in denominator" Tags="&lt;arithmetic&gt;" AnswerCount="5" CommentCount="9" />
  <row Id="3387" PostTypeId="2" ParentId="3382" CreationDate="2010-08-26T16:23:01.547" Score="6" ViewCount="" Body="&lt;p&gt;Another possibility: remember that the numerators and denominators of successive convergents of a continued fraction can be computed using a three term recurrence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a continued fraction&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$b_0+\cfrac{a_1}{b_1+\cfrac{a_2}{b_2+\dots}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with nth convergent $\frac{C_n}{D_n}$, the recurrence&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\begin{bmatrix}C_n\\D_n\end{bmatrix}=b_n\begin{bmatrix}C_{n-1}\\D_{n-1}\end{bmatrix}+a_n\begin{bmatrix}C_{n-2}\\D_{n-2}\end{bmatrix}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with starting values&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\begin{bmatrix}C_{-1}\\D_{-1}\end{bmatrix}=\begin{bmatrix}1\\0\end{bmatrix}$, $\begin{bmatrix}C_{0}\\D_{0}\end{bmatrix}=\begin{bmatrix}b_0\\1\end{bmatrix}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;holds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With $b_j=j+1$ and $a_j=j$, you now try to find a solution for those two difference equations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I leave to you to verify that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$C_n=\frac{(n+3)!}{n+2}\sum_{j=0}^{n+3}\frac{(-1)^j}{j!}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$D_n=\frac{(n+3)!}{n+2}\left(1-2\sum_{j=0}^{n+3}\frac{(-1)^j}{j!}\right)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;are solutions to the two difference equations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Divide $C_n$ by $D_n$ and take the limit as $n\to\infty$; you should get the expected result.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-08-26T16:23:01.547" CommentCount="6" />
  <row Id="3388" PostTypeId="1" AcceptedAnswerId="3390" CreationDate="2010-08-26T16:31:27.307" Score="1" ViewCount="177" Body="&lt;p&gt;Hello guys,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a problem I tried to solve, but couldn't because I don't know the method to solve it and I've never come across such problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's the problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(PQ)²=RSP&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where P, Q, R, S are distinct single digit natural numbers, then R=?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We need to find out the value of R, and given option are&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a) 1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;b) 2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;c) 3&lt;/p&gt;&#xA;&#xA;&lt;p&gt;d) 4&lt;/p&gt;&#xA;&#xA;&lt;p&gt;e) 5&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've tried solving it by option with newly created equation as Q²=RS/P and taking values of R from above options and other variables as distinct single digit numbers other than R, I've gone to a level after which I've got myself a lot confused and didn't know what to do. I think there is easier method to solve it which I don't know.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd appreciate if someone could give me a hand with this. Thanks in advance. :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS: The answer is c) 3&lt;/p&gt;&#xA;" OwnerUserId="712" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-08-26T19:43:50.210" LastActivityDate="2011-01-12T23:37:37.453" Title="(PQ)²=RSP, where P,Q,R,S are distinct single digit natural numbers, then R=?" Tags="&lt;arithmetic&gt;&lt;cryptarithm&gt;" AnswerCount="2" CommentCount="6" />
  <row Id="3389" PostTypeId="1" AcceptedAnswerId="3391" CreationDate="2010-08-26T16:39:31.910" Score="1" ViewCount="282" Body="&lt;p&gt;Let $X = \mathbb{R}^n$ and let $Z$ be a closed subspace of $X$.  The inclusion map $i: Z \hookrightarrow X$ induces a map $C_0(X) \to C_0(Z)$ (via pullback).  My question is simple: what is the kernel?  The answer should either be $C_0(X - Z)$ or $C(X - Z)$ and I think I know which one is correct.  But I need a sanity check, so I thought I would post the question here.&lt;/p&gt;&#xA;" OwnerUserId="1509" LastEditorUserId="1509" LastEditorDisplayName="" LastEditDate="2010-11-13T15:25:48.527" LastActivityDate="2011-02-09T18:55:46.297" Title="A silly question about C_0" Tags="&lt;topology&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="3390" PostTypeId="2" ParentId="3388" CreationDate="2010-08-26T16:40:51.550" Score="6" ViewCount="" Body="&lt;p&gt;The steps are as follows.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;A square number can only end in $1, 4, 5, 6, 9$.  To prove this one uses &lt;a href=&quot;http://en.wikipedia.org/wiki/Modular_arithmetic&quot; rel=&quot;nofollow&quot;&gt;modular arithmetic&lt;/a&gt;.  So $P = 1, 4, 5, 6, 9$.&lt;/li&gt;&#xA;&lt;li&gt;$40^2 &gt; 999$, so $P = 1$.  &lt;/li&gt;&#xA;&lt;li&gt;Since $(PQ)^2$ ends in $1$, it follows that $Q = 1, 9$.&lt;/li&gt;&#xA;&lt;li&gt;Since $P \neq Q$, it follows that $Q = 9$.&lt;/li&gt;&#xA;&lt;li&gt;So the digits are $19^2 = 361$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-08-26T16:40:51.550" CommentCount="1" />
  <row Id="3391" PostTypeId="2" ParentId="3389" CreationDate="2010-08-26T16:42:18.213" Score="2" ViewCount="" Body="&lt;p&gt;If $C_0$ denotes functions vanishing at infinity, then it is $C_0(X\setminus Z)$. The other candidate is much too big to be an ideal in $C_0(X)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If, on the other hand, $C_0$ denotes compactly supported functions, then Robin Chapman's answer applies.&lt;/p&gt;&#xA;" OwnerUserId="367" LastEditorUserId="367" LastEditorDisplayName="" LastEditDate="2011-02-09T18:55:46.297" LastActivityDate="2011-02-09T18:55:46.297" CommentCount="2" />
  <row Id="3392" PostTypeId="2" ParentId="3315" CreationDate="2010-08-26T16:44:37.667" Score="9" ViewCount="" Body="&lt;p&gt;Visually, the square root of a complex number (written in polar coordinates) $(\rho,\theta)$ is the number $(\sqrt\rho,\theta/2)$.&lt;img src=&quot;http://i.imgur.com/Gxqh7.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="354" LastEditorUserId="354" LastEditorDisplayName="" LastEditDate="2010-08-27T22:00:19.900" LastActivityDate="2010-08-27T22:00:19.900" CommentCount="2" />
  <row Id="3393" PostTypeId="2" ParentId="3389" CreationDate="2010-08-26T16:46:45.587" Score="2" ViewCount="" Body="&lt;p&gt;I presume $C_0(X)$ means compactly supported continuous&#xA;functions on $X$. So you are asking which compactly supported continuous&#xA;functions on $X$ which vanish on $Z$. Now these may not be compactly supported&#xA;on $X-Z$ for there may be a point $p$ in $Z$ and a sequence of points $(p_n)$&#xA;in converging to $X$ with $f(p_n)\ne0$ but $f(p)=0$. On the other hand there&#xA;will be continuous functions on $X-Z$ which don't converge to zero on the boundary.&#xA;So neither $C(X-Z)$ nor $C_0(X-Z)$ is correct!&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-08-26T16:46:45.587" CommentCount="1" />
  <row Id="3394" PostTypeId="2" ParentId="3386" CreationDate="2010-08-26T16:53:12.227" Score="3" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\;\quad\rm mod \; 97: \:\quad 100 \;\equiv\; 3 $  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence $\; 3007 \;\equiv\; 30 \cdot 100 + 7$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\quad\quad\quad\quad\quad\quad\;\; \;\equiv\; 30 \:\;\cdot\;\: 3 \;\: + \; 7 \;$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\quad\quad\quad\quad\quad\quad\;\; \;\equiv\; 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I.e. &lt;strong&gt;cast out 97's&lt;/strong&gt; in analogy to &lt;strong&gt;cast out nines&lt;/strong&gt;. See also &lt;a href=&quot;http://google.com/groups?selm=y8zwsvvxdmy.fsf@nestle.csail.mit.edu&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; where I discuss &lt;strong&gt;casting out 91's&lt;/strong&gt;.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-08-26T17:13:50.477" LastActivityDate="2010-08-26T17:13:50.477" CommentCount="4" />
  <row Id="3395" PostTypeId="1" AcceptedAnswerId="3396" CreationDate="2010-08-26T16:57:52.420" Score="1" ViewCount="84" Body="&lt;p&gt;Hello guys,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's one more question I need help with. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Samir has some amount of rupees with him, the difference between 45.45% and 36.36% of which is Rs. 789435689. Find the amount.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a) 8683792579&lt;/p&gt;&#xA;&#xA;&lt;p&gt;b) 8673792579&lt;/p&gt;&#xA;&#xA;&lt;p&gt;c) 7573781479&lt;/p&gt;&#xA;&#xA;&lt;p&gt;d) 88667292579&lt;/p&gt;&#xA;&#xA;&lt;p&gt;e) None of these&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, Samir is the name of Indian male, rupess is Indian currency and Rs. is symbol for Indian currency. You can try replacing those with $ and question will be the same. :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's what I've done. I've subtracted 45.45%-36.36%=9.09. I've multiplied 789435689*9.09/100. Which should be answer, but it isn't and I don't know why. Am I doing something wrong?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The actual answer is a) 8683792579.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone elaborate how its a) and if not, my method is correct or not?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance and really sorry for asking too many stupid questions. :(&lt;/p&gt;&#xA;" OwnerUserId="712" LastActivityDate="2010-08-26T17:19:25.260" Title="Difference between Percentage value calculation" Tags="&lt;arithmetic&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="3396" PostTypeId="2" ParentId="3395" CreationDate="2010-08-26T17:19:25.260" Score="2" ViewCount="" Body="&lt;p&gt;789435689 is 9.09% or 9.09/100 of the whole amount&amp;mdash;that is, $789435689=\frac{9.09}{100}x$ where $x$ is the whole amount&amp;mdash;so you want to divide by 9.09/100 rather than multiplying by it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It also may be worth noting that $0.\overline{36}=\frac{4}{11}$ and $0.\overline{45}=\frac{5}{11}$, so $0.\overline{09}=\frac{1}{11}$, so you are essentially multiplying by 11, which can be done by multiplying by 10 and adding the original number or by writing two copies of the number, shifted one place, and adding:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;   789435689&#xA;+ 789435689&#xA;------------&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-08-26T17:19:25.260" CommentCount="1" />
  <row Id="3397" PostTypeId="2" ParentId="3386" CreationDate="2010-08-26T17:20:27.413" Score="2" ViewCount="" Body="&lt;p&gt;This must be a multiple choice question that must be done under a time limit... so approximate!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here we go:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{43\cdot 93\cdot 47 \cdot 97}{3007} \approx \frac{50 \cdot 100 \cdot 50 \cdot 100}{3000}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can easily reduce this to $\frac{5 \cdot 100 \cdot5 \cdot 10}{3}$ which is simply $\frac{25000}{3} = 8333 \frac{1}{3}$.  &lt;em&gt;This is not all that satisfactory&lt;/em&gt; since the answer is $6063$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's be a little finer with our approximation (while being no more painful):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{43\cdot 93\cdot 47 \cdot 97}{3007} \approx \frac{40 \cdot 100 \cdot 50 \cdot 100}{3000}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, we are now looking at $\frac{40 \cdot 100 \cdot 5}{3}$ which is simply $\frac{20000}{3} = 6666 \frac{2}{3}$.  That should be close enough, but we already have more information than you may think, because the true answer is less than both of these estimates.  &lt;/p&gt;&#xA;" OwnerUserId="121" LastEditorUserId="121" LastEditorDisplayName="" LastEditDate="2010-08-26T18:54:54.070" LastActivityDate="2010-08-26T18:54:54.070" CommentCount="2" />
  <row Id="3398" PostTypeId="2" ParentId="3386" CreationDate="2010-08-26T17:45:15.867" Score="2" ViewCount="" Body="&lt;p&gt;(20:30)&#xA;First we notice that 43, 47 and 97 are all prime numbers and 93=3x31. Clearly 3 is not a factor of 3007, let's see about the rest using the good ol' Euclidean algorithm.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;43 is clearly not (as 43x7=280+21=301 therefore 300=43x6+42, and 427 is clearly not divisible by 43)&lt;/li&gt;&#xA;&lt;li&gt;in 47 case, 300=6x(47+3) therefore the reminder is 47-18=29, and again 297 is clearly not divisible by 47&lt;/li&gt;&#xA;&lt;li&gt;31 gives us 310=31x10 so 300=9x31+21, and 217 equals to 31x7. So we have one divisor, that is 97x31 = 3007&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Therefore the result is 43x47x3 which should be a simple calculation.&#xA;(20:38)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;8 minutes, would have taken 5 if I'd began with 31 :)&#xA;I tried to describe my thoughts into words and be not very formal. I hope it's clear and it gave you some insight about how I solved it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Addendum:&lt;/strong&gt;&lt;br&gt;&#xA;Say that it was the case where no factors are common, what's then? Then we have two options either use approximations by noticing how 97x93 is (95+2)(95-2) and same with 43x47=(45+2)(45-2) so it'd be simpler to try approximating 45x45x95x95/3000&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or you can take the numbers you've calculated in with the Euclidean algorithm when you checked for common factors and you'd have 3007/43, 3007/47, etc. So you can just multiply their inverses for the result.&lt;/p&gt;&#xA;" OwnerUserId="622" LastEditorUserId="622" LastEditorDisplayName="" LastEditDate="2010-08-26T17:50:35.273" LastActivityDate="2010-08-26T17:50:35.273" CommentCount="1" />
  <row Id="3399" PostTypeId="1" CreationDate="2010-08-26T17:49:36.227" Score="10" ViewCount="1056" Body="&lt;p&gt;A function $f : \mathbb{R} \to \mathbb{R}$ is &lt;b&gt;convex&lt;/b&gt; (or &quot;concave up&quot;) provided that for all $x,y \in \mathbb{R}$ and $t \in [0,1]$,&#xA;$$f(tx + (1-t)y) \le tf(x) + (1-t)f(y).$$&#xA;Equivalently, a line segment between two points on the graph lies &lt;em&gt;above&lt;/em&gt; the graph, the region &lt;em&gt;above&lt;/em&gt; the graph is convex, etc.  I want to know why the word &quot;convex&quot; goes with the inequality in this direction, and how I can remember it.  Every reason I have heard makes just as much sense applied to the opposite inequality (&quot;concave down&quot;).&lt;/p&gt;&#xA;" OwnerUserId="822" LastEditorUserId="121" LastEditorDisplayName="" LastEditDate="2010-08-26T18:15:49.333" LastActivityDate="2010-12-13T05:43:05.053" CommunityOwnedDate="2010-08-28T03:33:15.160" Title="Why does &quot;convex function&quot; mean &quot;concave *up*&quot;?" Tags="&lt;soft-question&gt;&lt;real-analysis&gt;&lt;terminology&gt;&lt;functions&gt;" AnswerCount="8" CommentCount="3" FavoriteCount="1" />
  <row Id="3400" PostTypeId="2" ParentId="3399" CreationDate="2010-08-26T18:00:14.660" Score="6" ViewCount="" Body="&lt;p&gt;One of my professors told me the following memorable line: &quot;A concave function looks like the roof of a cave.&quot; which helps me remember what is a concave and what is a convex function.&lt;/p&gt;&#xA;" LastActivityDate="2010-08-26T18:00:14.660" CommentCount="6" />
  <row Id="3401" PostTypeId="2" ParentId="3399" CreationDate="2010-08-26T18:00:58.597" Score="7" ViewCount="" Body="&lt;p&gt;Not sure why convex is defined that way, but one way to remember is that the derivative is monotonically increasing for some convex functions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or maybe just remember that $e^x$ is conv$e^x$. (I just thought of this one!)&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-08-26T18:00:58.597" CommentCount="3" />
  <row Id="3402" PostTypeId="2" ParentId="3363" CreationDate="2010-08-26T18:15:42.820" Score="7" ViewCount="" Body="&lt;p&gt;The number of loops is just the number of pairs of non-intersecting paths s.t. first one goes from (0,1) to (m-1,n) and the second one goes from (1,0) to (m,n-1).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Non-intersecting paths on a lattice are counted by some &lt;a href=&quot;http://qchu.wordpress.com/2009/11/17/the-lindstrom-gessel-viennot-lemma/&quot; rel=&quot;nofollow&quot;&gt;determinant formula&lt;/a&gt;. In this case it's just $\det\left(\begin{matrix}\binom{m+n-2}{m-1}&amp;amp;\binom{m+n-2}{m-2}\\\binom{m+n-2}{n-2}&amp;amp;\binom{m+n-2}{n-1}\end{matrix}\right)=\binom{m+n-2}{m-1}^2-\binom{m+n-2}{m-2}\binom{m+n-2}{n-2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's not hard to prove this formula directly: a pair (path from (0,1) to (m-1,n); path from (1,0) to (m,n-1)) either forms a loop without intersection or (if paths intersect) can be (canonically) identified with a pair (path from (0,1) to (m,n-1); path from (0,1) to (m,n-1).&lt;/p&gt;&#xA;" OwnerUserId="152" LastEditorUserId="152" LastEditorDisplayName="" LastEditDate="2010-08-26T18:22:44.583" LastActivityDate="2010-08-26T18:22:44.583" CommentCount="1" />
  <row Id="3403" PostTypeId="2" ParentId="3399" CreationDate="2010-08-26T18:16:42.187" Score="2" ViewCount="" Body="&lt;p&gt;A line is said to &quot;support&quot; the graph of a function (or indeed, any subset of the Cartesian plane) if it &quot;holds up&quot; the graph: that is, the graph lies entirely above or on the line.  (After all, gravity pulls downward!)  We might think of the union of all support lines as the &quot;ground&quot; on which the graph lies; everything else--its set-theoretic complement--is the &quot;sky&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A function of the real numbers is convex if and only if its graph is the boundary between the ground and the sky.  This is a special case of the more general idea of convexity that applies to arbitrary planar regions, the same as the familiar distinction between a convex and non-convex polygon, for example.  For arbitrary regions there is no definite &quot;up&quot; and &quot;down&quot; anymore, though, so we say that a line supports a region when the region lies entirely within one of the two closed half-planes bounded by that line.  (Thus, the interior and boundary of a convex polygon form its &quot;sky&quot; and everything outside is the &quot;ground.&quot;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In short, calling a &quot;concave upward&quot; function &quot;convex&quot; unites two closely related familiar concepts and is justified by the universal earthbound human experience that gravity usually pulls downward.&lt;/p&gt;&#xA;" OwnerUserId="1489" LastActivityDate="2010-08-26T18:16:42.187" CommentCount="2" />
  <row Id="3404" PostTypeId="2" ParentId="3356" CreationDate="2010-08-26T18:21:07.040" Score="5" ViewCount="" Body="&lt;p&gt;Here is a different way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f(x+y) = f(x) f(y) - g(x) g(y)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Differentiate wrt $y$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f'(x+y)  = f(x)f'(y) - g(x) g'(y)$ , put $y = 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f'(x) = -g(x)* g'(0)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly we get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$g'(x) = f(x) * g'(0)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus $f(x)*f'(x) + g(x)*g'(x) = 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus the function $f^{2}(x) + g^{2}(x)$ is constant, as it's derivative is zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f(0) = f^2(0) - g^{2}(0)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$g(0) = 2f(0)g(0)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Squaring and adding both we get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f^{2}(0) + g^{2}(0) = (f^{2}(0) + g^{2}(0))^2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now if $f^{2}(0) + g^{2}(0) = 0$ then because $f^{2}(x) + g^{2}(x)$ is a constant, we get $f(x) = 0$ which implies $f$ is constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus $f^{2}(0) + g^{2}(0) = 1$ and hence $f^{2}(x) + g^{2}(x) = 1$&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-08-26T18:21:07.040" />
  <row Id="3405" PostTypeId="1" AcceptedAnswerId="3517" CreationDate="2010-08-26T19:10:14.380" Score="34" ViewCount="5975" Body="&lt;p&gt;Some years ago I was interested in the following Markov chain&#xA;whose state space is the positive integers. The chain begins at state &quot;1&quot;, &#xA;and from state &quot;n&quot; the chain next jumps to a state uniformly&#xA;selected from {n+1,n+2,...,2n}.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As time goes on, this chain goes to infinity, with occasional&#xA;large jumps. In any case, the chain is quite unlikely to hit any&#xA;particular large n.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you define p(n) to be the probability that this chain&#xA;visits state &quot;n&quot;, then p(n) goes to zero like c/n for some &#xA;constant c. In fact, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$    np(n) \to c = {1\over 2\log(2)-1} = 2.588699. \qquad (1)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to prove this convergence, I recast it as an analytic &#xA;problem. Using the Markov property, you can see that the sequence&#xA;satisfies: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$   p(1)=1\quad\mbox{ and }\quad p(n)=\sum_{\lceil n/2\rceil}^{n-1} {p(j)\over j}\mbox{ for }n&gt;1.  \qquad (2)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For some weeks, using generating functions etc. I tried and failed to find&#xA;an analytic proof of the convergence in (1). Finally, at a conference&#xA;in 2003 Tom Mountford showed me a (non-trivial) probabilistic proof.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the result is true, but since then I've&#xA;continued to wonder if I missed something obvious. Perhaps there is&#xA;a standard technique for showing that (2) implies (1).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; Is there a direct (short?, analytic?) proof of (1)? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps someone who understands sequences better than I do could take a shot at this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update:&lt;/strong&gt; I'm digging through my old notes on this. I now remember that I had a proof (using generating functions) that &lt;em&gt;if&lt;/em&gt; $\ np(n)$ converges, then the limit is $1\over{2\log (2)-1}$. It was the convergence that eluded me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also found some curiosities like: $\sum_{n=1}^\infty {p(n)\over n(2n+1)}={1\over 2}.$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Another update&lt;/strong&gt;: Here is the conditional result mentioned above. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As in Qiaochu's answer, define $Q$ to be the generating function of $p(n)/n$, that is,&#xA; $Q(t)=\sum_{n=1}^\infty {p(n)\over n} t^n$ for $0\leq t&amp;lt;1$.&#xA; Differentiating gives&#xA; $$Q^\prime(t)=1+{Q(t)-Q(t^2)\over 1-t}.$$&#xA; This is slightly different from Qiaochu's expression because&#xA; $p(n)\neq \sum_{j=\lceil n/2\rceil}^{n-1} {p(j)\over j}$ when $n=1$,&#xA; so that $p(1)$ has to be treated  separately.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Differentiating again and multiplying by $1-t$, we get&#xA; $$(1-t)Q^{\prime\prime}(t)=-1+2\left[Q^\prime(t)-t Q^\prime(t^2)\right],$$&#xA; that is,&#xA; $$(1-t)\sum_{j=0}^\infty (j+1) p(j+2) t^j =  -1+2\left[\sum_{j=1}^\infty (jp(j)) {t^j-t^{2j}\over j}\right].$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume that $\lim_n np(n)=c$ exists. Letting $t\to 1$ above the left hand side gives $c$,&#xA; while the right hand side is $-1+2c\log(2)$ and hence $c={1\over 2\log(2)-1}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: $\sum_{j=1}^\infty {t^j-t^{2j}\over j}=\log(1+t).$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;New update: (Sept. 2)&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's an alternative proof of the conditional result that my colleague Terry Gannon&#xA; showed me in 2003.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Start with the sum $\sum_{n=2}^{2N}\ p(n)$, substitute the formula in the title,&#xA; exchange the variables $j$ and $n$, and rearrange to establish the identity:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$${1\over 2}=\sum_{j=N+1}^{2N} {j-N\over j}\ {p(j)}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $jp(j)\to c$, then&#xA;$1/2=\lim_{N\to\infty} \sum_{j=N+1}^{2N} {j-N\over j^2}\  c=(\log(2)-1/2)\  c,$ so that&#xA;$c={1\over 2\log(2)-1}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;New update: (Sept. 8)&lt;/strong&gt; Despite the nice answers and interesting discussion below, I am still holding out for an (nice?, short?) analytic proof of convergence. Basic Tauberian theory is allowed :) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;New update: (Sept 13)&lt;/strong&gt; I have posted a sketch of the probabilistic proof of convergence under &quot;A fun and frustrating recurrence sequence&quot; in the &quot;Publications&quot; section of my homepage.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Final Update:&lt;/strong&gt; (Sept 15th) The deadline is approaching, so I have decided to award the bounty to T.. &#xA;Modulo the details(!), it seems that the probabilistic approach is the most likely to lead to a proof.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My sincere thanks to everyone who worked on the problem, including those who tried it but &#xA;didn't post anything. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a sense, I &lt;em&gt;did&lt;/em&gt; get an answer to my question: there doesn't seem to be an easy, or standard&#xA;proof to handle this particular sequence.  &lt;/p&gt;&#xA;" OwnerUserId="940" LastEditorUserId="940" LastEditorDisplayName="" LastEditDate="2010-09-15T14:49:15.493" LastActivityDate="2010-11-25T23:04:30.593" Title="Convergence of $np(n)$ where $p(n)=\sum_{j=\lceil n/2\rceil}^{n-1} {p(j)\over j}$" Tags="&lt;sequences-and-series&gt;&lt;probability&gt;&lt;stochastic-processes&gt;" AnswerCount="7" CommentCount="6" FavoriteCount="5" />
  <row Id="3406" PostTypeId="2" ParentId="3399" CreationDate="2010-08-26T19:23:36.233" Score="0" ViewCount="" Body="&lt;p&gt;I don't know if this is the original reason for the terminology, but it makes a lot of sense when you look at optimization problems. Convex functions make minimization &quot;easy&quot;, in the sense that if you have a convex function on a convex domain, it has a unique local minimum which is also the global minimum, and gradient descent works.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you were typically looking at &lt;em&gt;maximization&lt;/em&gt; problems instead, it would make sense to switch the meanings of convex and concave functions. But since many problems of practical and/or theoretical interest are naturally expressed as minimization (of potential energy, cost, area, etc.), we have &quot;convex&quot; defined the way it is.&lt;/p&gt;&#xA;" OwnerUserId="856" LastActivityDate="2010-08-26T19:23:36.233" CommentCount="1" />
  <row Id="3407" PostTypeId="2" ParentId="3362" CreationDate="2010-08-26T19:39:44.927" Score="0" ViewCount="" Body="&lt;p&gt;The answer to your final question is no for arbitrary $\theta$.  Specifically, let $f(x) = 1/\sqrt(x)$ for $\epsilon &amp;lt; x &amp;lt;= 1$ and define $f$ however you like for $0 \le x &amp;lt; \epsilon$ &amp;lt;&amp;lt; .01, say.  (This can be done in a way that maintains the concavity and twice-differentiability of $f$.)  With $\theta = 3/4$, the sequence of Riemann sums begins (with $n=1$):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2., 2.04667, 2.056, 2.0579, 2.05757, 2.05647, 2.05511, 2.05368, ...,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;peaking at $n=4$.  (The sequence is monotonic after that.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem occurs because there is a section at the end of the graph of $f$, in this case near zero, that grows quickly enough to appreciably alter the very first term of the sum once $n$ grows large enough; this section is not sampled by the sum for smaller $n$.  By modifying $f$ appropriately within disjoint intervals converging to 0, you should be able to construct an example whose sequence of Riemann sums wiggles arbitrarily many times.  What this hand-waving argument does not settle is whether this process can be carried out &lt;em&gt;ad infinitum&lt;/em&gt;: the concern is that the resulting $f$ might not be continuous at 0, which conceivably is part of your concept of &quot;well-behaved&quot; (although it is not necessary for the theory of Riemann integration, which permits some points of discontinuity).&lt;/p&gt;&#xA;" OwnerUserId="1489" LastActivityDate="2010-08-26T19:39:44.927" CommentCount="1" />
  <row Id="3408" PostTypeId="1" CreationDate="2010-08-26T19:41:40.073" Score="11" ViewCount="372" Body="&lt;p&gt;Hi,&#xA;Is there a characterization of the nonconstant entire functions $f$ that satisfy $|f(z)|=1$ for all $|z|=1$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clearly, $f(z)=z^n$ works for all $n$. Also, it's not difficult to show that if $f$ is such an entire function, then $f$ must vanish somewhere inside the unit disk. What else can be said about those functions?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you&lt;/p&gt;&#xA;" OwnerUserId="1513" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-08-27T00:24:35.693" LastActivityDate="2010-08-27T17:57:19.797" Title="Characterizing nonconstant entire functions with modulus 1 on the unit circle" Tags="&lt;complex-analysis&gt;" AnswerCount="5" CommentCount="1" FavoriteCount="4" />
  <row Id="3409" PostTypeId="2" ParentId="3386" CreationDate="2010-08-26T19:54:27.973" Score="1" ViewCount="" Body="&lt;p&gt;The numerator is about $16000000$, so the quotient is about $5300$, say plus or minus a thousand.  Now I am going to play some tricks.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The numerator is $-9 \bmod 50$ and the denominator is $7 \bmod 50$, so $X \equiv 13 \bmod 50$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The numerator is $6 \bmod 9$ and the denominator is $2 \bmod 9$, so $X \equiv 3 \bmod 9$.  Hence $X \equiv 213 \bmod 450$.  So $X$ can really only be one of $4263, 4713, 5163, 5613, 6063$, and maybe $6513$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The numerator is $8 \bmod 11$ and the denominator is $4 \bmod 11$, so $X \equiv 2 \bmod 11$.  We now know the value of $X \bmod 4950$, which is more than enough.  Looking through the above list this gives $X = 6063$.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-08-26T19:54:27.973" CommentCount="3" />
  <row Id="3410" PostTypeId="2" ParentId="3386" CreationDate="2010-08-26T20:11:45.057" Score="1" ViewCount="" Body="&lt;p&gt;You can estimate it this way very quicly:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;43*93*47*97 = (50-7).(100-7).(50-3).(100-3) = (50-7).(50-3).(100-7).(100-3)&#xA;= (100²/4 - 10.100/2 + 21) * (100² - 10.100 + 21) =&#xA;= (100.(25-5)+21) * (100.(100-10)+21) = &#xA;= 2021 * 9021 ≈ 18 000 000 (if needed for aprox: greater error around 20/2000 =&amp;gt; 1%&#xA;&#xA;using 3000 instead of 3007 (error 7/3007 around 0,2% so less than 1% previous)&#xA;&#xA;18 000 000 / 3000 = 6000 if needed, you can reduce aproximation &#xA;correcting denominator error (1%) =&amp;gt; 6000 + 60 = 6060&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="1514" LastActivityDate="2010-08-26T20:11:45.057" CommentCount="3" />
  <row Id="3411" PostTypeId="2" ParentId="3405" CreationDate="2010-08-26T20:13:31.557" Score="2" ViewCount="" Body="&lt;p&gt;Here is an approach which should give a differential equation &lt;em&gt;without&lt;/em&gt; an $f(x^2)$ term.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider $s_{n} = \frac{p_n}{n}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have $s_{1} = 1$ and $s_{2} = 1/2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can easily see from the recursive formula for $p_{n}$ that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1)  $s_{2n+2} = s_{2n+1}$ for $n \ge 1$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2)  $s_{2n+1} = s_{2n} - \frac{s_n}{2n+1}$ for $n \ge 1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using these identities and putting $Q(x) = \sum {s_{i} x^{i}}$ gives us the generating function similar to what Qiaochu got (with the $Q(x^2)$ term), but we probably cannot deal with this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason the $Q(x^2)$ term appears is that the recurrence in (2)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$s_{2n+1} = s_{2n} - \frac{s_n}{2n+1}$ has both $s_{2n}$ and $s_{n}$, in order to avoid this, we aim to eliminate $s_{n}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$s_{4m+1} = s_{4m} - \frac{s_{2m}}{4m+1}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$s_{4m-1} = s_{4m-2} - \frac{s_{2m-1}}{4m-1}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $s_{2m} = s_{2m-1}$ we can eliminate that easily.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Multiply the first by $4m+1$, the second by $4m-1$ and subtract (also use $s_{4m} = s_{4m-1}$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(4m+1)s_{4m+1} = 8ms_{4m-1} - (4m-3)s_{4m-3}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly we get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(4m+3)s_{4m+3} = (8m+4)s_{4m+1} - (4m+1)s_{4m-1}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This I believe would be susceptible to standard generating function methods. (I believe it will give a differential equation of the form $A(x)*f''(x) + B(x)*f'(x) + C(x)*f(x) + D(x) = 0$ where $A(x),B(x),C(x),D(x)$ are polynomials).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please do recheck the algebra etc if you do choose to pursue this approach.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope that helps.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Incidentally, using the identities (1) and (2) above, we can show that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$s_{2} - s_{2n+1}$ =  $\sum_{j=1}^{n} {\frac{s_j}{2j+1}} = \sum_{j=1}^{n}{\frac{p_j}{j(2j+1)}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $s_{2n+1}$ tends to 0 (easy to prove), we have that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sum_{j=1}^{\infty}{\frac{p_j}{j(2j+1)}} = s_{2} = \frac{1}{2}$ which proves the curiosity you spoke of.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-08-27T16:40:18.890" LastActivityDate="2010-08-27T16:40:18.890" CommentCount="6" />
  <row Id="3412" PostTypeId="1" AcceptedAnswerId="3415" CreationDate="2010-08-26T20:22:09.897" Score="1" ViewCount="102" Body="&lt;p&gt;I read this fascinating article:&lt;br&gt;&#xA;&lt;a href=&quot;http://www.newscientist.com/article/mg20627550.200-enter-the-matrix-the-deep-law-that-shapes-our-reality.html&quot; rel=&quot;nofollow&quot;&gt;http://www.newscientist.com/article/mg20627550.200-enter-the-matrix-the-deep-law-that-shapes-our-reality.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately all the other papers I find googling are just not tangible to me :-(&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could anyone please point me to some material that bridges the gap from this popular science exposition to the hard core papers that seem to pile up in the Net?&lt;/p&gt;&#xA;" OwnerUserId="346" LastActivityDate="2010-08-26T20:43:16.093" Title="Accessible Intro to Random Matrix Theory (RMT)" Tags="&lt;linear-algebra&gt;&lt;random&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="1" />
  <row Id="3413" PostTypeId="2" ParentId="3315" CreationDate="2010-08-26T20:25:20.577" Score="1" ViewCount="" Body="&lt;p&gt;Gives a nice concise overview:&lt;br&gt;&#xA;&lt;a href=&quot;http://www.wolframalpha.com/input/?i=Sqrt(i&quot; rel=&quot;nofollow&quot;&gt;http://www.wolframalpha.com/input/?i=Sqrt(i&lt;/a&gt;)&lt;/p&gt;&#xA;" OwnerUserId="346" LastActivityDate="2010-08-26T20:25:20.577" />
  <row Id="3414" PostTypeId="2" ParentId="3408" CreationDate="2010-08-26T20:34:38.407" Score="3" ViewCount="" Body="&lt;p&gt;&lt;b&gt;Partial answer.&lt;/b&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $|f(z)|=1$ for all $|z|=1$ and $f$ is entire function, then $f:\mathbb D \to \mathbb D$, by the Maximum Modulus Theorem. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using Schwarz-Lemma we can characterize, at least, the conformal ones.&lt;br&gt;&#xA;( See Functions of one Complex Variable, Conway 2ed pag. 131 )&lt;/p&gt;&#xA;" OwnerUserId="633" LastActivityDate="2010-08-26T20:34:38.407" />
  <row Id="3415" PostTypeId="2" ParentId="3412" CreationDate="2010-08-26T20:43:16.093" Score="2" ViewCount="" Body="&lt;p&gt;I like very much and I think that the exposition is very good and clear in the Deift's book:&#xA;Orthogonal polynomials and random matrices: a Riemann-Hilbert approach.&lt;/p&gt;&#xA;" OwnerUserId="633" LastActivityDate="2010-08-26T20:43:16.093" />
  <row Id="3416" PostTypeId="2" ParentId="3336" CreationDate="2010-08-26T20:49:06.333" Score="1" ViewCount="" Body="&lt;p&gt;Try this:&lt;br&gt;&#xA;&lt;a href=&quot;http://www.wolframalpha.com/input/?i=sin&quot; rel=&quot;nofollow&quot;&gt;http://www.wolframalpha.com/input/?i=sin&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Click on all the &quot;more&quot;s - works of course for &quot;cos&quot; too!&lt;/p&gt;&#xA;" OwnerUserId="346" LastActivityDate="2010-08-26T20:49:06.333" />
  <row Id="3417" PostTypeId="2" ParentId="3408" CreationDate="2010-08-26T20:58:04.250" Score="5" ViewCount="" Body="&lt;p&gt;Edit: How frustrating, this is still not correct. It is not hard to make this proof work by generalizing to the functions holomorphic on unit disc and continuous on its closure, but such approach will be nothing else than proving the theorem cited by Jonas Meyer (see his answer). &lt;br&gt;&#xA;Second attempt:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Note that $f$ has finitely many zeros in $\mathbb{D}$ (identity principle + compactness).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Observe that if $F\in \mathcal{O}(\mathbb{D})\cap \mathcal{C}(\bar{\mathbb{D}}),$ the only zero of $F$ is in zero, and $|F|$ is constant on unit circle, then $F(z) = e^{it}z^{k}$ where $k = ord_{0}F$. &lt;br&gt;&#xA;(apply the maximum principle for $F(z)/z^{k}$ and $z^{k}/F(z)$)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Let $z_{1},\ldots, z_{n}$ be all pairwise distinct zeros of $f$ in $\mathbb{D}$. By induction on $n$ we will show that $n = 1$ and  $z_{1} = 0$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Let $n = 1$.&lt;br&gt;&#xA;We are taking $h \in Aut(\mathbb{D})$ s.t. $h(0) = z_{1}$ and $g:= f\circ h\in \mathcal{O}(\mathbb{D})\cap \mathcal{C}(\bar{\mathbb{D}})$.&lt;br&gt;&#xA;Then $|g|$ is constant on unit circle, $0$ is the only zero of $g$ in $\mathbb{D}$. So $g(z) = e^{it}z^{k}$.&lt;br&gt;&#xA;Therefore $f(z) = e^{it}\cdot \left(\frac{z - z_{1}}{1 - \bar{z_{1}}z}\right)^{k}$ in $\mathbb{D}$. &lt;br&gt;&#xA;From identity principle for meromorphic functions this equality holds in $\mathbb{C}$, and gives contradiction unless $z_{1} = 0$ (otherwise $f$ is not entire).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $n &gt; 1$, then we're reasoning by contradiction.&lt;br&gt; &#xA;W.l.o.g. $z_{n}\neq 0$.&lt;br&gt;&#xA;We are taking $h \in Aut(\mathbb{D})$ s.t. $h(0) = z_{n}$ and $g:= f\circ h\in \mathcal{O}(\mathbb{D})\cap \mathcal{C}(\bar{\mathbb{D}})$, and then $h^{-1}(z_{1}),\ldots, h^{-1}(z_{n-1}), 0$ are the all zeros of $g$ in $\mathbb{D}$.&lt;br&gt;&#xA;Let $k = ord_{0}g$, and $G(z) = g(z)/z^{k}\in \mathcal{O}(\mathbb{D})\cap \mathcal{C}(\bar{\mathbb{D}})$.&lt;br&gt;&#xA;$G$ has $n-1$ zeros on $\mathbb{D}$, $|G(z)| = 1$ on unit circle, &lt;del&gt;thus from inductive assumption $n-1 = 1$ or $G$ is constant&lt;/del&gt; (we cannot use inductive assumption since in general $G\not\in \mathcal{O}(\mathbb{C})$!).&lt;br&gt; &#xA;If $G$ is non-constant then $n-1 = 1$ and $G(z) = e^{it}z^{l}, \ l &gt; 0$ - a contradiction with $ord_{0}g = k$. Otherwise $n - 1 = 0$ - a contradiction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally 1. and 3. proves that all entire functions with desired property are of the form $f(z) = e^{it}z^{k}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The following reasoning is not correct (see the Malik Younsi's answer):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;del&gt;This should be the comment to the Leandro's answer, although I cannot add comments yet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;W.l.o.g. $f$ is non-constant. Then by maximum modulus $f$ has zero inside $\mathbb{D}$.&#xA;Denote this zero by $a$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Composing $f$ with an automorphism $h$ of disc s.t. $h(0) = a$ we obtain that $g = f\circ h$ satisfies $g(0)= 0$ and moreover $|g(z)|\le |z|^{k}$ inside $\mathbb{D}$, where $k$ is the order of zero of $g$ in $0$ (indeed $g(z)/z^{k}$ is entire and we may use the maximum modulus principle).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By Schwarz lemma there exists $\theta\in \mathbb{R}$ satisfying $g(z) = e^{i\theta}z^{k}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From classification of automorphisms of the disc, we have $h^{-1}(z) = e^{i\omega}\frac{z-a}{1-\bar{a}z}$ for some $\omega\in \mathbb{R}$, and from this the full classification easily follows.&lt;/del&gt;&lt;/p&gt;&#xA;" OwnerUserId="1454" LastEditorUserId="1454" LastEditorDisplayName="" LastEditDate="2010-08-27T10:09:19.460" LastActivityDate="2010-08-27T10:09:19.460" CommentCount="1" />
  <row Id="3418" PostTypeId="2" ParentId="3319" CreationDate="2010-08-26T21:11:30.640" Score="1" ViewCount="" Body="&lt;p&gt;The ubiquity of $e$ stems from the ubiquity of $e^x$. And $e^x$ is ubiquitous due to the fact that it permits one to solve any higher-order LDE with constant coefficients - either by factoring its differential operator over $\mathbb C$ into linear factors, or by converting it into linear system form and using matrix exponentials. See the exposition by Arnold below, from his beautiful textbook &quot;Ordinary differential equations&quot;. As Arnold says &lt;strong&gt;&quot;the exponential ... gives the solution of all differential equations quite generally&quot;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/gK98d.jpg&quot; alt=&quot;alt text&quot;&gt; &#xA;&lt;img src=&quot;http://i.imgur.com/PV2vn.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-08-28T23:31:39.980" LastActivityDate="2010-08-28T23:31:39.980" />
  <row Id="3419" PostTypeId="1" AcceptedAnswerId="3421" CreationDate="2010-08-26T21:14:46.133" Score="20" ViewCount="825" Body="&lt;p&gt;Consider the following expression. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;16313107343153908912074032799466965289077771751767944648966669091376847859711382649033004075188224&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a 98 decimal digit number. &#xA;This can be represented as $424^{37}$ which has just 5 digits.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or consider this number:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;16907351492333570491078177094338635851326626269508214301784177607286611534146052484795771275896190661372675631981127803129649521785142469703500691286617000071805893890889531804648801423948258750240509470456335529389117581957525380043352452775597911297901564395967893517511308057315467512494189332252686430935249121855914917866181252548011072665616976069886958296149475308550144566145651839224313331840075767830022374277939322452695654072920143693336239042875755246628767670638296599817906363150743450787176422650055877626&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This 200 decimal digit number can be simply expressed as $\log_e 56$ when we discard first 6 numbers and then consider first 200 digits.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the question is, is it possible to represent any and every huge random number using very few characters as possible, theoretically.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;...Also, is there any standard way to reduce it mathematically?&lt;/p&gt;&#xA;" OwnerUserId="1279" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2011-05-10T16:37:51.530" LastActivityDate="2011-05-10T16:37:51.530" CommunityOwnedDate="2010-08-26T21:14:46.133" Title="Is it possible to represent every huge number in abbreviated form?" Tags="&lt;number-theory&gt;&lt;big-numbers&gt;" AnswerCount="4" CommentCount="3" FavoriteCount="6" />
  <row Id="3420" PostTypeId="2" ParentId="3319" CreationDate="2010-08-26T21:23:21.227" Score="0" ViewCount="" Body="&lt;p&gt;Read this very, very good and accessible book!&lt;br&gt;&#xA;&lt;a href=&quot;http://books.google.com/books?id=eIsyLD_bDKkC&amp;amp;printsec=frontcover&quot; rel=&quot;nofollow&quot;&gt;http://books.google.com/books?id=eIsyLD_bDKkC&amp;amp;printsec=frontcover&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="346" LastEditorUserId="346" LastEditorDisplayName="" LastEditDate="2010-08-27T15:26:57.090" LastActivityDate="2010-08-27T15:26:57.090" CommentCount="1" />
  <row Id="3421" PostTypeId="2" ParentId="3419" CreationDate="2010-08-26T21:46:14.367" Score="60" ViewCount="" Body="&lt;p&gt;No.  The problem is very simple: there are more huge numbers than abbreviated forms.  Suppose that I allow you to use the numerals 0-9, the English alphabet, and spaces to describe whatever numbers you want; that's still only $37$ characters. There are only $37^{140}$ expressions you can write down using these characters which is $140$ characters or shorter (the length of a tweet!), so there are only $37^{140}$ numbers that you could conceivably tweet to someone else.  In particular, since $\log_{10} 37^{140} = 219.5$, this implies that there is at least one number with $220$ digits which is impossible to tweet to someone else without using something other than numerals and English.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;All right, I'll be even more generous: I'll let you use all $127$ ASCII characters. That means there are $127^{140}$ possible tweets, some of which mathematically describe some kind of number, and since $\log_{10} 127^{140} = 294.5$, there is at least one number with $295$ digits which is impossible to tweet to someone else (again, assuming that Twitter uses ASCII).  I could give you the entire Unicode alphabet and you still won't be able to tweet, for example, the kind of prime numbers that public key cryptosystems use.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These kinds of questions are studied in computer science and information theory; you will find the word &lt;a href=&quot;http://en.wikipedia.org/wiki/Kolmogorov_complexity&quot; rel=&quot;nofollow&quot;&gt;Kolmogorov complexity&lt;/a&gt; tossed around, as well as the word &lt;a href=&quot;http://en.wikipedia.org/wiki/Entropy_(information_theory)&quot; rel=&quot;nofollow&quot;&gt;entropy&lt;/a&gt;.  &lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="92" LastEditorDisplayName="" LastEditDate="2010-10-26T10:46:37.737" LastActivityDate="2010-10-26T10:46:37.737" CommunityOwnedDate="2010-08-26T21:46:14.367" CommentCount="7" />
  <row Id="3422" PostTypeId="2" ParentId="3419" CreationDate="2010-08-26T22:21:01.880" Score="15" ViewCount="" Body="&lt;p&gt;If follows from the pigeonhole principle that any lossless compression algorithm that decreases the size of at least one number must increase the size of another - else the compression wouldn't be invertible. For deeper results consult any textbook algorithmic information theory (Kolmogorov complexity), e.g. Li; Vitanyi.  An introduction to Kolmogorov complexity and its applications.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-08-28T23:28:30.923" LastActivityDate="2010-08-28T23:28:30.923" CommunityOwnedDate="2010-08-26T22:21:01.880" />
  <row Id="3423" PostTypeId="1" AcceptedAnswerId="3427" CreationDate="2010-08-26T22:24:13.840" Score="6" ViewCount="145" Body="&lt;p&gt;From a St. Petersburg school olympiad, 11th grade.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Prove or disprove: a non constant polynomial P with non-negative integer coefficients is uniquely determined by its values P(2) and P(P(2)).&lt;/p&gt;&#xA;" OwnerUserId="1313" LastActivityDate="2010-08-26T22:35:38.083" Title="A polynomial determined by two values." Tags="&lt;algebra-precalculus&gt;" AnswerCount="2" FavoriteCount="3" />
  <row Id="3424" PostTypeId="2" ParentId="3101" CreationDate="2010-08-26T22:28:00.110" Score="3" ViewCount="" Body="&lt;p&gt;The following is an attempt at an interpolation between (among?) the answers of Grigory M., Pierre-Yves Gaillard and Bill Dubuque (the last of which I confess that I do not completely understand, but this is one possible interpretation of it).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $J$ be an open interval in $\mathbb{R}$, and let $x_0 \in J$.  Following PYG's suggestion, let me begin by stating exactly what I will prove: if $f,g: J \rightarrow \mathbb{R}$ are differentiable at $x_0$, then so is their product, and &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(fg)'(x_0) = f'(x_0)g(x_0) + f(x_0) g'(x_0)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I will prove this &quot;congruentially&quot;, as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;let $R$ be the ring of all functions $f: J \rightarrow \mathbb{R}$ which are continuous at &#xA;$x_0$, under the operations of pointwise addition and multiplication.  Inside $R$, consider the set $I$ of all functions $f$ such that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\lim_{x \rightarrow x_0} \frac{f(x)}{x-x_0} = 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I claim that $I$ is an ideal of $R$.  This is easy to prove, but I note that it makes use &#xA;of the fact that every element of $R$ is continuous at $x_0$, hence bounded near $x_0$.  Now:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) For $f \in R$, $f$ lies in $I$ iff: $f(x_0) = 0$, $f$ is differentiable at $x_0$ and $f'(x_0) = 0$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) For $f \in R$, $f$ is differentiable at $x_0$ iff there exists $A \in \mathbb{R}$ such that $f \equiv f(x_0) + A(x-x_0)$.  If so, then necessarily $A = f'(x_0)$; in particular, it is uniquely determined.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) Thus, if $f$ and $g$ are both differentiable at $x_0$, then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$fg \equiv (f(x_0) + f'(x_0)(x-x_0))(g(x_0) + g'(x_0)(x-x_0))$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\equiv f(x_0)g(x_0) + (f'(x_0)g(x_0) + f(x_0)g'(x_0))(x-x_0) + f'(x_0)g'(x_0)(x-x_0)^2$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\equiv f(x_0)g(x_0) + (f'(x_0)g(x_0) + f(x_0)g'(x_0))(x-x_0) \pmod I$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using 2), it follows that $fg$ is differentiable at $x_0$ and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(fg)'(x_0) = f'(x_0)g(x_0) + f(x_0)g'(x_0)$.  &lt;/p&gt;&#xA;" OwnerUserId="299" LastActivityDate="2010-08-26T22:28:00.110" />
  <row Id="3425" PostTypeId="1" CreationDate="2010-08-26T22:28:48.893" Score="2" ViewCount="76" Body="&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Lanczos_resampling&quot; rel=&quot;nofollow&quot;&gt;Lanczos 1 function&lt;/a&gt; is defined as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\operatorname{Lanczos1}(x) = \begin{cases} \frac{(\sin(\pi x))^2}{\pi^2 x^2 } &amp;amp; -1 &amp;lt; x &amp;lt; 1, x \ne 0 \\&#xA; 1 &amp;amp; x = 0 \\ &#xA;0 &amp;amp; \text{otherwise} \end{cases}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If this function is to be implemented in a FIR filter with 4 taps, how do you calculate the filter coefficients, assuming that there are 16 phases?&#xA;Since the function is defined over the range -1 to +1, for 4 taps the coefficients will have to be shifted and spread to cover 0 to 3.&lt;/p&gt;&#xA;" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-08-27T03:47:41.540" LastActivityDate="2010-08-27T03:47:41.540" Title="Lanczos 1 function for FIR filter" Tags="&lt;signal-processing&gt;" CommentCount="2" />
  <row Id="3426" PostTypeId="2" ParentId="3423" CreationDate="2010-08-26T22:32:52.670" Score="7" ViewCount="" Body="&lt;p&gt;Look at $P(P(2))$ in base $P(2)$.  The nth place is the coefficient of $x^n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Steve&lt;/p&gt;&#xA;" OwnerUserId="641" LastActivityDate="2010-08-26T22:32:52.670" CommentCount="2" />
  <row Id="3427" PostTypeId="2" ParentId="3423" CreationDate="2010-08-26T22:33:14.270" Score="7" ViewCount="" Body="&lt;p&gt;True. If $P(x)=a_nx^n+\cdots+a_1x+a_0$ then each of the coefficients are less then $b\equiv P(2)$. Each of these coefficients can then be read off from the base-b expansion of $P(b)=P(P(2))$.&lt;/p&gt;&#xA;" OwnerUserId="1321" LastActivityDate="2010-08-26T22:33:14.270" CommentCount="5" />
  <row Id="3429" PostTypeId="1" CreationDate="2010-08-26T22:54:32.400" Score="1" ViewCount="135" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Lanczos_resampling&quot; rel=&quot;nofollow&quot;&gt;Lanczos-3 and Lanczos-4 functions&lt;/a&gt; are defined as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\operatorname{Lanczos3}(x) = \begin{cases} \frac{\sin(\pi x)\sin(\pi*x/3)}{(\pi*x)*(\pi*x/3)} &amp;amp; -3 &amp;lt; x &amp;lt; 3, x \ne 0 \\&#xA; 1 &amp;amp; x = 0 \\ &#xA;0 &amp;amp; \text{otherwise} \end{cases}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\operatorname{Lanczos4}(x) = \begin{cases} \frac{\sin(\pi x)\sin(\pi*x/4)}{(\pi*x)*(\pi*x/4)} &amp;amp; -4 &amp;lt; x &amp;lt; 4, x \ne 0 \\&#xA; 1 &amp;amp; x = 0 \\ &#xA;0 &amp;amp; \text{otherwise} \end{cases}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If a 4-tap FIR filter with 16 phases uses these functions, how do I calculate the coefficients for each of these functions for the filter?&lt;/p&gt;&#xA;" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-08-27T03:50:51.193" LastActivityDate="2010-08-27T03:50:51.193" Title="Lanczos 3 and Lanczos 4 functions" Tags="&lt;signal-processing&gt;" CommentCount="2" />
  <row Id="3430" PostTypeId="2" ParentId="3408" CreationDate="2010-08-26T22:59:40.940" Score="1" ViewCount="" Body="&lt;p&gt;This is meant as a comment to ifk's answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't see how you conclude from Schwarz's lemma that $g(z)=e^{iθ}z^k$, where k is the order of zero of g in 0. Moreover, the function $f$ you obtain is not entire : &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f=g \circ h^{-1}$, but $h^{-1}$ is not entire : it has a pole outside $\mathbb{D}$...&lt;/p&gt;&#xA;" OwnerUserId="1197" LastActivityDate="2010-08-26T22:59:40.940" CommentCount="1" />
  <row Id="3431" PostTypeId="2" ParentId="3352" CreationDate="2010-08-26T23:19:19.023" Score="1" ViewCount="" Body="&lt;p&gt;I can't fully characterize such sets, but can say one thing - the conjecture is false.&#xA;For a counterexample on one dimension, choose a closed and nowhere dense set X (such as the &lt;a href=&quot;http://en.wikipedia.org/wiki/Cantor_set&quot; rel=&quot;nofollow&quot;&gt;Cantor middle thirds set&lt;/a&gt;, which is not discrete) and let $g(x)=\min\{\vert x-y\vert\colon y\in X\}$ be the distance of a point x from X. Integrate it twice, $f(x)=\int_0^x\int_0^yg(z)\,dz\,dy$ to get a strictly convex function whose second derivative $f^{\prime\prime}=g$ vanishes precisely on X.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, you could take X to be the &lt;a href=&quot;http://en.wikipedia.org/wiki/Smith%E2%80%93Volterra%E2%80%93Cantor_set&quot; rel=&quot;nofollow&quot;&gt;Smith-Volterra-Cantor set&lt;/a&gt;, which has positive Lebesgue measure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This can be used to define a strictly convex function on $\mathbb{R}^2$, $\tilde f(x,y)=f(x)+f(y)$, whose Hessian vanishes completely on a set $X\times X$ of positive Lebesgue measure.&lt;/p&gt;&#xA;" OwnerUserId="1321" LastEditorUserId="1321" LastEditorDisplayName="" LastEditDate="2010-08-26T23:30:15.760" LastActivityDate="2010-08-26T23:30:15.760" CommentCount="1" />
  <row Id="3432" PostTypeId="2" ParentId="3405" CreationDate="2010-08-26T23:44:00.807" Score="5" ViewCount="" Body="&lt;p&gt;Here is the standard generating function gruntwork.  Let $Q(x) = \sum_{n \ge 1} \frac{p(n)}{n} x^n$.  Then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$Q'(x) = \sum_{n \ge 1} p(n) x^{n-1} = \sum_{n \ge 1}^{\infty} x^{n-1} \sum_{j = \lceil \frac{n}{2} \rceil}^{n-1} \frac{p(j)}{j}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Exchanging the order of summation gives&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$Q'(x) = \sum_{j=1}^{\infty} \frac{p(j)}{j} \sum_{n=j+1}^{2j} x^{n-1} = \sum_{j=1}^{\infty} \frac{p(j)}{j} \frac{x^{2j} - x^j}{x - 1}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So it follows that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$Q'(x) = \frac{Q(x^2) - Q(x)}{x - 1}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with initial conditions $Q(0) = 0, Q'(0) = 1$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, as I recall there are theorems in Flajolet and Sedgewick's &lt;a href=&quot;http://algo.inria.fr/flajolet/Publications/books.html&quot; rel=&quot;nofollow&quot;&gt;Analytic Combinatorics&lt;/a&gt; that allow us to deduce asymptotic behavior about the coefficients of $Q$ from these kinds of relations; I'm going to hunt one down now, and in the meantime others can see what they can do with this.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-08-26T23:44:00.807" CommentCount="2" />
  <row Id="3433" PostTypeId="1" AcceptedAnswerId="3435" CreationDate="2010-08-26T23:48:15.580" Score="1" ViewCount="193" Body="&lt;p&gt;I had asked this question sometime ago here: &lt;a href=&quot;http://math.stackexchange.com/questions/2536/finding-f-such-that-int-f-sum-f&quot;&gt;http://math.stackexchange.com/questions/2536/finding-f-such-that-int-f-sum-f&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now i have a question which i think is more or less related to it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $f$ be an increasing function (continuous of course!) with $f(1)=0$. Consider the sequence $s_{n}= ( \sum\limits_{k=1}^{n} f(k) - \int\limits_{1}^{n} f(x) dx )$. When does $s_{n}$ converge?&lt;/p&gt;&#xA;" LastEditorUserId="171" LastEditorDisplayName="" LastEditDate="2010-08-27T13:11:58.400" LastActivityDate="2010-12-02T15:19:00.770" Title="Convergence of $\sum_{k=1}^{n} f(k) - \int_{1}^{n} f(x) dx$" Tags="&lt;real-analysis&gt;&lt;sequences-and-series&gt;&lt;analysis&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="2" />
  <row Id="3434" PostTypeId="1" AcceptedAnswerId="3564" CreationDate="2010-08-27T00:56:48.797" Score="1" ViewCount="144" Body="&lt;p&gt;Suppose we have a tetrahedron defined by points (0,0,0),(1,1,0),(0,1,1),(1,0,1). Now define surface by (a,b,a + b - 2*a*b) for a,b between 0 and 1. Let E1 be the set of points inside the tetrahedron and E2 be the set of points on that surface, A the area of that surface and d be Euclidian distance&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it true that for all surfaces E3 with area not greater than A&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\max_{p \in E1} (\min_{q \in E2} (d(p,q))) \le \max_{p \in E1} (\min_{q \in E3} (d(p,q)))$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://yaroslavvb.com/upload/minimal-surface.png&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&#xA;g1 = Graphics3D[{Blue, Opacity[.1], &#xA;    GraphicsComplex[{{0, 0, 0}, {1, 1, 0}, {0, 1, 1}, {1, 0, 1}}, &#xA;     Polygon[{{2, 3, 4}, {3, 2, 1}, {4, 1, 2}, {1, 4, 3}}]]}];&#xA;g2 = Plot3D[a + b - 2*a*b, {a, 0, 1}, {b, 0, 1}, &#xA;   PlotStyle -&gt; Opacity[.5]];&#xA;Show[g1, g2]&#xA;&lt;/pre&gt;&#xA;" OwnerUserId="998" LastEditorUserId="998" LastEditorDisplayName="" LastEditDate="2010-08-28T17:47:58.447" LastActivityDate="2010-08-28T18:42:23.670" Title="Does this surface minimize maximum distance?" Tags="&lt;geometry&gt;&lt;minimal-surfaces&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="3435" PostTypeId="2" ParentId="3433" CreationDate="2010-08-27T01:50:13.057" Score="4" ViewCount="" Body="&lt;p&gt;Qiaochu was on the right track to use an integral-to-sum formula, but it sounds like you want the Abel-Plana summation formula:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\lim_{n\to\infty}\left(\sum_{k=m}^n f(k)-\int_m^n f(u)\mathrm{d}u\right)=\frac{f(m)}{2}-\int_{-\infty}^\infty \left(\frac{|t|}{\exp(|2\pi t|)-1}\right)\left(\frac{f(m+it)-f(m-it)}{2it}\right)\mathrm{d}t$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is used for instance to evaluate the Stieltjes constants. If the expression on the right hand side is convergent, then it is equivalent to the left hand side.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Adendum for Chandru:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Definitely $f(z)$ should be analytic, or at least analytic in the region where $\Re z\geq m$. Per Henrici's &quot;Applied and Computational Complex Analysis&quot;, the additional conditions are&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\lim_{t\to\infty}f(u\pm it)=0$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;uniformly in $u$, and that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\lim_{t\to\infty}|f(u\pm it)|\exp(\mp 2\pi t)=0$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;uniformly in $u$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For those scratching their head on just how Abel-Plana and Euler-Maclaurin are connected, the identity&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_{-\infty}^\infty \left(\frac{|t|}{\exp(|2\pi t|)-1}\right)|t|^{2n-2}\mathrm{d}t=\frac{|B_{2n}|}{2n}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;might be of interest.&lt;/p&gt;&#xA;" OwnerUserId="498" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-02T15:19:00.770" LastActivityDate="2010-12-02T15:19:00.770" CommentCount="4" />
  <row Id="3436" PostTypeId="2" ParentId="3408" CreationDate="2010-08-27T01:51:54.877" Score="3" ViewCount="" Body="&lt;p&gt;If a function $f$ is holomorphic in a neighborhood of the closed disk and has modulus 1 on the circle, then $f$ is a &lt;a href=&quot;http://en.wikipedia.org/wiki/Blaschke_product#Finite_Blaschke_products&quot; rel=&quot;nofollow&quot;&gt;finite Blaschke product&lt;/a&gt;.  You can prove this by taking all of the zeros inside the disk counted according to multiplicity, dividing by corresponding &lt;a href=&quot;http://planetmath.org/encyclopedia/AutomorphismsOfUnitDisk.html&quot; rel=&quot;nofollow&quot;&gt;holomorphic automorphisms of the disk&lt;/a&gt; that have those zeros, and showing that the result is constant.  (This quotient and its reciprocal are analytic and bounded by 1 on the disk...)  Thus, your $f$ is a finite Blaschke product, but it also has no poles in the plane.  The only such functions are constant multiples of power functions, as ifk already said.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2010-08-27T17:57:19.797" LastActivityDate="2010-08-27T17:57:19.797" CommentCount="2" />
  <row Id="3437" PostTypeId="1" CreationDate="2010-08-27T02:03:18.623" Score="3" ViewCount="123" Body="&lt;p&gt;I’m interested in some sequences which have no square factor.&#xA;$$&#xA;s_i =\begin{cases}&#xA;4 &amp;amp; i=0; \\&#xA;s_{i-1}^2 - 2 &amp;amp; \text{otherwise}&#xA;\end{cases}$$&#xA;This is Lucas–Lehmer primality test sequence.  &lt;a href=&quot;http://www.research.att.com/~njas/sequences/A003010&quot; rel=&quot;nofollow&quot;&gt;A003010 in OEIS&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;when ${s_i}$ has a square factor? I had checked that the sequence has no square factor for i=1,2,...,7&lt;/p&gt;&#xA;" OwnerUserId="1498" LastEditorUserId="171" LastEditorDisplayName="" LastEditDate="2010-08-27T06:18:48.657" LastActivityDate="2010-09-03T15:52:48.587" Title="Search square factors in Lucas–Lehmer sequence" Tags="&lt;number-theory&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="3438" PostTypeId="1" AcceptedAnswerId="4155" CreationDate="2010-08-27T02:16:01.887" Score="1" ViewCount="283" Body="&lt;p&gt;Prove that $\{n_k\}$ is a strictly increasing sequence of positive integers then the sum of the series $$\sum\limits_{k=1}^{\infty} \frac{2^{n_k}}{(n_{k})!}$$ is an irrational number. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is just a problem, given by a friend of mine, when we were doing some problems in Analysis. And if i remember correctly, he mentioned a theorem of &quot;Legendre&quot;, to be used for proving this result.  But this looks quite interesting. &lt;/p&gt;&#xA;" LastActivityDate="2010-09-06T18:27:17.693" Title="Nature of the series: $ \sum\limits_{k=1}^{\infty} \frac{2^{n_k}}{(n_{k})!}$" Tags="&lt;real-analysis&gt;&lt;analysis&gt;&lt;sequences-and-series&gt;" AnswerCount="1" CommentCount="7" FavoriteCount="0" />
  <row Id="3439" PostTypeId="1" CreationDate="2010-08-27T02:32:11.643" Score="0" ViewCount="75" Body="&lt;p&gt;&lt;strong&gt;Definition&lt;/strong&gt; A subgroup of a group is termed strongly potentially characteristic if there is an embedding of the bigger group in some group such that, in that embedding both the group and the subgroup become characteristic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One can refer this link: &lt;a href=&quot;http://groupprops.wiki-site.com/index.php/Strongly_potentially_characteristic_subgroup&quot; rel=&quot;nofollow&quot;&gt;http://groupprops.wiki-site.com/index.php/Strongly_potentially_characteristic_subgroup&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Although, i have somewhat understood the definition, i would like to have some examples of such subgroups. Moreover, i would also like to know, as to whats the essence of defining such subgroups in such a manner. Where are they applicable?&lt;/p&gt;&#xA;" LastActivityDate="2010-08-27T02:32:11.643" Title="Strongly Potentially Characteristic Subgroups." Tags="&lt;abstract-algebra&gt;&lt;group-theory&gt;" CommentCount="4" />
  <row Id="3440" PostTypeId="2" ParentId="3433" CreationDate="2010-08-27T03:09:21.327" Score="3" ViewCount="" Body="&lt;p&gt;The question implicitly asks for a &quot;simpler&quot; or &quot;more interesting&quot; criterion for convergence.  I doubt there is one.  Intuitively, f can do almost anything between (n, f(n)) and (n+1, f(n+1)) provided it is increasing.  Thus the terms of (s(n)) can be anything.  Convergence therefore is determined by fairly arbitrary properties of f(n) as n becomes arbitrarily large.  If you don't severely restrict f--e.g., require it to be analytic or bounded and concave or something like that--you shouldn't expect to find any simpler answer than &quot;(s(n)) converges when it converges.&quot;&lt;/p&gt;&#xA;" OwnerUserId="1489" LastActivityDate="2010-08-27T03:09:21.327" />
  <row Id="3441" PostTypeId="1" CreationDate="2010-08-27T04:02:58.813" Score="1" ViewCount="507" Body="&lt;p&gt;If I create a 10 digit password with the following requirements:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;At least one uppercase letter A-Z - 26&lt;/li&gt;&#xA;&lt;li&gt;At least one lowercase letter a-z - 26&lt;/li&gt;&#xA;&lt;li&gt;At least one digits 0-9 - 10&lt;/li&gt;&#xA;&lt;li&gt;At least one common symbol (#,$,%,etc) - 32&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;By inclusion-exclusion, I can calculate I have ~ 3.2333E+19 possible combinations&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if I change one of the requirements to at least TWO digits 0-9, how can I calculate the possible combinations?&lt;/p&gt;&#xA;" OwnerUserId="1524" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-08-27T04:10:02.783" LastActivityDate="2010-08-27T15:53:49.487" Title="How can I calculate the number of potential combinations in a password?" Tags="&lt;combinatorics&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="3442" PostTypeId="1" CreationDate="2010-08-27T04:19:45.673" Score="1" ViewCount="160" Body="&lt;p&gt;In particular, problems like $\lfloor\frac{10^{20000}}{3+10^{100}}\rfloor$. (From a putnam exam many years ago.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a general technique for attacking such problems?&lt;/p&gt;&#xA;" OwnerUserId="632" LastEditorUserId="632" LastEditorDisplayName="" LastEditDate="2010-08-27T04:56:19.760" LastActivityDate="2010-08-27T04:56:19.760" Title="Finding the units' digit of large quotients" Tags="&lt;number-theory&gt;&lt;soft-question&gt;" AnswerCount="1" CommentCount="0" />
  <row Id="3443" PostTypeId="2" ParentId="3442" CreationDate="2010-08-27T04:42:49.730" Score="1" ViewCount="" Body="&lt;p&gt;In the contest-style problems I have seen which ask you to do something like this, the trick is generally that the number in question is somehow very close to or otherwise closely related to a special integer, and you should compute this special integer's digits (using, probably, modular arithmetic) and then figure out if the number you're interested in is larger or smaller, and by how much.  For example, $\phi^n$ turns out to be very close to the Lucas number $L_n$; this is because $\phi^n + \varphi^n$ (the other root of $x^2 = x + 1$) is exactly equal to $L_n$, and $|\varphi| &amp;lt; 1$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem I have in mind, which I can't quite remember the details of, is a question which asks you to compute the first six digits after the decimal point of some seemingly complicated expression; it turns out to be an integer minus a tiny amount, so the answer is $999999$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this particular case, you don't seem to be asking about a solution to the problem itself, so I will suggest for now that you think about the identity $(10^{100} + 3)(10^{100} - 3) = 10^{200} - 9$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The best advice I can give you in general is that you should get a good sense for asymptotics.&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-08-27T04:47:56.487" LastActivityDate="2010-08-27T04:47:56.487" CommentCount="2" />
  <row Id="3444" PostTypeId="1" AcceptedAnswerId="3498" CreationDate="2010-08-27T05:01:25.447" Score="23" ViewCount="1072" Body="&lt;p&gt;In &lt;a href=&quot;http://terrytao.wordpress.com/2010/01/02/254a-notes-0a-stirlings-formula/&quot; rel=&quot;nofollow&quot;&gt;these notes by Terence Tao&lt;/a&gt; is a proof of Stirling's formula.  I really like most of it, but at a crucial step he uses the integral identity&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$n! = \int_{0}^{\infty} t^n e^{-t}  dt$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;coming from the Gamma function.  I have a mathematical confession to make: I have never &quot;grokked&quot; this identity.  Why should I expect the integral on the right to give me the number of elements in the symmetric group on $n$ letters?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(It's not that I don't know how to prove it.  It's quite fun to prove; my favorite proof observes that it is equivalent to the integral identity $\int_{0}^{\infty} e^{(x-1)t}  dt = \frac{1}{1 - x}$.  But if someone were to ask me, &quot;yes, but why, &lt;em&gt;really&lt;/em&gt;?&quot; I would have no idea what to say.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what are more intuitive ways of thinking about this identity?  Is there a probabilistic interpretation?  What kind of random variable has probability density function $\frac{t^n}{n!} e^{-t}$?  (What does this all have to do with Tate's thesis?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a rough measure of what I'm looking for, your answer should make it obvious that $t^n e^{-t}$ attains its maximum at $t = n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit:  The kind of explanation I'm looking for, as I described in the comments, is similar to &lt;a href=&quot;http://math.stackexchange.com/questions/3528/beta-function-derivation/3540#3540&quot;&gt;this explanation&lt;/a&gt; of the beta integral.&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-08-28T14:13:15.533" LastActivityDate="2011-02-03T13:41:30.760" Title="Intuition for the definition of the Gamma function?" Tags="&lt;probability-theory&gt;&lt;intuition&gt;&lt;gamma-function&gt;" AnswerCount="4" CommentCount="9" FavoriteCount="21" />
  <row Id="3445" PostTypeId="2" ParentId="2304" CreationDate="2010-08-27T06:14:08.463" Score="2" ViewCount="" Body="&lt;p&gt;To solve this you can use the function theoretic description of the invariant subspaces of the shift known as Beurling's theorem.  Identify your Hilbert space with the Hardy space $H^2$ on the disk such that $U$ is identified with &quot;multiplication by $z$&quot;.  Beurling's theorem says that each invariant subspace of $U$ has the form $\phi H^2$ for a so-called inner function $\phi$, i.e., a bounded analytic function on the disk whose radial (or non-tangential) limit function has modulus 1 a.e. on the circle. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your problem, there are inner functions $\phi_1$ and $\phi_2$ such that $M_1=\phi_1 H^2$ and $M_2=\phi_2 H^2$.  You have $\phi_1\in M_1\subset M_2$, so $\phi_1=\phi_2 f$ for some $f\in H^2$.  The modulus of $f$ on the circle is 1 a.e. because $f=\phi_1/\phi_2$ a.e., and thus $f$ is an inner function.  &lt;em&gt;Suppose&lt;/em&gt;, for the sake of argument, that you can write $f=gh$ for some nonconstant inner functions $g$ and $h$, and let $M=\phi_2 g H^2$.  Then $M_1\subset  M\subset M_2$.  I claim that the inclusions are strict, and more specifically that $\phi_2 g$ is in $M\setminus M_1$ and $\phi_2$ is in $M_2\setminus M$.  This amounts to the same thing as saying that $1/h$ and $1/g$ (respectively) are not in $H^2$.  Note that $1/h$ and $1/g$ have modulus 1 a.e. on the circle, so if they were in $H^2$ they would in fact be in $H^\infty$ and bounded by 1 on the disk.  But $h$ and $g$ are bounded by 1 on the disk and nonconstant, so this is impossible, showing that in fact $1/h$ and $1/g$ are not in $H^2$ as claimed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It remains to be seen why $f$ has such a factorization, and this is where the hypothesis about codimension is used.  Every inner function has a factorization into a Blaschke product times a singular inner function.  If the singular part of $f$ is nontrivial, it can be factored nontrivially by scaling the corresponding singular measure by numbers between 0 and 1.  If the Blaschke part of $f$ has more than one factor, then it factors. Given that $M_1\neq M_2$, $f$ is not constant, so the only other possibility is that $f$ is a Blaschke product with a single factor, a.k.a. a holomorphic automorphism of the disk.  I claim that this would imply that $M_1$ has codimension 1 in $M_2$, and more specifically $M_2=M_1 + \mathbb{C}\phi_2$.  To see this, let $\alpha$ be the zero of $f$, and let $G=\phi_2 H$ be an element of $M_2$.  Then $G=\phi_2 f\frac{H-H(\alpha)}{f}+\phi_2 H(\alpha)\in M_1+\mathbb{C}\phi_2$.  Q.E.D.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For more details on factorizations of inner functions and more, see Chapter 2 of the book named in the question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(As for the edit, you can't always take $M=UM_2$.  Let $0&amp;lt;|\alpha|&amp;lt;1$, and let $\phi_\alpha$ be the holomorphic automorphism of the disk that swaps 0 and $\alpha$, $\phi_\alpha(z)=\frac{\alpha-z}{1-\overline{\alpha}z}$.  Suppose that $M_2=H^2$ and $M_1=\phi_\alpha^2 H^2$.  Then your hypotheses are met, but $M_1$ is not contained in $UM_2$ because $\phi_\alpha^2$ is not in $UM_2$.)&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2010-08-27T08:03:53.797" LastActivityDate="2010-08-27T08:03:53.797" />
  <row Id="3446" PostTypeId="2" ParentId="3408" CreationDate="2010-08-27T06:20:47.470" Score="7" ViewCount="" Body="&lt;p&gt;The only such functions are $f(z)=az^n$ with $|a|=1$.&#xA;Let $f$ be such a function and define&#xA;$$g(z)=\overline{f(\overline{z}^{-1})}^{-1}.$$&#xA;Due to the two complex conjugations, $g$ is meromorphic&#xA;on $\mathbb{C}-\{0\}$. Also $g(z)=f(z)$ when $|z|=1$.&#xA;By analytic continuation, $f(z)=g(z)$ for all nonzero $z$.&#xA;Now $g$ has a pole of order $k$ at $\infty$ where $k$&#xA;is the order of $f$ at zero. Hence $f$ is a polynomial of degree $k$.&#xA;As $k$ is also the order at $0$ then $f(z)=az^k$; clearly&#xA;$|a|=1$.&lt;/p&gt;&#xA;" OwnerUserId="226" LastEditorUserId="226" LastEditorDisplayName="" LastEditDate="2010-08-27T16:55:22.003" LastActivityDate="2010-08-27T16:55:22.003" CommentCount="9" />
  <row Id="3447" PostTypeId="1" AcceptedAnswerId="3466" CreationDate="2010-08-27T06:33:29.233" Score="2" ViewCount="399" Body="&lt;p&gt;Motivated by &lt;a href=&quot;http://math.stackexchange.com/questions/3250/&quot;&gt;this&lt;/a&gt; question I would like to know whether there is an exact definition of modern mathematics. In which point in time  (century, decade) does one think, when speaking about modern mathematics. Does it refer to the Abstract Algebra?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; I got -1. If this is not a correct question, please state why. And it can be closed if the question is not proper.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;This is a genuine doubt and ignorance of mine!&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit 2:&lt;/strong&gt; In my humble opinion instead of closing, should be better to tag it as a community wiki. Anyway I &lt;strike&gt;do not&lt;/strike&gt; have &lt;strike&gt;any&lt;/strike&gt; no objections &lt;strike&gt;against&lt;/strike&gt; in closing it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit 3:&lt;/strong&gt; Now is a community wiki.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit 4.&lt;/strong&gt; I have learned from the answers and comments, including the explanation for closing!&lt;/p&gt;&#xA;" OwnerUserId="752" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2010-08-29T10:17:20.637" LastActivityDate="2010-08-29T10:17:20.637" CommunityOwnedDate="2010-08-27T12:42:06.947" ClosedDate="2010-08-27T15:41:16.057" Title="What is Modern Mathematics? Is this an exact concept with a clear meaning?" Tags="&lt;linear-algebra&gt;&lt;abstract-algebra&gt;&lt;terminology&gt;&lt;math-history&gt;" AnswerCount="3" CommentCount="12" FavoriteCount="1" />
  <row Id="3448" PostTypeId="1" AcceptedAnswerId="3449" CreationDate="2010-08-27T06:34:02.243" Score="4" ViewCount="267" Body="&lt;p&gt;I'm looking for clear mathematical rules on rounding a number to n decimal places.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Everything seems perfectly clear for positive numbers. Here is for example what I found on &lt;a href=&quot;http://math.about.com/od/arithmetic/a/Rounding.htm&quot; rel=&quot;nofollow&quot;&gt;math.about.com&lt;/a&gt; :&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Rule One&lt;/strong&gt; Determine what your rounding digit is and look to the right side of it. If that digit is 4, 3, 2, or 1, simply drop all digits to the right of it.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;Rule Two&lt;/strong&gt; Determine what your rounding digit is and look to the right side of it. If that digit is 5, 6, 7, 8, or 9 add one to the rounding digit and drop all digits to the right of it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But what about negative numbers ? Do I apply the same rules as above ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, what is the correct result when rounding -1.24 to 1 decimal place ? -1.3 or -1.2 ?&lt;/p&gt;&#xA;" OwnerUserId="1525" LastEditorUserId="1525" LastEditorDisplayName="" LastEditDate="2010-08-27T06:41:59.453" LastActivityDate="2010-08-27T07:09:16.280" Title="rules for rounding (positive and negative numbers)" Tags="&lt;real-number-computation&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="3449" PostTypeId="2" ParentId="3448" CreationDate="2010-08-27T06:51:57.240" Score="3" ViewCount="" Body="&lt;p&gt;$-1.24\approx -1.2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$-1.21,-1.22,-1.23$ and $-1.24$ all round to $-1.2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$-1.25,-1.26,-1.27,1.28$ and $-1.29$ all round to $-1.3$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the correct result in numerical methods.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rules are the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You take the absolute value of the number, round it, and then put the $+$ or $-$ sign.&lt;/p&gt;&#xA;" OwnerUserId="752" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2010-08-27T07:09:16.280" LastActivityDate="2010-08-27T07:09:16.280" CommentCount="2" />
  <row Id="3450" PostTypeId="1" AcceptedAnswerId="3451" CreationDate="2010-08-27T09:43:38.900" Score="5" ViewCount="389" Body="&lt;p&gt;Factorial is defined as &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$n! = n(n-1)(n-2)...1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But why mathematicians named this thing as &lt;strong&gt;FACTORIAL&lt;/strong&gt;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Has it got something to do with factors?&lt;/p&gt;&#xA;" OwnerUserId="705" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2011-05-10T15:37:32.490" LastActivityDate="2011-05-10T15:37:32.490" Title="Why the name 'FACTORIAL'?" Tags="&lt;combinatorics&gt;&lt;functions&gt;&lt;terminology&gt;&lt;math-history&gt;" AnswerCount="2" CommentCount="5" />
  <row Id="3451" PostTypeId="2" ParentId="3450" CreationDate="2010-08-27T10:08:58.140" Score="7" ViewCount="" Body="&lt;p&gt;Well, all positive integers smaller or equal to n are factors of n!&lt;/p&gt;&#xA;" OwnerUserId="1010" LastActivityDate="2010-08-27T10:08:58.140" CommentCount="1" />
  <row Id="3452" PostTypeId="1" AcceptedAnswerId="3458" CreationDate="2010-08-27T10:09:14.437" Score="3" ViewCount="476" Body="&lt;p&gt;Which one is bigger $2^{n!}$ or $(2^{n})!$? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;where n∈N.&lt;/p&gt;&#xA;" OwnerUserId="705" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-08-27T10:30:41.233" LastActivityDate="2010-08-27T20:53:34.373" Title="Which one is bigger $2^{n!}$ or $(2^{n})!$" Tags="&lt;asymptotics&gt;" AnswerCount="6" CommentCount="4" FavoriteCount="2" />
  <row Id="3453" PostTypeId="2" ParentId="3452" CreationDate="2010-08-27T10:43:13.620" Score="1" ViewCount="" Body="&lt;p&gt;Indefinite:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for $n=0$: $2^{0!}=2^1=2,$ $(2^0)!=1!=1$, so $2^{0!}&gt;(2^0)!$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for $n=2$: $2^{2!}=2^2=4,$ $(2^2)!=4!=12$, so $2^{2!}&amp;lt;(2^2)!$.&lt;/p&gt;&#xA;" OwnerUserId="163" LastActivityDate="2010-08-27T10:43:13.620" CommentCount="1" />
  <row Id="3454" PostTypeId="2" ParentId="3452" CreationDate="2010-08-27T10:44:34.617" Score="4" ViewCount="" Body="&lt;p&gt;It's enough to take this limit: $\lim_{n \rightarrow \infty} \frac{2^{n!}}{(2^n)!}$, by the Stirling formula we get $\lim_{n \rightarrow \infty} \frac{2^{(\frac{n}{e})^n \sqrt{2 \pi n}}}{(\frac{2^n}{e})^{2^n} \sqrt{2 \pi 2^n}}$, which is asymptotic (after elevating and taking logarithms) to $\lim_{n \rightarrow \infty} \frac{\exp(\frac{n^n}{e^n})}{\exp(2^n 2^{n/2})} = \lim_{n \rightarrow \infty} \frac{\exp(\frac{n^n}{e^n})}{\exp(2^{3/2 n})}$. Now, $\frac{n^n}{e^n}$ is bigger than $2^{3/2n}$ (and it is easy to verify). So the limit is infinite, which means that $2^{n!}$ is bigger than $(2^n)!$, if $n$ is big, of course.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Little note: I know that the asymptotic preseves constants and, even though I omitted them, it's good practice always to include them.&lt;/p&gt;&#xA;" OwnerUserId="1415" LastEditorUserId="1415" LastEditorDisplayName="" LastEditDate="2010-08-27T10:57:23.303" LastActivityDate="2010-08-27T10:57:23.303" CommentCount="3" />
  <row Id="3455" PostTypeId="2" ParentId="3447" CreationDate="2010-08-27T10:45:04.787" Score="8" ViewCount="" Body="&lt;p&gt;The opposite of &quot;modern&quot; is &quot;classical,&quot; and a common use of the word classical is &quot;whatever was known when I was in college.&quot;  That's a little tongue-in-cheek, but it really is how many people use the term.  See more discussion &lt;a href=&quot;http://www.johndcook.com/blog/2009/03/27/classical-science/&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="247" LastActivityDate="2010-08-27T10:45:04.787" CommunityOwnedDate="2010-08-27T16:05:48.460" CommentCount="5" />
  <row Id="3456" PostTypeId="2" ParentId="3452" CreationDate="2010-08-27T10:59:58.637" Score="4" ViewCount="" Body="&lt;p&gt;Actually, $(2^n)!$ is larger if 1 &amp;lt; n &amp;lt; &lt;a href=&quot;http://www.wolframalpha.com/input/?i=log_2%28%282%5en%29!%29%3Dn!&quot; rel=&quot;nofollow&quot;&gt;4.97399743597&lt;/a&gt;. See &lt;a href=&quot;http://www.wolframalpha.com/input/?i=plot+log%28log%282%5e%28n!%29%29%29+%2B+i+log%28log%28%282%5en%29!%29%29+for+n+%3D+0.1+to+10&quot; rel=&quot;nofollow&quot;&gt;this graph&lt;/a&gt; for detail.&lt;/p&gt;&#xA;" OwnerUserId="171" LastActivityDate="2010-08-27T10:59:58.637" CommentCount="1" />
  <row Id="3457" PostTypeId="2" ParentId="3452" CreationDate="2010-08-27T11:05:51.433" Score="4" ViewCount="" Body="&lt;p&gt;By proceeding straightforwardly we can answer without using Stirling's formula.&#xA;To start we show that $(n-1)!&gt;2^n$ for large n. [by induction]&#xA;First note that $5!&gt;2^6$.&#xA;Then we note that if we increase n by 1 we multiply the left hand side by n, and the right hand side by 2. Since n&gt;2 we conclude that the left hand side remains larger for bigger and bigger n.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Multiplying by n we get $n!&gt;n2^n$&#xA;Replacing n with $log(2^{n})$ [using base-2 logarithms] gets us: $n!&gt;2^{n}log(2^{n})$&#xA;Note that $2^{n}log(2^{n})$=&#xA;$log(2^{n})$+$log(2^{n})$+$log(2^{n})$+...$log(2^{n})$+$log(2^{n})$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[$2^n$ terms]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$&gt;log(2^{n})$+$log(2^{n}-1)$+$log(2^{n}-2)$+...log(2)+log(1) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;[since this has the same amount of terms, and some of them are smaller.]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;=$log(2^{n}!)$ [combining logs]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Putting together these inequalities gives us that n!&gt;$log(2^{n!})$, and raising 2 to these powers gives $2^{n!}$&gt;$(2^{n}!)$ [for n&gt;5 since our induction only told us that $(n-1)!&gt;2^n$ for n&gt;5]&lt;/p&gt;&#xA;" OwnerUserId="1149" LastEditorUserId="1149" LastEditorDisplayName="" LastEditDate="2010-08-27T11:23:17.223" LastActivityDate="2010-08-27T11:23:17.223" CommentCount="4" />
  <row Id="3458" PostTypeId="2" ParentId="3452" CreationDate="2010-08-27T11:09:26.857" Score="17" ViewCount="" Body="&lt;p&gt;We should expect $2^{n!}$ be larger. Note that $(2^n)! \leq (2^n)^{(2^n)} = 2^{(n 2^n)}$. &#xA;So we want to prove $n 2^n &amp;lt; n!$ eventually. (And this can be done very simply.)&lt;/p&gt;&#xA;" OwnerUserId="899" LastEditorUserId="899" LastEditorDisplayName="" LastEditDate="2010-08-27T11:14:47.690" LastActivityDate="2010-08-27T11:14:47.690" CommentCount="1" />
  <row Id="3459" PostTypeId="1" AcceptedAnswerId="3479" CreationDate="2010-08-27T11:23:38.270" Score="3" ViewCount="634" Body="&lt;p&gt;My textbook doesn't give any example of this kind of series. Could you provide some?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Trigonometric series is defined in wikipedia as :&#xA;$A_{0}+\sum_{n=1}^{\infty}(A_{n} \cos{nx} + B_{n} \sin{nx})$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$A_{n}=\frac{1}{\pi} \int^{2 \pi}_0 f(x) \cos{nx} dx\qquad (n=0,1,2,3 \dots)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$B_{n}=\frac{1}{\pi} \int^{2 \pi}_0 f(x) \sin{nx} dx\qquad (n=1,2,3, \dots)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is fourier series.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;thanks.&lt;/p&gt;&#xA;" OwnerUserId="1382" LastEditorUserId="1382" LastEditorDisplayName="" LastEditDate="2010-08-27T12:29:54.243" LastActivityDate="2010-08-29T20:19:50.293" Title="Example of a trigonometric series that is not fourier series?" Tags="&lt;calculus&gt;&lt;trigonometry&gt;&lt;fourier-series&gt;" AnswerCount="4" CommentCount="6" />
  <row Id="3460" PostTypeId="2" ParentId="3459" CreationDate="2010-08-27T11:33:53.637" Score="-1" ViewCount="" Body="&lt;p&gt;Take a look at &lt;a href=&quot;http://en.wikipedia.org/w/index.php?title=Trigonometric_series&amp;amp;oldid=370829487&quot; rel=&quot;nofollow&quot;&gt;this wikipedia stub&lt;/a&gt;. A trigonometric series, which is not a Fourier series, must therefore be one that is defined by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ f(x) = A_{0}+\displaystyle\sum_{n=1}^{\infty}(A_{n} \cos{nx} + B_{n} \sin{nx}) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for which the relations&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ A_{0}=\frac{1}{2\pi}\displaystyle\int^{2 \pi}_0 f(x)\,dx $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ A_{n}=\frac{1}{\pi}\displaystyle\int^{2 \pi}_0 f(x) \cos{nx} \,dx\qquad (n=1,2,3 \dots)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$B_{n}=\frac{1}{\pi}\displaystyle\int^{2 \pi}_0 f(x) \sin{nx}\, dx\qquad (n=0,1,2,3, \dots)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;do &lt;em&gt;not&lt;/em&gt; hold. But since {$\sin(nx),\cos(nx)$} forms a complete orthonormal basis, I cannot imagine one. &lt;strike&gt;Except for a non-convergent one perhaps...&lt;/strike&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;edit&lt;/strong&gt; As &lt;a href=&quot;http://math.stackexchange.com/users/498/j-mangaldan&quot;&gt;J. Mangaldan&lt;/a&gt; pointed out: &lt;a href=&quot;http://en.wikipedia.org/wiki/Lacunary_function#Lacunary_trigonometric_series&quot; rel=&quot;nofollow&quot;&gt;Lacunary trigonometric series&lt;/a&gt; might be the answer you search, i.e. series that cannot be analytically be continued.&lt;/p&gt;&#xA;" OwnerUserId="163" LastEditorUserId="163" LastEditorDisplayName="" LastEditDate="2010-08-27T11:43:55.100" LastActivityDate="2010-08-27T11:43:55.100" CommentCount="4" />
  <row Id="3461" PostTypeId="2" ParentId="3447" CreationDate="2010-08-27T11:34:51.943" Score="6" ViewCount="" Body="&lt;p&gt;Following in a tradition of similar purported definitions, one may say that &quot;modern mathematics is what modern mathematicians do&quot;. :-)&lt;/p&gt;&#xA;" OwnerUserId="688" LastActivityDate="2010-08-27T11:34:51.943" CommunityOwnedDate="2010-08-27T16:05:48.460" />
  <row Id="3462" PostTypeId="2" ParentId="3450" CreationDate="2010-08-27T12:54:41.790" Score="13" ViewCount="" Body="&lt;p&gt;Below is the etymology, from Jeff Miller's &lt;a href=&quot;http://jeff560.tripod.com/f.html&quot; rel=&quot;nofollow&quot;&gt;Earliest Known Uses of Some of the Words of Mathematics (F)&lt;/a&gt;. Perhaps a native French speaker can lend further insight.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;FACTORIAL&lt;/strong&gt;. The earlier term faculty was introduced around 1798 by Christian Kramp (1760-1826).&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;em&gt;Factorial&lt;/em&gt; was coined (in French as factorielle) by Louis François Antoine Arbogast (1759-1803).&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Kramp withdrew his term in favor of Arbogast's term. In the Preface, pp. xi-xii, of his &quot;Éléments d'arithmétique universelle,&quot; Hansen, Cologne (1808), Kramp remarks:&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;...je leur avais donné le nom de facultés. Arbogast lui avait substitué la nomination plus nette et plus française de factorielles; j'ai reconnu l'avantage de cette nouvelle dénomination; et en adoptant son idée, je me suis félicité de pouvoir rendre hommage à la mémoire de mon ami. [...I've named them facultes. Arbogast has proposed the denomination factorial, clearer and more French. I've recognised the advantage of this new term, and adopting its philosophy I congratulate myself of paying homage to the memory of my friend.] &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-08-27T12:54:41.790" CommentCount="2" />
  <row Id="3463" PostTypeId="1" AcceptedAnswerId="3470" CreationDate="2010-08-27T13:33:24.157" Score="3" ViewCount="136" Body="&lt;p&gt;I've been thinking about maps between sets. Injections, surjections and the rest. Often when thinking about some kind of map, it is interesting to say &quot;what about the maps from a set to &lt;em&gt;itself&lt;/em&gt;?&quot; Call these maps endomaps. Permutations of elements of the set are a special case of endomaps: they are bijective endomaps. Permutations have lots of interesting properties: they form groups and so on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But what about &lt;em&gt;general&lt;/em&gt; endomaps that are not necessarily bijective? Do they have any interesting properties that people study? They don't necessarily have inverses, which rules them out as forming groups, but composition of endomaps is associative, so they aren't totally devoid of interesting properties.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is also the case that for every endomap, there exists some subset of its domain (not necessarily unique) such that it is bijective on that subset. So these maps are permutations if restricted to a particular subset. Is this enough to make endomaps interesting in their own right, or are they only studied as a part of the study of maps between sets in general?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[It might be obvious that this question was motivated by thinking about category theory, but there's nothing particularly categorical about the question as such...]&lt;/p&gt;&#xA;" OwnerUserId="172" LastActivityDate="2010-08-27T15:32:28.523" Title="Do endomaps of sets have interesting properties?" Tags="&lt;set-theory&gt;&lt;category-theory&gt;" AnswerCount="3" CommentCount="4" FavoriteCount="1" />
  <row Id="3464" PostTypeId="2" ParentId="3419" CreationDate="2010-08-27T14:08:27.270" Score="0" ViewCount="" Body="&lt;p&gt;The way I understand this question, the answer is &quot;yes.&quot; Simply write the number in, say, hexadecimal notation, and it will always be shorter than in decimal (if it is &gt;=100000). If you are looking for the shortest representation, however, that is well defined only if you specify exactly what counts as a valid representation of a number. After all, every number you come up with can technically be represented in just one digit if you choose a sufficiently large base (at least one higher than the number itself).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;About the standard way to reduce it: I'd say the decimal system is the standard way of reducing numbers from just writing the appropriate number of dashes. :-) In the sense pointed out by Bill Dubuque, there is no representation that is more space efficient in principle, that is if you transfer the problem to a more abstract domain. If, instead, you are looking for the shortest representation using a given set of mathematical symbols, then no, there is no standard way to find it. I also suspect no algorithm can do better than brute force search (that is, first check all numbers representable using 1 character, then those representable in 2 characters, and so on). In any case, nobody would write a number as &quot;loge 56 (log 56 base e) when we discard first 6 numbers and then consider first 200 digits,&quot; so no conceivable set of standard symbols will account for that anyway.&lt;/p&gt;&#xA;" OwnerUserId="1386" LastActivityDate="2010-08-27T14:08:27.270" CommunityOwnedDate="2010-08-27T14:08:27.270" CommentCount="4" />
  <row Id="3465" PostTypeId="2" ParentId="3459" CreationDate="2010-08-27T14:21:41.163" Score="0" ViewCount="" Body="&lt;p&gt;When all the coefficients are 1, the series does not converge to a function.  (It does converge in the sense of a distribution, to the Dirac Delta.).&lt;/p&gt;&#xA;" OwnerUserId="1489" LastActivityDate="2010-08-27T14:21:41.163" />
  <row Id="3466" PostTypeId="2" ParentId="3447" CreationDate="2010-08-27T14:22:15.513" Score="5" ViewCount="" Body="&lt;p&gt;Further to the other answers, which are indeed correct: no technical definition exists- barks $\iff$ dog, frankly- but 'modern' &lt;em&gt;is&lt;/em&gt; a well defined concept outside of mathematics; and to a certain extent it is one to which the barkings of modern mathematics agree.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It was once the case that mathematicians believed proofs to uncover the neccessarily true- there must be numbers working thus, and those numbers must naturally have no zero divisors- there must be a geometry built thus, and this geometry must have angles in a triangle summing to $180^o$. With the exception of the work of Euclid (whose axioms were largely seen by others as immutable anyway), the theorems of mathematics were seen as universal truths, fouded in pure logic- facts about platonic ideals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Except none of this was true.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps the first chink in the armour of this classical mathematics came with the work of Bolyai and Gauss, constructing consistent geometries where triangles behaved unusually (turning, as we all know, into modern Hyperbolic geometry), that seeped from a change in the hitherto 'immutable' axioms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And from here the trickle began, which rushed and swelled with time, and burst the banks of mathematics as was: axioms became plastic, changeable at will, and with them the mathematics that followed from them. New concepts were created, and concepts of concepts, enriching and enlarging the mathematical landscape in ways that generations before could not have imagined.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Parallell to this explosion was the search for foundations for these axioms- the dying embers of platonism in the work of Frege, Russell and Whitehead; Hilbert's program, seeming at first promising, were spectacularly micturated on by Godel's incompleteness theorems. And it soon became (quite) clear, that any (provably) 'ultimate' description of mathematics was doomed to failure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Modernism outside of mathematics is characterised by a certain relativism- an understanding that different perspectives can lead to different (equally valid) conclusions. In modern mathematics one has the reals &lt;em&gt;and&lt;/em&gt; the p-adics, euclidean &lt;em&gt;and&lt;/em&gt;  non-euclidian geometries, topologies and metric spaces, groups, rings, algebras: sets and mereology- and we cannot claim one to be more valid than the others.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In modern mathematics, our truths are absolute but crucially contingent, the children of axioms in a pluralistic universe of possible postulates. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course some would say that 'modern' just means 'with categories', but that's not quite as neat- perhaps we can fit categories to 'post-modern' somehow....  &lt;/p&gt;&#xA;" OwnerUserId="160" LastEditorUserId="160" LastEditorDisplayName="" LastEditDate="2010-08-27T14:42:07.080" LastActivityDate="2010-08-27T14:42:07.080" CommunityOwnedDate="2010-08-27T14:22:15.513" CommentCount="10" />
  <row Id="3467" PostTypeId="2" ParentId="3459" CreationDate="2010-08-27T14:34:11.527" Score="2" ViewCount="" Body="&lt;p&gt;A standard example is &#xA;$$&#xA;f(t)= \sum_{n&gt;1} \frac{\sin(nt)}{\log(n)}$$&#xA;The conjugate of $f$ is a Fourier series but $f\not\in L^1(\mathbb{T})$ and hence is no Fourier series.&#xA;For further explanation see &lt;a href=&quot;http://books.google.com/books?id=gkpUE_m5vvsC&amp;amp;printsec=frontcover&amp;amp;dq=katznelson&amp;amp;hl=sv&amp;amp;ei=R8l3TN_tA5DLOIGhgBA&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=2&amp;amp;ved=0CCwQ6AEwAQ#v=snippet&amp;amp;q=conjugate&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;Katznelson's book page 85.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Edit: If $f$ is not in $L^1(\mathbb{T})$ it is hard to define Fourier coefficients.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added 29/8 - 2010&lt;/strong&gt;&#xA;Here is a screen dump from Zygmund's book &quot;Trigonometric Series Vol I&quot;&#xA;&lt;img src=&quot;http://i.imgur.com/BkVNH.png&quot; alt=&quot;A standard definition&quot;&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, in order to be a Fourier series the coefficients are to be Fourier coefficients of some function $f$, these are well defined if $f\in L^1(\mathbb{T})$ (because these are calculated using integration against a bounded measurable function). Thus it is natural to define a trigonometric series if the coefficients are Fourier coefficients of some $L^1$ function. Hence, all answers given to this problem are in fact right! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moreover, the Riemann-Lebesgue Lemma states that the Fourier coefficients of an $L^1$-function tends to $0$ as the index approach $\pm\infty$. Thus, it is easy to construct formal trigonometric series that are not Fourier series.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The example $f$ given above does in fact converge everywhere and hence this is a kind of non-trivial example of a trigonometric series that is not a Fourier series.&lt;/p&gt;&#xA;" OwnerUserId="1154" LastEditorUserId="1154" LastEditorDisplayName="" LastEditDate="2010-08-29T06:22:43.813" LastActivityDate="2010-08-29T06:22:43.813" CommentCount="2" />
  <row Id="3468" PostTypeId="2" ParentId="3463" CreationDate="2010-08-27T14:37:29.687" Score="2" ViewCount="" Body="&lt;p&gt;Given a set $X$, the collection of all maps $s\colon X\to X$ forms a &lt;em&gt;semigroup&lt;/em&gt; under the operation of composition, called the &quot;full transformation semigroup&quot; of $X$, often denoted $\mathcal{T}_X$. Full transformation semigroups play the same role in semigroup theory as symmetric groups do in group theory (the symmetric group is the group you get by considering the bijections from a set to itself, under composition). For example, the analogue of Cayley's Theorem (&quot;every group is a subgroup of a symmetric group&quot;) holds for semigroups (&quot;every semigroup is a subsemigroup of a full transformation semigroup&quot;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You claim that &quot;for every endomap, there exists some subset of its domain such that it is bijective on that subset&quot;. That is only true if you restrict yourself to finite sets or you allow restriction to the empty set: consider the map $f\colon\mathbb{N}\to\mathbb{N}$ given by $f(n)=n+1$. If $X$ is a nonempty subset of $\mathbb{N}$, then the minimum of $X$ is not in $f(X)$, so $f$ is not bijective on $X$.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-08-27T14:37:29.687" CommentCount="1" />
  <row Id="3470" PostTypeId="2" ParentId="3463" CreationDate="2010-08-27T14:41:47.543" Score="3" ViewCount="" Body="&lt;p&gt;One of the important properties of endomaps of an infinite set $X$ is that the expressive power of the logic that allows you to quantify over all endomaps of $X$ is the same as the expressive power of the full second order logic on $X.$ Roughly, this means that any relation on $X$ (a subset of $X^n$) can be modelled by means of endomaps. For instance, any equivalence relation $E$ on $X$ is modelled as follows: for every such equivalnce relation $E$ there is an endomap $f : X \to X$ such that&#xA;$$&#xA;E(x_1,x_2) \iff f(x_1)=f(x_2) \qquad (x_1,x_2 \in X). &#xA;$$&#xA;So endomaps of $X$ can tell you everything about $X,$ if you &quot;ask&quot; them properly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For other classes of structures, endomorphisms also carry lots and lots of information. For instance, the semigroup $\mathrm{End}(V)$ of all endomorphisms of a given vector space $V$ carries most essential information about $V,$ etc.&lt;/p&gt;&#xA;" OwnerUserId="1100" LastActivityDate="2010-08-27T14:41:47.543" />
  <row Id="3471" PostTypeId="2" ParentId="3433" CreationDate="2010-08-27T14:46:32.530" Score="0" ViewCount="" Body="&lt;p&gt;The following theorem (which I read in Number theory: algebraic numbers and functions&#xA; By Helmut Koch)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \sum\limits_{n=1}^{N} g(n) = g(1) + \int\limits_{1}^{N} g(x) \mathrm{d}x + \int\limits_{1}^{N} (x - [x]) g'(x) \mathrm{d}x $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;tells us that the difference converges when $$\int\limits_{1}^{N} (x - [x]) g'(x) \mathrm{d}x$$ does.&lt;/p&gt;&#xA;" LastActivityDate="2010-08-27T14:46:32.530" />
  <row Id="3472" PostTypeId="2" ParentId="3444" CreationDate="2010-08-27T15:01:26.480" Score="2" ViewCount="" Body="&lt;p&gt;You are really asking for a direct connection between some property of $n!$ and the integral.  This can be done from the recursion&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$S[n] = nS[n-1], S[0] = 1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $S[n]$ is the order of the symmetric group on $n$ elements.  The exponential generating function for this series equals $1/(1-t)$.  As in your favorite proof, replace that by $\int_{0}^{\infty} e^{(x-1)t}  dt$, expand $e^{xt}$ as a series, reverse the order of summation and integration, and you recover a power series in $t$ whose terms are precisely $\int_{0}^{\infty} t^n e^{-t}  dt/n!$.  The result follows upon term-by-term comparison.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I realize that's not a whole lot more insight, but it does show explicitly a connection between a defining property of $n!$ and the integral for the $\Gamma$ function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as your subsequent musings go: the integrand is the probability density of a Gamma variate, of course!  One useful relation is that the Gamma distribution with parameter $n$ is the sum (i.e., convolution) of $n$ independent exponential variates (i.e., Gamma variates with $n=1$).  The expectation of an exponential variate is 1 (this is easy), whence the expectation of a Gamma variate must be $n$ (because expectations add), strongly suggesting the mode of its pdf should be near $n$ (as justified for large $n$ by the Central Limit Theorem).&lt;/p&gt;&#xA;" OwnerUserId="1489" LastActivityDate="2010-08-27T15:01:26.480" CommentCount="5" />
  <row Id="3473" PostTypeId="2" ParentId="3452" CreationDate="2010-08-27T15:19:16.877" Score="1" ViewCount="" Body="&lt;p&gt;$ 2^{n!}=(2^n)^{(n-1)!} $, so we have to compare $(2^n)^{(n-1)!} $ and $ (2^n)! $. For $n\geq 6$, $(n-1)!&gt;2^n$. I may be wrong, but doesn't the result follow immediately from this? It's still true for $n=5$ though, but not for $n=4$.&lt;/p&gt;&#xA;" OwnerUserId="1313" LastActivityDate="2010-08-27T15:19:16.877" CommentCount="1" />
  <row Id="3474" PostTypeId="2" ParentId="3463" CreationDate="2010-08-27T15:32:28.523" Score="3" ViewCount="" Body="&lt;p&gt;The generalization of cycle decomposition to endomaps is quite interesting; rather than just cycles, endomaps break up into cycles in which each vertex is the root of a tree.  Counting endomaps (of which there are $n^n$) is therefore relevant to counting trees, and this is the basis of a beautiful proof due to Joyal (which I believe can be found in Bergeron, Labelle, and Leroux's &lt;a href=&quot;http://bergeron.math.uqam.ca/Species/especes.html&quot; rel=&quot;nofollow&quot;&gt;Combinatorial Species and Tree-like Structures&lt;/a&gt;) of &lt;a href=&quot;http://en.wikipedia.org/wiki/Cayleys_formula&quot; rel=&quot;nofollow&quot;&gt;Cayley's formula&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A set equipped with an endomap is just about the simplest kind of &lt;a href=&quot;http://en.wikipedia.org/wiki/Dynamical_system&quot; rel=&quot;nofollow&quot;&gt;dynamical system&lt;/a&gt;.  The category of sets equipped with endomaps is sometimes called the category of (discrete) dynamical systems and can be a useful example in category theory, e.g. it pops up in Lawvere and Schanuel's &lt;a href=&quot;http://books.google.com/books?id=o1tHw4W5MZQC&amp;amp;printsec=frontcover&amp;amp;dq=lawvere+and+schanuel&amp;amp;source=bl&amp;amp;ots=4u3nHsy70h&amp;amp;sig=PrsIRYFkI-CxO3AtnwLKf5dHP2E&amp;amp;hl=en&amp;amp;ei=ftp3TKumIIWBlAe7tcSzCg&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=2&amp;amp;ved=0CBsQ6AEwAQ#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;Conceptual Mathematics&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-08-27T15:32:28.523" />
  <row Id="3475" PostTypeId="2" ParentId="3441" CreationDate="2010-08-27T15:45:53.410" Score="2" ViewCount="" Body="&lt;p&gt;You have to choose 10 letters, and 2 of them must be digits. Furthermore, there must be one each of a lowercase letter, an upper case letter, and a common symbol. For the others, there are 5 choices to be made, and these are to be made from $26+26+10+32 = 94$ characters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This gives&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$94^5 \approx 7.339e9$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;choices for the the 6 other characters that do not have to be digits.&#xA;And for the digits, there are 2 choices from 10 characters. So this gives&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ 10^2 = 100$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;choices. And for the one each of lower-case letters, upper-case letters, and common symbols there are&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ 26 * 26 * 32  = 21632$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;choices.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now lastly, there are $10!$ permutations of these 10 characters, so the total number of combinations of these characters is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$26*26*32*100*94^5 * 10! \approx 5.76e22$&lt;/p&gt;&#xA;" OwnerUserId="1089" LastEditorUserId="1089" LastEditorDisplayName="" LastEditDate="2010-08-27T15:53:49.487" LastActivityDate="2010-08-27T15:53:49.487" CommentCount="1" />
  <row Id="3476" PostTypeId="1" CreationDate="2010-08-27T16:30:46.890" Score="4" ViewCount="123" Body="&lt;p&gt;In a very interesting &lt;a href=&quot;http://golem.ph.utexas.edu/category/2010/08/what_is_the_langlands_programm.html#comments&quot; rel=&quot;nofollow&quot;&gt;blog discussion&lt;/a&gt; at the $n$-category cafe, an anonymous poster made the following remark: &quot;... using the dictionary between number fields and function fields, Weil suggested that G-with-adelic-entries is analogous to the group of gauge transformations of a principal G-bundle over a Riemann surface.&quot; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to know how this particular piece of the number field / function field analogy is made precise. As a preliminary question, does anyone know a reference to where Weil (or someone else) explains it? I have a suspicion that it might be discussed in the notes from Weil's 1959-1960 lectures on &quot;Adeles and algebraic groups&quot;, but the copy in my local library is checked out so I'm not sure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My next question is for an explanation of the analogy in what I assume is the simplest case: holomorphic vector bundles on a compact Riemann surface. There are two obstacles here. First, I don't know what a gauge transformation is, and the &quot;Mathematical Formalism&quot; section &#xA;of the Wikipedia page seems nonsensical to me. What is being transformed, and what is the transformation itself? Can anyone provide a physics-free definition, purely in the language of geometry? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now given a curve $X$ over the complex numbers (or any field), we can form a topological ring $\mathbf{A}_X$ as the restricted product of the completed local rings at the closed points of $X$. The second obstacle is that when the ground field is infinite, these completions are not locally compact, so it seems unlikely that $\mathbf{A}_X$ is a good thing to consider. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nonetheless, can one put the constructions of the previous two paragraphs together to produce, for any rank $n$ holomorphic vector bundle $\mathcal{E}$ on $X$, a bijection as follows?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$${\text{gauge transformations of }\mathcal{E}}\stackrel{?}{\leftrightarrow} \mathrm{GL}_n(\mathbf{A}_X)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My final question is, what is the correct version of this when the complex numbers are replaced by  a finite field? Now we can consider algebraic vector bundles, and even better the adeles of $X$ are now a good thing to consider. So in trying to extend the putative bijection above to this case, the content seems to be in algebraizing the notion of gauge transform. Thus, I'm asking for a (second?) definition of gauge transformation which is native to algebraic geometry, if such exists.&lt;/p&gt;&#xA;" OwnerUserId="1265" LastActivityDate="2010-08-28T22:18:14.417" Title="How are gauge transformations of a $G$-bundle related to the adelic points of $G$?" Tags="&lt;number-theory&gt;&lt;algebraic-geometry&gt;&lt;differential-geometry&gt;&lt;riemann-surface&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="1" />
  <row Id="3478" PostTypeId="7" CreationDate="2010-08-27T19:02:16.537" Score="0" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;Mathematics - Stack Exchange&lt;/strong&gt; is for people studying mathematics at any level and professionals in related fields.  We welcome questions about:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Understanding mathematical concepts and theorems&lt;/li&gt;&#xA;&lt;li&gt;Hints on mathematical problems (but please read our &lt;a href=&quot;http://meta.math.stackexchange.com/questions/1803/how-to-ask-a-homework-question&quot;&gt;FAQ about homework questions&lt;/a&gt;)&lt;/li&gt;&#xA;&lt;li&gt;History and development of mathematics&lt;/li&gt;&#xA;&lt;li&gt;Solving mathematical puzzles&lt;/li&gt;&#xA;&lt;li&gt;Software that mathematicians use&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;There are certain subjects that, while still being on-topic here, you can get better response on our sister sites:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Research-level mathematics — &lt;a href=&quot;http://mathoverflow.net/&quot; rel=&quot;nofollow&quot;&gt;Math Overflow&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;Research level Algorithm design, complexity theory, etc — &lt;a href=&quot;http://cstheory.stackexchange.com/faq&quot;&gt;Theoretical Computer Science - Stack Exchange&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;Algorithm implementation/design, computer simulation and modelling, etc — &lt;a href=&quot;http://stackoverflow.com&quot;&gt;Stack Overflow&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;Statistical analysis  —  &lt;a href=&quot;http://stats.stackexchange.com&quot;&gt;Statistical Analysis - Stack Exchange&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;And some kinds of questions are considered off-topic:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Physics, engineering and financial questions — it is fine to ask the mathematical model and techniques, but concepts of the field would be off-topic.&lt;/li&gt;&#xA;&lt;li&gt;Typesetting equations — you can read &lt;a href=&quot;http://meta.math.stackexchange.com/questions/107/what-should-go-in-the-math-stackexchange-faq/117#117&quot;&gt;our guideline&lt;/a&gt; for writing equations on this site, or visit &lt;a href=&quot;http://tex.stackexchange.com&quot;&gt;TeX, LaTeX and Friends&lt;/a&gt; for questions about TeX&lt;/li&gt;&#xA;&lt;li&gt;Numerology&lt;/li&gt;&#xA;&lt;li&gt;Questions about the site itself — please ask on &lt;a href=&quot;http://meta.math.stackexchange.com/&quot;&gt;meta&lt;/a&gt; instead&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;If you want to get a more information of the site, you can browse our &lt;a href=&quot;http://meta.math.stackexchange.com/questions/107&quot;&gt;detailed list of FAQ&lt;/a&gt; on meta. &lt;/p&gt;&#xA;" OwnerUserId="-1" LastEditorUserId="1543" LastEditorDisplayName="" LastEditDate="2011-04-09T18:46:50.107" LastActivityDate="2011-04-09T18:46:50.107" />
  <row Id="3479" PostTypeId="2" ParentId="3459" CreationDate="2010-08-27T19:28:30.510" Score="5" ViewCount="" Body="&lt;p&gt;I don't know the answer (&lt;b&gt;I should have! -- see below&lt;/b&gt;), but in my opinion a lot of people are misinterpreting the question, so perhaps it is worth an answer to try to set this straight.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is an analogy: trigonometric series is to Fourier series as power series is to Taylor series.  In other words, a trigonometric series is just any series of the form&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$A_{0}+\sum_{n=1}^{\infty}(A_{n} \cos{nx} + B_{n} \sin{nx})$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This should be understood formally, i.e., this is a trigonometric series even if it doesn't converge anywhere.  A fourier series is the trigonometric series associated to an &#xA;$L^1$ function by taking $A_n$ and $B_n$ as above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just as a Taylor series need not need converge at any point except the central point, a Fourier series need not converge pointwise at any point.  Thus the Fourier series need not be a function and in particular the Fourier inversion theorem need not apply.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A theorem of Borel asserts that given any sequence $(a_n)$ of real numbers, there exists a $C^{\infty}$-function on the real line whose Taylor series at $0$ is $\sum_{n=0}^{\infty} a_n x^n$.  In particular, if the $a_n$'s grow too rapidly, the Taylor series will diverge away from zero and the function $f$ will not be analytic.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I interpret the question as asking whether the analogue of Borel's theorem is true for Fourier series: is every trigonometric series the Fourier series of some $L^1$ function (even if the trigonometric series does not converge to the function)?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S.: Boo to wikipedia for asserting the answer to this question without giving a reference.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Addendum&lt;/b&gt;: As Pierre-Yves Gaillard points out, the Fourier coefficients of any $L^1$ function $f$ are uniformly bounded (by $||f||_1$), so this answers the question as I have interpreted it.  &lt;/p&gt;&#xA;" OwnerUserId="299" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2010-08-29T20:19:50.293" LastActivityDate="2010-08-29T20:19:50.293" CommentCount="5" />
  <row Id="3480" PostTypeId="2" ParentId="3388" CreationDate="2010-08-27T19:46:08.087" Score="1" ViewCount="" Body="&lt;p&gt;Generalizing from decimal to arbitrary radix, a quick computer search seems to indicate  that there are only a handful of radices with such a unique solution. This leads to a proof of the following &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;THEOREM&lt;/strong&gt; $\;$ If in radix $\rm M$ notation we have $\rm (PQ)^2 = RSP$ with $\rm P,Q,R,S$ distinct digits and this solution is unique in radix $\:\rm M \:$ then it is one of the following, where $\;\rm A = 10,\; B = 11, \:\ldots,\: I = 18 \;$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{array}{|r|r|r|} &#xD;&#xA;\hline&#xD;&#xA;\rm M &amp;amp; \rm PQ &amp;amp; \rm RSP \\&#xD;&#xA;\hline&#xD;&#xA;6 &amp;amp; 15 &amp;amp; 321 \\&#xD;&#xA;7 &amp;amp; 23 &amp;amp; 562 \\&#xD;&#xA;8 &amp;amp; 17 &amp;amp; 341 \\&#xD;&#xA;9 &amp;amp; 18 &amp;amp; 351 \\&#xD;&#xA;10 &amp;amp; 19 &amp;amp; 361 \\&#xD;&#xA;11 &amp;amp; \rm 1A &amp;amp; 371 \\&#xD;&#xA;18 &amp;amp; \rm 1H &amp;amp; \rm 3E1 \\&#xD;&#xA;19 &amp;amp; \rm 1\:I &amp;amp; \rm 3F1 \\&#xD;&#xA;\end{array}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof&lt;/strong&gt;: $\;$ The two general solutions listed below prove that all radices $\;\rm M &amp;gt; 26 \;$ have nonunique solution (except possibly $\rm M = 44\:$). The remaining small number of exceptional cases were verified by computer.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\quad\quad\rm  (1\: M + M-1)^2 \;=\; \;\;\: 3\: M^2 + (M\;-\;\:4) M + 1, \quad\quad M &amp;gt; 7$   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\quad\quad\rm  (4\: M + M-2)^2 \;=\; 24\: M^2 + (M-20) M + 4, \quad\quad M&amp;gt;26,\;\; M \ne 44$ &lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2011-01-12T23:37:37.453" LastActivityDate="2011-01-12T23:37:37.453" CommentCount="2" />
  <row Id="3481" PostTypeId="1" AcceptedAnswerId="3487" CreationDate="2010-08-27T20:01:15.303" Score="3" ViewCount="200" Body="&lt;p&gt;In &lt;a href=&quot;http://math.stackexchange.com/questions/2395/what-are-the-recommended-textbooks-for-introductory-calculus/2446#2446&quot;&gt;an answer&lt;/a&gt; to &lt;a href=&quot;http://math.stackexchange.com/questions/2395/what-are-the-recommended-textbooks-for-introductory-calculus&quot;&gt;another question I asked&lt;/a&gt;, Isaac suggested a book that is the standard &quot;reform calculus&quot; book. In a comment, I asked what the phrase &quot;reform calculus&quot; means, and Isaac provided a link to &lt;a href=&quot;http://www.math.duke.edu/~das/essays/trends/summary.html&quot; rel=&quot;nofollow&quot;&gt;this summary page&lt;/a&gt;. It seems to be a new or revised teaching methodology. However, the page linked to doesn't provide a good explanation of what exactly the phrase &quot;reform calculus&quot; means or its history.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The summary describes things like courses being &quot;leaner in terms of the number of topics in the syllabus&quot; and says that &quot;technology alters the relative importance of specific techniques, and methodology because technology offers opportunities for creating new learning environments&quot;. But what do these things actually mean for people, especially the calculus student and the professor teaching a calculus course?&lt;/p&gt;&#xA;" OwnerUserId="91" LastActivityDate="2010-08-27T21:52:09.133" Title="What is &quot;reform calculus&quot;?" Tags="&lt;calculus&gt;&lt;terminology&gt;" AnswerCount="2" CommentCount="0" />
  <row Id="3483" PostTypeId="1" CreationDate="2010-08-27T20:08:54.687" Score="4" ViewCount="129" Body="&lt;p&gt;I'm brushing up on my calculus, and geometry and trigonometry keep coming up. However, these are probably my weakest areas mathematically. Are there any good sites or free PDFs/ebooks that I can use to quickly (in a weekend, preferably) work through geometry and trigonometry and refresh myself in topics such as areas and volumes of various geometric shapes, trigonometric functions, relationships between angles of a given shape or structure?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's been over 8 years since I took a geometric or trigonometric course (it was back in high school) and over 2 years since I took any course that even relied on geometry or trigonometry. I'm not looking for anything incredibly complex, either. Just something to cover the topics that I'm encountering in integral and differential calculus.&lt;/p&gt;&#xA;" OwnerUserId="91" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-09-11T06:26:46.183" LastActivityDate="2010-09-11T06:42:13.667" CommunityOwnedDate="2010-08-27T20:08:54.687" Title="What are some good resources for brushing up on geometry and trigonometry?" Tags="&lt;geometry&gt;&lt;algebra-precalculus&gt;&lt;trigonometry&gt;&lt;self-learning&gt;" AnswerCount="3" FavoriteCount="3" />
  <row Id="3484" PostTypeId="1" AcceptedAnswerId="3489" CreationDate="2010-08-27T20:24:03.847" Score="2" ViewCount="346" Body="&lt;p&gt;I know how to prove $\zeta (2)=\pi ^{2}/6$ by using the &lt;em&gt;trigonometric&lt;/em&gt; Fourier series expansion of $x^{2}/4$. How can one prove the same result using the &lt;em&gt;complex&lt;/em&gt; Fourier series of $f(x)=x$ for $0\leq x\leq 1$? Any suggestion?&lt;/p&gt;&#xA;" OwnerUserId="752" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2010-08-27T23:00:47.130" LastActivityDate="2011-06-05T09:55:37.400" Title="Hint on how to prove $\zeta ( 2) =\pi ^{2}/6$ using the complex Fourier series of $f(x)=x$" Tags="&lt;real-analysis&gt;&lt;sequences-and-series&gt;&lt;fourier-analysis&gt;&lt;fourier-series&gt;&lt;special-functions&gt;" AnswerCount="3" CommentCount="9" FavoriteCount="1" />
  <row Id="3487" PostTypeId="2" ParentId="3481" CreationDate="2010-08-27T21:22:35.727" Score="4" ViewCount="" Body="&lt;p&gt;It is hard to describe faithfully an entire movement or &quot;ism&quot;: suppose for instance you had asked for a description of Buddhism, or Marxism, or post-modernism.  For every principle that you put forward as a &quot;plank&quot; of the movement, there is someone to say that that's a misunderstanding/oversimplification, or there is a specific submovement formed out of the violation of that principle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a result, let me focus on a particular branch of calculus reform which is better defined (and which I have seen more of myself): the Harvard Calculus Consortium.  A very clear description, review and critique of this movement is given by Oliver Knill (who is well placed to comment on it, having been involved in the teaching and administration of calculus at Harvard for many years) here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.math.harvard.edu/~knill///pedagogy/harvardcalculus/index.html&quot; rel=&quot;nofollow&quot;&gt;http://www.math.harvard.edu/~knill///pedagogy/harvardcalculus/index.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I encourage you to read this first and then ask any further questions with at least one specific platform underneath your feet.&lt;/p&gt;&#xA;" OwnerUserId="299" LastActivityDate="2010-08-27T21:22:35.727" CommentCount="1" />
  <row Id="3488" PostTypeId="2" ParentId="3481" CreationDate="2010-08-27T21:52:09.133" Score="2" ViewCount="" Body="&lt;p&gt;Here's an excerpt from &lt;a href=&quot;http://ramanujan.math.trinity.edu/tumath/research/studpapers/s45.pdf&quot; rel=&quot;nofollow&quot;&gt;Reviewing Reformed Calculus&lt;/a&gt;, by Lisa Murphy, 2006.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The very beginnings of mathematics reform started in the 1960's, but the big push for Calculus&#xA;  reform started in earnest in 1989 with the publication of the National Council of Teachers of Mathematics'&#xA;  (NCTM) Principles and Standards for Mathematics Education. The NCTM published the&#xA;  Principles and Standards in response to the apathy of students towards math and the lack of academic&#xA;  success in the mathematics classroom. To combat these negative trends the NCTM outlined&#xA;  five goals for &quot;the processes of problem solving, reasoning and proof, connections, communication,&#xA;  and representation&quot; [1]. Through these goals it was hoped that students would be equipped with&#xA;  the basic skills and understanding that they would need to be successful.&#xA;  As the Principles and Standards inspired the reform of secondary mathematics education, thoughts&#xA;  of reform began to surface in the collegiate mathematics arena, especially with regard to Calculus.&#xA;  College Calculus courses were experiencing some of the same problems as secondary mathematics.&#xA;  Of the roughly 300,000 college students that are annually enrolled in an engineering-based Calculus&#xA;  course, only 140,000 earn a grade of D or higher [6]. Less than half of the students were performing&#xA;  &quot;well&quot; in their Calculus courses. Armed with statistics such as this, reform-minded professors set&#xA;  out to develop a new curriculum that would help raise the achievement level and stimulate student&#xA;  interest in mathematics.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;From the reform movement, numerous curricular designs have been generated. Calculus and&#xA;  Mathematica; Calculus, Concepts, Computers and Cooperative Learning (C4L); and The Calculus&#xA;  Consortium at Harvard (CCH) are a few of the commonly used curriculums. These new curriculums&#xA;  cover the entire spectrum of reform. Some are grounded in traditional techniques but incorporate&#xA;  snippets of reform, while others differ in most aspects from the traditional approach. Despite this&#xA;  vast array, there are some basic elements that are common to all reform curriculums in varying&#xA;  degrees that separate them from the traditional Calculus curriculum.&#xA;  One of the most noticeable differences of reformed Calculus is the use of graphing calculators&#xA;  and/or computers. The graphing calculator is a critical component in the reform classroom. Many&#xA;  reform classes include a weekly lab session where students meet in a computer lab. The students&#xA;  make use of calculators and math computer programs to investigate new topics and to graphically&#xA;  see what they are working on. Most reform textbooks urge students to read through the text with a&#xA;  calculator in hand to see directly what is discussed in the text. The idea behind the incorporation of&#xA;  calculators and computers is to alleviate the heavy algebraic manipulation that students typically do&#xA;  in a traditional Calculus setting. Reform supporters argue that the removal of manipulation allows&#xA;  students to move beyond the drudgery of computation and start learning the fundamental ideas of&#xA;  Calculus. They additionally argue that topics are discussed more fully with the use of graphical&#xA;  representations.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;A reformed Calculus class differs from a traditional course in methods of instruction. When&#xA;  walking into a reform classroom it is immediately clear that it is indeed a reform classroom. Most&#xA;  noticeably the teacher is no longer the central focus of the classroom experience. The lecture method&#xA;  of instruction, a standard of traditional curriculum, has a lesser place within a reform setting. The&#xA;  teacher still lectures occasionally and is available to answer questions from the students, but there is&#xA;  greater emphasis placed on cooperative learning. Reform students often work in groups to determine&#xA;  solutions or to explore concepts in a laboratory setting. This idea is rooted in the constructivist&#xA;  learning theory. Each student constructs their own meaning as they learn. Students are given the&#xA;  basic tools and from these discover how the pieces fit together to form the concept that they are&#xA;  studying. One of the primary goals of the C4L curriculum is to &quot;create situations which foster&#xA;  students to make the necessary mental constructions to learn mathematics concepts&quot; [10].&#xA;  Within the curriculum itself, the reformed method stresses the applications of Calculus. This&#xA;  emphasis hopes to justify the topics of study, which in theory raises interest in the material. In&#xA;  an effort to accomplish this, some of the mathematical rigor is removed from the curriculum. Most&#xA;  reform textbooks are void of a single proof. In the introduction to Calculus from Graphical, Numerical&#xA;  and Symbolic Point of View, the authors state that &quot;proving theorems in full generality is&#xA;  less valuable, we think, than helping students understand concretely what theorems say&quot; [9]. As&#xA;  a result of this change, a common question that arises from students new to reformed Calculus is&#xA;  &quot;Where is the math?&quot; [5].&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Accompanying this application heavy curriculum is a different method of assessment. Reformed&#xA;  Calculus courses emphasize the use of writing. Projects, reports and lengthy explanations of problem&#xA;  solutions are common place within the reform classroom. In some cases, the students are graded&#xA;  more on the thoroughness and completeness of written explanations as opposed to correctness of&#xA;  answer.&#xA;  [...]&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The emphasis on correct explanation rather than correct answer is seen explicitly in the directions&#xA;  for the midterm. The problem also provides an example of the type of application problems that&#xA;  reformed Calculus students are accustomed to working with. This midterm question additionally&#xA;  exhibits one of the flaws that traditional professors are quick to point out. The problem asks students&#xA;  to determine when the population becomes infinite, which is a misuse of the word infinite.&#xA;  The population may become uncontrollable but it will never become infinite. Traditional professors&#xA;  argue that the misuse of mathematical terms, such as infinite, teaches students the wrong meaning&#xA;  of or concept behind the term, which results in misunderstandings in future math work.&#xA;  More generally, there is a trend in reformed Calculus moving away from individual study and&#xA;  towards a social study of Calculus. The context of learning Calculus is now placed in a more social&#xA;  setting. Students work primarily in groups to gain knowledge both from a textbook and from each&#xA;  other.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-08-27T21:52:09.133" />
  <row Id="3489" PostTypeId="2" ParentId="3484" CreationDate="2010-08-27T23:09:53.373" Score="2" ViewCount="" Body="&lt;p&gt;Use the definition:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Say $f$ is defined on $[-\pi, \pi]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $f(z) = \sum_{-\infty}^{\infty} {c_{n} e^{inz}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$c_{n} = \frac{1}{2\pi}\int_{-\pi}^{\pi}{f(z)e^{-inz}} dz$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you put $f(z) = z$, can you work out what $c_{n}$ turns out to be?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To integrate, you can try integration by parts. &lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-08-28T02:41:51.660" LastActivityDate="2010-08-28T02:41:51.660" CommentCount="2" />
  <row Id="3490" PostTypeId="1" AcceptedAnswerId="3491" CreationDate="2010-08-27T23:26:05.400" Score="2" ViewCount="444" Body="&lt;p&gt;What's the simplest formula to calculate the optimal sample size from a population size? I've been reading a little about it and they ask me stuff like confidence level or confidence interval which I have no idea about. I need to test some HTML elements in web pages to get a better idea about general properties of HTML elements, and I have no idea about he results I'll get. So I just need the most neutral possible version of the sample size calculating formula when all I know is the population size.&lt;/p&gt;&#xA;" OwnerUserId="1542" LastActivityDate="2010-09-16T22:07:20.517" Title="Optimal sample size?" Tags="&lt;statistics&gt;" AnswerCount="2" CommentCount="5" />
  <row Id="3491" PostTypeId="2" ParentId="3490" CreationDate="2010-08-27T23:47:53.793" Score="1" ViewCount="" Body="&lt;p&gt;It depends on your population and the confidence that you need in your results. Typically, you should know something about what you are trying to achieve. If you know your confidence level, confidence interval, and population size, you can choose a sample size that will give you the properly confidence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your confidence level, typically around 90% (95% and 99% are also commonly used), tells you how sure you will be that your entire population reflects your results. The confidence interval describes the range of values that you are sure about within your confidence level.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Sample_size&quot; rel=&quot;nofollow&quot;&gt;Wikipedia provides an overview of sample size&lt;/a&gt; methodologies. That might get you started. But unless you know how sure you want to be of your results, you can't determine a sample size. Wikipedia also provides decent definitions of &lt;a href=&quot;http://en.wikipedia.org/wiki/Confidence_interval&quot; rel=&quot;nofollow&quot;&gt;confidence level and confidence interval&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From a statistical standpoint, if you don't have clearly defined questions, you really can't analyze your data, at least using statistical methods. Perhaps you should review data and try to formulate questions, or take an analytical instead of statistical approach to solving the problem/answering your question.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Assuming a normal distribution, you can use the equation $n \geq \left(\dfrac{z_{\alpha/2} \sigma}{\delta}\right)^2$ where z is found in a table, σ is your standard deviation (which you can use either a sample standard deviation or a population standard deviation, depending on what you know), and δ is your confidence level. The value for n is your population - be sure to round up if you get a fractional n.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the z-value is based on your confidence level. The value for α is a value between 1 - the decimal format of your confidence level (for a confidence level of 95%, α would be 1 - 0.95, or 0.05).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might also be interested in the answers to a few questions on the Statistical Analysis Stack Exchange:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://stats.stackexchange.com/questions/166/how-do-you-decide-the-sample-size-when-polling-a-large-population&quot;&gt;How do you decide the sample size when polling a large population?&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://stats.stackexchange.com/questions/276/how-large-should-a-sample-be-for-a-given-estimation-technique-and-parameters&quot;&gt;How large should a sample be for a given estimation technique and parameters?&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="91" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-09-16T22:07:20.517" LastActivityDate="2010-09-16T22:07:20.517" CommentCount="2" />
  <row Id="3492" PostTypeId="2" ParentId="3484" CreationDate="2010-08-28T00:08:33.340" Score="1" ViewCount="" Body="&lt;p&gt;This is not related but you would like to see this article: A Short Proof of ζ (2) = π2/6 T.H. Marshall American Math monthly April 2010.&lt;/p&gt;&#xA;" LastActivityDate="2010-08-28T00:08:33.340" />
  <row Id="3493" PostTypeId="2" ParentId="3483" CreationDate="2010-08-28T00:24:46.477" Score="1" ViewCount="" Body="&lt;p&gt;Hey there are lot of sources available on the internet. I enlist a few of them here. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) &lt;a href=&quot;http://www.free-ed.net/free-ed/Math/Trigonometry/trig02_SPK.asp&quot; rel=&quot;nofollow&quot;&gt;http://www.free-ed.net/free-ed/Math/Trigonometry/trig02_SPK.asp&lt;/a&gt; This website provides an nice introduction, which is very basic in nature. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) For geometry you can read this book by a very good Mathematician, which is available online : &lt;a href=&quot;http://math.mit.edu/~kedlaya/geometryunbound/&quot; rel=&quot;nofollow&quot;&gt;http://math.mit.edu/~kedlaya/geometryunbound/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am sure gigapedia.com will have lots of ebooks on this subjects. Try searching it.&lt;/p&gt;&#xA;" LastActivityDate="2010-08-28T00:24:46.477" CommunityOwnedDate="2010-08-28T00:24:46.477" CommentCount="1" />
  <row Id="3494" PostTypeId="2" ParentId="3024" CreationDate="2010-08-28T00:26:33.527" Score="1" ViewCount="" Body="&lt;p&gt;Your given metric is in the form of a &lt;a href=&quot;http://en.wikipedia.org/wiki/Warped_geometry&quot; rel=&quot;nofollow&quot;&gt;warped product&lt;/a&gt;, for which there are some well established methods of computing the curvature of the total space in terms of the curvature of the warp factors. See, for example, Barrett O'Neill's &lt;em&gt;Semi-Riemannian Geometry&lt;/em&gt;, Chapter 7. For the first factor (given by $u$ and $r$), it being two dimensional, you can just use the standard results for surfaces on its intrinsic curvature. &lt;/p&gt;&#xA;" OwnerUserId="1543" LastActivityDate="2010-08-28T00:26:33.527" />
  <row Id="3495" PostTypeId="2" ParentId="3483" CreationDate="2010-08-28T00:38:53.257" Score="0" ViewCount="" Body="&lt;p&gt;Although it's a little basic, I managed to find a &lt;a href=&quot;http://www.google.com/url?sa=t&amp;amp;source=web&amp;amp;cd=1&amp;amp;ved=0CBkQFjAA&amp;amp;url=http%3A%2F%2Fwww.videomathteacher.com%2Fuploaded%2Ffiles%2F02GeometryRefSheet.pdf&amp;amp;ei=4ll4TL7-IsP_lge_4v3rCw&amp;amp;usg=AFQjCNE44cwWueDYBTBPtf7RZcX2ZYgwQQ&amp;amp;sig2=aF6D3iZxDWn_99XSzI7lOQ&quot; rel=&quot;nofollow&quot;&gt;&quot;geometry reference sheet&quot; (direct PDF download)&lt;/a&gt;. It's got your standard two- and three-dimensional shapes, plus your law of sines, law of cosines, distance, and midpoint formulas. Although if it's been years since you had to find the volume of a sphere, this kind of thing might be useful.&lt;/p&gt;&#xA;" OwnerUserId="91" LastActivityDate="2010-08-28T00:38:53.257" CommunityOwnedDate="2010-08-28T00:38:53.257" />
  <row Id="3496" PostTypeId="2" ParentId="3335" CreationDate="2010-08-28T00:46:55.493" Score="2" ViewCount="" Body="&lt;p&gt;Since you like the writing style of Courant and John... for PDEs there are of course F. John's &lt;em&gt;Partial Differential Equations&lt;/em&gt;, and also Courant and Hilbert's &lt;em&gt;Methods of Mathematical Physics&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1543" LastActivityDate="2010-08-28T00:46:55.493" CommunityOwnedDate="2010-08-28T00:46:55.493" />
  <row Id="3497" PostTypeId="1" CreationDate="2010-08-28T00:48:36.883" Score="3" ViewCount="99" Body="&lt;p&gt;Sorry if this question is too basic. It is from Fröhlich and Taylor's &quot;Algebraic Number Theory&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $E/F$ be a finite Galois extension of fields, with $G=Gal(E/F)$, and let $K$ and $L$ be two subfields of $E$, containing $F$, such that $K/F$ and $L/F$ are both Galois. Let $M=Gal(E/K)$ and $N=Gal(E/L)$ be normal subgroups of $G$. Suppose ${\gamma_1,\ldots,\gamma_n}$ is a transversal for $MN$ in $G$, with $n=[G:MN]$. If $C$ is the compositum $KL$ in $E$, how can I show the map&#xA;$$ k\otimes l\mapsto (k^{\gamma_1}l,\ldots,k^{\gamma_n}l) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;induces an isomorphism between $K\otimes_F L$ and $\prod_{i=1}^n C$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is clear to me that this map is an $F$-algebra homomorphism, and that they both have the same dimension over $F$.  Thus surjectivity, or injectivity, would be enough.  I have not been able to figure out what the idempotents of $\prod_{i=1}^n C$ should look like in $K\otimes_F L$, so I have not been able to show surjectivity.  Meanwhile, I think injectivity should be easier to show, because if we have&#xA;$$ k_1\otimes l_1 + \cdots + k_m\otimes l_m\mapsto 0,$$&#xA;then we get a system of equations&#xA;$$ k_1l_1 + \cdots + k_ml_m=0$$&#xA;$$ \cdots$$&#xA;$$ k_1^{\gamma_n}l_1 + \cdots k_m^{\gamma_n}l_m=0.$$&#xA;Summing up the columns, I get&#xA;$$ \sum_{i=1}^m(\sum_{j=1}^n k_i^{\gamma_j})l_i=0 $$&#xA;and all this is happening in $L$.  But I can't seem to finish this argument.  Any help would be greatly appreciated.&lt;/p&gt;&#xA;" OwnerUserId="1544" LastActivityDate="2010-08-28T16:16:55.583" Title="Algebra structure of tensor product of two Galois extensions" Tags="&lt;number-theory&gt;&lt;abstract-algebra&gt;&lt;field-theory&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="3498" PostTypeId="2" ParentId="3444" CreationDate="2010-08-28T01:39:26.310" Score="11" ViewCount="" Body="&lt;p&gt;I haven't quite got this straight yet, but I think one way to go is to think about choosing points at random from the positive reals. This answer is going to rather longer than it really needs to be, because I'm thinking about this in a few (closely related) ways, which probably aren't all necessary, and you can decide to reject the uninteresting parts and keep anything of value. Very roughly, the idea is that if you &quot;randomly&quot; choose points from the positive reals and arrange them in increasing order, then the probability that the (n+1)th point is in a small interval (t,t+dt) is a product of probabilities of independent events, n factors of t for choosing n points in the interval [0,t], one factor of $e^{-t}$ as all the other points are in $[t,\infty)$, one factor of dt for choosing the point in $(t,t+dt)$, and a denominator of $n!$ coming from the reordering. At least, as an exercise in making a simple problem much harder, here goes...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll start with a bit of theory before trying to describe intuitively why the probability density $\frac{t^n}{n!}e^{-t}$ pops out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can look at the &lt;a href=&quot;http://en.wikipedia.org/wiki/Poisson_process&quot; rel=&quot;nofollow&quot;&gt;homogeneous Poisson process&lt;/a&gt; (with rate parameter 1). One way to think of this is to take a sequence on independent exponentially distributed random variables with rate parameter 1, $S_1,S_2,\ldots$, and set $T_n=S_1+\cdots+S_n$. As has been commented on already, $T_{n+1}$ has the probability density function $\frac{t^n}{n!}e^{-t}$. I'm going to avoid proving this immediately though, as It would just reduce to manipulating some integrals. Then, the Poisson process $X(t)$ counts the number of times $T_i$ lying in the interval [0,t].&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can also look at Poisson point processes (aka, &lt;a href=&quot;http://en.wikipedia.org/wiki/Poisson_random_measure&quot; rel=&quot;nofollow&quot;&gt;Poisson random measures&lt;/a&gt;, but that Wikipedia page is very poor). This is just makes rigorous the idea of randomly choosing unordered sets of points from a sigma-finite measure space $(E,\mathcal{E},\mu)$. Technically, it can be defined as a set of nonnegative integer-valued random variables $\{N(A)\colon A\in\mathcal{E}\}$ counting the number of points chosen from each subset A, such that $N(A)$ has the Poisson distribution of rate $\mu(A)$ and $N(A_1),N(A_2),\ldots$ are independent for pairwise disjoint sets $A_1,A_2,\ldots$. By definition, this satisfies&#xA;$$&#xA;\begin{array}{}\mathbb{P}(N(A)=n)=\frac{\mu(A)^n}{n!}e^{-\mu(A)}.&amp;amp;&amp;amp;(1)\end{array}&#xA;$$&#xA;The points $T_1,T_2,\ldots$ above defining the homogeneous Poisson process also define a Poisson random measure with respect to the Lebesgue measure $(\mathbb{R}_+,{\cal B},\lambda)$. Once you forget about the order in which they were defined and just regard them as a random set that is, which I think is the source of the $n!$. If you think about the probability of $T_{n+1}$ being in a small interval $(t,t+\delta t)$ then this is just the same as having $N([0,t])=n$ and $N((t,t+\delta t))=1$, which has probability $\frac{t^n}{n!}e^{-t}\delta t$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, how can we choose points at random so that each small set $\delta A$ has probability $\mu(\delta A)$ of containing a point, and why does (1) pop out? I'm imagining a hopeless darts player randomly throwing darts about and, purely by luck, hitting the board with some of them. Consider throwing a very large number $N\gg1$ of darts, independently, so that each one only has probability $\mu(A)/N$ of hitting the set, and is distributed according to the probability distribution $\mu/\mu(A)$. This is consistent, at least, if you think about the probability of hitting a subset $B\subseteq A$. The probability of missing with all of them is $(1-\mu(A)/N)^N=e^{-\mu(A)}$. This is a multiplicative function due to independence of the number hitting disjoint sets. To get the probability of one dart hitting the set, multiply by $\mu(A)$ (one factor of $\mu(A)/N$ for each individual dart, multiplied by N because there's N of them). For n darts, we multiply by $\mu(A)$ n times, for picking n darts to hit, then divide by n! because we have over-counted the subsets of size n by this factor (due to counting all n! ways of ordering them). This gives (1). I think this argument can probably be cleaned up a bit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Getting back to choosing points randomly on the positive reals, this gives a probability of $\frac{t^n}{n!}e^{-t}dt$ of picking n in the interval $[0,t]$ and one in $(t,t+dt)$. If we sort them in order as $T_1\lt T_2\lt\cdots$ then $\mathbb{P}(T_1\gt t)=e^{-t}$, so it is exponentially distributed. Conditional on this, $T_2,T_3,\ldots$ are chosen randomly from $[T_1,\infty)$, so we see that the differences $T_{i+1}-T_{i}$ are independent and identically distributed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is $\frac{t^n}{n!}e^{-t}$ maximized at t=n? I'm not sure why the mode should be a simple property of a distribution. It doesn't even exist except for unimodal distributions. As $T_{n+1}$ is the sum of n+1 IID random variables of mean one, the law of large numbers suggests that it should be peaked approximately around n. The central limit theorem goes further, and gives $\frac{t^n}{n!}e^{-t}\approx\frac{1}{\sqrt{2\pi n}}e^{-(t-n)^2/{2n}}$. Stirling's formula is just this evaluated at $t=n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What's this to do with Tate's thesis? I don't know, and I haven't read it (but intend to), but have a vague idea of what it's about. If there is anything to do with it, maybe it is something to do with the fact that we are relating the sums of independent random variables $S_1+\cdots+S_n$ distributed with respect to the Haar measure on the multiplicative group $\mathbb{R}_+$ (edit: oops, that's not true, the multiplicative Haar measure has cumulative distribution given by log, not exp) with randomly chosen sets according to the Haar measure on the additive group $\mathbb{R}$.&lt;/p&gt;&#xA;" OwnerUserId="1321" LastEditorUserId="1321" LastEditorDisplayName="" LastEditDate="2010-08-28T02:58:47.697" LastActivityDate="2010-08-28T02:58:47.697" CommentCount="2" />
  <row Id="3499" PostTypeId="1" AcceptedAnswerId="3524" CreationDate="2010-08-28T01:47:31.130" Score="0" ViewCount="91" Body="&lt;p&gt;Suppose $C \subset \mathbb{R}$ is of type $F_{\sigma}$. That is $C$ can be written as the union of $F_{n}$'s where each $F_{n}$'s are closed. Then can we prove that each point of $C$ is a point of discontinuity for some $f: \mathbb{R} \to \mathbb{R}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I refered this link on wiki : &lt;a href=&quot;http://en.wikipedia.org/wiki/Thomae%27s_function&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Thomae%27s_function&lt;/a&gt; and in the follow up subsection they given this result. I would like somebody to explain it more precisely.&lt;/p&gt;&#xA;" LastEditorUserId="171" LastEditorDisplayName="" LastEditDate="2010-08-28T07:48:01.743" LastActivityDate="2010-08-31T09:57:25.443" Title="$F_{\sigma}$ subsets of $\mathbb{R}$" Tags="&lt;real-analysis&gt;&lt;analysis&gt;&lt;general-topology&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="3501" PostTypeId="1" AcceptedAnswerId="3584" CreationDate="2010-08-28T02:23:59.140" Score="0" ViewCount="101" Body="&lt;p&gt;Hey guys,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's the question I've been thinking for 2 days and couldn't find an answer why.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If Y = √X, where both X and Y are real numbers then which of the following is true?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a) X&gt;=0; Y&gt;=0&lt;/p&gt;&#xA;&#xA;&lt;p&gt;b) X&amp;lt;=0; Y&gt;=0&lt;/p&gt;&#xA;&#xA;&lt;p&gt;c) X&amp;lt;=0; Y&amp;lt;=0&lt;/p&gt;&#xA;&#xA;&lt;p&gt;d) X&gt;=0; Y&amp;lt;=0&lt;/p&gt;&#xA;&#xA;&lt;p&gt;e) Either a) or b)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the answer is a) but I don't know why. I know a value of Y for which root of Y is negative.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;√4 = +2, -2&#xA;√9 = +3, -3&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the answer should be X&amp;lt;=0; Y&gt;=0, but its not. :(&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone explain me why? Thanks in advance. :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS: I think I messed up the under root symbol. (√ = Under Root / Square Root) I don't know how to get the proper symbol where I'll get a line of that symbol extended above numbers. :P&lt;/p&gt;&#xA;" OwnerUserId="712" LastEditorUserId="712" LastEditorDisplayName="" LastEditDate="2010-08-29T05:08:06.160" LastActivityDate="2010-08-29T05:11:59.420" Title="Problem related with equation consisting square root" Tags="&lt;arithmetic&gt;" AnswerCount="1" CommentCount="16" />
  <row Id="3502" PostTypeId="1" AcceptedAnswerId="3505" CreationDate="2010-08-28T02:39:42.697" Score="4" ViewCount="324" Body="&lt;p&gt;How does one sum the series, $$ S = a -\frac{2}{3}a^{3} + \frac{2 \cdot 4}{3 \cdot 5} a^{5} - \frac{ 2 \cdot 4 \cdot 6}{ 3 \cdot 5 \cdot 7}a^{7} + \cdots $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This was asked to me by a high school student, and i am embarrassed that i couldn't solve it. Can anyone give me a hint!!&lt;/p&gt;&#xA;" LastEditorUserId="171" LastEditorDisplayName="" LastEditDate="2010-08-28T13:58:02.783" LastActivityDate="2010-08-28T13:58:02.783" Title="Summing the series $(-1)^k \frac{(2k)!!}{(2k+1)!!} a^{2k+1}$" Tags="&lt;sequences-and-series&gt;&lt;analysis&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="3" />
  <row Id="3503" PostTypeId="2" ParentId="3502" CreationDate="2010-08-28T03:05:35.117" Score="8" ViewCount="" Body="&lt;p&gt;You can use the formula&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle \int_{0}^{\frac{\pi}{2}}{\sin^{2k+1}(x) dx} = \frac{2 \cdot 4 \cdot 6 \cdots 2k}{3\cdot 5 \cdots (2k+1)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is called Wallis's product.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we have $\displaystyle S(a) = \sum_{k=0}^{\infty} (-1)^k a^{2k+1} \int_{0}^{\frac{\pi}{2}}{\sin^{2k+1}(x) dx}$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Interchanging the sum and the integral&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle S(a) =   \int_{0}^{\frac{\pi}{2}}{\sum_{k=0}^{\infty}{(-1)^{k}(a\sin x)^{2k+1}} dx}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sum inside the integral is a geometric series of the form&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle x - x^3 + x^5 - \cdots = x(1 - x^2 + x^4 - \cdots)  = \frac{x}{1+x^2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle S(a) =   \int_{0}^{\frac{\pi}{2}}{\frac{a\sin x}{1 + (a\sin x)^2}}dx$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now substitute $\displaystyle t = a \cos x$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The integral becomes&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle \int_{0}^{a}{\frac{1}{1+a^2 - t^2}}dt = \frac{1}{2\sqrt{a^2+1}}\ln(\frac{\sqrt{a^2+1}+a}{\sqrt{a^2+1}-a})$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now $\displaystyle \ln(\frac{\sqrt{a^2+1}+a}{\sqrt{a^2+1}-a}) = \ln(\frac{(\sqrt{a^2+1}+a)^2}{(\sqrt{a^2+1}-a)(\sqrt{a^2+1}+a)}) = 2\ln(\sqrt{a^2+1}+a)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle S(a) = \frac{1}{\sqrt{a^2+1}}\ln(\sqrt{a^2+1}+a) = \frac{\sinh^{-1}(a)}{\sqrt{a^2+1}}$&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-08-28T13:42:44.377" LastActivityDate="2010-08-28T13:42:44.377" CommentCount="5" />
  <row Id="3504" PostTypeId="1" AcceptedAnswerId="3835" CreationDate="2010-08-28T03:25:23.280" Score="1" ViewCount="95" Body="&lt;p&gt;&lt;strong&gt;From a 2005 Russian olympiad. Prove that there do not exist four (pairwise) different quadratic polynomials, with leading coefficient 1, such that the sum of any two of them has a single root.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Optionally, this is what I've done so far:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $P_i(x)=x^2+b_ix+c_i$, then $P_i+P_j$ is single-rooted iff $(b_i+b_j)^2=8(c_i+c_j)$, with root $-(b_i+b_j)/2$. If two of the six roots are equal, we're done, so suppose they're different, and that $P_1+P_2$ is single-rooted, that is, $(b_1+b_2)^2=8(c_1+c_2)$. If we want $P_1+P_3$ and $P_2+P_3$ to be single-rooted as well, we need $(b_1+b_3)^2=8(c_1+c_3)$ and $(b_2+b_3)^2=8(c_2+c_3)$, from which, taking $b_1$ and $c_1$ as known, $b_2, c_2, b_3, c_3$ could be found.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This way looks cumbersome.)&lt;/p&gt;&#xA;" OwnerUserId="1313" LastEditorUserId="1313" LastEditorDisplayName="" LastEditDate="2010-09-02T01:55:30.690" LastActivityDate="2010-09-02T01:55:30.690" Title="Four polynomials with single-rooted sums." Tags="&lt;algebra&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" />
  <row Id="3505" PostTypeId="2" ParentId="3502" CreationDate="2010-08-28T03:31:51.030" Score="13" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\quad \:\;\;\rm (a^2+1) \: S' = 1 - a \: S \;\:$ by transmuting the coefficient recurrence to a differential equation.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\;\Rightarrow\; 1 = (a^2+1) \: S' + a \: S \; = \; f \: (f \; S)' \;\;$ for $\rm\;\; f = (a^2+1)^{1/2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\displaystyle\;\Rightarrow\; S = f^{-1} \int \; f^{-1} = \frac{\sinh^{-1}(a)}{(a^2+1)^{1/2}}$&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-08-28T04:36:16.937" LastActivityDate="2010-08-28T04:36:16.937" CommentCount="1" />
  <row Id="3506" PostTypeId="1" AcceptedAnswerId="3542" CreationDate="2010-08-28T04:00:36.603" Score="1" ViewCount="176" Body="&lt;p&gt;Is there a natural way to represent all the partitions of an integer set {1,2,3,...,n} as a matrix in the similar way permutations can be mapped to group of matrices?&lt;/p&gt;&#xA;" OwnerUserId="196" LastActivityDate="2010-08-28T12:58:51.107" Title="Matrix representation of a partition" Tags="&lt;combinatorics&gt;&lt;partitions&gt;" AnswerCount="1" CommentCount="8" />
  <row Id="3507" PostTypeId="1" AcceptedAnswerId="3511" CreationDate="2010-08-28T04:07:44.613" Score="10" ViewCount="212" Body="&lt;p&gt;At the beginning of &lt;a href=&quot;http://math.ucr.edu/home/baez/week300.html&quot; rel=&quot;nofollow&quot;&gt;Week 300 of John Baez's blog&lt;/a&gt;,&#xA;Baez gives a proof that the &quot;number&quot; of finite sets (more specifically, the cardinality of the groupoid of all finite sets, where an object in the groupoid counts as 1/n! if it has n! symmetries) equals e.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;He then says that this leads to a purely combinatorial proof that e^x is its own derivative.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone explain the purely combinatorial proof?&lt;/p&gt;&#xA;" OwnerUserId="1546" LastActivityDate="2010-08-28T05:20:07.213" Title="Purely combinatorial proof that d/dx(e^x) = e^x" Tags="&lt;combinatorics&gt;" AnswerCount="1" FavoriteCount="3" />
  <row Id="3508" PostTypeId="1" AcceptedAnswerId="3518" CreationDate="2010-08-28T04:08:01.697" Score="4" ViewCount="380" Body="&lt;p&gt;A matrix determinant (naively) can be computed in O(n!) steps, or with a proper LU decomposition O(n^3) steps. This assumes that all the matrix elements are constant. If, however the matrix elements are polynomials (say univariate of max order p) at each step of the LU decomposition an element is multiplied by another element producing (on average) ever larger polynomials. Each step therefore takes longer and longer - is the cost perform the decomposition still a polynomial?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;: To clarify my reasoning a bit, if polynomial (using FFT as Mangaldan suggests in the comments) takes O(m log m), and we must perform O(n^3) operations to get the LU decomposition, each step the polynomial could effectively double in degree at each multiplication*. The running time would look something like&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \approx O \left ( \sum_{k}^{n^3} (p \ln p)^{2k} \right ) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;* (it doesn't quite do that, and this is where I'm stuck)&lt;/p&gt;&#xA;" OwnerUserId="196" LastEditorUserId="196" LastEditorDisplayName="" LastEditDate="2010-08-28T05:28:02.023" LastActivityDate="2010-08-29T02:56:24.937" Title="Determinant of a polynomial matrix" Tags="&lt;linear-algebra&gt;&lt;matrices&gt;&lt;asymptotics&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="3509" PostTypeId="2" ParentId="3502" CreationDate="2010-08-28T04:08:59.450" Score="3" ViewCount="" Body="&lt;p&gt;Making my comments more explicit:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your sum of interest is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sum_{j=0}^\infty {(-1)^j \frac{(2j)!!}{(2j+1)!!} a^{2j+1}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $(2j)!!=2\cdot 4\cdot 6\cdots (2j)$ and $(2j+1)!!=3\cdot 5\cdot 7\cdots (2j+1)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To simplify things a bit, we rearrange the series a bit to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a\sum_{j=0}^\infty {\frac{(2j)!!}{(2j+1)!!}\left(-a^2\right)^j}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The double factorials can be also expressed as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(2j)!!=2^j j!=2^j (1)_j$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(2j+1)!!=2^j \left(\frac32\right)_j$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $(a)_j$ is a Pochhammer symbol.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Substitute both expressions into the series, and then note that the series now looks like a hypergeometric series. Now you can employ the formula &lt;a href=&quot;http://functions.wolfram.com/07.23.03.3098.01&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-08-28T04:08:59.450" CommentCount="5" />
  <row Id="3510" PostTypeId="1" AcceptedAnswerId="3513" CreationDate="2010-08-28T04:17:02.887" Score="11" ViewCount="2162" Body="&lt;p&gt;Could you provide a proof of Euler's formula: $\exp(i t)=\cos(t)+i\sin(t)$ ? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;thanks.&lt;/p&gt;&#xA;" OwnerUserId="1382" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-08-28T18:50:53.660" LastActivityDate="2011-05-17T04:46:15.120" Title="How to prove Euler's formula: $\exp(i t)=\cos(t)+i\sin(t)$ ?" Tags="&lt;complex-analysis&gt;" AnswerCount="9" CommentCount="3" FavoriteCount="7" />
  <row Id="3511" PostTypeId="2" ParentId="3507" CreationDate="2010-08-28T04:39:17.760" Score="8" ViewCount="" Body="&lt;p&gt;I am not quite sure how to translate this into groupoid cardinality language, but here is the standard proof.  Suppose $A(x) = \sum_{n \ge 0} a_n \frac{x^n}{n!}$ is an exponential generating function.  Then we should interpret $a_n$ as being the number of ways to put a certain structure on a set of size $n$.  For example, when $a_n = 1$ this is the structure of &quot;being a set.&quot;  When $a_n = n!$ this is the structure of &quot;being a totally ordered set.&quot;  And so forth.  We will call this an $A$-structure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then $A'(x) = \sum_{n \ge 0} a_{n+1} \frac{x^n}{n!}$ can be interpreted as having coefficients $b_n = a_{n+1}$ which count the number of ways to add an element to a set of size $n$, then put an $A$-structure on the resulting set of size $n+1$.  This is a purely combinatorial definition of differentiation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With this definition, the proof is quite obvious: there is exactly one way for a set to be a set, and there is also exactly one way to add an element to a set and then make the result a set.  So $\frac{d}{dx} e^x = e^x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This proof might seem contentless.  Try to see how it generalizes to show that $\frac{d}{dx} e^{ax} = ae^{ax}$ for any positive integer $a$, and if you're up for a challenge see if you can generalize it all the way to &lt;a href=&quot;http://qchu.wordpress.com/2009/06/24/gila-vi-the-cycle-index-polynomials-of-the-symmetric-groups/&quot; rel=&quot;nofollow&quot;&gt;this identity&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Vaguely the proof in groupoid cardinality language goes like this.  For a finite set $X$ the groupoid of finite sets equipped with a function to $X$ has cardinality $e^{|X|}$.  (The morphisms between two objects $A \to X, B \to X$ in this category are isomorphisms $A \simeq B$ such that the obvious triangle commutes.)  One way to think about this groupoid is as the groupoid of &quot;colored&quot; sets, where $X$ is the set of colors and an isomorphism must respect color.  Then it is easy to see that an isomorphism class of colored sets where there are $|X|$ colors is the same thing as a disjoint union of isomorphism classes of $|X|$ sets, one for each color.  One gets a direct interpretation of the terms in the expansion $\left( \sum_{n \ge 0}^{\infty} \frac{1}{n!} \right)^{|X|}$ this way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Differentiation replaces $|X|^n$ with $n|X|^{n-1}$, which means that we replace functions from an $n$-element set $S$ to $X$ with functions from $S - \{ s \}$ to $X$ where $s$ ranges over all elements of $S$.  The resulting groupoid is still the groupoid of finite sets equipped with a function to $X$; in particular, it has the same cardinality.  (Note that $X$ does not really have to be a finite set of a particular size for this argument to work; it can be a &quot;formal&quot; set in the same way that $x$ is a formal variable and the resulting groupoid cardinality is a generating function instead of a number.  I think this is what the formal theory of &quot;stuff types&quot; is for, but I am not familiar with it.)&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-08-28T05:20:07.213" LastActivityDate="2010-08-28T05:20:07.213" />
  <row Id="3512" PostTypeId="2" ParentId="3510" CreationDate="2010-08-28T04:48:13.147" Score="20" ViewCount="" Body="&lt;p&gt;Let $\mathbf{A}$ be an $n \times n$ matrix.  Recall that the system of differential equations&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathbf{x}' = \mathbf{Ax}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;has the unique solution $\mathbf{x} = e^{\mathbf{A}t} \mathbf{x}(0)$, where $\mathbf{x}$ is a vector-valued differentiable function and $e^{\mathbf{A}t}$ denotes the &lt;a href=&quot;http://en.wikipedia.org/wiki/Matrix_exponential&quot; rel=&quot;nofollow&quot;&gt;matrix exponential&lt;/a&gt;.  In particular, let $\mathbf{J} = \left[ \begin{array}{cc} 0 &amp;amp; 1 \\ -1 &amp;amp; 0 \end{array} \right]$.  Then the system of differential equations&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x' = y, y' = -x$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with initial conditions $x(0) = 1, y(0) = 0$ has the unique solution $\left[ \begin{array}{cc} x \\ y \end{array} \right] = e^{\mathbf{J}t} \left[ \begin{array}{cc} 1 \\ 0 \end{array} \right]$.  On the other hand, the above equations tell us that $x'' = -x$ and $y'' = -y$, and we know that the solutions to this differential equation are of the form $a \cos t + b \sin t$ for constants $a, b$.  By matching initial conditions we in fact find that $x = \cos t, y = \sin t$.  Now verify that on vectors multiplying by $\mathbf{J}$ has the same effect as multiplying a complex number by $i$, and you obtain Euler's formula.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This proof has the following attractive physical interpretation: a particle whose $x$- and $y$-coordinates satisfy $x' = y, y' = -x$ has the property that its velocity is always perpendicular to and has proportional magnitude to its displacement.  But from physics lessons you know that this uniquely describes particles which move in a circle. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another way to interpret this proof is as a description of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Exponential_map#Lie_theory&quot; rel=&quot;nofollow&quot;&gt;exponential map&lt;/a&gt; from the &lt;a href=&quot;http://en.wikipedia.org/wiki/Lie_algebra&quot; rel=&quot;nofollow&quot;&gt;Lie algebra&lt;/a&gt; $\mathbb{R}$ to the &lt;a href=&quot;http://en.wikipedia.org/wiki/Lie_group&quot; rel=&quot;nofollow&quot;&gt;Lie group&lt;/a&gt; $\text{SO}(2)$.  Euler's formula generalizes to quaternions, and this in turn can be thought of as describing the exponential map from the Lie algebra $\mathbb{R}^3$ (with the cross product) to $\text{SU}(2)$ (which can then be sent to $\text{SO}(3)$).  This is one reason it is convenient to use quaternions to describe 3-d rotations in computer graphics; the exponential map makes it easy to interpolate between two rotations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit:  whuber's answer reminded me of the following excellent graphic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://upload.wikimedia.org/wikipedia/commons/0/0e/ExpIPi.gif&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is what is happening geometrically in whuber's answer, and is essentially what happens if you apply Euler's method to the system of ODEs I described above.  Each step is a plot of the powers of $1 + \frac{i}{N}$ up to the $N^{th}$ power.&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-08-29T13:41:05.560" LastActivityDate="2010-08-29T13:41:05.560" CommentCount="7" />
  <row Id="3513" PostTypeId="2" ParentId="3510" CreationDate="2010-08-28T04:49:16.420" Score="13" ViewCount="" Body="&lt;p&gt;Assuming you mean $e^{ix}=\cos x+i\sin x$, one way is to use the MacLaurin series for sine and cosine, which are known to converge for all real x in a first-year calculus context, and the MacLaurin series for $e^z$, trusting that it converges for pure-imaginary z since this result requires complex analysis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The MacLaurin series:&#xA;$$\begin{align}&#xA;\sin x&amp;amp;=\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)!}x^{2n+1}=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\cdots&#xA;\\&#xA;\cos x&amp;amp;=\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n)!}x^{2n}=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\cdots&#xA;\\&#xA;e^z&amp;amp;=\sum_{n=0}^{\infty}\frac{z^n}{n!}=1+z+\frac{z^2}{2!}+\frac{z^3}{3!}+\cdots&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Substitute $z=ix$ in the last series:&#xA;$$\begin{align}&#xA;e^{ix}&amp;amp;=\sum_{n=0}^{\infty}\frac{(ix)^n}{n!}=1+ix+\frac{(ix)^2}{2!}+\frac{(ix)^3}{3!}+\cdots&#xA;\\&#xA;&amp;amp;=1+ix-\frac{x^2}{2!}-i\frac{x^3}{3!}+\frac{x^4}{4!}+i\frac{x^5}{5!}-\cdots&#xA;\\&#xA;&amp;amp;=1-\frac{x^2}{2!}+\frac{x^4}{4!}+\cdots +i\left(x-\frac{x^3}{3!}+\frac{x^5}{5!}-\cdots\right)&#xA;\\&#xA;&amp;amp;=\cos x+i\sin x&#xA;\end{align}$$&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-08-28T04:49:16.420" CommentCount="4" />
  <row Id="3514" PostTypeId="2" ParentId="3510" CreationDate="2010-08-28T04:56:41.357" Score="7" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\:$ Both $\:\rm e^{ix}\:$ and $\:\rm cos(x) + i \; sin(x) \:$ are solutions of $\;\rm y' = i \; y,\;\; y(0) = 1, \;$  so they are equal by the &lt;strong&gt;uniqueness theorem&lt;/strong&gt;. Alternatively, &lt;strong&gt;bisect&lt;/strong&gt; into even and odd parts the power series for $\:\rm e^{ix} \:,\;$  i.e.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xA;\rm f(x) \ \ &amp;amp;=\ \ \rm\frac{f(x)+f(-x)}{2} \;+\; \frac{f(x)-f(-x)}{2} \\ &#xA;\\&#xA;\Rightarrow\quad\quad \rm e^{ix} \ \ &amp;amp;=\ \ \rm\cos(x) \ +\ i \:\sin(x)  \end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;REMARKS 1.&lt;/strong&gt;$\;$ Uniqueness theorems provide powerful tools for proving equalities for functions that satisfy certain nice differential or difference (recurrence) equations. This includes a large majority of functions encountered in theory and practice. Such ideas lie behind algorithms implemented in computer algebra systems, e.g. search the computational literature using the terms &quot;D-finite&quot; and/or &quot;holonomic&quot; functions. &#xA;For a less trivial but still easy example of this technique see my &lt;a href=&quot;http://math.stackexchange.com/questions/3505&quot;&gt;recent post&lt;/a&gt; which proves the identity&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\rm \frac{sinh^{-1}(x)}{\sqrt{x^2+1}} \  = \ \ \sum_{k=0}^\infty\ (-1)^k \frac{(2k)!!}{(2k+1)!!} \: x^{2k+1}$$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a simple discrete example see my &lt;a href=&quot;http://google.com/groups?selm=y8zvdpvc099.fsf@nestle.csail.mit.edu&quot; rel=&quot;nofollow&quot;&gt;post here&lt;/a&gt; where I remark that $\rm\; 13 \:|\: 3^{n+1} + 4^{2n-1} =: f_n \;$ follows from $\;\rm f_2 = f_1 = 0 \;$ and the (obvious) fact that $\;\rm f_n \;$  satisfies a monic 2nd order linear recurrence. Namely let $\;\rm S\; f_n := f_{n+1} \;$ be the shift operator. Then $\;\rm S - 3 \;$ kills $\rm\; 3^{n+1} \;$ and $\;\rm S - 16 \;$ kills $\;\rm 4^{2n-1} \;$ therefore $\;\rm (S-3)\:(S-16) = S^2 - 19\: S + 48 \;$ kills their sum $\;\rm f_n \:,\;$ i.e. $\;\rm f_{n+2} = 19\; f_{n+1} - 48\; f_n\;$. Therefore $\:\rm mod\:13:\  f_2 = f_1 = 0  \;\Rightarrow\; f_3 = 19\: f_2 - 48\: f_1 = 0 \;\Rightarrow f_4 = 0 \Rightarrow f_5 = 0 \Rightarrow \cdots\Rightarrow\; f_n = 0\: $. Hence $\:0\:$ is the unique solution of the above recurrence that satisfies the intial conditions $\;\rm f_2 = f_1 = 0. \;$ This is simply an obvious special case of the uniqueness theorem for difference equations (recurrences). Once again, by invoking a uniqueness theorem, we have greatly simplified the deduction of an equality.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice that, above, we don't need to know the precise recurrence relation. Rather, we need only know a bound on its degree, so that we know how many initial terms are needed to determine the solution uniquely. In practice, as above, one can often easily derive simple upper bounds on the degree of the recurrence or differential equation - which makes the method even more practical.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;2.&lt;/strong&gt; $\;$ Generalizing the above bisection into even and odd parts, one can employ n'th roots of unity to take arbitrary n-part &lt;strong&gt;multisections&lt;/strong&gt; of power series and generating functions. They often prove handy, e.g.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EXERCISE&lt;/strong&gt; $\;$ Give elegant proofs of the following&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\quad\quad\rm\displaystyle sin(x)/e^{x} \quad\:$ has every $\rm 4k\;$'th term zero  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\quad\quad\rm\displaystyle cos(x)/e^{x} \quad$ has every $\rm 4k+2\;$'th term zero&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See the posts in &lt;a href=&quot;http://groups.google.com/group/sci.math/msg/642159e4f5db4e44&quot; rel=&quot;nofollow&quot;&gt;this thread&lt;/a&gt; for various solutions and more on multisections.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-10-25T16:46:06.767" LastActivityDate="2010-10-25T16:46:06.767" />
  <row Id="3515" PostTypeId="2" ParentId="3490" CreationDate="2010-08-28T06:11:33.497" Score="0" ViewCount="" Body="&lt;p&gt;what does it mean to &quot;test some HTML elements&quot; in a web page? what sort of [numerical] data do you expect to collect - and what, if anything, do you want to infer from what you get?&#xA;a particular example may help to clarify what you are driving at.&lt;/p&gt;&#xA;" OwnerUserId="1547" LastActivityDate="2010-08-28T06:11:33.497" />
  <row Id="3516" PostTypeId="2" ParentId="3476" CreationDate="2010-08-28T06:14:48.547" Score="1" ViewCount="" Body="&lt;p&gt;I'd like a see an answer to this question, and I hope someone here or on MO can provide a good one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I just want to address your question about gauge transformations.  Usually a gauge transformation on a principal G-bundle $P\to X$ is simply a G-equivariant map $P\to P$ which sends fibers to fibers (said another way, a gauge transformation is simply an automorphism of the principal G-bundle P).  A great place to read about the basics of gauge transformations is Atiyah and Bott's paper &quot;The Yang-Mills Equations over Riemann Surfaces&quot; (section 2 in particular).  &lt;/p&gt;&#xA;" OwnerUserId="1392" LastEditorUserId="1392" LastEditorDisplayName="" LastEditDate="2010-08-28T22:18:14.417" LastActivityDate="2010-08-28T22:18:14.417" CommentCount="3" />
  <row Id="3517" PostTypeId="2" ParentId="3405" CreationDate="2010-08-28T06:33:20.990" Score="12" ViewCount="" Body="&lt;p&gt;Update: the following probabilistic argument I had posted earlier shows only that $p(1) + p(2) + \dots + p(n) = (c + o(1)) \log(n)$ and not, as originally claimed, the convergence $np(n) \to c$.  Until a complete proof is available [edit: George has provided one in another answer] it is not clear whether $np(n)$ converges or has some oscillation, and at the moment there is evidence in both directions.  Log-periodic or other slow oscillation is known to occur in some problems where the recursion accesses many previous terms.  Actually, everything I can calculate about $np(n)$ is consistent with, and in some ways suggestive of, log-periodic fluctuations, with convergence being the special case where the bounds could somehow be strengthened and the fluctuation scale thus squeezed down to zero.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;$p(n) \sim c/n$ is [edit: only in average] equivalent to $p(1) + p(2) + \dots + p(n)$ being asymptotic to $c \log(n)$.  The sum up to $p(n)$ is the expected time the walk spends in the interval [1,n].  For this quantity there is a simple probabilistic argument that explains (and can rigorously demonstrate) the asymptotics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This Markov chain is a discrete approximation to a log-normal random walk.  If $X$ is the position of the particle, $\log X$ behaves like a simple random walk with steps $\mu \pm \sigma$ where $\mu = 2 \log 2 - 1 = 1/c$ and $\sigma^2 = (1- \mu)^2/2$.  This is true because &lt;em&gt;the Markov chain is bounded between two easily analyzed random walks with continuous steps&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Let X be the position of the particle and $n$ the number of steps; the walk starts at X=1, n=1.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lower bound walk $L$: at each step, multiply X by a uniform random number in [1,2] and replace n by (n+1).   $\log L$ increases, on average, by $\int_1^2 log(t) dt = 2 \log(2) - 1$ at each step.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Upper bound walk $U$: at each step, jump from X to uniform random number in [X+1,2X+1] and replace n by (n+1).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$L$ and $U$ have means and variances that are the same within $O(1/n)$, where the $O()$ constants can be made explicit.  Steps of $L$ are i.i.d and steps of $U$ are independent, asymptotically identical-distributed.  Thus, the Central Limit theorem shows that $\log X$ after $n$ steps is approximately a Gaussian with mean $n\mu + O(\log n)$ and variance $n\sigma^2 + O(\log n)$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The number of steps for the particle to escape the interval $[1,t]$ is therefore $({\log t})/\mu$ with fluctuations of size $A \sqrt{\log t}$ having probability that decays rapidly in A  (bounded by $|A|^p \exp(-qA^2)$ for suitable constants).   Thus, the sum p(1) + p(2) + ... p(n) is asymptotically equivalent to $(\log n)/(2\log (2)-1)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe this is equivalent to the 2003 argument from the conference.  If the goal is to get a proof from the generating function, it suggests that dividing by $(1-x)$ may be useful for smoothing the p(n)'s.&lt;/p&gt;&#xA;" OwnerUserId="467" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-11-12T17:23:20.473" LastActivityDate="2010-11-12T17:23:20.473" CommentCount="9" />
  <row Id="3518" PostTypeId="2" ParentId="3508" CreationDate="2010-08-28T06:33:39.540" Score="4" ViewCount="" Body="&lt;p&gt;Suppose each entry of your $n \times n$ matrix is a polynomial of degree $d$ in the variable $t$. Appealing to the cofactor expansion, we see that the determinant will be a polynomial in $t$ of degree at most $dn$; lets call it $D(t)$. So you can do the following: evaluate the determinant at $t=1,2, \ldots, dn$. You will thus know $D(1),D(2),\ldots,D(dn)$ and can find out the coefficients of $D$ by interpolation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This involves:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(i) $dn$ determinant evaluations, each of which takes $O(n^3)$ operations. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(ii) Interpolating a polynomial of degree $dn$. This can be done by solving a $dn \times dn$ system of equations (see &lt;a href=&quot;http://en.wikipedia.org/wiki/Polynomial_interpolation&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Polynomial_interpolation&lt;/a&gt; under &quot;Constructing the interpolating polynomial&quot;). There are &lt;a href=&quot;http://www.springerlink.com/content/q124843l4p368126/&quot; rel=&quot;nofollow&quot;&gt;algorithms&lt;/a&gt; which will do this in $O(dn \log^2 dn)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your final number of operations will be $O(d n^4 + dn \log^2 dn)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would be interested in knowing if its possible to do this faster. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Update: I corrected an error and incorporated a suggestion of J. Mangaldan in the comments. &lt;/p&gt;&#xA;" OwnerUserId="317" LastEditorUserId="317" LastEditorDisplayName="" LastEditDate="2010-08-29T02:56:24.937" LastActivityDate="2010-08-29T02:56:24.937" CommentCount="6" />
  <row Id="3519" PostTypeId="1" AcceptedAnswerId="3521" CreationDate="2010-08-28T06:34:51.013" Score="2" ViewCount="149" Body="&lt;p&gt;Liouville's Number is defined as $L = \sum_{n=1}^{\infty}(10^{-n!})$. Does it have other applications than just constructing a transcendental number?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Personally, I would have defined it (as &quot;Steven's Number&quot; :-)) as binary: $S = \sum_{n=1}^{\infty}(2^{-n!})$, since each digit can only be &quot;0&quot; or &quot;1&quot;: the corresponding power of 2 (instead of 10) included or not. Since according to Cantor most number are transcendental one can conjecture that this is also the case for Steven's Number. Can a proof for this be devised based on the proof for Liouville's number?)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not a mathematician, so please type slowly! :-)&lt;/p&gt;&#xA;" OwnerUserId="923" LastActivityDate="2010-09-11T07:36:33.910" Title="Liouville's number revisited" Tags="&lt;number-theory&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="1" />
  <row Id="3520" PostTypeId="2" ParentId="3483" CreationDate="2010-08-28T06:43:14.523" Score="0" ViewCount="" Body="&lt;p&gt;If you are starting trigonometry I'd suggest you to solve Plane trigonometry book by S.L. Loney&lt;/p&gt;&#xA;" LastActivityDate="2010-09-11T06:42:13.667" CommunityOwnedDate="2010-08-28T06:43:14.523" />
  <row Id="3521" PostTypeId="2" ParentId="3519" CreationDate="2010-08-28T06:43:22.343" Score="8" ViewCount="" Body="&lt;p&gt;the number $S$ is also transcendental, for precisely the same reason as $L$: they both have rational approximations that are far too good to hold for an algebraic number. See &lt;a href=&quot;http://en.wikipedia.org/wiki/Liouville_number&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; for more details. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the specific number $L$ is useful for anything else, I'm not aware of it. It's really just a simple example, one of many which can be dreamed up, of a number which fails to be algebraic since it defies Liouville's Lemma in diophantine approximation. &lt;/p&gt;&#xA;" OwnerUserId="308" LastActivityDate="2010-08-28T06:43:22.343" />
  <row Id="3522" PostTypeId="1" CreationDate="2010-08-28T07:04:58.937" Score="2" ViewCount="209" Body="&lt;p&gt;I have a standard linear programming problems I want to solve:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \min_x f^T x \text{ such that } \left\{ \begin{aligned} A\cdot x &amp;amp;\le b, \\ A_{eq}\cdot x &amp;amp;= b_{eq}, \\ lb \le x &amp;amp;\le ub. \end{aligned} \right. $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$F^T$ is a vector that doesn't involve $x$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem with the above minimization is that, I don't know what $n$, the number of variable $x$ is — it is also a part of the minimization. &#xA;There are a few constraints governing how the vector $x$ should behave. Additionally, there are constraints linking from $x_i$ to $x_{i+1}$. This means that given $x_i$, we know how to form the constraint for $x_{i+1}$. Also, even though we don't know about $n$, but in my problem I can easily construct the corresponding $f_i$ term for each $x_i$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The constraints are complicated in the sense that it is not easy to express the constraints in the following form:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ Ax \leq a $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ Bx = b $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I can do, at best, is to express $a$ and $b$ involving a first order of $x$ ( i.e., no $x^2$ and above). The matrix $A$ and $B$ are known values with no involvement of $x$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand that linear programming can be used to tackle problems such as this. But the two problems I mention above ( don't know what $n$ is, and cannot separate out the matrix/vector easily) stop me from proceeding.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any other techniques I can use to solve the problem?&lt;/p&gt;&#xA;" OwnerUserId="550" LastEditorUserId="550" LastEditorDisplayName="" LastEditDate="2010-08-28T08:59:25.493" LastActivityDate="2010-08-28T08:59:25.493" Title="How to Minimize A Function Where The Number of Variables is Unknown" Tags="&lt;optimization&gt;&lt;linear-programming&gt;" CommentCount="12" FavoriteCount="1" />
  <row Id="3523" PostTypeId="2" ParentId="3476" CreationDate="2010-08-28T07:08:12.473" Score="4" ViewCount="" Body="&lt;p&gt;If $X$ is a smooth projective curve over a finite field $k$, with field of rational functions $F$, then the quotient $GL_n(F)\backslash GL_n(\mathbf A_F)/GL_n(\mathcal O_F)$ (where&#xA;$\mathcal O_F$ denotes the subring of integral adeles in $\mathbf A_F$) is in natural bijection with the set of isomorphism classes of rank $n$ vector bundles on $X$, i.e. the &#xA;set $Bun_n(k)$ of $k$-valued points of the moduli stack of rank $n$ bundles on $X$.&#xA;(And the same should be true if we replace $GL_n$ by another reductive group $G$,&#xA;and replace &quot;rank $n$ vector bundles&quot; by principal $G$-bundles; recall that principal $GL_n$-bundles are &quot;the same&quot; as rank $n$ vector bundles; one goes from the latter to the former&#xA;by passing to the associate frame bundle.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The passage from a bundle to an a point in the double coset space is made as follows: given the vector bundle, one chooses an affine open subset $U = X \setminus {x_1,\ldots,x_r}$ over which&#xA;the bundle is trivialized, and also trivialiations in a n.h. $U_i$ of each of the points $x_i$.  The gluing data comparing the two trivializations on $U_i \cap U$ gives an element&#xA;$g_{x_i} \in GL_n(F_{x_i})$, where $F_{x_i}$ is the completion of $F$ at $x_i$.&#xA;If $x$ is a point distinct from the $x_i$, set $g_{x} = 1$.  We can then put all&#xA;the $g_x$ together into an adele, and its class in&#xA;the double coset space $GL_n(F)\backslash GL_n(\mathbf A_F)/GL_n(\mathcal O_F)$&#xA;is independent of all choices.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To see that this map is a bijection, one first observes&#xA;that any element in the double coset space can be represented as&#xA;$(g_x)$ with $g_x = 1$ for almost all $x$, and then one interprets these $g_x$ as formal&#xA;gluing data around the points $x_i$ at which $g_{x_i} \neq 1$ and uses them to &#xA;extend the trivial bundle on $X\setminus {x_1,\ldots,x_r}$ to a bundle on all of $X$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The relationship with gauge transformations is  that the $g_x$ are change of &#xA;basis matrices (gauge tranformations) in formal punctured n.h.s of the points $x$.&lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2010-08-28T07:08:12.473" CommentCount="1" />
  <row Id="3524" PostTypeId="2" ParentId="3499" CreationDate="2010-08-28T07:16:46.347" Score="2" ViewCount="" Body="&lt;p&gt;I believe you're looking for something like the construction mentioned &lt;a href=&quot;http://mathoverflow.net/questions/165/does-the-continuous-locus-of-a-function-have-any-nice-properties/174#174&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="308" LastActivityDate="2010-08-28T07:16:46.347" CommentCount="0" />
  <row Id="3525" PostTypeId="1" AcceptedAnswerId="3527" CreationDate="2010-08-28T07:48:50.807" Score="2" ViewCount="240" Body="&lt;p&gt;How to show the following identity holds? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle\sum_{n=1}^\infty\dfrac{2z}{z^2-n^2}=\pi\cot \pi z-\dfrac{1}{z}\qquad |z|&amp;lt;1$         &lt;/p&gt;&#xA;" OwnerUserId="752" LastActivityDate="2010-08-28T08:03:38.753" Title="A series expansion of $\cot (\pi z)$" Tags="&lt;sequences-and-series&gt;&lt;complex-analysis&gt;&lt;trigonometry&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="0" />
  <row Id="3526" PostTypeId="2" ParentId="3419" CreationDate="2010-08-28T08:00:29.070" Score="-1" ViewCount="" Body="&lt;p&gt;My guess is that for any sufficiently large n you can find integer numbers a, b and c, such that n = a^b + c and IntegerLength[N] &gt; IntegerLength[a] + IntegerLength[b] + IntegerLength[c]. Recurse for c (and a, b if they're large enough) and you'll find forms with even less digits.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;e.g. $n = 541^{547}  71^{73} + 29^{31} + 19^{23} + 997$ counts 1631 digits, its abbreviated form a total of 21, and probably even less is possible.&lt;/p&gt;&#xA;" OwnerUserId="923" LastEditorUserId="923" LastEditorDisplayName="" LastEditDate="2010-08-28T09:49:24.050" LastActivityDate="2010-08-28T09:49:24.050" CommunityOwnedDate="2010-08-28T08:00:29.070" CommentCount="0" />
  <row Id="3527" PostTypeId="2" ParentId="3525" CreationDate="2010-08-28T08:03:38.753" Score="1" ViewCount="" Body="&lt;p&gt;Hi, i have found a link which deals with this problem: &lt;a href=&quot;http://people.reed.edu/~jerry/311/cotan.pdf&quot; rel=&quot;nofollow&quot;&gt;people.reed.edu/~jerry/311/cotan.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" LastActivityDate="2010-08-28T08:03:38.753" CommentCount="2" />
  <row Id="3528" PostTypeId="1" AcceptedAnswerId="3540" CreationDate="2010-08-28T08:34:23.640" Score="5" ViewCount="421" Body="&lt;p&gt;How do I derive the Beta function using the definition of the beta function as the normalizing constant of the Beta distribution and only common sense random experiments?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm pretty sure this is possible, but can't see how.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can see that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sum_{a=0}^n {n \choose a} Beta(a+1, n-a+1) = 1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;because we can imagine that we are flipping a coin $n$ times.  The $2^n$ unique sequences of flips partition the probability space.  The Beta distribution with parameters $a$ and $n-a$ can be defined as the prior over the coin's bias probability $p$ given the observation of $a$ heads and $n-a$ tails.   Since there are ${n \choose a}$ such sequences for any $n$ and $a$, that explains the scaling factor, and we know that it all sums to unity since the sequences partition the probability space, which has total measure 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I can't figure out is why:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;${n \choose a} Beta(a+1, n-a+1) = \frac{1}{n+1} \qquad \forall n \ge 0,\quad a \in {0, \dots, n}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we knew that, we could easily see that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$Beta(a + 1,n - a + 1) = \frac{1}{(n+1){n \choose a}} = \frac{a!(n-a)!}{(n+1)!}$.&lt;/p&gt;&#xA;" OwnerUserId="774" LastEditorUserId="774" LastEditorDisplayName="" LastEditDate="2010-08-28T22:17:55.707" LastActivityDate="2011-01-20T18:35:31.860" Title="Beta function derivation" Tags="&lt;probability-theory&gt;&lt;statistics&gt;&lt;special-functions&gt;" AnswerCount="2" CommentCount="7" FavoriteCount="6" />
  <row Id="3529" PostTypeId="2" ParentId="1747" CreationDate="2010-08-28T08:48:11.187" Score="2" ViewCount="" Body="&lt;p&gt;This problem is a Bayesian probability problem.  You can never know whether or not the coin is fair because, as you point out yourself, even if the first n flips come up equally head and tails, you would merely have &quot;nothing to indicate [yet] that the coin is unfair&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can calculate the probability that the coin's bias is, say between 49% and 51%, by  evaluating the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$I_{0.51}(h+1,t+1) - I_{0.49}(h+1,t+1)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $I$ is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Beta_function&quot; rel=&quot;nofollow&quot;&gt;incomplete beta function&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This follows from the definition of the Beta distribution as the conjugate prior of the Bernoulli distribution and the definition of the incomplete beta function as its cdf.&lt;/p&gt;&#xA;" OwnerUserId="774" LastActivityDate="2010-08-28T08:48:11.187" />
  <row Id="3530" PostTypeId="2" ParentId="3504" CreationDate="2010-08-28T08:58:40.830" Score="2" ViewCount="" Body="&lt;p&gt;You problem statement is too vague.  I guess you mean all four polynomials and their pairwise sums are non-degenerate, otherwise we can simply pick any four straight lines whose slopes are nonzero and of different magnitudes.  There should be other restrictions as well, or else {x^2, 2x^2, 3x^2, 4x^2} will also be a solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The following transformation of the problem may help.  Let P_i(x) = a_ix^2 + b_ix + c_i.  If P_1+P_2 is a polynomial with a single root (actually, do you mean a double root?), it must be equal to some k_1(x-x_1)^2 with k_1 \not = 0 and x_1 \in \mathbb{R}.  Thus a_1+a_2 = k_1 and vice versa for other pairwise sums.  So we get the following matrix equation:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&#xA;[1 1 0 0] [a_1]   [k_1]&#xA;[1 0 1 0] [a_2]   [k_2]&#xA;[1 0 0 1] [a_3] = [k_3]&#xA;[0 1 1 0] [a_4]   [k_4]&#xA;[0 1 0 1]         [k_5]&#xA;[0 0 1 1]         [k_6].&#xA;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Call the leftmost matrix M.  Hence the vector u = (k_1, ..., k_6)^T must lie inside the the column space of M.  By inspecting the coefficients b_i and c_i, we see that v = (k_1x_1, ..., k_6x_6)^T and w = (k_1x_1^2, ..., k_6x_6^2)^T also lie inside this column space.  Now, if you can prove that this is impossible, you are done.&lt;/p&gt;&#xA;" OwnerUserId="1551" LastEditorUserId="1551" LastEditorDisplayName="" LastEditDate="2010-08-28T09:05:02.610" LastActivityDate="2010-08-28T09:05:02.610" />
  <row Id="3531" PostTypeId="1" AcceptedAnswerId="3557" CreationDate="2010-08-28T09:48:01.023" Score="6" ViewCount="334" Body="&lt;p&gt;Find the &lt;em&gt;equation&lt;/em&gt; of the ellipse circumscribing a right triangle whose lengths of it's sides are $3,4,5$ and such that its area is the minimum possible one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You may chose the origin and orientation of the $x,y$ axes as you want.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Motivation: It can be &lt;a href=&quot;http://www.math.purdue.edu/pow/fall2008/pdf/solution8.pdf&quot; rel=&quot;nofollow&quot;&gt;proved&lt;/a&gt; [&lt;em&gt;&lt;a href=&quot;http://www.math.purdue.edu/pow/fall2008/pdf/problem8.pdf&quot; rel=&quot;nofollow&quot;&gt;Problem of the Week, Problem No. 8 (Fall 2008 Series), Department of Mathematics, Purdue University&lt;/a&gt;&lt;/em&gt;] that the area of this ellipse is $8\pi /\sqrt{3}$, without the need of using its equation, but I am also interested in finding it.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Edit: picture from &lt;a href=&quot;http://math.stackexchange.com/questions/3531/minimal-ellipse-circumscribing-a-right-triangle/3588#3588&quot;&gt;this&lt;/a&gt; answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/EWwoy.png&quot; alt=&quot;alt text&quot;&gt; &lt;/p&gt;&#xA;" OwnerUserId="752" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2010-08-29T22:47:24.750" LastActivityDate="2010-08-29T22:47:24.750" CommunityOwnedDate="2010-08-29T22:47:24.750" Title="Minimal Ellipse Circumscribing A Right Triangle" Tags="&lt;geometry&gt;&lt;euclidean-geometry&gt;&lt;plane-curves&gt;&lt;conic-sections&gt;" AnswerCount="3" CommentCount="8" FavoriteCount="1" />
  <row Id="3532" PostTypeId="1" CreationDate="2010-08-28T10:25:24.843" Score="3" ViewCount="450" Body="&lt;p&gt;Mathematicians, and esp. number theorists, are used to working with big numbers. I have noted on several occasions that lots of people don't have a clear understanding of big numbers as far as the real world is concerned. I recall a &lt;a href=&quot;http://groups.google.com/group/sci.math/browse_thread/thread/5da7f8d21fa57d7d/29155a92aa828571?ie=utf-8&amp;amp;oe=utf-8&quot; rel=&quot;nofollow&quot;&gt;request&lt;/a&gt; for a list of all primes of less than 500 digits.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another example is homeopathic dilutions. I understand they use dilutions like 200C, which is 1 in $10^{400}$. An absurd number in view of the fact that the total number of particles in the universe is estimated (safe margin) to be less than a googol.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;How would you give people insight in big numbers?&lt;/strong&gt; I'm not talking about Skewes' Number or Graham's Number; for most practical purposes $10^{20}$ is equal to infinity.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;edit&lt;/strong&gt;&lt;br&gt;&#xA;To whoever voted me down: if you vote this down, please also tell me why. Thanks&lt;/p&gt;&#xA;" OwnerUserId="923" LastEditorUserId="923" LastEditorDisplayName="" LastEditDate="2010-08-28T12:28:55.447" LastActivityDate="2011-01-22T00:35:12.343" Title="How to explain Real Big Numbers?" Tags="&lt;teaching&gt;&lt;big-numbers&gt;" AnswerCount="2" CommentCount="6" />
  <row Id="3533" PostTypeId="1" CreationDate="2010-08-28T11:01:35.350" Score="1" ViewCount="143" Body="&lt;p&gt;how could i convert a non-uniform random variable distribution to a uniform distribution ? Is there a formula ? thanks .&lt;/p&gt;&#xA;" LastActivityDate="2010-08-28T11:51:53.550" Title="Converting non-uniform distribution to a uniform distribution" Tags="&lt;probability&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="3534" PostTypeId="1" AcceptedAnswerId="3538" CreationDate="2010-08-28T11:08:04.450" Score="3" ViewCount="97" Body="&lt;p&gt;Could you please explain how the math behind the first answer to this stack overflow questions works:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://stackoverflow.com/questions/3590308/testing-if-a-line-has-a-point-within-a-triangle&quot;&gt;http://stackoverflow.com/questions/3590308/testing-if-a-line-has-a-point-within-a-triangle&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="1554" LastActivityDate="2010-08-28T12:01:55.497" Title="Line triangle intersection." Tags="&lt;geometry&gt;" AnswerCount="1" />
  <row Id="3535" PostTypeId="1" AcceptedAnswerId="3536" CreationDate="2010-08-28T11:12:33.893" Score="0" ViewCount="385" Body="&lt;p&gt;Hi all, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that the product of two Gaussian functions is also Gaussian function. This is stated in &lt;a href=&quot;http://en.wikipedia.org/wiki/Gaussian_function&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;, but I might need to cite a classical (text)book/paper stating this result. A book containing the proof is welcome. Could you name such a book? &lt;/p&gt;&#xA;" OwnerUserId="1268" LastActivityDate="2010-08-28T11:37:34.607" CommunityOwnedDate="2010-08-28T11:12:33.893" Title="Texbook with proof that product of two Gaussian functions is also Gaussian" Tags="&lt;probability-theory&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" />
  <row Id="3536" PostTypeId="2" ParentId="3535" CreationDate="2010-08-28T11:24:08.583" Score="1" ViewCount="" Body="&lt;p&gt;The product of two normal (Gaussian) random variables is &lt;em&gt;not&lt;/em&gt; normal.  That would be convenient, but it's not true.  Here are a some related theorems that are true:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;The &lt;em&gt;sum&lt;/em&gt; of two normal random variables is normal.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The product of two &lt;em&gt;log normal&lt;/em&gt; random variables is log normal.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The ratio of two normal random variables is Cauchy.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;For more on distribution relationships, see this &lt;a href=&quot;http://www.johndcook.com/distribution_chart.html&quot; rel=&quot;nofollow&quot;&gt;chart&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you multiply two normal PDFs you get the joint PDF of a &lt;em&gt;multivariate&lt;/em&gt; normal, i.e. if f(x) is a normal PDF then g(x, y) defined by f(x)g(y) is a multivariate normal PDF, but in that case you're multiplying &lt;em&gt;densities&lt;/em&gt; and not &lt;em&gt;random variables&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update&lt;/strong&gt;: I looked back at your question and I see you said Gaussian &quot;function,&quot; rather than &quot;random variable&quot;, so perhaps you had in mind the PDF result I mentioned above. If so, just write down the definition of multivariate normal and there it is.&lt;/p&gt;&#xA;" OwnerUserId="247" LastEditorUserId="247" LastEditorDisplayName="" LastEditDate="2010-08-28T11:37:34.607" LastActivityDate="2010-08-28T11:37:34.607" CommunityOwnedDate="2010-08-28T11:24:08.583" />
  <row Id="3537" PostTypeId="2" ParentId="3533" CreationDate="2010-08-28T11:51:53.550" Score="1" ViewCount="" Body="&lt;p&gt;Use the &lt;a href=&quot;http://en.wikipedia.org/wiki/Probability_integral_transform&quot; rel=&quot;nofollow&quot;&gt;probability integral transform&lt;/a&gt;.&lt;/p&gt;&#xA;" LastActivityDate="2010-08-28T11:51:53.550" />
  <row Id="3538" PostTypeId="2" ParentId="3534" CreationDate="2010-08-28T12:01:55.497" Score="4" ViewCount="" Body="&lt;p&gt;Let's take the image from the anwer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/kKod9.png&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can think of the red line as a &lt;em&gt;two dimensional plane&lt;/em&gt;, where $4x − 3y + 2 = 0$ is it's coordinate equation with some normal vector&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\vec{n} = \begin{pmatrix} 4 \\ -3 \end{pmatrix}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;that is orthogonal to the line.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, by substituting a point into the equation, you get the distance of point and line along $\vec{n}$ (see. &lt;a href=&quot;http://en.wikipedia.org/wiki/Hesse_normal_form&quot; rel=&quot;nofollow&quot;&gt;Hesse's normal form&lt;/a&gt;). While we don't actually care for the distance itself, we do care for it's &lt;em&gt;sign&lt;/em&gt;. Positive sign indicates the point's lying on the side $\vec{n}$ points to, a negative one indicates the opposite side.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if all the values have the same sign, all triangle points lie on the same side of the line, i.e. there are no intersections. With different signs, the triangle must hit the line.&lt;/p&gt;&#xA;" OwnerUserId="917" LastActivityDate="2010-08-28T12:01:55.497" />
  <row Id="3539" PostTypeId="2" ParentId="3532" CreationDate="2010-08-28T12:23:54.560" Score="4" ViewCount="" Body="&lt;p&gt;Though I don't quite know that you actually want to hear - &lt;em&gt;what kind of numbers&lt;/em&gt; do you want to give people insight in, &lt;em&gt;whom&lt;/em&gt; and &lt;em&gt;why&lt;/em&gt;? - I'll give a few thoughts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I) &lt;strong&gt;Real cases&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's just understanding of natural sciences - numbers of particles in the universe, number of cells in a body ... Try to first of all break down the number by using smaller parts of the example - e.g. count bacteria in a drop of water and not in a whole lake.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;II) &lt;strong&gt;Thought experiments (explaining probabilities, complexity etc.)&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Extremely big numbers arise when you try to visualize probabilities or complexities, especially when exponential growth is involved. What about getting the jackpot ten times successively or trying to solve a TSP for 100 cities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you know people aren't comfortable with that big numbers, decide:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Is it really important to know the number? Maybe, extremely long or extremely improbable is just the important fact.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Can you find an easier to grasp example (special units)? Longer than the universe is old is better than &lt;em&gt;insert giant amount of milliseconds&lt;/em&gt;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Can you describe the growth differently? &lt;em&gt;If your problem with 999 cities can be solved in a certain amount of time and you take one additional city, you'll need &lt;strong&gt;1000&lt;/strong&gt; times longer&lt;/em&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;III) &lt;strong&gt;Data&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Especially in the context of CS / cryptography, numbers can often most accurately be explained as &lt;em&gt;some data you can calculate with&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;E.g. &lt;em&gt;RSA&lt;/em&gt; (as in your link) is of course a mathematical, number-based algorithm, but in fact, you're encrypting &lt;strong&gt;data&lt;/strong&gt;, so why not say &lt;em&gt;a 500 char key&lt;/em&gt; instead of explaining the giant number involved there. &lt;/p&gt;&#xA;" OwnerUserId="917" LastActivityDate="2010-08-28T12:23:54.560" CommentCount="3" />
  <row Id="3540" PostTypeId="2" ParentId="3528" CreationDate="2010-08-28T12:34:02.077" Score="10" ViewCount="" Body="&lt;p&gt;For non-negative integers $a, b$ and $t \in [0, 1]$, the expression $t^a (1 - t)^b$ describes the probability of randomly selecting $a+b$ real numbers in $[0, 1]$ such that the first $a$ are in $[0, t]$ and the last $b$ are in $[t, 1]$.  The integral $\int_0^{1} t^a (1 - t)^b$ then describes the probability of randomly selecting $a+b+1$ real numbers such that the first number is $t$, the next $a$ numbers are in $[0, t]$, and the next $b$ numbers are in $[t, 1]$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It follows that $ {a+b \choose b} \int_0^1 t^a (1 - t)^b$ describes the probability of randomly selecting $a+b+1$ real numbers such that the first number is $t$, some $a$ of the remaining numbers are in $[0, t]$, and some $b$ of the remaining numbers are in $[t, 1]$.  But this is the same as the probability that the first number happens to be $(a+1)^{st}$ in order, and this is just $\frac{1}{a+b+1}$.  Hence&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_0^1 t^a (1 - t)^b = \frac{a! b!}{(a+b+1)!}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;as desired.  I learned this proof through an exercise in a Putnam training seminar; the multidimensional generalization also works.&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-08-28T14:10:49.590" LastActivityDate="2010-08-28T14:10:49.590" CommentCount="1" />
  <row Id="3541" PostTypeId="1" AcceptedAnswerId="3546" CreationDate="2010-08-28T12:41:18.790" Score="3" ViewCount="191" Body="&lt;p&gt;Rotations in 3 dimensions are not commutative; however they are in the plane. In classical mechanics, are we allowed to say that angular momentum is a vector because particles only rotate along a single axis? Or do we make some kind of argument that you can assign angular velocity to an object because you can approximate it locally as being along a single rotation axis and so its &quot;locally commutative&quot; and hence appropriate to call it a vector? If that is the case what happens if we have a particle rotating in a way that isn't differentiable? Is there some physical reason that particles can't rotate in that sort of way?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm really confused by this and haven't taken physics since high school, any help would be appreciated. :)&lt;/p&gt;&#xA;" OwnerUserId="249" LastEditorUserId="249" LastEditorDisplayName="" LastEditDate="2010-08-28T13:21:50.010" LastActivityDate="2010-08-28T14:02:21.077" Title="How can angular velocity or angular momentum be a vector?" Tags="&lt;linear-algebra&gt;&lt;physics&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="3542" PostTypeId="2" ParentId="3506" CreationDate="2010-08-28T12:58:51.107" Score="1" ViewCount="" Body="&lt;p&gt;The important component of the question here seems to be: can we define a natural binary operation on the set P of partitions of $\{1,2,\ldots,n\}$ to form a group G?  &lt;a href=&quot;http://en.wikipedia.org/wiki/Cayley%27s_theorem&quot; rel=&quot;nofollow&quot;&gt;Cayley's Theorem&lt;/a&gt; tells us that if G exists, it is isomorphic to a subgroup of the symmetric group on G.  So G will be isomorphic to a group of permutation matrices.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, finding a &lt;em&gt;natural&lt;/em&gt; binary operation on P is going to be tricky.  Of course, P is just a set, and it would be possible to construct some highly-contrived binary operation on P, but typically it would not preserve the structure of the partitions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an off-the-top-of-my-head example of why I think it should be tricky: |P| is given by the &lt;a href=&quot;http://www.research.att.com/~njas/sequences/A000110&quot; rel=&quot;nofollow&quot;&gt;Bell Numbers&lt;/a&gt;, which can be a prime number, whence G must be a cyclic group and each non-identity element of G must somehow generate G.&lt;/p&gt;&#xA;" OwnerUserId="139" LastActivityDate="2010-08-28T12:58:51.107" />
  <row Id="3543" PostTypeId="1" AcceptedAnswerId="3544" CreationDate="2010-08-28T13:41:26.907" Score="1" ViewCount="287" Body="&lt;p&gt;One learns evaluating indefinite integrals of the form $$ \int \frac{f(x)}{p(x)\sqrt{q(x)}} \ dx $$ where $p(x)$ and $q(x)$ are linear polynomials. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How to evaluate these integrals if $p(x)$ and $q(x)$ are quadratic or cubic in nature. Is there any method for evaluating them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Examples of such integrals are: $\displaystyle \int \frac{1}{(x^{2}+a^{2})\sqrt{x^{2}+b^{2}}} \ dx$ or $\displaystyle \int\frac{1}{(x^{3}+a^{3})\sqrt{x^{3}+b^{3}}} \ dx$&lt;/p&gt;&#xA;" LastActivityDate="2010-08-31T17:10:32.993" Title="Evaluating Indefinite Integrals of the form $\frac{f(x)}{p(x) \sqrt{q(x)}}$" Tags="&lt;real-analysis&gt;&lt;analysis&gt;&lt;integral&gt;" AnswerCount="4" FavoriteCount="2" />
  <row Id="3544" PostTypeId="2" ParentId="3543" CreationDate="2010-08-28T13:46:50.493" Score="5" ViewCount="" Body="&lt;p&gt;If $q$ is quadratic, then these can be done by trigonometric/hyperbolic&#xA;substiuitions. If $q$ is cubic then they are are elliptic integrals,&#xA;not in general expressible in elementary terms, but expressible in terms&#xA;of elementary functions and the standard examples of elliptic integrals.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-08-28T13:46:50.493" CommentCount="1" />
  <row Id="3545" PostTypeId="2" ParentId="3543" CreationDate="2010-08-28T13:51:37.997" Score="2" ViewCount="" Body="&lt;p&gt;For the quadratic, first put $x = a \tan \theta$ then put $t = \sin \theta$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(First write $p(t)$ in the form $x^2 + a^2$ where $x = t+r$, $a$ could possibly be complex)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Don't know about the cubic.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-08-28T13:56:38.577" LastActivityDate="2010-08-28T13:56:38.577" CommentCount="2" />
  <row Id="3546" PostTypeId="2" ParentId="3541" CreationDate="2010-08-28T14:02:21.077" Score="6" ViewCount="" Body="&lt;p&gt;There are a couple of things at play here, which coincidentally(?) mirror the difference between mathematicians who say &quot;vector&quot; to mean anything can be added commutatively and multiplied by a scalar, and physicists who say &quot;vector&quot; to mean a 3- (or 4-)dimensional quantity that transforms properly under change of coordinates.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;While arbitrary rotations in 3 dimensions do not commute, &lt;em&gt;infinitesimal&lt;/em&gt; rotations do. (In fact any &quot;infinitesimal&quot; transformations commute, as you can see by multiplying $I + \epsilon A$ and $I + \epsilon B$ and ignoring second-order terms in $\epsilon$.) Since angular velocity can be thought of as &quot;infinitesimal rotation per infinitesimal time&quot;, it ends up being a vector (in the mathematical sense) even though rotation itself isn't. Similarly, angular momentum is the derivative of kinetic energy with respect to angular velocity, so it is a gradient, which is a (dual) vector (to angular velocity).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;There's an interesting mapping between these infinitesimal rotation matrices and vectors (in the physical sense) which only works in 3 dimensions. If you think of an infinitesimal rotation as $I + \epsilon A$, &lt;a href=&quot;http://mathworld.wolfram.com/InfinitesimalRotation.html&quot; rel=&quot;nofollow&quot;&gt;one can show that $A$ must be antisymmetric&lt;/a&gt;. This means its diagonal entries are 0, leaving only 3 degrees of freedom in the off-diagonal entries. Such a matrix can be associated with a vector using the cross product:&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;$$\begin{bmatrix}0 &amp;amp; -\omega_z &amp;amp; \omega_y \\ \omega_z &amp;amp; 0 &amp;amp; -\omega_x \\ -\omega_y &amp;amp; \omega_x &amp;amp; 0\end{bmatrix} \mathbf{r} = \mathbf{\omega} \times \mathbf{r}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So angular velocity isn't &quot;naturally&quot; a vector in physical space, but rather lives in a different 3-dimensional vector space. That's why once you interpret it as a physical vector, &lt;a href=&quot;http://en.wikipedia.org/wiki/Pseudovector&quot; rel=&quot;nofollow&quot;&gt;it turns out to transform slightly differently&lt;/a&gt;, as J. Mangaldan has pointed out in the comments.&lt;/p&gt;&#xA;" OwnerUserId="856" LastActivityDate="2010-08-28T14:02:21.077" CommentCount="4" />
  <row Id="3547" PostTypeId="1" AcceptedAnswerId="3548" CreationDate="2010-08-28T14:16:04.017" Score="2" ViewCount="287" Body="&lt;p&gt;Wikipedia has separate pages for symmetric group and permutation group, but I don't understand what the difference between them is. A symmetric group on a set is the set of all bijections from the set to itself with composition of functions as the group action. Permutation group on a set is the set of all permutations of elements on the set.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Aren't these two things the same thing?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On one of the discussion pages, someone suggested that permutation groups don't have to include all permutations: they just have to be collections of permutations on the set, closed under composition etc. But this seems weird. First, that's not how I was taught it, and second (thanks to Cayley's theorem) it looks like it's redundant: all groups are &quot;permutation groups&quot; on this reading.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there some subtle difference I'm missing?&lt;/p&gt;&#xA;" OwnerUserId="172" LastActivityDate="2010-08-28T19:47:14.960" Title="Permutation groups and symmetric groups" Tags="&lt;terminology&gt;&lt;group-theory&gt;" AnswerCount="2" />
  <row Id="3548" PostTypeId="2" ParentId="3547" CreationDate="2010-08-28T14:27:27.347" Score="6" ViewCount="" Body="&lt;p&gt;First, on the permutation group page, there's this line &quot;the term permutation group is usually restricted to mean a subgroup of the symmetric group.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second, Cayley's theorem doesn't really make the terminology &quot;permutation group&quot; redundant. When you talk about a permutation group, I think you are implicitly giving an action of the group on a set of some objects. This is an extra data other than the group structure.&lt;/p&gt;&#xA;" OwnerUserId="325" LastActivityDate="2010-08-28T14:27:27.347" CommentCount="1" />
  <row Id="3549" PostTypeId="2" ParentId="3101" CreationDate="2010-08-28T14:38:22.457" Score="2" ViewCount="" Body="&lt;p&gt;As per AndrejaKo's comment, I am posting my comment as an answer. Note that there are very interesting and useful answers to this question motivating the formula for the derivative of a product: my answer is simply a technical explanation of a typo in the argument provided in the question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You should add and subtract the quantity $f(x)g(x+h)$ from the left hand side of your equation (and the equality you wrote is indeed incorrect). Once you collect terms appropriately, everything should be clear!&lt;/p&gt;&#xA;" OwnerUserId="1028" LastActivityDate="2010-08-28T14:38:22.457" />
  <row Id="3550" PostTypeId="1" AcceptedAnswerId="3571" CreationDate="2010-08-28T14:51:53.123" Score="8" ViewCount="722" Body="&lt;p&gt;In my attempts to understand just how computer algebra systems &quot;do things&quot;, I tried to dig around a bit on Gröbner bases, which are described almost everywhere as &quot;a generalization of the Euclidean algorithm and Gaussian elimination&quot;. I've tried to look for examples of Gröbner bases in action, but have been unable to find any (that can be easily understood).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I could go ahead and just ask for an explanation with examples from people, but I'll go one step further.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;General plane conics can be represented by the Cartesian equation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ax^2+2bxy+cy^2+dx+fy+g=0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One common problem in dealing with conics is finding out if two conics intersect, and if so, where the intersection point(s) are. Usually one can do all this by eliminating variables accordingly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I ask: how would, say, Buchberger's method, proceed on determining if two given conics intersect, and then find where they intersect?&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-12-13T12:24:07.137" Title="Using Gröbner bases for solving polynomial equations." Tags="&lt;algebra&gt;&lt;polynomials&gt;" AnswerCount="5" CommentCount="1" FavoriteCount="6" />
  <row Id="3551" PostTypeId="2" ParentId="3532" CreationDate="2010-08-28T15:06:38.940" Score="5" ViewCount="" Body="&lt;p&gt;Very few (if any) mathematicians have significant insight regarding huge natural numbers (cf. various &lt;a href=&quot;http://en.wikipedia.org/wiki/Ultrafinitism&quot; rel=&quot;nofollow&quot;&gt;ultrafinitism&lt;/a&gt; arguments).  Perhaps the only exceptions are logicians who work with esoteric ordinal notations. This is one of the few ways one can gain any insight into arbitrarily large numbers - using various complicated inductions to show that some property holds for all naturals - thus lifting our intuition up from small naturals to arbitrarily large naturals. For example, see the &lt;a href=&quot;http://groups.google.com/groups?selm=WGD.95Dec11023450@martigny.ai.mit.edu&quot; rel=&quot;nofollow&quot;&gt;Goodstein sequence&lt;/a&gt; (or, more graphically: the Hercules vs. Hydra game) which encodes the ordinals below $\epsilon_0 = \omega^{\omega^{\omega^{\cdot^{\cdot^\cdot}}}} \;$ into huge natural numbers.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2011-01-22T00:35:12.343" LastActivityDate="2011-01-22T00:35:12.343" CommentCount="1" />
  <row Id="3552" PostTypeId="2" ParentId="3550" CreationDate="2010-08-28T16:09:52.510" Score="6" ViewCount="" Body="&lt;p&gt;You &lt;em&gt;have&lt;/em&gt; ro read through the book[David A. Cox, John B. Little and Don O'Shea, &lt;em&gt;Ideals, Varieties, Algorithms&lt;/em&gt;].&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In any case, if you start with a system of polynomial equations and compute a Groebner basis for the ideal they generate, you get a &quot;maximally triangular&quot; system of equations which is equivalent to the original one---that is why Groebner bases generalize Gaussian elimination.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me do a simple example using &lt;a href=&quot;http://www.math.uiuc.edu/Macaulay2/&quot; rel=&quot;nofollow&quot;&gt;Macaulay 2&lt;/a&gt;. Consider the ring $\mathbb Q[x,y,z]$:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;i1 : R = QQ[x, y, z, MonomialOrder =&amp;gt; Lex]&#xA;&#xA;o1 = R&#xA;&#xA;o1 : PolynomialRing&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;the Fermat quintic surface $x^5+y^5+z^5=1$, whose ideal is&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;i2 : surface = ideal (x^5 + y^5 + z^5 - 1)&#xA;&#xA;             5    5    5&#xA;o2 = ideal(x  + y  + z  - 1)&#xA;&#xA;o2 : Ideal of R&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;and the curve $x^2=y^2$, $y^3+1=z^3$:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;i3 : curve = ideal (x^2 - y^2, z^3 - y^3 - 1)&#xA;&#xA;             2    2     3    3&#xA;o3 = ideal (x  - y , - y  + z  - 1)&#xA;&#xA;o3 : Ideal of R&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The ideal of the intersection of the surface and the curve is the sum of the ideals of the intersecands:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;i4 : intersection = curve + surface&#xA;&#xA;              2    2     3    3       5    5    5&#xA;o4 = ideal (x  - y , - y  + z  - 1, x  + y  + z  - 1)&#xA;&#xA;o4 : Ideal of R&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The number of points on the intersection, counting multiplicities, is the degree of the ideal:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;i5 : degree intersection&#xA;&#xA;o5 = 30&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;if we now compute the degree of the radical of the ideal, we get a lower number, so not all the points are simple:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;i6 : degree radical intersection&#xA;&#xA;o6 = 25&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Now look at the lexicographic Groebner basis of the ideal of the intersection:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;i7 : ideal gens gb intersection&#xA;&#xA;              17      16      15      14      13      12       11       10      9      8       7       6      5      4      3  &#xA;o7 = ideal (9z   + 27z   + 54z   + 50z   + 15z   - 63z   - 104z   - 108z   - 35z  + 35z  + 108z  + 104z  + 63z  - 15z  - 50z  -&#xA;     ----------------------------------------------------------------------------------------------------------------------------&#xA;        2                 5            16      15      14      13      12       11      10      9       8      7       6       5&#xA;     54z  - 27z - 9, 20y*z  - 20y + 27z   + 18z   + 45z   - 75z   - 35z   - 164z   + 79z   - 10z  + 230z  - 10z  + 119z  - 164z &#xA;     ----------------------------------------------------------------------------------------------------------------------------&#xA;          4       3      2              3    3                                       16        15        14        13        12  &#xA;     - 35z  - 155z  + 45z  + 18z + 67, y  - z  + 1, 50x*z - 50x + 50y*z - 50y + 1242z   + 2718z   + 4770z   + 2220z   - 1235z   -&#xA;     ----------------------------------------------------------------------------------------------------------------------------&#xA;          11        10        9        8        7        6        5        4        3        2                 2    2&#xA;     7979z   - 7481z   - 6010z  + 1810z  + 5240z  + 9394z  + 5346z  + 1240z  - 4030z  - 4005z  - 2657z - 583, x  - y )&#xA;&#xA;o7 : Ideal of R&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The first generator depends only on $z$. The second one and the third, only on $z$ and $y$, and the fourth depends (linearly) on $x$ too. This is a &quot;triangular&quot; system of equations, from which you can compute the 30 points of intersection, assuming you get to solve the first equation for $z$, which is of degree 17...&lt;/p&gt;&#xA;" OwnerUserId="274" LastEditorUserId="274" LastEditorDisplayName="" LastEditDate="2010-12-13T12:24:07.137" LastActivityDate="2010-12-13T12:24:07.137" CommentCount="6" />
  <row Id="3554" PostTypeId="2" ParentId="3497" CreationDate="2010-08-28T16:16:55.583" Score="1" ViewCount="" Body="&lt;p&gt;To expand on Sam's awesome comment:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can write $K$ as $F(\alpha)$, where the minimal polynomial of $\alpha$ is $f(x)$.  Then we have a map from $L[x]$ to $K\otimes_F L$ given by&#xA;$$ a_0+a_1x+\cdots+a_nx^n\mapsto 1\otimes a_0 + \alpha\otimes a_1+\cdots + \alpha^n\otimes a_n.$$&#xA;This induces an isomorphism between $L[x]/fL[x]$ and $K\otimes_F L$. Composing with the map I've given above, we then want to show the map&#xA;$$ a(x)\mapsto (a(\alpha^{\gamma_1}),\ldots,a(\alpha^{\gamma_n}))$$&#xA;from $L[x]$ to $\prod_{i=1}^n C$, is surjective.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Set &#xA;$$ f_i=\prod_{n\in N} (x-\alpha^{\gamma_in});$$&#xA;this is a polynomial in $L[x]$; now let&#xA;$$ g_i=\prod_{j\neq i} f_j; $$&#xA;then $g_i$ maps to $(0,\ldots,g_i(\alpha^{\gamma_i}),0,\ldots)$.&#xA;Since $g_i(\alpha^{\gamma_i})$ is a non-zero element of $K$, we can write $g_i(\alpha^{\gamma_i})^{-1}=h_i(\alpha^{\gamma_i})$, and so $g_ih_i$ maps to $(0,\ldots,1,\ldots)$, where the $1$ is in the $i$th place. Since $C=L(\alpha)$, this map is surjective.&lt;/p&gt;&#xA;" OwnerUserId="1544" LastActivityDate="2010-08-28T16:16:55.583" />
  <row Id="3556" PostTypeId="2" ParentId="3550" CreationDate="2010-08-28T16:21:41.727" Score="2" ViewCount="" Body="&lt;p&gt;You might find Lecture 3 of James Carlson's CIMAT Lectures of value as well as other material on this page:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.math.utah.edu/~carlson/cimat/&quot; rel=&quot;nofollow&quot;&gt;http://www.math.utah.edu/~carlson/cimat/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1369" LastActivityDate="2010-08-28T16:21:41.727" CommentCount="1" />
  <row Id="3557" PostTypeId="2" ParentId="3531" CreationDate="2010-08-28T16:36:42.380" Score="5" ViewCount="" Body="&lt;p&gt;If you look at the proof that you linked, the proof is using an Affine transformation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now you can also see that the centroid of the triangles are mapped to each other in the affine transformation, and so do the centers of the ellipses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus &lt;code&gt;the ellipse you need has the same center as your triangle!&lt;/code&gt; This property uniquely defines the ellipse with the maximum area.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now consider the points $A = (-1,-4/3), B = (2, -4/3)$ and $C= (-1, 8/3)$. This is a 3-4-5 triangle whose center is origin which was gotten by starting with $(0,0), (3,0)$ and $(0,4)$ and translating so that the centroid is the origin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the equation of an ellipse whose center is the origin is given by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$Px^2 + Qxy + Ry^2 = 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus we must have that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) $P + 4Q/3 + 16R/9 = 1$ &lt;br/&gt;&#xA;(2) $4P - 8Q/3 + 16R/9 = 1$ &lt;br/&gt;&#xA;(3) $P -8Q/3 + 64R/9 = 1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solving these (see footnote) gives us the equation of the ellipse as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x^{2}/3 + xy/4 + 3y^{2}/16 = 1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to verify this, the area of $Px^2 + Qxy + Ry^2 = 1$ is given by $\displaystyle \frac{2\pi}{\sqrt{4PR - Q^2}}$ which comes out to $\displaystyle \frac{8\pi}{\sqrt 3}$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;You can ignore the below if you like. This is just manually solving the equations&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) $P + 4Q/3 + 16R/9 = 1$ &lt;br/&gt;&#xA;(2) $4P - 8Q/3 + 16R/9 = 1$ &lt;br/&gt;&#xA;(3) $P -8Q/3 + 64R/9 = 1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Subtracting (1) and (2) gives $3P = 4Q$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Subtracting (2) and (3) gives $3P = 48R/9$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus $Q = 3P/4$ and $R = 9P/16$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus using (3) we have that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$P - (8/3)*(3P/4) + 64/9 * (9P/16) = 1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i.e&#xA;$P - 2P + 4P = 1$ i.e $P = 1/3$.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-08-28T16:57:14.703" LastActivityDate="2010-08-28T16:57:14.703" CommentCount="7" />
  <row Id="3558" PostTypeId="2" ParentId="3444" CreationDate="2010-08-28T16:42:38.230" Score="4" ViewCount="" Body="&lt;p&gt;The geometric approach works.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let’s compute the volume of the $2n$ dimensional ball, $D^{2n}$, in two ways.  One way is extremely clever but has been known for centuries and provides interesting insights: it’s based on Liouville’s trick.  Specifically, we will compute two integrals in polar coordinates, one of which is the volume of the ball and the other of which reduces to a product of one-dimensional integrals.  Both integrands will depend (at most) on the radial coordinate $r$, which lets us separate out  the surface area of the boundary of the ball as a common factor.  Write this surface area as $S_{2n-1}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There’s essentially just one way to do this trick: integrate $exp(-r^2)$.  Its integral over $\mathbb{R}^{2n}$ equals&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$S_{2n-1} \int_0^\infty {exp\left(- r^2 \right) r^{2n-1} dr}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, because $r^2 = x_1^2 + x_2^2 + \ldots + x_{2n}^2$, the integrand (in &lt;em&gt;Cartesian&lt;/em&gt; coordinates $\left( x_1, x_2, \ldots, x_{2n} \right)$) factors as $exp\left(-r^2 \right) = exp\left(-x_1^2 \right) \cdots exp\left(-x_{2n}^2 \right)$, each of which must be integrated from $-\infty$ to $+\infty$.  Whence&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$S_{2n-1} \int_0^\infty {exp \left(- r^2 \right) r^{2n-1} dr} = \left( \int_{- \infty}^ \infty {exp \left( -x^2 \right) dx} \right) ^{2n}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I will call the left hand integral $\tfrac{1}{2} \Gamma \left(n \right)$, because that is what it is (as a simple change of variables shows).  In the same notation,  $\Gamma \left(1/2 \right) =  \int_{-\infty}^\infty {exp\left(-x^2 \right) dx}$.  Algebraic re-arrangement of the foregoing yields the volume of $D^{2n}$ as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$|D^{2n} | = S_{2n - 1} \int_0^1 {r^{2n - 1} dr}  = \frac{{S_{2n - 1} }}&#xA;{{2n}} = \frac {\Gamma \left(1/2 \right)^{2n}} { n \Gamma \left(n \right) }$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That was the first method: the result is a familiar one, but has been left expressed in a way that better reveals its origins in polar and Cartesian integration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The next way to compute the ball's volume is, I believe, new.  It is inspired by Cavalieri’s Principle: the idea that you can shift slices of a solid around without changing the volume of the solid.  The generalization is to move two-dimensional slices around &lt;em&gt;and to change their shapes while you do so,&lt;/em&gt; but in a way that does not change their areas.  It follows that the new solid has the same (hypervolume) as the original, although it might have a completely different shape.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We will compute the volume of a region $Q_n$ in $\mathbb{R}^{2n}$.  It is conveniently described by identifying $\mathbb{R}^{2n}$ with  $\mathbb{C}^{n}$, using coordinates $z_i = \left( x_{2i - 1}, x_{2i} \right)$, in terms of which&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$Q_n  = { \mathbf{z} \in \mathbb{C}^n :0 \leq \left| {z_1 } \right| \leq \left| {z_2 } \right| \leq  \cdots  \leq \left| {z_n } \right| \leq 1 }$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If these were &lt;em&gt;real&lt;/em&gt; variables, we could make the volume-preserving transformation $w_1 = z_1, w_2 = z_2 – z_1, \ldots , w_i = z_i – z_{i-1}, \ldots, w_n = z_n – z_{n-1}$, with the sole restriction that the sum of the $w_i$ (all of which are nonnegative) not exceed 1.  Because they are &lt;em&gt;complex&lt;/em&gt; variables, though, we have to consider the area of an annulus bounded by $z_{i-1}$ and $z_i$: it is proportional to $z_i^2 – z_{i – 1}^2$.  The circle of the same area has radius $w_i$ for which $w_i^2 = z_i^2 – z_{i – 1}^2$.  Therefore, &lt;em&gt;if we define new variables&lt;/em&gt; $w_i$ &lt;em&gt;according to this formula&lt;/em&gt;, we obtain a new region- - one of substantially different shape- - having the same volume.  This region is defined by $\left| {w_1 }^2 \right| + \cdots + \left| {w_n }^2 \right| \le 1$: that is, it’s our old friend $D^{2n}$.  Therefore, &lt;strong&gt;the volume of&lt;/strong&gt; $Q_n$ &lt;strong&gt;equals the volume of&lt;/strong&gt; $D^{2n}$ .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now for the punch line: $Q_n$ is a fundamental domain for the action of $S[n]$, the symmetric group, on the product of $n$ disks $T^{2n} = \left( D^2 \right) ^n$; $S[n]$ acts by permuting the Complex coordinates $z_1, \ldots, z_n$.  The volume of $T^{2n}$ equals $|D^2|^n = \pi ^n$.  Writing $|S[n]|$ for the number of permutations and equating our two completely different calculations of the volume of the $2n$ ball gives&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\pi ^ n / |S[n]| = \frac {\Gamma \left(1/2 \right)^{2n}} { n \Gamma \left(n \right) }$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;whence&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$|S[n]| = \frac{{\pi ^n n\Gamma \left( n \right)}}{{\Gamma \left( {1/2} \right)^{2n} }}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This simplifies: the volume formula for $n = 2$ must give the area of the unit circle, equal to $\pi$, whence $\Gamma \left( 1/2 \right)^2 = \pi$.  Finally, then,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$|S[n]| = n\Gamma \left( n \right)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I will finish by remarking that Liouville's method is a perfectly natural thing to encounter when working with the multivariate Normal distribution, so it's not really an isolated trick, but is rather a pretty basic result expressing a defining property of Normal (Gaussian) variates.  There are, of course, many other ways to compute the volume of $D^{2n}$, but this one gives us the Gamma function directly.&lt;/p&gt;&#xA;" OwnerUserId="1489" LastEditorUserId="1489" LastEditorDisplayName="" LastEditDate="2010-08-28T16:50:07.527" LastActivityDate="2010-08-28T16:50:07.527" CommentCount="3" />
  <row Id="3559" PostTypeId="2" ParentId="3531" CreationDate="2010-08-28T16:49:25.223" Score="3" ViewCount="" Body="&lt;p&gt;Given the proof to which you linked, as Rahul Narain said in a comment, there is an affine transformation that maps the unit circle and an inscribed equilateral triangle to the desired ellipse for a particular 3-4-5 right triangle.  Represent this affine transformation by&#xA;$$\begin{pmatrix}x'\\y'\\1\end{pmatrix}=\begin{pmatrix}a_{1,1} &amp;amp; a_{1,2} &amp;amp; b_1\\a_{2,1} &amp;amp; a_{2,2} &amp;amp; b_2\\0 &amp;amp; 0 &amp;amp; 1\end{pmatrix}\begin{pmatrix}x\\y\\1\end{pmatrix}.$$&#xA;Let the 3-4-5 right triangle have vertices (0,0), (0,3), and (4,0), and let the equilateral triangle have vertices (1,0), $(-\frac{1}{2},\frac{\sqrt{3}}{2})$, and $(-\frac{1}{2},-\frac{\sqrt{3}}{2})$.  Mapping these vertices in the order listed and solving for the parameters of the transformation gives the matrix&#xA;$$\begin{pmatrix}-\frac{4}{3} &amp;amp; -\frac{4}{\sqrt{3}} &amp;amp; \frac{4}{3}\\-1 &amp;amp; \sqrt{3} &amp;amp; 1\\0 &amp;amp; 0 &amp;amp; 1\end{pmatrix}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The unit circle has parametric representation $x=\cos t$, $y=\sin t$.  Applying the affine transformation to this unit circle gives the parametric representation $x=\frac{4}{3}(1-\cos t-\sqrt{3}\sin t)$, $y=1-\cos t+\sqrt{3}\sin t$ for the desired ellipse.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;edit&lt;/em&gt;: Looking at &lt;a href=&quot;http://math.stackexchange.com/questions/3531/minimal-ellipse-circumscribing-a-right-triangle/3557#3557&quot;&gt;Moron's answer&lt;/a&gt;, his choice of x and y coordinates are swapped relative to mine, and as he said, his right triangle is translated to have its centroid at the origin, but accounting for these differences, my parametric equations satisfy his equation.  That is, based on his equation, an equation in x and y alone for the ellipse with parametric equations as I specified is $9x^2+12xy+16y^2-36x-48y=0$.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;edit 2&lt;/em&gt;: I don't know too much of the detailed mechanics of the different representations of affine transformations.  The form that I used above is equivalent to&#xA;$$\begin{pmatrix}x'\\y'\end{pmatrix}=\begin{pmatrix}a_{1,1} &amp;amp; a_{1,2}\\a_{2,1} &amp;amp; a_{2,2}\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}+\begin{pmatrix}b_1\\b_2\end{pmatrix}$$&#xA;and the 1s and 0s in the third row in the earlier version are just the glue that allows an affine transformation to be represented with a single matrix multiplication.  My solution could have just as easily used this representation of the affine transformation.&lt;/p&gt;&#xA;" OwnerUserId="72" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-08-28T23:04:28.197" LastActivityDate="2010-08-28T23:04:28.197" CommentCount="7" />
  <row Id="3560" PostTypeId="2" ParentId="3250" CreationDate="2010-08-28T17:10:05.323" Score="0" ViewCount="" Body="&lt;p&gt;Perhaps these thoughts will shed some light on the discussion here. If one writes down a collection of axioms for Euclidean geometry, for example, Hilbert's axioms, the Pythagorean Theorem and discussion of a distance function do not explicitly appear. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.beva.org/math323/asgn7/dec12b.htm#hilb&quot; rel=&quot;nofollow&quot;&gt;http://www.beva.org/math323/asgn7/dec12b.htm#hilb&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A model of Euclidean geometry consists of taking points as (x,y), ordered pairs of real numbers and lines as linear equations ax + by + c = 0, (a, b, c real numbers) using the Euclidean distance function as the way to find the distance between pairs of points. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if one takes points as (x, y) (ordered pairs of real numbers) and lines as linear equations ax + by + c = 0 but uses |a-b| + |c- d| as the distance between the two points (a,b) and (c,d) and measures angles as one does in the Euclidean plane one gets a geometry, often called the Taxicab Plane, which obeys all of the axioms of the Euclidean Plane except for the axiom dealing with the congruence of triangles. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Details can be found in Eugene Krause's book Taxicab Geometry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Pyhtagorean Theorem does not hold here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For interesting subtleties, and lots about the taxicab plane, see Dawson's article available at this site:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://taxicabgeometry.net/research.html&quot; rel=&quot;nofollow&quot;&gt;http://taxicabgeometry.net/research.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1369" LastActivityDate="2010-08-28T17:10:05.323" />
  <row Id="3561" PostTypeId="1" CreationDate="2010-08-28T17:55:36.447" Score="3" ViewCount="285" Body="&lt;p&gt;Hello,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am going to be taking a first course in abstract algebra this fall and we are using Jacobson's Basic Algebra I for the course text. It looks like a good textbook, but it is lacking a lot of motivation (I am new to higher math), so I was wondering if there are other books to supplement my studies. Particularly historical or geometric motivations; Something to give me an idea about why we are learning about ideals, modules, etc. so I can &quot;picture&quot; what we are learning.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for any input!&lt;/p&gt;&#xA;" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-09-03T05:26:44.333" LastActivityDate="2010-09-03T05:26:44.333" CommunityOwnedDate="2010-08-28T17:55:53.997" Title="Supplement for Jacobson's Basic Algebra I" Tags="&lt;soft-question&gt;&lt;abstract-algebra&gt;&lt;math-history&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="3562" PostTypeId="2" ParentId="3550" CreationDate="2010-08-28T18:19:23.343" Score="4" ViewCount="" Body="&lt;p&gt;The example that you propose is a bit too general to serve as an introductory pedagogical example (compare e.g. the famous &lt;a href=&quot;http://portal.acm.org/citation.cfm?id=1088244&quot; rel=&quot;nofollow&quot;&gt;Kahan SIGSAM problem 9&lt;/a&gt; of determining conditions for an ellipse to lie inside the unit circle). Instead, begin with one of the earliest historical examples: Gauss's proof that every symmetric polynomial can be written uniquely as a polynomial in the elementary symmetric polynomials. This is one of the first and the simplest applications of rewriting reduction via a lexicographic order. For a nice presentation see Chapter 7 of Cox, Little, O'Shea: Ideals, Varieties and Algorithms. They also present generalizations to  the ring of invariants of a finite matrix group $\rm G \subset  GL(n,k)$. You can find online copies of said book via various ebook databases, e.g. &lt;a href=&quot;http://gigapedia.com/&quot; rel=&quot;nofollow&quot;&gt;gigapedia&lt;/a&gt; - which has links to over 20000 mathematics books.&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-08-28T18:19:23.343" CommentCount="1" />
  <row Id="3563" PostTypeId="2" ParentId="3550" CreationDate="2010-08-28T18:21:54.240" Score="2" ViewCount="" Body="&lt;p&gt;This is a simplistic version of the Grobner basis computation and complements Mariano's answer.  The idea behind the Buchberger algorithm is to build a basis so that you are to be able to do &quot;unique division&quot; between nonlinear polynomails.  This is done by computing (what is called the) S-polynomial between any 2 given polynomials until there are no more produced.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Decide some ordering (grlex, lex, revlex)&#xA;Initialize I = &amp;lt;f1, f2&amp;gt; &#xA;where we use the polynomials you gave above.&#xA;f_1=ax^2+2bxy+cy^2+dx+fy+g&#xA;f_2=ax^2+2bxy+cy^2+dx+fy+g&#xA;Set n=2&#xA;&#xA;Repeat the following in a loop&#xA;    Iteratively choose any two polynomials ( f_i, f_j ) from set I &#xA;    n = n + 1&#xA;    Compute f[n] = S-polynomial( f_i, f_j )&#xA;Until all S-polynomials return 0&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;At the end of the computation, you will have a set I = {f1,f2, ..., fk} such that one of the polynomials will be in only one term (say 'x').  Solve that equation to get the one coordinate of the intersection, and then work upwards through the rest of the polynomials.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The general question is related to Elimination theory, and AFAIK there are three basic methods ( I can't speak of their complexity )&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Grobner basis, as seen above&lt;/li&gt;&#xA;&lt;li&gt;Use Resultants  (example below)&lt;/li&gt;&#xA;&lt;li&gt;Wu's method of characteristic sets, which I can't say much about.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;If you wanted to use resultants to find the intersection of say &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;E={f1(x, y,z), f2(x, y, z), f3(x, y, z)},&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;you would first compute &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;E1 = {g1 = Resultant_in_x (f1, f2), g2 = Resultant_in_x(f2, f3)} &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;and then &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;E2 = { h1 = Resultant_in_y(g1, g2} }&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The last polynomial h1 would be in one variable and amenable to solution.&lt;/p&gt;&#xA;" OwnerUserId="1558" LastActivityDate="2010-08-28T18:21:54.240" CommentCount="4" />
  <row Id="3564" PostTypeId="2" ParentId="3434" CreationDate="2010-08-28T18:42:23.670" Score="2" ViewCount="" Body="&lt;p&gt;No. Your defined objective is the Hausdorff distance $d_H$ between a surface and the tetrahedron -- at least, it is as long as the surface is inside the tetrahedron. For the chosen surface, the distance is $d_H(E_1,E_2) = \frac{1}{2}$, attained at $p = \left(\frac{1}{2},\frac{1}{2},0\right), q = \left(\frac{1}{2},\frac{1}{2},\frac{1}{2}\right)$. You can have $d_H(E_1,E_3) \le \frac{1}{2}$ with the area of $E_3$ arbitrarily small by letting $E_3$ be a small sphere at $\left(\frac{1}{2},\frac{1}{2},\frac{1}{2}\right)$ with lots of very thin &quot;spikes&quot; reaching the boundary of the tetrahedron.&lt;/p&gt;&#xA;" OwnerUserId="856" LastActivityDate="2010-08-28T18:42:23.670" />
  <row Id="3565" PostTypeId="1" AcceptedAnswerId="3569" CreationDate="2010-08-28T18:44:25.117" Score="10" ViewCount="139" Body="&lt;p&gt;&lt;strong&gt;Question&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given some natural number, we can of course split it up into various sums of other naturals (e.g. $7 = 6 + 1 = 1 + 4 + 2 = \ldots$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More precisely, for $n \in \mathbb{N}$, we can a find distribution sequence (or &lt;em&gt;partition&lt;/em&gt;) $s_0,\ldots,s_u \in \mathbb{N}$ with&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{i=0}^{u}s_i = n$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now how can I find the partition $s$ for which the overall least common multiple of all elements is maximal.&#xA;Or differently formulated the maximum product of distinct prime factors of all elements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Example:&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ 7 \mapsto (3, 4); \Pi_{lcm} = 12 $$&#xA;$$ 7 \mapsto (1, 2, 4); \Pi_{lcm} = 4$$ &#xA;$$ \ldots $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here, the first solution is the desired one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Background:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The background of my question is: How can I split up a sequence/string into contiguous subsequences that, when repeated and &lt;a href=&quot;http://en.wikipedia.org/wiki/Zip_%28higher-order_function%29&quot; rel=&quot;nofollow&quot;&gt;zipped together&lt;/a&gt;, will yield the longest possible resulting sequence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the above example, I could split up a string of length $7$ in the following way:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;7: abcdefg 7:  abcdefg&#xA;&#xA;    I              II&#xA;1: aaaa    3: abcabcabcabc&#xA;2: bcbc    4: defgdefgdefg&#xA;4: defg &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Of course, using the second distribution, the resulting sequence has a much greater period.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;So:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What algorithm/approach can I use to solve this problem and maximize the product? Is this some known problem and how complex would calculating a solution be? It's not NP, I hope?!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit: Partial solution&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As @KennyTM pointed out in the comments, &lt;a href=&quot;http://en.wikipedia.org/wiki/Landau%27s_function&quot; rel=&quot;nofollow&quot;&gt;Landau's function $g$&lt;/a&gt; describes the maximum LCM, i.e. $g(7) = 12$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So this actually becomes: How to actually produce the partition? Does knowledge of $g(x)$ help here, maybe for a dynamic programming solution?&lt;/p&gt;&#xA;" OwnerUserId="917" LastEditorUserId="917" LastEditorDisplayName="" LastEditDate="2010-08-28T19:47:26.520" LastActivityDate="2010-08-28T20:56:51.250" Title="Split up $n \in \mathbb{N}$ into sum of naturals with maximum LCM" Tags="&lt;combinatorics&gt;&lt;sequences-and-series&gt;&lt;algorithms&gt;&lt;optimization&gt;&lt;extremal-combinatorics&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" />
  <row Id="3566" PostTypeId="2" ParentId="3543" CreationDate="2010-08-28T18:45:00.717" Score="3" ViewCount="" Body="&lt;p&gt;There are algorithms due to Abel, Liouville, Risch, Davenport, Trager et al. for solving the general problem of the &lt;strong&gt;algebraic&lt;/strong&gt; (vs. transcendental) case of integration in finite terms (elementary and algebraic functions). Here's an illuminating excerpt from the introduction of &lt;a href=&quot;http://hdl.handle.net/1721.1/15391&quot; rel=&quot;nofollow&quot;&gt;Barry Trager's 1984 MIT Thesis&lt;/a&gt; (coincidentally Barry was working right next door to me disproving Hardy's statement that no algorithm exists at the same time I was disproving an analogous Hardy statement while inventing algorithms for computing limits / asymptotics - while we were both student members of the MIT Mathlab Group - the research group that developed the Macsyma computer algebra system).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i36.tinypic.com/2mdnoef.jpg&quot; alt=&quot;alt text&quot;&gt;&#xA;&lt;img src=&quot;http://i38.tinypic.com/n1feac.jpg&quot; alt=&quot;alt text&quot;&gt;&#xA;&lt;img src=&quot;http://i34.tinypic.com/2z9ilac.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-08-28T19:52:20.247" LastActivityDate="2010-08-28T19:52:20.247" CommentCount="1" />
  <row Id="3567" PostTypeId="2" ParentId="3561" CreationDate="2010-08-28T19:21:06.743" Score="2" ViewCount="" Body="&lt;p&gt;That's a good question that - alas - does not really have a good answer - at least not a single textbook that one could point to. The best motivation for ideals and modules comes from   number theory. As such, you might find it helpful to attempt to learn some algebraic number theory in parallel, e.g. factorization theory in quadratic number fields. Unfortunately most of the algebraic number theory textbooks already assume solid algebra background, so you'll need to interpolate your own motivational examples. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One crucial point deserves emphasis. When first learning about new abstractions such as ideals and modules it is &lt;strong&gt;absolutely essential&lt;/strong&gt; to understand their motivating concrete instances. Thus when you are learning about ideals in algebra you should revisit your knowledge of elementary number theory and reinterpret results in terms of these new abstractions. For example, see my post &lt;a href=&quot;http://math.stackexchange.com/questions/2883/show-that-3p2q2-implies-3p-and-3q/2941#2941&quot;&gt;here&lt;/a&gt; and its links - which explains how various proofs of irrationality of sqrt's by descent on the size of a denominator are merely special cases of proofs that a denominator ideal - Dedekind's conductor ideal - is principal or invertible. From this general viewpoint the proof is a one-liner: the conductor ideal is invertible so cancelable. This shows that any Dedekind domain or PID is integrally closed - a monic generalization of the rational root test to a very wide class of rings. This is but one of many generalizations that you will meet when studying factorization theory in wider classes of rings. To better hone your intuition it is essential that you learn to easily move back-and-forth in this way between the concrete and abstract. Unfortunately most textbooks omit explicit discussion of such motivational material, so you'll need to develop it on-the-fly as need be. &lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-08-28T19:21:06.743" CommunityOwnedDate="2010-08-28T19:21:06.743" />
  <row Id="3568" PostTypeId="2" ParentId="3547" CreationDate="2010-08-28T19:47:14.960" Score="5" ViewCount="" Body="&lt;p&gt;&quot;Permutation group&quot; usually refers to a group that is acting (faithfully) on a set; this &lt;em&gt;includes&lt;/em&gt; the symmetric groups (which are the groups of all permutations of the set), but also every subgroup of a symmetric group. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Although all groups can be realized as permutation groups (by acting on themselves), this kind of action does not usually help in studying the group; special kinds of actions (irreducible, faithful, transitive, doubly transitive, etc), on the other hand, can give you a lot of information about a group. For example, Jordan proved that the only finite sharply five transitive groups are $A_7$, $S_6$, $S_5$, and the Mathieu group $M_{12}$. (A &quot;sharply five transitive group&quot; is a group $G$ acting on a set $X$ with five or more elements, such that for every ten elements $a_1,\ldots,a_5,b_1,\ldots,b_5\in X$, with $a_i\neq a_j$ for $i\neq j$ and $b_i\neq b_j$ for $i\neq j$, there exists one and only one $g\in G$ such that $g\cdot a_i = b_i$). (In fact, Jordan showed that the only finite sharply $k$-trasitive gruops for $k\geq 4$ are $S_k$, $S_{k+1}$, $A_{k+2}$, $M_{11}$, and $M_{12}$; see &lt;a href=&quot;http://en.wikipedia.org/wiki/Mathieu_group&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Mathieu_group&lt;/a&gt;.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You think of a permutation group as a group $G$, together with a faithful action $\sigma\colon G\times X\to X$ on a set $X$ (faithful here means that if $gx=x$ for all $x$, then $g=e$). Cayley's Theorem tells you that every group $G$ can be thought of as a permutation group, by taking $X$ to be the underlying set of $G$, and $\sigma$ to multiplication. But this gives you an embedding of $G$ into a very large symmetric group, because the set on which it is acting is large. You usually get more information if the set you are acting on is &quot;small&quot;-ish. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason for Cayley's Theorem is that, historically, people only considered permutation groups: collections of functions that acted on sets (the sets of roots of a polynomial, the points on the plane via symmetries, etc). Cayley was trying to abstract the notion of group; he then pointed out that his more abstract definition certainly included all the things that people were already considering, and that in fact it did not introduce any new ones in the sense that every abstract group could be considered as a permutation group. But, as he pointed out, it is sometimes more convenient or useful to consider the group abstractly, sometimes to consider it as a group of permutations. Having both viewpoints is better than having just one. I think Cayley's Theorem has more historical interest than practical interest these days, but your mileage may vary.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-08-28T19:47:14.960" />
  <row Id="3569" PostTypeId="2" ParentId="3565" CreationDate="2010-08-28T20:56:51.250" Score="4" ViewCount="" Body="&lt;p&gt;The blog post here &lt;a href=&quot;http://amca01.wordpress.com/2010/07/10/landaus-function/&quot; rel=&quot;nofollow&quot;&gt;describes&lt;/a&gt; somebody else's efforts at coding this up. He gets an improvement over the naive approach by using some nice internal structure of the problem. &lt;/p&gt;&#xA;" OwnerUserId="1560" LastActivityDate="2010-08-28T20:56:51.250" CommentCount="1" />
  <row Id="3570" PostTypeId="2" ParentId="187" CreationDate="2010-08-28T20:57:48.987" Score="1" ViewCount="" Body="&lt;p&gt;Let me discuss an analogous situation with regard to convex polyhedra.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Archimedes, generalizing what are often called the Platonic Solids or the convex regular polyhedra (there are five of them), seems to have discovered 13 polyhedra with the property that the pattern of faces around each vertex was the same for every vertex of the polyhedron, and all of the faces of the polyhedron were regular polygons. I say seems to have done this because the manuscript that describes what he did is lost. Pappus who lived many years after Archimedes describes these polyhedra explicitly and mentions 13 convex solids. Years latter many artists and mathematicians talked about these polyhedra. Kepler explicitly mentions two infinite families of polyhedra which have this property: the prisms (consisting of two regular n-gons and n squares) and the anti-prisms (consisting of two regular n-gons and 2n equilateral triangles). Kepler purports to give a proof a proof that there 13 such solids (with at least two types of faces). I say purports because there are in fact 14 convex polyhedra which meet the local symmetry condition described above. The one Archimedes, Pappus, Kepler and others missed is often called the pseudo-rhombicuboctahedron. (Kepler created confusion in referring to 14 solids in a place other than where he gave his &quot;proof.&quot;) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Elongated_square_gyrobicupola&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Elongated_square_gyrobicupola&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many books to this day continue to talk about 13 Archimedean convex solids (other than the prisms and antiprisms). With the definition that Archimedes almost certainly had in mind this is wrong - there are 14 such solids. However, if one considers convex polyhedra with at least two regular polygons as faces, and where the symmetry group of the solid is transitive on the vertices (that is, the symmetry group can take any vertex to any other vertex) then there are only 13 such solids. However, almost certainly Archimedes had no knowledge of the idea of a symmetry group in the modern sense. Depending on whether one uses a local symmetry notion or a global symmetry notion there are either 14 or 13 such convex solids (aside from the prisms and antiprisms).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Branko Grünbaum has a nice article about this called &quot;An enduring error.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://dlib.lib.washington.edu/dspace/bitstream/handle/1773/4592/An_enduring_error.pdf?sequence=1&quot; rel=&quot;nofollow&quot;&gt;https://dlib.lib.washington.edu/dspace/bitstream/handle/1773/4592/An_enduring_error.pdf?sequence=1&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1369" LastActivityDate="2010-08-28T20:57:48.987" CommentCount="1" />
  <row Id="3571" PostTypeId="2" ParentId="3550" CreationDate="2010-08-28T21:37:02.470" Score="3" ViewCount="" Body="&lt;p&gt;In order to find the solutions of the system of conics you mention, it suffices to give a procedure to find the projection of the simultaneous vanishing set of the two conics onto some axis, for instance, the $y$-axis: the coordinates of the projection onto the $y$-axis are the $y$-coordinates of the intersection points. Knowing them allows you to substitute back into the equations the values of $y$ and solve a system of equations in a single variable $x$, thus making the problem simpler. In terms of ideals, suppose that we can find a non-zero element $r$ in the ideal generated by the two conics, that depends only on a single variable, say $y$. This means that every solution of the system has $y$-coordinate satisfying the polynomial $r$. Thus we have severely limited the choices for the $y$-coordinates of the intersection (namely, they all have to satisfy the polynomial $r$) and, provided we can actually solve the polynomial $r$ in one variable, we can then substitute the various values of $y$ we found back into the initial equations and solve for $x$, again using our algorithm for solving polynomials in a single variable. So far, so good, I hope! The question is how to produce the element $r$. This is where Gröbner bases come in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to do the computation you ask, one would have to choose an appropriate monomial order. I will gloss over this, and simply &quot;do the obvious&quot;. It is an exercise for you to figure out which order to use so that the computation I am going to make is actually a Gröbner bases computation. Scattered throughout the computation there will be also a couple of special cases that I will not deal with: again, you can treat those as exercises for you!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, I am going to choose a generic basis, so that the first conic has the form &#xA;$$&#xA;x^2 + \alpha x + \beta ,&#xA;$$&#xA;where $\alpha$ and $\beta$ are polynomials in $y$ only (I am trying to simplify the notation as much as possible; the only &quot;assumption that I have made is that the coefficient of $x^2$ is non-zero, which can be arranged unless the &quot;conic&quot; is defined by a polynomial of degree at most one).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this basis, the second equation can be assumed to have no $x^2$ term (indeed, eliminating the $x^2$ term using the first equation would be the first step in any reasonable Gröbner basis computation in which the term $x^2$ is the highest term in sight). Thus I am going to write the second conic as &#xA;$$&#xA;\gamma x + \delta&#xA;$$&#xA;where, as before, $\gamma$ and $\delta$ are polynomials in $y$ only. Again, just to fix on a definite case, I am going to assume that the polynomial $\gamma$ is non-zero. (If $\gamma$ were zero, then we would have found a polynomial in the ideal generated by the two conics which only depends on $y$: this was our goal at the start anyway! Obviously, you should worry about the case $\gamma=\delta=0$, but I won't.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To compute a Gröbner basis, you would now compute S-polynomials: let's do it here as well. I am going to try to eliminate the $x^2$ term from the first equation, by using the second equation. This is easy: multiply the first equation by $\gamma$ and the second one by $x$ and take the difference: we are left with the equation&#xA;$$&#xA;(\alpha \gamma - \delta) x + \beta \gamma .&#xA;$$&#xA;Now we are going to eliminate the $x$ term from this last equation using again the second equation: multiply the second equation by $(\alpha \gamma - \delta)$, the last equation by $\gamma$ and subtract to obtain&#xA;$$&#xA;\delta^2 - \alpha \gamma \delta + \beta \gamma^2 .&#xA;$$&#xA;We found an expression independent of $x$!! We are done... provided this expression is not identically zero. You can figure out what this would mean and what happens in this case. Note also that the final expression is what we would have obtained if, at the very beginning, we had &quot;solved&quot; $x=-\frac{\delta}{\gamma}$ using the second equation, substituted in the first equation and cleared the denominators.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope that this &quot;hybrid&quot; computation explains what is going on: Gröbner bases and Buchberger's algorithm are a systematic way of &quot;solving&quot; systems of equations. You do not have to do any thinking, once you set up the problem. But you need to set up the problem so that it computes what you want. In this case, you could have used several shortcuts to get to the answer, without following all the steps. In more complicated situations, Buchberger's algorithm might be the best way of keeping track of all the steps to be taken.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me also comment that, except in the case in which you have a computer doing the computations for you, it is highly unlikely that Gröbner bases will help you with a specific question, unless you could have also found simple tricks to solve it right away.&lt;/p&gt;&#xA;" OwnerUserId="1028" LastActivityDate="2010-08-28T21:37:02.470" CommentCount="1" />
  <row Id="3572" PostTypeId="2" ParentId="3138" CreationDate="2010-08-28T22:33:49.803" Score="0" ViewCount="" Body="&lt;p&gt;One thing you might be interested in is Darboux's Theorem, which tells you that derivatives have the intermediate value property.  There's a decent  Wikipedia page  on the topic and there was also a nice proof given by Sam B. Nadler, Jr. in the Math Monthly a little while back (Feb. 2010). &lt;/p&gt;&#xA;" OwnerUserId="1392" LastActivityDate="2010-08-28T22:33:49.803" />
  <row Id="3573" PostTypeId="1" AcceptedAnswerId="3615" CreationDate="2010-08-28T23:16:29.683" Score="4" ViewCount="273" Body="&lt;p&gt;I came up with this problem after discussion of taxicab geometry in math class... I thought it was a simple problem, but still pretty neat; however, I am as of yet unsure of whether my answer is correct, or logical.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $[X]$ be the area of region $X$, and region $S_n$ be represented by the equation $|x-n|+|y-n|=k-n$ for all $n=0,1,2,\ldots,k-1$. Now let region $R_n$ be the region between $S_n$ and $S_{n+1}$ and $L=\displaystyle\sum_{n=0}^{k-2}{[R_n]}$. Find the smallest positive integer $k$ such that $L &gt; A$. &lt;strong&gt;($A$ is any number you can plug in)&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone else verify my result of $L=\frac{5k^2-k-4}{2}$?&lt;/p&gt;&#xA;" OwnerUserId="1562" LastEditorUserId="1562" LastEditorDisplayName="" LastEditDate="2010-08-29T20:12:38.220" LastActivityDate="2010-08-29T22:01:03.400" Title="Interesting Taxicab Problem?" Tags="&lt;geometry&gt;&lt;recreational-mathematics&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="2" />
  <row Id="3574" PostTypeId="1" AcceptedAnswerId="3575" CreationDate="2010-08-28T23:57:04.883" Score="2" ViewCount="141" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;How many &lt;strong&gt;unique&lt;/strong&gt; strings can be formed by the letters in &quot;aaamottu&quot;?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As a totally incompetent math student, I simply wrote a program to calculate it. And I ended up with 3360 unique strings:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;import itertools&#xA;unique = {}&#xA;for p in itertools.permutations(list(&quot;aaamottu&quot;), 8):&#xA;    unique[p] = True&#xA;&#xA;len(unique)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I'd like to know how to get that number using plain ol' math.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;nPr(8,8) gives 40,320. Obviously, lots of duplicates are being counted. How to I remove them?&lt;/p&gt;&#xA;" OwnerUserId="494" LastActivityDate="2010-08-29T00:17:56.540" Title="Remove duplicates from a permutation" Tags="&lt;combinatorics&gt;&lt;homework&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="3575" PostTypeId="2" ParentId="3574" CreationDate="2010-08-29T00:17:56.540" Score="2" ViewCount="" Body="&lt;p&gt;First pretend that the letters which are the same are different, then stop pretending.  Let us suppose that the string is $a_1 a_2 a_3 m o t_1 t_2 u$; then there really would be $8! = 40320$ different words we could form.  The key insight is that we can group these words according to how they use the $a$s and the $t$s; for example, we can group together the words&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$a_1 a_2 a_3 m o t_1 t_2 u, a_1 a_2 a_3 m o t_2 t_1 u$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;because the words are the same except for the order of the $t$s.  Since it is true that any word has a &quot;brother&quot; where the order of the $t$s is reversed, to account for the fact that we labeled the $t$s we have to divide by $2! = 2$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next we can group together according to how the $a$s are used; for example, we can group together the words&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$a_1 a_2 a_3 m o t_2 t_1, a_1 a_3 a_2 m o t_2 t_1, a_2 a_1 a_3 m o t_2 t_1, a_2 a_3 a_1 m o t_2 t_1, a_3 a_1 a_2 m o t_2 t_1, a_3 a_2 a_1 m o t_2 t_1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;because the words are the same except for the order of the $a$s.  Since it is true that any word has five &quot;partners&quot; where the order of the $a$s are different, to account for the fact that we labeled the $a$s we have to divide by $3! = 6$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the answer is $\frac{8!}{3! 2!} = \frac{40320}{12} = 3360$ as desired.  This function is an example of a &lt;a href=&quot;http://en.wikipedia.org/wiki/Multinomial_theorem&quot; rel=&quot;nofollow&quot;&gt;multinomial coefficient&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-08-29T00:17:56.540" />
  <row Id="3576" PostTypeId="1" AcceptedAnswerId="3577" CreationDate="2010-08-29T00:30:44.293" Score="0" ViewCount="129" Body="&lt;p&gt;In making up another problem today I came across something odd. I've been thinking it over and I can't exactly place why it's true, but after running a long Python script to check, I haven't yet found a counter example.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is $\sum_{n=1}^{m}{n^m}\equiv 0\mod m$ true for all odd $m \ge 3$? The script showed me that each term for odd $m$ is equivalent to $n$ when taken $\mod m$ (until term $m$), and so the sum would be $\frac{m(m-1)}{2}$ which is obviously $0 \mod m$. What I am unable to understand is why $n^m\equiv n \mod m$ only for odd $m$.&lt;/p&gt;&#xA;" OwnerUserId="1562" LastActivityDate="2010-08-29T00:45:46.527" Title="Why does this sum mod out to 0?" Tags="&lt;sequences-and-series&gt;&lt;modular-arithmetic&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="3577" PostTypeId="2" ParentId="3576" CreationDate="2010-08-29T00:45:46.527" Score="0" ViewCount="" Body="&lt;p&gt;If it is odd, each &lt;code&gt;n&lt;/code&gt; can be paired with &lt;code&gt;-n&lt;/code&gt;. So we get $n^m+(-n)^m=n^m-n^m=0$&lt;/p&gt;&#xA;" OwnerUserId="123" LastActivityDate="2010-08-29T00:45:46.527" CommentCount="2" />
  <row Id="3578" PostTypeId="1" CreationDate="2010-08-29T00:50:13.520" Score="2" ViewCount="104" Body="&lt;p&gt;Related to this question: &lt;a href=&quot;http://math.stackexchange.com/questions/2626/modelling-forces-acting-on-a-sail&quot;&gt;http://math.stackexchange.com/questions/2626/modelling-forces-acting-on-a-sail&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to create a simulation of the forces. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have done this before with cars:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;using a free physics engine&lt;/li&gt;&#xA;&lt;li&gt;a graphics library with OpenGL, Glut and glow&lt;/li&gt;&#xA;&lt;li&gt;and a 3D model of car. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;But I'm wondering if there is a better/more efficient way or your way :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks &lt;/p&gt;&#xA;" OwnerUserId="608" LastActivityDate="2010-08-29T00:50:13.520" Title="physics/mathematic simulations" Tags="&lt;learning&gt;&lt;self-learning&gt;" CommentCount="2" />
  <row Id="3579" PostTypeId="1" AcceptedAnswerId="3677" CreationDate="2010-08-29T01:19:15.317" Score="6" ViewCount="346" Body="&lt;p&gt;It can be seen &lt;a href=&quot;http://www-groups.dcs.st-and.ac.uk/~john/Zagier/Solution1.1.html&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; that the only numbers for which $n^{m+1}\equiv n \pmod{m}$ is true are 1, 2, 6, 42, and 1806. Through experimentation, it has been found that $\displaystyle\sum_{n=1}^{m}{n^m}\equiv 1 \bmod m$ is true for those numbers, and (as yet unproven) no others. Why is this true?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there is a simple relation between $n^{m+1} \bmod{m}$ and $n^m \bmod{m}$, that would probably make this problem make more sense. It is obvious that $n^m \equiv 1 \pmod{\frac{m}{d}}$ (dividing out $n$ from both sides gives this result) for all $n$ on the interval $[1,m]$ where $d$ is a divisor of $m$. As a result of this, $n^m \bmod{m}$ takes on only values of the form $1+k \frac{m}{d} \pmod m$ where $k = -1, 0, 1$. How can it be shown that the sum of those values is equivalent to $1 \bmod{m}$?&lt;/p&gt;&#xA;" OwnerUserId="1562" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-09-05T04:26:16.257" LastActivityDate="2010-09-05T04:26:16.257" Title="Why are the only numbers $m$ for which $n^{m+1}\equiv n \pmod{m}$ is true also unique for $\displaystyle\sum_{n=1}^{m}{n^m}\equiv 1 \bmod m$?" Tags="&lt;number-theory&gt;&lt;modular-arithmetic&gt;" AnswerCount="1" CommentCount="17" FavoriteCount="1" />
  <row Id="3580" PostTypeId="1" AcceptedAnswerId="3581" CreationDate="2010-08-29T01:26:59.140" Score="0" ViewCount="57" Body="&lt;p&gt;I had a question about an instantaneous z score.  The question reads as &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;P(z=2)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I think that this is 0 as z-score is the area under the curve and could be found by integrating the PDF over the range.  As the range is 0 the area also zero.  Is that correct?&lt;/p&gt;&#xA;" OwnerUserId="1564" LastActivityDate="2010-08-29T02:06:57.483" Title="Instantaneous Z-Score Value" Tags="&lt;statistics&gt;&lt;homework&gt;" AnswerCount="1" />
  <row Id="3581" PostTypeId="2" ParentId="3580" CreationDate="2010-08-29T01:36:39.840" Score="2" ViewCount="" Body="&lt;p&gt;Since $P(z=2)$ is the same as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac1{\sqrt{2\pi}}\int_2^2 \exp\left(-\frac{u^2}{2}\right)\mathrm{d}u$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;then yes, the result is zero; this can be interpreted also as saying&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac1{\sqrt{2\pi}}\int_{-\infty}^2 \exp\left(-\frac{u^2}{2}\right)\mathrm{d}u-\frac1{\sqrt{2\pi}}\int_{-\infty}^2 \exp\left(-\frac{u^2}{2}\right)\mathrm{d}u$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;via the Fundamental Theorem of Calculus.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Geometrically, the &quot;integral&quot; covers no area on the Gaussian curve, which agrees with the result of 0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;(previous version)&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I had to look at Abramowitz and Stegun just to check that you were saying what I thought you were saying; anyway, if by $P(z)$ you meant&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac1{\sqrt{2\pi}}\int_{-\infty}^z \exp\left(-\frac{u^2}{2}\right)\mathrm{d}u$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;then no, $P(2)$ is not zero. The integral&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac1{\sqrt{2\pi}}\int_{-\infty}^0 \exp\left(-\frac{u^2}{2}\right)\mathrm{d}u$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;by itself is equal to $\frac12$ (why? ;) ), and the other part&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac1{\sqrt{2\pi}}\int_0^2 \exp\left(-\frac{u^2}{2}\right)\mathrm{d}u$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;cannot be zero as well because it is the integral of a function whose graph does not cross the x-axis.&lt;/p&gt;&#xA;" OwnerUserId="498" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-08-29T02:06:57.483" LastActivityDate="2010-08-29T02:06:57.483" CommentCount="3" />
  <row Id="3582" PostTypeId="2" ParentId="2121" CreationDate="2010-08-29T02:05:01.930" Score="2" ViewCount="" Body="&lt;p&gt;Let $F_4 = Z/2[w]/w^2+w+1$ be the field of order 4.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $H = (F_4^+)^n = (Z/2)^{2n}$ for $n \geq 2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then $Aut(H) = GL(2n,F_2)$ which has order&#xA;$(4^n-1)(4^n-2)(4^n-4)...(4^n-2*4^(n-1))$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $G$ be $\langle H,a : a^3 = 1, axa^{-1} = wx \ \text{for} \ x \in H \rangle$.  So $G$ is the semi-direct product of $H$ with $Z/3$, where the action is given by multiplication by $H$.  An automorphism of $G$ must fix $H$ ($H$ consists of the set of elements of order 2).  It must also send a to one of the 2*4^n elements of order 3.  Let us consider only the automorphisms of $G$ which fix a (which is a subgroup of $Aut(G)$ of index $2*4^n$).  These are automorphisms of H which commute with multiplication by w.  Hence these are automorphisms of $F_4^n$ as an $F_4$ vector space, or elements of $GL(n,F_4)$.  Hence&#xA;$|Aut(G)| = 2*4^n* (4^n-1)(4^n-4)...(4^n-4^(n-1))$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $n \geq 2$, it is clear that $|Aut(H)| &gt; |Aut(G)|$.&lt;/p&gt;&#xA;" LastActivityDate="2010-08-29T02:05:01.930" CommentCount="1" />
  <row Id="3583" PostTypeId="1" CreationDate="2010-08-29T04:36:09.707" Score="4" ViewCount="365" Body="&lt;p&gt;A train with infinitely many seats, one for each rational number, stops in countably many villages, one for each positive integer, in increasing order, and then finally arrives at the city. At the first village, two women board the train. At the second village, one woman leaves the train to go visit her cousin, and two other women board the train. At the third village, one woman leaves the train to go visit her cousin, and two other women board the train. At the fourth village, and in fact at every later village, the same thing keeps happening: one woman off to visit her cousin, two new women on board the train. How many women arrive at the city?&lt;/p&gt;&#xA;" LastActivityDate="2010-08-30T14:12:06.803" Title="Infinite processes riddle" Tags="&lt;logic&gt;" AnswerCount="3" CommentCount="7" FavoriteCount="1" />
  <row Id="3584" PostTypeId="2" ParentId="3501" CreationDate="2010-08-29T05:11:59.420" Score="1" ViewCount="" Body="&lt;p&gt;As has been indicated in the comments, the radical symbol is generally taken to be the principal square root function, which for nonnegative real inputs gives nonnegative real outputs.  For negative real inputs, this function gives non-real outputs, so X must be nonnegative, so Y must be nonnegative, giving (a).&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-08-29T05:11:59.420" CommentCount="1" />
  <row Id="3586" PostTypeId="2" ParentId="3583" CreationDate="2010-08-29T06:09:19.830" Score="1" ViewCount="" Body="&lt;p&gt;This problem isn't well defined as it relies on infinity to actually exist. There are two ways of viewing this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we want the limit of the number of people on the train as we approach the infinitith stop, this series is +2-1+2-1..., so a number of people with the same cardinality as the integers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second and seemingly more correct interpretation is: &quot;How many women stay on the train throughout the whole journey?&quot;. If we make women leave the train in the same order they board, the answer is 0, as no-one remains on the train through the whole journey. If we make the last women to board leave, then the answer is infinite.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This behaviour is not strange if you have seen the behaviour of infinite series. For example, while +2-1+2-1... approached infinity, +2-1-1+2-1-1 will always be almost 0.&lt;/p&gt;&#xA;" OwnerUserId="123" LastEditorUserId="123" LastEditorDisplayName="" LastEditDate="2010-08-29T21:10:09.717" LastActivityDate="2010-08-29T21:10:09.717" CommentCount="6" />
  <row Id="3587" PostTypeId="2" ParentId="3583" CreationDate="2010-08-29T06:11:05.760" Score="10" ViewCount="" Body="&lt;p&gt;There is not enough information to uniquely determine the answer.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a situation in which $0$ women get to the city: we place the women in order according to when they boarded the train, and every time a new village is reached the woman who boarded the train farthest back in time leaves.  This procedure ensures that every woman eventually leaves the train.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a situation in which countably many women get to the city: every time a new village is reached one of the women who boarded at the previous village leaves.  This procedure ensures that the number of women who eventually reach the city increases by $1$ at each village.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One can in fact get any finite number of women to the city.  &lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-08-29T06:20:31.307" LastActivityDate="2010-08-29T06:20:31.307" CommentCount="4" />
  <row Id="3588" PostTypeId="2" ParentId="3531" CreationDate="2010-08-29T07:05:11.627" Score="2" ViewCount="" Body="&lt;p&gt;Not much to contribute at this point except this enhanced picture:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/OvSTr.png&quot; alt=&quot;Steiner circumellipse&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(dashed lines: axes; gray lines: medians)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and the &lt;a href=&quot;http://ifile.it/xse06ua&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;Mathematica&lt;/em&gt; notebook&lt;/a&gt; used to produce it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Thanks &lt;a href=&quot;http://math.stackexchange.com/users/72&quot;&gt;Isaac&lt;/a&gt;!)&lt;/p&gt;&#xA;" OwnerUserId="498" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-08-29T10:43:14.083" LastActivityDate="2010-08-29T10:43:14.083" CommentCount="9" />
  <row Id="3590" PostTypeId="1" CreationDate="2010-08-29T12:58:49.673" Score="7" ViewCount="327" Body="&lt;p&gt;(1) The Riemann hypothesis (RH) states that all non-trivial zeros of the zeta function have real part 1/2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) The zeta function is intimately connected with the Gamma function via the functional equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second fact suggests that there is an equivalent form of RH which is expressed solely in terms of the Gamma function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Question: What is the most natural form to translate RH as directly as possible (without mentioning the zeta function) into a hypothesis on the behaviour of the Gamma function?&lt;/p&gt;&#xA;" OwnerUserId="1484" LastActivityDate="2010-12-05T03:25:14.210" Title="How to express the Riemann hypothesis in terms of the Gamma function?" Tags="&lt;gamma-function&gt;&lt;riemann-hypothesis&gt;" AnswerCount="2" CommentCount="8" FavoriteCount="3" />
  <row Id="3592" PostTypeId="1" AcceptedAnswerId="3597" CreationDate="2010-08-29T13:29:16.830" Score="1" ViewCount="430" Body="&lt;p&gt;Calculus theory problem I'm belaboring to solve... Tried it for an hour only spinning my wheels. Any hints/nudges in the right direction would be greatly appreciated. Here's the problem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;f and g are functions that are differentiable for all real x. They have the following properties:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i) f'(x) = f(x) - g(x)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ii) g'(x) = g(x) - f(x)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;iii) f(0) = 5&lt;/p&gt;&#xA;&#xA;&lt;p&gt;iv) g(0) = 1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Prove that f(x) + g(x) = 6 for all x.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Good luck.&lt;/p&gt;&#xA;" LastActivityDate="2010-08-30T08:32:46.783" Title="Prove that f(x) + g(x) = 6" Tags="&lt;calculus&gt;" AnswerCount="5" CommentCount="5" />
  <row Id="3593" PostTypeId="2" ParentId="3510" CreationDate="2010-08-29T13:30:24.553" Score="11" ViewCount="" Body="&lt;p&gt;One could provide answers based on a wide range of definitions of $\exp$, $\cos$, and $\sin$ (&lt;em&gt;e.g.&lt;/em&gt;, via differential equations, power series, Lie theory, inverting integrals, infinite sums, infinite products, complex line integrals, continued fractions, circular functions, and even Euclidean geometry) as well as offering Euler's formula up as a tautology based on a definition.  But let's consider where in one's education this question arises: it's usually well before most of these other concepts are encountered.  The complex numbers have just been introduced; the Argand diagram is available; $\exp(x)$ is most likely defined as the limiting value of $(1 + x/n)^n$, and $\cos$ and $\sin$ are defined as circular functions (that is, as arclength parameters for coordinates of the unit circle).  So the question deserves a response in this context using mathematics accessible to someone at this level.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Accordingly, I propose that we interpret $\exp(i x)$ as the limit of $(1 + i x/n)^n$, because at least one understands how to compute the latter (as iterated multiplication of 1 by $1 + i x/n$), one at least vaguely intuits what the limit of a sequence of points in the complex plane might mean, and one has learned that $\cos(\theta) + i \sin(\theta)$ is quite literally the complex number plotted at $\left( \cos(\theta), \sin(\theta) \right)$ in the plane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems natural to evaluate the limit by looking at its modulus and argument separately.  The modulus is relatively easy: because $|1 + i x/n| = \sqrt {  1 + (x/n)^2  }$, the modulus of its $n$th power equals $\left( 1 + (x/n)^2 \right) ^{n/2}$.  We can exploit the one limit assumed to be known:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\left( 1 + (x/n)^2 \right) ^{n/2} = \left( \left( 1 + x^2 / n^2 \right) ^ {n^2} \right) ^ {1/{2n}} \simeq \left( \exp ( x^2 ) \right) ^ {1/{2n}} \to 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(There's some hand-waving in the last two steps.  This is characteristic of limiting arguments at this level.  Those of you who can see where the rigor is lacking also know exactly how to supply the missing steps.)  Whence, whatever $\exp( i x )$ might be, we deduce it should lie on the unit circle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we turn our attention to the argument of the limit.  The sequence of arguments&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\left( 1 + (x/n)^2 \right) ^{1/2}, \left( 1 + (x/n)^2 \right) ^{2/2}, \ldots, \left( 1 + (x/n)^2 \right) ^{k/2}, \ldots$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;obviously is non-decreasing, because it's a geometric sequence with multiplier of 1 or greater.  That is, each successive multiplication by $1 + i x / n$ is &lt;em&gt;expanding&lt;/em&gt; an original right triangle with vertices at (0,0), (1,0), and (1, $x/n$), but in the limit the amount of expansion is reduced to unity, as we have seen.  In the Argand diagram we're just laying out similar copies of this original right triangle, placing one leg of each new copy along the hypotenuse of the previous one.  In the limit, the length of the small leg (originally $x/n$) therefore remains constant.  These observations imply that near the limit, we can take all these $n$ little triangles to be essentially congruent, whence the length of the path traced out by the succession of images of the small leg must be extremely close to $n (x/n) = x$.  This pins down where on the circle $\exp( i x )$ must lie: it is the point reached by traveling (signed) distance $x$ counterclockwise along the circle beginning at (1,0).  To anyone exposed to the definition of $\sin$ and $\cos$ in terms of circular functions, it is immediate that the coordinates of this limiting location are $\left( \cos(x), \sin(x) \right)$ and we are done.&lt;/p&gt;&#xA;" OwnerUserId="1489" LastActivityDate="2010-08-29T13:30:24.553" />
  <row Id="3595" PostTypeId="2" ParentId="3592" CreationDate="2010-08-29T13:56:09.763" Score="0" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \frac{d}{dx} \bigl( f(x) + g(x) \bigr) = ? $$&lt;/p&gt;&#xA;" OwnerUserId="171" LastActivityDate="2010-08-29T13:56:09.763" />
  <row Id="3596" PostTypeId="2" ParentId="3592" CreationDate="2010-08-29T13:56:54.300" Score="2" ViewCount="" Body="&lt;p&gt;Let's introduce their sum function &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$s(x) := f(x) + g(x)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The statement to be proven thus becomes $s(x) = 6$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$s(0) = 5 + 1 = 6$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So now what derivative $s'$ do we need such that $s$ does never change it's constant value? &lt;/p&gt;&#xA;" OwnerUserId="917" LastActivityDate="2010-08-29T13:56:54.300" />
  <row Id="3597" PostTypeId="2" ParentId="3592" CreationDate="2010-08-29T14:09:55.280" Score="4" ViewCount="" Body="&lt;p&gt;As a follow up to Dario... If we let $s(x) := f(x) + g(x)$, $s'(x) = f'(x)+g'(x)$, which by (i) and (ii) is equal to $f(x)-g(x)+g(x)-f(x) = 0$. Since $s(0) = 6$ and $s'(x) = 0$, $s(x)$ is constant; $s(x) = 6$ for all $x$, therefore $f(x)+g(x) = 6$.&lt;/p&gt;&#xA;" OwnerUserId="1562" LastActivityDate="2010-08-29T14:09:55.280" CommentCount="1" />
  <row Id="3598" PostTypeId="2" ParentId="3592" CreationDate="2010-08-29T14:21:55.807" Score="1" ViewCount="" Body="&lt;p&gt;This problem should help you to understand and practice with two main facts about derivatives:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Derivative is a linear function&lt;/strong&gt; in fact if $f$ and $g$ are differentiable on $\mathbb{R}$ and $\alpha , \beta \in \mathbb{R}$ then $(\alpha f + \beta g)' = \alpha f' + \beta g'$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Derivative is a rate of change&lt;/strong&gt; this is more subtle, the essential fact is that if $f$ is differentiable on $\mathbb{R}$ and $f'(x) = 0$ for all $x \in \mathbb{R}$ then &lt;em&gt;the rate of change of $f$ is null&lt;/em&gt; on all $\mathbb{R}$ and then &lt;em&gt;the value of $f$ does not change&lt;/em&gt; on all $\mathbb{R}$ (pay attention, this is not always true if you break the domain of $f$ into pieces). This is a consequence of a more general and important result, the &lt;a href=&quot;http://en.wikipedia.org/wiki/Mean_value_theorem&quot; rel=&quot;nofollow&quot;&gt;mean value theorem&lt;/a&gt;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Observe that i) and ii) in the question give you &lt;em&gt;global information&lt;/em&gt; on $f$ and $g$ (i. e. they refer to all $x \in \mathbb{R}$), while iii) and iv) give you &lt;em&gt;local&lt;/em&gt; information (in the point $x = 0$). Your question is about a &lt;em&gt;global&lt;/em&gt; propriety of the function $f+g$, so you should try to combine local and global information to reach the solution.&lt;/p&gt;&#xA;" OwnerUserId="365" LastActivityDate="2010-08-29T14:21:55.807" />
  <row Id="3600" PostTypeId="1" AcceptedAnswerId="3603" CreationDate="2010-08-29T15:37:36.130" Score="4" ViewCount="256" Body="&lt;p&gt;So, I know that every abelian (commutative) group $G$ is such that, for any subgroup $H$, the left cosets of $H$ in $G$ are the right cosets. I guess this is true even if $G$ is not abelian, but $H$ is (but I'm not sure). Is this enough to characterise the subgroup as abelian, or are there examples of non-abelian subgroups with this property?&lt;/p&gt;&#xA;" OwnerUserId="172" LastActivityDate="2010-08-30T12:36:48.843" Title="Can a non-abelian subgroup be such that the right cosets equal the left cosets?" Tags="&lt;group-theory&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" />
  <row Id="3601" PostTypeId="2" ParentId="3600" CreationDate="2010-08-29T15:41:39.200" Score="8" ViewCount="" Body="&lt;p&gt;The left cosets of $H$ are the same as the right cosets&#xA;of $H$ if and only if $H$ is a &lt;a href=&quot;http://en.wikipedia.org/wiki/Normal_subgroup&quot; rel=&quot;nofollow&quot;&gt;normal subgroup&lt;/a&gt; of $G$.&#xA;There are non-abelian groups all subgroups of which are normal,&#xA;for example the quaternion group of order $8$.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-08-29T15:41:39.200" CommentCount="8" />
  <row Id="3602" PostTypeId="1" AcceptedAnswerId="3604" CreationDate="2010-08-29T15:50:12.367" Score="3" ViewCount="222" Body="&lt;p&gt;&lt;strong&gt;Motivation:&lt;/strong&gt; &lt;em&gt;Example&lt;/em&gt;. To solve a problem on evaluating the maximum of a product of $n$ real variables subject to an equality constraint on its sum $S$ ($=100$), I used the Lagrange multipliers method (which can be improved by adding the adequate conditions on the second order partial derivatives). Doing so I got a &quot;solution&quot; $x^*=(x_1,x_2,\dots ,x_n)\in \mathbb{R}^{n}$ with $x_1=x_2=\dots=x_n=100/n$. I proceeded by maximizing the single real variable function $u(t)=(100/t)^t$: the maximum is at $36\lt t^*\lt 37$. Finally I computed $u(36)\lt u(37)$. Hence, the optimum occurs at $n=37$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In spite of having come to the correct solution to the problem, I was told that &lt;em&gt;such an approach does not guarantee the correct solution, in the general case&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; In order to see some limitations of treating optimization problems as illustrated by the above example but with greater generality, a few examples in which this method fails would be appreciated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Remark&lt;/em&gt;: I hope this edited text improves the question. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt; The case I have in mind is that of finding the maximum/minimum of an objective function $f(x_1,x_2,\dots,x_n)$ subject to at least one constraint $g(x_1,x_2,\dots,x_n)=0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a kind  of generalization of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Simplex_algorithm&quot; rel=&quot;nofollow&quot;&gt;linear programming simplex problem&lt;/a&gt;, two differences being &lt;em&gt;a nonlinear optimization and one integer variable&lt;/em&gt;. &lt;/p&gt;&#xA;" OwnerUserId="752" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2010-12-23T23:47:46.083" LastActivityDate="2010-12-23T23:47:46.083" Title="A naive approach for nonlinear optimization on several real variables and one natural variable. Give examples of (potential) failure" Tags="&lt;calculus&gt;&lt;analysis&gt;&lt;optimization&gt;&lt;nonlinear-optimization&gt;&lt;multivariable-calculus&gt;" AnswerCount="2" CommentCount="6" FavoriteCount="0" />
  <row Id="3603" PostTypeId="2" ParentId="3600" CreationDate="2010-08-29T16:00:51.300" Score="3" ViewCount="" Body="&lt;p&gt;For a subgroup of a group, the condition of being abelian is neither necessary nor sufficient for being &lt;a href=&quot;http://en.wikipedia.org/wiki/Normal_subgroup&quot; rel=&quot;nofollow&quot;&gt;normal&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $G$ is non-abelian, then $H=G$ is a normal but non-abelian subgroup. As another example, the &lt;a href=&quot;http://en.wikipedia.org/wiki/Alternating_group&quot; rel=&quot;nofollow&quot;&gt;alternating group&lt;/a&gt; is a normal but non-abelian subgroup of the symmetric group.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a counterexample in the converse direction, consider the &lt;a href=&quot;http://en.wikipedia.org/wiki/Free_group&quot; rel=&quot;nofollow&quot;&gt;free group&lt;/a&gt; with two generators. The subgroup generated by one of the generators is abelian (isomorphic to $\mathbb Z$) but not normal. (This means that your guess above is wrong.)&lt;/p&gt;&#xA;" OwnerUserId="367" LastEditorUserId="367" LastEditorDisplayName="" LastEditDate="2010-08-30T12:36:48.843" LastActivityDate="2010-08-30T12:36:48.843" CommentCount="1" />
  <row Id="3604" PostTypeId="2" ParentId="3602" CreationDate="2010-08-29T17:10:36.493" Score="3" ViewCount="" Body="&lt;p&gt;The problem, if I understand it correctly, is best appreciated geometrically by drawing a picture of the level sets of the objective function and the lattice points satisfying the constraints: the optimum (over the reals) can occur at one location, but the nearby lattice points will necessarily lie on inferior contours.  Extending those contours identifies a border of almost-optimal locations where the objective is greater than at the lattice point you chose.  All you need for a spectacular failure to occur is for there to exist a distant lattice point somewhere within this marginal area: this would correspond to a dramatically different solution satisfying all the constraints and having a better objective value than yours.  (Your solution can have arbitrarily bad objective values, too, depending on the gradient of the objective function between the Lagrange multiplier solution and your rounded approximation to it.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have encountered exactly this difficulty with certain problems in statistics where the variables count things (and therefore must be integral) and the objective measures the precision of an estimate (and is to be minimized, but the change from maximization to minimization changes nothing essential).&lt;/p&gt;&#xA;" OwnerUserId="1489" LastActivityDate="2010-08-29T17:10:36.493" CommentCount="4" />
  <row Id="3605" PostTypeId="2" ParentId="3510" CreationDate="2010-08-29T17:14:39.880" Score="1" ViewCount="" Body="&lt;p&gt;I'll copy my answer to &lt;a href=&quot;http://math.stackexchange.com/questions/1133/natural-derivation-of-the-complex-exponential-function/1154#1154&quot;&gt;this question&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Let&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&#xA;f(x) = cos(x) + i*sin(x)&#xA;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Then&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&#xA;df/dx = -sin(x) + i*cos(x)&#xA;      = i*f(x)&#xA;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;So that&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&#xA;∫(1/f(x)) df = ∫i dx&#xA;ln(f(x)) = ix  + C&#xA;f(x) = e^(ix + C) = cos(x) + i*sin(x)&#xA;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Since &lt;code&gt;f(0) = 1&lt;/code&gt;, &lt;code&gt;C = 0&lt;/code&gt;, so&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&#xA;e^(ix) = cos(x) + i*sin(x)&#xA;&lt;/pre&gt;&#xA;" OwnerUserId="136" LastActivityDate="2010-08-29T17:14:39.880" CommentCount="1" />
  <row Id="3606" PostTypeId="1" CreationDate="2010-08-29T17:16:13.763" Score="3" ViewCount="182" Body="&lt;p&gt;I'm trying to express the following function more simply:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ \begin{align} f(x) = (\sum_{n=3}^{x} \prod_{k=2}^{\lceil \sqrt n \rceil} \lceil \dfrac {n}{k} \rceil - \lfloor \dfrac{n}{k} \rfloor ) + 1 \end{align} $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know this is kind of a vague request, but I'm trying to get away from the ceiling and floor functions and possibly the product notation, just because it feels like a hacky way of expressing what I'm trying to express.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any ideas?&lt;/p&gt;&#xA;" OwnerUserId="1574" LastActivityDate="2010-08-29T20:35:58.747" Title="Function simplification" Tags="&lt;number-theory&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="3607" PostTypeId="2" ParentId="3606" CreationDate="2010-08-29T17:20:23.020" Score="11" ViewCount="" Body="&lt;p&gt;$\left\lceil \frac{n}{k} \right\rceil - \left\lfloor \frac{n}{k} \right\rfloor$ is equal to $0$ or $1$ depending on whether $k$ does or does not divide $n$.  In other words,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\prod_{k=2}^{ \lceil \sqrt{n} \rceil} \left( \left\lceil \frac{n}{k} \right\rceil - \left\lfloor \frac{n}{k} \right\rfloor \right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is equal to $0$ or $1$ depending on whether $n$ is composite or prime.  So the function you are trying to describe is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Prime-counting_function&quot; rel=&quot;nofollow&quot;&gt;prime counting function&lt;/a&gt;.  There really is no simpler way to describe it than &quot;the prime counting function,&quot; although a lot is known about its asymptotic behavior.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit:  For functions like the prime counting function, you should replace the notion of &quot;nice formula&quot; with &quot;fast algorithm&quot; (the former being a special case of the latter).  The formula you wrote down is equivalent to a very slow algorithm: use trial division to test the primality of all the integers in your range.  There are much faster &lt;a href=&quot;http://en.wikipedia.org/wiki/Primality_test&quot; rel=&quot;nofollow&quot;&gt;primality tests&lt;/a&gt; which translate into much faster algorithms for computing $\pi(n)$ even though these tests cannot be easily translated into familiar-looking formulas.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit #2:  As Shreevatsar mentions in the comments, &lt;a href=&quot;http://en.wikipedia.org/wiki/Sieve_theory&quot; rel=&quot;nofollow&quot;&gt;sieve theory&lt;/a&gt; is also relevant to  counting or estimating $\pi(n)$ more directly, without having to do all that primality testing.  &lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-08-29T19:12:23.997" LastActivityDate="2010-08-29T19:12:23.997" CommentCount="5" />
  <row Id="3608" PostTypeId="1" CreationDate="2010-08-29T18:01:42.810" Score="1" ViewCount="148" Body="&lt;p&gt;In a recent (and to me, very beautiful) paper, entitled &quot;Relational EPR&quot;, &lt;a href=&quot;http://arxiv.org/abs/quant-ph/0604064&quot; rel=&quot;nofollow&quot;&gt;Smerlak and Rovelli&lt;/a&gt; present a way of thinking about EPR which relies upon Rovelli's previously published work on relational quantum mechanics (at see arxiv.org/abs/quant-ph/9609002 ).    In relational quantum mechanics, there is no non-locality, but the definition of when an event occurs is weakened from Einstein's strict definition and instead is localized to each observer-measurement apparatus, including subsequent observers.   There are (informal) coherence assumptions to ensure the consistency of reports from different subsequent observers.    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;All of this seems very similar to various results in modern categorical mathematics.    Is there a standard mathematical structure which well describes the structure of the space of localized measurements which Rovelli has envisioned?   I know of Isham's work on topos theory and quantum mechanics, but I think he is aiming at something a little different.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am asking the question here because I am only an interested amateur.&lt;/p&gt;&#xA;" OwnerUserId="1575" LastEditorUserId="1575" LastEditorDisplayName="" LastEditDate="2010-11-13T10:58:12.837" LastActivityDate="2010-11-13T10:58:12.837" Title="What categorical mathematical structure(s) best describe the space of &quot;localized events&quot; in &quot;relational quantum mechanics&quot;?" Tags="&lt;category-theory&gt;&lt;physics&gt;" CommentCount="3" />
  <row Id="3609" PostTypeId="1" CreationDate="2010-08-29T18:24:10.237" Score="6" ViewCount="247" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Goodstein%27s_theorem&quot; rel=&quot;nofollow&quot;&gt;Goodstein's Theorem&lt;/a&gt; is the statement that every Goodstein sequence eventually hits 0.  It is known to be independent of Peano Arithemtic (PA), and in fact, was the first such purely number theoretic result.  It is provable in ZFC.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One way of phrasing this is that the theory &quot;PA + Goodstein's theorem is false&quot; is consistent (assuming PA is).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By Godel's completeness theorem, there must exist a model of PA in which Goodstein's theorem fails.  In fact, applying the downward Lowenheim-Skolem theorem, we may assume this model is countable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, in the interest of speaking about this result to a group of grad students (of various interests), I'd like to run this backwards.  So,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;is there some known, obvious, or easy to construct countable nonstandard model of $PA$ in which Goodstein's theorem fails?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In answering this, I'm willing to accept the &quot;foundational&quot; first order logic theorems:  Godel's completeness and compactness results, the Lowenheim-Skolem theorem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is the kind of answer I'd really like:  There is an explicit countable collection $\Sigma = \{\phi_n\}$ of first order sentences (possibly in a slightly larger language) such that $\mathbb{N}$ is a model of $PA + \Sigma_0$ for any finite $\Sigma_0\subseteq \Sigma$, and such that $PA + \Sigma$ implies Goodstein's theorem is false.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;One approach I've thought of is to first enlarge the language by adding a constant symbol c.  Next, let $\phi_n$ be the statement &quot;The Goodstein sequence for $c$ takes longer than &quot;n&quot; steps to terminate&quot;.  (While I do not personally know how to encode &quot;the Goodstein sequence for $c$&quot; in first order language, I am confident it can be done, for otherwise, one could not even formulate &quot;PA proves the Goodstein sequence converges&quot;.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case, $\mathbb{N}$ is a model of any $PA + \Sigma_0$ by simply setting c = n+1, where n is the largest subscript of a $\phi_k$ in $\Sigma_0$ (which exists because $\Sigma_0$ is finite).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By Godel's and Lowenheim-Skolem theorems, $PA + \Sigma$ has a countable model $M$.  Then the interpretation of $c$ in this model satisfies $\phi_n$ for all $n$, and hence the Goodstein sequence doesn't terminate for $c$ in this model.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, since the independence of Goodstein's theorem was so difficult to prove, I'm quite certain there's a mistake in this line of reasoning (though I don't know where).  I'd love for someone to patch this up into something correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As always, please feel free to retag as neccesary, and thank you for the reponses!&lt;/p&gt;&#xA;" OwnerUserId="331" LastActivityDate="2010-09-14T05:40:24.700" Title="Is there a natural model of Peano Arithmetic where Goodstein's theorem fails?" Tags="&lt;number-theory&gt;&lt;logic&gt;&lt;model-theory&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="1" />
  <row Id="3610" PostTypeId="2" ParentId="3606" CreationDate="2010-08-29T18:30:40.307" Score="0" ViewCount="" Body="&lt;p&gt;It depends in what context you want to define that function (for computer implementations?), but following Qiaochu Yuan's post, you can e.g. use set-based notations to state the primality test more clearly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f(x) = \left| \lbrace x \in \lbrace 3,\ldots, x \rbrace \; |\; \forall k \in \lbrace 2,\ldots,\sqrt n \rbrace . k \not | \; n \rbrace \right| + 1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f(x) = \left| \mathbb{P} \cap \lbrace 3, \ldots, x \rbrace \right|$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\mathbb{P}$ denotes the set of prime numbers or simply &lt;em&gt;let $f$ be the prime counting function&lt;/em&gt; is even simpler.&lt;/p&gt;&#xA;" OwnerUserId="917" LastActivityDate="2010-08-29T18:30:40.307" />
  <row Id="3611" PostTypeId="2" ParentId="3592" CreationDate="2010-08-29T20:10:02.533" Score="5" ViewCount="" Body="&lt;p&gt;Dear Bob,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now that you've gotten the answer, here is a little bit of a post mortem clarifying the essential points, as may come up again in future problems.  (In general, once you've solved a problem, it's a really good idea to ask for the &quot;moral(s)&quot;.  If you can't identify one, how are you better off now than before you solved it?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case, I think the most important single step is the recognition that the conclusion&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;$f(x) + g(x) = 6$ for all $x$&quot; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;can be more usefully rephrased as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;$f(x) + g(x)$ is a constant function, with constant value $6$.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is much more suggestive, because a function $f$ defined on an interval is constant if and only if its dervative is constantly equal to zero.  (Stationarity is equivalent to identically zero velocity.)  Note that half of this statement is obvious, but the other half is not: it is consequence of the Mean Value Theorem that should be emphasized both in the text and by your instructor.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus you are clued in to the fact that the main thing you want to show is that $(f+g)' = f' + g' \equiv 0$ (i.e., constantly zero).  Since a constant function is determined by plugging in any point, once you know that $f+g$ is constant, seeing that it's constantly equal to SOMETHING -- in this case $6$ -- shouldn't be a problem.  So now you look and see how to get from what you're given to the conclusion that $f' + g' \equiv 0$, and you see that you're being given an expression for $f'$ and $g'$ separately.  So certainly you want to add them together and hope to get $0$; in this case, that hope is immediately fulfilled.&lt;/p&gt;&#xA;" OwnerUserId="299" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2010-08-30T08:32:46.783" LastActivityDate="2010-08-30T08:32:46.783" />
  <row Id="3612" PostTypeId="2" ParentId="3606" CreationDate="2010-08-29T20:22:16.507" Score="0" ViewCount="" Body="&lt;p&gt;Presuming that you seek a concise formula (versus a fast algorithm), then one can simplify it somewhat by replacing the product by a factorial, and by employing $\rm\;\; a\perp b \;\; := \; 1 \;\;\; if \;\;\; gcd(a,b)=1 \;\;\; else \;\;\; 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\rm \pi(x) \: = \: \sum_{n\;=\;2}^x \;\: n\perp {\lfloor \sqrt{n}\rfloor}!$$&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-08-29T20:35:58.747" LastActivityDate="2010-08-29T20:35:58.747" CommentCount="4" />
  <row Id="3613" PostTypeId="1" AcceptedAnswerId="3614" CreationDate="2010-08-29T21:27:24.127" Score="3" ViewCount="313" Body="&lt;p&gt;Hi guys - I was wondering if you could help...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have Math homework, I was hoping if you could check my answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Find the equation of the line perpendicular to 3x+2y-4=0 going through point (2,-3)&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;y = ( -3x + 4 ) /2&#xA;y = 2&#xA;&#xA;x = - ( 2 ( y - 2 ) / 3 )&#xA;x = -3&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Therefore my equation is correct?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;thanks for you help in advance guys.&lt;/p&gt;&#xA;" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-08-29T21:31:16.617" LastActivityDate="2010-08-30T11:34:52.533" Title="Algebra - Find the equation of the line perpendicular to 3x+2y-4=0 going through point (2,-3)" Tags="&lt;algebra-precalculus&gt;&lt;homework&gt;" AnswerCount="1" CommentCount="7" />
  <row Id="3614" PostTypeId="2" ParentId="3613" CreationDate="2010-08-29T21:37:12.580" Score="5" ViewCount="" Body="&lt;p&gt;If two lines are perpendicular, the product of their slopes is -1.  This is often restated as the slope of the line perpendicular to a given line is the opposite reciprocal of the slope of the given line.  For example, the line $6x-15y+3=0$ has slope $\frac{2}{5}$, so a line perpendicular to it will have slope $-\frac{5}{2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With that fact, you should be able to determine the slope of the line for which you are finding an equation, and you know a point on the line.  Those two pieces of information should be enough for you to write an equation of the line perpendicular to $3x+2y-4=0$ going through point $(2,-3)$.&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-08-29T21:37:12.580" CommentCount="1" />
  <row Id="3615" PostTypeId="2" ParentId="3573" CreationDate="2010-08-29T22:01:03.400" Score="2" ViewCount="" Body="&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/vRXKB.png&quot; alt=&quot;graph for k=10&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(diagram for k=10)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each region $S_n$ is a square with sides of slope $\pm 1$, center at $(n,n)$, and side length $\sqrt{2}(k-n)$.  Each pair of successive squares is positioned such that $S_{n+1}$ mostly overlaps $S_n$, but not quite.  As the top and right vertices of every square are, respectively, on the same horizontal and vertical line and 1 unit apart (the centers of successive squares are shifted 1 unit up and 1 unit to the right), the rectangular region of $S_{n+1}$ that is not inside $S_n$ sticks out by $\frac{1}{\sqrt{2}}$ and has length equal to the side length of $S_{n+1}$, which is $\sqrt{2}(k-(n+1))$, so this rectangular region has area $k-n-1$.  Thus, &#xA;$$\begin{align}&#xA;[R_n]&amp;amp;=(\text{area of }S_n)-(\text{area of }S_{n+1})+(\text{area of rectangular region})&#xA;\\&#xA;&amp;amp;=2(k-n)^2-2(k-n-1)^2+k-n-1&#xA;\\&#xA;&amp;amp;=5k-5n-3.&#xA;\end{align}$$&#xA;Now,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xA;\sum_{n=0}^{k-2}[R_n]&amp;amp;=\sum_{n=0}^{k-2}(5k-5n-3)&#xA;\\&#xA;&amp;amp;=5k\sum_{n=0}^{k-2}1-5\sum_{n=0}^{k-2}n-3\sum_{n=0}^{k-2}1&#xA;\\&#xA;&amp;amp;=5k(k-1)-5\left(\frac{(k-2)(k-1)}{2}\right)-3(k-1)&#xA;\\&#xA;&amp;amp;=\frac{10k^2-10k-5k^2+15k-10-6k+6}{2}&#xA;\\&#xA;&amp;amp;=\frac{5k^2-k-4}{2}.&#xA;\end{align}$$&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-08-29T22:01:03.400" CommentCount="3" />
  <row Id="3616" PostTypeId="2" ParentId="3602" CreationDate="2010-08-29T22:53:04.520" Score="2" ViewCount="" Body="&lt;p&gt;Here is a contrived example of the failure of the method.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Say you are given the function $f$ defined for $0 \leq x \leq 50$ as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle f(x) = 100 \sin\pi x, 0 \le x \le 1$ &lt;br/&gt;&#xA;$\displaystyle f(x) = x, 1 &amp;lt; x \le 50$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your task is to find the integer $n$ for which $f(n)$ is maximized. Your method gives $n=0$ or $n=1$ (max of function occurs at $x = 1/2$), while the correct value we need is $n=50$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason your method works for the example you have is that the function $\displaystyle S(t) = (100/t)^t$ is strictly increasing for $0 &amp;lt; t &amp;lt; 100/e$ and strictly decreasing for $100/e &amp;lt; t$.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-08-29T22:53:04.520" CommentCount="7" />
  <row Id="3617" PostTypeId="1" AcceptedAnswerId="3626" CreationDate="2010-08-30T00:12:15.020" Score="1" ViewCount="377" Body="&lt;p&gt;I am a statistician who wishes to learn real analysis in order to better understand the foundations of statistics. With that aim in mind I plan to go through Rudin's classic on &quot;Principles of Mathematical Analysis&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given the above context can I skip chapter 1? It seems to me that the material in chapter 1 is  not as important to someone with my goals. I understand that it may help in establishing the need for rigor in mathematical arguments but that is something I already appreciate. In particular, I wonder will skipping chapter 1 impede my understanding of the material in subsequent chapters?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any advice will be appreciated.&lt;/p&gt;&#xA;" OwnerUserId="1584" LastActivityDate="2010-08-30T11:03:54.400" Title="Can I skip the first chapter in Rudin's Principles of Mathematical Analysis?" Tags="&lt;soft-question&gt;&lt;real-analysis&gt;" AnswerCount="4" CommentCount="6" />
  <row Id="3618" PostTypeId="1" AcceptedAnswerId="3620" CreationDate="2010-08-30T00:51:48.890" Score="2" ViewCount="591" Body="&lt;p&gt;Is there any theory (analogous to Galois theory) for solving equations with irrational exponents like:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ x^{\sqrt{2}}+x^{\sqrt{3}}=1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;?&lt;/p&gt;&#xA;" OwnerUserId="1585" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2011-01-05T22:13:43.077" LastActivityDate="2011-01-05T22:13:43.077" Title="Solving an equation with irrational exponents" Tags="&lt;abstract-algebra&gt;&lt;irrational-numbers&gt;" AnswerCount="1" CommentCount="6" />
  <row Id="3619" PostTypeId="2" ParentId="3617" CreationDate="2010-08-30T00:57:32.447" Score="1" ViewCount="" Body="&lt;p&gt;I think you answered your own question. If you are willing to accept that the reals are the unique totally ordered archimedean field with the least upper bound property, then OK skip it.&#xA;But you may wish to refer back to it to know what all of these adjectives mean. &lt;/p&gt;&#xA;" OwnerUserId="722" LastActivityDate="2010-08-30T00:57:32.447" />
  <row Id="3620" PostTypeId="2" ParentId="3618" CreationDate="2010-08-30T01:05:55.390" Score="8" ViewCount="" Body="&lt;p&gt;The study of such equations is not &quot;abstract algebra&quot; as it is usually understood.  The reason is that to even define the function $x^{\sqrt{2}}$, for example, requires analysis; one has to prove certain properties of $\mathbb{R}$ to ensure that such a function exists.  This is in marked contrast to the case of integer or rational powers, where one has a purely algebraic definition and the background theory is equational.  To define the function $x^{\sqrt{2}}$ one has to either define $e^x$ and the logarithm or consider a limit of functions $x^{p_n}$ where $p_n$ form a sequence of rational approximations to $\sqrt{2}$, and this is irreducibly non-algebraic stuff.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, while polynomials can be studied in an absurdly general setting, transcendental equations like those you describe are more or less restricted to $\mathbb{R}$ (or $\mathbb{C}$ if you really want to pick a branch of the logarithm).  The LHS is an increasing function of $x$, so there is at most one root, which probably one can really only compute numerically if it exists.  (Its nonexistence can be ruled out by computing local minima in $(0, 1)$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is another question which touches on a theme which has come up several times on math.SE, which is that exponentiation should really not be thought of as one operation.  Instead, it is a collection of related operations with various degrees of generality and applicability which happen to share the same algebraic properties, and one should not infer too much about how similar these operations are.  &lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-08-30T01:50:09.997" LastActivityDate="2010-08-30T01:50:09.997" CommentCount="2" />
  <row Id="3621" PostTypeId="2" ParentId="3617" CreationDate="2010-08-30T02:28:20.500" Score="2" ViewCount="" Body="&lt;p&gt;It is probably less work to read through the first chapter than to ask if reading through it is necessary! Specially in the context of you planning to read through the whole book.&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2010-08-30T02:28:20.500" />
  <row Id="3622" PostTypeId="2" ParentId="3315" CreationDate="2010-08-30T03:32:35.663" Score="2" ViewCount="" Body="&lt;p&gt;As you can see, there are several ways to find the answer to your question, from jmoy's straightforward algebra calculation, to using Euler's formula $e^{r+i\theta} = e^r(\cos(\theta)+i \sin(\theta))$, to geometric interpretations of complex numbers.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One fact that hasn't been mentioned yet is that the complex numbers are &lt;strong&gt;algebraically closed.&lt;/strong&gt;  This means that every algebraic equation using complex numbers has all of its solutions in the complex numbers.  So, the equation $x^2=i$ has both of its solutions in the complex numbers; and the equations $x^4 = -7-12i$ and $x^4+(\pi -8i)x^3+x-\sqrt{5}=0$ each have all four of their solutions in the complex numbers.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The real numbers are not algebraically closed: the equations $x^2=-1$ and $x^4+7x^2+\pi=0$ cannot be solved unless we use complex numbers.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is one of the main reasons complex numbers are so important; they are the &lt;strong&gt;algebraic closure&lt;/strong&gt; of the real numbers.  You will never need &quot;higher levels&quot; of imaginary numbers or new mysterious square roots; numbers of the form $a+bi$ are all you need to find any root of real or complex polynomials.&lt;/p&gt;&#xA;" OwnerUserId="1461" LastActivityDate="2010-08-30T03:32:35.663" CommentCount="4" />
  <row Id="3623" PostTypeId="1" AcceptedAnswerId="3624" CreationDate="2010-08-30T05:31:48.753" Score="1" ViewCount="188" Body="&lt;p&gt;I would like to know something like what's the first PDE etc. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could you recommend book on the history of PDE/ODE?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;thanks.&lt;/p&gt;&#xA;" OwnerUserId="1382" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-08-30T05:59:23.760" LastActivityDate="2010-09-01T23:18:15.053" CommunityOwnedDate="2010-08-30T07:43:32.373" Title="Book recommendation on the history of PDE/ODE?" Tags="&lt;reference-request&gt;&lt;math-history&gt;&lt;differential-equations&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="3" />
  <row Id="3624" PostTypeId="2" ParentId="3623" CreationDate="2010-08-30T06:51:58.830" Score="1" ViewCount="" Body="&lt;p&gt;You can refer to these articles: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) &lt;a href=&quot;http://www.jstor.org/pss/2298771&quot; rel=&quot;nofollow&quot;&gt;http://www.jstor.org/pss/2298771&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) www.johnsasser.com/pdf/article13.pdf&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) &lt;a href=&quot;http://www.math.niu.edu/~rusin/known-math/index/34-XX.html#HIST&quot; rel=&quot;nofollow&quot;&gt;http://www.math.niu.edu/~rusin/known-math/index/34-XX.html#HIST&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bit of surfing yourself for things should help you out!&lt;/p&gt;&#xA;" LastActivityDate="2010-08-30T06:51:58.830" CommentCount="1" />
  <row Id="3625" PostTypeId="1" CreationDate="2010-08-30T09:59:38.337" Score="4" ViewCount="283" Body="&lt;p&gt;I want to fit to a dataset a power function ($y=Ax^B$). What is the best and easiest method to do this. I need the $A$ and $B$ parameters too.&#xA;I'm using in general financial data in my project, which looks like this:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;8553600 458.2&#xA;17193600    373.6&#xA;25833600    694.16&#xA;34646400    738.33&#xA;44064000    817.89&#xA;54259200    1040.67&#xA;67910400    1032.69&#xA;76291200    1222.1&#xA;84844800    1245.65&#xA;94089600    1217.44&#xA;102211200   1579.38&#xA;110592000   1859.24&#xA;118886400   1711.67&#xA;127612800   2303.62&#xA;136684800   2658.26&#xA;219196800   3669.23&#xA;225676800   3525.02&#xA;225763200   3749.27&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I need to implement the algorithm in a Java similar language called ActionScript.&lt;/p&gt;&#xA;" OwnerUserId="1541" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-19T00:55:44.507" LastActivityDate="2010-12-05T12:07:21.440" Title="easy to implement method to fit a power function (regression)" Tags="&lt;statistics&gt;&lt;algorithms&gt;&lt;regression&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="3626" PostTypeId="2" ParentId="3617" CreationDate="2010-08-30T10:05:05.417" Score="10" ViewCount="" Body="&lt;p&gt;I'll go through the paragraphs of the third edition, motivating why you should/shouldn't consider them&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;INTRODUCTION&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You should carefully read this. This paragraph is not so important for the subsequent development, but it's fundamental for your understanding of the utility of $\mathbb{R}$, it contains an enlightening example wich shows you (with some weird algebraic trick) that &lt;em&gt;$\mathbb{Q}$ contains gaps&lt;/em&gt; and so we really need to &quot;patch&quot; it in order to do interesting things.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;ORDERED SETS&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You should read only the definitions of &lt;em&gt;bound&lt;/em&gt; and &lt;em&gt;least upper bound&lt;/em&gt; (or &lt;em&gt;supremum&lt;/em&gt;). They really matter and are used, explicitly or not, in many theorems. In particular, the latter is subtle and you should practice with it, for example with exercises 4 and 5 at the end of the chapter. The rest of the paragraphs regards ordered fields, you can skip it if you are used to work with $\mathbb{R}$ and its ordering, and this practical experience should suffice to you. Perhaps, you could find strange and interesting that $\mathbb{C}$ cannot be ordered without destroying its algebraic properties.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;FIELDS&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can skip this. If your interest is real analysis, your only field will be $\mathbb{R}$ (perhaps $\mathbb{C}$ sometimes) and as said above, the practical properties of field and ordering should be enough for you (e.g. if $a,b,c \in \mathbb{R}$ and $a &amp;lt; b$ then $a + c &amp;lt; b + c$, you cannot divide by zero, etc.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;THE REAL FIELD&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are two ways to build $\mathbb{R}$. The first is &lt;em&gt;axiomatic&lt;/em&gt;: you say &quot;how I'd like to work in place that has such property&quot; and magic! You have it by axiom. The second way is &lt;em&gt;contructive&lt;/em&gt;: you take $\mathbb{Q}$, do something on it and come up with a mathematical structure that act as $\mathbb{R}$, has the properties of $\mathbb{R}$ and you eventually call it $\mathbb{R}$. It's very subtle and not practically useful, you should skip the latter method, reported in the appendix of the chapter, and know that when you follow the axiomatic method your are speaking of something that exists, in some mathematical sense. You should also consider theorem 1.20 (archimedean property and density of $\mathbb{Q}$ in $\mathbb{R}$) if you don't read the proof, at least carefully read the statement, it's very used and justify some mysterious things as: if $a \in \mathbb{R}$ and $0 \leq a &amp;lt; \epsilon$ for all $\epsilon &gt; 0$ then $a = 0$. Jump over the existence of the n-th root of a positive real, it's intuitive and you can prove it later in different (and simpler) ways.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;THE EXTENDED REAL NUMBER SYSTEM&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Not only it's not very useful, I think it's dangerous to introduce symbols for infinity when someone still isn't completely conscious of what infinity is and how it acts in many theorems of analysis. Skip.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;THE COMPLEX FIELD&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As in the case of the real field, if you know what $\mathbb{C}$ is and how to work with it, you can safely skip this, or read it later if you need. The only thing that you probably need are the &lt;a href=&quot;http://en.wikipedia.org/wiki/Triangular_inequality&quot; rel=&quot;nofollow&quot;&gt;trianglular inequality&lt;/a&gt; in theorem 1.33 and theorem 1.35, known as &lt;a href=&quot;http://en.wikipedia.org/wiki/Cauchy-Schwartz&quot; rel=&quot;nofollow&quot;&gt;Cauchy–Schwarz inequality&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EUCLIDEAN SPACES&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Read it, it's used in the following chapters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;APPENDIX&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As said above, skip it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EXERCISES&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As said above, 4 and 5 are very useful. I also suggest you to work on 6 and 7, they teach you what we mean when we say things like $3^{\pi}$ or when we talk about logarithms.&lt;/p&gt;&#xA;" OwnerUserId="365" LastEditorUserId="365" LastEditorDisplayName="" LastEditDate="2010-08-30T10:26:49.283" LastActivityDate="2010-08-30T10:26:49.283" CommentCount="3" />
  <row Id="3627" PostTypeId="1" AcceptedAnswerId="3634" CreationDate="2010-08-30T10:11:14.363" Score="3" ViewCount="172" Body="&lt;p&gt;I have a matrix $F=XDX^T$, where $D$ is $m\times m$ and diagonal and $X$ is $n\times m$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, I compute $F^{-1}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there an efficient method to update $F^{-1}$ if $D$ is updated by $D'=D+G$, where $G$ is a sparse diagonal matrix?&lt;/p&gt;&#xA;" OwnerUserId="1588" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-08-30T16:21:33.040" LastActivityDate="2010-10-02T13:32:10.003" Title="Diagonal update of the inverse of $XDX^T$" Tags="&lt;linear-algebra&gt;&lt;matrices&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="2" />
  <row Id="3628" PostTypeId="1" AcceptedAnswerId="3660" CreationDate="2010-08-30T10:58:22.460" Score="5" ViewCount="555" Body="&lt;p&gt;Hi! I'm practicing some problems from past exams and found this one: Find nth derivative of this function:&lt;br&gt;&#xA; $f(x)=\frac {x} {x^2-1}$.&lt;br&gt;&#xA; I have no idea how to start solving this problems. Is there any theorem for finding nth derivative?&lt;/p&gt;&#xA;" OwnerUserId="348" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-08-30T16:17:12.457" LastActivityDate="2010-08-30T19:34:42.577" Title="Nth derivative of a function: I don't know where to start." Tags="&lt;calculus&gt;" AnswerCount="5" CommentCount="3" FavoriteCount="2" />
  <row Id="3629" PostTypeId="2" ParentId="3617" CreationDate="2010-08-30T11:03:54.400" Score="2" ViewCount="" Body="&lt;p&gt;From a strategic point of view: given that mathematical terminology and formalism are not &lt;em&gt;totally&lt;/em&gt; standardised, it's worth at least reading through the chapter quickly, at least to stop yourself later having to refer back to what the various terms and symbols mean.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This advice goes for any book, I guess.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the point of view of content: no one but you is in a position to really decide whether you understand the material enough to skip the chapter.&lt;/p&gt;&#xA;" OwnerUserId="172" LastActivityDate="2010-08-30T11:03:54.400" CommentCount="1" />
  <row Id="3630" PostTypeId="2" ParentId="3628" CreationDate="2010-08-30T11:08:40.027" Score="8" ViewCount="" Body="&lt;p&gt;Split it into partial fractions then differentiate.&lt;/p&gt;&#xA;" OwnerUserId="1301" LastActivityDate="2010-08-30T11:08:40.027" CommentCount="2" />
  <row Id="3631" PostTypeId="2" ParentId="3609" CreationDate="2010-08-30T11:14:49.867" Score="3" ViewCount="" Body="&lt;p&gt;Here's the flaw in the compactness argument you sketched. In the model you construct, the interpretation of $c$ will indeed be a nonstandard number whose Goodstein sequence does not halt in a standard number of steps. But Goodstein's theorem has a universal quantifier for the number of steps, so nonstandard &quot;Goodstein sequences&quot; are also permitted. That is, when the theorem says &quot;there is a finite sequence&quot;, in a particular model that &quot;finite&quot; sequence might have the length of a nonstandard number, which cannot be excluded by the sequence of sentences in your argument. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One way to find an extension of PA where Goodstein's theorem is provable is to add enough transfinite induction to PA to formalize the usual proof of Goodstein's theorem. The details of how to do that are not so bad, although they may take too much time for an elementary class. &lt;/p&gt;&#xA;" OwnerUserId="630" LastEditorUserId="630" LastEditorDisplayName="" LastEditDate="2010-08-30T16:54:07.170" LastActivityDate="2010-08-30T16:54:07.170" CommentCount="1" />
  <row Id="3632" PostTypeId="2" ParentId="3628" CreationDate="2010-08-30T11:22:22.577" Score="6" ViewCount="" Body="&lt;p&gt;To add to Derek's hint: you will have to show the validity of the formula&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{\mathrm{d}^k}{\mathrm{d}x^k}\frac1{x}=\frac{(-1)^k k!}{x^{k+1}}$&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-08-30T11:22:22.577" CommentCount="1" />
  <row Id="3633" PostTypeId="1" CreationDate="2010-08-30T11:24:04.010" Score="10" ViewCount="614" Body="&lt;p&gt;There are some math quizzes like:&lt;blockquote&gt;&#xA;find a function $\phi:\mathbb{R}\rightarrow\mathbb{R}$ &#xA;such that $\phi(\phi(x)) = f(x) \equiv x^2 + 1.$&#xA;&lt;/blockquote&gt;&#xA;If such $\phi$ exists (it does in this example), $\phi$ can be viewed as a &quot;square root&quot; of $f$ in the sense of function composition because $\phi\circ\phi = f$.  Is there a general theory on the mathematical properties of this kind of square roots? (For instance, for what $f$ will a real analytic $\phi$ exist?)&lt;/p&gt;&#xA;" OwnerUserId="1551" LastEditorUserId="171" LastEditorDisplayName="" LastEditDate="2010-09-04T15:13:42.063" LastActivityDate="2011-01-02T20:44:33.180" Title="Square root of a function (in the sense of composition)" Tags="&lt;real-analysis&gt;&lt;functional-analysis&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="1" />
  <row Id="3634" PostTypeId="2" ParentId="3627" CreationDate="2010-08-30T11:24:09.673" Score="3" ViewCount="" Body="&lt;p&gt;Yes, there is a very efficient way to compute the updated $F^{-1}$. Begin by writing $X$ in terms of its columns: $X = [x_1 x_2 \dots x_m]$. It follows that:&#xA;$$&#xA;XGX^T = \sum_{i=1}^{m} G_{ii} x_ix_i^T = \sum_{i\in S} G_{ii} x_ix_i^T&#xA;$$&#xA;where $S \subset \{1,\dots,m\}$ is the set of indices for which $G_{ii}$ is nonzero. Suppose that $S$ contains $r$ indices. Since $G$ is sparse by assumption, $r$ is much smaller than $m$. If we assemble the columns of $X$ corresponding to indices in $S$, we obtain a new matrix $Y$, such that:&#xA;$$&#xA;XGX^T = YHY^T&#xA;$$&#xA;where $Y$ is $m \times r$, and $H$ is a diagonal matrix whose diagonal consists of all the nonzero entries along the diagonal of $G$. Now apply the &lt;a href=&quot;http://en.wikipedia.org/wiki/Woodbury_matrix_identity&quot; rel=&quot;nofollow&quot;&gt;Woodbury matrix identity&lt;/a&gt;:&#xA;$$&#xA;(F+YHY^T)^{-1} = F^{-1} - F^{-1}Y(H^{-1}+Y^TF^{-1}Y)^{-1}Y^TF^{-1}&#xA;$$&#xA;Computing the left-hand side directly is costly, as it requires inverting an $m\times m$ matrix. However, since $F^{-1}$ has already been computed, we can compute the right-hand side efficiently; the only inverse required is that of an $r\times r$ matrix, and $r$ is much smaller than $m$.&lt;/p&gt;&#xA;" OwnerUserId="1356" LastActivityDate="2010-08-30T11:24:09.673" CommentCount="1" />
  <row Id="3636" PostTypeId="2" ParentId="3625" CreationDate="2010-08-30T12:03:05.397" Score="5" ViewCount="" Body="&lt;p&gt;There's the obvious approach of proceeding based on taking the logarithm of both sides of your regression formula:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ln\;y=\ln\left(ax^b\right)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ln\;y=\ln\;a+b\ln\;x$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which can be seen to be an expression of the form&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$v=k+bu$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $v=\ln\;y$, $u=\ln\;x$, and $k=\ln\;a$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now a linear regression in the variables $v$ and $u$ applies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, formula numbers 16, 20, 27 and 28 of &lt;a href=&quot;http://mathworld.wolfram.com/LeastSquaresFitting.html&quot; rel=&quot;nofollow&quot;&gt;this page&lt;/a&gt; now apply.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you have the slope $b$ and the intercept $k$ of the linear transformation, $a$ is just the exponential of the intercept ($\exp\;k$), and $b$ is the slope of the transformed line.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I note here the possible caveat of the fit parameters being biased towards the data with ordinates small in magnitude. The &lt;em&gt;rigorous&lt;/em&gt; way of going about it would be to treat the parameters from the linear regression as provisional and then apply a nonlinear least-squares algorithm like Levenberg-Marquardt to the data, using the parameters from the linear regression as a starting point. This may or may not be needed though; it really depends on the data you have.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I'll polish off comments I gave earlier: again, the problem with using the logarithm to linearize your nonlinear function is that it tends to overemphasize the errors in the small values of y. Remember that the assumption of linear least squares is that the abscissas are accurate, but the ordinates are contaminated by error.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, the $y_i$ are actually of the form $\hat{y}_i\pm\sigma_i$ where the $\hat{y}_i$ are the &quot;true values&quot; (presumably unknown), and the $\sigma_i$ are inherent uncertainties. If one takes the logarithms of the $y_i$, the uncertainties are also transformed, and we have to take this into account.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The key formula is that if the $y_i$ are transformed by a function $f(y)$, the $\sigma_i$ are transformed according to $f&amp;#39;(y_i)\sigma_i$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the case at hand, the objective function we now have to minimize is of the form&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f(a,b)=\sum_i{y_i^2\;(\ln\;y_i-\ln\;a-b\ln\;x_i)^2}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and we have to modify the formulae for linear regression accordingly:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$m=\sum_i y_i^2$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\bar{x}=\frac{\displaystyle \sum_i y_i^2 \ln\;x_i}{m}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$t=\sum_i y_i^2 (\ln\;x_i-\bar{x})^2$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;whereupon&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$b=\frac{\displaystyle \sum_i y_i^2\ln\;y_i (\ln\;x_i-\bar{x})}{t}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$a=\exp\left(\frac{\displaystyle \sum_i y_i^2\ln\;y_i}{m}-b\bar{x}\right)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These should be better provisional values for later polishing.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;It turns out that for a separable nonlinear fit (linear in one of the parameters), the NLLS problem greatly simplifies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remember that the &lt;em&gt;actual&lt;/em&gt; quantity that we have to minimize is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$F(a,b)=\sum_i{(y_i-a x_i^b)^2}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we take the gradient $\nabla F(a,b)$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\nabla F(a,b)=(2a\sum_i x_i^{2b}-2\sum_i x_i^b y_i\quad 2a\sum_i \ln\left(x_i\right) x_i^{2b}-2\sum_i \ln\left(x_i\right) x_i^b y_i)^T$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;equate both components to 0, and then eliminate the linear parameter $a$, we get the univariate nonlinear equation in $b$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left(\sum_i x_i^b y_i\right)\left(\sum_i \ln\left(x_i\right) x_i^{2b}\right)-\left(\sum_i x_i^{2b}\right)\left(\sum_i \ln\left(x_i\right) x_i^b y_i\right)=0$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which can be attacked with standard techniques, e.g. the secant method or Newton's method. But, how do we start the iteration? Why, with the provisional $b$ we got from the (weighted) linear regression earlier!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Having gotten $b$, it is now a simple matter to get $a$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$a=\left(\sum_i x_i^b y_i\right) / \left(\sum_i x_i^{2b}\right)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and you now have your parameters.&lt;/p&gt;&#xA;" OwnerUserId="498" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-05T12:07:21.440" LastActivityDate="2010-12-05T12:07:21.440" CommentCount="8" />
  <row Id="3637" PostTypeId="1" AcceptedAnswerId="3638" CreationDate="2010-08-30T12:27:46.913" Score="10" ViewCount="347" Body="&lt;p&gt;I know, fundamental group of GL(n,C) is isomorphic to Z. It's written &lt;a href=&quot;http://en.wikipedia.org/wiki/General_linear_group#Complex_case&quot; rel=&quot;nofollow&quot;&gt;in wikipedia&lt;/a&gt;. Actually, I've succeed in proving this, but my proof is two pages long and very technical. I want&lt;br&gt;&#xA;1. to find some better proofs of this fact (in order to compare to mine);&lt;br&gt;&#xA;2. to find some book or article, using which I can learn, how to calculate fundamental groups and, more generally, connectedness components of maps from one space to another;&lt;br&gt;&#xA;3. to find something for the reference, which I can use in order to learn, how to write this proofs nicely, using standard terminology.&lt;/p&gt;&#xA;" OwnerUserId="695" LastActivityDate="2010-08-30T14:00:48.943" Title="Fundamental group of GL(n,C) is isomorphic to Z. How to learn to prove facts like this?" Tags="&lt;topology&gt;&lt;algebraic-topology&gt;" AnswerCount="3" FavoriteCount="3" />
  <row Id="3638" PostTypeId="2" ParentId="3637" CreationDate="2010-08-30T13:20:45.927" Score="11" ViewCount="" Body="&lt;p&gt;The first thing you have to do is to note that the inclusion $U(n)\to\mathrm{GL}(n,\mathbb C)$ induces an isomorphism on the fundamental groups. This can be done by noting that a loop in $\mathrm{GL}(n,\mathbb C)$ can be deformed to one in $U(n)$ by performing the Gram-Schmidt procedure at each point of the loop, and checking that this can be done continuously and so on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next, considering the beginning of the long exact sequence for the homotopy groups of the spaces appearing in the fibration $$U(n-1)\to U(n)\to S^{2n-1}$$ which arises from the transitive linear action of $U(n)$ on $S^{2n-1}\subseteq\mathbb C^{n}$ you can prove, by induction, that the inclusion $U(1)\to U(n)$ induces an isomorphism on fundamental groups.&#xA;Then you can explicitly describe $U(1)$ as a space homeomorphic to $S^1$.&lt;/p&gt;&#xA;" OwnerUserId="274" LastEditorUserId="274" LastEditorDisplayName="" LastEditDate="2010-08-30T13:34:42.150" LastActivityDate="2010-08-30T13:34:42.150" CommentCount="0" />
  <row Id="3639" PostTypeId="2" ParentId="3637" CreationDate="2010-08-30T13:34:03.417" Score="2" ViewCount="" Body="&lt;p&gt;For 2. and 3. I can recommend Allen Hatcher's Algebraic Topology book which can be accessed free from the author's webpage: &lt;a href=&quot;http://www.math.cornell.edu/~hatcher/AT/ATpage.html&quot; rel=&quot;nofollow&quot;&gt;http://www.math.cornell.edu/~hatcher/AT/ATpage.html&lt;/a&gt;&lt;/p&gt;&#xA;" LastActivityDate="2010-08-30T13:34:03.417" />
  <row Id="3640" PostTypeId="2" ParentId="3637" CreationDate="2010-08-30T14:00:48.943" Score="7" ViewCount="" Body="&lt;p&gt;The method Mariano discusses in his answer is absolutely the way that mathematicians compute fundamental groups (and also higher homotopy groups) of Lie groups.  Here I just want to mention how his first step applies in a more general context.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Concerning $\operatorname{GL}_n(\mathbb{C})$: the unitary group $U(n)$ is a maximal compact subgroup of $\operatorname{GL}_n(\mathbb{C})$, and moreover any maximal compact subgroup is conjugate to $U(n)$.  (This can be seen by considering Hermitian forms, c.f. e.g. Section 1 of &lt;a href=&quot;http://math.uga.edu/~pete/8410Chapter9.pdf&quot; rel=&quot;nofollow&quot;&gt;http://math.uga.edu/~pete/8410Chapter9.pdf&lt;/a&gt;.)  Moreover, the Gram-Schmidt process gives a &lt;a href=&quot;http://en.wikipedia.org/wiki/Deformation_retraction&quot; rel=&quot;nofollow&quot;&gt;deformation retraction&lt;/a&gt; from $\operatorname{GL}_n(\mathbb{C})$ to $U(n)$, hence these two spaces are homotopy equivalent.  And in fact even more is true: there exists a finite-dimensional Euclidean space $E$ such that $\operatorname{GL}_n(\mathbb{C})$ is homeomorphic to $U(n) \times E$: this is the &lt;a href=&quot;http://en.wikipedia.org/wiki/QR_decomposition#Square_matrix&quot; rel=&quot;nofollow&quot;&gt;QR decomposition&lt;/a&gt;.  Moreover:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Everything in 1) goes over verbatim for $\operatorname{GL}_n(\mathbb{R})$ with the unitary group $U(n)$ replaced by the orthogonal group $O(n)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) For any &lt;a href=&quot;http://en.wikipedia.org/wiki/Reductive_group&quot; rel=&quot;nofollow&quot;&gt;reductive group&lt;/a&gt; $G$ over $\mathbb{R}$ or $\mathbb{C}$, there exists a maximal compact subgroup $K$, any two such are conjugate in $G$, and $G$ is homeomorphic to the product of $K$ with a finite-dimensional Euclidean space.  This last fact is a consequence of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Iwasawa_decomposition&quot; rel=&quot;nofollow&quot;&gt;Iwasawa decomposition&lt;/a&gt;, a far-reaching generalization of the QR-decomposition.  &lt;/p&gt;&#xA;" OwnerUserId="299" LastActivityDate="2010-08-30T14:00:48.943" />
  <row Id="3641" PostTypeId="2" ParentId="3583" CreationDate="2010-08-30T14:12:06.803" Score="4" ViewCount="" Body="&lt;p&gt;The problem is not well posed, since you don't specify which woman leaves the train at each station. See &lt;a href=&quot;http://en.wikipedia.org/wiki/Ross%E2%80%93Littlewood_paradox&quot; rel=&quot;nofollow&quot;&gt;Ross–Littlewood paradox&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="89" LastActivityDate="2010-08-30T14:12:06.803" />
  <row Id="3643" PostTypeId="2" ParentId="3628" CreationDate="2010-08-30T14:36:11.403" Score="6" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\;\;\;$ Upon employing partial fractions it reduces to $\;\;\;\rm D^n \:\frac{1}{x+1}. \;\;$ Now employ the Taylor series&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\rm\;  f(t+x) = \sum_{k=0}^\infty \;\; D^n \: f(x) \; \frac{t^n}{n!}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and note that for this problem  we've  $\rm\; f(t+x) = \frac{1}{t+x+1}$ is a geometric series with known coefficients.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Such &quot;generating function&quot; approaches often work smoothly even in much more complicated problems. Indeed, there is a very powerful &lt;strong&gt;Umbral calculus&lt;/strong&gt; that frequently succeeds in computing such closed form expressions, e.g. see Steven Roman: The Umbral Calculus. 1984. For example, below is a small sample of derivative formulas for the countless number of polynomial sequences amenable to umbral calculus analysis&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{array}{|r|l|} &#xA;\hline&#xA;\rm Name &amp;amp; \rm Derivative \; formula \\&#xA;\hline\hline&#xA;\rm Laguerre  &amp;amp; \rm L_n^k(x) = (D+1)^{n+k}(-x)^n \\&#xA;\rm Exponential &amp;amp; \rm\;\; e_n(x) = e^{-x}(xD)^n e^x  \\&#xA;\rm Abel &amp;amp; \rm A_n^k(x) = x \; e^{-knD} x^{n-1} \\&#xA;\rm Hermite  &amp;amp; \rm H_n^k(x) = (-1)^n e^{x^2/(2n)} (kD)^n e^{-x^2/(2n)} \\&#xA;\rm Bernoulli &amp;amp; \rm B_n^k(x) = \left(\frac{D}{e^D-1}\right)^k  x^n \\&#xA;\rm Euler  &amp;amp; \rm E_n^k(x) = \left(\frac{2}{e^D+1}\right)^k  x^n \\&#xA;\end{array}$$&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-08-30T16:59:11.497" LastActivityDate="2010-08-30T16:59:11.497" CommentCount="1" />
  <row Id="3644" PostTypeId="1" CreationDate="2010-08-30T15:35:10.440" Score="1" ViewCount="172" Body="&lt;p&gt;Is there a easy way to compute the coefficients of the power series which represents $\frac{x - x^k + x^{k+1}}{1-2x + x^k - x^{k+1}}$. I am currently solving this by assuming it has the form $a_1x+a_2x^2+a_3x^3+\dots+a_nx^n+\dots$ and solving tedious recurrences obtained from $x - x^k + x^{k+1} = (1-2x + x^k - x^{k+1}) \times (a_1x+a_2x^2+a_3x^3+\dots+a_nx^n+\dots)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Motivation: If my calculations are right, the above is the generating function of the number of compositions of an integer which do not involve a given integer $k$. &lt;/p&gt;&#xA;" OwnerUserId="404" LastActivityDate="2010-08-30T17:40:49.340" Title="Computing the nth coefficient of the power series representing a given rational function" Tags="&lt;combinatorics&gt;&lt;complex-analysis&gt;" AnswerCount="2" CommentCount="8" FavoriteCount="1" />
  <row Id="3645" PostTypeId="1" AcceptedAnswerId="3689" CreationDate="2010-08-30T16:00:36.980" Score="3" ViewCount="153" Body="&lt;p&gt;Being motivated by &lt;a href=&quot;http://math.stackexchange.com/questions/3618/solving-an-equation-with-irrational-exponents&quot;&gt;this post&lt;/a&gt;, I was wondering if there is a proof (analogous to the case of Diophantine equations) that there is no general method for solving &lt;a href=&quot;http://en.wikipedia.org/wiki/Transcendental_equation&quot; rel=&quot;nofollow&quot;&gt;transcendental equations&lt;/a&gt;?  It seems pretty clear, intuitively, that there can be no general method; but the only reason I feel strongly about that is because I can't conceive that transcendental equations have a general method, but Diophantine equations don't.  I was never able to understand the proof for the case of Diophantine equations, so I am not in a position to even know where to begin thinking about this.  Has any work been done on this problem?&lt;/p&gt;&#xA;" OwnerUserId="1417" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-08-30T16:16:07.277" LastActivityDate="2010-08-31T04:32:55.990" Title="Is there a proof that there is no general method to solve transcendental equations?" Tags="&lt;analysis&gt;&lt;transcendence-theory&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" />
  <row Id="3646" PostTypeId="1" AcceptedAnswerId="3666" CreationDate="2010-08-30T16:20:00.250" Score="8" ViewCount="314" Body="&lt;p&gt;Since there is a 2-1 homomorphism from SU(2)xSU(2) to SO(4) there should be a way to recover the two SU(2) matrices given an SO(4) matrix.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe I could set this up as a system of equations using the map from above and solve for the coefficients of the SU(2) matrices.  However, I wonder if anyone has already done this or can point me to the forumlas.&lt;/p&gt;&#xA;" OwnerUserId="1596" LastActivityDate="2010-08-31T06:44:13.347" Title="Recovering the two SU(2) matrices from SO(4) matrix" Tags="&lt;matrices&gt;&lt;lie-groups&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="1" />
  <row Id="3647" PostTypeId="2" ParentId="3644" CreationDate="2010-08-30T16:42:50.167" Score="1" ViewCount="" Body="&lt;p&gt;There is an easier way to obtain the recurrences.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First rewrite in the more manageable form&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle g(x) = \frac{1-x}{1-2x+x^k-x^{k+1}} - 1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $\displaystyle f(x) = \frac{1-x}{1-2x+x^k-x^{k+1}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then we have that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle f(x)(1-2x+x^k-x^{k+1}) = 1-x$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now use Leibniz's formula for the $n^{th}$ derivative of a product.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sum_{r=0}^{n} {n \choose r} {f^{(n-r)'}(x)(1-2x+x^{k}-x^{k+1})^{r'}} = (1-x)^{n'}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;($r'$ is the $r^{th}$ derivative)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are interested in this when $x = 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only terms that matter are on the LHS would be for $r=0,1,k,k+1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The coefficient of the power series can be gotten by dividing the $n^{th}$ derivative at $0$ by $n!$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can probably also see all the above if you do a partial fraction reduction of the rational function.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-08-30T17:40:49.340" LastActivityDate="2010-08-30T17:40:49.340" CommentCount="2" />
  <row Id="3648" PostTypeId="2" ParentId="3644" CreationDate="2010-08-30T17:18:43.923" Score="2" ViewCount="" Body="&lt;p&gt;Depends on what you want to do with them.  If you want to prove things about them, there are theorems which tell you what such sequences look like in terms of the reciprocals of the roots of the denominator, but if you can't solve for the roots explicitly these are not useful for computation (although they tell you the asymptotic behavior rather precisely).  I wrote &lt;a href=&quot;http://web.mit.edu/~qchu/Public/TopicsInGF.pdf&quot; rel=&quot;nofollow&quot;&gt;some notes&lt;/a&gt; on this and other topics in generating functions if you're interested (Section 5).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you actually want to compute them for large indices, there is a way to do it faster than just running the recurrence.  The idea is that for sufficiently large indices $n$ there exists a matrix $A$ and vectors $v, w$ such that $a_n = v^T A^n w$ and one can use binary exponentiation to compute $A^n$ when $n$ is large.  In fact one can take $A$ to be the &lt;a href=&quot;http://en.wikipedia.org/wiki/Companion_matrix&quot; rel=&quot;nofollow&quot;&gt;companion matrix&lt;/a&gt; of the characteristic polynomial.  A special case of this idea includes the doubling formulas for the Fibonacci numbers.  For sequences with combinatorial interpretations it is often possible to choose $A$ to be the adjacency matrix of a graph and to choose particularly nice $v$ and $w$.  Stanley's &lt;em&gt;Enumerative Combinatorics&lt;/em&gt;, I think, discusses such issues; two keywords here are &quot;regular languages&quot; and &quot;finite state machines.&quot;&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-08-30T17:18:43.923" CommentCount="2" />
  <row Id="3649" PostTypeId="1" AcceptedAnswerId="3653" CreationDate="2010-08-30T17:26:03.807" Score="3" ViewCount="164" Body="&lt;p&gt;Let $S$ be some finite set. Let $Sym(S)$ be the symmetric group on $S$. Let $A$ be the subset of permutations $\sigma$ that are such that there exist $a,b,c \in S$ such that $\sigma(a) = b, \sigma (b) = c$ and $\sigma (c) =a$ and for every other element of $S$, the action of $\sigma$ is the identity. In short, let $A$ be the set of cycles of length 3.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do you show that the subgroup generated by $A$ is not the whole of $Sym(S)$?&lt;/p&gt;&#xA;" OwnerUserId="172" LastActivityDate="2010-08-30T17:54:17.950" Title="How do you prove cycles of length 3 don't generate the whole symmetric group?" Tags="&lt;group-theory&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="1" />
  <row Id="3650" PostTypeId="2" ParentId="3649" CreationDate="2010-08-30T17:28:27.637" Score="3" ViewCount="" Body="&lt;p&gt;Consider the parity.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-08-30T17:28:27.637" />
  <row Id="3651" PostTypeId="2" ParentId="3645" CreationDate="2010-08-30T17:29:13.200" Score="3" ViewCount="" Body="&lt;p&gt;Intuition may be misleading here. In fact transcendental cases are often much easier than the integer Diophantine case. For example below is a table listing the known decidability results in various rings for Hilbert's tenth problem and the full first order theory, excerpted from Bjorn Poonen's interesting paper &lt;a href=&quot;http://math.mit.edu/~poonen/papers/aws2003.pdf&quot; rel=&quot;nofollow&quot;&gt;Hilbert's tenth problem over rings of number-theoretic interest&lt;/a&gt;&lt;br&gt;&#xA;&lt;img src=&quot;http://i.imgur.com/yNEos.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-08-30T17:29:13.200" />
  <row Id="3652" PostTypeId="2" ParentId="3649" CreationDate="2010-08-30T17:30:48.270" Score="4" ViewCount="" Body="&lt;p&gt;Consider the action of $S_n$ on the ring of polynomials $R=\mathbb R[x_1,\dots,x_n]$ given by permutation of the variables, and consider the element $$\Delta=\prod_{1\leq i&amp;lt;j\leq n}(x_i-x_j)\in R.$$ Then every $3$-cycle fixes $\Delta$, so the whole subgroup of $S_n$ generated by $3$-cycles acts trivially on $\Delta$. Yet there are elements of $S_n$, like all transpositions, which do &lt;em&gt;not&lt;/em&gt; act trivially on $\Delta$ but map it to its opposite. Therefore $3$-cycles do not generate $S_n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One can alternatively use the action of $S_n$ on $\mathbb R^n$, again by permutation but now of the coordinates. Then the matrix corresponding to each $3$-cycle has determinant $1$, so every element of the subgroup generated by $3$-cycles is acts on $\mathbb R^n$ through a matrix with determinant $1$. Yet there are elements in $S_n$ whose corresponding matrix is of determinant $-1$: the same conclusion as before follows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;BN:&lt;/strong&gt; The point of doing this in this way is to avoid having to check first that the parity of elements of $S_n$ is well-defined; all answers so far depend on doing this (I wonder if it is at all side-steppable...). In fact, this can be massaged easily into becoming a proof of that very same fact.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a more or less canonical example of why we want to study representations of groups. &lt;/p&gt;&#xA;" OwnerUserId="274" LastEditorUserId="274" LastEditorDisplayName="" LastEditDate="2010-08-30T17:54:17.950" LastActivityDate="2010-08-30T17:54:17.950" />
  <row Id="3653" PostTypeId="2" ParentId="3649" CreationDate="2010-08-30T17:31:30.450" Score="5" ViewCount="" Body="&lt;p&gt;Look at the &lt;a href=&quot;http://en.wikipedia.org/wiki/Parity_of_a_permutation&quot; rel=&quot;nofollow&quot;&gt;sign&lt;/a&gt; of a permutation. If a permutation can be decomposed into a product of even number of transpositions (aka 2-cycles), we give it a sign of 1. Otherwise give it a sign of -1. It turns out that this is a well-defined notion. The group generated by the even permutations cannot cover the odd ones, thus cannot be the whole symmetric group.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S. The group generated by the even permutations is usually called the alternating group.&lt;/p&gt;&#xA;" OwnerUserId="325" LastActivityDate="2010-08-30T17:31:30.450" />
  <row Id="3655" PostTypeId="1" AcceptedAnswerId="3659" CreationDate="2010-08-30T19:01:29.603" Score="3" ViewCount="255" Body="&lt;p&gt;In one of my classes we discussed the ring of 2x2 matrices $M_{2}(\mathbb{Z})$. We said that its group of units was $SL_{2}(\mathbb{Z})$ which means that it is the set of 2x2 with determinant equal to $\pm$1.&#xA;Why can't we have a 2x2 matrix with entries a,b,c, and d such that $\frac{a}{ad-bc}$,$\frac{-b}{ad-bc}$,$\frac{-c}{ad-bc}$, and $\frac{d}{ad-bc}$ are all integers?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm sure its a simple contradiction argument, but I couldn't see it. So if anyone knows a quick elementary argument, it'd be greatly appreciated&lt;/p&gt;&#xA;" OwnerUserId="249" LastEditorUserId="664" LastEditorDisplayName="" LastEditDate="2011-01-30T18:45:05.010" LastActivityDate="2011-01-30T18:45:05.010" Title="Units of $M_2(Z)$" Tags="&lt;linear-algebra&gt;&lt;ring-theory&gt;" AnswerCount="4" CommentCount="9" />
  <row Id="3656" PostTypeId="2" ParentId="3628" CreationDate="2010-08-30T19:03:24.810" Score="1" ViewCount="" Body="&lt;p&gt;According to the Binomial Theorem (or using the usual formula for the sum of a geometric series with initial term 1 and common ratio $x^2$),&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f(x)=\frac {x} {x^2-1} = -x \frac {1} {1 - x^2} = -x \left( 1 + x^2 + x^4 + \cdots + x^{2n} + \cdots \right)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$= -x - x^3 - x^5 - \cdots - x^{2n+1} - \cdots$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because the right side converges absolutely for $|x^2| &amp;lt; 1$ you can differentiate it term by term, introducing a coefficient $(2n+1)(2n) \cdots (2n+1-k+1)$ for $x^{2n+1-k}$; in other words, the coefficient of $x^j$ is $(j+1)(j+2) \cdots (j+k)$.  Dividing the entire thing through by $k!$ gives a series you can easily relate to the binomial expansion of $( 1 - x^2 ) $ to a &lt;em&gt;negative&lt;/em&gt; integral value, yielding a closed form solution.&lt;/p&gt;&#xA;" OwnerUserId="1489" LastActivityDate="2010-08-30T19:03:24.810" />
  <row Id="3657" PostTypeId="2" ParentId="3655" CreationDate="2010-08-30T19:04:27.540" Score="4" ViewCount="" Body="&lt;p&gt;The determinant of the inverse is the inverse of the determinant, whence the determinant must be an integral unit.&lt;/p&gt;&#xA;" OwnerUserId="1489" LastActivityDate="2010-08-30T19:04:27.540" CommentCount="0" />
  <row Id="3658" PostTypeId="2" ParentId="3655" CreationDate="2010-08-30T19:04:50.833" Score="6" ViewCount="" Body="&lt;p&gt;The map $\det:M_2(\mathbb Z)\to\mathbb Z$ is multiplicative, and the determinant of the identity matrix is $1$. It follows from this that the determinant of any inversible element of $M_2(\mathbb Z)$ must be an inversible element of $\mathbb Z$. There are only two such elements in $\mathbb Z$, &lt;em&gt;v.g.&lt;/em&gt; $1$ and $-1$.&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2010-08-30T19:04:50.833" />
  <row Id="3659" PostTypeId="2" ParentId="3655" CreationDate="2010-08-30T19:20:35.763" Score="7" ViewCount="" Body="&lt;p&gt;As others have pointed out, the key here is that the determinant of $M^{-1}$ has to be equal to $\frac{1}{det(M)}$ on the one hand, by the properties of the determinant, and also that since $M$ and  $M^{-1}$ both have integer coefficients then their determinants must be integers. So you need an integer $e=det(M)$ such that both $e$ and $\frac{1}{e}$ are integers, and the only possibility is $e=\pm 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you don't consider arguments using the determinant to be &quot;elementary&quot;, perhaps the following will do: note that $ad-bc$ must divide each of $a$, $b$, $c$, and $d$. Let $D=ad-bc$; we can then write $a=Da'$, $b=Db'$, $c=Dc'$, $d=Dd'$; then $D = ad-bc=D^2(a'd'-b'c')$, so $D=D^2(a'd'-b'c')$. Therefore, since $D\neq 0$, we must have $D(a'd'-b'c')=1$, so $D|1$, hence $D=ad-bc=\pm 1$, which is what you wanted to show.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-08-31T13:24:47.480" LastActivityDate="2010-08-31T13:24:47.480" CommentCount="4" />
  <row Id="3660" PostTypeId="2" ParentId="3628" CreationDate="2010-08-30T19:34:42.577" Score="5" ViewCount="" Body="&lt;p&gt;Maybe we can add some more help -just in case you didn't succeed to find the answer yourself yet. Let &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\frac{x}{x^2 - 1} = \frac{A}{x-1} + \frac{B}{x+1}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;be the splitting into partial fractions. (I'm too lazy to compute the coeffitients $A$ and $B$.) Then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\frac{d}{dx} \frac{x}{x^2 - 1} = -\frac{A}{(x-1)^2} - \frac{B}{(x+1)^2}  \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Differentiating again,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\frac{d^2}{dx^2}\frac{x}{x^2 - 1}  = \frac{2A}{(x-1)^3} + \frac{2B}{(x+1)^3} \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One more time:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\frac{d^3}{dx^3} \frac{x}{x^2 - 1} = - \frac{3\cdot 2 A}{(x-1)^4} - \frac{3\cdot 2 B}{(x+1)^4} \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And sure enough you can find the general pattern now, can't you? Then, use induction to prove your guess.&lt;/p&gt;&#xA;" OwnerUserId="664" LastActivityDate="2010-08-30T19:34:42.577" CommentCount="1" />
  <row Id="3661" PostTypeId="2" ParentId="3655" CreationDate="2010-08-30T19:48:18.303" Score="3" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\;\;$ Multiplicative maps preserve units: $\rm\; MN = 1 \;\;\Rightarrow\;\; d(M)\:d(N) = 1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;NOTE&lt;/strong&gt; $\rm\;\; d(1) = 1\;$ via apply $\rm d$ to $1\cdot 1 = 1\:$ then cancel $\rm d(1)\ne 0,$ valid since $\mathbb Z$ has cancellation.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2011-01-30T05:59:33.360" LastActivityDate="2011-01-30T05:59:33.360" />
  <row Id="3662" PostTypeId="2" ParentId="3646" CreationDate="2010-08-30T20:27:08.237" Score="3" ViewCount="" Body="&lt;p&gt;Here's a start to figuring out the formulas. The first thing is to understand the map $SU(2)\times SU(2)\to SO(4)$. Let $std$ denote the standard 2-dimensional complex representation of $SU(2)$. Then the external tensor product $V=std\otimes std$ is a natural 4-dimensional complex representation of $G=SU(2)\times SU(2)$. Concretely, it's given by the homomorphism $G\to GL(V)=GL_4(\mathbf{C})$ sending $(g,h)\in G$ to the &quot;Kronecker product&quot; $g\otimes h$, which is just the matrix of $g\otimes h$ with respect to the basis for $V=\mathbf{C}^2\otimes \mathbf{C}^2$ given by $e_1\otimes e_1, e_1\otimes e_2, e_2\otimes e_1, e_2\otimes e_2$, with $e_1,e_2$ the standard basis for $\mathbf{C}^2$. Now the standard representation $std$ actually lands in $SL_2(\mathbf{C})$, since $SU(2)\subset SL_2(\mathbf{C})$. But $SL_2(\mathbf{C})=Sp_2$, which means that the standard representation $std$ is self-dual, and thus $\mathbf{C}^2$ has an $SU(2)$-invariant, anti-symmetric bilinear form. Concretely this takes a pair of column vectors $v,w\in \mathbf{C}^2$ to the determinant of the matrix with columns $v,w$. Call this bilinear form $\omega:std\otimes  std\to \mathbf{C}$. Then the external tensor product $\omega\otimes \omega:V\otimes V\to \mathbf{C}$ gives a $G$-invariant \textit{symmetric} bilinear form on $V$. In fact, this is an $SL_2(\mathbf{C})\times SL_2(\mathbf{C})$-invariant symmetric bilinear form on $V$. So we've shown that $SL_2(\mathbf{C})\times SL_2(\mathbf{C})$ actually lands in $SO(q)(\mathbf{C})$ for a certain 4-dimensional quadratic space $(V,q)$. (Here $q$ is the quadratic form associated to the sym. bilinear form $B=\omega\otimes\omega$.) By general Lie theory, this implies that the maximal compact $G$ of $SL_2(\mathbf{C})\times SL_2(\mathbf{C})$ lands in the maximal compact $SO(q)(\mathbf{R})$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, we really want a homomorphism $SU(2)\times SU(2)\to SO(4)$. To do this we must diagonalize $q$, over the complex number. It's not difficult to check that with respect to the tensor product basis for $V$ we were using above, the quadratic form $q$ is equal to (maybe a multiple of) $wz-xy$. That is, if $v\in V$ has coordinates $(w,x,y,z)$ with respect to the $e_i\otimes e_j$ basis, then $q(v)=wz-xy$. (Note that this is the &quot;determinant&quot; quadratic form!) You can diagonalize this form over $\mathbf{C}$ with the change of basis matrix &lt;/p&gt;&#xA;&#xA;&lt;p&gt;M= &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;[1   i    0   0]&#xA;&#xA;[0   0    i  -1]&#xA;&#xA;[0   0    i   1]&#xA;&#xA;[1  -i    0   0]'&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;unless I screwed up the arithmetic. So conjugating the representation $V$ by this matrix (thus not changing the isomorphism type of the repn), we obtain a homomorphism $SL_2(\mathbf{C})\times SL_2(\mathbf{C})\to SO(4,\mathbf{C})$ landing in the standard 4-dimensional special orthogonal group over the complex numbers. Now the maximal compact $G$ lands in the usual group $SO(4)$, and this gives the double cover you referred to in the question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In summary, I think that the 2-1 map $SU(2)\times SU(2)\to SO(4)$ is given explicitly in matrices by taking the Kronecker product and then conjugating by $M$. If I didn't make an error, one should find in particular that the condition of being in $SU(2)$ forces a pair $(g,h)$ to have real matrix entries under this map! I haven't bothered to check this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;OK, so how does that actually let you find the 2 preimages of each rotation matrix in $SO(4)$? I guess it amounts to solving a whole bunch of nonlinear equations, so that's totally disgusting and impractical. To get a nice answer to your question, you need to analyze the structure of $SU(2)$ and of $SO(4)$. The latter is described thoroughly on the Wikipedia page for $SO(4)$. The former is quite simple and is well described in M. Artin's book &lt;em&gt;Algebra&lt;/em&gt;. EDIT: I've deleted a bunch of stuff which Jason observed in a comment is totally bogus. Fortunately Robin gave another better answer which explains how to do this using quaternions.&lt;/p&gt;&#xA;" OwnerUserId="1265" LastEditorUserId="1265" LastEditorDisplayName="" LastEditDate="2010-08-30T22:43:44.523" LastActivityDate="2010-08-30T22:43:44.523" CommentCount="5" />
  <row Id="3663" PostTypeId="1" AcceptedAnswerId="3664" CreationDate="2010-08-30T20:49:04.227" Score="0" ViewCount="140" Body="&lt;p&gt;I can think of the set of bounded, continuous functions from R -&gt; R as a group, with composition as addition of functions.  {In other words, this group has the rule that the composition of two elements in the group, f(x) and g(x), is their point-wise sum f(x) + g(x)}  Since this set is path connected, the group is continuous, i.e. it is a lie group. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the lie algebra of this group?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am trying to use this approach to find a relationship between Lie theory and Fourier series, but I realize this might not work.  So I have decided to just ask the question in this form.  If anyone knows of any connections between Lie theory and Fourier series however, I am interested to hear about that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I realize I may have made some wrong/poorly stated assertions, please point these out in comments)&lt;/p&gt;&#xA;" OwnerUserId="1417" LastEditorUserId="1417" LastEditorDisplayName="" LastEditDate="2010-08-30T21:51:16.580" LastActivityDate="2010-08-30T21:51:16.580" Title="Lie algebra of the bounded continuous functions." Tags="&lt;analysis&gt;&lt;fourier-series&gt;&lt;lie-groups&gt;&lt;lie-algebra&gt;" AnswerCount="1" CommentCount="8" FavoriteCount="1" />
  <row Id="3664" PostTypeId="2" ParentId="3663" CreationDate="2010-08-30T20:58:45.247" Score="3" ViewCount="" Body="&lt;p&gt;Since your Lie group is infinite dimensional, one has to be careful about what exactly you mean by Lie algebra, and so on. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In any case, your «Lie group of bounded continuous functions $\mathbb R\to\mathbb R$ under composition» is not really a group... You'd have to restrict your attention to the group of homeomorphisms (or something along that line) to actually get a group. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The last sentence of your first paragraph is a bit misguided. You could do much worse that read Warner's book, say, on differentiable manifolds and lie groups, or Lang's, if you like infinite dimensional situations.&lt;/p&gt;&#xA;" OwnerUserId="274" LastEditorUserId="274" LastEditorDisplayName="" LastEditDate="2010-08-30T21:04:21.313" LastActivityDate="2010-08-30T21:04:21.313" CommentCount="6" />
  <row Id="3666" PostTypeId="2" ParentId="3646" CreationDate="2010-08-30T21:16:25.413" Score="11" ViewCount="" Body="&lt;p&gt;One way to think of the 2-to-1 map from $SU(2)\times SU(2)$ is via&#xA;quaternions. Regard $\mathbb{R}^4$ as $\mathbb{H}$, the set of quaternions,&#xA;and $SU(2)$ as the group of unit (that is, norm $1$) quaternions.&#xA;Then $SU(2)\times SU(2)$ acts on $\mathbf{H}$ by rotations in the following way:&#xA;$$h\mapsto q_1 hq_2^{-1}$$&#xA;is a rotation of $\mathbb{R}^4=\mathbb{H}$ for $q_1$, $q_2\in SU(2)$.&#xA;Then $(-q_1,-q_2)$ represents the same rotation as $(q_1,q_2)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose we have a rotation $\phi$ of $\mathbb{H}$. If $\phi$ fixes $1$,&#xA;it's effectively a rotation of the $\mathbb{R}^3$ spanned by $i$, $j$ and $k$&#xA;and is represented by the pair $(q,q)$ where $q=\cos(t/2)+h\sin(t/2)$&#xA;where $h=ai+bj+ck$ is a unit vector along the axis of rotation, and&#xA;$t$ is the angle of rotation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, let $q'$ be the unit quaternion defined by $\phi(1)$.&#xA;Then $\psi:h\mapsto q'^{-1}\phi(1)$ is a rotation fixing $1$ so that one&#xA;can find a quaternion $q$ such that $(q,q)$ represents $\psi$.&#xA;Then $(q'q,q)$ represents $\phi$.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-08-30T21:16:25.413" CommentCount="1" />
  <row Id="3667" PostTypeId="1" CreationDate="2010-08-30T21:33:12.253" Score="5" ViewCount="153" Body="&lt;p&gt;&lt;a href=&quot;http://math.stackexchange.com/questions/3649/how-do-you-prove-cycles-of-length-3-dont-generate-the-whole-symmetric-group&quot;&gt;Why don't 3-cycles generate the symmetric group?&lt;/a&gt; was asked earlier today. The proof is essentially that 3-cycles are even permutations, and products of even permutations are even.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So: do the 3-cycles generate the alternating group? Similarly, do the $k$-cycles generate the alternating group when $k$ is odd?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And do the $k$-cycles generate the symmetric group when $k$ is even? I know that transpositions ($2$-cycles) generate the symmetric group.&lt;/p&gt;&#xA;" OwnerUserId="173" LastActivityDate="2010-08-30T21:59:42.517" Title="What do all the $k$-cycles in $S_n$ generate?" Tags="&lt;permutations&gt;&lt;symmetric-groups&gt;" AnswerCount="2" />
  <row Id="3668" PostTypeId="1" AcceptedAnswerId="3674" CreationDate="2010-08-30T21:43:22.580" Score="8" ViewCount="592" Body="&lt;p&gt;What is the value of 1^i?&lt;/p&gt;&#xA;" OwnerUserId="796" LastActivityDate="2010-08-31T08:48:38.780" Title="What is the value of 1^i?" Tags="&lt;complex-numbers&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="2" />
  <row Id="3669" PostTypeId="2" ParentId="3668" CreationDate="2010-08-30T21:51:55.097" Score="0" ViewCount="" Body="&lt;p&gt;it's 1. 1 to the power of anything is 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; I'll elaborate. In defining $a^b$, we know intuitively what to do in certain cases. When $a$ is a positive integer and $b$ is an integer, for example. This definition is easily extended to when $a$ is real and positive, and $b$ is a real number. In the case where $a$ is negative, or complex, we run into trouble... one way to get around this is to define the power in terms of the logarithm, as $a^b = e^{b \log a}$. Then, we have the issue that the logarithm is multivalued, so we could get different answers depending on which branch we choose. This is the approach discussed in Carl's solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I do not believe this proposed method applies for the case where $a$ is positive and real, which is the relevant case being discussed. In the case where $a$ is positive and real, $a^b$ is unambiguously defined as $a^b = e^{b \ln a}$, where $\ln a$ is the unique real number $x$ satisfying $e^x = a$. This definition works even in the case where $b$ is complex. So we can apply this to the original problem: $1^i = e^{i\log(1)} = e^0 = 1$ (Myke, you got my +1!). In fact, $1^z = 1$ for any complex $z$. This viewpoint is shared by &lt;a href=&quot;http://mathworld.wolfram.com/Power.html&quot; rel=&quot;nofollow&quot;&gt;MathWorld&lt;/a&gt;, &lt;a href=&quot;http://tinyurl.com/25xjy9z&quot; rel=&quot;nofollow&quot;&gt;WolframAlpha&lt;/a&gt;, and &lt;a href=&quot;http://en.wikipedia.org/wiki/Power_%28mathematics%29&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;, for what that's worth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I guess at some level, it's a question of semantics and preference. Why not use multiple branches and all that jazz for the case where $a&gt;0$? Because (and this is my opinion), it's unnecessary, and doesn't fit with existing definitions. The exponential function $e^z$ is well-defined via a power series and I think everyone would agree that it is single-valued, even when $z$ is complex. It makes no sense to me that $a^z$ should be any different when $a$ is real and positive.&lt;/p&gt;&#xA;" OwnerUserId="1356" LastEditorUserId="1356" LastEditorDisplayName="" LastEditDate="2010-08-31T08:48:38.780" LastActivityDate="2010-08-31T08:48:38.780" CommentCount="8" />
  <row Id="3672" PostTypeId="2" ParentId="3667" CreationDate="2010-08-30T21:59:03.680" Score="1" ViewCount="" Body="&lt;p&gt;Yes, k-cycles generate the symmetric group when k is even and alternating group, when k is odd. As you've said, for k=2 you know the answer. Suppose k&gt;2.&lt;br&gt;&#xA;(1,2,...,k)(k,...,3,1,2)=(1,3,2)&lt;br&gt;&#xA;Similarly you can get any 3-cycle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose a is even element of Sn. Then, as you know, a is a product&lt;br&gt;&#xA;$(k_1,k_2)(k_3,k_4)...(k_{4l-1},k_{4l})$&lt;br&gt;&#xA;of 2l transpositions. But product of 2 transpositions can be written using 3-cycles:&lt;br&gt;&#xA;$(k_1,k_2)(k_2,k_3)=(k_1,k_2)(k_2,k_3)*(k_2,k_3)(k_3,k_4)$.&lt;br&gt;&#xA;Any of two products, separated by * in right hand side, is either 3-cycle or unity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If a is odd and k is even, than a multiplied by any k-cycle is even, so we can apply the previous algorithm to it.&lt;/p&gt;&#xA;" OwnerUserId="695" LastActivityDate="2010-08-30T21:59:03.680" />
  <row Id="3673" PostTypeId="2" ParentId="3667" CreationDate="2010-08-30T21:59:42.517" Score="11" ViewCount="" Body="&lt;p&gt;If $n\geq5$, then the only normal subgroups of the symmetric group $S_n$ are the trivial group, the alternating group and the symmetric group itself. Since the $k$-cycles form a full conjugacy class, it follows that the subgroup they generate is normal. This determines everything if $n \geq 5$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More specifically: the $k$-cycles in $S_n$ generate the alternating group if $k$ is odd and $k \ne 1$; they generate the full symmetric group if $k$ is even.&lt;/p&gt;&#xA;" OwnerUserId="1028" LastActivityDate="2010-08-30T21:59:42.517" />
  <row Id="3674" PostTypeId="2" ParentId="3668" CreationDate="2010-08-30T22:04:48.420" Score="26" ViewCount="" Body="&lt;p&gt;First, a concrete example of things that can happen with complex exponentiation if you aren't careful: $1 = e^{2\pi i}$, so we can naively try to compute $1^i = (e^{2\pi i})^i = e^{(2\pi i)i} = e^{-2\pi}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The formal moral of that example is that the value of $1^i$ depends on the branch of the complex logarithm that you use to compute the power. You may already know that $1=e^{0+2ki\pi}$ for every integer $k$, so there are many possible choices for $\log(1)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The textbook definition of complex exponentiation states that $1^i = e^{\log(1)i}$ where &quot;log&quot; is a branch of the complex logarithm. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now $e^{(2ki\pi)i} = e^{-2k\pi}$. If you take $k = 0$, which corresponds to using the principal branch of the logarithm, you get an answer of $1^i = e^0 = 1$. If you take $k = 2$ as in the example above, using the fact that $1=e^{2\pi i}$, you get $1^i = e^{-2\pi}$. There are infinitely many possible values for $1^i$, corresponding to different branches of the complex logarithm. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The confusing point here is that the formula $1^x = 1$ is not part of the definition of complex exponentiation, although it is an immediate consequence of the definition of natural number exponentiation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A second point that can be confusing is that the function $e^z$ used above is really the complex exponential function $\exp(z)$, which is defined by a power series. Otherwise, the definition of complex exponentiation would be circular. &lt;/p&gt;&#xA;" OwnerUserId="630" LastEditorUserId="630" LastEditorDisplayName="" LastEditDate="2010-08-31T02:10:10.303" LastActivityDate="2010-08-31T02:10:10.303" CommentCount="13" />
  <row Id="3675" PostTypeId="1" AcceptedAnswerId="3676" CreationDate="2010-08-30T22:14:44.757" Score="1" ViewCount="192" Body="&lt;p&gt;Let there be given some conditionally convergent infinite series S.  Then let R be some real number, and Q&lt;sub&gt;k&lt;/sub&gt; a rearrangement of S such that the sum is equal to R.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is Q&lt;sub&gt;k&lt;/sub&gt; unique?  In other words, is there some other rearrangement W&lt;sub&gt;k&lt;/sub&gt; for which the sum of S is also R?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe the answer is that Q&lt;sub&gt;k&lt;/sub&gt; is in fact unique, because the cardinality of the set of permutations of the natural numbers (i.e. the positions of each element of the series S), is the same as the cardinality of the real numbers; and therefore the mapping from the real numbers to the rearrangements should be 1-1.  Is this line of reasoning valid, or is this result well known?&lt;/p&gt;&#xA;" OwnerUserId="1417" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-08-30T22:44:06.913" LastActivityDate="2010-09-01T05:48:20.317" Title="Question on Riemann's rearrangement theorem" Tags="&lt;analysis&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="3676" PostTypeId="2" ParentId="3675" CreationDate="2010-08-30T22:40:40.413" Score="7" ViewCount="" Body="&lt;p&gt;The line of reasoning is not valid:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you find a particular $Q_k$, then I can find infinitely many other rearrangements that give the same value:  Pick any finite subset $X$ of $\mathbb{N}$ and let $\sigma$ be any permutation of $X$.  Then the re-rearrangement where we start with $Q_k$ and just mix up the subscripts in $X$ according to $\sigma$ will yield the same sum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The point is that only rearranging  a finite number of terms doesn't change the sum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your reasoning breaks down where you say &quot;and therefore the mapping should be 1-1&quot;.  Just because two sets have the same cardinality doesn't mean any function you think of should be 1-1.  Having the same cardinality just means &quot;there is some very specific choice of function which is 1-1 and onto&quot; - it says nothing at all about a generic function you pick between two sets.&lt;/p&gt;&#xA;" OwnerUserId="331" LastActivityDate="2010-08-30T22:40:40.413" CommentCount="3" />
  <row Id="3677" PostTypeId="2" ParentId="3579" CreationDate="2010-08-30T22:49:34.330" Score="2" ViewCount="" Body="&lt;p&gt;Well, I've made a full proof! Part 1 was solved &lt;a href=&quot;http://mathoverflow.net/questions/37097/37155#37155&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;, and Part 2 was solved &lt;a href=&quot;http://www-groups.dcs.st-and.ac.uk/~john/Zagier/Solution1.1.html&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Lemma 1:&lt;/strong&gt; Any integer $m$ which satisfies the original problem also satisfies $n^{m+1} \equiv n \bmod{m}$ for all $n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof:&lt;/strong&gt; Let $p$ be a prime dividing $m$. Then $\sum_{n=1}^mn^m\equiv1\pmod p$, so $(m/p)\sum_{n=1}^{p-1}n^m\equiv1\pmod p$, so $p^2$ doesn't divide $m$. Let $g$ be a primitive root mod $p$. Then $\sum_{n=1}^{p-1}n^m\equiv\sum_{r=0}^{p-2}g^{rm}$. That's a geometric series, it sums to $(1-g^{(p-1)m})/(1-g^m)$ which is zero mod $p$ - unless $g^m=1$, in which case it sums to $-1$ mod $p$. So we must have $p-1$ dividing $m$. Looking at  $n^{m+1}\equiv n\pmod m$ and letting $n=p$, we see that $p^2$ cannot divide $m$. Now looking mod $p$, we get $n^{m+1}\equiv n\pmod p$. This is equivalent to $m+1\equiv1\pmod{p-1}$ (if $a^x \equiv a^y \bmod{n}$, then $x \equiv y \bmod{\varphi(n)}$ by Euler's theorem, and $\varphi(p) = p-1$), that is, $p-1$ divides $m$, so any integer $n^{m+1} \equiv n \bmod{m}$ as $p-1|m$ for all $m$ if $p|m$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Lemma 2:&lt;/strong&gt; There are only a finite amount of integers $m$ which satisfy $n^{m+1} \equiv n \bmod{m}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof:&lt;/strong&gt; Since $p^2$ does not divide $m$, we may let $m = p_1 \ldots p_r$ with $p_1 &amp;lt; p_2 &amp;lt; \ldots &amp;lt; p_r$, with $p_i$ prime; as $p-1|m$ for all $p|m$, we say that $p_i-1|p_1 \ldots p_{i-1}$ for $i = 1, \ldots, r$. If we take $i = 1$, this forces $p_1-1|1$, so if $r \ge 1$, $p_1 = 2$. If $i = 2$, $p_2-1|2$, so if $r \ge 2$, $p_2 = 3$. Continuing, if $r \ge 3$, then $(p_3-1)|p_1 p_2 = 6$, so $p_3 = 7$; if $r \ge 4$, $(p_4 - 1)|p_1 p_2 p_3 = 42$, so $p_4 = 43$, as the numbers $d+1$ are not prime for other divisors $d$ of 42 larger than 6. If $r \ge 5$, then $(p_5 -1)|p_1 p_2 p_3 p_4 = 1806$, but 1, 2, 6 and 42 are the only divisors of 1806 with $d+1$ prime, so $p_5$ cannot exist. Therefore, $r \le 4$ and $m \in \{1, 2, 6, 42, 1806\}$.&lt;/p&gt;&#xA;" OwnerUserId="1562" LastEditorUserId="1562" LastEditorDisplayName="" LastEditDate="2010-08-30T22:55:58.090" LastActivityDate="2010-08-30T22:55:58.090" CommentCount="8" />
  <row Id="3678" PostTypeId="2" ParentId="3675" CreationDate="2010-08-30T23:01:52.917" Score="2" ViewCount="" Body="&lt;p&gt;A permutation $i \to \pi(i)$ preserves convergence of sums ($\Sigma a_{\pi(i)}$ converges whenever $\Sigma a_i$ does) if and only if it maps [1,N] to a bounded number of intervals of integers, with the bound allowed to depend on the permutation but not on N.  This places strong constraints on how much &quot;motion&quot; the permutation can create. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this respect a rearrangement has to be nearly trivial in order to never change the sum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(This is considering the problem for all convergent series simultaneously.  A single sum might be compatible with additional permutations.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: permutations satisfying the condition preserve the sum of the series and not only the convergence of the partial sums.  That is, $\Sigma a_i = \Sigma a_{\pi(i)}$.&lt;/p&gt;&#xA;" OwnerUserId="467" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-09-01T05:48:20.317" LastActivityDate="2010-09-01T05:48:20.317" />
  <row Id="3679" PostTypeId="1" AcceptedAnswerId="3682" CreationDate="2010-08-31T00:05:43.577" Score="5" ViewCount="237" Body="&lt;p&gt;To be more precise, I am interested in knowing &lt;strong&gt;if the intuition that &lt;em&gt;a Euclidean zero vector does not have a particular direction&lt;/em&gt; is actually correct, and if there is a rigorous formulation that would back it up.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Euclidean_vector#Null_vector&quot; rel=&quot;nofollow&quot;&gt;Wikipedia's entries&lt;/a&gt; on the &lt;a href=&quot;http://en.wikipedia.org/wiki/Null_vector&quot; rel=&quot;nofollow&quot;&gt;zero vector&lt;/a&gt; seems to agree with that intuition, but as always, one shouldn't blindly trust Wikipedia: &lt;a href=&quot;http://en.wikipedia.org/wiki/Null_vector&quot; rel=&quot;nofollow&quot;&gt;in one place&lt;/a&gt;, it is stated that a zero vector &quot;&lt;em&gt;is orthogonal to all other vectors with the same number of components,&lt;/em&gt;&quot; while &lt;a href=&quot;http://en.wikipedia.org/wiki/Inner-product&quot; rel=&quot;nofollow&quot;&gt;in another&lt;/a&gt;, it is stated that &quot;&lt;em&gt;two vectors can be considered orthogonal if and only if their dot product is zero, and they have non-null length.&lt;/em&gt;&quot; Correct me if I am wrong, but these two statements contradict each other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This question popped up in my head when I heard someone arguing that since a Euclidean vector is defined as a geometric entity that has both a magnitude and direction, and since a zero vector is a vector with length 0, then it is only fitting for a zero vector to have a &quot;&lt;em&gt;direction 0.&lt;/em&gt;&quot; I am personally inclined to say that a zero vector does not have a particular direction, but as I said, I would like to know if there is a rigorous formulation that would lead to this conclusion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And to put the question in a less &quot;discussion-inducing&quot; form, &lt;strong&gt;is there an agreement on the direction of a Euclidean zero vector?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="114" LastActivityDate="2010-08-31T22:32:49.823" Title="What is the Direction of a Zero (Null) Vector?" Tags="&lt;intuition&gt;&lt;euclidean-geometry&gt;&lt;definition&gt;" AnswerCount="3" />
  <row Id="3680" PostTypeId="1" AcceptedAnswerId="3681" CreationDate="2010-08-31T00:07:50.123" Score="3" ViewCount="90" Body="&lt;p&gt;Consider the following data points: 1, 1, 2, 3, 4&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand that...&#xA;the average is the total of the numbers divided by the count of numbers in the set, the median is the central value based on location in the set, the mode is the value occurring most often in the set, and the midrange is highest and lowest values divided by 2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a name for the calculation of the average based on the &lt;em&gt;unique&lt;/em&gt; values found in the set? So... (1 + 2 + 3 + 4) / 4 = 2.5? And is there a use for it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pardon a possibly elementary question, I'm a programmer but I'm not exactly a math-oriented person, and this has been bugging me recently.&lt;/p&gt;&#xA;" OwnerUserId="1603" LastActivityDate="2010-08-31T14:53:40.897" Title="Is there a name (and use) for an average based on the unique values of a set of data?" Tags="&lt;average&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="3681" PostTypeId="2" ParentId="3680" CreationDate="2010-08-31T00:19:20.283" Score="1" ViewCount="" Body="&lt;p&gt;From a practical viewpoint, I wouldn't think so.  For that example $ (1, 1, 2, 3, 4) $ your procedure gives $ 2.5 $.  If the data points had instead been $ (1, 1.01, 2, 3, 4) $, however, the result would jump to about $ 2.2 $.  The most common useful properties would seem to depend continuously upon the data, however.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since I don't know the application for which you're using that kind of average, though, it may well turn out to be helpful.  In this case, I'd recommend simply spelling it out as you did here.&lt;/p&gt;&#xA;" OwnerUserId="336" LastEditorUserId="336" LastEditorDisplayName="" LastEditDate="2010-08-31T00:24:55.170" LastActivityDate="2010-08-31T00:24:55.170" CommentCount="2" />
  <row Id="3682" PostTypeId="2" ParentId="3679" CreationDate="2010-08-31T00:20:58.113" Score="6" ViewCount="" Body="&lt;p&gt;The zero vector has no particular direction; this is consistent with the fact that it is orthogonal to every other vector.  (It doesn't really make sense to say it has &quot;&lt;em&gt;direction 0&lt;/em&gt;&quot;, since direction is not a magnitude; &quot;direction 0&quot; makes no more sense than &quot;direction 1&quot;&#xA;or &quot;direction 5.873&quot;.)   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively, you could say that it points in every direction, but with zero magnitude,&#xA;since if you take &lt;em&gt;any&lt;/em&gt; vector and multiply it by zero, you get the zero vector.  &quot;Every&#xA;direction&quot; is the same as &quot;no particular direction&quot;; it's just a different way of phrasing&#xA;things.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is one flaw with the traditional description of a vector as being a pair consisting of&#xA;a magnitude and a direction:  For the zero vector, the magnitude is zero, but the direction&#xA;is arbitrary.&lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2010-08-31T00:20:58.113" CommentCount="1" />
  <row Id="3683" PostTypeId="2" ParentId="3668" CreationDate="2010-08-31T01:24:40.413" Score="0" ViewCount="" Body="&lt;p&gt;$1^{i} = e^{log(1) i} = e^{0}=1$&lt;/p&gt;&#xA;" OwnerUserId="832" LastActivityDate="2010-08-31T01:24:40.413" />
  <row Id="3684" PostTypeId="1" AcceptedAnswerId="3686" CreationDate="2010-08-31T01:40:20.493" Score="2" ViewCount="215" Body="&lt;p&gt;I'm relatively new to ring theory so this is probably a simple question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Its easy to see that a finite ring that is commutative and has no-zero divisors (i.e. an integral domain) must have multiplicative inverses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am wondering if we can rearrange these properties and always get implication or produce finite rings with only 2 of the above properties.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Explicitly my questions are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Does a finite ring that is commutative with multiplicative inverses always have no-zero divisors?&#xA;(EDIT: this one is pretty easy too, let xy=0 and assume x not 0. then (x^-1)xy=(x^-1)0. Hence, y=0 as desired.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Is a finite ring that has multiplicative inverses and no zero divisors always commutative?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If these are actually simple exercises, I'd just like a hint to get started with the proof or any counterexamples.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks :)&lt;/p&gt;&#xA;" OwnerUserId="249" LastEditorUserId="249" LastEditorDisplayName="" LastEditDate="2010-08-31T02:40:58.157" LastActivityDate="2010-08-31T12:04:27.237" Title="When does a finite ring become a finite field?" Tags="&lt;abstract-algebra&gt;&lt;ring-theory&gt;" AnswerCount="2" />
  <row Id="3685" PostTypeId="2" ParentId="3684" CreationDate="2010-08-31T01:55:48.003" Score="1" ViewCount="" Body="&lt;p&gt;You might want to look at:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.springer.com/engineering/electronics/book/978-1-4020-7039-6&quot; rel=&quot;nofollow&quot;&gt;http://www.springer.com/engineering/electronics/book/978-1-4020-7039-6&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1369" LastActivityDate="2010-08-31T01:55:48.003" CommentCount="1" />
  <row Id="3686" PostTypeId="2" ParentId="3684" CreationDate="2010-08-31T02:23:42.300" Score="3" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt;  1) If $\;\rm R\;$ is finite then $\;\rm x\to r\:x\;$ is onto iff 1-1, so $\;\rm R\;$ is a field iff $\;\rm R\;$ is a domain. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) is &lt;a href=&quot;http://en.wikipedia.org/wiki/Wedderburn%27s_little_theorem&quot; rel=&quot;nofollow&quot;&gt;Wedderburn's little theorem&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-08-31T02:23:42.300" CommentCount="1" />
  <row Id="3687" PostTypeId="2" ParentId="2372" CreationDate="2010-08-31T03:43:57.950" Score="1" ViewCount="" Body="&lt;p&gt;There are 20!/4!5!8!3! arrangements of the 20 alphabets. Consider the action of $S_5$ on the arrangements by permuting the last 5 digits. You want to compute the number of orbits. You can then try to do it by &lt;a href=&quot;http://en.wikipedia.org/wiki/Burnside%27s_lemma&quot; rel=&quot;nofollow&quot;&gt;Burnside's formula&lt;/a&gt;. It is not difficult to do it at least for this case, since if a permutation of $S_5$ fixes an arrangement, it means that each cycle of the permutation corresponds to a color.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the general case, I suspect that Polya's enumeration theorem can do it, though I'm not certain.&lt;/p&gt;&#xA;" OwnerUserId="325" LastActivityDate="2010-08-31T03:43:57.950" CommentCount="1" />
  <row Id="3688" PostTypeId="1" CreationDate="2010-08-31T04:25:13.260" Score="1" ViewCount="167" Body="&lt;p&gt;Is anything known about these integrals? Textbook suggestions are welcome&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f(n,p)=\int_{x=-0.5}^p \frac{n!}{x!(n-x)!} dx$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$n&gt;0, p\le n+0.5$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, as n grows $2^n-f(n,n+1/2)$ seems to be positive and monotonically decreasing...how would I find the limit?&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&#xA;ListPlot[Table[2^n - NIntegrate[n!/(x! (n - x)!), {x, -1/2, n + 1/2}], {n, Range[30]}], PlotRange -&gt; All]&#xA;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://yaroslavvb.com/upload/gamma-integral.png&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="998" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-09-01T06:16:37.947" LastActivityDate="2010-09-01T06:16:37.947" Title="Gamma integrals" Tags="&lt;gamma-function&gt;&lt;special-functions&gt;" AnswerCount="1" CommentCount="7" FavoriteCount="1" />
  <row Id="3689" PostTypeId="2" ParentId="3645" CreationDate="2010-08-31T04:32:55.990" Score="3" ViewCount="" Body="&lt;p&gt;Yes.  The problem is that the term &quot;transcendental function&quot; is absurdly general.  For example, pick a computable bijection between the integers and the set of &lt;a href=&quot;http://en.wikipedia.org/wiki/Turing_machine&quot; rel=&quot;nofollow&quot;&gt;Turing machines&lt;/a&gt; and consider the function $f : \mathbb{R} \to \mathbb{R}$ which is $0$ on the integers corresponding to halting Turing machines, $1$ on the integers corresponding to non-halting Turing machines, and smoothly interpolated in between (for example via &lt;a href=&quot;http://en.wikipedia.org/wiki/Bump_function&quot; rel=&quot;nofollow&quot;&gt;bump functions&lt;/a&gt;) such that if $x$ is not an integer then $0 &amp;lt; f(x) &amp;lt; 1$.  Then the problem of determining the solutions to $f(x) = 1$ is equivalent to the &lt;a href=&quot;http://en.wikipedia.org/wiki/Halting_problem&quot; rel=&quot;nofollow&quot;&gt;halting problem&lt;/a&gt;.  (And $f$ is even &lt;em&gt;smooth&lt;/em&gt;.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So to get any reasonable answer to this question one must pick a much more specific definition of transcendental function.  One must also specify what it means to find a solution to an equation; is it enough to give an algorithm which will compute it to arbitrary accuracy, or must we get the answer in &quot;closed form&quot; (whatever that means)?  There are a lot of issues here.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-08-31T04:32:55.990" CommentCount="1" />
  <row Id="3690" PostTypeId="2" ParentId="2567" CreationDate="2010-08-31T04:42:53.600" Score="1" ViewCount="" Body="&lt;p&gt;As Chandru1 stated tentatively, the same arguments apply for $0&amp;lt; p&amp;lt;1$ as for $1\leq p&amp;lt;\infty$, so it is no surprise that a separate proof is hard to find.  For example, an introduction to $L^p$ spaces on subsets of $\mathbb{R}^n$ can be found in Chapter 8 of &lt;em&gt;Measure and integral&lt;/em&gt;  by Wheeden and Zygmund.  A proof of separability is outlined for $1\leq p&amp;lt;\infty$ in &lt;a href=&quot;http://books.google.com/books?id=YDkDmQ_hdmcC&amp;amp;lpg=PP1&amp;amp;dq=wheeden%20zygmund%20measure&amp;amp;pg=PA133#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;Theorem 8.15&lt;/a&gt;.  Part of Theorem 8.16 asserts separability for $0&amp;lt; p&amp;lt;1$, but for proof they simply refer to the proof of 8.15.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-08-31T04:42:53.600" CommentCount="2" />
  <row Id="3691" PostTypeId="2" ParentId="3688" CreationDate="2010-08-31T05:27:44.920" Score="1" ViewCount="" Body="&lt;p&gt;Not a full solution, but maybe enough for you to get a handle on the problem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using $\Gamma (x+1) = x \Gamma (x)$ and $\Gamma(x) \Gamma(1-x) = \frac{\pi}{\sin \pi x}$, one obtains&#xA;$$&#xA;\Gamma(x + 1) \Gamma(n - x + 1) = \frac{(-1)^n \pi}{\sin \pi x} \prod_{k=0}^{n} (x - k) .&#xA;$$&#xA;Thus,&#xA;$$&#xA;f(n, p) = \frac{(-1)^n n!}{\pi} \int_{-0.5}^p \frac{\sin \pi x}{\prod_{k=0}^n (x-k)} d x .&#xA;$$&#xA;Using partial fractions, you can now transform $f$ into a sum of values of the sine integral.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Addendum: Computing the sums yields&#xA;$$&#xA;f(n, n+\frac{1}{2}) = \frac{2}{\pi} \sum_{k = 0}^n \binom{n}{k} Si (\pi k + \frac{\pi}{2})&#xA;$$&#xA;Note, that $Si (x) = \frac{\pi}{2} + O(x^{- 2})$. However, I have no idea how to leverage this to find the limit.&lt;/p&gt;&#xA;" OwnerUserId="960" LastEditorUserId="960" LastEditorDisplayName="" LastEditDate="2010-08-31T07:10:29.103" LastActivityDate="2010-08-31T07:10:29.103" CommentCount="1" />
  <row Id="3692" PostTypeId="1" AcceptedAnswerId="3694" CreationDate="2010-08-31T06:40:55.103" Score="2" ViewCount="151" Body="&lt;p&gt;i. If it rains next Sunday, then Andy will only go to the football game if Bob does.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ii. If Andy will only go to the foot ball game on Sunday if it rains, then Bob will definitely go to the Sunday football game regardless of the weather.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My gut feeling is they are not equivalent. I chose the following three predicates: r: It rains next Sunday, b: Bob goes to the game, a: Andy goes to the game. And then it seems like the statements would then be:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i. $r \rightarrow (b \leftrightarrow a)$ &#xA;ii. $(a \leftrightarrow r) \rightarrow b$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then if I choose the case where it rains on Sunday, Bob goes to the game, but Andy doesn't, statement i is false, but statement ii is true. Have I translated these correctly?&lt;/p&gt;&#xA;" OwnerUserId="1583" LastActivityDate="2010-08-31T08:14:58.773" Title="Are these statements logically equivalent?" Tags="&lt;logic&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="3693" PostTypeId="2" ParentId="3646" CreationDate="2010-08-31T06:44:13.347" Score="3" ViewCount="" Body="&lt;p&gt;In terms of SU(2) = unit quaternions, the answer can be found in the sections &quot;Isoclinic decomposition&quot; and &quot;Relation to quaternions&quot; of &lt;a href=&quot;http://en.wikipedia.org/wiki/So%284%29#Isoclinic_decomposition&quot; rel=&quot;nofollow&quot;&gt;the Wikipedia article about SO(4)&lt;/a&gt;.&#xA;(This is equivalent to Robin Chapman's answer, I presume, but the formulas are a bit more explicit.)&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2010-08-31T06:44:13.347" />
  <row Id="3694" PostTypeId="2" ParentId="3692" CreationDate="2010-08-31T07:28:23.967" Score="1" ViewCount="" Body="&lt;p&gt;I don't think you've interpreted the end of (i) correctly: &quot;Andy will only go to the football game if Bob does.&quot;  This is not an &lt;em&gt;if-and-only-if&lt;/em&gt; statement; it is only an &lt;em&gt;only-if&lt;/em&gt; statement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have the same issue in your interpretation of (ii).&lt;/p&gt;&#xA;" OwnerUserId="1461" LastActivityDate="2010-08-31T07:28:23.967" CommentCount="1" />
  <row Id="3695" PostTypeId="2" ParentId="3692" CreationDate="2010-08-31T08:14:58.773" Score="0" ViewCount="" Body="&lt;p&gt;if we set R = &quot;Saturday it will rain&quot;, A = &quot;Andy goes to football match&quot;, B = &quot;Bob goes to football match&quot;, then your statement (i) says for triples (R,A,B) that (Y,Y,Y) and (Y,N,N) are true, while (Y,Y,N) and (Y,N,Y) are false. Nothing can be said about the other four triplets.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your statement (ii) is indeed different: the part &quot;Andy will only go to the foot ball game on Sunday if it rains&quot; says for couples (R,A) that (Y,Y) and (N,N) are true while (Y,N) and (N,Y) are false. But the statement &quot;If S then T&quot; is always true when S is false (Latins said &quot;ex falso quodlibet&quot;, from falsehood comes everything); therefore the triple (Y,N,Y) is true.&lt;/p&gt;&#xA;" OwnerUserId="89" LastActivityDate="2010-08-31T08:14:58.773" CommentCount="3" />
  <row Id="3696" PostTypeId="2" ParentId="3679" CreationDate="2010-08-31T08:17:22.677" Score="2" ViewCount="" Body="&lt;p&gt;Since adding the zero vector to any non-zero vector does not change the direction of the latter, it cannot have a proper direction because of the parallelogram rule :-)&lt;/p&gt;&#xA;" OwnerUserId="89" LastActivityDate="2010-08-31T08:17:22.677" />
  <row Id="3697" PostTypeId="1" AcceptedAnswerId="3714" CreationDate="2010-08-31T08:52:38.733" Score="3" ViewCount="534" Body="&lt;p&gt;I tried to find some reference on the net but couldn't find a good one. What is the advantage of pivoting in LU decomposition over regular LU decomposition? Is it something to do with matrix singular or not?&lt;/p&gt;&#xA;" OwnerUserId="957" LastActivityDate="2010-08-31T15:04:54.693" Title="pivoting in LU decomposition" Tags="&lt;linear-algebra&gt;&lt;matrices&gt;" AnswerCount="4" CommentCount="1" FavoriteCount="2" />
  <row Id="3698" PostTypeId="1" AcceptedAnswerId="3718" CreationDate="2010-08-31T08:59:10.580" Score="6" ViewCount="439" Body="&lt;p&gt;I know that 1 is not a prime number because $1\cdot\mathbb Z=\mathbb Z$ is, by convention, not a prime ideal in the ring $\mathbb Z$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, since $\mathbb Z$ is a domain, $0\cdot\mathbb Z=\{0\}$ &lt;em&gt;is&lt;/em&gt; a prime ideal in $\mathbb Z$. Isn't $(p)$ being a prime ideal the very definition of $p$ being a prime element?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I know that this would violate the &lt;a href=&quot;http://en.wikipedia.org/wiki/Fundamental_theorem_of_arithmetic&quot; rel=&quot;nofollow&quot;&gt;Fundamental Theorem of Arithmetic&lt;/a&gt;.)&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt;&#xA;Apparently the answer is that a prime element in a ring is, by convention a non-zero non-unit (see &lt;a href=&quot;http://en.wikipedia.org/wiki/Prime_element#Divisibility.2C_prime_and_irreducible_elements&quot; rel=&quot;nofollow&quot;&gt;wikipedia&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is strange because a prime ideal of a ring is, by convention, a proper ideal but not necessarily non-zero (see &lt;a href=&quot;http://en.wikipedia.org/wiki/Prime_ideal&quot; rel=&quot;nofollow&quot;&gt;wikipedia&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, my question is now: Why do we make this awkward convention?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit 2:&lt;/strong&gt; @Downvoter: Please explain your downvote in order to help me to improve my question.&lt;/p&gt;&#xA;" OwnerUserId="367" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-08-31T15:52:01.147" LastActivityDate="2011-03-15T22:45:37.660" Title="Why doesn't 0 being a prime ideal in Z imply that 0 is a prime number?" Tags="&lt;number-theory&gt;&lt;abstract-algebra&gt;&lt;terminology&gt;&lt;definition&gt;&lt;ring-theory&gt;" AnswerCount="3" CommentCount="8" FavoriteCount="2" />
  <row Id="3700" PostTypeId="2" ParentId="3697" CreationDate="2010-08-31T09:55:49.203" Score="2" ViewCount="" Body="&lt;p&gt;Pivoting is necessary to reduce the effect of rounding errors. In a world with exact arithmetic you wouldn't need pivoting. With a singular matrix it would just happen that you would not be able to proceed with computing the LU decomposition beyond a point. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But since finite computers must approximate reals with finite numbers of bits, every computation introduces a rounding error, and pivoting helps us control the accumulation of these errors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know of an online reference but most books on numerical analysis discuss this. An advanced source is Golub and van Loan's &lt;em&gt;Matrix Computations.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="1195" LastActivityDate="2010-08-31T09:55:49.203" CommentCount="1" />
  <row Id="3701" PostTypeId="2" ParentId="3499" CreationDate="2010-08-31T09:57:25.443" Score="0" ViewCount="" Body="&lt;p&gt;One can also see this article, &quot;S.S.Kim, Amer.Math.Monthly 106 (1999), 258-259&quot;.&lt;/p&gt;&#xA;" LastActivityDate="2010-08-31T09:57:25.443" />
  <row Id="3702" PostTypeId="2" ParentId="3697" CreationDate="2010-08-31T10:23:34.393" Score="7" ViewCount="" Body="&lt;p&gt;The matrix $\begin{bmatrix}0 &amp;amp; 1 \\ 1 &amp;amp; 0\end{bmatrix}$ is as far from singular as you can get, yet it cannot be factorized into an LU decomposition without pivoting.&lt;/p&gt;&#xA;" OwnerUserId="856" LastActivityDate="2010-08-31T10:23:34.393" />
  <row Id="3704" PostTypeId="1" AcceptedAnswerId="3720" CreationDate="2010-08-31T10:41:40.720" Score="3" ViewCount="381" Body="&lt;p&gt;Let's take a look at the following integrals :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) $\displaystyle \int\limits_{0}^{1} \frac{\log{x}}{1+x} \ dx = -\frac{\pi^{2}}{12} = -2 \sum\limits_{n=1}^{\infty} \frac{1}{n^2}= \zeta(2)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) For $c&amp;lt;1$ $\displaystyle \int\limits_{0}^{\frac{\pi}{2}} \arcsin(c \cos{x}) \ dx = \frac{c}{1^2} + \frac{c}{3^2} + \frac{c}{5^2} + \cdots $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) &lt;a href=&quot;http://math.stackexchange.com/questions/3502/summing-the-series-1k-frac2k2k1-a2k1&quot;&gt;http://math.stackexchange.com/questions/3502/summing-the-series-1k-frac2k2k1-a2k1&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are many infinite series which can be represented by integrals. I would like to know whether such type of representation is always possible for an infinite series?&lt;/p&gt;&#xA;" LastActivityDate="2010-08-31T20:01:37.980" ClosedDate="2010-08-31T14:51:33.513" Title="Integral Representation of Infinite series" Tags="&lt;analysis&gt;" AnswerCount="1" CommentCount="7" FavoriteCount="2" />
  <row Id="3705" PostTypeId="1" CreationDate="2010-08-31T10:49:25.463" Score="3" ViewCount="114" Body="&lt;p&gt;Say we have a set of symmetric $n \times n$ matrices $M_i$ for $1 \leq i \leq k$, elements in $\mathbb{R}$.  Suppose that for every $\boldsymbol{\lambda} = (\lambda_1, \dots , \lambda_k) \in \mathbb{R}^k$ we have that the kernel of $M_{\boldsymbol{\lambda}} = \sum_i \lambda_i M_i$ is nontrivial.  Does it follow that there exists some nonzero $n$ vector $\textbf{v}$ with $M_i \textbf{v} = 0$ for all $i$?&lt;/p&gt;&#xA;" OwnerUserId="1607" LastEditorUserId="1607" LastEditorDisplayName="" LastEditDate="2010-08-31T11:51:07.087" LastActivityDate="2010-08-31T15:26:58.350" Title="Sums of a set of symmetric matrices" Tags="&lt;linear-algebra&gt;" AnswerCount="2" FavoriteCount="2" />
  <row Id="3706" PostTypeId="2" ParentId="3697" CreationDate="2010-08-31T11:05:11.480" Score="1" ViewCount="" Body="&lt;p&gt;Most of the invertible square matrices admit an $LU$ decomposition with $L$ a lower triangular matrix with 1’s onthe diagonal and $U$ is an upper triangular matrix (with arbitrary nonzero diagonal entries). Such an expression is easily seen to be unique. The set of invertible matrices admitting decomposition is open and dense in the group of n by n invertible matrices $GL(n)$. A matrix belongs to this set if and only if all its leading principal minors are nonzero as mentioned in the &lt;a href=&quot;http://en.wikipedia.org/wiki/LU_decomposition&quot; rel=&quot;nofollow&quot;&gt;Wikipedia page&lt;/a&gt;)&#xA;If the matrix does not belong to this set it still admits an $LU$ decomposition with pivoting $A = wLU$, with $L$ and $U$ as before and $w$ is a permutaion matrix.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The geometric origin of these constructions is given by means of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Bruhat_decomposition&quot; rel=&quot;nofollow&quot;&gt;Bruhat cell decomposition&lt;/a&gt; of $GL(n)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$GL(n) = \coprod_{w \in W} BwB$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $B$ is the subgroup of invertibe upper triangular matrices, $W$ is the symmetric group, and $w$ is a permutation matrix.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When $w=w_0$ is the longest element of the symmetric group, given by an antidiagonal matrix with a unit antidiagonal, then: $ B w_0 = w_0 w_0^{-1} B w_0 = w_0 N $, where $N$ is the group of upper triangular matrices. Thus the set of matrices&#xA;admitting the regular LU decomposition is identified with the cell corresponding to the longest element. This cell which is dense in $GL(n)$ is called the big (Schubert) cell.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When, the matrix belongs to the big cell , but some of the leading principal minors are very small with respect to others, then the $LU$ decomposition will result big matrix elements, because the determinants of the principal minors appear in the denominators of the matrix elements of the $LU$ decomposition. In this case, it is customary to perform pivoting to move this matrix to another cell, to reduce the scale difference among the leading principal minors and gain numerical stability.&lt;/p&gt;&#xA;" OwnerUserId="1430" LastEditorUserId="1430" LastEditorDisplayName="" LastEditDate="2010-08-31T15:04:54.693" LastActivityDate="2010-08-31T15:04:54.693" />
  <row Id="3707" PostTypeId="2" ParentId="3543" CreationDate="2010-08-31T11:33:10.070" Score="1" ViewCount="" Body="&lt;p&gt;Here i have found one more method for evaluvating integrals of the form $$ I= \frac{1}{(x^{2}+a^{2})\sqrt{x^{2}+b^{2}}} \ dx $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $y = \displaystyle \sqrt{\frac{x^{2}+b^{2}}{x^{2}+a^{2}}}$, then by differentiating one can see that $$ \frac{dy}{dx} = \frac{(a^{2}-b^{2})x}{(x^{2}+a^{2})^{3/2} (x^{2}+b^{2})^{1/2}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus $I$ becomes $$\frac{(x^{2}+a^{2})^{1/2}}{(a^{2}-b^{2})x} \ dy$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also $(x^{2}+a^{2})y^{2}=x^{2}+b^{2}$ so that we have $\displaystyle x^{2}= \frac{b^{2}-a^{2}y^{2}}{y^{2}-1}$ and $$x^{2}+a^{2} = \frac{b^{2}-a^{2}}{y^{2}-1}$$ So $I$ further reduces to $$ I = \frac{1}{a^{2}-b^{2}} \int \frac{\sqrt{b^{2}-a^{2}}}{\sqrt{b^{2}-a^{2}y^{2}}} \ dy = \frac{1}{a\sqrt{b^{2}-a^{2}}} \cos^{-1}\frac{ay}{b} \ (a &amp;lt; b)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $a&gt;b$ one can rearrange $$I = \frac{1}{a^{2}-b^{2}} \int \frac{\sqrt{a^{2}-b^{2}}}{\sqrt{a^{2}y^{2}-b^{2}}} \ dy$$&lt;/p&gt;&#xA;" LastActivityDate="2010-08-31T17:10:32.993" CommentCount="1" />
  <row Id="3708" PostTypeId="1" CreationDate="2010-08-31T11:41:49.857" Score="1" ViewCount="119" Body="&lt;p&gt;Suppose $f$ is a continuous function, then can we find all $f$'s which satisfy $$\int\limits_{0}^{1} (1-(f(x))^{a})^{1/b} - (1-(f(x))^{b})^{1/a} \ dx =0 \quad (a&gt;b)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Motivation : &lt;a href=&quot;http://math.stackexchange.com/questions/316/why-is-int-01-1-x71-5-1-x51-7-dx0/345#345&quot;&gt;http://math.stackexchange.com/questions/316/why-is-int-01-1-x71-5-1-x51-7-dx0/345#345&lt;/a&gt;&lt;/p&gt;&#xA;" LastActivityDate="2010-08-31T11:41:49.857" Title="Finding all $f$'s such that $\int\limits_{0}^{1} (1-(f(x))^{a})^{1/b} - (1-(f(x))^{b})^{1/a} \ dx =0$" Tags="&lt;real-analysis&gt;&lt;analysis&gt;" FavoriteCount="1" />
  <row Id="3709" PostTypeId="2" ParentId="3705" CreationDate="2010-08-31T11:42:13.560" Score="3" ViewCount="" Body="&lt;p&gt;No. As a counterexample, consider the matrices&#xA;$M_1 = \pmatrix{1 &amp;amp; 0 \cr 0 &amp;amp; 0}$, $M_2 = \pmatrix{0 &amp;amp; 1 \cr 0 &amp;amp; 0}$. Any linear combination has a row of zeroes and therefore is not invertible, but the kernels of $M_1,M_2$ are disjoint (apart from 0).&lt;/p&gt;&#xA;" OwnerUserId="1609" LastActivityDate="2010-08-31T11:42:13.560" CommentCount="1" />
  <row Id="3710" PostTypeId="1" AcceptedAnswerId="3711" CreationDate="2010-08-31T11:54:16.687" Score="2" ViewCount="172" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have to get all combinations of a six digit number where each digit is unique. Its clear that this combination will not have a 0 at the start as it will not be a valid six digit number. Help me on how to attack this problem. &lt;/p&gt;&#xA;" OwnerUserId="1611" LastEditorUserId="1611" LastEditorDisplayName="" LastEditDate="2010-08-31T13:00:13.680" LastActivityDate="2010-09-24T11:58:01.933" Title="Help me out on permutation and combination" Tags="&lt;homework&gt;&lt;permutations&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="3711" PostTypeId="2" ParentId="3710" CreationDate="2010-08-31T11:58:27.207" Score="3" ViewCount="" Body="&lt;p&gt;First, if you're talking about a six-digit number, the order of the digits matters, so it's generally not referred to as a combination.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider how many possibilities there are for the first digit.  As you said, it cannot be zero, or you wouldn't have a six-digit number.  Once there is a first digit, how many possibilities are there for the second digit?  Then for the third?  And so on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The total number of six-digit numbers is the product of the various numbers of possibilities for each digit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(As the question feels like homework to me, this is more of a hint/outline than a complete answer.  If it actually is homework, please add the homework tag.  If it's not homework, or if this doesn't get you to an answer, please comment with more information.)&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-08-31T11:58:27.207" CommentCount="1" />
  <row Id="3713" PostTypeId="2" ParentId="3590" CreationDate="2010-08-31T12:06:38.267" Score="1" ViewCount="" Body="&lt;p&gt;The Wikipedia article gives a &#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Riemann_zeta_function#Mellin_transform&quot; rel=&quot;nofollow&quot;&gt; Mellin transform &lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Gamma(s)\zeta(s) =\int_0^\infty\frac{x^{s-1}}{e^x-1} dx.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Riemann_zeta_function#Reciprocal&quot; rel=&quot;nofollow&quot;&gt;Dirichlet series&lt;/a&gt; over the Möbius function gives the reciprocal &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \frac{1}{\zeta(s)} = \sum_{n=1}^{\infty} \frac{\mu(n)}{n^s} .$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus we may write&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Gamma(s) = \sum_{n=1}^{\infty} \frac{\mu(n)}{n^s} \int_0^\infty\frac{x^{s-1}}{e^x-1} dx .$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This holds true for every complex number s with real part &amp;gt; 1. Now let's try to enlarge the domain of validity of this representation. Riemann showed (see the book of H. M. Edwards, Riemann Zeta Function, for the details) that modifying the contour gives a formula valid for all complex s.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ 2\sin(\pi s)\Gamma(s)\zeta(s) = i \oint_C \frac{(-x)^{s-1}}{e^x-1}dx .$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This leads to &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \sin(\pi s) \Gamma(s) = \frac{i}{2} \sum_{n=1}^{\infty} \frac{\mu(n)}{n^s} &#xA;\oint_C \frac{(-x)^{s-1}}{e^x-1} dx . \qquad (*)  $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, this formula is again only valid for s with real part &amp;gt; 1 because of the use of the Dirichlet series. Wikipedia remarks: &amp;quot;The Riemann hypothesis is equivalent to the claim that [this representation of the reciprocal of the zeta function] is valid when the real part of s is greater than 1/2.&amp;quot; Thus a possible answer to my question is: &amp;quot;The representation (*) is valid for all s with real part &amp;gt; 1/2 if and only if the RH holds.&amp;quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps someone can elaborate further to give this relation a more geometric meaning? Where are the &lt;a href=&quot;http://en.wikipedia.org/wiki/Riemann_hypothesis&quot; rel=&quot;nofollow&quot;&gt;non-trivial zeros&lt;/a&gt; of the zeta function to be spotted in this setup? &lt;/p&gt;&#xA;" OwnerUserId="1484" LastActivityDate="2010-08-31T12:06:38.267" CommentCount="3" />
  <row Id="3714" PostTypeId="2" ParentId="3697" CreationDate="2010-08-31T13:06:19.230" Score="2" ViewCount="" Body="&lt;p&gt;To further generalize Rahul's answer, any matrix that has a singular leading block cannot have an LU decomposition. By allowing pivoting (or in matrix factorization terms, allowing the multiplication of your original matrix by an appropriate permutation matrix), all matrices admit an LU decomposition. This is the explanation for pivoting in exact arithmetic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In inexact arithmetic, the condition &quot;singular&quot; in the explanation above is replaced by the term &quot;ill-conditioned&quot;. Every matrix has an associated condition number, which is defined as the product of the norm of the matrix and the norm of its inverse. In attempting to proceed with the LU decomposition of a matrix with an ill-conditioned leading block, you will hit a point where you have to divide by a small number (resulting in a number that may be much larger in magnitude than the matrix's original entries), which causes all sorts of trouble in the succeeding additions/subtractions that have to be done to finish the LU decomposition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By pivoting, we avoid (or at the very least delay) the onset of encountering numbers much larger than the entries of the original matrix, which is one way precision is lost in the operations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, one could make the objection that the coefficients are badly scaled: for instance, if you have two equations in two unknowns, and you multiply both sides of any of the two equations by a number much smaller or much larger than the original coefficients, the solution to the system is still the same, but attempting to perform LU decomposition on the transformed system can be disastrous. This is where concepts like &quot;scaled pivoting&quot; comes in, where relative instead of absolute magnitudes are taken into account in the selection of pivots.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then there are applications where &quot;partial pivoting&quot; (swapping of rows) is not enough; rank determination of a matrix, for instance, requires &quot;complete pivoting&quot; (swapping of both rows and columns).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In short, LU decomposition behaves much better with pivoting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I have been intentionally vague in some parts; you would do well to read Golub and Van Loan, as already recommended by jmoy, or the books &quot;Matrix Decompositions&quot; by Stewart or &quot;Applied Numerical Linear Algebra&quot; by Demmel for more rigorous versions of my explanation.)&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-08-31T13:06:19.230" />
  <row Id="3715" PostTypeId="2" ParentId="3315" CreationDate="2010-08-31T13:10:23.783" Score="-2" ViewCount="" Body="&lt;p&gt;Fake answer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/iApFD.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1551" LastActivityDate="2010-08-31T13:10:23.783" CommentCount="2" />
  <row Id="3716" PostTypeId="2" ParentId="3698" CreationDate="2010-08-31T13:21:57.803" Score="5" ViewCount="" Body="&lt;p&gt;There are good reasons behind the convention of including $(0)$ as a prime ideal &#xA;but excluding $(1)\:.\ $ First, we include zero as a prime ideal because it facilitates many useful reductions. For example, in many ring theoretic problems involving an ideal $\rm\; I\;$, one can reduce to the case $\rm\;I = P\;$ prime, then reduce to $\rm\;R/P\;$, thus reducing to the case when the ring is a domain. In this case one simply says that we can &lt;strong&gt;factor out by the prime&lt;/strong&gt; $\rm\; P\;$, so w.l.o.g. assume $\rm\; P = 0\;$ is prime, so $\rm\:R\:$ is a domain. For example, I've appended to the end of this post an excerpt from Kaplansky's classic textbook &lt;em&gt;Commutative Rings&lt;/em&gt;, section 1-3: G-Ideals, Hilbert Rings, and the Nullstellensatz. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus we have solid evidence for the utility of the convention that the zero ideal is prime. So why don't we adopt the same convention for the unit ideal $(1)$ or, equivalently, why don't we permit the zero ring as a domain?  There are a number of reasons. First, in domains and fields it often proves very convenient to assume that one has a nonzero element available. This permits proofs by contradiction to conclude by deducing $\:1 = 0\:.\ $ More importantly, it implies that the  unit group is nonempty, so unit groups always exist. It'd be very inconvenient to have to always add the proviso (except if $\;\rm R = 0\;)$ to the many arguments involving units and unit groups. For a more general perspective it's worth emphasizing that the usual rules for equational logic are not complete for &lt;em&gt;empty structures&lt;/em&gt; so that is why groups and other algebraic structures are always axiomatized to prevent nonempty structures (see this &lt;a href=&quot;http://groups.google.com/group/sci.math/browse_frm/thread/3e8fac43081fcbf8&quot; rel=&quot;nofollow&quot;&gt;thread&lt;/a&gt; for details).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Below is the promised Kaplansky excerpt on reduction to domains by factoring out prime ideals. I've explicitly emphasized the reductions e.g. &lt;strong&gt;reduce to...&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Let $\rm\; I\;$ be any ideal in a ring $\rm\; R\;$. We write $\rm\; R^{*}\;$ for the quotient ring $\rm\; R/I\;$. In the polynomial ring $\rm\; R[x]\;$ there is a smallest extension $\rm\; IR[x]\;$ of $\rm\; I\;.\ $ The quotient ring $\rm\; R[x]/IR[x]\;$ is in a natural way isomorphic to $\rm\; R^*[x].\;$ In treating many problems, we can in this way &lt;strong&gt;reduce to the case&lt;/strong&gt; $\rm\; I = 0\;$, &#xA;and we shall often do so.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;THEOREM 28.&lt;/strong&gt;  $\:$ Let $\rm\: M\:$ be a maximal ideal in $\rm\: R[x]\:$ and suppose that the contraction $\rm\: M \cap R = N\:$ is maximal in $\rm\; R\:.\ $ Then $\rm\; M\;$ can be generated by $\rm\; N\;$ and one more element $\rm\; f\:.\ $ We can select $\rm\; f\;$ to be a monic polynomial which maps mod $\rm\; N\;$ into an irreducible polynomial over the field $\rm\; R/N\:.\ $ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof.&lt;/strong&gt; $\:$ We can &lt;strong&gt;reduce to the case&lt;/strong&gt; $\rm\; N = 0,\;$ i. e., $\rm\; R\;$ a field, and then &#xA;the statement is immediate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;THEOREM 31.&lt;/strong&gt; $\;$ A commutative ring $\rm\; R\;\;$ is a Hilbert ring if and only if the polynomial ring $\rm\; R[x] \;\;$ is a Hilbert ring. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof.&lt;/strong&gt; $\;$ If $\rm\; \;\rm R[x]\;$ is a Hilbert ring, so is its homomorphic image $\rm\; R\;$. &#xA;Conversely, assume that $\rm\; R\;$ is a Hilbert ring. Take a G-ideal $\rm\; Q\;$ in &#xA;$\rm\; R[x]\;$; we must prove that $\rm\; Q\;$ is maximal. Let $\rm\; P = Q \cap R\;$; we can &lt;strong&gt;reduce the problem to the case&lt;/strong&gt; $\rm\; P = 0\;$, which, incidentally, makes $\rm\; R\;$ a domain. &#xA;Let $\rm\; u\;$ be the image of $\rm\; x\;$ in the natural homomorphism $\rm\; R[x] \to R[x]/Q\;$. &#xA;Then $\rm\; R[u]\;$ is a G-domain. By Theorem 23, $\rm\:u\:$ is algebraic over $\rm\:R\:$ and $\rm\:R\:$ is a G-domain. Since $|rm\:R\:$ is both a G-domain and a Hilbert ring, $\rm\:R\:$ is a field. But this makes $\rm\; R[u] = R[x]/Q\;$ a field, proving $\rm\; Q\;$ to be maximal.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2011-03-15T22:45:37.660" LastActivityDate="2011-03-15T22:45:37.660" CommentCount="1" />
  <row Id="3717" PostTypeId="2" ParentId="3698" CreationDate="2010-08-31T13:41:11.140" Score="4" ViewCount="" Body="&lt;p&gt;Generally we make nice conventions because they make the statements of theorems nice.  The theorem relevant to prime ideals is that $P$ is a prime ideal of $R$ if and only if $R/P$ is an integral domain.  The theorem relevant to prime elements is prime factorization (when it holds).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;These two concepts &lt;em&gt;almost&lt;/em&gt; coincide for principal ideals, but we must distinguish between the generic point $(0)$ and closed points, and there are good reasons for doing this.  (The zero ideal, for example, can't occur in the factorization of a nonzero ideal in a Dedekind domain.)&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-08-31T13:41:11.140" CommentCount="3" />
  <row Id="3718" PostTypeId="2" ParentId="3698" CreationDate="2010-08-31T14:18:57.137" Score="13" ViewCount="" Body="&lt;p&gt;You have a point here: absolutely we want to count $(0)$ as a prime ideal in $\mathbb{Z}$ -- because $\mathbb{Z}$ is an integral domain -- whereas we do &lt;em&gt;not&lt;/em&gt; want to count $(1)$ as being a prime ideal -- because the zero ring is not an integral domain (which, to me, is much more a true fact than a convention: e.g., every integral domain has a field of fractions, and the zero ring does not).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think we do not want to call $0$ a prime element because, in practice, we never want to include $0$ in divisibility arguments.  Another way to say this is that we generally want to study factorization in integral domains, but once we have specified that a commutative ring $R$ is a domain, we know all there is to know about factoring $0$: $0 = x_1 \cdots x_n$ iff at least one $x_i = 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is one way to make this &quot;ignoring $0$&quot; convention look more natural: the notions of factorization, prime element, irreducible element, and so forth in an integral domain $R$ depend entirely on the multiplicative structure of $R$.  Thus we can think of factorization questions as taking place in the &lt;strong&gt;cancellative monoid&lt;/strong&gt; $(R \setminus 0,\cdot)$.  (Cancellative means: if $x \cdot y = x \cdot z$, then $y = z$.)  In this context it is natural to exclude zero, because otherwise the monoid would not be cancellative.  Contemporary algebraists often think about factorization as a property of monoids rather than integral domains &lt;em&gt;per se&lt;/em&gt;.  For a little more information about this, see e.g. Section 4.1 of &lt;a href=&quot;http://math.uga.edu/~pete/factorization2010.pdf&quot; rel=&quot;nofollow&quot;&gt;http://math.uga.edu/~pete/factorization2010.pdf&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="299" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2010-08-31T14:28:59.467" LastActivityDate="2010-08-31T14:28:59.467" CommentCount="6" />
  <row Id="3720" PostTypeId="2" ParentId="3704" CreationDate="2010-08-31T14:40:43.093" Score="10" ViewCount="" Body="&lt;p&gt;There is a &lt;em&gt;very powerful&lt;/em&gt; calculus of &lt;strong&gt;multidimensional residues&lt;/strong&gt; that accomplishes what you seek, see for example the book G. P. Egorychev. Integral representation and computation of combinatorial sums. AMS, Transl. of Math. Monogr. v. 59 Providence 2nd ed. 1989 - which can be located online via gigapedia.com. Below are two illustrative examples of Egorychev's &quot;method of coefficients&quot; excerpted from the survey by Egorychev and Zima in volume 5 of Hazewinkel's Handbook of Algebra (also available online via gigapedia)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/TiDQB.jpg&quot; alt=&quot;alt text&quot;&gt;&#xA;&lt;img src=&quot;http://i.imgur.com/xTpHh.jpg&quot; alt=&quot;alt text&quot;&gt;&#xA;&lt;img src=&quot;http://i.imgur.com/JMzAV.jpg&quot; alt=&quot;alt text&quot;&gt;&#xA;&lt;img src=&quot;http://i.imgur.com/MAZBo.jpg&quot; alt=&quot;alt text&quot;&gt;&#xA;&lt;img src=&quot;http://i.imgur.com/xOxhy.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-08-31T20:01:37.980" LastActivityDate="2010-08-31T20:01:37.980" CommentCount="0" />
  <row Id="3721" PostTypeId="2" ParentId="3680" CreationDate="2010-08-31T14:53:40.897" Score="1" ViewCount="" Body="&lt;p&gt;You could conceive of your result as being a weighted average.  In a batch of data consisting of $n_i \ge 1$ instances of $x_i$, $1 \le i \le n$, your average is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sum_{i=1}^{n} x_i / n = \sum_{i=1}^{n} \frac {1}{n_i} (n_i x_i ) / n$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$= \sum_{i=1}^{n} \sum_{j=1}^{n_j} \left( \frac {1}{n_i} x_i \right) /  \sum_{i=1}^{n} \sum_{j=1}^{n_j} \frac {1}{n_i}$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;exhibiting the weights as $\frac {1} {n_i}$.&lt;/p&gt;&#xA;" OwnerUserId="1489" LastActivityDate="2010-08-31T14:53:40.897" />
  <row Id="3722" PostTypeId="2" ParentId="3705" CreationDate="2010-08-31T15:26:58.350" Score="2" ViewCount="" Body="&lt;p&gt;Still no.  Counterexample:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$M = \lambda M_1 + \mu M_2 = \pmatrix{0 &amp;amp; \mu &amp;amp; \mu \cr \mu &amp;amp; \lambda &amp;amp; 0 \cr \mu &amp;amp; 0 &amp;amp; -\lambda}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously $\det(M)\equiv 0$ for all $\lambda$ and $\mu$.  However, the only $(x,y,z)$ that satisfies&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$M\pmatrix{x\cr y\cr z} = \pmatrix{0 &amp;amp; \mu &amp;amp; \mu \cr \mu &amp;amp; \lambda &amp;amp; 0 \cr \mu &amp;amp; 0 &amp;amp; -\lambda}\pmatrix{x\cr y\cr z} = \pmatrix{\mu(y+z)\cr \mu x + \lambda y\cr \mu x - \lambda z} = \mathbf{0}\quad \forall \lambda, \mu$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is the zero vector.&lt;/p&gt;&#xA;" OwnerUserId="1551" LastActivityDate="2010-08-31T15:26:58.350" CommentCount="0" />
  <row Id="3723" PostTypeId="1" AcceptedAnswerId="3836" CreationDate="2010-08-31T15:31:40.803" Score="4" ViewCount="414" Body="&lt;p&gt;Consider the piecewise constant function $\psi:I=[-1,1] \rightarrow \mathbb{R}$ given by &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\psi(x) = \begin{cases} \psi_1 &amp;amp; x \leq 0, \ \psi_2 &amp;amp; x &gt; 0 \end{cases}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for some constants $\psi_1, \psi_2 \in \mathbb{R}$.  I would like to evaluate the integral&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_I \delta(x) \psi(x) dx$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\delta$ is the Dirac delta distribution centered at $x=0$.  Of course, if we think about distributions in the usual way then you might say that $\delta(\psi) = \langle \delta, \psi \rangle = \psi(0) = \psi_1.$  But then the result depends on a fairly arbitrary choice when defining $\psi$: should the left- or right- half of the interval be closed?  This question doesn't seem to have a meaningful answer when dealing with a problem that arises from a physical system (say).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instead, consider a family of distributions $\phi_\epsilon(x)$ such that&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\phi_\epsilon(x)=\phi_\epsilon(-x)$,&lt;/li&gt;&#xA;&lt;li&gt;$\int_I \phi_\epsilon(x)=1$ for all $\epsilon$, and&lt;/li&gt;&#xA;&lt;li&gt;$\lim_{\epsilon \rightarrow 0} \phi_\epsilon = \delta$,&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;i.e., any family of even distributions with unit mass that approaches the Dirac delta distribution as $\epsilon$ approaches zero.  (For instance, you could use the family of Gaussians $\phi_\epsilon(x) = \frac{1}{\epsilon\sqrt{\pi}}e^{-x^2/\epsilon^2}$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can now think of my integral as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\lim_{\epsilon \rightarrow 0} \int_I \phi_\epsilon(x) \psi(x) dx.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For some $\epsilon &gt; 0$, the integral inside the limit can be expressed as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$u(\epsilon) = \psi_1 \int_{-1}^0 \phi_\epsilon(x) dx + \psi_2 \int_0^1 \phi_\epsilon(x) dx = \frac{1}{2}\left( \psi_1 + \psi_2 \right) = \bar{\psi},$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\bar{\psi}$ is the mean of the constant values.  Can we say, then, that $\lim_{\epsilon \rightarrow 0} u(\epsilon) = \bar{\psi}$?  It would seem so: for any $\mu &gt; 0$ there exists an $\epsilon_0$ such that $\epsilon &amp;lt; \epsilon_0$ implies $|u(\epsilon) - \bar{\psi}|&amp;lt;\mu$ (namely, $\epsilon$ is any positive constant!).  But clearly I've got a problem somewhere, because $\bar{\psi} \ne \psi_1$, i.e., this result does not agree with my earlier interpretation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what's the right thing to do here?  The latter answer ($\bar{\psi}$) agrees more with my &quot;physical&quot; intuition (because it's invariant with respect to deciding which half-interval is open), but I'm concerned about rigor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Edit:&lt;/em&gt; Since the problem as stated is not well-posed ($\delta$ cannot be evaluated on discontinuous functions), let me give some motivation.  Imagine that I have a pair of piecewise linear functions $f,g:I^2 \rightarrow \mathbb{R}$, which are again discontinuous only at $x=0$.  I would like to integrate the &lt;em&gt;wedge product&lt;/em&gt; of $df$ and $dg$ over the domain:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_{I^2} df \wedge dg = \int_I \int_I \frac{\partial f}{\partial x} \frac{\partial g}{\partial y} - \frac{\partial f}{\partial y}\frac{\partial g}{\partial x} dx dy.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider just the first term $(\partial f/\partial x)(\partial g/\partial y)$ and consider just the inner integral $\int_I \cdot dx$.  We now have (almost) the original problem: $\partial f/\partial x$ can be thought of as a $\delta$ (plus a piecewise constant), and $\partial g/\partial y$ is simply piecewise constant along the $x$-direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the problem could be restated as: how do I integrate the wedge product $df \wedge dg$ of piecewise linear 0-forms $f$ and $g$ defined over a planar region?  Formally this problem may again be ill-posed, yet it is a real problem that comes up in the context of finite element analysis where basis functions are nonsmooth or even discontinuous.&lt;/p&gt;&#xA;" OwnerUserId="1612" LastEditorUserId="1612" LastEditorDisplayName="" LastEditDate="2010-08-31T17:59:01.973" LastActivityDate="2010-09-02T01:59:08.120" Title="Integrate product of Dirac delta and discontinuous function?" Tags="&lt;calculus&gt;&lt;real-analysis&gt;&lt;analysis&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" />
  <row Id="3724" PostTypeId="1" AcceptedAnswerId="3732" CreationDate="2010-08-31T16:05:51.177" Score="4" ViewCount="672" Body="&lt;p&gt;Suppose you have 100 coins. 96 of them are heavy and 4 of them are light.  Nothing is known regarding the proportion of their weights. You want to find at least one genuine (heavy) coin. You are allowed to use a weight balance twice. How do you find it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assumptions: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Heavy coins all have the same weight; same for the light coins.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The weight balance compares the weight of two sides on the balance instead of giving numerical measurement of weights.&lt;/p&gt;&#xA;" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-09-02T01:25:52.793" LastActivityDate="2011-03-08T00:20:54.800" Title="Finding the Heavy Coin by weighing twice" Tags="&lt;combinatorics&gt;&lt;algorithms&gt;&lt;puzzle&gt;&lt;contest-math&gt;" AnswerCount="7" CommentCount="7" FavoriteCount="2" />
  <row Id="3726" PostTypeId="2" ParentId="3723" CreationDate="2010-08-31T17:25:28.610" Score="0" ViewCount="" Body="&lt;p&gt;It seems your question contains its own answer.  $\delta$ is not defined as evaluation-at-a-point &lt;em&gt;except&lt;/em&gt; for continuous functions.  (The inner product, strictly speaking, is defined in $L^p$ spaces of &lt;em&gt;equivalence classes&lt;/em&gt; of functions and the &quot;value&quot; of such a class at a point is undefined unless you limit the evaluation to continuous functions.)  Your second method is close to one standard method of creating (and calculating with) $\delta$ as a weak limit of integrals.  (Another standard method, based on integration by parts, requires the trial function to be differentiable, so applying that method won't work here.)  Thus, you have clearly demonstrated why evaluate-at-a-point is an unsuitable definition of $\delta$ for discontinuous functions.&lt;/p&gt;&#xA;" OwnerUserId="1489" LastActivityDate="2010-08-31T17:25:28.610" />
  <row Id="3728" PostTypeId="2" ParentId="3724" CreationDate="2010-08-31T17:28:58.667" Score="2" ViewCount="" Body="&lt;p&gt;If you pay me 95 of the coins in advance I'll show you how ;-).&lt;/p&gt;&#xA;" OwnerUserId="1489" LastActivityDate="2010-08-31T17:28:58.667" CommentCount="6" />
  <row Id="3729" PostTypeId="2" ParentId="3724" CreationDate="2010-08-31T17:45:48.853" Score="3" ViewCount="" Body="&lt;p&gt;To avoid wasting readers' time when posting puzzles, it is useful (and will improve the upvote/downvote ratio) to provide:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;an indication of the difficulty level&lt;/em&gt;. Was this puzzle in a math-is-fun book for children, or is it a generalization of a problem from the final round of a national olympiad?  Having schoolchildren attempt a question that (they were not told) is the $n=100$ specialization of a combinatorial lemma from a research paper, is a waste of time.  Having mathematicians spend time on (what might look like) a coin weighing puzzle from the literature, but is really an easy task the high school students could solve with some patience, is a waste of resources.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;the source, if known&lt;/em&gt;.  This helps those who want to evaluate the difficulty before taking time to attempt a solution, or who would like to look up hints or answers.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;indicators of whether the problem is correct or solvable as stated&lt;/em&gt;.  Problems from a book, a competition or a journal are likelier to be correct and unambiguously formulated.  Questions invented, circulated and (believed to be) solved among friends or students are more likely to have pitfalls that were missed in setting or solving the problem.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;clear definitions&lt;/em&gt;.  In this case, coin weighing problems involve at least two very different interpretations of &quot;use a weight balance&quot;.  Weighing can mean comparison or numerical measurement and what is possible is completely different in those settings.&#xA;Not specifying which is meant can waste the time of users who try to solve the problem in the wrong interpretation. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="467" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-08-31T17:51:14.767" LastActivityDate="2010-08-31T17:51:14.767" CommentCount="4" />
  <row Id="3730" PostTypeId="2" ParentId="2968" CreationDate="2010-08-31T17:49:19.060" Score="0" ViewCount="" Body="&lt;p&gt;It seems to me that this puzzle as stated can be resolved by a public counter that counts down from $n$ to $n - k$ in even time steps, each of which is longer than every player's reaction time.  Each player intends to stop the counter when it displays a value equal to or lower than the player's own card; whoever first stops the counter is the winner.  This does not seem to reveal anything that is not already available from $n$, $k$, and the value of the winner's card.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously this is impracticable when $n &gt;&gt; k$, but even then one could have the time count down in sufficiently large increments to be physically feasible.  When the number of steps is substantially larger than $k$ this would still have a good chance of identifying the winner.  Apparent ties in this procedure could be resolved by a quick run-off using the same technology (keeping the players in the runoff anonymous), giving an almost certain procedure to find a unique winner correctly.&lt;/p&gt;&#xA;" OwnerUserId="1489" LastActivityDate="2010-08-31T17:49:19.060" CommentCount="3" />
  <row Id="3731" PostTypeId="2" ParentId="3724" CreationDate="2010-08-31T17:49:19.420" Score="0" ViewCount="" Body="&lt;p&gt;[Edit] As pointed out by longway, there's a hole in this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My assumptions:&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Your heavy coins all have the same weight; same for the light coins.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The weight balance compares the weight of two sides on the balance instead of giving numerical measurement of weights.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Method:&lt;/strong&gt;&#xA;Divide the 100 coins into 2 groups A and B, each comprises 49 coins. The remaining 2 coins are labelled C and D.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Weight A and B.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;1.1 A &gt; B (which also handles symmetrically B &gt; A)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Divide A into 24 + 24 + 1 coins, and weigh 24 vs 24.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1.1.1 If the two sides are equal, all of them are genuine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1.1.2 Otherwise the heavier side contains all genuine coins.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1.2 A = B&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Divide A into 24 + 25 coins, add C to the 24 side, and weigh 25 vs 25. Call the side containing C the C-side.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1.2.1 If the C side is heavier, then C is genuine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1.2.2 If the C side is equal to the other side, then C side minus C are all genuine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1.2.3 If the C side is lighter, then the other side contains all genuine coins.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Explanation:&lt;/strong&gt;&#xA;Divide the 100 coins into 2 groups A and B, each of 49 coins, and label the remaining 2 coins as C and D.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Weight A and B.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;1.1 A &gt; B, it means that B has more fake coins than A. Clearly this means that A has either 0 or 1 fake coins.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Divide A into 24 + 24 + 1 coin, and weigh 24 coins against the other 24 coins.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1.1.1 If they are equal, then they must all be genuine since there's at most one fake coin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1.1.2 If they aren't equal, the heavier side contains all genuine coins.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The case where B &gt; A is symmetric.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1.2 A = B, it means that A and B have same number of fake coins. This number could be 1 or 2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Divide A into 24 + 25 coins, and add C to the 24 portion, and weigh these 25 coins against the other 25.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;If A and B both have 1 fake coin only, that means both C and D are fake. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If A and B both have 2 fake coins, that means both C and D are genuine.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;That means&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;if A has 1 fake coin, then the C side is &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;equal to the other side when C-side minus C are all genuine coins, and the other side     contains one fake coin.&lt;/li&gt;&#xA;&lt;li&gt;lighter than the other side when C-side contains the other fake coin as well. In this case the other side contains genuine coins.&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;if A has 2 fake coins, then the C side is &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;lighter than the other side when both fake coins are on the C-side. In this case the other side contains genuine coins.&lt;/li&gt;&#xA;&lt;li&gt;heavier than the other side when some fake coins are on the other side. In this case, C must be genuine anyway since A has 2 fake coins.&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;if the C side is heavier, then C is genuine.&lt;/li&gt;&#xA;&lt;li&gt;if the C side is equal to the other side, then C side minus C are all genuine coins.&lt;/li&gt;&#xA;&lt;li&gt;if the C side is lighter, then the other side contains all genuine coins.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="325" LastEditorUserId="325" LastEditorDisplayName="" LastEditDate="2011-03-08T00:20:54.800" LastActivityDate="2011-03-08T00:20:54.800" />
  <row Id="3732" PostTypeId="2" ParentId="3724" CreationDate="2010-08-31T17:54:53.823" Score="5" ViewCount="" Body="&lt;p&gt;I think this works.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Divide the coins into three groups: $A$ with $33$ coins, $B$ with $33$ coins and  $C$ with $34$ coins.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Weigh $A$ and $B$ against each other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now if $A$ is heavier than $B$, then $A$ cannot have two or more light coins, as in that case, $A$ would be lighter (or equal to $B$). Now split $A$ into groups of $16$ plus one odd coin. Weigh the groups of $16$ against each other. If they are the same, then any of those coins is heavy. If not, then any of the heavier 16 coins is heavy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the case when $A$ and $B$ are equal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The possibilities for $A$, $B$ and $C$ are:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&#xA;&#xA;+-----------+------------+-----------+&#xA;|  A        |   B        |  C        |&#xA;+-----------+------------+-----------+&#xA;|  33H      |  33H       |  30H + 4L |&#xA;|           |            |           |&#xA;|  32H + L  |  32H + L   |  32H + 2L |&#xA;|           |            |           |&#xA;|  31H + 2L |  31H + 2L  |  34H      |&#xA;+-----------+------------+-----------+&#xA;&#xA;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Now move one coin from $A$ to $B$ (call the resulting set $B'$) and weigh it against $C$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $B' &gt; C$, then the coin you moved from $A$ is a heavy coin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $B' = C$, then the coin you moved from $A$ is a light coin and the remaining coins in $A$ are heavy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $B' &amp;lt; C$, then all the coins in $C$ are heavy.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-10-11T20:53:51.037" LastActivityDate="2010-10-11T20:53:51.037" CommentCount="4" />
  <row Id="3734" PostTypeId="1" AcceptedAnswerId="3739" CreationDate="2010-08-31T18:53:13.827" Score="3" ViewCount="364" Body="&lt;p&gt;This is problem taken from this link Problem 4: &lt;a href=&quot;http://www.math.u-szeged.hu/~mmaroti/schweitzer/schweitzer-2001-eng.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.math.u-szeged.hu/~mmaroti/schweitzer/schweitzer-2001-eng.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I couldn't find the solution anywhere and i am curious to see a solution for this problem, as i can atleast comprehend the question and it seems, that the mechanism for the solution involved will be a somewhat understandable.&lt;/p&gt;&#xA;" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-09-05T04:25:29.857" LastActivityDate="2010-09-05T04:25:29.857" Title="Finding the Units in the Ring $\mathbb{Z}[t][\sqrt{t^{2}-1}]$" Tags="&lt;number-theory&gt;&lt;abstract-algebra&gt;&lt;ring-theory&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="1" />
  <row Id="3735" PostTypeId="2" ParentId="3399" CreationDate="2010-08-31T19:13:21.820" Score="2" ViewCount="" Body="&lt;p&gt;With the caveat that it's usually more helpful to devise your own mnemonics than follow someone else's&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;here are a couple of mine, poorly drawn (the second is same as Srikant Vadali's answer):&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://shreevatsa.files.wordpress.com/2009/03/convex.png&quot; alt=&quot;Convex function&quot;&gt;&#xA;&lt;img src=&quot;http://shreevatsa.files.wordpress.com/2009/03/concave1.png&quot; alt=&quot;Concave function&quot;&gt;  &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;convex: smiley face&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Another way of remembering them, if you recall the meanings of convex and concave outside mathematics (as in lenses, etc.), is that you &lt;strong&gt;look from below&lt;/strong&gt;: if the graph of the function viewed from below looks convex (i.e., bulging towards you) the function is convex, if it looks concave the function is concave.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Yet another way is to keep in mind the definition: &quot;a function is convex if its epigraph is a convex set&quot;. The &lt;a href=&quot;http://en.wikipedia.org/wiki/Epigraph_%28mathematics%29&quot; rel=&quot;nofollow&quot;&gt;epigraph&lt;/a&gt; is the set of points lying above the graph, and a &lt;a href=&quot;http://en.wikipedia.org/wiki/Convex_set&quot; rel=&quot;nofollow&quot;&gt;convex set&lt;/a&gt; is one in which every line segment between two points in the set lies within the set. [Actually, for me, this definition is more useful for remembering what epigraph means :-)]&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="205" LastActivityDate="2010-08-31T19:13:21.820" CommunityOwnedDate="2010-08-31T19:13:21.820" CommentCount="3" />
  <row Id="3736" PostTypeId="1" AcceptedAnswerId="3738" CreationDate="2010-08-31T19:37:27.057" Score="6" ViewCount="134" Body="&lt;p&gt;Hi,&#xA;Suppose $\Omega$ is a domain of the complex plane (i.e. an open and connected subset of the plane). Suppose $f$ is holomorphic on $\Omega$, and $f$ is not identically zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose $f$ has a holomorphic logarithm on $\Omega$, which means that there is a function $g$ holomorphic on $\Omega$ such that $e^g=f$. Then it is easy to show that $f$ has holomorphic $n$-th roots on $\Omega$ for each $n$, which means that for each integer $n$, there exist a function $g_n$ holomorphic on $\Omega$ such that $(g_n)^n = f$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is the converse true? i.e. if $f$ has holomorphic $n$-th roots on $\Omega$ for all $n$, then can we find a function $g$ holomorphic on $\Omega$ such that $f=e^g$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A few remarks :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One can prove that if $f$ has holomorphic $n$-th roots on $\Omega$ for all $n$, then $f$ does not vanish on $\Omega$. Therefore, we can define a holomorphic logarithm &lt;em&gt;locally&lt;/em&gt;, but is it possible to find a &lt;em&gt;global&lt;/em&gt; holomorphic logarithm?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore, notice that $\Omega$ is not supposed simply connected, in which case the answer to my question is yes.&lt;/p&gt;&#xA;" OwnerUserId="1197" LastActivityDate="2010-08-31T20:19:14.547" Title="A criterion for the existence of a holomorphic logarithm of a holomorphic function" Tags="&lt;complex-analysis&gt;" AnswerCount="1" />
  <row Id="3737" PostTypeId="2" ParentId="1748" CreationDate="2010-08-31T20:10:19.677" Score="3" ViewCount="" Body="&lt;p&gt;Hulek's «&lt;a href=&quot;http://www.ams.org/bookstore-getitem/item=stml-20&quot; rel=&quot;nofollow&quot;&gt;Elementary algebraic geometry&lt;/a&gt;» is a really nice text for undergraduates.  (It's part of the AMS's student mathematical library.)&lt;/p&gt;&#xA;" OwnerUserId="314" LastActivityDate="2010-08-31T20:10:19.677" CommunityOwnedDate="2010-08-31T20:10:19.677" />
  <row Id="3738" PostTypeId="2" ParentId="3736" CreationDate="2010-08-31T20:19:14.547" Score="8" ViewCount="" Body="&lt;p&gt;The condition that $f$ have a holomorphic logarithm is equivalent to&#xA;$df/f=f'(z)dz/f(z)$ being an exact differential. This is equivalent to&#xA;the integral of $df/f$ over all closed curves in $\Omega$ vanishing.&#xA;Let $C$ be a closed curve in $\Omega$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $f=g^n$ is an $n$-th power in $\Omega$ of a holomorphic $g$ then&#xA;$\int_C df/f=n\int_C dg/g$. But $\int_C dg/g$ is an integer multiple of $2\pi i$.&#xA;Hence $\int_C df/f$ is an integer multiple of $2\pi ni$. If this holds for&#xA;all $n$ then $\int_C df/f=0$. It follows that $f$ has a holomorphic logarithm.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-08-31T20:19:14.547" CommentCount="1" />
  <row Id="3739" PostTypeId="2" ParentId="3734" CreationDate="2010-08-31T20:21:27.620" Score="6" ViewCount="" Body="&lt;p&gt;For any non-negative integer $n$ the element $\pm (t \pm \sqrt{t^2 - 1})^n$ has norm $1$, and there are no elements $a(t) + b(t) \sqrt{t^2 - 1}$ of norm $-1$ because $a(1)^2 - b(1)^2 \sqrt{1^2 - 1} \ge 0$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose $a(t) + b(t) \sqrt{t^2 - 1}$ is a unit which is not one of the above units such that $\deg b$ is minimal.  Then it has norm $1$, so $a(t)^2 - b(t)^2 (t^2 - 1) = 1$.  It is not hard to see that the units $\pm 1$ are the only units for which $b = 0$, so WLOG $b$ is nonzero.  This implies that $\deg a = d+1, \deg b = d$ for some non-negative integer $d$, and moreover the leading terms of $a$ and $b$ must agree up to sign.  If the leading terms agree, then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$( a(t) + b(t) \sqrt{t^2 - 1})(t - \sqrt{t^2 - 1}) = (ta(t) - b(t) (t^2 - 1)) + (tb(t) - a(t)) \sqrt{t^2 - 1}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is a unit with the property that the coefficient of $\sqrt{t^2 - 1}$ has degree strictly less than that of $b$ which is not on the above list, which contradicts the assumption of minimality.  Similarly, if the leading terms of $a$ and $b$ are opposite in sign, then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$( a(t) + b(t) \sqrt{t^2 - 1})(t + \sqrt{t^2 - 1}) = (ta(t) + b(t) (t^2 - 1)) + (tb(t) + a(t)) \sqrt{t^2 - 1}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is a unit with the property that the coefficient of $\sqrt{t^2 - 1}$ has degree strictly less than that of $b$ which is not on the above list, again contradicting the assumption of minimality.  So no such units exist.&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-09-01T17:35:59.310" LastActivityDate="2010-09-01T17:35:59.310" CommentCount="3" />
  <row Id="3740" PostTypeId="2" ParentId="3561" CreationDate="2010-08-31T20:40:30.567" Score="0" ViewCount="" Body="&lt;p&gt;I highly recommend Artin's &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0130047635&quot; rel=&quot;nofollow&quot;&gt;Algebra&lt;/a&gt;, which is full of great motivation and originally kindled my love for algebra.&lt;/p&gt;&#xA;" OwnerUserId="314" LastActivityDate="2010-08-31T20:40:30.567" CommunityOwnedDate="2010-08-31T20:40:30.567" />
  <row Id="3741" PostTypeId="2" ParentId="3724" CreationDate="2010-08-31T20:43:24.707" Score="2" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Update:&lt;/strong&gt; It seems that &lt;a href=&quot;http://math.stackexchange.com/users/1102/moron&quot;&gt;Moron&lt;/a&gt; has a valid objection (&lt;em&gt;see the comments&lt;/em&gt;) that renders my solution incorrect:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Two sides are equal: You could have 2&#xA;  heavy and 2 light. Either all light or&#xA;  all heavy is incorrect.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In other words:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Yaser's answer misses one case. Where&#xA;  we get H+L vs H+L giving us equal on&#xA;  the first weighing.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;At first, I thought that even if this was the case, one could still find a heavy coin with only one more weighing. Alas, &lt;strong&gt;in every weighing scheme I could think of in the context of my solution, there was a situation where one weighing does not give the necessary information to guarantee selecting a heavy coin.&lt;/strong&gt; Perhaps there's some twisted weighing scheme I didn't think of that would guarantee a correct selection, but I really doubt that (don't take my word for it though.. it's left as an exercise to the reader.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://math.stackexchange.com/questions/3724/finding-the-heavy-coin-by-weighing-twice/3732#3732&quot;&gt;Moron's solution&lt;/a&gt; seems to be a correct one, especially that it is, according to &lt;a href=&quot;http://math.stackexchange.com/users/467/t&quot;&gt;T..&lt;/a&gt;, the same as the solution by the original proposer (Shapovalov). I still have to check it out, but it's getting late and I'm quite exhausted. For now, I think it's time for me to get some sleep and to accept the downvotes. Be gentle!&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Let me elaborate on &lt;a href=&quot;http://math.stackexchange.com/questions/3724/finding-the-heavy-coin-by-weighing-twice/3728#3728&quot;&gt;whuber's answer&lt;/a&gt; by putting it in a slightly different form.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's randomly choose 5 coins out of the 100. Let's denote this set of coins by . Since there are only 4 light coins, &lt;strong&gt;we are absolutely sure that  contains at least one heavy coin.&lt;/strong&gt; Now,  the problem is reduced to simply finding one heavy coin in  by using the weight balance twice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now stop for a moment and see if you can solve the new reduced problem on your own.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can't figure it out? Well, choose any 4 coins and put 2 on one side of the balance and 2 on the other. Now, what are the possible outcomes?&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;The two sides are equal.&lt;/strong&gt; This means that the 4 coins we put on the balance are &lt;em&gt;either all light or all heavy&lt;/em&gt; (&lt;strong&gt;Update: This is not true. See the update at the beginning&lt;/strong&gt;). In this case, pick any coin from the balance and compare it with the one we left out. If one is heavier, it would be trivial to select it as a heavy coin. If both are equal, then all of the coins in our initial  were actually heavy, so just select any!&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;One side is heavier.&lt;/strong&gt; This implies that &lt;em&gt;at least&lt;/em&gt; one of the 2 coins on this &quot;heavier&quot; side is a heavy coin. Take these 2 coins and compare them with each other. Now selecting a heavy coin is trivial (if they are of equal weight, then they are both heavy, so just select one!)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://math.stackexchange.com/questions/3724/finding-the-heavy-coin-by-weighing-twice/3728#3728&quot;&gt;Whuber's answer/hint&lt;/a&gt; (&lt;em&gt;look in the comments&lt;/em&gt;) put me on the right track to this solution, so don't forget to vote his answer up!&lt;/p&gt;&#xA;" OwnerUserId="114" LastEditorUserId="114" LastEditorDisplayName="" LastEditDate="2010-09-01T00:00:31.437" LastActivityDate="2010-09-01T00:00:31.437" CommentCount="2" />
  <row Id="3742" PostTypeId="1" AcceptedAnswerId="3743" CreationDate="2010-08-31T21:14:37.657" Score="3" ViewCount="319" Body="&lt;p&gt;Do you have any favorite open problem?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me mention one of my favorites. Let $A(\mathbb{T})$ be the Wiener algebra, that is, the linear space of absolutely convergent Fourier series on the unit circle $\mathbb{T}$ $(=\mathbb{R}/2\pi\mathbb{Z}=(-\pi,\pi])$ carrying the norm &#xA;$$ f\mapsto \|f\|=\sum_{n\in\mathbb{Z}}|\hat{f}(n)|&amp;lt;\infty$$&#xA;where $\hat{f}(n)=\int_{-\pi}^\pi f(t)e^{-int}dt/2\pi$ is the $n$:th Fourier coefficient of $f$. In fact $A(\mathbb{T})$ is a unitary commutative Banach algebra. By absolute convergence it follows that each $f\in A(\mathbb{T})$ is continuous on $\mathbb{T}$. Moreover, if $f(t)\not=0$ for all $t\in\mathbb{T}$ then obviously $1/f$ is also continuous on $\mathbb{T}$ - a famous theorem of Norbert Wiener (The Wiener Lemma) states that we also have $1/f\in A(\mathbb{T})$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next consider a possible quantitative refinement of the Wiener lemma: &#xA;Given $\delta&gt;0$  let $$C_\delta = \sup_{A_\delta}\|1/f\|$$&#xA;where $A_\delta={f\in A(\mathbb{T}):|f(t)|&gt;\delta,\ \|f\|\leq1}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Problem:&lt;/strong&gt; Find &#xA;$$\delta_\inf=\inf_{\delta&gt;0} \ C_\delta&amp;lt;\infty$$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remark 1: It is known that $\delta_\inf\leq 1/\sqrt{2}$ and that $\delta_\inf\geq 1/2$ (see [1,2]).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remark 2: The problem can be treated in any commutative Banach algebra we unit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[1] N. Nikolski, In search of the invisible spectrum, Annales de l'institut Fourier, 49 no. 6 (1999), p. 1925-1998&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[2] H.S. SHAPIRO, A counterexample in harmonic analysis, in Approximation Theory, Banach Center Publications, Warsaw (submitted 1975), Vol. 4 (1979), 233-236.&lt;/p&gt;&#xA;" OwnerUserId="1154" LastActivityDate="2010-09-01T14:31:30.020" CommunityOwnedDate="2010-08-31T21:21:43.353" Title="Favourite open problem?" Tags="&lt;big-list&gt;&lt;open-problem&gt;" AnswerCount="7" CommentCount="3" FavoriteCount="2" />
  <row Id="3743" PostTypeId="2" ParentId="3742" CreationDate="2010-08-31T21:32:35.137" Score="2" ViewCount="" Body="&lt;p&gt;I like this one that is simple to state but likely very difficult to prove or disprove: The irrationality of $\zeta (5)$.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The irrationality of $\zeta (3)$  was proved by Roger Apéry only in 1979. Despite considerable effort the picture is rather incomplete about $\zeta (s)$ for the other odd integers, $s=2t+1\gt 5$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;-- Martin Aigner and Günter Ziegler, Proofs from THE BOOK.&lt;/p&gt;&#xA;" OwnerUserId="752" LastActivityDate="2010-08-31T21:32:35.137" CommunityOwnedDate="2010-08-31T21:32:35.137" />
  <row Id="3744" PostTypeId="2" ParentId="3742" CreationDate="2010-08-31T21:46:40.883" Score="1" ViewCount="" Body="&lt;p&gt;It is open whether or not there exists an algorithm which will determine whether an integer sequence satisfying a linear recurrence with integer coefficients (such as the Fibonacci sequence) has only nonzero terms.  See Terence Tao's &lt;a href=&quot;http://terrytao.wordpress.com/2007/05/25/open-question-effective-skolem-mahler-lech-theorem/&quot; rel=&quot;nofollow&quot;&gt;excellent blog post&lt;/a&gt; on the subject.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-08-31T21:46:40.883" CommunityOwnedDate="2010-08-31T21:46:40.883" />
  <row Id="3745" PostTypeId="2" ParentId="3679" CreationDate="2010-08-31T21:57:16.167" Score="0" ViewCount="" Body="&lt;p&gt;Just a remark: there is some relevant basic algebraic topology here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As Matt E points out in his answer, the &quot;direction&quot; of a vector $v \in \mathbb{R}^n$ is not a number.  What then is it?  A natural answer is that a direction is an element of the unit sphere $S^{n-1} = {(x_1,\ldots,x_n) \in \mathbb{R}^n \ | \ x_1^2 + \ldots + x_n^2 = 1 }$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then the map, say $D$, which assigns each $v \in \mathbb{R}^n \setminus {0}$ to its direction to its direction $D(v) = \frac{v}{||v||} \in S^{n-1}$ is a &lt;a href=&quot;http://en.wikipedia.org/wiki/Deformation_retraction&quot; rel=&quot;nofollow&quot;&gt;deformation retraction&lt;/a&gt;.  It follows that $\mathbb{R}^n \setminus {0}$ has the same homotopy type as the sphere $S^{n-1}$.  In particular, $\mathbb{R}^n \setminus {0}$ does not have the same homotopy as $\mathbb{R}^n$ -- i.e., it is not contractible.  This can be taken as a precision of the idea that there is no natural way to extend $D$ so as to be defined at the zero vector.  (For that matter, there is evidently no &lt;em&gt;continuous&lt;/em&gt; extension of $D$ to $\mathbb{R}^n$, since $D$ is surjective on any deleted neighborhood of $0$.)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note also that any algebraic geometer will inevitably be reminded of projective space, in which the zero vector must be excluded for similar reasons.  In this context, (real) projective $n-1$-space is obtained by failing to distinguish between the directions of $v$ and $-v$.  Topologically, this amounts to taking a quotient of $S^{n-1}$ by identifying antipodal points.  &lt;/p&gt;&#xA;" OwnerUserId="299" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2010-08-31T22:32:49.823" LastActivityDate="2010-08-31T22:32:49.823" />
  <row Id="3746" PostTypeId="2" ParentId="3742" CreationDate="2010-08-31T22:04:28.907" Score="2" ViewCount="" Body="&lt;p&gt;The (ir)rationality of the Euler-Mascheroni constant has always intrigued me, as it's one of those constants that is &quot;obviously&quot; irrational and yet nobody has much of a clue on how to prove that. &lt;/p&gt;&#xA;" OwnerUserId="592" LastActivityDate="2010-08-31T22:04:28.907" CommunityOwnedDate="2010-08-31T22:04:28.907" />
  <row Id="3747" PostTypeId="2" ParentId="3734" CreationDate="2010-08-31T22:28:43.110" Score="2" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;PROOF&lt;/strong&gt; $\rm\;$ It's simply a typical unit group descent. Let $\rm\: U \:$ be the group generated by the units $\rm\: (\pm t,\: 1)\:$.&lt;br&gt;&#xA; If unit $\rm\; (a,\;b) \not\in U \;$ then $\rm\;  (a,\;b)(\pm t,\; 1) = (\cdots,\;a\pm bt) \;$ is a &quot;smaller&quot; unit $\rm\:\not\in U\;$  since one of $\rm\: a\pm bt \;$ has smaller degree than $\rm\: b\:,\:$ as is easily verified using $\rm\; (a-bt)(a+bt) = 1-b^2 \:. \:\;$ &lt;strong&gt;QED&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-09-01T13:57:33.720" LastActivityDate="2010-09-01T13:57:33.720" />
  <row Id="3748" PostTypeId="2" ParentId="3742" CreationDate="2010-08-31T22:35:55.217" Score="1" ViewCount="" Body="&lt;p&gt;Not that practical, but finding the smallest $\varepsilon$ in the statement &quot;Matrix multiplication/inversion takes $O(n^{2+\varepsilon})$ flops&quot; is still a nice open problem.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-08-31T22:35:55.217" CommunityOwnedDate="2010-08-31T22:35:55.217" />
  <row Id="3749" PostTypeId="1" AcceptedAnswerId="3755" CreationDate="2010-08-31T22:47:52.857" Score="7" ViewCount="567" Body="&lt;p&gt;When I first took linear algebra, we never learned about dual spaces. Today in lecture we discussed them and I understand what they are and everything, but I don't really understand why we want to study them within linear algebra.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was wondering if anyone knew a nice intuitive motivation for the study of dual spaces and whether or not they &quot;show up&quot; as often as other concepts in linear algebra? Is their usefulness something that just becomes more apparent as you learn more math and see them arise in different settings?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: I understand that dual spaces show up in functional analysis and multilinear algebra, but I still don't really understand the intuition/motivation behind their definition in the standard topics covered in a linear algebra course. (Hopefully, this clarifies my question)&lt;/p&gt;&#xA;" OwnerUserId="249" LastEditorUserId="249" LastEditorDisplayName="" LastEditDate="2010-09-01T00:26:00.073" LastActivityDate="2010-09-06T22:13:07.150" Title="Why do we care about dual spaces?" Tags="&lt;soft-question&gt;&lt;linear-algebra&gt;&lt;intuition&gt;" AnswerCount="5" CommentCount="2" FavoriteCount="8" />
  <row Id="3750" PostTypeId="1" AcceptedAnswerId="3788" CreationDate="2010-08-31T23:05:07.650" Score="12" ViewCount="973" Body="&lt;p&gt;This problem first came to me in high school, and a couple times since, and I even assigned it for extra credit in one of my calculus classes after I became a teacher.  So I know the solution.  What I am looking for is other WAYS to obtain the solution.  I've been told there exists a solution using only arithmetic, but have never figured it out.  Other solutions using ordinary calculus, trigonometry, algebra of conic sections, and so on are also possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is usually stated in the form of a letter from an Aunt and Uncle:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Dear niece/nephew, How are things&#xA;  going for you and your folks?  We hear&#xA;  you are doing quite well it school. &#xA;  Keep it up!  Given this success, we&#xA;  were hoping you could help us figure&#xA;  out a little dilemma.  As you know,&#xA;  our home is heated by fuel oil, and we&#xA;  have a big tank buried in the side&#xA;  yard.  The tank is a cylinder, 20 feet&#xA;  long and 10 feet in diameter, lying on&#xA;  its side five feet deep, with a narrow&#xA;  tube coming to a fill cap at ground&#xA;  level.  Your uncle has a 15 foot&#xA;  length of old pipe that we'd like to&#xA;  utilize as a dip stick in order to&#xA;  know when we are getting close to&#xA;  needing a fill-up.  We know that 0&#xA;  feet is empty, 5 feet is half full,&#xA;  and 10 feet is completely full. &#xA;  Trouble is, we don't know how to mark&#xA;  any other points. We are pretty sure&#xA;  they will not be uniformly spaced. &#xA;  What we really want is to know, within&#xA;  the nearest 0.01 foot, where to mark&#xA;  the dip stick for every multiple of&#xA;  10% from 0% to 100%.  Can you figure&#xA;  this out for us?  Of course, we will&#xA;  want to see details of your solution&#xA;  and check it ourselves, and it would&#xA;  especially help if you could draw us a&#xA;  scale model of the dip stick. Love,&#xA;  Auntie Flo and Uncle Jim&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That last sentence shows the teacher influence on the problem.&#xA;So, my challenge to this community is not to find any old solution, but to find the solution at the lowest possible grade level, so to speak.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;UPDATE:  To those who are focusing in on the .01 feet accuracy, I apologize.  The intent was merely to state, it is acceptable to estimate.  If the exact answer is sqrt(2)*pi/2 or some other silly thing, go ahead and just write 2.22 feet, for example.&lt;/p&gt;&#xA;" OwnerUserId="575" LastEditorUserId="575" LastEditorDisplayName="" LastEditDate="2010-09-01T23:40:29.073" LastActivityDate="2010-09-12T23:05:28.853" Title="Aunt and Uncle's fuel oil tank dip stick problem" Tags="&lt;calculus&gt;&lt;trigonometry&gt;&lt;arithmetic&gt;&lt;conic-sections&gt;" AnswerCount="4" CommentCount="5" FavoriteCount="2" />
  <row Id="3751" PostTypeId="2" ParentId="3749" CreationDate="2010-08-31T23:13:28.700" Score="3" ViewCount="" Body="&lt;p&gt;Absolutely yes to the last question.  There are already many applications listed in the &lt;a href=&quot;http://en.wikipedia.org/wiki/Dual_space&quot; rel=&quot;nofollow&quot;&gt;Wikipedia article&lt;/a&gt;.  I will give a few more.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First of all, together with tensor products, dual spaces can be used to talk about linear transformations; the vector space $\text{Hom}(V, W)$ of linear transformations from a vector space $V$ to a vector space $W$ is canonically isomorphic to the tensor product $V^{\ast} \otimes W$.  I assume you care about linear transformations, so you should care about tensor products and dual spaces as well.  (A really important feature of this decomposition is that it is true in considerable generality; for example, it holds for representations of groups and is a natural way to prove the orthogonality relations for characters.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dual spaces also appear in geometry as the natural setting for certain objects.  For example, a differentiable function $f : M \to \mathbb{R}$ where $M$ is a smooth manifold is an object that produces, for any point $p \in M$ and tangent vector $v \in T_p M$, a number, the directional derivative, in a linear way.  In other words, a differentiable function defines an element of the dual to the tangent space (the cotangent space) at each point of the manifold.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-08-31T23:13:28.700" CommentCount="3" />
  <row Id="3752" PostTypeId="2" ParentId="3749" CreationDate="2010-08-31T23:13:32.007" Score="0" ViewCount="" Body="&lt;p&gt;Dual spaces are important because they show up all over the place, and you'll see them arise in all sort of settings as you learn more math.&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2010-08-31T23:13:32.007" />
  <row Id="3753" PostTypeId="2" ParentId="3742" CreationDate="2010-08-31T23:51:18.870" Score="0" ViewCount="" Body="&lt;p&gt;Does favorite allow me two - both about convex 3-dimensional polyhedra?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Shephard's Conjecture)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it possible to cut along the edges of some spanning tree of any convex 3-dimensional polyhedron and unfold the cut open polyhedron so that the faces do not overlap (and one gets a simple polygon where the uncut edges indicate the faces of the original polyhedron in the resulting &quot;net&quot; in the plane)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Net_%28polyhedron%29&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Net_%28polyhedron%29&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Barnette's conjecture)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does every 3-valent convex 3-dimensional polyhedron all of whose faces have an even number of sides admit a hamiltonian circuit?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Barnette%27s_conjecture&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Barnette%27s_conjecture&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1369" LastActivityDate="2010-08-31T23:51:18.870" CommunityOwnedDate="2010-08-31T23:51:18.870" CommentCount="1" />
  <row Id="3754" PostTypeId="2" ParentId="3724" CreationDate="2010-08-31T23:54:55.537" Score="0" ViewCount="" Body="&lt;p&gt;Pick any four coins and weigh two against two.  If they are imbalanced then at least one of the two coins on the heavier side is heavy and they can now be weighed against each other to identify one or both as heavy.  Otherwise replace the two coins on either side of the balance with two other coins.  If they are now imbalanced then both coins on the heavier side are heavy, and if they are still balanced then all six coins are heavy.&lt;/p&gt;&#xA;" OwnerUserId="1284" LastActivityDate="2010-08-31T23:54:55.537" CommentCount="5" />
  <row Id="3755" PostTypeId="2" ParentId="3749" CreationDate="2010-09-01T00:44:27.447" Score="8" ViewCount="" Body="&lt;p&gt;Let $V$ be a vector space (over any field, but we can take it to be $\mathbb R$ if you like,&#xA;and for concreteness I will take the field to be $\mathbb R$ from now on;&#xA;everything is just as interesting in that case).  Certainly one of the interesting concepts &#xA;in linear algebra is that of a &lt;em&gt;hyperplane&lt;/em&gt; in $V$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, if $V = \mathbb R^n$, then a hyperplane is just the solution set to an equation&#xA;of the form&#xA;$$a_1 x_1 + \cdots + a_n x_n = b,$$&#xA;for some $a_i$ not all zero and some $b$.&#xA;Recall that solving such equations (or simultaneous sets of such equations) is one&#xA;of the basic motivations for developing linear algebra.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now remember that when a vector space is not given to you as $\mathbb R^n$,&#xA;it doesn't normally have a canonical basis, so we don't have a canonical way&#xA;to write its elements down via coordinates, and so we can't describe hyperplanes&#xA;by explicit equations like above.  (Or better, we can, but only after choosing&#xA;coordinates, and this is not canonical.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can we canonically describe hyperplanes in $V$?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For this we need a conceptual interpretation of the above equation.  And here linear&#xA;functionals come to the rescue.  More precisely, the map&#xA;$$\ell:(x_1,\ldots,x_n) \mapsto a_1 x_1 + \cdots a_n x_n$$&#xA;is a linear functional on $\mathbb R^n$, and so the above equation for the&#xA;hyperplane can be written as &#xA;$$\ell(v) = b.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More generally, if $V$ is any vector space, and $\ell: V \to \mathbb R$ is any&#xA;non-zero linear functional (i.e. non-zero element of the dual space), then&#xA;for any $b \in \mathbb R,$ the set&#xA;$$\{v \, | \, \ell(v) = b\}$$&#xA;is a hyperplane in $V$, and all hyperplanes in $V$ arise this way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So this gives a reasonable justification for introducing the elements of the dual&#xA;space to $V$; they generalize the notion of linear equation in several variables&#xA;from the case of $\mathbb R^n$ to the case of an arbitrary vector space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now you might ask: why do we make them a vector space themselves?  why do we want&#xA;to add them to one another, or multiply them by scalars?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are lots of reasons for this; here is one: Remember how important it is,&#xA;when you solve systems of linear equations, to add equations together, or&#xA;to multiply them by scalars (here I am referring to all the steps you typically&#xA;make when performing Gaussian elimination on a collection of simultaneous linear&#xA;equations)?  Well, under the dictionary above between linear equations&#xA;and linear functionals, these processes correspond precisely to adding together&#xA;linear functionals, or multiplying them by scalars.  If you ponder this for a bit,&#xA;you can hopefully convince yourself that making the set of linear&#xA;functionals a vector space is a pretty natural thing to do.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Summary: just as concrete vectors $(x_1,\ldots,x_n) \in \mathbb R^n$ are naturally&#xA;generalized to elements of vector spaces, concrete linear expressions&#xA;$a_1 x_1 + \ldots + a_n x_n$ in $x_1,\ldots, x_n$ are naturally generalized to linear functionals.&lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2010-09-01T00:44:27.447" CommentCount="5" />
  <row Id="3756" PostTypeId="2" ParentId="3749" CreationDate="2010-09-01T01:03:48.880" Score="3" ViewCount="" Body="&lt;p&gt;There are some very beautiful and easily accessible applications of duality, adjointness, etc. in Rota's modern reformulation of the Umbral Calculus. You'll quickly gain an appreciation for the power of such duality once you see how easily this approach unifies hundreds of diverse special-function identities, and makes their derivation essentially trivial. For a nice introduction see Steven Roman's book &quot;The Umbral Calculus&quot;.&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-09-01T01:03:48.880" CommentCount="2" />
  <row Id="3757" PostTypeId="2" ParentId="3750" CreationDate="2010-09-01T02:31:47.597" Score="6" ViewCount="" Body="&lt;p&gt;It is sufficient to consider the circular cross-section of the tank and volumes below 50% (marks for those above 50% are the reflection image of those below 50% across the 50% mark).  Consider the radius of the tank to be 1 unit.  For some amount of oil in the tank, consider the central angle formed by the points on the circular cross-section at the top of the oil and the center of the circle--call this $\alpha$, measured in radians.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/5pI7u.png&quot; alt=&quot;tank cross-section&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The area of the cross-section of the oil is the area of the sector determined by $\alpha$ ($\frac{\alpha}{2\pi}\pi r^2=\frac{\alpha}{2}$) minus the area of the triangle that is part of the sector but not part of the cross-section of oil ($\frac{1}{2}ab\sin C=\frac{\sin\alpha}{2}$), $\frac{1}{2}(\alpha-\sin\alpha)$.  The portion of the circular cross-section (and hence the portion of the volume) corresponding to this angle is $\frac{\alpha-\sin\alpha}{2\pi}$.  Setting this expression equal to 0.1, 0.2, 0.3, and 0.4 and solving for $\alpha$ will give the values of $\alpha$ corresponding to 10%, 20%, 30%, and 40% full--solving here is done numerically/graphically, as there is no algebraic method to solve these equations.  For each value of $\alpha$, the distance from the center of the circle to the oil level is $\cos\frac{\alpha}{2}$, so the depth of the oil is $1-\cos\frac{\alpha}{2}$.  (Note that these are for a radius of 1 unit and need to be rescaled for the original problem's specific numbers.)&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-09-01T02:31:47.597" CommentCount="6" />
  <row Id="3758" PostTypeId="1" AcceptedAnswerId="3773" CreationDate="2010-09-01T02:55:46.223" Score="1" ViewCount="519" Body="&lt;p&gt;When implicitly finding the derivative of:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;xy^3 - xy^3*sin(x) = 1&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;How do you find the implicit derivative of:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;xy^3*sin(x)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Is it using a &lt;em&gt;triple&lt;/em&gt; product rule of sorts?&lt;/p&gt;&#xA;" OwnerUserId="1602" LastActivityDate="2010-09-01T16:33:07.393" Title="Implicit Diferentiation... the triple product rule?" Tags="&lt;calculus&gt;" AnswerCount="5" CommentCount="6" FavoriteCount="1" />
  <row Id="3759" PostTypeId="2" ParentId="3734" CreationDate="2010-09-01T03:10:32.683" Score="5" ViewCount="" Body="&lt;p&gt;Here is another argument:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Write $x = t + \sqrt{t^2 - 1}$ and $y = t - \sqrt{t^2 - 1}$.    Then $x y = 1$, i.e. $y = x^{-1}$,&#xA;and $t = (x + y)/2$ , $\sqrt{t^2 - 1} = (x - y)/2$.&lt;br&gt;&#xA;Thus we see that&#xA;$\mathbb Z[t][\sqrt{t^2 - 1}] \subset \mathbb Z[1/2][x,x^{-1}].$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now it is easy to check that if $A$ is an integral domain, then the only units&#xA;in $A[x,x^{-1}]$ are of the form $a x^n,$ where $a \in A$ is a unit and $n$ is an integer.&#xA;Since the units in $\mathbb Z[1/2]$ are precisely the elements $\pm 2^m$ (for some $m$)&#xA;we see that the units in&#xA;$\mathbb Z[1/2][x,x^{-1}]$ are of the form $\pm 2^m x^n = \pm 2^m (t + \sqrt{t^2 - 1})^n.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is easy to check that if such an element and its inverse both actually lie in &#xA;$\mathbb Z[t][\sqrt{t^2 - 1}],$ then necessarily $m = 0$ (e.g. for norm reasons,&#xA;or just looking explicitly at their denominators), and so we get the answer that&#xA;the units are precisely the elements of the form $\pm (t + \sqrt{t^2 -1 })^n$.&#xA;(Here $n$ runs over $\mathbb Z$; this is the same set as $\pm (t \pm \sqrt{t^2 - 1})^n$,&#xA;where $n$ is now non-negative.)&lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2010-09-01T03:10:32.683" CommentCount="2" />
  <row Id="3760" PostTypeId="2" ParentId="3758" CreationDate="2010-09-01T03:13:41.847" Score="3" ViewCount="" Body="&lt;p&gt;It is possible to derive a 'triple product rule.' In fact, you can generalize it to an arbitrary number of products. You may ask why these rules aren't presented in these general forms. One reason for this is that writing the general formula out is a more complicated to memorize and more intimidating towards students just learning calculus. The other reason, is that having the product rule is enough to give you the general rule by a simple induction argument. Also, for practical purposes, problems can be solved with only the usual product rule being applied in multiple steps.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your case we can apply the product rule twice:&#xA;$\frac{d}{dx}xsin(x)y^{3}=y^{3}\frac{d}{dx}(xsin(x))+3y^2\frac{dy}{dx}(xsin(x))=y^{3}(sin(x)+xcos(x))+3y^{2}\frac{dy}{dx}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can also derive a 'triple product rule'&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{d}{dx}(f\cdot g\cdot h)=\frac{df}{dx}\cdot(h\cdot g)+\frac{dg}{dx}\cdot (f\cdot h)+\frac{dh}{dx}\cdot (f\cdot g)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But you see that you still have to apply the chain rule to that to get it to fit how you need to use it in this situation and so it would be a real mess to memorize all the different special cases. This is why we just provide the product rule with 2 functions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does my rambling make sense?&lt;/p&gt;&#xA;" OwnerUserId="249" LastEditorUserId="249" LastEditorDisplayName="" LastEditDate="2010-09-01T03:20:00.040" LastActivityDate="2010-09-01T03:20:00.040" CommentCount="2" />
  <row Id="3761" PostTypeId="2" ParentId="3399" CreationDate="2010-09-01T03:23:28.117" Score="8" ViewCount="" Body="&lt;p&gt;Lets say that you accept the definition of a convex set in higher dimensions, like a sphere in $\mathbb{R}^3$. The question I seek to provide insight into is why convex functions in one variable are defined as opening up instead of down, since this seems like an arbitrary definition. This is because, depending on how you look at the graph, you could naively view the function as bending outwards (like a convex set) or inwards (concave). However, there is a nice connection between these two things using metric spaces that I think can provide some meaning to the way it is defined.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most of the metrics that you are familiar with have open balls that are convex, such as the standard metric.  But some are actually non convex. A good example of this is $ d(x,y) = \sum \sqrt{|x_i - y_i |} $. (note that $\sqrt{x}$ is not a convex function)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is an interesting condition:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given a metric $d$. If for all $y,z\in E$ and $0\leq t\leq 1$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$d \left(x, \  t  y \; \, + \;  (1-t) z \right) \quad \leq \quad t  d(x,y) \; + \; (1-t)  d(x,z) $&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;then the open balls formed by $d$ are convex. [1] In other words, if you fix $x$ and $d(y):\mathbb{R}^n\rightarrow \mathbb{R}$ is a convex function, then the open balls are convex sets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Usually $d(x,y) = \sum f \, (x_i,y_i)$, for some $f:\mathbb{R}^2\rightarrow\mathbb{R}$.  If we fix $x$ and $f:\mathbb{R}\rightarrow \mathbb{R}$ fits the definition of a convex function, then $d$ will also be convex, and the condition will be satisfied, giving us convex balls. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So convex functions (if they can form a metric) will give you convex open balls.  A nice connection that makes the definition make more sense. Other conditions that guarantee convex open balls are discussed in the paper I reference. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;[1] Norfolk, T. (1991). When does a metric generate convex balls? www.math.uakron.edu/~norfolk/convex.ps&lt;/p&gt;&#xA;" OwnerUserId="565" LastEditorUserId="1457" LastEditorDisplayName="" LastEditDate="2010-12-12T04:00:23.837" LastActivityDate="2010-12-12T04:00:23.837" CommunityOwnedDate="2010-09-01T03:23:28.117" CommentCount="7" />
  <row Id="3762" PostTypeId="2" ParentId="1748" CreationDate="2010-09-01T03:48:41.877" Score="0" ViewCount="" Body="&lt;p&gt;Bix - Conics and Cubics is probably the most accessible. I think a high school student who did the BC calculus syllabus could read this and get a lot out of it. But I'd agree with the others about Cox, Little, O'Shea - Ideals Varieties and Algorithms being the 'best' undergraduate text.&lt;/p&gt;&#xA;" OwnerUserId="249" LastActivityDate="2010-09-01T03:48:41.877" CommunityOwnedDate="2010-09-01T03:48:41.877" />
  <row Id="3763" PostTypeId="1" AcceptedAnswerId="3775" CreationDate="2010-09-01T04:15:41.187" Score="1" ViewCount="528" Body="&lt;p&gt;How does one sum the given series: $$ \frac{1}{2n+1} + \frac{1}{2} \cdot \frac{1}{2n+3} + \frac{1 \cdot 3}{2 \cdot 4} \frac{1}{2n+5} + \frac{ 1 \cdot 3 \cdot 5}{2 \cdot 4 \cdot 6} \frac{1}{2n+7} + \cdots \ \text{ad inf}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given, such a series, how does one go about solving it. Getting an &lt;em&gt;Integral Representation&lt;/em&gt; seems tough for me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I thought of going along these lines, &lt;a href=&quot;http://math.stackexchange.com/questions/3502/summing-the-series-1k-frac2k2k1-a2k1&quot;&gt;http://math.stackexchange.com/questions/3502/summing-the-series-1k-frac2k2k1-a2k1&lt;/a&gt; but couldn't succeed.&lt;/p&gt;&#xA;" LastActivityDate="2010-09-02T01:18:09.083" Title="Summing the series $ \frac{1}{2n+1} + \frac{1}{2} \cdot \frac{1}{2n+3} + \cdots \ \text{ad inf}$" Tags="&lt;calculus&gt;&lt;real-analysis&gt;&lt;analysis&gt;&lt;sequences-and-series&gt;" AnswerCount="3" />
  <row Id="3764" PostTypeId="2" ParentId="3763" CreationDate="2010-09-01T04:28:45.190" Score="1" ViewCount="" Body="&lt;p&gt;FWIW, &lt;em&gt;Mathematica&lt;/em&gt; is able to sum this:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;In[1]:= Sum[1/(2 n + 2 k + 1) (2 k - 1)!!/(2 k)!!, {k, 0, \[Infinity]}]&#xA;&#xA;                        3 + 2 n&#xA;        Sqrt[Pi] Gamma[-------]&#xA;                           2&#xA;Out[1]= -----------------------&#xA;        (1 + 2 n) Gamma[1 + n]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="274" LastActivityDate="2010-09-01T04:28:45.190" />
  <row Id="3765" PostTypeId="1" CreationDate="2010-09-01T04:37:44.937" Score="1" ViewCount="93" Body="&lt;p&gt;Can someone please point me in the direction of any theory on graphs where the edge weights are not scalars but represent some relation between the nodes that is a simple function of a single variable (simple, say piecewise linear). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, I'm interested in various basic graph properties and also thinking of the graph as representing a network. So, for example, if the graph represented a communication network over time then the edge weights would be a function represented connectivity as a function of time, how do you find a valid path between nodes? I'm looking for help both on specific algorithms but also general theory if it exists?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm aware of time-extended networks where you explicitly expand out the dependence on the variable but, from what I've read, this incomplete and of limited applicability. &lt;/p&gt;&#xA;" OwnerUserId="1628" LastEditorUserId="1628" LastEditorDisplayName="" LastEditDate="2010-09-01T17:07:03.877" LastActivityDate="2010-09-01T17:07:03.877" Title="Generalized graph - edges weights are functions, not scalars" Tags="&lt;graph-theory&gt;" AnswerCount="1" CommentCount="9" />
  <row Id="3766" PostTypeId="2" ParentId="3763" CreationDate="2010-09-01T04:57:00.837" Score="1" ViewCount="" Body="&lt;p&gt;Consider the function&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f(a) = \sum_{k=0}^{\infty} \frac{1\cdot3\cdots(2k-1)}{2\cdot4\cdots(2k)}\cdot a^{2n+2k}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$|a| \leq 1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you want is $$\int_{0}^{1} f(a) da$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The $k^{th}$ coefficient can be written as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{2}{\pi}\int_{0}^{\frac{\pi}{2}}{\sin^{2k}x}dx$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(again Wallis's like product as in the other question).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f(a) = \frac{2a^{2n}}{\pi} \int_{0}^{\frac{\pi}{2}} \sum_{k=0}^{\infty} {(a\sin x)^{2k}} dx$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ = \frac{2a^{2n}}{\pi} \int_{0}^{\frac{\pi}{2}} {\frac{1}{1-(a\sin x)^2}}dx$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now $$\frac{1}{1-(a\sin x)^2} = \frac{1}{\cos^2 x + (\sqrt{1-a^2}\sin x)^2} = \frac{\sec^2 x}{1+(\sqrt{1-a^2}\tan x)^2}$$ which is the derivative of&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\arctan(\sqrt{1-a^2}\tan x)}{\sqrt{1-a^2}}$$ whose integral between $0$ and $\frac{\pi}{2}$ is $$\frac{\pi}{2\sqrt{1-a^2}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus we get $$f(a) = \frac{a^{2n}}{\sqrt{1-a^2}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(So in fact, we have shown the identity which Qiaochu uses in his answer).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence the required sum is $$\int_{0}^{1} \frac{a^{2n}}{\sqrt{1-a^2}}da$$ which can easily be found by the substitution $a = \sin x$ (like in Qiaochu's answer) and which gives us the result to be&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_{0}^{\frac{\pi}{2}} {\sin^{2n}x}dx$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now this integral for positive integer $n$ can easily be seen to be what Mariano got. I am unsure of whether that formula holds for an arbitrary real number $n$.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-09-02T01:18:09.083" LastActivityDate="2010-09-02T01:18:09.083" CommentCount="3" />
  <row Id="3767" PostTypeId="2" ParentId="3623" CreationDate="2010-09-01T05:01:02.167" Score="0" ViewCount="" Body="&lt;p&gt;If I were you I would start by looking for similar threads..  ;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://math.stackexchange.com/questions/3335/could-you-recommend-some-classic-textbooks-on-ordinary-partial-differential-equat&quot;&gt;Could you recommend some classic textbooks on ordinary/partial differential equation?&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://math.stackexchange.com/questions/2827/good-1st-pde-book-for-self-study&quot;&gt;Good 1st PDE book for self study&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1154" LastActivityDate="2010-09-01T05:01:02.167" CommunityOwnedDate="2010-09-01T05:01:02.167" />
  <row Id="3768" PostTypeId="1" AcceptedAnswerId="3770" CreationDate="2010-09-01T05:06:28.477" Score="9" ViewCount="440" Body="&lt;p&gt;From some reading, I've noticed that $\sup(\emptyset)=\min(S)$, but $\inf(\emptyset)=\max(S)$, given that $\min(S)$ and $\max(S)$ exist, where $S$ is the universe in which one is working. Is there some inherent reasoning/proof as to why this is? It seems strange to me that an upper bound of a set would be smaller than a lower bound of the same set.&lt;/p&gt;&#xA;" OwnerUserId="1583" LastActivityDate="2011-02-01T15:23:03.620" Title="Upper and Lower Bounds of $\emptyset$" Tags="&lt;analysis&gt;" AnswerCount="3" FavoriteCount="1" />
  <row Id="3769" PostTypeId="2" ParentId="3758" CreationDate="2010-09-01T05:09:35.083" Score="1" ViewCount="" Body="&lt;p&gt;You get a formula for the derivative of a product of $n$ factors from the formula for the product of $2$ factors by doing induction. Intuitively, you do it the same way as you go from $2$ to $3$: $(fgh)' = ((fg)h)' = (fg)'h + (fg)h' = (f'g + fg')h + (fg)h' = f'gh + fg'h + fgh'$. The pattern should now be clear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming you already know that $(f_1\cdots f_k)' = f_1'f_2\cdots f_k + f_1f_2'f_3\cdots f_k + \cdots + f_1\cdots f_{k-1}f_k'$, then inductively you have &#xA;$$(f_1\cdots f_kf_{k+1})' = (f_1\cdots f_k)'f_{k+1} + (f_1\cdots f_k)f_{k+1}'$$&#xA;and the formula follows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So for $xy^3\sin(x)$, the derivative will be &#xA;$$(x)'y^3\sin(x) + x(y^3)'\sin(x) + xy^3(\sin x)'$$&#xA;with the middle summand requiring implicit differentiation.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-09-01T05:09:35.083" />
  <row Id="3770" PostTypeId="2" ParentId="3768" CreationDate="2010-09-01T05:16:53.807" Score="8" ViewCount="" Body="&lt;p&gt;The supremum is defined to be the least upper bound. An upper bound for a set $K$ is defined to be an element $b$ such that for all $k\in K$, $k\leq b$. If $K$ is empty, then the condition &quot;for all $k\in K$, $k\leq b$&quot; is true by vacuity (you have an implication of the form &quot;if $k$ is an element of the empty set, then $\lt$something happens$\gt$&quot;, and the antecedent is always false so the implication is always true).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, every element of $S$ is an upper bound for $\emptyset$; since $\sup(\emptyset)$ is the least upper bound, that means that $\sup(\emptyset)=\min($upper bounds of $\emptyset) = \min(S)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, the infimum is the greatest lower bound, and every element of $S$ is a lower bound for $\emptyset$, so $\inf(\emptyset) = \max($lower bounds of $\emptyset) = \max(S)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yes, it is somewhat counterintuitive that you have $\sup(\emptyset)\leq\inf(\emptyset)$, but in fact the empty set is the only one for which you can have $\sup(\emptyset)\lt \inf(\emptyset)$. The empty set often causes counterintuitive results.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-09-01T05:16:53.807" CommentCount="1" />
  <row Id="3771" PostTypeId="2" ParentId="3742" CreationDate="2010-09-01T05:17:01.043" Score="1" ViewCount="" Body="&lt;p&gt;I am quite fond of the abc conjecture: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Define $q(n)$ to be the product of all distinct primes of n. The conjecture states, that for any $\epsilon &gt; 0$, there exist only finitely many triples of coprime positive integers $(a, b, c)$, satisfying $a + b = c$ for which $c &gt; q(a b c)^{1 + \epsilon}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This conjecture has an &lt;a href=&quot;http://en.wikipedia.org/wiki/Abc_conjecture#Some_consequences&quot; rel=&quot;nofollow&quot;&gt;impressive list of consequences&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="960" LastActivityDate="2010-09-01T05:17:01.043" CommunityOwnedDate="2010-09-01T05:17:01.043" />
  <row Id="3772" PostTypeId="2" ParentId="3768" CreationDate="2010-09-01T05:23:35.487" Score="12" ViewCount="" Body="&lt;p&gt;A quick counterpoint to Arturo's answer: yes, the fact that $\operatorname{sup}(\emptyset) = -\infty &amp;lt; \infty = \operatorname{inf}(\emptyset)$ seems strange at first.  But it is designed, among other things, to preserve an intuitive and useful property of infima and suprema:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $S$ and $T$ be subsets of $\mathbb{R}$ with $S \subset T$.  Then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\operatorname{inf}(T) \leq \operatorname{inf}(S)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\operatorname{sup}(T) \geq \operatorname{sup}(S)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Indeed, by adding elements to a set, its infimum can only get smaller and its supremum can only get larger.  Now if you take $S = \emptyset$ and let $T$ range through even all one-element subsets of $\mathbb{R}$, you see that in order for these inequalities to hold we &lt;em&gt;must&lt;/em&gt; define $\operatorname{inf}(\emptyset)$ and $\operatorname{sup}(\emptyset)$ as we did.&lt;/p&gt;&#xA;" OwnerUserId="299" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2010-09-01T05:29:56.220" LastActivityDate="2010-09-01T05:29:56.220" CommentCount="0" />
  <row Id="3773" PostTypeId="2" ParentId="3758" CreationDate="2010-09-01T05:33:48.730" Score="5" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\;$ &lt;a href=&quot;http://en.wikipedia.org/wiki/Logarithmic_differentiation&quot; rel=&quot;nofollow&quot;&gt;&lt;strong&gt;logarithmic differentiation&lt;/strong&gt;&lt;/a&gt; makes the n-ary generalization obvious:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\rm (abc)'= abc \;  log(abc)' = abc \;(log\; a + log\; b + log\; c)' = abc \; \bigg(\frac{a'}{a} + \frac{b'}{b} + \frac{c'}{c}\bigg) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously the same proof works for arbitrary length products yielding&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\rm (abc\cdots f)' = \: abc\cdots f\;\bigg(\frac{a'}{a} + \frac{b'}{b} + \frac{c'}{c} +\:\cdots\:+ \frac{f'}{f}\bigg) $$&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-09-01T06:09:19.707" LastActivityDate="2010-09-01T06:09:19.707" CommentCount="0" />
  <row Id="3775" PostTypeId="2" ParentId="3763" CreationDate="2010-09-01T06:09:07.147" Score="3" ViewCount="" Body="&lt;p&gt;Observe that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle \frac{1}{4^n} {2n \choose n} = \frac{(2n-1)(2n-3)(2n-5)...}{2n(2n-2)(2n-4)...}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and recall that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle \frac{1}{ \sqrt{1 - x^2} } = \sum_{k \ge 0} \frac{1}{4^k} {2k \choose k} x^{2k}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It follows that the desired quantity is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle \int_0^1 \frac{x^{2n}}{\sqrt{1 - x^2}} dx.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But letting $x = \sin \theta$ this is just&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle \int_0^{ \frac{\pi}{2} } \sin^{2n} \theta d \theta = \frac{\pi}{2} \frac{1}{4^n} {2n \choose n}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is equivalent to Mariano's closed form via the identities $\Gamma(x+1) = x \Gamma(x)$ and $\Gamma \left( \frac{1}{2} \right) = \sqrt{\pi}$.  I should mention that the integral at the end of Moron's answer is quite doable; write it in terms of a cosine and use the substitution $\cos \theta = \frac{1 - t^2}{1 + t^2}$ where $t = \tan \frac{\theta}{2}$ to reduce the problem to the integral of a rational function, and then one can use one of several related methods (partial fractions, contour integration).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Remark:&lt;/strong&gt;  The last integral identity above happens to be one of my favorite identities.  I describe a representation-theoretic and combinatorial proof of it &lt;a href=&quot;http://qchu.wordpress.com/2010/03/07/walks-on-graphs-and-tensor-products/&quot; rel=&quot;nofollow&quot;&gt;in this blog post&lt;/a&gt;.  Another approach implicitly occurs &lt;a href=&quot;http://qchu.wordpress.com/2009/10/07/extracting-the-diagonal/&quot; rel=&quot;nofollow&quot;&gt;in this blog post&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-09-01T06:21:16.963" LastActivityDate="2010-09-01T06:21:16.963" CommentCount="2" />
  <row Id="3776" PostTypeId="1" AcceptedAnswerId="3780" CreationDate="2010-09-01T07:22:25.440" Score="3" ViewCount="196" Body="&lt;p&gt;I was musing about a particular limit,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$L = \prod\limits_{n &gt; 0} \bigl(1 - 2^{-n}\bigr)$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;we may bound 0.288&amp;nbsp;&amp;lt;&amp;nbsp;L&amp;nbsp;&amp;lt;&amp;nbsp;0.308, which we may show by taking the logarithm:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ln(L) = \ln \bigl( \frac{315}{1024}\bigr) + \sum\limits_{n &gt; 4} \ln\bigl(1 - 2^{-n}\bigr) &gt; \ln\bigl(\frac{315}{1024}\bigr) - \sum\limits_{n &gt; 4} 2^{-n} =\; \ln\bigl(\frac{315}{1024} \cdot \mathrm e^{-1/16}\bigr)$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was wondering if this type of infinite product (or the corresponding sum of logarithms) has a name, and whether there are techniques for obtaining a closed form expression for the limit.&lt;/p&gt;&#xA;" OwnerUserId="439" LastEditorUserId="439" LastEditorDisplayName="" LastEditDate="2010-09-01T19:11:27.843" LastActivityDate="2010-09-01T19:11:27.843" Title="Limit of a particular variety of infinite product/series" Tags="&lt;analysis&gt;&lt;sequences-and-series&gt;&lt;limit&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="3777" PostTypeId="1" AcceptedAnswerId="3802" CreationDate="2010-09-01T07:22:33.540" Score="24" ViewCount="966" Body="&lt;p&gt;If memory serves, ten years ago to the week (or so), I taught first semester freshman calculus for the first time.  As many calculus instructors do, I decided I should ask some extra credit questions to get students to think more deeply about the material.  The first one I asked was this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Recall that a function $f: \mathbb{R} \rightarrow \mathbb{R}$ is said to have a &lt;strong&gt;removable discontinuity&lt;/strong&gt; at a point $x_0 \in \mathbb{R}$ if $\lim_{x \rightarrow x_0} f(x)$ exists but not does not equal $f(x_0)$.  Does there exist a function $f$ which has a removable discontinuity at $x_0$ &lt;em&gt;for every&lt;/em&gt; $x_0 \in \mathbb{R}$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Commentary: if so, we could define a new function $\tilde{f}(x_0) = \lim_{x \rightarrow x_0} f(x)$ and it seems at least that $\tilde{f}$ has a fighting chance to be continuous on $\mathbb{R}$.  Thus we have successfully &quot;removed the discontinuities&quot; of $f$, but in so doing we have changed the value at every point!  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remark: Lest you think this is too silly to even seriously contemplate, consider the function $f: \mathbb{Q} \rightarrow \mathbb{Q}$ given by $f(0) = 1$ and for a nonzero &#xA;rational number $\frac{p}{q}$, $f(\frac{p}{q}) = \frac{1}{q}$.  It is easy to see that this function has limit $0$ at every (rational) point!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I mentioned this problem to my students.  A week later, the only person who asked me about it at all was my Teaching Assistant, who was an older undergraduate, not even a math major, I think.  (I hasten to add that this was not in any sense an honors calculus class, i.e., I was pretty clueless back then.)  Thinking about it a bit, I asked him if he knew about uncountable sets, and he said that he didn't.  At that point I realized that I didn't have a solution in mind that he would understand (so still less so for the freshman calculus students) and I advised him to forget all about it.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;So &lt;strong&gt;my actual question is&lt;/strong&gt;: can you solve this problem using only the concepts in a non-honors freshman calculus textbook?  (In particular, without using notions of un/countability?)&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;[&lt;b&gt;Addendum&lt;/b&gt;: Let me say explicitly that I would welcome an answer that proceeds directly in terms of the least upper bound axiom.  Most freshman calculus books &lt;em&gt;do&lt;/em&gt; include this, albeit somewhere hidden from view of the casual readers, i.e., actual freshman calculus students.]&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you can't figure out how to answer the question at all, I think the following related question helps.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Define a function $f: \mathbb{R} \rightarrow \mathbb{R}$ to be &lt;strong&gt;precontinuous&lt;/strong&gt; if the limit exists at every point.  For such a function, we can define $\tilde{f}$ as above.  Prove/disprove that, as suggested above, $\tilde{f}$ is indeed continuous.  [Then think about $f - \tilde{f}$.]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now that I think about it, there is an entire little area here that I don't know anything about, e.g.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) The set of discontinuities of an arbitrary function is known -- any $F_{\sigma}$ set inside $\mathbb{R}$ can serve.  What can we say about the set of discontinuities of a &quot;precontinuous function&quot;?  [Edit: from the link provided in Chandru1's answer, we see that it is countable.  What &lt;em&gt;else&lt;/em&gt; can we say?  Note that taking the above example and extending by $0$ to the irrationals, we see that the set of points of discontinuity of a precontinuous function can be dense.]&lt;/p&gt;&#xA;" OwnerUserId="299" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2010-09-01T08:54:48.180" LastActivityDate="2010-09-03T02:18:35.137" Title="Is there a function with a removable discontinuity at every point?" Tags="&lt;calculus&gt;&lt;real-analysis&gt;" AnswerCount="3" CommentCount="11" FavoriteCount="10" />
  <row Id="3778" PostTypeId="2" ParentId="3758" CreationDate="2010-09-01T07:34:03.823" Score="1" ViewCount="" Body="&lt;p&gt;Here is a simple way to compute the $n$-th derivative of a product of $k$ functions. To ease the notation, stick to the case $k=3$. First follow your instinct, and then you'll see that the justification presents no difficulty. Write (with all summation indices varying from 0 to $\infty$)&#xA;$$\sum_n\frac{(fgh)^{(n)}}{n!}X^n&#xA;=\sum_i\frac{f^{(i)}}{i!}X^i\ \sum_j\frac{g^{(j)}}{j!}X^j\ \sum_k\frac{h^{(k)}}{k!}X^k$$&#xA;$$=\sum_{i,j,k}\frac{f^{(i)}g^{(j)}h^{(k)}}{i!\ j!\ k!}X^{i+j+k}&#xA;=\sum_n\ X^n\sum_{i+j+k=n}\frac{f^{(i)}g^{(j)}h^{(k)}}{i!\ j!\ k!}\quad.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In categorical language: To each $\mathbb Q$-algebra $A$ is attached the $\mathbb Q$-algebra $A':=A[[X]]$ equipped with the derivation $d/dX$ (the letter $X$ being an indeterminate). Then $A\mapsto A'$ is a &lt;strong&gt;right adjoint&lt;/strong&gt; to the forgetful functor from differential $\mathbb Q$-algebras to $\mathbb Q$-algebras. &lt;/p&gt;&#xA;" OwnerUserId="660" LastEditorUserId="660" LastEditorDisplayName="" LastEditDate="2010-09-01T07:41:21.887" LastActivityDate="2010-09-01T07:41:21.887" CommentCount="1" />
  <row Id="3779" PostTypeId="2" ParentId="3777" CreationDate="2010-09-01T07:38:06.377" Score="5" ViewCount="" Body="&lt;p&gt;No. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please see theorem 5.63 in this link: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://books.google.com/books?id=6l_E9OTFaK0C&amp;amp;pg=PA249&quot; rel=&quot;nofollow&quot;&gt;http://books.google.com/books?id=6l_E9OTFaK0C&amp;amp;pg=PA249&lt;/a&gt; &lt;/p&gt;&#xA;" LastActivityDate="2010-09-03T02:18:35.137" CommentCount="6" />
  <row Id="3780" PostTypeId="2" ParentId="3776" CreationDate="2010-09-01T08:22:57.220" Score="3" ViewCount="" Body="&lt;p&gt;The product $$\phi (x) = \prod_{n &gt; 0} (1 - x^n)$$ is called the Euler function, and is well studied. I don't know of any way to compute values at special points. The &lt;a href=&quot;http://en.wikipedia.org/wiki/Pentagonal_number_theorem&quot; rel=&quot;nofollow&quot;&gt;Euler identity&lt;/a&gt; can be used to compute numerical values.&lt;/p&gt;&#xA;" OwnerUserId="960" LastActivityDate="2010-09-01T08:22:57.220" CommentCount="5" />
  <row Id="3781" PostTypeId="2" ParentId="3765" CreationDate="2010-09-01T08:25:42.667" Score="1" ViewCount="" Body="&lt;p&gt;So, basically you have a graph G that varies over time. &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;    1    2    3&#xA;1  f11  f12  f13&#xA;2  f21  f22  f23&#xA;3  f31  f32  f33&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Evaluate $f_{ij}(t)$ for $t$ to get $G(t)$ graph at time t.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then you can use graph traversal algorithms like Depth-First Search,  Breadth-First Search etc. for finding out which nodes are reachable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However if you are allowing traveler to  wait at nodes then problem becomes &lt;a href=&quot;http://onlinelibrary.wiley.com/doi/10.1002/%28SICI%291097-0037%28199705%2929:3%3C141%3a%3aAID-NET2%3E3.0.CO;2-H/references&quot; rel=&quot;nofollow&quot;&gt;more difficult&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can create one big graph by connecting G(t) for various time values. You can keep that graph small if you can limit time to wait at each node. &#xA;If you know period of $f_{ij}(t)$ then also you can simplify this.&lt;/p&gt;&#xA;" OwnerUserId="705" LastEditorUserId="705" LastEditorDisplayName="" LastEditDate="2010-09-01T08:39:12.193" LastActivityDate="2010-09-01T08:39:12.193" CommentCount="1" />
  <row Id="3782" PostTypeId="1" AcceptedAnswerId="3819" CreationDate="2010-09-01T08:45:09.560" Score="11" ViewCount="924" Body="&lt;p&gt;I really like mathematics, but I am not good at learning it.  I find it takes me a long time to absorb new material by reading on my own and I haven't found a formula that works for me.  I am hoping a few people out there will tell me how they go about learning math so I can try out their systems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need to know basic things.  Should I use one book at a time or should I be reading many books on the same topic at once?  Do you stop reading when you hit on a fact that you don't understand or do you keep reading?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do you read all in one go or do you do a little bit and for how long (1 hr, 2 hr or more?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do you read all the chapters or do you do all the exercises before moving on from a chapter?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do you adjust your technique in Calculus(calculation heavy) vs. Analysis (proof heavy)? If so, how?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you make notes, what do you make notes about? Do you make notes while you read or after?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you think these decisions all depend, can you say what they depend on?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am really lost here.  I would appreciate any input.&lt;/p&gt;&#xA;" OwnerUserId="1004" LastActivityDate="2010-09-05T15:50:46.223" CommunityOwnedDate="2010-09-01T09:28:26.367" Title="How Do You Go About Learning Mathematics?" Tags="&lt;soft-question&gt;" AnswerCount="6" CommentCount="4" FavoriteCount="18" />
  <row Id="3783" PostTypeId="1" AcceptedAnswerId="3794" CreationDate="2010-09-01T08:51:46.267" Score="5" ViewCount="265" Body="&lt;p&gt;Can we find a polynomial $p(x) \in \mathbb{R}$ such that $\text{deg}\ p(x)&gt;1$ and which satisfies $$p^{2}(x)-1=p(x^{2}+1)$$ for all $x \in \mathbb{R}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This question can be very well identified with the one i have asked here: &lt;a href=&quot;http://math.stackexchange.com/questions/2346/continuous-function-satisfying-fkxfxk&quot;&gt;http://math.stackexchange.com/questions/2346/continuous-function-satisfying-fkxfxk&lt;/a&gt;&lt;/p&gt;&#xA;" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-09-01T11:15:18.370" LastActivityDate="2010-09-01T13:47:24.743" Title="Polynomials which satisfy $p^{2}(x)-1 = p(x^{2}+1)$" Tags="&lt;calculus&gt;&lt;real-analysis&gt;&lt;analysis&gt;&lt;functional-equations&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="1" />
  <row Id="3784" PostTypeId="2" ParentId="3782" CreationDate="2010-09-01T10:12:44.070" Score="0" ViewCount="" Body="&lt;p&gt;I fear that not only there is not a royal route to mathematics, but there is no single route which is good for everyone.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd suggest to start with a single book, maybe skimming at the first reading parts you are not confident with; in this way you could at least have an idea of what happens. Then you could either re-read the book more carefully, or read (carefully too) a different book whose approach maybe suits best to you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for exercises, it depends on you. I find them useful to set my mind on the themes, but too many exercises are boring.&lt;/p&gt;&#xA;" OwnerUserId="89" LastActivityDate="2010-09-01T10:12:44.070" CommunityOwnedDate="2010-09-01T10:12:44.070" />
  <row Id="3785" PostTypeId="1" CreationDate="2010-09-01T10:43:34.957" Score="3" ViewCount="167" Body="&lt;p&gt;For the equation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{\partial^2}{{\partial}x^2}(IE\frac{{\partial^2}u}{{\partial}x^2}) = \mu(\frac{{\partial^2}u}{{\partial}t^2})$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with $E$ a function of x, derive two ODE's by separation of variables.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I greatly appreciate your help/advice. I'm pretty unsure with how to handle derivatives like this. This is my best attempt, I've been working on this one for quite a while.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://mathbin.net/51398&quot; rel=&quot;nofollow&quot;&gt;http://mathbin.net/51398&lt;/a&gt;&lt;/p&gt;&#xA;" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-09-01T11:07:40.710" LastActivityDate="2010-09-01T13:35:17.563" Title="Seperating a PDE into two ODE's. Made an attempt but unsure if I'm correct." Tags="&lt;differential-equations&gt;" AnswerCount="0" CommentCount="7" FavoriteCount="1" />
  <row Id="3786" PostTypeId="2" ParentId="3782" CreationDate="2010-09-01T10:44:29.190" Score="7" ViewCount="" Body="&lt;p&gt;Von Neumann is credited with saying that you never understand mathematics, but merely get used to it. To an extent, I find this to be true. But what is it that you're getting used to when you learn mathematics?&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;You get used to the sorts of objects and properties-of-objects which are the subject of study. An excellent way to get used to them is to have examples or connections to material that you already know --- regardless of whether such examples or connections are supplied by the book itself.&lt;br&gt;&lt;br&gt;Therefore, as I read a book, I look for examples of the structures or properties which are the subject of a book. If I must, I will scan ahead in the book to hunt for them, or stop reading entirely to try and conjure one myself (or find one on the internet if I find this difficult). If I come across a theorem, I try to see how it holds for these examples --- and how attempts to construct counterexamples fail.&lt;br&gt;&lt;br&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;You get used to certain techniques for solving problems. Getting used to them involves getting used to &lt;em&gt;seeing&lt;/em&gt; them being used, getting used to &lt;em&gt;applying&lt;/em&gt; them, and getting used to the &lt;em&gt;circumstances&lt;/em&gt; in which they may be fruitfully applied.&lt;br&gt;&lt;br&gt; Therefore, in a book, if any of the material is new to me, I do the exercises. This holds regardless of whether the exercises consist of 'calculations' or 'proofs'.&lt;br&gt;&lt;br&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;You get used to the notation. That is, you get used to parsing the notation and translating it mentally into an alternative, often more explicit, description of what the new notation represents. Notation is important to me, in that I find a well-chosen notation an aid to understanding --- and conversely, a poorly-chosen notation a barrier. This is especially the case when the standard notation of one field looks a lot like the standard notation of a significantly different concept that I'm already familiar with.&lt;br&gt;&lt;br&gt; Therefore, as I go through a book, I carefully decide whether to adopt the notation of the book --- or to invent my own notation for the purposes of learning the concepts, without regard to the standard usage. Forcing a change of notation as you take notes has the side-effect of forcing you to reflect often as you do it, which helps to reinforce the new concepts; consciously choosing to reject the standard notation also helps you to be more conscious of the notation you are rejecting, and therefore paradoxiacally to interpret it properly (as you are continually conscious that you must distinguish it from conventions with which you're already familiar).&lt;br&gt;&lt;br&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;You get used to taking other people's ideas and revising them --- testing their boundaries, and perhaps even rediscovering a standard generalization --- to suit your purposes, or to arrive at a more general theory. This, of course, is something which a textbook cannot explicitly teach you. But especially as you see connections in other fields, or as you pick apart proofs in the textbook, you can ask yourself certain questions, such as:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Given a definition of a function or concept on the domain (0,1), is there a natural interpretation which may apply for the boundary points of 0 or 1?&lt;/li&gt;&#xA;&lt;li&gt;Given a definition of an isomorphism between two objects, and suspend the requirement of invertibility? What new relationships between objects do I obtain?&lt;/li&gt;&#xA;&lt;li&gt;Given a proof by contradiction of a result, can I obtain a direct proof of the same result --- perhaps even an efficient construction?&lt;br&gt;&lt;br&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Obviously, a lot of this process is facilitated by taking notes as one learns --- so I take notes, not just of the material in the book, but of the work I do and the reflections I make.&lt;/p&gt;&#xA;" OwnerUserId="439" LastActivityDate="2010-09-01T10:44:29.190" CommunityOwnedDate="2010-09-01T10:44:29.190" CommentCount="3" />
  <row Id="3787" PostTypeId="2" ParentId="3777" CreationDate="2010-09-01T10:57:03.030" Score="4" ViewCount="" Body="&lt;p&gt;&lt;del&gt;The uncountability of $\mathbb{R}$ must be used, but perhaps we can avoid mentioning it.&lt;/del&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Below is a sketch of proof.  By restriction of domain and subtraction of its limit from itself, we may consider $f:[0,1]\rightarrow\mathbb{R}$ such that $f$ has zero pointwise limit and removable discontinuity everywhere.  If we replace $f$ by $|f|$, we may further assume that $f$ is strictly positive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Denote the cardinality of a finite set $S$ by $|S|$.  Let&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$I_0 = \lbrace x: 1\le f(x)\rbrace$ and $I_n = \lbrace x: \frac{1}{n+1}\le f(x)\lt \frac{1}{n}\rbrace$ for $n\ge 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then each $I_n$ must be finite, or else it has an accumulation point $p\in[0,1]$ and this contradicts the assumption that the limit of $f$ at $p$ is zero.  Let&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$J_n = \lbrace x\in[0,1]: |x-y|&amp;lt;\frac{1/2}{3^{n+1}|I_n|} \textrm{ for some } y\in I_n\rbrace.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;($J_n$ is empty if $I_n$ is empty.)  Hence $[0,1] = \bigcup\limits_{n\ge 0} I_n \subseteq \bigcup\limits_{n\ge 0} J_n$.  But the latter has total length $\le \frac{1}{2}$.  So we arrive at a contradiction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[Edit] Alternatively, $\lbrace J_n \rbrace_{n=1,2,...}$ is an open cover of $[0,1]$.  So there is a finite subcover.  This contradicts that the total length of all $J_n$'s is at most $1/2$.&lt;/p&gt;&#xA;" OwnerUserId="1551" LastEditorUserId="1551" LastEditorDisplayName="" LastEditDate="2010-09-01T12:51:01.380" LastActivityDate="2010-09-01T12:51:01.380" CommentCount="3" />
  <row Id="3788" PostTypeId="2" ParentId="3750" CreationDate="2010-09-01T12:33:07.940" Score="8" ViewCount="" Body="&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/didog.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Upper figure: cylinder oil tank cross-section perpendicular to its horizontal axe. The vertical coordinate is the oil level in percentage.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lower figure: graph of oil volume/max. volume (in %) versus oil level $l$ (in feet). The horizontal straight lines represent the area/volume ratio $A(l)/A(10)=V(l)/V(10)$ (in %)  for every multiple of 10% from 0% to 100%.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;--&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the tank radius is $5$, the oil level with respect to the bottom of&#xA;the tank is given by $l=5-5\cos \frac{\theta }{2}$, where $\theta $ is the&#xA;central angle as shown in the figure. The area of the tank cross section&#xA;filled with oil is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$A(\theta )=\frac{25}{2}\theta -\frac{25}{2}\sin \theta $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$A(l)=25\arccos (\frac{5-l}{5})-\frac{25}{2}\sin (2\arccos (\frac{5-l}{5}))$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The area ratio $A(l)/A(10)=V(l)/V(10)$ where $V(l)$ is the oil volume.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $f(l)$ denote this area ratio in percentage:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f(l)=\frac{100}{\pi }\arccos \left( 1-\frac{1}{5}l\right) -\frac{50}{\pi }%&#xA;\sin \left( 2\arccos \left( 1-\frac{1}{5}l\right) \right) $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are the the sequence of $f(l)$ for $l=0,1,2,\ldots ,10$. The graph of $%&#xA;f(l)$ is shown above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f(0)=0$, $f(1)=5.2044$, $f(2)=14.238$, $f(3)=25.232$, $f(4)=37.353$, $f(5)=50$, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f(6)=62.647$, $f(7)=74.768$, $f(8)=85.762$, $f(9)=94.796$, $f(10)=100$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Edit&lt;/em&gt;: &lt;em&gt;One still needs to solve the nonlinear equation&lt;/em&gt; $f(l)-10k=0$ &lt;em&gt;for&lt;/em&gt; $k=1,2,3,4,6,7,8,9$, e. g. by the &lt;em&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Secant_method&quot; rel=&quot;nofollow&quot;&gt;Secant Method&lt;/a&gt;&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Edit&lt;/em&gt; 2: The problem of solving graphically, as shown in the secong figure, is that it would be very difficult, or impossible, to get the required accuracy of 0.01 (feet).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update:&lt;/strong&gt; The oil level marks (in feet) should be placed at &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$0,1.57,2.54,3.40,4.21,$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$5,5.79,6.60,7.46,8.44,10$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;corresponding to the oil volume percentage of&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$0,10,20,30,40,$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$50,60,70,80,90,100$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This calculation was based on the following $f$ function values:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f(0)=0.0$, $f(1.5648)=10.0$, $f(2.5407)=20.0$, $f(3.40155)=30.0$, $f(4.21135)=40.0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f(5)=50.0$, $f(5.7887)=60.0$, $f(6.59845)=70.000$, $f(7.4593)=80.000$, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f(8.4352)=90.000$, $f(10)=100.0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update 2&lt;/strong&gt; Figure of marks:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/j6vXn.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;[Rearranged to show the sequence of editions and updates.]&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="752" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2010-09-12T23:05:28.853" LastActivityDate="2010-09-12T23:05:28.853" CommentCount="5" />
  <row Id="3789" PostTypeId="2" ParentId="3782" CreationDate="2010-09-01T12:39:13.593" Score="0" ViewCount="" Body="&lt;p&gt;I think you best follow the curriculum a university publishes online. MIT has OCW for example. Your questions highly depend on yourself and what works best for you, I think you'll have to find that out on your own.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example: for me it is very hard to do the same thing for more than two hours, then I switch to something else because my mind just stops working otherwise.&lt;/p&gt;&#xA;" OwnerUserId="1120" LastActivityDate="2010-09-01T12:39:13.593" CommunityOwnedDate="2010-09-01T12:39:13.593" CommentCount="1" />
  <row Id="3790" PostTypeId="2" ParentId="783" CreationDate="2010-09-01T12:40:27.527" Score="3" ViewCount="" Body="&lt;p&gt;The important point here, I think, is about Hom(a,a) and Hom(b,b). nm is guaranteed to be an element of the first and mn an element of the second; to be an isomorphism, both of these maps must be the identity by definition. The definition of category requires that you have at least the identity in both endomorphism sets; if neither set has non-identity elements then you get an isomorphism, since the compositions don't have anything other than the identity to be, but this needn't hold in general.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maximally toy example: take the full subcategory of sets given by the one element set A and the two element set B. Hom(A,A) is only the identity, but Hom(B,B) has four elements, two of which are not isomorphisms.&lt;/p&gt;&#xA;" OwnerUserId="1646" LastActivityDate="2010-09-01T12:40:27.527" />
  <row Id="3793" PostTypeId="1" AcceptedAnswerId="3811" CreationDate="2010-09-01T13:35:29.013" Score="6" ViewCount="227" Body="&lt;p&gt;I have the following well-known exercise in cardinal arithmetics:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $\kappa, \lambda$ are cardinals such that $\lambda$ is infinite, then $\kappa^{&amp;lt;\lambda}$ equals the supremum of the $\kappa^{\theta}$, where $\theta &amp;lt; \lambda$ is a cardinal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Isn't this false for $\kappa=1$? The left side is $\lambda$ and the right side $1$. Remark that $\kappa^{&amp;lt;\lambda}$ is defined to be the cardinality of the disjoint(?) union of the function sets $map(\alpha,\kappa)$, where $\alpha &amp;lt; \lambda$ is an ordinal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Is this supremum understood as a cardinal supremum or an ordinal supremum? If the latter is the case, why should the supremum be a cardinal at all?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) Anyway, I don't know how to solve this exercise. I've already seen it several times, but 1) and 2) remain obscure in the literature I know.&lt;/p&gt;&#xA;" OwnerUserId="1650" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-09-02T18:43:52.797" LastActivityDate="2010-09-02T18:43:52.797" Title="cardinal exponentiation, $k^{&lt;\lambda}$" Tags="&lt;set-theory&gt;&lt;logic&gt;&lt;cardinals&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="3794" PostTypeId="2" ParentId="3783" CreationDate="2010-09-01T13:42:22.430" Score="10" ViewCount="" Body="&lt;p&gt;There is no solution. Here is the proof:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First note, that the highest coefficient of $p$ must be 1. In particular, $\lim_{x \to \infty} p(x) = \infty$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume, that there exists a real value $c$ such that $p (c) = 0$. Then&#xA;$$ 0 = (p(c)^2 - 1)^2 - 1 = p(c^2 + 1)^2 - 1 = p((c^2 + 1)^2 + 1)$$&#xA;But $(c^2 + 1)^2 + 1 &gt; c$, so any real root of $p$ would lead to another, higher root of $p$ which is impossible. So $p$ has no real root and thus $p(x) &gt; 0$ for all $x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But then, $p(x)^2 - 1 = p(x^2 + 1) &gt; 0$ and thus $p(x) &gt; 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Define a sequence $x_n$ by $x_0 = 0$ and $x_{n + 1}^2 + 1 = x_n$ with imaginary part of $x_n$ positive for $n &gt; 0$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We first show, that $p(x_n) \neq 0$ for all $x_n$:&#xA;clearly, $p(x_0) \neq 0$ and $p(i)^2 - 1 = p(0) &gt; 1$ and so $p(x_1) = p(i) \neq 0$.&#xA;Assume, that $p(x_{n+1}) = 0$ for some $n &gt; 1$. Then $p(x_n) = -1$ and so $p(x_{n - 1}) = 0$. This is impossible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the functional equation, we obtain by differentiating&#xA;$$&#xA;p(x) p'(x) = x p'(x^2 + 1)&#xA;$$&#xA;We now show by induction, that $p'(x_n) = 0$ for all $n$.&#xA;For $n = 0$, we obtain $p(0) p'(0) = 0$, and thus $p'(0) = 0$.&#xA;Assume, that $p'(x_n) = 0$, then&#xA;$$&#xA;p(x_{n+1}) p'(x_{n+1}) = x_{n + 1} p'(x_{n+1}^2+1) = x_{n+1} p'(x_n) = 0&#xA;$$&#xA;and so, $p'(x_{n+1}) = 0$ as required.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But $x_n$ is a sequence converging to $\frac{1 + i \sqrt{3}}{2}$, and in particular consists of infinitely many distinct points, all of which are roots of $p'$ which is impossible.&lt;/p&gt;&#xA;" OwnerUserId="960" LastEditorUserId="960" LastEditorDisplayName="" LastEditDate="2010-09-01T13:47:24.743" LastActivityDate="2010-09-01T13:47:24.743" CommentCount="1" />
  <row Id="3795" PostTypeId="1" AcceptedAnswerId="3797" CreationDate="2010-09-01T13:55:58.633" Score="1" ViewCount="77" Body="&lt;p&gt;Let $N \subseteq M$ be a subobject in an abelian category (say, modules). A complement of $N$ in $M$ is then defined to be a subobject $Q \subseteq M$ which is maximal with respect to the condition $Q \cap N = 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've got just a short question what maximality means here: Does it mean that every other $Q'$ with $Q' \cap N = 0$ is contained in $Q$, or does it mean that every $Q'$ with $Q' \cap N = 0$ and $Q \subseteq Q'$ satisfies $Q = Q'$? Usually the latter is meant by maximal, right? However, then I don't understand at all the proof (sketches) in the section on essential monomorphisms and injective hulls in Gabriel's thesis.&lt;/p&gt;&#xA;" OwnerUserId="1650" LastEditorUserId="16" LastEditorDisplayName="" LastEditDate="2010-09-04T03:34:04.163" LastActivityDate="2010-09-04T03:34:04.163" Title="Complement of a submodule" Tags="&lt;category-theory&gt;&lt;abelian-categories&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" />
  <row Id="3796" PostTypeId="2" ParentId="3742" CreationDate="2010-09-01T14:31:30.020" Score="2" ViewCount="" Body="&lt;p&gt;Can't believe no one spoke of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Collatz_conjecture&quot; rel=&quot;nofollow&quot;&gt;Collatz conjecture&lt;/a&gt; yet.&lt;/p&gt;&#xA;" OwnerUserId="1310" LastActivityDate="2010-09-01T14:31:30.020" CommunityOwnedDate="2010-09-01T14:31:30.020" />
  <row Id="3797" PostTypeId="2" ParentId="3795" CreationDate="2010-09-01T14:50:51.753" Score="1" ViewCount="" Body="&lt;p&gt;&quot;Maximal&quot; means the latter, i.e. if $Q'$ contains $Q$ and $Q'\cap N = 0$, then $Q' = Q$.&#xA;Since complements are very seldom unique (as one knows already from the case of subspaces&#xA;of vector spaces), the former condition is prohibitively strong, and would almost never&#xA;be realized in practice.&lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2010-09-01T14:50:51.753" />
  <row Id="3798" PostTypeId="1" AcceptedAnswerId="3800" CreationDate="2010-09-01T14:53:24.840" Score="6" ViewCount="193" Body="&lt;p&gt;The presentation of the dihedral group $D_{2n}$ is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$D_{2n}= \langle r,s \mid r^{n}=s^{2}=1, rs=sr^{-1} \rangle.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is it incorrect to conclude that $r=s=1$ ?   In other words why doesn't this group presentation describe the trivial group.&lt;/p&gt;&#xA;" OwnerUserId="832" LastEditorUserId="171" LastEditorDisplayName="" LastEditDate="2010-09-01T19:04:01.443" LastActivityDate="2010-09-02T20:50:36.947" Title="Presentation of $D_{2n}$" Tags="&lt;abstract-algebra&gt;&lt;group-theory&gt;" AnswerCount="3" CommentCount="3" />
  <row Id="3799" PostTypeId="1" AcceptedAnswerId="3803" CreationDate="2010-09-01T14:57:38.597" Score="4" ViewCount="118" Body="&lt;p&gt;Let $R$ be the coordinate ring of an affine complex variety (i.e. finitely generated, commutative, reduced $\mathbb{C}$ algebra) and $M$ be an $R$ module.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $s\in M$ be an element, such that $s\in \mathfrak{m}M$ for every maximal ideal $\mathfrak{m}$. Does this imply $s=0$?&lt;/p&gt;&#xA;" OwnerUserId="667" LastActivityDate="2010-09-01T16:07:46.627" Title="Does a section that vanishes at every point vanish?" Tags="&lt;algebraic-geometry&gt;&lt;commutative-algebra&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" />
  <row Id="3800" PostTypeId="2" ParentId="3798" CreationDate="2010-09-01T15:00:04.440" Score="6" ViewCount="" Body="&lt;p&gt;Call $G$ the group given by that presentation. Then you can show there exists an homomorphism $\phi:G\to\mathrm{GL}_2(\mathbb R)$ which maps $s$ to $S=\left(\begin{smallmatrix}-1&amp;amp;0\\0&amp;amp;1\end{smallmatrix}\right)$ and $r$ to $R=\left(&#xA;\begin{smallmatrix}&#xA; \cos (\theta ) &amp;amp; -\sin (\theta ) \\&#xA; \sin (\theta ) &amp;amp; \cos (\theta )&#xA;\end{smallmatrix}&#xA;\right)$ with $\theta=2\pi/n$; to check this, you have to use the characteristic property of group presentations, and verify that matrices $S$ and $R$ satisfy the same relations as the generators $s$ and $r$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, since $\phi(s)\neq I$ and $\phi(r)\neq I$, it follows of course that $s\neq 1_G$ and $r\neq 1_G$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(You can also easily find two non-identity permutations $\rho$, $\sigma\in S_n$ which satisfy the relations, and construct a map $g:G\to S_n$ such that $g(s)=\sigma$ and $g(r)=\rho$, giving another proof that $r\neq 1_G$ and $s\neq 1_G$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;NB&lt;/strong&gt; This is another canonical example of the things we want group representations for.&lt;/p&gt;&#xA;" OwnerUserId="274" LastEditorUserId="274" LastEditorDisplayName="" LastEditDate="2010-09-01T15:20:47.640" LastActivityDate="2010-09-01T15:20:47.640" CommentCount="6" />
  <row Id="3801" PostTypeId="1" CreationDate="2010-09-01T15:05:52.177" Score="0" ViewCount="146" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Possible Duplicate:&lt;/strong&gt;&lt;br&gt;&#xA;  &lt;a href=&quot;http://math.stackexchange.com/questions/656/how-can-i-write-an-equation-that-matches-any-sequence&quot;&gt;How can I write an equation that matches any sequence?&lt;/a&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;p&gt;I'm learning sequences right about now, and I'm having a really hard time finding the formula for a given sequence of numbers.  I'm worried that, at exam time I'll be spending as much time finding the formula as I am now.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, one problem gave me:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\langle a_n \rangle = \{2, 8, 18, 32, 50\}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Okay, it's giving me a Math Processing error. I'm sure the code is correct; I'm also seeing this error on other people's pages. If it's me, please let me know.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again, one problem gave me the sequence&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(a&lt;sub&gt;n&lt;/sub&gt;) = { 2, 8, 18, 32, 50 }&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I spent a good ten to fifteen minutes tinkering around, making some diagrams, plugging in the values. I'll post some examples:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a&lt;sub&gt;1&lt;/sub&gt; = 2     : +6    : +4&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a&lt;sub&gt;2&lt;/sub&gt; = 8     : +10   : +4&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a&lt;sub&gt;3&lt;/sub&gt; = 18    : +14   : +4&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a&lt;sub&gt;4&lt;/sub&gt; = 32    : +18   : +4&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a&lt;sub&gt;5&lt;/sub&gt; = 50   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I first checked to see by how much each values differs to the next. The +4 was the increment by how much &lt;em&gt;more&lt;/em&gt; it increased every time. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I therefore came up with the following solution (among many others):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2a&lt;sub&gt;n&lt;/sub&gt; + 4a&lt;sub&gt;n-1&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This worked for the first three values in the sequence. Then it broke. Damn.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Eventually I gave up and looked at the solution:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(a&lt;sub&gt;n&lt;/sub&gt;) = (2n&lt;sup&gt;2&lt;/sup&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Seriously? I would never have guessed this. So my question is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How are you supposed to find a pattern? Are there any good methods for doing so?&lt;/p&gt;&#xA;" OwnerUserId="44" LastActivityDate="2010-09-01T15:05:52.177" ClosedDate="2010-09-01T15:15:28.027" Title="How to Determine a Formula for a Sequence of Numbers?" Tags="&lt;discrete-mathematics&gt;" CommentCount="3" />
  <row Id="3802" PostTypeId="2" ParentId="3777" CreationDate="2010-09-01T15:18:17.817" Score="15" ViewCount="" Body="&lt;p&gt;I think the following works:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a sketch, I will fill in the details later if required.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $g(x) = \lim_{t\rightarrow x} f(t)$. Then we can show that $g(x)$ is continuous.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $h(x) = f(x) - g(x)$. Then $\lim_{t \rightarrow x} h(t)$ exists and is $0$ everywhere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We will now show that $h(c) = 0$ for some $c$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This will imply that $f(x)$ is continuous at $c$ as then we will have $f(c) = g(c) = \lim_{t-&gt;c} f(t)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider any point $x_0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By limit of $h$ at $x_0$ being $0$, there is a closed interval $I_0$ (of length &gt; 0) such that $|h(x)| &amp;lt; 1$ for all $x \in I_0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is because, given an $\epsilon &gt; 0$ there is a $\delta &gt; 0$ such that $|h(x)| &amp;lt; \epsilon$ for all $x$ such that $0 &amp;lt; |x - x_{0}| &amp;lt; \delta$. Pick $\epsilon = 1$ and pick $I_{0}$ to be any closed interval of non-zero length in $(x_{0}, x_{0} + \delta)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now pick any point $x_1$ in $I_0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By limit of $h$ at $x_1$ being $0$, there is a closed interval $I_1 \subset I_0$ (of length &gt; 0) such that $|h(x)| &amp;lt; 1/2$ for all $x \in I_1$, by argument similar to above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Continuing this way, we get a sequence of closed intervals $I_n$ such that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$|h(x)| &amp;lt; \frac{1}{n+1}$ for all $x \in I_n$. We also have that $I_{n+1} \subset I_n$ for each $n$, and that length $I_n$ &gt; 0. We could also arrange so that length $I_n \rightarrow 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now there is a point $c$ (by completeness of $\mathbb{R}$) such that $c \in \bigcap_{n=0}^{\infty}I_{n}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus we have that $|h(c)| &amp;lt; \frac{1}{n+1}$ for all $n$ and so $h(c) = 0$ and $f(c) = g(c)$.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-09-02T16:11:30.873" LastActivityDate="2010-09-02T16:11:30.873" CommentCount="8" />
  <row Id="3803" PostTypeId="2" ParentId="3799" CreationDate="2010-09-01T15:23:18.540" Score="7" ViewCount="" Body="&lt;p&gt;Not in general, no.  For example, if $R = \mathbb C[T]$ and $M$ is the field of fractions&#xA;of $R$, namely $\mathbb C(T)$, then (a) every maximal ideal of $R$ is principal; (b) every&#xA;element of $M$ is divisible by every non-zero element of $R$.  Putting (a) and (b) together&#xA;we find that $M = \mathfrak m M$ for every maximal ideal $\mathfrak m$ of $R$, but certainly&#xA;$M \neq 0.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a finitely generated example: again take $R = \mathbb C[T]$, and take $M = \mathbb C[T]/(T^2).$  Then $s = T \bmod T^2 \in \mathfrak m M$ for every $\mathfrak m$, because&#xA;$\mathfrak m M = M$ if $\mathfrak m$ is a maximal ideal other than $(T)$, and this&#xA;is clear from the choice of $s$ if $\mathfrak m = (T)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer &lt;em&gt;is&lt;/em&gt; yes if $M$ is finitely generated and torsion free.  For let $S$ be the total quotient ring of &#xA;$R$ (i.e. the product of functions fields $K(X)$ for each irreducible component $X$&#xA;of the variety attached to $R$).&lt;br&gt;&#xA;Then $M$ embeds into $S\otimes_R M$ (this is the torsion free condition), which in turn embeds into&#xA;a finite product of copies of $S$ (since it is finite type over $S$, which is just a product&#xA;of fields).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clearing denominators, we find that in fact $M$ then embeds into $R^n$ for some $n$.&#xA;Thus it suffices to prove the result for $M = R^n$, and hence for $R$, in which case&#xA;it follows from the Nullstellensatz, together with the fact that $R$ is reduced.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, note that for any finitely generated $R$-module, if $M = \mathfrak m M$ for all $\mathfrak m$&#xA;then $M = 0$ (since Nakayama then implies that $M_{\mathfrak m} = 0$ for all&#xA;$\mathfrak m$).  Thus if $M$ is non-zero it can't be that &lt;em&gt;every&lt;/em&gt; section lies&#xA;in $\mathfrak m M$ for all $\mathfrak m$.&lt;/p&gt;&#xA;" OwnerUserId="221" LastEditorUserId="221" LastEditorDisplayName="" LastEditDate="2010-09-01T15:49:16.953" LastActivityDate="2010-09-01T15:49:16.953" CommentCount="1" />
  <row Id="3804" PostTypeId="2" ParentId="3798" CreationDate="2010-09-01T15:29:08.167" Score="3" ViewCount="" Body="&lt;p&gt;A group given by presentation $G=\langle X|R\rangle$ is, by definition, the &lt;em&gt;most general&lt;/em&gt; group with a generating set $X$ and satisfying the relations $R$. Say, for simplicity, that $X=\{x_1,\ldots,x_n\}$ and $R=\{r_1(x_1,\ldots,x_n),\ldots,r_m(x_1,\ldots,x_n)\}$. Being the &quot;most general&quot; group with $n$ generators $x_1,\ldots,x_n$ that satisfy the $m$ relations $r_1,\ldots,r_m$ means that given &lt;em&gt;any&lt;/em&gt; group $H$ with elements $h_1,\ldots,h_n$ for which $r_i(h_1,\ldots,h_n)=1$ for $i=1,\ldots,m$, there will be a (unique) group homomorphism from $G$ to $H$ that maps $x_i$ to $h_i$.  It's like the &quot;free-est group&quot; with elements $x_1,\ldots,x_n$, subjected only to the relations in $R$. (The magic words are &lt;strong&gt;Van Dyck's Theorem&lt;/strong&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you are noting is that if we pick $H$ to be the trivial group, and $h_1,\ldots,h_n$ to be the trivial element, then all the relations are satisfied; but this does not tell you that the trivial group is the most general group that satisfies the given conditions, it only tells you that the trivial group is a &lt;em&gt;quotient&lt;/em&gt; of the most general group that satisfies the given conditions (as it must, since the trivial group is a quotient of everything).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of the dihedral group $D_{2n}$ you are looking for the most general group that has an element of order dividing $n$, an element of order dividing $2$, and such that those elements satisfy the relation $sr = r^{-1}s$ (which can be written as $srsr=1$). The trivial group certainly satisfies those relations, but it is not the most general one that does (e.g., if $n$ is even, you can look at $H=C_2\times C_2$, and identify $r$ with $(1,0)$ and $s$ with $(0,1)$). Every time you find a group and elements that satisfy the given relations, you have found a map from the group presented to the group you found, but you have not necessarily found the group presented.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To really know that you have found the &quot;most general group&quot; that satisfies the given conditions is not always trivial. One way is to show that the group you found (call it $G$, with elements $g_1,\ldots,g_n$ playing the roles of the $x_i$) has the property ascribed to the group presented by the presentation: given any group $H$, and any elements $h_1,\ldots,h_n$ in $H$ that satisfy $r_1,\ldots,r_m$ in $H$, there exists a unique group homomorphism from your group $G$ to $H$ that maps $g_i$ to $h_i$. The trivial group does not have this property for the presentation given in $D_{2n}$, so it cannot be $D_{2n}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another method that works sometimes is to find a &quot;normal form&quot; for the elements of the group presented, allowing you to count exactly how many elements there are. This is not hard to do in $D_{2n}$: every element of $D_{2n}$ can be written as a product of powers of $r$ times powers of $s$. Using the rules that say that $r^n=1$, $s^2=1$, and that $sr=r^{-1}s = r^{n-1}s$, you can &quot;move&quot; all the $r$s that occur in such a product all the way to the left, and all the $s$'s to the right, so that every element of $D_{2n}$ can be written in the form $r^is^j$, with $0\leq i\lt n$ and $0\leq j\lt 2$. We do not know if each element corresponds to one and only one such expression, but certainly each element corresponds to at least one such expression. That means that $D_{2n}$ has, at most, $2n$ elements. If you can find a group $G$, generated by elements $g_1$ and $g_2$ such that $g_1^n=1$, $g_2^2=1$, and $g_2g_1=g_1^{-1}g_2$, &lt;em&gt;and&lt;/em&gt; it turns out that $G$ has exactly $2n$ elements, then you know there is a homomorphism from $D_{2n}$ to $G$ that sends $r$ to $g_1$ and $s$ to $g_2$, so that $G$ will be a quotient of $D_{2n}$; since we are assuming that $G$ has exactly $2n$ elements, and we know that $D_{2n}$ has at most $2n$ elements, then you can conclude that the map has to be injective and the kernel is trivial, and so that $D_{2n}$ is in fact isomorphic to the group $G$ you found. This is one way of &quot;realizing&quot; a presentation as a group that you can get your hands on. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; Note that by the property ascribed to the group given by presentation, the group must be generated by $X$; otherwise, there is a map from $G$ onto $\langle X\rangle$ that sends each $x_i$ to itself, so you get two maps from $G$ to itself sending each $x_i$ to itself (the identity, and the composite map $G\to\langle X\rangle \hookrightarrow G$); but the definition requires that there be a unique homomorphism to accomplish this.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-09-01T15:49:21.970" LastActivityDate="2010-09-01T15:49:21.970" CommentCount="2" />
  <row Id="3805" PostTypeId="1" AcceptedAnswerId="3807" CreationDate="2010-09-01T15:36:29.613" Score="2" ViewCount="129" Body="&lt;p&gt;Hi everybody, first question on math.stackexchange :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm studying for a Cryptography - Communication Security exam, and it involves a certain quantity of number theory - finite field theory, so be warned: this is my first encounter with these topics, and you'll have to be extra-clear with me :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I thought I was doing pretty well with the questions and exercises in $GF(4)$… then I hit $GF(8)$ and realized I'm still missing the point :(&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand that to represent, let's say, $x^3$, the fourth element of $GF(8)$, I just take the result of the division of my irreducible polynomial of choice, let's say $x^3+x+1$, and I'm happy. I can then build addition and multiplication tables, and switch to and from a binary representation when it seems more convenient (e.g. when XORing with something else), and this also makes me happy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But why do I start dividing $x^0$, $x^1$, $x^2$, etc.?&lt;br&gt;&#xA;This sure seems like a sound idea when compared to “try some random polynomial“, but I can't figure why these are the candidates for being the members of the finite field? Why am I sure that this will generate all the elements?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bonus question: $GF(4)$ seems to have a number of properties that are quite useful for manipulation. It seems obvious that $x+x \equiv 0$ in any $GF(2^n)$, but do $x^2 \equiv x+1$ or $x^3\equiv x^2+x\equiv 1$ always hold? Is this in any way dependent on the irreducible polynomial chosen? Are these predictable?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please forgive me for any obvious error, and thanks in advance for any help.&lt;/p&gt;&#xA;" OwnerUserId="894" LastEditorUserId="894" LastEditorDisplayName="" LastEditDate="2010-09-01T16:39:03.310" LastActivityDate="2010-09-01T21:33:50.253" Title="Powers of $x$ as members of Galois Field and their representation as remainders" Tags="&lt;number-theory&gt;&lt;field-theory&gt;&lt;galois-theory&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="3806" PostTypeId="2" ParentId="3758" CreationDate="2010-09-01T16:24:36.397" Score="2" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;REMARK&lt;/strong&gt; $\ $ Pierre's answer is a generating function approach:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle\rm\quad\quad\quad\;\;  f(x+t) \;=\; \sum_{n=0}^\infty \;\;  f^{(n)}(x) \; \frac{t^n}{n!}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad (fgh)(x+t) \;=\; f(x+t)\: g(x+t)\: h(x+t)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\quad\quad\quad\quad\quad\quad\quad\;\:  \;=\; (f + f' t +\:\cdots) (g + g' t + \:\cdots) (h + h' t + \:\cdots) $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\quad\quad\quad\quad\quad\quad\quad\;\: \;=\; fgh + (f'gh + fg'h + fgh')\: t + \:\cdots $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad (fgh)(x+t) \;=\; fgh + (fgh)' t + \:\cdots $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus we conclude $\:\rm(fgh)' = f'gh + fg'h + fgh'$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This generalizes to n-th order derivatives, as Pierre mentioned, e.g.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\quad\displaystyle\rm\sum_{n=0}^\infty \: (fgh)^{(n)} \frac{t^n}{n!} \ =\ \sum_{n=0}^\infty \:\bigg(\ \sum_{i+j+k=n} \frac{f^{(i)}g^{(j)}h^{(k)}}{i!\ j!\ k!}\bigg) \frac{t^n}{n!} $&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-09-01T16:33:07.393" LastActivityDate="2010-09-01T16:33:07.393" CommentCount="1" />
  <row Id="3807" PostTypeId="2" ParentId="3805" CreationDate="2010-09-01T16:49:02.647" Score="5" ViewCount="" Body="&lt;p&gt;First: a Galois field is a finite field. A finite field will have $p^n$ elements for some prime $p$ and some positive integer $n$ (in fact, there is, up to isomorphism, one and only one finite field with $p^n$ elements). If a field has $p^n$ elements, then $x+x+\cdots+x=0$ ($p$ summands) for all $x$, and no smaller positive integer has that property; this is called the &lt;strong&gt;characteristic&lt;/strong&gt; of the field. For fields of order a power of $2$, you do indeed have $x+x=0$ for all $x$, but not for any other size Galois fields. However, in computer science and  cryptography there is some preference for working in fields of characteristic $2$, because they tend to be easier to represent and work with in computers (which themselves work in &quot;characteristic $2$&quot;). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure what you mean by &quot;why do I start dividing $x^0$, $x^2$, $x^2$, etc.?&quot;. As you have probably seen, the field $GF(2^n)$ can always be described in terms of a  monic polynomial $P(t)$ of degree $n$ that is irreducible over $GF(2)$. This amounts to constructing the field $GF(2)[t]/(P(t))$. What this tells you is that you have the smallest ring that contains $0$, $1$, and $x$, subject to the conditions $1+1=0$, $1\cdot 1=1$, and $P(x)=0$. Any element of this ring can be written as a polynomial expression in $x$ with coefficients $0$ or $1$, $q(x) = b_mx^m+\cdots + b_0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, because we are assuming that $P(x)=0$, using the division algorithm we can always write any polynomial $q(t)$ as $q(t)=P(t)b(t) + r(t)$, with $r(t)=0$ or $\deg(r)\lt \deg(P)=n$. Evaluating at $x$ and using the fact that $p(x)=0$, we get that $q(x)=r(x)$, so in fact every element in this ring can be written as a polynomial expression in $x$ of degree less than $n$. The expression is unique, because if $r(x)=s(x)$ with $r$ and $s$ of degree less than $n$, then $r-s$ would be a multiple of $P(t)$ (because the latter is irreducible), and degree considerations tell you that $r=s$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So each element of $GF(2^n)$ can be written uniquely as $q(x)$ where $q$ has degree less than $n$. Choosing different polynomials amounts to choosing different representations for the elements, and so different rules for what $x^n$ will mean.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So: in order for $x^2=x+1$ to hold in your Galois field $GF(p^n)$, you must have that $x^2-x-1=0$, so the polynomial $P$ you picked must divide $x^2-x-1$, and so you must be either in $GF(p)$ or in $GF(p^2)$. Similarly, for $x^3=x^2+x=1$ to hold you would need to be in $GF(p)$ or $GF(p^2)$ in order for $x^2+x-1=0$ to hold; and whether it holds depends on the polynomial $P$ that you picked to define how $x$ behaves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In summary: you divide by the polynomial so you can get a unique expression for each element in $GF(2^n)$ in terms of the distinguished element $x$ that helps you construct it (the behaviour of the element $x$ being determined by the polynomial). The condition $\alpha+\alpha=0$ will hold for every $\alpha$ in every Galois field of size $2^n$ for any $n$; but no power of $x$ smaller than $n$ will be expressible in terms of lower powers of $x$, and the expression of $x^n$ in terms of lower powers is determined by your choice of polynomial.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-09-01T16:49:02.647" CommentCount="3" />
  <row Id="3810" PostTypeId="2" ParentId="3776" CreationDate="2010-09-01T18:44:23.500" Score="4" ViewCount="" Body="&lt;p&gt;If you want to evaluate $$\prod_{n=1}^\infty (1-1/2^n)$$ numerically you will find that it's VERY close to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$2^{1/24}\sqrt{\frac{2\pi}{\log 2}}\exp{(-\pi^2/(6\log 2))}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which is about 0.2887880950866... And that's because $$\exp{(-4\pi^2/\log 2)}$$ is pretty small, approximately 1.839x10^(-25).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This can be seen from the fact that $$\Delta(-1/z) = z^{12}\Delta(z),$$ where $$\Delta(z)$$ is the cusp form of weight 12 defined by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Delta(z)=q\prod_{n=1}^\infty (1-q^n)^{24},$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $$q=e^{2\pi iz}$$ for Im(z)&gt;0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just put $$z=\frac{2\pi i}{\log 2}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can use this technique to greatly accelerate the convergence of your product for terms other than (1/2)^n.&lt;/p&gt;&#xA;" OwnerUserId="1301" LastActivityDate="2010-09-01T18:44:23.500" CommentCount="3" />
  <row Id="3811" PostTypeId="2" ParentId="3793" CreationDate="2010-09-01T18:52:04.667" Score="1" ViewCount="" Body="&lt;p&gt;First, (2): if you define a cardinal to be an ordinal that is not bijectable with any strictly smaller ordinal, then suppose that $\alpha$ is the ordinal supremum of $\kappa^{\theta}$, $\theta\lt\lambda$. Let $\beta$ be a strictly smaller ordinal than $\alpha$. By the definition of supremum, there must exist a $\theta\lt\lambda$ such that $\beta\lt\kappa^{\theta}\leq\alpha$; in particular, $\kappa^{\theta}$ cannot be bijected with $\beta$ (being a cardinal), and therefore neither can $\alpha$ (that would give an embedding of $\kappa^{\theta}$ into $\beta$, and Cantor-Bernstein would give you $|\beta|=\kappa^{\theta}$, contradicting that the latter is not bijectable with any strictly smaller ordinal). Thus, $\alpha$ is not bijectable with any strictly smaller ordinal, and so must be a cardinal. So whether you define the supremum as the ordinal-sup or the cardinal-sup, you will still get a cardinal (this holds for any set of cardinals).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second, (1): you are correct that the definition does not match this for $\kappa=1$ (or for $\kappa = 0$); as noted by Carl in the comments, this is likely an erratum or ommission; it should hold for any $\kappa&gt;1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; The definition as a sum is done running over all ordinals, rather than all cardinals, so I'm fixing this below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, (3): you are trying to show that the supremum of the $\kappa^{\theta}$ equals the sum over all ordinals $\alpha&amp;lt;\lambda$ of $|\kappa^{\alpha}$, assuming $\kappa\geq 2$ (to prevent the problems noted). This because $|\kappa^{\alpha}|=|map(\theta,\kappa)|$ by definition, and the cardinality of the disjoint union is the cardinal sum. I believe this can be shown by transfinite induction on $\lambda$ as I do below, but there probably is a simpler method.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the proposition we want to show is that for any infinite &lt;em&gt;ordinal&lt;/em&gt; $\lambda$, we have&#xA;$$\sup_{\alpha\lt\lambda}|\kappa^{\alpha}| = \sum_{\alpha\lt\lambda}|\kappa^{\alpha}|.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, the equality holds for $\lambda=\omega$: if $2\leq \kappa\lt\aleph_0$, then $\sup\{|\kappa^n|\,|\, n=0,1,2,3,\ldots\} = \aleph_0$ and $\sum_{n=0}^{\infty}|\kappa^n| = \aleph_0$; if $\aleph_0\leq\kappa$, then $|\kappa^n|=\kappa$ for all $n$, and $\sum_{n=0}^{\infty}|\kappa^n| = \sum_{n=0}^{\infty}\kappa = \kappa\aleph_0=\kappa$, so both sides agree.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume the result holds for $\lambda$; then the supremum of $|\kappa^{\alpha}|$ with $\alpha\lt\lambda^+$ is $|\kappa^{\lambda}|$; on the other hand, &#xA;$$\sum_{\alpha\lt\lambda^+}|\kappa^{\alpha}| = \left(\sum_{\alpha\lt\lambda}|\kappa^{\alpha}|\right) + \kappa^{\lambda} = \sup_{\alpha\lt\lambda}|\kappa^{\alpha}|+|\kappa^{\lambda}|=|\kappa^{\lambda}|,$$&#xA;where the last equality holds because $|\kappa^{\alpha}|\leq |\kappa^{\lambda}|$ for each $\alpha\lt \lambda$, so the supremum is at most $|\kappa^{\lambda}|$, and the sum of two infinite cardinals is equal to their maximum. So again the two expressions agree. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, we want to show that if $\lambda$ is a limit ordinal and the result holds for all $\beta\lt\lambda$, then it holds for $\lambda$. Then&#xA;$$\sup_{\alpha\lt\lambda}(|\kappa^{\alpha}|) = \sup_{\beta\lt\lambda}\left(\sup_{\alpha\lt\beta}(|\kappa^{\alpha}|\right) = \sup_{\beta\lt\lambda}\sum_{\alpha\lt\beta}|\kappa^{\alpha}| = \sum_{\alpha\lt\lambda}(|\kappa^{\alpha}|).$$&#xA;So the equality holds for $\lambda$ as well. This establishes the result by transfinite induction for all infinite ordinals $\lambda$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Further Edit, 2 Sep 2010:&lt;/strong&gt; Clarify how to finish it off.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the above shows that for any ordinal $\lambda$, $\sup_{\alpha\lt\lambda}|\kappa^{\alpha}| = \sum_{\alpha\lt\lambda}|\kappa^{\alpha}|$. To finish the exercise, we need to show that if $\lambda$ is a cardinal (that is, an ordinal that is not bijectable with any strictly smaller ordinal), then $\sup\{|\kappa^{\alpha}|\colon\alpha$ is an ordinal and $\alpha\lt\lambda\} = \sup\{|\kappa^{\theta}|\colon \theta$ is a cardinal and $\theta\lt\lambda\}$. To see this, note that $|\kappa^{\alpha}|=|\kappa|^{|\alpha|}$, and since $\lambda$ is assumed to be a cardinal, if $\alpha\lt\lambda$, then there exists a cardinal $\theta$, $\theta\lt\lambda$, such that $|\alpha|=|\theta|$, and hence $|\kappa^{\alpha}|=|\kappa^{\theta}|$.  Thus, the two sets are equal, so their suprema are equal as well. &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-09-02T18:43:06.160" LastActivityDate="2010-09-02T18:43:06.160" CommentCount="7" />
  <row Id="3812" PostTypeId="2" ParentId="3798" CreationDate="2010-09-01T18:55:25.270" Score="2" ViewCount="" Body="&lt;p&gt;In the question&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;$r^{n}=s^{2}=1, rs=sr^{-1}$ ... why is it incorrect to &lt;em&gt;conclude&lt;/em&gt; that $r=s=1$ ?  ... why doesn't this group presentation &lt;em&gt;describe&lt;/em&gt; the trivial group&quot;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;the key words are &quot;conclude&quot; and &quot;describe&quot;.  You are in effect asking what logic applies to statements about $r$ and $s$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer is that assertions about $r$ and $s$ are understood as valid in this context only if they hold in ALL groups.  The assertion &quot;if a pair of group elements $r$ and $s$ satisfies the defining relations of $D_{2n}$, then $r=s=1$&quot; is false in some groups and true in others.  It is true in groups with no subgroup of order $2n$, such as large cyclic groups of prime order.   It is false in other groups, such as the dihedral group of order 2n, or the group of order 2 where $r=1, s=-1$ is a pair satisfying the equations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Truth in all groups is a semantic definition and can be replaced by an equivalent, syntactic definition that gives the combinatorial rules for making all valid deductions from the defining relations.  $r=s=1$ is not a valid algebraic consequence of the algebraic rules.   As in all problems of logic, showing that something does NOT follow from some rules or axioms is a matter of constructing an example (a model) where the rules apply but the assertion is false.  Here the models are groups with a pair of elements $r,s$ satisfying the relations, so for showing what is not true in the logic of generators and relations the algebraic approach doesn't provide an alternative to constructing models.  For assertions that &lt;em&gt;do&lt;/em&gt; follow from the defining relations, the algebraic definition is useful, because instead of surveying a possibly infinite collection of models to show that the statement is correct, one can perform a finite combinatorial derivation that works for all models.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Added in response to comment from Matt E about normal forms: showing that any element can be placed in the normal form follows from the algebraic relations alone.  However, to use normal forms as a method of calculating in the group or understanding it, you need to show that distinct elements in the group have different normal forms. This is again the non-algebraic (or not purely algebraic) problem of showing that relations do not hold in a structure, which is done by constructing models such as actions of the group on particular spaces.)&lt;/p&gt;&#xA;" OwnerUserId="467" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-09-02T20:50:36.947" LastActivityDate="2010-09-02T20:50:36.947" CommentCount="2" />
  <row Id="3813" PostTypeId="2" ParentId="3805" CreationDate="2010-09-01T19:24:26.697" Score="3" ViewCount="" Body="&lt;p&gt;It appears that you may mistakenly believe that every element of your finite field is a power of $\rm x$. Here is a very simple counterexample: put $\rm\; f(x) = (x^5-1)/(x-1) = x^4 + x^3 + x^2 + x + 1 \;$, which is irreducible $\pmod 2$. Then $\rm\; (x-1) \: f(x) = x^5 - 1 \Rightarrow \rm x^5 \equiv 1 \pmod{f(x)}$. This implies that the powers of $\rm x$ generate only 5 elements, not all 15 nonzero elements of $\rm GF(2^4)$ constructed with $\rm f(x)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact there does exist an element whose powers generate all the nonzero elements of the finite field (said more technically: the multiplicative group is cyclic). Such generators are known as &lt;strong&gt;primitive elements&lt;/strong&gt;. However, generally there is no simple closed-form known for such generators.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason that you didn't see this in $\rm GF(4)$ or $\rm GF(8)$ is because their multiplicative groups have prime order $3$ and $7,$ so every element $\ne 1,0 \:$ is a generator (by Lagrange's theorem the order of an element divides the order of the group which, being a prime $\rm p,$ forces the order to be $\rm p\:$ for elements $\rm \ne 1,0\:$).&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-09-01T21:33:50.253" LastActivityDate="2010-09-01T21:33:50.253" />
  <row Id="3815" PostTypeId="1" AcceptedAnswerId="3817" CreationDate="2010-09-01T20:28:38.880" Score="5" ViewCount="547" Body="&lt;p&gt;Proposition: For a set $X$ and its power set $P(X)$, any function $f\colon P(X)\to X$ has at least two sets $A\neq B\subseteq X$ such that $f(A)=f(B)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can see how this would be true if $X$ is a finite set, since $|P(X)|\gt |X|$, so by the pidgeonhole principle, at least two of the elements in $P(X)$ would have to map to the same element. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does this proposition still hold for $X$ an infinite set? And if so, how does this show that the set of all singleton sets cannot exist?&lt;/p&gt;&#xA;" OwnerUserId="1583" LastEditorUserId="1583" LastEditorDisplayName="" LastEditDate="2010-09-11T02:11:33.977" LastActivityDate="2011-03-30T16:09:44.193" Title="Why does the set of all singleton sets not exist?" Tags="&lt;set-theory&gt;" AnswerCount="5" CommentCount="1" />
  <row Id="3817" PostTypeId="2" ParentId="3815" CreationDate="2010-09-01T20:43:50.437" Score="10" ViewCount="" Body="&lt;p&gt;If you had a function $f\colon P(X)\to X$ such that $f(A)=f(B)\Rightarrow A=B$, then $f$ would be one-to-one, which would imply $|P(X)|\leq |X|$; since $|X|\lt |P(X)|$ holds by Cantor's Theorem, that would give you $|P(X)|\lt|P(X)|$, which is impossible. This argument does not use finiteness of $X$, so it holds for $X$ infinite as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose that the collection of all singletons is a set $X$. Define a map $f\colon P(X)\to X$ by $f(A) = \{A\}$; this is well defined, since $\{A\}$ is a singleton. If $A\neq B$, then $\{A\}\neq\{B\}$, so $f$ would be one-to-one. Since, by the proposition, this is impossible, we conclude that our assumption that the collection of all singletons is a set must be false. Thus, there is no &quot;set of all singletons.&quot;&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-09-01T20:43:50.437" CommentCount="2" />
  <row Id="3818" PostTypeId="2" ParentId="3750" CreationDate="2010-09-01T21:16:48.140" Score="2" ViewCount="" Body="&lt;p&gt;Looking &lt;a href=&quot;http://math.stackexchange.com/questions/3750/aunt-and-uncles-fuel-oil-tank-dip-stick-problem/3757#3757&quot;&gt;@Isaac's answer&lt;/a&gt;, the appearance of &quot;$\alpha - \sin\alpha$&quot; and &quot;$1-\cos\frac{\alpha}{2}$&quot; made me think that power series would come in handy, with their respective initial terms vanishing. They didn't turn out to be as handy as I might have hoped.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Write $w$ for the ratio of oil to the tank's capacity, and $d$ for the ratio of the oil's depth to the radius of the tank. Then, by Isaac's work ...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$2\pi w = \alpha - \sin\alpha = \alpha-\sum_{m=0}^\infty\frac{(-1)^m}{(2m+1)!}\alpha^{2m+1}= \sum_{m=1}^\infty\frac{(-1)^{m+1}}{(2m+1)!}\alpha^{2m+1}$$&#xA;$$d = 1 - \cos\frac{\alpha}{2}=1-\sum_{m=0}^\infty\frac{(-1)^m}{(2m)!}\left(\frac{\alpha}{2}\right)^{2m}=\sum_{m=1}^\infty\frac{(-1)^{m+1}}{(2m)!}\left(\frac{\alpha}{2}\right)^{2m}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Write $w_n$ and $d_n$ for the values associated with lopping off the power series at the $n$-th term (that is, from $m=1$ to $m=n$). In particular, we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$2\pi w_1 = \frac{\alpha^3}{3!}$$&#xA;$$d_1 = \frac{\alpha^2}{2^2 2!}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Eliminating $\alpha$ yields the relation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$32 d_1^3 = 9 w_1^2 \pi^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This relation, though, isn't very accurate (as shown by the $n=1$ curve in the diagram).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/6aIcm.png&quot; alt=&quot;Ratio of depth to radius, vs ratio of volume to capacity&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we take $n=2$ ...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$2\pi w_2 = \frac{\alpha^3}{3!}-\frac{\alpha^5}{5!}$$&#xA;$$d_2 = \frac{\alpha^2}{2^2 2!}-\frac{a^4}{2^4 4!}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;... then eliminating $\alpha$ isn't quite so easy unless you do something like invoke Mathematica's &lt;code&gt;Resultant&lt;/code&gt; function, which yields&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$4718592 d_2^5 - 13762560 d_2^4 + 10035200 d_2^3 - 768000 d_2^2 w_2^2 \pi^2 + &#xA; 3024000 d_2 w_2^2 \pi^2 - 2822400 w_2^2 \pi^2 + 1875 w_2^4 \pi^4 = 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also not terribly accurate at we get close to $w=0.5$ (which should have $d=1.0$). With $n=3$ we're very close, and with $n=4$ and above, we've pretty much converged on our target (at least according to the graph). I won't give the $n=4$ equation (the coefficient on $d^9$ is $2^{71} 5^4 7^2$), but here are some data points:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(0.0, 0.000000)$$&#xA;$$(0.1, 0.312953)$$&#xA;$$(0.2, 0.508164)$$&#xA;$$(0.3, 0.680452)$$&#xA;$$(0.4, 0.842809)$$&#xA;$$(0.5, 1.001730)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, power series and resultants aren't exactly &quot;lowest possible grade level&quot;, but this is the best I can do at this point. &lt;/p&gt;&#xA;" OwnerUserId="409" LastActivityDate="2010-09-01T21:16:48.140" CommentCount="1" />
  <row Id="3819" PostTypeId="2" ParentId="3782" CreationDate="2010-09-01T23:06:25.530" Score="11" ViewCount="" Body="&lt;p&gt;Of course everybody has their own learning style.  Here are some general suggestions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Find a &lt;strong&gt;teacher&lt;/strong&gt;.  It is &lt;em&gt;hard&lt;/em&gt; to learn mathematics on your own until you have reached a certain level of mathematical sophistication; nobody is there to tell you what is important and what is unimportant.  Take courses at a university; as Agusti Roig mentioned, video lectures on MIT's &lt;a href=&quot;http://ocw.mit.edu/index.htm&quot; rel=&quot;nofollow&quot;&gt;OpenCourseWare&lt;/a&gt; are a good cheap alternative.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Read&lt;/strong&gt; as much mathematics as you possibly can, from as many sources as you possibly can.  This is not limited to textbooks but extends to popular math books, blogs, expository papers, MO, math.SE... doing this will get you used to not understanding things, which is important.  It will also expose you to many fascinating ideas that will fire up your curiosity enough for you to look at the material more seriously.  As &lt;a href=&quot;http://math.stanford.edu/~vakil/potentialstudents.html&quot; rel=&quot;nofollow&quot;&gt;Ravi Vakil&lt;/a&gt; says:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;...mathematics is so rich and infinite that it is impossible to learn it systematically, and if you wait to master one topic before moving on to the next, you'll never get anywhere. Instead, you'll have tendrils of knowledge extending far from your comfort zone. Then you can later backfill from these tendrils, and extend your comfort zone; this is much easier to do than learning &quot;forwards&quot;. &lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A specific way in learning backwards is easier than learning forwards is that instead of reading the proof of a theorem in a book, you might hear about a theorem without proof, but remember that someone on a blog said something vague about a crucial step, then gradually learn enough material that suddenly you can work out the proof independently.  I have done this a handful of times, and it is quite satisfying. For example, the theorem I proved in &lt;a href=&quot;http://qchu.wordpress.com/2010/04/27/the-mckay-correspondence-i/&quot; rel=&quot;nofollow&quot;&gt;this blog post&lt;/a&gt; is classical and extremely well-known, but I had never seen a proof of it.  I juggled around some ideas for about half a year until I figured out how to prove Lemma 6 (which I saw in a paper somewhere, again without proof), and I wrote down a proof.  Later I read a proof in an actual book, and although the second half of the proof was similar, it did not use Lemma 6.  I have yet to see a proof of Lemma 6 in print, although I am sure it is also well-known.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This might sound like more work.  But guess how well I &lt;em&gt;remember&lt;/em&gt; this theorem and its proof now!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Do&lt;/strong&gt; as much mathematics as you possibly can.  This is not limited to textbook exercises but includes competition problems, finding alternate proofs of theorems, working out concrete examples of abstract theorems, etc.  I try to do this as much as I can on &lt;a href=&quot;http://qchu.wordpress.com/&quot; rel=&quot;nofollow&quot;&gt;my blog&lt;/a&gt;; it keeps me sharp and is also, at least for me, much more fun than reading a textbook, which I can't do for long periods of time.  This is also why I post here so often.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question&lt;/strong&gt; everything.  There are a few aspects to this.  If something is unclear or unmotivated to you, ask yourself exactly where it becomes unclear or unmotivated.  Find someone to explain it to you (for example, on math.SE!).  Read a blog post about it.  &lt;em&gt;Write&lt;/em&gt; a blog post about it!  Ask yourself how things generalize and how they connect to other things you know.  (Again, math.SE is good for this.)  The worst thing you can do is to accept what a textbook tells you as the Word of God.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, &lt;strong&gt;teach&lt;/strong&gt; as much mathematics as you possibly can.  This is the other purpose of my blog, and is an amazing test of how well you understand something.  You would be surprised how much you can learn about something by teaching it.  &lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-09-01T23:14:49.710" LastActivityDate="2010-09-01T23:14:49.710" CommunityOwnedDate="2010-09-01T23:06:25.530" CommentCount="6" />
  <row Id="3820" PostTypeId="1" AcceptedAnswerId="3822" CreationDate="2010-09-01T23:13:34.313" Score="6" ViewCount="203" Body="&lt;p&gt;Must a polynomial function $f \in \mathbb{R}[x_1, \ldots, x_n]$ that's lower bounded by some $\lambda \in \mathbb{R}$ have a global minimum over $\mathbb{R}^n$?&lt;/p&gt;&#xA;" OwnerUserId="848" LastActivityDate="2010-09-02T02:29:43.950" Title="Does a polynomial that's bounded below have a global minimum?" Tags="&lt;optimization&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" />
  <row Id="3821" PostTypeId="2" ParentId="3623" CreationDate="2010-09-01T23:18:15.053" Score="1" ViewCount="" Body="&lt;p&gt;SIAM (The Society for Industrial and Applied Mathematics) has collected some oral interviews and other materials that have some bearing on the history of ordinary and partial differential equations:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://history.siam.org/&quot; rel=&quot;nofollow&quot;&gt;http://history.siam.org/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is also a lot of material on this topic that is related to more specialized historical studies:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;History of Hydraulics by Hunter Rouse and Simon Ince&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A History of the Theory of Elasticity and of the Strength of Materials ...&#xA;by Isaac Todhunter&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and work on the Euler equations and the Navier-Stokes equations.&lt;/p&gt;&#xA;" OwnerUserId="1369" LastActivityDate="2010-09-01T23:18:15.053" CommunityOwnedDate="2010-09-01T23:18:15.053" />
  <row Id="3822" PostTypeId="2" ParentId="3820" CreationDate="2010-09-01T23:25:02.963" Score="15" ViewCount="" Body="&lt;p&gt;No.  The polynomial $f(x, y) = (1 - xy)^2 + x^2$ is bounded below by $0$, cannot actually take the value $0$, but can take the value $\epsilon$ for any $\epsilon &gt; 0$.  I learned this example from &lt;a href=&quot;http://mathoverflow.net/questions/23478/examples-of-common-false-beliefs-in-mathematics/30979#30979&quot; rel=&quot;nofollow&quot;&gt;Richard Dore&lt;/a&gt; on MO.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-09-01T23:25:02.963" />
  <row Id="3823" PostTypeId="2" ParentId="2949" CreationDate="2010-09-01T23:49:43.943" Score="22" ViewCount="" Body="&lt;p&gt;$e^{i\pi} +1 = 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This still blows my mind.&lt;/p&gt;&#xA;" OwnerUserId="172" LastActivityDate="2010-09-01T23:49:43.943" CommunityOwnedDate="2010-09-01T23:49:43.943" CommentCount="2" />
  <row Id="3824" PostTypeId="2" ParentId="2949" CreationDate="2010-09-01T23:51:28.883" Score="11" ViewCount="" Body="&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Nine-point_circle&quot; rel=&quot;nofollow&quot;&gt;nine point circle&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Three sets of three points, each of which obviously determines a circle. That these three constructions &lt;em&gt;always&lt;/em&gt; give the &lt;strong&gt;same&lt;/strong&gt; circle!?&lt;/p&gt;&#xA;" OwnerUserId="172" LastActivityDate="2010-09-01T23:51:28.883" CommunityOwnedDate="2010-09-01T23:51:28.883" />
  <row Id="3825" PostTypeId="2" ParentId="3782" CreationDate="2010-09-01T23:54:22.107" Score="7" ViewCount="" Body="&lt;p&gt;I don't consider myself anywhere near an expert at this (there are a lot of things that I've tried to learn for a while but failed). However, I have spent a while trying to learn mathematics independently, so I will weigh in, but this should be taken with the usual truckload of salt.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Find a good book! It's easy to waste time with a poorly written textbook. I think this one is really important. Different people have different ideas about what constitutes a well-written textbook, though, so it may be desirable to look at books in a library and get a sense of them than simply following someone else's recommendation.&lt;/li&gt;&#xA;&lt;li&gt;Talk to people. I didn't realize how little I knew about algebraic geometry until I was having an email chat with someone and confused the universal properties of projective and affine space minus the origin.  That sorted out a few things in my mind.&lt;/li&gt;&#xA;&lt;li&gt;Don't worry about skipping over things that are unimportant. But try to ask someone what the important and unimportant things are.&lt;/li&gt;&#xA;&lt;li&gt;This may be just a personal preference, but I like to make copious amounts of notes when I learn something, starting from the beginning. This is one reason to keep a mathematical blog. However, creating large sets of notes is often not entirely suitable for blogging.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="536" LastActivityDate="2010-09-01T23:54:22.107" CommunityOwnedDate="2010-09-01T23:54:22.107" CommentCount="3" />
  <row Id="3826" PostTypeId="2" ParentId="2949" CreationDate="2010-09-01T23:56:34.847" Score="22" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Riemann_series_theorem&quot; rel=&quot;nofollow&quot;&gt;Riemann's rearrangement theorem.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is responsible for the counter-intuitive results of, for example &lt;a href=&quot;http://math.stackexchange.com/questions/107/paradox-increasing-sequence-that-goes-to-0&quot;&gt;this&lt;/a&gt; and &lt;a href=&quot;http://math.stackexchange.com/questions/3583/infinite-processes-riddle&quot;&gt;this&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="172" LastActivityDate="2010-09-01T23:56:34.847" CommunityOwnedDate="2010-09-01T23:56:34.847" CommentCount="2" />
  <row Id="3827" PostTypeId="2" ParentId="2949" CreationDate="2010-09-02T00:05:16.590" Score="11" ViewCount="" Body="&lt;p&gt;Fermat's &quot;two square theorem&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;G.H. Hardy's &lt;em&gt;A Mathematician's Apology&lt;/em&gt; is a book everyone should read, but for those who haven't here's something Hardy mentions that is rather surprising:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(If we ignore 2) All primes fit into two classes: those that leave remainder $1$ when divided by $4$ and those that leave remainder $3$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This much is obvious. The surprising thing is that all of the first class, and none of the second can be expressed as the sum of two integer squares.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is, for all prime $p$, if $p = 1 \mod 4$ then there exist $x,y$ integers such that $p = x^2 +y^2$ and if $p = 3 \mod 4$ there exists no such $x,y$&lt;/p&gt;&#xA;" OwnerUserId="172" LastActivityDate="2010-09-02T00:05:16.590" CommunityOwnedDate="2010-09-02T00:05:16.590" CommentCount="2" />
  <row Id="3828" PostTypeId="2" ParentId="3815" CreationDate="2010-09-02T00:06:16.827" Score="10" ViewCount="" Body="&lt;p&gt;If $x$ is a set containing all singletons, $\cup x$ is a set containing all sets, which leads to Russel's paradox. Thus, in $ZF$ there is no such $x$.&lt;/p&gt;&#xA;" OwnerUserId="1650" LastActivityDate="2010-09-02T00:06:16.827" CommentCount="3" />
  <row Id="3829" PostTypeId="2" ParentId="3815" CreationDate="2010-09-02T00:26:06.803" Score="1" ViewCount="" Body="&lt;p&gt;If you accept that the collection of all sets (denoted by $V$ usually) is not a set (but in fact a proper class) then if you look at the function $f(x) = \{ x \}$ it is 1-1 from $V$ into the sets of all singletons, thus resulting that $V$ would be a set, in contradiction to the fact that it is in fact a proper class.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But of course you'd have to accept that $V$ is a proper class as well, this is simpler and follows by the argument given in Arturo's answer (that the power-set of $V$ would be of the same cardinality).&lt;/p&gt;&#xA;" OwnerUserId="622" LastActivityDate="2010-09-02T00:26:06.803" />
  <row Id="3830" PostTypeId="1" AcceptedAnswerId="3885" CreationDate="2010-09-02T00:28:36.023" Score="2" ViewCount="103" Body="&lt;p&gt;Let $\kappa$ a cardinal of cofinality $\omega$; let $C \subseteq \kappa$ be a unbounded countable subset. Why is then $C$ closed (and thus a c.u.b.)? This means that if $\delta &amp;lt; \kappa$ is a limit ordinal, and $C \cap \delta$ is unbounded in $\delta$, then $\delta \in C$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I doubt that this is true; however, it's a short statement in Kunen's set theory book.&lt;/p&gt;&#xA;" OwnerUserId="1650" LastEditorUserId="1650" LastEditorDisplayName="" LastEditDate="2010-09-02T12:51:04.180" LastActivityDate="2010-09-02T17:15:02.703" Title="Example of a c.u.b. set" Tags="&lt;logic&gt;&lt;cardinals&gt;" AnswerCount="2" />
  <row Id="3832" PostTypeId="2" ParentId="3768" CreationDate="2010-09-02T00:58:02.623" Score="1" ViewCount="" Body="&lt;p&gt;A very natural explanation of the correct definition for the values of min and max on empty sets arises from their dual &lt;strong&gt;universal&lt;/strong&gt; definitions - analogous to the universal GCD, LCM definitions that I presented in a &lt;a href=&quot;http://math.stackexchange.com/questions/3149&quot;&gt;post here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First some notation. $\;$ Write $\ \ \rm x \le S \;\iff\; x \le s,\;\: \forall\: s \in S,\;$ and dually for $\rm\; x \ge S$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;DEFINITION&lt;/strong&gt; of $\:$ &lt;strong&gt;min&lt;/strong&gt; $\quad$ $\quad\rm x \le S \;\iff\; x \le min\ S$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;DEFINITION&lt;/strong&gt; of &lt;strong&gt;max&lt;/strong&gt; $\:\quad$ $\quad\rm x \ge S \;\iff\; x \ge max\ S$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For min, when $\;\rm S = \emptyset\;$ is empty, the first clause  $\;\rm x \le S\;$ in the min definition is vacuously true, hence the definition reduces to $\;\rm x \le \min \emptyset\;$ for all $\;\rm x\;$. Hence $\;\rm \min \emptyset = \infty\;$. Dually $\;\rm \max\emptyset = -\infty\;$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As I remarked in said GCD, LCM post, such universal definitions often facilitate slick proofs. For some nontrivial examples of min, max flavor consider the slick proofs of the integrality of various products of binomial coefficients by employing the floor function, e.g. see Joe Roberts: &lt;em&gt;Elementary Number Theory&lt;/em&gt;. Instead I close with an analogous slick GCD, LCM proof from my &lt;a href=&quot;http://math.stackexchange.com/questions/3149&quot;&gt;mentioned post&lt;/a&gt; (see it for further details).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Generally, in any domain, we have the following dual &lt;strong&gt;universal&lt;/strong&gt; definitions of LCM and GCD:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;DEFINITION&lt;/strong&gt; of &lt;strong&gt;LCM&lt;/strong&gt; $\quad$ If $\quad\rm a,b\ |\ c \;\iff\; [a,b]\ |\ c \quad$ then $\quad\rm [a,b] \;\;$ is an &lt;strong&gt;LCM&lt;/strong&gt; of $\;\rm a,b$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;DEFINITION&lt;/strong&gt; of &lt;strong&gt;GCD&lt;/strong&gt; $\quad$ If $\quad\rm c\ |\ a,b \;\iff\; c\ |\ (a,b) \quad$ then $\quad\rm (a,b) \;$ is an &lt;strong&gt;GCD&lt;/strong&gt; of $\;\;\rm a,b$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: that $\;\rm a,b\ |\ [a,b] \;$ follows by putting $\;\rm c = [a,b] \;$ in the definition. Dually $\;\rm (a,b)\ |\ a,b \;$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Such $\iff$ definitions provide slick unified proofs of both arrow directions, e.g. the fundamental&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;THEOREM&lt;/strong&gt;  $\rm\quad (a,b)\ =\ ab/[a,b] \;\;$  if $\;\rm\ [a,b] \;$  exists.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof&lt;/strong&gt;:  $\rm\quad\quad d\ |\ a,b \;\iff\; a,b\ |\ ab/d \;\iff\; [a,b]\ |\ ab/d \;\iff\; d\ |\ ab/[a,b] \quad\;\;$ &lt;strong&gt;QED&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The conciseness of this proof arises by exploiting to the hilt the $\iff$ definition of LCM and GCD.  Compare to less concise / general / illuminating proofs in many number theory textbooks.  &lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2011-02-01T15:23:03.620" LastActivityDate="2011-02-01T15:23:03.620" CommentCount="7" />
  <row Id="3833" PostTypeId="2" ParentId="3830" CreationDate="2010-09-02T00:59:35.310" Score="1" ViewCount="" Body="&lt;p&gt;If you set $\kappa = \aleph_{\omega}$ and let $C = \{0,,1,\aleph_1, 2, \aleph_2,...\}$ (so $C$ is the union of all natural numbers together with all $\aleph_n$ for $n&gt;0$) and you let $\delta = \omega$, I think you have a counterexample.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$C\cap\delta = \delta$, so is automatically unbounded in $\delta$.  But $\delta$ itself isn't in $C$ by construction.&lt;/p&gt;&#xA;" OwnerUserId="331" LastActivityDate="2010-09-02T00:59:35.310" CommentCount="1" />
  <row Id="3835" PostTypeId="2" ParentId="3504" CreationDate="2010-09-02T01:52:26.333" Score="1" ViewCount="" Body="&lt;p&gt;So we have the equations&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(b_1+b_2)^2=8(c_1+c_2)$ (i)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(b_1+b_3)^2=8(c_1+c_3)$ (ii)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(b_1+b_4)^2=8(c_1+c_4)$ (iii)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(b_2+b_3)^2=8(c_2+c_3)$ (iv)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(b_2+b_4)^2=8(c_2+c_4)$ (v)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(b_3+b_4)^2=8(c_3+c_4)$ (vi)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Substracting (ii) from (i) we get &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(b_2-b_3)(b_2+b_3+2b_1)=8(c_2-c_3)$ (1)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and substracting (vi) from (v) we get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(b_2-b_3)(b_2+b_3+2b_4)=8(c_2-c_3)$ (2)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, substracting (2) from (1) we get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$2(b_2-b_3)(b_1-b_4)=0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a similar fashion we arrive at&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$2(b_2-b_1)(b_3-b_4)=0$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$2(b_1-b_3)(b_2-b_4)=0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The last three equations show that $b_1=b_2=b_3=b_4$. Using (i) through (vi), we see that $c_1=c_2=c_3=c_4$. &lt;/p&gt;&#xA;" OwnerUserId="1313" LastActivityDate="2010-09-02T01:52:26.333" CommentCount="2" />
  <row Id="3836" PostTypeId="2" ParentId="3723" CreationDate="2010-09-02T01:59:08.120" Score="2" ViewCount="" Body="&lt;p&gt;The solution you have suggested is a perfectly good one. The only problem is that you are venturing outside the bounds of conventional Distribution theory. Consequently, the onus is on you to prove whatever properties of your definition that you use.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Actually, the &lt;a href=&quot;http://en.wikipedia.org/wiki/Heaviside_step#Zero_argument&quot; rel=&quot;nofollow&quot;&gt;Heaviside step function&lt;/a&gt; is more usually treated as a distribution itself. Bracewell's book, &quot;The Fourier Transform and its Applications&quot; considers the problem of defining a value at the discontinuity and concludes that it mostly doesn't matter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you are doing seems somewhat similar to the idea of density in physics. I'd be surprised if the physicists or the electrical engineers (signal processing) people hadn't already confronted this issue. However, after looking around a little bit I am unable to find anything specific.&lt;/p&gt;&#xA;" OwnerUserId="1123" LastActivityDate="2010-09-02T01:59:08.120" />
  <row Id="3837" PostTypeId="2" ParentId="3820" CreationDate="2010-09-02T02:29:43.950" Score="1" ViewCount="" Body="&lt;p&gt;@Agusti Roig. $f$ has no critical points in the interior of the disc, so you have to look for the minimum at the unit circle. One way would be to parametrize $x=\cos\theta, y=\sin\theta$, and minimize.&lt;/p&gt;&#xA;" OwnerUserId="1313" LastActivityDate="2010-09-02T02:29:43.950" CommentCount="2" />
  <row Id="3838" PostTypeId="1" AcceptedAnswerId="3843" CreationDate="2010-09-02T02:34:21.520" Score="2" ViewCount="121" Body="&lt;p&gt;Let $\Pi_1 : 7x-5y-2z=0, \Pi_2 : 5x-4y-z=0$ and $\mathbb{L}$ the line that passes through points $P=(-2,3,-3)$ and $Q=(-1,2,-1)$. Find all planes $\Pi$ that satisfy:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\Pi \cap \Pi_1 \cap \Pi_2 = \emptyset$&lt;/li&gt;&#xA;&lt;li&gt;$d(R,\Pi)=\sqrt{14} \forall R \in \mathbb{L}$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So far, I've found out that $\Pi_1\cap\Pi_2= { X : \lambda (1,1,1) } $ and $\mathbb{L} : X=\lambda(1,-1,2)+(-1,2,-1)$; given $\Pi : NX = NQ$, with $N$ being an orthogonal vector to the plane and $Q$ being any point in $\Pi$, it follows that $N \perp(1,1,1)$ and $N\perp(1,-1,2)$, so we can define $N=(1,1,1)\times(1,-1,2)=(3,-1,-2)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As $d(R,\Pi) = \frac{|N(R-Q)|}{||N||} = \sqrt{14}$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{|(3,-1,-2)((-3\lambda-1,\lambda+2,-4\lambda-1)-(x,y,z))|}{||(3,-1,-2)||} = \sqrt{14}$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$|(3,-1,-2)(-3\lambda-1-x,\lambda+2-y,-4\lambda-1-z)| = 1$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$-9\lambda-3-3x -\lambda-2+y +8\lambda+2+2z = \pm 1$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$-3x+y+2z-2\lambda-3 = \pm 1$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;and I simply don't know what to do next. Any help would be really appreciated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Solved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$d(R,\Pi) = \frac{|N(R-Q)|}{||N||} = \sqrt{14}$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(3,-1,-2)((-1,2,1)-(x,y,z)) = \pm 1$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$-3x+y+2z-7=\pm 1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we can take $Q$ as either $(0,4,2)$ or $(0,2,2)$ and we get two different planes:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Pi_3: 3x-y-2z=-8$ and $\Pi_4: 3x-y-2z=-6$. Once again, thanks a lot.&lt;/p&gt;&#xA;" OwnerUserId="1523" LastEditorUserId="1523" LastEditorDisplayName="" LastEditDate="2010-09-02T18:16:46.280" LastActivityDate="2010-09-02T18:16:46.280" Title="Finding all planes that satisfy two conditions." Tags="&lt;linear-algebra&gt;&lt;homework&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="3839" PostTypeId="2" ParentId="2949" CreationDate="2010-09-02T02:49:53.590" Score="4" ViewCount="" Body="&lt;p&gt;While not as surprising as, say, the countability of the rationals, and even fairly obvious to some people, the fundamental theorem of calculus joins two operations (differentiation and integration) which didn't look completely related to each other at first to me if you define them as the rate of change of a curve and the area beneath it.&lt;/p&gt;&#xA;" OwnerUserId="1523" LastActivityDate="2010-09-02T02:49:53.590" CommunityOwnedDate="2010-09-02T02:49:53.590" CommentCount="1" />
  <row Id="3841" PostTypeId="1" AcceptedAnswerId="3842" CreationDate="2010-09-02T03:26:51.857" Score="3" ViewCount="121" Body="&lt;p&gt;&lt;strong&gt;Question&lt;/strong&gt; If $p &gt; 3$ is a prime and $$ 1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{p} = \frac{a}{b}$$ then prove that $p^{4} \mid (ap-b)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is an exercise in Herstein which states, if $p &gt; 3$ is prime and if $$1 + \frac{1}{2} + \cdots + \frac{1}{p-1} = \frac{a}{b}$$ then $p^{2} \mid a$. (Page 116, Problem 2). But i am not sure whether this helps or not. Hints, suggestions would be of great help! &lt;/p&gt;&#xA;" LastActivityDate="2010-09-02T15:32:33.113" Title="If a prime $p$ satisfies given condition then $p^{4} \mid ap-b$" Tags="&lt;number-theory&gt;&lt;prime-numbers&gt;" AnswerCount="2" />
  <row Id="3842" PostTypeId="2" ParentId="3841" CreationDate="2010-09-02T03:42:39.203" Score="4" ViewCount="" Body="&lt;p&gt;$\frac{a}{b} = \frac{p^{2}\alpha}{\beta} + \frac{1}{p}$ where $(\alpha,\beta) = (p,\beta) = 1$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It follows immediately from this I believe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the exercise helps.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Related: &lt;a href=&quot;http://en.wikipedia.org/wiki/Wolstenholme%27s_theorem&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Wolstenholme%27s_theorem&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, a hint for the exercise: consider $f(x) = (x-1)(x-2)\cdots(x-p+1)$. Does $p^2$ divide $f'(0)$ (derivative of f at 0)? (This is a classic proof of that exercise and should be available in textbooks)&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-09-02T03:59:11.313" LastActivityDate="2010-09-02T03:59:11.313" CommentCount="0" />
  <row Id="3843" PostTypeId="2" ParentId="3838" CreationDate="2010-09-02T03:57:35.703" Score="2" ViewCount="" Body="&lt;p&gt;Martín, I think you can take any value of $\lambda$, because the distance between R and $\Pi$ doesn't change if R is a point of L, since you made sure L is parallel to $\Pi$. I'm not sure, though. (Btw, I suppose the first condition is actually $\Pi\cap(\Pi_1\cap\Pi_2)$.) &lt;/p&gt;&#xA;" OwnerUserId="1313" LastActivityDate="2010-09-02T03:57:35.703" CommentCount="3" />
  <row Id="3845" PostTypeId="2" ParentId="3841" CreationDate="2010-09-02T04:29:03.043" Score="0" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\rm\;\;\displaystyle p^4\:\big|\:ap-b = \:bp\bigg(\frac{a}{b} - \frac{1}{p}\bigg)\;\;$ via $\rm\;\; p^2|bp\;\;$ and $\rm\displaystyle\; p^2\:\big|\:\bigg(\frac{a}{b} - \frac{1}{p}\bigg)\;\;$ via &lt;a href=&quot;http://books.google.com/books?id=rey9wfSaJ9EC&amp;amp;pg=PA113&quot; rel=&quot;nofollow&quot;&gt;Wolstenholme&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-09-02T15:32:33.113" LastActivityDate="2010-09-02T15:32:33.113" />
  <row Id="3846" PostTypeId="2" ParentId="3815" CreationDate="2010-09-02T04:38:26.490" Score="0" ViewCount="" Body="&lt;p&gt;The question in the title is a consequence of the &lt;a href=&quot;http://books.google.com/books?id=sxr4LrgJGeAC&amp;amp;pg=PA53&amp;amp;dq=Axiom+of+regularity&amp;amp;hl=en&amp;amp;ei=lyh_TKLtJcPGswayz53lDw&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=1&amp;amp;ved=0CCkQ6AEwAA#v=onepage&amp;amp;q=Axiom%20of%20regularity&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;axiom of regularity&lt;/a&gt;.&#xA;In fact If there was such a set $X$, then $\{X\}\in X$. But obviously we always have $X\in\{X\}$. Hence, we would be able to construct an infinite downward chain of sets which contradicts the  axiom of regularity.&lt;/p&gt;&#xA;" OwnerUserId="1154" LastEditorUserId="1154" LastEditorDisplayName="" LastEditDate="2010-09-02T04:58:12.737" LastActivityDate="2010-09-02T04:58:12.737" CommentCount="3" />
  <row Id="3847" PostTypeId="1" AcceptedAnswerId="3848" CreationDate="2010-09-02T06:05:37.420" Score="0" ViewCount="248" Body="&lt;p&gt;How can $n!$ be represented in terms of $2^{n}$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I mean I want function $f$ such that $$n! = f(2^{n})$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, I would like to know $$\frac{n!}{2^n} = g(n) = ?$$&lt;/p&gt;&#xA;" OwnerUserId="705" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-09-05T04:25:00.143" LastActivityDate="2010-09-05T08:42:29.490" ClosedDate="2010-09-05T06:23:37.757" Title="How can $n!$ be represented in terms of $2^{n}$?" Tags="&lt;algebra-precalculus&gt;&lt;functions&gt;" AnswerCount="1" CommentCount="10" />
  <row Id="3848" PostTypeId="2" ParentId="3847" CreationDate="2010-09-02T06:10:39.773" Score="7" ViewCount="" Body="Try $$f(t) = \Gamma ( 1 + \log_2 t ), $$ where $\log_2$ is the base 2 logarithm." OwnerUserId="960" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2010-09-05T08:42:29.490" LastActivityDate="2010-09-05T08:42:29.490" CommentCount="3" />
  <row Id="3849" PostTypeId="1" AcceptedAnswerId="3851" CreationDate="2010-09-02T06:22:57.290" Score="1" ViewCount="157" Body="&lt;p&gt;Let $G$ be a finite group, and let $p^{\alpha} \mid |G|$, where $p$ is a prime. Now does this imply $p \mid |Aut(G)|$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clearly if $|G| \leq 2$, then the Automorphism group is the trivial group, so one can see that this need not be true for $\alpha =1$. I am curious to know for higher powers of $\alpha$.&lt;/p&gt;&#xA;" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-09-05T04:24:40.170" LastActivityDate="2010-09-05T04:24:40.170" Title="Let $p$ be a prime, then does $p^{\alpha} \mid |G| \Longrightarrow p \mid Aut(G)$?" Tags="&lt;abstract-algebra&gt;&lt;group-theory&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="1" />
  <row Id="3850" PostTypeId="2" ParentId="3849" CreationDate="2010-09-02T06:32:32.950" Score="3" ViewCount="" Body="&lt;p&gt;Cyclic.            &lt;/p&gt;&#xA;&#xA;&lt;p&gt;[15char]&lt;/p&gt;&#xA;" OwnerUserId="467" LastActivityDate="2010-09-02T06:32:32.950" CommentCount="2" />
  <row Id="3851" PostTypeId="2" ParentId="3849" CreationDate="2010-09-02T06:53:04.310" Score="6" ViewCount="" Body="&lt;p&gt;Let $P$ be a Sylow p-subgroup of $G$.  Now $G/Z(G)$ embeds in $Aut(G)$, so we can reduce to the case where $P\le Z(G)$, so $G=P\times H$, where $(|H|,p)=1$, and $P$ is abelian.  Now if $\alpha&gt;1$, then there will always be an element of order $p$ in $Aut(P)\subset Aut(G)$.  You can see this by looking at cyclic groups of order $p^m$ with $m&gt;1$, and elementary abelian p-groups of dimension greater than $1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Steve&lt;/p&gt;&#xA;" OwnerUserId="641" LastActivityDate="2010-09-02T06:53:04.310" CommentCount="4" />
  <row Id="3852" PostTypeId="1" AcceptedAnswerId="3860" CreationDate="2010-09-02T07:04:36.537" Score="22" ViewCount="2592" Body="&lt;p&gt;If A and B are square matrix such that AB = I where I is identity matrix. Show that BA = I. I do not understand anything more than the following.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Elementary row operations.&lt;/li&gt;&#xA;&lt;li&gt;Linear dependence.&lt;/li&gt;&#xA;&lt;li&gt;Row reduced forms and their relations with original matix.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;If Entries of matrix are not from a mathematical structure which supports commutativity, What can we say about this prob?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS : Using transform and/or inverse of a matrix MAY invite a death penalty.&lt;/p&gt;&#xA;" OwnerUserId="1674" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2011-02-22T23:54:44.987" LastActivityDate="2011-02-22T23:54:44.987" Title="If $AB = I$ then $BA = I$" Tags="&lt;linear-algebra&gt;" AnswerCount="15" CommentCount="2" FavoriteCount="8" />
  <row Id="3853" PostTypeId="2" ParentId="2949" CreationDate="2010-09-02T07:18:54.513" Score="0" ViewCount="" Body="&lt;p&gt;I was surprised to know that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$0! = 1$$&lt;/p&gt;&#xA;" OwnerUserId="705" LastActivityDate="2010-09-02T07:18:54.513" CommunityOwnedDate="2010-09-02T07:18:54.513" CommentCount="4" />
  <row Id="3854" PostTypeId="2" ParentId="3852" CreationDate="2010-09-02T07:20:24.273" Score="-8" ViewCount="" Body="&lt;p&gt;Suppose,&#xA;$AB = I$ and $BA = C \ne I$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$B(AB) = B(I)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(BA)B = B$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$CB = B$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;{Assuming B has a right inverse}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$C = I $$&#xA;so our assumption $C \ne I$ is wrong.&lt;/p&gt;&#xA;" OwnerUserId="705" LastActivityDate="2010-09-02T14:15:16.313" CommunityOwnedDate="2010-09-02T11:10:59.057" CommentCount="8" />
  <row Id="3855" PostTypeId="2" ParentId="3852" CreationDate="2010-09-02T07:35:03.933" Score="20" ViewCount="" Body="&lt;p&gt;So you want to find a proof of this well-known fact, which avoids the usual &quot;indirect&quot; proofs? I've also pondered over this some time ago.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have the following general assertion:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $M$ be a finite-dimensional $K$-algebra, and $a,b \in M$ such that $ab=1$, then $ba=1$. [For example, $M$ could be the algebra of $n \times n$ matrices]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proof: The sequence of subspaces $... \subseteq b^{k+1} M \subseteq b^k M \subseteq ... \subseteq M$ must be stationary, since $M$ is finite-dimensional. Thus there is some $k$ and some $c \in M$ such that $b^k = b^{k+1} c$. Now multiply with $a^k$ on the left to get $1=bc$. Then $ba=ba1 = babc=b1c=bc=1$. QED&lt;/p&gt;&#xA;&#xA;&lt;p&gt;No commutativity condition is needed. The proof shows more general that the claim holds in every left- or right-artinian ring $M$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remark that we needed, in a essential way, some finiteness condition. There is no purely algebraic manipulation with $a,b$, which shows $ab = 1 \Rightarrow ba=1$ (and shift operators provide a concrete counterexample). Every argument uses some argument of the type above. For example when you want to argue with linear maps, you have to know that every subspace of a finite-dimensional(!) vector space of the same dimension actually is the whole vector space, for which there is also no &quot;direct&quot; proof. I doubt that there is one.&lt;/p&gt;&#xA;" OwnerUserId="1650" LastEditorUserId="1650" LastEditorDisplayName="" LastEditDate="2010-09-02T11:09:47.677" LastActivityDate="2010-09-02T11:09:47.677" CommentCount="6" />
  <row Id="3858" PostTypeId="2" ParentId="1344" CreationDate="2010-09-02T08:01:22.617" Score="1" ViewCount="" Body="&lt;p&gt;Actually the result is used all the time in the basics of algebraic geometry. It provides &lt;em&gt;the&lt;/em&gt; formalization of the principle of gluing constructions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example if want to show that fibered products $X \times_S Y$, first do it for affine schemes $X,Y,S$ using the adjunction between Spec and global sections. Now, if $S$ is arbitrary, the functor $Sch/S \to Set, Z \mapsto Hom_S(Z,X) \times Hom_S(Z,Y)$ is a sheaf and locally representable on $S$, thus representable. Thus $X \times_S Y$ exists. Now if also $X$ is arbitrary, consider the functor $Sch/X \to Set, Z \mapsto Hom_S(Z',Y)$, where $Z' = Z \to X \to S$. This is a sheaf and locally representable on $X$, thus representable, which shows that $X \times_S Y$ exists. The usual &quot;ad hoc&quot; proofs for the existence of the fibered product actually just reprove the general representabilty result in the special case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is another example: If $A$ is a quasi-coherent sheaf of algebras on a scheme $X$, it is possible to construct the $X$-scheme $Spec(A)$. It is very laborious to check all the details of the gluing construction, well-definedness etc. when you just want to glue the $U$-schemes $Spec(A(U))$, $U \subseteq X$ affine, together. But instead, you could just consider the functor&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$Sch/X \to Set, (t : Z \to X) \mapsto Hom_{\mathcal{O}&lt;em&gt;X-Alg}(A,t&lt;/em&gt;* \mathcal{O}_Z)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and show that it is a sheaf (obvious) and locally on $X$ representable (usual adjunction with spectrum of a ring), so it is representable by an $X$-scheme $Spec(A)$ for which you also directly have a universal property. Again I want to emphasize: You get into a big mess when you want to construct this without using functors or universal properties. These rather abstract notions are very useful also in concrete situations, because they make you able to fix your ideas and make every construction fit together nicely. When you get more accustomed to these techniques, you stop thinking of specific functors, but you think in a &quot;functorial way&quot; and recognize, for example, why gluing constructions work.&lt;/p&gt;&#xA;" OwnerUserId="1650" LastActivityDate="2010-09-02T08:01:22.617" CommentCount="5" />
  <row Id="3859" PostTypeId="2" ParentId="3852" CreationDate="2010-09-02T08:09:36.297" Score="13" ViewCount="" Body="&lt;p&gt;Since there seem to be some lingering beliefs that an approach which does not make explicit use of the finite-dimensionality could be valid, here is a familiar counterexample in the infinite dimensional case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $V = \mathbb{R}[t]$ be the vector space of real polynomial functions.  Let $B: V \rightarrow V$ be differentiation: $p \mapsto p'$, and let $A: V \rightarrow V$ be anti-differentation with constant term zero: $\sum_{n=0}^{\infty} a_n t^n \mapsto \sum_{n=0}^{\infty} \frac{a_n}{n+1} t^{n+1}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These are both $\mathbb{R}$-linear maps and $B \circ A$ is the identity, but $A \circ B$ is not (the constant term is lost).  &lt;/p&gt;&#xA;" OwnerUserId="299" LastActivityDate="2010-09-02T08:09:36.297" CommentCount="4" />
  <row Id="3860" PostTypeId="2" ParentId="3852" CreationDate="2010-09-02T08:11:04.377" Score="14" ViewCount="" Body="&lt;p&gt;Dilawar says in 2. that he knows linear dependence! So I will give a proof, similar to that of TheMachineCharmer, which uses linear independence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose each matrix is $n$ by $n$. We consider our matrices to all be acting on some $n$-dimensional vector space with a chosen basis (hence isomorphism between linear transformations and $n$ by $n$ matrices).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then $AB$ has range equal to the full space, since $AB=I$. Thus the range of $B$ must also have dimension $n$. For if it did not, then a set of $n-1$ vectors would span the range of $B$, so the range of $AB$, which is the image under $A$ of the range of $B$, would also be spanned by a set of $n-1$ vectors, hence would have dimension less than $n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now note that $B=BI=B(AB)=(BA)B$. By the distributive law, $(I-BA)B=0$. Thus, since $B$ has full range, the matrix $I-BA$ gives $0$ on all vectors. But this means that it must be the $0$ matrix, so $I=BA$.&lt;/p&gt;&#xA;" OwnerUserId="873" LastActivityDate="2010-09-02T08:11:04.377" CommentCount="6" />
  <row Id="3861" PostTypeId="1" CreationDate="2010-09-02T08:51:28.307" Score="4" ViewCount="218" Body="&lt;p&gt;as a function of a real variable, apparently.&#xA;Part of the freedom in choosing a proof is that you get to choose what definition of $e^{ix}$ to start from -- do you use a differential equation? a power series? a definition in terms of trig functions?&#xA;Another bit of freedom is that you get to choose what definition of $\pi$ to start from.&lt;/p&gt;&#xA;" OwnerUserId="1678" LastActivityDate="2010-09-02T16:04:17.493" CommunityOwnedDate="2010-09-02T08:51:50.193" Title="What is your favorite proof that $e^{ix}$ has a period of $2\pi$?" Tags="&lt;calculus&gt;&lt;soft-question&gt;" AnswerCount="3" CommentCount="4" />
  <row Id="3862" PostTypeId="2" ParentId="3852" CreationDate="2010-09-02T09:11:45.910" Score="2" ViewCount="" Body="&lt;p&gt;You might have a look at the following note &quot;&lt;em&gt;Right inverse implies left inverse and vice versa&lt;/em&gt;&quot;: &lt;a href=&quot;http://www.lehigh.edu/~gi02/m242/08m242lrinv.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.lehigh.edu/~gi02/m242/08m242lrinv.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="55" LastActivityDate="2010-09-02T09:11:45.910" CommentCount="3" />
  <row Id="3863" PostTypeId="2" ParentId="978" CreationDate="2010-09-02T09:28:22.613" Score="2" ViewCount="" Body="&lt;p&gt;Once you have proved rank(AB) &amp;lt;= rank(B), you can obtain the other inequality by using transposition and the fact that it doesn't change the rank (see e.g. this &lt;a href=&quot;http://math.stackexchange.com/questions/2315/is-the-rank-of-a-matrix-the-same-of-its-transpose-if-yes-how-can-i-prove-it&quot;&gt;question&lt;/a&gt;). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Specifically, letting C=A^T and D=B^T, we have that rank(DC) &amp;lt;= rank(C) implies rank(C^TD^T) &amp;lt;= rank (C^T), which is rank(AB) &amp;lt;= rank(A).&lt;/p&gt;&#xA;" OwnerUserId="55" LastEditorUserId="55" LastEditorDisplayName="" LastEditDate="2010-09-02T10:03:24.290" LastActivityDate="2010-09-02T10:03:24.290" CommentCount="1" />
  <row Id="3865" PostTypeId="2" ParentId="3861" CreationDate="2010-09-02T09:55:14.410" Score="1" ViewCount="" Body="&lt;p&gt;$$e^{ix} = e^{i(x+T)} = e^{ix}e^{iT}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have to find $T$ for which $e^{iT} = 1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\rightarrow cos(T) + isin(T) = 1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\rightarrow sin(T) = 0$$ for all $$T = 2n\pi , n = 0,1,2,3...$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, period is $2\pi$.&lt;/p&gt;&#xA;" OwnerUserId="705" LastActivityDate="2010-09-02T09:55:14.410" CommunityOwnedDate="2010-09-02T09:55:14.410" CommentCount="1" />
  <row Id="3867" PostTypeId="2" ParentId="3749" CreationDate="2010-09-02T10:59:58.580" Score="1" ViewCount="" Body="&lt;p&gt;I still have very little mathematical experience (mainly linear algebra and a bit or real analysis) so the only answer I can give you is that dual spaces are already useful in linear algebra itself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;LEMMA dimension of the annihilator&lt;/strong&gt; Let $V$ be a vector space on the field $\mathbb{F}$ with $\operatorname{dim}V = n$ and $W \subseteq V$ a subspace, if $W^{0} = \{ \phi \in V^{*} | \phi_{|W} = 0 \}$ is the &lt;em&gt;annihilator&lt;/em&gt; of $W$, then $\operatorname{dim}W + \operatorname{dim}W^{0} = n$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $(w_1, \ldots, w_k)$ be a basis of $W$ and $(w_1,\ldots,w_k,v_{k+1},\ldots,v_n)$ a completion to basis of $V$, if $(w_1^{*},\ldots,w_k^{*},v_{k+1}^{*},\ldots,v_n*)$ is the corresponding dual base of $V^{*}$ then it's easy to prove by double inclusion that $W^{0}=\langle v_{k+1}^{*},...,v_n^{*}\rangle$ and thus $\operatorname{dim}W^{0}= n-k$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;THEOREM dimension of the orthogonal space&lt;/strong&gt; Let $V$ be a vector space on the field $\mathbb{F}$ with $\operatorname{dim}V = n$, endowed with a &lt;em&gt;nondegenerate&lt;/em&gt; scalar product $\langle , \rangle$, then if $W \subseteq V$ is a subspace $\operatorname{dim}W + \operatorname{dim}W^{\bot} = n$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, notice that the statement is not obvious. In fact if a scalar product is not positive-definite $V = W \oplus W^{\bot}$ could be false (you could have $v \in W$ such that $\langle v, v\rangle$ = 0 and therefore $W\cap W^{\bot}\neq \{ 0\}$). Now by mean of the scalar product you can define a very nice isomorphism $\psi : V \to V^{*}$ such that $\psi (v) = \langle \cdot , v\rangle$ for all $v \in V$, you can think about it as the &lt;em&gt;product by $v$&lt;/em&gt;. This is an isomorphism, in fact $\operatorname{dim}V = \operatorname{dim}V^{*}$ and $\operatorname{ker}\psi = 0$ because $\langle , \rangle$ is nondegenerate. Now observe that $\psi (W^{\bot}) = W^{0}$ and thus $\operatorname{dim}W^{\bot}=\operatorname{dim}W^{0}=n-\operatorname{dim}W$ by the lemma.&lt;/p&gt;&#xA;" OwnerUserId="365" LastEditorUserId="365" LastEditorDisplayName="" LastEditDate="2010-09-06T22:13:07.150" LastActivityDate="2010-09-06T22:13:07.150" CommentCount="2" />
  <row Id="3868" PostTypeId="2" ParentId="3861" CreationDate="2010-09-02T11:06:24.713" Score="5" ViewCount="" Body="&lt;p&gt;$$&#xA;e^{ix} = \cos x + i \sin x \ .&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="664" LastActivityDate="2010-09-02T11:06:24.713" CommunityOwnedDate="2010-09-02T11:06:24.713" CommentCount="4" />
  <row Id="3869" PostTypeId="1" AcceptedAnswerId="3871" CreationDate="2010-09-02T11:52:10.883" Score="8" ViewCount="546" Body="&lt;p&gt;Singular value decomposition (&lt;a href=&quot;http://en.wikipedia.org/wiki/Singular_value_decomposition&quot; rel=&quot;nofollow&quot;&gt;SVD&lt;/a&gt;) and principal component analysis (PCA) are two eigenvalue methods used to reduce a high-dimensional dataset into fewer dimensions while retaining important information. Articles online say that these methods are 'related' but never specify the exact relation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the intuitive relationship between PCA and SVD? As PCA uses the SVD in its calculation, clearly there is some 'extra' analysis done. What does PCA 'pay attention' to differently than the SVD? What kinds of relationships do each method utilize more in their calculations? Is one method 'blind' to a certain type of data that the other is not?&lt;/p&gt;&#xA;" OwnerUserId="1680" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2011-05-03T19:48:19.710" LastActivityDate="2011-05-11T11:23:29.617" Title="What is the intuitive relationship between SVD and PCA" Tags="&lt;linear-algebra&gt;&lt;statistics&gt;" AnswerCount="1" FavoriteCount="6" />
  <row Id="3870" PostTypeId="2" ParentId="3852" CreationDate="2010-09-02T12:06:16.607" Score="5" ViewCount="" Body="&lt;p&gt;I prefer to think in terms of linear operators rather than matrices. A function has a right inverse iff it is surjective and it has a left inverse iff it is injective. For a linear operator, this means that having a right inverse is equivalent to having range equal to the entire space and having a left inverse is equivalent to having trivial kernel. For a linear operator on a finite-dimensional space, the dimension of its kernel + the dimension of its range = the dimension of the entire space, so this does the job.&lt;/p&gt;&#xA;" OwnerUserId="687" LastActivityDate="2010-09-02T12:06:16.607" />
  <row Id="3871" PostTypeId="2" ParentId="3869" CreationDate="2010-09-02T12:31:53.333" Score="9" ViewCount="" Body="&lt;p&gt;(I assume for the purposes of this answer that the data has been preprocessed to have zero mean.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Simply put, the PCA viewpoint requires that one compute the eigenvalues and eigenvectors of the covariance matrix, which is the product $XX^T$, where X is the data matrix. Since the covariance matrix is symmetric, the matrix is diagonalizable, and the eigenvectors can be normalized such that they are orthonormal:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$XX^T=WDW^T$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, applying SVD to the data matrix $X$ as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$X=U \Sigma V^T$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and attempting to construct the covariance matrix from this decomposition gives&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$XX^T=(U \Sigma V^T)(U \Sigma V^T)^T$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$XX^T=(U \Sigma V^T)(V \Sigma U^T)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and since V is an orthogonal matrix ($V^T V=I$),&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$XX^T=U \Sigma^2 U^T$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and the correspondence is easily seen (the square roots of the eigenvalues of $XX^T$ are the singular values of $X$, etc.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, using the SVD to perform PCA makes much better sense numerically than forming the covariance matrix to begin with, since the formation of $XX^T$ can cause loss of precision. This is detailed in books on numerical linear algebra, but I'll leave you with an example of a matrix that can be stable SVD'd, but forming $XX^T$ can be disastrous, the Lauchli matrix:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\begin{pmatrix}1&amp;amp;1&amp;amp;1\\\\ \epsilon&amp;amp;0&amp;amp;0\\\\0&amp;amp;\epsilon&amp;amp;0\\\\0&amp;amp;0&amp;amp;\epsilon\end{pmatrix}^T$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\epsilon$ is a tiny number.&lt;/p&gt;&#xA;" OwnerUserId="498" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2011-05-11T11:23:29.617" LastActivityDate="2011-05-11T11:23:29.617" CommentCount="4" />
  <row Id="3872" PostTypeId="1" AcceptedAnswerId="3886" CreationDate="2010-09-02T12:49:22.513" Score="3" ViewCount="51" Body="&lt;p&gt;Let $\kappa$ ba weakly inaccessible cardinal. Why are there $\kappa$ regular cardinals $\lambda &amp;lt; \kappa$? I've tried a recursive construction, but I don't know what to do in the limit step. Supremum does not work, since then we loose regularity.&lt;/p&gt;&#xA;" OwnerUserId="1650" LastActivityDate="2010-09-02T17:21:05.037" Title="number of regular cardinals in a weakly inaccessible cardinal" Tags="&lt;logic&gt;&lt;cardinals&gt;" AnswerCount="1" />
  <row Id="3873" PostTypeId="1" AcceptedAnswerId="3874" CreationDate="2010-09-02T13:16:41.620" Score="3" ViewCount="86" Body="&lt;p&gt;Let us suppose to have a set of &lt;em&gt;N&lt;/em&gt; elements and to choose &lt;em&gt;m&lt;/em&gt; elements out of the set. I know I can do it in $ N \choose m $ ways; but I want to know how many of these combinations contain &lt;em&gt;k&lt;/em&gt; predefined elements (say 1,2,...,k). &lt;/p&gt;&#xA;" OwnerUserId="89" LastActivityDate="2010-09-02T13:38:20.583" Title="Probability of having k defined elements while choosing m elements from a set of N" Tags="&lt;combinatorics&gt;" AnswerCount="1" />
  <row Id="3874" PostTypeId="2" ParentId="3873" CreationDate="2010-09-02T13:19:52.267" Score="6" ViewCount="" Body="&lt;p&gt;If you fix k elements, you are left with m-k elements to choose among N-k elements, which can be done in $N-k \choose m-k$ ways. &lt;/p&gt;&#xA;" OwnerUserId="55" LastEditorUserId="55" LastEditorDisplayName="" LastEditDate="2010-09-02T13:38:20.583" LastActivityDate="2010-09-02T13:38:20.583" />
  <row Id="3875" PostTypeId="2" ParentId="3852" CreationDate="2010-09-02T14:01:40.353" Score="6" ViewCount="" Body="&lt;p&gt;It follows by the &lt;strong&gt;pigeonhole principle&lt;/strong&gt;. Here's an excerpt from my &lt;a href=&quot;http://groups.google.com/groups?selm=y8z4peo6ukm.fsf@nestle.csail.mit.edu&quot; rel=&quot;nofollow&quot;&gt;Dec 11 2007 sci.math post&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Recall (proof below) $\rm\; AB \:\:=\:\:\: I \:\;\Rightarrow\; BA \:\:=\:\: I\;\;\:$  easily reduces to:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;THEOREM&lt;/strong&gt; $\;$  $\rm\;\;B\;$ injective  $\rm\;\Rightarrow\:\:  B\;$ surjective, $\:$ for  linear $\rm\:B\:$ on a finite dim vector space $\rm\:V$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof&lt;/strong&gt;  $\rm\ \ \ B\;$ injective $\rm\;\Rightarrow\ B\;$ preserves injections:  $\rm\;R &amp;lt; S \;\Rightarrow\; BR &amp;lt; BS\;$&lt;br&gt;&#xA;Hence for   $\rm\;\;\; R \;\: &amp;lt; \;\; S &amp;lt; \cdots &amp;lt; \; V\;\;$  a chain of maximum length (= dim $\rm V\:$)&lt;br&gt;&#xA;its image   $\rm\;BR &amp;lt; BS &amp;lt; \cdots &amp;lt; BV \le V\;\;\:\;$   would have length greater&lt;br&gt;&#xA;if  $\rm\ BV &amp;lt; V\:,\: $  hence, instead   $\rm\:\:\:\ \ BV  = V\:,\;\:$  i.e. $\rm\; B \;$ is surjective. $\;$ &lt;strong&gt;QED&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice how this form of proof dramatically emphasizes the essence of the matter, namely that&lt;br&gt;&#xA;&lt;strong&gt;injective maps cannot decrease heights&lt;/strong&gt; (here = dimension = length of max subspace chain).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Below is said standard reduction to xxxjective form. See said &lt;a href=&quot;http://groups.google.com/groups?selm=y8z4peo6ukm.fsf@nestle.csail.mit.edu&quot; rel=&quot;nofollow&quot;&gt;sci.math post&lt;/a&gt; for much more,&lt;br&gt;&#xA;including references to folklore generalizations, e.g. work of Vasconcelos in the seventies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First,  notice that $\rm\;\;\ AB = I \;\Rightarrow\: B\:$ injective,  since  $\rm\;A\;$  times  $\rm\;B\:x = B\:y\;$  yields  $\rm\;x = y\:,\:$  and  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\ B\ $ surjective $\rm\ \Rightarrow\ BA = I \;\;$ since for all $\rm\;x\;$ exists $\rm\; y : \;\; x = B\:y = B(AB)\:y = BA \: x$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Combining them: $\rm\: AB = I \;\Rightarrow\: B\:$ injective $\rm\;\Rightarrow\; B\:$ surjective $\rm\;\Rightarrow\: BA = I$&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2011-02-22T17:36:15.830" LastActivityDate="2011-02-22T17:36:15.830" CommentCount="2" />
  <row Id="3876" PostTypeId="1" AcceptedAnswerId="3879" CreationDate="2010-09-02T14:32:54.373" Score="1" ViewCount="194" Body="&lt;p&gt;For a periodic ( and not so periodic) function, it is always possible to use fourier series to find out the frequencies contained in the function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But what about function that cannot be expressed in mathematical terms? For example, this graph ( accelerogram):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/pEwli.gif&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there anyway to apply a sort of fourier series, in order to find out the frequencies contained in it?&lt;/p&gt;&#xA;" OwnerUserId="550" LastActivityDate="2010-09-06T11:19:52.243" Title="Find the Frequency Components of a Time Series Graph" Tags="&lt;fourier-analysis&gt;" AnswerCount="4" CommentCount="3" FavoriteCount="1" />
  <row Id="3877" PostTypeId="2" ParentId="3876" CreationDate="2010-09-02T14:43:11.510" Score="2" ViewCount="" Body="&lt;p&gt;Er... that's what FFT is for. FFT your data, and take the absolute value of the result. If you want, take their phases too.&lt;/p&gt;&#xA;" OwnerUserId="498" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-09-02T14:48:58.213" LastActivityDate="2010-09-02T14:48:58.213" CommentCount="1" />
  <row Id="3878" PostTypeId="1" AcceptedAnswerId="3896" CreationDate="2010-09-02T14:43:36.777" Score="1" ViewCount="46" Body="&lt;p&gt;This queston can be regarded as a variant of this one:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://math.stackexchange.com/questions/3799/does-a-section-that-vanihes-at-every-point-vanish&quot;&gt;http://math.stackexchange.com/questions/3799/does-a-section-that-vanihes-at-every-point-vanish&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $X,Y$ be varieties, $M$ be an $\mathcal{O}_X$-Module and &#xA;$$\pi:X\times Y\rightarrow X$$&#xA;be the projection. Let $s$ be a local section of $M$, which vanishes along each &quot;horizontal line&quot;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\forall y: i_y^* s=0$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;where denotes the map $i_y:x\mapsto (x,y)$. Does it follow, that $s=0$?&lt;/p&gt;&#xA;" OwnerUserId="667" LastEditorUserId="667" LastEditorDisplayName="" LastEditDate="2010-09-02T17:54:16.667" LastActivityDate="2010-09-02T21:13:35.897" Title="Vanishing criterion for sections of module on a product" Tags="&lt;algebraic-geometry&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="3879" PostTypeId="2" ParentId="3876" CreationDate="2010-09-02T15:25:02.657" Score="4" ViewCount="" Body="&lt;p&gt;In short - yes. You can apply a discrete FFT and that should return the complex valued spectrum of your real data. Typically you can then plot out the magnitude of the value of each point in the spectrum (a.k.a. absolute value).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're going to take an FFT numerically (i.e. fftw routines, etc) it's also helpful to apply a window function (i.e. hamming). Your data doesn't look particularly periodic, so windowing it will help in removing the noise associated with that.&lt;/p&gt;&#xA;" OwnerUserId="1682" LastActivityDate="2010-09-02T15:25:02.657" CommentCount="1" />
  <row Id="3880" PostTypeId="2" ParentId="3861" CreationDate="2010-09-02T16:04:17.493" Score="5" ViewCount="" Body="&lt;p&gt;My favorite has always been Walter Rudin's proof in the prologue to his &quot;Real and Complex Analysis&quot; (2nd Ed.).  Here's a sketch:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Define $\exp$ in terms of the power series.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;By manipulating the series, deduce that $\exp$ is a homomorphism from the additive group to the group of complex units.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Show it satisfies the usual first order ODE.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;Define&lt;/em&gt; $\cos$ and $\sin$ as the real and imaginary parts of $\exp$, respectively.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;Define&lt;/em&gt; $\pi$ as twice the smallest positive real root of $\cos$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Deduce that $\exp( i \pi / 2) = i$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;By multiplying, conclude that $2 \pi i$ is a period of $\exp$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Show, by means of the preceding properties, that no smaller period exists.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1489" LastActivityDate="2010-09-02T16:04:17.493" CommunityOwnedDate="2010-09-02T16:04:17.493" />
  <row Id="3881" PostTypeId="2" ParentId="3852" CreationDate="2010-09-02T16:05:38.160" Score="15" ViewCount="" Body="&lt;p&gt;Let $x_1, x_2, \cdots, x_n$ be a basis of the space. At first we prove that $Bx_1, Bx_2, \cdots, Bx_n$ is also a basis. To do it we need to prove that those vectors are linearly independent. Suppose it's not true. Then there exist numbers $c_1, c_2, \cdots, c_n$ not all equal to zero such that $c_1 Bx_1 + c_2 Bx_2 + \cdots + c_n B x_n = 0$. Multiplying it by $A$ from the left, we get $c_1 ABx_1 + c_2 ABx_2 + \cdots + c_n ABx_n = 0$, hence $c_1 x_1 + c_2 x_2 + \cdots + c_n x_n = 0$ and so the vectors $x_1, x_2, \cdots, x_n$ are also linearly dependent. Here we get contradiction with assumption that the vectors $x_i$ form a basis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $Bx_1, Bx_2, \cdots, Bx_n$ is a basis, every vector $y$ can be represented as a linear combination of those vectors. This means that for any vector $y$ there exists some vector $x$ such that $Bx = y$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we want to prove that $BA = I$. It is the same as to prove that for any vector $y$ we have $BAy = y$. Now given any vector $y$ we can find $x$ such that $Bx = y$. Hence $BAy = BABx = Bx = y$ by associativity of matrix multiplication.&lt;/p&gt;&#xA;" OwnerUserId="334" LastEditorUserId="334" LastEditorDisplayName="" LastEditDate="2010-09-02T19:01:26.100" LastActivityDate="2010-09-02T19:01:26.100" CommentCount="7" />
  <row Id="3882" PostTypeId="2" ParentId="2079" CreationDate="2010-09-02T16:16:44.620" Score="0" ViewCount="" Body="&lt;p&gt;One can even look at this article for a general reference : &lt;a href=&quot;http://www.jstor.org/stable/2160835&quot; rel=&quot;nofollow&quot;&gt;http://www.jstor.org/stable/2160835&lt;/a&gt;&lt;/p&gt;&#xA;" LastActivityDate="2010-09-02T16:16:44.620" />
  <row Id="3883" PostTypeId="2" ParentId="3852" CreationDate="2010-09-02T16:47:49.057" Score="0" ViewCount="" Body="&lt;p&gt;Thanks guys for all the responses. Here, I was able to formulate this proof with all this material.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;0 : &lt;strong&gt;AB&lt;/strong&gt; = &lt;strong&gt;I&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;0.1 :  &lt;strong&gt;A&lt;/strong&gt; and &lt;strong&gt;B&lt;/strong&gt; are square matrix.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1 : &lt;strong&gt;AB&lt;/strong&gt; can be thought of as we are linearly combining columns of &lt;strong&gt;A&lt;/strong&gt; using the entries in a column of &lt;strong&gt;B&lt;/strong&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2 : Since &lt;strong&gt;AB = I&lt;/strong&gt; and &lt;strong&gt;I&lt;/strong&gt; is a matrix with the property that its columns are linearly independent (We can use &lt;em&gt;Row Reduced Echelon &lt;/em&gt; (RRE) form argument here to prove this - as done in &lt;EM&gt;Hoffman &amp;amp; Kunje's Linear Algebra&lt;/EM&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2.1 Every matrix have a unique RRE form. This can be proved from the fact that once you reach to a RRE form you can not modify it without disturbing its structure. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2.2 : it can be verified that row rank of a matrix is equal to its column rank.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2.2.1 : Row rank of a matrix &lt;strong&gt;A&lt;/strong&gt; = row rank of its RRE form. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2.2.2 : Row rank of RRE from = column rank of RRE form.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2.2.3 : Column rank of RRE form = column rank of Matrix &lt;strong&gt;A&lt;/strong&gt;. (I have a hazy proof since I have only defined elementary row operations and NOT the column operation. Anyone?). This proof, if elegant enough is different from the standard text-book proof which uses solution spaces. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;3 : From 2 it follows that matrix &lt;strong&gt;AB&lt;/strong&gt; has full rank i.e. its columns are linearly independent. Else, we'll not get matrix &lt;strong&gt;I&lt;/strong&gt; which is a RRE form. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;3.1 : From 2.2.2 it follows that its row rank is equal to its column rank.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4 : Rank of &lt;strong&gt;A&lt;/strong&gt; = row rank of &lt;strong&gt;AB&lt;/strong&gt; = row rank of &lt;strong&gt;B&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;5 : From 4 its follows that there exists a matrix &lt;strong&gt;C&lt;/strong&gt; such that &lt;strong&gt;BC&lt;/strong&gt; = &lt;strong&gt;I&lt;/strong&gt; i.e. we convert &lt;strong&gt;B&lt;/strong&gt; into its RRE form.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;This proves the existence of such a solution.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;6 :  No we use Davidac897 steps to find it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS : This is the first time I am writing a proof. If there is problem in its structure, please point out. &lt;/p&gt;&#xA;" OwnerUserId="1674" LastActivityDate="2010-09-02T16:47:49.057" CommunityOwnedDate="2010-09-02T16:47:49.057" />
  <row Id="3884" PostTypeId="2" ParentId="58" CreationDate="2010-09-02T17:07:44.020" Score="2" ViewCount="" Body="&lt;p&gt;From a physics perspective you can look at how Benford's law (or devations from it) can arise from the different stastical mechanics distributions (Boltzmann, Bose-Einstein, etc...). There is a good paper on the arVix that can get you started on this topic:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Abstract:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The occurrence of the nonzero leftmost digit, i.e., 1, 2, ..., 9, of numbers from many real world sources is not uniformly distributed as one might naively expect, but instead, the nature favors smaller ones according to a logarithmic distribution, named Benford's law. We investigate three kinds of widely used physical statistics, i.e., the Boltzmann-Gibbs (BG) distribution, the Fermi-Dirac (FD) distribution, and the Bose-Einstein (BE) distribution, and find that the BG and FD distributions both fluctuate slightly in a periodic manner around the Benford distribution with respect to the temperature of the system, while the BE distribution conforms to it exactly whatever the temperature is. Thus the Benford's law seems to present a general pattern for physical statistics and might be even more fundamental and profound in nature. Furthermore, various elegant properties of Benford's law, especially the mantissa distribution of data sets, are discussed.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://arxiv.org/abs/1005.0660&quot; rel=&quot;nofollow&quot;&gt;The Significant Digit Law in Statistical Physics&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="196" LastEditorUserId="196" LastEditorDisplayName="" LastEditDate="2010-09-02T17:27:29.593" LastActivityDate="2010-09-02T17:27:29.593" />
  <row Id="3885" PostTypeId="2" ParentId="3830" CreationDate="2010-09-02T17:15:02.703" Score="3" ViewCount="" Body="&lt;p&gt;What Kunen meant was that if $C$ is the range of a countable &lt;em&gt;increasing&lt;/em&gt; unbounded sequence in $\kappa$, that is, $C$ is unbounded and has order type $\omega$, then it will be vacuously closed, since it has no limit points below $\kappa$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This situation shows that the concept of club on cardinals of countable cofinality doesn't work as we want. &lt;/p&gt;&#xA;" OwnerUserId="413" LastActivityDate="2010-09-02T17:15:02.703" CommentCount="3" />
  <row Id="3886" PostTypeId="2" ParentId="3872" CreationDate="2010-09-02T17:21:05.037" Score="6" ViewCount="" Body="&lt;p&gt;Suppose that $\kappa$ is weakly inaccessible. Thus, it is a regular limit cardinal. So $\kappa=\aleph_\beta$ for some ordinal $\beta$. Since $\kappa$ is a limit cardinal, it must be that $\beta$ is a limit ordinal. Since $\kappa$ is regular, it cannot be that $\beta&amp;lt;\kappa$. So $\kappa=\aleph_\kappa$. Thus, there are $\kappa$ many cardinals below $\kappa$. All the successor cardinals $\aleph_{\beta+1}$ for $\beta&amp;lt;\kappa$ are regular, and there are $\kappa$ many of these.&lt;/p&gt;&#xA;" OwnerUserId="413" LastActivityDate="2010-09-02T17:21:05.037" CommentCount="1" />
  <row Id="3887" PostTypeId="2" ParentId="2949" CreationDate="2010-09-02T17:53:56.467" Score="1" ViewCount="" Body="&lt;p&gt;Similar to Thomae's function, I was impressed by the &lt;a href=&quot;http://mathworld.wolfram.com/DirichletFunction.html&quot; rel=&quot;nofollow&quot;&gt;Dirichlet function&lt;/a&gt;, which is not only discontinuous everywhere, but impossible to plot. The function is defined as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f(x)=\begin{cases}&#xA;  1 \mbox{ if } x\in\mathbb{Q} \\&#xA;  0 \mbox{ if }x\notin\mathbb{Q} &#xA;\end{cases}$&lt;/p&gt;&#xA;" OwnerUserId="1523" LastEditorUserId="1523" LastEditorDisplayName="" LastEditDate="2010-09-02T21:43:07.253" LastActivityDate="2010-09-02T21:43:07.253" CommunityOwnedDate="2010-09-02T17:53:56.467" />
  <row Id="3888" PostTypeId="1" AcceptedAnswerId="3889" CreationDate="2010-09-02T17:57:59.947" Score="4" ViewCount="430" Body="&lt;p&gt;Find all polynomials $p(x)$ such that for all $x$, we have $$(x-16)p(2x)=16(x-1)p(x)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried working out with replacing $x$ by $\frac{x}{2},\frac{x}{4},\cdots$, to have $p(2x) \to p(0)$ but then factor terms seems to create a problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Link: &lt;a href=&quot;http://web.mit.edu/rwbarton/Public/func-eq.pdf&quot; rel=&quot;nofollow&quot;&gt;http://web.mit.edu/rwbarton/Public/func-eq.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" LastActivityDate="2010-09-07T22:03:55.617" Title="Find polynomials such that $(x-16)p(2x)=16(x-1)p(x)$" Tags="&lt;real-analysis&gt;&lt;analysis&gt;&lt;functional-equations&gt;" AnswerCount="5" FavoriteCount="3" />
  <row Id="3889" PostTypeId="2" ParentId="3888" CreationDate="2010-09-02T18:08:51.287" Score="10" ViewCount="" Body="&lt;p&gt;Putting $x=1$ and $x=16$ we see that $p(2)=p(16) = 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $p(x) = (x-2)(x-16)g(x)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus we see that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(x-16)(2x-2)(2x-16)g(2x) = 16(x-1)(x-2)(x-16)g(x)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus $$ (x-8)g(2x) = 4(x-2)g(x)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We now see that $g(4) = g(8) = 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus $g(x) = (x-4)(x-8)h(x)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus&#xA;$$ (x-8)(2x-4)(2x-8)h(2x) = 4(x-2)(x-4)(x-8)h(x)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i.e&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ h(2x) = h(x) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This implies that $h(x)$ is a constant (as otherwise the non-identically zero polynomial $h(2x) - h(x)$ will have an infinite number of roots.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$p(x) = C(x-2)(x-4)(x-8)(x-16)$$&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-09-02T18:08:51.287" CommentCount="1" />
  <row Id="3890" PostTypeId="2" ParentId="58" CreationDate="2010-09-02T18:38:37.677" Score="3" ViewCount="" Body="&lt;p&gt;I recently attended a lecture on this. So I'll list a few interesting points that I remember.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The original discovery was by Newcomb who noticed that tables of logarithms were more worn out in the front than the back. It was rediscovered by Benford in 1938.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Areas of countries, areas of rivers and sizes of populations obey this rule. A more surprising example claims the numbers that appear in newspapers also follow this rule. In spite of the many observances of Benford's law, not a lot of instances are fully explained. I believe there have been proposals to provide simple models to associate an 'exponential-like' behavior of the areas of countries, but I don't really know anything about them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another note on the forensic accounting example mentioned by workmad3, I remember the lecturer pointed out that evidence pertaining to Benford's law is considered admissible in court.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are things that don't really obey Benford's law, for example: the number of pages in books.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and that's more or less what I can recall from the talk, hope that helps :)&lt;/p&gt;&#xA;" OwnerUserId="249" LastActivityDate="2010-09-02T18:38:37.677" />
  <row Id="3892" PostTypeId="2" ParentId="3852" CreationDate="2010-09-02T19:06:44.620" Score="4" ViewCount="" Body="&lt;p&gt;For a more elementary treatment ...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Fact.&lt;/b&gt; If the rows of $A$ are linearly dependent, then the rows of $A B$ are linearly dependent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Proof of fact.&lt;/em&gt; Consider the 3x3 case, where the linearly-dependent rows of $A$ are $\mathbf{a}_1$, $\mathbf{a}_2$, $\mathbf{a}_3 = h \mathbf{a}_1 + k \mathbf{a}_2$ (for some scalars $h$ and $k$):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$A = \begin{bmatrix}\mathbf{a}_1 \\ \mathbf{a}_2 \\ h\mathbf{a}_1 + k\mathbf{a}_2\end{bmatrix} = \begin{bmatrix}p &amp;amp; q &amp;amp; r \\ s &amp;amp; t &amp;amp; u \\ hp + ks &amp;amp; hq + kt &amp;amp; hr + ku \end{bmatrix}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Writing $\mathbf{b}_1$, $\mathbf{b}_2$, and $\mathbf{b}_3$ for the rows of $B$, we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$A B = \begin{bmatrix}p &amp;amp; q &amp;amp; r \\ s &amp;amp; t &amp;amp; u \\ hp + ks &amp;amp; hq + kt &amp;amp; hr + ku \end{bmatrix} \begin{bmatrix}\mathbf{b}_1 \\ \mathbf{b}_2 \\ \mathbf{b}_3\end{bmatrix} = \begin{bmatrix}p &#xA;\mathbf{b}_1+ q\mathbf{b}_2 + r\mathbf{b}_3 \\ s\mathbf{b}_1 + t\mathbf{b}_2 + u\mathbf{b}_3 \\ (hp + ks)\mathbf{b}_1 + (hq + kt)\mathbf{b}_2 + (hr + ku)\mathbf{b}_3 \end{bmatrix}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$= \begin{bmatrix}p &#xA;\mathbf{b}_1+ q\mathbf{b}_2 + r\mathbf{b}_3 \\ s\mathbf{b}_1 + t\mathbf{b}_2 + u\mathbf{b}_3 \\ h(p\mathbf{b_1}+q\mathbf{b}_2+r\mathbf{b}_3) + k(s\mathbf{b}_1 + t\mathbf{b}_2 + u\mathbf{b}_3) \end{bmatrix}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Generally, the linear dependence of the rows of $A$ carries over to the rows of the product, proving our Fact. (This reasoning actually shows the more-precise Fact that $rank(AB)\le rank(A)$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can restate the Fact this way:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Re-Fact.&lt;/b&gt; If the rows of $AB$ are linearly independent, then the rows of $A$ are linearly independent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To your question: If $A B = I$, then (by the Re-Fact) the rows of $A$ must be linearly independent. This implies that $A$ can be row-reduced to a diagonal matrix with no zero entries on that diagonal: the row-reduced form of $A$ must be the Identity matrix.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that row-reduction is &lt;em&gt;actually&lt;/em&gt; an application of matrix multiplication. (You can see this in the equations above, where (left-)multiplying $B$ by $A$ combined the rows of $B$ according to the entries in the rows of $A$.) This means that, if $R$ is the result of some row combinations of $A$, then there exists a matrix $C$ that &quot;performed&quot; the combinations:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$C A = R$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If (as in the case of your problem) we have determined that $A$ can be row-reduced all the way down to the Identity matrix, then the corresponding $C$ matrix must be a (&lt;em&gt;the&lt;/em&gt;) left-inverse of $A$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$C A = I$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's then straightforward to show that left and right inverses of $A$ must match. This has been shown in other answers, but for completeness ...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$A B = I \;\; \to \;\; C (A B) = C \;\; \to \;\; (C A) B = C \;\; \to \;\; I B = C \;\; \to \;\; B = C$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you start thinking (&lt;em&gt;ahem&lt;/em&gt;) &quot;outside the box (of numbers)&quot; to interpret matrices as linear transformations of vectors and such, you can interpret this result in terms of mapping kernels and injectivity-vs-surjectivity and all the kinds of sophisticated things other answers are suggesting. Nevertheless, it's worth noting that this problem is solvable within the realm of &lt;em&gt;matrix multiplication&lt;/em&gt;, plain and simple.&lt;/p&gt;&#xA;" OwnerUserId="409" LastActivityDate="2010-09-02T19:06:44.620" CommentCount="1" />
  <row Id="3894" PostTypeId="2" ParentId="3888" CreationDate="2010-09-02T19:28:29.997" Score="3" ViewCount="" Body="&lt;p&gt;I love Moron's approach.  Because it is usually interesting to look at other solutions, consider what happens if you just write $p(x) = p_n x^n + p_{n-1}x^{n-1} + \cdots + p_0$ and write out the equation.  Comparing the coefficients of $x^k$ ($k \gt 0$) yields the recursion&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$p_k= \frac{2^{k-5}-1}{2^k - 1} p_{k-1}$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;whence $p_5 = 0$, implying $p_6 = p_7 = \cdots = 0$, and the rest of the recursion yields $p_4, p_3, \ldots, p_0$ as multiples of $p_4$.  This approach, although pedestrian, also demonstrates that we need only assume $p$ is analytic near 0.&lt;/p&gt;&#xA;" OwnerUserId="1489" LastActivityDate="2010-09-02T19:28:29.997" />
  <row Id="3895" PostTypeId="2" ParentId="3852" CreationDate="2010-09-02T20:54:33.247" Score="18" ViewCount="" Body="&lt;p&gt;If $\rm\:B\:$ is a linear map on a finite dimensional vector space $\rm\: V\:$ over field $\rm\:K\:,\:$ then trivially by finite dimensionality (cf. Note below) there is a nonzero polynomial $\rm\:p(x)\in K[x]\;$ such that $\rm\:p(B) = 0\:$. We may assume $\rm\:p(0) \ne 0\;$ by canceling any factors of $\rm\:B\;$ from $\rm\;p(B)\;$ by left-multiplying by $\rm\:A\;$ and using $\rm\;AB = 1\;$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But $\rm\;\;\; AB=1  \ \ \Rightarrow\ \  (BA-1)\;\: B^n \;=\; 0\;$  for $\:\rm\;n&amp;gt;0\;$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;thus we have $\rm\;\; 0 \:=\: (BA-1)\ p(B)\ =\ (BA-1)\ p(0) \;\Rightarrow\;  BA=1 \quad\quad$&#xA; &lt;strong&gt;QED&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is essentially a special case of computing inverses by the Euclidean algorithm - see my &lt;a href=&quot;http://groups.google.com/groups?selm=y8zogksm5v0.fsf@berne.ai.mit.edu&quot; rel=&quot;nofollow&quot;&gt;Apr 13 1999 sci.math post&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Note&lt;/strong&gt;: The existence of $\rm\;p(x)\;$ follows simply from the fact that $\rm\:V\;$ finite-dimensional implies the same for the vector space $\rm\:L(V)\:$ of linear maps on $\rm\:V\:$ (indeed if $\rm\:V\;$ has dimension $\rm n$ then a linear map is uniquely determined by its matrix of $\:\rm n^2\:$ coefficients). So $\rm\: 1,\: B,\: B^2,\: B^3,\:\cdots\;$ are $\rm\:K$-linearly dependent in $\rm\: L(V)$ which yields the sought nonzero polynomial annihilating $\rm\:B\;$.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2011-02-22T17:47:57.890" LastActivityDate="2011-02-22T17:47:57.890" CommentCount="4" />
  <row Id="3896" PostTypeId="2" ParentId="3878" CreationDate="2010-09-02T21:13:35.897" Score="2" ViewCount="" Body="&lt;p&gt;If you take $X$ to be a point, then this reduces to your previous question: you are just taking fibres at each point $y \in Y$, and so the answer is &quot;no&quot; for the same reason.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just to see that this isn't just being pedantic, let me explain how to bootstrap the answer&#xA;to the previous question to give examples in this case which truly have a non-trivial product&#xA;structure.  For this, suppose that $Y = \mathbb A^1$ and that $X$ is the variety attached to&#xA;the ring $A$, so that $X \times Y$ corresponds to the ring $A[t]$.  Let $M$ be the module&#xA;$A[t]/t^2,$ and let $s$ be the section $t \bmod t^2$.  Then $s$ will vanish along each &quot;horizontal line&quot;, but is non-zero.  (The case $X = $ a point gives the counterexample to&#xA;your previous question.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again, if $M$ is torsion free and finitely generated then you should be okay.&lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2010-09-02T21:13:35.897" />
  <row Id="3897" PostTypeId="1" AcceptedAnswerId="3903" CreationDate="2010-09-02T21:15:33.480" Score="2" ViewCount="388" Body="&lt;p&gt;In Exercise 5 (f) of Angus Taylor's Advanced calculus (p. 659) one is asked to find the value of the following integral if convergent:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$I:=\underset{R}{\iiint}\dfrac{x^2 y^2 z^2}{r^{17/2}}\mathrm dV$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $R$ is the unit sphere $x^2+y^2+z^2\leq 1$ and $r^2=x^2+y^2+z^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Observing that $\dfrac{x^2 y^2 z^2}{r^{17/2}}\leq \dfrac{r^6}{r^{17/2}}=r^{-5/2}$ I proved that $I$ is convergent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using spherical co-ordinates $r$, $\theta $, $\phi $ i.e.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align*}x&amp;amp;=r\sin \phi \cos \theta\\y&amp;amp;=r\sin \phi \sin \theta\\z&amp;amp;=r\cos \theta\end{align*}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I transformed the integral $I$ into&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$I=\int_0^{2\pi }\left(\int_0^{\pi }\left(\lim_{\delta \to 0}\int_{\delta }^1\left(r^2 \sin \phi\right)\dfrac{x^2 y^2 z^2}{r^{17/2}}\;\mathrm dr\right)\;\mathrm d\phi \right)\;\mathrm d\theta$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$=\lim_{\delta \to 0}\left( \int_{\delta }^1 r^{-1/2}\mathrm dr\right)\int_0^{2\pi }\cos^4 \theta \sin^2 \theta \mathrm d\theta\int_0^{\pi }\sin^5 \mathrm d\phi $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$=2\cdot \dfrac18 \pi \cdot \dfrac{16}{15}=\dfrac4{15}\pi $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the solutions the answer is $\dfrac8{105}\pi$. Since sometimes there&#xA;are a few book typos (in the exercises) to prevent undue copying, I ask the&#xA;following&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Question: What is the correct solution, $\dfrac4{15}\pi $ or $\dfrac8{105}\pi $?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;UPDATE (Correction)&lt;/strong&gt;: instead of $z=r\cos \theta $ it is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$z=r\cos \phi $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;See a comment from whuber&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The integral $I$ is transformed into&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$I=\int_0^{2\pi }\left(\int_0^{\pi }\left(\lim_{\delta \to 0}\int_{\delta }^1\left(r^2 \sin \phi\right)\dfrac{x^2 y^2 z^2}{r^{17/2}}\;\mathrm dr\right)\;\mathrm d\phi \right)\;\mathrm d\theta$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(r^2 \sin \phi )\dfrac{x^2 y^2 z^2}{r^{17/2}}=(r^2\sin \phi )\dfrac1{r^{17/2}}\left( r\sin \phi \cos \theta \right)&#xD;&#xA;^{2}\left( r\sin \phi \sin \theta \right) ^{2}\left( r\cos \phi \right) ^{2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$=r^{-1/2}\cos ^{2}\theta \cdot\sin ^{2}\theta \cdot\cos ^{2}\phi \cdot\sin ^{5}\phi $,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the transformed integral becomes (if I am right):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$I=\left(\lim_{\delta \to 0} \int_{\delta&#xD;&#xA;}^1 r^{-1/2}\mathrm dr\right)\int_0^{2\pi }\cos^2\theta\cdot\sin^2\theta \;\mathrm d\theta&#xD;&#xA;\int_0^{\pi }\cos^2 \phi \cdot\sin^5 \phi \;\mathrm d\phi$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$=2\cdot \dfrac14 \pi \cdot \dfrac{16}{105}=\dfrac8{105}\pi$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The correct solution will be $\dfrac8{105}\pi $ as in the book. &lt;/p&gt;&#xA;" OwnerUserId="752" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2011-04-20T18:39:16.390" LastActivityDate="2011-04-20T18:39:16.390" Title="Evaluating a convergent improper triple integral over the unit sphere" Tags="&lt;calculus&gt;&lt;integral&gt;&lt;multivariable-calculus&gt;&lt;improper-integrals&gt;" AnswerCount="2" CommentCount="9" />
  <row Id="3898" PostTypeId="2" ParentId="3888" CreationDate="2010-09-02T21:33:02.127" Score="4" ViewCount="" Body="&lt;p&gt;Since $\rm\;p(2x)\;$ has roots being half the roots $\rm\;r_i\;$ of $\rm\;p(x)\;$, and $\rm\;\frac{p(x)}{p(2x)} = \frac{x-16}{16(x-1)}\;$, for the roots to all cancel like that leaving only one term in the numerator and denominator, we must have simply that the map $\rm\;r\to r/2\;$ is a left shift map on the (intermediate) roots. The numerator shows the largest root is 16, and the denomimator shows half the smallest root is 1. The other intermediate roots must obey the left shift $\rm\;r_i/2 = r_{i-1},\;$ so the roots are $\rm\;2,4,8,16,\;$ i.e. $\rm\;p(x) = c(x-2)(x-4)(x-8)(x-16)\;$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;$\:$ I purposely omitted a detail above since I thought it might be a homework problem. Based on the comments, this caused some confusion, so I will elaborate. Namely, notice that each linear linear factor of $\rm\;p(x)\;$ introduces a constant factor of $2\:$ in the denominator of $\rm\: p(x)/p(2x)$. But said denominator has constant factor $16$, so the $4$ factors found above are complete. In particular, $0$ cannot be a root of $\rm\:p(x).$&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-09-03T00:20:32.310" LastActivityDate="2010-09-03T00:20:32.310" CommentCount="12" />
  <row Id="3899" PostTypeId="1" AcceptedAnswerId="3902" CreationDate="2010-09-02T21:37:05.617" Score="1" ViewCount="273" Body="&lt;p&gt;Hi All, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am currently prepping for uni having been a few years out of the studying loop (programming as it happens). Anyway, I've been re-reading my A-level notes/exercises and looking through OpenCourseWare stuff and I noticed this assertation in my notes:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given $a \cos\theta \pm b \sin\theta = c$ then:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a \sin\theta \pm b\cos\theta = r\sin(\theta\pm\alpha)+c$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a \cos\theta \pm b\sin\theta = r\cos(\theta\pm\alpha)+c$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given $r \in \mathbb{N}$, $0 \leq \theta &amp;lt; 2\pi$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, I've heard of fourier series which have a very similar form to these equestions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, my question is, is there a relation between the two?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please bear in mind I know I'm stepping off a cliff and into &quot;unknown unknowns&quot; territory if they are - I have absolutely no idea what harmonic analysis is and I don't (yet) understand fourier series fully, although I grasp roughly how they work. I'm just interested to know if they are related and of course any further information / direction / reading I should follow up. I ask because this was one of those &quot;odd topics&quot; at A-level that we derived in class from a geometric argument then I've never seen again.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="892" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-09-05T04:24:00.237" LastActivityDate="2010-09-05T04:24:00.237" Title="Relation between &quot;harmonic form&quot; and fourier series?" Tags="&lt;algebra-precalculus&gt;&lt;trigonometry&gt;&lt;fourier-series&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="2" />
  <row Id="3900" PostTypeId="2" ParentId="3888" CreationDate="2010-09-02T22:22:37.647" Score="6" ViewCount="" Body="&lt;p&gt;Moron's solution and Dubuque's solution are essentilly the same, but I think their presentations can be made neater.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clearly, $p$ is not a constant polynomial (unless it is zero).  So we may write $p(x) = C(x-z_1)(x-z_2)\ldots(x-z_n)$.  Then the given equality implies that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(2x-z_1)(2x-z_2)\ldots(2x-z_n)(x-16) = 16(x-1)(x-z_1)(x-z_2)\ldots(x-z_n)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By comparing the leading coefficient on both sides, one sees $n=4$.  Thus,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(x-\frac{z_1}{2})(x-\frac{z_2}{2})(x-\frac{z_3}{2})(x-\frac{z_4}{2})(x-16)=(x-1)(x-z_1)(x-z_2)(x-z_3)(x-z_n)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence $\lbrace \frac{z_1}{2}, \frac{z_2}{2}, \frac{z_3}{2}, \frac{z_4}{2}, 16\rbrace = \lbrace 1, z_1, z_2, z_3, z_4\rbrace$.  Now one easily sees that, up to a relabelling of indices, we must have $z_i = 2^i$.&lt;/p&gt;&#xA;" OwnerUserId="1551" LastActivityDate="2010-09-02T22:22:37.647" CommentCount="10" />
  <row Id="3902" PostTypeId="2" ParentId="3899" CreationDate="2010-09-02T22:53:22.233" Score="5" ViewCount="" Body="&lt;p&gt;I can't really make sense of those equations, but it looks something like the statement that for any $ a, b \in \mathbb{R} $, then for $ r = \sqrt{a^2 + b^2 }$ and $ \phi = \arctan(b/a) $ we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ a \sin(\theta) + b \cos(\theta) = r \sin(\theta + \phi).$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you visualize x and y axes with $ cos(\theta) $ along the x-axis and $ sin(\theta) $ along the y-axis, then this says we can identify any sinusoidal, i.e. any function like $ r \sin(\theta + \phi) $, by its cartesian coordinates $(a,b)$ in this plane, or its polar coordinate $(r, \phi )$.  $r$ is called the magnitude of the sinusodial, and $ \phi $ the phase.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is directly related to Fourier Analysis.  The main idea in Fourier Analysis is to decompose a function into its sinusodial components.  For instance, if $f$ is a real-valued function on the interval $ [0, 2\pi] $ that is suitably regular, it turns out you can write $f$ uniquely as an infinite sum of sine and cosines, called the Fourier Series:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ f(x) = a_0 + \sum_{n=1}^{\infty} (a_n \cos(nx) + b_n \sin(nx)). $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The numbers $ a_n $ and $ b_n $ are called the Fourier coefficients of your function.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think this representation of Fourier series is sometimes difficult to grasp.  But using the relationship above, we see we could have also wrote:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ f(x) = a_0 + \sum_{n=1}^{\infty} r_n \sin(nx + \phi_n) $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this form I think you can see more clearly what's going on: For each $n$, there is only one sinusoidal component of $f$ of frequency $n$; $r_n = \sqrt{a_n^2 +b_n^2}$ tells you its magnitude and $\phi_n$ tells you its phase.&lt;/p&gt;&#xA;" OwnerUserId="1692" LastEditorUserId="1692" LastEditorDisplayName="" LastEditDate="2010-09-03T02:51:05.743" LastActivityDate="2010-09-03T02:51:05.743" CommentCount="3" />
  <row Id="3903" PostTypeId="2" ParentId="3897" CreationDate="2010-09-02T22:55:13.307" Score="6" ViewCount="" Body="&lt;p&gt;For this answer, I'll use $\rho$ instead of $r$, I tend to confuse things when dealing with both cylindrical and spherical coordinates&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Converting the original integrand into spherical coordinates gives&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{1}{16}\frac{(\sin(2\theta)\sin(2\phi)\sin(\phi))^2}{\rho^{5/2}}$$&#xD;&#xA;&#xD;&#xA;which we multiply by the Jacobian $\rho^2 \sin(\phi)$ for the triple integration to give the final integrand&#xD;&#xA;&#xD;&#xA;$$\frac{1}{16}\frac{(\sin(2\theta)\sin(2\phi))^2\sin(\phi)^3}{\sqrt{\rho}}$$&#xD;&#xA;&#xD;&#xA;As for setting up the limits of the integration, we can exploit symmetry and cut up the unit sphere into octants to simplify the mathematics, thus:&#xD;&#xA;&#xD;&#xA;$$8\int_{0}^{1}\int_{0}^{\pi/2}\int_{0}^{\pi/2}{\frac{1}{16}\frac{(\sin(2\theta)\sin(2\phi))^2\sin(\phi)^3}{\sqrt{\rho}}}\mathrm{d}\theta\mathrm{d}\phi\mathrm{d}\rho$$&#xD;&#xA;&#xD;&#xA;and then bring out the constant term to give&#xD;&#xA;&#xD;&#xA;$$\frac12\int_{0}^{1}\int_{0}^{\pi/2}\int_{0}^{\pi/2}{\frac{(\sin(2\theta)\sin(2\phi))^2\sin(\phi)^3}{\sqrt{\rho}}}\mathrm{d}\theta\mathrm{d}\phi\mathrm{d}\rho$$&#xD;&#xA;&#xD;&#xA;Let's separate things out:&#xD;&#xA;&#xD;&#xA;$$\frac12\int_{0}^{1}\frac{\mathrm{d}\rho}{\sqrt{\rho}}\int_{0}^{\pi/2}\sin(2\theta)^2 \mathrm{d}\theta\int_{0}^{\pi/2}{\sin(2\phi)^2\sin(\phi)^3}\mathrm{d}\phi$$&#xD;&#xA;&#xD;&#xA;The integral with respect to $\rho$ is 2, which cancels out the multiplicative factor, so we're left with&#xD;&#xA;&#xD;&#xA;$$\int_{0}^{\pi/2}\sin(2\theta)^2 \mathrm{d}\theta\int_{0}^{\pi/2}{\sin(2\phi)^2\sin(\phi)^3}\mathrm{d}\phi$$&#xD;&#xA;&#xD;&#xA;Evaluating the angular integrals gives:&#xD;&#xA;&#xD;&#xA;$$\left.\frac{\theta}{2}-\frac{\sin(4\theta)}{8}\right|_{\theta=0}^{\theta=\pi/2}=\frac{\pi}{4}$$&#xD;&#xA;&#xD;&#xA;$$\left.-\frac{5\cos(\phi)}{16}-\frac{\cos(3\phi)}{48}+\frac{3\cos(5\phi)}{80}-\frac{\cos(7\phi)}{112}\right|_{\phi=0}^{\phi=\pi/2}=\frac{32}{105}$$&#xD;&#xA;&#xD;&#xA;and we get the result $\frac{8\pi}{105}$.&lt;/p&gt;&#xA;" OwnerUserId="498" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-11T01:52:39.690" LastActivityDate="2010-11-11T01:52:39.690" CommentCount="2" />
  <row Id="3904" PostTypeId="1" AcceptedAnswerId="3907" CreationDate="2010-09-02T23:12:07.250" Score="3" ViewCount="107" Body="&lt;p&gt;Let me first start out by saying as you might have guessed, &lt;strong&gt;this is a homework problem&lt;/strong&gt;. Therefore I am &lt;strong&gt;not looking for an answer&lt;/strong&gt; to the question. I am looking for help in how to analyze it (My textbook is horrid).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a grid of 9 squares surrounded by a circle.&#xA;I can use three colors to color each region with no region touching any other region with the same color (corners don't count). **`&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I &lt;strong&gt;am asked to find the number of&#xA;  possible different colorings.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I drew a set of nodes with lines connecting neighboring regions, but I still have no idea how to figure out the number of combinations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also made a guess of three (which I know is wrong) since you can color the regions once with the three colors then map each color to a different one two times after the first coloring.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/B7R1O.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;No blue is touching any other blue regions, no white is touching any other white regions, no red is touching any other red regions.&lt;/p&gt;&#xA;" OwnerUserId="1670" LastEditorUserId="1670" LastEditorDisplayName="" LastEditDate="2010-09-03T00:16:09.403" LastActivityDate="2010-09-03T03:17:31.693" Title="How can I count the number of colored combinations in a set of regions?" Tags="&lt;combinatorics&gt;&lt;discrete-mathematics&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="1" />
  <row Id="3905" PostTypeId="1" AcceptedAnswerId="3957" CreationDate="2010-09-02T23:50:54.650" Score="0" ViewCount="215" Body="&lt;p&gt;I have a set of RGB colour values detected by a camera $C_{i_{RGB}}$ which are to be described by the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$C_{i_{RGB}} = X F_{i_{rgb}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $F_{i_{rgb}}$ is the component incident light at the detector, where the components $rgb$ are fixed at narrow bands and $X$, an $(n\times 3)$ matrix, is a description of the colour band overlap (cross talk if you will) for specific frequencies.  In a simplified RGB scheme, $X$ would be described:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\begin{align}&#xA;\mathsf X &amp;amp;= \begin{bmatrix}&#xA;  1 &amp;amp; R_{GR} &amp;amp; R_{BR} \\&#xA;  R_{RG} &amp;amp; 1 &amp;amp; R_{BG} \\&#xA;  R_{RB} &amp;amp; R_{G} &amp;amp; 1 &#xA;\end{bmatrix}\end{align}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is, the diagonal entries of $X$ are $1$'s, and the off-diagonals are rational.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;How can I solve for the values of $X$ only knowing $C_{i_{RGB}}$?&lt;/strong&gt; Eventually, given $C_i$ I need to find the corresponding $F_i$ using the constant $X$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Eventually added colour bands (band passes) will be added which will turn the above relation into something like this (and I fear the case where I have 5, 6, and 7 band passes to deal with):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\begin{bmatrix}&#xA;  C_{i_R} \\&#xA;  C_{i_G} \\&#xA;  C_{i_B} &#xA;\end{bmatrix} = \begin{align}\begin{bmatrix}&#xA;  X_{aR} &amp;amp; X_{bR} &amp;amp; X_{cR} &amp;amp; X_{dR} &amp;amp; X_{eR} &amp;amp; X_{fR} \\&#xA;  X_{aG} &amp;amp; X_{bG} &amp;amp; X_{cG} &amp;amp; X_{dG} &amp;amp; X_{eG} &amp;amp; X_{fG} \\&#xA;  X_{aB} &amp;amp; X_{bB} &amp;amp; X_{cB} &amp;amp; X_{dB} &amp;amp; X_{eB} &amp;amp; X_{fB} &#xA;\end{bmatrix}\end{align}\begin{bmatrix}&#xA;  F_{i_a} \\&#xA;  F_{i_b} \\&#xA;  F_{i_c} \\&#xA;  F_{i_d} \\&#xA;  F_{i_e} \\&#xA;  F_{i_f} &#xA;\end{bmatrix}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is for a graphics programming application and I lament that my first year algebra was 20 years ago.&lt;/p&gt;&#xA;" OwnerUserId="1067" LastEditorUserId="1067" LastEditorDisplayName="" LastEditDate="2010-09-03T12:34:11.863" LastActivityDate="2011-05-08T04:15:05.513" Title="How can I solve this matrix?" Tags="&lt;matrices&gt;" AnswerCount="4" CommentCount="5" />
  <row Id="3906" PostTypeId="2" ParentId="3876" CreationDate="2010-09-03T00:18:23.320" Score="3" ViewCount="" Body="&lt;p&gt;What you are after is known in statistics as spectral analysis. There is a whole literature on extracting frequency components from noisy data.  See &lt;a href=&quot;http://www.stat.pitt.edu/stoffer/tsa2/&quot; rel=&quot;nofollow&quot;&gt;Shumway and Stoffer (2006, chapter 4)&lt;/a&gt; for example.&lt;/p&gt;&#xA;" OwnerUserId="421" LastActivityDate="2010-09-03T00:18:23.320" />
  <row Id="3907" PostTypeId="2" ParentId="3904" CreationDate="2010-09-03T00:19:07.117" Score="3" ViewCount="" Body="&lt;p&gt;I'm not sure about this, maybe I don't understand the problem, but here it goes. If you colour the region outside the grid A, the 8 external cells of the grid must be coloured alternately with B and C, and the one in the center, with A. Thus you have two possible colourings. Same thing if you colour the region outside B, or C. Thus the total number of colourings would be 6.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update&lt;/strong&gt; Indeed, there are more. If the external cells are coloured B and C, then the one in the centre may be A, and either B or C, so there are 12 different colourings now.&lt;/p&gt;&#xA;" OwnerUserId="1313" LastEditorUserId="1313" LastEditorDisplayName="" LastEditDate="2010-09-03T00:54:18.047" LastActivityDate="2010-09-03T00:54:18.047" CommentCount="1" />
  <row Id="3908" PostTypeId="2" ParentId="3905" CreationDate="2010-09-03T00:37:26.793" Score="0" ViewCount="" Body="&lt;p&gt;If you don't know &lt;em&gt;both&lt;/em&gt; $X$ and $F_i$ then you have an infinite number of solutions.&lt;/p&gt;&#xA;" OwnerUserId="558" LastActivityDate="2010-09-03T00:37:26.793" CommentCount="1" />
  <row Id="3909" PostTypeId="2" ParentId="3904" CreationDate="2010-09-03T00:47:09.687" Score="4" ViewCount="" Body="&lt;p&gt;While it is valuable to attack such problems &quot;by hand,&quot; here are some generalities.  The general version of this problem is called &lt;a href=&quot;http://en.wikipedia.org/wiki/Graph_coloring&quot; rel=&quot;nofollow&quot;&gt;graph coloring&lt;/a&gt;.  (The graph being colored here is the graph whose vertices are the regions of the diagram where there is an edge between two regions if they are adjacent.)  Counting colorings is &lt;a href=&quot;http://en.wikipedia.org/wiki/Graph_coloring#Computational_complexity&quot; rel=&quot;nofollow&quot;&gt;known to be #P-hard&lt;/a&gt;, so it is not all that easy to do in general.  For small graphs there is a reasonable way to compute the number of colorings by any fixed number of colors $n$ by using the &lt;a href=&quot;http://en.wikipedia.org/wiki/Chromatic_polynomial#Deletion.E2.80.93contraction&quot; rel=&quot;nofollow&quot;&gt;deletion-contraction recurrence for the chromatic polynomial&lt;/a&gt;. &lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-09-03T03:17:31.693" LastActivityDate="2010-09-03T03:17:31.693" CommentCount="3" />
  <row Id="3910" PostTypeId="2" ParentId="3905" CreationDate="2010-09-03T01:11:38.343" Score="0" ViewCount="" Body="&lt;p&gt;If I understand correctly, you're trying to correlate $n$ separate equations for $n$ different values of $C_i$ and $F_i$ into a solution for $X$; is that correct?  If so, then the easiest way to go is to turn your set of vector equations into a matrix equation!  Simply write $C = [ C_1 C_2 C_3 \ldots C_n ]$ (assuming that the $C_n$ are being written as column vectors), $F = [ F_1 F_2 F_3 \ldots F_n ]$, and then by linearity you wind up with a single matrix equation $C = X F$ which can be solved by inverting F (if F isn't invertible then you have no guaranteed solution) and writing $X = C F^{-1}$.&lt;/p&gt;&#xA;" OwnerUserId="785" LastActivityDate="2010-09-03T01:11:38.343" CommentCount="3" />
  <row Id="3911" PostTypeId="2" ParentId="3904" CreationDate="2010-09-03T01:17:32.527" Score="1" ViewCount="" Body="&lt;p&gt;I've found that the simplest way to attack problems like this is really by straight enumeration.  Start by coloring the outer region, then see what colors that forces or eliminates; in this case, it means that none of the eight outer cells of your grid can share that color.  Now, pick another unforced cell (e.g., the top-left corner of the grid) and see how many ways that you can color it; in this case we have two separate colors that we can use, and whichever one we use then forces the colors on all of the outer cells of the grid, leaving us with just one un-forced color (the central square of the grid can be one of two colors), etc. While this is sort-of a graph coloring problem, it's really closer to a constraint satisfaction problem in disguise, and the methods used for solving a lot of the popular pen-and-paper puzzles like Sudoku/Kakuro/Picross/etc. are pretty closely related to this.&lt;/p&gt;&#xA;" OwnerUserId="785" LastActivityDate="2010-09-03T01:17:32.527" />
  <row Id="3912" PostTypeId="1" AcceptedAnswerId="3960" CreationDate="2010-09-03T02:06:57.783" Score="1" ViewCount="224" Body="&lt;p&gt;I'm trying to find out if this is a true statement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming asymptotic cases&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$n \sin(n \frac{\pi}{2}) + \log{n} = O(3)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How would you go about solving a problem like this systematically?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: O = &quot;Big O&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: lg = log base 2&lt;/p&gt;&#xA;" OwnerUserId="1630" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-09-03T02:42:17.873" LastActivityDate="2010-09-03T15:11:21.847" Title="Big O comparison in asymptotic cases" Tags="&lt;asymptotics&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="3913" PostTypeId="1" AcceptedAnswerId="3914" CreationDate="2010-09-03T02:39:38.483" Score="14" ViewCount="429" Body="&lt;p&gt;Back in the day, before approximation methods like splines became vogue in my line of work, one way of computing the area under an empirically drawn curve was to painstakingly sketch it on a piece of graphing paper (usually with the assistance of a French curve) along with the axes, &lt;strong&gt;painstakingly&lt;/strong&gt; cut along the curve, weigh the cut pieces, cut out a square equivalent to one square unit of the graphing paper, weigh that one as well, and reckon the area from the information now available.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One time, when we were faced with determining the area of a curve that crossed the horizontal axis thrice, I did the careful cutting of the paper, and made sure to separate the pieces above the horizontal axis from the pieces below the horizontal axis. Then, my boss suddenly scooped up all the pieces and weighed them all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I argued that the grouped pieces should have been weighed separately, and then subtract the weights of the &quot;below&quot; group from the &quot;above&quot; group, while my boss argued that we were calculating the total area, and thus, not separating the pieces was justifiable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which one of us was correct?&lt;/p&gt;&#xA;" OwnerUserId="498" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-09-03T07:00:14.310" LastActivityDate="2011-04-09T20:53:56.813" Title="The right &quot;weigh&quot; to do integrals." Tags="&lt;calculus&gt;&lt;approximation&gt;&lt;integral&gt;" AnswerCount="3" CommentCount="5" FavoriteCount="1" />
  <row Id="3914" PostTypeId="2" ParentId="3913" CreationDate="2010-09-03T03:01:06.763" Score="7" ViewCount="" Body="&lt;p&gt;you are correct. Actually, the integral is equal to: &quot;(total above weight) minus (total below weight)&quot;, which I'm sure is what you meant. What your boss is calculating is actually&#xA;$$&#xA;\int_a^b |f(x)| dx&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="1356" LastActivityDate="2010-09-03T03:01:06.763" />
  <row Id="3915" PostTypeId="2" ParentId="3912" CreationDate="2010-09-03T03:10:09.737" Score="2" ViewCount="" Body="&lt;p&gt;If $f(n)=O(g(n))$, with $g(n)&gt;0$ means that there is a a constant $M&gt;0$ that $|f(n)|\leq Mg(n)$ for every $n$, then (if you meant $O(n)$) we must show that there is a constant $M&gt;0$ that $|n\sin{(n\pi/2)}+\log{n}|\leq Mn$ for every $n$. But $|n\sin{(n\pi/2)}+\log{n}|\leq |n\sin{(n\pi/2)}| + \log{n}\leq n +\log{n}\leq 2n$ (for $n&gt; 1$), therefore $M=2$ will do if $n&gt;1$. For the case $n\to 0$, divide the inequality to prove by $n$ and note that $(\log{n})/n\to\infty$ when $n\to 0$, so it can't be true.&lt;/p&gt;&#xA;" OwnerUserId="1313" LastEditorUserId="1313" LastEditorDisplayName="" LastEditDate="2010-09-03T03:48:54.393" LastActivityDate="2010-09-03T03:48:54.393" CommentCount="2" />
  <row Id="3916" PostTypeId="2" ParentId="3913" CreationDate="2010-09-03T03:21:13.250" Score="3" ViewCount="" Body="&lt;p&gt;I guess we speak of the area &quot;under a curve&quot; when the piece of curve is above the $x$ axis. Sign does matter in the definition of the proper integral, so I would say you were right, and your boss was wrong. Doing integrals by &quot;weighing&quot; things should allow for negative weights.&lt;/p&gt;&#xA;" OwnerUserId="1313" LastActivityDate="2010-09-03T03:21:13.250" />
  <row Id="3917" PostTypeId="1" AcceptedAnswerId="3971" CreationDate="2010-09-03T03:43:29.130" Score="4" ViewCount="663" Body="&lt;p&gt;On my homework today, we had to find all the normal subgroups of $D_{n}$, the dihedral group of order 2n. I solved the problem by looking at how the conjugacy classes change based on whether n is even or odd and then constructed the normal subgroups as unions of the conjugacy classes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have 2 questions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(i) Is there a better way to approach the problem than looking at the conjugacy classes?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(ii) Could someone explain why we want to find all the normal subgroups of a particular group?   How does this provide us additional insight into the structure of the group we are studying? (Right now, this exercise feels more like a 'computation' to me than a way of understanding $D_{n}$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks :)&lt;/p&gt;&#xA;" OwnerUserId="249" LastActivityDate="2011-04-30T08:54:15.493" Title="Finding all normal subgroups of a group" Tags="&lt;abstract-algebra&gt;&lt;group-theory&gt;&lt;intuition&gt;" AnswerCount="4" CommentCount="2" FavoriteCount="1" />
  <row Id="3918" PostTypeId="2" ParentId="3917" CreationDate="2010-09-03T04:05:57.393" Score="1" ViewCount="" Body="&lt;p&gt;Very vague answer to (ii) (I'm taking this course myself): if you find all normal subgroups of a given group, then you know all its quotient groups, since quotient groups are defined through normal subgroups only - if you consider all left (or right) cosets of a given group, that isn't normal, they may not conform a group. And many theorems that follow in the theory are based on quotient groups.&lt;/p&gt;&#xA;" OwnerUserId="1313" LastActivityDate="2010-09-03T04:05:57.393" />
  <row Id="3919" PostTypeId="2" ParentId="3917" CreationDate="2010-09-03T04:30:42.387" Score="8" ViewCount="" Body="&lt;p&gt;(i) Conjugacy classes are certainly useful, since a normal subgroup must be a union of conjugacy classes. There are other ideas that can come into play: once you know a proper normal subgroup $N$, by taking the quotient you can obtain information about $G$ from information you may find in $G/N$, which will hopefully be easier than working directly in $G$ because $G/N$ will be smaller than $G$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(ii) In fact, that is one important reason for finding all normal subgroups: being able to look at the quotient, and thus obtain information about $G$ by looking at groups that are smaller than $G$. Another is that the normal subgroups are closely associated with all possible images of $G$ under homomorphisms. Both of these facts are part of the Isomorphism Theorems. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Right now you are probably engaged in computation and getting familiar with techniques for working with groups in general, and $D_n$ in particular; the importance of normal subgroups will likely emerge as you start using groups. For example, normal subgroups are extremely important when you are considering Galois groups (which help you study fields and field extensions). &lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-09-03T04:30:42.387" />
  <row Id="3920" PostTypeId="2" ParentId="3917" CreationDate="2010-09-03T04:32:09.093" Score="4" ViewCount="" Body="&lt;p&gt;Finding normal subgroups of a group is a first step towards understanding how it &quot;factors&quot; into simpler groups, analogous to finding prime factors of a number.  To understand a group it then suffices to understand i) the pieces it's built out of (the &lt;a href=&quot;http://en.wikipedia.org/wiki/Simple_group&quot; rel=&quot;nofollow&quot;&gt;simple groups&lt;/a&gt; in its &lt;a href=&quot;http://en.wikipedia.org/wiki/Composition_series&quot; rel=&quot;nofollow&quot;&gt;composition series&lt;/a&gt;) and ii) how those pieces fit together.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your case it is also just a good exercise in becoming familiar with groups and with basic constructions associated to groups.  Also, like Weltschmerz says, you should think of knowing the normal subgroups as the same thing as knowing all the quotient groups of a group, and the quotient groups of a group are like its &quot;shadows&quot; - they reveal some, but not all, of the structure of the group, and putting them together one gets a picture of the whole thing.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-09-03T04:32:09.093" />
  <row Id="3921" PostTypeId="1" AcceptedAnswerId="3922" CreationDate="2010-09-03T05:26:21.977" Score="6" ViewCount="333" Body="&lt;p&gt;Say I have a triangle with vertices &lt;code&gt;(0,0), (2,4), (4,0)&lt;/code&gt; that I want to rotate along the origin. Rotation by multiples of 90 is simple. However, I want to rotate by something a bit more complicated, such as 54 degrees. How do I figure out where the vertices would be then?&lt;/p&gt;&#xA;" OwnerUserId="1110" LastActivityDate="2010-09-03T05:38:19.880" Title="Rotations by degrees other than 90, 180, and 270." Tags="&lt;geometry&gt;&lt;euclidean-geometry&gt;&lt;transformation&gt;" AnswerCount="2" CommentCount="0" />
  <row Id="3922" PostTypeId="2" ParentId="3921" CreationDate="2010-09-03T05:29:42.237" Score="12" ViewCount="" Body="&lt;p&gt;One way is to use complex numbers.  Multiplying by $\cos\theta+i\sin\theta$ rotates by $\theta$ about 0, so you could multiply $(2+4i)(\cos 54^\circ+i\sin 54^\circ)$ to get the rotation image of (2,4).&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-09-03T05:29:42.237" />
  <row Id="3923" PostTypeId="1" AcceptedAnswerId="3940" CreationDate="2010-09-03T05:37:58.550" Score="3" ViewCount="294" Body="&lt;p&gt;I propose the following lemma and its proof. It is related to row-null and column-null matrices - i.e. matrices whose rows and columns both sum to zero. Could you please give your opinion on the plausibility of the lemma, and the validity of the proof?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Lemma:&lt;/strong&gt; Let $Z\in\text{GL}(n,\mathbb{R})$ be a general $n\times n$ real matrix, and let $Y\in\mathcal{S}(n,\mathbb{R})$, where $\mathcal{S}(n,\mathbb{R})$ is the space of row-null column-null $n\times n$ real matrices. Then $\text{Tr}(ZY)=0$ for all $Y$ in $\mathcal{S}(n,\mathbb{R})$ if and only if $Z$ has the form $$Z_{ij}=\left(p_{j}-p_{i}\right)+\left(q_{j}+q_{i}\right)$$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof:&lt;/strong&gt;&#xA;Consider the space of row-null and column-null matrices&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathcal{S}(n,\mathbb{R})= \left\{ Y_{ij}\in GL(n,\mathbb{R}):\sum_{i}Y_{ij}=0,\sum_{j}Y_{ij}=0 \right\} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Its dimension is&#xA;$$\text{dim}(S(n,\mathbb{R}))=N^{2}-2N+1$$&#xA;since the row-nullness and column-nullness are defined by $2N$ equations, only $2N-1$ of which are linearly independent.&#xA;Consider the following space&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathcal{G}(n,\mathbb{R})=\left\{ Z_{ij}\in GL(n,\mathbb{R}):Z_{ij}=\left(p_{j}-p_{i}\right)+\left(q_{j}+q_{i}\right)\right\}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Its dimension is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\text{dim}(\mathcal{G}(n,\mathbb{R}))=2N-1$$&#xA;where $N-1$ is the contribution from the antisymmetric part and $N$ is from the symmetric part.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume $Y\in\mathcal{S}$ and $Z\in\mathcal{G}$, then the Frobenius inner product of two such elements is&#xA;$$&#xA;\text{Tr}(ZY) =\sum_{ij}\left[\left(p_{j}-p_{i}\right)Y_{ji}+\left(q_{j}+q_{i}\right)Y_{ji}\right]&#xA;$$&#xA;$$&#xA;=\sum_{j}(q_{j}+p_{j})\sum_{i}Y_{ji}+\sum_{i}(q_{i}-p_{i})\sum_{j}Y_{ji}=0&#xA;$$&#xA;Since $\text{dim}(\mathcal{G})+\text{dim}(\mathcal{S})=\text{dim}(GL)$ and $\mathcal{G}\perp\mathcal{S}$, then $\mathcal{G}$ and $\mathcal{S}$ must be complementary in $GL$. Therfore, if $Y$ is orthogonal to all the matrices in $\mathcal{S}$, it must lie in $\mathcal{G}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS: How can I get the curly brackets {} to render in latex mode?&lt;/p&gt;&#xA;" OwnerUserId="1393" LastEditorUserId="1393" LastEditorDisplayName="" LastEditDate="2010-09-03T17:59:14.190" LastActivityDate="2010-09-03T17:59:14.190" Title="The orthogonal complement of the space of row-null and column-null matrices" Tags="&lt;linear-algebra&gt;" AnswerCount="2" CommentCount="7" FavoriteCount="1" />
  <row Id="3924" PostTypeId="2" ParentId="3921" CreationDate="2010-09-03T05:38:19.880" Score="6" ViewCount="" Body="&lt;p&gt;In the answer to &lt;a href=&quot;http://math.stackexchange.com/questions/2429&quot;&gt;this question&lt;/a&gt;, I mentioned the formula for the rotation matrix; one merely takes the product of the rotation matrix with the coordinates (treated as 2-vectors) to get the new rotated coordinates. Note that I gave the matrix for clockwise rotation; for anticlockwise rotation, negate the angle (thus switching the sign of the two sine components).&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-09-03T05:38:19.880" />
  <row Id="3925" PostTypeId="1" AcceptedAnswerId="3932" CreationDate="2010-09-03T06:12:35.230" Score="6" ViewCount="199" Body="&lt;p&gt;How does one prove that for a prime $p \geq 5$ the sum :  $$\sum\limits_{0 &amp;lt; k &amp;lt; \frac{2p}{3}} { p \choose k}$$ is divisible by $p^{2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since each term of $\displaystyle \sum\limits_{0 &amp;lt; k &amp;lt; \frac{2p}{3}} { p \choose k}$ is divisible by $p$, only thing remains is to prove that the sum $$\sum\limits_{ 0 &amp;lt; k &amp;lt; \frac{2p}{3}} \frac{1}{p} { p \choose k}$$ is divisible by $p$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How to evaluate this sum: $\displaystyle \frac{1}{p} { p \choose k} = \frac{(p-1)(p-2) \cdots (p-k+1)}{1 \cdot 2 \cdot 3 \cdots k}$&lt;/p&gt;&#xA;" LastActivityDate="2010-09-03T07:47:53.743" Title="Proving that this sum $\sum\limits_{0 &lt; k &lt; \frac{2p}{3}} { p \choose k}$ is divisible by $p^{2}$" Tags="&lt;number-theory&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" />
  <row Id="3926" PostTypeId="1" CreationDate="2010-09-03T07:01:05.730" Score="3" ViewCount="256" Body="&lt;p&gt;A 1-1 function is called injective. What is an n-1 function called ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm thinking about homomorphisms. So perhaps homojective ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Onto is surjective. 1-1 and onto is bijective.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What about n-1 and onto ? Projective ? Polyjective ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think n-m and onto should be hyperjective as in hypergroups.&lt;/p&gt;&#xA;" OwnerUserId="1613" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2010-11-24T20:33:50.990" LastActivityDate="2010-11-24T20:33:50.990" Title="A 1-1 function is called injective. What is an n-1 function called ?" Tags="&lt;abstract-algebra&gt;&lt;terminology&gt;&lt;elementary-set-theory&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="3927" PostTypeId="1" CreationDate="2010-09-03T07:24:55.703" Score="0" ViewCount="357" Body="&lt;p&gt;What are some interesting applications of the concept of homomorphism?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example: If there is a homorphism from a ring $R$ to a ring $r$ then a solution to a polynomial equation in $R$ gives rise to a solution in $r$. e.g. if $f:R \rightarrow r$ and $X^2+Y^2=0$ then $f(X^2+Y^2)=f(0), f(X^2)+f(Y^2)=0, f(X)^2+f(Y)^2=0, x^2+y^2=0$&lt;/p&gt;&#xA;" OwnerUserId="1613" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-09-03T07:56:42.363" LastActivityDate="2011-05-30T00:24:26.980" CommunityOwnedDate="2010-09-04T00:42:24.980" Title="Applications of the concept of homomorphism." Tags="&lt;abstract-algebra&gt;" AnswerCount="3" CommentCount="6" FavoriteCount="1" />
  <row Id="3929" PostTypeId="2" ParentId="3926" CreationDate="2010-09-03T07:32:59.803" Score="7" ViewCount="" Body="&lt;p&gt;n-1 + onto is sometimes called n-fold cover (by &lt;a href=&quot;http://en.wikipedia.org/wiki/Covering_space&quot; rel=&quot;nofollow&quot;&gt;analogy&lt;/a&gt;).&lt;/p&gt;&#xA;" OwnerUserId="152" LastActivityDate="2010-09-03T07:32:59.803" />
  <row Id="3930" PostTypeId="1" AcceptedAnswerId="3933" CreationDate="2010-09-03T07:42:10.253" Score="3" ViewCount="102" Body="&lt;p&gt;I was droodling a bit and a given moment I drew the following construction:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/XwnQg.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It appears that the three blue intersections are collinear (red line), no matter how I draw the construction lines. If this is always true, I assume that this a know fact [otherwise I have my first theorem! :-) erm conjecture, since I can't prove it :-(  ].&lt;br&gt;&#xA;What's the theorem called?&lt;br&gt;&#xA;TIA&lt;br&gt;&#xA;Steven  &lt;/p&gt;&#xA;" OwnerUserId="923" LastActivityDate="2010-09-03T07:52:34.150" Title="Collinear points theorem" Tags="&lt;geometry&gt;" AnswerCount="1" />
  <row Id="3931" PostTypeId="2" ParentId="3925" CreationDate="2010-09-03T07:42:27.643" Score="2" ViewCount="" Body="&lt;p&gt;From your last equation one gets&#xA;$$\frac1p{p\choose k}\equiv(-1)^k\frac 1k\pmod p.$$&#xA;So the problem is equivalent to&#xA;$$\sum_{0 &amp;lt; k &amp;lt; 2p/3}(-1)^k\frac1k\equiv0\pmod{p}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The case $p=1979$ was set as the first problem at the &lt;a href=&quot;http://www.imo-official.org/problems.aspx&quot; rel=&quot;nofollow&quot;&gt;1979 IMO&lt;/a&gt;.&#xA;The method used for this extends to all primes, and no doubt&#xA;you can find solutions for the IMO problem out on the interweb.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-09-03T07:42:27.643" />
  <row Id="3932" PostTypeId="2" ParentId="3925" CreationDate="2010-09-03T07:42:33.800" Score="6" ViewCount="" Body="&lt;p&gt;Since we are working in the field $\mathbb{F}_p$ we can write &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{(p-1)(p-2) \cdots (p-k+1)}{1 \cdot 2 \cdots k}$$ as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{(-1)(-2) \cdots (-(k-1))}{1 \cdot2 \cdots k}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;= $$\frac{(-1)^{k-1}}{k}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $N = [\frac{2p}{3}]$ and $M = [\frac{N}{2}]$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus what we need is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \sum_{k=1}^{N} \frac{(-1)^{k-1}}{k}$$  $$  = \sum_{k=1}^{N} \frac{1}{k} - 2 \sum_{k=1}^{M}\frac{1}{2k}$$  $$ =\sum_{k=M+1}^{N}\frac{1}{k}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now $N+M+1 =p$ so we can rewrite as &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{1}{N} + \frac{1}{M+1} + \frac{1}{N-1} + \frac{1}{M+2} + \cdots = $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{p}{N(M+1)} + \frac{p}{(N-1)(M+2)} + \cdots $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is $0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are $N-M$ terms, which is even, so each term gets paired off.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-09-03T07:47:53.743" LastActivityDate="2010-09-03T07:47:53.743" CommentCount="3" />
  <row Id="3933" PostTypeId="2" ParentId="3930" CreationDate="2010-09-03T07:52:34.150" Score="6" ViewCount="" Body="&lt;p&gt;Congratulations! Looks to me like you have rediscovered &lt;a href=&quot;http://mathworld.wolfram.com/PappussHexagonTheorem.html&quot; rel=&quot;nofollow&quot;&gt;Pappus' Hexagon Theorem&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The image from the link:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/SBcBj.gif&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-09-03T07:52:34.150" CommentCount="7" />
  <row Id="3934" PostTypeId="2" ParentId="58" CreationDate="2010-09-03T08:08:16.830" Score="3" ViewCount="" Body="&lt;p&gt;It may or may not be useful for detecting fraud in elections, for example the 2009 Iranian election.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some links:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://sbseminar.wordpress.com/2009/06/15/benfords-law-and-the-iranian-election&quot; rel=&quot;nofollow&quot;&gt;http://sbseminar.wordpress.com/2009/06/15/benfords-law-and-the-iranian-election&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.fivethirtyeight.com/2009/06/statistical-evidence-does-not-prove.html&quot; rel=&quot;nofollow&quot;&gt;http://www.fivethirtyeight.com/2009/06/statistical-evidence-does-not-prove.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="263" LastActivityDate="2010-09-03T08:08:16.830" />
  <row Id="3936" PostTypeId="2" ParentId="3913" CreationDate="2010-09-03T08:36:49.240" Score="7" ViewCount="" Body="&lt;p&gt;In my opinion, this is not really a math question.  Which procedure is correct depends what you're going to do with your calculation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a matter of definition, the integral indeed measures the signed area (positive area minus negative area), as you suggest.  So your approach is computing an approximation to the definite integral $\int_a^b f$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But maybe you want the total (unsigned) area.  E.g. if you're going to lay concrete along (some real-world space corresponding to) the region bounded by the curve and the x-axis then surely you want the total area -- there's no such thing as negative concrete.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Without knowing what you're using the calculation for, it's impossible to say.  (Essentially, you're asking us: &quot;Which of these is mathematically correct: $A-B+C$ or $A+B+C$?&quot;  Of course it depends upon what you're trying to do.) I would like to think that your boss knew what the point of it all was, so without further information I guess I would trust him.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, your story arouses my curiosity.  I suppose you're not putting us on, but weighing paper cutouts is just about the last method I would ever think of for computing area (aren't you going to need a very sensitive scale or an awfully big piece of paper to get anywhere with this?).  How long ago are we talking?  What was the job?  You don't have to answer these questions, but it would be interesting to know...&lt;/p&gt;&#xA;" OwnerUserId="299" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2011-04-09T20:53:56.813" LastActivityDate="2011-04-09T20:53:56.813" CommentCount="4" />
  <row Id="3937" PostTypeId="1" CreationDate="2010-09-03T09:10:48.437" Score="4" ViewCount="425" Body="&lt;p&gt;First there was arithmetic with numerical calculations (i.e. one unknown on one side of an equation). Then algebra with manipulations of variables (many unknowns anywhere in an equation). Then systems are studied that differ from ordinary arithmetic but share some of the same properties (equations where the unknowns represent all sorts of things - even functional equations) and then these properties are abstracted in abstract algebra and whole classes are studied such as groups and rings. Then category theory studies maps between structures (functorial equations), then n-category theory, then ...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where do we go now ? Is category theory the end of the road for the forseeable future ? &#xA;Is the only way forward to go backwards and generalize in a different direction ? (&quot;generalized equations&quot; of optimization)&lt;/p&gt;&#xA;" OwnerUserId="1613" LastActivityDate="2011-04-24T12:24:03.513" Title="What's more general than category theory?" Tags="&lt;category-theory&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="4" />
  <row Id="3938" PostTypeId="2" ParentId="3923" CreationDate="2010-09-03T09:16:43.543" Score="2" ViewCount="" Body="&lt;p&gt;I've checked your proof. It is correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to say obvious thing: I know two methods for checking the proof is correct.&lt;br&gt;&#xA;1. Check the answer, if you can get it from something else.&lt;br&gt;&#xA;1'. If you can't get the answer independently, check some of it's properties.&lt;br&gt;&#xA;2. Check the proof line by line.&lt;br&gt;&#xA;2'. Divide your proof on parts and check each of them, using one of the above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Applying second method is obvious. If you want, to apply first, you can try to prove the following.&lt;br&gt;&#xA;1. Every symmetric matrix, that is orthogonal to S(n,R), is of the form $q_i+q_j$.&lt;br&gt;&#xA;2. Every symmetric matrix, that is orthogonal to S(n,R), is of the form $q_i+q_j$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This can be done by using equations of the form Tr(ZY)=0 with matrices $Y\in S(n,\mathbb{R})$ of the form $Y_{ij}=a_i b_j$, where a and b are vectors of the form vectors with one coordinate equal to 1, another --- equal to -1, all other coordinates equal to 0 (like $(0,\dots,0,1,0,\dots,0,-1,0,\dots,0)$).&lt;/p&gt;&#xA;" OwnerUserId="695" LastActivityDate="2010-09-03T09:16:43.543" />
  <row Id="3940" PostTypeId="2" ParentId="3923" CreationDate="2010-09-03T10:39:13.047" Score="4" ViewCount="" Body="&lt;p&gt;Here is an alternate way of proving your Lemma. I'm not sure if its any simpler than your proof -- but it's different, and hopefully interesting to some.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $S$ be the set of $n\times n$ matrices which are row-null and column-null. We can write this set as:&#xA;$$&#xA;S = \left\{ Y\in \mathbb{R}^{n\times n} \,\mid\, Y1 = 0 \text{  and  }1^TY=0\right\}&#xA;$$&#xA;where $1$ is the $n\times 1$ vector of all-ones. The objective is the characterize the set $S^\perp$ of matrices orthogonal to every matrix in $S$, using the Frobenius inner product.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One approach is to &lt;em&gt;vectorize&lt;/em&gt;. If $Y$ is any matrix in $S$, we can turn it into a vector by taking all of its columns and stacking them into one long vector, which is now in $\mathbb{R}^{n^2\times 1}$. Then $\mathop{\mathrm{vec}}(S)$ is also a subspace, satisfying:&#xA;$$&#xA;\mathop{\mathrm{vec}}(S) = \left\{ y \in \mathbb{R}^{n^2\times 1} \,\mid\, (\mathbf{1}^T\otimes I)y = 0 \text{  and  } (I \otimes \mathbf{1}^T)y = 0 \right\}&#xA;$$&#xA;where $\otimes$ denotes the &lt;a href=&quot;http://en.wikipedia.org/wiki/Kronecker_product&quot; rel=&quot;nofollow&quot;&gt;Kronecker product&lt;/a&gt;. In other words,&#xA;$$&#xA;\mathop{\mathrm{vec}}(S) = \mathop{\mathrm{Null}}(A),\qquad\text{where: }&#xA;A = \left[ \begin{array}{c} \mathbf{1}^T\otimes I \\ I \otimes \mathbf{1}^T \end{array}\right]&#xA;$$&#xA;Note that vectorization turns the Frobenius inner product into the standard Euclidean inner product. Namely: $\mathop{\mathrm{Trace}}(A^T B) = \mathop{\mathrm{vec}}(A)^T \mathop{\mathrm{vec}}(B)$. Therefore, we can apply the range-nullspace duality and obtain:&#xA;$$&#xA;\mathop{\mathrm{vec}}(S^\perp) =&#xA;\mathop{\mathrm{vec}}(S)^\perp =&#xA;\mathop{\mathrm{Null}}(A)^\perp =&#xA;\mathop{\mathrm{Range}}(A^T)&#xA;$$&#xA;So every vector in $vec(S^\perp)$ is of the form $(\mathbf{1}\otimes I)a + (I\otimes \mathbf{1})b$ for some vectors $a$ and $b$ in $R^{n\times 1}$. It follows that every matrix in $S^\perp$ is of the form $a1^T + 1b^T$. This parametrization is equivalent to the one you presented if you set $a_i = q_i-p_i$ and $b_j = q_j + p_j$.&lt;/p&gt;&#xA;" OwnerUserId="1356" LastEditorUserId="1356" LastEditorDisplayName="" LastEditDate="2010-09-03T10:45:00.743" LastActivityDate="2010-09-03T10:45:00.743" />
  <row Id="3941" PostTypeId="1" AcceptedAnswerId="3954" CreationDate="2010-09-03T10:39:19.483" Score="12" ViewCount="492" Body="&lt;p&gt;Using trignometric identities one can see that $\cos{2x} = 2 \: \cos^{2}{x} - 1$ can be expressed as a polynomial of $\cos{x}$, where $p(\cos{x})=2 \: \cos^{2}{x}-1$. Then its natural to ask the same question for the sine function. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Can we express $\sin{2x}$ as a polynomial of $\sin{x}$?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="9413" LastEditorUserId="9413" LastEditorDisplayName="" LastEditDate="2011-06-01T18:14:27.740" LastActivityDate="2011-06-01T18:14:27.740" Title="Expressing $\sin(2x)$ as a polynomial of $\sin{x}$" Tags="&lt;real-analysis&gt;&lt;analysis&gt;&lt;trigonometry&gt;" AnswerCount="4" CommentCount="1" FavoriteCount="5" />
  <row Id="3942" PostTypeId="2" ParentId="3941" CreationDate="2010-09-03T10:41:04.540" Score="1" ViewCount="" Body="&lt;p&gt;No, you can't get any simpler than $2\sin\;x\;\cos\;x$. In general, only sines of odd angle multiples ($\sin((2k+1)x)$) are expressible as polynomials in $\sin(x)$, because any powers of cosines that may appear are raised to even exponents, and thus the identity $\sin^2(x)=1-\cos^2(x)$ applies, to turn it into an expression entirely made up of powers of sines.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-09-03T10:41:04.540" CommentCount="1" />
  <row Id="3943" PostTypeId="2" ParentId="3941" CreationDate="2010-09-03T10:42:05.770" Score="0" ViewCount="" Body="&lt;p&gt;you can express (sin2x)^2 as a polynomial of sinx&lt;/p&gt;&#xA;" OwnerUserId="1470" LastActivityDate="2010-09-03T10:42:05.770" />
  <row Id="3944" PostTypeId="2" ParentId="3937" CreationDate="2010-09-03T11:02:23.577" Score="14" ViewCount="" Body="&lt;p&gt;I don't really see a coherent logical progression in the branches of mathematics you're putting forward.  Mathematics isn't just about abstraction and generalizing, making things more and more general.  It's most often about solving particular problems.  Category theory was born out of algebraic topology, in many ways as a notational convenience, to make sense of the tremendous and difficult-to-follow messes algebraic topologists were producing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you've ever programmed in a language like &quot;C&quot; you know the concept of a &quot;macro&quot;.  This is an idea that is re-usable in many different contexts.  You plug in different objects and the macro continues to make sense.  That's much of the point of category theory, as there are so many ideas that are duplicated over and over again in mathematics that it's confusing to give them special names.  So we call these ideas by generic names that make sense in a wide-array of contexts, like &quot;the co-product (or whatever) in the category C (name your category)&quot;, etc.  It saves time and energy.   Moreover, once you've reduced the &quot;bulk&quot; of your notation sufficiently, there is a phenomena where the concepts are lighter and easier to play with.  So by using category theory you sometimes &quot;lighten the load&quot; a little, making other discoveries perhaps a little easier (if you're lucky). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In that regard what gets called &quot;category theory&quot; I think of as more of an attempt to find the natural language for certain types of ideas.  The general idea being that certain types of problems become easy when using appropriate notation. Not all, but some.  Some problems are just hard -- like the Poincare conjecture, or the Schoenflies problem, the classification of finite-simple groups, or existence and uniqueness of solutions to Navier-Stokes (and if you look at the work that's been done on these problems you will see almost no category theory at all, just a tiny little bit on the high-dimensional Poincare and Schoenflies problems). In programming category theory might be analogous to the study of data types, and how one structures memory efficiently. &lt;/p&gt;&#xA;" OwnerUserId="642" LastActivityDate="2010-09-03T11:02:23.577" CommentCount="2" />
  <row Id="3945" PostTypeId="2" ParentId="3937" CreationDate="2010-09-03T11:12:47.183" Score="8" ViewCount="" Body="&lt;p&gt;Generalization goes where problems lead it to. &lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2010-09-03T11:12:47.183" />
  <row Id="3946" PostTypeId="1" CreationDate="2010-09-03T11:34:07.223" Score="1" ViewCount="344" Body="&lt;p&gt;I've got a path that is made up of multiple points - i.e. 0,0  0,50  50,50  70,20&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I just draw this line on the screen it looks quite harsh as it puts a sharp angle at the joining of each point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence I was wondering what the bezier curves algorithm/method would look like which I could call that automatically change the sharp angles to &quot;tight&quot; curves?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wouldn't want the curve to be too big or generally effect the fall of main path, just soften the join out. If you have a look at the below here is a quick sample I have put together. The line on the left is what I have now, the one in the middle is the line I want. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The image on the right represents what I think I need the algorithm to do. Essentially, I add an additional point to each arc that makes up the join at a point which 10% away from the join, then I remove the join point and adjust the handles so that they are where the point was (not in the diagram they are slightly apart, this is just so that you can see). This is what I need to be able to do.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/EKY7U.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1700" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-09-03T12:29:32.183" LastActivityDate="2010-09-08T14:24:04.060" Title="Bezier curves algorithm complex yet simple" Tags="&lt;geometry&gt;&lt;algorithms&gt;&lt;bezier-curve&gt;" AnswerCount="3" />
  <row Id="3949" PostTypeId="2" ParentId="3946" CreationDate="2010-09-03T12:25:55.377" Score="0" ViewCount="" Body="&lt;p&gt;Well, Bézier curves are always guaranteed to lie within the convex hull of the control points; it is only a matter of taste/convenience for you to adjust the control points to get the shape you wish. Try finding a program that allows you to interactively move control points around, and experiment!&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-09-03T12:25:55.377" CommentCount="2" />
  <row Id="3950" PostTypeId="2" ParentId="3852" CreationDate="2010-09-03T13:32:10.677" Score="2" ViewCount="" Body="&lt;p&gt;Coincidentally, a totally unrelated MathSciNet search turned up the following article, which gives a result along the lines of (but slightly stronger than) the one in Martin Brandenburg's answer.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.math.uga.edu/~pete/Hill67.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.math.uga.edu/~pete/Hill67.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Theorem (Hill, 1967): Let $R$ be a ring satisfying the ascending chain condition on right ideals.  Then:&lt;br&gt;&#xA;a) For any $n \in \mathbb{Z}^+$, the matrix ring $M_n(R)$ also satisfies ACC on right ideals.&lt;br&gt;&#xA;b) If $R$ satisfies ACC on right ideals and $a,b \in R$ are such that $ab = 1$, then also $ba = 1$.&lt;/p&gt;&#xA;" OwnerUserId="299" LastActivityDate="2010-09-03T13:32:10.677" CommentCount="0" />
  <row Id="3951" PostTypeId="1" AcceptedAnswerId="3952" CreationDate="2010-09-03T13:35:25.853" Score="0" ViewCount="160" Body="&lt;p&gt;Can anyone give me an example of a vector space $V$ such that there is no norm which is complete on $V$? &lt;/p&gt;&#xA;" LastActivityDate="2010-09-03T14:02:55.607" Title="Vector spaces with no Complete norms" Tags="&lt;analysis&gt;&lt;functional-analysis&gt;" AnswerCount="1" CommentCount="7" />
  <row Id="3952" PostTypeId="2" ParentId="3951" CreationDate="2010-09-03T13:40:19.987" Score="1" ViewCount="" Body="&lt;p&gt;I assume you want $V$ to be a vector space over $\mathbb{R}$ or $\mathbb{C}$.  Since the latter can be viewed as a special case of the former, let's say $V$ is an $\mathbb{R}$-vector space to fix ideas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hints:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) A finite-dimensional subspace is always closed.&lt;br&gt;&#xA;2) A proper closed subspace is nowhere dense.&lt;br&gt;&#xA;3) There is a famous theorem about countable unions of nowhere dense subsets of a complete space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Addendum&lt;/b&gt;: I decided that my comment calling for a lack of other answers after having already given an answer myself was something of a conflict of interest, so I have made this answer CW.  &lt;/p&gt;&#xA;" OwnerUserId="299" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2010-09-03T14:02:55.607" LastActivityDate="2010-09-03T14:02:55.607" CommunityOwnedDate="2010-09-03T14:02:55.607" CommentCount="2" />
  <row Id="3953" PostTypeId="1" AcceptedAnswerId="3958" CreationDate="2010-09-03T13:43:40.270" Score="3" ViewCount="153" Body="&lt;p&gt;Define $R(0)=0=\emptyset, R(n+1)=P(R(n))$ and $R(\omega) = \cup_{n &amp;lt; \omega} R(n)$. Thus $R(\omega)$ is the set of all sets, which are build out of finitely many braces and $0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the following relation $E$ on $\omega$: If and only if the $n$th number in the binary representation of $m$ is $1$, then $n E m$. Now I want to construct an isomorphism $(R(\omega),\in) \cong (\omega,E)$ [This is an exercise in Kunen's set theory]. After some playing around I've come up with the following definition:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$g : R(\omega) \to \omega, g(x) = \sum_{y \in x} 2^{g(y)}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a well-defined recursion, since $rank(y) &amp;lt; rank(x)$. If $g$ is injective, then it is easy to see that $x \in y \Leftrightarrow g(x) E g(y)$. However I don't see this; neither why $g$ is surjective.&lt;/p&gt;&#xA;" OwnerUserId="1650" LastEditorUserId="1650" LastEditorDisplayName="" LastEditDate="2010-09-03T14:16:23.213" LastActivityDate="2010-09-04T04:02:30.310" Title="A nice enumeration of $R(\omega)$" Tags="&lt;set-theory&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="3954" PostTypeId="2" ParentId="3941" CreationDate="2010-09-03T14:01:05.510" Score="13" ViewCount="" Body="&lt;p&gt;Just for the sake of completeness, here is a proof that you cannot express it that way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(For a general proof, see the bottom of this answer).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose we could express $\sin 2x$ as a polynomial in $\sin x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i.e.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sin x \cos x  = P(\sin x)$ for some polynomial $P$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $\sin 0 = 0$, $x$ is a factor of $P(x)$. Hence we get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\cos x = Q(\sin x)$ for a polynomial $Q$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Put $x = \frac{\pi}{2}$ and $x=\frac{3\pi}{2}$ we see that $1,-1$ are roots of $Q(x)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus we see that for 0 &amp;lt; t &amp;lt; 1,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sqrt{1-t^2} = (t-1)(t+1)H(t)$ (because $Q(t) = (t-1)(t+1)H(t)$ for some polynomial $H$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By squaring we see that the degree of the left hand side polynomial is $2$ while the degree of the right hand side polynomial is $&gt;2$. This is not possible, as then their difference, which is not identically zero will have infinite roots.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus it is not possible to express $\sin 2x$ as a polynomial in $\sin x$.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;General Case:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a proof that in general $ \sin (2nx)$ cannot be written as a polynomial in $ \sin x$ based on the identity provided by J.M:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\sin(2nx)}{2\cos(x)}=(-1)^{\lfloor n-1/2\rfloor}\sum_{j=0}^{\lfloor n-1/2\rfloor}{(-1)^j \sin((2j+1)x)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First notice that $\sin ((2m+1)x)$ &lt;em&gt;can&lt;/em&gt; be written as a polynomial in $\sin x$ with &lt;em&gt;integer&lt;/em&gt; coefficients (follows easily from the DeMoivre's identity).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now suppose $ \sin (2nx)$ could be written as a polynomial $P$. Consider the $r^{th}$ derivative of that polynomial at $0$. This must be a rational number, as the $r^{th}$ derivative of $ \sin (2nx)$ at $0$ is rational. Thus the coefficients of $P$ must be rational.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus using the above identity we must have that $\cos x$ is of the form $\frac{P(\sin x)}{Q(\sin x)}$ where $P$ and $Q$ are polynomials with rational coefficients. Thus we must have that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sqrt{1-t^2} = \frac{P(t)}{Q(t)}$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ 0 &amp;lt; t &amp;lt; 1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Putting $t=\frac{1}{2}$ gives us a rational number on the RHS while an irrational number on the LHS.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence it is not possible to represent $ \sin (2nx)$ as a polynomial in $ \sin x$.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-09-03T15:07:14.007" LastActivityDate="2010-09-03T15:07:14.007" CommentCount="4" />
  <row Id="3955" PostTypeId="1" CreationDate="2010-09-03T14:15:19.793" Score="2" ViewCount="177" Body="&lt;p&gt;This is such a basic question I'm sure, but I've been trying to find a robust solution to it for a while.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lets say I have the numbers 0-100 in series.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the number increases, I want another number say 1000, to decrease. so that when you got to 100, it was at 0&lt;/p&gt;&#xA;&#xA;&lt;p&gt;0 = 1000&lt;/p&gt;&#xA;&#xA;&lt;p&gt;50 = 500&lt;/p&gt;&#xA;&#xA;&lt;p&gt;100 = 0&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is this kind of series called?&lt;/p&gt;&#xA;" OwnerUserId="1701" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-09-11T04:35:16.330" LastActivityDate="2010-09-11T04:35:16.330" Title="How to reduce lower one number, while another number increases goes up increments" Tags="&lt;algebra-precalculus&gt;&lt;terminology&gt;&lt;arithmetic&gt;" AnswerCount="4" CommentCount="1" />
  <row Id="3956" PostTypeId="2" ParentId="3926" CreationDate="2010-09-03T14:16:11.447" Score="7" ViewCount="" Body="&lt;p&gt;IMHO, an n to 1 function should be called &quot;an n to 1 function.&quot;&#xA;Simple, decriptive, to the point.  Adding a new term in this case just muddies the waters.&lt;/p&gt;&#xA;" OwnerUserId="575" LastActivityDate="2010-09-03T14:16:11.447" CommentCount="3" />
  <row Id="3957" PostTypeId="2" ParentId="3905" CreationDate="2010-09-03T14:32:51.310" Score="1" ViewCount="" Body="&lt;p&gt;Okay, as already mentioned the problem as posed is not tractable. However, reading between the lines, it looks to me a lot like Independent Components Analysis could work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically, assuming you have a sequence of such observations from the same sources, and its reasonable to assume that the individual sources are independent of each other, then the original signals can be recovered up to scaling and order.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a link to a nice tutorial paper by Aapo Hyvarinen:&#xA;&lt;a href=&quot;http://www.cs.helsinki.fi/u/ahyvarin/papers/NN00new.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.cs.helsinki.fi/u/ahyvarin/papers/NN00new.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1676" LastActivityDate="2010-09-03T14:32:51.310" CommentCount="4" />
  <row Id="3958" PostTypeId="2" ParentId="3953" CreationDate="2010-09-03T14:42:49.810" Score="1" ViewCount="" Body="&lt;p&gt;Take the minimal $n$ such that there are $x,y \in R(\omega)$ for which $x \not= y$ and $g(x)=g(y)=n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $g(x) = g(y)$ then we know the following: $|x| = |y|$ and $\forall u \in x \exists v \in y (g(u) = g(v))$ but since it is clear that $u \in v \Rightarrow g(u) &amp;lt; g(v)$ we have that there is some $u \in x, v\in y$ for which $g(u) = g(v) &amp;lt; g(y) = n$ which contradictions the assumption.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strike&gt;I've got no idea about the surjective part. Seems slightly tricky, but I'll get back to you when I have an answer.&lt;/strike&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know that $g(\emptyset) = 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $n$ be the minimal number that is not in the range of $g$, we can write $n = 2^{n_0}+\ldots +2^{n_k}$ where $n_0 &amp;lt; n_1 &amp;lt; \ldots &amp;lt; n_k$. Since $n_i &amp;lt; n$ we have that there is $x_i \in R(\omega)$ for which $g(x_i) = n_i$. So now we can take $x = \{x_i |0\le i\le k\}$ and it is a simple argument that $x \in R(\omega)$ and clearly $g(x) = n$.&lt;/p&gt;&#xA;" OwnerUserId="622" LastEditorUserId="622" LastEditorDisplayName="" LastEditDate="2010-09-03T14:56:33.820" LastActivityDate="2010-09-03T14:56:33.820" CommentCount="1" />
  <row Id="3959" PostTypeId="2" ParentId="3955" CreationDate="2010-09-03T14:43:38.377" Score="2" ViewCount="" Body="&lt;p&gt;The function you want is $y=10(100-x)$. This was obtained by attempting to compute the quadratic function that passes through the three points ($n+1$ points uniquely determine an nth-degree curve), and then noting that the interpolating polynomial is actually a linear function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively, computing the determinant&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\begin{vmatrix}1&amp;amp;0&amp;amp;1000\\1&amp;amp;50&amp;amp;500\\1&amp;amp;100&amp;amp;0\end{vmatrix}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(the area of the &quot;triangle&quot; formed by the three points)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;gives a value of 0, and thus the three points are collinear. Take any two of the three points, and you can derive the equation of the line passing through them.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-09-03T14:43:38.377" CommentCount="1" />
  <row Id="3960" PostTypeId="2" ParentId="3912" CreationDate="2010-09-03T15:11:21.847" Score="4" ViewCount="" Body="&lt;p&gt;Note that the notation--the use of $n$ and base-2 logs--indicates this problem comes from computer science, strongly suggesting that $n$ is an integer and we are being asked about the asymptotics as $n$ grows without bound.  When $n$ is even, the value is $\log(n)$, but when $n$ is odd, the value is either $n + \log(n)$ or $-n + \log(n)$.  Thus the growth equals $\max \left( O(\log(n)), O(-n + \log(n)), O(n + \log(n)) \right) = O(n)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is false that $O(3) \ge O(n)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;BTW, one wouldn't use an expression like $O(3)$, because the constant is implicit.  Conventionally, this is written $O(1)$.&lt;/p&gt;&#xA;" OwnerUserId="1489" LastActivityDate="2010-09-03T15:11:21.847" CommentCount="2" />
  <row Id="3962" PostTypeId="2" ParentId="3437" CreationDate="2010-09-03T15:52:48.587" Score="1" ViewCount="" Body="&lt;p&gt;It seems plausible that there may be no square factors other than $4$, or at least that there are only finitely many primes for which $p^2$ divides $s_n$. A sketchy argument follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Set $\tau=2 +\sqrt{3}$. Then $s_n = \tau^{2^n} + \tau^{-2^n}$. We work in the ring $R=\mathbb{Z}[\sqrt{3}]$, where $\tau$ is a unit. So $p^2$ divides $S_n$ if and only if $\tau^{2^{n+1}} \equiv 1 \mod p^2$, where our congruence is in $R$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $p$ be a prime $\geq 5$. Let $U_{p^2}$ be the unit group of $R/p^2 R$. You want to know whether the order of $\tau$ in $U_{p^2}$ is a power of $2$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are two cases. First, suppose there is a square root of $3$ mod $p$ (this happens when $p \equiv \pm 1 \mod 12$). This square root will lift to a solution $s$ of $s^2 \equiv 3 \mod p^2$ (we can lift by Hensel's lemma). In this case $R/p^2 \cong \mathbb{Z}/p^2 \oplus \mathbb{Z}/p^2$ and we can identify $\tau$ with the ordered pair $(2 + s, 2 - s)$. Notice that $2-s = (2+s)^{-1}$, so the two halves of the ordered pair have the same order. The unit group of $\mathbb{Z}/p^2$ is cyclic of order $p(p-1)$, and the number of elements whose orders are powers of $p$ is $2^{v_2(p-1)}$, where $v_2(k)$ is the exponent to which $2$ divides $k$. So, heuristically, I would expect the number of $p$ for which $p^2$ divides $s_n$ and there is a square root of $3$ mod $p$ to be something like&#xA;$$\sum_{p \equiv \pm 1 \mod 12} \frac{2^{v_2(p-1)}}{p(p-1)}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There case where where $p \equiv \pm 5 \mod 12$ is a little trickier but, if I haven't made any mistakes, the similar heuristic estimate is&#xA;$$\sum_{p \equiv \pm 5 \mod 12} \frac{2^{v_2(p+1)}}{p(p+1)}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, here is my point: those sums converge. In fact,&#xA;$$\sum_{n=1}^{\infty} \frac{2^{v_2(n)}}{n^2}$$&#xA;converges to $\pi^2/4$ (exercise!) and the above sums are less than this because they are restricted to just summing over primes. Moreover, the sum numerically looks pretty small. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the &quot;expected&quot; number of such primes is a small finite quantity, and it wouldn't surprise me if $2$ were the only one.&lt;/p&gt;&#xA;" OwnerUserId="448" LastActivityDate="2010-09-03T15:52:48.587" />
  <row Id="3963" PostTypeId="2" ParentId="3955" CreationDate="2010-09-03T15:54:22.283" Score="2" ViewCount="" Body="&lt;p&gt;There is no way to uniquely answer your question without knowing more about what class of functions you permit to interpolate these points. Notice that all of your points are divisible by 50, so we may simplify a bit by scaling down by 50, yielding the points $\rm (x,y) = (0,20), (1,10), (2,0)$. Since the intermediate slopes agree $(10-20)/(1-0) = -10 = (0-10)/(2-1)\;$ it follows that a linear function interpolates these points, viz. $\rm\;y = 20-10x,\;$ or $\rm\;Y = 1000 - 10X\;$ when scaled up by 50 to $\rm (X,Y) = (50x,50y)$. But one could easily interpolate many other curves between these points. Perhaps if you tell us more about your intended application we can elaborate further.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-09-11T03:25:41.893" LastActivityDate="2010-09-11T03:25:41.893" />
  <row Id="3964" PostTypeId="2" ParentId="164" CreationDate="2010-09-03T16:27:40.433" Score="2" ViewCount="" Body="&lt;p&gt;I wanted to post this as a comment to the answer by Justin L. but it is my first posting here so I couldn't. Archimedes was so proud of his proof that the volume of a sphere is two-thirds of the volume of a cylinder of the same height and diameter that he asked that a sculpture illustrating this would be placed at his tomb. The tomb was later found by the Roman orator Cicero who describes it as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;I managed to track down his grave. The Syracusians knew nothing about it, and indeed denied that any such thing existed. But there it was, completely surrounded and hidden by bushes of brambles and thorns. I remembered having heard of some simple lines of verse which had been inscribed on his tomb, referring to a sphere and cylinder modelled in stone on top of the grave. And so I took a good look round all the numerous tombs that stand beside the Agrigentine Gate. Finally I noted a little column just visible above the scrub: it was surmounted by a sphere and a cylinder.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The American painter Benjamin West imagined the scene in his 1797 painting &lt;a href=&quot;http://tinyurl.com/363fx3o&quot; rel=&quot;nofollow&quot;&gt;&quot;Cicero Discovering the Tomb of Archimedes&quot;.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1702" LastEditorUserId="1702" LastEditorDisplayName="" LastEditDate="2011-03-16T23:32:16.840" LastActivityDate="2011-03-16T23:32:16.840" />
  <row Id="3965" PostTypeId="2" ParentId="3927" CreationDate="2010-09-03T16:42:11.653" Score="5" ViewCount="" Body="&lt;p&gt;A professor of mine, Francisco Raggi, once said in class:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;There are two ways to study a ring. You can stare at it, and stare at it, and stare a it, until you can spot interesting features of the ring, and then you can start describing those interesting features. Or you can study its images and the way it acts on other structures, by considering homomorphic images and modules over the ring. The latter is usually easier and more fruitful.&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When you are studying a particular structure, you can study it in isolation: you can consider a vector space, and its subsets and subspaces, and say a lot of interesting things about that vector space; but in the end, it doesn't get you all that far. The same is true when you study a particular ring (and you can learn &lt;em&gt;a lot&lt;/em&gt; just by staring at that particular ring; think of classical Number Theory as the result of staring intently at the integers looking for interesting features), or groups, or topological spaces, etc. But it turns out that studying the &lt;em&gt;functions&lt;/em&gt; between similar objects that preserve the structure (linear transformations for vector spaces, ring and group homomorphisms, continuous maps, etc) leads to a much richer palette, with a lot more information and a lot more &lt;em&gt;structure&lt;/em&gt; to work with. In some cases, like vector spaces, the new objects have themselves the same structure that you were studying (the set of all linear transformations $T\colon\mathbf{V}\to\mathbf{W}$ can be given a natural structure of a vector space, so everything you know about vector spaces applies) or some other structure (if $A$ and $B$ are abelian groups, then the set of all group homomorphism $A\to B$ can be given a ring structure by pointwise addition and composition, so you can study &lt;em&gt;it&lt;/em&gt; as you would study a ring) which in turn gives you a wealth of information about the structures themselves. And of course, things like the Isomorphism Theorems can give you a lot of information about the structure you are looking at by looking at its images, which will often be &quot;simpler&quot; or &quot;smaller&quot; than the one you were originally concerned with. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even in the case were we have found a lot of good information by simply staring (as in classical number theory), just look at how big a leap forward was achieved with the introduction of congruences (homomorphisms from $\mathbb{Z}$ to $\mathbb{Z}/m\mathbb{Z}$), or with algebraic number theory (mapping to larger rings and then trying to import the information back into $\mathbb{Z}$).&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-09-03T16:42:11.653" CommunityOwnedDate="2010-09-04T00:42:24.980" />
  <row Id="3966" PostTypeId="1" CreationDate="2010-09-03T16:57:22.823" Score="1" ViewCount="209" Body="&lt;p&gt;Ok so I'm a programmer, I'm not a mathematician, I've got a minor in math but I didn't even do particularly well at it so please bear with some possibly really stupid thoughts.  Just please try to explain to me why what I'm asking is stupid so that I don't keep making a fool of myself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I recently found myself thinking along the following lines:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you take the stand that it's solipsism to talk about anything that cannot be put to use in  predicting an outcome then cannot you apply the same axiom to basic mathematical operations?  For example does addition really have any meaning outside of the context of placing two objects in a box and predicting how many are inside?  Yeah, I realize that's a dangerous question to ask a board full of mathematicians but I imagine you can ask this question about any axiom.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is that in reality if you put one thing and then another into a box it does not mean that there will be two things in the box.  There will likely be two things of course but if the objects are point particles like electrons there's a chance there will be one or three, or a million electrons there.  Heck, the same is even possible with apples, just incredibly unlikely.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't have the mathematical wherewithal to think through what this would mean but intuitively it would seem that this might not have many implications for addition, but might for subtraction (unless you could somehow have negative amounts of particles which I won't rule out), and certainly for things like integration and derivation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would assume that someone's done work along these lines before?  Has anyone actually created a system around it?  Was it useful?  Are there any accessible books or articles about it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just interested&lt;/p&gt;&#xA;" OwnerUserId="1703" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2010-09-04T05:01:28.840" LastActivityDate="2010-10-01T16:24:49.200" Title="Is there a Mathematical system which is stochastic at its core?" Tags="&lt;probability-theory&gt;" AnswerCount="3" CommentCount="7" FavoriteCount="1" />
  <row Id="3967" PostTypeId="2" ParentId="164" CreationDate="2010-09-03T16:59:07.623" Score="3" ViewCount="" Body="&lt;p&gt;To give you a larger perspective, the volume of an $n-1$-sphere in $\mathbb{R}^n$ is given by $C_n r^n$, where $r$ is the radius and $C_n=\frac{\pi^{n/2}}{\Gamma(\frac{n}{2} + 1)}$. For even $n$, this reduces to $C_n =\frac{\pi^{n/2}}{\left(\frac{n}{2}\right)!}$, and for odd $n$, it becomes $C_n =2^{(n+1)/2}\frac{\pi^{(n-1)/2}}{n!!}$, where $n!!$ denotes the double factorial.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus in the case $n=3$, the constant is $2^2 \frac{\pi^1}{3!!} = \frac{4 \pi}{3}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The point of this is to show you that the general formula also involves factors in the denominator, and that the formula for $n=3$ is not 'random' but rather fits into a general pattern.&lt;/p&gt;&#xA;" OwnerUserId="873" LastActivityDate="2010-09-03T16:59:07.623" />
  <row Id="3968" PostTypeId="1" AcceptedAnswerId="3974" CreationDate="2010-09-03T17:06:10.590" Score="3" ViewCount="153" Body="&lt;p&gt;I hope this question is suitable for the site.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I recently had to work with Mathematica, and the experience was, to put it kindly, unpleasing. I do not have much experience with similar programs, but I remember not liking much Matlab or Maple either. The result is that I am a mathematician who likes programming, but I never managed to learn how to work with a computer algebra system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does there exist a CAS which can be programmed using a standard language? I guess the best thing would be just an enormous library of mathematical algorithms implemented for C or Python or whatever.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know SAGE is based on Python, but as far as I understand (which is not much) it just collects preexisting open source software, so (I assume) one has to learn how to use a new tool for every different problem.&lt;/p&gt;&#xA;" OwnerUserId="233" LastActivityDate="2010-09-04T14:58:46.557" Title="CAS with a standard language" Tags="&lt;computer-algebra-systems&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="1" />
  <row Id="3969" PostTypeId="2" ParentId="3927" CreationDate="2010-09-03T17:29:03.960" Score="0" ViewCount="" Body="&lt;p&gt;Here are some striking examples from looking at the Lebesgue space $L^1(\mathbb{R})$ under Fourier transform. $L^1(\mathbb{R})$ is an &lt;em&gt;algebra&lt;/em&gt; under convolution, that is it is a vector space with a multiplication (here multiplication is the convolution). The Fourier transform $\mathcal{F}: L^1(\mathbb{R})\to C_0(\mathbb{R})$ is an &lt;em&gt;algebra homomorphism&lt;/em&gt; from $L^1$ into a sub-algebra $\mathcal(L^1(\mathbb{R}))$ of the space $C_0$ (continuous functions that vanish at $\pm\infty$). &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;In particular &#xA;$$\mathcal{F}(f*g) = \mathcal{F}(f)\cdot\mathcal{F}(g)$$ &#xA;and instead of studying convolution equations in $L^1$ we may study multiplicative equations in the Fourier image. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;An other example is &lt;a href=&quot;http://en.wikipedia.org/wiki/Wiener_tauberian_theorem&quot; rel=&quot;nofollow&quot;&gt;The Wiener Tauberian Theorem&lt;/a&gt;, that states that the translates of $f$ span a dense subspace of $L^1$ if and only if $\mathcal{F}(f)$ is non-zero.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="1154" LastActivityDate="2010-09-03T17:29:03.960" CommunityOwnedDate="2010-09-04T00:42:24.980" />
  <row Id="3970" PostTypeId="2" ParentId="3955" CreationDate="2010-09-03T17:34:47.683" Score="5" ViewCount="" Body="&lt;p&gt;As you mention “series” in the question I was wondering if something as simple as the term “Arithmetic Progression” is what you're after.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your first arithmetic progression being $$a_n=n$$ for n=0,1,...,100 and your second being&#xA;$$b_n=1000-10n=1000-10a_n.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given your chosen name, I was wondering if your intended application has anything to do with optics? (I ask as my wife is an optician.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Not really part of the answer but on a lighter note, here's a wonderful quote involving optics from the book “Littlewood's Miscellany”:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From time to time (since 1910) there were moves to get rid of the revolting optics and astronomy set in the Mathematical Tripos. It was discovered that over a period of years no wrangler attempted a question in either subject. An equivalent form of this proposition is that every attempt to do a question in optics or astronomy resulted in a failure to get a 1st. &lt;/p&gt;&#xA;" OwnerUserId="1301" LastActivityDate="2010-09-03T17:34:47.683" />
  <row Id="3971" PostTypeId="2" ParentId="3917" CreationDate="2010-09-03T17:41:47.910" Score="5" ViewCount="" Body="&lt;p&gt;I'm assuming you want to know more about group theory, so here is an answer to point you to some of the interesting things that you can learn if you are interested in computing with finite groups.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(i) If you have the conjugacy classes and there are not too many, then writing normal subgroups as unions of conjugacy classes is fast and easy.  This is often the case when you only have access to a group through its &lt;a href=&quot;http://en.wikipedia.org/wiki/Character_table&quot; rel=&quot;nofollow&quot;&gt;character table&lt;/a&gt;, such as when the Suzuki and the sporadic simple groups were being discovered, and we wanted to understand some of their (non-simple) subgroups.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a typical finite group given concretely as a permutation group, you use a special type of induction working along what is called a &lt;a href=&quot;http://en.wikipedia.org/wiki/Chief_series&quot; rel=&quot;nofollow&quot;&gt;chief series&lt;/a&gt;, where you find a maximal chain of normal subgroups.  It turns out there are some fairly easy ways to find these: for a &lt;a href=&quot;http://en.wikipedia.org/wiki/Solvable_group&quot; rel=&quot;nofollow&quot;&gt;solvable group&lt;/a&gt;, or any group G with an abelian quotient group, you can fairly easily and concretely find the &lt;a href=&quot;http://en.wikipedia.org/wiki/Derived_subgroup&quot; rel=&quot;nofollow&quot;&gt;derived subgroup&lt;/a&gt;, [G,G].  The quotient group is an abelian group, so every subgroup between the whole group and the derived subgroup is normal.  You then take the derived subgroup of the derived subgroup, but now you only take subgroups that are normalized by G/[G,G].  The action of G/[G,G] is relatively plain and easy to understand, so these &quot;invariant&quot; subgroups are pretty easy to find.  To finish the inductive step, you need to find the normal subgroups of G/[[G,G],[G,G]] that don't contain [G,G], and this can be done by a slightly easier version of the general subgroup lattice algorithm using what is called the first &lt;a href=&quot;http://en.wikipedia.org/wiki/Group_cohomology&quot; rel=&quot;nofollow&quot;&gt;cohomology group&lt;/a&gt;, or &quot;derivations&quot; (basically derivatives for groups, and related in a way to the derived subgroup).  When you reach the point where the group is its own derived subgroup, a &lt;a href=&quot;http://en.wikipedia.org/wiki/Perfect_group&quot; rel=&quot;nofollow&quot;&gt;perfect group&lt;/a&gt;, then a second algorithm begins, that identifies the simple groups involved in the top of the group, and then constructs the solvable group below them.  Sometimes you even have to repeat those last two steps, but not for groups of order less than 60&lt;sup&gt;24&lt;/sup&gt; ≈ 4.7e42 or so.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(ii) Often in mathematics, you want to use &quot;induction&quot;.  A normal subgroup is one of the two main ways to do induction in group theory.  Usually it is not necessary to find all normal subgroups, but rather a single (nice) chief series will do.  Some groups only have a single chief series (a fair number of dihedral groups are like this), and so finding the chief series and finding all normal subgroups is the same question.  You might try the symmetric group of degree 4 and order 24 as an example.&lt;/p&gt;&#xA;" OwnerUserId="583" LastActivityDate="2010-09-03T17:41:47.910" />
  <row Id="3972" PostTypeId="2" ParentId="3897" CreationDate="2010-09-03T18:32:52.033" Score="4" ViewCount="" Body="&lt;p&gt;As a double check, let's perform the integral in a completely different manner (so that my mistakes are unlikely to overlap yours!).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is handy to use a condensed notation.  I will write $[i,j,k;a]$ for the value of $x^{2 i}y^{2 j}z^{2 k} \rho^{2 a}$ integrated over the unit sphere, where $i,j,k$ are integral and $a$ is real.  These relations are easy to establish:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;$[i,j,k;a]$ is invariant under permutations of $(i,j,k)$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$6 [1,1,1;a] = [0,0,0;a+3] - 3[3,0,0;a] - 18[2,1,0;a]$ is a consequence of expanding $\rho^6 = (x^2+y^2+z^2)^3$ as a multinomial, using the first property to collect equal terms, and isolating $[1,1,1;a]$ on the lhs.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$[i,j,k;a+1] = [i+1,j,k;a] + [i,j+1,k;a] + [i,j,k+1;a]$ follows from $\rho^2 = x^2 + y^2 + z^2$.  Use this to compute $[2,1,0;a]$ in terms of $[3,0,0;a]$ and $[2,0,0;a+1]$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$[n,0,0;a] = \frac {4 \pi} {(2 n + 1) (2 n + 3 + 2 a)}$ can be obtained via integration by parts in cylindrical coordinates or by directly performing the integration of $ z^{2 n}  \rho ^ {2 a}$ over the sphere, which is very easy to do (because the angular part only involves $\phi$ and the integrand is exact).  (The numerator is the area of the unit sphere and the factors in the denominator come from integrating a $2 n$ power (in the angular part of $z$, which separates from the $\rho$ integral) and a $2 n + 2 a + 2$ power due to the $\rho^{2 n} \rho^{2 a} \rho^2 d \rho$ term in the integrand.)  Even when $a \lt 0$, this is justified whenever $2 n + 3 + 2 a \gt 0$, because the integral still converges at the origin.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;From these we algebraically obtain&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\eqalign{&#xD;&#xA;[1,1,1;a] = &amp;amp;\frac{1}{6} \left( [0,0,0;a+3] - 3 [3,0,0;a] - 18 [2,1,0;a] \right) \cr&#xD;&#xA;&#xD;&#xA;= &amp;amp;\frac{4 \pi} {6} \left( \frac{1}{2 a+9} - \frac{3}{7 (2 a+9)} - \frac{18}{35(2 a+9)} \right) \cr&#xD;&#xA;&#xD;&#xA;= &amp;amp;\frac{4 \pi}{105 (2 a+9)}.&#xD;&#xA;}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Setting $2 a = -17/2$ gives the textbook answer $\frac{8 \pi}{105}$.&lt;/p&gt;&#xA;" OwnerUserId="1489" LastEditorUserId="1489" LastEditorDisplayName="" LastEditDate="2010-12-24T19:33:58.383" LastActivityDate="2010-12-24T19:33:58.383" CommentCount="6" />
  <row Id="3973" PostTypeId="1" AcceptedAnswerId="4294" CreationDate="2010-09-03T19:15:12.087" Score="1" ViewCount="354" Body="&lt;p&gt;Hi guys, I was trying to obtain this result:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;integrate e^|x| dx&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.wolframalpha.com/input/?i=integrate+e%5e|x|+dx&quot; rel=&quot;nofollow&quot;&gt;The result&lt;/a&gt; is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \int e^{|x|} dx = \frac12 e^{-x} \left( (e^x-1)^2 \operatorname{sgn}(x) -2e^x +e^{2x} - 1 \right) \color{gray}{\,+ \text{constant}} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is fine but I would like to see the 2 part solution for $x&gt;0$ and $x&amp;lt;0$, I know you can see that because just after before the final output I can see it but then it switches to the one line solution above.&#xA;Is there a way to see the other solution?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it's not ok to ask about a tool used in mathematics I'm sorry. You can close it.&lt;/p&gt;&#xA;" OwnerUserId="1704" LastEditorUserId="171" LastEditorDisplayName="" LastEditDate="2010-09-04T15:25:30.147" LastActivityDate="2010-09-08T15:42:47.733" Title="Wolfram Alpha: How to show functions with abs() and sgn() as a piecewise function?" Tags="&lt;math-software&gt;&lt;integral&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="3974" PostTypeId="2" ParentId="3968" CreationDate="2010-09-03T19:24:44.093" Score="1" ViewCount="" Body="&lt;p&gt;If you're a big fan of C numerous libraries exist for doing certain tasks. MPIR/GMP are two libraries for doing arbitrary precision arithmetic. On top of these you'll find more specialised libraries like MPC which adds support for the complex field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Flint is a library that sits on top of MPIR, for example, providing an array of number theory functions. PARI/GP I believe provides a similar foundation but via the GMP library.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm sure if you googled &quot;library for linear algebra&quot; or &quot;library for x&quot; you would find it. However, a CAS unites all of this functionality.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we come to sage. From what I understand of the sage project, it is not so much a collection of open source libraries but an interface uniting them. For example, from the tutorial:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;sage: a = 5   # a is an integer&#xA;sage: type(a)&#xA;&amp;lt;type 'sage.rings.integer.Integer'&amp;gt;&#xA;sage: a = 5/3  # now a is a rational number&#xA;sage: type(a)&#xA;&amp;lt;type 'sage.rings.rational.Rational'&amp;gt;&#xA;sage: a = 'hello'  # now a is a string&#xA;sage: type(a)&#xA;&amp;lt;type 'str'&amp;gt;&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Is a sage session. As you're probably aware &lt;code&gt;sage.rings.integer.Integer&lt;/code&gt; is not a built-in python type.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would recommend giving sage a go. I think it is the closest thing out there to what you're after. I personally don't find Mathematica too bad.&lt;/p&gt;&#xA;" OwnerUserId="892" LastActivityDate="2010-09-03T19:24:44.093" />
  <row Id="3976" PostTypeId="1" AcceptedAnswerId="3986" CreationDate="2010-09-03T19:27:16.507" Score="4" ViewCount="328" Body="&lt;p&gt;I have been thinking a little bit about the binary decimal expansion of reciprocal prime numbers; and I have a few questions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found &lt;a href=&quot;http://brad.chalfan.net/binq/qtob.html&quot; rel=&quot;nofollow&quot;&gt;this neat table&lt;/a&gt; which lists the binary expansion of many fractions, and I was trying to find some patterns.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are my questions, for brevity I say a natural number has a period N if the binary decimal expansion of it's reciprocal has period N: (for example, 1/7 = .001001... has period 3)&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Given an arbitrary natural number N, does there exist a prime number of minimum period N?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;(By minimum period I mean to exclude the case that one prime has a period which is a multiple of the period of another prime.  For example, 1/3 = .0101... has period 2, and 1/5 = .00110011... has period 4; so while 1/3 has period 4, what I call it's &quot;minimum period&quot; is 2)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2.&#xA; Can two prime numbers have the same minimum period? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A useful result which I believe is well known, is that a natural number has a period N if and only if it is a factor of 2&lt;sup&gt;N&lt;/sup&gt; - 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anyone know of a good reference that describes some theory behind the relationship between the period of the reciprocal of a natural number, and the prime factorization of that number?&lt;/p&gt;&#xA;" OwnerUserId="1417" LastEditorUserId="1417" LastEditorDisplayName="" LastEditDate="2010-09-03T19:44:18.943" LastActivityDate="2010-09-04T00:24:40.890" Title="On the binary decimal expansion of the reciprocal prime's" Tags="&lt;prime-numbers&gt;&lt;decimal-expansion&gt;" AnswerCount="1" CommentCount="7" FavoriteCount="1" />
  <row Id="3977" PostTypeId="1" CreationDate="2010-09-03T20:10:17.607" Score="11" ViewCount="785" Body="&lt;p&gt;Does it exist a free interactive geometry software, like GeoGebra, which works for 3D geometry? I would be able to draw spheres, great circles, and so on.&lt;/p&gt;&#xA;" OwnerUserId="354" LastEditorUserId="1069" LastEditorDisplayName="" LastEditDate="2011-03-22T21:18:56.380" LastActivityDate="2011-03-22T21:40:13.513" CommunityOwnedDate="2010-09-03T20:10:17.607" Title="3D software like GeoGebra" Tags="&lt;geometry&gt;&lt;math-software&gt;&lt;teaching&gt;" AnswerCount="8" CommentCount="2" FavoriteCount="3" />
  <row Id="3978" PostTypeId="1" AcceptedAnswerId="4040" CreationDate="2010-09-03T20:15:22.570" Score="9" ViewCount="753" Body="&lt;p&gt;I happened to receive this from my friend.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $a,b \in \mathbb{N}$, such that $a^{n}+n \: \bigl|\: b^{n}+n$ for all $n \in \mathbb{N}$. Prove that $a=b$. How do we proceed?&lt;/p&gt;&#xA;" OwnerUserId="9413" LastEditorUserId="9413" LastEditorDisplayName="anonymous" LastEditDate="2011-06-02T16:46:48.740" LastActivityDate="2011-06-02T16:46:48.740" Title="$(a^{n}+n ) \mid (b^{n}+n) \Longrightarrow a=b$" Tags="&lt;number-theory&gt;&lt;elementary-number-theory&gt;" AnswerCount="5" CommentCount="13" FavoriteCount="3" />
  <row Id="3979" PostTypeId="2" ParentId="3977" CreationDate="2010-09-03T20:16:47.750" Score="2" ViewCount="" Body="&lt;p&gt;You can try pgftikz &lt;a href=&quot;http://www.texample.net/tikz/examples/&quot; rel=&quot;nofollow&quot;&gt;http://www.texample.net/tikz/examples/&lt;/a&gt;&lt;/p&gt;&#xA;" LastActivityDate="2010-09-03T20:24:43.840" CommunityOwnedDate="2010-09-03T20:16:47.750" CommentCount="1" />
  <row Id="3982" PostTypeId="2" ParentId="3977" CreationDate="2010-09-03T22:03:31.217" Score="2" ViewCount="" Body="&lt;p&gt;I think you can, to a limited extent, use &lt;a href=&quot;http://sketchup.google.com/&quot; rel=&quot;nofollow&quot;&gt;Google Sketchup&lt;/a&gt;, but that's not really the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.cabri.com/cabri-3d.html&quot; rel=&quot;nofollow&quot;&gt;Cabri3D&lt;/a&gt; is probably the most like what you want, but it's commercial.  &lt;a href=&quot;http://www.autograph-math.com/&quot; rel=&quot;nofollow&quot;&gt;Autograph&lt;/a&gt;, also commercial, may do some of that kind of stuff, but I'm not sure.&lt;/p&gt;&#xA;" OwnerUserId="72" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-09-04T01:28:37.913" LastActivityDate="2010-09-04T01:28:37.913" CommunityOwnedDate="2010-09-03T22:03:31.217" CommentCount="2" />
  <row Id="3983" PostTypeId="2" ParentId="3294" CreationDate="2010-09-03T22:36:41.800" Score="3" ViewCount="" Body="&lt;p&gt;You probably already know this, but let me explicitly point out:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;== exact match ==&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is not possible to construct a single cubic Bezier curve that exactly matches your new curve.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A single cubic Bezier curve has a constant &quot;curvature&quot; A along its length.&#xA;When you shrink it, the new curve has &quot;sharper&quot; curvature B (but still constant along the length of the new curve).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is not possible to construct a single cubic Bezier curve -- since, like all single cubic curves, it has constant curvature C along its length -- such that C=A along the part of the length, and C=B along the rest of the length, where A&amp;lt;&gt;B.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I'm mis-using the term &quot;curvature&quot; here for something that is more precisely the 3rd derivative of a curve, sometimes called the &quot;jerk&quot;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;== approximation ==&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps the simplest approximation is:&#xA;Use the initial starting point and first control knot -- p0 and p1,&#xA;and the final control knot and final ending point -- r'2 and r'3.&#xA;Use those 4 points as the start, control knots, and endpoint of a Bezier curve that approximates the one you want: a0, a1, a2, a3.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This approximation exactly hits the endpoints of your desired curve, and has the same initial and final slope at those endpoints, but it slightly diverges from your desired curve in the middle.&#xA;In particular, it probably doesn't go exactly through the cutpoint K.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are many other possible approximations you could make.&#xA;The &lt;a href=&quot;http://www.tinaja.com/cubic01.asp&quot; rel=&quot;nofollow&quot;&gt;&quot;Don Lancaster's Guru's Lair Cubic Spline Library&quot;&lt;/a&gt;&#xA;may have the details I'm leaving out:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;It is possible to nudge a1 along the line p0-p1, or to nudge a2 along the line r'2-r'3 -- or both -- such that the approximation curve not only starts and ends at the same points and slopes, but also passes through the cutpoint K.&lt;/li&gt;&#xA;&lt;li&gt;pick any 4 points along your new curve (perhaps the endpoints, the cutpoint K, and some other point) and generate a cubic Bezier curve that goes exactly through all 4 points.&lt;/li&gt;&#xA;&lt;li&gt;pick many points along your new curve -- bunched together in places where curve matching is important, perhaps near the endpoints and point K, and spaced further apart where curve matching is not so important -- and generate a cubic Bezier curve that is a least-squares best fit to those points.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1705" LastActivityDate="2010-09-03T22:36:41.800" CommentCount="3" />
  <row Id="3984" PostTypeId="2" ParentId="297" CreationDate="2010-09-03T22:55:10.810" Score="5" ViewCount="" Body="&lt;p&gt;Check out the&#xA;&lt;a href=&quot;http://stackoverflow.com/questions/tagged/pi&quot;&gt;questions tagged 'pi' at Stack Overflow&lt;/a&gt;.&#xA;Many of them discuss simple numerical methods for calculating any number of digits of pi.&lt;/p&gt;&#xA;" OwnerUserId="1705" LastActivityDate="2010-09-03T22:55:10.810" />
  <row Id="3985" PostTypeId="2" ParentId="297" CreationDate="2010-09-03T23:24:58.723" Score="1" ViewCount="" Body="&lt;p&gt;Check out the wiki page on approximations to pi, &lt;a href=&quot;http://en.wikipedia.org/wiki/Numerical_approximations_of_%CF%80#20th_century&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Numerical_approximations_of_%CF%80#20th_century&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I really love Ramanujan's method, which I recall converges extremely rapidly.&lt;/p&gt;&#xA;" OwnerUserId="1417" LastActivityDate="2010-09-03T23:24:58.723" />
  <row Id="3986" PostTypeId="2" ParentId="3976" CreationDate="2010-09-03T23:45:43.247" Score="2" ViewCount="" Body="&lt;p&gt;(In the below post there are several links with the apostrophes omitted; fill them in if the links don't work.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The phenomenon you are studying is a phenomenon in &lt;a href=&quot;http://en.wikipedia.org/wiki/Modular_arithmetic&quot; rel=&quot;nofollow&quot;&gt;modular arithmetic&lt;/a&gt;.  If a prime $p$ has period $n$, this means that there is some numerator $N$ such that $\frac{N}{2^n - 1} = \frac{1}{p}$.  This is equivalent to $Np = 10^n - 1$, or $p | 2^n - 1$, or $2^n \equiv 1 \bmod p$.  The smallest $n$ for which this is true is called the &lt;strong&gt;order&lt;/strong&gt; of $2 \bmod p$, sometimes denoted $\text{ord}_p(2)$ (although this is confusingly also used to denote the greatest power of $p$ which divides $2$...).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Fermats_little_theorem&quot; rel=&quot;nofollow&quot;&gt;Fermat's little theorem&lt;/a&gt; guarantees that $\text{ord}_p(2)$ always divides $p-1$; you have already observed this yourself.  However, predicting the exact order is very difficult to do in general.  For example, knowing that the order is actually equal to $p-1$ is equivalent to knowing that $2$ is a &lt;a href=&quot;http://en.wikipedia.org/wiki/Primitive_root_modulo_n&quot; rel=&quot;nofollow&quot;&gt;primitive root&lt;/a&gt;, and it is not currently even known whether this is true &lt;a href=&quot;http://en.wikipedia.org/wiki/Artins_conjecture_on_primitive_roots&quot; rel=&quot;nofollow&quot;&gt;infinitely often&lt;/a&gt;.  In any case, you should be able to find basic information about order in any good textbook on elementary number theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With that background out of the way...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer to question 1 is &lt;strong&gt;no&lt;/strong&gt;.  The only exception is $n = 6$ by &lt;a href=&quot;http://en.wikipedia.org/wiki/Zsigmondys_theorem&quot; rel=&quot;nofollow&quot;&gt;Zsigmondy's theorem&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer to question 2 is &lt;strong&gt;yes&lt;/strong&gt;.  If $p$ and $n$ are relatively prime, then $p$ has period $n$ if and only if $p$ divides $\Phi_n(2)$, where $\Phi_n(x)$ is the $n^{th}$ &lt;a href=&quot;http://en.wikipedia.org/wiki/Cyclotomic_polynomial&quot; rel=&quot;nofollow&quot;&gt;cyclotomic polynomial&lt;/a&gt;.  (This is more or less a restatement of the condition that $p | 2^n - 1$ but $p$ doesn't divide $2^k - 1$ for $k &amp;lt; n$.)  So it suffices to show that some number of this form has more than one prime factor relatively prime to $n$.  There are two cases here which are particularly classical:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;$n$ is a prime $q$.  In this case $\Phi_q(2) = 2^q - 1$ is a &lt;a href=&quot;http://en.wikipedia.org/wiki/Mersenne_prime&quot; rel=&quot;nofollow&quot;&gt;Mersenne number&lt;/a&gt;, and $2^{11} - 1 = 23 \cdot 89$ is the smallest composite Mersenne number, hence $23$ and $89$ both have period $11$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$n = 2^k$ for some $k$.  In this case $\Phi_{2^k}(2) = 2^{2^{k-1}} + 1$ is a &lt;a href=&quot;http://en.wikipedia.org/wiki/Fermat_number#Basic_properties&quot; rel=&quot;nofollow&quot;&gt;Fermat number&lt;/a&gt;, and $\Phi_{64}(2) = 2^{32} + 1 = 641 \cdot 6700417$ is the smallest composite Fermat number, hence $641$ and $6700417$ both have period $64$.  &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The answer to question 3 is the following.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Lemma:&lt;/strong&gt; If $n, m$ are relatively prime odd numbers, then $\text{ord}_{mn}(2) = \text{lcm}(\text{ord}_n(2), \text{ord}_m(2))$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Proof.&lt;/em&gt;  $\text{ord}_{mn}(2)$ is the order of the element $2$ in the &lt;a href=&quot;http://en.wikipedia.org/wiki/Multiplicative_group_of_integers_modulo_n&quot; rel=&quot;nofollow&quot;&gt;multiplicative group&lt;/a&gt; of $\mathbb{Z}/mn\mathbb{Z}$, which we will denote $U(mn)$.  By the &lt;a href=&quot;http://en.wikipedia.org/wiki/Chinese_remainder_theorem&quot; rel=&quot;nofollow&quot;&gt;Chinese remainder theorem&lt;/a&gt;, $U(mn)$ is isomorphic to the &lt;a href=&quot;http://en.wikipedia.org/wiki/Direct_product_of_groups&quot; rel=&quot;nofollow&quot;&gt;direct product&lt;/a&gt; $U(m) \times U(n)$, so the order of $2$ in $U(mn)$ must be the $\text{lcm}$ of the orders of $2$ in $U(m)$ and $U(n)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It follows that to compute $\text{ord}_m(2)$ for arbitrary $m$ it suffices to compute it for the odd prime power factors of $m$ and then to take the $\text{lcm}$ of the resulting numbers.  (Note that if $m$ is divisible by a power of $2$ this only contributes a leading string of zeroes to the binary expansion of $\frac{1}{m}$ and hence does not affect the computation of the period.)  Again, you can find a discussion of the Chinese remainder theorem in any good textbook on elementary number theory.  (I am particularly bad at recommending textbooks on elementary number theory because I learned mine through a summer program, not a textbook...)&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-09-04T00:24:40.890" LastActivityDate="2010-09-04T00:24:40.890" CommentCount="1" />
  <row Id="3987" PostTypeId="1" AcceptedAnswerId="4080" CreationDate="2010-09-03T23:51:48.490" Score="3" ViewCount="364" Body="&lt;p&gt;I was going to ask this on Stack Overflow, but finally decided this was more math than programming. I may still turn out to be wrong about that, but...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given a number represented in binary, it's fairly easy to derive a decimal representation using integer division and remainders to extract digits in reverse order. You can then simply count how many digits you extracted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I wondered about calculating an &lt;em&gt;exact&lt;/em&gt; number of decimal digits more efficiently by avoiding calculating the actual digits.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At first sight this is easy. First, count the binary digits needed (ie determine the position of the highest set bit). Many processors have an operation to do this directly, and there are bit-fiddling algorithms to do it in O(log n) time where n varies with the width of a machine register. In an arbitrary position integer, you normally know how many machine words you have, so you can usually jump directly to the most significant word.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you know the size in binary digits, you can scale by $\log 2 \over \log 10$ to get the number of decimal digits. Sort of.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is that this scale factor is (I think) an irrational number. If you have a maximum number of digits you need to worry about, you can use a rational approximation and you only need to worry about getting the rounding right.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is, therefore - is there a way to determine this number of decimal digits efficiently and precisely for numbers of any size? Is there (for example) a completely different approach to the calculation that I haven't thought of?&lt;/p&gt;&#xA;" OwnerUserId="510" LastEditorUserId="510" LastEditorDisplayName="" LastEditDate="2010-09-07T01:26:38.787" LastActivityDate="2010-09-07T01:26:38.787" Title="How to calculate the number of decimal digits for a binary number?" Tags="&lt;algorithms&gt;&lt;arithmetic&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="3989" PostTypeId="2" ParentId="297" CreationDate="2010-09-04T01:48:27.217" Score="1" ViewCount="" Body="&lt;p&gt;Since the AGM hasn't been mentioned yet (apart from my plug to read the Borwein and Borwein book), here's how to use the AGM to compute $\pi$:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;a←(1+1/√(2))/2&#xA;c←1/(8a)&#xA;s←a^2&#xA;f←1&#xA;repeat&#xA;a←(a+√((a-c)(a+c)))/2&#xA;c←(c^2)/(4a)&#xA;f←2f&#xA;s←s-fc^2&#xA;until |c| is small enough&#xA;(2a^2)/(2s-1)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;which exploits the Legendre relation.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-09-04T01:48:27.217" />
  <row Id="3990" PostTypeId="1" CreationDate="2010-09-04T02:44:33.170" Score="3" ViewCount="117" Body="&lt;p&gt;Assume you have a triangle in the plane defined by its three vertices at $(x_0,y_0)$, $(x_1,y_1)$, and $(x_2,y_2)$. Is there a general expression for the moments of the triangle, where the moments are defined as averages by the following expression:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\langle x^n y^m \rangle = \frac{1}{A}\int_\triangle x^n y^m \, dA $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Integrating manually to compute the first few moments, one can obtain:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\langle x \rangle = \frac{1}{3} \left(x_0+x_1+x_2\right) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\langle x^2 \rangle = \frac{1}{6} \left(x_0^2+x_1^2+x_2^2+x_0 x_1+x_0 x_2+x_1 x_2\right) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\langle x y \rangle = \frac{1}{12}\left(2 x_0 y_0+x_1 y_0+x_2 y_0+x_0 y_1+2 x_1 y_1+x_2 y_1+x_0 y_2+x_1 y_2+2 x_2 y_2 \right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it possible to obtain a closed form expression for general $n$ and $m$ using combinations of the coordinates?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can the same be done for line segments defined by their endpoints $(x_0,y_0)$ and $(x_1,y_1)$?&#xA;Again the first few moments in this case are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\langle x \rangle = \frac{1}{2} \left(x_0+x_1\right) $$&#xA;$$ \langle x^2 \rangle = \frac{1}{3} \left(x_0^2+x_0 x_1+x_1^2\right) $$&#xA;$$ \langle x^3 \rangle = \frac{1}{4} \left(x_0^3+x_0^2 x_1+x_0 x_1^2+x_1^3\right) $$&#xA;$$ \langle x y \rangle = \frac{1}{6} \left(2 x_0 y_0+x_1 y_0+x_0 y_1+2 x_1 y_1\right) $$&#xA;$$ \langle x^2 y \rangle  = \frac{1}{12} \left( 3 x_0^2 y_0+2 x_0 x_1 y_0+x_1^2 y_0+x_0^2 y_1+2 x_0 x_1 y_1+3 x_1^2 y_1\right)$$&lt;/p&gt;&#xA;" OwnerUserId="1393" LastEditorUserId="1393" LastEditorDisplayName="" LastEditDate="2010-09-04T02:55:54.703" LastActivityDate="2010-09-04T11:02:37.463" Title="Computing the moments of a triangle" Tags="&lt;geometry&gt;&lt;combinatorics&gt;&lt;euclidean-geometry&gt;&lt;integral&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="3991" PostTypeId="1" AcceptedAnswerId="3993" CreationDate="2010-09-04T03:02:22.437" Score="1" ViewCount="173" Body="&lt;p&gt;I'm working through a discrete maths text book and was stumped as to how to prove the closed form solution of the &lt;a href=&quot;http://mathworld.wolfram.com/HofstadterG-Sequence.html&quot; rel=&quot;nofollow&quot;&gt;Hofstadter G-Sequence&lt;/a&gt;&lt;br&gt;&#xA;$a(0) = 0$ and $a(n) = n - a(a(n-1)), n \geq 1$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The closed form is $a(n) = \lfloor(n+1)\mu\rfloor$  where $\mu$ is the inverse of the golden ratio.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The text book says to prove the following two statements on the way to proving this.&lt;br&gt;&#xA;Show that $n\mu -\lfloor n\mu \rfloor + n\mu^2 - \lfloor n\mu^2 \rfloor = 1$ for all n &gt; 0&lt;br&gt;&#xA;Show that $\lfloor(1+\mu)(1-\alpha)\rfloor + \lfloor\alpha+\mu\rfloor = 1$ for all $\alpha \in R, 0 \leq \alpha &amp;lt; 1\space and\space \alpha \ne 1-\mu$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm relatively certain I was able to confirm those those two statements but I'm not sure where to apply them when I start on the main proof. I think it's just my weakness with algebra, particularly how to deal with the floor that's holding me up.&lt;/p&gt;&#xA;" OwnerUserId="1455" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2011-06-03T20:11:49.927" LastActivityDate="2011-06-03T20:11:49.927" Title="Proof of closed form Hofstadter G-Sequence" Tags="&lt;sequences-and-series&gt;&lt;algebra-precalculus&gt;&lt;discrete-mathematics&gt;&lt;recurrence-relations&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="3992" PostTypeId="2" ParentId="3990" CreationDate="2010-09-04T03:07:59.690" Score="1" ViewCount="" Body="&lt;p&gt;It should be possible using &lt;a href=&quot;http://en.wikipedia.org/wiki/Green%27s_theorem&quot; rel=&quot;nofollow&quot;&gt;Green's Theorem&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/hnZe9.png&quot; alt=&quot;Green's Theorem Image from wikipedia&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Refer my earlier answer here for more details: &lt;a href=&quot;http://math.stackexchange.com/questions/3177/why-doesnt-a-simple-mean-give-the-position-of-a-centroid-in-a-polygon/3182#3182&quot;&gt;http://math.stackexchange.com/questions/3177/why-doesnt-a-simple-mean-give-the-position-of-a-centroid-in-a-polygon/3182#3182&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And especially the link given there is more relevant here: &lt;a href=&quot;http://www.worldserver.com/turk/computergraphics/Moments.pdf&quot; rel=&quot;nofollow&quot;&gt;Computing the Moments of a 2D polygon using Green's Theorem&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-09-04T03:17:10.493" LastActivityDate="2010-09-04T03:17:10.493" />
  <row Id="3993" PostTypeId="2" ParentId="3991" CreationDate="2010-09-04T04:00:41.780" Score="1" ViewCount="" Body="&lt;p&gt;Our goal is to prove $$\lfloor \mu  (n+1)\rfloor  = n-\lfloor \mu  \lfloor \mu  n+1\rfloor \rfloor. $$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you can prove  $\lfloor \mu  \lfloor \mu  n+1\rfloor \rfloor  = \lfloor (1-\mu ) (n+1)\rfloor $ then you should be able to prove the above without too much effort.&lt;/p&gt;&#xA;" OwnerUserId="899" LastActivityDate="2010-09-04T04:00:41.780" CommentCount="1" />
  <row Id="3994" PostTypeId="2" ParentId="3953" CreationDate="2010-09-04T04:02:30.310" Score="2" ViewCount="" Body="&lt;p&gt;It is easier to build the isomorphism in the other direction, and indeed we can see that the map is forced upon us. There is a unique isomorphism.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Specifically, you want a map $h:\mathbb{N}\to R(\omega)$ such that $n\mathrel{E} m\iff h(n)\in h(m)$. This very equivalence tells you that you must define &lt;code&gt;$h(m)=\{ h(n) \mid n\mathrel{E} m\}$&lt;/code&gt;. This function is defined by recursion, and obviously preserves $E$ to $\in$. Thus it is injective. It is surjective because your function $g$ is the inverse. QED&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If one knows about the Mostowski collapse, then you can see immediately that $h$ is precisely the Mostowski collapse of $(\mathbb{N},E)$, which is a well-founded extensional relation. This provides another way to see that $h$ is an isomorphism.&lt;/p&gt;&#xA;" OwnerUserId="413" LastActivityDate="2010-09-04T04:02:30.310" CommentCount="1" />
  <row Id="3995" PostTypeId="2" ParentId="3978" CreationDate="2010-09-04T04:38:29.900" Score="-2" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;UPDATE&lt;/strong&gt;: the proof seems incorrect as b=4, a=2, fail when n=4 (check the comments below).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, there is a counter example when n=1, where a=6 and b=13 (as $ (6+1) | (13+1) $ ). However I shall continue for all n&gt;1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $(a^n+n) | (b^n+n)$ then there is a natural number k $\geq 1$ such that $b^n+n=k(a^n+n)$. Clearly $a=b \mbox{ iff } k=1$. We shall proceed to prove that k is always 1 for all n&gt;1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Continuing from $b^n+n=k(a^n+n)$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$b^n+n \equiv 0 \quad (\mbox{mod } k)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we allow n to vary for the same a and b, then the only value for k that will make that equation hold for all n&gt;1 is k=1, which implies that a=b.&lt;/p&gt;&#xA;" OwnerUserId="558" LastActivityDate="2010-09-04T06:47:41.943" CommentCount="8" />
  <row Id="3996" PostTypeId="2" ParentId="3966" CreationDate="2010-09-04T05:00:00.780" Score="2" ViewCount="" Body="&lt;p&gt;Your question is undoubtedly philosophical in nature: it is predicated on the premise &quot;it's solipsism to talk about anything that cannot be put to use in predicting an outcome&quot;.  But &lt;a href=&quot;http://en.wikipedia.org/wiki/Solipsism&quot; rel=&quot;nofollow&quot;&gt;solipsism&lt;/a&gt; is a philosophical term (and more fundamentally, an inherently philosophical -- as opposed to mathematical or scientific -- idea)!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The philosophical view you advocate seems to be, roughly, that the truths of arithmetic are (only) empirical in nature.  As with many basic philosophical ideas, this has been propounded before and has gotten a lot of discussion.  Famously, it was the view of arithmetic espoused by &lt;a href=&quot;http://en.wikipedia.org/wiki/John_Stuart_Mill&quot; rel=&quot;nofollow&quot;&gt;John Stuart Mill&lt;/a&gt;.  For a pretty good quick summary of his positions, see&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://plato.stanford.edu/entries/mill/#GeoAri&quot; rel=&quot;nofollow&quot;&gt;http://plato.stanford.edu/entries/mill/#GeoAri&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mill's ideas on arithmetic and geometry have generally been poorly received by later philosophers.  Notably, much of the work of &lt;a href=&quot;http://en.wikipedia.org/wiki/Immanuel_Kant&quot; rel=&quot;nofollow&quot;&gt;Immanuel Kant&lt;/a&gt; is a reaction against Mill's ideas, especially his notion of the &quot;synthetic a priori&quot;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So far as I know, &lt;strike&gt;essentially no&lt;/strike&gt; very few mathematicians or philosophers of mathematics have accepted Mill's ideas.  Rather, we have followed the path of &lt;a href=&quot;http://en.wikipedia.org/wiki/Frege&quot; rel=&quot;nofollow&quot;&gt;Frege&lt;/a&gt; and viewed logic as being the foundation for mathematics.  (Later work by Russell, Godel and others have showed that there are some pitfalls and limitations of the logicist approach, but it has nevertheless been the one mathematicians have adopted for the last hundred years.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[As Qiaochu says, if you really want to talk mathematics you should remove the overtly philosophical lead to your question.  Instead, you should say more about the mathematical aspects of your question: what mathematical problem are you trying to solve?]&lt;/p&gt;&#xA;" OwnerUserId="299" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2010-09-04T13:46:05.657" LastActivityDate="2010-09-04T13:46:05.657" CommentCount="2" />
  <row Id="3997" PostTypeId="2" ParentId="3973" CreationDate="2010-09-04T07:52:00.100" Score="1" ViewCount="" Body="&lt;p&gt;You can use the following to get the two cases:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Assuming[t&amp;gt;0, integrate e^|x| dx from x=0 to t]&#xA;Assuming[t&amp;lt;0, integrate e^|x| dx from x=0 to t]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Or, if you don't care about choosing the constant of integration such that the two pieces fit together, simply:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Assuming[x&amp;gt;0, integrate e^|x| dx]&#xA;Assuming[x&amp;lt;0, integrate e^|x| dx]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;(although then it's so easy that I don't see why you would need Wolfram Alpha).&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2010-09-04T07:52:00.100" CommentCount="6" />
  <row Id="3998" PostTypeId="2" ParentId="3968" CreationDate="2010-09-04T08:04:50.717" Score="1" ViewCount="" Body="&lt;p&gt;If you don't want to install all of Sage (which is quite big), you could perhaps try one of its components, &lt;a href=&quot;http://wiki.sympy.org/wiki/Main_Page&quot; rel=&quot;nofollow&quot;&gt;SymPy&lt;/a&gt;, on its own to be begin with.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you use C++ maybe &lt;a href=&quot;http://www.ginac.de/&quot; rel=&quot;nofollow&quot;&gt;GiNaC&lt;/a&gt; is useful (I haven't tried it).&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2010-09-04T08:04:50.717" />
  <row Id="3999" PostTypeId="1" AcceptedAnswerId="4000" CreationDate="2010-09-04T08:58:14.113" Score="4" ViewCount="410" Body="&lt;p&gt;Is a prime number in the decimal system still a prime when converted to a different base?&lt;/p&gt;&#xA;" OwnerUserId="1409" LastActivityDate="2010-09-04T19:54:43.680" Title="Is a prime number still a prime when in a different base?" Tags="&lt;prime-numbers&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="1" />
  <row Id="4000" PostTypeId="2" ParentId="3999" CreationDate="2010-09-04T09:10:25.080" Score="7" ViewCount="" Body="&lt;p&gt;Here is the same question which has been asked at Mathforum: Here is the link&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://mathforum.org/library/drmath/view/55880.html&quot; rel=&quot;nofollow&quot;&gt;http://mathforum.org/library/drmath/view/55880.html&lt;/a&gt;&lt;/p&gt;&#xA;" LastActivityDate="2010-09-04T09:10:25.080" />
  <row Id="4001" PostTypeId="2" ParentId="3990" CreationDate="2010-09-04T09:42:03.227" Score="1" ViewCount="" Body="&lt;p&gt;If you don't mind a quadruple sum as a closed form (it was the best simplification of the integral I gave in the comments that I could manage, using repeated applications of the binomial theorem), then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\langle x^p y^q\rangle=\displaystyle 2\sum_{k=0}^p\sum_{\ell=0}^q \frac{(-1)^{k+\ell}\binom{p}{k}\binom{q}{\ell}\nu_{k\ell}}{k+\ell+2} x_0^{p-k} y_0^{q-\ell}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle \nu_{k\ell}=\sum_{i=0}^k\sum_{j=0}^\ell \frac{\binom{k}{i}\binom{\ell}{j}}{k-i+\ell-j+1}\left(x_0-x_1\right)^i \left(x_1-x_2\right)^{k-i} \left(y_0-y_1\right)^j \left(y_1-y_2\right)^{\ell-j}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is an explicit expression for $\langle x^p y^q\rangle$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;FWIW, &lt;em&gt;Mathematica&lt;/em&gt; is unable to evaluate this (as usual with such complicated multiple sums).&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;For the problem of finding the moments of a line segment with endpoints $\left(x_0, y_0\right)$ and $\left(x_1, y_1\right)$, the integral can be derived by noting that a set of parametric equations for the line passing through the two given points is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle \begin{align*}x&amp;amp;=x_0+t(x_1-x_0)\\y&amp;amp;=y_0+t(y_1-y_0)\end{align*}\qquad 0\leq t\leq1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Substituting the two equations into the expression $x^p y^q$ and integrating over the parameter range gives&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle \int_0^1\left(x_0+t\left(x_1-x_0\right)\right)^p\left(y_0+t\left(y_1-y_0\right)\right)^q\mathrm{d}t$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which can be expanded into a double sum (again by using the binomial theorem):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle \sum_{j=0}^p\sum_{k=0}^q\frac{\binom{p}{j} \binom{q}{k}}{j+k+1}\left(x_1-x_0\right)^j \left(y_1-y_0\right)^k x_0^{p-j} y_0^{q-k}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which can be &quot;simplified&quot; into the &lt;a href=&quot;http://mathworld.wolfram.com/AppellHypergeometricFunction.html&quot; rel=&quot;nofollow&quot;&gt;Appell hypergeometric function&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x_0^p y_0^q F_1\left(1;-p,-q;2;1-\frac{x_1}{x_0},1-\frac{y_1}{y_0}\right)$&lt;/p&gt;&#xA;" OwnerUserId="498" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-09-04T11:02:37.463" LastActivityDate="2010-09-04T11:02:37.463" />
  <row Id="4002" PostTypeId="2" ParentId="2735" CreationDate="2010-09-04T09:58:30.723" Score="0" ViewCount="" Body="&lt;p&gt;Searching for &quot;Out of core LU&quot; on google might help you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Essentially, there are two simple choices:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;An iterative method that merely depends on repeatedly computing A*v and perhaps A'*v for a given vector 'v'. Such a computation can be done without having the entire matrix A in RAM&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;A direct method, for this the &quot;out of core LU&quot; might help.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;In general, for both choices above, searching for external memory algorithms will be useful.&lt;/p&gt;&#xA;" OwnerUserId="1709" LastActivityDate="2010-09-04T09:58:30.723" CommentCount="1" />
  <row Id="4003" PostTypeId="1" AcceptedAnswerId="4006" CreationDate="2010-09-04T10:03:14.843" Score="1" ViewCount="383" Body="&lt;p&gt;We know that a group $G$ cannot be written as the set theoretic union of two of its proper subgroups. Also $G$ can be written as the union of 3 of its proper subgroups if and only if $G$ has a homomorphic image, a non-cyclic group of order 4?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this paper &lt;a href=&quot;http://www.jstor.org/stable/2695649&quot; rel=&quot;nofollow&quot;&gt;http://www.jstor.org/stable/2695649&lt;/a&gt; by M.Bhargava, it is shown that a group $G$ is the union of its proper normal subgroups if and only if its has a quotient that is isomorphic to $C_{p} \times C_{p}$ for some prime $p$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to make the condition more stringent on the subgroups. We know that Characteristic subgroups are normal. So can we have a group $G$ such that , $$G = \bigcup\limits_{i} H_{i}$$ where each $H_{i}$'s are Characteristic subgroups of $G$?&lt;/p&gt;&#xA;" LastActivityDate="2010-09-04T13:41:29.890" Title="Group as the Union of Subgroups" Tags="&lt;abstract-algebra&gt;&lt;group-theory&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="2" />
  <row Id="4004" PostTypeId="1" AcceptedAnswerId="4005" CreationDate="2010-09-04T11:47:03.397" Score="3" ViewCount="101" Body="&lt;h2&gt;Taken from : &lt;a href=&quot;http://answers.yahoo.com/question/index?qid=20100904034737AAfGkYF&quot; rel=&quot;nofollow&quot;&gt;answers.yahoo.com/question/index?qid=20100904034737AAfGkYF&lt;/a&gt;&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;A mall has sweep stakes where they give you a coupon for every X dollars you spend (amount is not important). You fill the coupon with your contact details and put it in the box. Every two weeks, they draw 10 coupons, give their owners prizes, and discard the rest of the coupons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, thanks to some very big purchase we made, we have about 300 coupons (going to hurt when filling them). To get the highest probability of winning, should we put all the coupons at once, or split them evenly between the three remaining draws?&lt;/p&gt;&#xA;" OwnerUserId="1711" LastActivityDate="2010-09-04T13:07:17.310" Title="Probability of winning sweep stakes, put all coupons in one box or split them between boxes?" Tags="&lt;probability&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="4005" PostTypeId="2" ParentId="4004" CreationDate="2010-09-04T13:07:17.310" Score="2" ViewCount="" Body="&lt;p&gt;Let's assume that in each box there are n other tickets besides yours. So if you put them all in the first box and 10 tickets are drawn your chance of winning something is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$q = 1 - \prod_{k=1}^{10} \left( 1 - \frac{300}{n+301-k} \right).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now if you split them, 100 each amongst three boxes (where, again, each box has n other tickets), your chance of winning something is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$p = 1 - \prod_{k=1}^{10} \left( 1 - \frac{100}{n+101-k} \right)^3 .$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now p&gt;q, but only slightly, for large n.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, with n=10000, $$p \approx 0.258176 \textrm{ and } q \approx 0.256004$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you're slightly better off if you split them up into three lots of 100. However, this assumes that the number of other tickets placed in the box every two weeks remains constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the difference between p and q decreases with increasing n.&lt;/p&gt;&#xA;" OwnerUserId="1301" LastActivityDate="2010-09-04T13:07:17.310" />
  <row Id="4006" PostTypeId="2" ParentId="4003" CreationDate="2010-09-04T13:16:00.683" Score="2" ViewCount="" Body="&lt;p&gt;One way to ensure this happens is to have every maximal subgroup be characteristic.  To get every maximal subgroup normal, it is a good idea to check p-groups first.  To make sure the maximal subgroups are characteristic, it makes sense to make sure they are simply not isomorphic.  To make sure there are not too many maximal subgroups, it makes sense to take p=2 and choose a rank 2 group.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact the &lt;a href=&quot;http://en.wikipedia.org/wiki/Quasidihedral_group&quot; rel=&quot;nofollow&quot;&gt;quasi-dihedral groups&lt;/a&gt; have this property.  Their three maximal subgroups are cyclic, dihedral, and quaternion, so each must be fixed by any automorphism.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So a specific example is QD16, the Sylow 2-subgroup of GL(2,3).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another small example is 4×S&lt;sub&gt;3&lt;/sub&gt;.  It has three subgroups of index 2, a cyclic, a dihedral, and a 4 acting on a 3 with kernel 2.  Since these are pairwise non-isomorphic, they are characteristic too.  It also just so happens (not surprisingly, by looking in the quotient 2×2) that every element is contained in one of these maximal subgroups.&lt;/p&gt;&#xA;" OwnerUserId="583" LastActivityDate="2010-09-04T13:16:00.683" CommentCount="1" />
  <row Id="4007" PostTypeId="1" AcceptedAnswerId="4013" CreationDate="2010-09-04T13:19:56.060" Score="3" ViewCount="126" Body="&lt;p&gt;One of my math professors gave me the following challenge. It isn't graded, it's just for fun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the function:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f_n(x)=x+3^3x^3+5^3x^5+...+(2n-1)^3x^{2n-1}, x \epsilon (0, 1)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to find which of the following functions $f_n$ is getting close to as $n$ gets larger:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ \displaystyle a)\frac{x(x+1)(x^4+22x^2+1)}{(x-1)^4(x+1)^4}$ $\displaystyle b) \frac{x(x^2+1)(x^4+22x^2+1)}{(x-1)^4(x+1)^4}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle c) \frac{x^2(x+1)(x^4+22x^2+1)}{(x-1)^4(x+1)^4}$ $\displaystyle d) \frac{x^2(x^2+1)(x^4+22x^2+1)}{(x-1)^4(x+1)^4}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Based on some tests i ran in mathematica by giving $n$ and $x$ values, it looks like $b)$ is the answer, but I am not sure. Can anyone confirm or deny this, and show how one might find the right answer, either with pen and paper or by using mathematica or maple or some other software?&lt;/p&gt;&#xA;" OwnerUserId="456" LastActivityDate="2010-09-05T19:24:10.970" Title="Finding a function that approaches another function" Tags="&lt;calculus&gt;&lt;math-software&gt;" AnswerCount="4" />
  <row Id="4008" PostTypeId="2" ParentId="4003" CreationDate="2010-09-04T13:31:06.323" Score="0" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Update:&lt;/strong&gt; While searching on the internet i found this paper, &quot;A remark on Hyperabelian groups&quot; by G.Baumslag where there is this remark &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Moreover so is any torsion abelian group of finite rank, since it is the union of finite characteristic subgroups.&lt;/strong&gt;  Would like to get a reference for this.&lt;/p&gt;&#xA;" LastActivityDate="2010-09-04T13:41:29.890" CommentCount="1" />
  <row Id="4009" PostTypeId="1" AcceptedAnswerId="4010" CreationDate="2010-09-04T13:32:07.540" Score="1" ViewCount="46" Body="&lt;p&gt;Hi! I'm doing some old exams and found following question:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Set $S={{1,2,3}}$ is given. Provide an&#xA;  example of binary operation in set S,&#xA;  binary relation in set S and a&#xA;  function $f:S\rightarrow R$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So, I'm thinking about $+_4$ as operation, but wouldn't it be a relation too? I can take ordered pair from $S$ and for every ordered pair associate an element form $S$ with every pair.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the third part, I'm thinking about $y=x*1.25$ or something similar, but that part isn't so problematic for me. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I did read the Wikipedia articles, but the difference between operations  and relations isn't clear to me.&lt;/p&gt;&#xA;" OwnerUserId="348" LastActivityDate="2010-09-04T13:40:15.120" Title="Confusion between operation and relation: Clarification needed." Tags="&lt;relations&gt;" AnswerCount="1" CommentCount="0" />
  <row Id="4010" PostTypeId="2" ParentId="4009" CreationDate="2010-09-04T13:40:15.120" Score="3" ViewCount="" Body="&lt;p&gt;A binary operation is a function from $S\times S \to S$ such as addition, multiplication or anything really.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A binary relation is just a subset of $S^2$, that is not necessarily a function and it doesn't have to include all the elements of $S$ in one way or another.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A function $f\colon S\to R$ is a relation, this time it's a subset of $S\times R$ however it satisfies a certain property, if you take some $s \in S$ then there is only a unique ordered pair with $s$ in it, so if you have $\langle s,r_1\rangle$ as well $\langle s,r_2\rangle$ then you can say that $r_1 = r_2$.&lt;/p&gt;&#xA;" OwnerUserId="622" LastActivityDate="2010-09-04T13:40:15.120" CommentCount="1" />
  <row Id="4011" PostTypeId="2" ParentId="3633" CreationDate="2010-09-04T13:42:42.280" Score="1" ViewCount="" Body="&lt;p&gt;([Update]:  Oh sorry; I'm new here. Didn't notice the much more substantial link to the mathoverflow. But perhaps it is interesting for some newbie anyway... [/update]) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the function is a polynomial (or powerseries) with constant term =/= zero this is difficult, but sometimes a meaningful solution can be given.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the function is as above, but the constant term is zero, then it is just the question of relatively simple manipulation of the formal powerseries/polynomial to construct a &quot;compositional&quot; (or &quot;iterative&quot;) root. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a very good deal in L.Comtet &quot;Advanced Combinatorics&quot; at pages around 130..150 (don't have it around).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also the keywords &quot;Bell-matrix&quot;, &quot;Carleman-matrix&quot; are helpful: these matrices transform the problem of funtion composition/iteration to matrix-products/matrix-powers. (The matrix-notation just implies the required formal powerseries-manipulations) . This is well established for functions f(x)  = a*x + b*x^2 + c*x^3 + ... &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the constant term does not vanish, f(x) = K + a*x + b*x^2 + c*x^3 + ... then things are usually much more difficult. But there is one -I think: again well established- workaround: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rewrite f(x) as some function g(x+x0) - x0 such that now g(x) has no constant term and then apply the above mentioned procedures (solving for iterative squarerroot and the like) to g(x) .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example: denote the iterative root of a some function f(x) by f°0.5(x) then solve&lt;/p&gt;&#xA;&#xA;&lt;p&gt;f°0.5(x) = g°0.5(x-x0)+x0&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where you first must find x0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a &lt;em&gt;very&lt;/em&gt; basic introduction you may try my small essay at &lt;/p&gt;&#xA;&#xA;&lt;p&gt;go.helms-net.de/math/tetdocs&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and look for the article &quot;countinuous functional iteration&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;HTH -&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gottfried Helms&lt;/p&gt;&#xA;" OwnerUserId="1714" LastEditorUserId="1714" LastEditorDisplayName="" LastEditDate="2010-09-04T15:11:09.747" LastActivityDate="2010-09-04T15:11:09.747" />
  <row Id="4012" PostTypeId="2" ParentId="4007" CreationDate="2010-09-04T14:09:57.463" Score="1" ViewCount="" Body="&lt;p&gt;Yes, b).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In Maple, if you do&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; map(factor, sum((2*i-1)^3*x^(2*i-1),i=1..n));&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The result will be 2 terms, one which is shape $(x^2)^{(n+1)}*r(n,x)$ with $r(n,x)$ polynomial in $n$ and rational in $x$, and the second term is (exactly) (b).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What this is is actually writing&#xA;$$\sum_{i=1}^{n} (2i-1)^3x^{2i-1} = \sum_{i=1}^{\infty} (2i-1)^3x^{2i-1} - \sum_{i=N-1}^{\infty} (2i-1)^3x^{2i-1}$$&#xA;(and the term order gets flipped in Maple's output).  Both the resulting sums are 'easy' to do by hand.&lt;/p&gt;&#xA;" OwnerUserId="620" LastActivityDate="2010-09-04T14:09:57.463" />
  <row Id="4013" PostTypeId="2" ParentId="4007" CreationDate="2010-09-04T14:19:01.903" Score="6" ViewCount="" Body="&lt;p&gt;Yes looks like it must be b)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is only one $x$ factor, so this eliminates choices c) and d).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The function is odd, i.e $f(-x) = -f(-x)$, so this eliminates a).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus b) must be the answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To find the answer by hand you can do the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Start with &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f(x) = x + x^3 + x^5 + \cdots = \frac{x}{1-x^2} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now differentiate &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f'(x)  = 1 + 3x^2 + 5x^4 + \cdots $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Multiply by $x$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f'(x)x = x + 3x^3 + 5x^5 + \cdots $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Differentiate again&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(f'(x)x)' = 1 + 3^2 x^2 + 5^2 x^4 + \cdots $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Multiply by $x$, differentiate again and then multiply by $x$ to give the answer.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-09-04T16:06:18.810" LastActivityDate="2010-09-04T16:06:18.810" CommentCount="2" />
  <row Id="4014" PostTypeId="2" ParentId="4007" CreationDate="2010-09-04T14:25:50.493" Score="0" ViewCount="" Body="&lt;p&gt;Here's yet another way to verify that it's (b). Let Mathematica do partial fractions decomposition:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Apart[x(x^2+1)(x^4+22x^2+1)/(x^2-1)^4]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This gives $\frac{1}{2 (x+1)}-\frac{7}{2(x+1)^2}+\frac{6}{(x+1)^3}-\frac{3}{(x+1)^4}+\frac{1}{2 (x-1)}+\frac{7}{2(x-1)^2}+\frac{6}{(x-1)^3}+\frac{3}{(x-1)^4}$, which can be expanded into power series using the geometric series and its derivatives.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course this is ugly, and requires you to know the answer in advance. Moron's answer is much better. ;-)&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2010-09-04T14:25:50.493" />
  <row Id="4016" PostTypeId="1" AcceptedAnswerId="4017" CreationDate="2010-09-04T15:26:52.140" Score="3" ViewCount="121" Body="&lt;p&gt;Consider the Banach space $\ell^1(\mathbb N)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sequence $(e_n)_{n\in\mathbb N}$ feels like a kind of basis because every element $a\in\ell^1(\mathbb N)$ can be written as an absolutely convergent infinite linear combination $\sum_{n\in\mathbb N}a(n)e_n$ in a unique way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Here $e_n$ denotes the vector whose $n$th entry is 1 and all of whose other entries vanish.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same is true for the Banach space $c_0(\mathbb N)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is the above property of the sequence $(e_n)_{n\in\mathbb N}$ appropriate in order to abstractly define a basis of a Banach space? Has this been considered?&lt;/p&gt;&#xA;" OwnerUserId="367" LastEditorUserId="171" LastEditorDisplayName="" LastEditDate="2010-09-04T15:34:21.130" LastActivityDate="2010-09-04T15:43:50.053" Title="Is there a notion of basis for Banach spaces?" Tags="&lt;functional-analysis&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" />
  <row Id="4017" PostTypeId="2" ParentId="4016" CreationDate="2010-09-04T15:43:50.053" Score="3" ViewCount="" Body="&lt;p&gt;As Soarer points out: Yes, it is called a &lt;a href=&quot;http://en.wikipedia.org/wiki/Schauder_basis&quot; rel=&quot;nofollow&quot;&gt;Schauder basis&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="367" LastActivityDate="2010-09-04T15:43:50.053" CommunityOwnedDate="2010-09-04T15:43:50.053" />
  <row Id="4018" PostTypeId="1" AcceptedAnswerId="4019" CreationDate="2010-09-04T15:47:43.173" Score="3" ViewCount="224" Body="&lt;p&gt;Let $X = \mathcal{C}([0,1],\mathbb{R})$, be the ring of all continuous valued functions $f:[0,1] \to \mathbb{R}$. For $x \in [0,1]$, let $M_{x} = \{ f \in M \ | \ f(x)=0\}$. One can show by using compactness of $[0,1]$ that every maximal ideal is of this form.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Extending the Question to Entire functions&lt;/strong&gt;: Let $\mathsf{C}(z)$ be the ring of complex entire functions. For $ \lambda \in \mathsf{C}$ let $M_{\lambda}$ denote the set of all entire functions which have a zero at $\lambda$. Then is $M_{\lambda}$ a maximal ideal in $\mathsf{C}(z)$, and does every ideal happen to be of this form? I don't know to prove this!&lt;/p&gt;&#xA;" LastActivityDate="2010-09-05T22:39:18.763" Title="Maximal Ideals in the Ring of Complex Entire Functions" Tags="&lt;analysis&gt;&lt;abstract-algebra&gt;&lt;complex-analysis&gt;" AnswerCount="2" FavoriteCount="4" />
  <row Id="4019" PostTypeId="2" ParentId="4018" CreationDate="2010-09-04T16:03:00.210" Score="11" ViewCount="" Body="&lt;p&gt;No.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $f(z)=1/\Gamma(z)$ where $\Gamma$ is the gamma function.&#xA;Then $f$ vanishes at $0$, $-1,-2,\ldots$ and nowhere else. For integers&#xA;$m$ let $f_m(z)=f(z+m)$. Then $f_m$ vanishes at $-m,-m-1,\ldots$.&#xA;Let $I$ be the ideal of the ring of the entire functions generated by the $f_m$.&#xA;Then $I$ is proper, since otherwise finitely many of the $f_m$&#xA;would generate an ideal containing the constant function $1$. But&#xA;any finite set of $f_m$ have a common zero, so this doesn't hold.&#xA;So $I\subseteq J$ for some maximal ideal $J$ (using the standard Zorn's&#xA;lemma argument). But $J\ne M_\lambda$ for any $\lambda$, as for each $\lambda$&#xA;there is some $m$ with $f_m(\lambda)\ne0$, and $f_m\in J$.&lt;/p&gt;&#xA;" OwnerUserId="226" LastEditorUserId="226" LastEditorDisplayName="" LastEditDate="2010-09-05T06:30:44.417" LastActivityDate="2010-09-05T06:30:44.417" CommentCount="2" />
  <row Id="4022" PostTypeId="1" CreationDate="2010-09-04T17:04:50.630" Score="4" ViewCount="125" Body="&lt;p&gt;I try to solve for the specific function&#xA;$f(x) = \frac{2-2a}{x-1} \int_0^{x-1} f(y) dy + af(x-1)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It looks similar to the function used to find the &lt;a href=&quot;http://mathworld.wolfram.com/RenyisParkingConstants.html&quot; rel=&quot;nofollow&quot;&gt;Renyi's parking constant&lt;/a&gt; because it came out from a simple generalization of that problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The skill I have gained in my differential class can't even solve&#xA;$f(x) = f'(x-1)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not looking for anyone to solve it. I just want to know the techniques for solving DE where functions and it's derivatives are evaluated at different points.(What's the terminology for this kind of DE?)&lt;/p&gt;&#xA;" OwnerUserId="96" LastActivityDate="2010-09-04T17:04:50.630" Title="How to solve DE that relate values of derivatives at different points?" Tags="&lt;differential-equations&gt;&lt;functional-equations&gt;" CommentCount="7" FavoriteCount="3" />
  <row Id="4023" PostTypeId="1" CreationDate="2010-09-04T17:07:04.957" Score="4" ViewCount="162" Body="&lt;p&gt;Here is a list of some applications of the exponential function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) The exponential mapping in Lie theory.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I put this first because my intuition tells me that this must be the most fundamental, or deep, way of thinking about the exponential function.  I have often been misled by my intuition however, and the main reason I feel strongly about this is because of how fundamental I consider Lie theory to be.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Fourier Series&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) Roots of Unity&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4) Gaussian Distribution&lt;/p&gt;&#xA;&#xA;&lt;p&gt;5) Boltzmann Distribution&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are certainly other applications, but it always kind of bothered me that I couldn't use symmetry methods to see how (all of) these applications are related.  Is it possible that there is no way to do this, i.e. that it's just a happy accident that the exponential function has these applications and it is unrelated to any continuous symmetries?&lt;/p&gt;&#xA;" OwnerUserId="1417" LastEditorUserId="1417" LastEditorDisplayName="" LastEditDate="2010-09-04T17:12:38.003" LastActivityDate="2011-01-05T16:44:36.070" Title="Is there a (deep) relationship between these various applications of the exponential function?" Tags="&lt;probability-theory&gt;&lt;fourier-series&gt;&lt;lie-groups&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" />
  <row Id="4024" PostTypeId="2" ParentId="3941" CreationDate="2010-09-04T17:07:48.370" Score="14" ViewCount="" Body="&lt;p&gt;Here's a slightly different way of proving that $\sin nx$ is not&#xA;a polynomial in $\sin x$ when $n$ is even. Indeed $\sin nx$&#xA;cannot be written as $f(\sin x)$ for any function $f$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To prove this all one has to do is to write down two numbers $a$&#xA;and $b$ such that $\sin a=\sin b$ but $\sin na\ne\sin nb$.&#xA;Let $a=\pi/(2n)$ and $b=\pi-a$. Then $\sin a=\sin b$. But&#xA;$na=\pi/2$ and $nb=n\pi-\pi/2$ so that $\sin na=1$ and $\sin nb=-1$&#xA;(recalling that $n$ is even).&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-09-04T17:07:48.370" CommentCount="5" />
  <row Id="4026" PostTypeId="2" ParentId="4023" CreationDate="2010-09-04T18:09:28.090" Score="2" ViewCount="" Body="&lt;p&gt;The most important traits of the exponential function that make $e^x$ occur in that much applications are&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;a)&lt;/strong&gt; &lt;a href=&quot;http://en.wikipedia.org/wiki/Euler%27s_identity&quot; rel=&quot;nofollow&quot;&gt;Euler's identity&lt;/a&gt; $e^{i\pi} + 1 = 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and its implications for the definition of complex trigonimetric functions, i.e.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$e^{i \cdot x} = \cos x + i \sin x$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(like Fourier transformation, unit roots)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;b)&lt;/strong&gt; the fact that $\frac{d}{dx}e^x = e^x$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;which lets $e^x$ naturally occur in many solutions of &lt;em&gt;differential equations&lt;/em&gt;, which are often the most basic formulation of natural laws (growth/decay, gaussian distribution).&lt;/p&gt;&#xA;" OwnerUserId="917" LastActivityDate="2010-09-04T18:09:28.090" CommentCount="1" />
  <row Id="4027" PostTypeId="2" ParentId="3999" CreationDate="2010-09-04T18:17:22.173" Score="2" ViewCount="" Body="&lt;p&gt;Alas, the accepted answer is misleading (and arguably incorrect). The reason that primality (or any other purely arithmetic property) is preserved in radix representation is simply that such representation faithfully preserves all of the arithmetic operations on integers. More precisely, $\:$ if $\rm\;n\to r(n)\;$ is the map from $\rm\:n\:$ to its radix $\rm\:d\;$ representation, then it preserves addition $\rm\;r(m+n) = r(m) + r(n),\;$ and multiplication $\rm\;r(mn) = r(m)\ r(n),\;$ and $\rm\;r\;$ has an inverse $\rm\;s\;$ that similarly also preserves addition and multiplication (technically: $\rm\:r\:$ is a ring isomorphism). This readily implies that the relation  of divisibility is faithfully preserved in radix representation, because the relation of divisibility can be expressed as an equation involving only arithmetic (ring) operations (namely multiplication), and such equations are necessarily preserved by the maps $\rm\;r\;$ and $\rm\;s\;$ - indeed these maps are defined precisely so to preserve these fundamental operations. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's instructive to examine more closely the preservation of divisibility. First, we recall the standard notation $\rm\;a|b\; := \: a\;$ divides $\rm\:b\:,\;\;$ i.e. $\rm\:\exists \:n\in\mathbb Z : \ an = b\:,\;$ i.e. there exists an integer $\rm\:n\:$ such that $\rm\;an = b\;$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;LEMMA&lt;/strong&gt; $\rm\;\quad a|b \iff r(a)|r(b)\quad\quad$ &lt;strong&gt;(Divisibility Preservation by Isomorphisms)&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof&lt;/strong&gt;: $\rm\;(\;\Rightarrow\;)\quad a|b \;\Rightarrow\; \exists\:n\in\mathbb Z: \: an = b \;\Rightarrow\; r(a)\ r(n) = r(an) = r(b) \;\Rightarrow\; r(a)|r(b)$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\;(\Leftarrow)\quad r(a)|r(b) \;\Rightarrow\;\exists\:c\in r(\mathbb Z): \: r(a)\: c = r(b)\;\Rightarrow\; r(a)\ r(n) = r(b) \;\Rightarrow\; a n = b\;$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The final $\;\Rightarrow\;$ above is by applying $\rm\;r\:$'s inverse $\rm\:s\:$ so to cancel the $\rm\;r\:$'s using $\rm\;\: sr = 1 =\;$ identity map.&lt;br&gt;&#xA;Note: this employs $\rm\:s\:$'s preservation of multiplication, $\:$ viz. $\rm\:s(r(a)\: r(n)) \:=\: sr(a)\: sr(n) \:=\: a\: n\;$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a corollary, we conclude that primes (i.e. irreducibles) are also preserved, since they are definable purely in terms of divisibilty, viz. $\rm\;p\;$ is prime $\;\: := \:\rm\;p = ab \;\Rightarrow\; p|a\;$ or $\rm\;p|b\;$ and $\rm\;p\;$ is not a unit, i.e. not $\rm\;p|1\;$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So your question reduces to the more fundamental  &lt;strong&gt;why is radix representation a ring isomorphism, i.e. why really does radix representation preserve the addition and multiplication operations?&lt;/strong&gt; This is a very good question that deserves a thoughtful answer. It's a serious pedagogical oversight that this topic is rarely discussed in algebra textbooks. Although most students understand this fact subconsciously, many have difficulty providing a rigorous proof (or, worse, they overlook the fact that it does require a rigorous proof). Your question would attract much more interest and receive much more interesting replies if you rephrase it in this manner. Thus I propose the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;SUGGESTION&lt;/strong&gt;: Edit the title of your question to say &quot;Why is radix representation a ring isomorphism, i.e. why does it preserve addition and multiplication?&quot;. Include in your question a brief description of your knowledge level so to help set the proper level for replies, e.g. do you already know any abstract algebra or elementary number theory?&#xA;Unaccept the currently accepted answer so that the software will continue to bump up the question till it gets a number of good answers. IMPORTANT: Wait at least a week before accepting any answer so that everyone has a chance to see the question. This helps serve to maximize its exposure and hence maximize your potential of receiving insightful answers. If all goes well this should result in an interesting thread that will help serve as a future reference for many similar frequently asked questions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;NOTE&lt;/strong&gt; to much more experienced readers: this problem is not as trivial as you might think at first glance (and certainly less trivial for novices). For example, the analogous problem for real numbers (or p-adics) is the subject of a famous paper [1]. For an introduction see e.g. &lt;a href=&quot;http://www.math.ou.edu/~dmccullough/teaching/slides/rotamath.pdf&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;. Its closing line is quite apropos:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;This is very much in keeping with Rota’s thinking that mathematics is not just a quest to solve problems, it is&#xA;  also a quest to understand the mathematical universe as&#xA;  clearly and as deeply as possible&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;[1] F. Faltin, N. Metropolis, B. Ross, G.-C. Rota,&lt;br&gt;&#xA;The real numbers as a wreath product.&lt;br&gt;&#xA;Advances in Math. 16 (1975), 278-304&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-09-04T19:54:43.680" LastActivityDate="2010-09-04T19:54:43.680" CommentCount="5" />
  <row Id="4028" PostTypeId="2" ParentId="3978" CreationDate="2010-09-04T18:23:12.220" Score="3" ViewCount="" Body="&lt;p&gt;Chandru1 asks how we might proceed, so in that spirit let me offer an idea that provides partial results and connects this problem to a long-standing one recently characterized as a &quot;frustrating question.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's aim to show that under the conditions assumed of $a$ and $b$, $a$ necessarily divides $b$.  This additional relation suffices to show that $a = b$.  (There does not seem to be a simple demonstration of this, but it is much easier than the original problem so I'll let it go for now as an &quot;exercise.&quot;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose $b &amp;lt; a^2$.  Then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\left( \frac{b}{a} \right) ^n - \frac{b^n + n}{a^n+n} = \frac{n (b^n - a^n)}{a^n (a^n+n)}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The right hand side, in the limit of large $n$, approaches zero from above.  The left hand side is the difference between $\left( \frac{b}{a} \right) ^n$ and an integer.  We conclude that eventually the fractional part of $\left( \frac{b}{a} \right) ^n$ approaches zero.  Results of Pisot, Vijayaraghavan and Andre Weil then imply that $\frac{b}{a}$ must be integral.  (See Akayama &lt;em&gt;et al.&lt;/em&gt;, &lt;em&gt;Powers of rationals modulo 1 and rational base number systems&lt;/em&gt;, &lt;a href=&quot;http://perso.telecom-paristech.fr/~jsaka/PUB/Files/RBNS-rev.pdf&quot; rel=&quot;nofollow&quot;&gt;http://perso.telecom-paristech.fr/~jsaka/PUB/Files/RBNS-rev.pdf&lt;/a&gt; .)  The intuition is that the fractional parts of powers of non-integers ought to fill the interval [0, 1) in a fairly &quot;random&quot; way.  Indeed, numerical experiments verify this for rational numbers (but not for all irrationals!): see &lt;a href=&quot;http://mathworld.wolfram.com/PowerFractionalParts.html&quot; rel=&quot;nofollow&quot;&gt;http://mathworld.wolfram.com/PowerFractionalParts.html&lt;/a&gt; .  So, convergence of the fractional part to zero--which is assured by the sequence of divisibility conditions in the problem--implies the ratio $b / a$ is not behaving like a proper fraction: it must actually be an integer.  That gives us enough leverage to show the equality $a = b$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suspect a similar approach should work for $b \ge a^2$, but I haven't found it.  Indeed some of the papers in the literature note changes in the behavior of powers of $b / a$ when $b$ exceeds $a^2$, so we should be cautious.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, note that elementary methods of number theory show that all primes dividing $a$ must also divide $b$.  That, however, doesn't seem to get us very far.&lt;/p&gt;&#xA;" OwnerUserId="1489" LastActivityDate="2010-09-04T18:23:12.220" CommentCount="1" />
  <row Id="4029" PostTypeId="2" ParentId="3978" CreationDate="2010-09-04T18:54:44.843" Score="5" ViewCount="" Body="&lt;p&gt;Claim if our hypothesis holds, $a \equiv b \ (\text{mod}\ p)$ for any prime $p$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proof:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Find $n$ so that $n \equiv -a \ (\text{mod}\ p)$ and $n \equiv 1 (\text{mod}\ p-1)$ ( we can do this by &lt;strong&gt;Chinese Remainder Theorem&lt;/strong&gt;).  Then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$a^{n} + n \equiv a^1 + n = a - a = 0 (\text{mod} \ p)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore since $a^n + n \mid b^n + n$, $b^n + n \equiv 0 (\text{mod}\ p)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$b^n + n \equiv b^1 + n \equiv b - a (\text{mod}\ p)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;therefore $b \equiv a \pmod p$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Our result now follows by picking any $p &gt; b$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;NOTE by BD&lt;/strong&gt; $\;$  This solution has been posted before in at least a few well-known math forums,  e.g. see my &lt;a href=&quot;http://groups.google.com/group/sci.math/browse_frm/thread/b04d0459db4c2cb5&quot; rel=&quot;nofollow&quot;&gt;sci.math post on April 4,2006&lt;/a&gt;, and see Rust's post on &lt;a href=&quot;http://www.artofproblemsolving.com/Forum/viewtopic.php?f=56&amp;amp;t=289683&quot; rel=&quot;nofollow&quot;&gt;AoPS, July 19, 2009&lt;/a&gt;. It also appeared in at least one other forum much more recently (alas, I can't recall which one). Almost surely, by now the problem and solution is listed in various problem collections, so it should be considered somewhat well-known.&lt;/p&gt;&#xA;" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-09-05T17:11:16.953" LastActivityDate="2010-09-05T17:11:16.953" CommentCount="17" />
  <row Id="4031" PostTypeId="1" CreationDate="2010-09-04T20:06:05.483" Score="2" ViewCount="266" Body="&lt;p&gt;Let $p_i$ denote the $i^{th}$ prime number. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Find the smallest positive integer $k$ such that the product $n = p_1 \cdot p_2 \cdots p_k$ satisfies $\sigma(n) &gt; 3n$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any positive integer $m &amp;lt; n$ satisfying $\sigma(m) &gt; 3m$?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Taken From&lt;/strong&gt; (Rosen), at the end of the chapter under challenge problems&lt;/p&gt;&#xA;" LastEditorUserId="1778" LastEditorDisplayName="" LastEditDate="2010-10-05T01:10:15.130" LastActivityDate="2010-10-05T01:10:15.130" Title="Sum of divisors and prime numbers, short proof" Tags="&lt;number-theory&gt;&lt;prime-numbers&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="4032" PostTypeId="1" AcceptedAnswerId="4041" CreationDate="2010-09-04T20:13:20.567" Score="4" ViewCount="285" Body="&lt;p&gt;I have to prove that $||A-B||=||A+B||\Leftrightarrow AB=0$ and I was wondering if this approach is correct, or if there's a better/more elegant way to prove this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given n-dimensional vectors A and B, we can write $||A-B||=||A+B||$ as:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sqrt{\sum\limits_{j=1}^{n}(a_j-b_j)^2}=\sqrt{\sum\limits_{j=1}^{n}(a_j+b_j)^2}$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Squaring both sides and expanding the binomials:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sum\limits_{j=1}^{n}a^2_j-2a_jb_j+b_j^2=\sum\limits_{j=1}^{n}a^2_j+2a_jb_j+b_j^2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Simplifying:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$-\sum\limits_{j=1}^{n}a_jb_j=\sum\limits_{j=1}^{n}a_jb_j$, which holds true if and only if $\sum\limits_{j=1}^{n}a_jb_j=0$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $AB$ is equivalent to $\sum\limits_{j=1}^{n}a_jb_j$, then $||A-B||=||A+B||\Leftrightarrow AB=0$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance.&lt;/p&gt;&#xA;" OwnerUserId="1523" LastActivityDate="2010-09-05T13:14:52.977" Title="Proving that $||A-B||=||A+B||\Leftrightarrow AB=0$" Tags="&lt;linear-algebra&gt;&lt;homework&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="1" />
  <row Id="4034" PostTypeId="2" ParentId="4031" CreationDate="2010-09-04T20:35:12.210" Score="1" ViewCount="" Body="&lt;p&gt;This should help you out: &lt;a href=&quot;http://www.mathhelpforum.com/math-help/f7/sum-divisors-function-question-154661.html&quot; rel=&quot;nofollow&quot;&gt;http://www.mathhelpforum.com/math-help/f7/sum-divisors-function-question-154661.html&lt;/a&gt;&lt;/p&gt;&#xA;" LastActivityDate="2010-09-04T20:35:12.210" CommentCount="1" />
  <row Id="4035" PostTypeId="1" AcceptedAnswerId="4037" CreationDate="2010-09-04T21:33:30.953" Score="3" ViewCount="420" Body="&lt;p&gt;Hello everyone,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've recently started making a small fractal app in Javascript using the famous Mandelbrot bulb (z = z^2 + c). I've been trying to find the best method of coloring the points on the complex plane, and I've come across some very interesting ideas:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://linas.org/art-gallery/escape/escape.html&quot; rel=&quot;nofollow&quot;&gt;http://linas.org/art-gallery/escape/escape.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikibooks.org/wiki/Fractals/Iterations_in_the_complex_plane/Mandelbrot_set#Real_Escape_Time&quot; rel=&quot;nofollow&quot;&gt;http://en.wikibooks.org/wiki/Fractals/Iterations_in_the_complex_plane/Mandelbrot_set#Real_Escape_Time&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I've been focusing on this &quot;normalized iteration count algorithm&quot; and I've run into a small mental hurdle sorting out what needs to be calculated. Basically, I'm confused about the first equation that shows up in that second link. Assuming we have a complex number &lt;code&gt;Z&lt;/code&gt; which is the end result of &lt;code&gt;n&lt;/code&gt; iterations. I can't quite figure out what &quot;Zx2&quot; or &quot;zy2&quot; is and how how one might add them together or take the square root of their sum. Is that just shorthand for some basic complex operators, or is there something else going on? Basically, my problem is that I'm not particularly good at reading mathematical notation in this area.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway...any directional guidance you can offer would be extremely helpful. Thanks in advance!&lt;/p&gt;&#xA;" OwnerUserId="1720" LastActivityDate="2010-09-04T22:36:03.963" Title="Continuous coloring of a Mandelbrot fractal" Tags="&lt;algorithms&gt;&lt;fractals&gt;" AnswerCount="2" />
  <row Id="4036" PostTypeId="2" ParentId="4035" CreationDate="2010-09-04T21:41:21.580" Score="2" ViewCount="" Body="&lt;p&gt;&lt;em&gt;edit&lt;/em&gt;: &quot;Zx2&quot; and &quot;Zy2&quot; are defined in the C code in the second code block &lt;a href=&quot;http://en.wikibooks.org/wiki/Fractals/Iterations_in_the_complex_plane/Mandelbrot_set#Optimisation&quot; rel=&quot;nofollow&quot;&gt;at this point in your wikibooks link&lt;/a&gt;.  They are the squares of the real and imaginary parts, respectively, of $z$.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I'm not sure if it's exactly the same, but what I've used is:&#xA;$$n-\log\left(\frac{\log(|z|)}{\log(r)}\right)$$&#xA;where $z$ is the result of $n$ iterations and has just escaped as noted by $|z|&gt;r$ (where $r$ is the escape radius).  This generates real numbers in $[0,n)$, so you could divide by $n$ to end up with a number in $[0,1)$.&lt;/p&gt;&#xA;" OwnerUserId="72" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-09-04T22:31:47.700" LastActivityDate="2010-09-04T22:31:47.700" CommentCount="1" />
  <row Id="4037" PostTypeId="2" ParentId="4035" CreationDate="2010-09-04T21:41:28.897" Score="4" ViewCount="" Body="&lt;p&gt;Based on the C code given there, &lt;code&gt;Zx&lt;/code&gt; is just the real part $\Re z$ and &lt;code&gt;Zy&lt;/code&gt; is just the imaginary part $\Im z$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remember that the complex iteration formula for the Mandelbrot set&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$z=z^2+c$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;when expanded to real variables looks like&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x=x^2-y^2+u$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$y=2xy+v$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $z=x+iy$ and $c=u+iv$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The variables with a 2 appended are the squares, as can be seen from how they were defined, so for instance $|z|=\sqrt{x^2+y^2}$ is &lt;code&gt;sqrt(Zx2+Zy2)&lt;/code&gt; which is used for determining how much the iteration diverges (escapes to infinity), which is used by the coloring functions.&lt;/p&gt;&#xA;" OwnerUserId="498" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-09-04T22:36:03.963" LastActivityDate="2010-09-04T22:36:03.963" CommentCount="2" />
  <row Id="4038" PostTypeId="2" ParentId="1647" CreationDate="2010-09-04T21:58:31.473" Score="2" ViewCount="" Body="&lt;p&gt;To expand on the suggestion to use the &lt;a href=&quot;http://en.wikipedia.org/wiki/Lambert_W_function&quot; rel=&quot;nofollow&quot;&gt;Lambert function&lt;/a&gt;, I'll show how it arises in your equation of interest.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Starting with&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$y=\frac1{2\pi v^2}\exp\left(-\frac{x^2}{2v^2}\right)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;we multiply both sides by $-\pi x^2$ to give&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$-\pi x^2 y=-\frac{x^2}{2v^2}\exp\left(-\frac{x^2}{2v^2}\right)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which can now be inverted to the Lambert function (recall that the Lambert function $W(z)$ is the inverse function of $z\exp(z)$, $W(z)\exp(W(z))=z$):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$-\frac{x^2}{2v^2}=W(-\pi x^2 y)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which we can now solve for $v$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$v=\frac{x}{\sqrt{-2W(-\pi x^2 y)}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The choice of sign for the square root is motivated by the fact that variances are conventionally taken to be positive.&lt;/p&gt;&#xA;" OwnerUserId="498" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-09-05T23:27:59.197" LastActivityDate="2010-09-05T23:27:59.197" CommentCount="3" />
  <row Id="4039" PostTypeId="2" ParentId="3977" CreationDate="2010-09-04T23:02:13.063" Score="2" ViewCount="" Body="&lt;p&gt;Geomview is fairly flexible, and allows you to work in Euclidean, spherical or hyperbolic geometry.  Depending on which operating system you use it might be easier or harder to get it installed.  It made it into the Ubuntu repositories recently, so if you're running Ubuntu it takes little effort to install. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That said, it's not the easiest software to &lt;em&gt;use&lt;/em&gt; -- quite a ways away from the interface of GeoGebra. More often than not, if I need a high-quality rendered image of something 3-dimensional, I code a PovRay script and render it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Blender is open-source and interactive.  It appears to be quite flexible -- it's used to produce models for 3d video games, for example.  I haven't really got comfortable using it, though. &lt;/p&gt;&#xA;" OwnerUserId="642" LastEditorUserId="642" LastEditorDisplayName="" LastEditDate="2010-09-05T22:39:55.250" LastActivityDate="2010-09-05T22:39:55.250" CommunityOwnedDate="2010-09-04T23:02:13.063" CommentCount="3" />
  <row Id="4040" PostTypeId="2" ParentId="3978" CreationDate="2010-09-04T23:17:46.360" Score="11" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;PROOF&lt;/strong&gt;$\;$ from my &lt;a href=&quot;http://groups.google.com/group/sci.math/browse_frm/thread/b04d0459db4c2cb5&quot; rel=&quot;nofollow&quot;&gt;sci.math post of April 4, 2006&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\begin{array}{}&#xA;p-1\:|\:n-1 \\&#xA;\quad\quad\;\rm p\:|\:a+n \\&#xA;\end{array} \;\; \Rightarrow\;\; p\;\;|\;\;a^n-a\;+\;a+n\;\;|\;\;b^n-b\;+\;b-a\;+\;a+n \;\;\Rightarrow\;\; p\:|\:b-a$&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-09-04T23:17:46.360" CommentCount="6" />
  <row Id="4041" PostTypeId="2" ParentId="4032" CreationDate="2010-09-04T23:25:57.187" Score="7" ViewCount="" Body="&lt;p&gt;What you've done is correct, but I think it's better to work without coordinates; just with the definition of norm in terms of the dot product:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\| A \| = +\sqrt{A\cdot A} \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then you may observe that, since $\|A \| \geq 0$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\|A+B\| =  \|A -B\| \  \Longleftrightarrow  \  \|A +B\|^2  =  \|A-B\|^2 .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, for instance, compute the difference&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA; \|A +B\|^2  -  \|A-B\|^2   &amp;amp;= (A+B)\cdot (A+B) - (A-B)\cdot(A-B)    \\&#xA;               &amp;amp;= A\cdot A + A\cdot B + B\cdot A + \cdots&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT. I forgot to point out an obvious geometric interpretation of this result: if you draw a parallelogram with sides $A$ and $B$, then $A+B$ and $A-B$ are the diagonals of the parallelogram, right? These diagonals are equal if and only if...?&lt;/p&gt;&#xA;" OwnerUserId="664" LastEditorUserId="664" LastEditorDisplayName="" LastEditDate="2010-09-05T13:14:52.977" LastActivityDate="2010-09-05T13:14:52.977" />
  <row Id="4042" PostTypeId="1" AcceptedAnswerId="4046" CreationDate="2010-09-05T00:56:04.653" Score="1" ViewCount="78" Body="&lt;p&gt;I am working through Theorem 6 in Chapter 8 of Hoffman and Kunze.  (The section is &lt;em&gt;Linear Functionals and Adjoints&lt;/em&gt;.)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The statement is &lt;blockquote&gt; Let $V$ be a finite-dimensional inner product space, and $f$ a linear functional on $V$.  Then there exists a unique vector $\beta$ in $V$ such that $f(\alpha) = (\alpha | \beta)$ for all $\alpha$ in $V$.&lt;/blockquote&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(&lt;strong&gt;EDIT:&lt;/strong&gt;  The following is written in terms of a &lt;em&gt;real&lt;/em&gt; vector space.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am fine with the fact that this essentially follows from what $f$ does to the elements of a basis for $V$, in other words:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For an &lt;em&gt;arbitrary&lt;/em&gt;* $\mathcal{B} = \{e_1, \ldots, e_{n} \}$ (where $n = \dim V\;\;$) we can assign $\xi_{i} = f(e_{i})$ for $i=1,\ldots, n$.  Then any $\alpha \in V$ can be written as $\alpha = x_{1}e_{1} + \cdots + x_{n}e_{n}$ so that &#xA;$$\begin{align}&#xA;f(\alpha) &amp;amp;= f(x_{1}e_{1} + \cdots + x_{n}e_{n}) \\ \\&#xA;          &amp;amp;= x_{1}f(e_{1}) + \cdots + x_{n}f(e_{n}) \\ \\&#xA;          &amp;amp;= x_{1} \xi_{1} + \cdots + x_{n}\xi_{n}.\end{align}$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;At this point we can identify the $\xi_{i}$'s as being independent of $\alpha \in V$, and only having been dependent on our choice of a basis.  With this, we can simply assert that $\beta = (\xi_{1}, \ldots, \xi_{n})$ (and is in fact unique).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Now&lt;/strong&gt;, this is apparently not the whole story!  *According to H. and K., the coordinates of our fixed vector $\beta$ are determined by the action of $f$ on an &lt;em&gt;orthonormal&lt;/em&gt; basis for $V$ rather than an arbitrary basis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt;  What role does the orthogonality (or even the &lt;em&gt;'orthonormality'&lt;/em&gt;) of the basis have in this argument?  It seems to work just fine for an arbitrary basis as far as I can tell.&lt;/p&gt;&#xA;" OwnerUserId="121" LastEditorUserId="121" LastEditorDisplayName="" LastEditDate="2010-09-05T01:58:55.693" LastActivityDate="2010-09-05T02:34:56.537" Title="Role of orthonormal basis for: Linear functional on a finite dimensional inner product space is 'inner product with a fixed vector in the space'" Tags="&lt;linear-algebra&gt;" AnswerCount="1" />
  <row Id="4043" PostTypeId="1" CreationDate="2010-09-05T01:57:01.843" Score="3" ViewCount="337" Body="&lt;p&gt;I'm trying to learn predicate logic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And I'm looking for some good resources on it:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've seen that I learn better when I can program&#xA;So I was wondering if there was a 'predicate logic' programning language?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe an interactive tutorial?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe lots of examples, something like: Predicate Logic by Example?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or at least some pretty basic books?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;TIA&lt;/p&gt;&#xA;" OwnerUserId="1724" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-07T02:13:54.573" LastActivityDate="2011-04-04T15:39:19.577" CommunityOwnedDate="2010-09-07T23:21:03.547" Title="What are good resources for learning predicate logic / predicate calculus?" Tags="&lt;logic&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" />
  <row Id="4044" PostTypeId="1" AcceptedAnswerId="4053" CreationDate="2010-09-05T02:15:29.800" Score="2" ViewCount="188" Body="&lt;p&gt;This question may smell like my older question &lt;a href=&quot;http://math.stackexchange.com/questions/1778/a-probability-game&quot;&gt;http://math.stackexchange.com/questions/1778/a-probability-game&lt;/a&gt;, but this time my intentions are different.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The above problem came from my friend. Or so I thought, it really came from me incorrectly remembering his problem. The problem he actually gave me was:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose you start with 1 dollar and flip a weighted coin. With probability 3/4 (call this heads) you win a dollar and with probability 1/4 (call this tails) you lose a dollar. What is the probability you eventually run out of money? (Suppose the house has as many dollar bills as needed).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know how to solve this problem in the following way:&#xA;First observe the number of tails must equal the number of heads plus one. If a string of Hs and Ts satisfy this property, call it Good. Now we want to count the number of Good strings that don't have Good proper prefixes. For a string of length 2n-1 this is just the nth Catalan number. Now we just take the appropriate summation and we arrive at the probability we eventually run out of money which turns out to be 1/3.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now when my friend told me this problem, he said he was able to apply a clever trick and solve it almost instantly. The thing is, he doesn't remember how he did it. He says he vaguely remembers using a recurrence relation and then telescoping a sum. (Which sounds very plausible) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So can you come up with a short and sweet solution?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On a side note, I am aware my title is poor at best. If you can think of a better one, please let me know.&lt;/p&gt;&#xA;" OwnerUserId="899" LastActivityDate="2010-09-06T05:34:08.040" Title="Finding a clever solution to a game of chance." Tags="&lt;combinatorics&gt;&lt;probability&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="1" />
  <row Id="4045" PostTypeId="1" CreationDate="2010-09-05T02:28:12.270" Score="4" ViewCount="148" Body="&lt;p&gt;I'm currently interested in the following result:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $f: X \to Y$ be a fpqc morphism of schemes. Then there is an equivalence of categories between quasi-coherent sheaves on $Y$ and &quot;descent data&quot; on $X$.  Namely, the second category consists of quasi-coherent sheaves $\mathcal{F}$ on $X$ with an isomorphism $p_{1}^*(\mathcal{F}) \simeq p_2^*(\mathcal{F})$, where $p_1, p_2: X \times_Y X \to X$ are the two projections.  &lt;strong&gt;Edit:&lt;/strong&gt; There is a further condition; a diagram involving an iterated fibered product is required to commute as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In Demazure-Gabriel's &lt;em&gt;Introduction to Algebraic Geometry and Algebraic Groups,&lt;/em&gt; it is proved (under the name ffqc (sic) descent theorem) that the sequence&#xA;$$ X \times_Y X \to^{p_1, p_2} X \to Y$$ is a coequalizer in the category of locally ringed spaces under the above hypotheses. If I am not mistaken, this is the same as the theorem that says that representable functors are sheaves in the fpqc topology.  On the other hand, D-G give a fairly explicit description of the quotient space.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Question: For a coequalizer diagram of (locally) ringed spaces, &#xA;$$A \to^{f,g} B \to C,$$&#xA;is there a descent diagram for quasi-coherent sheaves on $A,B,C$?  In particular, does the D-G form of the descent theorem directly, by itself, imply the more general one for quasi-coherent sheaves?&lt;/p&gt;&#xA;" OwnerUserId="536" LastEditorUserId="536" LastEditorDisplayName="" LastEditDate="2010-09-05T03:41:27.047" LastActivityDate="2010-09-05T03:41:27.047" Title="Do cokernels in RingSpc automatically lead to descent?" Tags="&lt;algebraic-geometry&gt;&lt;descent&gt;" CommentCount="3" />
  <row Id="4046" PostTypeId="2" ParentId="4042" CreationDate="2010-09-05T02:34:56.537" Score="2" ViewCount="" Body="&lt;p&gt;If the basis $\{e_1,\ldots,e_n\}$ is not orthonormal, then the inner product of the vectors $\alpha = x_1e_1 + \cdots + x_n e_n$ and $\beta = \xi_1 e_1 + \cdots + \xi_n e_n$ is not equal to $x_1\xi_1+\cdots+x_n\xi_n$.  Thus, while it is true that $f(\alpha) = x_1\xi_1+\cdots+x_n\xi_n$, it is not true that $f(\alpha)=(\alpha|\beta)$.&lt;/p&gt;&#xA;" OwnerUserId="1726" LastActivityDate="2010-09-05T02:34:56.537" CommentCount="3" />
  <row Id="4047" PostTypeId="2" ParentId="3987" CreationDate="2010-09-05T03:58:54.607" Score="0" ViewCount="" Body="&lt;p&gt;With no better answer, I'll just add a little to what I said in the question to get this closed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are series calculations for logarithms (http://en.wikipedia.org/wiki/Logarithm) so I could in principle calculate the base 10 log directly. This would be awkward and slow, though, as for arbitrary precision integers a machine float/double isn't precise enough. I'd probably need a rational library, and the rationals involved wouldn't be simple.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It may be possible to use an imprecise base-converting factor. With an underestimate factor, you'd get an underestimate log value. Divide through by 10^x for that log value, then repeat as often as necessary to top up your underestimate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is that dividing through by 10^x isn't trivial unless you happen to be working in base 10.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Probably best to stick with repeated division by 10 - or powers of 10, at least. Constructing a list of powers (10^1, 10^2, 10^4, 10^8 and so on) by squaring wouldn't be too hard. Stop when you exceed the target number, then do the divisions largest to smallest. Each power should only need to be considered once.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even then, it may be better to not build powers of 10 that won't fit in a machine word, and just do repeated division by (small powers of) 10.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, this sounds now more like programming than math, so sorry about that.&lt;/p&gt;&#xA;" OwnerUserId="510" LastActivityDate="2010-09-05T03:58:54.607" />
  <row Id="4048" PostTypeId="1" AcceptedAnswerId="4085" CreationDate="2010-09-05T04:09:27.013" Score="3" ViewCount="216" Body="&lt;p&gt;The only place I've encountered well-definition is with proving an operation defined on an equivalence class is independent of the choice of representative.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On my homework, it asks us to show that the ring of formal Laurent series is well-defined, and I don't understand what exactly I need to show. However, I don't understand what I'm trying to prove. I've read the wikipedia article and don't understand how there is anything to prove in this case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If anyone could either point me to some better references explaining well-definition or explain what I need to do to prove that a laurent series is well-defined, I would appreciate it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks, :)&lt;/p&gt;&#xA;" OwnerUserId="249" LastEditorUserId="249" LastEditorDisplayName="" LastEditDate="2010-09-05T04:20:45.633" LastActivityDate="2010-09-05T19:35:57.807" Title="How do you show the ring of formal laurent series is well-defined?" Tags="&lt;abstract-algebra&gt;&lt;terminology&gt;&lt;homework&gt;&lt;ring-theory&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" />
  <row Id="4049" PostTypeId="1" AcceptedAnswerId="4051" CreationDate="2010-09-05T04:31:50.393" Score="3" ViewCount="185" Body="&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; To prove that the Ring of Complex Entire functions is neither Artinian nor noetherian.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proof: Clearly $R$ is not Artinian because it is a commutative integral domain which is not a field, and $R$ is not noetherian because its not  a factorisation domain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a proof of this theorem using the Ascending Chain / Descending Chain condition for Artinian / Noetherian rings?&lt;/p&gt;&#xA;" LastActivityDate="2010-09-05T04:52:00.290" Title="Proving that Ring of Complex Entire functions is neither Artinian nor Noetherian." Tags="&lt;analysis&gt;&lt;abstract-algebra&gt;&lt;complex-analysis&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="4051" PostTypeId="2" ParentId="4049" CreationDate="2010-09-05T04:52:00.290" Score="4" ViewCount="" Body="&lt;p&gt;Let $J_n$ be the ideal of entire functions vanishing on the first $n$ positive integers, and let $I_n$ be the ideal of entire functions vanishing on positive integers greater than $n$.  Then $$J_1\supset J_2\supset J_3\supset\cdots,$$ $$I_1\subset I_2\subset I_3\subset\cdots,$$ and all of these containments are proper.  One way to see that the containments are proper is to use the fact that given any sequence of complex numbers $a_1,a_2,\ldots$, there is an entire function $f$ such that $f(n)=a_n$ for each positive integer $n$.  For more on this, see &lt;a href=&quot;http://mathoverflow.net/questions/20711/does-there-exist-a-holomorphic-function-which-takes-given-values-on-the-positive&quot; rel=&quot;nofollow&quot;&gt;these&lt;/a&gt; &lt;a href=&quot;http://mathoverflow.net/questions/2944/&quot; rel=&quot;nofollow&quot;&gt;MathOverflow&lt;/a&gt; &lt;a href=&quot;http://mathoverflow.net/questions/7328/interpolation-of-sequences-by-analytic-functions&quot; rel=&quot;nofollow&quot;&gt;questions&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-09-05T04:52:00.290" CommentCount="1" />
  <row Id="4052" PostTypeId="2" ParentId="3946" CreationDate="2010-09-05T04:52:02.053" Score="1" ViewCount="" Body="&lt;p&gt;It seems like you answered your own question in the last paragraph. You create the two new control points at each corner, and then replace the corner with a quadratic Bezier curve determined by the two new control points and the original corner point. So, it is not clear to me what you are asking. The &lt;a href=&quot;http://en.wikipedia.org/wiki/De_Casteljau%27s_algorithm&quot; rel=&quot;nofollow&quot;&gt;de Casteljau algorithm&lt;/a&gt; can be used to render the curves.&lt;/p&gt;&#xA;" OwnerUserId="1728" LastActivityDate="2010-09-05T04:52:02.053" />
  <row Id="4053" PostTypeId="2" ParentId="4044" CreationDate="2010-09-05T06:11:26.577" Score="3" ViewCount="" Body="&lt;p&gt;If $p$ is the probability that you eventually lose a dollar then we must have that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$p = \frac{1}{4} + \frac{3}{4}p^{2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(We lose a dollar, or win one and lose two eventually).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solving gives $p=1$ or $p=\frac{1}{3}$. I suppose we can eliminate $p=1$ somehow and get $p=\frac{1}{3}$ as the required answer.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-09-05T06:11:26.577" CommentCount="4" />
  <row Id="4054" PostTypeId="1" CreationDate="2010-09-05T08:39:40.427" Score="32" ViewCount="1462" Body="&lt;p&gt;There is a theorem that states that if $f$ is analytic in a domain $D$, and the closed disc {$  z:|z-\alpha|\leq r$} contained in $D$, and $C$ denotes the disc's boundary followed in the positive direction, then for every $z$ in the disc we can write:&#xA;$$f(z)=\frac{1}{2\pi i}\int\frac{f(\zeta)}{\zeta-z}d\zeta$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is:&#xA;&lt;strong&gt;What is the intuitive explanation of this formula?&lt;/strong&gt; (For example, but not necessary, geometrically.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Just to clarify - I know the proof of this theorem, I'm just trying to understand where does this exact formula come from.)&lt;/p&gt;&#xA;" OwnerUserId="766" LastEditorUserId="367" LastEditorDisplayName="" LastEditDate="2010-09-05T13:40:04.817" LastActivityDate="2011-04-14T05:18:23.210" Title="Intuitive explanation of Cauchy's Integral Formula in Complex Analysis" Tags="&lt;intuition&gt;&lt;complex-analysis&gt;" AnswerCount="12" CommentCount="4" FavoriteCount="25" />
  <row Id="4055" PostTypeId="2" ParentId="4054" CreationDate="2010-09-05T10:15:20.000" Score="2" ViewCount="" Body="&lt;p&gt;Perhaps you can see the Cauchy's integral formula as the &quot;Winding number&quot; of a curve $C$ around any complex number $a$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For more information, please see : &lt;a href=&quot;http://en.wikipedia.org/wiki/Winding_number&quot; rel=&quot;nofollow&quot;&gt;en.wikipedia.org/wiki/Winding_number&lt;/a&gt;&lt;/p&gt;&#xA;" LastActivityDate="2010-09-05T10:15:20.000" CommentCount="1" />
  <row Id="4056" PostTypeId="2" ParentId="1778" CreationDate="2010-09-05T10:17:19.317" Score="1" ViewCount="" Body="&lt;p&gt;Here is a completely probabilistic proof inspired by Moron in &lt;a href=&quot;http://math.stackexchange.com/questions/4044/finding-a-clever-solution-to-a-game-of-chance&quot;&gt;http://math.stackexchange.com/questions/4044/finding-a-clever-solution-to-a-game-of-chance&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let p be the probability that I end up with a net loss of 50 cents, then $p = \frac{1}{2} + \frac{1}{2} p^3$. That is to say, I either lose 50 cents, or gain a dollar and lose a dollar and 50 cents eventually. Solving for p I get 1, $(-1-\sqrt{5})/2$, and $(1-\sqrt{5})/2$ and it is not hard to see that we can dismiss the first two. Now the probability I end up with a net loss of one dollar is $p^2 = (3-\sqrt{5})/2$.&lt;/p&gt;&#xA;" OwnerUserId="899" LastActivityDate="2010-09-05T10:17:19.317" />
  <row Id="4057" PostTypeId="2" ParentId="67" CreationDate="2010-09-05T10:22:57.960" Score="0" ViewCount="" Body="&lt;p&gt;You can see this article as well: &lt;a href=&quot;http://www.cs.umd.edu/Honors/reports/ECCpaper.pdf&quot; rel=&quot;nofollow&quot;&gt;www.cs.umd.edu/Honors/reports/ECCpaper.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" LastActivityDate="2010-09-05T10:22:57.960" />
  <row Id="4058" PostTypeId="1" AcceptedAnswerId="4060" CreationDate="2010-09-05T10:25:59.877" Score="1" ViewCount="199" Body="&lt;p&gt;See this SO thread: &lt;a href=&quot;http://stackoverflow.com/questions/3634106/calculating-which-item-is-next-in-a-percentage-distribution&quot;&gt;&lt;strong&gt;Calculating which item is next in a percentage distribution&lt;/strong&gt;&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To summarize,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;you calculate&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;max ((rand1 * 10), (rand2 * 20), (rand3 * 30), (rand4 * 40))&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;for four random numbers in a range (say &lt;code&gt;1-100&lt;/code&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the &lt;code&gt;max&lt;/code&gt; is &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;rand1 * 10, you choose 'A'&#xA;rand2 * 20, you choose 'B'&#xA;rand3 * 30, you choose 'C'&#xA;rand4 * 40, you choose 'D'&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What is the expected distribution of choosing A, B, C and D?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How to derive it mathematically?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Using &lt;a href=&quot;http://codepad.org/F6VV6J3m&quot; rel=&quot;nofollow&quot;&gt;some C code&lt;/a&gt;, the percentages come out to be 1, 10, 31 and 56 respectively)&lt;/p&gt;&#xA;" OwnerUserId="324" LastActivityDate="2010-09-05T17:04:31.147" Title="How to calculate the expected distribution of results using random numbers?" Tags="&lt;calculus&gt;&lt;linear-algebra&gt;&lt;probability&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" />
  <row Id="4059" PostTypeId="2" ParentId="2581" CreationDate="2010-09-05T10:53:16.123" Score="2" ViewCount="" Body="&lt;p&gt;There are variations on the problem where the division is always possible and a proof using complete induction. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first variation is: [..original problem ...] where at least one of the sticks is &gt;=2n.&lt;br&gt;&#xA;The second variation is: [..original problem ...] where it is allowed to glue two of the sticks together and break once at an arbitrary position.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proof for the variations:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The case n=1 is trivial - one stick of length 1. Then we suppose the assumption holds for n and consider the case n+1. I.e. given a number of sticks of integral length &gt;= n+1 whose lengths add to (n+1)(n+2)/2 - can these be divided into sticks of lengths 1,2,3,…,n+1?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Break off from one of the sticks a length n+1. In the first variation we use the stick with lenght &gt;= 2n+2. This part will be the required stick of length n+1 in the solution. Because we broke off a length n+1 the remaining total length is (n+1)(n+2)/2 - (n+1) = n(n+1)/2. In the first variation the inducution step is valid, in the second variation we may not yet use the induction step because the other part of the broken stick may now be smaller then n+1. In that case the second variation allows us to glue together this part with one of the other sticks. Now the induction step is valid and we break the rest of the collection of sticks in {1,..n} and have succeeded in the division {1,.. n+1}.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;@downvoter: i.m.o. this try for an answer is useful, because it can show us the edges/restrictions of the original problem.&lt;/p&gt;&#xA;" OwnerUserId="706" LastEditorUserId="706" LastEditorDisplayName="" LastEditDate="2010-10-02T21:44:44.500" LastActivityDate="2010-10-02T21:44:44.500" CommentCount="6" />
  <row Id="4060" PostTypeId="2" ParentId="4058" CreationDate="2010-09-05T12:01:13.337" Score="2" ViewCount="" Body="&lt;p&gt;The probabilities are given by:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$p_i = \int_0^1 \prod_{j \in {1,2,3,4}, j\ne i} min(\frac{ix}{j},1)  dx, i = 1, 2, 3, 4$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;each factor in the integrand computes the probability density that the i-th random variable is larger than the j-th one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The exact result is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[1/96, 31/288 91/288 163/288]&lt;/p&gt;&#xA;" OwnerUserId="1430" LastEditorUserId="1430" LastEditorDisplayName="" LastEditDate="2010-09-05T12:10:32.463" LastActivityDate="2010-09-05T12:10:32.463" CommentCount="2" />
  <row Id="4061" PostTypeId="2" ParentId="4054" CreationDate="2010-09-05T12:16:19.607" Score="12" ViewCount="" Body="&lt;p&gt;As Chandru1 pointed out, a beautiful particular case of Cauchy's integral formula gives us a very particular &lt;em&gt;winding number&lt;/em&gt;. For the sake of simplicity, let us look only at closed paths on the boundary of the unit disc, $\gamma : [0,1] \longrightarrow S^1$. Then, the &lt;em&gt;winding number&lt;/em&gt; of $\gamma$ is the number of times $\gamma$ goes round the circumference. That is, if we write the complex number $\gamma (t)$ in its exponential form&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\gamma (t) = e^{i\theta (t)} \qquad \qquad \qquad  [1]&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;then you can prove that there is a &lt;em&gt;continuous&lt;/em&gt; choice for the argument $\theta (t)$ of $\gamma (t)$; that is, a continuous function $\theta : [0, 1] \longrightarrow \mathbb{R}&#xA;$ such that [1] holds for all $t$. In Algebraic Topology, $\theta$ is called a &quot;lifting&quot; of $\gamma$. Moreover, any two such continuous &quot;liftings&quot; of $\gamma$ differ necessarily by a constant integer multiple of $2\pi$: if $\widetilde{\theta}(t)$ is another (continuous) lifting of $ \gamma$, then there is a constant integer $k$, not depending on $t$, such that $\widetilde{\theta}(t) = \theta (t) + 2k\pi$ for all $t$. (This is called the &quot;lifting lemma&quot; in Algebraic Topology. For a proof in Complex Analysis, see theorem 7.1, in Stewart and Tall.) So the number of times $\gamma $ goes round the circumference (its &lt;em&gt;winding number&lt;/em&gt;) is well-defined as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;w (\gamma , 0) = \frac{\theta (1) - \theta (0)}{2\pi} \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then you can easily prove (see op.cit, section 7.5), that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;w (\gamma , 0 ) = \frac{1}{2\pi i}\int_\gamma \frac{1}{\zeta}d\zeta \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, Cauchy's integral formula for the constant function $f \equiv 1$, $z = 0$ (and $\gamma (t) = e^{i2\pi t}$, hence $\theta (t) =2 \pi t$) tells us that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\frac{1}{2\pi i}\int_\gamma \frac{1}{\zeta}d\zeta = 1 \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is, if you go round the circumference just one time, well, you are going round the circumference indeed exactly once. Isn't math amazing? :-)&lt;/p&gt;&#xA;" OwnerUserId="664" LastEditorUserId="664" LastEditorDisplayName="" LastEditDate="2010-09-05T18:04:13.617" LastActivityDate="2010-09-05T18:04:13.617" CommentCount="10" />
  <row Id="4062" PostTypeId="1" AcceptedAnswerId="4063" CreationDate="2010-09-05T13:24:02.937" Score="3" ViewCount="336" Body="&lt;p&gt;I've been struggling with figuring out how to add powers of $i$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, the result of $i^3 + i^4 + i^5$ is $1$. But how do I get the result of $i^3 + i^4 + ... + i^{50}$? Writing it all down would be pretty mundane...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It has to do something with division by 4, since the &quot;power cycle&quot; of $i$ repeats every fourth power.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you for any clues.&lt;/p&gt;&#xA;" OwnerUserId="267" LastActivityDate="2010-11-16T00:00:33.257" Title="Adding powers of i" Tags="&lt;complex-numbers&gt;" AnswerCount="4" CommentCount="3" />
  <row Id="4063" PostTypeId="2" ParentId="4062" CreationDate="2010-09-05T13:31:33.700" Score="6" ViewCount="" Body="&lt;p&gt;From $i^2=-1$ you get $i^{4n}=1$, $\quad i^{4n+1}=i$, $\quad i^{4n+2}=-1$ and $i^{4n+3}=-i$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then you just count your positive and negative multiples of $1$ and $i$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, $i^3+i^4+\cdots+i^{50}=0$.&lt;/p&gt;&#xA;" OwnerUserId="367" LastEditorUserId="367" LastEditorDisplayName="" LastEditDate="2010-09-05T13:42:38.977" LastActivityDate="2010-09-05T13:42:38.977" />
  <row Id="4064" PostTypeId="2" ParentId="4062" CreationDate="2010-09-05T13:35:44.510" Score="0" ViewCount="" Body="&lt;p&gt;Hi&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have $$i^{3} + i^{4} + i^{5} = 1 = i^{3} + i^{4} + i^{5} + i^{6} + i^{7} + i^{8} + i^{9} = i^{3} + i^{4} + \cdots +i^{4n+1}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now $i^{50}=1 \times -1$, therefore we have the sum is $0$.&lt;/p&gt;&#xA;" LastActivityDate="2010-09-05T13:49:06.583" />
  <row Id="4065" PostTypeId="2" ParentId="4062" CreationDate="2010-09-05T13:55:10.247" Score="17" ViewCount="" Body="&lt;p&gt;Observing that $i^{3}+i^{4}+\ldots +i^{50}$ is a &lt;em&gt;geometric progression&lt;/em&gt; with ratio $i$, first term $i^3$ and $50-3+1=48$ terms, we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$i^{3}+i^{4}+\ldots +i^{50}=i^{3}\times \dfrac{1-i^{50-3+1}}{1-i}=i^{3}\times &#xD;&#xA;\dfrac{1-i^{48}}{1-i}=i^{2}i\times \dfrac{1-(i^{2})^{24}}{1-i}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$=-i\dfrac{1-(-1)^{24}}{1-i}=-i\dfrac{1-1}{1-i}=0$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: &quot;arithmetic&quot; corrected to &quot;geometric&quot;&lt;/p&gt;&#xA;" OwnerUserId="752" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2010-11-16T00:00:33.257" LastActivityDate="2010-11-16T00:00:33.257" CommentCount="4" />
  <row Id="4066" PostTypeId="1" AcceptedAnswerId="4071" CreationDate="2010-09-05T15:25:07.790" Score="3" ViewCount="203" Body="&lt;p&gt;Consider the closed interval $[0,1]$, there is $\frac{2}{3} \in [0,1]$ where $p=2$ and $q=3$. Similarly consider $[2,3]$, one can have $\frac{5}{2} \in [2,3]$ where $p=5$ and $q=2$. Does every interval of the form $[a,b]$, where $a,b \in \mathbb{R}$ contain a rational of this kind. If yes, how can we prove it?&lt;/p&gt;&#xA;" LastActivityDate="2010-09-05T16:16:29.290" Title="Rationals of the form $\frac{p}{q}$ where $p,q$ are primes in $[a,b]$." Tags="&lt;number-theory&gt;&lt;real-analysis&gt;&lt;analysis&gt;&lt;prime-numbers&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" />
  <row Id="4067" PostTypeId="1" AcceptedAnswerId="4079" CreationDate="2010-09-05T15:26:02.213" Score="2" ViewCount="124" Body="&lt;p&gt;In the book &lt;em&gt;Elliptic Curves - McKean &amp;amp; Moll&lt;/em&gt; we are given the outline for a proof of Eulers addition theorem:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The (projective) quartic $\mathbf y^2 = (1-\mathbf x^2)(1-k^2 \mathbf x^2)$ has four intersections with the quadratic $\mathbf y = 1 + a \mathbf x + b \mathbf x^2$. One is located at $(\mathbf x, \mathbf y) = (0,1)$. Let the other three have covering parameters $x_1,x_2,x_3$. Check that their sum is independent of $a$ and $b$ and identify it, mod $\mathbb L$, as either $0$ or $2K$. Now compute $\mathbf x_1 + \mathbf x_2 + \mathbf x_3 = 2ab/c$ and $\mathbf x_1 \mathbf x_2 \mathbf x_3 = 2a/c$ with $c = k^2 - b^2$ and conclude that $\mathbf x_1 \mathbf y_2 + \mathbf x_2 \mathbf y_1 = \pm \mathbf x_3 (1 - k^2 \mathbf x_1^2 \mathbf x_2^2)$. What is the ambiguous sign?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It is too difficult for me to solve so I have some questions that I hope someone will be able to help me with:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;What are these covering parameters and how do we find information about their sum?&lt;/li&gt;&#xA;&lt;li&gt;Why does the information about the covering parameters matter?&lt;/li&gt;&#xA;&lt;li&gt;I have computed $\mathbf x_1 + \mathbf x_2 + \mathbf x_3$ and $\mathbf x_1 \mathbf x_2 \mathbf x_3$ but I don't see how to reach the conclusion from there. How is it done?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Thank you&lt;/p&gt;&#xA;" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-09-05T15:56:26.190" LastActivityDate="2010-09-05T17:31:11.483" Title="Deriving Eulers Addition Theorem for Elliptic Integrals" Tags="&lt;special-functions&gt;&lt;elliptic-curves&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="4068" PostTypeId="1" CreationDate="2010-09-05T15:50:23.323" Score="6" ViewCount="397" Body="&lt;p&gt;$$\ln2 = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \cdots&#xA;      = (1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \cdots) - 2(\frac{1}{2} + \frac{1}{4} + \cdots)$$&#xA;     $$= (1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \cdots) - (1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \cdots) = 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;thanks.&lt;/p&gt;&#xA;" OwnerUserId="1382" LastEditorUserId="1382" LastEditorDisplayName="" LastEditDate="2010-09-07T15:00:00.153" LastActivityDate="2010-09-07T20:02:28.423" Title="Explain why calculating this series could cause paradox?" Tags="&lt;sequences-and-series&gt;&lt;paradox&gt;" AnswerCount="3" CommentCount="11" FavoriteCount="3" />
  <row Id="4069" PostTypeId="2" ParentId="3782" CreationDate="2010-09-05T15:50:46.223" Score="-1" ViewCount="" Body="&lt;p&gt;With the Open Univeristy (http://www.open.ac.uk/) you can study at home, full or part-time, they send you reading material and regular tests to send back to be marked. It's cheaper than attending a bricks and mortar univeristy, but the fact that you are still paying for it is motivation to complete the tests the send you. After a few years you could have a degree in Mathematics from the OU. You can continue with them and study for a Masters degree.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They produce their own reading material which is tailored to the learning process. Much easier than reading most textbooks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is in my opinion, much better value for money than an attendance university although the latter may be suited to those who want to leave their parents' home for the first time for the social aspect and for the opportunity to rack up enormous debts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is based in the UK but many course are open to anyone.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If after giving the OU a go you still have difficulties then you might consider that there are 3 options in life:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a) be good&lt;/p&gt;&#xA;&#xA;&lt;p&gt;b) get good&lt;/p&gt;&#xA;&#xA;&lt;p&gt;c) give up&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you live in denial of this and choose d) none of the above then you are falling for the sunk cost fallacy.&lt;/p&gt;&#xA;" OwnerUserId="882" LastActivityDate="2010-09-05T15:50:46.223" CommunityOwnedDate="2010-09-05T15:50:46.223" />
  <row Id="4070" PostTypeId="2" ParentId="4068" CreationDate="2010-09-05T16:07:12.710" Score="8" ViewCount="" Body="&lt;p&gt;It's because the series for ln2 is &lt;a href=&quot;http://en.wikipedia.org/wiki/Conditional_convergence&quot; rel=&quot;nofollow&quot;&gt;conditionally convergent&lt;/a&gt;.  (see also &lt;a href=&quot;http://en.wikipedia.org/wiki/Riemann_series_theorem&quot; rel=&quot;nofollow&quot;&gt;Riemann's rearrangement theorem&lt;/a&gt;)&lt;/p&gt;&#xA;" OwnerUserId="1417" LastActivityDate="2010-09-05T16:07:12.710" CommentCount="14" />
  <row Id="4071" PostTypeId="2" ParentId="4066" CreationDate="2010-09-05T16:13:12.340" Score="11" ViewCount="" Body="&lt;p&gt;Roughly speaking this asks whether the quotients of two primes&#xA;are dense in the positive reals. The answer is yes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $0 &amp;lt; a &amp;lt; b$ and let $q$ be a prime.&#xA;Then there will a a prime $p$ with $a &amp;lt; p/q\le b$ if and only if&#xA;$\pi(bq) &gt; \pi(aq)$ where $\pi$ is the prime-counting function.&#xA;But by the prime number theorem, as $q\to\infty$,&#xA;$$\frac{\pi(bq)}{\pi(aq)}\sim\frac{b\log(aq)}{a\log(bq)}&#xA;=\frac{b(\log q+\log a)}{a(\log q+\log b)}\sim\frac ba&gt;1.$$&#xA;For all large enough $q$, $\pi(bq)/\pi(aq) &gt; 1$ as required.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-09-05T16:13:12.340" CommentCount="2" />
  <row Id="4072" PostTypeId="2" ParentId="4044" CreationDate="2010-09-05T16:14:34.773" Score="4" ViewCount="" Body="&lt;p&gt;Here is a more formal way to derive the answer: Define $q_n$ to be the probability that you eventually go broke when you start with $n$ dollars. Then $q_0 = 1$, $\lim_{n \to \infty} q_n = 0$, and $q_n = \frac{1}{4} q_{n-1} + \frac{3}{4} q_{n+1}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The last equation defines a recursive sequence with constant coefficients, whose general solution is&#xA;$q_n = c_1 + c_2 3^{- n}$ (the factors $1$ and $\frac{1}{3}$ come from solving the quadratic equation of Moron's answer, which is the fundamental equation of the recursion). To satisfy the other two conditions on $q_n$, we see that $c_1 = 1$ and $c_2 = -1$ i.e.&#xA;$q_n = 1 - 3^{-n}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, the probability to go broke starting with one dollar is $q_1 = \frac{2}{3}$.&lt;/p&gt;&#xA;" OwnerUserId="960" LastEditorUserId="960" LastEditorDisplayName="" LastEditDate="2010-09-05T18:18:30.160" LastActivityDate="2010-09-05T18:18:30.160" CommentCount="2" />
  <row Id="4073" PostTypeId="1" AcceptedAnswerId="4087" CreationDate="2010-09-05T16:30:53.760" Score="-1" ViewCount="282" Body="&lt;p&gt;How to build the real number system?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;thanks.&lt;/p&gt;&#xA;" OwnerUserId="1382" LastEditorUserId="622" LastEditorDisplayName="" LastEditDate="2010-09-05T16:44:56.553" LastActivityDate="2010-09-06T11:18:05.110" Title="How to build the real number system?" Tags="&lt;real-analysis&gt;" AnswerCount="6" CommentCount="3" />
  <row Id="4074" PostTypeId="2" ParentId="4073" CreationDate="2010-09-05T16:38:00.467" Score="1" ViewCount="" Body="&lt;p&gt;Have you learnt of dedikind cuts. If not please look at this: &lt;a href=&quot;http://en.wikipedia.org/wiki/Dedekind_cut&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Dedekind_cut&lt;/a&gt;&lt;/p&gt;&#xA;" LastActivityDate="2010-09-05T16:38:00.467" />
  <row Id="4075" PostTypeId="2" ParentId="4062" CreationDate="2010-09-05T16:40:11.807" Score="2" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\rm\quad\quad i^3 + \: i^4 \; + \:\;\cdots\;\: + \; i^k = 0\ \:\iff\: k\:\equiv\: 2 \:\pmod 4$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Generally, suppose that $\rm\: z \:$ has order $\rm m&gt;1\:.$ Therefore $\rm\; z^n = 1 \iff\ m\:|n\;\;\:$ hence:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;LEMMA&lt;/strong&gt; $\quad\rm z^j + z^{j+1} + \:\cdots + z^k = 0\;\; \iff \rm\: k \:\equiv\;\; j \:-\: 1 \:\pmod m $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof&lt;/strong&gt;: $\;\;\;\;\rm \displaystyle z^j \  (1+z+\cdots + z^{k-j}) \;=\; z^j \: \frac{1-z^{k-j+1}}{1-z} = 0 \;\iff\; \rm m\:|\:k-j+1\quad\;$&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-09-05T21:04:47.133" LastActivityDate="2010-09-05T21:04:47.133" />
  <row Id="4076" PostTypeId="1" CreationDate="2010-09-05T16:52:45.617" Score="1" ViewCount="135" Body="&lt;p&gt;Hello i am having some problems working out how to attack this assignment, and after spending hours on it, have i resolved to ask you guys here for help.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have been given the following differential equation&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;y^{4}-16y=u^{1}+u&#xA;y′′′′−16y=u′+u&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;And i have been asked to determine the complete real solution. I just completely stuck on this assignment and hope that one of you guys can give me a push in the right direction.&lt;/p&gt;&#xA;" OwnerUserId="1732" LastActivityDate="2010-09-05T18:25:00.547" Title="Determine the complete real solution to a homogeneous differential equation" Tags="&lt;homework&gt;&lt;differential-equations&gt;" CommentCount="5" FavoriteCount="1" />
  <row Id="4077" PostTypeId="2" ParentId="4058" CreationDate="2010-09-05T17:04:31.147" Score="1" ViewCount="" Body="&lt;p&gt;Suppose you only had three random numbers between 0 and 1. Point (x,y,z)=(rand1*10,rand2*20,rand3*30) falls uniformly inside parallelepiped with dimensions 10,20,30.&#xA;Sets x&gt;y&amp;amp;&amp;amp;x&gt;z, y&gt;x&amp;amp;&amp;amp;y&gt;z, z&gt;x&amp;amp;&amp;amp;z&gt;y partition this parallelepiped into 3 parts, as pictured below, and their relative volumes represent the fraction of the time that x, y or z are the maximum value, respectively. Working out expression for the volumes should give you the integral expression David gave above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://yaroslavvb.com/upload/extreme-values.gif&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="998" LastActivityDate="2010-09-05T17:04:31.147" CommentCount="2" />
  <row Id="4078" PostTypeId="2" ParentId="4073" CreationDate="2010-09-05T17:21:35.487" Score="2" ViewCount="" Body="&lt;p&gt;Alternatively, you can look at Cauchy sequences: &lt;a href=&quot;http://en.wikipedia.org/wiki/Construction_of_the_real_numbers&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Construction_of_the_real_numbers&lt;/a&gt; (and other approaches than can be found there).&lt;/p&gt;&#xA;" OwnerUserId="664" LastActivityDate="2010-09-05T17:21:35.487" />
  <row Id="4079" PostTypeId="2" ParentId="4067" CreationDate="2010-09-05T17:31:11.483" Score="2" ViewCount="" Body="&lt;p&gt;The elliptic curve is the image of $\mathbb{C}/\Lambda$ under a certain map.&#xA;Now, each point $(\mathbf{x}_0,\mathbf{y}_0)$ is the image of some point&#xA;$x$ of $\mathbb{C}/\Lambda$. This $x$ is the covering parameter. It's a complex&#xA;number unique up to translation by elements of $\Lambda$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each of $\mathbf{x}$ and $\mathbf{y}$ can be regarded as a function on&#xA;$\mathbb{C}/\Lambda$. We are interested in $f=\mathbf{y}&#xA;-1-a\mathbf{x}-b\mathbf{x}^2$. This has four zeros in each fundamental&#xA;region, and so four poles. If the zeros are $x_0,x_1,x_2,x_3$&#xA;and poles are $u_0,u_1,u_2,u_3$ then $x_0+x_1+x_2+x_3-u_0-u_1-u_2-u_3$&#xA;is an element of $\Lambda$. I suppose here, one must show that the poles&#xA;of $f$ are independent of $a$ and $b$, which leads to $x_0+x_1+x_2+x_3$&#xA;being independent of $a$ and $b$ (here $x_0$ is the covering parameter&#xA;for the point $(0,1)$).&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-09-05T17:31:11.483" />
  <row Id="4080" PostTypeId="2" ParentId="3987" CreationDate="2010-09-05T17:32:32.490" Score="4" ViewCount="" Body="&lt;p&gt;I would post this as a comment if I could, because it seems too simple for an answer. You already said that you can find the highest bit with a single instruction (if the number fits into a register). For every result $r$ you get (that is, your number is between $2^r$ and $2^{r+1}-1$), there are only two possible numbers $k$ and $k+1$ of decimal digits. So you could just use $r$ as an index into a lookup table that stores both $k$ and $10^k$, and output $k$ if your number is less than $10^k$, or $k+1$ otherwise.&lt;/p&gt;&#xA;" OwnerUserId="1386" LastActivityDate="2010-09-05T17:32:32.490" CommentCount="3" />
  <row Id="4082" PostTypeId="1" AcceptedAnswerId="4104" CreationDate="2010-09-05T19:00:15.490" Score="0" ViewCount="141" Body="&lt;p&gt;Given 3 pairs of coordinates, x1, y1, x2, y2, x3, y3, I need a function y(x) that will return the y coordinate of any x coordinate between x1 and x3 (it can be assumed that x1 &amp;lt; x2 &amp;lt; x3).&lt;br&gt;&#xA;y(x1) must equal y1, y(x2) must equal y2, and y(x3) must equal y3.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;However&lt;/strong&gt;, it must never return a y coordinate that is larger than the largest of y1, y2, and y3, or smaller than the smallest of y1, y2, and y3. This is the part which I have not yet figured out how to do.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(This is not schoolwork. I'm writing an audio creation library, and these curves will be used to define volume over time.)&lt;/p&gt;&#xA;" OwnerUserId="1734" LastEditorUserId="1734" LastEditorDisplayName="" LastEditDate="2010-09-05T19:17:14.653" LastActivityDate="2010-09-06T01:54:02.867" Title="equation of a curve given 3 points and additional (constant) requirements" Tags="&lt;functions&gt;&lt;algebraic-curves&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="4083" PostTypeId="2" ParentId="4054" CreationDate="2010-09-05T19:08:45.663" Score="16" ViewCount="" Body="&lt;p&gt;Irene, if you are looking for intution then let us &lt;em&gt;assume&lt;/em&gt; that we can expand $f(\zeta)$ into a power series around $z$: $f(\zeta) = \sum_{n \geq 0} c_n(\zeta - z)^n$. Note $c_0 = f(z)$.&#xA;If you plug this into the integral and interchange the order of integration &#xA;and summation then that integral on the right side of the formula &#xA;becomes $\sum_{n \geq 0} \int c_n(\zeta - z)^{n-1}d\zeta$.  Let us also assume that an integral along a contour doesn't change if we deform the contour continuously through a region where the function is &quot;nice&quot;. So let us take as our path of integration a circle going once around the point $z$ (counterclockwise). Then you are basically reduced to &#xA;showing that $\int (\zeta - z)^{m}d\zeta$ is 0 for $m \geq 0$ and is $2\pi i$ for $m = -1$.&#xA;These can be done by direct calculations using polar coordinates with $\zeta = z + e^{it}$.  Now divide by $2\pi i$ and you have the formula.  Of course this is a hand-wavy argument in places, but the question was not asking for a rigorous proof.  Personally, this is how I first came to terms with understanding how Cauchy's integral formula could be guessed.&lt;/p&gt;&#xA;" OwnerUserId="619" LastActivityDate="2010-09-05T19:08:45.663" />
  <row Id="4084" PostTypeId="1" AcceptedAnswerId="4086" CreationDate="2010-09-05T19:12:39.017" Score="1" ViewCount="61" Body="&lt;p&gt;Say you have a set of points and joining lines as show below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/xglug.png&quot; alt=&quot;t junctions are at D, F, and E&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this diagram of triangles, T-(or Y-like) junctions are at D, F, and E.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At vertex F for example, there is a triangle FDE, but no line FJ.  If there was a line FJ, then there would be no T junction at F. Two triangles would be formed instead out of triangle-ACJ as tri-AFC and tri-FCJ.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do you find the vertices that form &quot;T&quot; (or Y-like) junctions?&lt;/p&gt;&#xA;" OwnerUserId="1069" LastActivityDate="2010-11-04T12:59:47.517" Title="How do you find T-Junctions in an array of triangles?" Tags="&lt;geometry&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="4085" PostTypeId="2" ParentId="4048" CreationDate="2010-09-05T19:14:08.277" Score="4" ViewCount="" Body="&lt;p&gt;I'll make my comment into an answer so that it can be marked off.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The issue is probably not whether a formal Laurent series is well-defined by itself, but rather whether the operations you are defining on formal Laurent series (and in particular the operation of multiplication of two formal Laurent series) is well-defined, in the sense that if you take any two formal Laurent series, then the definition of &quot;product&quot; will in fact yield a formal Laurent series. When you work with formal power series (only nonnegative exponents), one usually defines the product by:&#xA;$$\left(\sum_{n=0}^{\infty}a_nx^n\right)\left(\sum_{n=0}^{\infty}b_nx^n\right) = \sum_{n=0}^{\infty}\left(\sum_{i+j=n}a_i+b_j\right)x^n.$$&#xA;In this case, since $a_k=b_k=0$ if $k&amp;lt;0$, then the definition makes sense, as each term on the right hand side is a finite sum, which makes sense. If you try doing the same thing with formal Laurent series, where the index runs from $-\infty$ to $\infty$, then it is not obvious that this definition always results in something that you can call a formal Laurent series. So one needs to check that it does in fact yield a formal Laurent series, and that the product so defined makes the set into a ring. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So here, the issue of &quot;&lt;em&gt;well-defined&lt;/em&gt;&quot;ness is not like the one when you define a function in terms of representatives (in terms of the &quot;name&quot; of an object when the object may have many different names), but rather in terms of whether the function actually makes sense for every input and yields an appropriate output. It is the same issue that arises if you try to define a map $f\colon(0,1)\to\mathbb{N}$ by taking a number in decimal expansion $0.a_1a_2a_3\ldots$ and &quot;defining&quot; $f(0.a_1a_2a_3\ldots) = \cdots a_3a_2a_1$; this map is not well-defined because for some inputs the output does not lie in the range or does not makes sense.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-09-05T19:35:57.807" LastActivityDate="2010-09-05T19:35:57.807" />
  <row Id="4086" PostTypeId="2" ParentId="4084" CreationDate="2010-09-05T19:15:36.903" Score="0" ViewCount="" Body="&lt;p&gt;My first stab at it is to consider each triangle in the diagram, as well as all its adjacent triangles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider tri-FEC first.  FEC is adjacent to ACJ, but only &lt;strong&gt;one&lt;/strong&gt; vertex is shared (C).  If there is to be no T-junction, then two vertices should have been shared between adjacent triangles.&lt;/p&gt;&#xA;" OwnerUserId="1069" LastEditorUserId="1069" LastEditorDisplayName="" LastEditDate="2010-11-04T12:59:47.517" LastActivityDate="2010-11-04T12:59:47.517" />
  <row Id="4087" PostTypeId="2" ParentId="4073" CreationDate="2010-09-05T19:16:25.460" Score="2" ViewCount="" Body="&lt;p&gt;Let's start by defining the &lt;em&gt;natural numbers&lt;/em&gt; (a version including 0) by &lt;a href=&quot;http://mathworld.wolfram.com/PeanosAxioms.html&quot; rel=&quot;nofollow&quot;&gt;Peano's Axioms&lt;/a&gt; (sometimes called the Peano Postulates):&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;0 is a number.&lt;/li&gt;&#xA;&lt;li&gt;If n is a number, the successor of n is a number.&lt;/li&gt;&#xA;&lt;li&gt;0 is not the successor of a number.&lt;/li&gt;&#xA;&lt;li&gt;Two numbers of which the successors are equal are themselves equal.&lt;/li&gt;&#xA;&lt;li&gt;(induction axiom.) If a set S of numbers contains 0 and also the successor of every number in S, then every number is in S.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Now, the &lt;em&gt;integers&lt;/em&gt; are the natural numbers and their additive inverses and the &lt;em&gt;rational numbers&lt;/em&gt; are numbers that can be expressed as a ratio of an integer to a non-zero integer.  The &lt;strong&gt;real numbers&lt;/strong&gt; are the set of numbers that are limits of &lt;a href=&quot;http://mathworld.wolfram.com/CauchySequence.html&quot; rel=&quot;nofollow&quot;&gt;Cauchy sequences&lt;/a&gt; of rational numbers (which is more or less a convenient way of saying that the sequence will converge in the reals).&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-09-05T19:16:25.460" />
  <row Id="4088" PostTypeId="2" ParentId="4082" CreationDate="2010-09-05T19:20:09.303" Score="0" ViewCount="" Body="&lt;p&gt;In general, no 2nd degree polynomial fits your requirements: If you fix its maximum/minimum at the highest/lowest of your three points, just one of the remaining two points would completely determine the curve due to its symmetry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if you want to use this to control volume, there are probably some additional side conditions that you need to consider anyway: If y2 is the highest point, you probably want the curve to have just one local maximum (at x2, then). I would also guess that the slope at x1 and x3 should be 0 (though that depends). In any case, you probably don't want a polynomial. The simplest solution might be to take pieces of sine curves between $0$ and $\pi$, stretch them according to your values, and attach them at x2. You could also use splines, but then you explicitly need to fix the slope at x2 to be 0.&lt;/p&gt;&#xA;" OwnerUserId="1386" LastActivityDate="2010-09-05T19:20:09.303" />
  <row Id="4089" PostTypeId="2" ParentId="4007" CreationDate="2010-09-05T19:24:10.970" Score="0" ViewCount="" Body="&lt;p&gt;This one might be another way of approaching the problem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f_n$ is a polynomial, so it is the Taylor expansion of itself at every point; then, we can pick any point in $[0,1]$, say 0, and see what the Taylor expansions of your four possibilities are. In this case you'll see that $b)$ has a Taylor expansion at $x =0$ which is exactly the same as you $f_n$, so that's your answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, since I did the calculations on my computer, I don't really know how messy it is to find the answer this way. (As Moron pointed out, also a quick check of the simmetry of the functions gives the right answer)&lt;/p&gt;&#xA;" OwnerUserId="1415" LastActivityDate="2010-09-05T19:24:10.970" />
  <row Id="4090" PostTypeId="1" CreationDate="2010-09-05T19:34:47.277" Score="2" ViewCount="109" Body="&lt;p&gt;&lt;a href=&quot;http://math.stackexchange.com/questions/4084/how-do-you-find-t-junctions-in-an-array-of-triangles&quot;&gt;Follow up to&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Say I have triangles as shown:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/xh4D0.png&quot; alt=&quot;triangles&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Its easy to see that ABC is adjacent to ACJ:  ABC and ACJ share 2 vertices (A and C).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However FEC and ACJ are &lt;strong&gt;also&lt;/strong&gt; adjacent, but only one vertex is shared.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I find if two triangles are adjacent when they differ in size?&lt;/p&gt;&#xA;" OwnerUserId="1069" LastActivityDate="2010-09-06T05:39:26.047" Title="How do I find if two triangles are adjacent, when they differ in size?" Tags="&lt;geometry&gt;" AnswerCount="3" CommentCount="4" />
  <row Id="4091" PostTypeId="2" ParentId="4090" CreationDate="2010-09-05T19:42:39.163" Score="0" ViewCount="" Body="&lt;p&gt;To find all adjacent triangles to ACJ for ex, my first stab at this is to consider all triangles that share &lt;em&gt;at least one vertex&lt;/em&gt; with ACJ.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, ACJ shares one vertex with ADF and FEC (adjacent because an entire side is shared).  It also shares one vertex with HBA and BIC (not adjacent because they don't share a side).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, for two triangle t1 and t2 to be adjacent then:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;At least one vertex must be shared between the two triangles t1 and t2;&lt;/li&gt;&#xA;&lt;li&gt;At least one vertex &lt;em&gt;other than the shared vertex&lt;/em&gt; from t2 must fall on a line that is drawn between the &lt;em&gt;shared vertex&lt;/em&gt; &lt;strong&gt;&lt;em&gt;and an unshared vertex on t1&lt;/em&gt;&lt;/strong&gt;.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So for t1=ACJ and t2=FEC, FEC is adjacent to ACJ because:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;C is shared between ACJ and FEC&lt;/li&gt;&#xA;&lt;li&gt;F (from t2) falls on the line between C and A (from t1).&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Thus ACJ and FEC are adjacent.&lt;/p&gt;&#xA;" OwnerUserId="1069" LastActivityDate="2010-09-05T19:42:39.163" CommentCount="1" />
  <row Id="4092" PostTypeId="2" ParentId="4054" CreationDate="2010-09-05T19:44:19.737" Score="3" ViewCount="" Body="&lt;p&gt;Expanding on my comment, this result can be translated into:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;A surface in R&lt;sup&gt;3&lt;/sup&gt; which satisfies the Maximum-Modulus principle is uniquely determined by specifying it's boundary&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For some more intuition imagine an arbitrary fixed boundary in R&lt;sup&gt;3&lt;/sup&gt;, and now try to imagine constructing a surface whose only local maximum's or minimum's occur at this boundary.  It makes sense &quot;intuitively&quot; (at least for me) that the no local maximum or minimum condition is strong enough to force the surface to be unique.  All that is left is to show that the real and imaginary parts of a holomorphic function satisfy this condition.&lt;/p&gt;&#xA;" OwnerUserId="1417" LastEditorUserId="1417" LastEditorDisplayName="" LastEditDate="2010-10-29T02:15:09.867" LastActivityDate="2010-10-29T02:15:09.867" CommentCount="2" />
  <row Id="4093" PostTypeId="2" ParentId="4054" CreationDate="2010-09-05T19:56:49.803" Score="3" ViewCount="" Body="&lt;p&gt;An exercise that was a big help to me was to compute the integral of 1/z on the closed path consisting of the square from (1,-1) to (1,1) to (-1,1) to (-1,-1) to (1,-1). It illustrates how the sides with varying real differ from the sides with varying imaginary values.&lt;/p&gt;&#xA;" OwnerUserId="1312" LastActivityDate="2010-09-14T19:18:43.203" />
  <row Id="4094" PostTypeId="1" AcceptedAnswerId="4122" CreationDate="2010-09-05T20:05:19.833" Score="3" ViewCount="177" Body="&lt;p&gt;I can't seem to work out the inequality $(\sum |x_n|^q)^{1/q} \leq (\sum |x_n|^p)^{1/p}$ for $p \leq q$ (which I'm assuming is the way to go about it).&lt;/p&gt;&#xA;" OwnerUserId="1736" LastActivityDate="2010-09-06T05:13:05.103" Title="How do you show that $l_p \subset l_q$ for $p \leq q$?" Tags="&lt;analysis&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" />
  <row Id="4095" PostTypeId="2" ParentId="4073" CreationDate="2010-09-05T20:07:28.650" Score="0" ViewCount="" Body="&lt;p&gt;You have a lot of possibilities, which dependes on what you want to achieve. See, for example &lt;a href=&quot;http://en.wikipedia.org/wiki/Construction_of_the_real_numbers&quot; rel=&quot;nofollow&quot;&gt;wikipedia&lt;/a&gt;, which also reports the following quote:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Few mathematical structures have&#xA;  undergone as many revisions or have&#xA;  been presented in as many guises as&#xA;  the real numbers. Every generation&#xA;  reexamines the reals in the light of&#xA;  its values and mathematical&#xA;  objectives.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="354" LastActivityDate="2010-09-05T20:07:28.650" />
  <row Id="4096" PostTypeId="2" ParentId="4094" CreationDate="2010-09-05T20:13:21.533" Score="3" ViewCount="" Body="&lt;p&gt;I don't think you need to prove the inequality you have in the question; that's a bit too strong. Note that $\{x_n\}\in\ell_p$ if and only if $\left(\sum|x_n|^p\right)^{1/p}$ is finite, if and only if $\sum|x_n|^p\lt\infty$. So you really just need to show that if $\sum|x_n|^p$ is finite, then $\sum|x_n|^q$ is finite, assuming $p\leq q$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You want to remember is two things:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;if $p\leq q$, then for $|x|&gt;1$ you have $|x|^p\leq|x|^q$, but if $|x|&amp;lt;1$, then $|x|^p \geq |x|^q$.&lt;/li&gt;&#xA;&lt;li&gt;$\sum_{n=1}^{\infty}a_n$ converges if and only if for every $m\geq 1$, $\sum_{n=m}^{\infty}a_n$ converges.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-09-05T20:25:06.690" LastActivityDate="2010-09-05T20:25:06.690" CommentCount="2" />
  <row Id="4097" PostTypeId="2" ParentId="4090" CreationDate="2010-09-05T20:32:15.033" Score="0" ViewCount="" Body="&lt;p&gt;I'm not so sure about this, but if you considered vectors FC and AC, the triangles are adjacent iff there exists some $x\in\mathbb{R}$ such that $FC = x(AC)$, and every point in AC as well as FC is contained in $L:X=\lambda(A-C)+Q$ given any $Q\in FC$.&lt;/p&gt;&#xA;" OwnerUserId="1523" LastActivityDate="2010-09-05T20:32:15.033" />
  <row Id="4098" PostTypeId="2" ParentId="4054" CreationDate="2010-09-05T21:24:34.693" Score="2" ViewCount="" Body="&lt;p&gt;Even though I am not totally satisfied by my argument, I would say it is about average values on the curve of integration that is dedicated to analytic functions. &#xA;Lets look at a circle $C$ around $z$: If we choose points $\zeta_1,\ldots,\zeta_n$ on $C$ such that they span a regular $n$-sided polygon the average of their sum is $z$, that is&#xA;$$\sum_{k=1}^n \zeta_k\cdot\frac{1}{n}= z.$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is far from Cauchy integral formula, however, since&#xA;$$1 = \frac{1}{2\pi i}\int_C \frac{1}{\zeta-z}d\zeta$$&#xA;we can write the Cauchy formula in the form &#xA;$$\frac{1}{2\pi i}\int_C\frac{f(\zeta)-f(z)}{\zeta-z}d\zeta=0.$$&#xA;Next, if we think of $f$ as a power series around $z$, we can interpret the formula in text as saying that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;If we remove the constant term $f(z)$ from $f(\zeta)$ and divide out the factor $\zeta-z$, the remaining will have average 0&quot;. &lt;/p&gt;&#xA;" OwnerUserId="1154" LastActivityDate="2010-09-05T21:24:34.693" />
  <row Id="4099" PostTypeId="1" AcceptedAnswerId="4100" CreationDate="2010-09-05T21:46:36.033" Score="1" ViewCount="147" Body="&lt;p&gt;I'm trying to show that all graphs with 5 vertices, each of degree 2, are isomorphic to each other.  Is there a more clever way than simply listing them all out?&lt;/p&gt;&#xA;" OwnerUserId="314" LastActivityDate="2010-09-05T22:23:29.500" Title="5-vertex graphs with vertices of degree 2" Tags="&lt;graph-theory&gt;" AnswerCount="2" />
  <row Id="4100" PostTypeId="2" ParentId="4099" CreationDate="2010-09-05T21:56:28.277" Score="3" ViewCount="" Body="&lt;p&gt;I'm assuming you do not allow multi-edges, as otherwise there is a trivial counterexample (the cycle of length $5$, versus a disconnected graph consisting of a triangle and two vertices joined by two edges).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pick one vertex $v_0$; it must be joined to two other distinct vertices, which we may call $v_{-1}$ and $v_1$. Each of those must be joined to another vertex; can they be joined to each other? Can they both be joined to the same vertex that is not yet listed? Consider the possibilities. Then see where each of them leads you. &lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-09-05T21:56:28.277" />
  <row Id="4101" PostTypeId="2" ParentId="4099" CreationDate="2010-09-05T22:17:54.967" Score="4" ViewCount="" Body="&lt;p&gt;Consider this simple, powerful theorem: all vertices of a graph have even degree if and only if the set of its edges can be partitioned in cycles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The number of edges of your graph $G$ is $e(G) = \frac 1 2 \sum_{x\in V(G)} d(x) = 5$. Observe that there are not isolated vertices and that every cycle has at least three edges. Therefore the only possible partition in cycles is $E(G)$ itself, i.e. all the graphs with the required property are 5-cycles and thus they are isomorphic.&lt;/p&gt;&#xA;" OwnerUserId="365" LastEditorUserId="365" LastEditorDisplayName="" LastEditDate="2010-09-05T22:23:29.500" LastActivityDate="2010-09-05T22:23:29.500" />
  <row Id="4102" PostTypeId="1" CreationDate="2010-09-05T22:30:02.247" Score="4" ViewCount="259" Body="&lt;p&gt;Given sample data $x_1, \ldots, x_n$ generated from a probability distribution $f(x|\theta)$ ($\theta$ being an unknown parameter), a statistic $T(x_1, \ldots, x_n)$ of the sample data is called sufficient if $f(x|\theta, t) = f(x|t)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I'm always kinda confused by this definition, since I think of a sufficient statistic as a function that gives just as much information about $\theta$ as the original data itself (which seems a little different from the definition above).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The definition of &lt;em&gt;Bayesian&lt;/em&gt; sufficiency, on the other hand, does mesh with my intuition: $T$ is a Bayesian sufficient statistic if $f(\theta|t) = f(\theta|x)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So why is the first definition of sufficiency important? What does it capture that Bayesian sufficiency doesn't, and how should I think about it? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;[Note: I believe that every sufficient statistic is also Bayesian sufficient, but not conversely (the reverse implication doesn't hold in the infinite-dimensional case, according to Wikipedia).]&lt;/p&gt;&#xA;" OwnerUserId="1546" LastActivityDate="2010-09-14T19:18:03.517" Title="Sufficient statistics vs. Bayesian sufficient statistics" Tags="&lt;probability&gt;&lt;statistics&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="3" />
  <row Id="4103" PostTypeId="2" ParentId="4018" CreationDate="2010-09-05T22:39:18.763" Score="2" ViewCount="" Body="&lt;p&gt;If I remember correctly from my student days, if D is a connected domain in the complex plane, then the ring &lt;em&gt;O&lt;/em&gt;(D) of holomorphic functions on D has the property that every finitely generated ideal is actually principal, but there are ideas tat cannot be finitely generated (such as in Robin's answer above)&lt;/p&gt;&#xA;" OwnerUserId="688" LastActivityDate="2010-09-05T22:39:18.763" />
  <row Id="4104" PostTypeId="2" ParentId="4082" CreationDate="2010-09-05T22:51:06.663" Score="0" ViewCount="" Body="&lt;p&gt;What you apparently want is a monotonic interpolation scheme. As mentioned already, fitting a single polynomial to your data will result in an interpolant that does not necessarily satisfy your constraints. Using a piecewise polynomial function (a function that has a different polynomial representation at each interval) should do a better job.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Classically, the most-used class of piecewise polynomial functions is the piecewise cubic Hermite interpolant, whose representation in an interval $[x_i,x_{i+1}]$ looks like&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$y_i+y_i^{\prime}\left(x-x_i\right)+c_i\left(x-x_i\right)^2+d_i\left(x-x_i\right)^3$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$c_i=\frac{3s_i-2y_i^\prime-y_{i+1}^\prime}{x_{i+1}-x_i}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$d_i=\frac{y_i^\prime+y_{i+1}^\prime-2s_i}{\left(x_{i+1}-x_i\right)^2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$s_i=\frac{y_{i+1}-y_i}{x_{i+1}-x_i}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and ${y_i^\prime,y_{i+1}^\prime}$ are the slopes (derivative values) of your interpolant at the corresponding points $(x_i,y_i)$, $(x_{i+1},y_{i+1})$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, you will rightly ask, &quot;how do I determine the $y_i^\prime$?&quot; There are a number of schemes to do this: one of the most popular is the cubic spline, which imposes the additional constraint that the interpolant have two continuous derivatives. The problem with this approach is that there can still be extraneous wiggles in your interpolant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One approach to ensure the monotonicity (i.e. the curve representing your data to fit against has no spurious wiggles) of your piecewise cubic Hermite interpolant is to use the Stineman scheme. The formula for computing slopes is a bit more elaborate, so I'll just have to refer you to the &lt;a href=&quot;http://ifile.it/x1yb0qt&quot; rel=&quot;nofollow&quot;&gt;paper&lt;/a&gt;. As an aside, Stineman also proposes a &lt;em&gt;rational&lt;/em&gt; interpolant you can use with the slope determination scheme in the linked paper; you can try that one instead if for some reason using a piecewise Hermite cubic is unsatisfactory.&lt;/p&gt;&#xA;" OwnerUserId="498" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-09-06T01:54:02.867" LastActivityDate="2010-09-06T01:54:02.867" />
  <row Id="4105" PostTypeId="1" CreationDate="2010-09-05T23:37:52.187" Score="1" ViewCount="240" Body="&lt;p&gt;What's a good way of solving the Diffie–Hellman problem when those exchanging the message have chosen a low primitive root $g$ (e.g. $g=3$)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course you could brute force it but I'm interested in knowing whether there is a formula for solving it when you know $g^a \pmod{p}$ and $g^b \pmod{p}$ as well as $p$ and $g$ of course.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; For those unfamiliar with the Diffie–Hellman problem the integers $g$ and $p$ (with $1 &amp;lt; g &amp;lt; p$ and $p$ being prime), $g^a \pmod{p}$ and $g^b \pmod{p}$ are public. The integers $a$ and $b$ are private integers and we want to calculate the secret key $s = g^{ab} \pmod{p}$.&lt;/p&gt;&#xA;" OwnerUserId="1737" LastEditorUserId="1737" LastEditorDisplayName="" LastEditDate="2010-09-06T09:04:13.473" LastActivityDate="2010-09-06T09:04:13.473" Title="Solving Diffie–Hellman problem for low primitive root" Tags="&lt;number-theory&gt;&lt;modular-arithmetic&gt;&lt;cryptography&gt;" CommentCount="7" />
  <row Id="4106" PostTypeId="1" AcceptedAnswerId="4960" CreationDate="2010-09-05T23:51:00.007" Score="4" ViewCount="336" Body="&lt;p&gt;&lt;b&gt;Bounty update&lt;/b&gt;: this can be solved by change of basis, but I'm intrigued by David's solution relying on Fourier Transform of Dirac Delta function, so the bounty is for whoever finds a way to fix his solution to give the right result.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose I have a non-negative real-valued function over $d$-dimensional real vectors as follows&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f(\mathbf{x})=\exp(-\mathbf{x}' A \mathbf{x})$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $A$ is some symmetric positive definite $d\times d$ matrix. What is the normalization factor to turn this into a valid density over the following set?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$S_d=\{(x_1,\ldots,x_d)\in \mathbf{R}^d | \sum_i x_i=0 \}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Below, David Bar Moshe gives a general solution to computing that integral over space orthogonal to some vector $v$, but I suspect it has a mistake because the answer depends on the norm of $v$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, suppose $A$ is $d$-by-$d$ identity matrix. Let $v$ be a vector of all ones. Because of symmetry, integrating over space orthogonal to $v$ should be the same as as $d-1$ dimensional Gaussian integral, ie $\pi^{(d-1)/2}$, whereas David's solution gives&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\pi^{(d-1)/2}}{d}$$&lt;/p&gt;&#xA;" OwnerUserId="998" LastEditorUserId="998" LastEditorDisplayName="" LastEditDate="2010-09-12T21:05:32.623" LastActivityDate="2010-09-18T23:16:59.507" Title="normalization factor for restricted density" Tags="&lt;integral&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="3" />
  <row Id="4107" PostTypeId="1" AcceptedAnswerId="4142" CreationDate="2010-09-05T23:52:36.430" Score="9" ViewCount="539" Body="&lt;p&gt;What are some examples of theories stronger than Presburger Arithmetic but weaker than Peano Arithmetic?  Are all such theories decidable?  If not, by what methods other than Gödelization can undecidability be established?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Both answers have indicated Robinson Arithmetic as an intermediate theory, but I don't understand how it can be &quot;stronger&quot; than Presburger Arithmetic in any reasonable way, because Robinson Arithmetic cannot prove commutativity of addition.  I have attempted to define &quot;stronger&quot; in the comments.  I would appreciate any suggestions for how to clarify the question and this definition.&lt;/p&gt;&#xA;" OwnerUserId="1284" LastEditorUserId="1284" LastEditorDisplayName="" LastEditDate="2010-09-07T20:57:18.857" LastActivityDate="2010-09-08T04:11:15.663" Title="What are some examples of theories stronger than Presburger Arithmetic but weaker than Peano Arithmetic?" Tags="&lt;number-theory&gt;&lt;logic&gt;" AnswerCount="2" CommentCount="14" FavoriteCount="2" />
  <row Id="4108" PostTypeId="2" ParentId="4102" CreationDate="2010-09-06T00:47:27.640" Score="0" ViewCount="" Body="&lt;p&gt;Suppose, we can factorize the pdf as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f(x|\theta) = h(x) g(t(x),\theta)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we can achieve the above factorization then it is obvious that the estimate of $\theta$ will depend only on the value of $t(x)$. In other words, our estimate of $\theta$ will be the same for two sets of data ($x_1$, $x_2$) if $t(x_1) = t(x_2)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The above feature of $t(x)$ is conceptualized by calling $t(x)$ as a sufficient statistic. It is sufficient in the sense that once we know $t(x)$ we implicitly have an estimate of $\theta$ and is termed as a statistic as it is a function of the data $x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am a bit unsure about the relationship with bayesian sufficiency but will update my answer if I have some relevant thoughts on that issue.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the way your definition of sufficient statistic does not seem to make sense to me as the right handside should depend on $\theta$.&lt;/p&gt;&#xA;" LastActivityDate="2010-09-06T00:47:27.640" CommentCount="3" />
  <row Id="4109" PostTypeId="1" CreationDate="2010-09-06T00:52:00.360" Score="0" ViewCount="180" Body="&lt;p&gt;I have been working on the following lemma for a few days. Any hints would be appreciated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Lemma&lt;/strong&gt;&lt;br&gt;&#xA;&lt;em&gt;Let $G$ be a finite group&lt;/em&gt;, $H \le G$ &lt;em&gt;and&lt;/em&gt; $P \in Syl_{p}(H)$. &lt;em&gt;If&lt;/em&gt; $N_{G}(P) \subseteq H$ &lt;em&gt;then&lt;/em&gt; $P$ &lt;em&gt;is a Sylow $p$-subgroup of $G$&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Partial Proof (Contrapostive).&lt;/em&gt;&lt;br&gt;&#xA;First assume that $P$ is not a Sylow $p$-subgroup of $G$. Therefore $P$ is properly contained in some Sylow $p$-subgroup of $G$, say $Q$. Next let $x \in Q \setminus P$. Therefore $P^{x} \subseteq Q^{x}=Q.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I would like to show that infact $P^{x} = P$ so that $x \in N_{G}(P)$. Therefore $N_{G}(P) \not \subseteq H$. However I am not sure if this is true. One interesting fact is that Q contains only one Sylow $p$-subgroup of $H$. So if $P^{x} \subset H$ then the proof is complete. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have the feeling that I may be missing something very trivial. Maybe a contradiction proof would work better. &lt;/p&gt;&#xA;" OwnerUserId="1738" LastActivityDate="2010-09-06T02:04:28.817" Title="A Nice Sylow Theory Problem" Tags="&lt;group-theory&gt;" AnswerCount="1" />
  <row Id="4110" PostTypeId="2" ParentId="4109" CreationDate="2010-09-06T01:26:45.463" Score="3" ViewCount="" Body="&lt;p&gt;Instead of taking an &lt;em&gt;arbitrary&lt;/em&gt; $x\in Q\setminus P$, note that $N_Q(P)$ must properly contain $P$ (since $Q$ is a $p$-group, and you are assuming that $P\neq Q$). So pick $x\in N_Q(P)\setminus P$. It cannot lie in $H$, as $\langle P,x\rangle$ is a subgroup of $Q$.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-09-06T02:04:28.817" LastActivityDate="2010-09-06T02:04:28.817" />
  <row Id="4113" PostTypeId="2" ParentId="4073" CreationDate="2010-09-06T02:46:08.337" Score="0" ViewCount="" Body="&lt;p&gt;This is a great article that constructs the real numbers as classes of rational Cauchy sequences:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://math.mit.edu/~tkemp/18.024/R.pdf&quot; rel=&quot;nofollow&quot;&gt;http://math.mit.edu/~tkemp/18.024/R.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1605" LastActivityDate="2010-09-06T02:46:08.337" />
  <row Id="4114" PostTypeId="1" AcceptedAnswerId="4116" CreationDate="2010-09-06T02:55:42.890" Score="5" ViewCount="416" Body="&lt;p&gt;In the definition of topological space, we require the intersection of a finite number of open sets to be open while we require the arbitrary union of open sets to be open. why is this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm assuming this has something to do with the following observation: $\cap_{n=1}^{\infty} (-\frac{1}{n},\frac{1}{n}) = \{0\}$ and there is some reason we don't want singletons to be considered open, I am wondering what this reason is. Am I thinking in the right direction here?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks :)&lt;/p&gt;&#xA;" OwnerUserId="249" LastEditorUserId="249" LastEditorDisplayName="" LastEditDate="2010-09-06T04:05:37.717" LastActivityDate="2010-09-06T14:50:15.403" Title="Why do we require a topological space to be closed under finite intersection?" Tags="&lt;intuition&gt;&lt;topology&gt;&lt;definition&gt;" AnswerCount="5" CommentCount="8" FavoriteCount="1" />
  <row Id="4115" PostTypeId="2" ParentId="4114" CreationDate="2010-09-06T03:16:07.613" Score="2" ViewCount="" Body="&lt;p&gt;&lt;strike&gt;Minor correction: $[0,1/n]$ is not open in the usual topology of the real line.  But you could get the same thing with&lt;/strike&gt; The example $\cap_{n=1}^\infty (-1/n,1/n)=${0} is a good one.  And yes, points are not open in the usual topology on the line.  If they were, then every set, being the union of one-point sets, would be open.  This is called the discrete topology.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think this definition can be fairly well motivated by what happens in metric spaces, of which the real line is a special case.  Call a set $U$ open if around each point in $U$ there is a ball (with respect to the metric) centered at that point and contained in $U$.  Then you can show as an exercise that intersections of finite collections of open sets are open, and that arbitrary unions of open sets are open.  However, the example above shows that you can't take arbitrary intersections.  (In the case of the real line, &quot;ball&quot; just means open interval.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But why does it work so well to define general topological spaces to satisfy this property of metric spaces?  I don't know.  For more on the motivation of defining topologies in terms of &quot;open sets&quot;, you may be interested in &lt;a href=&quot;http://mathoverflow.net/questions/19152/why-is-a-topology-made-up-of-open-sets&quot; rel=&quot;nofollow&quot;&gt;this MathOverflow question&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2010-09-06T03:42:00.263" LastActivityDate="2010-09-06T03:42:00.263" CommentCount="0" />
  <row Id="4116" PostTypeId="2" ParentId="4114" CreationDate="2010-09-06T03:25:16.147" Score="7" ViewCount="" Body="&lt;p&gt;You need to think about what the intuition behind open sets are. One way to think about it is through neighborhoods: an open set is a set which is a neighborhood of each of its points. What is a neighborhood of a point? A neighborhood of a point $x$ is a set that contains of all points that are &quot;sufficiently close&quot; to $x$ (what does &quot;sufficiently close&quot; mean? It depends on the situation; you think of different neighborhoods perhaps specifying different degrees of closeness). In particular, any set that contains a neighborhood of $x$ is itself a neighborhood of $x$. And specifying two degrees of closeness specifies another degree of closeness that makes sense (the smaller of the two at any given place, say).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So: if you think about open sets as sets that are neighborhoods of all of the points they contain, then it is natural that the union of any family of pen sets will be open: each point in the union is one of the open sets, and that open set is a neighborhood, and the union contains that neighborhood and so is itself a neighborhood. So the arbitrary union of open sets should still be open.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What about intersection? Well, if you take two open sets $O_1$ and $O_2$, and you consider a point $x$ in $O_1\cap O_2$, then $O_1$ contains all points that are $1$-sufficiently close to $x$, and $O_2$ contains all points that are $2$-sufficiently close to $x$ (with &quot;$1$-sufficiently&quot; and &quot;$2$-sufficiently&quot; describing the two degrees of closeness required), so $O_1\cap O_2$ will contains all points that are both $1$-sufficiently and $2$-sufficiently close to $x$, and so it contains all points that are &quot;sufficiently close&quot; to $x$ for some meaning of &quot;sufficiently close&quot;, so it is also an open set. This gives you, inductively, any finite intersection.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But what about arbitrary intersections? Then you run into trouble, because specfying two degrees of &quot;closeness&quot; gives you a degree of closeness (the smaller one), but an infinite number of degrees of closeness may end up excluding everything! (Just as in your example, taking the intersection of all $(-\frac{1}{n},\frac{1}{n})$, which specify all points that are $\frac{1}{n}$-close to $0$, but the intersection excludes everything). So we don't want to require that an arbitrary intersection of neighborhoods be a neighborhood, and so we don't want to require that an arbitrary intersection of open sets be an open set. &lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-09-06T03:25:16.147" />
  <row Id="4119" PostTypeId="2" ParentId="4114" CreationDate="2010-09-06T04:48:29.960" Score="3" ViewCount="" Body="&lt;p&gt;If you think of it in the case of the topology on the real line - as soon as we leave the restriction from finite intersections we would get singletons as you have seen. &#xA;However, any set of real numbers is a union of singletons that is&#xA;$$A= \bigcup_{x\in A} \{ x \}. $$&#xA;It would follow that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;each&lt;/strong&gt; set would be open &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;each&lt;/strong&gt; set would be closed &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;only finite&lt;/strong&gt; sets would be compact &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;each&lt;/strong&gt; function $f:\mathbb{R}\to X$ would be continuous, etc...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Thanks Jonas!)&lt;/p&gt;&#xA;" OwnerUserId="1154" LastEditorUserId="1154" LastEditorDisplayName="" LastEditDate="2010-09-06T07:14:50.543" LastActivityDate="2010-09-06T07:14:50.543" CommentCount="4" />
  <row Id="4120" PostTypeId="1" AcceptedAnswerId="4151" CreationDate="2010-09-06T04:56:35.983" Score="0" ViewCount="62" Body="&lt;p&gt;Given $C=XF$, and only data for $C$, solving for $X$ and $F$ has been described as &lt;a href=&quot;http://math.stackexchange.com/questions/3905/how-can-i-solve-this-matrix/3908#3908&quot;&gt;not tractable&lt;/a&gt; (which I'm content to believe, but others aren't).  If that's the case, where does the following argument fall apart?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose $C^i=XF^i$ is written:&#xA;$\begin{bmatrix}&#xA;  C_1^i \\&#xA;  \vdots \\&#xA;  C_m^i &#xA;\end{bmatrix} = \begin{align}\begin{bmatrix}&#xA;  X_{1,1} &amp;amp; \dots &amp;amp; X_{1,n} \\&#xA;  \vdots &amp;amp;  &amp;amp; \vdots \\&#xA;  X_{m,1} &amp;amp; \dots &amp;amp; X_{m,n} &#xA;\end{bmatrix}\end{align}\begin{bmatrix}&#xA;  F_1^i \\&#xA;  \vdots \\&#xA;  F_n^i &#xA;\end{bmatrix}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One instance of data for a given $i$ of $C^i$ would produce $m$ equations and $mn+n$ unknowns. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$q$ instances of data for $C^i$ would produce $qm$ equations and $mn+qn$ unknowns.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;If&lt;/em&gt; the supposition that you need more equations than unknowns is sufficient, why can I contradict others statements of not tractability  by choosing whole number values to satisfy the inequality:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$qm &gt;= mn+qn$? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using say $q=6,m=3,n=2$.&lt;/p&gt;&#xA;" OwnerUserId="1067" LastEditorUserId="1067" LastEditorDisplayName="" LastEditDate="2010-09-06T05:14:41.060" LastActivityDate="2010-09-06T16:23:05.307" Title="Looking for a proof of non tractability (or perhaps a definition of tractable) ..." Tags="&lt;linear-algebra&gt;&lt;matrices&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="4121" PostTypeId="2" ParentId="4090" CreationDate="2010-09-06T05:05:24.403" Score="1" ViewCount="" Body="&lt;p&gt;First, we need a definition of &quot;adjacent plane figures&quot;.  If we take &quot;plane figure&quot; to mean a closed plane curve (the boundary) along with its interior, then one definition might be that two plane figures are adjacent when (a) the intersection of their boundaries is a curve but not a point; (b) their interiors are disjoint.  For polygons (boundaries consist of line segments) any subtle topological issues disappear, so this will certainly do for triangles. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given this, then I think it's probably not too hard to prove this hypothesis:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Theorem: Triangles ABC and DEF are adjacent iff all of these hold:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(a) At least one side of ABC (say AB) and one side of DEF (say DE) are identical as lines (not line segments).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(b) At least one of A and B is strictly between D and E, or at least one of D and E is strictly between A and B.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(c) C and F are on different sides of line AB (which is the same as line DE).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The nice thing about this theorem, if true, is that it should not be hard to come up with a simple algorithm to test for adjacency in the Cartesian plane. I suspect the image processing folks have done this and much much more along the same lines.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are some of the trickier cases...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://djlewis.smugmug.com/photos/996266339_m4xKQ-XL.jpg&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="271" LastEditorUserId="271" LastEditorDisplayName="" LastEditDate="2010-09-06T05:39:26.047" LastActivityDate="2010-09-06T05:39:26.047" />
  <row Id="4122" PostTypeId="2" ParentId="4094" CreationDate="2010-09-06T05:13:05.103" Score="2" ViewCount="" Body="&lt;p&gt;You are right @user1736.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $a\leq1$ then &#xA;$$\left(\sum |a_n|\right)^a\leq \sum|a_n|^a.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence if $p\leq q$ we have $p/q\leq1$&#xA;$$\left(\sum_n |x_n|^q\right)^{1/q}=\left(\sum_n |x_n|^q\right)^{p/qp}\leq \left(\sum_n |x_n|^{q(p/q)}\right)^{1/p}=\left(\sum|x_n|^p\right)^{1/p}$$&lt;/p&gt;&#xA;" OwnerUserId="1154" LastActivityDate="2010-09-06T05:13:05.103" />
  <row Id="4124" PostTypeId="2" ParentId="4044" CreationDate="2010-09-06T05:34:08.040" Score="1" ViewCount="" Body="&lt;p&gt;the assertion $q_n \to 0$ as $n \to \infty$ in michael ulm's response is perhaps not entirely obvious - and is related to the remark in moron's response: &quot;I suppose we can eliminate $p=1$ somehow..&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i'll elaborate a bit more on this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;suppose $\alpha$ is the probability of winning $\$1$ and $\beta = 1-\alpha$ is the chance of losing $\$1$. two facts are clear:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(1)\kern{15pt} q_0=1\kern{10pt} {\rm and\ for}\kern{10pt} n\ge 1,\kern{10pt}  q_n = \alpha q_{n+1} + \beta q_{n-1},$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\kern{-1.2in}(2) \kern{1.5in}  q_n = q_1^n,\qquad n\ge 1.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;taking $n = 1$ in (1) and writing $q_1 = q$ gives&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\kern{-1.7in}(3) \kern{1.5in}  q = \alpha q^2 + \beta,$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which means $q = 1$ or $q = {\beta\over\alpha}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so if $\alpha \le \frac{1}{2},\ q = 1$ [because one root is $\ge 1$], while if $\alpha &gt; \frac{1}{2}$, both roots are admissible answers.&lt;br&gt;&#xA;[this is what moron also concluded.]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;to rule out the case $q = 1$, we can argue as follows: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;suppose $q = 1$ and let $S_n$ be the player's fortune at time $n$. then &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$T = T_1 = in\kern{-1pt}f \{ n \ge 1 :  S_n = 0\}$ is finite ($wp1$).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;now consider just the random walk $\{S_n : n \ge 1\}$ [forgetting about the gambling context]. we can recursively define $T_{k+1} = in\kern{-1pt}f \{n &gt; T_k : S_n = -k\} $. then $T_k$ is finite ($wp1$) for all $k\ge 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;then $\{S_{T_k} = 1-k : k \ge 1\}$ is a nonpositive subsequence of $\{S_n\}_1^\infty$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but this contradicts the SLLN which $\mathit inter\ alia$ implies that $S_n\to \infty$ when&#xA;$\alpha &gt; \frac{1}{2}$.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;so the case $q = 1$ is ruled out, and therefore $q = \frac{\beta}{\alpha} = \frac{1}{3}$ when $\alpha = \frac{3}{4}$. &lt;/p&gt;&#xA;" OwnerUserId="1547" LastActivityDate="2010-09-06T05:34:08.040" />
  <row Id="4125" PostTypeId="1" AcceptedAnswerId="4132" CreationDate="2010-09-06T06:54:04.803" Score="26" ViewCount="1254" Body="&lt;p&gt;In the comments to the question: &lt;a href=&quot;http://math.stackexchange.com/questions/3978/ann-mid-bnn-longrightarrow-ab&quot;&gt;$a^n+n|b^n+n \Longrightarrow a=b$&lt;/a&gt;, there was a claim that $5^n+n$ is never prime (for integer $n&gt;0$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It does not look obvious to prove, nor have I found a counter example.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this really true?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update&lt;/strong&gt;: $5^{7954} + 7954$ has been found to be prime by a computer: &lt;a href=&quot;http://www.mersenneforum.org/showpost.php?p=233370&amp;amp;postcount=46&quot; rel=&quot;nofollow&quot;&gt;http://www.mersenneforum.org/showpost.php?p=233370&amp;amp;postcount=46&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks to Douglas (and lavalamp)!&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-10-14T16:24:20.287" LastActivityDate="2010-10-14T21:29:43.933" Title="$5^n+n$ is never prime?" Tags="&lt;number-theory&gt;" AnswerCount="3" CommentCount="9" FavoriteCount="6" />
  <row Id="4127" PostTypeId="2" ParentId="4106" CreationDate="2010-09-06T07:53:05.780" Score="2" ViewCount="" Body="&lt;p&gt;First, one may assume that $\mathbf{A}$ is a symmetric matrix, since its antisymmetric part doesn't contribute to the density value. The normalization factor is equal to the reciprocal of the integral over constraint surface. This integral&#xA;can be converted into an integral over the whole of $\mathbb{R}^n$ as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$N^{-1}=\int_{\mathbf{x} \in S_d} \exp(-\mathbf{x}^T \mathbf{A}\mathbf{x}) = \frac1{2\pi}\int_{-\infty}^{\infty}\mathrm{d}k \int_{\mathbf{x} \in \mathbb{R}^d} \exp(-\mathbf{x}^T \mathbf{A}\mathbf{x} + \sqrt{-1}k \mathbf{v}^T \mathbf{x})$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $\mathbf{v}$ is a $d$-dimensional vector having all components equal to one. This construction can be understood by exchanging the integration order, then the integration over $k$ generates a Dirac measure concentrated on the constraint surface. The change of order is possible because the integrand is bounded. Performing a square completion of the exponent, we obtain:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$N^{-1}=  \frac1{2\pi}\int_{-\infty}^{\infty} \mathrm{d}k \int_{\mathbf{x} \in \mathbb{R}^d} \exp(-(\mathbf{x}+\frac{\sqrt{-1}}{2}k\mathbf{A}^{-1}\mathbf{v})^T \mathbf{A}(\mathbf{x}+\frac{\sqrt{-1}}{2}k\mathbf{A}^{-1}\mathbf{v}) - \frac{k^2}{4} \mathbf{v}^T \mathbf{A}^{-2}\mathbf{v})$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solving the Gaussian integral over $\mathbb{R}^d$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ N^{-1}= \frac{(2\pi)^{\frac{d-3}{2}}}{2\det(\mathbf{A})^{\frac{1}{2}}}\int_{-\infty}^{\infty} \mathrm{d}k \exp(-\frac{k^2}{4} \mathbf{v}^T \mathbf{A}^{-1}\mathbf{v}))$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solving the Gaussian integral over $k$, we obtain the final result:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ N^{-1} = \frac{(\pi)^{\frac{(d-1)}{2}}}{\det(\mathbf{A})^{\frac{1}{2} }(\mathbf{v}^T \mathbf{A}^{-1}\mathbf{v})^{\frac{1}{2}}}$&lt;/p&gt;&#xA;" OwnerUserId="1430" LastEditorUserId="1430" LastEditorDisplayName="" LastEditDate="2010-09-08T08:21:26.127" LastActivityDate="2010-09-08T08:21:26.127" CommentCount="12" />
  <row Id="4129" PostTypeId="2" ParentId="4125" CreationDate="2010-09-06T08:17:01.243" Score="4" ViewCount="" Body="&lt;p&gt;If n is odd, then 5^n + n is always even because LSD of 5^n is always 5 for n &gt; 0. Hence, for odd n ( n &gt; 0), 5 ^n + n is composite.&lt;/p&gt;&#xA;" OwnerUserId="1744" LastActivityDate="2010-09-06T08:17:01.243" CommentCount="2" />
  <row Id="4130" PostTypeId="2" ParentId="4125" CreationDate="2010-09-06T10:18:41.600" Score="6" ViewCount="" Body="&lt;p&gt;After reading Douglas S. Stones comment I asked mathematica to check if 5^(2*3977) + (2*3977) is prime and after about 27 seconds, found that it is indeed prime. So the claim $5^n +n$ is never prime is false.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: It turns out the function I used in mathematica is not a deterministic algorithm. However we can still say the claim $5^n +n$ is never prime is false is most likely true.&lt;/p&gt;&#xA;" OwnerUserId="899" LastEditorUserId="899" LastEditorDisplayName="" LastEditDate="2010-09-06T11:17:45.353" LastActivityDate="2010-09-06T11:17:45.353" CommentCount="4" />
  <row Id="4132" PostTypeId="2" ParentId="4125" CreationDate="2010-09-06T10:35:33.550" Score="27" ViewCount="" Body="&lt;p&gt;A general rule-of-thumb for &quot;is there a prime of the form f(n)?&quot; questions is, unless there exists a set of small divisors D, called a &lt;em&gt;covering set,&lt;/em&gt; that divide every number of the form f(n), then there will eventually be a prime.  See, e.g. &lt;a href=&quot;http://en.wikipedia.org/wiki/Sierpinski_number&quot; rel=&quot;nofollow&quot;&gt;Sierpinski numbers&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Running WinPFGW (it should be available from the primeform yahoo group &lt;a href=&quot;http://tech.groups.yahoo.com/group/primeform/&quot; rel=&quot;nofollow&quot;&gt;http://tech.groups.yahoo.com/group/primeform/&lt;/a&gt;), it found that $5^n+n$ is &lt;a href=&quot;http://en.wikipedia.org/wiki/Probable_prime&quot; rel=&quot;nofollow&quot;&gt;3-probable prime&lt;/a&gt; when n=7954.  Moreover, for every n less than 7954, we have $5^n+n$ is composite.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To actually certify that $5^{7954}+7954$ is a prime, you could use Primo (available from &lt;a href=&quot;http://www.ellipsa.eu/public/misc/downloads.html&quot; rel=&quot;nofollow&quot;&gt;http://www.ellipsa.eu/public/misc/downloads.html&lt;/a&gt;).  I've begun running it (so it's passed a few more pseudo-primality tests), but I doubt I will continue until it's completed -- it could take a long time (e.g. a few months).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: $5^{7954}+7954$ is officially prime.  A proof certificate was given by lavalamp at &lt;a href=&quot;http://www.mersenneforum.org/showpost.php?p=233370&amp;amp;postcount=46&quot; rel=&quot;nofollow&quot;&gt;mersenneforum.org&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="139" LastEditorUserId="139" LastEditorDisplayName="" LastEditDate="2010-10-14T21:29:43.933" LastActivityDate="2010-10-14T21:29:43.933" CommentCount="12" />
  <row Id="4133" PostTypeId="2" ParentId="566" CreationDate="2010-09-06T10:44:28.233" Score="0" ViewCount="" Body="&lt;p&gt;Relatively simple, but still fun...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unknown source, other than my friend telling me yesterday&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a room with 50 doors in it, each door leading to a cell that is completely sound-proof and light-proof, and each cell can hold a maximum of 1 person.  There are 50 prisoners.  There also exists a light in the middle of the main room, so that when a prisoner is pulled out of their cell, they could see the light.  There is also a light switch inside the main room which controls the light.  The 50 prisoners are the only people allowed to touch the light switch that controls the light (even the prison guard cannot touch it)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Before any prisoner is placed in the cells, the prison guard tells them that he will pull out, at random, any one of them at any time, and then put them back in their cell after they get a chance to turn the light on or off or do nothing at all to it. (Only one prisoner can be out of their cell at any given time)  He also guarantees them that he will pull at least one prisoner out at least once a day.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The goal is, when any one prisoner knows when all 50 prisoners have been pulled out of their cells at least one time, and he tells the prison guard this fact correctly, all prisoners will be let free.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The prisoners are allowed one short meeting before they all get placed in their cells.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They also have a (reasonably) infinite amount of time to let the prison guard know when all prisoners have been let out of their cells at least one time.  (say 5 years or something to that effect)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is only one guess allowed ever, so it better be the right answer, or else they will all be fed to the cannibals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What method did they come up with to know exactly when all 50 prisoners have been out of their cells at least one time?&lt;/p&gt;&#xA;" OwnerUserId="1630" LastActivityDate="2010-09-06T10:44:28.233" CommunityOwnedDate="2010-09-06T10:44:28.233" CommentCount="1" />
  <row Id="4134" PostTypeId="2" ParentId="4114" CreationDate="2010-09-06T10:57:31.373" Score="3" ViewCount="" Body="&lt;p&gt;I think that one should have played around with metric spaces before dealing with topological spaces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The finite intersection of open balls centered at $x$ is again an open ball; the radius is just the minimum of the finitely many radia. But if you take infinitely many balls, the radius is the infimum of infinitely many positive numbers, which will be $0$ in many situations (for example if the radia are $1/n$). A similar argument shows that you cannot prove in general that an infinite intersection of open subsets in a metric space is open again.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since topological spaces are the natural generalization of metric spaces (replacing concrete distances by the concept of neighborhoods), the same is true for topological spaces and in the definition only finite intersections of open sets are allowed.&lt;/p&gt;&#xA;" OwnerUserId="1650" LastActivityDate="2010-09-06T10:57:31.373" />
  <row Id="4135" PostTypeId="1" CreationDate="2010-09-06T10:59:16.043" Score="2" ViewCount="92" Body="&lt;p&gt;The following is the start of basic corollary in my logic text:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For any set $\Gamma$ of sentences, $\Gamma\subseteq\text{Th}(\text{Mod}(\Gamma))$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What happens when $\text{Mod}(\Gamma)$ is empty? Suppose $\Gamma={p_1,\neg p_1}$, so that there is no truth value $V$ which models $\Gamma$, or otherwise $V(p_1)=\text{T}=V(\neg{p_1})$, which doesn't respect logical connectives. Then I would have something like $\Gamma\subseteq\text{Th}(\emptyset)$. If I'm working under the definition that $\text{Th}(\emptyset)$ is the set of all tautologies, then this doesn't make any sense.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Should there be an extra condition that $\text{Mod}(\Gamma)$ be nonempty, or am I missing something?&lt;/p&gt;&#xA;" OwnerUserId="1583" LastActivityDate="2010-09-14T05:39:48.133" Title="Does this proposition hold if $\text{Mod}(\Gamma)=\emptyset$?" Tags="&lt;logic&gt;&lt;model-theory&gt;" AnswerCount="1" />
  <row Id="4136" PostTypeId="2" ParentId="3145" CreationDate="2010-09-06T11:02:47.323" Score="0" ViewCount="" Body="&lt;p&gt;$U_i \setminus (Z \cap U_i)$ is open in $U_i$, thus also open in $X$. Then $X \setminus Z = \cup_ {i \in I} U_i \setminus (Z \cap U_i)$ is open in $X$, i.e. $Z$ is closed in $X$.&lt;/p&gt;&#xA;" OwnerUserId="1650" LastActivityDate="2010-09-06T11:02:47.323" />
  <row Id="4137" PostTypeId="2" ParentId="4135" CreationDate="2010-09-06T11:05:42.363" Score="2" ViewCount="" Body="&lt;p&gt;$Th(\emptyset)$ consists of every sentence: $S \models \phi$ means that $M \models \phi$ for every model $M$ in $S$, but in the case of $S=\emptyset$ this is an empty condition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Besides, the proof of $\Gamma \subseteq Th(Mod(\Gamma))$ is straight forward with the definitions and does not use $Mod(\Gamma) \neq \emptyset$.&lt;/p&gt;&#xA;" OwnerUserId="1650" LastActivityDate="2010-09-06T11:05:42.363" CommentCount="2" />
  <row Id="4138" PostTypeId="2" ParentId="4073" CreationDate="2010-09-06T11:18:05.110" Score="0" ViewCount="" Body="&lt;p&gt;The construction of $\mathbb{R}$ depends on your theory from which you want to start. For example it is reasonable for analysis to start with the axiom &quot;There is a complete ordered field&quot;, show that this field is actually unique up to unique isomorphism and call it $\mathbb{R}$. Then you may define the other number systems, too. Or we start from $\mathbb{N}$ and use Grothendieck constructions and completions to get $\mathbb{R}$ (see the other answers).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, we use $ZFC$ to make all these constructions work. There we have no chance to define(!) the natural numbers by means of the Peano axioms (because we have to prove the Peano axioms from $ZFC$ for $\omega$, which may be defined as the smallest inductive set). So take care of Isaac's first sentence. Also, if we start with a postulated complete ordered field, this actually follows from $ZFC$, but for the basics of analysis, this does not play any role and may be postponed to the next algebra or logic course.&lt;/p&gt;&#xA;" OwnerUserId="1650" LastActivityDate="2010-09-06T11:18:05.110" />
  <row Id="4139" PostTypeId="2" ParentId="3876" CreationDate="2010-09-06T11:19:52.243" Score="1" ViewCount="" Body="&lt;p&gt;Quinn and Hannan's &lt;a href=&quot;http://books.google.com.au/books?id=mTluO2TZ8V8C&amp;amp;dq=estimation+and+tracking+of+frequency&amp;amp;printsec=frontcover&amp;amp;source=bn&amp;amp;hl=en&amp;amp;ei=4s2ETMuYIYqHcYa-pNAL&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=4&amp;amp;ved=0CCoQ6AEwAw#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;The estimation and tracking of frequency&lt;/a&gt; is dedicated to this topic. I can highly recommend it.&lt;/p&gt;&#xA;" OwnerUserId="567" LastActivityDate="2010-09-06T11:19:52.243" />
  <row Id="4140" PostTypeId="2" ParentId="4054" CreationDate="2010-09-06T12:16:54.660" Score="6" ViewCount="" Body="&lt;p&gt;The trick is to remember that differentiable is really a way of saying 'locally linear', that is, close to $z$ we have $$f( \zeta )= f(z) + (\zeta -z)f'(z) +g( \zeta )$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $g(\zeta)$ is an error term that vanishes as we get close to $z$ (in fact it is &lt;a href=&quot;http://en.wikipedia.org/wiki/Little_o#Little-o_notation&quot; rel=&quot;nofollow&quot;&gt;$o(\zeta -z)$&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, dividing through by $(\zeta -z)$ as in the integral formula gives expression under the integral sign as: $$\frac{f( \zeta )}{(\zeta -z)}= \frac{f(z)}{(\zeta -z)} + f'(z) +\frac{g( \zeta )}{(\zeta -z)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We now integrate this termwise:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The last term still vanishes as $\zeta \to z$ (by the definition of little o), which allows us to ignore it by way of a homotopy argument. You see, since we can shrink the loop $\gamma$ around $z$ arbitrarily small without changing the value of the integral (an old theorem of Cauchy's says that the integral along a closed curve is homotopy invariant $(*)$) and, as the loop gets smaller, we see that the value of the last term on $\gamma$ gets closer to vanishing- and so, morally, its integral over $\gamma$ may be bounded above in modulus by successively smaller $\epsilon$. But, because of homotopy invariance, these $\epsilon$ bound the integral over all such loops- and $\epsilon$ is arbitrary- so the integral of the last term vanishes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second term is easy- this integrates to $[\zeta f'(z)]_{\gamma (0)}^{\gamma (1)}$, with $\gamma (0)=\gamma (1)$. Plug it in to see this vanishes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first term is the interesting one and others have covered it well in their posts, but I'll finish the argument anyways. Having eliminated the other terms (and observing that $f(z)$ is a constant) we have a pair of statements equivalent to the original formula:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{1}{2\pi i}\int\frac{f(z)}{\zeta-z}d\zeta =f(z) \iff \int\frac{1}{\zeta-z}d\zeta =2\pi i$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This last one we can use substitution $u=\zeta-z$ and homotopy to make about the integral over a circle $\gamma (t)= e^{2\pi it}$ around the origin giving:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_{\gamma} \frac{1}{u} du (**)= \int_0^1 \frac{\gamma '(t)}{\gamma (t)}dt= = \int_0^1 \frac{2 \pi i e^{2\pi it}}{e^{2\pi it}}dt$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The last one is a total gift after cancellation and we all go home smiling. We could of course have morally calculated $(**)$ without resorting to technicalities by observing that the primitive of $\frac{1}{u}= log(u)$ and looking at the integrand $[log(u)]_{e^0}^{e^{(2\pi -\epsilon) i}}$ and letting $\epsilon \to 0$. But the 'right' intuition is certainly that of Augusti's answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for $(*)$, the intuition is another story altogether, but I may add it later if it looks worth it...&lt;/p&gt;&#xA;" OwnerUserId="160" LastEditorUserId="160" LastEditorDisplayName="" LastEditDate="2010-09-06T15:37:32.027" LastActivityDate="2010-09-06T15:37:32.027" />
  <row Id="4141" PostTypeId="1" AcceptedAnswerId="4143" CreationDate="2010-09-06T12:17:11.820" Score="2" ViewCount="165" Body="&lt;p&gt;According to the definitions of &lt;a href=&quot;http://en.wikipedia.org/wiki/Algebra_%28ring_theory%29&quot; rel=&quot;nofollow&quot;&gt;algebra over a ring&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/Algebra_over_a_field&quot; rel=&quot;nofollow&quot;&gt;algebra over a field&lt;/a&gt;, are &lt;a href=&quot;http://en.wikipedia.org/wiki/Elementary_algebra&quot; rel=&quot;nofollow&quot;&gt;Elementary Algebra&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/Boolean_algebra_%28logic%29&quot; rel=&quot;nofollow&quot;&gt;Boolean Algebra&lt;/a&gt; examples of algebra over a ring or algebra over a field? How can one show that? What are the ring or field that the algebras are with respect to?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks and regards!&lt;/p&gt;&#xA;" OwnerUserId="1281" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-09-06T15:33:49.940" LastActivityDate="2010-09-06T15:33:49.940" Title="Are Elementary Algebra and Boolean Algebra Algebras over a Ring (or Field)?" Tags="&lt;abstract-algebra&gt;" AnswerCount="2" FavoriteCount="0" />
  <row Id="4142" PostTypeId="2" ParentId="4107" CreationDate="2010-09-06T12:32:31.550" Score="6" ViewCount="" Body="&lt;p&gt;Peano arithmetic is relatively strong, but that standards of arithmetics, and decidable theories of arithmetic are fairly unusual.  Here's a &quot;tower&quot; of three theories between the two, of increasing strength:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Robinson_arithmetic&quot; rel=&quot;nofollow&quot;&gt;Robinson's Q&lt;/a&gt;, which is an induction-free axiomatisation of arithmetic, that gives addition and mutliplication as its only function constant.  It cannot prove exponentiation total;&lt;/li&gt;&#xA;&lt;li&gt;Exponential function arithmetic (EFA) allows exponentiation and a very limited form of induction.  It cannot prove iterated exponetiation (a.k.a. &lt;a href=&quot;http://en.wikipedia.org/wiki/Tetration&quot; rel=&quot;nofollow&quot;&gt;tetration&lt;/a&gt;) total;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Primitive_recursive_arithmetic&quot; rel=&quot;nofollow&quot;&gt;Primitive recursive arithmetic&lt;/a&gt; is sometimes axiomatised as Peano arithmetic but with induction restricted to formulae without universal quantification (it is more usually given as a purely equational theory).  It cannot prove the Ackermann function total.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Note how I've distinguished them by which arithmetic operators thay have, and which they cannot prove total.  This is a very natural measure of  the strength of theories of arithmetic: an essential part of proving the incompleteness theorem is providing a diagonalisation operation, which we can do only if we have multiplication.  These theories have this, and so all of these theories admit the incompleteness theorems.&lt;/p&gt;&#xA;" OwnerUserId="100" LastActivityDate="2010-09-06T12:32:31.550" CommentCount="0" />
  <row Id="4143" PostTypeId="2" ParentId="4141" CreationDate="2010-09-06T14:33:09.603" Score="4" ViewCount="" Body="&lt;p&gt;This should be a comment, I guess...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, a silly remark: every algebra over a field is an algebra over a ring, because fields are rings!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;A&lt;/em&gt; boolean algebra (in the technical sense of the term, more or less explained in the page &lt;a href=&quot;http://en.wikipedia.org/wiki/Boolean_algebra_(structure&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Boolean_algebra_(structure&lt;/a&gt;) ) is an algebra over the field of two elemetns $\mathbb F_2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, &lt;em&gt;Boolean algebra&lt;/em&gt; which is the subject of the Wikipedia page you linked to, is not an algebra in that sense. Likewise, &lt;em&gt;Elementary algebra&lt;/em&gt;, the subject of the other page you linked to, is also not an algebra in the technical sense.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Boolean algebra&lt;/em&gt; and &lt;em&gt;Elementary algebra&lt;/em&gt;, as the terms are used in the Wikipedia pages you linked to, are &lt;em&gt;subjects&lt;/em&gt;, not algebraic structures.&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2010-09-06T14:33:09.603" CommentCount="1" />
  <row Id="4144" PostTypeId="2" ParentId="4141" CreationDate="2010-09-06T14:40:54.620" Score="1" ViewCount="" Body="&lt;p&gt;Disclaimer: I still don't have a great grasp on a lot of abstract algebra, but I'd like to try and help when I can.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: I think Mariano Suárez-Alvarez's comment was more to the point by addressing the issues of structures vs. subjects&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The trouble with the boolean algebra you linked to is that there aren't inverse operations and neutral elements to union and intersection because with intersection and union each element is its own neutral element while the axioms of a field or ring state that there is one neutral element that works for your whole ring or field. Without neutral elements, you can't define inverses because the result of 'multiplying' or 'adding' an element with its inverse element is supposed to be the multiplicatively neutral or additively neutral element. So boolean algebra can't be an algebra over a ring or an algebra over a field since it is neither a ring nor field since there are no neutral elements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand elementary algebra (assuming you are working with either the real or complex numbers) is a ring and also a field. We can make elementary algebra into an algebra over a field by letting the vector space be the field itself and the axioms will be satisfied; however, this is technically adding additional structure to elementary algebra (the vector space) and so its probably no technically considered 'elementary algebra.'&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please let me know if there are any unclear points or errors :)&lt;/p&gt;&#xA;" OwnerUserId="249" LastEditorUserId="249" LastEditorDisplayName="" LastEditDate="2010-09-06T14:50:31.693" LastActivityDate="2010-09-06T14:50:31.693" />
  <row Id="4145" PostTypeId="2" ParentId="4114" CreationDate="2010-09-06T14:50:15.403" Score="2" ViewCount="" Body="&lt;p&gt;It is not that we &lt;em&gt;require&lt;/em&gt; open sets to be closed under finite intersection...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Definitions are not invented out of thin air: people had topological spaces long before anyone thought of coming up with the definition of topological spaces and in the examples they had available, open sets were closed under finite intersection and that property was in fact quite useful to get things done. In other words, the definition of topological spaces (and the requirement you mention in the title along with it) was abstracted from examples, not imposed on them.&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2010-09-06T14:50:15.403" CommentCount="3" />
  <row Id="4146" PostTypeId="1" CreationDate="2010-09-06T15:25:53.557" Score="4" ViewCount="333" Body="&lt;p&gt;Hello everybody,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for somebody having a quite strong background in Mathematics, which are some good books for the domain of Operations research? I guess there are textbooks covering topics like linear and nonlinear optimization, convex optimization and quadratic programming, dynamic programming, multicriterial optimizations (did I miss something?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks,&#xA;Lucian &lt;/p&gt;&#xA;" OwnerUserId="1268" LastActivityDate="2011-02-01T21:13:26.937" CommunityOwnedDate="2010-09-06T15:25:53.557" Title="Operations research book to start with" Tags="&lt;books&gt;&lt;optimization&gt;&lt;convex-analysis&gt;&lt;nonlinear-optimization&gt;" AnswerCount="4" CommentCount="1" FavoriteCount="3" />
  <row Id="4147" PostTypeId="2" ParentId="4054" CreationDate="2010-09-06T15:28:06.283" Score="7" ViewCount="" Body="&lt;p&gt;Cauchy's Formula has a remarkable interpretation in terms of hyperbolic geometry. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To understand it, you need to know very little about hyperbolic geometry. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, you only need to know that if you define the &quot;plane&quot; as the open unit disk $D$ in $\mathbb C$, and the &quot;lines&quot; as those arcs of circle inside $D$ which are orthogonal to the unit circle $\partial D$, you get a model for the hyperbolic plane. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;[There is a first miracle: the orientation preserving isometries of $D$ into itself are precisely the holomorphic automorphisms of $D$.] &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $f$ is holomorphic in a neighborhood of the closure of $D$, and $z$ is in $D$, then Cauchy's Formula says that &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$f(z)$ is the average of $f$ on $\partial D$ &quot;as you see it from $z$&quot;. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;By this I mean that $f(z)$ is the integral of $f$ on $\partial D$ with respect to the measure which assigns to an arc in $\partial D$ the number $\theta/2\pi$ where $\theta\in[0,2\pi]$ is the angle between the (hyperbolic) half-lines going from $z$ to the end points of the arc. (That's the measure you think the arc has, as a part of your horizon, if you look at it from $z$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT 1 OF NOV 1, 2010.&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is to clarify the relationship between the Cauchy's and Poisson's Formulas. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $|z|&amp;lt;1$ and $\theta$ real define the &lt;strong&gt;Cauchy kernel&lt;/strong&gt; by &#xA;$$C(\theta,z):=\frac{1}{2\pi}\ \frac{e^{i\theta}}{e^{i\theta}-z}\quad,$$&#xA;and define the &lt;strong&gt;Cauchy transform&lt;/strong&gt; of a continuous function $g$ on the unit circle $\partial D$ by &#xA;$$(Cg)(z):=\int_0^{2\pi}g(e^{i\theta})\ C(\theta,z)\ d\theta$$ &#xA;for $|z|&amp;lt;1$. In particular Cauchy's Formula says that if $g$ is the boundary value of a &lt;strong&gt;holomorphic&lt;/strong&gt; function $f$ on $D$, then $Cg=f$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $|z|&amp;lt;1$ and $\theta$ real define the &lt;strong&gt;Poisson kernel&lt;/strong&gt; by &#xA;$$P(\theta,z):=C(\theta,z)+\overline{C(\theta,z)}-\frac{1}{2\pi}\quad,$$&#xA;and define the &lt;strong&gt;Poisson transform&lt;/strong&gt; of a continuous function $g$ on the unit circle $\partial D$ by &#xA;$$(Pg)(z):=\int_0^{2\pi}g(e^{i\theta})\ P(\theta,z)\ d\theta$$ &#xA;for $|z|&amp;lt;1$. In particular Poisson's Formula says that if $g$ is the boundary value of a &lt;strong&gt;harmonic&lt;/strong&gt; function $f$ on $D$, then $Pg=f$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT 2 OF NOV 1, 2010.&lt;/strong&gt; I stole this from Bill Thurston: go to page 180 of (or search for &quot;visual&quot; in) &lt;a href=&quot;http://www.math.unl.edu/~mbrittenham2/classwk/990s08/public/thurston.notes.pdf/8a.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.math.unl.edu/~mbrittenham2/classwk/990s08/public/thurston.notes.pdf/8a.pdf&lt;/a&gt; . &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT OF NOV 2, 2010.&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Stricto sensu&lt;/em&gt;, there is no geometric interpretation of the Cauchy kernel, because it is not invariant. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Indeed, if $G$ denotes the group of biholomorphic transforms of the open unit disk $D$, then the &lt;strong&gt;Cauchy kernel&lt;/strong&gt; $C(z,\theta)d\theta$, viewed as a 1-form on $D\times\partial D$, is &lt;strong&gt;not&lt;/strong&gt; $G$-invariant. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, its restriction to $\{0\}\times\partial D$ &lt;strong&gt;is&lt;/strong&gt; invariant under the stabilizer of $0\in D$ in $G$ (which is the circle group). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a result, this restriction extends in a unique way to a $G$-invariant 1-form on $D\times\partial D$, and this 1-form is the &lt;strong&gt;Poisson kernel&lt;/strong&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;More precisely, the (complex) vector space of $G$-invariant $(0,1)$-forms on $D\times\partial D$ is one dimensional, generated by the Poisson kernel. [This is because the action of $G$ on $D\times\partial D$ is simply transitive, and the $(0,1)$-forms on $D\times\partial D$ are the sections of a homogeneous line bundle over $D\times\partial D$.] &lt;/p&gt;&#xA;" OwnerUserId="660" LastEditorUserId="660" LastEditorDisplayName="" LastEditDate="2010-11-02T05:45:36.757" LastActivityDate="2010-11-02T05:45:36.757" CommentCount="9" />
  <row Id="4148" PostTypeId="1" AcceptedAnswerId="4149" CreationDate="2010-09-06T15:36:59.347" Score="5" ViewCount="176" Body="&lt;p&gt;I have a set containing numbers $1$ to $49$ $\{1,2,3, \cdots, 49\}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, I want to divide the set into $7$ subsets such that each subset should contain $7$ elements and sum of the elements of each subset should be $175$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it possible to prove that such subsets exist?&lt;/p&gt;&#xA;" OwnerUserId="1353" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-09-06T15:54:19.063" LastActivityDate="2010-09-06T15:54:19.063" Title="finding seven disjoint seven element subsets of {1,2, ..., 49} with same sum" Tags="&lt;combinatorics&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="4149" PostTypeId="2" ParentId="4148" CreationDate="2010-09-06T15:45:47.597" Score="10" ViewCount="" Body="&lt;p&gt;Lookup &lt;a href=&quot;http://en.wikipedia.org/wiki/Magic_square&quot; rel=&quot;nofollow&quot;&gt;Magic Squares&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which has the following:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;22  47  16  41  10  35  4&#xA;5    23  48  17  42  11  29&#xA;30  6   24  49  18  36  12&#xA;13  31  7   25  43  19  37&#xA;38  14  32  1   26  44  20&#xA;21  39  8   33  2   27  45&#xA;46  15  40  9   34  3   28&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-09-06T15:45:47.597" />
  <row Id="4150" PostTypeId="2" ParentId="3888" CreationDate="2010-09-06T16:19:07.153" Score="8" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\:$ Note the equation is $\displaystyle \rm\; \frac{\sigma\:f}f \:=\; \frac{\sigma^4 g}g\ $ for $\rm\ g = x-16 \:,\:\;$ shift  $\rm\;\;\sigma\:f(x) = f(\sigma\:x) = f(2\:x)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or, written additively $\rm\  (\sigma-1) \; f \;=\: (\sigma^4 - 1) \; g \quad\;$ [see Note below on this additive notation] &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore  $\rm\;\; f \:=\: (1+\sigma+\sigma^2+\:\sigma^3) \: g \;=\: (x-16)\:(2x-16)\:(4x-16)\:(8x-16) \;$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;unique upto a factor of $\rm\;h \in ker(\sigma-1) = \{h: \sigma\:h = h\:\} = \;$  constants, i.e. $\rm deg\;h = 0 \quad$ &lt;strong&gt;QED&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;REMARK&lt;/strong&gt; $\;$ This reformulation of my prior answer is intended to dramatically illustrate the innate symmetry. Its striking simplicity arises precisely from the simple structure of the orbits of the shift automorphism $\rm\: \sigma\;$.  Namely, by orbit decomposition the problem reduces to one on the single orbit of an irreducible polynomial. By exploiting polynomial structure there, we reduce the problem to a trivial polynomial division by $\rm\:\sigma - 1\:$. This nicely illustrates the essence of the ideas employed to solve general difference equations (recurrences) over rational function fields - ideas at the foundation of algorithms employed in computer algebra systems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hopefully the symmetry is clearer in the above reformulation. Based on prior comments, and someone reposting my prior solution stripped of the symmetry, I fear my prior answer didn't succeed in explicitly emphasizing the beautiful innate symmetry at the heart of this problem  - a germ of the Galois theory of difference fields. Probably the problem was devised to help spur one to discover this beautiful structure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;NOTE&lt;/strong&gt;$\;\;$ The point of the additive notation is to exploit the natural polynomial structure of the action of $\rm\:\sigma\:$ on the multiplicative group generated by the elements $\rm\: \sigma^n \:f\;$ in the orbit of $\rm\: f \;$ under $\rm\:\sigma\:$. This action is best comprehended by examining it on a specific example.$\:$ Namely, recalling that $\rm\;\sigma\;f(x) \:=\; f(2\:x)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\quad\quad\begin{align}{}&#xA;\rm \sigma\:(\:f(2^{-2}\: x)^a \; f(2^3 x)^b \;\: f(2^5 x)^c)\; =&amp;amp; \rm\;\;\: f(\:2^{-1} x)^a \;\: f(2^4 x)^b \;\: f(2^6 x)^c \\&#xA;\rm \iff \quad\;\;\: \sigma\;\:(\:(\:\sigma^{-2}\: f\;)^a \;\; (\:\sigma^3\: f\:)^b \;\; (\:\sigma^5\: f\:)^c)\; =&amp;amp; \rm\;\; \;\; (\:\sigma^{-1} \: f\:)^a \;\; (\:\sigma^4 \: f\:)^b \;\; (\:\sigma^6 \: f\;)^c \\&#xA;\rm \iff \quad\;\; \sigma\; (\:a\:\sigma^{-2} \;+\:\;\; b\:\sigma^3 \;+\;\; c\:\sigma^5)\:\; f\quad\; =&amp;amp; \rm\; (\:a\:\sigma^{-1} \: + \:\;\: b\: \sigma^4 \; +\;\;  c\: \sigma^6)\; f \\&#xA;\end{align}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the Galois theory literature the above action is frequently written in a highly suggestive exponential form. To illustrate this, below is the key identity in th problem at hand, expressed in this exponential notation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\quad\quad\begin{align}{}&#xA;\rm g^{\:\sigma^4-1} \;=\;&amp;amp; \rm g^{\:(1\:+\;\sigma\:+\;\sigma^2\:+\;\:\sigma^3)\:(\sigma-1)} \\&#xA;\iff\quad\quad\rm  \frac{\sigma^4 g}g \;=\;&amp;amp; \rm (g \;\: \sigma\:g \;\:\sigma^2 g \;\:\sigma^3 g)^{\sigma - 1} \;=\; \frac{\phantom{g\;\;\:} \sigma\:g \;\;\: \sigma^2 g \;\;\:\sigma^3 g \;\;\:\sigma^4 g}{g \;\;\:\sigma\:g \;\;\:\sigma^2 g \;\;\:\sigma^3 g\phantom{\;\;\:\sigma^4 g}} \\&#xA;\end{align}$&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-09-07T22:03:55.617" LastActivityDate="2010-09-07T22:03:55.617" CommentCount="6" />
  <row Id="4151" PostTypeId="2" ParentId="4120" CreationDate="2010-09-06T16:23:05.307" Score="1" ViewCount="" Body="&lt;p&gt;I don't think there's a general definition of tractability in the sense you probably mean, other than &quot;unsolvable under the given conditions&quot;; however &quot;given conditions&quot; are very important.&#xA;In practice conditions on a theorem aren't set beforehand but typically arise from working on the problem - you want to show X, but find out that X doesn't hold unless you also make assumption Y, and so then you ask yourself if Y is a reasonable thing to assume in the problem setting you are interested in. In the end you (hopefully) prove a theorem that has as few assumptions as you can get away with, and still is useful for the problem you are interested in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, there is a clear distinction between the two situations &quot;can I very likely get a practically satisfactory solution to (some impossible problem) if I add some extra constraints by applying domain knowledge&quot; and &quot;can I solve a problem that can be proved to be impossible in general&quot;. In your previous post, the problem you gave was impossible in general, but a good enough solution was tractable under mild assumptions. So, tractability is a function of how you choose to work on a problem, and specifically of which conditions or constraints you assume to hold/are happy to accept in the first place.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can recommend Imre Lakatos's &quot;Proofs and Refutations&quot;, which makes this point better than I could ever hope to and, IMHO, is one of those books that every intelligent person should read at sometime in their life. &lt;/p&gt;&#xA;" OwnerUserId="1676" LastActivityDate="2010-09-06T16:23:05.307" CommentCount="2" />
  <row Id="4152" PostTypeId="1" AcceptedAnswerId="4153" CreationDate="2010-09-06T17:58:27.650" Score="2" ViewCount="123" Body="&lt;p&gt;How can we show that the number of pairs (where $(a, b)$ is the same as $(b, a)$) with $LCM(a, b) = n$ is equal to:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{(2e_1+1)(2e_2+1)...(2e_k+1)+1}{2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$n=p_1^{e_1}\cdot p_2^{e_2}\cdot ...\cdot p_k^{e_k}, p_i\space prime\space \forall 1\leq i\leq k$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I managed to guess this formula for a programming problem I was working on, but I'd be interested in a deduction or proof.&lt;/p&gt;&#xA;" OwnerUserId="456" LastActivityDate="2010-09-06T18:10:17.293" Title="Pairs of numbers with a given LCM" Tags="&lt;number-theory&gt;&lt;combinatorics&gt;&lt;counting&gt;" AnswerCount="1" />
  <row Id="4153" PostTypeId="2" ParentId="4152" CreationDate="2010-09-06T18:10:17.293" Score="4" ViewCount="" Body="&lt;p&gt;First count the pairs with $\gcd(a,b)=n$ counting $(a,b)$ and $(b,a)$ as&#xA;distinct. You have $a=\prod p_i^{r_i}$ and $b=\prod p_i^{s_i}$&#xA;and need $\max(r_i,s_i)=e_i$ for all $i$. So, given a positive integer $e$,&#xA;how many pairs $(r,s)$ of nonnegative integers are there with $\max(r,s)=e$?&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-09-06T18:10:17.293" CommentCount="2" />
  <row Id="4154" PostTypeId="2" ParentId="3852" CreationDate="2010-09-06T18:10:36.187" Score="1" ViewCount="" Body="&lt;p&gt;here is a try. we will use  AB = I to show that the columns of $B$ are linearly independent and then use that to show $BA$ is identity on the range of $B$ which is all of the space due to linear independence of the columns of $B$. This implies that $BA$ is identity. linear independence of the columns of  follows if we can show Bx = 0 implies x = 0. assume $Bx = 0, x = Ix = ABx = A0 = 0$. now to show that  $BA$ is an identity on range of  $B$ we have $(BA)Bx = B(AB)x = Bx $ and and we are done.&lt;/p&gt;&#xA;" LastActivityDate="2010-09-07T12:09:01.310" />
  <row Id="4155" PostTypeId="2" ParentId="3438" CreationDate="2010-09-06T18:27:17.693" Score="0" ViewCount="" Body="&lt;p&gt;I received this solution via email. I am TeXing it out here.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;We can write $n!= 2^{\alpha(n)}\beta(n)$, where $\beta(n)$ is odd. From a theorem of Legendre we have $\alpha(n)=n -\gamma(n)$ where $\gamma(n)$ denotes the sum of ones in the binary representation of $n$. Moreover, have $$ \sum\limits_{k=1}^{\infty} \frac{2^{n_k}}{n_{k}!} = \sum\limits_{n=1}^{\infty} \delta_{n} \frac{2^n}{n!}$$ where $\delta_{n}=1$ if $n=n_{k}$ and $\delta_{n}=0$, otherwise. Suppose, contrary to our claim, if $$\sum\limits_{n=1}^{\infty} \delta_{n} \frac{2^n}{n!}= \frac{p}{q}$$ Write $q=2^{s}t$, where $t$ is odd. Take $N=2^{r} &gt; \max \{t,2^{s+2}\}$, it then follows that $\frac{\beta(n)}{n} \in \mathbb{N}$. Therefore $2^{s}\beta(N)\frac{p}{q}=\frac{\beta(N)}{t} p \in \mathbb{N}$. A simple calculation shows that $N!=2^{N-1}\beta(N)$. Multiplying the equality $$ \frac{p}{q} = \sum\limits_{n=1}^{N} \delta_{n} \frac{2^n}{n!} + \sum\limits_{n=N+1}^{\infty} \delta_{n} \frac{2^n}{n!}$$ by $2^{s}\beta(N)$, we get $$ 2^{s}\beta(N)\frac{p}{q} = 2^{s}\beta(N) \sum\limits_{n=1}^{N} \delta_{n} \frac{2^n}{n!} + 2^{s}\beta(N) \sum\limits_{n=N+1}^{\infty} \delta_{n} \frac{2^n}{n!} \quad (\star) $$ Note that $$ 2^{s} \beta(N) \sum\limits_{n=1}^{N} \delta_{n} \frac{2^n}{n!} = 2^{s} \sum\limits_{n=}^{N} \delta_{n} \frac{\beta(N)2^{n}}{2^{\alpha(n)}\beta(n)}$$ Since $\beta(n)$ divides $\beta(N)$ we see that the first term on the right hand side of $(\star)$ is an integer. To obtain contradiction we will show that, $$0 &amp;lt; 2^{s}\beta(N) \sum\limits_{n=N+1}^{\infty} \delta_{n} \frac{2^{n}}{n!} &amp;lt;1$$ Indeed, we have $$2^{s}\beta(N) \sum\limits_{N+1}^{\infty} \delta_{n} \frac{2^n}{n!} = 2^{s-N+1}N! \sum\limits_{n=N+1}^{\infty} \frac{2^n}{n!}$$ $$= 2^{s+2} \sum\limits_{n=N+1}^{\infty} \delta_{n}2^{n-N-1} \frac{N!}{n!} &amp;lt; \frac{2^{s+2}}{N+1} \sum\limits_{n=N+1}^{\infty} \Bigl(\frac{2}{N+2}\Bigr)^{n-N-1}$$ $$=\frac{2^{s+2}}{N+1} \frac{N+2}{N} &amp;lt; \frac{2^{s+3}}{N+1} &amp;lt; 1$$&lt;/p&gt;&#xA;" LastActivityDate="2010-09-06T18:27:17.693" />
  <row Id="4156" PostTypeId="2" ParentId="4146" CreationDate="2010-09-06T19:13:08.237" Score="2" ViewCount="" Body="&lt;p&gt;Stephen Boyd and Lieven Vandenberghe's book is popular, and available free online:&#xA;&lt;a href=&quot;http://www.stanford.edu/~boyd/cvxbook/&quot; rel=&quot;nofollow&quot;&gt;http://www.stanford.edu/~boyd/cvxbook/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1676" LastActivityDate="2010-09-06T19:13:08.237" CommunityOwnedDate="2010-09-06T19:13:08.237" />
  <row Id="4157" PostTypeId="2" ParentId="4107" CreationDate="2010-09-06T19:27:59.217" Score="6" ViewCount="" Body="&lt;p&gt;There are infinitely many different theories stronger than Presburger arithmetic and weaker than Peano arithmetic. The classic examples of theories weaker than Peano arithmetic are determined by adding different amounts of induction to Robinson arithmetic.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;The $\Sigma^0_n$ induction axioms say that any set of numbers definable by a $\Sigma^0_n$ formula that contains 0 and is closed under successor contains every natural number.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The $\Sigma^0_n$ bounding axioms say that any there is no $\Sigma^0_n$ definable function on a bounded initial segment of the natural numbers with an unbounded range.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The theory Q of &quot;Robinson arithmetic&quot; consists of a small number of axioms about the arithmetical operations. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The theory I$\Sigma^0_n$ is Q plus the $\Sigma^0_n$ induction axioms.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The theory B$\Sigma^0_n$ is Q plus the $\Sigma^0_n$ bounding axioms.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;It is known that, for each $n$, I$\Sigma^0_{n+1}$ implies B$\Sigma^0_{n+1}$, which in turn implies I$\Sigma^0_n$, and the implications are all strict.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;None of these theories is decidable, nor is any other consistent extension of Robinson arithmetic. The first incompleteness theorem applies to all these theories. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The standard proof of the second incompleteness theorem applies to all the theories I$\Sigma^0_1$ and stronger.  The second incompleteness theorem can be proved for some weaker systems as well, but things get very technical for systems weaker than I$\Sigma^0_1$. &lt;/p&gt;&#xA;" OwnerUserId="630" LastActivityDate="2010-09-06T19:27:59.217" CommentCount="7" />
  <row Id="4158" PostTypeId="1" CreationDate="2010-09-06T19:30:50.797" Score="2" ViewCount="91" Body="&lt;p&gt;&lt;sup&gt;(This question is inspired by the &lt;a href=&quot;http://stackoverflow.com/questions/3653475/converting-prime-numbers&quot;&gt;StackOverflow question &lt;em&gt;Converting prime numbers&lt;/em&gt;&lt;/a&gt; by &lt;a href=&quot;http://stackoverflow.com/users/158163/manas&quot;&gt;@Manas&lt;/a&gt;, which turns out to be the &lt;a href=&quot;http://www.spoj.pl/problems/PPATH/&quot; rel=&quot;nofollow&quot;&gt;problem &lt;code&gt;PPATH&lt;/code&gt; of SPOJ&lt;/a&gt;.)&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's construct a graph &lt;em&gt;G&lt;/em&gt; of all primes &lt;em&gt;p&lt;/em&gt; having &lt;em&gt;L&lt;/em&gt; digits in base-&lt;em&gt;n&lt;/em&gt;. Two primes &lt;em&gt;p&lt;/em&gt; and &lt;em&gt;q&lt;/em&gt; form an edge in &lt;em&gt;G&lt;/em&gt; iff they only differ in 1 digit (e.g. 1&lt;b&gt;0&lt;/b&gt;09 and 1&lt;b&gt;1&lt;/b&gt;09 form an edge).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there some (approximated?) condition on &lt;em&gt;n&lt;/em&gt; and &lt;em&gt;L&lt;/em&gt; that &lt;em&gt;G&lt;/em&gt; is connected? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, it &lt;em&gt;seems&lt;/em&gt; that when &lt;em&gt;n&lt;/em&gt; is odd &lt;em&gt;G&lt;/em&gt; is usually disconnected, while when &lt;em&gt;n&lt;/em&gt; is even (&gt;2) &lt;em&gt;G&lt;/em&gt; is usually connected.&lt;/p&gt;&#xA;" OwnerUserId="171" LastActivityDate="2010-09-07T20:11:35.073" Title="Condition for Connectedness of Graph of Prime Numbers Differing only 1 Digit." Tags="&lt;number-theory&gt;&lt;prime-numbers&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="4160" PostTypeId="2" ParentId="4146" CreationDate="2010-09-06T19:54:16.820" Score="2" ViewCount="" Body="&lt;p&gt;&lt;em&gt;&lt;a href=&quot;http://www.math.ucla.edu/~tom/LP.pdf&quot; rel=&quot;nofollow&quot;&gt;Linear Programming&lt;/em&gt; -- &lt;em&gt;A Concise Introduction&lt;/a&gt;&lt;/em&gt; by Thomas S. Ferguson and other ebooks/lecture notes on &lt;em&gt;Optimization&lt;/em&gt; listed in &lt;a href=&quot;http://stochastix.wordpress.com/books/&quot; rel=&quot;nofollow&quot;&gt;Rod Carvalho's web notebook&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Addendum:  The classic and complete book by Hillier and Lieberman &lt;em&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0073017795&quot; rel=&quot;nofollow&quot;&gt;Introduction to Operations Research&lt;/a&gt;&lt;/em&gt; &lt;/p&gt;&#xA;" OwnerUserId="752" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2010-09-06T21:45:45.530" LastActivityDate="2010-09-06T21:45:45.530" CommunityOwnedDate="2010-09-06T19:54:16.820" />
  <row Id="4161" PostTypeId="2" ParentId="29" CreationDate="2010-09-06T20:49:31.687" Score="1" ViewCount="" Body="&lt;p&gt;In case you happen to be an Emacs user, the &lt;a href=&quot;http://www.gnu.org/software/emacs/calc.html&quot; rel=&quot;nofollow&quot;&gt;Emacs Calc&lt;/a&gt; package is a very good programmable stack-based calculator/mini-CAS.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(&quot;M-x calc&quot; to start, then &quot;i&quot; to open the manual in info format.)&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2010-09-06T20:49:31.687" CommunityOwnedDate="2010-09-06T20:49:31.687" />
  <row Id="4162" PostTypeId="1" AcceptedAnswerId="4164" CreationDate="2010-09-06T20:53:05.990" Score="5" ViewCount="125" Body="&lt;p&gt;This question isn't about the actual substance of math so much as it is about style.  I'm not going to school, so my feedback on these sorts of problems is limited.  How would you improve this proof's readability?  And if this isn't the sort of question that's suitable here, where would be a more appropriate venue?  The problem is pretty straightforward (from little Spivak):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $f:\mathbb{R}^2\rightarrow \mathbb{R}$ and $D_2f=0$, show that $f$ is independent of the second variable.  If $D_1f=D_2f=0$, show that $f$ is constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The definition of &quot;independent of the second variable&quot; that we are using is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A function $f:\mathbb{R}^2 \rightarrow \mathbb{R}$ is independent of the second variable if for each $x\in \mathbb{R}$ we have $f(x,y_1)=f(x,y_2)$ for all $y_1,y_2\in\mathbb{R}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My solution as I have written it is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume $f:\mathbb{R}^2\rightarrow \mathbb{R}$ and $D_2f=0$.  Let $x$ be arbitrary.  By integration, $\int f_2(x,y)=\int 0=C$ for some constant $C$.  It follows that $f(x,y_1)=f(x,y_2)=C$ for all $y_1,y_2\in\mathbb{R}$.  Since $x$ was arbitrary, it follows from the definition that $f$ is independent of the second variable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To show that $f$ is constant when $D_1f=D_2f=0$ we proceed by contradiction.  Assume that $f$ is not constant.  Then there exists some $x_1,y_1,x_2,y_2$ such that $f(x_1,y_1) \ne f(x_2,y_2)$.  Since $f$ is independent of the second variable, it follows that $f(x_1,y_1)=f(x_1,y_2)$ and $f(x_1,y_2) \ne f(x_2,y_2)$.  By the mean value theorem there exists $x\in(x_1,x_2)$ such that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$Df_1(x,y_2)=\displaystyle\frac{f(x_2,y_2)-f(x_1,y_2)}{x_2-x_1} \ne 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is the desired contradiction.&lt;/p&gt;&#xA;" OwnerUserId="326" LastActivityDate="2010-09-06T21:57:34.613" Title="How can I write this proof about partial derivatives better?" Tags="&lt;calculus&gt;&lt;real-analysis&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="4163" PostTypeId="1" AcceptedAnswerId="4166" CreationDate="2010-09-06T21:08:02.197" Score="0" ViewCount="78" Body="&lt;p&gt;If F and G are functors between two arbitrary categories C and D, does a &lt;a href=&quot;http://en.wikipedia.org/wiki/Natural_transformation&quot; rel=&quot;nofollow&quot;&gt;natural transformation&lt;/a&gt; η from F to G always exists? What is the condition for its existence?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks and regards!&lt;/p&gt;&#xA;" OwnerUserId="1281" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-09-06T21:21:40.160" LastActivityDate="2010-09-06T21:21:40.160" Title="Existence of Natural Transformation between Functors" Tags="&lt;category-theory&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="4164" PostTypeId="2" ParentId="4162" CreationDate="2010-09-06T21:16:51.407" Score="4" ViewCount="" Body="&lt;p&gt;I've marked this as &quot;community wiki&quot; for myself...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your argument is a bit vague in the first paragraph; you have an indefinite integral, but what you actually want is the function; that is, you want $f(x,y_1)=C$. &lt;strong&gt;Edit:&lt;/strong&gt; The problem with this is that, if you recall, the indefinite integral does not produce a function, but produces a family of functions, namely all the antiderivatives of your function, and not a value of a particular function. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It would be better to use the Fundamental Theorem of Calculus, which you can use to get specific values of a function (or to be more precise, the difference between two values of a function), and write something along these lines:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Let $f\colon\mathbb{R}^2\to\mathbb{R}$ be a function such that $D_2f=0$. Fix an arbitrary $x$; then for any $y$ we have:&#xA;  $$f(x,y)-f(x,0) = \int_0^y D_2f(x,t)dt = \int_0^y 0dt = 0,$$&#xA;  so for  fixed $x$, $f(x,y)=f(x,0)$ for all $y$. Therefore, $f$ is independent of the second variable.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You may, of course, replace $0$ with another point above. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the second part, why not use the first part? From $D_2f=0$ you know $f$ is independent of the second variable; the same argument, by exchanging $x$ and $y$, shows that from $D_1f=0$ you get that $f$ is independent of the first variable. So you have&#xA;$$f(x_1,y_1) = f(x_1,y_2) = f(x_2,y_2)$$&#xA;for any $(x_1,y_1)$ and $(x_2,y_2)$; first equality because $f$ is independent of the second variable, second because it is independent of the first. Then you don't have to invoke something like the Mean Value Theorem. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But in any case, let me point out that you seem to be assuming that $(x_1,y_1)\neq(x_2,y_2)$ implies both that $x_1\neq x_2$ and $y_1\neq y_2$, but this is not true.  You do &lt;em&gt;not&lt;/em&gt;  know a priori that $x_2\neq x_1$; all you know is that $(x_1,y_1)\neq(x_2,y_2)$, and therefore &lt;em&gt;either&lt;/em&gt; $x_1\neq x_2$ or $y_1\neq y_2$ (or both). Your argument breaks down if $x_1=x_2$, and you need to consider that case separately. I would also add at some point, in the case $x_1\neq x_2$, that &quot;we may assume without loss of generality that $x_1\lt x_2$&quot;, so that &quot;$x\in(x_1,x_2)$&quot; makes sense.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-09-06T21:57:34.613" LastActivityDate="2010-09-06T21:57:34.613" CommunityOwnedDate="2010-09-06T21:16:51.407" CommentCount="2" />
  <row Id="4165" PostTypeId="2" ParentId="4146" CreationDate="2010-09-06T21:19:56.967" Score="2" ViewCount="" Body="&lt;p&gt;If you want books that give a broad introduction to operations research (including optimization, queuing theory etc.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The classics are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Operations Research - Ronald Rardin&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Operations Research - Wayne Winston&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They are excellent from pedagogical point of view. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to get into linear programming, this book is widely regarded to be one of the best:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Linear Programming - Vasek Chvatal.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For nonlinear programming:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Nonlinear programming - Dimitri Bertsekas&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="1124" LastActivityDate="2010-09-06T21:19:56.967" CommunityOwnedDate="2010-09-06T21:19:56.967" />
  <row Id="4166" PostTypeId="2" ParentId="4163" CreationDate="2010-09-06T21:21:11.397" Score="3" ViewCount="" Body="&lt;p&gt;For a natural transformation $\eta$ to exist between $F$ and $G$, you need for each object $C$ of C a morphism in $D$ $\eta(C)\colon F(C)\to G(C)$. So for an easy example in which no natural transformation exists, take D to be a category with two objects, $A$ and $B$, and in which the only arrows are $1_A\colon A\to A$ and $1_B\colon B\to B$ (the two identity arrows). Take your favorite category C with at least one object, and let $F$ be the functor that maps every object of C to $A$ and every arrow of C to $1_A$, and take $G$ to be the functor that maps every object of C to $B$ and every arrow of C to $1_B$. Then there can be no natural transformation form $F$ to $G$, since there are no morphisms from $F(C)$ to $G(C)$ for any $C$ in C.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-09-06T21:21:11.397" />
  <row Id="4167" PostTypeId="2" ParentId="4163" CreationDate="2010-09-06T21:21:35.730" Score="1" ViewCount="" Body="&lt;p&gt;have you seen this:  &lt;a href=&quot;http://mathoverflow.net/questions/35731/conditions-for-natural-transformations-to-exist&quot; rel=&quot;nofollow&quot;&gt;http://mathoverflow.net/questions/35731/conditions-for-natural-transformations-to-exist&lt;/a&gt;&lt;/p&gt;&#xA;" LastActivityDate="2010-09-06T21:21:35.730" />
  <row Id="4168" PostTypeId="2" ParentId="3750" CreationDate="2010-09-06T21:41:55.050" Score="3" ViewCount="" Body="&lt;p&gt;I will give the calculus-based solution myself just for the sake of argument, taking note that I am still hoping to obtain a wide variety of other solutions, if possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For this, I am going to mentally rotate the tank (or the oil within it) $90^\circ$ clockwise, cut it in half, and center it at $(0,0)$, so that the upper half of the tank is represented by $y=\sqrt{25-x^2}$ and the volume of the oil by $\int_{-5}^h \sqrt{25-x^2}\mathrm{d}x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know from simple $A=\pi r^2$ that the total cross-section area of the tank is $25\pi$ and thus for the upper half from -5 to +5 is $12.5\pi$.  I will set the result of the integral to the various 10% proportions of this value, knowing that 10% of the upper half will occur at the same position as 10% of the entire circle etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The integral is $\frac12\left(x\sqrt{25-x^2} + 25\arcsin\left(\frac{x}{5}\right)\right)$ evaluated from -5 to h, so the equation we need to solve is:&#xA; $\frac12\left(h\sqrt{25-h^2}+25\arcsin(\frac{h}{5})\right)+\frac{25\pi}{4}=12.5\pi P$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Substituting values of .1, .2, .3, .4, and .5 successively for $P$ and using various tools to estimate $h$, my results are -3.43424, -2.45931, -1.59846, -0,788681, 0.  Adding 5 to account for the central displacement, rounding off, then reflecting these values for the right (upper) values, confirms the values given earlier by Americo.&lt;/p&gt;&#xA;" OwnerUserId="575" LastEditorUserId="575" LastEditorDisplayName="" LastEditDate="2010-09-07T11:44:57.350" LastActivityDate="2010-09-07T11:44:57.350" CommentCount="2" />
  <row Id="4169" PostTypeId="1" AcceptedAnswerId="4172" CreationDate="2010-09-06T21:50:41.957" Score="0" ViewCount="315" Body="&lt;p&gt;Let $f$ be continuous on $\mathbb{R}$. Then how to find all continuous functions satisfying $f(f(x))=f(x)+x$&lt;/p&gt;&#xA;" LastActivityDate="2010-10-22T02:47:52.827" Title="Solving the functional Equation $f(f(x))=f(x)+x$" Tags="&lt;real-analysis&gt;&lt;analysis&gt;&lt;functional-equations&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="1" />
  <row Id="4170" PostTypeId="1" CreationDate="2010-09-06T22:06:31.337" Score="5" ViewCount="570" Body="&lt;p&gt;I just started to learn mathematical logic. I'm a graduate student. I need a book with relatively more examples. Any recommendation?&lt;/p&gt;&#xA;" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-26T01:13:12.783" LastActivityDate="2011-02-14T19:01:34.110" CommunityOwnedDate="2010-09-06T22:55:47.140" Title="Good books on mathematical logic?" Tags="&lt;logic&gt;&lt;big-list&gt;&lt;books&gt;" AnswerCount="8" CommentCount="2" FavoriteCount="4" />
  <row Id="4171" PostTypeId="2" ParentId="4102" CreationDate="2010-09-06T22:07:12.740" Score="1" ViewCount="" Body="&lt;p&gt;this response addresses part of grautur's question: how to think about the property of being &quot;sufficient&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ra fisher suggested the terminology &quot;sufficient&quot; for a statistic $t$ that satisfies the [heuristic] requirement $$\kern{-2.5in}(1)\kern{2.5in} f(x|\theta, t) = f(x|t).$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;[there is nothing heuristic about this way of writing the requirement if $x$ is discrete. but if $x$ is continuous, the &lt;em&gt;pdf&lt;/em&gt;  notation $f(x|t)$ is usually heuristic. an example of the latter is when $x = (x_1,\dots x_n)$, where the $\{x_i\}$ are $iid$ N($\theta$,1). here the sample mean $\bar x$ is a sufficient statistic and $(1)$ requires considering the conditional &lt;em&gt;pdf&lt;/em&gt; $$\kern{-.5in} (2)\kern{.5in} f(x_1,\dots, x_n|\bar x) \equiv f(x_1 - \bar x,\dots, x_n - \bar x | \bar x) = f(x_1 - \bar x,\dots, x_n - \bar x).$$ the last equality in $(2)$ holds since the sample deviations are independent of the sample mean in the normal case. (this fact gives an instant proof that the sample mean and variance are independent in the normal case.) here the &lt;em&gt;&quot;pdf&quot;&lt;/em&gt; on the RHS of (2) does not exist as an $n-$dimensional &lt;em&gt;pdf&lt;/em&gt; since the joint distribution of the sample deviations is singular - they sum to zero. replacing &quot;$f$&quot; in (1) by &quot;dist&quot;, removes its heuristic nature.]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;at any rate, the idea of (1) is that, as the conditional distribution of $x|t$ does not depend on $\theta$, one can [in principle] generate a new $x$ - call it $x^*$, say, from the known conditional distribution of $x|t$, so that $x^* \sim x$ for all $\theta$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;as an illustration, consider the normal example above. here, obtaining an $x^*$  is particularly easy since the deviations are independent of $\bar x$. thus one can generate $n$ &lt;em&gt;iid&lt;/em&gt; N(0,1) variables $z_1,\dots, z_n$ and let $x^* = (x_1^*,\dots. x_n^*)$, where $x_i^* = z_i - \bar z + \bar x : 1\le i\le n$.&#xA;clearly $x \sim x^*$ for all values of $\theta$, so that $x^*$ is just as &quot;good&quot; a sample from the population as the original $x$ for learning about $\theta$. clearly no one would actually want to use the sample deviations for $x^*$ in addition to its sufficient statistic $\bar x^* = \bar x$, as those deviations, the $\{z_i - \bar z\}$, were obtained by a completely extraneous random experiment having nothing to do with the actual data process. one then should agree that the original sample deviations $\{x_i - \bar x\}$ should also not be used, since they can be considered as being generated in the same way, by an extraneous random experiment, where now nature, rather than the statistician, did the generating.&lt;/p&gt;&#xA;" OwnerUserId="1547" LastActivityDate="2010-09-06T22:07:12.740" />
  <row Id="4172" PostTypeId="2" ParentId="4169" CreationDate="2010-09-06T23:39:25.487" Score="7" ViewCount="" Body="&lt;p&gt;This one is a problem from a journal or from competitions at the level of the Putnam contest (see reference below).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint:  $g(x) = x + Af(x)$ satisfies $g(f^n(x))=A^ng(x)$ when $A^2 = A + 1$; consider the cases $n \to \pm \infty$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Source for a similar problem, with solution:  &lt;a href=&quot;http://books.google.com/books?id=-CNbGp2ZFXUC&amp;amp;pg=PA21&quot; rel=&quot;nofollow&quot;&gt;http://books.google.com/books?id=-CNbGp2ZFXUC&amp;amp;pg=PA21&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="467" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-10-22T02:47:52.827" LastActivityDate="2010-10-22T02:47:52.827" />
  <row Id="4173" PostTypeId="1" AcceptedAnswerId="4177" CreationDate="2010-09-06T23:55:24.823" Score="3" ViewCount="157" Body="&lt;p&gt;I asked this question on mathoverflow, but it was deemed too simple, so I'm posting here instead -- &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a nice way to characterize an orthonormal basis of eigenvectors of the following $d\times d$ matrix?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathbf{I}-\frac{1}{d} \mathbf{v}\mathbf{v}'$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $\mathbf{v}$ is a $d\times 1$ vector of 1's. This is similar to the &lt;a href=&quot;http://en.wikipedia.org/wiki/Householder_transformation&quot; rel=&quot;nofollow&quot;&gt;Householder matrix&lt;/a&gt;, except the $v's$ are not normalized. One eigenvector is $\mathbf{v}$ with corresponding eigenvalue 0, remaining eigenvalues should be 1. I'm looking for an expression in terms of unknown d.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Motivation: this is covariance matrix of uniform multinomial distribution, so expression for orthonormal basis produces a linear transformation that will make variables uncorrelated for large n&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example: below are 5 orthonormal eigenvectors vectors I get from Gram-Schmidt for d=5...what is the expression for general d? An even bigger example -- columns of &lt;a href=&quot;http://yaroslavvb.com/upload/bigmatrix.png&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; form orthonormal basis for d=20&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$-\frac{1}{\sqrt{2}},0,0,0,\frac{1}{\sqrt{2}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$-\frac{1}{\sqrt{6}},0,0,\sqrt{\frac{2}{3}},-\frac{1}{\sqrt{6}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$-\frac{1}{2 \sqrt{3}},0,\frac{\sqrt{3}}{2},-\frac{1}{2 \sqrt{3}},-\frac{1}{2 \sqrt{3}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$-\frac{1}{2 \sqrt{5}},\frac{2}{\sqrt{5}},-\frac{1}{2 \sqrt{5}},-\frac{1}{2 \sqrt{5}},-\frac{1}{2 \sqrt{5}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{1}{\sqrt{5}},\frac{1}{\sqrt{5}},\frac{1}{\sqrt{5}},\frac{1}{\sqrt{5}},\frac{1}{\sqrt{5}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Update 09/08&lt;/b&gt;&#xA;I came across another interesting characterization, when d=2^k, for some k, then Walsh Functions  form orthogonal basis for this matrix. In particular, let {$\mathbf{x_i}$} represent the list of vectors of binary expansion of integers 1 to d, ie {(0,0,0),(0,0,1),(0,1,0)...}. Then, rows (and columns) of $M$ define the orthonormal basis of matrix in question, where&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$M_{ij}=(-1)^{x_i \cdot x_j}$$&lt;/p&gt;&#xA;" OwnerUserId="998" LastEditorUserId="998" LastEditorDisplayName="" LastEditDate="2010-09-13T18:02:19.070" LastActivityDate="2010-09-13T18:02:19.070" Title="Characterizing orthonormal basis of I-c v v'" Tags="&lt;linear-algebra&gt;" AnswerCount="4" CommentCount="2" FavoriteCount="1" />
  <row Id="4174" PostTypeId="1" AcceptedAnswerId="4449" CreationDate="2010-09-07T00:19:05.313" Score="3" ViewCount="194" Body="&lt;p&gt;I'm trying to construct a simple proof of the Propositional Interpolation Theorem. For the following, let $At(\phi)$ be the set of sentence symbols that occur in a sentence $\phi$. Suppose that $\psi$ is a tautological consequence of $\phi$, but neither $\neg\phi$ nor $\psi$ is a tautology, and so $\psi$ is not a tautological consequence of $\phi$ for trivial reasons. I want to show that there is some sentence $\gamma$ such that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;i) $\gamma$ is a tautological consequence of $\phi$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ii) $\psi$ is a tautological consequence of $\gamma$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;iii) $At(\gamma)\subseteq At(\phi)\cap At(\psi)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From iii) $\gamma$ must be constructed from the sentence symbols found both in $\phi$ and $\psi$, so I first showed that $At(\phi)\cap At(\psi)\neq\emptyset$. Since neither $\neg\phi$ nor $\psi$ is a tautology, there must exist truth assignments $V,W$ such that $V(\neg\phi)=F$ and $W(\psi)=F$ and from this $V(\phi)=T$ and $W(\phi)=F$. If $At(\phi)\cap At(\psi)=\emptyset$ we may define a truth assignment for sentence symbols $p_i$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$U(p_i)=V(p_i)$ if $p_i\in At(\phi)$, $U(p_i)=W(p_i)$ if $p_i\in At(\psi)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But then $U(\phi)=V(\phi)=T$, but $U(\psi)=W(\psi)=F$, which contradicts the fact that $\psi$ is a tautological consequence of $\phi$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question now is, is there some sort of method to construct $\gamma$ such that conditions i) and ii) are satisfied?&lt;/p&gt;&#xA;" OwnerUserId="1583" LastEditorUserId="1583" LastEditorDisplayName="" LastEditDate="2010-09-07T01:13:47.133" LastActivityDate="2011-05-14T18:37:50.600" Title="Constructing an example that supports the Propositional Interpolation Theorem" Tags="&lt;logic&gt;" AnswerCount="3" CommentCount="4" FavoriteCount="1" />
  <row Id="4175" PostTypeId="1" AcceptedAnswerId="4187" CreationDate="2010-09-07T00:21:33.297" Score="11" ViewCount="602" Body="&lt;p&gt;Let $m,n\ge 1$ be two integers. Prove that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{k=m}^n (-1)^{k-m} \binom{k}{m} \binom{n}{k} = \delta_{mn}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\delta_{mn}$ stands for the &lt;em&gt;Kronecker's delta&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: I put the tag &quot;linear algebra&quot; because i think there is an elegant way to attack the problem using a certain type of matrices.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope you will enjoy. :)&lt;/p&gt;&#xA;" OwnerUserId="685" LastEditorUserId="152" LastEditorDisplayName="" LastEditDate="2010-09-07T06:10:15.277" LastActivityDate="2010-10-14T04:12:32.343" Title="Beautiful identity: $\sum_{k=m}^n (-1)^{k-m} \binom{k}{m} \binom{n}{k} = \delta_{mn}$" Tags="&lt;linear-algebra&gt;&lt;combinatorics&gt;&lt;binomial-coefficients&gt;" AnswerCount="5" FavoriteCount="1" />
  <row Id="4176" PostTypeId="2" ParentId="4170" CreationDate="2010-09-07T00:45:11.897" Score="1" ViewCount="" Body="&lt;p&gt;Ebbinghaus, Flum and Thomas. &lt;em&gt;Mathematical Logic&lt;/em&gt; (&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0387942580&quot; rel=&quot;nofollow&quot;&gt;Amazon&lt;/a&gt;)&lt;/p&gt;&#xA;" OwnerUserId="1195" LastActivityDate="2010-09-07T00:45:11.897" CommunityOwnedDate="2010-09-07T00:45:11.897" CommentCount="1" />
  <row Id="4177" PostTypeId="2" ParentId="4173" CreationDate="2010-09-07T00:54:02.997" Score="3" ViewCount="" Body="&lt;p&gt;If you write $P= I - \frac{1}{d}vv^t$ (I assume your $v'$ means the transpose of $v$), then you certainly have a symmetric matrix&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;P^t = \left( I- \frac{1}{d}vv^t\right)^t = I -\frac{1}{d}v^{tt}v^t = I - \frac{1}{d}vv^t = P \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So it diagonalizes and has an orthonormal basis of eigenvectors. That is, there exists an orthogonal matrix $S$, ($S^{-1} = S^t$) and a diagonal matrix $D $ such that $S^tP S = D$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moreover, since $P^2 = P$, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;P^2 &amp;amp;= \left( I- \frac{1}{d}vv^t\right) \left( I- \frac{1}{d}vv^t\right) \\&#xA;    &amp;amp;= I - 2\frac{1}{d}vv^t  + \frac{1}{d^2}vv^tvv^t \\&#xA;    &amp;amp;= I - 2\frac{1}{d}vv^t  + \frac{1}{d}vv^t \\&#xA;    &amp;amp;=I- \frac{1}{d}vv^t = P &#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and $P \neq 0,I$, its minimal polynomial is $x^2-x = x(x-1)$. So the only eigenvalues of $P$ are $0,1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for eigenvectors, as you say, $v$ is one of them, of $0$ eigenvalue:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;Pv = \left( I- \frac{1}{d}vv^t\right)v = v - \frac{1}{d}vv^tv = v - v = 0 \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, you can take $\frac{v}{\|v\|} = \frac{1}{\sqrt{d}} (1, \dots ,1)$ as the first vector of your orthonormal basis. The other vectors are an orthonormal basis of the orthogonal complement of $v$, $[v]^\bot$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;\left( I- \frac{1}{d}vv^t\right)w = w &amp;amp;\quad\Longleftrightarrow \quad w - \frac{1}{d}vv^tw = w \\&#xA;           &amp;amp;\quad\Longleftrightarrow \quad vv^tw = 0  \\&#xA;           &amp;amp;\quad\Longleftrightarrow \quad v^tw = 0&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, you only need to compute a basis for $[v]^\bot$ and orthonormalize it. In coordinates you must find the solutions of the linear equation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;x_1 + \dots + x_d = 0 \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;(-1, 1, 0, \dots ,0), (-1, 0, 1, 0, \dots ,0), \dots , (-1, 0, \dots , 0,1) \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And now you apply the Gram-Schmidt process &lt;a href=&quot;http://en.wikipedia.org/wiki/Gram_schmidt&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Gram_schmidt&lt;/a&gt; to these vectors.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;EDIT. I think Unkz intuition is right. Before normalizing, the general pattern looks as follows. You have $d-1$ vectors of the form&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;(-1, 0,\dots,0, i, -1,\dots, -1), \qquad \text{for} \quad i=1,\dots,d-1 \ ,&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $i$ is in the $d-i+1$ coordinate (so there are $i-1$ coordinates with $-1$ after it) and one more last vector&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;(d,\dots, d) \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's check that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(a) Those first $d-1$ vectors belong to $\ker (P-I) = [(1,\dots, 1)]^\bot$ (so they are eigenvectors corresponding to the eigenvalue $1$):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;(1,\dots, 1)\cdot (-1, 0,\dots,0, i, -1,\dots, -1) = -1 + i + (i-1)(-1)= 0 \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(b) Those first $d-1$ are mutually orthogonal vectors. We may assume $i&gt;j$, for instance:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;(-1, 0,\dots,0, i, -1,\dots, -1) \cdot (-1, 0,\dots,0, j, -1,\dots, -1) = 1 - i +(i-1)(-1)^2 = 0 \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, you quotient out their norms&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\sqrt{1 + i^2 + (i-1)} = \sqrt{i^2 + i} = \sqrt{i (i+1)} \qquad \text{for} \quad i= 1, \dots, d-1&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\sqrt{d^2d}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and you are done.&lt;/p&gt;&#xA;" OwnerUserId="664" LastEditorUserId="664" LastEditorDisplayName="" LastEditDate="2010-09-07T03:32:17.257" LastActivityDate="2010-09-07T03:32:17.257" CommentCount="1" />
  <row Id="4179" PostTypeId="2" ParentId="4170" CreationDate="2010-09-07T01:05:03.783" Score="1" ViewCount="" Body="&lt;p&gt;Shoenfield's &quot;Mathematical Logic&quot;.  The notation is a bit dated, but the exercises are great.&lt;/p&gt;&#xA;" OwnerUserId="851" LastActivityDate="2010-09-07T01:05:03.783" CommunityOwnedDate="2010-09-07T01:05:03.783" />
  <row Id="4180" PostTypeId="2" ParentId="4170" CreationDate="2010-09-07T01:15:14.367" Score="0" ViewCount="" Body="&lt;p&gt;François G. Dorais and others made some &lt;a href=&quot;http://mathoverflow.net/questions/15951/propositional-logic-first-order-logic-and-higher-order-logics/15954#15954&quot; rel=&quot;nofollow&quot;&gt;great recommendations to me&lt;/a&gt; some time ago over on MathOverflow. They're fairly high-level (not exactly introductory courses) but they're good reads.&lt;/p&gt;&#xA;" OwnerUserId="56" LastActivityDate="2010-09-07T01:15:14.367" CommunityOwnedDate="2010-09-07T01:15:14.367" />
  <row Id="4182" PostTypeId="2" ParentId="4068" CreationDate="2010-09-07T01:54:13.063" Score="2" ViewCount="" Body="&lt;p&gt;In this chain of 4 equations,  #1, #3 and #4 are correct, and no.2 is the mistake. The equations are assertions about (limits of) some finite sums. Let $H_n = \Sigma_{j=1}^n 1/j$ and $A_n = \Sigma_{i=1}^n (-1)^{i-1} 1/i$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The correct formula is $A_n = H_n - H_{[n/2]}$ (which is approximately $\log(n) - \log(n/2) \sim \log(2) = A_{\infty}$) , but the second equation cut off at $n$ terms is claiming $A_n = H_n - H_n$ (which is zero).&lt;/p&gt;&#xA;" OwnerUserId="467" LastActivityDate="2010-09-07T01:54:13.063" CommentCount="2" />
  <row Id="4183" PostTypeId="2" ParentId="4173" CreationDate="2010-09-07T02:25:34.227" Score="2" ViewCount="" Body="&lt;p&gt;While I haven't taken the time to prove it, if you look at the numbers I think you'll see a very nice pattern if you multiply out all your irrational denominators.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(-1,0,0,0,1) norm=$\sqrt{1 \cdot 2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(-1,0,0,2,-1) norm=$\sqrt{2 \cdot 3}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(-1,0,3,-1,-1) norm=$\sqrt{3 \cdot 4}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(-1,4,-1,-1,-1) norm=$\sqrt{4 \cdot 5}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(5,5,5,5,5) norm=$\sqrt{5^2 \cdot 5}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This appears to work for a few other random choices of d, where you have -1 in the first column, (1..d) down the backwards diagonal, and -1 under the backwards diagonal, along with 1,1,...,1 on the bottom row.&lt;/p&gt;&#xA;" OwnerUserId="326" LastActivityDate="2010-09-07T02:25:34.227" />
  <row Id="4184" PostTypeId="2" ParentId="4175" CreationDate="2010-09-07T02:35:44.010" Score="8" ViewCount="" Body="&lt;p&gt;The vector space of polynomials in one variable has two bases $\{1, x, x^2, ... \}$ and $\{1, (x+1), (x+1)^2, ... \}$ and I believe what you've written down is equivalent to the statement that the change-of-basis matrices between these two bases multiply to the identity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am still thinking about an inclusion-exclusion argument.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-09-07T02:35:44.010" CommentCount="1" />
  <row Id="4185" PostTypeId="1" AcceptedAnswerId="4190" CreationDate="2010-09-07T03:42:37.597" Score="3" ViewCount="136" Body="&lt;p&gt;consider the following game: 10 dice are tossed and those showing 3 are more are retained. [those showing 2 or less are discarded.] the remaining dice are tossed again and those showing 4 or more are retained. finally, the remaining dice are tossed once more and those showing 5 or more are retained. &#xA;if $Z$ denotes the number of dice retained after the third round of play, what is the probability that&#xA;$Z = 0$?&lt;/p&gt;&#xA;" OwnerUserId="1547" LastActivityDate="2010-09-07T07:49:10.740" Title="a sequential game of dice" Tags="&lt;probability&gt;&lt;recreational-mathematics&gt;" AnswerCount="2" />
  <row Id="4186" PostTypeId="2" ParentId="29" CreationDate="2010-09-07T04:03:57.357" Score="1" ViewCount="" Body="&lt;p&gt;If you want a quick RPN calculator with print capabilities for standard calculations, I use &lt;code&gt;free42&lt;/code&gt; which is compatible with HP42S (RPN). There is a print key to print when you need and you can configure the file used to save those prints. You can use skins to make it look the way you want, including the original HP calculator. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://thomasokken.com/free42/&quot; rel=&quot;nofollow&quot;&gt;free42&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1514" LastActivityDate="2010-09-07T04:03:57.357" CommunityOwnedDate="2010-09-07T04:03:57.357" />
  <row Id="4187" PostTypeId="2" ParentId="4175" CreationDate="2010-09-07T04:05:49.773" Score="10" ViewCount="" Body="&lt;p&gt;This follows easily from the &lt;a href=&quot;http://en.wikipedia.org/wiki/Multinomial_theorem&quot; rel=&quot;nofollow&quot;&gt;Multinomial Theorem&lt;/a&gt;, I believe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ 1 = 1^n = (1 - x + x)^n$$&#xA;$$ = \sum_{a+b+c=n} {n \choose a,b,c} 1^a \cdot (-x)^b \cdot x^c$$&#xA; $$ = \sum_{m=0}^{n} \sum_{k=m}^{n} {n \choose m,k-m,n-k} 1^{m}  \cdot (-x)^{k-m} \cdot x^{n-k} $$&#xA;  $$ = \sum_{m=0}^{n} (\sum_{k=m}^{n} (-1)^{k-m} {k \choose m}{n \choose k}) x^{n-m}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Comparing coefficients now gives the result immediately.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-09-07T04:05:49.773" />
  <row Id="4188" PostTypeId="2" ParentId="4173" CreationDate="2010-09-07T04:24:16.157" Score="0" ViewCount="" Body="&lt;p&gt;a slight retouching of unkz's idea is the following orthogonal basis for $v^\perp$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$u_1 = (1,-1,0,\dots,0),\&#xA;u_2 =  (1,1,-2,0,\dots,0),\&#xA;   \dots, \&#xA;u_{d-1} = (1,\dots,1,-d+1)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;in this form it is pretty easy to see that the $\{u_i\}$ are in $v^\perp$ and are mutually orthogonal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i don't quite understand the remark that normalizing the $\{u_i\}$ to get an ONB is the hardest part - since $d$ is unknown.  what sort of expression &quot;in terms of [the] unknown $d$&quot; &lt;em&gt;would&lt;/em&gt; be suitable?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ps: i would post this as a comment - if this @!*&amp;amp;*%#! site would let me.&lt;/p&gt;&#xA;" OwnerUserId="1547" LastActivityDate="2010-09-07T04:24:16.157" />
  <row Id="4189" PostTypeId="2" ParentId="4175" CreationDate="2010-09-07T05:08:29.670" Score="5" ViewCount="" Body="&lt;p&gt;Here's another way to look at Moron's proof: the sum counts all the partitions of $[n]$ into three sets $A,B,C$ satisfying $|C|=m$, weighted according to $(-1)^{|A|}$. The identity just says that if $n \neq m$, the number of partitions with $|A|$ even is the same as those with $|A|$ odd.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The latter fact is proved by the following sign-changing involution: pick the first element which is not in $C$ (there must be one since $n \neq m$), and flip it from $A$ to $B$ or vice versa.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-09-07T05:08:29.670" />
  <row Id="4190" PostTypeId="2" ParentId="4185" CreationDate="2010-09-07T05:30:59.203" Score="8" ViewCount="" Body="&lt;p&gt;The probability that a given die survives all three rounds is $(4/6)(3/6)(2/6) = 1/9$. Thus the probability that no dice survive is $(8/9)^{10} \approx 31\%$.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-09-07T05:30:59.203" CommentCount="1" />
  <row Id="4191" PostTypeId="2" ParentId="4170" CreationDate="2010-09-07T05:39:25.800" Score="0" ViewCount="" Body="&lt;p&gt;You should take a look at &lt;a href=&quot;http://books.google.com/books?id=4u9gQ6pctuIC&amp;amp;printsec=frontcover&amp;amp;dq=van+dalen+logic&amp;amp;source=bl&amp;amp;ots=koMWzUiwUR&amp;amp;sig=Noea7aYSfL54r63KwnunQOB1qRE&amp;amp;hl=en&amp;amp;ei=Vc-FTNL1EMjNswbKq-yaBQ&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=5&amp;amp;ved=0CCoQ6AEwBA#v=onepage&amp;amp;q=van%20dalen%20logic&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;D. Van Dalen: Logic and Structure&lt;/a&gt;. &lt;/p&gt;&#xA;" OwnerUserId="1154" LastActivityDate="2010-09-07T05:39:25.800" CommunityOwnedDate="2010-09-07T05:39:25.800" />
  <row Id="4192" PostTypeId="1" AcceptedAnswerId="4196" CreationDate="2010-09-07T06:25:10.123" Score="1" ViewCount="230" Body="&lt;p&gt;How does one find the value of the sum : $$\sum\limits_{n=1}^{p-1} [\sqrt{np} \ ]$$ where $p$ is a prime such that $p \equiv 1 (\text{mod} \ 4)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If i remember correctly, i got this sometime back, when i was an undergrad.&lt;/p&gt;&#xA;" LastActivityDate="2010-09-07T07:07:19.977" Title="Finding the Value of $\sum\limits_{n=1}^{p-1} [\sqrt{np} \ ]$" Tags="&lt;number-theory&gt;&lt;prime-numbers&gt;" AnswerCount="2" />
  <row Id="4193" PostTypeId="1" AcceptedAnswerId="4247" CreationDate="2010-09-07T06:54:19.977" Score="3" ViewCount="200" Body="&lt;p&gt;In an introductory algebra course, one proves the Structure theorem for finitely generated modules over a principal ideal domain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Structure_theorem_for_finitely_generated_modules_over_a_principal_ideal_domain&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Structure_theorem_for_finitely_generated_modules_over_a_principal_ideal_domain&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In every book that I have looked in, the uniqueness part of the proof is a mess.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone give me a reference for a nicer proof of the uniqueness part? I am especially interested in a proof of uniqueness when the underlying ring is the ring of polynomials in one variable over a field.&lt;/p&gt;&#xA;" OwnerUserId="1725" LastEditorUserId="1725" LastEditorDisplayName="" LastEditDate="2010-09-07T08:02:57.600" LastActivityDate="2010-09-08T18:41:49.100" Title="Proving uniqueness in the structure theorem for finitely generated modules over a principal ideal domain" Tags="&lt;abstract-algebra&gt;&lt;reference-request&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="4194" PostTypeId="2" ParentId="4192" CreationDate="2010-09-07T06:56:09.037" Score="4" ViewCount="" Body="&lt;p&gt;This is the number of pairs $(a,b)$ of integers with $0 &amp;lt; a &amp;lt; p$, $b &gt; 0$&#xA;and $b^2 &amp;lt; ap$. Reversing the implied summation gives&#xA;$\sum_{b=1}^{p-1}(p-1-\lfloor b^2/p\rfloor)$. This will reduce to&#xA;summing the fractional parts of $b^2/p$, which I'll leave you to think about.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-09-07T06:56:09.037" />
  <row Id="4195" PostTypeId="1" CreationDate="2010-09-07T07:00:17.280" Score="2" ViewCount="93" Body="&lt;p&gt;Given complete fan $\Delta$ defining a projective toric variety (so that $\Delta$ is the normal fan of some polytope). How do one go on to find a defining ideal of the toric variety in projective space?&#xA; Or, going the other way, given binomial equations defining a projective toric variety, is there some way to recover the fan?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In my case, I am trying to recover the fan of the hypersurface $Z(x_0x_3-x_1x_2)$ in $\mathbb{P}^n$ ($n&gt;3$).&lt;/p&gt;&#xA;" OwnerUserId="1764" LastEditorUserId="1764" LastEditorDisplayName="" LastEditDate="2010-09-07T17:36:21.443" LastActivityDate="2010-09-07T17:36:21.443" Title="Equations of a projective toric variety" Tags="&lt;algebraic-geometry&gt;" />
  <row Id="4196" PostTypeId="2" ParentId="4192" CreationDate="2010-09-07T07:07:19.977" Score="3" ViewCount="" Body="&lt;p&gt;If you don't feel like thinking too much, here are all the details for Robin's answer...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's begin with the formula $\lfloor\sqrt{np}\rfloor = \sharp[C \geq 1 : np \geq C^2]$. Therefore the original sum is equal to $\sharp[n &amp;lt; p, C \geq 1 : np \geq C^2]$. Now we can change the order of summation to $\sum_{C=1}^{p-1} (p - \lceil C^2/p \rceil)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next, we have $\lceil C^2/p \rceil = C^2/p + (-C^2 \mod p)/p$. Substituting, the sum reduces to $\sum_{C=1}^{p-1} (p - C^2/p) - \sum_{C=1}^{p-1} (-C^2 \mod p)/p$. We can calculate the first summand explicitly using the formula for sum of squares as $p(p-1) - (p-1)(2p-1)/6 = (p-1)(4p+1)/6$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to calculate the second summand, notice that we are summing over all quadratic residues, twice ($-1$ is a square for our $p$). Since $p \equiv 1 \pmod{4}$, the sum of all quadratic residues equals the sum of all quadratic non-residues, and so the second summand is $-\sum_{k=1}^{p-1} k/p = -(p-1)/2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Putting it all together, we get $(p-1)(4p+1)/6 - (p-1)/2 = (p-1)(2p-1)/3$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In case you were wondering, if $p$ were a prime of the other residue class, then the answer would depend on some class number.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-09-07T07:07:19.977" />
  <row Id="4197" PostTypeId="2" ParentId="4185" CreationDate="2010-09-07T07:49:10.740" Score="1" ViewCount="" Body="&lt;p&gt;For a much lengthier approach, the problem can be decomposed into three smaller problems: the chance of discarding all the dice in one round, two rounds, or three rounds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The chance of discarding all the dice in the first round is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(2/6)^{10}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your chance of having $n$ dice remaining after the first round is $(2/6)^{10-n} (4/6)^n$, and your chance of discarding all the dice at this stage are&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{i=1}^9 (10)choose(n)(2/6)^i (4/6)^{10-i} (3/6)^{10-i}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $i$ is the number of dice eliminated in the first pass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, your chance of having $m$ dice remaining after the second round is $(10)choose(n)(2/6)^{10-n} (4/6)^n (n)choose(m)(3/6)^{n-m} (3/6)^m=(2/6)^{10-n} (4/6)^n (3/6)^n$, and your chance of discarding the remaining dice at this stage is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{i=1}^9 \sum_{j=1}^{9-i} (10)choose(i)(2/6)^i (4/6)^{10-i} (10-i)choose(10-i-j)(3/6)^i (4/6)^{10-i-j}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $i$ is the number of dice eliminated in the first pass and $j$ is the number of dice eliminated in the second pass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sum of those three probabilities gives you the answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For me, it works out to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Removing all dice in first: 0.00001694&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Removing all dice in second: 0.01732459&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Removing all dice in third: 0.29060462&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Removing all dice in any round: 0.30794615&lt;/p&gt;&#xA;" OwnerUserId="326" LastActivityDate="2010-09-07T07:49:10.740" CommentCount="1" />
  <row Id="4198" PostTypeId="1" AcceptedAnswerId="4210" CreationDate="2010-09-07T10:04:23.990" Score="2" ViewCount="195" Body="&lt;p&gt;I am reading Jech's Set Theory, in particular the chapter about large cardinals. After discussing measurable cardinals he moves on to weakly-compact cardinals, which have been discussed far earlier in the book.&lt;br&gt;&#xA;I went back to the chapter dealing with weakly-compact cardinals and began retracing the definition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Eventually, it came into this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We denote $[k]^n = \{X \subset \kappa | |X| = n\}$. If $\lambda$ is a cardinal, we denote $\kappa \to (\lambda)^2$ when for every partition of $[\kappa]^2$ into $2$ we have $H \subset \kappa$ that is of cardinality $\lambda$, and for which $[H]^2$ is strictly in one part.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And we say that $\kappa$ is weakly-compact if it satisfies the property $\kappa \to (\kappa)^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is that I'm a bit lost in all those definitions, and not even sure about the $\kappa \to (\lambda)^2$ notation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My questions are, if so, can someone help me make some sense into those definitions, and is there an equivalent definition for weakly-compact cardinals which can help me understand their properties better?&lt;/p&gt;&#xA;" OwnerUserId="622" LastActivityDate="2010-09-07T19:28:18.887" Title="Weakly-compact cardinals" Tags="&lt;set-theory&gt;&lt;cardinals&gt;" AnswerCount="2" />
  <row Id="4199" PostTypeId="1" AcceptedAnswerId="4334" CreationDate="2010-09-07T10:04:56.257" Score="6" ViewCount="186" Body="&lt;p&gt;Usually, applying the conversion formulae&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$r^2=x^2+y^2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\cos\;\theta=\frac{x}{r}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sin\;\theta=\frac{y}{r}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;to transform an equation in polar coordinates to an implicit Cartesian equation is quite straightforward for an equation of the form&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$r=f(\cos(n\theta),\sin(n\theta))$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with $n$ an integer, thanks to multiple angle formulae. Polar equations of the form&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$r=f\left(\cos\left(\frac{p}{q}\theta\right),\sin\left(\frac{r}{s}\theta\right)\right)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $p$, $q$, $r$ and $s$ are integers, and $\frac{p}{q}$ and $\frac{r}{s}$ are in lowest terms, are more difficult to handle, but the case where $q$, $s$ are powers of 2 is slightly eased by the existence of identities like $\tan\left(\frac{\theta}{2}\right)=\frac{\sin\;\theta}{1+\cos\;\theta}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The really difficult ones to handle are the cases like &lt;a href=&quot;http://mathworld.wolfram.com/CayleysSextic.html&quot; rel=&quot;nofollow&quot;&gt;Cayley's sextic&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$r=4a\cos^3\left(\frac{\theta}{3}\right)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and in general, the cases with fractions whose denominators are not powers of 2. In particular for Cayley's sextic, I can't seem to find a quick way to exploit the appropriate multiple angle identities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One &quot;cheat&quot; I have seen is to instead represent the polar equation as a pair of parametric equations, and then make the substitution $\theta=3\arctan\;t$ so that everything is represented algebraically. The problem is that apparently it takes some insight to recognize how to remove the new parameter $t$ easily. For even tougher cases like&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$r=\cos\left(\frac{2\theta}{3}\right)-\sin\left(\frac{3\theta}{5}\right)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the appropriate substitution after transforming to parametric equations is $\theta=\mathrm{LCM}(3,5)\arctan\;t$, but the expressions, though algebraic, look even more nightmare-ish to manipulate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Mathematica&lt;/em&gt; has no trouble figuring out the implicit Cartesian equation, through a judicious use of Gröbner bases:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Factor /@ GroebnerBasis[Append[Thread[{x, y} == TrigExpand[{4a Cos[t/3]^3 Cos[t], 4a Cos[t/3]^3 Sin[t]}]],&#xA;Cos[t/3]^2 + Sin[t/3]^2 == 1], {x, y}, {Cos[t/3], Sin[t/3]}, MonomialOrder -&amp;gt; EliminationOrder]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;but I'm pretty sure the Cartesian equations have already been determined way before Buchberger was born.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How would one determine the Cartesian equation of such a curve using only classical techniques?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Apropos to this question, is there a quick way to determine the degree of an algebraic curve represented in polar coordinates without having to do a conversion?&lt;/p&gt;&#xA;" OwnerUserId="498" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-09-09T12:21:50.400" LastActivityDate="2010-09-09T19:08:05.050" Title="Deriving an implicit Cartesian equation from a polar equation with fractional multiples of the angle" Tags="&lt;geometry&gt;&lt;plane-curves&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="4201" PostTypeId="1" AcceptedAnswerId="4209" CreationDate="2010-09-07T11:20:05.537" Score="3" ViewCount="284" Body="&lt;p&gt;Let X = complex manifold of dimension n. Thus, it's a real manifold of dimension 2n.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now cohomology is a topological concept so it should not depend upon the structure given on a topological space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know that k'th Singular cohomology of X is 0 for k &gt; 2n. We can also define a sheaf cohomology on that space using derived functor approach of Grothendieck. Then (by a result of Grothendieck) we know that k'th sheaf cohomology is 0 for k &gt; n. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, for constant sheaves [say &lt;strong&gt;R&lt;/strong&gt;], the sheaf cohomology agrees with singular cohomology [with coefficient &lt;strong&gt;R&lt;/strong&gt;]. Does this means that even the k'th singular cohomology of X vanishes for k &gt; n ??&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[Edited] I now feel that the result which says that sheaf and singular agrees is actually this that k'th sheaf cohomology [of a complex manifold and constant sheaf] will agree with 2k'th singular cohomology [of the underlying real manifold]. Is this correct?? I would still like others to comment. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="1767" LastEditorUserId="1767" LastEditorDisplayName="" LastEditDate="2010-09-07T11:27:52.820" LastActivityDate="2010-09-08T07:28:28.167" Title="Singular and Sheaf Cohomology" Tags="&lt;algebraic-geometry&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" />
  <row Id="4202" PostTypeId="1" AcceptedAnswerId="4204" CreationDate="2010-09-07T11:39:16.947" Score="19" ViewCount="992" Body="&lt;p&gt;Hi&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of my Fellows asked me whether total induction is applicable to real numbers, too ( or at least all real numbers ≥ 0) . We only used that for natural numbers so far. &#xA;Of course you have to change some things in the inductive step, when you want to use it on real numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I guess that using induction on real numbers isn't really possible, since $[r,r+\epsilon]$ with $\epsilon &gt; 0$, $r \in \mathbb R$ is never empty.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you either give a good reason, why it isn't possible or provide an example where it is used?&lt;/p&gt;&#xA;" OwnerUserId="1768" LastEditorUserId="1768" LastEditorDisplayName="" LastEditDate="2010-09-07T18:11:44.733" LastActivityDate="2010-09-19T22:20:54.500" Title="Induction on Real Numbers" Tags="&lt;induction&gt;" AnswerCount="4" CommentCount="1" FavoriteCount="7" />
  <row Id="4204" PostTypeId="2" ParentId="4202" CreationDate="2010-09-07T12:38:05.140" Score="21" ViewCount="" Body="&lt;p&gt;Okay, I can't resist: here is a quick answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am construing the question in the following way: &quot;Is there some criterion for a subset of $[0,\infty)$ to be all of $[0,\infty)$ which is (a) analogous to the principle of mathematical induction on $\mathbb{N}$ and (b) useful for something?&quot;&#xA;The answer is yes, at least to (a).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me work a little more generally: let $(X,\leq)$ be a totally ordered set which has&lt;br&gt;&#xA;$\bullet $a least element, called $0$, and no greatest element.&lt;br&gt;&#xA;$\bullet$ The &lt;strong&gt;greatest lower bound property&lt;/strong&gt;: any nonempty subset $Y$ of $X$ has a greatest lower bound.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Principle of Induction on $(X,\leq)$: Let $S \subset X$ satisfy the following properties:&lt;br&gt;&#xA;(i) $0 \in S$.&lt;br&gt;&#xA;(ii) For all $x$ such that $x \in S$, there exists $y &gt; x$ such that $[x,y] \subset S$.&lt;br&gt;&#xA;(iii) If for any $y \in X$, the interval $[0,y) \subset S$, then also $y \in S$.&lt;br&gt;&#xA;Then $S = X$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Indeed, if $S \neq X$, then the complement $S' = X \setminus S$ is nonempty, so has a least upper bound, say $y$.  By (i), we cannot have $y = 0$, since $y \in S$.  By (ii), we cannot have $y \in S$, and by (iii) we cannot have $y \in S'$.  Done!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that in case $(X,\leq)$ is a well-ordered set, this is equivalent to the usual statement of transfinite induction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It also applies to an interval in $\mathbb{R}$ of the form $[a,\infty)$.  It is not hard to adapt it to versions applicable to any interval in $\mathbb{R}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that I believe that some sort of converse should be true: i.e., an ordered set with a principle of induction should have the GLB / LUB property.  [&lt;b&gt;Added&lt;/b&gt;: yes, this is true.  A totally ordered set with minimum element $0$ satisfies the principle of ordered induction as stated above iff every nonempty subset has an infimum.]  &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Added: as for usefulness, one can use &quot;real induction&quot; to prove the three basic &lt;strong&gt;Interval Theorems&lt;/strong&gt; of honors calculus / basic real analysis.  These three theorems assert three fundamental properties of any continuous function $f: [a,b] \rightarrow \mathbb{R}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First Interval Theorem: $f$ is bounded.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Inductive Proof: Let $S = \{x \in [a,b] \ | \ f|_{[a,x]} \text{is bounded} \}$.  It suffices to show that $S = [a,b]$, and we prove this by induction.&lt;br&gt;&#xA;(i) Of course $f$ is bounded on $[a,a]$.&#xA;(ii) Suppose that $f$ is bounded on $[a,x]$.  Then, since $f$ is continuous at $x$, $f$ is bounded near $x$, i.e., there exists some $\epsilon &gt; 0$ such that $f$ is bounded on &#xA;$(x-\epsilon,x+\epsilon)$, so overall $f$ is bounded on $[0,x+\epsilon)$.&lt;br&gt;&#xA;(iii) If $f$ is bounded on $[0,y)$, of course it is bounded on $[0,y]$.  Done!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Corollary: $f$ assumes its maximum and minimum values.&lt;br&gt;&#xA;Proof: Let $M$ be the least upper bound of $f$ on $[a,b]$.  If $M$ is not a value of $f$, &#xA;then $f(x)-M$ is never zero but takes values arbitrarily close to $0$, so $g(x) = \frac{1}{f(x)-M}$ is continuous and unbounded on $[a,b]$, contradiction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Unfortunately in the proof I said &quot;least upper bound&quot;, and I suppose the point of proofs by induction is to remove explicit appeals to LUBs.  Perhaps someone can help me out here.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second Interval Theorem (Intemediate Value Theorem): Suppose that $f(a) &amp;lt; 0$ and $f(b) &gt; 0$.  Then there exists $c \in (a,b)$ such that $f(c) = 0$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proof: Define $S = {x \in [a,b] \ | \ f(x) \leq 0}$.  In this case we are given that $S \neq [a,b]$, so at least one of the hypotheses of real induction must fail.  But which?&lt;br&gt;&#xA;(i) Certainly $a \in S$.&lt;br&gt;&#xA;(iii) If $f(x) \leq 0$ on $[a,y)$ and $f$ is continuous at $y$, then we must have $f(y) \leq 0$ as well: otherwise, there is a small interval about $y$ on which $f$ is positive.&lt;br&gt;&#xA;So it must be that (ii) fails: there exists some $x \in (a,b)$ such that $f \leq 0$ on &#xA;$[a,x]$ but there is no $y &gt; x$ such that $f \leq 0$ on $[a,y]$.  As above, since $f$ is continuous at $x$, we must have $f(x) = 0$!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Third Interval Theorem: $f$ is uniformly continuous.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Proof left to the interested reader.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moreover, one can give even easier inductive proofs of the following basic theorems of analysis: that the interval $[a,b]$ is connected, that the interval $[a,b]$ is compact (Heine-Borel), that every infinite subset of $[a,b]$ has an accumulation point (Bolzano-Weierstrass).&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Acknowledgement&lt;/b&gt;: My route to thinking about real induction was the following paper by I. Kalantari:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.math.uga.edu/~pete/Kalantari07.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.math.uga.edu/~pete/Kalantari07.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;His setup is slightly different from mine -- instead of (ii) and (iii), he has the single axiom that $[0,x) \subset S$ implies there exists $y &gt; x$ such that $[0,y) \subset S$ -- and I am sorry to say that I was initially confused by it and didn't think it was correct.  But I corresponded with Prof. Kalantari this morning and he kindly set me straight on the matter. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For that matter, several other papers exist in the literature doing real induction (mostly in the same setup as Kalantari's, but also with some other variants such as assuming (ii) and that the subset $S$ be closed).  The result goes back at least to a 1922 paper of Khinchin, who in fact later used real induction as a basis for an axiomatic treatment of real analysis.  It is remarkable to me that this appealing concept is not more widely known -- there seems to be a serious PR problem here.&lt;/p&gt;&#xA;" OwnerUserId="299" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2010-09-08T08:41:43.393" LastActivityDate="2010-09-08T08:41:43.393" CommentCount="7" />
  <row Id="4205" PostTypeId="1" AcceptedAnswerId="4207" CreationDate="2010-09-07T13:35:18.387" Score="2" ViewCount="126" Body="&lt;p&gt;I'm a beginner to maths and have trouble simplyfying the following function:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{p^y \cdot (pq)^o}{p^{2y+o} \cdot q^{o-2}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The final answer is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$p^{-y} \cdot q^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I'm not sure how to get there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any help is appreciated.&lt;/p&gt;&#xA;" OwnerUserId="1771" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-09-07T13:41:57.787" LastActivityDate="2010-09-07T13:52:44.417" Title="How to expand and simplify (p^y * (pq)^o) / (p^(2y+o) * q^(o-2))?" Tags="&lt;algebra-precalculus&gt;" AnswerCount="1" />
  <row Id="4206" PostTypeId="2" ParentId="530" CreationDate="2010-09-07T13:44:30.737" Score="2" ViewCount="" Body="&lt;p&gt;Note: for more on Bézout domains, see e.g.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Section 8.2 of &lt;a href=&quot;http://math.uga.edu/~pete/factorization2010.pdf&quot; rel=&quot;nofollow&quot;&gt;http://math.uga.edu/~pete/factorization2010.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;or&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Section 12.4 of &lt;a href=&quot;http://math.uga.edu/~pete/integral.pdf&quot; rel=&quot;nofollow&quot;&gt;http://math.uga.edu/~pete/integral.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="299" LastActivityDate="2010-09-07T13:44:30.737" />
  <row Id="4207" PostTypeId="2" ParentId="4205" CreationDate="2010-09-07T13:46:39.623" Score="4" ViewCount="" Body="&lt;p&gt;Here's the method in general, without actually working out your example. You should do that yourself to seal the concepts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The intermediate goal is to get all the powers of p and q separated in both the numerator and denominator.  In this case, it's almost there, with the exception of (pq)^o. So expand that first, using the principle (x*y)^a = x^a * y^a.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then gather the p's and q's using the properties of multiplication and exponentiation, x^a * x^b = x^(a+b). Finally, match the p's in the numerator and denominator, likewise the q's, and using the principle x^a / x^b = x^(a-b) calculate the ultimate powers of p and q. Note that there are two ways you could handle the power of p in the final answer, since it is negative.&lt;/p&gt;&#xA;" OwnerUserId="271" LastEditorUserId="271" LastEditorDisplayName="" LastEditDate="2010-09-07T13:52:44.417" LastActivityDate="2010-09-07T13:52:44.417" CommentCount="4" />
  <row Id="4208" PostTypeId="2" ParentId="530" CreationDate="2010-09-07T13:54:59.220" Score="2" ViewCount="" Body="&lt;p&gt;The easiest example I know is the ring of all algebraic integers (roots of monic polynomials with integer coefficients). As noted, it is a Bezout domain, so every finitely generated ideal is principal, and in particular for every two algebraic integers $a$ and $b$ there exist algebraic integers $\alpha$ and $\beta$ such that $\alpha a+\beta b = d$, where $d$ is a gcd for $a$ and $b$. However, the ideal $(2, 2^{1/2}, 2^{1/4}, 2^{1/8}, \ldots, 2^{1/2^{n}},\ldots)$ is not principal, so the ring is not a PID.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-09-07T13:54:59.220" CommentCount="6" />
  <row Id="4209" PostTypeId="2" ParentId="4201" CreationDate="2010-09-07T14:00:55.417" Score="6" ViewCount="" Body="&lt;p&gt;The cohomological dimension of a &lt;em&gt;real&lt;/em&gt; $n$-manifold $M$ is $n$: this means that $H^i(M,\mathscr F)=0$ for each sheaf $\mathscr F$ of abelian groups on $M$ if $i&gt;n$, and that there exist sheaves $\mathcal F$ on $M$ with $H^n(M,\mathscr F)\neq0$. You'll find this proved in Bredon's book on &lt;em&gt;Sheaf theory&lt;/em&gt;, §II.16. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It follows that the cohomological dimension of a &lt;em&gt;complex&lt;/em&gt; $n$-manifold is $2n$. For example, you reach the maximum, at least for compact ones, for the constant sheaf $\mathbb R$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer to your «Does this means that even the k'th singular cohomology of X vanishes for k &gt; n??» question is &lt;em&gt;No&lt;/em&gt; (You can answer it without determining the cohomological dimension: just consider a compact complex $n$-manifold, which is automatically oriented: what is it $2n$-th cohomology group?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question in you [Edited] paragraph also has a negative answer. Consider examples to see that it is so.&lt;/p&gt;&#xA;" OwnerUserId="274" LastEditorUserId="274" LastEditorDisplayName="" LastEditDate="2010-09-07T16:06:24.777" LastActivityDate="2010-09-07T16:06:24.777" CommentCount="1" />
  <row Id="4210" PostTypeId="2" ParentId="4198" CreationDate="2010-09-07T14:48:47.730" Score="4" ViewCount="" Body="&lt;p&gt;There are many ways to think of these definitions. Here's a way one can understand for example what $[\kappa]^2$ means and what it means to have a homogeneous subset, that I find intuitive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume that you have a complete undirected graph with $\kappa$ many nodes. That means that you have $\kappa$ points and you connect every pair of them with a line. Now assume that we have two colours, red and blue, and that every line between two nodes is coloured either red or blue. A subset of these $\kappa$ nodes is called homogeneous if the lines between each and every of its nodes have the same colour (that's the same as saying that it has a complete subgraph whose lines are of one colour only).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we say that $\kappa\to(\lambda)^2$ is true if no matter how we paint those lines using two colours, we can find a homogeneous set of cardinality $\lambda$. That is, for every way we colour the lines we can find $\lambda$ many nodes that every line between them is of the same colour.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, for every function $f:[\kappa]^2\to 2$ (you can think of this as a function that sends every two elements of $\kappa$ to one of the 2 colours) we can find a subset of $\kappa$ (let's call it$H$) that has cardinality $\lambda$ such that for every $x,y,z,w\in H$ we have $f({ x,y} )=f({ z,w} )$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To generalize this if for every function $f:[\kappa]^n\to\mu$ (again you can see this as a function that sends every $n$ elements of $\kappa$ to one of the $\mu$ colours, or that it partitions the subsets of $\kappa$ of cardinality $n$ into $\mu$ partitions) we can find a set $H$ such that $\left|H\right|=\lambda$ and for every $x_1,\ldots,x_n,y_1,\ldots,y_n\in H$ we have $f({ x_1,\ldots,x_n} )=f({ y_1,\ldots,y_n} )$ then we say that $\kappa\to(\lambda)^n_\mu$ is true.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The arrow notation, even though appears weird at first is used because the property remains true if we replace the cardinal in the left side of the arrow with a larger cardinal or if we replace any cardinal in the right side of the arrow with a smaller cardinal (if the subscript in the left side of the arrow is omitted then it is assumed that it is 2). It should be obvious that the notation only has meaning if $\lambda&amp;lt;\kappa$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are many equivalent definitions of weakly compact cardinals. Jech's book mentions most of them, but here are some. If $\kappa$ is a cardinal then it is weakly compact iff:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\kappa\to(\kappa)^2$ is true&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\kappa\to(\kappa)^n_\lambda$ where $\lambda&amp;lt;\kappa$ is true&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\kappa$ is inaccessible and has the tree property, that is, if you have a tree with height $\kappa$ and each level of the tree has less than $\kappa$ elements then it has a branch of length $\kappa$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\kappa$ is $\prod^1_1\textrm{-indescribable}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\kappa$ is inaccessible and the languages $\mathcal{L}_{\kappa,\omega}$ satisfy the weak compactness theorem (hence the name). This means that in a language where we allow $\lambda$ conjunctions and disjunctions (where $\lambda&amp;lt;\kappa$), if we have a set of sentences $\Sigma$ where$\left|\Sigma\right|=\kappa$ and every subset of $\Sigma$ with less than $\kappa$ sentences is satisfiable then $\Sigma$ is satisfiable.&lt;/p&gt;&#xA;" OwnerUserId="1772" LastEditorUserId="1772" LastEditorDisplayName="" LastEditDate="2010-09-07T15:59:59.743" LastActivityDate="2010-09-07T15:59:59.743" />
  <row Id="4211" PostTypeId="1" AcceptedAnswerId="4213" CreationDate="2010-09-07T15:37:21.660" Score="2" ViewCount="143" Body="&lt;p&gt;Motivation: In S.J. Patterson's &lt;em&gt;An introduction to the theory of the Riemann&#xA;Zeta-Function&lt;/em&gt; it is proved (p.132) that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle -\Gamma ^{\prime }(t)/\Gamma (t)=\gamma +t^{-1}+\underset{j\geq 1}{\sum }[(j+t)^{-1}-j^{-1}]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In Exercise A4.1 (p.135) one is asked to show that if $|t|&amp;lt;1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle -\Gamma ^{\prime }(t)/\Gamma (t)=\gamma +t^{-1}+\underset{k\geq 1}{\sum }\zeta (k+1)(-t)^{k}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; How does one prove that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle\underset{j\geq 1}{\sum }[(j+t)^{-1}-j^{-1}]=\underset{k\geq 1}{\sum }\zeta (k+1)(-t)^{k}\qquad $for $|t|&amp;lt;1$?&lt;/p&gt;&#xA;" OwnerUserId="752" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2010-11-15T18:11:01.060" LastActivityDate="2010-11-15T18:11:01.060" Title="Proving identity $\displaystyle\sum_{j\geq 1}[(j+t)^{-1}-j^{-1}]=\displaystyle\sum_{k\geq 1}\zeta (k+1)(-t)^{k}$" Tags="&lt;sequences-and-series&gt;&lt;special-functions&gt;&lt;zeta-function&gt;" AnswerCount="2" />
  <row Id="4212" PostTypeId="2" ParentId="4211" CreationDate="2010-09-07T16:08:49.557" Score="4" ViewCount="" Body="&lt;p&gt;Use&#xA;$$&#xA;\frac{1}{j + t} = \frac{1}{j} \frac{1}{1 + \frac{t}{j}} = \frac{1}{j} \sum_{k \geq 0} (-t)^k j^{- k}&#xA;$$&#xA;to obtain&#xA;$$&#xA;\sum_{j \geq 1} [ (j + t)^{-1} - j^{-1} ] =&#xA;\sum_{j \geq 1} \sum_{k \geq 1} (-t)^k j^{-k-1} = \sum_{k \geq 1} \zeta (k + 1) (-t)^k .&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="960" LastActivityDate="2010-09-07T16:08:49.557" />
  <row Id="4213" PostTypeId="2" ParentId="4211" CreationDate="2010-09-07T16:09:10.307" Score="5" ViewCount="" Body="&lt;p&gt;Just apply $\sum_{n=0}^\infty x^n = \frac1{1-x}$. For $|t| &amp;lt; \min j = 1$, we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ (j + t)^{-1} = j^{-1} \left( 1 + \frac tj \right)^{-1} = j^{-1} \sum_{k=0}^\infty \left( -\frac tj\right)^k, $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;thus the LHS is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \sum_{j=1}^\infty \sum_{k=1}^\infty \frac{(-t)^k}{j^{k+1}}&#xA;= \sum_{k=1}^\infty \sum_{j=1}^\infty \frac{(-t)^k}{j^{k+1}}&#xA;= \sum_{k=1}^\infty (-t)^k \left( \sum_{j=1}^\infty \frac1{j^{k+1}} \right) &#xA;= \sum_{k=1}^\infty (-t)^k \zeta(k+1) $$&lt;/p&gt;&#xA;" OwnerUserId="171" LastActivityDate="2010-09-07T16:09:10.307" />
  <row Id="4214" PostTypeId="1" CreationDate="2010-09-07T16:34:27.840" Score="7" ViewCount="390" Body="&lt;p&gt;In b&lt;sup&gt;a&lt;/sup&gt; = x, b is the base, a is the exponent and x is the result of the operation. But in its logarithm counterpart, log&lt;sub&gt;b&lt;/sub&gt;(x) = a, b is still the base, and a is now the result. What is x called here? The exponent?&lt;/p&gt;&#xA;" OwnerUserId="1773" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2011-05-11T02:49:15.470" LastActivityDate="2011-05-11T04:21:49.437" Title="What is the x in $\log_b x$ called?" Tags="&lt;terminology&gt;&lt;logarithms&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="1" />
  <row Id="4215" PostTypeId="1" CreationDate="2010-09-07T17:08:21.957" Score="4" ViewCount="134" Body="&lt;p&gt;I suspect it is impossible to split a (any) 3d solid into two, such that each of the pieces is identical in shape (but not volume) to the original.  How can I prove this?&lt;/p&gt;&#xA;" OwnerUserId="1775" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-09-07T17:47:41.687" LastActivityDate="2010-09-07T18:09:15.317" Title="Can we slice an object into two pieces similar to the original?" Tags="&lt;geometry&gt;" AnswerCount="2" CommentCount="5" />
  <row Id="4216" PostTypeId="2" ParentId="4193" CreationDate="2010-09-07T17:18:19.717" Score="4" ViewCount="" Body="&lt;p&gt;Let $M$ be the finitely generated module, and $M_{tor}$ its torsion submodule.&#xA;As part of the proof of the structure theorem, one shows that $M/M_{tor}$ is free&#xA;of finite rank, and that $M$ is isomorphic to $M_{tor} \oplus M/M_{tor}$,&#xA;so it suffices to show uniqueness for free modules and torsion modules separately.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a free module, its free rank can be computed as the maximal number of linearly independent&#xA;elements (just using the usual linear algebra definition, but applied to the free module).&#xA;So the free rank of $M/M_{tor}$ is uniquely determined.  (In fact, one can apply this&#xA;directly to $M$ --- the free part of $M$ in the decomposition given by the structure theorem&#xA;is just the maximal number of linearly independent elements of $M$ --- and so you&#xA;don't need to explicitly consider $M/M_{tor}$ if you don't want to.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We are now reduced to the case when $M = M_{tor}$, i.e. $M$ is torsion.  The structure theorem then gives us that $M$ is a direct sum of modules of the&#xA;form $A/\wp_j^{i_j} A$ for various non-zero prime ideals $\wp_j$ of $A$, and various indices $i_j$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine now that we have another such decomposition, into a direct sum of $A/\wp'_j^{i'_j}.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we fix a non-zero prime ideal $\wp$ of $M$, and some $n \geq 0,$&#xA;then let $M[\wp^n]$ denote the submodule of $M$ consisting of elements annihilated by&#xA;$\wp^n$.  A direct computation with each of the decompositions in turn shows that&#xA;$M[\wp^n]/M[\wp^{n-1}]$ is an $A/\wp$-vector space of dimension equal to the number&#xA;of $j$ for which $\wp_j = \wp$ and $i_j \geq n,$ and also equal to the number of &#xA;$j$ for which $\wp'_j = \wp$ and $i'_j \geq n.$  Since $\wp$ and $n$ were arbitrary,&#xA;this shows that the two decompositions coincide.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One can also make the following variant of this argument, in which we break it into two steps.&#xA;Namely, first fix a non-zero prime ideal of $A$, and consider $M[\wp^n]$.  If $n$ is larger than&#xA;any of the $i_j$ or $i'_j$, then this is the direct sum of just those $A/\wp_j^{i_j}$&#xA;for which $\wp_j = \wp$, and also is the direct sum of just those&#xA;$A/\wp'_j^{i'_j}$ for which $\wp'_j = \wp$.&#xA;Thus, replacing $M$ by $M[\wp^n]$ as $\wp$ runs over all the $\wp_j$ and $\wp'_j$, we may assume that all the $\wp_j$ and $\wp'_j$ coincide with some fixed prime ideal $\wp$.&#xA;We can then repeat the above argument, but just varying $n$, and keeping $\wp$ equal&#xA;to our given choice of prime ideal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is also the following variant, using quotients, rather than subobjects, namely,&#xA;fix a non-zero prime ideal of $A$, and consider $M/\wp^n M$.  Again, if $n$ is larger than&#xA;any of the $i_j$ or $i_j'$, then this is the direct sum of just those $A/\wp_j^{i_j}$&#xA;for which $\wp_j = \wp$, and also is the direct sum of just those&#xA;$A/\wp'_j^{i'_j}$ for which $\wp'_j = \wp$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At this point, we could consider submodules of prescribed torsion, as above,&#xA;but, since we have started by using quotients, we can finish by using quotients as well,&#xA;by considering the quotients of the form $\wp^n M/\wp^{n+1}M$ just as in&#xA;Pierre-Yves Gaillard's comment above.  As he explains, this also gives the desired uniqueness.&lt;/p&gt;&#xA;" OwnerUserId="221" LastEditorUserId="221" LastEditorDisplayName="" LastEditDate="2010-09-08T07:31:41.870" LastActivityDate="2010-09-08T07:31:41.870" />
  <row Id="4217" PostTypeId="2" ParentId="4215" CreationDate="2010-09-07T17:22:52.970" Score="3" ViewCount="" Body="&lt;p&gt;It seems that Puppe and others proved that this is impossible for &lt;em&gt;any&lt;/em&gt; strictly convex solid. See [B. L van den Waerden, Aufgabe Nr 51, &lt;em&gt;Elem. Math.&lt;/em&gt; &lt;strong&gt;4&lt;/strong&gt; (1949) &lt;strong&gt;18&lt;/strong&gt;, 140]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reference comes from &lt;em&gt;Unsolved problems in geometry&lt;/em&gt; by Hallard T. Croft, K. J. Falconer and Richard K. Guy.&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2010-09-07T17:22:52.970" CommentCount="2" />
  <row Id="4218" PostTypeId="2" ParentId="4214" CreationDate="2010-09-07T18:00:05.650" Score="14" ViewCount="" Body="&lt;p&gt;Argument (as you call the $x$ in any $f(x)$ argument of the function)&lt;/p&gt;&#xA;" OwnerUserId="917" LastActivityDate="2010-09-07T18:00:05.650" />
  <row Id="4219" PostTypeId="2" ParentId="4215" CreationDate="2010-09-07T18:09:15.317" Score="8" ViewCount="" Body="&lt;p&gt;You can certainly take a rectangular box, $2^{1/3} \times 2^{2/3} \times 2$ and slice it into two boxes of size $1 \times 2^{1/3} \times 2^{2/3}$.&lt;/p&gt;&#xA;" OwnerUserId="448" LastActivityDate="2010-09-07T18:09:15.317" CommentCount="5" />
  <row Id="4220" PostTypeId="1" AcceptedAnswerId="4248" CreationDate="2010-09-07T18:35:56.797" Score="6" ViewCount="408" Body="&lt;p&gt;In the course of analyzing a certain Markov chain, I once had to prove the following algebraic identity. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Is there a slick or known proof?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $n$-tuples $(x_1,x_2,\dots, x_n)$ of positive real numbers define&#xA;$$\mu(x_1,x_2,\dots, x_n)=\prod_{j=1}^n {x_j\over x_j+x_{j+1}+\cdots+x_n}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then if $x^\ast$ is another positive real, and $1\leq k\leq n+1$, then &#xA;define $x^*_k$ to be the $(n+1)$-tuple $(x_1,x_2,\dots, x^*,\dots, x_n)$ where &#xA;$x^*$ is in the $k$th place. The identity is  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{k=1}^{n+1}\ \mu(x^\ast_k)=\mu(x_1,x_2,\dots, x_n).$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example,&#xA;$$ {xyz\over(x+y+z)(y+z)z}&#xA; + {yxz\over(y+x+z)(x+z)z}&#xA; + {yzx\over(y+z+x)(z+x)x}={yz\over(y+z)z}.$$&lt;/p&gt;&#xA;" OwnerUserId="940" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-09-08T05:47:50.403" LastActivityDate="2010-09-08T05:48:44.363" Title="Is this a known algebraic identity?" Tags="&lt;algebra-precalculus&gt;&lt;probability&gt;&lt;polynomials&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="4221" PostTypeId="2" ParentId="4158" CreationDate="2010-09-07T18:50:18.430" Score="0" ViewCount="" Body="&lt;p&gt;This question seems fairly advanced. You may want to consider asking it on &lt;a href=&quot;http://mathoverflow.net/&quot; rel=&quot;nofollow&quot;&gt;MathOverflow&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1776" LastActivityDate="2010-09-07T18:50:18.430" />
  <row Id="4222" PostTypeId="2" ParentId="4198" CreationDate="2010-09-07T19:28:18.887" Score="2" ViewCount="" Body="&lt;p&gt;Let me add to Ivan's answer another equivalent&#xA;characterization of weak compactness, which might appeal to&#xA;you, as it makes them resemble miniature measurable&#xA;cardinals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Namely, if $\kappa$ is a cardinal and&#xA;$\kappa^{&amp;lt;\kappa}=\kappa$, then $\kappa$ is weakly compact&#xA;if and only if for every transitive set $M$ of size&#xA;$\kappa$ with $\kappa\in M$, there is another transitive&#xA;set $N$ and an elementary embedding $j:M\to N$ having&#xA;critical point $\kappa$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This embedding characterization admits myriad forms. For&#xA;example, one can insist that $M\models ZF^-$ or even&#xA;$M\models ZFC$, and that $M^{&amp;lt;\kappa}\subset M$, or that&#xA;every $A\subset \kappa$ can be placed into such an $M$, and so&#xA;on. One can even insist that $j\in N$, a property known as&#xA;the &lt;em&gt;Hauser&lt;/em&gt; property.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These various embedding formulations of weak compactness&#xA;allow one to borrow many of the methods and techniques from&#xA;much larger large cardinals, which are most often described&#xA;in terms of embeddings, and apply them with weakly compact&#xA;cardinals. For example, using Easton support forcing&#xA;iterations, one can control the value of $2^\kappa$ while&#xA;preserving the weak compactness of $\kappa$.&lt;/p&gt;&#xA;" OwnerUserId="413" LastActivityDate="2010-09-07T19:28:18.887" />
  <row Id="4223" PostTypeId="2" ParentId="4068" CreationDate="2010-09-07T20:02:28.423" Score="1" ViewCount="" Body="&lt;p&gt;In Matt's answer I began discussing with Matt E over several comments, which I think should be written out as an answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As Matt pointed out, this is a rearrangement of this conditionally convergent series which is why you have this sort of paradox.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However it was unclear about how this is exactly a rearrangement, as the equities seems perfectly legal - even for a conditionally convergent series.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \ldots = 1 + (\frac{1}{2} - 1) + \frac{1}{3} + (\frac{1}{4} - \frac{1}{2})$ is the first step, which is legal as you simply replace the negative terms by pairs of a positive and negative terms, but you don't change the order of summation from the original series which makes this exchange legit.&lt;/li&gt;&#xA;&lt;li&gt;$1 + (\frac{1}{2} - 1) + \frac{1}{3} + (\frac{1}{4} - \frac{1}{2}) = 1 + \frac{1}{2} + \frac{1}{3} + \ldots - 1 - \frac{1}{2} - \ldots$ &lt;strong&gt;this&lt;/strong&gt; is where things break, you've taken a conditionally convergent series and changed the order, basically we've performed infinitely many commutations in order to rearrange the series into this order, and that is what breaks the summation.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The rearrangement wasn't very obvious, but it was hiding there with its big sharp pointy teeth... and when you stepped too close to its cave - it jumped out at you and bit your head off.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The series in the question is closely reminding me of the one my calculus teacher used when he first showed us what changing conditionally convergent series can do, although his was even less obvious.&lt;/p&gt;&#xA;" OwnerUserId="622" LastActivityDate="2010-09-07T20:02:28.423" CommentCount="4" />
  <row Id="4224" PostTypeId="2" ParentId="4202" CreationDate="2010-09-07T20:09:42.483" Score="3" ViewCount="" Body="&lt;p&gt;There's a phenomenal paper by Don Zagier which uses a sort of induction over the real numbers to give an explicit solution to some recurrence relations (which, it just so happens, calculate the Betti numbers of the moduli space of central Yang-Mills connections on bundles over Riemann surfaces).  But you don't need to understand those words to appreciate Zagier's paper.  The reference is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Zagier, Don&#xA;Elementary aspects of the Verlinde formula and of the Harder-Narasimhan-Atiyah-Bott formula. (English summary) Proceedings of the Hirzebruch 65 Conference on Algebraic Geometry (Ramat Gan, 1993), 445--462,&#xA;Israel Math. Conf. Proc., 9, Bar-Ilan Univ., Ramat Gan, 1996. &lt;/p&gt;&#xA;" OwnerUserId="1392" LastActivityDate="2010-09-07T20:09:42.483" CommentCount="3" />
  <row Id="4225" PostTypeId="2" ParentId="4158" CreationDate="2010-09-07T20:11:35.073" Score="1" ViewCount="" Body="&lt;p&gt;I think Terry Tao's paper might help here: &lt;a href=&quot;http://arxiv.org/abs/0802.3361&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/0802.3361&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It shows that the isolated primes are of positive relative density.  I'm not sure if the method can be extended to show that there are large connected blocks.&lt;/p&gt;&#xA;" OwnerUserId="1778" LastActivityDate="2010-09-07T20:11:35.073" />
  <row Id="4226" PostTypeId="2" ParentId="1870" CreationDate="2010-09-07T20:35:49.157" Score="1" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://www.mathxpert.com/&quot; rel=&quot;nofollow&quot;&gt;MathXpert&lt;/a&gt; supposedly fits your first 3 criteria (though I have never tried it).&lt;/p&gt;&#xA;" OwnerUserId="1386" LastActivityDate="2010-09-07T20:35:49.157" CommunityOwnedDate="2010-09-07T20:35:49.157" />
  <row Id="4227" PostTypeId="2" ParentId="262" CreationDate="2010-09-07T20:52:10.283" Score="10" ViewCount="" Body="&lt;p&gt;Proofs from the Book, &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/3540636986&quot; rel=&quot;nofollow&quot;&gt;http://www.amazon.com/Proofs-BOOK-Martin-Aigner/dp/3540636986&lt;/a&gt;&lt;/p&gt;&#xA;" LastActivityDate="2010-09-07T20:52:10.283" CommunityOwnedDate="2010-09-07T20:52:10.283" />
  <row Id="4228" PostTypeId="1" AcceptedAnswerId="4347" CreationDate="2010-09-07T21:40:48.790" Score="2" ViewCount="239" Body="&lt;p&gt;Say P is a convex Euclidean polytope, where the origin is not contained in any bounding hyperplane containing a facet of P, with n facets given by f ◦ x = 1 and m vertices v. (That is, each facet equation has been multiplied through as needed to obtain 1 on the right side.)&#xA;Is it true that the polytope P’ with m facets given by v ◦ x = 1 and n vertices f is combinatorially equivalent to the polar dual P* of P?&lt;/p&gt;&#xA;" OwnerUserId="1782" LastEditorUserId="1782" LastEditorDisplayName="" LastEditDate="2010-09-08T13:23:22.503" LastActivityDate="2010-09-13T09:20:17.983" Title="“Cartesian” dual vs. polar dual of convex polytope" Tags="&lt;geometry&gt;" AnswerCount="1" />
  <row Id="4229" PostTypeId="2" ParentId="4220" CreationDate="2010-09-07T22:23:12.113" Score="3" ViewCount="" Body="&lt;p&gt;Here is a proof by induction on the number of elements in the tuple.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Hopefully, I got the algebra right. In any case, it would be easy to verify).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First a Lemma.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Lemma&lt;/strong&gt;: For $n \ge 2$ we have that $\displaystyle \mu(x_1, ..., x_n) = \frac{x_{n-1}}{x_{n-1}+x_n} \mu(x_1, .., x_{n-2}, x_{n-1} + x_n)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof&lt;/strong&gt;: Easy Algebra.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $y = x^{*}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Base Case&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $n=1$ it is easy to see that $ \displaystyle \mu(y, x) + \mu(x, y) = \mu(x)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Induction Step&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now suppose $n \ge 2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have that (I have changed your notation slightly, I start from 0 instead of 1)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{k=0}^{n-2} \mu(x_1, x_2, \dots , x_k, y, x_{k+1}, \dots,x_{n-1}, x_{n})$$&#xA;$$ = \sum_{k=0}^{n-2}\frac{x_{n-1}}{x_{n-1}+x_{n}} \mu(x_1, x_2, ..., x_k, y, \dots, x_{n-1} + x_{n})$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now it is easy to see that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mu(x_1, \dots, x_{n-1}, y, x_n) + \mu(x_1, \dots, x_{n-1}, x_n, y)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$  = \frac{x_{n-1}}{x_{n-1}+x_n} \mu(x_1, x_2, ..., x_{n-2}, x_{n-1} + x_n, y)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus we have that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \sum_{k=0}^{n} \mu(x_{k}^{*}) $$ &#xA;$$ = \sum_{k=0}^{n-1} \frac{x_{n-1}}{x_{n-1}+x_{n}} \mu (z_{k}^{*})$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ z = (x_1, \dots, x_{n-2}, x_{n-1}+x_{n})$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus by induction hyptothesis, this is the same as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \frac{x_{n-1}}{x_{n-1}+x_{n}} \mu (x_1, \dots, x_{n-2}, x_{n-1}+x_{n}) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;= $$ \mu(x_1, x_2, \dots, x_{n-1}, x_n)$$ &lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-09-07T22:23:12.113" CommentCount="5" />
  <row Id="4230" PostTypeId="2" ParentId="4214" CreationDate="2010-09-07T22:24:53.630" Score="3" ViewCount="" Body="&lt;p&gt;antilogarithm (noun): a compound of Greek anti-&quot;against&quot; and logarithm (qq. v.). In algebra if $log_{b}(x) = y$, y is the logarithm of x, and x is said to be the antilogarithm of y. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Source: The Words of Mathematics: An Etymological Dictionary of Mathematical Terms Used in English, Steven Schwartzman &lt;/p&gt;&#xA;" OwnerUserId="1746" LastActivityDate="2010-09-07T22:24:53.630" CommentCount="5" />
  <row Id="4231" PostTypeId="2" ParentId="566" CreationDate="2010-09-07T22:45:17.960" Score="0" ViewCount="" Body="&lt;p&gt;Quite simply, a monkey's mother is twice as old as the monkey will be when the monkey's father is twice as old as the monkey will be when the monkey's mother is less by the difference in ages between the monkey's mother and the monkey's father than three times as old as the monkey will be when the monkey's father is one year less than twelve times as old as the monkey is when the monkey's mother is eight times the age of the monkey, notwithstanding the fact that when the monkey is as old as the monkey's mother will be when the difference in ages between the monkey and the monkey's father is less than the age of the monkey's mother by twice the difference in ages between the monkey's mother and the monkey's father, the monkey's mother will be five times as old as the monkey will be when the monkey's father is one year more than ten times as old as the monkey is when the monkey is less by four years than one seventh of the combined ages of the monkey's mother and the monkey's father.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If in a number of years equal to the number of times a monkey's mother is as old as the monkey, the monkey's father will be as many times as old as the monkey as the monkey is now, and assuming no &lt;em&gt;a priori&lt;/em&gt; knowledge of the monkeys' longevity, find their respective ages. :-D&lt;/p&gt;&#xA;" OwnerUserId="1770" LastActivityDate="2010-09-07T22:45:17.960" CommunityOwnedDate="2010-09-07T22:45:17.960" CommentCount="2" />
  <row Id="4232" PostTypeId="2" ParentId="4201" CreationDate="2010-09-07T22:53:15.210" Score="3" ViewCount="" Body="&lt;p&gt;The statement is the following: If you take a variety with the &lt;em&gt;analytic&lt;/em&gt; topology (or a complex manifold), then yes, sheaf cohomology of the constant sheaf agrees with singular cohomology. More generally, sheaf cohomology of a constant sheaf on a &lt;em&gt;locally contractible space&lt;/em&gt; agrees with singular cohomology of that space. I think there's an explanation of this somewhere in Warner's book &quot;Foundations of Differentiable Manifolds and Lie Groups&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, if you have an irreducible variety with the &lt;em&gt;Zariski&lt;/em&gt; topology, then sheaf cohomology $H^i$ of the constant sheaf is zero for $i &gt; 0$, because the constant sheaf is flasque. (I'm not sure -- does it make sense to talk about Zariski topology on a (non-algebraic) complex manifold?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Further remarks: If what you want is to be able to deal with singular cohomology in a more &quot;algebraic&quot; way, you can do so by using the Hodge decomposition $H^n(X;\mathbb{C}) = \bigoplus_{p+q=n}H^q(X,\Omega^p)$ which expresses singular cohomology in terms of sheaf cohomologies of $\Omega^p$'s. This works if your $X$ is a compact Kaehler manifold, e.g., a smooth projective variety. If your $X$ is an algebraic variety, you can also use etale cohomology for an &quot;algebraic&quot; way of dealing with singular cohomology; see Milne's book/notes on etale cohomology for more.&lt;/p&gt;&#xA;" OwnerUserId="263" LastEditorUserId="263" LastEditorDisplayName="" LastEditDate="2010-09-08T07:28:28.167" LastActivityDate="2010-09-08T07:28:28.167" CommentCount="3" />
  <row Id="4233" PostTypeId="2" ParentId="4043" CreationDate="2010-09-07T22:59:44.917" Score="2" ViewCount="" Body="&lt;p&gt;Introduction to Logic: Predicate Logic by Prospesel is the text probably most frequently associated with the topic for introductory students from a purely academic perspective.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you really want to code in languages that are based around predicate logic (logic programming is the more formal term), there are a few options:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Prolog is easily the best known and without a doubt the best documented logic programming language&lt;/li&gt;&#xA;&lt;li&gt;Oz is a lesser known alternative that incorporates a number of more modern programming language features like concurrency and object orientation (modern Prolog implementations typically add some of these to the language, but it never seems a natural fit to the language)&lt;/li&gt;&#xA;&lt;li&gt;Castor is a fairly common library for C++ that builds a logic model within the syntax of C++&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1781" LastActivityDate="2010-09-07T22:59:44.917" CommunityOwnedDate="2010-09-07T23:21:03.547" CommentCount="1" />
  <row Id="4234" PostTypeId="1" AcceptedAnswerId="4235" CreationDate="2010-09-07T23:33:24.873" Score="1" ViewCount="136" Body="&lt;p&gt;Suppose {x_t} is a random walk with mean zero. (either discrete or continuous time)&#xA;Fix a time T. &#xA;What is: Probability[x_t &amp;lt; 0, for all t &amp;lt;= T]?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;in words, what's the probability the random walk from 0 to T has never been above 0?&#xA;Thank you.&lt;/p&gt;&#xA;" OwnerUserId="1784" LastActivityDate="2010-09-08T00:38:04.520" Title="Random walk &lt; 0" Tags="&lt;stochastic-processes&gt;" AnswerCount="1" />
  <row Id="4235" PostTypeId="2" ParentId="4234" CreationDate="2010-09-08T00:38:04.520" Score="3" ViewCount="" Body="&lt;p&gt;Here is a solution for the discrete time, simple (steps are $\pm 1$), symmetric random walk $(X_t)$&#xA;that uses the reflection principle.&#xA;You didn't specify your starting point, but presumably it is negative, say $-b$ for some $b&gt;0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm going to consider the equivalent problem where the random walk starts at zero, and investigate&#xA; when it hits $b&gt;0$, that is, $T_b:= \inf (t&gt;0: X_t =b )$.&#xA;Then you should be able to use equation $(*)$ to find what you are looking for. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $t,b$ have opposite parity. Then&#xA;$$P(X_t &gt; b) = P(X_t&gt;b \mid  T_b &amp;lt; t) P(T_b &amp;lt; t)={1\over 2} P(T_b &amp;lt; t).$$&#xA;The conditional probability is exactly $1/2$ because $X_t=b$ is impossible.&#xA;Given $T_b &amp;lt; t$, the sample paths satisfy $X_{T_b}=b$ at some time $T_b$ prior to $t$.&#xA;The reflection principle shows that&#xA;these paths divide equally into those with $X_t &gt; b$ and those with $X_t &amp;lt; b$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This gives us the formula  $P(T_b &amp;lt; t)=2P(X_t &gt; b)$ or $P(T_b\geq t)=P(-b\leq X_t\leq b)$.&#xA;Since the random walk is symmetric around the origin, the random variables&#xA;$T_b$ and $T_{-b}$ have the same  distribution. Thus, we can combine the two cases as follows,&#xA; taking into account that $X_t=\pm b$ is impossible,&#xA;$$P(T_b\geq t)=P(-|b|&amp;lt; X_t &amp;lt; |b|),\quad b\in {\mathbb Z}\setminus \lbrace 0\rbrace ,\ t+b\mbox{ odd}.\qquad \qquad (*)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, in the special case where $b=1$, with the even time point $2t$,&#xA;we get $$P(T_1\geq 2t)=P(-1&amp;lt; X_{2t} &amp;lt; 1)=P(X_{2t}=0)={2t\choose t}\left({1\over 2}\right)^{2t}.$$   &lt;/p&gt;&#xA;" OwnerUserId="940" LastActivityDate="2010-09-08T00:38:04.520" CommentCount="1" />
  <row Id="4236" PostTypeId="1" AcceptedAnswerId="4237" CreationDate="2010-09-08T00:41:52.957" Score="0" ViewCount="85" Body="&lt;p&gt;&lt;strong&gt;Note&lt;/strong&gt;: logs below are base 2.  (Not sure how to do subscripts here)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wondering if the below equation is true when thinking asymptotically (Computer Science)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$log_2((n!)^n) = O(n \sin(n \frac{\pi}{2}) + \log_2{n})$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I'm not sure how to compute this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm guessing we need to take the log of both sides of the following equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$log_2((n!)^n) &amp;lt; n sin(n (pi/2)) + log_2(n)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;getting us:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$log_2(log_2((n!)^n)) &amp;lt; log_2 (n  sin(n (pi/2)) + log_2(n) )$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Not sure where to go from there.&lt;/p&gt;&#xA;" OwnerUserId="1630" LastEditorUserId="1630" LastEditorDisplayName="" LastEditDate="2010-09-08T04:17:09.247" LastActivityDate="2010-09-08T04:17:09.247" Title="Big O boundary condition truth value" Tags="&lt;asymptotics&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="4237" PostTypeId="2" ParentId="4236" CreationDate="2010-09-08T01:37:20.163" Score="1" ViewCount="" Body="&lt;p&gt;If you are interested in computer science, a very useful thing to remember is that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \log n! = \theta(n \log n) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now given this, can you tell what $f(n)$ is, if $ \log (n!^{n}) = \theta(f(n))$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How would that compare to the right hand side?&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-09-08T01:37:20.163" CommentCount="19" />
  <row Id="4238" PostTypeId="1" AcceptedAnswerId="4320" CreationDate="2010-09-08T03:12:13.693" Score="3" ViewCount="205" Body="&lt;p&gt;The two element group has a centre (of course it does, it's abelian) but no outer automorphisms. I can come up with a (convoluted) countably infinite non-abelian example. Is there a finite one? If so what is the smallest? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a name for this sort of group? (&quot;Outerless?&quot; Since a group without a centre is &quot;centreless&quot;, or &quot;complete&quot; if it has no outermorphisms either.)&lt;/p&gt;&#xA;" LastActivityDate="2010-09-09T00:19:34.470" Title="Finite non-abelian group with centre but no outer automorphism" Tags="&lt;group-theory&gt;" AnswerCount="2" />
  <row Id="4239" PostTypeId="1" AcceptedAnswerId="4240" CreationDate="2010-09-08T03:18:05.003" Score="2" ViewCount="63" Body="&lt;p&gt;Say we have a partially ordered set $(S,\preceq)$, and some subset $E\subseteq S$ such that $E$ is bounded below and $\inf E$ exists. My question is, since $S$ is not totally ordered is it possible to have two lower bounds that are not comparable, and hence be unable to say whether one lower bound is really the greatest lower bound since you cannot compare to every other lower bound? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If such a thing is not possible, (which I believe is the case since it is given $\inf E$ exists) would it also be fair to say that the existence of $\inf E$ implies that the set $L(E)$ of lower bounds of $E$ is a totally ordered subset of $S$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="1583" LastActivityDate="2010-09-08T03:32:22.013" Title="Does the existence of an infimum imply that the set of lower bounds of a set is totally ordered?" Tags="&lt;analysis&gt;&lt;set-theory&gt;" AnswerCount="1" />
  <row Id="4240" PostTypeId="2" ParentId="4239" CreationDate="2010-09-08T03:32:22.013" Score="1" ViewCount="" Body="&lt;p&gt;No, because even if a partially ordered set is not totally ordered, there can be a &quot;largest&quot; element.  For a simple example, take a set $X$ with at least 2 elements, let $S$ be the power set of $X$ and let the partial order be inclusion.  Then $\inf\{X\}=X$, but the set of lower bounds of $\{X\}$ is $S$, which is not totally ordered because $X$ has more than one element.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-09-08T03:32:22.013" CommentCount="2" />
  <row Id="4241" PostTypeId="1" CreationDate="2010-09-08T03:43:29.700" Score="2" ViewCount="312" Body="&lt;p&gt;I was preparing for an area exam in analysis and came across a problem in the book &lt;em&gt;Real Analysis&lt;/em&gt; by Haaser &amp;amp; Sullivan. From p.34 Q 2.4.3, If the field &lt;em&gt;F&lt;/em&gt; is isomorphic to the subset &lt;em&gt;S'&lt;/em&gt; of &lt;em&gt;F'&lt;/em&gt;, show that &lt;em&gt;S'&lt;/em&gt; is a subfield of &lt;em&gt;F'&lt;/em&gt;. I would appreciate any hints on how to solve this problem as I'm stuck, but that's not my actual question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand that for finite fields this implies that two sets of the same cardinality must have the same field structure, if any exists. The classification of finite fields answers the above question in a constructive manner.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What got me curious is the infinite case. Even in the finite case it's surprising to me that the field axioms are so &quot;restrictive&quot;, in a sense, that alternate field structures are simply not possible on sets of equal cardinality. I then started looking for examples of fields with characteristic zero while thinking about this problem. I didn't find many. So far, I listed the rationals, algebraic numbers, real numbers, complex numbers and the p-adic fields. What are other examples? Is there an analogous classification for fields of characteristic zero?&lt;/p&gt;&#xA;" OwnerUserId="1761" LastActivityDate="2010-09-08T09:55:44.503" Title="Examples of fields of characteristic 0?" Tags="&lt;soft-question&gt;&lt;abstract-algebra&gt;&lt;reference-request&gt;" AnswerCount="4" />
  <row Id="4243" PostTypeId="2" ParentId="4202" CreationDate="2010-09-08T04:22:53.383" Score="1" ViewCount="" Body="&lt;p&gt;If you assume a particular limit ordinal for the cardinality of the reals, you can prove new theorems about them using transfinite induction with a bounded limit case!  But I doubt these theorems have any application.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other answers are very interesting but it is hard for me to believe in a process truly analogous to induction on a set which cannot be well-ordered.  The ability to well-order the reals depends on the Axiom of Choice, and the precise consequences of such an ordering depend on the Continuum Hypothesis.&lt;/p&gt;&#xA;" OwnerUserId="1284" LastEditorUserId="1284" LastEditorDisplayName="" LastEditDate="2010-09-08T04:45:36.660" LastActivityDate="2010-09-08T04:45:36.660" CommentCount="0" />
  <row Id="4244" PostTypeId="2" ParentId="4241" CreationDate="2010-09-08T04:35:23.093" Score="2" ViewCount="" Body="&lt;p&gt;If $F$ is any field, the rational functions over it form a field $F(t)$ with the same characteristic (and cardinality, if $F$ is infinite). This field consists of all rational functions $P(t)/Q(t)$ (considered as equivalence classes, i.e. if $P_1(t) Q_2(t) = P_2(t) Q_1(t)$ then $P_1(t)/Q_1(t)$ and $P_2(t)/Q_2(t)$ are identified).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can also replace polynomials with formal power series to get a different field. And you can iterate the construction or just consider rational functions in several variables.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-09-08T04:35:23.093" />
  <row Id="4245" PostTypeId="2" ParentId="4241" CreationDate="2010-09-08T05:04:03.750" Score="2" ViewCount="" Body="&lt;p&gt;Yuval just posted the function field construction, which is an important example for your question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You mentioned the algebraic numbers, but just to make it explicit, there are many algebraic number fields (often called just &quot;number fields&quot;).  We have $ \bar Q $, the field of &lt;em&gt;all&lt;/em&gt; algebraic numbers, but we also have $Q(\sqrt{2})$, $Q(\sqrt[5]{7})$, $Q(i,\sqrt{2},\sqrt[5]{7})$, and any other field you wish to generate over $Q$ by some tower of finite extensions.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(If we take $Q(\pi)$, we get a field isomorphic to the function field $Q(t)$, since $\pi$ is not algebraic.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can also take infinitely many different algebraic extensions of the p-adic fields, like $Q_3(i)$ or $Q_3(\sqrt[4]{2})$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you like big constructions, you can take the algebraic closure $\bar {Q_p}$of a p-adic field (which turns out not to be p-adically complete), then take the completion $\hat { \bar {Q_p}}$ of that to get a p-adcially complete and algebraically complete field which is isomorphic to the field of complex numbers (but not in any canonical way).&lt;/p&gt;&#xA;" OwnerUserId="1461" LastActivityDate="2010-09-08T05:04:03.750" />
  <row Id="4246" PostTypeId="2" ParentId="4214" CreationDate="2010-09-08T05:24:03.990" Score="12" ViewCount="" Body="&lt;p&gt;Another name (that I've only ever seen when someone else asked this &lt;a href=&quot;http://en.wikipedia.org/wiki/Wikipedia%3aReference_desk/Archives/Mathematics/2010_February_17#Follow_up%3a_the_name_for_the_argument_of_the_logarithmic_function&quot; rel=&quot;nofollow&quot;&gt;question&lt;/a&gt;) is &quot;logarithmand&quot;.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;From &lt;a href=&quot;http://books.google.com/books?id=9uoDAAAAQAAJ&amp;amp;vq=logarithmand&amp;amp;pg=PA36&quot; rel=&quot;nofollow&quot;&gt;page 36&lt;/a&gt; of &lt;em&gt;&lt;a href=&quot;http://books.google.com/books?id=9uoDAAAAQAAJ&amp;amp;printsec=frontcover&quot; rel=&quot;nofollow&quot;&gt;The Spirit of Mathematical Analysis&lt;/a&gt;&lt;/em&gt; by &lt;a href=&quot;http://en.wikipedia.org/wiki/Martin_Ohm&quot; rel=&quot;nofollow&quot;&gt;Martin Ohm&lt;/a&gt;, translated from the German by &lt;a href=&quot;http://en.wikipedia.org/wiki/Alexander_John_Ellis&quot; rel=&quot;nofollow&quot;&gt;Alexander John Ellis&lt;/a&gt;, 1843:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/fSgzo.png&quot; alt=&quot;page from The Spirit of Mathematical Analysis&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1461" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2011-05-11T04:21:49.437" LastActivityDate="2011-05-11T04:21:49.437" CommentCount="2" />
  <row Id="4247" PostTypeId="2" ParentId="4193" CreationDate="2010-09-08T05:32:50.650" Score="3" ViewCount="" Body="&lt;p&gt;As a complement to Matt E's nice answer, let $A$ be your PID, and $p\in A$ a prime. It suffices to prove the uniqueness in the case where your finitely generated module $M$ is the product of a finite family of modules of the form $$M_i:=A/p^{i+1}A.$$ For each $j$ the quotient $p^jM/p^{j+1}M$ is an $A/pA$ vector space of finite dimension $n_j$. The multiplicity of $A/p^{i+1}A$ is then $n_i-n_{i+1}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a way to see this. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Form the polynomial $M(X):=\sum\ n_j\ X^j$. We have &#xA;$$M_i(X)=\frac{X^{i+1}-1}{X-1}=1+X+X^2+\cdots+X^i,$$ &#xA;and we must solve $\sum\ m_i\ M_i(X)=\sum\ n_j\ X^j$ for the $m_i$, where the $n_j$ are considered as known quantities (almost all equal to zero). Multiplying throughout by $X-1$ we get $$\sum\ m_{i-1}\ X^i-\sum\ m_i=\sum\ (n_{i-1}-n_i)\ X^i,$$ &#xA;whence the formula.&lt;/p&gt;&#xA;" OwnerUserId="660" LastEditorUserId="660" LastEditorDisplayName="" LastEditDate="2010-09-08T18:41:49.100" LastActivityDate="2010-09-08T18:41:49.100" CommentCount="2" />
  <row Id="4248" PostTypeId="2" ParentId="4220" CreationDate="2010-09-08T05:48:44.363" Score="10" ViewCount="" Body="&lt;p&gt;Consider the following experiment: there is an interval of length $x_1 + \cdots + x_n$ divided into $n$ segments of sizes $x_1,\ldots,x_n$. We sample an infinite sequence of points from the interval, and write out the segments corresponding to the points. Then, we look at the order in which the segments are &quot;discovered&quot;, i.e. which one was hit first, which was the next one, and so on. Then $\mu(x_1,\ldots,x_n)$ is the probability that the segments will be discovered in the order $x_1,\ldots,x_n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now suppose we add an extra segment $x^* $, and repeat the same experiment. If we just forget about this extra segment, the new experiment is just like the old one. This is the same as running the new experiment and deleting $x^* $ from the order of discovery. Your identity follows.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-09-08T05:48:44.363" CommentCount="4" />
  <row Id="4249" PostTypeId="2" ParentId="4238" CreationDate="2010-09-08T05:50:10.607" Score="4" ViewCount="" Body="&lt;p&gt;$C_2\times M_{11}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Steve&lt;/p&gt;&#xA;" OwnerUserId="641" LastActivityDate="2010-09-08T05:50:10.607" CommentCount="1" />
  <row Id="4250" PostTypeId="1" AcceptedAnswerId="4251" CreationDate="2010-09-08T06:00:57.767" Score="1" ViewCount="63" Body="&lt;p&gt;This is a follow up question to one I posted earlier. I'm trying to decide that if for $(S,\preceq)$ a partially ordered set and $E\subseteq S$, one has $L(E)=\langle s]$ for some $s\in S$ iff $\inf E$ exists, and in particular, $L(E)=\langle\inf E]$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm simply curious about showing that $L(E)\subseteq\langle\inf E]$ when $\inf E$ exists. Taking some $x\in L(E)$, if $x$ is comparable to $\inf E$, then $x\preceq\inf E$, and so $x\in\langle\inf E]$. But what if $x$ and $\inf E$ are not comparable? Is it still possible to show such an inclusion?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: $\langle s]$ is the set {$x\in S \ | \ x\preceq s$}.&lt;/p&gt;&#xA;" OwnerUserId="1583" LastEditorUserId="1583" LastEditorDisplayName="" LastEditDate="2010-09-08T06:30:17.353" LastActivityDate="2010-09-08T06:31:45.597" Title="Can one show that $L(E)=\langle s]$ on a partially ordered set?" Tags="&lt;set-theory&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="4251" PostTypeId="2" ParentId="4250" CreationDate="2010-09-08T06:31:45.597" Score="2" ViewCount="" Body="&lt;p&gt;If I correctly understand your question, it boils down to whether $\inf E$ is &quot;larger&quot; than or equal to every lower bound of $E$.  Isn't this the definition?  A greatest lower bound of a set must be, by definition, comparable to and &quot;larger&quot; than (or equal to) every lower bound.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the confusion arises because, as seen in your last question, having an inf does not imply that the set of lower bounds is totally ordered.  It does imply, however, that all lower bounds are comparable to the inf.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the answer is yes: If $E$ has an inf, then $L(E)=\langle \inf E]$.  Anything smaller than a lower bound is also a lower bound, which gives the right to left inclusion.  The other inclusion follows from the fact that $\inf E$ is the &quot;greatest&quot; lower bound.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-09-08T06:31:45.597" CommentCount="2" />
  <row Id="4252" PostTypeId="2" ParentId="3852" CreationDate="2010-09-08T06:55:08.603" Score="5" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;MOTIVATION&lt;/strong&gt; $\ $ If vector space $\rm V\:$ has a one-one map $\rm B$ that's not onto, i.e. $\rm V &amp;gt; BV\:,\:$ then this yields an $\infty$ descending chain of subspaces by  $\rm V &amp;gt; \: BV &amp;gt; \;\cdots\: &amp;gt; B^i V$ by repeatedly shifting by $\rm B\:$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;THEOREM&lt;/strong&gt; $\rm\;\;\;\: AB = 1 \;\Rightarrow\; BA=1 \;\;$ for linear maps $\rm\; A,\:B \;$ on a finite dimensional vector space $\rm\: V $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof&lt;/strong&gt; $\rm\;\; V &amp;gt; BV, \;$ times $\rm\; B^i\:$ yields $\rm\ B^i V &amp;gt; B^{i+1} V \;$ (else $\rm\; B^i V = B^{i+1} V, \;$ times $\rm\; A^i \Rightarrow V = BV)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\ \ \Rightarrow\rm\;\;\; V &amp;gt; BV &amp;gt; B^2 V &amp;gt; \;\cdots \;\:$ is an $\infty$ descending chain $\rm\; \Rightarrow\; dim\: V = \infty\:,\:$ contra hypothesis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence $\rm\;\; V = BV \;\Rightarrow\; (BA-1)V = (BA-1)BV = B(AB-1)V = 0 \quad\quad$ &lt;strong&gt;QED&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;REMARK&lt;/strong&gt; $\;\;$ Hence vector space $\rm\:V\;$ &lt;strong&gt;has infinite dimension&lt;/strong&gt; $\rm\;\iff V\:$ &lt;strong&gt;is Dedekind infinite&lt;/strong&gt;, i.e. $\rm\:V\:$ has an isomorphic proper subspace $\rm BV\:,\:$ viz. the theorem proves $(\Leftarrow)$ and the converse follows immediately by taking $\rm B\:$ to be a basis shift map $\rm\; (v_1,v_2,v_3,\cdots\:)\:\to\: (0,v_1,v_2,v_3,\cdots\:)\:,\;$  i.e. said simply, a vector space is infinite iff it has a subspace chain model of &lt;strong&gt;Hilbert's infinite hotel&lt;/strong&gt;. $\:$ That is the simple essence of the matter.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2011-02-22T17:42:59.130" LastActivityDate="2011-02-22T17:42:59.130" CommentCount="2" />
  <row Id="4253" PostTypeId="1" AcceptedAnswerId="4292" CreationDate="2010-09-08T08:00:06.240" Score="1" ViewCount="159" Body="&lt;p&gt;Could someone give me some tips?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let e1,e2,... be iid normal mean 0 variance 1. Let Xt := e1+...+et, for t=1,2.. and X0 := 0. (So we have a discrete-time random walk whose steps are iid N(0,1))  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Define first-hitting time T0 := inf{t&gt;0 : Xt &amp;lt;0}.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the CDF of T0, i.e. what is Prob(T0 &amp;lt;= t) ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And second (probably more difficult) question: what is ExpectedValue( X_t | T0 &gt; t )?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you.&lt;/p&gt;&#xA;" OwnerUserId="1784" LastEditorUserId="1784" LastEditorDisplayName="" LastEditDate="2010-09-08T08:55:23.643" LastActivityDate="2010-09-08T15:29:13.577" Title="Ruin-time in Gambler's Ruins" Tags="&lt;probability-theory&gt;&lt;probability&gt;&lt;stochastic-processes&gt;" AnswerCount="1" FavoriteCount="2" />
  <row Id="4254" PostTypeId="2" ParentId="907" CreationDate="2010-09-08T08:36:29.900" Score="0" ViewCount="" Body="&lt;p&gt;Mathematics is generally seen as a precise and specific language.  Surely if there is ambiguity in the phrase &quot;in the sequel&quot; it should be avoided.  There are far more accurate ways of expressing both the meanings that phrase could carry so why create needless confusion?&lt;/p&gt;&#xA;" LastActivityDate="2010-09-08T08:36:29.900" />
  <row Id="4255" PostTypeId="2" ParentId="4241" CreationDate="2010-09-08T09:04:38.537" Score="0" ViewCount="" Body="&lt;p&gt;Yes, you are right in saying that field axioms are restrictive. Some other examples of the restriction [for finite fields] are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) You can't have field of arbitrary order. Only of the order $p^n$ are possible. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Non-zero elements of a field form a multiplicative group. When the field is finite, this group is cyclic [it's not straightforward to prove this starting with field axioms but you should try it]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) Any finite division ring is a field [Wedderburn's theorem]. This is a very surprising result as a division ring need not by commutative. But finiteness imposes it.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the question that you mentioned [from that book], you should note that when some set with some algebraic structure [like group, ring, field] is isomorphic [as that algebraic structure] to some other set, then that under that isomorphism, we are giving an algebraic structure to that other set. Hence in your case S' is a field. &lt;/p&gt;&#xA;" OwnerUserId="1767" LastActivityDate="2010-09-08T09:04:38.537" CommentCount="2" />
  <row Id="4256" PostTypeId="1" AcceptedAnswerId="4258" CreationDate="2010-09-08T09:17:32.093" Score="4" ViewCount="341" Body="&lt;p&gt;&lt;em&gt;This problem is taken from Vojtěch Jarník International Mathematical Competition 2010, &lt;a href=&quot;http://vjimc.osu.cz/j20/j20problems1.pdf&quot; rel=&quot;nofollow&quot;&gt;Category I, Problem 1&lt;/a&gt;. &lt;sub&gt;&lt;sup&gt;— edit by KennyTM&lt;/sup&gt;&lt;/sub&gt;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;On going through this post &lt;a href=&quot;http://math.stackexchange.com/questions/2120/does-there-exist-an-f-mathbbn-to-mathbbn-such-that-sum-fn-n2-conve&quot;&gt;http://math.stackexchange.com/questions/2120/does-there-exist-an-f-mathbbn-to-mathbbn-such-that-sum-fn-n2-conve&lt;/a&gt; i happened to get the following 2 problems into my mind:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $f: \mathbb{N} \to \mathbb{N}$ be a bijection. Then does the series $$\sum\limits_{n=1}^{\infty} \frac{1}{nf(n)}$$ converge?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next, consider the series $$\sum\limits_{n=1}^{\infty} \frac{1}{n+f(n)}$$ where $f: \mathbb{N} \to \mathbb{N}$ is a bijection. Clearly by taking $f(n)=n$ we see that the series is divergent. Does there exist a bijection such that the sum above is convergent?&lt;/p&gt;&#xA;" LastEditorUserId="171" LastEditorDisplayName="" LastEditDate="2010-09-08T17:32:27.090" LastActivityDate="2010-09-09T09:23:46.643" Title="Convergence of $\sum\limits_{n=1}^{\infty} \frac{1}{nf(n)}$" Tags="&lt;real-analysis&gt;&lt;analysis&gt;&lt;sequences-and-series&gt;" AnswerCount="3" CommentCount="6" />
  <row Id="4257" PostTypeId="2" ParentId="4256" CreationDate="2010-09-08T09:32:56.567" Score="2" ViewCount="" Body="&lt;p&gt;A &lt;a href=&quot;http://www.scienzematematiche.it/forum/viewtopic.php?f=7&amp;amp;t=2322&quot; rel=&quot;nofollow&quot;&gt;couple of proofs&lt;/a&gt; in italian.&lt;/p&gt;&#xA;" OwnerUserId="685" LastActivityDate="2010-09-08T09:32:56.567" CommentCount="1" />
  <row Id="4258" PostTypeId="2" ParentId="4256" CreationDate="2010-09-08T09:33:53.490" Score="3" ViewCount="" Body="&lt;p&gt;Hints. For the first series, prove that&#xA;$$\sum_{n=1}^N\frac1{nf(n)}\le\sum_{n=1}^N\frac1{n^2}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the second, what would happen if $f$ were an involution,&#xA;and in each pair $(n,f(n))$ one term was much bigger than the other?&lt;/p&gt;&#xA;" OwnerUserId="226" LastEditorUserId="226" LastEditorDisplayName="" LastEditDate="2010-09-08T10:12:09.643" LastActivityDate="2010-09-08T10:12:09.643" />
  <row Id="4259" PostTypeId="2" ParentId="4241" CreationDate="2010-09-08T09:55:44.503" Score="2" ViewCount="" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Is there an analogous classification for fields of characteristic zero?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes, but it is somewhat useless and nobody would call it a classification.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Every field of characteristic zero has the form $Quot(\mathbb{Q}[X]/S)$, where $X$ is a set of variables and $S$ is a set of polynomials in $\mathbb{Q}[X]$ (which you may replace by the ideal generated by $S$, which must be prime). This may be improved by the existence of transcendence bases: Every field of characteristic zero has the form $Quot(\mathbb{Q}[X])[T]/S$, where $X$ and $T$ are sets of variables and $S$ consists of polynomials, which have each only one variable of $T$.&lt;/p&gt;&#xA;" OwnerUserId="1650" LastActivityDate="2010-09-08T09:55:44.503" CommentCount="1" />
  <row Id="4260" PostTypeId="1" AcceptedAnswerId="4281" CreationDate="2010-09-08T09:58:17.100" Score="2" ViewCount="72" Body="&lt;p&gt;Hello everyone,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have have a question which should be pretty basic commutative algebra, but I can't find a reference and I' m stuck on proving a result myself, so here it goes:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $\varphi: S \to R$ be a morphism of commutative rings and let $M$ be a finitely generated $S$-module. Then, we can &quot;extend the scalars of $M$&quot; by considering $M \otimes_S R$, which is naturally an $R$-module. I would like to get a description of $\text{supp}(M \otimes_S R)$ (as an $R$-module) involving $\text{supp}(M)$ and $\varphi$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe this helps: by Eisenbud: Commutative algebra with a view toward algebraic geometry, Corollary 2.7, the problem can probably be solved by finding a despcription of $\text{Ann}(M \otimes_S R)$ and it is not difficult to see that $\varphi(\text{Ann}(M)) \cdot R \subset \text{Ann}(M \otimes_S R)$, but I fail to prove the other inclusion.&lt;/p&gt;&#xA;" OwnerUserId="1766" LastActivityDate="2010-09-08T13:14:19.080" Title="Support of a module with extended scalars" Tags="&lt;commutative-algebra&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="4261" PostTypeId="1" AcceptedAnswerId="4265" CreationDate="2010-09-08T10:08:15.933" Score="5" ViewCount="405" Body="&lt;p&gt;Mathematica tells me that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum _{k=0}^n { n \choose k} \frac{(-1)^k}{2k+1} = \frac{(2n)!!}{(2n+1)!!}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Although I have not been able to come up with a proof.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proofs, hints, or references are all welcome.&lt;/p&gt;&#xA;" OwnerUserId="899" LastActivityDate="2010-09-09T14:23:03.063" Title="Proving a binomial sum identity" Tags="&lt;combinatorics&gt;&lt;complex-analysis&gt;&lt;generating-functions&gt;" AnswerCount="5" CommentCount="0" FavoriteCount="1" />
  <row Id="4263" PostTypeId="2" ParentId="4261" CreationDate="2010-09-08T10:20:08.917" Score="6" ViewCount="" Body="&lt;p&gt;Sums of the form&#xA;$$\sum_{k=0}^n(-1)^k{n\choose k}f(k)$$&#xA;can often be attacked via the &#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Finite_difference&quot; rel=&quot;nofollow&quot;&gt;Calculus of finite differences&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-09-08T10:20:08.917" />
  <row Id="4264" PostTypeId="2" ParentId="4261" CreationDate="2010-09-08T10:21:21.307" Score="3" ViewCount="" Body="&lt;p&gt;I can't offer specific help, but I'd recommend thumbing through &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0201558025&quot; rel=&quot;nofollow&quot;&gt;Concrete Mathematics&lt;/a&gt; looking for techniques.  It has many sums that look similar, though of course the difficulty is in the details.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's also the book &lt;a href=&quot;http://www.math.upenn.edu/~wilf/AeqB.html&quot; rel=&quot;nofollow&quot;&gt;A=B&lt;/a&gt;, but Concrete Mathematics gives an introduction to the content of A=B and in my opinion is easier to read, so I'd start with Concrete Mathematics.&lt;/p&gt;&#xA;" OwnerUserId="247" LastEditorUserId="247" LastEditorDisplayName="" LastEditDate="2010-09-09T14:23:03.063" LastActivityDate="2010-09-09T14:23:03.063" />
  <row Id="4265" PostTypeId="2" ParentId="4261" CreationDate="2010-09-08T10:32:48.580" Score="12" ViewCount="" Body="&lt;p&gt;You could consider the integral&#xA;$$ \int_{0}^{1} (1-x^2)^n dx .$$&lt;/p&gt;&#xA;" OwnerUserId="1301" LastActivityDate="2010-09-08T10:32:48.580" CommentCount="1" />
  <row Id="4266" PostTypeId="1" AcceptedAnswerId="4282" CreationDate="2010-09-08T10:49:44.020" Score="1" ViewCount="116" Body="&lt;p&gt;In my previous question, &lt;a href=&quot;http://math.stackexchange.com/q/4198/622&quot;&gt;Weakly-compact cardinals&lt;/a&gt;, I was asking about weakly-compact cardinals and equivalent definitions to the basic one, which is $\kappa \to (\kappa)^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of which was that $\kappa$ is inaccessible and has the &lt;em&gt;tree property&lt;/em&gt; (that is if any tree of cardinality $\kappa$ for which every level is of cardinality $&amp;lt;\kappa$ then it has a branch (i.e. a maximal chain) of cardinality $\kappa$). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can understand the property itself and what it means. However, since $\aleph_1$ or $\aleph_\omega$ are clearly not weakly-compact cardinals, there should be a tree which contradicts this property.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do you build this sort of tree?&lt;/p&gt;&#xA;" OwnerUserId="622" LastEditorUserId="622" LastEditorDisplayName="" LastEditDate="2010-09-08T14:25:42.047" LastActivityDate="2010-09-08T14:25:42.047" Title="The tree property for non-weakly compact $\kappa$" Tags="&lt;set-theory&gt;&lt;cardinals&gt;&lt;counter-examples&gt;" AnswerCount="1" />
  <row Id="4267" PostTypeId="2" ParentId="4261" CreationDate="2010-09-08T10:54:31.803" Score="3" ViewCount="" Body="&lt;p&gt;Your identity is a special case of the &lt;a href=&quot;http://mathworld.wolfram.com/Chu-VandermondeIdentity.html&quot; rel=&quot;nofollow&quot;&gt;Chu-Vandermonde identity&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;${}_2 F_1(-n,b;c;1)=\frac{(c-b)_n}{(c)_n}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with $b=\frac12$ and $c=\frac32$. More info on it is in A=B, as mentioned already by John.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-09-08T10:54:31.803" CommentCount="1" />
  <row Id="4268" PostTypeId="1" AcceptedAnswerId="4269" CreationDate="2010-09-08T11:17:01.167" Score="3" ViewCount="516" Body="&lt;p&gt;I know the definition of prime number when dealing with integers, but I can't understand why the following definition also works:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A prime is a quantity $p$ such that whenever $p$ is a factor of some product $a\cdot b$, then either $p$ is a factor of $a$ or $p$ is a factor of $b$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For example, take $4$ (which clearly is not a prime): it is a factor of $16=8\cdot 2$, so I should check that either $4\mid 8$ or $4\mid 2$. But $4\mid 8$ is true. So $4$ is a prime, which is absurd.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please note that English is not my first language, so I may have easily misunderstood the above definition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: Let me try formalize the definition as I understood it: $p$ is prime if and only if $\exists a\exists b(p\mid a\cdot b)\rightarrow p\mid a\lor p\mid b$.&lt;/p&gt;&#xA;" LastActivityDate="2010-09-19T22:34:53.240" Title="Alternate definition of prime number" Tags="&lt;prime-numbers&gt;&lt;definition&gt;" AnswerCount="4" CommentCount="2" FavoriteCount="1" />
  <row Id="4269" PostTypeId="2" ParentId="4268" CreationDate="2010-09-08T11:23:41.937" Score="11" ViewCount="" Body="&lt;p&gt;Note that this is &lt;strong&gt;whenever&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$4|4 = 2\cdot 2$ however $4\not{|}2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To the edited question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;No! The word &quot;whenever&quot; comes to say that $p$ is prime if $\forall a \forall b (p | a\cdot b \Rightarrow (p | a \vee p | b))$&lt;/p&gt;&#xA;" OwnerUserId="622" LastEditorUserId="622" LastEditorDisplayName="" LastEditDate="2010-09-08T12:13:18.727" LastActivityDate="2010-09-08T12:13:18.727" />
  <row Id="4270" PostTypeId="1" AcceptedAnswerId="4275" CreationDate="2010-09-08T11:35:28.740" Score="1" ViewCount="206" Body="&lt;p&gt;I was wondering if a field is an algebra over itself (http://en.wikipedia.org/wiki/Algebra_over_a_field)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also is a ring an algebra over itself (http://en.wikipedia.org/wiki/Algebra_(ring_theory)? If not, does the ring require to be commutative?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks and regards!&lt;/p&gt;&#xA;" OwnerUserId="1281" LastEditorUserId="1281" LastEditorDisplayName="" LastEditDate="2010-09-08T12:01:18.037" LastActivityDate="2010-09-08T18:24:30.387" Title="Is a field (ring) an algebra over itself?" Tags="&lt;abstract-algebra&gt;" AnswerCount="3" CommentCount="3" />
  <row Id="4272" PostTypeId="1" CreationDate="2010-09-08T11:46:19.083" Score="2" ViewCount="264" Body="&lt;p&gt;I'm trying to express the integral&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$I = \int_{-\infty}^{+\infty} \int_{-\infty}^{+\infty} f(x_1, x_2) \; g(x_1, x_2) \; dx_1 dx_2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;of two real valued functions $f(x_1,x_2)$ and $g(x_1,x_2)$ in terms of their Fourier transforms $\tilde{f}(\omega_1, \omega_2)$ and $\tilde{g}(\omega_1, \omega_2)$. However, as $f(x_1,x_2)$ and $g(x_1,x_2)$ are real-valued functions and $I$ is not defined as the the product of $f(x_1,x_2)$ and $g(x_1,x_2)^\ast$ I'm not working with the complex conjugate)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f(x_1,x_2) = \int_{-\infty}^{+\infty} \int_{-\infty}^{+\infty} \tilde{f}(\omega_1, \omega_2) \; e^{2 \pi i ( \;\omega_1 x_1 \;+ \;\omega_2 x_2 \;) } \; d\omega_1 d\omega_2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$g(x_1,x_2) = \int_{-\infty}^{+\infty} \int_{-\infty}^{+\infty} \tilde{g}(\omega_1, \omega_2) \; e^{2 \pi i ( \;\omega_1 x_1 \;+ \;\omega_2 x_2 \;) } \; d\omega_1 d\omega_2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The integral becomes&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$I = \int_{-\infty}^{+\infty} \int_{-\infty}^{+\infty} \; \left[ \int_{-\infty}^{+\infty} \int_{-\infty}^{+\infty} \tilde{f}(\omega_1, \omega_2) \; e^{2 \pi i ( \;\omega_1 x_1 \;+ \;\omega_2 x_2 \;) } \; d\omega_1 d\omega_2 \right] \; \left[ \int_{-\infty}^{+\infty} \int_{-\infty}^{+\infty} \tilde{g}(\omega_1, \omega_2) \; e^{2 \pi i ( \;\omega_1 x_1 \;+ \;\omega_2 x_2 \;) } \; d\omega_1 d\omega_2 \right] dx_1 dx_2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is where it starts to get confusing. I'm not sure if I should write&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$I = \int_{-\infty}^{+\infty} \int_{-\infty}^{+\infty} d\omega_1 \; \tilde{f}(\omega_1, \omega_2) \; \int_{-\infty}^{+\infty} \int_{-\infty}^{+\infty} d\omega_2 \tilde{g}(\omega_1, \omega_2) \; \; \int_{-\infty}^{+\infty} \; e^{2 \pi i x_1 ( \; \omega_1 \;+ \;\omega_2 \;) } \; dx_1 \; \int_{-\infty}^{+\infty} \; e^{2 \pi i x_2 ( \; \omega_1 \;+ \;\omega_2 \;) } \; dx_2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and extract two one-dimensional dirac deltas, or write&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$I = \int_{-\infty}^{+\infty} \int_{-\infty}^{+\infty} d\omega_1 \; \tilde{f}(\omega_1, \omega_2) \; \int_{-\infty}^{+\infty} \int_{-\infty}^{+\infty} d\omega_2 \tilde{g}(\omega_1, \omega_2) \; \;\int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty} e^{2 \pi i x_1 ( \; \omega_1 \;+ \;\omega_2 \;) + x_2 ( \; \omega_1 \;+ \;\omega_2 \;) } \; dx_1 \; dx_2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with the hope of extracting the two-dimensional dirac delta. (I'm not even sure what the inverse Fourier transform of the two dimensional dirac delta is).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd appreciate advise on how to proceed. (Thanks.)&lt;/p&gt;&#xA;" OwnerUserId="1752" LastActivityDate="2010-10-09T20:53:43.363" Title="Trying to derive two dimensional version of Parseval's theorem (for real valued functions)" Tags="&lt;fourier-analysis&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="4273" PostTypeId="1" AcceptedAnswerId="4377" CreationDate="2010-09-08T11:49:05.373" Score="1" ViewCount="100" Body="&lt;p&gt;Consider a situation where a person has partial knowledge, but we have a more complete picture. For example, suppose that we want to know the probability that a fish is red. Suppose that the person with partial information knows 1/3 of all fish are red, but we know that the particular species is actually red 2/3 of the time. Do these two separate probabilities have special names? If they don't have any standard names, what would you call them?&lt;/p&gt;&#xA;" OwnerUserId="123" LastEditorUserId="123" LastEditorDisplayName="" LastEditDate="2010-09-10T08:55:29.000" LastActivityDate="2010-09-10T14:21:50.153" Title="Terminology for handling probabilities with partial knowledge" Tags="&lt;terminology&gt;&lt;probability&gt;" AnswerCount="2" CommentCount="9" />
  <row Id="4274" PostTypeId="1" AcceptedAnswerId="4293" CreationDate="2010-09-08T12:08:34.743" Score="1" ViewCount="167" Body="&lt;p&gt;What is the area of the region bounded by $y = f_n(x)$ and the $x$-axis as $n$ gets large?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f_0=1-|x|$&lt;br&gt;&#xA;$f_n=1-|1-2f_{n-1}(x)|$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume the domain of $f$ to be the real numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By trying various values for $n$ and $x$  I got that $f_n(x) = 2^{n}\cdot f_0(x)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By integrating this I got that the area is given by $2^{n-1}\cdot x\cdot (2-|x|)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I was also told that the area is in fact $1$ as $n$ approaches infinity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The formula I came up with (guessed) seems to check out if I integrate $2^{n-1}\cdot f_0(x)$ between certain values and for a certain $n$ and compare it to manually expanding that certain $f_n$ and integrating it between the same values.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My questions are:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Is my formula wrong?&lt;/li&gt;&#xA;&lt;li&gt;How can we prove the area is 1?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="456" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2011-06-03T20:11:18.663" LastActivityDate="2011-06-03T20:11:18.663" Title="Area of a recursive function" Tags="&lt;calculus&gt;&lt;functions&gt;" AnswerCount="1" CommentCount="7" FavoriteCount="1" />
  <row Id="4275" PostTypeId="2" ParentId="4270" CreationDate="2010-09-08T12:38:19.897" Score="3" ViewCount="" Body="&lt;p&gt;A field K is an algebra over a field (the field being K). A commutative ring R is an algebra over a ring (the ring being R). &lt;/p&gt;&#xA;" OwnerUserId="1794" LastActivityDate="2010-09-08T12:38:19.897" CommentCount="1" />
  <row Id="4276" PostTypeId="1" AcceptedAnswerId="4277" CreationDate="2010-09-08T12:50:07.077" Score="2" ViewCount="402" Body="&lt;p&gt;We can divide $7^{17} - 7^{15}$ by?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer is $6$, but how?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance.&lt;/p&gt;&#xA;" OwnerUserId="1795" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-09-08T17:49:41.067" LastActivityDate="2010-09-08T21:20:09.317" Title="We can divide $7^{17} - 7^{15}$ by?" Tags="&lt;algebra-precalculus&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="4277" PostTypeId="2" ParentId="4276" CreationDate="2010-09-08T12:53:19.317" Score="14" ViewCount="" Body="&lt;p&gt;Note that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$7^{17}-7^{15}=(49-1)*7^{15}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and 48 is divisible by 6.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-09-08T12:53:19.317" CommentCount="23" />
  <row Id="4279" PostTypeId="1" AcceptedAnswerId="4289" CreationDate="2010-09-08T13:06:42.420" Score="2" ViewCount="179" Body="&lt;p&gt;How do I evaluate:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\lim_{\tau\rightarrow \infty} \int_{-c/\tau}^0 e^{-\beta x}(\tau x+c)^{(\lambda/\tau) -1}dx$$ to give $$1/\lambda$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suspect I need to convert to gamma function but cannot simplify it to get the answer.&lt;/p&gt;&#xA;" OwnerUserId="626" LastEditorUserId="56" LastEditorDisplayName="" LastEditDate="2010-09-10T16:13:16.397" LastActivityDate="2010-09-10T16:13:16.397" Title="evaluating $\lim_{\tau\rightarrow \infty} \int_{-c/\tau}^0 e^{-\beta x}(\tau x+c)^{(\lambda/\tau) -1}dx$" Tags="&lt;calculus&gt;&lt;integral&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="4280" PostTypeId="1" AcceptedAnswerId="4285" CreationDate="2010-09-08T13:13:17.363" Score="3" ViewCount="102" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm currently getting confused about indices in some spectral sequences. Assume we work in the category of modules for simplicity. Let $A^\cdot$ be a (bounded on the right) complex and let $B^\cdot$ (I don't think we have to assume anything about the boundedness of $B$). I want to compute $Ext^n(A^\cdot,B^\cdot)$, which is classically called hyperext (and sometimes denoted by $\mathbb{E}xt$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, (perhaps assuming $B^\cdot$ to be bounded on the right), there exists a spectral sequence&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$E^{p,q}_2 = Ext^p(A^\cdot,H^q(B^\cdot)) \Rightarrow Ext^{p+q}(A^\cdot,B^\cdot).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There should be an analogous by switching A and B, but I'm unsure of the indices, so my question is&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;is $$ E^{p,q}_2 = Ext^q(A^\cdot,H^{-p}(B^\cdot)) \Rightarrow Ext^{q-p}(A^\cdot,B^\cdot) $$&#xA;  the right thing?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="1007" LastEditorUserId="263" LastEditorDisplayName="" LastEditDate="2010-09-08T21:09:09.433" LastActivityDate="2010-09-08T21:09:09.433" Title="Contravariant Grothendieck Spectral Sequence" Tags="&lt;abstract-algebra&gt;&lt;homological-algebra&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="4281" PostTypeId="2" ParentId="4260" CreationDate="2010-09-08T13:14:19.080" Score="2" ViewCount="" Body="&lt;p&gt;The support of $M$ is the set $\{ \mathfrak q \in \text{ Spec } S \, | \, \kappa(q)\otimes_S M&#xA;\neq 0\}$, where $\kappa(\mathfrak q)$ denotes the fraction field of the domain $S/\mathfrak q$.  (Since $M$ is finitely generated, one sees by Nakayama's lemma that the stalk $M_{\mathfrak q}$ is non-zero if and only if $\kappa(\mathfrak q)\otimes_S M$ is non-zero).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The support of $R\otimes_S M$ is defined analogously.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now if $\mathfrak p$ is an element of Spec $R$, then&#xA;$$\kappa(\mathfrak p)\otimes_R (R\otimes _S M) = \kappa(\mathfrak p)\otimes_S M&#xA;= \kappa(\mathfrak p)\otimes_{\kappa(\mathfrak q)} (\kappa(\mathfrak q)\otimes_S M),$$&#xA;where $\mathfrak q$ is the preimage of $\mathfrak p$ in $S$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $\kappa(\mathfrak q) \to \kappa(\mathfrak p)$ is just an inclusion of fields,&#xA;we see that $\kappa(\mathfrak p)\otimes_R (R\otimes_S M) \neq 0$ if and only if&#xA;$\kappa(q)\otimes_S M \neq 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In conclusion, the support of $R\otimes_S M$ is the preimage in Spec $R$ of the support&#xA;of $M$ in Spec $S$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All of this has a geometric interpretation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first description of the support says that, when $M$ is finitely generated, we can&#xA;check that the stalks of $M$ are non-zero by instead checking if the fibres are non-zero.&#xA;Then the computation relating the fibre of $R\otimes_S M$ at $\mathfrak p$ to&#xA;the fibre of $M$ at $\mathfrak q$ just says that the fibre of the pull-back is the pull-back of the fibre.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, note that if you think geometrically, the result makes intuitive sense:&#xA;we are pulling back a sheaf (i.e. applying $\varphi^*$), and we can think of the sheaf as being a subset of&#xA;Spec $R$ (namely its support) with extra decoration (i.e. each point has a certain fibre&#xA;of the sheaf lying over it).  When we pull back the sheaf, we pull back the subset &#xA;(i.e. we pull back the support), and then at each point of the pull back we also pull back&#xA;the extra decoration.  Thinking about supports just involves forgetting the extra decoration.&lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2010-09-08T13:14:19.080" CommentCount="1" />
  <row Id="4282" PostTypeId="2" ParentId="4266" CreationDate="2010-09-08T13:22:55.863" Score="3" ViewCount="" Body="&lt;p&gt;What you are looking for is the concept of &lt;a href=&quot;http://en.wikipedia.org/wiki/Aronszajn_tree&quot; rel=&quot;nofollow&quot;&gt;Aronszajn tree&lt;/a&gt;. You can read about constructions of Aronszajn trees in any graduate level set theory text, and meanwhile, the Wikipedia page lists a summary of the basic facts:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;König's lemma states that $\aleph_0$-Aronszajn trees do not exist.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The existence of Aronszajn trees ($=\aleph_1$-Aronszajn trees) was proven by Nachman Aronszajn, and implies that the analogue of König's lemma does not hold for uncountable trees.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The existence of $\aleph_2$-Aronszajn trees is undecidable (assuming a certain large cardinal axiom): more precisely, the continuum hypothesis implies the existence of an $\aleph_2$-Aronszajn tree, and Mitchell and Silver showed that it is consistent (relative to the existence of a weakly compact cardinal) that no $\aleph_2$-Aronszajn trees exist.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Jensen proved that $V=L$ implies that there is a κ-Aronszajn tree (in fact a κ-Suslin tree) for every infinite successor cardinal κ.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Cummings &amp;amp; Foreman (1998) showed (using a large cardinal axiom) that it is consistent that no $\aleph_n$-Aronszajn trees exist for any finite n other than 1.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If κ is weakly compact then no κ-Aronszajn trees exist. Conversely if κ is inaccessible and no κ-Aronszajn trees exist then κ is weakly compact.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Finally, let me point out a small inaccuracy in your question. The equivalence is that $\kappa$ is weakly compact iff it is inaccessible and has the tree property. It is not correct to drop the inaccessibility part, as you did, since it is consistent that $\aleph_2$ has the tree property.&lt;/p&gt;&#xA;" OwnerUserId="413" LastEditorUserId="413" LastEditorDisplayName="" LastEditDate="2010-09-08T13:35:06.503" LastActivityDate="2010-09-08T13:35:06.503" />
  <row Id="4283" PostTypeId="2" ParentId="4175" CreationDate="2010-09-08T13:26:54.670" Score="2" ViewCount="" Body="&lt;p&gt;The given quantity is the constant term in $ (-1)^m  \times { \sum_{ k \geq 0 } (-1)^{k} \binom{n}{k} (\frac{1}{x})^k } \times \sum_{k \geq 0} \binom{k}{m} x^k $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or the constant term in $(-1)^m \times (1-\frac{1}{x})^n \times x^{m} \times (1-x)^{-(m+1)}$ =  $(-1)^{m+n} x^{m-n} \times (1-x)^{n-m-1} $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $m &gt;n$ clearly the constant term is 0, if $m &amp;lt; n$ then writing the above as $(-1)^{m+n} \frac{(1-x)^{n-m-1}}{x^{n-m}}$ and noting the maximum exponent of $x$ in numerator is $n-m-1$ we again see the constant term is 0. If $n=m$ then the constant term is clearly 1.&lt;/p&gt;&#xA;" OwnerUserId="404" LastActivityDate="2010-09-08T13:26:54.670" />
  <row Id="4284" PostTypeId="1" CreationDate="2010-09-08T13:30:38.307" Score="4" ViewCount="156" Body="&lt;p&gt;On a lark, I wanted to know if one can use &lt;a href=&quot;http://mathworld.wolfram.com/PadeApproximant.html&quot; rel=&quot;nofollow&quot;&gt;Padé approximants&lt;/a&gt; to compute the exponential $\exp(z)$ of a quaternion $z=a+b\mathbf{i}+c\mathbf{j}+d\mathbf{k}$. Since &lt;em&gt;Mathematica&lt;/em&gt; has a package meant for manipulating quaternions, as well as computing the exponential of a quaternion, I decided to try things out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The numerator $N_{pq}(z)$ and denominator $D_{pq}(z)$ of the $(p,q)$ Padé approximant for the exponential function can be computed using the following formulae:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle N_{pq}(z)=\sum_{j=0}^p \frac{(p+q-j)!p!}{(p+q)!j!(p-j)!}z^j$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle D_{pq}(z)=\sum_{j=0}^q \frac{(p+q-j)!q!}{(p+q)!j!(q-j)!}(-z)^j$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;after which the Padé approximant is $R_{pq}(z)=\frac{N_{pq}(z)}{D_{pq}(z)}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For simplicity (and since they are the ones most used in practice), I will only consider here the &quot;diagonal&quot; case, $p=q$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The case where $z$ is a matrix is well-studied, and it is known that $R_{pq}(z)=N_{pq}(z)\cdot D_{pq}(z)^{-1}=D_{pq}(z)^{-1}\cdot N_{pq}(z)$. Banking on the isomorphism of quaternions to specially constructed matrices, I expected that the order of multiplication won't matter in the quaternion case, and experimentation seems to show that this holds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What was mystifying in my experiments is that, although in general $|R_{pp}(z)-\exp(z)|$ can be made small enough by taking $p$ large enough, one pattern I saw seems to indicate that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$R_{pp}(a-b\mathbf{i}+c\mathbf{j}+d\mathbf{k})\approx \overline{\exp(a-b\mathbf{i}+c\mathbf{j}+d\mathbf{k})}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for $b&gt;0$, where the bar denotes conjugation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can't see an immediate equivalent of this phenomenon in the theory of the matrix exponential, so I'm wondering why the Padé approximants to the quaternion exponential behave this way.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;For those who want to retry my &lt;em&gt;Mathematica&lt;/em&gt; experiments:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;&amp;lt;&amp;lt;Algebra`Quaternions`&#xA;&#xA;z=Quaternion[1.,-1.,1.,1.](*Quaternion[2 Random[] - 1, 2 Random[] - 1, 2 Random[] - 1, 2 Random[] - 1]*)&#xA;&#xA;p=q=8;&#xA;&#xA;num=Sum[(((p+q-j)!p!)/((p+q)!j!(p-j)!)) z^j, {j, 0, p}];&#xA;&#xA;den=Sum[(((p+q-j)!q!)/((p+q)!j!(q-j)!)) (-z)^j, {j, 0, q}];&#xA;&#xA;{num**(den^-1),(den^-1)**num,Exp[z],Conjugate[Exp[z]]}&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="498" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-09-08T21:51:06.697" LastActivityDate="2010-09-08T21:51:06.697" Title="Using Padé approximants for the quaternion exponential" Tags="&lt;approximation&gt;&lt;quaternions&gt;" />
  <row Id="4285" PostTypeId="2" ParentId="4280" CreationDate="2010-09-08T13:33:18.003" Score="2" ViewCount="" Body="&lt;p&gt;I think that you will want $B^{\bullet}$ to be bounded below (i.e. on the left), so that&#xA;after replacing either $A^{\bullet}$ by a projective resolution or $B^{\bullet}$ by an&#xA;injective resolution, the complex $Hom(\text{complex},\text{complex})$ that you compute&#xA;will be bounded below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In any case, the spectral sequence you are looking for comes about by taking this  $Hom$&#xA;complex (the one you get after replacing $A^{\bullet}$ or $B^{\bullet}$ by a projective&#xA;or injective resolution), writing it as the total complex of a double complex, and applying the standard&#xA;machine.  If you do this, you will find that the second spectral sequence that you want is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$E_2^{p,q} = Ext^p(H^{-q}(A^{\bullet}),B^{\bullet}) \Rightarrow Ext^{p+q}(A^{\bullet},B^{\bullet}).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Your first spectral sequence comes from replacing $A^{\bullet}$ by a projective resolution.&#xA;This one comes from replacing $B^{\bullet}$ by an injective resolution.)&lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2010-09-08T13:33:18.003" />
  <row Id="4286" PostTypeId="1" AcceptedAnswerId="4290" CreationDate="2010-09-08T13:58:32.343" Score="1" ViewCount="149" Body="&lt;p&gt;Suppose I have a 4-index, real-valued tensor $\lambda_{ij}^{ab}$ where $i$ and $j$ are from some subspace $O$ of the whole numbers, and $a$ and $b$ are from some other subspace $V,$ which is disjunct from $O.$  Suppose also that we know that $\lambda$ is antisymmetric.  If we impose the constraint that $\lambda$ is Hermitian (i.e. $\lambda_{ij}^{ab} = (\lambda_{ab}^{ij})^*=\lambda_{ab}^{ij}$ since $\lambda$ is real), and we have some function:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$E = \sum_{ij\in O, ab \in V} k_{ab}^{ij}\lambda_{ij}^{ab} + \sum_{ij\in O, ab \in V} k_{ij}^{ab}\lambda_{ab}^{ij}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(where $k_{ij}^{ab}$ is some antisymmetric, Hermitian tensor independent of $\lambda$) and then we take the partial derivative of that function with respect to some specific $\lambda_{a^\prime b^\prime}^{i^\prime j^\prime}$, what do we get?  What I don't understand is that if you make the substitution for the Hermitian constraint before differentiating, you can combine the two summations into one, but then differentiating gives you the same answer with a different constant, right?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer that the derivation I am reading gives is $k_{ij}^{ab}$, rather than $2k_{ij}^{ab}$.  Could someone explain why?&lt;/p&gt;&#xA;" OwnerUserId="1590" LastActivityDate="2010-09-08T14:56:33.960" Title="Partial Derivatives and Dependent Variables" Tags="&lt;calculus&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="4287" PostTypeId="2" ParentId="3946" CreationDate="2010-09-08T14:24:04.060" Score="1" ViewCount="" Body="&lt;p&gt;What exactly do you want?&#xA;If you want step-by-step software algorithm to draw a Bezier curve,&#xA;and the &lt;a href=&quot;http://www.tinaja.com/cubic01.asp&quot; rel=&quot;nofollow&quot;&gt;&quot;Don Lancaster's Guru's Lair Cubic Spline Library&quot;&lt;/a&gt; doesn't have the answer,&#xA;then the StackOverflow.com website is a better place to ask such questions.&lt;/p&gt;&#xA;" OwnerUserId="1705" LastActivityDate="2010-09-08T14:24:04.060" />
  <row Id="4288" PostTypeId="2" ParentId="4279" CreationDate="2010-09-08T14:34:57.997" Score="3" ViewCount="" Body="&lt;p&gt;Try the subsitution $s=\beta(x+\frac{c}{\tau})$. Then use: &#xA;\begin{equation}&#xA;\int_0^a t^{b-1}e^{-t}dt=a^be^{-a}\sum_0^{\infty}\frac{a^n}{b(b+1)\ldots(b+n)}&#xA;\end{equation}&lt;/p&gt;&#xA;" OwnerUserId="1794" LastEditorUserId="1794" LastEditorDisplayName="" LastEditDate="2010-09-08T14:43:15.930" LastActivityDate="2010-09-08T14:43:15.930" CommentCount="1" />
  <row Id="4289" PostTypeId="2" ParentId="4279" CreationDate="2010-09-08T14:42:14.090" Score="3" ViewCount="" Body="&lt;p&gt;A sketch of the proof, as I'm in a hurry - I hope I've done all the calculations correctly:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Set $y = \tau x + c$ to obtain&#xA;$$&#xA;\frac{1}{\tau} \int_0^c e^{\frac{\beta}{\tau} (c - y)} y^{\lambda / \tau - 1} d y&#xA;$$&#xA;Now, partial integration yields&#xA;$$&#xA;\frac{1}{\lambda} c^{\lambda / \tau} + \frac{\beta}{\lambda \tau} \int_0^c e^{\frac{\beta}{\tau} (c - y)} y^{\lambda / \tau} dy&#xA;$$&#xA;The limit of the first summand is $\frac{1}{\lambda}$, and the integrand in the second part is bounded (maximum is obtained for $y = \frac{\lambda}{\beta}$), so the second summand will vanish when $\tau \to \infty$.&lt;/p&gt;&#xA;" OwnerUserId="960" LastActivityDate="2010-09-08T14:42:14.090" CommentCount="1" />
  <row Id="4290" PostTypeId="2" ParentId="4286" CreationDate="2010-09-08T14:56:33.960" Score="1" ViewCount="" Body="&lt;p&gt;It turns out that $2k_{ij}^{ab}$ is, in fact, the correct answer; it's just that when the derivation was done, the next step was to set the derivative equal to zero, which allows you to divide through by 2, and so the original deriver still got the right answer.  Sorry for the confusion!&lt;/p&gt;&#xA;" OwnerUserId="1590" LastActivityDate="2010-09-08T14:56:33.960" />
  <row Id="4291" PostTypeId="2" ParentId="4273" CreationDate="2010-09-08T15:26:49.897" Score="4" ViewCount="" Body="&lt;p&gt;I would call the second of these a &lt;em&gt;conditional&lt;/em&gt; probability when trying to emphasise the difference:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The probability of a chosen fish being red is 1/3.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The conditional probability of a fish being red, given that it is species X, is 2/3.&lt;/p&gt;&#xA;" OwnerUserId="1750" LastActivityDate="2010-09-08T15:26:49.897" CommentCount="2" />
  <row Id="4292" PostTypeId="2" ParentId="4253" CreationDate="2010-09-08T15:29:13.577" Score="2" ViewCount="" Body="&lt;p&gt;Here is an answer for your first question. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is easier than dealing with simple symmetric random walk.&#xA;When the steps are diffuse,&#xA;you don't have to worry about &lt;em&gt;ties&lt;/em&gt; or &lt;em&gt;parity&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are looking at a Brownian motion, evaluated at integer times $t$.&#xA;Then defining  $$T_0=\inf(t &gt; 0, t\in {\mathbb Z}: B_t &amp;lt; 0)$$ and integer $t &gt; 0$ &#xA;we have $$ P(T_0 &gt; t)=P(B_1 &gt; 0,B_2 &gt; 0,\dots, B_t &gt; 0)={2t\choose t}/4^t.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It turns out that the answer is the same for any random walk with&#xA;symmetric, diffuse increments.&#xA;This follows from Theorem 9.11 of Kallenberg's &lt;em&gt;Foundations of Modern Probability (2nd edition)&lt;/em&gt;, for example.&#xA;(A great book, by the way!)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The key insight is that you should not to try to solve the&#xA;probability exactly, but rather set up a recursion that the probabilities solve.&#xA;Setting $p(t)=P(T_0&gt;t)$ for integer $t\geq 0$, you can show that&#xA;for any $t$ we have $\sum_{j=0}^t p(t-j)p(j)=1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The probabilistic justification for the recursion formula is that&#xA;every path of a diffuse, symmetric, discrete-time random walk can be broken into two&#xA;&lt;em&gt;positive&lt;/em&gt; paths at the time $j$  where it achieves its minimum value. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You may also find Konrad Jacob's book &lt;em&gt;Discrete Stochastics&lt;/em&gt; to be useful. &lt;/p&gt;&#xA;" OwnerUserId="940" LastActivityDate="2010-09-08T15:29:13.577" />
  <row Id="4293" PostTypeId="2" ParentId="4274" CreationDate="2010-09-08T15:36:23.693" Score="4" ViewCount="" Body="&lt;p&gt;You must be integrating between -1 and 1; otherwise, the integral diverges.  Within this region, $f_n$ is a periodic sawtooth.  This is obvious by following the steps geometrically for any one of the teeth: $-2f_{n-1}$ doubles the height of $f_{n-1}$ and flips the graph over the x-axis, which is then raised by $1$.  The absolute value creates a new &quot;tooth&quot;, and then subtracting all that from $1$ flips everything over.  The process converts each tooth into a pair of teeth of the same height but half the width, whence the new teeth have half the original area of the teeth they have replaced.  Two of them occupy the original area.  Therefore, by induction, the integrals of &lt;em&gt;all&lt;/em&gt; the $f_n$ between -1 and 1 equal the integral of $f_0$, which is 1.&lt;/p&gt;&#xA;" OwnerUserId="1489" LastActivityDate="2010-09-08T15:36:23.693" CommentCount="4" />
  <row Id="4294" PostTypeId="2" ParentId="3973" CreationDate="2010-09-08T15:42:47.733" Score="4" ViewCount="" Body="&lt;p&gt;Apparently, giving the input&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;Integrate[Exp[Piecewise[{{x, x &amp;gt; 0}}, -x]],x]&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;to WolframAlpha returns a piecewise result, which may be what you're expecting.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-09-08T15:42:47.733" />
  <row Id="4295" PostTypeId="2" ParentId="4261" CreationDate="2010-09-08T15:58:14.510" Score="1" ViewCount="" Body="&lt;p&gt;There are a powerful &lt;strong&gt;algorithms generalizing telescopy&lt;/strong&gt; (Gosper, Zeilberger et al.) that easily tackle this case of the &lt;strong&gt;Chu-Vandermonde identity&lt;/strong&gt; and much more complicated sums. For example, see &lt;a href=&quot;http://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.71.3348&quot; rel=&quot;nofollow&quot;&gt;this paper&lt;/a&gt; which gives as an application a very interesting q-analogy - namely that L. J. Rogers' classical finite version of &lt;strong&gt;Euler's pentagonal number theorem&lt;/strong&gt; is simply the dual of a special case of a q-Chu-Vandermonde.&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-09-08T15:58:14.510" />
  <row Id="4296" PostTypeId="2" ParentId="4270" CreationDate="2010-09-08T16:09:04.717" Score="2" ViewCount="" Body="&lt;p&gt;[This should be a comment, I guess]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the kind of question you can answer yourself by simply checking the conditions in the definition are satisfied. You &lt;em&gt;really&lt;/em&gt; should do it, independently of what anyone answers here or elsewhere!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You should know, though, that a field can be an algebra over itself in many different ways, as soon as you have a field $K$ such that there are at least two ring homomorphisms $K\to K$ (there is always one such ring homomorphism, the identity map, so to get something interesting, we need two). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is easy to give examples of fields which do have such non-identity homomorphisms. For example, every Galois extension of $\mathbb Q$, or the field of rational functions $\mathbb C(x)$ (there is a non-identity homomorphism $\mathbb C(x)\to \mathbb C(x)$ which maps $x$ to $x^2$, and in fact many similar other ones), &amp;amp;c.&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2010-09-08T16:09:04.717" CommentCount="2" />
  <row Id="4297" PostTypeId="1" CreationDate="2010-09-08T16:26:44.983" Score="0" ViewCount="174" Body="&lt;p&gt;Let $n \in \mathbb{N}$. How to find the least positive integer $\gamma$ such that $$ \sum\limits_{i=1}^{n} \cos{\theta_{n}} \leq \gamma$$ provided $$\prod\limits_{i=1}^{n} \tan{\theta_{i}} = 2^{\frac{n}{2}}$$ for any $\theta_{i} \in \Bigl(0,\frac{\pi}{2}\Bigr)$, $i=1,2,...,n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I had trouble solving this problem, some time ago. If i am not mistaken, i had seen this in Loney's trigonometry book. (Although i don't remember it correctly!).&lt;/p&gt;&#xA;" LastActivityDate="2010-09-08T20:34:11.083" Title="Finding the least postive integer such that $\sum \cos{\theta_{i}} \leq \gamma$" Tags="&lt;trigonometry&gt;" AnswerCount="1" CommentCount="0" />
  <row Id="4298" PostTypeId="1" AcceptedAnswerId="4299" CreationDate="2010-09-08T16:43:58.943" Score="1" ViewCount="213" Body="&lt;p&gt;Suppose $K$ is the splitting field of $x^3-2$ over $\mathbb{Q}$, and let $\mathcal{O}_K$ denote its ring of integers.  I want to show that for any prime number $p$, $p\mathcal{O}_K$ is not a prime ideal of $\mathcal{O}_K$. The hint I was given was to consider the factorization of $x^3-2$ over $F_p$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My question(s)&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My first instinct was to use Kummer's criterion, which guarantees that $p\mathcal{O}_K$ is not prime as long as $x^3-2$ is reducible (has a root) in $F_p$ [and $p&gt;3$].  But how to handle the case when $p=2,3$ (I suspect it ramifies, but I can't seem to show this).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My second instinct was that if $p\mathcal{O}_K$ is prime in $\mathcal{O}_K$, then $\mathcal{O}_K/p\mathcal{O}_K$ would be the splitting field of $x^3-2$ over $F_p$, and so I just need to check that the degree is always less than $[K:\mathbb{Q}]=6$.  But is it always true in this case that $\mathcal{O}_K/p\mathcal{O}_K$ is the splitting field of $x^3-2$ over $F_p$?&lt;/p&gt;&#xA;" OwnerUserId="1544" LastActivityDate="2010-09-09T03:44:18.263" Title="Factoring prime ideals in a Galois extension of Q" Tags="&lt;algebraic-number-theory&gt;" AnswerCount="2" />
  <row Id="4299" PostTypeId="2" ParentId="4298" CreationDate="2010-09-08T16:50:38.300" Score="2" ViewCount="" Body="&lt;p&gt;Let's work inside $L=\mathbb{Q}(\sqrt[3]2)$.&#xA;It really should be easy to see that $2$ ramifies (there an ideal $\mathfrak{p}$&#xA;with $\mathfrak{p}^3=2\mathfrak{O}_L$). Proving that $3$ ramifies is a little trickier.&#xA;If we could find an element of $\mathfrak{O}_L$ whose minimal polynomial is Eisenstein&#xA;at $3$, that would do. An Eisenstein polynomial at $3$ would be&#xA;$$x^3+ax^2+bx+c$$&#xA;where $a$, $b$ and $c$ are divisible by $3$ and $c$ isn't divisible by $9$.&#xA;Obviously $\sqrt[3]2$ doesn't work - but is there something similar which does?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added&lt;/strong&gt;&#xA;I was implicitly arguing that since the primes $2$ and $3$ aren't inert in&#xA;$L=\mathbb{Q}(\sqrt[3]2)$ then &lt;em&gt;a fortiori&lt;/em&gt; they aren't inert in its Galois&#xA;closure $K$. But no prime remains inert in a non-cyclic Galois extension&#xA;of a number field. For a prime to remain inert, its decomposition group&#xA;is the whole Galois group and its inertia group is trivial. But the&#xA;quotient of the decomposition group by the inertia group must be cyclic&#xA;(its the Galois group of the extension of residue fields)....&lt;/p&gt;&#xA;" OwnerUserId="226" LastEditorUserId="226" LastEditorDisplayName="" LastEditDate="2010-09-08T18:43:29.873" LastActivityDate="2010-09-08T18:43:29.873" CommentCount="5" />
  <row Id="4300" PostTypeId="2" ParentId="4270" CreationDate="2010-09-08T17:04:15.667" Score="0" ViewCount="" Body="&lt;p&gt;The answer is clear if you know the universal view of $\rm\:R\:$-algebras, namely they are simply those rings $\rm\:A\:$ into which one can always evaluate $\rm R[x]$ for every $\rm\:a \in A\:$. More directly,  an $\rm\:R$-algebra is just a ring $\rm A\:$ containing a central subring $\rm R'$ that's a ring image of $\:\rm R\:,\:$ i.e. $\rm\: R'\:$ is either an embedding of $\rm\:R\:$ or $\rm\:R/I\:$ for some ideal $\rm\;I\in R\:.\;$ Being central is precisely the condition needed for elts in $\rm\:R'\:$ to serve as &quot;coefficients&quot; in the sense that this makes polynomial rings $\rm\: R[x]\:$ be a &lt;strong&gt;universal&lt;/strong&gt; $\rm\:R\:$-algebras. Namely, the fact that the coefficients commute with all elts of $\rm\:A\:$ is precisely what is required to make the evaluation map be a ring homomorphism $\rm\: R[x]\to A\:,\:$&#xA;viz $\rm\;\; r\; x = x\: r\;$ in $\rm\:R[x]\;\Rightarrow\;  r\: a = a\:r\;\;$ by evaluation $\rm\: x\to a\in A,\:$ i.e. by definition polynomial multiplication assumes that the coefficients commute with the indeterminates, so this property must remain true at values of indeterminates if evaluation is to be a ring homomorphism.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-09-08T18:24:30.387" LastActivityDate="2010-09-08T18:24:30.387" />
  <row Id="4301" PostTypeId="2" ParentId="4298" CreationDate="2010-09-08T17:07:08.013" Score="1" ViewCount="" Body="&lt;p&gt;In general, if the ring of integers of $\mathcal{O}_K$ is of the form $\mathbb{Z}[\alpha]$, and $q(x)$ is the irreducible polynomial of $\alpha$ over $\mathbb{Q}$, then there is a natural bijection between the primes in $\mathcal{O}_K$ that lie above the rational prime $p$ and the irreducible factors of $\overline{q}(x)$ in $\mathbb{F}_p$, where $\overline{q}$ is the image of $q(x)$ under reduction modulo $p$. The bijection is that a prime $\mathfrak{p}$ of $\mathcal{O}_K$ lies above $(p)$ corresponds to the factor $\overline{Q}(x)$ if and only if $\mathfrak{p}$ is the kernel of the map $\mathbb{Z}[\alpha]\to\mathbb{F}_p[\alpha]$. So the factorization of $(p)$ in $\mathcal{O}_K$ exactly mimicks the factorization of $\overline{q}(x)$ over $\mathbb{F}_p$: same number of factors, same repetitions, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you want to factor $x^3-2$ over $\mathbb{F}_p$; this suffices here because if $\alpha$ is a root of $x^3-2$, then $\mathcal{O}_K = \mathbb{Z}[\alpha]$ (see for example Problem 41(a)-(d) in Chapter 2 of Marcus's &lt;em&gt;Number Fields&lt;/em&gt;). &lt;strong&gt;Edit:&lt;/strong&gt; I was working in $\mathbb{Q}(\sqrt[3]{2})$ instead of the splitting field; sorry. But from the fact that they ramify in this subextension, you know they will ramify in $K$; the only question is whether they ramify as $\mathfrak{p}^6$ or as $\mathfrak{p_1}^3\mathfrak{p_2}^3$, depending on whether the prime above $p$ splits or ramifies, or is inert in $K$. Since you can go from $$\mathbb{Q}(\sqrt[3]{2})$ to $K$ by adjoining $\omega$, and the ring of integers will be, if I'm not mistaken, $\mathbb{Z}[\alpha][i]$, you can use the same theorem above (which holds for arbitrary integral extensions, not just over $\mathbb{Z}$) to get that next step, looking at $x^2+x+1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With $p=2$, the polynomial factors over $\mathbb{F}_p$ as $x^3$, so $(2) = \mathfrak{p}^3$ for some prime $\mathfrak{p}$ of $\mathcal{O}_K$; for $p=3$, you have $x^3 + 2 = x^3 - 1 = (x-1)^3$, so again $(3)=\mathfrak{q}^3$ for some prime $\mathfrak{q}$ of $\mathcal{O}_K$. &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-09-09T03:44:18.263" LastActivityDate="2010-09-09T03:44:18.263" CommentCount="3" />
  <row Id="4302" PostTypeId="2" ParentId="4276" CreationDate="2010-09-08T17:35:22.310" Score="7" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\rm\quad X^{n+2} - X^n \;=\; (X^2 - 1)\: X^n \;=\; (X-1)\: (X+1)\: X^n$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Often number identities are more perceptively viewed as special cases of function or polynomial identities. For example, &lt;a href=&quot;http://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.114.548&quot; rel=&quot;nofollow&quot;&gt;Aurifeuille, Le Lasseur and Lucas&lt;/a&gt; discovered so-called &lt;a href=&quot;http://www.ams.org/journals/mcom/2006-75-253/S0025-5718-05-01766-7/S0025-5718-05-01766-7.pdf&quot; rel=&quot;nofollow&quot;&gt;&lt;strong&gt;Aurifeuillian factorizations&lt;/strong&gt;&lt;/a&gt; of cyclotomic polynomials $\rm\;\Phi_n(x) = C_n(x)^2 - n\ x\  D_n(x)^2\;$. These play a role in factoring numbers of the form $\rm\; b^n \pm 1\:$, cf. the &lt;a href=&quot;http://homes.cerias.purdue.edu/~ssw/cun/&quot; rel=&quot;nofollow&quot;&gt;Cunningham Project&lt;/a&gt;. Below are some simple examples of such factorizations:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{array}{rl}&#xA;x^4 + 2^2 \quad=&amp;amp;  (x^2 + 2x + 2)\;(x^2 - 2x + 2) \\&#xA;\frac{x^6 + 3^2}{x^2 + 3} \quad=&amp;amp;  (x^2 + 3x + 3)\;(x^2 - 3x + 3) \\&#xA;\frac{x^{10} - 5^5}{x^2 - 5} \quad=&amp;amp;  (x^4 + 5x^3 + 15x^2 + 25x + 25)\;(x^4 - 5x^3 + 15x^2 - 25x + 25) \\&#xA;\frac{x^{12} + 6^6}{x^4 + 36} \quad=&amp;amp;  (x^4 + 6x^3 + 18x^2 + 36x + 36)\;(x^4 - 6x^3 + 18x^2 - 36x + 36) \\&#xA;\end{array}$$&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-09-08T21:20:09.317" LastActivityDate="2010-09-08T21:20:09.317" CommentCount="3" />
  <row Id="4303" PostTypeId="2" ParentId="4256" CreationDate="2010-09-08T18:03:16.837" Score="0" ViewCount="" Body="&lt;p&gt;For question 2, consider any f(n) which is 2^n for n which are not powers of 2. This lets us devide our sum into two parts, both of which converge.&lt;/p&gt;&#xA;" OwnerUserId="796" LastActivityDate="2010-09-08T18:03:16.837" />
  <row Id="4304" PostTypeId="1" AcceptedAnswerId="4331" CreationDate="2010-09-08T18:33:42.550" Score="1" ViewCount="453" Body="&lt;p&gt;I am trying to understand &lt;a href=&quot;http://en.wikipedia.org/wiki/All-pairs_testing&quot; rel=&quot;nofollow&quot;&gt;&lt;strong&gt;pairwise testing&lt;/strong&gt;&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How many combinations of tests would be there for example, if&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;code&gt;a&lt;/code&gt; can take values from 1 to m&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;code&gt;b&lt;/code&gt; can take values from 1 to n&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;code&gt;c&lt;/code&gt; can take values from 1 to p&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;a&lt;/code&gt;, &lt;code&gt;b&lt;/code&gt; and &lt;code&gt;c&lt;/code&gt; can take m, n and p distinct values respectively. &lt;strong&gt;What are the total number of pairwise combinations possible?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;With a pairwise testing tool that I am testing, I am getting 40 results for m = n = p = 6. I am trying to mathematically understand how I get 40 values. &lt;/p&gt;&#xA;" OwnerUserId="324" LastEditorUserId="324" LastEditorDisplayName="" LastEditDate="2010-09-08T18:53:12.003" LastActivityDate="2010-09-09T11:51:05.527" Title="How to derive the number of pairwise combinations of a set of factors?" Tags="&lt;calculus&gt;&lt;linear-algebra&gt;&lt;combinatorics&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="4305" PostTypeId="2" ParentId="4272" CreationDate="2010-09-08T18:33:58.773" Score="1" ViewCount="" Body="&lt;p&gt;I don't see what you're doing. Are you trying to show that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_{\mathbb R^2} f(x) g(x) \, dx = \int_{\mathbb R^2} \widehat{f(\xi)} \overline{\widehat{g(\xi)}} \, d\xi$$?&#xA;This is usually called Parseval's identity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can prove it as follows (even in $\mathbb R^n$), first prove by Fubini (very simple) that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_{\mathbb R^2} f(x) \widehat{g(x)} \, dx = \int_{\mathbb R^2} \widehat{g(x)} f(x) \, dx$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then let $g = \overline{\hat{h}}$ and note then that $\hat{g} = \overline{h}$ (by Fourier inversion), then we get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_{\mathbb R^2} f(x) h(x) \, dx = \int_{\mathbb R^2} \widehat{g(x)} \overline{\widehat{h(x)}} \, dx$$&#xA;(assuming $h$ is real, otherwise we get the conjugate).&lt;/p&gt;&#xA;" OwnerUserId="1120" LastEditorUserId="1120" LastEditorDisplayName="" LastEditDate="2010-09-09T19:56:42.063" LastActivityDate="2010-09-09T19:56:42.063" CommentCount="3" />
  <row Id="4306" PostTypeId="2" ParentId="4304" CreationDate="2010-09-08T18:45:30.167" Score="1" ViewCount="" Body="&lt;p&gt;If each parameter had $10$ choices you'd be testing $300$ vs $1000$ combinations, namely hold $\rm a$ constant and vary $\rm b,c$ through $10\cdot 10 = 100$ values. Similarly hold, $\rm b$ constant; then $\rm c$. As the number of variables $\rm k$ increases you get better savings, roughly $\rm (k N)^2$ vs. $\rm N^k$, where $\rm N =$ max domain size. For QA purposes usually such rough upper bounds suffice. Do you have an intended application where you need something more precise? If so perhaps you should reveal some further details, e.g. the distribution of the sizes of the domains, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: After reviewing your latest revision, it appears that the following web pages may be of interest: &lt;a href=&quot;http://www.developsense.com/pairwiseTesting.html&quot; rel=&quot;nofollow&quot;&gt;Pairwise Testing&lt;/a&gt;, which refers to various &lt;a href=&quot;http://en.wikipedia.org/wiki/Taguchi_methods&quot; rel=&quot;nofollow&quot;&gt;Taguchi methods&lt;/a&gt; such as those &lt;a href=&quot;http://www.freequality.org/html/tools.html&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;. See also these links to introductions to &lt;a href=&quot;http://www.combinatorialtesting.com/clear-introductions-1&quot; rel=&quot;nofollow&quot;&gt;combinatorial testing&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-09-09T02:14:11.153" LastActivityDate="2010-09-09T02:14:11.153" />
  <row Id="4308" PostTypeId="2" ParentId="4297" CreationDate="2010-09-08T19:20:14.317" Score="3" ViewCount="" Body="&lt;p&gt;Try this. Find the maximum of $$\sum_{k=1}^n \frac{1}{\sqrt{1+{x_k}^2}},$$&#xA;where $\prod_{k=1}^n {x_k}^2=2^n,$ and $x_k = \tan\theta_k$, using Lagrange multipliers. Each $x_k$ satisfies the same equation involving the multiplier $\lambda,$&#xA;$$\lambda 2^{n+1} = \frac{{x_k}^2}{(1+{x_k}^2)^{3/2}}.$$&#xA;And so all the $x_k$ are equal and hence $x_k = \sqrt{2}.$&#xA;Therefore the maximum value of our sum is $n/\sqrt{3}$ and so the required smallest integer is&#xA;$$\lceil \frac{n}{\sqrt{3}} \rceil.$$&lt;/p&gt;&#xA;" OwnerUserId="1301" LastEditorUserId="1301" LastEditorDisplayName="" LastEditDate="2010-09-08T20:34:11.083" LastActivityDate="2010-09-08T20:34:11.083" CommentCount="2" />
  <row Id="4312" PostTypeId="1" AcceptedAnswerId="5484" CreationDate="2010-09-08T20:18:51.057" Score="4" ViewCount="228" Body="&lt;p&gt;Given a sequence of row sums $r_1, \ldots, r_m$ and column sums $c_1, \ldots, c_n$, all positive, I'd like to find a matrix $A_{m\times n}$ consistent with the given row and column sums that has the fewest nonzero entries. That is, we want to minimize $|\{a_{ij} \ne 0\}|$ under the constraint that $\sum_j a_{ij} = r_i$ and $\sum_i a_{ij} = c_j$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One can also think of this as a network flow problem, with $m$ sources and $n$ sinks that must be connected with the minimum number of edges to achieve the specified flow out of each source ($r_i$) or into each sink ($c_j$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this a known problem, or does it fit into a well-known class of problems? What is the computational complexity of finding the optimal solution? According to the comments on &lt;a href=&quot;http://www.reddit.com/r/math/comments/darwh/accounting_help_with_a_dash_of_linear_algebra/&quot; rel=&quot;nofollow&quot;&gt;this Reddit discussion&lt;/a&gt;, it could be related to the &lt;a href=&quot;http://en.wikipedia.org/wiki/Cutting_stock_problem&quot; rel=&quot;nofollow&quot;&gt;cutting stock problem&lt;/a&gt;, but it's not exactly the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The motivation for this problem comes from settling debts. Suppose you have a set of people $P$ with debts between them. These debts can be settled by (1) finding the net debt $d(p)$ owed to each person $p \in P$, (2) dividing $P$ into two disjoint sets of people who are owed, $P_+ = \{p : d(p) &gt; 0\}$, and people who owe, $P_- = \{p : d(p) &amp;lt; 0\}$, and (3) having the people in $P_-$ pay the ones in $P_+$. Then the problem of resolving debts with the fewest transactions is equivalent to the above optimization problem, with $a_{ij}$ being the amount that the $i$th person in $P_-$ must give the $j$th person in $P_+$.&lt;/p&gt;&#xA;" OwnerUserId="856" LastEditorUserId="856" LastEditorDisplayName="" LastEditDate="2010-09-26T06:38:40.120" LastActivityDate="2010-09-26T06:38:40.120" Title="Sparsest matrix with specified row and column sums" Tags="&lt;combinatorics&gt;&lt;optimization&gt;" AnswerCount="2" />
  <row Id="4313" PostTypeId="2" ParentId="2655" CreationDate="2010-09-08T20:28:19.610" Score="2" ViewCount="" Body="&lt;p&gt;Andrew Booker's &lt;a href=&quot;http://primes.utm.edu/nthprime/&quot; rel=&quot;nofollow&quot;&gt;Nth prime&lt;/a&gt; page is excellent... but it can't handle your example number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have custom code that can calculate values up to about 2^64, but your number is larger than that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks to Dusart [1], we can say that its rank is somewhere between 24244547260299402427 and 24247918127257270377.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the Riemann Hypothesis is true, then we know by Schoenfeld [2] that its rank is somewhere between 24245911027060346607 and 24245911157987206331.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[1] Pierre Dusart, 'Estimates of Some Functions Over Primes without R.H.', preprint (2010), &lt;a href=&quot;http://arxiv.org/abs/1002.0442&quot; rel=&quot;nofollow&quot;&gt;arXiv:1002.0442&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[2] Lowell Schoenfeld, 'Sharper Bounds for the Chebyshev Functions theta(x) and psi(x). II'. Mathematics of Computation, Vol 30, No 134 (Apr 1976), pp. 337-360.&lt;/p&gt;&#xA;" OwnerUserId="1778" LastActivityDate="2010-09-08T20:28:19.610" />
  <row Id="4314" PostTypeId="2" ParentId="1257" CreationDate="2010-09-08T20:39:47.683" Score="6" ViewCount="" Body="&lt;p&gt;Far better than sieving in the large range ShreevatsaR suggested (which, for the 10^15-th prime, has 10^15 members and takes about 33 TB to store in compact form), take a good first guess like Riemann's R and use one of the advanced methods of computing pi(x) for that first guess.  (If this is far off for some reason -- it shouldn't be -- estimate the distance to the proper point and calculate a new guess from there.)  At this point you can sieve the small distance, perhaps just 10^8 or 10^9, to the desired number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is about 100,000 times faster for numbers around the size I indicated.  Even for numbers as small as 10 to 12 digits this is faster if you don't have a precomputed table large enough to contain your answer.&lt;/p&gt;&#xA;" OwnerUserId="1778" LastActivityDate="2010-09-08T20:39:47.683" CommentCount="2" />
  <row Id="4315" PostTypeId="2" ParentId="4312" CreationDate="2010-09-08T21:03:18.290" Score="4" ViewCount="" Body="&lt;p&gt;This is linear optimization in the $L_0$ (Hamming) metric which in full generality is NP-hard, so either there is a solution adapted to the specific structure of the debt settlement equations, or it is intractable. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One structure the debt settlement system has is sparsity.  It can be written as $m+n$ equations in $mn$ variables, with only $2mn$ (out of $(mn)^2$) nonzero coefficients. This is a very sparse system to which $L_0$-$L_1$ transference (so-called &quot;compressed sensing&quot;) methods may apply, reducing the problem to linear programming.&lt;/p&gt;&#xA;" OwnerUserId="467" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-09-08T21:36:00.543" LastActivityDate="2010-09-08T21:36:00.543" CommentCount="2" />
  <row Id="4316" PostTypeId="1" CreationDate="2010-09-08T22:34:25.857" Score="0" ViewCount="169" Body="&lt;p&gt;Please explain your reasoning too. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is for a lucky draw, I estimate around 240, with about 12 candies in 20 layers. The size of candies is about standard candy. I am sorry I have no more info to share.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/xx3Zk.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1802" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-09-08T22:37:39.593" LastActivityDate="2010-09-08T22:37:39.593" ClosedDate="2010-09-08T23:06:29.800" Title="How many candies are in this picture?" Tags="&lt;geometry&gt;" CommentCount="4" />
  <row Id="4317" PostTypeId="1" AcceptedAnswerId="4318" CreationDate="2010-09-08T23:31:32.027" Score="13" ViewCount="455" Body="&lt;p&gt;For computing the present worth of an infinite sequence of equally spaced payments $(n^{2})$ I had the need to evaluate&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle\sum_{n=1}^{\infty}\frac{n^{2}}{x^{n}}=\dfrac{x(x+1)}{(x-1)^{3}}\qquad x&gt;1.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The method I used was based on the geometric series $\displaystyle\sum_{n=1}^{\infty}x^{n}=\dfrac{x}{1-x}$ differentiating each side followed by a multiplication by $x$, differentiating a second time and multiplying again by $x$. There is at least a second (more difficult) method that is to compute the series partial sums and letting $n$ go to infinity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; Is there a closed form for&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle\sum_{n=1}^{\infty }\dfrac{n^{p}}{x^{n}}\qquad x&gt;1,p\in\mathbb{Z}^{+}\quad ?$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the sketch of its proof in case it exists?&lt;/p&gt;&#xA;" OwnerUserId="752" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2010-10-05T10:19:02.307" LastActivityDate="2010-10-05T10:19:02.307" Title="Generalizing $\sum_{n=1}^{\infty }n^{2}/x^{n}$ to $\sum_{n=1}^{\infty }n^{p}/x^{n}$" Tags="&lt;calculus&gt;&lt;sequences-and-series&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="2" />
  <row Id="4318" PostTypeId="2" ParentId="4317" CreationDate="2010-09-08T23:48:13.087" Score="11" ViewCount="" Body="&lt;p&gt;The general closed form is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle \sum_{k=1}^{\infty} k^n x^k = \frac{1}{(1 - x)^{n+1}} \left( \sum_{m=0}^{n} A(n, m) x^{m+1} \right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $A(n, m)$ are the &lt;a href=&quot;http://en.wikipedia.org/wiki/Eulerian_number&quot; rel=&quot;nofollow&quot;&gt;Eulerian numbers&lt;/a&gt;.  When I have time I will edit with a few more details.  If you only want the answer for a particular small value of $n$ then see Section 3 of &lt;a href=&quot;http://web.mit.edu/~qchu/Public/TopicsInGF.pdf&quot; rel=&quot;nofollow&quot;&gt;my notes on generating functions&lt;/a&gt;.  I will also mention that for a particular value of $n$ one can deduce the answer by using the identity&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle \sum_{k=0}^{\infty} {k+n \choose n} x^k = \frac{1}{(1 - x)^{n+1}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and writing $k^n$ as a linear combination of the polynomials ${k+r \choose r}$ (in $k$), for example using a finite difference table.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-09-08T23:48:13.087" CommentCount="6" />
  <row Id="4319" PostTypeId="2" ParentId="4317" CreationDate="2010-09-08T23:54:33.407" Score="3" ViewCount="" Body="&lt;p&gt;According to &lt;a href=&quot;http://mathworld.wolfram.com/GeometricDistribution.html&quot; rel=&quot;nofollow&quot;&gt;mathworld&lt;/a&gt; The $n^{th}$ moment of the geometric distribution with parameter $p$, which is $ \sum p (1-p)^n n^k $ can be expressed in terms of the polylogarithm function.&lt;/p&gt;&#xA;" OwnerUserId="404" LastActivityDate="2010-09-08T23:54:33.407" CommentCount="1" />
  <row Id="4320" PostTypeId="2" ParentId="4238" CreationDate="2010-09-09T00:19:34.470" Score="3" ViewCount="" Body="&lt;p&gt;The smallest example is 2×AΓL(1,8) of order 336.  A fairly standard permutation representation is generated by (+,-), (∞,0)(1,3)(2,6)(4,5), (1,2,3,4,5,6,7), and (1,2,4)(3,6,5), or you can use more GAP friendly points to get (9,10), (8,7)(1,3)(2,6)(4,5), (1,2,3,4,5,6,7), and (1,2,4)(3,6,5). AΓL(1,8) is a complete, solvable group with no normal subgroup of index 2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I just checked the groups ordered by size, skipping nilpotent groups since they always have outer automorphisms.  Probably one could be a bit smarter, looking for complete groups with no 2-quotients, but I wasn't sure the smallest example would be of the form 2×Complete.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are examples like 2.Sz(8) which have no outer automorphisms, have a non-trivial center, but where the quotient by the center is not complete (so it is definitely not 2×Complete).&lt;/p&gt;&#xA;" OwnerUserId="583" LastActivityDate="2010-09-09T00:19:34.470" CommentCount="1" />
  <row Id="4321" PostTypeId="2" ParentId="4304" CreationDate="2010-09-09T03:52:39.740" Score="2" ViewCount="" Body="&lt;p&gt;After reading &lt;a href=&quot;http://www.satisfice.com/tools.shtml&quot; rel=&quot;nofollow&quot;&gt;this page&lt;/a&gt;, it seems that pairwise testing requires a set of test cases in which every pair of values from any two of the n categories occurs at least once among the test case n-tuples. In the present case, the problem is to find a minimal subset of the 6x6x6 = 216 total triples (a,b,c) such that&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;each pair of values for a and b&lt;br&gt;&#xA;occurs at least once, i.e. (a,b,*),&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;each pair of a and c values occurs&lt;br&gt;&#xA;at least once, i.e. (a,*,c)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;each pair of b and c values occurs&#xA;at least once, i.e. (*,b,c)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Any subset satisfying these requirements must have at least 36 elements just to satisfy the (a,b,*) requirement. In the present case I think 36 test cases are also sufficient, as in the following set of triples:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1, 1, 1), (1, 2, 2), (1, 3, 3), (1, 4, 4), (1, 5, 5), (1, 6, 6)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2, 1, 6), (2, 2, 1), (2, 3, 2), (2, 4, 3), (2, 5, 4), (2, 6, 5)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(3, 1, 5), (3, 2, 6), (3, 3, 1), (3, 4, 2), (3, 5, 3), (3, 6, 4)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(4, 1, 4), (4, 2, 5), (4, 3, 6), (4, 4, 1), (4, 5, 2), (4, 6, 3)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(5, 1, 3), (5, 2, 4), (5, 3, 5), (5, 4, 6), (5, 5, 1), (5, 6, 2)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(6, 1, 2), (6, 2, 3), (6, 3, 4), (6, 4, 5), (6, 5, 6), (6, 6, 1)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this example each of the three kinds of pairs occurs once and only once, i.e. there is no overlap. I don't think this will be possible in general, so it might not always be easy to come up with minimal subsets that cover all the cases.&lt;/p&gt;&#xA;" OwnerUserId="1123" LastActivityDate="2010-09-09T03:52:39.740" CommentCount="1" />
  <row Id="4322" PostTypeId="1" AcceptedAnswerId="4328" CreationDate="2010-09-09T06:25:46.773" Score="2" ViewCount="405" Body="&lt;p&gt;I have a 3D plane defined by three points: $P_0$ , $P_1$ and $P_2$. How to check whether a point $P$ is located right on and &lt;strong&gt;&lt;em&gt;inside&lt;/em&gt;&lt;/strong&gt; the 3D triangle?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, for example, if I have a plane defined by $({0,0,0})$, $({10,0,0})$ and $({0,10,0})$, then the point $({50,0,0})$ is considered &lt;strong&gt;&lt;em&gt;not&lt;/em&gt;&lt;/strong&gt; located on the plane, whereas the point  $({5,0,0})$ is. &lt;/p&gt;&#xA;" OwnerUserId="550" LastEditorUserId="550" LastEditorDisplayName="" LastEditDate="2010-09-09T07:20:53.350" LastActivityDate="2011-03-25T23:05:31.370" Title="Check whether a point is within a 3D Triangle" Tags="&lt;geometry&gt;" AnswerCount="3" CommentCount="10" FavoriteCount="1" />
  <row Id="4323" PostTypeId="1" AcceptedAnswerId="4327" CreationDate="2010-09-09T06:28:02.247" Score="14" ViewCount="640" Body="&lt;p&gt;If we have an algebraic number &amp;#945; with (complex) absolute value 1, it does not follow that &amp;#945; is a root of unity (i.e., that &amp;#945;&lt;sup&gt;&lt;em&gt;n&lt;/em&gt;&lt;/sup&gt;=1 for some &lt;em&gt;n&lt;/em&gt;).  For example, (3/5 + 4/5 &lt;em&gt;i&lt;/em&gt;) is not a root of unity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if we assume that &amp;#945; is an algebraic integer with absolute value 1, does it follow that &amp;#945; is a root of unity?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I know that if all conjugates of &amp;#945; have absolute value 1, then &amp;#945; is a root of unity by the argument below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The minimal polynomial of &amp;#945; over &lt;em&gt;Z&lt;/em&gt; is $\prod_{i=1}^d (x-\alpha_i)$, where the $\alpha_i$ are just the conjugates of &amp;#945;.  Then $\prod_{i=1}^d (x-\alpha_i^n)$ is a polynomial over &lt;em&gt;Z&lt;/em&gt; with &amp;#945;&lt;sup&gt;n&lt;/sup&gt; as a root.  It also has degree d, and all roots have absolute value 1.  But there can only be finitely many such polynomials (since the coefficients are integers with bounded size), so we get that &amp;#945;&lt;sup&gt;n&lt;/sup&gt;=&amp;#963;(&amp;#945;) for some Galois conjugation &amp;#963;. If &amp;#963;&lt;sup&gt;m&lt;/sup&gt;(&amp;#945;)=&amp;#945;, then &amp;#945;&lt;sup&gt;n&lt;sup&gt;m&lt;/sup&gt;&lt;/sup&gt;=&amp;#945;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus &amp;#945;&lt;sup&gt;n&lt;sup&gt;m&lt;/sup&gt;-1&lt;/sup&gt;=1.&lt;/p&gt;&#xA;" OwnerUserId="1461" LastActivityDate="2010-09-10T15:40:24.017" Title="Are all algebraic integers with absolute value 1 roots of unity?" Tags="&lt;algebraic-number-theory&gt;" AnswerCount="3" CommentCount="8" FavoriteCount="5" />
  <row Id="4326" PostTypeId="2" ParentId="4322" CreationDate="2010-09-09T06:44:24.010" Score="0" ViewCount="" Body="&lt;p&gt;There may well be a more efficient algorithm, but here's a way to check that &lt;em&gt;P&lt;/em&gt; is on the triangle defined by the three points&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Compare the cross products $\vec {P_0 P_1} \times \vec {P_0 P_2} = \vec a$ and $\vec {P P_1} \times \vec {P P_2} = \vec b$.  If $\vec b = k_1 \vec a$ with $k_1 \geq 0$, then P is on the plane and is on the correct side of the line through &lt;em&gt;P&lt;sub&gt;1&lt;/sub&gt;&lt;/em&gt; and &lt;em&gt;P&lt;sub&gt;2&lt;/sub&gt;&lt;/em&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, compute $\vec {P P_2} \times \vec {P P_0} = \vec c$ and $\vec {P P_0} \times \vec {P P_1} = \vec d$.  If $\vec c = k_2 \vec a$ and $\vec d = k_3 \vec a$ with $k_2, k_3 \geq 0$, then &lt;em&gt;P&lt;/em&gt; lies within the triangle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, there's a lot of redundancy in this calculation.  Once we know &lt;em&gt;P&lt;/em&gt; is on the plane, we could just check one coordinate of the $\vec c$ and $\vec d$ to see that they have the correct sign.  This is just projecting the problem onto one of the coordinate planes.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;My old answer below explains how to check that &lt;em&gt;P&lt;/em&gt; is on the infinite plane defined by the first three points, not how to check if it lies inside the triangle defined by the first three points, which is what the question intended.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;A general plane equation is &lt;em&gt;Ax + By + Cz = D&lt;/em&gt;.  The 3-dimensional vector &amp;lt;&lt;em&gt;A,B,C&lt;/em&gt;&gt; is perpendicular to the plane.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Note that there is not a unique equation of this form; you can multiply the equation by any nonzero number and get another equation of the plane.  --So you can try to find a solutions with &lt;em&gt;D=1&lt;/em&gt;, and if that doesn't work, use &lt;em&gt;D=0&lt;/em&gt;.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you find an equation of the plane defined by your first three points, you can just plug in the fourth point to check if it satisfies the equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can solve for the plane equation by hand by setting up equation from the first three points; but a more efficient method is to take the cross product of the vector $\vec {P_0 P_1}$ and the vector $\vec {P_0 P_2}$, and to use the resulting vector as &amp;lt;&lt;em&gt;A,B,C&lt;/em&gt;&gt;.  You'll find it very useful to understand the cross product if you like working in 3-D space.&lt;/p&gt;&#xA;" OwnerUserId="1461" LastEditorUserId="1461" LastEditorDisplayName="" LastEditDate="2010-09-09T07:22:27.317" LastActivityDate="2010-09-09T07:22:27.317" CommentCount="1" />
  <row Id="4327" PostTypeId="2" ParentId="4323" CreationDate="2010-09-09T06:46:01.717" Score="14" ViewCount="" Body="&lt;p&gt;No. There are algebraic integers on the unit circle which aren't roots of unity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://ramanujan.math.trinity.edu/rdaileda/research/papers/p1.pdf&quot; rel=&quot;nofollow&quot;&gt;This paper&lt;/a&gt; by Ryan Daileda provides some useful information and references.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also see &lt;a href=&quot;http://en.wikipedia.org/wiki/Salem_number&quot; rel=&quot;nofollow&quot;&gt;Salem numbers&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-09-09T06:46:01.717" CommentCount="2" />
  <row Id="4328" PostTypeId="2" ParentId="4322" CreationDate="2010-09-09T07:18:24.993" Score="3" ViewCount="" Body="&lt;p&gt;Try to solve the system&#xA;$$&#xA;x * (P_1 - P_0) + y * (P_2 - P_0) = P - P_0&#xA;$$&#xA;If it is solveable, the point $P$ lies on the plane. If in addition $x \geq 0$, $y \geq 0$, and $x + y \leq 1$, then $P$ lies inside the triangle.&lt;/p&gt;&#xA;" OwnerUserId="960" LastEditorUserId="960" LastEditorDisplayName="" LastEditDate="2010-09-09T08:32:03.627" LastActivityDate="2010-09-09T08:32:03.627" CommentCount="7" />
  <row Id="4329" PostTypeId="2" ParentId="4173" CreationDate="2010-09-09T10:31:47.873" Score="1" ViewCount="" Body="&lt;p&gt;One way is to find the householder matrix Q that maps v to a multiple of e_1 (first coordinate basis vector). Then (since Q is symmtric and orthogonal) w_2=Q*e_2 ... will be a basis of the orthogonal complement of v. &#xA;Explicitly, I get&#xA;w_k = e_k - u&#xA;where u = (v+sqrt(d)*e_1)/(d+sqrt(d))&lt;/p&gt;&#xA;" LastActivityDate="2010-09-09T10:31:47.873" />
  <row Id="4330" PostTypeId="2" ParentId="4268" CreationDate="2010-09-09T11:11:21.380" Score="6" ViewCount="" Body="&lt;p&gt;If you're willing to accept the &lt;a href=&quot;http://math.stackexchange.com/questions/4268/alternate-definition-of-prime-number&quot;&gt;fundamental theorem of arithmetic&lt;/a&gt; (which can be proved by elementary -i.e., &quot;ideal free&quot;- methods, see the link) that states that any integer different from $\pm 1$ can be written as a unique product (up to ordering of the factors) of prime numbers, then the equivalence of both definitions of prime numbers&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;(\ m \vert p \ \Longrightarrow \ m = \pm 1 \ \text{or} \ \pm p \ ) &#xA;\qquad \Longleftrightarrow \qquad&#xA;(\ p \vert ab \ \Longrightarrow \ p \vert a \ \text{or} \ p \vert b \ )&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is clear:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[Warning. If you feel more comfortable working only with positive integer numbers, then just drop those $\pm$ all over this answer.]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(\Longrightarrow)$ If $p\vert ab$, then $ab = pq$ for some integer $q$. Since both $ab$ and $pq$ can be &lt;em&gt;uniquely&lt;/em&gt; written as a product of prime numbers, $p$ must be a prime factor of $ab$. But prime factors of $ab$ are necessarily prime factors of $a$ or $b$, again by the uniqueness of the fundamental theorem of arithmetic. So, $p$ is necessarily a prime factor of $a$ or $b$; that is, $p\vert a$ or $p\vert b$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(\Longleftarrow)$ Again, if $m\vert p$, all prime factors of $m$ must be prime factors of $p$. But the only prime factors of $p$ are $\pm p$. So, either $m = \pm p$, or $m=\pm 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, there is little (big?) circular argument here, because one already uses implication $(\Longrightarrow)$ in the proof of the fundamental theorem of arithmetic. This implication is called &lt;a href=&quot;http://en.wikipedia.org/wiki/Euclid%27s_lemma&quot; rel=&quot;nofollow&quot;&gt;Euclid's lemma&lt;/a&gt;. But it can be proved by elementary  methods too, that rely on &lt;a href=&quot;http://en.wikipedia.org/wiki/B%C3%A9zout%27s_identity&quot; rel=&quot;nofollow&quot;&gt;Bézout's identity&lt;/a&gt; (whith elementary proof too: see the last link).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, as you can see, everything is elementary, but, perhaps, too long for a single answer and to remember in one go. While reasoning using the fundamental theorem of arithmetic may be seen as little short cut easier to remember the next time you have any doubt about the equivalence of both definitions of prime number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(In fact, a number satisfying the left hand side condition is called &quot;irreducible&quot;, while $p$ on the right hand side is called &quot;prime&quot;. So &quot;prime = irreducible&quot;, at least for integer numbers.)&lt;/p&gt;&#xA;" OwnerUserId="664" LastActivityDate="2010-09-09T11:11:21.380" CommentCount="2" />
  <row Id="4331" PostTypeId="2" ParentId="4304" CreationDate="2010-09-09T11:51:05.527" Score="3" ViewCount="" Body="&lt;p&gt;Pairwise testing tests for all possible 2-way interactions efficiently -- I gave a quick overview here: &lt;a href=&quot;http://cstheory.stackexchange.com/questions/891/&quot;&gt;http://cstheory.stackexchange.com/questions/891/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are looking for strength 2 &lt;a href=&quot;http://math.nist.gov/coveringarrays/coveringarray.html&quot; rel=&quot;nofollow&quot;&gt;covering arrays&lt;/a&gt;.  In each pair of columns every pair of symbols occur -- this ensures all 2-way interactions are observed in some way.  Here's a very simple example of a covering array of strength 2 with 2 columns:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;11&#xA;12&#xA;21&#xA;22&#xA;12&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;What castel has drawn is essentially the &lt;a href=&quot;http://en.wikipedia.org/wiki/Latin_square&quot; rel=&quot;nofollow&quot;&gt;Latin square&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;123456&#xA;612345&#xA;561234&#xA;456123&#xA;345612&#xA;234561&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;If you look at each entry and write the list (r,c,s), where r is the row index, c is the column index, and s is the symbol, you will construct an orthogonal array (as depicted below) -- a covering array of strength 2 with the minimum number of rows (36).&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;111&#xA;122&#xA;133&#xA;...&#xA;661&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;In fact, Latin squares exist for all orders n.  So if you have three columns (e.g. three variables) and n symbols for each variable, then you can always find a strength 2 covering array with n&lt;sup&gt;2&lt;/sup&gt; rows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many combinatorial designs give rise to particularly efficient covering arrays.  Strength 2 covering arrays with more than three columns and n&lt;sup&gt;2&lt;/sup&gt; rows are equivalent to sets of &lt;a href=&quot;http://designtheory.org/library/encyc/mols/a/&quot; rel=&quot;nofollow&quot;&gt;mutually orthogonal Latin squares&lt;/a&gt; (the reference shows the construction).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your case, if you have 40 results, then you are not using the most efficient covering array.&lt;/p&gt;&#xA;" OwnerUserId="139" LastActivityDate="2010-09-09T11:51:05.527" />
  <row Id="4332" PostTypeId="2" ParentId="4323" CreationDate="2010-09-09T12:00:58.340" Score="12" ViewCount="" Body="&lt;p&gt;Let x be an algebraic number with absolute value 1. Then x and its complex conjugate x* = 1/x have the same minimal polynomial. Writing f(T) for the minimal polynomial of x over Q, with degree n, the polynomials T^nf(1/T) and f(T) are irreducible over Q with root x*, so the polynomials are equal up to a scaling factor: T^nf(1/T) = cf(T). Setting T = 1, f(1) = cf(1). &#xA;Assuming x is not rational (i.e., x is not 1 or -1), f has degree greater than 1, so &#xA;f(1) is nonzero and thus c = 1. Therefore T^nf(1/T) = f(T), so f(T) has symmetric coefficients. In particular, its constant term is 1. Moreover, the roots of f(T) come in reciprocal pairs (since 1 and -1 are not roots), so n is even.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Partial conclusion: an algebraic number other than 1 or -1 which has absolute value 1 &#xA;has even degree over Q and its minimal polynomial has constant term 1. In particular, &#xA;if x is an algebraic integer then it must be a unit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are no examples of algebraic integers with degree 2 and abs. value 1 that are not roots of unity since a real quadratic field has no elements on the unit circle besides 1 and -1 and the units in an imaginary quadratic field are all roots of unity (and actually are only 1 and -1 except for Q(i) and Q(w)). Thus the smallest degree x could have over Q is 4 and there are examples with degree 4: the polynomial x^4 - 2x^3 - 2x + 1 has two roots on the unit circle and two real roots (one between 0 and 1 and the other greater than 1). &lt;/p&gt;&#xA;" OwnerUserId="619" LastActivityDate="2010-09-09T12:00:58.340" CommentCount="1" />
  <row Id="4333" PostTypeId="2" ParentId="368" CreationDate="2010-09-09T12:48:34.330" Score="5" ViewCount="" Body="&lt;p&gt;Contrast these two methods for calculating n!:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Counting permutations of a n-element set one-by-one (this is what n! counts), vs.&lt;/li&gt;&#xA;&lt;li&gt;Multiplying together the numbers 1,2,...,n.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Therefore n! represents an amazingly efficient method for counting permutations!  So I think (and I don't think I'm the only one) that n! should probably be considered a closed form (unless you have some other clear definition of what constitutes a &quot;closed&quot; form).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For further reading, I recommend:  H. S. Wilf, What is an answer?, Amer. Math. Monthly, 89 (1982), pp. 289–292.&lt;/p&gt;&#xA;" OwnerUserId="139" LastActivityDate="2010-09-09T12:48:34.330" CommentCount="1" />
  <row Id="4334" PostTypeId="2" ParentId="4199" CreationDate="2010-09-09T14:21:45.030" Score="3" ViewCount="" Body="&lt;p&gt;We could do something simple, since we have:&#xA;\begin{equation}&#xA;\cos\left(\frac{p}{q}\theta\right)=\frac{1}{2}\left(e^{i\frac{p}{q}\theta}+e^{-i\frac{p}{q}\theta}\right)=\frac{1}{2}\left(\left(\frac{x}{r}+i\frac{y}{r}\right)^{\frac{p}{q}}+\left(\frac{x}{r}-i\frac{y}{r}\right)^{\frac{p}{q}}\right)&#xA;\end{equation}&#xA;That is &#xA;\begin{equation}&#xA;\cos\left(\frac{p}{q}\theta\right)=\frac{1}{2}\left(\frac{1}{x^2+y^2}\right)^{\frac{p}{2q}}\left(\left(x+iy\right)^{\frac{p}{q}}+\left(x-iy\right)^{\frac{p}{q}}\right)&#xA;\end{equation}&#xA;and similarly we have&#xA;\begin{equation}&#xA;\sin\left(\frac{p}{q}\theta\right)=\frac{1}{2i}\left(\frac{1}{x^2+y^2}\right)^{\frac{p}{2q}}\left(\left(x+iy\right)^{\frac{p}{q}}-\left(x-iy\right)^{\frac{p}{q}}\right)&#xA;\end{equation}&#xA;Now we can use these to express your equations in terms of $x$ and $y$. Is this any use?&lt;/p&gt;&#xA;" OwnerUserId="1794" LastEditorUserId="1794" LastEditorDisplayName="" LastEditDate="2010-09-09T18:45:05.497" LastActivityDate="2010-09-09T18:45:05.497" CommentCount="4" />
  <row Id="4335" PostTypeId="1" AcceptedAnswerId="4341" CreationDate="2010-09-09T14:51:35.530" Score="4" ViewCount="500" Body="&lt;p&gt;I'm starting a very long quest to learn about math so I can program games. I'm mostly a coorporate developer and it's somewhat boring and non exiting. When I began my career I chose it because I wanted to create games.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm told Linear Algebra is the best place to start - where should I go?&lt;/p&gt;&#xA;" LastActivityDate="2010-09-11T07:20:06.867" CommunityOwnedDate="2010-09-09T16:15:04.960" Title="Where to start learning Linear Algebra?" Tags="&lt;linear-algebra&gt;" AnswerCount="5" CommentCount="1" FavoriteCount="6" />
  <row Id="4336" PostTypeId="1" AcceptedAnswerId="4340" CreationDate="2010-09-09T14:52:53.860" Score="5" ViewCount="615" Body="&lt;p&gt;If the matrix is positive definite, then all its eigenvalues are strictly positive. Is the converse also true? That is, if the eigenvalues are strictly positive, then matrix is positive definite? Can you give example of 2x2 matrix with 2 positive eigenvalues but is not positive definite?&lt;/p&gt;&#xA;" OwnerUserId="957" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-09-09T16:14:39.923" LastActivityDate="2011-04-23T05:01:45.897" Title="if eigenvalues are positive, is the matrix positive definite?" Tags="&lt;linear-algebra&gt;&lt;matrices&gt;" AnswerCount="4" CommentCount="4" FavoriteCount="2" />
  <row Id="4337" PostTypeId="2" ParentId="4336" CreationDate="2010-09-09T14:57:07.710" Score="1" ViewCount="" Body="&lt;p&gt;True. If you consider a diagonal matrix $A = diag(d_1,\cdots,d_n)$ where each diagonal entry $d_i$ is positive, then clearly it is positive definite, since $x^TAx = \sum d_ix_i^2 &gt; 0$ unless $x = 0$ ($x_i$ are the components of vector $x$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now apply spectral theorem for symmetric matrix to reduce to the diagonal case.&lt;/p&gt;&#xA;" OwnerUserId="325" LastActivityDate="2010-09-09T14:57:07.710" CommentCount="0" />
  <row Id="4338" PostTypeId="2" ParentId="4335" CreationDate="2010-09-09T15:00:48.857" Score="8" ViewCount="" Body="&lt;p&gt;Hi Sergio. MIT has a complete online course for linear algebra, complete with video lectures, lecture notes, and assignments.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://ocw.mit.edu/courses/mathematics/18-06-linear-algebra-spring-2005/&quot; rel=&quot;nofollow&quot;&gt;http://ocw.mit.edu/courses/mathematics/18-06-linear-algebra-spring-2005/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure how much of the course is applicable to what you'll be doing in game development, but it's a solid foundation.&lt;/p&gt;&#xA;" OwnerUserId="592" LastActivityDate="2010-09-09T15:00:48.857" CommunityOwnedDate="2010-09-09T16:15:04.960" />
  <row Id="4340" PostTypeId="2" ParentId="4336" CreationDate="2010-09-09T15:39:51.320" Score="10" ViewCount="" Body="&lt;p&gt;I think this is false. Let $A = \begin{pmatrix} 1 &amp;amp; -3 \\ 0 &amp;amp; 1 \end{pmatrix}$ be a 2x2 matrix, in the canonical basis of $\mathbb R^2$. Then A has a double eigenvalue b=1. If $v=\begin{pmatrix}1\\1\end{pmatrix}$, then $\langle v, Av \rangle &amp;lt; 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The point is that the matrix can have all its eigenvalues strictly positive, but it does not need to be symmetric. &lt;/p&gt;&#xA;" OwnerUserId="1178" LastEditorUserId="171" LastEditorDisplayName="" LastEditDate="2010-09-09T15:59:34.003" LastActivityDate="2010-09-09T15:59:34.003" CommentCount="3" />
  <row Id="4341" PostTypeId="2" ParentId="4335" CreationDate="2010-09-09T17:43:56.283" Score="6" ViewCount="" Body="&lt;p&gt;You are right: Linear Algebra is not just the &quot;best&quot; place to start. It's THE place to start.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Among all the books cited in &lt;a href=&quot;http://en.wikipedia.org/wiki/Linear_algebra&quot; rel=&quot;nofollow&quot;&gt;Wikipedia - Linear Algebra&lt;/a&gt;, I would recommend:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;* Strang, Gilbert, Linear Algebra and Its Applications (4th ed.) &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Strang's book has at least two reasons for being recomended. First, it's extremely easy and short. Second, it's the book they use at MIT for the extremely good video Linear Algebra course you'll find in the link of Unreasonable Sin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a view towards applications (though maybe not necessarily your applications) and still elementary:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;* B. Noble &amp;amp; J.W. Daniel: Applied Linear Algebra, Prentice-Hall, 1977&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Linear algebra has two sides: one more &quot;theoretical&quot;, the other one more &quot;applied&quot;. Strang's book is just elementary, but perhaps &quot;theoretical&quot;. Noble-Daniel is definitively &quot;applied&quot;. The distinction from the two points of view relies in the emphasis they put on &quot;abstract&quot; vector spaces vs specific ones such as $\mathbb{R}^n$ or $\mathbb{C}^n$, or on matrices vs linear maps.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe because my penchant towards &quot;pure&quot; maths, I must admit that sometimes I find matrices somewhat annoying. They are funny, specific, whereas linear maps can look more &quot;abstract&quot; and &quot;ethereal&quot;. But, for instance: I can't stand the proof that the matrix product is associative, whereas the corresponding associativity for the composition of (linear or non linear) maps is true..., well, just because it can't help to be true the first moment you write it down.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, at a more advanced level in the &quot;theoretical&quot; side you can use:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;* Greub, Werner H., Linear Algebra, Graduate Texts in Mathematics (4th ed.), Springer&#xA;&#xA;* Halmos, Paul R., Finite-Dimensional Vector Spaces, Undergraduate Texts in Mathematics, Springer&#xA;&#xA;* Shilov, Georgi E., Linear algebra, Dover Publications &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;In the &quot;applied&quot; (?) side, a book that I love and you'll appreciate if you want to study, for instance, the exponential of a matrix is &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0486445542&quot; rel=&quot;nofollow&quot;&gt;Gantmacher&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And, at any time, you'll need to do a lot of exercises. Lipschutz's is second to none in this:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;* Lipschutz, Seymour, 3,000 Solved Problems in Linear Algebra, McGraw-Hill&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Enjoy!   :-)&lt;/p&gt;&#xA;" OwnerUserId="664" LastEditorUserId="664" LastEditorDisplayName="" LastEditDate="2010-09-09T18:30:27.973" LastActivityDate="2010-09-09T18:30:27.973" CommunityOwnedDate="2010-09-09T17:43:56.283" CommentCount="2" />
  <row Id="4342" PostTypeId="2" ParentId="368" CreationDate="2010-09-09T17:57:48.713" Score="1" ViewCount="" Body="&lt;p&gt;This is a riff on some of the comments about what might constitute an &quot;answer&quot; and what &quot;closed form&quot; might mean: although it's somewhat facetious, it's intended to prompt thoughts about these issues.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Our base-10 number system interprets a string ${a_n}{a_{n-1}} \cdots {a_0}$ as the sum $\sum_{i=0}^{n} a_i 10^i $ (which can be, and is, computed recursively as $a_0 + 10 \left( a_1 + 10 \left( \cdots + 10 a_n \right) \cdots \right)$).  If you take the former to be an acceptable &quot;closed form&quot; representation, then why not use a slight modification of this number system?  Specifically, interpret the same string as equal to $a_0 + 2 \left( a_1 + 3 \left( \cdots + (n+1) a_n \right) \cdots \right)$ and require that $0 \le a_0 \le 1, 0 \le a_1 \le 2, \ldots, 0 \le a_n \le n$.  In this &quot;factorial&quot; number system, $n! = 10 \cdots 0$ is represented as a simple $n$-digit string: it's &quot;closed&quot;!&lt;/p&gt;&#xA;" OwnerUserId="1489" LastActivityDate="2010-09-09T17:57:48.713" />
  <row Id="4343" PostTypeId="2" ParentId="4323" CreationDate="2010-09-09T18:38:21.673" Score="4" ViewCount="" Body="&lt;p&gt;The answer to your question is no - not all algebraic integers with absolute value 1 are roots of unity - as you may have learned by now from the &lt;a href=&quot;http://ramanujan.math.trinity.edu/rdaileda/research/papers/p1.pdf&quot; rel=&quot;nofollow&quot;&gt;2005 paper of Daileda&lt;/a&gt; that was referenced in the answer supplied by Robin Chapman. However, you should be aware that &lt;strong&gt;Daileda has in fact rediscovered&lt;/strong&gt; these simple folklore results on unimodular units. These results are probably at least a half-century old - if not much older. Indeed, I recall reading similar results in papers published  by Iwasawa in the fifties or sixties. I don't have the time now to locate said Iwasawa reference, but here is another reference that suffices to prove my point. This &lt;a href=&quot;http://ci.nii.ac.jp/naid/110000010754/en&quot; rel=&quot;nofollow&quot;&gt;1979 paper of Nakahata&lt;/a&gt; includes the same results and the same simple proofs found in Daileda's paper. Moreover, Nakahata places these results naturally into a more general context. Thus I highly recommend that you consult Nakahata's paper in addition to Daileda's. I've appended its Zbl review below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/lofUp.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-09-10T15:40:24.017" LastActivityDate="2010-09-10T15:40:24.017" CommentCount="1" />
  <row Id="4344" PostTypeId="2" ParentId="4317" CreationDate="2010-09-09T18:50:11.550" Score="5" ViewCount="" Body="&lt;p&gt;Here is a different look:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The differentiating and multiplying by $x$ gives rise to &lt;a href=&quot;http://en.wikipedia.org/wiki/Stirling_numbers_of_the_second_kind&quot; rel=&quot;nofollow&quot;&gt;Stirling Numbers of the Second Kind&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Say you denote the operator of differentiating and multiplying by $x$ as $D_{x}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then we have that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(D_{x})^{n}f(x) = \sum_{k=1}^{n} s(n,k) f^{k}(x) x^{k}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $s(n,k)$ is the stirling number of the second kind and $f^k(x)$ is the $k^{th}$ derivative of $f(x)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This can easily be proven using the identity $$s(n,k) = s(n-1,k-1) + k \cdot s(n-1,k)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a table for the Stirling numbers of the second kind (from the wiki page):&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&#xA;n/k   0     1     2     3     4      5      6      7      8      9&#xA;0     1&#xA;1     0     1&#xA;2     0     1     1&#xA;3     0     1     3     1&#xA;4     0     1     7     6     1&#xA;5     0     1     15    25    10     1&#xA;6     0     1     31    90    65     15     1&#xA;7     0     1     63    301   350    140    21     1&#xA;8     0     1     127   966   1701   1050   266    28     1&#xA;9     0     1     255   3025  7770   6951   2646   462    36     1&#xA;&#xA;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;So in your case, we can start with $ f(x) = \frac{1}{1-x}$ and obtain that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \sum_{k=0}^{\infty} k^{n} x^k = \sum_{r=1}^{n} r! \cdot s(n,r) \frac{x^{r}}{(1-x)^{r+1}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, in your case for $n=3$ we get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{k=0}^{\infty} k^3 x^k = \frac{1! \cdot 1 \cdot x}{(1-x)^2} + \frac{2! \cdot 3 \cdot x^2}{(1-x)^3} + \frac{3! \cdot 1 \cdot x^3}{(1-x)^4}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ = \frac{x(1-x)^2 + 6x^{2}(1-x) + 6x^3}{(1-x)^4} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ = \frac{x^3 + 4x^2 + x}{(1-x)^4} $$&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-09-09T22:16:49.427" LastActivityDate="2010-09-09T22:16:49.427" CommentCount="2" />
  <row Id="4345" PostTypeId="2" ParentId="4199" CreationDate="2010-09-09T19:08:05.050" Score="0" ViewCount="" Body="&lt;p&gt;Here's a systematic method, but not one recommended for paper-and-pen&#xA;calculation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $n$ be a common denominator for all rationals $r$ such that $\sin r\theta$&#xA;or $\cos r\theta$ arises in the formula.&#xA;Let $u=\cos(\theta/n)$ and $v=\sin(\theta/n)$. There are polynomials&#xA;$\phi_n$ and $\psi_n$ such that $\cos\theta=\phi_n(u,v)$ and&#xA;$\sin\theta=\psi_n(u,v)$. The curve has a polar equation which can be written&#xA;as $f(r,u,v)=0$. Now consider the system of equations:&#xA;$$f(s^2,u,v)=0,\ \ \ x=s^2\phi_n(u,v),\ \ \ y=s^2\psi_n(u,v),\ \ \ u^2+v^2=1.$$&#xA;Yes, four equations in five variables. Using some systematic&#xA;elimination procedure, for instance &lt;a href=&quot;http://en.wikipedia.org/wiki/Gr%C3%B6bner_basis&quot; rel=&quot;nofollow&quot;&gt;Groebner bases&lt;/a&gt;, eliminate the&#xA;three variables $s$, $u$ and $v$ to get (one hopes) one equation in $x$ and $y$.&#xA;It certainly helps to have some computer package to do this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an example, for $r=4a\cos^3(\theta/3)$, the system of equations becomes&#xA;$$s^2=4au^3,\ \ \ x=s^2(u^3-3uv^2),\ \ \ y=s^2(3u^2v-v^3),\ \ \ u^2+v^2=1.$$&#xA;Well, I'm not going to do the elimination, but $s^2$ disappears readily enough....&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-09-09T19:08:05.050" CommentCount="1" />
  <row Id="4346" PostTypeId="1" AcceptedAnswerId="4348" CreationDate="2010-09-09T19:21:49.167" Score="10" ViewCount="254" Body="&lt;p&gt;In &lt;a href=&quot;http://math.stackexchange.com/questions/4169/solving-the-functional-equation-ffxfxx&quot;&gt;this&lt;/a&gt; thread, the question was to find a $f: \mathbb{R} \to \mathbb{R}$ such that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f(f(x)) = f(x) + x$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(which was revealed in the comments to be solved by $f(x) = \varphi x$ where $\varphi$ is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Golden_ratio&quot; rel=&quot;nofollow&quot;&gt;golden ratio&lt;/a&gt; $\frac{1+\sqrt{5}}{2}$)&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Having read about &lt;a href=&quot;http://en.wikipedia.org/wiki/Function_iteration&quot; rel=&quot;nofollow&quot;&gt;iterated functions&lt;/a&gt; shortly before though, I came up with this train of thought:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f(f(x)) = f(x) + x$$&#xA;$$\Leftrightarrow f^2 = f^1 + f^0$$&#xA;$$f^2 - f - f^0 = 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $f^n$ denotes the $n$'th iterate of $f$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I solved the resulting quadratic equation much as I did with plain numbers&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f = \frac{1}{2} \pm \sqrt{\frac{1}{4} + 1}$$&#xA;$$f = \frac{1 \pm \sqrt{1+4}}{2} = \frac{1 \pm \sqrt{5}}{2}\cdot f^0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And finally the solution&#xA;$$f(x) = \frac{1 \pm \sqrt{5}}{2} x$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now my question is: &lt;em&gt;Is it somehow allowed to work with functions in that way?&lt;/em&gt;* I know that in the above, there are denotational ambiguities as $1$ is actually treated as $f^0 = id$ ... But since the result is correct, there seems to be some correct thing in this approach.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So can I actually solve certain functional equations like this? And if true, &lt;strong&gt;how would the correct notation of the above be&lt;/strong&gt;?&lt;/p&gt;&#xA;" OwnerUserId="917" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-09-09T23:10:28.813" LastActivityDate="2010-09-11T21:42:37.693" Title="Solving (quadratic) equations of iterated functions, such as f(f(x))=f(x)+x" Tags="&lt;real-analysis&gt;&lt;notation&gt;&lt;algebra&gt;&lt;functional-equations&gt;" AnswerCount="3" FavoriteCount="3" />
  <row Id="4347" PostTypeId="2" ParentId="4228" CreationDate="2010-09-09T19:54:31.250" Score="2" ViewCount="" Body="&lt;p&gt;Yes. Your polytope $P'$ is exactly the polar $P^*$ of $P$ with respect to the standard unit sphere. Choosing another conic instead of the unit sphere will in general give another polar. See &lt;a href=&quot;http://en.wikipedia.org/wiki/Pole_and_polar&quot; rel=&quot;nofollow&quot;&gt;this wikpedia page&lt;/a&gt; for the two-dimensional case of polarity with respect to other conics. So your &quot;Cartesian&quot; dual corresponds to the unit sphere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Polarity/duality of polytopes or more generally, convex bodies, is treated in many books on convexity. Unfortunately, many of them leave a lot of detail to the reader. I recommend &lt;a href=&quot;http://books.google.co.uk/books?id=SWQmykCwabwC&amp;amp;lpg=PP1&amp;amp;dq=webster%20convexity&amp;amp;pg=PP1#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;Webster's &lt;em&gt;Convexity&lt;/em&gt;&lt;/a&gt;, &lt;a href=&quot;http://books.google.co.uk/books?id=rvrYsRREzSYC&amp;amp;lpg=PA56&amp;amp;dq=polar%20of%20polytope&amp;amp;pg=PP1#v=onepage&amp;amp;q=polar%20of%20polytope&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;Brønsted's &lt;em&gt;An Introduction to Convex Polytopes&lt;/em&gt;&lt;/a&gt; or &lt;a href=&quot;http://books.google.co.uk/books?id=dH5DK1URwQgC&amp;amp;lpg=PP1&amp;amp;dq=matousek&amp;amp;pg=PP1#v=onepage&amp;amp;q=matousek&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;Matousek's &lt;em&gt;Lectures on Discrete Geometry&lt;/em&gt;&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1021" LastEditorUserId="1021" LastEditorDisplayName="" LastEditDate="2010-09-13T09:20:17.983" LastActivityDate="2010-09-13T09:20:17.983" CommentCount="1" />
  <row Id="4348" PostTypeId="2" ParentId="4346" CreationDate="2010-09-09T20:37:06.800" Score="6" ViewCount="" Body="&lt;p&gt;One way to think about this is that you are assuming that $f(x) = cx$ and then solving for the value of $c$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But there is something much more interesting going on; you have started doing some &lt;a href=&quot;http://en.wikipedia.org/wiki/Abstract_algebra&quot; rel=&quot;nofollow&quot;&gt;abstract algebra&lt;/a&gt; without knowing it.  $f$ is not a real number, but it does live in something called an &lt;a href=&quot;http://en.wikipedia.org/wiki/Algebra_over_a_field&quot; rel=&quot;nofollow&quot;&gt;algebra over a field&lt;/a&gt;, which is a special type of &lt;a href=&quot;http://en.wikipedia.org/wiki/Ring_(mathematics)&quot; rel=&quot;nofollow&quot;&gt;ring&lt;/a&gt;.  In particular, $f$ lives in the algebra of continuous functions $\mathbb{R} \to \mathbb{R}$.  In this algebra there is a multiplicative identity $f(x) = x$ which plays the role of the zeroth power, an addition, a multiplication, and a scalar multiplication by real numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many manipulations which are possible with ordinary real numbers are possible in rings and algebras; in particular, the first half of the proof of the quadratic formula carries through totally abstractly (the part where you complete the square).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, the second half does not.  In other words, it is not true that there are only two solutions to the equation $f^2 = a$ in a general algebra.  This is because algebras are not in general &lt;a href=&quot;http://en.wikipedia.org/wiki/Integral_domain&quot; rel=&quot;nofollow&quot;&gt;integral domains&lt;/a&gt;.  There may be none or infinitely many!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, in this special case $a$ is a positive real multiple of the identity, so we know it has &lt;em&gt;at least&lt;/em&gt; two square roots (even though there may be more).  These are the solutions that you found, and this method for finding them is perfectly valid. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a very important technique.  It is often used in the case where $f$ is a differential operator as a concise way to solve linear homogeneous ODEs.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-09-09T20:37:06.800" CommentCount="1" />
  <row Id="4350" PostTypeId="1" AcceptedAnswerId="4352" CreationDate="2010-09-09T21:57:38.147" Score="7" ViewCount="209" Body="&lt;p&gt;I'm not sure if this a proper recurance relation &lt;em&gt;per se&lt;/em&gt; but I'd be interested in the methodology in solving a recurrence relation of the following form:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$Z_0 = 1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$Z_1 = x_1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$Z_2 = x_1Z_1 + x_2 = x_1^2 + x_2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$Z_3 = x_1Z_2 + x_2Z_1 + x_3 = x_1^3 + x_1x_2 + x_1x_2 + x_3$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$Z_n = x_1 Z_{n-1} + x_2 Z_{n-2} + ... + x_n Z_0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As written, each term requires the knowledge of the previous terms. Is it possible to write down a closed form for $Z_n$? For this &lt;em&gt;particular&lt;/em&gt; recurrence, I can write down the result  of $Z_n$  but I would be very interested in seeing how one can derive this from the recurrence relation itself. My gut feeling is that a generating function is lurking underneath all this.&lt;/p&gt;&#xA;" OwnerUserId="196" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2011-06-03T20:09:36.510" LastActivityDate="2011-06-03T20:09:36.510" Title="Solving recurrence relations that involve all previous terms" Tags="&lt;recurrence-relations&gt;&lt;generating-functions&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="4351" PostTypeId="2" ParentId="4336" CreationDate="2010-09-09T22:01:03.530" Score="11" ViewCount="" Body="&lt;p&gt;This question does a great job of illustrating the problem with thinking about these things in terms of coordinates.  The thing that is positive-definite is not a matrix $M$ but the &lt;em&gt;quadratic form&lt;/em&gt; $x \mapsto x^T M x$, which is a very different beast from the linear transformation $x \mapsto M x$.  For one thing, the quadratic form does not depend on the antisymmetric part of $M$, so using an asymmetric matrix to define a quadratic form is redundant.  And there is &lt;em&gt;no reason&lt;/em&gt; that an asymmetric matrix and its symmetrization need to be at all related; in particular, they do not need to have the same eigenvalues.  &lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-09-09T22:01:03.530" CommentCount="5" />
  <row Id="4352" PostTypeId="2" ParentId="4350" CreationDate="2010-09-09T22:08:06.407" Score="7" ViewCount="" Body="&lt;p&gt;Your gut feeling is right, but let me change your $x$s to $a$s to make the answer easier to read.  Consider the generating functions&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$Z(x) = \sum_{n \ge 0} Z_n x^n$$&#xA;$$A(x) = \sum_{n \ge 1} a_n x^n.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then the recurrence relation you have written down is equivalent to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$Z(x) = Z(x) A(x) + 1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which gives&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$Z(x) = \frac{1}{1 - A(x)}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if you know the generating function $A(x)$ in closed form, you then know the generating function $Z(x)$ in closed form.  This is one of the most basic manipulations to do with generating functions, and techniques like this are thoroughly covered in, for example, Wilf's &lt;a href=&quot;http://www.math.upenn.edu/~wilf/DownldGF.html&quot; rel=&quot;nofollow&quot;&gt;generatingfunctionology&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-09-09T22:08:06.407" />
  <row Id="4354" PostTypeId="2" ParentId="4350" CreationDate="2010-09-10T00:01:28.557" Score="4" ViewCount="" Body="&lt;p&gt;To answer the general question in the title (&quot;solving recurrence relations that involve all previous terms&quot;), such relations can be thought of as &lt;em&gt;path- or history-dependent&lt;/em&gt; calculations and one tries to replace them by equivalent &lt;em&gt;state-dependent&lt;/em&gt; (Markovian) calculations that retain a finite amount of state information and keep transforming the state, but not remembering any additional information. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, if the recurrence is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$S_n = S_0 + S_1 + \dots S_{n-1}$     &lt;/p&gt;&#xA;&#xA;&lt;p&gt;this is equivalent, for $n&gt;1$,  to $S_n = S_{n-1} + S_{n-1} = 2S_{n-1}$, and remembering the previous term is enough to reproduce the recurrence as iterated doubling (after the first two terms).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The example in the question is also an iteration of a single transformation, but on an unbounded-dimensional state space.&lt;/p&gt;&#xA;" OwnerUserId="467" LastActivityDate="2010-09-10T00:01:28.557" />
  <row Id="4355" PostTypeId="1" CreationDate="2010-09-10T01:16:08.810" Score="1" ViewCount="95" Body="&lt;p&gt;I'm having a hard time solving this problem. Would appreciate any hint! (and special thanks to  Byron Schumuland for answering my previous 2 questions. This 3rd Q is different.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let et e1,e2,... be iid normal mean 0 variance 1. Let Xt := e1+...+et, for t=1,2.. and X0 := 0. (So we have a discrete-time random walk whose steps are iid N(0,1))&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Define passage time T* := inf{t&gt;0 : X_t &amp;lt; -t}.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the CDF of T*, i.e. what is Prob(T* &amp;lt;= t) ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Equivalently: let e'_t ~ N(1,1), and T* above is simply inf{t&gt;0 : X'_t &amp;lt; 0}.)&#xA;Is there a closed form solution? (Remember we have zero correlation, just non-zero mean)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some reference: &#xA;&lt;a href=&quot;http://dl.dropbox.com/u/4260685/orthant.pdf&quot; rel=&quot;nofollow&quot;&gt;http://dl.dropbox.com/u/4260685/orthant.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;THANK YOU!&lt;/p&gt;&#xA;" OwnerUserId="1784" LastActivityDate="2010-09-10T01:16:08.810" Title="Orthant probability / passage-time" Tags="&lt;probability&gt;&lt;statistics&gt;&lt;stochastic-processes&gt;" CommentCount="1" />
  <row Id="4356" PostTypeId="1" AcceptedAnswerId="4359" CreationDate="2010-09-10T02:32:55.873" Score="1" ViewCount="269" Body="&lt;p&gt;I have a constant area $A$   , and I can mold that area into a &lt;a href=&quot;http://en.wikipedia.org/wiki/Regular_polygon&quot; rel=&quot;nofollow&quot;&gt;&lt;strong&gt;regular&lt;/strong&gt;&lt;/a&gt; $n$ -side polygon, where $n&gt;=3$. The issue now is how to find the $n$  such that it can contain the most number of circles, each with a constant radius $R$ ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit to clarify the question: the area must be a regular $n$ -side polygon. Sorry for not making this clear&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="550" LastEditorUserId="550" LastEditorDisplayName="" LastEditDate="2010-09-10T07:04:47.900" LastActivityDate="2010-09-15T15:55:25.090" Title="Find the regular $n$ -side polygon of A Constant Area that can contain Most Number of Circles" Tags="&lt;geometry&gt;&lt;optimization&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="4357" PostTypeId="2" ParentId="4346" CreationDate="2010-09-10T02:41:09.587" Score="1" ViewCount="" Body="&lt;p&gt;Assume $f$ is a linear transformation, then linear algebra lets us give it a matrix representation $f(x) = F \cdot x$. It is also the case that $f \circ g = F G$, function composition is represented by matrix multiplication.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your derivation you do not make any distinction between $f$ and $F$, this is fine as long as you know you are doing it. Also the matrix $F$ is a 1x1 because $\mathbb{R}$ is a 1-dimensional vector space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is why we have:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \begin{eqnarray}&#xA;f(f(x)) &amp;amp;= f(x) + x \\&#xA; \iff \\&#xA;F^2 &amp;amp;= F + I \\&#xA; \iff \\&#xA;\varphi^2 &amp;amp;= \varphi + 1&#xA;\end{eqnarray} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Where $f(x) = F \cdot x = \varphi x$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only problem is that we cannot be sure $f$ is a linear transform just based on what was told. So it is not known whether this gives all solutions.&lt;/p&gt;&#xA;" LastActivityDate="2010-09-10T02:41:09.587" CommentCount="1" />
  <row Id="4358" PostTypeId="1" AcceptedAnswerId="4361" CreationDate="2010-09-10T03:27:33.737" Score="2" ViewCount="130" Body="&lt;p&gt;I've been reading a bit about how the set of bounds changes for a set depending on what superset one works with. I considered the sets $S\subseteq T\subseteq\mathbb{Q}$ and worked out a few contrived examples:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $S=T=$ {$x\in\mathbb{Q}\ | \ x^2\lt 2$}, so here $S$ is not bounded above in $T$, but it is bounded about in $\mathbb{Q}$, with $2$ being a possibility.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, if $S=$ {$x\in\mathbb{Q}\ | \ x^2\lt 1$} and $T=$ {$x\in\mathbb{Q}\ | \ x\lt 2 \ \text{and}\ x\neq 1$}, then $S$ is bounded in $T$ and $\sup_\mathbb{Q} S=1$ exists, but $\sup_T S$ does not exist.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is, is it possible for $S$ to be bounded in $T$ where $\sup_T S$ exists, but $\sup_\mathbb{Q} S$ does not? And moreover, can both $\sup_T S$ and $\sup_\mathbb{Q} S$ exist, but not be equal? Any example of this would be much appreciated.&lt;/p&gt;&#xA;" OwnerUserId="1583" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2010-11-26T15:27:13.790" LastActivityDate="2010-11-26T15:27:13.790" Title="For $S\subseteq T\subseteq\mathbb{Q}$, it is possible for $\sup_T S$ to exist, but $\sup_\mathbb{Q} S$ does not?" Tags="&lt;analysis&gt;&lt;elementary-set-theory&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="0" />
  <row Id="4359" PostTypeId="2" ParentId="4356" CreationDate="2010-09-10T03:35:06.723" Score="5" ViewCount="" Body="&lt;p&gt;If $n$ is unconstrained, as it seems to be in the question, then the polygon can be taken to be an $\epsilon$-accurate approximation to the optimal figure, and we can ask directly what that figure (or figures) look like. &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;If you don't assume the polygon is convex, the answer is trivial, in that you can get as close as desired to a collection of $k$ disjoint circles connected by very thin tubes. In this version of the problem the only constraint is that $A &gt; k \pi R^2$ from which the maximum $k$ is easily determined.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If convexity is required, finding the tightest configuration of circles -- the one whose convex hull is a minimum-area figure containing $k$ or more discs of given size -- is a hard nonlinear optimization problem which is a variant of the classical problem of &lt;em&gt;circle packing&lt;/em&gt; (finding the smallest circle enclosing $k$ unit discs).   Recent experiments indicate that the optima, even for large $k$, do not have a connectivity pattern approximating the optimal lattice packing.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If the polygon is convex and you are satisfied with an asymptotically optimal solution, start from a hexagonal packing of circles of radius $R$ in the whole plane, and draw a circle of area $A$ that encloses as many of these as possible, then take the convex hull of the circles inside, then approximate the convex hull closely enough by a polygon.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;(added: for asymmetry in high density finite packings up to n=348, see &lt;a href=&quot;http://arxiv.org/abs/1002.0604&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/1002.0604&lt;/a&gt; and a long series of theory papers by the same authors.  Best known packings of small numbers of disks in circles, hexagons, squares, and other shapes are displayed at: &lt;a href=&quot;http://www2.stetson.edu/~efriedma/packing.html&quot; rel=&quot;nofollow&quot;&gt;http://www2.stetson.edu/~efriedma/packing.html&lt;/a&gt;  .)&lt;/p&gt;&#xA;" OwnerUserId="467" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-09-10T07:32:52.603" LastActivityDate="2010-09-10T07:32:52.603" CommentCount="14" />
  <row Id="4360" PostTypeId="2" ParentId="4358" CreationDate="2010-09-10T03:50:19.883" Score="3" ViewCount="" Body="&lt;p&gt;For the first question, let $S$ be the set of all rationals less than $\pi$.  Then $S$ has no least upper bound in $\mathbb{Q}$.  On the other hand, if $T = S\cup {4}$, then 4 is the least upper bound of $S$ inside $T$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the second question, let $S$ be the set of all rationals strictly less than 1, and let $T = \mathbb{Q}\cap \left([-\infty, 1)\cup [2, \infty)\right)$.  Then the least upper bound of $S$ inside $T$ is 2.  But of course the least upper bound of $S$ in $\mathbb{Q}$ is 1.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This shows the answer to your second question is &quot;Yes.&quot;&lt;/p&gt;&#xA;" OwnerUserId="1392" LastActivityDate="2010-09-10T03:50:19.883" CommentCount="1" />
  <row Id="4361" PostTypeId="2" ParentId="4358" CreationDate="2010-09-10T03:50:54.620" Score="5" ViewCount="" Body="&lt;p&gt;Yes, it is possible for $\sup_T(S)$ to exist, but $\sup_\mathbb{Q}(S)$ not to exist. Yes, it's possible for both to exist and be distinct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is an example of the latter. Take $S=\{x\in\mathbb{Q}|0\leq x\lt 1\}$. Let $T=S\cup{{2}}$. Then $S$ is bounded above in $T$, and has a supremum, namely $2$. Indeed, for every $t\in T$, if $t\lt 2$, then there exists $s\in S$ such that $t\lt s\leq 2$, so $2$ is the supremum of $S$. And of course, the supremum of $S$ in $\mathbb{Q}$ is $1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the former, take $S=\{x\in\mathbb{Q}|0\leq x \lt \sqrt{2}\}$ (or any irrational number you please), and take $T=S\cup\{2\}$ (or any number greater than $\sqrt{2}$). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, if you have partially ordered sets $P\subseteq Q$, and a subset $A$ of $P$, you can have that $A$ has a supremum in $P$ but not in $Q$; has a supremum in $Q$ but not in $P$; has suprema in neither; has the same supremum in both; or has suprema in both but they are distinct. The one thing you can say is that if both suprema exist, then $\sup_Q(A)\leq\sup_P(A)$. So in your situation, you do know that if they both exist you will have $\sup_{\mathbb{Q}}(S)\leq\sup_{T}(S)$, but you can have strict inequality.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-09-10T04:02:12.033" LastActivityDate="2010-09-10T04:02:12.033" CommentCount="1" />
  <row Id="4362" PostTypeId="1" AcceptedAnswerId="4364" CreationDate="2010-09-10T04:05:02.650" Score="8" ViewCount="243" Body="&lt;p&gt;Consider the following sequence $a_n$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a_1 = 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a_n = 1 + \frac{1}{2^n-2} \sum_{i=1}^{n-1} \binom{n}{i} a_i$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first few terms are $0,1,\frac{3}{2},\frac{13}{7},\frac{15}{7}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sequence comes out of the analysis of a certain process, whose asymptotics we expect to be about $\log_2 n$. Numerical computations show that indeed $a_n \approx \log_2 n - C$, where $C \approx 0.318$. It should be easy to show that $a_n = \log_2 n + O(1)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt; Is it true that $a_n = \log_2 n - C + o(1)$ for some constant $C$? &lt;/b&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are some sequences defined by similar recurrence relations for which the &quot;constant&quot; part actually wiggles (unfortunately the reference escapes me).&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-29T22:48:46.400" Title="A recurrence that wiggles?" Tags="&lt;asymptotics&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="4363" PostTypeId="1" CreationDate="2010-09-10T04:21:49.203" Score="4" ViewCount="354" Body="&lt;p&gt;If $\varphi:U\subset \mathbb{R}^n \to \mathbb{R}^m$ is $C^1$, let $\mathrm{T}\varphi:\mathrm{T}U \to \mathrm{T}R^m$ be its tangent map. The inverse function theorem tells us that if $\ker(\mathrm{T}\varphi(x))$ is zero, $\varphi$ is injective in some neighborhood of $x$. If the kernel is non-zero, what can we say about $\varphi$ near $x$ provided we know the kernel? In particular, can we say anything about curves through $x$ whose tangents belong to this kernel? &lt;/p&gt;&#xA;" OwnerUserId="1320" LastEditorUserId="6296" LastEditorDisplayName="" LastEditDate="2011-03-20T14:27:49.813" LastActivityDate="2011-03-20T14:27:49.813" Title="Kernel of the tangent map" Tags="&lt;differential-geometry&gt;" AnswerCount="2" CommentCount="8" />
  <row Id="4364" PostTypeId="2" ParentId="4362" CreationDate="2010-09-10T07:14:27.373" Score="10" ViewCount="" Body="&lt;p&gt;This is the Erdos-Renyi &quot;20 Questions&quot; problem.  You want to guess an integer in a set of N elements by selecting subsets and asking questions of the form &quot;and is it in &lt;em&gt;this&lt;/em&gt; subset?&quot;.  For the question to provide information the subset cannot be empty or the entire set of uneliminated possibilities, and if an informative question (subset) is chosen at random one gets the above recursion for the expected length of the game.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Erdos and Renyi showed that the expected number of questions is $\log_2(n)$ plus a non-constant periodic function of $\log_2(n)$.  I don't remember if they allowed uninformative questions, so the problem could be slightly different, but similar log-periodic behavior should appear in both versions of the problem.  (added:  the reason it should be the same is that the dominant $\log_2(n)$ term comes from the concentration of the size of the random subset around $k/2$ where $k$ is the number of remaining possibilities, and the log-periodic part comes from the $O(\sqrt{k})$ dispersion of the subset size near $k/2$.  What happens at the very extremes of the distribution is much less influential.)&lt;/p&gt;&#xA;" OwnerUserId="467" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-09-10T07:42:23.283" LastActivityDate="2010-09-10T07:42:23.283" CommentCount="1" />
  <row Id="4365" PostTypeId="1" AcceptedAnswerId="4366" CreationDate="2010-09-10T07:17:48.323" Score="5" ViewCount="124" Body="&lt;p&gt;A number $n$ is perfect if it's equal to the sum of its divisors (smaller than itself). A well known theorem by Euler states that every even perfect number is of the form $2^{p-1}(2^p-1)$ where $2^p-1$ is prime (this is what is called a &lt;strong&gt;Mersenne&lt;/strong&gt; prime).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How is this theorem proved? The converse (showing that every number of that form is perfect) is simple (the divisors of such a number are powers of 2 and powers of 2 times the Mersenne prime, and it's an easy sum to calculate) but I couldn't find a proof for the theorem itself.&lt;/p&gt;&#xA;" OwnerUserId="1818" LastActivityDate="2010-09-10T07:28:01.350" Title="How to show that all even perfect numbers are obtained via Mersenne primes?" Tags="&lt;number-theory&gt;" AnswerCount="1" />
  <row Id="4366" PostTypeId="2" ParentId="4365" CreationDate="2010-09-10T07:28:01.350" Score="6" ViewCount="" Body="&lt;p&gt;Many elementary number theory texts have the proof, e.g., Hardy &amp;amp; Wright's.&#xA;&lt;a href=&quot;http://primes.utm.edu/notes/proofs/EvenPerfect.html&quot; rel=&quot;nofollow&quot;&gt;Here&lt;/a&gt; is an online proof from Chris Caldwell's &lt;em&gt;Prime Pages&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-09-10T07:28:01.350" />
  <row Id="4367" PostTypeId="1" CreationDate="2010-09-10T07:58:40.090" Score="1" ViewCount="53" Body="&lt;p&gt;In the attached paper the writer proves the isomorphism between potential and congestion games by building a congestion game from a given potential game (theorem 1, subsection 2.7). He defines the groups M_1 and M_2 as facilities in the congestion games. If there are only 2 players in the given potential game, then M_1 and M_2 are sets of matrices. Can I look at an element from either of them as a payment matrix for a potential game? A&lt;a href=&quot;http://ie.technion.ac.il/~dov/ijcai_multi6.pdf&quot; rel=&quot;nofollow&quot;&gt;link text&lt;/a&gt;nd if so - why?&lt;/p&gt;&#xA;" LastActivityDate="2010-09-10T07:58:40.090" Title="The isomorphism between congestion games and potential games" Tags="&lt;game-theory&gt;" CommentCount="1" />
  <row Id="4368" PostTypeId="1" AcceptedAnswerId="4369" CreationDate="2010-09-10T08:01:17.977" Score="5" ViewCount="326" Body="&lt;p&gt;I am trying to compute the asymptotic growth-rate in a specific combinatorial problem depending on a parameter w, using the Transfer-Matrix method. This amounts to computing the largest eigenvalue of the corresponding matrix.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For small values of w, the corresponding matrix is small and I can use the so-called power method - start with some vector, and multiply it by the matrix over and over, and under certain conditions you'll get the eigenvector corresponding to the largest eigenvalue. However, for the values of w I'm interested in, the matrix becomes to large, and so the vector becomes too large - $n&gt;10,000,000,000$ entries or so, so it can't be contained in the computer's memory anymore and I need extra programming tricks or a very powerful computer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the matrix itself, I don't need to store it in memory - I can access it as a black box, i.e. given $i,j$ I can return $A_{ij}$ via a simple computation. Also, the matrix has only 0 and 1 entries, and I believe it to be sparse (i.e. only around $\log n$ of the entries are 1's, $n$ being the number of rows/columns). However, the matrix is &lt;strong&gt;not&lt;/strong&gt; symmetric.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there some method more space-effective for computation of eigenvalues for a case like this?&lt;/p&gt;&#xA;" OwnerUserId="1818" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-09-10T10:31:04.830" LastActivityDate="2010-09-12T10:19:35.053" Title="Computing the largest Eigenvalue of a very large sparse matrix?" Tags="&lt;linear-algebra&gt;&lt;matrices&gt;&lt;numerical-methods&gt;" AnswerCount="3" CommentCount="4" FavoriteCount="1" />
  <row Id="4369" PostTypeId="2" ParentId="4368" CreationDate="2010-09-10T09:00:57.040" Score="5" ViewCount="" Body="&lt;p&gt;You could use the Arnoldi Iteration algorithm. This algorithm only requires the matrix $A$ for matrix-vector multiplication. I'm expecting that you will be able to black-box the function $v\rightarrow Av$. What you generate is an upper Hessenberg matrix $H$ whose eigenvalues whose can be computed cheaply (by a direct method or Rayleigh quotient iteration) and which approximate the eigenvalues of $A$. Arnoldi Iteration will give the best approximation to the dominant eigenvalue so I suspect you won't have to do many iterations before you have a good estimate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An excellent introduction to this is: &quot;Numerical Linear Algebra&quot; by Trefethen and Bau. (p250)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The basic algorithm can be found here: &lt;a href=&quot;http://en.wikipedia.org/wiki/Arnoldi_iteration&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Arnoldi_iteration&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the only thing that is required to make this a fully functional algorithm is a termination condition. Since you don't seem to need the dominant eigenvalue to a high degree of accuracy I would not worry and just stop when the dominant eigenvalue estimate doesn't change too much.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have Matlab you can always use the built in function eigs(Afun,n,...) where Afun is the black-box function handle that computes $Av$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Good luck!&lt;/p&gt;&#xA;" OwnerUserId="1794" LastEditorUserId="1794" LastEditorDisplayName="" LastEditDate="2010-09-10T09:29:00.427" LastActivityDate="2010-09-10T09:29:00.427" CommentCount="3" />
  <row Id="4370" PostTypeId="1" CreationDate="2010-09-10T11:02:52.957" Score="0" ViewCount="250" Body="&lt;p&gt;If $h(x_{t\wedge \tau})$ is a submartingale such that $h(0)=1$ and $\tau$ is a stopping time. Why is it that $h(x_\tau)=1_{{\tau&amp;lt;\infty}}$?. Also $h(x)$ is decreasing.&lt;/p&gt;&#xA;" OwnerUserId="626" LastEditorUserId="626" LastEditorDisplayName="" LastEditDate="2010-09-10T14:26:50.317" LastActivityDate="2010-09-10T14:26:50.317" Title="stopping time and martingale" Tags="&lt;stochastic-processes&gt;" CommentCount="11" />
  <row Id="4371" PostTypeId="1" AcceptedAnswerId="4374" CreationDate="2010-09-10T11:48:22.973" Score="3" ViewCount="480" Body="&lt;p&gt;I would like to learn Graph Theory from the beginning. It seems to me that one does not need to be familiar with many abstract type subjects to be able to understand the more basic concepts of graphs.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Which subjects should one know prior to learn Graph Theory at the introductory level?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;And which book or lecture notes would you advise to learn it? &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="752" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2010-09-10T18:17:40.333" LastActivityDate="2010-09-14T08:41:39.037" Title="Prerequisites for learning (basic) Graph Theory" Tags="&lt;books&gt;&lt;graph-theory&gt;" AnswerCount="5" FavoriteCount="1" />
  <row Id="4372" PostTypeId="1" CreationDate="2010-09-10T11:54:39.127" Score="1" ViewCount="158" Body="&lt;p&gt;If $f$ is a positive function, the intuitive interpretation of the Riemann integral&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\int_a^b f(x) dx$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is the area under the curve $f$ between $a$ and $b$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose $f$ and $g$ are smooth positive functions. Is it correct to interpret the Riemann-Stieltjes integral&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\int_a^b f(x) d g(x)$ as the volume under a &quot;ribbon&quot;, where the height of the ribbon at a point $u$ between $a$ and $b$ is determined by $f$ and the thickness is determined by $g'$?&lt;/p&gt;&#xA;" OwnerUserId="1820" LastActivityDate="2010-09-10T15:55:48.090" Title="Is this interpretation of Stieltjes integration correct?" Tags="&lt;real-analysis&gt;&lt;intuition&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="1" />
  <row Id="4373" PostTypeId="2" ParentId="4371" CreationDate="2010-09-10T12:11:23.777" Score="1" ViewCount="" Body="&lt;p&gt;You don't need more than knowledge of basic notations in Mathematics to read a &lt;strong&gt;basic&lt;/strong&gt; book on Graph Theory. However, some experience in mathematics is helpful, even if the material is not used directly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My favorite books for &quot;pure&quot; graph theory is &quot;Graph Theory&quot; by Harary and &quot;Modern Graph Theory&quot; by Bollobas. They also delve into algebraic graph theory in later chapters and for that you'll need some basic linear algebra and group theory. Also, graph theory is widely used in computer science and so, for example, many chapters in the &quot;Introduction to algorithms&quot; by Corman and co. deal with algorithms on graphs (an interesting subject by itself but maybe not what you have in mind).&lt;/p&gt;&#xA;" OwnerUserId="1818" LastActivityDate="2010-09-10T12:11:23.777" CommentCount="2" />
  <row Id="4374" PostTypeId="2" ParentId="4371" CreationDate="2010-09-10T12:13:41.563" Score="8" ViewCount="" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;(a)&lt;/strong&gt; Basic logic + set operations almost goes without saying (&lt;em&gt;e.g.&lt;/em&gt; logical conjunction&amp;nbsp;/&amp;nbsp;set intersections; also equivalence classes, sets and relations obtained by modding out subsets, &lt;em&gt;etc&lt;/em&gt;).&lt;br&gt;&lt;br&gt;&lt;strong&gt;(b)&lt;/strong&gt; Depending on how 'basic' you mean, you may or may not derive benefit from linear algebra (for playing with adjacency matrices), probability theory (for applying the probabilistic method, natch), elementary number theory (&lt;em&gt;e.g.&lt;/em&gt; arithmetic modulo &lt;em&gt;N&lt;/em&gt;) and experience with asymptotic analysis (knowledge of what O(&lt;em&gt;N&lt;/em&gt;), o(&lt;em&gt;N&lt;/em&gt;), &amp;omega;(&lt;em&gt;N&lt;/em&gt;), etc. refer to and how to play with them).&lt;br&gt;&lt;br&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I found Diestel's text quite nice. (Furthermore, it is freely available from &lt;a href=&quot;http://diestel-graph-theory.com/&quot; rel=&quot;nofollow&quot;&gt;his website&lt;/a&gt;). I also found that undoing his proofs-by-contradiction provides good exercise.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="439" LastActivityDate="2010-09-10T12:13:41.563" CommentCount="3" />
  <row Id="4377" PostTypeId="2" ParentId="4273" CreationDate="2010-09-10T14:21:50.153" Score="4" ViewCount="" Body="&lt;p&gt;They are called &quot;personal probabilities&quot; in the literature on Bayesian statistics, which is the field where the possibility of different probability assessments among different observers is considered.  You could check out the Wikipedia article on &lt;a href=&quot;http://en.wikipedia.org/wiki/Bayesian_probability#Personal_probabilities_and_objective_methods_for_constructing_priors&quot; rel=&quot;nofollow&quot;&gt;Bayesian probability&lt;/a&gt;, for instance.&lt;/p&gt;&#xA;" OwnerUserId="1489" LastActivityDate="2010-09-10T14:21:50.153" CommentCount="4" />
  <row Id="4378" PostTypeId="2" ParentId="4362" CreationDate="2010-09-10T15:03:45.297" Score="4" ViewCount="" Body="&lt;p&gt;Not really an answer, but too long for a comment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;A recurrence with a wiggle&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The following was proposed by Lennart Rade&#xA;  in a 1991 issue of the &lt;em&gt;American Mathematical Monthly&lt;/em&gt;, and the&#xA;   solution appears in the January 1994 issue, on page 78.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Begin with $n$ fair coins all showing tails.&#xA;  Toss them all and pick up only the coins with tails.&#xA;  Repeat this procedure until all coins show heads. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The number of coins showing tails is a Markov chain that&#xA;starts in state $n$ and ends up at 0. The transition probabilities&#xA;are binomial: $p_{n,i}={n \choose i}\left({1\over 2}\right)^n \mbox{ for }0\leq i\leq n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sequence $u_n=P_{n+1}(\mbox{chain hits }1)$ satisfies  $u_0=1$ and for $n\geq1$&#xA; $$u_n=\sum_{j=0}^n{n+1\choose j}2^{-(n+1)} u_{n-j},$$&#xA; or $$u_n=\sum_{j=1}^n{n+1\choose j}{2^{-(n+1)}\over 1-2^{-(n+1)}} u_{n-j}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, this problem can be solved explicitly giving&#xA;$$u_n={n+1\over2}\sum_{k=0}^\infty 2^{-k}(1-2^{-k})^n.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Analysis of this expression shows that, despite appearances, the sequence $u_n$ does&#xA; &lt;strong&gt;not&lt;/strong&gt; converge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It slowly oscillates about the pseudo limit $${1\over2\ln(2)}=.72134$$&#xA;with an amplitude of about $${|\Gamma(1-2\pi i/\ln(2))|\over2\ln(2)}=.00000356.$$&lt;br&gt;&#xA;It's a fun example to show students as the oscillations are invisible graphically. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For related examples see &#xA;&lt;em&gt;The asymptotic probability of a tie for first place&lt;/em&gt;,&#xA;by B. Eisenberg, G. Stengle, and G. Strang, &lt;strong&gt;Annals of Applied Probability 3&lt;/strong&gt;, 731-745 (1993). &lt;/p&gt;&#xA;" OwnerUserId="940" LastActivityDate="2010-09-10T15:03:45.297" CommentCount="1" />
  <row Id="4379" PostTypeId="2" ParentId="4372" CreationDate="2010-09-10T15:55:48.090" Score="1" ViewCount="" Body="&lt;p&gt;You are effectively replacing $\int_a^b f(x) dg(x)$ with $\int_a^b f(x) g'(x) dx$. &lt;a href=&quot;http://en.wikipedia.org/wiki/Riemann%E2%80%93Stieltjes_integral#Properties_and_relation_to_the_Riemann_integral&quot; rel=&quot;nofollow&quot;&gt;A quick trip to Wikipedia&lt;/a&gt; reveals that this is perfectly fine when $g$ is absolutely continuous, or when it has a continuous derivative, but may not be valid otherwise:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&lt;p&gt;If $g$ should happen to be everywhere differentiable, then the integral may still be different from the Riemann integral&#xA;&#xA;    $$\int_a^b f(x) g'(x) \, dx,$$&#xA;&#xA;for example, if the derivative is unbounded. But if the derivative is continuous, they will be the same. This condition is also satisfied if $g$ is the (Lebesgue) integral of its derivative; in this case $g$ is said to be absolutely continuous.&#xA;&#xA;&lt;p&gt;However, $g$ may have jump discontinuities, or may have derivative zero almost everywhere while still being continuous and increasing (for example, $g$ could be the Cantor function), in either of which cases the Riemann–Stieltjes integral is not captured by any expression involving derivatives of $g$.&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Edit:&lt;/em&gt; I didn't notice the word &quot;smooth&quot; in your question. In this light, the answer to your question is yes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, &lt;a href=&quot;http://mathworld.wolfram.com/StieltjesIntegral.html&quot; rel=&quot;nofollow&quot;&gt;MathWorld is probably a better reference&lt;/a&gt;: the two expressions are equal &quot;if $f$ is continuous and $g'$ is Riemann integrable over the specified interval&quot;.&lt;/p&gt;&#xA;" OwnerUserId="856" LastActivityDate="2010-09-10T15:55:48.090" />
  <row Id="4380" PostTypeId="1" AcceptedAnswerId="4382" CreationDate="2010-09-10T20:06:17.260" Score="3" ViewCount="280" Body="&lt;p&gt;If you look at &lt;a href=&quot;http://www.wolframalpha.com/input/?i=world+population&quot; rel=&quot;nofollow&quot;&gt;Wolfram Alpha&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/x0ILQ.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or this Wikipedia page &lt;a href=&quot;http://en.wikipedia.org/wiki/List_of_countries_by_median_age&quot; rel=&quot;nofollow&quot;&gt;List of countries by median age&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/QBtjT.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clearly &lt;a href=&quot;http://en.wikipedia.org/wiki/Median&quot; rel=&quot;nofollow&quot;&gt;&lt;strong&gt;median&lt;/strong&gt;&lt;/a&gt; seems to be the statistic of choice when it comes to ages.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not able to explain to myself why &lt;a href=&quot;http://en.wikipedia.org/wiki/Arithmetic_mean&quot; rel=&quot;nofollow&quot;&gt;&lt;strong&gt;arithmetic mean&lt;/strong&gt;&lt;/a&gt; would be a worse statistic. Why is it so?&lt;/p&gt;&#xA;" OwnerUserId="324" LastEditorUserId="324" LastEditorDisplayName="" LastEditDate="2010-09-10T20:23:15.127" LastActivityDate="2010-09-12T22:17:48.857" Title="Why is median age a better statistic than mean age?" Tags="&lt;statistics&gt;&lt;average&gt;" AnswerCount="2" CommentCount="6" />
  <row Id="4381" PostTypeId="1" AcceptedAnswerId="4383" CreationDate="2010-09-10T20:16:48.073" Score="3" ViewCount="267" Body="&lt;p&gt;Hi! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was doing a physics problem today and I needed to find when some movement is accelerating and when it is decelerating. So after some calculations, I got the equation I needed. I know that it is correct because when plotted, I can see from graph that I got the expected solution. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately I don't know how to actually start solving the inequality. It looks like &#xA;$$k \sin(\frac {\pi x} {2})\cos(\frac {\pi x} {2})&gt;0$$ where $k$ is some huge sausage which is constant and positive.&lt;/p&gt;&#xA;" OwnerUserId="348" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-09-11T06:21:53.403" LastActivityDate="2010-09-11T06:21:53.403" Title="How to solve this trigonometric inequality: $k \sin(\frac {\pi x} {2})\cos(\frac {\pi x} {2})&gt;0$?" Tags="&lt;algebra-precalculus&gt;&lt;trigonometry&gt;" AnswerCount="3" CommentCount="9" FavoriteCount="1" />
  <row Id="4382" PostTypeId="2" ParentId="4380" CreationDate="2010-09-10T20:19:13.060" Score="4" ViewCount="" Body="&lt;p&gt;Median is what many people actually have in mind when they say &quot;mean.&quot;  It's easier to interpret the median: half the population is above this age and half are below.  Mean is a little more subtle. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;People look for symmetry and sometimes impose symmetry when it isn't there.  The age distribution in a population is far from symmetric, so the mean could be misleading. Age distributions are something like a pyramid.  Lots of children, not many elderly.  (Or at least that's how it is in a sort of steady state.  In the US, for example, the post-WWII baby boom generation has distorted the distribution as they age.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With an asymmetrical distribution, it may be better to report the median because it is a symmetrical statistic in the sense that it splits the population in half.  Said another way, the median is symmetrical even if the distribution isn't.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update&lt;/strong&gt;: I got my logic backward when I first answered and said the mean would be lower than the median.  I meant the opposite.&lt;/p&gt;&#xA;" OwnerUserId="247" LastEditorUserId="247" LastEditorDisplayName="" LastEditDate="2010-09-12T22:17:48.857" LastActivityDate="2010-09-12T22:17:48.857" CommentCount="3" />
  <row Id="4383" PostTypeId="2" ParentId="4381" CreationDate="2010-09-10T20:21:39.463" Score="6" ViewCount="" Body="&lt;p&gt;You can use the identity&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sin 2x = 2 \sin x \cos x$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus you only need to determine when $$ \sin \pi x  &gt; 0$$ and this happens if and only if $$ x \in (2m, 2m+1), m \in \mathbb{Z}$$&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-09-10T20:21:39.463" />
  <row Id="4384" PostTypeId="2" ParentId="4381" CreationDate="2010-09-10T20:24:25.693" Score="3" ViewCount="" Body="&lt;p&gt;Since $k$ is positive, it is immaterial. You want to know the values of $x$ for which $\sin(\frac{\pi x}{2})$ and $\cos(\frac{\pi x}{2})$ have the same sign.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each function has period $4$, so we can just look at what happens on $[0,4]$. $\sin(\frac{\pi x}{2})$ is positive when $0\lt\frac{\pi x}{2}\lt\pi$ (that is, when $0\lt x\lt 2$), and negative when $2\lt x \lt 4$. $\cos(\frac{\pi x}{2})$ is positive when $0\leq \frac{\pi x}{2}\lt \frac{\pi}{2}$ and when $\frac{3\pi}{2}\lt \frac{\pi x}{2}\lt 4$ (that is, when $0\leq x\lt 1$ and when $3\lt x\lt 4$). So the two have the same sign on $(0,1)$, and on $(2,3)$. Lather, rinse, and repeat. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Of course, this is easier to do using the identity that others have suggested; the approach of looking at the signs of the different factors is what you would use in general, or when there is no easy way of reducing the expression to a single item). &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-09-10T20:32:54.247" LastActivityDate="2010-09-10T20:32:54.247" />
  <row Id="4385" PostTypeId="2" ParentId="4371" CreationDate="2010-09-10T20:40:17.037" Score="3" ViewCount="" Body="&lt;p&gt;Graph Theory is indeed a very quick starting subject in the sense that one does need to have studied calculus and other mathematical subjects to get started. However, there are lots of books that are specialized towards particular purposes: applications in general, network applications, distances in graphs, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One book that I think has nice balance is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Introduction to Graph Theory (Second edition) by Douglas West, Prentice-Hall, 2001.&lt;/p&gt;&#xA;" OwnerUserId="1369" LastEditorUserId="1369" LastEditorDisplayName="" LastEditDate="2010-09-11T00:59:56.030" LastActivityDate="2010-09-11T00:59:56.030" />
  <row Id="4386" PostTypeId="1" CreationDate="2010-09-10T20:47:45.427" Score="4" ViewCount="79" Body="&lt;p&gt;Hi, I'm a noob, and I'm not a mathematician (Although I will be a Math major next semester).  My question is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have 68 maps I would like to use as posters on my wall at home.  They are all rectangles, and all different dimensions.  Is there an algorithm (or better yet, a website or other simple computer program I can use) to take the dimensions of all these posters as input and then compute the optimal tiling arrangement on the wall so there is little as possible &quot;wasted&quot; empty space?&lt;/p&gt;&#xA;" OwnerUserId="1825" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-09-10T23:23:42.210" LastActivityDate="2010-09-10T23:23:42.210" Title="Tiling Posters on a Wall" Tags="&lt;geometry&gt;&lt;tiling&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="1" />
  <row Id="4387" PostTypeId="2" ParentId="4386" CreationDate="2010-09-10T21:00:47.747" Score="1" ViewCount="" Body="&lt;p&gt;You are attempting to solve what is referred to as a &quot;packing problem&quot;. It's generally a hard problem but for your case I think that the existing methods will work. Try looking here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://pollinimini.net/blog/rectangle-packing/&quot; rel=&quot;nofollow&quot;&gt;http://pollinimini.net/blog/rectangle-packing/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems he has already coded something. If it's not what you're after, google &quot;rectangle packing&quot; and I'm sure you'll run into something.&lt;/p&gt;&#xA;" OwnerUserId="1818" LastActivityDate="2010-09-10T21:00:47.747" />
  <row Id="4388" PostTypeId="2" ParentId="4386" CreationDate="2010-09-10T21:02:12.937" Score="2" ViewCount="" Body="&lt;p&gt;This problem is known as 2-dimensional bin packing, and some algorithms for this problem can be found here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.csc.liv.ac.uk/~epa/surveyhtml.html&quot; rel=&quot;nofollow&quot;&gt;http://www.csc.liv.ac.uk/~epa/surveyhtml.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1369" LastActivityDate="2010-09-10T21:02:12.937" />
  <row Id="4390" PostTypeId="2" ParentId="4380" CreationDate="2010-09-10T22:28:44.567" Score="1" ViewCount="" Body="&lt;p&gt;The best statistic to summarize a distribution depends upon the distribution and what you want to use it for.  For distributions that are nicely bell-shaped, the mean, median, and mode are close together and it doesn't matter.  For skew distributions the mean is out on the skew side from the median, but it still represents the expected value of the average of a large number of samples.  The median is closer to more of the individuals than the mean.  For stranger distributions no one number can provide a useful summary.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-09-10T22:28:44.567" CommentCount="1" />
  <row Id="4391" PostTypeId="2" ParentId="4381" CreationDate="2010-09-10T22:31:35.077" Score="1" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Slighty simpler&lt;/strong&gt;:$\;$ note $\rm\; ab &gt; 0 \iff a/b &gt; 0\;\;$ so $\;\;\sin(z)\cos(z)&gt;0 \iff \tan(z) &gt; 0$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rest is trivial using $\;\;\tan(-z) = -\tan(z)\;\;\:$ and $\:\;\;\tan(z+\pi) = \tan(z)$&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-09-10T22:42:48.923" LastActivityDate="2010-09-10T22:42:48.923" />
  <row Id="4392" PostTypeId="2" ParentId="4268" CreationDate="2010-09-11T00:05:04.600" Score="4" ViewCount="" Body="&lt;p&gt;As far as I know, your definition&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A prime is an element p such that whenever p divides ab, then either p divides a or p divides b,&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;is the true definition of &quot;prime&quot;. The usual one, &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;... an element p which cannot be expressed as a product of non-unit elements,&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;is the definition of an &lt;em&gt;irreducible&lt;/em&gt; element. Now, in every ring all primes are also irreducible, but the converse is in general not true. In other words, there are rings where the two definitions are not equivalent. One case in which they are indeed equivalent is when the ring is a &lt;em&gt;unique factorization domain&lt;/em&gt; (the statement of the so-called &quot;fundamental theorem of arithmetic&quot; is that the integers form one such ring).&lt;/p&gt;&#xA;" OwnerUserId="1828" LastEditorUserId="1828" LastEditorDisplayName="" LastEditDate="2010-09-11T11:39:25.117" LastActivityDate="2010-09-11T11:39:25.117" CommentCount="3" />
  <row Id="4393" PostTypeId="2" ParentId="4335" CreationDate="2010-09-11T00:11:10.983" Score="2" ViewCount="" Body="&lt;p&gt;I'll add another title (it's a bit on the theoretical side, but still at the introductory level, very readable and definitely worth):&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Serge Lang, &lt;em&gt;Linear algebra&lt;/em&gt; 3rd. ed., Springer&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1828" LastActivityDate="2010-09-11T00:11:10.983" CommunityOwnedDate="2010-09-11T00:11:10.983" />
  <row Id="4394" PostTypeId="2" ParentId="2395" CreationDate="2010-09-11T00:32:10.590" Score="0" ViewCount="" Body="&lt;p&gt;If you can get a copy (which is improbable, I'm afraid),&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Vladimir Smirnov, &lt;em&gt;A Course of Higher Mathematics, VOLUME 1&lt;/em&gt;, Pergamon press 1964.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This is a translation of a beautiful treatise on mathematical analysis which served as a textbook for generations of engineering students in the USSR. The first volume, in particular, is one of the best introductions to calculus I've ever seen.&lt;/p&gt;&#xA;" OwnerUserId="1828" LastActivityDate="2010-09-11T00:32:10.590" CommunityOwnedDate="2010-09-11T00:32:10.590" />
  <row Id="4395" PostTypeId="2" ParentId="4335" CreationDate="2010-09-11T00:48:32.547" Score="0" ViewCount="" Body="&lt;p&gt;In my opinion, Schaum's Linear Algebra is very straight forward, has a nice balance between pure and applied, and has the advantage that it is cheap. Definitely, if you are auto-didactic, it will be easier to work through. &lt;/p&gt;&#xA;" OwnerUserId="722" LastActivityDate="2010-09-11T00:48:32.547" CommunityOwnedDate="2010-09-11T00:48:32.547" />
  <row Id="4396" PostTypeId="2" ParentId="3955" CreationDate="2010-09-11T01:25:02.193" Score="1" ViewCount="" Body="&lt;p&gt;Here is how you can construct a linear relation for general bounds:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If $x$ goes from $a$ to $b$,&lt;/li&gt;&#xA;&lt;li&gt;then $x-a$ goes from $0$ to $b-a$,&lt;/li&gt;&#xA;&lt;li&gt;then $\frac{x-a}{b-a}$ goes from $0$ to $1$,&lt;/li&gt;&#xA;&lt;li&gt;then $(x-a) \frac{d-c}{b-a}$ goes from $0$ to $d-c$,&lt;/li&gt;&#xA;&lt;li&gt;then $y := c + (x-a) \frac{d-c}{b-a}$ goes from $c$ to $d$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;If you plug $a=0$, $b=100$, $c=1000$, $d=0$ as in your example, you obtain $y = 1000 + (x-0) \frac{0-1000}{100-0} = 1000 - 10x$ as in the other answers.&lt;/p&gt;&#xA;" OwnerUserId="1828" LastActivityDate="2010-09-11T01:25:02.193" />
  <row Id="4397" PostTypeId="2" ParentId="250" CreationDate="2010-09-11T01:38:29.773" Score="4" ViewCount="" Body="&lt;p&gt;&quot;The natural numbers are as many as the even natural numbers&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This statement is trivial and not worth to be named &quot;theorem&quot;, but it is rather counterintuitive if you don't know the meaning of &quot;as many&quot;. At least, it was for Galileo :)&lt;/p&gt;&#xA;" OwnerUserId="1828" LastEditorUserId="1828" LastEditorDisplayName="" LastEditDate="2010-09-11T01:44:31.540" LastActivityDate="2010-09-11T01:44:31.540" CommunityOwnedDate="2010-09-11T01:38:29.773" CommentCount="2" />
  <row Id="4398" PostTypeId="2" ParentId="1819" CreationDate="2010-09-11T02:22:37.813" Score="3" ViewCount="" Body="&lt;p&gt;For $X_1$, $X_2$, $\ldots$, i.i.d., with finite variance $\sigma ^{2}$, the &lt;strong&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Central_limit_theorem#Classical_central_limit_theorem&quot; rel=&quot;nofollow&quot;&gt;Central Limit Theorem&lt;/a&gt;&lt;/strong&gt; (convergence in distribution) says that if we wait long enough, that is, take a large enough sample, then $$ \frac{\sqrt n(\overline{X}_n-\mu)}{\sigma}$$ will have a probability distribution which is arbitrarily close to a &lt;em&gt;norm&lt;/em&gt;(0,1) distribution.  Notice the CLT doesn't say anything about the actual behavior of any particular $\overline{X}$ for large $n$, only that if we observed a whole bunch of them (all at that large $n$) and made a histogram then it would be approximately bell-shaped. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;strong&gt;Weak Law of Large Numbers&lt;/strong&gt; (convergence in probability) says that if we wait long enough (&lt;em&gt;i.e.&lt;/em&gt;, take a large enough sample) then the chance that $$ \frac{\overline{X}_n-\mu}{\sigma}$$ is nearby zero can be made arbitrarily high.  Note a couple of things: &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;the $\sqrt{n}$ is gone, and&lt;/li&gt;&#xA;&lt;li&gt;again, nothing is said about any particular observed $\overline{X}$, only that it would be a pretty safe bet that the above is close to zero.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The &lt;strong&gt;Strong Law of Large Numbers&lt;/strong&gt; (convergence almost surely) says that $$ \frac{\overline{X}_n-\mu}{\sigma}$$ is 100% sure to be arbitrarily close to 0, provided we wait long enough (take a large enough sample).  That is &lt;em&gt;strong&lt;/em&gt;, indeed. Again, please note a couple of things:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;the $\sqrt{n}$ is still gone. In fact, this is the same fellow that was in the WLLN, and&lt;/li&gt;&#xA;&lt;li&gt;this time, something is said about the &lt;em&gt;particular&lt;/em&gt; sequence of $\overline{X}_n$'s that we are watching go by.  It's still possible that it would be far away (more than $\epsilon$), but all sequences for which $\overline{X}_n$ &lt;em&gt;never&lt;/em&gt; converges to $\mu$ have probability zero.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Of course, nothing is ever guaranteed in probability; we can do no better than 100% sure.  Also, the SLLN doesn't say anything about &lt;em&gt;how long&lt;/em&gt; we would have to wait to be within $\epsilon$, we would need the &lt;strong&gt;Law of the Iterated Logarithm&lt;/strong&gt; for something like that.  Finally, this discussion is about convergence (in distribution/probability/a.s.) to a constant, while the general definition is about convergence to another random variable (or even convergence of two sequences).  But we can regain the intuition if we think about differences.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't see anything wrong with the other two answers given for this question, I just thought I'd offer another perspective.&lt;/p&gt;&#xA;" OwnerUserId="1830" LastEditorUserId="123" LastEditorDisplayName="" LastEditDate="2010-09-11T06:47:32.997" LastActivityDate="2010-09-11T06:47:32.997" />
  <row Id="4399" PostTypeId="2" ParentId="4268" CreationDate="2010-09-11T03:10:58.760" Score="3" ViewCount="" Body="&lt;p&gt;You have encountered the general ring-theoretic definition of a prime (versus irreducible) element. For general rings one distinguishes between the inequivalent properties of being &lt;strong&gt;irreducible&lt;/strong&gt;, i.e. having no nontrivial factors, and being &lt;strong&gt;prime&lt;/strong&gt;, i.e a nonunit such that if it divides a product then it divides some factor of the product. The latter property is key to unique factorization since one easily proves by induction that products of primes factor uniquely in any domain, i.e. a ring such that $\rm\; ab = 0 \;\Rightarrow\; a=0\ \  or\ \ b=0$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are the precise definitions. Let $\rm\; a,b,p \;$ be elements of a domain $\rm\:Z\:$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;DEFINITION&lt;/strong&gt; $\rm\quad\; a\;$ is a &lt;strong&gt;unit&lt;/strong&gt; $\ \ $ if $\rm\ \ a\: |\: 1\quad\quad\quad\ \ \:\; (\:a\ |\ b\ :=\ a\ \ divides\ \ b\:)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;DEFINITION&lt;/strong&gt; $\quad$ nonunit $\rm\; p\;$ is $\quad\;$ &lt;strong&gt;prime&lt;/strong&gt; $\quad\;$ if $\;\;\;\rm p \;\;\:|\:\;\; ab \;\;\Rightarrow\;\; p|a \;\;\; or \;\;\; p|b$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;DEFINITION&lt;/strong&gt; $\quad$ nonunit $\rm\; p\;$ is $\:$ &lt;strong&gt;irreducible&lt;/strong&gt;$\;$ if $\;\;\:\rm p = ab \;\;\Rightarrow\;\; p|a \;\;\; or \;\;\; p|b\quad\;\;$ (a.k.a.$\:$ &lt;strong&gt;atom&lt;/strong&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;COROLLARY&lt;/strong&gt; $\quad\;$ prime $\;\;\Rightarrow\;\;$ irreducible, since $\;\;\;\rm p = ab \;\;\Rightarrow\;\; p|ab$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Conversely&lt;/strong&gt; $\:\;$ irreducible $\;\Rightarrow\;\;$ prime $\;\;$ iff $\;$ factorizations into irreducibles are unique (upto order and units). Indeed, it is very easy to prove that a factorization into primes is unique, i.e. any other factorization into irreducibles is the same (upto order and units). &lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-09-19T22:34:53.240" LastActivityDate="2010-09-19T22:34:53.240" />
  <row Id="4400" PostTypeId="1" AcceptedAnswerId="4401" CreationDate="2010-09-11T05:30:01.867" Score="7" ViewCount="632" Body="&lt;p&gt;The following probability question appeared in an &lt;a href=&quot;http://math.stackexchange.com/questions/250/a-challenge-by-r-p-feynman-give-counter-intuitive-theorems-that-can-be-transl/346#346&quot;&gt;earlier thread&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I have two children. One is a boy born on a Tuesday. What is the probability I have two boys?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The claim was that it is not actually a mathematical problem and it is only a language problem.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;If one wanted to restate this problem formally the obvious way would be like so:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Definition&lt;/strong&gt;: &lt;em&gt;Sex&lt;/em&gt; is defined as an element of the set $\{\text{boy},\text{girl}\}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Definition&lt;/strong&gt;: &lt;em&gt;Birthday&lt;/em&gt; is defined as an element of the set $\{\text{Monday},\text{Tuesday},\text{Wednesday},\text{Thursday},\text{Friday},\text{Saturday},\text{Sunday}\}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Definition&lt;/strong&gt;: A &lt;em&gt;Child&lt;/em&gt; is defined to be an ordered pair: (sex $\times$ birthday).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $(x,y)$ be a pair of children,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Define an auxiliary predicate $H(s,b) :\!\!\iff s = \text{boy} \text{ and } b = \text{Tuesday}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Calculate $P(x \text{ is a boy and } y \text{ is a boy}|H(x) \text{ or } H(y))$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I don't see any other sensible way to formalize this question.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;To actually solve this problem now requires no thought (infact it is thinking which leads us to guess incorrect answers), we just compute&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{align*}&#xA;&amp;amp; P(x \text{ is a boy and } y \text{ is a boy}|H(x) \text{ or } H(y)) \\&#xA;=&amp;amp; \frac{P(x\text{ is a boy and }y\text{ is a boy and }(H(x)\text{ or }H(y)))}&#xA;        {P(H(x)\text{ or }H(y))} \\&#xA;=&amp;amp; \frac{P((x\text{ is a boy and }y\text{ is a boy and }H(x))\text{ or }(x\text{ is a boy and }y\text{ is a boy and }H(y)))}&#xA;        {P(H(x)) + P(H(y)) - P(H(x))P(H(y))} \\&#xA;=&amp;amp; \frac{\begin{align*} &amp;amp;P(x\text{ is a boy and }y\text{ is a boy and }x\text{ born on Tuesday}) \\&#xA;   + &amp;amp;P(x\text{ is a boy and }y\text{ is a boy and }y\text{ born on Tuesday}) \\&#xA;   - &amp;amp;P(x\text{ is a boy and }y\text{ is a boy and }x\text{ born on Tuesday and }y\text{ born on Tuesday}) \\&#xA;   \end{align*}}&#xA;   {P(H(x)) + P(H(y)) - P(H(x))P(H(y))} \\&#xA;=&amp;amp; \frac{1/2 \cdot 1/2 \cdot 1/7 + 1/2 \cdot 1/2 \cdot 1/7 - 1/2 \cdot 1/2 \cdot 1/7 \cdot 1/7}&#xA;        {1/2 \cdot 1/7 + 1/2 \cdot 1/7 - 1/2 \cdot 1/7 \cdot 1/2 \cdot 1/7} \\&#xA;=&amp;amp; 13/27&#xA;\end{align*}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Now what I am wondering is, does this refute the claim that this puzzle is just a language problem or add to it? Was there a lot of room for misinterpreting the questions which I just missed? &lt;/p&gt;&#xA;" LastEditorUserId="1543" LastEditorDisplayName="anon" LastEditDate="2011-04-18T12:47:39.557" LastActivityDate="2011-05-05T07:25:02.153" Title="Boy Born on a Tuesday - is it just a language trick?" Tags="&lt;probability&gt;&lt;faq&gt;" AnswerCount="9" CommentCount="3" FavoriteCount="7" />
  <row Id="4401" PostTypeId="2" ParentId="4400" CreationDate="2010-09-11T05:42:25.577" Score="8" ViewCount="" Body="&lt;p&gt;There are even trickier aspects to this question. For example, what is the strategy of the guy telling you about his family? If he always mentions a boy first and not a daughter, we get one probability; if he talks about the sex of the first born child, we get a different probability. Your calculation makes a choice in this issue - you choose the version of &quot;if the father has a boy and a girl, he'll mention the boy&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I'm aiming to is this: the question is not well-defined mathematically. It has several possible interpretations, and as such the &quot;problem&quot; here is indeed of the language; or more correctly, the fact that a simple statement in English does not convey enough information to specify the precise model for the problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's look at a simplified version without days. The probability space for the make-up of the family is {BB, GB, BG, GG} (GB means &quot;an older girl and a small boy&quot;, etc). We want to know what is $P(BB|A)$ where A is determined by the way we interpret the statement about the boys. Now let's look at different possible interpretations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) If there is a boy in the family, the statement will mention him. In this case A={BB,BG,GB} and so the probability is $1/3$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) If there is a girl in the family, the statement will mention her. In this case, since the statement talked about a boy, there are NO girls in the family. So A={BB} and so the probability is 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) The statement talks about the sex of the firstborn. In this case A={BB,BG} and so the probability is $1/2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The bottom line: The statement about the family looks &quot;constant&quot; to us, but it must be looked as a function from the random state of the family - and there are several different possible functions, from which you must choose one otherwise no probabilistic analysis of the situation will make sense.&lt;/p&gt;&#xA;" OwnerUserId="1818" LastEditorUserId="1818" LastEditorDisplayName="" LastEditDate="2010-09-11T06:39:56.287" LastActivityDate="2010-09-11T06:39:56.287" CommentCount="24" />
  <row Id="4402" PostTypeId="2" ParentId="4400" CreationDate="2010-09-11T06:53:52.183" Score="0" ViewCount="" Body="&lt;p&gt;The Tuesday is a red herring. It's stated as a fact, thus the probability is 1. Also, it doesn't say &quot;&lt;em&gt;only&lt;/em&gt; one boy is born on a Tuesday&quot;. But indeed, this could be a language thing.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;With 2 children you have the following possible combinations:&lt;br&gt;&#xA;    1. two girls&lt;br&gt;&#xA;    2. a boy and a girl&lt;br&gt;&#xA;    3. a girl and a boy&lt;br&gt;&#xA;    4. two boys  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If at least 1 is a boy we only have to consider the last three combinations. That gives us one in three that both are boys.&lt;br&gt;&#xA;The error which is often made is to consider 2. and 3. as a single combination.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;edit&lt;/strong&gt;&lt;br&gt;&#xA;I find it completely counter-intuitive that the outcome is influenced by the day, and I simulated the problem for one million families with 2 kids. And lo and behold, the outcome is 12.99 in 27. I was wrong.&lt;/p&gt;&#xA;" OwnerUserId="923" LastEditorUserId="923" LastEditorDisplayName="" LastEditDate="2010-09-12T07:16:19.863" LastActivityDate="2010-09-12T07:16:19.863" CommentCount="4" />
  <row Id="4403" PostTypeId="1" AcceptedAnswerId="4434" CreationDate="2010-09-11T07:18:28.213" Score="3" ViewCount="53" Body="&lt;p&gt;I've been struggling with the following problem I found for a while now:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose $(T,\preceq)$ is a partially ordered subset of $(U,\preceq)$ and $S\subseteq T$. If $\inf_T S$ and $u=\inf_U S$ both exist, and there exists a subset $S'\subseteq T$ such that $u=\sup_U S'$, then $\inf_U S=\inf_T S$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;First I noted that $\inf_T S\preceq\inf_U S$, since $\inf_T S\in U$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also saw that $u=\inf_U([u)\cap T)$ by the following: For $x\in S$, $u\preceq x$ and $x\in T$, so $x\in [u)\cap T$, and so $S\subseteq [u)\cap T$. So $u$ is a lower bound of $[u)\cap T$. Now suppose there exists some $y\in U$ such that $u\prec y$, but $y\preceq x$ for all $x\in [u)\cap T$. Since $S\subseteq [u)\cap T$, $y$ would be a lower bound of $S$ larger than $u$, contradicting the fact that $u=\inf_U S$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was hoping to show that $\inf_U S\preceq\inf_T S$ in order to show equality, but I can't piece it together. Can someone explain how to show this?&lt;/p&gt;&#xA;" OwnerUserId="1583" LastActivityDate="2010-09-11T22:32:45.570" Title="For some sets $S\subseteq T\subseteq U$, when is $\inf_T S=\inf_U S$?" Tags="&lt;analysis&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="4404" PostTypeId="2" ParentId="4335" CreationDate="2010-09-11T07:20:06.867" Score="1" ViewCount="" Body="&lt;p&gt;Can't comment, so this will have to be a post. As well as seconding Gilbert Strang's lively lectures on the open MIT site, his book, and Artin's, which are all excellent and all already recommended, I think you could also try Paul Dawkins' online math notes:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://tutorial.math.lamar.edu/Classes/LinAlg/LinAlg.aspx&quot; rel=&quot;nofollow&quot;&gt;http://tutorial.math.lamar.edu/Classes/LinAlg/LinAlg.aspx&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which are a fair bit shorter and, given your ultimate aim, the online OpenGL tutorials at NeHe are also probably useful (because you will see lots of examples of the application of linear algebra in the area of your interest):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://nehe.gamedev.net/&quot; rel=&quot;nofollow&quot;&gt;http://nehe.gamedev.net/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1676" LastActivityDate="2010-09-11T07:20:06.867" CommunityOwnedDate="2010-09-11T07:20:06.867" />
  <row Id="4405" PostTypeId="2" ParentId="3519" CreationDate="2010-09-11T07:36:33.910" Score="1" ViewCount="" Body="&lt;p&gt;Liouville's Number makes it's next appearance in &lt;em&gt;Making Transcendence Transparent - Edward B. Burger, Robert Tubbs&lt;/em&gt;. The transcendence of $e + L$ is proved. (This might be the end of it's career though).&lt;/p&gt;&#xA;" LastActivityDate="2010-09-11T07:36:33.910" />
  <row Id="4406" PostTypeId="1" CreationDate="2010-09-11T07:40:51.017" Score="2" ViewCount="307" Body="&lt;p&gt;What is the precise difference or relation between these terms in logic: &#xA;Inference, Reasoning, Deduction, and Induction?&lt;/p&gt;&#xA;" OwnerUserId="1831" LastActivityDate="2010-10-11T17:16:40.700" Title="Difference or relation between Inference, Reasoning, Deduction, and Induction?" Tags="&lt;logic&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="2" />
  <row Id="4407" PostTypeId="2" ParentId="4400" CreationDate="2010-09-11T08:19:15.810" Score="2" ViewCount="" Body="&lt;p&gt;There is always room for misinterpreting a question when one does not fully understand the language in which it is written. I think that the way mathematics and mathematicians use conditional probability is clear:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$P(A|B)=P(A \cap B)/P(B).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I believe that this is the interpretation that one should take, and thus arrive at your answer of 13/27, and not search for further nuances, which are not too difficult to find.&lt;/p&gt;&#xA;" OwnerUserId="1301" LastActivityDate="2010-09-11T08:19:15.810" CommentCount="2" />
  <row Id="4408" PostTypeId="2" ParentId="4400" CreationDate="2010-09-11T08:31:26.883" Score="9" ViewCount="" Body="&lt;p&gt;It is actually &lt;em&gt;impossible&lt;/em&gt; to have a unique and unambiguous answer to the puzzle without explicitly articulating a probability model for how the information on gender and birthday is generated.  The reason is that (1) for the problem to have a unique answer some random process is required, and (2) the answer is a function of which random model is used.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The problem assumes that a unique probability can be deduced as the answer. This requires that the set of children described is chosen by a random process, otherwise the number of boys is a deterministic quantity and the probability would be 0 or 1 but with no ability to determine which is the case.  More generally one can consider random processes that produce the complete set of information referenced in the problem: choose a parent, then choose what to reveal about the number, gender, and birth days of its children.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The answer depends on which random process is used.  If the Tuesday birth is disclosed only when there are two boys, the probability of two boys is 1.    If Tuesday birth is disclosed only when there is a sister, the probability of two boys is 0.  The answer could be any number between 0 or 1 depending on what process is assumed to produce the data. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;There is also a linguistic question of how to interpret &quot;one is a boy born on Tuesday&quot;.  It could mean that the number of Tuesday-born males is exactly one, or at least one child.&lt;/p&gt;&#xA;" OwnerUserId="467" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-09-13T08:58:03.033" LastActivityDate="2010-09-13T08:58:03.033" CommentCount="1" />
  <row Id="4409" PostTypeId="2" ParentId="4400" CreationDate="2010-09-11T08:35:12.623" Score="4" ViewCount="" Body="&lt;p&gt;I guess the following two versions of the experiment provide two different answers:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Dave has two children. Is atleast one of them a boy who is born on Tuesday? Dave answers Yes.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Dave has two children. I ask him to first choose and fix one child at random, and tell me if it is a boy who was born on Tuesday. Dave answers yes he is a boy born on Tuesday.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;For 1st the answer is 13/27, while the second has answer 1/2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The way in which the question is asked, it's in line with 1st, hence the answer should be 13/27.&lt;/p&gt;&#xA;" OwnerUserId="1310" LastActivityDate="2010-09-11T08:35:12.623" CommentCount="2" />
  <row Id="4410" PostTypeId="2" ParentId="154" CreationDate="2010-09-11T09:15:10.803" Score="7" ViewCount="" Body="&lt;p&gt;Here is an example of a physical quantity which comes naturally as a complex number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Electrical_impedance&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;impedance&lt;/em&gt;&lt;/a&gt; between two nodes of a linear electrical circuit working in alternate current, which is the analogous of the resistance of a resistor and is also measured in Ohms, is a complex quantity. For instance, the impedance of an (ideal) capacitor is imaginary.&lt;/p&gt;&#xA;" OwnerUserId="1828" LastEditorUserId="1828" LastEditorDisplayName="" LastEditDate="2010-09-11T09:30:21.907" LastActivityDate="2010-09-11T09:30:21.907" CommunityOwnedDate="2011-04-15T20:45:29.080" />
  <row Id="4411" PostTypeId="2" ParentId="4363" CreationDate="2010-09-11T10:04:00.793" Score="2" ViewCount="" Body="&lt;p&gt;As Wikipedia says: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;The inverse function theorem (and the implicit function theorem) can be seen as a special case of the constant rank theorem, which states that a smooth map with locally constant rank near a point can be put in a particular normal form near that point.&quot; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;See &lt;a href=&quot;http://en.wikipedia.org/wiki/Derivative_rule_for_inverses#Constant_rank_theorem&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Derivative_rule_for_inverses#Constant_rank_theorem&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Constant Rank Theorem is stated as Theorem (7.1) p. 47 of &lt;strong&gt;An Introduction to Differentiable Manifolds and Riemannian Geometry&lt;/strong&gt;, Revised Second Edition, William M. Boothby, Academic Press. (This is the reference given by Wikipedia.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is, for the reader's convenience, a statement of the Constant Rank Theorem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $k,n$ and $r$ be positive integers, let $a$ be in $\mathbb R^n$, let $b$ be in $\mathbb R^k$, let $f$ be a smooth map from a neighborhood of $a$ to $\mathbb R^k$ sending $a$ to $b$, and let $\ell$ be the linear map from $\mathbb R^n$ to $\mathbb R^k$ sending $x$ to $(x_1,\dots,x_r,0,\dots,0)$. Assume that the rank of the tangent map to $f$ at $x$ is equal to $r$ for all $x$ in our neighborhood of $a$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then there is a diffeomorphism $g$ from a neighborhood of $a$ to a neighborhood of 0 in $\mathbb R^n$, and a diffeomorphism $h$ from a neighborhood of 0 in $\mathbb R^k$ to a neighborhood of $b$, such that the equality $f=h\circ\ell\circ g$ holds in some neighborhood of $a$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT OF MARCH 19, 2011&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a statement and a proof of the Constant Rank Theorem. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Theorem.&lt;/strong&gt; Let $U$ be open in $\mathbb{R}^n$, let $a$ be a point in $U$, and let $f$ be $C^p$ map ($1\le p\le\infty$) of rank $r$ from $U$ to $\mathbb{R}^k$. Then there are open sets $U_1,U_2\subset\mathbb{R}^n$, $U_3\subset\mathbb{R}^k$ and $C^p$ diffeomorphisms $\varphi:U_1\to U_2$, $\psi:U_3\to U_3$ such that $a\in U_1$ and $(\psi\circ f\circ\varphi^{-1})(x)=(x_1,\dots,x_r,0,\dots,0)$ for all $x$ in $U_2$. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof.&lt;/strong&gt; For $$x\in\mathbb{R}^r,\quad y\in\mathbb{R}^{n-r},\quad(x,y)\in U$$ write &#xA;$$f(x,y)=(f_1(x,y),f_2(x,y)),\quad f_1(x,y)\in\mathbb{R}^r,\quad f_2(x,y)\in\mathbb{R}^{k-r}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can assume that $\partial f_1(x,y)/\partial x$ is invertible for all $(x,y)\in U$. Define $$\varphi:U\to\mathbb{R}^n,\quad(x,y)\mapsto(f_1(x,y),y).$$ By the Inverse Function Theorem, there are open sets &#xA;$$U_1\subset\mathbb{R}^n,\quad U_4\subset\mathbb{R}^r,\quad U_5\subset\mathbb{R}^{n-r}$$ such that $a\in U_1\subset U$, $\varphi$ is a $C^p$ diffeomorphism from $U_1$ onto $U_2:=U_4\times U_5$, and $U_5$ is connected. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then $f(\varphi^{-1}(x,y))=(x,g(x,y))$ for some $C^p$ map $g$ from $U_2$ to $\mathbb{R}^{k-r}$. As $\partial g/\partial y=0$, we can write $g(x)$ for $g(x,y)$, and it suffices to set $U_3:=U_4\times\mathbb{R}^{k-r}$ and $\psi(u,v):=(u,v-g(u))$ for $ u\in U_4$ and $v\in\mathbb{R}^{k-r}.$&lt;/p&gt;&#xA;" OwnerUserId="660" LastEditorUserId="660" LastEditorDisplayName="" LastEditDate="2011-03-20T12:11:16.847" LastActivityDate="2011-03-20T12:11:16.847" CommentCount="4" />
  <row Id="4412" PostTypeId="1" AcceptedAnswerId="4413" CreationDate="2010-09-11T10:35:49.663" Score="6" ViewCount="239" Body="&lt;p&gt;We know that if $f : \mathbb{R} \to \mathbb{R}$ is a continuous function, then $f$ carries connected sets to connected sets and compact sets to compact sets. That is if $A \subset \mathbb{R}$ is connected then $f(A)$ is connected, and if $A$ is compact then $f(A)$ is compact.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; Suppose $f: \mathbb{R} \to \mathbb{R}$ is a function such that for every connected, compact subsets $A \subset \mathbb{R}$, $f(A)$ is connected, compact, then is $f$ continuous? If yes, i would like to see a proof.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update&lt;/strong&gt;: Does this result remain true if $f: \mathbb{R}^{2} \to \mathbb{R}$, or from any $f: \mathbb{R}^{m} \to \mathbb{R}^{n}$.&lt;/p&gt;&#xA;" LastActivityDate="2010-09-13T04:52:37.927" Title="Characterising Continuous functions" Tags="&lt;real-analysis&gt;&lt;analysis&gt;" AnswerCount="2" FavoriteCount="2" />
  <row Id="4413" PostTypeId="2" ParentId="4412" CreationDate="2010-09-11T10:42:50.813" Score="12" ViewCount="" Body="&lt;p&gt;No. See &lt;a href=&quot;http://en.wikipedia.org/wiki/Conway_base_13_function&quot; rel=&quot;nofollow&quot;&gt;Conway's base 13&lt;/a&gt; function. If we call this function $f$,&#xA;while $f$ itself is not a counterexample to your question, $g(x)=\sin f(x)$&#xA;is. The connected compact subsets of $\mathbb{R}$ are the empty set, the&#xA;one-point sets and the closed intervals $[a,b]$ with $a &amp;lt; b$. The Conway&#xA;function $f$ takes such closed intervals to $\mathbb{R}$ and so $g$&#xA;takes them to $[-1,1]$. The functions $f$ and $g$ are discontinuous everywhere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;ADDED&lt;/strong&gt;&#xA;For functions with domain $\mathbb{R}^n$ the answer is still no. Just&#xA;take&#xA;$$(x_1,\ldots,x_n)\mapsto g(x_1)$$&#xA;where $g$ is the function above.&lt;/p&gt;&#xA;" OwnerUserId="226" LastEditorUserId="226" LastEditorDisplayName="" LastEditDate="2010-09-11T11:54:28.047" LastActivityDate="2010-09-11T11:54:28.047" />
  <row Id="4414" PostTypeId="1" AcceptedAnswerId="4420" CreationDate="2010-09-11T11:08:06.980" Score="5" ViewCount="211" Body="&lt;p&gt;In a proof or deduction system, there are some axioms and some inference rules. To have a small proof system, axioms are defined for the primitives (for example for negation and implication in propositional logic). Now, we want to have a proof system that proves all provable formulae including the formulae contain non-primitive logical connectors (such as OR in propositional logic). The question is should we add definitions for non-primitives as axioms to the proof system, or we should do something else? In other words, do logicians accept definitions as axioms?&lt;/p&gt;&#xA;" OwnerUserId="1831" LastActivityDate="2011-02-10T01:43:20.323" Title="Can we define definitions as axioms in logic?" Tags="&lt;logic&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="1" />
  <row Id="4415" PostTypeId="2" ParentId="154" CreationDate="2010-09-11T12:34:46.353" Score="1" ViewCount="" Body="&lt;p&gt;I'm not a mathematician and for a long time I've struggled with the question if numbers exist.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mathematics is a language. Numbers are part of its vocabulary, like &quot;apple&quot; is part of the English vocabulary. Just like you can build a sentence with the word &quot;apple&quot; you can build equations with numbers, and just like you have to follow certain rules to make a correct sentence you have to follow certain rules for your equation to hold true.&lt;br&gt;&#xA;Now, and I'm trying not go into semiotics (of which I know nothing), the word &quot;apple&quot; is not an apple: you can't eat the word &quot;apple&quot; for one thing. But it exists in the language. You can invent things in a language to represent things in the real world. Does the verb &quot;to walk&quot; exist? It represents an action in the real world, but in itself it is nothing.&lt;br&gt;&#xA;I think the same is true with numbers: numbers exist because we defined them in the language of mathematics, but they only make sense if you can connect them to something real. Natural numbers often represent the cardinality of sets: 10 may the number of real world apples in a real world basket. Rational numbers come in handy when comparing things: one apple mat be 1.2 times as large as another one. To non-mathematicians irrational numbers make less sense, for everyday non-mathematical use even pi is rational: 22/7.&lt;br&gt;&#xA;Now for complex numbers same thing: we define them in the language of mathematics, but just like irrational numbers it's much harder to make them represent something real, although especially physicists are very good at describing the real world with them.  What about other numbers like quaternions? They're part of the language, and are often required to make the grammar fit.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, are complex numbers real? (mind the pun! :-))&lt;br&gt;&#xA;They obviously exist in the language of mathematics, and may be used to describe real world things and events, but in themselves I don't think they exist.&lt;/p&gt;&#xA;" OwnerUserId="923" LastActivityDate="2010-09-11T12:34:46.353" CommunityOwnedDate="2011-04-15T20:45:29.080" CommentCount="2" />
  <row Id="4416" PostTypeId="2" ParentId="4400" CreationDate="2010-09-11T13:41:30.877" Score="0" ViewCount="" Body="&lt;p&gt;This is discussed also on &lt;a href=&quot;http://johncarlosbaez.wordpress.com/2010/08/24/probability-puzzles-from-egan/#comment-1313&quot; rel=&quot;nofollow&quot;&gt;http://johncarlosbaez.wordpress.com/2010/08/24/probability-puzzles-from-egan/#comment-1313&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="875" LastActivityDate="2010-09-11T13:41:30.877" />
  <row Id="4417" PostTypeId="2" ParentId="4400" CreationDate="2010-09-11T14:36:56.830" Score="0" ViewCount="" Body="&lt;p&gt;This, in my opinion, is why the intuitive approach fails:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One has a tendency to think that &#xA;the probability of 7*P(b AND d1) = P(b AND d1) + P(b AND d2) + ... + P(b AND d7) = P((b AND d1) OR (b AND d2) OR ... OR (b AND d7)) = P(b AND (d1 OR d2 OR ... OR d7)) = P(b).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the flaw here is that, in reality, P(b AND d1) + P(b AND d2) + ... + P(b AND d7) is NOT equal to P((b AND d1) OR (b AND d2) OR ... OR (b AND d7)). This means that mentioning independent (and one might think irrelevant) information alongside with relevant information actually changes the resulting probabilities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One interesting consequence: if I say something like&#xA;&quot;I have two children. One of them is a boy who was born at 10:24 PM on February 10th,&quot;&#xA;The probability that I have two boys is now almost exactly the same as as the probability that I have a girl and a boy. Adding a unique or almost unique piece of information makes the stuff I want to know about the other child independent of the information I have on the first child. If I took this to the extreme and said that I have a firstborn boy, won't know anything additional about the other child.&lt;/p&gt;&#xA;" OwnerUserId="796" LastActivityDate="2010-09-11T14:36:56.830" CommentCount="1" />
  <row Id="4418" PostTypeId="2" ParentId="4412" CreationDate="2010-09-11T14:55:53.767" Score="12" ViewCount="" Body="&lt;p&gt;Much more general results are known: no statement of this form can characterize continuous functions, derivatives, Baire class 1 functions, Borel functions, measurable functions, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;THEOREM&lt;/strong&gt; $\;$ There do not exist families of sets of reals $\rm\: \cal A\:,\:\cal B\;$ such that the following statement is true:   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\quad\quad$ for every function $\rm\; f : \mathbb R\to \mathbb R\:,\;\;\; f\:$ is continuous $\iff$ for every $\: A\in {\cal A}\:,\;\; {\rm f}\:(A) \in \cal B$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the elementary proof see this &lt;a href=&quot;http://www.jstor.org/stable/2974580&quot; rel=&quot;nofollow&quot;&gt;1997 Monthly paper&lt;/a&gt; by Velleman and for more general results on classes of functions characterizable by images of sets see &lt;a href=&quot;http://at.yorku.ca/p/a/a/f/12.htm&quot; rel=&quot;nofollow&quot;&gt;this paper&lt;/a&gt;, whose abstract I have appended below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For non-empty topological spaces $X$ and $Y$ and arbitrary&#xA;families $\cal{A}\subseteq\cal{P}(X)$ and $\cal{B}\subseteq\cal{P}(Y)$ we put&#xA;$\cal{C}_{\cal{A},\cal{B}}=\{f\in Y^X\colon(\forall A\in\cal{A})(f[A]\in\cal{B})\}$.&#xA;In this paper we will examine&#xA;which classes of functions $\cal{F}\subseteq Y^X$&#xA;can be represented as $\cal{C}_{\cal{A},\cal{B}}$. We will be mainly&#xA;interested in the case&#xA;when&#xA;$\cal{F}=\cal{C}(X,Y)$ is the class of all continuous functions from $X$ into $Y$.&#xA;We prove that for non-discrete Tychonoff space $X$ the class&#xA;$\cal{F}=\cal{C}(X,\mathbb{R})$&#xA;is not equal to $\cal{C}_{\cal{A},\cal{B}}$ for any&#xA;$\cal{A}\subseteq \cal{P}(X)$ and $\cal{B}\subseteq\cal{P}(\mathbb{R})$. Thus, $\cal{C}(X,\mathbb{R})$&#xA;&lt;strong&gt;cannot be characterized by images of sets&lt;/strong&gt;.&#xA;We also show that none of the&#xA;following classes of real functions can be represented as&#xA;$\cal{C}_{\cal{A},\cal{B}}$:&#xA;upper (lower) semicontinuous functions,&#xA;derivatives, approximately  continuous functions,&#xA;Baire class 1 functions, Borel functions, and measurable functions.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-09-13T04:52:37.927" LastActivityDate="2010-09-13T04:52:37.927" CommentCount="3" />
  <row Id="4419" PostTypeId="2" ParentId="4406" CreationDate="2010-09-11T15:28:52.267" Score="0" ViewCount="" Body="&lt;p&gt;If your question is about philosophical matters of logic, then I think that you may find very interesting the following paper authored by a leading logician (see also other papers in the same handbook).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wilfrid Hodges, &lt;strong&gt;The scope and limits of logic&lt;/strong&gt;&lt;br&gt;&#xA;pp. 41-63 in Handbook of the Philosophy of Science: Philosophy of Logic,&lt;br&gt;&#xA;ed. Dale Jacquette, 2007, Elsevier, Amsterdam&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-09-11T15:28:52.267" />
  <row Id="4420" PostTypeId="2" ParentId="4414" CreationDate="2010-09-11T15:45:55.513" Score="8" ViewCount="" Body="&lt;p&gt;Given a theory $T$, you can enlarge it with definitions in two steps. First you enlarge the alphabet $L(T)$ to a new alphabet $L'$ which will contain the new symbols of constants, functions and relations that you need. Then you enlarge $T$ itself by adding axioms of the following form.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $\phi(x_1, \dots, x_n)$ be a formula of $T$ with free variables $x_1, \dots, x_n$ and $\psi$ a new symbol of $n$-ary relation. Then you can add an axiom $$\phi(x_1, \dots, x_n) \leftrightarrow \psi(x_1, \dots, x_n).$$ This axiom &quot;defines the meaning of $\psi$&quot;. Similarly you can add new defininig axioms for functions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The theory $T'$ with language $L'$ that you obtain in this way is called an &lt;em&gt;extension by definitions&lt;/em&gt; of $T$. It is not difficult to prove that $T'$ is a conservative extension of $T$; that is, $T$ and $T'$ prove the same formulas of $L$. In this sense one can make precise the fact that definitions do not add nothing new to a theory.&lt;/p&gt;&#xA;" OwnerUserId="233" LastActivityDate="2010-09-11T15:45:55.513" />
  <row Id="4421" PostTypeId="2" ParentId="1257" CreationDate="2010-09-11T17:39:07.933" Score="1" ViewCount="" Body="&lt;p&gt;One of India's prestigious computer scientists developed this test called as the AKS primality test. Although this doesn't answer your question, it shows as to how much progress has been made to determine the primes: Please see this article: &lt;a href=&quot;http://en.wikipedia.org/wiki/AKS_primality_test&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/AKS_primality_test&lt;/a&gt;&lt;/p&gt;&#xA;" LastActivityDate="2010-09-11T17:39:07.933" CommentCount="2" />
  <row Id="4422" PostTypeId="1" CreationDate="2010-09-11T17:57:48.697" Score="4" ViewCount="323" Body="&lt;p&gt;I am taking a fourier analysis course at the graduate level and I am unhappy with the textbook (Stein and Shakarchi).  What I am looking for is a book that is less conversational and more to the point.  Further, I am not terribly interested in applications and would rather be exposed to how Fourier Analysis fits into the broader framework of analysis.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For background, I used Baby Rudin for a one-year course in advanced calculus, I am currently taking a course from Kolmogorov and Fomin's Introductory Real Analysis and I have taken complex analysis (using Conway's text, Functions of One Complex Variable) as well as topology (using Munkres as well as Engelking) at the graduate level, but I have not yet been introduced to the Lebesgue integral.&lt;/p&gt;&#xA;" OwnerUserId="121" LastActivityDate="2011-03-07T09:54:41.060" CommunityOwnedDate="2010-09-11T17:57:48.697" Title="Fourier Analysis textbook recommendation" Tags="&lt;reference-request&gt;&lt;fourier-analysis&gt;" AnswerCount="5" CommentCount="2" FavoriteCount="3" />
  <row Id="4424" PostTypeId="2" ParentId="4202" CreationDate="2010-09-11T18:23:05.610" Score="4" ViewCount="" Body="&lt;p&gt;Such results on continuous induction are folklore dating back over a century - much older than  the references in the Kalantari paper referenced by Pete L. Clark. I once had a large file of papers on this topic but, alas, it's long lost. But I can still recall one &lt;a href=&quot;http://projecteuclid.org/DPubS?service=UI&amp;amp;version=1.0&amp;amp;verb=Display&amp;amp;handle=euclid.bams/1183425067&quot; rel=&quot;nofollow&quot;&gt;old paper [1]&lt;/a&gt;  because the author was the world-renowned multi-talented Chinese linguist Yuen Ren Chao. If memory serves correct he did his thesis under Sheffer at Harvard on aspects of the continuum, so his BAMS paper was probably an offshoot of his thesis. Since Project Euclid seems to be having intermittent problems I have appended the short paper below. For some biographical information on Chao see &lt;a href=&quot;http://content.cdlib.org/view?docId=hb8779p27v&amp;amp;brand=calisphere&amp;amp;doc.view=entire_text&quot; rel=&quot;nofollow&quot;&gt;here1&lt;/a&gt; and &lt;a href=&quot;http://www.cuhk.edu.hk/lin/cbrc/interview/eng.pdf&quot; rel=&quot;nofollow&quot;&gt;here2&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[1] Yuen Ren Chao. &lt;a href=&quot;http://projecteuclid.org/DPubS?service=UI&amp;amp;version=1.0&amp;amp;verb=Display&amp;amp;handle=euclid.bams/1183425067&quot; rel=&quot;nofollow&quot;&gt;A note on “Continuous mathematical induction&quot;.&lt;/a&gt;&lt;br&gt;&#xA;Bull. Amer. Math. Soc. Volume 26, Number 1 (1919), 17-18. &#xA;&lt;img src=&quot;http://i.imgur.com/whVt9.jpg&quot; alt=&quot;alt text&quot;&gt;&#xA;&lt;img src=&quot;http://i.imgur.com/LD8M4.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-09-19T22:20:54.500" LastActivityDate="2010-09-19T22:20:54.500" CommentCount="2" />
  <row Id="4425" PostTypeId="2" ParentId="4422" CreationDate="2010-09-11T18:39:40.483" Score="2" ViewCount="" Body="&lt;p&gt;Here are the ones which i would recommend:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Fourier series by R.Bhatia. Link: &lt;a href=&quot;http://www.maa.org/reviews/BhatiaFourierSeries.html&quot; rel=&quot;nofollow&quot;&gt;http://www.maa.org/reviews/BhatiaFourierSeries.html&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Korner, T. W. Fourier Analysis. Cambridge. 1990&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The second one is very good and you can comprehend it if you are familiar with the analysis which you have mentioned. I would also like to advice you to read &quot;A radical approach to Lebesgue theory of Integration&quot; by D.M.Bressoud, link &lt;a href=&quot;https://www.maa.org/EbusPPRO/DynamicSearch/ProductDetailsAdvancedSearch/tabid/176/ProductId/1498/Default.aspx&quot; rel=&quot;nofollow&quot;&gt;https://www.maa.org/EbusPPRO/DynamicSearch/ProductDetailsAdvancedSearch/tabid/176/ProductId/1498/Default.aspx&lt;/a&gt; this is a beautiful book to learn Measure theory&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update&lt;/strong&gt; You might be interested to see this link: &lt;a href=&quot;http://www.cargalmathbooks.com/#FourierAnalysis&quot; rel=&quot;nofollow&quot;&gt;http://www.cargalmathbooks.com/#FourierAnalysis&lt;/a&gt;&lt;/p&gt;&#xA;" LastActivityDate="2010-09-11T19:07:12.847" CommunityOwnedDate="2010-09-11T18:39:40.483" CommentCount="2" />
  <row Id="4426" PostTypeId="1" AcceptedAnswerId="4456" CreationDate="2010-09-11T18:46:04.340" Score="3" ViewCount="180" Body="&lt;p&gt;In classical real analysis I've only seen absolute continuity defined for functions on compact interval $[a,b]$, where the two equivalent definitions are: $f:[a,b]\rightarrow\mathbb{R}$ is AC if&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1)  Given $\epsilon &gt; 0$ there is a $\delta &gt; 0$ such that $\sum_{i=1}^n |f(y_i)-f(x_i)|&amp;lt; \epsilon$ for every finite collection of nonoverlapping intervals $( (x_i,y_i) )_{i=1}^n $ each contained in $[a,b]$ with $\sum_{i=1}^n |y_i-x_i|&amp;lt; \delta$. Or,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) $f'$ exists a.e. on $[a,b]$, $f'$ is integrable on $[a,b]$, and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f(x) = \int_a^x f'(y) dy + f(a)$ for all $x \in [a,b]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there an accepted definition for absolute continuity of a function on an open, possibly unbounded, interval $(a,b)$ where $-\infty \leq a &amp;lt; b \leq \infty$? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems that definition (1) extends easily to this case if we replace $[a,b]$ by $(a,b)$.  If we call this condition (1'), &lt;strike&gt;then it's easy to show (1') is equivalent to&lt;/strike&gt; (see answer by Jonas below).  A natural extension of (2) to this case would be&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2') $f$ is AC on an open set $U$ if, for all compact intervals $[c,d] \subset (a,b)$, $f$ is AC in the sense of (2) on $[c,d]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which of these is the best extension of the definition?  I don't know enough about the notion of absolute continuity of measures to know if my extended definitions are consistent with that generalization as well.&lt;/p&gt;&#xA;" OwnerUserId="1692" LastEditorUserId="1692" LastEditorDisplayName="" LastEditDate="2010-09-13T15:59:05.423" LastActivityDate="2010-09-13T15:59:05.423" Title="Absolute continuity on an open interval of the real line?" Tags="&lt;real-analysis&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="4427" PostTypeId="2" ParentId="4422" CreationDate="2010-09-11T18:57:44.763" Score="1" ViewCount="" Body="&lt;p&gt;I'd like to suggest &lt;em&gt;Fourier Series and Integrals&lt;/em&gt; by Dym and McKean.&#xA;It's old, but still an excellent book. Chapters 3 and 4 show how Fourier analysis fits in with some other parts of mathematics. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the Preface:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The level of preparation expected is a thorough knowledge of advanced calculus.&#xA;To this must be added a willingness to believe in (or to study up on) the Lebesgue integral. &lt;/p&gt;&#xA;" OwnerUserId="940" LastEditorUserId="940" LastEditorDisplayName="" LastEditDate="2010-09-12T16:30:40.530" LastActivityDate="2010-09-12T16:30:40.530" CommunityOwnedDate="2010-09-11T18:57:44.763" />
  <row Id="4428" PostTypeId="1" AcceptedAnswerId="4429" CreationDate="2010-09-11T19:19:49.593" Score="0" ViewCount="83" Body="&lt;p&gt;What's the difference between a bar chart and a histogram?&lt;/p&gt;&#xA;" LastActivityDate="2010-09-11T21:57:01.193" Title="Bar chart or histogram" Tags="&lt;statistics&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="4429" PostTypeId="2" ParentId="4428" CreationDate="2010-09-11T19:23:19.257" Score="0" ViewCount="" Body="&lt;p&gt;With bar charts, each column represents a group defined by a categorical variable; and with histograms, each column represents a group defined by a quantitative variable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See this link: &lt;a href=&quot;http://wiki.answers.com/Q/What_is_the_difference_between_Bar_chart_and_a_frequency_histogram&quot; rel=&quot;nofollow&quot;&gt;http://wiki.answers.com/Q/What_is_the_difference_between_Bar_chart_and_a_frequency_histogram&lt;/a&gt;&lt;/p&gt;&#xA;" LastActivityDate="2010-09-11T19:23:19.257" />
  <row Id="4430" PostTypeId="1" CreationDate="2010-09-11T19:23:50.987" Score="1" ViewCount="248" Body="&lt;p&gt;&lt;strong&gt;Question&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do you know any application or in which field this trajectories can be applied?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Comment&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you know any book or any site that can show me the applications, please let me know. Thanks for your time&lt;/p&gt;&#xA;" OwnerUserId="1833" LastEditorUserId="8246" LastEditorDisplayName="" LastEditDate="2011-05-10T20:04:20.077" LastActivityDate="2011-05-10T20:04:20.077" Title="Applications of orthogonal trajectories" Tags="&lt;reference-request&gt;&lt;differential-equations&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="4432" PostTypeId="5" CreationDate="2010-09-11T19:36:12.317" Score="0" ViewCount="0" Body="" OwnerUserId="-1" LastEditorUserId="171" LastEditorDisplayName="" LastEditDate="2010-09-11T19:36:12.317" LastActivityDate="2010-09-11T19:36:12.317" />
  <row Id="4433" PostTypeId="4" CreationDate="2010-09-11T19:36:12.317" Score="0" ViewCount="0" Body="[reference-request] are for questions that the poster wants to seek for references (books, articles, etc) in a subject." OwnerUserId="-1" LastEditorUserId="171" LastEditorDisplayName="" LastEditDate="2010-09-11T19:36:12.317" LastActivityDate="2010-09-11T19:36:12.317" />
  <row Id="4434" PostTypeId="2" ParentId="4403" CreationDate="2010-09-11T19:48:40.353" Score="1" ViewCount="" Body="&lt;p&gt;So, you have shown that $\inf_T(S)\leq \inf_U(S)=u$. You want to show that $u\leq\inf_T(S)$. Clearly, you need to use the existence of $S'$ somehow, because without the existence of $S'$ you can trivially show this is false. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You know that $\inf_T(S)\leq\inf_U(S)$. You want to show that you do not have $\inf_T(S)\lt\inf_U(S)=u$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; (added case) If $S'=\emptyset$, then $u=\inf_U(\emptyset)=\sup(U)$, so in particular $\inf_T(S)\leq u$ and we are done. So we may assume that $S'\not= \emptyset$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Added later&lt;/em&gt;: Actually, this is not needed strictly speaking, but it may be a good point to mention explicitly; if $t\lt u$, then $t$ is not an upper bound for $S'$; the only way that can occur in the first place is if $S'$ is not empty, since everything is an upper bound for the empty set, so this case is covered by the assumption being made.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strike&gt;If $t\in T$ is such that $t\lt u$, then since $t\in U$ and $u=\sup_U(S')$, you know that there exists $s'\in S'$ such that $t\lt s'\leq u$.&lt;/strike&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; (implicitly assumed certain things were comparable, which may not be the case in general; fixed below). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that for every $s'\in S'$ and $s\in S$, you have $s'\leq u\leq s$ (since $u=\inf_U(S)$), and so in particular $s'\leq \inf_T(S)$, since $S'\subseteq T$. So $\inf_T(S)$ is comparable to every element of $S'$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose that $t\in T$ is such that $t\lt u$. Since $u=\sup_U(S')$, it follows that $t$ cannot be an upper bound for $S'$ in $U$. In particular, there exists $s'\in S'$ such that $s'\not\leq t$; so either $t\lt s'$, or else $t$ and $s'$ are incomparable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In summary, for every $t\in T$, if $t\lt u$ and $t$ is comparable to every element of $S'$, then $t\lt\inf_T(S)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $\inf_T(S)\in T$, $\inf_T(S)$ is comparable to every element of $S'$ (as noted above), and $\inf_T(S)\not\lt\inf_T(S)$, you can can conclude by contrapositive that $\inf_T(S)\not\lt u$. Since you already know that $\inf_T(S)\leq u$, you are done.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-09-11T22:32:45.570" LastActivityDate="2010-09-11T22:32:45.570" CommentCount="6" />
  <row Id="4435" PostTypeId="1" AcceptedAnswerId="4448" CreationDate="2010-09-11T19:49:05.417" Score="0" ViewCount="126" Body="&lt;p&gt;Problem taken from the link:  &lt;a href=&quot;http://web.mit.edu/rwbarton/Public/func-eq.pdf&quot; rel=&quot;nofollow&quot;&gt;http://web.mit.edu/rwbarton/Public/func-eq.pdf&lt;/a&gt; I am stating the question here&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For which $\alpha$ does there exists a nonconstant function $f: \mathbb{R} \to \mathbb{R}$ such that $f(\alpha(x+y))=f(x)+f(y)$ for all $x,y \in \mathbb{R}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clearly for $\alpha=1$ we see that this case is satisfied, by taking the identity function. But are there other values of $\alpha$ for which this condition is satisfied. &lt;/p&gt;&#xA;" LastActivityDate="2010-09-11T22:42:08.930" Title="Finding $\alpha$ such that $f(\alpha(x+y))=f(x)+f(y)$" Tags="&lt;real-analysis&gt;&lt;analysis&gt;&lt;functional-equations&gt;" AnswerCount="1" />
  <row Id="4436" PostTypeId="1" CreationDate="2010-09-11T19:59:37.853" Score="1" ViewCount="70" Body="&lt;p&gt;Doing a little practice exercise, and came across the following question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are three functions: $f(x)$, $g(x)$, and $h(x)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Find $f(x)$ and $g(x)$, such that $h(x) = f \circ g(x)$ and $g(x) = \sqrt{4} + 8$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The value of $h(x)$ is defined as $h(x) = (\sqrt{4} + 8) ^ 4$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not really sure how to approach the problem. I know I'm going to have to solve this algebraically, but I just can't seem to wrap my mind around the manipulation of functions as opposed to variables and numbers.&lt;/p&gt;&#xA;" OwnerUserId="1836" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-09-11T21:15:09.247" LastActivityDate="2010-09-11T21:15:09.247" Title="Solving Compositions of Functions Algebraically" Tags="&lt;algebra-precalculus&gt;" AnswerCount="1" CommentCount="7" />
  <row Id="4437" PostTypeId="1" AcceptedAnswerId="4443" CreationDate="2010-09-11T20:08:18.757" Score="11" ViewCount="548" Body="&lt;p&gt;To prove the &lt;a href=&quot;http://en.wikipedia.org/wiki/Cayley%E2%80%93Hamilton_theorem&quot; rel=&quot;nofollow&quot;&gt;Cayley-Hamilton&lt;/a&gt; theorem in linear algebra, my professor said that a lemma was necessary:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Lemma&lt;/strong&gt;: Let $A \in M_n(\mathbb{K})$ be an $n\times n$ matrix over a field $\mathbb{K}$, let $b(t) \in M_n(\mathbb{K})[t]$ and $P(t) = b(t)[A-tI]$, then $P(A) = 0$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The theorem (which says that if $f$ is an endomorphism of V, then $f$ is a solution to its characteristic polynomial), was then proven thus:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;let $B(t) = \text{adj}[A-tI]$ and $P(t) = B(t)[A-tI]$, then $P(A)=0$ but also $P(t) = \delta I$ (where $\delta = \det(A-tI)$). Since $\delta = \chi_f(t)$, so $P(A) = 0 \Rightarrow \chi_f(A) = 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is: since we interpret the $P(t)$ of the theorem as a polynomial with matrix coefficients, isn't the whole thing kind of obvious for the properties of a polynomial ring? (Assuming we all know how to switch between matrices and endomorphisms)&lt;/p&gt;&#xA;" OwnerUserId="1415" LastEditorUserId="1415" LastEditorDisplayName="" LastEditDate="2010-09-12T22:08:06.633" LastActivityDate="2010-09-17T23:16:20.873" Title="Is the proof of this lemma really necessary?" Tags="&lt;linear-algebra&gt;&lt;proof-strategy&gt;" AnswerCount="4" CommentCount="6" FavoriteCount="8" />
  <row Id="4438" PostTypeId="2" ParentId="4436" CreationDate="2010-09-11T20:34:11.837" Score="1" ViewCount="" Body="&lt;p&gt;The question, as currently posted (even after I edited it) is a mess. Are $g(x)$ and $h(x)$ meant to be &lt;em&gt;constant&lt;/em&gt; functions? Was that supposed to be $g(4) = \sqrt{4} + 8$ and $h(4)=(\sqrt{4}+8)^4$, or was it supposed to be $g(x)=\sqrt{x}+8$ and $h(x)=(\sqrt{x}+8)^4$? Something else?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In any case: a composition is a function that is obtained by first applying one function, and then taking the output of that function and plugging it into the next function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take the expression $(\sqrt{4}+8)^4$. Suppose you were trying to actually compute this number. How would you go about it? First, we would take the number $4$ and plug it into the square root to obtain $\sqrt{4}$. The we would take the ouput of that, and add $8$. Finally, we would take the result we got from doing all that, and raise it to the fourth power.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you can think of this as the process of doing three things, one after another, where the input to each of the steps is the output of the previous step:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Take the square root of what you have;&lt;/li&gt;&#xA;&lt;li&gt;Add $8$ to what you have;&lt;/li&gt;&#xA;&lt;li&gt;Raise what you have to the fourth power.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Now, you are told to try to express this as the result of doing &lt;em&gt;two&lt;/em&gt; things in succession: the first one is $g(x)$, and the second one will be $f(x)$. And you are told that the first thing you are doing is $g(x)=\sqrt{4}+8$. So, what else do you need to do to the output of this process in order to get to the final result? You need to raise what you are given to the fourth power. So, whatever $f(x)$ is, it should take the output of $g(x)$ and raise it to the fourth power.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, what is $f(x)$?&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-09-11T20:34:11.837" CommentCount="6" />
  <row Id="4439" PostTypeId="2" ParentId="4346" CreationDate="2010-09-11T20:47:15.340" Score="1" ViewCount="" Body="&lt;p&gt;Whether or not $f$ is linear, the equation can in fact be rewritten as $F^2 = F + 1$, in a suitable interpretation where to a function $f$ is associated a linear operator $F$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $V$ be the vector space of sums $aX + bf(X)$. Denote by $F$ the operator that takes an element of $V$ to its composition with $f$, that is,  $F (h(x)) = h(f(x))$.  Let $1$ denote the identity operator on $V$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then $F$ is linear (from the definition), and satisfies $F^2=F+1$ as one can check on basis elements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have been temporarily but deliberately unclear as to whether $V$ is the 2-dimensional space of formal sums with basis vectors $x$ and $f(x)$, or the at most 2-dimensional space of functions of that form.   The question of whether $f$ is linear is the same as asking whether the second space is one-dimensional, which is no longer an algebraic question but one of analysis using regularity assumptions on $f$. (In particular, seeing $f$ as an operator or 2x2 matrix $F$, illuminates but does not quite trivialize the problem of showing that linear functions are the only continuous solutions to the functional equation. Additional arguments are needed.)   The equation $F^2 = F + 1$ holds in both interpretations.&lt;/p&gt;&#xA;" OwnerUserId="467" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-09-11T21:42:37.693" LastActivityDate="2010-09-11T21:42:37.693" />
  <row Id="4440" PostTypeId="2" ParentId="4371" CreationDate="2010-09-11T21:37:08.233" Score="1" ViewCount="" Body="&lt;p&gt;I am learning some graph theory myself as an independent study in college.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I started with a very simple, but informative text, Introductory Graph Theory by Chatrand. It is a Dover book, and can be bought for very cheap on Amazon.&#xA;I then went to my university library and took out Modern Graph Theory by Bollobas. It is not an easy book, but it is incredibly clear and informative.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are also great introductions to be found online.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a rather lengthly sample of Bollobas's &lt;I&gt;Modern Graph Theory&lt;/I&gt;:&#xA;&lt;a href=&quot;http://books.google.com/books?id=SbZKSZ-1qrwC&amp;amp;printsec=frontcover&amp;amp;source=gbs_slider_thumb#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;http://books.google.com/books?id=SbZKSZ-1qrwC&amp;amp;printsec=frontcover&amp;amp;source=gbs_slider_thumb#v=onepage&amp;amp;q&amp;amp;f=false&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1838" LastEditorUserId="1838" LastEditorDisplayName="" LastEditDate="2010-09-12T03:32:20.460" LastActivityDate="2010-09-12T03:32:20.460" CommentCount="1" />
  <row Id="4441" PostTypeId="2" ParentId="4437" CreationDate="2010-09-11T21:41:03.873" Score="5" ViewCount="" Body="&lt;p&gt;It's not clear to me what the exact problem is here, but there is a common pitfall&#xA;concerning polynomials with matrix coefficients.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given a polynomial $P(t)=\sum_i B_i t^i$ we can certainly substitute a matrix&#xA;$A$ for $t$ to get $P(A)=\sum_i B_i A^i$. But there is a catch when we multiply&#xA;polynomials. If $Q(t)$ is another polynomial with matrix coefficients, then we may&#xA;have&#xA;$$(PQ)(A)\ne P(A)Q(A).$$&#xA;A sufficient condition for $(PQ)(A)=P(A)Q(A)$ to hold is that $A$ commutes&#xA;with the coefficients of $Q$. But when it doesn't then in general&#xA;$(PQ)(A)\ne P(A)Q(A)$.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-09-11T21:41:03.873" CommentCount="3" />
  <row Id="4442" PostTypeId="1" AcceptedAnswerId="4454" CreationDate="2010-09-11T21:46:34.420" Score="3" ViewCount="332" Body="&lt;p&gt;I've been reading Ravi Vakil's notes on algebraic geometry &lt;a href=&quot;http://math.stanford.edu/~vakil/216blog/FOAGsept1010public.pdf&quot; rel=&quot;nofollow&quot;&gt;notes&lt;/a&gt; and am having a hard time connecting the standard definition of the localization of a module to the definition in terms of universal properties (page 24). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anybody either explain the connection or provide some kind of 'roadmap' of the material necessary to understand this connection?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I apologize if this question is vague or unclear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: This is an attempt to reformulate my original question to make it more precise:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose we did not know the definition of the localization of a module and we just saw the diagram presented on page 24, how do we recover the usual definition and how do we determine that anything else satisfying those properties is isomorphic to the usual definition?&lt;/p&gt;&#xA;" OwnerUserId="249" LastEditorUserId="664" LastEditorDisplayName="" LastEditDate="2010-09-21T19:29:16.827" LastActivityDate="2010-09-21T19:29:16.827" Title="Learning to think categorically (localization of rings and modules)" Tags="&lt;category-theory&gt;&lt;modules&gt;" AnswerCount="2" CommentCount="6" FavoriteCount="2" />
  <row Id="4443" PostTypeId="2" ParentId="4437" CreationDate="2010-09-11T21:56:20.730" Score="13" ViewCount="" Body="&lt;p&gt;The reason we need the lemma is that from $P(t)=b(t)(A-tI)$ one cannot directly conclude that $P(A)=b(A)(A-AI)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $R$ is a &lt;em&gt;commutative&lt;/em&gt; ring, then there is a natural map $R[t]\to R^R$ which is a ring homomorphism (we endow $R^R$ with the pointwise ring structure: $(f+g)(r) = f(r)+g(r)$, and $fg(r) = f(r)g(r)$ for every $r\in R$). If $p(t)=q(t)s(t)$, then for every $r\in R$ you have that $p(r)=q(r)s(r)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But this doesn't work if $R$ is not commutative. For example, taking $p(t) = at$, $q(t) = t$ and $s(t)=a$, you have $p(t)=q(t)s(t)$ in $R[t]$ (since $t$ is central in $R[t]$ even when $R$ is not commutative), but $p(r) = ar$ while $q(r)s(r) = ra$. So you get $p(r)=q(r)s(r)$ if and only if $a$ and $r$ commute. Thus, while you can certainly define a map $\psi\colon R[t]\to R^R$ by&#xA;$$\psi(a_0+a_1t+\cdots+a_nt^n)(r) =  a_0 + a_1r + \cdots + a_nr^n,$$&#xA;this map is not a ring homomorphism when the ring is not commutative. This is the situation we have here, where the ring $R$ is the ring $n\times n$ matrices over $\mathbb{K}$, which is not commutative when $n\gt 1$. In particular, from $P(t) = B(t)(A-tI)$ one cannot simply conclude that $P(A)=B(A)(A-AI)$. This implicitly assumes that your map $M_n(\mathbb{K})[t]\to M_n(\mathbb{K})^{M_n(\mathbb{K})}$ is multiplicative, which it is not in this case. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If your $A$ happens to be central in $M_n(\mathbb{K})$, then it is true that the induced map $M_n(\mathbb{K})[t]\to M_n(\mathbb{K})$ is a homomorphism. But then you would be assuming that your $A$ is a scalar multiple of the identity. It would also be true if the coefficients of the polynomial $b(t)$ centralize $A$, but you are not assuming that.  So you do need to prove that in this case you have $P(A)=b(A)(A-AI)$, since it does not follow from the general set-up (the way it would in a commutative setting).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S. In fact, this is the subtle point where the proof that a polynomial over a field of degree $n$ has at most $n$ roots breaks down for skew fields/division rings. If $K$ is a division ring, then the division algorithm holds for polynomials with coefficients over $K$, so one can show that for every $p(t)\in K[t]$ and $a(t)\in K[t]$, $a(t)\neq 0$, there exist unique $q(t)$ and $r(t)$ such that $p(t)=q(t)a(t) + r(t)$ and $r(t)=0$ or $\deg(r)\lt \deg(a)$. From this, we can deduce that for every polynomial $p(t)$ and for every $a\in K$, we can write $p(t) = q(t)(t-a) + r$, where $r\in K$. But the proof of the Remainder and Factor Theorems no longer goes through, because we cannot go from $p(t)=q(t)(t-a)+r$ to $p(a)=q(a)(a-a)+r$; and you cannot get the recursion argument to work, because from $p(t)=q(t)(t-a)$, and $p(b)=0$ with $b\neq a$, you cannot deduce that $q(b)=0$. For instance, over the real quaternions, we have $p(t)=t^2+1=(t+i)(t-i)$, but $p(j)=j^2+1\neq 2k = ij-ji = (j+i)(j-i)$.  I remember when I first learned the corresponding theorems for polynomial rings, the professor challenging us to identify all the field axioms used in the proofs of the Remainder and Factor Theorem; none of us spotted the use of commutativity in the evaluation map.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-09-12T17:51:10.830" LastActivityDate="2010-09-12T17:51:10.830" CommentCount="2" />
  <row Id="4444" PostTypeId="2" ParentId="4428" CreationDate="2010-09-11T21:57:01.193" Score="1" ViewCount="" Body="&lt;p&gt;I disagree with wiki answers, which although it's partially correct, completely misses the main point.  The &lt;em&gt;heights&lt;/em&gt; in a bar chart represent counts.  You can use a bar chart for this purpose with categorical data or with binned &quot;continuous&quot; variates; either is fine.  The &lt;em&gt;areas&lt;/em&gt; in a histogram represent &lt;em&gt;relative frequencies&lt;/em&gt; or &lt;em&gt;proportions per unit value&lt;/em&gt; of a continuous variate.  A histogram thus is a discrete approximation to a frequency or probability density function, whereas a bar chart has no such meaning (except accidentally when it's really serving as a histogram!).  The distinction comes to the fore when the bars in a histogram have varying widths.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reference: Freedman, Pisani, Purves, &lt;em&gt;Statistics&lt;/em&gt;.  (Any of the first through fifth editions should be fine.)&lt;/p&gt;&#xA;" OwnerUserId="1489" LastActivityDate="2010-09-11T21:57:01.193" />
  <row Id="4445" PostTypeId="1" AcceptedAnswerId="4446" CreationDate="2010-09-11T22:02:05.623" Score="2" ViewCount="146" Body="&lt;p&gt;From any textbook on fourier analysis: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;It is easily shown that for $f$ and $g$, both $2 \pi$-periodic functions on $[-\pi,\pi]$, we have $$(f \ast g)(x) = \int_{-\pi}^{\pi}f(x-y)g(y)\;dy = \int_{-\pi}^{\pi}f(z)g(x-z)\;dz = (g \ast f)(x),$$ by using the substitution $z = x-y.$&quot;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;I don't doubt that this is true, but I cannot figure out what happened to the negative sign coming from $dy = -dz\;$ after the change of variable $z = x - y$.  In particular, after the change of variable $z = x-y,\;$ I am coming up with $ -\int_{-\pi}^{\pi}f(z)g(x-z)\;dz$. What am I missing here?&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="121" LastActivityDate="2010-09-11T22:17:44.833" Title="Proving commutativity of convolution $(f \ast g)(x) = (g \ast f)(x)$" Tags="&lt;fourier-analysis&gt;&lt;convolution&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" />
  <row Id="4446" PostTypeId="2" ParentId="4445" CreationDate="2010-09-11T22:14:20.233" Score="7" ViewCount="" Body="&lt;p&gt;You need to check the bounds on your integral. since $y$ ranges from $-\pi$ to $\pi$, you'll have $z = x-y$ ranging from $x+\pi$ to $x-\pi$. Therefore:&#xA;$$&#xA;\int_{-\pi}^{\pi}f(x-y)g(y)dy = -\int_{x+\pi}^{x-\pi}f(z)g(x-z)dz =&#xA;\int_{x-\pi}^{x+\pi}f(z)g(x-z)dz = \int_{-\pi}^{\pi}f(z)g(x-z)dz&#xA;$$&#xA;in the second-last step, I swapped the two bounds on the integral (this changes the sign). In the final step, I shifted both bounds on the integral by $-x$, which does not change the value because we are integrating over an interval of length $2\pi$ and the function is $2\pi$-periodic.&lt;/p&gt;&#xA;" OwnerUserId="1356" LastActivityDate="2010-09-11T22:14:20.233" CommentCount="1" />
  <row Id="4447" PostTypeId="2" ParentId="4445" CreationDate="2010-09-11T22:17:44.833" Score="1" ViewCount="" Body="&lt;p&gt;Here is something I've sometimes wondered about. If $f,g$ are both nonnegative proving &#xA;commutativity of convolution can be done without a tedious change of variable. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Indeed, let $X$ be a random variable with density $f$ and let $Y$ be a random variable with density $g$. Its easy to see that $f$ convolved with $g$ is the density of $X+Y$ (or in your case $X+Y ~{\rm mod} ~2 \pi$). By commutativity of addition, the density of $X+Y$ is the same as the density of $Y+X$ and we are done! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wonder if an argument of this sort can work in the general case. &lt;/p&gt;&#xA;" OwnerUserId="317" LastActivityDate="2010-09-11T22:17:44.833" />
  <row Id="4448" PostTypeId="2" ParentId="4435" CreationDate="2010-09-11T22:19:08.367" Score="5" ViewCount="" Body="&lt;p&gt;$f(0+0)=f(0) + f(0)$ implies $f(0)=0$. Set $y=0$ to get $f(\alpha x)=f(x)$ so that $f(\alpha (x+y))=f(x+y)$ and the function is additive.  If $\alpha \neq 1$ this means $f((\alpha - 1)x)=0$ and hence $f=0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is too easy to be on rwbarton's list, and in fact I cannot find this problem in the file that you linked.  Which problem number is it in the list? &lt;/p&gt;&#xA;" OwnerUserId="467" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-09-11T22:42:08.930" LastActivityDate="2010-09-11T22:42:08.930" CommentCount="2" />
  <row Id="4449" PostTypeId="2" ParentId="4174" CreationDate="2010-09-11T22:36:56.227" Score="5" ViewCount="" Body="&lt;p&gt;Induction on the number of sentence symbols is the kind of proof Craig gave originally, and is what logicians call a syntactic proof. It's also possible to give a semantic or model-theoretic proof. Bell and Slomson give one for first-order logic in Chapter 7, section 4 of their text &lt;a href=&quot;http://amzn.com/0486449793&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;Models and Ultraproducts&lt;/em&gt;.&lt;/a&gt; I haven't seen a model-theoretic proof for propositional logic.  (Submitted as an answer because I don't have enough rep yet to comment.)&lt;/p&gt;&#xA;" OwnerUserId="1813" LastEditorUserId="1813" LastEditorDisplayName="" LastEditDate="2010-09-11T22:42:03.903" LastActivityDate="2010-09-11T22:42:03.903" CommentCount="1" />
  <row Id="4450" PostTypeId="2" ParentId="4442" CreationDate="2010-09-11T22:44:58.487" Score="3" ViewCount="" Body="&lt;p&gt;In my opinion, the most enlightening (and the simplest) way to present the universal construction of localizations (and fractions) is to use instead of the pair construction the natural presentation in terms of generators and relations. This allows one to exploit the universal properties of quotient rings and polynomial rings to quickly construct and derive the basic properties of localizations (and to avoid the many tedious verifications required in the pair approach). Moreover, this approach is much more conceptual. Indeed, the pairs in the pair construction are nothing but normal forms for the polynomial terms in the presentation based approach.  For details of this approach see e.g. the exposition in section 11.1 of Rotman's Advanced Modern Algebra, and Voloch's: &lt;a href=&quot;http://www.ma.utexas.edu/users/voloch/Homework/fractions.pdf&quot; rel=&quot;nofollow&quot;&gt;Rings of fractions the hard way&lt;/a&gt;. Note: presumably Voloch's title is a joke - since the presentation based approach is actually the easiest way - in fact both Rotman's and Voloch's expositions can be simplified.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're just beginning to understand universal constructions then I highly recommend that you peruse the beautiful exposition in George Bergman's &lt;a href=&quot;http://www.math.berkeley.edu/~gbergman/245&quot; rel=&quot;nofollow&quot;&gt;An Invitation to General Algebra and Universal Constructions&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might also find illuminating Paul Cohn's historical article &lt;a href=&quot;http://www.maths.ed.ac.uk/~aar/books/nlat.pdf#page=15&quot; rel=&quot;nofollow&quot;&gt;Localization in general rings, a historical survey&lt;/a&gt; - as well as other papers in that &lt;a href=&quot;http://www.maths.ed.ac.uk/~aar/books/nlat.pdf&quot; rel=&quot;nofollow&quot;&gt;volume [1]&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[1] Ranicki, A.(ed). &lt;a href=&quot;http://www.maths.ed.ac.uk/~aar/books/nlat.pdf&quot; rel=&quot;nofollow&quot;&gt;Noncommutative localization in algebra and topology&lt;/a&gt;. ICMS 2002&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-09-19T22:14:03.513" LastActivityDate="2010-09-19T22:14:03.513" CommentCount="1" />
  <row Id="4451" PostTypeId="2" ParentId="4437" CreationDate="2010-09-11T23:21:58.943" Score="1" ViewCount="" Body="&lt;p&gt;The problem is quite simple: the definition of polynomial multiplication assumes that the coefficients commute with the indeterminates since e.g. $\rm\; r\ x = x\ r\:.\:$ So for evaluation to preserve polynomial arithmetic, i.e. for it to be a ring homomorphism, one can only evaluate polynomials in rings where the coefficient ring is central, i.e. where coefficients commute with all other ring elements. This concept is so important that it has a name, viz. $\rm\:R\:$-algebras are those rings $\rm\:A\:$ in which polynomials in $\rm R[x]$ can be evaluated. More precisely, this means that $\rm\:A\:$ contains an image $\rm\:R'$ of $\rm\:R\:$ and that for all $\rm\;a\in A\:$ the evaluation map $\rm\:x\to a\:$ is a ring homomorphism $\rm\:R[x]\to A\:$. In particular, evaluating $\rm\;\; r\: x = x\: r\;\;$ at $\rm\:x = a\in A\:$ implies that $\rm\;R'$ is central, i.e. it commutes with all elements of $\rm\:A\:$. That's precisely all that's required for the existence of a unique evaluation homomorphism $\rm\:\epsilon_a : x\to a\:$ for every $\rm\:a\in A\:.\:$ This permits one to view the polynomial ring $\rm R[x]$ as the universal $\rm\:R$-algebra generated by $\rm\:x\:.$&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-09-17T23:16:20.873" LastActivityDate="2010-09-17T23:16:20.873" CommentCount="2" />
  <row Id="4452" PostTypeId="2" ParentId="4400" CreationDate="2010-09-11T23:58:43.577" Score="0" ViewCount="" Body="&lt;p&gt;Well, given the unstated assumption that the writer is a mathematician and therefore not using regular english, then I agree with the 13/27 answer. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But in everyday english, from &quot;there are two fleems, one is a glarp&quot; we all infer that the other is not a glarp. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From &quot;there are two fleems, one is a glarp, which is snibble&quot; we would still infer that the other is not a glarp. Whereas from &quot;there are two fleems, one is a glarp which is snibble&quot; (absence of comma, or when spoken, difference in intonation) we would infer that the other is not a snibble glarp, but it could still be an unsnibble glarp.&lt;/p&gt;&#xA;" LastActivityDate="2010-09-11T23:58:43.577" CommentCount="2" />
  <row Id="4453" PostTypeId="1" CreationDate="2010-09-12T00:20:11.347" Score="3" ViewCount="122" Body="&lt;p&gt;This is a simple terminology question.  Let $S$ be a (let's say smooth)&#xA;surface in $\mathbb{R}^3$, and $p$ a point&#xA;on $S$.  Suppose the principle curvatures $\kappa_1$ and $\kappa_2$ at $p$ are both negative.&#xA;I am imagining $p$ sitting at the bottom of a dent in the surface.&#xA;Is there an accepted term to describe such a point?&#xA;The difficulty is that the Gaussian curvature $\kappa_1 \kappa_2$ is positive, so intrinsically&#xA;$p$ is no different than if it were on a bump rather than a dent. &#xA;I could make up my own term of course, e.g., &lt;em&gt;valley point&lt;/em&gt; or &lt;em&gt;cup point&lt;/em&gt;, but I'd rather follow convention.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="237" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-09-12T01:53:46.720" LastActivityDate="2011-01-23T05:31:51.747" Title="Terminology for point in dent in surface?" Tags="&lt;terminology&gt;&lt;differential-geometry&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="4454" PostTypeId="2" ParentId="4442" CreationDate="2010-09-12T01:15:06.697" Score="2" ViewCount="" Body="&lt;p&gt;As the Rotman's, or Voloch's, constructions of Dubuque's answers show, you &lt;em&gt;can't&lt;/em&gt; recover &lt;em&gt;exactly&lt;/em&gt; &quot;the usual definition&quot; of the localization of a ring from its universal property. That is, a universal property won't tell you, in general, how to construct an object that verifies it (as you can see, there are at least, two constructions for the localization, both verifying the universal property: how are you going to choose among them?). What a universal property will tell you is that all the guys verifying it are necessarily isomorphic: for instance, the construction in Vakil's notes, and Rotman's are necessarily isomorphic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Precisely, let's stay with rings and let $\gamma' : A \longrightarrow B$ be some guy that satisfies the universal property of the localization. Let $\gamma: A \longrightarrow S^{-1}A$ denote the construction of the localization explained in Vakil's notes, 2.3.3. Then, since both $S^{-1}A$ and $B$ satisfy the universal property, you'll have morphisms $\widetilde{\gamma} : S^{-1}A \longrightarrow B$ and $\widetilde{\gamma'}: B \longrightarrow S^{-1}A$ such that $\widetilde{\gamma'}\gamma = \gamma'$ and $\widetilde{\gamma}\gamma' = \gamma$. Let's show that both $\widetilde{\gamma}$ and $\widetilde{\gamma'}$ are inverses of each other. For instance,  $\widetilde{\gamma'} \widetilde{\gamma} = \mathrm{id}_B$ because&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;(\widetilde{\gamma'} \widetilde{\gamma})\gamma' = \widetilde{\gamma'}\gamma = \gamma' \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But $\mathrm{id}_B$ verifies the same identity clearly:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\mathrm{id}_B \gamma'= \gamma' \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, because of the uniqueness of the universal property, you must have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\widetilde{\gamma'} \widetilde{\gamma} = \mathrm{id}_B \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Analogously,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\widetilde{\gamma}\widetilde{\gamma'} = \mathrm{id}_{S^{-1}A} \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you can (and must) do is to verify that the &lt;em&gt;usual&lt;/em&gt; definition (particular construction) of the localization verifies the universal property. That is, given some morphism of rings $\varphi : A \longrightarrow B$ such that $\varphi s \in B$ is a unit for every $s \in S$, then there is a unique morphism of rings $\widetilde{\varphi} : S^{-1}A \longrightarrow B$ such that $\widetilde{\varphi}\gamma = \varphi$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint: you can easily check that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\widetilde{\varphi} \left( \frac{a}{s}  \right) = \frac{\varphi (a)}{\varphi (s)}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is well-defined (that is, if $\frac{a}{s} = \frac{b}{t}$, then $\widetilde{\varphi}\left( \frac{a}{s}  \right) = \widetilde{\varphi}\left( \frac{b}{t}  \right) $), is a morphism of rings, verifies $\widetilde{\varphi} \gamma = \varphi$ and is unique (that is, if you had another $\psi : S^{-1}A \longrightarrow B$ such that $\psi\gamma = \varphi$, then $\psi = \widetilde{\varphi}$).&lt;/p&gt;&#xA;" OwnerUserId="664" LastEditorUserId="664" LastEditorDisplayName="" LastEditDate="2010-09-12T06:55:16.627" LastActivityDate="2010-09-12T06:55:16.627" CommentCount="1" />
  <row Id="4455" PostTypeId="1" AcceptedAnswerId="4457" CreationDate="2010-09-12T02:13:38.003" Score="3" ViewCount="87" Body="&lt;p&gt;I fear I over simplified the following problem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For any partially ordered set $(A,\leq)$, let $A^* = A- ${$ maxA,minA $} if $maxA$ and $minA$ exist. Show the inclusion $(A^*,\leq)\hookrightarrow (A,\leq)$ is sup-continuous.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I took any set $B\subseteq A^*$ such that $supB$ exists. Then $\iota(B)=B$, and $\iota(supB)=supB$. Then $\iota(supB)=supB=sup(\iota(B))$, and so $\iota$ is sup-continuous.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This seems too simple so I'm sure I've misinterpreted something. Can someone point out the source of error?&lt;/p&gt;&#xA;" LastActivityDate="2010-09-12T03:44:46.230" Title="Showing the Inclusion is sup-continuous." Tags="&lt;set-theory&gt;" AnswerCount="1" />
  <row Id="4456" PostTypeId="2" ParentId="4426" CreationDate="2010-09-12T02:58:27.513" Score="1" ViewCount="" Body="&lt;p&gt;(1') is not equivalent to (2').  For example, $f(x)=x^2$ satisfies (2') on $(-\infty,\infty)$ but not (1'). It is not even uniformly continuous.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Condition (2'), being AC on all compact subintervals, is a condition I have at least seen used, and it is the right one if you want to extend the equivalence to being an indefinite integral.  Namely, it is equivalent to:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(3) If $a\lt c \lt b$, then $f(x)=f(c)+ \int_c^x f'$ for all $x\in(a,b)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is only a slight modification of (2), which must be made because $f$ may be unbounded or otherwise undefined at $a$ and $f'$ may not be integrable on $(a,x)$, even if $f$ satisfies the stronger condition (1') (e.g., $f(x)=x$ on $(-\infty,\infty)$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But you asked if this is the &quot;best&quot; definition or if there is an &quot;accepted&quot; definition, and of that I am not sure.  I have not seen anyone write &quot;$f$ is AC on $(a,b)$&quot; when they mean (2') holds.  For the case $(-\infty,\infty)$, I have seen simply &quot;f is AC on bounded intervals&quot;.  On the Wikipedia page, they use the phrase &quot;locally absolutely continuous&quot; in the section on the relationship to measures.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2010-09-12T03:15:06.287" LastActivityDate="2010-09-12T03:15:06.287" CommentCount="2" />
  <row Id="4457" PostTypeId="2" ParentId="4455" CreationDate="2010-09-12T03:38:49.213" Score="2" ViewCount="" Body="&lt;p&gt;The set of upper bounds depends on whether you are in A or in A*, so technically you have to check that sup(B) actually is the least upper bound of B in A.  It would still be an upper bound of B, but there are 2 new contenders, max(A) and min(A).  Of course sup(B) &amp;lt; max(A), so max(A) is a larger upper bound and presents no problem.  On the other hand, min(A) &amp;lt; sup(B), so if min(A) is an upper bound for B in A, then it must in fact be the sup of B in A.  This could only happen if B is empty, and it could happen.  For example, let A={0,1,2} with the usual order and let B be the empty set.  Then &quot;sup(B)&quot; in A*={1} is 1, but the sup of B in A is 0.  This problem occurs only because A* has a least element, thus making the empty set have a sup in A*.  Everything is fine if B is nonempty.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I've correctly hit on the point you were stumbling over, perhaps it would help to adjust notation to keep in mind that sups are conditional, e.g., $\sup_A (B)$ for the sup of $B$ in $A$.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2010-09-12T03:44:46.230" LastActivityDate="2010-09-12T03:44:46.230" CommentCount="4" />
  <row Id="4458" PostTypeId="1" CreationDate="2010-09-12T03:56:11.690" Score="5" ViewCount="248" Body="&lt;p&gt;My goal here is to make my own custom &lt;a href=&quot;http://www.progonos.com/furuti/MapProj/Normal/ProjPoly/projPoly.html&quot; rel=&quot;nofollow&quot;&gt;&quot;polyhedral map&quot;&lt;/a&gt; of Earth. If you print out something from the &lt;a href=&quot;http://www.progonos.com/furuti/MapProj/Normal/ProjPoly/Foldout/foldout.html&quot; rel=&quot;nofollow&quot;&gt;&quot;Map Fold-outs&quot;&lt;/a&gt; page, you will have something almost exactly like what I'm trying to make.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have 2 triangles. One is a spherical triangle drawn on a 3D globe. By definition, each edge of a spherical triangle is part of a great circle. When you look at that 3D globe, there are a bunch of cities, coastlines, etc. that are (hopefully) accurately plotted on that 3D globe, inside that spherical triangle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other triangle is a flat, plane (2D), straight-edged, Euclidean triangle. On paper. At the moment the interior of that triangle is blank white paper, but eventually I want to draw a copy of all those cities, coastlines, etc. into that area.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Every map projection will map that 3D spherical triangle to a 2D image. Then it's easy to (in 2D) rotate and slide and shrink, and perhaps skew, that image until the 3 corners exactly line up with the 3 corners of my plane triangle. If I stack the results of a bunch of different map projections on top of each other, even though I've forced the 3 corners to exactly line up, each projection will put the cities in a slightly different location. Unfortunately, many projections take cities that are slightly inside the spherical triangle and draw them slightly outside the flat 2D triangle. (One symptom of this problem is that the sides of the spherical triangle are mapped to plane (2D) curves whose endpoints match the corners properly, but they bulge outward slightly from perfectly straight lines drawn between the corners). That leads to the city being completely missing from my polyhedral map.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd rather not run the risk of some Polish mathematician getting upset that Warsaw is inexplicably missing from the flat, plane (2d), paper map that I've made :-).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I'm looking for map projections that &quot;keep it inside the lines&quot;. I need 3 specific great circles (the edges of the spherical triangle) to be mapped to straight lines on paper. I don't care about other great circles -- straight, non-straight, whatever.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What map projections meet that criteria? And can you give me a link to the (x,y) = f(lat, long) equations for that projection?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hear that the map projection used for the Dymaxion map meets that criteria; and someone told me that he thought it used the Chamberlin trimetric projection.&#xA;Alas, when I use my (extremely rough and probably buggy) implementation of Chamberlin trimetric projection to map the spherical triangle formed by its 3 base points to the plane, I seem to be getting a shape that is almost a triangle, but the three &quot;lines&quot; curve and bulge out. Is that a bug in my code, or is it supposed to do that?&#xA;So apparently either (a) I'm using the wrong equations -- so where can I find the right equations? Or (b) Dymaxion actually uses some other projection -- so where can I find the equations for that projection?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm looking for answers of the form &quot;The gnomonic projection meets your criteria. The &lt;a href=&quot;http://en.wikipedia.org/wiki/Gnomonic_projection&quot; rel=&quot;nofollow&quot;&gt;gnomonic projection equations&lt;/a&gt;.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm trying to get a list of several map projections that meet that critera.&#xA;I know that the gnomonic projection is not the only one, because&#xA;the Collignon projection and the &lt;a href=&quot;http://en.wikipedia.org/wiki/Peirce_quincuncial_projection&quot; rel=&quot;nofollow&quot;&gt;Peirce quincuncial projection&lt;/a&gt; can map an octant (a spherical triangle with 90 degree corners) to a straight-edged plane triangle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Forgive me for reposting &lt;a href=&quot;http://stackoverflow.com/questions/3305056/how-do-i-map-a-spherical-triangle-to-a-plane-triangle&quot;&gt;this question from StackOverflow&lt;/a&gt;. I hope that a few people here will know a few answers to this that no one at StackOverflow seems to know).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Several responses have mentioned that the &lt;em&gt;only&lt;/em&gt; way to make every straight line on the plane map correspond to a great circle on the globe is to use the gnomonic projection (or some linear affine transformation of it).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's technically correct, but irrelevant to my question.&#xA;I'm not asking for a projection that maps &lt;em&gt;every&lt;/em&gt; straight line on the plane map correspond to a great circle on the globe.&#xA;I'm asking for projections that map 3 particular straight lines -- the boundaries of one facet of my map -- to 3 particular great circles on the globe -- the great circles connecting a few particular carefully-chosen locations on Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While certainly the gnomonic projection is &lt;em&gt;one&lt;/em&gt; way to map the cities, coastlines, etc. in a spherical triangle to a plane triangle, it is not the only way. As an existence proof, as I mentioned before, the Collignon projection and the Peirce quincuncial projection also map the octant -- a spherical triangle -- to a straight-edged plane triangle.&#xA;(What are the equations for the &lt;a href=&quot;http://www.grunch.net/synergetics/map/dymax.html&quot; rel=&quot;nofollow&quot;&gt;Dymaxion map projection&lt;/a&gt;?&#xA;I hear that it is yet another way to map a spherical triangle to a plane triangle).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alas, some people do not like the area distortion, length distortion, and the angle distortion of the gnomonic projection.&#xA;I'm hoping that by allowing a little distortion in the paths of &lt;em&gt;some&lt;/em&gt; great circles (mapping them to slightly curved paths on the plane), I will end up with less area distortion, length distortion, and angle distortion.&#xA;(I see that the Peirce quincuncial projection has zero angle distortion at most points).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;there are infinitely many smooth mappings from the sphere to the plane that satisfy that criterion. ... you can always take any such mapping and &quot;distort&quot; it a little in the interior.&quot;&#xA;Yes, but most of them make area distortion, length distortion, and angle distortion &lt;em&gt;worse&lt;/em&gt;, and have horrifically complicated equations describing them.&#xA;I'm looking for projections that are &lt;em&gt;better&lt;/em&gt; than the gnomonic projection, preferably with relatively simple equations describing the projection.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hesitate to say that I'm looking for the projection that &lt;em&gt;minimizes&lt;/em&gt; distortion, because I don't want several pages of horrifically complicated equations that happen to minimize some specific arbitrarily-weighted average of area distortion, length distortion, and angle distortion.&lt;/p&gt;&#xA;" OwnerUserId="1705" LastEditorUserId="1705" LastEditorDisplayName="" LastEditDate="2010-09-13T15:52:36.827" LastActivityDate="2010-09-13T15:52:36.827" Title="How do I map a spherical triangle to a plane triangle?" Tags="&lt;spherical-geometry&gt;" AnswerCount="2" CommentCount="5" />
  <row Id="4459" PostTypeId="2" ParentId="4437" CreationDate="2010-09-12T05:00:26.510" Score="0" ViewCount="" Body="&lt;p&gt;I'd just like to make the following three observations as a minor complement to the other answers. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;OBSERVATION 1.&lt;/strong&gt; What the Cayley-Hamilton Theorem says is $$\det&#xA;\begin{pmatrix}&#xA;a_{11}-A &amp;amp; a_{12}&amp;amp;\cdots&amp;amp;a_{1n}\\ &#xA;a_{21}&amp;amp;a_{22}-A&amp;amp;\cdots&amp;amp;a_{2n}\\ &#xA;\vdots&amp;amp;\vdots&amp;amp;\vdots&amp;amp;\vdots\\ &#xA;a_{n1}&amp;amp;a_{n2}&amp;amp;\cdots&amp;amp;a_{nn}-A&#xA;\end{pmatrix}=0.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;OBSERVATION 2.&lt;/strong&gt; The proof of the Cayley-Hamilton Theorem I like best (among the ones I know) is on &lt;a href=&quot;http://www.iecn.u-nancy.fr/~gaillard/atiyah_macdonald_p_21.pdf&quot; rel=&quot;nofollow&quot;&gt;page 21&lt;/a&gt; (proof of Proposition 2.4) of &lt;strong&gt;Introduction to Commutative Algebra&lt;/strong&gt; by Atiyah and MacDonald. The argument can be phrased as follows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $K$ be a commutative ring; let $n$ be a positive integer; let $A=(a_{ij})\in M_n(K)$ be an $n$ by $n$ matrix with entries in $K$; let $\chi$ be its characteristic polynomial; define $B=(b_{ij})\in M_n(K[A])$ by $b_{ij}:=\delta_{ij}\,A-a_{ij}$; observe $$\sum_i\ \ b_{ij}\ e_i=0,\quad\det B=\chi(A);$$ and write $(c_{ij})$ for the adjugate of $B$. Applying (a trivial case of) Fubini's Theorem to the double sum $\sum_{i,j}\ c_{jk}\ b_{ij}\ e_i$, we get $\chi(A)=0$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;OBSERVATION 3.&lt;/strong&gt; It's easy to define the ring $R[X]$ of polynomials in the indeterminate $X$ with coefficients in the &lt;strong&gt;noncommutative&lt;/strong&gt; $R$. But when you think in terms of universal properties, you see that this construction is not very natural. So, it's better, I think, not to introduce it just to prove the Cayley-Hamilton Theorem. &lt;/p&gt;&#xA;" OwnerUserId="660" LastActivityDate="2010-09-12T05:00:26.510" />
  <row Id="4460" PostTypeId="2" ParentId="2969" CreationDate="2010-09-12T07:23:21.410" Score="1" ViewCount="" Body="&lt;p&gt;I realized about 80% through that this answer may have too many parts to implement, so I'm also going to leave a shorter one.&lt;br&gt;&lt;br&gt;&#xA;If you prioritize filling some cells over others, you can at least prune the computer's search tree. I suggest filling the center cube, followed by the six face centers, followed by the eight corners, finishing with everything else.&lt;br&gt;&lt;br&gt;&#xA;Here, let me number the squares of a Y pentomino:&lt;br&gt;1&lt;br&gt;23&lt;br&gt;4&lt;br&gt;5&lt;br&gt;By starting with the Y filling the center cube, the search space can be reduced by a factor of 12 by holding this piece's orientation constant. Note that only cells 2, 3, and 4 can occupy the center of the cube. If 2 or 4 is the center cell, then five face centers will remain to be filled; if it's 3, then all six will remain.&lt;br&gt;&lt;br&gt;&#xA;For the five or six centers, the Y occupying each center can be placed in one of 36 ways, disregarding overlap, as follows:&lt;br&gt;&lt;br&gt;&#xA;-The &quot;spine&quot; (the cubes 1245) can point in any of four directions&lt;br&gt;-Again, only 2, 3, or 4 can fill these cubes&lt;br&gt;-The full Y can lie on the face in three ways for each cube: on either Y face for all three, plus either along the spine for 2 and 4 or alongthe furthest opposing face for 3&lt;br&gt;&lt;br&gt;&#xA;Now for the eight corners. Only 1 and 5 can occupy these cubes, so for each of these corners we can work on partial placements for just the spines along the edges. Now, each corner is adjacent to three three-square edges, although some of the twelve edges are certainly part-occupied by Y's placed in the previous phase. The corners to be filled need to be prioritized by fewest available edges.&lt;br&gt;&lt;br&gt;&#xA;-If, at any point, an unfilled corner has 0 free edges, then either the last &quot;temporary edge placement&quot; or the entire face-center configuration needs to be advanced to the next one&lt;br&gt;-Whenever a corner piece is laid along an edge and the other corner sharing that edge is not yet filled, that other corner should be re-prioritized for having one fewer free edge&lt;br&gt;-If some unfilled corner has only one free edge, a spine is placed along that corner and edge&lt;br&gt;-And if no unfilled corner has fewer than two free edges, pick and note a &quot;temporary edge placement&quot; to fix if it leads to a contradiction. (I guarantee at least two corners will have no more than two edges. If the first placement is wrong, the second placement should be used but not noted as temporary.)&lt;br&gt;&lt;br&gt;&#xA;This gives eight partial placements and six or seven total placements. This leaves eleven or ten Y's to place within a volume of 63 or 58, respectively, which includes at least 26 cubes that are absolutely not filled regardless of how exactly the corner Y's are placed (each one still has up to four possibilities for the placement of the 3 cube), a much smaller search space.&lt;/p&gt;&#xA;" LastActivityDate="2010-09-12T07:23:21.410" />
  <row Id="4461" PostTypeId="2" ParentId="2969" CreationDate="2010-09-12T07:32:13.777" Score="0" ViewCount="" Body="&lt;p&gt;For a simpler solution path, I again suggest filling the center cube first to reduce symmetries, but make sure cube 1 or 5 is on the top layer (or bottom, depending on which you want to fill first). Then you can fill the first layer (possibly prioritizing the corners), followed by the subsequent layers.&lt;/p&gt;&#xA;" LastActivityDate="2010-09-12T07:32:13.777" />
  <row Id="4462" PostTypeId="1" AcceptedAnswerId="6225" CreationDate="2010-09-12T09:08:12.540" Score="4" ViewCount="183" Body="&lt;p&gt;&lt;a href=&quot;http://www.mathreference.com/ag,fgaf.html&quot; rel=&quot;nofollow&quot;&gt;This&lt;/a&gt; page contains a result which it refers to as the UFD field lemma. I was wondering if anybody knew of any other references which discuss this result--this page is the only place I've seen it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The UFD field lemma appears to assert that if $R$ is a unique factorization domain containing infinitely many prime elements, if $F$ is the field of fractions of $R$, and if $A$ is a finitely generated $R$-algebra which is a field and is algebraic over $F$, then $A$ does not contain $F$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm looking for other sources because I find the exposition on that page a little hard to follow.&lt;/p&gt;&#xA;" OwnerUserId="251" LastEditorUserId="251" LastEditorDisplayName="" LastEditDate="2010-10-06T20:40:03.507" LastActivityDate="2010-10-06T21:12:40.120" Title="The UFD field lemma" Tags="&lt;abstract-algebra&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" />
  <row Id="4463" PostTypeId="2" ParentId="4368" CreationDate="2010-09-12T09:49:10.053" Score="0" ViewCount="" Body="&lt;p&gt;Please search for &lt;strong&gt;external memory linear algebra&lt;/strong&gt; or &lt;strong&gt;out of core&lt;/strong&gt; computation to see if you find implementations / ideas for how to manage these huge 'v' vectors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moreover, since computing the largest eigenvalue + corresponding eigenvector are in a way core parts of Google's pagerank algorithm, you might find useful leads on how to perform this huge computation by using ideas similar to those used for pagerank.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, if you are content with approximate answers, other options like random projections open up.&lt;/p&gt;&#xA;" OwnerUserId="1709" LastActivityDate="2010-09-12T09:49:10.053" />
  <row Id="4465" PostTypeId="2" ParentId="4368" CreationDate="2010-09-12T10:19:35.053" Score="0" ViewCount="" Body="&lt;p&gt;For a computer package that solves the eigenvalue for large sparse matrix problem. Use &lt;a href=&quot;http://www.caam.rice.edu/software/ARPACK/&quot; rel=&quot;nofollow&quot;&gt;ARPACK&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For &lt;a href=&quot;http://en.wikipedia.org/wiki/ARPACK&quot; rel=&quot;nofollow&quot;&gt;wiki&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The package is designed to compute a&#xA;  few eigenvalues and corresponding&#xA;  eigenvectors of large sparse or&#xA;  structured matrices, using the&#xA;  Implicitly Restarted Arnoldi Method&#xA;  (IRAM) or, in the case of symmetric&#xA;  matrices, the corresponding variant of&#xA;  the Lanczos algorithm.the Lanczos algorithm.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="550" LastActivityDate="2010-09-12T10:19:35.053" />
  <row Id="4466" PostTypeId="1" AcceptedAnswerId="4493" CreationDate="2010-09-12T10:57:24.120" Score="3" ViewCount="175" Body="&lt;p&gt;I'm working my way through Niven's Introduction to Number Theory, and the wording of the following problem is making me unsure of my answer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Show that there is a one-to-one correspondence between twin primes and numbers $n$ such that $n^2-1$ has just four positive divisors. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I felt an obvious bijection would be $f\colon A\rightarrow B\colon (p,p+2)\mapsto p+1$, where $A$ is the set of all pairs of twin primes, and $B$ is the set of all positive $n$ such that $n^2-1$ has only four positive divisors. $f$ is injective, and for any such $n$, if $n^2-1=(n-1)(n+1)$, has only four positive divisors, they must be $1,(n-1),(n+1),n^2-1$, implying that $n-1$ and $n+1$ are twin primes, and thus $(n-1,n+1)$ would be a suitable preimage. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I assumed the wording of the problem meant I should show a bijection from unordered pairs of twin primes to positive integers, but I'm not sure if the problem intended for me to find a bijection that takes a single prime that happens to be a twin prime, to any such $n$, not necessarily positive which is problematic since both $n$ and $-n$ give the same $n^2-1$. Have I interpreted this problem correctly? If not, do other bijections exist with the differing domains and ranges I mentioned above?&lt;/p&gt;&#xA;" OwnerUserId="1583" LastActivityDate="2010-09-15T10:42:42.653" Title="Bijection between twin primes and numbers $n$ such that $n^2-1$ has exactly four positive divisors" Tags="&lt;number-theory&gt;&lt;prime-numbers&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="4467" PostTypeId="1" AcceptedAnswerId="4478" CreationDate="2010-09-12T11:24:41.593" Score="3" ViewCount="1127" Body="&lt;p&gt;I got this interesting question in my mind:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do we prove that if $a \in  \mathbb{N}$ then $\sqrt{a}$ is an integer or an irrational number?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can we extend this result?  That is, can it be shown that if $a,b \in \mathbb{N}$, then $a^{1/b}$ is an integer or an irrational?&lt;/p&gt;&#xA;" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2011-01-05T00:37:02.193" LastActivityDate="2011-03-13T21:48:36.523" Title="$a^{1/2}$ is either an integer or an irrational number." Tags="&lt;number-theory&gt;&lt;algebra&gt;&lt;irrational-numbers&gt;" AnswerCount="9" CommentCount="11" FavoriteCount="4" />
  <row Id="4468" PostTypeId="2" ParentId="4467" CreationDate="2010-09-12T11:32:07.030" Score="10" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Theorem&lt;/strong&gt;: If $a$ and $b$ are positive integers, then $a^{1/b}$ is either irrational or an integer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $a^{1/b}=x/y$ where $y$ does not divide $x$, then $a=(a^{1/b})^b=x^b/y^b$ is not an integer (since $y^b$ does not divide $x^b$), giving a contradiction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I subsequently found a variant of this proof on Wikipedia, under &lt;a href=&quot;http://en.wikipedia.org/wiki/Square_root_of_2#Proofs_of_irrationality&quot; rel=&quot;nofollow&quot;&gt;Proof by unique factorization&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The bracketed claim is proved below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Lemma&lt;/strong&gt;:  If $y$ does not divide $x$, then $y^b$ does not divide $x^b$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Fundamental_theorem_of_arithmetic&quot; rel=&quot;nofollow&quot;&gt;Unique prime factorisation&lt;/a&gt; implies that there exists a prime $p$ and positive integer $t$ such that $p^t$ divides $y$ while $p^t$ does not divide $x$.  Therefore $p^{bt}$ divides $y^b$ while $p^{bt}$ does not divide $x^t$.  Hence $y^b$ does not divide $x^b$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[OOC: This answer has been through several revisions (some of the comments below might not relate to this version)]&lt;/p&gt;&#xA;" OwnerUserId="139" LastEditorUserId="139" LastEditorDisplayName="" LastEditDate="2010-09-13T10:40:59.017" LastActivityDate="2010-09-13T10:40:59.017" CommentCount="10" />
  <row Id="4469" PostTypeId="2" ParentId="4467" CreationDate="2010-09-12T11:47:59.613" Score="4" ViewCount="" Body="&lt;p&gt;The general theorem is that a natural number $a$ has a rational square root if and only if the multiplicity of every prime factor of $a$ is even.  For example $2^43^611^2$ has a rational square root but $5^411^3$ does not. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moreover, if a natural number has a rational square root, that square root is always obtained by halving the multiplicity of each prime factor, and so the square root is also a natural number. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same principle works for $n$th roots, $n &gt; 1$.&lt;/p&gt;&#xA;" OwnerUserId="630" LastActivityDate="2010-09-12T11:47:59.613" CommentCount="1" />
  <row Id="4470" PostTypeId="1" AcceptedAnswerId="4473" CreationDate="2010-09-12T12:08:16.930" Score="1" ViewCount="107" Body="&lt;p&gt;In &lt;a href=&quot;http://en.wikipedia.org/wiki/Parallelepiped&quot; rel=&quot;nofollow&quot;&gt;Wikipedia it is stated that the volume of the parallelepiped&lt;/a&gt; given its edge  lengths $a,b,c$, and the internal angles between the edges $\alpha ,\beta ,\gamma $ is: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$V=abc\sqrt{1+2\cos \alpha \cos \beta \cos \gamma -\cos ^{2}\alpha&#xA;- \cos^{2}\beta - \cos ^{2}\gamma }\qquad$(*).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was not able to derive it by using the determinant formula and expressing $\cos \alpha ,\cos \beta ,\cos \gamma $ in terms of $a,b,c$ and $\alpha&#xA;,\beta ,\gamma $. For instance&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a_{1}b_{1}+a_{2}b_{2}+a_{3}b_{3}=ab\cos \alpha $.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; Could you give a hint on how can the formula (*) be proved?&lt;/p&gt;&#xA;" OwnerUserId="752" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2010-09-12T13:35:24.507" LastActivityDate="2010-09-12T13:35:24.507" Title="Deriving volume of parallelepiped as a function of edge lengths and angles between the edges" Tags="&lt;geometry&gt;&lt;linear-algebra&gt;&lt;trigonometry&gt;&lt;euclidean-geometry&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="4471" PostTypeId="2" ParentId="4467" CreationDate="2010-09-12T12:10:41.543" Score="2" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Definition&lt;/strong&gt;: An &lt;em&gt;algebraic integer&lt;/em&gt; is a solution of a monic polynomial with integer coefficients. This set is closed by sums and products and such.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Theorem&lt;/strong&gt;: If an algebraic integer is rational, then it is an integer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof&lt;/strong&gt;: Apply rational roots theorem.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;This theorem proves that $\sqrt[b]{a}$ is irrational unless $a$ is a $b$-th power.&lt;/p&gt;&#xA;" LastActivityDate="2010-09-12T12:10:41.543" />
  <row Id="4472" PostTypeId="2" ParentId="4470" CreationDate="2010-09-12T12:55:29.160" Score="1" ViewCount="" Body="&lt;p&gt;Please see this link, under trigonometric approach: &lt;a href=&quot;http://www.ehow.com/how_6545076_prove-formula-volume-cuboids.html&quot; rel=&quot;nofollow&quot;&gt;http://www.ehow.com/how_6545076_prove-formula-volume-cuboids.html&lt;/a&gt;&lt;/p&gt;&#xA;" LastActivityDate="2010-09-12T12:55:29.160" />
  <row Id="4473" PostTypeId="2" ParentId="4470" CreationDate="2010-09-12T13:01:33.887" Score="4" ViewCount="" Body="&lt;p&gt;The volume satisfies $V=|\det D|$ where $D$ is the matrix&#xA;$$\pmatrix{a_1&amp;amp;a_2&amp;amp;a_3\\&#xA;b_1&amp;amp;b_2&amp;amp;b_3\\&#xA;c_1&amp;amp;b_2&amp;amp;c_3}$$&#xA;where $\mathbf{a}=(a_1,a_2,a_3)$ is the vector corresponding to $a$ etc.&#xA;Thus&#xA;$$V^2=\det(DD^t)=\det&#xA;\pmatrix{\mathbf{a}\cdot\mathbf{a}&#xA;&amp;amp;\mathbf{a}\cdot\mathbf{b}&#xA;&amp;amp;\mathbf{a}\cdot\mathbf{c}\\&#xA;\mathbf{b}\cdot\mathbf{a}&#xA;&amp;amp;\mathbf{b}\cdot\mathbf{b}&#xA;&amp;amp;\mathbf{b}\cdot\mathbf{c}\\&#xA;\mathbf{c}\cdot\mathbf{a}&#xA;&amp;amp;\mathbf{c}\cdot\mathbf{b}&#xA;&amp;amp;\mathbf{c}\cdot\mathbf{c}}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now you can express $V^2$ in terms of $a$, $b$, $c$ and the various&#xA;cosines. &lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-09-12T13:01:33.887" />
  <row Id="4474" PostTypeId="2" ParentId="4430" CreationDate="2010-09-12T13:34:24.417" Score="0" ViewCount="" Body="&lt;p&gt;Perhaps what you mean is how differential equations can be applied to find othogonal trajectories?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.sosmath.com/diffeq/first/orthogonal/orthogonal.html&quot; rel=&quot;nofollow&quot;&gt;http://www.sosmath.com/diffeq/first/orthogonal/orthogonal.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1369" LastActivityDate="2010-09-12T13:34:24.417" />
  <row Id="4475" PostTypeId="2" ParentId="4458" CreationDate="2010-09-12T13:47:51.423" Score="1" ViewCount="" Body="&lt;p&gt;You might want to look at the relatively recent book:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Portrait of the Earth: A Mathematician Looks at Maps by Timothy Freeman&lt;/p&gt;&#xA;&#xA;&lt;p&gt;American Mathematical Society, Providence, 2002&lt;/p&gt;&#xA;" OwnerUserId="1369" LastActivityDate="2010-09-12T13:47:51.423" />
  <row Id="4476" PostTypeId="1" AcceptedAnswerId="4507" CreationDate="2010-09-12T13:48:47.790" Score="1" ViewCount="197" Body="&lt;p&gt;The Fourier transform can be defined on $L^1(\mathbb{R}^n) \cap L^2(\mathbb{R}^n)$, and we can extend this to $X:=L^2(\mathbb{R}^n)$ by a density argument.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, by Plancherel we know that $\|\widehat{f}\|_2 = \|f\|_2$, so the Fourier transform is an isometry on this space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question now is, what is a theorem that guarantees that the Fourier transform has a fixed point on $L^2$? I know the Gaussian is a fixed point, but I'm also interested in other integral transforms, but I just take the Fourier transform as an example.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Banach Fixed Point Theorem does not work here since we don't have a contraction (operator norm $&amp;lt; 1$). Can we apply the Tychonoff fixed point theorem? Then we would need to show that there exists a non-empty compact convex set $C \subset X$ such that the Fourier transform restricted to $C$ is a mapping from $C$ to $C$. Is this possible?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we have a fixed point, what would be a way to show it is unique? By linearity we obviously have infinitely many fixed points of we have at least two of them.&lt;/p&gt;&#xA;" OwnerUserId="1120" LastActivityDate="2010-10-26T16:14:35.587" Title="Fixed point Fourier transform (and similar transforms)." Tags="&lt;fourier-analysis&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" />
  <row Id="4477" PostTypeId="2" ParentId="2418" CreationDate="2010-09-12T14:13:50.823" Score="1" ViewCount="" Body="&lt;p&gt;This is not intended as an answer, but clearly is too long for comments.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are some approaches I have been playing around with; I would appreciate any comments, advice, or suggestions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clearly the condition s&lt;sub&gt;i&lt;/sub&gt; = s&lt;sub&gt;j&lt;/sub&gt; + s&lt;sub&gt;k&lt;/sub&gt; is the key thing about this problem.  It seems like in order to state this as a problem (instead of as a conjecture), it must have been solved by someone, otherwise where does |T| $\leq$ 7 come from?  Since we don't know for sure if the problem has been solved, I am assuming that |T| $\leq$ 7 is possibly wrong and just trying to work out the consequences of the condition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This condition is very constrictive, and it's consequences are not at all readily apparent.  One question I have is, &lt;strong&gt;how many independent elements can a set which satisfies the condition have?&lt;/strong&gt;  The idea of an independent element is not rigorous for a set like this, but the idea is to figure out how to formulate the idea of independent element so it is rigorous.  Given a set of integers H, with |H| less than 15, is it possible to add numbers (by combining elements of H) so that H has 15 elements and satisfies the condition; if so, can we say something about the original size of H?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this direction I have made the following (slightly trivial) observations:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A set which satisfies the condition contains:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;at least two (distinct) positive numbers&lt;/li&gt;&#xA;&lt;li&gt;at least two (distinct) negative numbers&lt;/li&gt;&#xA;&lt;li&gt;at least two (distinct) numbers which are the sum of a positive and a negative number&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This follows from the fact that each such set will contain a largest positive number which must be the sum of only positive numbers (similarly for negative); and the smallest positive number must be the sum of a positive and negative (similar for smallest negative number).  I'm not sure how this relates to the (poorly defined) notion of &quot;independent element&quot;, but it feels like a start in that direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another idea is that the condition s&lt;sub&gt;i&lt;/sub&gt; = s&lt;sub&gt;j&lt;/sub&gt; + s&lt;sub&gt;k&lt;/sub&gt; can be &quot;unpacked&quot; by applying it recursively to s&lt;sub&gt;j&lt;/sub&gt; and s&lt;sub&gt;k&lt;/sub&gt;.  My hope is that I can show that there will always be one element which can be &quot;unpacked&quot; so that it is equal to 8 elements in S, one of which is itself.  This would of course show that the sum of the other 7 is zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;s&lt;sub&gt;i&lt;/sub&gt; = s&lt;sub&gt;j&lt;/sub&gt; + s&lt;sub&gt;k&lt;/sub&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;= s&lt;sub&gt;l&lt;/sub&gt; + s&lt;sub&gt;m&lt;/sub&gt; + s&lt;sub&gt;n&lt;/sub&gt; + s&lt;sub&gt;p&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;= s&lt;sub&gt;q&lt;/sub&gt; + s&lt;sub&gt;r&lt;/sub&gt; + s&lt;sub&gt;s&lt;/sub&gt; + s&lt;sub&gt;t&lt;/sub&gt; + s&lt;sub&gt;u&lt;/sub&gt; + s&lt;sub&gt;v&lt;/sub&gt; + s&lt;sub&gt;w&lt;/sub&gt; + s&lt;sub&gt;i&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The next idea is to show that any set satisfying the condition will contain what I call a &quot;7 cycle&quot;, which is inspired from the totally anti-symmetric unit tensor.  A &quot;7 cycle&quot; is a kind of symmetry for the index's of s&lt;sub&gt;i&lt;/sub&gt; = s&lt;sub&gt;j&lt;/sub&gt; + s&lt;sub&gt;k&lt;/sub&gt;, I will just write down what a 7 cycle looks like, because that is more clear than explaining it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;s&lt;sub&gt;1&lt;/sub&gt; = s&lt;sub&gt;2&lt;/sub&gt; + s&lt;sub&gt;3&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;s&lt;sub&gt;2&lt;/sub&gt; = s&lt;sub&gt;3&lt;/sub&gt; + s&lt;sub&gt;4&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;s&lt;sub&gt;3&lt;/sub&gt; = s&lt;sub&gt;4&lt;/sub&gt; + s&lt;sub&gt;5&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;s&lt;sub&gt;4&lt;/sub&gt; = s&lt;sub&gt;5&lt;/sub&gt; + s&lt;sub&gt;6&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;s&lt;sub&gt;5&lt;/sub&gt; = s&lt;sub&gt;6&lt;/sub&gt; + s&lt;sub&gt;7&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;s&lt;sub&gt;6&lt;/sub&gt; = s&lt;sub&gt;7&lt;/sub&gt; + s&lt;sub&gt;1&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;s&lt;sub&gt;7&lt;/sub&gt; = s&lt;sub&gt;1&lt;/sub&gt; + s&lt;sub&gt;2&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it's possible to show that any set satisfying the condition contains a 7 cycle, then it's clear that the sum  of s&lt;sub&gt;1&lt;/sub&gt;...s&lt;sub&gt;7&lt;/sub&gt; will be zero (since it is equal to twice itself).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another idea is a combinatorial graph theory approach.  Clearly, a set satisfying the condition generates a directed graph (as pointed out in the MO link).  For clarity, each number in S corresponds to a vertex, and if for example, s&lt;sub&gt;1&lt;/sub&gt; = s&lt;sub&gt;2&lt;/sub&gt; + s&lt;sub&gt;3&lt;/sub&gt;, then there will be edges directed from s&lt;sub&gt;2&lt;/sub&gt; to s&lt;sub&gt;1&lt;/sub&gt; and from s&lt;sub&gt;3&lt;/sub&gt; to s&lt;sub&gt;1&lt;/sub&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A question I have which I can't seem to make progress on, is: &lt;strong&gt;what is the minimum possible number of vertex's which have outgoing edges?&lt;/strong&gt; (a vertex has an outgoing edge if it can be added to an element of S to obtain another element of S, also the assumptions used in the previously given &quot;weaker statement&quot; led to the conclusion that each vertex has an outgoing edge, and this was used to show |T| $\leq$ 7).  My hope is that once I know the minimum number of vertex's with outgoing edges, I can use some sort of combinatorial reasoning to show that there will be a closed cycle with a maximum length.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Googling &quot;zero sum subset&quot; quickly leads to references for the &lt;a href=&quot;http://en.wikipedia.org/wiki/Zero-sum_problem&quot; rel=&quot;nofollow&quot;&gt;EGZ theorem&lt;/a&gt;, and by thinking about the problem backwards, it makes sense that since this is one of the best known results for zero sum problems; that if this problem has already been solved this result might be involved in the solution somehow.  I can't quite wrap my head around exactly how it will be involved, but since it is a result dealing with the integers mod N, the idea would be to somehow show that the condition forces some (possibly elaborate) modular arithmetic structure on the set S which satisfies it.  (my understanding is that EGZ says that given an arbitrary set of 2n - 1 integers in $\mathbb{Z}$ mod n, there will be a subset of n elements whose sum is congruent to 0 mod n)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yet another idea is to reformulate the problem in terms of polynomials.  Given a set S which satisfies the condition, consider the 15th degree polynomial whose roots are the elements of S.  Then these roots possess the symmetry that every root is the sum of two other roots; these types of polynomials might have some property which forces the sum of at most 7 elements to be zero. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't feel strongly that any of these approaches will ultimately prove successful, but hopefully by making this post I can help keep this question alive and inspire others to work on it.  &lt;/p&gt;&#xA;" OwnerUserId="1417" LastEditorUserId="1417" LastEditorDisplayName="" LastEditDate="2010-09-12T20:33:13.723" LastActivityDate="2010-09-12T20:33:13.723" CommentCount="2" />
  <row Id="4478" PostTypeId="2" ParentId="4467" CreationDate="2010-09-12T14:47:42.780" Score="5" ViewCount="" Body="&lt;p&gt;These (standard) results are discussed in detail in &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://math.uga.edu/~pete/4400irrationals.pdf&quot; rel=&quot;nofollow&quot;&gt;http://math.uga.edu/~pete/4400irrationals.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the second handout for a first course in number theory at the advanced undergraduate level.  Three different proofs are discussed:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) A generalization of the proof of irrationality of $\sqrt{2}$, using the decomposition of any positive integer into a perfect $k$th power times a $k$th power-free integer, followed by Euclid's Lemma.  (For some reason, I don't give all the details of this proof.  Maybe I should...)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) A proof using the functions $\operatorname{ord}_p$, very much along the lines of the one Carl Mummert mentions in his answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) A proof by establishing that the ring of integers is integrally closed.  This is done directly from unique factorization, but afterwards I mention that it is a special case of the Rational Roots Theorem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me also remark that every proof I have ever seen of this fact uses the Fundamental Theorem of Arithmetic (existence and uniqueness of prime factorizations) in some form.  [&lt;b&gt;Edit&lt;/b&gt;: I have now seen Robin Chapman's answer to the question, so this is no longer quite true.]  However, if you want to prove any particular case of the result, you can use a brute force case-by-case analysis that avoids FTA.  &lt;/p&gt;&#xA;" OwnerUserId="299" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2010-09-13T07:30:29.880" LastActivityDate="2010-09-13T07:30:29.880" CommentCount="4" />
  <row Id="4479" PostTypeId="1" AcceptedAnswerId="4487" CreationDate="2010-09-12T15:11:52.683" Score="4" ViewCount="233" Body="&lt;p&gt;Hello,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $\mathcal{M}$ be an n-dimensional manifold endowed with an affine connection $\nabla$. Let $\gamma_1:[a,b]\rightarrow M$ and $\gamma_2:[c,d]\rightarrow \mathcal{M}$ be two curves with the same initial and final points, that is,&#xA;$p=\gamma_1(a)=\gamma_2(c), q=\gamma_1(b)=\gamma_2(d)$. Take  $X\in T_p\mathcal{M}$. Parallelly propagating $X$ along $\gamma_1$ and $\gamma_2$ we obtain two vectors $X_1, X_2\in T_q\mathcal{M}$, respectively. Let $R$ be the curvature tensor of the connection, $R(X,Y)Z=\nabla_X\nabla_Y Z - \nabla_Y\nabla_X Z -\nabla_{[X,Y]}Z$, and $\tau$ its torsion, $\tau(X,Y)=\nabla_X Y-\nabla_Y X -[X,Y]$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is: How can I compare the two vectors $X_1$ and $X_2$? Can I write the difference $(X_2-X_1)$ in terms of $R,\tau$ and the curves?&lt;/p&gt;&#xA;" OwnerUserId="1178" LastActivityDate="2010-09-12T17:52:45.653" Title="Parallel transport of a vector along two distinct curves" Tags="&lt;differential-geometry&gt;&lt;differential-topology&gt;" AnswerCount="2" FavoriteCount="2" />
  <row Id="4480" PostTypeId="1" AcceptedAnswerId="4499" CreationDate="2010-09-12T15:21:08.620" Score="8" ViewCount="391" Body="&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/vzhVr.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Find the radius of the circles if the size of the larger square is 1x1.&lt;br&gt;&#xA;Enjoy!  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(read about the &lt;a href=&quot;http://mathworld.wolfram.com/SangakuProblem.html&quot; rel=&quot;nofollow&quot;&gt;origin of sangaku&lt;/a&gt;)&lt;/p&gt;&#xA;" OwnerUserId="923" LastEditorUserId="923" LastEditorDisplayName="" LastEditDate="2010-09-13T14:44:59.953" LastActivityDate="2010-09-13T14:44:59.953" Title="sangaku - a geometrical puzzle" Tags="&lt;geometry&gt;&lt;puzzle&gt;" AnswerCount="3" FavoriteCount="2" />
  <row Id="4481" PostTypeId="2" ParentId="4422" CreationDate="2010-09-12T16:58:17.337" Score="0" ViewCount="" Body="&lt;p&gt;Oops I misread the question, but I actually like Shakarchi and Stein volume 1!  In my opinion the second volume is a lot better than the first one however.&lt;/p&gt;&#xA;" OwnerUserId="1417" LastActivityDate="2010-09-12T16:58:17.337" CommunityOwnedDate="2010-09-12T16:58:17.337" />
  <row Id="4482" PostTypeId="2" ParentId="4479" CreationDate="2010-09-12T17:05:46.370" Score="1" ViewCount="" Body="&lt;p&gt;The difference depends on the homotopy classes of the paths, not only local data like the curvature and torsion.  I don't think there is a simpler answer than &quot;the difference in the parallel transport along the two paths&quot;.  But see the (superb) Wikipedia page on holonomy, especially the sections on the Ambrose-Singer theorem and affine connections.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Holonomy&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Holonomy&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="467" LastActivityDate="2010-09-12T17:05:46.370" />
  <row Id="4483" PostTypeId="1" AcceptedAnswerId="4486" CreationDate="2010-09-12T17:06:05.153" Score="4" ViewCount="515" Body="&lt;p&gt;Any linear map between two finite-dimensional vector spaces can be represented as a matrix under the bases of the two spaces. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if one or all of the vector spaces is infinite dimensional, is the linear map still represented as a matrix under their bases?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there is matrix of infinite dimension, what is it used for if not used as a representation of a linear map between vector spaces?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks and regards!&lt;/p&gt;&#xA;" OwnerUserId="1281" LastActivityDate="2011-01-20T05:42:12.077" Title="Matrix of Infinite Dimension" Tags="&lt;abstract-algebra&gt;&lt;matrices&gt;&lt;vector-space&gt;" AnswerCount="2" FavoriteCount="3" />
  <row Id="4485" PostTypeId="2" ParentId="4480" CreationDate="2010-09-12T17:42:13.457" Score="4" ViewCount="" Body="&lt;p&gt;Let $r$ be the length the radius of the circles, and let $\theta$ be the measure of the (smaller) angle made at the corner of the big square.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/9Gurs.png&quot; alt=&quot;Labeled image&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The width of the square is equal to two radii and the projection of a double diameter (a quadruple-radius), so that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(1)\hspace{1.0in}4r\cos\theta=1-2r$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Looking at the four right triangles, we see that the center circle's diameter is equal to the difference in the lengths of the legs; since the hypotenuse has length $1$, we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(2)\hspace{1.0in}2r = \cos\theta - \sin\theta$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From here, we simply need to eliminate $\theta$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Multiplying (2) through by $4r$ and substituting in from (1) ...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$8 r^2 = 4r\cos\theta - 4r \sin\theta = 1 - 2r - 4r \sin\theta$$&#xA;$$4r \sin\theta = 1 - 2r - 8 r^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{eqnarray}16r^2 &amp;amp;=&amp;amp; (4r \cos\theta)^2 + (4 r \sin\theta)^2 \\ &#xA;&amp;amp;=&amp;amp; ( 1 - 2r )^2 + ( 1 - 2r - 8 r^2 )^2 \\&#xA;&amp;amp;=&amp;amp; 2 - 8 r - 8 r^2 + 32r^3 + 64 r^4 \end{eqnarray}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$0 = 32 r^4 + 16 r^3 - 12 r^2 - 4 r + 1 = (2r+1)(2r-1)(8 r^2 + 4 r - 1)$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The roots of the polynomial are $\pm1/2$ and $(-1\pm\sqrt{3})/4$. We can eliminate three of them from consideration to conclude that $r = (-1+\sqrt{3})/4$.&lt;/p&gt;&#xA;" OwnerUserId="409" LastActivityDate="2010-09-12T17:42:13.457" CommentCount="3" />
  <row Id="4486" PostTypeId="2" ParentId="4483" CreationDate="2010-09-12T17:46:45.500" Score="8" ViewCount="" Body="&lt;p&gt;Sure, if $T:V\to W$ is a linear transformation between vector spaces $V$ and $W$ with bases $B$ and $C$, respectively, then $T$ can be described in terms of the coordinates with respect to these bases, thus yielding a &quot;matrix&quot;.  How closely this relates to the usual notion of matrix depends on the nature of $B$ and $C$.  In the usual notion, you take bases that are not only finite, but ordered, so that it makes sense to talk about the 1st row, etc., of the matrix; that is, you make all bases indexed by sets of the form $\{1,2,\ldots,n\}$.  The closest to this in the infinite dimensional setting would be to have bases indexed by the positive integers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More generally, suppose $B=\{v_j\}_{j\in J}$ and $C=\{w_i\}_{i\in I}$, where $I$ and $J$ are sets.  Then the matrix of $T$ can be described as a function $M:I\times J\to F$, where $F$ is the base field, by taking $M(i,j)$ to be the coefficient of $w_i$ in the $C$-expansion of $Tv_j$.  Such matrices are &lt;em&gt;column-finite&lt;/em&gt;, in the sense that for each $j\in J$, the set of $i\in I$ such that $M(i,j)\neq0$ is finite.  Conversely, each column-finite matrix, in this sense, corresponds uniquely to a linear transformation between $V$ and $W$.  Coordinate-wise addition of such matrices corresponds to addition of the linear transformations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can also extend multiplication.  Suppose that $S:W\to X$ is a linear transformation and that $X$ has basis $D=\{x\_k\}\_{k\in K}$.  Let $N:K\times I\to F$ denote the $C$-$D$ matrix of $S$.  Then $ST:V\to X$ has $B$-$D$ matrix $NM:K\times J\to F$ defined by &#xA;$$(NM)(k,j)=\sum_{i\in I}N(k,i)M(i,j).$$  In particular, note that this sum is always finite because $M$ is column-finite.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Motivated by Calle's answer, I decided to add a little on a different kind of matrix for continuous linear transformations on Banach spaces with Schauder bases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $X$ is an infinite dimensional separable Banach space, then a sequence $(e_n)_{n=1}^\infty$ in $X$ is called a Schauder basis for $X$ if every $x\in X$ has a unique representation $x=\sum_{n=1}^\infty a_ne_n$, the $a_n$ being scalar and the sum being norm convergent.  If $X$ and $Y$ are Banach spaces with Schauder bases $(e_n)$ and $(f_n)$ respectively, and if $T:X\to Y$ is a bounded linear operator, then $T$ can be described by a matrix $(a_{ij})_{i,j=1}^\infty$, with $a_{ij}$ being the coefficient of $f_i$ in the $(f_n)$ expansion of $Te_j$.  The map from bounded operators to matrices in one-to-one and preserves algebraic structure, but there is typically not any nice description of which matrices correspond to bounded operators.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, in a separable Hilbert space any orthonormal basis is a Schauder basis.  For maps between Hilbert spaces the coefficients are found as $a_{ij}=\langle Te_j,f_i\rangle$.  In $c_0$, the space of sequences converging to $0$ with sup norm, and in $\ell^p$, the sequence space with norm $\|(x_n)_{n=1}^\infty\|_p=(\sum_{n=1}^\infty|x_n|^p)^{1/p}$, the sequence $(e_n)$ such that the $n^\text{th}$ component of $e_n$ is $1$ and all other components are $0$ forms a Schauder basis.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $c$ is the space of convergent sequences with sup norm, then this will no longer be a Schauder basis, and in particular it is clear that $\sum_{n=1}^\infty x_n e_n$ is not norm convergent unless $\lim_{n\to\infty}x_n=0$.  A Schauder basis for $c$ can be obtained by adding $e_0=(1,1,1,\ldots)$.  If $(x_n)\in c$ and $x=\lim_n x_n$, then $(x_n)=xe_0 +\sum_{n=1}^\infty(x_n-x)e_n$ is the basis representation.  As in Calle's answer, suppose that $T:c\to c$ is defined by $T(x_1,x_2,x_3,\ldots)=(x,0,0,\ldots)$.  Then $T$ has a matrix representation with repsect to $(e_0,e_1,\ldots)$ (but not with respect to $(e_1,e_2,\ldots)$), namely $a_{10}=1$ and all other components are $0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similar to Olod's warning, such matrices typically play only a marginal role, even in cases where they are guaranteed to exist, like on Hilbert space.  Not every separable Banach space has a Schauder basis.  &lt;a href=&quot;http://en.wikipedia.org/wiki/Per_Enflo&quot; rel=&quot;nofollow&quot;&gt;Enflo&lt;/a&gt; first &lt;a href=&quot;http://www.springerlink.com/content/g124318rp4588310/&quot; rel=&quot;nofollow&quot;&gt;gave an example&lt;/a&gt; of a separable Banach space without the &lt;a href=&quot;http://en.wikipedia.org/wiki/Approximation_property&quot; rel=&quot;nofollow&quot;&gt;approximation property&lt;/a&gt;, which guarantees that it has no Schauder basis.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2011-01-20T05:42:12.077" LastActivityDate="2011-01-20T05:42:12.077" CommentCount="8" />
  <row Id="4487" PostTypeId="2" ParentId="4479" CreationDate="2010-09-12T17:52:45.653" Score="3" ViewCount="" Body="&lt;p&gt;If your two paths are homotopic you can make a comparison between your two vectors. And yes the comparison involves an integral over the homotopy of a function of curvature. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;See for example Theorem 13.6.4 in Pressley's &quot;Elementary Differential Geometry&quot; (Google books will bring up the statement of the theorem). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If your paths are not homotopic you're out of luck, as T describes. There are things you can say of course but it's not clear what you're looking for.  You should think of an example of a Riemann manifold you're interested in, to get a sense for how bad it can get. &lt;/p&gt;&#xA;" OwnerUserId="642" LastActivityDate="2010-09-12T17:52:45.653" CommentCount="3" />
  <row Id="4488" PostTypeId="2" ParentId="4467" CreationDate="2010-09-12T18:03:03.590" Score="2" ViewCount="" Body="&lt;p&gt;As muad points out, you can also obtain this as an easy consequence of the Rational Root Theorem: if $a_nx^n+\cdots+a_0$ is a polynomial with integer coefficients, and $\frac{p}{q}$ is a rational root with $\gcd(p,q)=1$, then $p|a_0$ and $q|a_n$ (plug in, clear denominators, factor out). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if you look at the polynomial $x^b-a$, with $b$ and $a$ positive integers, then a rational root must be of the form $\frac{p}{q}$, with $\gcd(p,q)=1$, and $q|1$. Thus, it must be an integer. So if it has a rational root, then the root is integral. &lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-09-12T18:03:03.590" />
  <row Id="4490" PostTypeId="2" ParentId="4483" CreationDate="2010-09-12T20:10:39.423" Score="5" ViewCount="" Body="&lt;p&gt;What, however, must be understood that the role of matrices when one works with linear operators of infinite-dimensional vector spaces is a (very) marginal one. Familiar techniques such as the use of determinants, traces etc. no longer work. For instance, any determinant-like function on $\mathrm{GL}(V)$ where $V$ is an infinite-dimensional vector space over a field is necessarily the trivial one, since any element of $\mathrm{GL}(V)$ is a product of commutators (the group $\mathrm{GL}(V)$ is perfect, in other words; a result by Alex Rosenberg of 1958.)&lt;/p&gt;&#xA;" OwnerUserId="1100" LastEditorUserId="1100" LastEditorDisplayName="" LastEditDate="2010-09-12T20:20:40.280" LastActivityDate="2010-09-12T20:20:40.280" />
  <row Id="4491" PostTypeId="2" ParentId="4467" CreationDate="2010-09-12T20:25:03.167" Score="6" ViewCount="" Body="&lt;p&gt;Here is a simple &lt;em&gt;conceptual&lt;/em&gt; $\:$ proof of the irrationality of sqrts - completely from first principles.&lt;br&gt;&#xA;The key idea is that the least positive denominator of a fraction $\rm\:r\:$ divides &lt;em&gt;every&lt;/em&gt;  denominator of $\rm\:r\:$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;THEOREM&lt;/strong&gt; $\ \ \rm r = \sqrt{a}\ $ is integral if rational,$\:$ for $\:\rm a\in\mathbb{N}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof&lt;/strong&gt; $\quad$  Put $\ \ \displaystyle\rm r = \frac{c}d ,\;$ least $\rm\; d&amp;gt;0\:.\;$ $\  \displaystyle\rm\sqrt{a}\; = \frac{a}{\sqrt{a}} \ \Rightarrow\ \frac{c}d = \frac{ad}c \ \Rightarrow\ d\:$ divides $\rm\: c\ $ via this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;LEMMA&lt;/strong&gt; $\;$ The least positive denominator of a fraction $\:\rm r\:$ divides every denominator of $\rm\:r\;$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof&lt;/strong&gt; $\rm\ \ \ n,m\ $ denominators $\: \Rightarrow\: $ so is $\rm\ n-m\ $ via $\;\rm n\:r,\: m\:r\in \mathbb Z \ \Rightarrow\: (n-m)\:r\in \mathbb Z\:,\:$ now apply&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Lemma&lt;/strong&gt; $\ \ $ If a set of positive integers $\rm\: M\:$ satisfies $\rm n &amp;gt; m \in M \ \Rightarrow\ \: n-m \in M$ &lt;br/&gt; then every element of $\rm\:M\:$ is a multiple of the least element $\rm\:m \in M$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof&lt;/strong&gt; $\:$ If not there'd be a least nonmultiple $\rm\:n\in M\:,$ contra  $\rm\:n-m \in M\:$ is a nonmultiple of $\rm\:m\:$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;REMARK&lt;/strong&gt; $\:$ The above LEMMA is quite fundamental to factorization. I frequently  refer to it by the suggestive moniker &lt;strong&gt;unique fractionization&lt;/strong&gt;, since it's equivalent to  uniqueness of factorizations into irreducibles (indeed one easily sees that it is equivalent to Euclid's Lemma, which implies that irreducibles are prime).&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-12-02T19:31:11.163" LastActivityDate="2010-12-02T19:31:11.163" CommentCount="3" />
  <row Id="4492" PostTypeId="2" ParentId="4480" CreationDate="2010-09-12T20:39:04.157" Score="3" ViewCount="" Body="&lt;p&gt;Somewhat related to Don's solution: From the figure, we see that the four triangles are 1: congruent, and 2: right triangles. The hypotenuse of one triangle has length 1, and if we let $\theta$ be the smaller of the two angles of the right triangle, and use $r$ to denote the radius of one circle, then the Pythagorean relation is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\cos^2\;\theta+(\cos\;\theta-2r)^2=1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This can now be solved as a simultaneous equation with any of the other two equations Don obtained, or we can use another equation, the expression for the &lt;a href=&quot;http://mathworld.wolfram.com/Incircle.html&quot; rel=&quot;nofollow&quot;&gt;inradius&lt;/a&gt; $r$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$r^2=\frac{(s-1)(s-\cos\;\theta)(s+2r-\cos\;\theta)}{s}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $s=\frac{1+\cos\;\theta+(\cos\;\theta-2r)}{2}$ is the semiperimeter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we eliminate $\cos\;\theta$ and solve the two equations here for $r$, we find that the roots of the resulting quartic equation are&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$r=\frac{\pm 1\pm\sqrt{3}}{4}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we carry out Don's approach as well, we find that only one &lt;strong&gt;positive&lt;/strong&gt; value of $r$ is consistent with both systems, and thus has to be the correct answer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$r=\frac{-1+\sqrt{3}}{4}$$&lt;/p&gt;&#xA;" OwnerUserId="498" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-09-12T20:46:24.930" LastActivityDate="2010-09-12T20:46:24.930" CommentCount="1" />
  <row Id="4493" PostTypeId="2" ParentId="4466" CreationDate="2010-09-12T20:59:35.247" Score="0" ViewCount="" Body="&lt;p&gt;As Yuval points out, (F(n-1=p, n+1=q) := n)^2-1 = pq is the obvious interpretation.  If you want to interpret the problem as specifying as a domain for the bijection a single prime p for which either p+2 or p-2 is also prime, or two primes which are part of the same twin pair but given in either order, or if n is allowed to be negative while p and q are conventionally positive, then (in any of those cases) a bijection exists if and only if twin primes are infinite in number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you extend &lt;em&gt;both&lt;/em&gt; the domain and range in this way then it still works: F(-n+1=p, -n-1=q) := n.  Also if you extend the range and extend the domain by allowing negative primes then it still works without adding to the definition of F.&lt;/p&gt;&#xA;" OwnerUserId="1284" LastEditorUserId="1284" LastEditorDisplayName="" LastEditDate="2010-09-12T21:48:36.947" LastActivityDate="2010-09-12T21:48:36.947" />
  <row Id="4495" PostTypeId="1" AcceptedAnswerId="4567" CreationDate="2010-09-12T21:00:51.220" Score="6" ViewCount="443" Body="&lt;p&gt;An ordinary eigenvector can be viewed as a vector on which the operator acts by only stretching (without rotating) it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a similar intuition behind generalized eigenvectors?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: By generalized eigenvectors I'm referring to vectors in the kernel of $(T-\lambda I)^k$ for some $k$.&lt;/p&gt;&#xA;" OwnerUserId="766" LastEditorUserId="766" LastEditorDisplayName="" LastEditDate="2010-09-13T06:23:00.797" LastActivityDate="2010-09-13T18:47:28.640" Title="The intuition behind generalized eigenvectors" Tags="&lt;linear-algebra&gt;&lt;intuition&gt;" AnswerCount="2" CommentCount="6" FavoriteCount="2" />
  <row Id="4496" PostTypeId="2" ParentId="4495" CreationDate="2010-09-12T21:09:56.267" Score="7" ViewCount="" Body="&lt;p&gt;Consider first eigenvectors associated to $0$; these are elements of the nullspace. Now, it is possible for a vector to not lie in the nullspace, but for its image to lie in the nullspace (for example, consider the linear transformation $T\colon \mathbb{R}^2\to\mathbb{R}^2$ given by $T(x,y) = (y,0)$; then $T(0,1) = (1,0)$, and $T(T(0,1))=(0,0)$. Or for the image of the image to lie in the nullspace... etc. If you want to think about all the vectors that will &lt;em&gt;eventually&lt;/em&gt; map to zero if you keep applying $T$ (which seems like a sensible thing to think about in many circumstances), then you are looking for the vectors that lie in union of the nullspaces of $T^n$, $n=1,2,\ldots$. Notice that $\mathbf{N}(T^n)\subseteq\mathbf{N}(T^{n+1})$, so this is an increasing union. Once a vector maps to $0$ under a power of $T$, it stays there. These vectors are &lt;em&gt;precisely&lt;/em&gt; the generalized eigenvectors of $T$ associated to $\lambda=0$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, what about arbitrary eigenvalues? Instead of thinking of an eigenvector associated to $\lambda$ as a vector on which $T$ acts by stretching/compressing, think of an eigenvector associated to $\lambda$ as an element of the nullspace of $T-\lambda I$. The generalized eigenvector associated to $\lambda$ are then the vectors that lie in $\mathbf{N}((T-\lambda I)^n)$ for some positive integer $n$. &lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-09-12T21:09:56.267" CommentCount="3" />
  <row Id="4497" PostTypeId="1" CreationDate="2010-09-12T23:01:39.370" Score="2" ViewCount="163" Body="&lt;p&gt;What is a conditional double limit and how to compute it for $\lim_{(x,y)\rightarrow\infty,x\leq y}\frac{2x-1}{x-1}+\frac{x}{y}$?&lt;/p&gt;&#xA;" OwnerUserId="1856" LastEditorUserId="1856" LastEditorDisplayName="" LastEditDate="2010-09-12T23:17:13.473" LastActivityDate="2010-09-13T22:22:43.270" CommunityOwnedDate="2010-09-13T22:26:13.870" Title="What is a conditional double limit and how to compute it?" Tags="&lt;classical-analysis&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="1" />
  <row Id="4498" PostTypeId="1" CreationDate="2010-09-12T23:11:20.007" Score="3" ViewCount="249" Body="&lt;p&gt;I take it there isn't a classification of finite complete groups yet. Has someone put together a classification of small complete groups? I.e. $S_4$, $\text{Aut}(G)$ for $G$ simple, $\text{Hol}(C_p)$ for $p$ odd prime are complete. The smallest complete group not of one of these forms is $H$ of order $\left|H\right|$, which generalises to a class $P$. The smallest complete group not of one of these forms is $J$, and so on, until we run out of ideas.&lt;/p&gt;&#xA;" LastActivityDate="2010-10-01T12:43:32.170" Title="classification of small complete groups" Tags="&lt;group-theory&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="2" />
  <row Id="4499" PostTypeId="2" ParentId="4480" CreationDate="2010-09-13T01:14:31.620" Score="10" ViewCount="" Body="&lt;p&gt;Here's a nicer approach ...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/fBUgX.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given two segments emanating from a point and ending at points of tangency with a circle, we know that those segments are congruent. In the figure, the three vertices of the right triangle give rise to three fundamental lengths; note that, since the &quot;green&quot; angle is a right angle, the lengths of the tangent segments are equal to the radius of the circle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pythagoras tells us that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(b+r)^2 + ( r + a )^2 = ( a + b )^2 $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But we also have that $a+b=1$ (the side of the square), and that $b=a+2r$ (via tangent circles on the &quot;outside&quot; of the right triangle). From these relations, we find that $a=1/2-r$ and $b=1/2+r$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(1/2+2r)^2 + ( 1/2 )^2 = ( 1 )^2$$&#xA;$$1/4+2r+4r^2+1/4=1$$&#xA;$$8r^2+4r-1=0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The roots are $(-1\pm\sqrt{3})/4$, and we select the positive value: $r = (-1+\sqrt{3})/4$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As Américo noted: The sides of the triangle have lengths $r+a=1/2$, $3r+a=\sqrt{3}/2$, and $1$, so that we have a 30-60-90 triangle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I like that there's but the one extraneous value this time, rather than the three in my first attempt. Is there yet a better approach that yields the answer directly, with &lt;em&gt;no&lt;/em&gt; extraneous values?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Edit.&lt;/b&gt; There is. Immediately after we have deduced that $a=1/2-r$, we know that the short leg of the triangle has length $1/2$, so that its longer leg is $\sqrt{3}/2$. Since that longer leg is also $a+3r=1/2+2r$, we have that $r=(-1+\sqrt{3})/4$.)&lt;/p&gt;&#xA;" OwnerUserId="409" LastEditorUserId="409" LastEditorDisplayName="" LastEditDate="2010-09-13T01:25:43.373" LastActivityDate="2010-09-13T01:25:43.373" CommentCount="3" />
  <row Id="4500" PostTypeId="2" ParentId="4497" CreationDate="2010-09-13T01:42:47.590" Score="1" ViewCount="" Body="&lt;p&gt;The limit depends on the exact way $x,y$ approach infinity. The first summand $(2x-1)/(x-1)$ actually tends to $2$ (always). The second summand can tend to any limit in $[0,1]$, or to no limit at all (in which case the limit of the entire sum does not exist). To get a limit of $0$, use $y = x^2$. To get a limit of $0 &amp;lt; a\leq 1$, use $x = ay$. To get no limit, you can use $x = y(sin(y)+2)/3$ [thanks AD!], as well as many other options.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suppose you could define $\lim \sup$ as the supremum of all limits attainable for some sequence $(x_n,y_n)$, and then $\lim \sup_{(x,y)\rightarrow\infty, x\leq y} x/y = 1$ (and similarly $\lim \inf_{(x,y)\rightarrow\infty, x\leq y} x/y = 0$).&lt;/p&gt;&#xA;" OwnerUserId="1277" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2010-09-13T07:27:34.767" LastActivityDate="2010-09-13T07:27:34.767" CommentCount="3" />
  <row Id="4501" PostTypeId="1" AcceptedAnswerId="4502" CreationDate="2010-09-13T04:09:54.390" Score="4" ViewCount="172" Body="&lt;p&gt;In a inner product space with inner product $\langle\ ,\ \rangle$ and real or complex line as its base field, for each point $x$ in the space, is $\langle x,-\rangle$ continuous function on the second argument, and is $\langle - ,x\rangle$ continous function on the first argument? &quot;Continuous&quot; is defined respect to the topology induced by the inner product.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks and regards!&lt;/p&gt;&#xA;" OwnerUserId="1281" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-09-13T04:46:29.767" LastActivityDate="2010-09-13T14:11:13.623" Title="Is Inner product continous when one arg is fixed?" Tags="&lt;inner-product-space&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="3" />
  <row Id="4502" PostTypeId="2" ParentId="4501" CreationDate="2010-09-13T04:50:42.973" Score="6" ViewCount="" Body="&lt;p&gt;Yes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fix x in the inner product space, and let $f(y) = \langle y, x \rangle$ denote the inner product function.  Note that this is a linear functional -- that is, it is linear in y, and maps vectors to scalars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is a well-known theorem that linear functionals are continuous (on the entire space) if and only if they are bounded.  Here, &quot;bounded&quot; means that there exists a constant M such that $|f(y)| \leq M|y|$ for all y in the space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That the inner product functional is bounded now follows from the Cauchy-Schwarz Inequality: $|f(y)| \leq |x||y|.$&lt;/p&gt;&#xA;" OwnerUserId="640" LastActivityDate="2010-09-13T04:50:42.973" CommentCount="1" />
  <row Id="4503" PostTypeId="2" ParentId="4501" CreationDate="2010-09-13T04:51:20.257" Score="4" ViewCount="" Body="&lt;p&gt;If we define $T_x v = \langle v,x\rangle$ then $T_x$ is a linear functional. Since an inner product induces a norm, and thus it is continuous if and only if it is bounded on the unit circle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And so $|T_x v | = |\langle v,x\rangle | \le ||v||\cdot ||x||$ and for $v$ on the unit circle, i.e. $||v|| = 1$ we have that $T_x$ is indeed bounded and therefore continuous.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The proof is similar if you fix the left argument, however it may not be a linear functional if it is a complex vector space, but rather antilinear (that is linear up to conjugation). The norm is unaffected by that, though, so it's not a big step to overcome.&lt;/p&gt;&#xA;" OwnerUserId="622" LastEditorUserId="622" LastEditorDisplayName="" LastEditDate="2010-09-13T05:12:15.050" LastActivityDate="2010-09-13T05:12:15.050" />
  <row Id="4504" PostTypeId="2" ParentId="4501" CreationDate="2010-09-13T04:51:59.490" Score="3" ViewCount="" Body="&lt;p&gt;Fix $x$. We have $|\langle x,y\rangle - \langle x,z\rangle| = |\langle x,y-z\rangle|\leq ||x||||y-z||$ by the Cauchy-Schwartz inequality. This gives you easily that $\langle x,-\rangle\colon \mathbf{V}\to\mathbb{F}$ is not only continuous, but uniformly continuous. Similarly for $\langle -,x\rangle$. &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-09-13T14:11:13.623" LastActivityDate="2010-09-13T14:11:13.623" />
  <row Id="4505" PostTypeId="2" ParentId="4458" CreationDate="2010-09-13T04:53:16.880" Score="0" ViewCount="" Body="&lt;p&gt;The uniqueness of gnomonic projections is a consequence of the uniqueness of line-preserving transformations of the plane.  If you have map projections $P_1$ and $P_2$ that take all arcs of great circles to segments of lines, then $P_1 \circ P_2^{-1}$ is a transformation of the plane (or of one part of the plane onto another) that sends line segments to line segments. [In case the $\circ$ notation is not clear, I mean the transformation that results from taking the target of $P_2$, un-projecting it back to the sphere, and projecting that pre-target from the sphere back to the plane using $P_1$.]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is known that, within the class of topology-preserving (continuous and one-to-one) transformations between connected regions in the plane, any collinearity-preserving transformation is the restriction of a single linear transformation.   If you allow points at infinity (for the gnomonic projection these are the projection of points on the equator of the sphere) then it is also true that any collinearity-preserving transformation is the restriction of a single projective transformation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The consequence for gnomonic projections is that if you consider the projection plane to be a rigid object, with an x-y coordinate system drawn on it, but that coordinatized plane can be moved around in space, then any map projection from a sphere (or parts of the sphere) to the plane that takes arcs of great circles to segments or lines, is a gnomonic projection onto &lt;em&gt;some&lt;/em&gt; position of the plane.  &lt;/p&gt;&#xA;" OwnerUserId="467" LastActivityDate="2010-09-13T04:53:16.880" CommentCount="2" />
  <row Id="4506" PostTypeId="1" AcceptedAnswerId="4509" CreationDate="2010-09-13T06:47:12.453" Score="3" ViewCount="246" Body="&lt;p&gt;We, know that there are infinitely many primes of the form $4n-1,4n+1,5n-1,\cdots, \text{etc}$. I saw these things in Apostol's &lt;em&gt;Introduction to Analytic Number theory&lt;/em&gt; textbook. I would like to have an argument working for $n^{2}$. The first expression which came to my mind was $4n^{2}+3$, which gives a prime for $n=1,2,3,4,5$. For $n=6$ it gives $147$ which is divisible by 3. For $n=7$ it gives $199$ which is again a prime. Then for $n=11$, it gives $487$ which is again a prime. I would like to know whether there are &lt;em&gt;infinitely many primes of the form $4n^{2}+3$&lt;/em&gt;? If yes, then a proof!&lt;/p&gt;&#xA;" LastActivityDate="2010-09-13T07:11:29.300" Title="Are there infinitely many primes of the form $4n^{2}+3$?" Tags="&lt;number-theory&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="4507" PostTypeId="2" ParentId="4476" CreationDate="2010-09-13T06:48:26.437" Score="0" ViewCount="" Body="&lt;p&gt;My Functional Analysis Fu has gotten bit weak lately, but I think the following should work:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Schauder fixed point theorem says, that a continuous function on a compact convex set in a topological vector space has a fixed point. Because of isometry, the Fourier transform maps the unit ball in $L^2$ to itself. Owing to the Banach Alaoglu theorem, the unit ball in $L^2$ is compact with respect to the weak topology. The Fourier transform is continuous in the weak topology, because if $( f_n, \phi ) \to (f, \phi)$ for all $\phi \in L^2$, then&#xA;$$&#xA;(\hat{f}_n, \phi) = (f_n, \hat{\phi}) \to (f, \hat{\phi}) = (\hat{f}, \phi).&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="960" LastActivityDate="2010-09-13T06:48:26.437" CommentCount="5" />
  <row Id="4508" PostTypeId="2" ParentId="4467" CreationDate="2010-09-13T06:48:26.500" Score="2" ViewCount="" Body="&lt;p&gt;To answer Pete's comment as to how to prove integral&#xA;closure of $\mathbb{Z}$ without using the UFD property.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $a/b$ be a rational ($a$, $b\in\mathbb{Z}$) which is&#xA;integral over $\mathbb{Z}$. Let $R=\mathbb{Z}[a/b]$. Then&#xA;$R$ is a finitely-generated $\mathbb{Z}$-module. It follows that&#xA;$b^n R\subseteq\mathbb{Z}$ for some $n$. We reduce to proving&#xA;the lemma that if $R$ is a ring with &#xA;$\mathbb{Z}\subseteq R\subseteq N^{-1}\mathbb{Z}$ for some&#xA;nonzero integer $N$ then $R=\mathbb{Z}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are various ways of proving this. For instance if&#xA;$R\ne\mathbb{Z}$ there is an element of $R$ strictly between two&#xA;consecutive integers (this is the division algorithm) and so&#xA;an element $x$ of $R$ strictly between $0$ and $1$. If $y$ is the least&#xA;such number then considering $y^2$ gives a contradiction.&#xA;Alternatively, $M=xN$ is an integer and&#xA;$R\subseteq M^{-1}\mathbb{Z}$ so we can always replace $N$ by a smaller integer etc.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-09-13T06:48:26.500" CommentCount="7" />
  <row Id="4509" PostTypeId="2" ParentId="4506" CreationDate="2010-09-13T07:11:29.300" Score="5" ViewCount="" Body="&lt;p&gt;Many people &quot;would like to have an argument working for $n^2$&quot;, but what is available at the moment (and for the last two centuries) are conjectures.  For any list of integer polynomials there is a conjecture on how often all polynomials on the list are prime:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Bateman%E2%80%93Horn_conjecture&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Bateman%E2%80%93Horn_conjecture&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is extremely hard to prove that any natural set of integers of density 0 contains infinitely many primes.  It is known for the set of values of $x^2 + y^4$ but not for the values of any single-variable polynomial of degree higher than one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The asymptotic formula in the Bateman-Horn conjecture isn't necessarily the most general expression of what people in the field believe to be true (and it is probably a lot older than Bateman and Horn's article that formally codified it), but it does subsume many earlier conjectures on primes of the form $n^2+1$, prime twins and k-tuplets, Schinzel's Hypothesis and Buniakowsy's conjecture.  You can calculate from the formula the predicted frequency of $n$ such that $4n^2 + 3$ is prime.&lt;/p&gt;&#xA;" OwnerUserId="467" LastActivityDate="2010-09-13T07:11:29.300" CommentCount="2" />
  <row Id="4510" PostTypeId="1" AcceptedAnswerId="4566" CreationDate="2010-09-13T09:02:07.990" Score="5" ViewCount="379" Body="&lt;p&gt;In many combinatorial enumeration problems it is possible to find a rational generating function (i.e. the quotient of two polynomials) for the sequence in question. The question is - given the generating function, how can we find (algorithmically) the values of the sequence, i.e. the coefficients of the corresponding power series?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that for a rational generating function, the sequence satisfies a recurrence relation given by the coefficients of the polynomial in the denominator, so it's really just the question of finding the finite initial values.&lt;/p&gt;&#xA;" OwnerUserId="1818" LastActivityDate="2010-09-14T18:21:33.887" Title="Finding the power series of a rational function" Tags="&lt;real-analysis&gt;&lt;combinatorics&gt;&lt;commutative-algebra&gt;" AnswerCount="3" CommentCount="5" FavoriteCount="2" />
  <row Id="4511" PostTypeId="1" AcceptedAnswerId="4519" CreationDate="2010-09-13T10:23:05.620" Score="2" ViewCount="134" Body="&lt;p&gt;I found the following in the wikipedia page for &lt;a href=&quot;http://en.wikipedia.org/wiki/Euler_class#Squares_to_top_Pontryagin_class&quot; rel=&quot;nofollow&quot;&gt;Euler class&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;If the rank r is even, then this cohomology class $e(E) \cup e(E)$ equals the top Pontryagin class $p_{r/2}(E)$.&#xA;Under the splitting principle, this corresponds to the square of the Vandermonde polynomial equaling the discriminant: the Euler class corresponds to the Vandermonde polynomial, the basic alternating polynomial, while the top Pontryagin class corresponds to the discriminant, a symmetric polynomial.&#xA;More formally, the Euler class of a direct sum of line bundles is the Vandermonde polynomial (orientation determines the order of the line bundles up to sign), while top Pontryagin class is the discriminant.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anybody clarify me this? In particular I don't understand the sense of &quot;the Euler class of a direct sum of line bundles is the Vandermonde polynomial&quot;.  I checked the references but there is nothing about that.&lt;/p&gt;&#xA;" OwnerUserId="1860" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-09-13T10:27:50.697" LastActivityDate="2010-09-13T11:26:56.987" Title="euler class and vandermonde polynomial" Tags="&lt;algebraic-topology&gt;" AnswerCount="1" FavoriteCount="2" />
  <row Id="4512" PostTypeId="1" AcceptedAnswerId="4549" CreationDate="2010-09-13T10:23:45.590" Score="3" ViewCount="76" Body="&lt;p&gt;I came across the following theorem that for any class $K$ of truth assignments, $K\subseteq Mod(Th(K))$, and if $K$ is axiomatizable, then $K=Mod(Th(K))$. Together these seem to me to imply that there is some class $K$ such that $K\neq Mod(Th(K))$, however, I can't find any class $K$ that would verify my assumption.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I do know that $K$ would have to contain more than one truth assignment, since if $K$ were to have only one assignment $V$, $Th(V)$ has a unique model $V$, so then $K=V=Mod(Th(K))$. Does anyone know of any example where $K\neq Mod(Th(K))$, or at least how to construct one?&lt;/p&gt;&#xA;" OwnerUserId="1583" LastActivityDate="2010-09-13T17:03:52.947" Title="Is it possible for $Mod(Th(K))\neq K$ for some class $K$ of truth assignments?" Tags="&lt;logic&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="4513" PostTypeId="1" CreationDate="2010-09-13T10:44:29.463" Score="0" ViewCount="373" Body="&lt;p&gt;I would like to find the center of arc from the starting point, the endpoint, and i,j values&lt;/p&gt;&#xA;" OwnerUserId="1861" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-09-13T11:08:07.073" LastActivityDate="2011-05-18T20:58:56.040" Title="How to calculate center of arc from startpoint, endpoint and i,j values?" Tags="&lt;geometry&gt;&lt;algorithms&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" />
  <row Id="4514" PostTypeId="2" ParentId="4497" CreationDate="2010-09-13T10:48:59.640" Score="1" ViewCount="" Body="&lt;p&gt;Just adding some extra comments.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First $$\frac{2x-1}{x-1}=\frac{2-1/x}{1-1/x}\to\frac{2-0}{1-0}=2$$&#xA;as $x\to\infty$ (this is independent of how the limit is taken).&#xA;Secondly, we note that&#xA;$$0\leq\frac{x}{y}\leq1$$&#xA;the first inequality comes from that both $x,y$ are $&gt;0$ (since they should tend to $\infty$), the second comes from $x\leq y$ (divide by $y$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At this point we have &#xA;$$2\leq\liminf_{(x,y)\to\infty,x\leq y} \frac{2x-1}{x-1} + \frac{x}{y}\leq\limsup_{(x,y)\to\infty,x\leq y} \frac{2x-1}{x-1} + \frac{x}{y}\leq3$$&#xA;At the boundary of the limit, that is when $x=y$ we would get 3. To reach the liminf we only have to find a way to press $x/y$ as low as we can (recall that we already know it is $x/y\ge0$) - at this point you have do some testing, you want $x$ to tend slower to infinity than $y$, $x=\log y$ or $x=\sqrt{y}$ works fine! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, the linear choice $x=ay$, earlier suggested by Yuval Filmus above works fine to see that all limits between 2 and 3 are possible. The example in the the comment of Yuval Filmus shows that there are choices of $x$'s such that no limit is attained.&lt;/p&gt;&#xA;" OwnerUserId="1154" LastActivityDate="2010-09-13T10:48:59.640" CommentCount="0" />
  <row Id="4515" PostTypeId="1" AcceptedAnswerId="4516" CreationDate="2010-09-13T10:51:46.270" Score="4" ViewCount="864" Body="&lt;p&gt;I've been using the sentence:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If a series converges then the limit of the sequence is zero&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;as a criterion to prove that a series diverges (when lim != 0) and I can understand the rationale behind it, but I can't find a &lt;strong&gt;formal proof&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you help me?&lt;/p&gt;&#xA;" OwnerUserId="1862" LastActivityDate="2010-09-13T12:24:41.803" Title="Proof: Series converges =&gt; the limit of the sequence is zero" Tags="&lt;sequences-and-series&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="4516" PostTypeId="2" ParentId="4515" CreationDate="2010-09-13T10:57:18.580" Score="5" ViewCount="" Body="&lt;p&gt;Yes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\lim_{n \to \infty} \left ( \sum_{k = 1}^{n + 1} a_k - \sum_{k = 1}^{n} a_k \right ) = \lim_{n \to \infty} a_{n + 1} $$&#xA;And both sums will converge to the same number so the limit is zero. This is by far the easiest proof I know.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the Cauchy criterion in disguise by the way, so you could use that too.&lt;/p&gt;&#xA;" OwnerUserId="1120" LastEditorUserId="1120" LastEditorDisplayName="" LastEditDate="2010-09-13T12:24:41.803" LastActivityDate="2010-09-13T12:24:41.803" CommentCount="6" />
  <row Id="4518" PostTypeId="2" ParentId="4515" CreationDate="2010-09-13T11:16:13.927" Score="2" ViewCount="" Body="&lt;p&gt;If we know that the sequence converges and merely wish to show it converges to zero, then a proof by contradiction gives a little more intuition here (although the direct proofs are simple and beautiful). Assume $a_n\to a$ with $a&gt;0$, then for all $n&gt;N$ for some large enough $N$ we have $a_n &gt; a/2$ (take $\varepsilon = a/2$ in the definition of the limit). Now the sum diverges: $\sum_{n&gt;N}a_n &gt; \sum_{n&gt;N}a/2 = \infty$. A similar argument works when $a&amp;lt;0$.&lt;/p&gt;&#xA;" OwnerUserId="1818" LastEditorUserId="1818" LastEditorDisplayName="" LastEditDate="2010-09-13T11:53:11.037" LastActivityDate="2010-09-13T11:53:11.037" CommentCount="2" />
  <row Id="4519" PostTypeId="2" ParentId="4511" CreationDate="2010-09-13T11:26:56.987" Score="2" ViewCount="" Body="&lt;p&gt;According to the &lt;a href=&quot;http://en.wikipedia.org/wiki/Chern%E2%80%93Weil_homomorphism&quot; rel=&quot;nofollow&quot;&gt;Chern-Weil theory&lt;/a&gt; (see for example these &lt;a href=&quot;http://data.imf.au.dk/publications/ln/2003/imf-ln-2003-69.pdf&quot; rel=&quot;nofollow&quot;&gt;lecture notes&lt;/a&gt; by Johan Dupont) , characteristic classes such as the Euler and Pontryagin classes correspond to invariant polynomials (under the bundle structure group) of the curvature form of the vector bundle. Basically, what the &lt;a href=&quot;http://en.wikipedia.org/wiki/Splitting_principle&quot; rel=&quot;nofollow&quot;&gt;splitting principle&lt;/a&gt; tells us is that these polynomials can be &quot;formally&quot; written in terms of pullbacks of curvature forms of line bundles which correspond to the &quot;eigenvalues&quot; of the curvature form (viewed as an antisymmetric endomorphism).The invariant polynomial corresponding to the Euler class is the Vandermonede polynomial. For a given rank of the vector bundle, one can always write these polynomials in terms of the vector bundle curvature form and the use of the line bundle curvature forms is for simplifying the algebraic work.&lt;/p&gt;&#xA;" OwnerUserId="1430" LastActivityDate="2010-09-13T11:26:56.987" CommentCount="2" />
  <row Id="4520" PostTypeId="1" AcceptedAnswerId="4522" CreationDate="2010-09-13T12:11:29.667" Score="4" ViewCount="88" Body="&lt;p&gt;Let $X$ be a contractible space. If $x_0 \in X$, it is not necessarily true that the pointed space $(X,x_0)$ is contractible (i.e., it is possible that any contracting homotopy will move $x_0$).  An example is given in 1.4 of Spanier: the comb space.  However, this space is contractible as a pointed space if the basepoint is in the bottom line.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a contractible space which is not contractible as a pointed space for &lt;em&gt;any&lt;/em&gt; choice of basepoint?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My guess is that this will have to be some kind of pathological space, because for CW complexes, we have the &lt;a href=&quot;http://en.wikipedia.org/wiki/Whitehead_theorem&quot; rel=&quot;nofollow&quot;&gt;Whitehead theorem&lt;/a&gt;. (So I'm not completely sure that the Whitehead theorem is actually a statement about the &lt;em&gt;pointed&lt;/em&gt; homotopy category, but hopefully I'm right.)&lt;/p&gt;&#xA;" OwnerUserId="536" LastEditorUserId="536" LastEditorDisplayName="" LastEditDate="2010-09-13T12:18:37.577" LastActivityDate="2010-09-13T12:21:19.947" Title="Does contractibility imply contractibility with a basepoint?" Tags="&lt;algebraic-topology&gt;&lt;homotopy-theory&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="4521" PostTypeId="1" AcceptedAnswerId="4523" CreationDate="2010-09-13T12:16:08.380" Score="2" ViewCount="303" Body="&lt;p&gt;$$\arccos\left(\frac{1-x^2}{1+x^2}\right) = 2\arctan{x}$$ for $x \geq 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not even sure what kind of math to try? :(&lt;/p&gt;&#xA;" OwnerUserId="1862" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-09-13T12:52:59.240" LastActivityDate="2010-10-04T18:26:36.017" Title="How can I prove this trigonometric equality?" Tags="&lt;trigonometry&gt;" AnswerCount="4" FavoriteCount="1" />
  <row Id="4522" PostTypeId="2" ParentId="4520" CreationDate="2010-09-13T12:21:19.947" Score="3" ViewCount="" Body="&lt;p&gt;Yes.  See exercise 7 &lt;a href=&quot;http://books.google.com/books?id=tru-TWQFDAIC&amp;amp;lpg=PP1&amp;amp;dq=using%20the%20borsuk%20ulam%20theorem&amp;amp;pg=PA7#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot; title=&quot;here&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="641" LastActivityDate="2010-09-13T12:21:19.947" CommentCount="1" />
  <row Id="4523" PostTypeId="2" ParentId="4521" CreationDate="2010-09-13T12:23:52.773" Score="5" ViewCount="" Body="&lt;p&gt;Put $t=\arctan x$. Then $x=\tan t$.&#xA;You have to prove that $2t=\arccos(1-x^2)/(1+x^2)$.&#xA;This is more-or-less the same as $(1-x^2)/(1+x^2)=\cos2t$. If you put&#xA;$x=\tan t$ into $(1-x^2)/(1+x^2)$ what do you get?&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-09-13T12:23:52.773" />
  <row Id="4524" PostTypeId="2" ParentId="4521" CreationDate="2010-09-13T12:24:55.407" Score="1" ViewCount="" Body="&lt;p&gt;One way would be to use the complex logarithm representations of the inverse trigonometric functions. For this answer, I'll be more pedestrian.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take the cosines of both sides, like so:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\cos\left(\arccos\left(\frac{1-x^2}{1+x^2}\right)\right)=\cos\left(2\arctan\left(x\right)\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;to obtain&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{1-x^2}{1+x^2}=\cos\left(2\arctan\left(x\right)\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;use the double angle formula,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{1-x^2}{1+x^2}=2\cos^2\left(\arctan\left(x\right)\right)-1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;switch to secants&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{1-x^2}{1+x^2}=\frac{2}{\mathrm{sec}^2\left(\arctan(x)\right)}-1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;use a Pythagorean identity&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{1-x^2}{1+x^2}=\frac{2}{1+\tan^2\left(\arctan(x)\right)}-1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and now you have a function and its inverse composed&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{1-x^2}{1+x^2}=\frac{2}{1+x^2}-1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and this last one is a definite identity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, as already noted, the left side (of the original identity) is even; the right side odd. We check that the two expressions agree at $x=0$, and both have the same limiting behavior as $x\to\infty$. The identity is thus only valid for nonnegative $x$.&lt;/p&gt;&#xA;" OwnerUserId="498" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-09-13T21:04:30.483" LastActivityDate="2010-09-13T21:04:30.483" CommentCount="3" />
  <row Id="4525" PostTypeId="2" ParentId="4521" CreationDate="2010-09-13T12:50:21.960" Score="4" ViewCount="" Body="&lt;p&gt;Setting $x=\tan a$ your equality&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\arccos \dfrac{1-x^{2}}{1+x^{2}}=2\arctan x\qquad (1)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;becomes&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\cos 2a=\dfrac{1-\tan ^{2}a}{1+\tan ^{2}a}.\qquad (2)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To prove (2) which is &lt;a href=&quot;http://en.wikipedia.org/wiki/List_of_trigonometric_identities&quot; rel=&quot;nofollow&quot;&gt;listed in the wikipedia&lt;/a&gt; we can take the duplication formula&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\cos 2a=\cos ^{2}a-\sin ^{2}a$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dividing by $\cos ^{2}a+\sin ^{2}a=1$, we establish&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\cos 2a=\dfrac{\cos ^{2}a-\sin ^{2}a}{\cos ^{2}a+\sin ^{2}a}=\dfrac{1-\tan&#xA;^{2}a}{1+\tan ^{2}a}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Edit:&lt;/em&gt; One must note that $\arctan x$ is an &lt;a href=&quot;http://en.wikipedia.org/wiki/Even_and_odd_functions&quot; rel=&quot;nofollow&quot;&gt;odd function&lt;/a&gt;, while $\arccos \frac{1-x^{2}}{1+x^{2}}$ is an &lt;a href=&quot;http://en.wikipedia.org/wiki/Even_and_odd_functions&quot; rel=&quot;nofollow&quot;&gt;even&lt;/a&gt; one. Only for $x\geq 0$ are both sides of $(1)$ equal. Identity $(2)$ is valid for both positive and negative values of $a$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/UZHEW.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Blue: $y=\arccos((1-x^2)/(1+x^2))$ for $x\ge 0$;  Red: $y=2\arctan x$ &lt;/p&gt;&#xA;" OwnerUserId="752" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2010-10-04T18:26:36.017" LastActivityDate="2010-10-04T18:26:36.017" CommentCount="3" />
  <row Id="4526" PostTypeId="1" AcceptedAnswerId="4538" CreationDate="2010-09-13T13:07:23.073" Score="10" ViewCount="467" Body="&lt;p&gt;I would like to collect a list of mathematical concepts that have been named after mathematicians, which are now used in lowercase form (such as &quot;abelian&quot;).  This question is partly motivated by my former supervisor, who joked (something like):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;You know you've made it as a mathematician when they start using your name in lowercase.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is a &lt;em&gt;community wiki&lt;/em&gt;, please:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;One mathematician per answer.  Please update already-existing answers if that mathematician already exists.&lt;/li&gt;&#xA;&lt;li&gt;Please give a brief description of the concept and the person (ideally with links to wikipedia).&lt;/li&gt;&#xA;&lt;li&gt;This is not about whether or not you agree with the capitalisation.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Expect your answer to be edited otherwise.&lt;/p&gt;&#xA;" OwnerUserId="139" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-09-13T14:22:07.607" LastActivityDate="2010-09-13T20:45:24.727" CommunityOwnedDate="2010-09-13T13:07:23.073" Title="Mathematical concepts named after mathematicians that have become acceptable to spell in lowercase form (e.g. abelian)?" Tags="&lt;big-list&gt;&lt;terminology&gt;&lt;mathematicians&gt;" AnswerCount="12" CommentCount="5" FavoriteCount="1" />
  <row Id="4527" PostTypeId="2" ParentId="4526" CreationDate="2010-09-13T13:13:48.567" Score="8" ViewCount="" Body="&lt;p&gt;I've seen &quot;&lt;a href=&quot;http://en.wikipedia.org/wiki/Jacobian_matrix_and_determinant&quot; rel=&quot;nofollow&quot;&gt;jacobian&lt;/a&gt;&quot; in lower case in a few books.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the uninitiated: it refers to either a matrix of partial derivatives, or its determinant. This is named after &lt;a href=&quot;http://en.wikipedia.org/wiki/Carl_Gustav_Jacob_Jacobi&quot; rel=&quot;nofollow&quot;&gt;Carl Gustav Jacob Jacobi&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sources using lowercase &quot;jacobi&quot;:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Claus Bendtsen, Preliminary Experiences with Extrapolation Methods for Parallel Solution of Differential, in Parallel scientific computing, 1994.&lt;/li&gt;&#xA;&lt;li&gt;Foundations of Modern Analysis, By J. Dieudonne.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="498" LastEditorUserId="139" LastEditorDisplayName="" LastEditDate="2010-09-13T14:36:20.377" LastActivityDate="2010-09-13T14:36:20.377" CommunityOwnedDate="2010-09-13T13:13:48.567" />
  <row Id="4528" PostTypeId="2" ParentId="4526" CreationDate="2010-09-13T13:18:54.437" Score="11" ViewCount="" Body="&lt;p&gt;For something even more common, &quot;boolean&quot; (e.g. &quot;&lt;a href=&quot;http://en.wikipedia.org/wiki/Boolean_data_type&quot; rel=&quot;nofollow&quot;&gt;boolean type&lt;/a&gt;&quot;, &quot;&lt;a href=&quot;http://en.wikipedia.org/wiki/Boolean_algebra_%28structure%29&quot; rel=&quot;nofollow&quot;&gt;boolean algebra&lt;/a&gt;&quot;) is increasingly not being capitalized as well. &lt;a href=&quot;http://en.wikipedia.org/wiki/George_Boole&quot; rel=&quot;nofollow&quot;&gt;George Boole&lt;/a&gt; formalized the manipulation of logical statements, and thus his name crops up frequently in connection to logic.&lt;/p&gt;&#xA;" OwnerUserId="498" LastEditorUserId="139" LastEditorDisplayName="" LastEditDate="2010-09-13T13:39:33.170" LastActivityDate="2010-09-13T13:39:33.170" CommunityOwnedDate="2010-09-13T13:18:54.437" />
  <row Id="4529" PostTypeId="2" ParentId="4526" CreationDate="2010-09-13T13:48:53.747" Score="2" ViewCount="" Body="&lt;p&gt;Noetherian rings:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Noetherian_ring&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Noetherian_ring&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;in honor of Emmy Noether:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Emmy_Noether&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Emmy_Noether&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example sources:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Lectures on modules and rings, By Tsit-Yuen Lam&lt;/li&gt;&#xA;&lt;li&gt;Commutative algebra, By Oscar Zariski, Pierre Samuel, Irvin Sol Cohen&lt;/li&gt;&#xA;&lt;li&gt;Model theory and modules, By Mike Prest&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;(plus many others)&lt;/p&gt;&#xA;" OwnerUserId="1369" LastEditorUserId="139" LastEditorDisplayName="" LastEditDate="2010-09-13T14:41:29.983" LastActivityDate="2010-09-13T14:41:29.983" CommunityOwnedDate="2010-09-13T13:48:53.747" CommentCount="1" />
  <row Id="4530" PostTypeId="2" ParentId="4526" CreationDate="2010-09-13T13:50:56.303" Score="2" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Euclid&quot; rel=&quot;nofollow&quot;&gt;Euclid&lt;/a&gt;'s name has been attached to &lt;a href=&quot;http://en.wikipedia.org/wiki/Euclidean_space&quot; rel=&quot;nofollow&quot;&gt;euclidean space&lt;/a&gt; and the &lt;a href=&quot;http://en.wikipedia.org/wiki/Euclidean_algorithm&quot; rel=&quot;nofollow&quot;&gt;euclidean algorithm&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="247" LastEditorUserId="139" LastEditorDisplayName="" LastEditDate="2010-09-13T14:08:33.927" LastActivityDate="2010-09-13T14:08:33.927" CommunityOwnedDate="2010-09-13T13:50:56.303" CommentCount="1" />
  <row Id="4531" PostTypeId="2" ParentId="4526" CreationDate="2010-09-13T13:51:39.810" Score="9" ViewCount="" Body="&lt;p&gt;&lt;code&gt;cartesian&lt;/code&gt;. Many things are described as cartesian, after &lt;a href=&quot;http://en.wikipedia.org/wiki/Ren%C3%A9_Descartes&quot; rel=&quot;nofollow&quot;&gt;René Descartes&lt;/a&gt;. In philosophy things attributed to Descartes tend to keep the capital when described as &quot;Cartesian&quot;, but in mathematics, I've seen &quot;cartesian product&quot; lower case a few times. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What's weird is that this isn't more prevalent, given that Descartes is older than Abel by almost 200 years, and I conjecture that cartesian products, cartesian planes and the like come up at least as much as abelian groups... And since the adjectival form of &quot;Descartes&quot; isn't just the name+ending, it's strange how common the capitalised form has remained...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Google books reveals that lowercase &quot;&lt;a href=&quot;http://en.wikipedia.org/wiki/Cartesian_coordinate_system&quot; rel=&quot;nofollow&quot;&gt;cartesian plane&lt;/a&gt;&quot; was used in the following books (for example):&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Algebra Through Practice: Volume 2, Matrices and Vector Spaces,  By T. S. Blyth, E. F. Robertson&lt;/li&gt;&#xA;&lt;li&gt;Cracking the MCAT with CD-ROM, By James L. Flowers&lt;/li&gt;&#xA;&lt;li&gt;Symmetry And Spectroscopy Of Molecules, By K Veera Reddy&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="172" LastEditorUserId="139" LastEditorDisplayName="" LastEditDate="2010-09-13T14:26:12.987" LastActivityDate="2010-09-13T14:26:12.987" CommunityOwnedDate="2010-09-13T13:51:39.810" />
  <row Id="4532" PostTypeId="2" ParentId="4526" CreationDate="2010-09-13T13:52:31.240" Score="8" ViewCount="" Body="&lt;p&gt;Many things have been named gaussian after &lt;a href=&quot;http://en.wikipedia.org/wiki/Gauss&quot; rel=&quot;nofollow&quot;&gt;Carl Friedrick Gauss&lt;/a&gt;: &lt;a href=&quot;http://en.wikipedia.org/wiki/Normal_distribution&quot; rel=&quot;nofollow&quot;&gt;gaussian distribution&lt;/a&gt;, &lt;a href=&quot;http://en.wikipedia.org/wiki/Gaussian_integer&quot; rel=&quot;nofollow&quot;&gt;gaussian integers&lt;/a&gt;, etc.&lt;/p&gt;&#xA;" OwnerUserId="247" LastEditorUserId="139" LastEditorDisplayName="" LastEditDate="2010-09-13T14:28:37.853" LastActivityDate="2010-09-13T14:28:37.853" CommunityOwnedDate="2010-09-13T13:52:31.240" />
  <row Id="4533" PostTypeId="1" AcceptedAnswerId="4535" CreationDate="2010-09-13T13:56:32.420" Score="1" ViewCount="213" Body="&lt;p&gt;I've got two questions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1). I would like to know if by definition the Sobolev space norm is defined only by differentials with respect to one variable i.e.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\parallel f \parallel \; = \; \left[ \int_{\mathbb{R}} \left( \left| f \right|^p \; + \; \left| \frac{\partial f}{\partial x} \right|^p \; + \ldots + \; \left| \frac{\partial^m f}{\partial x^m} \right|^p \right) dx \right]^{\frac{1}{p}} $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What would the corresponding Sobolev space norm be if $f$ were a function of two variables e.g. $f(x,y)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2). If the Sobolev space norm exists for multivariate functions, is it correct to say that the Beppo-Levi space norm a special case of the Sobolev space norm? From the little I've read, it appears that the Beppo-Levi space norm is given by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\parallel f \parallel \; = \; \int_{\mathbb{R}^2} \sum_{i = 0}^m {m \choose i} \left( \frac{\partial^m f}{\partial x^i \partial y^{m - i}} \right)^2 dx dy$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What if the partial derivative is raised to the power of 3. Would this still be a Beppo-Levi space norm?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: based on Robin's answer and the paper &quot;Spline functions and stochastic filtering&quot; by Christine Thomas-Agnan, here's my understanding of the &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sobolev norm:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\parallel f \parallel^p \; = \; \sum_{ 0 \: \le \: ( \: \mid \alpha \mid \: = \: \alpha_1 \: + \: \cdots \: + \: \alpha_n \: ) \: \le \: p}\int_{\mathbb{R}^n} \; \left| \frac{\partial^{\mid \alpha \mid} f}{\partial x_1^{\alpha_1} \: \cdots \: \partial x_n^{\alpha_n}} \right|^p \: dx$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Beppo-Levi norm:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\parallel f \parallel^m \; = \; \int_{\mathbb{R}^n} \; \sum_{\alpha_i \: + \: \cdots \: + \: \alpha_n \: = \: m } {m \choose {\alpha_i! \; \cdots \; \alpha_n!}} \left| \frac{\partial^m f(x)}{\partial x_1^{\alpha_1} \cdots \partial x_n^{\alpha_n} } \right|^2 dx$&lt;/p&gt;&#xA;" OwnerUserId="1752" LastEditorUserId="1752" LastEditorDisplayName="" LastEditDate="2010-09-15T20:35:01.523" LastActivityDate="2010-09-15T20:35:01.523" Title="Sobolev Space norm and Beppo-Levi Space norm" Tags="&lt;functional-analysis&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="4534" PostTypeId="2" ParentId="4526" CreationDate="2010-09-13T13:58:53.970" Score="2" ViewCount="" Body="&lt;p&gt;The algebraic structure formed by a set with a commutative binary operation is described as &lt;a href=&quot;http://en.wikipedia.org/wiki/Abelian_group&quot; rel=&quot;nofollow&quot;&gt;abelian&lt;/a&gt;.  If G is a group, then the quotient group G/[G,G] (where [G,G] is the commutator subgroup) is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Abelianization#Abelianization&quot; rel=&quot;nofollow&quot;&gt;abelianization&lt;/a&gt; of G.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Niels_Henrik_Abel&quot; rel=&quot;nofollow&quot;&gt;Niels Henrik Abel&lt;/a&gt; wikipedia page suggests that he pretty much invented group theory.  [[citation needed]]&lt;/p&gt;&#xA;" OwnerUserId="139" LastActivityDate="2010-09-13T13:58:53.970" CommunityOwnedDate="2010-09-13T13:58:53.970" />
  <row Id="4535" PostTypeId="2" ParentId="4533" CreationDate="2010-09-13T14:06:04.017" Score="2" ViewCount="" Body="&lt;p&gt;The Sobolev norm is also defined for regions in $\mathbb{R}^n$. You would typically&#xA;have&#xA;$$\|f\|^p=\sum_{(i_1,\ldots,i_n)}\int\left|&#xA;\frac{\partial^I f}{\partial x_1^{i_1}\cdots\partial x_n^{i_n}}&#xA;\right|^pd\mathbf{x}$$&#xA;where $I=i_1+\cdots+i_n$ and the sum is over all tuples with $0\le I\le m$.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-09-13T14:06:04.017" CommentCount="7" />
  <row Id="4536" PostTypeId="1" AcceptedAnswerId="4564" CreationDate="2010-09-13T14:37:53.260" Score="9" ViewCount="260" Body="&lt;p&gt;I'm looking at the random sequence $x_n$ with $x_0=x_1=1$ and &#xA;\begin{equation}&#xA;x_{n+1}=2x_n\pm x_{n-1}&#xA;\end{equation}&#xA;where we choose the $\pm$ sign independently with equal probability. Now considering the recurrence relations $x_{n+1}=2x_n+x_{n-1}$ and $x_{n+1}=2x_n-x_{n-1}$ separately, it is clear that $x_n\rightarrow\infty$ as $n\rightarrow\infty$. However, the sequence $x_n^{1/n}$ seems to tend to a limit near 1.91 (got this from numerical computation and some brute force by a Monte Carlo simulating the $\pm$ sign). Thus the sequence $x_n^{1/n}$ seems to convenge almost surely. I was wondering if anyone could show that the sequence was indeed almost surely converging and/or work out the limit. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update:&lt;/strong&gt;&lt;br&gt;&#xA;This comment shows that $\lim_{n\rightarrow\infty} |x_n|^{1/n}$ converges almost surely. Let $y_n=\frac{x_n}{2^n}$ then $2^{n+1}y_{n+1}=2^{n+1}y_n\pm 2^{n-1}y_{n-1}$. Hence&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$y_{n+1}=y_n\pm \frac{1}{4}y_{n-1}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Embree-Trefethen showed that $\lim_{n\rightarrow\infty} |y_n|^{1/n}$ converges almost surely. See Embree, M.; Trefethen, L. N. (1999), &quot;Growth and decay of random Fibonacci sequences&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, finding the almost sure limit accurately or analytically is proving difficult at the moment. &lt;/p&gt;&#xA;" OwnerUserId="1794" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2011-06-03T20:08:39.863" LastActivityDate="2011-06-03T20:08:39.863" Title="Solving randomized recurrence relation" Tags="&lt;probability&gt;&lt;probability-theory&gt;&lt;approximation&gt;&lt;recurrence-relations&gt;" AnswerCount="3" />
  <row Id="4537" PostTypeId="1" AcceptedAnswerId="4547" CreationDate="2010-09-13T14:39:25.920" Score="1" ViewCount="269" Body="&lt;p&gt;In India, generally during the graduate years, we follow a course work pattern, unlike many places in the U.S where students are exposed to research during their undergraduate years itself. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As, a graduate student, we have basic courses like, Algebra, Analysis, Rings and Modules, Measure theory, Topology, functional analysis etc.. Generally topology is one subject, which i don't find that much of interest. But in some universities, students are forced to take Topology, and Algebraic topology, during their graduate years, and i have seen many students facing trouble, as they have to study a subject which is not of their interests. My question, would be, for a student whose research are in Analytic and Algebraic Number theory, does he needs to know Algebraic Topology?&lt;/p&gt;&#xA;" LastActivityDate="2010-09-13T16:13:09.910" CommunityOwnedDate="2010-09-13T14:39:25.920" Title="Algebraic Topology Need!" Tags="&lt;soft-question&gt;" AnswerCount="1" CommentCount="10" />
  <row Id="4538" PostTypeId="2" ParentId="4526" CreationDate="2010-09-13T14:40:26.240" Score="6" ViewCount="" Body="&lt;p&gt;Like &quot;cartesian&quot;, &quot;&lt;a href=&quot;http://en.wikipedia.org/wiki/Algorithm&quot; rel=&quot;nofollow&quot;&gt;algorithm&lt;/a&gt;&quot; is not only a lowercased but a derivative form of the name of a mathematician (&lt;a href=&quot;http://en.wikipedia.org/wiki/Al-Khw%C4%81rizm%C4%AB&quot; rel=&quot;nofollow&quot;&gt;al-Khwarismi&lt;/a&gt;).&lt;/p&gt;&#xA;" LastEditorUserId="139" LastEditorDisplayName="" LastEditDate="2010-09-13T14:44:11.460" LastActivityDate="2010-09-13T14:44:11.460" CommunityOwnedDate="2010-09-13T14:40:26.240" CommentCount="2" />
  <row Id="4539" PostTypeId="2" ParentId="4526" CreationDate="2010-09-13T14:41:14.677" Score="2" ViewCount="" Body="&lt;p&gt;I have found the term &quot;karoubianization&quot; in the literature (although google only finds for me instances where it is capitalized...). It is a synonym for &quot;pseudo-abelianization&quot; or &quot;idempotent-completion&quot;, and refers to the process whereby one enlarges a category universally so that all its idempotents split. The word refers to Max Karoubi.&lt;/p&gt;&#xA;" OwnerUserId="274" LastEditorUserId="274" LastEditorDisplayName="" LastEditDate="2010-09-13T14:48:46.953" LastActivityDate="2010-09-13T14:48:46.953" CommunityOwnedDate="2010-09-13T14:41:14.677" CommentCount="1" />
  <row Id="4541" PostTypeId="2" ParentId="4526" CreationDate="2010-09-13T15:29:58.107" Score="3" ViewCount="" Body="&lt;p&gt;David Rusin compiled &lt;a href=&quot;http://www.math.niu.edu/~rusin/known-math/98/MSC.names&quot; rel=&quot;nofollow&quot;&gt;a list of mathematicians whose names appear in the 1991 MSC&lt;/a&gt;. (That's the &quot;mathematics subject classification&quot;.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are 357. He claims Abel's name is the only one among these which appears noncapitalized in the MSC. The other people who have been named in this question (Noether, Descartes, Euclid, Karoubi, Gauss, Boole, Jacobi) all appear in the MSC.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, the 1991 MSC is nearly twenty years old, and is likely to reflect more-formal-than-average usage.&lt;/p&gt;&#xA;" OwnerUserId="173" LastActivityDate="2010-09-13T15:29:58.107" CommunityOwnedDate="2010-09-13T15:29:58.107" CommentCount="1" />
  <row Id="4542" PostTypeId="2" ParentId="4521" CreationDate="2010-09-13T15:32:46.893" Score="2" ViewCount="" Body="&lt;p&gt;Differentiate the difference $$\arccos\left(\frac{1-x^2}{1+x^2}\right) - 2\arctan{x},$$ simplify the resulting expression and find $0$ when $x&gt;0$, concluding that the difference is constant. Now evaluate at $1$, to see that the difference is in fact constantly zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;N.B.:&lt;/strong&gt; This is of course quite unenlightenling, but you always know that if two things are equal, their difference is constant so you never lose much by trying to show that it is constant and zero. Specially if you have a computer algebra system at hand that can do the derivatives and simplifications for you!&lt;/p&gt;&#xA;" OwnerUserId="274" LastEditorUserId="274" LastEditorDisplayName="" LastEditDate="2010-09-13T15:46:10.360" LastActivityDate="2010-09-13T15:46:10.360" />
  <row Id="4544" PostTypeId="1" AcceptedAnswerId="4548" CreationDate="2010-09-13T15:53:18.607" Score="11" ViewCount="613" Body="&lt;p&gt;Hi,&lt;br&gt;&#xA;the fact that Ramanujan's Constant $e^{\pi \sqrt{163}}$ is almost integer ($262 537 412 640 768 743.99999999999925...$) doesn't seem to be a coincidence, but has to do with the $163$ appearing in it. Can you explain why it's almost-but-not-quite integer in layman's terms (I'm not a mathematician)?&lt;br&gt;&#xA;TIA&lt;br&gt;&#xA;Steven&lt;/p&gt;&#xA;" OwnerUserId="923" LastActivityDate="2010-09-13T16:23:36.007" Title="Why is $e^{\pi \sqrt{163}}$ almost integer?" Tags="&lt;number-theory&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="2" />
  <row Id="4545" PostTypeId="2" ParentId="4544" CreationDate="2010-09-13T16:05:10.260" Score="0" ViewCount="" Body="&lt;p&gt;As QY pointed out this cannot be explained in layman terms: Here are the papers which you would like to see: &lt;a href=&quot;http://www.isibang.ac.in/~sury/ramanujanday.pdf&quot; rel=&quot;nofollow&quot;&gt;www.isibang.ac.in/~sury/ramanujanday.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.isibang.ac.in/~sury/episq163.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.isibang.ac.in/~sury/episq163.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" LastActivityDate="2010-09-13T16:05:10.260" />
  <row Id="4546" PostTypeId="2" ParentId="4544" CreationDate="2010-09-13T16:05:34.727" Score="17" ViewCount="" Body="&lt;p&gt;This is quite a challenge to express in &quot;layman's terms&quot;, but the&#xA;reason is that&#xA;$$j\left(\frac{1+\sqrt{-163}}{2}\right)$$&#xA;is an integer where $j$ is the &lt;a href=&quot;http://en.wikipedia.org/wiki/J-invariant&quot; rel=&quot;nofollow&quot;&gt;$j$-function&lt;/a&gt;. When you substitute&#xA;$(1+\sqrt{-163})/2$ into the $q$-expansion (see the wikipedia page)&#xA;of $j$, all terms save the first two are small, and the first two equal&#xA;$$-\exp(\pi\sqrt{163})+744.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason that this $j$-value is an integer is due to the quadratic&#xA;field $\mathbb{Q}(\sqrt{-163})$ having&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Class_number_%28number_theory%29&quot; rel=&quot;nofollow&quot;&gt;class number&lt;/a&gt; one, or equivalently that all positive-definite&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Binary_quadratic_form&quot; rel=&quot;nofollow&quot;&gt;integer binary quadratic forms&lt;/a&gt; of discriminant $-163$ are equivalent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added&lt;/strong&gt;&#xA;I'll try to explain the connection with binary quadratic forms. Consider&#xA;a quadratic form&#xA;$$Q(x,y)=ax^2+bxy+cy^2$$&#xA;with $a$, $b$ and $c$ integers. I'll only consider forms $Q$ which are&#xA;&lt;em&gt;primitive&lt;/em&gt;, so that $a$, $b$ and $c$ have no common factor $ &gt; 1$,&#xA;and &lt;em&gt;positive-definite&lt;/em&gt;, that is $a &gt; 0$ and the &lt;em&gt;discriminant&lt;/em&gt;&#xA;$D=b^2-4ac &amp;lt; 0$. There is a notion of equivalence of quadratic forms,&#xA;and two primitive positive-definite forms $Q$ and $Q'(x,y)=a'x^2+b'xy+c'y^2$&#xA;(necessarily also of discriminant $D$) are equivalent if and only if&#xA;$$j\left(\frac{b+\sqrt{-D}}{2a}\right)&#xA;=j\left(\frac{b'+\sqrt{-D}}{2a'}\right).$$&#xA;For each possible discriminant there are only finitely many equivalence&#xA;classes. Thus we get a finite set of $j$-values for each discriminant, and&#xA;the big theorem is that they are the solutions of a monic algebraic equation&#xA;with integer coefficients. When there is only one class the equation has the&#xA;form $x-k=0$ where $x$ is an integer, and the $j$-value must be an integer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My recommended reference for this is David Cox's book&#xA;&lt;a href=&quot;http://books.google.co.uk/books?id=pSMlAQAAIAAJ&amp;amp;q=cox+primes&amp;amp;dq=cox+primes&amp;amp;hl=en&amp;amp;ei=0E2OTIKIHZOCswb3ysmAAg&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=1&amp;amp;ved=0CCgQ6AEwAA&quot; rel=&quot;nofollow&quot;&gt;Primes of the form $x^2+ny^2$&lt;/a&gt;. But these results appear towards&#xA;the end of this 350-page book.&lt;/p&gt;&#xA;" OwnerUserId="226" LastEditorUserId="226" LastEditorDisplayName="" LastEditDate="2010-09-13T16:23:36.007" LastActivityDate="2010-09-13T16:23:36.007" CommentCount="1" />
  <row Id="4547" PostTypeId="2" ParentId="4537" CreationDate="2010-09-13T16:13:09.910" Score="2" ViewCount="" Body="&lt;p&gt;Well, I'm not an expert in Number Theory and I don't know if your interests may include in the future things like &lt;a href=&quot;http://en.wikipedia.org/wiki/Weil_conjectures&quot; rel=&quot;nofollow&quot;&gt;Weil conjectures&lt;/a&gt; and $\zeta$-functions for algebraic varieties, or on the contrary, you'll try to avoid any contact with Algebraic Geometry and Homological Algebra.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, just in case, your interests lead you towards the first issues, then you'll have a nasty encounter with something called $\ell$-adic and étale cohomologies. I wouldn't like to be in your shoes at that moment, without having seen before any other simpler cohomology (as the singular cohomology you're going to learn in Algebraic Topology) and the classical Lefchstez fixed-point theorem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More generally, except you're not going to use Algebraic Geometry at all in your research, nor Homological Algebra, or your use of the first is limited to the most classical aspects, you'll have to be familiar with sheaf cohomology and derived functors such as $\mathrm{Ext}$ and $\mathrm{Tor}$. It isn't impossible to learn sheaf cohomology without knowing a word of singular cohomology, and the same applies for derived functors, but you'll clearly have a tremendous gap at that point.&lt;/p&gt;&#xA;" OwnerUserId="664" LastActivityDate="2010-09-13T16:13:09.910" CommunityOwnedDate="2010-09-13T16:13:09.910" />
  <row Id="4548" PostTypeId="2" ParentId="4544" CreationDate="2010-09-13T16:14:10.200" Score="14" ViewCount="" Body="&lt;p&gt;I do not think &quot;why&quot; has a reasonable layman's-terms answer, but let me at least explain &quot;how&quot; with a simpler example of such a numerical coincidence.  If one takes powers of the golden ratio $\phi = \frac{1 + \sqrt{5}}{2}$ it is not hard to see that they are close to integers.  For example, $\phi^{20} = 15126.999934...$.  One might ask an analogous question about why these numbers are close to integers.  The answer is that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\phi^n + \varphi^n = L_n$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $L_n$ is the $n^{th}$ &lt;a href=&quot;http://en.wikipedia.org/wiki/Lucas_number&quot; rel=&quot;nofollow&quot;&gt;Lucas number&lt;/a&gt; (an integer), and where $\varphi = \frac{1 - \sqrt{5}}{2}$ has absolute value less than $1$.  As $n$ gets larger, $\varphi^n$ gets smaller, so $\phi^n$ becomes a better and better approximation to the integer $L_n$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A similar, but much more complicated, phenomenon is happening here.  The reason $e^{\pi \sqrt{163}}$ is so close to an integer is that it, plus a small error term, gives a special formula for that integer.  But &quot;why&quot; this formula exists is a rather &lt;a href=&quot;http://en.wikipedia.org/wiki/Complex_multiplication&quot; rel=&quot;nofollow&quot;&gt;long and complicated story&lt;/a&gt; (as Robin Chapman hints at) and I do not think there is any reasonable way to talk about it in layman's terms.  &lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-09-13T16:14:10.200" CommentCount="2" />
  <row Id="4549" PostTypeId="2" ParentId="4512" CreationDate="2010-09-13T17:03:52.947" Score="3" ViewCount="" Body="&lt;p&gt;Your use of the term &quot;truth assignments&quot; indicates that you are asking this question in the context of propositional logic, rather than the context of model theory and first order logic, where it often arises. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer is Yes, there are $K$ for which $K\neq Mod(Th(K))$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the propositional logic context, we have a collection $K$ of truth assignments of a fixed set of propositional variables to {true,false}. That is, $K$ is a set of rows in the truth table. In this case, $Th(K)$ is the set of propositional assertions true in all those rows, and $Mod(Th(K))$ is the set of all models of all those assertions. Since the rows appearing in $K$ do model all those assertions, we see that immediately that $K\subset Mod(Th(K))$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But here is a counterexample showing that $K\neq Mod(Th(K))$ is possible. Suppose that there are infinitely many propositional variables $p_0,p_1,\ldots$, and let $K$ be the rows for which at most finitely many $p_n$ are true. In this case, I claim $K\neq Mod(Th(K))$. The reason is that any statement $\varphi$ involves only finitely many variables, and so if it is not tautological, that is, if it fails anywhere, then it will fail in a model for which only finitely many variables are true (just ignore all variables not in $\varphi$). Thus, $Th(K)$ includes only the tautologies, and so $Mod(Th(K))$ includes all truth assignments. So this is a counterexample. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is not difficult to see that any counterexample will involve infinitely many propositional variables, since if there are only finitely many variables, then the assingments in $K$ can be explicitly described and the theory axiomatized. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the context of first order logic, we take $K$ to be a set of first order structures, and there are numerous examples of $K$ not forming an elementary class, as in the link provided by Asaf. One example is the collection $K$ of all countably infinite structures (of a fixed consistent theory in a countable language). This will not be $Mod(Th(K))$, since by Lowenheim-Skolem, there are arbitrary large structures of any consistent theory.&lt;/p&gt;&#xA;" OwnerUserId="413" LastActivityDate="2010-09-13T17:03:52.947" CommentCount="2" />
  <row Id="4550" PostTypeId="2" ParentId="4536" CreationDate="2010-09-13T17:08:04.447" Score="0" ViewCount="" Body="&lt;p&gt;To start with: the sequence can be written as &#xA;\begin{equation}&#xA;    x_{n+1} = 2 x_{n} + e_{n+1} x_{n-1}   \; [1]&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $e_n$ is a sequence of iid Bernoully variables taking the values $(1,-1)$ with equal probabilities. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, taking expectations, and because $e_{n+1}$ and $x_{n-1}$ are independent, it results that $E(x_n) = 2^{n-1}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, I'd say that, if the sequence $x_n^{1/n}$ converges, it should converge to 2. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;== continued ==&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's see if we can compute the variance.&#xA;Multiplying by $x_{n+1}$ and taking expectations:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;  E(x^2_{n+1})=2E(x_n x_{n+1}) + E(e_{n+1} x_{n+1} x_{n-1})  \; [2]&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To compute the first term we multiply [1]  by  $x_{n}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;  E(x_n x_{n+1})= 2 E(x_n^2) + E(e_{n+1} x_{n} x_{n-1}) = 2 E(x_n^2)  \; [3]&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To compute the second term we multiply [1] by  $e_{n+1} x_{m-1}$ and recall that $e_{n+1}^2 =1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;   E(e_{n+1} x_{n+1} x_{n-1}) =  2 E(e_{n+1} x_{n-1} x_{n}) + E(e_{n+1}^2) = E(e_{n+1}^2)  \; [4]&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;   E(x_{n+1}^2) = 4 E(x^2_n) + E(x^2_{n-1}) \; [5]&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;====&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Updated: this doesnt lead where I expected to ... I thought that the linearly-normalized sequence (say, $x_n / 2^n$ ) would have an asymtotically vanishing variance, but this does not appear to be so.&lt;/p&gt;&#xA;" OwnerUserId="312" LastEditorUserId="312" LastEditorDisplayName="" LastEditDate="2010-09-13T18:33:08.490" LastActivityDate="2010-09-13T18:33:08.490" CommentCount="5" />
  <row Id="4551" PostTypeId="1" AcceptedAnswerId="4552" CreationDate="2010-09-13T17:21:45.143" Score="1" ViewCount="1001" Body="&lt;p&gt;If A,B non empty, upper bounded sets and &lt;strong&gt;A+B={a+b}&lt;/strong&gt;, how can I prove that &lt;strong&gt;sup(A+B)=supA+supB&lt;/strong&gt;?&lt;/p&gt;&#xA;" OwnerUserId="1862" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2010-11-21T23:08:43.327" LastActivityDate="2010-11-21T23:08:43.327" Title="How can I prove sup(A+B)=supA+supB if A+B={a+b}" Tags="&lt;elementary-set-theory&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="4552" PostTypeId="2" ParentId="4551" CreationDate="2010-09-13T17:25:11.050" Score="5" ViewCount="" Body="&lt;p&gt;Show that $\sup(A+B)$ is less than or equal to $\sup(A)+\sup(B)$ by showing that the latter is an upper bound for $A+B$. Then show that $\sup(A)+\sup(B)$ is less than or equal to $\sup(A+B)$ by showing that $\sup(A+B)$ is an upper bound for $A+\sup(B)$ and that $\sup(A+\sup(B)) = \sup(A)+\sup(B)$. &lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-09-13T17:25:11.050" />
  <row Id="4553" PostTypeId="2" ParentId="4551" CreationDate="2010-09-13T17:27:31.970" Score="1" ViewCount="" Body="&lt;p&gt;Set x=supA, y=supB. Given $a\in A, b\in B$ we have $a\le x, b\le y$ so $a+b\le x+y$ so it's an upper bound. Now take $\varepsilon &gt; 0$ and find $a,b$ such that $a&gt;x-\varepsilon/2, b&gt;y-\varepsilon/2$ and you have $a+b &gt; x+y - \varepsilon$. That suffices (since it means that every potential &quot;smaller upper bound&quot; $x+y-\varepsilon$ is not really an upper bound).&lt;/p&gt;&#xA;" OwnerUserId="1818" LastActivityDate="2010-09-13T17:27:31.970" />
  <row Id="4554" PostTypeId="2" ParentId="4526" CreationDate="2010-09-13T17:33:27.877" Score="4" ViewCount="" Body="&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Wronskian&quot; rel=&quot;nofollow&quot;&gt;wronskian&lt;/a&gt;, named after Józef Hoene-Wroński, sometimes appears uncapitalised (though it doesn't on Wikipedia, it does in several of the first hits on Google Scholar).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The wronskian of a set of n functions, as a function of x, is defined as the determinant of the matrix whose (i,j)'th entry is the (i-1)'th derivative of the j'th function, evaluated at x.&lt;/p&gt;&#xA;" OwnerUserId="1835" LastActivityDate="2010-09-13T17:33:27.877" CommunityOwnedDate="2010-09-13T17:33:27.877" CommentCount="1" />
  <row Id="4556" PostTypeId="1" AcceptedAnswerId="4559" CreationDate="2010-09-13T17:42:43.110" Score="2" ViewCount="177" Body="&lt;p&gt;Let $c$ be a postive integer that is not prime. Show that there is some positive integer $b$ such that $b|c$ and $b \leq \sqrt{c}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know this can be proved by contradiction, but I'm not sure how to approach it. Usually I write the proof in the form $P \to Q$, and then if we can prove $P \land \neg Q$ is false, $P \to Q$ must be true.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case, I wrote it as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $c$ is a composite, positive integer, &lt;b&gt;then&lt;/b&gt; $b|c$ and $b \leq \sqrt{c}$, for some positive integer $b$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm guessing that as long as I assume that $b \not| c$ or $b &gt; \sqrt{c}$, then this is still valid as $\neg Q$- that is, I don't have to assume the converse of both parts of Q?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moving on, if $b &gt; \sqrt{c}$, and $b|c$, then $br=c$ for some integer $r$, which means $r &amp;lt; \sqrt{c}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And this is where I get stuck.&lt;/p&gt;&#xA;" OwnerUserId="1868" LastActivityDate="2010-09-13T17:57:54.647" Title="Understanding proof by contradiction" Tags="&lt;number-theory&gt;&lt;proof-strategy&gt;&lt;contradiction&gt;" AnswerCount="3" CommentCount="5" FavoriteCount="2" />
  <row Id="4557" PostTypeId="2" ParentId="4556" CreationDate="2010-09-13T17:52:06.207" Score="1" ViewCount="" Body="&lt;p&gt;What you want is to assume that &lt;strong&gt;every&lt;/strong&gt; b that divides c is &quot;too large&quot; and derive a contradiction; however, you don't need proof by contradiction here at all. If b divides c and is too large, then it's easy to show directly that $c/b$ also divides c and is small enough.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This also can be phrased with contradiction: assume $c/b &gt; \sqrt{c}$ and $b &gt; \sqrt{c}$ then $c=(c/b)\cdot b &gt; \sqrt{c}\cdot\sqrt{c}=c$ - contradiction.&lt;/p&gt;&#xA;" OwnerUserId="1818" LastActivityDate="2010-09-13T17:52:06.207" />
  <row Id="4558" PostTypeId="2" ParentId="4556" CreationDate="2010-09-13T17:55:28.150" Score="0" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;I:&lt;/strong&gt; That there exists some $b$ with $b|c$ is true by the definition (c is not prime).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we assume that - as &lt;strong&gt;II&lt;/strong&gt; - there is no $b \leq \sqrt c$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since there &lt;strong&gt;has&lt;/strong&gt; to be &lt;em&gt;some&lt;/em&gt; $b$ due to &lt;strong&gt;I&lt;/strong&gt;, this $b$ now has to meet $b &gt; \sqrt c$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now $z = \frac{c}{b}$ surely is an integer that divides $c$ too, and, furthermore, $z \leq \sqrt c$. Therefore, by assuming &lt;strong&gt;II&lt;/strong&gt;, we disprove &lt;strong&gt;II&lt;/strong&gt; - Contradiction, q.e.d.&lt;/p&gt;&#xA;" OwnerUserId="917" LastActivityDate="2010-09-13T17:55:28.150" />
  <row Id="4559" PostTypeId="2" ParentId="4556" CreationDate="2010-09-13T17:57:54.647" Score="5" ViewCount="" Body="&lt;p&gt;When you negate statements with quantifiers, you need to flip the quantifiers.  For example, the opposite of &quot;every dog is black&quot; is not &quot;every dog is not black,&quot; but &quot;there exists some dog which is not black.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case, the correct negation of &quot;there exists some $b$ such that $b | c$ and $b \le \sqrt{c}$,&quot; if you want to keep the condition that $b$ is a divisor, is &quot;for all $b$ such that $b | c$, $b &gt; \sqrt{c}$.&quot;  This is the statement from which you can prove a contradiction: for every $b$ that satisfies this condition, $\frac{c}{b}$ must also satisfy this condition, giving the contradiction $c &gt; c$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(And everywhere you want the condition that $b &gt; 1$ or else the problem is trivial as KennyTM says in the comments.)&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-09-13T17:57:54.647" CommentCount="1" />
  <row Id="4560" PostTypeId="2" ParentId="2969" CreationDate="2010-09-13T18:05:53.573" Score="2" ViewCount="" Body="&lt;p&gt;While you talk about pruning the search tree through symmetry, generally the best way I've seen to approach these problems is by breaking cells down into classes or layers; for instance, many of the classic polyomino non-tiling problems can be solved with suitable colorings, and constraints provided by colorings can help reduce the search space immensely (e.g., the Soma cube result that the 'T' piece must have its spine on an edge of the cube).  The thing that jumps out at first glance is the 'intermediate' layer of cells, the 26 cells that are neither on the outermost face of the cube nor the center cell.  Whatever piece occupies the center must also occupy at least (and in fact, exactly) three of these cells, and any piece that's not wholly on an outer layer occupies at least one, with many placements forcing several cells to be occupied; it seems intuitively like they might be at a premium and that that's a constraint worth investigating.  The section on the Soma cube in the last volume of &lt;em&gt;Winning Ways For Your Mathematical Plays&lt;/em&gt; has some discussion of these sorts of colorings, and that might be a good place to start.&lt;/p&gt;&#xA;" OwnerUserId="785" LastActivityDate="2010-09-13T18:05:53.573" />
  <row Id="4562" PostTypeId="1" AcceptedAnswerId="4563" CreationDate="2010-09-13T18:18:45.433" Score="0" ViewCount="382" Body="&lt;p&gt;If A, B two non empty sets of real numbers and for every a from A and b from B&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;a &amp;lt; b&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;how can I prove that &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;supA ≤ infB&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;?&lt;/p&gt;&#xA;" OwnerUserId="1862" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2010-11-25T16:59:19.413" LastActivityDate="2010-11-25T16:59:19.413" Title="If a &lt; b =&gt; sup A ≤ inf B" Tags="&lt;real-analysis&gt;&lt;elementary-set-theory&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="1" />
  <row Id="4563" PostTypeId="2" ParentId="4562" CreationDate="2010-09-13T18:23:33.673" Score="4" ViewCount="" Body="&lt;p&gt;I don't think this is true. Let $A=(0,1)$ and $B=(1,2)$ then $supA=1$ and $infB=1$ but $A$ and $B$ satisfy your conditions. &lt;/p&gt;&#xA;" OwnerUserId="1794" LastActivityDate="2010-09-13T18:23:33.673" CommentCount="3" />
  <row Id="4564" PostTypeId="2" ParentId="4536" CreationDate="2010-09-13T18:29:37.443" Score="3" ViewCount="" Body="&lt;p&gt;If you replace 2 with 1 in your recurrence, you get the &lt;a href=&quot;http://en.wikipedia.org/wiki/Random_Fibonacci_sequence&quot; rel=&quot;nofollow&quot;&gt;random Fibonacci sequence&lt;/a&gt;. &lt;a href=&quot;http://www.ams.org/journals/mcom/2000-69-231/S0025-5718-99-01145-X/home.html&quot; rel=&quot;nofollow&quot;&gt;Viswanath has proven&lt;/a&gt; that the random Fibonacci sequence almost surely has exponential growth (and gives the growth constant).  I haven't carefully read the proof, though, so I can't comment on how it could be adapted to the case you're interested in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.jstor.org/stable/2237962&quot; rel=&quot;nofollow&quot;&gt;Kesten and Furstenberg&lt;/a&gt; showed quite a while ago that the limit you say is around 1.91 at least exists, by restating the problem as one about multiplication of random matrices.&lt;/p&gt;&#xA;" OwnerUserId="173" LastActivityDate="2010-09-13T18:29:37.443" CommentCount="1" />
  <row Id="4565" PostTypeId="2" ParentId="350" CreationDate="2010-09-13T18:42:45.433" Score="0" ViewCount="" Body="&lt;p&gt;For quick review Schaum's Discrete Mathematics is good.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have time, read &lt;strong&gt;Foundations of Computer Science&lt;/strong&gt; By Aho-Ullman. It's free and available online. Around 800 pages.&lt;/p&gt;&#xA;" OwnerUserId="1869" LastActivityDate="2010-09-13T18:42:45.433" CommunityOwnedDate="2010-09-13T20:09:02.853" />
  <row Id="4566" PostTypeId="2" ParentId="4510" CreationDate="2010-09-13T18:43:31.310" Score="3" ViewCount="" Body="&lt;p&gt;If $$f(x) = \frac{P(x)}{Q(x)}$$ where $P,Q$ are polynomials, then we have that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f(x)Q(x) = P(x)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now use Leibniz's formula for differentiating a product&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{r=0}^{n} {n \choose r} f^{r}(x) Q^{n-r}(x) = P^{n}(x)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which you can evaluate at $0$ for $n=1,2,\dots$ and you can obtain $(n+1)^{th}$ derivative of $f$ at $0$ from the first $n$ derivatives using the above formula.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The coefficient you need would be $\frac{f^{n}(0)}{n!}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This should be easily doable by a computer. No integration etc needed.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-09-13T18:43:31.310" />
  <row Id="4567" PostTypeId="2" ParentId="4495" CreationDate="2010-09-13T18:47:28.640" Score="2" ViewCount="" Body="&lt;p&gt;Some intuition is afforded by the Jordan Normal Form, and even more is obtained by understanding a proof.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Jordan form of an endomorphism decomposes it as the direct sum of &quot;Jordan blocks&quot;; if you have an intuition for what a block is doing, you can understand the entire direct sum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A Jordan block geometrically is the sum of two operations.  One rescales everything by a constant, the eigenvalue.  The other essentially collapses the module onto a codimension one submodule.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The collapsing is the nilpotent part.  A geometric intuition is obtained by considering the simplest nontrivial example, the linear map defined (in coordinates) by $(x,y) \to (y,0)$: two dimensions are collapsed onto one essentially by forgetting the first dimension.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a Jordan block, the nilpotent operation is generally $(x,y, \ldots, z) \to (y, \ldots, z, 0)$.  This establishes a hierarchy in the module: the last dimension generates the kernel of $T$, the last two dimensions are killed by $T^2$, and so on.  Iterating $T$ enough times eventually produces the zero map--that's what it means to be nilpotent.  Thus the kernel of $(T-\lambda I)^k$ picks up all the Jordan blocks associated with eigenvalue $\lambda$ and, speaking somewhat loosely, each generalized eigenvector gets rescaled by $\lambda$, &lt;em&gt;up to some &quot;error&quot; term&lt;/em&gt; generated by certain of the other generalized eigenvectors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In two dimensions, then, a Jordan block effects a transformation that can be written in suitable coordinates (given by two of the generalized eigenvectors) as $(x,y) \to (\lambda x + y, \lambda y)$: an isothety and skew transformation together.  The geometric picture isn't really any different for higher-dimensional blocks.&lt;/p&gt;&#xA;" OwnerUserId="1489" LastActivityDate="2010-09-13T18:47:28.640" />
  <row Id="4568" PostTypeId="2" ParentId="153" CreationDate="2010-09-13T18:56:55.120" Score="2" ViewCount="" Body="&lt;p&gt;It seems you want to give it a try and play around a while before you jump into, if at all, seriously. I recommend you learn Python. Then use Sage. You'll have lot of fun, I bet. It's awesome. Learning python not only will lead you to sage, it'll lead you to several other python based computing environments. See python's website for further reference.&lt;/p&gt;&#xA;" OwnerUserId="1869" LastActivityDate="2010-09-13T18:56:55.120" />
  <row Id="4569" PostTypeId="2" ParentId="2723" CreationDate="2010-09-13T19:00:08.430" Score="2" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://www.khanacademy.org/&quot; rel=&quot;nofollow&quot;&gt;Khan Academy&lt;/a&gt; offers dozens of videos on particular Algebra topics, among thousands of videos on other academic topics.&lt;/p&gt;&#xA;" OwnerUserId="1869" LastEditorUserId="1995" LastEditorDisplayName="" LastEditDate="2010-09-27T02:52:58.960" LastActivityDate="2010-09-27T02:52:58.960" CommunityOwnedDate="2010-09-13T19:00:08.430" />
  <row Id="4571" PostTypeId="2" ParentId="4497" CreationDate="2010-09-13T19:40:22.727" Score="1" ViewCount="" Body="&lt;p&gt;It seems to me that the limit of the function does not exist, since the function approaches to different limits along different ways. However, we can compute iterated limits:&#xA;\begin{equation}&#xA;\lim_{x\rightarrow\infty}\lim_{y\rightarrow\infty}\frac{2x−1}{x−1}+\frac{x}{y}=2&#xA;\end{equation}&#xA;and&#xA;\begin{equation}&#xA;\lim_{y\rightarrow\infty}\lim_{x\rightarrow y}\frac{2x−1}{x−1}+\frac{x}{y}=3.&#xA;\end{equation}&#xA;The non-existence of the double limit is also implicated by the non-equality of the iterated limits.&lt;/p&gt;&#xA;" OwnerUserId="1856" LastEditorUserId="1856" LastEditorDisplayName="" LastEditDate="2010-09-13T22:22:43.270" LastActivityDate="2010-09-13T22:22:43.270" CommentCount="4" />
  <row Id="4572" PostTypeId="1" AcceptedAnswerId="4574" CreationDate="2010-09-13T19:47:37.310" Score="6" ViewCount="695" Body="&lt;p&gt;In the question here: &lt;a href=&quot;http://math.stackexchange.com/questions/4467/a1-2-is-either-an-integer-or-an-irrational-number&quot;&gt;http://math.stackexchange.com/questions/4467/a1-2-is-either-an-integer-or-an-irrational-number&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There was an answer by Douglas S Stones:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;&#xA;If $\sqrt{a}=x/y$ where $y$ does not divide $x$, then $a=(\sqrt{a})^2=x^2/y^2$ is not an integer (since $y^2$ does not divide $x^2$), giving a contradiction.&#xA;&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Two people whose opinions I respect claimed that this proof approach was &quot;totally bogus/circular&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't really see how this is circular, or bogus for that matter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Doesn't the result follow immediately from unique factorization?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So my question is: What is wrong with that answer?&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-09-15T20:37:15.570" LastActivityDate="2010-09-15T20:37:15.570" ClosedDate="2010-09-15T23:04:46.297" Title="Is this proof of $a^{1/2}$ being either integer or irrational circular/incorrect?" Tags="&lt;number-theory&gt;" AnswerCount="3" CommentCount="38" FavoriteCount="4" />
  <row Id="4573" PostTypeId="2" ParentId="4572" CreationDate="2010-09-13T19:56:58.343" Score="2" ViewCount="" Body="&lt;p&gt;The result does follow from unique factorization, but the point being made here is that the stated claim - that if $y$ doesn't divide $x$, it's also true that $y^2$ doesn't divide $x^2$ - is &lt;strong&gt;equivalent&lt;/strong&gt; to the problem statement, so it's circular to use it to prove the problem statement (or in any case it doesn't address the meat of the problem).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit:  Let me try to make it clearer why I think this proof is not a proof in the sense that no mathematical work has been done.  We want to show that if $a$ is a positive integer, $\sqrt{a}$ is either an integer or irrational.  What does that mean?  That means if $\sqrt{a} = \frac{p}{q}$ where $p, q$ are positive integers, then $q | p$.  Equivalently, if $a = \frac{p^2}{q^2}$ where $p, q$ are positive integers, then $q | p$.  Equivalently, if $q^2 | p^2$, then $q | p$.  Equivalently, if $q$ does not divide $p$, then $q^2$ does not divide $p^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have done no mathematical work so far.  All I have done is unpack definitions.  The statement that I have ended up with is 1) exactly as hard to prove as the statement I started with, and 2) true in all of the same rings as the statement I started with.  A crucial part of the problem - that we are working in $\mathbb{Z}$ - has not yet been used.  To claim that the statement is &quot;obvious&quot; from here is to ignore an essential nontrivial property of $\mathbb{Z}$, namely that it is &lt;a href=&quot;http://en.wikipedia.org/wiki/Integrally_closed&quot; rel=&quot;nofollow&quot;&gt;integrally closed&lt;/a&gt; (which follows from unique factorization).  There is a reason this property has a name.&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-09-13T20:49:07.357" LastActivityDate="2010-09-13T20:49:07.357" CommentCount="24" />
  <row Id="4574" PostTypeId="2" ParentId="4572" CreationDate="2010-09-13T20:12:27.690" Score="3" ViewCount="" Body="&lt;p&gt;It sounds like we only need to be clear about what is being asserted here, because there shouldn't be any dispute.  Evidently you want to say that when $a$ is an integer, $\sqrt{a}$ cannot be a non-integral rational number.  Assume the contrary in order to achieve a contradiction; that is, suppose there exists a rational number $x/y$ whose square equals $a$ but which itself is non-integral.  Because $x/y$ is not an integer, $y$ does not divide $x$.  Then $y^2$ does not divide $x^2$, whence $x^2/y^2 = (x/y)^2 = a^2$ is nonintegral, the contradiction.  &lt;em&gt;Ergo,&lt;/em&gt; this is a valid argument.  (Like all logical arguments, it's a complete tautology.  But that's not a circularity!)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In response to a request in the comments below the main question, I am happy to point out that one possible justification for the final step in this demonstration is the unique factorization property of the integers.  (UF is not strictly necessary, though: the assertion that for all $x, y$ in a ring $x^2 \mid y^2 \Rightarrow x \mid y$ in and of itself does not guarantee unique factorization.)&lt;/p&gt;&#xA;" OwnerUserId="1489" LastEditorUserId="1489" LastEditorDisplayName="" LastEditDate="2010-09-15T18:22:49.877" LastActivityDate="2010-09-15T18:22:49.877" CommentCount="19" />
  <row Id="4575" PostTypeId="2" ParentId="4562" CreationDate="2010-09-13T20:24:59.137" Score="1" ViewCount="" Body="&lt;p&gt;With the edited problem: if $a\lt b$ for all $a\in A$ and $b\in B$, why is it that $\sup(A)\leq\inf(B)$? Is every element of $A$ a lower bound for $B$? If so, what does that tell you about the elements of $A$ relative to $\inf(B)$? And given that, what does &lt;em&gt;that&lt;/em&gt; tell you about $\sup(A)$?&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-09-13T20:24:59.137" />
  <row Id="4576" PostTypeId="1" CreationDate="2010-09-13T20:29:02.237" Score="3" ViewCount="171" Body="&lt;p&gt;&lt;strong&gt;The problem:&lt;/strong&gt; Let X be a product of equidimensional spheres of arbitrary dimension, say k, and G a finite group acting freely on X. Assume that the induced G-action on the Z_2-cohomology ring of X is trivial. We then have a Serre spectral sequence with the E_2-page given by the expression H*(G; H*(X; Z_2)) and converging to H*(X/G; Z_2).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first non-zero differential is certainly d^{k+1}. Let M be the image of the transgression (hence M is a subgroup of H^{k+1}(G; Z_2)). Now suppose that M = 0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do I deduce that E_{\infty} = E_{k+2} = E_2? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Background:&lt;/strong&gt; I've been trying to learn some spectral sequences these past few days. After a while I had decided I kind of have a grasp of what is going on and have been trying to understand some research-level examples. The one described above comes from the paper &lt;em&gt;Free compact group actions on products of spheres&lt;/em&gt; by R. Oliver. In fact, he takes G to be the alternating group on four letters, but I don't think this has anything to do with my question. (My guess is that it &lt;em&gt;somehow&lt;/em&gt; follows from the multiplicative structure of the spectral sequence.)&lt;/p&gt;&#xA;" OwnerUserId="1875" LastActivityDate="2010-10-28T16:44:08.287" Title="Why does this particular Serre spectral sequence collapse at the E_{k+2}-page?" Tags="&lt;algebraic-topology&gt;" AnswerCount="1" FavoriteCount="2" />
  <row Id="4577" PostTypeId="1" CreationDate="2010-09-13T20:34:38.340" Score="0" ViewCount="109" Body="&lt;p&gt;The following is an example given by my professor, but there is an equality that I don't understand. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let a partially ordered set $(S,\preceq)$ is the union of three sets such that $S=X\cup Y\cup Z$ such that $x\preceq y$ and $x\preceq z$ for all $x\in X, y\in Y, z\in Z$, but no $y\in Y$ and no $z\in Z$ are comparable. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let $T=S\cup$ {$\nu,\eta$} and $T'=S\cup$ {$\nu$} where&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x\prec\nu\prec y$ for all $x\in X$ and all $y\in Y$,&#xA;$x\prec\eta\prec z$ for all $x\in X$ and all $z\in Z$,&#xA;but $\nu$ and $\eta$ are not comparable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It continues to say that $L(Y)=(\nu]\cap S=X=L(Z)$. However, I don't see how the first equality holds. Since $\nu=\inf_{T'}Y$, it would follow that $\nu\in L(Y)$, but isn't $\nu\notin (\nu]\cap S$, since $\nu\notin S$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps I have mistook $L(Y)$ as $L_{T'}(Y)$, and so it is actually not the case that $\nu\in L(Y)$? Thanks for any answer to my doubts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Also, is there a different way to TeX { and } on this site? { and \left{ don't seem to translate.)&lt;/p&gt;&#xA;" LastActivityDate="2011-01-12T04:44:44.183" Title="Set Equality between Lower Bounds and and an Intersection of a Subset and Interval" Tags="&lt;set-theory&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="4578" PostTypeId="2" ParentId="4526" CreationDate="2010-09-13T20:45:24.727" Score="3" ViewCount="" Body="&lt;p&gt;Here they talk about &lt;a href=&quot;http://en.wikipedia.org/wiki/Abelian_and_tauberian_theorems&quot; rel=&quot;nofollow&quot;&gt;abelian and tauberian theorems&lt;/a&gt; after (Niels-Henrik Abel and Alfred Tauber), however it is also common to use &lt;a href=&quot;http://www.britannica.com/EBchecked/topic/1084/Abelian-theorem&quot; rel=&quot;nofollow&quot;&gt;Abelian theorem&lt;/a&gt; and &lt;a href=&quot;http://books.google.com/books?id=_G4k2onFlQwC&amp;amp;pg=PA65&amp;amp;lpg=PA65&amp;amp;dq=Adjective+tauberian+Lennart+carleson&amp;amp;source=bl&amp;amp;ots=n7eubjLKk7&amp;amp;sig=xIDZ82wz-eugYyDGox7ipnDHILs&amp;amp;hl=en&amp;amp;ei=U4qOTOqrK4nHswafh6WVAg&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=1&amp;amp;ved=0CBIQ6AEwAA#v=onepage&amp;amp;q=Adjective%20tauberian%20Lennart%20carleson&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;Tauberian theorem&lt;/a&gt;. Also, in the last paper, by Lennart Carleson, we can read about &quot;this thread&quot; -- that is about people that becomes adjectives.&lt;/p&gt;&#xA;" OwnerUserId="1154" LastActivityDate="2010-09-13T20:45:24.727" CommunityOwnedDate="2010-09-13T20:45:24.727" />
  <row Id="4579" PostTypeId="2" ParentId="4577" CreationDate="2010-09-13T20:48:38.043" Score="1" ViewCount="" Body="&lt;p&gt;I do not think you know that $\nu=\inf_{T'}(Y)$. Since $S\subset T'$, you know that if $\inf_{T'}(Y)$ and $\inf_S(Y)$ both exist, then $\inf_S(Y)\preceq\inf_{T'}(Y)$; but you could have equality (e.g., if $Y$ has a minimum). If you have equality, then $\nu$ is not the infimum of $Y$ in $T'$. You do know that $\nu\preceq\inf_{T'}(Y)$, because $\nu$ is a lower bound for $Y$ in $T'$, but the  inequality could be strict. I would argue that you do not know that $L(Y)=(\nu]\cap S$, because it excludes the possibility that there is a $y\in Y$ that is a lower bound for $Y$. For the same reason, $L_{T'}(Y)$ cannot equal $(\nu]\cap S$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, presumably you are also assuming that $Y$ has no minimum. If that's the case, then I think that $L(Y)$ is meant to be $L_S(Y)$, rather than $L_{T'}(Y)$ or $L_T(Y)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S. I'm assuming that $L(A)$ means &quot;lower bounds of $A$&quot;; it's possible it is supposed to mean &quot;strict lower bounds of $A$&quot; (that is, $L(A) = \{x|x\prec y$ for all $y\in Y\}$), in which case my objection above vanishes. In any case, I suspect that $L(Y)$ is meant to be $L_S(Y)$, $S$ being the set that we started with so the one that it makes sense to take as understood from context.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-09-14T04:19:34.840" LastActivityDate="2010-09-14T04:19:34.840" />
  <row Id="4580" PostTypeId="1" CreationDate="2010-09-13T20:53:44.400" Score="0" ViewCount="74" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Possible Duplicate:&lt;/strong&gt;&lt;br&gt;&#xA;  &lt;a href=&quot;http://math.stackexchange.com/questions/4561/show-process-of-solution&quot;&gt;Show process of solution? &lt;/a&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;p&gt;Show process of solution? [reposted]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clarification: no python program or programming needed/necessary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;cryptogram - two variations - most likely by substitution - please help me solve:&#xA;Explain the process I need to take. Thanks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;cfcfcf&lt;br&gt;&#xA;adgddlb&lt;br&gt;&#xA;cfcicmc&lt;br&gt;&#xA;gddo&lt;br&gt;&#xA;bcfcfc&lt;br&gt;&#xA;dgddg&lt;br&gt;&#xA;cfcjcfc&lt;br&gt;&#xA;dgdhgo&lt;br&gt;&#xA;cfcfb&lt;br&gt;&#xA;ehk&lt;br&gt;&#xA;cb  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(or)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;ab&lt;br&gt;&#xA;cdccdcdce&lt;br&gt;&#xA;fgfgfgfgf&lt;br&gt;&#xA;cdcdcdcdchc&lt;br&gt;&#xA;fifjfb&lt;br&gt;&#xA;cdcdcdchk&lt;br&gt;&#xA;flmgfgn&lt;br&gt;&#xA;bcoco  &lt;/p&gt;&#xA;" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-09-13T21:13:51.697" LastActivityDate="2010-09-13T21:13:51.697" ClosedDate="2010-09-13T21:39:22.320" Title="Show process of solution? [reposted]" Tags="&lt;cryptography&gt;" CommentCount="7" />
  <row Id="4581" PostTypeId="1" AcceptedAnswerId="4595" CreationDate="2010-09-13T21:05:57.657" Score="10" ViewCount="258" Body="&lt;p&gt;The following is a problem in an algebra textbook, probably a well-known fact, but I just don't know how to google it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $K/k$ be a finite field extension. Then $K(X)/k(X)$ is also finite with the same degree as $K/k$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously if $v_1,...,v_n$ is a $k$-Basis of $K$, any polynomial in $K(X)$ can be written as a $k(X)$-linear combination of $v_1,...,v_n$, but I have no idea what to do with a nontrivial denominator.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there perhaps a more elegant way of proving this?  &lt;/p&gt;&#xA;" OwnerUserId="1878" LastActivityDate="2010-09-16T11:14:54.187" Title="Why does K-&gt;K(X) preserve the degree of field extensions?" Tags="&lt;abstract-algebra&gt;&lt;field-theory&gt;" AnswerCount="5" CommentCount="1" FavoriteCount="5" />
  <row Id="4582" PostTypeId="1" AcceptedAnswerId="4583" CreationDate="2010-09-13T22:45:16.783" Score="4" ViewCount="361" Body="&lt;p&gt;I am reviewing Calc 2 material and I came across a problem which asked me to explain why $x^\pi$  does not have a Taylor Series expansion around x=0. To me it seems that it would have an expansion but it would just be 0, so maybe it's not a suitable expansion. It doesn't have any holes and it is infinitely differentiable so I don't know why it couldn't have an expansion. &lt;/p&gt;&#xA;" OwnerUserId="1880" LastActivityDate="2010-09-14T01:19:00.630" Title="Explanation of Maclaurin Series of x^pi" Tags="&lt;sequences-and-series&gt;&lt;taylor-expansion&gt;&lt;analytic&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="1" />
  <row Id="4583" PostTypeId="2" ParentId="4582" CreationDate="2010-09-13T22:58:35.483" Score="5" ViewCount="" Body="&lt;p&gt;Consider what happens for higher powered derivatives. $\frac{d}{d^{4}x}x^{\pi}=(\pi)(\pi -1)(\pi -2)(\pi -3)x^{\pi -4}$ You can think of this as $c \frac{1}{x^{k}}$ for some positive real number k which is not defined at 0.&lt;/p&gt;&#xA;" OwnerUserId="249" LastActivityDate="2010-09-13T22:58:35.483" CommentCount="3" />
  <row Id="4584" PostTypeId="1" AcceptedAnswerId="4608" CreationDate="2010-09-13T23:25:16.033" Score="2" ViewCount="85" Body="&lt;p&gt;The [infinite] Ramsey theorem states that &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Let $n$ and $k$ be natural numbers. Every partition $\{X_1,\ldots ,X_k\}$ of $[\omega]^n$ into $k$ pieces has an infinite homogeneous set. Equivalently, for every $F\colon [\omega]^n\to \{1, . . . , k\}$ there exists an infinite $H \subseteq \omega$ such that $F$ is constant on $[H]^n$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Where $[X]^k := \{ Y \subset X | |Y| = k\}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, when $k=2$ we can interpret this as coloring edges of a complete graph. But what happens when $k&gt;2$, is there some geometrical or graphic, in a similar sense, to which we can turn this state into?&lt;/p&gt;&#xA;" OwnerUserId="622" LastActivityDate="2010-09-14T07:02:56.020" Title="A &quot;geometrical&quot; representation for Ramsey's theorem" Tags="&lt;combinatorics&gt;&lt;set-theory&gt;&lt;intuition&gt;" AnswerCount="1" />
  <row Id="4585" PostTypeId="2" ParentId="4467" CreationDate="2010-09-13T23:31:58.853" Score="4" ViewCount="" Body="&lt;p&gt;Below is a simple proof of irrationality of square-roots that I discovered as a teenager (inspired by a proof of Dedekind). It employs the &lt;a href=&quot;http://en.wikipedia.org/wiki/Bezouts_identity&quot; rel=&quot;nofollow&quot;&gt;Bezout identity&lt;/a&gt; for the gcd, i.e. the gcd $\rm\:(a,b)\:$ of two integers $\rm\:a,b\:$ may be expressed as an integral linear combination of the given integers: $\rm\:\ (a,b)\ =\ a\ d - b\ c\:.\ $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;THEOREM&lt;/strong&gt; $\quad \rm r = \sqrt{n}\;\;$ is integral if rational, $\:$ for $\:\rm n\in\mathbb{N}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof&lt;/strong&gt; $\quad\ \rm r = a/b,\ \ \gcd(a,b) = 1\ \Rightarrow\ ad-bc \;=\; 1\;$  for some $\:\rm c,d \in \mathbb{Z}\;\;$ by Bezout.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus $\;\;\rm 0\ =\ (a-br)\: (c+dr) \ =\ ac-bdn \:+\: r \ \Rightarrow\ r \in \mathbb{Z}\quad\quad$ &lt;strong&gt;QED&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nowadays my favorite proof is the $1$-line gem using Dedekind's &lt;strong&gt;conductor ideal&lt;/strong&gt;. As I explained at length &lt;a href=&quot;http://mathoverflow.net/questions/30313&quot; rel=&quot;nofollow&quot;&gt;elsewhere&lt;/a&gt;, it beautifully encapsulates the denominator descent in ad-hoc &quot;elementary&quot; irrationality proofs. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;See also my &lt;a href=&quot;http://math.stackexchange.com/questions/4491&quot;&gt;other post&lt;/a&gt; here. That post concisely proves the theorem after first proving from scratch the fundamental lemma that the least positive denominator $\rm\:b\:$ of a fraction divides every other denominator, i.e. $\rm\ a/b = c/d\ \Rightarrow\ b\ |\ d\ $ if $\rm\ a/b\ $ is in lowest terms. The irrationality proof follows immediately from this principal denominator ideal theorem (&quot;unique fractionization&quot;), namely  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof&lt;/strong&gt; $\ \ $  Suppose $\displaystyle\;\rm \sqrt{n} \:= \frac{a}b\:,\;\;$ least $\rm b&amp;gt;0\:.\ \ $ $\displaystyle\rm\sqrt{n}\; = \frac{n}{\sqrt{n}} \ \Rightarrow\ \frac{a}b = \frac{nb}a \ \Rightarrow\ b\:|\:a\ \Rightarrow\ \sqrt{n}\ \in\ \mathbb Z$&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2011-03-13T21:48:36.523" LastActivityDate="2011-03-13T21:48:36.523" />
  <row Id="4586" PostTypeId="2" ParentId="4582" CreationDate="2010-09-13T23:44:32.417" Score="6" ViewCount="" Body="&lt;p&gt;I decided to elaborate on my comment above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is $x^\pi$ if $x$ is negative?  Trying to approximate with $x^r$ for rational $r$ is problematic, depending on which rational numbers you use to approximate; should the answer be positive, negative, imaginary?  Perhaps more reasonable would be to take $x^\pi=e^{\pi \log(x)}$ for some suitably chosen branch of the logarithm on $(-\infty,0)$, the most common choice being $\log x=\ln(-x)+\pi i$.  Your function then becomes &#xA; $$&#xA;   f(x) = \left\{&#xA;     \begin{array}{lr}&#xA;       e^{\pi \ln x} &amp;amp; : x&gt;0 \\&#xA;       0 &amp;amp; : x=0 \\&#xA;       e^{i\pi^2}e^{\pi \ln(-x)} &amp;amp; :x&amp;lt;0&#xA;     \end{array}&#xA;   \right.&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, it is not real-valued.  You can still go ahead and try to take its derivatives, but this piecewise representation may make it less surprising that there is going to be a singularity at zero.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-09-13T23:44:32.417" CommentCount="2" />
  <row Id="4587" PostTypeId="1" AcceptedAnswerId="4588" CreationDate="2010-09-13T23:50:36.443" Score="7" ViewCount="140" Body="&lt;p&gt;If I want to take the derivative of $ax^n$, I will get $anx^{n-1}$. If I were to take the derivative again, I get $an(n-1)x^{n-2}$.&lt;br&gt;&#xA;We can generalize this for integer k easily to get the kth derivative $a\frac{n!}{(n-k)!} x ^{n-k}$. But what about for a more general k?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does this have some name? Has it been widely studied? If so, can you show how to generalize this formula for kth derivative of $ax^n$, and explain how it works? If not, is there a good reason it is impossible?&lt;/p&gt;&#xA;" OwnerUserId="1881" LastActivityDate="2010-09-14T00:36:10.343" Title="Is it meaningful to take the derivative of a function a non-integer number of times?" Tags="&lt;calculus&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="4588" PostTypeId="2" ParentId="4587" CreationDate="2010-09-14T00:36:10.343" Score="5" ViewCount="" Body="&lt;p&gt;To expand on Jonas's comment: Yes, it makes sense. For the case of the power function, one can consider&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\Gamma(n+1)}{\Gamma(n-\alpha+1)}x^{n-\alpha}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;as the $\alpha$-th derivative of the power function $x^n$, where $\Gamma(z)$ is the gamma function, the generalization of the factorial to the complex plane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, one has a number of definitions for so-called &quot;fractional derivatives&quot;, or, as &lt;a href=&quot;http://www.sciencedirect.com/science/bookseries/00765392&quot; rel=&quot;nofollow&quot;&gt;Spanier and Oldham&lt;/a&gt; prefer to call it, the &quot;differintegral&quot;. Negative values of $\alpha$ in expressions like the one given above correspond to integration, positive values correspond to differentiation, and in general $\alpha$ can be complex.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's a lot of things to look at (Caputo derivatives, Riemann-Lioville integrals, Grunwald-Lednikov series), and I suggest you look at the book I linked to first, and then search around the web. Have fun!&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-09-14T00:36:10.343" CommentCount="1" />
  <row Id="4589" PostTypeId="1" CreationDate="2010-09-14T00:43:13.930" Score="1" ViewCount="165" Body="&lt;p&gt;I am seeking methods to see if a polynomial $f \in O(\mathbb{C^2},0)$ is irreducible. The subject is really new to my and I am studying for myself, for which I don't see about this subject.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;would thank them a lot they could provide me some bibliography where to read about this topic&lt;/p&gt;&#xA;" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-09-14T00:47:32.017" LastActivityDate="2010-09-14T02:48:01.320" Title="Irreducible Polynomials in two complex variables" Tags="&lt;reference-request&gt;&lt;algebra&gt;&lt;polynomials&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" />
  <row Id="4590" PostTypeId="2" ParentId="4582" CreationDate="2010-09-14T01:19:00.630" Score="3" ViewCount="" Body="&lt;p&gt;As a graphical supplement to Jonas's and WWright's answers:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/gK1wy.png&quot; alt=&quot;real and imaginary parts of x^\pi&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a plot of the real and imaginary parts of $(x+iy)^\pi$ in the complex plane. Note the cut running across the negative real axis. This cut is precisely the reason why you cannot have a Maclaurin expansion; polynomials cannot exhibit cuts, and a Maclaurin expansion amounts to approximating your function with a sequence of polynomials.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-09-14T01:19:00.630" CommentCount="5" />
  <row Id="4591" PostTypeId="2" ParentId="4467" CreationDate="2010-09-14T01:34:51.157" Score="2" ViewCount="" Body="&lt;p&gt;So... we can prove something pretty general here: the only rational roots of polynomials $x^n + \cdots + a_0$ with integer coefficients are integers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Indeed, suppose $p/q$ is a rational root, in lowest form, then $p^n = -q(a_0q^{n-1} + a_1pq^{n-2} + \cdots + a_{n-1}p^{n-1})$. Now, if $q&gt;1$, then any prime divisor of $q$ also divides $p^n$, and hence $p$. But this contradicts our assumption that $p/q$ is in lowest form, so we conclude that $q=1$, so the root is integral.&lt;/p&gt;&#xA;" OwnerUserId="423" LastEditorUserId="423" LastEditorDisplayName="" LastEditDate="2010-09-14T02:07:13.703" LastActivityDate="2010-09-14T02:07:13.703" CommentCount="5" />
  <row Id="4592" PostTypeId="2" ParentId="4589" CreationDate="2010-09-14T02:48:01.320" Score="2" ViewCount="" Body="&lt;p&gt;Here's one idea: if $f(x, y)$ is reducible, then the projective closure of $f(x, y) = 0$ has at least two components, which intersect by Bezout's theorem.  These intersections are singularities and therefore can be found by setting all partial derivatives to zero.  If no singularities exist then $f$ is irreducible.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-09-14T02:48:01.320" CommentCount="1" />
  <row Id="4593" PostTypeId="1" CreationDate="2010-09-14T03:01:06.940" Score="1" ViewCount="109" Body="&lt;p&gt;I want to know the tools and some source codes for generating these amazing pictures.&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Portal%3aMathematics/Featured_picture_archive&quot; rel=&quot;nofollow&quot;&gt;Featured_picture_archive in wiki&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1498" LastActivityDate="2010-09-14T03:01:06.940" ClosedDate="2010-09-14T05:40:43.403" Title="How to generate these mathematical (animated) pictures" Tags="&lt;math-software&gt;" CommentCount="3" FavoriteCount="1" />
  <row Id="4594" PostTypeId="1" AcceptedAnswerId="26122" CreationDate="2010-09-14T03:11:31.470" Score="3" ViewCount="216" Body="&lt;p&gt;It seems to be an accepted belief based on decades of experience that naive algorithms are not &#xA;adequate to solve NP-complete problems in a reasonable amount of time. Even those who believe P = NP seem to be looking for an algorithm with a very clever representation, so far without success.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can the observations above be formalized? Let's use the full, unrestricted Clique problem as an example. Suppose it could be shown that no clever representation exists for the Clique problem. That is, suppose it could be proven for every algorithm that either:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;the algorithm is correct and uses an internal representation such that no two distinct input cliques    lead to the same internal representation,&#xA;or &lt;/li&gt;&#xA;&lt;li&gt;the algorithm is incorrect. &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Would this be a worthwhile result? How important would it be? Or is it wrong?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there already a proof that CLIQUE or SAT, for example, cannot be solved in polynomial time by performing operations on cliques or Boolean expressions respectively?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;References with your answer would be helpful. Thanks&lt;/p&gt;&#xA;" OwnerUserId="1884" LastActivityDate="2011-03-10T03:46:12.637" Title="Question about the P versus NP Problem" Tags="&lt;computer-science&gt;&lt;np-complete&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="4595" PostTypeId="2" ParentId="4581" CreationDate="2010-09-14T03:38:20.003" Score="7" ViewCount="" Body="&lt;p&gt;The missing step is to show that $v_1,\ldots,v_n$ is linearly independent over $k(x)$.  So suppose that there exist rational functions $r_1(x),\ldots,r_n(x)$, not all the zero function, such that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$r_1(x) v_1 + \ldots + r_n(x) v_n = 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We may write $r_i(x) = n_i(x)/d(x)$, i.e., let $d(x)$ be a common denominator of the rational functions.  Multiplying through by $d(x)$, we get a polynomial dependence relation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$n_1(x) v_1 + \ldots + n_n(x) v_n = 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let $N(x)$ be the greatest common divisor of $n_1(x),\ldots,n_r(x)$.  (Such things exist since the polynomial ring $k[x]$ has a division algorithm, hence is a Unique Factorization Domain.)  Dividing through by $N(x)$, we get, say,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$N_1(x) v_1 + \ldots + N_n(x) v_n = 0$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with $\operatorname{gcd}(N_1,\ldots,N_n) = 1$.  In particular, not all of the $N_i$'s are divisible by $x$, so plugging in $x = 0$ gives a nontrivial linear dependence relation &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$N_1(0) v_1 + \ldots + N_n(0) v_n = 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;over $k$, a contradiction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In more sophisticated language, we are showing that the extensions $K/k$ and $k(x)/k$ are &lt;strong&gt;linearly disjoint&lt;/strong&gt;.  This can be rephrased in terms of tensor products, for  instance...but in the end the above proof is the simplest I can think of.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Added&lt;/b&gt;: Robin Chapman and Steve D are right: I misread the question and thought that the OP had already worked out that a basis $v_1,\ldots,v_n$ of $K/k$ also spans $K(x)$ over $k(x)$, but in fact this is the hardest part of the argument.  As the OP says, it is easy to see that the set of all $k(x)$-linear combinations of the $v_1,\ldots,v_n$ contains all elements of $K[x]$.  As Robin says, it is enough to show that this span also contains all elements $\frac{1}{Q(x)}$ with $Q(x) \in K[x] \setminus {0}$, and a nice way to see this is to show that every $Q \in K[x]$ divides some nonzero polynomial $q \in k[x]$, for if $Q(x) g(x) = q(x)$, $\frac{1}{Q} = = \frac{1}{q} \cdot g(x)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Robin gives a nice argument for this: essentially he extends the norm map from a finite dimensional field extension to polynomials.  I might as well complete my answer, and I might as well do it in a different way, so here goes:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is enough to assume that $Q$ is irreducible in the UFD $K[x]$, i.e., that $\mathcal{P} = (Q)$ is a prime ideal.  Put $\mathfrak{p} = \mathcal{P} \cap k[x]$.  In great generality, the restriction of a prime ideal to a subring is again a prime ideal (this is even true for the preimage of a prime ideal under an arbitrary ring homomorphism, and is very easy to show).  What we want to show is that $\mathfrak{p} \neq 0$.  But $K[x]$ is a free $k[x]$-module of dimension $n = [K:k]$.  In particular, the extension $K[x] / k[x]$ is finitely generated as a module, hence an &lt;strong&gt;integral&lt;/strong&gt; extension, and in such an extension any maximal ideal pulls back to a maximal ideal.  So $\mathfrak{p}$ is maximal, hence nonzero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These facts follow immediately from the definition of integral extensions: see e.g. Proposition 160 and Corollary 168 of &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://math.uga.edu/~pete/integral.pdf&quot; rel=&quot;nofollow&quot;&gt;http://math.uga.edu/~pete/integral.pdf&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that a possible virtue of this argument is that one does not need to treat the separable and inseparable cases differently.&lt;/p&gt;&#xA;" OwnerUserId="299" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2010-09-14T08:01:11.157" LastActivityDate="2010-09-14T08:01:11.157" CommentCount="1" />
  <row Id="4597" PostTypeId="1" AcceptedAnswerId="4598" CreationDate="2010-09-14T04:31:26.257" Score="7" ViewCount="987" Body="&lt;p&gt;Coming from an understanding of Fermat's primality test, I'm looking for a clear explanation of the Miller-Rabin primality test.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Specifically: I understand that for some reason, having non-trivial square roots of 1 mod &lt;code&gt;p&lt;/code&gt; means that &lt;code&gt;p&lt;/code&gt; is definitely composite; and I gather that you can find these non-trivial square roots by squaring &lt;code&gt;x&lt;/code&gt;, but I don't really understand what these reasons are. Specific examples of non-trivial roots of a composite number would be helpful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cheers!&lt;/p&gt;&#xA;" OwnerUserId="1885" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-09-14T06:00:27.883" LastActivityDate="2011-05-27T12:47:37.170" Title="Simple explanation and examples of the Miller-Rabin Primality Test" Tags="&lt;number-theory&gt;&lt;algorithms&gt;&lt;prime-numbers&gt;" AnswerCount="4" CommentCount="4" />
  <row Id="4598" PostTypeId="2" ParentId="4597" CreationDate="2010-09-14T05:14:20.197" Score="7" ViewCount="" Body="&lt;p&gt;Suppose $p$ was prime, and $y$ was a non-trivial square root of $1$ mod $p$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then we must have that $y^2 = 1 \mod p$ and so $(y-1)(y+1) = 0 \mod p$. This implies that either $y = 1 \mod p$ or $y = -1 \mod p$, which implies that $y$ is a trivial square root.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, if there is a non-trivial square root of $1$ mod $p$, then $p$ has to be composite.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For an example of a non trivial square root of a composite, consider $p = 15$. We have that $4^2 = 16 = 1 \mod 15$. Thus $15$ is composite.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the witness in the primality test is not necessarily a non-trivial square root of $1$ mod $p$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fact about non-trivial square roots can be used to prove that if $p$ is prime, then for any $a$ relatively prime to $p$, some power of $a$ from a given set of powers (the powers are based on the even factors of $p-1$) must be $-1$ or a specific odd power of $a$ (again based on factor of $p-1$) must be $1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If for some $a$ none of the above set of powers is $-1$ and the specific odd power is not $1$, then it must be the fact that $p$ is composite.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It can also be shown that for composite $p$, the chances of finding such $a$ is atleast $3/4$. This $a$ is the witness in the primality test and is not necessarily a non-trivial square root of $1$ mod $p$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The squaring that is done is to get the powers described above which are based on the factors of $p-1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The wiki page has really got a lot of good information (including the exact powers of $a$ that need to be taken): &lt;a href=&quot;http://en.wikipedia.org/wiki/Miller%E2%80%93Rabin_primality_test&quot; rel=&quot;nofollow&quot;&gt;Miller Rabin Primality Test&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-09-14T05:57:43.760" LastActivityDate="2010-09-14T05:57:43.760" CommentCount="2" />
  <row Id="4599" PostTypeId="2" ParentId="4597" CreationDate="2010-09-14T05:19:37.370" Score="2" ViewCount="" Body="&lt;p&gt;If $p$ is prime then $\mathbb{Z}_p$ (integers modulo $p$) is a &lt;strong&gt;field&lt;/strong&gt;. It is a basic result in algebra that in a field, a polynomial of degree $n$ has at most $n$ roots, and so the polynomial $x^2-1$ has exactly two roots: $1$ and $-1$ (which exist in every field).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $n$ is composite, then $\mathbb{Z}_n$ is never a field because not all elements have an inverse; it it well known that $a\in \mathbb{Z}_n$ has an inverse if and only if $a$ is relatively prime to $n$. Let's look at the case $n$ is the product of two primes, $n=pq$. In this case we can do arithmetic in $\mathbb{Z}_n$ by doing arithmetic in $\mathbb{Z}_p$ and $\mathbb{Z}_q$ and combining the results using the Chinese remainder theorem (which basically states that $\mathbb{Z}_n\cong\mathbb{Z}_p\times\mathbb{Z}_q$. Since $\mathbb{Z}_p$ and $\mathbb{Z}_q$ are both fields, $1$ has two roots in each of them. For every combination of a root from $\mathbb{Z}_p$ and a root from $\mathbb{Z}_q$ we'll get a root of 1 in $\mathbb{Z}_n$, meaning we'll get 4 roots of 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The major challenge of the Miller-Rabin test is to show that there is a &quot;large&quot; chance to stumble upon a non-trivial root while squaring random elements of $\mathbb{Z}_n$, and the proof, although not difficult, is not immediate either.&lt;/p&gt;&#xA;" OwnerUserId="1818" LastActivityDate="2010-09-14T05:19:37.370" CommentCount="6" />
  <row Id="4600" PostTypeId="1" AcceptedAnswerId="5210" CreationDate="2010-09-14T05:22:47.953" Score="1" ViewCount="77" Body="&lt;p&gt;I'm just starting to introduce myself to first-order logic, and I find it very confusing. As such, I'm not entirely sure how well I constructed the following. For $L$ a first-order language, let $L$ be equipped with the following nonlogical symbols: a unary function $\dot{+}$, a binary function $\dot{1}$, and a 3-ary relation $\dot{&amp;lt;}$, give an example of an $L-structure$ with an infinite universe. (Here the functions have nothing to do with $+$, $1$ and $\lt$ as they typically are.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I decided to let $\mathbb{R}$ be the universe, and for $r,s,t\in\mathbb{R}$, I defined the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\dot{+}\colon\mathbb{R}\to\mathbb{R}\colon \dot{+}(r)\mapsto r$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\dot{1}\colon\mathbb{R}\times\mathbb{R}\to\mathbb{R}\colon \dot{1}(r,s)\mapsto rs$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\dot{&amp;lt;}\subseteq\mathbb{R}\times\mathbb{R}\times\mathbb{R}\colon (r,s,t)\in \dot{&amp;lt;}$ iff $r\lt s\lt t$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is $(\mathbb{R},+,-,\times)$ then a valid $L-structure$ with an infinite universe? $\mathbb{R}$ is infinite and both functions are closed, and $\dot{&amp;lt;}$ is in the correct cartesian product, so I can't see any problems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If this is correct, I was surprised by the amount of freedom in defining the nonlogical symbols. This led me to ask if there's a particular limit to $L-structures$ depending on the universe and the functions in the language. For example, suppose I had a finite universe $U=\{a,b,c,d\}$, with a unary function $-$, and two binary functions $+,\times$. In this case, would there be $16^4$ possible $L-structures$? My thinking was that for the binary functions, there are $16$ elements in the domain, and $4$ elements in the codomain, and for $-$, there are 4 in each for a total of $(16)(4)(16)(4)(4)(4)$ ways to define all the functions, which accounts for all possible functions, relations, and constants. Or am I totally off base? Thanks for any input.&lt;/p&gt;&#xA;" OwnerUserId="1583" LastActivityDate="2010-09-22T13:21:54.113" Title="Constructing an L-structure with an Infinite Universe, and Counting L-structures on a Finite Universe." Tags="&lt;logic&gt;&lt;functions&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="4601" PostTypeId="1" AcceptedAnswerId="4602" CreationDate="2010-09-14T05:30:07.387" Score="6" ViewCount="195" Body="&lt;p&gt;The word &quot;quad&quot; generally means 4. Quadratics don't have 4 of anything. Can anyone explain where the name comes from?&lt;/p&gt;&#xA;" OwnerUserId="123" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-09-14T05:36:07.857" LastActivityDate="2010-09-14T05:36:07.857" Title="Why are quadratic equations called quadratics?" Tags="&lt;algebra-precalculus&gt;&lt;terminology&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="2" />
  <row Id="4602" PostTypeId="2" ParentId="4601" CreationDate="2010-09-14T05:35:21.427" Score="9" ViewCount="" Body="&lt;p&gt;From &lt;a href=&quot;http://mathworld.wolfram.com/Quadratic.html&quot; rel=&quot;nofollow&quot;&gt;MathWorld&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The Latin prefix quadri- is used to indicate the number 4, for example, quadrilateral, quadrant, etc. However, it also very commonly used to denote objects involving the number 2. This is the case because quadratum is the Latin word for square, and since the area of a square of side length $x$ is given by $x^2$, a polynomial equation having exponent two is known as a quadratic (&quot;square-like&quot;) equation. By extension, a quadratic surface is a second-order algebraic surface.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-09-14T05:35:21.427" CommentCount="1" />
  <row Id="4603" PostTypeId="1" AcceptedAnswerId="4605" CreationDate="2010-09-14T05:45:55.633" Score="5" ViewCount="430" Body="&lt;p&gt;I'm studying for qualifying exams and ran into this problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Show that if ${a_n}$ is a nonincreasing sequence of positive real&#xA;numbers such that $\sum_n{a_n}$ converges, then $\lim_{n \rightarrow \infty}{n a_n} = 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the definition of the limit, this is equivalent to showing&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;\forall \epsilon &gt; 0 \; \exists n_0 \; \mbox{such that} \;&#xA;|n a_n| &amp;lt; \epsilon \; \forall n &gt; n_0&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;\forall \epsilon &gt; 0 \; \exists n_0 \; \mbox{such that} \;&#xA;a_n &amp;lt; \frac{\epsilon}{n} \; \forall n &gt; n_0&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically, the terms must be bounded by the harmonic series. Thanks, I'm really stuck on this seemingly simple problem!&lt;/p&gt;&#xA;" OwnerUserId="1761" LastEditorUserId="1761" LastEditorDisplayName="" LastEditDate="2010-09-14T06:03:22.813" LastActivityDate="2010-09-14T14:03:56.783" Title="Series converges implies $\lim{n a_n} = 0$" Tags="&lt;sequences-and-series&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="1" />
  <row Id="4604" PostTypeId="2" ParentId="4597" CreationDate="2010-09-14T06:00:58.723" Score="3" ViewCount="" Body="&lt;p&gt;Below is little-known general yet simple answer to your question about nontrivial sqrts $\rm (mod\ m)\:$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;THEOREM&lt;/strong&gt; $\;$ One may quickly factor $\rm m&amp;gt;1\:$ given a polynomial with more roots mod $\rm\: m\:$ than its degree, viz. suppose that,  modulo $\rm m,\;$ the polynomial $\rm\: f(x)\ne 0\:$ has degree $\rm\:n\:$ but has $\rm\:n+1 \:$ distinct roots $\rm\:r_{\:i}\:$.  Then one of $\rm\;gcd(m\;,\:r_{\:i} - r_{\:j}\:),\; i\ne j \:$ must yield a proper factor of $\rm\:m\;$. For if that failed, then all of the gcds must be improper so $1\:,\:$ not $\rm\;m \;$ since $\rm\; i\ne j\;\Rightarrow\; r_{\:i}\: \not\equiv r_{\:j}\ (mod\ m) \:$. Now an induction using the Factor Theorem yields $\rm\;f(x) = (x-r_1)\cdots(x-r_{n+1})\; g(x),\;\; g(x) \ne 0 \;\;$ contra $\rm\;\:\deg\: f = n \;$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EXAMPLE&lt;/strong&gt; $\rm\;(deg\ f = 1)\;\:$ Suppose, modulo $\rm\:m\:$, that $\rm\; 0 \:\ne\: f \:=\: a\:x\;$ has a &quot;nontrivial&quot; root $\rm\: b\ne 0\:$. Then $\:\rm gcd(m,b)\:$ is a proper factor of $\rm\: m\:.\ $ More directly: $\rm\; m\:|\:ab\:,\;$ not $\rm\; m|a\;\:$ nor $\rm\: m|b \;\Rightarrow\; gcd(m,b)\;$ is a proper factor of $\rm\:m\:,\:$ i.e. if $\rm\: m\:$ fails the Prime Divisor Property then $\rm\: m\:$ is &quot;constructively&quot; composite.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EXAMPLE&lt;/strong&gt; $\rm\;(deg\ f = 2)\;\:$ Suppose, modulo $\rm\:m\:$, that $\rm\; x^2 = 1\;$ has a &quot;nontrivial&quot; square root $\rm\: b\ne \pm1\:$. Then $\;\rm gcd(m,b\pm 1)\;$ yields a proper factor of $\rm\: m\:$ when $\rm\: m\:$ is odd. This idea lies at the heart of many integer factorization algorithms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The above proof easily extends to yield a proof of the following&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;THEOREM&lt;/strong&gt; $\ $ Ring $\rm\: R\:$ is a domain $\iff\:$ every $\rm 0 \ne f(x)\in R[x]\:$ has at most $\rm\: deg\ f(x)\:$ roots in $\rm R$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof&lt;/strong&gt; $\ $ $(\Rightarrow)\;$ Employ induction on the polynomial degree, as in the proof of the above Theorem. $(\Leftarrow)\ \ $ If $\rm\; a \ne 0\;$ then  the polynomial $\rm\: a\:x\:$ has only the root $\rm\; x = 0\:,\:$ hence $\rm\ ab = 0 \ \Rightarrow\ b = 0\:,\:$ therefore $\rm\: R\:$ has no zero divisors.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2011-03-09T22:32:08.850" LastActivityDate="2011-03-09T22:32:08.850" CommentCount="3" />
  <row Id="4605" PostTypeId="2" ParentId="4603" CreationDate="2010-09-14T06:08:26.923" Score="6" ViewCount="" Body="&lt;p&gt;Some hints:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $S_{n} = \sum_{k=1}^{n} a_{k}$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;then what is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\lim_{n \to \infty} S_{2n} - S_{n}$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now can you use the fact that $a_{n}$ is non-increasing to upper bound a certain term of the sequence $na_{n}$ with a multiple of $S_{2n} - S_{n}$?&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-09-14T14:01:52.847" LastActivityDate="2010-09-14T14:01:52.847" CommentCount="2" />
  <row Id="4607" PostTypeId="2" ParentId="4581" CreationDate="2010-09-14T06:24:46.793" Score="8" ViewCount="" Body="&lt;p&gt;This is quite a subtle problem, and there are important differences between&#xA;the finite extension and infinite extension cases. The key is to show&#xA;that $v_1,\ldots,v_n$ form a $k(X)$-basis of $K(X)$. Pete has dealt with&#xA;their linear indepdendence but it remains to show that their $k(X)$-span is&#xA;$K(X)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $f\in K(X)$. Then $f=g/h$ where $g$ and $h$ are polynomials over $K$. The key&#xA;is that there is a non-zero polynomial $h_1$  with coefficients in $k$&#xA;such that $h\mid h_1$. In the separable case one can take $h_1$ be&#xA;the product of Galois conjugates of $h$; in general one may need to take&#xA;a positive power of that. The upshot is that $f=g_1/h_1$ where $h_1$&#xA;has coefficients in $k$. This reduces the problem to showing that $g_1$&#xA;is in the $k(X)$-span of the $v_i$, which you have already done.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The subtlety is that this breaks down for infnite non-algebraic extensions.&#xA;If $a\in K$ is transcendental over $k$ then $1/(X+a)$ is not in the $k(X)$-linear&#xA;span of the elements of $K$. In short, even though $K$ and $k(X)$ are&#xA;linearly disjoint over $k$, $K(X)\ne K\cdot k(X)$ in this case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added&lt;/strong&gt; (15/9/2010)&#xA;The real distinction isn't between finite and infinite extensions&#xA;(as I said above)&#xA;but rather between algebraic and transcendental extensions. To summarize:&#xA;the natural map&#xA;$$K\otimes_k k(X)\to K(X)$$&#xA;is always injective, and is surjective if and only if $K/k$ is&#xA;an algebraic extension.&lt;/p&gt;&#xA;" OwnerUserId="226" LastEditorUserId="226" LastEditorDisplayName="" LastEditDate="2010-09-15T17:31:28.907" LastActivityDate="2010-09-15T17:31:28.907" />
  <row Id="4608" PostTypeId="2" ParentId="4584" CreationDate="2010-09-14T06:56:55.233" Score="1" ViewCount="" Body="&lt;p&gt;Partitioning $[\omega]^{(n)}$ into $k$ pieces can be interpreted as colouring the complete $n$-uniform hypergraph, with vertex set $\omega$, in $k$ colours. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When $n=2$ the complete $2$-uniform hypergraph is the complete graph with vertex set $\omega$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a definition of hypergraph see &lt;a href=&quot;http://en.wikipedia.org/wiki/Hypergraph&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Hypergraph&lt;/a&gt;. Briefly it is the generalisation of a graph where edges can join more than two vertices. And by complete $n$-uniform I mean all the edges contain exactly $n$ vertices and every possible edge of size $n$ is in the hypergraph. &lt;/p&gt;&#xA;" OwnerUserId="1794" LastEditorUserId="1794" LastEditorDisplayName="" LastEditDate="2010-09-14T07:02:56.020" LastActivityDate="2010-09-14T07:02:56.020" CommentCount="1" />
  <row Id="4609" PostTypeId="1" AcceptedAnswerId="4664" CreationDate="2010-09-14T07:06:11.427" Score="1" ViewCount="119" Body="&lt;p&gt;I have data modelled to be Gaussian-distributed, and then displayed on a logarithmic scale and like to know some properties of the displayed data.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Effektively, I ignore that a Gaussian $X$ (with positive mean) may be negative and need mean, standard deviation and full width at half maximum of $\log X$. Unfortunately, the probability density is sufficiently ugly to prevent me from calculating this myself. Has this distribution a special name that I might look for?&lt;/p&gt;&#xA;" OwnerUserId="505" LastEditorUserId="505" LastEditorDisplayName="" LastEditDate="2010-09-14T07:30:05.237" LastActivityDate="2010-09-14T16:29:19.703" Title="Characteristics of a Gaussian on a logarithmic scale?" Tags="&lt;probability-theory&gt;&lt;normal-distribution&gt;" AnswerCount="2" CommentCount="0" />
  <row Id="4611" PostTypeId="1" AcceptedAnswerId="4613" CreationDate="2010-09-14T07:19:19.770" Score="1" ViewCount="81" Body="&lt;p&gt;Given a quadratic form with discriminant $D$, what does the class number of $\mathbb{Q}(\sqrt{D})$ tell us?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(This question is inspired by a comment on the question &lt;a href=&quot;http://math.stackexchange.com/questions/4489/proving-that-the-given-diophantine-equation-has-a-solution&quot;&gt;here&lt;/a&gt;)&lt;/p&gt;&#xA;" LastActivityDate="2010-09-14T07:34:27.720" Title="What does the class number tell us about a quadratic form?" Tags="&lt;number-theory&gt;" AnswerCount="1" />
  <row Id="4612" PostTypeId="2" ParentId="4510" CreationDate="2010-09-14T07:23:21.070" Score="2" ViewCount="" Body="&lt;p&gt;If one is computing by hand then the following small observation is useful: If $f(x) = P(x)/Q(x)$ and $Q(x) = x^d(1 - g(x))$ for some $d \geq 0$ and some polynomial $g(x)$ (which we may assume after rescaling&#xA;$P(x)$ and $Q(x)$), then using the formula for a geometric series we find that&#xA;$$f(x) = x^{-d}P(x)(1 + g(x) + g(x)^2 + \cdots ).$$&#xA;For explicit pen and paper computations, I normally find this is the simplest way to&#xA;compute the Taylor series for $f(x)$.&lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2010-09-14T07:23:21.070" />
  <row Id="4613" PostTypeId="2" ParentId="4611" CreationDate="2010-09-14T07:34:27.720" Score="3" ViewCount="" Body="&lt;p&gt;I guess you mean a binary quadratic form, i.e. of the form $a x^2 + b x y + c y^2$,&#xA;with $a,b,c$ integers and of discrmimant $D$ (i.e. $b^2 - 4 a c = D$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $\mathbb Q(\sqrt{D})$ has class number one, then the conditions for solving&#xA;$a x^2 + b x y + c y^2 = p$ ($p$ a prime not dividing $D$) depend only on the residue class&#xA;of $p$ mod $D$, in fact only on the value of the Kronecker symbol of $p$ mod $D$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the class group is just a product of cyclic groups of order 2 (note that this can't&#xA;be detected by the class number alone, which e.g. can't distinguish between $C_2 \times C_2$&#xA;and $C_4$) then the condition for solving $a x^2 + b x y + c y^2 = p$ depends only on the&#xA;residue class of $p$ mod $D$, but one has to consider not just the Kronecker symbol mod $D$,&#xA;but  other Kronecker symbols modulo various divisors of $D$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the class group is not a product of cyclic groups of order 2 (e.g. if the class number&#xA;is not a power of two) then there is no congruence condition on $p$ which guarantees being able to solve $a x^2 + b x y + c y^2$.&lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2010-09-14T07:34:27.720" />
  <row Id="4617" PostTypeId="2" ParentId="4371" CreationDate="2010-09-14T08:41:39.037" Score="1" ViewCount="" Body="&lt;p&gt;For easy read:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;&lt;p&gt;Combinatorics and Graph Theory by Harris&lt;/p&gt;&lt;/li&gt;&#xA;  &lt;li&gt;&lt;p&gt;Introduction to Graph Theory by Wilson&lt;/p&gt;&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1869" LastActivityDate="2010-09-14T08:41:39.037" />
  <row Id="4618" PostTypeId="2" ParentId="4603" CreationDate="2010-09-14T08:56:07.027" Score="3" ViewCount="" Body="&lt;p&gt;Try to stay away from quantifier-laden formulas, which make the problem harder to understand, and draw a picture.   There is an equivalent problem for decreasing functions (as in the Integral Test for convergence) and the picture makes it obvious what is true in that case.  Having seen the continuous proof, run the same argument for the sequence, or specialize the function to a sequence by using step functions or approximation thereof. I won't spoil the &quot;aha!&quot; proof-by-picture experience by posting more details, but it is quite easy once you draw the graph.&lt;/p&gt;&#xA;" OwnerUserId="467" LastActivityDate="2010-09-14T08:56:07.027" />
  <row Id="4619" PostTypeId="1" AcceptedAnswerId="4645" CreationDate="2010-09-14T09:20:53.783" Score="3" ViewCount="560" Body="&lt;p&gt;Please mention ONE problem that is NP-Hard but not NP-Complete? And, explain why. I see some papers assert &lt;strong&gt;Degree Constrained Minimum Spanning Tree&lt;/strong&gt; is an NP-Hard problem and some say it's NP-Complete. Why so? Is it something that we don't have clear idea about?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="1869" LastActivityDate="2010-11-24T12:01:14.003" Title="Mention 1 problem that is NP Hard but not NP Complete and why" Tags="&lt;np-complete&gt;" AnswerCount="4" CommentCount="3" FavoriteCount="1" />
  <row Id="4620" PostTypeId="2" ParentId="4594" CreationDate="2010-09-14T09:27:23.250" Score="3" ViewCount="" Body="&lt;p&gt;There are formal notions of Natural Proofs (Razborov-Rudich) and of &lt;em&gt;proofs&lt;/em&gt; that relativize (Baker-Gill-Solovay) and both are known to not work for showing $P \neq NP$ .  You are asking about relativizable &lt;em&gt;algorithms&lt;/em&gt;, because any algorithm that works only on the &quot;external shell&quot; or &quot;superficial interface&quot; of SAT or Max-Clique will probably work equally in environments where Turing machines are connected to an oracle, and P=NP can be true or false depending on the oracle.  At least for oracles with P &amp;lt; NP the algorithm will not run in polynomial time, and I think this is true for random oracles, so possibilities for SAT solvers that use only externally visible features to work in polynomial time are fairly limited.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are also problems where algorithms that use only the external interface are outperformed by algorithms that take apart the internal representation of the data. For example, sorting by comparisons requires $n \log n$ operations compared to the linear-time radix sorting.&lt;/p&gt;&#xA;" OwnerUserId="467" LastActivityDate="2010-09-14T09:27:23.250" />
  <row Id="4621" PostTypeId="1" AcceptedAnswerId="4622" CreationDate="2010-09-14T09:30:34.807" Score="1" ViewCount="148" Body="&lt;p&gt;Below are the common set of curves that are used in a DJ mixer on the crossfader. I have a software equivalent and am using the &quot;Transition&quot; type curve but would like to include some different curve types. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On both the x and y axis' my range is from 0-1. What I usually do is have a function that gives me one side of the curve. (i.e. just the red from the diagrams below) Then as I need two levels (the red and the blue) for a given x value I invert the x value and feed it to the same formula (something like invertedX = x * -1 + 1)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need the formula for the Dipped curve in the diagram below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Extra credit goes to those who can give me the formulas to the other curves &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Intermediate&lt;/li&gt;&#xA;&lt;li&gt;Constant Power&lt;/li&gt;&#xA;&lt;li&gt;I'm pretty sure Slow Fade, Slow Cut and Fast Cut are all the same formula with just a parameter difference or two.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I have Transition (the easiest of them all)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/IO7KF.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="676" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-09-14T12:43:07.867" LastActivityDate="2011-03-10T07:48:53.223" Title="Simple Formula for Curve of DJ Crossfader volume : Dipped" Tags="&lt;algebra-precalculus&gt;&lt;functions&gt;&lt;algebraic-curves&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="4622" PostTypeId="2" ParentId="4621" CreationDate="2010-09-14T09:49:23.603" Score="1" ViewCount="" Body="&lt;p&gt;Intermediate is clearly  $y = 1 - x$ and $y = x$; should be similarly related to transition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I &lt;em&gt;want&lt;/em&gt; to say that Dipped is a parabola (not sure it is; hard to tell); in which case it would be&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ y = (x-1)^2 $ and $ y = x^2 $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But there are many parabolas that fit to the points (0,0) and (1,1), or (0,1) and (1,0).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The name &quot;Power&quot; seems to imply &lt;a href=&quot;http://en.wikipedia.org/wiki/Sound_power&quot; rel=&quot;nofollow&quot;&gt;Power in Sound&lt;/a&gt;, which would mean logarithms.  I could imagine fitting logarithms into that, but perhaps it would be easier to just use 1-dipped.  That is, $ y = 1 - (x-1)^2 $ and $ y = 1 - x^2 $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fade/cut/cut could be cubic formulas, shifted upwards +0.5.  So they'd be variations of $x^3 + 0.5$ stretched vertically/horizontally.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll get back to this when I'm less sleepy, heh.&lt;/p&gt;&#xA;" OwnerUserId="51" LastActivityDate="2010-09-14T09:49:23.603" CommentCount="1" />
  <row Id="4624" PostTypeId="2" ParentId="4621" CreationDate="2010-09-14T09:59:48.427" Score="1" ViewCount="" Body="&lt;p&gt;For the slow fade may be you can take the Gaussian curve $y = e^{-x^{2}}$ and here is the diagram below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/gdbEu.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And for the transition you could try $y=-|x|$.&lt;/p&gt;&#xA;" LastActivityDate="2010-09-14T09:59:48.427" CommentCount="2" />
  <row Id="4625" PostTypeId="2" ParentId="4619" CreationDate="2010-09-14T10:01:39.670" Score="5" ViewCount="" Body="&lt;p&gt;The difference is that NP-complete means both NP-hard and in NP. Sometimes it's not important to mention that something is in NP even if it is, so NP-hard is said instead. I don't think there's some sophisticated motive behind it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The halting problem is a classical example of NP-hard but not in NP problem; it can't be in NP since it's not even decidable, and it's NP-hard since given any NP-language L and an NP machine M for it, then the reduction from L to HP goes like this: reduce the input x to the input (M',x) where M' is a machine that runs M on x in all the possible ways (M is nondeterministic so there are several such ways, but a finite number since M is polynomial) and halts if M accepts in at least one of these computations. Of course, M' is not polynomial, but the reduction (producing (M',x) given x) is polynomial, so this is a polynomial-time reduction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sometimes people are not satisfied with HP since it's undecidable. It's much more difficult to give a concrete example of a decidable problem, since most of the everyday problems are in PSPACE and it is yet unknown if PSPACE is different than P (if it is, than every PSPACE-complete language, such as TQBF (The language of all true quantified boolean formulas) will do. For now we need to go as far as NEXP-complete problems (since the hierarchy theorems show that NP is different than NEXP). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Complete problems for NEXP can be constructed from complete graph problems for NP (such as Hamiltonian cycle) if we change the representation of the graph - instead of being given directly, it is given via a black-box algorithm that takes two vertices as input and outputs whether there is an edge between them or not (so the size of the representation of the graph is exponentially smaller than it would be for if the edges were given directly).&lt;/p&gt;&#xA;" OwnerUserId="1818" LastEditorUserId="1818" LastEditorDisplayName="" LastEditDate="2010-09-14T11:58:48.733" LastActivityDate="2010-09-14T11:58:48.733" CommentCount="2" />
  <row Id="4626" PostTypeId="1" AcceptedAnswerId="5374" CreationDate="2010-09-14T10:13:18.280" Score="8" ViewCount="684" Body="&lt;p&gt;&lt;em&gt;Notation&lt;/em&gt;: $b_{0}+\underset{n=1}{\overset{\infty }{\mathbb{K}}}\left( a_{n}/b_{n}\right) $ is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Generalized_continued_fraction&quot; rel=&quot;nofollow&quot;&gt;Gauss Notation for generalized continued fractions&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Description of the &lt;a href=&quot;http://mathworld.wolfram.com/Bauer-MuirTransformation.html&quot; rel=&quot;nofollow&quot;&gt;Bauer-Muir transformation&lt;/a&gt;&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Based on &lt;a href=&quot;http://calcauxprobteor.files.wordpress.com/2010/09/cfwa-bmtpdf.pdf&quot; rel=&quot;nofollow&quot;&gt;pp. 76-77&lt;/a&gt; of Lisa Lorentzen and Haakon Waadeland's &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0444892656&quot; rel=&quot;nofollow&quot;&gt;book&lt;/a&gt; (A), chapter II, theorems 11, and Theorem 7;  also section 5 of J. Mc Laughlin, and Nancy J. Wyshinski's online &lt;a href=&quot;http://arxiv.org/PS_cache/math/pdf/0402/0402462v2.pdf&quot; rel=&quot;nofollow&quot;&gt;paper&lt;/a&gt; (C), and section 5 of Bruce C. Berndt, Sen-Shan Huang, Jaebum Sohn, and Seung Hwan Son's online &lt;a href=&quot;http://www.math.uiuc.edu/~berndt/articles/bhss.pdf&quot; rel=&quot;nofollow&quot;&gt;paper&lt;/a&gt; (B) ). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given a convergent c.f. $\underset{n=1}{\overset{\infty }{\mathbb{K}}}(a_{n}/b_{n})=\lim_{n\rightarrow \infty }A_{n}/B_{n}$ and a sequence ${w_{n}}$ we can construct a new c.f $\underset{n=1}{\overset{\infty }{\mathbb{K}}}({c_{n}/d_{n}})=\lim_{n\rightarrow \infty }C_{n}/D_{n}$ (if convergent) which is its &lt;em&gt;Bauer-Muir transform with respect to&lt;/em&gt; ${w_{n}}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;By theorem 11, chapter II, of Lisa Lorentzen and Haakon Waadeland's &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0444892656&quot; rel=&quot;nofollow&quot;&gt;book&lt;/a&gt; (A), &lt;a href=&quot;http://calcauxprobteor.files.wordpress.com/2010/09/cfwa-bmtpdf.pdf&quot; rel=&quot;nofollow&quot;&gt;pp.76-77&lt;/a&gt;, the relations between $A_{n},B_{n}$ and $C_{n},D_{n}$ are given by:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$C_{n}=A_{n}+A_{n-1}w_{n}$, $D_{n}=B_{n}+B_{n-1}w_{n}$, with the initial conditions $C_{-1}=1,D_{-1}=0.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If for $n\geq 1$,  $\lambda_{n}=a_{n}-w_{n-1}(b_{n}+w_{n})\neq 0$, then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$c_{n}=a_{n-1}\lambda_{n}/\lambda_{n-1},$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$d_{n}=b_{n}+w_{n}-w_{n-2}\lambda_{n}/\lambda _{n-1},$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for $n\geq 2,$ and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\underset{n=1}{\overset{\infty }{\mathbb{K}}}(a_{n}/b_{n})=w_{0}+\dfrac{\lambda_{1}}{b_{1}+w_{1}+\underset{n=1}{\overset{\infty }{\mathbb{K}}}(c_{n}/d_{n})}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The elements of $\underset{n=1}{\overset{\infty }{\mathbb{K}}}(a_{n}/b_{n})$ are computed by an application of Theorem 7, chapter II, of Lisa Lorentzen and Haakon Waadeland's &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0444892656&quot; rel=&quot;nofollow&quot;&gt;book&lt;/a&gt; (A), which transforms a sequence into a continued fraction. &lt;em&gt;I was able to derive&lt;/em&gt; $c_{n}$ &lt;em&gt;but not&lt;/em&gt; $d_{n}$. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Example:&lt;/strong&gt; Application to the $\log (1+t)$ expansion. By choosing $w_{0}=w_{1}=0,w_{n}=(n-1)t$, for $n\geq 2$, one can derive&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\underset{n=1}{\overset{\infty }{\mathbb{K}}}\left( n^{2}t/\left( \left(&#xA;n+1\right) -kt\right) \right) =\dfrac{t}{2+\underset{n=3}{\overset{\infty }{%&#xA;\mathbb{K}}}\left( \left( n-2\right) ^{2}t/\left( n-\left( n-3\right)&#xA;t\right) \right) }.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence (see &lt;a href=&quot;http://en.wikipedia.org/wiki/Natural_logarithm&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;) the expansion&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\log (1+t)=\displaystyle\sum_{n=1}^{\infty }\dfrac{(-1)^{n-1}t^{n}}{n}=\dfrac{t}{1+%&#xA;\underset{n=1}{\overset{\infty }{\mathbb{K}}}\left( n^{2}t/\left( \left(&#xA;n+1\right) -nt\right) \right) },$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;can be improved with respect to the convergence speed to this one &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\log (1+t)=\dfrac{t}{1+\dfrac{t}{2+\underset{n=3}{\overset{\infty }{\mathbb{%&#xA;K}}}\left( \left( n-2\right) ^{2}t/\left( n-\left( n-3\right) t\right)&#xA;\right) }}.$$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Derivation of $c_n$&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is how I got $c_n$. In page 77 of reference A it is proved that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$C_{n-1}D_n-D_{n-1}C_n=(A_{n-1}B_{n-2}-A_{n-2}B_{n-1})\lambda_n.$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$C_{n-2}D_{n-1}-D_{n-2}C_{n-1}=(A_{n-2}B_{n-3}-A_{n-3}B_{n-2})\lambda_{n-1}.$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $n\ge 2$ Theorem 7 of reference A (derived from the fundamental c.f. recurrence) gives  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$c_n=\dfrac{C_{n-1}D_n-D_{n-1}C_n}{-(C_{n-2}D_{n-1}-D_{n-2}C_{n-1})}=\dfrac{(A_{n-1}B_{n-2}-A_{n-2}B_{n-1})\lambda_n}{-(A_{n-2}B_{n-3}-A_{n-3}B_{n-2})\lambda_{n-1}}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the determinant formula we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$A_{n-1}B_{n-2}-A_{n-2}B_{n-1}=-a_{n-1}(A_{n-2}B_{n-3}-A_{n-3}B_{n-2}).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$c_n=a_{n-1}\dfrac{\lambda_n}{\lambda_{n-1}}.$$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;QUESTION 1: How does one prove&lt;/strong&gt; &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$$d_{n}=b_{n}+w_{n}-w_{n-2}\lambda_{n}/\lambda_{n-1}$$&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;from&lt;/strong&gt; &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$$d_{n}=\dfrac{C_{n}D_{n-2}-D_{n}C_{n-2}}{C_{n}D_{n-1}-D_{n}C_{n-1}}\qquad\text{for}\quad n\ge 2$$&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;and&lt;/strong&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$\lambda_{n}=a_{n}-w_{n-1}(b_{n}+w_{n})$,&#xA;  $C_{n}=A_{n}+A_{n-1}w_{n}$,&#xA;  $D_{n}=B_{n}+B_{n-1}w_{n}$ &lt;strong&gt;?&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;References&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A - Lisa Lorentzen and Haakon Waadeland, &lt;em&gt;&lt;a href=&quot;http://calcauxprobteor.files.wordpress.com/2010/09/cfwa-bmtpdf.pdf&quot; rel=&quot;nofollow&quot;&gt;Continued Fractions and Applications&lt;/a&gt;&lt;/em&gt;, North-Holland, Amsterdam, 1992. (&lt;a href=&quot;http://calcauxprobteor.files.wordpress.com/2010/09/cfwa-bmtpdf.pdf&quot; rel=&quot;nofollow&quot;&gt;pdf file of pp. 76-77&lt;/a&gt;) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;B - Bruce C. Berndt, Sen-Shan Huang, Jaebum Sohn, and Seung Hwan Son, &lt;em&gt;&lt;a href=&quot;http://www.math.uiuc.edu/~berndt/articles/bhss.pdf&quot; rel=&quot;nofollow&quot;&gt;A Transformation Formula in Rogers--Ramanujan Continued Fraction&lt;/a&gt;&lt;/em&gt;. (section 5) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;C - J. Mc Laughlin, and Nancy J. Wyshinski, &lt;em&gt;&lt;a href=&quot;http://arxiv.org/PS_cache/math/pdf/0402/0402462v2.pdf&quot; rel=&quot;nofollow&quot;&gt;Real numbers with polynomial continued fraction expansions&lt;/a&gt;&lt;/em&gt;, arXiv, 2004. &lt;/p&gt;&#xA;" OwnerUserId="752" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2010-10-05T09:31:16.683" LastActivityDate="2011-01-13T20:22:43.467" CommunityOwnedDate="2010-09-18T10:50:23.373" Title="How do I prove the partial denominators formula of the Bauer-Muir transformation of a generalized continued fraction ?" Tags="&lt;number-theory&gt;&lt;sequences-and-series&gt;&lt;continued-fractions&gt;&lt;computational-mathematics&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="0" />
  <row Id="4627" PostTypeId="1" AcceptedAnswerId="5203" CreationDate="2010-09-14T10:28:27.603" Score="2" ViewCount="203" Body="&lt;p&gt;Hey, I'm looking for a guide on how I find Q given the following, where $a$ and $b$ are constants:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;\frac{dQ}{dt} = \frac{a + Q}{b}&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have the answer and working for specific case I'm trying to solve but do not understand the steps involved. A guide on how I can solve this, with an explanation of each step would be much appreciated.&lt;/p&gt;&#xA;" OwnerUserId="1888" LastEditorUserId="664" LastEditorDisplayName="" LastEditDate="2010-09-21T19:27:43.467" LastActivityDate="2010-09-22T07:25:38.223" Title="How do I find a function from a differential equation?" Tags="&lt;calculus&gt;&lt;integral&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="2" />
  <row Id="4628" PostTypeId="2" ParentId="4627" CreationDate="2010-09-14T10:58:10.750" Score="1" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; Your particular differential equation can be solved without any calculation if you know the solutions of $$\frac{dQ}{dt} = Q$$, i.e. $C\exp(t), C\in\mathbb{R}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that, if $f$ is a solution to $\frac{dQ}{dt} = \frac 1b Q$, then $f-a$ is a solution to your equation, therefore is suffices to only look at $a=0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The chain rule can be used to see, that if $f$ solves $\frac{dQ}{dt} = Q$, then $f(t/b)$ solves $\frac{dQ}{dt} = \frac 1b Q$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore all your solutions are of the form $C\exp(t/b) - a, C\in\mathbb{R}$.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Old Post:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think these differential equations can be solved by &quot;Separation of Variables&quot;. See &lt;a href=&quot;http://en.wikipedia.org/wiki/Separation_of_variables&quot; rel=&quot;nofollow&quot;&gt;the Wikipedia&lt;/a&gt; for a guide. It's got examples! =)&lt;/p&gt;&#xA;" OwnerUserId="505" LastEditorUserId="505" LastEditorDisplayName="" LastEditDate="2010-09-14T13:23:52.040" LastActivityDate="2010-09-14T13:23:52.040" CommentCount="5" />
  <row Id="4630" PostTypeId="2" ParentId="4363" CreationDate="2010-09-14T11:33:59.377" Score="0" ViewCount="" Body="&lt;p&gt;Here, I think is what the OP is after- we can say:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The image of the curve under the map will be '$C^1$' (that is as a given parametrisation from an interval to $\mathbb{R}^m$) if the original curve is (under similar caveats of what that means): to see this, choose a $C^1$ parameterisation $\gamma :I \to U$ and observe that $\varphi \circ \gamma$ must also be $C^1$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;However:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\varphi \circ \gamma$ need not be injective: let $m=n=2$ and let $\psi$ be a bump function of height one centred around the x- axis, then the map $(x,y) \mapsto (x, (1- \psi )y )$ makes any parametrisation of the y-axis non-injective&lt;/li&gt;&#xA;&lt;li&gt;The preimage of a curve in $\mathbb{R}^m$ in $U$, even if it exists and is unique (that is, the preimage gives you something that looks like a bona fide curve rather than, say, several curves meeting at a point, or worse), need not be smooth: take $\varphi(x,y)= (x, y^2)$ and $\gamma=${$(t,t^3)| t \in [0, \infty)$}&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Indeed, I am fairly sure that the first bullet point is *the only interesting thing we can say about such curves. Fun excercise/ research practice: try to come up with more statements about such curves and see if you can find a counter example to each...&lt;/p&gt;&#xA;" OwnerUserId="160" LastActivityDate="2010-09-14T11:33:59.377" CommentCount="2" />
  <row Id="4631" PostTypeId="1" AcceptedAnswerId="4633" CreationDate="2010-09-14T11:42:43.140" Score="2" ViewCount="171" Body="&lt;p&gt;Why are differential equations called differential equations?&lt;/p&gt;&#xA;" OwnerUserId="705" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-09-14T14:36:57.943" LastActivityDate="2010-09-14T14:36:57.943" Title="Why are differential equations called differential equations?" Tags="&lt;calculus&gt;&lt;terminology&gt;&lt;differential-equations&gt;" AnswerCount="1" />
  <row Id="4632" PostTypeId="1" AcceptedAnswerId="4652" CreationDate="2010-09-14T11:44:50.143" Score="3" ViewCount="287" Body="&lt;p&gt;I am trying to identify the general case algorithm for counting the different ways dice can add to a given number.  For instance, there are &lt;a href=&quot;http://wizardofodds.com/gambling/dice.html&quot; rel=&quot;nofollow&quot;&gt;six ways to roll a seven&lt;/a&gt; with two 6-dice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've spent quite a bit of time working on this (for a while my friend and I were using Figurate numbers, as the early items in the series match up) but at this point I'm tired and stumped, and would love some assistance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So far we've got something to this effect (apologies for the feeble attempt at mathematical notation - I usually reside on StackOverflow):&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;count(x):&#xA;   x = min(x,n*m-x+n)&#xA;   if x = n&#xA;      1&#xA;   else&#xA;      some sort of (recursive?) operation&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The first line simplifies the problem to just the lower numbers - where the count is increasing.  Then, if we're looking for the count of the minimum possible (which is also now the max because of the previous line) there is only one way to do that so it is 1, no matter the n or m.&lt;/p&gt;&#xA;" OwnerUserId="1887" LastActivityDate="2010-09-14T15:06:16.907" Title="How can I (algorithmically) count the number of ways n m-sided dice can add up to a given number?" Tags="&lt;algorithms&gt;&lt;probability&gt;&lt;probability-theory&gt;&lt;counting&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="2" />
  <row Id="4633" PostTypeId="2" ParentId="4631" CreationDate="2010-09-14T11:49:20.940" Score="6" ViewCount="" Body="&lt;p&gt;Because they are equations (with the variable being a function, not a number) that involve a function and its derivatives (the functions obtained by differentiating it).&lt;/p&gt;&#xA;" OwnerUserId="1818" LastActivityDate="2010-09-14T11:49:20.940" />
  <row Id="4634" PostTypeId="1" AcceptedAnswerId="4659" CreationDate="2010-09-14T11:52:09.570" Score="3" ViewCount="218" Body="&lt;p&gt;I know the &lt;a href=&quot;http://en.wikipedia.org/wiki/Four_color_theorem&quot; rel=&quot;nofollow&quot;&gt;four colour theorem&lt;/a&gt; was solved by a computer checking a large number of cases. What I don't understand is why there are only a large but finite number of cases. It seems like there should be infinitely many planar graphs... What is the intuition behind cutting this down to a finite number of configurations?&lt;/p&gt;&#xA;" OwnerUserId="172" LastEditorUserId="172" LastEditorDisplayName="" LastEditDate="2010-09-15T10:45:34.653" LastActivityDate="2010-09-15T10:45:34.653" Title="Intuition behind the &quot;large but finite search space&quot; of the proof of the four colour theorem?" Tags="&lt;graph-theory&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="1" />
  <row Id="4635" PostTypeId="2" ParentId="4619" CreationDate="2010-09-14T11:53:02.537" Score="1" ViewCount="" Body="&lt;p&gt;As noted in the earlier answers, &lt;strong&gt;NP&lt;/strong&gt;-hard means that any problem in &lt;strong&gt;NP&lt;/strong&gt; can be reduced to it. This means that any complete problem for a class (e.g. &lt;strong&gt;PSPACE&lt;/strong&gt;) which contains &lt;strong&gt;NP&lt;/strong&gt; is also &lt;strong&gt;NP&lt;/strong&gt;-hard.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to get a problem which is &lt;strong&gt;NP&lt;/strong&gt;-hard but not &lt;strong&gt;NP&lt;/strong&gt;-complete, it suffices to find a computational class which (&lt;em&gt;a&lt;/em&gt;)&amp;nbsp;has complete problems, (&lt;em&gt;b&lt;/em&gt;)&amp;nbsp;provably contains &lt;strong&gt;NP&lt;/strong&gt;, and (&lt;em&gt;c&lt;/em&gt;)&amp;nbsp;is provably different from &lt;strong&gt;NP&lt;/strong&gt;. So:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;While &lt;strong&gt;PSPACE&lt;/strong&gt; contains &lt;strong&gt;NP&lt;/strong&gt;, and has complete problems, the containment is not yet known to be strict. Therefore,&lt;strong&gt;PSPACE&lt;/strong&gt;-complete problems do not suffice.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The same goes for any complete problem for any class in the polynomial hierarchy. If the hierarchy &lt;em&gt;does not&lt;/em&gt; collapse, any complete problem for any class in the hierarchy (aside from &lt;strong&gt;P&lt;/strong&gt;, &lt;strong&gt;NP&lt;/strong&gt;, or &lt;strong&gt;coNP&lt;/strong&gt;) would suffice; but without a proof that the PH doesn't collapse, we can't use such problems&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;We &lt;em&gt;can&lt;/em&gt; prove that &lt;strong&gt;NEXP&lt;/strong&gt; (the class of problems solvable in &lt;em&gt;exponential time&lt;/em&gt; on a nondeterministic Turing Machine) strictly contains &lt;strong&gt;NP&lt;/strong&gt;, and has complete problems. So any &lt;strong&gt;NEXP&lt;/strong&gt;-complete problem is &lt;strong&gt;NP&lt;/strong&gt;-hard but not &lt;strong&gt;NP&lt;/strong&gt;-complete.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="439" LastEditorUserId="439" LastEditorDisplayName="" LastEditDate="2010-09-14T14:46:24.167" LastActivityDate="2010-09-14T14:46:24.167" />
  <row Id="4636" PostTypeId="1" AcceptedAnswerId="4650" CreationDate="2010-09-14T12:06:35.343" Score="1" ViewCount="107" Body="&lt;p&gt;A polyomino is a connected subset of $\mathbb{Z}^2$ - a set of squares joined along their edges such that the resulting form is connected (or, more shortly, a generalized form of Tetris cube). A polyomino can be easily thought of as graph with vertices being elements of $\mathbb{Z}^2$ and edges between two vertices of hamming distance 1 (i.e. with one coordinate equal and the other different by 1).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Hamiltonian path/cycle problem is the problem of determining for a given graph whether it contains a path/cycle that visits every vertex exactly once. It is known to be NP-complete for general graphs and even for planar graphs. My question is whether it's still NP-complete when restricting the graphs to be polyominoes (and if it is NP-complete, how it is shown).&lt;/p&gt;&#xA;" OwnerUserId="1818" LastActivityDate="2010-09-14T13:46:50.203" Title="The Hamiltonian problem on Polyominoes" Tags="&lt;computer-science&gt;&lt;graph-theory&gt;&lt;np-complete&gt;" AnswerCount="1" />
  <row Id="4637" PostTypeId="2" ParentId="4627" CreationDate="2010-09-14T12:09:52.217" Score="3" ViewCount="" Body="&lt;p&gt;Yet, another method for solving this differential equation is to look at it as a &lt;em&gt;linear differential equation&lt;/em&gt;, whose general form is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;y'(x) = a(x) y(x) + b(x) \ , \qquad\qquad\qquad [1]&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $a(x), b(x)$ are arbitrary functions depending on the variable $x$. In your case:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;x = t \ , y(x) = Q(t) \ , b(x) = \frac{a}{b} \quad \text{and}\quad a(x) = \frac{1}{b} \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A general procedure for solving [1] is the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;1.&lt;/strong&gt; First, try to solve the associated &lt;em&gt;homogeneous&lt;/em&gt; linear differential equation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;y' = a(x)y \ .  \qquad\qquad\qquad    [2]&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is easy: the general solution is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;y = K e^{A(x)} \ ,   \qquad\qquad\qquad   [3]&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $K\in \mathbb{R}$ is an arbitrary constant and $A(x) = \int a(x)dx$ is a primitive function of $a(x)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;2.&lt;/strong&gt; Once you have the general solution [3] of [2], you apply &lt;em&gt;variation of constants&lt;/em&gt;; that is, you look for solutions of [1] of the following kind:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;y = K(x) e^{A(x)} \ .   \qquad \qquad \qquad  [4]&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here, we have replaced the arbitrary constant $K$ by an arbitrary unkown function $K(x)$ (hence the name &quot;variation of constants&quot;) to be determined. How? Imposing that we want [4] to be a solution of our first differential equation [1]. It goes like this: if you replace $y$ in [4] into [1], you get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;K'(x) e^{A(x)} + K(x) A'(x) e^{A(x)} = a(x) K(x) e^{A(x)} + b(x) \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $A'(x) = a(x) $, this is the same as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;K'(x) e^{A(x)} = b(x) \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;K(x) = \int b(x)e^{-A(x)}dx + C \ ,&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $C \in \mathbb{R}$ is an arbitrary constant. Now you put this $K(x)$ into [4] and get the general solution of your differential equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;y(x) = Ce^{A(x)} + e^{A(x)}\int b(x) e^{-A(x)}dx \ . \qquad\qquad\qquad   [5]&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since I've never could remember formula [5], I use to repeat the whole process for each particular linear differential equation, which is not hard and you can do it for yours.&lt;/p&gt;&#xA;" OwnerUserId="664" LastActivityDate="2010-09-14T12:09:52.217" CommentCount="1" />
  <row Id="4638" PostTypeId="1" CreationDate="2010-09-14T12:11:31.153" Score="2" ViewCount="173" Body="&lt;p&gt;A at 45,10 ,  B at 10,20,  AB=AC and angle C=20 degree find the coordinates of C .suggest the formula so i can write code in Perl.&lt;/p&gt;&#xA;" OwnerUserId="1890" LastActivityDate="2010-09-14T19:00:53.433" Title="Find the coordinates in an isosceles triangle if the triangle it self is in positive axis" Tags="&lt;geometry&gt;&lt;trigonometry&gt;" AnswerCount="2" CommentCount="0" />
  <row Id="4639" PostTypeId="2" ParentId="4638" CreationDate="2010-09-14T12:27:25.813" Score="1" ViewCount="" Body="&lt;p&gt;You have the coordinates of A and B, so you can compute the distance AB.  AB=AC, so you then know the distance AC.  Let the coordinates of C be (x,y).  Apply the distance formula to A and C and set the result equal to the distance you already computed.  This equation guarantees that AB=AC.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the angle at C is determined by the vectors $\overrightarrow{CA}$ and $\overrightarrow{CB}$.  These vectors can be found by subtracting the coordinates of C from A and B (respectively).  $(CA)(CB)\cos C = \overrightarrow{CA}\cdot\overrightarrow{CB}$, and you know the distances CA = AC = AB and the measure of C, and you can compute CB and the dot product in terms of (x,y).  This equation guarantees that the measure of C is 20&amp;deg;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solve the system resulting from the two equations above to find the coordinates of C.  There are almost certainly 2 solutions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(See also &lt;a href=&quot;http://math.stackexchange.com/questions/3120/find-the-coordinates-in-an-isosceles-triangle&quot;&gt;this question&lt;/a&gt;.)&lt;/p&gt;&#xA;" OwnerUserId="72" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-09-14T16:19:14.913" LastActivityDate="2010-09-14T16:19:14.913" />
  <row Id="4640" PostTypeId="1" AcceptedAnswerId="4642" CreationDate="2010-09-14T12:33:21.410" Score="1" ViewCount="91" Body="&lt;p&gt;When given an algrebraic expression of the form&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \ 6A + 7{A^2}\ $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and asked to simplify, would the correct answer be&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a)&#xA;$$ \ A(6 + 7A)\ $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or&lt;/p&gt;&#xA;&#xA;&lt;p&gt;b)&#xA;That this cannot be simplified further&lt;/p&gt;&#xA;" OwnerUserId="1409" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-09-14T12:39:44.953" LastActivityDate="2010-09-14T15:55:58.170" Title="Simplification or Alternate form" Tags="&lt;algebra-precalculus&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="4642" PostTypeId="2" ParentId="4640" CreationDate="2010-09-14T12:39:33.503" Score="3" ViewCount="" Body="&lt;p&gt;&quot;Simplify&quot; is not particularly well-defined.  For the expression by itself, I think (b) might be the more common textbook answer; if the expression were the denominator of a rational expression, (a) is more likely to be useful.&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-09-14T12:39:33.503" />
  <row Id="4643" PostTypeId="1" AcceptedAnswerId="4654" CreationDate="2010-09-14T12:58:51.490" Score="1" ViewCount="153" Body="&lt;p&gt;There are a number of ways to describe this problem. I shall name a few.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Submultisets&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $(M, f)$ be a multiset where $M = {x_1, ... x_k}, |M| = k$ and $f(x_i) = m_i$, i ranging from 0 to k inclusive. Find the number of submultisets $(S,g)$ of $(M,f)$ of size n, $|(S, g)| = n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Integer Partitions&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $n, x_1, ... x_k, m_1, ... m_k \in \mathbb{N}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Find the number of solutions of the equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x_1 + x_2 + ... x_k = n$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;such that $0 \le x_i \le m_i$, i of course ranging from 0 to k inclusive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Unordered combinations of size n&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let a collection of objects M have k types of objects, where there are $m_i$ identical objects for each type, i from 0 to k inclusive. How many ways can I pick n elements from M, given the restriction that I cannot pick more than $m_i$ from each type?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unless I've made a mistake somewhere, each of these problems should be identical. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the ordinary undergrad combinatorics literature, there is a method for picking the number of ways to solve this, without the $m_i$ restrictions I mentioned (or at least $m_i \ge n$. That is call the multiset coefficient. There is also the number of ways of ordering all of the elements of M, provided you use all of them. That is called the multinomial coefficient. This, of course, uses only n elements from M, and does not include ordering.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only solution I've come across, given the 5 measly hours I've researched the problem, was in Combinatorics and Graph Theory by Harris et. al, which mentions &lt;a href=&quot;http://books.google.com/books?id=C-0y7LI2FFMC&amp;amp;printsec=frontcover&amp;amp;dq=harris+combinatorics&amp;amp;hl=en&amp;amp;ei=6nCPTOGyLcGB8gazx_WhDQ&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=1&amp;amp;ved=0CCUQ6AEwAA#v=onepage&amp;amp;q=almaalabama&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;a crude method&lt;/a&gt; for counting the number of strings of length 4 of a given set of letters &quot;Alma, Alabama&quot;. They enumerate 4 different patterns by hand, and use well known equations on those patterns. It works when n and k are small, but is inefficient otherwise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sorry, I should have made my question a bit more explicit. I'm looking for an efficient method to do it by hand. The recurrence equations are helpful, but it's not quite what I was looking for.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't think there is a simple equation that provides the answers. &lt;em&gt;A recurrence equation is likely to catch all of the cases, but may not be efficient for computing by hand. I don't think there is an efficient way to do this, if k is large and n is small. Or is there?&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update 2&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Forgive me for bringing up the concept of a multiset. I thought it was in the standard literature. The size of a multiset $|(M, f)|$ is just the summation of the multiplicities $|(M, f)| = \sum_{i} f(x_i)$. It's not the same as the size of the support $|M|$. They are only equal when $f(x_i) = 1$ for all of the elements, just a regular set.&lt;/p&gt;&#xA;" OwnerUserId="1892" LastEditorUserId="1892" LastEditorDisplayName="" LastEditDate="2010-09-14T20:03:33.847" LastActivityDate="2010-09-14T21:20:49.373" Title="An efficient method for computing the number of submultisets of size n, of a given multiset" Tags="&lt;combinatorics&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="4645" PostTypeId="2" ParentId="4619" CreationDate="2010-09-14T13:31:16.480" Score="6" ViewCount="" Body="&lt;p&gt;I will answer the following part of the question:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;And, explain why. I see some papers assert Degree Constrained Minimum Spanning Tree is an NP-Hard problem and some say it's NP-Complete. Why so?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Some people say “the Degree Constrained Minimum Spanning Tree problem (DCMST) is NP-hard” for a reason, other people say “DCMST is NP-complete” for a different reason.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As is explained in the other answers, the word “NP-complete” means that a problem belongs to NP and is NP-hard.  Note that NP is a class of decision problems (that is, problems whose answers are yes/no).  Some people avoid saying “DCMST is NP-complete” and choose to say “DCMST is NP-hard” because DCMST is a function problem (that is, not a decision problem) and therefore cannot belong to NP by definition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, this is not the end of the story.  Other people &lt;em&gt;do&lt;/em&gt; say “DCMST is NP-complete.”  When they say this, they consider the &lt;em&gt;decision version&lt;/em&gt; of the problem implicitly:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The function version of DCMST: Given a graph G with edge weights and a positive integer d, find a minimum-weight spanning tree of G whose maximum degree is at most d.&lt;/li&gt;&#xA;&lt;li&gt;The decision version of DCMST: Given a graph G with edge weights and positive integers d and k, decide whether G has a spanning tree whose maximum degree is at most d and whose total weight is at most k.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Therefore, if you accept the convention to refer to the decision version as DCMST, it is perfectly fine to say that DCMST is NP-complete.  In computational complexity theory, this convention is often used.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the decision version of DCMST can be solved in polynomial time if and only if the function version of DCMST can be (“if” is easy, “only if” is slightly more difficult).  This is why calling the function version and the decision version by the same name causes little confusion.&lt;/p&gt;&#xA;" OwnerUserId="926" LastActivityDate="2010-09-14T13:31:16.480" CommentCount="1" />
  <row Id="4646" PostTypeId="2" ParentId="4609" CreationDate="2010-09-14T13:38:49.803" Score="0" ViewCount="" Body="&lt;p&gt;You can work out the distribution using the transformation theorem.  If you have a normal density truncated to positive values, the log of that distribution has PDF&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{2}{\sqrt{2\pi}} \exp(x -\exp(2x)/2)$&lt;/p&gt;&#xA;" OwnerUserId="247" LastActivityDate="2010-09-14T13:38:49.803" CommentCount="1" />
  <row Id="4648" PostTypeId="1" AcceptedAnswerId="4661" CreationDate="2010-09-14T13:40:32.997" Score="5" ViewCount="148" Body="&lt;p&gt;I made an assertion in &lt;a href=&quot;http://math.stackexchange.com/questions/4107/what-are-some-examples-of-theories-stronger-than-presburger-arithmetic-but-weaker&quot;&gt;What are some examples of theories stronger than Presburger Arithmetic but weaker than Peano Arithmetic?&lt;/a&gt; that Q has higher consistency strength than Pres, Presburger arithmetic; i.e., Q proves the consistency sentence for Pres.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But in fact, I only know something weaker, that Q can formalise the provability predicate for Hilbert systems, and so prove, say, that Peano arithmetic proves Pres consistent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a direct proof of consistenct of Pres in Q?&lt;/p&gt;&#xA;" OwnerUserId="100" LastActivityDate="2010-09-14T15:45:12.443" Title="Can Robinson's Q prove Presburger arithmetic consistent?" Tags="&lt;logic&gt;&lt;proof-theory&gt;&lt;incompleteness&gt;" AnswerCount="1" />
  <row Id="4649" PostTypeId="2" ParentId="4634" CreationDate="2010-09-14T13:45:56.010" Score="2" ViewCount="" Body="&lt;p&gt;Perhaps this expository account of an &quot;improved&quot; proof of the 4-color theorem will help:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://people.math.gatech.edu/~thomas/FC/fourcolor.html&quot; rel=&quot;nofollow&quot;&gt;http://people.math.gatech.edu/~thomas/FC/fourcolor.html&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="1369" LastActivityDate="2010-09-14T13:45:56.010" />
  <row Id="4650" PostTypeId="2" ParentId="4636" CreationDate="2010-09-14T13:46:50.203" Score="4" ViewCount="" Body="&lt;p&gt;This problem (both the path version and the circuit version) was shown to be NP-complete by Itai, Papadimitriou and Szwarcfiter [IPS82] by a reduction from a special case of the Hamiltonian circuit problem for planar graphs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[IPS82] Alon Itai, Christos H. Papadimitriou and Jayme Luiz Szwarcfiter.  Hamilton paths in grid graphs.  &lt;em&gt;SIAM Journal on Computing&lt;/em&gt;, 11(4):676–686, 1982.  &lt;a href=&quot;http://dx.doi.org/10.1137/0211056&quot; rel=&quot;nofollow&quot;&gt;http://dx.doi.org/10.1137/0211056&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="926" LastActivityDate="2010-09-14T13:46:50.203" />
  <row Id="4651" PostTypeId="2" ParentId="4603" CreationDate="2010-09-14T14:03:56.783" Score="1" ViewCount="" Body="&lt;p&gt;You might also do this the other way around. What if &#xA;$$\lim_{n\to\infty}na_n \not=0,$$&#xA;that is if the limit does not exist or if it is positive, what does this tell you about $a_n$? What about sub-sequences of $(a_n)$?&lt;/p&gt;&#xA;" OwnerUserId="1154" LastActivityDate="2010-09-14T14:03:56.783" />
  <row Id="4652" PostTypeId="2" ParentId="4632" CreationDate="2010-09-14T14:12:25.033" Score="3" ViewCount="" Body="&lt;p&gt;Since the order matters i.e 2+3+4 is different from 3+4+2, you can use generating functions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The number of ways to get a sum $S$ by rolling $n$ dice (with numbers $1,2,\dots,m$) is the coefficient of $x^S$ in&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ (x+x^2+\dots+x^m)^n = x^n(\frac{1-x^{m}}{1-x})^n$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;= $$ x^n(1-x^{m})^{n}(\sum_{k=0}^{\infty} {n+k-1 \choose k} x^k)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus the number of ways is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \sum_{rm+k=S-n} {n \choose r} {n+k-1 \choose k} (-1)^{r}$$&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-09-14T15:06:16.907" LastActivityDate="2010-09-14T15:06:16.907" CommentCount="8" />
  <row Id="4653" PostTypeId="1" CreationDate="2010-09-14T14:15:18.113" Score="-1" ViewCount="286" Body="&lt;p&gt;HI All,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want a funny maths story that must be logically high to understand and funny enough with the use of maths concept .  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The targeted audience are students.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit:&lt;br&gt;&#xA;nothing particular any topic from cloud of maths is acceptable provided that it must be understood by children :P&lt;/p&gt;&#xA;" OwnerUserId="1894" LastEditorUserId="1894" LastEditorDisplayName="" LastEditDate="2010-09-14T14:25:55.443" LastActivityDate="2010-09-14T14:53:04.023" ClosedDate="2010-09-14T15:24:23.227" Title="Maths funny Story needed for a presentation" Tags="&lt;math-history&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="4654" PostTypeId="2" ParentId="4643" CreationDate="2010-09-14T14:23:31.043" Score="1" ViewCount="" Body="&lt;p&gt;I don't really understand the description of the SubMultiset problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think I do understand the description of the integer partition problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This can be solved by generating functions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The number of solutions to &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x_1 + x_2 + \dots + x_k = n$ with $0 \le x_i \le m_i$ is the coefficient of $x^n$ in&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ (1+x+\dots+x^{m_1})(1+x+\dots+x^{m_2})\dots(1+x+\dots+x^{m_k}) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ = \frac{\prod_{i}^{k}(1-x^{m_i+1})}{(1-x)^{k}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ = (\sum_{j=0}^{\infty}{k+j-1 \choose j} x^j) \prod_{i}^{k}(1-x^{m_i+1})$$ &lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-09-14T14:23:31.043" CommentCount="2" />
  <row Id="4655" PostTypeId="2" ParentId="4632" CreationDate="2010-09-14T14:23:42.873" Score="0" ViewCount="" Body="&lt;p&gt;Continuing Moron's answer, you can break the final sum as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \sum_{r=0}^{\lfloor S/(m+1) \rfloor} (-1)^r \binom{n-1+S-r(m+1)}{n-1} \binom{n}{r} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, for your example, $m=6$, $n=2$, $S=7$, and the formula gives&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \binom{8}{1} \binom{2}{0} - \binom{1}{1} \binom{2}{1} = 8 - 2 = 6 $$&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-09-14T14:23:42.873" CommentCount="2" />
  <row Id="4656" PostTypeId="2" ParentId="4581" CreationDate="2010-09-14T14:40:38.417" Score="3" ViewCount="" Body="&lt;p&gt;There is also a more elementary argument explaining why $v_{1},\ldots, v_{n}$ spans $K(X)/k(X)$. To understand it one doesn't need to know separable field extensions nor integral ring extensions, only a little bit of linear algebra:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $E = span_{k(X)} \lbrace v_{1},\ldots, v_{n}\rbrace$. Note that $K[X]\subset E$ implies $K[X]E\subset E,\ 1\in E$, as $v_{1},\ldots, v_{n}$ is a basis of $K/k$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take an arbitrary $h\in K[X]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\cdot h:K(X)\ni a\mapsto a\cdot h\in K(X)$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;is an isomorphism of linear spaces over $k(X)$ ($h^{-1}$ defines inverse linear mapping $\cdot h^{-1}$). Furthermore $\cdot h(E)\subset E$ as $K[X]\subset E$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\cdot h|_{E}: E\to E$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is a monomorphism of finite dimensional linear spaces over $k(X)$, so in fact an isomorphism.&#xA;Therefore $\cdot h^{-1}|_{E}:E\to E$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular $(\cdot h^{-1})(1) = h^{-1}\in E$, and $gh^{-1}\in E$ for every $g\in K[X]$.&lt;/p&gt;&#xA;" OwnerUserId="1454" LastActivityDate="2010-09-14T14:40:38.417" CommentCount="0" />
  <row Id="4657" PostTypeId="2" ParentId="4653" CreationDate="2010-09-14T14:53:04.023" Score="3" ViewCount="" Body="&lt;p&gt;You can borrow from the idea Futurama presented in one of their latest episodes:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://theinfosphere.org/The_Prisoner_of_Benda&quot; rel=&quot;nofollow&quot;&gt;http://theinfosphere.org/The_Prisoner_of_Benda&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the idea can be understood by children and its a nice problem which can be presented in a funny way.&lt;/p&gt;&#xA;" OwnerUserId="1818" LastActivityDate="2010-09-14T14:53:04.023" CommentCount="1" />
  <row Id="4658" PostTypeId="1" CreationDate="2010-09-14T14:53:33.890" Score="7" ViewCount="977" Body="&lt;p&gt;If you were to flip a coin 150 times, what is the probability that it would land tails 7 times in a row?  How about 6 times in a row?  Is there some forumula that can calculate this probability?&lt;/p&gt;&#xA;" OwnerUserId="1895" LastActivityDate="2011-01-25T17:17:05.057" Title="What is the probability of a coin landing tails 7 times in a row in a series of 150 coin flips?" Tags="&lt;probability&gt;&lt;statistics&gt;" AnswerCount="4" CommentCount="4" FavoriteCount="6" />
  <row Id="4659" PostTypeId="2" ParentId="4634" CreationDate="2010-09-14T15:19:51.157" Score="14" ViewCount="" Body="&lt;p&gt;I'm not sure whether or not you understand how the proof of the $4$ color map theorem worked. I'll sketch that, and then make some philosophical comments.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's start by proving the $6$ color map theorem. Without loss of generality, we may assume that all the faces of our graph are triangular, because adding edges just makes a graph harder to color. (We are coloring vertices.) Let $V$, $E$ and $F$ be the number of vertices, edges and faces. Every edge is in two faces, and every face has three edges, so $F=(2/3)E$. We have $V-E+F=2$ by Euler's relation, so $V=E/3 +2$. Since the sum of all vertex degrees is $2E$, the average degree of a vertex is $2E/V = (2E)/(E/3 + 2) &amp;lt; 6$. So the average degree is less than $6$ and there must be a vertex $v$ of degree $\leq 5$. Delete $v$, $6$-color the rest of the graph (by induction), and add $v$ back. Since the vertex has degree $\leq 5$, there is some color which does not border $v$, color $v$ that color.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To prove the $5$ color map theorem requires one new idea. Find the vertex $v$ as before, color the rest of the graph and add it back. But we have a problem if the $v$ has degree $5$ and its neighbors all have different colors. The idea to get around this is to consider the subgraphs $G_{ij}$ consisting of vertices of colors $i$ and $j$, for different pairs $(i,j)$. Because $G_{ij}$ and $G_{k \ell}$ can not cross, for $i$, $j$, $k$ and $\ell$ distinct, some of these graphs must be disconnected. By considering the finitely possible many arrangements of these graphs, one sees that, in each case, one can take some component of $G_{ij}$ and switch the colors $i$ and $j$ on that component in order to make $v$ have only $4$ distinct colors of neighbor. See, for example, &lt;a href=&quot;http://en.wikipedia.org/wiki/Five_color_theorem#Outline_of_the_proof_by_contradiction&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt; for details.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The $4$ color map theorem requires the recoloring idea of the previous paragraph, plus one more idea. Instead of showing that there is always a vertex of degree $3$, $4$ or $5$, one shows that one of a finite list of larger configurations is unavoidable. This list tends to have 600-2000 members, depending on exactly which copy of the proof you read. The idea to showing that this list is unavoidable is an averaging argument like the one I used; the technical term to search for is &lt;a href=&quot;http://en.wikipedia.org/wiki/Discharging_method_%28discrete_mathematics%29&quot; rel=&quot;nofollow&quot;&gt;&quot;discharging&quot;&lt;/a&gt;. For each of these hundreds of configurations, one removes it from the graph, colors the rest, and analyzes the possible positions of the graphs $G_{ij}$. In each case one can recolor some of the components of the $G_{ij}$ in such a way that, when the configuration is put back in, the whole graph will be colorable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As I understand it, computers are used in two ways. The less interesting one is that there is a routine which verifies that the discharging argument is correct, finds the possible positions of the $G_{ij}$ and checks that they can be recolored. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The more interesting one is that, when the routine in the previous paragraph fails, the computer perturbs the discharging argument, finds a new resulting list of configurations and tries again.&#xA;It is basically a genetic process, where the computer tweaks the discharging algorithm until it finds a list of configurations which works.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The leap of faith which Appel and Haken took was to believe that there was some sufficiently complex discharging algorithm, and some list of configurations for which this procedure would work, and they just had to search long enough. Since then, this search has been implemented many times, with the details of the search algorithm different in each case, and many different solutions have been found.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To my mind, the whole thing reminds me of &lt;a href=&quot;http://en.wikipedia.org/wiki/Parallel_evolution&quot; rel=&quot;nofollow&quot;&gt;parallel evolution&lt;/a&gt;. There are lots of ways to make a wing, and if you leave an organism for long enough in an environment that rewards flight then it will develop one. But all of those ways are so complex that a human intellegence couldn't engineer them. Similarly, there are lots of discharging algorithms that prove the $4$ color map theorem, but they are all too bizarre for a human mathematician to find directly.&lt;/p&gt;&#xA;" OwnerUserId="448" LastActivityDate="2010-09-14T15:19:51.157" CommentCount="6" />
  <row Id="4661" PostTypeId="2" ParentId="4648" CreationDate="2010-09-14T15:45:12.443" Score="4" ViewCount="" Body="&lt;p&gt;I believe that Theorem 1 of Bezboruah and Shepherdson 1976 [1] covers your question, at least in spirit. Their theory $T_0$ is a finite theory extending $Q$. Quoting their paper:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Theorem 1.&lt;/strong&gt; Let $L$ be any formal system with a recursive set of axioms, a finite number of finitary and recursive rules of inference including modus ponens and having $A \to A$ as a theorem for all sentences $A$. Let&#xA;  $$Con_L =_{df}\quad \lnot(\exists y,z)(Th_L(y) \land Th_L(z) \land neg(z,y))$$&#xA;  where $\text{Th}_L$, $\text{neg}$ are given in Definition 3 below. Then $\text{Con}_L$ is not provable in $T_0$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The authors, however, express the common doubt that consistency proofs in Q are philosophically meaningful. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;We must agree with Kreisel that this is devoid of any philosophical interest and that in such a weak system this formula cannot be said to express consistency but only an algebraic property which in a stronger system (e.g. Peano arithmetic P) could reasonably be said to express the consistency of Q.&quot;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The (well known?) difficulty here is that Q can formalize the provability predicate but cannot verify the Hilbert-Bernays derivability conditions for it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1: A. Bezboruah and J. C. Shepherdson, &quot;Gödel's Second Incompleteness Theorem for Q&quot;, &lt;em&gt;The Journal of Symbolic Logic&lt;/em&gt; Vol. 41, No. 2 (Jun., 1976), pp. 503-512, &lt;a href=&quot;http://www.jstor.org/stable/info/2272251&quot; rel=&quot;nofollow&quot;&gt;JStor&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="630" LastActivityDate="2010-09-14T15:45:12.443" CommentCount="3" />
  <row Id="4662" PostTypeId="2" ParentId="4640" CreationDate="2010-09-14T15:55:58.170" Score="1" ViewCount="" Body="&lt;p&gt;There can be no meaningful answer without a prior definition of &quot;simpler&quot;. What is simplest or most useful is highly context dependent. For example, in some particular contexts it might prove &quot;simpler&quot; to employ $\rm\;A^2 + 6A(A+1)\;$ instead of either of the above forms.&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-09-14T15:55:58.170" />
  <row Id="4663" PostTypeId="2" ParentId="4638" CreationDate="2010-09-14T16:10:26.273" Score="3" ViewCount="" Body="&lt;p&gt;We have $A(45,10),B(10,20),C(x_c,y_c)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$AB=AC=\sqrt{(45-10)^{2}+(10-20)^{2}}=5\sqrt{53}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$C=B=20^{%&#xA;%TCIMACRO{\U{ba}}%&#xA;%BeginExpansion&#xA;{{}^o}%&#xA;%EndExpansion&#xA;}=\pi /9$ rad&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$A=180^{%&#xA;%TCIMACRO{\U{ba}}%&#xA;%BeginExpansion&#xA;{{}^o}%&#xA;%EndExpansion&#xA;}-40^{%&#xA;%TCIMACRO{\U{ba}}%&#xA;%BeginExpansion&#xA;{{}^o}%&#xA;%EndExpansion&#xA;}=140^{%&#xA;%TCIMACRO{\U{ba}}%&#xA;%BeginExpansion&#xA;{{}^o}%&#xA;%EndExpansion&#xA;}=7\pi /9$ rad&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/ziKPQ.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's make the following change of variables: $x=X+45,y=Y+10$ (translation of axes). Then $A$ becomes the origin of the $XY$ referential.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The vector $\overrightarrow{AB}$ can be written in this $XY$ referential as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\overrightarrow{AB}=(5\sqrt{53}\cos \left( \pi -\arctan \frac{2}{7}\right)&#xA;,5\sqrt{53}\sin \left( \pi -\arctan \frac{2}{7}\right) )=(-35,10)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and the vector $\overrightarrow{AC}$ as &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\overrightarrow{AC}=(5\sqrt{53}\cos \left( \pi -\frac{7\pi }{9}-\arctan &#xA;\frac{2}{7}\right) ,5\sqrt{53}\sin \left( \pi -\frac{7\pi }{9}-\arctan \frac{%&#xA;2}{7}\right) )$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore in the original referencial $xy$, we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x_{C}=5\sqrt{53}\cos \left( -\arctan \frac{2}{7}-\frac{7\pi }{9}+\pi&#xA;\right) +45\approx 78.239$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$y_{C}=5\sqrt{53}\sin \left( -\arctan \frac{2}{7}-\frac{7\pi }{9}+\pi&#xA;\right) +10\approx 24.837$&lt;/p&gt;&#xA;" OwnerUserId="752" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2010-09-14T19:00:53.433" LastActivityDate="2010-09-14T19:00:53.433" />
  <row Id="4664" PostTypeId="2" ParentId="4609" CreationDate="2010-09-14T16:29:19.703" Score="3" ViewCount="" Body="&lt;p&gt;No, it does not have a standard name.  By analogy to the lognormal distribution it might be dubbed the &quot;exponential-normal&quot; distribution.  But the analogy is imperfect, because you have to truncate the normal part at zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The integrals to compute the moments (or, equivalently, the characteristic function) do not have closed forms.  You will need to compute them numerically.  They are obtained by integrating&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{1}{\sqrt{2 \pi } \sigma } x^k e^{x-\frac{\left(e^x-\mu \right)^2}{2 \sigma ^2}} dx$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;over the entire real line.  (This comes from the pdf for a Normal distribution of $y$ with mean $\mu$ and standard deviation $\sigma$ and substituting $y = exp(x)$.)  $k = 1$ yields the mean; $k = 2$ gives the second moment; subtracting the square of the mean from that is the variance; and the square root of the variance is the standard deviation.  Finding the width at half maximum also requires numerical methods (but convergence should be rapid).&lt;/p&gt;&#xA;" OwnerUserId="1489" LastActivityDate="2010-09-14T16:29:19.703" CommentCount="1" />
  <row Id="4665" PostTypeId="1" AcceptedAnswerId="4673" CreationDate="2010-09-14T16:49:26.483" Score="2" ViewCount="243" Body="&lt;p&gt;Hi,&#xA;It is a well known theorem that any doubly connected domain can be conformally mapped onto an annulus.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the simpler version :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose $D$ is a bounded domain whose boundary is two non-intersecting circles. Then $D$ can be conformally mapped onto an annulus.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe that the proof of this should be easier, that the conformal map would be just a linear fractional transformation. However, I'm having trouble constructing it. Could someone help?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you&lt;/p&gt;&#xA;" LastActivityDate="2010-09-14T18:44:31.647" Title="Conformal mapping of a doubly connected domain onto an annulus" Tags="&lt;complex-analysis&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="4666" PostTypeId="1" AcceptedAnswerId="4677" CreationDate="2010-09-14T17:05:08.740" Score="4" ViewCount="217" Body="&lt;p&gt;In measure theory, we have &quot;lambda systems&quot; and &quot;pi systems&quot;.  Pearl's message passing algorithm has &quot;lambda messages&quot; and &quot;pi messages&quot;.  Is there a reason that lambda and pi go together?&lt;/p&gt;&#xA;" OwnerUserId="774" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2011-04-16T18:02:39.123" LastActivityDate="2011-04-16T18:02:39.123" Title="Why do lambda and pi go together?" Tags="&lt;notation&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="4667" PostTypeId="1" CreationDate="2010-09-14T17:30:49.437" Score="3" ViewCount="109" Body="&lt;p&gt;Is it possible to quantify the number of dimensions in combinatorial spaces.  The space I am particularly interested is all partitions of a set, bounded by the Bell number, where objects in this space are particular partitions.&lt;/p&gt;&#xA;" OwnerUserId="1453" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2010-11-24T02:31:34.547" LastActivityDate="2010-11-24T02:31:34.547" Title="Quantifying dimensionality of combinatorial space" Tags="&lt;combinatorics&gt;&lt;graph-theory&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" />
  <row Id="4668" PostTypeId="2" ParentId="4643" CreationDate="2010-09-14T18:15:02.590" Score="0" ViewCount="" Body="&lt;p&gt;Counting submultisets is the same as asking for the number of divisors of $\prod p_i^{m_i}$ (where the $p_i$ are distinct but otherwise arbitrary), so it's just $\prod(m_i+1)$.&lt;/p&gt;&#xA;" OwnerUserId="1778" LastActivityDate="2010-09-14T18:15:02.590" CommentCount="4" />
  <row Id="4669" PostTypeId="1" AcceptedAnswerId="4672" CreationDate="2010-09-14T18:18:32.560" Score="5" ViewCount="215" Body="&lt;p&gt;Could anyone give an example of a sequence of &lt;em&gt;differentiable&lt;/em&gt; (real) functions that uniformly converges to a &lt;em&gt;differentiable&lt;/em&gt; function, but the derivatives of which don't converge to the derivative of the limit function?&lt;/p&gt;&#xA;" OwnerUserId="766" LastActivityDate="2010-09-14T19:17:06.103" Title="Example: Function sequence uniformly converges, its derivatives don't." Tags="&lt;calculus&gt;&lt;real-analysis&gt;&lt;intuition&gt;" AnswerCount="2" />
  <row Id="4670" PostTypeId="2" ParentId="4510" CreationDate="2010-09-14T18:21:33.887" Score="2" ViewCount="" Body="&lt;p&gt;Sometimes partial fractions decomposition is useful. Especially if there are only simple poles, since then each partial fraction can be expanded in a geometric series.&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2010-09-14T18:21:33.887" />
  <row Id="4671" PostTypeId="2" ParentId="4669" CreationDate="2010-09-14T18:25:45.673" Score="3" ViewCount="" Body="&lt;p&gt;Consider the function $f_{n}: [0,2\pi] \to \mathbb{R}$ defined by $f_{n}(x)=n^{-1/2}\sin{nx}$ and let $f:[0,2\pi] \to \mathbb{R}$ be the zero function, that is $f(x)=0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ok, here i work out the details. Since $\sin{x}$ oscillates between -1 and 1, we have $d_{\infty}(f_{n},f) \leq n^{-1/2}$, where $d_{\infty}$ is the uniform metric defined as $d_{\infty}(f,g)= \sup_{x \in [0, 2\pi]} |f(x)-g(x)|$. Since $\frac{1}{\sqrt{n}} \to 0$, using squeeze test we see that $f_{n} \to f$ uniformly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whereas $f_{n}'(x)= \sqrt{n}\cos{nx}$ so we have $|f_{n}'(0-f'(0)|=\sqrt{n}$, which says that $f_{n}'$ does not converge pointwise to $f'$. &lt;/p&gt;&#xA;" LastActivityDate="2010-09-14T19:17:06.103" CommentCount="0" />
  <row Id="4672" PostTypeId="2" ParentId="4669" CreationDate="2010-09-14T18:26:07.623" Score="8" ViewCount="" Body="&lt;p&gt;This is from the book, Counterexamples in Analysis by Gelbaum and Olmstead. (Google books link: &lt;a href=&quot;http://books.google.com/books?id=cDAMh5n4lkkC&quot; rel=&quot;nofollow&quot;&gt;http://books.google.com/books?id=cDAMh5n4lkkC&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is under:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;A sequence of infinitely differentiable functions converging uniformly to zero, the sequence of whose derivatives diverges everywhere.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f_{n}(x) = \frac{\sin nx}{\sqrt{n}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is on page 76, Chapter 7.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-09-14T18:26:07.623" />
  <row Id="4673" PostTypeId="2" ParentId="4665" CreationDate="2010-09-14T18:44:31.647" Score="3" ViewCount="" Body="&lt;p&gt;Suppose your outer circle is the unit circle $C_1=\{z:|z|=1\}$.&#xA;All you have to do is to get a linear fractional transformation&#xA;taking $C_1$ to itself and the inner circle $C_2$ to a circle centred&#xA;at the origin. You might as well rotate $C_2$ so that its centre&#xA;is on the real axis. Then $C_2$ meets the real axis in points $a$ and $b$&#xA;with $-1 &amp;lt; a &amp;lt; b &amp;lt; 1$. The required linear fractional transformation&#xA;will have real coefficients and take $-1$, $a$, $b$, $1$ to $-1$,&#xA;$-c$, $c$, $1$ where $0 &amp;lt; c &amp;lt; 1$. The typical linear fractional transformation&#xA;fixing $-1$ and $1$ has the form&#xA;$$f_u:z\mapsto\frac{z+u}{uz+1}.$$&#xA;We are considering $u$ real and for $u$ to take the unit disc to itself,&#xA;$|u| &amp;lt; 1$. We need $f_u(a)=-f_u(b)$. This is a quadratic equation in $u$ with two&#xA;real roots, one of which satsifies $|u| &amp;lt; 1$.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-09-14T18:44:31.647" CommentCount="1" />
  <row Id="4674" PostTypeId="2" ParentId="4658" CreationDate="2010-09-14T19:06:24.917" Score="10" ViewCount="" Body="&lt;p&gt;I'll sketch a solution; details are left to you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you flip your coin, think about what data you would want to keep track of to see whether $7$ heads have come up yet. You'd want to know: Whether you have already won and what the number of heads at the end of your current sequence was. In other words, there are $8$ states:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$A$: We have not flipped $7$ heads in a row yet, and the last flip was $T$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$B$: We have not flipped $7$ heads in a row yet, and the last two flips was $TH$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$C$: We have not flipped $7$ heads in a row yet, and the last three flips were $THH$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ldots$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$G$: We have not flipped $7$ heads in a row yet, and the last seven flips were $THHHHHH$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$H$: We've flipped $7$ heads in a row!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we are in state $A$ then, with probability $1/2$ we move to state $B$ and with probability $1/2$ we stay in state $A$. If we are in state $B$ then, with probability $1/2$ we move to state $C$ and with probability $1/2$ we move back to state $A$. $\ldots$ If we are in state $G$, with probability $1/2$ we move forward to state $H$ and with probability $1/2$ we move back to state $A$. Once we are in state $H$ we stay there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In short, define $M$ to be the matrix&#xA;$$\begin{pmatrix} &#xA;1/2 &amp;amp; 1/2 &amp;amp; 1/2 &amp;amp; 1/2 &amp;amp; 1/2 &amp;amp; 1/2 &amp;amp; 1/2 &amp;amp; 0 \\&#xA;1/2 &amp;amp; 0   &amp;amp; 0   &amp;amp; 0   &amp;amp; 0   &amp;amp; 0   &amp;amp; 0   &amp;amp; 0 \\&#xA;0   &amp;amp; 1/2 &amp;amp; 0   &amp;amp; 0   &amp;amp; 0   &amp;amp; 0   &amp;amp; 0   &amp;amp; 0 \\&#xA;0   &amp;amp; 0   &amp;amp; 1/2 &amp;amp; 0   &amp;amp; 0   &amp;amp; 0   &amp;amp; 0   &amp;amp; 0 \\&#xA;0   &amp;amp; 0   &amp;amp; 0   &amp;amp; 1/2 &amp;amp; 0   &amp;amp; 0   &amp;amp; 0   &amp;amp; 0 \\&#xA;0   &amp;amp; 0   &amp;amp; 0   &amp;amp; 0   &amp;amp; 1/2 &amp;amp; 0   &amp;amp; 0   &amp;amp; 0 \\&#xA;0   &amp;amp; 0   &amp;amp; 0   &amp;amp; 0   &amp;amp; 0   &amp;amp; 1/2 &amp;amp; 0   &amp;amp; 0 \\&#xA;0   &amp;amp; 0   &amp;amp; 0   &amp;amp; 0   &amp;amp; 0   &amp;amp; 0   &amp;amp; 1/2 &amp;amp; 1 &#xA;\end{pmatrix}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then the entries of $M^n$ give the probability of transitioning from one given state to another in $n$ coin flips. (Please, please, please, do not go on until you understand why this works! This is one of the most standard uses of matrix multiplication.) You are interested in the lower left entry of $M^{150}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, a computer algebra system can compute this number for you quite rapidly. Rather than do this, I will discuss some interesting math which comes out of this.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt; &#xA;&#xA;&lt;p&gt;(1) The &lt;a href=&quot;http://en.wikipedia.org/wiki/Perron%E2%80%93Frobenius_theorem&quot; rel=&quot;nofollow&quot;&gt;Perron-Frobenius theorem&lt;/a&gt; tells us that $1$ is an eigenvalue of $M$ (with corresponding eigenvector $(0,0,0,0,0,0,0,1)^T$, in this case) and all the other eigenvalues are less then $1$. Let $\lambda$ be the largest eigenvalue less than $1$, then probability of getting $7$ heads in a row, when we flip $n$ times, is approximately $1-c \lambda^n$ for some constant $c$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) You might wonder what sort of configurations of coin flips can be answered by this method. For example, could we understand the case where we flip $3$ heads in a row before we flip $2$ tails in a row? (Answer: Yes.) Could we understand the question of whether the first $2k$ flips are a palindrome for some $k$? (Answer: No, not by this method.) In general, the question is which properties can be recognized by &lt;a href=&quot;http://en.wikipedia.org/wiki/Finite-state_machine&quot; rel=&quot;nofollow&quot;&gt;finite state automata&lt;/a&gt;, also called the regular languages. There is a lot of study of this subject. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(3) See chapter $8.4$ of &lt;em&gt;Concrete Mathematics&lt;/em&gt;, by Graham, Knuth and Patashnik, for many more coin flipping problems.&lt;/p&gt;&#xA;" OwnerUserId="448" LastActivityDate="2010-09-14T19:06:24.917" />
  <row Id="4675" PostTypeId="2" ParentId="4658" CreationDate="2010-09-14T19:11:06.607" Score="12" ViewCount="" Body="&lt;p&gt;Here are some details; I will only work out the case where you want $7$ tails in a row, and the general case is similar.  I am interpreting your question to mean &quot;what is the probability that, at least once, you flip at least 7 tails in a row?&quot;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $a_n$ denote the number of ways to flip $n$ coins such that at no point do you flip more than $6$ consecutive tails.  Then the number you want to compute is $1 - \frac{a_{150}}{2^{150}}$.  The last few coin flips in such a sequence of $n$ coin flips must be one of $H, HT, HTT, HTTT, HTTTT, HTTTTT$, or $HTTTTTT$.  After deleting this last bit, what remains is another sequence of coin flips with no more than $6$ consecutive tails.  So it follows that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$a_{n+7} = a_{n+6} + a_{n+5} + a_{n+4} + a_{n+3} + a_{n+2} + a_{n+1} + a_n$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with initial conditions $a_k = 2^k, 0 \le k \le 6$.  Using a computer it would not be very hard to compute $a_{150}$ from here, especially if you use the matrix method that David Speyer suggests.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In any case, let's see what we can say approximately.  The asymptotic growth of $a_n$ is controlled by the largest positive root of the characteristic polynomial $x^7 = x^6 + x^5 + x^4 + x^3 + x^2 + x + 1$, which is a little less than $2$.  Rearranging this identity gives $2 - x = \frac{1}{x^7}$, so to a first approximation the largest root is $r \approx 2 - \frac{1}{128}$.  This means that $a_n$ is approximately $\lambda \left( 2 - \frac{1}{128} \right)^n$ for some constant $\lambda$, which means that $\frac{a_{150}}{2^{150}}$ is roughly&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\lambda \left( 1 - \frac{1}{256} \right)^{150} \approx \lambda e^{ - \frac{150}{256} } \approx 0.56 \lambda$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;although $\lambda$ still needs to be determined.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt;  So let's approximate $\lambda$.  I claim that the generating function for $a_n$ is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$A(x) = 1 + \sum_{n \ge 1} a_{n-1} x^n = \frac{1}{1 - x - x^2 - x^3 - x^4 - x^5 - x^6 - x^7}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is because, by iterating the argument in the second paragraph, we can decompose any valid sequence of coin flips into a sequence of one of seven blocks $H, HT, ...$ uniquely, except that the initial segment does not necessarily start with $H$.  To simplify the above expression, write $A(x) = \frac{1 - x}{1 - 2x + x^8}$.  Now, the partial fraction decomposition of $A(x)$ has the form&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$A(x) = \frac{\lambda}{r(1 - rx)} + \text{other terms}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\lambda, r$ are as above, and it is this first term which determines the asymptotic behavior of $a_n$ as above.  To compute $\lambda$ we can use l'Hopital's rule; we find that $\lambda$ is equal to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\lim_{x \to \frac{1}{r}} \frac{r(1 - rx)(1 - x)}{1 - 2x + x^8} = \lim_{x \to \frac{1}{r}} \frac{-r(r+1) + 2r^2x}{-2 + 8x^7} = \frac{r^2-r}{2 - \frac{8}{r^7}} \approx 1.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So my official guess at the actual value of the answer is $1 - 0.56 = 0.44$.  Anyone care to validate it?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Sequences like $a_n$ count the number of words in objects called &lt;a href=&quot;http://en.wikipedia.org/wiki/Regular_language&quot; rel=&quot;nofollow&quot;&gt;regular languages&lt;/a&gt;, whose enumerative behavior is described by &lt;a href=&quot;http://en.wikipedia.org/wiki/Recurrence_relation#Linear_homogeneous_recurrence_relations_with_constant_coefficients&quot; rel=&quot;nofollow&quot;&gt;linear recurrences&lt;/a&gt; and which can also be analyzed using &lt;a href=&quot;http://en.wikipedia.org/wiki/Finite-state_machine&quot; rel=&quot;nofollow&quot;&gt;finite state machines&lt;/a&gt;.  Those are all good keywords to look up if you are interested in generalizations of this method.  I discuss some of these issues in my &lt;a href=&quot;http://web.mit.edu/~qchu/Public/TopicsInGF.pdf&quot; rel=&quot;nofollow&quot;&gt;notes on generating functions&lt;/a&gt;, but you can find a more thorough introduction in the relevant section of Stanley's &lt;a href=&quot;http://www-math.mit.edu/~rstan/ec/&quot; rel=&quot;nofollow&quot;&gt;Enumerative Combinatorics&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-09-14T19:51:28.350" LastActivityDate="2010-09-14T19:51:28.350" CommentCount="10" />
  <row Id="4676" PostTypeId="2" ParentId="4102" CreationDate="2010-09-14T19:12:04.467" Score="0" ViewCount="" Body="&lt;p&gt;I think I see where your confusion is rooted.  The notation f(x|θ) is chosen in the first place because there's a implication of a causal relationship between theta (the inferred cause) and x (the resulting observations.)  And also because we usually have simple pdf representing the density on x given θ. This function is necessarily invertible, so nothing forces us to write things this way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, in the problem you presented, we are given x and inferring θ.  So, f(x|θ) is read backwards.  It could have been written L(θ|x) (the likelihood of θ given x).  And then your intuition is almost prefect, except for one flaw discussed below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can write that a statistic t is sufficient iff L(θ|x) = L(θ|t,x), or equivalently, L(θ|t,x) = L(θ|t).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The flaw with your equation is that you need to condition on both t and x. Because even if L(θ|x) = L(θ|t), t could be based on information other than (just) x.  For example, if x comprises n examples, θ could be the result of a different n examples.  In that case, the combination of t and x would be like seeing 2n examples, and would almost always yield a different result.&lt;/p&gt;&#xA;" OwnerUserId="774" LastEditorUserId="774" LastEditorDisplayName="" LastEditDate="2010-09-14T19:18:03.517" LastActivityDate="2010-09-14T19:18:03.517" />
  <row Id="4677" PostTypeId="2" ParentId="4666" CreationDate="2010-09-14T20:01:25.120" Score="3" ViewCount="" Body="&lt;p&gt;The measure theory ideas of $\pi$-system and $\lambda$-system were introduced by Dynkin in his book &lt;em&gt;Die Grundlagen der Theorie der Markoffschen Prozesse&lt;/em&gt; (1961 German translation of 1959 Russian original); a note at the end of the book mentions they are new, but doesn't explain why they are so called.  My guess has been that $\pi$ is for &quot;product&quot; and $\lambda$ is for &quot;limit,&quot; or some Russian cognates thereof; I'm not sure there's any connection between the letters themselves.  Although Professor Dynkin has recently retired, he still has an office here at Cornell; if I see him and I think of it, I may ask him about it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately I don't know anything about message passing, so I can't say whether Pearl's terminology is related or a coincidence.&lt;/p&gt;&#xA;" OwnerUserId="822" LastActivityDate="2010-09-14T20:01:25.120" />
  <row Id="4678" PostTypeId="1" CreationDate="2010-09-14T20:08:52.257" Score="1" ViewCount="205" Body="&lt;p&gt;Let $G$ be a &lt;a href=&quot;http://en.wikipedia.org/wiki/Directed_graph&quot; rel=&quot;nofollow&quot;&gt;directed graph&lt;/a&gt; which contains N vertex's, and which satisfies the following condition: each vertex of $G$ has at least one incoming edge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(For clarity, a vertex $V_1$ has an &quot;incoming edge&quot; if there is an edge $E$ which connects $V_1$ to some other vertex on the graph, $V_2$, for which $E$ is directed from $V_2$ to $V_1$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then I believe it follows that $G$ contains a closed cycle.  The reason I am posting this is because the result seems straightforward/obvious, and yet it would apparently solve an &lt;a href=&quot;http://mathoverflow.net/questions/16857/existence-of-a-zero-sum-subset&quot; rel=&quot;nofollow&quot;&gt;open problem on MO regarding zero sum subsets&lt;/a&gt;, so I am concerned that I may have a mistake.  (To be frank, I am not comfortable posting on MO)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The basic idea is to show that starting with a set of N vertex's, and no edges.  That it is impossible to construct a directed graph with no closed cycles by adding one (directed) edge at a time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let a set of N vertex's be given, and let us attempt to construct a directed graph with no closed cycle by adding one directed edge at a time.  Choose some arbitrary vertex $V_1 \in G$, then by the defining condition of $G$, $V_1$ must have an incoming edge, say from $V_2$.  It's clear that $V_2$ has an incoming edge, if this edge originates from $V_1$, then the graph would have a closed cycle, so we choose some other vertex which connects to $V_2$: and in this way we are forced to create a chain of N-1 vertex's, $V_1$, has an edge coming from $V_2$, which has an edge coming from $V_3$, ...$V_{N-2} $, which has an edge coming from $V_{N-1}$.  It's clear that if any of the vertex's in this chain has an incoming edge originating from another vertex on the chain, that $G$ will contain a closed cycle.  So once we construct a chain of N-1 distinct vertex's, there is one vertex left, and since this vertex must have an incoming edge originating from another vertex which is already on the chain, $G$ must contain a closed cycle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let me explain why I think this fairly trivial result solves the &lt;a href=&quot;http://mathoverflow.net/questions/16857/existence-of-a-zero-sum-subset&quot; rel=&quot;nofollow&quot;&gt;open zero sum subset problem&lt;/a&gt; on MO.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This question asks: given a finite set of real numbers $S$ with the property that every number in the set can be written as the sum of two numbers in the set (not necessarily different), if there exists a subset $S$ which sums to zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, clearly a set $S$ which satisfies the condition of the problem stated above corresponds to a directed graph for which each vertex has at least one incoming edge (each real number in the set $S$ is associated with a vertex, and if given three numbers $a,b,c \in S$, with $a = b + c$, then the vertex associated with $a$ will have incoming edges from the vertex's associated with $b$ and $c$).   Therefore the graph corresponding to an arbitrary S which satisfies the condition contains a closed cycle.  By adding the numbers in the closed cycle, we obtain a sum of the form:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a + b + c + .... = a$, hence it follows that $b + c + ... = 0$, i.e. a set $S$ which satisfies the condition must contain a zero sum subset.  Have I made a mistake somewhere, or does this solve the problem? &lt;/p&gt;&#xA;" OwnerUserId="1417" LastEditorUserId="1417" LastEditorDisplayName="" LastEditDate="2010-09-14T20:29:47.717" LastActivityDate="2010-09-14T20:42:03.677" Title="Does this solve the zero sum subset problem?" Tags="&lt;graph-theory&gt;" AnswerCount="0" CommentCount="4" />
  <row Id="4680" PostTypeId="1" AcceptedAnswerId="4681" CreationDate="2010-09-14T21:00:16.327" Score="6" ViewCount="320" Body="&lt;p&gt;As an example, consider the polynomial f(x) = x^3 + x - 2 = (x - 1)(x^2 + x + 2) which clearly has a root x = 1.&#xA;But we can also find the roots using Cardano's method, which leads to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x = \sqrt[3]{\sqrt{28/27} + 1} - \sqrt[3]{\sqrt{28/27} - 1}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and 2 other roots.&#xA;It's easy to check numerically that this expression is really equal to 1, but is there a way to derive it algebraically which isn't equivalent to showing that this expression satisfies f(x) = 0?&lt;/p&gt;&#xA;" OwnerUserId="1470" LastEditorUserId="4290" LastEditorDisplayName="" LastEditDate="2011-02-19T01:32:35.580" LastActivityDate="2011-02-19T23:08:28.830" Title="Simplification of expressions containing radicals" Tags="&lt;algebra&gt;&lt;roots&gt;" AnswerCount="3" FavoriteCount="1" />
  <row Id="4681" PostTypeId="2" ParentId="4680" CreationDate="2010-09-14T21:12:14.693" Score="6" ViewCount="" Body="&lt;p&gt;Yes.  The first thing to try is to guess that $\sqrt[3]{ \left( \sqrt{ \frac{28}{27} } \pm 1 \right) } = \pm \frac{1}{2} + \sqrt{a}$ for some $a$.  Cubing both sides then gives&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{2}{9} \sqrt{21} \pm 1 = \pm \frac{1}{8} + \frac{3}{4} \sqrt{a} \pm \frac{3}{2} a + a \sqrt{a}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Setting $1 = \frac{1}{8} + \frac{3a}{2}$ gives $a = \frac{7}{12}$, and we can verify that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{3}{4} \sqrt{a} + a \sqrt{a} = \frac{1}{8} \sqrt{21} + \frac{7}{72} \sqrt{21} = \frac{2}{9} \sqrt{21}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;as desired.  If this method doesn't work then the problem becomes harder.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-09-14T21:12:14.693" CommentCount="4" />
  <row Id="4682" PostTypeId="1" AcceptedAnswerId="4688" CreationDate="2010-09-14T21:12:57.730" Score="2" ViewCount="107" Body="&lt;p&gt;I'm a software engineer and have been asked to research a Voronoi implementation in four dimensions. I'm not asking for &quot;teh codez&quot; but &lt;em&gt;am&lt;/em&gt; interested in approachable tutorials on Voronoi decomposition and/or implementations in any dimension. Any help here would be greatly appreciated!&lt;/p&gt;&#xA;" OwnerUserId="124" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-09-14T22:08:10.453" LastActivityDate="2010-09-15T05:38:59.310" Title="Voronoi decomposition implementation in four dimensions?" Tags="&lt;algorithms&gt;&lt;reference-request&gt;&lt;discrete-geometry&gt;" AnswerCount="2" />
  <row Id="4683" PostTypeId="1" AcceptedAnswerId="4684" CreationDate="2010-09-14T21:46:05.027" Score="4" ViewCount="426" Body="&lt;p&gt;I know that a continuous function which is a BV may not be absolutely continuous.  Is there an example of such a function?   I was looking for a BV whose derivative is not Lebesgue integrable but I couldn't find one.&lt;/p&gt;&#xA;" OwnerUserId="832" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2010-09-14T23:01:42.567" LastActivityDate="2010-09-14T23:01:42.567" Title="Continuous and bounded variation does not imply absolutely continuous." Tags="&lt;real-analysis&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="4684" PostTypeId="2" ParentId="4683" CreationDate="2010-09-14T22:01:46.720" Score="4" ViewCount="" Body="&lt;p&gt;The Devil's staircase function does the trick.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Its derivative is almost surely zero with respect to Lebesgue measure,&#xA;so the function is not absolutely continuous.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;See &lt;a href=&quot;http://mathworld.wolfram.com/DevilsStaircase.html&quot; rel=&quot;nofollow&quot;&gt;http://mathworld.wolfram.com/DevilsStaircase.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="940" LastActivityDate="2010-09-14T22:01:46.720" CommentCount="1" />
  <row Id="4685" PostTypeId="2" ParentId="4683" CreationDate="2010-09-14T22:10:53.583" Score="3" ViewCount="" Body="&lt;p&gt;Byron already answered your main question, but your last sentence is another matter.  You want a BV function whose derivative is not integrable, but such things don't exist.  In particular, if $f$ is monotone on $[a,b]$, then $f'$ exists a.e., is Lebesgue integrable, and $\int_a^b f' \leq f(b)-f(a)$.  Thus half of the fundamental theorem of calculus holds, so to speak.  General BV functions are differences of monotone functions, so their derivatives are also Lebesgue integrable.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-09-14T22:10:53.583" CommentCount="2" />
  <row Id="4687" PostTypeId="1" CreationDate="2010-09-14T23:15:33.803" Score="6" ViewCount="169" Body="&lt;p&gt;My frustration started after hours of searching failed to turn up a formula for the vertex of a parabola in the general form&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$Ax^2+Bxy+Cy^2+Dx+Ey+F=0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As is already well known, the discriminant $\Delta=B^2-4AC$ can be used to diagnose the type of conic one has if it it given in the general form. Other useful formulae I found in the midst of searching include the formula for the slope of the principal axis:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\tan\;\theta=\frac{B}{A-C+(\mathrm{sgn}\;B)\sqrt{B^2+(A-C)^2}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the eccentricity (intended only for elliptic or hyperbolic cases)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\varepsilon=\left(\sqrt{\frac12-\frac{(\mathrm{sgn}\;\Delta)(A+C)}{2\sqrt{B^2+(A-C)^2}}}\right)^{-1}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and the coordinates of the center for a central conic (I won't list it here so you can figure it out or find it yourself ;) ), but there were no formulae given for finding the coordinates of the vertex of a parabola.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, for the parabola case, I could rotate axes, find the vertex through completing the square, and then rotate back, but I was hoping somebody already went through the trouble of deriving a formula so I don't have to reinvent the wheel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On to my query: is there a comprehensive compilation (book, article) somewhere of formulae related to dealing with conic sections? The implicit Cartesian form is what I'm currently dealing with, but lists relating to other forms (polar, parametric) are welcome as well.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-09-14T23:15:33.803" Title="Comprehensive compilation of conic section formulae" Tags="&lt;geometry&gt;&lt;reference-request&gt;&lt;conic-sections&gt;&lt;plane-curves&gt;" CommentCount="8" FavoriteCount="1" />
  <row Id="4688" PostTypeId="2" ParentId="4682" CreationDate="2010-09-14T23:24:44.577" Score="4" ViewCount="" Body="&lt;p&gt;The standard survey is &lt;a href=&quot;http://www.pi6.fernuni-hagen.de/publ/tr198.pdf&quot; rel=&quot;nofollow&quot;&gt;Aurenhammer and Klein&lt;/a&gt;.  Have you looked at that?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[This would be a comment, but I don't have enough rep on this site yet]&lt;/p&gt;&#xA;" OwnerUserId="826" LastActivityDate="2010-09-14T23:24:44.577" />
  <row Id="4689" PostTypeId="1" CreationDate="2010-09-14T23:47:54.910" Score="1" ViewCount="43" Body="&lt;p&gt;In signal detection, an observer is assigned the task of discerning the presence (or absence) of some signal with accompanying noise. There are four possible outcomes: a hit, a miss, a false alarm, or a correct rejection. An iso-sensitivity function is the function that describes the relation between correct and false reports of a difference, with p(H) on the Y axis and p(FA) on the X axis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If p(H) = p + (1-p)FA&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and p(FA) = g  (guess)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and p, g both lie between 0 and 1,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;can someone walk me through how to derive the prediction for the iso-sensitivity function in ROC space? (Or point me to some source that can clearly explain this)?&lt;/p&gt;&#xA;" OwnerUserId="1902" LastActivityDate="2010-09-14T23:47:54.910" Title="Deriving Isosensitivity Functions in ROC space from elementary signal detection parameters" Tags="&lt;calculus&gt;&lt;functions&gt;" CommentCount="1" />
  <row Id="4690" PostTypeId="2" ParentId="3405" CreationDate="2010-09-15T00:02:09.473" Score="0" ViewCount="" Body="&lt;p&gt;Not a proof, but a conjecture.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $$s_{n} = \frac{p(n)}{n}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Conjecture 1&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems like the sequence $$\{(n+2)(n+3)s_{2n}\}$$ is monotonically decreasing! I verified this for the first 5000 terms or so, using a computer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The statement that the above sequence is monotonically decreasing is equivalent to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Conjecture 1.1&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ (n+4) s_{2n+1} &amp;lt; (n+2) s_{2n}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I haven't been able to prove that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe someone else can.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Others, feel free to edit this to add your observations if you like.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is some python code&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;from fractions import Fraction&#xA;&#xA;s = []&#xA;s.append(0)&#xA;s.append(Fraction(1,1))&#xA;s.append(Fraction(1,2))&#xA;&#xA;for i in xrange(3,1000):&#xA;    if (i %2 == 0):&#xA;        s.append(s[i-1])&#xA;    else:&#xA;        s.append(s[i-1] - s[(i-1)/2]/i )&#xA;&#xA;&#xA;for i in xrange(2,499):&#xA;    if (s[2*i]*(i+3) &amp;gt;= s[2*i-2]*(i+1)):&#xA;        print &quot;Failed&quot;&#xA;        print i&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-09-15T04:16:10.193" LastActivityDate="2010-09-15T04:16:10.193" CommunityOwnedDate="2010-09-15T00:02:09.473" CommentCount="1" />
  <row Id="4691" PostTypeId="1" CreationDate="2010-09-15T02:04:14.107" Score="2" ViewCount="815" Body="&lt;p&gt;Is there a way to derive the Maclaurin series for $\frac{1}{(1-x)}$ after finding the Maclaurin series for $(1+x)^n$ which is $\displaystyle\sum\limits_{k=0}^\infty \frac{f^k(0)}{k!}*x^k$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the original equation I could substitute -1 for n and -x for x but I don't see how I could do that in the summation. The intention of this is getting the next series without having to do the expansion from scratch.&lt;/p&gt;&#xA;" OwnerUserId="1880" LastActivityDate="2010-09-15T11:20:35.807" Title="Maclaurin Series of 1/(1-x) derived from maclaurin series of (1+x)^n" Tags="&lt;calculus&gt;&lt;sequences-and-series&gt;&lt;taylor-expansion&gt;" AnswerCount="4" CommentCount="6" FavoriteCount="1" />
  <row Id="4692" PostTypeId="2" ParentId="4691" CreationDate="2010-09-15T02:08:44.560" Score="1" ViewCount="" Body="&lt;p&gt;Try division? 1/(1-x) = 1 + x + x^2 + x^3 ...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(edit: I'm not sure why you would try to derive it from the series for (1+x)^n ) &lt;/p&gt;&#xA;" OwnerUserId="120" LastActivityDate="2010-09-15T02:08:44.560" CommentCount="1" />
  <row Id="4693" PostTypeId="2" ParentId="4691" CreationDate="2010-09-15T02:15:09.933" Score="6" ViewCount="" Body="&lt;p&gt;The &lt;a href=&quot;http://mathworld.wolfram.com/BinomialSeries.html&quot; rel=&quot;nofollow&quot;&gt;binomial series&lt;/a&gt; works for any real $n$.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-09-15T02:15:09.933" CommentCount="2" />
  <row Id="4694" PostTypeId="1" AcceptedAnswerId="4709" CreationDate="2010-09-15T02:20:53.777" Score="3" ViewCount="323" Body="&lt;p&gt;While teaching the concept of vector spaces, my professor mentioned that addition and multiplication aren't necessarily what we &lt;em&gt;normally&lt;/em&gt; call addition and multiplication, but any other function that complies with the eight axioms needed by the definition of a vector space (for instance, associativity, commutativity of addition, etc.). Is there any widely used vector space in which alternative functions are used as addition/multiplication?&lt;/p&gt;&#xA;" OwnerUserId="1523" LastActivityDate="2011-02-25T11:40:05.273" Title="What are some alternative definitions of vector addition and scalar multiplication?" Tags="&lt;linear-algebra&gt;&lt;vector-space&gt;" AnswerCount="4" CommentCount="1" FavoriteCount="2" />
  <row Id="4695" PostTypeId="2" ParentId="4694" CreationDate="2010-09-15T02:49:29.250" Score="2" ViewCount="" Body="&lt;p&gt;Many of the operations commonly encountered in vector spaces, such as modular arithmetic, functional convolution, or even p-adic arithmetic, ultimately are related to the addition and multiplication with which you are familiar.  One interesting exception is Conway's field of &lt;em&gt;nimbers&lt;/em&gt; (which is effectively a one-dimensional vector space).  I wouldn't say it's &quot;widely used,&quot; but it's well enough known to qualify as a genuine example.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;References: &lt;a href=&quot;http://mathoverflow.net/questions/6455/nimber-multiplication&quot; rel=&quot;nofollow&quot;&gt;http://mathoverflow.net/questions/6455/nimber-multiplication&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Surreal_number&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Surreal_number&lt;/a&gt; (nimbers form a subfield of the surreal numbers).&lt;/p&gt;&#xA;" OwnerUserId="1489" LastActivityDate="2010-09-15T02:49:29.250" CommentCount="8" />
  <row Id="4696" PostTypeId="2" ParentId="4694" CreationDate="2010-09-15T02:50:19.483" Score="3" ViewCount="" Body="&lt;p&gt;This example is certainly not &quot;widely used&quot;, but I think it's worth thinking about anyway.  This answer comes from an MO post by John Goodrick: &lt;a href=&quot;http://mathoverflow.net/questions/9402/pedagogical-question-about-linear-algebra&quot; rel=&quot;nofollow&quot;&gt;http://mathoverflow.net/questions/9402/pedagogical-question-about-linear-algebra&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll quote the entirety of his post here in case you don't feel like clicking on the link (and since I've done no work on my own, I'll make this post CW)&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;You could try giving the following example: the set of all positive  real numbers, considered as a vector space over the field R, with vector addition given by multiplication and scalar multiplication given by taking exponents.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;As a first step, you could verify that this satisfies a few of the vector-space axioms, and then let students check the rest of them (say, as homework). Then, you could ask questions like, &quot;what is the dimension of this vector space?&quot; or, &quot;give an example of a (nontrivial) linear transformation from this space into R^3.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="331" LastActivityDate="2010-09-15T02:50:19.483" CommunityOwnedDate="2010-09-15T02:51:22.770" />
  <row Id="4697" PostTypeId="2" ParentId="4680" CreationDate="2010-09-15T03:03:43.943" Score="10" ViewCount="" Body="&lt;p&gt;There are very general algorithms known for radical denesting. Below is the structure theorem which lies at the foundation of these algorithms. It widely generalizes the heuristic employed by Qiaochu in his answer. It may be employed heuristically - in a similar manner as Qiaochu - to perform complicated denestings, without requiring much comprehension of the underlying theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In Bloemer's FOCS '91, '92 papers you'll find his polynomial-time algorithms for radical denesting. Informally, the Denesting Structure Theorem says that if a radical $\rm\; r^{1/d} \;$ denests in any radical extension $\rm\; F' \;$ of its base field $\rm\; F \;$, then a suitable multiple $\rm\; q b\: r \;$ of &#xA;the radicand $\rm\; r \;$ must already denest in the field $\rm\; F' \;$ defined by the &#xA;radicand. More precisely &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;DENESTING STRUCTURE THEOREM&lt;/strong&gt;$\;\; \;$  Let $\rm\; F \;$ be a real field and &#xA;$\rm\; F' = F(q_1^{1/d1},\ldots,q_k^{1/dk}) \;$ be a real radical extension of $\rm\; F \;$ &#xA;of degree $\rm\; n \;$. By $\rm\; B = \{b_0,\ldots, b_{n-1}\}$ denote the standard &#xA;basis of $\rm\; F' \;$ over $\rm\; F \;$. If $\rm\; r \;$ is in $\rm\; F' \;$ and $\rm\; d \;$ is a positive integer such &#xA;that $\rm\; r^{1/d} \;$ denests over $\rm\; F \;$ using only real radicals, that is, &#xA;$\rm\; r^{1/d} \;$ is in $\rm\; F(a_1^{1/t_1},\ldots,a_m^{1/t_m}) \;$ for some positive integers &#xA;$\rm\; t_i \;$ and positive $\rm\; a_i \in F \;$, then there exists a nonzero $\rm\; q \in F \;$ and a &#xA;$\rm\; b \in B \;$ such that $\rm\; (q b r)^{1/d} \in F' \;$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I.e. multiplying the radicand by a $\rm\; q \;$ in the base field $\rm\; F \;$ &#xA;and a power product $\rm\; b \;=\; q_1^{e_1/d_1}\cdots q_k^{e_k/d_k} \;$ we can &#xA;normalize any denesting so that it denests in the field defined &#xA;by the radicand. E.g. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \sqrt{\sqrt[3]5 - \sqrt[3]4} \;\;=\; \frac{1}3 (\sqrt[3]2 + \sqrt[3]{20} - \sqrt[3]{25})$$&#xA;normalises to $$ \sqrt{18\ (\sqrt[3]10 - 2)} \;\;=\; 2 + 2\ \sqrt[3]{10} - \sqrt[3]{10}^2 $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An example with nontrivial $\rm\:b$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$  \sqrt{12 + 5\ \sqrt 6} \;\;=\; (\sqrt 2 + \sqrt 3)\ 6^{1/4} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;normalises to &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \sqrt{\frac{1}3 \sqrt{6}\: (12 + 5\ \sqrt 6)} \;\;=\; 2 + \sqrt{6} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here $\rm\; F=\mathbb Q,\ F' = \mathbb Q(\sqrt 6),\ n=2,\ B = \{1,\sqrt 6\},\ d=2,\ q=1/3,\ b= \sqrt 6\:$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The structure theorem also hold for complex fields except that &#xA;in this case one has to assume that $\rm\; F \;$ contains enough roots of &#xA;unity (which may be computationally expensive in practice, to &#xA;wit doubly-exponential complexity). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the complexity of even simpler problems involving &#xA;radicals is currently unknown. For example, no polynomial time &#xA;algorithm is known for determining the sign of a sum of real &#xA;radicals $\rm\; \sum{c_i\: q_i^{1/r_i}} \;$ where $\rm\; c_i,\: q_i \;$ are rational numbers and &#xA;$\rm\; r_i \;$ is a positive integer. Such sums play an important role in &#xA;various geometric problems (e.g. Euclidean shortest paths and &#xA;traveling salesman tours). Even though testing whether such a sum &#xA;of radicals is zero can be decided in polynomial time, this is &#xA;of no help in determining the sign, it only shows that if sign &#xA;testing is in $\rm\; NP \;$ then it is already in $\rm\; NP \cap \text{co-NP} \;$. &lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-09-15T03:55:47.930" LastActivityDate="2010-09-15T03:55:47.930" CommentCount="2" />
  <row Id="4698" PostTypeId="2" ParentId="4694" CreationDate="2010-09-15T03:07:04.753" Score="4" ViewCount="" Body="&lt;p&gt;My favorite example is the set of subsets of a set under the operation of &lt;a href=&quot;http://en.wikipedia.org/wiki/Symmetric_difference&quot; rel=&quot;nofollow&quot;&gt;symmetric difference&lt;/a&gt; (otherwise known as bitwise XOR).  This forms a vector space over the &lt;a href=&quot;http://en.wikipedia.org/wiki/Finite_field&quot; rel=&quot;nofollow&quot;&gt;finite field&lt;/a&gt; $\mathbb{F}_2$.  This example is important in computer science, coding theory, combinatorics, ...&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-09-15T03:07:04.753" />
  <row Id="4699" PostTypeId="1" CreationDate="2010-09-15T03:53:04.560" Score="1" ViewCount="108" Body="&lt;p&gt;Among the smooth 1-manifolds (with or without boundary) which embed into $\mathbb{R}^2$, which ones can be represented by a single parametrization $z = (x,y) = f(t)$, for $t \in I$, where $I$ is an interval (not necessarily open or closed), and $f$ is smooth (i.e. infinitely differentiable)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason I ask is that in my reading, I've come across two definitions for &quot;line integral&quot; (which I'm sure turn out to be equivalent):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first is the standard definition given in most multivariable calculus and complex analysis classes, which relies on defining a &quot;curve&quot; as a (sufficiently nice) function (or image set) $z = \gamma(t)$.  That is: $\int_\gamma f(z) dz = \int_a^b f(g(t))g'(t) dt$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second is the more high-powered definition involving 1-manifolds and differential 1-forms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So my question is not really about whether or not these two definitions are equivalent per se, but rather about how much generality is lost by looking at only the &quot;special&quot; 1-manifolds which admit representations as $z = \gamma(t)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(My own thoughts: (1) Do these 1-manifolds end up being the 1-manifolds with an atlas consisting of one chart?  I'm thinking not, because the circle cannot be given an atlas with one chart, yet can still be parametrized by a single function. (2) Does the Implicit Function Theorem have any role to play?)&lt;/p&gt;&#xA;" OwnerUserId="640" LastEditorUserId="640" LastEditorDisplayName="" LastEditDate="2010-09-15T06:46:28.080" LastActivityDate="2010-09-15T06:46:28.080" Title="Which smooth 1-manifolds can be represented by a single smooth parametrization?" Tags="&lt;complex-analysis&gt;&lt;differential-topology&gt;" CommentCount="7" FavoriteCount="1" />
  <row Id="4700" PostTypeId="1" AcceptedAnswerId="4712" CreationDate="2010-09-15T04:26:54.693" Score="3" ViewCount="217" Body="&lt;p&gt;Here's the problem (from little Spivak):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Define $f:\mathbb{R}\rightarrow\mathbb{R}$ by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f(x)=\begin{cases}&#xA;e^{-x^{-2}} &amp;amp; x\ne 0 \\&#xA;0 &amp;amp; x=0 &#xA;\end{cases}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Show that $f$ is a $C^\infty$ function, and $f^{(i)}(0)=0$ for all $i$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyhow, here's my proof.  It gets pretty ugly at the end and I do a little handwaving (every term is of the form ... pretty much &quot;because I say so&quot;, and the last step should be a better form of induction instead of &quot;successive applications of the rule...&quot; but the exact way to say it correctly escapes me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For arbitrary $p,q:\mathbb{R}\rightarrow\mathbb{R}$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(e^{q(x)}p(x))'=e^{q(x)} q'(x)p(x) + e^{q(x)} p'(x)=e^{q(x)}(q'(x)p(x)+p'(x))$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $q(x)=-x^{-2}$, $p_0(x)=1$, and $p_{n+1}(x)=q'(x)p_{n}(x)+p_n'(x)=\frac{2}{x^3} p_n(x) + p_n'(x)$.  Every term of $p_n(x)$ is therefore of the form $a x^{-n}$ for some $a,n\in\mathbb{R}$ and $n&gt;0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then $f^{(n)}(x)=e^{q(x)} p_n(x)=e^{-x^{-2}}p_n(x)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the definition of derivative,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align*}&#xA;f^{(n+1)}(0) &amp;amp;= \lim_{h\rightarrow 0}\frac{f^{(n)}(h)}{h}\&#xA;&amp;amp;= \lim_{h\rightarrow 0}\frac{e^{-x^{-2}} p_n(h)}{h}\&#xA;&amp;amp;= \lim_{h\rightarrow 0}\frac{h^{-1} p_n(h)}{e^{h^{-2}}}\&#xA;\end{align*}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since every term of $p_n(h)$ is of the form $a h^{-m}$ for some $a,m\in\mathbb{R},n&gt;0$, it follows that every term of $h^{-1} p_n(h)$ is also of the same form.  Thus if we can show that $\lim_{h\rightarrow 0}\frac{ah^{-m}}{e^{h^{-2}}}=0$, the result follows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Applying L'Hopital's rule results in&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\lim_{h\rightarrow 0}\frac{-m h^{-m-1}}{e^{h^{-2}} (-\frac{2}{h^3})}=\lim_{h\rightarrow 0}\frac{2m h^{-m+2}}{e^{h^{-2}}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Successive applications of L'Hopital's rule result in multiplying numerator by some constant and increasing the exponent of $h$ by 2.  Thus apply the rule $m/2+1$ times will give a positive exponent, and the limit will be clearly 0.&lt;/p&gt;&#xA;" OwnerUserId="326" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-09-15T17:30:07.710" LastActivityDate="2010-09-15T17:30:07.710" Title="Improve my proof about this $C^\infty$ function." Tags="&lt;calculus&gt;&lt;real-analysis&gt;&lt;proof-strategy&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="1" />
  <row Id="4701" PostTypeId="2" ParentId="4682" CreationDate="2010-09-15T05:38:59.310" Score="2" ViewCount="" Body="&lt;p&gt;Even though you didn't ask for 'teh codez' :), &lt;a href=&quot;http://www.qhull.org/html/qvoronoi.htm&quot; rel=&quot;nofollow&quot;&gt;qvoronoi&lt;/a&gt; is a pretty good implementation of Voronoi diagrams for low dimensions (i.e 4). It actually computes the Delaunay triangulation, which in turn is constructed by a convex hull algorithm (qhull) one dimension higher. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The convex hull algorithm itself &lt;a href=&quot;http://www.qhull.org/html/index.htm#algorithm&quot; rel=&quot;nofollow&quot;&gt;is a combination of&lt;/a&gt; &quot;the 2-d Quickhull algorithm with the n-d beneath-beyond algorithm [c.f., Preparata &amp;amp; Shamos '85]&quot;&lt;/p&gt;&#xA;" OwnerUserId="865" LastActivityDate="2010-09-15T05:38:59.310" CommentCount="1" />
  <row Id="4702" PostTypeId="2" ParentId="4581" CreationDate="2010-09-15T08:48:59.140" Score="2" ViewCount="" Body="&lt;p&gt;Here is what I believe I understand of the other answers. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $K/k$ be an extension and $X$ an indeterminate. We identify $k(X)\otimes_kK$ with $k(X)\otimes_{k[X]}K[X]$, and denote this domain by $A$. Everything will follow from the &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;OBSERVATION.&lt;/strong&gt; Any element of $A$ can be written as &#xA;  $$\frac{1}{p(X)}\otimes P(X)$$ with $p(X)$ in $k[X]$, $p(X)\not=0$, $P(X)$ in $K[X]$. Its image in $K(X)$ is $P(X)/p(X)$. In particular, the natural map from $A$ to $K(X)$ is injective. Thus we can (and will) view $A$ as the subdomain of $K(X)$ formed by the elements which can be written, in the above notation, as $P(X)/p(X)$. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The fraction field of $A$, regarded as a subfield of $K(X)$, is called the &lt;a href=&quot;http://en.wikipedia.org/wiki/Tensor_product_of_fields#Compositum_of_fields&quot; rel=&quot;nofollow&quot;&gt;compositum&lt;/a&gt; of $k(X)$ and $K$. It is denoted $k(X)K$, and coincides with $K(X)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $a$ is in $K$ and $1/(X-a)$ is in $A$, then $a$ is algebraic over $k$. So, $A$ is &lt;strong&gt;not&lt;/strong&gt; a field if $K/k$ is transcendental. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $K/k$ has finite degree, the domain $A$, being finite dimensional over $k(X)$, is a field, and is thus equal to $K(X)$. The same conclusion holds if $K$ is a union of finite degree extensions of $k$, that is, if $K/k$ is algebraic. &lt;/p&gt;&#xA;" OwnerUserId="660" LastEditorUserId="660" LastEditorDisplayName="" LastEditDate="2010-09-16T08:59:44.957" LastActivityDate="2010-09-16T08:59:44.957" />
  <row Id="4703" PostTypeId="2" ParentId="4691" CreationDate="2010-09-15T09:15:34.800" Score="2" ViewCount="" Body="&lt;p&gt;The expression $\displaystyle \sum_{k=0}^{\infty} \frac{f^k(0)}{k!} x^k$ is not just the Maclaurin series for functions of the form $f(x)=(1+x)^n$; it is a formula for the Maclaurin series of any function infinitely differentiable at $0$.  It gives us the power series in $x$ with all derivatives at $x=0$ matching the derivatives of $f(x)$ at $x=0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to use this formula to find the Maclaurin series for $f(x)=\frac{1}{1-x}$, you need to compute all of the derivatives of $f(x)$.  --Fortunately, there is a simple pattern you can observe and prove (by induction).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you know the derivatives, you can plug in $x=0$, simplify your summation formula, and you'll have your answer.  &lt;/p&gt;&#xA;" OwnerUserId="1461" LastActivityDate="2010-09-15T09:15:34.800" />
  <row Id="4704" PostTypeId="2" ParentId="4466" CreationDate="2010-09-15T10:42:42.653" Score="0" ViewCount="" Body="&lt;p&gt;I did this same problem for a class on number theory, the book omits a constraint that n&gt;3 since 3^2 -1 = 8 which is divisible by 1,2,4,8 and 4 is not prime it does however hold for all n&gt;3. Intuitively this is a consequence of the fact that 2 is the only even prime and that the product of three consecutive integers is divisible by 6 (n-1 being divisible by two would typically eliminate its candidacy for being prime EXCEPT for the n-1 = 2 case). Proving that a one-to-one correspondence exists is done by proving a that there is an iff statement relating (n-1),(n+1) being prime and (n-1)(n+1) having 4 dividers. Since this would imply that the set of all n such that n^2-1 has 4 dividers is equal to the set of n s.t. (n-1),(n+1) are prime and the function mapping a set back to itself is a trivial bijective map.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One direction is pretty easy, (n-1),(n+1) prime =&gt; s = (n-1)(n+1) = p1*p2 therefore s has 4 dividers, 1,s,p1, and p2. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other direction is left for you to figure out,&lt;/p&gt;&#xA;" OwnerUserId="1906" LastActivityDate="2010-09-15T10:42:42.653" />
  <row Id="4705" PostTypeId="1" AcceptedAnswerId="4747" CreationDate="2010-09-15T11:00:54.413" Score="2" ViewCount="602" Body="&lt;p&gt;I am looking for an intuitive explanation of $\liminf$ and $\limsup$ for sequence of sets and how it corresponds to $\liminf$ and $\limsup$ for sets of real numbers. I researched online but cannot find a good comparison. Any link, reference or answer very much appreciated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, what is $\liminf$ and $\limsup$ of real number sequences $a_n=(-1)^n$ and $b_n=1/n$. Corresponding to this, what is the $\liminf$ and $\limsup$ of sequence of sets $A_n=\{(−1)^n\}$ and $B_n=\{1/n\}$?&lt;/p&gt;&#xA;" OwnerUserId="957" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2010-11-24T20:34:11.740" LastActivityDate="2010-11-24T20:34:11.740" Title="limit inferior and superior for sets vs real numbers" Tags="&lt;measure-theory&gt;&lt;elementary-set-theory&gt;&lt;limit&gt;" AnswerCount="3" CommentCount="5" FavoriteCount="1" />
  <row Id="4706" PostTypeId="1" CreationDate="2010-09-15T11:11:32.410" Score="1" ViewCount="103" Body="&lt;p&gt;Under the condition that the spaces (or maybe just the total) are connected and locally path connected, is then the a covering the same as a homeomorphism? &lt;/p&gt;&#xA;" OwnerUserId="1907" LastActivityDate="2010-09-15T11:28:49.493" Title="Covering for connected and locally path connected spaces" Tags="&lt;topology&gt;" CommentCount="4" />
  <row Id="4707" PostTypeId="2" ParentId="4691" CreationDate="2010-09-15T11:20:35.807" Score="5" ViewCount="" Body="&lt;p&gt;Elaborating on J.M. answer, you can procede this way:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(1+x)^n = 1+ n x + \frac{1}{2}(n-1) n x^2 + \frac{1}{6}(n-2)(n-1)n x^3 + \ldots$$, more generally,&#xA;$$(1+x)^\alpha = 1 +  \sum_{k = 1}^{n} \binom{\alpha}{k} x^k + o(x^n)$$ for any real $\alpha$, where $\binom{\alpha}{k} = \frac{(\alpha -1)(\alpha-2)\ldots (\alpha - k +1)}{k!}$ is the binomial coefficient extended to any real number. Then, you get that $$(1-x)^n = 1 - nx + \frac{1}{2}(n-1)n x^2 - \frac{1}{6}(n-2)(n-1)n x^3 + \ldots$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;then, you only have to plug in $n = -1$ to get $$ \frac{1}{1-x} = 1 + x + x^2 + x^3 + \ldots$$&lt;/p&gt;&#xA;" OwnerUserId="1415" LastActivityDate="2010-09-15T11:20:35.807" />
  <row Id="4709" PostTypeId="2" ParentId="4694" CreationDate="2010-09-15T11:33:54.470" Score="4" ViewCount="" Body="&lt;p&gt;What do you &quot;normally&quot; call addition and multiplication?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Just&lt;/em&gt; those operations with real numbers, or &lt;em&gt;all&lt;/em&gt; kind of addition and multiplication &quot;derived&quot; from the well-known operations with real numbers, or that &quot;look like&quot; these operations?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because, in the first case, you have plenty of elementary and widely used vector spaces with operations which are &lt;em&gt;not&lt;/em&gt; those of real numbers:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$\mathbb{R}^2$, the set of ordered pairs of real numbers $(x,y)$, is a real vector space, with addition and multiplication defined as $(x,y) + (u,v) = (x+u, y+v)$ and $\lambda (x,y) = (\lambda x , \lambda y)$. These operations are defined using the &quot;normal&quot; addition and multiplication of real numbers, but are &lt;em&gt;not&lt;/em&gt; the &quot;normal&quot; addition and multiplication of real numbers just because $(x,y)$ is &lt;em&gt;not&lt;/em&gt; a real number.&lt;/li&gt;&#xA;&lt;li&gt;${\cal C}^0 (\mathbb{R}, \mathbb{R})$, the set of continuous functions $f:\mathbb{R} \longrightarrow \mathbb{R}$, is a real vector space, with addition and multiplication defined point-wise; that is $(f+g)(x) = f(x) + g(x)$ and $(\lambda f)(x) = \lambda f(x)$. Again, these addition and multiplication are defined using the &quot;normal&quot; addition and multiplication of real numbers, but are not the &quot;normal&quot; addition and multiplication of real numbers for the same reason.&lt;/li&gt;&#xA;&lt;li&gt;$\mathbb{Z}/2\mathbb{Z}$, the set of integers mod 2, is a $\mathbb{Z}/2\mathbb{Z}$-vector space, with addition and multiplication $\widetilde{m} + \widetilde{n} = \widetilde{n+m}$ and $\widetilde{\lambda}\widetilde{m} = \widetilde{\lambda m}$, where $\widetilde{m}$ denotes the class of $m$ mod 2. Ditto.&lt;/li&gt;&#xA;&lt;li&gt;$\mathbb{R}(x)$, the field of rational functions $\frac{p(x)}{q(x)}$, where $p(x), q(x) \in \mathbb{R}[x]$ are polynomials, $q(x) \neq 0$, is a $\mathbb{R}(x)$-vector space, with addition and multiplication $\frac{p(x)}{q(x)} + \frac{r(x)}{s(x)} = \frac{p(x) s(x) + r(x) q(x)}{q(x)s(x)} $ and $\frac{p(x)}{q(x)}  \frac{r(x)}{s(x)} = \frac{p(x)r(x)}{q(x)s(x)}$. Ditto.&lt;/li&gt;&#xA;&lt;li&gt;$\mathbb{C}$, the set of complex numbers, is a $\mathbb{C}$-vector space, whit the addition and multiplication of complex numbers. Ditto.&lt;/li&gt;&#xA;&lt;li&gt;$\mathbb{K}^n$, the set of ordered families $(x_1, \dots , x_n)$ of elements of any field $\mathbb{K}$, is a $\mathbb{K}$-vector space, with addition and multiplication defined as in example 1. Examples 3, 4 and 5 are particular cases of this one with $n=1$ and $\mathbb{K} =$ $\mathbb{Z}/2\mathbb{Z}$, $\mathbb{R}(x)$ and $\mathbb{C}$, respectively. Example 1 is also a particular case, with $n=2$ and $\mathbb{K} = \mathbb{R}$. Addition and multiplication in $\mathbb{K}$ may have nothing in common with the operations with real numbers.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="664" LastActivityDate="2010-09-15T11:33:54.470" />
  <row Id="4710" PostTypeId="2" ParentId="4705" CreationDate="2010-09-15T11:43:40.493" Score="0" ViewCount="" Body="&lt;p&gt;As I know the limits of sequences of sets, the limit superior consists of all points contained in infinitely many of these sets, the limit inferior are all points contained in almost all sets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\limsup A_n = \{x | x\in A_n \text{ for infinitely many }n\}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\liminf A_n = \{x | x\in A_n \text{ for all but finitely many }n\}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as I know, they do not correspond with the limits of sequences of real numbers in an obvious way. I've come to the conclusion that they are poorly named, but I may lack insight here. =)&lt;/p&gt;&#xA;" OwnerUserId="505" LastEditorUserId="505" LastEditorDisplayName="" LastEditDate="2010-09-15T12:05:23.763" LastActivityDate="2010-09-15T12:05:23.763" CommentCount="4" />
  <row Id="4711" PostTypeId="1" AcceptedAnswerId="4777" CreationDate="2010-09-15T12:08:49.927" Score="3" ViewCount="303" Body="&lt;p&gt;If I have an arbitrary shape, I would like to fill it only with 45-45-90&#xA;triangles. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The aim is to get a Tangram look, so it's related to &lt;a href=&quot;http://stackoverflow.com/questions/3717178/how-can-i-fill-an-outline-with-predefined-tangram-shapes&quot;&gt;this question&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Starting with 45-45-90 triangles would be an amazing start. After&#xA;the shape if filled I imagine I could pick adjacent triangles, either&#xA;2 or 4 and draw squares and parallelograms instead, but just getting&#xA;the outline estimated with packed 45-45-90 triangles would be great. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do I get started ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;&#xA;@J.M.'s comment makes perfect sense, which means I should first make sure my shape is suited for this. Here's a sketch I did to illustrate this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/ROZGP.gif&quot; alt=&quot;hand&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The black shape is the 'arbitrary shape', the blue path is the path I should be filling.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So far, I see the first step is to 'estimate' arbitrary paths with lines at right or 45 degree angles. The second step would be the initial question, packing 45-45-90 triangles into the shape.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hints for estimating random angles lines with 45/90 degrees angled lines or 45-45-90 triangle packing ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;UPDATE2&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've gone ahead with a naive approach to estimate an arbitrary path(outline) using only straight or diagonal(45 degrees) lines.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;function estimate45(points:Vector.&amp;lt;Point&amp;gt;):Vector.&amp;lt;Point&amp;gt; {&#xA;    var result:Vector.&amp;lt;Point&amp;gt; = new Vector.&amp;lt;Point&amp;gt;();&#xA;    var pNum:int = points.length,angle:Number,pi:Number = 3.141592653589793,qpi:Number = pi*.25,d:Number = 0;&#xA;    for(var i:int = 0 ; i &amp;lt; pNum ; i++){&#xA;        if(i == 0) angle = Math.atan2(points[i].y,points[i].x);&#xA;        else {&#xA;            angle = Math.atan2(points[i].y-points[i-1].y,points[i].x-points[i-1].x);&#xA;            d = Math.sqrt((points[i].x-points[i-1].x)*(points[i].x-points[i-1].x)+(points[i].y-points[i-1].y)*(points[i].y-points[i-1].y))&#xA;        }&#xA;        //constraint to 45 (1. scale to values between 0,1 (/qpi) 2. round, 3. do whatever(convert to degrees/radians as needed) &#xA;        angle = (Math.round(angle/qpi)) * 45 / 57.2957795;&#xA;        if(i == 0) result.push(new Point(Math.cos(angle)*d,Math.s(angle)*d));&#xA;        else       result.push(new Point(result[i-1].x+Math.cos(angle)*d,result[i-1].y+Math.s(angle)*d));&#xA;    }&#xA;    return result;&#xA;}&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I loop trough the the path (an ordered list of points) and I calculate the angle and radius of each line(cartesian to polar I think). Then I 'round' the angle to 45 degrees,&#xA;and draw a line to a 45 degrees constrained version of the original angle, and keep the same length for the line.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This isn't very good way to do it, especially for consecutive lines with similar angles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are some tests:&#xA;&lt;img src=&quot;http://i.imgur.com/gFFAK.gif&quot; alt=&quot;crescent&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The faded red is the original, the green is the estimation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/Vgiec.gif&quot; alt=&quot;hand 45&quot;&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;@Américo Tavares's suggestion is great though. I could use this approach for bitmap graphics too, not just vector graphics. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I would go with this approach, I imagine I would do something like:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;get a mosaic(create a grid of boxes to cover the size of the shape)&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;boxes_xnum = floor(w/box_size)&#xA;boxes_ynum = floor(h/box_size)&#xA;for(y to boxes_ynum):&#xA;    for(x to boxes_xnum):&#xA;        grid.addBitmap(copyPixels(source,x*box_size,y*box_size,box_size,box_size));//copypixels(source,x,y,width,height)&#xA;&#xA;for box in grid:&#xA;   if(box.nonAlphaPixels/box.totalPixels &amp;gt; .75): fullBox&#xA;   else:&#xA;        checkDiagonalType()//is it like this / or like this \&#xA;        checkFillSide()//which of the two sides should be filled&#xA;        //which means I should check for something constrained to 45 degrees angles \| or _\ or |/ or /_&#xA;        //in the case of halfs go for random diagonal ? &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;If I think about this better,&lt;br&gt;&#xA;when I loop though the pixels of a box, keep a pixel count per box 'quadrant'(top-left).&#xA;If the non transparent pixels in one quadrant in larger than .5 or .75 it's marked as used.&#xA;Based on how many and which of the 4 quadrants are used in a box, a diagonal with direction is used.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/2CLf7.jpg&quot; alt=&quot;solve&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does this make sense, or am I over complicating this ?&lt;/p&gt;&#xA;" OwnerUserId="304" LastEditorUserId="304" LastEditorDisplayName="" LastEditDate="2010-09-16T14:46:59.817" LastActivityDate="2010-09-16T16:29:51.100" Title="How can I pack 45-45-90 triangles inside an arbitrary shape ?" Tags="&lt;geometry&gt;&lt;algorithms&gt;&lt;euclidean-geometry&gt;" AnswerCount="1" CommentCount="7" FavoriteCount="1" />
  <row Id="4712" PostTypeId="2" ParentId="4700" CreationDate="2010-09-15T13:16:58.637" Score="3" ViewCount="" Body="&lt;p&gt;You seem to have two main observations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;statement $A_{n}$ (for $n \ge 1$):&#xA;$$f^{n}(x) = e^{q(x)}p_{n}(x), x \neq 0$$&#xA;where $$p_{n}(x) = \sum \frac{a_{r}}{x^r}, x \neq 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;statement $B_{n}$ (for $n \ge -1$):&#xA;$$\lim_{h \to 0} \frac{1}{h^{n} e^{h^{-2}}}  = 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The result you have is that $A_{n-1}$ implies $A_{n}$ and that $B_{n-2}$ implies $B_{n}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you rewrite those formally using the right form of induction, with proper base cases?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you have those, you finally can prove the third statement $C_{n}$  (for $n \ge 1$) :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ f^{n}(0) = 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You haven't explicitly spelled this out in your proof, but you are showing that $C_{n}$ (combined with the $A_{i}$ and $B_{j}$) implies $C_{n+1}$ which can again be formalized by induction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance the statement:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f^{(n+1)}(0) = \lim_{h\rightarrow 0}\frac{f^{(n)}(h)}{h}$$ is implicitly using the fact that $f^{n}(0) = 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope that helps.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-09-15T13:16:58.637" CommentCount="1" />
  <row Id="4713" PostTypeId="1" AcceptedAnswerId="4715" CreationDate="2010-09-15T13:29:47.237" Score="1" ViewCount="259" Body="&lt;p&gt;I've just come up with this question as I'm studying for a number theory midterm. If $p$ and $q$ are different prime numbers, and it's known that $2^p \equiv 1 \bmod{q}$, then $q\equiv 1 \bmod{p}$. I've tried a few cases and it seems to be true. How can it be proved?&lt;/p&gt;&#xA;" OwnerUserId="1313" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-09-17T15:26:13.887" LastActivityDate="2010-09-17T15:26:13.887" Title="If the order divides a prime P then the order is P (or 1)" Tags="&lt;number-theory&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" />
  <row Id="4715" PostTypeId="2" ParentId="4713" CreationDate="2010-09-15T13:57:24.333" Score="7" ViewCount="" Body="&lt;p&gt;$\rm\displaystyle\;\ 2^p,\;  2^{q-1} \equiv 1 \;\Rightarrow\; 2^{gcd(p,q-1)} \equiv 1\ (mod\  q)\;\Rightarrow\; gcd(p,q-1) = p\; $ (not $1$ else $\rm \:q|2^1-1\:$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Said in group theory language: if an element $\ne 1$ has order dividing a prime $\:\rm p\;$ then it must have order $\rm p\:$. In ring theory language: if a principal ideal$\;\ne 1$ contains an irreducible element then that element generates the ideal (which is why the minimal polynomial is sometimes called the &quot;irreducible polynomial&quot;). Here the group / ideal is simply the so-called &lt;strong&gt;order ideal&lt;/strong&gt; $\rm\; \{n : x^n = 1 \}$ which, being nonempty and closed under subtraction, comprises a subgroup / ideal of $\:\mathbb Z\:$. The Euclidean algorithm implies that ideals in $\mathbb Z$ are principal, so every element of a nonzero ideal is a multiple of the least positive element. For an order ideal this simply says that every &quot;possible&quot; order is a multiple of the least possible order $\rm m\:$, $\:$ i.e. $\rm\; x^n = 1 \;\Rightarrow\; m\:|\:n\;$. Compare this ring-theoretic proof to the ubiquitous group-theoretic proof using Lagrange's theorem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a simple additive example: the order ideal of a fraction is the &lt;strong&gt;denominator ideal&lt;/strong&gt; $\rm\; \{n : n\: x \in \mathbb Z \:\}\:$. Here the analogous result is that if a proper fraction can be written with a prime denominator then it is the least possible denominator (in the multiplicative sense), i.e. it divides ever other possible denominator.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Denominator ideals and fractional ideals can yield significant number-theoretical simplifications - in much the same way that fractional arithmetic often serves to simplify integer arithmetic in elementary number theory. For example, see my post on &lt;a href=&quot;http://math.stackexchange.com/questions/4491&quot;&gt;irrationality proofs using the denominator ideal&lt;/a&gt; or, more generally, see my proof employing Dedekind's &lt;a href=&quot;http://mathoverflow.net/questions/30313&quot; rel=&quot;nofollow&quot;&gt;conductor ideal&lt;/a&gt; - a sort of universal denominator for an entire extension ring. Follow the links there to my sci.math posts for much further discussion on this and related topics.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-09-17T15:19:44.253" LastActivityDate="2010-09-17T15:19:44.253" />
  <row Id="4717" PostTypeId="1" CreationDate="2010-09-15T14:05:37.410" Score="2" ViewCount="79" Body="&lt;p&gt;Is there a reliable method for testing how invertible a Hermitian Toeplitz matrix is without going through the work of actually inverting it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A determinant is obviously easy to compute, but I'm not sure what threshold to compare it against.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The algorithm I am using is Levinson recursion.  The function can detect the divide-by-zero condition, but I am interested in also detecting/scoring the matrices that are &lt;em&gt;almost&lt;/em&gt; singular.&lt;/p&gt;&#xA;" OwnerUserId="1908" LastActivityDate="2010-09-15T14:20:13.547" Title="fast and reliable method for testing invertibility of Hermitian Toeplitz matrix" Tags="&lt;linear-algebra&gt;&lt;matrices&gt;&lt;numerical-methods&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="4718" PostTypeId="2" ParentId="4717" CreationDate="2010-09-15T14:20:13.547" Score="2" ViewCount="" Body="&lt;p&gt;It's the condition number $\kappa$ you should be looking at, not the determinant. Since Levinson is an O(n²) method, you can hook it up to the &lt;a href=&quot;http://doi.acm.org/10.1145/50063.214386&quot; rel=&quot;nofollow&quot;&gt;Hager-Higham condition estimator&lt;/a&gt; for determining how near to singularity your matrix is. The classical heuristic is that one loses $\log_{b}\kappa$ base-$b$ significant figures in solving a linear system.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-09-15T14:20:13.547" CommentCount="1" />
  <row Id="4719" PostTypeId="1" AcceptedAnswerId="4724" CreationDate="2010-09-15T14:50:44.543" Score="4" ViewCount="127" Body="&lt;p&gt;Apologies for the simple question. Is it possible to analytically minimize a function such as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f(x) = a^x + c x (x-1) \qquad x &gt; 0, \quad 0 &amp;lt; a,c &amp;lt; 1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nothing else can be said about the constants $a$ or $c$. &lt;/p&gt;&#xA;" OwnerUserId="1910" LastEditorUserId="1910" LastEditorDisplayName="" LastEditDate="2010-09-15T15:22:08.600" LastActivityDate="2010-09-15T15:45:25.863" Title="Minimization of a function including quadratic and exponential terms." Tags="&lt;optimization&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="2" />
  <row Id="4720" PostTypeId="1" AcceptedAnswerId="4723" CreationDate="2010-09-15T14:53:26.083" Score="7" ViewCount="685" Body="&lt;p&gt;I'm taking a course in differential geometry, and have here been introduced to the wedge product of to vectors defined (in &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0132125897&quot; rel=&quot;nofollow&quot;&gt;Differential Geometry of Curves and Surfaces by Manfredo Perdigão do Carmo&lt;/a&gt;) by:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $\mathbf{u}$, $\mathbf{v}$ be in $\mathbb{R}^3$. $\mathbf{u}\wedge\mathbf{v}$ in $\mathbb{R}^3$ is the unique vector that satisfies:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(\mathbf{u}\wedge\mathbf{v})\cdot\mathbf{w} = \det\;(\mathbf{u}\;\mathbf{v}\;\mathbf{w})$ for all $\mathbf{w}$ in $\mathbb{R}^3$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And to clarify, $(\mathbf{u}\;\mathbf{v}\;\mathbf{w})$ is the 3×3 matrix with $\mathbf{u}$, $\mathbf{v}$ and $\mathbf{w}$ as its columns, in that order.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question: is there &lt;em&gt;any&lt;/em&gt; difference between this and the regular cross product or vector product of two vectors, as long as we stay in $\mathbb{R}^3$? And if there is no difference, then why introduce the wedge?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cheers!&lt;/p&gt;&#xA;" OwnerUserId="1911" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-09-17T01:35:13.070" LastActivityDate="2010-09-17T01:35:13.070" Title="Wedge product and cross product - any difference?" Tags="&lt;linear-algebra&gt;&lt;terminology&gt;&lt;differential-geometry&gt;&lt;cross-product&gt;&lt;exterior-algebra&gt;" AnswerCount="4" CommentCount="6" FavoriteCount="3" />
  <row Id="4721" PostTypeId="2" ParentId="4719" CreationDate="2010-09-15T14:58:20.430" Score="1" ViewCount="" Body="&lt;p&gt;I think there is no analytical solution to $f^{\prime }(x)=0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f^{\prime }(x)=2cx-c+\left( \ln a\right) a^{x}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Update: I mean no solution in elementary functions.&lt;/p&gt;&#xA;" OwnerUserId="752" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2010-09-15T15:18:25.763" LastActivityDate="2010-09-15T15:18:25.763" />
  <row Id="4722" PostTypeId="2" ParentId="3405" CreationDate="2010-09-15T15:27:50.337" Score="2" ViewCount="" Body="&lt;p&gt;Here is an observation: It looks like $(2n) p(2n) - c$ converges to $0$ much faster than $(2n+1) p(2n+1)-c$ does. Here is a plot of the points $(\log n, n p(n) -c)$, with blue points for$n$ even and red points for $n$ odd.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/R6asV.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It might be easiest to first prove the result for $n$ even, then use the fact that $(2n) p(2n) - (2n-1) p(2n-1) = p(2n) + p(2n-1)$ to get that the odd and even points must have the same limit. &lt;/p&gt;&#xA;" OwnerUserId="448" LastActivityDate="2010-09-15T15:27:50.337" CommentCount="2" />
  <row Id="4723" PostTypeId="2" ParentId="4720" CreationDate="2010-09-15T15:39:08.383" Score="7" ViewCount="" Body="&lt;p&gt;No difference at all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've been trying to write a little proof, but the software on this page seems to have forgotten how to write maths.  :-(   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway: I assume that by &quot;regular cross/vector product&quot; you mean the definition with coordinates as in &lt;a href=&quot;http://en.wikipedia.org/wiki/Cross_product&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Try to compute both sides of your equation $(u\wedge v ) \cdot w = \det (u, v, w)$ with your definition using coordinates for the cross / vector product.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since this equation defines a &lt;em&gt;unique&lt;/em&gt; vector, has just &lt;em&gt;one&lt;/em&gt; solution, once you've checked your definition with coordinates verifies it, that means that your definition with coordinates agrees with the present one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT. I forgot. The adventage of this &quot;new&quot; definition is that it is &quot;coordinate-free&quot;. &lt;em&gt;So what?&lt;/em&gt; -I can hear someone whispering at the end of the classroom. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;So, if you ever wonder if your &quot;former&quot; definition could be affected just in case you change the (positive orthonormal) basis in which you're writing your vectors, the answer, you betcha, is &quot;no&quot;.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="664" LastEditorUserId="664" LastEditorDisplayName="" LastEditDate="2010-09-15T17:57:13.470" LastActivityDate="2010-09-15T17:57:13.470" CommentCount="1" />
  <row Id="4724" PostTypeId="2" ParentId="4719" CreationDate="2010-09-15T15:45:25.863" Score="6" ViewCount="" Body="&lt;p&gt;Alright, here's my expansion of my comment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's rearrange the equation in VFG's answer like so:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\ln\;a}{2c}=\left(\frac12-x\right)a^{-x}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and multiply both sides by $\sqrt{a}$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\sqrt{a}\ln\;a}{2c}=\left(\frac12-x\right)a^{\frac12-x}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We're close to a form where we can now use the Lambert function, but let's change bases first:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\sqrt{a}\ln\;a}{2c}=\left(\frac12-x\right)\exp\left(\frac{\ln\;a}{2}-x\ln\;a\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We're much closer now, so we multiply both sides by a factor of $\ln\;a$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\sqrt{a}(\ln\;a)^2}{2c}=\left(\frac{\ln\;a}{2}-x\ln\;a\right)\exp\left(\frac{\ln\;a}{2}-x\ln\;a\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and then perform the inversion:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$W\left(\frac{\sqrt{a}(\ln\;a)^2}{2c}\right)=\frac{\ln\;a}{2}-x\ln\;a$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is now easily solved for $x$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x=\frac12-\frac1{\ln\;a}W\left(\frac{\sqrt{a}(\ln\;a)^2}{2c}\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the Lambert function has two real branches $W(z)$ (principal branch) and $W_{-1}(z)$; one is real for $z\geq-e^{-1}$, and the other is real only in the interval $[-e^{-1},0)$. If both $a$ and $c$ are in the interval $(0,1)$, the argument of the Lambert function would be much greater than 1, and thus we can dispose of the $W_{-1}$ branch, as well as consider $W(z)$ only for $z\geq 0$, which is true for $a$ and $c$ in the interval $(0,1)$. Thus, the Lambert solution using the principal branch is valid.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-09-15T15:45:25.863" CommentCount="1" />
  <row Id="4725" PostTypeId="1" AcceptedAnswerId="4727" CreationDate="2010-09-15T16:01:02.187" Score="3" ViewCount="90" Body="&lt;p&gt;I have a stocastic event (say a weighted coin toss) that produces a positive outcome (heads) according to some unknown probability P.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given N total events (coin tosses) and n positive outcomes (heads), how can I measure the likelihood that n/N is a good approximation of P?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously, as N grows, it becomes more likely than n/N is close to the value P, but how can this convergence be described mathematically?&lt;/p&gt;&#xA;" OwnerUserId="19" LastActivityDate="2010-09-17T19:29:53.720" Title="Deriving the probability of an event from limited trial data" Tags="&lt;probability&gt;" AnswerCount="3" />
  <row Id="4726" PostTypeId="2" ParentId="4725" CreationDate="2010-09-15T16:13:34.967" Score="1" ViewCount="" Body="&lt;p&gt;The wiki on &lt;a href=&quot;http://en.wikipedia.org/wiki/Convergence_of_random_variables&quot; rel=&quot;nofollow&quot;&gt;convergence of random variables&lt;/a&gt; is what you probably want to read. It discusses various forms of convergence and in particular &lt;a href=&quot;http://en.wikipedia.org/wiki/Convergence_of_random_variables#Convergence_in_probability&quot; rel=&quot;nofollow&quot;&gt;convergence in probability&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/Convergence_of_random_variables#Almost_sure_convergence&quot; rel=&quot;nofollow&quot;&gt;almost sure convergence&lt;/a&gt; are what you need to take a look at.&lt;/p&gt;&#xA;" LastActivityDate="2010-09-15T16:13:34.967" />
  <row Id="4727" PostTypeId="2" ParentId="4725" CreationDate="2010-09-15T16:15:42.640" Score="3" ViewCount="" Body="&lt;p&gt;While I realise this isn't a full answer, the convergence you discuss is guaranteed by the weak and strong laws of large numbers. You might also be interested in Chebychev's inequality, which would say $$\mathbb{P}(|n/N - p| &gt; \alpha) \leq \frac{\sigma^2}{\alpha^2}$$ where $\sigma^2$ is the variance of your estimator, $n/N$. Finally the Central Limit Theorem (CLT) proves that as $N \to \infty$ the estimator becomes normally distributed about $p$ with variance $\frac{\sigma^2}{N}$. &lt;/p&gt;&#xA;" OwnerUserId="1910" LastActivityDate="2010-09-15T16:15:42.640" />
  <row Id="4728" PostTypeId="2" ParentId="4705" CreationDate="2010-09-15T16:32:11.570" Score="2" ViewCount="" Body="&lt;p&gt;Here are two facts that may help with the analogy.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Let $x_n$ be a real sequence; let $A_n = (-\infty, x_n)$.  Then $\limsup A_n = (-\infty, \limsup x_n)$ or $(-\infty, \limsup x_n]$.  Similarly, $\liminf A_n = (-\infty, \liminf x_n)$ or $(-\infty, \liminf x_n]$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Let $A_n$ be subsets of some set $\Omega$, and let $1_{A_n} : \Omega \to \mathbb{R}$ be the function which is 1 on $A_n$ and 0 on $A_n^c$.  Then $\limsup 1_{A_n}(\omega) = 1_{\limsup A_n}(\omega)$ for all $\omega \in \Omega$, and similarly $\liminf 1_{A_n}(\omega) = 1_{\liminf A_n}(\omega)$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="822" LastActivityDate="2010-09-15T16:32:11.570" CommentCount="2" />
  <row Id="4730" PostTypeId="2" ParentId="4720" CreationDate="2010-09-15T17:07:16.000" Score="6" ViewCount="" Body="&lt;p&gt;I would have left this as a comment, but apparently I can't comment yet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To answer your final question, why introduce the wedge, the point is that the wedge product (as explained in the &lt;a href=&quot;http://en.wikipedia.org/wiki/Exterior_algebra#Formal_definitions_and_algebraic_properties&quot; rel=&quot;nofollow&quot;&gt;Wikipedia article&lt;/a&gt;) is a notion that generalizes to R^n and indeed any vector space---in general the output is what is called a &quot;bivector&quot;. Now, it just so happens that in R^3 there is a natural identification between bivectors and vectors, and the vector that corresponds to the (bivector) wedge product is indeed the cross product, but this does not work in other vector spaces. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, to sum up, the notion of cross product is specific to R^3, whereas wedge product makes sense in any vector space.&lt;/p&gt;&#xA;" OwnerUserId="1914" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-09-15T17:24:17.300" LastActivityDate="2010-09-15T17:24:17.300" CommentCount="1" />
  <row Id="4731" PostTypeId="1" CreationDate="2010-09-15T18:03:22.457" Score="2" ViewCount="353" Body="&lt;p&gt;For functions on $\mathbb{R}^2$, it lets you classify extreme points. When function is log-likelihood, it gives volume of the region where maximum likelihood estimates tend to fall. Are there other places it comes up?&lt;/p&gt;&#xA;" OwnerUserId="998" LastEditorUserId="998" LastEditorDisplayName="" LastEditDate="2010-09-16T20:50:26.483" LastActivityDate="2010-09-16T21:46:51.393" Title="Determinant of the Hessian, what is it good for?" Tags="&lt;calculus&gt;&lt;matrices&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" />
  <row Id="4732" PostTypeId="1" AcceptedAnswerId="4740" CreationDate="2010-09-15T18:34:15.633" Score="4" ViewCount="222" Body="&lt;p&gt;I am  trying to solve a ODE that arises from a Hamilton-Jacobi-Bellman (HJB) equation. The equation is $$\frac{1}{2}b^2(1-\rho_s^2)\psi''-\frac{1}{2}\left(\frac{\mu-r}{\sigma}\right)^2\frac{(\psi')^2}{\psi''}+[ru+\theta a+b\rho_s(\mu-r)(1-\frac{2}{\sigma})]\psi'=0,$$ where $\mu, r, \sigma, \theta, a, \rho_s, b$ are constant. I want to determine $\psi'(u)$ (so that I get an integral form for $\psi(u)$). I have tried guessing (trial and error method) forms of the solution but didn't get far. I also tried the Legendre transform, but could not get the linear form. These are the method that I have seen being used in with these problems. &lt;/p&gt;&#xA;" OwnerUserId="626" LastEditorUserId="626" LastEditorDisplayName="" LastEditDate="2010-09-15T21:17:12.803" LastActivityDate="2010-09-15T22:03:08.357" Title="Solution of a Hamilton-Jacobi-Bellman (HJB) equation" Tags="&lt;differential-equations&gt;&lt;stochastic-processes&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" />
  <row Id="4733" PostTypeId="1" AcceptedAnswerId="4748" CreationDate="2010-09-15T18:48:49.570" Score="4" ViewCount="377" Body="&lt;p&gt;Let $f$ be defined as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f(x)=\begin{cases} x \sin(\frac{1}{x}) &amp;amp; x\ne 0 \\ 0 &amp;amp; x=0 \end{cases}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f(x)$  is not absolutely continuous so it &lt;strike&gt;cannot&lt;/strike&gt; &lt;em&gt;might not&lt;/em&gt; satisfy the Lusin N condition. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a direct proof of that it does not?  i.e. I wanted to know how to construct a set of zero measure which does not satisfy Lusin's Condition for this function.&lt;/p&gt;&#xA;" OwnerUserId="832" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-09-15T20:54:00.430" LastActivityDate="2010-09-16T02:24:28.570" Title="Direct proof that f(x)=x sin(1/x) does not satisfy Lusin N condition." Tags="&lt;real-analysis&gt;" AnswerCount="1" CommentCount="11" />
  <row Id="4734" PostTypeId="2" ParentId="4572" CreationDate="2010-09-15T18:55:10.053" Score="7" ViewCount="" Body="&lt;p&gt;It seems to me that maybe the best way to describe the situation is this: Douglas Stone's original answer to the original question consisted of rephrasing the question in such a way as to make it accessible to a proof using basic properties on the integers (specifically, unique factorization).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In my opinion, one thing which is being lost in all this discussion is just how important it can be to rephrase a question! Sure, the process of rephrasing contains &quot;no math&quot; as Qiaochu has pointed out.  But that doesn't make it useless (and I wouldn't use the word circular here either).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finding ways to rephrase questions so that they become accessible to the methods available is a basic skill beginning students of mathematics need to learn.  For example, much of the material in the early chapters of modern linear algebra books consists of teaching students how to rephrase questions in linear algebra so that they can be solved by row reduction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wouldn't accept Douglas Stone's answer as a complete solution to the problem if it were turned in by a student in an elementary number theory class, just as in my linear algebra class, reducing a problem to a question of row reduction isn't a complete solution.  But if a student came to me and said he or she was stuck on the problem, the first thing I'd try to do is get them to rephrase the question in precisely the way Douglas Stone did.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pleasantly, the community has pointed out (both here and at the original question) exactly how to finish the proof after rephrasing it in this useful way.&lt;/p&gt;&#xA;" OwnerUserId="1392" LastActivityDate="2010-09-15T18:55:10.053" CommentCount="5" />
  <row Id="4736" PostTypeId="2" ParentId="4731" CreationDate="2010-09-15T19:41:41.533" Score="1" ViewCount="" Body="&lt;p&gt;According to &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0132125897&quot; rel=&quot;nofollow&quot;&gt;Do Carmo&lt;/a&gt;, exercise 22 in page 173, the Hessian gives you the second fundamental form of a surface, as John pointed out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Precisely, let $S$ be a surface, $p\in S$ a point and $h: S \longrightarrow \mathbb{R}$ be the height function of $S$ relative to $T_p S$; that is, $h(q) = (q-p)\cdot N(p)$, where $N(p)$ is the normal vector to the surface $S$ at the point $p$. Then the Hessian of the function $h$ at $p$ is the second fundamental form of $S$ at $p$.&lt;/p&gt;&#xA;" OwnerUserId="664" LastActivityDate="2010-09-15T19:41:41.533" CommentCount="1" />
  <row Id="4737" PostTypeId="1" CreationDate="2010-09-15T19:51:58.437" Score="10" ViewCount="284" Body="&lt;p&gt;Does there exists a family of discs $\lbrace D_{n}\rbrace_{n=1}^{\infty}$ in euclidean plane such that:&lt;br&gt;&#xA;-the radius of $D_{n}$ is $\frac{1}{n+1}$&lt;br&gt;&#xA;-each $D_{n}$ is contained in the unit disc&lt;br&gt;&#xA;and&lt;br&gt;&#xA;-$D_{n}\cap D_{m} = \emptyset$ for each $n\neq m$ ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I'm not sure what tags are appropriate for this kind question, so if You have any suggestions, You're welcome to inform me about it via comments)&lt;/p&gt;&#xA;" OwnerUserId="1454" LastActivityDate="2010-10-03T15:46:37.793" Title="Packing disjoint family of discs with radii $\tfrac{1}{2}, \tfrac{1}{3}, \tfrac{1}{4},\ldots$ inside the unit disc" Tags="&lt;geometry&gt;&lt;puzzle&gt;" AnswerCount="2" CommentCount="8" />
  <row Id="4738" PostTypeId="1" AcceptedAnswerId="4790" CreationDate="2010-09-15T20:18:03.397" Score="0" ViewCount="231" Body="&lt;p&gt;Suppose $X$ is a metric space. A real valued function $f$ defined on a metric space $X$ is said to be of first baire class, if $f$ is a pointwise, limit of a sequence of continuous functions on $X$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A famous well know theorem asserts that if &lt;em&gt;$f$ is in the first baire class, then the set of points of discontinuity is of the first category&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strike&gt;Using this definition, can we characterize for what &lt;em&gt;Metric Spaces&lt;/em&gt;, will the points of continuity of a real valued function, be dense in itself.&lt;/strike&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ok, since we are talking about metric spaces, lets consider the 2 famous ones, complete and Compact. A compact metric space, is &quot;Complete and Totally Bounded&quot; by definition, so lets not look into it. So lets look at complete metric spaces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the question basically, is if, $X$ is a complete metric space, and $f$ is as above(1st Baire class), then is the points of continuity of $f$ dense in $X$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are some exercises related, but i didn't take it from here. So dont misunderstand me. &lt;a href=&quot;http://books.google.com/books?id=CoSIe7h5mTsC&amp;amp;pg=PA166&amp;amp;dq=f+is+of+first+baire+class,+points+of+continuity&quot; rel=&quot;nofollow&quot;&gt;http://books.google.com/books?id=CoSIe7h5mTsC&amp;amp;pg=PA166&amp;amp;dq=f+is+of+first+baire+class,+points+of+continuity&lt;/a&gt;&lt;/p&gt;&#xA;" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-09-16T15:29:25.780" LastActivityDate="2010-09-16T15:29:25.780" Title="Points of continuity of a function being dense" Tags="&lt;real-analysis&gt;&lt;analysis&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="4740" PostTypeId="2" ParentId="4732" CreationDate="2010-09-15T22:03:08.357" Score="1" ViewCount="" Body="&lt;p&gt;Assume $\psi' = f$, then the equation is of the form:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$Af' + B\frac{f^2}{f'} + (cu+d)f = 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By putting $$g = \frac{f'}{f} = (\log f)'$$ we see that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$Ag + \frac{B}{g} + (cu+d) = 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a quadratic in $g$ and can be easily solved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We get $$ \psi' = f = e^{\int g}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope that helps.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-09-15T22:03:08.357" CommentCount="1" />
  <row Id="4741" PostTypeId="1" AcceptedAnswerId="4792" CreationDate="2010-09-15T22:07:56.313" Score="6" ViewCount="266" Body="&lt;p&gt;Suppose $\alpha: \mathbb{R} \rightarrow \mathbb{R}^3$ is a $C^\infty$&#xA;curve, parameterized by arc length ($\left\|\alpha'(t)\right\| = 1$),&#xA;and with $\alpha(0) = \alpha(\ell)$.  Show that&#xA;there exists a $t_0 \in [0,\ell]$ such that&#xA;$&#xA;\left\| \alpha''(t_0)\right\| \geq \frac{2\pi}{\ell}.&#xA;$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/WD65E.png&quot; alt=&quot;loop&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we had that $\alpha'(0) = \alpha'(\ell)$, then the result would&#xA;follow from Fenchel's theorem, but we don't. Actually there is a nice&#xA;extension to Fenchel's theorem described in Toponogov's book&#xA;``Differential Geometry of Curves and Surfaces: A Concise Guide'',&#xA;under the name Fenchel-Reshetnyak, which gives lower bounds on the&#xA;integral curvature in an open curve. This can be adapted to our&#xA;situation to show that $\left\| \alpha''(t_0)\right\| \geq&#xA;\frac{\pi}{\ell}$, but that bound is only half as good as it&#xA;presumably could be. It makes sense that the integral curvature bound&#xA;will be only half as good if we don't constrain $\alpha'(0)$&#xA;w.r.t. $\alpha'(\ell)$ (in particular, one may be the negative of the&#xA;other), but nonetheless it seems that the original bound on the&#xA;&lt;i&gt;maximum&lt;/i&gt; curvature in the loop should still stand. How can we&#xA;recover it?&lt;/p&gt;&#xA;" OwnerUserId="1728" LastEditorUserId="1728" LastEditorDisplayName="" LastEditDate="2010-09-15T22:13:02.567" LastActivityDate="2010-09-16T23:47:45.373" Title="A good lower bound on the maximum curvature in a loop" Tags="&lt;differential-geometry&gt;" AnswerCount="1" />
  <row Id="4742" PostTypeId="1" CreationDate="2010-09-15T22:36:08.850" Score="2" ViewCount="445" Body="&lt;p&gt;Hey guys,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My professor wants us to do this problem to refresh ourselves with substitution. We have to solve:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$$\int\sqrt{1 + \sqrt{x}}\,\mathrm dx$$&lt;br&gt;&#xA;  $$\int\sqrt{1 + \sqrt{1 + \sqrt{x}}}\,\mathrm dx$$&lt;br&gt;&#xA;  ...etc...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If someone could point me in the right direction with the first one, I think I can handle the others.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for the help guys!&lt;/p&gt;&#xA;" OwnerUserId="1917" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2011-04-30T03:46:40.770" LastActivityDate="2011-04-30T03:46:40.770" Title="Integral of $\sqrt{1 + \sqrt{x}}$" Tags="&lt;calculus&gt;&lt;homework&gt;&lt;integral&gt;" AnswerCount="3" CommentCount="9" />
  <row Id="4744" PostTypeId="1" AcceptedAnswerId="4750" CreationDate="2010-09-16T02:11:30.943" Score="5" ViewCount="380" Body="&lt;p&gt;Given that I have two formal power series:&#xA;$$ A(x) = \sum_{k \ge 0} a_k x^k $$&#xA;$$ B(x) = \sum_{k \ge 0} b_k x^k $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Cauchy Product gives a series&#xA;$$ C(x) = \sum_{k \ge 0} c_k x^k $$&#xA;$$ c_k  = \sum_{n=0}^k a_n b_{k-n} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which comes from taking the product of the two series $C(x)=A(x)B(x)$. What then, in terms of $A(x)$ and $B(x)$, is this series?&#xA;$$Y(x) = \sum_{k \ge 0} a_k b_k x^k $$&lt;/p&gt;&#xA;" OwnerUserId="196" LastEditorUserId="196" LastEditorDisplayName="" LastEditDate="2010-09-16T02:27:41.210" LastActivityDate="2010-09-18T17:08:01.813" Title="Formal power series coefficient multiplication" Tags="&lt;generating-functions&gt;&lt;power-series&gt;" AnswerCount="4" CommentCount="2" FavoriteCount="1" />
  <row Id="4745" PostTypeId="2" ParentId="4742" CreationDate="2010-09-16T02:11:45.247" Score="2" ViewCount="" Body="&lt;p&gt;One technique usually worth trying in integration is to be extremely optimistic: ask yourself what is &quot;hard&quot; about the integrand and then make a substitution that simplifies it, even if the substitution looks awful.  In your case, the &quot;hard&quot; part about $\sqrt{1 + \sqrt{x}}$ is that the argument of the outer root is not evidently a perfect square, so it doesn't simplify.  So substituting $u^2$ for $1 + \sqrt{x}$ would be worth considering.  (You will obtain an integrand that is a polynomial in $u$.)&lt;/p&gt;&#xA;" OwnerUserId="1489" LastActivityDate="2010-09-16T02:11:45.247" CommentCount="0" />
  <row Id="4746" PostTypeId="2" ParentId="4744" CreationDate="2010-09-16T02:21:25.970" Score="-1" ViewCount="" Body="&lt;p&gt;Stupid answer: $Y\left(x\right)=\delta_{ik}A\left(x\right)B\left(1\right)$, for $\delta_{ik}$ the Kronecker Delta, and $i$ being the index on $B\left(x\right)$.&lt;/p&gt;&#xA;" OwnerUserId="16" LastActivityDate="2010-09-16T02:21:25.970" CommentCount="3" />
  <row Id="4747" PostTypeId="2" ParentId="4705" CreationDate="2010-09-16T02:21:31.497" Score="2" ViewCount="" Body="&lt;p&gt;Let's deal with the examples first.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The $\liminf$ of the real sequence $\{a_n\}$ with $a_n = (-1)^n$ is $-1$; the $\limsup$ of the same sequence is $1$. For the sequence $\{b_n\}$, with $b_n=\frac{1}{n}$, since the sequence converges to $0$, every subsequence converges to $0$ so both $\liminf$ and $\limsup$ are equal to $0$. You can think of the $\liminf$ as the infimum of the all the limits of all converging subsequences of the sequence, and the $\limsup$ is the supremum of all the limits of all converging subsequences of the sequences.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the definitions given by Jens, for $A_n=\{(-1)^n\}$ the $\limsup$ is $\{-1, 1\}$, because both elements show up in infinitely many of the $A_n$, and the $\liminf$ is empty, because no element occurs in all but finitely many of the $A_n$. For $B_n$, they are both empty because no element occurs in infinitely many of the sets, nor in all but finitely many (each element occurs in only one $B_n$). Again, $\liminf$ is the collection of all points that are in all but finitely many of the sets, while $\limsup$ is the collection of all points that are in infinitely many of the sets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But you are looking at the &lt;em&gt;wrong&lt;/em&gt; sets if you want your sets to be related to your sequences. As Nate Eldredge points out, what you should be looking at is the set $A_n = (-\infty,a_n)$, or $A_n = (-\infty,(-1)^n)$. Using &lt;em&gt;that&lt;/em&gt; definition, you have that $\limsup A_n = (-\infty,1)$ (as expected, since $\limsup a_n = 1$; each of these numbers occurs in infinitely many of the $A_n$), and $\liminf A_n=(-\infty,-1)$ because those are the only ones that occur in all but finitely many of the sets (in fact, in all; every other number that occurs in any $A_n$ occurs only in the $A_n$ with even $n$, so is missing from infinitely many $A_n$); while if you let $B_n = (-\infty,\frac{1}{n})$, then $\liminf B_n=\limsup B_n = (-\infty,0]$ (again, as expected, since the limit inferior and limit superior of $b_n$ are both equal to $0$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the reason you seem to be getting hung up is that there seems to be little relation between the limits inferior and superior of a $a_n$, and the limits inferior and superior of the sequence of sets $\{a_n\}$. But the point that Nate Eldredge made is that these are not the sets you want to associate with the sequence $a_n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You may remember that a sequence $\{a_n\}$ converges to $L$ if and only if every subsequence $\{a_{n_k}\}$ converges to $L$. Also, every sequence contains a monotone sequence, so if we allow $\infty$ and $-\infty$ as &quot;limits&quot;, it follows that every sequence will necessarily have a converging subsequence. So one can ask: &quot;what are all the points $M$ for which there is a subsequence of $\{a_n\}$ that converges to $M$?&quot; One can view the limits inferior and superior in terms of this set: the limit inferior of the sequence is the smallest number $\ell$ (including possibly $\infty$ or $-\infty$) for which there is a subsequence of $\{a_n\}$ converging to $\ell$. The limit superior is the largest number $L$ for which there is a subsequence of $\{a_n\}$ that converges to $L$. As it happens, the limit exists if and only if $\ell=L$. The limits inferior and superior can also be defined by&#xA;$$\liminf a_n = \lim_{n\to\infty}(\inf\{a_m|m\geq n\}) = \sup_n\left(\inf\{a_m|m\geq n\}\right)$$&#xA;and&#xA;$$\limsup a_n = \lim_{n\to\infty}(\sup\{a_m|m\geq n\}) = \inf_n\left(\sup\{a_m|m\geq n\}\right).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Viewed like this, you can perhaps see a bit more of a connection with the limits inferior and superior of a sequence of sets. If $\{A_n\}$ is a sequence of sets, then the limits inferior and superior are defined to be:&#xA;$$\liminf A_n = \cup_{n=1}^{\infty}\left(\cap_{m=n}^{\infty} A_m\right)$$&#xA;and&#xA;$$\limsup A_n = \cap_{n=1}^{\infty}\left(\cup_{m=n}^{\infty} A_m\right).$$&#xA;Think of an intersection as taking &quot;smallest&quot; thing in common (so like an infimum), and think of union as taking &quot;largest&quot; thing in common (so like a supremum). The limit inferior is the supremum of the infima, while the limit superior is the infimum of the suprema. It is now a nice exercise to verify that $\liminf A_n$ is the collection of all things which are in all but finitely many of the $A_i$, while $\limsup A_n$ is the collection of all things which are in infinitely many of the $A_i$ (as described by Jens). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, how do you connect a sequence $\{a_n\}$ to sets so that the limits inferior and superior correspond in some way? You can't just take $A_n = \{a_n\}$, because then each $A_n$ knows nothing about what came before or after; you lose all information that could tell you something about subsequences. You could try letting $A_n =\{a_m|m\geq n\}$, and that will even work in some instances, but the problem here is that the information you are losing is that in the real numbers, a sequence may converge to a number even if &lt;em&gt;no term&lt;/em&gt; in the sequence equals the limit; then the limit is never going to show up in any of the sets, and it's not going to show up in the limits inferior nor superior of the sets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What's the solution? The limit inferior of a sequence is going to be a lower bound for all but finitely many of the terms of the sequence (if there were infinitely many terms of the sequence strictly smaller than $\liminf a_n$, then you would be able to get a subsequence from among them that converges to something strictly smaller than $\liminf a_n$, a contradiction). This suggests that what you want to do is let $A_n$ be the collection of all lower bounds to $a_n$; then the limit inferior of the $A_n$ will be the collection of all things that are lower bounds to all but finitely many terms of the sequence, exactly the set you want to consider to find $\liminf a_n$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the limit superior of $a_n$? You can define it dually, as the smallest of all numbers that upper bounds for all but finitely many terms of the sequence (this will lead to the formula that says that $\limsup a_n = -\liminf(-a_n)$). Or you can try to define it in terms of the lower bounds again: any number $k$ strictly smaller than the limit superior must have infinitely many terms of the sequence larger than $k$ (otherwise, no subsequence could converge to something larger, so no subsequence could converge to the limit superior). That is: look at the collection of all things that are lower bounds for &lt;em&gt;infinitely many&lt;/em&gt; of the $a_n$, and the supremum of that will be the limit superior.  So we again look at $A_n = (-\infty,a_n)$ (the set of all lower bounds to $a_n$), and consider the limit superior of the $A_n$; this is the collection of all numbers that are lower bounds to infinitely many of the $a_n$, and so its supremum will be $\limsup a_n$. So that's why you consider the set $A_n(-\infty,a_n)$ instead of the set $\{a_n\}$, and where they come from.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are other ways of associating to each $a_n$ an appropriate set; in this situation, note that $\sup A_n = a_n$ for each $n$, that $\sup(\liminf A_n) = \liminf (\sup A_n) = \liminf a_n$, and $\sup(\limsup A_n) = \limsup(\sup A_n) = \limsup a_n$, which makes this association pretty nice. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope this helps clarify it further.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-09-16T05:02:01.010" LastActivityDate="2010-09-16T05:02:01.010" CommentCount="5" />
  <row Id="4748" PostTypeId="2" ParentId="4733" CreationDate="2010-09-16T02:24:28.570" Score="2" ViewCount="" Body="&lt;p&gt;The function sends each set of measure zero to a set of measure zero.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $A\subset\mathbb R$ be a null set (i.e. $A$ has measure 0).  Then $f(A)=f(A\cap \{0\})\cup f(A\cap(\mathbb{R}\setminus \{0\}))$.  The first set in the union has at most one point, so we need only worry about the second.  The set $A\cap(\mathbb{R}\setminus \{0\})$ can be expressed as a countable union of sets of the form $A\cap [a,b]$ with $0\lt a$ or $b\lt 0$, and therefore, since the image of a union is the union of the images, $f(A\cap(\mathbb{R}\setminus \{0\}))$ can be expressed as a countable union of sets of the form $f(A\cap [a,b])$ with $0\lt a$ or $b\lt 0$.  For each such $a$ and $b$, $f$ is continuously differentiable in a neighborhood of $[a,b]$, and hence the restriction of $f$ to $[a,b]$ is absolutely continuous (e.g. by the fundamental theorem of calculus for $C^1$ functions).  Therefore the image of the null set $A\cap[a,b]$ under $f$ is null.  Countable unions of null sets are null, so this shows that $f(A)$ is null.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-09-16T02:24:28.570" CommentCount="3" />
  <row Id="4749" PostTypeId="1" AcceptedAnswerId="4752" CreationDate="2010-09-16T02:28:19.010" Score="1" ViewCount="142" Body="&lt;p&gt;For plane curves, twice the quadratic term of the Taylor expansion at a point on that curve is precisely the curvature. I know of one proof as featured in &lt;a href=&quot;http://www.math.cornell.edu/~hubbard/vectorcalculus.html&quot; rel=&quot;nofollow&quot;&gt;Hubbard and Hubbard&lt;/a&gt;, but I was wondering if there was another more elegant solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="16" LastEditorUserId="1489" LastEditorDisplayName="" LastEditDate="2010-09-16T02:45:13.877" LastActivityDate="2010-09-16T04:56:46.857" Title="A nice proof of the fact that the Quadratic Term of the Taylor polynomial is Curvature" Tags="&lt;calculus&gt;&lt;differential-geometry&gt;&lt;curvature&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="2" />
  <row Id="4750" PostTypeId="2" ParentId="4744" CreationDate="2010-09-16T02:33:44.670" Score="8" ViewCount="" Body="&lt;p&gt;It is known as the &lt;strong&gt;Hadamard product&lt;/strong&gt; $\rm\; f \star g\;$. There is no closed form for general hadamard products, but some classes of functions are known to be closed under such products, e.g. rational power series. However, algebraic series are not generally closed under Hadamard products, the standard example being $\rm\quad f = g = (1-4x)^{-1/2} \;=\; \sum \binom{2n}{n}\ x^n\:.\quad$ However, $\rm\: f\:$ rational, $\rm\: g\:$ algebraic $\rm\;\Rightarrow\; f\star g\;$ algebraic. Also &lt;a href=&quot;http://www-math.mit.edu/~rstan/pubs/pubfiles/45.pdf&quot; rel=&quot;nofollow&quot;&gt;D-finite power series&lt;/a&gt; are closed under Hadamard products, i.e. power series satisfying a linear differential equation with polynomial coefficients or, equivalently, series whose coefficients satisfy a linear recursive equation with polynomial coefficients.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hadamard products can also be defined in terms of diagonals of multivariate series (and vice versa), e.g. see this &lt;a href=&quot;http://pictor.math.uqam.ca/~plouffe/OEIS/archive_in_pdf/005_a084.pdf&quot; rel=&quot;nofollow&quot;&gt;paper&lt;/a&gt;, where you'll find some interesting connections with finite automata which, e.g. help one to easily observe that algebraic series over $\rm\mathbb{F}_p$ &lt;em&gt;are&lt;/em&gt; closed under Hadamard product (which fails in characteristic 0)&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-09-16T03:40:03.010" LastActivityDate="2010-09-16T03:40:03.010" />
  <row Id="4751" PostTypeId="2" ParentId="4744" CreationDate="2010-09-16T02:33:49.990" Score="6" ViewCount="" Body="&lt;p&gt;$Y$ is known as the Hadamard product of $A$ and $B$, and there is no simple way to find it from $A$ and $B$ in general.  (If you don't believe me, set $A = B = e^x$.)  The closest thing I know to a formula is (with all the analytic caveats that makes everything converge)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$Y(r) = \frac{1}{2\pi } \int_0^{2\pi} A(e^{i \theta}) B(re^{-i \theta}) d \theta$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which follows by &lt;a href=&quot;http://en.wikipedia.org/wiki/Parsevals_theorem&quot; rel=&quot;nofollow&quot;&gt;Parseval's theorem&lt;/a&gt;.  If $A$ and $B$ are sufficiently simple (for example if they are rational functions) then it is possible to evaluate this integral, but in general there's not much you can do; $Y$ can be much more complicated than $A$ or $B$.  (Another idea is to write the above as a contour integral, and if the integrand ends up being meromorphic you can try to use the residue theorem.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Computing Hadamard products is a special case of computing the diagonal of a two-variable generating function, a problem which I describe with a few examples &lt;a href=&quot;http://qchu.wordpress.com/2009/10/07/extracting-the-diagonal/&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; using the residue theorem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Of course for very special $A$ it is possible to say more, e.g. when $a_k$ is a polynomial in $k$.)&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-09-16T02:51:24.297" LastActivityDate="2010-09-16T02:51:24.297" CommentCount="1" />
  <row Id="4752" PostTypeId="2" ParentId="4749" CreationDate="2010-09-16T02:43:52.113" Score="2" ViewCount="" Body="&lt;p&gt;I'm unfamiliar with your reference, so please forgive me if this might be reminiscent of its solution.  I'll use the Serret-Frenet formulae.  For a twice differentiable curve $\gamma$ parameterized by arclength, one of them asserts that the normal component of the derivative of the (unit) tangent vector is the curvature $\kappa$.  (This is often taken as the definition of curvature.)  To simplify things, but wlg, let's use coordinates in which the point in question is the origin, so that the Taylor expansion is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\gamma(t) = \left(a_1 t + \frac{a_2}{2}t^2 + \cdots, b_1 t + \frac{b_2}{2}t^2 + \cdots \right)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whence the unit tangent vector at the origin is $T = (a_1, b_1)$ and its derivative is $(a_2, b_2)$.  To obtain the component normal to $T$, it suffices to compute the inner product of this derivative with the rotated tangent vector $(-b_1, a_1)$ (because the tangent vector has unit length), yielding&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\kappa = -a_2 b_1 + b_2 a_1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This general result, specialized to the case where $\gamma$ is of the form $(t, f(t))$ for a function $f(t) = 0 + b_1 t + \frac{b_2}{2}t^2 + \cdots$, reduces (because $a_1 = 1$ and $a_2 = 0$) to $b_2$, &quot;twice the quadratic term in the Taylor expansion&quot; at this point.&lt;/p&gt;&#xA;" OwnerUserId="1489" LastActivityDate="2010-09-16T02:43:52.113" CommentCount="3" />
  <row Id="4753" PostTypeId="2" ParentId="4742" CreationDate="2010-09-16T03:13:22.113" Score="0" ViewCount="" Body="&lt;p&gt;This is the answer from &lt;a href=&quot;http://www.wolframalpha.com/input/?i=integrate+of+sqrt%281+%2B+sqrt%28x%29%29&quot; rel=&quot;nofollow&quot;&gt;Wolframalpha&lt;/a&gt;: note that you have to use the substitution&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$u=\sqrt{x}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;first. Note that Wolframalpha able to give you the full step by step solution.&lt;/p&gt;&#xA;" OwnerUserId="550" LastEditorUserId="550" LastEditorDisplayName="" LastEditDate="2010-09-16T09:21:16.290" LastActivityDate="2010-09-16T09:21:16.290" />
  <row Id="4754" PostTypeId="2" ParentId="4749" CreationDate="2010-09-16T04:56:46.857" Score="2" ViewCount="" Body="&lt;p&gt;Here's a completely different answer inspired by J.M.'s description of the text.  It might give a bit more geometric insight.  It essentially shows that (a) twice the second coefficient in the Taylor series describes a parabola closely approximating the curve at the origin, (b) the curvature is determined by a circle closely approximating the curve at the origin, and (c) the curvature of the Taylor series parabola is twice that coefficient.  In other words, when we're talking about curvature we're really talking about the second-order behavior of the curve and that's exactly what the quadratic term in the Taylor series describes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The relevant definition of the curvature here is the reciprocal of the radius of the osculating circle (taken to be negative if the circle lies below the curve, with 1/infinity = 0).  An osculating (&quot;kissing&quot;) circle is found by taking three points close to the point under consideration (which again might as well be the origin), fitting a circle to them, letting the points independently converge to the origin, and seeing what the circle settles down to.  Assuming the circle exists, we can find it by considering some special configurations of points.  Choose the three consisting of the origin itself, with coordinates (0,0), and the points at nonzero distance $x$ to both sides of the origin.  Their coordinates are given by the Taylor series.  One is is at $(x, \frac{b}{2}x^2 + O(x^3))$ and the second is at $(-x, \frac{b}{2}(-x)^2 + O((-x)^3))$.  The center of the circle is equidistant from all three.  If we choose $x$ small enough (witness some geometric hand-waving here that can be made rigorous at the expense of making the following algebra a little harder), the situation is essentially symmetric around the origin and we expect the center to be located directly above (or below) the origin at a point $(0,a)$.  Consider the distances from $(0,a)$ to the origin and to the point corresponding to $x$: their squares must be equal.  Whence&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a^2 = (x-0)^2 + (\frac{b}{2}x^2 - a)^2 + O(x^3)$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;at least to an excellent approximation when $x$ is small.  Some terms cancel and an easy algebraic simplification gives (remember, we're solving for $a$ in terms of $x$ and $b$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$0 = 1 - a b + \frac{b^2}{4}x^4 + O(x)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;implying $1/a = b$ in the limit as $x \to 0$.  Since $1/a$ is the curvature, our task is done.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yes, as I have warned, this lacks rigor, but the rigor can be supplied (I recall Michael Spivak does this early in Volume II of his Differential Geometry series) and what it lacks in sophistication may be compensated by the elementary geometric intuition it affords of both curvature and the Taylor series.&lt;/p&gt;&#xA;" OwnerUserId="1489" LastActivityDate="2010-09-16T04:56:46.857" CommentCount="1" />
  <row Id="4755" PostTypeId="2" ParentId="4742" CreationDate="2010-09-16T05:25:19.160" Score="0" ViewCount="" Body="&lt;p&gt;Suprisingly useful hint: $x = 1+(x-1)$.&lt;/p&gt;&#xA;" OwnerUserId="467" LastActivityDate="2010-09-16T05:25:19.160" />
  <row Id="4757" PostTypeId="1" CreationDate="2010-09-16T06:17:34.643" Score="1" ViewCount="224" Body="&lt;p&gt;In a sense, I did just ask this question.  However, since the question is about how to write a beautiful looking proof and my proof is entirely rewritten, it seems like it should be a new question.  What I'm looking for here are (aside from any actual mathematical problems in the proof of course) is advice on how to better structure what I'm writing or to better communicate the ideas.  Anything from overall structure, word choice, what letters I give my variables, anything at all.  Anyhow, the question from Spivak:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Define $f:\mathbb{R}\rightarrow\mathbb{R}$ by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f(x)=\begin{cases}&#xA;e^{-x^{-2}}&amp;amp; x\ne 0\\&#xA;0 &amp;amp; x=0\\&#xA;\end{cases}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Show that $f$ is a $C^\infty$ function, and $f^{(i)}(0)=0$ for all $i$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, here is my proof.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, we show that $f\in C^\infty$ for $x\ne 0$ by induction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $-g(x)=\frac{1}{x^2}$.  Then we can write $f$ as $f(x)=e^{g(x)}$.  Let $P={p:\mathbb{R}\rightarrow\mathbb{R} : p(x)=\sum \frac{a_n}{x^n},a_n\in\mathbb{R}}$.  Now we can show by induction that $f^{(n)}(x)=e^{g(x)}p_n(x)$ for some $p_n \in P$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $p_0=\frac{1}{x^0}=1$.  Then $p\in P$ and $f(x)=e^{g(x)}p_0(x)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume $p_n\in P$ such that $p_n=\sum_{i=0}^\infty\frac{a_i}{x^i}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align*}&#xA;(e^{g(x)}p_n(x)) &amp;amp;= e^{g(x)}g'(x) p_n(x) + e^{g(x)} p_n'(x)\\&#xA;&amp;amp;= e^{g(x)}(g'(x) p_n(x) + p_n'(x))\\&#xA;&amp;amp;= e^{g(x)}(\frac{2}{x^3} p_n(x) + p_n'(x))\\&#xA;&amp;amp;= e^{g(x)}(\frac{2}{x^3} \sum \frac{a_i}{x^i} + \sum \frac{-ia_i}{x^{i+1}})\\&#xA;&amp;amp;= e^{g(x)}(\sum \frac{2a_i}{x^{i+3}} + \sum \frac{-ia_i}{x^{i+1}})\\&#xA;\end{align*}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $\sum \frac{2a_i}{x^{i+3}} + \sum \frac{-ia_i}{x^{i+1}}\in P$, it follows that $p_n\in P \Rightarrow p_{n+1}\in P$.  Thus by induction, $f^{(n)}=e^{g(x)}p_n(x)$ where $p_n\in P$ and $f\in C^\infty$ for $x\ne 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we show by induction that $f^{(n)}(0)=0$ for all $n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, we prove that for all $n \ge -1$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$m(n)=\lim_{h\rightarrow 0} \frac{1}{h^n e^{\frac{1}{h^2}}}=0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $n=-1$ we have $\lim_{h\rightarrow 0}\frac{1}{h^{-1} e^{\frac{1}{h^2}}}=\lim_{h\rightarrow 0}\frac{h}{e^{\frac{1}{h^2}}}=0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $n=0$ we have $\lim_{h\rightarrow 0}\frac{1}{h^{0} e^{\frac{1}{h^2}}}=\lim_{h\rightarrow 0}\frac{1}{e^{\frac{1}{h^2}}}=0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now assume that $m(n-1)=0$ and $m(n-2)=0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align*}&#xA;m(n) &amp;amp;= \lim_{h\rightarrow 0} \frac{1}{h^n e^{\frac{1}{h^2}}}\\&#xA;&amp;amp;= \lim_{h\rightarrow 0} \frac{h^{-n}}{e^{\frac{1}{h^2}}}\\&#xA;&amp;amp;= \lim_{h\rightarrow 0} \frac{-nh^{-n-1}}{e^{\frac{1}{h^2}} (-2 h^{-3})}\tag{L'H\^{o}pital's rule}\\&#xA;&amp;amp;= \frac{n}{2} \lim_{h\rightarrow 0} \frac{h^{-n-1+3}}{e^{\frac{1}{h^2}}}\\&#xA;&amp;amp;= \frac{n}{2} \lim_{h\rightarrow 0} \frac{1}{h^{n-2}e^{\frac{1}{h^2}}}\\&#xA;&amp;amp;= \frac{n}{2} m(n-2)=0\\&#xA;\end{align*}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus by induction, $m(n)=0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f^{(n)}(x)=e^{g(x)}(\sum \frac{2a_i}{x^{i+3}} + \sum \frac{-ia_i}{x^{i+1}})=\sum 2a_i m(i+3)+\sum-im(i+1)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;it follows that $f^{(n)}=0$ for all $n$.&lt;/p&gt;&#xA;" OwnerUserId="326" LastEditorUserId="326" LastEditorDisplayName="" LastEditDate="2010-09-17T01:46:40.937" LastActivityDate="2010-10-17T02:45:36.573" Title="Improve my proof about this $C^\infty$ function even more!" Tags="&lt;calculus&gt;&lt;real-analysis&gt;&lt;proof-strategy&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="4758" PostTypeId="1" AcceptedAnswerId="4760" CreationDate="2010-09-16T06:59:57.377" Score="0" ViewCount="234" Body="&lt;p&gt;I first noticed this is true for the integers of the sequence $9, 99, 999, 9999,\dots$, since for some term $a_n=10^n-1$ in the sequence and $p$ a prime other than $2$ or $5$, we have $a_n\equiv 0 (mod\ p)\Leftrightarrow 10^n\equiv 1 (mod\ p)$ and we can find infinitely many $n=k(p-1)$ so that $10^n=(10^{p-1})^k\equiv 1^k\equiv 1 (mod\ p)$ by Fermat's little theorem. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm curious, is there a similar way to show this is true for repunits $1, 11, 111,\dots$?&lt;/p&gt;&#xA;" OwnerUserId="1583" LastActivityDate="2010-09-19T21:30:05.707" Title="Why do primes other than 2 and 5 divide infinitely many repunits?" Tags="&lt;number-theory&gt;&lt;prime-numbers&gt;" AnswerCount="2" />
  <row Id="4759" PostTypeId="1" CreationDate="2010-09-16T07:03:46.947" Score="1" ViewCount="282" Body="&lt;p&gt;Hi guys, I wasnt sure where to ask this but since this is an algorithmic question here it goes. I've come face to face with a math problem and can't seem to get over it for the last couple of days. It goes like this:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;You are given an adding machine that sums a set of N+1 digits&#xA;  consisting of positive integers 1 to N&#xA;  as it's given the numbers (e.g. the&#xA;  machine is given 3 as the first number&#xA;  and outputs 3. It's then given 6 as&#xA;  the second number and outputs 9. It's&#xA;  given 11 as the third number and&#xA;  outputs 20. Etcetera until it has&#xA;  processed N+1 numbers). One (and only&#xA;  one) of the digits is repeated. How do&#xA;  you determine which number is&#xA;  repeated?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It seems like a trick question and I'd be really pissed off it is just that a question to which the answer is 'not possible' - any ideas here?&lt;/p&gt;&#xA;" OwnerUserId="1922" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-09-16T13:50:43.397" LastActivityDate="2010-09-16T13:50:43.397" Title="Need help with a math trick question" Tags="&lt;algebra-precalculus&gt;&lt;average&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="4760" PostTypeId="2" ParentId="4758" CreationDate="2010-09-16T07:08:00.653" Score="7" ViewCount="" Body="&lt;p&gt;Note that the $n$-th repunit is $r_n=(10^n-1)/9$.&#xA;If $p$ is a prime and $p\ne 3$ then $p\mid r_n$ iff $p\mid(10^n-1)$.&#xA;(This is due to Euclid's lemma that $p\mid ab$ implies $p\mid a$&#xA;or $p\mid b$.) So when $p\notin\{2,3,5\}$ the problem reduces to the one&#xA;you've done.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Only the case $p=3$ remains. Now remember that $3\mid n$ iff $3$ divides&#xA;the &lt;em&gt;digital sum&lt;/em&gt; of $n$ (the sum of the decimal digits of $n$)....&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-09-16T07:08:00.653" CommentCount="1" />
  <row Id="4761" PostTypeId="2" ParentId="4759" CreationDate="2010-09-16T07:14:47.423" Score="5" ViewCount="" Body="&lt;p&gt;We know that $1+2+\ldots +N = \frac{1}{2}N(N+1)$. Suppose the end amount is $T$ then the repeated number is $T-\frac{1}{2}N(N+1)$.&lt;/p&gt;&#xA;" OwnerUserId="1794" LastActivityDate="2010-09-16T07:14:47.423" CommentCount="3" />
  <row Id="4762" PostTypeId="2" ParentId="4720" CreationDate="2010-09-16T08:03:57.763" Score="-1" ViewCount="" Body="&lt;p&gt;There is a difference. Both products take two vectors in $\mathbb{R}^3$. The cross product gives a vector in the same $\mathbb{R}^3$ and the wedge product gives a vector in a different $\mathbb{R}^3$. The two output vector spaces are indeed isomorphic and if you choose an isomorphism you can identify the two products. However this isomorphism is a choice, or to put it another way depends on fixing a convention.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In higher dimensions the wedge product gives a vector in a vector space of higher dimension and so no identification is possible.&lt;/p&gt;&#xA;" OwnerUserId="638" LastActivityDate="2010-09-16T08:03:57.763" CommentCount="1" />
  <row Id="4763" PostTypeId="2" ParentId="4720" CreationDate="2010-09-16T08:05:43.583" Score="2" ViewCount="" Body="&lt;p&gt;Yes, this definition is chosen for a reason, as the unique solution to a pedagogical problem. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do Carmo's definition is awkward and redundant in 3 dimensions, but it is the only one among the usual definitions for the cross product that when generalized to $n$ dimensions (there is a cross-product of $n-1$ vectors in $R^n$) is rigorous, visibly basis independent, and produces a vector in the same space --- all without a long digression on multilinear algebra, duality, and other sophisticated theory.  Consider the other definitions:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The right hand rule does work in $n$ dimensions, and is coordinate-independent, but it is hard to describe without a formula which hand is the &quot;right&quot; one, and verifying that the result is multilinear can be awkward if one does not have a reliable formalism of $n$-dimensional Euclidean geometry.   I doubt most students would consider this approach rigorous even if, in principle, it can be made so.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The exterior algebra relies on theory not known by all audiences of Do Carmo's book (e.g, engineers).  It does not produce a vector in $R^n$ without even more theory.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The definition as a determinant of the vector coordinates can be confusing, with vectors and scalars mixed together as entries of one matrix.  It is also geometrically meaningful only when proven independent of the choice of (orthonormal) basis for the coordinates. In 3-dimensions this is known by the equivalence with the right hand rule and by drawing pictures, techniques that require a long development to perform in $n$ dimensions.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="467" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-09-16T08:12:24.537" LastActivityDate="2010-09-16T08:12:24.537" />
  <row Id="4764" PostTypeId="1" AcceptedAnswerId="4767" CreationDate="2010-09-16T08:53:14.743" Score="3" ViewCount="413" Body="&lt;p&gt;We know that the sine function takes it values between $[-1,1]$. So is the set $$A = \{ \sin{n} \ : \ n \in \mathbb{N}\}$$ dense in $[-1,1]$. Generally, for showing the set is dense, one proceeds, by finding out what is $\overline{A}$ of this given set. And if $\overline{A} = [-1,1]$, we are through with the proof, but i having trouble here!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly can one do this with cosine function also, that is proving $B= \{ \cos{n} \ : \ n \in \mathbb{N}\}$ being dense in $[-1,1]$&lt;/p&gt;&#xA;" LastActivityDate="2010-09-16T11:37:04.880" Title="Sine function dense in $[-1,1]$" Tags="&lt;real-analysis&gt;&lt;analysis&gt;" AnswerCount="3" FavoriteCount="2" />
  <row Id="4765" PostTypeId="1" AcceptedAnswerId="4766" CreationDate="2010-09-16T08:56:02.620" Score="1" ViewCount="171" Body="&lt;p&gt;I have a grid made up of overlapping 3x3 squares like so:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/BaY9s.png&quot; alt=&quot;Grid&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The numbers on the grid indicate the number of overlapping squares. Given that we know the maximum number of overlapping squares (9 at the middle), and the size of the squares (3x3), is there a simple way to calculate the rest of the number of overlaps?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;e.g. I know the maximum number of overlaps is 9 at point (2,2) and the square size is 3x3. So given point (3,2) how can I calculate that there are 6 overlaps at that point?&lt;/p&gt;&#xA;" OwnerUserId="1588" LastActivityDate="2010-09-16T11:05:58.773" Title="Grid of overlapping squares" Tags="&lt;geometry&gt;" AnswerCount="2" />
  <row Id="4766" PostTypeId="2" ParentId="4765" CreationDate="2010-09-16T09:15:06.143" Score="1" ViewCount="" Body="&lt;p&gt;The maximum number of overlaps is always $k^{2}$ (where $k$ is the small square's side length) (assuming the larger square is always of side length $2k-1$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The number of overlaps at any other point $(i,j)$ is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\min(i,2k-i) \cdot \min(j,2k-j)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a pointer towards a proof consider the first row and the first column of the larger square which must always be the sequence $(1,2,\ldots,k,\ldots,2,1)$.  Then look at why the other rows and columns are the products of their intersection with the edge row and column.&lt;/p&gt;&#xA;" OwnerUserId="1750" LastActivityDate="2010-09-16T09:15:06.143" CommentCount="1" />
  <row Id="4767" PostTypeId="2" ParentId="4764" CreationDate="2010-09-16T09:18:13.813" Score="4" ViewCount="" Body="&lt;p&gt;The hard part is to show that for any $x$ such that $0 \le x \le 2\pi$, and any $\epsilon&gt;0$ there exists a real number $y$ and two integers $m$ and $n$ such that $|y-x|&amp;lt;\epsilon$ and $n=2\pi m+y$. Hint: break up $[0,2\pi]$ into small subintervals, remember that $\pi$ is irrational and apply the pigeonhole principle.&lt;/p&gt;&#xA;" OwnerUserId="1195" LastActivityDate="2010-09-16T09:18:13.813" />
  <row Id="4768" PostTypeId="1" CreationDate="2010-09-16T10:09:04.857" Score="3" ViewCount="571" Body="&lt;p&gt;How we can show every permutation is either even or odd,but not both......I can't arrive at a proof for this ..... Can anybody give me the proof...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance...&lt;/p&gt;&#xA;" OwnerUserId="1924" LastActivityDate="2010-09-16T15:35:07.063" Title="every permutation is either even or odd,but not both......" Tags="&lt;permutations&gt;" AnswerCount="4" CommentCount="4" FavoriteCount="1" />
  <row Id="4769" PostTypeId="1" AcceptedAnswerId="4771" CreationDate="2010-09-16T10:13:05.400" Score="0" ViewCount="120" Body="&lt;p&gt;I have a suspicion that the following expression is true, however my algebra skills aren't brilliant, so any help would be appreciated:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is it possible to infer $x = \frac{y + z}{2}$ from $a^{\frac{1}{x}} = b^{\frac{1}{y}} + c^{\frac{1}{z}}$ and $a = b + c$&lt;/p&gt;&#xA;" OwnerUserId="1923" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-09-16T13:49:34.157" LastActivityDate="2010-09-16T13:49:34.157" Title="algebraic puzzle" Tags="&lt;algebra-precalculus&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="4770" PostTypeId="2" ParentId="4764" CreationDate="2010-09-16T10:13:43.853" Score="2" ViewCount="" Body="&lt;p&gt;Jmoy's is a good hint, but I think an even nicer approach is to&#xA;prove that the points $(\cos n,\sin n)$ are dense on the unit circle,&#xA;and then deduce the results for the individual terms.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-09-16T10:13:43.853" CommentCount="2" />
  <row Id="4771" PostTypeId="2" ParentId="4769" CreationDate="2010-09-16T10:20:44.077" Score="4" ViewCount="" Body="&lt;p&gt;Let $a=2$, $b=c=1$. Also let $x=1$ then $a=b+c$ and $a^{\frac{1}{x}}=2$. Regardless of the values of $y$ and $z$ then $1^{\frac{1}{y}}+1^{\frac{1}{z}}=2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus we cannot infer that $x=\frac{y+z}{2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Counterexample:&#xA;Let $a=2$, $b=c=1$, $x=1$ and $y=z=2$ then $b+c=2=a$ and $b^{\frac{1}{y}}+c^{\frac{1}{z}}=2=a^{\frac{1}{x}}$. But $x\neq \frac{y+z}{2}=2$. &lt;/p&gt;&#xA;" OwnerUserId="1794" LastEditorUserId="1794" LastEditorDisplayName="" LastEditDate="2010-09-16T10:31:21.043" LastActivityDate="2010-09-16T10:31:21.043" CommentCount="2" />
  <row Id="4772" PostTypeId="2" ParentId="4768" CreationDate="2010-09-16T10:20:47.133" Score="3" ViewCount="" Body="&lt;p&gt;One way is to define the sign of a permutation $\sigma$ using the polynomial $\Delta = \Pi (x_i-x_j)$ with $1\le i &amp;lt; j \le n$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is easy to see that $\sigma(\Delta) = \Pi (x_{\sigma(i)}-x_{\sigma(j)})$ satisfies $\sigma(\Delta)=\pm\Delta$. Now define the sign by $sign(\sigma)=\frac{\Delta}{\sigma(\Delta)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With a little more work you can show that this function is a homomorphism of groups, and that on transpositions it return -1. Therefore, if $\sigma=\tau_1\cdots\tau_k$ is a way to write $\sigma$ as a product of transpositions, we have $sign(\sigma)=(-1)^k$ and so for even permutations (permutations whose sign is 1) k must always be even, and for odd permutations (whose sign is -1) it must always be odd.&lt;/p&gt;&#xA;" OwnerUserId="1818" LastActivityDate="2010-09-16T10:20:47.133" CommentCount="3" />
  <row Id="4773" PostTypeId="1" AcceptedAnswerId="4782" CreationDate="2010-09-16T10:35:48.917" Score="0" ViewCount="417" Body="&lt;p&gt;A sequence of sets is defined as $A_n=\{x \in [0,1] : |\sum_{i=0}^{n-1} 1_{[\frac{i}{2n},\frac{2i+1}{4n})} - 1_{[\frac{2i+1}{4n},\frac{i+1}{2n})}| \geq p\}$ for some positive $p\geq0$. What is $\limsup_{n\to\infty}A_n(p)$ and $\liminf_{n\to\infty}A_n(p)$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For any $x$, the function $f_n(x) = \sum_{i=0}^{n-1} 1_{[\frac{i}{2n},\frac{2i+1}{4n})} - 1_{[\frac{2i+1}{4n},\frac{i+1}{2n})}$ is $1$ or $-1$ based on number of paritions $n$. So, $|f_n(x)|$ is always $1$ for any given $x$. Hence, if $p &gt; 1$, $A_n$ is empty set for all $n$. Hence both $\limsup$ and $\liminf$ is empty. If $p \leq 1$, then $A_n = [0,1]$ for all $n$. Therefore, both $\limsup$ and $\liminf$ is $[0,1]$. Is this correct?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I experimented further by changing $f_n(x)$ definition as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f_n(x) = \sum_{i=0}^{n-1} 1_{[\frac{i}{2n},\frac{2i+1}{4n})} - 1_{[\frac{2i+1}{4n},\frac{i+1}{2n})} ~~~~\mbox{if } 0 \leq x &amp;lt; 1/2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f_n(x) = x^n ~~~~\mbox{if } 1/2 \leq x \leq 1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is $\limsup_{n\to\infty}A_n(p)$ and $\liminf_{n\to\infty}A_n(p)$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am still trying to solve this.&lt;/p&gt;&#xA;" OwnerUserId="957" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2010-11-21T23:07:22.160" LastActivityDate="2010-12-22T00:44:27.000" Title="limit superior and limit inferior of the given sequence of sets" Tags="&lt;measure-theory&gt;&lt;limit&gt;&lt;elementary-set-theory&gt;" AnswerCount="1" CommentCount="9" FavoriteCount="1" />
  <row Id="4774" PostTypeId="2" ParentId="4765" CreationDate="2010-09-16T11:05:58.773" Score="1" ViewCount="" Body="&lt;p&gt;If you are just considering $3\times 3$ squares then the number of overlapping squares at the $(i,j)$ is the number of $1\times 1$ squares (including itself) which are internal neighbours. i.e. neighbouring squares which are not on the edge.&lt;/p&gt;&#xA;" OwnerUserId="1794" LastActivityDate="2010-09-16T11:05:58.773" />
  <row Id="4775" PostTypeId="2" ParentId="4581" CreationDate="2010-09-16T11:08:27.053" Score="3" ViewCount="" Body="&lt;p&gt;Let $K/k$ be a field extension. Then $K \otimes_k k(X)$ is a localization of the integral domain $K \otimes_k k[X] = K[X]$, thus the natural map $K \otimes_k k(X) \to K(X)$ is injective. Now we have the following equivalence:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a) $K/k$ is algebraic&lt;/p&gt;&#xA;&#xA;&lt;p&gt;b) Every polynomial over $K$ divides some polynomial over $k$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;c) $K \otimes_k k(X) = K(X)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a \Rightarrow b$: Because of uniqueness of the division algorithm, it suffices to prove divisibilty in an algebraic extension. So take a splitting field of the polynomial and reduce to the case of linear factors $x-a$. But they divide the minimal polynomial over $k$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$b \Rightarrow c$: $K \otimes_k k(X)$ is the localization of $K[X]$ at all nontrivial polynomials over $k$. Since we have $b$, we localize at all nontrivial polynomials  over $K$, i.e. we get $K(X)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$c \Rightarrow a$: Take $u \in K$. Then $1/(X-u) = p/q$ for $p \in K[X], q \in k[X]$, i.e. $p (X-u)$ is a polynomial over $k$ with root $u$. QED&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, if $K/k$ is algebraic, we have $[K(X):k(X)] = [K:k]$.&lt;/p&gt;&#xA;" OwnerUserId="1650" LastEditorUserId="226" LastEditorDisplayName="" LastEditDate="2010-09-16T11:14:54.187" LastActivityDate="2010-09-16T11:14:54.187" CommentCount="2" />
  <row Id="4776" PostTypeId="2" ParentId="4764" CreationDate="2010-09-16T11:37:04.880" Score="1" ViewCount="" Body="&lt;p&gt;While searching on google, i got this JSTOR link: &lt;a href=&quot;http://www.jstor.org/stable/2688681&quot; rel=&quot;nofollow&quot;&gt;http://www.jstor.org/stable/2688681&lt;/a&gt; which answers the question in an intricate way.&lt;/p&gt;&#xA;" LastActivityDate="2010-09-16T11:37:04.880" />
  <row Id="4777" PostTypeId="2" ParentId="4711" CreationDate="2010-09-16T11:44:40.270" Score="3" ViewCount="" Body="&lt;p&gt;I show here my interpretation of your question and comments until now:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/Y6yoF.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One has to define the triangle sides lengths (one is of course enough) depending on the resolution one needs. In the figure I approximated the red shape by the green right triangles with two equal angles. The shape boundary determines the triangles over or closed to it, the criteria being that at least 50% of the triangle area lies inside the shape. The inner triangles can be chosen more freely.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remark: there are a few mistakes in the sketch.  &lt;/p&gt;&#xA;" OwnerUserId="752" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2010-09-16T16:29:51.100" LastActivityDate="2010-09-16T16:29:51.100" CommentCount="1" />
  <row Id="4778" PostTypeId="1" AcceptedAnswerId="4786" CreationDate="2010-09-16T11:51:08.257" Score="4" ViewCount="195" Body="&lt;p&gt;Assume $M$ is a set, in which all axioms of $ZF - P + (V=L)$ hold. Does then $M$ believe that there exists an uncountable ordinal? I mean, why should the class of all countable ordinal numbers be a set?&lt;/p&gt;&#xA;" OwnerUserId="1650" LastEditorUserId="1650" LastEditorDisplayName="" LastEditDate="2010-09-16T12:55:42.130" LastActivityDate="2010-09-16T13:35:24.383" Title="Uncountable ordinals without power set axiom" Tags="&lt;set-theory&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" />
  <row Id="4779" PostTypeId="2" ParentId="4768" CreationDate="2010-09-16T12:15:49.947" Score="4" ViewCount="" Body="&lt;p&gt;I claim all three &quot;proofs&quot; of this result (currently) on &lt;a href=&quot;http://en.wikipedia.org/wiki/Parity_of_a_permutation&quot; rel=&quot;nofollow&quot;&gt;wikipedia&lt;/a&gt; are incomplete.  Let's begin with the observation that, assuming that identity permutation is not an odd permutation, then the result follows fairly easily.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Theorem&lt;/strong&gt;: Assuming the identity permutation is not an odd permutation, then all permutations are either even xor odd.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Proof&lt;/em&gt;:  Let $\sigma$ be both an even and an odd permutation.  Then there exists transpositions $t_i$ and $s_j$ such that \[\sigma=t_1 \circ t_2 \circ \cdots \circ t_k=s_1 \circ s_2 \circ \cdots \circ s_m\] where $k$ is even and $m$ is odd.  Note that \[\sigma^{-1}=s_m \circ s_{m-1} \circ \cdots \circ s_1.\]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then the identity permutation $\sigma \circ \sigma^{-1}$ is the odd permutation \[t_1 \circ t_2 \circ \cdots \circ t_k \circ s_m \circ s_{m-1} \circ \cdots \circ s_1,\] giving a contradiction.  Thus completing the proof of the theorem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is that we cannot just assume that the identity permutation not an odd permutation (yes, it's an even permutation, but what's to stop it being an odd permutation also?), it does not follow from the definition and it is the base case of the &quot;induction&quot;.  To prove it, we need to show that the identity permutation cannot be decomposed into an odd number of transpositions (without using the theorem itself).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, we can deduce from the above theorem that either (a) all permutations are both even and odd or (b) all permutations are either even xor odd.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On wikipedia: Proof 1 states that the identity permutation is an even permutation (which it is) then assumes that the identity permutation is not also an odd permutation (this is analogous to assuming that a closed set is not an open set).  Proof 2 essentially says that we can switch signs by multiplying by a transposition (which is fine, if you know &lt;em&gt;a priori&lt;/em&gt; that there exists some permutation that is not even or not odd).  Proof 3 neglects the possibility that an even-length word might be equal to an odd-length word.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same problem arises in:  &lt;a href=&quot;http://www.jstor.org/stable/2316272&quot; rel=&quot;nofollow&quot;&gt;An Historical Note on the Parity of Permutations&lt;/a&gt;, T. L. Barlow, American Mathematical Monthly Vol. 79, No. 7 (Aug. - Sep., 1972), pp. 766-769.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;EDIT&lt;/em&gt;:  Keith Conrad gave a proof that the identity permutation is not an odd permutation &lt;a href=&quot;http://www.math.uconn.edu/~kconrad/blurbs/grouptheory/sign.pdf&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.  It is almost a page long (and is the majority of the proof of the result in question).&lt;/p&gt;&#xA;" OwnerUserId="139" LastEditorUserId="139" LastEditorDisplayName="" LastEditDate="2010-09-16T13:23:27.850" LastActivityDate="2010-09-16T13:23:27.850" CommentCount="6" />
  <row Id="4780" PostTypeId="2" ParentId="4768" CreationDate="2010-09-16T12:16:37.730" Score="1" ViewCount="" Body="&lt;p&gt;This is overkill, but it follows from general facts about Coxeter groups as outlined &lt;a href=&quot;http://qchu.wordpress.com/2010/06/26/coxeter-groups/&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.  In particular, it follows from the fact that $S_n$ has presentation $s_i^2 = 1, (s_i s_{i+1})^3 = 1, s_i s_j = s_j s_i, |i - j| \ge 2$ (which follows from the faithfulness of the geometric representation) that the homomorphism $s_i \to -1$ is well-defined.  &lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-09-16T12:16:37.730" CommentCount="1" />
  <row Id="4781" PostTypeId="1" AcceptedAnswerId="13306" CreationDate="2010-09-16T12:58:18.613" Score="3" ViewCount="105" Body="&lt;p&gt;The complexity class RL is described at the complexity zoo as: Has the same relation to L as RP does to P. The randomized machine must halt with probability 1 on any input. It must also run in polynomial time (since otherwise we would just get NL).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is - how can we get NL when omitting the demand for polynomial time? I have a solution of my own but it seems strange to me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My solution: If suffices to solve ST-CON. We can use the same NL algorithm for ST-CON (guessing a path) with one major difference - we count the number of steps we make, and if it suppresses the number of vertices in the graph, we restart the computation, without remembering ANYTHING.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means we can play this game indefinitely. If the graph is not ST-connected, then we'll never halt, but if it's ST-connected we'll halt in probability 1 (this is the same as saying that a geometric random variable obtains a finite value in probability 1). However, since we do not halt for NO-instances, this solution &quot;feels wrong&quot; to me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there another solution? And is my solution correct?&lt;/p&gt;&#xA;" OwnerUserId="1818" LastEditorUserId="1818" LastEditorDisplayName="" LastEditDate="2010-09-16T13:14:08.263" LastActivityDate="2010-12-06T22:55:46.200" Title="Complexity - why is RL=NL when omitting the demand for polynomial run-time?" Tags="&lt;algorithms&gt;&lt;probability&gt;&lt;computer-science&gt;" AnswerCount="1" CommentCount="7" />
  <row Id="4782" PostTypeId="2" ParentId="4773" CreationDate="2010-09-16T12:59:37.720" Score="1" ViewCount="" Body="&lt;p&gt;I believe you have got the first example correct now. For the second example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $p&gt;1$ then since $f_n(x)\leq 1$ for all $x\in(\frac{1}{2},1]$ we have $A_n=\emptyset$. If $p\leq 1$ then $f_n(x)\geq p$ if and only if $x\geq p^{\frac{1}{n}}$. Thus $A_n=[0,\frac{1}{2})\cup [p^{\frac{1}{n}},1]$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For clarity for $p=1$: I take the set $[1,1]=\{1\}$ so this agrees with what you suspected.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $p\geq 1$, $A_n$ doesn't depend on $n$ so the $\liminf$ and $\limsup$ are easy. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $p&amp;lt;1$ we have $A_{n+1}\subset A_n$. So $\liminf=\limsup=\lim=[0,\frac{1}{2})\cup\{1\}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can always get this from $\liminf A_n=\bigcup_{n=1}^{\infty}\bigcap_{m=n}^{\infty} A_m$ and similarly $\limsup A_n=\bigcap_{n=1}^{\infty}\bigcup_{m=n}^{\infty} A_m$&lt;/p&gt;&#xA;" OwnerUserId="1794" LastEditorUserId="1794" LastEditorDisplayName="" LastEditDate="2010-09-16T18:52:05.473" LastActivityDate="2010-09-16T18:52:05.473" CommentCount="8" />
  <row Id="4783" PostTypeId="1" AcceptedAnswerId="4784" CreationDate="2010-09-16T13:20:23.217" Score="4" ViewCount="401" Body="&lt;p&gt;We know that $$ 2^{n}= (1+1)^{n} = \sum\limits_{k=0}^{n} {n \choose k}$$ I was asked to solve this limit, $$\lim_{n \to \infty} \ \sum\limits_{k=0}^{n} {n \choose k}^{-1}=? \quad \text{for} \ n \geq 1$$&lt;/p&gt;&#xA;" LastActivityDate="2010-10-12T16:13:56.180" Title="Finding $\displaystyle \lim_{n \to \infty} \sum\limits_{k=0}^{n} { n \choose k}^{-1}$" Tags="&lt;analysis&gt;&lt;sequences-and-series&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="2" />
  <row Id="4784" PostTypeId="2" ParentId="4783" CreationDate="2010-09-16T13:26:55.173" Score="8" ViewCount="" Body="&lt;p&gt;The terms from $k = 2$ to $k = n-2$ are all at most $\frac{1}{ {n \choose 2} }$ by unimodality, hence their sum is at most $\frac{2}{n-1}$.  So the entire sum is at most $2 + \frac{2}{n} + \frac{2}{n-1}$.  The limit is $2$.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-09-16T13:26:55.173" CommentCount="3" />
  <row Id="4785" PostTypeId="1" CreationDate="2010-09-16T13:27:46.853" Score="2" ViewCount="94" Body="&lt;p&gt;For a sequence of positive integers $a_1, \ldots, a_n$ and a base ring $R$ there is a graded ring $R[x_1,\ldots, x_n]$ where $x_i$ is in degree $a_i$.  We can then apply Proj and get a scheme, and this is usually called a weighted projective space; if all of the $a_i$ are 1, then the resulting scheme really is projective space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the way that this arises is as the quotient of $\mathbb{A}^n \setminus 0$ by an action of the multiplicative group, given by $(x_1,\ldots,x_n) \simeq (\lambda^{a_1} x_1, \ldots, \lambda^{a_n} x_n)$ for all $\lambda$.  This is a &quot;coarse&quot; group quotient.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is an alternative version where one instead takes the associated quotient stack/orbifold, and this has a number of nice properties (including possession of a line bundle $\mathcal{O}(1)$); this is true more generally of a graded ring.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the appropriate terminology for the stack-theoretic version of this construction if &quot;weighted projective space&quot; is taken?&lt;/p&gt;&#xA;" OwnerUserId="1925" LastActivityDate="2010-09-16T13:27:46.853" Title="Terminology for weighted projective spaces" Tags="&lt;terminology&gt;&lt;algebraic-geometry&gt;" CommentCount="2" />
  <row Id="4786" PostTypeId="2" ParentId="4778" CreationDate="2010-09-16T13:35:24.383" Score="4" ViewCount="" Body="&lt;p&gt;No. One cannot prove that $\omega_1$ exists in ZF - P, with or without V = L.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The set $HC$ of hereditarily countable sets always satisfies ZF - P. (This is straightforward to check axiom per axiom.) Of course, $\omega_1 \notin HC$ but moreover every set in $HC$ is countable or finite as witnessed by a function in $HC$. Therefore, $HC$ knows that every set in $HC$ is countable and hence $HC$ is a model of ZF - P + &quot;every set is at most countable.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Throwing in V = L into the mix doesn't help. Indeed, $HC^L = L_{\omega_1^L}$ (see my answer to your &lt;a href=&quot;http://mathoverflow.net/questions/38570/inclusions-of-constructible-sets-are-not-definable/38576#38576&quot; rel=&quot;nofollow&quot;&gt;recent MO question&lt;/a&gt; for a proof) is a model of ZF - P + V = L + &quot;every set is at most countable.&quot;&lt;/p&gt;&#xA;" OwnerUserId="25" LastActivityDate="2010-09-16T13:35:24.383" CommentCount="4" />
  <row Id="4787" PostTypeId="1" AcceptedAnswerId="4796" CreationDate="2010-09-16T13:42:39.973" Score="5" ViewCount="293" Body="&lt;p&gt;Let's take the numbers 0-10.  Their mean is 5, and the individual deviations from 5 are&lt;br&gt;&#xA;&lt;code&gt;-5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5&lt;/code&gt;&lt;br&gt;&#xA;And so the average (magnitude of) deviation from the mean is $30/11 \approx 2.72$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, this is not the standard deviation.  The standard deviation is $\sqrt{10} \approx 3.16$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first mean-deviation is a simpler and by far more intuitive definition of the &lt;em&gt;&quot;standard-deviation&quot;&lt;/em&gt;, so I'm sure it's the first definition statisticians worked with.  However, for some reason they decided to adopt the second definition instead.  What is the reasoning behind that decision?&lt;/p&gt;&#xA;" OwnerUserId="136" LastEditorUserId="136" LastEditorDisplayName="" LastEditDate="2010-09-16T13:48:19.807" LastActivityDate="2010-09-16T16:15:24.087" Title="Motivation behind standard deviation?" Tags="&lt;intuition&gt;&lt;statistics&gt;&lt;standard-deviation&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" />
  <row Id="4788" PostTypeId="1" AcceptedAnswerId="4794" CreationDate="2010-09-16T14:20:22.483" Score="11" ViewCount="631" Body="&lt;p&gt;Despite its name, its often claimed that the fundamental theorem of algebra (which shows that the Complex numbers are algebraically closed - this is not to be confused with the claim that a polynomial of degree n has &lt;strong&gt;at most&lt;/strong&gt; n roots) is not considered fundamental by algebraists as it's not needed for the development of modern algebra. My question is - what &lt;strong&gt;are&lt;/strong&gt; the major uses of the theorem, and to which extent can they justify the claim that the theorem is fundamental for &lt;strong&gt;something&lt;/strong&gt;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An example I think of is the Jordan canonical form for matrices, but I don't think it suffices.&lt;/p&gt;&#xA;" OwnerUserId="1818" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-09-17T16:58:23.690" LastActivityDate="2010-09-17T16:58:23.690" Title="How fundamental is the fundamental theorem of algebra?" Tags="&lt;terminology&gt;&lt;algebra&gt;&lt;complex-numbers&gt;" AnswerCount="5" CommentCount="3" FavoriteCount="2" />
  <row Id="4789" PostTypeId="2" ParentId="4787" CreationDate="2010-09-16T14:28:19.333" Score="2" ViewCount="" Body="&lt;p&gt;Squaring is nicer than taking the absolute value, e.g. it is smooth.  It also leads to a definition of &lt;a href=&quot;http://en.wikipedia.org/wiki/Variance&quot; rel=&quot;nofollow&quot;&gt;variance&lt;/a&gt; which has nice mathematical properties, e.g. it is additive.  But for me the theorem that really justifies using standard deviation over the mean absolute error is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Central_limit_theorem&quot; rel=&quot;nofollow&quot;&gt;central limit theorem&lt;/a&gt;.  The central limit theorem is at work whenever we measure the mean and standard deviation of a distribution we assume to be normal (e.g. heights in a population) and use that to make predictions about the entire distribution, since a normal distribution is completely specified by its mean and standard deviation.&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-09-16T14:48:34.323" LastActivityDate="2010-09-16T14:48:34.323" CommentCount="6" />
  <row Id="4790" PostTypeId="2" ParentId="4738" CreationDate="2010-09-16T14:33:34.057" Score="1" ViewCount="" Body="&lt;p&gt;Yes.  The Baire category theorem asserts that in a complete metric space, a set of first category not only is not the whole space, but actually has dense complement.&lt;/p&gt;&#xA;" OwnerUserId="822" LastActivityDate="2010-09-16T14:33:34.057" />
  <row Id="4791" PostTypeId="2" ParentId="4788" CreationDate="2010-09-16T15:03:33.453" Score="4" ViewCount="" Body="&lt;p&gt;Well, there is something called &quot;Algebraic Geometry&quot;. You may check the first pages of most of the books on that subject and you'll see something like: &quot;Let $k$ be an algebraically closed field...&quot; [for instance, $\mathbb{C}$]. Indeed, this is the beginning of &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0387902449&quot; rel=&quot;nofollow&quot;&gt;Hartshorne's book&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason? It's much harder to do geometry over the real numbers, where all your theorems about the set of solutions of some system of polynomial equations (the primary object of Algebraic Geometry) can miserably fail, just because there are equations like $x^2 + 1 = 0$ with no solution at all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, the existence of polynomial equations like $x^2 + 1 = 0$ with no solution over the reals, ruins the basic bijection between prime  ideals in $\mathbb{R}[x_1, \dots , x_n]$ and irreducible algebraic sets  in the affine space $\mathbb{R}^n$, which is central for Algebraic Geometry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This doesn't mean you cannot do Algebraic Geometry over the reals, only it's another business.&lt;/p&gt;&#xA;" OwnerUserId="664" LastEditorUserId="664" LastEditorDisplayName="" LastEditDate="2010-09-16T17:44:59.567" LastActivityDate="2010-09-16T17:44:59.567" CommentCount="5" />
  <row Id="4792" PostTypeId="2" ParentId="4741" CreationDate="2010-09-16T15:04:32.990" Score="3" ViewCount="" Body="&lt;p&gt;This smells like a problem in the Calculus of Variations and can be solved as such.  We therefore make a slight reformulation: among all $C^3$ curves of length $l$ originating and terminating at the same point, find the one(s) for which the maximum curvature is smallest.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's begin by developing some intuition.  In general, a minimax problem like this has a solution where the objective (the curvature in this case) is as constant as possible.  A circle is an obvious candidate for a solution and indeed a circle of length $l$ has constant radius $l / (2 \pi)$ and constant curvature of $2 \pi / l$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(As a tiny simplification, by choosing our linear units of measurement let's henceforth assume $l=1$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The technique of the Calculus of Variations is to assume we have a solution and perturb it a little, showing that any perturbation, no matter how small, cannot decrease the value of the objective.  To motivate this in the present case, though, I find it more appealing to consider how one might go about reducing the maximum curvature of any unit-length loop.  A point of maximum curvature is part of a relatively sharp &quot;bump&quot; on the curve.  If we push that bump inwards a little, we should be able to decrease its sharpness and at the same time we slightly decrease the total length of the curve.  To make up for the latter, uniformly dilate the entire curve relative to its point of origin until the dilated curve again has unit length.  This uniformly &lt;em&gt;decreases&lt;/em&gt; all curvatures along the loop.  In this fashion the maximum curvature has strictly decreased.  The only way this operation can fail is when there is no &quot;bump&quot;: the curvature everywhere is constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To make this go through rigorously, use an adapted orthonormal frame for the curve: $T(t) = \alpha'(t)$ is the tangent, $N(t)$ is the inward-pointing unit normal with $T'(t) = \kappa(t)N(t)$, and $B(t)$ is the unit binormal, with $N'(t) = -\kappa(t)T(t) + \tau(t)B(t)$ and $B'(t) = -\tau(t)N(t)$.  ($\tau$ is the &lt;em&gt;torsion&lt;/em&gt;.  These Serret-Frenet formulae generalize readily to higher dimensions and in two dimensions just forget about $B$.)  Notice that $\kappa \ge 0$.  Let $\epsilon &gt; 0$ be small and let $\delta(t)$ be a smooth non-negative function in an open neighborhood of $[0,1]$, vanishing at $0$ and $1$ (in order to keep the endpoints of the new curve fixed).  Set the &quot;pushed&quot; curve to be&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\psi(t) = \alpha(t) + \epsilon \delta(t) N(t).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Compute the curvature of $\psi$ to first order in $\epsilon$ using the Serret-Frenet formulae.  I find that its square becomes&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\kappa^2 + 2 \kappa \left( \delta'' - \delta \tau^2 \right) \epsilon + O(\epsilon^2).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose the curvature is not constant.  Then there is a closed interval of maximum curvatures (perhaps reducing to a point) and within some neighborhood of that interval all curvatures are strictly less than the maximum.  We can make $\delta$ zero outside this larger neighborhood, let it increase slowly enough (and choose $\epsilon$ sufficiently small) so that $\delta''$ keeps the new curvature less than the maximum within the outer neighborhood, and make $\delta''$ strictly negative at all points within the interval of maximum curvature.  This procedure strictly decreases the maximum curvature within the outer interval.  You can also check, again to second order in $\epsilon$, that the length of $\psi'$ decreases by $\epsilon \delta \kappa \ge 0$, so we will be able to apply our dilation trick in order to keep the total loop length constant after this perturbation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A little more formally (and using less work, actually), the preceding equation shows that when $\kappa$ is constant and $\tau = 0$, we would need $\delta'' \lt \delta \tau^2 = 0$ everywhere in order to decrease the curvature, which implies $\delta$ is identically zero in the unit interval.  Otherwise (namely, if $\kappa$ is not constant or $\tau \gt 0$), it will be possible to decrease the curvature subject to maintaining the unit length of the curve.  Curves of constant curvature and zero torsion are circular arcs, and the only circular arc that returns to its starting point is a full circle.  Therefore the maximum curvature in &lt;em&gt;any&lt;/em&gt; sufficiently smooth unit-length loop can be no less than $2 \pi$ and is strictly greater than that if the loop is not a circle, QED.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe the same technique works in higher dimensions, too, although I have not done the calculations.  (There are additional vectors in the adapted frame, and therefore additional torsion terms, but this doesn't seem to present any obstacle to carrying out the same program.)&lt;/p&gt;&#xA;" OwnerUserId="1489" LastEditorUserId="1489" LastEditorDisplayName="" LastEditDate="2010-09-16T23:47:45.373" LastActivityDate="2010-09-16T23:47:45.373" CommentCount="7" />
  <row Id="4793" PostTypeId="2" ParentId="4768" CreationDate="2010-09-16T15:35:07.063" Score="0" ViewCount="" Body="&lt;p&gt;It is amazing. It seems no one (including the Wikipedia contributors) is aware of the link between the parity of a permutation and the determinant of the associated permutation matrix :)&lt;/p&gt;&#xA;" LastActivityDate="2010-09-16T15:35:07.063" CommentCount="1" />
  <row Id="4794" PostTypeId="2" ParentId="4788" CreationDate="2010-09-16T15:59:11.423" Score="8" ViewCount="" Body="&lt;p&gt;I can immediately think of four important areas for which the FTA is actually quite fundamental. Maybe other people may come up with more contributions.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Algebraic Geometry (as already touched in Agusti Roig's answer). In particular, we should couple the FTA with Lefschetz's Principle which basically says that the complex projective space is sort of &quot;universal&quot; environment for algebraic geometry in characteristic 0.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Classification of algebraic structures over local or global fields, where one uses constantly that $\Bbb C$ has no non trivial finite extensions (a trivial consequence of the FTA)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The theory of representations of groups, where the TFA plays a big role making the theory of complex representations simpler.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Galois theory and algebraic number theory, where the FTA allows to realize $\Bbb C$ as a good environment for studying finite extensions of $\Bbb Q$ (such in the basic resut that a number field of degree $n$ always admits $n$ independent embedings into $\Bbb C$) and allowing the use of the &quot;geometry of numbers&quot; (Minkowski's theorem) to prove important arithmetic results such as the finiteness of the class number and the structure of the units (Dirichlet's theorem)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="688" LastActivityDate="2010-09-16T15:59:11.423" />
  <row Id="4795" PostTypeId="2" ParentId="4788" CreationDate="2010-09-16T16:08:30.833" Score="2" ViewCount="" Body="&lt;p&gt;Perhaps it is worth expanding my answer.  (Forgive me if I get any details wrong.)  If you are interested in Lie groups, you are interested in their classification.  A connected Lie group is the quotient of its universal cover by a discrete subgroup of its center, so to classify connected Lie groups it suffices to classify simply connected Lie groups, then compute their centers.  This classification is essentially equivalent to the classification of finite-dimensional real Lie algebras.  The &lt;a href=&quot;http://en.wikipedia.org/wiki/Levi_decomposition&quot; rel=&quot;nofollow&quot;&gt;Levi decomposition&lt;/a&gt; says that to classify such Lie algebras it is more or less enough to classify solvable and semisimple Lie algebras.  The former classification problem, as I understand it, is wild; however the latter is tractable.  The basic idea is first to study &lt;em&gt;complex&lt;/em&gt; semisimple Lie algebras and then to study their real forms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A big reason this is possible is that the complex numbers are algebraically closed.  This means that eigenvalues always exist, and from there the entire classification is possible.  &lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-09-16T16:08:30.833" />
  <row Id="4796" PostTypeId="2" ParentId="4787" CreationDate="2010-09-16T16:15:24.087" Score="8" ViewCount="" Body="&lt;p&gt;Your guess is correct: least absolute deviations was the method tried first historically.  The first to use it were astronomers who were attempting to combine observations subject to error.  Boscovitch in 1755 published this method and a geometric solution.  It was used later by Laplace in a 1789 work on geodesy.  Laplace formulated the problem more mathematically and described an analytical solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Legendre appears to be the first to use least squares, doing so as early as 1798 for work in celestial mechanics.  However, he supplied no probabilistic justification.  A decade later, Gauss (in an 1809 treatise on celestial motion and conic sections) asserted axiomatically that the arithmetic mean was the best way to combine observations, invoked the maximum likelihood principle, and then showed that a probability distribution for which the likelihood is maximized at the mean must be proportional to $\exp(-x^2 / (2 \sigma^2))$ (now called a &quot;Gaussian&quot;) where $\sigma$ quantifies the precision of the observations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The likelihood (when the observations are statistically independent) is the product of these Gaussian terms which, due to the presence of the exponential, is most easily maximized by minimizing the negative of its logarithm.  Up to an additive constant, the negative log of the product is the sum of the squares (all divided by a constant $2 \sigma^2$, which will not affect the minimization).  Thus, even historically, the method of least squares is intimately tied up with likelihood calculations and averaging.  There are plenty of other modern justifications for least squares, of course, but this derivation by Gauss--with the almost magical appearance of the Gaussian, which had first appeared some 70 years early in De Moivre's work on sums of Bernoulli variables (the Central Limit Theorem)--is memorable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This story was researched, and is ably recounted, by Steven Stigler in his &lt;em&gt;The History of Statistics - The Measurement of Uncertainty before 1900&lt;/em&gt; (1986).  Here I have merely given the highlights of parts of chapters 1 and 4.&lt;/p&gt;&#xA;" OwnerUserId="1489" LastActivityDate="2010-09-16T16:15:24.087" CommentCount="2" />
  <row Id="4797" PostTypeId="1" AcceptedAnswerId="4798" CreationDate="2010-09-16T17:24:57.733" Score="1" ViewCount="149" Body="&lt;p&gt;Suppose A is a &lt;a href=&quot;http://en.wikipedia.org/wiki/Hadamard_matrix&quot; rel=&quot;nofollow&quot;&gt;Hadamard matrix&lt;/a&gt; of size $d$. Let A be in normalized in a sense that first row and first column are all ones. What is the sum of rows? I tried random Hadamard matrices and seem to get $d,0,0,0,0,\ldots$&lt;/p&gt;&#xA;" OwnerUserId="998" LastActivityDate="2010-09-23T17:18:23.430" Title="What is sum of rows of Hadamard matrix" Tags="&lt;combinatorics&gt;" AnswerCount="1" CommentCount="0" />
  <row Id="4798" PostTypeId="2" ParentId="4797" CreationDate="2010-09-16T17:29:39.057" Score="5" ViewCount="" Body="&lt;p&gt;Follows immediately from the property of Hadamard Matrices: Any two distinct rows are orthogonal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since all the other rows are orthogonal to the all ones row, the sum of the elements in each of those rows must be zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: To answer Qiaochu's query:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;MUltiplying a row or column of a Hadamard matrix by -1 gives yet another Hadamard matrix.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using this operation, we can easily normalize a Hadamard matrix to make the elements of the first row to be all ones.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-09-23T17:18:23.430" LastActivityDate="2010-09-23T17:18:23.430" CommentCount="3" />
  <row Id="4799" PostTypeId="2" ParentId="4788" CreationDate="2010-09-16T17:49:36.060" Score="12" ViewCount="" Body="&lt;p&gt;One cannot over-emphasize the point that one of the primary benefits of working over an algebraically closed field such as $\mathbb C$ is the immense simplification gained by &lt;strong&gt;linearizing&lt;/strong&gt; what would otherwise be much more complicated &lt;strong&gt;nonlinear&lt;/strong&gt; phenomena. The ability to factor polynomials completely into linear factors over $\mathbb C$ enables widespread linearization simplifications of diverse problems. An example familiar to any calculus student is the fact that integration of rational functions is much simpler over $\mathbb C$ (vs. $\mathbb R$) since partial fraction decompositions involve at most linear (vs. quadratic) polynomials in the denominator. Analogously, one may reduce higher-order constant coefficient differential and difference equations (i.e. recurrences) to linear (first-order) equations by factoring them as linear operators over $\mathbb C$ (i.e. &quot;operator algebra&quot;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More generally, such &lt;strong&gt;simplification by linearization&lt;/strong&gt; was at the heart of the development of abstract algebra. Namely, Dedekind, by abstracting out the essential linear structures (ideals and modules) in number theory, greatly simplified the prior nonlinear Gaussian theory (based on quadratic forms). This enabled him to exploit to the hilt the power of linear algebra. Examples abound of the revolutionary breakthroughs that this brought to number theory and algebra - e.g. it provided the methods needed to generalize the law of quadratic reciprocity to higher-order reciprocity laws - a longstanding problem that motivated much of the early work in number theory and algebra. &lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-09-17T00:09:12.653" LastActivityDate="2010-09-17T00:09:12.653" CommentCount="10" />
  <row Id="4800" PostTypeId="1" CreationDate="2010-09-16T18:22:11.177" Score="2" ViewCount="210" Body="&lt;p&gt;Suppose I have an number of distances from an unknown location to a known location. I can use these distances and the known locations to draw a number of circles. The point where all the circles intersect is my unknown location. This is easy to solve. However, if my distance measurements have some error, I will get a cluster of points around my unknown location. The question is, for each pair of intersection points, how can I tell which solution is closer to my unknown point?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Every solution I have come up with involves many special cases, such as when all the known locations are inside one circle, or all the known locations are co-linear, and so on, buy I am hoping to come up with a more elegant solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lets make this more simple. I have 2 circles that have at most 2 intersection points. If they don't intersect, I can easily choose the point between them. If they do intersect at 2 points, how can I use a third circle to choose which one of these 2 intersections is the correct one?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My current plan is to do a comparison between the distance between the radius of the third circle and the intersection points and choose the smaller one. The problem with this is that I don't know how much error this approach can tolerate before you choose the wrong side. If I knew that, I could put a goal on my measurement method to try to reduce the error by X%&lt;/p&gt;&#xA;" LastActivityDate="2010-09-17T12:56:14.577" Title="Choosing solutions for the intersections of n number of circles" Tags="&lt;geometry&gt;&lt;algorithms&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" />
  <row Id="4802" PostTypeId="2" ParentId="4800" CreationDate="2010-09-16T19:07:58.613" Score="4" ViewCount="" Body="&lt;p&gt;One handles error with a probability model.  In this case, the typical error in a distance measurement is likely proportional to the distance itself.  The errors themselves are often thought of as accumulated small nearly independent errors, allowing one to invoke the Central Limit Theorem and suppose, at least as a reasonable hypothesis, that the errors are normally distributed.  We usually assume there's no systematic bias in the errors: they should average out to zero in principle.  A simple model is obtained by supposing the errors are independent.  (Other suppositions can be treated, depending on how the measurements were made, but it quickly gets complicated.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This leaves us with just three unknowns to estimate: the true coordinates $(x,y)$ of your location and the precision of the relative errors, usually expressed as their (common) standard deviation $\sigma$.  Your data consist of $n$ measured distances to the known locations $(x_i, y_i)$, say $d_i, i = 1 \ldots n$.  Mathematically, these assumptions translate to the following.  The probability of observing $d_i$ equals&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{1}{\sqrt{2 \pi} \sigma \delta_i} \exp \left( -\frac{(d_i - \delta_i)^2}{2 \sigma^2 \delta_i^2)} \right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\delta_i = \delta_i(x,y) = \sqrt{(x - x_i)^2 + (y - y_i)^2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The probability of your data $(d_1, d_2, \ldots, d_n)$ is the product of these probabilities.  This is the &lt;em&gt;likelihood&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One reasonably tractable estimator maximizes the likelihood (as a function of the unknown parameters $x$, $y$, and $\sigma$).  This is usually done by maximizing the negative logarithm of the likelihood, which will be a sum of terms, one for each data value.  It's nonlinear and a little nasty, but we know geometrically that there will be at least one global optimum and, quite likely, exactly one.  Many methods exist to solve this, available in statistical packages, numerical optimization routines, and general-purpose math software like &lt;em&gt;Mathematica&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course the optimal arguments $(x,y)$ tell you where your location likely is.  The optimal value of $\sigma$ estimates the typical relative error in the distance measurement: you can check that it's a reasonable value.  There are ways to extract confidence intervals for all three parameters (from the Hessian of the log likelihood).  A joint confidence interval for $(x,y)$ gives you an ellipse in which the true point is likely to be.  Statistical packages will give you this information.  If you are doing this by hand, you can analyze any other candidate solution $(x',y')$, such as the intersection of a few circles, by evaluating the likelihood there and comparing it to the optimum likelihood.  If the logarithms differ by less than $2$, the candidate solution is consistent with the data.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;BTW, as you've seen, you're practically forced to take a statistical approach.  With real data you find that some circles don't even intersect and that overall the set of data has little or no internal consistency.  You &lt;em&gt;have&lt;/em&gt; to model the error somehow.&lt;/p&gt;&#xA;" OwnerUserId="1489" LastEditorUserId="1489" LastEditorDisplayName="" LastEditDate="2010-09-16T19:19:00.463" LastActivityDate="2010-09-16T19:19:00.463" />
  <row Id="4803" PostTypeId="1" CreationDate="2010-09-16T19:13:24.393" Score="3" ViewCount="194" Body="&lt;p&gt;If $n$ is an integer, is there a nice way to write the partial fraction expansion of $\frac{1}{x^n - 1}$? I figure that if $\zeta$ is the $n$-th root of unity, then for some coefficients $a_0, a_1, \ldots, a_{n-1}$ we may write &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\frac{1}{x^n - 1} = \frac{a_0}{x - 1} + \frac{a_1}{x - \zeta} + \frac{a_2}{x - \zeta^2} + \ldots + \frac{a_{n-1}}{x - \zeta^{n-1}}.&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then for $0 \leq i \leq n -1$,&#xA;$$&#xA;a_i = \lim_{x \to \zeta^{i}} \frac{x - \zeta^{i}}{x^n - 1} = \frac{1}{(\zeta^{i} - 1) \ldots (\zeta^{i} - \zeta^{i-1}) (\zeta^{i} - \zeta^{i + 1}) \ldots (\zeta^{i} - \zeta^{n - 1})}.&#xA;$$&#xA;Is there a simpler expression for this and if so, how could I see it easily?&lt;/p&gt;&#xA;" LastActivityDate="2010-09-16T19:34:55.040" Title="The partial fraction expansion of $\frac{1}{x^n - 1}$." Tags="&lt;special-functions&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="4804" PostTypeId="2" ParentId="4803" CreationDate="2010-09-16T19:29:30.823" Score="7" ViewCount="" Body="&lt;p&gt;Use l'Hopital's rule.  $a_i = \lim_{x \to \zeta^i} \frac{1}{nx^{n-1}} = \frac{\zeta^i}{n}$.  Note that the identity I am claiming is true is equivalent to the identity&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{i=0}^{n-1} (\zeta^i)^k = \begin{cases} n \text{ if } n | k \\ 0 \text{ otherwise} \end{cases}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;by taking the generating function of both sides.  More generally, the coefficient of $\frac{1}{x - a}$ in the partial fraction decomposition of $\frac{P(x)}{Q(x)}$ is (as long as $a$ is a simple root of $Q$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\lim_{x \to a} \frac{(x - a) P(x)}{Q(x)} = \frac{P(a)}{Q'(a)}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This formula is tremendously useful (for example I recently used it on &lt;a href=&quot;http://math.stackexchange.com/questions/4658/what-is-the-probability-of-a-coin-landing-tails-7-times-in-a-row-in-a-series-of-1/4675#4675&quot;&gt;this math.SE question&lt;/a&gt;) but does not seem to be widely known.&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-09-16T19:34:55.040" LastActivityDate="2010-09-16T19:34:55.040" CommentCount="2" />
  <row Id="4805" PostTypeId="1" AcceptedAnswerId="4809" CreationDate="2010-09-16T19:51:20.197" Score="0" ViewCount="79" Body="&lt;p&gt;What do you call this function?&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;f(x) = x * (1.0 - x)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Does it have a name?  It’s cropped up in at least two statistical formulas I've been asked to program so far.  (In both cases, &lt;code&gt;x&lt;/code&gt; was something called a &lt;em&gt;p-value&lt;/em&gt; (for what it's worth).)&lt;/p&gt;&#xA;" OwnerUserId="984" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-09-16T21:51:11.370" LastActivityDate="2010-09-16T21:51:11.370" Title="If x is a p-value, what do we call x*(1-x)?" Tags="&lt;terminology&gt;&lt;statistics&gt;&lt;functions&gt;" AnswerCount="1" CommentCount="0" />
  <row Id="4806" PostTypeId="1" CreationDate="2010-09-16T19:53:36.750" Score="1" ViewCount="128" Body="&lt;p&gt;While learning about limits and derivatives, I came accross the following problem on one of Stewart's exercises book. I've been trying to wrap my head around it but I haven't got anywhere useful:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$ \forall x f''(x) $ exists&lt;/li&gt;&#xA;&lt;li&gt;$ \exists c \in \mathbb{R} $ / $\forall x \neq c, f'(x) &gt; 0 \wedge f'(c) = 0 $&lt;/li&gt;&#xA;&lt;li&gt;Then, $(c, f(c))$ is an &lt;strong&gt;inflection point&lt;/strong&gt;.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;From this I can gather that:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$f$ is continuous&lt;/li&gt;&#xA;&lt;li&gt;$f$ is increasing $\forall x \neq c$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;My intuition tells me that $(3)$ is false since I might be able to come up with a function defined by parts that contradicts the statement, but I haven't found a way to prove this. Any pointers would be greatly appreciated.&lt;/p&gt;&#xA;" OwnerUserId="1935" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-09-16T21:38:48.490" LastActivityDate="2010-09-16T21:38:48.490" Title="Finding the inflection point for a function" Tags="&lt;calculus&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" />
  <row Id="4807" PostTypeId="2" ParentId="4806" CreationDate="2010-09-16T20:01:57.357" Score="2" ViewCount="" Body="&lt;p&gt;Because the second derivative exists everywhere, you also know that $f'$ is continuous everywhere. Now think about the function $f'(x)$; it is positive to the left and to the right of $c$, and is $0$ at $c$, and it is continuous. Does that tell you something about $f''$, and hence about concavity?&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-09-16T20:01:57.357" CommentCount="3" />
  <row Id="4808" PostTypeId="1" AcceptedAnswerId="4816" CreationDate="2010-09-16T20:02:03.207" Score="7" ViewCount="503" Body="&lt;p&gt;In this wikipedia, article &lt;a href=&quot;http://en.wikipedia.org/wiki/Circle#Area_enclosed&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Circle#Area_enclosed&lt;/a&gt; its stated that the circle is the closed curve which has the maximum area for a given arc length. First, of all i would like to see different proofs, for this result. (If there are any elementary ones!)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One, interesting observation, which one can think while seeing this problem, is: How does one propose such type of problem? Does, anyone take all closed curves, and calculate their area to come this conclusion. I don't think thats the right intuition.&lt;/p&gt;&#xA;" LastEditorUserId="2937" LastEditorDisplayName="" LastEditDate="2011-03-10T17:58:53.540" LastActivityDate="2011-03-10T17:58:53.540" Title="Why Circle encloses largest Area?" Tags="&lt;calculus&gt;&lt;geometry&gt;&lt;analysis&gt;&lt;calculus-of-variations&gt;" AnswerCount="6" CommentCount="1" FavoriteCount="3" />
  <row Id="4809" PostTypeId="2" ParentId="4805" CreationDate="2010-09-16T20:23:21.037" Score="5" ViewCount="" Body="&lt;p&gt;If &lt;code&gt;Z&lt;/code&gt; can take values 1 or 0 with probability &lt;code&gt;x&lt;/code&gt; and &lt;code&gt;1-x&lt;/code&gt;, respectively, we usually call &lt;code&gt;x*(1-x)&lt;/code&gt; the &lt;strong&gt;variance&lt;/strong&gt; of Z.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a special case of the more general &lt;a href=&quot;http://en.wikipedia.org/wiki/Variance#Definition&quot; rel=&quot;nofollow&quot;&gt;variance&lt;/a&gt; formula, and works because Z is what we call an &lt;a href=&quot;http://en.wikipedia.org/wiki/Indicator_function#Mean.2C_Variance_and_covariance&quot; rel=&quot;nofollow&quot;&gt;indicator variable&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="73" LastActivityDate="2010-09-16T20:23:21.037" />
  <row Id="4810" PostTypeId="2" ParentId="4758" CreationDate="2010-09-16T20:43:35.940" Score="4" ViewCount="" Body="&lt;p&gt;The essence of the matter is both strikingly simple and extremely general. Namely, there are infinitely many solutions to your congruence simply because the solutions lie in the orbit of an invertible map on a finite set. But such permutations decompose into finite &lt;strong&gt;cycles&lt;/strong&gt; - so they repeat infinitely often upon iteration.  Below I discuss your simple linear example and then I show how the technique generalizes to nonlinear recursions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First we convert $\rm\;f_n = 10^n-1 \;$ to shift form. Note $\rm\;f_{n+2}-f_{n+1} = 9\cdot 10^{n+1} = 10\ (f_{n+1}-f_n) \;$ which yields the recurrence relation $\rm\;\: f_{n+2} \;=\; 11\ f_{n+1}-10\ f_n \:$.  This is simpler as a shift map on pairs: $\rm\;S : (f_n,f_{n+1})\to (f_{n+1},f_{n+2}) =\: (f_{n+1},11\ f_{n+1} - 10\ f_n) \;$. Extended to all pairs it yields the  map $\rm\;S : (a,b)\to (b,\:11\:b-10\:a = c)\:$  with the obvious inverse $\rm\;(b,c)\to (a = (11\:b-c)/10,\:b)\:$ which clearly exists in any ring $\rm\:R\:$ where $10\:$ has an inverse. In particular, let $\rm\;R =\mathbb Z\pmod m \;$ be the ring of integers modulo $\rm\:m&gt;1\:$, for any $\rm\:m\:$ coprime to $10$. Since $\rm\:R \:$ is finite, so too is its set of pairs $\rm  R^2$. But on such pairs the shift map $\rm\;S \;$ is an invertible map on a finite set, so it is a permutation. Therefore its orbit decomposition consists of finite cycles. Thus the values $\rm f_n$ all occur in the orbit of the initial condition pair $\rm (f_0,f_1) = (0,9)$, and each value necessarily occurs infinitely many times since the shift map successively steps along the finite circular orbit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The method also works for nonlinear recurrences with invertible shift map, e.g. from one of my old posts&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EXERCISE&lt;/strong&gt; $\;\;$ A sequence $\rm f_n$ satisfies the recurrence $\rm\; f_{n+2} = f_{n+1}^{\:2} - f_n\:,\:$ with $\rm\; f_1 = 39,\; f_2 = 45 \;$.&lt;br&gt;&#xA;Prove that $\: 1986 \:$ divides infinitely many terms of the sequence. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\;\;$ The shift map is $\rm\; (a,b)\to (b,\;b^2-a = c)\;$ with inverse $\rm\; (b,c)\to (a = b^2-c,\; b)\;$&lt;br&gt;&#xA;and maps $\;(39,45)\to (45,0) \;\pmod {1986}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;NOTE&lt;/strong&gt; $\;\;$ When the recurrence is linear the shift map has a matrix representation, which may be used to quickly compute the terms of the recurrence by fast exponentiation algorithms (e.g. repeatedly squaring). However, neither the matrix nor any other consequences of linearity are required to deduce the cyclicity properties above. Often many students (and even some professors) don't realize this, which can lead to obfuscated solutions (compared to the simple approach above). I've even encountered some solutions that completely overlook the innate symmetry and, as a result, end up completely reinventing the wheel, i.e. discovering the innate cycle structure completely from scratch - not realizing that it is simply a special case of the cycle structure of permutations, i.e. orbit decomposition. The moral of the story is: always look for any innate symmetries of a problem before diving in head-first to apply more brute-force techniques.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;NOTE 2&lt;/strong&gt; $\;\;$ For another very nice application of orbit decompositions see my &lt;a href=&quot;http://math.stackexchange.com/questions/4150&quot;&gt;recent post&lt;/a&gt; which illustrates how they prove key for finding polynomial and rational function solutions of recurrences.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-09-19T21:30:05.707" LastActivityDate="2010-09-19T21:30:05.707" />
  <row Id="4811" PostTypeId="2" ParentId="4731" CreationDate="2010-09-16T20:44:14.473" Score="4" ViewCount="" Body="&lt;p&gt;Yep. The condition that the Hessian is nonsingular at every critical point means that the function in question is suitably &quot;nice,&quot; and is called a Morse function. Morse functions are useful because their inverse images $f^{-1}((-\infty, a])$ behave very nicely.  Basically, as $a$ varies, you get things that are (up to homotopy equivalence) obtained by attaching cells when $a$ passes through a critical value (the type of cell being determined by the type of the critical point).  So since every compact manifold admits a Morse function (the &quot;typical&quot; function is, in fact, Morse--I believe this is a standard transversality idea in differential topology), it follows that every compact manifold is homotopy equivalent to a CW complex. &lt;/p&gt;&#xA;" OwnerUserId="536" LastEditorUserId="536" LastEditorDisplayName="" LastEditDate="2010-09-16T21:46:51.393" LastActivityDate="2010-09-16T21:46:51.393" />
  <row Id="4812" PostTypeId="2" ParentId="4808" CreationDate="2010-09-16T20:55:39.897" Score="2" ViewCount="" Body="&lt;p&gt;As for the second question, the result is quite intuitive.  First of all one can readily see that we can suppose WLOG that the curve is convex, so already it cannot be too far from being circle-like.  Second, it's not hard to see that stretching the curve out so as to make it non-uniform causes it to enclose less area, e.g. consider the analogous problem for rectangles (or even try it for $n$-gons).  The intuition here is that a kink in the boundary of an area does not create much extra area but creates extra arc length.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course the hard part is to justify these intuitions.  Probably the proof that comes closest proceeds via Steiner symmetrization; there is a link at the Wikipedia article.  There is also a neat Fourier-analytic proof.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-09-16T20:55:39.897" />
  <row Id="4813" PostTypeId="2" ParentId="3405" CreationDate="2010-09-16T21:04:49.547" Score="12" ViewCount="" Body="&lt;p&gt;After getting some insight by looking at some numerical calculations to see what np(n)-c looks like for large n, I can now describe the convergence in some detail. First, the results of the simulations strongly suggested the following.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;for n odd, np(n) converges to c from below at rate O(1/n).&lt;/li&gt;&#xA;&lt;li&gt;for n a multiple of 4, np(n) converges to c from above at rate O(1/n&lt;sup&gt;2&lt;/sup&gt;).&lt;/li&gt;&#xA;&lt;li&gt;for n a multiple of 2, but not 4, np(n) converges to c from below at rate O(1/n&lt;sup&gt;2&lt;/sup&gt;).&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This suggests the following asymptotic form for p(n).&#xA;$$&#xD;&#xA;p(n)=\frac{c}{n}\left(1+\frac{a_1(n)}{n}+\frac{a_2(n)}{n^2}\right)+o\left(\frac{1}{n^3}\right)&#xD;&#xA;$$&#xD;&#xA;where a&lt;sub&gt;1&lt;/sub&gt;(n) has period 2, with a&lt;sub&gt;1&lt;/sub&gt;(0)&amp;nbsp;=&amp;nbsp;0, a&lt;sub&gt;1&lt;/sub&gt;(1)&amp;nbsp;&amp;lt;&amp;nbsp;0 and a&lt;sub&gt;2&lt;/sub&gt;(n) has period 4 with a&lt;sub&gt;2&lt;/sub&gt;(0)&amp;nbsp;&amp;gt;&amp;nbsp;0 and a&lt;sub&gt;2&lt;/sub&gt;(2)&amp;nbsp;&amp;lt;&amp;nbsp;0. In fact, we can expand p(n) to arbitrary order [Edit: Actually, not quite true. See below]&#xA;$$&#xD;&#xA;\begin{array}&#xD;&#xA;{}p(n)=c\sum_{r=0}^s\frac{a_r(n)}{n^{r+1}}+O(n^{-s-2})&amp;amp;&amp;amp;(1)&#xD;&#xA;\end{array}&#xD;&#xA;$$&#xD;&#xA;and the term a&lt;sub&gt;r&lt;/sub&gt;(n) is periodic in n, with period 2&lt;sup&gt;r&lt;/sup&gt;. Here, I have normalized a&lt;sub&gt;0&lt;/sub&gt;(n) to be equal to 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can compute all of the coefficients in (1). As p satisfies the recurrence relation&#xA;$$&#xD;&#xA;\begin{array}&#xD;&#xA;\displaystyle p(n+1)=(1+1/n)p(n)-1_{\lbrace n\textrm{ even}\rbrace}\frac2np(n/2) -1_{\lbrace n=1\rbrace}.&amp;amp;&amp;amp;(2)&#xD;&#xA;\end{array}&#xD;&#xA;$$&#xD;&#xA;we can simply plug (1) into this, expand out the terms $(n+1)^{-r}=\sum_s\binom{-r}{s}n^{-r-s}$ on the left hand side, and compare coefficients of 1/n.&#xA;$$&#xD;&#xA;\begin{array}&#xD;&#xA;\displaystyle a_r(k+1)-a_r(k)=a_{r-1}(k)-\sum_{s=0}^{r-1}\binom{-s-1}{r-s}a_{s}(k+1)-1_{\lbrace k\textrm{ even}\rbrace}2^{r+1}a_{r-1}(k/2).&amp;amp;&amp;amp;(3)&#xD;&#xA;\end{array}&#xD;&#xA;$$&#xD;&#xA;Letting &amp;#257;&lt;sub&gt;r&lt;/sub&gt; be the average of a&lt;sub&gt;r&lt;/sub&gt;(k) as k varies, we can average (3) over k to get a recurrence relation for &amp;#257;&lt;sub&gt;r&lt;/sub&gt;. Alternatively, the function f(n)&amp;nbsp;=&amp;nbsp;&amp;Sigma;&lt;sub&gt;r&lt;/sub&gt;&amp;#257;&lt;sub&gt;r&lt;/sub&gt;n&lt;sup&gt;-r-1&lt;/sup&gt; must satisfy f(n+1)&amp;nbsp;=&amp;nbsp;(1+1/n)f(n)&amp;nbsp;-&amp;nbsp;f(n/2)/n which is solved by f(n)&amp;nbsp;=&amp;nbsp;1/(n+1)&amp;nbsp;=&amp;nbsp;&amp;Sigma;&lt;sub&gt;r&amp;ge;0&lt;/sub&gt;(-1)&lt;sup&gt;r&lt;/sup&gt;n&lt;sup&gt;-r-1&lt;/sup&gt;, so we get &amp;#257;&lt;sub&gt;r&lt;/sub&gt;&amp;nbsp;=&amp;nbsp;(-1)&lt;sup&gt;r&lt;/sup&gt;. Then, (3) can be applied iteratively to obtain a&lt;sub&gt;r&lt;/sub&gt;(k+1)-a&lt;sub&gt;r&lt;/sub&gt;(k) in terms of a&lt;sub&gt;s&lt;/sub&gt;(k) for s&amp;nbsp;&amp;lt;&amp;nbsp;r. Together with &amp;#257;&lt;sub&gt;r&lt;/sub&gt;, this gives a&lt;sub&gt;r&lt;/sub&gt;(k) and it can be seen that the period of a&lt;sub&gt;r&lt;/sub&gt;(k) doubles at each step. Doing this gives a&lt;sub&gt;r&lt;/sub&gt;&amp;nbsp;&amp;equiv;&amp;nbsp;(a&lt;sub&gt;r&lt;/sub&gt;(0),...,a&lt;sub&gt;r&lt;/sub&gt;(2&lt;sup&gt;r&lt;/sup&gt;-1)) as follows&#xA;$$&#xD;&#xA;\begin{align}&#xD;&#xA;&amp;amp; a_0=(1),\\\\&#xD;&#xA;&amp;amp; a_1=(0,-2),\\\\&#xD;&#xA;&amp;amp; a_2=(7,-3,-9,13)/2&#xD;&#xA;\end{align}&#xD;&#xA;$$&#xD;&#xA;These agree with the numerical simulation mentioned above.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Update: I tried another numerical simulation to check these asymptotics, by successively subtracting out the leading order terms. You can see it converges beautifully to the levels a&lt;sub&gt;0&lt;/sub&gt;, a&lt;sub&gt;1&lt;/sub&gt;, a&lt;sub&gt;2&lt;/sub&gt; but, then...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/ZDNzb.gif&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;... it seems that after the a&lt;sub&gt;2&lt;/sub&gt;n&lt;sup&gt;-3&lt;/sup&gt; part, there is an oscillating term! I wasn't expecting that, but it there is an asymptotic of the form cn&lt;sup&gt;-r&lt;/sup&gt;sin(alogn+&amp;theta;), where you can solve to leading order to obtain r&amp;nbsp;&amp;asymp;&amp;nbsp;3.54536, a&amp;nbsp;&amp;asymp;&amp;nbsp;10.7539.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Update 2: I was re-thinking this question a few days ago, and it suddenly occured how you can not only prove it using analytic methods, but give a full asymptotic expansion to arbitrary order. The idea involves some very cool maths! (if I may say so myself). Apologies that this answer has grown and grown, but I think it's worth it. It is a very interesting question and I've certainly learned a lot by thinking about it. The idea is that, instead of using a power series generating function as in Qiaochu's answer, you use a Dirichlet series which can be inverted with &lt;a href=&quot;http://en.wikipedia.org/wiki/Perron%27s_formula&quot; rel=&quot;nofollow&quot;&gt;Perron's formula&lt;/a&gt;. First, the expansion is as follows,&#xA;$$&#xD;&#xA;\begin{array}{ccc}&#xD;&#xA;\displaystyle&#xD;&#xA;p(n)=\sum_{\Re[r]+k\le \alpha}a_{r,k}(n)n^{-r-k}+o(n^{-\alpha}),&amp;amp;&amp;amp;(4)&#xD;&#xA;\end{array}&#xD;&#xA;$$&#xD;&#xA;for any real &amp;alpha;. The sum is over nonnegative integers k and complex numbers r with real part at least 1 and satisfying r+1&amp;nbsp;=&amp;nbsp;2&lt;sup&gt;r&lt;/sup&gt; (the leading term being r=1), and a&lt;sub&gt;r,k&lt;/sub&gt;(n) is a periodic function of n, with period 2&lt;sup&gt;k&lt;/sup&gt;. The reason for such exponents is that the difference equation (2) has the continuous-time limit p'(x)&amp;nbsp;=&amp;nbsp;p(x)/x&amp;nbsp;-&amp;nbsp;p(x/2)/x which has solutions p(x)&amp;nbsp;=&amp;nbsp;x&lt;sup&gt;-r&lt;/sup&gt; for precisely such exponents. Splitting into real and imaginary parts r&amp;nbsp;=&amp;nbsp;u+iv, all solutions to r+1&amp;nbsp;=&amp;nbsp;2&lt;sup&gt;r&lt;/sup&gt; lie on the line (1+u)&lt;sup&gt;2&lt;/sup&gt;+v&lt;sup&gt;2&lt;/sup&gt;&amp;nbsp;=&amp;nbsp;4&lt;sup&gt;u&lt;/sup&gt; and, other than the leading term u=1, v=0, there is precisely one complex solution with imaginary part 2n&amp;pi;&amp;nbsp;&amp;le;&amp;nbsp;vlog2&amp;nbsp;&amp;le;2n&amp;pi;+&amp;pi;/2 (positive integer n) and, together with the complex conjugates, this lists all possible exponents r.&#xA;Then, a&lt;sub&gt;r,k&lt;/sub&gt;(n) is determined (as a multiple of a&lt;sub&gt;r,0&lt;/sub&gt;) for k&amp;nbsp;&gt;&amp;nbsp;0 by substituting this expansion back into the difference equation as I did above. I arrived at this expansion after running the simulations plotted above (and T..'s mention of complex exponents of n helped). Then, the Dirichlet series idea nailed it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Define the Dirichlet series with coefficients p(n)/n&#xA;$$&#xD;&#xA;L(s)=\sum_{n=1}^\infty p(n)n^{-1-s},&#xD;&#xA;$$&#xD;&#xA;which converges absolutely for the real part of s large enough (greater than 0, since p is bounded by 1). It can be seen that L(s)&amp;nbsp;-&amp;nbsp;1 is of order 2&lt;sup&gt;-1-s&lt;/sup&gt; in the limit as the real part of s goes to infinity. Multiplying (2) by n&lt;sup&gt;-s&lt;/sup&gt;, summing and expanding n&lt;sup&gt;-s&lt;/sup&gt; in terms of (n+1)&lt;sup&gt;-s&lt;/sup&gt; on the LHS gives the functional equation&#xA;$$&#xD;&#xA;\begin{array}&#xD;&#xA;\displaystyle&#xD;&#xA;(s-1+2^{-s})L(s)=s+\sum_{k=1}^\infty(-1)^k\binom{-s}{k+1}(L(s+k)-1).&amp;amp;&amp;amp;(5)&#xD;&#xA;\end{array}&#xD;&#xA;$$&#xD;&#xA;This extends L to a meromorphic function on the whole complex plane with simple poles precisely at the points -r with real part at least one and r+1&amp;nbsp;=&amp;nbsp;2&lt;sup&gt;r&lt;/sup&gt;, and then at -r-k for nonnegative integers k. The pole with largest real component is at s&amp;nbsp;=&amp;nbsp;-1 and has residue&#xA;$$&#xD;&#xA;{\rm Res}(L,-1)={\rm Res}(s/(s-1+2^{-s}),-1)=\frac{1}{2\log2-1}.&#xD;&#xA;$$&#xD;&#xA;If we &lt;em&gt;define&lt;/em&gt; p'(n) by the truncated expansion (4) for some suitably large &amp;alpha;, then it will satisfy the recurrence relation (2) up to an error term of size O(n&lt;sup&gt;-&amp;alpha;-1&lt;/sup&gt;) and its Dirichlet series will satisfy the functional equation (5) up to an error term which will be an analytic function over R[s]&amp;nbsp;&amp;gt;&amp;nbsp;-&amp;alpha; (being a Dirichlet series with coefficients o(n&lt;sup&gt;-&amp;alpha;-1&lt;/sup&gt;)). It follows that p' has simple poles in the same places as p on the domain R[s]&amp;nbsp;&amp;gt;&amp;nbsp;-&amp;alpha; and, by choosing a&lt;sub&gt;r,0&lt;/sub&gt; appropriately, it will have the same residues. Then, the Dirichlet series associated with q(n)&amp;nbsp;=&amp;nbsp;p'(n)&amp;nbsp;-&amp;nbsp;p(n) will be analytic on R[s]&amp;nbsp;&amp;gt;&amp;nbsp;-&amp;alpha; We can use &lt;a href=&quot;http://en.wikipedia.org/wiki/Perron%27s_formula&quot; rel=&quot;nofollow&quot;&gt;Perron's formula&lt;/a&gt; to show that q(n) is of size O(n&lt;sup&gt;&amp;beta;&lt;/sup&gt;) for any &amp;beta;&amp;nbsp;&amp;gt;&amp;nbsp;-&amp;alpha; and, by making &amp;alpha; as large as we like, this will prove the asymptotic expansion (4). Differentiated, Perron's formula gives&#xA;$$&#xD;&#xA;dQ(x)/dx = \frac{1}{2\pi i}\int_{\beta-i\infty}^{\beta+i\infty}x^sL(s)\,ds&#xD;&#xA;$$&#xD;&#xA;where Q(x)&amp;nbsp;=&amp;nbsp;&amp;Sigma;&lt;sub&gt;n&amp;le;x&lt;/sub&gt;q(n) and &amp;beta;&amp;nbsp;&amp;gt;&amp;nbsp;-&amp;alpha;. This expression is intended to be taken in the sense of distributions (i.e., multiply both sides by a smooth function with compact support in (0,&amp;infin;) and integrate). If &amp;theta; is a smooth function of compact support in (0,&amp;infin;) then&#xA;$$&#xD;&#xA;\begin{array}&#xD;&#xA;\displaystyle\sum_{n=1}^\infty q(n)\theta(n/x)/x&amp;amp;\displaystyle=\int_0^\infty\theta(y)\dot Q(xy)\,dy\\\\&#xD;&#xA;&amp;amp;\displaystyle=\frac{1}{2\pi i}\int_{\beta-i\infty}^{\beta+i\infty}x^sL(s)\int\theta(y)y^s\,dy\,ds=O(x^\beta)\ \ (6)&#xD;&#xA;\end{array}&#xD;&#xA;$$&#xD;&#xA;We obtain the final bound, because by integration by parts, the integral of &amp;theta;(y)y&lt;sup&gt;s&lt;/sup&gt; can be shown to go to zero faster than any power of 1/s, so the integrand is indeed integrable and the x&lt;sup&gt;&amp;beta;&lt;/sup&gt; term can be pulled outside.&#xA;This is enough to show that q(n) is itself of O(n&lt;sup&gt;&amp;beta;&lt;/sup&gt;). Trying to finish this answer off without too much further detail, the argument is as follows. If q(n)n&lt;sup&gt;-&amp;beta;&lt;/sup&gt; was unbounded then would keep exceeding its previous maximum and, by the recurrence relation (2), it would take time at least &lt;a href=&quot;http://en.wikipedia.org/wiki/Big-omega_notation&quot; rel=&quot;nofollow&quot;&gt;&amp;Omega;(n)&lt;/a&gt; to get back close to zero. So, if &amp;theta; has support in [1,1+&amp;epsilon;] for small enough &amp;epsilon;, the integral $\int\theta(y)\dot Q(ny)\,dy$ will be of order q(n) at such times and, as this happens infinitely often, it would contradict (6). Phew! I knew that this could be done, but it took some work! Possibly not as simple or direct as you were asking for, but Dirichlet series are quite standard (more commonly in analytic number theory, in my experience). However, maybe not really more difficult than the probabilistic method and you do get a whole lot more. This approach should also work for other types of recurrence relations and differential equations too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, I added a much more detailed writeup on my blog, fleshing out some of the details which I skimmed over here. See &lt;a href=&quot;http://almostsure.wordpress.com/2010/10/06/asymptotic-expansions-of-a-recurrence-relation/&quot; rel=&quot;nofollow&quot;&gt;Asymptotic Expansions of a Recurrence Relation&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1321" LastEditorUserId="1321" LastEditorDisplayName="" LastEditDate="2010-11-19T19:04:11.270" LastActivityDate="2010-11-19T19:04:11.270" CommunityOwnedDate="2010-11-19T18:58:39.777" CommentCount="10" />
  <row Id="4814" PostTypeId="1" AcceptedAnswerId="4824" CreationDate="2010-09-16T21:14:45.120" Score="2" ViewCount="165" Body="&lt;p&gt;I know that many types graphs have unique names, ie. a directed graph where each node has exactly one outbound edge is known as a &lt;code&gt;functional&lt;/code&gt; graph. Do the graphs with exactly 2 inbound/outbound edges have a name?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Bonus&lt;/strong&gt;: If the 2 inbound/outbound graph is named, is the nth inbound/outbound graph named?&lt;/p&gt;&#xA;" OwnerUserId="196" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-09-17T01:28:31.090" LastActivityDate="2010-09-17T19:56:26.913" Title="What are the names of graphs with 2 in/outbound edges?" Tags="&lt;terminology&gt;&lt;graph-theory&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="4815" PostTypeId="1" AcceptedAnswerId="4817" CreationDate="2010-09-16T21:16:38.457" Score="4" ViewCount="175" Body="&lt;p&gt;Given an infinite dimensional normed linear space, how would one show that it is possible to fit an infinite collection of non-overlapping balls of radius 1/4 in the unit ball?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I guess one can immediately reduce the problem to a normed linear space of countably infinite dimensions. The solution seems clear if the concept of orthogonality exists, but not every normed linear space has an inner product so it's not possible to apply something like Gram-Schmidt to produce an orthogonal basis. Is there any way around this, or there another approach that can be used?&lt;/p&gt;&#xA;" OwnerUserId="1736" LastEditorUserId="1736" LastEditorDisplayName="" LastEditDate="2010-09-16T21:33:39.653" LastActivityDate="2010-09-16T23:23:22.190" Title="Fitting an infinite collection of balls in an infinite dimensional unit ball" Tags="&lt;functional-analysis&gt;" AnswerCount="1" />
  <row Id="4816" PostTypeId="2" ParentId="4808" CreationDate="2010-09-16T22:27:00.530" Score="4" ViewCount="" Body="&lt;p&gt;There are relatively simple proofs in textbooks on calculus of variations. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In more elementary approaches a convex figure is deformed, in discrete steps or through a continuous unbending process, toward a circle, and two things need to be proved: convergence to the circle, and increase of the isoperimetric ratio throughout the flow. Usually one step is easy and the other is difficult, requiring non-elementary methods to make rigorous.   It is also necessary to make explicit what class of curves is considered: rectifiable, piecewise smooth, or something else.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The simplest argument I know that is elementary and rigorous is to prove the finite-dimensional approximation, that for fixed edge lengths of a polygon, there is a maximum area (by compactness) and (by elementary geometry or Lagrange multipliers) it is the one where all vertices are on a circle. Then, use this to prove that any smooth curve, if it beats the circle, has a finite polygonal approximation that beats the inscribed polygon. &lt;/p&gt;&#xA;" OwnerUserId="467" LastActivityDate="2010-09-16T22:27:00.530" />
  <row Id="4817" PostTypeId="2" ParentId="4815" CreationDate="2010-09-16T23:03:28.723" Score="3" ViewCount="" Body="&lt;p&gt;Here is a standard lemma you could use. If $M$ is a closed proper subspace of a normed linear space $E$, then for all $\epsilon\gt0$ there is an $x\in E$ of norm 1 whose distance to $M$ is greater than $1-\epsilon$.  (E.g., &lt;a href=&quot;http://books.google.com/books?id=maRLP-UBquYC&amp;amp;lpg=PP1&amp;amp;pg=PA56#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;here's a proof&lt;/a&gt; in Lemma 3-6.10 of Tsoy-Wo Ma's &lt;em&gt;Classical analysis on normed spaces&lt;/em&gt;.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's how you could use it.  Let $B$ denote the unit ball of an infinite dimensional normed space $X$.  Let $x_1\in X$ have norm 1, and let $M_1$ be the span of $\{x_1\}$.  By the lemma there is an $x_2\in X$ of norm 1 whose distance to $M_1$ is greater than $\frac{2}{3}$.  Let $M_2$ be the span of $\{x_1,x_2\}$, and let $x_3\in X$ have norm 1 and distance greater than $\frac{2}{3}$ to $M_2$.  Repeat countably infinitely many times to obtain a sequence $x_1,x_2,\ldots$ of elements of $X$ of norm 1 with pairwise distances greater than $\frac{2}{3}$.  Note that the lemma will always apply, because each $M_k=\operatorname{span}\{x_1,x_2,\ldots,x_k\}$ is finite dimensional, hence closed and proper.  Then the balls of radius $\frac{1}{4}$ centered at the points $\frac{3}{4}x_1,\frac{3}{4}x_2,\ldots$ are disjoint and contained in $B$.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2010-09-16T23:23:22.190" LastActivityDate="2010-09-16T23:23:22.190" CommentCount="3" />
  <row Id="4818" PostTypeId="2" ParentId="4808" CreationDate="2010-09-16T23:34:06.463" Score="3" ViewCount="" Body="&lt;p&gt;As Qiaochu Yuan pointed out, this is a consequence of the isoperimetric inequality that relates the length $L$ and the area $A$ for any closed curve $C$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;4\pi A \leq L^2 \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Taking a circumference of radius $r$ such that $2\pi r = L$, you obtain&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;A \leq \frac{L^2}{4\pi} = \frac{4 \pi^2 r^2}{4\pi} = \pi r^2 \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is, the area $A$ enclosed by the curve $C$ is smaller than the area enclosed by the circumference of the same length.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;As for the proof of the isoperimetric inequality&lt;/strong&gt;, here is the one I've learnt as undergraduate, which is elementary and beautiful, I think.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Go round your curve $C$ counterclockwise. For a plane vector field $(P,Q)$, &lt;a href=&quot;http://en.wikipedia.org/wiki/Green%27s_theorem&quot; rel=&quot;nofollow&quot;&gt;Green's theorem&lt;/a&gt; says&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\oint_{\partial D}(Pdx + Qdy) = \int_D \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right)  dxdy\ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Apply it for the vector field $(P,Q) = (-y,x)$ and when $D$ is the region enclosed by your curve $C = \partial D$. You obtain&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;A = \frac{1}{2} \oint_{\partial D} (-ydx + xdy) \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, parametrize $C= \partial D$ with arc length: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\gamma : [0,L] \longrightarrow \mathbb{R}^2 \ ,\qquad \gamma (s) = (x(s), y(s)) \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Taking into account that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;0= xy \vert_0^L = \int_0^L x'yds + \int_0^L xy'ds \ ,&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;we get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;A = \int_0^L xy'ds = -\int_0^L x'yds \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;So enough for now with our curve $C$. Let's look for a nice circumference to compare with!&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First of all, $[0,L]$ being compact, the function $x: [0,L] \longrightarrow \mathbb{R}$ will have a global maximum and a global minimum. Changing the origin of our parametrization if necessary, me may assume the minimum is attained at $s=0$. Let the maximum be attained at $s=s_0 \in [0,L]$. Let $q = \gamma (0)$ and $p = \gamma (s_0)$. (If there are more than one minimum and more than one maximum, we choose one of each: the ones you prefer.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $x'(0) = x'(s_0) = 0$, we have vertical tangent lines at both points $p,q$ of our curve $C$. Draw a circumference between these parallel lines, tangent to both of them (a little far away of $C$ to avoid making a mess). So the radius of this circumference will be $r = \frac{\| pq \|}{2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's take the origin of coordinates at the center of this circumference. We parametrize it with the same $s$, the arc length of $C$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\sigma (s) = (\overline{x}(s), \overline{y}(s)) \ , \quad s \in [0, L] \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, $\overline{x}(s)^2 + \overline{y}(s)^2 = r^2$ for all $s$. If we choose $\overline{x}(s) = x(s)$, this forces us to take $ \overline{y}(s) = \pm \sqrt{r^2 - \overline{x}(s)^2}$. In order that $\sigma (s)$ goes round all over our circumference counterclockwise too, we choose the minus sign if $0\leq s \leq s_0$ and the plus sign if $s_0 \leq s \leq L$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;We are almost done, just a few computations left.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $\overline{A}$ denote the area of our circumference. So, we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;A = \int_0^L xy'ds = \int_0^L \overline{x}y'ds \qquad \text{and} \qquad \overline{A}= \pi r^2 = -\int_0^L\overline{y}\overline{x}'ds = -\int_0^L\overline{y} x'ds \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{align}&#xA;A + \pi r^2  &amp;amp;= A + \overline{A} = \int_0^L (\overline{x}y' - \overline{y}x')ds \\&#xA;            &amp;amp;\leq \int_0^L \vert \overline{x}y' - \overline{y}x'\vert ds     \\&#xA;            &amp;amp;= \int_0^L \vert (\overline{x}, \overline{y})\cdot (y', -x')\vert ds   \\&#xA;            &amp;amp;\leq \int_0^L \sqrt{\overline{x}^2 + \overline{y}^2} \cdot \sqrt{(y')^2+ (-x')^2}ds \\&#xA;            &amp;amp;= \int_0^L rds = rL \ .&#xA;\end{align}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The last inequality is Cauchy-Schwarz's one and the last but one equality is due to the fact that $s$ is the arc-length of $C$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Summing up:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;A + \pi r^2 \leq rL \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, since the geometric mean is always smaller than the arithmetic one,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\sqrt{A\pi r^2} \leq \frac{A + \pi r^2}{2} \leq \frac{rL}{2} \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;A \pi r^2 \leq \frac{r^2L^2}{4}  \qquad \Longrightarrow \qquad 4\pi A \leq L^2 \ .&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="664" LastEditorUserId="664" LastEditorDisplayName="" LastEditDate="2010-09-17T02:33:49.903" LastActivityDate="2010-09-17T02:33:49.903" />
  <row Id="4819" PostTypeId="2" ParentId="4757" CreationDate="2010-09-16T23:52:12.517" Score="1" ViewCount="" Body="&lt;p&gt;I think the first part of your proof is a little clumsy. For $x\neq 0$ you don't need to use induction: note that what you have for $x\neq 0$ is a composite of $C^\infty$ functions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you write  $(e^{g(x)}p_n(x))'=e^{g(x)}g′(x)p_n(x)+e^{g(x)}p_n′(x)$ you are utilizing the chain rule. Thus, you are assuming that the composite $(e^{g(x)}p_n(x))$ is differentiable. Of course this function is differentiable, but this is what you want to prove: that $f^{(n)}$ is differentiable for $x\neq 0$. The point is that you are proving even more: you're giving the general form of the derivatives.  &lt;/p&gt;&#xA;" OwnerUserId="1178" LastActivityDate="2010-09-16T23:52:12.517" CommentCount="3" />
  <row Id="4820" PostTypeId="1" AcceptedAnswerId="4831" CreationDate="2010-09-17T00:29:17.893" Score="6" ViewCount="114" Body="&lt;p&gt;If we know the rational homology of X is 0, can we get some information about the rational homology of X/G, where G is a finite group? Thank you very much for the answers!&lt;/p&gt;&#xA;" OwnerUserId="1944" LastActivityDate="2010-09-17T16:13:10.693" Title="Is there any relation about rational homology of X and X/G" Tags="&lt;algebraic-topology&gt;&lt;cohomology&gt;&lt;homological-algebra&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" />
  <row Id="4821" PostTypeId="1" CreationDate="2010-09-17T00:39:55.013" Score="2" ViewCount="170" Body="&lt;p&gt;I've somewhat got this question down but I'm only half way. &#xA;Q: How many 7-digit even numbers less than 3 000 000 can be formed using all the digits 1,2,2,3,5,5,6?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I figured that there's about 4 possible approaches&#xA;1 _ _ _ _ _ 6&#xA;1 _ _ _ _ _ 2&#xA;2 _ _ _ _ _ 6&#xA;2 _ _ _ _ _ 2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do I fill in the middle? I tried 5! factorial and dividing out the similar factorials but I didn't get the right answer &lt;/p&gt;&#xA;" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-09-17T08:03:07.690" LastActivityDate="2010-09-18T00:32:16.700" Title="Permutations with identical items" Tags="&lt;combinatorics&gt;" AnswerCount="3" />
  <row Id="4822" PostTypeId="2" ParentId="4808" CreationDate="2010-09-17T00:43:03.727" Score="0" ViewCount="" Body="&lt;p&gt;First you can propose this problem as: The area $A$ encompassed by any simple closed rectifiable curve $C$, of length $L$, satisfacts the inequality $A\geq \frac{L^2}{4\pi}$, and equality occurs, if and only if, $C$ is a circle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only proof I have done for this was using Parseval's identity (and therefore Fourier series), so it's not elementary (but it's rather simple if you know the aforementioned identity). Thought if you want I can post that proof.&lt;/p&gt;&#xA;" OwnerUserId="1605" LastActivityDate="2010-09-17T00:43:03.727" />
  <row Id="4823" PostTypeId="2" ParentId="4821" CreationDate="2010-09-17T00:56:34.037" Score="1" ViewCount="" Body="&lt;p&gt;Try the &lt;a href=&quot;http://en.wikipedia.org/wiki/Multinomial_theorem&quot; rel=&quot;nofollow&quot;&gt;multinomial coefficient.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1892" LastActivityDate="2010-09-17T00:56:34.037" />
  <row Id="4824" PostTypeId="2" ParentId="4814" CreationDate="2010-09-17T01:10:29.217" Score="3" ViewCount="" Body="&lt;p&gt;This is known as a $d$-regular graph (at least if the graph is undirected). If your graph is a sub-graph of a bigger graph, then it is a $d$-factor. Finally, for $d=2$ it's also called a &lt;i&gt;cycle cover&lt;/i&gt;, since it's a collection of cycles (if it's finite). For $d=1$ it's called a &lt;i&gt;matching&lt;/i&gt;. For $d=3$ it is known as a &lt;i&gt;cubic graph&lt;/i&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For directed graphs it's usually not clear what is supposed to be regular, the in-degree, the out-degree or both.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2010-09-17T19:56:26.913" LastActivityDate="2010-09-17T19:56:26.913" CommentCount="0" />
  <row Id="4825" PostTypeId="2" ParentId="4821" CreationDate="2010-09-17T01:19:16.293" Score="3" ViewCount="" Body="&lt;p&gt;We actually have only two cases: 1... and 2... .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the first case, we have to fit the numbers 2,2,3,5,5,6 somehow. Imagine that we tag each of the repeated elements somehow, e.g. 2',2&quot;,3,5',5&quot;,6. If we care about the tags, then there are $6!$ possible ways to arrange the numbers. Now each untagged arrangement corresponds to $2!2!$ different tagged arrangements (consider the order in which the tags on the $2$s and $5$s appear). In total, this gives $6!/2!2!= 180$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the second case, we have only one repeated element $5$ so there are $6!/2! = 360$ possibilities, for a total of $540$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, if we have $k$ unique elements repeated $t_1,\ldots,t_k$ times (respectively), then the answer is going to be $(t_1+\cdots+t_k)!/t_1!\cdots t_k!$, using the same reasoning. This is known as a multinomial coefficient since it appears in the multinomial theorem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(x_1 + \cdots + x_k)^n = \sum_{t_1 + \cdots + t_k = n} n!/t_1!\cdots t_k! x_1^{t_1} \cdots x_k^{t_k}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where the sum is over all non-negative integers summing to $n$. The multinomial coefficient is sometimes denoted $\binom{n}{t_1\ldots t_k}$, although one of these numbers is really redundant. When $k=2$ we get the binomial coefficient, and we usually omit $t_2$.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-09-17T01:19:16.293" CommentCount="1" />
  <row Id="4826" PostTypeId="1" AcceptedAnswerId="5009" CreationDate="2010-09-17T01:37:56.247" Score="5" ViewCount="217" Body="&lt;p&gt;Here's the setup: Let $U,V$ open $\subset \mathbb{R}^n$, and let $u:V\rightarrow \mathbb{R}$ be harmonic, and $v:U\rightarrow V$ be conformal, i.e. $v$ is $C^1$ and the Jacobian $J_v(x)$ is a scalar multiple of an orthogonal transformation for all [; x\in U ;].  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm trying to prove $u\circ v$ is harmonic. [I've seen this stated as a fact in a few places without reference, namely here: &lt;a href=&quot;http://en.wikipedia.org/wiki/Conformal_map#Uses&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Conformal_map#Uses&lt;/a&gt; , but maybe my hypotheses are slightly different and this is not true at all]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've seen a proof that if $u$ is $C^2$ and $T$ is an orthogonal transformation then &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Delta (u \circ T) = \Delta(u) \circ T$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I'm thinking that to show $u\circ v$ is harmonic, we can use the fact that $v$ acts locally as its Jacobian, which is an orthogonal transformation, and move the Laplacian onto $u$ and conclude $u\circ v$ is harmonic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I'm having trouble making this idea precise.  After glancing at my copy of baby Rudin, my hunch is to use the inverse function theorem or constant rank theorem, but I'm unsure how to apply those.  Any suggestions?&lt;/p&gt;&#xA;" OwnerUserId="1692" LastEditorUserId="1692" LastEditorDisplayName="" LastEditDate="2010-09-17T19:12:12.670" LastActivityDate="2010-09-19T19:02:10.397" Title="Harmonic function composed with conformal map is harmonic (in R^n)" Tags="&lt;real-analysis&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="4827" PostTypeId="1" AcceptedAnswerId="4829" CreationDate="2010-09-17T01:41:36.120" Score="3" ViewCount="69" Body="&lt;p&gt;I came across the two congruences that for $p$ an odd prime,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$1^2\cdot 3^2\cdot 5^2\cdots (p-2)^2\equiv (-1)^{(p+1)/(2)}(mod\ p)$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;and &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$2^2\cdot 4^2\cdot 6^2\cdots (p-1)^2\equiv (-1)^{(p+1)/(2)}(mod\ p)$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the reason for this? It seems to me to be closely related to Wilson's Theorem, and possibly Fermat's little theorem since $p-1=2k$ for some $k$, and then $j^{2k}\equiv 1(mod\ p)$ for all the evens and odds $j$ less than $p$, but I wasn't able to complete this train of thought.&lt;/p&gt;&#xA;" OwnerUserId="1583" LastActivityDate="2010-09-17T03:21:42.447" Title="Why is the square of all odds less than an odd prime $p$ congruent to $(-1)^{(p+1)/(2)}(mod\ p)$?" Tags="&lt;number-theory&gt;" AnswerCount="1" />
  <row Id="4829" PostTypeId="2" ParentId="4827" CreationDate="2010-09-17T01:56:02.783" Score="3" ViewCount="" Body="&lt;p&gt;Modulo $p$, you have $k=-(p-k)$ and so both products are equal to $(p-1)!(-1)^{(p-1)/2} \equiv (-1)^{(p+1)/2} \pmod{p}$ by Wilson's theorem.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2010-09-17T03:21:42.447" LastActivityDate="2010-09-17T03:21:42.447" CommentCount="3" />
  <row Id="4830" PostTypeId="2" ParentId="4788" CreationDate="2010-09-17T02:40:25.903" Score="5" ViewCount="" Body="&lt;p&gt;It's easy to be jaded about FTA after mathematical history has run for another two centuries.  Sure, it is not the number one most important result any more, or the center of any research program (though understanding the algebraic closure of Q could be considered as half of number theory).  But consider the situation around 1800.  In addition to solution of algebraic equations one had new methods of constructing numbers, using power series, integrals and other limits.  Algebra and number theory dealt with the first situation, to a limited extent, and analysis showed that the second type of construction could be iterated but still stay within the same realm of numbers.  There was still the possibility that solving equations with $\pi$ and $e$ as coefficients could require an entirely new type of super-transcendental analysis.  Fundamental Theorem of Algebra is self-defeating in this sense: it shows that nothing more was needed than complex numbers.  But this is not clear in a world where you don't know that FTA is true.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To get an idea what algebraic geometry looked like without complex numbers, look up Newton's classification of degree 3 algebraic curves in the plane, $P_3(x,y)=0$, using real coordinates.  The reason this work is obscure today is that there are many dozens of cases compared to the complex projective version.   As in Lie groups and topology, looking from the universal covering (C) downward, modulo some Galois-theoretic details (R), is usually easier than working from the bottom up. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose you want to evaluate the integral, from $- \infty$ to $+\infty$, of a rational function (one with integer coefficients would illustrate the point).  The answer will involve $\pi$ and the usual method for finding it will use the specific location of the roots in the complex plane, so is more specific to complex numbers than the existence of roots in an algebraic closure.  There are some non-usual methods that stay entirely within the real numbers, but they are nonstandard because they are more complicated, and harder to understand and adapt to other problems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Algebraic geometry in general has a transcendental part -- periods, Hodge theory, uniformization, etc -- that, in the present state of knowledge, cannot be fully substituted by algebraic methods over fields of characteristic 0 (or p).  Sometimes Lefschetz principle or reduction to positive characteristic can be used, sometimes not, or the theory is unknown.&lt;/p&gt;&#xA;" OwnerUserId="467" LastActivityDate="2010-09-17T02:40:25.903" CommentCount="1" />
  <row Id="4831" PostTypeId="2" ParentId="4820" CreationDate="2010-09-17T02:49:24.020" Score="5" ViewCount="" Body="&lt;p&gt;When $G$ is finite, the rational cohomology of $X/G$ is the fixed point set $H^*(X;\mathbb{Q})^G$.  This is proven in Grothendieck's Tohoku paper (Theorem 5.3.1 and the Corollary to Proposition 5.3.2).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if the rational cohomology of $X$ is trivial, the same is true for $X/G$. And rationally the cohomology and homology are isomorphic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For paracompact Hausdorff spaces, these cohomology groups can be taken to be the Cech cohomology groups.  Note that if $X$ is homotopy equivalent to a CW complex, then Cech cohomology agrees with singular cohomology.  You might also want to look at Oscar Randall-Williams comments here: &lt;a href=&quot;http://mathoverflow.net/questions/18898/grothendiecks-tohoku-paper-and-combinatorial-topology/30015#30015&quot; rel=&quot;nofollow&quot;&gt;http://mathoverflow.net/questions/18898/grothendiecks-tohoku-paper-and-combinatorial-topology/30015#30015&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1392" LastEditorUserId="1392" LastEditorDisplayName="" LastEditDate="2010-09-17T16:13:10.693" LastActivityDate="2010-09-17T16:13:10.693" CommentCount="5" />
  <row Id="4832" PostTypeId="2" ParentId="4737" CreationDate="2010-09-17T03:04:53.763" Score="2" ViewCount="" Body="&lt;p&gt;I think Apollonian circle packing is relevant here.  It's certainly not identical with the question's topic, but it's enough to convince me that the answer is yes-- you can fit disks of radii 1/2, 1/3, 1/4,... inside a disk of radius 1.  Hopefully someone more knowledgeable can make a rigorous argument, and solve the question for a disk of radius 5/6.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found a &lt;a href=&quot;http://arxiv.org/abs/0811.2236&quot; rel=&quot;nofollow&quot;&gt;paper&lt;/a&gt; on arXiv by Alex Kontorovich and Hee Oh demonstrating that for a bounded Apollonian circle packing, the number of circles with curvature at most N is asymptotic to cN&lt;sup&gt;a&lt;/sup&gt; where a=1.30...  This is good news for our problem!  Apollonian packing greedily fits the largest possible circle in the gap created by any 3 mutually tangent circles, and the number of circles with radius at least 1/N grows as N&lt;sup&gt;a&lt;/sup&gt;, where we just need it to grow as N.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a picture from Wikipedia of an Apollonian packing, labeled by curvature (1/r):&#xA;&lt;img src=&quot;http://i.imgur.com/pGCwR.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The outer circle has radius 1, and we want to find space for disks with curvature 2,3,4,5,...  Compare the list of curvatures in the picture with the list we want to find space for:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2,2,3,3,6,6,6,6,11,11,11,11,14,14,14,14,15,15,18,18,18,18,23,23,23,23,26,26,26,26,27,27,&#xA;2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;27,27,30,30,30,30,35,35,35,35,35,35,38,38,38,38,38,38,38,38,...&#xA;34,35,36,37,38,39,40,41,42,43,44,45,46,47,48,49,50,51,52,53,...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Everywhere the curvature in the lower list is $\geq$ the curvature in the upper list, we can just fit our disk inside the corresponding circle in the Apollonian packing.  The only case in the list that doesn't fit is 10-- but since we've left unused space several times already, it's clear we can make room for it.  Of course, shifting things to make room breaks the Apollonian packing (though we can leave three-quarters of it intact); but it sure looks like we have plenty of space left, and the asymptotic formula also suggests we are safe.  I haven't looked closely at the paper for any explicit bounds that would make this rigorous. I realize it is just a sketch, but hopefully a convincing one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second Apollonian packing picture is analogous to the radius=5/6 case, scaled up 5x.&lt;br&gt;&#xA;&lt;img src=&quot;http://i.imgur.com/SRTtF.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here, we want to fit disks with curvature 10,15,20,25,30,35,40,45,50,...  We run into trouble already with 30 (the disk of radius 1/6), and we'll have to shift things around quite a bit to fit our disks in.  The Apollonian packing does not lead to a solution as easily as it seems to in the radius 1 case.  Nonetheless, the asymptotic formula suggests that if we can pack the first N disks (for N perhaps not very large), that there will be plenty of room for the rest.  One would have to work out some details to find the value of N.&lt;/p&gt;&#xA;" OwnerUserId="1461" LastEditorUserId="1461" LastEditorDisplayName="" LastEditDate="2010-09-17T03:47:30.980" LastActivityDate="2010-09-17T03:47:30.980" CommentCount="1" />
  <row Id="4833" PostTypeId="1" CreationDate="2010-09-17T03:22:11.793" Score="3" ViewCount="74" Body="&lt;p&gt;I forget for a while, we don't need the compactness condition here right?&lt;/p&gt;&#xA;" OwnerUserId="1944" LastActivityDate="2010-09-17T07:10:04.400" Title="Does noncompact manifold or orbifold have the homotopy type,of CW complex?" Tags="&lt;algebraic-topology&gt;&lt;homotopy-theory&gt;" AnswerCount="1" FavoriteCount="2" />
  <row Id="4834" PostTypeId="1" AcceptedAnswerId="4857" CreationDate="2010-09-17T03:40:37.583" Score="2" ViewCount="166" Body="&lt;p&gt;If I want to find the number of all possible groupings of a set of numbers, including all groupings with only a single number and groupings will all numbers is there a better method than this. For this use I don't want to count empty groups and order is important.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take the summation of the number of permutations with repetition of size i where i goes from 1 to the number of items in my collection?&lt;br&gt;&#xA;s = number of elements in the collection&lt;br&gt;&#xA;i = number I'm summing over from one to s&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{i=1}^{s} s^i$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;basically I want to know this:&#xA;1. will this count all possible groupings except for the empty group?&#xA;2. how do you get math equations to show up in questions for future use?&#xA;3. is there a better way to express this (if you can understand what I'm trying to say)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ok small example. if I have a collection (1,2,3) I want to count the possible collections that could be made from is such as: &#xA;(1),(2),(3)&#xA;(1,1),(1,2),(1,3),(2,1),(2,2),(2,3),(3,1),(3,2),(3,3),&#xA;(1,1,1),(1,1,2),(1,1,3),(1,2,1),(1,2,2),(1,2,3),(1,3,1),(1,3,2),(1,3,3),(2,1,1),(2,1,2),(2,1,3),(2,2,1),(2,2,2),(2,2,3),(2,3,1),(2,3,2),(2,3,3),(3,1,1),(3,1,2),(3,1,3),(3,2,1),(3,2,2),(3,2,3),(3,3,1),(3,3,2),(3,3,3).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so for the collection (1,2,3) I can make 39 collections from it not exceeding the length of the original collection.&lt;/p&gt;&#xA;" OwnerUserId="1582" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-09-17T17:24:59.643" LastActivityDate="2010-09-17T17:24:59.643" Title="Given a collection of items, how many ordered subsets can I make?" Tags="&lt;combinatorics&gt;" AnswerCount="1" CommentCount="6" />
  <row Id="4836" PostTypeId="1" AcceptedAnswerId="4837" CreationDate="2010-09-17T05:37:22.933" Score="2" ViewCount="106" Body="&lt;p&gt;Let $M$ be a 2n dimensional manifold, and let $A$ be a n-dimensional cycle on M. I want to compute the self-intersection $(A.A)$ of A with itself. Let $\eta_A$ be the form in $H^n(M, \mathbb{R})$ given by the Poincare duality isomorphism, i.e. $\int_B \eta_A=(A.B)$ for an n-cycle B. Then by the rule that intersection of cycles corresponds to taking wedge product, we have $\int_M \eta_A \wedge \eta_B=(A.B)$. Taking A=B, since $\eta_A \wedge \eta_A =0$, I get $(A.A)=0$, which I don't think is the case. Where am I going wrong? &lt;/p&gt;&#xA;" LastActivityDate="2010-09-17T05:48:51.757" Title="computing intersection number via differential forms" Tags="&lt;differential-topology&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="4837" PostTypeId="2" ParentId="4836" CreationDate="2010-09-17T05:48:51.757" Score="6" ViewCount="" Body="&lt;p&gt;$\omega \wedge \eta = (-1)^{ab}\eta \wedge \omega$ where $a$ is the dimension of $\omega$ and $b$ is the dimension of $\eta$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So your &quot;since $\eta_A \wedge \eta_A=0$&quot; comment is only true when $n$ is odd.&lt;/p&gt;&#xA;" OwnerUserId="642" LastActivityDate="2010-09-17T05:48:51.757" CommentCount="4" />
  <row Id="4839" PostTypeId="2" ParentId="4833" CreationDate="2010-09-17T07:10:04.400" Score="4" ViewCount="" Body="&lt;p&gt;According to &lt;em&gt;The Topology of CW-Complexes&lt;/em&gt; by Lundell and Weingram&#xA;(Van Nostrand Reinhold, 1969) the answer is yes for (separable) manifolds.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-09-17T07:10:04.400" CommentCount="1" />
  <row Id="4840" PostTypeId="1" CreationDate="2010-09-17T07:23:18.387" Score="6" ViewCount="109" Body="&lt;p&gt;I have a terrain, which is represented by  &lt;a href=&quot;http://en.wikipedia.org/wiki/Polygon_mesh&quot; rel=&quot;nofollow&quot;&gt;one mesh with a lot of polygons&lt;/a&gt; as shown below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/jZRSC.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This terrain will be cut by a plane at a certain level. So there are volumes of the terrains that are located above the plane ( cut volume), and volumes that are located below the plane ( fill volume). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is, how do I obtain the cut/ fill volume? My current approach is simply take one mesh at a time, and then form a tetrahedron with the plane, and compute the volume. &lt;strong&gt;But this is slow&lt;/strong&gt;. Is there other better approach?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One approach that I have in mind, is to try to form Bezier surface for the terrain, and then try to use integration to compute the volume. But I don't know how to proceed with this. Any idea?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit: Terminology updated&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="550" LastEditorUserId="550" LastEditorDisplayName="" LastEditDate="2010-09-19T08:32:53.333" LastActivityDate="2010-09-19T08:32:53.333" Title="Find the Volume Enclosed by Terrain and a Certain Sea Level" Tags="&lt;geometry&gt;&lt;computational-geometry&gt;" AnswerCount="1" CommentCount="9" FavoriteCount="1" />
  <row Id="4841" PostTypeId="1" CreationDate="2010-09-17T08:05:46.290" Score="7" ViewCount="2144" Body="&lt;p&gt;How can we prove that inverse of upper(lower) triangular matrix is upper(lower) triangular...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anybody answer to this question.......&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance.........&lt;/p&gt;&#xA;" OwnerUserId="1924" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-09-17T08:10:34.963" LastActivityDate="2010-09-18T17:06:21.803" Title="Inverse of a triangular matrix( both upper &amp; lower ) is triangular" Tags="&lt;linear-algebra&gt;&lt;matrices&gt;" AnswerCount="4" CommentCount="2" FavoriteCount="2" />
  <row Id="4842" PostTypeId="2" ParentId="4841" CreationDate="2010-09-17T08:16:49.233" Score="0" ViewCount="" Body="&lt;p&gt;Suppose that $U$ is upper. The $i$th column $x_i$ of the inverse is given by $Ux_i=e_i$ where $e_i$ is the $i$th unit vector. By backward subsitution you can see that $(x_i)_j=0$ for $i+1\leq j\leq n$. I.e all the entries in the $i$th column of the inverse below the diagonal are zero. This is true for all $i$ and hence the inverse $U^{-1}=[x_1|\ldots|x_n]$ is upper triangular. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same thing works for lower triangular using forward subsitution.&lt;/p&gt;&#xA;" OwnerUserId="1794" LastActivityDate="2010-09-17T08:16:49.233" />
  <row Id="4843" PostTypeId="2" ParentId="4841" CreationDate="2010-09-17T09:07:27.647" Score="2" ViewCount="" Body="&lt;p&gt;I'll add nothing to alext87 answer, or J.M. comments. Just &quot;display&quot; them.   :-)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remeber that you can compute the inverse of a matrix by reducing it to &lt;a href=&quot;http://math.stackexchange.com/questions/1822/is-reducing-a-matrix-to-row-echelon-form-useful-at-all/1831#1831&quot;&gt;row echelon form&lt;/a&gt; and solving the simultaneous systems of linear equations $ (A \vert I)$, where $A$ is the matrix you want to invert and $I$ the unit matrix. When you have finished the process, you'll get a matrix like $(I\vert A^{-1})$ and the matrix on the right, yes!, is the inverse of $A$. (Why?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your case, half of the work is already done:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{pmatrix}&#xA;a^1_1 &amp;amp; a^1_2 &amp;amp; \cdots &amp;amp; a^1_{n-1}     &amp;amp; a^1_n     &amp;amp; 1 &amp;amp; 0      &amp;amp; \cdots &amp;amp; 0 &amp;amp; 0 \\&#xA;      &amp;amp; a^2_2 &amp;amp; \cdots &amp;amp; a^2_{n-1}     &amp;amp; a^2_n     &amp;amp;   &amp;amp; 1      &amp;amp; \cdots &amp;amp; 0 &amp;amp; 0  \\&#xA;      &amp;amp;       &amp;amp; \ddots &amp;amp; \vdots        &amp;amp; \vdots    &amp;amp;   &amp;amp;        &amp;amp; \ddots &amp;amp; \vdots  &amp;amp; \vdots \\&#xA;      &amp;amp;       &amp;amp;        &amp;amp; a^{n-1}_{n-1} &amp;amp; a^{n-1}_n &amp;amp;   &amp;amp;        &amp;amp;        &amp;amp; 1 &amp;amp; 0 \\&#xA;      &amp;amp;       &amp;amp;        &amp;amp;               &amp;amp; a^n_n     &amp;amp;   &amp;amp;        &amp;amp;        &amp;amp;   &amp;amp; 1&lt;br&gt;&#xA;\end{pmatrix}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, what happens when you do back substitution starting with $a^n_n$ and then continuing with $a^{n-1}_{n-1}$...?&lt;/p&gt;&#xA;" OwnerUserId="664" LastActivityDate="2010-09-17T09:07:27.647" />
  <row Id="4844" PostTypeId="1" CreationDate="2010-09-17T09:24:22.037" Score="10" ViewCount="224" Body="&lt;p&gt;&lt;strong&gt;Problem&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$N$ point charges are distributed in the unit ball in $\mathbb{R}^k$, $k=2,3$. Given locations of the particles $x_1,\ldots,x_N$ the potential energy is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$E=\sum_{j=1}^{N-1}\sum_{k=j+1}^N |x_j-x_k|^{-1}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $|x_j-x_k|$ is Euclidean distance between $x_j$ and $x_k$. I'm interested in both the minimal value of $E$ over all possible locations of the particles in the unit ball and what this configuration looks like. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;On the Unit Interval&lt;/strong&gt;&#xA;For $k=1$ the $N$ charges are distributed on the interval $[-1,1]$ according to the roots of the $N+1$th Chebshev polynomial. See: &lt;a href=&quot;http://en.wikipedia.org/wiki/Chebyshev_polynomials#Roots_and_extrema&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Chebyshev_polynomials#Roots_and_extrema&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1794" LastEditorUserId="550" LastEditorDisplayName="" LastEditDate="2010-09-28T15:33:49.910" LastActivityDate="2010-10-18T02:40:32.613" Title="Distributions of point charges" Tags="&lt;numerical-methods&gt;&lt;physics&gt;" AnswerCount="4" CommentCount="3" FavoriteCount="2" />
  <row Id="4845" PostTypeId="1" AcceptedAnswerId="4847" CreationDate="2010-09-17T10:46:40.253" Score="2" ViewCount="206" Body="&lt;p&gt;This paper &lt;a href=&quot;http://arxiv.org/pdf/0911.1933&quot; rel=&quot;nofollow&quot;&gt;arxiv.org/pdf/0911.1933&lt;/a&gt; discusses, regarding the irrationality of certain trigonometric functions. Recently, i encountered this problem which says states the given function, $$ f(n)=\frac{1}{\pi} \arccos{\frac{1}{\sqrt{n}}}$$ is irrational for every odd $n \geq 3$. But i couldn't find the proof anywhere. Can anyone provide me with the proof.&lt;/p&gt;&#xA;" LastActivityDate="2011-05-16T04:05:42.487" Title="Irrationality of $ \frac{1}{\pi} \arccos{\frac{1}{\sqrt{n}}}$" Tags="&lt;real-analysis&gt;&lt;analysis&gt;" AnswerCount="2" FavoriteCount="3" />
  <row Id="4846" PostTypeId="1" AcceptedAnswerId="4856" CreationDate="2010-09-17T11:02:26.057" Score="5" ViewCount="705" Body="&lt;p&gt;Everyone in this community i think would be familiar with International Mathematical Olympiad, which is an International Mathematics Competition held for high school students, with many countries participating from around the world. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whats interesting to note is many of the IMO participants have gone to win the Fields Medal. Notable personalities would be &quot;Terence Tao (2006)&quot;, &quot;Ngo Bau Chau (2010)&quot;, Grigori Perelman..etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to know, what advantage does an IMO student possess over a normal student in terms of Mathematical Research. Does, the IMO competition help the student in becoming a good research Mathematician. &lt;/p&gt;&#xA;" LastActivityDate="2010-09-17T17:46:17.557" CommunityOwnedDate="2010-09-17T11:02:26.057" Title="Advantages of IMO students in Mathematical Research" Tags="&lt;soft-question&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="4" />
  <row Id="4847" PostTypeId="2" ParentId="4845" CreationDate="2010-09-17T11:05:12.800" Score="8" ViewCount="" Body="&lt;p&gt;You could find the proof in &lt;a href=&quot;http://books.google.com/books?id=CuTpgZaLahkC&amp;amp;pg=PA40&quot; rel=&quot;nofollow&quot;&gt;this book&lt;/a&gt;: &lt;i&gt;Proofs from the Book&lt;/i&gt; by Martin Aigner, Günter M. Ziegler, Karl H. Hofmann, pages 40-41&lt;/p&gt;&#xA;" OwnerUserId="823" LastActivityDate="2010-09-17T11:05:12.800" CommunityOwnedDate="2010-09-17T11:05:12.800" CommentCount="4" />
  <row Id="4848" PostTypeId="2" ParentId="213" CreationDate="2010-09-17T11:37:12.163" Score="0" ViewCount="" Body="&lt;p&gt;Not unpopular, but I wish I had studied the theory of &lt;strong&gt;Rings and Fields&lt;/strong&gt;, and basic &lt;strong&gt;Topology&lt;/strong&gt; earlier, because in my opinion both these branches appear in many interesting subjects studied relatively early during one's undergraduate studies.&#xA;For example, the whole concept of a minimal polynomial of a linear transformation $T$ is more intuitive (at least for me) when viewed as the generator of the ideal of polynomials such that $P(t)=0$.&#xA;Metric Spaces (studied in the introductory Topology course at my university) appear as early as in calculus, and are generally a basis to many definitions there. Also, many algebraic structures can be viewed as topologies, which can sometimes give a new insight or assist in proofs (a favourite of mine is a topological proof of the infinity of the primes, which can be found in &lt;em&gt;Proofs from the Book&lt;/em&gt;).&lt;/p&gt;&#xA;" OwnerUserId="766" LastActivityDate="2010-09-17T11:37:12.163" CommunityOwnedDate="2010-09-17T11:37:12.163" CommentCount="1" />
  <row Id="4849" PostTypeId="2" ParentId="4846" CreationDate="2010-09-17T11:46:22.357" Score="22" ViewCount="" Body="&lt;p&gt;I would say that olympiads build some, but far from all, of the skills needed to excel at mathematical research. I'd compare it to running 100 meters versus playing soccer. Usain Bolt is probably a better soccer player than the vast majority of the population, because he could outsprint anyone and because he's generally in fantastic shape. But that doesn't mean he's going to be able to play on a top team. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Being a successful researcher requires the ability to learn new fields of mathematics, and develop ways of thinking about them that others haven't. It requires the discipline to spend months or years returning to a problem and trying new angles on it. It requires, or at least is strongly aided by, the ability to communicate and &quot;sell&quot; one's results, in writing and in talks. It requires the ability to write good definitions, that will be useful and cover the boundary cases correctly. It requires the ability to form an intelligent guess as to which unproven statements are true and which are false. It requires the ability to hold a complex argument in one's head and play with it. It requires, or at least is strongly aided by, the ability to find clever technical arguments.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would say that olympiads are very helpful in developing the last skill, somewhat helpful in developing the fifth and sixth, and not at all in developing the first four. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I definitely, at some points in my research, find myself needing lemmas which would be fair to put on an IMO or a Putnam exam. And when that I happens I feel myself relaxing, because I know I can do that. But I also spend a lot of my time trying to learn how to think about a subject, or figuring out what to prove, or trying to figure out how broadly a phenomenon holds. And those are not skills which I found olympiad training helpful in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the record, I was very competitive in my High School years, but not a world level competitor; I was the first alternate to the US team in 1998 and, during my senior year, I regularly came in somewhere in the top 10 spots in national contests. I've heard similar views expressed by Kiran Kedlaya and Ravi Vakil, who were much more impressive competitors.&lt;/p&gt;&#xA;" OwnerUserId="448" LastActivityDate="2010-09-17T11:46:22.357" CommunityOwnedDate="2010-09-17T11:46:22.357" />
  <row Id="4850" PostTypeId="2" ParentId="4846" CreationDate="2010-09-17T12:02:29.173" Score="9" ViewCount="" Body="&lt;p&gt;Keep in mind that what you have here is a correlation, not a causation. While doubtlessly Olimpiad Training would help develop some skills necessary for research, I think it is likely that many of the strongest mathematicians participate in these competitions when they are in high school and go on to do research later.&lt;/p&gt;&#xA;" OwnerUserId="796" LastActivityDate="2010-09-17T12:02:29.173" CommunityOwnedDate="2010-09-17T12:02:29.173" CommentCount="9" />
  <row Id="4851" PostTypeId="1" AcceptedAnswerId="5208" CreationDate="2010-09-17T12:25:36.453" Score="2" ViewCount="88" Body="&lt;p&gt;Let $p : E \rightarrow B$ be a cover s.t. Aut(E) acts transitively on $p^{-1}(b)$ for some fix $b \in B$. Then $E/Aut(E)$ is homeomorphic to $B$ where $Aut(E) \subset Cov(E,E)$. The problem is that I don't even understand how the map E/Aut(E) -&gt; B is defined, would someone please give me this definition? Apparently it is &quot;clear&quot; but I fail to see this. &lt;/p&gt;&#xA;" OwnerUserId="1907" LastActivityDate="2010-09-22T15:07:40.383" Title="Cover E -&gt; B gives a homeomorphism E/Aut(E) -&gt; B" Tags="&lt;topology&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" />
  <row Id="4853" PostTypeId="2" ParentId="250" CreationDate="2010-09-17T12:59:33.820" Score="5" ViewCount="" Body="&lt;p&gt;One that Feynman would have rejected: There are always two antipodal points on the Earth that are the same temperature. (This follows from the fact that a continuous scalar field on a circle has a diameter whose endpoints have the same value)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One that Feynman might have preferred:&#xA;Draw a triangle. Draw a circle through the midpoints of the circle's sides. This circle also passes through the the foot of each altitude and the midpoint of the line segment from the orthcentre to each vertex. &lt;a href=&quot;http://en.wikipedia.org/wiki/Nine-point_circle&quot; rel=&quot;nofollow&quot;&gt;The nine point circle&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="172" LastActivityDate="2010-09-17T12:59:33.820" CommunityOwnedDate="2010-09-17T12:59:33.820" />
  <row Id="4854" PostTypeId="1" AcceptedAnswerId="4877" CreationDate="2010-09-17T13:03:03.767" Score="6" ViewCount="144" Body="&lt;p&gt;I am obsessed with &lt;a href=&quot;http://en.wikipedia.org/wiki/Nine-point_circle&quot; rel=&quot;nofollow&quot;&gt;the nine point circle&lt;/a&gt;. I was thinking, is there a generalisation to aribtrary tetrahedra and spheres? What about higher dimensions? For each face of the tetrahedron, there is a nine point circle. Do these circles all lie on a sphere?&lt;/p&gt;&#xA;" OwnerUserId="172" LastActivityDate="2010-09-17T19:00:29.907" Title="Does the nine point circle generalise to some theorem about n-spheres and n-simplices?" Tags="&lt;geometry&gt;&lt;euclidean-geometry&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="4855" PostTypeId="2" ParentId="4854" CreationDate="2010-09-17T13:11:08.130" Score="4" ViewCount="" Body="&lt;p&gt;Take a look at:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.springerlink.com/content/mj55u41710w625j8/&quot; rel=&quot;nofollow&quot;&gt;http://www.springerlink.com/content/mj55u41710w625j8/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Tetrahedron&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Tetrahedron&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is also this paper that deals with sphere analogues:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.jstor.org/pss/3617834&quot; rel=&quot;nofollow&quot;&gt;http://www.jstor.org/pss/3617834&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1369" LastEditorUserId="1369" LastEditorDisplayName="" LastEditDate="2010-09-17T13:28:58.697" LastActivityDate="2010-09-17T13:28:58.697" />
  <row Id="4856" PostTypeId="2" ParentId="4846" CreationDate="2010-09-17T14:04:43.390" Score="21" ViewCount="" Body="&lt;p&gt;Training for competitions will help you solve competition problems - that's all. These are not the sort of problems that one typically struggles with later as a professional mathematician - for many different reasons. First, and foremost, the problems that one typically faces at research level are not problems carefully crafted so that they may be solved in certain time limits. Indeed, for problems encountered &quot;in the wild&quot;, one often does not have any inkling whether or not they are true. So often one works simultaneously looking for counterexamples and proofs. Often solutions require discovering fundamentally new techniques - as opposed to competition problems - which typically may be solved by employing variations of methods from a standard toolbox of &quot;tricks&quot;. Moreover, there is no artificial time limit constraint on solving problems in the wild. Some research level problems require years of work and immense persistence (e.g. Wiles proof of FLT). Those are not skills that can be measured by competitions. While competitions might be used to encourage students, they should never be used to discourage them. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a great diversity among mathematicians. Some are prolific problem solvers (e.g. Erdos) and others are grand theory builders (e.g. Grothendieck). Most are somewhere between these extremes. All can make significant, surprising contributions to mathematics. History is a good teacher here. One can learn from the masters not only from their mathematics, but also from the way that they learned their mathematics. You will find much interesting advice in the (auto-)biographies of eminent mathematicians. Time spent perusing such may prove much more rewarding later in your career than time spent learning yet another competition trick. Strive to aim for a proper balance of specialization and generalization in your studies.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-09-17T17:46:17.557" LastActivityDate="2010-09-17T17:46:17.557" CommunityOwnedDate="2010-09-17T14:04:43.390" CommentCount="2" />
  <row Id="4857" PostTypeId="2" ParentId="4834" CreationDate="2010-09-17T14:20:50.137" Score="3" ViewCount="" Body="&lt;p&gt;Seems like you need&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{i=1}^{s} s^i$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a geometric progression with the sum&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \frac{s(s^{s} -1)}{s-1} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $s=3$ this gives $\frac{3(3^3 -1)}{3-1} = 39$.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-09-17T14:20:50.137" CommentCount="1" />
  <row Id="4858" PostTypeId="1" CreationDate="2010-09-17T14:22:03.873" Score="1" ViewCount="115" Body="&lt;p&gt;Let $S$ be a (n-1)-submanifold of a n-manifold $M$ and that be a submanifold of (n+1)-manifold V. (All of the manifolds are assumed orientable) Let $V$ carry a metric of signature $(1,n)$. Using terminology as in Einstein's Theory let $n$ be a future directed unit normal to $M$ and $m$ be an unit normal to $S$ pointing outward which is tangent to $M$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $h$ be a metric on $S$ , $g'$ be a metric on $M$ and $g$ be a metric on $V$ such that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$g_{\mu \nu}' = g_{\mu \nu} + n_\mu n_\nu$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$h_{\mu \nu} = g_{\mu \nu} + n_\mu n_\nu - m_\mu m _\nu$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(In each of the cases $\mu$ and $\nu$ run over the dimensions of the lowest dimensional manifold whose metric is involved) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to know as to what is the motivation for the above choice of metrics that is often made. What makes the above &quot;natural&quot;? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To pull-back a metric on a submanifold, one needs to fix an embedding first. &#xA;Here what is the implicit embedding? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;{Now the LaTeX is compiling} &lt;/p&gt;&#xA;" OwnerUserId="965" LastEditorUserId="965" LastEditorDisplayName="" LastEditDate="2010-09-18T11:29:27.783" LastActivityDate="2010-09-18T11:29:27.783" Title="A particular method of pulling back a metric on a submanifold" Tags="&lt;differential-geometry&gt;" CommentCount="6" />
  <row Id="4859" PostTypeId="1" AcceptedAnswerId="4864" CreationDate="2010-09-17T15:21:25.000" Score="5" ViewCount="128" Body="&lt;p&gt;Let $C$ be a small pre-additive category. Let $R(C)$ denote its category ring, that is,&#xA;$$&#xA;R(C)=\bigoplus_{a,b\in \mathrm{Ob}(C)} C(a,b)&#xA;$$&#xA;as Abelian group, where the direct sum runs over all object $a$, $b$ of $C$. The multiplication in $R(C)$ is given by composition of composable morphisms and 0 for uncomposable morphisms (extended by bilinearity).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This constuction is functorial: An additive functor $C\to D$ between small pre-additive categories induces a ring homomorphism $R(C)\to R(D)$ between the corresponding category rings in a canocical way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence we have a functor $R$ from the category of small pre-additive categories (with additive functors as morhisms) to the category of rings (with ring homomorphisms as morphisms).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the category of small pre-additive categories has a 2-categorical structure given by natural transformations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence my question: Does this 2-categorical structure have a counterpart in the category of rings? More precisely, &lt;strong&gt;is there a natural notion of 2-morphisms between ring homomorphisms turning $R$ into a 2-functor?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="367" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-09-17T20:25:02.957" LastActivityDate="2010-09-19T09:02:35.137" Title="Can we turn the functor &quot;category ring&quot; into a 2-functor in a natural way?" Tags="&lt;abstract-algebra&gt;&lt;category-theory&gt;" AnswerCount="2" CommentCount="7" FavoriteCount="2" />
  <row Id="4860" PostTypeId="2" ParentId="4841" CreationDate="2010-09-17T15:44:04.247" Score="3" ViewCount="" Body="&lt;p&gt;Another method is as follows. An invertible upper triangular matrix has the form&#xA;$A=D(I+N)$ where $D$ is diagonal (with the same diagonal entries as $A$)&#xA;and $N$ is upper triangular with zero diagonal. Then $N^n=0$ where $A$&#xA;is $n$ by $n$. Both $D$ and $I+N$ have upper triangular inverses:&#xA;$D^{-1}$ is diagonal, and $(I+N)^{-1}=I-N+N^2-\cdots +(-1)^{n-1}N^{n-1}$.&#xA;So $A^{-1}=(I+N)^{-1}D^{-1}$ is upper triangular.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-09-17T15:44:04.247" CommentCount="1" />
  <row Id="4861" PostTypeId="2" ParentId="4808" CreationDate="2010-09-17T15:46:08.957" Score="2" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Assuming that such a closed curve exists&lt;/strong&gt;, it is fairly straightforward to show that it must be a circle. For example, check out this &lt;a href=&quot;http://www.cut-the-knot.org/do_you_know/isoperimetric.shtml&quot; rel=&quot;nofollow&quot;&gt;page&lt;/a&gt; for a proof.  It is interesting that the existence hypothesis is the &lt;em&gt;hard&lt;/em&gt; part of the proof.  &lt;/p&gt;&#xA;" OwnerUserId="964" LastActivityDate="2010-09-17T15:46:08.957" />
  <row Id="4862" PostTypeId="1" CreationDate="2010-09-17T16:19:58.933" Score="3" ViewCount="699" Body="&lt;p&gt;Ok this is a bit of an unanswerable question, but hopefully someone will answer. As I have been going through college &amp;amp; high school there has been a kind of &quot;path&quot; through which you learn math. For example I went through basic math, prealgebra, algebra, precalc, calculus 1-3. My question is where to go after calculus three?(which is where I am at now btw). It looks to me like math really starts branching after calculus. I worried that I'll end up in classes that I can't fully understand because I didn't have the prerequisite knowledge. Do I go to linear algebra, real analysis, number theory, another branch ...? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm working towards a degree in mathematics and would like to get the most out of every class, So if anybody would tell me what worked/didn't work for them I would be grateful.&lt;/p&gt;&#xA;" OwnerUserId="1582" LastActivityDate="2010-09-22T20:21:51.227" CommunityOwnedDate="2010-09-17T17:15:13.250" Title="Where to go after calculus?" Tags="&lt;soft-question&gt;&lt;education&gt;&lt;learning&gt;" AnswerCount="8" CommentCount="18" FavoriteCount="2" />
  <row Id="4863" PostTypeId="2" ParentId="4862" CreationDate="2010-09-17T16:31:33.200" Score="1" ViewCount="" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;One of the disappointments experienced by most mathematics students is that they never get a course in mathematics. They get courses in calculus, algebra, topology, and so on, but the divison of labor in teaching seems to prevent these different topics from being combined into a whole. In fact, some of the most important and natural questions are stifled because they wall on the wrong side of topic boundary lines. ... Thus, if students are to feel they really know mathematics by the time they graduate, there is a need to unify the subject.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;from Stillwell - Mathematics and Its History&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;So you could take this time to study some of the history of mathematics. Another thing which is useful is to study some formal logic (the stuff that makes proofs) - since (at least in my experience) this is not taught very much in classes.&lt;/p&gt;&#xA;" LastActivityDate="2010-09-17T16:31:33.200" CommunityOwnedDate="2010-09-17T17:15:13.250" CommentCount="1" />
  <row Id="4864" PostTypeId="2" ParentId="4859" CreationDate="2010-09-17T16:33:00.127" Score="2" ViewCount="" Body="&lt;p&gt;Rings are precisely those pre-additive categories with exactly one object. Thus the category of rings (noncommutative here) is a $2$-category. If $f,g : R \to S$ are $1$-morphisms, then a $2$-morphism $f \to g$ is an element $s \in S$ such that for all $r \in R$, we have $s f(r) = g(r) s$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So take a natural transformation $\eta : F \to G$ between additive functors $F,G : C \to D$. Then we may take $s = \sum_{x \in C} \eta(x)$ as an element of $R(D)$. But wait, this sum does not have to be finite. Then let's define $R(-)$ simply as the product and not the direct sum. This causes other problems, see the answer of Agusti Roig.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the desired equation $s F(r) = G(r) s$ for some morphism $r : x \to y$ in $C$ is equivalent to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sum_{u : F(u)=F(x)} \eta(u) F(r) = \sum_{v : G(v)=G(x)} G(r) \eta(v)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This seems to be true only if $F$ and $G$ are injective on objects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So my answer would be: No, unfortunately $R$ cannot be made into a $2$-functor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But somehow, $R$ &lt;em&gt;should&lt;/em&gt; be a $2$-functor, and perhaps we can modify the whole setting a but, so that it works.&lt;/p&gt;&#xA;" OwnerUserId="1650" LastEditorUserId="1650" LastEditorDisplayName="" LastEditDate="2010-09-19T09:02:35.137" LastActivityDate="2010-09-19T09:02:35.137" />
  <row Id="4865" PostTypeId="2" ParentId="4862" CreationDate="2010-09-17T16:44:57.550" Score="4" ViewCount="" Body="&lt;p&gt;A good History of Math course would probably be enjoyable, and give you a good idea of what things are and where they lie.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the mathematical point of view, Linear algebra is often a good &quot;first advanced mathematics course&quot;, a good jumping-off point: it is still concrete enough that you won't get lost in a sea of abstraction (a possible issue with abstract algebra depending on how it is taught), cover entirely new ideas (&quot;advanced calculus&quot; can feel like you are just re-treading the same ground you already know, and depending on the specific topics analysis might also feel like it), but it should make you work through proofs and concepts in a way with which you probably have not done so far. In addition, linear methods will show up all over the place later on, so it would prove useful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In that same vein, Number Theory can be a really good &quot;first abstract course in mathematics&quot;, while sticking close to things you are very familiar with (the integers and rationals) while also probably delivering some exciting surprises. It often surprises a lot of people just how &lt;em&gt;much&lt;/em&gt; of mathematics arises out of number theory (complex analysis and abstract algebra, to name just two). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to stick to the applied side, differential equations is a good place to go as well. Linear algebra would be useful there, though.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, I would suggest linear algebra or number theory first (if you can also get a good history of math course, do that as well), then decide if you want to go towards abstraction (in which case, head to abstract algebra, mathematical analysis, or whichever of linear algebra or number theory you did not take) or more towards applications (differential equations, a good advanced probability/statistics course, or a discrete mathematics course). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is indeed the case that mathematics starts branching out, but some of the most interesting things happen where the branches meet; it would be ideal to be able to take a good one semester or one year sequence in the major areas (analysis, algebra, differential equations, topology, logic/set theory, number theory), then go on to more advanced courses in whichever area(s) you find interesting. But the truth is that this is very hard to do: not only would such a wide choice not be available except in the largest universities, but it would also mean a &lt;em&gt;lot&lt;/em&gt; of your time. I did my undergraduate in Mexico, where all I did in college was mathematics courses, and it basically took six semesters before that had been covered (in addition to the calculus sequence, a linear algebra sequence, an abstract algebra sequence, a mathematical analysis sequence, differential equations, discrete mathematics, complex analysis, probability and statistics, plus some other stuff to &quot;fill in the corners&quot;; it would be barely possible to do it in two years if you are not taking the calculus sequence, but not if you are also taking other coursework as you would be in most institutions in the United States).&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-09-17T16:44:57.550" CommunityOwnedDate="2010-09-17T17:15:13.250" CommentCount="5" />
  <row Id="4866" PostTypeId="2" ParentId="4862" CreationDate="2010-09-17T16:47:11.060" Score="2" ViewCount="" Body="&lt;p&gt;Large parts of mathematics as it has evolved in the 20th century are quite independent of calculus.  Game theory, graph theory, combinatorics, discrete mathematics, etc. are often courses which are taught without a specific Calculus prerequisite. The order in which one studies mathematics is to some extent related to what one's goals and &quot;natural talents&quot; are. For example, there are parts of geometry which don't require calculus (directly) while there are other parts of geometry where using Calculus methods are very central. If one majors in mathematics typically there is a &quot;recommended route&quot; of courses at the school one is studying at. &lt;/p&gt;&#xA;" OwnerUserId="1369" LastActivityDate="2010-09-17T16:47:11.060" CommunityOwnedDate="2010-09-17T17:15:13.250" CommentCount="2" />
  <row Id="4867" PostTypeId="2" ParentId="4862" CreationDate="2010-09-17T17:13:43.583" Score="2" ViewCount="" Body="&lt;p&gt;If you're interested in building on your calculus knowledge in a way that builds links to many other areas of mathematics, a natural direction to go would be to take some linear algebra and a differential geometry course.  From there you could take some analysis and then a manifolds course.  At that point the subject of Lie groups acts as something of a unifying picture for all of the above said courses -- it fits everything together in a very pleasant way and gives you many pleasant connections between group theory, linear algebra and analysis.  Stillwell's book &quot;Naive Lie Theory&quot; is something you could pick up right now and start reading. When reading the book you'll likely be able to spot what you're comfortable with and what you aren't, and go from there. &lt;/p&gt;&#xA;" OwnerUserId="642" LastActivityDate="2010-09-17T17:13:43.583" CommunityOwnedDate="2010-09-17T17:15:13.250" />
  <row Id="4868" PostTypeId="1" AcceptedAnswerId="4892" CreationDate="2010-09-17T17:15:27.010" Score="4" ViewCount="302" Body="&lt;p&gt;Let $A \subset \mathbb{C} $ be an open set containing the closed unit disc.&#xA;Let $f$ be an analytic function from $A$ to $\mathbb{C}$ such that $|f(z)|=1$ if $|z|=1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does it follow that $f(z) = a z^{n} \frac{cz^{m}-b}{1-cz^{m}\bar{b}} $ for some $a,b,c \in \mathbb{C}$ s.t. $|c|=1$ $|a|=1$, $|b|&amp;lt;1$ and some $n,m\ge 0 $?&lt;/p&gt;&#xA;" OwnerUserId="832" LastActivityDate="2010-09-19T05:47:47.120" Title="Mapping that takes unit circle to unit circle" Tags="&lt;complex-analysis&gt;" AnswerCount="2" CommentCount="9" />
  <row Id="4869" PostTypeId="1" AcceptedAnswerId="4938" CreationDate="2010-09-17T17:16:59.183" Score="6" ViewCount="513" Body="&lt;p&gt;In real analysis, we can define a line integral, but we also define (earlier) the regular definite integral.&#xA;Why is it that in complex analysis we are interested only in a line integral?&lt;/p&gt;&#xA;" OwnerUserId="766" LastActivityDate="2010-10-20T23:53:13.067" Title="Why is an integral of a complex function defined as a line integral?" Tags="&lt;complex-analysis&gt;" AnswerCount="2" CommentCount="8" FavoriteCount="2" />
  <row Id="4870" PostTypeId="1" AcceptedAnswerId="4873" CreationDate="2010-09-17T17:22:34.210" Score="1" ViewCount="123" Body="&lt;p&gt;Let's assume X,Y are arbitrary sets, and f is a function between them. What must those sets, and f, satisfy, in order for us to be able to define an integral of f in a way that &quot;makes sense&quot;?&lt;/p&gt;&#xA;" OwnerUserId="766" LastActivityDate="2010-09-17T21:36:19.200" Title="$f:X \rightarrow Y$. What are the conditions on $f$,$X$,$Y$ in order for an integral of $f$ to be defined?" Tags="&lt;analysis&gt;&lt;functions&gt;&lt;integral&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" />
  <row Id="4871" PostTypeId="2" ParentId="4862" CreationDate="2010-09-17T17:26:11.257" Score="0" ViewCount="" Body="&lt;p&gt;I would strongly recommend you to read the book, &quot;Differential Calculus For Normed Linear Banach Spaces&quot; by Kalyan Mukherjee. Its published by the Hindustan Book Agency. I think it is one of those rare wonderful books which bridges the gap between calculus and manifold theory. It has beautiful sections on differentiating and Taylor expanding and integrating functions between arbitrary finite dimensional euclidean planes. He masterfully uses all this to motivate Lie Groups and calculates tangents to curves in Lie Groups. He further explains subtle things like differentiating matrix multiplication and determinant maps. &lt;/p&gt;&#xA;" OwnerUserId="965" LastActivityDate="2010-09-17T17:26:11.257" CommunityOwnedDate="2010-09-17T17:26:11.257" />
  <row Id="4872" PostTypeId="1" AcceptedAnswerId="4879" CreationDate="2010-09-17T18:26:14.560" Score="2" ViewCount="251" Body="&lt;p&gt;I've read through the elementary proof of why there exists a solution $x$ to $x^2\equiv -1\pmod p$ iff $p\equiv 1 \pmod 4$ for $p$ an odd prime. Is there a group theory generalization for this fact as well? &lt;/p&gt;&#xA;" OwnerUserId="1583" LastActivityDate="2010-09-18T00:14:11.493" Title="Group theory proof of existence of a solution to $x^2\equiv -1\pmod p$ iff $p\equiv 1 \pmod 4$" Tags="&lt;number-theory&gt;&lt;algebraic-number-theory&gt;" AnswerCount="3" FavoriteCount="1" />
  <row Id="4873" PostTypeId="2" ParentId="4870" CreationDate="2010-09-17T18:39:34.167" Score="5" ViewCount="" Body="&lt;p&gt;First of all, the question is not &quot;what conditions must $X$ and $Y$ satisfy&quot; but &quot;what &lt;em&gt;structures&lt;/em&gt; must $X$ and $Y$ have?&quot;  Second of all, there is not one most general definition of the integral; depending on what you consider an &quot;integral,&quot; there are several definitions which generalize in different directions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Probably the most basic requirement is that the integral is a linear operator.  For this to make sense, as Akhil says, $Y$ needs some kind of linear structure.  If $Y$ is an &lt;a href=&quot;http://en.wikipedia.org/wiki/Abelian_group&quot; rel=&quot;nofollow&quot;&gt;abelian group&lt;/a&gt; and $X$ a finite set, then a perfectly sensible definition of the integral of $f$ is $ \sum_{x \in X} f(x)$, which is in fact a linear operator from the abelian group of functions $X \to Y$ to $Y$ (in fact the unique one, up to scaling, which is invariant under the symmetry group of $X$).  But of course the finite case is not very interesting.  (Note that in order for the sum to make sense above without an ordering on $X$ it is necessary that $Y$ be abelian.)  More generally one can stipulate that the only &quot;integrable functions&quot; are the finitely supported ones, in which case this definition alwaysm makes sense.  But again, a pretty boring definition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the infinite case, for example if $X = \mathbb{N}$, we would like to define the integral as some kind of limit, for example maybe $\lim_{n \to \infty}  \sum_{k=1}^{n} f(k)$, because we would intuitively like to &quot;sum over $X$&quot; (but can't, at least not directly).  For this to make sense $Y$ needs a notion of convergence and $f$ needs to satisfy a &quot;smallness&quot; condition.  At a bare minimum, then, it needs to be a topological abelian group (say complete with respect to a norm), but there are some additional technicalities and I think they are usually resolved by requiring that $Y$ is a &lt;a href=&quot;http://en.wikipedia.org/wiki/Banach_space&quot; rel=&quot;nofollow&quot;&gt;Banach space&lt;/a&gt; over a &lt;a href=&quot;http://en.wikipedia.org/wiki/Complete_field&quot; rel=&quot;nofollow&quot;&gt;complete field&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another feature of the infinite case is that, if indicator functions of &quot;small&quot; subsets of $X$ are to be integrable, then a notion of integral on functions out of $X$ implies a notion of &quot;size&quot; or measure on $X$.  The usual way people handle this is to require that $X$ be a &lt;a href=&quot;http://en.wikipedia.org/wiki/Measure_(mathematics)&quot; rel=&quot;nofollow&quot;&gt;measure space&lt;/a&gt;, in which case (together with the structures on $Y$ described above) the usual construction of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Lebesgue_integration&quot; rel=&quot;nofollow&quot;&gt;Lebesgue integral&lt;/a&gt; goes through.&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-09-17T21:36:19.200" LastActivityDate="2010-09-17T21:36:19.200" />
  <row Id="4877" PostTypeId="2" ParentId="4854" CreationDate="2010-09-17T19:00:29.907" Score="5" ViewCount="" Body="&lt;p&gt;An &lt;em&gt;orthocentric&lt;/em&gt; tetrahedron is one for which the altitudes from the vertices to the opposite faces are concurrent (this is not true for all tetrahedrons).  For an orthocentric tetrahedron, there exists a sphere (the 24-point sphere) that intersects each face of the tetrahedron in its 9-point circle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See also the summary of a talk by Steve West (&quot;Discovering Theorems Using Cabri 3-D&quot;) in &lt;a href=&quot;http://mmcchicago.org/dinners08-09/November08.pdf&quot; rel=&quot;nofollow&quot;&gt;the November 2008 issue of &lt;em&gt;Points &amp;amp; Angles&lt;/em&gt; (PDF)&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-09-17T19:00:29.907" />
  <row Id="4878" PostTypeId="2" ParentId="4872" CreationDate="2010-09-17T19:11:11.813" Score="3" ViewCount="" Body="&lt;p&gt;Presumably you refer to the following proof, which contradicts Lagrange's or Fermat's little Theorem&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose $\rm\; p = 4n+3\;$ is prime. Then $\rm\ x^2 = -1 \;\Rightarrow\; x^{p-1} = (x^2)^{2n+1} = -1 \pmod{p}\;\Rightarrow\Leftarrow$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A finite field of order $\rm\:4n+3\:$ has subgroup of squares of order $\rm\:2n+1\:$. But one easily shows that in any group of odd order $\rm\; 2n+1\:$ the equation $\rm\; x^2 = a\;$ has the solution $\rm\:  a^{n+1}\;$ (due to Lagrange in 1769). Specializing $\rm\; a = -1\:$ we conclude that $\rm\; x^2 = -1\:$ has no roots. On the other hand, the equation always has a root in a finite field of order $\rm 4n+1$ by the case $\rm k=4$ of &lt;a href=&quot;http://www.math.wisc.edu/~jensen/frobtalk.pdf&quot; rel=&quot;nofollow&quot;&gt;Frobenius's  theorem (1907)&lt;/a&gt; - which says that if $\rm k$ divides the order of a finite group $\rm G$ then $\rm k$ also divides the number of $\rm k$'th roots of $1$ in $\rm G\:$.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-09-17T21:04:49.020" LastActivityDate="2010-09-17T21:04:49.020" />
  <row Id="4879" PostTypeId="2" ParentId="4872" CreationDate="2010-09-17T19:18:32.203" Score="6" ViewCount="" Body="&lt;p&gt;I would sketch the group theoretic fact as follows: Let $G$ be a finite abelian group. Then the following are equivalent:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) For every $d$, there are at most $d$ solutions to $g^d=1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) $G$ is cyclic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(3) For $n$ dividing $|G|$, there are precisely $\phi(n)$ elements of order $n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only place that I know of where hypothesis (1) shows up naturally is considering the multiplicative groups of finite fields. From this we get:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is an element of order $d$ in $\mathbb{F}_{p^r}$ if and only if $d| p^r-1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure there is any more to say.&lt;/p&gt;&#xA;" OwnerUserId="448" LastActivityDate="2010-09-17T19:18:32.203" />
  <row Id="4880" PostTypeId="2" ParentId="4725" CreationDate="2010-09-17T19:29:53.720" Score="1" ViewCount="" Body="&lt;p&gt;This is the problem of &lt;em&gt;confidence intervals&lt;/em&gt; in statistics.  Your question is about the original problem of that type, the  &quot;confidence interval for a proportion in binomial sampling&quot; (wikipedia &lt;a href=&quot;http://en.wikipedia.org/wiki/Binomial_proportion_confidence_interval&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Binomial_proportion_confidence_interval&lt;/a&gt; ) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The asymptotic theory, when $Np(1-p)$ is very large, is described by the Central Limit Theorem, which says that $n = pN + A\sqrt{Np(1-p)}$ where the probability distribution of $A$ is the famous Gaussian (normal, &quot;bell shaped&quot;) curve.  Qualitatively, fluctuations of $n$ around the expected value are usually within a small factor of $\sqrt{Np(1-p)}$.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For limited data, the story is more complicated, with no perfect solution, many proposals, and a large literature.&lt;/p&gt;&#xA;" OwnerUserId="467" LastActivityDate="2010-09-17T19:29:53.720" />
  <row Id="4881" PostTypeId="1" AcceptedAnswerId="4887" CreationDate="2010-09-17T19:42:51.613" Score="4" ViewCount="173" Body="&lt;p&gt;Let $T$ be the operator from $L^2(\mathbb R^n)$ to $L^2(\mathbb R^n)$ that is given by $Tf := f * g$ where $g$ is in $L^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do I now find that the spectrum of $T$ is equal to the essential range of $\hat{g}$? How is the spectrum of $T$ related to the invertibility of the operator $G\hat{f} = \hat{f}\hat{g}$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The hat denotes the Fourier transform.&lt;/p&gt;&#xA;" OwnerUserId="1120" LastActivityDate="2011-01-29T11:25:39.050" Title="Spectrum of a convolution operator." Tags="&lt;functional-analysis&gt;&lt;fourier-analysis&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" />
  <row Id="4882" PostTypeId="1" AcceptedAnswerId="4883" CreationDate="2010-09-17T19:42:58.150" Score="1" ViewCount="159" Body="&lt;p&gt;Define a binary step-order set as a finite set S of positive integers s&lt;sub&gt;i&lt;/sub&gt;, i = 0 to n-1, where s&lt;sub&gt;i&lt;/sub&gt; ≡ 2&lt;sup&gt;i&lt;/sup&gt; (mod 2&lt;sup&gt;i+1&lt;/sup&gt;) for each i from 0 to n-1. So, i is the power of 2 appearing in the prime factorization of s&lt;sub&gt;i&lt;/sub&gt;. It’s elementary that binary step-order sets have distinct subset sums.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the least upper bound on the number of pairs of disjoint subsets of an n-element binary step-order set S whose sums differ by 1?&lt;/p&gt;&#xA;" OwnerUserId="1782" LastActivityDate="2010-09-17T20:43:14.167" Title="What is the least upper bound on the number of pairs of disjoint subsets of a binary step-order set S whose sums differ by 1?" Tags="&lt;combinatorics&gt;" AnswerCount="1" />
  <row Id="4883" PostTypeId="2" ParentId="4882" CreationDate="2010-09-17T20:26:01.663" Score="1" ViewCount="" Body="&lt;p&gt;The answer is F&lt;sub&gt;n+1&lt;/sub&gt;, where {F&lt;sub&gt;i&lt;/sub&gt;} is the Fibonacci sequence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lemma – Let S be a binary step-order set with n elements. For any positive integer k, F&lt;sub&gt;n+1&lt;/sub&gt; is an upper bound for the number of pairs of disjoint subsets of S whose sums differ by k, where {F&lt;sub&gt;i&lt;/sub&gt;} is the Fibonacci sequence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proceeding by math induction, first prove the cases n = 1 and n = 2. For n = 1, the only possible pair of disjoint subsets is (S, φ); thus, F&lt;sub&gt;2&lt;/sub&gt; = 1 is an upper bound for the number of pairs of disjoint subsets with sums differing by k. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For n ≥ 2, observe that finding a pair of disjoint subsets whose sums differ by k is equivalent to finding an n-tuple of coefficients &lt;strong&gt;d&lt;/strong&gt; ∊ {-1, 0, 1}&lt;sup&gt;n&lt;/sup&gt; such that ∑ d&lt;sub&gt;i&lt;/sub&gt;s&lt;sub&gt;i&lt;/sub&gt; = k – the elements of the higher-sum subset have coefficients 1, the elements of the lower-sum subset (if any – this may be φ) have coefficients -1, and the elements in neither subset have coefficients 0. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For n = 2, we have s&lt;sub&gt;0&lt;/sub&gt; odd and s&lt;sub&gt;1&lt;/sub&gt; even. So, there are two cases; s&lt;sub&gt;0&lt;/sub&gt; &gt; s&lt;sub&gt;1&lt;/sub&gt; and s&lt;sub&gt;0&lt;/sub&gt; &amp;lt; s&lt;sub&gt;1&lt;/sub&gt;.&#xA;For s&lt;sub&gt;0&lt;/sub&gt; &gt; s&lt;sub&gt;1&lt;/sub&gt;, the only possible coefficient ordered pairs d are (1, 0), (0, 1), (1, -1) and (1, 1); otherwise ∑ d&lt;sub&gt;i&lt;/sub&gt;s&lt;sub&gt;i&lt;/sub&gt; is not positive. Furthermore, these all yield distinct values of ∑ d&lt;sub&gt;i&lt;/sub&gt;s&lt;sub&gt;i&lt;/sub&gt;, because s&lt;sub&gt;0&lt;/sub&gt; + s&lt;sub&gt;1&lt;/sub&gt; &gt; s&lt;sub&gt;0&lt;/sub&gt; &gt; s&lt;sub&gt;1&lt;/sub&gt;, and because s&lt;sub&gt;1&lt;/sub&gt; ≠ s&lt;sub&gt;0&lt;/sub&gt; – s&lt;sub&gt;1&lt;/sub&gt; because the former is even and the latter is odd. So, in this case, 1 is an upper bound for the number of disjoint subset pairs whose sums differ by k.&#xA;For s&lt;sub&gt;1&lt;/sub&gt; &gt; s&lt;sub&gt;0&lt;/sub&gt;, the only possible coefficient ordered pairs d are (1, 0), (0, 1), (-1, 1) and (1, 1); otherwise ∑ d&lt;sub&gt;i&lt;/sub&gt;s&lt;sub&gt;i&lt;/sub&gt; is not positive. We have s&lt;sub&gt;0&lt;/sub&gt; + s&lt;sub&gt;1&lt;/sub&gt; &gt; s&lt;sub&gt;1&lt;/sub&gt; &gt; s&lt;sub&gt;0&lt;/sub&gt;, so the only possible way to have ∑ d&lt;sub&gt;i&lt;/sub&gt;s&lt;sub&gt;i&lt;/sub&gt; = k for two distinct &lt;strong&gt;d&lt;/strong&gt; is if s&lt;sub&gt;0&lt;/sub&gt; = s&lt;sub&gt;1&lt;/sub&gt; – s&lt;sub&gt;0&lt;/sub&gt;; i.e., s&lt;sub&gt;1&lt;/sub&gt; = 2s&lt;sub&gt;0&lt;/sub&gt;.  So, in this case, F&lt;sub&gt;3&lt;/sub&gt; = 2 is an upper bound for this case and the general n = 2 case of the number of disjoint subset pairs whose sums differ by k.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, assume that the lemma is true for all positive integers from 1 through n ≥ 2. Consider an n+1 element binary step-order set S&lt;sub&gt;n+1&lt;/sub&gt;. Let S&lt;sub&gt;n&lt;/sub&gt; be the n-element set obtained by omitting the odd element of S&lt;sub&gt;n+1&lt;/sub&gt; and dividing all the other elements by 2 – clearly S&lt;sub&gt;n&lt;/sub&gt; is a binary step-order set. Let S&lt;sub&gt;n-1&lt;/sub&gt; be the n-1 element set obtained by omitting the odd element of S&lt;sub&gt;n&lt;/sub&gt; and dividing all the other elements by 2 – clearly S&lt;sub&gt;n-1&lt;/sub&gt; is also a binary step-order set. So s&lt;sub&gt;n,i&lt;/sub&gt; = s&lt;sub&gt;n+1,i+1&lt;/sub&gt; / 2 and s&lt;sub&gt;n-1,i&lt;/sub&gt; = s&lt;sub&gt;n,i+1&lt;/sub&gt; / 2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the case where k is even. The coefficient of s&lt;sub&gt;n+1,0&lt;/sub&gt; must be 0 for any &lt;strong&gt;d&lt;/strong&gt; in the solution set; otherwise ∑ d&lt;sub&gt;i&lt;/sub&gt;s&lt;sub&gt;i&lt;/sub&gt; is odd and ≠ k.  Each solution &lt;strong&gt;d&lt;/strong&gt; = (0, d&lt;sub&gt;n+1,1&lt;/sub&gt;, …, d&lt;sub&gt;n+1,n&lt;/sub&gt;) corresponds to a solution d&lt;sub&gt;n&lt;/sub&gt; for k/2 in S&lt;sub&gt;n&lt;/sub&gt; via bijection, where d&lt;sub&gt;n,i&lt;/sub&gt; = d&lt;sub&gt;n+1,i&lt;/sub&gt; for i = 0 through n – 1. So, F&lt;sub&gt;n+1&lt;/sub&gt; is an upper bound for the number of disjoint subset pairs of S&lt;sub&gt;n+1&lt;/sub&gt; whose sums differ by an even k.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the case where k is odd. The coefficient d&lt;sub&gt;n+1,0&lt;/sub&gt; of s&lt;sub&gt;n+1,0&lt;/sub&gt; must be either 1 or -1 for any &lt;strong&gt;d&lt;/strong&gt; in the solution set; otherwise ∑ d&lt;sub&gt;i&lt;/sub&gt;s&lt;sub&gt;i&lt;/sub&gt; is even and ≠ k.  Now, for both of k and s&lt;sub&gt;n+1,0&lt;/sub&gt;, we have cases ≡ 1 (mod 4) and ≡ 3 (mod 4); as a consequence, exactly one of k + s&lt;sub&gt;n+1,0&lt;/sub&gt; and k - s&lt;sub&gt;n+1,0&lt;/sub&gt; is ≡ 2 (mod 4), and the other is ≡ 0 (mod 4).  Say k + s&lt;sub&gt;n+1,0&lt;/sub&gt; ≡ 0 (mod 4), and k - s&lt;sub&gt;n+1,0&lt;/sub&gt; ≡ 2 (mod 4); the other case is similar &amp;amp; will be omitted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each solution &lt;strong&gt;d&lt;/strong&gt; = (1, d&lt;sub&gt;n+1,1&lt;/sub&gt;, …, d&lt;sub&gt;n+1,n&lt;/sub&gt;) corresponds to a solution &lt;strong&gt;d&lt;sub&gt;n&lt;/sub&gt;&lt;/strong&gt; for |( k - s&lt;sub&gt;n+1,0&lt;/sub&gt;)/2| in S&lt;sub&gt;n&lt;/sub&gt; via bijection, where&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-If k &gt; s&lt;sub&gt;n+1,0&lt;/sub&gt;, d&lt;sub&gt;n,i&lt;/sub&gt; = d&lt;sub&gt;n+1,i+1&lt;/sub&gt; for i = 0 through n – 1. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;-If k &amp;lt; s&lt;sub&gt;n+1,0&lt;/sub&gt;, d&lt;sub&gt;n,i&lt;/sub&gt; = -d&lt;sub&gt;n+1,i+1&lt;/sub&gt; for i = 0 through n – 1. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, F&lt;sub&gt;n+1&lt;/sub&gt; is an upper bound for the number of disjoint subset pairs of S&lt;sub&gt;n+1&lt;/sub&gt; whose sums differ by k with d&lt;sub&gt;n+1,0&lt;/sub&gt; = 1. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the solutions &lt;strong&gt;d&lt;/strong&gt; where d&lt;sub&gt;n+1,0&lt;/sub&gt; = -1.  We have ∑ d&lt;sub&gt;i&lt;/sub&gt;s&lt;sub&gt;i&lt;/sub&gt; = k, so ∑ d&lt;sub&gt;i&lt;/sub&gt;s&lt;sub&gt;i&lt;/sub&gt; + s&lt;sub&gt;n+1,0&lt;/sub&gt; ≡ 0 (mod 4). Hence, for the ordered n+1 tuple &lt;strong&gt;d’&lt;/strong&gt; where d’&lt;sub&gt;n+1,0&lt;/sub&gt; = 0 and d’&lt;sub&gt;n+1,i&lt;/sub&gt; = d&lt;sub&gt;n+1,i&lt;/sub&gt; for i = 1 through n, we have   ∑ d’&lt;sub&gt;i&lt;/sub&gt;s&lt;sub&gt;i&lt;/sub&gt; ≡ 0 (mod 4). Therefore, d’&lt;sub&gt;n+1,1&lt;/sub&gt; = d&lt;sub&gt;n+1,1&lt;/sub&gt; = 0; otherwise ∑ d’&lt;sub&gt;i&lt;/sub&gt;s&lt;sub&gt;i&lt;/sub&gt; ≡ 2 (mod 4).  Each solution &lt;strong&gt;d&lt;/strong&gt; = (-1, 0, d&lt;sub&gt;n+1,2&lt;/sub&gt;, …, d&lt;sub&gt;n+1,n&lt;/sub&gt;) corresponds to a solution d&lt;sub&gt;n-1&lt;/sub&gt; for( k + s&lt;sub&gt;n+1,0&lt;/sub&gt;)/4 in S&lt;sub&gt;n-1&lt;/sub&gt; via bijection, where d&lt;sub&gt;n-1,i&lt;/sub&gt; = d&lt;sub&gt;n+1,i+2&lt;/sub&gt; for i = 0 through n – 2. So, F&lt;sub&gt;n&lt;/sub&gt; is an upper bound for the number of disjoint subset pairs of S&lt;sub&gt;n+1&lt;/sub&gt; whose sums differ by k with d&lt;sub&gt;n+1,0&lt;/sub&gt; = -1. As we have exhausted the two cases for d&lt;sub&gt;n+1,0&lt;/sub&gt;, F&lt;sub&gt;n+2&lt;/sub&gt; = F&lt;sub&gt;n+1&lt;/sub&gt; + F&lt;sub&gt;n&lt;/sub&gt; is an upper bound for the total number of disjoint subset pairs of S&lt;sub&gt;n+1&lt;/sub&gt; whose sums differ by an odd k, or any positive integer k./&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Applying the lemma to k = 1, conclude that F&lt;sub&gt;n+1&lt;/sub&gt; is an upper bound on the number of pairs of disjoint subsets of an n-element binary step-order set S whose sums differ by 1. To show that F&lt;sub&gt;n+1&lt;/sub&gt; is the least upper bound, we’ll construct a sequence of binary step-order sets where this upper bound is realized.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let S&lt;sub&gt;1&lt;/sub&gt; = {1}, S&lt;sub&gt;2&lt;/sub&gt; = {1,2}, and S&lt;sub&gt;3&lt;/sub&gt; = {3,2,4}. Given S&lt;sub&gt;n&lt;/sub&gt; for n ≥ 2, S&lt;sub&gt;n+1&lt;/sub&gt; is constructed by setting s&lt;sub&gt;n+1,0&lt;/sub&gt; = 3, and s&lt;sub&gt;n+1,i&lt;/sub&gt; = 2s&lt;sub&gt;n,i-1&lt;/sub&gt;for i = 1 to n. (The set elements are listed in the order of increasing power of 2 in the prime factorization of the element.) There are 2 solutions for &lt;strong&gt;d&lt;/strong&gt; for S&lt;sub&gt;2&lt;/sub&gt;; namely (1,0) and (-1,1). There are 3 solutions for &lt;strong&gt;d&lt;/strong&gt; for S&lt;sub&gt;3&lt;/sub&gt;; namely (1,-1,0), (1,1,-1) and (-1,0,1). (Because the 0 term of the sum is either 3 or -3, the subtotal of the other terms must be either -2 or 4 to have the sum of 1.) The F&lt;sub&gt;n+2&lt;/sub&gt; solutions for S&lt;sub&gt;n+1&lt;/sub&gt; are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-The F&lt;sub&gt;n+1&lt;/sub&gt; n+1 tuples d defined by setting d&lt;sub&gt;n+1,0&lt;/sub&gt; = 1 and  d&lt;sub&gt;n+1,i&lt;/sub&gt; = -d&lt;sub&gt;n,i-1&lt;/sub&gt; for i = 1 to n from all the solutions d&lt;sub&gt;n&lt;/sub&gt; for S&lt;sub&gt;n&lt;/sub&gt;; plus&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-The F&lt;sub&gt;n&lt;/sub&gt; n+1 tuples d defined by setting d&lt;sub&gt;n+1,0&lt;/sub&gt; = -1, d&lt;sub&gt;n+1,1&lt;/sub&gt; = 0 and  d&lt;sub&gt;n+1,i&lt;/sub&gt; = d&lt;sub&gt;n-1,i-1&lt;/sub&gt; for i = 1 to n from all the solutions d&lt;sub&gt;n-1&lt;/sub&gt; for S&lt;sub&gt;n-1&lt;/sub&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: I couldn't work out fully how to do sigma notation; hopefully the sum limits are clearly implied.&lt;/p&gt;&#xA;" OwnerUserId="1782" LastEditorUserId="1782" LastEditorDisplayName="" LastEditDate="2010-09-17T20:43:14.167" LastActivityDate="2010-09-17T20:43:14.167" CommentCount="2" />
  <row Id="4884" PostTypeId="1" AcceptedAnswerId="4885" CreationDate="2010-09-17T20:49:45.077" Score="1" ViewCount="149" Body="&lt;p&gt;In the book Introduction to Linear Optimization by Bertsimas Dimitri, a polyhedron is defined as a set $ \lbrace x \in \mathbb{R^n} | Ax \geq b \rbrace $, where A is an m x n matrix and b is a vector in $\mathbb{R^m}$. What it means is that a polyhedron is the intersection of several halfspaces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A ball can also be viewed as the intersection of infinitely many halfspaces. So I was wondering if a ball is also a polyhedron by that definition or by any other definition that you might use?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks and regards!&lt;/p&gt;&#xA;" OwnerUserId="1281" LastEditorUserId="1281" LastEditorDisplayName="" LastEditDate="2010-09-17T23:02:37.393" LastActivityDate="2010-09-17T23:02:37.393" Title="Is a ball a polyhedron?" Tags="&lt;geometry&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="4885" PostTypeId="2" ParentId="4884" CreationDate="2010-09-17T20:59:39.307" Score="3" ViewCount="" Body="&lt;p&gt;No a ball is not a polyhedron, even by this definition. In your definition the matrix $A$ is of size $m\times n$, where $m\in\mathbb{N}$ thus the matrix is finite. The integer $m$ is an upper bound on the number of halfspaces which intersect to form the polyhedron. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason $m$ is an upper bound is because suppose $A$ has two rows identical. Then there are two hyperspaces which are parallel so at least one of them does not form any part of the polyhedron.&lt;/p&gt;&#xA;" OwnerUserId="1794" LastActivityDate="2010-09-17T20:59:39.307" />
  <row Id="4886" PostTypeId="2" ParentId="4884" CreationDate="2010-09-17T21:00:19.213" Score="2" ViewCount="" Body="&lt;p&gt;The usual definition of a polyhedron requires that either one intersects a finite number of half-spaces, or one takes the convex hull of a finite set of points.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See the book Convex Polytopes by Branko Grünbaum (either the first or second edition).&lt;/p&gt;&#xA;" OwnerUserId="1369" LastActivityDate="2010-09-17T21:00:19.213" CommentCount="2" />
  <row Id="4887" PostTypeId="2" ParentId="4881" CreationDate="2010-09-17T21:06:31.267" Score="5" ViewCount="" Body="&lt;p&gt;The point is that the Fourier transform $\mathcal{F}$&#xA;conjugates $T$ to the multiplication&#xA;operator $S:h\mapsto\hat g h$ (i.e. $\mathcal{F}(f\ast g)=\hat g\mathcal{F}(f)$).&#xA;Hence $T$ and $S$ have the same spectrum.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-09-17T21:06:31.267" CommentCount="2" />
  <row Id="4889" PostTypeId="1" CreationDate="2010-09-17T21:19:10.037" Score="2" ViewCount="444" Body="&lt;p&gt;In many elementary (and not-so-elementary) Euclidean geometry texts, a (simple) polygon is said to be &lt;em&gt;tangential&lt;/em&gt;&amp;nbsp; if it is convex and has an inscribed circle (i.e., a circle that intersects and is tangent to each side of the polygon).  The assumption of convexity is not needed: I've come up with a rather laborious proof that every polygon with an inscribed circle is convex.  But I'd like to find either a simple elementary proof or a reference to a proof in the literature.  (By &quot;elementary,&quot; I mean using only standard facts of axiomatic Euclidean geometry.)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anyone know of a reference for a proof of this fact (elementary or not)?  Or can anyone think of a straightforward elementary proof?  You can use any definition of &quot;convex polygon&quot; that you like, but the easiest one to work with is that for each edge, the vertices not on that edge lie on one side of the line through that edge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Interestingly, the corresponding fact for circumscribed circles--i.e., that every polygon with a circumscribed circle is convex--is quite easy to prove: If P has a circumscribed circle, any two nonadjacent sides of P are non-intersecting chords of the circle; it is easy to show that both endpoints of each chord lie on the same side of the line through the other, and from there it is an easy matter to prove that P is convex.)&lt;/p&gt;&#xA;" OwnerUserId="1421" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-09-17T23:54:25.870" LastActivityDate="2010-09-18T01:42:20.337" Title="Proof that every polygon with an inscribed circle is convex?" Tags="&lt;geometry&gt;&lt;euclidean-geometry&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="4890" PostTypeId="2" ParentId="4870" CreationDate="2010-09-17T21:28:41.303" Score="2" ViewCount="" Body="&lt;p&gt;In Stefan Rolewicz book &lt;a href=&quot;http://books.google.com/books?id=qVmHL2M5I9oC&amp;amp;dq=rolewicz+Metric+linear+spaces&amp;amp;printsec=frontcover&amp;amp;source=bn&amp;amp;hl=en&amp;amp;ei=1teTTMnnIouPswaolsX3CQ&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=4&amp;amp;sqi=2&amp;amp;ved=0CCcQ6AEwAw#v=onepage&amp;amp;q=riemann&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;Metric Linear Spaces&lt;/a&gt; there are different kinds of general and useful results regarding integration on &quot;more abstract spaces&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example  $f:[0,1]\to X$ where $X$ is an &lt;a href=&quot;http://en.wikipedia.org/wiki/F-space&quot; rel=&quot;nofollow&quot;&gt;$F$-space&lt;/a&gt;, he discuss the Riemann-integral and if, in addition, $X$ is &lt;a href=&quot;http://en.wikipedia.org/wiki/Locally_convex&quot; rel=&quot;nofollow&quot;&gt;locally convex&lt;/a&gt; then the Bochner-Lebesgue integral is more fruitful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, though I like your question I believe that in the end I think it is better to have some application in mind and then, whenever needed, try to look for some integration methods for the situation. Sometimes it is good to be open to alternative methods in other areas of mathematics that might be more handy, for example &quot;abstract maximum principles&quot;. &lt;/p&gt;&#xA;" OwnerUserId="1154" LastActivityDate="2010-09-17T21:28:41.303" />
  <row Id="4891" PostTypeId="2" ParentId="4889" CreationDate="2010-09-17T22:05:04.180" Score="0" ViewCount="" Body="&lt;p&gt;&lt;em&gt;edit 2&lt;/em&gt;: Suppose that a simple polygon has an inscribed circle.  Without loss of generality, pick a &quot;first&quot; edge and let the circle be on the &quot;right&quot; side of that edge.  The angle that is on the same side as the circle--that is, to the right--between the first edge and the second edge must have measure less than 180&amp;deg;.  Similarly, the angle between the second and third edges that is also on the right must also have measure less than 180&amp;deg;, and so on, so that all of the angles on the right side of the polygon's perimeter (whether it is the inside or the outside) must have measure less than 180&amp;deg; and all of the angles on the left side must have measure greater than 180&amp;deg;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the sum of the interior angles of a simple polygon with &lt;em&gt;n&lt;/em&gt; sides is 180&amp;deg;(&lt;em&gt;n&lt;/em&gt;&amp;nbsp;&amp;ndash;&amp;nbsp;2), the average measure of an interior angle of a simple polygon with &lt;em&gt;n&lt;/em&gt; sides is 180&amp;deg;&amp;nbsp;&amp;ndash;&amp;nbsp;360&amp;deg;/&lt;em&gt;n&lt;/em&gt;, which is strictly less than 180&amp;deg;.  So, since the angles on the left side of the perimeter all have measure greater than 180&amp;deg;, their average is greater than 180&amp;deg;, so the left cannot be the interior of the polygon and the right side must be the interior, so the inscribed circle must be in the interior of the polygon and the internal angles all have measure less than 180&amp;deg;.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;original answer&lt;/em&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming the polygon is non-self-intersecting, then two consecutive sides of the polygon correspond to two consecutive points on tangency on the circle, with the angle formed by the two sides subtending the minor arc between the points of tangency.  The measure of the angle of the polygon is half the difference between the measure of the major and minor arcs between the points of tangency.  The greatest possible difference would be the degenerate case where the minor arc has measure 0&amp;deg; and the major arc has measure 360&amp;deg;, giving the angle measure 180&amp;deg;; for non-degenerate cases, the difference in the arc measures must be less than 360&amp;deg;, so the angle measure must be less than 180&amp;deg;.  This applies to every pair of consecutive sides, so every interior angle of the polygon has measure less than 180&amp;deg;, so the polygon is convex.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;edit&lt;/em&gt;: Alternately, and blatantly assuming that the inscribed circle must be in the interior of the polygon, suppose that a polygon is not convex, so that there is a vertex at which the interior angle has measure greater than 180&amp;deg;.  For a circle to be tangent to the two sides that meet at that vertex, the circle must be exterior to the polygon at that vertex (so as to be in the non-reflex side of that angle) and thus cannot be the inscribed circle.&lt;/p&gt;&#xA;" OwnerUserId="72" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-09-18T01:31:56.330" LastActivityDate="2010-09-18T01:31:56.330" CommentCount="8" />
  <row Id="4892" PostTypeId="2" ParentId="4868" CreationDate="2010-09-17T22:37:12.357" Score="2" ViewCount="" Body="&lt;p&gt;Quoting part of my answer to &lt;a href=&quot;http://math.stackexchange.com/questions/3408/characterizing-nonconstant-entire-functions-with-modulus-1-on-the-unit-circle&quot;&gt;this other question&lt;/a&gt; is my answer to yours (and Robin basically already said it in the comments):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If a function $f$ is holomorphic in a neighborhood of the closed disk and has modulus 1 on the circle, then $f$ is a &lt;a href=&quot;http://en.wikipedia.org/wiki/Blaschke_product#Finite_Blaschke_products&quot; rel=&quot;nofollow&quot;&gt;finite Blaschke product&lt;/a&gt;.  You can prove this by taking all of the zeros inside the disk counted according to multiplicity, dividing by corresponding &lt;a href=&quot;http://planetmath.org/encyclopedia/AutomorphismsOfUnitDisk.html&quot; rel=&quot;nofollow&quot;&gt;holomorphic automorphisms of the disk&lt;/a&gt; that have those zeros, and showing that the result is constant.  (This quotient and its reciprocal are analytic and bounded by 1 on the disk...)  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Of course I'm assuming $A$ is connected.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-09-17T22:37:12.357" CommunityOwnedDate="2010-09-17T22:38:35.357" CommentCount="1" />
  <row Id="4893" PostTypeId="1" AcceptedAnswerId="4895" CreationDate="2010-09-17T23:27:17.700" Score="3" ViewCount="215" Body="&lt;p&gt;Why is saying $\frac{dy}{dx}\frac{dy}{dx}=y\frac{d^2y}{(dx)^2 }$ not valid? Does Leibniz notation (and thinking of it as an infinitesimal quotient) not work for higher-order derivatives?&lt;/p&gt;&#xA;" OwnerUserId="1958" LastActivityDate="2010-09-18T11:43:30.850" Title="Can Leibniz Notation Be Treated As a Quotient?" Tags="&lt;calculus&gt;&lt;notation&gt;" AnswerCount="3" />
  <row Id="4894" PostTypeId="1" CreationDate="2010-09-17T23:39:21.200" Score="3" ViewCount="338" Body="&lt;p&gt;It is not quite intuitive , at least from its origin. Could any one can give me an intuitive explanation?Thank you!&lt;/p&gt;&#xA;" OwnerUserId="1944" LastEditorUserId="8575" LastEditorDisplayName="" LastEditDate="2011-05-23T16:46:56.257" LastActivityDate="2011-05-23T16:46:56.257" Title="Why does symplectic geometry have many applications in mathematics" Tags="&lt;geometry&gt;&lt;soft-question&gt;&lt;symplectic-geometry&gt;" AnswerCount="4" />
  <row Id="4895" PostTypeId="2" ParentId="4893" CreationDate="2010-09-17T23:48:21.867" Score="4" ViewCount="" Body="&lt;p&gt;Consider $y=3x$ then $\frac{dy}{dx} =3$. So the left hand side of your equation would be 9. However, $\frac{d^{2}y}{dx^{2}} =0$, so the right hand side of your equation would be 0 for this example, but $0\neq 9$. Thus, the function $y=3x$ provides a counterexample. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Leibniz notation does work for higher derivatives, the nth derivative of y is denoted by $\frac{d^{n}y}{dx^{n}}$. However, treating the derivative as a quotient is called &lt;a href=&quot;http://en.wikipedia.org/wiki/Abuse_of_notation#Derivative&quot; rel=&quot;nofollow&quot;&gt;abuse of notation&lt;/a&gt; and is considered more as a convenient tool to help students remember certain rules like the chain rule. There is actually a way of treating infinitesimals rigorously is treated in what is known as &lt;a href=&quot;http://en.wikipedia.org/wiki/Non-standard_analysis&quot; rel=&quot;nofollow&quot;&gt;non-standard analysis&lt;/a&gt;, but this isn't really the most common approach.&lt;/p&gt;&#xA;" OwnerUserId="249" LastActivityDate="2010-09-17T23:48:21.867" />
  <row Id="4896" PostTypeId="1" AcceptedAnswerId="4911" CreationDate="2010-09-17T23:54:03.710" Score="2" ViewCount="137" Body="&lt;p&gt;Say we have a undirected, connected graph that represents a network of streets.  How would you prove that there always exists a tour of the network, where one ``drives'' in both directions on every street exactly once?&lt;/p&gt;&#xA;" OwnerUserId="1960" LastActivityDate="2010-09-18T10:52:04.190" Title="Network of streets" Tags="&lt;graph-theory&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="4897" PostTypeId="2" ParentId="4896" CreationDate="2010-09-18T00:09:43.980" Score="3" ViewCount="" Body="&lt;p&gt;What you are looking to prove is that the graph is &lt;a href=&quot;http://en.wikipedia.org/wiki/Eulerian_path&quot; rel=&quot;nofollow&quot;&gt;Eulerian&lt;/a&gt;, which is true as each vertex has an even degree.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-09-18T00:09:43.980" CommentCount="4" />
  <row Id="4898" PostTypeId="2" ParentId="4896" CreationDate="2010-09-18T00:12:02.480" Score="0" ViewCount="" Body="&lt;p&gt;How do you represent the fact that you want to traverse each edge twice?  How do you know whether you can trace a graph with a single continuous path?&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-09-18T00:12:02.480" CommentCount="2" />
  <row Id="4899" PostTypeId="2" ParentId="4872" CreationDate="2010-09-18T00:14:11.493" Score="1" ViewCount="" Body="&lt;p&gt;Probably the simplest way to state a corresponding group theory result is the generalization of Frobenius's theorem mentioned by Bill Dubuque. The generalization says that if $G$ is a finite group, then the number of solutions in $G$ to $x^n=1$ is a multiple of $\gcd(|G|,n)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of the congruence $x^2\equiv -1\pmod{p}$, the solutions to this congruence are exactly the solutions to $x^4\equiv 1 \pmod{p}$ that are not solutions to $x^2\equiv 1\pmod{p}$. The order of the group here is $p-1$, so the number of solutions to the first is a multiple of $\gcd(4,p-1)$; the number of solutions of the second is a multiple of $2$, and we know the two: $1$ and $-1$. Since every solution to $x^4\equiv 1\pmod{p}$ is a solution to $x^2\equiv 1 \pmod{p}$, there are solutions to $x^2\equiv -1\pmod{p}$ if and only if $\gcd(4,p-1) = 4$. &lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-09-18T00:14:11.493" CommentCount="1" />
  <row Id="4900" PostTypeId="1" AcceptedAnswerId="4906" CreationDate="2010-09-18T00:21:52.723" Score="2" ViewCount="167" Body="&lt;p&gt;From an olympic book:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $m$ be an odd number and $a_1, ... , a_m$ integers. $x=(x_1,...,x_m)$ is a permutation of the integers $1, ..., m$ and $f$ at $x$ is defined by $f(x)=x_1a_1 + ... + x_m a_m$. Prove that there exist two permutations $x$ and $y$ such that $f(x)-f(y)$ is divisible by $m!$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm aware of a few facts. For instance, if all $a_i$ are odd (or even), any permutations of $x$ results in values of $f$ of the same parity, but there are only $m!/2$ (for $m&gt;1$) residues module $m!$ of the same parity, and an application of the Pigeon Hole Principle yields the conclusion. So in the general case I'd like to say something like there's a greater number of permutations that give a value of $f$ of one parity than the other. I don't know how to show this.&lt;/p&gt;&#xA;" OwnerUserId="1313" LastActivityDate="2010-09-18T07:14:57.937" Title="pigeon-hole and parity" Tags="&lt;number-theory&gt;&lt;combinatorics&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="4901" PostTypeId="2" ParentId="4821" CreationDate="2010-09-18T00:32:16.700" Score="2" ViewCount="" Body="&lt;p&gt;It appears that both previous answers missed that the number was to be even.  So the real count is 5!/2! + 5!/2!/2! + 5!/2! + 5!/2! = 210&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-09-18T00:32:16.700" CommentCount="1" />
  <row Id="4902" PostTypeId="2" ParentId="4889" CreationDate="2010-09-18T01:42:20.337" Score="2" ViewCount="" Body="&lt;p&gt;The definition of &lt;em&gt;simple&lt;/em&gt; establishes that the polygon itself equals the intersection of the half-planes tangent to the circle at the points where the polygon's sides contact the circle.  Any nonempty intersection of halfplanes is convex.&lt;/p&gt;&#xA;" OwnerUserId="1489" LastActivityDate="2010-09-18T01:42:20.337" CommentCount="6" />
  <row Id="4903" PostTypeId="2" ParentId="4894" CreationDate="2010-09-18T02:47:28.480" Score="1" ViewCount="" Body="&lt;p&gt;Symplectic geometry is applicable in lot of areas such as Classical mechanics, Hamiltonian mechanics. While browsing through the web i have found a reason as to why SG is good for classical mechanics. You may read it here: &lt;a href=&quot;http://research.microsoft.com/en-us/um/people/cohn/thoughts/symplectic.html&quot; rel=&quot;nofollow&quot;&gt;http://research.microsoft.com/en-us/um/people/cohn/thoughts/symplectic.html&lt;/a&gt;&lt;/p&gt;&#xA;" LastActivityDate="2010-09-18T02:47:28.480" CommentCount="2" />
  <row Id="4904" PostTypeId="2" ParentId="4841" CreationDate="2010-09-18T06:04:04.877" Score="4" ViewCount="" Body="&lt;p&gt;Personally, I prefer arguments which are more geometric to arguments rooted in matrix algebra. With that in mind, here is a proof.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, two observations on the geometric meaning of an upper triangular invertible linear map. &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Define $S_k = {\rm span} (e_1, \ldots, e_k)$, where $e_i$ the standard basis vectors. &#xA;Clearly, the linear map $T$ is upper triangular if and only if $T S_k \subset S_k$. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If $T$ is in addition invertible, we must have the stronger relation $T S_k = S_k$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Indeed, if $T S_k$ was a strict subset of $S_k$, then $Te_1, \ldots, Te_k$ are $k$ vectors in a space of dimension strictly less than $k$, so they must be dependent: $\sum_i \alpha_i Te_i=0$ for some $\alpha_i$ not all zero. This implies that $T$ sends the &lt;b&gt;nonzero&lt;/b&gt; vector $\sum_i \alpha_i e_i$ to zero, so $T$ is not invertible. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;With these two observations in place, the proof proceeds as follows. Take any $s \in S_k$. Since $TS_k=S_k$ there exists some $s' \in S_k$ with $Ts'=s$ or $T^{-1}s = s'$. In other words, $T^{-1} s$ lies in $S_k$, so $T^{-1}$ is upper triangular. &lt;/p&gt;&#xA;" OwnerUserId="317" LastEditorUserId="317" LastEditorDisplayName="" LastEditDate="2010-09-18T17:06:21.803" LastActivityDate="2010-09-18T17:06:21.803" CommentCount="3" />
  <row Id="4905" PostTypeId="1" CreationDate="2010-09-18T06:41:39.513" Score="6" ViewCount="312" Body="&lt;p&gt;The following seems to hold in numerical simulations, is it true?&#xA;$$\lim_{n\to \infty} \int_0^1 dx \frac{n! 2^{-n} n}{(n x)!(n-n x)!\sqrt{x(1-x)}}=2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's a combination of two known integrals&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\lim_{n\to \infty} \int_0^1 dx \frac{n! 2^{-n} n}{(n x)!(n-n x)!}=1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_0^1 dx \frac{1}{\sqrt{x(1-x)}}=\pi$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First one follows from asymptotic &lt;a href=&quot;http://www.physicsforums.com/attachment.php?attachmentid=27909&amp;amp;d=1283240321&quot; rel=&quot;nofollow&quot;&gt;expansion&lt;/a&gt; of Gamma and the second is done using trig substitution&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Edit&lt;/b&gt;: $x!$ is short for $\Gamma(x+1)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Limit convergence is apparent for $n$ between 10 and $10^6$, $\log_{10} n$ gives an almost perfect fit to the log of deviation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://yaroslavvb.com/upload/error-convergence.png&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Motivation -- the last integral is an instance of &quot;Information Volume&quot;, an estimate of number of statistically distinguishable distributions in the Bernoulli model, used as a measure of model complexity . A major annoyance is that this integral fails to exist for some popular models, like the geometric distribution family. Perhaps we can fix this by instead estimating the number of observation sequences that are well fit by some distribution in the model? That's the first integral.&lt;/p&gt;&#xA;" OwnerUserId="998" LastEditorUserId="998" LastEditorDisplayName="" LastEditDate="2010-09-18T20:50:47.637" LastActivityDate="2010-09-19T08:47:07.610" Title="Limit of Gamma integrals" Tags="&lt;gamma-function&gt;&lt;integral&gt;" AnswerCount="2" CommentCount="7" />
  <row Id="4906" PostTypeId="2" ParentId="4900" CreationDate="2010-09-18T07:14:57.937" Score="2" ViewCount="" Body="&lt;p&gt;Let&#xA;$$S=\{(x_1,\ldots,x_m):x_1,\ldots,x_m\textrm{ is a permutation of }1,\ldots m\}.$$&#xA;So for a vector $a=(a_1,\ldots,a_m)\in\mathbb{Z}^n$ we need to show there are&#xA;distinct $x$, $y\in S$ with $x\cdot a\equiv y\cdot a$ (mod $m!$). If this&#xA;wasn't true, $x\cdot a$ runs through all congruence classes modulo $m!$&#xA;as $x$ runs through $S$. Therefore&#xA;$$\left(\sum_{x\in S}x\right)\cdot a\equiv\sum_{k=0}^{m!-1}k\pmod{m!}.$$&#xA;Each coordinate of $X=\sum_{x\in S}x$ equals $(m-1)!\sum_{j=1}^m j$.&#xA;But since $m$ is odd, this is divisible by $m!$. Hence $X\cdot a\equiv0$&#xA;(mod $m!$) but $\sum_{k=0}^{m!-1}k\not\equiv0$ (mod $m!$).&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-09-18T07:14:57.937" CommentCount="2" />
  <row Id="4907" PostTypeId="2" ParentId="4893" CreationDate="2010-09-18T08:23:06.837" Score="2" ViewCount="" Body="&lt;p&gt;Your intuition is correct that the derivative OPERATOR can be multiplied together in such a manner i.e. $\frac{d}{dx}\frac{d}{dx} = \frac{d^2}{dx^2}$. However, the expression $\frac{dy}{dx} = \frac{d}{dx}(y)$ is an evaluated derivative on the function $y$ we write the $y$ in the numerator with $d$ as a notational convenience. When you evaluate a derivative, it no longer behaves like the derivative operator and is now simply a function. Another way to look at it is if you consider $\frac{d}{dx}$ as a &quot;function&quot; then the product $\frac{d}{dx}\frac{d}{dx}$ is simply the composition of two derivative functions, conversely $\frac{dy}{dx} \frac{dy}{dx}$ is the product of two evaluated functions. So we can restructure your question of follows : does $f(x)*f(x) = f(f(x))$ ? the answer is obviously no. However, $(f\circ f)(x) = f(f(x))$ is true by the definition of composition of functions. &lt;/p&gt;&#xA;" OwnerUserId="1906" LastEditorUserId="1906" LastEditorDisplayName="" LastEditDate="2010-09-18T08:48:54.713" LastActivityDate="2010-09-18T08:48:54.713" />
  <row Id="4908" PostTypeId="1" AcceptedAnswerId="4909" CreationDate="2010-09-18T09:01:11.907" Score="5" ViewCount="341" Body="&lt;p&gt;Sometime back i watched the documentary of Andrew Wiles proving the Fermat's Last theorem. A truly inspiring video and i still watch it whenever i am in a depressed mood. There are certain things(infact many) which i couldn't follow and i would like it to be explained here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The Taniyama-Shimura conjecture. In the video it's said that that an elliptic curve is a modular form in disguise. I would want someone to explain this statement. I have seen the definition of a Modular form in Wikipedia, but i can't correlate this with an elliptic curve. The definition of an elliptic curve is simply a cubic equation of the form $y^{2}=x^{3}+ax+b$. How can it be a modular form?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Next, there was a mention of this Mathematician named &quot;Gerhard Frey&quot; who seems to considered this question of what could happen if there was a solution to the equation, $x^{n}+y^{n}=z^{n}$, and by considering this he constructed a curve which is not modular, contradicting Taniyama-Shimura. If he had constructed such a curve then what was the need for Prof. Ribet to actually prove the Epsilon conjecture.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lastly, here i would like to know this answer: How many of you agree with Wiles, that possibly Fermat could have fooled himself by saying that he had a proof of this result? I certainly disagree with his statement. Well, the reason, is just instinct!&lt;/p&gt;&#xA;" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-22T02:12:57.537" LastActivityDate="2010-11-22T02:12:57.537" Title="A question on FLT and Taniyama Shimura" Tags="&lt;number-theory&gt;&lt;soft-question&gt;&lt;algebraic-number-theory&gt;&lt;elliptic-curves&gt;&lt;modular-forms&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="1" />
  <row Id="4909" PostTypeId="2" ParentId="4908" CreationDate="2010-09-18T09:18:34.297" Score="6" ViewCount="" Body="&lt;p&gt;The elliptic curve is not a modular form. The idea that there is&#xA;a modular form $f$ associated to $E$. It satisfies&#xA;$$f(z)=\sum_{n=1}^\infty c_n q^n$$&#xA;where $q=\exp(2\pi i z)$, $c_1=1$ and (with finitely many exceptions)&#xA;for prime $p$, the equation $y^2=x^3+ax=b$ has $p-c_p$ solutions $(x,y)$&#xA;considered modulo $p$. It also satisfies various other conditions&#xA;that I won't spell out (that it's a &quot;newform&quot; for a modular group $\Gamma_0(N)$&#xA;etc.)&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-09-18T09:18:34.297" CommentCount="2" />
  <row Id="4911" PostTypeId="2" ParentId="4896" CreationDate="2010-09-18T10:52:04.190" Score="2" ViewCount="" Body="&lt;p&gt;This is essentially the same as Isaac’s comment on &lt;a href=&quot;http://math.stackexchange.com/questions/4896/network-of-streets/4897#4897&quot;&gt;Moron’s answer&lt;/a&gt;, but I’ll post it as an answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I assume that by “one ‘drives’ in both directions on every street exactly once,” you mean “one ‘drives’ in &lt;em&gt;each&lt;/em&gt; direction on every street exactly once.”&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A directed graph has an Eulerian circuit (i.e. a circuit which uses every edge exactly once) if and only if it is strongly connected and each vertex has equal in-degree and out-degree.  This fact can be proved in the same way as the well-known fact that an undirected graph has an Eulerian circuit if and only if it is connected and each vertex has even degree.  See a textbook on graph theory for a proof.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Back to your question, let &lt;em&gt;G&lt;/em&gt; be the connected undirected graph representing the network of streets.  You are looking for an Eulerian circuit (or an Eulerian path if you do not require the tour to end at the starting point) in the directed graph &lt;em&gt;G&lt;/em&gt;′ obtained by replacing each edge in &lt;em&gt;G&lt;/em&gt; by a pair of directed edges in both directions.  It is easy to see that the directed graph &lt;em&gt;G&lt;/em&gt;′ satisfies the two conditions above, and therefore &lt;em&gt;G&lt;/em&gt;′ has an Eulerian circuit.&lt;/p&gt;&#xA;" OwnerUserId="926" LastActivityDate="2010-09-18T10:52:04.190" />
  <row Id="4912" PostTypeId="2" ParentId="4859" CreationDate="2010-09-18T11:38:49.393" Score="1" ViewCount="" Body="&lt;p&gt;This is not an answer, but an elementay question about this category ring $R({\cal C})$. Because of Martin's remark, I'll use the product notation: elements of $R({\cal C})$ are tuples &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;f= (f_{ab}) \in \prod_{a, b \in {\mathrm ob} {\cal C}} {\cal C}(a,b) \ ,&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;in which every $f_{ab} : a \longrightarrow b$ is a morphism of ${\cal C}$. Right?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(By the way: are we still in our universe when we make such a product? I mean: this won't be a &quot;ring&quot; at all, that is: a small ring.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the problem that Martin has pointed out comes from the fact that, when you multiply two of these tuples $f=(f_{ab})$ and $g= (g_{ab})$ a sum appears in the $ab$-component of $g\cdot f$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\sum_x g_{xb}\circ f_{ax} \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it so?&lt;/p&gt;&#xA;" OwnerUserId="664" LastEditorUserId="664" LastEditorDisplayName="" LastEditDate="2010-09-18T20:05:55.780" LastActivityDate="2010-09-18T20:05:55.780" CommentCount="1" />
  <row Id="4913" PostTypeId="2" ParentId="4893" CreationDate="2010-09-18T11:43:30.850" Score="2" ViewCount="" Body="&lt;p&gt;Since&#xA;$$f''(a)=\lim_{h\to0}\frac{f(a+2h)-2f(a+h)+f(a)}{h^2}$$&#xA;under suitable conditions, I suppose one could conceptualize&#xA;the $d^2 f$ in $d^2 f/dx^2$ as a &quot;second-order&quot; infinitesimal difference:&#xA;i.e., as $f(x+2dx)-2f(x+dx)+f(x)$. Mind you, I've never liked thinking&#xA;in terms of &quot;infinitesimals&quot; :-)&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-09-18T11:43:30.850" />
  <row Id="4914" PostTypeId="1" AcceptedAnswerId="4924" CreationDate="2010-09-18T11:54:21.823" Score="1" ViewCount="368" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Find the Cartesian equation for the perpendicular bisector of the line joining A(2,3) and B(0,6)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;How do I do this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you! &lt;/p&gt;&#xA;" OwnerUserId="1968" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-09-18T20:30:28.980" LastActivityDate="2010-09-18T20:30:28.980" Title="Cartesian Equation for the perpendicular bisector of a line." Tags="&lt;geometry&gt;&lt;algebra-precalculus&gt;&lt;analytic-geometry&gt;" AnswerCount="5" />
  <row Id="4915" PostTypeId="2" ParentId="4914" CreationDate="2010-09-18T11:57:45.883" Score="0" ViewCount="" Body="&lt;p&gt;Here is a hint. The mid point of the line joining $A$ and $B$ is $(1, \frac{9}{2})$. Use this to find the equation of the line passing through this point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As, J.M says the product of the slopes of the perpendiculars is $-1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Slope of the line passing through $A$ and $B$ is $\displaystyle m_{1}= \frac{6-3}{0-2} = -\frac{3}{2}$. Therefore the slope of the line perpendicular to it is $m_{2}= \frac{2}{3}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence the required equation of the line is $(y-\frac{9}{2})=\frac{2}{3}(x-1)$.&lt;/p&gt;&#xA;" LastActivityDate="2010-09-18T12:17:31.120" />
  <row Id="4916" PostTypeId="2" ParentId="4914" CreationDate="2010-09-18T11:59:00.220" Score="2" ViewCount="" Body="&lt;p&gt;Hints:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Get the slope and the midpoint of the segment joining your two given points.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Recall the relationship of the slopes of two perpendicular lines&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Use the point-slope form of the equation of a line.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-09-18T11:59:00.220" />
  <row Id="4917" PostTypeId="2" ParentId="4908" CreationDate="2010-09-18T13:04:03.790" Score="2" ViewCount="" Body="&lt;p&gt;The starting point of the link between elliptic curves and modular forms is the following.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From a topological point of view, elliptic curves are just (2-dimensional) tori, i.e. products $S^1\times S^1$, where $S^1$ is a circle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A torus has always an invariant never vanishing tangent field. Dually, one can find a non-vanishing invariant differential form $\omega$ on every elliptic curve $E$. It is a useful exercise to write $\omega$ in terms of the coordinates $x$ and $y$ when $E$ is given as a Weierstrass cubic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have a modular parametrization $\pi:X_0(N)\rightarrow E$ you can pull-back the form $\omega$ to $X_0(N)$ and in terms of the coordinate $z$ in the complex upper halfplane $\pi^*(\omega)=f(z)dz$ for some holomorphic function $f(z)$. It is basically immediate that $f(z)$ is a modular form of weight $2$.&lt;/p&gt;&#xA;" OwnerUserId="688" LastActivityDate="2010-09-18T13:04:03.790" CommentCount="12" />
  <row Id="4919" PostTypeId="2" ParentId="4914" CreationDate="2010-09-18T13:24:42.747" Score="2" ViewCount="" Body="&lt;p&gt;The perpendicular bisector of the segment $AB$ is the locus of points&#xA;$P$ equidistant from $A$ and $B$, that is $|AP|=|BP|$. It's&#xA;easier to consider the equation $|AP|^2=|BP|^2$ which, when $A=(a,b)$,&#xA;$B=(c,d)$ and $P=(x,y)$ becomes&#xA;$$(x-a)^2+(y-b)^2=(x-c)^2+(y-d)^2$$&#xA;and can be simplified further....&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-09-18T13:24:42.747" CommentCount="2" />
  <row Id="4920" PostTypeId="1" AcceptedAnswerId="4933" CreationDate="2010-09-18T14:52:03.653" Score="3" ViewCount="249" Body="&lt;p&gt;According to my textbook, the second derivative of $y^{2}+xy-x^{2}=9$ is $\frac{90}{(2y+x)^{3}}$. The problem states &quot;Express $\frac{d^{2}y}{dx^{2}}$ in terms of $x$ and $y$.&quot; I've tried for two days straight now, and I can't get that answer. I am convinced the book has a typo.&lt;/p&gt;&#xA;" OwnerUserId="1958" LastActivityDate="2010-09-18T18:15:41.697" Title="Second Derivatives Using Implicit Differentiation" Tags="&lt;calculus&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="4921" PostTypeId="2" ParentId="4840" CreationDate="2010-09-18T15:02:21.430" Score="1" ViewCount="" Body="&lt;p&gt;How close do you have to be?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you just find the center of the each triangle in space then divide the cutting plane into a 2d grid and do a series of rectangular volume calculations using the length and width of the grid section and the average height of the triangle midpoint above/below that grid section?&lt;/p&gt;&#xA;" OwnerUserId="1582" LastEditorUserId="1582" LastEditorDisplayName="" LastEditDate="2010-09-18T15:22:57.477" LastActivityDate="2010-09-18T15:22:57.477" CommentCount="5" />
  <row Id="4922" PostTypeId="2" ParentId="4920" CreationDate="2010-09-18T15:07:27.587" Score="1" ViewCount="" Body="&lt;p&gt;Hint: treat $y$ as a function $y(x)$, so differentiating $xy(x)$ should give something like $x y^{\prime}(x)+y(x)$. Differentiate expressions twice, and solve for $y^{\prime\prime}(x)$&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-09-18T15:07:27.587" CommentCount="4" />
  <row Id="4923" PostTypeId="2" ParentId="4869" CreationDate="2010-09-18T16:13:12.483" Score="3" ViewCount="" Body="&lt;p&gt;Let f:C to C be a complex-valued function. If you want to do ''regular'' integration, by which I assume you mean integrating over 2-dimensional regions in the complex plane, then you would have that the integral of f(x+iy)=u(x,y)+iv(x,y) over some region A in R&lt;sup&gt;2&lt;/sup&gt; equals the integral of the real component over A plus i*integral of the imaginary component of f (by sensible definition). I suspect it turns out that the theory of ''regular'' integration of complex-valued functions of the plane reduces quite literally to the theory of ''regular'' integration of real-valued functions functions of the plane, and hence is of not much interest and not much discussed as a theory of complex integration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now line integrals are different.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can associate with your complex-valued function f a vector-field on the plane Pdx+Qdy given by f(x+iy)=P(x,y)+iQ(x,y) where P and Q are the real-valued real and imaginary parts of f as function of the real plane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The condition that f be differentiable (i.e. the Cauchy-Riemann equations) translates to requiring that the associated vector field have vanishing curl, which means that the vector field Pdx+Qdy is conservative, which in turn means that it has path-invariant line integrals, which are precisely the line integrals of the complex-valued function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This though only works on simply connected domains, but it gives enough reason to consider line integrals of (differentiable) complex-valued functions as something special and worthy of interest.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Question: if we can think of complex-valued functions as vector-fields, why do we bother with having a theory of complex-valued functions with its own terminology? The answer is that the language of vector fields is only applicable to the simplest bootstrapping case of the complex plane; as soon as you leave that haven (for, say, meromorphic functions that have poles, multiple variables, etc) it all becomes much different and then the vector-fields simply do not reflect what is going on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ADDED:&#xA;For a rigorous exposition of the above sketch, check out the papers on Polya vector fields on Paul Fishback's web-site: &lt;a href=&quot;http://faculty.gvsu.edu/fishbacp/complex/complex.htm&quot; rel=&quot;nofollow&quot;&gt;http://faculty.gvsu.edu/fishbacp/complex/complex.htm&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="400" LastEditorUserId="400" LastEditorDisplayName="" LastEditDate="2010-09-18T18:33:03.293" LastActivityDate="2010-09-18T18:33:03.293" />
  <row Id="4924" PostTypeId="2" ParentId="4914" CreationDate="2010-09-18T16:30:50.033" Score="2" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\;$ The equation is  $\rm\;\: 2\ (A-B)\cdot (x,y) \;=\; |A|-|B|\;\;\;$ where $\rm\;\;\; |(a,b)| \ =\  a^2 + b^2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which, if worked out, yields $\rm\;\: (-4,6)\cdot (x,y) \;=\; \;36 \;- 13\;\;\:$ for $\rm\; A = (0,6),\;\; B = (2,3)$   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;which, after simplifying, yields the equation $\rm\; 6y =\; 4x+23$&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-09-18T17:55:35.957" LastActivityDate="2010-09-18T17:55:35.957" CommentCount="3" />
  <row Id="4925" PostTypeId="2" ParentId="250" CreationDate="2010-09-18T16:50:06.530" Score="4" ViewCount="" Body="&lt;p&gt;Morely's trisection theorem -- not discovered until 1899.  &lt;a href=&quot;http://www.mathpages.com/home/kmath376/kmath376.htm&quot; rel=&quot;nofollow&quot;&gt;http://www.mathpages.com/home/kmath376/kmath376.htm&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Trisect the three angles of a triangle. Take the intersection of each trisector with its nearest trisector of the nearest other vertex. These three points form an equilateral triangle!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With bisectors, they all meet in a point.  That trisectors should give an equilateral triangle is pretty surprising.&lt;/p&gt;&#xA;" OwnerUserId="271" LastActivityDate="2010-09-18T16:50:06.530" CommunityOwnedDate="2010-09-18T16:50:06.530" CommentCount="1" />
  <row Id="4926" PostTypeId="2" ParentId="4744" CreationDate="2010-09-18T17:08:01.813" Score="2" ViewCount="" Body="&lt;p&gt;For more on Hadamard Products or Hadamard Multiplication Theorem look for papers by Louis R. Bragg in American Mathematical Monthly, Jan. 1999, pp 36-42 or SIAM J. Math. Anal., Vol. 17, 1986, pp 220-230 for starters.&lt;/p&gt;&#xA;" LastActivityDate="2010-09-18T17:08:01.813" />
  <row Id="4927" PostTypeId="1" AcceptedAnswerId="4929" CreationDate="2010-09-18T17:24:27.460" Score="1" ViewCount="240" Body="&lt;p&gt;How to &quot;show&quot;, that if x=1, then its powers will end with 1, if x=5, it's powers end with 5.&lt;/p&gt;&#xA;" OwnerUserId="1795" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-09-18T20:47:05.410" LastActivityDate="2010-09-22T13:48:47.127" Title="If x=1, then its powers end with 1" Tags="&lt;number-theory&gt;&lt;algebra-precalculus&gt;" AnswerCount="4" CommentCount="0" />
  <row Id="4928" PostTypeId="2" ParentId="4914" CreationDate="2010-09-18T17:24:44.900" Score="1" ViewCount="" Body="&lt;p&gt;Generally, &lt;a href=&quot;http://math.stackexchange.com/questions/4914/cartesian-equation-for-the-perpendicular-bisector-of-a-line/4916#4916&quot;&gt;J.M.'s answer&lt;/a&gt; is what I'd suggest at the high school level.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, this is a somewhat common problem on timed math contests (or part of a problem), and in that settings, I'd take advantage of the fact that in the form $ax+by=c$, $\langle a,b\rangle$ is a vector perpendicular to the line:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Find the vector $\overrightarrow{AB}=\langle x_a,y_a\rangle$, which is perpendicular to the line you want.&lt;/li&gt;&#xA;&lt;li&gt;The midpoint of $\overline{AB}$ is $A+\frac{1}{2}\overrightarrow{AB}=\langle x_m,y_m\rangle$, which is a point on the line.&lt;/li&gt;&#xA;&lt;li&gt;An equation for the line is $\langle x_a,y_a\rangle\cdot\langle x,y\rangle=\langle x_a,y_a\rangle\cdot\langle x_m,y_m\rangle$ or $x_ax+y_ay=x_ax_m+y_ay_m$.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-09-18T17:24:44.900" />
  <row Id="4929" PostTypeId="2" ParentId="4927" CreationDate="2010-09-18T17:28:35.923" Score="5" ViewCount="" Body="&lt;p&gt;I am presuming you are talking about the last digit of a number in base 10 being 1 and 5 implying the powers of that number will also end with 1 and 5.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I will give you a hint:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Give two numbers whose last digits are 1, what does their product end with?&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-09-18T17:28:35.923" CommentCount="11" />
  <row Id="4930" PostTypeId="2" ParentId="4927" CreationDate="2010-09-18T17:29:36.980" Score="11" ViewCount="" Body="&lt;p&gt;Use mathematical induction:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Show that $x^1$ ends in 1 (or 5).&lt;/li&gt;&#xA;&lt;li&gt;Show that if $x^n$ ends in 1 (or 5), then $x^{n+1}$ also ends in 1 (or 5).  (Think about multiplying a number than ends in 1 by 1 (or ends in 5 by 5).)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-09-18T17:29:36.980" />
  <row Id="4931" PostTypeId="1" AcceptedAnswerId="4936" CreationDate="2010-09-18T18:14:20.677" Score="0" ViewCount="271" Body="&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/qBfas.png&quot; alt=&quot;http://img683.imageshack.us/img683/7001/mathw.png&quot;&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Show that BE is the perpendicular bisector to AC&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried to prove this through Pythagoras, but the answer I got did not prove it was at a right angle, and therefore said it was not the perpendicular bisector, I'm probably making a silly mistake, so any advice would be helpfull!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks a lot :)!&lt;/p&gt;&#xA;" OwnerUserId="1968" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-09-19T03:59:12.470" LastActivityDate="2010-09-19T04:37:25.237" Title="Perpendicular bisector." Tags="&lt;geometry&gt;&lt;algebra-precalculus&gt;&lt;analytic-geometry&gt;" AnswerCount="7" CommentCount="1" />
  <row Id="4932" PostTypeId="2" ParentId="4927" CreationDate="2010-09-18T18:14:41.033" Score="1" ViewCount="" Body="&lt;p&gt;If you know that addition and multiplication are well-defined modulo $n$, then just work (mod 10).&lt;br&gt;&#xA;Let $x_0 \in$ {1,5} be the ones digit of $x$.  Then for any $k$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x^k \equiv x_0^k \equiv x_0$ (mod 10).&lt;/p&gt;&#xA;" OwnerUserId="964" LastActivityDate="2010-09-18T18:14:41.033" />
  <row Id="4933" PostTypeId="2" ParentId="4920" CreationDate="2010-09-18T18:15:41.697" Score="3" ViewCount="" Body="&lt;p&gt;Based on &lt;a href=&quot;http://imgur.com/4DDa7.png&quot; rel=&quot;nofollow&quot;&gt;your work&lt;/a&gt; as linked in the comments on &lt;a href=&quot;http://math.stackexchange.com/questions/4920/second-derivatives-using-implicit-differentiation/4922#4922&quot;&gt;J.M.'s answer&lt;/a&gt;, you've very nearly got it (except for a typo in differentiating *: there's a dy/dx where there should be a d/dx, but the mathematics that follows is correct as if it were d/dx).  The numerator you have is &#xA;$$\begin{align}&#xA;-2 ((2 x - y)^2 &amp;amp;+ (2 x - y) (2 y + x) - (2 y + x)^2)&#xA;\\&#xA;&amp;amp;=-10x^2+10xy+10y^2&#xA;\\&#xA;&amp;amp;=10(y^2+xy-x^2)&#xA;\\&#xA;&amp;amp;=10\cdot9&#xA;\\&#xA;&amp;amp;=90.&#xA;\end{align}$$&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-09-18T18:15:41.697" CommentCount="3" />
  <row Id="4934" PostTypeId="2" ParentId="4927" CreationDate="2010-09-18T18:19:05.497" Score="5" ViewCount="" Body="&lt;p&gt;If you know modular arithmetic: $\rm\; e\equiv 1,5\;\Rightarrow\; e^2\equiv e \;\Rightarrow\; e^k\equiv e\pmod{10}\;$ for all $\rm\; k&gt;0$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If not, use induction: suppose $\rm\;\; e^2 = e + 10j\;$ and  $\rm\; (e + 10 m)^k = e + 10n$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;then $\rm\; (e + 10 m)^{k+1} = (e + 10n) (e + 10m) = e + 10(j+em+en+10mn)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;REMARK&lt;/strong&gt; Note how the use of modular arithmetic has allowed us to rephrase the problem in a completely arithmetical form that makes the induction step utterly trivial - namely that $\rm 1^k \equiv 1$ or, more generally that $\rm\; e^2 \equiv e\;\Rightarrow e^k \equiv e\;$ for $\rm\: k&gt;0$. This is but one very simple illustration of how one may often reduce problems about integers to simpler problems about integers modulo $\rm\:m\:$. Because the integers modulo $\rm\:m\:$ satisfy many of  the same arithmetical laws as the integers (associative, commutative, distributive, etc) - i.e. they form a ring - we can reuse our strong intuition about integers to help discover proofs in this simpler structure integers modulo $\rm\:m\:$. And, if we're lucky, we can sometimes lift these simple solutions from the integers modulo $\rm\:m\:$ up to the integers. If you study abstract algebra you'll see many more examples of problems solved by passing to such simpler structures (technically they are known as &lt;a href=&quot;http://en.wikipedia.org/wiki/Quotient_ring&quot; rel=&quot;nofollow&quot;&gt;quotient objects&lt;/a&gt; - structures modulo a congruence, i.e. a structure-compatible equivalence relation). Said in an informal way, such &lt;strong&gt;reduction to quotient objects&lt;/strong&gt; is a method to &lt;strong&gt;divide and conquer&lt;/strong&gt; mathematically.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-09-22T13:48:47.127" LastActivityDate="2010-09-22T13:48:47.127" CommentCount="2" />
  <row Id="4935" PostTypeId="2" ParentId="4931" CreationDate="2010-09-18T18:19:44.653" Score="1" ViewCount="" Body="&lt;p&gt;To show that the segments are perpendicular, consider their slopes.  The product of the slopes of perpendicular lines is -1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To show that BE bisects AC, show that the distance AE is equal to the distance EC.&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-09-18T18:19:44.653" />
  <row Id="4936" PostTypeId="2" ParentId="4931" CreationDate="2010-09-18T18:21:49.947" Score="1" ViewCount="" Body="&lt;p&gt;Consider the slopes of AE and EB. AE has a slope of -1, whereas EB has a slope of 1. Two lines on a plane are perpendicular if their slopes multiply to -1, so since -1*1=-1, the lines are in fact perpendicular.&lt;/p&gt;&#xA;" OwnerUserId="796" LastActivityDate="2010-09-18T18:21:49.947" CommentCount="1" />
  <row Id="4937" PostTypeId="2" ParentId="4931" CreationDate="2010-09-18T18:39:39.277" Score="1" ViewCount="" Body="&lt;p&gt;BE is perpendicular to AC, that is correct. You don't need any analytic geometry to see that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Observe that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;AE = EC, so E is the midpoint of AC.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;AB = CB, so ABC is isosceles triangle with B being the the vertex opposite the base.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus BE is perpendicular to AC, as the line joining the vertex opposite the base (B) to the midpoint of base (E) is perpendicular to the base (AC).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, analytic geometry is probably more useful in similar situations.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-09-18T19:00:09.837" LastActivityDate="2010-09-18T19:00:09.837" />
  <row Id="4938" PostTypeId="2" ParentId="4869" CreationDate="2010-09-18T18:41:43.833" Score="9" ViewCount="" Body="&lt;p&gt;Let's suppose, imitating the real case, the we want to integrate the expression&#xA;$f(z) dz$, where $f$ is a function of the complex variable $z$.  What would this mean?&#xA;Well, imitating the real case, we find that we have to form Riemann sums&#xA;$\sum_{i = 1}^{n-1} f(z_i) (z_{i+1} - z_i)$, and then let the absolute value of the differences$| z_{i+1} - z_i|$ tend to zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This sum is easily seen to be the line integral along the polygonal arc joining $z_1$ to&#xA;$z_n$ via $z_2, z_3, \ldots ,$ of the piecewise constant function whose value is $f(z_i)$ on&#xA;the arc joining $z_i$ to $z_{i+1}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we can form the limit in any reasonable sense, we will get a line integral of $f(z) dz$&#xA;along the curve which is the limit of these polygonal arcs as&#xA;$n \to \infty$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, the limit we get will depend not just on the end points $a$ and $b$,&#xA;but (at least &lt;em&gt;a priori&lt;/em&gt;)  on the particular path that arises as the limit of these polygonal arcs.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Seeing this, we now see that to define the integral $\int_a^b f(z) dz$, we should&#xA;choose a path joining $a$ to $b$, and choose the $z_i$ to lie along this path.&#xA;The resulting integral will depend on this choice of path &lt;em&gt;a priori&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What we are seeing is that although passing from the real to the complex numbers involves&#xA;passing from a one-dimensional to a two-dimensional situation, taking an integral&#xA;of $f(z) dz$ still involves choosing a finite sequence of points and then passing to the&#xA;limit.  In the real case, these points fill out an interval (in the limit), while in the complex case they will fill out a curve.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you wanted to write an integral over an area in terms of complex variables, you would&#xA;need to do integrals like $\int f(z) dzd\overline{z}$; this now involves adding up Riemann sums in which the terms involve products $\Delta z\Delta\overline{z}$ of a small change in $z$ and an independent small change in $\overline{z}$.  One computes that this&#xA;is the same as the integral $-2i\int f(z) dxdy$, which is a usual double integral in the plane.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In summary: If we try to generalize the one real variable case to the one complex variable case,&#xA;we find that we are forced to choose a path along which to take the integral, because&#xA;the complex numbers lie in a two-dimensional plane, while the idea of forming Riemann sums to compute the integral requires us to choose some arc along which the points $z_i$ lie.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fundamentally, adding up small changes $\Delta z$ (weighted by values of the function $f$)&#xA;requires looking at changes in $z$ along some one-dimensional object.  Thus $\int f(z) dz$ has to be a line integral.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Adding up products of indepenent small&#xA;changes $\Delta z$ and $ \Delta \overline{z}$ (weighted by vaues of the function $f$)  requires moving in two independent directions, and so takes place over a two-dimensional&#xA;region.  Thus $\int f(z) dz d \overline{z}$ has to be a double integral.&lt;/p&gt;&#xA;" OwnerUserId="221" LastEditorUserId="221" LastEditorDisplayName="" LastEditDate="2010-10-20T23:53:13.067" LastActivityDate="2010-10-20T23:53:13.067" />
  <row Id="4939" PostTypeId="2" ParentId="4931" CreationDate="2010-09-18T18:57:22.877" Score="0" ViewCount="" Body="&lt;p&gt;You can also use the simple general formula that I mentioned in my &lt;a href=&quot;http://math.stackexchange.com/questions/4914/cartesian-equation-for-the-perpendicular-bisector-of-a-line/4924#4924&quot;&gt;prior post&lt;/a&gt;, namely  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\;$ The equation is  $\rm\;\: 2\ (A-B)\cdot (x,y) \;=\; |A|-|B|\;\;\;$ where $\rm\;\;\; |(a,b)| \ =\  a^2 + b^2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which, if worked out, yields $\rm\;\: (-8,8)\cdot (x,y) \;=\; \;17 \;- 9\;\;\:$ for $\rm\; A = (4,-1),\;\; B = (0,3)$   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;which, after simplifying, yields the equation $\rm\;\;\: y \;=\; x-1$&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-09-18T18:57:22.877" />
  <row Id="4941" PostTypeId="1" AcceptedAnswerId="4943" CreationDate="2010-09-18T19:08:10.340" Score="3" ViewCount="97" Body="&lt;p&gt;I was reviewing my class notes and found the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;The name 'torsion' comes from topology and refers to spaces that are twisted, ex. Mobius band&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In our notes we used the following definition for torsion element and torsion module:&#xA;An element m of an R-module M is called a torsion element if $rm=0$ for some $r\in R$. &#xA;A torsion module is a module which consists solely of torsion elements&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the relationship between torsion modules and twisted spaces? Was the definition of torsion module somehow motivated from topological considerations of twisted spaces?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't really see any obvious connection. I'm taking my first topology class this semester, so I apologize if this is something you learn about later in courses like algebraic topology, but I haven't been able to find any explanation of this.&lt;/p&gt;&#xA;" OwnerUserId="249" LastEditorUserId="249" LastEditorDisplayName="" LastEditDate="2010-09-18T19:17:51.037" LastActivityDate="2010-09-18T19:17:51.037" Title="Relationship between torsion modules and topology" Tags="&lt;soft-question&gt;&lt;terminology&gt;&lt;intuition&gt;&lt;topology&gt;&lt;modules&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="1" />
  <row Id="4942" PostTypeId="1" CreationDate="2010-09-18T19:08:15.440" Score="0" ViewCount="161" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;The perpendicular bisector of the line&#xA;  joining A(0,1) and C(-4,7) intersects&#xA;  the x-axis at B and the y-axis at D.&#xA;  Find the area of the quadrilateral&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Thank you in advance!&lt;/p&gt;&#xA;" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-09-19T03:48:57.737" LastActivityDate="2010-09-19T06:33:39.527" Title="Area of a quadrilateral" Tags="&lt;geometry&gt;&lt;algebra-precalculus&gt;&lt;analytic-geometry&gt;" AnswerCount="4" CommentCount="7" />
  <row Id="4943" PostTypeId="2" ParentId="4941" CreationDate="2010-09-18T19:14:21.010" Score="6" ViewCount="" Body="&lt;p&gt;When you compute the homology groups of &quot;twisted&quot; spaces (which are abelian groups), you (sometimes) find that they contain non-zero torsion elements; furthermore, the presence of these particular elements in the homology is due to the twisting (in that, when you compute the homology groups, you see that is the twisting in the space that causes the calculation to give rise to torsion elements).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since you haven't studied algebraic topology yet, I won't say more here.  Hopefully, while necessarily vague, the above description gives you some feeling for the meaning of the remark in your notes.&lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2010-09-18T19:14:21.010" />
  <row Id="4944" PostTypeId="2" ParentId="4941" CreationDate="2010-09-18T19:15:31.693" Score="5" ViewCount="" Body="&lt;p&gt;The definition of torsion in modules is a generalization of the definition of torsion in $\mathbb{Z}$-modules, e.g. abelian groups.  Torsion in abelian groups refers to elements of finite order, and this in turn relates to topology because to any topological space we can associate abelian groups called (integral) &lt;a href=&quot;http://en.wikipedia.org/wiki/Homology_(mathematics)&quot; rel=&quot;nofollow&quot;&gt;homology groups&lt;/a&gt;, and torsion in these groups is suggestive of a kind of &quot;twistedness&quot; in the space.  The simplest example of this is in the first integral homology of closed surfaces; the group has torsion if and only if the surface is non-orientable, such as the &lt;a href=&quot;http://en.wikipedia.org/wiki/Klein_bottle&quot; rel=&quot;nofollow&quot;&gt;Klein bottle&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-09-18T19:15:31.693" />
  <row Id="4945" PostTypeId="2" ParentId="4942" CreationDate="2010-09-18T19:17:41.793" Score="1" ViewCount="" Body="&lt;p&gt;Tag the question as homework and show what you've already tried, otherwise it will be (rightfully) closed soon.&lt;/p&gt;&#xA;&#xA;&lt;hr/&gt;&#xA;&#xA;&lt;p&gt;Now, we just had &lt;a href=&quot;http://math.stackexchange.com/questions/4931/perpendicular-bisector&quot;&gt;this question&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;find middle of $\overrightarrow{AC}$&lt;/li&gt;&#xA;&lt;li&gt;create a line's equation $b: y = m\cdot x + c$ for the bisector&lt;/li&gt;&#xA;&lt;li&gt;compute the intersections of $b$ and the coordinate axis&lt;/li&gt;&#xA;&lt;li&gt;compute the &lt;a href=&quot;http://en.wikipedia.org/wiki/Quadrilateral#Area_of_a_convex_quadrilateral&quot; rel=&quot;nofollow&quot;&gt;area&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="917" LastActivityDate="2010-09-18T19:17:41.793" CommentCount="1" />
  <row Id="4946" PostTypeId="2" ParentId="4942" CreationDate="2010-09-18T19:19:26.537" Score="3" ViewCount="" Body="&lt;p&gt;The perpendicular bisector of the segment $AC$, by definition, passes through the midpoint of the segment. Look at your drawing. The segment $BD$ is a base to the triangles $BCD$ and $BAD$. Note also that the heights of the triangles are equal.&lt;/p&gt;&#xA;" OwnerUserId="1313" LastActivityDate="2010-09-18T19:19:26.537" />
  <row Id="4947" PostTypeId="1" AcceptedAnswerId="4954" CreationDate="2010-09-18T19:34:37.793" Score="0" ViewCount="123" Body="&lt;p&gt;I have to show that any line parallel to Ax+ By + c =0 is of the form Ax + By + k =0&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do I show this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you! &lt;/p&gt;&#xA;" OwnerUserId="1968" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-09-18T20:29:49.010" LastActivityDate="2010-09-19T11:34:27.870" Title="Equation of a line" Tags="&lt;geometry&gt;&lt;algebra-precalculus&gt;&lt;analytic-geometry&gt;" AnswerCount="4" />
  <row Id="4948" PostTypeId="1" AcceptedAnswerId="4951" CreationDate="2010-09-18T19:35:35.967" Score="1" ViewCount="145" Body="&lt;p&gt;&lt;strong&gt;The Questions&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;70% of all vehicles pass inspection. Assuming vehicles pass or fail independently. What is the probability:&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;a) exactly one of the next 3 vehicles  passes&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;b) at most 1 of the next 3 vehicles  passes&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The answer to a) is .189. The way I calculated it was:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;P(success) * P(fail) * P(fail) + &#xA;P(fail) * P(success) * P(fail) + &#xA;P(fail) * P(fail) * P(success) = .7*.3*.3 + .3*.7*.3 + .3*.3*.7 = .189&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I summed the 3 possible permutations of 1 success and 2 failures. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For b) the answer is .216. To get that answer you take your answer to a) and add the probability of exactly 0 successes which is P(fail) * P(fail) * P(fail) =&gt; &lt;code&gt;.189 + .3*.3*.3 = .216&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I don't understand is why the probability of exactly 0 successes doesn't follow the pattern of exactly 1 success. Why doesn't the &quot;formula&quot; work:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;P(fail) * P(fail) * P(fail) + &#xA;P(fail) * P(fail) * P(fail) + &#xA;P(fail) * P(fail) * P(fail) = .3*.3*.3+.3*.3*.3+.3*.3*.3 = .081&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;=&amp;gt; .189 + .081 = .27 (not .216)&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I'm wondering if I calculated the answer to a) the wrong way, and it was merely a coincidence that I got the right answer!&lt;/p&gt;&#xA;" OwnerUserId="1977" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-09-19T01:57:52.823" LastActivityDate="2010-09-19T01:57:52.823" Title="What is wrong with my reasoning?" Tags="&lt;probability&gt;" AnswerCount="3" />
  <row Id="4949" PostTypeId="2" ParentId="4947" CreationDate="2010-09-18T19:38:46.163" Score="3" ViewCount="" Body="&lt;p&gt;Examples don't show it in any case. You cannot prove that something holds for infinitely many things by doing nothing except checking finitely many examples.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you need to do is remember that two lines are parallel if and only if they have the same slope or they are both vertical (of the form $x=k$ for some constant $k$). Then figure out what the slope of an arbitrary line $\alpha x+\beta y+\gamma=0$ would be in terms of $\alpha$, $\beta$, and $\gamma$ (careful when dividing by constants to consider what happens if the constant you are dividing by is equal to zero). Then see what this means if you want to find all lines $\alpha x+\beta y+\gamma = 0$ that are parallel to a given line $Ax+By+c=0$. &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-09-18T19:46:31.370" LastActivityDate="2010-09-18T19:46:31.370" CommentCount="2" />
  <row Id="4950" PostTypeId="2" ParentId="4947" CreationDate="2010-09-18T19:44:09.873" Score="0" ViewCount="" Body="&lt;p&gt;You can solve the following system of equations:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{cases}&#xA;Ax+By+c=0,\\&#xA;Ax+By+k=0&lt;br&gt;&#xA;\end{cases}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you subtract the second line from the firs, you obtain $c-k=0$. Then, if $c\ne k$, you have no solution: the lines are parallel.&lt;/p&gt;&#xA;" OwnerUserId="354" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-09-18T19:47:31.967" LastActivityDate="2010-09-18T19:47:31.967" CommunityOwnedDate="2010-09-18T19:44:43.477" CommentCount="2" />
  <row Id="4951" PostTypeId="2" ParentId="4948" CreationDate="2010-09-18T19:44:42.060" Score="6" ViewCount="" Body="&lt;p&gt;The case that the &lt;em&gt;first&lt;/em&gt; car passes and the next two fail is &lt;em&gt;different&lt;/em&gt; from the case that the first car fails, the second passes, and the third fails. But there is only one case in which all cars fail: namely, &lt;em&gt;all cars fail&lt;/em&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Think of it with coin tosses: if you flip a penny, a nickel, and a dime (different coins, in case you are not in the U.S. or Canada), then there is only &lt;em&gt;one&lt;/em&gt; way for all three coins to come out tails,  namely, the penny comes out tails, the nickel comes out tails, and the dime comes out tails. But there are three ways in which exactly one coin comes out heads: you can have the penny come out heads and the nickel and dime come out tails; you can have the nickel come out heads and the penny and dime come out tails; or you can have the dime come out heads and the penny and nickel come out tails. These are three different outcomes, but you do not have three different ways in which the three can come out tails.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-09-18T19:54:25.133" LastActivityDate="2010-09-18T19:54:25.133" />
  <row Id="4952" PostTypeId="2" ParentId="4948" CreationDate="2010-09-18T19:46:02.347" Score="1" ViewCount="" Body="&lt;p&gt;I'm not too confident with my probability, but I'll try my hand at an explanation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the first part, you are considering the next 3 trials as successive events so you can have a success in either the first spot, second spot, or third spot.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In part b, you require all 3 events to be failures so there is only 1 way that the 3 failures can occur in the next 3 events.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does that make any sense? I never felt probability was my strong point :)&lt;/p&gt;&#xA;" OwnerUserId="249" LastActivityDate="2010-09-18T19:46:02.347" />
  <row Id="4953" PostTypeId="2" ParentId="4948" CreationDate="2010-09-18T19:48:54.787" Score="3" ViewCount="" Body="&lt;p&gt;In part a, there are exactly three ways for one out of three cars to pass, they are the three possibilities that you added. But there is only one way for all the cars to fail! The first car must fail, the second car must fail, and the third car must fail. Since there is only one way for this to happen you only consider this one probability. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Btw there are also three ways for exactly two cars to pass and only one way for all three of them to pass. Hope this helps. &lt;/p&gt;&#xA;" OwnerUserId="1352" LastActivityDate="2010-09-18T19:48:54.787" />
  <row Id="4954" PostTypeId="2" ParentId="4947" CreationDate="2010-09-18T20:16:31.697" Score="1" ViewCount="" Body="&lt;p&gt;(Based on &lt;a href=&quot;http://math.stackexchange.com/questions/4947/equation-of-a-line/4950#4950&quot;&gt;zar's answer&lt;/a&gt; and Moron's comment there.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If two lines $Ax+By+C=0$ and $Dx+Ey+F=0$ are parallel, then they have no points of intersection (or infinitely many if you allow a line to be parallel to itself).  In either case, the system of equations &#xA;$$\begin{cases}&#xA;Ax+By+C=0,\\&#xA;Dx+Ey+F=0 &#xA;\end{cases}$$&#xA;will not have a single, unique solution.  The system of equations is equivalent to&#xA;$$\begin{bmatrix}&#xA;A &amp;amp; B\\ &#xA;D &amp;amp; E&#xA;\end{bmatrix}\cdot\begin{bmatrix}&#xA;x\\&#xA;y&#xA;\end{bmatrix}+\begin{bmatrix}&#xA;C\\&#xA;F&#xA;\end{bmatrix}=0$$&#xA;or&#xA;$$\begin{bmatrix}&#xA;A &amp;amp; B\\ &#xA;D &amp;amp; E&#xA;\end{bmatrix}\cdot\begin{bmatrix}&#xA;x\\&#xA;y&#xA;\end{bmatrix}=\begin{bmatrix}&#xA;-C\\&#xA;-F&#xA;\end{bmatrix}.$$&#xA;This matrix equation has a single unique solution if and only if the coefficient matrix $\begin{bmatrix}&#xA;A &amp;amp; B\\ &#xA;D &amp;amp; E&#xA;\end{bmatrix}$ is invertible, which is true if and only if its determinant is nonzero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, if the lines are parallel, then &#xA;$$\begin{vmatrix}&#xA;A &amp;amp; B\\ &#xA;D &amp;amp; E&#xA;\end{vmatrix}=AE-BD=0$$&#xA;or&#xA;$$\frac{A}{D}=\frac{B}{E}$$&#xA;which means that $\frac{A}{D}(Dx+Ey+F)=Ax+By+\frac{FA}{D}=0$, so $Dx+Ey+F=0$ can be written in the form $Ax+By+k=0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;edit&lt;/em&gt;: from Arturo Magidin's comment below, to handle the cases where $D=0$ or $E=0$ (in which $A/D = B/E$ makes no sense):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If $D=0$ then $A=0$ or $E=0$, but you cannot have both $D$ and $E$ zero. So $D=0$ implies $A=0$, and $B\neq 0\neq E$. Similarly, if $E=0$, then $B=0$ and $A\neq 0 \neq D$. And you get the similar conclusion without forgetting about horizontal and vertical lines.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="72" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-09-19T03:49:55.520" LastActivityDate="2010-09-19T03:49:55.520" CommentCount="2" />
  <row Id="4955" PostTypeId="2" ParentId="4942" CreationDate="2010-09-18T20:21:36.047" Score="1" ViewCount="" Body="&lt;p&gt;If the quadrilateral is ABCD--that is, if AC and BD are its diagonals--then the quadrilateral has perpendicular diagonals, so its area is half the product of the lengths of the diagonals (which you can find from the coordinates of A, B, C, and D).&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-09-18T20:21:36.047" />
  <row Id="4957" PostTypeId="2" ParentId="4905" CreationDate="2010-09-18T20:52:55.890" Score="2" ViewCount="" Body="&lt;p&gt;Limit (and even asymptotic series) of your integral can be calculated using &lt;a href=&quot;http://en.wikipedia.org/wiki/Laplace%27s_method&quot; rel=&quot;nofollow&quot;&gt;Laplace's method&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/Stirling%27s_approximation&quot; rel=&quot;nofollow&quot;&gt;Stirling's approximation&lt;/a&gt;. The solution can be done in 5 steps:&lt;br&gt;&#xA;1. Prove, that integral over $[0,1/n]$ can be neglected. In order to do this, use Stirling's approximation for all factorials except $(nx)!$, which can be approximated by a constant.&lt;br&gt;&#xA;2. Prove, that integral over $[1/n,\delta]$ can be neglected (for $\delta&amp;lt;1/2$). In order to do this, use Stirling's approximation for all factorials (note, that $z!/(\sqrt{2 \pi z} (z/e)^z))$ is bounded from both sides by positive constants if $z&gt;=1$).&lt;br&gt;&#xA;3. Using symmetry you know, that your limit is equal to $\lim_{n\to\infty}\int_{\delta}^{1-\delta}dx(\dots)$.&lt;br&gt;&#xA;4. Use Stirling's approximation to approximate factorials. You will get the following:&lt;br&gt;&#xA;$$\int_{\delta}^{1-\delta}dx\frac{n!2^{-n}n}{(xn)!(n-xn)!\sqrt{x(1-x)}}=(1+O(n^{-1}))\sqrt{\frac{n}{2\pi}}\cdot$$&#xA;$$\int_{\delta}^{1-\delta} x^{-1}(1-x)^{-1}exp\bigl(-n(x\ln x+(1-x)\ln(1-x)+\ln 2)\bigr)dx$$&lt;br&gt;&#xA;5. Apply Laplace's method to the last integral and get it's asymptotic.&lt;/p&gt;&#xA;" OwnerUserId="695" LastActivityDate="2010-09-18T20:52:55.890" CommentCount="2" />
  <row Id="4959" PostTypeId="1" AcceptedAnswerId="4991" CreationDate="2010-09-18T21:26:36.597" Score="1" ViewCount="83" Body="&lt;p&gt;In the area of convex analysis and the area of optimization in their general sense, are convex subsets assumed to be in vector spaces or topological vector spaces? Are convex functions defined to be from convex subsets in vector spaces or topological vector spaces to general field or some special field such as $\mathbb{R}$ or $\mathbb{C}$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now instead of considering the general sense, what are the special cases that have been extensively studied and have significance in application?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks and regards!&lt;/p&gt;&#xA;" OwnerUserId="1281" LastActivityDate="2010-09-19T13:27:55.917" Title="study objects of convex analysis and optimization" Tags="&lt;optimization&gt;&lt;convex-analysis&gt;&lt;convex-optimization&gt;" AnswerCount="1" />
  <row Id="4960" PostTypeId="2" ParentId="4106" CreationDate="2010-09-18T22:31:21.533" Score="1" ViewCount="" Body="&lt;p&gt;This is a correction of the answer of David Bar Moshe &lt;strong&gt;edit:&lt;/strong&gt; and generalizing it for the case, when $\mathbf{A}$ is degenerate. Using formulas for Fourier and inverse Fourier transforms we can write&#xA;$$f(0)=\frac{1}{2\pi}\int_{-\infty}^{\infty}\mathrm{d}k\int_{-\infty}^{\infty} \mathrm{d}y f(y)e^{iky}. (1)$$&#xA;Suppose $S_d(y)=$&quot;$S_d$ shifted by $(y,0,\dots,0)$&quot; and&#xA;$f(y)=\int_{\mathbf{x} \in S_d(y)} \mathrm{d}\mathbf{x} \exp(-\mathbf{x}^T \mathbf{A}\mathbf{x}).$&#xA;Then using the formula (1) we will get&#xA;$$N^{-1}=f(0) = \frac{1}{2\pi}\int_{-\infty}^{\infty}\mathrm{d}k\int_{-\infty}^{\infty} \mathrm{d}y f(y)e^{iky}  =$$&#xA;$$\frac{1}{2\pi}\int_{-\infty}^{\infty}\mathrm{d}k\int_{-\infty}^{\infty} \mathrm{d}y \int_{\mathbf{x} \in S_d(y)}\mathrm{d}\mathbf{x} \exp(-\mathbf{x}^T \mathbf{A}\mathbf{x} + iky) =$$&#xA;$$\frac{\|\mathbf{v}\|}{2\pi}\int_{-\infty}^{\infty}\mathrm{d}k \int_{\mathbf{x} \in \mathbb{R}^d}\mathrm{d}\mathbf{x} \exp(-\mathbf{x}^T \mathbf{A}\mathbf{x} + ik \mathbf{v}^T \mathbf{x}),$$&#xA;where $\mathbf{v}$ is vector, orthogonal to the plane we are interested in. For example, we can take $\mathbf{v}=(1,\dots,1)$. Note $\|\mathbf{v}\|$ in the last integral, appearing from change of coordinates from $y$ and coordinates in $S_d$ to coordinates in $\mathbb{R}^d$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now $N^{-1}=$&#xA;$$\frac{\|\mathbf{v}\|}{2\pi}\int_{-\infty}^{\infty} \mathrm{d}k \int_{\mathbf{x} \in \mathbb{R}^d}\mathrm{d}\mathbf{x} \exp(-(\mathbf{x}+\frac{i}{2}k\mathbf{A}^{-1}\mathbf{v})^T \mathbf{A}(\mathbf{x}+\frac{i}{2}k\mathbf{A}^{-1}\mathbf{v}) - \frac{k^2}{4} \mathbf{v}^T \mathbf{A}^{-1}\mathbf{v})=$$&#xA;$$\frac{\|\mathbf{v}\|}{2\pi} \frac{\pi^{d/2}}{\sqrt{\det{A}}} \frac{2\sqrt{\pi}}{\sqrt{\mathbf{v}^T \mathbf{A}^{-1}\mathbf{v}}}=\frac{\|\mathbf{v}\| \pi^{(d-1)/2}}{\sqrt{\det{A}}\sqrt{\mathbf{v}^T \mathbf{A}^{-1}\mathbf{v}}}=\frac{\|\mathbf{v}\| \pi^{(d-1)/2}}{\sqrt{\mathbf{v}^T \mathbf{C}\mathbf{v}}},$$&#xA;where $\mathbf{C}$ is &lt;a href=&quot;http://en.wikipedia.org/wiki/Adjugate_matrix&quot; rel=&quot;nofollow&quot;&gt;adjugate matrix&lt;/a&gt; to $\mathbf{A}$. Initial integral and the answer both are continuous in $\mathbf{A}$ when $A$ restricted to $S_d$ is not degenerate. From this we can conclude, that equation&#xA;$N^{-1}=\frac{\|\mathbf{v}\| \pi^{(d-1)/2}}{\sqrt{\mathbf{v}^T \mathbf{C}\mathbf{v}}}$&#xA;is true whenever $\mathbf{A}$ restricted to $S_d$ is not degenerate (i.e. even if $\mathbf{A}$ is degenerate).&lt;/p&gt;&#xA;" OwnerUserId="695" LastEditorUserId="695" LastEditorDisplayName="" LastEditDate="2010-09-18T23:16:59.507" LastActivityDate="2010-09-18T23:16:59.507" />
  <row Id="4961" PostTypeId="1" AcceptedAnswerId="4971" CreationDate="2010-09-18T23:54:21.547" Score="19" ViewCount="1076" Body="&lt;p&gt;I was just looking at a calculus textbook preparing my class for next week on complex numbers. I found it interesting to see as an exercise a way to calculate the usual freshman calculus integrals $\int e^{ax}\cos{bx}\ dx$ and $\int e^{ax}\sin{bx}\ dx$ by taking the real and imaginary parts of the &quot;complex&quot; integral $\int e^{(a + bi)x} \ dx$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So my question is if you know of other &quot;relatively interesting&quot; results that can be obtained easily by using complex numbers. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It may be something that one can present to engineering students taking the usual calculus sequence, but I'm also interested in somewhat more advanced examples (if they're available under the condition that the process to get them is somewhat easy, or not too long). Thank you all.&lt;/p&gt;&#xA;" OwnerUserId="900" LastActivityDate="2010-10-27T04:38:14.523" CommunityOwnedDate="2010-09-19T00:13:12.210" Title="Interesting results easily achieved using complex numbers" Tags="&lt;complex-analysis&gt;&lt;complex-numbers&gt;" AnswerCount="17" CommentCount="5" FavoriteCount="16" />
  <row Id="4962" PostTypeId="2" ParentId="4961" CreationDate="2010-09-19T00:16:20.763" Score="3" ViewCount="" Body="&lt;p&gt;If you feel like picking up a good book on this topic, &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0691118221&quot; rel=&quot;nofollow&quot;&gt;&quot;Dr. Euler's Fabulous Formula&quot; by Nahin&lt;/a&gt; provides many beautiful applications of complex numbers to other parts of mathematics (he is an electrical engineer).&lt;/p&gt;&#xA;" OwnerUserId="1352" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-09-20T23:56:06.797" LastActivityDate="2010-09-20T23:56:06.797" CommunityOwnedDate="2010-09-19T00:16:20.763" CommentCount="2" />
  <row Id="4963" PostTypeId="1" AcceptedAnswerId="5582" CreationDate="2010-09-19T00:51:02.560" Score="7" ViewCount="243" Body="&lt;p&gt;Suppose $A$ is a commutative algebra over $\mathbb{R}$ with unity. $\mathbb{R}$-linear map $\xi\colon A\to A$ is a &lt;a href=&quot;http://en.wikipedia.org/wiki/Derivation_%28abstract_algebra%29&quot; rel=&quot;nofollow&quot;&gt;derivation&lt;/a&gt; of $A$ iff $\xi(ab)=a\xi(b)+\xi(a)b$ for any $a,b\in A$.&lt;br&gt;&#xA;If $\gamma\colon \mathbb{R}\to A$, $a\in A$, then we say, that $a=\frac{\partial}{\partial t}| _ {t=\tau} \gamma(t)$ iff $h(a) =\frac{\partial}{\partial t}|_{t=\tau} h(\gamma(t))$ for any $\mathbb{R}$-linear map $h\colon A\to\mathbb{R}$.&lt;br&gt;&#xA;Suppose $\xi$ is a derivation of $A$. Then $\Phi\colon A\times\mathbb{R}\to A$ is it's flow iff $\Phi(a,0)=a$ for any $a\in A$ and&lt;br&gt;&#xA;$$\frac{\partial}{\partial t} \Phi(a,t) = \xi \Phi(a,t).\tag{1}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; I like algebras $A$, such that any derivation of $A$ possesses a flow. Is there any simple sufficient condition for them?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Examples:&lt;/strong&gt; (with sketches of the proofs)&lt;br&gt;&#xA;1. Algebra $C^\infty(M)$ of smooth functions on a &lt;a href=&quot;http://en.wikipedia.org/wiki/Closed_manifold&quot; rel=&quot;nofollow&quot;&gt;closed manifold&lt;/a&gt; $M$ --- yes (if I haven't made a mistake), any derivation possesses a flow. This, I believe, can be checked using &lt;a href=&quot;http://en.wikipedia.org/wiki/Picard%E2%80%93Lindel%C3%B6f_theorem&quot; rel=&quot;nofollow&quot;&gt;Picard-Lindelof theorem&lt;/a&gt;.&lt;br&gt;&#xA;2. Algebra $C^\infty((0,1))$ of smooth functions on an interval --- no, $\frac{\partial}{\partial x}$ does not possess a flow.&lt;br&gt;&#xA;3. Algebra $C^\infty([0,1])$ of smooth functions on a segment --- yes, any derivation possesses a flow, but it's not always unique (for example, for $\frac{\partial}{\partial x}$ it is not). In order to prove this, one can consider an embedding of $[0,1]$ to some closed manifold $N$ and prolong any function from $[0,1]$ to $N$. Then use example 1.&lt;br&gt;&#xA;4. Algebra $C^\infty(\mathbb{R})$ --- no, because it is isomorphic to the algebra from example 2.&lt;br&gt;&#xA;5. Algebra $\mathbb{R}[x]$ --- no. In order to prove this one can consider derivation $\xi=x^2\frac{\partial}{\partial x}$ and manually solve equation (1) for $a=x$. Any solution locally should be of the form $\frac{x}{1+xt}$. It is not in $\mathbb{R}[x]$.&lt;br&gt;&#xA;6. Algebra $\mathbb{R}[x,y]/(x^2+y^2-1)$ --- no. In order to prove this take $\xi = y (x\frac{\partial}{\partial y} - y \frac{\partial}{\partial x})=\sin(\varphi)\frac{\partial}{\partial \varphi}$ and solve equation (1) manually (locally) in polar coordinates (take, for example, $a=y$). Check that the answer is not a polynomial.&lt;/p&gt;&#xA;" OwnerUserId="695" LastEditorUserId="695" LastEditorDisplayName="" LastEditDate="2010-09-19T19:54:08.407" LastActivityDate="2010-09-27T18:24:31.163" Title="In which commutative algebras does any derivation possess a flow?" Tags="&lt;algebraic-geometry&gt;&lt;differential-geometry&gt;&lt;differential-equations&gt;&lt;commutative-algebra&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="1" />
  <row Id="4964" PostTypeId="2" ParentId="4961" CreationDate="2010-09-19T01:01:09.213" Score="5" ViewCount="" Body="&lt;p&gt;You might get some ideas from this MathOverflow thread:&#xA;&lt;a href=&quot;http://mathoverflow.net/questions/30156/demystifying-complex-numbers/30185&quot; rel=&quot;nofollow&quot;&gt;http://mathoverflow.net/questions/30156/demystifying-complex-numbers/30185&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="940" LastEditorUserId="1778" LastEditorDisplayName="" LastEditDate="2010-09-21T03:59:19.373" LastActivityDate="2010-09-21T03:59:19.373" CommunityOwnedDate="2010-09-19T01:01:09.213" />
  <row Id="4965" PostTypeId="2" ParentId="4961" CreationDate="2010-09-19T01:01:49.913" Score="26" ViewCount="" Body="&lt;p&gt;Without complex numbers, it's a mystery why the power series for $1/(1+ x^2)$ centered at the origin has radius of convergence 1.  The function is infinitely differentiable, no strange behavior at 1 or -1, etc.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But in the complex domain, there are singularities at $\pm i$ which explains why the radius of convergence is 1.&lt;/p&gt;&#xA;" OwnerUserId="247" LastActivityDate="2010-09-19T01:01:49.913" CommunityOwnedDate="2010-09-19T01:01:49.913" CommentCount="3" />
  <row Id="4966" PostTypeId="2" ParentId="4961" CreationDate="2010-09-19T01:54:08.453" Score="4" ViewCount="" Body="&lt;p&gt;The prime number theorem can be proved without complex analysis, but the elementary proof is generally considered much more difficult than the standard one (which deduces it from analytic properties of the zeta-function).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know if Dirichlet's theorem on primes in arithmetic progressions can be proved without complex analysis (one special case can using cyclotomic polynomials), but the analytic proof is the only one I've seen.  &lt;strong&gt;Edit:&lt;/strong&gt; Adrian below says that you can avoid complex analysis, but it won't look as nice. KCd has given references to specific papers. In either case, note that the analytic proof chronologically preceded the elementary one.&lt;/p&gt;&#xA;" OwnerUserId="536" LastEditorUserId="536" LastEditorDisplayName="" LastEditDate="2010-09-19T13:54:06.360" LastActivityDate="2010-09-19T13:54:06.360" CommunityOwnedDate="2010-09-19T01:54:08.453" CommentCount="4" />
  <row Id="4967" PostTypeId="2" ParentId="4961" CreationDate="2010-09-19T01:58:33.013" Score="2" ViewCount="" Body="&lt;p&gt;Since one of my interests is the geometry of plane curves, the examples I'll be giving reflect this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, a lot of manipulative advantages occur when using the Argand form of a plane curve $z=f(t)+ig(t)$ instead of the parametric form $(x\;\;y)=(f(t)\;\;g(t))$: translation is the addition of an appropriate complex quantity, rotation corresponds to a multiplication by a factor $\exp(i\theta)$, and compare the formula for the curvature of a parametric curve (formula 13 &lt;a href=&quot;http://mathworld.wolfram.com/Curvature.html&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;) with the formula for the curvature for a curve in Argand form:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\kappa=\frac{\Im(\bar{z}^{\prime}z^{\prime\prime})}{|z^{\prime}|^3}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll stop here, because at this point I should refer you to Zwikker's &quot;Advanced Plane Geometry&quot;, it's an oldie but goodie.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-09-19T01:58:33.013" CommunityOwnedDate="2010-09-19T01:58:33.013" CommentCount="1" />
  <row Id="4968" PostTypeId="1" AcceptedAnswerId="4969" CreationDate="2010-09-19T02:02:47.307" Score="6" ViewCount="189" Body="&lt;p&gt;We know this is true for commutative ring, but if $S\subset R$ is a left and right Ore set, and $S^{-1}R$ its localization by this Ore set, is this always a flat $R$-module?&lt;/p&gt;&#xA;" OwnerUserId="16" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-09-26T05:54:41.077" LastActivityDate="2010-09-27T15:35:52.190" Title="Are localized rings always flat as R-modules?" Tags="&lt;abstract-algebra&gt;&lt;modules&gt;&lt;homological-algebra&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="4969" PostTypeId="2" ParentId="4968" CreationDate="2010-09-19T02:08:13.267" Score="6" ViewCount="" Body="&lt;p&gt;This is Proposition 2.1.16 in McConnel+Robson's book on &lt;em&gt;Noncommutative Noetherian Rings&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2010-09-19T02:08:13.267" CommentCount="2" />
  <row Id="4970" PostTypeId="2" ParentId="4961" CreationDate="2010-09-19T02:13:45.607" Score="3" ViewCount="" Body="&lt;p&gt;Yet another example close to my heart: one of the &lt;a href=&quot;http://www.siam.org/books/100digitchallenge/&quot; rel=&quot;nofollow&quot;&gt;hundred-dollar hundred-digit problems&lt;/a&gt; asked for the value of&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$I=\lim_{\varepsilon\to 0} \int_{\varepsilon}^1 \frac1{t}\cos\left(\frac{\ln\;t}{t}\right) \mathrm{d}t$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One solution was to transform it into an oscillatory integral with infinite limit, but the slickest solution to this problem was to use a contour in the complex plane that side-stepped the numerical difficulties with having an oscillatory integrand:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$I=\Re\left(\int_0^1 t^{i/t-1} \mathrm{d}t\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and by using an appropriate contour, even the humble trapezoidal rule suffices to handle the integral.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-09-19T02:13:45.607" CommunityOwnedDate="2010-09-19T02:13:45.607" CommentCount="2" />
  <row Id="4971" PostTypeId="2" ParentId="4961" CreationDate="2010-09-19T02:34:26.107" Score="26" ViewCount="" Body="&lt;p&gt;There are too many examples to count.  Let me just mention one that is particularly concrete: how many subsets of an $n$-element set have a cardinality divisible by $3$ (or any positive integer $k$)?  In other words, how do we evaluate&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{k=0}^{\lfloor \frac{n}{3} \rfloor} {n \choose 3k}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;in closed form?  Although the statement of this problem does not involve complex numbers, the answer does: the key is what is known in high school competition circles as the roots of unity filter and what is known among real mathematicians as the discrete Fourier transform.  Starting with the generating function&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(1 + x)^n = \sum_{k=0}^{n} {n \choose k} x^k$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;we observe that the identity&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$1 + \omega^k + \omega^{2k} = \begin{cases} 3 \text{ if } 3 | k \\ 0 \text{ otherwise} \end{cases}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\omega = e^{ \frac{2 \pi i}{3} }$ is a primitive third root of unity implies that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{k=0}^{ \lfloor \frac{n}{3} \rfloor} {n \choose 3k} = \frac{(1 + 1)^n + (1 + \omega)^n + (1 + \omega^2)^n}{3}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $1 + \omega = -\omega^2$ and $1 + \omega^2 = - \omega$, this gives&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{k=0}^{ \lfloor \frac{n}{3} \rfloor} {n \choose 3k} = \frac{2^n + (-\omega)^n + (-\omega^2)^n}{3}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This formula can be stated without complex numbers (either by using cosines or listing out cases) but both the statement and the proof are much cleaner with it.  More generally, complex numbers make their presence known in combinatorics in countless ways; for example, they are pivotal to the theory of asymptotics of combinatorial sequences.  See, for example, Flajolet and Sedgewick's &lt;a href=&quot;http://algo.inria.fr/flajolet/Publications/books.html&quot; rel=&quot;nofollow&quot;&gt;Analytic Combinatorics&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-09-19T02:34:26.107" CommunityOwnedDate="2010-09-19T02:34:26.107" CommentCount="2" />
  <row Id="4973" PostTypeId="2" ParentId="4931" CreationDate="2010-09-19T04:12:27.713" Score="1" ViewCount="" Body="&lt;p&gt;I don't understand why do you say &quot;I tried to prove this through Pythagoras, but the answer I got did not prove it was at a right angle&quot;, because, to me at least, Pythagoras tells me that it &lt;em&gt;is&lt;/em&gt; a right angle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To see this, you must know that Pythagoras' theorem actually is an &quot;if and only if&quot;. That is, if you have a triangle with two sides named $u$ and $v$, then the third one is $u-v$ and Pythagoras says:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;u\cdot v = 0 \qquad \Longleftrightarrow \qquad \| u \|^2 + \|v\|^2 = \|u-v\|^2 \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is, if sides (cathetuses) $u$ and $v$ are perpendicular to each other, &lt;em&gt;then&lt;/em&gt; the sum of their squares is equal to the square of the third side (hypotenuse) $u-v$. But the reciprocal is also true, if the equality on the right hand side of the displayed formula above holds, &lt;em&gt;then&lt;/em&gt; $u$ and $v$ are perpendicular to each other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The proof is easy if you have already seen some Linear Algebra (if not, you may skip what follows, and go for the actual computations for your problem down below). So, the proof:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\|u\|^2+  \|v\|^2 - \|u-v\|^2 = u\cdot u + v\cdot v - (u-v)\cdot (u-v) = -2u\cdot v \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for your problem, let's compute the three sides:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\|be\| = \sqrt{3^2 + 3^2} = \sqrt{18}     \ , \quad \|ce\|= \sqrt{2^2 + 2^2} = \sqrt{8}   \quad \text{and} \quad \|bc\|= \sqrt{1^2 + 5^2} = \sqrt{26} \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And the quantities appearing in the right hand side of the &quot;extended&quot; Pythagoras' theorem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\|be\|^2 + \|ce\|^2 = 18 + 8 = 26 \qquad \text{and} \qquad \|bc\|^2 = 26 \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, $be$ and $ce$ are perpendicular to each other.&lt;/p&gt;&#xA;" OwnerUserId="664" LastActivityDate="2010-09-19T04:12:27.713" />
  <row Id="4974" PostTypeId="2" ParentId="4931" CreationDate="2010-09-19T04:20:30.987" Score="2" ViewCount="" Body="&lt;p&gt;A meta-analysis: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The calculation with Pythagoras' theorem should in fact work for the points indicated.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If done correctly it has to be consistent with any correct method. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope the result of the Q &amp;amp; A is not to treat the posted solutions as correct (maybe because there are several of them that all support the problem statement, sound authoritative, are posted by high-ranked math.SE users, or you checked the reasoning in detail, or whatever) but simultaneously disregard the original computational evidence that another equally valid principle seems to give a different answer.  Tracing back to locate the source of the inconsistency should be worthwhile.  &lt;/p&gt;&#xA;" OwnerUserId="467" LastActivityDate="2010-09-19T04:20:30.987" />
  <row Id="4975" PostTypeId="1" CreationDate="2010-09-19T04:28:05.337" Score="2" ViewCount="177" Body="&lt;p&gt;For context I am developing a ray-tracer for a computer science class, and want to implement some more advanced shapes than just spheres.  So while this is related to schoolwork, I'm not asking you to do my work for me, the work is implementing the programming, and it's the math I don't understand, so I'm just looking for help understanding how the math works.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am trying to understand how to calculate the intersection point, and the normal vector from that point, of several &lt;a href=&quot;http://homepage.univie.ac.at/herwig.hauser/bildergalerie/gallery.html&quot; rel=&quot;nofollow&quot;&gt;algebraic surfaces&lt;/a&gt;.  I am at the very frustrating point of knowing what I need to do, and how it is theoretically done, but not really grasping how to actually do it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that I need to take the equation for the line and substitute the x, y, and z variables in the surface equation for the equivalent portions of the line equation, but as soon as I sit down to do that, I immediately hit a mental brick wall.  As for the normal calculations, I'm really lost, I'm not even sure there is a general case way to calculate the normals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, I'd love some help on how to calculate the intersection and normal of some of these shapes, and any sort of general case rules for these calculations would be fantastic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update&lt;/strong&gt;&#xA;While real general case solutions would be super awesome, it's ok to assume the shapes are in their standard orientation, not rotated or transformed at all - just positioned and (maybe) scaled. This make the problem much simpler, I believe. If there are other limitations you can use to make the problem even simpler, that's likely fine.&lt;/p&gt;&#xA;" OwnerUserId="1887" LastEditorUserId="1887" LastEditorDisplayName="" LastEditDate="2010-09-19T10:36:52.527" LastActivityDate="2010-09-20T00:03:34.130" Title="Calculating Intersections of Lines and Algebraic Surfaces" Tags="&lt;geometry&gt;&lt;algebra-precalculus&gt;&lt;algebraic-geometry&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="4976" PostTypeId="2" ParentId="4931" CreationDate="2010-09-19T04:37:25.237" Score="2" ViewCount="" Body="&lt;p&gt;To expand on Agusti's point further, here's a strategy that only makes use of Pythagoras (remember that the distance formula for two points &lt;em&gt;is&lt;/em&gt; a result of the Pythagorean relation!): You have two adjacent triangles $\triangle AEB$ and $\triangle BEC$; compute the distances of all relevant segments, show that the two triangles are right triangles (thus proving the &quot;perpendicular&quot; part) and then show that $\overline{AE}\cong\overline{EC}$ (thus proving the &quot;bisector&quot; part).&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-09-19T04:37:25.237" />
  <row Id="4977" PostTypeId="2" ParentId="4961" CreationDate="2010-09-19T04:37:48.623" Score="17" ViewCount="" Body="&lt;p&gt;As Dylan Wilson comments in Byron's link, you can derive most of &lt;a href=&quot;http://en.wikipedia.org/wiki/Trigonometric_identity&quot; rel=&quot;nofollow&quot;&gt;trigonometric identities&lt;/a&gt; thanks to &lt;a href=&quot;http://en.wikipedia.org/wiki/Euler%27s_formula&quot; rel=&quot;nofollow&quot;&gt;Euler's formula&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;e^{i\theta} = \cos \theta + i \sin \theta \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, I never could learn by heart the formula for the cosine or the sine of the sum of two angles, but on one hand&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;e^{i(\alpha + \beta)} = e^{i\alpha} \cdot e^{i\beta} = (\cos\alpha \cos\beta -\sin\alpha\sin\beta) + i (\sin\alpha\cos\beta + \cos\alpha \sin\beta) \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And on the other hand:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;e^{i(\alpha + \beta)} = \cos(\alpha + \beta) + i\sin(\alpha + \beta) \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\cos(\alpha + \beta) = \cos\alpha \cos\beta -\sin\alpha\sin\beta&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\sin(\alpha + \beta) = \sin\alpha\cos\beta + \cos\alpha \sin\beta \ .&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="664" LastEditorUserId="664" LastEditorDisplayName="" LastEditDate="2010-09-19T07:48:42.670" LastActivityDate="2010-09-19T07:48:42.670" CommunityOwnedDate="2010-09-19T04:37:48.623" CommentCount="3" />
  <row Id="4978" PostTypeId="2" ParentId="4975" CreationDate="2010-09-19T04:45:44.783" Score="1" ViewCount="" Body="&lt;p&gt;The &quot;normal vector&quot; to a surface is easy enough: compute the partial derivatives with respect to both dependent variables (with respect to $x$ and $y$ if you have $z=f(x,y)$; with respect to parameters $u$ and $v$ if you have a parametric representation $(x\;\;y\;\;z)=(f(u,v)\;\;g(u,v)\;\;h(u,v))$), and then take the cross product of the vectors obtained, (with the option of normalizing it to have unit length).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In any event, I think you're better off trying to do surfaces with &quot;easy&quot; parametric representations: the manipulations are much easier than if you have them in implicit Cartesian form.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;For polyhedra, finding normals to the faces is an even easier task: take the barycenter/centroid of the face under consideration, subtract the components of that from the components of any two selected vertices in the face, and take the cross product of the two resulting vectors (with again the option to normalize to unit length afterward).&lt;/p&gt;&#xA;" OwnerUserId="498" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-09-20T00:03:34.130" LastActivityDate="2010-09-20T00:03:34.130" CommentCount="6" />
  <row Id="4979" PostTypeId="2" ParentId="3852" CreationDate="2010-09-19T05:00:56.167" Score="0" ViewCount="" Body="&lt;p&gt;i agree to &quot; TheMachineCharmer &quot; good concern.&lt;/p&gt;&#xA;" LastActivityDate="2010-09-19T05:00:56.167" />
  <row Id="4980" PostTypeId="2" ParentId="4868" CreationDate="2010-09-19T05:47:47.120" Score="0" ViewCount="" Body="&lt;p&gt;The answer is No. Recall the question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $A\subset\mathbb{C}$ be an open set containing the closed unit disc. Let $f$ be an analytic function from $A$ to $\mathbb{C}$ such that $|f(z)|=1$ if $|z|=1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does it follow that &#xA;$$f(z)=az^{n}\,\frac{cz^{m}-b}{1-\bar{b}cz^{m}}$$ &#xA;for some $a,b,c\in\mathbb{C}$ s.t. $|c|=1$, $|a|=1$, $|b|\le1$ and some $n,m\ge0$? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $B$ be the set of analytic functions from $A$ to $\mathbb{C}$ such that $|f(z)|=1$ if $|z|=1$, and let $C$ the set of rational fractions of the indicated form.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $f$ is an element of $C$ with $0&amp;lt;|b|&amp;lt;1$ and $m\ge1$, then $z\mapsto f(z)^2$ is in $B$, but, having a multiple zero $z_0\not=0$, is not in $C$. &lt;/p&gt;&#xA;" OwnerUserId="660" LastActivityDate="2010-09-19T05:47:47.120" />
  <row Id="4981" PostTypeId="2" ParentId="4942" CreationDate="2010-09-19T06:13:29.510" Score="1" ViewCount="" Body="&lt;p&gt;Making Weltschmerz's answer explicit, and very much related to Isaac's answer: note that the quadrilateral can be split into two triangles. The area of one triangle is half the product of the length of $\overline{BD}$ (the base) and half the length of $\overline{AC}$ (the height). Doubling that gives an area expression that is exactly what Isaac stated.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;As for computing the coordinates  of $B$ and $D$, here's a twofer method: once you can compute the equation of the line from the point-slope form, transform the equation you have into the &quot;two-intercept&quot; form&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{x}{a}+\frac{y}{b}=1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;whence your x- and y-intercepts are (a,0) and (0,b).&lt;/p&gt;&#xA;" OwnerUserId="498" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-09-19T06:33:39.527" LastActivityDate="2010-09-19T06:33:39.527" />
  <row Id="4982" PostTypeId="2" ParentId="4905" CreationDate="2010-09-19T08:37:21.843" Score="7" ViewCount="" Body="&lt;p&gt;This is nothing mysterious, it is the continuous analogue of $\Sigma 2^{-n} \binom{n}{k} =  1$ with factorials interpolated by Gamma functions that vary quite smoothly in the &lt;em&gt;middle range&lt;/em&gt; where the sum is concentrated and has approximately Gaussian distribution, $k = n/2 + O(\sqrt{n})$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The ratio between the discrete step function and continuous interpolation is $1+O(1/n)$ in the middle range (using the first few terms of Stirling asymptotic expansion for $\Gamma$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The $1/\sqrt{x(1-x)}$ is irrelevant because it is effectively a constant factor $2 + O(1/n)$ in the middle range. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The middle range has width less than $n^{1/2 + \epsilon}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, writing the integral in terms of $u = nx$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$I_n = \int_0^n \binom{n}{u} 2^{-n} du \frac {1} {\sqrt{x(1-x)}} \sim 2 \int_0^n \binom{n}{u} 2^{-n} du \sim 2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with an additive error of $O(n^{\epsilon - 1/2})$. The $\epsilon$ can be a small positive constant or an infinitesimal depending on how you dissect the integral into middle and tails (we have to choose a cut for each $n$, such that the entire middle range is covered as $n \to \infty$; any set of choices will prove convergence, but the rate of convergence depends on the choices).&lt;/p&gt;&#xA;" OwnerUserId="467" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-09-19T08:47:07.610" LastActivityDate="2010-09-19T08:47:07.610" CommentCount="6" />
  <row Id="4983" PostTypeId="1" AcceptedAnswerId="4985" CreationDate="2010-09-19T10:31:50.380" Score="3" ViewCount="522" Body="&lt;p&gt;My (maybe wrong) impression is that while probability is widely used in science (for example, in statistical mechanics), it is rarely seen in pure mathematics. Which leads me to the question - &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Are there some interesting application of Probability Theory in pure mathematics, outside Probability Theory itself?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="766" LastEditorUserId="822" LastEditorDisplayName="" LastEditDate="2010-09-19T16:55:36.493" LastActivityDate="2010-09-27T17:19:39.553" CommunityOwnedDate="2010-09-19T18:23:53.840" Title="Applications of Probability Theory in pure mathematics" Tags="&lt;big-list&gt;&lt;probability&gt;&lt;probability-theory&gt;&lt;applications&gt;" AnswerCount="8" CommentCount="1" />
  <row Id="4984" PostTypeId="2" ParentId="4983" CreationDate="2010-09-19T10:41:09.070" Score="1" ViewCount="" Body="&lt;p&gt;Hi--&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please see this link: &lt;a href=&quot;http://en.wikipedia.org/wiki/Probabilistic_number_theory&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Probabilistic_number_theory&lt;/a&gt;&lt;/p&gt;&#xA;" LastActivityDate="2010-09-19T10:41:09.070" CommentCount="0" />
  <row Id="4985" PostTypeId="2" ParentId="4983" CreationDate="2010-09-19T10:58:49.080" Score="8" ViewCount="" Body="&lt;p&gt;The topic of Probabilistic Combinatorics. See &lt;a href=&quot;http://en.wikipedia.org/wiki/Probabilistic_method&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Probabilistic_method&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a powerful way of giving non-constuctive existence proofs for lots of different (finite) mathematical structures and determining their properties. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found this the highlight of all my undergraduate probability courses. If you are interested in learning more I would recommend: &quot;The Probabilistic Method&quot; by Alon and Spencer.&lt;/p&gt;&#xA;" OwnerUserId="1794" LastActivityDate="2010-09-19T10:58:49.080" CommentCount="1" />
  <row Id="4986" PostTypeId="2" ParentId="4983" CreationDate="2010-09-19T10:59:23.567" Score="1" ViewCount="" Body="&lt;p&gt;One of my favorite examples is a proof of the Weierstrass approximation theorem, the fact that polynomials are dense in the space of continuous functions over an interval.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I like the proof because it's elementary and the conclusion comes as a surprise.  &lt;/p&gt;&#xA;" OwnerUserId="247" LastActivityDate="2010-09-19T10:59:23.567" CommentCount="3" />
  <row Id="4987" PostTypeId="2" ParentId="4947" CreationDate="2010-09-19T11:34:27.870" Score="1" ViewCount="" Body="&lt;p&gt;It is a preliminary and useful exercise to show that the line $Ax+By+C=0$ and the vector $v=(A,B)$ are perpendicular (one says that $v$ is the normal vector of the line). With these, all amounts to observe that two lines are parallel if and only if their normal vectors are parallel.&lt;/p&gt;&#xA;" OwnerUserId="688" LastActivityDate="2010-09-19T11:34:27.870" />
  <row Id="4988" PostTypeId="1" AcceptedAnswerId="5123" CreationDate="2010-09-19T12:24:45.797" Score="2" ViewCount="144" Body="&lt;p&gt;I have a problem, which has $n$ variables. Let's call these variables as $a_1$, $a_2$, $a_3$ ... $a_n$. Each of these variables have their own range (i.e, $a_{imin} \leq a_i \leq a_{imax}$ for all $i$).  And most importantly, each $a_n$ is ranked in terms of importance. Without a loss of generality one can assume that that $a_1$ is more important than $a_2$, which in turn is more important than $a_3$, and so on. This means that as long as the constraint is satisfied, it is better to have a smaller $a_1$ compare to have a smaller $a_2$, and a smaller $a_3$ is better compare to a smaller $a_4$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These variables are required to fulfill one ( yes, just one!!) constraint function&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$F_{min} \leq f(a_1,a_2,a_3, .... a_n) \leq F_{max}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question now is, how to construct a minimizing function that satisfy the above condition? If I have two variables ($a_1$ and $a_2$), and both of them are between or equal to 0 and 9, then I can easily write the minimizing function as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$L=10a_1 + a_2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;given that $a_1$ is more important than $a_2$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But for complicated case, I am a bit confused. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: Maybe I should clarify further. My point is, for the above case, if $a_1=3, a_2=4$ and $a_1=4, a_2=3$ both satisfy the constraint equation, $a_1=3, a_2=4$ should be selected. &lt;strong&gt;So how to construct $L$ in this case to reflect the situation?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit 2 : Referring to &lt;a href=&quot;http://math.stackexchange.com/questions/4988/formulate-a-minimizing-function-that-prefers-a-minimal-variable-over-another/5044#5044&quot;&gt;the answer&lt;/a&gt; by Rahul, it seems that it is not possible to formulate such a minimizing function. Is there other methods of attack I can use?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit 3: &lt;a href=&quot;http://math.stackexchange.com/questions/4988/formulate-a-minimizing-function-that-prefers-a-minimal-variable-over-another/4992#4992&quot;&gt;alext87&lt;/a&gt; answer is a good start, it contains general guidelines on how to proceed. But I want a hard, quantitative rule that tells me how to select for $lambda_i$ for all $i$. &lt;/p&gt;&#xA;" OwnerUserId="550" LastEditorUserId="550" LastEditorDisplayName="" LastEditDate="2010-09-21T01:57:19.580" LastActivityDate="2010-09-21T05:06:34.000" Title="Formulate A Minimizing Function that Prefers A Minimal Variable Over Another" Tags="&lt;optimization&gt;" AnswerCount="3" CommentCount="4" />
  <row Id="4989" PostTypeId="1" AcceptedAnswerId="4994" CreationDate="2010-09-19T12:44:31.903" Score="1" ViewCount="88" Body="&lt;p&gt;Hello! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This question may seem simple, but I have no idea where to start. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm in design phase of a hobby electronics project and need your help. Basically, I'm going to have two platforms with electronics on both of them. I'm planning to have a &lt;a href=&quot;http://en.wikipedia.org/wiki/Stepper_motor&quot; rel=&quot;nofollow&quot;&gt;stepper motor&lt;/a&gt; on the bottom platform which will be connected to top platform by an axle. I plan to use a microcontroller to precisely control the intensity of angular velocity of the motor and the angle by which it will rotate. I will also need to use wires to connect the bottom platform to the top platform. My idea is to have wires run on the outside of the axle and be long enough that the axle will be able to rotate by some angle of $±θ$. In normal operation the angle shouldn't be grater than 2π, so I want to avoid constructing some complex mechanism which will transfer current irrelevant of the axle's rotation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Image (I apologize for my bad drawing skills) shows the two states which are interesting (at least in my opinion) here. The top is when the axle is in neutral position and angle by which it is rotated is 0. While the image does not show it, the wire will be loose. The bottom image shows what I expect will happen after the axle has been rotated by a certain angle. &#xA;The two black rectangles are soldering joints by which the cable will be connected to the bottom and the top platforms. The red line is wire.&lt;br&gt;&#xA;&lt;a href=&quot;http://i.imgur.com/OSeFP.png&quot; rel=&quot;nofollow&quot;&gt;Here&lt;/a&gt;'s direct link to the image. It is more clear there.&#xA;&lt;img src=&quot;http://i.imgur.com/OSeFP.png&quot; alt=&quot;Image of the problem&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's my question: Is I have an axle with length L and radius R, how would I calculate length l of a cable running form the top of the axle to the bottom after axle has rotated by a certain angle θ? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, I have no idea which tags would be appropriate, so do feel free to edit them.&lt;/p&gt;&#xA;" OwnerUserId="348" LastEditorUserId="348" LastEditorDisplayName="" LastEditDate="2010-09-19T13:20:41.003" LastActivityDate="2010-09-20T01:14:37.283" Title="Calulating length of cable running along exterior of an axle" Tags="&lt;geometry&gt;&lt;trigonometry&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="4990" PostTypeId="1" AcceptedAnswerId="4996" CreationDate="2010-09-19T12:57:24.710" Score="2" ViewCount="1284" Body="&lt;p&gt;The birational change of variables $(u,v) = (\frac{36+y}{6x},\frac{36-y}{6x})$ maps $u^3+v^3=1$ to $y^2 = x^3 - 432$ which has discriminant $-2^{12}\cdot 3^9$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using pari/gp we can compute the torsion subgroup:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;? elltors(ellinit([0,0,0,0,-432]))&#xA;%1 = [3, [3], [[12, 36]]]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This says the torsion subgroup has order 3, is $\mathbf{Z}/3\mathbf{Z}$ and is generated by $(12,36)$ (which corresponds to $1^3+0^3=1^3$). The reason it has order 3 is because this also includes the projective solution $[0:0:1]$ of $X^3+Y^3=Z^3$.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;: By Nagell-Lutz one only needs to solve $y^2 = x^3 - 432$ in integers for $y=0$ and $y^2|2^{12}\cdot 3^9$ (which is a simple generate and test) to compute the elements of the torsion subgroup 'on paper'.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The group of rational points for this curve is then (by Mordell's Theorem) of the form $\mathbf{Z}^r \times \mathbf{Z}/3\mathbf{Z}$ where $r$ is the rank of the curve. If we can show the rank is 0 then this would prove fermats last theorem for $n = 3$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can it be shown directly the rank of this curve is 0?&lt;/p&gt;&#xA;" LastActivityDate="2010-10-18T00:01:51.407" Title="How could I calculate the rank of this elliptic curve?" Tags="&lt;number-theory&gt;&lt;elliptic-curves&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="4991" PostTypeId="2" ParentId="4959" CreationDate="2010-09-19T13:16:58.720" Score="2" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Q1:&lt;/strong&gt; Are convex subsets assumed to be in vector spaces or topological vector spaces?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Convex sets are easily defined in topological vector spaces over a subfield of $\mathbb{C}$, following the usual definition: $A$ is a convex set if $\forall x,y\in A$ we have $tx+(1-t)y\in A$ for $t\in[0,1]$. Basically means we can draw a line between any two points and the line segment is contained in $A$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For non-euclidean spaces convexity generalises to geodesically convex set see &quot;Convex functions and optimization methods on Riemannian manifolds&quot; by Constantin.  There are a few other generalisations. For the most abstract there are Convexity spaces which is an axiomatic description but I can't remember much about this. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Q2:&lt;/strong&gt; Are convex functions defined to be from convex subsets in vector spaces or topological vector spaces to general field or some special field such as R or C?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Firstly I have seen a convex function defined without the need for its domain to be a convex set at all! It required the function to only be convex on any convex subset of its domain. (I don't think is standard though.)&#xA;Secondly, let $f:I\rightarrow K$ be a convex function where $I$ is a convex subset. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $K=\mathbb{R}$ then $f$ is convex iff $f(tx+(1-t)y)\leq tf(x)+(1-t)f(y)$, $t\in[0,1]$. Is suspect this could be generalised to any ordered field. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $K=\mathbb{C}$ then we can't use this definition as $\mathbb{C}$ is not a totally ordered field. Instead we have $\mathbb{C}\sim\mathbb{R}^2$ and use the hessian: $f$ is convex iff $\nabla^2f$ is positive semi-definite. However this requires $f\in\mathcal{C}^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Q3:&lt;/strong&gt; What are the special cases that have been extensively studied and have significance in application?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The majority of the applications in convex optimisation are with a real vector space and a convex real-valued function. There are so many applications for this the mind boggles. A short memory dump is:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Least squares problems&lt;/li&gt;&#xA;&lt;li&gt;Linear programming: network flow problems, mircoeconomics.&lt;/li&gt;&#xA;&lt;li&gt;Quadratic programming: This forms the basis of so many optimization algorithms&lt;/li&gt;&#xA;&lt;li&gt;Entropy maximisation&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1794" LastEditorUserId="1794" LastEditorDisplayName="" LastEditDate="2010-09-19T13:27:55.917" LastActivityDate="2010-09-19T13:27:55.917" />
  <row Id="4992" PostTypeId="2" ParentId="4988" CreationDate="2010-09-19T13:25:05.347" Score="1" ViewCount="" Body="&lt;p&gt;You can turn this into an optimization problem. For each $a_i$ set up a parameter $\lambda_i$, where $\lambda_i$ you determine to be the importance of the variable being small. For example as $a_i$ is more important to be small than $a_{i+1}$ you will select $\lambda_i&gt;\lambda_{i+1}$. Then solve the following problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\min \lambda_1a_1+\ldots+\lambda_na_n$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;subject to &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$F_{\text{min}}\leq f(a_1,a_2,\ldots,a_n)\leq F_{\text{max}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a_{imin}\leq a_i\leq a_{imax}$ for all i&lt;/p&gt;&#xA;" OwnerUserId="1794" LastActivityDate="2010-09-19T13:25:05.347" CommentCount="5" />
  <row Id="4993" PostTypeId="2" ParentId="4989" CreationDate="2010-09-19T13:28:48.473" Score="3" ViewCount="" Body="&lt;p&gt;Looks like you can approximate the cable's equation with a helix, $(x\quad y\quad z)=(r\cos(m t)\quad r\sin(m t)\quad \frac{h t}{2\pi})$ where $m$ is the number of turns, $h$ is the height of the cylinder, $r$ is the radius of the cylinder, and the parameter $t$ runs from $0$ to $2\pi$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we compute the arclength of this one, we get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sqrt{4\pi^2 r^2 m^2+h^2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This should be a nice starting point, unless there's a better way to approximate the behavior of the cable.&lt;/p&gt;&#xA;" OwnerUserId="498" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-09-20T01:14:37.283" LastActivityDate="2010-09-20T01:14:37.283" CommentCount="1" />
  <row Id="4994" PostTypeId="2" ParentId="4989" CreationDate="2010-09-19T13:36:41.717" Score="5" ViewCount="" Body="&lt;p&gt;This is a total and utter guess, but suppose you fix one endpoint of the wire and unroll the axle to make a rectangle? This is the same trick used to justify the rectangle in the surface area of a cylinder. Then the wire is the diagonal of a rectangle of width $L$ and height $r\theta$. So by the Pythagorean theorem, $l = \sqrt{L^2+r^2\theta^2}$.&lt;/p&gt;&#xA;" OwnerUserId="854" LastActivityDate="2010-09-19T13:36:41.717" CommentCount="3" />
  <row Id="4996" PostTypeId="2" ParentId="4990" CreationDate="2010-09-19T14:00:39.650" Score="2" ViewCount="" Body="&lt;p&gt;I don't know much about this, but there is a theorem due to combined work of Zagier, Kolyvagin, Gross, etc, that says that if the analytic rank of the elliptic curve $E$ is 0 then it's rank is in fact 0. So you can compute the analytic rank for instance using SAGE and then by this theorem you will have shown that the rank is 0. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course this is like cheating since I think there ought to be some simpler way to do this, but you'll have to wait for the experts to respond you. For the time being, you can look at Theorem 5.8 and the definition preceding it in the following paper &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.ams.org/bull/2002-39-04/S0273-0979-02-00952-7/S0273-0979-02-00952-7.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.ams.org/bull/2002-39-04/S0273-0979-02-00952-7/S0273-0979-02-00952-7.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="900" LastActivityDate="2010-09-19T14:00:39.650" CommentCount="3" />
  <row Id="4997" PostTypeId="1" AcceptedAnswerId="5000" CreationDate="2010-09-19T14:19:15.433" Score="0" ViewCount="90" Body="&lt;p&gt;Let $I$ be a finite nonempty set, and $D$ a nonvoid compact subset of an N-dimensional Euclidean space. The functions $V:D\rightarrow\mathbb{R}$ and $u_i:D\rightarrow\mathbb{R}$ are given. They are all continuous, and, moreover, each $u_i$ satisfies $u_i(\lambda p + (1-\lambda) q) = \lambda u_i( p) + (1-\lambda) u_i(q)$ when $\lambda \in (0,1)$ (I don't believe this last piece of information is important, though).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$V$ and $u_i$ also satisfy: (P) if $u_i(p) = u_i(q)$ for all $i\in I$, then $V(p) = V(q)$. This condition allows us to construct a mapping $W:L\rightarrow\mathbb{R}$, where $L =$ {$ (u_1(p),...,u_I(p)): p \in D$}, and $W$ is defined as $W(u) = V(p)$ for any $p \in D$ such that $u = (u_1(p),...,u_I(p))$. Condition (P) ensures W is well defined.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question&lt;/strong&gt;: Is $W$ continuous?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Motivation&lt;/strong&gt;: I am sketching a &quot;proof&quot; from a (published) paper in &quot;applied math&quot; I found online, but that &lt;em&gt;I cannot verify&lt;/em&gt;. Let $(u^n)$ be a sequence in $L$ converging to $u \in L$. Let $(p_n)$ be a sequence in $D$ such that $u^n = (u_1(p_n),...,u_I(p_n))$ for all $n$, and $p \in D$ satisfy $u = (u_1(p),...,u_I(p))$. Because $D$ is compact, there exists a convergent subsequence $p_{n_{k}}\rightarrow p^\prime$. Because $V$ is continuous, we obtain $V(p_{n_{k}})\rightarrow V(p^\prime)$. Moreover, since each $u_i$ is also continuous, we have $\lim u^n = \lim u^{n_{k}} $, and hence $(u_1(p),...,u_I(p)) = (u_1(p^\prime),...,u_I(p^\prime))$. By property (P), we have $V(p) = V(p^\prime)$. Finally, &lt;em&gt;the article goes on and concludes&lt;/em&gt; that $W(u^n) = V(p_n) \rightarrow V(p^\prime) = V(p) = W(u)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't think the conclusion follows from the previous arguments. In particular, nothing guarantees that $V(p_n)$ converges to $p$ in the first place, right? Nevertheless, I cannot think of an easy counter-example...&lt;/p&gt;&#xA;" LastActivityDate="2010-09-19T15:23:43.323" Title="Continuity of an induced function" Tags="&lt;real-analysis&gt;&lt;counter-examples&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="4998" PostTypeId="2" ParentId="4983" CreationDate="2010-09-19T14:44:40.490" Score="1" ViewCount="" Body="&lt;p&gt;You may find some interesting examples on this MathOverFlow thread:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://mathoverflow.net/questions/9218/probabilistic-proofs-of-analytic-facts&quot; rel=&quot;nofollow&quot;&gt;http://mathoverflow.net/questions/9218/probabilistic-proofs-of-analytic-facts&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="940" LastActivityDate="2010-09-19T14:44:40.490" CommunityOwnedDate="2010-09-19T16:58:00.910" />
  <row Id="4999" PostTypeId="2" ParentId="4968" CreationDate="2010-09-19T14:51:20.470" Score="4" ViewCount="" Body="&lt;p&gt;As I mentioned in a prior &lt;a href=&quot;http://math.stackexchange.com/questions/4442/learning-to-think-categorically-localization-of-rings-and-modules/4450#4450&quot;&gt;post here&lt;/a&gt; there is a wealth of information on noncommutative localizations in Ranicki, A.(ed). &lt;a href=&quot;http://www.maths.ed.ac.uk/~aar/books/nlat.pdf&quot; rel=&quot;nofollow&quot;&gt;Noncommutative localization in algebra and topology. ICMS 2002&lt;/a&gt;. In particular, there you will find an interesting paper on this very topic by Beachy: &quot;On flatness and the Ore condition&quot;. Below is general reference information for flatness in the commutative case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a very nice treatment of flatness in Bourbaki's &quot;Commutative Algebra&quot; - which begins with an excellent chapter on flat modules before turning to localizations in Chapter 2 (see Theorem 2.41. p. 68 for the result you seek). Also perhaps of interest is the following motivational remark from the introduction&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The study of the passage from a ring $\rm A$ to a local ring $\rm A$, or to a completion &#xA;  $\rm \hat A$ brings to light a feature common to these two operations, the property of &#xA;  flatness of the $\rm A$-modules $\rm A$, and $\rm \hat A$, which allows amongst other things &#xA;  the use of tensor products of such $\rm A$-modules with arbitrary $\rm A$-modules somewhat &#xA;  similar to that of tensor products of vector spaces, that is, without all the precautions &#xA;  surrounding their use in the general case. The properties associated with this notion, which &#xA;  are also applicable to modules over non-commutative rings, are  the object of study in Chapter I. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;See also Atiyah and Macdonald, Corollary 3.6 and Proposition 3.10 pp. 40-41.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-09-27T15:35:52.190" LastActivityDate="2010-09-27T15:35:52.190" CommentCount="4" />
  <row Id="5000" PostTypeId="2" ParentId="4997" CreationDate="2010-09-19T15:23:43.323" Score="0" ViewCount="" Body="&lt;p&gt;You can think of your $u_i$ as the components of a continuous&#xA;map $U:D\to\mathbb{R}^n$ for some $n$. So you have the condition&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;if $U(p)=U(q)$ then $V(p)=V(q)$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;and so $V=W\circ U$ for some map from $L=U(D)$ to $\mathbb{R}$.&#xA;To show $W$ is continuous, it suffices to prove that $U:D\to L$&#xA;is a quotient map, that is $X\subseteq L$ is closed in $L$ iff $U^{-1}(X)$&#xA;is closed in $D$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By continuity if $X$ is closed in $L$ then $U^{-1}(X)$ is closed&#xA;in $D$, so suppose that $Y=U^{-1}(X)$ is closed in $D$. As $D$ is compact,&#xA;then $Y$ is compact (Heine-Borel), and as $U$ is continuous, $U(Y)$ is a compact&#xA;subspace of $\mathbb{R}^n$. Thus $U(Y)$ is a closed subset of $\mathbb{R}^n$&#xA;and &lt;em&gt;a fortiori&lt;/em&gt; of $L$. But $U(Y)=X$, as $U$ maps $D$ onto $L$.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-09-19T15:23:43.323" CommentCount="1" />
  <row Id="5002" PostTypeId="2" ParentId="4961" CreationDate="2010-09-19T16:16:07.843" Score="2" ViewCount="" Body="&lt;p&gt;The Cauchy integral formula's are pretty easy to prove from the Cauchy-Riemann equation, especially if your class already knows Green's theorem.  I think these are my favorite &quot;easily achieved results&quot;; because these formula's are central to complex analysis, the proof's are rather easy, and the applications are extremely powerful.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are discussing complex numbers in a calculus class, it seems natural to discuss applications to integration theory in my opinion.&lt;/p&gt;&#xA;" OwnerUserId="1417" LastActivityDate="2010-09-19T16:16:07.843" CommunityOwnedDate="2010-09-19T16:16:07.843" />
  <row Id="5003" PostTypeId="1" CreationDate="2010-09-19T16:21:10.127" Score="0" ViewCount="212" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Possible Duplicate:&lt;/strong&gt;&lt;br&gt;&#xA;  &lt;a href=&quot;http://math.stackexchange.com/questions/3633/square-root-of-a-function-in-the-sense-of-composition&quot;&gt;Square root of a function (in the sense of composition)&lt;/a&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;p&gt;I'm interested in solving equations of the form $f(f(x))=g(x)$ for $x\in\mathbb{R}$ where $g(x)$ is a known function. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example: if $g(x)=x$ then we $f(f(x))=x$ so one solution is $f(x)=x$, $\forall x\in\mathbb{R}$. However, another is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f(x)=x+1$ if $x\in(0,1]$,  $f(x)=x-1$ if $x\in(1,2]$ and $f(x)=x$ otherwise. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence there are infinitely many solutions to this equation when $g(x)=x$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Problem:&lt;/strong&gt; Find an $f$ where $f(f(x))=e^x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Extra:&lt;/strong&gt; Are there extra constraints that could be placed on $f$ so the solution is unique?&lt;/p&gt;&#xA;" OwnerUserId="1794" LastActivityDate="2010-09-19T16:21:10.127" ClosedDate="2010-09-19T22:15:29.917" Title="Solving $f(f(x))=g(x)$ equations." Tags="&lt;functional-equations&gt;" CommentCount="6" />
  <row Id="5004" PostTypeId="2" ParentId="4983" CreationDate="2010-09-19T16:54:50.140" Score="2" ViewCount="" Body="&lt;p&gt;There is a strong connection between probability theory and areas of real analysis such as linear PDE and potential theory.  If $X_t$ is a continuous time stochastic process on a state space $S$, the function $u(x,t) = E_x[f(X_t)]$ tends to satisfy an equation like $\partial_t u + Lu = 0$, where $L$ is some interesting operator.  ($E_x$ denotes that the process should start at the point $x \in S$, and $f$ is a real-valued function on $S$.)  The classical example is when $X_t$ is Brownian motion on $\mathbb{R}^n$; then $L$ is the Laplace operator.  So one can study heat equations, harmonic functions, spectral theory, functional inequalities, and lots of other &quot;analysis&quot; topics by thinking about a corresponding stochastic process.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The theory of Dirichlet forms makes this correspondence quite precise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the state space $S$ is a manifold, then the study of the operator $L$ tends to be closely related to the geometry of $S$, so one can also study differential geometry through probability theory.&lt;/p&gt;&#xA;" OwnerUserId="822" LastActivityDate="2010-09-19T16:54:50.140" CommunityOwnedDate="2010-09-19T16:55:05.467" />
  <row Id="5005" PostTypeId="1" AcceptedAnswerId="5006" CreationDate="2010-09-19T18:31:11.157" Score="3" ViewCount="196" Body="&lt;p&gt;In my daughter's class of 23, three students and the teacher all share the same birthday. Of course, there are 365 days in the year, and the first case of the shared birthday is not counted in the probability. But is the likelihood of this just (1/365)*3 = .0082? For my curiosity, leap years can be omitted.&lt;/p&gt;&#xA;" OwnerUserId="1990" LastActivityDate="2010-09-19T19:06:34.193" Title="Birthday Probability" Tags="&lt;probability&gt;" AnswerCount="1" CommentCount="7" FavoriteCount="1" />
  <row Id="5006" PostTypeId="2" ParentId="5005" CreationDate="2010-09-19T18:42:23.453" Score="1" ViewCount="" Body="&lt;p&gt;No, 1/365^3 is (roughly) the probability that, given a date and 3 people, those 3 people have a birthday on that date.  This is much smaller than the probability that, given a date and 23 people, 3 of those people have their birthday on that date.   It seems now that you want, given 24 people (and no date fixed beforehand), the probability that 4 of those people have their birthday on the same date.  A generalization of this question can be found at &lt;a href=&quot;http://stats.stackexchange.com/questions/1308/extending-the-birthday-paradox-to-more-than-2-people&quot;&gt;http://stats.stackexchange.com/questions/1308/extending-the-birthday-paradox-to-more-than-2-people&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2010-09-19T19:06:34.193" LastActivityDate="2010-09-19T19:06:34.193" CommunityOwnedDate="2010-09-19T18:55:55.683" />
  <row Id="5007" PostTypeId="1" AcceptedAnswerId="5047" CreationDate="2010-09-19T18:51:01.750" Score="5" ViewCount="208" Body="&lt;p&gt;Seems straight-forward but I've been unable to get it right. Here are my steps:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$y'(x) = \sqrt{-2y(x) + 28},\hspace{20 pt} y(-4)=-4$$&#xA;$$\int {1 \over \sqrt{28-2y} }\hspace{2 pt}\text{d}y = \int \text{d}x$$&#xA;$$-\sqrt{28-2y} = x + c$$&#xA;$$(28-2y) = (x+c)^2$$&#xA;$$y = -1/2x^2 - cx - c^2/2 + 14$$&#xA;$$c = {-2, 10}$$&#xA;$$\Rightarrow y = -1/2x^2 - 10x - 36$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've checked it over many countless times but for the life of me I can't figure out why it won't work. I've tried plugging the result back into the original equation and it seems to me like it checks out if you take the negative of the square root..&lt;/p&gt;&#xA;" OwnerUserId="1989" LastEditorUserId="1989" LastEditorDisplayName="" LastEditDate="2010-09-19T19:41:12.673" LastActivityDate="2011-04-03T09:27:56.940" Title="Solving separable differential equation" Tags="&lt;differential-equations&gt;&lt;homework&gt;" AnswerCount="3" CommentCount="4" FavoriteCount="1" />
  <row Id="5008" PostTypeId="1" AcceptedAnswerId="5011" CreationDate="2010-09-19T18:58:11.280" Score="1" ViewCount="175" Body="&lt;p&gt;What does the following term mean?&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;category of geometrically reduced schemes of finite type over some field&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(I know what a category and a field is but I cannot translate any of the middle bits)&lt;/p&gt;&#xA;" LastActivityDate="2010-11-06T08:31:51.333" Title="What is the meaning of &quot;reduced schemes of finite type&quot;?" Tags="&lt;terminology&gt;" AnswerCount="2" />
  <row Id="5009" PostTypeId="2" ParentId="4826" CreationDate="2010-09-19T19:02:10.397" Score="1" ViewCount="" Body="&lt;p&gt;Here is an attempt at a solution: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $x_0\in U$, since $U$ is open there is a $\delta&gt;0$ such that $B(x_0,\delta)\subset U$ and by Talyor's theorem (using $v\in\mathcal{C}^1$):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$v(x) = v(x_0)+\nabla v(x_{\epsilon})^T(x-x_0)$  for some $x_{\epsilon}\in B(x_0,\delta)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We want to show that $\Delta u(v(x_0)+\nabla v(x_{\epsilon})^T(x-x_0))=0$ for $\forall x\in B(x_0,\delta)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the fact that $\nabla v(x_{\epsilon})^T$ is orthogonal and the theorem stated in the question we have &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Delta u(\nabla v(x_{\epsilon})^T(x-x_0))=\Delta (u)\nabla v(x_{\epsilon})^T(x-x_0)=0$  $\forall x\in B(x_0,\delta)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $v(x_0)$ is the zero vector then we are done. Else $v(x_0)+\nabla v(x_{\epsilon})^T(x-x_0)$ is a translation of $\nabla v(x_{\epsilon})^T(x-x_0)$ by a constant vector $v(x_0)$ which is an orthogonal transformation. Let this translation be denoted as the linear orthogonal operator $T_{v(x_0)}$ then we have &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Delta u(T_{v(x_0)}[\nabla v(x_{\epsilon})^T(x-x_0)]) = \Delta (u\circ T_{v(x_0)})\nabla v(x_{\epsilon})^T(x-x_0)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;as $v(x_{\epsilon})^T$ is orthogonal and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Delta (u\circ T_{v(x_0)})\nabla v(x_{\epsilon})^T(x-x_0)= \Delta (u)\circ T_{v(x_0)}(\nabla v(x_{\epsilon})^T(x-x_0))=0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;as $T_{v(x_0)}$ is orthogonal. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $x_0$ was any point in $U$ we have showed that $\Delta(u\circ v)=0$ in $U$ and thus is a harmonic function. &lt;/p&gt;&#xA;" OwnerUserId="1794" LastActivityDate="2010-09-19T19:02:10.397" CommentCount="2" />
  <row Id="5010" PostTypeId="2" ParentId="5007" CreationDate="2010-09-19T19:03:01.003" Score="2" ViewCount="" Body="&lt;p&gt;Your second line should be&#xA;$$\int {1 \over \sqrt{28-2y} }\hspace{2 pt}\text{d}y = \int \text{d}x.$$&lt;/p&gt;&#xA;" OwnerUserId="1301" LastActivityDate="2010-09-19T19:03:01.003" CommentCount="1" />
  <row Id="5011" PostTypeId="2" ParentId="5008" CreationDate="2010-09-19T19:12:26.003" Score="7" ViewCount="" Body="&lt;p&gt;You need to consult a text on scheme theory, for instance&#xA;Hartshorne's &lt;em&gt;Algebraic Geometry&lt;/em&gt;. A scheme of finite type&#xA;over a field $k$ is one with a finite cover of spectra of&#xA;rings of the form $k[x_1,\ldots,x_m]/I$ where $I$ is an ideal.&#xA;A reduced scheme is a scheme where for each open set $U$ the ring&#xA;of functions defined over $U$ has no nonzero nilpotent elements.&#xA;It's geometrically reduced if it remains reduced when the base&#xA;field is changed to its algebraic closure.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-09-19T19:12:26.003" CommentCount="1" />
  <row Id="5012" PostTypeId="2" ParentId="4990" CreationDate="2010-09-19T20:22:32.287" Score="2" ViewCount="" Body="&lt;p&gt;For some of the programs that output a rank, such as MAGMA, when they do so it is a rigorous certificate that the rank is the one displayed, and not only an apparent rank calculated from the L-function assuming the Birch and Swinnerton-Dyer conjecture.  So although for some cases such as $L(E,1) \neq 0$ or using parity etc, the state of knowledge on BSD is enough to conclude the rank from accurate approximate calculation on the L-function and other invariants, these tricks are not needed given the program output.&lt;/p&gt;&#xA;" OwnerUserId="467" LastActivityDate="2010-09-19T20:22:32.287" />
  <row Id="5013" PostTypeId="2" ParentId="4983" CreationDate="2010-09-19T20:50:28.710" Score="1" ViewCount="" Body="&lt;p&gt;You might find interesting various probabilistic plausibility arguments for the $\rm\:3n+1$ conjecture. For an introduction see &lt;a href=&quot;http://www.cecm.sfu.ca/organics/papers/lagarias/&quot; rel=&quot;nofollow&quot;&gt;Lagarias's survey&lt;/a&gt; excerpted below, and for more sophisticated arguments see his paper &lt;a href=&quot;http://g4gardner.pbworks.com/f/mm-lagarias3XP1.pdf&quot; rel=&quot;nofollow&quot;&gt;How random are the 3x+1 function iterates?&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;1. Introduction.&lt;/strong&gt;$\;$ The $\rm\:3x + 1\:$ problem, also known as the Collatz problem, the Syracuse problem,&#xA;Kakutani's problem, Hasse's algorithm, and Ulam's problem, concerns the behavior of the &#xA;iterates of the function which takes odd integers $\rm\:n\:$ to $\rm\:3n + 1\:$ and even integers $\rm\:n\:$ to $\rm\:n/2\:$. The &#xA;$\rm\: 3x + 1\:$ Conjecture asserts that, starting from any positive integer $\rm\:n\:$, repeated iteration of this &#xA;function eventually produces the value $\rm\:1\:$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The $\rm\:3x + 1\:$ Conjecture is simple to state and apparently intractably hard to solve. It shares &#xA;these properties with other iteration problems, for example that of aliquot sequences (see Guy &#xA;[36], Problem B6) and with celebrated Diophantine equations such as Fermat's last theorem. Paul &#xA;Erdos commented concerning the intractability of the $\rm\:3x + 1\:$ problem: &quot;Mathematics is not yet &#xA;ready for such problems.&quot; Despite this doleful pronouncement, study of the $\rm\:3x + 1\:$ problem has &#xA;not been without reward. It has interesting connections with the Diophantine approximation of &#xA;$\log_2 3\:$ and the distribution (mod $\rm\:1\:$) of the sequence $\rm\:\{(3/2)^k : k = 1,2,\ldots\}\:$, with questions of ergodic theory on the $\rm\:2\:$-adic integers $\rm\:\mathbb Z_2\:$, and with computability theory--a generalization of the $\rm\:3x + 1\:$ problem has been shown to be a computationally unsolvable problem. In this paper I &#xA;describe the history of the $\rm\:3x + 1\:$ problem and survey all the literature I am aware of about this &#xA;problem and its generalizations. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;2.1. A heuristic argument.&lt;/strong&gt; $\;$ The following heuristic probabilistic argument supports the $\rm\:3x + 1\:$ &#xA;Conjecture (see [28]). Pick an odd integer $\rm\:n_0\:$ at random and iterate the function $\rm\:T(n) = n/2\ \text{if n is even}\:$ $\rm\:\text{else}\ (3n+1)/2\:$ until another  odd integer $\rm\:n_1\:$ occurs. Then $\rm\:1/2\:$ of the time $\rm\:n_1 = (3 n_o + 1)/2,\ 1/4\:$ of the time, $\rm\:n_1 = (3 n_o + 1)/4,\ 1/8\:$ of the time $\rm\:n_1 = (3 n_o + 1)/8\:$ and so on. If one supposes that the function $\rm\:T\:$ is sufficiently &quot;mixing&quot; that successive odd integers in the trajectory of $\rm\:n\:$ behave as though they were drawn at &#xA;random (mod $\rm\:2^k\:$) from the set of odd integers (mod $\rm\:2^k\:$) for all $\rm\:k\:$, then the expected growth in size &#xA;between two consecutive odd integers in such a trajectory is the multiplicative factor &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{3}{2}^{1/2}\frac{3}{4}^{1/2}\frac{3}{8}^{1/2}\cdots \ = \ \frac{3}{4} &amp;lt; 1\:.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consequently this heuristic argument suggests that on average the iterates in a trajectory tend to &#xA;shrink in size, so that divergent trajectories should not exist. Furthermore it suggests that the total &#xA;stopping time $\rm\:\sigma_x(n)\:$ is (in some average sense) a constant multiple of $\rm\:\log n\:$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the viewpoint of this heuristic argument, the central difficulty of the $\rm\:3x + 1\:$ problem lies &#xA;in understanding in detail the &quot;mixing&quot; properties of iterates of the function $\rm\:T(n)\ (mod\ 2^k)\:$ for &#xA;all powers of $\rm\:2\:$. The function $\rm\:T(n)\:$ does indeed have some &quot;mixing&quot; properties given by &#xA;Theorems B and K below; these are much weaker than what one needs to settle the $\rm\:3x + 1\:$ Conjecture. &lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-09-19T20:50:28.710" CommunityOwnedDate="2010-09-19T20:50:28.710" />
  <row Id="5014" PostTypeId="2" ParentId="4983" CreationDate="2010-09-19T21:10:38.803" Score="1" ViewCount="" Body="&lt;p&gt;In addition to examples, the overall &quot;gestalt&quot; answer to the question is: probability is used everywhere in mathematics.  It is a basic idea on the level of algorithm, algebraic structure, geometry, calculus, or other very ubiquitous things.  It has become a very popular source of questions and intuitions in research, in all fields.   Knowing that this is true, it is not surprising that many examples of theoretical uses of probability can be posted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Being a basic language, it is also true that many of the uses of probability are basic, and do not go far beyond the idea of a probability distribution, frequencies of events, expectations, related combinatorics and so on.  But in some fields, advanced results in probability theory are constantly being used.&lt;/p&gt;&#xA;" OwnerUserId="467" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-09-19T21:18:28.140" LastActivityDate="2010-09-19T21:18:28.140" CommunityOwnedDate="2010-09-19T21:10:38.803" />
  <row Id="5015" PostTypeId="1" CreationDate="2010-09-19T21:52:30.600" Score="1" ViewCount="173" Body="&lt;p&gt;I'm sure this is a solved problem but I don't know what to search for:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a set of positive numbers (256 at the moment but it could get bigger) and I need to find another set of positive numbers such that for each number in the original set, there is a sub set of the new set that sums to within some error bound of the number:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Give $S\subset\mathbb{R}^+$&lt;/li&gt;&#xA;&lt;li&gt;Find a small set $P\subset\mathbb{R}^+$&lt;/li&gt;&#xA;&lt;li&gt;Sutch that $\forall i\in S, (\exists P_i\subset P,|i-\sum P_i|\le\epsilon)$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The best solution I have found so far is itterative:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;find $M = max(S_n)$.&lt;/li&gt;&#xA;&lt;li&gt;find $m = min(x | x \in S_n \wedge x &gt; M/2)$&lt;/li&gt;&#xA;&lt;li&gt;$P_{n+1} = P_n \cup m$&lt;/li&gt;&#xA;&lt;li&gt;$S_{n+1} = (x | x \in S_n \wedge x &amp;lt; m) \cup (x - m | x \in S_n \wedge x \ge m)$&lt;/li&gt;&#xA;&lt;li&gt;Stop when the approximation is good enough (i.e. $\forall x\in P_n,x\le\epsilon $)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1171" LastEditorUserId="1171" LastEditorDisplayName="" LastEditDate="2010-09-23T14:56:29.940" LastActivityDate="2010-09-23T14:56:29.940" Title="Minimal set with subsets that sum to given values" Tags="&lt;combinatorics&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="5016" PostTypeId="1" AcceptedAnswerId="5054" CreationDate="2010-09-19T21:54:32.000" Score="2" ViewCount="136" Body="&lt;p&gt;Suppose $\mathcal{F}$ is a famliy of analytic functions of the unit disc.   Suppose also that&#xA;$( Re(f(z)) )^2 \ne ( Im(f(z)) ) $for all $|z|&amp;lt;1$ and all $f \in \mathcal{F}$.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It follows from the &lt;a href=&quot;http://en.wikipedia.org/wiki/Fundamental_normality_test&quot; rel=&quot;nofollow&quot;&gt;Fundamental Normality test&lt;/a&gt;   that $\mathcal{F}$ is a normal family. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a for elementary way of showing $\mathcal{F}$ is a normal family without invoking the Fundamental Normality Test?&lt;/p&gt;&#xA;" OwnerUserId="832" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-09-19T23:05:05.237" LastActivityDate="2010-09-20T10:31:29.050" Title="Normal Families" Tags="&lt;complex-analysis&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="5018" PostTypeId="1" AcceptedAnswerId="5024" CreationDate="2010-09-19T22:26:24.410" Score="7" ViewCount="817" Body="&lt;p&gt;Hello,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I fell in love with mathematics a bit too late when I've already taken decisions regarding my future, career-wise. Now I would like to learn math on my own but I'm a bit confused as where to start. My knowledge of mathematics is comparable to that of a 15-16 year old highschool freshman. I would like to know how would you (if you were in my position) start your learning adventure on your own. There are a lot of resources online, of that I'm sure but I would like to follow a path. I'm pretty sure I can learn mathematics on my own and been thinking about this decision for almost a year. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you!&lt;/p&gt;&#xA;" LastActivityDate="2010-09-25T11:37:29.643" CommunityOwnedDate="2010-09-19T23:19:34.053" Title="How to start with mathematics?" Tags="&lt;learning&gt;&lt;self-learning&gt;" AnswerCount="8" CommentCount="2" FavoriteCount="5" />
  <row Id="5019" PostTypeId="1" AcceptedAnswerId="5041" CreationDate="2010-09-19T22:30:09.340" Score="3" ViewCount="289" Body="&lt;p&gt;Why are row reductions so useful in linear algebra? It is easy to get lost in the mechanical solving of equations. I know I can get a matrix into reduced row echelon form. But what are the outcomes of this? What can this mean?&lt;/p&gt;&#xA;" OwnerUserId="1992" LastActivityDate="2010-09-20T13:33:19.623" Title="What are we solving with row reduction?" Tags="&lt;linear-algebra&gt;" AnswerCount="5" CommentCount="1" FavoriteCount="1" />
  <row Id="5020" PostTypeId="2" ParentId="4975" CreationDate="2010-09-19T22:44:09.240" Score="1" ViewCount="" Body="&lt;p&gt;Perhaps this more elementary description could help.&#xA;Let $e$ be the eye/camera, and $v$ a line-of-sight vector.&#xA;You want to solve simultaneously $e + t v$ with the surface you want to view, solving for $t$.&#xA;If you have two or more surfaces, don't try to intersect them with one another,&#xA;which can be algebraically complex, but rather let the ray tracing (effectively) do it for you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose you have a surface patch $S$ (perhaps a Bezier surface) parametrized by $a$ and $b$.&#xA;So now you want to solve simultaneously for $(t, a, b)$.  If $S$ is a sphere or cylinder,&#xA;this amounts to quadratic equations.  If $S$ is a cubic patch, it will reduce to solving&#xA;cubic equations.  If $S$ is a torus, degree-4 equations. Once you have $(a,b)$, you can get the normal vector at that point from your parametric equations, as&#xA;J.M. describes.&lt;/p&gt;&#xA;" OwnerUserId="237" LastActivityDate="2010-09-19T22:44:09.240" />
  <row Id="5021" PostTypeId="2" ParentId="5019" CreationDate="2010-09-19T22:46:29.707" Score="4" ViewCount="" Body="&lt;p&gt;So you have a matrix $A$ and vectors $x,b$, to solve $Ax=b$ you apply row reductions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So considering a concrete example, if we wanted to solve&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left(\begin{array}{cc} 1 &amp;amp; 1 \\ 1 &amp;amp; 2\end{array}\right) x = \left(\begin{array}{c} 5 \\ 8\end{array}\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can use the row that changes the bottom row into bottom row minus top:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left(\begin{array}{cc} 1 &amp;amp; 1 \\ 0 &amp;amp; 1\end{array}\right) x = \left(\begin{array}{c} 5 \\ 3\end{array}\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which tells us that $x = \left(\begin{array}{c} 2 \\ 3\end{array}\right)$.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The secret is that the second line is actually&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left(\begin{array}{cc} 1 &amp;amp; 0 \\ -1 &amp;amp; 1\end{array}\right) \left(\begin{array}{cc} 1 &amp;amp; 1 \\ 1 &amp;amp; 2\end{array}\right) x = \left(\begin{array}{cc} 1 &amp;amp; 0 \\ -1 &amp;amp; 1\end{array}\right) \left(\begin{array}{c} 5 \\ 8\end{array}\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So a row operation corresponds to multiplying both sides by a matrix - Why is this useful for solving a system? Well it's just the same as when you solve any equation, for example&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$2x+3=7x-4$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can add 4 to both sides and then subtract 2x from both to get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$7=5x$$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;That is the meaning of row operations, and why they help to solve a system - it's also interesting to note that if we start out with $AX=I$ and use row operations until it's in the form $IX=B$ then the inverse of $A$ has been found. Viewing the row operations as matrices, lets say it took 3 operations $R_1, R_2, R_3$ then we have $R_3 R_2 R_1 A X = I X = R_3 R_2 R_1 I$ so $A^{-1} = X = R_3 R_2 R_1$.&lt;/p&gt;&#xA;" LastActivityDate="2010-09-19T22:46:29.707" />
  <row Id="5023" PostTypeId="2" ParentId="5018" CreationDate="2010-09-19T23:11:56.337" Score="0" ViewCount="" Body="&lt;p&gt;Pardon me if I'm understating your current knowledge, but are you familiar with geometry, elementary algebra and pre-calculus? Those are pretty standard courses for high school students, and I'm sure most standard textbooks is fine to get the basics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are already familiar with that, computational calculus and computational linear algebra is usually the next step, and books like Tom Apostol's Calculus Vol. 1-2 are a nice introduction to single and multivariable calculus with some basic linear algebra. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you're comfortable with computational mathematics, it might be nice to move on to more theoretical mathematics, like linear algebra, abstract algebra, and real analysis. Some good texts in my opinion are Friedberg, Insel and Spence for Linear algebra, Dummit and Foote for Abstract Algebra, and Rudin for Real Analysis. Those are just some suggestions that I would pick and choose parts which I would want to learn.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm currently self-teaching myself as well, so I don't have as much experience as many others here, so they can probably be more helpful.&lt;/p&gt;&#xA;" OwnerUserId="1583" LastEditorUserId="1583" LastEditorDisplayName="" LastEditDate="2010-09-19T23:17:33.603" LastActivityDate="2010-09-19T23:17:33.603" />
  <row Id="5024" PostTypeId="2" ParentId="5018" CreationDate="2010-09-19T23:13:24.627" Score="5" ViewCount="" Body="&lt;p&gt;I hope this is useful, I am just describing what worked well for me in the past. I recommend using real physical books (for example, I used &lt;em&gt;Mathematics for the Million - Lancelot Hogben&lt;/em&gt; at this stage and I would recommend this particular book also because I am very fond of it).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also you can a very good grip of the basic mathematical concepts by programming them (e.g. algorithms to convert between bases, factoring quadratics, finding square roots by newtons method when you get to calculus) - if you are just programming for programmings sake you will probably not learn very much mathematics (or at least, most of that which you do learn is by accident), but if you do it as an act of &lt;em&gt;detailed explanation&lt;/em&gt; it can be an excellent process to find much clearer understanding of whatever it is you're learning about (and as a bonus you can run them to automatically solve problems that you already know how to do!).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In finding a path it would be useful for you to make a list of the more advanced theorems or applications which fascinate you (you will have to choose your own but I mean things like why are the orbits of planets in newtons universe ellipses, why you cannot square the circle or how raytracing works etc.) - then find out which setting they live in, then study the building blocks needed to get started working on the background - it doesn't even matter if you give up after a couple months extremely hard work by discovering that you cannot possibly study this stuff yet because you will still have got a lot out of it.&lt;/p&gt;&#xA;" LastActivityDate="2010-09-19T23:13:24.627" />
  <row Id="5025" PostTypeId="1" AcceptedAnswerId="5038" CreationDate="2010-09-19T23:20:32.243" Score="1" ViewCount="147" Body="&lt;p&gt;Let $e(n)$ be the number of partitions of $n$ with even number of even parts and let $o(n)$ denote the number of partitions with odd number of even parts. In Enumerative Combinatorics 1, it is claimed that it is easy to see that $\sum_{n \geq 0} (e(n) - o(n)) x^n = \frac{1}{(1-x) \times (1+x^2) \times (1 - x^3 ) \times (1+x^4) \times ... }$. I have been racking my head over this for the past few hours, and I can't see any light. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I noticed, that $e(n) - o(n) = 2e(n) - p(n)$ where $p(n)$ is the number of partitions of $n$, so the above claim is equivalent to showing $\sum_{ n \geq 0} e(n)x^n = \frac{1}{2} \frac{1}{(1-x)(1-x^3)(1-x^5)...}( \frac{1}{(1-x^2)(1-x^4)....} + \frac{1}{(1+x^2)(1+x^4)........})$, and similarly, it is equivalent to  $\sum_{ n \geq 0} o(n)x^n = \frac{1}{2} \frac{1}{(1-x)(1-x^3)(1-x^5)...}( \frac{1}{(1-x^2)(1-x^4)....} - \frac{1}{(1+x^2)(1+x^4)........})$, but these identities appear more difficult than the original one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any hints and suggestions appreciated.&lt;/p&gt;&#xA;" OwnerUserId="404" LastActivityDate="2010-09-20T03:28:20.987" Title="partitions with even number of even parts - partitions with odd number of even parts" Tags="&lt;combinatorics&gt;" AnswerCount="1" />
  <row Id="5026" PostTypeId="2" ParentId="5018" CreationDate="2010-09-19T23:23:10.687" Score="1" ViewCount="" Body="&lt;p&gt;I am a work in progress myself, but I think I can give some advice.  I think tackling mathematics as an older person has some advantages because you have a lot more life experience than a 15-16 year old person.  You know yourself better.  This can be helpful in terms of keeping a consistent work ethic.  I base these observations on tutoring a wide number of age groups (at a local community college).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, unfortunately, your question is a bit broad and I can't really tell from your description what a 15-16 year old high school freshman might know in your country.  In North America, preparation can vary greatly between school districts.  So my advice has to be general.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would set myself a goal of learning the basic topics that are needed at the high school level in the local school system: read the books and do the exercises.  Ensuring that you don't have major gaps in the topics covered in your school system has the advantage of making it easier to enter the formal system if at some point later you find that you want to take a class perhaps at a local community college.  Such classes are likely to be tuned to the educational background of the typical high school student.  As an independent learner myself, I occasionally have the problem of weird gaps in my background. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mathematics at the high school level is straightforward I think.  If you read the book and do all the exercises, you can get through most of the curriculum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think it's important to have social feedback.  I have a few ideas in this area:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;You can participate in online mathematics community such as this site.  In addition to this site, I like &lt;a href=&quot;http://www.artofproblemsolving.com/Forum/index.php&quot; rel=&quot;nofollow&quot;&gt;Art of Problem Solving&lt;/a&gt;.  I think it's worth it to interact with others and keep your enthusiasm high.  I also like the Mathematics Category on Yahoo Answers.  There are a large variety of questions and you can compare your answers to others.  The questions are quite often in the high school to early college range.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;When you get to a level where you feel confident enough, you might tutor either as a volunteer or as a way to make some extra money.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;You can start your own blog.  I haven't tried it, but a few people have reported that it's a great way to learn new things.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="1004" LastActivityDate="2010-09-19T23:23:10.687" CommunityOwnedDate="2010-09-19T23:23:10.687" />
  <row Id="5027" PostTypeId="2" ParentId="5019" CreationDate="2010-09-19T23:44:10.017" Score="6" ViewCount="" Body="&lt;p&gt;The main point of row operations is that they do not change the solution set of the underlying linear system.  So when you take a system of linear equations, write down its (augmented) coefficient matrix, and row reduce that matrix, you get a new system of equations that has the same solutions as the original system.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the coefficient matrix is in reduced row echelon form, it is particularly easy to understand the solution set of the linear system.  In particular, you can readily tell whether the system is consistent or inconsistent, and when the system is consistent, you can readily see how a choice for the values of the free (non-pivotal) variables leads to a solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I recommend David Lay's book as a textbook for this material.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One thing that sometimes gets lost in the process of Gaussian elimination (the usual algorithm for putting a matrix into reduced row echelon form) is just how similar this process is to the approach, often taught in high school algebra, of solving for one variable in a particular equation and then eliminating its occurrences from all the other equations.  When you produce a pivot in a row of the coefficient matrix, this corresponds to solving the equation for that variable.  When you add or subtract (multiples of) this pivotal row from the other rows to produce zeros in the pivotal column, the new rows represent the equations you get by substituting in for the pivotal variable.  In a sense, Gaussian elimination is just an excellent way to keep track of the process of solving for variables and then eliminating them from the other equations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's an example.  We start with the equations&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$3x + 6y + 3z = 3$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$2x + y  + 7z  = 6$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The corresponding matrix is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\left[\begin{array}{rrr|r} 3 &amp;amp; 6 &amp;amp; 3 &amp;amp; 3\\&#xA; 2 &amp;amp; 1 &amp;amp; 7 &amp;amp; 6\end{array}\right]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dividing the first row/equation by 2 gives you the equation $x + 2y + z = 1$, which you can think of as $x = 1 - 2y - z$.  You can plug that expression into the second equation to get&#xA;$2(1-2y -z) + y + 7z = 6$, which simpifies to $-3y + 5z = 4$.  Notice that when you row reduce by dividing the first row by 2 and then subtracting twice row 1 from row 2, you get the same thing:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\left[\begin{array}{rrr|r} 3 &amp;amp; 6 &amp;amp; 3 &amp;amp; 3\\&#xA; 2 &amp;amp; 1 &amp;amp; 7 &amp;amp; 6\end{array}\right] \to  \left[\begin{array}{rrr|r} 1 &amp;amp; 2 &amp;amp; 1 &amp;amp; 1\\&#xA; 2 &amp;amp; 1 &amp;amp; 7 &amp;amp; 6 \end{array}\right] \to  \left[\begin{array}{rrr|r} 1 &amp;amp; 2 &amp;amp; 3 &amp;amp; 1\\&#xA; 0 &amp;amp; -3 &amp;amp; 5 &amp;amp; 4\end{array}\right].$&lt;/p&gt;&#xA;" OwnerUserId="1392" LastEditorUserId="1392" LastEditorDisplayName="" LastEditDate="2010-09-20T03:07:27.073" LastActivityDate="2010-09-20T03:07:27.073" CommentCount="2" />
  <row Id="5029" PostTypeId="2" ParentId="5018" CreationDate="2010-09-20T00:12:16.573" Score="3" ViewCount="" Body="&lt;p&gt;A good place to sample quick starting parts of mathematics is through the books of Martin Gardner, who sadly, died recently:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.york.cuny.edu/~malk/biblio/martin-gardner-biblio.html&quot; rel=&quot;nofollow&quot;&gt;http://www.york.cuny.edu/~malk/biblio/martin-gardner-biblio.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another suggestion is to sample the essays in the wonderful book (1034 pages):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Princeton Companion to Mathematics (2008)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edited by Timothy Gowers&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Princeton U. Press, Princeton, NJ.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These essays cover a wide swath of mathematics and there are suggestions about learning more about areas that may catch your interest.&lt;/p&gt;&#xA;" OwnerUserId="1369" LastActivityDate="2010-09-20T00:12:16.573" CommunityOwnedDate="2010-09-20T00:12:16.573" />
  <row Id="5030" PostTypeId="2" ParentId="5018" CreationDate="2010-09-20T00:27:17.597" Score="3" ViewCount="" Body="&lt;p&gt;I can only speculate on your background, so I would recommend only that in addition to the stuff you picked up from school, learn how to program (insert language/environment of choice here). It can greatly help to have another tool apart from pen and paper to try out whatever piece of mathematics you're trying to learn.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After that little prerequisite... it's open season for problems! Math problems don't have to be hard, advanced, or a &quot;hot topic&quot;; all they have to be is to be captivating and fascinating to you! Look around encyclopedias like MathWorld or Wikipedia so you can get a feel of what's out there (sometimes the &quot;Random Entry&quot; function is helpful in this regard if you don't know where to start), and if you find something you like, you can start from there! In addition to that, the bibliographies should give you ideas on what to type into the search engine should the encyclopedia entry prove inadequate in giving you what you want. If you find your background is less than adequate in understanding fully what you saw, that's okay too; this is one good thing about being in the Internet age, the ability to search for information without having to leave your seat.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a further supplement, go to the nearest public library and take a look at the math section. Since not all the great books are already online, so to speak, a look at the printed pages can be helpful.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-09-20T00:27:17.597" CommunityOwnedDate="2010-09-20T00:27:17.597" CommentCount="2" />
  <row Id="5031" PostTypeId="1" AcceptedAnswerId="5032" CreationDate="2010-09-20T00:53:07.813" Score="4" ViewCount="169" Body="&lt;p&gt;I was reading about isomorphisms and homomorphisms on general structures, and first came across the definition of an injective homomorphism, or an embedding. This made me curious, is it possible for two structures $A$ and $B$ to be embeddable in each other, yet no isomorphism exists between them?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After some looking around, I let the structures be $A=\mathbb{R}$ and $B=[-1,1]$ with $f\colon [-1,1]\to\mathbb{R}\colon r\mapsto r$ and $g\colon\mathbb{R}\to [-1,1]\colon r\mapsto \frac{2}{\pi}\arctan(r)$. If I refrain from defining any relations, functions, or distinguished elements in the universes of $A$ and $B$, then it is vacuously true that $f$ and $g$ are homomorphisms. Also, $[-1,1]$ and $\mathbb{R}$ would not be isomorphic since $[-1,1]$ has a maximum and minimum element. (Or would this require me to define $\lt$ on $[-1,1]$?).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there some other structures, even contrived ones, where such embeddings $f$ and $g$ exist, but $A$ and $B$ are still not isomorphic?&lt;/p&gt;&#xA;" OwnerUserId="1583" LastEditorUserId="1583" LastEditorDisplayName="" LastEditDate="2010-09-20T02:19:09.570" LastActivityDate="2010-09-20T03:38:37.437" Title="Can two structures be embeddable in each other, but not isomorphic?" Tags="&lt;abstract-algebra&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="1" />
  <row Id="5032" PostTypeId="2" ParentId="5031" CreationDate="2010-09-20T00:54:00.837" Score="7" ViewCount="" Body="&lt;p&gt;Yes. This was discussed on MO and many examples were given.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See &lt;a href=&quot;http://mathoverflow.net/questions/1058/when-does-cantor-bernstein-hold&quot; rel=&quot;nofollow&quot;&gt;http://mathoverflow.net/questions/1058/when-does-cantor-bernstein-hold&lt;/a&gt; and the links thereon.&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2010-09-20T00:54:00.837" CommentCount="2" />
  <row Id="5033" PostTypeId="2" ParentId="5015" CreationDate="2010-09-20T01:50:35.353" Score="1" ViewCount="" Body="&lt;p&gt;First, from your description, it appears that P=S is a valid, but likely non-optimal[1], solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Begin with this. Sort the set from smallest to largest. For every member of the set, if there exists a subset that sums to that member, remove that member. Do not repeat.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[1] For S=[1,2,4,8] then optimal P=[1,2,4,8]&lt;/p&gt;&#xA;" OwnerUserId="1995" LastEditorUserId="1995" LastEditorDisplayName="" LastEditDate="2010-09-21T04:36:17.883" LastActivityDate="2010-09-21T04:36:17.883" CommentCount="5" />
  <row Id="5034" PostTypeId="2" ParentId="5019" CreationDate="2010-09-20T01:52:26.740" Score="2" ViewCount="" Body="&lt;p&gt;Well, Dan and muad have already told you everything, but let me stress the two main points (in my opinion):&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Row reduction is just the same as solving one equation for one unkown in terms of the others and then plugging the obtained expression into the remaining equations. (But without having to write down the unkowns all the time.) In a nutshell, &lt;strong&gt;row reduction = solving your system&lt;/strong&gt;.&lt;/li&gt;&#xA;&lt;li&gt;Row reduction doesn't change the solution of the system of linear equations. (So the target in performing row reduction is to obtain a simpler system, with the same solutions as the original one, but such that solutions can be read directly from it.)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The second statement can be proved formally, but I think that, if I can convince you that &lt;strong&gt;(1)&lt;/strong&gt; is true, you won't need this formal proof, will you? (Because, when you solve your system, you're sure that, if you don't make mistakes, the solution you arrive at is, well, &lt;em&gt;the solution&lt;/em&gt;. You don't change the solution of a system when you perform your operations in order to solve it, do you? So, if row reduction = solving your system and you don't change the solutions of your system while solving it, then row reduction doesn't change the solutions of your system.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So let's see how row reduction = solving your system. Say you're trying to solve the following system:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{align}&#xA;5x - 3y &amp;amp;= 3 \\&#xA;2x + 4y &amp;amp;= 2&#xA;\end{align}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You realize that the second equation looks nicer than the first one, so you decide to start with it and &lt;strong&gt;interchange&lt;/strong&gt; both equations:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{align}&#xA;2x + 4y &amp;amp;= 2    \\&#xA;5x - 3y &amp;amp;= 3 &#xA;\end{align}&#xA;\qquad \qquad \qquad \qquad \textbf{Step one}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then you solve the first equation for $x$. You &lt;strong&gt;divide by two&lt;/strong&gt; your, now, first equation &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;x + 2y = 1&#xA;\qquad \qquad \qquad \qquad \textbf{Step two}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;solve it for $x$, $x = 1 -2y$, and &lt;strong&gt;plug it into the second one&lt;/strong&gt;, that is $5(1-2y) - 3y = 3$. You obtain:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;-13y = -2&#xA;\qquad \qquad \qquad \qquad \textbf{Step three}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, you solve the second equation for $y$, &lt;strong&gt;dividing by&lt;/strong&gt; $-13$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;y = \frac{2}{13}&#xA;\qquad \qquad \qquad \qquad \textbf{Step four}&lt;br&gt;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and you perform &lt;strong&gt;back subtitution&lt;/strong&gt;, that is you plug this $y$ into the first equation, $x = 1- 2\frac{2}{13}$, getting&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;x = \frac{9}{13}&#xA;\qquad \qquad \qquad \qquad \textbf{Step five}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you know &lt;strong&gt;the solution&lt;/strong&gt; of your system is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{align}&#xA;x &amp;amp;= \frac{9}{13} \\&#xA;y &amp;amp;= \frac{2}{13}&#xA;\end{align}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, we are going to do &lt;strong&gt;exactly the same&lt;/strong&gt;, but with &lt;strong&gt;row reduction&lt;/strong&gt;. Our system of equations is the same as its augmented matrix:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\left(&#xA;\begin{array}{rr|r}&#xA;5 &amp;amp; -3 &amp;amp; 3 \\&#xA;2 &amp;amp; 4  &amp;amp; 2&#xA;\end{array}&#xA;\right)&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In &lt;strong&gt;Step one&lt;/strong&gt;, we've interchanged both equations. Now, we interchange the two rows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\left(&#xA;\begin{array}{rr|r}&#xA;2 &amp;amp; 4  &amp;amp; 2 \\&#xA;5 &amp;amp; -3 &amp;amp; 3&#xA;\end{array}&#xA;\right)&#xA;\qquad \qquad \qquad \qquad \textbf{Step one}&lt;br&gt;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In &lt;strong&gt;Step two&lt;/strong&gt;, we divided by two the first equation. Now, we divide by two the first row:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\left(&#xA;\begin{array}{rr|r}&#xA;1 &amp;amp; 2  &amp;amp; 1 \\&#xA;5 &amp;amp; -3 &amp;amp; 3&#xA;\end{array}&#xA;\right)&#xA;\qquad \qquad \qquad \qquad \textbf{Step two}&lt;br&gt;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In &lt;strong&gt;Step three&lt;/strong&gt;, we plugged $x = 1 -2y$ into the second equation. Now, we substract five times the first equation from the second:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\left(&#xA;\begin{array}{rr|r}&#xA;1 &amp;amp; 2  &amp;amp; 1 \\&#xA;0 &amp;amp; -13 &amp;amp; -2&#xA;\end{array}&#xA;\right)&#xA;\qquad \qquad \qquad \qquad \textbf{Step three}&lt;br&gt;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In &lt;strong&gt;Step four&lt;/strong&gt;, we divided the second equation by $-13$. Now, we divide the second row by $-13$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\left(&#xA;\begin{array}{rr|r}&#xA;1 &amp;amp; 2  &amp;amp; 1 \\&#xA;0 &amp;amp; 1 &amp;amp; \frac{2}{13}&#xA;\end{array}&#xA;\right)&#xA;\qquad \qquad \qquad \qquad \textbf{Step four}&lt;br&gt;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In &lt;strong&gt;Step five&lt;/strong&gt;, we performed back substitution. Now, we substract $2$ times the second row from the first one:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\left(&#xA;\begin{array}{rr|r}&#xA;1 &amp;amp; 0  &amp;amp; \frac{9}{13} \\&#xA;0 &amp;amp; 1  &amp;amp; \frac{2}{13}&#xA;\end{array}&#xA;\right)&#xA;\qquad \qquad \qquad \qquad \textbf{Step five}&lt;br&gt;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, &lt;strong&gt;the solution&lt;/strong&gt; of your system is in the third column, because the system that corresponds to this matrix is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{align}&#xA;x &amp;amp;= \frac{9}{13} \\&#xA;y &amp;amp;= \frac{2}{13}&#xA;\end{align}&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="664" LastEditorUserId="664" LastEditorDisplayName="" LastEditDate="2010-09-20T07:20:56.887" LastActivityDate="2010-09-20T07:20:56.887" />
  <row Id="5035" PostTypeId="1" CreationDate="2010-09-20T02:17:59.813" Score="5" ViewCount="1040" Body="&lt;p&gt;Is there any formula for this series?&lt;/p&gt;&#xA;" OwnerUserId="1997" LastActivityDate="2010-09-20T14:11:32.997" Title="1 + 1/2 + 1/3 + ... + 1/n = ?" Tags="&lt;sequences-and-series&gt;" AnswerCount="5" CommentCount="2" FavoriteCount="1" />
  <row Id="5036" PostTypeId="1" CreationDate="2010-09-20T02:22:51.703" Score="4" ViewCount="182" Body="&lt;p&gt;For the definition of moduli space of Riemann surface, it seems that it is not based on the definiton in terms of functor(representable...) but rather put a topology on the set of isormophism classes. Is there any difference between these? Or if I miss something? Is there any reference ? Thank you!&lt;/p&gt;&#xA;" OwnerUserId="1944" LastActivityDate="2010-09-20T04:17:20.900" Title="moduli space of Riemann surface" Tags="&lt;topology&gt;&lt;riemann-surface&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" />
  <row Id="5037" PostTypeId="2" ParentId="5035" CreationDate="2010-09-20T02:43:30.307" Score="12" ViewCount="" Body="&lt;p&gt;There is no formula for the nth partial sum of the harmonic series, only approximations. A well known approximation by Euler is that the nth partial sum is approximately $$ln(n) + \gamma $$ where $\gamma$ is the Euler–Mascheroni constant and is close to 0.5772. The amount of error in this approximation gets arbitrarily small for sufficiently large values of n.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A well known fact in mathematics is that the harmonic series is divergent which means that if you add up enough terms in the series you can make their sum as large as you wish.&lt;/p&gt;&#xA;" OwnerUserId="1352" LastActivityDate="2010-09-20T02:43:30.307" CommentCount="2" />
  <row Id="5038" PostTypeId="2" ParentId="5025" CreationDate="2010-09-20T03:28:20.987" Score="3" ViewCount="" Body="&lt;p&gt;Now,&#xA;$$ \frac{1}{(1-x)(1+x^2)(1-x^3)\dots} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ = (1+x+x^2+\dots)(1-x^2+x^4-\dots)(1+x^3+x^6+\dots)(1-x^4+x^8-\dots)\dots $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider a term which provides a negative coefficient to $x^n$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we pick $x^{2n_2}$ from the second term, $x^{4n_4}$ from the fourth and so on,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The coefficient of $-1$ is $$(-1)^{n_2 + n_4 + \dots}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;this will be negative if and only if $n_2 + n_4 + \dots$ is odd which comes from $o(n)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The positive ones come from $e(n)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus we must have that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum (e(n) - o(n))x^n = \frac{1}{(1-x)(1+x^2)(1-x^3)\dots}$$&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-09-20T03:28:20.987" CommentCount="1" />
  <row Id="5039" PostTypeId="2" ParentId="5036" CreationDate="2010-09-20T03:29:02.857" Score="2" ViewCount="" Body="&lt;p&gt;From the mention of representability I assume you mean Riemann surfaces as algebraic curves, not the complex-analytic or conformal moduli problems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Representing the moduli functor would mean you have object in a &quot;reasonable&quot; category (e.g., a variety, scheme, or algebraic space) that is a universal family of curves of whatever kind you are considering.  But this doesn't exist for algebraic curves without further restrictions (eg., stability) and rigidification data (such as marked points) to get a &quot;fine moduli space&quot;.  However, if by topology you mean a Grothendieck topology then you can define such a thing, i.e., a suitable category of coverings associated to the problem, as a virtual moduli object that you can consider as a space in the sense that cohomology and other geometrical invariants can be associated to it.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The canonical and much admired exposition that explained this to the world is &quot;Picard Groups of Moduli Problems&quot; by David Mumford.   Among other achievements it shows how to work with moduli stacks, without using the word &quot;stack&quot; (which may have been coined later, I don't know).&lt;/p&gt;&#xA;" OwnerUserId="467" LastActivityDate="2010-09-20T03:29:02.857" CommentCount="4" />
  <row Id="5040" PostTypeId="2" ParentId="5031" CreationDate="2010-09-20T03:38:37.437" Score="4" ViewCount="" Body="&lt;p&gt;There are unenlightening examples if you don't ask for the image of the embedding to be large.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, let A be a square and B an annulus, considered as topological spaces.  A small copy of each one can be placed inside the other but they are not isomorphic in the topological category. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bidirectional embedding is an equivalence relation, as is bidirectional dense (epimorphic) embedding, so at least linguistically the more natural problem in a given category is to ask what is the same about A and B if they are equivalent in this sense.&lt;/p&gt;&#xA;" OwnerUserId="467" LastActivityDate="2010-09-20T03:38:37.437" />
  <row Id="5041" PostTypeId="2" ParentId="5019" CreationDate="2010-09-20T03:51:09.947" Score="3" ViewCount="" Body="&lt;p&gt;The point of row reduction is to transform the system into an equivalent form that makes it simpler to deduce key information. In particular the reduced row echelon form immediately reveals fundamental information such as the  &lt;strong&gt;column space&lt;/strong&gt;, spanned by the pivot columns, the &lt;strong&gt;rank&lt;/strong&gt;, the &lt;strong&gt;nullspace&lt;/strong&gt;, etc. Generally such information can not be so simply deduced for a matrix in general form. Here is a worked example from Strang's Linear Algebra, 4ed p. 84:&#xA;&lt;img src=&quot;http://i.imgur.com/R8yqm.jpg&quot; alt=&quot;alt text&quot;&gt;&#xA;&lt;img src=&quot;http://i.imgur.com/6LjBp.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-09-20T03:51:09.947" />
  <row Id="5042" PostTypeId="2" ParentId="5036" CreationDate="2010-09-20T04:17:20.900" Score="3" ViewCount="" Body="&lt;p&gt;In the complex analytic setting one can also work in terms of representing functors and so on; unfortunately, just as in the algebraic setting, it will turn out that the natural functor is not representable without either imposing level structure or passing to a larger category of stack-like objects.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the complex analytic study of Riemann surfaces and their moduli has its own traditions, coming from Teichmuller theory, the theory of the mapping class group, the analytic theory of Ahlfors and Bers, and so on, and these traditions come with their own view-points, in which emphasizing the representability of the functor by a certain stack-like object may not be of paramount importance.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just to see why this might be, consider that if $f: X \to S$ is a smooth proper map of complex analytic manifolds (e.g. a family of Riemann surfaces over a smooth base), then by Ehresmann's theorem, $f$ is a fibre-bundle in the $\mathcal C^{\infty}$-category.  So one sees that&#xA;the fibres of $f$ don't vary topologically, and one can try to consider directly how the complex structure varies.  (This is the Kodaira--Spencer view-point on deformation theory.)  The fact that one has these vivid geometric view-points available means that there is less pressure to use categorical language to describe the situation.&lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2010-09-20T04:17:20.900" CommentCount="3" />
  <row Id="5043" PostTypeId="2" ParentId="5035" CreationDate="2010-09-20T04:57:39.863" Score="6" ViewCount="" Body="&lt;p&gt;Let $H_n=1+1/2+\cdots+1/n$, as usual. Then, for $n&gt;1$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$H_n =\log(n+1/2) + \gamma + \varepsilon(n)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$0 &amp;lt; \varepsilon(n) &amp;lt; 1/(24n^2)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so if you can use a high-precision estimate rather than an exact answer this may be good enough.  For example, $1+1/2+1/3+\cdots+1/10^{100}$ is between&lt;/p&gt;&#xA;&#xA;&lt;p&gt;230.83572496430610126240565755851882319115230819881722120213855733164212869451291269453666757225157658376140985147843194582191305052276721850285291090752309248454422116629840542211766342541591511108644544&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;230.83572496430610126240565755851882319115230819881722120213855733164212869451291269453666757225157658376140985147843194582191305052276721850285291090752309248454422116629840542211766342541591511108644546&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where the numbers are identical up to their last digit.  If you need higher precision, a series expansion will yield formulas that are increasingly precise for large values of $n$.  (For small values of $n$, calculate directly...)&lt;/p&gt;&#xA;" OwnerUserId="1778" LastEditorUserId="1778" LastEditorDisplayName="" LastEditDate="2010-09-20T05:04:19.720" LastActivityDate="2010-09-20T05:04:19.720" CommentCount="1" />
  <row Id="5044" PostTypeId="2" ParentId="4988" CreationDate="2010-09-20T04:58:59.010" Score="2" ViewCount="" Body="&lt;p&gt;I think my point in the comments was not taken, so I'll elaborate here and make this a proper answer. As far as I understand, you want to find the feasible point with the smallest $a_1$. If there is more than one such point, you want to find the one among them with the smallest $a_2$, and so on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is no such objective function that will work for all constraints $f$. Taking your example $L = 10x + y$ (I've renamed the variables as it'll make things easier later), I can choose the constraint to be $f(x,y) = 100x + y \in [1,2]$, making the feasible set to be a convex quadrilateral with vertices at $(0.01,0)$, $(0.02,0)$, $(0,1)$, and $(0,2)$. Now by your criteria you would like to choose the point $(0,1)$ as it minimizes $x$, but if you work out the values of $L$, you'll find that it is instead minimized at $(0.01,0)$. So the objective function you could &quot;easily write&quot; does not give you the desired solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, suppose $L$ is a function that gives you the desired solution for all constraints. Pick two values $x_1 &amp;lt; x_2$ and consider two corresponding &quot;slices&quot; of the domain $S_1 = \{(x_1,y) : y_{\min} \le y \le y_{\max}\}$ and $S_2 = \{(x_2,y) : y_{\min} \le y \le y_{\max}\}$. Then $L$ attains strictly lower values on $S_1$ than on $S_2$, because all points in $S_1$ are preferred over all points in $S_2$. On the other hand, $L$ must increase by a nonzero amount over $S_1$ itself, because $L(x_1,y_{\max}) &gt; L(x_1,y_{\min})$; call the difference $\Delta L(x_1)$. So, $L(x_2,y_{\min}) &gt; L(x_1,y_{\max}) = L(x_1,y_{\min}) + \Delta L(x_1)$. This is the key point — if $x_2 &gt; x_1$, $L(x_2,y_{\min}) &gt; L(x_1,y_{\min}) + \Delta L(x_1)$ — that is, keeping $y$ fixed, moving along $x$ by &lt;em&gt;any&lt;/em&gt; arbitrarily small amount must cause a finite increase in $L$ independent of how little you move.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From here, the proof is easy to wrap up. There are uncountably many $x$s in $[x_{\min},x_{\max}]$, yet $L(x,y_{\min})$ must increase by a positive amount $\Delta L(x)$ at each $x$. So $L(x_{\max},y_{\min}) \ge L(x_{\min},y_{\min}) + \sum_x \Delta L(x)$, but this is impossible because &lt;a href=&quot;http://mathforum.org/library/drmath/view/66534.html&quot; rel=&quot;nofollow&quot;&gt;the sum of uncountably many positive numbers must diverge&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="856" LastActivityDate="2010-09-20T04:58:59.010" CommentCount="9" />
  <row Id="5046" PostTypeId="2" ParentId="5018" CreationDate="2010-09-20T06:28:24.687" Score="9" ViewCount="" Body="&lt;p&gt;I'm also an older person trying to teach myself math. I took math till high school and since then I have taught myself to beginning-graduate level in one area (analysis) and bits and pieces of other areas at an undergraduate level. My suggestions,&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;Time is limited, don't waste it on popular books.&lt;/strong&gt; Study high-school and undergraduate textbooks and do the problems.  It may be a good idea to     go to an older generation of textbooks since they are less chatty than the current ones.      The &lt;a href=&quot;http://www.ocf.berkeley.edu/~abhishek/chicmath.htm&quot; rel=&quot;nofollow&quot;&gt;Chicago Undergradute Mathematics Bibliography&lt;/a&gt; is useful. But don't obsess too much on which book to read. At this level it is hard to get things very wrong. Pick one book and persevere with it.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Work in sprints.&lt;/strong&gt; It takes    time to get into a&#xA;productive frame    of mind. So trying to do&#xA;math for a little    while each day&#xA;does not lead to any    progress and&#xA;results in frustration.    Instead,   dedicate&#xA;a week of your vacation or at least a &#xA;whole weekend for learning something &#xA;new. Between these sprints revise what you have learned by doing more problems.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Get it right the first time.&lt;/strong&gt; While studying a text if you feel that you don't understand something fully, stop and work on your difficulty. Look at other books. Ask for help on forums like this. Go and brush up on the prerequisites if need be. But never leave a gap. Don't skip more than one exercise in ten.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Be    patient.&lt;/strong&gt; A good undergraduate     training takes 4 years of     work with teachers and peers. My      guess is it will take at least a      dozen years to achieve the same level   of mathematical maturity working      part-time on your own. But at the end your understanding will be better than an undergraduate's since you will have had the benefit of greater intellectual maturity and more time to reflect.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Is it worth it?&lt;/strong&gt; Math is no doubt beautiful. But the time you give to it comes at the cost of your career, personal life and other interests. Are you willing to make the necessary sacrifices? If you are then it might be a good idea to make the sacrifices upfront and go back to school to learn math. The chances of fulfillment would be much higher.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1195" LastActivityDate="2010-09-20T06:28:24.687" CommunityOwnedDate="2010-09-20T06:28:24.687" CommentCount="3" />
  <row Id="5047" PostTypeId="2" ParentId="5007" CreationDate="2010-09-20T06:31:51.513" Score="2" ViewCount="" Body="&lt;p&gt;As Robin Chapman has pointed out, the correct choice is $c=-2$, not $c=10$. This gives&#xA;$$y(x)=\frac{28-(x-2)^2}{2} = -\frac{x^2}{2} + 2x + 12.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, this formula describes a parabola $y(x)$ which hits its highest point $y=14$ at $x=-2$, and this is a decreasing function for $x&gt;-2$, which disagrees with the ODE. (Since the square root in the right-hand side of the ODE can't be negative, the solution $y(x)$ can't be decreasing.) Moreover, the right-hand side of the ODE is undefined for $y&gt;14$, so the solution can't continue to increase beyond 14. The conclusion is that the only possible continuation is $y(x)=14$ for all $x&gt;-2$. Hence, the solution is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$y(x)=\frac{28-(x-2)^2}{2} = -\frac{x^2}{2} + 2x + 12, \quad -4 \le x \le -2,$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$y(x)=14, x&gt;-2.$$&lt;/p&gt;&#xA;" OwnerUserId="1242" LastEditorUserId="1242" LastEditorDisplayName="" LastEditDate="2010-09-20T08:26:11.113" LastActivityDate="2010-09-20T08:26:11.113" CommentCount="1" />
  <row Id="5048" PostTypeId="2" ParentId="5035" CreationDate="2010-09-20T06:33:58.260" Score="0" ViewCount="" Body="&lt;p&gt;Well there are some formulae for the harmonic numbers. For example, one of the most well known is&#xA;$ H_n = \int_0^1 {1 - x^n \over 1 - x} dx$, which has certain useful properties. However, it is not in any way simpler than the original definition. And if you're looking for numerical approximations rather than exact identities, you certainly are better off with the approximation $H_n= \ln (n+1/2) +\gamma + \epsilon (n)$&lt;/p&gt;&#xA;" LastActivityDate="2010-09-20T06:33:58.260" />
  <row Id="5049" PostTypeId="1" AcceptedAnswerId="5051" CreationDate="2010-09-20T07:59:14.853" Score="4" ViewCount="213" Body="&lt;p&gt;According this problem/solution set from an MIT class (http://ocw.mit.edu/courses/mathematics/18-100c-analysis-i-spring-2006/exams/exam1_sol.pdf), the assertion:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Every collection of disjoint intervals in R is countable.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is True, because &quot;every interval contains a rational number&quot;, and the rationals are countable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems to me this should be False, with possible counterexample:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;{ [x,x] | x is an element of R}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ie the set of all singelton intervals on R. Why isn't this a valid counterexample? &lt;/p&gt;&#xA;" OwnerUserId="2003" LastEditorUserId="2003" LastEditorDisplayName="" LastEditDate="2010-09-20T11:00:42.273" LastActivityDate="2010-09-20T11:00:42.273" Title="Countability of disjoint intervals" Tags="&lt;real-analysis&gt;&lt;analysis&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="5050" PostTypeId="2" ParentId="5049" CreationDate="2010-09-20T08:28:51.477" Score="1" ViewCount="" Body="&lt;p&gt;Because &quot;singleton interval&quot; is usually not considered to be an interval.&lt;/p&gt;&#xA;" OwnerUserId="695" LastActivityDate="2010-09-20T08:28:51.477" CommentCount="1" />
  <row Id="5051" PostTypeId="2" ParentId="5049" CreationDate="2010-09-20T08:29:37.840" Score="5" ViewCount="" Body="&lt;p&gt;Your thinking is correct; the set of all singleton sets of &lt;strong&gt;R&lt;/strong&gt; is certainly uncountable.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems that the question meant something like &quot;Every collection of disjoint &lt;em&gt;open&lt;/em&gt; intervals in &lt;strong&gt;R&lt;/strong&gt; is countable.&quot;  (In this case, the claim that each interval contains a rational number is valid.)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe there was some convention in the course that &quot;interval&quot; meant open interval, or excluded singleton sets; perhaps it's simply a mistake.  Either way, it's good that you noticed this detail!&lt;/p&gt;&#xA;" OwnerUserId="1461" LastActivityDate="2010-09-20T08:29:37.840" CommentCount="1" />
  <row Id="5052" PostTypeId="1" AcceptedAnswerId="5339" CreationDate="2010-09-20T09:39:39.077" Score="1" ViewCount="232" Body="&lt;p&gt;I have heard that Legendre functions are important in number theory. Can any one tell me how?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Legendre function of the first kind $P_s$ is defined by&#xA;\begin{eqnarray*}P_s(x) =&amp;amp; \frac{1}{2\pi}\int_{-\pi}^{\pi}\left(x+\sqrt{x^2-1}\cos\theta\right)^s d\theta&lt;br&gt;&#xA;\newline&#xA;=&amp;amp;\frac{1}{\pi}\int_0^1\left(x+\sqrt{x^2-1}(2t-1)\right)^s\frac{dt}{\sqrt{t(1-t)}},&amp;amp; s\in\mathbb{C}, x\ge 1.&#xA;\end{eqnarray*}&lt;/p&gt;&#xA;" OwnerUserId="1154" LastActivityDate="2010-11-15T21:16:22.587" Title="Legendre functions in number theory" Tags="&lt;number-theory&gt;&lt;special-functions&gt;" AnswerCount="2" CommentCount="8" />
  <row Id="5053" PostTypeId="1" AcceptedAnswerId="5055" CreationDate="2010-09-20T10:21:45.337" Score="5" ViewCount="175" Body="&lt;p&gt;The German mathematician &lt;a href=&quot;http://en.wikipedia.org/wiki/Martin_Eichler&quot; rel=&quot;nofollow&quot;&gt;Martin Eichler&lt;/a&gt; once stated that there were five fundamental operations of mathematics: addition, subtraction, multiplication, division, and modular forms. This was also mentioned in Simon Singh's book on Fermat's last theorem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know much about modular forms beyond the &lt;a href=&quot;http://en.wikipedia.org/wiki/Modular_forms&quot; rel=&quot;nofollow&quot;&gt;Wikipedia entry&lt;/a&gt;, and am quite confused by this remark. The usual operations seem to be different entities (two numbers input, one number output) than a modular form (one number input, one number output).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anyone have some deeper insight into Martin Eichler's assertion?&lt;/p&gt;&#xA;" OwnerUserId="960" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-22T02:10:57.960" LastActivityDate="2010-11-22T02:10:57.960" Title="How are modular forms fundamental operations?" Tags="&lt;modular-forms&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="5054" PostTypeId="2" ParentId="5016" CreationDate="2010-09-20T10:31:29.050" Score="2" ViewCount="" Body="&lt;p&gt;Your condition means that  if $\gamma$ is the parabola $x^2-y=0$ then for any $f\in\mathcal{F}$ we have $f(\mathbb{D})\cap\gamma=\emptyset$. Thus &#xA;$\mathcal{F}=\mathcal{G}\cup \mathcal{H}$ where each $g$ in $\mathcal{G}$ maps the disk in the open set $x^2-y&gt;$0 and each $h$ in $\mathcal{H}$ maps the disk in the open set $x^2-y&amp;lt;0$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now just note that these two open sets are biholomorphic to the unit disk by uniformization theorem, thus $\mathcal{G}$ and $\mathcal{H}$ are normal families.  If you don't want to use uniformization theorem try to prove by hand that the open sets are biholomorphic to the disk.&lt;/p&gt;&#xA;" OwnerUserId="1860" LastActivityDate="2010-09-20T10:31:29.050" CommentCount="3" />
  <row Id="5055" PostTypeId="2" ParentId="5053" CreationDate="2010-09-20T10:44:14.377" Score="0" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://mathoverflow.net/questions/6955/modular-forms-eichler-quote&quot; rel=&quot;nofollow&quot;&gt;http://mathoverflow.net/questions/6955/modular-forms-eichler-quote&lt;/a&gt;&lt;/p&gt;&#xA;" LastActivityDate="2010-09-20T10:44:14.377" CommunityOwnedDate="2010-09-20T14:34:30.470" CommentCount="1" />
  <row Id="5057" PostTypeId="2" ParentId="5035" CreationDate="2010-09-20T13:24:30.727" Score="2" ViewCount="" Body="&lt;p&gt;This question made me recall this wonderful formula for $H_n$ due to &lt;a href=&quot;http://en.wikipedia.org/wiki/Euler-Mascheroni_constant&quot; rel=&quot;nofollow&quot;&gt;Gregory Fontana&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$H_n = \gamma + \log n + {1 \over 2n} - \sum_{k=2}^\infty { (k-1)! C_k \over n(n+1)\ldots(n+k-1)},&#xA;\qquad \textrm{ for } n=1,2,3,\ldots,$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where the coefficients $C_k$ are the Gregory coefficients given by&#xA;$${ z \over \log(1-z)} = \sum_{k=0}^\infty C_k z^k \qquad \textrm{ for } |z|&amp;lt;1.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Although I remember proving this as a student, about 30 years ago, I've just spent a frustrating 90 minutes trying to recover the proof without success, so unfortunately I cannot sketch it here. PS: I'll post a question shortly, and perhaps someone can beat me to it and put me out of my misery :-)&lt;/p&gt;&#xA;" OwnerUserId="1301" LastActivityDate="2010-09-20T13:24:30.727" />
  <row Id="5058" PostTypeId="1" AcceptedAnswerId="5234" CreationDate="2010-09-20T13:33:17.043" Score="14" ViewCount="531" Body="&lt;p&gt;My question was inspired by &lt;a href=&quot;http://math.stackexchange.com/questions/5035/1-1-2-1-3-1-n&quot;&gt;this&lt;/a&gt; stackexchange question. For the last 90 minutes I have been trying to prove this formula due to &lt;a href=&quot;http://en.wikipedia.org/wiki/Euler-Mascheroni_constant&quot; rel=&quot;nofollow&quot;&gt;Gregorio Fontana&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$H_n = \gamma + \log n + {1 \over 2n} - \sum_{k=2}^\infty { (k-1)! C_k \over n(n+1)\ldots(n+k-1)},&#xA;\qquad \textrm{ for } n=1,2,3,\ldots,$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $H_n = \sum_{k=1}^n 1/k$ and the coefficients $C_k$ are the Gregory coefficients given by&#xA;$${ z \over \log(1-z)} = \sum_{n=0}^\infty C_k z^k \qquad \textrm{ for } |z|&amp;lt;1.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's a bit frustrating as it's something I recall proving as a student many years ago. I have a vague recollection that I began with something like:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$H_n = \int_0^1 {1-(1-x)^n \over x } \textrm{d}x,$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but my attempts to follow on from there have failed. Can you help?&lt;/p&gt;&#xA;" OwnerUserId="1301" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-09-22T22:16:34.700" LastActivityDate="2010-09-22T22:16:34.700" Title="Formula for the harmonic series $H_n = \sum_{k=1}^n 1/k$ due to Gregorio Fontana." Tags="&lt;sequences-and-series&gt;" AnswerCount="1" FavoriteCount="3" />
  <row Id="5059" PostTypeId="2" ParentId="5019" CreationDate="2010-09-20T13:33:19.623" Score="1" ViewCount="" Body="&lt;p&gt;i think of gaussian elimination or row reduction of a matrix A as giving you a simpler basis for the row space. it is only a small step from there by looking at the reduction of augmented matrix to find the compatibility condition on the right hand side b and solve ax = b&#xA;in that case.&lt;/p&gt;&#xA;" LastActivityDate="2010-09-20T13:33:19.623" />
  <row Id="5060" PostTypeId="2" ParentId="5035" CreationDate="2010-09-20T14:11:32.997" Score="3" ViewCount="" Body="&lt;p&gt;Here is a way to interpret the harmonic numbers combinatorially:&#xA;$$H_n=\dfrac {\genfrac{[}{]}{0pt}{}{n+1}{2}}{n!}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\genfrac{[}{]}{0pt}{}{n+1}{2}$ is the absolute value of the Stirling number of the first kind, namely, the number of permutations of ${1,2,\dots,n,n+1}$ that have exactly $2$ cycles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These satisfy the following recurrence:&#xA;$$\genfrac{[}{]}{0pt}{}{n}{k}=\genfrac{[}{]}{0pt}{}{n-1}{k-1}+(n-1)\genfrac{[}{]}{0pt}{}{n-1}{k}$$ which makes it algebraically obvious that they are related to the Harmonic numbers the way they are, though there is also a purely combinatorial proof. The virtue of this interpretation is that you can prove a whole host of crazy identities involving the Harmonic numbers by just translating the natural identities for the Stirling numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Stirling numbers of the first kind are also the coefficients of $x(x-1)(x-2)\cdots(x-(n-1))$, so their absolute values are the coefficients of $x(x+1)(x+2)\cdots (x+n-1)$, and so in particular&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The n&lt;sup&gt;th&lt;/sup&gt; Harmonic number is the coefficient of $x^2$ in $\frac1{n!}x(x+1)(x+2)\dots(x+n)$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;All of this can be found in Chapter 7 of Benjamin &amp;amp; Quinn's wonderful book Proofs That Really Count.&lt;/p&gt;&#xA;" OwnerUserId="400" LastActivityDate="2010-09-20T14:11:32.997" />
  <row Id="5061" PostTypeId="1" AcceptedAnswerId="5067" CreationDate="2010-09-20T15:34:56.363" Score="2" ViewCount="158" Body="&lt;p&gt;I managed to prove that if $\phi(0)$ and $\forall n(n\in\mathbb{N},\phi(n)\rightarrow\phi(n+1))$ then $\forall n(n\in\mathbb{N}\rightarrow\phi(n))$. The proof is based on the fact that $\{m\in\mathbb{N}:\phi(m)\}$ is an inductive set. Now I'd like to show (but I can't) that if you take a $k\in\mathbb{N}$, if $\phi(k)$ and $\forall n(n\in\mathbb{N},k\leq n,\phi(n)\rightarrow\phi(n+1))$ then $\forall n(n\in\mathbb{N},k\leq n\rightarrow\phi(n))$. The problem is I cannot show that $\{m\in\mathbb{N}:k\leq m,\phi(m)\}$ is an inductive set, since it is obvious, for example, that it doesn't contain zero as an element (except if $k=0$). What can I do instead? Thanks.&lt;/p&gt;&#xA;" LastActivityDate="2010-09-20T16:06:38.437" Title="Induction principle" Tags="&lt;induction&gt;" AnswerCount="2" />
  <row Id="5062" PostTypeId="2" ParentId="5061" CreationDate="2010-09-20T15:39:34.700" Score="0" ViewCount="" Body="&lt;p&gt;You could define $\psi$ on $\mathbb{N}$ by $\psi(n)=\phi(n+k)$, apply your first result to $\psi$, and then reinterpret for $\phi(n)=\psi(n-k)$.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-09-20T15:39:34.700" CommentCount="5" />
  <row Id="5063" PostTypeId="1" AcceptedAnswerId="5066" CreationDate="2010-09-20T15:46:37.340" Score="2" ViewCount="234" Body="&lt;p&gt;Ok this is for a HW but I'm not looking for a handout...just a hint to get me on the right track.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have no idea where to begin proving this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Show that if A is a symmetric matrix, then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x^TAy = y^TAx$$&lt;/p&gt;&#xA;" OwnerUserId="2014" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-09-20T17:07:09.550" LastActivityDate="2010-09-20T18:43:00.450" Title="How do I prove that $x^TAy = y^TAx$ if A is symmetric?" Tags="&lt;linear-algebra&gt;&lt;homework&gt;" AnswerCount="4" />
  <row Id="5064" PostTypeId="2" ParentId="5063" CreationDate="2010-09-20T15:48:12.963" Score="3" ViewCount="" Body="&lt;p&gt;Prove that whenever $A$ and $B$ are matrices for which you can compute the product $AB$, then $$(AB)^t=B^tA^t$$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next apply $(\mathord-)^t$ to the left hand side of your equation, and compare the result to the right hand side, keeping in mind that both sides are actually numbers (well, $1$-by-$1$ matrices)&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2010-09-20T15:48:12.963" />
  <row Id="5065" PostTypeId="1" AcceptedAnswerId="5127" CreationDate="2010-09-20T16:02:23.313" Score="4" ViewCount="230" Body="&lt;p&gt;For a continuous scalar field on a circle, there is a diameter of the circle such that the endpoints of the diameter have the same value. If you think of the scalar field as &quot;temperature&quot;, then what this says is that there are points on opposite sides of the circle that are the same temperature: isothermic antipodes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, for a continuous scalar field on a sphere, the same is true: there are isothermic antipodes. (Just consider any great circle.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, is there more you can say? Can you say for example that there is a closed loop on the surface of the sphere such that every point on the loop has the same value as the other endpoint of its diameter?&lt;/p&gt;&#xA;" OwnerUserId="172" LastActivityDate="2010-09-24T15:44:58.387" Title="Does a continuous scalar field on a sphere have continuous loop of &quot;isothermic antipodes&quot;" Tags="&lt;geometry&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="5066" PostTypeId="2" ParentId="5063" CreationDate="2010-09-20T16:05:48.257" Score="10" ViewCount="" Body="&lt;p&gt;$x^tAy$ is a scalar. So $x^tAy=(x^tAy)^t$. Can you continue from here?&lt;/p&gt;&#xA;" OwnerUserId="1794" LastActivityDate="2010-09-20T16:05:48.257" CommentCount="0" />
  <row Id="5067" PostTypeId="2" ParentId="5061" CreationDate="2010-09-20T16:06:38.437" Score="2" ViewCount="" Body="&lt;p&gt;Prove instead that the set $\{m\in\mathbb{N}\colon m\lt k\}\cup\{m\in\mathbb{N}\colon\phi(m)\}$ is inductive, and thus equals $\mathbb{N}$. Then use this to deduce what you want. (That is, take every element for which $\phi(m)$ holds, and then &quot;throw in for free&quot; every number strictly less than $k$, since you do not care what happens to them, and show that this is the set of all natural numbers). &lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-09-20T16:06:38.437" />
  <row Id="5068" PostTypeId="2" ParentId="5065" CreationDate="2010-09-20T16:12:09.417" Score="0" ViewCount="" Body="&lt;p&gt;As for what more you can say in higher dimensions, take a look at the Borsuk-Ulam Theorem: &lt;a href=&quot;http://en.wikipedia.org/wiki/Borsuk%E2%80%93Ulam_theorem&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Borsuk%E2%80%93Ulam_theorem&lt;/a&gt;.  &lt;/p&gt;&#xA;" OwnerUserId="1392" LastActivityDate="2010-09-20T16:12:09.417" CommentCount="5" />
  <row Id="5069" PostTypeId="2" ParentId="4961" CreationDate="2010-09-20T16:14:06.070" Score="5" ViewCount="" Body="&lt;p&gt;A result that might be particularly suitable to students in a calculus sequence is complex partial fraction decomposition.  Over the real numbers, you get denominators with both linear and quadratic factors, but over the complex numbers, algebraic closure means you only have to think about linear factors.  For example, since&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{1}{1 + x^2} = \frac{1}{2} \left( \frac{1}{1 + ix} + \frac{1}{1 - ix} \right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;it follows that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int \frac{1}{1 + x^2}  dx = \frac{1}{2} \left( i \log(1 - ix) - i \log(1 + ix) \right).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(It turns out that this function is just the arctangent in disguise.)  So complex numbers give you a simple algorithm which you can use, in principle, to integrate any rational function.  For example, in &lt;a href=&quot;http://math.stackexchange.com/questions/4803/the-partial-fraction-expansion-of-frac1xn-1&quot;&gt;this recent math.SE thread&lt;/a&gt; it is shown that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{1}{1 - x^n} = \frac{1}{n} \left( \sum_{k=0}^{n-1} \frac{1}{1 - \zeta^k x} \right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\zeta = e^{ \frac{2 \pi i}{n} }$ is a primitive $n^{th}$ root of unity.  This implies that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int \frac{1}{1 - x^n} dx = \frac{1}{n} \left( \sum_{k=0}^{n-1} -\zeta^{-k} \log (1 - \zeta^k x) \right).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Not an integral most students would suspect is expressible in closed form!  The analogous computation of $\int \frac{1}{1 + x^n} dx$ would even allow you to express the sum $\sum_{k=0}^{\infty} \frac{(-1)^k}{kn+1}$ in closed form (which is just the definite integral from $0$ to $1$).  These examples illustrate another application of the discrete Fourier transform, but the method I'm describing here is totally general.&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-09-20T21:46:49.423" LastActivityDate="2010-09-20T21:46:49.423" CommunityOwnedDate="2010-09-20T16:14:06.070" CommentCount="1" />
  <row Id="5070" PostTypeId="2" ParentId="4961" CreationDate="2010-09-20T16:43:04.623" Score="8" ViewCount="" Body="&lt;p&gt;I'd like to mention another combinatorial result, this time about asymptotics.  Let $a_n$ denote the number of ways that $n$ horses can finish in a race (with ties); in other words, $a_n$ is the number of &quot;lists of (nonempty) sets&quot; with $n$ total members.  Generating function methods allow you to deduce that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$A(z) = \sum_{n=0}^{\infty} \frac{a_n}{n!} z^n = \frac{1}{1 - (1 - e^z)} = \frac{1}{2 - e^z}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This function is meromorphic, so the asymptotic behavior of $a_n$ is controlled by its poles.  The dominant pole occurs at $z = \ln 2$ with residue $-\frac{1}{2}$, which means that the asymptotic behavior of $a_n$ is given by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{a_n}{n!} \approx \frac{1}{2 (\ln 2)^{n+1}}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So far, so real-variable.  Where do complex numbers come in?  The &lt;strong&gt;error&lt;/strong&gt; in this approximation is controlled by the remaining poles of $A(z)$, all of which are complex.  The next two most dominant poles are at $\ln 2 \pm 2 \pi i$ with the same residue, which means that the asymptotic behavior of the error is given by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{a_n}{n!} - \frac{1}{2 (\ln 2)^{n+1}} \approx \frac{1}{2 (\ln 2 + 2 \pi i)^{n+1}} + \frac{1}{2 (\ln 2 - 2 \pi i)^{n+1}}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Letting $\frac{1}{\ln 2 + 2 \pi i} = r e^{i \theta}$ where $r = \frac{1}{\sqrt{(\ln 2)^2 + 4 \pi^2}}$ and $\theta = -\arctan \frac{2 \pi}{\ln 2}$, it follows that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{a_n}{n!} - \frac{1}{2(\ln 2)^{n+1}} \approx r^{n+1} \cos (n+1)\theta.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, the error in the above approximation is quasi-periodic.  (Of course there are infinitely many poles, each pair of which also contributes quasi-periodic terms, but as $n$ becomes large these terms become less and less important, most of the time.)  This is a phenomenon you can easily see for yourself by actually computing the error for several consecutive values of $n$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So think about this: even if you correctly guessed the asymptotic behavior of $a_n$ (for instance by making a table of the values $\frac{a_n}{n!}$ (or its inverse, if you are interested in the probability that there are no ties in the race) and noticing that the number of digits grows linearly), and even if you computed experimentally that the error in the approximation is quasi-periodic, how on earth could you possibly have deduced the value of either $r$ or $\theta$ without complex numbers?  &lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-09-20T19:44:14.303" LastActivityDate="2010-09-20T19:44:14.303" CommunityOwnedDate="2010-09-20T16:43:04.623" />
  <row Id="5071" PostTypeId="2" ParentId="4961" CreationDate="2010-09-20T16:59:42.500" Score="5" ViewCount="" Body="&lt;p&gt;I know of several ways to prove the Fundamental Theorem of Algebra using only basic tools from complex analysis. Suppose $P(z)$ is a non-constant polynomial. $|P(z)|$ is large outside of a large disk, and inside that disk since $1/P(z)$ is analytic, $1/|P(z)|$ is bounded by the maximum it has on the boundary. Thus $1/P(z)$ is a bounded analytic function on the whole plane so by Liouville's Theorem it is constant, a contradiction. Thus $P(z)$ has a zero somewhere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This really only uses the fact that polynomials are analytic, and analytic functions satisfy the maximum principle. Liouville's Theorem takes a little more work, but should be covered in any first complex analysis course.&lt;/p&gt;&#xA;" OwnerUserId="2000" LastActivityDate="2010-09-20T16:59:42.500" CommunityOwnedDate="2010-09-20T16:59:42.500" />
  <row Id="5072" PostTypeId="2" ParentId="5063" CreationDate="2010-09-20T17:04:00.743" Score="3" ViewCount="" Body="&lt;p&gt;This condition can actually be used as an equivalent definition of $A$ being symmetric.  The key point is that both sides are bilinear in $x$ and $y$, so it suffices to prove the result when $x$ and $y$ are basis vectors, say $e_i$ and $e_j$.  What does the condition say then?&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-09-20T17:04:00.743" />
  <row Id="5074" PostTypeId="2" ParentId="5063" CreationDate="2010-09-20T17:56:36.643" Score="0" ViewCount="" Body="&lt;p&gt;compute $[x_1, x_2]A[y_1 y_2]^T$ for a 2 x 2 symmetric matrix $A = [[a, b], [b, c]]$.&#xA;you get $ax_1y_1 + b(x_1y_2 + x_2y_1) + cx_2y_2$. this expression is clearly symmetric&#xA;in $x$ and $y$ i.e., does not change when $x$ and $y$ are exchanged. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;for an n x n matrix you may want to use sigma notation.  &lt;/p&gt;&#xA;" LastActivityDate="2010-09-20T18:43:00.450" />
  <row Id="5076" PostTypeId="1" AcceptedAnswerId="5082" CreationDate="2010-09-20T18:17:06.570" Score="2" ViewCount="1814" Body="&lt;p&gt;Given $f(x) = \frac{1}{2}x^TAx + b^Tx + \alpha $ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;where A is an nxn symmetric matrix, b is an n-dimensional vector, and alpha a scalar. Show that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\bigtriangledown _{x}f(x) = Ax + b$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$H = \bigtriangledown ^{2}_{x}f(x) = A$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this simply a matter of taking a derivative with respect to X, how would you attack this one?&lt;/p&gt;&#xA;" OwnerUserId="2014" LastEditorUserId="2014" LastEditorDisplayName="" LastEditDate="2010-09-20T18:30:52.767" LastActivityDate="2010-09-20T18:56:13.823" Title="what does ∇ (upside down triangle) symbol mean in this problem" Tags="&lt;linear-algebra&gt;" AnswerCount="2" CommentCount="7" />
  <row Id="5077" PostTypeId="1" AcceptedAnswerId="5078" CreationDate="2010-09-20T18:24:33.607" Score="0" ViewCount="60" Body="&lt;p&gt;Hi&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any rectangle, that if you divide it into another rectangle (or any other quadrilateral), the &lt;strong&gt;relation&lt;/strong&gt; between the two different edges is the same in both rectangles?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example - &lt;strong&gt;Orginal rectangle:&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/WF3f0.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Divided Rectangle:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/raRto.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that these illustrations aren't really exact, it's just to clear the point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="1445" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-09-20T18:26:08.280" LastActivityDate="2010-09-20T18:31:46.527" Title="Relation of different edges in rectangles" Tags="&lt;geometry&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="5078" PostTypeId="2" ParentId="5077" CreationDate="2010-09-20T18:31:41.593" Score="1" ViewCount="" Body="&lt;p&gt;This amounts to solving an equation. For the case you illustrate (where we divide the longer side in half), you fall into one of two cases: if $b\gt 2a$, so that in the second rectangle the horizontal sides are still the longer sides, you want $\frac{b/2}{a}=\frac{b}{a}$. For this to occur, you would need $ba = 2ba$, which is impossible since both $a$ and $b$ are assumed positive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $b\leq 2a$, then in the second rectangle it is the vertical side that are longest, so you want $\frac{a}{b/2} = \frac{b}{a}$. This gives you $2a^2 = b^2$, or $\sqrt{2}a=b$ (since $a$ and $b$ are lengths, they will be positive). So, for example, if $a$ is $1$ in the first figure, you need $b=\sqrt{2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, it will all depend on what proportions you want. For example, one can consider the case in which you &quot;cut off&quot; a square of side length $a$ from your original rectangle, leaving you with a rectangle with sides $a$ and $b-a$. Then you would want this latter rectangle to have the same proportions as the original one, which means you want $\frac{b}{a}=\frac{a}{b-a}$ (turns out you cannot do it if $b-a\gt a$). This leads to an equation that you can solve for $b$ in terms of $a$ (or solve for $a$ in terms of $b$), giving you the proportions that the original rectangle must have. Etc.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-09-20T18:31:41.593" />
  <row Id="5079" PostTypeId="2" ParentId="5077" CreationDate="2010-09-20T18:31:46.527" Score="3" ViewCount="" Body="&lt;p&gt;I presume you want $b/a=a/(b/2)$, that is $(b/a)^2=2$ so the ratio of sides&#xA;is $\sqrt 2$. The &lt;a href=&quot;http://en.wikipedia.org/wiki/A4_paper_size#The_A_series&quot; rel=&quot;nofollow&quot;&gt;A series&lt;/a&gt; of paper sizes is based on this ratio.&#xA;To the nearest millimetre the ratio of the long to the short size&#xA;is $\sqrt2$, so tearing a sheet of A4 along a line through its centre&#xA;parallel to its short sides gives two sheets of A5 etc.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-09-20T18:31:46.527" />
  <row Id="5080" PostTypeId="2" ParentId="4961" CreationDate="2010-09-20T18:32:32.130" Score="3" ViewCount="" Body="&lt;p&gt;There are tons of applications of complex numbers in places you would not expect. Number theory is one, for example. This example is from &lt;em&gt;The Art and Craft of Problem Solving&lt;/em&gt; by Zeitz.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proposition: If $m$ and $n$ are integers that can be written as the sum of two squares, then $mn$ can also be written as the sum of two squares.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proof: Let $m=a^2+b^2$ and $n=c^2+d^2$ and define $z=(a+bi)(c+di)$. Then $|z|^2=|a+bi|^2 |c+di|^2=mn$. Since $\Re(z)=ac-bd$ and $\Im(z)=bc+ad$ are both integers and $|z|^2$ equals the sum of the squares of its real and imaginary components, then $|z|^2=mn$ is the sum of two squares.&lt;/p&gt;&#xA;" LastActivityDate="2010-09-20T19:13:23.457" CommunityOwnedDate="2010-09-20T18:32:32.130" />
  <row Id="5081" PostTypeId="2" ParentId="5076" CreationDate="2010-09-20T18:47:49.380" Score="0" ViewCount="" Body="&lt;p&gt;$\bigtriangledown f$ finds the direction of maximal change in f.&lt;/p&gt;&#xA;" LastActivityDate="2010-09-20T18:47:49.380" />
  <row Id="5082" PostTypeId="2" ParentId="5076" CreationDate="2010-09-20T18:56:13.823" Score="5" ViewCount="" Body="&lt;p&gt;$\nabla f = (\partial f/\partial x_1, \ldots, \partial f/\partial x_n)^t$ denotes the vector of partial derivatives of $f$ and is a completely standard notation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, $\nabla^2 f$ seems to be used here in an unusual way, namely to denote the Hessian (the matrix of all second order partial derivatives), $(\partial^2 f/\partial x_i \partial x_j)_{i,j=1}^n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(The usual meaning of $\nabla^2 f$ is the Laplacian, $\partial^2 f/\partial x_1^2 + \ldots + \partial^2 f/\partial x_n^2$.)&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2010-09-20T18:56:13.823" CommentCount="3" />
  <row Id="5083" PostTypeId="1" AcceptedAnswerId="5086" CreationDate="2010-09-20T19:10:45.387" Score="1" ViewCount="98" Body="&lt;p&gt;Hello all,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know how to partition given $p(x)$ using a generating function as my textbook on discrete mathematics explains it in detail. However, I want to know if it is possible to restrict the source elements of the partition? So, given I want to find the partition of $x$ is it possible to do it using only elements from $S={y_1, y_2, y_3, ..., y_p : y_i \in \mathbb{Z} \forall i}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;An example might be partition any number using only 4 and 9.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If so, how?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a homework question of sorts but this is &lt;strong&gt;not&lt;/strong&gt; the actual homework question (because I can solve that - I just can't find a general answer for it yet and this is a way I think I can generalise (possibly)).&lt;/p&gt;&#xA;" OwnerUserId="892" LastActivityDate="2010-09-20T19:54:15.903" Title="Partitioning Integers using only elements from a specific set" Tags="&lt;number-theory&gt;&lt;discrete-mathematics&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="5084" PostTypeId="1" AcceptedAnswerId="5085" CreationDate="2010-09-20T19:22:14.977" Score="3" ViewCount="98" Body="&lt;p&gt;Given a set of vectors $\{x_1,\ldots,x_n\}$, how can I find a set of orthonormal vectors $\{y_1,\ldots,y_m\colon m\leq n\}$ spanning the same space spanned by $\{x_i\}$?&lt;/p&gt;&#xA;" OwnerUserId="2014" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-09-20T19:27:13.807" LastActivityDate="2010-09-20T19:28:01.827" Title="Orthogonal Vectors" Tags="&lt;linear-algebra&gt;" AnswerCount="1" />
  <row Id="5085" PostTypeId="2" ParentId="5084" CreationDate="2010-09-20T19:28:01.827" Score="6" ViewCount="" Body="&lt;p&gt;Look up &lt;a href=&quot;http://en.wikipedia.org/wiki/Gram-Schmidt_Process&quot; rel=&quot;nofollow&quot;&gt;the Gram-Schmidt process&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-09-20T19:28:01.827" />
  <row Id="5086" PostTypeId="2" ParentId="5083" CreationDate="2010-09-20T19:48:26.263" Score="2" ViewCount="" Body="&lt;p&gt;It's the same generating function method.  If $p_S(n)$ denotes the number of partitions of $n$ using only positive integers in some set $S \subset \mathbb{N}$, then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{n \ge 0} p_S(n) x^n = \prod_{s \in S} \frac{1}{1 - x^s}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A popular choice is $S = \{ 1, 5, 10, 25 \}$ (the problem of making change).  When $S$ is finite the above function is rational and it is possible to give a closed form for $p_S$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the question is about whether such a partition exists at all, for sufficiently large $n$ this is possible if and only if the greatest common divisor of the elements of $S$ divides $n$.  For small $n$ there are greater difficulties; see the Wikipedia article on the &lt;a href=&quot;http://en.wikipedia.org/wiki/Coin_problem&quot; rel=&quot;nofollow&quot;&gt;Frobenius problem&lt;/a&gt;.  When $|S| = 2$ (say $S = \{ a, b \}$) and the elements of $S$ are relatively prime, the largest $n$ for which no such partition exists is known to be $ab - a - b$.  When $a = 4, b = 9$ this gives $23$, so for any $n \ge 24$ such a partition always exists.&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-09-20T19:54:15.903" LastActivityDate="2010-09-20T19:54:15.903" CommentCount="1" />
  <row Id="5087" PostTypeId="2" ParentId="5018" CreationDate="2010-09-20T19:57:12.203" Score="4" ViewCount="" Body="&lt;p&gt;Khan Academy has excellent short videos you can take at your own pace, from basic arithmetic to advanced calculus. He has a very conversational style as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.khanacademy.org/&quot; rel=&quot;nofollow&quot;&gt;http://www.khanacademy.org/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2022" LastActivityDate="2010-09-20T19:57:12.203" CommunityOwnedDate="2010-09-20T19:57:12.203" />
  <row Id="5088" PostTypeId="1" AcceptedAnswerId="5090" CreationDate="2010-09-20T20:06:43.977" Score="3" ViewCount="122" Body="&lt;p&gt;&lt;em&gt;Just out of pure curiosity ...&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose I want to connect the two points $(0|0)$ and $(1|1)$ with the graph of some continuous and differentiable function &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f : [0; 1] \to [0; 1]$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and let $s$ be the &lt;a href=&quot;http://en.wikipedia.org/wiki/Arc_length#Finding_arc_lengths_by_integrating&quot; rel=&quot;nofollow&quot;&gt;arc length&lt;/a&gt; of that function in $[0; 1]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, the function with minimum $s$ that satisfies the above conditions is $f(x) = x$ with $s = \sqrt 2$. So for $s = \sqrt 2$, exactly one matching function can be found.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But &lt;strong&gt;what happens to the number of these functions if $s$ increases?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Surely, more functions can be found to match the given arc length - uncountably many more I suppose due to the nature of the real numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But intuitively, I'd think that the &lt;em&gt;number of such functions grows even more the greater $s$ gets&lt;/em&gt;, since there is more &quot;space&quot; the graph can use.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, despite continuum cardinality, &lt;strong&gt;are there any means of measuring the number of such functions&lt;/strong&gt; against $s$ or is it all the same once that minimal way of $f(x) = x$ as been taken?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And would this change if we limited the ways of constructing such functions to e.g. some elementary ones?&lt;/p&gt;&#xA;" OwnerUserId="917" LastActivityDate="2010-09-20T20:48:46.797" Title="Number of continuous $[0; 1] \to [0; 1]$ functions for given arc length" Tags="&lt;functional-analysis&gt;&lt;functions&gt;&lt;cardinals&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="5089" PostTypeId="2" ParentId="5065" CreationDate="2010-09-20T20:25:00.070" Score="1" ViewCount="" Body="&lt;p&gt;For any continuous scalar field $f$ on the sphere, you can define a continuous scalar field $g$ as $g(p) = f(p) - f(\bar{p})$, where $\bar{p}$ is the antipodal point of $p$. Unless $g$ is zero everywhere, there is a point $q$ for which $g(p) &gt; 0$ and $g(\bar{q}) = -g(q) &amp;lt; 0$. As $g$ is continuous, a closed contour separating $q$ from $\bar{q}$ exists on which $g$ is zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Edit:&lt;/em&gt; The last statement &quot;seems obvious&quot; to me, but I don't have enough knowledge of compact topological spaces or manifolds to make it rigorous. Perhaps a counterexample exists.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Edit 2:&lt;/em&gt; See &lt;a href=&quot;http://math.stackexchange.com/questions/5065/does-a-continuous-scalar-field-on-a-sphere-have-continuous-loop-of-isothermic-an/5127#5127&quot;&gt;T..'s answer&lt;/a&gt; for a sketch of a counterexample. Apparently I was thinking about the existence of a &lt;em&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Continuum_%28topology%29#Definitions&quot; rel=&quot;nofollow&quot;&gt;curve&lt;/a&gt;&lt;/em&gt; while the question asked for a &lt;em&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Loop_%28topology%29&quot; rel=&quot;nofollow&quot;&gt;loop&lt;/a&gt;&lt;/em&gt;?&lt;/p&gt;&#xA;" OwnerUserId="856" LastEditorUserId="856" LastEditorDisplayName="" LastEditDate="2010-09-21T06:39:25.573" LastActivityDate="2010-09-21T06:39:25.573" CommentCount="4" />
  <row Id="5090" PostTypeId="2" ParentId="5088" CreationDate="2010-09-20T20:29:28.300" Score="3" ViewCount="" Body="&lt;p&gt;There is probably a sophisticated answer to this; let me just mention an unsophisticated one.  Several possible discrete analogues of the question you ask exhibit the desired behavior.  Consider, for example, the problem of finding a path from $(0, 0)$ to $(n, n)$ in the lattice $\mathbb{Z}^2$ where the only allowable steps are between diagonally adjacent lattice points.  Let the length of such a path be the total number of steps.  Then, of course, there is only one path of length $n$, which is the obvious diagonal one.  However, there are ${n+2 \choose 2} + n$ paths of length $n+2$, already a big leap.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The general answer can be computed as follows.  In a path of length $n+2k$ we need to choose $2j$ of the steps to be up-left or down-right (for some $j$ between $0$ and $k$) and we need to choose $n+2k-2j$ of the steps to be up-right or down-left.  Of the first set of steps, half need to be of one type and half need to be the other, and of the second set of steps, $n+k-j$ need to be of one type and $k-j$ need to be the other.  This gives the total number of paths of length $n+2k$ as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{j=0}^{k} {n+2k \choose 2j} {2j \choose j} {n+2k-2j \choose k-j}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The dominant term when $n$ is large compared to $k$ is the term associated to $j = k$, which is ${n+2k \choose 2k}$ and hence grows asymptotically like $\frac{n^{2k}}{(2k)!}$.  So, again when $n$ is large compared to $k$, it is indeed true that the number of allowable paths grows with the length of the path.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-09-20T20:29:28.300" />
  <row Id="5091" PostTypeId="2" ParentId="5088" CreationDate="2010-09-20T20:48:46.797" Score="1" ViewCount="" Body="&lt;p&gt;Here's my take on an intuitive answer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since there are $2^{\aleph_0}$ continuous functions from the unit interval to itself, it is clear that there can be at most that many functions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, given $s &gt; \sqrt{2}$ (which as you well observed the minimal distance), we define $t = s - \sqrt{2}$ and for any $x\in (0,1)$ we can construct a function that has the wanted arc length, simply by &quot;pulling&quot; the value of $f(x)$ smoothly in a way that extends the length of the arc by $t$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(The construction is very clear when you don't request $f$ to be differentiable, as you can just stick a triangle in the proper length, however the idea should be clear)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you have at least $|(0,1)| = 2^{\aleph_0}$ functions as such, which means that you have exactly that many.&lt;/p&gt;&#xA;" OwnerUserId="622" LastActivityDate="2010-09-20T20:48:46.797" />
  <row Id="5092" PostTypeId="1" AcceptedAnswerId="5093" CreationDate="2010-09-20T21:47:22.200" Score="7" ViewCount="367" Body="&lt;p&gt;I studied applied math, so each course (except abstract algebra) was dedicated to solution of a similar problems. After those courses it seems that every branch of mathematics has a developed theory with a few unsolved problems. I believe that is wrong. I found out that there is no general approach to Diophantine equations (the 10th Hilbert problem), to problems like &quot;Collatz conjecture&quot; and to nonlinear differential equations. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And so the question is: &lt;strong&gt;what is the other branches of mathematics (collections of similar problems) without a general method to solve them?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="2016" LastEditorUserId="2016" LastEditorDisplayName="" LastEditDate="2010-09-21T12:14:37.597" LastActivityDate="2011-04-01T19:30:08.773" CommunityOwnedDate="2010-09-20T21:59:25.957" Title="Branches of mathematics not having a general method to solve" Tags="&lt;diophantine-equations&gt;&lt;open-problem&gt;&lt;conjectures&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="1" />
  <row Id="5093" PostTypeId="2" ParentId="5092" CreationDate="2010-09-20T21:58:43.243" Score="9" ViewCount="" Body="&lt;p&gt;Diophantine equations are not special; any branch of mathematics in which it is possible to ask sufficiently strong questions will suffer a similar fate due to the existence of problems like the &lt;a href=&quot;http://en.wikipedia.org/wiki/Halting_problem&quot; rel=&quot;nofollow&quot;&gt;halting problem&lt;/a&gt; and the possibility of asking a question equivalent to the halting problem (or something similar).  For example, group theory has such problems: it is generally impossible to decide from a presentation of a group whether it is trivial or not.  See &lt;a href=&quot;http://mathoverflow.net/questions/11540/what-are-the-most-attractive-turing-undecidable-problems-in-mathematics&quot; rel=&quot;nofollow&quot;&gt;this list of problems&lt;/a&gt; on MathOverflow.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-09-20T21:58:43.243" CommunityOwnedDate="2010-09-20T21:59:25.957" CommentCount="1" />
  <row Id="5094" PostTypeId="1" AcceptedAnswerId="5096" CreationDate="2010-09-20T22:45:53.223" Score="0" ViewCount="143" Body="&lt;p&gt;I think this is obvious and I am just missing something but:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Given some finitely presentable algebra, can we always reduce the relations to commutator relations?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And moreover, if yes, then is there a reasonable algorithm?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As I said, I feel like I should know this, so I apologize if it is obvious.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks, &lt;/p&gt;&#xA;" OwnerUserId="16" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-09-20T23:04:10.703" LastActivityDate="2010-09-21T03:27:25.343" Title="Reducing Relations to Commutator Relations" Tags="&lt;abstract-algebra&gt;&lt;algorithms&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="5095" PostTypeId="1" AcceptedAnswerId="5103" CreationDate="2010-09-20T22:59:42.003" Score="6" ViewCount="196" Body="&lt;p&gt;I know the limits in the categories of groups, abelian groups and topological spaces and was wondering about the same thing.&lt;/p&gt;&#xA;" OwnerUserId="2026" LastActivityDate="2010-09-21T00:53:31.440" Title="what are the product and coproduct in the category of topological groups" Tags="&lt;category-theory&gt;" AnswerCount="1" FavoriteCount="3" />
  <row Id="5096" PostTypeId="2" ParentId="5094" CreationDate="2010-09-20T23:01:23.320" Score="4" ViewCount="" Body="&lt;p&gt;(Apologies for putting this in the form of an answer - I haven't got enough rep yet to make it a comment)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What about elements of finite order?  It seems like even a simple algebra such as the one generated by $a^m = b^n = e$ couldn't possibly have a commutator presentation, unless I'm missing something immensely obvious...&lt;/p&gt;&#xA;" OwnerUserId="785" LastActivityDate="2010-09-20T23:01:23.320" CommentCount="1" />
  <row Id="5097" PostTypeId="2" ParentId="4961" CreationDate="2010-09-20T23:31:32.633" Score="3" ViewCount="" Body="&lt;p&gt;I just remembered something that at the moment seemed pretty nice to me. Usually in an elementary number theory course one is taught that the integer solutions to the diophantine equation $x^2 + y^2 = z^2$ are given by the formulas $x = a^2 - b^2$, $y = 2ab$ and $z = a^2 + b^2$. By working in a naive way in the ring of Gaussian integers $\mathbb{Z}[i] = \{ a + bi \mid a, b \in \mathbb{Z} \}$ one can get these formulas without much effort. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We factor the equation as &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ (x + iy)(x - iy) = z^2 $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then since the right hand side is a square, the two factors on the left must be squares of numbers in $\mathbb{Z}[i]$ so for instance&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ x + iy = (a + bi)^2 $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and squaring this gives&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ x + iy = a^2 - b^2 + 2abi $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now equating real and imaginary parts shows that $x = a^2 - b^2$ and $y = 2ab$. From this one easily gets $z = a^2 + b^2$. Of course all of these manipulations need some more details but it would make for a nice example after the formulas have been derived in the usual way (without complex numbers).&lt;/p&gt;&#xA;" OwnerUserId="900" LastEditorUserId="900" LastEditorDisplayName="" LastEditDate="2010-09-20T23:38:37.350" LastActivityDate="2010-09-20T23:38:37.350" CommunityOwnedDate="2010-09-20T23:31:32.633" />
  <row Id="5098" PostTypeId="1" CreationDate="2010-09-20T23:57:08.490" Score="0" ViewCount="226" Body="&lt;p&gt;For the equation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f(x,y) = x^2/y$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;given that $x$ and $y$ are integers&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the easiest/fastest way to find at least one tuple $(x_0,y_0)$ solution to $f(x,y)=4$ or any other number really. I know that $(2,1)$ would work, but is there a process for finding the solution to any integer that is equal to $f(x,y)$? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any easy way to just enter this in and calculate it in Mathematica, MATLAB, Sage, etc.? Is there any way to work this out by hand besides just guessing which values would work?&lt;/p&gt;&#xA;" LastEditorUserId="1778" LastEditorDisplayName="" LastEditDate="2010-09-21T03:29:58.507" LastActivityDate="2010-09-21T03:29:58.507" Title="Finding positive integer solutions to functions with only positive integer input" Tags="&lt;math-software&gt;&lt;mathematica&gt;&lt;matlab&gt;&lt;diophantine-equations&gt;" AnswerCount="4" />
  <row Id="5099" PostTypeId="2" ParentId="5098" CreationDate="2010-09-21T00:03:12.930" Score="0" ViewCount="" Body="&lt;p&gt;In &lt;em&gt;Mathematica&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;FindInstance[x^2/y == 4, {x, y}, Integers, n]&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where &lt;code&gt;n&lt;/code&gt; is an integer returns &lt;code&gt;n&lt;/code&gt; possible instances.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-09-21T00:03:12.930" CommentCount="4" />
  <row Id="5100" PostTypeId="2" ParentId="5098" CreationDate="2010-09-21T00:11:16.983" Score="1" ViewCount="" Body="&lt;p&gt;This could help: $x^2/y=a$, where $x,y,a$ are integers. If $a$ is a perfect square, $y=1$ and $x=\sqrt{a}$. If it isn't a perfect square, suppose $a=p_1^{\alpha_1}...p_k^{\alpha_k}$. Then any $y$ of the form $p_1^{\beta_1}...p_k^{\beta_k}$ such that $\alpha_i +\beta_i$ is even will do, because then $ay$ will be a perfect square. Moreover, $ay$ times any other perfect square is good too.&lt;/p&gt;&#xA;" OwnerUserId="1313" LastActivityDate="2010-09-21T00:11:16.983" />
  <row Id="5101" PostTypeId="1" AcceptedAnswerId="5102" CreationDate="2010-09-21T00:23:59.827" Score="1" ViewCount="483" Body="&lt;p&gt;I'm learning about proof by contrapositive and by mathematical induction in a computer science class. I'm banging my head trying to solve this problem and would like some help:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Prove that the sum of the first n even numbers is n^2 + n&#xA;(a) indirectly by assuming that the sum of the first n odd numbers is n^2&#xA;(b) directly by mathematical induction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have no problem doing (b). But I can't figure out how to do (a) using indirect proof. I can only come up with this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sum of first n numbers is n^2 + n =&gt; numbers are even&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Contrapositive: numbers are odd =&gt; sum of first n numbers is not n^2 + n&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This just doesn't seem logical to me. Any hints would be greatly appreciated.&lt;/p&gt;&#xA;" OwnerUserId="2028" LastActivityDate="2010-09-21T00:46:04.013" Title="Indirect proof that sum of first n even numbers is n^2 + n" Tags="&lt;sequences-and-series&gt;&lt;homework&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="5102" PostTypeId="2" ParentId="5101" CreationDate="2010-09-21T00:32:42.227" Score="3" ViewCount="" Body="&lt;p&gt;Using the contrapositive isn't the only indirect way of proving a theorem, and I highly doubt it's what your instructors had in mind here.  As a hint for part (a): what can you say about the first $n$ even numbers (individually) compared to the first $n$ odd numbers?  Can you find some convenient correspondence between, say, the 5th even number and the 5th odd number?&lt;/p&gt;&#xA;" OwnerUserId="785" LastActivityDate="2010-09-21T00:32:42.227" CommentCount="4" />
  <row Id="5103" PostTypeId="2" ParentId="5095" CreationDate="2010-09-21T00:44:40.223" Score="10" ViewCount="" Body="&lt;p&gt;The product of topological groups is simply the product of the underlying groups with the product topology. The universal property is easily verified. The coproduct topology is more complicated. Again, the underlying group is just the coproduct of the underlying groups, which is also called the free product. Here is a description of the topology:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $(G_i) (i \in I)$ our family of topological groups. Then consider the class of all topological groups $H$, such that there is a surjective group homomorphism $G:=\coprod_{i \in I} G_i \to H$, such that the composites $G_i \to H$ are continuous. Notice that this class is essentially a set. Namely, consider the set of all quotient groups $G / N$ endowed with some topology. In particular, we may define the topology on $G$ to be the initial one with respect to all maps $G \to H$ as above. Then $G$ is a topological group. For example, the inversion map $G \to G$ is continuous since the composite with every $G \to H$ as above is the composite with $G \to H$ and the inversion map $H \to H$. The universal property is also easy to verify.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This works in general: If $\tau$ is some type of algebraic structures, then the category of topological $\tau$-algebras is complete and cocomplete, and the forgetful functor to $\tau$-algebras preserves and creates limits and colimits. You can also prove this by applying Freyd's representabilty theorem. In the above example, I have written down a solution set.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, it is hard to describe this topology explicitely (and thus, for instance, proving separation axioms). See for example &lt;a href=&quot;http://linkinghub.elsevier.com/retrieve/pii/S0166864101000852&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; or &lt;a href=&quot;http://linkinghub.elsevier.com/retrieve/pii/S0166864103000890&quot; rel=&quot;nofollow&quot;&gt;that&lt;/a&gt; article.&lt;/p&gt;&#xA;" OwnerUserId="1650" LastEditorUserId="1650" LastEditorDisplayName="" LastEditDate="2010-09-21T00:53:31.440" LastActivityDate="2010-09-21T00:53:31.440" CommentCount="2" />
  <row Id="5104" PostTypeId="1" AcceptedAnswerId="5113" CreationDate="2010-09-21T00:45:06.603" Score="1" ViewCount="82" Body="&lt;p&gt;When studying group theory, I was interested to learn that a the quotient set $G/N$ of a group $G$ is again a group with $\pi\colon G\to G/N$ a homomorphism when $N$ is a normal subgroup of $G$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a more general set of criteria for this situation outside of group theory? As in, for any set $U$ and an equivalence relation $\sim$ on $U$, is there a similar set of criteria for $\sim$ such that $\pi\colon U\to U/\sim$ is a homomorphism in that it respects all functions and relations on $U$? Where by this I mean that $R^{U}(a_1,\dots,a_{\nu_L(R)})\Leftrightarrow R^{U/\sim}(\pi(a_1),\dots,\pi(a_{\nu_L(R)}))$ and $\pi(f^U(a_1,\dots,a_{\nu_L(f)}))=f^{U/\sim}(\pi(a_1),\dots,\pi(a_{\nu_L(f)}))$.&lt;/p&gt;&#xA;" OwnerUserId="1583" LastActivityDate="2010-09-21T04:03:16.997" Title="Quotienting a Set by an Equivalence Relation such that the Natural Projection is a Homomorphism." Tags="&lt;abstract-algebra&gt;&lt;logic&gt;" AnswerCount="2" />
  <row Id="5105" PostTypeId="2" ParentId="5101" CreationDate="2010-09-21T00:46:04.013" Score="2" ViewCount="" Body="&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/8UmRp.png&quot; alt=&quot;grid&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This diagram should help you see the proper approach to proving this.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-09-21T00:46:04.013" />
  <row Id="5106" PostTypeId="2" ParentId="5104" CreationDate="2010-09-21T00:47:27.913" Score="1" ViewCount="" Body="&lt;p&gt;Any homomorphism $A \to B$ of structured sets which is surjective has this property.  This is because any function $f : A \to B$ defines an equivalence relation on $A$ whose equivalence classes are the subsets $f^{-1}(b), b \in B$.  So write down all the conditions necessary for $f$ to be a morphism and you'll get all the conditions you need on the corresponding equivalence relation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, for a surjective function $f : G \to H$ between groups we require that $f(ab) = f(a) f(b)$.  This implies that the preimage of the identity in $H$ is a subgroup $F$ of $G$, and from there it follows that the preimage of $h \in H$ is the coset $g F = F g$ for any $g \in f^{-1}(h)$, hence that $F$ is normal.&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-09-21T02:34:09.403" LastActivityDate="2010-09-21T02:34:09.403" CommentCount="8" />
  <row Id="5107" PostTypeId="2" ParentId="5098" CreationDate="2010-09-21T00:59:30.733" Score="0" ViewCount="" Body="&lt;p&gt;One obvious solution to f(x,y)=a is x=y=a, since a^2/a = a.&lt;/p&gt;&#xA;" OwnerUserId="796" LastActivityDate="2010-09-21T00:59:30.733" />
  <row Id="5108" PostTypeId="1" AcceptedAnswerId="5111" CreationDate="2010-09-21T01:01:49.050" Score="16" ViewCount="595" Body="&lt;p&gt;I'm taking a course in Complex Analysis,&#xA;and the teacher mentioned that if we do not&#xA;restrict our attention to analytic functions,&#xA;we would just be looking at functions from $\mathbb{R}^2$ to $\mathbb{R}^2$.&lt;br&gt;&#xA;What I don't understand is why this is not true&#xA;when we do restrict our attention to analytic functions.&#xA;I understand that complex analytic functions&#xA;have different properties than real functions on $\mathbb{R}^2$,&#xA;but I don't understand why this is so.&lt;br&gt;&#xA;If I look at a complex number $z$ as a vector in $\mathbb{R}^2$,&#xA;then isn't differentiability of $w=f(z)$ in $\mathbb{C}$ defined the same way&#xA;as differentiability of $(u,v)=F(x,y)$ in $\mathbb{R}^2$ ?&lt;/p&gt;&#xA;" OwnerUserId="2029" LastActivityDate="2010-09-21T10:50:49.930" Title="How is $\mathbb{C}$ different than $\mathbb{R}^2$?" Tags="&lt;complex-analysis&gt;&lt;complex-variables&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="10" />
  <row Id="5109" PostTypeId="2" ParentId="5108" CreationDate="2010-09-21T01:10:20.613" Score="3" ViewCount="" Body="&lt;p&gt;The essential point of complex analysis is that the answer to this question is a resounding no.  Differentiability of a function $f$ in the complex sense requires taking a limit in which the limiting variable is a complex number:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f'(z) = \lim_{w\to z} \frac{f(w) - f(z)}{w-z}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For this limit to exist, one needs convergence uniformly in all directions (think through the $\epsilon$-$\delta$ definition of this limit).  On the other hand, a function from $\mathbb{R}^2\to \mathbb{R}^2$ (or more generally $\mathbb{R}^n\to \mathbb{R}^m$) is differentiable (in the sense explained in Ryan's answer) at all points in an open neighborhood if and only if all the partial derivatives exist at every point and are continuous on the given neighborhood.  Of course that forces all the directional derivatives to exist, but doesn't require the same sort of uniformity from different directions.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This Wikipedia page is a good place to learn some more: &lt;a href=&quot;http://en.wikipedia.org/wiki/Holomorphic_function&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Holomorphic_function&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Cauchy-Riemann equations, as Ryan's answer explains, relate the directional derivatives of the real and complex parts of a holomorphic function and give rise to all sorts of interesting properties.&lt;/p&gt;&#xA;" OwnerUserId="1392" LastEditorUserId="1392" LastEditorDisplayName="" LastEditDate="2010-09-21T03:37:31.913" LastActivityDate="2010-09-21T03:37:31.913" CommentCount="7" />
  <row Id="5110" PostTypeId="2" ParentId="2865" CreationDate="2010-09-21T01:16:32.377" Score="2" ViewCount="" Body="&lt;p&gt;You can construct colimits by a rather general procedure. See my answer &lt;a href=&quot;http://math.stackexchange.com/questions/5095/what-are-the-product-and-coproduct-in-the-category-of-topological-groups&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The topologies mentioned by Agusti are not correct. The continuity of the group laws does not hold in general. The problem is that quotient topologies do not commute with product topologies.&lt;/p&gt;&#xA;" OwnerUserId="1650" LastActivityDate="2010-09-21T01:16:32.377" CommentCount="2" />
  <row Id="5111" PostTypeId="2" ParentId="5108" CreationDate="2010-09-21T01:27:57.953" Score="18" ViewCount="" Body="&lt;p&gt;For a function $f : \mathbb R^2 \to \mathbb R^2$ &quot;differentiable&quot; at a point $x \in \mathbb R^2$ means you have a linear approximation $f'_x$ which satisfies&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\lim_{y \to x} \frac{f(x)-f(y)-f'_x(x-y)}{|x-y|} = 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Saying that $f$ is complex analytic is the constraint that $f'_x$ is a complex linear function for all $x$.  &quot;Complex linear&quot; means that not only is it true that $f'_x(av+bw)=af'_x(v)+bf'_x(w)$ for $a, b \in \mathbb R$, but it also holds for $a,b \in \mathbb C$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One way to say $L$ is &quot;Complex linear&quot; is that $L$ is a regular (real linear) function plus $L(iv)=iL(v)$ for all vectors $v$.   Stated in terms of the derivative, the directional derivative of $f$ in the direction $(0,1)$ is $i$ times the directional derivative in the direction $(1,0)$.  In component notation these are the &quot;Cauchy-Riemann&quot; equations. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \frac{\partial f}{\partial y} = i \frac{\partial f}{\partial x}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where if you write $f(z) = f(x+iy) = u(x,y) + iv(x,y)$ translates to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x}, &#xA;   \frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The formula:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \frac{\partial f}{\partial y} = i \frac{\partial f}{\partial x}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;gives the key picture.  A complex linear map means that the map looks like the composite of a rotation together with a re-scaling map $v \longmapsto av$ where $a \in \mathbb R$.  These maps are called &quot;conformal&quot;.  The nice things about conformal linear maps is they preserve all angles -- they do not always preserve length.  So the &quot;nice&quot; thing about complex differentiable maps is that if you have any collection of curves in the plane, and you apply your complex differentiable function to them, it preserves the angles of intersection of your curves.  That's a very special property. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;edit: An instructive example would be to think through two different functions from $\mathbb R^2$ to $\mathbb R^2$.  The first function:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(x,y) \longmapsto x(\cos(y),\sin(y))$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and the second function&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(x,y) \longmapsto e^x(\cos(y),\sin(y))$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first function preserves the angles between the coordinate grid lines -- curves like $x=a$ and $y=b$ in the coordinate plane.  The first function is not complex differentiable but the second is!  So this means that the second function preserves all angles (not just the coordinate lines $x=a, y=b$).  Can you spot curves in the domain which intersect in some angle $\theta$, but when after you compose them with the 1st function, their angle of intersection is not $\theta$? &lt;/p&gt;&#xA;" OwnerUserId="642" LastEditorUserId="642" LastEditorDisplayName="" LastEditDate="2010-09-21T01:53:54.667" LastActivityDate="2010-09-21T01:53:54.667" CommentCount="1" />
  <row Id="5112" PostTypeId="2" ParentId="5098" CreationDate="2010-09-21T02:57:03.617" Score="0" ViewCount="" Body="&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Mathematica is great. Mathematica is a crutch. If you depend on a calculator, you won't get the meaning of your answer and you won't know what to do with it.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If you know nothing about the RHS, to get a 'minimal' answer by generate and test, go in rings around the origin:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;{0,0}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;{1,0},{0,1},{-1,0},{0,-1}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;{2,0},{1,1},{0,2},{-1,1},{-2,0},{-1,-1},{0,-2},{1,-1}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;...&#xA;(is that how Mathematics's FindInstance works?)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If the RHS is $x^2/y$, to understand the problem, what is the thing  that causes immediate problems with integers (getting integer solutions)? It's the division. If $x$ divides $y^2$ (in order to make f an integer) $x$ has to divide $y$. That is, beyond yrudoy's answer, you can make $x$ be any divisor of $y$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1919" LastActivityDate="2010-09-21T02:57:03.617" />
  <row Id="5113" PostTypeId="2" ParentId="5104" CreationDate="2010-09-21T03:07:41.240" Score="2" ViewCount="" Body="&lt;p&gt;Assuming I've understood your question correctly and you are only interested in equivalence relations that respect &lt;em&gt;some&lt;/em&gt; given operations and/or relations (as opposed to &quot;all&quot; operations and relations at the same time, which I suspect would lead you to only the identity or the total relation), let me give you the punchline first: an equivalence relation will &quot;respect the operations and the relations&quot; if and only if it is a substructure when considered as a subset of $U\times U$, where $U\times U$ is given the &quot;coordinatewise&quot; structure in the obvious way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The concept you are looking for is that of a &lt;a href=&quot;http://en.wikipedia.org/wiki/Congruence_relation#Universal_algebra&quot; rel=&quot;nofollow&quot;&gt;&lt;strong&gt;congruence&lt;/strong&gt;&lt;/a&gt;, and is studied in universal algebra.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose that you have a set $U$ with operators $\{f_i\}_{i\in I}$, with $f_i$ of arity $n_i$; and relations $\{R_j\}_{j\in J}$ with $R_j$ and $m_j$-ary relation. Let $\sim$ be an equivalence relation which &quot;respects the operations&quot;, so that if $a_1,\ldots,a_{n_i},b_1,\ldots,b_{n_i}\in U$, $a_k\sim b_k$, then $f_i(a_1,\ldots,a_{n_i})\sim f_i(b_1,\ldots,b_{n_i})$, and and if $c_1,\ldots,c_{m_j},d_1,\ldots,d_{m_j}\in U$ with $c_k\sim d_k$ and $(c_1,\ldots,c_{m_j})\in R_j$, then $(d_1,\ldots,d_{m_j})\in R_j$. Then $\sim$, considered as subset of $U\times U$ is closed under each of $f_i\times f_i$, and the restrictions of $R_j\times R_j$ work just as needed to make $\sim$ into a substructure of $U\times U$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conversely, if $\sim$ is an equivalence relation that is also a substructure of $U\times U$, then if $a_k\sim b_k$ you have $(a_1,b_1),\ldots,(a_{n_i},b_{n_i})\in \sim$, so $$f_i\times f_i\bigl((a_1,b_1),\ldots,(a_{n_i},b_{n_i})\bigr) = \bigl(f_i(a_1,\ldots,a_{n_i}),f_i(b_1,\ldots,b_{n_i})\bigr)\in\sim,$$&#xA;hence $f_i(a_1,\ldots,a_{n_i})\sim f_i(b_1,\ldots,b_{n_i})$; similarly with the relations. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So an equivalence relation &quot;respects the operations and relations&quot; if and only if it is a substructure of $U\times U$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a theorem that is completely analogous to the usual Isomorphism Theorems for groups and rings, for congruences (in fact, the theorems for groups and rings are special cases). If you are interested in this, you can see Gratzer's book (titled &quot;Universal Algebra&quot;), or George Bergman's &lt;a href=&quot;http://math.berkeley.edu/~gbergman/245/&quot; rel=&quot;nofollow&quot;&gt;An Invitation to General Algebra and Universal Constructions&lt;/a&gt;. (Though the material on universal algebra is in Chapter 8; Gratzer will get to this much quicker). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S. You need to be careful with relational algebras; in general, defining a substructure is a bit tricky; if you think of functions as special kinds of relations, you can see that you need some kind of &quot;relational closure&quot; for your set to be a substructure; but if you define substructures adequately, then the condition on $\sim$ that it be a substructure of $U\times U$ will take care of it.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-09-21T04:03:16.997" LastActivityDate="2010-09-21T04:03:16.997" CommentCount="2" />
  <row Id="5114" PostTypeId="2" ParentId="5094" CreationDate="2010-09-21T03:27:25.343" Score="4" ViewCount="" Body="&lt;p&gt;No, you cannot even do it in groups. For if you could, consider the identity $x^n=e$ with $n&gt;1$. If this were equivalent to a set of commutator words, then every abelian group would satisfy the relation, but of course this is false: only those abelian groups that are of exponent $n$ satisfy it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More generally, in groups every identity is equivalent to a pair of identities, one a commutator word and one a power word; equivalently, if $w$ is a word in the free group $F$ on countably many generators $x_1,\ldots,x_n,\ldots$, then there exist words $w_1$ and $w_2$, with $w_1=x_1^a$ for some positive integer $a$, and $w_2\in [F,F]$, such that for all groups $G$, $G$ satisfies $w$ as an identity (every evaluation of $w$ in $G$ is trivial) if and only if $G$ satisfies $w_1$ and $w_2$ as identities. This is due to Bernhard Neumann; write $w=x_1^{n_1}\cdots x_m^{n_m}c$, with $c$ a commutator word, by using the collection process. Then let $a=\gcd(n_1,\ldots,n_m)$, and $w_2=c$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So at least in groups you also need &quot;power identities&quot;. &lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-09-21T03:27:25.343" />
  <row Id="5115" PostTypeId="1" CreationDate="2010-09-21T03:45:23.580" Score="3" ViewCount="1293" Body="&lt;p&gt;So for the formula 1/x, If you were to add up all y values from x=1 to x=∞, wouldn't the sum approach a number because even though you are always adding, aren't you just adding smaller and smaller numbers? Wouldn't this mean that it approached a certain number?&lt;/p&gt;&#xA;" OwnerUserId="2033" LastEditorUserId="1778" LastEditorDisplayName="" LastEditDate="2010-09-21T14:39:27.730" LastActivityDate="2010-10-02T20:03:09.010" Title="Why does 1/x diverge?" Tags="&lt;calculus&gt;&lt;sequences-and-series&gt;&lt;integral&gt;" AnswerCount="4" CommentCount="2" FavoriteCount="1" />
  <row Id="5116" PostTypeId="2" ParentId="4414" CreationDate="2010-09-21T03:46:49.333" Score="0" ViewCount="" Body="&lt;p&gt;Yes, absolutely you can introduce definitions as axioms.  That's what Metamath does:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://us.metamath.org/mpegif/mmset.html#definitions&quot; rel=&quot;nofollow&quot;&gt;http://us.metamath.org/mpegif/mmset.html#definitions&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See also&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.hss.cmu.edu/philosophy/techreports/181_Avigad.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.hss.cmu.edu/philosophy/techreports/181_Avigad.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which defines a formal system with definitions:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;We extend the foundational framework in two ways. First, we allow for explicit deﬁnitions of new predicates and functions on the universe of sets. And,&#xA;  second, we allow function symbols to denote functions that are only partially de-&#xA;  ﬁned, using a logic of partial terms. We call the resulting formal system DZFC .&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1778" LastActivityDate="2010-09-21T03:46:49.333" />
  <row Id="5117" PostTypeId="2" ParentId="5115" CreationDate="2010-09-21T03:54:19.810" Score="14" ViewCount="" Body="&lt;p&gt;Ignoring the 1, if you group together the first term, then the next two terms, then the next four terms, and so on, you get:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(1/2) + (1/3 + 1/4) + (1/5 + 1/6 + 1/7 + 1/8) + \cdots$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is greater than&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(1/2) + (1/4 + 1/4) + (1/8 + 1/8 + 1/8 + 1/8) + \cdots$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where now each group is exactly equal to 1/2.  This shows that the sum of the first $2^n$ terms is at least $1 + 1/2 \cdot n$, and so the sum of all the terms is unbounded.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you know a little bit of calculus, $\int dx/x=\log x$, so $1+1/2+1/3+\cdots+1/n\ge\int_1^ndx/x=\log n-\log1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See &lt;a href=&quot;http://math.stackexchange.com/questions/5035&quot;&gt;http://math.stackexchange.com/questions/5035&lt;/a&gt; which discusses precisely how fast the series diverges.&lt;/p&gt;&#xA;" OwnerUserId="1778" LastActivityDate="2010-09-21T03:54:19.810" CommentCount="1" />
  <row Id="5118" PostTypeId="2" ParentId="5115" CreationDate="2010-09-21T03:59:58.100" Score="4" ViewCount="" Body="&lt;p&gt;Yes it is true that the numbers you are adding are getting smaller and smaller. The key is that they do not get small quick enough. There are many proofs that can be found easily online (search for proof that the harmonic series diverges) that show that you can add up enough terms of the harmonic series to make its sum as large as you wish. The key is that the rate of growth of the (partial sums of the) harmonic series is logarithmic. Even though $\ln(x)$ grows &lt;strong&gt;extremely&lt;/strong&gt; slowly, it can still be made larger than any fixed value given sufficiently large $x$.&lt;/p&gt;&#xA;" OwnerUserId="1352" LastActivityDate="2010-09-21T03:59:58.100" />
  <row Id="5119" PostTypeId="1" AcceptedAnswerId="5121" CreationDate="2010-09-21T04:26:32.180" Score="1" ViewCount="618" Body="&lt;p&gt;If $\mathcal{F}_1 \subset \mathcal{F}_2 \subset \dotsb$ are sigma algebras, what is wrong with claiming that $\cup_i\mathcal{F}_i$ is a sigma algebra?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems closed under complement since for all $x$ in the union, $x$ has to belong to some F_i, and so must its complement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems closed under countable union, since for any countable unions of $x_i$ within it, each of the $x_i$ must be in some $F_j$, and so we can stop the sequence at any point and take the highest $j$ and we know that all the $x_i$'s up to that point are in $F_j$, and thus so must be their union.  There must be some counterexample, but I don't see it.&lt;/p&gt;&#xA;" OwnerUserId="774" LastActivityDate="2010-10-07T03:38:17.343" Title="Sigma algebra question" Tags="&lt;probability-theory&gt;&lt;measure-theory&gt;" AnswerCount="3" CommentCount="4" FavoriteCount="2" />
  <row Id="5120" PostTypeId="2" ParentId="5119" CreationDate="2010-09-21T04:37:12.093" Score="0" ViewCount="" Body="&lt;p&gt;If $x_1\in F_1$, $x_2\in F_2$, and so forth, then the infinite union $x_1 \cup x_2 \cup \cdots$ does not lie in any one $F_j$, and therefore does not lie in the union of the $F_j$'s.&lt;/p&gt;&#xA;" OwnerUserId="1726" LastActivityDate="2010-09-21T04:37:12.093" CommentCount="1" />
  <row Id="5121" PostTypeId="2" ParentId="5119" CreationDate="2010-09-21T04:42:48.320" Score="6" ViewCount="" Body="&lt;p&gt;The problem arises in the countable union; your argument is correct as far as it goes, but from the fact that $\cup_{i=1}^n x_i\in \cup_{i=1}^{\infty}F_i$ for each $n$ you cannot conclude that $\cup_{i=1}^{\infty} x_i$  lies in $\cup_{i=1}^{\infty} F_i$: the full union must be in one of the $F_j$ in order to be in $\cup_{i=1}^{\infty}F_i$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For an explicit example, take $X=\mathbb{N}$; let $F_n$ be the sigma algebra that consists of all subsets of $\{1,\ldots,n\}$ and their complements in $X$. Now let $x_i=\{2i\}$. Then each $x_i$ is in $\cup F_i$, but the union does not lie in any of the $F_k$, hence does not lie in $\cup F_i$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Added:&lt;/em&gt; In this example, $\cup_{i=1}^{\infty}F_n$ is the algebra of subsets of $X$ consisting of all subsets that are either finite or cofinite, so any infinite subset with infinite complement will not lie in the union, and such a set can always be expressed as a countable union of elements of $\cup F_i$.  &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-09-21T19:46:24.373" LastActivityDate="2010-09-21T19:46:24.373" />
  <row Id="5122" PostTypeId="1" AcceptedAnswerId="5133" CreationDate="2010-09-21T05:03:41.010" Score="0" ViewCount="101" Body="&lt;p&gt;I was considering the group of automorphisms on a vector space $\mathbb{Q}^3$ with the binary operation of addition. Is there a general description of the elements of this group? At first, it seemed to be that it would consist of bijective linear transformations, but I'm not sure how informative that is. Is there a specific category that this automorphism group would fall under?&lt;/p&gt;&#xA;" OwnerUserId="1583" LastActivityDate="2010-09-21T08:28:51.533" Title="Is there a particular characterization of $Aut(\mathbb{Q}^3)$?" Tags="&lt;linear-algebra&gt;&lt;abstract-algebra&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="5123" PostTypeId="2" ParentId="4988" CreationDate="2010-09-21T05:06:34.000" Score="5" ViewCount="" Body="&lt;p&gt;The kind of ordering you specify is known to economists as 'lexicographic preferences'. There is no objective function which can represent these preferences. Here is an outline of the proof for two variables:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume on the contrary that there is such a function $f(x,y)$. It must be the case for any $x$ that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f(x,0) &amp;lt; f(x,1),$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so we can find a rational number, say $q(x)$ such that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f(x,0) &amp;lt; q(x) &amp;lt; f(x,1) $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now if $x &amp;lt; y$, then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$q(x) &amp;lt; f(x,1) &amp;lt; f(y,0) &amp;lt; q(y)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, $q(\cdot)$ is a one-to-one mapping from the set of real numbers into the set of rational numbers which is impossible since the set of real numbers is uncountable while the set of rational numbers is countable.&lt;/p&gt;&#xA;" OwnerUserId="1195" LastActivityDate="2010-09-21T05:06:34.000" CommentCount="4" />
  <row Id="5124" PostTypeId="1" AcceptedAnswerId="5125" CreationDate="2010-09-21T05:12:27.760" Score="1" ViewCount="331" Body="&lt;p&gt;I'm actually looking for more intuition behind what's going on than a simple answer.  I understand that the Borel sets are the smallest sigma algebra that contain all of the open sets.  I can see how open intervals and their unions are in this set, and how it makes sense to measure them.  It's hard to see why you'd want to measure rational numbers (the probability that a normally distributed variable takes on a rational number?), and more importantly, why the sigma algebra should have to contain rational numbers (why can't we get away with a sigma algebra that doesn't contain the set of rational numbers?)&lt;/p&gt;&#xA;" OwnerUserId="774" LastActivityDate="2010-09-29T06:54:30.900" Title="Why are the sets of rational and irrational numbers Borel sets (over the reals)?" Tags="&lt;probability-theory&gt;&lt;measure-theory&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="5125" PostTypeId="2" ParentId="5124" CreationDate="2010-09-21T05:17:06.120" Score="1" ViewCount="" Body="&lt;p&gt;If you want your sigma algebra to contain all of the one-point sets, so in particular if it's the Borel sigma-algebra generated by a $T_1$ topology, then taking countable unions yields all of the countable sets, including the set of rational numbers in your case.  There are sigma-algebras of subsets of the real numbers that don't contain the rational numbers, but if you're OK with open intervals, then you have to be OK with points.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-09-21T05:17:06.120" CommentCount="1" />
  <row Id="5126" PostTypeId="1" CreationDate="2010-09-21T05:31:10.720" Score="1" ViewCount="245" Body="&lt;p&gt;Can anyone recommend good, free online articles or books to learn Discrete mathematics? When I google'd for them, I came across few resources..but don't know whether they are good to start learning with. It's better to know a tutorial or a book that get you started in Discrete Mathematics. Can readers of SE Mathematics can suggest a good starting point?&lt;/p&gt;&#xA;" LastActivityDate="2010-09-21T11:14:46.630" CommunityOwnedDate="2010-09-21T08:16:45.550" Title="Free resources to start learning Discrete Mathematics" Tags="&lt;books&gt;&lt;discrete-mathematics&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="5127" PostTypeId="2" ParentId="5065" CreationDate="2010-09-21T05:51:03.533" Score="4" ViewCount="" Body="&lt;p&gt;[edit: as predicted, there is a counterexample with no continuous loop. See addition below.]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think that you may not quite get a loop, only a topological continuum (a compact connected subset) on the sphere whose complement contains multiple components.  The continuum can be gotten by elaborating Rahul's answer to choose a suitable component of the $g=0$ locus.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The existence of topologically wild continua such as the &quot;Warsaw circle&quot; suggests that you can draw such a creature on the sphere or projective plane and then extend to a continuous function that would give a counterexample.  Or you could take a field that has the equator as the locus of isothermal antipodes ($g=0$) and try to perform a (antisymmetric) bending construction that modifies parts of the equator, turning it into a wild curve that cannot be traced by a continuous loop.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[added:  the extension construction would work as follows.  Take two opposite points on the equator.  Join them with a wild continuum in one hemisphere, and the antipode of that continuum in the opposite hemisphere. Define $f(x)$ to be the distance to the wild thing, in one hemisphere, and the negative of distance to the wild thing, in the opposite hemisphere.   Hence $f(x) = -f(-x)$ on the whole sphere, and $f=0$ only on the wild construction that cannot be traversed continuously by a path.] &lt;/p&gt;&#xA;" OwnerUserId="467" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-09-24T15:44:58.387" LastActivityDate="2010-09-24T15:44:58.387" CommentCount="2" />
  <row Id="5128" PostTypeId="2" ParentId="5122" CreationDate="2010-09-21T06:21:56.903" Score="2" ViewCount="" Body="&lt;p&gt;The distinction between preserving addition and also preserving scalar multiplication is illusory in this case, because any vector can be divided by $n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The distinction resurfaces, however, for the same problem with $Q$ replaced by (almost) any other field, such as the real numbers.  There preserving scalar multiplication is an additional property that does not follow from addition being conserved.&lt;/p&gt;&#xA;" OwnerUserId="467" LastActivityDate="2010-09-21T06:21:56.903" CommentCount="5" />
  <row Id="5129" PostTypeId="1" AcceptedAnswerId="5130" CreationDate="2010-09-21T06:26:58.610" Score="3" ViewCount="243" Body="&lt;p&gt;Hi - I have the Cartesian coordinates of the hypotenuse 'corners' in a right angle triangle. I also have the length of the sides of the triangle. What is the method of determining the coordinates of the third vertex where the opposite &amp;amp; adjacent sides meet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks, Kevin.&lt;/p&gt;&#xA;" OwnerUserId="1529" LastEditorUserId="1529" LastEditorDisplayName="" LastEditDate="2010-09-21T06:35:40.807" LastActivityDate="2010-09-21T12:48:56.120" Title="Given coordinates of hypotenuse, how can I calculate coordinates of other vertex?" Tags="&lt;geometry&gt;&lt;trigonometry&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="5130" PostTypeId="2" ParentId="5129" CreationDate="2010-09-21T06:37:13.393" Score="5" ViewCount="" Body="&lt;p&gt;You have two points $A=(a,b)$ and $B=(c,d)$ and want a point $P$&#xA;at given distances from $A$ and $B$, say $l$ and $m$. Then $|PA|^2=l^2$&#xA;and $|PB|^2=m^2$ that is&#xA;$$(x-a)^2+(y-b)^2=l^2\qquad\qquad(1)$$&#xA;and&#xA;$$(x-c)^2+(y-d)^2=m^2.\qquad\qquad(2)$$&#xA;Subtracting (2) from (1) gives a linear equation. Use this to eliminate&#xA;one variable from (1). This yields a quadratic equation in the other variable.&#xA;Solving this will give the two possible positions for $P$.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-09-21T06:37:13.393" CommentCount="1" />
  <row Id="5131" PostTypeId="1" AcceptedAnswerId="5135" CreationDate="2010-09-21T07:30:17.017" Score="1" ViewCount="184" Body="&lt;p&gt;Here is a question that I can't find the answer to in my notes or my textbook.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any relation between $\text{Sym}_n$ and $\text{Sym}_k$ where $k &amp;lt; n$? Is $\text{Sym}_k \subset \text{Sym}_n$? Since $k! |  n!\text{ } \forall  k &amp;lt; n $ they satisfy Lagrange's theorem, but of course that doesnt guarantee the existence of a subgroup&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm trying to use this result to show that that there is a transitive action of a finite group $G$ on a smaller finite set $S$ by showing that $G$ or a subgroup of $G$ is isomorphic to $\Sigma (S)$ (The permutation group of $S$) which has a natural transitive action. &lt;/p&gt;&#xA;" OwnerUserId="1906" LastEditorUserId="1906" LastEditorDisplayName="" LastEditDate="2010-09-21T08:30:45.137" LastActivityDate="2010-09-21T11:50:45.277" Title="Relationships between Symmetric Groups" Tags="&lt;abstract-algebra&gt;&lt;group-theory&gt;&lt;homework&gt;&lt;symmetric-groups&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="5132" PostTypeId="2" ParentId="5131" CreationDate="2010-09-21T07:37:57.587" Score="2" ViewCount="" Body="&lt;p&gt;Yes. $Sym_k &amp;lt; Sym_n$ for $n &amp;lt; k$ and it corresponds to a subgroup of all permutations with the same $k-n$ elements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(That is, fix $k-n$ elements first, then look at all the transpositions that keep them in place. This subgroup of $S_k$ is isomorphic to $S_n$.)&lt;/p&gt;&#xA;" OwnerUserId="622" LastActivityDate="2010-09-21T07:37:57.587" />
  <row Id="5133" PostTypeId="2" ParentId="5122" CreationDate="2010-09-21T08:28:51.533" Score="2" ViewCount="" Body="&lt;p&gt;A vector space over $\mathbb{Q}$ is the same as a torsionfree divisible abelian group. The product $p/q * v$ is the unique vector so that $q * p/q v = p v$. In particular, every group homomorphism between vector spaces over $\mathbb{Q}$ is automatically $\mathbb{Q}$-linear. Thus, for example, the group automorphisms of $\mathbb{Q}^n$ may be described by $GL_n(\mathbb{Q})$.&lt;/p&gt;&#xA;" OwnerUserId="1650" LastActivityDate="2010-09-21T08:28:51.533" />
  <row Id="5135" PostTypeId="2" ParentId="5131" CreationDate="2010-09-21T10:30:18.830" Score="5" ViewCount="" Body="&lt;p&gt;$S_n$ is not contained in $S_m$ when $n&amp;lt;m$ if you define $S_k$ to be the set of all bijections $\{1,\dots,k\}\to\{1,\dots,k\}$, simply because an element of $S_n$ has domain $\{1,\dots,n\}$ and elements of $S_m$ have domain $\{1,\dots,m\}$, and the two sets are &lt;em&gt;different&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now this is a silly problem, so surely there is some way out...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose again that $n&amp;lt;m$. There is a map $\phi:S_n\to S_m$ such that whenever $\pi\in S_n$, so that $\pi:\{1,\dots,n\}\to\{1,\dots,n\}$ is a bijection, then $\phi(\pi):\{1,\dots,m\}\to\{1,\dots,m\}$ is the map given by $$\phi(\pi)(i)=\begin{cases}\pi(i),&amp;amp;\text{if $i\leq n$;}\\i,&amp;amp;\text{if $i&amp;gt;n$.}\end{cases}$$ You can easily check that $phi$ is well-defined (that is, that $\phi(\pi)$ is a bijection for all $\pi\in S_n$) and that moreover $\phi$ is a group homomorphism which is &lt;em&gt;injective&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It follows from this that the image of $\phi$ is a subgroup of $S_m$ which is isomorphic to $S_n$. It is usual to &lt;em&gt;identify&lt;/em&gt; $S_n$ with its image $\phi(S_n)\subseteq S_m$, and then we can say that «$S_n$ is a subgroup of $S_m$», but this is nothing but a &lt;em&gt;façon de parler&lt;/em&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It should be noted, though, that there are &lt;em&gt;many&lt;/em&gt; injective homomorphisms $S_n\to S_m$, so there are many ways to identify a subgroup of $S_m$ with $S_n$. The one I described above is nice because it looks very natural.&lt;/p&gt;&#xA;" OwnerUserId="274" LastEditorUserId="274" LastEditorDisplayName="" LastEditDate="2010-09-21T11:50:45.277" LastActivityDate="2010-09-21T11:50:45.277" CommentCount="1" />
  <row Id="5137" PostTypeId="2" ParentId="5108" CreationDate="2010-09-21T10:50:49.930" Score="6" ViewCount="" Body="&lt;p&gt;Analytic functions map tiny disks to tiny disks.  (Of course that's not rigorous, but you could make it rigorous by putting in the right limit language.)  Analytic functions can shift, stretch, and rotate disks, but they can't flip disks over. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Smooth functions of two real variables can map disks to ellipses.  That is, they can stretch a disk more in one direction than in another.  Complex analytic functions can't do that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Complex conjugation is not analytic because it flips disks over.  &lt;/p&gt;&#xA;" OwnerUserId="247" LastActivityDate="2010-09-21T10:50:49.930" />
  <row Id="5138" PostTypeId="2" ParentId="5126" CreationDate="2010-09-21T11:14:46.630" Score="1" ViewCount="" Body="&lt;p&gt;Try MIT Open Courseware -- Course 6.042J (Mathematics for Computer Science) on &lt;a href=&quot;http://ocw.mit.edu/courses/electrical-engineering-and-computer-science/&quot; rel=&quot;nofollow&quot;&gt;http://ocw.mit.edu/courses/electrical-engineering-and-computer-science/&lt;/a&gt; would be a good place to start. &lt;/p&gt;&#xA;" OwnerUserId="759" LastActivityDate="2010-09-21T11:14:46.630" CommunityOwnedDate="2010-09-21T11:14:46.630" />
  <row Id="5139" PostTypeId="1" AcceptedAnswerId="5181" CreationDate="2010-09-21T11:56:36.070" Score="2" ViewCount="275" Body="&lt;p&gt;This an Exercise from Apostol's analytic number theory. I have been struggling with this problem for quite some time. Looks elementary though. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is to prove that this quantity &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Biggl[\frac{8n+13}{25}\Biggr] - \Biggl[\frac{n-12 - \Bigl[\frac{n-17}{25}\Bigr]}{3}\Biggr]$$ is independent of $n$.&lt;/p&gt;&#xA;" LastActivityDate="2010-09-21T21:59:50.207" Title="Proving $\Bigl[\frac{8n+13}{25}\Bigr] - \Bigl[\frac{n-12 - \Bigl[\frac{n-17}{25}\Bigr]}{3}\Bigr]$ is independent of $n$." Tags="&lt;analysis&gt;&lt;functions&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="5140" PostTypeId="2" ParentId="5139" CreationDate="2010-09-21T12:09:28.623" Score="8" ViewCount="" Body="&lt;p&gt;Write $n=25m+r$ with $0\leq r&amp;lt;25$, and replace in the equation.You'll see that the result only depends on $r$. Now make the $25$ computations corresponding to each of the possible values of $r$, and check that the result is $4$ in all those finitely many cases.&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2010-09-21T12:09:28.623" CommentCount="4" />
  <row Id="5141" PostTypeId="2" ParentId="5129" CreationDate="2010-09-21T12:48:56.120" Score="2" ViewCount="" Body="&lt;p&gt;While I'd use the same algebra as &lt;a href=&quot;http://math.stackexchange.com/questions/5129/given-coordinates-of-hypotenuse-how-can-i-calculate-coordinates-of-other-vertex/5130#5130&quot;&gt;Robin Chapman's solution&lt;/a&gt;, my first thought on this problem yields a third circle equation.  The circumcenter of a right triangle is at the midpoint of its hypotenuse.  Given the endpoints of the hypotenuse, $A=(a,b)$ and $B=(c,d)$, and letting $h$ be the length of the hypotenuse, the circumcircle has equation&#xA;$$\left(x-\frac{a+c}{2}\right)^2+\left(y-\frac{b+d}{2}\right)^2=\left(\frac{h}{2}\right)^2.$$&#xA;This may look a bit intimidating in symbols, but isn't really any different to work with for solving than the other two circles.  I don't know that this offers any advantages, though.&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-09-21T12:48:56.120" CommentCount="2" />
  <row Id="5143" PostTypeId="1" AcceptedAnswerId="5189" CreationDate="2010-09-21T13:36:22.633" Score="1" ViewCount="108" Body="&lt;p&gt;I have a series of inequalities:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$y_{1min} \leq y_{1}(x) \leq y_{1max}$$&#xA;$$y_{2min} \leq y_{2}(x) \leq y_{2max}$$&#xA;$$..$$&#xA;$$y_{nmin} \leq y_{n}(x) \leq y_{nmax}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that $x\in\mathbb{R}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is, is there a &lt;strong&gt;general method&lt;/strong&gt; that allows me to find the range of $x$ that satisfies the above inequalities?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Things are easy if there is just one inequality, and $y_1(x)$ is at most a quadratic function. But I am looking for a general solution here. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: My $y$ function can roughly be written as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$y=\Biggl[\frac{(x^2+bx+c)^f}{(dx+e)^g}\Biggr]$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $c$ and $e$ are small comparatively.$(dx+e)^g$ must be positive. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In one of the $y_i(x)$, $f=5/3, g=2/3$, but in another, $f=1, g=1$. It is safe to assume that both $f,g\in\mathbb{Q}$, that is they are &lt;a href=&quot;http://en.wikipedia.org/wiki/Rational_number&quot; rel=&quot;nofollow&quot;&gt;rational numbers&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$b,c,d,e\in\mathbb{R}$ , and they vary from one $y_i(x)$ to another. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope this helps.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strike&gt;Edit:Let's make this question a little general; instead of letting $y(x)$ depends on one variable, what if $y$ depending on multiple variables? That is, how to find the range for $(x_1,x_2,x_3,..., x_i)$, given that  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$y_{1min} \leq y_{1}(x_1,x_2,x_3,..., x_i) \leq y_{1max}$$&#xA;$$y_{2min} \leq y_{2}(x_1,x_2,x_3,..., x_i) \leq y_{2max}$$&#xA;$$..$$&#xA;$$y_{nmin} \leq y_{n}(x_1,x_2,x_3,..., x_i) \leq y_{nmax}$$&#xA;&lt;/strike&gt;&lt;/p&gt;&#xA;" OwnerUserId="550" LastEditorUserId="550" LastEditorDisplayName="" LastEditDate="2010-09-22T00:47:47.083" LastActivityDate="2010-09-22T00:47:47.083" Title="Find the range of $x$, given $y_{min} \leq y(x) \leq y_{max}$, where $y(x) $ can be any function ( Updated)" Tags="&lt;optimization&gt;&lt;numerical-methods&gt;&lt;nonlinear-optimization&gt;" AnswerCount="2" CommentCount="28" />
  <row Id="5144" PostTypeId="2" ParentId="5092" CreationDate="2010-09-21T13:44:45.603" Score="5" ViewCount="" Body="&lt;p&gt;There is no general method for solving differential equations.  There are general methods for proving existence and uniqueness of solutions, but not for computing solutions (if that's even possible) or for studying detailed properties of solutions.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can make a tiny change to a differential equation and it can go from being well understood to being an area of research.  Or it can go from something one group of mathematicians study to something another group studies.&lt;/p&gt;&#xA;" OwnerUserId="247" LastActivityDate="2010-09-21T13:44:45.603" CommunityOwnedDate="2010-09-21T13:44:45.603" CommentCount="1" />
  <row Id="5145" PostTypeId="1" AcceptedAnswerId="5155" CreationDate="2010-09-21T13:52:11.043" Score="4" ViewCount="340" Body="&lt;p&gt;The CASIO fx-7400G calculator is very quick at doing $(1+1/1000000)^{1000000}$, giving an output of 2.718280469.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mathematica (v7) takes over a minute, running on an Intel Core2 Duo processor, and its output is this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/dP0ac.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With n=10000000, my patience ran out. The calculator is quick again with an output of 2.718281693.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anybody know why this happens?&lt;/p&gt;&#xA;" OwnerUserId="1313" LastEditorUserId="1313" LastEditorDisplayName="" LastEditDate="2010-09-21T16:12:17.713" LastActivityDate="2010-09-21T16:30:21.757" Title="Casio hand calculator vs Mathematica at (1+1/n)^n" Tags="&lt;math-software&gt;" AnswerCount="3" CommentCount="5" FavoriteCount="1" />
  <row Id="5146" PostTypeId="2" ParentId="5143" CreationDate="2010-09-21T13:54:22.603" Score="1" ViewCount="" Body="&lt;p&gt;For a start, suppose we can find the interval $I_i\subseteq\mathbb{R}$ such that $y_{imin}\leq y_i(x)\leq y_{imax}$. Then the interval $I$ which satisfies all the constraints is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$I=\bigcap_{i=1}^n I_i$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus we can concentrate on finding $I_i$ separately for each $1\leq i\leq n$. Here I'm sure we will need some conditions on $y_i(x)$. Do your functions have compact support? Are they continuous? Do they oscillate wildly in any region?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;In the general case - finding a feasible point&lt;/strong&gt;&#xA;You mentioned that $y_i(x)$ are continuous. Lets suppose that in addition they are continuously differentiable. I'll do the problem with $1$ constraint for easy of notion but the general case is exactly the same though more difficult to write down.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rewrite the constraints as &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$y_1(x)\geq y_{1min}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$-y_1(x)\geq -y_{1max}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let $A(x) = \left(\nabla y_1(x),\nabla(-y_1(x))\right)^T$ (the jacobian matrix of the constraints). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also let $b=\left(y_{1min},-y_{1max}\right)^T$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then we need to find a $x_0$ such that $Ax_0\geq b$. We do this by solving the following (phase-1) problem. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;let $x_{guess}\in\mathbb{R}^n$. Let $r=\min(b-Ax_{guess},0)$ and solve the problem&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\min_{x\in\mathbb{R}^n,\mu\in\mathbb{R}}$  $\mu$ subject to $Ax + \mu r\geq b$ and $\mu\geq 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, you may not think that we have come very far but because $(x,\mu)=(x_{guess},1)$ is an initial feasible point for this minimisation problem it can be solved by standard software. If the optimal point gives an optimal value $&gt;0$ then the orignal problem is not feasible. If the optimal value is $0$ then the orignal problem is feasible and the optimal point is a feasible point for the orignal problem. &lt;/p&gt;&#xA;" OwnerUserId="1794" LastEditorUserId="1794" LastEditorDisplayName="" LastEditDate="2010-09-21T14:40:11.597" LastActivityDate="2010-09-21T14:40:11.597" CommentCount="4" />
  <row Id="5147" PostTypeId="2" ParentId="5145" CreationDate="2010-09-21T14:11:08.300" Score="2" ViewCount="" Body="&lt;p&gt;I think your calculator might use $$ a^b = \exp(b\ln a)$$ together with precalculated $\exp$ and $\ln$ tables to speed things up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know I would =)&lt;/p&gt;&#xA;" OwnerUserId="505" LastActivityDate="2010-09-21T14:11:08.300" />
  <row Id="5148" PostTypeId="1" AcceptedAnswerId="5151" CreationDate="2010-09-21T14:19:19.650" Score="1" ViewCount="70" Body="&lt;p&gt;An affine transformation is a linear transformation followed by a translation. They are &lt;a href=&quot;http://en.wikipedia.org/wiki/Morphism&quot; rel=&quot;nofollow&quot;&gt;morphism&lt;/a&gt; between affine spaces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A rigid transformation consists of a rotation and a translation. I was wondering what are the spaces between which the rigid transformations can become morphisms?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;what are the spaces between which the rotation transformations can become morphisms?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Same question for similarity transformation and projective transformation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks and regards!&lt;/p&gt;&#xA;" OwnerUserId="1281" LastEditorUserId="1281" LastEditorDisplayName="" LastEditDate="2010-09-21T15:16:47.767" LastActivityDate="2010-09-21T15:36:04.607" Title="What are the most special spaces between which rigid transformations preserve the structures of the spaces" Tags="&lt;rigid-transformation&gt;&lt;morphism&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="5149" PostTypeId="2" ParentId="5145" CreationDate="2010-09-21T14:19:43.443" Score="7" ViewCount="" Body="&lt;p&gt;Try&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;N[1 + 1/100000000000000, 2000]^100000000000000&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;which makes &lt;em&gt;Mma&lt;/em&gt; do the computation with floating point numbers (but using 2000 exact digits, waaaaay more that your calculator!) In my computer,&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;In[41]:= Timing[N[1 + 1/100000000000000, 2000]^100000000000000]&#xA;&#xA;Out[41]= {0.001999, &#xA;2.71828182845903164395114517625107361345759538109318401368108700081263&#xA;8880450911203511655259489457817904923933478503365902065203002443715271&#xA;2825230612837327582882359132201752486932090411441742152459981748434424&#xA;6796125955942676709724311384945123182525138185474706841748184641523835&#xA;6399319640050121555981826495805200410019220302352025402844917823115708&#xA;4890277526840712036588689004850449217720498524694546241483265385120337&#xA;3036215908323704165098770345249059641531345794163630084215403881301114&#xA;0222776646139077637586730556318554208747513123989912904288599990729397&#xA;4105912301684777926066523131379164170616992617301710309080427339172694&#xA;8652209704120914116266089477162936814414049552884109177338585344418934&#xA;4664761359706737137907335308204793693031274512475102742160130609557028&#xA;0606249997104370287581131479088448270923696582800846475772654584603126&#xA;9574917252623375481387197279590503661010907784866738170370137382154502&#xA;5239391867955746777742097465819430570361920416259681558416618161699424&#xA;5237267230224583604177051168028000345047412788448526660107314695031108&#xA;5918243291907671190858585383905903649083888525646705126721235172922992&#xA;7474539583108858881111281802649615453982269311194004892438663488977893&#xA;2481678450112257387961011092938028584251024526367562391996230685117325&#xA;0436239706780584781622006285082275199325910403142256578320132681839729&#xA;1241283875634170613592405031777334496908146321259758762276254422018659&#xA;2501711558155518458153097538288499261588998555672920722850605105301158&#xA;3203321297849315979775985916474141976627081937753054953188751346134290&#xA;6581791976723248211158139510858187401773833843382750617470712252506403&#xA;3501718283248275223376526908879174360214195321187111170739858908536480&#xA;0084276450965142494861141929264441784847261049017482036450996057648280&#xA;0994695573343998546256052750841029336112669690421548936514156835398676&#xA;3390097789579643086512481978170502348966487726085752931492687574582562&#xA;2555386293362518511636564659786083245494967887013590756572099854766124&#xA;625831647757374695456737571}&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;takes 0.002 seconds.&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2010-09-21T14:19:43.443" CommentCount="4" />
  <row Id="5151" PostTypeId="2" ParentId="5148" CreationDate="2010-09-21T14:29:46.070" Score="1" ViewCount="" Body="&lt;p&gt;If by &quot;rotation&quot; you mean &quot;orientation-preserving isometry of $\mathbb{R}^n$ fixing the origin,&quot; then &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;the spaces between which rigid transformations are morphisms are the &quot;affine oriented (finite-dimensional) inner product spaces&quot; over $\mathbb{R}$, by which I mean &lt;a href=&quot;http://math.ucr.edu/home/baez/torsors.html&quot; rel=&quot;nofollow&quot;&gt;torsors&lt;/a&gt; over a finite-dimensional &lt;a href=&quot;http://en.wikipedia.org/wiki/Orientation_(mathematics)&quot; rel=&quot;nofollow&quot;&gt;oriented&lt;/a&gt; inner product space over $\mathbb{R}$,&lt;/li&gt;&#xA;&lt;li&gt;the spaces between which rotations are morphisms are the oriented finite-dimensional inner product spaces over $\mathbb{R}$, and&lt;/li&gt;&#xA;&lt;li&gt;the spaces between which projective transformations are morphisms are the &lt;a href=&quot;http://en.wikipedia.org/wiki/Projective_space&quot; rel=&quot;nofollow&quot;&gt;projective spaces&lt;/a&gt; over $\mathbb{R}$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I cannot off the top of my head think of a good name for the spaces between which similarities are morphisms.  I &lt;em&gt;think&lt;/em&gt; you are looking for &quot;conformal affine spaces,&quot; e.g. torsors over $\mathbb{R}^n$ equipped with a notion of oriented angle (but not the full inner product).&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-09-21T15:36:04.607" LastActivityDate="2010-09-21T15:36:04.607" CommentCount="1" />
  <row Id="5152" PostTypeId="1" CreationDate="2010-09-21T14:43:15.710" Score="0" ViewCount="133" Body="&lt;p&gt;I would like to know what is an eulerian approach and an Lagragian approach.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Someone can recommend some references?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks,&lt;/p&gt;&#xA;" OwnerUserId="2045" LastActivityDate="2010-09-21T15:39:40.733" Title="Eulerian vs Lagrangian approach" Tags="&lt;calculus&gt;" AnswerCount="0" CommentCount="0" FavoriteCount="0" />
  <row Id="5153" PostTypeId="1" AcceptedAnswerId="5171" CreationDate="2010-09-21T14:46:18.590" Score="3" ViewCount="257" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Goursat%27s_lemma&quot; rel=&quot;nofollow&quot;&gt;Goursat's Lemma&lt;/a&gt; characterizes the subgroups of direct products. Is there a similar characterization for the subgroups of semidirect products? What about if I'm only interested in the normal subgroups?&lt;/p&gt;&#xA;" OwnerUserId="759" LastActivityDate="2010-09-21T19:47:50.883" Title="What are the subgroups of a semidirect product?" Tags="&lt;group-theory&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="5154" PostTypeId="2" ParentId="5115" CreationDate="2010-09-21T14:55:23.287" Score="3" ViewCount="" Body="&lt;p&gt;Your logic fails much more obviously for the series&lt;br&gt;&#xA;$1.1 + 1.01 + 1.001 + 1.0001 + \cdots$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Though each term is smaller than the last, it is clearly divergent (think of $1 + 1 + 1 + \cdots$)&lt;/p&gt;&#xA;" OwnerUserId="136" LastActivityDate="2010-09-21T14:55:23.287" CommentCount="1" />
  <row Id="5155" PostTypeId="2" ParentId="5145" CreationDate="2010-09-21T15:05:55.810" Score="5" ViewCount="" Body="&lt;p&gt;General purpose computer algebra systems, unlike most calculators, are able to perform both exact/symbolic computations (e.g. exact integer arithmetic) in addition to approximate numerical computations (e.g. floating point). Thus when you present the problem in exact terms - as you do above - they will preserve that form - so as not to lose any information. Thus your problem will be interpreted as a computation in exact rational arithmetic, i.e. raising a rational number to an integer power. If instead you desire a real approximation of the result then you must explicitly specify such, e.g. using N[...] in Mma. This is but one of the many complexities that arise when one passes from special-purpose computation systems (e.g. calculators) to general purpose computer algebra systems (speaking as one who has often tried to tackle less trivial such complexities while working as a developer of the Macsyma computer algebra system). &lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-09-21T16:30:21.757" LastActivityDate="2010-09-21T16:30:21.757" />
  <row Id="5157" PostTypeId="2" ParentId="4536" CreationDate="2010-09-21T15:30:09.140" Score="0" ViewCount="" Body="&lt;p&gt;I've been testing a new math search engine and found a similar discussion on the mathoverflow, there is link to search results (http://uniquation.com/en/solutions.aspx?query=F_{n%2B1}%3D2F_n%2BF_{n-1})&lt;/p&gt;&#xA;" OwnerUserId="2016" LastActivityDate="2010-09-21T15:30:09.140" />
  <row Id="5158" PostTypeId="2" ParentId="5160" CreationDate="2010-09-21T15:39:40.733" Score="1" ViewCount="" Body="&lt;p&gt;There is a Eulerian vs. Lagrangian distinction in fluid mechanics:&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Lagrangian_and_Eulerian_specification_of_the_flow_field&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Lagrangian_and_Eulerian_specification_of_the_flow_field&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1195" LastActivityDate="2010-09-21T15:39:40.733" />
  <row Id="5159" PostTypeId="1" CreationDate="2010-09-21T16:16:21.723" Score="3" ViewCount="142" Body="&lt;p&gt;Let $C$ be the curve in $\mathbb{C}^2$ given by $f(x,y)=x^3+y^3-y=0$ and let g:$C \to \mathbb{C}$ be the projection $g(x,y)=x$. I want to find where this map $g$ is ramified, and do this by computing the sheaf of relative differentials. I have $(3y^2-1)dy=-3x^2dx$. Now, by the definition of relative differentials I thought I set $dx=0$, and since ramified points are where this sheaf is nonzero, I get that dy is not killed when $3y^2-1=0$ but I don't think that gives the right answer. On the other hand, the derivative $dy/dx=\frac{-3x^2}{3y^2-1}$ vanishes when x=0, and that gives the right answer. Apparently I should be setting $dy=0$ but that doesn't make sense since the target of $g$ is $Spec\ \mathbb{C}[x]$. I'm confused why my sheaf of differentials approach isn't working - clearly I just have a basic misunderstand about it and I was hoping someone could clear if up for me.&lt;/p&gt;&#xA;" LastActivityDate="2010-12-24T21:44:25.883" Title="stuck on computing points of ramification of a map of curves via sheaf of relative differentials" Tags="&lt;algebraic-geometry&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" />
  <row Id="5160" PostTypeId="1" CreationDate="2010-09-21T16:24:49.670" Score="1" ViewCount="92" Body="&lt;p&gt;I am reading about active contours models. It is said that this can be broadly classified as either parametric active contour models and geometric active contour models. The first one are represented explicitly as parameterized curves in a Lagragian framework, while the geometric active contours are represented implicitly as level sets of a 2-dimensional function that evolves in a Eulerian framework.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is an Eulerian approach and an Lagragian approach?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks,&lt;/p&gt;&#xA;" OwnerUserId="2045" LastActivityDate="2010-10-21T17:01:54.407" Title="Active Contour Models - Lagrangian/Eulerian approches" Tags="&lt;calculus&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="0" />
  <row Id="5161" PostTypeId="1" AcceptedAnswerId="5162" CreationDate="2010-09-21T16:43:06.410" Score="3" ViewCount="351" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd like to know the name of this kind of polynomial&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$p(x)=x^n+a_{1}x^{n-1}+\ldots+a_{n-1}x+1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where the $a_{i}\in\lbrace0,1\rbrace$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="1747" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-30T01:32:14.937" LastActivityDate="2010-11-30T01:32:14.937" Title="What is the name of this kind of polynomial?" Tags="&lt;terminology&gt;&lt;polynomials&gt;" AnswerCount="1" CommentCount="8" />
  <row Id="5162" PostTypeId="2" ParentId="5161" CreationDate="2010-09-21T17:21:29.123" Score="11" ViewCount="" Body="&lt;p&gt;They're known as &lt;strong&gt;Newman polynomials&lt;/strong&gt;. They are often studied in contexts where one is interested in learning what interesting consequences result from placing such restrictions on the coefficients, for example &lt;a href=&quot;http://www.ams.org/journals/mcom/2009-78-265/S0025-5718-08-02138-8/S0025-5718-08-02138-8.pdf&quot; rel=&quot;nofollow&quot;&gt;see here&lt;/a&gt;.  Erdos and Littlewoood posed several questions about the effects this has on the minimum modulus of the polynomial on the unit circle, e.g. see &lt;a href=&quot;http://www.ams.org/journals/proc/1992-116-02/S0002-9939-1992-1101984-4/S0002-9939-1992-1101984-4.pdf&quot; rel=&quot;nofollow&quot;&gt;this paper&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-09-21T17:35:28.707" LastActivityDate="2010-09-21T17:35:28.707" CommentCount="2" />
  <row Id="5163" PostTypeId="1" AcceptedAnswerId="5169" CreationDate="2010-09-21T17:34:00.923" Score="1" ViewCount="245" Body="&lt;p&gt;So first off all, this WAS homework. Submitted it about 3 hours ago.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $A$ be an $m \times m$ matrix. then show that if rank A = m, then Ax = 0 has a unique solution. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My roommate said this is wrong but here is how I solved it. &#xA;very simply, i stated that if the rank of A is m that means the number of leading ones are the same as the columns. Hence there exists a determinant. According to theorms, if a matrix has a determinant then it has an inverse. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So keeping that in mind, we do this &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A * A^(-1) x = 0 * A^(-1)  &amp;lt;--- multipling both sides by the inverse. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I x = 0 where I is the identity matrix and hence showing that if rank A = m then Ax = 0 has a unique solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So is this completely wrong? is there an alternative?&lt;/p&gt;&#xA;" OwnerUserId="287" LastActivityDate="2010-09-23T02:56:15.583" Title="Finding a unique solution given rank A = m" Tags="&lt;linear-algebra&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="5164" PostTypeId="2" ParentId="5163" CreationDate="2010-09-21T17:42:42.327" Score="0" ViewCount="" Body="&lt;p&gt;Yes, the rank is the number of pivots in Gaussian elimination. But you must explicitly state how you utilize this to conclude that the determinant is nonzero - viz. by pigeonhole, all diagonal elements must be 1, not 0. For further insight see this &lt;a href=&quot;http://en.wikipedia.org/wiki/Invertible_matrix&quot; rel=&quot;nofollow&quot;&gt;Wikipedia page&lt;/a&gt; for at least 17 properties that are equivalent to $\rm A$ being invertible. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's one alternative solution: since the matrix has full column rank, the columns are linearly independent. Therefore any solution, being a linear combination of the columns, is necessarily unique. Said slightly more generally: $\rm\ A\:$ has full column rank $\iff$ $\rm\:A$ is injective (1-1) as a linear map $\rm\iff\ ker\ A = \:0$&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-09-21T19:20:23.707" LastActivityDate="2010-09-21T19:20:23.707" />
  <row Id="5165" PostTypeId="2" ParentId="5163" CreationDate="2010-09-21T17:46:12.803" Score="4" ViewCount="" Body="&lt;p&gt;Your reasoning is not quite right. Every square matrix has a determinant regardless if it has an inverse or not. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The correct statement is: A square matrix has an inverse iff its determinant is non-zero. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the easiest way to solve your problem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $A=[a_1|\ldots|a_m]$    i.e. $a_i$ is the $i^{\text{th}}$ column of $A$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;[Recall: $\text{rank}(A) = \text{dim}(\text{Span}(a_1,\ldots,a_m))$]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;rank($A$)=m iff Span$(a_1,\ldots,a_m)=\mathbb{R}^m$ This comes from the definition of rank. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have $m$ vectors $a_1,\ldots,a_m$ that span an $m$ dimension space $\mathbb{R}^m$ and thus $a_1,\ldots,a_m$ are linearly independent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$0=Ax=x_1a_1+\ldots + x_ma_m$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and by definition of linearly independence, $x_i=0$ for $1\leq i\leq m$. Hence the zero vector is the unique solution.&lt;/p&gt;&#xA;" OwnerUserId="1794" LastEditorUserId="1794" LastEditorDisplayName="" LastEditDate="2010-09-21T18:43:29.647" LastActivityDate="2010-09-21T18:43:29.647" />
  <row Id="5166" PostTypeId="1" CreationDate="2010-09-21T19:02:42.603" Score="2" ViewCount="468" Body="&lt;p&gt;Hi all!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have all points where my curve pass through, but I need to get the coordinates of the control points to be able to draw the curve. How can I do to calculate this points?&lt;/p&gt;&#xA;" OwnerUserId="2050" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-09-22T01:48:57.900" LastActivityDate="2010-11-02T18:37:10.093" Title="algorithm to calculate the control points of a cubic Bezier curve" Tags="&lt;geometry&gt;&lt;bezier-curve&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" />
  <row Id="5167" PostTypeId="1" AcceptedAnswerId="5176" CreationDate="2010-09-21T19:06:15.070" Score="2" ViewCount="359" Body="&lt;p&gt;I was reading this book &quot;Rational points on Elliptic curves&quot; by J.Silverman, and J.Tate, 2 prominent figures in Number theory and was very intrigued after reading the first couple of pages.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The connection between Algebra and Geometry is displayed in a professional manner in that book. One can generally ask this question:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;It's clear that the Mathematicians had to develop some laws in order to make the points on an Elliptic curve to a group. The first axioms of adding 2 points on an elliptic curve is clear. But i would like to know how does one come up with the mechanism of the remaining axioms(like associativity) which are actually intricate in nature. What is the motivation behind it. Can we define an axiom in an other way so that the point on the elliptic curve still forms a group? Also why to define the group axioms only on for this type of curves. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A very much related question was asked in MO link: &lt;a href=&quot;http://mathoverflow.net/questions/6870/why-is-an-elliptic-curve-a-group&quot; rel=&quot;nofollow&quot;&gt;http://mathoverflow.net/questions/6870/why-is-an-elliptic-curve-a-group&lt;/a&gt; but i actually expect more about the mechanism of the intricate axioms. &lt;/p&gt;&#xA;" LastActivityDate="2010-09-21T20:41:51.347" Title="Group Law for an Elliptic curve" Tags="&lt;number-theory&gt;&lt;soft-question&gt;&lt;elliptic-curves&gt;" AnswerCount="3" CommentCount="15" />
  <row Id="5168" PostTypeId="2" ParentId="5153" CreationDate="2010-09-21T19:10:19.390" Score="0" ViewCount="" Body="&lt;p&gt;Hi--&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While searching on the net i found this article. See whether it helps or not: &lt;a href=&quot;http://www.springerlink.com/content/l272110h87u05667/fulltext.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.springerlink.com/content/l272110h87u05667/fulltext.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If one can't access this then try: &lt;a href=&quot;http://dx.doi.org/10.1007/BF01058705&quot; rel=&quot;nofollow&quot;&gt;dx.doi.org/10.1007/BF01058705&lt;/a&gt; or &lt;a href=&quot;http://ams.org/mathscinet-getitem?mr=1148867&quot; rel=&quot;nofollow&quot;&gt;ams.org/mathscinet-getitem?mr=1148867&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks to Jack.&lt;/p&gt;&#xA;" LastActivityDate="2010-09-21T19:47:50.883" CommentCount="4" />
  <row Id="5169" PostTypeId="2" ParentId="5163" CreationDate="2010-09-21T19:12:05.420" Score="2" ViewCount="" Body="&lt;p&gt;There are several misstatements and one error in your solution, though the &lt;em&gt;idea&lt;/em&gt; is essentially correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, it is false that if the matrix has rank $m$ then &quot;the number of leading ones are he same as the columns.&quot; The matrix $A$ may have no &quot;leading ones&quot; whatsoever. What you meant, presumably, is that the *&lt;em&gt;row-echelon form&lt;/em&gt; (or the row-reduced form) of $A$ will have as many leading $1$s as there are columns.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second: determinants exist for square matrices regardless of the form of the row-echelon form; the determinant of a square matrix is nonzero if and only if the matrix is invertible, if and only if the number of leading $1$s in the row-echelon form of $A$ equals the number of columns. This has to be proven in some way; for example: the elementary row operations will either multiply the determinant by $-1$ in the case of row exchanges, will multiply it by a nonzero constant $\alpha$ in the case of multiplying a row by $\alpha\neq 0$, or will leave the determinant without change in the case that you add the multiple of one row to another row. Thus, performing elementary row operations will not change whether the determinant is or is not equal to $0$, in the sense that if you obtain $B$ from $A$ by a finite sequence of elementary row operations, then $\det(A)=0$ if and only if $\det(B)=0$. Since the determinant of an upper triangular matrix is just the product of the diagonal entries, the determinant of a square matrix in row echelon form will be $1$ if the leading $1$s are all in the diagonal, and $0$  otherwise. So in your case, the row echelon form of $A$ has nonzero determinant, and therefore (by what we saw before) the determinant of $A$ is nonzero, so $A$ has an inverse.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That said, your next line is incorrect. You go from&#xA;\[ A\mathbf{x}=\mathbf{0}\]&#xA;to&#xA;\[ AA^{-1}\mathbf{x} = \mathbf{0}A^{-1}.\]&#xA;This not only does not follow, it does not even make sense! First, because if $A$ is an $m\times m$ matrix, then $\mathbf{x}$ and $\mathbf{0}$ are $m\times 1$ matrices; but for $\mathbf{0}A^{-1}$ to make sense, the number of columns of $\mathbf{0}$ (namely, 1) has to equal the number of rows in $A^{-1}$ (namely $m$); and this is not true unless you are dealing with $1\times 1$ matrices. Second, even if you &lt;em&gt;could&lt;/em&gt; multiply both sides by $A^{-1}$ on the right, then you would not get $AA^{-1}\mathbf{x}$ on the left hand side of the equation, you would get $A\mathbf{x}A^{-1}$, and then you need to remember that matrix multiplication is not commutative, so in general you cannot go from $\mathbf{x}A^{-1}$ to $A^{-1}\mathbf{x}$ (as noted above, it may not even make sense to write one of them).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, you multiplied by $A^{-1}$ on the wrong side, and you did it incorrectly. Rather, you want to multiply by $A^{-1}$ on the &lt;em&gt;left&lt;/em&gt; of each side of the equality, not on the right. That is, you want to go from&#xA;\[ A\mathbf{x}=\mathbf{0}\]&#xA;to&#xA;\[ A^{-1}(A\mathbf{x}) = A^{-1}\mathbf{0}.\]&#xA;And from here you can proceed as you did.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So... partial, but definitely not full credit.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-09-23T02:56:15.583" LastActivityDate="2010-09-23T02:56:15.583" CommentCount="2" />
  <row Id="5170" PostTypeId="1" AcceptedAnswerId="5172" CreationDate="2010-09-21T19:20:33.360" Score="1" ViewCount="132" Body="&lt;p&gt;So I have some circles that look kinda like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://img80.imageshack.us/img80/792/mmmcircles.png&quot; width=&quot;800&quot; height=&quot;400&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm given the radius of the circle with center point A which is also the distance AB, the distance AB between the two center points on the x axis (they share same y values for the centers), and the distance CD which is the height of the shape created by the intersection.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can also assume that despite my drawing the circle ARE perfect circles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm looking for a way to find the distance BD which is also the radius of the circle centered on point B.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance for any help.&#xA;&lt;br&gt;--AvatarOfChronos&lt;/p&gt;&#xA;" OwnerUserId="1582" LastActivityDate="2010-09-21T19:27:20.497" Title="Find the radius of a circle based off of its intersection with another." Tags="&lt;geometry&gt;&lt;circle&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" />
  <row Id="5171" PostTypeId="2" ParentId="5153" CreationDate="2010-09-21T19:26:29.230" Score="4" ViewCount="" Body="&lt;p&gt;The short answer is that there is nothing nearly so nice as Goursat's Lemma. You can certainly reduce easily to the case where $\pi_K(H)=K$, much like you can reduce Goursat's Lemma to the case of a subdirect product, but after that it gets complicated. To give you an idea, here are three references.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A theorem of Rosenbaum (&lt;em&gt;Die Untergruppen von halbdirekten Produkten&lt;/em&gt;, Rostock. Math. Kolloq. No. 35 (1988), 21-30) gives (from the MathScieNet Review MR991728 (90c:20032)):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Theorem.&lt;/strong&gt; A set $U$ of elements of the semidirect product $G=NK$ with $N\triangleleft G$ is a subgroup of $G$ if and only if&lt;/p&gt;&#xA;  &#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;$UN\cap K$ and $U\cap K$ are subgroups of $G$;&lt;/li&gt;&#xA;  &lt;li&gt;$U\cap N$ is a subgroup and $UK\cap N$ is a collection of $U\cap N$-cosets in $N$; and&lt;/li&gt;&#xA;  &lt;li&gt;There is a mapping $\varphi$ defined for all $g\in UK\cap N$ mapping $(U\cap K)g$ onto some coset $n(U\cap N)$, with $n\in N$, satisfying $\varphi(g_1g_2)=g_2^{-1}\varphi(g_1)g_2\varphi(g_2)$.&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The criterion was then used by Gutiérrez-Barrios to develop a criterion for a set of elements to be a normal subgroup of the semidirect product (&lt;em&gt;Die Normalteiler von halbdirekten Produkten.&lt;/em&gt; Wiss. Z. Pädagog. Hochsch. Erfurt/Mühlhausen Math.-Natur. Reihe &lt;strong&gt;25&lt;/strong&gt; (1989), no. 2, 108-114. MR1044548 (91b:20029))&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Usenko (&lt;em&gt;Subgroups of semidirect products&lt;/em&gt;, english translation in &#xA;Ukrainian Math. J. &lt;strong&gt;43&lt;/strong&gt; (1991), no. 7-8, 982-988 (1992), MR1148867 (92k:20045)) uses crossed homomorphisms to study subgroups of semidirect products. &lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-09-21T19:26:29.230" />
  <row Id="5172" PostTypeId="2" ParentId="5170" CreationDate="2010-09-21T19:27:20.497" Score="2" ViewCount="" Body="&lt;p&gt;First, note that segment AB bisects segment CD.  Call their point of intersection E, which is also the midpoint of CD.  Since you know the length of CD, you know the length of CE.  The measure of angle BAC (which could also be called EAC) is $\sin^{-1}\left(\frac{CE}{AC}\right)$.  Apply the Law of Cosines to triangle ABC to find BC=BD: &#xA;$$BC^2=AC^2+AB^2-2\cdot AC\cdot AB\cdot\cos\left(\sin^{-1}\left(\frac{CE}{AC}\right)\right).$$&#xA;Fill in the known lengths and solve.&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-09-21T19:27:20.497" CommentCount="2" />
  <row Id="5173" PostTypeId="2" ParentId="5167" CreationDate="2010-09-21T19:52:23.680" Score="5" ViewCount="" Body="&lt;p&gt;I agree that from the chord-and-tangent definition of the group law it is far from clear why associativity should hold.  This definition is the one presented in Silverman and Tate because all the other ways you might define the group law require &lt;strong&gt;much&lt;/strong&gt; more background; however, these are the definitions from which associativity makes more sense.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In one approach an elliptic curve is a group of the form $\mathbb{C}/\Lambda$ where $\Lambda$ is a &lt;strong&gt;lattice&lt;/strong&gt;, e.g. a discrete subgroup of $\mathbb{C}$ isomorphic to $\mathbb{Z}^2$.  (The motivation for looking at such quotients comes basically from the &lt;a href=&quot;http://en.wikipedia.org/wiki/Uniformization_theorem&quot; rel=&quot;nofollow&quot;&gt;uniformization theorem&lt;/a&gt;).  Such a quotient forms a compact Riemann surface, topologically a torus, hence it admits no nonconstant holomorphic functions; therefore we would like to describe the meromorphic functions.  A natural way to do this is to write down a function and then try to average over $\Lambda$, which (if you do it right) will lead you to the definition of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Weierstrass_elliptic_function&quot; rel=&quot;nofollow&quot;&gt;Weierstrass elliptic function&lt;/a&gt; $\wp(z)$.  By examining its poles and the poles of its derivative $\wp'(z)$ we can deduce the differential equation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\wp'(z)^2 = 4 \wp(z)^3 - g_2 \wp(z) - g_3$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $g_2, g_3$ are certain constants associated to $\Lambda$.  In fact the map from $\mathbb{C}/\Lambda$ to the elliptic curve $y^2 = 4x^3 - g_2 x - g_3$ given by $z \mapsto (\wp(z), \wp'(z))$ is an isomorphism.  This forms the connection between the complex-analytic picture and the algebraic picture in terms of Weierstrass normal forms.  The group law here is just regular addition, e.g. the sum of the points $(\wp(a), wp'(a))$ and $(\wp(b), wp'(b))$ is just $(\wp(a+b), wp'(a+b))$, and so it is obvious that associativity holds.  The fact that if $a + b + c = 0$ then the corresponding points are collinear is then a relatively simple computation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Historically, this is not how the first group laws on elliptic curves were discovered.  For the elliptic-function perspective I recommend Stevenhagen's &lt;a href=&quot;http://websites.math.leidenuniv.nl/algebra/ellipticfunctions.pdf&quot; rel=&quot;nofollow&quot;&gt;notes on the subject&lt;/a&gt;.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In another approach an elliptic curve (over $\mathbb{C}$, for simplicity) is a smooth projective curve of genus $1$.  The category of smooth projective curves over $\mathbb{C}$ is known to be equivalent to the category of compact Riemann surfaces, which is part of the connection between this definition and the above definition.  To any smooth projective curve $C$ one can associate its &lt;a href=&quot;http://en.wikipedia.org/wiki/Jacobian_variety&quot; rel=&quot;nofollow&quot;&gt;Jacobian variety&lt;/a&gt;, which is a group of the form $\mathbb{C}^g/\Lambda$ where $g$ is the genus which naturally occurs when one thinks about integration over such a curve.  The &lt;a href=&quot;http://en.wikipedia.org/wiki/Abel-Jacobi_theorem&quot; rel=&quot;nofollow&quot;&gt;Abel-Jacobi theorem&lt;/a&gt; asserts that the Jacobian variety of a curve is isomorphic to its &lt;a href=&quot;http://en.wikipedia.org/wiki/Picard_group&quot; rel=&quot;nofollow&quot;&gt;Picard group&lt;/a&gt; (the group of &lt;a href=&quot;http://en.wikipedia.org/wiki/Divisor_(algebraic_geometry)&quot; rel=&quot;nofollow&quot;&gt;divisors&lt;/a&gt; modulo principal divisors), and the points of the Picard group of a curve of genus $1$ can be put into natural bijection with the curve itself (once one chooses an identity), so the group law on the Jacobian variety (which is perfectly natural) then gives a group law on the original curve (again once one chooses an identity).  The connection to collinear points comes from the fact that lines determine principal divisors, which are zero in the Picard group.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for your last question, elliptic curves are, as it turns out, the only projective curves which can be given the structure of algebraic groups; see the Wikipedia article on &lt;a href=&quot;http://en.wikipedia.org/wiki/Abelian_variety&quot; rel=&quot;nofollow&quot;&gt;abelian varieties&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-09-21T19:59:55.597" LastActivityDate="2010-09-21T19:59:55.597" />
  <row Id="5174" PostTypeId="1" AcceptedAnswerId="5187" CreationDate="2010-09-21T19:55:51.983" Score="0" ViewCount="132" Body="&lt;p&gt;I hope you will excuse my vague title. Let's define&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\varsigma \ \colon\ \mathbb N\to \mathbb Z\ \colon\ k\mapsto (-1)^{\lfloor \frac{k-1}{2}\rfloor}$$&#xA;and&#xA;$$S(m,n)=\displaystyle\sum_{k=m}^n k\cdot \varsigma(k)$$&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Find every integer $n\ge 1$ such that $S(1,\lfloor n/2\rfloor)=S(\lfloor n/2\rfloor+1,n)$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Show that $-(n+1)\le \delta(n)\le 2(n+1)$, when $\delta(n)=S(1,\lfloor n/2\rfloor)-S(\lfloor n/2\rfloor+1,n)$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="685" LastEditorUserId="685" LastEditorDisplayName="" LastEditDate="2010-09-21T22:35:47.367" LastActivityDate="2010-09-21T23:59:49.583" Title="About a certain signed sum of consecutive integers ($\sum_{k=1}^n k\cdot \varsigma(k)=\sum_{k=n}^{2n} k\cdot \varsigma(k)$)" Tags="&lt;arithmetic&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="5175" PostTypeId="2" ParentId="5167" CreationDate="2010-09-21T20:41:17.853" Score="1" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;This is a failed attempt at constructing the group law with associativity built in&lt;/strong&gt;, It is still interesting to consider though - I will explain why at the end.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Every line intersects 3 points on an elliptic curve, call them $P,Q,R$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can pick some designated point $\mathcal O$ as the identity on the curve and then set the ternary relation $P+Q+R = \mathcal O$ to hold for all triples which lie on a line (N.B. I could just write this $C(P,Q,R)$ but I use the suggestive notation of $+$ and $=$ instead).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Inspired by the group axioms we have $P + \mathcal O + -P = \mathcal O$, so we know how to compute the inverse of a point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given any two points $P,Q$ we could also find a third $Q$ easily, so that $P + Q + R = \mathcal O$. We can now define the binary operation of addition as $P + Q = -R$.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Now it is possible to reduce the proof of associativity of the group law to a purely combinatoric statement.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;$(P + Q) + R$ means that $P + Q + -PQ = \mathcal O$ and $PQ + R + -PQR = \mathcal O$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$P + (Q + R)$ means that $Q + R + -QR = \mathcal O$ and $P + QR + -PQR' = \mathcal O$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;(and also the extra relations between positive and negative terms)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These relations define an &lt;a href=&quot;http://en.wikipedia.org/wiki/Incidence_structure&quot; rel=&quot;nofollow&quot;&gt;incidence structure&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;{(P,Q,-PQ),(PQ,O,-PQ),(PQ,R,-PQR),(PQR,O,-PQR),(Q,R,-QR),(QR,O,-QR),(P,QR,-PQR'),(PQR',O,-PQR')}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now for the interesting bit! We would hope that a Fano plane style situation would force PQR to be equal to PQR' but it doesn't happen. So we didn't get any proof out of this, but we do get a sense of the difficulty of this associativity problem: There is something special about the &lt;em&gt;shape&lt;/em&gt; of the elliptic curve which forces these points to be equal. That implies that any proof or construction of this group law &lt;strong&gt;must&lt;/strong&gt; use the geometric properties of the curve in an essential way.&lt;/p&gt;&#xA;" LastActivityDate="2010-09-21T20:41:17.853" CommentCount="3" />
  <row Id="5176" PostTypeId="2" ParentId="5167" CreationDate="2010-09-21T20:41:51.347" Score="6" ViewCount="" Body="&lt;p&gt;Dear Chandru, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The group law on an elliptic curve was not discovered in a vacuum.  It came up in the context&#xA;of abelian integrals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $y^2 = f(x)$, where $f(x)$ is a cubic in $x$ be an elliptic curve; call it $E$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Elliptic integrals are integrals of the form $$\int_{a}^x dx/y = \int_a^x \frac{d x}{\sqrt{&#xA;f(x)}}.$$  (Here $a$ is some fixed base-point.) They come up (in a slightly transformed manner) when computing the arclength of an ellipse (whence their name).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If was realized at some point (in the 1600s or 1700s) (at least in special cases) that if &#xA;you apply certain substitutions to $x$, you can double the value of the integral, or that&#xA;if you apply certain substitusions of the form $x = \phi(x_1,x_2)$, the integral you compute is the sum of the individual integrals for $x_1$ and $x_2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Real understanding came from the work of Abel and Jacobi.  (Unfortunately I don't know&#xA;the precise history or attributions here.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What they realized (in modern terms) is that, if we fix the base-point $a$ and let $b$&#xA;vary, then the elliptic integral is giving a multivalued map from the elliptic curve $E$&#xA;to $\mathbb C$,&#xA;and that the formula $\phi(x_1,x_2)$ mentioned above shows us a way to add points on the&#xA;elliptic curve, so that this map is a (multi-valued) group homomorphism.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Taking the inverse of this multi-valued map gives a single-valued map (which is how we are&#xA;more used to thinking about it) &#xA;$$\mathbb C \to E,$$&#xA;which is a homomorphism when we give $\mathbb C$ its addivite structure and $E$ the&#xA;group law coming from $(x_1,x_2)\mapsto \phi(x_1,x_2)$.   The kernel of this map turns out&#xA;to be a lattice $\Lambda$, so that we get an isomorphism $\mathbb C/\Lambda \cong E.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The formula $\phi(x_1,x_2)$ turns out to be precisely the formula describing addition&#xA;on the elliptic curve via chords and tangents, and there are lots of theoretical explanations for it, as you can find on the linked MO page.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a higher genus curve $C$, it turns out that there is not just the one holomorphic differential $dx/y$, but $g$ linearly independent such (if the curve has genus $g$),&#xA;say $\omega_1, \ldots,\omega_g$.&#xA;Furthermore, if we fix a particular differential $\omega_i$, then there is &lt;em&gt;no&lt;/em&gt; formula&#xA;$\phi(x_1,x_2)$ such that the sum of the integrals of $\omega$ for $x_1$ and $x_2$ is&#xA;equal to the integral of $\omega$ for $\phi(x_1,x_2)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, what Abel and Jacobi found is that, if we consider the map &#xA;$$(x_1,\ldots,x_g) \mapsto (\sum_{i = 1}^g \int_a^{x_i} \omega_1,&#xA;\ldots,\sum_{i = 1}^g \int_a^{x_i} \omega_g),$$&#xA;which gives a multi-valued map $$Sym^g C \to \mathbb C$$&#xA;(here $Sym^g C$ denotes the $g$th symmetric power of $C$, so it is the product of&#xA;$g$ copies of $C$, modulo the action of the symmetric group on the $g$ factors),&#xA;then we can find a formula $\phi(x_{1,1},\ldots,x_{1,g},x_{2,1},\ldots,x_{2,g})$&#xA;such that $\phi$ defines a group law on $Sym^g C$ (at least generically) and&#xA;such that this map is a homomorphism.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again, this map&#xA; becomes well-defined and (generically) single valued if we quotient out the target by an appropriate lattice $\Lambda$ (the period lattice), to get a birational map&#xA;$$Sym^g C \to \mathbb C^g /\Lambda.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The target here is called the Jacobian of $C$, and can be identified with $Pic^0(C)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In summary, to generalize the addition law on an elliptic curve to higher genus curves,&#xA;you have to consider unordererd $g$-tuples of points (where $g$ is the genus), and add those (not just individual&#xA;points).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(The relationship with $Pic^0(C)$ is that if $x_1,\ldots,x_g$ are $g$ points on $C$,&#xA;then $x_1 + \ldots + x_g - g a$ is a degree zero divisor on $C$, and every degree zero&#xA;divisor is linearly equivalent to a divisor of this form, and, generically, to a unique&#xA;such divisor.)&lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2010-09-21T20:41:51.347" />
  <row Id="5177" PostTypeId="1" CreationDate="2010-09-21T21:11:53.667" Score="2" ViewCount="104" Body="&lt;p&gt;Do generalized hypergeometric functions &#xA;$${}_p F_q(a_1,\ldots,a_p; b_1, \ldots,b_q; z) $$&#xD;&#xA;with $p = q+1$ always possess a singularity at $z=1$, independent of the their parameters $a_1,\ldots,a_p$ and $b_1,\ldots,b_q$ under the provision that all cases of finite polynomials are excluded? Can someone mention a reliable source for such a statement or provide a counterexample?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance.&lt;/p&gt;&#xA;" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-20T23:50:41.000" LastActivityDate="2010-12-20T23:50:41.000" Title="A singularity of hypergeometric functions" Tags="&lt;special-functions&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="5178" PostTypeId="2" ParentId="250" CreationDate="2010-09-21T21:18:16.207" Score="5" ViewCount="" Body="&lt;p&gt;It is possible for a group of people to hold a secret ballot election in which all communication is done publicly. This is one of the many surprising consequences of the existence of &lt;a href=&quot;http://en.wikipedia.org/wiki/Secure_multi-party_computation&quot; rel=&quot;nofollow&quot;&gt;secure multiparty computation&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Of course, the ballots are only &quot;secret&quot; under some reasonable cryptographic assumptions. I guess Feynman may have objected to this.)&lt;/p&gt;&#xA;" OwnerUserId="1776" LastActivityDate="2010-09-21T21:18:16.207" CommunityOwnedDate="2010-09-21T21:18:16.207" />
  <row Id="5179" PostTypeId="1" AcceptedAnswerId="5183" CreationDate="2010-09-21T21:34:59.040" Score="3" ViewCount="211" Body="&lt;p&gt;What are the benefits of using a conjugate linear inner product in a complex vector space vs a simple linear inner product?  That is, why do we demand that (y,x) = (x,y)* as opposed to (y,x)=(x,y)?  Of course, this ensures that (x,x) is real and thus makes an easy definition of norm, but is that necessary?  &lt;/p&gt;&#xA;" OwnerUserId="948" LastActivityDate="2010-09-21T22:14:27.290" Title="Is complex conjugation needed for valid inner product?" Tags="&lt;inner-product-space&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="2" />
  <row Id="5180" PostTypeId="2" ParentId="5159" CreationDate="2010-09-21T21:39:34.387" Score="0" ViewCount="" Body="&lt;p&gt;This should not under any circumstances be considered rigorous, but intuitively here is what is happening in my mind. If you project onto the &quot;$x$-axis&quot;, then think of the tangent space of the curve $C$ at some point $(a,b)$. If the tangent direction is &quot;vertical&quot;, then any vector in that space will project to 0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So after projecting, there is no &quot;$y$-component&quot; to any of the tangent vectors, and hence the $dy$ always evaluates to $0$. So in some sense $g$ is transverse to the projection precisely when there is some &quot;$x$-direction&quot;. This is precisely when $-3x^2dx\neq 0$.&lt;/p&gt;&#xA;" OwnerUserId="2000" LastActivityDate="2010-09-21T21:39:34.387" />
  <row Id="5181" PostTypeId="2" ParentId="5139" CreationDate="2010-09-21T21:59:50.207" Score="5" ViewCount="" Body="&lt;p&gt;There is a simpler way (in terms of manual computations) than the other answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We use the following facts:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $A$ is an integer then $$ A -[x] = [A-x + \delta] $$ where $\delta = 0$ if $x$ is an integer, and $\delta = 1$ otherwise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $n &gt; 1$ is a positive integer then $$ [\frac{[x]}{n}] = [\frac{x}{n}]$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using this we see that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$[\frac{n-12 - [\frac{n-17}{25}]}{3}] = [\frac{n-12-\frac{(n-17)}{25}+\delta}{3}]$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\delta = 0$ iff $n=17 \mod 25$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ = [\frac{24n-283+25\delta}{75}]$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $24n+39 = r \mod 75$ where $0 \leq r &amp;lt; 75$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice that $r$ must be divisible by $3$. Thus $0 \leq r \leq 72$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose $24n+39 = 75m+r$, then we have that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$[\frac{8n+13}{25}] = [\frac{24n+39}{75}] = m$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and $$[\frac{24n-283+25\delta}{75}] = [\frac{75m+r-375+53+25\delta}{75}]$$&#xA;= $$m-5 + [\frac{r + 53 + 25\delta}{75}]$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now $\delta = 0$ iff $n=17 \mod 25$ iff $r = 72$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus for $r &amp;lt; 72$, $\delta = 1$, for which we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$[\frac{r + 53 + 25\delta}{75}] = [\frac{r + 53 + 25}{75}]  = 1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $r = 72$, we have $\delta=0$ for which we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$[\frac{r + 53 + 25\delta}{75}] = [\frac{r + 53}{75}] = 1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus the whole expression is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ m - ((m-5)+1) = 4 $$&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-09-21T21:59:50.207" />
  <row Id="5182" PostTypeId="2" ParentId="5179" CreationDate="2010-09-21T22:01:00.577" Score="3" ViewCount="" Body="&lt;p&gt;You need to define the inner product this way (in a complex vector space) to have a good definition of the norm of a vector. Let $V$ be this vector space, and take $X\in V$. Then, the norm of X is defined as $|X|=\sqrt{(X,X)}$. This definition is good ($|X|$ is well defined for all $X$) if $(y,x)=(x,y)^*$. If this is not the case, for example if $(y,x)=(x,y)$, then $ (x,x) $ need not be positive. More: $(x,x)$ need not be real. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Aother thing, related to the above one: in metric spaces, given an inner product we define a norm (the same way we did above) and, from the norm, a distance function. This distance function determines the topology of the space (the topology induced by the inner product). To do this, we need to be able to define in a good way the norm of a vector, and the problem appears again.&lt;/p&gt;&#xA;" OwnerUserId="1178" LastActivityDate="2010-09-21T22:01:00.577" />
  <row Id="5183" PostTypeId="2" ParentId="5179" CreationDate="2010-09-21T22:14:27.290" Score="6" ViewCount="" Body="&lt;p&gt;It is in fact necessary. The inner product axioms without the conjugation are inconsistent:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Here $u$, $v$, $w$ are vectors and $c$ is a scalar)&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$\langle cu, v\rangle = c\langle u, v\rangle$&lt;/li&gt;&#xA;&lt;li&gt;$\langle u,v\rangle = \langle v,u\rangle$&lt;/li&gt;&#xA;&lt;li&gt;If $u \neq 0$, then $\langle u,u\rangle$ is a positive real number&lt;/li&gt;&#xA;&lt;li&gt;$\langle u+v,w\rangle = \langle u,w\rangle + \langle v,w\rangle$&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;In fact, 1-3 alone are inconsistent. Indeed, let $u$ be any nonzero vector, so $\langle u,u\rangle &gt; 0$ by condition 3. But if $i = \sqrt{-1}$, then $\langle iu, iu\rangle = i\langle u, iu\rangle$ (by 1) $ = i\langle iu, u\rangle$ (by 2) $ = i^2\langle u, u\rangle$ (by 1) $ = -\langle u, u\rangle &amp;lt; 0$, contradicting condition 3. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The upshot is that you can choose: either conjugate one side of condition 2, giving you the axioms for an inner product, or get rid of condition 3, giving you the axioms for a symmetric bilinear form. You could also consider a weaker version of 3, like requiring that if $u\neq 0$, then $\langle u, v\rangle \neq 0$ for some $v$. That gives you nondegenerate symmetric bilinear forms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that there's nothing wrong with bilinear forms on complex vector spaces; they're just not inner products. They're disjoint concepts, unlike in real vector spaces, where inner products are just special symmetric bilinear forms. In some ways, bilinear forms are nicer than inner products, since you don't have to worry about complex conjugation. However, bilinear forms over the complex numbers do not give rise to norms, which means they don't endow vector spaces with good geometry. Inner products do, hence their ubiquity. &lt;/p&gt;&#xA;" OwnerUserId="2032" LastActivityDate="2010-09-21T22:14:27.290" />
  <row Id="5184" PostTypeId="2" ParentId="5177" CreationDate="2010-09-21T22:31:36.307" Score="1" ViewCount="" Body="&lt;p&gt;For the $p=q+1$ case, taking the branch cut from 1 to ∞ is conventional so that you can speak of the &lt;em&gt;principal branch&lt;/em&gt; of ${}_{q+1} F_q$ ; the function is multivalued outside the unit disk, and 1 and ∞ are branch points.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For unit argument, the key word &quot;Saalschützian&quot; is what you might take into account; see &lt;a href=&quot;http://www.ams.org/journals/proc/1992-114-01/S0002-9939-1992-1068116-2/S0002-9939-1992-1068116-2.pdf&quot; rel=&quot;nofollow&quot;&gt;this article&lt;/a&gt; by Bühring for more information.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-09-21T22:31:36.307" />
  <row Id="5185" PostTypeId="1" AcceptedAnswerId="5186" CreationDate="2010-09-21T22:36:10.747" Score="8" ViewCount="244" Body="&lt;p&gt;Prompted by today's Minute Math question on the MAA site (&lt;a href=&quot;http://amc.maa.org/mathclub/5-0,problems/T-problems/T-web,ia/2005web/tb05-12-ia.shtml&quot; rel=&quot;nofollow&quot;&gt;http://amc.maa.org/mathclub/5-0,problems/T-problems/T-web,ia/2005web/tb05-12-ia.shtml&lt;/a&gt;), I started thinking about the probability that the sum of the numbers rolled on a set of $n$ dice is prime, particularly the asymptotics as $n\rightarrow\infty$.  Heuristics strongly suggest that this is proportional to $1/\mathrm{ln}\ n$, and in fact, that it's $1/\mathrm{ln}\ n - O(1/\mathrm{ln}^2n)$, but I'm wondering how I would go about getting better asymptotics on the second term.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the record, the heuristic argument goes something like this: assume for concreteness' sake that we're rolling 6-sided dice.  Then the sum of the dice is closely approximated by a normal variable with mean $\mu=7n/2$ and variance $\sigma^2=35n/12$, and since the PNT says that the 'probability' of an integer $n$ being prime is roughly $1/\mathrm{ln}\ n$, we should be able to integrate that probability with respect to the normal distribution:&#xA;$$p = {1\over \sqrt{2\pi\sigma^2}}\int_n^{6n} e^{-\left({(t-\mu)^2\over 2\sigma^2}\right)} {1\over\mathrm{ln}\ t} dt$$&#xA;And since $1/\mathrm{ln}\ t$ is monotonic, the value of the integral is bounded by the values we get by replacing its term in the integral with its maximum and minimum values on the integration interval:&#xA;$${1\over\mathrm{ln}\ 6n} {1\over \sqrt{2\pi\sigma^2}}\int_n^{6n} e^{-\left({(t-\mu)^2\over 2\sigma^2}\right)} dt &amp;lt; p &amp;lt; {1\over\mathrm{ln}\ n} {1\over \sqrt{2\pi\sigma^2}}\int_n^{6n} e^{-\left({(t-\mu)^2\over 2\sigma^2}\right)} dt$$&#xA;Both of the integrals in the latter formula are essentially 1 (by definition), so we get ${1\over\mathrm{ln}\ 6n} &amp;lt; p &amp;lt; {1\over\mathrm{ln}\ n}$; replacing $\mathrm{ln}\ 6n$ by $(\mathrm{ln}\ 6 + \mathrm{ln}\ n)$ and using the binomial formula gives the heuristic approximation I alluded to above.  This leads me to a couple of questions:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&lt;li&gt;How safe is the heuristic argument above?  I know that the PNT gives good bounds on the number of primes in an interval (on the order of $n^{1/2}$ here, which in particular means that the error from the prime-counting would be $O(n^{-1/2})$ and so much smaller than the inverse-log terms above), but my analytic number theory isn't good enough to know whether 'weighting' by the normal distribution would throw off the classical proofs.&#xA;&lt;li&gt;How would I go about evaluating the integral above?  Obviously the bounds I use bring it in to a fairly small range, but it seems as though to get a second term in my asymptotics I'd need to be able to at least approximate the integral, and there aren't any obvious tricks that look like they'd handle it well...&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="785" LastActivityDate="2010-09-21T23:05:33.727" Title="What's the probability that a sum of dice is prime?" Tags="&lt;number-theory&gt;&lt;probability&gt;&lt;prime-numbers&gt;&lt;recreational-mathematics&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="5186" PostTypeId="2" ParentId="5185" CreationDate="2010-09-21T22:57:45.520" Score="2" ViewCount="" Body="&lt;p&gt;For (1.), Rosser and Schoenfeld (sp?) non-asymptotic estimates bound $\pi(n)$ between functions of the form $x/(\log x + C)$ and this should be enough.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For (2) your integrals are rapid decaying and incredibly close to the integrals on the whole real line.  O($n$) standard deviations from the average is quite an unlikely event.&lt;/p&gt;&#xA;" OwnerUserId="467" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-09-21T23:05:33.727" LastActivityDate="2010-09-21T23:05:33.727" CommentCount="3" />
  <row Id="5187" PostTypeId="2" ParentId="5174" CreationDate="2010-09-21T23:59:49.583" Score="1" ViewCount="" Body="&lt;p&gt;The easiest way to attack this is elementary; if we define $S_0(n) = S(1,n)$, then $S(m,n) = S_0(n)-S_0(m)$, so all that's needed is to analyze $S_0$.  That most easily happens by cases; a bit of quick calculation and some simple algebra shows that $S_0(4k+1) = 1$, $S_0(4k+3) = 0$, $S_0(4k+2) = 4k+3$, and $S_0(4k) = -4k$.  From there it's straightforward to find the possible cases for $\delta(n)$ based on $n$ mod $8$; in particular, it seems like $\delta(n)$ will be zero whenever $n = 8k$ (where $S_0(n/2) = S_0(n)/2$) or $n=8k+7$ (where $S_0(n) = S_0(\lfloor n/2 \rfloor) = 0$).&lt;/p&gt;&#xA;" OwnerUserId="785" LastActivityDate="2010-09-21T23:59:49.583" />
  <row Id="5188" PostTypeId="2" ParentId="5018" CreationDate="2010-09-22T00:01:24.820" Score="3" ViewCount="" Body="&lt;p&gt;I want to point out that there are, topicwise, different paths from the most suggested standard one:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As noted by &quot;jmoy&quot;, time is limited - the comments telling you to learn some physics or  programming first or on the way seem to ignore this somewhat. Sure it is good and will pay off to have these things as background, but it is not indispensable. Apparently you already have discovered your fascination for mathematics, so you don't have to back it up with these things that surround mathematics. Physics and programming can give you good intuition, but so can doing mathematics itself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, going through the whole standard curriculum with lots of calculus and coordinate geometry is by no means the only way to come into mathematics!! Certainly a basic knowledge of these things will be indispensable &lt;em&gt;at some point&lt;/em&gt;. If you actually enjoy studying calculus, then do it, it's an excellent start! But don't just do it because it is the standard curriculum in US colleges. Often people who went through this can't imagine having learned things in a different way, but that is perfectly possible, and normal in many countries (I went through a German school and university and learned much less calculus on the way than an average US student - it didn't do me any harm).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After all you are wanting to study mathematics for having a rewarding experience, and while there will be some frustration on the way, it should feel somewhat rewarding right from the beginning.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So here are three alternative routes - in case they pique your curiosity more:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Number Theory&lt;/em&gt; is a field which seems perfect for self study: It is concerned with objects that you know very well already - the natural numbers - yet it is a deep and difficult field and connects to virtually every other part of mathematics. This is mathematics of an entirely different flavour than you would see following the calculus path, and it can bring you just as far. In particular it will bring you to a point where you want to study calculus because it tells you things about number theory.&#xA;For this road, pick up a book on Elementary Number Theory (e.g. &lt;a href=&quot;http://modular.math.washington.edu/ent/&quot; rel=&quot;nofollow&quot;&gt;(1)&lt;/a&gt;, free for download -  but if it's too hard going, there are more basic ones also) and work your way through that. Accompany it with the book &quot;Fearless Symmetry&quot; &lt;a href=&quot;http://books.google.com/books?id=h7oUr88S7JQC&amp;amp;printsec=frontcover&amp;amp;dq=fearless+symmetry&amp;amp;source=bl&amp;amp;ots=MzULIGz4SJ&amp;amp;sig=pI76wlCbxjZok0-1psJhclj4Hd8&amp;amp;hl=de&amp;amp;ei=7xyZTKHhNcH7OZOirNMP&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=2&amp;amp;ved=0CCEQ6AEwAQ#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;(2)&lt;/a&gt; an excellent popular math book, and not a waste of time! It leads you very far into things which connect to current research (the so-called &quot;Langlands program&quot;) but is very friendly written. It gives you a path to follow; you get intuitive ideas for example about &quot;group theory&quot;, and whenever you get to such a point where a new concept (like &quot;group&quot;) is introduced, you should accompany it with a more formal treatment, where you can see the basic theorems about that structure and get some exercises to solve. I suggest Michael Artin's &quot;Algebra&quot; &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0130047635&quot; rel=&quot;nofollow&quot;&gt;(3)&lt;/a&gt; which contains all you will need to know for a long time on this way. It starts at a low level, which either is already ok for you, or will take only little intermediate reading to bridge the gap.&#xA;(If you start wondering when the calculus comes in, pick up Serre's &quot;Course in Arithmetic&quot; &lt;a href=&quot;http://books.google.com/books?id=Ohj9NNk6btUC&amp;amp;printsec=frontcover&amp;amp;dq=course+in+arithmetic&amp;amp;hl=en&amp;amp;ei=Y0OZTJ3rFY-MOOPN9KsP&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=1&amp;amp;ved=0CDIQ6AEwAA#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;(4)&lt;/a&gt; and jump to part II, but only after you got some feeling for what's in the first half of Artin's &quot;Algebra&quot;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Combinatorics&lt;/em&gt; (e.g. with the book &quot;Proofs that really count&quot; &lt;a href=&quot;http://books.google.com/books?id=FXLGzXwbwIAC&amp;amp;dq=proofs+that+really+count&amp;amp;printsec=frontcover&amp;amp;source=bn&amp;amp;hl=en&amp;amp;ei=d0CZTJaSKcvKOMPd1JgP&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=4&amp;amp;ved=0CCQQ6AEwAw#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;(5)&lt;/a&gt;) - this is also a topic which needs no motivation from physics or any other area about which you would have to learn things first. As in number theory you can see how one develops mathematical reasoning to answer questions that make sense immediately. And again it can lead you the point where you want to study other mathematics, e.g. calculus, because it makes sense to pursue those questions in the context of what you already know...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Logic&lt;/em&gt;, e.g. starting with Carnielli, Epstein: &quot;Computability&quot; &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0534546447&quot; rel=&quot;nofollow&quot;&gt;(6)&lt;/a&gt; - this is a basic introduction into logic and computabiliy theory, written for philosophers, and therefore starting very gently. It also paves a way into mathematics via a motivating question - When should we call something &quot;computable&quot;? - and leads you all the way to a proof of Gödel's famous undecidability theorems. Again this is connected to a lot of mathematics. Maybe you would after that like to go on studying non-classical logics, which will lead you to algebra, lattice theory and topology...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whatever route you choose (and I have to emphasize again that the calculus route is very good as well, if you feel like it), talk to people about the things you study, ask and answer questions, e.g. work as a tutor, or spend time here in the forum! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Enjoy your journey to mathematics!&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;" OwnerUserId="2048" LastEditorUserId="2048" LastEditorDisplayName="" LastEditDate="2010-09-25T11:37:29.643" LastActivityDate="2010-09-25T11:37:29.643" CommunityOwnedDate="2010-09-22T00:01:24.820" CommentCount="2" />
  <row Id="5189" PostTypeId="2" ParentId="5143" CreationDate="2010-09-22T00:43:30.657" Score="2" ViewCount="" Body="&lt;p&gt;If $f$ and $g$ are rational the problem is very easy.  Multiply out by the denominator, take the $d^{th}$ power of both sides where $d$ is the least common denominator of $f$ and $g$, and you've reduced to the polynomial case, which is easy (check the extrema and so forth).  This is why it is important to provide details.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-09-22T00:43:30.657" CommentCount="1" />
  <row Id="5190" PostTypeId="2" ParentId="5115" CreationDate="2010-09-22T00:51:17.320" Score="1" ViewCount="" Body="&lt;p&gt;Let's try a series that diverges more quickly.  The sequence $b_n = \sqrt{n}$ clearly grows to infinity, but slowly.  The sequence &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$a_n = b_{n+1} - b_n = \sqrt{n+1} - \sqrt{n} = \frac{1}{\sqrt{n+1} + \sqrt{n}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;goes to zero.  Nevertheless, it is clear that the series $a_1 + a_2 + a_3 + ... $ diverges.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-09-22T00:51:17.320" />
  <row Id="5191" PostTypeId="1" CreationDate="2010-09-22T01:41:11.400" Score="22" ViewCount="654" Body="&lt;p&gt;A common way to define a group is as the group of structure-preserving transformations on some structured set.  For example, the symmetric group on a set $X$ preserves no structure: or, in other words, it preserves only the structure of being a set.  When $X$ is finite, what structure can the &lt;em&gt;alternating&lt;/em&gt; group be said to preserve?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a way of making the question precise, is there a natural definition of a category $C$ equipped with a faithful functor to $\text{FinSet}$ such that the skeleton of the underlying groupoid of $C$ is the groupoid with objects $X_n$ such that $\text{Aut}(X_n) \simeq A_n$?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt;  I've been looking for a purely combinatorial answer, but upon reflection a geometric answer might be more appropriate.  If someone can provide a convincing argument why a geometric answer is more natural than a combinatorial answer I will be happy to accept that answer (or Omar's answer).&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-09-28T15:37:36.763" LastActivityDate="2011-05-27T14:48:13.317" Title="What structure does the alternating group preserve?" Tags="&lt;combinatorics&gt;&lt;group-theory&gt;&lt;category-theory&gt;" AnswerCount="5" CommentCount="4" FavoriteCount="3" />
  <row Id="5192" PostTypeId="2" ParentId="5166" CreationDate="2010-09-22T02:28:15.560" Score="1" ViewCount="" Body="&lt;p&gt;When what you already have is a set of points where your curve must pass through, Bézier is not what you want; you should be using a parametric (cubic) spline.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming that your points $\mathbf{P}_i=(x_i,y_i)$, $i=1\dots n$ represent a general curve (as opposed to a function $f(x)$, for which simply applying the usual cubic spline algorithm on your points suffices), Eugene Lee proposed &lt;a href=&quot;http://www.cs.bgu.ac.il/~leonid/na105/Splines/Lee.pdf&quot; rel=&quot;nofollow&quot;&gt;centripetal parametrization&lt;/a&gt; to generate suitable parameter values $t_i$ associated with your $\mathbf{P}_i$. The prescription for generating the $t_i$ (in its most general form) is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/ffWgx.gif&quot; alt=&quot;centripetal&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\left\| \cdot \right\|$ is the (Euclidean) length, $e$ is an adjustable exponent in the interval $[0,1]$, and $t_1=0$. (A usual value for $e$ is 0.5, but $e=1$ is sometimes used as well.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From this, one applies the usual cubic spline algorithm to the sets $(t_i,x_i)$ and $(t_i,y_i)$, from which you now have your parametric spline. (The periodic spline is recommended for closed curves, and the &quot;not-a-knot&quot; spline for all other cases.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The MATLAB Spline Toolbox has support for this parametrization scheme, though it shouldn't be too hard to write your own implementation.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-09-22T02:28:15.560" />
  <row Id="5193" PostTypeId="1" CreationDate="2010-09-22T02:59:52.830" Score="6" ViewCount="170" Body="&lt;p&gt;The torsion of a curve in $\mathbb{R}^3$ indicates how much it twists around. The curl of a vector field indicates how much the vector field twists around. Is there a relation between the curl of a vector field and the torsion of a curve through that vector field at a given point?&lt;/p&gt;&#xA;" OwnerUserId="873" LastActivityDate="2010-09-22T03:59:19.987" Title="Relation between torsion of a curve and the curl of a vector field" Tags="&lt;calculus&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" />
  <row Id="5194" PostTypeId="1" AcceptedAnswerId="5195" CreationDate="2010-09-22T03:34:19.700" Score="1" ViewCount="54" Body="&lt;p&gt;Fix a finite group $G$, and look at all its irreducible representations/$\mathbb{C}$. It is said in Serre's book that &quot;there cannot be any $\mathbb{C}$-linear relation between the matrix coefficients of these irreducible representations (of course we fix a set of bases first), because of the orthogonality formulas&quot;. I don't understand the quoted part. Can anyone explain that? I think I must be missing something obvious here.&lt;/p&gt;&#xA;" OwnerUserId="325" LastActivityDate="2010-09-22T03:43:23.977" Title="No linear relation between matrix coefficients of all the irreducible repn of finite group" Tags="&lt;group-theory&gt;&lt;representation-theory&gt;" AnswerCount="1" />
  <row Id="5195" PostTypeId="2" ParentId="5194" CreationDate="2010-09-22T03:43:23.977" Score="3" ViewCount="" Body="&lt;p&gt;Essentially because a set of pairwise orthgonal nonzero vectors must be&#xA;linearly independent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here orthogonality is with respect to a complex (sesquilinear)&#xA;positive-definite inner product, and we can regard each matrix entry&#xA;as a vector of length $|G|$.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-09-22T03:43:23.977" CommentCount="1" />
  <row Id="5196" PostTypeId="2" ParentId="5193" CreationDate="2010-09-22T03:59:19.987" Score="5" ViewCount="" Body="&lt;p&gt;The paper &quot;&lt;a href=&quot;http://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.13.9738&quot; rel=&quot;nofollow&quot;&gt;Curvature Measures of 3D Vector Fields and Their Applications&lt;/a&gt;&quot; by Weinkauf et al. describes how to compute the curvature and torsion of the tangent curves in terms of the spatial derivatives of the underlying vector field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Going the other way and recovering any information about the vector field from its tangent curves is not necessarily possible. As a simple counterexample, consider that all vector fields of the form $\mathbf{v}(r,\theta,z) = f(r)\mathbf{e}_\theta$ in cylindrical coordinates have exactly the same tangent curves (circles around the $z$-axis) but completely different curls.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, the difficulty is that the properties of the tangent curve can only tell you about the derivatives of the vector field along the parallel direction; you get no information about how the vector field varies in the plane normal to the curve.&lt;/p&gt;&#xA;" OwnerUserId="856" LastActivityDate="2010-09-22T03:59:19.987" />
  <row Id="5197" PostTypeId="2" ParentId="5191" CreationDate="2010-09-22T04:22:08.603" Score="4" ViewCount="" Body="&lt;p&gt;The alternating group preserves orientation, more or less by definition. I guess you can take C to be the category of simplices together with an orientation. I.e., the objects of C are  affinely independent sets of points in some R^n together with an orientation and the morphisms are affine transformations taking the vertices of one simplex to the vertices of another (we don't require anything special for the orientation: some morphisms preserve it and some invert it). Of course this is cheating since if you actually try to define orientation you'll probably wind up with something like &quot;coset of the alternating group&quot; as the definition. On the other hand, some people find orientations of simplices to be a geometric concept, so this might conceivably be reasonable to you.&lt;/p&gt;&#xA;" OwnerUserId="1070" LastActivityDate="2010-09-22T04:22:08.603" CommentCount="2" />
  <row Id="5198" PostTypeId="1" AcceptedAnswerId="5200" CreationDate="2010-09-22T05:02:19.277" Score="3" ViewCount="208" Body="&lt;p&gt;How does one find a simple expression of the one below applying the binomial theorem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{k=1}^n k \cdot 2^k{ n \choose k}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{d}{dx}(x^k)=kx^{k-1}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(1 + x)^n = \sum_{k=0}^n { n \choose k}x^k = \sum_{k=0}^n { n \choose k}\frac{d}{dx}(x^k) = \sum_{k=0}^n { n \choose k}kx^{k-1} $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$n(1+x)^{n-1} = \sum_{k=1}^n k{ n \choose k}x^{k-1}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit 2:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$nx(1+x)^{n-1} = \sum_{k=1}^n k{ n \choose k}2^{k}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;if substitute &lt;strong&gt;x = 2&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$2n(1+2)^{n-1} = \sum_{k=1}^n k{ n \choose k}2^{k} $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this correct?&lt;/p&gt;&#xA;" LastActivityDate="2010-09-22T15:17:46.350" Title="Finding a simple expression - Binomial Theorem" Tags="&lt;homework&gt;&lt;binomial-coefficients&gt;" AnswerCount="1" CommentCount="9" />
  <row Id="5200" PostTypeId="2" ParentId="5198" CreationDate="2010-09-22T05:09:42.863" Score="3" ViewCount="" Body="&lt;p&gt;Since this looks like homework, I won't tell you how to do it with the binomial theorem and will tell you how to do it bijectively instead.  How many ways are there to choose a committee of $k$ people out of $n$ people, split the committee into Democrats and Republicans, then elect a President from that committee, for some $k$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the one hand, the answer is $\displaystyle \sum_{k=1}^{n} k \cdot 2^k {n \choose k}$.  On the other hand, one can first choose the President, then split the non-Presidents into Democrats, Republicans, and people not on the committee, then choose a party for the President, which can be done in $2n \cdot 3^{n-1}$ ways.  &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;As for how to do it through the binomial theorem, here is a different but related application.  Recall the geometric series formula&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{1}{1 - x} = \sum_{k=0}^{\infty} x^k.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What happens if we take the derivative of both sides?  We get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{1}{(1 - x)^2} = \sum_{k=1}^{\infty} kx^{k-1}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Substituting $x = \frac{1}{2}$, we then conclude that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{k=1}^{\infty} \frac{k}{2^{k-1}} = 4.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This problem requires essentially the same trick.&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-09-22T05:49:01.817" LastActivityDate="2010-09-22T05:49:01.817" CommentCount="1" />
  <row Id="5201" PostTypeId="1" AcceptedAnswerId="5202" CreationDate="2010-09-22T05:19:17.853" Score="1" ViewCount="134" Body="&lt;p&gt;Before I begin, note that this is significantly outside what I've studied, so if it's a load of crap just let me know.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Up till now, everything I've covered assumes the same domain for all variables.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm trying to define an arbitrary one-to-one relationship between elements in two different sets, and want each of the variables x and y in the formula below to be in the domain of its particular set:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;X = {X&lt;sub&gt;1&lt;/sub&gt;, X&lt;sub&gt;2&lt;/sub&gt;, X&lt;sub&gt;3&lt;/sub&gt;}&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Y = {Y&lt;sub&gt;1&lt;/sub&gt;, Y&lt;sub&gt;2&lt;/sub&gt;, Y&lt;sub&gt;3&lt;/sub&gt;}&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That is, I want:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;x ∈ X, y ∈ Y&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If possible, how would I go about specifying this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, I've phrased the formula&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;∀x ∃y [ A(x, y) ∧ ¬A(¬x, y) ]&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Where A(x, y) indicates an relationship between x and y. Is this statement well formed? Particularly the ¬x in ¬A(¬x, y)? And does it mean what I think it means - That A(x, y) for all x ∈ X except the x in question, and for the y in question, must be false in order for the statement ot be true? That is, does it define a one-to-one relationship?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks a lot,&#xA;Wyatt&lt;/p&gt;&#xA;" OwnerUserId="2058" LastActivityDate="2010-09-22T20:28:58.890" Title="different domains for each variable in predicate logic" Tags="&lt;logic&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="5202" PostTypeId="2" ParentId="5201" CreationDate="2010-09-22T05:53:00.740" Score="5" ViewCount="" Body="&lt;p&gt;No, I don't think it makes sense to write $\neg x$ in this context. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The simplest way to specify that $x\in X$ and $y\in Y$ is to make them clauses in your statement. For example, to write that for every $x\in X$ there should exist a $y\in Y$ such that $A(x,y)$ holds, you would usually write:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$$\forall x \left(x\in X\rightarrow \exists y(y\in Y\wedge A(x,y)\right).$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For the second part: to say that no other $z\in X$ will also be associated to $y$, the easiest why is to say that anyone who is associated to $y$ is going to be equal to $x$; so you would want a clause $A(z,y)\rightarrow x=z$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To put it all together:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$$\forall x\left( x\in X\rightarrow \exists y(y\in Y\wedge A(x,y)\wedge \forall z(z\in X\wedge A(z,y)\rightarrow x=z))\right).$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It is also possible to use common (but perhaps slightly abusive) notation and write&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$$\forall x\in X\Bigl(\exists y\in Y\Bigl(\forall z\in X\Bigl(A(x,y)\wedge (A(z,y)\rightarrow x=z)\Bigr)\Bigr).$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Added:&lt;/em&gt; As noted in several comments, the above does not completely describe a one-to-one &lt;em&gt;function&lt;/em&gt;; it only describes a relation with domain $X$, codomain $Y$, such that for each $y\in Y$ there is at most one $x\in X$ such that $A(x,y)$ holds; this does not fully describe a function (unless you happen to know some other things about $X$ and $Y$, e.g., that they are both finite and of equal size, in which case the other condition will follow), nor a bijective function. In general, the clause you want for &quot;function-ness&quot; is the dual of the condition you have: for all $x\in X$, and for all $y\in Y$, $A(x,y)\wedge A(x,y')\rightarrow y=y'$ (so each input has at most one output). For bijectivity (so you have a true correspondence) you also need that for all $y\in Y$ there exists $x\in X$ such that $A(x,y)$. &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-09-22T15:46:56.413" LastActivityDate="2010-09-22T15:46:56.413" CommentCount="1" />
  <row Id="5203" PostTypeId="2" ParentId="4627" CreationDate="2010-09-22T07:25:38.223" Score="1" ViewCount="" Body="&lt;p&gt;To deal well with substitution of variables in differential equations you should know when there is a differential operator and when there is a derivative. In the first case a variable is linked with differential operator and changing it you should also change a differential operator, in the second a derivative is just a function so you can simultaneously change the same argument in all equation's functions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's see on your equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{dQ}{dt}=\frac{a+Q}{b}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And rewrite it using a differential operator notation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{d}{dt}[Q(t)]=\frac{a+Q(t)}{b}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's define a functional map $F_1(f)(x)=f(x)-a$, obvious this map is bijection and let $g$ be a solution of $F_1(g)=Q$. So $Q(t)=F_1(g)(t)=g(t)-a$ and we rewrite the equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{d}{dt}[g(t)-a]=\frac{a+g(t)-a}{b}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reduce it and calculate a derivative:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$g'(t)=\frac{1}{b}g(t)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And back to a differential operator notation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{d}{dt}[g(t)]=\frac{1}{b}g(t)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Repeat the trick with a map $F_2(f)(x)=f(\frac{1}{b}x)$, let $w$ be a solution of $F_2(w)=g$, so $g(t)=F_2(w)(t)=w(\frac{1}{b}t)$ and substitute it into the equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{d}{dt}[w(\frac{1}{b}t)]=\frac{1}{b}w(\frac{1}{b}t)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Calculate a derivative and reduce:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$w'(\frac{1}{b}t)=w(\frac{1}{b}t)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But there is no a differential operator, so we change an argument $\frac{1}{b}t$ to $t$ in whole equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$w'(t)=w(t)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's known that a solution is $w(t)=C_1e^t$ and $Q=F_1(F_2(w))$ so $Q(t)=C_1e^{\frac{1}{b}t}-a$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a very verbose solution, but then understand it you can just use rules of substitution of variables and solve an equation in a few lines.&lt;/p&gt;&#xA;" OwnerUserId="2016" LastActivityDate="2010-09-22T07:25:38.223" CommunityOwnedDate="2010-09-22T07:25:38.223" />
  <row Id="5205" PostTypeId="1" CreationDate="2010-09-22T09:36:14.377" Score="2" ViewCount="372" Body="&lt;p&gt;I apologise for the really basic question. This didn't really fit on any other StackExchange website so the Maths one was the closest one where I could ask.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Really Basic Question- What are 4 sided shapes called again? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Like how triangles are 3 sided shapes, octagons are 8 sided shapes, ... What are the 4 sided ones called then?&lt;/p&gt;&#xA;" OwnerUserId="2062" LastEditorUserId="171" LastEditorDisplayName="" LastEditDate="2010-09-22T14:39:18.223" LastActivityDate="2010-09-22T14:39:18.223" Title="What Are 4 Sided Shapes Called Again?" Tags="&lt;geometry&gt;&lt;terminology&gt;" AnswerCount="4" CommentCount="4" FavoriteCount="1" />
  <row Id="5206" PostTypeId="2" ParentId="5205" CreationDate="2010-09-22T09:36:33.927" Score="1" ViewCount="" Body="&lt;p&gt;Wait, never mind. It's quadrilateral. :)&lt;/p&gt;&#xA;" OwnerUserId="2062" LastActivityDate="2010-09-22T09:36:33.927" />
  <row Id="5207" PostTypeId="2" ParentId="5205" CreationDate="2010-09-22T10:24:15.247" Score="5" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Quadrilateral&quot; rel=&quot;nofollow&quot;&gt;The word &lt;strong&gt;quadrilateral&lt;/strong&gt; is made of the words &lt;strong&gt;quad (meaning &quot;four&quot;)&lt;/strong&gt; and &lt;strong&gt;lateral (meaning &quot;of sides&quot;)&lt;/strong&gt;.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="705" LastActivityDate="2010-09-22T10:24:15.247" />
  <row Id="5208" PostTypeId="2" ParentId="4851" CreationDate="2010-09-22T10:44:53.467" Score="2" ViewCount="" Body="&lt;p&gt;Let's remove this question from the list of unanswered questions by up-voting this community wiki answer stating that Agusti Roig's comment above contains the right answer.&lt;/p&gt;&#xA;" OwnerUserId="367" LastActivityDate="2010-09-22T10:44:53.467" CommunityOwnedDate="2010-09-22T10:44:53.467" />
  <row Id="5209" PostTypeId="2" ParentId="5201" CreationDate="2010-09-22T11:06:05.337" Score="0" ViewCount="" Body="&lt;p&gt;There are lots of ways to write a formula A(x,y) that defines a bijection between two sets. If you're trying to keep the depth of quantifiers to a minimum, you could use&#xA;$$&#xA;\begin{align}&#xA;&amp;amp;  (\forall x)(\exists y) [ x \in X \to y \in Y \land A(x,y)] \\&#xA;&amp;amp; {} \land (\forall y)(\exists x) [ y \in Y \to x \in X \land A(x,y)]\\&#xA;&amp;amp; {} \land (\forall x,x',y,y') ([x \in X \land y \in Y) \\&#xA;&amp;amp; \qquad {} \to [(A(x,y) \land A(x',y) \to x=x'] &#xA; \land [A(x,y) \land A(x,y') \to y = y'])]\\&#xA;\end{align}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This can be written more elegantly with the abbreviations $(\forall x \in X) $ and $(\exists x \in X)$:&#xA;$$&#xA;\begin{align}&#xA;(\forall x \in X)\Phi &amp;amp;\equiv (\forall x) [ x \in X \to \Phi] \\&#xA;(\exists x \in X)\Phi &amp;amp;\equiv (\exists x) [ x \in X \land \Phi] \\&#xA;\end{align}&#xA;$$&#xA;like so: &#xA;$$&#xA;\begin{align}&#xA;&amp;amp;  (\forall x \in X)(\exists y\in Y) A(x,y) \land (\forall y \in Y )(\exists x \in X) A(x,y)\\&#xA;&amp;amp; {} \land (\forall x \in X)(\forall x' \in X)(\forall y \in Y)(\forall y' \in Y)([(A(x,y) \land A(x',y) \to x=x'] \\&#xA;&amp;amp; \qquad \qquad {} \land [A(x,y) \land A(x,y') \to y = y'])\\&#xA;\end{align}&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="630" LastEditorUserId="630" LastEditorDisplayName="" LastEditDate="2010-09-22T12:14:16.893" LastActivityDate="2010-09-22T12:14:16.893" />
  <row Id="5210" PostTypeId="2" ParentId="4600" CreationDate="2010-09-22T13:08:04.003" Score="2" ViewCount="" Body="&lt;p&gt;You are absolutely right with your understanding. The notion of L-structure is very loose, it is just a way of attaching (any) meaning to the symbols. It is only after this that it makes sense to ask whether a given formal statement is true in a given L-structure - i.e. whether the L-structure is a &lt;em&gt;model&lt;/em&gt; of your statement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Very roughly, the area of model theory called Classification theory looks how many isomorphism classes of models for a given set of statements there are - of course you have to fix a cardinality, otherwise two models can't be isomorphic. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could continue your exercise of classifying all L-structures of cardinality 4 (which you did correctly) by classifying all isomorphism classes of such L-structures. An isomorphism of L-structures would be a bijection which preserves the relation and is compatible with the functions (see any book for a formal definition). You will see that lots of them are non-isomorphic - but maybe you won't want to make a complete list, it seems quite a lot of work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This was then the classification of models of the empty set of sentences. If you now require your L-structures to render a certain set of sentences true, you get a subset of your L-structures (usually easier to see) and can ask the same about this subset - your classification from before is of course still valid. This gives you a small taste of Classification theory (which of course in reality proceeds by very advanced and more effective methods, which don't feel like this exercise at all).&lt;/p&gt;&#xA;" OwnerUserId="2048" LastActivityDate="2010-09-22T13:08:04.003" />
  <row Id="5211" PostTypeId="2" ParentId="4862" CreationDate="2010-09-22T13:18:34.913" Score="0" ViewCount="" Body="&lt;p&gt;I have suggested three alternative paths to the usual calculus one in &lt;a href=&quot;http://math.stackexchange.com/questions/5018/how-to-start-with-mathematics/5188#5188&quot;&gt;this&lt;/a&gt; answer!&lt;/p&gt;&#xA;" OwnerUserId="2048" LastActivityDate="2010-09-22T13:18:34.913" CommunityOwnedDate="2010-09-22T13:18:34.913" />
  <row Id="5212" PostTypeId="2" ParentId="4600" CreationDate="2010-09-22T13:21:54.113" Score="1" ViewCount="" Body="&lt;p&gt;The reason that you have so much freedom is that you have not required that the structure satisfy any axioms at all. This is somewhat like saying &quot;Let $f$ be an arbitrary function&quot; - of course there are lots of examples.  If you added a list of axioms that the structure has to satisfy, it would reduce the freedom you have in constructing the structure. &lt;/p&gt;&#xA;" OwnerUserId="630" LastActivityDate="2010-09-22T13:21:54.113" />
  <row Id="5213" PostTypeId="2" ParentId="5205" CreationDate="2010-09-22T13:57:14.117" Score="1" ViewCount="" Body="&lt;p&gt;Shorter version is &quot;quadrangle&quot;.&lt;/p&gt;&#xA;" OwnerUserId="817" LastActivityDate="2010-09-22T13:57:14.117" CommentCount="3" />
  <row Id="5214" PostTypeId="2" ParentId="5205" CreationDate="2010-09-22T14:11:12.273" Score="0" ViewCount="" Body="&lt;p&gt;I have a draft article where I 'should' use 'quadrilateral' dozens of times. I refuse, stating up front that I will use 'quad' instead.&lt;/p&gt;&#xA;" OwnerUserId="1782" LastActivityDate="2010-09-22T14:11:12.273" CommunityOwnedDate="2010-09-22T14:11:12.273" CommentCount="2" />
  <row Id="5215" PostTypeId="1" CreationDate="2010-09-22T14:22:38.240" Score="1" ViewCount="59" Body="&lt;p&gt;Given $X_i, Y_i \in SO_3, i=1..n, n&gt;2$, find $A,B$ that satisfies $AX_iB=Y_i$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $n=1$, there are an infinite number of solutions, a trivial one is:&lt;br&gt;&#xA;$B=I$&lt;br&gt;&#xA;$A=Y_1X_1^T$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $n=2$, there should be exactly one solution. One could do the following:&lt;br&gt;&#xA;Compute $A$ from the equation where $i=1$: $A=Y_1B^{-1}X_1^{-1}$,&lt;br&gt;&#xA;Substituting this into equation where $i=2$: $Y_1B^{-1}X_1^{-1}X_2B=Y_2$,&lt;br&gt;&#xA;Now, let $C=X_1^{-1}X_2$ and $D=Y_1^{-1}Y_2$,&lt;br&gt;&#xA;We get: $B^{-1}CB=D$,&lt;br&gt;&#xA;Which can be rewritten as: $CB=BD$&lt;br&gt;&#xA;I'm already lost at this point. How do I solve for $B$, given $D,C$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And now for $n&gt;2$, how can I find a solution that minimizes the least squares error? Is it possible to rearrange the $n$ equations into a linear system at all? Or do I need to look into using some kind of iterative approach or linear programming instead?&lt;/p&gt;&#xA;" OwnerUserId="2063" LastEditorUserId="171" LastEditorDisplayName="" LastEditDate="2010-09-22T14:30:13.770" LastActivityDate="2010-09-22T14:48:06.623" Title="Looking for closed-form least squares solution of a simplified inverse kinematics problem" Tags="&lt;linear-algebra&gt;" AnswerCount="1" />
  <row Id="5216" PostTypeId="1" AcceptedAnswerId="5242" CreationDate="2010-09-22T14:32:45.077" Score="2" ViewCount="132" Body="&lt;p&gt;I have an array of &quot;lines&quot; each defined by 2 points. I am working with only the line segments lying between those points. I need to search lines that could continue one another (relative to some angle) and lie on the same line (with some offset)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I mean I had something like 3 lines&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/4uzcj.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I solved some mathematical problem (formulation of which is my question) and got understanding that there are lines that could be called relatively one line (with some angle K and offset J)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/9YEFd.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And of course by math formulation I meant some kind of math formula like&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/NsKKl.gif&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so I know the algorithm I used such simple approach like one presented by Victor Liu &lt;a href=&quot;http://stackoverflow.com/questions/3769885/how-to-formulate-such-problem-mathematicaly-line-continuation-search/3770275#3770275&quot;&gt;http://stackoverflow.com/questions/3769885/how-to-formulate-such-problem-mathematicaly-line-continuation-search/3770275#3770275&lt;/a&gt; but its Pusedo solution code and what I need is problem formulation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My problem is its formalisation into math using some Matrix and other math words (to write into some paper=) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;guys - I am a programmer - sorry=) &lt;/p&gt;&#xA;" OwnerUserId="2065" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-09-22T17:17:36.690" LastActivityDate="2010-09-22T17:42:06.953" Title="How to formulate such problem mathematicaly? (line continuation search)" Tags="&lt;geometry&gt;&lt;algorithms&gt;" AnswerCount="1" CommentCount="9" FavoriteCount="0" />
  <row Id="5217" PostTypeId="1" AcceptedAnswerId="5227" CreationDate="2010-09-22T14:47:45.047" Score="4" ViewCount="128" Body="&lt;p&gt;The problem asks us to calculate:&lt;br&gt;&#xA;$ \sum_{i = 0}^{n}(-1)^i \binom{n}{i} \binom{n}{n-i}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The way I tried solving is:&lt;br&gt;&#xA;The given sum is the coefficient of $x^n$ in $ (1+x)^n(1-x)^n $, which is $ (1 - x^2)^n $.&#xA;The coefficient of $x^n$ in $(1 -x^2)^n$ is $ (-1)^{n/2} \binom{n}{n/2} $.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Am I doing it right?&lt;/p&gt;&#xA;" OwnerUserId="813" LastEditorUserId="813" LastEditorDisplayName="" LastEditDate="2010-09-22T15:57:08.537" LastActivityDate="2010-09-22T15:57:08.537" Title="a combinatorial exercise" Tags="&lt;combinatorics&gt;&lt;binomial-coefficients&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="5218" PostTypeId="2" ParentId="5215" CreationDate="2010-09-22T14:48:06.623" Score="1" ViewCount="" Body="&lt;p&gt;$\mathbf{C}\mathbf{B}-\mathbf{B}\mathbf{D}=\mathbf{0}$ is a &lt;a href=&quot;http://en.wikipedia.org/wiki/Sylvester_equation&quot; rel=&quot;nofollow&quot;&gt;Sylvester equation&lt;/a&gt;, for which algorithms like &lt;a href=&quot;http://doi.acm.org/10.1145/361573.361582&quot; rel=&quot;nofollow&quot;&gt;Bartels-Stewart&lt;/a&gt; or &lt;a href=&quot;http://ieeexplore.ieee.org/xpls/abs_all.jsp?arnumber=1102170&quot; rel=&quot;nofollow&quot;&gt;Hessenberg-Schur&lt;/a&gt; apply. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I haven't seen your more general problem in my notes, but I'll update this answer when I find something.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-09-22T14:48:06.623" CommentCount="1" />
  <row Id="5219" PostTypeId="1" AcceptedAnswerId="5228" CreationDate="2010-09-22T14:57:18.170" Score="2" ViewCount="460" Body="&lt;p&gt;Assume $k,n$ are positive integers. The sum&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\dfrac{1}{k+1}+\dfrac{1}{k+2}+\dfrac{1}{k+3}+\ldots +\dfrac{1}{k+n-1}+\dfrac{1}{k+n}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is not an integer. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The case $k=0$ is proved in this question &quot;&lt;a href=&quot;http://math.stackexchange.com/questions/2746/is-there-an-elementary-proof-that-k1n-1-k-is-never-an-integer&quot;&gt;Is there an elementary proof that $\sum_{k=1}^{n}1/k$ is never an integer?&lt;/a&gt;&quot;&lt;/p&gt;&#xA;" OwnerUserId="752" LastActivityDate="2010-09-22T16:43:28.570" Title="How do I prove this sum is not an integer" Tags="&lt;number-theory&gt;&lt;sequences-and-series&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="0" />
  <row Id="5220" PostTypeId="1" AcceptedAnswerId="5222" CreationDate="2010-09-22T15:05:14.607" Score="2" ViewCount="152" Body="&lt;h3&gt;Background:&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;I was studying Theorem 2.3.3 from Introduction to Graph Theory by W. B. West. The main idea of his proof is as follows:&#xA;&lt;br&gt;&#xA;&lt;b&gt;T&lt;/b&gt;, resulting tree.&#xA;&lt;br&gt;&#xA;&lt;b&gt;T&lt;sup&gt; *&lt;/sup&gt;&lt;/b&gt;, spanning tree of minimum weight.&#xA;&lt;br&gt;&#xA;Let, T $\neq$ T&lt;sup&gt; *&lt;/sup&gt;, then there are edges in T&lt;sup&gt; *&lt;/sup&gt; that are not in T.&#xA;We first consider only 1 edge. Let's name it e. Hence, T&lt;sup&gt; *&lt;/sup&gt; has all the edges that are in T, except e. Now, if we add e to T&lt;sup&gt; *&lt;/sup&gt;, we should get a cycle, say 'C'. This implies, C has an edge that is not in T. Let's name it e'. Now we get spanning tree T&lt;sup&gt; *&lt;/sup&gt;+ e - e'.&#xA;Now, let us think what happened when Kruskal ran and produced T. It examined all edges in G. Including e and e'. But, it included only e. So, w(e)$\leq$w(e').&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Thus T&lt;sup&gt; *&lt;/sup&gt;+ e - e' is a spanning tree with weight at most T&lt;sup&gt; *&lt;/sup&gt; that agrees with T for a &lt;em&gt;longer initial list&lt;/em&gt; of edges than T&lt;sup&gt; *&lt;/sup&gt; does.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;What does it mean actually?&lt;/li&gt;&#xA;&lt;li&gt;Anybody knows any other proof?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1869" LastActivityDate="2010-09-22T15:16:29.410" Title="Prove : In a connected weighted Graph G, Kruskal's Algorithm constructs a minimum spanning tree" Tags="&lt;graph-theory&gt;&lt;proof-strategy&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="5221" PostTypeId="2" ParentId="4851" CreationDate="2010-09-22T15:07:40.383" Score="-1" ViewCount="" Body="&lt;p&gt;Let's honor Rasmus' creative handling of the software restrictions by downvoting &lt;a href=&quot;http://math.stackexchange.com/questions/4851/cover-e-b-gives-a-homeomorphism-e-aute-b/5221#5221&quot;&gt;this&lt;/a&gt; self-referential community wiki answer.&lt;/p&gt;&#xA;" OwnerUserId="2048" LastActivityDate="2010-09-22T15:07:40.383" CommunityOwnedDate="2010-09-22T15:07:40.383" CommentCount="1" />
  <row Id="5222" PostTypeId="2" ParentId="5220" CreationDate="2010-09-22T15:16:29.410" Score="2" ViewCount="" Body="&lt;p&gt;If T is a tree and one joins two vertices of this tree that are not already joined by an edge with a new edge, one gets a unique circuit C. Now, what happens if one removes an edge from C different from the edge just added to T? One gets a new tree.&lt;/p&gt;&#xA;" OwnerUserId="1369" LastActivityDate="2010-09-22T15:16:29.410" CommentCount="2" />
  <row Id="5224" PostTypeId="2" ParentId="4862" CreationDate="2010-09-22T15:19:11.307" Score="1" ViewCount="" Body="&lt;p&gt;I do not know what &quot;Calculus 1-3&quot; entails, but if the terms &quot;complete ordered field&quot; and &quot;least upper bound property&quot; don't mean anything to you, I'd start with a first course in real analysis. The usual stuff: Cauchy sequences, absolutely and conditionally convergent series, the $\varepsilon$-$\delta$ definition of continuity, the intermediate value theorem, a rigorous construction of the Riemann integral using upper and lower sums$\ldots$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All (at least non-nonstandard) analysis flows from this, up to and including metric spaces, topology (to a lesser degree admittedly), differential geometry, complex analysis, functional analysis and meassure theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Paraphrasing the late Randy Pausch: Fundamentals. Fundamentals. Fundamentals. You need to have the basics straight or the fancy stuff ain't gonna work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From there I'd look at linear algebra. As has been mentioned earlier, it's nice and concrete but still very useful and shows up in a lot of places. For example in abstract algebra when studying field extensions(a).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Speaking of abstract algebra, it takes up an interesting place in undergraduate mathematics. I find it hard to explain, but it's like it's sitting in a corner off by itself doing its own thing - at least until you get to for example algebraic topology. But algebraic topology isn't usually taught to undergraduates (at least that I know of), so algebra feels rather disconnected from the rest of an undergraduate mathematics education.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what it boils down to is that I'd stay away from abstract algebra beyond a first course (because all undergraduate math students need to at least know groups, rings and fields) unless you either 1) loved the first course, 2) find out that algebraic tools are needed for something else you like, or 3) the rules say you must do more algebra.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(a) Though probably not in a first course in abstract algebra since Galois theory can't really be taught until after you've done the basics of groups, rings and fields which take up at least a semester.&lt;/p&gt;&#xA;" OwnerUserId="1269" LastActivityDate="2010-09-22T15:19:11.307" CommunityOwnedDate="2010-09-22T15:19:11.307" />
  <row Id="5225" PostTypeId="2" ParentId="4498" CreationDate="2010-09-22T15:19:57.477" Score="2" ViewCount="" Body="&lt;p&gt;I'm not sure you've asked a question with an answer, but I'll try to address what I can.  Surely people have looked at small complete groups, but people do not tend to publish such things.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The smallest complete group that is not S4, Aut(simple), nor Hol(p) is S3 x Hol(5).  Direct products of complete groups are often but certainly not always complete.  You might try to find reasonable conditions on when the direct product of completes is complete.  Other people have written down some conditions, but I can't recall where.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An infinite family omitted are the AΓL(1,q) of semi-linear transformations of a one-dimensional vector space.  If q is prime, this is just Hol(q).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The next two groups not covered might be interesting to understand.  They are of the form G⋉V where G is complete and V is an absolutely irreducible G-module (G being AΓL(1,9) or Hol(5), V being smallest non-trivial dimension and rational).  I don't think all such groups are complete, but it would be interesting to know conditions on G,V to make them complete.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't think one will get any good sort of classification as there are just weird complete groups.  &lt;s&gt;One group V of order 128 has an automorphism G of order 7 (and several other outer automorphisms), but the semi-direct product G⋉V of order 128*7 is complete.&lt;/s&gt; (SmallGroup(432,520) and SG(1344,816) are almost as weird).  I at least find it a little disturbing to have small subgroups of Hol(V) be complete when neither V nor Hol(V) is complete.  The other G⋉V examples are like this too, but perhaps I find it less disturbing when V is abelian.&lt;/p&gt;&#xA;" OwnerUserId="583" LastEditorUserId="583" LastEditorDisplayName="" LastEditDate="2010-09-30T15:23:13.437" LastActivityDate="2010-09-30T15:23:13.437" CommentCount="3" />
  <row Id="5226" PostTypeId="1" AcceptedAnswerId="13394" CreationDate="2010-09-22T15:35:46.090" Score="4" ViewCount="102" Body="&lt;p&gt;Call a finite group A &lt;strong&gt;affine&lt;/strong&gt; if it has a normal, self-centralizing, complemented, elementary abelian subgroup V.  Such a group A is a semi-direct product G⋉V where V is a vector space of dimension n over GF(p) and G is a group of matrices in GL(n,p).  The elements of A can be written as matrices [ g, v ; 0, 1 ] where g in G, v in V and 0,1 are row vectors of the appropriate length.  Conversely, given G ≤ GL(V), V a vector space over GF(p), A=G⋉V is an affine group.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An example is the full affine group, AGL(n,p) where G is GL(n,p) and V is GF(p)^n, that is A is all (n+1)×(n+1) matrices of the form [ g, v ; 0, 1 ] where g is n×n invertible, v is anything, 0 is a zero vector, and 1 is just a 1×1 identity matrix.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;V becomes a G-module and its G-module structure has a large influence on the group theoretic structure of A.  In particular, V contains no &quot;trivial&quot; (central) summand as a G-module iff A is centerless.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Supposing A is centerless, what conditions on V ensure A is a complete group, that is, so that A is also &quot;outerless&quot;?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;See the previous questions:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://math.stackexchange.com/questions/4238/finite-non-abelian-group-with-centre-but-no-outer-automorphism&quot;&gt;4238: Finite non-abelian group with centre but no outer automorphism&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://math.stackexchange.com/questions/4498/classification-of-small-complete-groups&quot;&gt;4498: Classification of small complete groups&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="583" LastActivityDate="2010-12-07T16:53:38.637" Title="When is an affine group complete?" Tags="&lt;group-theory&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="5227" PostTypeId="2" ParentId="5217" CreationDate="2010-09-22T15:36:32.957" Score="4" ViewCount="" Body="&lt;p&gt;Your solution is correct for even $n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $n$ is odd then your last sentence should read &quot;The coefficient of $x^n$ in $(1-x^2)^n$ is $0$.&quot; This is because only even powers of $x$ occur when expanding $(1-x^2)^n$.&lt;/p&gt;&#xA;" OwnerUserId="367" LastEditorUserId="367" LastEditorDisplayName="" LastEditDate="2010-09-22T15:42:07.033" LastActivityDate="2010-09-22T15:42:07.033" CommentCount="1" />
  <row Id="5228" PostTypeId="2" ParentId="5219" CreationDate="2010-09-22T15:38:13.810" Score="6" ViewCount="" Body="&lt;p&gt;The proof that &lt;a href=&quot;http://math.stackexchange.com/questions/2752&quot;&gt;I gave in that thread&lt;/a&gt; works just as well here. It depends only on the fact that in any contiguous sequence of integers (here denominators) the maximal power $\rm 2^k$ that divides any element occurs in precisely one element. Indeed, after the first (necessarily odd) multiple of $\rm 2^k$, the next multiple would, by contiguity, be an even multiple, so a multiple of $\rm 2^{k+1}$, contra maximality. Here is said proof:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\;$ Since there is a unique denominator $\rm 2^k$ having maximal power of $2$, upon multiplying all terms through by $\rm 2^{k-1}$ one deduces the contradiction that $\rm\; a/2 = b/c \;$ with $\rm \; a,\ c \:$ odd. As an example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\quad\quad\quad\quad\quad\quad m = \frac{1}{2} + \frac{1}{3} +\; \frac{1}{4} \;+\; \frac{1}{5} + \frac{1}{6} + \frac{1}{7} $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\quad\quad\Rightarrow\quad\;\; 2m = \; 1 + \frac{2}{3} +\; \frac{1}{2} \;+\; \frac{2}{5} + \frac{1}{3} + \frac{2}{7} $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\quad\quad\Rightarrow\quad -\frac{1}{2}  = \; 1 + \frac{2}{3} - 2m  + \frac{2}{5} + \frac{1}{3} + \frac{2}{7}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The prior sum has all odd denominators so reduces to a fraction with odd denominator $\rm d\:|3\cdot 5\cdot 7$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Note&lt;/strong&gt; $\:$ I purposely avoided any use of valuation theory because Anton requested an &quot;elementary&quot; solution. The above proof can easily be made comprehensible to a high-school student. The proof is trivial to anyone who knows valuation theory: namely the sum has a lone dominant term with 2-adic value $\rm v_2&amp;lt;0\:$ so, by the &lt;strong&gt;domination principle&lt;/strong&gt;, the sum has 2-adic value $\rm v_2&amp;lt;0\:,\:$ i.e. the sum has even denominator in lowest terms, so it is nonintegral.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-09-22T16:43:28.570" LastActivityDate="2010-09-22T16:43:28.570" CommentCount="1" />
  <row Id="5229" PostTypeId="2" ParentId="5219" CreationDate="2010-09-22T15:38:55.540" Score="2" ViewCount="" Body="&lt;p&gt;{Differences of harmonic numbers are not integers}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider $\sum_{k=m}^n {1\over k}$. Find the&#xA;largest power of $2^r$ that divides one of the denominators between $m$ and $n$.&#xA;There can be no even multiples of $2^r$ between $m$ and $n$, hence there&#xA; is only &lt;strong&gt;one&lt;/strong&gt; odd multiple.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore $2^{r-1}(1/m+\cdots+1/n)$ is equal to $1/2o_1$ plus a bunch of&#xA;fractions with odd denominator. Adding them gives one fraction with&#xA;an odd denominator, write it as $x/o_2$.&#xA;Here $o_1$ and $o_2$ are odd integers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have $2^{r-1}(1/m+\cdots+1/n)=1/2o_1+x/o_2$ so that&#xA;$${1\over m}+\cdots+{1\over n}={2xo_1+o_2\over 2^r o_1 o_2}.$$&#xA;Therefore $2^r$ divides the denominator.&#xA;If $n\not=m$, then $r\geq1$ and the sum is not an integer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cf. Exercise 6.21 (page 311) of Graham, Knuth, and Patashnik.&lt;/p&gt;&#xA;" OwnerUserId="940" LastActivityDate="2010-09-22T15:38:55.540" />
  <row Id="5230" PostTypeId="1" AcceptedAnswerId="5237" CreationDate="2010-09-22T15:55:06.450" Score="2" ViewCount="149" Body="&lt;p&gt;Suppose $\mu$ is a measure. Is there any difference in meaning between the notation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\int f(x)d\mu$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and the notation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\int f(x) \mu(dx)$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many thanks.&lt;/p&gt;&#xA;" OwnerUserId="1820" LastEditorUserId="1820" LastEditorDisplayName="" LastEditDate="2010-09-22T16:27:14.247" LastActivityDate="2010-09-22T16:55:27.000" Title="Notation question: Integrating against a measure" Tags="&lt;notation&gt;&lt;integral&gt;&lt;measure-theory&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="5231" PostTypeId="1" AcceptedAnswerId="5961" CreationDate="2010-09-22T16:05:06.420" Score="1" ViewCount="239" Body="&lt;p&gt;I have coordinates for 4 vertices/points that define a plane and the normal/perpendicular.&#xA;The plane has an arbitrary rotation applied to it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I 'un-rotate'/translate the points so that the plane has rotation 0 on x,y,z ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've tried to get the plane rotation from the plane's normal:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;rotationX = atan2(normal.z,normal.y);&#xA;rotationY = atan2(normal.z,normal.x);&#xA;rotationZ = atan2(normal.y,normal.x);&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Is this correct ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do I apply the inverse rotation to the position vectors ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've tried to create a matrix with those rotations and multiply it with the vertices,&#xA;but it doesn't look right.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the moment, I've wrote a simple test using &lt;a href=&quot;http://processing.org/&quot; rel=&quot;nofollow&quot;&gt;Processing&lt;/a&gt; and can be seen &lt;a href=&quot;http://lifesine.eu/so/vertex_rotation/&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;float s = 50.0f;//scale/unit&#xA;PVector[] face = {new PVector(1.08335042,0.351914703846,0.839020013809),&#xA;new PVector(-0.886264681816,0.69921118021,0.839020371437),&#xA;new PVector(-1.05991327763,-0.285596489906,-0.893030643463),&#xA;new PVector(0.909702301025,-0.63289296627,-0.893030762672)};&#xA;PVector n = new PVector(0.150384, -0.500000, 0.852869);&#xA;PVector[] clone;&#xA;&#xA;void setup(){&#xA;  size(400,400,P3D);&#xA;  smooth();&#xA;  clone = unRotate(face,n,true);&#xA;}&#xA;&#xA;void draw(){&#xA;  background(255);&#xA;  translate(width*.5,height*.5);&#xA;  if(mousePressed){&#xA;    rotateX(map(mouseY,0,height,0,TWO_PI));&#xA;    rotateY(map(mouseX,0,width,0,TWO_PI));&#xA;  }&#xA;  stroke(128,0,0);&#xA;  beginShape(QUADS);&#xA;    for(int i = 0 ; i &amp;lt; 4; i++) vertex(face[i].x*s,face[i].y*s,face[i].z*s);&#xA;  endShape();&#xA;  stroke(0,128,0);&#xA;  beginShape(QUADS);&#xA;    for(int i = 0 ; i &amp;lt; 4; i++) vertex(clone[i].x*s,clone[i].y*s,clone[i].z*s);&#xA;  endShape(); &#xA;}&#xA;//get rotation from normal&#xA;PVector getRot(PVector loc,Boolean asRadians){&#xA;  loc.normalize();&#xA;  float rz = asRadians ? atan2(loc.y,loc.x) : degrees(atan2(loc.y,loc.x));&#xA;  float ry = asRadians ? atan2(loc.z,loc.x) : degrees(atan2(loc.z,loc.x));&#xA;  float rx = asRadians ? atan2(loc.z,loc.y) : degrees(atan2(loc.z,loc.y));&#xA;  return new PVector(rx,ry,rz);&#xA;}&#xA;//translate vertices&#xA;PVector[] unRotate(PVector[] verts,PVector no,Boolean doClone){&#xA;  int vl = verts.length;&#xA;  PVector[] clone;&#xA;  if(doClone) {&#xA;    clone = new PVector[vl];&#xA;    for(int i = 0; i&amp;lt;vl;i++) clone[i] = PVector.add(verts[i],new PVector());&#xA;  }else clone = verts;&#xA;  PVector rot = getRot(no,false);&#xA;&#xA;  PMatrix3D rMat = new PMatrix3D();&#xA;  rMat.rotateX(-rot.x);rMat.rotateY(-rot.y);rMat.rotateZ(-rot.z);&#xA;  for(int i = 0; i&amp;lt;vl;i++) rMat.mult(clone[i],clone[i]);&#xA;  return clone;&#xA;}&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Any syntax/pseudo code or explanation is useful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What trying to achieve is this:&#xA;If I have a rotated plane:&#xA;&lt;img src=&quot;http://i.imgur.com/bZ1fn.png&quot; alt=&quot;rotated plane&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can move the vertices to have something that would have no rotation:&#xA;&lt;img src=&quot;http://i.imgur.com/ogFR0.png&quot; alt=&quot;plane with no rotations&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;UPDATE:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;@muad&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure I understand. I thought I was using matrices for rotations.&#xA;PMatrix3D's rotateX,rotateY,rotateZ calls should done the rotations for me.&#xA;Doing it manually would be declaring 3d matrices and multiplying them.&#xA;Here's a little snippet to illustrate this:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; PMatrix3D rx = new PMatrix3D(1,          0,          0,  0,&#xA;                               0, cos(rot.x),-sin(rot.x),  0,&#xA;                               0, sin(rot.x),cos(rot.x) ,  0,&#xA;                               0,          0,          0,  1);&#xA;  PMatrix3D ry = new PMatrix3D(cos(rot.y), 0,sin(rot.y),  0,&#xA;                                        0, 1,0         ,  0,&#xA;                              -sin(rot.y), 0,cos(rot.y),  0,&#xA;                                        0, 0,0         ,  1);&#xA;  PMatrix3D rz = new PMatrix3D(cos(rot.z),-sin(rot.z), 0, 0,&#xA;                               sin(rot.z), cos(rot.z), 0, 0,&#xA;                               0         ,          0, 1, 0,&#xA;                               0         ,          0, 0, 1);&#xA;  PMatrix3D r = new PMatrix3D();&#xA;  r.apply(rx);r.apply(ry);r.apply(rz);&#xA;&#xA;  //test&#xA;  PMatrix rmat = new PMatrix3D();rmat.rotateX(rot.x);rmat.rotateY(rot.y);rmat.rotateZ(rot.z);&#xA;  float[] frmat = new float[16];rmat.get(frmat);&#xA;  float[] fr    = new float[16];r.get(fr);&#xA;  println(frmat);println(fr);&#xA;&#xA;/*&#xA;Outputs:&#xA;[0] 0.059300933&#xA;[1] 0.09312407&#xA;[2] -0.99388695&#xA;[3] 0.0&#xA;[4] 0.90466285&#xA;[5] 0.41586864&#xA;[6] 0.09294289&#xA;[7] 0.0&#xA;[8] 0.42198166&#xA;[9] -0.9046442&#xA;[10] -0.059584484&#xA;[11] 0.0&#xA;[12] 0.0&#xA;[13] 0.0&#xA;[14] 0.0&#xA;[15] 1.0&#xA;[0] 0.059300933&#xA;[1] 0.09312407&#xA;[2] -0.99388695&#xA;[3] 0.0&#xA;[4] 0.90466285&#xA;[5] 0.41586864&#xA;[6] 0.09294289&#xA;[7] 0.0&#xA;[8] 0.42198166&#xA;[9] -0.9046442&#xA;[10] -0.059584484&#xA;[11] 0.0&#xA;[12] 0.0&#xA;[13] 0.0&#xA;[14] 0.0&#xA;[15] 1.0&#xA;&#xA;*/&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="304" LastEditorUserId="304" LastEditorDisplayName="" LastEditDate="2010-09-22T22:28:31.207" LastActivityDate="2010-11-24T23:35:50.863" Title="How can I remove rotations from points defining a plane ?" Tags="&lt;geometry&gt;&lt;euclidean-geometry&gt;&lt;matrices&gt;" AnswerCount="3" CommentCount="4" FavoriteCount="0" />
  <row Id="5232" PostTypeId="1" CreationDate="2010-09-22T16:17:23.533" Score="4" ViewCount="99" Body="&lt;p&gt;I spent a little time looking around the Wikipedia and Wolfram articles on Penrose Tiling, the Domino Problem, Wang Tiles, etc., but I'm having a little trouble getting into them. A lot of these resources are fairly non-linear in that they often have circular dependencies of concepts that you need to know. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, I can tell that aperiodicity is important somehow, but I don't know why and I'm not quite sure about the actual definition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I keep seeing tiling linked in unexpectedly (e.g. L-systems, the Entscheidungsproblem) or in non-mathematical places (games, geometric art... even Neal Stephenson's &lt;em&gt;Anathem&lt;/em&gt;) that it seems important and I'd really like to know more.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, with that said: does anyone have any resources they'd recommend? I'm looking for something fairly complete or at least introductory, not just an article about a certain facet. Books are fine, visual or interactive resources are great. Would love descriptions of practical implications and relationships with other fields.&lt;/p&gt;&#xA;" OwnerUserId="2059" LastActivityDate="2010-09-22T17:46:27.240" Title="Getting started with aperiodic tiling" Tags="&lt;tiling&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" />
  <row Id="5233" PostTypeId="1" CreationDate="2010-09-22T16:34:50.537" Score="7" ViewCount="433" Body="&lt;p&gt;When teaching abstract vector spaces for the first time, it is handy to have some really weird examples at hand, or even some really weird non-examples that may illustrate the concept. For example, a physicist friend of mine uses &quot;color space&quot; as a (non) example, with two different bases given essentially {red, green, blue} and {hue, saturation and brightness} (see &lt;a href=&quot;http://en.wikipedia.org/wiki/Color_space&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Color_space&lt;/a&gt;). I say this is a non-example for a number of reasons, the most obvious being the absence of &quot;negative color&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyhow, what are some bizarre and vivid examples of vector spaces you've come across that would be suitable for a first introduction?&lt;/p&gt;&#xA;" OwnerUserId="941" LastActivityDate="2011-05-27T12:32:53.567" CommunityOwnedDate="2011-02-02T17:46:15.680" Title="Vivid examples of vector spaces?" Tags="&lt;soft-question&gt;&lt;education&gt;" AnswerCount="7" CommentCount="3" FavoriteCount="3" />
  <row Id="5234" PostTypeId="2" ParentId="5058" CreationDate="2010-09-22T16:43:40.970" Score="10" ViewCount="" Body="&lt;p&gt;Let's rewrite the formula, using $C_1=1/2$ as&#xA;$$S_n=\gamma+\log n-H_{n-1}$$&#xA;where&#xA;$$S_n=\sum_{k=1}^\infty\frac{(k-1)!C_k}{n(n+1)\cdots(n+k-1)}.$$&#xA;Use the formula&#xA;$$\frac{(k-1)!}{n(n+1)\cdots(n+k-1)}dx=\int_0^1 x^{n-1}(1-x)^{k-1}$$&#xA;which is a special case of the beta integral.&#xA;Therefore&#xA;$$S_n=\int_0^1x^{n-1}\sum_{k=1}^\infty C_k(1-x)^k&#xA;=\int_0^1\left(\frac{1}{1-x}+\frac{1}{\log x}\right)x^{n-1}dx.$$&#xA;When $n=1$ we get&#xA;$$S_1=\int_0^1\left(\frac{1}{1-x}+\frac{1}{\log x}\right)dx=\gamma$$&#xA;by a well-known &lt;a href=&quot;http://en.wikipedia.org/wiki/Euler-Mascheroni_constant&quot; rel=&quot;nofollow&quot;&gt;formula&lt;/a&gt; which can be derived from the aymptotic&#xA;$$\int_t^\infty\frac{e^{-x}}{x}dx=-\log t-\gamma+O(t)$$&#xA;as $t\to0$, for the exponential integral.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By induction it suffices to consider the difference&#xA;$$S_n-S_{n+1}=\int_0^1\left(x^{n-1}+\frac{x^{n-1}-x^n}{\log x}\right)dx&#xA;=\frac1n-\int_0^\infty\frac{e^{-ny}-e^{-(n+1)y}}{y}dy.$$&#xA;Using the identity&#xA;$$\int_0^\infty\frac{e^{-ay}-e^{-by}}{y}=\log\frac ba$$&#xA;for $0 &amp;lt; a &amp;lt; b$ &#xA;which follows by integrating $e^{-xy}$ over the region $[a,b]\times[0,\infty)$&#xA;we get&#xA;$$S_n-S_{n+1}=\frac1n-\log\frac{n+1}{n}.$$&#xA;The desired formula for $S_n$ now follows by induction.&lt;/p&gt;&#xA;" OwnerUserId="226" LastEditorUserId="226" LastEditorDisplayName="" LastEditDate="2010-09-22T22:09:09.710" LastActivityDate="2010-09-22T22:09:09.710" CommentCount="1" />
  <row Id="5235" PostTypeId="2" ParentId="5232" CreationDate="2010-09-22T16:44:32.807" Score="1" ViewCount="" Body="&lt;p&gt;Try &lt;a href=&quot;http://www.ams.org/bookstore-getitem/item=STML-1&quot; rel=&quot;nofollow&quot;&gt;Miles of Tiles&lt;/a&gt; by Charles Radin.&lt;/p&gt;&#xA;" OwnerUserId="589" LastActivityDate="2010-09-22T16:44:32.807" />
  <row Id="5236" PostTypeId="2" ParentId="5233" CreationDate="2010-09-22T16:47:32.023" Score="3" ViewCount="" Body="&lt;p&gt;I like the color example.  It shows how the idea of a basis is useful, even though it's not a vector space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Barycentric coordinates are another example of something like a vector space but not a vector space.&lt;/p&gt;&#xA;" OwnerUserId="247" LastActivityDate="2010-09-22T16:47:32.023" CommunityOwnedDate="2011-02-02T17:46:15.680" CommentCount="2" />
  <row Id="5237" PostTypeId="2" ParentId="5230" CreationDate="2010-09-22T16:55:27.000" Score="7" ViewCount="" Body="&lt;p&gt;There is no difference in meaning. I grabbed five books at random from my shelf and four out of five use the form $\int f(x) \mu(dx)$, while one uses the form  $\int f(x) d\mu(x)$. The form suppressing the variable of integration $\int f d\mu$ is also very common. I have never seen $\int f(x) d\mu$, which seems to me a weird hybrid. I work in probability so there may be some bias.&lt;/p&gt;&#xA;" OwnerUserId="940" LastActivityDate="2010-09-22T16:55:27.000" CommunityOwnedDate="2010-09-22T16:55:27.000" CommentCount="3" />
  <row Id="5238" PostTypeId="1" AcceptedAnswerId="5245" CreationDate="2010-09-22T17:27:40.383" Score="1" ViewCount="158" Body="&lt;p&gt;For my homework, I have been asked to rationalise and simplify this surd;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{11}{3\sqrt{3}+7}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each time I do this I get the wrong answer. The method I am using is;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \frac{11}{3\sqrt3-7} \times \frac{3\sqrt3-7}{3\sqrt3-7} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I ended up with $\frac{33+11\sqrt3-77}{9+3+21+7\sqrt3-21-7\sqrt3}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This ends up no where near the right answer, even once it is simplified, can someone tell me where i'm going wrong? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many thanks!&lt;/p&gt;&#xA;" OwnerUserId="1968" LastEditorUserId="171" LastEditorDisplayName="" LastEditDate="2010-09-22T17:34:29.293" LastActivityDate="2010-09-23T03:01:01.600" Title="Rationalising the denominator" Tags="&lt;algebra-precalculus&gt;&lt;homework&gt;" AnswerCount="4" CommentCount="2" />
  <row Id="5239" PostTypeId="2" ParentId="5238" CreationDate="2010-09-22T17:36:44.000" Score="1" ViewCount="" Body="&lt;p&gt;After multiplying the numerator and denominator by $3\sqrt3-7$&#xA;the new denominator is &#xA;$$(3\sqrt 3+7)(3\sqrt3-7)=(3\sqrt3)^2-7^2=27-49=-22$$&#xA;a nice integer to divide by.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-09-22T17:36:44.000" />
  <row Id="5240" PostTypeId="2" ParentId="5238" CreationDate="2010-09-22T17:37:09.383" Score="1" ViewCount="" Body="&lt;p&gt;Do you mean rationalise&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\frac{11}{3\sqrt{3}-7} \qquad \text{?}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And are you sure you're trying&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\frac{11}{3\sqrt{3}-7} \cdot \frac{3\sqrt{3} + 7}{3\sqrt{3} + 7}\qquad \text{?}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, &lt;a href=&quot;http://en.wikipedia.org/wiki/Rationalisation_%28mathematics%29&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt; is almost always of great help too.&lt;/p&gt;&#xA;" OwnerUserId="664" LastEditorUserId="664" LastEditorDisplayName="" LastEditDate="2010-09-23T03:01:01.600" LastActivityDate="2010-09-23T03:01:01.600" />
  <row Id="5241" PostTypeId="2" ParentId="5238" CreationDate="2010-09-22T17:38:19.957" Score="1" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT:&lt;/strong&gt; The general trick is&#xA;$$&#xA;\frac{1}{\sqrt{a}+b}=\frac{\sqrt{a}-b}{(\sqrt{a}+b)(\sqrt{a}-b)}=\frac{\sqrt{a}-b}{a-b^2}.&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="367" LastActivityDate="2010-09-22T17:38:19.957" />
  <row Id="5242" PostTypeId="2" ParentId="5216" CreationDate="2010-09-22T17:42:06.953" Score="2" ViewCount="" Body="&lt;p&gt;Write one procedure that takes seg1 and seg2 as input, and returns how close their closest endpoints are.  Write another procedure that computes the minimum angle between the lines containing seg1 and seg2.&#xA;Finally, mix the two with some weights to determine a &lt;em&gt;continuation score&lt;/em&gt;: how good seg2 serves as a continuation of seg1.  Then select the seg2 that has the best continuation score for seg1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(If you have, say, $10^5$ segments, then you will need to make it more efficient, which is an entirely different topic.)&lt;/p&gt;&#xA;" OwnerUserId="237" LastActivityDate="2010-09-22T17:42:06.953" />
  <row Id="5243" PostTypeId="2" ParentId="5233" CreationDate="2010-09-22T17:43:45.657" Score="2" ViewCount="" Body="&lt;p&gt;I like the example $C([0,1])$ of continuous functions on the interval (or something similar). It is familiar-looking but shows that there is not always a natural choice of basis.&lt;/p&gt;&#xA;" OwnerUserId="367" LastActivityDate="2010-09-22T17:43:45.657" CommunityOwnedDate="2011-02-02T17:46:15.680" CommentCount="5" />
  <row Id="5244" PostTypeId="2" ParentId="5232" CreationDate="2010-09-22T17:46:27.240" Score="2" ViewCount="" Body="&lt;p&gt;Chaim Goodman-Strauss has quite a few papers on his web page about this topic, including this specific one:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://comp.uark.edu/~strauss/papers/newsmall.pdf&quot; rel=&quot;nofollow&quot;&gt;http://comp.uark.edu/~strauss/papers/newsmall.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Goodman-Strass's publications page:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://comp.uark.edu/~strauss/papers/index.html&quot; rel=&quot;nofollow&quot;&gt;http://comp.uark.edu/~strauss/papers/index.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1369" LastActivityDate="2010-09-22T17:46:27.240" />
  <row Id="5245" PostTypeId="2" ParentId="5238" CreationDate="2010-09-22T17:47:58.703" Score="2" ViewCount="" Body="&lt;p&gt;You're mistakenly multiplying $\rm\; a * b\sqrt{3} \ =\ ab + a\sqrt{3}\:\,\;$ but $\rm\; ab\:\sqrt{3}\;$ is correct.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words $\rm\; b\:\sqrt{3}\;$ means $\rm b * \sqrt{3}\:,\;$ not $\rm\; b + \sqrt{3}\:.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, to rationalize the denominator use $\rm\; (a+b\sqrt 3)\:(a-b\sqrt 3)\ =\ a^2 - 3 b^2$&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-09-22T17:47:58.703" />
  <row Id="5246" PostTypeId="2" ParentId="2837" CreationDate="2010-09-22T18:06:33.650" Score="-2" ViewCount="" Body="&lt;p&gt;Dear God you guys over complicated things...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Not being a math expert along the lines of you guys, I set the problem down and came back to it later. Upon looking it over again, I realized I could really simplify this by using the given data to look at it as different geometry. Instead of a line segment and a circle and testing for intersection, I used the radius of the circle to convert the line to a rectangle and then checked if the center of the circle (as just a point, no longer as a circle) is inside the new rectangle.&#xA;&lt;img src=&quot;http://i.imgur.com/NRO80.png&quot; alt=&quot;Line Circle Intersect Using New Geometry&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please excuse my rectangle not having perfect dimensions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Although my answer meets my needs for my specific application, to better answer my own question, about the semi circles at the end of the rectangle (to test if an endpoint of the line segment is contained in the circle), I can compare the distance between an endpoint and the center of the circle, if that distance is less than the radius of the circle, the circle contains the end point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My application was for a Tower Defense game. The line represented some sort of laser stretched across the trail that enemies (the circle) would walk past, I needed to know when the laser contacted the enemy. The laser endpoints are attached to towers that are not positioned on the path the enemy walks down, so the enemy's circle could not contain an endpoint. I realize I was unclear during the original question about some of these details.&lt;/p&gt;&#xA;" OwnerUserId="757" LastEditorUserId="757" LastEditorDisplayName="" LastEditDate="2010-09-25T20:12:43.303" LastActivityDate="2010-09-25T20:12:43.303" CommentCount="2" />
  <row Id="5247" PostTypeId="2" ParentId="5231" CreationDate="2010-09-22T18:10:55.087" Score="1" ViewCount="" Body="&lt;p&gt;Try to represent rotations using matrices instead of angles - then finding the inverse is easy.&lt;/p&gt;&#xA;" LastActivityDate="2010-09-22T18:10:55.087" CommentCount="5" />
  <row Id="5248" PostTypeId="1" AcceptedAnswerId="5322" CreationDate="2010-09-22T19:09:03.090" Score="11" ViewCount="1455" Body="&lt;p&gt;A famous exercise which one encounters while doing Complex Analysis(Residue theory): is to prove that the given integral $$\int_{0}^{\infty} \frac{\sin{x}}{x} \ dx = \frac{\pi}{2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, can anyone prove this without using Residue theory. I actually thought of doint this: $$\int_{0}^{\infty} \frac{\sin{x}}{x} \ dx = \lim_{t \to \infty} \int_{0}^{t} \frac{1}{t} \Bigl( t - \frac{t^3}{3!} + \frac{t^5}{5!} + \cdots \Bigr) \ dt$$ but i don't see how $\pi$ comes here, since we need the answer to be equal to $\frac{\pi}{2}$.&lt;/p&gt;&#xA;" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-10-31T02:06:54.593" LastActivityDate="2011-04-26T08:05:57.303" Title="Solving the integral $\int_{0}^{\infty} \frac{\sin{x}}{x} \ dx = \frac{\pi}{2}$?" Tags="&lt;analysis&gt;&lt;sequences-and-series&gt;&lt;complex-analysis&gt;" AnswerCount="8" CommentCount="2" FavoriteCount="9" />
  <row Id="5249" PostTypeId="2" ParentId="5248" CreationDate="2010-09-22T19:19:06.853" Score="6" ViewCount="" Body="&lt;p&gt;See &lt;a href=&quot;http://en.wikipedia.org/wiki/Dirichlet_integral&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Dirichlet_integral&lt;/a&gt; for a proof using differentiation under the integral sign.&lt;/p&gt;&#xA;" OwnerUserId="367" LastActivityDate="2010-09-22T19:19:06.853" CommunityOwnedDate="2010-09-22T19:19:06.853" CommentCount="1" />
  <row Id="5250" PostTypeId="1" CreationDate="2010-09-22T20:01:31.330" Score="3" ViewCount="210" Body="&lt;p&gt;I am trying to find a formula that will enable me to calculate the illumination of the moon down to one thousandth of a percent, given the gregorian year, month, day, and hour is known. I am unsure if this is the place to ask this sort of a question, but can anyone help me on this formula?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you!&lt;/p&gt;&#xA;" OwnerUserId="2070" LastEditorUserId="630" LastEditorDisplayName="" LastEditDate="2010-10-31T20:40:36.337" LastActivityDate="2010-10-31T20:40:36.337" Title="Illumination calculations?" Tags="&lt;physics&gt;&lt;applications&gt;" CommentCount="7" />
  <row Id="5251" PostTypeId="2" ParentId="5248" CreationDate="2010-09-22T20:07:13.260" Score="3" ViewCount="" Body="&lt;p&gt;In my &lt;a href=&quot;http://problemasteoremas.files.wordpress.com/2008/06/tayloradvancedcalculus1.jpg?w=351&amp;amp;h=490&quot; rel=&quot;nofollow&quot;&gt;Advanced Calculus book (Angus Taylor)&lt;/a&gt; it is shown that, if $a\gt 0$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle\int_0^{\infty}\dfrac{e^{-at}\sin xt}{t}dt=\arctan\dfrac{x}{a}\qquad (1)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $x&gt;0$, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle\int_0^{\infty}\dfrac{\sin xt}{t}dt=\dfrac{\pi}{2}\qquad (2)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;follows from $(1)$, observing that the integrand is $G(0)$ for&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$G(a)=\displaystyle\int_0^{\infty}\dfrac{e^{-at}\sin xt}{t}dt\qquad (3)$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$G$ is uniformly convergent when $a\ge 0$, and $G(a)$ approaches $G(0)$ as $a$ tends to $0^+$.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Answer to Qiaochu: $(1)$ is proved as an application of the following theorem to $F(x)=\displaystyle\int_0^{\infty}\dfrac{e^{-at}\sin xt}{t}dt$: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Let $F(x)=\displaystyle\int_c^{\infty}f(t,x)dt$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;be convergent when $a\le x\le b$. Let $\dfrac{\partial f}{\partial x}$ be continuous in $t,x$ for $c\lt t$,$a\lt x\lt b$, and let $\displaystyle\int_c^{\infty}\dfrac{\partial f}{\partial x}dt$ converge uniformly on $[a,b]$. Then $F'(x)=\displaystyle\int_c^{\infty}\dfrac{\partial f}{\partial x}dt$&quot;.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;PS. It was a surprise to me to see the evaluation of $F(x)=\displaystyle\int_0^{\infty}\dfrac{e^{-at}\sin xt}{t}dt$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$F(x)=\dfrac{\pi}{2}\qquad$ if $x&gt;0$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$F(x)=-\dfrac{\pi}{2}\qquad$ if $x&amp;lt;0$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$F(x)=0\qquad\qquad$ if $x=0$.&lt;/p&gt;&#xA;" OwnerUserId="752" LastEditorUserId="136" LastEditorDisplayName="" LastEditDate="2010-09-22T22:11:15.680" LastActivityDate="2010-09-22T22:11:15.680" CommentCount="5" />
  <row Id="5252" PostTypeId="2" ParentId="4862" CreationDate="2010-09-22T20:21:51.227" Score="0" ViewCount="" Body="&lt;p&gt;Many of the suggestions here are good ones, if perhaps a little advanced. A course on multivarible calculus and a course on ordinary differential equations seem like the logical progression to me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In an undergrad maths course, this would typically be accompanied by courses on linear algebra and abstract algebra.&lt;/p&gt;&#xA;" OwnerUserId="1820" LastActivityDate="2010-09-22T20:21:51.227" CommunityOwnedDate="2010-09-22T20:21:51.227" />
  <row Id="5254" PostTypeId="2" ParentId="5201" CreationDate="2010-09-22T20:28:58.890" Score="0" ViewCount="" Body="&lt;p&gt;It seems to me what you're trying to do is formulate a first-order theory. This has non-logical axioms. Your theory uses two one-place predicates, say &quot;F&quot; for the Xs and &quot;G&quot; for the Ys. It has one non-logical axiom, perhaps, which says there are Fs and there are Gs. Then it has another which says nothing is both F and G. Then you have a two-place predicate &quot;R&quot; to represent the relationship, and another non-logical axiom for that, one that uses &quot;F&quot; and &quot;G&quot; as well as &quot;R&quot;. From the point of view of logic, there's still only one domain; it's your non-logical axioms that divide it up. -- It's the predicates and axioms that are important, not the variables &quot;x&quot; and &quot;y&quot;. Shoenfield's text &lt;em&gt;Mathematical Logic&lt;/em&gt; is good for examples of first-order theories.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: by non-logical axiom I mean an axiom that's not itself already provable in first-order logic.&lt;/p&gt;&#xA;" OwnerUserId="1813" LastActivityDate="2010-09-22T20:28:58.890" />
  <row Id="5255" PostTypeId="1" AcceptedAnswerId="5256" CreationDate="2010-09-22T20:36:34.023" Score="3" ViewCount="100" Body="&lt;p&gt;Simplicial sequences generalize the familiar &quot;linear&quot;, &quot;triangular&quot;, and &quot;tetrahedral&quot; number sequences. (A line segment is a 1-simplex, a triangle is a 2-simplex, a tetrahedron is a 3-simplex, and so on, so I'm calling these simplicial sequences.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, the n-th d-simplicial number is the binomial coefficient C(n+d-1,d).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These pictures below graphically show the n-th element (modulo 512) of 512 simplicial sequences side by side. The white dots in a given frame have coordinates (d,C(n+d-1,d) mod 512), where d runs from 0 through to 511, and the value of n is displayed at the bottom of the diagram.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, if you are using a &quot;modern browser&quot; (i.e. not IE6,7,8), you can see an animation running through all values of n &lt;a href=&quot;http://www.brainjam.ca/stackoverflow/simplicialNumbers.htm&quot; rel=&quot;nofollow&quot;&gt;by following this link&lt;/a&gt;.  The animation allows you to start and stop, and even save individual frames.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you watch the animation, you'll see that binary visual patterns emerge, in ways that are threaded through the image and increase and decrease in strength.  The images below for n=5 and n=256 indicate the extremes .. for n=5 the points look randomly distributed, for n=256 they are strongly patterned.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So my question is whether there are any explanations for the emergence and disappearance of visual binary patterns as we iterate through n.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/eur9v.png&quot; alt=&quot;alt text&quot;&gt;&#xA;&lt;img src=&quot;http://i.imgur.com/WIZ0x.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1257" LastEditorUserId="1257" LastEditorDisplayName="" LastEditDate="2010-09-22T21:37:41.203" LastActivityDate="2010-09-22T21:37:41.203" Title="Is there an explanation for the patterns formed by the binomial coefficients C(n+d-1,d) mod 512?" Tags="&lt;sequences-and-series&gt;&lt;binomial-coefficients&gt;&lt;experimental-mathematics&gt;" AnswerCount="1" />
  <row Id="5256" PostTypeId="2" ParentId="5255" CreationDate="2010-09-22T20:48:42.823" Score="4" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Theorem (Kummer):&lt;/strong&gt;  The greatest power of $p$ dividing ${m+n \choose n}$ is the number of carries it takes to add $m$ and $n$ in base $p$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular the greatest power of $2$ dividing ${n+d-1 \choose d}$ is the number of carries it takes to add $n-1$ and $d$ in base $2$, and so depends entirely on the binary digits of both $n-1$ and $d$.  (When $n = 256$, for example, there are likely to be a lot of carries, which is why the position of the dots is so restricted in that case.  More generally the same happens when $n-1$ has lots of $1$s in its binary representation.)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The above theorem places a strong restriction on where the white dots can be, which I think will be clearer visually if you plot things in the shape of &lt;strong&gt;Pascal's triangle&lt;/strong&gt; instead of whatever it is you're doing now (for example $\bmod 2$ you will get the &lt;a href=&quot;http://en.wikipedia.org/wiki/Sierpinski_triangle&quot; rel=&quot;nofollow&quot;&gt;Sierpinski triangle&lt;/a&gt;.  More generally I think you should try plotting the entries of Pascal's triangle $\bmod 512$ where the modulus determines the color of a block.  This won't be an animation but will allow you to consolidate all of the patterns you're seeing into one image which I think will ultimately be clearer).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;See also &lt;a href=&quot;http://en.wikipedia.org/wiki/Lucas_theorem&quot; rel=&quot;nofollow&quot;&gt;Lucas' theorem&lt;/a&gt; for a partial explanation of patterns in the exact remainder $\bmod 512$.&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-09-22T20:57:08.937" LastActivityDate="2010-09-22T20:57:08.937" CommentCount="2" />
  <row Id="5257" PostTypeId="2" ParentId="5248" CreationDate="2010-09-22T21:36:29.747" Score="30" ViewCount="" Body="&lt;p&gt;I believe this can also be solved using double integrals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is possible (if I remember correctly) to justify switching the order of integration to give the equality:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_{0}^{\infty} \Bigg(\int_{0}^{\infty} e^{-xy} \sin x \,dy \Bigg)\, dx = \int_{0}^{\infty} \Bigg(\int_{0}^{\infty} e^{-xy} \sin x \,dx \Bigg)\,dy$$&#xA;Notice that&#xA;$$\int_{0}^{\infty} e^{-xy} \sin x\,dy = \frac{\sin x}{x}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This leads us to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_{0}^{\infty} \Big(\frac{\sin x}{x} \Big) \,dx = \int_{0}^{\infty} \Bigg(\int_{0}^{\infty} e^{-xy} \sin x \,dx \Bigg)\,dy$$&#xA;Now the right hand side can be found easily, using integration by parts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align*}&#xD;&#xA;I &amp;amp;= \int e^{-xy} \sin x \,dx = -e^{-xy}{\cos x} + y \int e^{-xy} \cos x \, dx\\&#xD;&#xA;&amp;amp;= -e^{-xy}{\cos x} + y \Big(e^{-xy}\sin x - y \int e^{-xy} \sin x \,dx \Big)\\&#xD;&#xA;&amp;amp;= \frac{ye^{-xy}\sin x - e^{-xy}\cos x}{1+y^2}.&#xD;&#xA;\end{align*}$$&#xA;Thus $$\int_{0}^{\infty} e^{-xy} \sin x \,dx = \frac{1}{1+y^2}$$&#xA;Thus $$\int_{0}^{\infty} \Big(\frac{\sin x}{x} \Big) \,dx = \int_{0}^{\infty}\frac{1}{1+y^2}\,dy = \frac{\pi}{2}.$$&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="8476" LastEditorDisplayName="" LastEditDate="2011-04-17T04:30:41.777" LastActivityDate="2011-04-17T04:30:41.777" CommentCount="3" />
  <row Id="5259" PostTypeId="1" AcceptedAnswerId="5287" CreationDate="2010-09-22T22:36:01.287" Score="2" ViewCount="288" Body="&lt;p&gt;I'm trying to approximate $\pi$ using Monte Carlo integration; I am approximating the integral&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int\limits_0^1\!\frac{4}{1+x^2}\;\mathrm{d}x=\pi$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is working fine, and so is estimating the error (variance), $\sigma$. However, when I then try to use importance sampling with a Cauchy(0,1) distribution, things start to go wrong:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{1}{n}\sum\limits_{i=0}^n\frac{f(x_i)}{p(x_i)}=\frac{1}{n}\sum\limits_{i=0}^n\frac{\frac{4}{1+x^2}}{\frac{1}{\pi(1+x^2)}}=\frac{1}{n}\sum\limits_{i=0}^n\frac{4\pi(1+x^2)}{1+x^2}=\frac{1}{n}\sum\limits_{i=0}^n4\pi=4\pi$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously something's wrong, since the mean is computed independently of the random variables I generate. Where is this going wrong? Is the distribution too close to $f$? &lt;/p&gt;&#xA;" OwnerUserId="708" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-09-23T14:16:00.327" LastActivityDate="2010-09-23T14:16:00.327" Title="Approximating $\pi$ using Monte Carlo integration" Tags="&lt;integral&gt;&lt;approximation&gt;&lt;numerical-methods&gt;&lt;monte-carlo&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="5261" PostTypeId="2" ParentId="5191" CreationDate="2010-09-22T22:46:22.590" Score="1" ViewCount="" Body="&lt;p&gt;Here is one idea, although I do not find it very satisfying.  An object of $C$ is a finite set $X$ equipped with $\frac{|X|!}{2}$ (or $1$ if $|X| = 1$) total orders, all of which are &lt;strong&gt;even&lt;/strong&gt; with respect to each other (in other words, basically a coset of $A_n$ in $S_n$).  A morphism between two objects in $X$ is a map of sets preserving these orders (in other words, take one of the orderings on $X$ and apply a function $f : X \to Y$ to its elements.  The result, after throwing out repeats, must be compatible with an ordering on $Y$.)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is more or less a discretization of Omar's answer.  Again, I would like to do better than this, or at least see the data described above packaged in a more satisfying way.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-09-22T22:46:22.590" />
  <row Id="5262" PostTypeId="1" AcceptedAnswerId="5264" CreationDate="2010-09-22T23:18:55.690" Score="2" ViewCount="154" Body="&lt;p&gt;I have 2 numbers (a &amp;amp; b) but I need a formula (or a how to) to find which 2 numbers (c &amp;amp; d) will add together to give a and times together to give b. &#xA;So&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;c + d = a&#xA;c * d = b &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="2075" LastActivityDate="2010-09-23T02:55:40.663" Title="Formula(/How) to find 2 numbers that add together to give one number and times to give another" Tags="&lt;recreational-mathematics&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="5263" PostTypeId="2" ParentId="5233" CreationDate="2010-09-22T23:21:58.090" Score="2" ViewCount="" Body="&lt;p&gt;The real vector space of all fibonacci sequences (the first two values are arbitrary) is quite instructive. Or the subspace of all smooth functions satisfying the differential equation $f'=f$. It is quite illuminating that elementary linear algebra has a interplay with analysis.&lt;/p&gt;&#xA;" OwnerUserId="1650" LastActivityDate="2010-09-22T23:21:58.090" CommunityOwnedDate="2011-02-02T17:46:15.680" />
  <row Id="5264" PostTypeId="2" ParentId="5262" CreationDate="2010-09-22T23:31:58.390" Score="2" ViewCount="" Body="&lt;p&gt;Consider the quadratic&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(x-c)(x-d) = x^2 -(c+d)x + cd = x^2 - ax +b$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus the roots of $$x^2 -ax + b = 0$$ are your numbers $c$ and $d$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The roots are given by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{ a \pm \sqrt{a^2 -4b}}{2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Check this page out for more information: &lt;a href=&quot;http://en.wikipedia.org/wiki/Quadratic_equation&quot; rel=&quot;nofollow&quot;&gt;Quadratic Equation&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-09-22T23:31:58.390" CommentCount="1" />
  <row Id="5265" PostTypeId="2" ParentId="5262" CreationDate="2010-09-22T23:32:58.130" Score="2" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\rm\quad (X - c)\: (X - d) \; =\; X^2 - a\ X + b $&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-09-22T23:32:58.130" />
  <row Id="5266" PostTypeId="1" AcceptedAnswerId="5284" CreationDate="2010-09-22T23:36:08.967" Score="1" ViewCount="191" Body="&lt;p&gt;In my statistical signal modelling class we were given the following task:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For simplicity, let's assume E(y) = E(x) = 0. Show that if the real variables x and y are normally distributed, then the equation E(y|x) = corr(y,x)*x/var(x) holds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We were also given a hint: write the assisting variable s = y - corr(x,y)*x/var(x) and show that it's uncorrelated with x, meaning that E(sx) = E(s)E(x) = 0. Because s and x together are normally distributed, it follows that s and x are uncorrelated, ie. the conditional probability of s doesn't depend on x. Now you can easily write y using s, and calculate E(y|x).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've been hacking at this for 3 hours without much progress so any help is appreciated&lt;/p&gt;&#xA;" OwnerUserId="2076" LastActivityDate="2010-09-23T06:05:59.897" Title="Proving the conditional probability E(y|x) = corr(y,x)*x/(stdev(x)^2)" Tags="&lt;probability&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="5267" PostTypeId="1" AcceptedAnswerId="5268" CreationDate="2010-09-22T23:46:35.733" Score="3" ViewCount="310" Body="&lt;p&gt;Hi there!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've got a programming exercise I need to do, but I just can't figure out the math part. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need to check if 3 of 6 lines intersect in the same point. I am given the equation ax+by=c, and I input random numbers in a,b,c. How would I go about solving this mathematically?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="2078" LastEditorUserId="2078" LastEditorDisplayName="" LastEditDate="2010-09-22T23:58:53.790" LastActivityDate="2010-09-23T01:48:20.000" Title="How to find where 3 lines intersect." Tags="&lt;geometry&gt;" AnswerCount="3" FavoriteCount="1" />
  <row Id="5268" PostTypeId="2" ParentId="5267" CreationDate="2010-09-22T23:52:29.170" Score="3" ViewCount="" Body="&lt;p&gt;You're trying to check for the condition of &lt;a href=&quot;http://mathworld.wolfram.com/Concurrent.html&quot; rel=&quot;nofollow&quot;&gt;concurrency&lt;/a&gt; of three lines; the last determinant formula involving the coefficients of your three lines in that link may be of help.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you've shown that your three lines are concurrent, take any two of the three and solve as simultaneous equations (two equations in two unknowns) to get the coordinates of the intersection point.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-09-22T23:52:29.170" CommentCount="3" />
  <row Id="5269" PostTypeId="2" ParentId="5267" CreationDate="2010-09-23T00:46:18.733" Score="0" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT:&lt;/strong&gt;$\;$ &lt;strong&gt;Divide and conquer&lt;/strong&gt;. $\:$ Split the lines into two sets of three lines. Compute the intersection points of each set (at most three). If either intersection set has cardinality one then you're done. Else check if any intersection point from one set lies on any line in the other set. Total operation count: 6 line intersections and 6 line evaluations.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: if the lines need not be distinct then you need to make slight modifications to the above.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-09-23T01:12:26.247" LastActivityDate="2010-09-23T01:12:26.247" CommentCount="3" />
  <row Id="5270" PostTypeId="2" ParentId="5233" CreationDate="2010-09-23T00:58:21.423" Score="3" ViewCount="" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The positive real numbers, where 1 is the &quot;zero vector,&quot; &quot;scalar multiplication&quot; is really numerical exponentiation, and &quot;addition&quot; is really numerical multiplication. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Simplicial complexes, as in algebraic topology, are another good example, but perhaps this is even &lt;em&gt;too&lt;/em&gt; weird. Still, it might be fun to throw out the idea that mathematicians like to add triangles to each other to get quadrilaterals and negate them to reverse orientation, but they'll probably have to take it on faith that this is actually useful. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="2032" LastActivityDate="2010-09-23T00:58:21.423" CommunityOwnedDate="2011-02-02T17:46:15.680" CommentCount="4" />
  <row Id="5271" PostTypeId="2" ParentId="5267" CreationDate="2010-09-23T01:48:20.000" Score="1" ViewCount="" Body="&lt;p&gt;The first step to write a program is to break down a problem into small parts.  I would do the following:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Write a function to decide whether given three lines intersect at one point.  &lt;a href=&quot;http://math.stackexchange.com/questions/5267/how-to-find-where-3-lines-intersect/5268#5268&quot;&gt;J. M.’s answer&lt;/a&gt; gives some material to do this.&lt;/li&gt;&#xA;&lt;li&gt;Iterate over all combinations of three lines out of six.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;If the program does not work, try to test the two parts separately.  Smaller, simpler programs are easier to verify.&lt;/p&gt;&#xA;" OwnerUserId="926" LastActivityDate="2010-09-23T01:48:20.000" CommentCount="2" />
  <row Id="5272" PostTypeId="2" ParentId="5262" CreationDate="2010-09-23T02:42:14.620" Score="2" ViewCount="" Body="&lt;p&gt;Since I'm not so smart to imagine that the solution has anything to do with Vieta formulae or roots of polynomials,  ;-)  I'd try the following. First, let me call your unknowns $c,d$ with letters that look like unkowns (I've tried to do the problem with $c,d$ and at a certain point I forgot which among $a,b,c,d$ were the unkowns and which the data). So let $x = c$ and $y =d$. Your system becomes&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{align}&#xA;x + y &amp;amp;= a \\&#xA;xy    &amp;amp;= b   \ .&#xA;\end{align}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, do what you would do in those cases. For instance, solve the first equation for $x$ in terms of $y$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;x = a -y  \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then substitute this expression for $x$ into the second equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;(a-y)y = b \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And, surprise!, you've got a polynomial equation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;y^2 -ay + b = 0&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;that looks like the ones I've been told. (So this Vieta must have been a really smart guy, but you've rediscovered his trick.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And, of course, once you've got the solutions $y$ of this second degree equation, you should use $x = a-y$ to find $x$... (Just to find out that if $y$ is one of the roots of the last polynomial equation, $x$ is the other one.)&lt;/p&gt;&#xA;" OwnerUserId="664" LastEditorUserId="664" LastEditorDisplayName="" LastEditDate="2010-09-23T02:55:40.663" LastActivityDate="2010-09-23T02:55:40.663" CommentCount="7" />
  <row Id="5273" PostTypeId="2" ParentId="227" CreationDate="2010-09-23T03:09:14.133" Score="2" ViewCount="" Body="&lt;p&gt;Some advice from the eminent macroeconomist Thomas Sargent&#xA;&lt;a href=&quot;http://homepages.nyu.edu/~ts43/math_courses.html&quot; rel=&quot;nofollow&quot;&gt;http://homepages.nyu.edu/~ts43/math_courses.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1195" LastActivityDate="2010-09-23T03:09:14.133" CommunityOwnedDate="2010-09-23T03:09:14.133" CommentCount="1" />
  <row Id="5274" PostTypeId="1" AcceptedAnswerId="5276" CreationDate="2010-09-23T03:19:56.110" Score="2" ViewCount="529" Body="&lt;p&gt;While working on a proof showing that all functions limited to the domain of real numbers can be expressed as a sum of their odd and even components, I stumbled into a troublesome roadblock; namely, I had no clue how one divides the function into these even and odd parts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Looking up a solution for the proof, I found these general formulas for the even and odd parts of a function f(n):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/Sl6Hk.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While I understand that in an even function f(n) = f(-n) and that in an odd function f(-n) = -f(n), I still don't get how these general formulas for the even and odd parts were obtained. Can someone guide me through the logic?&lt;/p&gt;&#xA;" OwnerUserId="1956" LastActivityDate="2010-09-23T04:06:16.363" Title="How do I divide a function into even and odd sections?" Tags="&lt;functions&gt;" AnswerCount="2" CommentCount="5" />
  <row Id="5275" PostTypeId="2" ParentId="5274" CreationDate="2010-09-23T03:31:05.070" Score="3" ViewCount="" Body="&lt;p&gt;Suppose $f(x)=g(x)+h(x)$ with $g$ even and $h$ odd.  Then $f(-x)=g(-x)+h(-x)=g(x)-h(x)$.  Think of $f(x)=g(x)+h(x)$ and $f(-x)=g(x)-h(x)$ as a system of two equations in two unknowns, $g(x)$ and $h(x)$, to solve for $g$ and $h$.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-09-23T03:31:05.070" CommentCount="1" />
  <row Id="5276" PostTypeId="2" ParentId="5274" CreationDate="2010-09-23T03:32:02.367" Score="12" ViewCount="" Body="&lt;p&gt;Suppose you could write a function $f(x)$ as the sum of an even and an odd function; call them $E(x)$ and $O(x)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, you would have&#xA;\[f(x) = E(x)+O(x)\]&#xA;and you would also have&#xA;\[f(-x) = E(-x) + O(-x) = E(x) - O(x)\]&#xA;with the latter equation because we are assuming $E$ is even and $O$ is odd, so $E(x)=E(-x)$ and $O(-x) = -O(x)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Adding both equations you get $f(x)+f(-x) = 2E(x)$. Subtracting the second equation from the first gives you $f(x)-f(-x)=2O(x)$. Now solve for $E(x)$ and $O(x)$, and you get the formulas you see in the solution. Then you check that the answer does indeed work (that is, you check that the formulas you found do give you an even and an odd function in all cases).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words: pretend you already know the answer, and try to deduce conditions that the answer must satisfy (these will be &lt;em&gt;necessary&lt;/em&gt; conditions); if things go well, you'll get enough information about what they &lt;em&gt;must&lt;/em&gt; be like to figure out &lt;em&gt;what&lt;/em&gt; they are.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-09-23T04:06:16.363" LastActivityDate="2010-09-23T04:06:16.363" CommentCount="1" />
  <row Id="5277" PostTypeId="1" AcceptedAnswerId="5279" CreationDate="2010-09-23T05:10:47.343" Score="4" ViewCount="602" Body="&lt;p&gt;How do I mathematically determine if a number is prime?  I'm writing a program that the user inputs any integer and from that the program determines whether the number is prime, but how do I go about that?&lt;/p&gt;&#xA;" OwnerUserId="1602" LastEditorUserId="1778" LastEditorDisplayName="" LastEditDate="2010-09-27T19:26:26.430" LastActivityDate="2010-09-27T19:26:26.430" Title="Determine whether a number is prime" Tags="&lt;number-theory&gt;&lt;algorithms&gt;&lt;prime-numbers&gt;" AnswerCount="5" CommentCount="3" FavoriteCount="1" />
  <row Id="5278" PostTypeId="1" AcceptedAnswerId="5285" CreationDate="2010-09-23T05:16:47.097" Score="8" ViewCount="319" Body="&lt;p&gt;This question is taken from Dummit and Foote(14.9 #6). Any help will be appreciated:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Show that if $t$ is transcendental over $\mathbb{Q}$, then $\mathbb{Q}(t,\sqrt{t^3-t})$ is not a purely transcendental extension of $\mathbb{Q}$.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's what I've got so far:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Abbreviate $\sqrt{t^3-t}$ as $u$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've shown that the transcendence degree is 1, so the problem boils down to showing that $\mathbb{Q}(t,u) \supset \mathbb{Q}(f(t,u)/g(t,u))$ strictly, for all polynomials $f,g$ in two variables.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose for contradiction that $\mathbb{Q}(t,u) = \mathbb{Q}(f(t,u)/g(t,u))$. Look at this field as $\mathbb{Q}(t)[x]/(x^2-(t^3-t))$, with $\bar{x}=u$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then since $t$ and $u$ are generated by $f/g$, we have that for some polynomials $a,b,c,d$ in 1 variable, $a(\frac{f(t,x)}{g(t,x)})/b(\frac{f(t,x)}{g(t,x)})-t \in (x^2-(t^3-t))$, and $c(\frac{f(t,x)}{g(t,x)})/d(\frac{f(t,x)}{g(t,x)}) - x \in (x^2-(t^3-t))$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I then tried playing with degrees, but I haven't found a contradiction.&lt;/p&gt;&#xA;" OwnerUserId="2080" LastActivityDate="2010-09-23T18:00:35.540" Title="Why is $\mathbb{Q}(t,\sqrt{t^3-t})$ not a purely transcendental extension of $\mathbb{Q}$?" Tags="&lt;abstract-algebra&gt;&lt;field-theory&gt;" AnswerCount="2" FavoriteCount="4" />
  <row Id="5279" PostTypeId="2" ParentId="5277" CreationDate="2010-09-23T05:19:22.370" Score="5" ViewCount="" Body="&lt;p&gt;Have the program find the remainder when dividing the input (say n) by 2, 3, 4, ..., $\sqrt n$  (or the following integer if $\sqrt n$ is not an integer.) If this value ever leaves a remainder of zero then your number is composite and you can stop checking divisors. If the remainder is non-zero for all of these values then your number is prime.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are more efficient ways to check primeness but this is probably the easiest way to program it and I imagine will suffice for your purposes. Unless you are checking very very large numbers? &lt;/p&gt;&#xA;" OwnerUserId="1352" LastEditorUserId="1352" LastEditorDisplayName="" LastEditDate="2010-09-23T05:36:26.947" LastActivityDate="2010-09-23T05:36:26.947" CommentCount="2" />
  <row Id="5280" PostTypeId="2" ParentId="5277" CreationDate="2010-09-23T05:23:11.573" Score="6" ViewCount="" Body="&lt;p&gt;There are many different algorithms for primality testing. See the &lt;a href=&quot;http://en.wikipedia.org/wiki/Primality_test&quot; rel=&quot;nofollow&quot;&gt;Wikipedia page&lt;/a&gt; for an introduction and see Henri Cohen's book &quot;A course in computational algebraic number theory&quot; for further details. See also Caldwell's &lt;a href=&quot;http://primes.utm.edu/prove/index.html&quot; rel=&quot;nofollow&quot;&gt;Prime Pages&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-09-23T05:36:27.897" LastActivityDate="2010-09-23T05:36:27.897" />
  <row Id="5281" PostTypeId="2" ParentId="5277" CreationDate="2010-09-23T05:32:48.977" Score="15" ViewCount="" Body="&lt;p&gt;For very small numbers (less than a million), trial division is the best way: divide by 2, 3, 5, and so on until the square root of the number.  If you find a factor, the number is composite; otherwise, the number is prime.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For larger numbers there are better methods, but choosing which one depends on how much work you're willing to put into the program.  It is now known that there are no BPSW-pseudoprimes below $2^{64}$, so if you can write that test (see &lt;a href=&quot;http://www.trnicely.net/misc/bpsw.html&quot; rel=&quot;nofollow&quot;&gt;here for details&lt;/a&gt;) then you have a very quick test for primality.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you only need to test up to $2^{32}$, you can simply check if the number is a 2-strong pseudoprime.  If so, test if it's one of 2314 exceptions (this can be done in 12 or 13 steps with a binary search); if the test fails or it's an exception, the number is composite, otherwise prime.  (You can go higher than $2^{32}$ if you're willing to build an appropriate table of exceptions.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For larger numbers, the work is usually split into two parts: determining with high probability (say, 99.99999999%) that the number is prime, then actually proving that it is.  What type of proof depends on the form and size of the number.&lt;/p&gt;&#xA;" OwnerUserId="1778" LastActivityDate="2010-09-23T05:32:48.977" CommentCount="2" />
  <row Id="5282" PostTypeId="2" ParentId="5277" CreationDate="2010-09-23T05:47:36.980" Score="1" ViewCount="" Body="&lt;p&gt;Algorithm posted by &lt;a href=&quot;http://math.stackexchange.com/questions/5277/determine-whether-a-number-is-prime/5279#5279&quot;&gt;jericson&lt;/a&gt; is the best for basic purposes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;IMHO, for &lt;em&gt;programming competitions&lt;/em&gt; and for &lt;em&gt;practical purposes&lt;/em&gt; &lt;strong&gt;randomized algorithms&lt;/strong&gt; are best. Rabin-Miller is my favorite. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take a look at &lt;a href=&quot;http://www.topcoder.com/tc?module=Static&amp;amp;d1=tutorials&amp;amp;d2=primalityTesting&quot; rel=&quot;nofollow&quot;&gt;&lt;strong&gt;Rabin-Miller primality testing algorithm code @ TopCoder&lt;/strong&gt;&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As primes are in P, there is deterministic, polynomial time algorithm called &lt;strong&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/AKS_primality_test&quot; rel=&quot;nofollow&quot;&gt;AKS primality test.&lt;/a&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="705" LastEditorUserId="705" LastEditorDisplayName="" LastEditDate="2010-09-23T12:52:46.370" LastActivityDate="2010-09-23T12:52:46.370" CommentCount="3" />
  <row Id="5283" PostTypeId="2" ParentId="5278" CreationDate="2010-09-23T06:01:30.980" Score="5" ViewCount="" Body="&lt;p&gt;By definition a purely transcendental extension of $\mathbf{Q}$ is the function field of a rational variety over $\mathbf{Q}$, namely the fraction field of a polynomial ring $\mathbf{Q}[x_1,\ldots, x_n]$ for some $n$. Call your field $K$. Obviously the transcendence degree of $K$ over $\mathbf{Q}$ is 1, so it's enough to show that $K$ is not isomorphic to a rational function field in 1 variable, i.e. to $\mathbf{Q}(t)$. Now by inspection $K$ is the function field of the elliptic curve $E:y^2 = x^3-x$ over $\mathbf{Q}$. It's a standard fact, shown in various books concerned with algebraic curves, that $K\cong \mathbf{Q}(t)$ if and only if $E$ is birational to the projective line $\mathbf{P}^1_{\mathbf{Q}}$. This is not the case, because (for example) the genus of $E$ is 1 (say by the degree-genus formula for plane curves), the genus of $\mathbf{P}^1$ is zero, and the genus is a birational invariant of a curve. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know how to solve this problem with &quot;elementary&quot; (non-geometric) methods off the top of my head; hopefully someone else can explain how to do this.&lt;/p&gt;&#xA;" OwnerUserId="1265" LastActivityDate="2010-09-23T06:01:30.980" />
  <row Id="5284" PostTypeId="2" ParentId="5266" CreationDate="2010-09-23T06:05:59.897" Score="2" ViewCount="" Body="&lt;p&gt;Note that $corr(x,y) = E(xy)-E(x)E(y) = E(xy)$ and $var(x) = E(x^2)-E(x)^2=E(x^2)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$E(sx) = E(yx) - \frac{corr(y,x)}{var(x)}E(x^2) = E(yx) -\frac{E(xy)}{E(x^2)}E(x^2) = 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then also $y=s+\frac{corr(x,y)x}{var(x)}$ so &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$E(y|x) = E(s+\frac{corr(x,y)x}{var(x)}|x) = E(s|x) + E(\frac{corr(x,y)x}{var(x)}|x) = E(s) + \frac{corr(x,y)x}{var(x)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but $E(s)=0$ giving the result.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You will need to justify all these formulas and steps.&lt;/p&gt;&#xA;" OwnerUserId="1794" LastActivityDate="2010-09-23T06:05:59.897" CommentCount="2" />
  <row Id="5285" PostTypeId="2" ParentId="5278" CreationDate="2010-09-23T06:45:32.530" Score="13" ViewCount="" Body="&lt;p&gt;For convenience change the base field to $k=\mathbb{C}$&#xA;(if the extension were purely transcendental before&#xA;then it would be purely transcendental after).&#xA;As the extension has transcendance degree 1, then&#xA;if the extension were purely transcendental it would equal $k(x)$&#xA;for some $x$. Then there are nonconstant rational functions&#xA;$f(x)$ and $g(x)$ such that&#xA;$$g(x)^2=f(x)^3-f(x).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's easy to see we can write $f(x)=u(x)/w(x)^2$&#xA;and $g(x)=v(x)/w(x)^3$ in lowest terms&#xA;where $u$, $v$ and $w$ are polynomials.&#xA;We find that&#xA;$$\phi(x)=\frac{g'(x)}{3f(x)^2-1}=\frac{f'(x)}{2g(x)}.$$&#xA;If fact $\phi(x)$ is a polynomial. Otherwise the denominator&#xA;of $\phi$ has a factor $x-a$. Expressing $\phi$ in terms of $u$, $v$&#xA;and $w$ (details?) we see that this implies that&#xA;$$2g(a)=3f(a)^2-1=0$$&#xA;as well as&#xA;$$g(a)^2=f(a)^3-f(a).$$&#xA;This is impossible. As $f$ and $g$ are nonconstant, $\phi$ is a nonzero&#xA;polynomial.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now replace $f(x)$ and $g(x)$ by $f(1/x)$ and $g(1/x)$. Then $\phi(x)$&#xA;is replaced by $-x^{-2}\phi(1/x)$ so that is also a polynomial, which&#xA;it isn't. So we get the required contradiction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added&lt;/strong&gt; The above was composed in a rush, and I cut a few corners,&#xA;so some extra details. We have $f'(x)=\textrm{polynomial}/w(x)^3$ and so&#xA;$$\phi(x)=\frac{f'(x)}{g(x)}=\frac{\textrm{polynomial}}{v(x)}.$$&#xA;Similarly&#xA;$$\phi(x)=\frac{\textrm{polynomial}}{3u(x)^2-w(x)^4}.$$&#xA;If $x-a$ divides the denominator of $\phi(x)$ then $v(a)=0=u(a)^2-w(a)^4$.&#xA;We can't have $w(a)=0$ as then $x-a$ would divide $u(x)$ and $w(x)$&#xA;contrary to $u(x)/v(x)^2$ being in lowest terms. Hence $f(a)$ and $g(a)$&#xA;make sense and we get $g(a)=3f(a)^2-1=0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is really a geometric argument. On the elliptic curve&#xA;$$E:\qquad\qquad z^2=t^3-t$$&#xA;the &quot;invariant differential&quot;&#xA;$$\omega=\frac{dt}{2z}=\frac{dz}{3t^2-1}$$&#xA;would have no poles on the (projective curve) $E$. But were $E$ rational&#xA;then $\omega=\phi(x)dx$ has no poles on the projective line; but every&#xA;nonzero differential on the projective line has a pole. The above is just&#xA;a naive version of this argument, which works for all elliptic curves.&lt;/p&gt;&#xA;" OwnerUserId="226" LastEditorUserId="226" LastEditorDisplayName="" LastEditDate="2010-09-23T18:00:35.540" LastActivityDate="2010-09-23T18:00:35.540" CommentCount="2" />
  <row Id="5286" PostTypeId="1" CreationDate="2010-09-23T07:24:51.937" Score="5" ViewCount="295" Body="&lt;p&gt;I'm developing some instructional material for a Calculus 1 class and I wanted to know from experience for yourself, tutoring others, and/or helping people on this site where is the most difficulty in Calculus? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you had any good methods of helping people that would be very helpful. &lt;/p&gt;&#xA;" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-09-23T14:38:11.700" LastActivityDate="2010-09-23T14:44:28.830" CommunityOwnedDate="2010-09-23T11:15:58.493" Title="What concepts were most difficult for you to understand in Calculus?" Tags="&lt;calculus&gt;&lt;soft-question&gt;&lt;big-list&gt;&lt;teaching&gt;" AnswerCount="5" CommentCount="2" FavoriteCount="1" />
  <row Id="5287" PostTypeId="2" ParentId="5259" CreationDate="2010-09-23T07:33:27.643" Score="2" ViewCount="" Body="&lt;p&gt;This is quite a common error when doing Monte Carlo integration. The support of the random variable you choose must be equal to the range of integration. Though the Cauchy distribution has support on $\mathbb{R}$ we can adapt it slightly to make it work here. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that: $\int_0^1 \frac{1}{\pi(1+x^2)} dx = \frac{1}{\pi}\tan^{-1}(1)=\frac{1}{4}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus $g(x) = \frac{4}{\pi(1+x^2)}$ for $x\in(0,1)$ is a probability density with support on $(0,1)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;using this we have &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{1}{n}\sum_{i=0}^n \frac{f(x_i)}{g(x_i)} = \pi$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is not a problem! Since $g(x) = \frac{1}{\pi}f(x)$ you have found exactly the right probability distribution to use to evaluate $\int_0^1f(x)dx$! The error is zero for any sample, regardless of the size. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note however, I needed to know how to integrate $\int_0^1\frac{1}{\pi(1+x^2)}$ in the first place to form the probability distribution. Thus for arbitrary functions it is impossible to get this situation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also note the Monte Carlo is not a very good way of approximating integrals in general. Far better deterministic methods are quadrature rules. &lt;/p&gt;&#xA;" OwnerUserId="1794" LastActivityDate="2010-09-23T07:33:27.643" CommentCount="6" />
  <row Id="5288" PostTypeId="2" ParentId="5286" CreationDate="2010-09-23T07:35:51.060" Score="0" ViewCount="" Body="&lt;p&gt;The area formula, namely, how could&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{i=1}^{\infty} y_i \delta x_i=\int y dx$$&lt;/p&gt;&#xA;" OwnerUserId="550" LastActivityDate="2010-09-23T07:35:51.060" CommunityOwnedDate="2010-09-23T11:15:58.493" />
  <row Id="5289" PostTypeId="1" CreationDate="2010-09-23T07:48:12.340" Score="1" ViewCount="102" Body="&lt;p&gt;This is a similar question with &lt;a href=&quot;http://math.stackexchange.com/questions/5143/find-the-range-of-x-given-y-min-leq-yx-leq-y-max-where-yx-can&quot;&gt;this one&lt;/a&gt;. But with a bit of twist.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have two inequalities:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$y_{1min} \leq y_{1}(x_1,x_2) \leq y_{1max}$$&#xA;$$y_{2min} \leq y_{2}(x_1,x_2) \leq y_{2max}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$y_i(x_1,x_2)=\Biggl[\frac{(x_{1}^2x_{2}+b_{i}x_{1}x_{2}+c_{i})^{f_i}}{(x_{1}(2+x_{2})+e_{i})^{g_i}}\Biggr]$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $f_1=g_1=1, f_2=5, g_2=2.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$b_{i},c_{i},e_{i},x_1,x_2\in\mathbb{R}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question: How to find the range of $x_{1},x_{2}$ that satisfy the above inequalities?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;em&gt;range&lt;/em&gt; of $x_{1},x_{2}$ is defined as all the $x_1,x_2$ pairs that satisfy the above inequalities. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit: Maybe I'll relax the question a bit: let's assume that $b_{1}=b_{2}, c_{1}=c_{2}, e_{1}=e_{2}$.&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="550" LastEditorUserId="550" LastEditorDisplayName="" LastEditDate="2010-09-24T06:47:51.483" LastActivityDate="2010-09-24T06:47:51.483" Title="Find Range of $x_1,x_2$, where $y_{min} \leq y(x_1, x_2) \leq y_{max}$" Tags="&lt;optimization&gt;" CommentCount="7" />
  <row Id="5290" PostTypeId="2" ParentId="5286" CreationDate="2010-09-23T07:53:51.670" Score="8" ViewCount="" Body="&lt;p&gt;The hardest thing for me was to understand what is meant when someone writes $\mathrm d x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I still don't know...&lt;/p&gt;&#xA;" LastActivityDate="2010-09-23T07:53:51.670" CommunityOwnedDate="2010-09-23T11:15:58.493" CommentCount="2" />
  <row Id="5291" PostTypeId="1" AcceptedAnswerId="5292" CreationDate="2010-09-23T08:32:35.283" Score="3" ViewCount="235" Body="&lt;p&gt;Hi &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am having trouble answering the following question:&lt;br&gt;&#xA;&lt;br&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Problem&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By integrating the function:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f(z) = \frac{R + z}{z(R - z)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;round a suitable contour, prove that, for $0 \leq r &amp;lt; R$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{1}{2\pi}\int_0^{2\pi} \frac{R^2-r^2}{R^2 - 2Rr\cos(\theta) + r^2}\mathrm{d}\theta = 1   $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My Attempt&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I &lt;em&gt;think&lt;/em&gt; they want me to do is to use the equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_C f(z) \mathrm{d}z  =  \int_{z_1}^{z_2} f(z) \mathrm{d}z = \int_a^bf(z(t))\cdot z'(t)\mathrm{d}t  $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where C is a contour from $z_1$ to $z_2$ and $z(a) = z_1$ and $z(b) = z_2$. Then I should find a contour $z(t)$ in terms of $r$ and $t$ such that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f(z(t))\cdot z'(t) = \frac{R^2-r^2}{R^2 - 2Rr\cos(t) + r^2}  $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And then with this contour I could find values for $z_1$ and $z_2$ meaning I could evaluate:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \int_{z_1}^{z_2} f(z) \mathrm{d}z  $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which should equal 2$\pi$ or something. However so far I have not been able to find a suitable z(t). The closest I have come is $z(t) = re^{i\theta}$ which yields:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f(z(t))\cdot z'(t) = \frac{R^2\cdot i - 2Rr\sin(t) + r - r^2\cdot i}{R^2 - 2Rr\cos(t) + r^2}  $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However this entire approach could be wrong. I've got to the stage when you start to question your whole approach and feel like resorting to violence...According to Einstein, only insane people do the same thing over and over again and expect different results......Oh wait that's me...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please help me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit 1&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;First of all a big thank you to everyone who has helped me here - your hints are like a life raft to a drowning man! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To Mariano: I have not done anything on residues at this stage in the course - so the question &lt;em&gt;should&lt;/em&gt; be answerable without using them.  That technique does look very powerful though and I hope I will get a good understanding of it in the future...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To Hans Lundmark: Your link helped me to see the bigger picture put this equation in context (wow its more that a figment of my prof's imagination!)...Thanks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To J. M: Your note helped a lot to remove the log jam in my head but there are still some issues that remain with my answer. I would very much appreciate your comments on my 2nd attempt. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;2nd Attempt&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Step 1)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The function $f(z)$ can be decomposed using partial fractions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f(z) = \frac{R + z}{z(R - z)}  = \frac{1}{z} + \frac{2}{R - z}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and then integrated with respect to $z$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int f(z)dz = log(z) - 2log(R - z) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Step 2)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With this information we can evaluate:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_C f(z)dz $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where C is the positively oriented circle $z = re^{i\theta}$. Because log(z) is not defined at its branch cut we cannot integrate around the whole circle but must split the circle into halves $C_1$ and $C_2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $C_1$ denote the right half of the circle where $ -\frac{\pi}{2} &amp;lt;= \theta &amp;lt;= \frac{\pi}{2}$ and consider the branch: $log(z) = ln(r) + i\theta$ where $-\pi &amp;lt;= \theta &amp;lt;= \pi$. Therefore:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \int_{C_1} f(z) = [log(z) - 2log(R - z)]_{-ri}^{ri} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which equals:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \int_{C_1} f(z)dz = \pi{i} - 2log(R - ri) + 2log(R + ri) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $C_2$ denote the left half of the circle where $\frac{\pi}{2} &amp;lt;= \theta &amp;lt;= \frac{3\pi}{2}$ and consider the branch: $log(z) = ln(r) + i\theta$ where $0 &amp;lt;= \theta &amp;lt;= 2\pi$. Therefore:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \int_{C_2} f(z) = [log(z) - 2log(R - z)]_{ri}^{-ri} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which equals:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \int_{C_2} f(z)dz = \pi{i} - 2log(R + ri) + 2log(R - ri) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and since:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \int_{C} f(z)dz = \int_{C_1} f(z)dz + \int_{C_2} f(z)dz  $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;it follows that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \int_{C} f(z)dz = 2\pi{i}  $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Step 3)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the part I am currently having difficulty with:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does the equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_C f(z) \mathrm{d}z  =  \int_a^bf(z(t))\cdot f'(t) dt $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;imply that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Im\int_C f(z) \mathrm{d}z  =  \int_a^b\Im[{f(z(t))}\cdot f'(t) dt] $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because if it did I could say that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \Im(2\pi{i}) = \int_0^{2\pi} \frac{R^2-r^2}{R^2 - 2Rr\cos(\theta) + r^2}\mathrm{d}\theta   $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and the proof would be complete...But I dont know if I can do this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can some one help please?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit 2&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Thanks to J.M step 3 should be:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the integral operator is linear:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \int z(t) \mathrm{d}t=\int \Re z(t) \mathrm{d}t+i\int \Im z(t) \mathrm{d}t$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;therfore:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \int_C f(z) \mathrm{d}z  =  \int_a^b\Re{f(z(t))}\cdot f'(t) dt + i\int_a^b\Im{f(z(t))}\cdot f'(t) dt $$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;also we know that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \int_{0}^{2\pi}\Re[{f(z(t))}\cdot f'(t) dt] = 0 $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \int_C f(z) \mathrm{d}z = 2{\pi}i $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \Im[f(z(t))\cdot z'(t)] = \frac{R^2-r^2}{R^2 - 2Rr\cos(t) + r^2}  $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \int_C f(z) \mathrm{d}z  =  i\int_0^{2\pi}\Im[{f(z(t))}\cdot f'(t)] dt $$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ 2{\pi}i  =  i\int_0^{2\pi}\Im[{f(z(t))}\cdot f'(t)] dt $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ 2{\pi}  =  \int_0^{2\pi}\Im[{f(z(t))}\cdot f'(t)] dt $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ 2{\pi}  =  \int_0^{2\pi}\frac{R^2-r^2}{R^2 - 2Rr\cos(t) + r^2}dt  $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and the proof is complete. Awesome!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope this helps some other dude out there who is trying to understand this stuff...&lt;/p&gt;&#xA;" OwnerUserId="2081" LastActivityDate="2010-09-25T17:40:36.797" Title="How to find a suitable contour to integrate round?" Tags="&lt;complex-analysis&gt;&lt;integral&gt;" AnswerCount="1" CommentCount="8" FavoriteCount="1" />
  <row Id="5292" PostTypeId="2" ParentId="5291" CreationDate="2010-09-23T08:58:53.410" Score="3" ViewCount="" Body="&lt;p&gt;Note that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Im\left(\frac{R+r\exp(i\theta)}{r\exp(i\theta)(R-r\exp(i\theta))}(i r\exp(i\theta))\right)=\frac{R^2-r^2}{R^2+r^2-2Rr\cos\theta}$$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;As for your second attempt, remember that the integral is a &lt;em&gt;linear&lt;/em&gt; operator, so&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int z(t) \mathrm{d}t=\int \Re z(t) \mathrm{d}t+i\int \Im z(t) \mathrm{d}t$$&lt;/p&gt;&#xA;" OwnerUserId="498" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-09-24T07:37:25.970" LastActivityDate="2010-09-24T07:37:25.970" CommentCount="3" />
  <row Id="5293" PostTypeId="1" AcceptedAnswerId="5299" CreationDate="2010-09-23T09:09:58.437" Score="1" ViewCount="388" Body="&lt;p&gt;An orthogonal matrix is a matrix $A$ over the reals such that $A^t=A^{-1}$ (its transpose is its inverse). The Frobenius norm over $n\times n$ real matrices is given by $\|A\| = \sqrt{trace(A^tA)}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have come across the following claim: The distance (induced by the Frobenius norm) between any two (non equal) orthogonal matrices is $\sqrt{n}$. I can't find a proof for this claim, but no refutation either (of course, if the difference between two orthogonal matrices is itself an orthogonal matrix the claim is clear, but I don't know if that's true either).&lt;/p&gt;&#xA;" OwnerUserId="1818" LastActivityDate="2010-09-23T12:37:42.073" Title="The distance between orthogonal matrices induced by the Frobenius norm" Tags="&lt;linear-algebra&gt;&lt;matrices&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="5295" PostTypeId="2" ParentId="5293" CreationDate="2010-09-23T09:23:55.657" Score="5" ViewCount="" Body="&lt;p&gt;I don't think this is true. Let $A=I_2$ and $B$ be 2 by 2 matrix with $b_{11}=b_{22}=0$ and $b_{12}=b_{21}=1$. Then both these matrices are orthogonal but &lt;/p&gt;&#xA;&#xA;&lt;p&gt;and $||A-B||_F = 2\neq \sqrt{2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An even easier counterexample: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $A=1$ and $B=-1$ the 1 by 1 orthogonal matrices. Then $A-B=2$ so $||A-B||_F=2\neq 1$. &lt;/p&gt;&#xA;" OwnerUserId="1794" LastEditorUserId="1794" LastEditorDisplayName="" LastEditDate="2010-09-23T09:29:08.910" LastActivityDate="2010-09-23T09:29:08.910" />
  <row Id="5297" PostTypeId="1" CreationDate="2010-09-23T09:55:21.410" Score="10" ViewCount="484" Body="&lt;p&gt;In &lt;a href=&quot;http://math.stackexchange.com/questions/2865/is-a-direct-limit-of-topological-groups-always-a-topological-group&quot;&gt;Direct limit&lt;/a&gt;, Martin rightly pointed out that my naive construction (now deleted) of the colimit (direct limit) of topological abelian groups was wrong. He shows how to do it properly (at least the coproduct) &lt;a href=&quot;http://math.stackexchange.com/questions/5095/what-are-the-product-and-coproduct-in-the-category-of-topological-groups&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since then, I've been lurking some of the literature about the subject and this problem of the colimit of topological groups (about which I had previously no idea) seems at the same time classical and topical. For instance, &lt;a href=&quot;http://projecteuclid.org/DPubS?service=UI&amp;amp;version=1.0&amp;amp;verb=Display&amp;amp;handle=euclid.kjm/1250518067&quot; rel=&quot;nofollow&quot;&gt;this 1998's paper&lt;/a&gt; points out that the Encyclopedic Dictionary of Mathematics, second edition, MIT (1987), article 210, has made the same mistake that I did, stating that the direct limit of topological groups, with the &lt;em&gt;inductive limit of topologies&lt;/em&gt; (my naive attempt) has always a continuous multiplication. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The authors of this paper show a counter-example (example 1.2, page 553) and here is my question: I must be absolutely dumb, but don't understand it. Could anyone help me?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For those who don't have access to the paper, here is the example.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $G_n = \mathbb{Q} \times \mathbb{R}^n$ with the usual topology. Imbed $G_n \hookrightarrow G_{n+1}$ as $x \mapsto (x,0)$. Then, as a plain abelian group, $G = \varinjlim_{n} G_n= \mathbb{Q} \times \prod'\mathbb{R}$, where $\prod'\mathbb{R}$ denotes the &lt;em&gt;weak&lt;/em&gt; or &lt;em&gt;restricted&lt;/em&gt; product, which is the way guys in this area call the &lt;em&gt;direct sum&lt;/em&gt;; that is, elements of $\prod'\mathbb{R}$ are infinite tuples $( x_1, \dots , x_n, \dots )$, in which all of its components $x_n \in \mathbb{R}$ are zero, except a finite number of them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;em&gt;inductive limit topology&lt;/em&gt; is the finest one that makes all the inclusions $G_n \hookrightarrow G$ continuous. That is, $U \subset G$ is open if and only if $U\cap G_n$ is open for all $n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's see how they show that, with this inductive topology, the &quot;product&quot; (in fact, addition) $\mu : G \times G \longrightarrow G$ is &lt;strong&gt;not&lt;/strong&gt; continuous ($\mu$ is the operation induced as an honest colimit of groups -no topologies- by the operations $\mu_n : G_n \times G_n \longrightarrow G_n$, which I assume are the usual additions of those linear spaces). In plain English:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;(x_0, x_1, \dots , x_n , \dots ) + (y_0, y_1, \dots , y_n , \dots ) = (x_0+y_0, x_1 + y_1 , \dots , x_n + y_n , \dots ) \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, in this situation it's enough to produce an open neighbourhood $U \subset G$ of the neutral element $e \in G$ such that $V^2$ is not in $U$, for any open neighbourhood $V$ of $e$ -where, I assume, $V^2$ means $V + V$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ok, here is the guy that is supposed to ruin (and sure it does) my naive attempt:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;U = \left\{ &#xA;x = (x_0, x_1, \dots , x_n , \dots ) \quad \vert \quad \vert x_j\vert &amp;lt; \vert \cos (jx_0) \vert \ , 1 \leq j&#xA;\right\}  \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This guy is an open set of $G$ because $x_0$ being a rational number guarantees $\cos (jx_0) \neq 0$ for all $j$. Assume there is an open neighbourhood $V$ of $e$ such that $V^2 \subset U$. Then, $V \cap G_j$ contains an open interval $(-\varepsilon_j , \varepsilon_j)$ in $\mathbb{R}$ with $\epsilon_j &gt; 0$ such that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;(-\varepsilon_0 , \varepsilon_0) \times (-\varepsilon_j , \varepsilon_j) \subset &#xA;\left\{&#xA;(x_0 , x_j) \in \mathbb{Q} \times \mathbb{R} \quad \vert \quad \vert x_j\vert &amp;lt; \vert \cos (jx_0) \vert&#xA;\right\}  \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And here come the two final sentences of the example I don't understand: &lt;strong&gt;This is impossible if $2j\varepsilon_0 &gt; \pi$. A contradiction.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any hints or remarks (even humiliating ones) will be welcome.&lt;/p&gt;&#xA;" OwnerUserId="664" LastEditorUserId="664" LastEditorDisplayName="" LastEditDate="2010-09-24T05:29:07.693" LastActivityDate="2010-10-03T05:53:24.397" Title="Colimit of topological groups (again)" Tags="&lt;topology&gt;&lt;general-topology&gt;" AnswerCount="3" CommentCount="14" FavoriteCount="5" />
  <row Id="5298" PostTypeId="2" ParentId="5286" CreationDate="2010-09-23T10:16:45.010" Score="1" ViewCount="" Body="&lt;p&gt;I think the entire concept behind integration is hard to grasp for students who are not familiar with analysis. They tend to think of it only as the &quot;inverse operation of derivatives&quot;, which is quite restrisctive, in my opinion.&lt;/p&gt;&#xA;" OwnerUserId="1415" LastActivityDate="2010-09-23T10:16:45.010" CommunityOwnedDate="2010-09-23T11:15:58.493" />
  <row Id="5299" PostTypeId="2" ParentId="5293" CreationDate="2010-09-23T10:40:17.970" Score="4" ViewCount="" Body="&lt;p&gt;I don't know where you found your claim, but it seems that the distance induced by the Frobenius norm between any two orthogonal matrices can be any real number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{align}&#xA;\|A - B\|^2 &amp;amp;= \mathrm{tr} \left( (A-B)^t (A-B)    \right) \\&#xA;            &amp;amp;= \mathrm{tr} \left( (A^t -B^t) (A-B)  \right)  \\&#xA;            &amp;amp;= \mathrm{tr} (A^tA -A^tB - B^tA + B^tB)    \\&#xA;            &amp;amp;= \mathrm{tr} (2I) - \mathrm{tr}(2A^tB)  \\&#xA;            &amp;amp;= 2n - 2\mathrm{tr}(A^tB)&#xA;\end{align}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The last but one equality is due to the fact that $A^tA = B^tB = I$ and $(B^tA)^t = A^tB$ and a matrix and its transpose have the same trace.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, take $A = I$ and we've got&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\| I - B\|^2 = 2n - 2\mathrm{tr}(B)&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for any orthogonal matrix $B$. So, if you take as $B$ the family of orthogonal matrices&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{pmatrix}&#xA;\cos\theta  &amp;amp;  -\sin\theta \\&#xA;\sin\theta  &amp;amp;  \cos\theta&#xA;\end{pmatrix} \ ,&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\theta \in [0, 2\pi]$, their traces $\mathrm{tr} (B) = 2\cos\theta$ can be any real number between $-2$ and $2$. So, their Frobenius distances to the unit matrix $I$ can be any real number from $0$ to $\sqrt{8}$.&lt;/p&gt;&#xA;" OwnerUserId="664" LastEditorUserId="664" LastEditorDisplayName="" LastEditDate="2010-09-23T12:37:42.073" LastActivityDate="2010-09-23T12:37:42.073" />
  <row Id="5300" PostTypeId="1" AcceptedAnswerId="5301" CreationDate="2010-09-23T10:42:06.343" Score="0" ViewCount="245" Body="&lt;p&gt;a. For every real number c, neither f nor g is continuous at c&lt;/p&gt;&#xA;&#xA;&lt;p&gt;b. For every real number c, the sum f + g is continuous at c&lt;/p&gt;&#xA;&#xA;&lt;p&gt;c. For every real number c, the product fg is continous at c&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint: consider functions defined one way on rationals and another on irrationals&lt;/p&gt;&#xA;" OwnerUserId="1958" LastActivityDate="2010-09-23T17:30:14.290" Title="Give examples of functions f and g so that:" Tags="&lt;calculus&gt;&lt;limit&gt;" AnswerCount="2" CommentCount="5" />
  <row Id="5301" PostTypeId="2" ParentId="5300" CreationDate="2010-09-23T10:49:42.680" Score="3" ViewCount="" Body="&lt;p&gt;Take $f=1_\mathbb Q$ (that is, $f(x)=1$ if $x$ is rational and $f(x)=0$ otherwise) and $g=1-f$. Then neither $f$ nor $g$ is continuous at any point but $f+g$ and $fg$ are very simple.&lt;/p&gt;&#xA;" OwnerUserId="367" LastActivityDate="2010-09-23T10:49:42.680" />
  <row Id="5302" PostTypeId="2" ParentId="5286" CreationDate="2010-09-23T11:13:59.747" Score="4" ViewCount="" Body="&lt;p&gt;I really struggled with the $\epsilon-\delta$ definition of limits, especially for non-linear functions.  This was also my first exposure to proof, as in: Prove that $$\lim_{x \to 2} (x^2 + 3) = 7$$ and I had a hard time with it at first.  To be clear, computing these limits was no problem, but using the definition to prove they were correct really confused me.&lt;/p&gt;&#xA;" OwnerUserId="506" LastActivityDate="2010-09-23T11:13:59.747" CommunityOwnedDate="2010-09-23T11:15:58.493" CommentCount="5" />
  <row Id="5303" PostTypeId="1" AcceptedAnswerId="5304" CreationDate="2010-09-23T12:48:32.990" Score="8" ViewCount="794" Body="&lt;p&gt;The motivation for this question is : &lt;a href=&quot;http://math.stackexchange.com/questions/4066/rationals-of-the-form-fracpq-where-p-q-are-primes-in-a-b&quot;&gt;http://math.stackexchange.com/questions/4066/rationals-of-the-form-fracpq-where-p-q-are-primes-in-a-b&lt;/a&gt; and some other problems in Mathematics which looks as if they are elementary but their proofs are very much sophisticated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to consider two famous questions: First the &quot;Fermat's Last Theorem&quot; and next the unproven &quot;Goldbach conjecture&quot;. These questions appear elementary in nature, but require a lot of Mathematics for even comprehending the solution. Even the problem, which i posed in the link is so elementary but i don't see anyone even giving a proof without using the prime number theorem. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the question is: Why is this happening? If i able to understand the question, then i should be able to comprehend the solution as well. A Mathematician once quoted: Mathematics is the understanding of how nature works. Is nature so complicated that a common person can't understand as to how it works, or is it we are making it complicated. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the same time, i appreciate the beauty of Mathematics also: Paul Erdos proof of Bertrands postulate is something which i admire so much because, of its elementary nature. But at the same time i have my scepticism about FLT and other theorems.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I have stated 2 examples of questions which appear elementary, but the proofs are intricate. I know some other problems, in number theory which are of this type. Are there any other problems of this type, which are not Number Theoritical? If yes, i would like to see some of them.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" LastEditorUserId="468" LastEditorDisplayName="anon" LastEditDate="2011-05-27T01:45:58.683" LastActivityDate="2011-05-27T01:45:58.683" Title="Why do statements which appear elementary have complicated proofs?" Tags="&lt;soft-question&gt;&lt;logic&gt;&lt;philosophy&gt;&lt;proof-theory&gt;" AnswerCount="6" CommentCount="13" FavoriteCount="0" />
  <row Id="5304" PostTypeId="2" ParentId="5303" CreationDate="2010-09-23T13:57:51.133" Score="16" ViewCount="" Body="&lt;p&gt;Simple questions often have complex answers, or no answers at all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you were to ask me why the sky was blue, to give a complete answer I would have to describe the heliocentric model, the earth's atmosphere, and the electromagnetic spectrum. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you asked me how my computer connected to the internet, the answer would take a considerable amount of time to explain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you asked my whether there was life on other planets, I wouldn't be able to give an answer; we just don't know.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Often, it is impossible to give a simple answer to a simple question. This is true in any field you might encounter, ranging from the sciences to the humanities. And it is true in mathematics,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mathematics allows us to formalize our questions within an axiomatic structure. It lets us ask our questions more precisely. But it in no way guarantees that the simplicity of the question would translate into simplicity of the answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some other simple problems which cannot be answered simply:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Four color theorem states that any map made up of continuous regions can be colored with 4 colors such that each region gets 1 color and no two adjacent regions get the same color. The proof is quite complex, requiring the use of a computer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Scheinerman's conjecture is the conjecture that &quot;every planar graph is the intersection graph of a set of line segments in the plane&quot; (sourced from &lt;a href=&quot;http://en.wikipedia.org/wiki/Scheinerman%27s_conjecture&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Scheinerman%27s_conjecture&lt;/a&gt;). However, the proof was only completed in 2009, and is fairly difficult.&lt;/p&gt;&#xA;" OwnerUserId="796" LastActivityDate="2010-09-23T13:57:51.133" CommentCount="1" />
  <row Id="5306" PostTypeId="2" ParentId="5286" CreationDate="2010-09-23T14:44:28.830" Score="2" ViewCount="" Body="&lt;p&gt;At the school I was taught to look at the derivative as the instantaneous rate of change and that fit well with applications in physics. But later, when I was learning Economics in college, I had to learn to look at the derivative as the best linear (affine) approximation, and a differentiable function as a function which had 'good' linear approximations. That is also the intuition that generalizes to many variables. I wish it had been discussed in my early calculus classes.&lt;/p&gt;&#xA;" OwnerUserId="1195" LastActivityDate="2010-09-23T14:44:28.830" CommunityOwnedDate="2010-09-23T14:44:28.830" CommentCount="1" />
  <row Id="5307" PostTypeId="1" CreationDate="2010-09-23T15:01:34.520" Score="2" ViewCount="161" Body="&lt;p&gt;I'm trying to find a percentage to describe how close a generated frequency distribution is to an ideal distribution, but am having a total blank. Basically I need to rate distributions&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So say I have the below frequency distributions for a sample size of say 100:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;A   B&#xA;1   0&#xA;2   0&#xA;3   0&#xA;4   100&#xA;5   0&#xA;6   0&#xA;7   0&#xA;8   0&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The above is my ideal distribution and would receive a rating of 100%&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;A   B&#xA;1   1&#xA;2   4&#xA;3   15&#xA;4   50&#xA;5   20&#xA;6   6&#xA;7   3&#xA;8   1&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The above would also be considered ideal as well with the spike falling on the same frequency and the count is sufficiently  larger then the surrounding values so a rating of 100% (or close to)&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;A   B&#xA;1   15&#xA;2   50&#xA;3   20&#xA;4   8&#xA;5   4&#xA;6   2&#xA;7   1&#xA;8   0&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This while having a sufficient sized spike the spiking is happening on the wrong frequency so I want it to be weighted more towards spike happening on the right frequency then the size of the spike. So this might get rated at 50%&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;A   B&#xA;1   5&#xA;2   30&#xA;3   10&#xA;4   4&#xA;5   6&#xA;6   10&#xA;7   30&#xA;8   5&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Then something like this with two large spikes, but one being close to the required frequency might receive a rating of 20%&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically I'm after a formula to rate a distribution to how close it is to ideal with more weighting towards getting the frequency right (ie the closer the spike is to the ideal frequency the better the score, but the score will also be effected by the the size of the spike and other spikes etc etc.)&lt;/p&gt;&#xA;" OwnerUserId="2089" LastActivityDate="2010-09-23T15:13:12.867" Title="Quantifying difference between two frequency distributions" Tags="&lt;normal-distribution&gt;" AnswerCount="1" />
  <row Id="5308" PostTypeId="2" ParentId="5307" CreationDate="2010-09-23T15:06:09.973" Score="2" ViewCount="" Body="&lt;p&gt;Maybe the statistics stack exchange site will give you better answers. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://stats.stackexchange.com/questions/4/assessing-the-significance-of-differences-in-distributions&quot;&gt;http://stats.stackexchange.com/questions/4/assessing-the-significance-of-differences-in-distributions&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This link seems to give the answers you seek. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Good luck!&lt;/p&gt;&#xA;" OwnerUserId="1794" LastEditorUserId="1794" LastEditorDisplayName="" LastEditDate="2010-09-23T15:13:12.867" LastActivityDate="2010-09-23T15:13:12.867" CommentCount="2" />
  <row Id="5309" PostTypeId="1" AcceptedAnswerId="5310" CreationDate="2010-09-23T15:18:21.037" Score="3" ViewCount="83" Body="&lt;p&gt;Let G be a topological group and let $s_1$ and $s_2$ be loops in G (both loops are based at the identity e of G). Is it true that the loop $s_1s_2$ (where the multiplication is the one of the group structure of G) is equal, in $\pi_1(G,e)$, to the loop $s_1*s_2$ where this product is given by first going around $s_1$ and then $s_2$ (i.e., do we have $[s_1s_2] = [s_1*s_2]$)? If yes, what is the proof for this? &lt;/p&gt;&#xA;" OwnerUserId="1907" LastActivityDate="2010-09-23T15:27:18.733" Title="Topological group: Multiplying two loops is homotopic to linking these paths?" Tags="&lt;algebraic-topology&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="2" />
  <row Id="5310" PostTypeId="2" ParentId="5309" CreationDate="2010-09-23T15:27:18.733" Score="2" ViewCount="" Body="&lt;p&gt;Yes and this is known as the Eckmann-Hilton argument. (&lt;a href=&quot;http://en.wikipedia.org/wiki/Eckmann%E2%80%93Hilton_argument&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;)&lt;/p&gt;&#xA;" OwnerUserId="1650" LastActivityDate="2010-09-23T15:27:18.733" CommentCount="1" />
  <row Id="5311" PostTypeId="2" ParentId="227" CreationDate="2010-09-23T15:32:47.960" Score="0" ViewCount="" Body="&lt;p&gt;I think game theory is becoming increasingly important for economists. In addition to classical models such as matrix games (zero-sum and examples like Prisoner's Dilemma and Chicken) and the work of John Nash on equilibria and bargaining, there is the Gale-Shapley model for two-markets, mechanism design (a recent Nobel Memorial Prize was awarded for this work), auctions, fair division methods, group decision making, etc.. &lt;/p&gt;&#xA;" OwnerUserId="1369" LastActivityDate="2010-09-23T15:32:47.960" CommunityOwnedDate="2010-09-23T15:32:47.960" />
  <row Id="5312" PostTypeId="1" AcceptedAnswerId="5313" CreationDate="2010-09-23T16:27:49.820" Score="3" ViewCount="262" Body="&lt;p&gt;Suppose we have a 3x3x3 cube of cubes (i. e. 27 cubes) and consider the graph whose vertices are the cubes, and let an edge connect two iff they have touching faces.  I'm quite sure there is no Hamilton path that starts at the middle cube and ends in a corner cube, though I'm not sure how to prove it without tedious casework.&lt;/p&gt;&#xA;" OwnerUserId="1960" LastActivityDate="2010-09-24T20:40:01.887" Title="3-dimensional graph of cubes" Tags="&lt;graph-theory&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="5313" PostTypeId="2" ParentId="5312" CreationDate="2010-09-23T16:30:39.640" Score="4" ViewCount="" Body="&lt;p&gt;If you give 3D coordinates to the cube nodes $(x,y,z)$. Two nodes are adjacent iff they differ by exactly $1$ in exactly one of the coordinates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider $x+y+z$. The parity(whether it is even or odd) of this flips each time you move to an adjacent node.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The parity of the center cube and the corner cube are different. In 26 moves, you will end up at a node with the same parity as the initial node.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So no Hamiltonian Path can exist which starts at the center cube and ends at a corner cube.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-09-23T16:30:39.640" CommentCount="2" />
  <row Id="5314" PostTypeId="1" CreationDate="2010-09-23T16:58:23.503" Score="5" ViewCount="146" Body="&lt;p&gt;Consider the following problem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are given a multiset (a set with repetitions allowed) of $2n+1$ real numbers, say $S = \{r_1, \dots, r_{2n+1}\}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;These numbers are such that for every $k$, the multiset $S - \{r_k\}$ can be split into two multisets of size $n$ each, such that the sum of numbers in one multiset is same as the sum of numbers in the other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Show that all the numbers must be equal.( i.e. $r_{i} = r_{j}$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please stop reading further if you want to try and solve this problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Spoiler:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Now this problem can easily be solved using Linear Algebra. We have a set of $2n+1$ linear equations, which corresponds to a matrix equation $Ar = 0$. It can be shown that $A$ has rank at least $2n$ which implies the result.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The question is, is there any solution to this problem which does not involve any linear algebra? &lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2011-03-07T04:28:59.537" LastActivityDate="2011-03-07T04:28:59.537" Title="Odd number of reals with equal partitions" Tags="&lt;combinatorics&gt;" AnswerCount="1" CommentCount="6" FavoriteCount="0" />
  <row Id="5315" PostTypeId="1" AcceptedAnswerId="5320" CreationDate="2010-09-23T16:59:56.350" Score="4" ViewCount="263" Body="&lt;p&gt;Let $X$ be a set, and $d,d'$ two metrics on $X$.  Consider the identity map $i : (X,d) \to (X,d')$ as a map of metric spaces.  There are (at least) three reasonable notions of equivalence for $d$ and $d'$, in increasing order of strength:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;$i$ is a homeomorphism, i.e. $d$ and $d'$ induce the same topology on $X$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$i$ and $i^{-1}$ are uniformly continuous.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$i$ is bilipschitz, i.e. $C_1 d' \le d \le C_2 d'$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I would like to know what terms are used for these notions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, Mathworld says that the term &quot;equivalent&quot; refers to sense 1.  This seems counterintuitive since, for instance, sense 1 does not preserve completeness.  Munkres's &lt;em&gt;General Topology&lt;/em&gt; uses &quot;metrically equivalent&quot; for sense 2.  Does this agree with people's experience of standard usage?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Edit:&lt;/b&gt; I will point out that 3 implies 2 implies 1 (since Lipschitz implies uniformly continuous implies continuous) but converses are false.  Let $X = \mathbb{R}$, let $d_1(x,y) = |x-y|$, $d_2(x,y) = |x-y| \wedge 1$, $d_3(x,y) = |\phi(x)-\phi(y)|$, where $\phi : \mathbb{R}\to (0,1)$ is your favorite homeomorphism.  In sense 1 all three are equivalent, in sense 2 $d_1 \sim d_2 \not\sim d_3$, and in sense 3 all are inequivalent.  In particular note that $d_1, d_2$ are complete but $d_3$ is not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Edit:&lt;/b&gt; Further confusing the matter is the fact that if $X$ is a vector space and $d,d'$ are induced by norms $||\cdot||, ||\cdot||'$, then all three senses coincide, and sense 3 is usually taken as the definition of &quot;equivalent norms.&quot;&lt;/p&gt;&#xA;" OwnerUserId="822" LastEditorUserId="822" LastEditorDisplayName="" LastEditDate="2010-09-23T18:40:30.957" LastActivityDate="2010-09-23T18:40:30.957" Title="Notions of equivalent metrics" Tags="&lt;terminology&gt;&lt;topology&gt;&lt;metric-spaces&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="5316" PostTypeId="2" ParentId="5277" CreationDate="2010-09-23T17:08:19.857" Score="3" ViewCount="" Body="&lt;p&gt;&lt;em&gt;How do I mathematically determine if a number is prime?&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the number is n, then dividing it by every prime number less than or equal to sqrt(n) and showing that there is a remainder.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are a number of different &lt;a href=&quot;http://en.wikipedia.org/wiki/Generating_primes&quot; rel=&quot;nofollow&quot;&gt;sieve&lt;/a&gt; solutions for finding prime numbers, the oldest and most famous of which is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Sieve_of_Eratosthenes&quot; rel=&quot;nofollow&quot;&gt;Sieve of Eratosthenes&lt;/a&gt;.  These are generally easy to programme and can, for example, find all of the primes below 100,000 in a few milliseconds on a modern processor.&lt;/p&gt;&#xA;" OwnerUserId="1032" LastEditorUserId="1032" LastEditorDisplayName="" LastEditDate="2010-09-27T09:42:35.480" LastActivityDate="2010-09-27T09:42:35.480" />
  <row Id="5317" PostTypeId="2" ParentId="5300" CreationDate="2010-09-23T17:30:14.290" Score="2" ViewCount="" Body="&lt;p&gt;This isn't specific to the real numbers.  The domain, range, sum and product could be specified (almost) arbitrarily.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $S= f+g$ and $P=fg$.  These are continuous functions.  $f$ and $g$ are roots of a quadratic equation with continuous coefficients, $z^2 - Sz + P$.  There is a formula expressing the solution continuously in terms of $S$ and $P$, using one extraction of a square root.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only possible source of discontinuity is in the selection of which sign of the square root is used to define $f$ and which one for $g$.   To attain discontinuity everywhere, one needs to have $S^2 \neq 4P$ everywhere (so that $f,g$ are always distinct) and for the set where the plus sign is used for determining $f$ to be dense with dense complement.  Hence, a splitting of the domain into two dense sets (such as rational and irrational) is not an artificial feature of the answer posted earlier, it is necessary and sufficient for the solution. &lt;/p&gt;&#xA;" OwnerUserId="467" LastActivityDate="2010-09-23T17:30:14.290" />
  <row Id="5318" PostTypeId="2" ParentId="5314" CreationDate="2010-09-23T17:57:23.190" Score="6" ViewCount="" Body="&lt;p&gt;You can't avoid some sort of algebra, because the statement is false in a commutative group where $nx = 0$ has nontrivial solutions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you allow use of the linear algebra fact that rank is the same over any field containing the coefficients of the equations, it is enough to consider rational (and thus integer) solutions and extra structure is available.  One can then avoid use of determinants or matrices:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $\Sigma$ is the sum of all elements, $\Sigma - r_i$ is even and thus all $r_i$ have the same parity.  We can replace each $r_i$ by $(r_i-r_k)/2$ and get a smaller solution, where $r_k$ is the smallest of the numbers.  This process ends at the zero solution, and is reversible, so the original solution has all numbers equal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(added:  you can consider this a use of either the real or 2-adic metric on integers, so this must correspond to a linear algebra argument using inequalities or reduction mod 2^(2n+1) on the system of equations or its determinant.)&lt;/p&gt;&#xA;" OwnerUserId="467" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-09-23T18:11:29.450" LastActivityDate="2010-09-23T18:11:29.450" CommentCount="4" />
  <row Id="5319" PostTypeId="2" ParentId="5248" CreationDate="2010-09-23T18:01:22.667" Score="5" ViewCount="" Body="&lt;p&gt;Here is a sketch of another elementary solution based on a proof in Bromwich's Theory of Infinite Series.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using $\sin(2k+1)x-\sin(2k-1)x = 2\cos2kx\sin x$ and summing from k=1 to k=n we have&#xA;$$\sin(2n+1)x = \sin x \left( 1+ 2 \sum_{k=1}^n \cos 2kx \right),$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and hence&#xA;$$ \int_0^{\pi/2} {\sin(2n+1)x \over \sin x} dx = \pi/2. \qquad (1)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $y=(2n+1)x$ and this becomes&#xA;$$ \int_0^{(2n+1)\pi/2} {\sin y \over (2n+1) \sin (y/(2n+1))} dy = \pi/2.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and since $\lim_{n \to \infty} (2n+1) \sin { y \over 2n+1} = y$ it suggests that there is a proof lurking in there somewhere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So let's put&#xA;$$\begin{align}&#xA;I_n &amp;amp;=  \int_0^{n\pi/(2n+1)} {\sin(2n+1)x \over \sin x} dx&#xA;\ &amp;amp;= \sum_{k=0}^{n-1} \int_{k\pi/(2n+1)}^{(k+1)\pi/(2n+1)} {\sin(2n+1)x \over \sin x} dx.&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence we have $I_n = u_0 – u_1 + u_2 \cdots + (-1)^{n-1}u_{n-1},$ where $u_k$ is a decreasing sequence of positive terms. We can see this from the shape of the curve&#xA;$y = \sin(2n+1)x / \sin x,$ which crosses the x-axis at $\pi/(2n+1), 2\pi/(2n+1),\ldots,n\pi/(2n+1).$ (I said that this is just a sketch, you have to check the details.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence the sequence $I_n$ converges, and by (1) it converges to $\pi/2.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now if we make the substitution $y=(2n+1)x$ we see that&#xA;$$u_k = \int_{k\pi}^{(k+1)\pi} {\sin y \over (2n+1) \sin (y/(2n+1))} dy,$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and since $I_n$ can be written as an alternating sequence of decreasing positive terms we can truncate the sequence wherever we like and the value of $I_n$ lies between two successive partial summations. Hence&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \int_{0}^{2m\pi} {\sin y \over (2n+1) \sin (y/(2n+1))} dy &amp;lt; I_n &amp;lt;&#xA; \int_{0}^{(2m+1)\pi} {\sin y \over (2n+1) \sin (y/(2n+1))} dy. \qquad (2)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for any m such that $2m+1 \le n.$ (Take $m=[\sqrt{n}],$ say, $n \ge 6.$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now $$\left| { \sin y \over y} - {\sin y \over (2n+1) \sin(y/(2n+1))} \right| &#xA;&amp;lt; { \pi^2(2m+1)^2 \over 3(2n+1)^2}$$&#xA;and so this difference tends to zero uniformly in the interval $0 \le y \le (2m+1)\pi$&#xA;and so by taking the $\lim_{n \to \infty}$ in (2) we obtain&#xA;$$\int_0^{\infty} { \sin x \over x } dx = { \pi \over 2}.$$&lt;/p&gt;&#xA;" OwnerUserId="1301" LastActivityDate="2010-09-23T18:01:22.667" />
  <row Id="5320" PostTypeId="2" ParentId="5315" CreationDate="2010-09-23T18:08:45.897" Score="1" ViewCount="" Body="&lt;p&gt;I don't know if there is a standard, but I can provide two sample points.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Burago, Burago, and Ivanov in &quot;A course in metric geometry&quot;, p. 9&#xA;call definition (3) Lipschitz equivalent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dugundji in &quot;Topology&quot; calls those metrics satisfying (1) equivalent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Burago et al. is available online: www.math.psu.edu/petrunin/papers/alexandrov/bbi.pdf&lt;/p&gt;&#xA;" OwnerUserId="1728" LastEditorUserId="1728" LastEditorDisplayName="" LastEditDate="2010-09-23T18:16:53.353" LastActivityDate="2010-09-23T18:16:53.353" CommentCount="1" />
  <row Id="5321" PostTypeId="2" ParentId="5315" CreationDate="2010-09-23T18:14:56.790" Score="0" ViewCount="" Body="&lt;p&gt;I would call (1) topologically equivalent or though as yasmar mentions some just use equivalent. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would call (3) strongly equivalent but in numerical analysis we often use just equivalent (as in your norm equivalent example) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm fairly confident that (1) iff (2) and certain that (3) implies (1). &lt;/p&gt;&#xA;" OwnerUserId="1794" LastActivityDate="2010-09-23T18:14:56.790" CommentCount="2" />
  <row Id="5322" PostTypeId="2" ParentId="5248" CreationDate="2010-09-23T18:18:28.747" Score="11" ViewCount="" Body="&lt;p&gt;Here's another way of finishing off Derek's argument. He proves&#xA;$$\int_0^{\pi/2}\frac{\sin(2n+1)x}{\sin x}dx=\frac\pi2.$$&#xA;Let&#xA;$$I_n=\int_0^{\pi/2}\frac{\sin(2n+1)x}{x}dx=&#xA;\int_0^{(2n+1)\pi/2}\frac{\sin x}{x}dx.$$&#xA;Let&#xA;$$D_n=\frac\pi2-I_n=\int_0^{\pi/2}f(x)\sin(2n+1)x\ dx$$&#xA;where&#xA;$$f(x)=\frac1{\sin x}-\frac1x.$$&#xA;We need the fact that if we define $f(0)=0$ then $f$ has a continuous&#xA;derivative on the interval $[0,\pi/2]$. Integration by parts yields&#xA;$$D_n=\frac1{2n+1}\int_0^{\pi/2}f'(x)\cos(2n+1)x\ dx=O(1/n).$$&#xA;Hence $I_n\to\pi/2$ and we conclude that&#xA;$$\int_0^\infty\frac{\sin x}{x}dx=\lim_{n\to\infty}I_n=\frac\pi2.$$&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-09-23T18:18:28.747" CommentCount="1" />
  <row Id="5323" PostTypeId="2" ParentId="5315" CreationDate="2010-09-23T18:28:51.380" Score="1" ViewCount="" Body="&lt;p&gt;The first one is simply the definition of topological equivalence - it verbatim extends to general topological spaces (not necessarily metric spaces). I don't have a name for the second one. The third, which is used often, is called &quot;Lipschitz equivalence&quot;.&lt;/p&gt;&#xA;" OwnerUserId="2093" LastActivityDate="2010-09-23T18:28:51.380" />
  <row Id="5324" PostTypeId="2" ParentId="5297" CreationDate="2010-09-23T18:53:41.657" Score="2" ViewCount="" Body="&lt;p&gt;This means that $j\epsilon_0 &gt; \frac{\pi}{2}$. Hence $-j\epsilon_0 &amp;lt; -\frac{\pi}{2}$. By the density of rationals, this tells you that there exists a sequence $q_1, q_2,\dots,q_n$ of rational numbers in $(-\epsilon_0,\epsilon_0)$ such that $jq_n\rightarrow \frac{\pi}{2}$. Hence $|\cos(jq_n)|\rightarrow 0$. Combining this with $|x_j| &amp;lt; |\cos(jq_n)|$ for all $n$ gives the contradiction.&lt;/p&gt;&#xA;" OwnerUserId="2093" LastActivityDate="2010-09-23T18:53:41.657" CommentCount="2" />
  <row Id="5325" PostTypeId="1" AcceptedAnswerId="5327" CreationDate="2010-09-23T19:07:55.263" Score="0" ViewCount="105" Body="&lt;p&gt;I have a series of terms $\{t_n : t_n = a_n x_n\}$, and I want to talk about the product of each term except $t_j$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would any of these be an appropriate way to say that?  I like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\prod_{i \ne j} t_i$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or maybe this&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\prod_{i=1,i \ne j}^n t_i$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it a normal thing to do, to put qualifiers on an index like that, or is there a better way to do it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or would it be better to make a subset, $\{b_n : a_n, n \ne j \}$ and use that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\prod_{i=1}^n b_i$$&lt;/p&gt;&#xA;" OwnerUserId="326" LastActivityDate="2010-09-23T20:03:17.657" Title="What is the correct way to write product or sum in capital pi or capital sigma notation when you wish to exclude an index?" Tags="&lt;soft-question&gt;&lt;terminology&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="5326" PostTypeId="2" ParentId="4844" CreationDate="2010-09-23T19:08:16.377" Score="0" ViewCount="" Body="&lt;p&gt;This problem seems related to evenly distributing points across the surface of a sphere (specifically, that is very similar to your k=2 case). That problem is addressed in a programming competition challenge named &lt;a href=&quot;http://www.spoj.pl/problems/PSPHERE/&quot; rel=&quot;nofollow&quot;&gt;PSPHERE&lt;/a&gt; at &lt;a href=&quot;http://www.spoj.pl/&quot; rel=&quot;nofollow&quot;&gt;SPOJ&lt;/a&gt;. The solutions there are not public, but perhaps approaching the leading contestants on that particular challenge could prove helpful.&lt;/p&gt;&#xA;" OwnerUserId="1995" LastActivityDate="2010-09-23T19:08:16.377" />
  <row Id="5327" PostTypeId="2" ParentId="5325" CreationDate="2010-09-23T19:21:40.023" Score="4" ViewCount="" Body="&lt;p&gt;It is normal to write&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\prod_{\begin{matrix}i=1 \\ i \ne j \end{matrix}}^n t_i$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but another good notation is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\prod_{t \in T \setminus \{t_j\}} t$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(with $T = \{t_n : t_n = a_n x_n\}$, also note that it is only defined up to permutation)&lt;/p&gt;&#xA;" LastActivityDate="2010-09-23T20:03:17.657" CommentCount="4" />
  <row Id="5328" PostTypeId="1" CreationDate="2010-09-23T19:38:23.503" Score="2" ViewCount="310" Body="&lt;p&gt;I have the standard logical equivalence:&#xA;$(p\rightarrow q)\wedge(q\rightarrow r)\Leftrightarrow p\rightarrow (q\wedge r)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using several distributive laws I was able to get it down to:&#xA;$(\neg p\wedge\neg q) \vee (\neg p\wedge r) \vee  (q\wedge r)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I must be missing some manipulation I can do to reduce this.&lt;/p&gt;&#xA;" OwnerUserId="1880" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-09-23T19:46:45.363" LastActivityDate="2010-09-23T21:47:09.907" Title="Proof of logical equivalence" Tags="&lt;logic&gt;&lt;propositional-calculus&gt;" AnswerCount="1" CommentCount="7" />
  <row Id="5329" PostTypeId="1" AcceptedAnswerId="5345" CreationDate="2010-09-23T20:00:24.023" Score="2" ViewCount="125" Body="&lt;p&gt;I have to prove that if $P$ is a $R$-module , $P$ is projective $\Leftrightarrow$ there is a family $\{x_i\}$ in $P$ and morphisms $f_i\colon P\to R$ such that for all $x\in P$&#xA;$$ x= \sum_{i\in I} f_i(x)x_i$$&#xA;where for each $x\in P$, $f_i(x)=0$ for almost all $i\in I$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any help?&lt;/p&gt;&#xA;" OwnerUserId="2095" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-09-23T20:03:22.763" LastActivityDate="2010-09-23T21:58:44.937" Title="Projective modules" Tags="&lt;abstract-algebra&gt;&lt;modules&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="5330" PostTypeId="1" AcceptedAnswerId="5341" CreationDate="2010-09-23T20:15:24.113" Score="1" ViewCount="62" Body="&lt;p&gt;I am reading the book Stochastic Simulation, Algorithms and Analysis by Asmussen. On page 39, when talking about inversion method for sampling a distribution, he gave an example:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;an r.v. distributed as the overshoot X − a given X &gt; a can be generated as $F^{-1} (U (1-F (a)) − a$,&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;where $U$ is a r.v. uniform in $[0,1]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I think the distribution is $F^{-1} (U (1-F (a)) + F(a)) - a$. I was wondering which is correct and why?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks and regards!&lt;/p&gt;&#xA;" OwnerUserId="1281" LastEditorUserId="1281" LastEditorDisplayName="" LastEditDate="2010-09-23T20:58:54.077" LastActivityDate="2010-09-23T21:28:32.360" Title="A problem of inversion method for sampling" Tags="&lt;probability&gt;&lt;monte-carlo&gt;&lt;simulation&gt;" AnswerCount="1" />
  <row Id="5331" PostTypeId="1" AcceptedAnswerId="17435" CreationDate="2010-09-23T20:30:33.460" Score="6" ViewCount="279" Body="&lt;p&gt;A Blaschke product is a function of the form&#xA;$$B(z):=z^k\prod_{n=1}^{\infty}\frac{a_n-z}{1-\overline{a_n}z}\frac{|a_n|}{a_n}$$&#xA;where the $a_n$ are the non-zero zeros of $B$, and satisfie $\sum_{n=1}^{\infty}(1-|a_n|) &amp;lt; \infty$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Blashke products are holomorphic and bounded by 1 on the unit disk. A well known theorem asserts that $B$ has radial limits almost everywhere on the unit circle, i.e. that the limit&#xA;$$\lim_{r \rightarrow 1} B(re^{i \theta})$$ exist for almost every $\theta$. I'm looking for an example of Blashke product such that the radial limit does not exist at a certain point, say $1$ for example. In particular, a Blaschke product with zeros in $(0,1)$ such that &#xA;$$\limsup_{r \rightarrow 1}|B(r)| =1$$ would work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anyone have a construction or reference?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you,&#xA;Malik&lt;/p&gt;&#xA;" OwnerUserId="1197" LastActivityDate="2011-01-13T22:21:20.940" Title="On radial limits of Blaschke Products" Tags="&lt;complex-analysis&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="5333" PostTypeId="1" AcceptedAnswerId="5336" CreationDate="2010-09-23T20:38:49.070" Score="2" ViewCount="136" Body="&lt;p&gt;The modular group is the group $G$ consisting of all linear fractional transformations $\phi$ of the form&#xA;$$\phi(z)=\frac{az+b}{cz+d}$$&#xA;where $a,b,c,d$ are integers and $ad-bc=1$. I have read that $G$ is generated by the transformations $\tau(z)=z+1$ and $\sigma(z)=-1/z$. Is there an easy way to prove this? In particular, is there a proof that uses the relation between linear fractional transformations and matrices? Any good reference would be helpful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you,&#xA;Malik &lt;/p&gt;&#xA;" OwnerUserId="1197" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-09-23T20:50:22.003" LastActivityDate="2010-09-23T20:55:25.240" Title="On the generators of the Modular Group" Tags="&lt;linear-algebra&gt;&lt;group-theory&gt;" AnswerCount="1" />
  <row Id="5334" PostTypeId="1" AcceptedAnswerId="5353" CreationDate="2010-09-23T20:39:53.030" Score="4" ViewCount="160" Body="&lt;p&gt;So, I &lt;a href=&quot;http://math.stackexchange.com/questions/5267/how-to-find-where-3-lines-intersect&quot;&gt;asked a question&lt;/a&gt; about how to find if 3 lines are concurrent. I built the algorithm I needed,and it was working well, until I started doubting my power of judgement. So my question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Are 2 equal lines also intersecting?&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For an example, 9x+9y=9;7x+7y+7 ? Because now, if I would draw these lines and add &lt;em&gt;any&lt;/em&gt; line to the comparison, these 3 lines would be concurrent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;help?&lt;/p&gt;&#xA;" OwnerUserId="2078" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-09-24T05:52:25.167" LastActivityDate="2010-09-24T05:52:25.167" Title="Do 2 equal lines also count as intersecting?" Tags="&lt;geometry&gt;" AnswerCount="1" CommentCount="7" />
  <row Id="5335" PostTypeId="2" ParentId="5331" CreationDate="2010-09-23T20:41:19.550" Score="0" ViewCount="" Body="&lt;p&gt;If you have the zeros on (0,1), as you suggest, then certainly &#xA;$$\liminf_{r\to 1}|B(r)|=0$$&#xA;(as long as you have infinite numbers of zeros of course). &lt;/p&gt;&#xA;" OwnerUserId="1154" LastActivityDate="2010-09-23T20:41:19.550" CommentCount="1" />
  <row Id="5336" PostTypeId="2" ParentId="5333" CreationDate="2010-09-23T20:49:28.917" Score="8" ViewCount="" Body="&lt;p&gt;Yes; this statement is essentially equivalent to the Euclidean algorithm.  I discuss these issues in &lt;a href=&quot;http://www.artofproblemsolving.com/blog/10588&quot; rel=&quot;nofollow&quot;&gt;this old blog post&lt;/a&gt;.  (A very brief sketch: by applying the generators and the inverses to an arbitrary element of the modular group it is possible to perform the Euclidean algorithm on $a$ and $c$ (or maybe it's $a$ and $b$).  The rest is casework.)  You can think of this as a form of row reduction, which is generalized by the notion of &lt;a href=&quot;http://en.wikipedia.org/wiki/Smith_normal_form&quot; rel=&quot;nofollow&quot;&gt;Smith normal form&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is also a geometric proof using the action on the upper half plane which is given, for example, in the relevant section of Serre's &lt;em&gt;Course in Arithmetic.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-09-23T20:55:25.240" LastActivityDate="2010-09-23T20:55:25.240" CommentCount="2" />
  <row Id="5338" PostTypeId="1" AcceptedAnswerId="5342" CreationDate="2010-09-23T21:19:23.557" Score="2" ViewCount="111" Body="&lt;p&gt;Let φ(w)ψ(x) be a solution to the heat problems. Given that ψ(−1) = ψ(1) = 0 &#xA;prove that it is not possible for ψ(x) and ψ′′ (x) to be strictly positive for all x such that &#xA;−1 &amp;lt; x &amp;lt; 1. You may use any facts, such as the mean value theorem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So yes, this is homework. But we have been at it for hours now,and could use a little nudge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume $ψ(x), ψ''(x)$ are positive.&#xA;then the mean value theorem states $ψ'(c) = 0$ according to:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ψ'(c) = \frac{ψ(1)-ψ(−1)}{1+(-1)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $ψ(c) = 0$, then we know there exists a tangent parallel to x-axis. This is the extrema which is strictly positive. &lt;em&gt;Here is where we get stuck&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since we assumed $ψ''(c)$ is positive then we know that there exists a minimum due to the the derivative test. But how does this help us prove that the range from -1 to 1 is NOT strictly positive. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A little background info:&#xA;This is for real analysis. Both me and roommate have not taken a prior proofs course (which is acutally the pre-req). So we are not experts are proving so if any of you take the time to answer, please dumb it down as much as you can. &lt;/p&gt;&#xA;" OwnerUserId="287" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-09-23T21:26:36.077" LastActivityDate="2010-09-23T21:37:36.713" Title="Proving a function is below zero using the mean value theorm." Tags="&lt;real-analysis&gt;&lt;homework&gt;" AnswerCount="1" />
  <row Id="5339" PostTypeId="2" ParentId="5052" CreationDate="2010-09-23T21:19:26.577" Score="2" ViewCount="" Body="&lt;p&gt;Perhaps this has something for you: &lt;a href=&quot;http://wis.kuleuven.be/analyse/walter/coimbra.pdf&quot; rel=&quot;nofollow&quot;&gt;http://wis.kuleuven.be/analyse/walter/coimbra.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This uses the Legendre polynomials in proving irrationality of certain numbers.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-09-23T21:19:26.577" />
  <row Id="5340" PostTypeId="1" CreationDate="2010-09-23T21:23:02.227" Score="2" ViewCount="379" Body="&lt;p&gt;If the probability of an event is $\frac{1}{\infty}$ and $\infty$ trials are conducted, how many times will the event occur — $0$, $1$, or $\infty$?&lt;/p&gt;&#xA;" OwnerUserId="2094" LastEditorUserId="2736" LastEditorDisplayName="" LastEditDate="2011-05-11T03:50:07.170" LastActivityDate="2011-05-11T03:50:07.170" Title="Probability and Infinity" Tags="&lt;probability&gt;&lt;infinity&gt;" AnswerCount="3" CommentCount="8" />
  <row Id="5341" PostTypeId="2" ParentId="5330" CreationDate="2010-09-23T21:28:32.360" Score="1" ViewCount="" Body="&lt;p&gt;Let $Y=X$ with support on $[a,\infty)$. If $f_X$ is the p.d.f of $X$ and $f_Y$ is the pd.f. of $Y$ then &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f_Y(y) = \frac{f_X(y)}{1-F(a)}$  for $y&gt;a$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$F_Y(y) = \frac{F_X(y)-F_X(a)}{1-F_X(a)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $U=F_Y(Y)$ then as usual we have $U\sim Uni[0,1]$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$U=\frac{F_X(Y)-F_X(a)}{1-F_X(a)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rearranging gives us &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$Y=F_X^{-1}(U(1-F_X(a))+F_X(a))$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus the over shoot r.v. $X-a$ for $X&gt;a$ can be generated as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$F_X^{-1}(U(1-F_X(a))+F_X(a))-a$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I agree with you. &lt;/p&gt;&#xA;" OwnerUserId="1794" LastActivityDate="2010-09-23T21:28:32.360" />
  <row Id="5342" PostTypeId="2" ParentId="5338" CreationDate="2010-09-23T21:37:36.713" Score="1" ViewCount="" Body="&lt;p&gt;Hint:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $\psi(x)$ is strictly positive, what does it say about the trend of $\psi'(x)$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $\psi''(x)$ is strictly positive, what does it say about the trend of $\psi'(x)$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It might be useful to draw a graph.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-09-23T21:37:36.713" CommentCount="10" />
  <row Id="5343" PostTypeId="1" AcceptedAnswerId="5348" CreationDate="2010-09-23T21:37:45.247" Score="2" ViewCount="152" Body="&lt;p&gt;I'm looking at the quantum group $SU_q(2)$ (over ${\mathbb C}$) and can't see why it has no zero divisors. It's clear that $M_q(2)$, the quantum $2 \times 2$ matrices have no zero divisors, but I can't seem to see why/if this extends to the quotient over $det_q - 1$. I think one can argue it from the grading on $M_q(2)$, but that's just a guess.&lt;/p&gt;&#xA;" OwnerUserId="2099" LastActivityDate="2010-11-23T03:44:19.833" Title="Zero divisors in $SU_q(2)$" Tags="&lt;abstract-algebra&gt;&lt;zero-divisor&gt;&lt;quantum-groups&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" />
  <row Id="5344" PostTypeId="2" ParentId="5328" CreationDate="2010-09-23T21:47:09.907" Score="6" ViewCount="" Body="&lt;p&gt;Umm... maybe I am missing something, but&lt;/p&gt;&#xA;&#xA;&lt;p&gt;if $p$ is false, $q$ is true and $r$ is false, then we have that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(p\rightarrow q)\wedge(q\rightarrow r)$ is false&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$p\rightarrow (q\wedge r)$ is true.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I don't see how you can prove the equivalence.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-09-23T21:47:09.907" />
  <row Id="5345" PostTypeId="2" ParentId="5329" CreationDate="2010-09-23T21:58:44.937" Score="4" ViewCount="" Body="&lt;p&gt;Put the f&lt;sub&gt;i&lt;/sub&gt; together to form one giant f from P to R&lt;sup&gt;(I)&lt;/sup&gt;, the direct sum of I copies of the ring R.  The condition that x = Sum f_i(x) x_i just means that there is some g:R&lt;sup&gt;(I)&lt;/sup&gt;→P such that g(f(x)) = x, namely g((r1,r2,...)) = r1*x1 + r2*x2 + .... In other words, P is a direct summand of the free module R&lt;sup&gt;(I)&lt;/sup&gt;.&lt;/p&gt;&#xA;" OwnerUserId="583" LastActivityDate="2010-09-23T21:58:44.937" CommentCount="4" />
  <row Id="5347" PostTypeId="1" AcceptedAnswerId="5360" CreationDate="2010-09-23T23:58:16.693" Score="3" ViewCount="196" Body="&lt;p&gt;In Stochastic Simulation: Algorithms and Analysis by Søren Asmussen, on Page 38&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A Poisson r.v. $N$ with rate $\lambda$ ($P(X = n) = e^{−\lambda} \frac{\lambda^n}{n!}$) can be constructed using the relation between the exponential distribution and the Poisson process: generate $X_1 , X_2 , ...$ as i.i.d. exponentials with rate $\lambda$ one at a time until the random time at which the sum exceeds 1, say $X_1 + · · ·+ X_N &amp;lt; 1 &amp;lt; X_1 + · · ·+ X_{N +1}$ . Then N is the desired Poisson r.v. To reduce the number of evaluations of special functions, use&#xA;                      $N = max \lbrace n \geq 0 : \prod_{i=1}^n U_i &gt; e^{-\lambda} \rbrace$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;where $U_i$ is r.v. uniform in $[0,1]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was wondering why to &quot;reduce the number of evaluations of special functions&quot;, the $N$ in the last equation has Poisson distribution with rate λ? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks and regards!&lt;/p&gt;&#xA;" OwnerUserId="1281" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-09-24T05:49:11.787" LastActivityDate="2010-09-24T05:49:11.787" Title="Sample Poisson Distribution" Tags="&lt;probability&gt;&lt;monte-carlo&gt;&lt;simulation&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="5348" PostTypeId="2" ParentId="5343" CreationDate="2010-09-24T00:13:14.040" Score="1" ViewCount="" Body="&lt;p&gt;The algebra can be constructed as an iterated Ore extension starting from the base field. By standard properties of such extensions, it is then a domain.&lt;/p&gt;&#xA;" OwnerUserId="274" LastEditorUserId="274" LastEditorDisplayName="" LastEditDate="2010-09-24T00:49:11.097" LastActivityDate="2010-09-24T00:49:11.097" CommentCount="2" />
  <row Id="5349" PostTypeId="1" CreationDate="2010-09-24T00:35:45.427" Score="5" ViewCount="173" Body="&lt;p&gt;I'm looking for different proofs (references or sketch of main ideas) of the structure theorem for finitely generated modules over a PID. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If possible, a comparison in terms of clarity, elegance or usefulness would be appreciated.&lt;/p&gt;&#xA;" OwnerUserId="1746" LastEditorUserId="1746" LastEditorDisplayName="" LastEditDate="2010-09-24T03:00:09.187" LastActivityDate="2011-05-21T06:59:56.340" Title="Proofs of the structure theorem for finitely generated modules over a PID" Tags="&lt;abstract-algebra&gt;&lt;modules&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="5350" PostTypeId="2" ParentId="5340" CreationDate="2010-09-24T00:35:57.423" Score="1" ViewCount="" Body="&lt;p&gt;Sorry byoogle, I misinterpreted the point of your question. So now I feel compelled to try to answer. I don't know much about probability, but by chance I stumbled across something today that might be interesting to you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll try to frame a question that I think is close to what you are asking, and is also answered by this nice one page paper I found.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I'm looking for $x \in [0,1]$, and each &quot;instant&quot; I &quot;look&quot;, I select a set $A_\alpha \subset [0,1]$. Unfortunately, the probability that $x \in A_\alpha$ is always $0$. But, I have at least this: my set from instant $\alpha$ includes all the previous sets I selected. It seems little consolation though, because at any given $\alpha$ the probability that I have $x \in A_\alpha$ is still 0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, if I am allowed to go on looking indefinitely, is the probability that $x$ is &quot;found&quot; zero? Well, it seems that the probability can in fact be 1, if we choose well. At least that is how I interpret this:&#xA;&lt;a href=&quot;http://www.math.harvard.edu/~elkies/Misc/quickie.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.math.harvard.edu/~elkies/Misc/quickie.pdf&lt;/a&gt; &lt;br&gt;&#xA;When you read that, you can read &quot;measure zero&quot; or &quot;negligible&quot; as &quot;probability zero&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a fair chance that my analogy is broken, and hopefully that will be pointed out if that's the case. Our &quot;instants&quot; can't be countable, in other words we need to be continously looking, rather than just looking every microsecond or so. Also, my analogy seems to be using the ordering on the real numbers, and maybe that is a logical stumble.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope that helps, or is at least amusing.&lt;/p&gt;&#xA;" OwnerUserId="1728" LastEditorUserId="1728" LastEditorDisplayName="" LastEditDate="2010-09-24T00:58:23.260" LastActivityDate="2010-09-24T00:58:23.260" CommentCount="3" />
  <row Id="5352" PostTypeId="2" ParentId="5340" CreationDate="2010-09-24T01:37:56.520" Score="2" ViewCount="" Body="&lt;p&gt;Here's one possible answer.  Say each event occurs with probability $\varepsilon&gt;0$ and there are $1/\varepsilon$ total events.  Taking the limit as $\varepsilon\to0^+$, the event occurs 0 times with probability $1/e\approx36.8\%$, once with probability $1/e\approx36.8\%$, twice with probability $1/2e\approx18.4\%$, three times with probability $1/6e\approx6.13\%$, and so on: the Poisson distribution with $\lambda=1$.&lt;/p&gt;&#xA;" OwnerUserId="1778" LastActivityDate="2010-09-24T01:37:56.520" CommentCount="2" />
  <row Id="5353" PostTypeId="2" ParentId="5334" CreationDate="2010-09-24T01:39:51.163" Score="3" ViewCount="" Body="&lt;p&gt;In the analytical geometry version of Euclidean Geometry we need a convenient way of thinking of parallelism and there is also the interest in constructing the real projective plane (where there are no parallel lines) from  Euclidean Geometry. (The key idea is that we think of all the lines which have the same slope as having a special additional point on them at &quot;infinity&quot; (thereby creating new lines in the projective plane) and all these new points, one for each slope, lie on a new line, the line at &quot;infinity.&quot;) In the analytical geometry framework of Euclidean Geometry lines are parallel when they have the same slope but there is the problem that a nonzero multiple of an equation looks like a  different equation but consists of all of the same points. Furthermore, one would like parallelism to be an equivalence relation, which means it obeys being reflexive, symmetric and transitive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Equivalence_relation&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Equivalence_relation&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So it is CONVENIENT to think of a line as being parallel to itself. Now all lines with the same slope can be thought of as parallel. (All the lines with undefined slope are also parallel to each other; they are the vertical lines.) Furthermore, we can think of the Euclidean plane's lines as being partitioned in classes by this parallelism relation.&lt;/p&gt;&#xA;" OwnerUserId="1369" LastActivityDate="2010-09-24T01:39:51.163" />
  <row Id="5354" PostTypeId="2" ParentId="5340" CreationDate="2010-09-24T02:18:49.220" Score="6" ViewCount="" Body="&lt;p&gt;There are meaningful ways to work out probabilities on infinite spaces, but your question is not well-defined.  We could define situations matching your question with the answer 0, 1, $\infty$, or any other finite answer.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Probability distributions on finite sets are pretty easy to write down and work with.  For rolling a single die, the probability distribution is {p(1)=1/6, p(2)=1/6, p(3)=1/6, p(4)=1/6, p(5)=1/6, p(6)=1/6}.  From this, we can ask and answer many different questions-- for example, the probability of rolling a 3 three times in a row is 1/216.  However, the probability of rolling a 7 is 0; and it doesn't matter how many times you roll, you'll never get a 7.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Probability distributions on infinite sets require a lot more background to define and work with.  You need a &lt;a href=&quot;http://en.wikipedia.org/wiki/Probability_measure&quot; rel=&quot;nofollow&quot;&gt;probability measure&lt;/a&gt; and you can compute integrals to find probabilities.  (Measure theory is usually a graduate-level topic, and it requires very careful, abstract thought to get the details straight.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An instructive and fun problem to read about is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Random_walk&quot; rel=&quot;nofollow&quot;&gt;Random Walk&lt;/a&gt;.  Suppose I start at (0,0) on an infinite 2-dimensional grid; and each second, I go up, down, left, or right one unit.  (Each direction has probability 1/4 every second.)  The probability space is the set of &lt;em&gt;all&lt;/em&gt; possible walks, which is an infinite set.  The probability measure on this space is determined by the fact that each direction has equal probability every second.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even though there are many walks in the probability space that never return to (0,0), the subset of walks that &lt;em&gt;do&lt;/em&gt; return to (0,0) has measure 1.  This implies that on a 2-dimensional random walk, the expected number of returns to (0,0) is infinite.  However, the expected number of times I take a diagonal step is 0 (because, by my definition of the problem, it never happens).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can also consider the 3-dimensional random walk.  Interestingly, on a 3-dimensional random walk, the expected number of returns to (0,0,0) is about 0.516385.  (You can think of this as the average number of returns if you sample many 3-dimensional random walks.)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;These results requires some work to obtain, and you always have to start with a precisely defined problem.  If you think about them and read other examples on your own, it should help you gain some understanding of infinite probability spaces, probability 0 events, and probability 1 events.&lt;/p&gt;&#xA;" OwnerUserId="1461" LastEditorUserId="1461" LastEditorDisplayName="" LastEditDate="2010-09-24T02:48:05.317" LastActivityDate="2010-09-24T02:48:05.317" />
  <row Id="5356" PostTypeId="2" ParentId="5297" CreationDate="2010-09-24T04:03:48.703" Score="1" ViewCount="" Body="&lt;p&gt;I would have added a comment, but I fear the answer would overlap the allowed character limit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, maybe I'm missing something, but I think I see a clear contradiction. Here we go...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $S_j = \{(x_0, x_j)\in\mathbb{Q}\times\mathbb{R}: |x_j| &amp;lt; |\cos(jx_0)|\}$ Let $V_j = (-\epsilon_0,\epsilon_0)\times(-\epsilon_j,\epsilon_j)$. The claim is that for all $j$, we have $V_j\subset S_j$. Fix $j$ and assume this inclusion holds. On the other hand, if $2j\epsilon_0 &gt; \pi$, there is a sequence $q_1,q_2,\dots$ of rationals in $(-\epsilon_0,\epsilon_0)$, such that $jq_n\rightarrow \frac{\pi}{2}$. Let $x_j\in (-\epsilon_j, \epsilon_j)$ not equal to zero. Since the inclusion $V_j\subset S_j$ holds, we have $|x_j| &amp;lt; |\cos(jq_n)|$ for all $n$. Clearly impossible, since $x_j &gt; 0$. &lt;/p&gt;&#xA;" OwnerUserId="2093" LastActivityDate="2010-09-24T04:03:48.703" CommentCount="1" />
  <row Id="5357" PostTypeId="1" AcceptedAnswerId="5369" CreationDate="2010-09-24T04:55:03.190" Score="2" ViewCount="169" Body="&lt;p&gt;I know that the product rule is generalised by Leibniz's general rule and the chain rule by Faà di Bruno's formula, but what about the quotient rule? Is there a generalisation for it analogous to these? Wikipedia mentions both  Leibniz's general rule and Faà di Bruno's formula for the product and the chain rule, but rather nothing for the quotient rule. &lt;/p&gt;&#xA;" OwnerUserId="2104" LastEditorUserId="2104" LastEditorDisplayName="" LastEditDate="2010-09-24T05:01:11.980" LastActivityDate="2010-09-24T11:45:43.443" Title="What's the generalisation of the quotient rule for higher derivatives?" Tags="&lt;calculus&gt;&lt;analysis&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="0" />
  <row Id="5358" PostTypeId="2" ParentId="5357" CreationDate="2010-09-24T05:00:12.950" Score="1" ViewCount="" Body="&lt;p&gt;Quotient Rule is actually Product Rule.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$D(u/v) = D(uw)$ where $w = 1/v$.&lt;/p&gt;&#xA;" OwnerUserId="705" LastActivityDate="2010-09-24T05:00:12.950" CommentCount="1" />
  <row Id="5359" PostTypeId="2" ParentId="4961" CreationDate="2010-09-24T05:17:34.757" Score="5" ViewCount="" Body="&lt;p&gt;The useful identity&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \frac{1}{2} + \cos x + \cos 2x + \cdots + \cos nx = \frac{\sin (n+x/2)}{2 \sin (x/2)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which appears in the study of Fourier series is most easily proven by replacing every $\cos(kx)$ on the left-hand side with $$\cos(kx)=\frac{e^{ikx}+e^{-ikx}}{2}$$ and applying the formula for the sum of the geometric series. &lt;/p&gt;&#xA;" OwnerUserId="317" LastActivityDate="2010-09-24T05:17:34.757" CommunityOwnedDate="2010-09-24T05:17:34.757" />
  <row Id="5360" PostTypeId="2" ParentId="5347" CreationDate="2010-09-24T05:36:13.623" Score="5" ViewCount="" Body="&lt;p&gt;Note that by inverse sampling we can generate exponential r.v. with rate $\lambda$ by &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$X=F^{-1}(U)$  where $F(x) = 1-e^{-\lambda x}$, is the c.d.f of exponential r.v. with rate $\lambda$ and $U\sim Uni[0,1]$. Thus $F^{-1}(U)=\frac{-\log(1-U)}{\lambda}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But $1-U\sim Uni[0,1]$ so if $U_i\sim Uni[0,1]$ then  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$X_i=\frac{-\log(U_i)}{\lambda}$ is exponential with rate $\lambda$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$X_1+\ldots +X_n= -\frac{1}{\lambda}(\log(U_1)+\ldots +\log(U_n)) = -\frac{1}{\lambda}\log(U_1U_2\ldots U_n)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is $X_1+\ldots + X_n&amp;lt;1$ iff $\prod_{i=1}^nU_i&gt;e^{\lambda}$. Clearly in both cases $N$ is actually the same random variable. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Why reduce the evaulation of special functions?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose $\lambda=10^{-3}$ (small). In the first method, where you generate each $X_i$ in turn add them up and check if totals less than one, you would expect to generate $10^3$ exponentials.  Each one requires you to evaluate the logarithm (with argument in $(0,1)$) and simulate a uniform r.v.. So you have 1 thousand evaluations of a special function, in the region where it is hard to approximate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the second method, you expect to generate $10^3$ uniform r.v. and check if the product is greater than $e^{-10^3}$. This method requires you to calculate $e^{-10^3}$ but only once. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both methods generate what you desired but the second method will be much more efficient on a computer. &lt;/p&gt;&#xA;" OwnerUserId="1794" LastActivityDate="2010-09-24T05:36:13.623" />
  <row Id="5361" PostTypeId="2" ParentId="4961" CreationDate="2010-09-24T06:42:40.913" Score="4" ViewCount="" Body="&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/2AaSE.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Theorem:&#xA;&lt;strong&gt;The line segments joining the centers of opposite squares are perpendicular and of equal length.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $2a,2b,2c,2d$ be complex numbers running along the edges of the quadrilateral.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Condition for quadrilateral to close up is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a + b + c + d = 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let origin be at vertex where $2a$ begins &lt;/p&gt;&#xA;&#xA;&lt;p&gt;so&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$p = a + ia$ i.e. $a$ + ($a$ rotated though 90 degrees) &lt;br /&gt;&#xA;$ q = 2a + (1+i)b$&lt;br /&gt;&#xA; $r = 2a + 2b + (1+i)c$&lt;br /&gt;&#xA;$ s = 2a + 2b + 2c + (1+i)d$&lt;br /&gt;&lt;br /&gt;&#xA; $A = s-q$  and $B = r - p$ &lt;br /&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;we want to show that A and B are perpendicular and equal i.e. $B = iA$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$A + iB = (a+b+c+d) + i(a+b+c+d) = 0 $ since $a + b + c + d = 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This image, theorem, proof and everything is taken from &lt;strong&gt;&lt;a href=&quot;http://usf.usfca.edu/vca//&quot; rel=&quot;nofollow&quot;&gt;Visual Complex Analysis by Tristan Needham.&lt;/a&gt;&lt;/strong&gt;[One of the Best Books. Highly Recommended.]&lt;/p&gt;&#xA;" OwnerUserId="705" LastActivityDate="2010-09-24T06:42:40.913" CommunityOwnedDate="2010-09-24T06:42:40.913" CommentCount="2" />
  <row Id="5362" PostTypeId="1" AcceptedAnswerId="5417" CreationDate="2010-09-24T09:07:45.970" Score="3" ViewCount="120" Body="&lt;p&gt;Let $A$ be a set of infinite cardinals. Assume that for every regular $\lambda$, the subset $A \cap \lambda$ of $\lambda$ is not stationary. Then I want to prove that there is an injective function $g$ on $A$ which satisfies $g(\alpha) &amp;lt; \alpha$ for all $\alpha \in A$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is an exercise in Kunen's set theory. I dont' understand if $g$ should be a map from $A \to A$. But then the claim would be false if $A = \{\kappa\}$. But if $g$ may have arbitrary ordinal values, then the claim is somehow trivial, also without the assumptions. So I don't know what this exercise about.&lt;/p&gt;&#xA;" OwnerUserId="1650" LastActivityDate="2010-09-25T05:10:16.097" Title="set of infinite cardinals admits an injective regressive function" Tags="&lt;set-theory&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="5363" PostTypeId="1" AcceptedAnswerId="5387" CreationDate="2010-09-24T09:52:58.310" Score="1" ViewCount="276" Body="&lt;p&gt;There is something in the definition of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Free_product&quot; rel=&quot;nofollow&quot;&gt;free product&lt;/a&gt; of two groups that annoys me, and it's this &quot;word&quot; thing:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If G and H are groups, a word in G and H is a product of the form&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$$&#xA;     s_1 s_2 \dots s_m,&#xA;  $$&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;where each $s_i$ is either an element of G or an element of H. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So what is this &quot;word&quot; guy? Does it come out of the blue? Does it come from some sort of new operation that I can perform with the two sets $G$ and $H$ -in addition to the well-known ones of union, intersection, Cartesian product...?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fortunatelly, I think there is nothing new under the sun of set operations: it's easy to realise that words can be identified with elements of some Cartesian product (see below):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;(s_1, s_2, \dots , s_m )  \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And Cartesian product is a well-established set-theoretical operation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I tried to translate the rest of Wikipedia's definition &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Such a word may be &lt;em&gt;reduced&lt;/em&gt; using the following operations:&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Remove an instance of the identity element (of either G or H).&#xA;  Replace a pair of the form $g_1g_2$ by its product in $G$, or a pair $h_1h_2$ by its product in $H$.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Every reduced word is an alternating product of elements of G and elements of H, e.g.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$$&#xA;  g_1 h_1 g_2 h_2 \dots g_r h_r.&#xA;  $$&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The free product $G ∗ H$ is the group whose elements are the reduced words in G and H, under the operation of concatenation followed by reduction.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;in an elementary set setting. First, consider the set of &quot;unreduced&quot; tuples of elements of $G$ and $H$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;U =  G \sqcup H \sqcup (G\times G) \times (G\times H) \sqcup (H\times G) \sqcup (H\times H) \sqcup (G\times G \times G) \sqcup \dots &#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More concisely:&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;EDIT:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the following formula may be less messier than the one I wrote previously:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;U = \bigsqcup_{r \geq 1} (S_1 \times \cdots \times S_r)&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $S_i = G$ or $S_i = H$.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;So, elements of $U$ are ordered tuples (&lt;em&gt;unreduced&lt;/em&gt; ones)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;(s_1, s_2, \dots , s_m)&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where each $s_i$ is either an element of $G$ or an element of $H$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The product of two unreduced tuples is defined by concatenation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;(s_1, \dots , s_m) \cdot (t_1,  \dots , t_n) = (s_1, \dots , s_m, t_1 , \dots , t_n) \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, consider the following equivalence relation in the set of unreduced tuples $U$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;(s_1, s_2, \dots , s_{i-1}, 1, s_{i+1}, \dots , s_n) \sim (s_1, s_2, \dots, s_{i-1}, s_i, \dots , s_n) \ ,&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $1$ is either the unit element of $G$ or the one of $H$. And&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;(s_1, s_2, \dots , s_i,s_{i+1}, \dots , s_r) \sim (s_1, s_2, \dots , s_is_{i+1}, \dots , s_r )&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;whenever two adjacent $s_i, s_{i+1} \in G$ or $s_i, s_{i+1} \in H$ at the same time. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want, you may call the equivalence class of a tuple under this equivalence relation a &lt;em&gt;reduced&lt;/em&gt; tuple. So every reduced tuple is an alternating one, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;(g_1, h_1, \dots , g_r , h_r) \ ,&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with $g_i \in G$ and $h_i \in H$ for all $i = 1, \dots , r$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Define the free product of $G$ and $H$ as the quotient:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;G*H = U/\sim \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, one verifies that concatenation is well-defined on unreduced tuples and gives $G*H$ a group structure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After performing this elementary exercise I understand perfectly well why nobody defines the free product in this way, but I still wanted to ask:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Is this correct?&lt;/li&gt;&#xA;&lt;li&gt;Is it written somewhere?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="664" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-09-26T16:28:17.277" LastActivityDate="2010-09-26T16:28:17.277" Title="Set-theoretical description of the free product?" Tags="&lt;abstract-algebra&gt;" AnswerCount="5" CommentCount="2" FavoriteCount="1" />
  <row Id="5364" PostTypeId="2" ParentId="5363" CreationDate="2010-09-24T10:09:52.663" Score="1" ViewCount="" Body="&lt;p&gt;Well, I think that the usual definition of the free product is just a slightly less formal version of what you've written, which, by the way, is certainly correct. I haven't seen it completely written out like that before but maybe others have.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Meanwhile, one can avoid the whole messy construction and prove (using, for example, a version of the adjoint functor theorem) that the category Grp has small colimits. The free product is just the coproduct in Grp, so that would suffice for all practical purposes, and it is a rigorous formulation, if that's what you're worried about.&lt;/p&gt;&#xA;" OwnerUserId="423" LastActivityDate="2010-09-24T10:09:52.663" CommentCount="5" />
  <row Id="5365" PostTypeId="2" ParentId="5363" CreationDate="2010-09-24T11:19:25.500" Score="1" ViewCount="" Body="&lt;p&gt;This is not an answer to your questions, but I guess I would point out that the “words” in the definition in Wikipedia can be given a structure: consider the free monoid freely generated by the set &lt;i&gt;G&lt;/i&gt;⊔&lt;i&gt;H&lt;/i&gt;, and just call its elements &lt;em&gt;words in G and H&lt;/em&gt;.  Now you can introduce an equivalence relation and define the free product as the quotient (similarly to what you did), verifying that the operation on the quotient is well-defined and makes it a group.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, this is not at all shorter than your written-down definition if we write down the definition of the free monoid and the definition of the equivalence relation.  I just wanted to point out that the “words” are a natural concept and that this two-step construction may be slightly tidier than constructing the free product of groups in one big step (based on sets).&lt;/p&gt;&#xA;" OwnerUserId="926" LastActivityDate="2010-09-24T11:19:25.500" CommentCount="1" />
  <row Id="5366" PostTypeId="1" CreationDate="2010-09-24T11:24:50.330" Score="4" ViewCount="285" Body="&lt;p&gt;I apologise for this question, as it is no doubt very simple, but I've never been very confident with proofs. Our lecturer today (in a course related to maths but not mathematical itself) was playing with doing the modulus of powers, and used the above fact - $(x.y)\; mod\; m\; ==\; ((x\; mod\; m).y)\; mod\; m$ - to show a quick way to do it. She mentioned offhand that the proof was short and easy, and that we should try it. Well, I did. And I'm not really sure that what I've done is mathematically sound. So I'd very much appreciate it if anyone could give me some help with how to do this proof correctly - googling for it turned up many pages which just state it as simple fact. I would not call my proof short or easy, so I reckon I've just got completely the wrong end of the stick.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My proof:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When considering any number, denoted n, in regards to another number m, we can write it as $n_b$ + $n_r$, where the former is some multiple of m, and the latter is the remainder when you take n mod m.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we expand the original formula, $(x.y)\; mod\; m$ like this, we get $((x_b + x_r)(y_b + y_r))\; mod\; m$. This then expands to $(x_b y_b + x_b y_r + x_r y_b + x_r y_r)\; mod\; m$. The first three terms in the brackets are all some multiple of m, and so when they are taken modulus m they are 0. The final term we do not know, so we are left with: $(x_r y_r)\; mod\; m$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we can consider the left hand side. $((x\; mod\; m).y)\; mod\; m$ is equivalent to $((x_r).y)\; mod\; m$. Expanding y gives us: $((x_r)(y_b + y_r))\; mod\; m$. Multiplying out then gives $(x_r y_b + x_r y_r)\; mod\; m$. As before, the first term in the brackets are multiples of m and so are 0 when taken mod m, and the last one we do not know, so we are left with: $(x_r y_r)\; mod\; m$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since both the LHS and RHS boil down to the same thing, the equality holds.&lt;/p&gt;&#xA;" LastActivityDate="2011-01-02T23:00:02.660" Title="Very simple question, but what is the proof that x.y mod m == ((x mod m).y) mod m?" Tags="&lt;modular-arithmetic&gt;" AnswerCount="2" CommentCount="0" />
  <row Id="5368" PostTypeId="2" ParentId="5366" CreationDate="2010-09-24T11:43:57.320" Score="2" ViewCount="" Body="&lt;p&gt;I guess you already know how to prove that (&lt;i&gt;s&lt;/i&gt;+&lt;i&gt;t&lt;/i&gt;) mod &lt;i&gt;m&lt;/i&gt; = ((&lt;i&gt;s&lt;/i&gt; mod &lt;i&gt;m&lt;/i&gt;)+(&lt;i&gt;t&lt;/i&gt; mod &lt;i&gt;m&lt;/i&gt;)) mod &lt;i&gt;m&lt;/i&gt;.  If not, you should first prove it, because you are using it in your proof.  Other than this, your proof is correct except for a tiny point that you have not stated what &lt;i&gt;y&lt;/i&gt;&lt;sub&gt;b&lt;/sub&gt; and &lt;i&gt;y&lt;/i&gt;&lt;sub&gt;r&lt;/sub&gt; are.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I call your proof easy, but slightly verbose.  Most importantly, you do not have to break &lt;i&gt;y&lt;/i&gt; into &lt;i&gt;y&lt;/i&gt;&lt;sub&gt;b&lt;/sub&gt; and &lt;i&gt;y&lt;/i&gt;&lt;sub&gt;r&lt;/sub&gt;.&lt;/p&gt;&#xA;" OwnerUserId="926" LastActivityDate="2010-09-24T11:43:57.320" CommentCount="4" />
  <row Id="5369" PostTypeId="2" ParentId="5357" CreationDate="2010-09-24T11:45:43.443" Score="1" ViewCount="" Body="&lt;p&gt;As others have already said, you just apply the product rule to $f.g^{-1}.$ However, the is an&#xA;&lt;a href=&quot;http://www.jstor.org/pss/2324425&quot; rel=&quot;nofollow&quot;&gt;American Mathematical Monthly article&lt;/a&gt; on how NOT to do it, which you may find instructive.&lt;/p&gt;&#xA;" OwnerUserId="1301" LastActivityDate="2010-09-24T11:45:43.443" />
  <row Id="5370" PostTypeId="1" AcceptedAnswerId="5372" CreationDate="2010-09-24T11:48:43.340" Score="4" ViewCount="293" Body="&lt;p&gt;Please suggest a suitable approach for this problem.&lt;/p&gt;&#xA;" OwnerUserId="2109" LastEditorUserId="2109" LastEditorDisplayName="" LastEditDate="2010-11-11T15:13:55.503" LastActivityDate="2010-11-11T15:13:55.503" Title="How many number of three digit numbers lying between 100 and 999 (inclusive) and having only two consecutive digits identical ?" Tags="&lt;combinatorics&gt;" AnswerCount="2" CommentCount="5" />
  <row Id="5371" PostTypeId="2" ParentId="3710" CreationDate="2010-09-24T11:58:01.933" Score="2" ViewCount="" Body="&lt;p&gt;This seems simple : The digits will be from the set {0,1,2,3,4,5,6,7,8,9} &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Number of possibility for first digit (Most significant digit) = &lt;strong&gt;9&lt;/strong&gt; (since we can't use 0)&#xA;Number of possibility for second digit = &lt;strong&gt;9&lt;/strong&gt; &#xA;Number of possibility for third digit  = &lt;strong&gt;8&lt;/strong&gt;&#xA;...&#xA;Number of possibility for sixth digit (Least significant digit)  = &lt;strong&gt;5&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence total possibility = &lt;strong&gt;9*9*8*7*6*5&lt;/strong&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2109" LastActivityDate="2010-09-24T11:58:01.933" />
  <row Id="5372" PostTypeId="2" ParentId="5370" CreationDate="2010-09-24T12:08:32.873" Score="4" ViewCount="" Body="&lt;p&gt;Ok. Ask yourself first how many contain 00. Next, how many contain 11 (it's 17, you work it out). Then consider the number that contain 22, etc. You will get 9+17*9=162. I hope this helps.&lt;/p&gt;&#xA;" OwnerUserId="1301" LastActivityDate="2010-09-24T12:08:32.873" CommentCount="1" />
  <row Id="5373" PostTypeId="2" ParentId="2293" CreationDate="2010-09-24T12:10:49.987" Score="1" ViewCount="" Body="&lt;p&gt;So I just stumbled across a very cool, easy to read paper by Eckmann called &quot;Social Choice and Topology A Case of Pure and Applied Mathematics.&quot; It's a survey of a result that Eckmann proved a long time ago about spaces that have &quot;n-means,&quot; and the application of this result to social theory. Basically, if you have some function that assigns a single preference when given the preferences of a population, and this function is &quot;continuous&quot; and fair (you can switch the order of the population and it won't change the output, and if everyone has the same preference then that's what you get out), then this says a LOT about the space of preferences. Given a few assumptions about the space of preferences (I dunno how reasonable they are or realistic, since I have no idea what it would mean for a space of preferences to be a cell complex...), one can show right off the bat that the fundamental group is abelian (we have an H-space structure), and later on he basically classifies all possible spaces that admit social choice functions. Anyway, I thought this was pretty cool as an example of beautiful mathematics and a quirky &quot;application.&quot;&lt;/p&gt;&#xA;" OwnerUserId="423" LastActivityDate="2010-09-24T12:10:49.987" CommunityOwnedDate="2010-09-24T12:10:49.987" CommentCount="1" />
  <row Id="5374" PostTypeId="2" ParentId="4626" CreationDate="2010-09-24T12:53:21.430" Score="1" ViewCount="" Body="&lt;p&gt;Answer to Question 1: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Look at &quot;Die Transformation von Bauer und Muir&quot;, §7 of Oskar Perron, &lt;em&gt;Die Lehre von den Kettenbrüchen&lt;/em&gt;, Band II, 1957.&lt;/p&gt;&#xA;" OwnerUserId="752" LastActivityDate="2010-09-24T12:53:21.430" CommunityOwnedDate="2010-09-24T12:53:21.430" CommentCount="0" />
  <row Id="5375" PostTypeId="2" ParentId="2293" CreationDate="2010-09-24T13:29:51.577" Score="5" ViewCount="" Body="&lt;p&gt;This isn't a specific application but an area of application.  Robert Ghrist and others use algebraic topology as a way of integrating local data about sensor networks into global information.  For example, you may want to determine whether there are any holes in your sensor coverage.  Here's an &lt;a href=&quot;http://www.johndcook.com/blog/2010/09/13/applied-topology-and-dante-an-interview-with-robert-ghrist/&quot; rel=&quot;nofollow&quot;&gt;interview&lt;/a&gt; with Ghrist where he elaborates on this idea.&lt;/p&gt;&#xA;" OwnerUserId="247" LastActivityDate="2010-09-24T13:29:51.577" CommunityOwnedDate="2010-09-24T13:29:51.577" />
  <row Id="5376" PostTypeId="2" ParentId="5366" CreationDate="2010-09-24T14:07:04.347" Score="4" ViewCount="" Body="&lt;p&gt;Here's the &lt;strong&gt;product rule&lt;/strong&gt; proof expressed in both divisibility form and congruence form, using the standard notation: $\rm\ \ \ a\ |\ b \ :=\ a\:$ divides $\rm\: b\:,\;$ and $\rm\ \; a\equiv b\ \ (mod\ m)\: \iff\: m\:|\:a-b$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;LEMMA&lt;/strong&gt; $\rm\quad\: m\ |\ X-x\ $ and $\rm\ m\:|\:Y-y \ \Rightarrow\ m\:|\:XY - \: xy$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof&lt;/strong&gt;: $\rm\quad\ \ \ \  m\:|\:(X-x)\:y\ +\ X\:(Y-y) \ \ \ =\ \ \ XY - \: xy \quad$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;LEMMA&lt;/strong&gt; $\rm\quad\quad\quad\ \: X\equiv x\quad\:$ and $\rm\quad\: Y\equiv y \ \ \ \ \Rightarrow\:\ \ XY \equiv xy\ \ \ (mod\ m)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof&lt;/strong&gt;: $\rm\ \ \ 0 \ \equiv\ (X-x)\:y\ +\ X\:(Y-y) \ \ \: \equiv\ \ \ XY - \: xy \ \ \ (mod\ m)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice how the congruence notation eliminates the cumbersome manipulation of relations (divisibility). Indeed, the relations are replaced by a generalized equality (congruence) which, being compatible with multiplication (as above) and addition (similar proof), enables us to exploit our well-honed intuition manipulating integer equations - which immediately generalizes to manipulating congruences (mod m). When you study abstract algebra you'll learn that this is a very special case of a quotient or residue ring. This product rule arises in many analogous contexts, e.g. see my post on the &lt;a href=&quot;http://math.stackexchange.com/questions/3178&quot;&gt;product rule for derivatives&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2011-01-02T23:00:02.660" LastActivityDate="2011-01-02T23:00:02.660" CommentCount="3" />
  <row Id="5377" PostTypeId="1" AcceptedAnswerId="5405" CreationDate="2010-09-24T14:24:55.983" Score="8" ViewCount="664" Body="&lt;p&gt;I came across the following &quot;proof&quot; for the inconsistency of ZFC and can't find the flaw in it (if there is one...):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Construct a Turing machine A which sequentially runs on all proofs in ZFC and checks if the claim &quot;A does not halt or ZFC is inconsistent&quot; is proved. If such a proof is found, A halts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that A can know its encoding - see Kleene's recursion theorem - and even if it can't this &quot;obstacle&quot; can be overcome. So it seems to me such A is constructible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if A halts then it is because it has found a proof for &quot;A does not halt or ZFC is inconsistent&quot;. Since A halted, it is implied that ZFC is inconsistent. So either A does not halt, or ZFC is inconsistent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now (and my guess is that this is the part where this proof &quot;cheats&quot;) we have just &lt;strong&gt;proved&lt;/strong&gt; the claim &quot;A does not halt or ZFC is inconsistent&quot;, so such a proof exists in ZFC. So A must halt on this proof - and we have seen that this implies that ZFC is inconsistent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where is the mistake?&lt;/p&gt;&#xA;" OwnerUserId="1818" LastEditorUserId="1818" LastEditorDisplayName="" LastEditDate="2010-09-24T14:37:39.790" LastActivityDate="2010-12-24T03:53:16.367" Title="&quot;Proof&quot; that ZFC is inconsistent using Turing machines" Tags="&lt;logic&gt;&lt;fake-proofs&gt;&lt;computability&gt;" AnswerCount="7" CommentCount="7" FavoriteCount="1" />
  <row Id="5378" PostTypeId="1" AcceptedAnswerId="5379" CreationDate="2010-09-24T14:30:15.703" Score="2" ViewCount="425" Body="&lt;p&gt;Hi,&lt;br&gt;&#xA;I understand that there are different types of infinity: one can (even intuitively) understand that the infinity of the reals is different from the infinity of the natural numbers. Or that the infinity of the even numbers is the same as that of the natural numbers. How many types of infinity are there? Or is this infinite itself?&lt;br&gt;&#xA;TIA&lt;br&gt;&#xA;Steven&lt;/p&gt;&#xA;" OwnerUserId="923" LastActivityDate="2010-09-25T22:14:30.547" Title="Types of infinity" Tags="&lt;infinity&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="2" />
  <row Id="5379" PostTypeId="2" ParentId="5378" CreationDate="2010-09-24T14:33:04.053" Score="3" ViewCount="" Body="&lt;p&gt;Not only infinite - it's &quot;so big&quot; that there is no infinite set so large as the collection of all types of infinity (in Set Theoretic terms, the collection of all types of infinity is a class, not a set).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can easily see that there are infinite types of infinity via Cantor's theorem which shows that given a set A, its power set P(A) is strictly larger in terms of infinite size (the technical term is &quot;cardinality&quot;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Cantor%27s_theorem&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Cantor%27s_theorem&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1818" LastActivityDate="2010-09-24T14:33:04.053" CommentCount="1" />
  <row Id="5380" PostTypeId="2" ParentId="5378" CreationDate="2010-09-24T14:40:01.110" Score="1" ViewCount="" Body="&lt;p&gt;well, there is no definite answer, in my opinion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some mathematicians contend that the concept of (actual) infinite itself is ill-defined, and you may have only a potential infinite, the one you have when you say that a straight line segment may be extended at will, but you do not ave a whole straight line.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you accept Cantor theory, the cardinal numbers (the ones which count how many objects there are in a set) are infinite; you start from &amp;alefsym;&lt;sub&gt;0&lt;/sub&gt; which corresponds to the integers, and go on. With the usual axioms, you cannot decide whether there are or not infinite cardinal numbers strictly greater than &amp;alefsym;&lt;sub&gt;0&lt;/sub&gt; and strictly less than the cardinality of the real numbers, but anyway the infinite numbers are infinite.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You may also consider ordinal numbers (the ones which consider the ordered elements of a set); they are many more than the cardinal numbers ;-)&lt;/p&gt;&#xA;" OwnerUserId="89" LastActivityDate="2010-09-24T14:40:01.110" />
  <row Id="5381" PostTypeId="2" ParentId="5363" CreationDate="2010-09-24T14:43:55.670" Score="2" ViewCount="" Body="&lt;p&gt;In the question above, it seems that the description is far too complicated as it actually is. I will elaborate on the construction of coproducts of groups or monoids a bit. This will go beyond the scope of the specific question, but I hope it's helpful in other situations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First assume $M,N$ are sets. What is their coproduct? Obviously it's the disjoint union, which I will denote by $M + N$. Now assume $M,N$ are monoids. What is their coproduct? Well we have to introduce products of elements in $M + N$. Thus consider $(M + N)^{&amp;lt;\omega}$, the set of finite sequences with entries in $M + N$. This is a monoid by concatenation, but the inclusions $M,N \to (M + N)^{&amp;lt;\omega}$ are no homomorphisms. Well, then let's force it! Mod out the smallest congruence relation $\sim$ on our monoid, which satisfies $1 \sim (1_M), (m,m') \sim (mm')$ for all $m,m' \in M$ and similarily for elements of $N$. Then the quotient $(M + N)^{&amp;lt;\omega} / \sim$ is obviously a monoid (since $\sim$ is a congruence relation), and the universal property of the coproduct is also verified easily.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now what about groups? The correct definition of a group involves the operations of the underlying monoid and the inversion. But in practice, this inversion can be reconstructed from the rest of the data ($x=y^{-1}$ iff $xy=1$), and actually the category of groups is a full subcategory of the category of monoids. If the monoids $M,N$ above happen to be groups, their coproduct turns out to be a group, and clearly the universal property then also holds with respect to groups. Thus we have constructed the coproduct in the category of groups.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's turn back to monoids. The above construction of $M \coprod N$ is rather general, it shows the existence of colimits in every finitary algebraic category, but what can be said about the &lt;em&gt;elements&lt;/em&gt;? Do they have a canonical representation? Now there are two ways of doing it:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a) The elegant, short, &quot;geometric&quot; one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;b) The long, tedious one. This one is preferred in textbooks ...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In our case, b) means that you write down the set of reduced words, endow it with a terribly complicated monoid structure, show that every monoid axiom is satisfied, and finally check the universal property. Good luck.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a) you just use that $M \coprod N$ exists. We have constructed it, but the construction does not answer such simple questions as: Do $M$ and $N$ intersect in $M \coprod N$ trivially? Therefore we just use the existence of the coproduct together with the structure maps $i : M \to M \coprod N, j : N \to M \coprod N$. The idea is now to define an action of $M \coprod N$ on another object. This may be a geometric one, but in our case it's our desired set of reduced words.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Observe that the elements in $M \coprod N$, which are products of elements, which are in the image of $i$ or $j$, constitute a submonoid, which verifies the same universal property. In other words, every element in $M \coprod N$ is such a product. In such a product we may replace $i(m) i(m')$ by $i(mm')$ and cancel $i(1)$, similarily for $j$. Thus every element is in the image of the canonical map $X \to M \coprod N$, where&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$X := \{(... ,m_1,n_1,m_2,n_2, ...)  : m_i \in M - \{1\}, n_i \in N - \{1\}\}.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we prove that this map $X \to M \coprod N$ is a bijection, i.e. every element of the coproduct has a &lt;em&gt;unique&lt;/em&gt; representation as $... i(m_1) j(n_1) i(m_2) ...$. To do this, we define an action of the monoid $M \coprod N$ on $X$, i.e. a monoid homomorphism $M \coprod N \to Sym(X), (m \mapsto (x \mapsto mx))$, which should imitate the usual multiplication. By the universal property, it is enough to construct this homomorphism on $M$ and on $N$. If $m \in M$ and $x \in X$, define $mx$ as follows: If $x$ starts with an element in $N$, just concatenate with $(m)$. If $x$ starts with an element of $M$, which is not inverse to $m$, then multiply $m$ in the first entry. Otherwise delete the first entry. Similarily the homomorphism $N \to Sym(X)$ is defined. The resultung homomorphism $M \coprod N \to Sym(X)$ can be composed with the evaluation at the empty sequence to get a map $M \coprod N \to X$, which turns out to be a left inverse to $X \to M \coprod N$. Thus, $X \to M \coprod N$ is a bijection.&lt;/p&gt;&#xA;" OwnerUserId="1650" LastEditorUserId="1650" LastEditorDisplayName="" LastEditDate="2010-09-24T14:51:02.093" LastActivityDate="2010-09-24T14:51:02.093" />
  <row Id="5382" PostTypeId="2" ParentId="5377" CreationDate="2010-09-24T14:49:06.107" Score="3" ViewCount="" Body="&lt;p&gt;You've just proved that &quot;either A does not halt or ZFC is inconsistent&quot; must be true, not that this statement is provable in ZFC. There exist unprovable true statements in ZFC (if it is consistent). See Gödel's first incompleteness theorem.&lt;/p&gt;&#xA;" OwnerUserId="519" LastActivityDate="2010-09-24T14:49:06.107" CommentCount="3" />
  <row Id="5383" PostTypeId="1" AcceptedAnswerId="5385" CreationDate="2010-09-24T14:54:40.403" Score="1" ViewCount="151" Body="&lt;p&gt;Consider the following problem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/XsLEN.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I got $g'(1) = f'(1) = 2$, which is correct according to the expected answer. However, I get that $a$ can be any value, as it doesn't affect anything and $g$ is differentiable no matter what $a$ is (am I missing something here?). The expected answer is that $a$ must be $2$ however. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can someone explain why $a$ needs to be $2$?&lt;/p&gt;&#xA;" OwnerUserId="456" LastActivityDate="2010-09-24T14:57:45.340" Title="Conditions for differentiability" Tags="&lt;calculus&gt;&lt;homework&gt;" AnswerCount="2" />
  <row Id="5384" PostTypeId="2" ParentId="5383" CreationDate="2010-09-24T14:56:42.663" Score="2" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Hint&lt;/strong&gt;: if $g$ is supposed to be differentiable at $x=1$, the left and right derivatives of $g$ should be identical. Now, find the value of $a$ that would make it so.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-09-24T14:56:42.663" />
  <row Id="5385" PostTypeId="2" ParentId="5383" CreationDate="2010-09-24T14:57:45.340" Score="2" ViewCount="" Body="&lt;p&gt;The derivative of g from the right at $x=1$ is a, and from the left it's $f^\prime(1) = 2$. They must be equal.&lt;/p&gt;&#xA;" OwnerUserId="1818" LastActivityDate="2010-09-24T14:57:45.340" CommentCount="2" />
  <row Id="5386" PostTypeId="2" ParentId="5377" CreationDate="2010-09-24T15:23:19.603" Score="2" ViewCount="" Body="&lt;p&gt;For such questions, it is hard to get an intuitive understanding. I think the best way to intuitively understand it is by formalizing it using model-theory. So lets look at it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Say you define your A as a set in ZFC, using an $\cal{L}_1$-Formula, and denote the property of &quot;halting&quot; as $\cal{H}$, which can also be expressed in ZFC. So by your assumption, you have $ZFC\models{\cal H}(A)\leftrightarrow(ZFC\models\bot \vee ZFC\models\lnot{\cal H}(A))$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With a little more theory, you can imply therefore $ZFC+{\cal H}(A)\models (ZFC\models\bot\vee ZFC\models\lnot{\cal H}(A))$, and therefore, $ZFC+{\cal H}(A)\models (ZFC\models\bot)$. That is your one direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other direction goes with $ZFC+\lnot{\cal H}(A)\models \lnot (ZFC\models\bot \vee ZFC\models\lnot{\cal H}(A))$ (which follows from the first assumption), i.e. $ZFC+\lnot{\cal H}(A)\models (ZFC\not\models\bot) \wedge (ZFC\not\models\lnot{\cal H}(A))$. That is, in every model of ZFC, in which $\lnot{\cal H}(A)$ holds, every &quot;sub-model&quot; which can be created doint meta-theory inside that one, does neither model $\bot$ nor model $\lnot{\cal H}(A)$.&lt;/p&gt;&#xA;" OwnerUserId="1858" LastActivityDate="2010-09-24T15:23:19.603" />
  <row Id="5387" PostTypeId="2" ParentId="5363" CreationDate="2010-09-24T15:35:56.497" Score="5" ViewCount="" Body="&lt;p&gt;In answer to your question about whether this is written somewhere: the construction of free products in my book &lt;em&gt;Introduction to Topological Manifolds&lt;/em&gt; proceeds very much along these lines. (I'm not the only one who's written it down, but I don't have other references handy.) I don't go through the set-theoretic description of the set of ordered tuples in as much detail as you do, but it can be described more simply as just the disjoint union of the sets $(G\amalg H)^n$ over all n.  I like the concreteness of this construction as a way of giving one a handle on what the free product looks like.  Once the group is constructed, it's a relatively easy matter to prove after the fact that it is uniquely determined up to isomorphism as the coproduct in the category of groups.&lt;/p&gt;&#xA;" OwnerUserId="1421" LastActivityDate="2010-09-24T15:35:56.497" CommentCount="7" />
  <row Id="5388" PostTypeId="2" ParentId="5370" CreationDate="2010-09-24T15:42:54.137" Score="6" ViewCount="" Body="&lt;p&gt;In your comment, you said that the given solution was 9 &amp;bull; 9 + 1 &amp;bull; 9 + 8 &amp;bull; 9 = 162.  I'll attempt to explain a logic that yields that calculation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the 3-digit numbers that start with two identical digits.  There are 9 choices of the first digit (and inherently the second digit): 11, 22, 33, 44, 55, 66, 77, 88, and 99 (not 00 because the number is in the range 100-999).  For each of these, there are 9 choices of the final digit (0-9, except whatever digit was already chosen for the first two).  So, there are 9 &amp;bull; 9 such numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, suppose that the number ends with two identical digits.  There are 10 choices for the last digit (and inherently the second-to-last digit): 00, 11, 22, 33, 44, 55, 66, 77, 88, and 99, but we need to treat 00 separately from the rest.  If the number ends with 00, then the first digit can be 1-9, so 9 choices, so 1 &amp;bull; 9.  If the number ends with 11-99, there are 8 choices of first digit (1-9 except the digit already chosen), so 9 &amp;bull; 8.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While I have the 8 and 9 transposed in the final term, this is term-by-term the same expression as in the solution you gave.&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-09-24T15:42:54.137" CommentCount="1" />
  <row Id="5389" PostTypeId="2" ParentId="5377" CreationDate="2010-09-24T15:52:58.030" Score="3" ViewCount="" Body="&lt;p&gt;You might find &lt;a href=&quot;http://www.scottaaronson.com/democritus/lec3.html&quot; rel=&quot;nofollow&quot;&gt;Scott Aaronson's lecture&lt;/a&gt; on this useful. Search for the phrase &quot;Talk to the axioms!&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In summary, suppose ZFC is consistent. I'll work with Alex's simplification, where $A$ searches for a ZFC proof of &quot;A does not halt.&quot; Then there is a model $M$ of ZFC+&quot;$A$ halts&quot;. In $M$, there is a set $S$ which satisfies the whatever sentence of ZFC you have cooked up to encode &quot;a halting sequence of states for $A$&quot;. There is a set $P$ which satisfies the first order sentence you have cooked up to encode &quot;a ZFC proof that $A$ does not halt.&quot; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the perspective of the model $M$, there is no contradiction. The model sees that $P$ encodes a ZFC proof that $A$ doesn't halt, the model sees that $A$ does halt, so the model believes that ZFC is inconsistent. Since Con(ZFC) is not an axiom of ZFC, the model sees nothing wrong with this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From an outside perspective, $S$ does not actually look like a sequence of states for the Turing machine $A$, and $P$ does not look like a proof. They are just objects of the model which the model thinks have those features, because it has some strange interpretation of set theory. Even if you could construct $M$ (and I think there are obstacles to $M$ being computable in any reasonable way), you could not convert $P$ into an actual ZFC proof that $A$ does not halt, nor could you convert $S$ into a halting sequence for $A$. Any attempt to convert them into an actual proof or actual halting sequence will run into the same difficulty Scott describes in trying to extract a proof of NOT(Con(PA)) from a model of PA+NOT(Con(PA))&lt;/p&gt;&#xA;" OwnerUserId="448" LastActivityDate="2010-09-24T15:52:58.030" CommentCount="0" />
  <row Id="5390" PostTypeId="2" ParentId="5378" CreationDate="2010-09-24T15:58:03.450" Score="4" ViewCount="" Body="&lt;p&gt;A very nice introduction to the many different notions of infinity in mathematics is Rudy Rucker's book: &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0691001723&quot; rel=&quot;nofollow&quot;&gt;Infinity and the Mind&lt;/a&gt;. Unlike many other popularizations, this is written by someone who did a Ph.D. on the topic. Moreover, Rucker has gone to great lengths to make the presentation faithful to the mathematics but still accessible to an educated layperson. Below is an excerpt on the Alephs.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;stevenh: your recent comment &lt;a href=&quot;http://math.stackexchange.com/questions/201/is-there-possibly-a-largest-prime-number/2653#2653&quot;&gt;to this answer&lt;/a&gt; is based upon a misunderstanding. It is not only correct, but also beautiful, to reformulate Euclid's proof in the form: $\rm N(N+1)$ has more prime factors than does $\rm N$. For further details see my comments there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/z7owE.jpg&quot; alt=&quot;alt text&quot;&gt;&#xA;&lt;img src=&quot;http://i.imgur.com/kcYlt.jpg&quot; alt=&quot;alt text&quot;&gt;&#xA;&lt;img src=&quot;http://i.imgur.com/zfeyJ.jpg&quot; alt=&quot;alt text&quot;&gt;&#xA;&lt;img src=&quot;http://i.imgur.com/kH2we.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-09-25T22:14:30.547" LastActivityDate="2010-09-25T22:14:30.547" />
  <row Id="5391" PostTypeId="1" AcceptedAnswerId="5414" CreationDate="2010-09-24T16:06:53.790" Score="3" ViewCount="128" Body="&lt;p&gt;If we have a covering $p:H\rightarrow G$ where $G$ is a topological group, then $H$ is also a topological group. The multiplication function can be defined as follows. Consider the map $f:H\times H \rightarrow G$ which is a composition of the map $p\times p$ and the multiplication function on $G$. Choose $h\in p^{-1}(e)$ where $e$ is the identity element of $G$. If $$f_* (\pi_1(H\times H,(h,h))) \subset p_*(\pi_1(H,h)),$$ then $f$ can be lifted to a map $g:H\times H \rightarrow H$  such that $p\circ g = f$ and $g(h,h) = h$. Suppose we have shown the &quot;if&quot; part, then $g$ should function as our multiplication map on $H$. But given any $x\in H$, why do we know that $g(x,h) = x$ and that $g(x,h)$ does &lt;strong&gt;not&lt;/strong&gt; equal any other element of $p^{-1}(p(x))$?&lt;/p&gt;&#xA;" OwnerUserId="1907" LastEditorUserId="1907" LastEditorDisplayName="" LastEditDate="2010-09-24T19:01:34.190" LastActivityDate="2010-09-25T01:43:46.533" Title="Covering of a topological group is a topological group" Tags="&lt;topology&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="1" />
  <row Id="5392" PostTypeId="1" AcceptedAnswerId="5404" CreationDate="2010-09-24T17:14:00.647" Score="5" ViewCount="391" Body="&lt;p&gt;While I understand the intuition behind expectation, I don't really understand the meaning of variance and moment. &#xA;What is a good way to think of those two terms?&lt;/p&gt;&#xA;" OwnerUserId="766" LastActivityDate="2010-09-25T23:22:57.000" Title="Intuitive explanation of variance and moment in Probability." Tags="&lt;probability&gt;&lt;intuition&gt;&lt;probability-theory&gt;" AnswerCount="4" FavoriteCount="5" />
  <row Id="5393" PostTypeId="2" ParentId="5392" CreationDate="2010-09-24T17:17:39.800" Score="2" ViewCount="" Body="&lt;p&gt;The variance is the expected squared deviation between the random variable and its expectation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It measures how much the random variable scatters or spreads around its expectation.&lt;/p&gt;&#xA;" OwnerUserId="367" LastActivityDate="2010-09-24T17:17:39.800" CommentCount="3" />
  <row Id="5394" PostTypeId="2" ParentId="5363" CreationDate="2010-09-24T17:21:45.837" Score="4" ViewCount="" Body="&lt;p&gt;You can see essentially the same construction in two different ways in &lt;a href=&quot;http://math.berkeley.edu/~gbergman/245/&quot; rel=&quot;nofollow&quot;&gt;George Bergman's &lt;em&gt;An Invitation to General Algebra and Universal Constructions&lt;/em&gt;&lt;/a&gt; in &lt;a href=&quot;http://math.berkeley.edu/~gbergman/245/Ch.2.ps&quot; rel=&quot;nofollow&quot;&gt;Chapter 2&lt;/a&gt; (link is to a postscript file) for the free group. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, you define &lt;em&gt;&quot;the set of all terms in the elements of the set $X$ under the formal group operations $\mu$, $i$, $e$&quot;&lt;/em&gt; to mean a set which is given with functions symb${}_T\colon X\to T$, $\mu_T\colon T^2\to T$, $i_T\colon T\to T$, and $e_T\colon T^0\to T$, such that each of these maps is one-to-one, their images are disjoint, and $T$ is the union of the images, and $T$ is generated by symb${}_T(X)$ under the operations $\mu_T$, $i_T$, and $e_T$. Such a set exists (it can be constructed inductively with enough care; given in &lt;a href=&quot;http://math.berkeley.edu/~gbergman/245/Ch.1.ps&quot; rel=&quot;nofollow&quot;&gt;Chapter 1&lt;/a&gt; of the same notes). Then one defines an apropriate equivalence relation $\sim$ on $T$; the set $T/\sim$ gives the underlying set of the free group, and one defines the operations in the free group via representatives in the natural way. Bergman labels this &quot;the logician's approach&quot; (section 2.2). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;An alternative construction (&quot;the classical construction&quot;, section 2.4) gives &quot;free groups as groups of words&quot;. Again, you start with a set $X$, and let $T$ be the set of all group-theoretic terms of $X$; identify $X$ with its image under symb, and one defines a subset $T_{red}$ of &quot;reduced terms&quot; (defining what this means appropriately) and then defining operations $\otimes$, ${}^{(-)}$, and $e_T$ on this set to make it into a group. Proving it is a group can be done either in the straightforward but tedious way, or by using &quot;van der Waerden's trick&quot; (embed the set $T_{red}$ into a group of permutations, and check that the operations you defined correspond to the operations in the image, so that &quot;group&quot;-ness gets inherited). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To get the free product, you let $X$ be the disjoint union of the underlying sets of $G$ and $H$, and either adds to the equivalence relation (in the &quot;logician's approach&quot;), or restricts the definition of &quot;reduced words&quot; (in the &quot;classical approach&quot;), in essentially the way you did. &lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-09-24T17:21:45.837" CommentCount="1" />
  <row Id="5396" PostTypeId="1" CreationDate="2010-09-24T17:36:48.840" Score="2" ViewCount="152" Body="&lt;p&gt;Let $v^i$ be a vector on a Riemannian 3-manifold with metric $g_{ij}$ embedded inside a 3+1 space-time such that for some constant $N_M$ it satisfies the inequality $g_{ij}v^iv^j \leq  N_M ^2$. Let $K$ be a symmetric rank-2 tensor on the 3-manifold. Then apparently the following holds:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\vert K_{ij} v^i v^j \vert \leq \vert K \vert _g N_M ^2.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This looks like some sort of a Cauchy-Schwartz inequality but given that $K$ is a tensor as described, I don't understand what the notation on the RHS means. For a rank-2 symmetric tensor $K$ what does $\vert K \vert _g$ mean?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If one knows that for some function $N$, $g_{ij}v^iv^j \leq  N^2,$ where the function $N$ is itself bounded between constants&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$N_m \leq N \leq N_M,$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;then using inequalities like the above one can apparently show the following bound:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int _{t_1} ^t \frac{1}{N} \Big(-v^i \partial _i N - \frac{dN}{dt} + K_{ij}v^iv^j\Big) dt&#xD;&#xA; \leq -2\log N_m + \frac{1}{N_m} \int _{t_1}^t (\vert \nabla N \vert _g N_M + \vert K \vert _g N_M ^2 )dt,$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for some fixed $t_1$ and $t$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can't understand that first &quot;log&quot; term in the above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also once the above bound is shown does it follow that the integral can be unbounded above or below depending solely on the property of the function $N$? If yes then what would be needed of $N$ to make the integral unbounded above or below? &lt;/p&gt;&#xA;" OwnerUserId="965" LastEditorUserId="4322" LastEditorDisplayName="" LastEditDate="2011-05-02T13:07:01.313" LastActivityDate="2011-05-02T13:07:01.313" Title="What is the modulus of a tensor on a Riemannian 3-manifold?" Tags="&lt;differential-geometry&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="5397" PostTypeId="1" AcceptedAnswerId="5401" CreationDate="2010-09-24T17:46:26.540" Score="0" ViewCount="245" Body="&lt;p&gt;How do I find a (longitude, latitude) point (any will do) on a circle where the only info I have is a (longitude, latitude) center point, and a radius measured in Feet (ft.)?&lt;/p&gt;&#xA;" OwnerUserId="2113" LastEditorUserId="2113" LastEditorDisplayName="" LastEditDate="2010-09-24T18:33:15.820" LastActivityDate="2010-09-24T18:33:15.820" Title="How to find a (longitude, latitude) point on a circle when given only the center (longitude, latitude) point and radius measured in Feet" Tags="&lt;circle&gt;&lt;geodesy&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="1" />
  <row Id="5398" PostTypeId="2" ParentId="5392" CreationDate="2010-09-24T17:47:12.987" Score="4" ViewCount="" Body="&lt;p&gt;Variance of a r.v. $X$ is $Var(X) = \mathbb{E}[(X-\mathbb{E}(X))^2]$.  Variance is used as a measure for the dispersion of a random variable. If you are interested in how far a random variable is from what you expect; perhaps you would consider $\mathbb{E}(|X-\mathbb{E}(X)|)$ but see: &lt;a href=&quot;http://stats.stackexchange.com/questions/118/standard-deviation-why-square-the-difference-instead-of-taking-the-absolute-val&quot;&gt;http://stats.stackexchange.com/questions/118/standard-deviation-why-square-the-difference-instead-of-taking-the-absolute-val&lt;/a&gt; for why this we use the square instead.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For all except the first moment (i.e. the mean) the central moments (centred about the mean) are far more interesting. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the $r$th (central) moment of $X$, that is $\mathbb{E}[(X-\mathbb{E}(X))^r]$.  The lower moments ($r&amp;lt;5$) are related to properties of the distribution:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$r=0$ is $1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$r=1$ is $0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$r=2$ is the variance which is a measure of the spread from the mean. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$r=3$ is related to skewness which is a measure of the asymmetry of the distribution. See: &lt;a href=&quot;http://en.wikipedia.org/wiki/Skewness&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Skewness&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$r=4$ is related to kurtois which is a measure of the 'peakedness' of a distribution. See: &lt;a href=&quot;http://en.wikipedia.org/wiki/Kurtosis&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Kurtosis&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $r\geq5$ they become slightly harder to compute. The $5th$ central moment is a measure for the asymmetry of the distributions tails, but I would use skewness instead. &lt;/p&gt;&#xA;" OwnerUserId="1794" LastEditorUserId="1794" LastEditorDisplayName="" LastEditDate="2010-09-24T19:08:56.037" LastActivityDate="2010-09-24T19:08:56.037" CommentCount="5" />
  <row Id="5399" PostTypeId="1" AcceptedAnswerId="5403" CreationDate="2010-09-24T17:55:20.583" Score="2" ViewCount="153" Body="&lt;p&gt;I'm trying to understand how the performance of financial investments are measuring when the asset has multiple points of investment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, say I invest &lt;code&gt;$1000&lt;/code&gt; in a stock. After 6 months, it's worth &lt;code&gt;$1200&lt;/code&gt;. At this point, I invest an additional &lt;code&gt;$2000&lt;/code&gt;, for a total value of &lt;code&gt;$3200&lt;/code&gt;. If after an additional 6 months, the entire investment is worth &lt;code&gt;$4000&lt;/code&gt;, how would I calculate the overall yield or effective interest rate? Normally, you'd do (final-initial)/initial, but in this example there are multiple initial values, but only one final. How would I reconcile this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Note, some values are supposed to be dollar amounts, but for some reason the site's markup is interpreting the dollar sign as the start of italics...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Escaped dollar amounts (thanks KennyTM). &lt;/p&gt;&#xA;" OwnerUserId="2114" LastEditorUserId="2114" LastEditorDisplayName="" LastEditDate="2010-09-24T19:06:41.017" LastActivityDate="2010-09-25T15:09:11.527" Title="Measuring Financial Investment Performance" Tags="&lt;finance&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="2" />
  <row Id="5400" PostTypeId="2" ParentId="5377" CreationDate="2010-09-24T17:57:12.420" Score="0" ViewCount="" Body="&lt;p&gt;The problem is that if the machine manages to prove it - it halts. and therefore ZFC inconsistent.&lt;/p&gt;&#xA;" OwnerUserId="2116" LastActivityDate="2010-09-24T17:57:12.420" CommentCount="1" />
  <row Id="5401" PostTypeId="2" ParentId="5397" CreationDate="2010-09-24T18:24:25.353" Score="4" ViewCount="" Body="&lt;p&gt;Assuming you are using a spherical model of the earth, a simple way is to proceed due north or due south, for then each degree of latitude has a constant length (with value extremely close to 10000000/90 meters, because the French revolutionaries originally defined the meter as 10^-7 times the distance from the equator to the North Pole along the Paris meridian!).  For large radii or points close to poles you will have to adjust the latitude appropriately to account for passing over the pole (and add 180 degrees to the longitude).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, starting at (lon, lat) = (-90, 40) with a radius of 10,000 feet, you would convert that to 10,000 / (39.37/12) = 3048.006 meters, then convert that to 3048.006 * 90 / 10000000 = 0.0274321 degrees, landing you at (-90, 40.0274321).  (Notice I write the longitude first, because it's the horizontal coordinate.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you need more accuracy, &lt;em&gt;don't&lt;/em&gt; look up the approximate spherical radius of the earth: use the geoid appropriate for the datum of your coordinates.  Appropriate formulae appear in books on geodesy and earth projections.  A good resource is the USGS Professional Paper 1395, &lt;em&gt;Map Projections--A Working Manual&lt;/em&gt;, by John Snyder.&lt;/p&gt;&#xA;" OwnerUserId="1489" LastActivityDate="2010-09-24T18:24:25.353" />
  <row Id="5402" PostTypeId="2" ParentId="5331" CreationDate="2010-09-24T18:26:14.623" Score="1" ViewCount="" Body="&lt;p&gt;Let $c_n$ be a sequence dense in $S^1$ and define $a_n = \left ( 1 + \frac{1}{n^2} \right ) c_n$. Then $1 - |z_n| = \frac{1}{n^2}$ so this are the zeros of a Blaschke product.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, let $c_n = r e^{i \phi_n}$ and fill this in in the Blaschke product. Then we if $B_n(z,r)$ is the term inside the Blaschke product, then we can try to evaluate if $\sum 1 - B_n(z,r)$ converges. We can evaluate the convergence of this with the integral test and then we see the limit does not exist as $r \to 1^{-}$.&lt;/p&gt;&#xA;" OwnerUserId="1120" LastEditorUserId="1120" LastEditorDisplayName="" LastEditDate="2010-09-26T20:16:59.990" LastActivityDate="2010-09-26T20:16:59.990" />
  <row Id="5403" PostTypeId="2" ParentId="5399" CreationDate="2010-09-24T18:41:24.773" Score="2" ViewCount="" Body="&lt;p&gt;Excel has a function for this called IRR.  To roll your own you just need to find the rate of return that produces the final value.  You have investments i1, i2, ... that have been in the account for m1, m2, ... months.  For a given return per month you should have i1*(1+r)^m1 + i2*(1+r)^m2 + ...  Now just search for r that makes this equal to the final value.  For some patterns of cash flow, there may be multiple values of r that satisfy the equation.  You can use your favorite root finder, but bisection is quite robust.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-09-24T18:41:24.773" CommentCount="4" />
  <row Id="5404" PostTypeId="2" ParentId="5392" CreationDate="2010-09-24T18:43:48.710" Score="3" ViewCount="" Body="&lt;p&gt;There are many good answers possible, depending on how one's intuition works and one's experience with related things like averages, $L^p$ norms, etc.  One answer that builds on your intuition of an expectation is to think of the higher central moments as &lt;em&gt;weighted&lt;/em&gt; expectations.  The central third moment, for example, is defined as the expectation of the third powers of deviations from the mean.  Split a third power into two parts: one is the deviation from the mean (let's call it, after common statistical parlance, the &quot;residual&quot;) and the other is the squared deviation from the mean (&lt;em&gt;ergo&lt;/em&gt;, the squared residual).  Treat the latter as a weight.  As the residual gets larger in size, these weights get larger (in a quadratic fashion).  The third central moment is the weighted &quot;average&quot; of the residuals.  (I have to put &quot;average&quot; in quotes because, strictly speaking, to deserve being called an average one would require the weights themselves to average out to 1, but they don't usually do that.)  Thus, the third central moment gives disproportionately greater emphasis to larger residuals compared to smaller ones.  (It is possible for that emphasis to be so great that the third central moment doesn't even exist: if the probability doesn't decay rapidly enough with the size of the residual, the weights can swamp the probability, with the net effect of creating an infinitely great third central moment.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can understand, and even analyze, all other central moments in the same way, even fractional moments and absolute moments.  It should be immediately clear, for instance, that the higher the moment, the more the weights are emphasized at extreme values: higher central moments measure (average) behavior of the probability distribution at greater distances from the mean.&lt;/p&gt;&#xA;" OwnerUserId="1489" LastActivityDate="2010-09-24T18:43:48.710" />
  <row Id="5405" PostTypeId="2" ParentId="5377" CreationDate="2010-09-24T19:13:44.730" Score="7" ViewCount="" Body="&lt;p&gt;Here's my take on it. Use $(A \uparrow)$ to mean $A$ does not halt, and $(A \downarrow)$ to mean $A$ does halt. Let $\Phi$ be any sentence; the question uses ~Con(ZFC) but this is not material. Let $\mathrm{Pvbl}(b)$ be the standard ZFC-formalized provability predicate, which says that there is a coded proof of the formula with number $b$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is right that, by the recursion theorem, we can create a specific machine $A$ such that&#xA;$$&#xA;(A \downarrow) \Leftrightarrow \mathrm{Pvbl}((A\uparrow) \lor \Phi)&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moreover, because of the specific form of the formula $(A \downarrow)$, ZFC is able to prove&#xA;$$&#xA;(A \downarrow) \to \mathrm{Pbvl}(A\downarrow)&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Working in ZFC, assume $(A \downarrow)$. Then we know $\mathrm{Pvbl}(A \downarrow)$ and $\mathrm{Pvbl}((A \uparrow) \lor \Phi)$.  ZFC is able to prove enough about the Pvbl predicate to ensure that&#xA;$$\mathrm{Pvbl}(\psi) \land \mathrm{Pvbl}(\lnot \psi \lor \theta) \rightarrow \mathrm{Pvbl}(\theta)&#xA;$$&#xA;for all $\psi$ and $\theta$. So we can obtain $\mathrm{Pvbl}(\Phi)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now: what we have obtained is: &quot;A does not halt or Pvbl($\Phi$)&quot;. In the fourth paragraph of the question, it instead claims we can obtain &quot;A does not halt or $\Phi$&quot;. That is a stronger statement, and this is the first error I see in the proof. In the fourth paragraph, it is silently assumed that provability implies truth, but this is not correct for formalized provability. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;ZFC does not prove $\mathrm{Pvbl}(\Phi) \to \Phi$ in general. In particular, ZFC does not prove $\mathrm{Pvbl}(0=1) \to (0=1)$ because there are models of ZFC in which both $\mathrm{Pvbl}(0=1)$ and $0 \not = 1$ are satisfied.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Addendum: Löb's theorem addresses this exact question. Applied to ZFC, it states that if ZFC proves $\mathrm{Pvbl}(\Phi) \to \Phi$ then ZFC already proves $\Phi$. So, in particular, ZFC does not prove that Pvbl(~Con(ZFC)) implies ~Con(ZFC). &lt;/p&gt;&#xA;" OwnerUserId="630" LastEditorUserId="630" LastEditorDisplayName="" LastEditDate="2010-09-25T02:19:39.400" LastActivityDate="2010-09-25T02:19:39.400" CommentCount="4" />
  <row Id="5407" PostTypeId="2" ParentId="5312" CreationDate="2010-09-24T19:26:58.917" Score="2" ViewCount="" Body="&lt;p&gt;Here is a more general framework for problems of this kind.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A bipartite graph is one for which the vertices can be divided into two sets so that any edges only join vertices which are not in the same same one of these two sets. Bipartite graphs can be vertex colored with two colors and any circuits they have must have even length. If the two color sets of the bipartite graph differ in size then the graph can not have a hamiltonian circuit and if the two color sets differ by two the graph can not have a hamiltonian path. If the two color sets differ by 1 any hamiltonian path must start and end at vertices of the same color. If a bipartite graph has color sets of equal size it may or may not have a hamiltonian circuit. For example the 1x1x1 cube has a hamiltonian circuit but the rhombic dodecahedron does not. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A famous conjecture of David Barnette suggests that every planar 3-connected 3-valent bipartite graph has a hamiltonian circuit. Another way of putting this is that the graphs of 3-valent convex polyhedra all of whose faces have an even number sides have hamiltonian circuits. &lt;/p&gt;&#xA;" OwnerUserId="1369" LastEditorUserId="1369" LastEditorDisplayName="" LastEditDate="2010-09-24T20:40:01.887" LastActivityDate="2010-09-24T20:40:01.887" />
  <row Id="5408" PostTypeId="1" AcceptedAnswerId="5409" CreationDate="2010-09-24T19:55:58.140" Score="2" ViewCount="241" Body="&lt;p&gt;Suppose $x_{1}&gt;0$, and consider the sequence, $\{x_{n}\}$ defined as follows: $$x_{n+1}=\log(1+x_{n}) \quad n\geq 1 $$ Find the value of $\displaystyle \lim_{n \to \infty} nx_{n}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am having trouble solving it. One thing is clear, that since $x_{n}&gt;0$ and $x_{n+1} &amp;lt; x_{n}$, we can have a sequence which converges an $f$ which satisfies $f=\log(1+f)$, so that $f=0$. Any way as to how we can proceed from here.&lt;/p&gt;&#xA;" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-09-24T22:31:40.427" LastActivityDate="2010-09-24T22:31:40.427" Title="Limit of the sequence $nx_{n}$ where $x_{n+1} = \log (1 +x_{n})$" Tags="&lt;real-analysis&gt;&lt;analysis&gt;&lt;sequences-and-series&gt;&lt;asymptotics&gt;" AnswerCount="1" CommentCount="0" />
  <row Id="5409" PostTypeId="2" ParentId="5408" CreationDate="2010-09-24T20:19:48.617" Score="7" ViewCount="" Body="&lt;p&gt;Using the methods similar to the answer by David Speyer here: &lt;a href=&quot;http://math.stackexchange.com/questions/3215/convergence-of-sqrtnx-n-where-x-n1-sinx-n&quot;&gt;http://math.stackexchange.com/questions/3215/convergence-of-sqrtnx-n-where-x-n1-sinx-n&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I suggest you read that first)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe we get the answer to be $2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Use the fact that $$\log (1+x) = x - \frac{x^2}{2} + O(x^3)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $$y_{n}  = \frac{1}{x_n}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then we have that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$y_{n+1} = \frac{1}{x_n(1 - x_{n}/2 + O((x_{n})^2))} = y_{n} +  \frac{1}{2} + O(x_{n})$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This gives rise to (again similar to that answer)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{y_{n}}{n} = \frac{1}{2} + \frac{1}{n}O(\sum_{k=1}^{n-1} x_{k})$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $x_{n} \rightarrow 0$ we have that $$\frac{y_{n}}{n} \rightarrow \frac{1}{2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and thus $$nx_{n} \rightarrow 2$$&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-09-24T20:19:48.617" />
  <row Id="5410" PostTypeId="2" ParentId="4844" CreationDate="2010-09-24T21:18:12.523" Score="1" ViewCount="" Body="&lt;p&gt;You can look at one particular configuration at&#xA;&quot;Animated (Java) Illustrations {of} &lt;a href=&quot;http://mathpages.com/home/polyjava.htm&quot; rel=&quot;nofollow&quot;&gt;24 Electrons on a Sphere&lt;/a&gt;&quot;&#xA;and a few more at&#xA;&lt;a href=&quot;http://mathpages.com/home/kmath005/kmath005.htm&quot; rel=&quot;nofollow&quot;&gt;&quot;Min-Energy Configurations of Electrons On A Sphere&quot;&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1705" LastEditorUserId="1705" LastEditorDisplayName="" LastEditDate="2010-09-27T22:03:00.287" LastActivityDate="2010-09-27T22:03:00.287" CommentCount="2" />
  <row Id="5411" PostTypeId="1" CreationDate="2010-09-24T21:58:21.090" Score="2" ViewCount="329" Body="&lt;p&gt;I would like to recover the probability mass function (pmf) from the characteristic function (CF) of a discrete probability distribution using &lt;i&gt;Mathematica&lt;/i&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;p&gt; Ideally, I'd like to do calculations like this example (not necessarily this simple). To compute the pmf f&lt;sub&gt;X+X&lt;/sub&gt;[x] of the sum of two iid discrete uniform distributions X with support S={1,...,6}, it seems reasonable to try something like this:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&#xA;Subscript[\[CurlyPhi], X + X][t_] := &#xA; CharacteristicFunction[DiscreteUniformDistribution[{1, 6}], t]^2&#xA;&#xA;Subscript[f, X + X][x_] := &#xA; InverseFourierTransform[Subscript[\[CurlyPhi], X + X][t], t, x]&#xA;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;but &lt;b&gt;InverseFourierTransform&lt;/b&gt; doesn't recover the pmf from the CF of discrete distributions (it does recover pdfs from the CFs of continuous distributions). Am I using the wrong function? (There are other candidates, but &lt;b&gt;InverseFourier&lt;/b&gt; only works on lists of numbers, and &lt;b&gt;InverseZTransform&lt;/b&gt; doesn't seem to work here either.) Am I forgetting to set some necessary options to the inverse function? Or is there just no built-in to recover the pmf from the CF of a discrete probability distribution?&lt;/p&gt;&#xA;" LastActivityDate="2010-09-27T20:50:46.520" Title="Recovering the probability mass function from the characteristic function of a discrete probability distribution using Mathematica" Tags="&lt;probability&gt;&lt;fourier-analysis&gt;&lt;mathematica&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" />
  <row Id="5412" PostTypeId="2" ParentId="5399" CreationDate="2010-09-24T22:09:13.207" Score="4" ViewCount="" Body="&lt;p&gt;My interpretation of your question, my hand calculations and explanations are as follows. For periods of six months you have:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;At the beginning of period 1 you invest $C_{0}=1000$ (dollars). &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;At the end of period 1 $C_{0}$ worths $F_{1}=1200$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;At the beginning of period 2 you invest the additional capital of $%&#xA;C_{1}=2000$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;At the end of period 2 your investement worths $F_{2}=4000$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/gk7Gh.jpg&quot; alt=&quot;alt text&quot;&gt; &lt;strong&gt;Figure: Cash Flows&lt;/strong&gt; of -1000, -2000 and +4000 dollars&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let us denote by $r$ the anual nominal interest rate of your investement. In &#xA;$n$ periods of six months the investement worths $(1+r/2)^{n}$ per currency unit of invested capital.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The initial investement $C_{0}$ worths $C_{0}(1+r/2)^{2}$ at the end of&#xA;period 2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The additional capital $C_{1}$ worths $C_{1}(1+r/2)$ at the end of period 2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$C_{0}(1+r/2)^{2}+C_{1}(1+r/2)=F_{2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$1000(1+r/2)^{2}+2000(1+r/2)=4000$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(1+r/2)^{2}+2(1+r/2)=4$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$2r+\dfrac{1}{4}r^{2}+3=4.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The solution is: $r=2\sqrt{5}-4\approx 0.47214\approx 47.214\%$ anual&#xA;nominal rate&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To determine the effective interest rate, you could find how much would you&#xA;need to invest so that at a nominal rate of $r$ you yould have $4000$ in 1&#xA;year (2 periods):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$P\cdot 1.2361^{2}=4000$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$P=4000/1.2361^{2}=2617.9$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The anual effective interest rate is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$i_{eff}=\dfrac{4000}{2617.9}-1=0.52794\approx 52,794\%$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remark: The future value $F_{n}$ at the end of period $n$ of a present value &#xA;$P$ is given by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$F_{n}=P\left( 1+\dfrac{r}{m}\right) ^{n}$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $r=i_N$ is the anual nominal interest rate compounded $m$ periods per year. In your case $m=2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reference: Rigg, Bedworth and Randhawa, &lt;em&gt;Engineering Economics&lt;/em&gt;, Mc Graw Hill, 4th ed., 1996.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;All errors and omissions are of mine, of course.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="752" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2010-09-25T15:09:11.527" LastActivityDate="2010-09-25T15:09:11.527" />
  <row Id="5413" PostTypeId="2" ParentId="5362" CreationDate="2010-09-25T00:18:37.227" Score="0" ViewCount="" Body="&lt;p&gt;I'm no specialist, but see how this goes.  When you say &quot;the subset A∩λ of λ is not stationary&quot;, stationary in what?  Presumably not stationary in λ.  So there is a closed and unbounded subset of λ that is disjoint from A.  Pick any element of this subset greater than sup{g(α):α&amp;lt;λ}&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-09-25T00:18:37.227" CommentCount="1" />
  <row Id="5414" PostTypeId="2" ParentId="5391" CreationDate="2010-09-25T01:43:46.533" Score="3" ViewCount="" Body="&lt;p&gt;Consider the map $k: H\to H$ given by $k(x) = g(x,h)$.  Then for any $x\in H$, we have $$p\circ k(x)  = p\circ g(x,h) = m\circ (p\times p)(x,h) = m(p(x),e) = p(x),$$ &#xA;which implies that $k$ is an automorphism of the covering $p$ (which you can also think of as a lift of $p$).  Note also that $k(h)  = g(h,h) = h$.  Thus $k$ and the identity are both automorphisms of $p$ that agree at a point, so they are equal. This implies $g(x,h)=x$ for all $x$.&lt;/p&gt;&#xA;" OwnerUserId="1421" LastActivityDate="2010-09-25T01:43:46.533" CommentCount="3" />
  <row Id="5415" PostTypeId="2" ParentId="201" CreationDate="2010-09-25T03:16:39.083" Score="3" ViewCount="" Body="&lt;p&gt;Here's a really nice proof that there are an infinite number of primes:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\prod(1-p_i^{-2})^{-1} = \zeta(2) = \pi^2/6$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $\pi$ is transcendental, the RHS is irrational. Therefore there must be an infinite number of terms in the product on the left.&lt;/p&gt;&#xA;" OwnerUserId="1820" LastEditorUserId="1820" LastEditorDisplayName="" LastEditDate="2010-09-25T04:34:12.230" LastActivityDate="2010-09-25T04:34:12.230" />
  <row Id="5416" PostTypeId="1" AcceptedAnswerId="5518" CreationDate="2010-09-25T04:58:05.933" Score="4" ViewCount="532" Body="&lt;p&gt;The derivative of a function $f: C \rightarrow C$ with respect to $\overline{z}$ is defined as&#xA;$$ \frac{1}{2} \left( \frac{\partial}{\partial x} + i \frac{\partial}{\partial y} \right) $$&#xA;where $\frac{\partial}{\partial x}$ is defined as the operator which maps $u+iv$ to $u_x + iv_x$, and $\frac{\partial}{\partial y}$ is defined likewise. I am wondering whether there is a geometric way to think about this quantity. &lt;/p&gt;&#xA;" OwnerUserId="2120" LastActivityDate="2010-09-27T08:46:24.137" Title="geometric meaning of differentiation with respect to the complex conjugate of $z$" Tags="&lt;complex-analysis&gt;" AnswerCount="7" FavoriteCount="1" />
  <row Id="5417" PostTypeId="2" ParentId="5362" CreationDate="2010-09-25T05:10:16.097" Score="2" ViewCount="" Body="&lt;p&gt;Hi Martin. The exercise is asking for a map from $A$ to the ordinals, not to itself. Why do you think the exercise is trivial? If $\lambda$ is regular and $A\cap\lambda$ is non-stationary (in $\lambda$) then there is a club subset of $\lambda$ disjoint from $A$. You can then pick $g(\lambda)$ in this club. You need to ensure, however, that this element is not $g(\tau)$ for some prior $\tau$, and this requires some argument: If, as suggested in another answer, you are to pick something larger than $\sup(g(\alpha)\mid\alpha\in A\cap\lambda)$, you need to argue that this supremum is strictly below $\lambda$. But this is not automatic (or true in general), as $A$ may contain all cardinals up to and including some inaccessible. Also, $A$ may contain singular cardinals, and you need to define $g$ on them as well. &lt;/p&gt;&#xA;" OwnerUserId="462" LastActivityDate="2010-09-25T05:10:16.097" CommentCount="4" />
  <row Id="5418" PostTypeId="1" CreationDate="2010-09-25T05:17:25.233" Score="1" ViewCount="132" Body="&lt;p&gt;in constrained Lagrangian optimization what is a general way to figure out how the optimal point varies with respect to parameters of the constraints? For example maximize x*y when x + y &amp;lt; k, and I wanted to find dx*/dk and dy*/dk where (x*,y*) is the optimal point. How would I do that?&lt;/p&gt;&#xA;" OwnerUserId="2121" LastActivityDate="2010-09-25T05:49:29.387" Title="Lagrangian (constrained) optimization - rate of change of optimal values." Tags="&lt;optimization&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" />
  <row Id="5419" PostTypeId="2" ParentId="5418" CreationDate="2010-09-25T05:49:29.387" Score="2" ViewCount="" Body="&lt;p&gt;The optimal point may not be differentiable at a value of $k$ where a constraint goes from binding to non-binding. At any other value of $k$ just differentiate with respect to $k$ the (equality) first order conditions for the constraints which bind and solve for $\partial x^*/\partial k$ etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take a look at any standard optimization book, for eg. Luenberger &lt;em&gt;Linear and Non-linear Programming&lt;/em&gt; or Sundaram's &lt;em&gt;A First Course in Optimization Theory&lt;/em&gt;,&lt;/p&gt;&#xA;" OwnerUserId="1195" LastActivityDate="2010-09-25T05:49:29.387" CommentCount="1" />
  <row Id="5420" PostTypeId="1" CreationDate="2010-09-25T05:54:40.267" Score="10" ViewCount="374" Body="&lt;p&gt;I'm working on a multiple-precision library. I'd like to make it possible for users to ask for higher precision answers for results already computed at a fixed precision. My $\mathrm{sqrt}(x)$ can pick up where it left off, even if $x$ changes a bit, because it uses Newton-Raphson. But $\exp(x)$ computed using a Maclaurin series or continued fraction has to be computed from scratch.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there an iterative refinement (i.e. Newton-Raphson, gradient descent) method for computing $\exp(x)$ that uses only arithmetic and integer roots?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I know Newton-Raphson can solve $\log(y)-x=0$ to compute $\exp(x)$. I am specifically not asking for that. Newton-Raphson can also solve $\exp(y)-x=0$ to compute $\log(x)$. Note that each requires the other. I have neither right now as an arbitrary-precision function. I have arithmetic, integer roots, and equality/inequality tests.)&lt;/p&gt;&#xA;" OwnerUserId="2122" LastEditorUserId="2122" LastEditorDisplayName="" LastEditDate="2010-09-25T16:30:28.770" LastActivityDate="2010-11-17T20:57:10.527" Title="Iterative refinement algorithm for computing exp(x) with arbitrary precision" Tags="&lt;algorithms&gt;&lt;numerical-methods&gt;&lt;special-functions&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="4" />
  <row Id="5422" PostTypeId="2" ParentId="5416" CreationDate="2010-09-25T09:33:58.763" Score="1" ViewCount="" Body="&lt;p&gt;I think the hypothesis that f is a totally differentiable function is necessary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A basic first result is that, f is holomorphic if and only if $\frac{\partial f}{\partial \overline{z}}=0$, and in that case $f' = \frac{\partial f}{\partial z}$, i.e &quot;Holomorphic functions are independent of  $\overline{z}$ and depend only on z&quot;, by Cauchy-Riemann.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Interpreting f as a real function of two independent real variables x, y, where z = x + i y. Hence $x = \frac{z + \overline{z}}{2}$, and $y = \frac{z - \overline{z}}{2i}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Minimizing f with respect to $\overline{z}$ and applying the chain rule&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{\partial f}{\partial \overline{z}} = \frac{\partial x}{\partial \overline{z}}_z\frac{\partial f}{\partial x}_y +  \frac{\partial y}{\partial \overline{z}}_z\frac{\partial f}{\partial y}_x$&#xA;$= \frac{1}{2} ( \frac{\partial f}{\partial x}_x + i \frac{\partial f}{\partial y}_y)$&#xA;and to minimize over all z = x + iy (since minimizing w.r.t x (resp. y) is to have the partial derivative = 0), then $\frac{\partial f}{\partial \overline{z}} = 0$.&#xA;see &lt;a href=&quot;http://people.ccmr.cornell.edu/~muchomas/P480/Notes/dft/node10.html&quot; rel=&quot;nofollow&quot;&gt;http://people.ccmr.cornell.edu/~muchomas/P480/Notes/dft/node10.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It should be mentioned that both $\frac{\partial f}{\partial \overline{z}}$ along with $\frac{\partial f}{\partial z}$ are known as Wirtinger derivatives(operators). I'm not sure if geometric intuition is necessary, but these derivatives give the Wirtinger calculus. One builds a theory (a &quot;symbolical tool&quot;?!!) that's analogous to the real case that is applied even to nonholomorphic functions (see &lt;a href=&quot;http://books.google.com.eg/books?id=CC0dQxtYb6kC&amp;amp;pg=PA67&amp;amp;lpg=PA67&amp;amp;dq=Wirtinger+calculus&amp;amp;source=bl&amp;amp;ots=70YeqX9-k6&amp;amp;sig=asCVvTFAvG_y3GrPXSnHnNphl9Q&amp;amp;hl=en&amp;amp;ei=i5mdTJ6yNIjH4AakrrjIDg&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=1&amp;amp;ved=0CBIQ6AEwADge#v=onepage&amp;amp;q=Wirtinger%20calculus&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;&quot;Theory of complex functions&quot;, Remmert, page 67-69&lt;/a&gt; for a quick development), or the beginnings of &quot;&lt;a href=&quot;http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.86.6515&amp;amp;rep=rep1&amp;amp;type=pdf&quot; rel=&quot;nofollow&quot;&gt;The Complex Gradient Operator and the CR-Calculus, ECE275A – Lecture Supplement&lt;/a&gt;&quot;, which is more elementary but could be useful. &lt;/p&gt;&#xA;" OwnerUserId="1746" LastEditorUserId="1746" LastEditorDisplayName="" LastEditDate="2010-09-25T09:47:14.573" LastActivityDate="2010-09-25T09:47:14.573" />
  <row Id="5424" PostTypeId="1" CreationDate="2010-09-25T11:08:13.333" Score="2" ViewCount="201" Body="&lt;p&gt;Without strenuous arithmetic. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a programme I can download to do so?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are the quadratic residues modulo 5^4 or 5^5?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="2127" LastEditorUserId="622" LastEditorDisplayName="" LastEditDate="2010-09-25T11:21:35.063" LastActivityDate="2011-04-02T17:16:41.017" Title="How can I find out whether a number is a quadratic residue in a large modulo?" Tags="&lt;number-theory&gt;&lt;math-software&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="5425" PostTypeId="2" ParentId="5416" CreationDate="2010-09-25T11:30:46.813" Score="3" ViewCount="" Body="&lt;p&gt;For getting geometric intuition about complex analysis Tristram Needham's book &lt;a href=&quot;http://books.google.com/books?id=ogz5FjmiqlQC&amp;amp;dq=visual+complex+analysis&amp;amp;printsec=frontcover&amp;amp;source=bn&amp;amp;hl=en&amp;amp;ei=j9qdTJ2kDMrKOLDomZ8M&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=6&amp;amp;ved=0CC8Q6AEwBQ#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;&quot;Visual Complex Analysis&quot;&lt;/a&gt; seems to be the best place you can turn to! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There, reading from page 216 on (and you can safely just jump there, if you already have arrived at questions like the one you are asking here), you will learn that holomorphic functions are the ones which look locally like compositions of rotations, translations and rescalings (I mean if you really draw the complex plane as a plane), i.e. they are the ones which carry disks to disks (and I mean the real round disks, with boundary a circle :-). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So since the derivation of a function $f$ with respect to $\bar{z}$ is zero iff $f$ is holomorphic, you could say that it measures how far $f$ is from being holomorphic or, in view of the above characterization, how much a disk can get deformed (e.g. into an ellipsoid) when you apply $f$.&lt;/p&gt;&#xA;" OwnerUserId="2048" LastEditorUserId="2048" LastEditorDisplayName="" LastEditDate="2010-09-25T11:49:50.690" LastActivityDate="2010-09-25T11:49:50.690" />
  <row Id="5426" PostTypeId="2" ParentId="5424" CreationDate="2010-09-25T11:31:51.597" Score="4" ViewCount="" Body="&lt;p&gt;If you want to know the squares modulo $p^m$ where $p$ is an odd prime,&#xA;they are congruent to $0$ or to numbers $a p^{2k}$ where $0\le k &amp;lt; m/2$&#xA;and the Legendre symbol $\left(\frac ap\right)=1$. For large primes $p$,&#xA;Legendre symbols can be calculated by regarding them as&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Jacobi_symbol&quot; rel=&quot;nofollow&quot;&gt;Jacobi symbols&lt;/a&gt; and using the law of quadratic reciprocity&#xA;for Jacobi symbols.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't regard $5$ as a large prime. :-)&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-09-25T11:31:51.597" CommentCount="2" />
  <row Id="5427" PostTypeId="1" CreationDate="2010-09-25T13:42:11.600" Score="1" ViewCount="149" Body="&lt;p&gt;I'm studying on K. Hrbacek and T. Jech, Introduction to Set Theory. In the third chapter, they prove the usual properties of the strict ordering on natural numbers in the following way:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;They prove transitivity using induction.&lt;/li&gt;&#xA;&lt;li&gt;They prove irreflexivity using induction and transitivity.&lt;/li&gt;&#xA;&lt;li&gt;They prove asymmetry by contradiction and using transitivity and irreflexivity.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;There is nothing wrong with this, but I'd like, if possible, to prove the previous properties (in particular irreflexivity and asymmetry) independently and using induction only.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, let $\phi(x)$ be $x\not\in x$. Of course $\phi(0)$ is true ($0\not\in 0$ since $0$ is the empty set). Now I should suppose that $\phi(x)$ is true and show that $\phi(x+1)$ is true, too. $\phi(x+1)$ stands for $x+1\not\in x+1$, that is $x+1\not\in x$ and $x+1\not=x$. But here I cannot go on, mainly because $x+1$ is at the left of the relations symbols.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any hint? Thank you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: $x+1$ is defined as $x\cup\{x\}$.&lt;/p&gt;&#xA;" OwnerUserId="2128" LastEditorUserId="2128" LastEditorDisplayName="" LastEditDate="2010-09-25T17:48:53.300" LastActivityDate="2010-09-25T20:57:35.340" Title="Strict ordering on natural numbers" Tags="&lt;set-theory&gt;&lt;induction&gt;&lt;natural-number&gt;" AnswerCount="2" />
  <row Id="5428" PostTypeId="2" ParentId="5297" CreationDate="2010-09-25T13:43:33.377" Score="1" ViewCount="" Body="&lt;p&gt;This is a new version of the answer. The comments refer to previous versions. Martin Brandenburg's comment was especially helpful. A comment of Dan Ramras, somewhere else in this thread, drew my attention to Lemma 5.5 &lt;a href=&quot;http://www.iecn.u-nancy.fr/~gaillard/milnor_64.pdf&quot; rel=&quot;nofollow&quot;&gt;page 64&lt;/a&gt; in Milnor and Stasheff's &lt;strong&gt;Characteristic Classes&lt;/strong&gt;. Thank you also to Agusti Roig for his wonderful question. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me state the Lemma of Milnor and Stasheff mentioned above: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) Let $A_1\subset A_2\subset\cdots$ and $B_1\subset B_2\subset\cdots$ be sequences of locally compact spaces with inductive limits $A$ and $B$ respectively. Then the product topology on $A\times B$ coincides with the inductive limit topology which is associated with the sequence $A_1\times B_1\subset A_2\times B_1\subset\cdots$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The proof (which can be found &lt;a href=&quot;http://www.iecn.u-nancy.fr/~gaillard/milnor_64.pdf&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;) shows in fact that the following more technical statement also holds: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) Let $A_1\subset A_2\subset\cdots$ and $B_1\subset B_2\subset\cdots$ be as above. For each $i$ let $C_i\subset A_i$ and $D_i\subset B_i$ be subspaces. Assume $C_i\subset C_{i+1}$ and $D_i\subset D_{i+1}$ for all $i$, and call $C$ and $D$ the respective inductive limits. Then the product topology on $C\times D$ coincides with the inductive limit topology which is associated with the sequence $C_1\times D_1\subset C_2\times D_2\subset\cdots$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, if the $C_i$ are topological groups, then so is $C$ (assuming, of course, that the topological group structure on $C_i$ is induced by that of $C_{i+1}$). This seems to indicate that the alleged counterexample of Tatsuuma et al. is not really a counterexample. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know no example of an inductive limit of topological groups which is not a topological group. (I think that such examples do exist. It would be interesting to know if $(\mathbb R^\infty)^\infty$ is a topological group.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's prove (2). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $W$ be a subspace of $C\times D$ which is open in the inductive limit topology. For each $i$ let $W_i$ be an open subset of $A_i\times B_i$ which has the same intersection with $C_i\times D_i$ as $W$. Let $(c,d)$ be in $W\cap(C_i\times D_i)$. There is a compact neighborhood $K_i$ of $c$ in $A_i$, and a compact neighborhood $L_i$ of $d$ in $B_i$, such that $K_i\times L_i\subset W_i$. There is also a compact neighborhood $K_{i+1}$ of $K_i$ in $A_{i+1}$, and a compact neighborhood $L_{i+1}$ of $L_i$ in $B_{i+1}$, such that $K_{i+1}\times L_{i+1}\subset W_{i+1}$. Then $K_{i+1}\cap C_{i+1}$ is a neighborhood of $K_i\cap C_i$ in $C_{i+1}$, and $L_{i+1}\cap D_{i+1}$ is a neighborhood of $L_i\cap D_i$ in $D_{i+1}$. The union $U$ of the $K_i\cap C_i$ is open in $C$, and the union $V$ of the $L_i\cap D_i$ is open in $D$, the spaces $C$ and $D$ being equipped with the inductive limit topology. Moreover we have $(c,d)\in U\times V\subset W$.&lt;/p&gt;&#xA;" OwnerUserId="660" LastEditorUserId="660" LastEditorDisplayName="" LastEditDate="2010-10-03T05:53:24.397" LastActivityDate="2010-10-03T05:53:24.397" CommunityOwnedDate="2010-10-03T05:29:23.220" CommentCount="2" />
  <row Id="5429" PostTypeId="2" ParentId="213" CreationDate="2010-09-25T14:33:17.487" Score="0" ViewCount="" Body="&lt;p&gt;My answer is: fundamental concepts and methods of both first order logic and set theory. I really wish I learned them much earlier, since all mathematics is based on them.&lt;/p&gt;&#xA;" OwnerUserId="2128" LastActivityDate="2010-09-25T14:33:17.487" CommunityOwnedDate="2010-09-25T14:33:17.487" />
  <row Id="5430" PostTypeId="1" AcceptedAnswerId="5435" CreationDate="2010-09-25T16:23:39.410" Score="2" ViewCount="106" Body="&lt;p&gt;This question stumped me on a recent calculus test... I would really like an explanation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The region $\mathbf{R}$ is bounded by $h(x) = -0.8(x-2)^2 + 8$ and $p(x) = x\cos(x) - \ln(2x+2)$. Find the lateral surface area of the solid formed when $\mathbf{R}$ is revolved around the line $y=11$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried to convert both functions into terms of y so that I could get the arc length, but that didn't get me very far. Any help would be greatly appreciated.&lt;/p&gt;&#xA;" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-09-25T16:26:43.407" LastActivityDate="2010-09-25T17:16:43.820" Title="Lateral surface area of a region" Tags="&lt;calculus&gt;&lt;integral&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="5431" PostTypeId="1" AcceptedAnswerId="5434" CreationDate="2010-09-25T16:38:45.680" Score="4" ViewCount="172" Body="&lt;p&gt;Say I have a function $f(x) = ax^3 + bx^2 + cx + d$ where $a &gt; 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's clear that for a high enough value of $x$, the $x^3$ term will dominate and I can say $f(x) \in O(x^3)$, but this doesn't seem very formal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The formal definition is $f(x) \in O(g(x))$ if constants $k, x_0 &gt; 0$ exist, such that $0 \le f(x) \le kg(x)$ for all $x &gt; x_0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is, what are appropriate values for $k$ and $x_0$? It's easy enough to find ones that apply (say $k = a + b + c + d$). By the formal definition, all I have to do is show that these numbers exist, so does it actually matter which numbers I use? For some value of $x$, $k$ could be anywhere from $1$ to $a + b + c + d + ... $. From my understanding, it doesn't matter what numbers I pick as long as they 'work', but is this right? It seems too easy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="1957" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-09-25T22:21:09.597" LastActivityDate="2010-09-25T22:21:09.597" Title="Formally proving that a function is $O(x^n)$" Tags="&lt;algorithms&gt;&lt;asymptotics&gt;" AnswerCount="3" CommentCount="7" FavoriteCount="1" />
  <row Id="5432" PostTypeId="1" AcceptedAnswerId="5515" CreationDate="2010-09-25T16:53:58.793" Score="1" ViewCount="289" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The well known formula for perfect numbers is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$P_n=2^{n-1}(2^{n}-1)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;this formula is obtained by observing some patterns on the sum of the perfect number's divisors. Take for example $496$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$496=1+2+4+8+16+31+62+124+248$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;one can see that the first pattern is a sequence of powers of $2$ that stops at $16$, the second pattern starts with a prime number, in this case $31$, the rest of them are multiples of $31$, e.g. $2\cdot31, 4\cdot31$ and $8\cdot31$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I found that the pattern proceeds after $31$, for example $31=2^5-2^0$, $62=2^6-2^1$, $124=2^7-2^2$ and finally $248=2^8-2^3$, so the perfect number can be written as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$496=1+2+4+8+16+(2^5-2^0)+(2^6-2^1)+(2^7-2^2)+(2^8-2^3)$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;or &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$496=(1+2+4+8+16+32+64+128+256) -(1+2+4+8)$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the formula follows very naturally from this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've searched but didn't find this formulation anywhere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, is this something new? Have anyone seen this somewhere?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks. &lt;/p&gt;&#xA;" OwnerUserId="1747" LastActivityDate="2010-09-26T23:41:41.400" Title="Perfect numbers, the pattern continues..." Tags="&lt;number-theory&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="5433" PostTypeId="2" ParentId="5424" CreationDate="2010-09-25T17:03:07.043" Score="3" ViewCount="" Body="&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Euler%27s_criterion&quot; rel=&quot;nofollow&quot;&gt;Euler criterion&lt;/a&gt; is easily generalized to yield the following test for squareness (mod $\rm n\:$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;THEOREM&lt;/strong&gt; $\ $ Let $\rm\ a,\:n\:$ be integers, with $\rm\:a\:$ coprime to $\rm\:n\ =\ 2^e \:p_1^{e_1}\cdots p_k^{e_k}\:,\ \ p_i\:$ primes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\quad \ x^2\ =\ a\ \ (mod\ n)\ $ is solvable for $\rm\:x\:$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\quad \: \iff\ \ \: a^{(p_i\ -\ 1)/2} \ \ \equiv\ \ 1\ \ (mod\ p_i)\quad\quad\ \ $  for all $\rm\ i\le k$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\quad\quad\ $ and $\rm\quad\ \ e&amp;gt;1 \:\Rightarrow\: a\equiv 1\ \ (mod\ 2^{2+\delta}\:),\ \ \ \delta = 1\ \ if\ \ e\ge 3\ \ else\ \ \delta = 0$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof&lt;/strong&gt;: See Ireland and Rosen,  A Classical Introduction to Modern Number Theory, Proposition 5.1.1 p.50.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The above criterion is practical if one knows a full factorization of $\rm\:n\:$, since the exponentiations may be quickly computed by repeated squaring.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;BEWARE&lt;/strong&gt; $\ $ The criterion cannot be expressed equivalently as a simple &lt;a href=&quot;http://en.wikipedia.org/wiki/Jacobi_symbol&quot; rel=&quot;nofollow&quot;&gt;Jacobi symbol&lt;/a&gt; calculation. For example we have $\rm(8|15) = 1\ $ but $8$ is not a square (mod $15$).&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2011-04-02T17:16:41.017" LastActivityDate="2011-04-02T17:16:41.017" />
  <row Id="5434" PostTypeId="2" ParentId="5431" CreationDate="2010-09-25T17:16:38.247" Score="1" ViewCount="" Body="&lt;p&gt;The argument you are getting at as I understand is is, roughly: $x^n \in O(x^n)$ and thus $kx^n \in O(x^n)$, so $f(x)$ acts like $O(x^3) + O(x^2) + O(x) + O(1)$ which can be reduced to $O(x^3)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the theorem we would like to prove now is that for $n\geq m$: $f \in O(x^n)$ and $g \in O(x^m)$ implies $f + g \in O(x^n)$. Once we have this you just add up the monomials of the polynomial and that proves the result.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Look at what we have, from the definitions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f \in O(x^n) \Rightarrow \exists x_0, k,\;\; \forall x&gt;x_0,\;\; f(x) \leq kx^n$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$g \in O(x^m) \Rightarrow \exists x_1, k',\;\; \forall x&gt;x_1,\;\; g(x) \leq kx^m$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $x_2$ be the maximum of $x_0$ and $x_1$, $k''$ be the maximum of $k$ and $k'$ and add these inequalities:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\forall x&gt;x_2,\;\; (f+g)(x) \leq kx^n + k'x^m \leq k''(x^n + x^m) \leq 2 k'' x^n $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the pair of values $(x_2,2k'')$ prove that $f+g \in O(x^n)$.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;To consider abstract functions like this makes the proof very easy, but it is clear that the values we exhibit to prove the existential may not be the &lt;em&gt;best&lt;/em&gt;, although we still prove the theorem in an effective way. In particular, you could do a very detailed analysis of the functions in specific cases to get very tight bounds - in this lucky case it's not needed which is why the theorem is easier to prove in the abstract case.&lt;/p&gt;&#xA;" LastActivityDate="2010-09-25T17:16:38.247" />
  <row Id="5435" PostTypeId="2" ParentId="5430" CreationDate="2010-09-25T17:16:43.820" Score="3" ViewCount="" Body="&lt;p&gt;First, let's shift the functions down so that instead of revolving about $y=11$, we're revolving about the x-axis, $y=0$: $h_1(x)=h(x)-11$, $p_1(x)=p(x)-11$.  Let $a&amp;lt; b$ be the x-coordinates of the points of intersection of $h_1(x)$ and $p_1(x)$ (in the vicinity of -2 and 5)--that is, the least and greatest x-coordinates of any points in $\mathbf{R}$.  There are two lateral surface areas to compute: the &quot;inner&quot; part bounded by $h_1$, and the &quot;outer&quot; part bounded by $p_1$.  For each, the surface area is given by $\int_{a}^{b}2\pi y(x)\sqrt{1+(y'(x))^2}dx$, replacing $y$ with $h_1$ or $p_1$ (and $y'$ with its derivative) as appropriate.&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-09-25T17:16:43.820" CommentCount="1" />
  <row Id="5436" PostTypeId="2" ParentId="5432" CreationDate="2010-09-25T17:29:28.050" Score="1" ViewCount="" Body="&lt;p&gt;You've observed that $P_5 = 2^4(2^5-1) = 496$ can also be written as the sum of the first 9 powers of two minus the sum of the first four powers of two.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Sums of powers of two&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Powers of two written in binary look like $1, 10, 100, 1000, \cdots$ but you can also write them like this $1, 1+1, 11+1, 111+1, \cdots$. This explains why (now in decimal notation) $1 + 2 + 4 + 8 = 15$ and the general sums of this form.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Nine and four&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well obvious $4 = 5-1$, and $9 = 2\cdot 5 - 1$, but where do these come from? Well just multiply out the formula! $2^{n-1}(2^n-1) = 2^{2n-1}-2^{n-1}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This explains why the form $(1 + 2 + 4 + 8 + \cdots) - (1 + 2 + \cdots)$ works.&lt;/p&gt;&#xA;" LastActivityDate="2010-09-25T17:29:28.050" CommentCount="3" />
  <row Id="5437" PostTypeId="2" ParentId="5427" CreationDate="2010-09-25T17:37:15.960" Score="1" ViewCount="" Body="&lt;p&gt;Assuming that $0 = \{\}$ and $x + 1 = \{x,\{x\}\}$ are the definitions you use, the recursive induction step would be to prove:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x \notin x \Rightarrow \{x,\{x\}\} \notin \{x,\{x\}\}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By extensionality what we really need to show is the two cases&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\{x,\{x\}\} \neq \{x\}$&lt;/li&gt;&#xA;&lt;li&gt;$\{x,\{x\}\} \neq x$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The first is immediate since one is a singleton and the other is not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the second part we notice that if it were true, we would have $x \in x$ but that contradicts the induction hypothesis.&lt;/p&gt;&#xA;" LastActivityDate="2010-09-25T17:37:15.960" CommentCount="5" />
  <row Id="5438" PostTypeId="2" ParentId="5432" CreationDate="2010-09-25T17:46:51.900" Score="2" ViewCount="" Body="&lt;p&gt;The patterns you're after tell us nothing about the perfect(ness) of a number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since they hold even if the number is not perfect, for example, take n = 6. $2^5 (2^6-1) = 2016 = 2^{10} + 2^9 + ... + 2^5$. Which is valid for any n, since in binary, 2^6-1 is 6-1=5 1's from left, and multiplication by 2^5 is adding 5 zeros from right.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moreover, the following also hold for any n,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$2^5 (2^6-1) = 2^0 + ... + 2^{6-1} + (2^6-2^0) + (2^{6+1} + 2^1) + ... + (2^{2*6-2} - 2^{6-2})$.&#xA;$= 1 + ... + 32 + (2^6 - 2^0) + ... + (2^10 - 2^4)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but 2016 is not perfect. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here, $2^n-1$ must be prime so that $2^{n-1}(2^n-1)$ is the unique prime factorization of the number, and in which case, the terms in  $\sum_{k=0}^{n-1} 2^k + \sum_{k=0}^{n-2} 2^k(2^n - 1)$ are precisely the divisors of $2^{n-1}(2^n-1)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Theorem: &lt;strong&gt;If P is an even perfect number, then P = $2^{k-1} (2^k - 1)$ for some k&gt;1 with (2^k - 1) prime.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proof: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose P is even perfect. Then we can find k&gt;1 such that $P = 2^{k-1}m$, for m odd. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, $2^{k-1} - 1 = 1 + ... + 2^{k-2}$ (as explained earlier, 2^{k-1} - 1 has k-2 ones). &#xA;i.e explaining why 31 = 1 + 2 + ... + 16.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Write $P = (2^{k-1}-1 + 1)m$. Then $P = (1 + ... + 2^{k-2})m + m$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The case m is prime:&lt;/strong&gt; The proper divisors of P are then $1,2, ..., 2^{k-1}, m, 2m, ..., 2^{k-2}m$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since P is assumed perfect, $P = 1 + 2 + ... + 2^{k-1} + m + 2m + ... + 2^{k-2}m$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But $P = (1 + ... + 2^{k-2})m + m$ as shown above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, $m = 1 + 2 + ... + 2^{k-1} = 2^{k}-1$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;i.e $P = 2^{k-1} (2^k - 1)$ (note 2^{k}-1 prime).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The case for m is not prime:&lt;/strong&gt; Assume without loss of generality $m = p_1 p_2$, neither a unit or even. Then the divisors (hopefully I didn't miss a divisor) of P are:&#xA;$1,2, ..., 2^{k-1}, p_1, 2p_1, ... 2^{k-1}p_1, p_2, 2p_2, ... 2^{k-1}p_2, 2m, ..., 2^{k-2}m$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Taking the sum of the divisors and equating with $P = (1 + ... + 2^{k-2})m + m$. Then &#xA;$m = 1 + 2 + ... + 2^{k-1} + (1 + ... + 2^{k-1})p_1 + (1 + ... + 2^{k-1})p_2 = (2^{k} - 1) (1 + p_1 + p_2)$, but then $p_1 = (1 + p_1 + p_2)$, i.e p_2=-1 or $p_2 = (1 + p_1 + p_2)$, in either case a contradiction.&lt;/p&gt;&#xA;" OwnerUserId="1746" LastEditorUserId="1746" LastEditorDisplayName="" LastEditDate="2010-09-25T23:58:50.547" LastActivityDate="2010-09-25T23:58:50.547" CommentCount="3" />
  <row Id="5439" PostTypeId="1" CreationDate="2010-09-25T18:07:44.677" Score="1" ViewCount="110" Body="&lt;p&gt;I'm trying to learn about the Selberg zeta function, but it seems like introductory texts assume more knowledge of Riemannian geometry than I'm comfortable with.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have some basic questions that someone might be able to help with:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is the composition of two closed geodesics itself a closed geodesic?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is composition of geodesics a commutative operation?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can all geodesics be decomposed into compositions of primitive closed geodesics?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If anyone has comments or references, I would appreciate them.&lt;/p&gt;&#xA;" OwnerUserId="1820" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-10-01T16:28:10.980" LastActivityDate="2010-10-01T16:28:10.980" Title="Some basic questions about the Selberg zeta function" Tags="&lt;geometry&gt;&lt;zeta-function&gt;" CommentCount="4" />
  <row Id="5440" PostTypeId="2" ParentId="5431" CreationDate="2010-09-25T18:18:25.783" Score="1" ViewCount="" Body="&lt;p&gt;Suppose you have a function $f(x)$ and $g(x)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A really simple method (that works when $f(x)$ and $g(x)$ are polynomials) of determining a constant that works is the following. Consider&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\lim_{x\rightarrow\infty}\frac{f(x)}{g(x)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the limit exists and if a constant $0\leq C&amp;lt; \infty$. Then $C+\epsilon$ for any $\epsilon&gt;0$ is a constant that works. To see this just apply the definition of the limit. $\forall \epsilon&gt;0$ there exists a $x_0(\epsilon)$ such that $\forall x&gt;x_0(\epsilon)$ we have &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\left|\frac{f(x)}{g(x)}-C\right|&amp;lt;\epsilon$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{f(x)}{g(x)} &amp;lt; C+\epsilon$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now you know the constant from calculating the limit and know the existence of $x_0$. Since you are always considering asymptotics when using this definition you are never concerned with the value of $x_0$ (only that it exists)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It does not matter what constants you use and someone could easily use different constants to get $f(x)=O(g(x))$. This notation is to compare the growth rate of two functions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $\lim_{x\rightarrow \infty} \frac{f(x)}{g(x)}$ does not exist or is hard to calculate then as long as you can bound it above you still have $f(x)=O(g(x))$&lt;/p&gt;&#xA;" OwnerUserId="1794" LastEditorUserId="1794" LastEditorDisplayName="" LastEditDate="2010-09-25T18:35:41.037" LastActivityDate="2010-09-25T18:35:41.037" />
  <row Id="5441" PostTypeId="1" CreationDate="2010-09-25T18:37:31.767" Score="2" ViewCount="231" Body="&lt;p&gt;Can you advise some good books covering areas which are required for serious probability theory studies (e.g. measure theory, functional analysis)? Preferably this book should have some problem sets to work on. Thanks!&lt;/p&gt;&#xA;" OwnerUserId="1463" LastActivityDate="2010-09-30T15:22:57.440" CommunityOwnedDate="2010-09-26T11:14:44.290" Title="Supplementary reading for probability theory studies" Tags="&lt;soft-question&gt;&lt;probability&gt;&lt;books&gt;&lt;self-learning&gt;" AnswerCount="7" CommentCount="3" />
  <row Id="5442" PostTypeId="2" ParentId="5431" CreationDate="2010-09-25T18:43:02.207" Score="2" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\quad\rm  ax^3 + bx^2 + cx + d\  \le \ (|a|+|b|+|c|+|d|)\ x^3 \ $  for $\rm\ x &gt; 1$&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-09-25T18:43:02.207" />
  <row Id="5443" PostTypeId="2" ParentId="5441" CreationDate="2010-09-25T18:55:21.170" Score="2" ViewCount="" Body="&lt;p&gt;Take a look at &quot;Measure Theory and Probability&quot; by A.K. Basu. This book has lots of problem sets in them and contains some serious probabilty theory. It developed measure theory with probability as the main application. Also through-out the book there are examples which were really helpful.&lt;/p&gt;&#xA;" OwnerUserId="1794" LastActivityDate="2010-09-25T18:55:21.170" CommunityOwnedDate="2010-09-28T07:16:29.983" CommentCount="1" />
  <row Id="5444" PostTypeId="2" ParentId="5441" CreationDate="2010-09-25T18:56:04.320" Score="3" ViewCount="" Body="&lt;p&gt;For a broad overview in one book at a graduate level, it is hard to beat &lt;em&gt;Foundations of Modern Probability (Second edition)&lt;/em&gt; by Olav Kallenberg. His presentation is classroom tested, and many standard complicated proofs have been made very concise. I'm a big fan of this book!    &lt;/p&gt;&#xA;" OwnerUserId="940" LastEditorUserId="940" LastEditorDisplayName="" LastEditDate="2010-09-30T15:22:57.440" LastActivityDate="2010-09-30T15:22:57.440" CommunityOwnedDate="2010-09-25T18:56:04.320" />
  <row Id="5445" PostTypeId="2" ParentId="201" CreationDate="2010-09-25T19:05:21.913" Score="4" ViewCount="" Body="&lt;p&gt;$$\sum_p \frac{1}{p}=\infty$$&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Proof_that_the_sum_of_the_reciprocals_of_the_primes_diverges&quot; rel=&quot;nofollow&quot;&gt;Here is a link&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; If there was finitely many primes, then the sum would be finite.&lt;/p&gt;&#xA;" OwnerUserId="1154" LastEditorUserId="1154" LastEditorDisplayName="" LastEditDate="2010-10-15T19:40:24.540" LastActivityDate="2010-10-15T19:40:24.540" />
  <row Id="5446" PostTypeId="2" ParentId="5303" CreationDate="2010-09-25T20:02:58.737" Score="12" ViewCount="" Body="&lt;p&gt;I suppose there is a way to formalize your question into a relatively precise one.   Let $P$ be the set of all propositions in the language of Zermelo-Frankel set theory for which there exist proofs. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given a natural number $k$ we want to have an upper bound $l(k)$ on the minimal-length of the proof of any proposition $p$ from $P$ such that the length of the proposition $p$ is at most $k$. Here I'm using &quot;length&quot; in the sense of the number of ASCII characters it would take to write the proposition (or proof of the proposition, respectively). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Presumably $l(k)$ is a non-computable function. Since if $l(k)$ were computable, given a proposition $p$ of length $k$ you'd have a &quot;simple&quot; procedure to find a proof of any statement in Zermelo-Frankel set theory, provided one exists.  The idea would be to iterate through all propositions of length at most $l(k)$ and verify they're proofs of $p$.  :) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I imagine such a statement is known to logicians.  If it is known, I would call it the &quot;rationalization is hard theorem&quot;.&lt;/p&gt;&#xA;" OwnerUserId="642" LastEditorUserId="642" LastEditorDisplayName="" LastEditDate="2010-09-25T20:08:08.670" LastActivityDate="2010-09-25T20:08:08.670" CommentCount="5" />
  <row Id="5447" PostTypeId="1" AcceptedAnswerId="5448" CreationDate="2010-09-25T20:46:04.593" Score="1" ViewCount="226" Body="&lt;p&gt;A problem that is an example of this is $x^2+16$. I have to know how to factor this over the set of complex numbers. How do you do that? I used to know it's just been a long time.&lt;/p&gt;&#xA;" OwnerUserId="2135" LastEditorUserId="367" LastEditorDisplayName="" LastEditDate="2010-09-26T16:52:18.107" LastActivityDate="2010-09-27T10:00:27.323" Title="How do you factor an equation over the set of complex numbers?" Tags="&lt;algebra-precalculus&gt;" AnswerCount="3" />
  <row Id="5448" PostTypeId="2" ParentId="5447" CreationDate="2010-09-25T20:50:07.613" Score="3" ViewCount="" Body="&lt;p&gt;Notice that $16 = -(4i)^2$, so $x^2 + 16$ is a difference of two squares and can be written as $(x - 4i)(x + 4i)$.&lt;/p&gt;&#xA;" LastActivityDate="2010-09-25T20:50:07.613" CommentCount="2" />
  <row Id="5449" PostTypeId="2" ParentId="5427" CreationDate="2010-09-25T20:57:35.340" Score="1" ViewCount="" Body="&lt;p&gt;As the induction hypothesis is clearly not strong enough, how about strengthening your proof obligation? Just adding $x+1\not\in x$ will mean you'll also need $x+2\not\in x$, so I'd suggest $y\not\in x$ for all $y$ such that $x\le y$.&lt;/p&gt;&#xA;" OwnerUserId="2061" LastActivityDate="2010-09-25T20:57:35.340" CommentCount="2" />
  <row Id="5450" PostTypeId="1" AcceptedAnswerId="5453" CreationDate="2010-09-25T21:00:50.127" Score="2" ViewCount="300" Body="&lt;p&gt;I'm stuck trying to solve an inequality of the form $\left|\frac{f(x)}{g(x)}\right| \geq 1$; specifically, $\left|\frac{2x + 5}{x + 1}\right| \geq 1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried the approach that was taught for solving inequalities in general, and solved first for x when the part enclosed in absolute value signs is  $\geq 1$, and then solved for when it is $\leq -1$, but plugging in numbers for the resulting domain $[-4, -2]$ back into the original inequality resulted in some answers that were not $\geq 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where have I gone wrong in trying to solve this inequality? How is this different from solving an inequality like $\left|\frac{3}{x + 1}\right| \geq 1$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;: Okay, since I'm clearly messing this up horribly(and now realize that I was absent-mindedly taking the intersection and not the union of my results in spite of having not mixed up the two many times before), I'll edit with the specifics of the solution I was trying and which had me confused before I asked the question. So:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\left|\frac{2x + 5}{x + 1}\right| \geq 1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Meaning 1: $\frac{2x + 5}{x + 1} \geq 1$ or 2: $\frac{2x + 5}{x + 1} \leq -1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Tackling the first: multiplied 1 by $x + 1$, forgetting to account for the fact that $x + 1$ could potentially be negative: $2x + 5 \geq x + 1$, then rearrange for $x \geq -4$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the other inequality: multiplied -1 by $x + 1$, again forgetting to consider $x + 1$ being negative: $2x + 5 \leq -x - 1$ , rearrange and divide by 3 for $x \leq -2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Domain is the union, so $[-4, \infty] \cup [-\infty, -2]$, so $R - { 1 }$. This is obviously not right, as plugging in -3 shows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, in inequalities where the denominator is some function g(x), if I split it and choose to multiply, will I have to solve each part of the inequality twice(once assuming g(x) is positive, a second time assuming it's negative) and then check which solution makes sense by plugging in numbers?&lt;/p&gt;&#xA;" OwnerUserId="1956" LastEditorUserId="1956" LastEditorDisplayName="" LastEditDate="2010-09-25T23:05:00.250" LastActivityDate="2010-09-26T00:34:45.693" Title="How do I solve inequalities of the form $\left|\frac{f(x)}{g(x)}\right| \geq 1$?" Tags="&lt;algebra-precalculus&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="5451" PostTypeId="2" ParentId="5450" CreationDate="2010-09-25T21:07:40.203" Score="4" ViewCount="" Body="&lt;p&gt;The solution set of the inequality $$|f(x)/g(x)|\geq1$$ is equal to the union of the solutions sets of the inequalities $$f(x)/g(x)\leq-1$$ and $$f(x)/g(x)\geq1,$$ so solve these two and compute the union.&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2010-09-25T21:07:40.203" CommentCount="11" />
  <row Id="5452" PostTypeId="2" ParentId="5447" CreationDate="2010-09-25T21:25:47.457" Score="2" ViewCount="" Body="&lt;p&gt;In general, it is true that every polynomial splits (i.e., factors into linear factors) over the complex number field. However, finding this factorization for polynomials of degree up to four is manageable, and factoring something of degree higher than 4 is impossible (in the sense that there does not exist, for a general polynomial of degree 5 or higher, an algorithm for finding its roots). Well, actually, this isn't entirely true. I should say that there is no formula for the roots of a general quintic expressed by radicals. There are other methods: for instance, the so called ultraradicals, approximations, etc.&lt;/p&gt;&#xA;" OwnerUserId="2093" LastActivityDate="2010-09-25T21:25:47.457" />
  <row Id="5453" PostTypeId="2" ParentId="5450" CreationDate="2010-09-25T21:36:27.947" Score="4" ViewCount="" Body="&lt;p&gt;Your answer of $[-4,-2]$ appears to be the solution set to $\left|\frac{2x + 5}{x + 1}\right| \leq 1$.  Without seeing all of your work, I can't be sure what happened.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, for inequalities of this type (actually, pretty much any inequalities involving a quotient with variables in the denominator), the method I'd suggest is the following:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Split it into two inequalities to solve separately: $\frac{f(x)}{g(x)} \geq 1$ or $-1\geq \frac{f(x)}{g(x)}$.  Once these are solved, take the union of their solution sets (only one of these inequalities must be true for your original inequality to hold).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;For each of these two inequalities, manipulate them so that the comparison is to 0 instead of 1 or -1.  For example, $\frac{f(x)}{g(x)} \geq 1$ can be rewritten as $\frac{f(x)}{g(x)}-1\geq 0$ or $\frac{f(x)-g(x)}{g(x)} \geq 0$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;When solving an inequality that compares an expression of the form $\frac{a(x)}{b(x)}$ to zero, the solution set will be the union of intervals whose endpoints are at the zeros of $a(x)$ or $b(x)$.  That is:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;determine the zeros of $a(x)$ and $b(x)$&lt;/li&gt;&#xA;&lt;li&gt;plot the zeros on a number line&#xA;&lt;ul&gt;&#xA;&lt;li&gt;use open circles (excluded values) for the zeros of $b(x)$&lt;/li&gt;&#xA;&lt;li&gt;if the inequality is $\leq$ of $\geq$, use closed circles (included values) for the zeros of $a(x)$&lt;/li&gt;&#xA;&lt;li&gt;if the inequality is $&amp;lt;$ or $&amp;gt;$, use open circles for the zeros of $a(x)$&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;for each region between two zeros, test a value in the inequality you are solving&lt;/li&gt;&#xA;&lt;li&gt;if that value makes the inequality true, then the region is part of the solution set&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This is sometimes called the &quot;boundary algorithm&quot; or the &quot;test-point method&quot; for solving inequalities.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&lt;strong&gt;edit&lt;/em&gt;&lt;/strong&gt;:  Regarding &quot;if I split it and choose to multiply, will I have to solve each part of the inequality twice,&quot; well...  sort of.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Tackling the first: multiplied 1 by $x + 1$, forgetting to account for the fact that $x + 1$ could potentially be negative: $2x + 5 \geq x + 1$, then rearrange for $x \geq -4$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here, you should have $(2x+5\geq x+1\text{ and }x\geq -1)\text{ or }(2x+5\leq x+1\text{ and }x&amp;lt;-1)$.  In solving, remember that the &quot;and&quot; becomes an intersection of solution sets and the &quot;or&quot; becomes a union of solution sets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As I'd indicated in the comments, I use the method I originally described in order to avoid this sort of casework.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;edit 2&lt;/em&gt;:  As an example of the method that I describe above, let's solve $\frac{2x+5}{x+1}\geq 1$.  Subtract 1 from both sides, get a common denominator on the left side, and combine the terms to get $\frac{x+4}{x+1}\geq 0$.  The only zero of the numerator is -4; the only zero of the denominator is -1.  Graphing these on a number line gives:&#xA;$$\leftarrow\!\!\underset{-4}{-\!\!\bullet\!\!-}\!\!-\!\!\underset{-1}{-\!\!\circ\!\!-}\!\!\rightarrow$$&#xA;(open circle at -1 because $x=-1$ makes the quotient undefined, closed circle at -4 because $x=-4$ makes the numerator 0, so the quotient is 0, and the inequality is $\geq 0$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, there are 3 intervals to test: $(-\infty,-4]$, $[-4,-1)$, and $(-1,\infty)$.  Relatively simple-to-test values in the intervals are -5, -2, and 0, respectively. $\frac{-5+4}{-5+1}=\frac{-}{-}=+\geq 0$; $\frac{-2+4}{-2+1}=\frac{+}{-}=-\not\geq 0$; $\frac{0+4}{0+1}=\frac{+}{+}=+\geq 0$.  So, the solution is $(-\infty,-4]\cup(-1,\infty)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(To complete the original problem, you still need to solve $\frac{2x+5}{x+1}\leq -1$ and take the union of that solution with the set we just found.)&lt;/p&gt;&#xA;" OwnerUserId="72" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-09-26T00:34:45.693" LastActivityDate="2010-09-26T00:34:45.693" />
  <row Id="5454" PostTypeId="1" AcceptedAnswerId="5456" CreationDate="2010-09-25T21:46:33.617" Score="1" ViewCount="136" Body="&lt;p&gt;How do we go from $$\frac{1}{1+(x/2)+(x^2/3)+(x^3/4)+\dots}$$ to $$1-(x/2)-(x^2/12)-\dots$$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I could consolidate the series in denominator in some form, then I could use binomial expansion in numerator by raising it to power -1, but I don't see the one in denominator in any known form other than converting it to log.&lt;/p&gt;&#xA;" OwnerUserId="754" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-09-25T22:48:19.407" LastActivityDate="2010-09-25T22:48:19.407" Title="How is this series in denominator converted to a series in numerator?" Tags="&lt;sequences-and-series&gt;" AnswerCount="2" CommentCount="10" />
  <row Id="5455" PostTypeId="2" ParentId="5450" CreationDate="2010-09-25T21:56:12.230" Score="3" ViewCount="" Body="&lt;p&gt;You can also rewrite in equivalent form as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f^2(x) \ge g^2(x) \wedge g(x) \neq 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i.e.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ (f(x)+g(x))(f(x)-g(x)) \ge 0 \wedge g(x) \neq 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your particular case, we get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ (3x+6)(x+4) \ge 0 \wedge x+1 \ne 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which is equivalent to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ (x \ge -2 \vee x \le -4) \wedge x \ne -1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The $\wedge$ stand for AND and $\vee$ stands for OR, so the solution set is $x$ is not -1 AND either $x \ge -2$ OR $x \le -4$, i.e $$\mathbb{R} - ((-4,-2) \cup \{1\})$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since you seem to have the answer reversed, my guess it that you multiplied one of you inequalities by a negative number at some stage and forgot to reverse the inequality.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance $$ x &amp;lt; 1$$&#xA;Mulitplying by $-1$ gives&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ -x &gt; -1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if you forget to reverse the inequality it gives $$ -x &amp;lt; -1$$ which is incorrect.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-09-25T21:56:12.230" CommentCount="2" />
  <row Id="5456" PostTypeId="2" ParentId="5454" CreationDate="2010-09-25T22:14:02.963" Score="2" ViewCount="" Body="&lt;p&gt;Using $$-\log(1-x) = x + \frac{x^2}{2} + \frac{x^3}{3} + \dots$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ - \frac{x}{\log(1-x)} = \frac{1}{1 + \frac{x}{2} + \frac{x^2}{3} + \dots}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the series for $$\frac{x}{\log(1-x)}$$ is well known.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See the question asked on this very site here: &lt;a href=&quot;http://math.stackexchange.com/questions/5058/formula-for-the-harmonic-series-h-n-sum-k1n-1-k-due-to-gregorio-fontana&quot;&gt;http://math.stackexchange.com/questions/5058/formula-for-the-harmonic-series-h-n-sum-k1n-1-k-due-to-gregorio-fontana&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And the page here: &lt;a href=&quot;http://en.wikipedia.org/wiki/Euler-Mascheroni_constant&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Euler-Mascheroni_constant&lt;/a&gt;. (search the page for Gregory).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The series expansion is given by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{x}{\log(1-x)} = \sum_{k=0}^{\infty} C_{k} x^{k} = -1 + \frac{x}{2} + \frac{x^2}{12} + \frac{x^3}{24} + \dots$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The $C_{k}$ are called as Gregory coefficients. The wiki page I linked above tells you how they can be calculated using a recursive formula.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So to answer your question, we get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{1}{1 + \frac{x}{2} + \frac{x^2}{3} + \dots} = - \sum_{k=0}^{\infty} C_{k} x^k = 1 - \frac{x}{2} - \frac{x^2}{12} - \frac{x^3}{24} - \dots$$&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-09-25T22:14:02.963" CommentCount="2" />
  <row Id="5457" PostTypeId="2" ParentId="5454" CreationDate="2010-09-25T22:20:25.113" Score="1" ViewCount="" Body="&lt;p&gt;Let $f(x) = \sum f_n x^n$ be a formal power series.  To compute the inverse of $1 - x f(x)$, write&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{1}{1 - x f(x)} = \sum_{n \ge 0} x^n f^n(x).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To compute the coefficient of $x^k$ it suffices to compute the contributions from the first $k$ terms on the RHS, so this results in a finite algorithm which will let you compute any particular coefficient of the inverse.  If $f$ has special properties (for example it is a polynomial or more generally meromorphic) then often one can even find a nice closed form.  &lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-09-25T22:20:25.113" />
  <row Id="5458" PostTypeId="2" ParentId="5416" CreationDate="2010-09-25T23:00:31.773" Score="1" ViewCount="" Body="&lt;p&gt;This goes along with the question of how to interpret&#xA;$$&#xA;\frac{\partial}{\partial z} \;=\; \frac{1}{2}\left(\frac{\partial}{\partial x} - i \frac{\partial}{\partial y}\right)&#xA;$$&#xA;for non-holomorphic functions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One possible answer is that $\partial f / \partial z$ is the average value of &#xA;$$&#xA;\frac{f(w) - f(z)}{w - z}&#xA;$$&#xA;as $w$ takes values in a small circle centered at $z$.  To be precise:&#xA;$$&#xA;\frac{\partial f}{\partial z}(z) \;=\; \lim_{r\to 0}\frac{1}{2\pi}\int_0^{2\pi} \frac{f(z+re^{it}) - f(z)}{re^{it}} dt.&#xA;$$&#xA;Similarly, $\partial f / \partial \overline{z}$ is the average value of&#xA;$$&#xA;\frac{f(w) - f(z)}{\overline{w} - \overline{z}}&#xA;$$&#xA;as $w$ takes values in a small circle centered at $z$.&lt;/p&gt;&#xA;" OwnerUserId="1726" LastActivityDate="2010-09-25T23:00:31.773" />
  <row Id="5459" PostTypeId="2" ParentId="201" CreationDate="2010-09-25T23:18:23.443" Score="2" ViewCount="" Body="&lt;p&gt;Another proof is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the numbers $$9^{2^n} + 1, \ \  n = 1,2,\dots$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now if $$9^{2^n} + 1 = 0 \mod p$$ then we have that, for $ m &gt; n$ that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$9^{2^m} + 1 = (9^{2^n})^{2^{m-n}} + 1 = (-1)^{2^{m-n}} + 1 = 1+1 = 2 \mod p$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus if one term of the sequence is divisible by a prime, none of the next terms are divisible by that prime, i.e. if you write out the factors of the terms of the sequence, each term of this sequence gives rise to a prime not seen before!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a curiosity, it can be shown that each number in the sequence has at least one prime factor &gt; 40. See this question on this very site: &lt;a href=&quot;http://math.stackexchange.com/questions/3243/does-92n-1-always-have-a-prime-factor-larger-than-40/&quot;&gt;http://math.stackexchange.com/questions/3243/does-92n-1-always-have-a-prime-factor-larger-than-40/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-09-25T23:18:23.443" CommentCount="2" />
  <row Id="5460" PostTypeId="2" ParentId="5392" CreationDate="2010-09-25T23:22:57.000" Score="2" ViewCount="" Body="&lt;p&gt;The localization and the dispersion probabilistic measures can be &quot;seen&quot; as the corresponding momentums of mechanical systems of &quot;probabilistic masses&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The expectation has the following mechanical interpretation. Given that $F(x)$ is the &quot;probalilistic mass&quot; contained in the interval $0\le X\lt x$ (in one dimention), the mathematical expectation of the random variable $X$ is the static momentum with respect to the origin of the &quot;probabilistic masses&quot; system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The variance is the mechanical analog of the inertia momentum of the &quot;probabilistic masses&quot; system with respect to the center of masses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The variance of the random variable $X$ is the 2nd order momentum of $X-m$ where $m$ is the expectation of $X$, i.e. $m=E(X)=\displaystyle\int x dF(x)$ ($F(x)$ is the repartion function).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The $k$-order momentum is the expectation of $(X-m)^k$.&lt;/p&gt;&#xA;" OwnerUserId="752" LastActivityDate="2010-09-25T23:22:57.000" />
  <row Id="5461" PostTypeId="2" ParentId="213" CreationDate="2010-09-26T00:02:24.270" Score="1" ViewCount="" Body="&lt;p&gt;I wish I'd learned about special functions earlier.  The subject is a treasure trove of results that were commonly known a century ago but now few people know.&lt;/p&gt;&#xA;" OwnerUserId="247" LastActivityDate="2010-09-26T00:02:24.270" CommunityOwnedDate="2010-09-26T00:02:24.270" />
  <row Id="5462" PostTypeId="2" ParentId="5441" CreationDate="2010-09-26T01:05:01.477" Score="2" ViewCount="" Body="&lt;p&gt;Pollard, &lt;em&gt;A User's Guide to Measure Theoretic Probability&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="1195" LastActivityDate="2010-09-26T01:05:01.477" CommunityOwnedDate="2010-09-28T07:16:29.983" />
  <row Id="5463" PostTypeId="2" ParentId="5441" CreationDate="2010-09-26T01:06:01.020" Score="2" ViewCount="" Body="&lt;p&gt;Dudley, &lt;em&gt;Real Analysis and Probability&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="1195" LastActivityDate="2010-09-26T01:06:01.020" CommunityOwnedDate="2010-09-28T07:16:29.983" />
  <row Id="5464" PostTypeId="2" ParentId="5377" CreationDate="2010-09-26T02:18:55.970" Score="1" ViewCount="" Body="&lt;p&gt;Here's a simplified form of your original argument:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Construct a Turing machine A which sequentially runs on all proofs in ZFC and checks if the claim &quot;A does not halt&quot; is proved. If such a proof is found, A halts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that A can know its encoding - see Kleene's recursion theorem - and even if it can't this &quot;obstacle&quot; can be overcome. So it seems to me such A is constructible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if A halts then it is because it has found a proof for &quot;A does not halt&quot;. This is impossible, since A halted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we have just proved the claim &quot;A does not halt&quot;, so such a proof exists in ZFC. So A must halt on this proof - a contradiction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Hopefully this version makes the error more clear.  The proof that &quot;A does not halt&quot; assumes the consistency of ZFC, and is therefore a proof in ZFC + Con(ZFC), not ZFC.  In fact, the above argument is essentially a proof that Con(ZFC) is not provable in ZFC.)&lt;/p&gt;&#xA;" OwnerUserId="1726" LastActivityDate="2010-09-26T02:18:55.970" CommentCount="1" />
  <row Id="5465" PostTypeId="1" CreationDate="2010-09-26T03:09:51.650" Score="6" ViewCount="271" Body="&lt;p&gt;I'm talking about the shape made up of a rectangle with a semi-circle at each end.  Does it have a particular name?&lt;/p&gt;&#xA;" OwnerUserId="2139" LastActivityDate="2010-09-26T12:07:00.697" Title="What is the proper geometric description of a the oval used for a horse racetrack?" Tags="&lt;geometry&gt;&lt;definition&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="1" />
  <row Id="5466" PostTypeId="1" AcceptedAnswerId="5475" CreationDate="2010-09-26T03:15:00.497" Score="0" ViewCount="247" Body="&lt;p&gt;I was told to write the 3 cube roots of 27 in rectangular form how is this to be done? I know how to do cube roots on something like 5+0i in polar form, but how do you do it with just a number?&lt;/p&gt;&#xA;" OwnerUserId="2135" LastActivityDate="2010-09-26T09:25:09.030" Title="How do you find the cube roots of just a number?" Tags="&lt;algebra-precalculus&gt;" AnswerCount="3" FavoriteCount="1" />
  <row Id="5468" PostTypeId="2" ParentId="5465" CreationDate="2010-09-26T03:34:07.780" Score="3" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://en.wiktionary.org/wiki/obround&quot; rel=&quot;nofollow&quot;&gt;Obround&lt;/a&gt;, apparently.  I don't know Wiktionary's source.  This definition of obround does not appear in OED, for example.  Googling indicates that this definition is commonly used for machine parts having this shape.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2010-09-26T04:27:43.150" LastActivityDate="2010-09-26T04:27:43.150" CommentCount="1" />
  <row Id="5469" PostTypeId="2" ParentId="3724" CreationDate="2010-09-26T03:34:23.287" Score="1" ViewCount="" Body="&lt;p&gt;The answer from Soarer has a hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The following is not correct:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1.2.2 If the C side is equal to the other side, then C side minus C are all genuine&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is possible that group A has 2 fake coins. In that case we could have 1 fake at the C size (within the 24 coins) and another fake coin on the other size (within 25 coins). C is no fake when A has two fakes. This will make C side equal to the other side also and we could not claim that C side minus C are all genuine.&lt;/p&gt;&#xA;" OwnerUserId="2140" LastActivityDate="2010-09-26T03:34:23.287" />
  <row Id="5470" PostTypeId="2" ParentId="5465" CreationDate="2010-09-26T03:42:31.557" Score="5" ViewCount="" Body="&lt;p&gt;I'm pretty sure they actually use four &lt;a href=&quot;http://mathworld.wolfram.com/CornuSpiral.html&quot; rel=&quot;nofollow&quot;&gt;clothoid&lt;/a&gt; arcs joined together in practice, e.g. &lt;a href=&quot;http://dx.doi.org/10.1111/j.2042-3306.1975.tb03240.x&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt;. This has a lot to do with the fact that the clothoid is the curve whose curvature is directly proportional to its arclength; an abrupt variation in curvature would equate to an abrupt variation in centripetal force, which can be bad for the racehorses (or even racecars, for that matter).&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Here's a simulated clothoid track drawn in &lt;em&gt;Mathematica&lt;/em&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/XfWQ8.png&quot; alt=&quot;clothoid track&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just to show that the bends are honest-to-goodness clothoids, I drew the clothoid corresponding to the lower right portion of the track in full (the dashed gray one).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The parametrization used is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(x\qquad y)=\left(\sqrt{\frac{\pi}{2}}C\left(\sqrt{\frac{2}{\pi}}s\right)\qquad \sqrt{\frac{\pi}{2}}S\left(\sqrt{\frac{2}{\pi}}s\right)\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $C(x)$ and $S(x)$ are the &lt;a href=&quot;http://dlmf.nist.gov/7.2.iii&quot; rel=&quot;nofollow&quot;&gt;Fresnel integrals&lt;/a&gt;; I leave you to verify using those expressions that the curvature of the clothoid is indeed directly proportional to the arclength.&lt;/p&gt;&#xA;" OwnerUserId="498" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-09-26T05:31:53.197" LastActivityDate="2010-09-26T05:31:53.197" CommentCount="7" />
  <row Id="5471" PostTypeId="2" ParentId="5466" CreationDate="2010-09-26T03:44:11.583" Score="2" ViewCount="" Body="&lt;p&gt;$z^\frac{1}{n} = +\sqrt[n]{\vert z\vert} e^{ i \frac{\arg (z) + 2k\pi }{n} } $, where $\arg(z)$ is the principal argument of the complex number $z$, and $k = {0, 1, ... n-1}$. In this case, since the complex number $z$ is purely real, $\arg(z) = 0$. Therefore we are left with:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$27^\frac{1}{3} = \sqrt[3]{27} e^{i\frac{2k\pi}{3} }$, which, iterating over $k$, gives us the three roots $3, 3e^{i\frac{2\pi}{3}}$, and $3e^{i\frac{4\pi}{3}} $. Now, to express them in rectangular form, simply use Euler's formula; $e^{ix} =  \cos(x) + i \sin(x)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your final answers should then be $3, \frac{3}{2}(-1+i\sqrt{3})$, and $\frac{3}{2}(-1-i\sqrt{3})$.&lt;/p&gt;&#xA;" LastEditorUserId="664" LastEditorDisplayName="" LastEditDate="2010-09-26T09:25:09.030" LastActivityDate="2010-09-26T09:25:09.030" CommentCount="2" />
  <row Id="5472" PostTypeId="2" ParentId="5466" CreationDate="2010-09-26T03:44:40.397" Score="1" ViewCount="" Body="&lt;p&gt;Yes, a number has three cube roots. Think about how you would factor the polynomial $x^3-27$.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-09-26T03:44:40.397" CommentCount="1" />
  <row Id="5473" PostTypeId="1" AcceptedAnswerId="5474" CreationDate="2010-09-26T03:46:19.667" Score="1" ViewCount="143" Body="&lt;p&gt;First of all, I had no idea what to title this question, so feel free to change it to something more appropriate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a set defined as such:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\{a\frown b\mid a \in A \wedge b \in \{c\frown d\mid c \in C \wedge d \in D\}\}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is that logically equivalent to:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\{a\frown (c\frown d)\mid a \in A \wedge c \in C \wedge d \in D\}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming that $\frown$ is an associative relation.&lt;/p&gt;&#xA;" OwnerUserId="2141" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2010-11-24T20:35:16.637" LastActivityDate="2010-11-24T20:35:16.637" Title="Expansion of terms in set definition?" Tags="&lt;elementary-set-theory&gt;" AnswerCount="1" />
  <row Id="5474" PostTypeId="2" ParentId="5473" CreationDate="2010-09-26T03:55:27.147" Score="1" ViewCount="" Body="&lt;p&gt;The sets are equal, regardless of whether $\frown$ is associative.  For instance, a straightforward &quot;element chase&quot; shows this. More directly, note that $x=a\frown b$ for some $b$ of the form $c\frown d$ means precisely the same thing as $x=a\frown(c\frown d)$ for some $c$ and $d$.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-09-26T03:55:27.147" CommentCount="2" />
  <row Id="5475" PostTypeId="2" ParentId="5466" CreationDate="2010-09-26T03:59:49.687" Score="2" ViewCount="" Body="&lt;p&gt;If you know how to find the cube roots of $5+0i$ in polar form, you're most of the way there already.  $27=27+0i$, so you can use the same techniques on it as on $5+0i$ to get the polar form of the cube roots.  To get to rectangular form from polar form, evaluate the polar form&amp;mdash;that is, $$re^{i\theta}=r(\cos\theta+i\sin\theta)=r\cos\theta+(r\sin\theta)i.$$&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-09-26T03:59:49.687" />
  <row Id="5476" PostTypeId="2" ParentId="5416" CreationDate="2010-09-26T04:11:06.040" Score="4" ViewCount="" Body="&lt;p&gt;The dee zee bar operator in the secret blogging seminar: &lt;a href=&quot;http://sbseminar.wordpress.com/2009/01/08/what%C2%B4s-up-with-dee-zee-bar/&quot; rel=&quot;nofollow&quot;&gt;What´s up with dee zee bar&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="776" LastActivityDate="2010-09-26T04:11:06.040" />
  <row Id="5477" PostTypeId="1" CreationDate="2010-09-26T04:49:16.450" Score="3" ViewCount="97" Body="&lt;p&gt;I have attempted to google for this, but the searching is marred by the close relationship of Jordan Algebras and Quantum Mechanics. I have been passively thinking of this question for ages, tonight I finally post it. :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strike&gt;My question is fairly simple. Similar to the universal enveloping algebra for Lie algebras, we should be able to form a universal enveloping algebra for Jordan algebras. So question one:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Can we indeed construct this &quot;Universal enveloping algebra&quot;?&lt;/strike&gt;(Thanks Matt)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We can form a Universal enveloping algebra for Jordan algebras. So;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Are there some sort of Serre relations for Jordan algebras?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and then as we do for Lie algebras,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Can we find &quot;quantum Jordan 'Serre' relations&quot; for quantum Jordan algebras?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;or just;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Can we deform the Jordan Universal Enveloping algebra in a similar way that we deform Lie Enveloping algebras to their quantized versions? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I would be happy with an explanation or a reference. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance!&lt;/p&gt;&#xA;" OwnerUserId="16" LastEditorUserId="16" LastEditorDisplayName="" LastEditDate="2010-10-01T06:04:02.560" LastActivityDate="2010-10-01T06:04:02.560" Title="Quantum Jordan Algebras(Still wondering!)" Tags="&lt;abstract-algebra&gt;&lt;reference-request&gt;" AnswerCount="1" />
  <row Id="5478" PostTypeId="2" ParentId="5477" CreationDate="2010-09-26T05:31:48.860" Score="3" ViewCount="" Body="&lt;p&gt;Yes, there is a notion of universal enveloping algebra for Jordan algbera; it is left adjoint&#xA;to the functor that associates a Jordan algebra to an associative algebra.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, there is no analogue of the Poincare--Birkhoff--Witt theorem for Jordan algebras,&#xA;and so the map from the Jordan algebra to its universal enveloping algebra need not be injective.  (I forget where I learned this, but it should be discussed in most texts that discuss Jordan algebras; indeed, there is an adjective --- which I also forget --- used to separate those Jordan algebras that do embed into their enveloping algebra from those that&#xA;don't.)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for deformation quantization, I don't know.&lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2010-09-26T05:31:48.860" />
  <row Id="5480" PostTypeId="1" AcceptedAnswerId="5492" CreationDate="2010-09-26T05:40:06.160" Score="4" ViewCount="404" Body="&lt;p&gt;Or, in other words, why are all functions not surjective? Isn't any subset of the codomain which isn't part of the image rather arbitrary?&lt;/p&gt;&#xA;" OwnerUserId="2058" LastActivityDate="2010-09-26T14:04:16.200" Title="How is the codomain for a function defined?" Tags="&lt;functions&gt;" AnswerCount="6" CommentCount="1" FavoriteCount="1" />
  <row Id="5481" PostTypeId="1" AcceptedAnswerId="5491" CreationDate="2010-09-26T05:46:10.890" Score="1" ViewCount="193" Body="&lt;p&gt;I am looking at Perturbation Theory by E.J. Hinch. The author introduces Van Dyke's matching rule: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(m term inner)(n term outer) = (n term outer)(m term inner)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To be used to match the terms in an inner solution to the the terms in the outer solution for some perturbed differential equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hinch when comparing with introducing intermediate variables says: 'Van Dyke's matching rule usually works and is more convenient.'&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anyone know an example where this matching rule doesn't work?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;&#xA;In section $5.2.6$ of the book above, the author gives an example of where the matching rule is meant to fail. Is this an example where the matching rule actually fails or just shows some care in comparing terms is required? In particular, would the intermediate variables approach have any problems?&lt;/p&gt;&#xA;" OwnerUserId="1794" LastEditorUserId="1794" LastEditorDisplayName="" LastEditDate="2010-09-26T08:38:38.417" LastActivityDate="2010-09-26T09:11:16.160" Title="Van Dyke's matching rule" Tags="&lt;differential-equations&gt;&lt;perturbation-theory&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="5482" PostTypeId="1" CreationDate="2010-09-26T06:01:38.737" Score="2" ViewCount="188" Body="&lt;p&gt;Could someone please help me understand this proof given in an article by William Miller &lt;/p&gt;&#xA;&#xA;&lt;p&gt;its supposed to follow from the prime number theorem that given,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$A(x)$ which is the sum of all primes less than or equal to $x$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and $\theta(x)$ which is the sum of the logarithm of all primes less than or equal to x &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$A(x)\sim \frac{x^2}{2\log x}$ and $\theta(x) \sim x$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the following identity is used: $\theta(x) = \int_1^x \log(t)\mathrm{d}(\pi(t))$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\pi(t)$ is the prime counting function. I don't understand why this is. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here $\sim$ means asymptotic to i.e. $\lim_{n\to\infty} \frac{f(x)}{g(x)}=1$&lt;/p&gt;&#xA;" OwnerUserId="2143" LastEditorUserId="226" LastEditorDisplayName="" LastEditDate="2010-09-26T06:14:03.830" LastActivityDate="2010-09-26T06:45:00.243" Title="Prime Number theorem and the prime counting function" Tags="&lt;number-theory&gt;&lt;prime-numbers&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="5483" PostTypeId="2" ParentId="5480" CreationDate="2010-09-26T06:28:31.547" Score="4" ViewCount="" Body="&lt;p&gt;This is something that trips people up the more they dwell on it.  And depending on how you like to think about mathematical objects, you have different preferred answers.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;More often than not, I tend to think of a function as consisting of three things: a domain $A$, a range $B$ and a subset $S$ of $A \times B$ which satisfies this condition, that for any $a \in A$ there is a unique $b \in B$ such that $(a,b) \in S$.  To be precise, a &quot;function&quot; $f$ is a triple $(A,B,S)$ where $S \subset A\times B$ satisfies the above criterion. We say $f : A \to B$ and $f(a) = b$ provided $(a,b) \in S$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the above formalism, you can't describe a function &lt;em&gt;without&lt;/em&gt; specifying the domain and range before hand. To be specific, here are two different ways of specifying the function that in a calculus class would just be called $x^2$:  $f=(\mathbb R, \mathbb R, \{(x,x^2) : x \in \mathbb R\})$.   $g=(\mathbb R, [0,\infty), \{(x,x^2) : x \in \mathbb R\})$.  So in this formalism, $f$ is not an onto function, while $g$ is an onto function.  Although $f(x)=g(x)$ for all $x$ in the domain of $f$ and $g$, $f \neq g$ since they have different ranges. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you don't use a formalism like the above, yes, the range (or co-domain as you say) is just any arbitrary set containing the image.  Frequently this is taken care of by considering the range to be a typical &quot;universal&quot; set. In a standard calculus course what happens is people never really mention the range, they only talk about the image. &lt;/p&gt;&#xA;" OwnerUserId="642" LastActivityDate="2010-09-26T06:28:31.547" CommentCount="7" />
  <row Id="5484" PostTypeId="2" ParentId="4312" CreationDate="2010-09-26T06:36:47.443" Score="1" ViewCount="" Body="&lt;p&gt;It turns out that, as T.. had guessed, the problem is indeed NP-hard. I realised this after participating in some discussions in the comments of &lt;a href=&quot;http://www.toastandtesla.com/engineering/maths/good-friendship-requires-good-accounting/&quot; rel=&quot;nofollow&quot;&gt;this blog post&lt;/a&gt; where the same problem turned up again, and &lt;a href=&quot;http://www.reddit.com/r/math/comments/dioxr/resolving_debts_to_your_friends_with_minimal/c10huhd&quot; rel=&quot;nofollow&quot;&gt;the corresponding Reddit discussion&lt;/a&gt;. The solution I am posting here is based on the above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, a lemma: The optimal solution for $m+n$ people cannot have more than $m+n-1$ nonzero entries.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The proof is by inductive construction of a solution with $m+n-1$ entries. For $m = n = 1$, the solution trivially has 1 entry. When $m+n &gt; 2$, let $a_{11} = \min(r_1,c_1)$. This satisfies one row or column (or both, if $r_1 = c_1$), leaving a subproblem of size $m+n-1$ whose solution has at most $m+n-2$ entries. This completes the proof of the lemma.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose a solution exists with fewer than $m+n-1$ entries. Consider the network flow view of the problem, with $m+n$ nodes connected by edges. If there are fewer than $m+n-1$ edges, the underlying undirected graph is disconnected. Then each connected component is &quot;self-sufficient&quot; in that it has no net in- or out-flow. That means that the sum of the corresponding subset of $\{r_i\} \cup \{-c_j\}$ is zero. So, this problem includes the &lt;a href=&quot;http://en.wikipedia.org/wiki/Subset_sum_problem&quot; rel=&quot;nofollow&quot;&gt;subset sum problem&lt;/a&gt;, which is NP-complete.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More precisely, suppose an instance of the subset sum problem is given as a set of integers $X = \{x_1,\ldots,x_k\}$, and the problem is to determine whether there is a nonempty subset of $X$ whose sum is zero. Let these be the debts $d(p)$ among a set of people $P$ as in the question, introducing an additional person with $d(p) = -\sum X$ to balance the books. Then there is a solution with fewer than $|P|-1$ entries if and only if there is a subset of $X$ with sum zero. (The above lemma is useful for proving the &quot;if&quot; part.) Therefore the subset sum problem can be reduced to this problem.&lt;/p&gt;&#xA;" OwnerUserId="856" LastActivityDate="2010-09-26T06:36:47.443" />
  <row Id="5485" PostTypeId="2" ParentId="5482" CreationDate="2010-09-26T06:45:00.243" Score="2" ViewCount="" Body="&lt;p&gt;The notation&#xA;$$\int \log(t)\ d(\pi(t))$$&#xA;is an example of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Stieltjes_integral&quot; rel=&quot;nofollow&quot;&gt;Stieltjes integral&lt;/a&gt;. Some like this (I don't)&#xA;when discussing summation, as integrating with a step function $f(t)$&#xA;inside the $d$ notation is essentially a summation. Here $\pi(t)$&#xA;is constant, save at prime values of $t$ where it jumps by $1$. This means&#xA;that&#xA;$$\int_a^b g(t)\ d(\pi(t))=\sum_{p\ \mathrm{prime}}g(p)\times 1.$$&#xA;Here the summation is over all primes in the interval from $a$ to $b$&#xA;(I can't remember whay you do about the endpoints; one reason I don't&#xA;like the notation). If you like you can think that&#xA;$$\int_a^b g(t)\ d(\pi(t))=\int_a^b g(t)\pi'(t)\ dt$$&#xA;where $\pi'$ is seen as a generalized function, so as a bunch of delta-functions&#xA;at the primes. Again one has to do the right thing (or is it the left thing? :-))&#xA;at the endpoints.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-09-26T06:45:00.243" />
  <row Id="5486" PostTypeId="2" ParentId="5480" CreationDate="2010-09-26T06:50:11.617" Score="1" ViewCount="" Body="&lt;p&gt;Yes it is arbitrary, and in a sense this is exactly the problem. When we have a function we would like to be able to talk about the inverse of the function (assuming it is 1-1.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the following: I have some function $f$ that maps $\mathbb{N} \rightarrow \mathbb{R}$ such that $$f(x) = (e^\sqrt{3.7x}+137.2)^3$$ The point here is that it is much easier to say that my codomain is $\mathbb{R}$ and my function is not surjective than to explicitly list the image of my function and say that $f$ is surjective on this subset of $\mathbb{R}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that every output will be a real number, but it is hard to say exactly which numbers I will get out of $f$ and which real numbers I can never get. To get to the point, if my function was surjective I know that I could plug any real number into $f^{-1}$. But I know it's not and I therefore have to be very careful with my inputs to $f^{-1}$, restricting my attention only to $f$'s image (otherwise the inverse is undefined.) In this case, we say that $f$ does not have the 'nice' property of surjectivity.&lt;/p&gt;&#xA;" OwnerUserId="1352" LastEditorUserId="1352" LastEditorDisplayName="" LastEditDate="2010-09-26T07:07:46.933" LastActivityDate="2010-09-26T07:07:46.933" CommentCount="0" />
  <row Id="5487" PostTypeId="1" AcceptedAnswerId="5488" CreationDate="2010-09-26T07:49:39.227" Score="-1" ViewCount="114" Body="&lt;p&gt;What is the k(e) and r in Coulomb's law (scalar form)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Coulomb%27s_law&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Coulomb%27s_law&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2145" LastActivityDate="2010-09-26T08:13:13.377" ClosedDate="2010-09-26T11:24:53.393" Title="What is the k(e) and r in Coulomb's law?" Tags="&lt;physics&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="5488" PostTypeId="2" ParentId="5487" CreationDate="2010-09-26T08:13:13.377" Score="0" ViewCount="" Body="&lt;p&gt;$k$ is the columb constant, and $r$ is the distance between two charges. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What's the thing that you really don't understand?&lt;/p&gt;&#xA;" OwnerUserId="550" LastActivityDate="2010-09-26T08:13:13.377" />
  <row Id="5489" PostTypeId="2" ParentId="5480" CreationDate="2010-09-26T08:30:59.407" Score="1" ViewCount="" Body="&lt;p&gt;In the same sense you could ask why all functions aren't total functions. The real question is perhaps &quot;what is the use of writing $f:\ S\rightarrow T$&quot; or even more general &quot;what is the use of writing &quot;$v \in S$&quot;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd expect that in most cases where it is mentioned explicitly it's a way to say &quot;this object belongs to this class of objects&quot; and is done because it belonging to that class is somehow considered relevant or useful at that point. This might sound a bit obvious, but I don't think there is much more to it.&lt;/p&gt;&#xA;" OwnerUserId="2061" LastActivityDate="2010-09-26T08:30:59.407" CommentCount="6" />
  <row Id="5490" PostTypeId="2" ParentId="5480" CreationDate="2010-09-26T08:53:49.270" Score="0" ViewCount="" Body="&lt;p&gt;As far as I know, relations can be defined in two ways:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Given two sets $A$ and $B$, any subset $R$ of $A\times B$ is a relation. The domain is $\{x\in A:\exists y(y\in B,(x,y)\in R)\}$. The codomain is $\{y\in B:\exists x(x\in A,(x,y)\in R)\}$. Obviously the domain is a subset of $A$ and the codomain is a subset of $B$. Not every relation is surjective.&lt;/li&gt;&#xA;&lt;li&gt;Every set that only contains ordered pairs is a relation. The domain is $\{x\in\bigcup\bigcup R:\exists y((x,y)\in R)\}$. The codomain is $\{y\in\bigcup\bigcup R:\exists x((x,y)\in R)\}$. Every relation is surjective.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;My preferred definition is the first one, but you may like the second one, instead.&lt;/p&gt;&#xA;" OwnerUserId="2128" LastActivityDate="2010-09-26T08:53:49.270" />
  <row Id="5491" PostTypeId="2" ParentId="5481" CreationDate="2010-09-26T09:11:16.160" Score="0" ViewCount="" Body="&lt;p&gt;Okay I have found an example: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f_{xx} + \frac{1}{x}f_x+f_x^2 +\epsilon ff_x=0$ for $x&gt;1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f=0$ on $x=1$ and $f\rightarrow 1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The solution is something close to $\log(1+\log(x)/\log(1/\epsilon))$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Van Dyke's rule fails because when expanding an infinite number terms are of the same order (in size). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, with the intermediate variable approach since we are scaling by $\epsilon^{-\alpha}$ with $\alpha \in[0,1]$ we are able to choose $\alpha$ so that there are only a finite number of terms the same order. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The example is completely contrived though, I would feel much better with a more natural example. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you for your effort J.Meyer and J.M.&lt;/p&gt;&#xA;" OwnerUserId="1794" LastActivityDate="2010-09-26T09:11:16.160" />
  <row Id="5492" PostTypeId="2" ParentId="5480" CreationDate="2010-09-26T09:46:45.360" Score="7" ViewCount="" Body="&lt;p&gt;Often what is important is not functions but &lt;em&gt;collections&lt;/em&gt; of functions, especially subsets of the set of functions between two sets $X$ and $Y$, some of which will be surjective and some of which won't.  What also doesn't get emphasized a lot at the level of elementary set theory is the compositional structure of functions, e.g. a function $f : X \to Y$ can be composed with a function $g : Y \to Z$ to give a function $fg : X \to Z$.  For the purposes of studying this compositional structure (for example if $X = Y = Z$) it is generally important not to require that $f$ and $g$ be surjective or you will miss out on structure.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, a simple way to define a dynamical system is just as a function $f : X \to X$.  Whether $f$ is surjective or not is an important aspect of classifying the dynamics of $f$, or in other words of classifying the behavior of the sequences $\{ x, f(x), f^2(x), f^3(x), ... \}$ for various $x$.  This sequence is not well-defined if you pretend that the domain and codomain of $f$ are different just because the range and the codomain aren't equal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is another way of saying that a function really consists of three pieces of data (a domain, a codomain, and the mapping from one to the other), but the reason these three pieces of data are all important is really the compositional structure.  &lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-09-26T09:46:45.360" />
  <row Id="5493" PostTypeId="2" ParentId="5465" CreationDate="2010-09-26T12:07:00.697" Score="1" ViewCount="" Body="&lt;p&gt;It's called a &lt;em&gt;stadium&lt;/em&gt;. &lt;/p&gt;&#xA;" OwnerUserId="875" LastActivityDate="2010-09-26T12:07:00.697" CommentCount="3" />
  <row Id="5494" PostTypeId="1" AcceptedAnswerId="5517" CreationDate="2010-09-26T12:10:38.490" Score="1" ViewCount="285" Body="&lt;p&gt;How can one build a 3x3 linear equation system (3 equations with 3 variables) if the variables are known, for example, a=1, b=2 and c=3?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you in advance!&lt;/p&gt;&#xA;" OwnerUserId="2145" LastActivityDate="2010-12-10T18:06:59.877" Title="How to build a linear equation system?" Tags="&lt;linear-algebra&gt;" AnswerCount="3" CommentCount="5" FavoriteCount="1" />
  <row Id="5495" PostTypeId="2" ParentId="5494" CreationDate="2010-09-26T12:12:55.787" Score="2" ViewCount="" Body="&lt;p&gt;It's ill-posed; you can construct any &lt;strong&gt;nonsingular&lt;/strong&gt; matrix $\mathbf A$ to be multiplied with $(1\;2\;3)^T$ for the corresponding right hand side.&lt;/p&gt;&#xA;" OwnerUserId="498" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-10T18:06:59.877" LastActivityDate="2010-12-10T18:06:59.877" CommentCount="0" />
  <row Id="5496" PostTypeId="2" ParentId="329" CreationDate="2010-09-26T12:24:39.960" Score="2" ViewCount="" Body="&lt;p&gt;Manin and Panchishkin's &lt;a href=&quot;http://lib.org.by/info/M_Mathematics/MT_Number%20theory/Manin%20I.,%20Panchishkin%20A.%20Introduction%20to%20modern%20number%20theory..%20fundamental%20problems,%20ideas%20and%20theories%20%28ISBN%203540203648%29%282ed.,%20Springer,%202005%29%28520s%29_MT_.pdf&quot; rel=&quot;nofollow&quot;&gt;Introduction to Modern Number Theory&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2048" LastActivityDate="2010-09-26T12:24:39.960" CommunityOwnedDate="2010-09-26T12:24:39.960" CommentCount="3" />
  <row Id="5499" PostTypeId="2" ParentId="5480" CreationDate="2010-09-26T14:04:16.200" Score="3" ViewCount="" Body="&lt;p&gt;Here is an answer which directly addresses the question in the title:  the codomain has to be given as part of the information telling you what the functions is; it can't be deduced (or in the language of question, &quot;defined&quot;) if all you know is the domain and values of&#xA;the function.  It is an extra piece of data.  (This is why it seems arbitrary to you; you are thinking about how to determine the codomain from the other data, which can't be done!&#xA;You have to be told what it is as part of the initial description of the function.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First note that this is &lt;em&gt;incompatible&lt;/em&gt; with one traditional definition of a function as being a set of ordered pairs.  The set of ordered pairs definition determines the domain and values of the function, but not the codomain.   (I guess that some people do use this definition of function; for them, a function doesn't have a codomain separate from its image.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To define a function which has a domain and a codomain, one should instead use the scheme&#xA;described by Ryan: a function is a triple (domain $A$, codomain $B$, set of elements of&#xA;$A\times B$ determining its values).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As to why we introduce the concept of codomain, Qiaochu's answer describes this.&lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2010-09-26T14:04:16.200" />
  <row Id="5500" PostTypeId="1" CreationDate="2010-09-26T14:26:12.770" Score="1" ViewCount="171" Body="&lt;p&gt;Can we represent all algebraic structures in First-Order logic?&lt;/p&gt;&#xA;" OwnerUserId="1831" LastActivityDate="2010-09-26T21:52:53.773" Title="Can we represent all algebraic structures in First-Order logic?" Tags="&lt;logic&gt;&lt;algebra&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="5501" PostTypeId="2" ParentId="5441" CreationDate="2010-09-26T14:48:51.350" Score="2" ViewCount="" Body="&lt;p&gt;Billingsley, &lt;em&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0471007102&quot; rel=&quot;nofollow&quot;&gt;Probability and Measure&lt;/a&gt;&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="367" LastActivityDate="2010-09-26T14:48:51.350" CommunityOwnedDate="2010-09-26T14:48:51.350" CommentCount="1" />
  <row Id="5502" PostTypeId="1" AcceptedAnswerId="5509" CreationDate="2010-09-26T14:50:12.323" Score="0" ViewCount="81" Body="&lt;p&gt;In relation to this geometric construction:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/AoQDP.png&quot; alt=&quot;construction&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where D is the center of both circles, if the inner radius (&lt;em&gt;x&lt;/em&gt; = length of line segments DA and DE), the angle &lt;em&gt;φ&lt;/em&gt; = ∠CAB, and the length Δ&lt;i&gt;g&lt;/i&gt; of line segment AB are given, what is the length Δ&lt;i&gt;x&lt;/i&gt; of line segment AC?&lt;/p&gt;&#xA;" OwnerUserId="184" LastEditorUserId="184" LastEditorDisplayName="" LastEditDate="2010-09-26T15:28:01.870" LastActivityDate="2010-09-26T15:42:36.620" Title="What is the difference in radii of two concentric circles given an angle and length of a triangle that is inscribed in the annulus?" Tags="&lt;geometry&gt;&lt;geometric-construction&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="5503" PostTypeId="2" ParentId="5441" CreationDate="2010-09-26T14:50:41.923" Score="2" ViewCount="" Body="&lt;p&gt;A quite new and german one: Klenke, &lt;em&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/3540255451&quot; rel=&quot;nofollow&quot;&gt;Wahrscheinlichkeitstheorie&lt;/a&gt;&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="367" LastActivityDate="2010-09-26T14:50:41.923" CommunityOwnedDate="2010-09-26T14:50:41.923" CommentCount="1" />
  <row Id="5504" PostTypeId="2" ParentId="5500" CreationDate="2010-09-26T15:17:53.973" Score="1" ViewCount="" Body="&lt;p&gt;I am not sure I understand your question. If, by definition, an algebraic structure consists of a set $A\ne\emptyset$ (or a family of such sets) together with a collection of finitary functions on $A$ (functions $f:A^n\to A$ or even ${\mathbf f}:A^n\to A^m$), finitary relations (subsets of $A^k$), and constants (elements of $A$), then 'algebraic structures' are precisely the structures that first-order concerns itself with (you have to use a multi-sorted version of first-order if you want to appropriately work in the setting of &quot;families of sets&quot;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any such structure can be axiomatized in a first-order language as long as you are willing to allow enough function, relation, and constant symbols into your language.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, the Lowenheim-Skolem theorems will prevent you from uniquely characterizing infinite structures unless you impose additional conditions that are not first-order (like &quot;completeness&quot;, for the reals).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But perhaps you mean something else. If you are interested in &lt;em&gt;families&lt;/em&gt; of such structures, which is more natural, then the answer is no in general (simple groups, for example). And many interesting &quot;structures&quot; that appear in algebra are second (or higher) order in nature, and typically first-order falls short to capture them explicitly.&lt;/p&gt;&#xA;" OwnerUserId="462" LastActivityDate="2010-09-26T15:17:53.973" />
  <row Id="5505" PostTypeId="1" AcceptedAnswerId="5506" CreationDate="2010-09-26T15:24:22.467" Score="5" ViewCount="210" Body="&lt;p&gt;If you look at &lt;a href=&quot;http://math.stackexchange.com/questions/3573/interesting-taxicab-problem&quot;&gt;this previous problem&lt;/a&gt;, I asked how to find the sum of all the areas between two taxicab geometry circles. However, upon learning about &lt;a href=&quot;http://en.wikipedia.org/wiki/Lp_space#Motivation&quot; rel=&quot;nofollow&quot;&gt;$\ell^p$ norms&lt;/a&gt;, I thought it would be pretty interesting to extend the problem to all $\ell^p$ norm circles, not just $\ell^1$ (taxicab).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $p=1$, then the result has already been found (the total area is $\frac{5k^2-k-4}{2}$). If $p = \infty$, then each &quot;circle&quot; is just a square, and the area is also easily found (I'm too tired to think about it, but I think it would just be $4(k^2-1)$). Is there, however, a general formula for the area of each circle and the total area of the regions between circles in terms of $k$ and $p$; that is, what is the equation for the area of each overlapping region?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The area of an individual circle, if I did it correctly, is the area of a Lamé curve with $r = p$ and a radius of $k-n$ (see the &lt;a href=&quot;http://math.stackexchange.com/questions/3573/interesting-taxicab-problem&quot;&gt;linked problem&lt;/a&gt;), which equals $\displaystyle 4(k-n)^2\frac{(\Gamma(1+\frac{1}{p})^2)}{\Gamma(1+\frac{2}{p})}$. This can be reduced to $\displaystyle 2(k-n)^2 \frac{\Gamma(\frac{1}{p})^2}{p \Gamma(\frac{2}{p})}$ (see equations 41 and 42 &lt;a href=&quot;http://mathworld.wolfram.com/GammaFunction.html#eqn41&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are some explanatory pictures:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$k=5, p=1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://imgur.com/lic2e.png&quot; alt=&quot;k=5, p=1&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$k=5, p=2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://imgur.com/KYOs7.png&quot; alt=&quot;k=5, p=2&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$k=5, p=3$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://imgur.com/0mFHv.png&quot; alt=&quot;k=5, p=3&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1562" LastEditorUserId="1562" LastEditorDisplayName="" LastEditDate="2010-10-02T18:57:36.037" LastActivityDate="2010-10-05T04:03:42.103" Title="Extension of previous problem, involving $\ell^p$ norm circles" Tags="&lt;calculus&gt;&lt;real-analysis&gt;&lt;functional-analysis&gt;&lt;algebra&gt;" AnswerCount="1" CommentCount="7" FavoriteCount="1" />
  <row Id="5506" PostTypeId="2" ParentId="5505" CreationDate="2010-09-26T15:29:17.017" Score="1" ViewCount="" Body="&lt;p&gt;The &lt;a href=&quot;http://mathworld.wolfram.com/Superellipse.html&quot; rel=&quot;nofollow&quot;&gt;superellipses&lt;/a&gt; would seem to fit your bill, as long as $p &amp;lt; \infty$.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-09-26T15:29:17.017" CommentCount="5" />
  <row Id="5507" PostTypeId="2" ParentId="329" CreationDate="2010-09-26T15:31:55.163" Score="8" ViewCount="" Body="&lt;p&gt;It depends on the level.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For an undergraduate interested in &lt;strong&gt;algebraic&lt;/strong&gt; number theory, I would strongly suggest (parts of) Serre's &lt;em&gt;Cours d'arithmetique&lt;/em&gt; and also Samuel's &lt;em&gt;Théorie algébriques des nombres&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a graduate student aiming at a future of research work in number theory, Cassels &amp;amp; Fröhlich is a must.&lt;/p&gt;&#xA;" OwnerUserId="688" LastActivityDate="2010-09-26T15:31:55.163" CommunityOwnedDate="2010-09-26T15:31:55.163" CommentCount="1" />
  <row Id="5509" PostTypeId="2" ParentId="5502" CreationDate="2010-09-26T15:42:36.620" Score="2" ViewCount="" Body="&lt;p&gt;Consider the triangle DAB. You know two sides and an angle. The length of the third side is $x + \Delta x$.&lt;/p&gt;&#xA;" OwnerUserId="856" LastActivityDate="2010-09-26T15:42:36.620" CommentCount="1" />
  <row Id="5510" PostTypeId="1" AcceptedAnswerId="5608" CreationDate="2010-09-26T17:44:57.837" Score="2" ViewCount="880" Body="&lt;p&gt;I was maths student but now I'm a software engineer and almost all my knowledge about maths formulas is perished.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of my client wants to calculate optimal price for perishable products. He has come across an article named &lt;strong&gt;Optimal Pricing Policies for Perishable Products&lt;/strong&gt; by Miguel F. Anjos¤ Russell C.H. Cheng Christine S.M. Currie.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It has a formula which is called Karush-Kuhn-Tucker condition. I'm completely baffled by looking at it, may be my basics are completely vanished that's why I became completely blank. I understood the basic requirement and has come up with a proper solution for that but I'm not sure whether my solution is completely addressing the condition given in this article. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anybody explain me in plain english about this condition so that I can convert it to algorithm written in a programming language. Or if there is any programming library available for this condition?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm sorry to ask without giving full information about this condition. I'm assuming someone will know about this condition, if not I'll have to post that entire formula which is of around 2-3 pages. :(&lt;/p&gt;&#xA;" OwnerUserId="2151" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-09-27T16:18:38.750" LastActivityDate="2010-12-17T21:19:40.990" Title="Karush-Kuhn-Tucker condition - Lagrange multiplier" Tags="&lt;differential-equations&gt;&lt;integral&gt;&lt;optimization&gt;&lt;linear-programming&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="1" />
  <row Id="5511" PostTypeId="2" ParentId="5441" CreationDate="2010-09-26T17:46:53.417" Score="2" ViewCount="" Body="&lt;p&gt;Jacod and protter is great.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Probability Theory: A Comprehensive Course by Klenke is worth looking at too.&lt;/p&gt;&#xA;" OwnerUserId="2152" LastActivityDate="2010-09-26T17:46:53.417" CommunityOwnedDate="2010-09-26T17:46:53.417" />
  <row Id="5512" PostTypeId="2" ParentId="5510" CreationDate="2010-09-26T18:19:38.540" Score="3" ViewCount="" Body="&lt;p&gt;The KKT conditions are necessary conditions for a constrainted optimization problem to have a local minimum. For complex problems these conditions normally motivate the development of minimization algorithms rather than trying to find the values that satisfy the KKT conditions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is black-boxed optimization software and it is advised to take advantages of these. Perhaps a SQP (Sequential quadratic programming) would be ideal for this problem. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The optimization toolbox in Matlab is a really easy to use. The function fmincon is likely to solve your problem. See: &lt;a href=&quot;http://www.mathworks.com/products/optimization/&quot; rel=&quot;nofollow&quot;&gt;http://www.mathworks.com/products/optimization/&lt;/a&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are alternatives though I personally have never used them.  See: &lt;a href=&quot;http://wwwasdoc.web.cern.ch/wwwasdoc/minuit/minmain.html&quot; rel=&quot;nofollow&quot;&gt;http://wwwasdoc.web.cern.ch/wwwasdoc/minuit/minmain.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1794" LastEditorUserId="1794" LastEditorDisplayName="" LastEditDate="2010-09-26T18:35:13.687" LastActivityDate="2010-09-26T18:35:13.687" CommentCount="2" />
  <row Id="5513" PostTypeId="1" AcceptedAnswerId="5514" CreationDate="2010-09-26T19:19:26.007" Score="0" ViewCount="186" Body="&lt;p&gt;I have a series of problems that involve events that happen one after another, and where the usual formulas like P(A/B)=P(AB)/P(A) don't seem to help.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In a horse race there are 3 horses. E12&#xA;  means horse 1 finishes before horse 2.&#xA;  E123 means horse 1 finishes before 2,&#xA;  which finishes before 3, etc.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;If &lt;strong&gt;P(E13)=2/3&lt;/strong&gt; and &lt;strong&gt;P(E23)=1/2&lt;/strong&gt; then&#xA;  &lt;strong&gt;P(E123)=?&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;How can I go about solving this kind of problems?&lt;/p&gt;&#xA;" OwnerUserId="1862" LastActivityDate="2010-09-26T19:46:00.677" Title="How to solve this kind of Conditional Probability problems?" Tags="&lt;probability&gt;&lt;probability-theory&gt;" AnswerCount="1" />
  <row Id="5514" PostTypeId="2" ParentId="5513" CreationDate="2010-09-26T19:46:00.677" Score="3" ViewCount="" Body="&lt;p&gt;This is &lt;strong&gt;not&lt;/strong&gt; a conditional probability question.&#xA;Your particular example has insufficient information to solve it.&#xA;Here $E123$ is the event of the horses finishing in the order 1, 2, 3.&#xA;So (assuming the horses can't tie for a position)&#xA;$$P(E13)=P(E132)+P(E123)+P(E213),$$&#xA;$$P(E23)=P(E123)+P(E213)+P(E231)$$&#xA;and of course&#xA;$$1=P(E123)+P(E132)+P(E213)+P(E231)+P(E312)+P(E321).$$&#xA;Knowing $P(E13)$ and $P(E23)$ isn't enough to determine any one&#xA;of the $P(Eijk)$s.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-09-26T19:46:00.677" />
  <row Id="5515" PostTypeId="2" ParentId="5432" CreationDate="2010-09-26T19:56:43.197" Score="3" ViewCount="" Body="&lt;p&gt;What you have discovered is the special case $\rm\ a = 2,\ m = n-1\ $ of the following simple identity&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\displaystyle\ a^{m}\:\frac{a^n-1}{a-1}\: =\: \frac{a^{m+n}-1}{a-1} - \frac{a^m-1}{a-1}\: =\: 1+a+\cdots+a^{m+n-1} - (1+a+\cdots+a^{m-1})$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, it yields $\ 11111000 = 11111111 - 111\ $ for $\rm\ a = 10,\ m = 3,\ n = 5\:$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, except in the special case when the left hand side is an even perfect number, the right hand side is not the (rearranged) sum of the divisors of the left hand side, so it bears little relation to perfect numbers.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-09-26T23:41:41.400" LastActivityDate="2010-09-26T23:41:41.400" CommentCount="2" />
  <row Id="5516" PostTypeId="2" ParentId="5494" CreationDate="2010-09-26T20:06:49.030" Score="3" ViewCount="" Body="&lt;p&gt;If this is homework, it's a pretty strange one.   :-?  Anyway, do you like the following system?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;\begin{align}&#xD;&#xA;x  &amp;amp;=  1   \\\&#xD;&#xA; y &amp;amp;=  2  \\\&#xD;&#xA; z &amp;amp;=  3&#xD;&#xA;\end{align}&#xD;&#xA;$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-Too much easy, I see.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, try this: your system must fulfil the following equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;A &#xD;&#xA;\begin{pmatrix}&#xD;&#xA;1 \\\&#xD;&#xA;2 \\\&#xD;&#xA;3&#xD;&#xA;\end{pmatrix}  = b&#xD;&#xA;$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $A$ is a $3\times 3$ matrix and $b$ is a column vector with $3$ coordinates (a $3\times 1$ matrix). Write $a_1, a_2, a_3$ for the columns of $A$. Then this equation is the same as the following equality between vectors in $\mathbb{R}^3$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;a_1 + 2a_2 + 3a_3 = b \ .&#xD;&#xA;$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, pick your three (linearly independent) favourite vectors in $\mathbb{R}^3$, put in the places of $a_1, a_2, a_3$, compute the resulting $b$ and you have another system of linear equations with solution $x=1, y=2$, and $z=3$. For instance, if you take $a_1 = (1\ 0\ 0)^t$, $a_2 = (0\ 1 \ 0)^t$ and $a_3 = (0\ 0\ 1)^t$, you'll obtain my first example.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;EDIT. Maybe I should add something to explain the necessity of my misterious condition that $a_1, a_2, a_3$ must be &quot;linearly independent&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason is the following: if you pick your three vectors randomly (that is to say, you put no matter what coefficients into the matrix $A$), you could obtain systems like&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;x + y + z = 6 &#xD;&#xA;$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(take $a_1 = a_2 = a_3 = (1\ 0\ 0)^t$). Of course this system has the desired solution $x=1, y=2,$ and $ z=3$, but also an infinite number of them. In fact, the solution set is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;(x, y, z) = (1, 2, 3) + \lambda (1, 0, -1) + \mu (0, 1, -1)&#xD;&#xA;$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for all values of $\lambda, \mu$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, if you fill in the matrix $A$ randomly, the probability of &lt;em&gt;not&lt;/em&gt; obtaining three linearly independent vectors $a_1, a_2, a_3$ is zero, but, it won't take you too much time and effort to verify that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;\det (A) \neq 0&#xD;&#xA;$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;before submitting your homework.&lt;/p&gt;&#xA;" OwnerUserId="664" LastEditorUserId="664" LastEditorDisplayName="" LastEditDate="2010-12-10T18:05:15.563" LastActivityDate="2010-12-10T18:05:15.563" />
  <row Id="5517" PostTypeId="2" ParentId="5494" CreationDate="2010-09-26T20:12:48.813" Score="2" ViewCount="" Body="&lt;p&gt;If I understand correctly, you have the desired solution for a system of three equations in 3 unknowns and want to know how to construct a system with those solutions.  This is a fairly common task for people who teach high-school algebra.  The way that I typically do it is to make up the coefficients of the variables first, so something like:&#xA;$$\begin{align}&#xA;3a+4b+5c&amp;amp;=&#xA;\\&#xA;8a-5b+7c&amp;amp;=&#xA;\\&#xA;-5a+b+c&amp;amp;=&#xA;\end{align}$$&#xA;then substitute in the desired solutions to get the constant terms on the right side of each equation (using your $a=1$, $b=1$, and $c=3$):&#xA;$$\begin{align}&#xA;3\cdot 1+4\cdot 2+5\cdot 3&amp;amp;=26&#xA;\\&#xA;8\cdot 1-5\cdot 2+7\cdot 3&amp;amp;=19&#xA;\\&#xA;-5\cdot 1+2+3&amp;amp;=0&#xA;\end{align}$$&#xA;so that one system with your given solutions would be:&#xA;$$\begin{align}&#xA;3a+4b+5c&amp;amp;=26&#xA;\\&#xA;8a-5b+7c&amp;amp;=19&#xA;\\&#xA;-5a+b+c&amp;amp;=0&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;edit&lt;/em&gt;: I should probably note (similar to the edited note in &lt;a href=&quot;http://math.stackexchange.com/questions/5494/how-to-build-a-linear-equation-system/5516#5516&quot;&gt;Agusti Roig's answer&lt;/a&gt;) that this method only guarantees that the given values are a solution to the system, not that they are the only solution to the system.  As with nearly any problem, it is worthwhile to check that the resulting system has only the solution(s) you want by solving it.&lt;/p&gt;&#xA;" OwnerUserId="72" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-09-27T14:07:35.747" LastActivityDate="2010-09-27T14:07:35.747" />
  <row Id="5518" PostTypeId="2" ParentId="5416" CreationDate="2010-09-26T20:35:48.577" Score="3" ViewCount="" Body="&lt;p&gt;Geometrically one can see what's going on by analyzing $f$'s induced map on tangent vectors.  You'll see that the operator $\frac{\partial f}{\partial \overline{z}}$ is measuring the failure of a function to respect multiplication by $i$ between domain tangent vectors and range tangent vectors.  It is helpful to think of $f$ mapping from a domain $C$ ~ $R^2$ to a distinct range $C$ ~ $R^2$.  Any differentiable function $f$ will induce a map from tangent vectors in the domain to tangent vectors in the range, denoted here $v \rightarrow f'(v)$.  This is the derivative of $f$ along the vector $v$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both our domain and range are complex and thus have a concept of multiplication by $i$ on tangent vectors.&#xA;Multiplying by $i$ is like rotating a vector counter-clockwise 90 degrees, e.g. at a point $z=(x,y)$, the vector $\frac{\partial}{\partial x}$ gets sent to $\frac{\partial}{\partial y}$ under multiplication by $i$.  The beauty of holomorphic functions is that they respect this 90 degree rotation between the domain and range.  That is, if $f$ is holomorphic, then $f'(iv) = i f'(v)$ where the first $i$ multiplication is in the domain and the second represents $i$ multiplication in the range.  So, from the image of a vector in the domain (the derivative in one direction at some point), you can infer the image of $f'$ in every direction at that point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We write this as, for all $v$, $f'(iv) = if'(v)$.  The failure to achieve holomorphicity is thus $f'(iv)-if'(v)$ being different from $0$ for some $v$.  The operator $\frac{\partial f}{\partial \overline{z}}$ is essentially this measurement for a specific v.  Let v be the unit vector $\frac{-\partial}{\partial y}$.  Then $i v$ is $\frac{\partial}{\partial x}$.  The above formula $f'(iv)-if'(v)$ becomes $f'(\frac{\partial}{\partial x}) - i f'(-\frac{\partial}{\partial y})$ = $\frac{\partial f}{\partial x} + i (\frac{\partial f}{\partial y})$ = $2 \frac{\partial f}{\partial \overline{z}}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course there's a certain symmetry here.  While functions where $\frac{\partial f}{\partial \overline{z}}=0$ are holomorphic, functions where $\frac{\partial f}{\partial z}=0$ are anti-holomorphic; $f'$ sends the vector $i v$ to $- i f'(v)$ for all $v$.&lt;/p&gt;&#xA;" OwnerUserId="948" LastActivityDate="2010-09-26T20:35:48.577" />
  <row Id="5520" PostTypeId="2" ParentId="5500" CreationDate="2010-09-26T21:52:53.773" Score="2" ViewCount="" Body="&lt;p&gt;The subject of &lt;a href=&quot;http://en.wikipedia.org/wiki/Universal_algebra&quot; rel=&quot;nofollow&quot;&gt;universal algebra&lt;/a&gt; is the study of algebraic structures as classes of first-order structures in languages having only function symbols as opposed to relation symbols. This subject is regarded by many as much closer to algebra than the rest of model theory, allowing one to combine methods from model theory with a stronger category-theoretic perspective. A central concern are the classes of algebras, called &lt;a href=&quot;http://en.wikipedia.org/wiki/Variety_%28universal_algebra%29&quot; rel=&quot;nofollow&quot;&gt;varieties&lt;/a&gt;, defined by a system of algebraic equations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, &lt;a href=&quot;http://en.wikipedia.org/wiki/Variety_(universal_algebra)#Birkhoff.27s_theorem&quot; rel=&quot;nofollow&quot;&gt;Birkoff's theorem&lt;/a&gt; states that a class of algebraic structures is a variety of algebras if and only if it is closed under homomorphisms, subalgebras and products.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A similar corresponding theorem in first order logic is due to Kiesler and Shelah: two structures are elementary equivalent if and only if they have a common ultrapower, and one can use this to characterize which collections of algebras form an &lt;a href=&quot;http://en.wikipedia.org/wiki/Elementary_class&quot; rel=&quot;nofollow&quot;&gt;elementary class&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="413" LastActivityDate="2010-09-26T21:52:53.773" />
  <row Id="5521" PostTypeId="2" ParentId="5416" CreationDate="2010-09-26T22:30:25.000" Score="3" ViewCount="" Body="&lt;p&gt;This is a rephrasing of aspects of what several people above have said. In particular Weaam's 3rd paragraph shows you how to prove:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A $\mathbb R$-linear function $f : \mathbb R^2 \to \mathbb R^2$ can be decomposed uniquely into a sum $f = Cf+Af$ of two $\mathbb R$-linear functions $Cf, Af : \mathbb R^2 \to \mathbb R^2$, where $Cf$ is &quot;complex linear&quot; meaning that $Cf(av+bw)=aCf(v)+bCf(w)$ for all complex numbers $a,b$ and vectors $v,w\in \mathbb R^2$.   $Af$ is complex anti-linear $Af(av+bw)=\overline{a}Af(v)+\overline{b}Af(w)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So given an arbitrary Frechet ($\mathbb R$-differentiable) function $f$ defined on an open subset of $\mathbb R^2$,&#xA;$\partial f/\partial z$ is the complex-linear part of the derivative of $f$.  $\partial f/\partial \overline{z}$ is the complex anti-linear part of the derivative of $f$.   So this explains why $f$ is complex differentiable if and only if $\partial f/\partial \overline{z} = 0$.  In the notation of the previous paragraph, $\partial f/\partial \overline{z} = Af'$, and $\partial f/\partial z = Cf'$. &lt;/p&gt;&#xA;" OwnerUserId="642" LastActivityDate="2010-09-26T22:30:25.000" />
  <row Id="5522" PostTypeId="1" CreationDate="2010-09-26T22:34:19.713" Score="1" ViewCount="258" Body="&lt;p&gt;What are some examples of cutting-edge research involving fractals or self-similar structures?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Who's actively contributing high-quality research in this field?&lt;/p&gt;&#xA;" LastActivityDate="2010-09-27T03:38:32.643" Title="Where have fractals gone since Mandelbrot?" Tags="&lt;fractals&gt;" AnswerCount="4" CommentCount="2" />
  <row Id="5523" PostTypeId="1" AcceptedAnswerId="5612" CreationDate="2010-09-27T00:20:49.510" Score="2" ViewCount="148" Body="&lt;p&gt;It's said on Wikipedia page about &lt;a href=&quot;http://en.wikipedia.org/wiki/Post%27s_lattice&quot; rel=&quot;nofollow&quot;&gt;Post's lattice&lt;/a&gt; that DM clone(all monotone self-dual functions) has majority function(xy + xz + yz) as one of its bases. What is the proof of this fact? &lt;/p&gt;&#xA;" LastActivityDate="2010-09-28T01:50:16.483" Title="Monotone self-dual boolean functions clone" Tags="&lt;logic&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="5524" PostTypeId="2" ParentId="5522" CreationDate="2010-09-27T00:54:25.027" Score="0" ViewCount="" Body="&lt;p&gt;Fractal &lt;a href=&quot;http://en.wikipedia.org/wiki/Fractal_antenna&quot; rel=&quot;nofollow&quot;&gt;antennae&lt;/a&gt; lend theory to the design of fractal solar panels.&lt;/p&gt;&#xA;" OwnerUserId="1995" LastActivityDate="2010-09-27T00:54:25.027" />
  <row Id="5525" PostTypeId="2" ParentId="2581" CreationDate="2010-09-27T01:17:23.687" Score="1" ViewCount="" Body="&lt;p&gt;It is possible to brute force a list of possible solutions for every possible combination of sticks. Ignoring permutations... For n=1 there is a single solution. Ditto for n=2. For n=3 there are two solutions for one combination of sticks. For n=4 there are three solutions for two combinations of sticks. For n=5 I think there are ten solutions for four combinations of sticks. For n=6 I think there are 24 solutions. A completely baseless and uniformed search of the OEIS suggests this sequence might continue with 130 solutions for n=7 and 504 solutions for n=8. This trend &lt;em&gt;suggests&lt;/em&gt; that there are increasingly more solutions for any such n, and a non-decreasing number of solutions for the &quot;hardest&quot; combination of sticks for a given n, and the pattern by which such solutions can be enumerated may suggest a proof of the original question.&lt;/p&gt;&#xA;" OwnerUserId="1995" LastActivityDate="2010-09-27T01:17:23.687" CommentCount="3" />
  <row Id="5526" PostTypeId="2" ParentId="5522" CreationDate="2010-09-27T01:28:24.273" Score="-1" ViewCount="" Body="&lt;p&gt;Behold the mandelbulb &lt;a href=&quot;http://www.skytopia.com/project/fractal/mandelbulb.html&quot; rel=&quot;nofollow&quot;&gt;http://www.skytopia.com/project/fractal/mandelbulb.html&lt;/a&gt;  !!!!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;oddly enough the mandelbulb looks almost organic.&lt;/p&gt;&#xA;" OwnerUserId="1582" LastActivityDate="2010-09-27T01:28:24.273" />
  <row Id="5527" PostTypeId="1" AcceptedAnswerId="5530" CreationDate="2010-09-27T02:32:41.370" Score="1" ViewCount="110" Body="&lt;p&gt;This is mostly related to doing large modular exponentiation by hand. For example, a problem I was doing was to find the last 3 digits of $7^{9729}$; that is, find $7^{9729}\bmod{1000}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the simplest concept for Euler's theorem, I found that $7^{400}\equiv 1\pmod{1000}$, since $\varphi(1000)=400$. Using Carmichael's theorem, I found a smaller number, $7^{100}\equiv 1\pmod{1000}$, as $\lambda(1000)=100$. Now, by manually multiplying it out, I found that $7^{20}\equiv 1 \pmod{1000}$, and that is the first $n$ for which that is true, meaning I just need to find $7^9 \bmod{1000}$, making the answer 607.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a way to arrive at this answer without multiplying it out each time for every number I get? For example, could I do something like $13^{12937}\bmod{1000}$ without sitting around modding out multiples of $13^4$? (I know that the first $n$ for 13 would be 100, so no less than using Carmichael's theorem, but I want to know if there are other ways to find numbers lower than those given by Carmichael's theorem)&lt;/p&gt;&#xA;" OwnerUserId="1562" LastActivityDate="2010-09-27T04:13:32.837" Title="How do I find the lowest $n$ for which $a^n \equiv 1 \pmod{b}$?" Tags="&lt;algebra&gt;&lt;modular-arithmetic&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" />
  <row Id="5528" PostTypeId="2" ParentId="5527" CreationDate="2010-09-27T03:05:04.210" Score="1" ViewCount="" Body="&lt;p&gt;As I discuss in &lt;a href=&quot;http://math.stackexchange.com/questions/3976/on-the-binary-decimal-expansion-of-the-reciprocal-primes&quot;&gt;my answer to this question&lt;/a&gt;, this is a hard quantity to compute in general.  For practical purposes you should use &lt;a href=&quot;http://en.wikipedia.org/wiki/Exponentiation_by_squaring&quot; rel=&quot;nofollow&quot;&gt;exponentiation by squaring&lt;/a&gt; to test the possibilities (and you should definitely use exponentiation by squaring for computing remainders in general; this is how computer algebra systems do it).  Note that using Carmichael's theorem is not trivial, as it requires that you know the prime factorization of $b$.  &lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-09-27T03:05:04.210" CommentCount="5" />
  <row Id="5529" PostTypeId="2" ParentId="5522" CreationDate="2010-09-27T03:17:03.313" Score="0" ViewCount="" Body="&lt;p&gt;render 3D fractals based on the original 2D mandelbrot formula, such as mandelbox and mandelbulb + many more - using free open source software called &quot;&lt;a href=&quot;http://sourceforge.net/projects/mandelbulber/&quot; rel=&quot;nofollow&quot;&gt;mandelbulber&lt;/a&gt;&quot;&lt;/p&gt;&#xA;" LastActivityDate="2010-09-27T03:29:42.487" />
  <row Id="5530" PostTypeId="2" ParentId="5527" CreationDate="2010-09-27T03:28:58.353" Score="4" ViewCount="" Body="&lt;p&gt;If $\rm\: gcd(a,10)=1\:$ then order of $\rm\: a\:$ in $\:\mathbb Z/1000 \:$ must be a divisor of $100 = \lambda(1000)$. You can compute the order simply and quickly by computing in order  $\rm a^2, a^4, a^5, a^{10}, a^{20}, a^{25}, a^{50}\:$ by squaring or multiplying previous entries. This requires at most 5 squaring and 2 multiplication operations $\rm (mod\ 1000)$. Obviously the same sort of optimized divisor lattice searching works for any modulus.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively one can use the following well-known simple order algorithm for groups&#xA;&lt;img src=&quot;http://i.imgur.com/77uZH.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://groups.csail.mit.edu/cis/theses/sutherland-phd.pdf&quot; rel=&quot;nofollow&quot;&gt;This thesis&lt;/a&gt; is a good reference on order algorithms in generic groups. Here's the abstract:&#xA;&lt;img src=&quot;http://i.imgur.com/VAw3T.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-09-27T04:13:32.837" LastActivityDate="2010-09-27T04:13:32.837" CommentCount="3" />
  <row Id="5531" PostTypeId="2" ParentId="5522" CreationDate="2010-09-27T03:29:46.897" Score="3" ViewCount="" Body="&lt;p&gt;As pointed out in the comments, this is a tremendously broad area and there are many people doing a lot of very good research, so I can only say a little bit about the small part of it that I'm familiar with.  When dealing with &quot;chaotic&quot; dynamical systems (a notion which can be made precise in a few different ways), one often finds that the chaotic behaviour of the dynamics is intimately connected to the presence of various geometric structures in the phase space that are best characterised as fractals -- the Julia sets found in complex dynamics are examples of this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To study these structures, one uses various dimensional quantities, such as Hausdorff dimension.  It turns out that some fundamental dynamical quantities such as topological entropy can also be defined as &quot;dimensions&quot; of a sort, so that there are deep connections between fractal geometry and chaotic dynamics.  One of the standard (advanced) references on this is &quot;Dimension Theory in Dynamical Systems&quot;, by Yakov Pesin.  A more introductory exposition (with apologies for self-advertising) is &quot;Lectures on Fractal Geometry and Dynamical Systems&quot;, by Pesin and Climenhaga.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an example of the sort of thing that occurs, suppose you have a dynamical system with phase space $X$ and an observable function $\phi\colon X\to \mathbb{R}$.  You take measurements of $\phi$ as time goes along, and calculate its average value as time goes to infinity.  This average value is a function of your starting position $x\in X$.  Let $K_\alpha$ denote the set of points for which that average value is equal to $\alpha$; then typically speaking, there is a range of values of $\alpha$ for which $K_\alpha$ is a fractal with quite an intricate structure.  This is called a &lt;em&gt;multifractal decomposition&lt;/em&gt;.  One can define a &lt;em&gt;multifractal spectrum&lt;/em&gt; by $B(\alpha) = \dim_H(K_\alpha)$, and it turns out (somewhat miraculously) that the function $\alpha\mapsto B(\alpha)$ is actually concave and analytic in a great many (important) examples!  This multifractal analysis has deep connections to thermodynamic formalism and statistical properties of chaotic dynamical systems; a good reference is &quot;Thermodynamics of Chaotic Systems&quot;, by Beck and Schl&amp;ouml;gl.&lt;/p&gt;&#xA;" OwnerUserId="876" LastActivityDate="2010-09-27T03:29:46.897" />
  <row Id="5533" PostTypeId="2" ParentId="2581" CreationDate="2010-09-27T03:55:25.067" Score="0" ViewCount="" Body="&lt;p&gt;This is not a solution, but a potential restatement of the problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider a single stick of length n(n+1)/2 that can be broken in any single place, leaving two pieces of length &gt;=n. The resulting two sticks can be broken into 1...n easily, possibly in multiple ways. Can you show that for any position of that break that leaves a stick of length &gt;=2n, there exists a subdivision of the stick such that the longer stick can be broken into any possible two pieces of length &gt;=n? If so, I think that you may have the desired proof.&lt;/p&gt;&#xA;" OwnerUserId="1995" LastActivityDate="2010-09-27T03:55:25.067" />
  <row Id="5534" PostTypeId="1" AcceptedAnswerId="5548" CreationDate="2010-09-27T04:23:19.910" Score="5" ViewCount="301" Body="&lt;p&gt;Let random variable $X\sim\text{Binomial}\left(a+b,\frac{a}{a+b}\right)$, where $a$ and $b$ are positive integers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm trying to prove that $\mathbb{P}[X\geq\mathbb{E}[X]]\leq\frac{3}{4}$, which appears to be true numerically.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anyone have a suggestion on how to proceed?&#xA;The complication in this particular problem is that the condition is not a strict inequality &quot;$&gt;$&quot;.&#xA;I've tried the Chernoff bound but it's not tight enough.&lt;/p&gt;&#xA;" OwnerUserId="1023" LastEditorUserId="1023" LastEditorDisplayName="" LastEditDate="2010-10-24T17:54:31.377" LastActivityDate="2010-10-24T17:54:31.377" Title="Proving an upper bound for Prob[X&gt;=E[X]]" Tags="&lt;probability&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" />
  <row Id="5535" PostTypeId="2" ParentId="227" CreationDate="2010-09-27T05:29:23.870" Score="1" ViewCount="" Body="&lt;p&gt;Looks like you're set on linear algebra and real analysis, both of which are great choices. You can never know enough linear algebra, and if you haven't taken an analysis class yet, it'll definitely get you thinking more &quot;mathematically&quot; than in many of your previous classes, which like Katie said, is definitely one of the main goals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might also want to take a look into probability theory. You mentioned that you talked to a couple of finance professors, so I'm guessing that finance might be a possibility in your potential field of research. In that case, probability will be a pretty key component. It's also useful to know it in a lot of other areas of economics as well. &lt;/p&gt;&#xA;" OwnerUserId="1736" LastActivityDate="2010-09-27T05:29:23.870" CommunityOwnedDate="2010-09-27T05:29:23.870" />
  <row Id="5537" PostTypeId="1" AcceptedAnswerId="5576" CreationDate="2010-09-27T06:15:09.690" Score="8" ViewCount="160" Body="&lt;p&gt;This question is motivated by the following homework problem.  I'm trying to explicitly compute the homeomorphism $f:S^2 \rightarrow \mathbb{CP}^1$ by using stereographic projection and considering $\mathbb{CP}^1 = \mathbb{C}\cup {\infty}$.  I'll want to prove that this is an isometry, where $S^2$ has the standard angle metric and $\mathbb{CP}^1$ has the Fubini-Study metric given by $d(\overline{x},\overline{y})=2\cos^{-1}|(x,y)|$, where $x,y\in \mathbb{C}^2$ are unit vectors (and presumably $(-,-)$ is the usual Hermitian inner product).  Later, I'll use this to explicitly compute the Lie group homomorphism $U(2)\rightarrow SO(3)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My stereographic projection is from the north pole, takes the equator to the unit circle, and puts the south pole at the origin.  What I've gotten so far is that for $z\not= 1$, \begin{equation*} f(x,y,z)=\left( \frac{x}{1-z} , \frac{y}{1-z} \right) = \frac{x+iy}{1-z} = [x+iy : 1-z ], \end{equation*} where these are coordinates in $\mathbb{R}^2$, $\mathbb{C}$, and $\mathbb{C}\subseteq \mathbb{CP}^1$ respectively.  This is troublesome, because philosophically I'd expect that I should be able to define this for $(x,y,z)\not= (0,0,1)$ and then end up with a function to projective space that extends continuously over the north pole; that's sort of the point of projective space, to make $\infty$ into just another point.  However, it is not immediately obvious that this works, although luckily \begin{equation*} \left| \frac{x+iy}{1-z} \right| = \sqrt{ \frac{|x+iy|^2}{(1-z)^2} } = \sqrt{ \frac{1-z^2}{(1-z)^2}}, \end{equation*} and the limit of this expression as $z\rightarrow 1^-$ is indeed $\infty$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, fair enough.  This ends up extending to a continuous function after all.  But: &lt;strong&gt;Am I wrong in my philosophical understanding of projective space?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(For what it's worth, I tried using my calculations to verify that $f$ is an isometry, and it didn't look like it was going to work out.  So maybe I really am just doing something wrong.)&lt;/p&gt;&#xA;" OwnerUserId="2159" LastEditorUserId="2159" LastEditorDisplayName="" LastEditDate="2010-09-27T15:40:24.537" LastActivityDate="2010-09-28T07:31:38.220" Title="Does projectivizing always fix problems at infinity? (Or, am I making a mistake somewhere?)" Tags="&lt;differential-geometry&gt;&lt;lie-groups&gt;&lt;projective-space&gt;" AnswerCount="2" CommentCount="5" FavoriteCount="1" />
  <row Id="5538" PostTypeId="1" CreationDate="2010-09-27T07:05:40.413" Score="2" ViewCount="194" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is one of the question asked in a written test conducted by a company. The question sounded stupid to me. May be its not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Given the area of the coin to be 'A'. If the probability of getting a tail, head and the edge are same, what is the thickness of the coin?&lt;/p&gt;&#xA;" OwnerUserId="1611" LastActivityDate="2010-09-27T10:05:11.777" Title="Find thickness of a coin" Tags="&lt;probability&gt;" AnswerCount="4" FavoriteCount="1" />
  <row Id="5539" PostTypeId="2" ParentId="5538" CreationDate="2010-09-27T07:38:21.937" Score="1" ViewCount="" Body="&lt;p&gt;If you assume that the probability of getting tail, head or edge only depends on the surface area that those regions of the coin have, then you see that the area of the edge must also equal $A$. If you denote by $T$ the thickness of the coin, then the area of the edge is given by $T \cdot C$, where $C$ is the circumference of the coin. Now, as you are given $A$, you can determine $C = 2 \sqrt{A \pi}$ and thus you get $T = \frac{A}{C} = \frac{A}{2 \sqrt{A \pi}} = \frac{\sqrt{A\pi}}{2\pi}$&lt;/p&gt;&#xA;" OwnerUserId="1766" LastEditorUserId="1766" LastEditorDisplayName="" LastEditDate="2010-09-27T09:59:23.747" LastActivityDate="2010-09-27T09:59:23.747" CommentCount="2" />
  <row Id="5540" PostTypeId="1" AcceptedAnswerId="5546" CreationDate="2010-09-27T08:14:13.247" Score="4" ViewCount="341" Body="&lt;p&gt;Generally one can see that there are infinite number of solutions for this equation $$a^{2}+b^{2}=c^{2}$$ by taking multiples of the solution $3,4$ and $5$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can i use this as a fact to prove, that $k^{2} + (k+1)^{2}$ is a perfect square for infinitely many $k \in \mathbb{N}$? Any hints, suggestions would be helpful. If not, then how do i prove this fact!&lt;/p&gt;&#xA;" LastEditorUserId="1778" LastEditorDisplayName="" LastEditDate="2010-09-27T15:51:39.387" LastActivityDate="2011-04-04T05:50:21.750" Title="$k^{2}+(k+1)^{2}$ being a perfect square for infinitely many $k$." Tags="&lt;number-theory&gt;&lt;diophantine-equations&gt;" AnswerCount="3" FavoriteCount="1" />
  <row Id="5541" PostTypeId="2" ParentId="4576" CreationDate="2010-09-27T08:14:17.583" Score="2" ViewCount="" Body="&lt;p&gt;To see that $E_{\infty}=E_{k+2}$: your $E_2$ page is supported in the 0th and kth rows, so any differentials beyond $d^{k+1}$ must be zero because they start and/or end at entries that are 0 (and have been since the $E_2$ page).  Remember that the differential $d^j$ goes $j$ to the right and $j-1$ downwards.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You're right that $E_{k+2}=E_2$ follows from the fact that the transgression is 0.  In fact, since the coefficients are trivial (and the cohomology rings of $G$ and $X$ are both finitely generated in each degree), then $E_2 = H^&lt;em&gt;(G;H^&lt;/em&gt;(X)) \cong H^&lt;em&gt;(G) \otimes H^&lt;/em&gt;(X)$, and under this identification all the differentials must satisfy a signed Leibniz rule.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm trying to make sense of your &quot;Serre SS&quot;, since it's not the one I'm familiar with.  The one I'm used to has $E_2^{p,q} = H^p(B;H^q(F;\mathbb{Z})) \Rightarrow H^{p+q}(E;\mathbb{Z})$; here, $F \hookrightarrow E \rightarrow B$ is a fibration, and we assume either that $\pi_1(B)=0$ or at least that $\pi_1(B)$ acts trivially on $H^q(F)$.  (Otherwise you get twisted coefficients, which are awesome but harder to deal with.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the setup you've got, it sounds like you have a fiber bundle with discrete fiber (i.e., a covering space) $G \hookrightarrow X \rightarrow X/G$, where by $G$ I actually mean the underlying set of the group $G$.  So $H^0(G;\mathbb{Z}) = \mathbb{Z}^{\oplus |G|}$ and $H^q(G;\mathbb{Z})=0$ for $q&gt;0$.  I'd imagine that the assumption of triviality of the induced $G$-action on $H^*(X)$ would guarantee that the coefficients are untwisted, although I'm not sure this is right.  If it is, the $E_2$ term of my Serre SS for this would only have nonzero entries along the bottom row; this reflects the close connection between the (co)homology of a base space and a covering space.  So the SS has &lt;em&gt;no&lt;/em&gt; nonzero differentials, again because all possible differentials $d^i$ ($i\geq 2$) either start at a trivial group or end at a trivial group or both.  So $E_2 = E_\infty$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I guess probably that paper is using some fancier SS for group actions that I've never heard of.  Since usually the condition for untwisted coefficients is that $\pi_1(B)$ acts trivially on $H^q(F)$, I'm inclined to think that perhaps your cohomology is actually group cohomology or something.  I don't know much about this, but I'm pretty sure that $H^q_{group}(G)=H^q_{sing}(BG)$, where $BG=K(G,1)$ is the classifying space of $G$.  This would mesh with what I've said so far, because $\pi_1(K(G,1))=G$.&lt;/p&gt;&#xA;" OwnerUserId="2159" LastEditorUserId="2159" LastEditorDisplayName="" LastEditDate="2010-09-28T15:44:43.150" LastActivityDate="2010-09-28T15:44:43.150" CommentCount="1" />
  <row Id="5542" PostTypeId="2" ParentId="5538" CreationDate="2010-09-27T08:22:17.747" Score="-1" ViewCount="" Body="&lt;p&gt;I think that the question is stupid, because it does not take into account the fact that even with a really thick coin it's easier to tilt it when it lands on the edge than when it lands on a face.&lt;br&gt;&#xA;This said, as Sebastian wrote down in formulas you should think at the edge of the coin as a surface, and not at the single line where the coin ends up.&lt;/p&gt;&#xA;" OwnerUserId="89" LastActivityDate="2010-09-27T08:22:17.747" CommentCount="2" />
  <row Id="5543" PostTypeId="1" AcceptedAnswerId="5556" CreationDate="2010-09-27T08:44:07.763" Score="7" ViewCount="142" Body="&lt;p&gt;Again I have trouble with some exercises in Kunen's set theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the following, let $\kappa &gt; \omega$ a cardinal. Then I want to show that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) $|H(\kappa)| = 2^{&amp;lt;\kappa}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) $H(\kappa)=R(\kappa)$ if and only if $\kappa = \beth_{\kappa}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In 1) the inequality $\geq$ is easy, but I cannot prove $\leq$. In 2) I have no idea. It is always true that $H(\kappa) \subseteq R(\kappa)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you give me a hint?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS: It seems that LaTeX here does not have the Beth symbol.&lt;/p&gt;&#xA;" OwnerUserId="1650" LastActivityDate="2011-03-22T16:35:52.587" Title="Cardinality of H(k)" Tags="&lt;set-theory&gt;" AnswerCount="2" />
  <row Id="5544" PostTypeId="2" ParentId="5416" CreationDate="2010-09-27T08:46:24.137" Score="0" ViewCount="" Body="&lt;p&gt;The formulas&#xA;$$\frac{\partial f}{\partial z} \;=\; \frac{1}{2}\left(\frac{\partial f}{\partial x} - i \frac{\partial f}{\partial y}\right)$$ $$\frac{\partial f}{\partial \bar{z}} \;=\; \frac{1}{2}\left(\frac{\partial f}{\partial x} + i \frac{\partial f}{\partial y}\right)$$&#xA;are just the trigonometric-to-exponential Fourier coefficient transformations&#xA;$$c_1=\frac{1}{2}(a_1-ib_1)$$ $$c_{-1}=\frac{1}{2}(a_1+ib_1)$$&lt;/p&gt;&#xA;" LastActivityDate="2010-09-27T08:46:24.137" />
  <row Id="5545" PostTypeId="2" ParentId="1519" CreationDate="2010-09-27T08:47:56.927" Score="3" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Note that you're only attaching the &lt;em&gt;boundary&lt;/em&gt; of the 4-cell.&lt;/strong&gt;  This is a map $S^3 \rightarrow S^2$.  The resulting topological space has a completely intact copy of the interior of the 4-cell.  This is similar to making $\mathbb{RP}^2$ by gluing a disk to a circle by the degree-2 map $\partial D^2 = S^1 \rightarrow S^1$ (given, say, in complex coordinates by $z\mapsto z^2$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the issue of having high-dimensional spheres &quot;wrap around&quot; low-dimensional cells: up to homotopy, this can be phrased as a question about &lt;em&gt;homotopy groups&lt;/em&gt;, that is, maps from spheres into some space $X$.  (This doesn't address the question of singularities, but I think it is still helpful for trying to think about these gluing constructions.)  These are groups $\pi_k(X)$ for all $k\geq 1$; when $k=1$, this just the fundamental group.  In general, they are quite poorly understood.  Even for the simplest possible CW complexes $X=S^n$ (except $n=1$), we don't know all the homotopy groups!  This might sound like an easy problem, but it's really....REALLY...nontrivial.  Now, when you make CW complexes, you're gluing on cells $e^k$ one by one; these are really just maps from $\partial e^k = S^{k-1}$ to the stuff you already had (the &quot;$(k-1)$-skeleton&quot;).  People often only care about the homotopy types of these maps.  In summary, while it's certainly a good idea to try and visualize these things, it's also good to be able to deal abstractly with gluing constructions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The wikipedia article on &lt;a href=&quot;http://en.wikipedia.org/wiki/Homotopy_groups_of_spheres&quot; rel=&quot;nofollow&quot;&gt;homotopy groups of spheres&lt;/a&gt; is really interesting, and contains a sort of picture of the Hopf fibration.  Basically, you want to think of it as a bunch of circles that somehow fill out $S^3$, which is just $\mathbb{R}^3$ with one extra point at $\infty$.  Interestingly, any two of these circles are linked.  So you can kind of think of it locally as the unit circle in the xy-plane, and then you can move that circle in 2 dimension's worth of directions, and in every one of those directions that you push the circle off itself you're going to get a new, nearby circle that's linked to the original one.  (Of course, there is precisely an $S^2$'s worth of circles!)  You can take the circle going through $\infty$ to be the z-axis.  So by what I just said, all of these other circles need to wind once around the z-axis too.&lt;/p&gt;&#xA;" OwnerUserId="2159" LastActivityDate="2010-09-27T08:47:56.927" />
  <row Id="5546" PostTypeId="2" ParentId="5540" CreationDate="2010-09-27T08:54:13.960" Score="8" ViewCount="" Body="&lt;p&gt;You can find infinitely many by considering solutions of Pell's equation $ m^2 - 2a^2=-1.$&#xA;Your (3,4,5) solution comes from$$(7-5\sqrt{2})(7+5\sqrt{2})=-1,$$&#xA;where $m=7$ and $a=5$.&#xA;For example $$(7-5\sqrt{2})^3$$ yields&#xA;$$696^2+697^2 = 985^2.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just to make it a bit more clear note that&#xA;$$\left( {m-1 \over 2} \right)^2 + \left( {m+1 \over 2} \right)^2 = a^2.$$&lt;/p&gt;&#xA;" OwnerUserId="1301" LastEditorUserId="1301" LastEditorDisplayName="" LastEditDate="2010-09-27T09:01:39.200" LastActivityDate="2010-09-27T09:01:39.200" />
  <row Id="5547" PostTypeId="2" ParentId="5538" CreationDate="2010-09-27T09:10:46.327" Score="4" ViewCount="" Body="&lt;p&gt;This problem is also considered in the book &lt;a href=&quot;http://books.google.co.uk/books?id=QiuqPejnweEC&amp;amp;pg=PA10&amp;amp;lpg=PA10&amp;amp;dq=%22how+thick+should+a+coin+be%22&amp;amp;source=bl&amp;amp;ots=uI9vTy3Egy&amp;amp;sig=S7SjnQTJ-SSwygsV2fTfm4lEOTo&amp;amp;hl=en&amp;amp;ei=516gTP_rHM-TjAfyvtWBDQ&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=1&amp;amp;ved=0CBkQ6AEwAA#v=onepage&amp;amp;q=%22how%20thick%20should%20a%20coin%20be%22&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;Fifty challenging problems in probability with solutions&lt;/a&gt; by Frederick Mosteller.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a more in depth study on bias in coin tossing you could read:&#xA;&lt;a href=&quot;http://comptop.stanford.edu/preprints/heads.pdf&quot; rel=&quot;nofollow&quot;&gt;http://comptop.stanford.edu/preprints/heads.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1301" LastActivityDate="2010-09-27T09:10:46.327" />
  <row Id="5548" PostTypeId="2" ParentId="5534" CreationDate="2010-09-27T09:22:31.547" Score="4" ViewCount="" Body="&lt;p&gt;Here is an answer giving the outline of a proof.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The Bound is tight&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the bound is true then it is tight: If $a=1$, $b=1$ then &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\mathbb{P}[X\geq \mathbb{E}[x]] = \mathbb{P}[X\geq 1] =1-\mathbb{P}[X=0] = 1-\frac{1}{2}\frac{1}{2}=\frac{3}{4}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Getting a feel for the problem&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a rule of thumb (though can be made precise) if $a&gt;5$ and $b&gt;5$ then we can approximate the Binomial by normal random variable by the central limit theorem. That is $X\sim N(a,Var(X))$ approximately.  The error in this approximation becomes increasing small as $a$ and $b$ increase. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence $\mathbb{P}[X\geq \mathbb{E}[x]] \approx \frac{1}{2}&amp;lt;\frac{3}{4}$ for $a&gt;5$ and $b&gt;5$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To make precise, first read this: &lt;a href=&quot;http://en.wikipedia.org/wiki/Binomial_distribution#Normal_approximation&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Binomial_distribution#Normal_approximation&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and then: Box, Hunter and Hunter (1978). Statistics for experimenters. Wiley. p. 130&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We are left with three cases: a and b both small, a large and b small, b large and a small. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;$b$ large and $a\leq 5$&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $b+a&gt;100$ then a poisson approximation to the Binomial r.v. becomes appealling. That is $X\sim Po(a)$ approximately. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then $\mathbb{P}(X\geq a)=1-\sum_{i=0}^{a-1}\frac{e^{-a}a^i}{i!}\leq 0.63&amp;lt;\frac{3}{4}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For explicit error bounds look in A bound on the Poisson-binomial relative error by Teerapabolarn (2007). Theorem 2.1 contains a good error bound.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;$b\leq 5$ and $a$ large&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $Y=a+b-X\sim  \text{Bin}(a+b,\frac{b}{a+b})$ now if $b+a&gt;100$ $Y$ is approximately Poisson and &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\mathbb{P}(X\geq a)\approx\mathbb{P}(Y\leq b) &amp;lt;0.45 \leq \frac{3}{4}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Remaining cases&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the cases above are not quite mutually exclusive, but there a finite number of cases left. The cases left are &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) $a\leq 5$ and $b\leq 100$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) $b\leq 5$ and $a\leq 100$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And this can be checked on a computer. To give the bound. &lt;/p&gt;&#xA;" OwnerUserId="1794" LastEditorUserId="1794" LastEditorDisplayName="" LastEditDate="2010-09-28T12:52:11.903" LastActivityDate="2010-09-28T12:52:11.903" CommentCount="3" />
  <row Id="5549" PostTypeId="1" AcceptedAnswerId="5554" CreationDate="2010-09-27T09:48:02.227" Score="0" ViewCount="269" Body="&lt;p&gt;If $\displaystyle \frac{a^n + b^n}{a^{n-1} + b^{n-1}}$ , ( a != b and a,b,n are real numbers ) is the Arithmetic mean &lt;strong&gt;or&lt;/strong&gt; Geometric mean between a and b then find the values of n for both means respectively.&lt;/p&gt;&#xA;" OwnerUserId="2109" LastEditorUserId="2109" LastEditorDisplayName="" LastEditDate="2010-09-27T10:10:05.670" LastActivityDate="2010-10-06T15:45:41.487" Title="A simple algebra problem" Tags="&lt;algebra-precalculus&gt;" AnswerCount="2" CommentCount="14" />
  <row Id="5550" PostTypeId="1" AcceptedAnswerId="5553" CreationDate="2010-09-27T09:57:45.970" Score="2" ViewCount="141" Body="&lt;p&gt;I want to show that if $F:\mathbb{R^n}\to\mathbb{R^m}$ has a derivative of rank = r at a point $p$, then there is a neighbourhood of that point where the rank of the function doesn't decrease, i.e. $\exists N=B(p, \delta)$ such that $rk(DF_x) \geq rk(DF_p) ~ \forall x\in N$  &lt;/p&gt;&#xA;" OwnerUserId="2162" LastEditorUserId="2162" LastEditorDisplayName="" LastEditDate="2010-09-27T10:03:23.590" LastActivityDate="2010-09-27T12:13:18.800" Title="Why doesn't small perturbations of a matrix decrease its rank?" Tags="&lt;calculus&gt;&lt;linear-algebra&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" />
  <row Id="5551" PostTypeId="2" ParentId="5447" CreationDate="2010-09-27T10:00:27.323" Score="0" ViewCount="" Body="&lt;p&gt;The easiest way to factor polynomials is by finding roots: if $p(a)=0$ then $p(x)=(x-a)q(x)$ with $q(x)$ being a polynomial of smaller degree than $p(x)$ and you can repeat the process by finding a root of $q(x)$. So if you have a root $a$, you simply compute $q(x)=p(x)/(x-a)$ and continue recursively. This works over any field - the power of the Complex numbers stems from the fact that this process &lt;strong&gt;always&lt;/strong&gt; works in the complex number - every polynomial has a root.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Still, finding such a root is not always an easy task. For polynomials of degree 2 it is trivial using the well known formula $z_{1,2} = \frac{-b\pm\sqrt{b^2-4ac}}{2a}$ which gives the roots of the polynomial $az^2+bz+c$. In simpler cases, such as $x^2+16$ you gave above, one can solve directly ($x^2=-16$ and then taking roots you find $\pm 4i$). However, for higher degrees the task becomes more difficult. There is a formula for polynomials of degree 3 and 4 but it is more complex; and for degree 5 you don't have a simple formula anymore (with &quot;simple&quot; meaning &quot;involving only the four operations of arithmetic and taking roots&quot;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For those cases, it is simpler to find an approximate solution and use the division method described above. A typical method for finding approximate roots of polynomials is the Newton-Rapson method.&lt;/p&gt;&#xA;" OwnerUserId="1818" LastActivityDate="2010-09-27T10:00:27.323" />
  <row Id="5552" PostTypeId="2" ParentId="5538" CreationDate="2010-09-27T10:05:11.777" Score="1" ViewCount="" Body="&lt;p&gt;I assumed that the probability of getting a head, tail or edge depended on the angle from the centre of the coin that the side lies in.  So the head, tail and edge must each occupy 120 degrees when viewed along the axis of rotation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/dw3GQ.gif&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the diagram above the angles at the centre are all (meant to be) 60 degrees and the radius of each face is $\sqrt{A/\pi}$.  A small amount of trigonometry later and I found the edge length to be $\sqrt{\frac{4A}{3\pi}}$.&lt;/p&gt;&#xA;" OwnerUserId="1750" LastActivityDate="2010-09-27T10:05:11.777" />
  <row Id="5553" PostTypeId="2" ParentId="5550" CreationDate="2010-09-27T10:11:25.013" Score="9" ViewCount="" Body="&lt;p&gt;Use the continuity of the determinant. If the rank is $r$ the Jacobian matrix&#xA;has an $r$-by-$r$ submatrix $A$ with nonzero determinants. Peturbing the point&#xA;will perturb the Jacobian matrix and so $A$. As the determinant is a continuous&#xA;function of its entries a small enough perturbation will leave the new $A$&#xA;having nonzero determinant, and so the new Jacobian matrix will have rank $\ge r$.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-09-27T10:11:25.013" CommentCount="1" />
  <row Id="5554" PostTypeId="2" ParentId="5549" CreationDate="2010-09-27T10:16:17.500" Score="3" ViewCount="" Body="&lt;p&gt;For the current version of the problem the answer is $n=1$ (for arithmetic mean) and $n=1/2$ (for geometric mean).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider first the case&#xA;$$&#xA;\frac{a^n + b^n}{a^{n-1} + b^{n-1}} = \frac{a + b}{2}&#xA;$$&#xA;This equality implies $2a^n + 2b^n = a^{n} + ab^{n-1} + b^{n} + ba^{n-1}$. Hence $a^n + b^n = ab^{n-1} + ba^{n-1}$ and so $(a^{n-1} - b^{n-1})(a-b) = 0$. Hence (since $a \ne b$) we have $n=1$. It's easy to check that $n=1$ indeed satisfies the conditions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now consider the case&#xA;$$&#xA;\frac{a^n + b^n}{a^{n-1} + b^{n-1}} = \sqrt{ab}&#xA;$$&#xA;This equality implies $a^n + b^n = a^{n-1/2}\sqrt{b} + \sqrt{a}b^{n-1/2}$. Hence $(a^{n-1/2} - b^{n-1/2})(\sqrt{a}-\sqrt{b}) = 0$. So (since $a \ne b$) we have $n=1/2$. It's easy to check that $n=1/2$ indeed satisfies the conditions.&lt;/p&gt;&#xA;" OwnerUserId="334" LastActivityDate="2010-09-27T10:16:17.500" CommentCount="2" />
  <row Id="5555" PostTypeId="2" ParentId="5537" CreationDate="2010-09-27T10:45:25.247" Score="0" ViewCount="" Body="&lt;p&gt;&quot;Am I wrong in my philosophical understanding of projective space?&quot;&lt;br&gt;&#xA;I think your understanding is incomplete in some ways.&lt;br&gt;&#xA;E.g. your remark &quot;resolve singularities by replacing a point with a copy of projective space&quot; doesn't make much sense to me.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: I think I didn't understand the question correctly. But I couldn't ask any clarification in comments to the question like others do, because of lack of reputation!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It may help in your understanding of &quot;projective spaces&quot; to realize that their resolvement of &lt;code&gt;\infty&lt;/code&gt; singularities uses homogene coordinates, e.g. in &lt;code&gt;\mathbb{CP}^1 = \mathbb{C}\cup {\infty}&lt;/code&gt; one uses coordiantes \rho\cdot (z, 1) for elements in &lt;code&gt;\mathbb{C}&lt;/code&gt; and &lt;code&gt;\rho (1, 0)&lt;/code&gt; for &lt;code&gt;{\infty}&lt;/code&gt;. Then the mapping should involve both coordinates and account for the free factor &lt;code&gt;$\rho != 0$&lt;/code&gt;.&lt;/p&gt;&#xA;" OwnerUserId="706" LastEditorUserId="706" LastEditorDisplayName="" LastEditDate="2010-09-28T07:31:38.220" LastActivityDate="2010-09-28T07:31:38.220" CommentCount="4" />
  <row Id="5556" PostTypeId="2" ParentId="5543" CreationDate="2010-09-27T11:20:38.220" Score="3" ViewCount="" Body="&lt;p&gt;1) To show that $|H(\kappa)| \leq 2^{&amp;lt;\kappa}$, we need to code elements of $H(\kappa)$ using subsets of ordinals $&amp;lt; \kappa$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $x \in H(\kappa)$ and let $a$ be the transitive closure of $\{x\}$. Fix a bijection $f:|a|\to a$ with $f(0) = x$ and let $E \subseteq |a|^2$ be defined by $\xi \mathrel{E} \zeta$ iff $f(\xi) \in f(\zeta)$. Then $x$ is recovered as the value of $0$ in the transitive collapse of $(|a|,E)$. The pair $(|a|,E)$ can be encoded as a subset of $|a|^2 &amp;lt; \kappa$. This encoding process shows that $|H(\kappa)| \leq 2^{&amp;lt;\kappa}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) To see that $H(\kappa) \subseteq R(\kappa)$, note that the rank of an element $x \in H(\kappa)$ is smaller than $|a|^+$, where $a$ is once again the transitive closure of $\{x\}$. Since $|R(\kappa)| = \beth_\kappa$, we see that $H(\kappa) = R(\kappa)$ can only hold when $2^{&amp;lt;\kappa} = \beth_\kappa$. Note that $2^{\lambda} &amp;lt; \beth_\kappa$ for every $\lambda &amp;lt; \beth_\kappa$. Therefore, $H(\kappa) = R(\kappa)$ entails that $\kappa = \beth_\kappa$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To see that $\kappa = \beth_\kappa$ entails that $H(\kappa) = R(\kappa)$ it suffices to check that every element of $R(\kappa)$ has transitive closure of size less than $\beth_\kappa$. In fact, this holds for every limit ordinal $\kappa$: if $x \in R(\kappa)$ then $x \in R(\alpha)$ for some $\alpha &amp;lt; \kappa$ and hence the transitive closure of $x$ has size at most $|R(\alpha)| \leq \beth_\alpha &amp;lt; \beth_\kappa$.&lt;/p&gt;&#xA;" OwnerUserId="25" LastEditorUserId="25" LastEditorDisplayName="" LastEditDate="2010-09-27T19:03:13.797" LastActivityDate="2010-09-27T19:03:13.797" CommentCount="3" />
  <row Id="5557" PostTypeId="2" ParentId="5550" CreationDate="2010-09-27T11:41:01.770" Score="2" ViewCount="" Body="&lt;p&gt;The &quot;why doesn't ... decrease its rank&quot; is because it is hard to decrease rank.  The rank is almost everywhere at its highest possible value, with special exceptions that are destroyed by perturbation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There will be a maximum rank $r(f) \quad$ that holds at almost all points, and possibly some lower-rank points contained in lower-dimensional pieces of the domain of $f$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lower rank is an unstable property. Perturbation away from a low-rank point destroys low rank and replaces it by maximum rank, unless you perturb in special directions that remain inside the lower-dimensional locus where the rank is below maximum.   At the &quot;generic point&quot;, the rank is the highest one possible, $r(f)$, and for any point, all or (in the case of low-rank points, almost all) of the nearby points to which one can perturb, will be generic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As in Robin Chapman's answer this is because low rank requires some functions (determinants) computed from $Df$ to equal zero, while maximum rank requires them to be nonzero.  The first property is unstable and the second is stable with respect to perturbation.  It is also true that the rank cannot decrease through a small perturbation, because the set of determinants that are nonzero can only expand as one moves to a close enough nearby point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Sard%27s_theorem&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Sard%27s_theorem&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="467" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-09-27T12:13:18.800" LastActivityDate="2010-09-27T12:13:18.800" />
  <row Id="5558" PostTypeId="1" AcceptedAnswerId="5570" CreationDate="2010-09-27T11:44:31.777" Score="6" ViewCount="354" Body="&lt;p&gt;Please suggest an approach for this task.&lt;/p&gt;&#xA;" OwnerUserId="2109" LastEditorUserId="2109" LastEditorDisplayName="" LastEditDate="2010-11-11T15:13:18.873" LastActivityDate="2010-11-11T15:13:18.873" Title="Find the sum to n terms of the series $\displaystyle \frac{1} {1.2.3.4} + \frac{1} {2.3.4.5} + \frac{1} {3.4.5.6} $..." Tags="&lt;algebra-precalculus&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="2" />
  <row Id="5559" PostTypeId="1" CreationDate="2010-09-27T11:45:57.837" Score="0" ViewCount="90" Body="&lt;p&gt;Hi. I'm looking for some FOSS/GPL programs (or Python libraries) for dealing with sparse matrices. I haven't found much online about these. Can someone please point me in the right direction?&lt;/p&gt;&#xA;" OwnerUserId="2163" LastActivityDate="2010-09-27T14:12:25.760" Title="Good computer programs for dealing with sparse matrices" Tags="&lt;matrices&gt;&lt;math-software&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="5560" PostTypeId="2" ParentId="5559" CreationDate="2010-09-27T11:59:15.413" Score="2" ViewCount="" Body="&lt;p&gt;I haven't really seen any serious numerical computing being done in Python; FORTRAN is still standard, and &lt;a href=&quot;http://www.netlib.org/&quot; rel=&quot;nofollow&quot;&gt;Netlib&lt;/a&gt; has a number of (free!) libraries for solving sparse linear systems, sparse eigenproblems, ... etc. You are of course welcome to construct an interface from Python to these, or probably even better, translate these into Python. Good luck!&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-09-27T11:59:15.413" />
  <row Id="5561" PostTypeId="2" ParentId="5559" CreationDate="2010-09-27T12:13:18.613" Score="2" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://www.caam.rice.edu/software/ARPACK/&quot; rel=&quot;nofollow&quot;&gt;&lt;strong&gt;ARPACK&lt;/strong&gt;&lt;/a&gt; is the most common and effective solution to problems involving large matrices (both sparse and dense). It's a FORTRAN library (with C++ bindings) and a very stable/mature one at that. I know personally that it is used extensively in academia, so I strongly recommend you give it a go.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In terms of Python, I believe &lt;a href=&quot;http://www.scipy.org/&quot; rel=&quot;nofollow&quot;&gt;SciPy&lt;/a&gt;/NumPY actually includes wrappers for ARPACK, though you'll need to investigate this a bit more. The actual implementation is written in C, so efficiency is not at all a problem.&lt;/p&gt;&#xA;" OwnerUserId="56" LastEditorUserId="56" LastEditorDisplayName="" LastEditDate="2010-09-27T14:12:25.760" LastActivityDate="2010-09-27T14:12:25.760" CommentCount="5" />
  <row Id="5563" PostTypeId="1" CreationDate="2010-09-27T12:34:55.677" Score="0" ViewCount="151" Body="&lt;p&gt;hi everybody could you people give some of shortcut methods to do algebra calculation&lt;/p&gt;&#xA;" OwnerUserId="1328" LastActivityDate="2010-09-27T12:34:55.677" ClosedDate="2010-09-27T15:14:29.307" Title="easy way to do algebra calculations?" Tags="&lt;algebra&gt;" CommentCount="4" FavoriteCount="1" />
  <row Id="5564" PostTypeId="1" AcceptedAnswerId="5692" CreationDate="2010-09-27T12:37:06.477" Score="8" ViewCount="396" Body="&lt;p&gt;I am reading a Set Theory book by Kunen. He presents first-order logic and claims that if a set of sentences in inconsistent, then it proves every possible sentence. Since he does not explicitly specify the inference rules, I became curious as to how fundamental this property of inconsistent systems is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So my question is what is the simplest proof, with the least use of assumptions, of the vague claim that &quot;inconsistent systems can prove anything&quot; - in particular I'm interested in the assumptions about the system needed to prove this - is it true only for first order logic? Only for first order logic with the &quot;standard&quot; rules of inference (Modus ponens and GEN)? Or is it such a basic truth that it can be proved for every &quot;reasonable&quot; proof system (and what is &quot;reasonable&quot;)?&lt;/p&gt;&#xA;" OwnerUserId="1818" LastActivityDate="2010-09-29T07:40:45.423" Title="Why an inconsistent formal system can prove everything?" Tags="&lt;logic&gt;&lt;proof-theory&gt;" AnswerCount="4" CommentCount="7" FavoriteCount="2" />
  <row Id="5565" PostTypeId="2" ParentId="5564" CreationDate="2010-09-27T12:51:36.213" Score="2" ViewCount="" Body="&lt;p&gt;From my recollection, it goes something like this: if $K$ is a first-order inconsistent theory, there exists, by definition, a formula $C$ such that $\vdash_{K} C$ and $\vdash_{K}\neg C$. If $D$ is an arbitrary formula in $K$ then we have the following chain (proof):&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$C$&lt;/li&gt;&#xA;&lt;li&gt;$\neg C$&lt;/li&gt;&#xA;&lt;li&gt;$\neg C \Rightarrow (C\Rightarrow D)$ (Tautology)&lt;/li&gt;&#xA;&lt;li&gt;$C \Rightarrow D$ (2,3, Modus Ponens)&lt;/li&gt;&#xA;&lt;li&gt;$D$ (1,4, Modus Ponens).&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Now, if inconsistency is defined the same way for logics of any other order, I think this proof wouldn't change. I don't see anything explicit about first-orderness in it, but correct me if I'm wrong.&lt;/p&gt;&#xA;" OwnerUserId="1313" LastActivityDate="2010-09-27T12:51:36.213" CommentCount="2" />
  <row Id="5566" PostTypeId="2" ParentId="5564" CreationDate="2010-09-27T13:18:40.877" Score="5" ViewCount="" Body="&lt;p&gt;If $T$ is an inconsistent set of first order theorems (or axioms for the proof), then it is possible to prove from $T$ for some $\alpha$ both $\alpha$ and $\neg \alpha$. So without the loss of generality we can assume that $T$ includes $\alpha$ and $\neg\alpha$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now suppose that $\beta$ is whatever first order sentence that you want to prove.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$\alpha$ (in $T$)&lt;/li&gt;&#xA;&lt;li&gt;$\beta \to \alpha$ (easily verified to be true since $\alpha$ is an axiom of $T$)&lt;/li&gt;&#xA;&lt;li&gt;$\neg \alpha \to \neg \beta$ &lt;/li&gt;&#xA;&lt;li&gt;$\neg \alpha$ (axiom of $T$)&lt;/li&gt;&#xA;&lt;li&gt;$\neg \beta$ (inferred from 3 &amp;amp; 4)&lt;/li&gt;&#xA;&lt;li&gt;$\neg \beta \to \alpha$ (holds for the same reason in 2, further more we have $\neg\beta$)&lt;/li&gt;&#xA;&lt;li&gt;$\neg \alpha \to \neg\neg\beta$&lt;/li&gt;&#xA;&lt;li&gt;$\neg \neg \beta$ (inferred from 4,7)&lt;/li&gt;&#xA;&lt;li&gt;$\neg \neg \beta \to \beta$ (tautology)&lt;/li&gt;&#xA;&lt;li&gt;$\beta$ (inferred from 8,9)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;So you see, you can prove pretty much anything you want from $\{\alpha ,\neg\alpha\}$ for some first order sentence $\alpha$.&lt;/p&gt;&#xA;" OwnerUserId="622" LastEditorUserId="622" LastEditorDisplayName="" LastEditDate="2010-09-27T14:17:08.203" LastActivityDate="2010-09-27T14:17:08.203" CommentCount="5" />
  <row Id="5567" PostTypeId="2" ParentId="5540" CreationDate="2010-09-27T13:44:48.540" Score="7" ViewCount="" Body="&lt;p&gt;This is obvious by ascent in the &lt;a href=&quot;http://mathoverflow.net/questions/33697/assistance-with-understanding-parent-child-relationships-in-pythagorean-triples/33726#33726&quot; rel=&quot;nofollow&quot;&gt;ternary tree of Pythagorean triples&lt;/a&gt;. Explicitly, ascent  yields this formula:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm (x,x+1,z) \to (X,X+1,Z),\ \ X = 3x+2z+1,\ \ Z = 4x+3z+2\:.\ \ $  For example  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm (3,4,5)\to (20, 21, 29)\to (119, 120, 169)\to (696, 697, 985)\to (4059, 4060, 5741)\to\cdots$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This corresponds to always taking the middle branch in the tree, see below&#xA;&lt;img src=&quot;http://i27.tinypic.com/ohimfs.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;br&gt;&#xA;The reflection that yields the descent in the triples tree is the following&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\quad\quad (x,y,z)\; \mapsto (x,y,z) - 2 \dfrac{(x,y,z)\cdot(1,1,1)}{(1,1,1)\cdot(1,1,1)} (1,1,1)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\quad\quad\quad\quad\quad\quad = (x,y,z) - 2 \; (x+y-z) \; (1,1,1)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\quad\quad\quad\quad\quad\quad = (-x-2y+2z, \; -2x-y+2z, \; -2x-2y+3z)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We ascend the tree by inverting this reflection, combined with trivial sign-changing reflections:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\quad\quad (-3,+4,5) \mapsto (-3,+4,5) - 2 \; (-3+4-5) \; (1,1,1) = ( 5,12,13)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\quad\quad (-3,-4,5) \mapsto (-3,-4,5) - 2 \; (-3-4-5) \; (1,1,1) = (21,20,29)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\quad\quad (+3,-4,5) \mapsto (+3,-4,5) - 2 \; (+3-4-5) \; (1,1,1) = (15,8,17)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Continuing in this way enables one to reflectively generate the entire tree of primitive Pythagorean triples. This has a beautiful geometric interpretation in terms of refelections - see my said &lt;a href=&quot;http://mathoverflow.net/questions/33697/assistance-with-understanding-parent-child-relationships-in-pythagorean-triples/33726#33726&quot; rel=&quot;nofollow&quot;&gt;MathOverflow post&lt;/a&gt; and &lt;a href=&quot;http://math.stackexchange.com/questions/15977/proofs-that-every-natural-number-is-a-sum-of-four-squares/15999#15999&quot;&gt;see here&lt;/a&gt; for sums of four squares.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2011-04-04T05:50:21.750" LastActivityDate="2011-04-04T05:50:21.750" CommentCount="0" />
  <row Id="5568" PostTypeId="1" AcceptedAnswerId="5575" CreationDate="2010-09-27T14:02:17.930" Score="4" ViewCount="338" Body="&lt;p&gt;&lt;a href=&quot;http://mathworld.wolfram.com/CatalansConstant.html&quot; rel=&quot;nofollow&quot;&gt;Catalan's Constant&lt;/a&gt; and quite a few other mathematical constants are known to have an infinite continued fraction (see the bottom of that webpage). On  wikipedia (I'm sorry, I can't post anymore hyperlinks because of my low rep.), a condition for irrationality of that continued fraction is given (see 'generalized continued fractions'). It is said that a given continued fraction converges to an irrational limit if $ b_n &gt; a_n $ in the continued fraction $b_0 + a_0/(b_1 + a_1/(b_2 + a_2/(...b_n+a_n)))$ for some sufficiently large $n$. In the webpage I provided you with, however, the degree of the polynomial $a_n$ of the continued fractions is bigger than the one of $b_n$. Therefore, all values of $b_n$ will never exceed all values of $a_n$, even after some large $n$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is: Why does the degree of $a_n$ need to be smaller than the degree of $b_n$ in order for a continued fraction representation of a constant to be irrational? I think I read somewhere it had to do with something like 'Tietschzes criterion' (but I'm not sure). (bonus question: Does anyone know where a proof of this 'criterion for irrationality' can be found?) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Max Muller&lt;/p&gt;&#xA;" OwnerUserId="1055" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2011-01-06T18:55:06.253" LastActivityDate="2011-01-06T18:55:06.253" Title="Why are some mathematical constants irrational by their continued fraction while others aren't ?" Tags="&lt;number-theory&gt;&lt;reference-request&gt;&lt;continued-fractions&gt;&lt;irrational-numbers&gt;&lt;approximation-theory&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="5569" PostTypeId="2" ParentId="5540" CreationDate="2010-09-27T14:02:39.833" Score="1" ViewCount="" Body="&lt;p&gt;Just to add to Derek's excellent answer,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pell's equation becomes essential in getting &lt;em&gt;all&lt;/em&gt; the solutions to $$k^2 + (k+1)^2 = j^2$$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $gcd(k,k+1) =1 $ we may assume that (by the formula for all primitive pythagorean triples) that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$m^2 - n^2 = 2mn \pm 1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a quadratic in $m$ and solving for $m$ gives us&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ m = n \pm \sqrt{2n^2 \pm 1}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus any solution of $$k^2 + (k+1)^2 = j^2$$ can be used to get a solution to the equation $$2n^2 \pm 1 = b^2$$ and vice versa.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-09-27T14:02:39.833" />
  <row Id="5570" PostTypeId="2" ParentId="5558" CreationDate="2010-09-27T14:03:43.960" Score="8" ViewCount="" Body="&lt;p&gt;$\dfrac{1}{k(k+1)(k+2)(k+3)} = \dfrac{1}{3} (\dfrac{k+3}{k(k+1)(k+2)(k+3)} - \dfrac{k}{k(k+1)(k+2)(k+3)})$&#xA;$ = \dfrac{1}{3}(\dfrac{1}{k(k+1)(k+2)} - \dfrac{1}{(k+1)(k+2)(k+3)})$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sum_{k=1}^{\infty}\dfrac{1}{k(k+1)(k+2)(k+3)} = \dfrac{1}{3} \dfrac{1}{2*3} = \dfrac{1}{18}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;@moron Yes, you are right. For the first n terms:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sum_{k=1}^{n}\dfrac{1}{k(k+1)(k+2)(k+3)} = \dfrac{1}{3} (\dfrac{1}{1*2*3} - \dfrac{1}{(n+1)(n+2)(n+3)})$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[edit] more details&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sum_{k=1}^{n}\dfrac{1}{k(k+1)(k+2)(k+3)} = \sum_{k=1}^{n} \dfrac{1}{3} (\dfrac{1}{k(k+1)(k+2)} - \dfrac{1}{(k+1)(k+2)(k+3)})$&#xA;$= \dfrac{1}{3} [\sum_{k=1}^{n} \dfrac{1}{k(k+1)(k+2)} - \sum_{k=1}^{n} \dfrac{1}{(k+1)(k+2)(k+3)}]$&#xA;$= \dfrac{1}{3} [\sum_{k=0}^{n-1} \dfrac{1}{(k+1)(k+2)(k+3)} - \sum_{k=1}^{n} \dfrac{1}{(k+1)(k+2)(k+3)}]$&#xA;$= \dfrac{1}{3} (\dfrac{1}{1*2*3} - \dfrac{1}{(n+1)(n+2)(n+3)})$&#xA;&lt;/p&gt;&#xA;" OwnerUserId="2164" LastEditorUserId="2164" LastEditorDisplayName="" LastEditDate="2010-09-27T16:29:40.460" LastActivityDate="2010-09-27T16:29:40.460" CommentCount="8" />
  <row Id="5571" PostTypeId="2" ParentId="5549" CreationDate="2010-09-27T14:18:46.397" Score="1" ViewCount="" Body="&lt;p&gt;Put $\rm\ A = a^{n-1},\ B = b^{n-1}\:$ below and the results follow immediately&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\displaystyle\quad \frac{a\:A+b\:B}{A+B}\ =\ \frac{a+b}2\ \iff\ (a-b)\ (A-B)\ =\ 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\displaystyle\quad \frac{a\:A+b\:B}{A+B}\ = \ \ \sqrt{ab}\ \ \ \:\iff\ (\sqrt a - \sqrt b)\ (A\sqrt{a}-B\sqrt{b})\ =\ 0$&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-10-06T15:45:41.487" LastActivityDate="2010-10-06T15:45:41.487" />
  <row Id="5572" PostTypeId="2" ParentId="5558" CreationDate="2010-09-27T14:49:50.117" Score="7" ViewCount="" Body="&lt;p&gt;This is similar to what has been said by Branimir, but shows how we can extend the result to&#xA;$$\sum_{k=1}^n {1 \over k(k+1) \cdots (k+m)}, \qquad m \in \mathbb{N}.$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can build up the result from the identities&#xA;$${1 \over k(k+1)} = {1 \over k} - { 1 \over k+1}, \qquad (1)$$&#xA;$${1 \over k(k+1)(k+2)} = {1 \over 2} \left( {1 \over k(k+1)} - { 1 \over (k+1)(k+2)} \right),$$&#xA;$${1 \over k(k+1)(k+2)(k+3)} = &#xA;{1 \over 3} \left( {1 \over k(k+1)(k+2)} - { 1 \over (k+1)(k+2)(k+3)} \right), \quad \textrm{ etc...}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Write $S_1 = \sum_{k=1}^n {1 \over k(k+1)},$&#xA;$S_2 = \sum_{k=1}^n {1 \over k(k+1)(k+2)},$ etc&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Summing for $S_1$ using (1) all terms on RHS cancel to get the classic&#xA;$$S_1 = \sum_{k=1}^n {1 \over k(k+1)} = 1 – {1 \over n+1} = {n \over n+1}.$$&#xA;We then sum the series for $S_2$ using this result obtained for $S_1,$ and so on.&lt;/p&gt;&#xA;" OwnerUserId="1301" LastActivityDate="2010-09-27T14:49:50.117" CommentCount="1" />
  <row Id="5573" PostTypeId="2" ParentId="5558" CreationDate="2010-09-27T14:54:16.097" Score="4" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\rm\displaystyle\ \frac{1}{(k+1)(k+2)(k+3)(k+4)} = \frac{1}{6(k+1)} - \frac{1}{2(k+2)}+\frac{1}{2(k+3)}-\frac{1}{6(k+4)}$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\ f(k+1)-f(k)\: = $ above $\rm\displaystyle\  \Rightarrow\  f(k) \:=\: c_0 + \frac{c_1}{k+1}\ \:+\:\ \frac{c_2}{k+2}\ \:+\:\ \frac{c_3}{k+3}$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Calculating yields $\rm\ c_0,c_1,c_2,c_3 \ =\ 1/18,\ -1/6,\ 1/3,\ -1/6$.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For remarks on the group theory behind rational indefinite summation see &lt;a href=&quot;http://math.stackexchange.com/questions/4150&quot;&gt;my post here&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-09-27T14:54:16.097" />
  <row Id="5574" PostTypeId="1" AcceptedAnswerId="5580" CreationDate="2010-09-27T15:19:08.110" Score="3" ViewCount="156" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $d(n)$ represent the divisor function as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$d(n)=\displaystyle\sum\limits_{k|n}1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and the divisor summatory function as &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$D(x)=\displaystyle\sum\limits_{n \leq x}d(n)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found the following triangular representation for the values of $D(n)$&#xA;&lt;img src=&quot;http://i.imgur.com/J0ZFW.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The values on the right are the sum of all elements in a row.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT 1&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The above picture is the result of the following observation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $v_{m}(n)$ be the greatest power of $m$ that divides $n$ with $ m,n \in  \mathbb{N}$  , so we get that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$D(n)=\displaystyle\sum\limits_{m=2}^{\infty}v_{m}(p^{n}), p \in  \mathbb{P}$ where $p$ is a fixed prime number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I didn't try to prove this. I don't know how to do it, but hopefully some one will have some idea on how to prove or disprove this conjecture.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I'd like to know if this is a known fact. I don't have a proof but I've tested lots of values and woks all the time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="1747" LastEditorUserId="1747" LastEditorDisplayName="" LastEditDate="2010-09-27T20:37:39.393" LastActivityDate="2010-09-27T20:37:39.393" Title="A triangular representation for the divisor summatory function, $D(x)$" Tags="&lt;number-theory&gt;" AnswerCount="1" CommentCount="10" />
  <row Id="5575" PostTypeId="2" ParentId="5568" CreationDate="2010-09-27T15:22:45.500" Score="5" ViewCount="" Body="&lt;p&gt;You can find the &lt;a href=&quot;http://www.maths.unsw.edu.au/~angell/5535/chapter7.ps&quot; rel=&quot;nofollow&quot;&gt;proofs here&lt;/a&gt;, taken from Chrystal's Algebra - which is one of the best references on continued fractions. I suspect that if you read Chrystal then almost all of your questions will be answered. See also chapter 9 of Fowler's &quot;The mathematics of Plato's Academy&quot; &lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-09-27T16:13:29.007" LastActivityDate="2010-09-27T16:13:29.007" CommentCount="4" />
  <row Id="5576" PostTypeId="2" ParentId="5537" CreationDate="2010-09-27T15:28:44.587" Score="4" ViewCount="" Body="&lt;p&gt;One complication in your situation is that you are mixing real and complex coordinates.&lt;br&gt;&#xA;If you were considering a map from a complex curve to $\mathbb{CP}^1$, then the kind of computation you are trying to make would work out more straightforwardly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because you are looking at a map of real analytic manifolds, not complex analytic ones&#xA;(concretely, you are working with the variables $x,y,z$, which are real coordinates), the&#xA;point of view you have adopted is perhaps not quite as natural.  Nevertheless, it can be&#xA;made to work, as follows:&#xA;$$[x+iy:1-z] \text{ (which is where you finished) }&#xA;= [x^2 + y^2: (1-z)(x - i y)]$$&#xA;$$ = [ 1 - z^2: (1-z)(x-iy)] = [1+z:(x-iy)].$$&#xA;This rewriting of your map to $\mathbb{CP}^1$ is now well-defined in a neighbourhood of $(0,0,1)$ on the sphere.  (The fact that I introduced a complex conjugate of $x + i y$ to&#xA;facilitate the computation is related to the real vs. complex issue mentioned above.  This is also essentially the same computation you made to check that your map tends to $\infty$ as $z \to 1$, just rewritten in homogeneous coordinates.)&lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2010-09-27T15:28:44.587" CommentCount="2" />
  <row Id="5577" PostTypeId="1" AcceptedAnswerId="5585" CreationDate="2010-09-27T15:57:58.893" Score="2" ViewCount="171" Body="&lt;p&gt;According &lt;a href=&quot;http://oeis.org/A180926&quot; rel=&quot;nofollow&quot;&gt;A180926&lt;/a&gt;, the elements of the set&#xA;{$a:\exists m,n|60a=5n^2+5n=3m^2+3m$} satisfy the following recurrence relation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$a_{n}=\frac{62a_{n-1}+1+\sqrt{(48a_{n-1}+1)(80a_{n-1}+1)}}{2}$$ &#xA;$$a_1=0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any idea how this can be derived?&lt;/p&gt;&#xA;" OwnerUserId="1778" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-09-27T16:42:25.387" LastActivityDate="2010-09-28T15:50:39.207" Title="Recurrence relation for $a_{n}$ where $6a_{n}$ and $10a_{n}$ are both triangular." Tags="&lt;number-theory&gt;&lt;diophantine-equations&gt;" AnswerCount="2" CommentCount="7" />
  <row Id="5578" PostTypeId="2" ParentId="1519" CreationDate="2010-09-27T16:30:14.270" Score="3" ViewCount="" Body="&lt;p&gt;Tom, I'm not sure I see how it's getting any harder in passing from a torus to a projective space.  In your $\mathbb CP^2$ case, you have $\mathbb CP^1$ sitting inside of it, and the boundary of a regular (tubular) neighbourhood of the $\mathbb CP^1$ is $S^3$.  And $D^4$ has $S^3$ as its boundary, so the attaching map is tautological.  The normal bundle is the missing data and that's what your CW-decomposition is ignoring. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is essentially what always happens.  Perhaps the conceptual hump you're dealing with is that you're asking for CW-decompositions of manifolds.  Morse functions generically only build homotopy-equivalences to CW-complexes, they do not put CW-structures on the manifold without some significant work. Moreover, CW-decompositions ignore some of the most essential properties of the manifold, like smooth structures.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If instead you work with handle decompositions, what I state in my first paragraph is basically a generality -- critical points amount to handle attachments and the gluing instructions are always given in a direct way from the flow lines of the Morse function's (suitably normalized) gradient. So the handle decomposition is &lt;em&gt;on&lt;/em&gt; the given manifold -- unlike the CW-case where you only have a homotopy-equivalence to a CW-complex. &lt;/p&gt;&#xA;" OwnerUserId="642" LastActivityDate="2010-09-27T16:30:14.270" />
  <row Id="5579" PostTypeId="1" CreationDate="2010-09-27T16:30:23.987" Score="1" ViewCount="103" Body="&lt;p&gt;In blind experiments subjects are randomly assigned to one of groups.  The most commonly used solution is to use (equivalent of) a coin toss, with the same probability to be assigned to each group.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have read about two modifications of a procedure of random assignment in blind tests:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;fair / just coin&lt;/strong&gt;, where probability of assignment to group A is proportional to the number of subjects assigned to other group (group B) divided by number of all assigned subjects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This solution is meant to have more equal distribution of subjects to groups (as truly random coin can result in one group to be over-represented).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is known as &lt;strong&gt;covariate-minimization randomization&lt;/strong&gt;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;result-based probability&lt;/strong&gt;, where probability of assignment to given group is proportional to how well this group performs in experiment (e.g. in drug test).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is known as &lt;strong&gt;response-adaptive randomization&lt;/strong&gt;, or &lt;strong&gt;outcome-adaptive randomization&lt;/strong&gt;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;How based on results of such not fully random assignment method can one recover (calculate) the same estimates and error bounds as for random sampling?&lt;/p&gt;&#xA;" OwnerUserId="2167" LastActivityDate="2010-09-27T17:12:50.193" Title="Random assignment in blind experiments and &quot;fair / just coin&quot;" Tags="&lt;probability-theory&gt;&lt;statistics&gt;&lt;random&gt;" AnswerCount="1" />
  <row Id="5580" PostTypeId="2" ParentId="5574" CreationDate="2010-09-27T16:42:24.107" Score="4" ViewCount="" Body="&lt;p&gt;Yes, this is true.  Write $D(x) = \sum_{n \le x} d(n) = \sum_{n \le x} \sum_{d | n} 1 = \sum_{d \le x} \lfloor \frac{x}{d} \rfloor$; this is equivalent to the pattern you observe.  The last step is exchanging the order of summation together with the observation that the number of times a number $d$ appears in the double sum is the number of numbers less than or equal to $x$ it divides.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-09-27T16:42:24.107" CommentCount="1" />
  <row Id="5581" PostTypeId="1" CreationDate="2010-09-27T16:46:08.483" Score="0" ViewCount="304" Body="&lt;p&gt;(Wade's rewording)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a phrase that contains 4 words and 22 total letters (not counting the three spaces between the words). One letter appears four times, no letter appears three times, three letters appear twice, and 12 letters appear once each. What is this phrase?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;======================================&lt;/p&gt;&#xA;&#xA;&lt;p&gt;@@$$##&amp;amp;&amp;amp;&amp;amp;%%%%????????????&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are 4 words and the &amp;amp; represents a whitespace.&lt;br&gt;&#xA;@,#,$,% represent one letter each (4 total letters).&lt;br&gt;&#xA;Each ? represents one of the other 22 letter of the alphabet (12 total letters).&lt;br&gt;&#xA;16 different letters used.&lt;br&gt;&#xA;ciphertext also needs to be rearranged - it is 'scrambled.'  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do I approach this?&lt;br&gt;&#xA;Not looking for the answer really.&lt;br&gt;&#xA;I just need a tool to unscramble the phrase and decrypt the ciphertext simultaneously.&lt;br&gt;&#xA;Decrypto only does the later. Thank you.    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;@Noldorin: This is not homework. From elsewhere. &#xA;@Kenny: How are encryption/decryption methodologies not math? This is an applied math problem, whereas your preferences tend to be theoretical/ pure math?&lt;/p&gt;&#xA;" LastActivityDate="2010-09-27T21:12:34.590" Title="Solve cryptogram - ciphertext given" Tags="&lt;cryptography&gt;" AnswerCount="2" CommentCount="6" FavoriteCount="0" />
  <row Id="5582" PostTypeId="2" ParentId="4963" CreationDate="2010-09-27T16:57:56.233" Score="1" ViewCount="" Body="&lt;p&gt;@ fiktor: due to the low amount of reputation points I have at the moment, I can't comment yet. I just wanted to point out that, since I suspect this is a research-level question, there's a website called www.mathoverflow.com which might be an (even) better platform to launch this question from.  &lt;/p&gt;&#xA;" OwnerUserId="1055" LastEditorUserId="1055" LastEditorDisplayName="" LastEditDate="2010-09-27T18:24:31.163" LastActivityDate="2010-09-27T18:24:31.163" CommentCount="1" />
  <row Id="5583" PostTypeId="2" ParentId="5579" CreationDate="2010-09-27T17:12:50.193" Score="2" ViewCount="" Body="&lt;p&gt;Here's a paper that may be helpful regarding your second point: &lt;a href=&quot;http://www.johndcook.com/power_bias.pdf&quot; rel=&quot;nofollow&quot;&gt;Power and bias in adaptively randomized clinical trials&lt;/a&gt;.  When looked at from the Bayesian standpoint, you simply compute posterior probabilities.  The randomization mechanism doesn't matter.  If you want to get frequentist operating characteristics, you can run simulations.&lt;/p&gt;&#xA;" OwnerUserId="247" LastActivityDate="2010-09-27T17:12:50.193" />
  <row Id="5584" PostTypeId="2" ParentId="4983" CreationDate="2010-09-27T17:19:39.553" Score="2" ViewCount="" Body="&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Erd%C5%91s%E2%80%93Kac_theorem&quot; rel=&quot;nofollow&quot;&gt;Erdős–Kac theorem&lt;/a&gt; shows that the (log of the log of the) prime factors of a number are Poisson/normally distributed.&lt;/p&gt;&#xA;" OwnerUserId="1778" LastActivityDate="2010-09-27T17:19:39.553" CommunityOwnedDate="2010-09-27T17:19:39.553" />
  <row Id="5585" PostTypeId="2" ParentId="5577" CreationDate="2010-09-27T17:22:00.967" Score="5" ViewCount="" Body="&lt;p&gt;Since we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ 5(n^2 +n) = 3(m^2 +m)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;multiplying by $4$ gives us&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$5(4n^2 + 4n + 1) - 5 = 3(4m^2 + 4m + 1) -3$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;if $x = 2n+1$ and $y = 2m+1$ we get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$5x^2 - 3y^2 = 2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a generalized Pell's equation of the form $ax^2 - by^2 = c$ and I believe this is a solved problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, see this: &lt;a href=&quot;http://books.google.com/books?id=YD6UlfcRlGQC&amp;amp;pg=PA81&quot; rel=&quot;nofollow&quot;&gt;http://books.google.com/books?id=YD6UlfcRlGQC&amp;amp;pg=PA81&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This should give us all the solutions, and so make the recurrence easy to verify.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-09-27T17:22:00.967" CommentCount="8" />
  <row Id="5586" PostTypeId="1" AcceptedAnswerId="5589" CreationDate="2010-09-27T18:22:43.087" Score="6" ViewCount="145" Body="&lt;p&gt;Suppose I embed a manifold-with-boundary $M$ in some $\mathbb{R}^n$.  Are there conditions (necessary, sufficient, or both) that can help determine when the topological boundary of $M$ is equal to the manifold boundary?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By &quot;topological boundary,&quot; I'm referring to $\text{Bd } M$, which is the closure minus the interior (relative to $\mathbb{R}^n$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By &quot;manifold boundary,&quot; I mean the boundary $\partial M$ that is specified in the definition of &quot;manifold-with-boundary.&quot;&lt;/p&gt;&#xA;" OwnerUserId="640" LastEditorUserId="640" LastEditorDisplayName="" LastEditDate="2011-01-02T06:27:46.397" LastActivityDate="2011-01-02T06:27:46.397" Title="When does the topological boundary of an embedded manifold equal its manifold boundary?" Tags="&lt;topology&gt;&lt;differential-geometry&gt;&lt;differential-topology&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="5587" PostTypeId="1" AcceptedAnswerId="5590" CreationDate="2010-09-27T18:25:08.720" Score="1" ViewCount="71" Body="&lt;p&gt;The title and tags are a little goofy since I'm not quite sure what to call this.  This may be more of a computer science question, but I thought I'd ask here too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have two integers n and m that are arbitrarily (logically) related with integers x and y, respectively, where all values are finite and &gt; 0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd like to know if there's a way to combine n and x, and m and y, into two numbers (integer, decimal, whatever) such that neither will overlap.  Additionally, I need to be able to perform some kind of operation to reverse the previous process to once again obtain n and x, and m and y.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me know if this isn't clear.  Thanks in advance!&lt;/p&gt;&#xA;" OwnerUserId="2170" LastActivityDate="2010-09-27T18:39:56.247" Title="Related Integer Combination" Tags="&lt;computer-science&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="5588" PostTypeId="1" AcceptedAnswerId="5592" CreationDate="2010-09-27T18:25:52.523" Score="4" ViewCount="229" Body="&lt;p&gt;I recently learned about the Classification Theorem for compact 2-manifolds.  Is there a similar classification theorem for ALL 2-manifolds, not just the compact ones?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moreover, is there a theorem which classifies the 2-manifolds with boundary?&lt;/p&gt;&#xA;" OwnerUserId="640" LastActivityDate="2010-09-28T05:00:44.507" Title="Classification Theorem for Non-Compact 2-Manifolds? 2-Manifolds With Boundary?" Tags="&lt;topology&gt;&lt;differential-topology&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="1" />
  <row Id="5589" PostTypeId="2" ParentId="5586" CreationDate="2010-09-27T18:31:50.703" Score="5" ViewCount="" Body="&lt;p&gt;If you embed a smooth $m$-manifold $M$ smoothly in $\mathbb{R}^n$&#xA;then the &quot;topological boundary&quot; of $M$ is the closure of $M$. As&#xA;locally each point of $M$ has a neighbourhood in $\mathbb{R}^n$&#xA;where $M$ looks like $\mathbb{R}^m$, then no point of $M$ is interior.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When $m=n$ and $M$ is compact then, yes (at least in the smooth case)&#xA;the topological and manifold boundaries coincide.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I expect the above hold for topological embeddings but won't swear&#xA;to it; they need not be locally flat (see nasties like the Alexander&#xA;horned sphere),&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-09-27T18:31:50.703" CommentCount="2" />
  <row Id="5590" PostTypeId="2" ParentId="5587" CreationDate="2010-09-27T18:39:23.933" Score="3" ViewCount="" Body="&lt;p&gt;You could employ a pairing function, e.g. &lt;a href=&quot;http://en.wikipedia.org/wiki/Pairing_function&quot; rel=&quot;nofollow&quot;&gt;Cantor's&lt;/a&gt; or &lt;a href=&quot;http://szudzik.com/ElegantPairing.pdf&quot; rel=&quot;nofollow&quot;&gt;others&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-09-27T18:39:23.933" />
  <row Id="5592" PostTypeId="2" ParentId="5588" CreationDate="2010-09-27T18:56:03.307" Score="6" ViewCount="" Body="&lt;p&gt;Yes, there's a classification theorem for non-compact 2-manifolds.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This paper gives the classification for triangulable 2-manifolds:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.jstor.org/stable/1993768&quot; rel=&quot;nofollow&quot;&gt;http://www.jstor.org/stable/1993768&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That an arbitrary (2nd countable, Hausdorff) topological 2-manifold admits a triangulation is fairly classical.  Ahlfors book &quot;Riemann Surfaces&quot; has a proof.  There are others available, see for example this list:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://mathoverflow.net/questions/17578/triangulating-surfaces&quot; rel=&quot;nofollow&quot;&gt;http://mathoverflow.net/questions/17578/triangulating-surfaces&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If all you're interested in is compact manifolds with boundary, you get that classification immediately from the closed manifold case.  Because if you have a compact manifold with boundary, its boundary is a disjoint union of circles.  So cap those circles off with discs to produce a closed manifold.  So compact manifolds with boundary are classified by the closed manifold you get by &quot;capping off&quot; and the number of boundary circles you started with. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Non-compact manifolds have a more delicate classification -- think for example about the complement of a Cantor set in a compact surface. &lt;/p&gt;&#xA;" OwnerUserId="642" LastEditorUserId="642" LastEditorDisplayName="" LastEditDate="2010-09-28T05:00:44.507" LastActivityDate="2010-09-28T05:00:44.507" CommentCount="2" />
  <row Id="5593" PostTypeId="2" ParentId="5581" CreationDate="2010-09-27T19:10:17.413" Score="6" ViewCount="" Body="&lt;p&gt;Captain Kirk was wondering why his journal was not more excited to hear about his exploits. He said&lt;/p&gt;&#xA;&#xA;&lt;p&gt;LOG, YOU SEEM FAINTHEARTED!&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;More seriously, I dumped the letters EEEETTAAOOINSHRDLUGFMY into a multiword anagrammer, and it immediately found more than 24000 answers, of which this was one of the more amusing ones. What I deduce from this experience is that, without the knowledge that you are looking for a common English phrase, this puzzle is dramatically under constrained. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I recommend you look for a file of common English phrases which you can test for this property. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This isn't math, but there is a bit of a mathematical lesson here -- it is worth doing a quick test to see whether you have enough data to answer a question before working too hard on it.&lt;/p&gt;&#xA;" OwnerUserId="448" LastEditorUserId="448" LastEditorDisplayName="" LastEditDate="2010-09-27T19:36:28.547" LastActivityDate="2010-09-27T19:36:28.547" CommunityOwnedDate="2010-09-27T19:10:17.413" CommentCount="2" />
  <row Id="5594" PostTypeId="2" ParentId="5581" CreationDate="2010-09-27T19:59:23.460" Score="3" ViewCount="" Body="&lt;p&gt;As David Speyer notes, the problem is too unconstrained for a solution as it stands. You can make some guesses using frequency tables for letters in English, but that's about the only place where some math will come in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it were a &quot;straight cryptogram&quot; (the letters are placed in the correct order, whether with the correct word spacing or in blocks), then you are a little closer to a problem in which mathematics can come in. You would have frequency tables, plus also tables on digraphs or trigraphs (blocks of 2 or 3 letters, when certain letters tend to occur next to each other); the two together can give you a lot of information (though the shorter the text, the less useful information they give: you need larger cryptograms to avoid the information getting lost in the &quot;noise&quot;). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to see some of the general procedures to attack &quot;classic&quot; cryptographic systems like monoalphabetic ciphers (where each letter in the original is replaced by another letter, the same one throughout, and the order of the letters is not modified), and polyalphabetic ciphers (where the substitutions depend on the location of the letter; these range from simple &lt;a href=&quot;http://en.wikipedia.org/wiki/Vigen%C3%A8re_cipher&quot; rel=&quot;nofollow&quot;&gt;Vigenère substitution&lt;/a&gt; all the way to substitutions like the &lt;a href=&quot;http://en.wikipedia.org/wiki/Enigma_machine&quot; rel=&quot;nofollow&quot;&gt;Engima cipher&lt;/a&gt;), or transposition systems (where the order of the letters is changed, but the letters themselves are not), then I can recommend &lt;em&gt;Elementary Cryptanalysis: A mathematical approach&lt;/em&gt; by Abraham Sinkov (Mathematical Association of America, New Mathematical Library, ISBN 0883856220). Pretty much all of them are obsolete as far as modern cryptography goes, but it can give you a taste of how (and when) mathematics comes in. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For your problem as stated, however, the role of mathematics is so limited as to become nearly useless. &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-09-27T21:12:34.590" LastActivityDate="2010-09-27T21:12:34.590" CommentCount="6" />
  <row Id="5595" PostTypeId="1" AcceptedAnswerId="5596" CreationDate="2010-09-27T20:15:47.413" Score="11" ViewCount="458" Body="&lt;p&gt;This is a neat little problem that I was discussing today with my lab group out at lunch. Not particularly difficult but interesting implications nonetheless&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine there are a 100 people in line to board a plane that seats 100. The first person in line realizes he lost his boarding pass so when he boards he decides to take a random seat instead. Every person that boards the plane after him will either take their &quot;proper&quot; seat, or if that seat is taken, a random seat instead.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Question: What is the probability that the last person that boards will end up in his/her proper seat.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moreover, and this is the part I'm still pondering about. Can you think of a physical system that would follow this combinatorial statistics? Maybe a spin wave function in a crystal etc...&lt;/p&gt;&#xA;" OwnerUserId="1906" LastActivityDate="2011-04-17T19:09:35.813" Title="Taking Seats on a Plane" Tags="&lt;combinatorics&gt;&lt;probability&gt;" AnswerCount="5" FavoriteCount="3" />
  <row Id="5596" PostTypeId="2" ParentId="5595" CreationDate="2010-09-27T20:30:13.373" Score="13" ViewCount="" Body="&lt;p&gt;This is a classic puzzle!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer is that the probability that the last person ends in up in his proper seat is exactly $\frac{1}{2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reasoning goes as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First observe that the fate of the last person is determined the moment either the first or the last seat is selected! This is because the last person will either get the first seat or the last seat. Any other seat will necessarily be taken by the time the last guy gets to 'choose'.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since at each choice step, the first or last is equally probable to be taken, the last person will get either the first or last with equal probability: $\frac{1}{2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sorry, no clue about a physical system.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-09-27T20:30:13.373" CommentCount="1" />
  <row Id="5597" PostTypeId="2" ParentId="5411" CreationDate="2010-09-27T20:50:46.520" Score="1" ViewCount="" Body="&lt;p&gt;Here's one way to solve the problem. The trick is to substitute &lt;code&gt;-I*Log[x]&lt;/code&gt; for t in the characteristic function $\varphi[t]$. This turns it into a generating function. Then extract the coefficient of $x^n$ from the resulting series using the &lt;b&gt;SeriesCoefficient&lt;/b&gt; built-in. The coefficient of $x^n$ is the value of the pmf f at n. Here's a simple example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;p&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&#xA;        (* Characteristic function (CF) of the discrete probability&#xA;           distribution for summing two 6-sided dice *)&#xA;In[1]:= \[CurlyPhi][t_] :=&#xA;          CharacteristicFunction[DiscreteUniformDistribution[{1, 6}], t]^2&#xA;&#xA;        (* Recover the probability mass function f from this CF *)&#xA;In[2]:= f[n_] := SeriesCoefficient[\[CurlyPhi][-I*Log[x]], {x, 0, n}]&#xA;&#xA;        (* Show the range of the pmf f over its domain, {2,..,12} *)&#xA;In[3]:= f[#] &amp; /@ Range[2, 12]&#xA;Out[3]= {1/36, 1/18, 1/12, 1/9, 5/36, 1/6, 5/36, 1/9, 1/12, 1/18, 1/36}&#xA;&#xA;        (* Note that applying f to a number outside its domain (like 1)&#xA;           produces 0 *)&#xA;In[4]:= f[1]&#xA;Out[4]= 0&#xA;&#xA;        (* Now use this technique to show the range of the pmf of summing three&#xA;           6-sided dice *)&#xA;In[5]:= Map[&#xA;          Function[n, &#xA;            SeriesCoefficient[&#xA;              CharacteristicFunction[&#xA;                DiscreteUniformDistribution[{1, 6}], -I*Log[x]]^3, {x, 0, n}]], &#xA;                  Range[3, 18]]&#xA;Out[5]= {1/216, 1/72, 1/36, 5/108, 5/72, 7/72, 25/216, 1/8, 1/8, 25/216, 7/72,&#xA;          5/72, 5/108, 1/36, 1/72, 1/216}&#xA;&#xA;        (* And verify that the values in the range of this pmf sum to 1 *)&#xA;In[6]:= Fold[Plus, 0, %]&#xA;Out[6]= 1&#xA;&lt;/pre&gt;&#xA;" LastActivityDate="2010-09-27T20:50:46.520" />
  <row Id="5601" PostTypeId="2" ParentId="5568" CreationDate="2010-09-27T22:42:04.467" Score="2" ViewCount="" Body="&lt;p&gt;The continued fraction expansion of any possible real number falls in one of 2 possible categories:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;the continued fraction expansion has a finite length if and only if the number is rational.&lt;/li&gt;&#xA;&lt;li&gt;the continued fraction expansion never ends if and only if the number is irrational.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&quot;quite a few other mathematical constants are known to have an infinite continued fraction&quot;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yes, and all such constants are therefore all irrational.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&quot;Catalan's Constant ... known to have an infinite continued fraction&quot;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do you have a proof? My understanding is that we don't yet know whether or not Catalan's Constant has an finite continued fraction -- therefore we don't yet know whether or not it is rational.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Why does the degree of an need to be smaller than the degree of bn in order for a continued fraction representation of a constant to be irrational?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is required for the sequence to converge on some specific value.&#xA;If the degree of an is &lt;em&gt;larger&lt;/em&gt;, that doesn't mean the result is rational -- it indicates that it doesn't converge to any real number, and therefore isn't a &quot;constant&quot; at all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a way to make the &lt;a href=&quot;http://en.wikipedia.org/wiki/Generalized_continued_fraction#Conditions_for_irrationality&quot; rel=&quot;nofollow&quot;&gt;Wikipedia &quot;generalized continued fraction&quot; article&lt;/a&gt; a little less confusing?&lt;/p&gt;&#xA;" OwnerUserId="1705" LastActivityDate="2010-09-27T22:42:04.467" CommentCount="2" />
  <row Id="5602" PostTypeId="1" AcceptedAnswerId="5605" CreationDate="2010-09-27T23:13:35.370" Score="3" ViewCount="103" Body="&lt;p&gt;I'm reading Jech's Set Theory, and in the chapter about measurable cardinals there is a theorem that if $\kappa$ is real-measurable but not measurable then it is $\le 2^{\aleph_0}$ and so and so. (Corollary 10.10)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can a cardinal number be real-measurable without being measurable? Can't a measure be &quot;destructed&quot; (as in opposite of constructed) into a trivial $0,1$ measure?&lt;/p&gt;&#xA;" OwnerUserId="622" LastActivityDate="2010-09-28T00:00:14.313" Title="Real-measurable cardinals that are not measurable ones" Tags="&lt;set-theory&gt;&lt;cardinals&gt;" AnswerCount="2" />
  <row Id="5603" PostTypeId="2" ParentId="5577" CreationDate="2010-09-27T23:13:57.980" Score="8" ViewCount="" Body="&lt;p&gt;Here's a complete solution, including a derivation of the recurrence relation in &lt;a href=&quot;http://oeis.org/A180926&quot; rel=&quot;nofollow&quot;&gt;A180926&lt;/a&gt;. First, notice&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\ 60\:a \:=\: 5\:(n^2+n)\ =\ 3\:(m^2+m)\ \ $  is easily transformed into the following &lt;a href=&quot;http://en.wikipedia.org/wiki/Pell%27s_equation&quot; rel=&quot;nofollow&quot;&gt;Pell equation&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\quad\quad\quad\ \: 5\:x^2-3\:y^2 \:= \ 2\ \ \ $  for $\rm\ \ \ (x,\:y)\: =\: (2\:n+1,\:2\:m+1)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence $\rm\ \ \ \: x^2 - 1\ =\ 4\: (n^2+n)\ =\ 48\:a,\quad\ y^2 - 1\ =\ 4\: (m^2+m)\ =\ 80\:a\quad\quad\quad\ \ (*)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By &lt;a href=&quot;http://www.alpertron.com.ar/QUAD.HTM&quot; rel=&quot;nofollow&quot;&gt;standard Pell theory&lt;/a&gt; $\rm\ \: (x,y) \:\to\: (X,Y)\:=\: (4\:x+3\:y\ \ ,\ \ 5\:x+4\:y)\ $ is a solution too, since&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\phantom{\quad\Rightarrow\quad}\ X+Y\omega \: =\: (4+5\:\omega)\:(x+y\:\omega) \: = \: (4\:x+3\:y + (5\:x+4\:y)\:\omega) \: ,\ \ \ \omega \:=\: \sqrt{3/5}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\Rightarrow\quad  X- Y\omega \: =\: (4-5\:\omega)\:(x-y\:\omega)\quad\quad\:$ via conjugate prior equation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\Rightarrow\quad  5\: X^2 - 3\: Y^2 \: =\  5\: x^2 - 3\: y^2\: =\: 2\ \ $ via multiply the prior two equations, then scale by 5&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This yields a recurrence to generate a new solution $\rm (X,Y) = (x_{n+1},\:y_{n+1})$ from a known solution $\rm\: (x,\:y) = (x_n,\:y_n)\:.\:$ We square this recurrence to obtain a recurrence for $\rm (A,\:a) = (a_{n+1},\:a_n)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Namely $\rm\ \: (4\:x+3\:y)^2 =\ X^2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\quad\quad\quad \Rightarrow\quad\ 24\ x\:y\ =\ X^2-1 - 16\:(x^2-1) - 9\:(y^2-1) - 24$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\quad\quad\quad\quad \phantom{\Rightarrow\ 24\ x\:y}\ =\ 48\:A - 16\:(48\:a) - 9\:(80\:a) - 24\quad$ by $\ (*)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\quad\quad\quad\quad \phantom{\Rightarrow\ 24\ x\:y}\ =\ 24\ (2\:A - 62\: a - 1)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\quad\quad\quad \Rightarrow\quad\ x^2\ y^2\ =\ (2\:A - 62\: a - 1)^2 $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But also $\rm\quad\quad\quad\: x^2\ y^2\ =\ (48\: a + 1)\ (80\: a + 1)\quad$ by $\ (*)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence $\rm\ \ (2\ a_{n+1} - 62\ a_n - 1)^2 \ = \ (48\ a_n+1)\ (80\ a_n+1)\ \ $  as was to be proved.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-09-28T15:50:39.207" LastActivityDate="2010-09-28T15:50:39.207" CommentCount="2" />
  <row Id="5604" PostTypeId="2" ParentId="5602" CreationDate="2010-09-27T23:36:29.937" Score="2" ViewCount="" Body="&lt;p&gt;It is a good question. The answer is that &lt;a href=&quot;http://en.wikipedia.org/wiki/Measurable_cardinal#Real-valued_measurable&quot; rel=&quot;nofollow&quot;&gt;real-valued measurable cardinal&lt;/a&gt; need not be strongly inaccessible, whilst every measurable cardinal is strongly inaccessible. Indeed, it is consistent that the continuum itself is a real-valued measurable cardinal, but the continuum can never be a measurable cardinal, since every measurable cardinal is strongly inaccessible. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nevertheless, part of what you claim is true: Solovay proved that every real-valued measurable cardinal $\kappa$ is fully measurable (with a two-valued measure) in an inner model of the universe. That is, if $\kappa$ is a real-valued measurable cardinal, then there is a definable transitive class $W$ satisfying ZFC in which $\kappa$ is an actual measurable cardinal. The class $W$ is defined directly from the real-valued measure on $\kappa$, and this provides a sense in which the measure is deconstructed to form a 2-valued measure.&lt;/p&gt;&#xA;" OwnerUserId="413" LastActivityDate="2010-09-27T23:36:29.937" />
  <row Id="5605" PostTypeId="2" ParentId="5602" CreationDate="2010-09-28T00:00:14.313" Score="2" ViewCount="" Body="&lt;p&gt;Two-valued measures behave very differently from real-valued measures. For example, suppose $\mathcal{U}$ is a countably complete ultrafilter on a set $X$ and suppose that $f:X\to2^\omega$ is an injection. There is a $b \in 2^\omega$ such that&#xA;$$B_n = \{ a \in X : f(a)(n) = b(n) \} \in \mathcal{U}$$&#xA;for every $n &amp;lt; \omega$. By countable completeness, $B = \bigcap_{n&amp;lt;\omega} B_n \in \mathcal{U}$. But $B$ contains exactly one element (namely $f^{-1}(b)$) since $f$ is an injection. Therefore, $\mathcal{U}$ is a principal ultrafilter. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This argument shows that the first measurable cardinal is larger than $2^{\aleph_0}$. Indeed, a slightly more general argument can be used to show that a measurable cardinal must be inaccessible. However, this argument cannot be carried out with a real-valued measure. In fact, it is possible (assuming the consistency of a measurable cardinal) for Lebesgue measure to be extended to a measure defined on all subsets of $\mathbb{R}$.&lt;/p&gt;&#xA;" OwnerUserId="25" LastActivityDate="2010-09-28T00:00:14.313" CommentCount="1" />
  <row Id="5606" PostTypeId="1" AcceptedAnswerId="5610" CreationDate="2010-09-28T00:09:26.330" Score="1" ViewCount="98" Body="&lt;p&gt;Hi people. Can you help me realize why this is true? I can tell you that $P_i$ and $P_j$ are probabilities, i.e. $0 \leq P_i, P_j \leq 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle \sum_{i=1}^\infty \sum_{j=1}^\infty ijP_iP_j \leq \sum_{i=1}^\infty \sum_{j=1}^\infty j^2P_jP_i$.&lt;/p&gt;&#xA;" OwnerUserId="2174" LastActivityDate="2010-09-28T01:38:05.997" Title="Help me realize this inequality of infinite sums." Tags="&lt;sequences-and-series&gt;&lt;probability-theory&gt;" AnswerCount="1" />
  <row Id="5607" PostTypeId="1" AcceptedAnswerId="5620" CreationDate="2010-09-28T00:33:48.630" Score="3" ViewCount="105" Body="&lt;p&gt;Suppose one wants to prove that a 1-dimensional integral proper scheme over an algebraically closed field is projective. This is a step in how Hartshorne has you prove that any 1-dim proper scheme (over an algebraically closed field) is projective.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The method is that first note the normalization is non-singular and hence projective. Consider $f: \tilde{X}\to X$, and let $\mathcal{L}$ be a very ample sheaf on $\tilde{X}$. The goal is to prove that there is an effective divisor $D=\sum P_i$ such that $\mathcal{L}(D)\simeq \mathcal{L}$ and such that each $f(P_i)$ are non-singular points.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rest of the proof follows from the fact that there merely exists some very ample sheaf with that property, but the exercise seems to imply that any very ample sheaf satisfies this property.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think I'm missing something really obvious, but take a proper curve over $k$, say $C$, that has a single singularity. Blow-up that singularity, $\pi: \tilde{C}\to C$, this is the normalization. Take a point $P\in \tilde{C}$ such that $\pi(P)$ is the singularity. It seems to me that $D=P$ is an effective divisor, so $\mathcal{L}(D)$ is an invertible sheaf. If the above is correct, then $\mathcal{L}(D)$ cannot be very ample. Does anyone have a simple reason for this?&lt;/p&gt;&#xA;" OwnerUserId="2000" LastActivityDate="2010-09-28T08:14:21.607" Title="Very ample sheaf on a blowup" Tags="&lt;algebraic-geometry&gt;" AnswerCount="2" />
  <row Id="5608" PostTypeId="2" ParentId="5510" CreationDate="2010-09-28T00:59:06.213" Score="5" ViewCount="" Body="&lt;p&gt;I've glanced at the paper you linked to. I have a few remarks:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;It looks the authors of the paper managed to derive explicit closed-form expressions for the stationarity conditions for a special class of functions. This is sometimes possible for very specific problems. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If you remember your calculus, for a univariate function $f(x)$, the minimum or maximum is found when $f'(x) = 0$. You can think of the KKT as a generalization of this idea for an optimization problem with constraints. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The reason I (and the authors) say &quot;stationarity&quot; conditions instead of &quot;optimality&quot; conditions is because KKT conditions are necessary but insufficient conditions for optimality. For a true guarantee of optimality, you need to test your solution against SSOCs (Sufficient Second Order Conditions). Again, the second order conditions are analogous to the Calculus-101 idea of determining whether $f''(x)$ is positive or negative, which tells you whether the solution of $f'(x) = 0$ is a minimum or maximum.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Nevertheless in this case, under the assumptions in the paper, in the absence of saddle points, we may presume the solution to be optimal.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;This means you &lt;strong&gt;do not have to use any optimization algorithm/library in this particular case&lt;/strong&gt;. For a given pricing function, all you have to do is evaluate a function based on the results given in the paper. This is a very nice property -- it means the optimization is implicitly done for you. See Example 1 in the paper for a worked example. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In general, the KKT does not provide an algorithm for finding an optimum; it's only a test for stationarity. However, in very specific cases (i.e. with regularity conditions assumed, convex, no inequalities, etc.), the KKT conditions can provide closed-form expressions for the optimal solution of a problem. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact... this is the idea behind the Method of Lagrange Multipliers. The Lagrange Multiplier method is simply a special case of the KKT conditions with no inequality constraints.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Side Note&lt;/em&gt;: one of the reasons behind the difficulty in using the KKT as a practical algorithm to find stationary/optimal points is due to the &quot;complementarity conditions&quot; in the KKT system (see Wikipedia article).  when you have inequality constraints, you have to satisfy a set of constraints known as &quot;complementarity constraints&quot;. There can be nasty, nonconvex, degenerate bilinear terms in these constraints, and special techniques are required to handle them -- enumerative methods are required for obtaining global optimality, and therefore the problem is NP-hard. There is a whole new field of research in burgeoning area of complementarity problems (and its related problem class, MPECs = mathematical programs with equilibrium constraints)... currently there's a lot of work being done on determining different types of stationarity conditions for this class of problems. It is outside the scope of the question, but I thought I'd just let you know what's out there.&lt;/p&gt;&#xA;" OwnerUserId="1124" LastEditorUserId="1124" LastEditorDisplayName="" LastEditDate="2010-09-28T01:22:31.307" LastActivityDate="2010-09-28T01:22:31.307" CommentCount="3" />
  <row Id="5609" PostTypeId="1" AcceptedAnswerId="5611" CreationDate="2010-09-28T01:32:41.710" Score="1" ViewCount="240" Body="&lt;p&gt;I'm trying to figure out the limit of this sequence. I feel like I'm close to understanding it, I just need a little push.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\lbrace\sum_{n=1}^{k}\frac{1}{\sqrt{k^2 + n}}\rbrace_{k=1}^\infty$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand that I need to show that this sequence lies between $\lbrace \frac{k}{\sqrt{k^2 + k}}\rbrace_{k=1}^\infty$ and $\lbrace \frac{k}{\sqrt{k^2 + 1}}\rbrace_{k=1}^\infty$, and that both of those sequences converge to 1. However, I'm not quite sure how to show that both of them converge to 1, or why the first sequence would lie between them. The summation and the square roots are confusing me, I think.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: Okay, so I have the notion that I should divide everything in the convergent sequences by k, but I don't think I'm doing the algebra right. For the first sequence, dividing by k seems to get me $\frac{1}{\sqrt{k + 1}}$, which seems to converge to 0 as k approaches $\infty$ since $\frac{1}{\sqrt{k + 1}} \leq \frac{1}{k}$. Similarly, dividing everything in the latter sequence seems to give $\frac{1}{\sqrt{k + \frac{1}{k}}}$, which would also converge to zero as k approaches $\infty$. I don't feel like I'm distributing the k properly over the square root.&lt;/p&gt;&#xA;" OwnerUserId="2175" LastEditorUserId="2175" LastEditorDisplayName="" LastEditDate="2010-09-28T04:29:15.193" LastActivityDate="2010-09-28T04:29:15.193" Title="Finding the limit of $\lbrace\sum_{n=1}^{k}\frac{1}{\sqrt{k^2 + n}}\rbrace_{k=1}^\infty$" Tags="&lt;sequences-and-series&gt;&lt;homework&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="5610" PostTypeId="2" ParentId="5606" CreationDate="2010-09-28T01:38:05.997" Score="7" ViewCount="" Body="&lt;p&gt;For a given pair $(m,n)$ with $m\ne n$, consider the sum of the two terms corresponding to $(i,j) = (m,n)$ and $(i,j)=(n,m)$.  For the summation on the left, this will be&#xA;$$&#xA;mnP_mP_n + nmP_nP_m = 2mnP_mP_n,&#xA;$$&#xA;while for the sum on the right this will be&#xA;$$&#xA;n^2P_n P_m + m^2 P_m P_n = (n^2+m^2)P_mP_n.&#xA;$$&#xA;Since $2mn \le m^2+n^2$ for all $m$ and $n$, it follows that&#xA;$$&#xA;\sum_{i \ne j} ijP_iP_j \le \sum_{i\ne j} j^2 P_j P_i.&#xA;$$&#xA;But&#xA;$$&#xA;\sum_{i = j} ijP_iP_j = \sum_{i = j} j^2 P_j P_i,&#xA;$$&#xA;so this proves the given inequality.&lt;/p&gt;&#xA;" OwnerUserId="1726" LastActivityDate="2010-09-28T01:38:05.997" />
  <row Id="5611" PostTypeId="2" ParentId="5609" CreationDate="2010-09-28T01:42:18.963" Score="2" ViewCount="" Body="&lt;p&gt;Since this looks like homework, I will give a hint.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which is the largest among $\sqrt{k^2+1}, \sqrt{k^2+2},\dots, \sqrt{k^2+k}$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For your next step:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the mistake you are doing is substituting&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{\sqrt{k^2+k}}{k}$ with $\sqrt{\frac{k^2+k}{k}}$ instead of $\sqrt{\frac{k^2+k}{k^2}}$.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-09-28T03:49:19.593" LastActivityDate="2010-09-28T03:49:19.593" CommentCount="6" />
  <row Id="5612" PostTypeId="2" ParentId="5523" CreationDate="2010-09-28T01:50:16.483" Score="2" ViewCount="" Body="&lt;p&gt;I proved this a few years ago as a homework assignment. Even if it's not &lt;em&gt;your&lt;/em&gt; homework assignment, that makes me reluctant to give a complete solution; however, the hints below should be helpful. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Method 1:&lt;/strong&gt; The proof I submitted is along the following lines: given a monotone self-dual function $f$, show by induction that for all sufficiently small nonnegative integers $l$, there is another function $g_l$ (of the same arity as $f$) such that $||f-g_l||_1 = 2l$ and $g_l$ is in the clone. (Here $||f-g||_1$ is the &quot;$l_1$-norm,&quot; which in this case is just the number of $x$ for which $f(x)\neq g(x)$.) The basis case is obvious: pick an arbitrary function you know is in the clone. For the inductive step, you have to figure out how to take $g_l$ and produce $g_{l-1}$, which I'll leave out, but a hint is that there is a $g_{l-1}$ with $||g_l - g_{l-1}||_1 = 2$, i. e., $g_{l-1}$ is almost the same as $g_l$. The conclusion (by induction) is the case $l=0$, but $g_0$ has to be $f$, so $f$ is in the clone. (Informally: you can start at some function in the clone and &quot;walk&quot; by simple steps to any other monotone self-dual function of the same arity, and adjacent functions on the walk can be obtained from one another using the majority function.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Method 2:&lt;/strong&gt; However, there's a much neater argument by induction on the arity of $f$. To simplify future notation, let $\bar{x} = (x_4,\ldots,x_n)$, so $f(x_1,\ldots,x_n) = f(x_1,x_2,x_3,\bar{x})$. Observe that if&#xA;$f:\{0,1\}^n\to \{0,1\}$ is monotone and self-dual and $n&gt; 3$, then the&#xA;three functions of $n-1$ arguments $$f_1(x,y,\bar{x}) =&#xA;f(x,y,y,\bar{x}),$$ $$f_2(x,y,\bar{x}) = f(y,x,y,\bar{x}),$$&#xA;$$f_3(x,y,\bar{x}) = f(y,y,x,\bar{x}),$$ are monotone and self-dual&#xA;as well, and moreover, they determine $f$. (Indeed, out of $x_1,x_2,x_3$,&#xA;at least 2 must be equal, so for example, if $x_1 = x_2$, then&#xA;$f(x_1,\ldots,x_n) = f_3(x_3,x_2,\bar{x})$.) All you have to do now is reconstruct $f$ from $f_1,f_2,f_3$ using only the majority function. Thankfully,  the construction is very simple, although it may not be obvious at first why it works: $$f(x_1,\ldots,x_n) = M(f_1(x_1,x_3,\bar{x}), f_2(x_2,x_1,\bar{x}), f_3(x_3,x_2,\bar{x}))$$ (where $M$ is the majority function). (Hint: prove the case $x_1=1,x_2=x_3=0$, and the rest should become obvious. You need monotonicity here, but surprisingly not self-duality.) The basis case is $n\leq 3$ (where self-duality is needed). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Oh, well; I guess that's almost a complete solution anyway, but I don't feel like going back and removing steps from the argument at the risk of making it cryptic and useless. :-)&lt;/p&gt;&#xA;" OwnerUserId="2032" LastActivityDate="2010-09-28T01:50:16.483" />
  <row Id="5613" PostTypeId="2" ParentId="5609" CreationDate="2010-09-28T02:13:47.860" Score="0" ViewCount="" Body="&lt;p&gt;You want to take the limits of your two sequences as they approach infinity. if you are in a calculus class, you can use L'Hôpital's rule (http://en.wikipedia.org/wiki/L%27H%C3%B4pital%27s_rule). Otherwize, you should devide both the numerator and denominator by k, and then bring the k in the denominator into the radical. Then, as you go to infinity the stuff inside the radical goes to 1, and since 1/sqrt(1)=1, that becomes your answer.&lt;/p&gt;&#xA;" OwnerUserId="796" LastActivityDate="2010-09-28T02:13:47.860" CommentCount="1" />
  <row Id="5614" PostTypeId="1" AcceptedAnswerId="5638" CreationDate="2010-09-28T04:15:14.757" Score="0" ViewCount="185" Body="&lt;p&gt;Let $N$ be a normal subgroup of a finite group $G$. Let $S$ belonging to $G$ be a conjugacy class of elements in $G$, and assume that $S$ belongs to $N$. Prove that $S$ is a union of $n$ conjugacy classes in $N$, all having the same cardinality, where $n$ equals the index $[G : NCent(x)]$ of the group generated by $N$ and the centralizer $Cent(x)$ in $G$ of and element $x$ belonging to $S$.&lt;/p&gt;&#xA;" OwnerUserId="2177" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-09-28T14:45:56.877" LastActivityDate="2010-09-28T14:51:48.860" Title="a question about group" Tags="&lt;group-theory&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="5615" PostTypeId="2" ParentId="5588" CreationDate="2010-09-28T04:33:17.677" Score="4" ViewCount="" Body="&lt;p&gt;Amazingly, the complete classification of noncompact 2-manifolds with boundary was not completed until 2007.  Here's the reference:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A. O. Prishlyak and K. I. Mischenko, &lt;a href=&quot;http://www.ams.org/mathscinet-getitem?mr=2308580&quot; rel=&quot;nofollow&quot;&gt;Classification of noncompact surfaces with&#xA;boundary&lt;/a&gt;, Methods Funct. Anal. Topology 13 (2007), no. 1, 62–66.  &lt;/p&gt;&#xA;" OwnerUserId="1421" LastActivityDate="2010-09-28T04:33:17.677" />
  <row Id="5616" PostTypeId="1" AcceptedAnswerId="5623" CreationDate="2010-09-28T05:35:49.283" Score="1" ViewCount="148" Body="&lt;p&gt;If  S1, S2, and S are the sums of n terms, 2n terms and to infinity of a G.P. Then, find the value of S1(S1-S).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS:Nothing is given about the common ratio.&lt;/p&gt;&#xA;" OwnerUserId="2109" LastActivityDate="2010-09-28T22:41:46.897" Title="Geometric Progression" Tags="&lt;algebra-precalculus&gt;" AnswerCount="2" CommentCount="10" FavoriteCount="2" />
  <row Id="5617" PostTypeId="1" CreationDate="2010-09-28T05:50:49.190" Score="4" ViewCount="580" Body="&lt;p&gt;Recently, I came across this problem in my textbook. Any hints on how to solve this using dynamic programming?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Professor Stewart is consulting for the president of a corporation that is planning a company party. The company has a hierarchical structure; that is, the supervisor relation forms a tree rooted at the president. The personnel office has ranked each employee with a conviviality rating, which is a real number. In order to make the party fun for all attendees, the president does not want both an employee and his or her immediate supervisor to attend. Professor Stewart is given the tree that describes the structure of the corporation, using the left-child, right-sibling representation. Each node of the tree holds, in addition to the pointers, the name of an employee and that employee’s conviviality ranking. Describe an algorithm to make up a guest list that maximizes the sum of the conviviality ratings of the guests.&quot;&lt;/p&gt;&#xA;" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-09-28T08:21:45.810" LastActivityDate="2010-10-28T14:44:08.577" Title="Party planning problem" Tags="&lt;algorithms&gt;&lt;puzzle&gt;" AnswerCount="1" CommentCount="9" FavoriteCount="3" />
  <row Id="5618" PostTypeId="2" ParentId="5616" CreationDate="2010-09-28T06:05:23.293" Score="1" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\quad\:$  In $\rm\  \ (1-X)\ (1-(1-X))\ =\ 1-X^2-(1-X)\ \ \ $ put $\rm\ \ \ X = x^n\ $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;then multiply both sides by $\rm\  1/(1-x)^2\ =\ S/(1-x)\:.\ \ $  More generally one has&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\ \ (1-x^a)\:(1-x^b)\ =\ (1-x^a) + (1-x^b) - (1-x^{a+b})$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\quad\quad\ \Rightarrow\quad\quad S_a\ S_b\ =\ S\ (S_a + S_b - S_{a+b})\:,\quad S_n = \displaystyle\frac{1-x^n}{1-x},\quad S = S_\infty = \frac{1}{1-x}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This generalizes to arbitrary products $\rm\: S_{a}\: S_b\: S_c\cdots S_k\:$ using the &lt;a href=&quot;http://en.wikipedia.org/wiki/Inclusion%E2%80%93exclusion_principle&quot; rel=&quot;nofollow&quot;&gt;Inclusion–exclusion principle.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-09-28T22:41:46.897" LastActivityDate="2010-09-28T22:41:46.897" />
  <row Id="5619" PostTypeId="2" ParentId="3364" CreationDate="2010-09-28T06:18:20.320" Score="0" ViewCount="" Body="&lt;p&gt;These examples may be silly, but just to add to your list:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Sequential compactness (i.e. every sequence has a convergent subsequence)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Countable compactness (i.e. every &lt;em&gt;countable&lt;/em&gt; open cover has a finite subcover)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) $\sigma$-compactness (i.e. the space is a countable union of compact sets)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Less trivially, as mentioned in the Wikipedia article, the Lindelof property and separability are both preserved.&lt;/p&gt;&#xA;" OwnerUserId="640" LastEditorUserId="640" LastEditorDisplayName="" LastEditDate="2010-09-28T06:23:21.680" LastActivityDate="2010-09-28T06:23:21.680" />
  <row Id="5620" PostTypeId="2" ParentId="5607" CreationDate="2010-09-28T06:48:58.417" Score="2" ViewCount="" Body="&lt;p&gt;A very ample invertible sheaf $\mathcal L(D)$ is characterized by there being an embedding $\tilde{C} \hookrightarrow \mathbb P^n$ so that $D$ is the intersection of $\tilde{C}$&#xA;with a hyperplane $H$ in $\mathbb P^n$.  If a hyperplane intersects $\tilde{C}$ in the point&#xA;$P$ with multiplicity one, then $\tilde{C}$ has to be a degree one curve in $\mathbb P^n$, which is to say, a line.  Thus $\tilde{C}$ must have genus zero.  Thus all points on $\tilde{C}$ must be linearly equivalent, and so in this case $\mathcal L(P) = \mathcal L(P')$ for any&#xA;point $P'$ on $\tilde{C}$ (just to say the same thing a different way, we have&#xA;that $\tilde{C} = \mathbb P^1$, and that $\mathcal L(D) = \mathcal O(1)$ on this&#xA;$\mathbb P^1$).&#xA;In particular, we can take $P'$ to be a point whose image in $C$ is non-singular.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another way to phrase the argument, which avoids this precise intersection theoretic analysis, is to say that if $\mathcal L$ is very ample on $\tilde{C}$, then it comes from an embedding&#xA;$\tilde{C} \hookrightarrow \mathbb P^n$ by pulling back $\mathcal O(1)$, and so is of the&#xA;form $\mathcal L(D)$ for any intersection of $\tilde{C}$ with a hyperplane in $\mathbb P^n$.  But we can always choose this hyperplane so that its intersection with $\tilde{C}$ misses any given finite set of points.  So we can always choose this hyperplane so that &#xA;$D$ consists only of points whose images in $C$ are non-singular, which is what we want.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, when thinking about what very ample means, you want to relate it to the corresponding projective embedding, and then use geometric facts about hyperplane intersections.  (Primarily: that there are a lot of hyperplanes (!), so you can hope to get a lot of control over the $D := \tilde{C} \cap H$ for which your very ample invertible&#xA;sheaf equals $\mathcal L(D)$.  Although it isn't relevant here, because you are just working with curves, a typical tool that you might apply to understand the possible such $D$ is Bertini's theorem.)&lt;/p&gt;&#xA;" OwnerUserId="221" LastEditorUserId="221" LastEditorDisplayName="" LastEditDate="2010-09-28T08:14:21.607" LastActivityDate="2010-09-28T08:14:21.607" CommentCount="1" />
  <row Id="5621" PostTypeId="2" ParentId="3364" CreationDate="2010-09-28T07:03:30.743" Score="0" ViewCount="" Body="&lt;p&gt;Following from muad's comment one could say that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) $A \subseteq X_2$ is open, then $f^{-1}(A) \subseteq X_1$ is open &lt;br /&gt;&#xA;2) same thing for closed sets&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which I think are more simple conditions for characterizing a continuous function as opposed to those in your list, since the concept of open and closed sets are &quot;simpler&quot; than compact of connected sets.&lt;/p&gt;&#xA;" OwnerUserId="1415" LastActivityDate="2010-09-28T07:03:30.743" CommentCount="2" />
  <row Id="5622" PostTypeId="2" ParentId="5607" CreationDate="2010-09-28T07:42:50.917" Score="2" ViewCount="" Body="&lt;p&gt;Notice that in your 2nd paragraph you say the goal is to prove that «there is an effective divisor $D=\sum P_i$ such that $\mathcal{L}(D)\simeq \mathcal{L}$ and such that each $f(P_i)$ are non-singular points», but in your 4th paragraph you are picking a non singular point and asking if there is an effective divisor with a specific support. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The two things have quantifiers in different places. In particular, the claim in your 2nd paragraph does not say that you can pick the support of the divisor as you want!&lt;/p&gt;&#xA;" OwnerUserId="274" LastEditorUserId="274" LastEditorDisplayName="" LastEditDate="2010-09-28T07:50:47.420" LastActivityDate="2010-09-28T07:50:47.420" />
  <row Id="5623" PostTypeId="2" ParentId="5616" CreationDate="2010-09-28T08:57:42.210" Score="4" ViewCount="" Body="&lt;p&gt;I change your notation from S1, S2 and S to $S_{n},S_{2n}$ and $S$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sum of $n$ terms of a geometric progression of ratio $r$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$u_{1},u_{2},\ldots ,u_{n}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is given by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$S_{n}=u_{1}\times \dfrac{1-r^{n}}{1-r}\qquad (1)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore the sum of $2n$ terms of the same progression is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$S_{2n}=u_{1}\times \dfrac{1-r^{2n}}{1-r}\qquad (2)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming that the sum $S$ exists, it is given by &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$S=\lim S_{n}=u_{1}\times \dfrac{1}{1-r}\qquad (3)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the &quot;answer is S(S1-S2)&quot;, we have to prove this identity &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$S_{n}(S_{n}-S)=S(S_{n}-S_{2n})\qquad (4).$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Plugging $(1)$, $(2)$ and $(3)$ into $(4)$ we have to prove the following equivalent algebraic identity:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$u_{1}\times \dfrac{1-r^{n}}{1-r}\left( u_{1}\times \dfrac{1-r^{n}}{1-r}%&#xA;-u_{1}\times \dfrac{1}{1-r}\right) $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$=u_{1}\times \dfrac{1}{1-r}\left( u_{1}\times \dfrac{1-r^{n}}{1-r}-u_{1}\times \dfrac{1-r^{2n}}{1-r}\right)&#xA;\qquad (5)$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which, after simplifying $u_1$ and the denominator $1-r$, becomes: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\dfrac{1-r^{n}}{1}\left( \dfrac{1-r^{n}}{1}-\dfrac{1}{1}\right) =\left( \dfrac{%&#xA;1-r^{n}}{1}-\dfrac{1-r^{2n}}{1}\right) \qquad (6)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is equivalent to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\left( 1-r^{n}\right) \left( -r^{n}\right) =-r^{n}+r^{2n}\iff 0=0\qquad (7)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given that $(7)$ is true, $(5)$ and $(4)$ are also true.&lt;/p&gt;&#xA;" OwnerUserId="752" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2010-09-28T10:31:24.270" LastActivityDate="2010-09-28T10:31:24.270" CommentCount="2" />
  <row Id="5624" PostTypeId="1" AcceptedAnswerId="5629" CreationDate="2010-09-28T09:23:32.537" Score="2" ViewCount="60" Body="&lt;p&gt;for a finite group G and a trivial abelian G module A, there is the short exact sequence&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$0 \rightarrow Ext^1 (G_{ab},A) \rightarrow H^2(G,A) \rightarrow Hom (H_2 (G,Z), A) \rightarrow 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm looking for a description for the connecting maps. Specially, I want to use the representation of $H_2(G,Z)$ as $M(G)=[F,F]\cap R/ [F,R]$ where $G\cong F/R$ and F is free.&lt;/p&gt;&#xA;" OwnerUserId="2125" LastActivityDate="2010-09-28T12:25:33.993" Title="connecting maps in the universal coefficients theorem" Tags="&lt;cohomology&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="5625" PostTypeId="1" CreationDate="2010-09-28T10:00:42.893" Score="1" ViewCount="207" Body="&lt;p&gt;Greetings, dear Mathematicians!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How to implement base-9 to base-3 conversion in O(log(n)) on 4-tuple ((stateFrom, encounterWhat, Command, StateTo), where Command can be &amp;lt; (turn left), &gt; (turn right), # (stop execution), or symbol from an alphabet) Turing machine, finite from the left side and infinite from the right side?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you for your answers!&lt;/p&gt;&#xA;" OwnerUserId="2181" LastActivityDate="2010-10-15T03:58:15.300" Title="Turing machine to convert base-9 number to base-3 in O(log(n))" Tags="&lt;algorithms&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="5626" PostTypeId="2" ParentId="5614" CreationDate="2010-09-28T10:16:29.237" Score="0" ViewCount="" Body="&lt;p&gt;for $x \in S$ define a function from $G/ NCent(x)$ to conjugacy classes in N by $gNCent(x) \mapsto [gxg^{-1}]$. This is well defined because changing by an element of Cent(x) does nothing to x, and changing by an element of N changes $gxg^{-1}$ to another element in the same conjugacy class of N.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is easy to show that this is onto the conjugacy classes in S. also this is injective, because if $[hxh^{-1}]=[gxg^{-1}]$ then there is an $m\in N$ such that $(g^{-1}mh)x=x(g^{-1}mh)$, so you have that $g^{-1}h (h^{-1}mh) \in Cent(x)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;let $m_i\in N$ be the elements such that ${ m_i x m_i ^{-1} }$ are all the elements in the conjugacy class of x in N. From this we get that $(g_i m_i g_i^{-1}) (g_i x g_i^{-1}) (g_i m_i g_i^{-1})^{-1} $ are different elements in the conjugacy class of $(g_i x g_i^{-1})$ so $ | [x] | \leq | [gxg^{-1}] | $, but there is nothing special about taking x from S so we have the opposite inequality as well.&lt;/p&gt;&#xA;" OwnerUserId="2125" LastActivityDate="2010-09-28T10:16:29.237" />
  <row Id="5627" PostTypeId="1" AcceptedAnswerId="5628" CreationDate="2010-09-28T11:37:43.690" Score="2" ViewCount="217" Body="&lt;p&gt;Helping a buddy with his &quot;intro to math&quot; course for CompSci. I'm afraid my 'leet math skills are already giving out, so any help would be appreciated!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If &lt;em&gt;f&lt;/em&gt; and &lt;em&gt;g&lt;/em&gt; are both surjective functions, then so is the composition &lt;em&gt;f&lt;/em&gt; o &lt;em&gt;g&lt;/em&gt;. (so far so good).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would expect the converse to be true too: If &lt;em&gt;f&lt;/em&gt; o &lt;em&gt;g&lt;/em&gt; is a surjective function, does that mean &lt;em&gt;f&lt;/em&gt; and &lt;em&gt;g&lt;/em&gt; are surjective?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From my research so far it seems that in this case &lt;em&gt;f&lt;/em&gt; is indeed surjective but &lt;em&gt;g&lt;/em&gt; is not necessarily so. Can anybody explain why, please? An example would rock!&lt;/p&gt;&#xA;" OwnerUserId="2183" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2010-11-24T02:30:44.593" LastActivityDate="2010-11-24T07:19:41.617" Title="Surjection on composed function?" Tags="&lt;functions&gt;&lt;elementary-set-theory&gt;" AnswerCount="4" />
  <row Id="5628" PostTypeId="2" ParentId="5627" CreationDate="2010-09-28T12:04:17.360" Score="4" ViewCount="" Body="&lt;p&gt;It is important to consider the codomain when talking about the `surjectivity' of a function. Let us give a real-valued example of functions for now.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $g(x)$ be any function that maps the real number line to the interval $(-1, 1)$ (eg. sin) and let $f(x) = \frac{x}{x^2-1}$ be considered only over the interval $(-1, 1)$. Notice that $f(x)$ is continuous over this interval, and the range of f is the real line (You can graph x/(x^2-1) at wolframalpha.com to see this as well).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus it is not necessary that $g(x)$ be surjective. In fact, the real number line is isomorphic to any open interval.&lt;/p&gt;&#xA;" OwnerUserId="557" LastEditorUserId="557" LastEditorDisplayName="" LastEditDate="2010-09-28T12:11:42.830" LastActivityDate="2010-09-28T12:11:42.830" CommentCount="1" />
  <row Id="5629" PostTypeId="2" ParentId="5624" CreationDate="2010-09-28T12:25:33.993" Score="3" ViewCount="" Body="&lt;p&gt;By «connecting maps» you mean the maps that appear in the short exact sequence you wrote?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In any case, you will find the details of the construction of the sequence in Hilton-Stammbach's book &lt;em&gt;A Course on homological algebra&lt;/em&gt;, GTM4. The result is theorem VI.15.1, but the meat of the construction is in section §V.3.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want the expressions in terms of the isomorphisms between $H_2(G,\mathbb Z)$ and $M(G)$, well, then you have to write down both and compose them :) I do not have time now to do the computation, but you should &lt;em&gt;really&lt;/em&gt; do it yourself!&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2010-09-28T12:25:33.993" />
  <row Id="5630" PostTypeId="2" ParentId="5627" CreationDate="2010-09-28T12:28:14.370" Score="1" ViewCount="" Body="&lt;p&gt;Consider $f: X \rightarrow Y$, $g: Y \rightarrow Z$, then $g \circ f: X \rightarrow Z$. If it is surjective, it means that for any $z \in Z$ there exists $x \in X$ such that $(g \circ f)(x) = g(f(x)) = z$; thus, if $y = f(x)$, then $g(y) = z$, which shows that $g$ is surjective&lt;/p&gt;&#xA;" OwnerUserId="1415" LastActivityDate="2010-09-28T12:28:14.370" />
  <row Id="5631" PostTypeId="2" ParentId="5564" CreationDate="2010-09-28T12:34:21.673" Score="0" ViewCount="" Body="&lt;p&gt;If you assume natural deduction, this is easy to explain. Let $\Sigma$ be an inconsistent set of axioms. This means you can prove both $\phi$ and $\lnot\phi$ for some statement $\phi$: in symbols $\Sigma\vdash\phi$ and $\Sigma\vdash\lnot\phi$. Now you suppose the negation of whatever statement $\psi$ you like to prove, but since, by monotonicity, $\Sigma\cup\{\lnot\psi\}\vdash\phi$ and $\Sigma\cup\{\lnot\psi\}\vdash\lnot\phi$, you can apply negation elimination obtaining $\Sigma\vdash\psi$.&lt;/p&gt;&#xA;" OwnerUserId="2128" LastActivityDate="2010-09-28T12:34:21.673" />
  <row Id="5632" PostTypeId="1" AcceptedAnswerId="5633" CreationDate="2010-09-28T12:54:57.133" Score="0" ViewCount="206" Body="&lt;p&gt;I think this is really weird. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to solve the following equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(2x^2+3x+5)^3+(2x^2+3x+5)^2=0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a polynomial equation of 6 degrees. It should have 6 roots.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But &lt;a href=&quot;http://www.wolframalpha.com&quot; rel=&quot;nofollow&quot;&gt;wolframalpha&lt;/a&gt; only returns 4 roots! Anything I miss?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the command I use:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;solve ((2x^2+3x+5)^3)+((2x^2+3x+5)^2)=0&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit: Ah, Wolframalpha doesn't return repeated roots. So now case solved.&lt;/strong&gt; &lt;/p&gt;&#xA;" OwnerUserId="550" LastEditorUserId="550" LastEditorDisplayName="" LastEditDate="2010-09-28T13:06:09.000" LastActivityDate="2010-09-28T13:15:43.557" Title="WolframAlpha Returns only 2 Roots for a Polynomial Equation of 6 Degree" Tags="&lt;wolframalpha&gt;" AnswerCount="1" CommentCount="7" />
  <row Id="5633" PostTypeId="2" ParentId="5632" CreationDate="2010-09-28T13:15:43.557" Score="3" ViewCount="" Body="&lt;p&gt;Trying&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Solve[(2x^2 + 3x + 5)^3 + (2x^2 + 3x + 5)^2 == 0, x]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;on WolframAlpha returns four roots; the truth of the matter is that the equation has two roots of multiplicity two, so that you have six roots in total.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not entirely sure why the multiplicities aren't indicated...&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-09-28T13:15:43.557" />
  <row Id="5634" PostTypeId="1" CreationDate="2010-09-28T13:24:34.177" Score="2" ViewCount="155" Body="&lt;p&gt;I'm trying to convey to some folks some logical and bitwise operators such as AND, OR, XOR, &amp;amp;c in programming languages such as C &amp;amp;| Java.  Though these I guess could be considered &quot;boolean&quot; operators there is no &quot;Boolean Algebra&quot; in here, just the results of various operation, yes?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've never heard of Boolean &lt;strong&gt;Arithmetic&lt;/strong&gt;, but would one consider these simple questions&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;5 AND 8 = ?&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;boolean arithmetic (operating bitwise) or rather &lt;em&gt;logical&lt;/em&gt; or &lt;em&gt;bitwise&lt;/em&gt; arithmetic?&lt;/p&gt;&#xA;" OwnerUserId="2184" LastActivityDate="2010-09-28T15:02:02.490" Title="Arithmetic? Logical/Boolean/Bitwise operations?" Tags="&lt;arithmetic&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="5635" PostTypeId="2" ParentId="5634" CreationDate="2010-09-28T13:40:49.277" Score="4" ViewCount="" Body="&lt;p&gt;The terms &lt;em&gt;boolean&lt;/em&gt;, &lt;em&gt;binary&lt;/em&gt;, and &lt;em&gt;bitwise&lt;/em&gt; arithmetic are all equivalent in my mind. However, it is most commonly referred to as &lt;em&gt;binary arithmetic&lt;/em&gt; in my experience.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Boolean_algebra_%28logic%29&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;Boolean algebra&lt;/em&gt;&lt;/a&gt; however is a more general field (encompassing more than just arithmetic), and is really synonymous for &lt;em&gt;classical propositional calculus&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="56" LastActivityDate="2010-09-28T13:40:49.277" />
  <row Id="5636" PostTypeId="2" ParentId="5627" CreationDate="2010-09-28T13:49:36.963" Score="3" ViewCount="" Body="&lt;p&gt;The example to keep in mind is the following: take $X=\{x\}$ be a one element set, $Y=\{1,2\}$, a two elements set; and $Z=\{a\}$ a one-element set. Let $f\colon X\to Y$ be given by $f(x)=1$, and $g\colon Y\to Z$ by $g(1)=g(2)=a$. Then $g\circ f\colon X\to Y$ is bijective (one-to-one and onto), $f$ is one-to-one but not onto, and $g$ is onto but not one-to-one. So even if you know &lt;em&gt;more&lt;/em&gt; about the composition, still the best you can conclude is that the second function is onto, but not necessarily the first.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So: if a composition $g\circ f$ is one-to-one, the best you can say is that $f$ is one-to-one; if $g\circ f$ is onto, then the best you can say is that $g$ is onto. Intuitively, in the one-to-one case, the second function can be ill-behaved outside the image of $f$, but the composition &quot;will never know&quot;. And in the onto-case, $g$ can have a lot of redundancy, so that $f$ does not have to get to &lt;em&gt;every&lt;/em&gt; point in the domain of $g$, just to &quot;enough&quot; of them.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-09-28T13:49:36.963" />
  <row Id="5637" PostTypeId="2" ParentId="5617" CreationDate="2010-09-28T14:02:42.817" Score="3" ViewCount="" Body="&lt;p&gt;I'll give you a hint without referring to the representation of the graph:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given a tree structure of a corporation, you can either have the root (president) attend the party or not. Say the root has n children. If the root is not in the party, you can have a party which is the &quot;sum&quot; of n different parties, each representing the subtree at each of the n children (this is the recursive step). If the root is in the party, then you again have the sum of n different parties, but this time each &quot;subparty&quot; &lt;em&gt;cannot&lt;/em&gt; have its root included (this is another recursive step).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can reformat this recursion into a dynamic programming solution, or solve it recursively with memoization.&lt;/p&gt;&#xA;" OwnerUserId="26" LastActivityDate="2010-09-28T14:02:42.817" CommentCount="0" />
  <row Id="5638" PostTypeId="2" ParentId="5614" CreationDate="2010-09-28T14:35:18.560" Score="3" ViewCount="" Body="&lt;p&gt;The fact that $S$ is a union of conjugacy classes of $N$ should be immediate; also, the sizes are the same: let $x\in S$, and let $U$ be the $N$-conjugacy class of $x$. If $T$ is any other $N$-conjugacy class contained in $S$, then let $y\in T$. There exists $g\in G$ such that $gxg^{-1}=y$. Then $|T|=|gUg^{-1}|=|U|$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the real meat of the problem is showing that the &lt;em&gt;number&lt;/em&gt; of conjugacy classes is the index of $NC_G(x)$ in $G$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whenever you see a problem in which you need to show that the number of something or other in a group equals the index of a subgroup $H$ in $G$, this should suggest finding a map from $G$ to the set you want, and then showing that the images of two elements coincide if and only if they lie in the same coset of $H$ in $G$. This will then give you that the number of distinct images is the index of $H$ in $G$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So here what you want to do is find a map from $G$ to the conjugacy classes of $N$ contained in $S$, and arrange matters so that two elements of $G$ map to the same thing if and only if they lie in the same coset of $NC_G(x)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So... how to map it? The obvious thing to do is to take a conjugacy class of $N$ contains in $S$, represented by some $x\in S$, and map it to the conjugacy class of $gxg^{-1}$ (which lies in $N$ because $N$ is assumed to be normal). If $x\sim_N y$ (that is, $x$ is conjugate to $y$ in $N$) then $y=nxn^{-1}$, so $gyg^{-1} = gnxn^{-1}g^{-1} = (gng^{-1})x(gng^{-1})$, and since $gng^{-1}\in N$, then $gyg^{-1}\sim_N gxg^{-1}$. So this map is well-defined on $N$-conjugacy classes. Since $S$ is a conjugacy class, $gxg^{-1}\in S$ as well, so the image of the $N$-conjugacy class of $x$ is also an $N$-conjugacy class contained in $S$. That is: conjugation maps $N$-conjugacy classes contained in $S$ to $N$-conjugacy classes contained in $S$. And note that this map is onto: if you pick &lt;em&gt;any&lt;/em&gt; $x\in S$, then every element of $S$ is of the form $gxg^{-1}$ for some $g\in G$, so you can pick your favorite $x\in S$, and use it as a basis: you will &quot;hit&quot; every $N$-conjugacy class that makes up $S$ this way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, when will $g$ and $h$ do the exact same thing to every $N$-conjugacy class contained in $N$? Suppose that $gxg^{-1}\sim_N hxh^{-1}$ for some $x\in S$. Then there exists $n\in N$ such that $ngxg^{-1}n^{-1} = hxh^{-1}$, hence $h^{-1}ngxg^{-1}n^{-1}h = x$; that is, there exists $n\in N$ such that $h^{-1}ng\in C_G(x)$ (we're getting somewhere... we've gotten to the centralizer of $x$ in $G$). This means that $ng C_G(x)=hC_G(x)$. Not quite what we want: that $n$ is  &quot;on the wrong side&quot; so to speak. Ah, but $N$ is normal. So $ng = (gg^{-1})ng = g(g^{-1}ng) = gn'$ for some $n'\in N$, so we have $gn'C_G(x)=hC_G(x)$. Thus, of $gxg^{-1}\sim_N hxh^{-1}$, then there exists $n'\in N$ such that $gn'C_G(x)=hC_G(x)$. And this is equivalent to $gNC_G(x) = hNC_g(x)$. (Almost there!) Conversely, suppose that $gNC_G(x) = hNC_g(x)$; then there exists $n\in N$ and $c\in C_G(x)$ such that $h = gnc$. Then $hxh^{-1} = (gnc)x(gnc)^{-1} = gn(cxc^{-1})n^{-1}g^{-1} = gnxn^{-1}g^{-1}$ (since $cx=xc$); since $n$ is normal, $gn = gn(g^{-1}g) = (gng^{-1})g = n'g$ for some $n'\in N$, so $hxh^{-1} = gnxn^{-1}g^{-1} = n'gxg^{-1}n'^{-1}$, hence $hxh^{-1}\sim_N gxg^{-1}$. That is: $g$ and $h$ map the $N$-conjugacy class of $x$ to the same $N$-conjugacy class if and only if $g$ and $h$ are in the same left coset of $NC_G(x)$ in $G$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, since every element of $S$ is $G$-conjugate to $x$, this is enough: we can get to &lt;em&gt;any&lt;/em&gt; $N$-conjugacy class contained in $N$ by conjugating $x$ by some element of $G$. And we have just seen that two elements of $G$ have the same image if and only if they are in the same left cosets of $NC_G(x)$. So the number of distinct images, that is, the number of $N$-conjugacy classes that make up $S$ (and here we can use &quot;number&quot; to mean cardinality, even in the infinte case!), is exactly the same as the number of left cosets of $NC_G(x)$ in $G$. That is, the number of conjugacy classes is $[G:NC_G(x)]$, as claimed. &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-09-28T14:51:48.860" LastActivityDate="2010-09-28T14:51:48.860" />
  <row Id="5639" PostTypeId="1" AcceptedAnswerId="5642" CreationDate="2010-09-28T14:47:08.623" Score="6" ViewCount="420" Body="&lt;p&gt;A simple question here, which likely demands a somewhat complex answer... Or rather, a set of related questions.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;What are the advantages of typed lambda calculus over untyped lambda calculus in terms of proof theory?&lt;/li&gt;&#xA;&lt;li&gt;Specifically, Church's original Lambda Calculus was untyped and allows arbitrarily high-order functions. What are the limitations with respect to constructing a proof calculus from it?&lt;/li&gt;&#xA;&lt;li&gt;Are not untyped and typed lambda calculi inherently higher-order formal systems?&lt;/li&gt;&#xA;&lt;li&gt;What are the reasons for using complex (e.g. polymorphic/dependent) type theories over the simple type theory in lambda calculus? Are they more 'powerful' in some sense; if so, how exactly?&lt;/li&gt;&#xA;&lt;li&gt;Do semantics (interpretation) have anything to say here, with respect to typed and untyped theories, especially in terms of soundness and completeness?&lt;/li&gt;&#xA;&lt;li&gt;The well-known proof verifier &lt;em&gt;Coq&lt;/em&gt;, which (I believe) uses a language of higher-order complex-typed lambda calculus to represent proofs in constructive (intuistionistic) mathematics. I have read that the theory behind it (the &lt;em&gt;Calculus of Constructions&lt;/em&gt;) is essentially an extension of the Curry-Howard isomorphism to higher-order logic. Are there any elaborations/clarifications I should be aware of here?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="56" LastActivityDate="2010-09-28T15:52:16.553" Title="The Power of Lambda Calculi" Tags="&lt;logic&gt;&lt;proof-theory&gt;&lt;lambda-calculus&gt;&lt;formal-systems&gt;" AnswerCount="1" FavoriteCount="4" />
  <row Id="5640" PostTypeId="2" ParentId="5634" CreationDate="2010-09-28T15:02:02.490" Score="6" ViewCount="" Body="&lt;p&gt;But there is boolean algebra here. Namely, you can consider an $\rm n$-bit integer as a characteristic function for a subset of $\rm\: n = \{0,1,\ldots,n-1\}$, viz. bit $\rm i$ of $\rm k$ is $1$ iff $\rm i$ lies in the subset represented by $\rm k$. The bitwise boolean operations suffice to define all of the boolean algebra operations in this powerset boolean algebra.&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-09-28T15:02:02.490" />
  <row Id="5641" PostTypeId="1" AcceptedAnswerId="5643" CreationDate="2010-09-28T15:38:11.287" Score="7" ViewCount="316" Body="&lt;p&gt;I'm going to ask a very silly question, so I'm begging you to be understanding if it is absolutely trivial, or if it's an exercise in some Bourbaki. I'm afraid of asking you, because the question entails this particular case: is the addition of (a finite, but variable quantity of) real numbers a continuous function?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ok, let me try to explain what I mean.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm insisting in doing exercices about topological groups, see if I can understand something about those previous &lt;a href=&quot;http://math.stackexchange.com/questions/2865/is-a-direct-limit-of-topological-groups-always-a-topological-group&quot;&gt;question 1&lt;/a&gt;, &lt;a href=&quot;http://math.stackexchange.com/questions/5095/what-are-the-product-and-coproduct-in-the-category-of-topological-groups&quot;&gt;question 2&lt;/a&gt; and &lt;a href=&quot;http://math.stackexchange.com/questions/5297/colimit-of-topological-groups-again&quot;&gt;question 3&lt;/a&gt;. This one should be more elementary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $G$ be a topological Abelian group, written additively, and let $I$ be an arbitrary set. I don't mind if you take $G = \mathbb{R}$, with the usual Euclidian topology and the usual addition of real numbers, but you &lt;strong&gt;cannot&lt;/strong&gt; assume $I$ is finite, nor countable. Let &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;G^I = \prod G = \prod_\alpha G_\alpha , \qquad \text{where}\ G_\alpha = G \qquad \text{for all}\ \alpha \in I \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let us denote  $(x_\alpha)$ the elements of $\prod G$. This is an Abelian topological group with the usual &lt;a href=&quot;http://en.wikipedia.org/wiki/Product_topology&quot; rel=&quot;nofollow&quot;&gt;product topology&lt;/a&gt; and operations defined component-wise:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{align}&#xA;(x_\alpha) + (y_\alpha) &amp;amp;= (x_\alpha + y_\alpha)  \\&#xA;-(x_\alpha)              &amp;amp;= (-x_\alpha)&#xA;\end{align}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider also the &lt;em&gt;weak product&lt;/em&gt; $\prod' G$ (aka, &lt;em&gt;direct sum&lt;/em&gt;). That is, elements of $\prod' G$ are those tuples $(x_\alpha) \in \prod  G$ such that $x_\alpha = 0$, except for a finite number of indexes $\alpha \in I$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\prod'G $ is of course a subgroup of $\prod G$, $\prod' G \subset \prod G$. Let's consider the &lt;a href=&quot;http://en.wikipedia.org/wiki/Subspace_topology&quot; rel=&quot;nofollow&quot;&gt;subspace topology&lt;/a&gt; on $\prod' G$, induced from the product topology on $\prod G$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then we have a well-defined addition map that we hadn't in $\prod G$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\sum : \prod ' G \longrightarrow G \ , \qquad  (x_\alpha ) \mapsto \sum_\alpha x_\alpha \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This makes sense, despite the fact $I$ is not necessarily an ordered set because, given an element $(x_\alpha) \in \prod'G$, if $x_{\alpha_1}, \dots , x_{\alpha_n}$ are its non-zero components, we don't have a canonical choice in order to perform the addition $x_{\alpha_1} + \cdots + x_{\alpha_n}$. Nevertheless, since the sum is associative and commutative, we may do it as we please: the result will always be the same. Hence $\sum_\alpha x_\alpha$ is well-defined.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question.&lt;/strong&gt; Is this map $\sum$ continuous?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is, if you don't like too much abstraction in your life, but prefer to be very specific: is the sum of real numbers&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\sum: \prod' \mathbb{R} \longrightarrow \mathbb{R} \ , \qquad (x_\alpha) \mapsto \sum_\alpha x_\alpha&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;continuous?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Remark 1.&lt;/strong&gt; The question seems silly, doesn't it? Well, maybe it is, but the first answer that came to my mind is wrong, as far as I can see: you cannot say &quot;this $\sum$ is continuous because it is a composition of continuous maps&quot;; namely, the iteration of the addition of $G$. Yeah, but: &lt;em&gt;which&lt;/em&gt; composition? Notice that there is no canonical choice for doing the addition in an specific order -there could be no order at all in $I$. So every time you compute $\sum_\alpha x_\alpha$ you can change the order in which you perform the (continuous) sums $x_\alpha + x_\beta$. So, $\sum$ is &lt;em&gt;not&lt;/em&gt; a composition of a finite number of specific maps. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Edit. Previous Remark 2 was wrong.)&lt;/p&gt;&#xA;" OwnerUserId="664" LastEditorUserId="664" LastEditorDisplayName="" LastEditDate="2010-09-30T07:30:10.030" LastActivityDate="2010-09-30T07:30:10.030" Title="Is addition continuous?" Tags="&lt;topology&gt;&lt;general-topology&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" />
  <row Id="5642" PostTypeId="2" ParentId="5639" CreationDate="2010-09-28T15:52:16.553" Score="3" ViewCount="" Body="&lt;ol&gt;&#xA;&lt;li&gt;What are the advantages of typed lambda calculus over untyped lambda calculus in terms of proof theory? &lt;em&gt;and also&lt;/em&gt;&lt;/li&gt;&#xA;&lt;li&gt;Specifically, Church's original Lambda Calculus was untyped and allows arbitrarily high-order functions. What are the limitations with respect to constructing a proof calculus from it?&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Simply, typed lambda calculus has a proof theory, and untyped lambda calculus doesn't because it lacks a normal-form theorem.&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;Are not untyped and typed lambda calculi inherently higher-order formal systems?&#xA;&lt;ul&gt;&#xA;&lt;li&gt;This is tricky, because there are two ways of looking at higher-orderness.  They are higher-order when you look at the terms, since the definition of higher order is abstraction over higher-order entities like functions, and lambda abstraction is abstraction.  But note, under the formulae-as-types correspondence, the proposaitions in logic are associated with the types of the lambda terms, and in, say, the simply-typed lambda calculus, there is no abstraction over anything.  This point confused me when I first studied Church's simple theory of types, because it is a higher-order calculus based on simply-typed lambda calculus, where the propositions are formed using the lambda terms, rather than, as with the formulae-as-types corespondence, being the types.&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;What are the reasons for using complex (e.g. polymorphic/dependent) type theories over the simple type theory in lambda calculus? Are they more 'powerful' in some sense; if so, how exactly?&#xA;&lt;ul&gt;&#xA;&lt;li&gt;They do add power.  Under the formulae-as-types correspondence, the simply-typed lambda calculus has as its matching logic propositional calculus, adding dependent types adds universal and existential quantification at higher finite types to its logic.  Polymorphic types allow types allowing all usual mathematical entities to be constructed (although without the usual theory of provability) and has high proof-theoretic consistency: Strong normalisation of system F has (over a base theory such as RCL0) the same strength as consistency second order arithmetic, even though it lacks an induction principle.&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;Do semantics (interpretation) have anything to say here, with respect to typed and untyped theories, especially in terms of soundness and completeness?&#xA;&lt;ul&gt;&#xA;&lt;li&gt;They are much less useful with these sorts of theories than model theory is with classical logic, although there are good applications of category theory to the semantics of type theory.&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;The well-known proof verifier Coq, which (I believe) uses a language of higher-order complex-typed lambda calculus to represent proofs in constructive (intuistionistic) mathematics. I have read that the theory behind it (the Calculus of Constructions) is essentially an extension of the Curry-Howard isomorphism to higher-order logic. Are there any elaborations/clarifications I should be aware of here?&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Yes, that polymorphic types are tricky, and non-conservative over the base theory.  I'd recommend sarting an exploration of formulae-as-types with Martin-Löf's dependent type theory.  If you want to work with a proof assistant, there is &lt;a href=&quot;http://wiki.portal.chalmers.se/agda/pmwiki.php&quot; rel=&quot;nofollow&quot;&gt;Agda&lt;/a&gt;, which is a functional programming language whose type system is Martin-Löf's type theory.&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="100" LastActivityDate="2010-09-28T15:52:16.553" CommentCount="9" />
  <row Id="5643" PostTypeId="2" ParentId="5641" CreationDate="2010-09-28T16:47:07.040" Score="6" ViewCount="" Body="&lt;p&gt;In the case where $I$ is infinite and $G$ has a non trivial topology, then the answer is no.&#xA;let $(x_\alpha) \in \prod 'G_\alpha$ and let $U$ be an open set containing $( x_\alpha)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;U comes from an open set in the regular product topology, so we can assume that it is $\prod U_\alpha \cap \prod'G_\alpha$ where only for finite number $U_\alpha \neq G$. the assumption that $I$ is infinite gives at least one $U_\beta=G$ so for all $(y_\alpha)$ such that $y_\alpha = 0 \forall \alpha\neq \beta$  we have $(x_\alpha)+(y_\alpha) \in U$ so $\sum (U) = G$, because we can choose $y_\beta$ to be any element of G.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means that if $V\neq G$ is an open neighborhood of $\sum (x_\alpha)$ then there is no open neighborhood U of $(x_\alpha)$ such that $\sum (U) \subseteq V$. &lt;/p&gt;&#xA;" OwnerUserId="2125" LastActivityDate="2010-09-28T16:47:07.040" CommentCount="1" />
  <row Id="5644" PostTypeId="2" ParentId="2837" CreationDate="2010-09-28T17:01:14.923" Score="0" ViewCount="" Body="&lt;p&gt;This dude have some nice info about this topic!&#xA;Click his name to see more math stuff.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://local.wasp.uwa.edu.au/~pbourke/geometry/sphereline/&quot; rel=&quot;nofollow&quot;&gt;http://local.wasp.uwa.edu.au/~pbourke/geometry/sphereline/&lt;/a&gt;&lt;/p&gt;&#xA;" LastActivityDate="2010-09-28T17:01:14.923" />
  <row Id="5645" PostTypeId="1" AcceptedAnswerId="7905" CreationDate="2010-09-28T17:05:24.147" Score="6" ViewCount="258" Body="&lt;p&gt;I want to show that there is a finite conjunction $\phi$ of axioms of $ZF$, such that every transitive proper class $M$, which satisfies $\phi$, is already a model of $ZF$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is an exercise in Kunen's set theory. There is a hint, it seems to be useful to apply the Reflection principle to the union $M = \cup_{\alpha} M \cap R(\alpha)$. But I don't know with which axioms we can do that (we can only use finitely many!), and why this yields an ordinal which is independent from $M$. Please give me only a hint, because basically I want to solve this on my own, but I don't know how to start with the hint above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, what is the &quot;philosophical&quot; reason that we cannot deduce from this, that $ZF$ is finitely axiomatizable (which is wrong)? I mean I cannot prove that this $\phi$ above proves every axiom, but is there also a deeper reason for this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: There was an answer with some hints, but it was deleted...&#xA;I still don't know how to produce this strange sentence $\phi$.&lt;/p&gt;&#xA;" OwnerUserId="1650" LastEditorUserId="1650" LastEditorDisplayName="" LastEditDate="2010-10-26T11:52:22.077" LastActivityDate="2010-10-26T12:26:52.290" Title="ZF is almost finitely axiomatizable" Tags="&lt;set-theory&gt;&lt;logic&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="5646" PostTypeId="1" AcceptedAnswerId="5654" CreationDate="2010-09-28T19:48:05.417" Score="3" ViewCount="418" Body="&lt;p&gt;I've been trying to understand how  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;${x^3-12x+9}$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;factors to &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(x-3) (x^2+3 x-3)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What factoring rule does this follow? The net result seems to be similar to what is attained through the sum/difference of cubes factoring pattern, but the signs are different.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Additionally, what type of problem is this, so I can make better and more relevant searches for help on future questions. Is it a cubic trinomial?&lt;/p&gt;&#xA;" OwnerUserId="2185" LastActivityDate="2010-09-28T23:27:48.137" Title="Factoring a Cubic Polynomial" Tags="&lt;algebra-precalculus&gt;" AnswerCount="3" FavoriteCount="1" />
  <row Id="5647" PostTypeId="1" AcceptedAnswerId="5651" CreationDate="2010-09-28T19:53:21.637" Score="2" ViewCount="158" Body="&lt;p&gt;How would you find the image of ${ x+yi : -\pi &amp;lt; x &amp;lt; \pi, y &gt; 0 }$ under $\sin(z)$?  I see that $\text{Re } \sin(x+yi) = \sin x \cosh y$ and $\text{Im } \sin(x+yi) = \cos x \sinh y$, but I'm not sure what to do next since $\sinh$ and $\cosh$ are unbounded.&lt;/p&gt;&#xA;" OwnerUserId="1960" LastActivityDate="2010-09-28T20:24:30.960" Title="Image of sin(z) over a complex set" Tags="&lt;complex-analysis&gt;" AnswerCount="2" />
  <row Id="5648" PostTypeId="1" AcceptedAnswerId="5722" CreationDate="2010-09-28T19:57:36.457" Score="5" ViewCount="226" Body="&lt;p&gt;If $h(x,t)$ is measurable and the measures involved are $\sigma$-finite, does there exist a sequences of functions &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$h_n(x,t) = \sum_{j=1}^{N_n} f_{j,n}(x) 1_{F_{j,n}}(t)$$ &#xA;where the sets are pairwise disjoint (in $j$) such that $h_n \to h$ pointwise almost everywhere?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that if we fix $x$ we can get a sequence like that but we cannot just make the coefficients depending on $x$ since the sets $F_{j,n}$ will be different. Can I somehow combine them? Do I need the $\sigma$-finiteness?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the answer is positive, we can use this to prove Minkowski's inequality for integrals quite easily.&lt;/p&gt;&#xA;" OwnerUserId="1120" LastEditorUserId="1120" LastEditorDisplayName="" LastEditDate="2010-09-29T15:28:12.087" LastActivityDate="2010-09-29T18:17:02.123" Title="Sequence of measurable functions." Tags="&lt;real-analysis&gt;&lt;measure-theory&gt;" AnswerCount="2" CommentCount="7" FavoriteCount="2" />
  <row Id="5649" PostTypeId="2" ParentId="5646" CreationDate="2010-09-28T20:00:03.967" Score="2" ViewCount="" Body="&lt;p&gt;One way to see that is that $x=3$ is a root of $x^3 - 12x  + 9$. So $(x-3)$ will be a factor (by the &lt;a href=&quot;http://en.wikipedia.org/wiki/Factor_theorem&quot; rel=&quot;nofollow&quot;&gt;Factor Theorem&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now you can try to get $x-3$ somehow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One way you can do that is to rewrite&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x^3-12x + 9 = x^3 - (3^3 - 3^3) - 12(x - 3+3)  + 9$$&#xA;$$ = x^3 - 27 + 27 - 12(x - 3) - 36 + 9 = x^3 - 27 - 12(x-3) + (9 -36 + 27)$$&#xA;$$ = (x-3)(x^2+3x+9) - 12(x-3) = (x-3)(x^2+3x-3) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here we used the fact that $x^3 - a^3 = (x-a)(x^2 + ax + a^2)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, if $r$ is a root of $f(x) = a_{n}x^n + a_{n-1}x^{n-1} + \dots + a_{0}$, then $f(x) - f(r) = f(x)$ gives us a way to factorize $f(x)$ as $(x-r)g(x)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f(x) = a_{n}x^n + a_{n-1}x^{n-1} + \dots + a_{0} - (a_{n}r^{n} + a_{n-1}r^{n-1} + \dots +a_{0})$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ = a_{n}(x^n - r^n) + a_{n-1}(x^{n-1} - r^{n-1}) + \dots + a_{1}(x-r)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just like $x^3 - a^2 = (x-a)(x^2 + ax + a^2)$ we have that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x^n - r^n = (x-r)(x^{n-1} + rx^{n-1} + \dots + r^{n-1})$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and so&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f(x) = (x-r) (a_{n}(x^{n-1} + rx^{n-2} + \dots + r^{n-1}) + a_{n-1}(x^{n-2} + \dots +r^{n-1}) + \dots + a_1)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once we know a root, we can also try using &lt;a href=&quot;http://www.sosmath.com/algebra/factor/fac01/fac01.html&quot; rel=&quot;nofollow&quot;&gt;Polynomial Long Division&lt;/a&gt; to get the other factor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For cubics, the roots can be found without the need to guess. Check this out: &lt;a href=&quot;http://en.wikipedia.org/wiki/Cubic_function#Cardano.27s_method&quot; rel=&quot;nofollow&quot;&gt;Cardano's Method&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does that help?&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-09-28T22:17:37.967" LastActivityDate="2010-09-28T22:17:37.967" CommentCount="4" />
  <row Id="5650" PostTypeId="1" AcceptedAnswerId="5667" CreationDate="2010-09-28T20:14:32.190" Score="2" ViewCount="187" Body="&lt;p&gt;Right, this is an exercise in Apostol, which i am not being able to solve. I was able to prove this result for a small case, that is the case when $n=2$, $[x] + \Bigl[x + \frac{1}{2}\Bigr]=[2x]$, but i am struggling with the generalization.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Prove that $$\sum\limits_{k=0}^{n-1} \Biggl[x + \frac{k}{n}\Biggr] = [nx]$$, where $[ \ ]$ denotes the greatest integer function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can this be proved via induction, i ask this because i have shown it to bealready true for $n=2$. &lt;/p&gt;&#xA;" LastActivityDate="2010-09-29T21:12:36.650" Title="Proving $\sum\limits_{k=0}^{n-1} \Bigl[x + \frac{k}{n}\Bigr] = [nx]$" Tags="&lt;functions&gt;" AnswerCount="2" />
  <row Id="5651" PostTypeId="2" ParentId="5647" CreationDate="2010-09-28T20:17:25.490" Score="3" ViewCount="" Body="&lt;p&gt;What shape do you get when you fix y and consider all values of x? Then see how the shape depends on y. Of course, you could try the reverse as well (fix x and vary y first), but my way, you don't have to deal with the &quot;y &gt; 0&quot; condition until the end. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, draw a picture showing the images of the lines x=constant and y=constant. You get two families of curves that are orthogonal to one another (since sin(z) is holomorphic and thus generally preserves angles). I promise that this will be very enlightening. &lt;/p&gt;&#xA;" OwnerUserId="2032" LastActivityDate="2010-09-28T20:17:25.490" />
  <row Id="5652" PostTypeId="2" ParentId="5650" CreationDate="2010-09-28T20:24:09.220" Score="4" ViewCount="" Body="&lt;p&gt;It is enough to prove it for $0 &amp;lt; x &amp;lt; 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, let $M$ be an integer such that, $\frac{M}{n} \le x &amp;lt; \frac{M+1}{n}$ where $0 \le M &amp;lt; n$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus $[nx] = M$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $0 \le k \le n-M-1$, we have that $[x+\frac{k}{n}] = 0$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $n-1 \ge k &gt; n-M-1$ we have that $[x + \frac{k}{n}] = 1$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The result follows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't think induction can be used here.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-09-28T20:31:00.237" LastActivityDate="2010-09-28T20:31:00.237" />
  <row Id="5653" PostTypeId="2" ParentId="5647" CreationDate="2010-09-28T20:24:30.960" Score="2" ViewCount="" Body="&lt;p&gt;The image is the whole complex plane minus the interval $[-1,1]$ (on the real axis) and the negative imaginary axis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.mai.liu.se/~halun/complex/domain_coloring-unicode.html#fig%3asin&quot; rel=&quot;nofollow&quot;&gt;Here is a figure&lt;/a&gt; illustrating the mapping. (The color-coding is explained at the top of that page.)&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2010-09-28T20:24:30.960" CommentCount="2" />
  <row Id="5654" PostTypeId="2" ParentId="5646" CreationDate="2010-09-28T20:29:57.350" Score="4" ViewCount="" Body="&lt;p&gt;By the &lt;a href=&quot;http://mathworld.wolfram.com/RationalZeroTheorem.html&quot; rel=&quot;nofollow&quot;&gt;Rational Zero Theorem&lt;/a&gt; all the rational roots of $x^{3}-12x+9$ must have a numerator which is a factor of $9$ and a denominator which is a factor of $1$. Therefore they have to be of the form $\frac{9}{1}=9$ or $\frac{3}{1}=3$. Let $f(x)=x^{3}-12x+9$. Since $f(9)=630$ and $f(3)=0$, $3$ is a root of $f(x)$. So it can be factored as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x^{3}-12x+9=(x-3)\left( ax^{2}+bx+c\right) =ax^{3}+\left( b-3a\right)&#xA;x^{2}+\left( c-3b\right) x-3c$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Comparing coefficients we get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a=1,b-3=0\iff b=3,-3c=9\iff c=-3$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x^{3}-12x+9=(x-3)\left( x^{2}+3x-3\right) $.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS.  Or we could apply &lt;a href=&quot;http://en.wikipedia.org/wiki/Ruffini%27s_rule&quot; rel=&quot;nofollow&quot;&gt;Ruffini's rule&lt;/a&gt; to find the coefficients of $ax^{2}+bx+c$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PPS. As commented by user1827 &quot;The rational zero theorem&quot; also permits &quot;$1/1=1$ and $-9,-3,-1$. But they are not zeros.&quot; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, since $f(3)=0$ we can factor $f(x)$ rightaway, without taking into consideration all the remaining possibilities.&lt;/p&gt;&#xA;" OwnerUserId="752" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2010-09-28T23:27:48.137" LastActivityDate="2010-09-28T23:27:48.137" CommentCount="4" />
  <row Id="5655" PostTypeId="2" ParentId="5646" CreationDate="2010-09-28T20:38:04.710" Score="3" ViewCount="" Body="&lt;p&gt;It's not a special case of any general factorization rule (except perhaps the &lt;a href=&quot;http://en.wikipedia.org/wiki/Rational_root_theorem&quot; rel=&quot;nofollow&quot;&gt;Rational Root Test&lt;/a&gt;).&lt;br&gt;&#xA;However, one easy ad-hoc way to derive the factorization is to notice&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\quad\quad\  f(x) + 3\:(x-3)\ =\ x\: (x^2 - 9)  $  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus $\rm\ \ f(x)\ =\ (x-3)\ (x\: (x+3) - 3)$ &lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-09-28T20:38:04.710" />
  <row Id="5656" PostTypeId="1" AcceptedAnswerId="5674" CreationDate="2010-09-28T20:46:44.590" Score="1" ViewCount="197" Body="&lt;p&gt;Looking at a 2&lt;sup&gt;nd&lt;/sup&gt;-order Taylor series approximation of the function $f$, I have this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f(t_1) = f(t_0) + hf'(t_0) + {h^2\over 2}f''(t_0) + O(h^3)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now say I approximate $f''(t0)$ with a 2&lt;sup&gt;nd&lt;/sup&gt;-order central difference method:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f''(t) = {1\over 2}{f'(t+h) - f'(t-h)\over h} + O(h^2)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What's the resulting error of this method?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The naive approach would be to substitute the central difference equation into the Taylor series, giving something like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f(t_1) = f(t_0) + hf'(t_0) + {h\over 4}(f'(t_0+h)-f'(t_0-h)) + {1\over 2}O(h^4) + O(h^3)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is that plausible? Would error actually &lt;em&gt;decrease&lt;/em&gt; (go from 2&lt;sup&gt;nd&lt;/sup&gt;-order to 4&lt;sup&gt;th&lt;/sup&gt;-order)?&lt;/p&gt;&#xA;" OwnerUserId="2192" LastEditorUserId="2192" LastEditorDisplayName="" LastEditDate="2010-09-28T21:01:43.040" LastActivityDate="2010-09-29T15:35:36.130" Title="How do I combine &quot;error of order&quot; terms in numerical analysis?" Tags="&lt;numerical-methods&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" />
  <row Id="5657" PostTypeId="1" AcceptedAnswerId="5673" CreationDate="2010-09-28T20:54:48.890" Score="3" ViewCount="137" Body="&lt;p&gt;I made a picture of the problem here:&#xA;&lt;img src=&quot;http://i.imgur.com/phRsj.png&quot; alt=&quot;(statement of the problem, rendered as an image)&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the link does not work, read this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $a$ be a non-zero vector in $\mathbb{R}^n$. Let S be the set of all orthogonal vectors to $a$ in $\mathbb{R}^n$. I.e., for all $x \in \mathbb{R}^n$, $a\cdot x = 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Prove that the interior of S is empty.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I show that for every point in S, all &quot;close&quot; points are either in the complement of S or in S itself?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is what I attempted:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $u\in B(r,x) = \{ v \in \mathbb{R}^n : |v - x| &amp;lt; r \} $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So $|u - x| &amp;lt; r$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, $|a||u - x| &amp;lt; |a|r$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By Cauchy-Schwarz, $|a\cdot(u-x)| \leq |a||u - x|$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, $|a\cdot u - a\cdot x| &amp;lt; |a|r$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $u\in S$, then either $u\in S^{\text{int}}$ or $u\in \delta S$.  ($\delta$ denotes boundary).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $u\in S^{\text{int}}$, then $B(r,x) \subset S$, and $a\cdot u = 0$. But then the inequality becomes $|a|r &gt; 0$ which implies $B(r,x) \subset S \forall r &gt; 0$, but this is impossible since it would also imply that $S = \mathbb{R}^n$ and $S^c$ is empty, which is false. Therefore, if $u\in S$, then $u\in\delta S$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, $\forall u\in B(r, x)$ such that $u\in S$, $u\in\delta S$. Thus $S^{\text{int}}$ is empty.&lt;/p&gt;&#xA;" OwnerUserId="2193" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-09-28T23:28:03.907" LastActivityDate="2010-09-29T04:13:52.067" Title="Prove that the interior of the set of all orthogonal vectors to &quot;a&quot; is empty." Tags="&lt;topology&gt;" AnswerCount="4" CommentCount="3" FavoriteCount="2" />
  <row Id="5659" PostTypeId="2" ParentId="5191" CreationDate="2010-09-28T21:24:13.810" Score="1" ViewCount="" Body="&lt;p&gt;You can do with something very slightly weaker than an ordering: an identification of each $n$-element set with a single &quot;universal&quot; (unordered) $n$-set.  This data canonically identifies any two $n$-element sets and thus associates a permutation to any bijection of such sets.  Even bijections can then be defined in terms of the cycle structure of the permutations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is not much of an improvement, but you are in effect asking for a lifting of the alternating group to a groupoid of &lt;em&gt;maps&lt;/em&gt; between finite sets.  It is hard to see how to determine whether a bijection of finite sets is even (reducing to the usual notion when the sets are the same, and also &quot;transporting structure&quot; along the whole category) without having a coordinatization of the sets.&lt;/p&gt;&#xA;" OwnerUserId="467" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-09-28T22:21:54.793" LastActivityDate="2010-09-28T22:21:54.793" />
  <row Id="5660" PostTypeId="2" ParentId="5657" CreationDate="2010-09-28T21:29:46.407" Score="2" ViewCount="" Body="&lt;p&gt;Consider the function $\phi:x\in\mathbb R^n\mapsto x\cdot a\in\mathbb R$, which is clearly linear. If $x_0$ is a point in the interior of your set, compute the derivatives of $\phi$ at $x_0$. Now look at the Taylor development of $\phi$ at $x_0$...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A more geometric way: suppose that $x$ is orthogonal to $a$ (and that $a\neq0$) and consider the sequence $(x_n)_{n\geq1}$ with $x_n=x+\tfrac1na$ for each $n$. Show that $x_n\to x$ when $n\to\infty$ and then see why this is useful for you.&lt;/p&gt;&#xA;" OwnerUserId="274" LastEditorUserId="274" LastEditorDisplayName="" LastEditDate="2010-09-28T21:54:57.497" LastActivityDate="2010-09-28T21:54:57.497" CommentCount="0" />
  <row Id="5661" PostTypeId="1" AcceptedAnswerId="5663" CreationDate="2010-09-28T22:11:03.640" Score="2" ViewCount="236" Body="&lt;p&gt;I was reading about Euler's totient function on wikipedia, and it eventually led me to this book on google:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://books.google.com/books?id=1MTcYrbTdsUC&amp;amp;pg=PA74&quot; rel=&quot;nofollow&quot;&gt;Page 74 of the book, Prime numbers: the most mysterious figures in math&#xA; By David G. Wells&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, the books lists many assertions without proof, only references which I can't find. One I could not solve for myself said that not all even numbers are values of $\phi(n)$. The sequence of even non-values of $\phi(n)$ starts:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$14, 26, 34, 38, 50, 62,\dots$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After thinking about it for a while, I've made little headway. Looking at 14 specifically, I suppose if such a solution $a$ to $\phi(x)=14$ were to exist, $(a,m_i)=1$ for $1\leq i\leq 14$, for some $m_i\lt a$ and so there must also exist inverses $\overline{a_i}$ such that $\overline{a_i}a\equiv 1 \pmod {m_i}$ for each $i$. I'm looking for some contradiction, possibly of the Chinese Remainder Theorem, to show such $a$ cannot exist. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there some way to generalize which values are not taken by $\phi$, or at least explain why this is the case? I was hoping to see why $14$ is the least such integer such that this is true, but values $1$ to $13$ must be indeed taken. I suppose this would also explain why 26 is the next such value that is not taken, while $15$ to $25$ are.&lt;/p&gt;&#xA;" OwnerUserId="1583" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-09-28T22:20:29.707" LastActivityDate="2010-09-28T23:09:58.773" Title="Generalizing values which Euler's-totient function does not take." Tags="&lt;number-theory&gt;&lt;prime-numbers&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="5663" PostTypeId="2" ParentId="5661" CreationDate="2010-09-28T22:47:38.607" Score="3" ViewCount="" Body="&lt;p&gt;It's just casework on the prime factorization of $n$.  If $\phi(n) = 14$ then $n$ can't be divisible by any primes larger than $7$ (because $p-1$ cannot divide $14$ for such primes), and it can't be divisible by any of $5, 7$ because $4, 6$ don't divide $14$.  So it can only be divisible by $2$ or $3$.  Since $14$ is not divisible by $3$, $3$ can only divide $n$ at most once, so $n$ is of the form $2^k 3$.  But then $\phi(n) = 2^k$; contradiction.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, if $\phi(n) = 26$ then $n$ can't be divisible by any primes larger than $13$, and it can't be divisible by any of $5, 7, 11, 13$ because $4, 6, 10, 12$ don't divide $26$.  So it can only be divisible by $2$ or $3$, and then the same argument as above works.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For any particular potential value of $\phi(n)$ a similar, but longer, argument should work; in particular it should be possible to test in finite time whether a particular candidate works.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-09-28T22:47:38.607" CommentCount="1" />
  <row Id="5664" PostTypeId="2" ParentId="5661" CreationDate="2010-09-28T22:50:54.370" Score="3" ViewCount="" Body="&lt;p&gt;From the comments to &lt;a href=&quot;http://www.research.att.com/~njas/sequences/?q=A005277&amp;amp;sort=0&amp;amp;fmt=0&amp;amp;language=english&amp;amp;go=Search/%22A005277%22&quot; rel=&quot;nofollow&quot;&gt;A005277&lt;/a&gt; :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If p is prime then the following two statements are true. I. 2p is in the sequence iff 2p+1 is composite (p is not a Sophie Germain prime). II. 4p is in the sequence iff 2p+1 and 4p+1 are composite. - Farideh Firoozbakht (mymontain(AT)yahoo.com), Dec 30 2005 &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This covers most of your cases, 50 is covered by the next comment.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-09-28T22:50:54.370" CommentCount="2" />
  <row Id="5666" PostTypeId="1" AcceptedAnswerId="5669" CreationDate="2010-09-28T23:09:53.210" Score="3" ViewCount="339" Body="&lt;p&gt;I was in the middle of writing the same old geographic distance calculation using the &lt;a href=&quot;http://en.wikipedia.org/wiki/Haversine_formula&quot; rel=&quot;nofollow&quot;&gt;Haversine formula&lt;/a&gt; when it occurred to me: shouldn't there be simpler way to do this? Haversine is of course derived from the Law of Cosines. But in thinking about this problem, I ran across &lt;a href=&quot;http://en.wikipedia.org/wiki/Spherical_trigonometry&quot; rel=&quot;nofollow&quot;&gt;Napier's Rules&lt;/a&gt; for right-angled spherical triangles. It seems like Napier's Rules should apply, after all Latitude small circles and Longitude great circles always intersect at right angles, so you should be able to draw a right-angled spherical triangle where the hypotenuse connects any two points (unless the case devolves into a line or single point).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if I'm applying Napier's Rules correctly, if our delta latitude and delta longitude in radians are $a$ and $b$ respectively, the angle $c$ should trace the arc between the two points. So Napier says:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sin(\pi-c) = \cos(a) \cos(b)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which should then simplify to:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$c = \arcsin(\cos(a) \cos(b))$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But when I try to verify this with a couple test points, the result doesn't match the Haversine formula. Is there a mistake in my (admittedly rusty) algebra or is my mistake in assuming I can apply Napier's Rules to this problem?&lt;/p&gt;&#xA;" OwnerUserId="2195" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-09-28T23:22:48.820" LastActivityDate="2010-09-28T23:22:48.820" Title="Napier's Rules applied to spherical distance calculations" Tags="&lt;geometry&gt;&lt;trigonometry&gt;&lt;spherical-geometry&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="5667" PostTypeId="2" ParentId="5650" CreationDate="2010-09-28T23:16:08.783" Score="5" ViewCount="" Body="&lt;p&gt;Both sides are equal since they count the same set: the RHS counts naturals $\rm\:\le n\:x\:$. The LHS counts them in a unique mod $\rm\ n\ $ representation, $\:$ viz. $\rm\ \: j \:\le\: x+k/n\: \iff \ j\:n-k \:\le\: n\:x\:,\ \ j&gt;0 \le k &amp;lt; n\:$.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;REMARK&lt;/strong&gt; $\:$ That every natural has a unique representation of form $\rm \: j\:n-k \ \ \:$ where $\rm\ \ \: j&gt;0 \le k &amp;lt; n\ \ \ $ is simply a slight variant of the &lt;strong&gt;Division Algorithm&lt;/strong&gt; where one utilizes negative (vs. positive) remainders.$\ \ $ To derive this negative form, simply  perform the following transformation on the positive remainder form $\rm\ q\: n + r\ \to\ (q+1)\:n + r-n\ $ if $\rm\ r\ne 0\:$, i.e. inc the quotient, dec the remainder by the dividend. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus the result is equivalent to the Division Algorithm, whose normal proof is indeed by &lt;em&gt;induction&lt;/em&gt;. One could give a direct inductive proof of the result if, instead of invoking the Division Algorithm by name, one unwinds or inlines this inductive proof directly into the proof of the result - much as the same way that the classic &lt;a href=&quot;http://www.math.uiuc.edu/~dan/ShortProofs/uf.html&quot; rel=&quot;nofollow&quot;&gt;Lindenmann - Zermelo direct proof of unique factorization of naturals&lt;/a&gt; inlines a division / Euclidean algorithm based descent proof of the fundamental Prime Divisor property $\rm\ p|ab\ \Rightarrow\ p|a\ \ or\ \ p|b\:$.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-09-29T21:12:36.650" LastActivityDate="2010-09-29T21:12:36.650" />
  <row Id="5668" PostTypeId="2" ParentId="5191" CreationDate="2010-09-28T23:20:45.783" Score="4" ViewCount="" Body="&lt;p&gt;$A_n$ is the symmetry group of the chamber of the Tits building of $\mathbb{P}GL_n$. The shape of this chamber is independent of what coefficients you insert into the group scheme $\mathbb{P}GL_n$, just the number and configuration of chambers changes. If you insert the finite fields $\mathbb{F}_p$ then you get finite simplicial complexes as buildings, and the smaller $p$ gets, the fewer chambers you have. You can analyse and even reconstruct the group in terms of its action on this building. The natural limit case would be just having one chamber and the symmetry group of this chamber - the Weyl group - is $A_n$. This is how Tits first thought that there is a limit case to the sequence of finite fields - which he called the field with one element.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe somewhat more algebraically you can think in terms of Lie algebras - as I said the shape of the chamber does not change with different coefficients. The reason is that it is determined just by the Lie algebra of the group and thus describable by a Dynkin diagram or by a root system (ok, geometry creeps in again). The Wikipedia page about &lt;a href=&quot;http://en.wikipedia.org/wiki/Weyl_group&quot; rel=&quot;nofollow&quot;&gt;Weyl groups&lt;/a&gt; tells you that the Weyl group of the Lie algebra $sl_n$ is $S_n$. If have no experience with Lie algebras, but maybe you can get $A_n$ the same way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you can get hold of it, you can read Tits' original account, it's nice to read (but geometric) see the reference on this &lt;a href=&quot;http://en.wikipedia.org/wiki/Field_with_one_element&quot; rel=&quot;nofollow&quot;&gt;Wikipedia page&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: Aha, I found a link now: Lieven Le Bruyn's &lt;a href=&quot;http://matrix.cmi.ua.ac.be/fun/index.php/kapranov-smirnov-on-f_un.html&quot; rel=&quot;nofollow&quot;&gt;F_un page&lt;/a&gt; is back online. You can look there under &quot;papers&quot; and find Tits' article. And, since you are picking up the determinant ideas, you should definitely take a look at Kapranov/Smirnov!&lt;/p&gt;&#xA;" OwnerUserId="2048" LastEditorUserId="2048" LastEditorDisplayName="" LastEditDate="2010-09-29T17:43:34.493" LastActivityDate="2010-09-29T17:43:34.493" CommentCount="5" />
  <row Id="5669" PostTypeId="2" ParentId="5666" CreationDate="2010-09-28T23:22:29.867" Score="4" ViewCount="" Body="&lt;p&gt;Latitude &quot;lines&quot; are not spherical-geometry lines, as they are not great circles, so the &quot;triangles&quot; you are getting are not spherical triangles.&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-09-28T23:22:29.867" CommentCount="7" />
  <row Id="5670" PostTypeId="1" CreationDate="2010-09-29T00:27:28.717" Score="4" ViewCount="236" Body="&lt;p&gt;There is a lemma in my introduction to analysis book for which the proof eludes me. I was hoping to get some clarification. Here goes:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lemma: If $\lbrace b_n \rbrace_{n=1}^\infty$ converges to B and B $\not=$ 0, then there is a positive real number $M$ and a positive integer $N$ such that if $n \geq N$, then $\mid b_n \mid \geq M$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proof: Since $B \not= 0, \frac{\mid B \mid}{2} = \epsilon &gt; 0.$ There is $N$ such that if $n \geq N, \mid b_n - B \mid &amp;lt; \epsilon.$ Let $M = \frac{\mid B \mid}{2}.$ Thus for $n \geq N$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\mid b_n \mid = \mid b_n - B + B \mid \geq \mid B \mid - \mid b_n - B \mid \geq \mid B \mid - \frac{\mid B \mid}{2} = \frac{\mid B \mid}{2} = M$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The lemma makes intuitive sense to me because of course if a sequence converges to a number other than 0, its absolute value will always be above some number $M$ for $n \geq N$. However, I get lost in the inequalities of the proof. I don't understand how the author came up with $\mid b_n - B + B \mid \geq \mid B \mid - \mid b_n - B \mid$. I understand that he added and subtracted $B$ to $b_n$ for the left side, but I'm not so clear on the how he came up with $\mid B \mid - \mid b_n - B \mid$. It looks like an iteration of the triangle inequality, but I don't quite see it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an aside, the book had an exercise for a variation on this lemma that involved proving $b_n \geq M$ for all $n$. The proof for that involved setting $M = min \lbrace b_1, b_2, ..., b_n \rbrace$, which made a lot more sense to me than the inequalities for this proof.&lt;/p&gt;&#xA;" OwnerUserId="2175" LastEditorUserId="2175" LastEditorDisplayName="" LastEditDate="2010-09-30T03:53:22.440" LastActivityDate="2010-09-30T03:53:22.440" Title="Clarification of a sequence convergence proof." Tags="&lt;sequences-and-series&gt;&lt;proof-strategy&gt;" AnswerCount="2" />
  <row Id="5671" PostTypeId="2" ParentId="5657" CreationDate="2010-09-29T00:34:24.367" Score="0" ViewCount="" Body="&lt;p&gt;I would do it this way. First of all, by choosing $\frac{a}{\|a\|}$ as the last vector of an orthonormal basis of $\mathbb{R}^n$, you may assume that $a = (0,\dots , 0,\alpha), \alpha \in \mathbb{R}, \alpha \neq 0$. Hence &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;S =\left\{ (x_1 , \dots , x_n) \in \mathbb{R}^n\ \vert \ x_n = 0    \right\} \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, pick any $x = (x_1 , \dots , x_{n-1} , 0) \in S$ and let's show that, for any $r&gt;0$, $B(r,x) $ is not included in $S$, so the interior of $S$ will be empty.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To this end it's enough to produce a single vector in $B(r,x)$ which is not in $S$, right? Ok, take a look at $v = (x_1, \dots , x_{n-1}, \frac{r}{2})$ and compute (draw a picture in $\mathbb{R}^3$ too to convince yourself).&lt;/p&gt;&#xA;" OwnerUserId="664" LastActivityDate="2010-09-29T00:34:24.367" />
  <row Id="5672" PostTypeId="2" ParentId="5670" CreationDate="2010-09-29T00:51:25.863" Score="3" ViewCount="" Body="&lt;p&gt;You are right, it's just the triangle inequality. $B=b_n+(B-b_n)$. So $|B|\leq |b_n| + |B-b_n|$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the way, shouldn't it be $n \geq N$ in your second para?&lt;/p&gt;&#xA;" OwnerUserId="1195" LastActivityDate="2010-09-29T00:51:25.863" CommentCount="2" />
  <row Id="5673" PostTypeId="2" ParentId="5657" CreationDate="2010-09-29T01:00:08.310" Score="1" ViewCount="" Body="&lt;p&gt;It is a tautology to say &quot;every point in S, all 'close' points are either in the complement of S or in S itself&quot;. Also I am not sure how you use $a|r|&gt;0$ to conclude that  $B(r,x) \subset S \, \forall r&gt;0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway you don't need to think of all &quot;close&quot; points. You just need to prove that arbitrarily close to any $x \in S$ there is at least one point which is not in $S$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint: think of points of the form $x+\lambda a$.&lt;/p&gt;&#xA;" OwnerUserId="1195" LastEditorUserId="1195" LastEditorDisplayName="" LastEditDate="2010-09-29T03:25:21.597" LastActivityDate="2010-09-29T03:25:21.597" CommentCount="1" />
  <row Id="5674" PostTypeId="2" ParentId="5656" CreationDate="2010-09-29T01:03:50.563" Score="2" ViewCount="" Body="&lt;p&gt;If we start with the Taylor expansion (I'll change variables here, too many subscripts confuse me):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f(x+h)=f(x)+h f^{\prime}(x)+\frac{h^2}{2}f^{\prime\prime}(x)+\frac{h^3}{3!}f^{\prime\prime\prime}(x)+O(h^4)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and the derivative of this w.r.t. $h$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f^{\prime}(x+h)=f^{\prime}(x)+h f^{\prime\prime}(x)+\frac{h^2}{2}f^{\prime\prime\prime}(x)+O(h^3)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and the version of this with $h$ replaced by negative $h$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f^{\prime}(x-h)=f^{\prime}(x)-h f^{\prime\prime}(x)+\frac{h^2}{2}f^{\prime\prime\prime}(x)+O(h^3)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;subtracting the third expression from the second expression gives&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f^{\prime}(x+h)-f^{\prime}(x-h)=2h f^{\prime\prime}(x)+O(h^3)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and we see that the even powers drop out of this error expansion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we solve for $f^{\prime\prime}(x)$ like so:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f^{\prime\prime}(x)=\frac{f^{\prime}(x+h)-f^{\prime}(x-h)}{2h}+O(h^2)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and substitute in the first expression,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f(x+h)=f(x)+h f^{\prime}(x)+\frac{h^2}{2}\left(\frac{f^{\prime}(x+h)-f^{\prime}(x-h)}{2h}+O(h^2)\right)+\frac{h^3}{3!}f^{\prime\prime\prime}(x)+O(h^4)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;we can take the $O(h^2)$ within the parentheses out as an $O(h^4)$ term:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f(x+h)=f(x)+h f^{\prime}(x)+\frac{h}{2}\left(\frac{f^{\prime}(x+h)-f^{\prime}(x-h)}{2}\right)+\frac{h^3}{3!}f^{\prime\prime\prime}(x)+O(h^4)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the leading term after the replaced portion is $O(h^3)$, thus simplifying to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f(x+h)=f(x)+h f^{\prime}(x)+\frac{h}{2}\left(\frac{f^{\prime}(x+h)-f^{\prime}(x-h)}{2}\right)+O(h^3)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and we see that the formula has $O(h^3)$ error: cutting $h$ in half decreases the error by a factor of $2^3=8$.&lt;/p&gt;&#xA;" OwnerUserId="498" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-09-29T15:35:36.130" LastActivityDate="2010-09-29T15:35:36.130" CommentCount="7" />
  <row Id="5675" PostTypeId="1" CreationDate="2010-09-29T02:29:54.573" Score="3" ViewCount="157" Body="&lt;p&gt;Me and a team of programmers are programming a robot to play a game in a competition. We did not create the hardware, and all teams will use the same type of robot.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;hr /&gt;&#xA;&lt;strong&gt;GAME DESCRIPTION&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The game has two players opposing each other and are both trying to achieve an objective. This objective is to move to a position to pick up a payload then return to another position to return the payload. Each team has their own payload and their own target to bring them to, and it is impossible to pick up the other team's payload. The team that brings their payload to the target first wins and the game is ended. There is a time limit of 210 seconds, and if the games times out, the team who held the payload the longest wins.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, it is a little more complicated than that. The robots also have the ability to face the opponent and &quot;push them back&quot; from any range (the closer they are the more&#xA;forcefully they push). Also, if a robot is pushed out of the bounds of the playing field, they drop their payload and it is moved back inside the playing field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;hr /&gt;&#xA;&lt;strong&gt;GAME THEORY QUESTIONS&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First of all, is this a zero-sum game or not? I am new to game theory math and I am not completely sure if it is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, how does the minimax theorem apply to this specific game? I understand how it works, but I do not know what the values to account for would be in this game (would it be the difference in how long each team has held the payload somehow combined with how close they are to bringing to the target?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I really am not sure at all how to calculate any of this, but if anyone can at least point me in the right direction for coming up with an effective strategy system for this I would be very appreciative.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you so much for your time, this project is very important to me and your help means a lot. If I need to clarify anything please ask, I am not sure if I included all the information needed for this problem.&lt;/p&gt;&#xA;" LastActivityDate="2010-09-29T03:43:02.943" Title="Zero-Sum Game Theory" Tags="&lt;game-theory&gt;" AnswerCount="2" />
  <row Id="5676" PostTypeId="1" CreationDate="2010-09-29T02:34:58.383" Score="10" ViewCount="443" Body="&lt;p&gt;By looking at an integral and bounding the error?&lt;/p&gt;&#xA;" OwnerUserId="1838" LastActivityDate="2010-09-30T17:44:02.330" Title="How closely can we estimate $\sum_{i=0}^n \sqrt{i}$" Tags="&lt;sequences-and-series&gt;&lt;integral&gt;" AnswerCount="6" CommentCount="6" FavoriteCount="2" />
  <row Id="5677" PostTypeId="2" ParentId="5675" CreationDate="2010-09-29T02:40:56.580" Score="2" ViewCount="" Body="&lt;p&gt;Yes, this is a zero sum game, since both players are fighting over a single &quot;win&quot;. What that means in terms of strategy is that hurting your opponent is as effective as helping yourself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't see how you would use minmax theory in this situation. As far as I know, game theory tools are geared towards games with a relatively small number of possible situations and moves, whereas in your game, you have a very large number of possible situations (because the position of the robots is variable).&lt;/p&gt;&#xA;" OwnerUserId="796" LastActivityDate="2010-09-29T02:40:56.580" />
  <row Id="5678" PostTypeId="2" ParentId="5676" CreationDate="2010-09-29T02:49:06.793" Score="2" ViewCount="" Body="&lt;p&gt;Since square root is a monotonically increasing function, that summation will be between the integral of sqrt(i) from 0 to n and the integral of sqrt(i) from 1 to n-1.&lt;/p&gt;&#xA;" OwnerUserId="796" LastActivityDate="2010-09-29T02:49:06.793" />
  <row Id="5679" PostTypeId="2" ParentId="5676" CreationDate="2010-09-29T03:01:32.453" Score="3" ViewCount="" Body="&lt;p&gt;The standard approach should be along the lines of $\sum_{i\le n}\sqrt i=\sqrt n\sum_{i\le n}\sqrt{\frac{i}{n}}$, so $\frac1{n\sqrt n}\sum_{i\le n}\sqrt i=\sum_{i\le n}\sqrt{\frac{i}{n}}\frac1n\to\int_0^1\sqrt x{} dx=\frac23$, or $\sum_{i\le n}\sqrt i\sim\frac23 n\sqrt n$. What techniques do you know to estimate the error between integrals and their Riemann sums?&lt;/p&gt;&#xA;" OwnerUserId="462" LastActivityDate="2010-09-29T03:01:32.453" CommentCount="1" />
  <row Id="5680" PostTypeId="2" ParentId="5675" CreationDate="2010-09-29T03:43:02.943" Score="0" ViewCount="" Body="&lt;p&gt;This is a zero sum game &lt;a href=&quot;http://en.wikipedia.org/wiki/Differential_game&quot; rel=&quot;nofollow&quot;&gt;differential game&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your strategy is a &lt;em&gt;function&lt;/em&gt; from the state of the board (everyone's position and velocity and how long they have held their payload) to your control inputs. Your payoff is 1 if you win and 0 if you lose. [Which is why this is a zero sum game, since the sum of payoffs over all players equals 1]. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Having said that, the actual solved examples of differential games I have seen have been for simple two player problems. You may be better off using some kind of heuristic for the contest.&lt;/p&gt;&#xA;" OwnerUserId="1195" LastActivityDate="2010-09-29T03:43:02.943" />
  <row Id="5681" PostTypeId="2" ParentId="5676" CreationDate="2010-09-29T03:59:57.267" Score="5" ViewCount="" Body="&lt;p&gt;The &quot;sophisticated&quot; way to do this is using the &lt;a href=&quot;http://en.wikipedia.org/wiki/Euler%E2%80%93Maclaurin_formula&quot; rel=&quot;nofollow&quot;&gt;Euler-Maclaurin formula&lt;/a&gt; - look under the heading &quot;Asymptotic expansion of sums&quot;. You'll want to start your sum at 1, not at 0, to avoid dividing by zero.&lt;/p&gt;&#xA;" OwnerUserId="173" LastActivityDate="2010-09-29T03:59:57.267" />
  <row Id="5682" PostTypeId="2" ParentId="5657" CreationDate="2010-09-29T04:13:52.067" Score="0" ViewCount="" Body="&lt;p&gt;The condition that $x$ be orthogonal to $a$, i.e. that $x$ lies in $S$, is that $x \cdot a = 0$.  Imagine perturbing $x$ by a small amount, say to $x'$.  If $x$ were in the interior,&#xA;than one would have $x' \cdot a = 0$ as well, provided that $x'$ is very close to $x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Think about whether this is possible for every $x'$.  (Hint: $x'$ has to be close to $x$,&#xA;i.e. $x - x'$ has to be small.  But it can point in any direction!)&lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2010-09-29T04:13:52.067" />
  <row Id="5683" PostTypeId="2" ParentId="5670" CreationDate="2010-09-29T04:24:56.963" Score="2" ViewCount="" Body="&lt;p&gt;Just a remark (that is too long to put in a comment box): The idea of the proof is just this, that if $x_n$ approaches $B$,&#xA;then eventually (i.e. for large enough $n$) it has to be within distance $|B/2|$ of $B$, &#xA;and once this happens we can be sure that $x_n \geq B/2$ (if $B$ is positive) or that&#xA;$x_n \leq B/2$ (if $B$ is negative).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If this isn't clear just from how I've written it, draw  picture: label $B$,&#xA;draw the interval of points within distance $|B/2|$ of $B$, which is the interval&#xA;$[B/2, 3B/2]$ (or switch the endpoints if $B$ is negative), and then just randomly&#xA;draw in a sequence of points getting closer and closer to $B$.  You'll see that eventually&#xA;you have to put them all inside this interval, so they will all be at least $B/2$&#xA;(or at most $B/2$ if $B$ is negative).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The manipulations with absolute values are just a slightly tortuous verification that&#xA;if $x$ is within distance $|B/2|$ of $B$, then $x$ is at least $B/2$ (or at most $B/2$&#xA;is $B$ is negative).  It is easy to write down statements that are obvious when you say&#xA;them out like this, or when you draw a picture, but which become a bit of a battle when&#xA;you try to derive them formally from the triangle inequality.  The fact that the author&#xA;is treating the case when $B$ is negative at the same time adds another small but distracting layer of complexity, because you have to write $|B/2|$ rather than $B/2$ for the distance under consideration.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you become more familiar with these sorts of manipulations, when you come to a point&#xA;like this in a text where you recognize some kind of triangle inequality manipulations, it is often easier (and good training!) just to draw a picture and figure out the manipulations for yourself, rather than working through somebody else's slightly painful manipulations on the page.  &lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2010-09-29T04:24:56.963" CommentCount="3" />
  <row Id="5684" PostTypeId="1" AcceptedAnswerId="5687" CreationDate="2010-09-29T05:21:05.217" Score="3" ViewCount="189" Body="&lt;p&gt;An exercise from Herstein asks to prove that the number of elements of order $p$, $p$ a prime in $S_{p}$ is $(p-1)!+1$. I would like somebody to help me out on this and also i would like to know whether can we prove Wilson's theorem which says $(p-1)! \equiv -1 \ (\text{mod} \ p)$ using this result.&lt;/p&gt;&#xA;" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-09-29T23:17:03.257" LastActivityDate="2010-09-29T23:17:03.257" Title="Number of Elements of order $p$ in $S_{p}$" Tags="&lt;abstract-algebra&gt;&lt;group-theory&gt;" AnswerCount="2" />
  <row Id="5686" PostTypeId="2" ParentId="5648" CreationDate="2010-09-29T06:20:37.350" Score="1" ViewCount="" Body="&lt;p&gt;The answer here is no. I am assuming you're trying to prove the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given $\sigma$-finite measure spaces $\Omega_1, \Omega_2$, and $\Omega = \Omega_1\times\Omega_2$ carries the product measure, then if $h: \Omega\rightarrow \mathbb{R}$ is a measurable function, then $h$ is point-wise approximated almost everywhere by the sequence $h_n$ given by:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$h_n(x,t) = \sum_{j=1}^{N_n}f_{j,n}(x)1_{F_{j,n}}(t)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, in general this is not the case since the sets $F_{j,n}$ will generally depend on $x$. If $\Omega_1$ is countable, the situation is not too bad: the family of sets $\{F_{j,n}(x): x\in \Omega_1\}$ is countable. In general, however, the situation is hopeless. I think using $\Omega_1 = \Omega_2 = \mathbb{R}_{\geq 0}$ and $h(x,t) = x^2 + e^t$, or something to that effect, should do as a counter example. Play with it.&lt;/p&gt;&#xA;" OwnerUserId="2093" LastActivityDate="2010-09-29T06:20:37.350" CommentCount="3" />
  <row Id="5687" PostTypeId="2" ParentId="5684" CreationDate="2010-09-29T06:35:25.243" Score="4" ViewCount="" Body="&lt;p&gt;Maybe you mean the number of elements of order dividing $p$ (so that you are including the identity)?  (Think about the case $p = 3$ --- there are two three cycles, not three of them.)  For the general question, think about the possible cycle structure of an element of order $p$&#xA;in $S_p$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can go from the formula in your question to Wilson's theorem by counting the number of $p$-Sylow subgroups (each contains $p-1$ elements of order $p$), and then appealing to Sylow's theorem.  (You will find that there are $(p-2)!$ $p$-Sylow subgroups, and by Sylow's theorem this number is congruent to $1$ mod $p$.  Multiplying by $p-1$, we find that&#xA;$(p-1)!$ is congruent to $-1$ mod $p$.)&lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2010-09-29T06:35:25.243" />
  <row Id="5688" PostTypeId="2" ParentId="5124" CreationDate="2010-09-29T06:54:30.900" Score="0" ViewCount="" Body="&lt;p&gt;It isn't that we &quot;Want them to be,&quot; but that they must be simply from definition: Single point sets are necessarily closed in the topology of $\mathbb{R}$, hence are in $\mathcal{B}$ - the Borel $\sigma$-algebra. Since $\mathbb{Q}$ is a countable union of single point sets, it must be in $\mathcal{B}$. Hence so is $\mathbb{Q}^c$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If nothing else, this at least allows some nice examples. For instance, define $f(x)$ to be identically $1$ on $\mathbb{Q}$ and identically $0$ on $\mathbb{Q}^c$. $f$ is not Riemann integrable, however, $f$ is Lebesgue measurable, and integrates to zero in the Lebesgue sense (which is something one would expect).&lt;/p&gt;&#xA;" OwnerUserId="2093" LastActivityDate="2010-09-29T06:54:30.900" />
  <row Id="5689" PostTypeId="1" CreationDate="2010-09-29T07:19:26.897" Score="2" ViewCount="94" Body="&lt;p&gt;What is the general solution to the differential equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{\partial^2 u}{\partial y^2}=\frac{\partial u}{\partial x}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm a little stuck because all the techniques I know are unable to solve it.&lt;/p&gt;&#xA;" OwnerUserId="2148" LastActivityDate="2010-09-29T07:23:53.117" Title="Solution to 2nd order PDE" Tags="&lt;differential-equations&gt;" AnswerCount="1" FavoriteCount="2" />
  <row Id="5690" PostTypeId="2" ParentId="5689" CreationDate="2010-09-29T07:23:53.117" Score="4" ViewCount="" Body="&lt;p&gt;This is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Heat_equation&quot; rel=&quot;nofollow&quot;&gt;heat equation&lt;/a&gt;. Don't expect a single formula for the general solution, because the problem is too complicated for that. There are, however, fairly general formulas which apply under special circumstances (Fourier series solution for the initial–boundary value problem, etc.). &lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2010-09-29T07:23:53.117" />
  <row Id="5691" PostTypeId="2" ParentId="5676" CreationDate="2010-09-29T07:23:58.463" Score="1" ViewCount="" Body="&lt;p&gt;You will find some really nice approximations at &lt;a href=&quot;http://www.mathkb.com/Uwe/Forum.aspx/undergrad/3288/sequence-sum&quot; rel=&quot;nofollow&quot;&gt;MathKB&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS: You'll also find a couple of exact formulas there, though not written with elementary functions of course.&lt;/p&gt;&#xA;" OwnerUserId="1301" LastEditorUserId="1301" LastEditorDisplayName="" LastEditDate="2010-09-29T07:40:17.620" LastActivityDate="2010-09-29T07:40:17.620" />
  <row Id="5692" PostTypeId="2" ParentId="5564" CreationDate="2010-09-29T07:40:45.423" Score="5" ViewCount="" Body="&lt;p&gt;It doesn't have to: logics which don't are called paraconsistent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most important paraconsistent logic is relevance logic, which repudiates the K axiom:&#xA;$$\alpha \rightarrow \beta \rightarrow \alpha$$&#xA;and replaces it by axioms that do not allow there to be unused assumptions.  This is equivalent to saying weakening, the principle that if $\Gamma \vdash \alpha$ then $\Gamma'\vdash \alpha$ for $\Gamma\subset\Gamma'$.  This blocks derivations such as Weltschmertz's, which appeals to the K axiom once, Asaf's which uses it twice; Francesco appeals to monotonicity in his proof, which is another name for weakening.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's not difficult to see that this also blocks proofs of everything from a contradictory pair of propositions in a logic satisfying compactness, since one can prove inductively about such proof systems that if $\alpha\rightarrow\beta$, then all positive atoms in $\beta$ must occur either negatively in $\beta$ or positively in $\alpha$.  So if our contradictory pair (over an assumption) takes the form $\alpha\rightarrow\beta$ and $\alpha\rightarrow\neg\beta$, we need to prove for any $\gamma$ that $\alpha\rightarrow\gamma$.  But if we choose $\gamma$ to be any positive atom not occuring in $\alpha$, our inductive proof tells us this cannot be done.  We need compactness here, to be ensure that the basis for all contradictory pairs can be expressed by a finitary formula.&lt;/p&gt;&#xA;" OwnerUserId="100" LastActivityDate="2010-09-29T07:40:45.423" />
  <row Id="5693" PostTypeId="1" CreationDate="2010-09-29T08:03:47.577" Score="1" ViewCount="304" Body="&lt;p&gt;first show you only need to consider squares of functions as f.g = 1/4 [(f+g)sqr - (f-g)sqr]. show then that you only need to consider only positive valued functions becuase f(x).g(x)=|f(x)|sqr. then , if 0 &amp;lt;=f(x) &amp;lt;= M on [a,b] show that f sqr(x) - f sqr(y) &amp;lt;= 2M (f(x)-f(y)).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;does anyone know how i would answer this ??&lt;/p&gt;&#xA;" LastActivityDate="2010-09-29T14:37:29.703" Title="proof : product of two riemann integrals are integrable." Tags="&lt;calculus&gt;&lt;analysis&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" />
  <row Id="5694" PostTypeId="2" ParentId="5693" CreationDate="2010-09-29T08:14:09.137" Score="5" ViewCount="" Body="&lt;p&gt;This follows from Lebesgue's characterization of Riemann integrable functions&#xA;as bounded functions continuous outside a set of Lebesgue measure zero.&#xA;This characterization is usually the swiftest way of deciding on the&#xA;Riemannn integrability of a function.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-09-29T08:14:09.137" CommentCount="4" />
  <row Id="5695" PostTypeId="2" ParentId="5684" CreationDate="2010-09-29T08:21:17.313" Score="2" ViewCount="" Body="&lt;p&gt;Every element of order $p$ in $S_p$ is a $p$-cycle. The symmetric group $S_{p-1}$ acts transitively on these $p$ cycles.&lt;/p&gt;&#xA;" OwnerUserId="1650" LastActivityDate="2010-09-29T08:21:17.313" CommentCount="2" />
  <row Id="5696" PostTypeId="1" AcceptedAnswerId="5704" CreationDate="2010-09-29T08:25:08.957" Score="1" ViewCount="179" Body="&lt;p&gt;I think Cauchy's integral formula and the Hilbert transform can be used to prove one direction, but is this an equivalence or only an implication?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;edit&lt;/strong&gt; for clarification:&#xA;Is a function $f : \mathbb C \to \mathbb C, z\mapsto f(z)$ analytical $\Leftrightarrow$ The Fourier-Transform $\mathcal F\{f\}(\omega) = N \int_{\mathbb R} f(z) e^{i\omega z} dz$ (choose whatever normalization $N$ you like, I prefer symmetry $N=\sqrt{2\pi}$) is zero for all $\omega&amp;lt;0$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or shorter: Is the following true? $f$ analytical $\Leftrightarrow$ $\mathrm{supp}_{\mathcal F\{f\}}=\mathbb R^+$&lt;/p&gt;&#xA;" OwnerUserId="163" LastEditorUserId="367" LastEditorDisplayName="" LastEditDate="2010-09-29T10:54:29.867" LastActivityDate="2010-09-29T18:19:17.247" Title="Is a function analytical on C iff its Fourier-transform vanishes for negative frequencies?" Tags="&lt;analysis&gt;&lt;functional-analysis&gt;&lt;fourier-analysis&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="5697" PostTypeId="1" AcceptedAnswerId="5732" CreationDate="2010-09-29T08:48:42.770" Score="7" ViewCount="356" Body="&lt;p&gt;I will restate the 3-D version of the problem. In how many ways can you color a regular cube with 2 colors up to a rotational isometry. The answer is of course a special case of Burnsides Lemma which can be used to show that the number of distinct face permutations is $\frac{1}{24}(N^6 + 3N^4 + 12N^3 + 8N^2)$ where $N$ is the number of colors used, $2$ in this case, which gives us an answer of $10$ distinct permutations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is how can you expand this to a tesseract, and then more generally, to any hypercube. The rotational isometries of a cube are somewhat simple to comprehend, but the rotational isometries o hypercube are difficult to grasp (even after an hour of playing with the 4d rubiks cube app)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My initial thought was to consider the expansion of a tesseract as 8 interconnected cubelets. For the two color case each one of these cubelets has 10 distinct states. Which gives us $10^8$ non-distinct permutations of the hypercube. Or more generally, this reduces to how many distinct ways one can color a 8 faced 3-dimensional figure using 10 colors. So the complicated question of 4-dimensional isometries reduces to 3 dimensional rotational isometries of a octahedron.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I'm a physicist so what the hell do I know :D&lt;/p&gt;&#xA;" OwnerUserId="1906" LastEditorUserId="1906" LastEditorDisplayName="" LastEditDate="2010-09-29T09:04:51.830" LastActivityDate="2010-09-30T18:04:50.063" Title="Coloring the faces of a hypercube" Tags="&lt;geometry&gt;&lt;combinatorics&gt;&lt;abstract-algebra&gt;&lt;group-theory&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="1" />
  <row Id="5698" PostTypeId="1" AcceptedAnswerId="5701" CreationDate="2010-09-29T09:04:24.670" Score="1" ViewCount="184" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Possible Duplicate:&lt;/strong&gt;&lt;br&gt;&#xA;  &lt;a href=&quot;http://math.stackexchange.com/questions/548/explanation-of-method-for-showing-that-0-0-is-undefined&quot;&gt;Explanation of method for showing that 0 / 0 is undefined&lt;/a&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;p&gt;always i wonder this question why we cant calculate 0/0 in real numbers. where is my mind. i cant think anything logically&lt;/p&gt;&#xA;" OwnerUserId="2202" LastActivityDate="2010-09-29T10:02:56.333" ClosedDate="2010-09-29T14:06:04.673" Title="why we cant calculate 0/0 why is it undefinable?" Tags="&lt;zeros&gt;&lt;zero-divisor&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="5699" PostTypeId="2" ParentId="5696" CreationDate="2010-09-29T09:10:31.603" Score="2" ViewCount="" Body="&lt;p&gt;I'm not sure if I got your question right, but as I understand it the answer is no. Let's stick with the fourier transform as an operator $L^2(\mathbb{R}) \to L^2(\mathbb{R})$.&#xA;One of the basic facts about the fourier transform is that compactly supported functions transform into analytic functions. So as a counterexample just pick the characteristic function of $[-1,1]$. Its fourier transform is analytical but it certainly has &quot;negative frequencies&quot;, namely, it has frequencies almost everywhere in the interval $[-1,1]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: To make this more explicit: Let $f := \hat \chi_{[-1,1]}$. Then $f$ is analytic (in fact, we use the definition $\hat u(\xi) = \int e^{i\xi x} f(x) d x$ then $f(\xi)=2 \sin \xi / \xi$ (and $f(0)=2$), which is analytic). By the Fourier inversion theorem, $\hat f(x)=\hat{\hat \chi}_{[-1,1]} (x)=2\pi \chi _{[-1,1]}(-x)$ , whose support is $[-1,1]$. So $f$ is a counterexample to your statement.&lt;/p&gt;&#xA;" OwnerUserId="1609" LastEditorUserId="1609" LastEditorDisplayName="" LastEditDate="2010-09-29T11:38:04.110" LastActivityDate="2010-09-29T11:38:04.110" CommentCount="7" />
  <row Id="5700" PostTypeId="2" ParentId="5697" CreationDate="2010-09-29T09:12:33.813" Score="1" ViewCount="" Body="&lt;p&gt;I presume you want to colour the &lt;em&gt;faces&lt;/em&gt; of the cube?&#xA;If so there are $2^6$ colourings as there are six faces.&#xA;But do you want two colourings which are equivalent under&#xA;rotations (or rotations and reflections) to count as as the same?&#xA;If so then this is a classic application of&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Burnside%27s_lemma&quot; rel=&quot;nofollow&quot;&gt;Burnside's lemma&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added&lt;/strong&gt;&#xA;If you want to do the same in higher dimensions then you need a good&#xA;handle on the symmetries of an $n$-cube. Its convenient to centre the&#xA;cube at the origin and take its vertices to be all points $(\pm1,\ldots,\pm1)$.&#xA;Then all symmetries of the cube fix the origin and their matrices&#xA;are &quot;signed&quot; permutation matrices: each row and column is all zero save for&#xA;one entry which is $\pm1$. The rotations correspond to signed permutation&#xA;matrices of determinant 1. The faces are the sets defined by $x_i=1$ and&#xA;$x_i=-1$. Again use Burnside's lemma but the book-keeping gets fiddlier.&lt;/p&gt;&#xA;" OwnerUserId="226" LastEditorUserId="226" LastEditorDisplayName="" LastEditDate="2010-09-29T12:06:39.490" LastActivityDate="2010-09-29T12:06:39.490" CommentCount="1" />
  <row Id="5701" PostTypeId="2" ParentId="5698" CreationDate="2010-09-29T09:17:13.343" Score="1" ViewCount="" Body="&lt;p&gt;The simple answer is that division by zero is not well defined. As Andrea pointed out the fraction $\frac{0}{0}$ can be restated as $x$ s.t. $x*0 = 0$ which is true for all x in the real numbers. However, we still are motivated to find some kind of resolution to this issue, what exactly $ does $ $\frac{0}{0}$ equal to?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For this we have to turn to the world of calculus, $\frac{0}{0}$ is a standard limit case that we can examine. The answer in that case depends on the function you are taking the limit of. And somewhat surprisingly (or not), it can be equal to any real number. Just to illustrate this, I want to give you a simple example for the three cases, where the limit is equal to $0$, $N$, or $\pm\infty$ .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f_1(x) = \frac{Nx}{x}$$&#xA;$$f_2(x) = \frac{x^2}{x}$$&#xA;$$f_3(x) = \frac{x}{x^2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In each of these three cases, you see that plugging in zero gives you the undefined expression of $\frac{0}{0}$. However, if you take the limit of $f_n(x)$ as $x\rightarrow0$ you see that the three functions tend to completely different values. One tends to $N$ the second tends to $0$ and the third tends to $\pm\infty$ depending on the direction you are coming from. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, we resolve the issue in some sense, because now we can understand how this beast behaves, but on the other hand we still have the ambiguity that we had in the beginning.&lt;/p&gt;&#xA;" OwnerUserId="1906" LastEditorUserId="1906" LastEditorDisplayName="" LastEditDate="2010-09-29T10:02:56.333" LastActivityDate="2010-09-29T10:02:56.333" CommentCount="4" />
  <row Id="5702" PostTypeId="2" ParentId="1730" CreationDate="2010-09-29T09:33:16.630" Score="1" ViewCount="" Body="&lt;p&gt;A function is something that takes a number , twists it around and spits out another number&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once they grasp that, you can talk about how you can have functions that work with things besides numbers. And how generally, a functions can twist around any object, tangible or mathematical and spit out another object.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then you can move on to explaining the concept of domain and range. Domain is type of objects that your function can accept, and range is the type of object that your function can conceivably spit out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A good concrete example. A soda machine is a function that maps (don't use that word though, it scares students at their first encounter with it) the domain of coins to the range of soda's. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once they have the image in their head and some intuition THEN you go back and discuss the formal definitions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At least that what I do with my algebra students&lt;/p&gt;&#xA;" OwnerUserId="1906" LastActivityDate="2010-09-29T09:33:16.630" />
  <row Id="5703" PostTypeId="2" ParentId="5698" CreationDate="2010-09-29T09:47:00.633" Score="1" ViewCount="" Body="&lt;p&gt;Perhaprs the simplest way to understand this is to use &quot;words from the elementary school&quot;. In that time the teacher asked me how much is 12/4 and why. So my answer was 3, bacause $3\cdot 4 = 12$. Imagine now that the teacher asks you how much is 0/0? You can say for example 1 and by the previous argument you are right, because $1\cdot 0 = 0$. But on the other side you could say 0 or -1 or any real number. In this case many answers seem to be correct, so in such situations in mathematics the expression cannot be well-defined.&lt;/p&gt;&#xA;" OwnerUserId="2035" LastActivityDate="2010-09-29T09:47:00.633" />
  <row Id="5704" PostTypeId="2" ParentId="5696" CreationDate="2010-09-29T11:44:28.857" Score="1" ViewCount="" Body="&lt;p&gt;So, you are only interested in the Fourier transform of $f$&#xA;restricted to the real line. Typically the restriction&#xA;of such a function won't be in $L^2(\mathbb{R})$ (or any $L^p(\mathbb{R})$&#xA;for that matter) and won't have polynomial growth, so won't be a tempered&#xA;distribution. I don't know a way in which such a function (e.g. $f(z)=\exp(z)$)&#xA;could be said to have a Fourier transform.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But any compactly supported distribution on $\mathbb{R}$ will have&#xA;a Fourier transform that is an entire analytic function.&#xA;Some other tempered distributions also have this property, notably&#xA;the Gaussian $g(z)=\exp(-z^2)$. As Florian points out, that even when&#xA;an entire function has a Fourier transform, it need not be supported&#xA;on the positive reals; there is absolutely no bias towards positivity&#xA;or negativity in its support.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are interested in Fourier transforms of analytic functions, you should&#xA;look at the &lt;a href=&quot;http://en.wikipedia.org/wiki/Paley%E2%80%93Wiener_theorem&quot; rel=&quot;nofollow&quot;&gt;Paley-Wiener theorem&lt;/a&gt; which translates between properties&#xA;of the one and of the other.&lt;/p&gt;&#xA;" OwnerUserId="226" LastEditorUserId="226" LastEditorDisplayName="" LastEditDate="2010-09-29T18:19:17.247" LastActivityDate="2010-09-29T18:19:17.247" CommentCount="1" />
  <row Id="5705" PostTypeId="2" ParentId="1099" CreationDate="2010-09-29T11:54:18.223" Score="6" ViewCount="" Body="&lt;p&gt;Starting with the very well known iteration formula that creates the &lt;em&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Mandelbrot_set&quot; rel=&quot;nofollow&quot;&gt;Mandelbrot set&lt;/a&gt;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$z_{n+1}=z_n^2+c$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $c=z_0=x_0+iy_0=\text{Re}(c)+i\text{Im}(c)$ is a complex constant which is the starting point of the trajectory generated in the complex plane by the iteration, in &lt;em&gt;&lt;a href=&quot;http://fraktalwelten.wordpress.com/2010/01/11/doppelpot/&quot; rel=&quot;nofollow&quot;&gt;Doppelpot&lt;/a&gt;&lt;/em&gt; , from the German blog &lt;em&gt;&lt;a href=&quot;http://fraktalwelten.wordpress.com/&quot; rel=&quot;nofollow&quot;&gt;Fraktale Welten&lt;/a&gt; &lt;/em&gt;by &lt;em&gt; &lt;a href=&quot;http://tamagothi.wordpress.com/&quot; rel=&quot;nofollow&quot;&gt;Nachtwaechter&lt;/a&gt;&lt;/em&gt;, it is explained that, when we have two complex exponentiations the fractals generated by the recursive formula are normally beautiful, particularly those of  Julia sets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I &lt;a href=&quot;http://fraktalwelten.wordpress.com/2010/01/11/doppelpot/#comment-3186&quot; rel=&quot;nofollow&quot;&gt;suggested&lt;/a&gt;  to the author to try the following formulae:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$z_{n+2}=z_{n+1}^{2}+z_{n}+c$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$z_{n+1}=z_{n}^{z_{n}c}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$z_{n+1}=z_{n}^{z_{n}+c}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$z_{n+2}=z_{n+1}^{3}+c^{z_{n}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$z_{n+1}=z_{n}^{c}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In &lt;a href=&quot;http://fraktalwelten.wordpress.com/2010/02/22/funf-formeln/&quot; rel=&quot;nofollow&quot;&gt;Fünf Formeln&lt;/a&gt; the fractals are shown.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I reproduced some with the author's permission in this &lt;a href=&quot;http://problemasteoremas.wordpress.com/2010/02/25/alguns-dos-fractais-mostrados-no-fraktalen-welten-gerados-por-formulas-de-iteracao-que-sugeri/&quot; rel=&quot;nofollow&quot;&gt;entry&lt;/a&gt; of my blog. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;This one is generated by  $z_{n+2}=z_{n+1}^{2}+z_{n}+c$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/LNuSp.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and this by $z_{n+2}=z_{n+1}^{3}+c^{z_{n}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/P3Ysd.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="752" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2010-09-30T12:18:43.050" LastActivityDate="2010-09-30T12:18:43.050" />
  <row Id="5706" PostTypeId="1" AcceptedAnswerId="5708" CreationDate="2010-09-29T12:48:00.587" Score="0" ViewCount="121" Body="&lt;p&gt;Let $G$ be a (not necessarily finite) group with the property that for each subgroup $H$ of $G$, there exists a `retraction' of $G$ to $H$ (that is, a group homomorphism from $G$ to $H$ which is identity on $H$). Then, we claim :&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;$G$ is abelian.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Each element of $G$ has finite order.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Each element of $G$ has square-free order.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Let $g$ be a nontrivial element of $G$ and consider a retraction $T : G \to \langle{g\rangle}$ which is identity on $\langle{g\rangle}$. As $G/Ker(T)$ is isomorphic to $\text{Img}\langle{g\rangle}$, it is cyclic and so, it is abelian. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other than this i don't know how to prove the other claims of the problem. Moreover, a similar problem was asked in Berkeley Ph.D exam, in the year 2006, which actually asks us to prove that:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If $G$ is finite and there is a retraction for each subgroups $H$ of $G$, then $G$ is the products of groups of prime order.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" LastActivityDate="2010-09-29T14:48:53.463" Title="Showing $G$ is the product of groups of prime order." Tags="&lt;abstract-algebra&gt;&lt;group-theory&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="5707" PostTypeId="1" CreationDate="2010-09-29T12:55:33.773" Score="2" ViewCount="149" Body="&lt;p&gt;If I have an equation &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$p(x)\frac{\partial^2u}{\partial x^2} + r(x)\frac{\partial^2u}{\partial x\partial y} + q(x)\frac{\partial^2 u}{\partial y^2}=f(x,y,u)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $f$ maybe contains first partial derivatives for $u$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone give me a worked example of how to solve this using the method of characteristics. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance. &lt;/p&gt;&#xA;" OwnerUserId="2148" LastActivityDate="2010-09-30T19:51:11.527" Title="Characteristics for 2nd order differential equations" Tags="&lt;differential-equations&gt;" AnswerCount="1" FavoriteCount="2" />
  <row Id="5708" PostTypeId="2" ParentId="5706" CreationDate="2010-09-29T13:20:03.193" Score="3" ViewCount="" Body="&lt;p&gt;Let $H$ be a subgroup of $K$ which is a subgroup of $G$.&#xA;If there's a retraction of $G$ onto $H$, it restricts to&#xA;a retraction of $K$ onto $H$. So if a group $G$ has this&#xA;property (let's say it's &quot;retractible&quot;) then each subgroup of $G$&#xA;is retractible. Which cyclic groups are retractible?&lt;/p&gt;&#xA;" OwnerUserId="226" LastEditorUserId="226" LastEditorDisplayName="" LastEditDate="2010-09-29T14:45:18.367" LastActivityDate="2010-09-29T14:45:18.367" CommentCount="3" />
  <row Id="5709" PostTypeId="2" ParentId="5693" CreationDate="2010-09-29T14:37:29.703" Score="4" ViewCount="" Body="&lt;p&gt;This is not a problem I would assign as homework (at least, not without substantial guidance).  Rather, it is one of the fundamental results of the subject -- the subject being advanced calculus / elementary real analysis -- and as such I would expect any instructor / textbook to supply a proof.  For instance, Rudin's &lt;em&gt;Principles of Mathematical Analysis&lt;/em&gt; covers this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For those readers who do not have access to any basic analysis textbooks, see Corollary 8 of&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://math.uga.edu/~pete/243integrals2.pdf&quot; rel=&quot;nofollow&quot;&gt;http://math.uga.edu/~pete/243integrals2.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As Robin says, the result also follows from Lebesgue's criterion of Riemann integrability: now &lt;em&gt;that&lt;/em&gt;'s something I would leave as an exercise, since finding this short argument on one's own helps to drive home the power of the Lebesgue criterion.  &lt;/p&gt;&#xA;" OwnerUserId="299" LastActivityDate="2010-09-29T14:37:29.703" />
  <row Id="5710" PostTypeId="2" ParentId="5706" CreationDate="2010-09-29T14:48:53.463" Score="0" ViewCount="" Body="&lt;p&gt;Let $g$ be a nontrivial element of $G$ and consider a retraction $T : G \to \langle{g\rangle}$ which is identity on $\langle{g\rangle}$. As $G/Ker(T)$ is isomorphic to $\text{Img}\langle{g\rangle}$, it is cyclic and so, it is abelian. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus $[G,G]$ is contained in $Ker(T)$. Since $g \notin Ker(T)$, $g \notin [G,G]$. As $g$ is an arbitrary nontrivial element of $G$, this means that $[G,G] = {e}$; that is, $G$ is abelian.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Look at any element $g \in G$ and consider a retraction $T:G \to \langle{g^2 \rangle}$.&#xA;$T(g)$ is in $\langle{g^2 \rangle}$ means $T(g) = g^{2r}$ for some $r$. Also, $T(g^2)=g^2$ means then that&#xA;$g^{4r}=g^2$; that is, $g^{4r-2} = e$. As $4r-2$ is not zero, we get that $g$ has finite order.&lt;/p&gt;&#xA;" LastActivityDate="2010-09-29T14:48:53.463" />
  <row Id="5711" PostTypeId="2" ParentId="5707" CreationDate="2010-09-29T14:54:35.983" Score="1" ViewCount="" Body="&lt;p&gt;See these &lt;a href=&quot;http://research.nianet.org/~grossman/lectnotes3_09-ch5.pdf&quot; rel=&quot;nofollow&quot;&gt;lecture notes&lt;/a&gt; on the method of characteristics for second-order PDEs, where the method is applied to steady isentropic flow (gasdynamics)&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-09-29T15:00:48.453" LastActivityDate="2010-09-29T15:00:48.453" />
  <row Id="5712" PostTypeId="1" AcceptedAnswerId="5714" CreationDate="2010-09-29T16:24:58.593" Score="6" ViewCount="220" Body="&lt;p&gt;I've recently spent some time learning (the very basics of) K-theory for $C^*$-algebras and topological K-theory. Actually, my main fields of interest are PDEs and related topics, in particular functional calculus for unbounded operators, Sobolev/Bessel potential/Besov spaces, interpolation theory, semigroups and so on. You get the idea.&#xA;Now I would like to deepen (and broaden) my knowledge about the machinery of K-theory, ideally by learning some interesting connection to said PDE-related topics such as for example some PDE-relevant result admitting a proof with a k-theoretic flavour. Or some general idea how K-theory might provide insight into (or an interesting point of view on) some PDE-related results or concepts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I'd be very thankful, and I hope this request is not too broad, if you could provide me with some examples of interesting relations, if they exist, between K-theory and mentioned PDE-related topics.&lt;/p&gt;&#xA;" OwnerUserId="1403" LastEditorUserId="1543" LastEditorDisplayName="" LastEditDate="2011-04-18T12:16:26.370" LastActivityDate="2011-04-18T19:39:02.427" Title="Connections between K-Theory and PDEs?" Tags="&lt;topology&gt;&lt;algebraic-geometry&gt;&lt;algebraic-topology&gt;&lt;differential-equations&gt;&lt;pde&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="5713" PostTypeId="1" AcceptedAnswerId="5715" CreationDate="2010-09-29T16:31:41.140" Score="3" ViewCount="208" Body="&lt;p&gt;This is (a slight paraphrase) of question 1.3.14 in Chang and Keisler's Model Theory book.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Show that for each natural number $n$, there is a language $L_n$ and finite model $M_n$ of $L$ such that $M_n$ has precisely $n$ undefinable elements.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here, an element $x\in M$ is definable if there is a (first order) formula in $L$, called $\phi$, such that $x$ is the unique element of $M$ satisfying $\phi$.  Of course, &quot;undefinable&quot; means &quot;not definable&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is starred, indicating that it is more difficult than a standard problem in that book.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Chang and Keisler remark that $n=1$ is the only difficult case.  In that spirit, here is the proof for all $n\neq 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $L_n$ have a single 2 place predicate symbol (I'm thinking of $L_n = \{ &amp;lt; \}$).  Let $M_n$ be the partial order with minimum a and with elements $b_1,...,b_{n}$ with $a &amp;lt; b_i$ for all $i$ and the $b_i$ pairwise incomparable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First note that a is definable:  it uniquely satisfies $\phi(x) =$ for all y, $x\leq y$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if $n =0$, there are no $b_i$, and hence in this model, we have 0 undefinable elements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $n &gt; 1$, then I claim that all the $b_i$ are undefinable.  The short answer is that any permutation of the $b_i$ can be extended to a unique isomorphism of $M_n$.  Hence, for any formula $\phi$, we have $\phi(b_i)$ iff $\phi(b_j)$ for all $b_j$.  Thus, no $\phi$ can single out any particular $b_i$, so each $b_i$ is undefinable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This proof fails completely for $n=1$, for then $b_1$ is the unique element that satisfies &quot;b_1 is not a&quot;.  Or more in the spirit of first order logic, $b_1$ is the unique element satisfying $\phi(x) =$ there is a $y$ such that for all $z$, $y&amp;lt; z$ and $x$ is not equal to $y$.&quot;  Incidentally, this proves that any such model that works for $n=1$ must have at least 3 elements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, my question is: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What is an example of a language with finite model having precisely one undefinable element?  Is the smallest cardinality of such a model known?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance!&lt;/p&gt;&#xA;" OwnerUserId="331" LastEditorUserId="331" LastEditorDisplayName="" LastEditDate="2010-09-29T16:38:47.560" LastActivityDate="2010-11-23T12:00:56.430" Title="What is an example of a finite model in first order logic having a unique undefinable element?" Tags="&lt;logic&gt;&lt;model-theory&gt;" AnswerCount="3" FavoriteCount="1" />
  <row Id="5714" PostTypeId="2" ParentId="5712" CreationDate="2010-09-29T16:35:54.333" Score="7" ViewCount="" Body="&lt;p&gt;I would suggest the &lt;a href=&quot;http://en.wikipedia.org/wiki/Atiyah%E2%80%93Singer_index_theorem&quot; rel=&quot;nofollow&quot;&gt;Atiyah–Singer index theorem&lt;/a&gt; if this counts as a PDE-related topic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a theorem about elliptic differential operators on compact manifolds. The original proof uses K-theory.&lt;/p&gt;&#xA;" OwnerUserId="367" LastActivityDate="2010-09-29T16:35:54.333" CommentCount="2" />
  <row Id="5715" PostTypeId="2" ParentId="5713" CreationDate="2010-09-29T16:54:21.850" Score="6" ViewCount="" Body="&lt;p&gt;If your languages necessarily have the = relation, which is a common assumption, then the case n=1 is impossible in a finite model. The reason is that if a model $M$ has $k$ elements $x_1$, $x_2$, ... $x_k$ for finite $k\gt 1$ and each $x_i$ is defined by $\varphi_i(x)$ for $i\lt k$, then the remaining element $x_k$ is defined by the formula $\neg\varphi_1(x)\wedge\cdots\wedge \neg\varphi_{k-1}(x)$. If $M$ has only one element, then it is defined by the formula $x=x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if we do not insist that $=$ is in the language, then let $L$ be the empty language, having no relations at all. In this case, there are no atomic formulas and hence no well-formed formulas to define elements, so a one-point model has exactly one undefinable element.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we allow infinite models, then we can easily arrange to have exactly one undefinable element, even in a language with $=$. For example, consider the lanuage with infinitely many constant symbols, and have a model where all these constants are interpreted by different elements, and there is one extra un-named object. That extra object will be the only non-definable element, even when $=$ is in the language.&lt;/p&gt;&#xA;" OwnerUserId="413" LastActivityDate="2010-09-29T16:54:21.850" CommentCount="1" />
  <row Id="5716" PostTypeId="2" ParentId="5713" CreationDate="2010-09-29T16:58:13.757" Score="5" ViewCount="" Body="&lt;p&gt;You can't have equality as a relation in the language that is interpreted as equality in the model, otherwise you can define the undefinable element as that (one) which is not equal to any of the definable ones.  Most languages in common use have equality so it is hard to think of a natural example that describes, e.g., algebraic structures or combinatorial objects. &lt;/p&gt;&#xA;" OwnerUserId="467" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-09-29T17:03:41.610" LastActivityDate="2010-09-29T17:03:41.610" CommentCount="1" />
  <row Id="5717" PostTypeId="1" AcceptedAnswerId="5719" CreationDate="2010-09-29T17:10:19.397" Score="0" ViewCount="295" Body="&lt;p&gt;Looking for a hint on show to show convexity in a set....&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $f\colon \mathbb{R}^n \rightarrow \mathbb{R}$ be a convex function and let $c$ be some constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Show that the set $s=\{x \in \mathbb{R}^n \mid f(x) \le c \}$ is convex.&lt;/p&gt;&#xA;" OwnerUserId="2014" LastEditorUserId="367" LastEditorDisplayName="" LastEditDate="2010-09-30T18:06:54.723" LastActivityDate="2010-09-30T18:06:54.723" Title="How to show a set is convex" Tags="&lt;optimization&gt;&lt;linear-programming&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="5718" PostTypeId="1" AcceptedAnswerId="5751" CreationDate="2010-09-29T17:12:34.920" Score="5" ViewCount="124" Body="&lt;p&gt;Let $K$ be a number field and $L$ an infinite algebraic extension of $K$. Fix a non-trivial absolute value $v$ on $K$ (so $v$ is induced either by an embedding into the complex numbers or by a prime ideal in the ring of integers of $K$). If $K_v$ is the corresponding completion and $\overline{K}_v$ a choice of algebraic closure of $K_v$, then the extensions of $v$ to an absolute value on $K$ are in bijection with the $Gal(\overline{K}_v/K_v)$-orbits of $Hom_{K-alg}(L,\overline{K}_v)$ (this is described in detail in, e.g., Neukirch's &lt;em&gt;Algebraic Number Theory&lt;/em&gt;). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question regards the case when $v$ is non-Archimedean, arising from a prime ideal $\mathfrak{p}$ with residue characteristic $p$. In this case, &lt;strong&gt;is there a bijection between prime ideals of $\mathscr{O}_L$ lying above $\mathfrak{p}$ and places of $L$ above $v$?&lt;/strong&gt; Since $\mathscr{O}_L$ is not generally going to be Dedekind (though it is a one-dimensional, integrally closed domain), we don't get an (additive) discrete valuation in the usual way from a prime ideal of $\mathscr{O}_L$, but if $w$ is a non-Archimedean absolute value on $L$ extending $v$, then by restricting to finite sub-extensions $L_i$ of $L/K$ we get sort of a coherent sequence of prime ideals $\mathfrak{p}_i$ in the integer rings of the $L_i$. Perhaps there is a way to convert this sequence of primes into a single prime ideal of $\mathscr{O}_L$ lying above each of the $\mathfrak{p}_i$? I thought maybe some kind of compactness\inverse limit type argument might work, but I'm not sure anymore...even so, if such an argument were to work, it would probably just show the existence of such a prime, as opposed to determining it uniquely. Alternatively, if I look at the maximal ideal of the valuation ring of $w$ in $L$ and intersect it with $\mathscr{O}_L$, I should get a (hopefully) non-zero prime ideal of $\mathscr{O}_L$ that maybe fits the bill. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This might be the wrong approach entirely (and maybe the answer to my question is just &quot;no&quot;). The reason I'm interested is because, for example, in Washington's book on cyclotomic fields, he (in an appendix) defines the decomposition group of a prime ideal in an infinite Galois extension of number fields, but makes no mention of the decomposition group of a &lt;em&gt;place&lt;/em&gt; of an infinite algebraic extension of a number field. When one starts considering objects like the maximal unramified abelian $p$-extension of a $\mathbb{Z}_p$-extension of a number field, surely this (possibly more general) notion becomes relevant. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would greatly appreciate if anyone could set me straight on this issue or point me in the direction of a reference where it's discussed. Thanks. &lt;/p&gt;&#xA;" OwnerUserId="628" LastActivityDate="2010-09-30T07:53:57.700" Title="What is the relationship between non-Archimedean places of infinite extensions of number fields and primes in the ring of integers?" Tags="&lt;algebraic-number-theory&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="5719" PostTypeId="2" ParentId="5717" CreationDate="2010-09-29T17:13:00.920" Score="7" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Hint:&lt;/strong&gt; Well, just write down a convex combination of elements in s and verify that it belong to s. You will find the convexity of f useful for this.&lt;/p&gt;&#xA;" OwnerUserId="367" LastActivityDate="2010-09-29T17:13:00.920" CommentCount="5" />
  <row Id="5720" PostTypeId="1" AcceptedAnswerId="5726" CreationDate="2010-09-29T18:12:31.190" Score="0" ViewCount="105" Body="&lt;p&gt;What are linear combinations of a column space?   What equation to they satisfy? Why are they useful in leaner algebra? &lt;/p&gt;&#xA;" OwnerUserId="1992" LastActivityDate="2010-09-29T19:16:18.000" Title="What are linear combinations of a column space?" Tags="&lt;linear-algebra&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="5721" PostTypeId="1" AcceptedAnswerId="5724" CreationDate="2010-09-29T18:13:53.370" Score="0" ViewCount="130" Body="&lt;p&gt;What is the homogeneous problem?   What is the purpose of null space of a vector in this context? &lt;/p&gt;&#xA;" OwnerUserId="1992" LastEditorUserId="1992" LastEditorDisplayName="" LastEditDate="2010-09-29T18:29:01.080" LastActivityDate="2010-09-29T19:03:36.190" Title="What is the homogeneous problem?" Tags="&lt;linear-algebra&gt;&lt;differential-equations&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="5722" PostTypeId="2" ParentId="5648" CreationDate="2010-09-29T18:17:02.123" Score="2" ViewCount="" Body="&lt;p&gt;In contrast to William, I think the answer is yes.  I had to prove something similar once; here is a suitably adapted argument.  It might be more complicated than necessary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, reduce to the case that $(\Omega_1, \mu_1), (\Omega_2, \mu_2)$ are finite measure spaces and $h$ is bounded.  Let $\mathcal{P}$ be the set of all functions on $\Omega_1 \times \Omega_2$ of the form $F(x,y) = f(x) g(y)$ where $f$ is bounded and measurable and $g$ is simple; such functions can be written in the form you seek.  Let $\mathcal{Q}$ be the linear span of $\mathcal{P}$; these functions can also be written in the desired form.  Let $\mathcal{L}$ be the closure of $\mathcal{Q}$ in $L^1(\mu_1 \times \mu_2)$ and let $\mathcal{L}_b$ be the bounded functions from $\mathcal{L}$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now $\mathcal{L}_b$ is a vector space which is closed under bounded convergence (by the dominated convergence theorem), contains the constants and contains $\mathcal{P}$.  $\mathcal{P}$ is closed under multiplication and contains all functions of the form $1_{A \times B}$ with $A \subset \Omega_1$, $B \subset \Omega_2$ measurable; the collection of such sets $A \times B$ generates the product $\sigma$-algebra.  By the functional version of the Dynkin $\pi$-$\lambda$ theorem (references below), $\mathcal{L}_b$ contains all bounded measurable functions; in particular it contains $h$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $h$ is in $\mathcal{L}$ which is the $L^1$ closure of $\mathcal{Q}$, there is a sequence $\mathcal{Q} \ni h_n \to h$ in $L^1$.  Then some subsequence converges almost everywhere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the functional $\pi$-$\lambda$ theorem, see Theorem 8.2 of &lt;a href=&quot;http://www.math.ucsd.edu/~bdriver/Cornell%20Summer%20Notes%202010/Lecture_Notes/Probability%20Tools%20with%20Examples.pdf&quot; rel=&quot;nofollow&quot;&gt;Bruce Driver's probability notes&lt;/a&gt;. A reference is also given to C. Dellacherie, &lt;i&gt;Capacit&amp;eacute;s et processus stochastiques&lt;/i&gt;, page 14.&lt;/p&gt;&#xA;" OwnerUserId="822" LastActivityDate="2010-09-29T18:17:02.123" CommentCount="2" />
  <row Id="5723" PostTypeId="2" ParentId="155" CreationDate="2010-09-29T18:29:31.810" Score="8" ViewCount="" Body="&lt;p&gt;The techniques used for indefinite integration of elementary functions are actually quite simple in the transcendental (vs. algebraic) case, i.e. the case where the integrand lies in a purely transcendental extension of the field of rational functions $\rm\mathbb C(x)$. Informally this means that the integrand lies in some tower of fields $\rm\mathbb C(x) = F_0 &amp;lt; \cdots &amp;lt; F_n = \mathbb C(x,t_1,\ldots,t_n)$ which is built by adjoining an exponential or logarithm of an element from the prior field, i.e $\rm\  t_{i+1} =\: exp(f_i)\ $ or $\rm\ t_{i+1} =\: log(f_i)\ $ for $\rm\ f_i \in F_i$ where $\rm t_{i+1}$ is transcendental over $\rm F_i\:.\ $ For example $\rm\ exp(x),\ log(x)\ $ are transcendental over $\rm\mathbb C(x)$ but $\rm\ exp(2\ log(x)) = x^2\ $ is not. Now, because $\rm\ F_{i} = F_{i-1}(t_{i})$ is transcendental it has particularly simple structure, viz. it is isomorphic to the field of rational functions in one indeterminate $\rm\:t_i\:$ over $\rm\ F_{i-1}\ $. In particular, this means that one may employ well-known rational function integration, techniques such as expansions into partial fractions. This, combined with a simple analysis of the effect of differentiation on the degree of polynomials $\rm\  p(t_i)$, quickly leads to the fundamental result of Liouville on the structure of antiderivatives, namely they must lie in the same field $\rm F$ as the integrand except possibly for the addition of constant multiples of log's over $\rm F$. With this structure theorem in hand, the transcendental case reduces to elementary computations in rational function fields. This transcendental case of the algorithm is so simple that it may be  easily comprehended by anyone who has mastered a first course in abstract algebra.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, the full-blown algebraic case of the algorithm requires nontrivial results from the theory of algebraic functions. Although there are some simple special case algorithms for sqrt's and cube-roots (Davenport, Trager) the general algorithm requires deep results about points of finite order on abelian varieties over finitely generated ground fields. This algebraic case of the integration algorithm was discovered by Robert Risch in 1969 - who did his Berkeley Ph.D. on this topic (under Max Rosenlicht).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a very nice introduction to the theory see Max Rosenlicht's Monthly paper, available &lt;a href=&quot;http://www.jstor.org/stable/2318066&quot; rel=&quot;nofollow&quot;&gt;from JSTOR&lt;/a&gt; and &lt;a href=&quot;http://vigo.ime.unicamp.br/~fismat/Integration.pdf&quot; rel=&quot;nofollow&quot;&gt;also here&lt;/a&gt;. This exposition includes a complete proof of the Liouville structure theorem along with a derivation of Liouville's classic criterion for $\rm\int f(z)\: e^{g(z)}\: dz\ $ to be elementary, for $\rm\: f(z),\: g(z)\in \mathbb C(x)$. For algorithms see &lt;a href=&quot;http://dspace.mit.edu/handle/1721.1/15391&quot; rel=&quot;nofollow&quot;&gt;Barry Trager's 1984 MIT thesis&lt;/a&gt; and Manual Bronstein: &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/3540214933&quot; rel=&quot;nofollow&quot;&gt;Symbolic Integration I: Transcendental Functions.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Disclaimer: I implemented the integration algorithm in Macsyma (not the much older Maxima) so, perhaps due to this experience,  my judgment of simplicity might be slightly biased. However, the fact that the basic results are derived in a handful of pages in Rosenlicht's paper yields independent evidence for such claims.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-09-29T18:36:49.347" LastActivityDate="2010-09-29T18:36:49.347" />
  <row Id="5724" PostTypeId="2" ParentId="5721" CreationDate="2010-09-29T18:47:51.987" Score="5" ViewCount="" Body="&lt;p&gt;The answer to what I think you're asking:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The homogeneous problem is: &quot;Given a matrix $A$, find all solutions to the problem $Ax = 0$, where $x$ is a vector of appropriate dimension, and '0' is the vector of all zeros&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Nullspace of $A$ is precisely the set of all such solutions. There is always at least one solution to the homogeneous problem. Indeed, $x=0$ (the vector of all zeros) is always a solution. In the case where $A$ is a square and invertible matrix, $x=0$ is the only solution. In general, there can be other solutions, and the set of all solutions (the Nullspace) is actually a subspace. To see why, notice that if $x_1$ and $x_2$ both satisfy $Ax=0$, then so does $a_1 x_1 + a_2 x_2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope this answers your question. Best of luck!&lt;/p&gt;&#xA;" OwnerUserId="1356" LastActivityDate="2010-09-29T18:47:51.987" CommentCount="2" />
  <row Id="5725" PostTypeId="2" ParentId="1099" CreationDate="2010-09-29T18:50:41.613" Score="4" ViewCount="" Body="&lt;p&gt;Isaac, the auxiliary question that we should ask ourselves before answering your question is: &lt;strong&gt;what is the &quot;right&quot; (i. e., most mathematically relevant/interesting) generalization of the Mandelbrot set to other parametrized families of functions?&lt;/strong&gt; Of course, you can take some arbitrary holomorphic family of functions $f_c(z)$ that depends holomorphically on the parameter $c$, and for each $c$, pick some starting value $z_0$ (which depends on $c$ in an essentially arbitrary way) and color the plane somehow based on the long-term behavior of the iterates. However, as some of the other posts show, you typically end up with objects that don't really resemble the Mandelbrot set, so the status of this construction as the &quot;right&quot; generalization of the M-set is very dubious.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's what I think is the right answer to my auxiliary question: the boundary of the standard Mandelbrot set is the &lt;em&gt;bifurcation locus&lt;/em&gt; for the holomorphic family $f_c(z) = z^2 + c$: it's the set of $c$ at which the long-term dynamical behavior of $f_c$ is not continuous. (For example, if you cross the boundary and move from the inside of the M-set to the outside, you go from having a finite-valued attracting orbit to not having one (with only the fixed point at infinity attracting)). This concept applies equally well to other holomorphic families, and this is what I think is the right generalization of the M-set: the bifurcation locus of a family $f_c(z)$ of holomorphic functions that depends holomorphically on $c$. More generally, we can think of not just the bifurcation locus, but a more complete &quot;catalog&quot; of the dynamics of the family: a partition of the complex plane into the bifurcation set plus the union of a bunch of open sets in which the long-term dynamical behavior varies continuously (which for the M-set, would be a partition of the plane into the outside, the boundary, and all of the cardiods and bulbs (including those in smaller copies of the M-set found near the boundary)).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, to answer &lt;em&gt;your&lt;/em&gt; question: yes, these objects have been studied in some generality, but I'm not an expert in the field and thus can't tell you everything about them. However, for example, it is known that every nonempty bifurcation locus contains copies of the boundary of the standard M-set (or the bifurcation locus of the family $z\mapsto = z^d + c$ for some $d$, the &quot;degree-$d$ M-set&quot;), and moreover, that these copies are dense in every bifurcation locus. &lt;/p&gt;&#xA;" OwnerUserId="2032" LastActivityDate="2010-09-29T18:50:41.613" />
  <row Id="5726" PostTypeId="2" ParentId="5720" CreationDate="2010-09-29T18:59:11.673" Score="3" ViewCount="" Body="&lt;p&gt;A linear combination of vectors, say ${ x_1, \ldots , x_n}$, of a vector space on some field $\mathbb{K}$ is $$\sum_{i = 1}^{n} \lambda_i x_i$$, where $\lambda_i \in \mathbb{K}$ and not all of them are zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These kind of combinations are important in linear algebra because you can define &quot;linear independence&quot; and &quot;generators&quot;, which, if combined, give you a &quot;basis&quot; for the vector space (you can find all the definitions here: &lt;a href=&quot;http://en.wikipedia.org/wiki/Vector_space&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, ${x_1, \ldots , x_n }$ vectors are said to be independent, if $$\sum_{i = 1}^{n} \lambda_i x_i = 0 \Rightarrow \lambda_i = 0 \quad \forall i$$. This definition, as you can see, uses the concept of linear combination. On the same note, you can say that hose vectors generate the space $V$ if $\forall x \in V \quad x = \sum_{i = 1}^{n} \lambda_i x_i$ for appropriate $\lambda_i \in \mathbb{K}$. As you can see, also this definition uses the concept of linear combination.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the rest of your question, I can't really grasp its meaning.&lt;/p&gt;&#xA;" OwnerUserId="1415" LastEditorUserId="1415" LastEditorDisplayName="" LastEditDate="2010-09-29T19:16:18.000" LastActivityDate="2010-09-29T19:16:18.000" CommentCount="2" />
  <row Id="5727" PostTypeId="2" ParentId="5721" CreationDate="2010-09-29T19:03:36.190" Score="2" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\ $ For $\rm A\:$ linear, $\rm\ A\:X_1 = B = A\:X_2 \ \iff\ 0 \:=\: A\:X_1 - A\:X_2 = A\:(X_1-X_2)$   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This implies that the general solution of $\rm\ \ \ \:A\:X = B\ $ is the sum of any particular solution plus a solution of the associated &quot;homogeneous&quot; equation $\rm\ A\:X = 0\:$. This property holds true for every linear operator, e.g. for matrices, linear differential equations, linear recurrences, etc.&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-09-29T19:03:36.190" />
  <row Id="5728" PostTypeId="2" ParentId="5676" CreationDate="2010-09-29T19:07:48.493" Score="15" ViewCount="" Body="&lt;p&gt;This was an interesting challenge, to try and come up with an estimate for this sum in an elementary way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The estimate I got was&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$1 + \sqrt{2} + \dots + \sqrt{n} \sim \frac{2}{3}n^{3/2} + \frac{\sqrt{n}}{2} + C$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for some constant $C$ which appears to be close to $-0.207$ (which I guess will be $\zeta(-1/2)$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(By $a_{n} \sim b_{n}$ I mean $\lim_{n \rightarrow \infty} (a_{n}-b_{n}) = 0$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe here is a completely elementary proof of that fact:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First consider the inequality for $x &gt; 0$ and $k &gt; 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \sqrt{x} \le \frac{x}{2\sqrt{k}} + \frac{\sqrt{k}}{2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This follows easily by the arithmetic mean $\ge$ geometric mean inequality.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus we have that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_{k}^{k+1} \sqrt{x} \ dx \le \int_{k}^{k+1} (\frac{x}{2\sqrt{k}} + \frac{\sqrt{k}}{2}) \ dx$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ = \frac{(k+1)^2 - k^2}{4\sqrt{k}} + \frac{\sqrt{k}}{2} = \sqrt{k} + \frac{1}{4\sqrt{k}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus $$\sum_{k=1}^{n-1} \int_{k}^{k+1} \sqrt{x} \ dx \le \sum_{k=1}^{n-1} (\sqrt{k} + \frac{1}{4\sqrt{k}})$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i.e.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_{1}^{n} \sqrt{x} \ dx \le \sum_{k=1}^{n-1} (\sqrt{k} + \frac{1}{4\sqrt{k}})$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and so&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \frac{2}{3} n^{3/2} - \frac{2}{3} \le \sum_{k=1}^{n-1} (\sqrt{k} + \frac{1}{4\sqrt{k}})$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we have inequality&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{1}{2\sqrt{k}} &amp;lt; \frac{1}{\sqrt{k} + \sqrt{k-1}} = \sqrt{k} -\sqrt{k-1}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And so, we have that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \sum_{k=1}^{n-1} \frac{1}{4\sqrt{k}} &amp;lt; \frac{\sqrt{n-1}}{2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \frac{2}{3} n^{3/2} - \frac{2}{3} \le \frac{\sqrt{n-1}}{2} + \sum_{k=1}^{n-1} \sqrt{k} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if $$S_{n} = \sum_{k=1}^{n} \sqrt{k}$$ we have that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$S_{n} \ge \frac{2}{3} n^{3/2} - \frac{2}{3} + \sqrt{n} - \frac{\sqrt{n-1}}{2} \ge \frac{2}{3} n^{3/2} - \frac{2}{3} + \frac{\sqrt{n}}{2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let $$G_{n} = S_{n} - \frac{2}{3} n^{3/2} - \frac{\sqrt{n}}{2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have that $$G_{n} \ge -\frac{2}{3}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can easily show that (using tedious but not too complicated algebra&lt;sup&gt;*&lt;/sup&gt;) $$G_{n+1} &amp;lt; G_{n}$$ and so $G_{n}$ is a convergent sequence, as it is bounded below and monotonically decreasing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus there exists a constant $C$ (the limit of $G_{n}$) such that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$1 + \sqrt{2} + \dots + \sqrt{n} \sim \frac{2}{3}n^{3/2} + \frac{\sqrt{n}}{2} + C$$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;*&lt;/sup&gt; For the sake of completeness, we show that $G_{n+1} &amp;lt; G_{n}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider $$6(G_{n}-G_{n+1}) = 4(n+1)\sqrt{n+1} + 3\sqrt{n+1} - 6\sqrt{n+1} - 4n\sqrt{n} - 3\sqrt{n}$$&#xA;$$ = \sqrt{n+1} + 4n(\sqrt{n+1} -\sqrt{n}) - 3\sqrt{n}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Multiplying by $\sqrt{n+1} + \sqrt{n}$ does not change the sign, so we look at&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \sqrt{n+1}(\sqrt{n+1} + \sqrt{n}) + 4n - 3\sqrt{n}(\sqrt{n+1} + \sqrt{n})$$&#xA;$$ = 2n+1 - 2\sqrt{n^2 + n}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now $$ (2n+1)^2 = 4n^2 + 4n + 1 &gt; 4n^2 + 4n = (2\sqrt{n^2+n}) ^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ 2n+1 &gt; 2\sqrt{n^2+n}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and so&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$G_{n} &gt; G_{n+1}$$&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-09-29T23:04:12.800" LastActivityDate="2010-09-29T23:04:12.800" CommentCount="2" />
  <row Id="5729" PostTypeId="2" ParentId="5720" CreationDate="2010-09-29T19:12:42.697" Score="2" ViewCount="" Body="&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Column_space&quot; rel=&quot;nofollow&quot;&gt;column space&lt;/a&gt; of a matrix is simply the space spanned by all linear combinations of the columns of the matrix. If $\rm \:A_i\:$ denotes the $\rm i$'th column of $\rm \:A\:$ then $\rm \ b = Ax = A_i\: x_1 + \cdots + A_n\: x_n$ has a solution iff $\rm\:b\:$ lies in the column space. Said in geometric language, the column space spans the image (range) of the linear map corresponding to the matrix $\rm\: A\:$.&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-09-29T19:12:42.697" />
  <row Id="5731" PostTypeId="2" ParentId="5697" CreationDate="2010-09-29T19:53:45.247" Score="2" ViewCount="" Body="&lt;p&gt;First let's state the special case of Burnside's lemma that is relevant here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Lemma:&lt;/strong&gt;  Let $G$ be a finite group acting on a finite set $X$.  The number of ways to color the elements of $X$ with $z$ different colors, up to the action of $G$, is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{1}{|G|} \sum_{g \in G} z^{c(g)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $c(g)$ is the number of cycles in the cycle decomposition of $g$ acting on $X$.  (&lt;a href=&quot;http://qchu.wordpress.com/2009/06/16/gila-iii-the-orbit-counting-lemma-and-baby-polya/&quot; rel=&quot;nofollow&quot;&gt;Proof&lt;/a&gt;.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here $X$ is the set of faces of a hypercube.  In $n$ dimensions there are $2n$ such faces.  $G$ is the subgroup of index $2$ in the &lt;a href=&quot;http://en.wikipedia.org/wiki/Hyperoctahedral_group#Subgroups&quot; rel=&quot;nofollow&quot;&gt;hyperoctahedral group&lt;/a&gt; with determinant $1$, also known as the Coxeter group $D_n$.  So our job now is to count, for each $k$, the number of elements of $D_n$ with $k$ cycles in the action on $X$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, note that to analyze the action of $D_n$ on the faces it suffices to analyze the action of $D_n$ on the midpoints of the faces.  But these are precisely the $2n$ points $(0, 0, ... \pm 1, ..., 0, 0)$, so writing the elements of $D_n$ as signed permutation matrices is very well-suited to analyzing their action on these points; in particular, it suffices to figure out the answer for a single signed cycle.  But this turns out to be very simple: there are either one or two cycles depending on whether the product of the signs is $-1$ or $+1$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(It might be helpful here to play with a specific example.  Consider $\left[ \begin{array}{cc} 0 &amp;amp; 1 &amp;amp; 0 \\ 0 &amp;amp; 0 &amp;amp; -1 \\ -1 &amp;amp; 0 &amp;amp; 0 \end{array} \right]$ acting on the six points $(\pm 1, 0, 0), (0, \pm 1, 0), (0, 0, \pm 1)$ to get a feel for what's going on in the general case.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From here I think it's easiest to work with generating functions because the combinatorics get a little messy.  Begin with the identity&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{m \ge 0} Z(S_n) t^n = \exp \left( z_1 t + \frac{z_2 t^2}{2} + \frac{z_3 t^3}{3} + ... \right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $Z(S_n)$ is the &lt;a href=&quot;http://qchu.wordpress.com/2009/06/21/gila-v-the-polya-enumeration-theorem-and-applications/&quot; rel=&quot;nofollow&quot;&gt;cycle index polynomial&lt;/a&gt; for the action of $S_n$ on $\{ 1, 2, ... n \}$.  Each $z_i$ is the term that controls cycles of length $i$.  We want to modify this generating function so that it tells us how the cycles in $D_n$ work.  There are $2^i$ signed cycles of length $i$ which come in two flavors: half of them have positive sign product (two unsigned cycles) and half of them have negative sign product (one unsigned cycle), so to keep track of the total number of unsigned cycles we should replace $z_i$ with $2^{i-1} z^2 + 2^{i-1} z$.  We also have to keep in mind that the determinant of a signed cycle is its sign product multiplied by $(-1)^{i+1}$, and we only want permutations with determinant $1$.  So the generating function we want is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{n \ge 0} f_n(z) \frac{t^n}{n!} = \frac{1}{2} \left( \exp \left( \sum_{i \ge 1} \frac{2^{i-1} z^2 + 2^{i-1} z) t^i}{i} \right) + \exp \left( \sum_{i \ge 1} (-1)^{i+1} \frac{2^{i-1} z^2 - 2^{i-1} z) t^i}{i} \right) \right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $f_n(z) = \sum_{g \in D_n} z^{c(g)}$.  After some simplification the above becomes&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{n \ge 0} \frac{1}{|D_n|} f_n(z) t^n = \frac{1}{(1 - t)^{(z^2+z)/2}} + (1+t)^{(z^2-z)/2}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Substituting $z = 2$ gives, at last, the answer&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{n \ge 0} \frac{1}{|D_n|} f_n(2) t^n = \frac{1}{(1 - t)^3} + 1 + t.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, for $n \ge 2$ we simply have $\frac{1}{|D_n|} f_n(2) = {n+2 \choose 2}$.  This is such a simple answer that there should be a direct proof of it.  I'll keep working on it.  (There is a rather straightforward proof of the corresponding result with &quot;hypercube&quot; replaced by &quot;simplex,&quot; so my guess is something along those lines is possible here.)&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-09-29T20:00:18.270" LastActivityDate="2010-09-29T20:00:18.270" CommentCount="1" />
  <row Id="5732" PostTypeId="2" ParentId="5697" CreationDate="2010-09-29T20:30:38.770" Score="3" ViewCount="" Body="&lt;p&gt;Once someone else has done the hard work and come up with the answer,&#xA;as Qiaochu has done, then it's easy to prove it on an &lt;em&gt;ad hoc&lt;/em&gt;&#xA;basis as this interloper (me) will do.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider an $n$-cube with faces coloured red and blue. Consider a set&#xA;of $n$ balls. Colour the $j$-th ball red, blue or green according to the&#xA;colours of the $j$-th pair of faces in the cube, that is if they both have&#xA;the same colour give the ball that colour, else if they have different&#xA;colours colour the ball green. Now a symmetry operation on the cube&#xA;will shuffle the balls so the number of each colour of balls will be the same.&#xA;Conversely given the $n$ coloured balls there will be various coloured&#xA;cubes giving rise to the colour distribution of the balls, but these&#xA;will all be equivalent under the full symmetry group of the cube.&#xA;However for $n\ge 2$ each $2$-coloured cube has a non-rotational symmetry,&#xA;so any two coloured cubes giving rise to the same colour distribution&#xA;of balls are related by a rotation. Hence the number of $2$-coloured $n$-cubes&#xA;up rotation is the same as the number of triples $(r,b,g)$ of nonnegative&#xA;integers adding to $n$ and there are ${n+2\choose 2}$ of these.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One can repeat this with $k$-colours for the cube. This time one needs&#xA;${k+1\choose 2}$ colours for the balls, and the argument ensuring that&#xA;rotations and the full symmetry group give the same answer requires&#xA;$n&gt;{k\choose 2}$ (why?).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added&lt;/strong&gt; (30/9/2010)&#xA;One can get the general result using these considerations. For&#xA;$n&gt;{k\choose 2}$ one gets&#xA;$${n+(k^2+k)/2-1\choose (k^2+k)/2-1}$$&#xA;colourings up to rotations. For $n\le {k\choose 2}$ there are $k$-colourings&#xA;having no improper (determinant $-1$) symmetries. But any such colouring&#xA;has no nontrivial symmetries at all. If two opposite faces have&#xA;the same colour, one can reflect through a hyperplane parallel to them.&#xA;So assume that no colour is opposite itself. Then if two pairs of opposite&#xA;faces have the same two colours between them, one can reflect in a plane&#xA;$x_i=x_j$or $x_i=-x_j$. Hence in these &quot;special&quot; colourings each pair&#xA;of opposite faces has a distinct pair of distinct colours. Up to symmetry&#xA;there are&#xA;$${(k^2-k)/2\choose n}$$&#xA;of these special colourings. So for $n\le{k\choose 2}$ there are&#xA;$${(k^2-k)/2\choose n}+{n+(k^2+k)/2-1\choose (k^2+k)/2-1}$$&#xA;up to rotational symmetries.&lt;/p&gt;&#xA;" OwnerUserId="226" LastEditorUserId="226" LastEditorDisplayName="" LastEditDate="2010-09-30T18:04:50.063" LastActivityDate="2010-09-30T18:04:50.063" CommentCount="1" />
  <row Id="5734" PostTypeId="1" AcceptedAnswerId="5750" CreationDate="2010-09-29T21:31:58.827" Score="1" ViewCount="88" Body="&lt;p&gt;Suppose $G$ is a compact topological group.  We can construct the profinite completion of $G$; let's call this $\Gamma$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My questions are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Assuming that we know nothing about the (original) topology of $G$ other than that it is compact, is there anything we can say linking the topology of $G$ to the topology of $\Gamma$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I assume the answer to this question is &quot;no&quot; since it seems to me that we usually regard $G$ as an abstract group when thinking about constructing its profinite completion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) If not (and, like I said, I assume the answer to (1) is &quot;no&quot;) is there a way to construct a profinite completion (or something like this) of $G$ that takes into account the topology we already have?&lt;/p&gt;&#xA;" OwnerUserId="2209" LastActivityDate="2010-09-30T05:31:12.057" Title="Relationship between topology of a compact group and the topology of its profinite completion." Tags="&lt;group-theory&gt;&lt;topology&gt;" AnswerCount="1" />
  <row Id="5735" PostTypeId="2" ParentId="120" CreationDate="2010-09-29T21:34:28.633" Score="3" ViewCount="" Body="&lt;p&gt;It's worth emphasizing that, in addition to unique factorization, there are also somewhat deeper structural reasons underlying the convention that $\:1\:$ is neither prime nor a prime ideal, but $\:0\:$ is. Below I discuss the motivations for these differing conventions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One important motivation for including the zero ideal as prime is that this facilitates powerful reductions. For example, in many ring theoretic problems involving an ideal $\rm\; I\;$, one can reduce to the case $\rm\;I = P\;$ prime, and then reduce to $\rm\;R/P\;$, therefore reducing to the case when the ring is a domain. In this case one simply says that one can factor out $\rm\; P\;$, so wlog assume $\rm\; P = 0\;$ is prime, hence the ring is a domain. For example, at the end of this post is an excerpt from Kaplansky's classic textbook &quot;Commutative Rings&quot;, section 1-3: G-Ideals, Hilbert Rings, and the Nullstellensatz, where I've explicitly highlighted a few prototypical examples of such reductions - cf. &lt;strong&gt;reduce to...&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus we have solid evidence for the utility of the convention that the zero ideal is prime.&#xA;So why don't we adopt the same convention for the unit ideal $1$ or, equivalently, why don't we permit the zero ring as a domain?  There are a number of reasons. First, in domains and fields it often proves very convenient to assume that one has a nonzero element available. This permits proofs by contradiction to conclude by deducing $1 = 0\:$. More importantly, it implies that the  unit group is nonempty, so unit groups always exist. It'd be very inconvenient to have to always add the proviso (except if $\;\rm R = 0)$ to the ubiquitous arguments involving units and unit groups. More generally it's worth emphasizing that the usual rules for equational logic are not complete for &lt;em&gt;empty structures&lt;/em&gt;. That is why groups and other algebraic structures are always axiomatized to prevent nonempty structures (see this &lt;a href=&quot;http://groups.google.com/group/sci.math/browse_frm/thread/3e8fac43081fcbf8&quot; rel=&quot;nofollow&quot;&gt;thread&lt;/a&gt; for further details).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Below is the mentioned Kaplansky excerpt on reduction to domains by factoring out prime ideals. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Let $\rm\; I\;$ be any ideal in a ring $\rm\; R\;$. We write $\rm\: R^{*}\:$ for the quotient ring $\rm\; R/I\;$. &#xA;  In the polynomial ring $\rm\; R[x]\;$ there is a smallest extension $\rm\; IR[x]\;$ of $\rm\; I\;$. The quotient ring $\rm\; R[x]/IR[x]\;$ is in a natural way isomorphic to $\rm\; R^*[x].\;$ In treating many problems, we can in this way &lt;strong&gt;reduce to the case&lt;/strong&gt; $\rm\; I = 0\;$, and we shall often do so.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;THEOREM 28.&lt;/strong&gt; $\;$ Let $\rm\; M\;$ be a maximal ideal in $\rm\; R[x]\;$ and suppose that &#xA;  the contraction $\rm\; M \cap R\ =\ N\;$ is maximal in $\rm\; R\;.\ $ Then $\rm\; M\;$ can be generated by $\rm\; N\;$ and one more element $\rm\; f\:.\ $ We can select $\rm\; f\;$ to be a monic polynomial which maps mod $\rm\; N\;$ into an irreducible polynomial over the field $\rm\; R/N\;$. &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;Proof.&lt;/strong&gt; $\;$ We can &lt;strong&gt;reduce to the case&lt;/strong&gt; $\rm\; N = 0,\;$ i. e., $\rm\; R\;$ a field, and then &#xA;  the statement is immediate.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;THEOREM 31.&lt;/strong&gt; $\;$ A commutative ring $\rm\; R\;\;$ is a Hilbert ring if and only if &#xA;  the polynomial ring $\rm\; R[x] \;\;$ is a Hilbert ring. &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;Proof.&lt;/strong&gt; $\;$ If $\rm\; \;\rm R[x]\;$ is a Hilbert ring, so is its homomorphic image $\rm\; R\;$. Conversely, assume that $\rm\; R\;$ is a Hilbert ring. Take a G-ideal $\rm\: Q\:$ in $\rm\; R[x]\;$; we must prove that $\rm\: Q\:$ is maximal. Let $\rm\: P = Q \cap R\:$; we can &lt;strong&gt;reduce the problem to the case&lt;/strong&gt; $\rm\ P = 0\:,\ $ which, incidentally, makes $\rm\ R\ $ a domain. Let $\rm\; u\;$ be the image of $\rm\; x\;$ in the natural homomorphism $\rm\ R[x] \ \to\  R[x]/Q\ .\ \ $ Then $\rm\; R[u]\;$ is a G-domain. $\ $ By Theorem 23, $\rm\ u\ $ is algebraic over $\rm\ R\ $ and $\rm\ R\ $ is a G-domain. Since $\rm\:R\:$ is both a G-domain and a Hilbert ring, $\rm\ R\ $ is a field. $\ $ But this makes $\rm\ R[u] = R[x]/Q\ $ a field, $\ $ proving $\rm\ Q\ $ to be maximal.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-12-07T05:25:50.567" LastActivityDate="2010-12-07T05:25:50.567" />
  <row Id="5736" PostTypeId="1" AcceptedAnswerId="5737" CreationDate="2010-09-29T21:58:52.043" Score="1" ViewCount="39" Body="&lt;p&gt;Hello&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can someone shed a light on how to solve something like this?&#xA;I've been looking for permutations but so far I found it very confusing.&#xA;As I'm not at home atm, I can't reach for my math books either for an answer.&#xA;Maybe you can help? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;a 5x5 grid&lt;/li&gt;&#xA;&lt;li&gt;each cell can have 2 possible values: 0 or 1&lt;/li&gt;&#xA;&lt;li&gt;each resulting grid &quot;layout&quot; must be bilateral-symmetric ( on the vertical axis )&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Question:&#xA;What's the number of possible variations?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks a lot!&lt;/p&gt;&#xA;" OwnerUserId="1969" LastEditorUserId="1969" LastEditorDisplayName="" LastEditDate="2010-09-29T22:06:30.000" LastActivityDate="2010-09-29T22:06:30.000" Title="Possible variations in a bilateral symmetric system" Tags="&lt;permutations&gt;" AnswerCount="1" />
  <row Id="5737" PostTypeId="2" ParentId="5736" CreationDate="2010-09-29T22:05:27.367" Score="1" ViewCount="" Body="&lt;p&gt;You only need to consider all possible $3\times 5$ grids, from which you can get a symmetrical $5\times 5$ grid by mirroring the first two columns about the third. Each cell in the $3\times 5$ grid has 2 possibilities, independent of all the others, so the total number of possibilities is $2^{3\times 5} = 32768$.&lt;/p&gt;&#xA;" OwnerUserId="856" LastActivityDate="2010-09-29T22:05:27.367" CommentCount="1" />
  <row Id="5738" PostTypeId="1" AcceptedAnswerId="5741" CreationDate="2010-09-29T22:18:31.353" Score="1" ViewCount="193" Body="&lt;p&gt;This question is from the book How to Prove It and I'm having trouble getting started with it.  The book provides the hint &quot;first prove a &amp;lt; 0&quot; however I can't figure out how to get that far with the provided givens.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given a≠0 and b≠0 and a &amp;lt; 1/a &amp;lt; b &amp;lt; 1/b prove a &amp;lt; -1&lt;/p&gt;&#xA;" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-09-30T10:19:41.543" LastActivityDate="2010-10-01T03:08:07.393" Title="help with a proof: Given a≠0 and b≠0 and a &lt; 1/a &lt; b &lt; 1/b prove a &lt; -1" Tags="&lt;algebra-precalculus&gt;&lt;proof-strategy&gt;" AnswerCount="6" CommentCount="0" />
  <row Id="5739" PostTypeId="2" ParentId="5738" CreationDate="2010-09-29T22:25:16.040" Score="1" ViewCount="" Body="&lt;p&gt;Hint: Try to prove $a &amp;lt; 0$ by contradiction, i.e. assume $a &gt; 0$ and try to derive a contradiction.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-09-29T22:25:16.040" />
  <row Id="5740" PostTypeId="2" ParentId="5738" CreationDate="2010-09-29T22:30:08.330" Score="1" ViewCount="" Body="&lt;p&gt;Hint: What does a&amp;lt;1/a say about a? What does b&amp;lt;1/b say about b?&lt;/p&gt;&#xA;" OwnerUserId="1149" LastActivityDate="2010-09-29T22:30:08.330" />
  <row Id="5741" PostTypeId="2" ParentId="5738" CreationDate="2010-09-29T22:32:23.673" Score="2" ViewCount="" Body="&lt;p&gt;Suppose that $a$ were positive. Then, multiplying the string of ineqs. (doesn't change their direction) by $a$ gives $a^2 &amp;lt;1 &amp;lt; ba &amp;lt; a/b$. From here we see that $b&gt;1$ (why?), but then, from the last ineq., $b^2 a &amp;lt; a$ which is impossible (why?). Therefore $a&amp;lt;0$. Now multiply again by $a$ and conclude.&lt;/p&gt;&#xA;" OwnerUserId="1313" LastActivityDate="2010-09-29T22:32:23.673" CommentCount="2" />
  <row Id="5742" PostTypeId="2" ParentId="5738" CreationDate="2010-09-29T22:58:42.047" Score="4" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\ \ \ $ Multiplying $\rm\ a &amp;lt; 1/a\ \:$ by $\rm\ a^2\ $ yields $\rm\:\ a^3 &amp;lt; a\ $. By symmetry $\rm\ b^3 &amp;lt; b\ $. So both $\rm\: a\:$ and $\rm \:b\:$ must lie in the intervals where the graph of $\ x^3 - x &amp;lt; 0\:,\ $ i.e. either in  $(-\infty,-1)$ or in $(0,1)\:$. But $\rm\ a \in (0,1)\ \Rightarrow\ 1/a &gt; 1\ $ contra $\rm\ 1/a &amp;lt; b &amp;lt; 1\ $.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/CXxGs.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-09-29T23:05:34.013" LastActivityDate="2010-09-29T23:05:34.013" CommentCount="2" />
  <row Id="5743" PostTypeId="1" AcceptedAnswerId="5744" CreationDate="2010-09-29T23:16:52.600" Score="1" ViewCount="99" Body="&lt;p&gt;In the literature it is stated that to each quadratic irrational $\gamma=\frac{P+\sqrt{D}}{Q}$ there is a corresponding ideal $I=[|Q|/\sigma , (P+\sqrt{D})/\sigma]$, where $\sigma=1$, if $\Delta \equiv0$ mod $4$ and $\sigma=2$, otherwise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, in the case of $\frac{2+\sqrt{13}}{3}$ the associated ideal must be $I=[3/2, (2+\sqrt{13})/2]$ which makes no sense, as $N(I)=3/2$ is supposed to be a rational integer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What am I doing wrong here? &lt;/p&gt;&#xA;" OwnerUserId="2212" LastActivityDate="2010-09-30T02:19:15.367" Title="Ideals in a Quadratic Number Fields" Tags="&lt;algebraic-number-theory&gt;" AnswerCount="1" />
  <row Id="5744" PostTypeId="2" ParentId="5743" CreationDate="2010-09-29T23:41:35.470" Score="1" ViewCount="" Body="&lt;p&gt;Below is a proof of the standard equivalences between forms, ideals and numbers, excerpted from section 5.2, p. 225 of Henri Cohen's book &quot;A course in computational algebraic number theory&quot;. Note that your quadratic number is not of the form specified in this equivalence, viz. $\rm\ \tau = (-b+\sqrt{D})/(2a)\:,\:$ and $\rm\: 4\:a\:|\:(D-b^2)\:,\:\:$ i.e. $\rm\ a\:|\:N(a\tau)\:,\:$ a condition equivalent to the $\rm\mathbb Z$-module $\rm\ a\:\mathbb Z + a\tau\ \mathbb Z\ $ being an ideal when $\rm\:D\:$ and $\rm\:b\:$ have the same parity, e.g. see &lt;a href=&quot;http://www.fen.bilkent.edu.tr/~franz/ant/ant02.pdf&quot; rel=&quot;nofollow&quot;&gt;Proposition 2.8 p.18 in Franz Lemmermeyer's notes.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/eIFwF.jpg&quot; alt=&quot;alt text&quot;&gt;&#xA;&lt;img src=&quot;http://i.imgur.com/bs8E6.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-09-30T02:19:15.367" LastActivityDate="2010-09-30T02:19:15.367" />
  <row Id="5745" PostTypeId="1" AcceptedAnswerId="5771" CreationDate="2010-09-30T00:27:56.233" Score="-1" ViewCount="234" Body="&lt;p&gt;In Quaternion mathematics, normally $$i ^ {2} = j ^ {2} = k ^ {2} = i*j*k = {-1}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I am looking at a different type of Formula where $$i ^ {j ^ k} = {-1} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(edit again, since somehow this remains elusive to explain...)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With functions, $f$ and all permutations of the subset variables $N$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $i$ is used in cases of non-real $f(x)=N$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $i ^ j$ is used in cases of non-real $f(x,y)=N$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $i ^ {j ^ k}$ is used in cases of non-real $f(x,y,z) = N$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This of course must obey other permutations of $x$, $y$, and $z$ such as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $k ^ i$ is used in cases of non-real $f(z,x) = N$; etc...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(&lt;em&gt;EDIT NOT A RING&lt;/em&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have given a bit more thought about this, and since $x$, $y$ and $z$ are possibly inter-changable depending on the orientation in real space, consider the possibility:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$i ^ {j ^ k} = j ^ {k ^ i} = k ^ {i ^ j} = -1$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would be an &quot;exponential-type ring&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(&lt;em&gt;EDIT MAYBE IT IS A RING?&lt;/em&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are some important fundamental differences (or formulae) I should be looking at in how i, j and k compare? (more importantly is the second formula even possible?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT TO FURTHER ELABORATE&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider in the second example a hypothetical anti-DeSitter space, where the real spatial directions &lt;code&gt;X, Y, and Z&lt;/code&gt; are real components, and &lt;code&gt;i, j and k&lt;/code&gt; are imaginary spatial directions that associate to each one. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$i^2 = -1$; for $x$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$j^2 = -1$; for $y$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$k^2 = -1$; for $z$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because, I am looking at a &lt;em&gt;different&lt;/em&gt; approach to &lt;code&gt;Riemann Tensors&lt;/code&gt;, my approach is to use exponentiation, rather than multiply. I am hoping someone that has experience with this type of thing can point out glaring flaws in this type of approach, and apologies in advance if I have errors in the reasoning here, its been some time since I've worked out anything this complex. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have further details &lt;a href=&quot;http://physicsmathforums.com/showthread.php?t=2343&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; &lt;a href=&quot;http://physicsmathforums.com/showthread.php?t=2416&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; and &lt;a href=&quot;http://physicsmathforums.com/showthread.php?t=2417&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; regarding my problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT FOR SIMPLE CASE EXAMPLE&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What values are possible or valid for $a$, $b$, and $c$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(ai)^{(bj)^{(ck)}} = -1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;if $i^2 = j^2 = k^2 = -1$&lt;/p&gt;&#xA;" OwnerUserId="2213" LastEditorUserId="2213" LastEditorDisplayName="" LastEditDate="2010-09-30T10:54:05.580" LastActivityDate="2010-10-01T03:33:23.723" Title="Differences in Quaternion properties?" Tags="&lt;complex-numbers&gt;&lt;quaternions&gt;" AnswerCount="1" CommentCount="16" />
  <row Id="5746" PostTypeId="1" CreationDate="2010-09-30T03:29:43.493" Score="1" ViewCount="315" Body="&lt;p&gt;How do I use second order Taylor method to solve a system of non-linear equations? Is there a good reference that gives details? I found mentions of it in a dozen of numerical analysis books, but no examples&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Specifically, $f:\mathbb{R}^n \to \mathbb{R}^m$, solve $f(\mathbf{x})=\mathbf{0}$ using second order Taylor expansion of $f$ around initial guess $\mathbf{x_0}$&lt;/p&gt;&#xA;" OwnerUserId="998" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-09-30T04:00:18.637" LastActivityDate="2010-09-30T16:18:11.627" Title="Second order Taylor method to solve system of equations" Tags="&lt;sequences-and-series&gt;&lt;numerical-methods&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="5747" PostTypeId="1" AcceptedAnswerId="5748" CreationDate="2010-09-30T04:28:16.190" Score="1" ViewCount="167" Body="&lt;p&gt;A variation of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Roots_of_unity&quot; rel=&quot;nofollow&quot;&gt;Root of Unity&lt;/a&gt; problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to find all possible answers to this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$z^n = i$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $$i^2 = -1$$&lt;/p&gt;&#xA;" OwnerUserId="2213" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-09-30T05:26:20.187" LastActivityDate="2010-09-30T09:45:10.913" Title="How can you find the complex roots of i?" Tags="&lt;complex-numbers&gt;&lt;roots&gt;" AnswerCount="3" CommentCount="5" />
  <row Id="5748" PostTypeId="2" ParentId="5747" CreationDate="2010-09-30T04:38:09.700" Score="4" ViewCount="" Body="&lt;p&gt;If the polar form of $z$ is $z=r(\cos\theta + i\sin\theta)$, there are $n$ distinct solutions to the equation $w^n = z$: $w=\sqrt[n]{r}(\cos\frac{\theta +2\pi k}{n}+ i \sin\frac{\theta +2\pi k}{n})$, where $k=0,1,...,n-1$. In your case, $z=i$, whose polar form is given by $r=1$, $\theta = \pi /2$.&lt;/p&gt;&#xA;" OwnerUserId="1313" LastActivityDate="2010-09-30T04:38:09.700" />
  <row Id="5749" PostTypeId="2" ParentId="5747" CreationDate="2010-09-30T04:41:44.857" Score="2" ViewCount="" Body="&lt;p&gt;Generally, the answers would be of the form&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sqrt[n]{i}\omega_n^j$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\omega_n=\exp\left(\frac{2i\pi}{n}\right)$ is a root of unity, and $j=0\dots n-1$.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-09-30T04:41:44.857" />
  <row Id="5750" PostTypeId="2" ParentId="5734" CreationDate="2010-09-30T05:31:12.057" Score="4" ViewCount="" Body="&lt;p&gt;In principle, the answer to question (1) should be &quot;no&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for question (2), it seems that a basic construction would be to form a &quot;profinite completion&quot; of the topological group $G$ by taking the inverse limit of all quotients $G/N$, where $N$ ranges over all finite-index &lt;em&gt;open&lt;/em&gt; normal subgroups of $G$.  The resulting completion $\hat{G}$ will have the property that the canonical function $G\to\hat{G}$ is continuous, and it should be somehow universal with respect to this property.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, let $\mathbb{Z}_2^\infty$ be the direct sum of infinitely many copies of $\mathbb{Z}_2$, which is a subspace of the infinite product $\mathbb{Z}_2^\omega$.  Then $\mathbb{Z}_2^\infty$ has many finite-index subgroups that are not open (e.g. the subgroup of elements with an even number of $1$'s), but it seems to me that every finite-index open subgroup should be the intersection of an open subgroup of $\mathbb{Z}_2^\omega$ with $\mathbb{Z}_2^\infty$.  Then the resulting topological profinite completion of $\mathbb{Z}_2^\infty$ ought to be $\mathbb{Z}_2^\omega$.&lt;/p&gt;&#xA;" OwnerUserId="1726" LastActivityDate="2010-09-30T05:31:12.057" CommentCount="2" />
  <row Id="5751" PostTypeId="2" ParentId="5718" CreationDate="2010-09-30T07:53:57.700" Score="3" ViewCount="" Body="&lt;p&gt;Since $\mathcal O_L$ is the union of $\mathcal O_{L_i}$, where $L_i$ runs over all the finite subextensions, giving a prime ideal $\mathfrak p$ in $\mathcal O_L$ is the same as giving a compatible collection of primes ideals $\mathfrak p_i$ in each $\mathcal O_{L_i}$.  (We set&#xA;$\mathfrak p_i := \mathfrak p \cap \mathcal O_{L_i}$, and $\mathfrak p = \cup_i \mathfrak p_i$.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $L$ is the union of the $L_i$, giving an absolute value $v$ on $L$ is the same as giving compatible absolute values $v_i$ on the various $L_i$.  (Take $v_i$ to be the restriction to $L_i$ of $v$.)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Combining the preceding two remarks, we see that the bijection between primes ideals and non-Archimedean valuations in the case of finite extensions extends to a corresponding bijection in the case of infinite extensions.  &lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2010-09-30T07:53:57.700" CommentCount="1" />
  <row Id="5752" PostTypeId="2" ParentId="5747" CreationDate="2010-09-30T09:45:10.913" Score="1" ViewCount="" Body="&lt;p&gt;Also, observe that if $z^n=i$ then $z^{4n}=1$. Thus, the complex numbers you're looking for are particular $4n$-th roots of $1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you know that the $m$-th roots of 1 (any $m$) can be written as powers of a single well-chosen one (a primitive root), it shouldn't be too hardto find exactly which $4n$-th roots have the desired property.&lt;/p&gt;&#xA;" OwnerUserId="688" LastActivityDate="2010-09-30T09:45:10.913" />
  <row Id="5753" PostTypeId="2" ParentId="5738" CreationDate="2010-09-30T09:48:53.903" Score="0" ViewCount="" Body="&lt;p&gt;Robin-presumably that is ' not equal to' as opposed to factorial!&lt;/p&gt;&#xA;" LastActivityDate="2010-09-30T09:48:53.903" CommunityOwnedDate="2010-09-30T09:48:53.903" />
  <row Id="5755" PostTypeId="1" AcceptedAnswerId="5843" CreationDate="2010-09-30T10:32:50.683" Score="5" ViewCount="130" Body="&lt;p&gt;I'm looking for a convenient upper bound on the integral $\int_y^\infty x^k \exp(-(x-\mu)^2/2) dx$ for (possibly large) positive integer k.  This is equivalent to finding higher moments of a truncated normal distribution. A bound that works for non-integer k as well would be even better.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course &quot;convenient&quot; is in the eye of the beholder, but I'd like some sort of fairly simple expression that I can use in further calculations.  For example, an upper bound of the form $f(x) \exp( -g(x))$ where where $f$ and $g$ are low-degree polynomials would be great. I'm more interested in simplicity of form than in obtaining the tightest possible bound.&lt;/p&gt;&#xA;" OwnerUserId="247" LastEditorUserId="247" LastEditorDisplayName="" LastEditDate="2010-09-30T14:48:56.203" LastActivityDate="2010-10-01T20:09:45.113" Title="Bounding higher moments of truncated normal" Tags="&lt;real-analysis&gt;&lt;probability-theory&gt;" AnswerCount="2" />
  <row Id="5756" PostTypeId="1" AcceptedAnswerId="5776" CreationDate="2010-09-30T10:35:01.987" Score="5" ViewCount="74" Body="&lt;p&gt;What is the largest (acyclic, connected) transitively reduced acyclic graph on $n$ vertices, for every $n$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How many edges $e$ does it have?  How does $e$ grow as a function of $n$?  Faster than $n^2/4$?&lt;/p&gt;&#xA;" OwnerUserId="2217" LastEditorUserId="2217" LastEditorDisplayName="" LastEditDate="2010-09-30T11:20:40.093" LastActivityDate="2010-09-30T18:52:55.323" Title="How big are transitively reduced graphs?" Tags="&lt;graph-theory&gt;" AnswerCount="2" CommentCount="5" FavoriteCount="1" />
  <row Id="5757" PostTypeId="2" ParentId="5755" CreationDate="2010-09-30T11:09:43.610" Score="1" ViewCount="" Body="&lt;p&gt;Let $I_k=\int_y^\infty x^k\mathrm{exp}(-(x-\mu)^2/2)dx$ and $a_k=y^{k}\mathrm{exp}(-(y-\mu)^2/2)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Integration by parts gives $I_k=a_{k-1}+(k-1)I_{k-2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, for even $n$,&#xA;\begin{multline}&#xA;I_n=a_{n-1}+(n-1)a_{n-3}+(n-1)(n-3)a_{n-5}+\cdots\newline+(n-1)(n-3)\cdots 1 a_1+(n-1)(n-3)\cdots 1 I_0.&#xA;\end{multline}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Feel free to expand this answer and deduce an estimate from the above.&lt;/p&gt;&#xA;" OwnerUserId="367" LastEditorUserId="367" LastEditorDisplayName="" LastEditDate="2010-09-30T11:15:56.733" LastActivityDate="2010-09-30T11:15:56.733" CommunityOwnedDate="2010-09-30T11:19:34.993" CommentCount="3" />
  <row Id="5758" PostTypeId="1" AcceptedAnswerId="5759" CreationDate="2010-09-30T12:14:30.857" Score="2" ViewCount="169" Body="&lt;p&gt;If you start with an infinite set, you can have a sequence of nested sets which converge to a single point. (ie Intersection of (-1/n, 1/n) as n-&gt;infinity))&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, at no time during the sequence is there a first element with only one point. In fact, for any finite (but unbounded) n, the number of points in the interval is uncountably large. So, how do you understand the idea that this infinite intersection contains just one point? How do you make sense of it?&lt;/p&gt;&#xA;" OwnerUserId="2003" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2010-11-24T02:31:12.190" LastActivityDate="2010-11-24T02:31:12.190" Title="Understanding of convergence of intersections of sets" Tags="&lt;real-analysis&gt;&lt;intuition&gt;&lt;elementary-set-theory&gt;&lt;infinity&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="1" />
  <row Id="5759" PostTypeId="2" ParentId="5758" CreationDate="2010-09-30T12:28:52.460" Score="3" ViewCount="" Body="&lt;p&gt;A point is in the intersection if it's in &lt;em&gt;every&lt;/em&gt; set of the sequence. Focus on that. No matter how close to $0$ a point is (excluding $0$ itself), there will be a sufficiently large $n$ such that the set $(-1/n,1/n)$ does not contain it, and therefore that point will not be in the (infinite) intersection, because, again, a point is in the intersection iff and it is in every set specified by the intersection.&lt;/p&gt;&#xA;" OwnerUserId="1313" LastActivityDate="2010-09-30T12:28:52.460" />
  <row Id="5760" PostTypeId="1" AcceptedAnswerId="5764" CreationDate="2010-09-30T13:02:10.917" Score="1" ViewCount="68" Body="&lt;p&gt;Suppose $x_1, x_2, y_1, y_2$ are known integers which satisfy&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$y_1 \equiv \gamma x_1 \pmod c$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$y_2 \equiv \gamma x_2 \pmod c$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\gamma$ and $c$ are unknown integers.   Is there a way to determine the values of $\gamma$ and $c$? &lt;/p&gt;&#xA;" OwnerUserId="832" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-09-30T13:08:59.750" LastActivityDate="2010-09-30T13:56:39.927" Title="System of Congruences" Tags="&lt;number-theory&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="5761" PostTypeId="2" ParentId="5758" CreationDate="2010-09-30T13:02:17.637" Score="4" ViewCount="" Body="&lt;p&gt;I think the source of the confusion is this: taking the cardinality of a set seems like such a simple operation that we subconsciously expect it to be continuous.  We might think that simple limiting operations like countable nested intersections should give convergent cardinalities.  But the simple fact is that they don't; our subconscious guess is just wrong, and this example proves it.  Cardinality is not continuous in this sense.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is related to the standard &quot;paradox&quot; of the supertask where at time $1-2^{-n}$ you add to a box two balls numbered $2n-1$ and $2n$, and remove the one numbered $n$.  Despite the fact that the cardinality of the balls in the box is always increasing, in the limit at time 1 the box is empty.  Again, the paradox arises only if you feel that cardinality should be continuous.  If you realize that it isn't, then you no longer see such examples as paradoxical.&lt;/p&gt;&#xA;" OwnerUserId="822" LastActivityDate="2010-09-30T13:02:17.637" CommentCount="2" />
  <row Id="5762" PostTypeId="2" ParentId="5756" CreationDate="2010-09-30T13:25:23.663" Score="2" ViewCount="" Body="&lt;p&gt;Bipartite graph is the largest.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof:&lt;/strong&gt; Suppose G is a graph, v is it's vertex. Let $L(v)$ is equal to the maximal length of a directed path, which starts in $v$. Let $L(G)=max_{v\in G} L(v)$. If $L(G)=1$, the graph is bipartite. In order to prove, that bipartite graph is the largest, I will prove, that for any graph $G$ such that $L(G)&gt;1$ there exists graph $H$ with the same number of vertices and edges, such that $L(G)&gt;L(H)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note, that any vertex with $L(v)=0$ has only inbound connections. This connections are from vertices w with $L(w)=1$. We will do graph $H$ in two steps.&lt;br&gt;&#xA;Step 1. Remove all vertices with $L(v)=0$, and edges to them. Note that $L(G)$ has been decreased.&#xA;Step 2. Add all the vertices, removed on step 1. Add edges, removed on step 1, but reverse their direction. Call the new graph $H$. Note, that it has the same number of vertices and edges and $L(H) &amp;lt; L(G)$.&lt;/p&gt;&#xA;" OwnerUserId="695" LastActivityDate="2010-09-30T13:25:23.663" CommentCount="1" />
  <row Id="5763" PostTypeId="1" AcceptedAnswerId="5780" CreationDate="2010-09-30T13:29:45.010" Score="3" ViewCount="258" Body="&lt;p&gt;I had asked this question: &lt;a href=&quot;http://math.stackexchange.com/questions/4412/characterising-continuous-functions&quot;&gt;http://math.stackexchange.com/questions/4412/characterising-continuous-functions&lt;/a&gt; some time back, and this question is more or less related to that question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose we have a function $f: \mathbb{R} \to \mathbb{R}$ and suppose the set  $G = \{ (x,f(x) : x \in \mathbb{R}\}$ is connected and closed in $\mathbb{R}^{2}$, then does it imply $f$ is continuous?&lt;/p&gt;&#xA;" LastActivityDate="2010-09-30T21:35:51.343" Title="Characterizing continuous functions based on the graph of the function." Tags="&lt;real-analysis&gt;&lt;analysis&gt;&lt;general-topology&gt;" AnswerCount="4" FavoriteCount="1" />
  <row Id="5764" PostTypeId="2" ParentId="5760" CreationDate="2010-09-30T13:50:45.917" Score="2" ViewCount="" Body="&lt;p&gt;No, because if c=pq, for primes p and q, then both congruences hold mod p and mod q as well. So c is not uniquely determined.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, if $y_1=y_2=11$ and $x_1=x_2=1$ and $\gamma=5$, c could be 1,2,3 or 6.&lt;/p&gt;&#xA;" OwnerUserId="1301" LastEditorUserId="1301" LastEditorDisplayName="" LastEditDate="2010-09-30T13:56:39.927" LastActivityDate="2010-09-30T13:56:39.927" />
  <row Id="5766" PostTypeId="2" ParentId="5763" CreationDate="2010-09-30T15:40:17.687" Score="0" ViewCount="" Body="&lt;p&gt;suppose you have a function $f:\mathbb{R} \rightarrow \mathbb{R}$ which is not continuous and its graph is closed and connected. wlog assume f is not continuous at zero and f(0)=0. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;first, I want to show that when x approaches zero then either f(x) approaches 0 or goes to infinity. &#xA;for every b&gt;a&gt;0 let $ X_{a,b} = \{ x&gt;0 \mid a &amp;lt; f(x) &amp;lt; b \}  $ . if this set is not empty, then its infimum cannot be 0, because then the closure of the graph will give us that $ a \leq f(0) \leq b $ (you can find a series $a_n \rightarrow 0 \subseteq X_{a,b}$ and a sub-series such that $f(a_{n_i})$ is monotone and thus converges in $R^2$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;now I want to use the connectedness to show that f(x) cannot go to infinity when x goes to zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f$ isn't continuous at zero, so wlog there is an $\varepsilon &gt; 0$ such that $ X_{2\varepsilon, \infty}$ is not empty and its infimum is 0. $X_{\varepsilon, 2\varepsilon}$ has infimum &gt; 0 (if not empty) - denote it by $\delta$. if you have $0 &amp;lt; x_0 &amp;lt; \delta$ such that $ x_0 \notin X_{2 \varepsilon, \infty} $ (and so $f(x_0) &amp;lt; \varepsilon $ ) then you have the open sets $ A = (0, x_0) \times (1.5 \varepsilon, \infty) $ and $ R^2 - \bar A$ which separates the garph of the function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;otherwise, if $ (0,\delta) \subseteq X_{2\varepsilon, \infty} $ then you can take the sets $A = \{x\leq 0 \} \cup (0, \delta/2) \times (-\infty, 1.5 \varepsilon)$ and $R^2 - \bar A$.&lt;/p&gt;&#xA;" OwnerUserId="2125" LastActivityDate="2010-09-30T15:40:17.687" />
  <row Id="5767" PostTypeId="2" ParentId="5763" CreationDate="2010-09-30T15:57:21.707" Score="0" ViewCount="" Body="&lt;p&gt;The answer is yes.  Here's one way to prove it.  (There might be a slicker way, but this seems to work.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume $G$ is connected and closed. Let $a\in\mathbb R$ be arbitrary, and let $\epsilon&gt;0$ be given.  Because $(a,f(a)-\epsilon)\notin G$, the fact that the complement of $G$ is open implies that there is a product neighborhood of the form $(a-\delta,a+\delta)\times (f(a)-\epsilon-c, f(a)-\epsilon+c)$ contained in the complement of $G$.  This means that  $|x-a|&amp;lt;\delta$ implies that one of the following two inequalities holds:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$f(x)\ge f(a)-\epsilon+c&gt;f(a)-\epsilon$, or&lt;/li&gt;&#xA;&lt;li&gt;$f(x)\le f(a)-\epsilon-c$.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;If there is any $x\in (a-\delta,a+\delta)$ so that the second inequality holds, say $f(b)\le f(a)-\epsilon-c$ (without loss of generality, we may assume $b\lt a$), then the graph of $f$ does not intersect the following set:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\{(b,y): y\ge f(a)-\epsilon \}\cup \{(x,f(a)-\epsilon): b\le x \le a\} \cup \{(a,y): y\le f(a)-\epsilon\}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(See the diagram below.)  This broken line disconnects $G$, contradicting the assumption that $G$ is connected.  Therefore inequality (1) holds when $|x-a|&amp;lt;\delta$.&lt;br&gt;&#xA;A similar argument shows that $f(x)\lt f(a)+\epsilon$ when $|x-a|&amp;lt;\delta$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Putting these together, we conclude that $|x-a|&amp;lt;\delta$ implies $|f(x)-f(a)|&amp;lt;\epsilon$, so $f$ is continuous at $a$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/OrP64.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1421" LastActivityDate="2010-09-30T15:57:21.707" />
  <row Id="5768" PostTypeId="2" ParentId="5763" CreationDate="2010-09-30T16:00:40.263" Score="0" ViewCount="" Body="&lt;p&gt;Yes, I think so.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, observe that such $f$ must have the intermediate value property.  For suppose not; then there exist $a &amp;lt; b$ with (say) $f(a) &amp;lt; f(b)$ and $y \in (f(a),f(b))$ such that $f(x) \ne y$ for all $x \in (a,b)$.  Then $A = (-\infty,a) \times \mathbb{R} \cup (-\infty,b) \times (-\infty,y)$ and $B = (b, +\infty) \times \mathbb{R} \cup (a,+\infty) \times (y,+\infty)$ are disjoint nonempty open subsets of $\mathbb{R}^2$ whose union contains $G$, contradicting connectedness.  (Draw  a picture.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now take some $x \in \mathbb{R}$, and suppose $f(x) &amp;lt; y &amp;lt; \limsup_{t \uparrow x} f(t) \le +\infty$. Then there is a  sequence $t_n \uparrow x$ with $f(t_n) &gt; y$ for each $n$.  By the intermediate value property, for each $n$ there is $s_n \in (t_n, x)$ with $f(s_n) = y$.  So $(s_n, y) \in G$ and $(s_n,y) \to (x,y)$, so since $G$ is closed $(x,y) \in G$ and $y = f(x)$, a contradiction.  So $\limsup_{t \uparrow x} f(t) \le f(x)$.  Similarly, $\limsup_{t \downarrow x} f(t) \le f(x)$, so $\limsup_{t \to x} f(t) \le f(x)$.  Similarly, $\liminf_{t \to x} f(t) \ge f(x)$, so that $\lim_{t \to x} f(t) = f(x)$, and $f$ is continuous at $x$.&lt;/p&gt;&#xA;" OwnerUserId="822" LastActivityDate="2010-09-30T16:00:40.263" CommentCount="2" />
  <row Id="5769" PostTypeId="2" ParentId="5746" CreationDate="2010-09-30T16:18:11.627" Score="1" ViewCount="" Body="&lt;p&gt;One way to solve f(x) = y is to minimize g(x) = (f(x) - y)^2.  You could do this by taking the quadratic approximation to g(x) that comes from the second order Taylor series centered at your initial guess, finding its minimum, and letting that minimum be the starting guess for the next iteration. This isn't necessarily the best algorithm, but it's easy to understand and implement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For high-dimensional problems, the work is in solving the system of equations to minimize the quadratic approximation.  A great deal of research has gone into doing that step cleverly, taking advantage of sparse matrix structure etc.&lt;/p&gt;&#xA;" OwnerUserId="247" LastActivityDate="2010-09-30T16:18:11.627" CommentCount="4" />
  <row Id="5770" PostTypeId="2" ParentId="5738" CreationDate="2010-09-30T16:42:18.513" Score="0" ViewCount="" Body="&lt;p&gt;$\ $ If $\rm\ a &gt; 0\ $ then $\rm\ a &amp;lt; b\ $ times $\rm\ 1/a &amp;lt; 1/b\ \Rightarrow\ 1 &amp;lt; 1$   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So $\rm\ a &amp;lt; 0\ \:$ and $\rm\:\ a &amp;lt; 1/a\ \:\Rightarrow\:\ a^2 &gt; 1 \ \:\Rightarrow\:\ a &amp;lt; -1$&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-10-01T03:08:07.393" LastActivityDate="2010-10-01T03:08:07.393" />
  <row Id="5771" PostTypeId="2" ParentId="5745" CreationDate="2010-09-30T16:48:32.307" Score="2" ViewCount="" Body="&lt;p&gt;This may be a complete stupid thing, for which I'll lose some hundred points of reputation, but here it is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't understand anything at all about your functionss $f$ &quot;and all permutations of the subset variables N...&quot;, nor the rest of your question. But I think that expressions like&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;i^{j^k} \ , \qquad j^{k^i} \ , \qquad \text{and}\qquad   k^{i^j} &#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;could make sense. However, they don't seem to be equal to $-1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's remember how we define $x^y$ for real numbers $x, y$, $x&gt;0$. We put&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;x^y = e^{y\log x} \ ,&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;don't we? After choosing a determination for the complex logarithm, we define also&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;i^z = e^{z \log i }\ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we need to compute $\log i$. For this we can rely on Euler's formula, which, for a real number $\theta$, says:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;e^{i\theta} = \cos\theta + i \sin\theta \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we ask a complex number $z = x + iy$ to be a logarithm for $i$, we must have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;i = e^z = e^{x + iy} = e^x e^{iy} = e^x (\cos y + i \sin y) \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;0 = e^x \cos y \qquad \text{and} \qquad 1 = e^x\sin y \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we may take $y = \frac{\pi}{2}$ and then $x= 0$. Hence a logarithm for $i$ is $\log i = i\frac{\pi}{2}$ and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;i^z = e^{zi\frac{\pi}{2}} \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So far so good: what about quaternions? Well, we don't need to define exponentials and logarithms for them, but just notice that Euler's formula only needs two things to be true: (a) the Taylor series for $\exp z$, $\cos z$ and $\sin z$ and (b) &quot;something&quot;, called $i$, that can be added with real numbers and itself, multyplied by real numbers and such that $i^2 = -1$. Right? Okay. So, for the same reasons as for $i$, we may take &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\log i = i\frac{\pi}{2} \ , \qquad &#xA;\log j = j\frac{\pi}{2} \ , \qquad \text{and} \qquad&#xA;\log k = k\frac{\pi}{2} \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, we can compute $j^k$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;j^k = e^{k\log j} = e^{k j\frac{\pi}{2}} = e^{-i \frac{\pi}{2}} = \cos \left( \frac{-\pi}{2} \right) +i \sin \left( \frac{-\pi}{2} \right) = -i \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;i^{j^k} = i^{-i} = e^{-i \log i} = e^{-i i \frac{\pi}{2}} = e^{\frac{\pi}{2}} \ .&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="664" LastEditorUserId="664" LastEditorDisplayName="" LastEditDate="2010-10-01T03:33:23.723" LastActivityDate="2010-10-01T03:33:23.723" CommentCount="5" />
  <row Id="5772" PostTypeId="1" AcceptedAnswerId="5822" CreationDate="2010-09-30T17:09:45.773" Score="5" ViewCount="327" Body="&lt;p&gt;I've proved $\neg \neg \beta \rightarrow \beta$ is a theorem but I can't figure out a way to do the same for $\beta \rightarrow \neg \neg \beta$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems the proof would use Axiom 2 and the deduction theorem (which allows $\beta$ to be an axiom) - but i've endlessy tried values to no avail.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Axiom 1: $A \rightarrow ( B \rightarrow A )$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Axiom 2: $( A \rightarrow ( B \rightarrow C ) ) \rightarrow ( ( A \rightarrow B ) \rightarrow (A \rightarrow C) ) $.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Axiom 3: $( \neg B \rightarrow \neg A) \rightarrow ( ( \neg B \rightarrow A) \rightarrow B )$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So to clarify A, B, C $\alpha$ and $\beta$ are propositions (i.e. assigned True or False). $\rightarrow$ and $\neg$ have the standard logical meanings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: $\TeX$ification does not work in the IE9 beta.&lt;/p&gt;&#xA;" OwnerUserId="2222" LastEditorUserId="2222" LastEditorDisplayName="" LastEditDate="2010-10-01T16:05:59.743" LastActivityDate="2011-02-09T20:40:41.433" Title="Prove that $\beta \rightarrow \neg \neg \beta$ is a theorem using standard axioms 1,2,3 and MP." Tags="&lt;logic&gt;" AnswerCount="2" CommentCount="7" FavoriteCount="0" />
  <row Id="5773" PostTypeId="2" ParentId="5676" CreationDate="2010-09-30T17:44:02.330" Score="4" ViewCount="" Body="&lt;p&gt;In case you were wondering, like me, Moron's excellent proof adapts easily to show that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$1 + \sqrt[3]{2} + \dots + \sqrt[3]{n} \sim \frac{3}{4}n^{4/3} + \frac{\sqrt[3]{n}}{2} + C,$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for some constant $C.$ In this case $C = \zeta(-1/3) \approx -0.277343.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where, as before, $a_n \sim b_n$ means $\lim_{n \rightarrow \infty} (a_{n}-b_{n}) = 0.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similar to the previous proof, we use the AM-GM inequality to show&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sqrt[3]{x} \le \frac{x}{3k^{2/3}} + \frac{2k^{1/3}}{3}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Summing from $k=1$ to $n-1$ and integrating we arrive at&#xA;$$ \frac{3}{4}n^{4/3} - \frac{3}{4} \le \sum_{k=1}^{n-1}\sqrt[3]{n} +&#xA;\frac{1}{6}\sum_{k=1}^{n-1} \frac{1}{k^{2/3}}, \qquad n&gt;1.$$&#xA;And so,&#xA;$$\sum_{k=1}^{n}\sqrt[3]{n} \ge \frac{3}{4}n^{4/3} + n^{1/3} - \frac{3}{4} - \frac{1}{6}\sum_{k=1}^{n-1} \frac{1}{k^{2/3}}, \qquad n&gt;1.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using $\sum_{k=1}^{n-1} \frac{1}{k^{2/3}} \le 1+ \int_1^n x^{-2/3} dx,&#xA;\qquad n&gt;1,$ we obtain&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{1}{6}\sum_{k=1}^{n-1} \frac{1}{k^{2/3}} \le \frac{1}{2}n^{1/3} - \frac{1}{3}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And hence&#xA;$$\sum_{k=1}^{n}\sqrt[3]{n} \ge \frac{3}{4}n^{4/3} + \frac{1}{2}n^{1/3} - \frac{5}{12}, \qquad n&gt;1.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As in the previous argument, set&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$G_n = \sum_{k=1}^{n}\sqrt[3]{n} - \frac{3}{4}n^{4/3} - \frac{1}{2}n^{1/3}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then $G_n \ge -5/12,$ and we can show $G_n – G_{n+1} &gt; 0$ by showing that&#xA;$$\frac{(n+1)^{4/3} – n^{4/3}}{(n+1)^{1/3} + n^{1/3}} &gt; \frac{2}{3}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, as before, $G_n$ is convergent, since it is bounded below and monotonically decreasing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suspect this argument also adapts easily to the more general case&#xA;$\sum_{k=1}^{n}\sqrt[r]{n},$ for $r \in \mathbb{N},$ where I'm guessing, we'll find&#xA;$$1 + \sqrt[r]{2} + \dots + \sqrt[r]{n} \sim \frac{r}{r+1}n^{(r+1)/r} + \frac{\sqrt[r]{n}}{2} + C,$$ &#xA;where $C = \zeta(-1/r).$ However, I confess to not having checked the details of this further generalisation.&lt;/p&gt;&#xA;" OwnerUserId="1301" LastActivityDate="2010-09-30T17:44:02.330" />
  <row Id="5774" PostTypeId="1" AcceptedAnswerId="6019" CreationDate="2010-09-30T17:54:20.380" Score="3" ViewCount="140" Body="&lt;p&gt;I have recently been working out the full game tree for tic-tac-toe, just for fun.  I am using the well known equivalence relation of rotations/reflections to simplify this tree in the standard way (which begins by noting that there are only 3 opening moves: edge, corner, and center).  I actually could not find an image of the full game tree for tic-tac-toe, so if anyone can provide a link I'd appreciate that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This work motivates the following;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let there be given an N x N grid; and let $m$ be a natural number.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $I$ be the set of all possible ways to place $m$ copies of the letter $X$ in the grid, and $m$ copies of the letter $O$ in the grid (with the restriction that we can only place one letter per space of the grid; in other words, just imagine playing a game of tic-tac-toe on an N x N board for an even number of moves).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Problem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How many ways are there to completely break the rotational/reflection symmetry of an N x N grid by placing $m$ copies of $X$ and $m$ copies of $O$?&lt;/p&gt;&#xA;" OwnerUserId="1417" LastEditorUserId="1417" LastEditorDisplayName="" LastEditDate="2010-10-01T01:36:26.670" LastActivityDate="2011-05-20T15:18:31.803" Title="Counting the number of ways to break the symmetry of an N x N grid by placing one X and one O at a time" Tags="&lt;combinatorics&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="2" />
  <row Id="5775" PostTypeId="1" AcceptedAnswerId="5782" CreationDate="2010-09-30T18:37:53.607" Score="5" ViewCount="197" Body="&lt;p&gt;Given is a sequence $\langle a_1,a_2,\ldots,a_n\rangle$ over the alphabet $\{1,2,\ldots,m\}$ chosen uniformly at random among the $m^n$ possibilities. What is the expected size of the &lt;em&gt;set&lt;/em&gt; $\{a_1,a_2,\ldots,a_n\}$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $m=n$ it &lt;a href=&quot;http://rgrig.blogspot.com/2010/09/expected-size-of-set.html&quot; rel=&quot;nofollow&quot;&gt;seems&lt;/a&gt; the answer tends to $(1-1/e)n$ as $n\to\infty$, but I don't know why.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I bumped into this while benchmarking some code for hashtables, so I wouldn't be surprised if it is a standard result in the hash world.&lt;/p&gt;&#xA;" OwnerUserId="512" LastActivityDate="2011-01-20T21:09:33.727" Title="How many bins do random numbers fill?" Tags="&lt;combinatorics&gt;&lt;probability&gt;&lt;discrete-mathematics&gt;" AnswerCount="2" />
  <row Id="5776" PostTypeId="2" ParentId="5756" CreationDate="2010-09-30T18:52:55.323" Score="3" ViewCount="" Body="&lt;p&gt;Since the transitively reduced graph $G$ is acyclic, there is a corresponding undirected simple graph $G'$, which can be formed by ignoring the direction of the edge. The number of edges in $G$ is same as the number of edges in $G'$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $G$ to be transitively reduced, a necessary condition is the $G'$ must not contain any triangles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is well known (&lt;a href=&quot;http://en.wikipedia.org/wiki/Tur%C3%A1n%27s_theorem&quot; rel=&quot;nofollow&quot;&gt;Turan's/Mantel's theorem&lt;/a&gt;) that any simple undirected graph on $n$ vertices and more than $n^2/4$ edges has a triangle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is also known that the undirected graph on $n$ vertices with maximum edges and no triangles is the complete bipartite graph $K_{[n/2],[n/2]}$ (for instance check out: &lt;a href=&quot;http://books.google.com/books?id=SbZKSZ-1qrwC&amp;amp;pg=PA28&quot; rel=&quot;nofollow&quot;&gt;http://books.google.com/books?id=SbZKSZ-1qrwC&amp;amp;pg=PA28&lt;/a&gt;, exercise 4)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From your comments, it looks like you found a $G$ whose corresponding $G'$ is $K_{[n/2],[n/2]}$ and so that proves it.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-09-30T18:52:55.323" CommentCount="1" />
  <row Id="5777" PostTypeId="1" AcceptedAnswerId="5781" CreationDate="2010-09-30T18:55:15.780" Score="4" ViewCount="794" Body="&lt;p&gt;I know i'm supposed to do sin(3x) - sin x = 0 but beyond that im stuck.. i tryed expanding sin(3x) but that didn't help  &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;i want the value of x in the interval [0, 2π)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="2227" LastEditorUserId="2227" LastEditorDisplayName="" LastEditDate="2010-09-30T19:09:28.747" LastActivityDate="2010-09-30T20:04:32.573" Title="sin(3x) = sin(x)" Tags="&lt;trigonometry&gt;&lt;homework&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="5778" PostTypeId="1" CreationDate="2010-09-30T18:58:20.143" Score="1" ViewCount="239" Body="&lt;p&gt;This is probably a very simple question to all of you, but I have never taken a physics class, so I am lost for what I need to do for a programming problem. I will need to specify a force by inputting the newtons x, newtons y, and newtons z.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I calculate these 3 forces based on the desired angle and desired force for that direction?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you for your time, this project is important to me.&lt;/p&gt;&#xA;" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-11-19T16:46:29.377" LastActivityDate="2010-11-19T16:46:29.377" Title="Physics: Calculating the Forces Needed from an Angle and a Force" Tags="&lt;physics&gt;" AnswerCount="1" CommentCount="8" />
  <row Id="5779" PostTypeId="2" ParentId="4658" CreationDate="2010-09-30T19:00:50.207" Score="4" ViewCount="" Body="&lt;p&gt;With respect to the exact answer, the recursion is related to the  &lt;a href=&quot;http://mathworld.wolfram.com/Fibonaccin-StepNumber.html&quot; rel=&quot;nofollow&quot;&gt;Fibonacci n-Step Numbers&lt;/a&gt;  (look at the phrase below the table).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With respect to an approximate/asympotic solution: it can also be obtained by probabilistic reasoning.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instead of throwing N (150) coins, lets consider the (thought) experiment of throwing M alternate &lt;em&gt;runs&lt;/em&gt; (a ‘run’ is a sequence of consecutive tails/heads). That is, instead of throwing N iid Bernoulli random variables (two values with prob=1/2), we throw M iid geometric random variables ( p(1)=1/2 p(2)=1/4 p(3)=1/8 ...) which we interpret as the length of each alternate run of consecutive tails/head in the coins sequence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The expected value of the the runs (in both experiments) is 2. If we choose M=N/2, then the expected total number of coins (in the seconde experiment) will be N, and so we can expect (informally) that the two experiments are asympotically equivalent (in the original experiment the number of coins is fixed, the number of runs is a random variable; in the second, the reverse; this could be related to the use of different ensembles in statistical physics).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, in the modified experiment it’s easy to compute the probability that no run exceeds a length L: it’s just an  (1-(1/2)^(L-1))^(N/2)&#xA;If we consider just tails, the number of &quot;trials&quot; would be N/4 instead. This, in our case, (L=150, N=6) gives P=0.44160&lt;/p&gt;&#xA;" OwnerUserId="312" LastActivityDate="2010-09-30T19:00:50.207" />
  <row Id="5780" PostTypeId="2" ParentId="5763" CreationDate="2010-09-30T19:03:12.147" Score="9" ViewCount="" Body="&lt;p&gt;This &lt;a href=&quot;http://www.jstor.org/stable/2324521&quot; rel=&quot;nofollow&quot;&gt;Monthly paper&lt;/a&gt; has short simple proofs of the following&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;THEOREM&lt;/strong&gt;$\ $ TFAE if $\rm\ f: \mathbb R\to \mathbb R\ $ has a closed graph in $\:\mathbb R^2$&lt;br&gt;&#xA;(a)$\rm\ \ f\ $ is continuous.&lt;br&gt;&#xA;(b)$\rm\ \  f\ $ is locally bounded.&lt;br&gt;&#xA;(c)$\rm\ \ f\ $ has the intermediate value property.&lt;br&gt;&#xA;(d)$\rm\ \ f\ $ has a connected graph in $\rm\mathbb R^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More generally the result is merely a special case of R. L. Moore's 1920 characterization of a topological line as a locally compact metric space that is separated into two connected sets by each of its points.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Per request, I've appended the proof of the theorem below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/vnoai.jpg&quot; alt=&quot;alt text&quot;&gt;&#xA;&lt;img src=&quot;http://i.imgur.com/UyIrJ.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-09-30T21:35:51.343" LastActivityDate="2010-09-30T21:35:51.343" CommentCount="8" />
  <row Id="5781" PostTypeId="2" ParentId="5777" CreationDate="2010-09-30T19:03:21.460" Score="6" ViewCount="" Body="&lt;p&gt;We have $\sin{3x} = 3\sin{x} - 4\sin^{3}{x}$ which says that we have to solve the equation $$3\sin{x} - 4\sin^{3}{x} - \sin{x}=0$$, that is $2 \sin{x} - 4\sin^{3}{x}=0$. Take $y = \sin{x}$ and so you have $$2y-4y^{3}=0 \Longrightarrow 2y(1-2y^{2})=0$$ and then see what happens. I hope this helps you out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or you can even try this $$\sin{3x} - \sin{x} = 2 \cos\Bigl(\frac{3x+x}{2}\Bigr) \cdot \sin\Bigl(\frac{3x-x}{2}\Bigr) = 2\cos{2x} \cdot \sin{x}$$&lt;/p&gt;&#xA;" LastActivityDate="2010-09-30T20:04:32.573" CommentCount="9" />
  <row Id="5782" PostTypeId="2" ParentId="5775" CreationDate="2010-09-30T19:10:24.803" Score="14" ViewCount="" Body="&lt;p&gt;Define the indicator random variable $I_i$ for $1 \leq i \leq m$ as $1$ if alphabet $i$ is present in the set ${a_1,\dots,a_n}$. Then the size of the set is simply $\sum_{i=1}^m I_i$. The expectation of this can be easily computed by linearity of expectation. The probability that $I_i$ equals $1$ is given by $1-\left( \frac{m-1}{m} \right)^n$ and therefore the expected size of the set is $m \left[ 1- \left( 1 - \frac{1}{m} \right)^n \right]$. For $m=n$, the limiting value is indeed as you mentioned in the question.&lt;/p&gt;&#xA;" OwnerUserId="618" LastActivityDate="2010-09-30T19:10:24.803" />
  <row Id="5783" PostTypeId="1" CreationDate="2010-09-30T19:37:11.613" Score="1" ViewCount="108" Body="&lt;p&gt;please help.. In my notes, it is the V itself. I just want to confirm this. Can you also explain and give an example if that is so...&lt;/p&gt;&#xA;" OwnerUserId="2231" LastActivityDate="2010-09-30T19:40:39.770" Title="What is the center of V, the Klein 4 group?" Tags="&lt;abstract-algebra&gt;&lt;group-theory&gt;" AnswerCount="1" />
  <row Id="5784" PostTypeId="2" ParentId="5783" CreationDate="2010-09-30T19:40:39.770" Score="4" ViewCount="" Body="&lt;p&gt;Every abelian group is its own center.  If you look at the definitions you will see this.&lt;/p&gt;&#xA;" OwnerUserId="822" LastActivityDate="2010-09-30T19:40:39.770" CommentCount="1" />
  <row Id="5785" PostTypeId="2" ParentId="5777" CreationDate="2010-09-30T19:47:23.353" Score="8" ViewCount="" Body="&lt;p&gt;You could use the fact that $\sin x=\sin y$ if and only if either&#xA;$x-y$ is an even integer times $\pi$ or $x+y$ is an odd integer times $\pi$.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-09-30T19:47:23.353" />
  <row Id="5787" PostTypeId="2" ParentId="3966" CreationDate="2010-09-30T20:56:21.630" Score="0" ViewCount="" Body="&lt;p&gt;The first thing that comes to mind is Quantum Electrodynamics. I don't really know much more than the name, but the first chapters of Richard Feynmans's Six Easy Pieces (I think that was the one) are pretty accessible. They raise the problems of trying to keep track of single photons, and how they behave completely differently from regular solid objects. With a bunch of photons you can only keep track of the distribution of photons (more or less). Look for Feynmann vidoes on the web. He was entertaining and you won't regret the time.&lt;/p&gt;&#xA;" OwnerUserId="1312" LastActivityDate="2010-09-30T20:56:21.630" />
  <row Id="5788" PostTypeId="1" AcceptedAnswerId="5790" CreationDate="2010-09-30T21:03:57.143" Score="2" ViewCount="120" Body="&lt;p&gt;it has been years since I have done logs, I remember something like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x^{\log_z(y)} = y^{\log_z(x)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(where $z$ is the base)&#xA;Is that correct?  It doesn't seem so, since&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$3^{\log_2(4)} \neq 4^{\log_2(3)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Am I right in that assumption?&#xA;If I am, does&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x^{\log_z(y)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;convert to anything easily?&lt;/p&gt;&#xA;" OwnerUserId="2232" LastActivityDate="2010-10-01T01:31:55.180" Title="simple/dumb logarithmic conversion question" Tags="&lt;algebra-precalculus&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="5789" PostTypeId="2" ParentId="5778" CreationDate="2010-09-30T22:01:50.540" Score="1" ViewCount="" Body="&lt;p&gt;Isaac's comment asking how the angles are given is a good one.  If you specify that and the coordinate system the problem will be well posed.  Guessing that it is in azimuth $(\phi )$ and elevation $(\theta )$, with $\phi =\theta =0$ along the positive x axis and positive z up,&#xA;$F_x=F\cos(\theta )*\cos(\phi ), F_y=F\cos(\theta )*\sin(\phi ), F_z=F\sin(\theta )$&lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2010-11-19T01:50:41.683" LastActivityDate="2010-11-19T01:50:41.683" />
  <row Id="5790" PostTypeId="2" ParentId="5788" CreationDate="2010-09-30T22:27:08.903" Score="3" ViewCount="" Body="&lt;p&gt;We do have &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x^{\log _{z}y}=y^{\log _{z}x}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;because &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(\log_{z}y)\log_{z}x=(\log_{z}x)\log _{z}y$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The numerical relation is an equality&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$3^{\log _{2}4}=4^{\log _{2}3}=9$.&lt;/p&gt;&#xA;" OwnerUserId="752" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2010-09-30T22:35:03.687" LastActivityDate="2010-09-30T22:35:03.687" />
  <row Id="5791" PostTypeId="1" CreationDate="2010-09-30T22:27:30.120" Score="0" ViewCount="258" Body="&lt;p&gt;Here is the dot&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/aQBA2.jpg&quot; alt=&quot;image&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And here is how I arrived at it »&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/4LBJi.jpg&quot; alt=&quot;image&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pretty simple IFS using a 2 x 2 grid as the base for the iterations. Is there a way to describe this point as well as its siblings, all of which lie on the dotted red line(pic 2), using an equation?&lt;/p&gt;&#xA;" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-09-30T23:24:03.020" LastActivityDate="2011-05-30T14:41:27.140" Title="How do I plot this red dot?" Tags="&lt;sequences-and-series&gt;&lt;fractals&gt;" AnswerCount="2" CommentCount="6" />
  <row Id="5792" PostTypeId="2" ParentId="5775" CreationDate="2010-09-30T22:30:51.203" Score="0" ViewCount="" Body="&lt;p&gt;This is dealt with in depth at &lt;a href=&quot;http://www.math.uah.edu/stat/urn/Birthday.xhtml&quot; rel=&quot;nofollow&quot;&gt;http://www.math.uah.edu/stat/urn/Birthday.xhtml&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1789" LastEditorUserId="1789" LastEditorDisplayName="" LastEditDate="2011-01-20T21:09:33.727" LastActivityDate="2011-01-20T21:09:33.727" />
  <row Id="5793" PostTypeId="1" CreationDate="2010-09-30T23:44:46.070" Score="-2" ViewCount="136" Body="&lt;p&gt;In matlab, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;i have 12 variables, i want to calculate the each columns mean and subtract from each data dimension (every variable in column) and get a new matrix.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for your help&lt;/p&gt;&#xA;" OwnerUserId="2235" LastActivityDate="2010-10-01T10:47:35.380" Title="How can i subtract the mean from each of the data dimension in matlab?" Tags="&lt;linear-algebra&gt;&lt;matrices&gt;&lt;matlab&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="5794" PostTypeId="2" ParentId="5788" CreationDate="2010-10-01T00:42:53.970" Score="2" ViewCount="" Body="&lt;p&gt;If $\rm\:\ \ell\:X\ =\ \log_Z{X}\ \ $ then $\rm\ \ell(X^{\:\ell\:{Y}})\ =\ \ell X\ \ell Y\ =\ \ell Y\ \ell X\ =\ \ell(Y^{\:\ell\: X})$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or, without taking logarithms $\rm\displaystyle\ \ \: X^{\:\ell\:{Y}}\ \ =\ Z^{\:\ell\: X\ \ell\: Y}\ =\ Z^{\:\ell\: Y\ \ell\: X}\ =\ Y^{\:\ell\: X}$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your final example is a special case $\rm\ \ Y\ =\ Z^n\:$, $\rm \ $ but it also has a simple direct proof,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;namely $\rm\ \ \: X^{\:\ell\: Z^{\:n}}\ =\ X^n\ =\ Z^{\:\ell\: X^n}\ =\ Z^{n\:\ell\: X}\:$.$\:$ Yours is $\rm\: X = 3,\ Z = 2 = n$&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-10-01T01:31:55.180" LastActivityDate="2010-10-01T01:31:55.180" />
  <row Id="5795" PostTypeId="1" AcceptedAnswerId="5798" CreationDate="2010-10-01T01:18:33.547" Score="5" ViewCount="132" Body="&lt;p&gt;Suppose I have a weakly sequentially continuous linear operator T between two normed linear spaces X and Y (i.e. $x_n \stackrel {w}{\rightharpoonup} x$ in $X$ $\Rightarrow$ $T(x_n) \stackrel {w}{\rightharpoonup} T(x)$ in $Y$). Does this imply that my operator T must be bounded?   &lt;/p&gt;&#xA;" OwnerUserId="1736" LastEditorUserId="1736" LastEditorDisplayName="" LastEditDate="2010-10-01T02:59:39.523" LastActivityDate="2010-10-01T03:26:16.743" Title="weak sequential continuity of linear operators" Tags="&lt;functional-analysis&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" />
  <row Id="5796" PostTypeId="2" ParentId="4498" CreationDate="2010-10-01T02:03:53.727" Score="0" ViewCount="" Body="&lt;p&gt;This is not a classification, but it is interesting to notice that every finite complete group can be obtained by the following process: start with any finite centerless group $G$ and iterate the automorphism group operations to build the automorphism tower:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$G\to\mathop{Aut}(G)\to \mathop{Aut}(\mathop{Aut}(G))\to\cdots$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Each group maps into the next by the inner automorphisms. If $G$ is centerless, then so is $\mathop{Aut}(G)$ and all the groups in the tower are centerless. The tower terminates at a complete group, a centerless group that is isomorphic to $\mathop{Aut}(G)$ by the canonical map. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is a beautiful theorem of Wielandt (1939) that every finite centerless group has an automorphism tower that terminates in finitely many steps. Thus, every finite complete group arises in this fashion. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This question is also considered at &lt;a href=&quot;http://mathoverflow.net/questions/5635/does-autaut-autg-stabilize&quot; rel=&quot;nofollow&quot;&gt;this classic MO question&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Truly interesting things occur if one is willing to push things harder by continuing the iteration transfinitely. That is, having built the finite part of the automorphism tower, which has an associated system of mappings, one may simply take the direct limit to produce the limit group $G_\omega$ at stage $\omega$ and continue the iteration transfinitely with &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$G_0\to G_1\to G_2\to\cdots G_\omega\to G_{\omega+1}\to\cdots\to G_\alpha\to\cdots$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;where one takes the automorphism group at successor ordinal stages and direct limits of the system at limit ordinal stages. Simon Thomas proved the wonderful theorem that every centerless group leads in this transfinite manner to a complete group, where the process stops (see &lt;a href=&quot;http://www.jstor.org/pss/2044505&quot; rel=&quot;nofollow&quot;&gt;Proceedings AMS article&lt;/a&gt;). I extended this result by proving:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Theorem.&lt;/b&gt; Every group $G$ has a terminating transfinite automorphism tower. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The proof proceeds by showing that every $G$ leads eventually to a centerless group in the transfinite tower, whose subsequent tower then terminates by Thomas' theorem.  You can see &lt;a href=&quot;http://arxiv.org/abs/math/9808014&quot; rel=&quot;nofollow&quot;&gt;my paper at the ArXiv&lt;/a&gt;, published in the &lt;a href=&quot;http://www.jstor.org/stable/119135&quot; rel=&quot;nofollow&quot;&gt;Proceedings of the AMS&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is an open question how tall the automorphism tower of a finite group can be. Examples are known with nontrivial centers that do not terminate at any finite stage and make it out to $\omega+n$ for any desired $n$, but the smallest upper bound is on the order of the least inaccessible cardinal. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Simon Thomas is writing a book on the subject of the automorphism tower problem, and the preliminary versions that I have seen are excellent.&lt;/p&gt;&#xA;" OwnerUserId="413" LastEditorUserId="413" LastEditorDisplayName="" LastEditDate="2010-10-01T12:43:32.170" LastActivityDate="2010-10-01T12:43:32.170" />
  <row Id="5798" PostTypeId="2" ParentId="5795" CreationDate="2010-10-01T02:21:31.977" Score="2" ViewCount="" Body="&lt;p&gt;In my original answer I only mentioned that it works for $Y$ complete, but as Nate pointed out in a comment, I never actually used completeness of $Y$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer is yes. Weakly convergent sequences in a normed space are bounded, as a consequence of the uniform boundedness principle applied to the dual space (which is a Banach space) and the fact that a convergent sequence of real (or complex) numbers is bounded. If $T$ is unbounded, then there is a sequence $x_1,x_2,\ldots$ in $X$ converging in norm (and hence weakly) to 0 such that $\|T(x_n)\|\to\infty$, so by the previous sentence this implies that $T(x_1),T(x_2),\ldots$ does not converge weakly. &lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2010-10-01T03:26:16.743" LastActivityDate="2010-10-01T03:26:16.743" CommentCount="5" />
  <row Id="5799" PostTypeId="1" AcceptedAnswerId="5811" CreationDate="2010-10-01T02:56:04.147" Score="2" ViewCount="160" Body="&lt;p&gt;Consider a canonical eliptical cone C with its vertex at the origin, with height h, and with a base given by:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;(x/a)&lt;sup&gt;2&lt;/sup&gt;+(y/b)&lt;sup&gt;2&lt;/sup&gt; = 1; z = h&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;a, b not equal to 0.  Given a point p = (p&lt;sub&gt;x&lt;/sub&gt;, p&lt;sub&gt;y&lt;/sub&gt;, p&lt;sub&gt;z&lt;/sub&gt;) how would you determine if p is interior to C?  I realize I can first check to see if p is outside of the elliptical right cylinder where one base the base of the elliptical cone and the other is on the XY plane, so for grins, assume that check has already been made, and p is in fact inside that cylinder.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most obvious solution I see is to linearly scale a and b by p&lt;sub&gt;z&lt;/sub&gt;/h and see if p is inside the ellipse:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;(h/p&lt;sub&gt;z&lt;/sub&gt;)&lt;sup&gt;2&lt;/sup&gt;[(x/a)&lt;sup&gt;2&lt;/sup&gt;+(y/b)&lt;sup&gt;2&lt;/sup&gt;]=1; z = p&lt;sub&gt;z&lt;/sub&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;p&gt;Are there other approaches?  Thanks.&lt;/p&gt;&#xA;" OwnerUserId="1287" LastActivityDate="2010-10-01T04:39:12.180" Title="How do I determine if a point is interior to an elliptical cone?" Tags="&lt;geometry&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="5801" PostTypeId="1" AcceptedAnswerId="5813" CreationDate="2010-10-01T03:22:36.873" Score="3" ViewCount="292" Body="&lt;p&gt;This is an old exam problem I found online. For any positive integers $a,b$, one has $a^4|b^3$ implies $a|b$. Clearly if $a^4|b^3$, then $a|b^3$. It seemed simple on first reading, but I can't figure out how to show that $a|b$ follows, since $a$ is not necessarily prime. Is there some observation I'm missing?&lt;/p&gt;&#xA;" OwnerUserId="1583" LastActivityDate="2010-10-08T16:27:11.213" Title="For any positive integers $a,b$, one has $a^4|b^3$ implies $a|b$?" Tags="&lt;number-theory&gt;" AnswerCount="4" FavoriteCount="1" />
  <row Id="5802" PostTypeId="2" ParentId="5801" CreationDate="2010-10-01T03:37:36.380" Score="1" ViewCount="" Body="&lt;p&gt;Suppose $p$ is prime and $p^k$ divides $a$ (and no higher power of $p$ divides $a$).  Then $p^{4k}$ divides $a^4$ and hence $p^{4k}$ divides $b^3$.  It follows that $p^{4k/3}$ divides $b$.  In particular, $p^k$ must divide $b$ since $4k/3 \geq k$.  Doing this for all $p$ shows that $a|b$.&lt;/p&gt;&#xA;" OwnerUserId="331" LastActivityDate="2010-10-01T03:37:36.380" CommentCount="3" />
  <row Id="5804" PostTypeId="1" AcceptedAnswerId="13355" CreationDate="2010-10-01T03:48:41.257" Score="7" ViewCount="200" Body="&lt;p&gt;I think this is asked as a standard exercise in books about wavelets (e.g. exercise 7.2 in &lt;a href=&quot;http://books.google.com/books?id=hbVOfWQNtB8C&amp;amp;pg=370&quot; rel=&quot;nofollow&quot;&gt;Mallat's book&lt;/a&gt;), but I couldn't find a proof.  Let $\phi$ be a scaling function (see definition below).  I would like to learn why &#xA;$$\sum_{k\in\mathbb Z} \phi(x-k) = 1 $$&lt;br&gt;&#xA;almost everywhere.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Definition.&lt;/strong&gt; A sequence of subspaces $\{V_j: j\in \mathbb{Z}\}$ of $L^2(\mathbb R)$ is&#xA;called a &lt;em&gt;multiresolution analysis&lt;/em&gt;  if it satisfies the following:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$V_j \subset V_{j+1}$&#xD;&lt;/li&gt;&#xA;&lt;li&gt;$\bigcap_{j}V_j = \{0\}$&#xD;&lt;/li&gt;&#xA;&lt;li&gt;$\overline{\bigcup_jV_j} = L^2(\mathbb R)$&#xD;&lt;/li&gt;&#xA;&lt;li&gt;$f(x)\in V_j$ if and only if $f(2x) \in V_{j+1}$&#xD;&lt;/li&gt;&#xA;&lt;li&gt;There exists a function $\phi \in V_0$ such that&#xA;$\{\phi(x-k)\}_{k\in\mathbb Z}$ is an orthogonal basis for $V_0$&#xD;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The function $\phi$ here is called as a &lt;em&gt;scaling function&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="145" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-09T10:27:48.813" LastActivityDate="2010-12-09T10:27:48.813" Title="Integer translates of a scaling function" Tags="&lt;real-analysis&gt;&lt;functional-analysis&gt;" AnswerCount="1" FavoriteCount="2" />
  <row Id="5805" PostTypeId="1" AcceptedAnswerId="5808" CreationDate="2010-10-01T04:07:22.047" Score="2" ViewCount="224" Body="&lt;p&gt;Suppose I have a 8x6 rectangle, with its lower left corner at the origin (0, 0). I want to scale this rectangle by 0.5 at an anchor point (3, 3). So the resulting rectangle is 4x3, but I cannot figure out how to compute the distance from the origin to the lower left corner of the new rectangle. Help is appreciated.&lt;/p&gt;&#xA;" OwnerUserId="2239" LastActivityDate="2010-10-01T04:36:09.453" Title="Scaling at an arbitrary point and figuring out the distance from origin" Tags="&lt;geometry&gt;&lt;matrices&gt;&lt;transformation&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="5806" PostTypeId="2" ParentId="5801" CreationDate="2010-10-01T04:08:41.013" Score="9" ViewCount="" Body="&lt;p&gt;It's simple: $\rm\ a^3|b^3\ \Rightarrow\ (b/a)^3\in\mathbb Z\ \Rightarrow\ b/a\in \mathbb Z\ \ $  by the &lt;a href=&quot;http://en.wikipedia.org/wiki/Rational_root_theorem&quot; rel=&quot;nofollow&quot;&gt;Rational Root Test&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To elaborate, the &lt;strong&gt;Rational Root Test&lt;/strong&gt; implies, as a special case, that if $\rm\ r\ $ is a rational root of an integer coefficient polynomial that is &lt;strong&gt;monic&lt;/strong&gt; (i.e. has leading coefficient equal to 1), then $\rm\:r\:$ is necessarily an integer. Hence, in the case above, since $\rm\ r = b/a\ $ is a root of the monic polynomial $\rm\ x^3 - n\ $ for some $\rm n\in \mathbb Z\:$, we deduce that $\rm\ r = b/a\ $ is an integer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is merely a special case of results equivalent to the irrationality of $\rm\:n\:$'th roots. Namely, any domain $\rm Z$ satisfying the following equivalent conditions is called &lt;strong&gt;root-closed&lt;/strong&gt;. This is a weaker condition than being integrally-closed, i.e. satisfying the monic Rational Root Test.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;THEOREM&lt;/strong&gt; $\:$ TFAE for $\rm\: a,b\: $ in domain $\rm\:Z\:,\ \: r \in Q \:=\:$ fraction field of $\rm\: Z\:,\ n\in \mathbb N$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) $\rm\ \ r = \sqrt[n]a  \ \Rightarrow\ r \in Z$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) $\rm\ \ r^n \in \:Z \:\ \Rightarrow\  r \in Z$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(3) $\rm\ \ \ a^n\:|\:b^n \:\ \Rightarrow\:\  a\:|\:b$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(4) $\rm\ \ (a^n) =  (b^n,\: c^n) \ \Rightarrow\  (a) = (b,c)\ $  as ideals in $\rm\:Z$&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-10-01T15:30:56.830" LastActivityDate="2010-10-01T15:30:56.830" CommentCount="1" />
  <row Id="5808" PostTypeId="2" ParentId="5805" CreationDate="2010-10-01T04:20:49.670" Score="3" ViewCount="" Body="&lt;p&gt;Scaling by 0.5 about the center (3,3) is the same as translating by (-3,-3), scaling by 0.5 about (0,0), then translating by (3,3).  So, to find the coordinates of the lower left corner of the new rectangle, take the coordinates of the lower left corner of the original rectangle, subtract 3 from each coordinate, multiply each coordinate by 0.5, and add 3 to each coordinate.  Once you have the point's coordinates, you can find its distance from the origin.&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-10-01T04:20:49.670" CommentCount="8" />
  <row Id="5809" PostTypeId="2" ParentId="5801" CreationDate="2010-10-01T04:21:37.420" Score="0" ViewCount="" Body="&lt;p&gt;As you said, $a\vert b$. Hence every prime factor of $a$ is a prime factor of $b$. So we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;a = p_1^{m_1} \dots p_r^{m_r} \qquad \text{and} \qquad b = p_1^{n_1} \dots p_r^{n_r} q_1&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $q_1$ is a positive integer number such that none of the $p_i$ divides it. So in order to see that $a\vert b$ it's enough to prove that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;n_i \geq m_i \qquad \text{for all} \ i = 1, \dots , r \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume $n_i &amp;lt; m_i$ for some $i$. Now,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;a^4 \vert b^3 \qquad \Longleftrightarrow \qquad b^3 = a^4q_2&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for some positive integer $q_2$. Thus&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;p_1^{3n_1} \dots p_r^{3n_r} q_1^3 = p_1^{4m_1} \dots p_r^{4m_r}q_2 \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, $q_2$ may have, or may have not, some of the $p_i$ as prime factors (but $q_1$ hasn't). In any case, we must have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;3n_i = 4m_i + t_i \qquad \text{for all} \ i= 1, \dots , r \ ,&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $t_i \geq 0$ for all $i$. But we have assumed that there is some $i$ such that $n_i &amp;lt; m_i$. So, for this $i$, we would have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;4m_i + t_i = 3n_i &amp;lt; 3m_i \qquad \Longleftrightarrow \qquad m_i + t_i &amp;lt; 0 \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But this is impossible, since both $m_i, t_i \geq 0$. A contradiction. Hence $n_i \geq m_i$ for all $i$ and so $a\vert b$.&lt;/p&gt;&#xA;" OwnerUserId="664" LastActivityDate="2010-10-01T04:21:37.420" />
  <row Id="5810" PostTypeId="2" ParentId="5805" CreationDate="2010-10-01T04:36:09.453" Score="1" ViewCount="" Body="&lt;p&gt;If you want to scale by .5 with anchor (a,b), then (x,y) will be sent to ((a+x)/2,((b+y)/2), the midpoint of the segment connecting the two points.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-10-01T04:36:09.453" />
  <row Id="5811" PostTypeId="2" ParentId="5799" CreationDate="2010-10-01T04:39:12.180" Score="5" ViewCount="" Body="&lt;p&gt;Write down the equation for the cone:  it is &#xA;$$(x/a)^2 + (y/b)^2 - (z/h)^2 = 0.$$&#xA;Now the interior of the cone will be defined by the inequalities&#xA;$$(x/a)^2 + (y/b)^2 - (z/h)^2 &amp;lt; 0 \text{ and } 0 &amp;lt; z &amp;lt; h.$$&#xA;So given a point $(x,y,z)$, assuming that $0 &amp;lt; z &amp;lt; h$ (so that at least the $z$-coordinate&#xA;is compatible with lying in the interior of the cone), you evaluate $(x/a)^2 + (y/b)^2 - (z/h)^2$ and consider whether or not it is negative.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is of course just a rephrasing of the description you give in your post,&#xA;but it is the most practical test that I can think of, and should be easy to program.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;NB: I am writing $(x,y,z)$ rather than $(p_x,p_y,p_z)$ just to make the notation simpler.&lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2010-10-01T04:39:12.180" />
  <row Id="5812" PostTypeId="1" AcceptedAnswerId="5814" CreationDate="2010-10-01T05:25:00.387" Score="2" ViewCount="154" Body="&lt;p&gt;Hello,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope I'm at right place, as I asked this question at stackoverflow.com where the question was closed and suggested I go ask here. I'll try to be more comprehensive also, looking at couple of replies I got there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is simple for those who know how algorithms work, and as I'm physics student I'm not sure about the answer. I had exam in Statistics where professor suggested that formula I used (red outlined) in determining &quot;Least-Squares Regression Line&quot; was computationally inferior to the one derived below (green outline). My simple thinking was that first one uses n squares while derived (n+1) squares, but supposedly that's not enough. Can someone point why could derived formula be computationally more effective?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/cd9ex.gif&quot; alt=&quot;variance&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2011-05-10T10:24:15.847" LastActivityDate="2011-05-10T10:24:15.847" Title="Simple algorithm effectiveness (Least-Squares Regression Line in Terms of Sample Variances)" Tags="&lt;statistics&gt;&lt;algorithms&gt;" AnswerCount="1" />
  <row Id="5813" PostTypeId="2" ParentId="5801" CreationDate="2010-10-01T05:57:43.487" Score="6" ViewCount="" Body="&lt;p&gt;This may be a situation where it is enlightening to ask (and answer) a more general question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Problem: let $m$ and $n$ be positive integers.  Show that the following are equivalent:&lt;br&gt;&#xA;(i) For all positive integers $a,b,$ $a^m \ | \ b^n$ implies  $a \ | \ b$.&lt;br&gt;&#xA;(ii) $m \geq n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All of the other answers given should generalize to this situation.  As people who are familiar with my number theory notes probably know, I like arguments using the functions $\operatorname{ord}_p(n)$, defined to be the largest number $a$ such that $p^a \ | \ n$.  The &quot;local to global principle&quot; for divisibility in the integers is: for $x,y \in \mathbb{Z}^+$, $x \ | \ y$ iff for all primes $p$, $\operatorname{ord}_p(x) \leq \operatorname{ord}_p(y)$.  [More generally, this is true in any unique factorization domain.  &lt;b&gt;Added&lt;/b&gt;: in fact, with appropriate modifications it is true in any &lt;a href=&quot;http://en.wikipedia.org/wiki/Krull_ring&quot; rel=&quot;nofollow&quot;&gt;Krull domain&lt;/a&gt;.  This actually came up in my own research recently...]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the present situation, $a^m \ | \ b^n$ implies that for all prime numbers $p$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$m \cdot \operatorname{ord}_p(a) = \operatorname{ord}_p(a^m) \leq \operatorname{ord}_p(b^n) = n \cdot \operatorname{ord}_p(b)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if $m \geq n$, we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$n \cdot  \operatorname{ord}_p(a) \leq m \cdot \operatorname{ord}_p(a) \leq n \cdot \operatorname{ord}_p(b)$, so&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\operatorname{ord}_p(a) \leq \operatorname{ord}_p(b)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This shows that (ii) implies (i).  I'll leave the other direction to the reader as a simple exercise in understanding what's going on here, with the hint that one should prove the contrapositive: assume $m &amp;lt; n$ and take $a$ and $b$ to be powers of the same prime number. &lt;/p&gt;&#xA;" OwnerUserId="299" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2010-10-08T16:27:11.213" LastActivityDate="2010-10-08T16:27:11.213" CommentCount="1" />
  <row Id="5814" PostTypeId="2" ParentId="5812" CreationDate="2010-10-01T06:09:13.433" Score="3" ViewCount="" Body="&lt;p&gt;You're comparing the two variance formulae&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$s_x^2=\frac1{n}\sum_{i=1}^n \left(x_i-\bar{x}\right)^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$s_x^2=\frac1{n}\left(\sum_{i=1}^n x_i^2\right)-\bar{x}^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With respect to the computational effort needed, the second formula looks better: it only requires one loop to pass through your data, as opposed to the two needed for the first one (first for computing the mean, and then for adding up the squared deviations).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The truth of the matter, however, is that the first formula makes much more computational sense than the second one. In practice, the two terms being subtracted in the second form can be very large; if the variance is small, you lose a lot of significant figures in the subtraction of two nearly-equal quantities (&lt;em&gt;subtractive cancellation&lt;/em&gt;). Worst comes to worst, you may even end up with a negative variance with the second formula!&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-10-01T06:09:13.433" CommentCount="5" />
  <row Id="5815" PostTypeId="2" ParentId="370" CreationDate="2010-10-01T07:15:09.930" Score="1" ViewCount="" Body="&lt;p&gt;&lt;em&gt;MATH 4135/5135: Introduction to Category Theory&lt;/em&gt; by Peter Selinger&#xA;(17pp). Concise course outline. Only wish it covered more topics. Available in PS or PDF format.&#xA;&lt;br&gt; &lt;a href=&quot;http://www.mscs.dal.ca/~selinger/4135/&quot; rel=&quot;nofollow&quot;&gt;http://www.mscs.dal.ca/~selinger/4135/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Handbook of Categorical Algebra&lt;/em&gt; (Encyclopedia of Mathematics and its Applications) by Francis Borceux. Rigorous. Comprehensive. This is NOT free, but you can see the contents/index/excerpts at the publisher's web site, listed below. This is a three volume set: &lt;br&gt;&#xA;(v. 1) Basic Category Theory, 364pp. (ISBN-13: 9780521441780)&#xA;&lt;br&gt;http://www.cambridge.org/catalogue/catalogue.asp?isbn=9780521441780&#xA;&lt;br&gt;&#xA;(v. 2) Categories and Structures, 464pp. (ISBN-13: 9780521441797)&#xA;&lt;br&gt;http://www.cambridge.org/catalogue/catalogue.asp?isbn=9780521441797&#xA;&lt;br&gt;&#xA;(v. 3) Sheaf Theory, 544pp. (ISBN-13: 9780521441803)&#xA;&lt;br&gt;http://www.cambridge.org/catalogue/catalogue.asp?isbn=9780521441803&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Reprints in Theory and Applications of Categories (TAC)&lt;/em&gt;. This site has 18 books and articles on category theory in PDF, including several by F.W. Lawvere.&#xA;&lt;br&gt;http://www.tac.mta.ca/tac/reprints/index.html&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Abstract and Concrete Categories-The Joy of Cats&lt;/em&gt; by Jirı Adamek, Horst Herrlich, and George E. Strecker (524pp). Free PDF. Published under the GNU Free Documentation License. Mentioned already by Seamus in reference to Wikipedia's external links for Category Theory, but worth repeating, because it's very readable.&#xA;&lt;br&gt;http://katmat.math.uni-bremen.de/acc&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;A Gentle Introduction to Category Theory (the calculational approach)&lt;/em&gt; by Maarten M. Fokkinga (80pp).&#xA;&lt;br&gt;http://wwwhome.cs.utwente.nl/~fokkinga/mmf92b.html&lt;/p&gt;&#xA;" LastActivityDate="2010-10-01T07:15:09.930" CommunityOwnedDate="2010-10-01T07:15:09.930" />
  <row Id="5816" PostTypeId="2" ParentId="5793" CreationDate="2010-10-01T10:47:35.380" Score="0" ViewCount="" Body="&lt;p&gt;Something like this should do the trick:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;myNewArray  =  myOldArray - mean(myOldArray,1);&lt;/p&gt;&#xA;" OwnerUserId="2244" LastActivityDate="2010-10-01T10:47:35.380" CommentCount="1" />
  <row Id="5817" PostTypeId="1" AcceptedAnswerId="5873" CreationDate="2010-10-01T10:50:34.227" Score="3" ViewCount="198" Body="&lt;p&gt;Given a differential equation of the form &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\epsilon \frac{d^ny}{dx^n} + \sum_{k=0}^{n-1} a_k(x)\frac{d^ky}{dx^k}=0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then the WKB method says to choose the ansatz $y\sim exp({\frac{i\phi(x)}{\epsilon}})A(x,\epsilon)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $A(x,\epsilon) = \sum_{k=0}^{\infty} A_k(x)\epsilon^n$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wondered what is the motivation for this educated guess. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance. &lt;/p&gt;&#xA;" OwnerUserId="1794" LastActivityDate="2010-10-02T09:47:07.130" Title="The WKB method: motivation" Tags="&lt;differential-equations&gt;&lt;perturbation-theory&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="5818" PostTypeId="1" AcceptedAnswerId="5829" CreationDate="2010-10-01T11:37:35.233" Score="2" ViewCount="519" Body="&lt;p&gt;This is an exercise from Apostol, which i have been struggling for a while.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given a prime $p$, how does one show that $${n \choose p} \equiv \biggl[\frac{n}{p}\biggr] \ (\text{mod} \ p)$$ Note that $\Bigl[\frac{n}{p}\Bigr]$ denotes the integral part of $\frac{n}{p}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would also like to know as to how does one try to solve this problem. Well, what we need is to show is whenever one divides ${n \choose p}$ by a prime $p$ the remainder is the integral part of $\frac{n}{p}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, $${ n \choose p} = \frac{n!}{p! \cdot (n-p)!}$$ Now $n!$ can be written as $$n!= n \cdot (n-1) \cdot (n-2) \cdots (n-p) \cdots 2 \cdot 1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But i am really struggling in getting the integral part.&lt;/p&gt;&#xA;" LastActivityDate="2011-03-31T07:00:41.160" Title="Proving ${n \choose p} \equiv \Bigl[\frac{n}{p}\Bigr] \ (\text{mod} \ p)$" Tags="&lt;number-theory&gt;" AnswerCount="7" FavoriteCount="2" />
  <row Id="5819" PostTypeId="1" AcceptedAnswerId="5821" CreationDate="2010-10-01T11:50:04.010" Score="8" ViewCount="268" Body="&lt;p&gt;Let's say we have some topological space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Axiom $T_1$ states that for any two points $y \neq x$, there is an open neighborhood $U_y$ of $y$ such that $x \notin U_y$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then we say that a topological space is $T_4$ if it is $T_1$ and also satisfies that for any two closed, non-intersecting sets $A,B$, there are open neighborhoods $U_A,U_B$ respectively, such that $U_A\cap U_B = \emptyset$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could anyone give an example of a topological space which satisfies the second condition of $T_4$, but which is not $T_1$?&lt;/p&gt;&#xA;" OwnerUserId="766" LastActivityDate="2010-12-01T19:35:26.107" Title="Why is $T_1$ required for a topological space to be $T_4$?" Tags="&lt;topology&gt;&lt;general-topology&gt;&lt;counter-examples&gt;&lt;examples&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="2" />
  <row Id="5821" PostTypeId="2" ParentId="5819" CreationDate="2010-10-01T12:03:06.167" Score="14" ViewCount="" Body="&lt;p&gt;Consider a topological space $X$ with two (different) points and the trivial topology (that is, only $\emptyset$ and $X$ are open sets).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This space satisfies the second condition of $T_4$: the only two closed non-intersecting sets are $\emptyset$ and $X$ and you can take as open neighborhoods $U_\emptyset = \emptyset$ and $U_X = X$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But $X$ is not $T_1$, as you can easily check.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$T_1$ (also called &lt;em&gt;Fréchet&lt;/em&gt; space) is equivalent to the fact that every point is a closed set (exercise: prove this!). So, if you had a &lt;em&gt;regular&lt;/em&gt; space ($T_3$) $X$ that was not $T_1$, you could not say that $X$ is $T_2$ (or &lt;em&gt;Hausdorff&lt;/em&gt;); that is, you couldn't say that $T_3 \Rightarrow T_2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, it is necessary to add the $T_1$ condition to $T_3$ and $T_4$ in order to have the beautiful sequence of implications that makes all of us happy :-)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;T_4 \ \text{(normal)} \ \Longrightarrow \&#xA;T_3 \ \text{(regular)} \ \Longrightarrow \&#xA;T_2 \ \text{(Hausdorff)} \ \Longrightarrow \&#xA;T_1 \ \text{(Fréchet)}&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="664" LastEditorUserId="664" LastEditorDisplayName="" LastEditDate="2010-10-01T17:46:03.360" LastActivityDate="2010-10-01T17:46:03.360" />
  <row Id="5822" PostTypeId="2" ParentId="5772" CreationDate="2010-10-01T12:08:14.550" Score="6" ViewCount="" Body="&lt;p&gt;I'll use the deduction theorem, so I'll assume $\beta$ and need to prove $\neg\neg\beta$.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;$\beta$ (assumption)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$\beta\to (\neg\neg\neg\beta\to\beta)$ (axiom 1)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$\neg\neg\neg\beta\to\beta$ (modus ponens using 1 and 2)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$\neg\neg\neg\beta\to\neg\beta$ (you have proved that $\neg\neg\beta\to\beta$ is a theorem)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$(\neg\neg\neg\beta\to\neg\beta)\to((\neg\neg\neg\beta\to\beta)\to\neg\neg\beta)$ (axiom 3)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$(\neg\neg\neg\beta\to\beta)\to\neg\neg\beta$ (modus ponens using 4 and 5)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$\neg\neg\beta$ (modus ponens using 3 and 6)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="1772" LastActivityDate="2010-10-01T12:08:14.550" CommentCount="2" />
  <row Id="5823" PostTypeId="2" ParentId="5818" CreationDate="2010-10-01T12:09:31.810" Score="1" ViewCount="" Body="&lt;p&gt;Consider ${n \choose p}$ as $(n)(n-1)...(p+2)(p+1)/(n-p)!$. That can be written as $(n)/(n-p) \cdot (n-1)/(n-p-1) \cdot \cdots \cdot (p+1)/(1)$. Notice that these are all of the form $(x+p)/x$. Thus, terms where $x$ is  not divisible by $p$ can be reduced modulo $p$ to $x/x=1$. The rest of the terms are going to be &#xA;$(p [n/p])/((p [n/p]-1))) \cdot \cdots \cdot (3p)/(2p) \cdot (2p)/(p)$. This product telescopes to $(p [n/p])/p = [n/p]$.&lt;/p&gt;&#xA;" OwnerUserId="796" LastActivityDate="2010-10-01T16:04:02.720" CommentCount="5" />
  <row Id="5824" PostTypeId="1" CreationDate="2010-10-01T12:30:39.897" Score="7" ViewCount="246" Body="&lt;p&gt;Let us say that a Rubik's cube in a particular configuration is in a particular &quot;state&quot;. All other configurations of this cube (other &quot;states&quot;), which can be achieved by rotations of the cube can be thought to be connected to each other... rotations are like walking the edges of the graph where the different states are the vertices.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if our Rubik's cube is of the type where we can peel off the stickers easily and put them back too, then we can swap 2 of the stickers in a corner.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand that the cube cannot be solved any more. Our cube has moved to a new state which is disconnected from the previous graph built above. Also, all the states reachable from this new state are also disconnected from the other graph. (otherwise the cube would be solvable). So now we have 2 graphs which are disconnected from each other. A third set of states might be disconnected from both the above states.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For all possible reasonable color assignments ( by reasonable, I mean to preserve the number of tiles for each color=9), how many such disconnected graphs exist for a 3x3x3 Rubik's cube?&lt;/p&gt;&#xA;" OwnerUserId="2246" LastEditorUserId="2246" LastEditorDisplayName="" LastEditDate="2010-10-02T02:43:29.877" LastActivityDate="2010-10-02T02:43:29.877" Title="How many disconnected graphs of the Rubik's cube exist?" Tags="&lt;combinatorics&gt;&lt;graph-theory&gt;&lt;recreational-mathematics&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="5825" PostTypeId="2" ParentId="5824" CreationDate="2010-10-01T12:55:23.830" Score="2" ViewCount="" Body="&lt;p&gt;You can calculate this using Burnside's Lemma, &lt;a href=&quot;http://en.wikipedia.org/wiki/Burnside%27s_lemma&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Burnside%27s_lemma&lt;/a&gt;, once you know the group of transformations for Rubik's cube.&lt;/p&gt;&#xA;" OwnerUserId="1470" LastActivityDate="2010-10-01T12:55:23.830" />
  <row Id="5826" PostTypeId="1" CreationDate="2010-10-01T13:54:16.557" Score="1" ViewCount="168" Body="&lt;p&gt;How to prove connect sum of two manifolds doesn't depend on the choices of balls(which would be cutted) and different gluing of boundary spheres? Is it still true in differential category? Thanks!&lt;/p&gt;&#xA;" LastActivityDate="2010-10-04T16:37:27.990" Title="a basic question about connect sum of two manifolds" Tags="&lt;geometry&gt;&lt;topology&gt;" AnswerCount="2" CommentCount="5" FavoriteCount="1" />
  <row Id="5827" PostTypeId="1" AcceptedAnswerId="5841" CreationDate="2010-10-01T15:37:55.637" Score="1" ViewCount="277" Body="&lt;p&gt;This is a problem which appeared in one of my tests, which i wasn't able to solve.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $\Omega$ be a uncountable set. Let $S$ be the collection of subsets of $\Omega$ given by: $A \in S$ if and only if $A$ is countable or $A^{c}$ is countable. Suppose $f: \Omega \to \mathbb{R}$ is a real measurable function. Prove that there exists a $y \in \mathbb{R}$ and a countable set $B$ such that the $f(x)=y$ is on $B^{c}$.&lt;/p&gt;&#xA;" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-10-01T18:06:28.700" LastActivityDate="2010-10-01T23:15:45.617" Title="Measurable function remaining constant" Tags="&lt;analysis&gt;&lt;measure-theory&gt;" AnswerCount="2" CommentCount="9" />
  <row Id="5828" PostTypeId="1" AcceptedAnswerId="5836" CreationDate="2010-10-01T15:51:24.840" Score="2" ViewCount="130" Body="&lt;p&gt;(Apologies for the vague way the question is stated.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We define the function like this: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) $f(x) = x$ for $0\leq x \leq a$ (for some $a \geq 15 \in \mathbb{R}$) $\land$ $f(x) = a$ if $ x &gt; a$ . For $x&amp;lt;0$, it doesn't really matter what $f(x)$ looks like. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Has someone studied this type of function? Does it have a 'nicer' definition than the one stated above? Any references on the subject? It is not absolutely necessary for the function to have &lt;em&gt;all&lt;/em&gt; of the above characteristics. If you know a function that is similar to it, please let me know.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Motivation:&#xA;I'm looking for a function that assigns a constant to 'infinite values' of x and does not 'deform' finite values (too much). So the function should satisfy $f(\zeta(1))=a$ for some finite value of $a \in \mathbb{R}$ and $f(3)=3$, for example. It doesn't really matter whether $f(b)=b$ or any multiple of b or that $f(x)$ is some finite polynomial, as long as it's 'easy' to find the value of $f(b)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Max Muller &lt;/p&gt;&#xA;" OwnerUserId="1055" LastActivityDate="2010-10-01T18:12:46.290" Title="Does the following function have a name and what properties does it have?" Tags="&lt;reference-request&gt;&lt;functions&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="5829" PostTypeId="2" ParentId="5818" CreationDate="2010-10-01T16:18:03.913" Score="5" ViewCount="" Body="&lt;p&gt;It's easy: $\:$ if $\rm\ n\equiv j\ \: (mod\ p)\:\: $ and $\rm\: \bigl[\frac{n}{p}\bigr] = k\;, \: $ then pairing factors top $\equiv$ bottom $\rm\:(mod\ p)\:$ we have  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\displaystyle\quad\quad {n \choose p}\ =\ \frac{n\:(n-1)\:\cdots\:(n-p+1)}{p\:(p-1)\:\cdots\: 1} $  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\displaystyle\quad\quad \phantom{{n \choose p}}\ =\ \frac{n}{j}\ \frac{n-1}{j-1}\cdots\frac{kp+1}{1}\ \frac{kp}p\ \frac{kp-1}{p-1}\:\cdots\frac{n-p+2}{j+2}\ \frac{n-p+1}{j+1}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $\rm\:n\equiv j\:,\: $ all of the above fractions are $\rm\: \equiv 1\:$ except the middle-shown one $\rm\: (kp)/p = k$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a very special case of much more general arithmetical results on binomial coefficients. For example, see Andrew Granville's very interesting survey &lt;a href=&quot;http://www.cecm.sfu.ca/organics/papers/granville/&quot; rel=&quot;nofollow&quot;&gt;The Arithmetic Properties of Binomial Coefficients&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;REMARK&lt;/strong&gt; The above proof is a special case of a very simple &lt;a href=&quot;http://math.stackexchange.com/questions/2192&quot;&gt;purely arithmetical proof&lt;/a&gt; that I devised to show that binomial coefficients are integral. Namely, the same idea of exploiting the innate symmetry by aligning the numerators and denominators $\rm\:(mod\ p)\:$ extends to yield a simple algorithm that, given a binomial coefficient and a prime $\rm p$, rewrites the binomial coefficient as a product of fractions whose denominators are all coprime to $\rm p$. This implies that no prime divides the lowest-terms denominator, so that we may therefore conclude that the binomial coefficient is integral. For an example and further discussion see my &lt;a href=&quot;http://math.stackexchange.com/questions/2192&quot;&gt;prior post&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-10-01T19:50:52.380" LastActivityDate="2010-10-01T19:50:52.380" />
  <row Id="5830" PostTypeId="2" ParentId="3966" CreationDate="2010-10-01T16:24:49.200" Score="1" ViewCount="" Body="&lt;p&gt;You are looking at the problem of mathematics without &lt;em&gt;knowing&lt;/em&gt; mathematics. Most of what you write applies to Chinese, provided you do not understand Chinese. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For one thing: A question like «does addition really have any meaning outside of the context of placing two objects in a box and predicting how many are inside?» is meaningless to anyone who understands arithmetic and, better, &quot;higher arithmetic&quot;: it is obvious to him/her that addition is meaningful!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More generally, your &quot;axiom&quot; regarding solipsism is quite against observation.&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2010-10-01T16:24:49.200" />
  <row Id="5831" PostTypeId="2" ParentId="5818" CreationDate="2010-10-01T16:56:34.653" Score="5" ViewCount="" Body="&lt;p&gt;A useful result in such problems is &lt;a href=&quot;http://planetmath.org/encyclopedia/LucassTheorem.html&quot; rel=&quot;nofollow&quot;&gt;Lucas's Theorem&lt;/a&gt; which states that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $p$ is a prime and if $a = \sum_{i=0}^{k} a_{i} p^{i},  \  0 \le a_i &amp;lt; p \ $ i.e. the $a_i$ are digits of $a$ in base $p$ and similarly $b = \sum_{i=0}^{k} b_{i}p^{i}$ (pad with zeroes if required) then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$${a \choose b} = \prod_{i=0}^{k} {a_i \choose b_i} \mod p$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In our case $n=a$ and $b=p$. Since $b = p$ we have $b_1 = 1$ and rest of the $b_i = 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$${a \choose p} = {a_1 \choose 1} = a_1 \mod p$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is easy to see that $a_1 = [\frac{a}{p}] \mod p$.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-10-01T16:56:34.653" />
  <row Id="5832" PostTypeId="2" ParentId="5595" CreationDate="2010-10-01T17:03:20.077" Score="4" ViewCount="" Body="&lt;p&gt;This analysis is correct, but not complete enough to convince me.  For example, why is the fate of the last person settled as soon as the first person's seat chosen?  Why will any other seat but the first person's or the last person's be taken by the time the last person boards?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I had to fill in the holes for myself this way...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The last person's fate is decided as soon as anybody chooses the first person's seat (nobody is now in a wrong seat, so everybody else gets their assigned seat, including the last person) or the last person's seat (the last person now won't get their correct seat).  Any other choice at any stage doesn't change the probabilities at all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rephrasing... at each stage, either the matter gets settled and there is a 50/50 chance it gets settled each way for the last person's seat, or the agony is just postponed.  The matter can thus be settled at any stage, and the probabilities at that stage are the only ones that matter -- and they are 50/50 no matter what stage. Thus, the overall probability is 50/50.&lt;/p&gt;&#xA;" OwnerUserId="271" LastActivityDate="2010-10-01T17:03:20.077" CommentCount="0" />
  <row Id="5833" PostTypeId="2" ParentId="5818" CreationDate="2010-10-01T17:28:48.993" Score="4" ViewCount="" Body="&lt;p&gt;Here's another way to look at it. Suppose $n=kp+j,$ for $0 \le j \le p-1.$ Then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ (p-1)! {n \choose p} = \frac{n(n-1) \cdots (n-p+1)}{p} = \left( {n-j \over p} \right)&#xA;\prod_{i=0,i\ne j }^{p-1} (n-i)$$&#xA;$$ \equiv k(p-1)! (\textrm{ mod } p). $$&#xA;Since the product runs through a complete set of non-zero residues mod p.&lt;/p&gt;&#xA;" OwnerUserId="1301" LastEditorUserId="1301" LastEditorDisplayName="" LastEditDate="2010-10-01T17:39:26.450" LastActivityDate="2010-10-01T17:39:26.450" CommentCount="10" />
  <row Id="5834" PostTypeId="1" CreationDate="2010-10-01T17:49:19.413" Score="2" ViewCount="95" Body="&lt;p&gt;I'm looking for good online resources for 2nd order system dynamics. Any recommendations? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm looking for stuff that ideally includes discussion of Q, damping ratio, overshoot, bode plots, for systems with transfer functions  of&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ H_1(s) = \frac{1}{\tau_2^2 s^2 + \tau_1 s + 1} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ H_2(s) = \frac{\tau_1 s + 1}{\tau_2^2 s^2 + \tau_1 s + 1} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ H_3(s) = \frac{s}{\tau_2^2 s^2 + \tau_1 s + 1} $$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;edit:&lt;/strong&gt; I've done change of variables to rewrite as &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ H_1(\sigma) = \frac{1}{\sigma^2 + 2\zeta\sigma + 1} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ H_2(\sigma) = \frac{2\zeta\sigma + 1}{\sigma^2 + 2\zeta\sigma + 1} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ H_3(\sigma) / \omega_0 = \frac{\sigma}{\sigma^2 + 2\zeta\sigma + 1} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\sigma = \tau_2 s = s / \omega_0$, $\omega_0 = 1 / \tau_2$, and $\zeta = \tau_1 / 2\tau_2$ to normalize out the time scaling factor and end up with a transfer function of the single variable $\zeta$ (damping factor) which I think is the conventional treatment.&lt;/p&gt;&#xA;" OwnerUserId="120" LastEditorUserId="120" LastEditorDisplayName="" LastEditDate="2010-10-01T18:18:57.637" LastActivityDate="2011-05-29T22:05:45.053" CommunityOwnedDate="2010-10-01T18:19:29.117" Title="good online resources for 2nd-order system dynamics" Tags="&lt;control-theory&gt;&lt;online-resources&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="5835" PostTypeId="2" ParentId="5834" CreationDate="2010-10-01T18:03:16.463" Score="0" ViewCount="" Body="&lt;p&gt;Searching in Google itself gives you a great deal of information. I got this link which explains has the content which you need.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.facstaff.bucknell.edu/mastascu/econtrolhtml/SysDyn/SysDyn2.html&quot; rel=&quot;nofollow&quot;&gt;http://www.facstaff.bucknell.edu/mastascu/econtrolhtml/SysDyn/SysDyn2.html&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Next, [redacted: inappropriate link to pirated content] is an excellent source for those who are in need of books online. &lt;/p&gt;&#xA;" LastEditorUserId="1543" LastEditorDisplayName="" LastEditDate="2011-03-30T20:55:56.277" LastActivityDate="2011-03-30T20:55:56.277" CommunityOwnedDate="2010-10-01T18:03:29.587" CommentCount="1" />
  <row Id="5836" PostTypeId="2" ParentId="5828" CreationDate="2010-10-01T18:12:46.290" Score="2" ViewCount="" Body="&lt;p&gt;f(x)=min(x,a)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can rewrite this in terms of the absolute value function as (x+a)/2-|x-a|/2, the average minus half the difference. If you prefer this can be seen as getting the graph of the absolute value function, and manipulating it to fit the function you want. This is possible since they are both formed of two straight lines meeting at a point.&lt;/p&gt;&#xA;" OwnerUserId="1149" LastActivityDate="2010-10-01T18:12:46.290" CommentCount="0" />
  <row Id="5837" PostTypeId="1" CreationDate="2010-10-01T18:14:29.650" Score="3" ViewCount="261" Body="&lt;p&gt;A student of mine has cooked up a new graphical notation for computing with knots on surfaces. The trouble is, writing up his results is difficult due to his new notation. Is there a good &quot;drawing tool&quot; for mathematics that anyone can suggest?&lt;/p&gt;&#xA;" OwnerUserId="941" LastActivityDate="2010-10-03T00:01:40.123" Title="A good, free, graphics package for mathematics?" Tags="&lt;math-software&gt;" AnswerCount="5" />
  <row Id="5838" PostTypeId="2" ParentId="5837" CreationDate="2010-10-01T18:24:42.500" Score="0" ViewCount="" Body="&lt;p&gt;LaTeX is a good way to write all mathematical formulas and graphical notations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is a lot more than that, but personnaly, I use it frequently only for that.&lt;/p&gt;&#xA;" OwnerUserId="2252" LastActivityDate="2010-10-01T18:24:42.500" />
  <row Id="5839" PostTypeId="1" AcceptedAnswerId="5850" CreationDate="2010-10-01T18:45:17.327" Score="2" ViewCount="550" Body="&lt;p&gt;Given that I know the point coordinates of A and B on segment AB and the expected length of a perpendicular segment CD crossing the middle of AB, how do I calculate the point coordinates of segment CD? &lt;/p&gt;&#xA;" OwnerUserId="2250" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2010-10-02T18:40:52.153" LastActivityDate="2011-05-14T12:12:07.673" Title="Finding point coordinates of a perpendicular bisector" Tags="&lt;geometry&gt;&lt;trigonometry&gt;&lt;analytic-geometry&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="2" />
  <row Id="5840" PostTypeId="1" AcceptedAnswerId="5845" CreationDate="2010-10-01T18:47:34.920" Score="2" ViewCount="154" Body="&lt;p&gt;I have read a few weeks ago that some mathematical researchers have discover that the minimum number of movements to solve any initial configuration of a Rubik's cube has been downsized to 20.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do they calculate such a number ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="2252" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-10-02T01:43:52.543" LastActivityDate="2010-10-02T05:52:22.320" Title="How is done the calculation of the minimum number of movement to solve any configuration of Rubik's Cube ?" Tags="&lt;combinatorics&gt;&lt;recreational-mathematics&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="5841" PostTypeId="2" ParentId="5827" CreationDate="2010-10-01T18:48:20.660" Score="3" ViewCount="" Body="&lt;p&gt;Note, that the countable intersection of co-countable sets (i.e., sets who complement is countable) is co-countable ( since its complement is a countable union of countable sets).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the union of the inverse image of the sets $[n,n+1]$ under as $n$ varies over all integers is all of $\Omega$. Since the inverse image of each $[n,n+1]$ is either countable or co-countable, so at least one of them must be co-countable (since $\Omega$ is uncountable).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Say $[n_1,n_1+1] = [a_1,b_1] $ is a set with co-countable (and hence uncountable) inverse image. Clearly, one of $[n_1, n_1 + 1/2]$ and $[n_1+1/2,n_1+1]$ must have a co-countable inverse image, call it $[a_2,b_2]$, ... proceeding in this manner we get a sequence of nested intervals $[a_n,b_n]$ each of whose inverse image is co-countable and $\lim_{n\to\infty} b_n - a_n = 0$, their intersection consists of a single point, say $y$, and  $f^{-1}({y}) = \cap f^{-1}( [a_n,b_n] )$ being an intersection of co-countable sets is co-countable. Call $B^{c} = f^{-1}({y})$, we are done.&lt;/p&gt;&#xA;" OwnerUserId="404" LastEditorUserId="404" LastEditorDisplayName="" LastEditDate="2010-10-01T23:15:45.617" LastActivityDate="2010-10-01T23:15:45.617" CommentCount="1" />
  <row Id="5842" PostTypeId="1" AcceptedAnswerId="5844" CreationDate="2010-10-01T18:49:18.760" Score="3" ViewCount="234" Body="&lt;p&gt;I'd like to know if the function $ y = f(x) : [0,1] \rightarrow [0,1]$ defined implicitly by the trascendental equation $$\displaystyle y = e^{(y-1)/x}$$ is &quot;well known&quot; (name, properties) or is related to some other well known function.&lt;/p&gt;&#xA;" OwnerUserId="312" LastEditorUserId="1543" LastEditorDisplayName="" LastEditDate="2011-04-18T12:06:58.280" LastActivityDate="2011-04-18T12:06:58.280" Title="Implicit function y = e^((y-1)/x)" Tags="&lt;special-functions&gt;&lt;functional-equations&gt;" AnswerCount="2" />
  <row Id="5843" PostTypeId="2" ParentId="5755" CreationDate="2010-10-01T18:54:48.807" Score="2" ViewCount="" Body="&lt;p&gt;I assume $y \gt 0$ and $y \gg \mu$.  If you replace $x^k$ by $y^k \exp {\left( (k/y)x - k \right) }$ you will overestimate the $x^k$ term (because this is the exponential of the first two terms of the MacLaurin series of $\log(x^k)$ expanded around $x=y$, the series is alternating, and the remainder term is negative).  Completing the square yields a closed-form formula for an upper bound, one of whose factors is a Gaussian integral:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sqrt{2\pi }y^k \exp \left( {\frac{k (k+2 y (-y+\mu ))}{2 y^2}} \right) \Phi \left(\frac{k}{y}-y+\mu \right)$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This will work extremely well when $k$ is large compared to $y$ and $\mu$ because then most of the mass of the integral is concentrated at its lower limit where the exponential upper bound to $x^k$ is a good approximation.  To avoid exponential overflow, use logarithms to compute the product.&lt;/p&gt;&#xA;" OwnerUserId="1489" LastEditorUserId="1489" LastEditorDisplayName="" LastEditDate="2010-10-01T20:09:45.113" LastActivityDate="2010-10-01T20:09:45.113" CommentCount="1" />
  <row Id="5844" PostTypeId="2" ParentId="5842" CreationDate="2010-10-01T19:02:00.540" Score="3" ViewCount="" Body="&lt;p&gt;It is solvable employing the &lt;a href=&quot;http://en.wikipedia.org/wiki/Lambert_W_function&quot; rel=&quot;nofollow&quot;&gt;Lambert W function&lt;/a&gt; as follows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your equation is $\rm\quad\quad\ Y \ \: =\ e^{\:Z-YZ}\ \: $ for $\rm\: \ Z\ =\: -\frac{1}X$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\quad\iff\ \ Y\:Z\ e^{Y\:Z}\ \:=\:\ Z\ e^Z$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\quad\iff\ \  Y\:Z\ \ \:=\:\ \ W(Z\ e^Z) $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\quad\iff\ \ Y\: =\: -X\ W(-\frac{1}X e^{-1/X}) $&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-10-02T02:46:31.823" LastActivityDate="2010-10-02T02:46:31.823" />
  <row Id="5845" PostTypeId="2" ParentId="5840" CreationDate="2010-10-01T19:07:56.737" Score="2" ViewCount="" Body="&lt;p&gt;See the discussion and references on the associated Wikipedia page &lt;a href=&quot;http://en.wikipedia.org/wiki/Optimal_solutions_for_Rubik%27s_Cube&quot; rel=&quot;nofollow&quot;&gt;Optimal solutions for Rubik's Cube&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-10-01T19:07:56.737" CommentCount="1" />
  <row Id="5846" PostTypeId="2" ParentId="5839" CreationDate="2010-10-01T19:13:05.830" Score="3" ViewCount="" Body="&lt;p&gt;Knowing the coordinates of A and B, you can find the slope of the line AB and the coordinates of the midpoint M of segment AB.  The slopes of perpendicular lines have product -1, so you can use the slope of AB to find the slope of CD.  With this slope and the point M, you can write an equation for the line CD.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem here is that the points C and D could be anywhere on that line, so long as they are the given length apart, so I suspect there is a bit of information missing.  If M is the midpoint of CD as well (so that AB and CD are the perpendicular bisectors of each other), then C and D are each half the given length from M, so you can write an equation for the circle with radius half the length and center M, and find where the line CD intersects the circle (solve the system of the two equations).&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-10-01T19:13:05.830" />
  <row Id="5847" PostTypeId="2" ParentId="5826" CreationDate="2010-10-01T19:16:56.157" Score="1" ViewCount="" Body="&lt;p&gt;First of all, you need your surfaces to be connected, otherwise the result is not true: let $X = \mathbb{T}^2 \sqcup \mathbb{S}^2$ and $Y = \mathbb{T}^2$. ($\mathbb{T}^2$ a torus, $\mathbb{S}^2$ a sphere.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then the connected sum $X \oplus Y$ &lt;em&gt;does&lt;/em&gt; depend on the choice of the disks. It doesn't matter which disk you choose on $Y$, of course, but if you choose your disk on $X$ belonging to $\mathbb{T}^2$, then you get $2\mathbb{T}^2 \sqcup \mathbb{S}^2$ and if you choose your disk belonging to $\mathbb{S}^2$, you obtain $\mathbb{T}^2 \sqcup \mathbb{T}^2$.&lt;/p&gt;&#xA;" OwnerUserId="664" LastEditorUserId="664" LastEditorDisplayName="" LastEditDate="2010-10-04T16:37:27.990" LastActivityDate="2010-10-04T16:37:27.990" />
  <row Id="5848" PostTypeId="1" AcceptedAnswerId="5852" CreationDate="2010-10-01T20:13:52.973" Score="0" ViewCount="336" Body="&lt;p&gt;This is the question which i am attempting to solve, and it seems to difficult to get rid of the exponents.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Show that a the two cubic curves $Y^3 = X^2 + X^3$ and $X^3 = Y^2 + Y^3$ intersect in nine points.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Any help would be appreciated. By the way i would also like to know whether is there any general method for solving such problems. &lt;/p&gt;&#xA;" LastEditorUserId="8575" LastEditorDisplayName="" LastEditDate="2011-05-26T15:23:31.827" LastActivityDate="2011-05-26T15:23:31.827" Title="Intersection of Cubic curves." Tags="&lt;geometry&gt;&lt;algebraic-curves&gt;&lt;intersection&gt;" AnswerCount="3" CommentCount="5" FavoriteCount="2" />
  <row Id="5849" PostTypeId="2" ParentId="5848" CreationDate="2010-10-01T20:20:22.343" Score="1" ViewCount="" Body="&lt;p&gt;Like your earlier question about how to prove the associativity of the group law on elliptic curves, this too is a consequence of general results on intersection theory, esp. (variants of) &lt;a href=&quot;http://en.wikipedia.org/wiki/Bezouts_theorem&quot; rel=&quot;nofollow&quot;&gt;Bezout's theorem&lt;/a&gt;. For computational purposes you can employ various constructive elimination techniques, e.g Grobner bases, various triangularization methods, etc, which are available in most computer algebra systems&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-10-01T20:26:29.420" LastActivityDate="2010-10-01T20:26:29.420" />
  <row Id="5850" PostTypeId="2" ParentId="5839" CreationDate="2010-10-01T20:29:45.717" Score="0" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Remark&lt;/strong&gt;: Let $M$ be the midpoint of $[CD]$. I assume that $M$ belongs to $[AB]$, as shown above. If it is not the case, then this is only a particular solution.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/KxkHD.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $A\left( x_{A},y_{A}\right) ,B(x_{B},y_{B}),C(x_{C},y_{C}),D(x_{D},y_{D})&#xD;&#xA;$ be the four points. By translation and rotation of the $x,y$ axes as shown&#xA;in the figure, we can center $[A,B,C,D]$. The translation corresponds to the&#xA;change of variables&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x=\overline{x}+X,\qquad y=\overline{y}+Y,$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\overline{x}=\dfrac{x_{A}+x_{B}}{2}\qquad\overline{y}=\dfrac{y_{A}+y_{B}}{2},$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and the rotation corresponds to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$X=x^{\prime }\cos \theta -y^{\prime }\sin \theta,\qquad Y=x^{\prime }\sin \theta +y^{\prime }\cos \theta ,$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\theta =\arctan \dfrac{y_{B}-y_{A}}{x_{B}-x_{A}}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Combining both transformations, we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x=\overline{x}+x^{\prime }\cos \theta -y^{\prime }\sin \theta,\qquad y=\overline{y}+x^{\prime }\sin \theta +y^{\prime }\cos \theta .$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Denoting the distance between the points $C$ and $D$ by $\overline{CD}=d_{CD}&#xD;&#xA;$, we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x_{C}^{\prime }=x_{D}^{\prime }=0\qquad y_{C}^{\prime }=\dfrac{d_{CD}}{2}\qquad y_{D}^{\prime }=-\dfrac{d_{CD}}{2}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the above transformations we get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$C=\left( x_{C},y_{C}\right) =\left( \overline{x}-\dfrac{d_{CD}}{2}\sin&#xD;&#xA;\theta ,\overline{y}+\dfrac{d_{CD}}{2}\cos \theta \right) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$D=\left( x_{D},y_{D}\right) =\left( \overline{x}+\dfrac{d_{CD}}{2}\sin&#xD;&#xA;\theta ,\overline{y}-\dfrac{d_{CD}}{2}\cos \theta \right) ,$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\overline{x}$ and $\theta $ are given above. &lt;/p&gt;&#xA;" OwnerUserId="752" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2011-05-14T12:12:07.673" LastActivityDate="2011-05-14T12:12:07.673" CommentCount="3" />
  <row Id="5851" PostTypeId="2" ParentId="5818" CreationDate="2010-10-01T20:32:00.997" Score="2" ViewCount="" Body="&lt;p&gt;The binomial theorem together with the identity $(x+y)^p = x^p + y^p \mod p$ is useful for this type of problem. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here we want the the $x^p$ coefficient of $(1+x)^{Ap+B} = (1+x^p)^A (1+x)^B = (1 + Ax^p + \dots)(1+X)^B$ where $0 \leq B &amp;lt; p$. This coefficient comes only from multiplication of $(Ax^p)$ by $(1)$.  So the result mod p is $A$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another solution is to use combinatorics.  Partition the set of integers from 1 to $n$ into $A$ blocks of size $p$ and a (perhaps empty) remainder of size $B$, as above.  Any subset is either a complete block or it contains a proper nonempty subset of at least one block.  Rotating the elements in the first partially filled block organizes the $\binom{n}{p}$ p-subsets into collections of size $p$, plus a &quot;remainder&quot; consisting of complete $p$-blocks, and there are $A$ of the those.  &lt;/p&gt;&#xA;" OwnerUserId="467" LastActivityDate="2010-10-01T20:32:00.997" CommentCount="1" />
  <row Id="5852" PostTypeId="2" ParentId="5848" CreationDate="2010-10-01T20:43:15.100" Score="4" ViewCount="" Body="&lt;p&gt;In addition to the usual generalities on Bezout's theorem and resultants, notice that for this system, $X^2 + Y^2 = 0$, so that there are 6 finite solutions (3 on each line $X = \pm iY$, organized as 1 transverse intersection point per line, plus one double point at (0,0) on each line, leading to a quadruple intersection of the curves at 0)  and accounting for the other 3 solutions, the solutions &quot;at infinity&quot; with $X^3=Y^3$, i.e., the asymptotic directions $(X/Y) \to \omega, \omega^3=1, |X| \to \infty$.  In other words, this example is built to show the need for complex projective geometry, and the counting of points according to their multiplicities, as the environment where the Bezout 3x3=9 calculation works.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(the generalities are: if f(x,y)=0 and g(x,y)=0 are the polynomials, you can write down a basis for k[x,y]/(f=g=0), and check that as a vector space it is of dimension 9. If you meant that there are 9 distinct intersection points, you can check that by writing down the action of the multiply-by-x operator on this vector space and seeing whether it has repeated roots. This can all be done by hand but is not as efficient as what is done in computer algebra systems. )&lt;/p&gt;&#xA;" OwnerUserId="467" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-10-06T18:16:28.757" LastActivityDate="2010-10-06T18:16:28.757" CommentCount="3" />
  <row Id="5853" PostTypeId="2" ParentId="123" CreationDate="2010-10-01T21:03:01.563" Score="1" ViewCount="" Body="&lt;p&gt;Most curves that look parabolic are actually Catenaries, which is based in the hyperbolic cosine function. A good example of a Catenary would be the Gateway Arch in Saint Louis, Missouri. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The tractrix, which is based in the hyperbolic secant, is also known as the pursuit curve, which models objects like cargo trucks turning corners, or a dog on a porch starting to chase a car in the street.&lt;/p&gt;&#xA;" OwnerUserId="2254" LastActivityDate="2010-10-01T21:03:01.563" />
  <row Id="5854" PostTypeId="2" ParentId="5837" CreationDate="2010-10-01T21:15:23.110" Score="0" ViewCount="" Body="&lt;p&gt;Hi--&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is also something called as TeXmacs which you might be interested in. As Xavier quote LaTeX is the best way to start. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In case you are in need of a drawing package i list the following 3 packages which are commonly used:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;pstricks&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;PgfTikZ&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Gnuplot&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Please google the names and you shall find lot of interesting things.&lt;/p&gt;&#xA;" LastActivityDate="2010-10-01T21:15:23.110" />
  <row Id="5856" PostTypeId="2" ParentId="5827" CreationDate="2010-10-01T21:30:36.700" Score="3" ViewCount="" Body="&lt;p&gt;Here's a fun way to write the solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Define on $(\Omega, \mathcal{S})$ the probability measure $P(A) = 0$ if $A$ is countable, $P(A)=1$ if $A^c$ is countable.  Then $f$ can be seen as a random variable $X$.  Since all events have probability $0$ or $1$, all events, and hence all random variables, are independent.  Now there must be some $N$ with $P(|X| \le N) &gt; 0$ (since $\Omega = \bigcup_{N=1}^\infty \{|X| \le N\}$ and $P$ is countably additive), hence $P(|X| \le N) = 1$.  So $X$ is a.s. bounded and in particular is $L^2$.  But $Var(X) = Cov(X,X) = 0$ since $X$ is independent of itself!  So $X = EX$ a.s., i.e. except on a countable set.&lt;/p&gt;&#xA;" OwnerUserId="822" LastActivityDate="2010-10-01T21:30:36.700" />
  <row Id="5857" PostTypeId="2" ParentId="5848" CreationDate="2010-10-01T21:31:15.850" Score="3" ViewCount="" Body="&lt;p&gt;Suppose $(x,y)$ satisfies both equations. Substituting one equation into the other, we obtain $x^2+y^2=0$. The only real solution to this equation are $x=y=0$, as pointed out by Moron.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now suppose $x\neq 0$, and so $x$ is a complex number. Then $y = \pm i x$. Eliminating $y$ from the other equation yields two possibilities:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;$y = ix$, and $-ix^3 = x^2 + x^3$. We assume $x\neq 0$, so we have: $-ix = 1 + x$. This yields the solution: $x = (-1+i)/2, y=(-1-i)/2$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$y = -ix$, and $ix^3 = x^2 + x^3$. Similarly to before, we have: $ix = 1+x$. This yields the solution: $x = (-1-i)/2, y=(-1+i)/2$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;As far as I can see, these are the only possible solutions... There are &lt;strong&gt;three&lt;/strong&gt; solutions, not nine.&lt;/p&gt;&#xA;" OwnerUserId="1356" LastEditorUserId="1356" LastEditorDisplayName="" LastEditDate="2010-10-02T05:13:03.113" LastActivityDate="2010-10-02T05:13:03.113" CommentCount="1" />
  <row Id="5858" PostTypeId="1" AcceptedAnswerId="5865" CreationDate="2010-10-01T21:39:33.073" Score="1" ViewCount="184" Body="&lt;p&gt;I have a two state dynamical system. The two state variables are $P$ and $Z$ and $a,b,c,d$ are parameters. The system equations are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{dP}{dt}=a\cdot P-b\cdot PZ=P\left(a-bZ\right)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{dZ}{dt}=c\cdot PZ-d\cdot Z=Z\left(cP-d\right)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Multiplying both sides by $\frac{1}{PZ}$ :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{dP}{dt}\frac{1}{PZ}=a\frac{1}{Z}-b$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{dZ}{dt}\frac{1}{PZ}=c-d\frac{1}{P}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Multiplying with the original system equation gives:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{1}{PZ}\frac{dP}{dt}\frac{dZ}{dt}=\frac{dZ}{dt}\left(a\frac{1}{Z}-b\right)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{1}{PZ}\frac{dP}{dt}\frac{dZ}{dt}=\frac{dP}{dt}\left(c-d\frac{1}{P}\right)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Taking the difference of the last two equations gives:&#xA;$0=\frac{dZ}{dt}\left(a\frac{1}{Z}-b\right)+\frac{dP}{dt}\left(d\frac{1}{P}-c\right)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the chain rule, the total time derivative of my conserved quantity is:&#xA;$\frac{d}{dt}E\left(P,Z\right)=\frac{dZ}{dt}\frac{\partial}{\partial Z}E\left(P,Z\right)+\frac{dP}{dt}\frac{\partial}{\partial P}E\left(P,Z\right)$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;so then the following should hold:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{\partial}{\partial Z}E\left(P,Z\right)=\left(a\frac{1}{Z}-b\right)$ and&#xA;$\frac{\partial}{\partial P}E\left(P,Z\right)=\left(d\frac{1}{P}-c\right)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$&#xA;E\left(P,Z\right)=\int\left(a\frac{1}{Z}-b\right)\partial Z=\int\left(d\frac{1}{P}-c\right)\partial P&#xA;$&#xA;$=a\ln{Z}-bZ+C_{Z}\left(P\right)=d\ln{P}-cP+C_{P}\left(Z\right)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally: &#xA;$E\left(P,Z\right)=a\ln{Z}+dln{P}-bZ-cP$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt; and this quantity is conserved! I had a bad calculus mistake.&lt;/p&gt;&#xA;" OwnerUserId="2256" LastEditorUserId="2256" LastEditorDisplayName="" LastEditDate="2010-10-02T00:03:43.957" LastActivityDate="2010-10-02T00:04:48.137" Title="Determine a conserved quantity in a dynamical system Lotka-Volterra" Tags="&lt;calculus&gt;&lt;dynamical-systems&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="5859" PostTypeId="2" ParentId="3977" CreationDate="2010-10-01T22:01:46.423" Score="4" ViewCount="" Body="&lt;p&gt;There will be a public preview of GeoGebra 3D soon. Keep an eye on the GeoGebra forums :)&lt;/p&gt;&#xA;" LastActivityDate="2010-10-01T22:01:46.423" CommunityOwnedDate="2010-10-01T22:01:46.423" CommentCount="1" />
  <row Id="5860" PostTypeId="2" ParentId="2158" CreationDate="2010-10-01T22:07:06.190" Score="5" ViewCount="" Body="&lt;p&gt;Here is another proof:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We use Legendre's formula for the exact power of a prime $p$ which divides $n!$ which is given by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \sum_{k=1}^{\infty} [\frac{n}{p^k}]$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $[x]$ is the integer part of $x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Coupled with $$ \sum_{i=1}^{n}  [x_i] \le [\sum_{i=1}^{n} x_i]$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;we get that if $\sum_{i=1}^{n} a_i = N$ then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{i=1}^{n} [\frac{a_i}{p^k}] \le [\frac{N}{p^k}] $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And so any prime power which divides $(a_1)! \dots (a_n)!$ divides $N!$ and so $\frac{N!}{(a_1)! \dots (a_n)!}$ is an integer.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-10-01T22:07:06.190" />
  <row Id="5861" PostTypeId="1" CreationDate="2010-10-01T22:24:58.737" Score="7" ViewCount="195" Body="&lt;p&gt;Let $(\Omega, \mathcal{F}, P)$ be a probability space, $X$ an&#xA;integrable random variable, $\mathcal{G} \subset \mathcal{F}$ a&#xA;$\sigma$-field.  The &lt;b&gt;conditional expectation&lt;/b&gt; of $X$ given&#xA;$\mathcal{G}$ is by definition the unique random variable $Y$ which is&#xA;$\mathcal{G}$-measurable and satisfies $E[Y;A] = E[X;A]$ for all $A&#xA;\in \mathcal{G}$.  Proving the uniqueness of $Y$ is easy, but&#xA;existence is harder.  I am looking for a nice existence proof with&#xA;minimal prerequisites.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The traditional proof is to invoke the Radon-Nikodym theorem: the&#xA;signed measure $\nu(A) = E[X;A]$ on $(\Omega, \mathcal{G})$ is&#xA;absolutely continuous to $\mu = P|_\mathcal{G}$, so take $Y$ to be the&#xA;Radon-Nikodym derivative, and it clearly has the desired properties.&#xA;But the proofs I know of the Radon-Nikodym theorem, while elementary,&#xA;are somewhat involved (at least 2 pages, even if you only do the&#xA;absolutely continuous case).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another proof is to first take $X$ with finite variance, and note that&#xA;$K = L^2(\Omega, \mathcal{G}, P)$ is a closed subspace of the Hilbert&#xA;space $H = L^2(\Omega, \mathcal{F}, P)$; then take $Y$ to be the&#xA;orthogonal projection of $X$ onto $K$.  Again, it is then easy to see&#xA;that $Y$ has the desired properties.  But this is not as suitable&#xA;for students with no functional analysis background.  You can develop&#xA;the necessary facts from scratch but it's a little tedious.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I am wondering if anyone knows of a simple proof, preferably using&#xA;only basic measure theory and probability facts.&lt;/p&gt;&#xA;" OwnerUserId="822" LastActivityDate="2010-10-06T15:55:10.333" Title="Slick constructions of conditional expectation" Tags="&lt;probability-theory&gt;&lt;reference-request&gt;&lt;proof-strategy&gt;" AnswerCount="1" CommentCount="0" />
  <row Id="5862" PostTypeId="2" ParentId="5818" CreationDate="2010-10-01T22:26:41.750" Score="1" ViewCount="" Body="&lt;p&gt;Below is a sketch of the generating-function-based proof of Lucas's theorem - which is mentioned in passing in a number of other posts here. Below we will see that the original problem is just a special linear case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\quad\quad\quad\ (1 + X)^{\:a + b P + c P^2} \ \ \ (mod\ P)\quad\quad$ with $\rm\quad\quad 0 \le a,\: b,\: c &amp;lt; P$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\quad  =\ \ \: (1 + X)^a \ (1 + X^P)^b \ (1 + X^{P^2})^c \ \ \ (mod\ P)\quad\quad$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\quad  =\ \ \:  (1 + \:\binom{\:a\:}1 \ X\ \ \ \:+ \ \:\binom{\:a\:}2\ X^2\ \ \  +\  \binom{a}3\ X^3 \ \ \ + \:\cdots\: + \ X^{a}\ )$&lt;br&gt;&#xA;$\rm\quad\quad\ \ *\   (1 + \binom{b}1 \ X^P \ + \ \binom{b}2\: X^{2P} \ \: +\ \binom{b}3\ X^{3P} \ \:+\: \cdots\: + \ X^{bP}) $&#xA;$\rm\quad\quad\ \ * \ (1 +\: \binom{\:c\:}1 \ X^{P^2} + \ \binom{\:c\:}2\ X^{2P^2} +\ \binom{c}3\ X^{3P^2} +\: \cdots\: + \ X^{cP^2}) \quad\ \ (mod\ P)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\quad =\ \ \sum\ \binom{a}A \binom{b}B \binom{c}C\  X^{\: A + BP + CP^2}\quad\quad\quad\: (mod\ P)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore $\rm\quad \ \ \binom{a}A \binom{b}B \binom{c}C = \binom{a + bP + cP^2}{A + BP + CP^2}\ \ \ (mod\ P)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\ \Rightarrow\quad\ b\ \ =\ \ \binom{b}1 \quad\ \: =\quad\ \: \binom{a+bp}p\ \ \ $ for $\rm\ \  B = 1,\ A = C = c = 0,\ P = p$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is the result sought with $\rm\ n = a+bp\ \Rightarrow\ [n/p] = b\:$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For other proofs see Granville's delightful survey &lt;a href=&quot;http://www.cecm.sfu.ca/organics/papers/granville/&quot; rel=&quot;nofollow&quot;&gt;The Arithmetic Properties of Binomial Coefficients&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-10-04T02:41:07.190" LastActivityDate="2010-10-04T02:41:07.190" />
  <row Id="5863" PostTypeId="1" AcceptedAnswerId="5871" CreationDate="2010-10-01T22:43:35.017" Score="1" ViewCount="365" Body="&lt;p&gt;If you are given of a ring, how do you find its ideals? &lt;/p&gt;&#xA;" OwnerUserId="6838" LastActivityDate="2011-04-04T15:43:10.890" Title="If you are given of a ring, how do you find its ideals?" Tags="&lt;ring-theory&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="1" />
  <row Id="5864" PostTypeId="2" ParentId="5837" CreationDate="2010-10-01T23:51:08.280" Score="6" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://asymptote.sourceforge.net/&quot; rel=&quot;nofollow&quot;&gt;Asymptote&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="822" LastActivityDate="2010-10-01T23:51:08.280" />
  <row Id="5865" PostTypeId="2" ParentId="5858" CreationDate="2010-10-02T00:04:48.137" Score="1" ViewCount="" Body="&lt;p&gt;I have edited the original post to include the correct derivation. My mathematical weakness showed when I confused integration with differentiation!&lt;/p&gt;&#xA;" OwnerUserId="2256" LastActivityDate="2010-10-02T00:04:48.137" />
  <row Id="5866" PostTypeId="2" ParentId="5837" CreationDate="2010-10-02T00:13:15.560" Score="1" ViewCount="" Body="&lt;p&gt;On unix or Mac environments try xfig. There are windows simulators. Picture environments can be used with LaTeX to draw, but they are a pain (we used this in &quot;The Classical and Quantum 6j symbols&quot;). We used xfig for both &quot;Knotted Surfaces and Their Diagrams&quot; and &quot;Surfaces in 4-Space.&quot; Otherwise, get a good set of Rapidograph pens, some blue pencil, and lots of white-out. &lt;/p&gt;&#xA;" OwnerUserId="722" LastActivityDate="2010-10-02T00:13:15.560" CommentCount="1" />
  <row Id="5867" PostTypeId="2" ParentId="5842" CreationDate="2010-10-02T01:10:59.630" Score="5" ViewCount="" Body="&lt;p&gt;To give a more detailed solution, we start by rearranging the original equation as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$y\exp\left(\frac{1-y}{x}\right)=1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$y\exp\left(\frac1{x}\right)\exp\left(-\frac{y}{x}\right)=1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$-y\exp\left(-\frac{y}{x}\right)=-\exp\left(-\frac1{x}\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$-\frac{y}{x}\exp\left(-\frac{y}{x}\right)=-\frac1{x}\exp\left(-\frac1{x}\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which we can now invert to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$-\frac{y}{x}=W\left(-\frac1{x}\exp\left(-\frac1{x}\right)\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and finally we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$y=-xW\left(-\frac1{x}\exp\left(-\frac1{x}\right)\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Though the Lambert function has &lt;em&gt;two&lt;/em&gt; real branches, only the &quot;principal branch&quot; gives a nontrivial solution; the solution corresponding to the $W_{-1}(\cdot)$ branch is $y=1$&lt;/p&gt;&#xA;" OwnerUserId="498" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-10-02T12:00:13.993" LastActivityDate="2010-10-02T12:00:13.993" />
  <row Id="5868" PostTypeId="1" AcceptedAnswerId="5872" CreationDate="2010-10-02T04:03:12.450" Score="1" ViewCount="140" Body="&lt;p&gt;Given lines $\mathbb{L}_1 : \lambda(1,3,2)+(-1,3,1)$, $\mathbb{L}_2 : \lambda(-1,2,3)+(0,0,-1)$ and $\mathbb{L}_3 : \lambda(1,1,-2)+(2,0,1)$, find a line $\mathbb{L}$ such that $\mathbb{L}$ is parallel to $\mathbb{L}_1$, $\mathbb{L}\cap\mathbb{L}_2 \neq \emptyset$ and $\mathbb{L}\cap\mathbb{L}_3 \neq \emptyset$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $\mathbb{L}$ must be parallel to $\mathbb{L}_1$, then $\mathbb{L}:\lambda(1,3,2)+(x,y,z)$ but I can't figure out how to get that (x,y,z) point. I'd like to be given just a slight nod because I'm sure the problem is really easy. Thanks a lot!&lt;/p&gt;&#xA;" OwnerUserId="1523" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-10-02T04:20:04.417" LastActivityDate="2010-10-03T13:12:35.357" Title="Finding a line that satisfies three conditions." Tags="&lt;geometry&gt;" AnswerCount="4" CommentCount="1" FavoriteCount="1" />
  <row Id="5869" PostTypeId="2" ParentId="5868" CreationDate="2010-10-02T04:20:51.297" Score="1" ViewCount="" Body="&lt;p&gt;$\mathbb L \cap \mathbb L_2 \ne \emptyset$ implies that for some value of $\lambda$, your equation for $\mathbb L$ is on $\mathbb L_2$.  Since substituting $\lambda' = \lambda - c$ doesn't change the line $\mathbb L$, you can take $\lambda = 0$ without loss of generality.  That is, $(x,y,z)$ is on $\mathbb L_2$, or $(x,y,z) = \lambda_2(-1,2,3) + (0,0,-1)$. ...&lt;/p&gt;&#xA;" OwnerUserId="2236" LastActivityDate="2010-10-02T04:20:51.297" CommentCount="3" />
  <row Id="5870" PostTypeId="2" ParentId="5840" CreationDate="2010-10-02T05:52:22.320" Score="2" ViewCount="" Body="&lt;p&gt;See &lt;a href=&quot;http://www.cube20.org/&quot; rel=&quot;nofollow&quot;&gt;http://www.cube20.org/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1770" LastActivityDate="2010-10-02T05:52:22.320" />
  <row Id="5871" PostTypeId="2" ParentId="5863" CreationDate="2010-10-02T06:15:29.217" Score="6" ViewCount="" Body="&lt;p&gt;What Bill say in the comments is that there is no method to determine all ideals of a ring. Even if you specialize and look just for the maximal ideals in a given ring it might imply hard work! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A great example is to look at $H^\infty(\mathbb{D})$ which is the collection of all power series $$f(z)=\sum_{n\ge0}a_nz^n$$ such that &#xA;$$\sup_{|z|&amp;lt;1}|f(z)|&amp;lt;\infty.$$&#xA;In other words, the space of all bounded analytic functions on the unit disc $\mathbb{D}$.&#xA;It is fairly easy to see that $R=H^\infty(\mathbb{D})$ is a ring under point-wise operations. &#xA;However, for long it was unknown how the maximal ideals ideals looks like. That problem was solved in 1962 by L. Carleson and is known as the &lt;a href=&quot;http://en.wikipedia.org/wiki/Corona_theorem&quot; rel=&quot;nofollow&quot;&gt;Corona Theorem&lt;/a&gt;. I recommend reading this popular text regarding the &lt;a href=&quot;http://www.abelprisen.no/en/prisvinnere/2006/lennartcarleson/koronateoremet.html&quot; rel=&quot;nofollow&quot;&gt;Corona theorem&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this day no one knows for sure how the the maximals looks like in $H^\infty(\mathbb{D}\times\mathbb{D})$, that is the space of bounded analytic functions in the bi-disc in $\mathbb{C}^2$.&lt;/p&gt;&#xA;" OwnerUserId="1154" LastActivityDate="2010-10-02T06:15:29.217" CommentCount="7" />
  <row Id="5872" PostTypeId="2" ParentId="5868" CreationDate="2010-10-02T07:32:12.837" Score="5" ViewCount="" Body="&lt;p&gt;Your first step is correct. Let's write the vector equations of your line $\mathbb{L}$ and the remaining ones:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{align}&#xA;\mathbb{L}_{\ \ } \qquad (x,y,z) &amp;amp;= (x_0, y_0, z_0) + \lambda (1, 3,2) \\&#xA;\mathbb{L}_2 \qquad (x,y,z) &amp;amp;= (0,0,-1) + \lambda_2 (-1, 2,3) \\&#xA;\mathbb{L}_3 \qquad (x,y,z) &amp;amp;= (2,0,1) + \lambda_3 (1,1,-2)&#xA;\end{align}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $\mathbb{L}\cap\mathbb{L}_2 \neq \emptyset$, you may take the point $(x_0, y_0, z_0) \in \mathbb{L}_2$. So the equation for $\mathbb{L}$ must look like&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\mathbb{L} \qquad (x,y,z) = (-\lambda_2 , 2\lambda_2, -1 + 3\lambda_2 ) + \lambda (1, 3,2)    \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for $\mathbb{L}\cap\mathbb{L}_3 \neq \emptyset$, you should find the common point thanks to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;(-\lambda_2 , 2\lambda_2, -1 + 3\lambda_2 ) + \lambda (1, 3,2) = (2,0,1) + \lambda_3 (1,1,-2)&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Somewhat in disguise, this is a $3\times 3$ system of linear equations, with unkowns $\lambda, \lambda_2$ and $\lambda_3$. Solve it. In particular, find $\lambda_2$ and you have your line $\mathbb{L}$.&lt;/p&gt;&#xA;" OwnerUserId="664" LastEditorUserId="664" LastEditorDisplayName="" LastEditDate="2010-10-02T08:15:36.193" LastActivityDate="2010-10-02T08:15:36.193" CommentCount="3" />
  <row Id="5873" PostTypeId="2" ParentId="5817" CreationDate="2010-10-02T09:47:07.130" Score="2" ViewCount="" Body="&lt;p&gt;Look at the introduction to Chapter 10 in &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0387989315&quot; rel=&quot;nofollow&quot;&gt;Bender &amp;amp; Orszag&lt;/a&gt;. (&quot;Click to look inside&quot;, then search for &quot;WKB&quot; and go to p.484 in the search results.)&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2010-10-02T09:47:07.130" CommentCount="2" />
  <row Id="5874" PostTypeId="1" AcceptedAnswerId="5894" CreationDate="2010-10-02T11:39:46.653" Score="2" ViewCount="162" Body="&lt;p&gt;Given a set $E$ which is bounded, measurable and $m^{\ast}(E)=x &gt;0$, then for each $y \in (0,x)$ we can find a measurable set $A \subset E$, such that $m^{\ast}(A)=y$. To see this one considers measure as a continuous function and applies the Intermediate Value property.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we extend the question in the following manner: Suppose $E$ is a set which has finite measure, then for each positive $x &amp;lt; m^{\ast}(E)$ prove that there exists a perfect set $A \subset E$, such that $m^{\ast}(A)=x$.&lt;/p&gt;&#xA;" LastActivityDate="2010-10-02T21:25:18.457" Title="Existence of a perfect measurable set" Tags="&lt;measure-theory&gt;" AnswerCount="1" CommentCount="9" />
  <row Id="5875" PostTypeId="2" ParentId="5868" CreationDate="2010-10-02T11:58:59.940" Score="3" ViewCount="" Body="&lt;p&gt;I'd like to add the following to Agustí Roig's splendid answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The equation of the general line passing through both $\mathbb{L}_2$ and $\mathbb{L}_3$ is given by $\underline{r} = \underline{a}+ \mu(\mathbb{L}_2 - \mathbb{L}_3),$ where $\underline{a}$ is some point on this line and its direction is $\mathbb{L}_2 - \mathbb{L}_3.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we know that the direction is (1,3,2), so set&#xA;$\lambda (1,3,2) = \mathbb{L}_2 - \mathbb{L}_3.$&#xA;That is&#xA;$$\lambda (1,3,2) = (0,0,-1)+\lambda_2(-1,2,3) – (2,0,1) - \lambda_3(1,1,-2).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence we obtain $\lambda_2 = -2$, $\lambda_3 = 2$ and $\lambda = -2.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Putting $\lambda_2 = -2$ in the equation of $\mathbb{L}_2$ gives the point (2,-4,-7), which is equivalent to w=0 in my comment to Agustí's answer. I hope that this adds some value for you.&lt;/p&gt;&#xA;" OwnerUserId="1301" LastEditorUserId="1301" LastEditorDisplayName="" LastEditDate="2010-10-02T12:08:54.683" LastActivityDate="2010-10-02T12:08:54.683" />
  <row Id="5876" PostTypeId="2" ParentId="3627" CreationDate="2010-10-02T13:32:10.003" Score="0" ViewCount="" Body="&lt;p&gt;What if rank(G) is m? Is there any efficient way?&lt;/p&gt;&#xA;" LastActivityDate="2010-10-02T13:32:10.003" CommentCount="1" />
  <row Id="5877" PostTypeId="1" AcceptedAnswerId="5883" CreationDate="2010-10-02T14:07:17.043" Score="9" ViewCount="602" Body="&lt;p&gt;The web is littered with any number of pages (&lt;a href=&quot;http://planetmath.org/encyclopedia/ProofOfThuesLemma.html&quot; rel=&quot;nofollow&quot;&gt;example&lt;/a&gt;) giving an existence and uniqueness proof that a pair of squares can be found summing to primes congruent to 1 mod 4 (and also that there are no such pairs for primes congruent to 3 mod 4).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, none of the stuff I've read on the topic offers any help with actually efficiently finding (ie other than a straight search up to sqrt(p)) the concrete values of such squares.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What's the best way to actually find them ?&lt;/p&gt;&#xA;" OwnerUserId="2266" LastActivityDate="2010-10-06T13:33:13.530" Title="Efficiently finding two squares which sum to a prime" Tags="&lt;number-theory&gt;&lt;algorithms&gt;&lt;prime-numbers&gt;" AnswerCount="5" FavoriteCount="2" />
  <row Id="5878" PostTypeId="2" ParentId="531" CreationDate="2010-10-02T14:19:56.223" Score="2" ViewCount="" Body="&lt;p&gt;Greg Egan has some fun with this idea in one of his best short stories, &quot;Luminous&quot; (published in the collection of the same name).  A pair of researchers are exploring an apparent &quot;defect&quot; in mathematics:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;You still don't get it, do you,&#xA;  Bruno? You're still thinking like a&#xA;  Platonist. The universe has only been&#xA;  around for fifteen billion years. It&#xA;  hasn't had time to create infinities.&#xA;  The far side can't go on&#xA;  forever-because somewhere beyond the&#xA;  defect, there are theorems that don't&#xA;  belong to any system. Theorems that&#xA;  have never been touched, never been&#xA;  tested, never been rendered true or&#xA;  false.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Terrific stuff!&lt;/p&gt;&#xA;" OwnerUserId="2266" LastEditorUserId="2266" LastEditorDisplayName="" LastEditDate="2010-10-03T22:58:53.233" LastActivityDate="2010-10-03T22:58:53.233" />
  <row Id="5879" PostTypeId="1" AcceptedAnswerId="5897" CreationDate="2010-10-02T14:27:39.830" Score="0" ViewCount="114" Body="&lt;p&gt;I am having a little trouble understanding some results from a program i wrote. I think i have the correct results but i need to understand the theory behind it a little more (i am a programmer, not mathematician).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My program generates symmetric k-regular graphs (single undirected edges). It generates an adjacency matrix and i take that adjacency matrix and find it's canonical isomorph. I compare the canonical isomorph to the list of all previous ones i have found to determine it if's a new k-regular graph.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So here is an example and i think it is correct but i just need to understand the result a little more.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;n = 6&lt;br /&gt;&#xA;k = 2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Adjacency Matrix 1: &lt;br /&gt;&#xA;&lt;code&gt;&lt;br /&gt;&#xA;0 1 1 0 0 0 &lt;br /&gt;&#xA;1 0 1 0 0 0 &lt;br /&gt;&#xA;1 1 0 0 0 0 &lt;br /&gt;&#xA;0 0 0 0 1 1 &lt;br /&gt;&#xA;0 0 0 1 0 1 &lt;br /&gt;&#xA;0 0 0 1 1 0 &lt;br /&gt;&#xA;&lt;/code&gt;&#xA;&lt;br /&gt;&lt;br /&gt;&#xA;Canonical Isomorph:&#xA;&lt;code&gt;&lt;br /&gt;&#xA;0 1 1 0 0 0 &lt;br /&gt;&#xA;1 0 1 0 0 0 &lt;br /&gt;&#xA;1 1 0 0 0 0 &lt;br /&gt;&#xA;0 0 0 0 1 1 &lt;br /&gt;&#xA;0 0 0 1 0 1 &lt;br /&gt;&#xA;0 0 0 1 1 0 &lt;br /&gt;&#xA;&lt;/code&gt;&#xA;&lt;br /&gt;&#xA;The relabeling Label:&lt;br /&gt; &#xA;0 1 2 3 4 5&lt;br /&gt;&#xA;(which is the identity)&#xA;&lt;br /&gt;&lt;br /&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Adjacency Matrix 2: &#xA;&lt;code&gt;&lt;br /&gt;&#xA;0 1 1 0 0 0 &lt;br /&gt;&#xA;1 0 0 1 0 0 &lt;br /&gt;&#xA;1 0 0 0 1 0 &lt;br /&gt;&#xA;0 1 0 0 0 1 &lt;br /&gt;&#xA;0 0 1 0 0 1 &lt;br /&gt;&#xA;0 0 0 1 1 0 &lt;br /&gt;&#xA;&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Canonical Isomorph: &#xA;&lt;code&gt;&lt;br /&gt;&#xA;0 1 1 0 0 0 &lt;br /&gt;&#xA;1 0 0 0 1 0 &lt;br /&gt;&#xA;1 0 0 0 0 1 &lt;br /&gt;&#xA;0 0 0 0 1 1 &lt;br /&gt;&#xA;0 1 0 1 0 0 &lt;br /&gt;&#xA;0 0 1 1 0 0 &lt;br /&gt;&#xA;&lt;/code&gt;&#xA;Relabeling: &lt;br /&gt;&#xA;0 1 2 5 3 4&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does this look correct? Labeling means each vertex is labeled from 0-5.&lt;/p&gt;&#xA;" OwnerUserId="43" LastActivityDate="2010-12-30T21:26:24.213" Title="Canonical Isomorphic Graphs" Tags="&lt;combinatorics&gt;&lt;graph-theory&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="5880" PostTypeId="2" ParentId="5877" CreationDate="2010-10-02T14:33:27.897" Score="5" ViewCount="" Body="&lt;p&gt;You can try using the &lt;a href=&quot;http://www.numbertheory.org/courses/MP313/lectures/lecture13/page4.html&quot; rel=&quot;nofollow&quot;&gt;Hermite-Serret&lt;/a&gt; algorithm, which is briefly as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Find a $z$ such that $z^2 = -1 \mod p$. Apply the Euclidean algorithm to $p$ and $z$ stopping when the first pair of remainders $(x,y)$ go below $\sqrt{p}$. Hermite proved that $x^2 + y^2 = p$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To find $z^2 = -1 \mod p$, you can probably guess a $z$ and start with $z^{p-1} = 1 \mod p$ and start taking square roots or as Robin noted (in a now deleted comment/his answer) guess an $a$ and see if $z = a^{\frac{p-1}{4}}$ is a candidate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There have been improvements to this by BrillHart. Derek was kind enough to hunt down the reference to that here: &lt;a href=&quot;http://jstor.org/pss/2005889&quot; rel=&quot;nofollow&quot;&gt;http://jstor.org/pss/2005889&lt;/a&gt; and a free link (thanks to J.M) &lt;a href=&quot;http://www.ams.org/journals/mcom/1972-26-120/S0025-5718-1972-0314745-6/S0025-5718-1972-0314745-6.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.ams.org/journals/mcom/1972-26-120/S0025-5718-1972-0314745-6/S0025-5718-1972-0314745-6.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-10-05T13:54:35.157" LastActivityDate="2010-10-05T13:54:35.157" CommentCount="6" />
  <row Id="5881" PostTypeId="1" AcceptedAnswerId="5882" CreationDate="2010-10-02T15:28:16.143" Score="2" ViewCount="107" Body="&lt;p&gt;Hi all,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have the following assertion in my notes from last year that I'm trying hard to digest, but I think it isn't true:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If $p$ is prime $\Leftrightarrow$ if $p | ab$ then either $p | a$ or $p | b$ or both.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A valid proof must prove both directions, so:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;if $p$ is prime, let $a_p$ be the prime factorisation of $a$ and likewise $b_p$ be the prime factorisation of $b$. By the fundamental theorem of arithmetic, if $p|ab$ then $p|a$ or $p|b$ or both.&lt;/li&gt;&#xA;&lt;li&gt;However, assuming $p|ab$ and that $p|a$ or $p|b$ or both, I can find a case that doesn't work. Let $p=4$, $a=7$, $b=8$. Then $4|56$ does imply $4|8$ ($p|b$) but $4$ is not prime.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Am I missing something? The rest of my notes use this result quite a lot (albeit mostly from the point of view of $p$ prime which is clearly true). Unfortunately I'm not sure I'll be happy to progress unless I'm happy this statement is true...&lt;/p&gt;&#xA;" OwnerUserId="892" LastActivityDate="2010-10-02T17:28:29.477" Title="Prime divisibility" Tags="&lt;number-theory&gt;&lt;prime-numbers&gt;&lt;divisibility&gt;" AnswerCount="2" />
  <row Id="5882" PostTypeId="2" ParentId="5881" CreationDate="2010-10-02T15:36:01.467" Score="5" ViewCount="" Body="&lt;p&gt;What you are missing is the following: The criterion &quot;$p|ab$ implies that $p|a$ or $p|b$&quot; has to hold &lt;strong&gt;for all natural numbers $a$ and $b$&lt;/strong&gt; for $p$ to be prime.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, $p=4$ is not prime, because $4|36$ but not $4|6$.&lt;/p&gt;&#xA;" OwnerUserId="367" LastActivityDate="2010-10-02T15:36:01.467" CommentCount="1" />
  <row Id="5883" PostTypeId="2" ParentId="5877" CreationDate="2010-10-02T15:46:07.847" Score="12" ViewCount="" Body="&lt;p&gt;In practice this comes to the same algorithm as Moron's, but I'll give&#xA;my two penn'orth anyway. There are two stages: (i) find $z$ with $z^2\equiv -1$&#xA;(mod $p$), and (ii) use $z$ to find $x$ and $y$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Stage (i). If $a$ is a quadratic nonresidue modulo $p$ then $a^{(p-1)/2}&#xA;\equiv-1$ (mod $p$) so we can take $z\equiv a^{(p-1)/4}$ (mod $p$)&#xA;(and that is easily computed by &lt;a href=&quot;http://en.wikipedia.org/wiki/Modular_exponentiation&quot; rel=&quot;nofollow&quot;&gt;modular exponentiation&lt;/a&gt;).&#xA;How does one find quadratic nonresidues? Well just over half of all&#xA;numbers $a$ such that $1 &amp;lt; a &amp;lt; p-1$ &#xA;are quadratic nonresidues, so taking $a$ at random&#xA;will need at most two goes on average.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Stage (ii). Compute the greatest common divisor of $p$ and $z+i$ using the&#xA;Euclidean algorithm for the &lt;a href=&quot;http://en.wikipedia.org/wiki/Gaussian_integer&quot; rel=&quot;nofollow&quot;&gt;Gaussian integers&lt;/a&gt;. The answer will&#xA;be $x+yi$ where $x^2+y^2=p$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $p\equiv 1$ (mod $2^k$) where $k\ge3$ one can speed up stage (i) a bit&#xA;on average. Compute $w=a^{(p-1)/2^k}$ modulo $p$. If $w\equiv\pm1$ we lose&#xA;but otherwise keep squaring $w$. Then we eventually get to $-1$ and before&#xA;that we have the required $z$. This sometimes wins even if $a$ is a quadratic&#xA;residue.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In effect the Gaussian gcd above and Moron's Hermite-Serret algorithm amount&#xA;to the same thing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Below is some rough-and-ready Python code to compute $a$ and $b$ such that&#xA;$p=a^2+b^2$. The function that does this is 2sq. It will behave erratically&#xA;if fed p not a prime congruent to 1 modulo 4.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;def mods(a, n):&#xA;    if n &amp;lt;= 0:&#xA;        return &quot;negative modulus&quot;&#xA;    a = a % n&#xA;    if (2 * a &amp;gt; n):&#xA;        a -= n&#xA;    return a&#xA;&#xA;def powmods(a, r, n):&#xA;    out = 1&#xA;    while r &amp;gt; 0:&#xA;        if (r % 2) == 1:&#xA;            r -= 1&#xA;            out = mods(out * a, n)&#xA;        r /= 2&#xA;        a = mods(a * a, n)&#xA;    return out&#xA;&#xA;def quos(a, n):&#xA;    if n &amp;lt;= 0:&#xA;        return &quot;negative modulus&quot;&#xA;    return (a - mods(a, n))/n&#xA;&#xA;def grem(w, z):&#xA;    # remainder in Gaussian integers when dividing w by z&#xA;    (w0, w1) = w&#xA;    (z0, z1) = z&#xA;    n = z0 * z0 + z1 * z1&#xA;    if n == 0:&#xA;        return &quot;division by zero&quot;&#xA;    u0 = quos(w0 * z0 + w1 * z1, n)&#xA;    u1 = quos(w1 * z0 - w0 * z1, n)&#xA;    return(w0 - z0 * u0 + z1 * u1,&#xA;           w1 - z0 * u1 - z1 * u0)&#xA;&#xA;def ggcd(w, z):&#xA;    while z != (0,0):&#xA;        w, z = z, grem(w, z)&#xA;    return w&#xA;&#xA;def root4(p):&#xA;    # 4th root of 1 modulo p&#xA;    if p &amp;lt;= 1:&#xA;        return &quot;too small&quot;&#xA;    if (p % 4) != 1:&#xA;        return &quot;not congruent to 1&quot;&#xA;    k = p/4&#xA;    j = 2&#xA;    while True:&#xA;        a = powmods(j, k, p)&#xA;        b = mods(a * a, p)&#xA;        if b == -1:&#xA;            return a&#xA;        if b != 1:&#xA;            return &quot;not prime&quot;&#xA;        j += 1&#xA;&#xA;def sq2(p):&#xA;    a = root4(p)&#xA;    return ggcd((p,0),(a,1))&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="226" LastEditorUserId="226" LastEditorDisplayName="" LastEditDate="2010-10-06T08:21:18.800" LastActivityDate="2010-10-06T08:21:18.800" CommentCount="3" />
  <row Id="5884" PostTypeId="2" ParentId="5877" CreationDate="2010-10-02T16:22:30.517" Score="7" ViewCount="" Body="&lt;p&gt;Just for amusement, here is an exact formula for finding $a$ so that $a^2 + b^2 = p$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a  = \dfrac{1}{2}\sum_{x = 0}^{p-1} \bigl( \dfrac{x^3 - x}{p} \bigr),$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\bigl( \dfrac{a}{p} \bigr)$ denotes the Legendre symbol of $a$ mod $p$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Added: At Bill Dubuque's suggestion, I am noting here that this exact formula is not&#xA;an &lt;em&gt;efficient&lt;/em&gt; means to compute the value of $a$ (hence my prefatory &quot;Just for amusement ...&quot;).&lt;/p&gt;&#xA;" OwnerUserId="221" LastEditorUserId="221" LastEditorDisplayName="" LastEditDate="2010-10-04T05:32:38.230" LastActivityDate="2010-10-04T05:32:38.230" CommentCount="6" />
  <row Id="5885" PostTypeId="2" ParentId="5881" CreationDate="2010-10-02T17:28:29.477" Score="1" ViewCount="" Body="&lt;p&gt;You could also take a look at this &lt;a href=&quot;http://math.stackexchange.com/questions/4268/alternate-definition-of-prime-number&quot;&gt;previous question&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="664" LastActivityDate="2010-10-02T17:28:29.477" CommentCount="1" />
  <row Id="5886" PostTypeId="2" ParentId="5868" CreationDate="2010-10-02T17:49:49.377" Score="1" ViewCount="" Body="&lt;p&gt;Notation can help us see what's going on.  Let's systematically parametrize the lines: let $\mathbb{L_2}$ start at $\mathbf{a} = (0,0,-1)$ and have a direction vector $\mathbf{u} = (-1,2,3)$ and similarly let $\mathbf{b} = (2,0,1)$, $\mathbf{v} = (1,1,-2)$ determine $\mathbb{L_3}$.  All we need to know about $\mathbb{L_1}$ is its direction vector, $\mathbf{w} = (1,3,2)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there exists a solution then we can start at the origin $\mathbf{a}$ of $\mathbb{L_2}$ and reach the origin $\mathbf{b}$ of $\mathbb{L_3}$ via a path that takes us along $\mathbb{L_2}$, thence in the direction $\mathbf{w}$ of $\mathbb{L_1}$ to a point on $\mathbb{L_3}$, and finally back along $\mathbb{L_3}$ to $\mathbf{b}$.  Let $\lambda$, $\rho$, and $\mu$ be the multiples of the direction vectors $\mathbf{u}$,  $\mathbf{w}$, and  $\mathbf{v}$ that are traveled, respectively.  Whence&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathbf{b - a} = \lambda \mathbf{u} + \rho \mathbf{w} + \mu \mathbf{v}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and the solution line $\mathbb{L}$ passes through $\mathbf{a} +  \lambda \mathbf{u}$ in the $\mathbf{w}$ direction.  All we need to do, then, is find $\lambda$.  A simple way is to find a linear form that kills both $\mathbf{w}$ and $\mathbf{v}$ but not $\mathbf{u}$ and apply it to both sides.  The obvious one is to take a dot product with some vector that is mutually perpendicular to $\mathbf{w}$ and $\mathbf{v}$.  All vectors are multiples of the cross product $\mathbf{v} \times \mathbf{w}$, so let's use it.  Thus&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(\mathbf{b - a}) \cdot (\mathbf{v} \times \mathbf{w}) =(\lambda \mathbf{u} + \rho \mathbf{w} + \mu \mathbf{v}) \cdot (\mathbf{v} \times \mathbf{w}) = \lambda \mathbf{u} \cdot (\mathbf{v} \times \mathbf{w}).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dividing gives the solution (if it exists!)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\lambda = \frac{(\mathbf{b - a}) \cdot (\mathbf{v} \times \mathbf{w})}{\mathbf{u} \cdot (\mathbf{v} \times \mathbf{w})} = -2.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A parametrization of the line $\mathbb{L}$ therefore is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathbf{a} + \mathbf{u} \frac{(\mathbf{b - a}) \cdot (\mathbf{v} \times \mathbf{w})}{\mathbf{u} \cdot (\mathbf{v} \times \mathbf{w})} + t \mathbf{w}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for a real parameter $t$.  To check the solution, reverse the roles of the lines $\mathbb{L_2}$ and $\mathbb{L_3}$, thereby finding $\mu$ and the point along $\mathbb{L_3}$.  (We don't have to recompute the triple product in the denominator; only its sign changes.)  The corresponding point (on $\mathbb{L_3}$ now) indeed lies on the solution line (with $t = 2$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This approach immediately shows a solution does exist and is unique if and only if the triple product $\mathbf{u} \cdot (\mathbf{v} \times \mathbf{w})$ is nonzero; i.e., the three direction vectors $\mathbf{u}$, $\mathbf{v}$, and $\mathbf{w}$ are not coplanar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course this all is just another way of reproducing the previous solutions.  Its possible advantages are its geometric clarity and the straightforward formulas it provides for the answer.&lt;/p&gt;&#xA;" OwnerUserId="1489" LastEditorUserId="1489" LastEditorDisplayName="" LastEditDate="2010-10-03T13:12:35.357" LastActivityDate="2010-10-03T13:12:35.357" CommentCount="1" />
  <row Id="5887" PostTypeId="2" ParentId="5839" CreationDate="2010-10-02T18:09:03.357" Score="0" ViewCount="" Body="&lt;p&gt;Well, I'm not too good at proving things, but here goes my attempt. This is a general result for when we don't know one of the endpoints of $\overline{CD}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have points $A(a_x,a_y)$ and $B(b_x,b_y)$, and length of the perpendicular bisector $d$. Let the midpoint of $\overline{AB}$ be $\displaystyle M\left(\frac{a_x+b_x}{2},\frac{a_y+b_y}{2}\right)$, and let the point $Q$ be the point $d$ units away from $M$ along the perpendicular bisector $\overline{QM}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The slope of $\overline{AB}$ is $\displaystyle\frac{b_y-a_y}{b_x-a_x}$, so the slope of the perpendicular bisector is $\displaystyle -\frac{b_x-a_x}{b_y-a_y}$. The equation of the perpendicular bisector in vector form is the collection of all points $P$ where $P(t)=M+\langle a_y-b_y,b_x-a_x \rangle t$, $t \in \mathbb{R}$ ($M$ being the midpoint of $\overline{AB}$). Now, making the vector into a unit vector is much easier, and since the magnitude of the vector ($\sqrt{(a_y-b_y)^2+(b_x-a_x)^2}$) is simply the length $AB$, we can say that the point equation is $P(t)=M+\frac{1}{AB}\langle a_y-b_y,b_x-a_x \rangle t$, and thus the point $Q$ would be the point returned by the equation when $t=d$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All possibilities for the points $C$ and $D$ can be found by finding the right number for $t$; namely, the first point $P(t)$ on the range of $t \in [-d,0]$ ($t = 0$ making it the midpoint of $\overline{AB}$), and the second point then being $P(t+d)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Correct me in any way necessary! I feel like that $P(t)$ equation will give you all the points possible if you use the correct range.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Note&lt;/strong&gt;: I think an advantage of this method is that if the question were posed something like finding the endpoints of the perpendicular bisector with a certain length, where $C$ was, say, $\frac{2}{7}$ths of the distance along the line away, all it would take is plugging in $t=\pm\frac{2d}{7},\mp\frac{5d}{7}$.&lt;/p&gt;&#xA;" OwnerUserId="1562" LastEditorUserId="1562" LastEditorDisplayName="" LastEditDate="2010-10-02T18:28:31.227" LastActivityDate="2010-10-02T18:28:31.227" CommentCount="2" />
  <row Id="5888" PostTypeId="2" ParentId="5861" CreationDate="2010-10-02T18:19:29.450" Score="1" ViewCount="" Body="&lt;p&gt;For the basic case:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume that $X$ and $Y$ are random variables on a probability space $(\Omega, \mathcal{F}, P)$ where $E[|Y|] &amp;lt; \infty$. Further assume that $X$ and $Y$ have joint probability distribution $f_{X, Y}(x,y)$. Define:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$g(x) = \int_{\mathbb{R}} \frac{f_{X,Y}(x,y)}{f_X(x)} \, dy$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $f_X$ is the marginal density of $X$. now $g$ is the conditional expectation $E[Y|X = x]$ from elementary probability theory. Now we can see that $E[Y|X] = g(X)$. Now $g$ is $\sigma(X)$-measurable so now we need to check:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_A g(X) \, dP = \int_A Y \, dP \textrm{ for $A$ in $\sigma(X)$}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the partial-averaging-property, so we get the conditional expectation. Well, this is just some syntax-manipulation so I'll skip that. I can add it if you want.&lt;/p&gt;&#xA;" OwnerUserId="1120" LastEditorUserId="1120" LastEditorDisplayName="" LastEditDate="2010-10-06T15:55:10.333" LastActivityDate="2010-10-06T15:55:10.333" CommentCount="2" />
  <row Id="5889" PostTypeId="1" CreationDate="2010-10-02T19:00:20.743" Score="11" ViewCount="204" Body="&lt;p&gt;Let $G = A *_C B$ be an amalgamated free product of groups.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is: suppose $C$ and $G$ are finitely generated, can we prove that so is $A$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've been trying to prove it by contradiction. Any suggestion?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks...&lt;/p&gt;&#xA;" OwnerUserId="2270" LastEditorUserId="664" LastEditorDisplayName="" LastEditDate="2010-10-02T19:27:43.247" LastActivityDate="2010-10-06T19:24:47.760" Title="Finite generation in amalgamated free products" Tags="&lt;group-theory&gt;&lt;category-theory&gt;&lt;algebraic-topology&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" />
  <row Id="5892" PostTypeId="1" CreationDate="2010-10-02T20:12:27.387" Score="8" ViewCount="506" Body="&lt;p&gt;As it is required for most students who wish to do a Ph.D in maths in the US to sit the GRE subject specific mathematics exam, I hope this question will be of interest to the mathematical community and will not be closed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Essentially, the exam was &quot;rescaled&quot; (made more difficult) in 2001 and I have only been able to find 2 past &quot;rescaled&quot; papers, one of which is available on the official website, the other available &lt;a href=&quot;http://www.math.ucsb.edu/mathclub/GRE/GRE9768.pdf&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;. Are other past papers available elsewhere? Thanks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: In actual fact, the exam was not made more difficult (see comments), so any past paper would be of interest.&lt;/p&gt;&#xA;" OwnerUserId="2272" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-10-05T09:06:28.970" LastActivityDate="2010-10-31T11:49:06.070" CommunityOwnedDate="2010-10-03T16:28:14.450" Title="GRE past papers" Tags="&lt;reference-request&gt;&lt;education&gt;" AnswerCount="1" CommentCount="7" FavoriteCount="6" />
  <row Id="5894" PostTypeId="2" ParentId="5874" CreationDate="2010-10-02T21:25:18.457" Score="3" ViewCount="" Body="&lt;p&gt;Suppose first that $E$ is contained in the closed interval $[a,b]$ and let $E' = [a,b] \setminus E$. By definition of Lebesgue measure, we can find an open set $G$ containing $E'$ such that $m(G) = m(E') + \varepsilon = (b - a) - m(E) + \varepsilon$, where $\varepsilon$ is small enough. Then $F = [a,b] \setminus G$ is a compact set contained in $E$ with measure $m(F) = m(E) - \varepsilon$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose we have chosen $\varepsilon$ small enough that $x \leq m(E)-\varepsilon = m(F)$. Your intermediate value argument proves the existence of some point $c \in [a,b]$ such that $m(F \cap [a,c]) = x$. By the Cantor-Bendixson Theorem, we have a decomposition $F \cap [a,c] = K \cup H$ where $K$ is perfect and $H$ is countable. Since $m(H) = 0$, it follows that $m(K) = m(F \cap [a,c]) = x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $E$ is unbounded, then pick $[a,b]$ such that $m(E \cap [a,b]) = m(E) - \delta$, where $\delta$ is small enough, and apply the above argument to $E \cap [a,b]$.&lt;/p&gt;&#xA;" OwnerUserId="25" LastActivityDate="2010-10-02T21:25:18.457" />
  <row Id="5895" PostTypeId="2" ParentId="5877" CreationDate="2010-10-02T21:28:50.797" Score="4" ViewCount="" Body="&lt;p&gt;One may quickly compute a representation of a prime $\rm\:p\equiv 1\ (mod\ 4)\:$ as a sum of two squares by employing the Euclidean GCD algorithm in $\rm\mathbb Z[\:i\:]\:$ and an algorithm for computing square roots $\rm\:(mod\ p)\:$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;THEOREM&lt;/strong&gt; $\ $  Let $\rm\ \: c = \sqrt{-1}\ \:(mod\ p)\ $ and $\rm\ gcd(p,\:i-c)\: = \: a+b\:i\ $.$\ $ Then $\rm\ p = a^2 + b^2\:$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof&lt;/strong&gt; $\ $ We shall show:  $\:$(1)$\ $ Representing $\rm\: p\:$ as a sum of squares is equivalent to finding a proper splitting $\rm\: p = \alpha\:\beta\ $ in $\rm\: \mathbb Z[\:i\:]\:$. $\ $(2)$\ $ Since $\rm\: \mathbb Z[\:i\:]\:$ has a Euclidean algorithm, a proper splitting of $\rm\: p\:$ can be computed by GCD with a suitable splitting of some multiple of $\rm\: p\ $. (3)$\ \:$ A suitable splitting of some multiple of $\rm\: p\:$ arises by factoring $\rm\ \: x^2+1\ \ (mod\ p\:)\:$,$\ $ i.e.$\ $ by computing $\rm\ \ \sqrt{-1}\ \ (mod\ p\:)\:$. $\:$  Below are the proofs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) $\:$ If $\rm\:\alpha|p\ $ properly then, conjugating, $\rm\:\alpha'|p\ $ properly. Multiplying them yields $\rm\:\alpha\:\alpha'|p^2\:$ properly in $\:\mathbb Z\:$.  But  in $\rm\mathbb{Z}\:$ the only proper factor$&gt;0$ of $\rm\:p^2\:$ is $\rm\: p\:$, thus $\rm\: p = \alpha\:\alpha' = (a+b\:i)(a-b\:i) = a^2 + b^2\:$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) $\ $ If $\rm\ p\ \gamma \:=\: \alpha\:\beta\ $  and $\rm\:p\:$ does not divide $\rm\:\alpha\:$ nor $\rm\:\beta\:$,$\ $ then $\rm\: gcd(\alpha,p)\:$ is a $\:$ &lt;em&gt;proper&lt;/em&gt;$\ $ factor of $\rm\:p\ $.&lt;br&gt;&#xA;Else $\rm\ gcd(\alpha,p) = p\:$ or $1$. If the gcd is $\rm p$ then $\rm\: p\:|\:\alpha\:$ contra hypothesis.  Otherwise, if $\rm\ gcd(\alpha,p)=1\ $ then by Euclid's Lemma, $\rm\ \alpha\:|\:p\:\gamma \ \Rightarrow\ \alpha\:|\:\gamma\:$,$\ $ so $\rm\  \gamma/\alpha = \beta/p\ \Rightarrow\ p\:|\:\beta\:$,$\ $ again contra hypothesis. Note: generally, in rings, GCDs are unique only up to unit multiples. Here the units are $\rm\ i^n = \pm 1,\:\pm i\:$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(3) $\rm\ \ x^2 + 1 \ =\ (x-c)\:(x+c) + p\  f(x)\ $ in $\rm\mathbb\ Z[x]\ \: \Rightarrow\:  -p\ f(i)\ =\ (i-c)\:(i+c)\ $ in $\rm\mathbb\ Z[\:i\:]\ \ $ by evaluation at $\rm\: x = i\:$. This splitting is suitable to split $\rm\:p\:$ by (2) since $\rm\:p\:$ doesn't divide $\rm\: i\pm c\ \:$ in $\rm\ \: Z[\:i\:]\:$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;REMARK&lt;/strong&gt; $\ $ There are many variations on the Euclidean algorithm in $\rm\mathbb \ Z[\:i\:]\ $ that are employed in practice, e.g. employing continued fractions, binary quadratic forms, etc. Also, there are also at least a few algorithms for computing sqrts $\rm\:(mod\ p)\:$, e.g. by factoring polynomials $\rm\:(mod\ p)\:$, by elliptic curves (Schoof), and and algorithm of Tonelli and Shanks. For much further information see Henri Cohen's book &quot;A course in computational algebraic number theory&quot;. Here is an excerpt on the beautiful algorithm of Cornacchia:&#xA;&lt;img src=&quot;http://i.imgur.com/fG9OE.jpg&quot; alt=&quot;alt text&quot;&gt;&#xA;&lt;img src=&quot;http://i.imgur.com/rdIY5.jpg&quot; alt=&quot;alt text&quot;&gt;&#xA;&lt;img src=&quot;http://i.imgur.com/FjAUY.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-10-04T05:46:51.930" LastActivityDate="2010-10-04T05:46:51.930" />
  <row Id="5896" PostTypeId="1" AcceptedAnswerId="5903" CreationDate="2010-10-02T22:17:43.243" Score="6" ViewCount="296" Body="&lt;p&gt;$\sum_{n=1}^{\infty} \frac{\varphi(n)}{n}$ where $\varphi(n)$ is 1 if the variable $\text n$ has the number $\text 7$ in its typical base-$\text10$ representation, and $\text0$ otherwise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am supposed to find out if this series converges or diverges. I think it diverges, and here is why.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can see that there is a series whose partial sums are always below our series, but which diverges. Compare some of the terms of each sequence&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{1}{7} &gt; \frac{1}{8}$&lt;br/&gt;&#xA;$\frac{1}{70} &gt; \frac{1}{80}$&lt;br/&gt;&#xA;$\frac{1}{71} &gt; \frac{1}{80}$&lt;br/&gt;&#xA;$\frac{1}{72} &gt; \frac{1}{80}$&lt;br/&gt;&#xA;$\text ... $&lt;br/&gt;&#xA;$\frac{1}{79} &gt; \frac{1}{80}$&lt;br/&gt;&#xA;$\text ... $&lt;br/&gt;&#xA;$\frac{1}{700} &gt; \frac{1}{800}$&lt;br/&gt;&#xA;$\text ... $&lt;br/&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And continue in this way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously some terms are left out of the sequence on the left, which is fine since our sequence of terms on the left is already greater than the right side. Notice the right side can be grouped into&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{1}{8} + \frac{1}{8} + ... $ because we will have $10$ $\frac{1}{80}$s, $100$ $\frac{1}{800}$s, etc etc. Thus we are adding up infinitely many 1/8s. This is similar to the idea of the divergence of the harmonic series. So, my conclusion is that it diverges. A bunch of other students in my real analysis class have come to the conclusion that is, in fact, convergent, and launched into a detailed verbal explanation about comparison with a geometric series that I couldn't follow without seeing their work. Is my reasoning, like they suspect, flawed? I can't see how.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sorry about the poor format, I'm new to TeX and couldn't figure out how to format a piecewise function (it was telling me a my \left delimiter wasn't recognized).&lt;/p&gt;&#xA;" LastActivityDate="2010-10-03T23:13:03.293" Title="An interesting series." Tags="&lt;real-analysis&gt;&lt;sequences-and-series&gt;" AnswerCount="3" CommentCount="5" FavoriteCount="1" />
  <row Id="5897" PostTypeId="2" ParentId="5879" CreationDate="2010-10-02T22:43:13.530" Score="1" ViewCount="" Body="&lt;p&gt;Looks okay to me.  The first graph is two disjoint copies of C_3 whereas the second graph is C_6.  They are not isomorphic and they receive a different canonical label (which is correct).  And there's no other 2-regular simple graphs on 6 vertices.  The permutation listed is applied to the rows and columns of the matrix.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there anything more specific you want to clarify?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could also compare your output with that of &lt;a href=&quot;http://cs.anu.edu.au/~bdm/nauty/&quot; rel=&quot;nofollow&quot;&gt;nauty&lt;/a&gt;.  It also has a function &lt;em&gt;geng&lt;/em&gt; that can be used to generate graphs with various properties.&lt;/p&gt;&#xA;" OwnerUserId="139" LastActivityDate="2010-10-02T22:43:13.530" CommentCount="2" />
  <row Id="5898" PostTypeId="2" ParentId="5896" CreationDate="2010-10-02T23:11:36.427" Score="3" ViewCount="" Body="&lt;p&gt;Your argument seems fine to me...  so I'll give the argument that popped into my head when I read the question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sum phi(n)/n for n congruent to 7 (mod 10) and multiply by 10 (which does not affect divergence/convergence).  Note that&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;10/7  &amp;gt; 1/7 +1/8 +...+1/16&#xA;10/17 &amp;gt; 1/17+1/18+...+1/26&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;and so on.  The result is &quot;larger&quot; than the harmonic series minus the first six terms (which diverges).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Although, I actually prefer the OP's proof since it's self-contained (i.e. doesn't require the harmonic series).&lt;/p&gt;&#xA;" OwnerUserId="139" LastActivityDate="2010-10-02T23:11:36.427" CommentCount="1" />
  <row Id="5899" PostTypeId="1" AcceptedAnswerId="5907" CreationDate="2010-10-02T23:26:08.467" Score="1" ViewCount="99" Body="&lt;p&gt;If the order of a group is pqr with p q r primes, then there exist three elements A B C with order p q r. Is it possible that the order of the subgroup generated by A and B has the order of pqr ?&lt;/p&gt;&#xA;" OwnerUserId="2177" LastActivityDate="2010-10-03T03:32:53.533" Title="a question about finite group" Tags="&lt;group-theory&gt;" AnswerCount="1" />
  <row Id="5900" PostTypeId="2" ParentId="5837" CreationDate="2010-10-03T00:01:40.123" Score="3" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://inkscape.org/&quot; rel=&quot;nofollow&quot;&gt;Inkscape&lt;/a&gt; is a good open source drawing package, and it &lt;a href=&quot;http://www.johndcook.com/blog/2009/12/22/including-latex-in-inkscape/&quot; rel=&quot;nofollow&quot;&gt;integrates with LaTeX&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="247" LastActivityDate="2010-10-03T00:01:40.123" />
  <row Id="5901" PostTypeId="1" AcceptedAnswerId="5920" CreationDate="2010-10-03T00:09:03.130" Score="6" ViewCount="263" Body="&lt;p&gt;Suppose $V, V'$ are subspaces of dimension $d$ of a vector space $X$. Then there is a subspace $W$ of $X$ of codimension $d$ such that $W \cap V = W \cap V' = { 0 }$. &#xA;This can be proved by choosing an explicit basis for $X$ which contains a basis for $V$ and a basis for $V'$ and a basis for $V \cap V'$.  On the other hand, there should be a nice way to do this without choosing a basis. Can anyone explain this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Disclosure: This came up when I was doing a homework problem. However, I'm just going to use the non-basis-free approach when I write up my answer.&lt;/p&gt;&#xA;" OwnerUserId="536" LastActivityDate="2010-10-03T06:34:25.707" Title="How to show that you can find two subspaces that don't intersect" Tags="&lt;linear-algebra&gt;" AnswerCount="2" CommentCount="6" FavoriteCount="2" />
  <row Id="5902" PostTypeId="1" AcceptedAnswerId="5912" CreationDate="2010-10-03T01:42:35.153" Score="2" ViewCount="200" Body="&lt;p&gt;I was wondering what are the differences and relations: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;between geometry and topology; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;between differential geometry and differential topology;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;between algebraic geometry and algebraic topology?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are they studying different objects? Such as different mathematical structures/spaces? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;if they study the same object, but study different aspects/properties of the same object?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reading their wikipedia pages really confuses me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks and regards!&lt;/p&gt;&#xA;" OwnerUserId="1281" LastActivityDate="2010-10-03T05:12:38.343" Title="geometry and topology" Tags="&lt;geometry&gt;&lt;topology&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="5903" PostTypeId="2" ParentId="5896" CreationDate="2010-10-03T02:23:50.617" Score="4" ViewCount="" Body="&lt;p&gt;Yes your argument seems fine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another argument is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Divide the integers into blocks of 10&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$[1 \dots 10] [11 \dots 20] [21 \dots 30] \dots$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each block will have an number with digit $7$ in it and so the series has sum at least&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{1}{10} + \frac{1}{20} + \dots + \frac{1}{10n} + \dots $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ = \frac{1}{10}(1 + \frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{n} + \dots)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is divergent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yet another argument which uses the following useful result (the main reason for posting my answer):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $A = \{a_i\}$ is a sequence of natural numbers such that $\sum_{i=1}^{\infty} \frac{1}{a_i}$ converges then the &lt;a href=&quot;http://en.wikipedia.org/wiki/Natural_density&quot; rel=&quot;nofollow&quot;&gt;natural density&lt;/a&gt; of $A$ is zero. (This is an exercise in Ivan Niven's book on Number theory and for a proof see here: &lt;a href=&quot;http://math.stackexchange.com/questions/5932/theorem-on-natural-density/5967#5967&quot;&gt;http://math.stackexchange.com/questions/5932/theorem-on-natural-density/5967#5967&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In our case the density of the numbers under consideration is at least $\frac{1}{10}$ and thus the sum cannot be convergent.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-10-03T23:13:03.293" LastActivityDate="2010-10-03T23:13:03.293" CommentCount="0" />
  <row Id="5904" PostTypeId="1" CreationDate="2010-10-03T02:35:58.657" Score="2" ViewCount="150" Body="&lt;p&gt;Does the basis of an Euclidean space have to be ordered by definition? Or can be left unordered?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was also wondering about what is the morphism (i.e. the mapping that can preserve all the structures) on Euclidean spaces? Is it &lt;a href=&quot;http://planetmath.org/encyclopedia/Rotate.html&quot; rel=&quot;nofollow&quot;&gt;Euclidean transformation&lt;/a&gt; (rigid transformation), consisting of rotation, translation and reflection? Or the reflection is not part of it, because the basis of a Euclidean space must be ordered.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks and regards!&lt;/p&gt;&#xA;" OwnerUserId="1281" LastEditorUserId="1281" LastEditorDisplayName="" LastEditDate="2010-10-03T04:19:27.420" LastActivityDate="2010-10-03T12:06:21.923" Title="Must Basis of an Euclidean Space Be Ordered" Tags="&lt;euclidean-geometry&gt;" AnswerCount="4" CommentCount="2" FavoriteCount="2" />
  <row Id="5905" PostTypeId="2" ParentId="5902" CreationDate="2010-10-03T02:55:27.563" Score="3" ViewCount="" Body="&lt;p&gt;Your question is quite vague.  There are two parallel dynamics that go on when you learn mathematics.  On one level, mathematics is extremely specific so as you learn one subject in detail it appears as if all you know is that subject and there appears to be no relation to any other subject. But once you get to understand something you start noticing patterns -- standard examples that appear in many different fields, but in different guises, standard constructions many of which fit into categorical or other natural frameworks that are beyond the specifics of one field.  So to some extent there are broad unifying themes between subjects in mathematics.  In that regard there's many connections between subjects labelled by names where you combine two of the words from the set {geometry(ic), topology, algebra(ic)}.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But at its most coarse, primitive level, there are some big differences.  Algebraic geometry is about the study of algebraic varieties -- solutions to things like polynomial equations. Geometric topology is largely about the study of manifolds -- which are like varieties but with no singularities, i.e. homogeneous objects.  Algebraic topology you could say is more about the study of homotopy-type or &quot;holes in spaces&quot;.  These are all inaccurate descriptions as in some sense subjects definitions are shaped by their histories.  I'd say for example that Algebraic topology is more defined by the nature of the tools it employs. While geometric topology is more motivated by objects it wants to prove theorems about.  That can seem like an artificial distinction, too, since isn't a &quot;tool&quot; an &quot;object&quot;?  Geometric topology is very much motivated by low-dimensional phenomena -- and the very notion of low-dimensional phenomena being special is due to the existence of a big tool called the Whitney Trick, which allows one to readily convert certain problems in manifold theory into (sometimes quite complicated) algebraic problems.  The thing is the Whitney trick fails in dimensions $4$ and lower.  In that regard, geometric topology has some characteristics of a grumpy old man who is really set on figuring out something specific. And algebraic topology in some sense has more of the air of the person that follows the natural lay-of-the-land from some formal perspective. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, it's much more common in the geometric topology community to go to a talk that's an illustration of an idea, or a problem illustrated entirely by examples: generating tables of knots or a census of manifolds, or testing a hypothesis by computer experiement, etc.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But after all that is said, there are many links between these fields and others, so it frequently difficult to disambiguate them except in rather pat, artificial ways.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure if that helps at all.  But that's a generic 1st response. &lt;/p&gt;&#xA;" OwnerUserId="642" LastActivityDate="2010-10-03T02:55:27.563" CommentCount="1" />
  <row Id="5906" PostTypeId="1" AcceptedAnswerId="5940" CreationDate="2010-10-03T02:58:34.453" Score="5" ViewCount="166" Body="&lt;p&gt;Is a triangle (its sides and the region enclosed by its sides) in a 2D Euclidean space $\mathbb{E}^2$ a manifold? I was thinking to use the identity mapping as its charts, but for each point on the sides of the triangle, there is no neighborhood of it can be mapped to an open subset in $\mathbb{E}^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="1281" LastActivityDate="2010-10-03T19:16:11.357" Title="is triangle a manifold ?" Tags="&lt;manifold&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="2" />
  <row Id="5907" PostTypeId="2" ParentId="5899" CreationDate="2010-10-03T03:32:53.533" Score="4" ViewCount="" Body="&lt;p&gt;Yes. Consider the group of 2x2 matrices whose entries come from the field of 7 elements.  Take A = [ -1, 0 ; 0, 1 ], B = [ 2, 1 ; 0, 1 ], C = [ 1, 1 ; 0, 1 ].  Then A,B,C generate a group of order 42=2*3*7 called Hol(7) = AGL(1,7).  A has order 2, B has order 3, C has order 7, but the subgroup generated by A and B also has order 42.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is basically the same trick used to generate the the non-abelian group of order 6 using two elements of order 2.  You can't take them from the same Sylow 2-subgroup, but if you choose from two Sylow 2-subgroups things work.  For p=2, q=3, r=7, we cannot choose two from the same Hall {p,q}-subgroup, lest we get a subgroup of order pq, so we choose from two different Hall 6-subgroups, and generate the whole group.&lt;/p&gt;&#xA;" OwnerUserId="583" LastActivityDate="2010-10-03T03:32:53.533" />
  <row Id="5909" PostTypeId="1" AcceptedAnswerId="5917" CreationDate="2010-10-03T04:01:19.087" Score="2" ViewCount="205" Body="&lt;p&gt;Consider a simple random walk on a undirected, connected graph. This is the random walk which, at every time step, moves to a random neighbor, with all neighbors being equally likely. Lets assume every node has a self-loop to avoid issues associated with periodicity. Then, the following (surprising to me) fact is true: &lt;b&gt;under the stationary distribution, every edge is traversed with the same probability&lt;/b&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm trying to get some intuition for why this holds, and I am wondering whether someone &#xA;can provide an explanation for this phenomenon. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Naturally, I know the standard proof which&#xA;observes that $\pi_i = d_{i}/\sum_k d_k$ satisfies the equations for the stationary distribution, and so $\pi_i p_{ij} = 1/\sum_k d_k$ whenever the edge $(i,j)$ is present in the graph. This proof, however, feels more like a lucky calculation to me than an explanation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the fact in boldface is false for directed graphs. Its also false for the corresponding two-step chain, i.e. the Markov chain with probability transition matrix $P^2$, where $P$ is the transition matrix of the simple random walk. &lt;/p&gt;&#xA;" OwnerUserId="2120" LastEditorUserId="2120" LastEditorDisplayName="" LastEditDate="2010-10-03T04:15:35.890" LastActivityDate="2010-10-03T06:06:45.550" Title="simple random walks on undirected graphs" Tags="&lt;probability&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" />
  <row Id="5910" PostTypeId="2" ParentId="5904" CreationDate="2010-10-03T04:02:30.953" Score="4" ViewCount="" Body="&lt;p&gt;Bases must be ordered, otherwise you couldn't speak of coordinates as &lt;em&gt;ordered&lt;/em&gt; tuples. And this could make things pretty messy. For instance, if the standard basis for $\mathbb{R}^2$ could be either $e_1, e_2$ or $e_2, e_1$ simultaneously, then it would be difficult to speak of &quot;the point of coordinates $(2,1)$&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reflexions send bases to bases, because they are isomorphisms. But this doesn't mean they necessarily send a particular base to the &lt;em&gt;same&lt;/em&gt; particular base. So they can change the order of the vectors of the basis. They don't preserve &lt;a href=&quot;http://en.wikipedia.org/wiki/Orientation_%28mathematics%29&quot; rel=&quot;nofollow&quot;&gt;orientation&lt;/a&gt;, but this is an extra piece of structure, not included in the definition of Euclidean space.&lt;/p&gt;&#xA;" OwnerUserId="664" LastEditorUserId="664" LastEditorDisplayName="" LastEditDate="2010-10-03T10:34:30.693" LastActivityDate="2010-10-03T10:34:30.693" CommentCount="2" />
  <row Id="5911" PostTypeId="2" ParentId="5896" CreationDate="2010-10-03T04:11:03.563" Score="3" ViewCount="" Body="&lt;p&gt;Here's another proof that your sum diverges. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the sum $\sum_{n-1}^\infty (1-\phi(n))/n$.  This is the sum of the reciprocals of integers which &lt;i&gt;don't&lt;/i&gt; have a 7 in their decimal expansion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The number of integers $n$ with $1-\phi(n)=1$ and $1 \le n &amp;lt; 10^k$ is $9^k - 1$. (We can choose each of $k$ decimal digits to be anything but 7, except we don't want to choose all zeroes.) Thus the numbers of integers $n$ with $1-\phi(n) = 1$ and $10^{k-1} \le n &amp;lt; 10^k$ is $(9^k - 1) - (9^{k-1}-1) = 8 \times 9^{k-1}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore&#xA;$$ S_k = \sum_{n=10^{k-1}}^{10^k - 1} {1-\phi(n) \over n} $$&#xA;has $8 \times 9^{k-1}$ nonzero terms; they are each at most $10^{k-1}$. So $S_k \le 8 \times (9/10)^{k-1}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The infinite sum is then&#xA;$$ \sum_{k=1}^\infty S_k \le 8 \times \sum_{k=1}^\infty (9/10)^{k-1} = 8 \times 10 = 80 $$&#xA;and in particular it's finite.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the harmonic series $\sum_{n=1}^\infty 1/n$ diverges; removing terms whose sum converges (like the sum that I just showed to converge) won't change that. So your series diverges.&lt;/p&gt;&#xA;" OwnerUserId="173" LastEditorUserId="173" LastEditorDisplayName="" LastEditDate="2010-10-03T04:57:16.833" LastActivityDate="2010-10-03T04:57:16.833" CommentCount="1" />
  <row Id="5912" PostTypeId="2" ParentId="5902" CreationDate="2010-10-03T04:24:49.217" Score="6" ViewCount="" Body="&lt;p&gt;As a supplement to Ryan's answer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Differential geometry typically studies Riemannian metrics on manifolds, and properties of them.  A typical differential geometry result is the &lt;em&gt;sphere theorem&lt;/em&gt;, stating that if $M$ is a closed manifold equipped with a Riemannian metric for which the sectional curvatures lie in the half-open interval $(1/4, 1]$, then $M$ is a sphere.  Note that the basic object is a manifold equipped with a Riemannian metric (a &lt;em&gt;Riemannian manifold&lt;/em&gt;), and the &lt;em&gt;curvature&lt;/em&gt; of the metric plays a key role in the statement of the theorem.  These are typical features of problems/theorems in differential geometry.  Note though that the conclusion of the theorem involves a statement about the topology of $M$; so there is certainly overlap between differential geometry and the concerns of topology.  (One might say that the sphere theorem is a global result, using geometric hypotheses to draw topological conclusions.  One can also have local results, in which topology plays no role in the hypothesis or conclusions: e.g. that a Riemanninan manifold with everywhere zero curvature is locally isometric to Euclidean space; one can also have global results that begin with topology and conclude with geometry: e.g. that any compact orientable surface of genus 2 or higher admits a Riemannian metric with constant curvature $-1$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Differential topology refers to results about manifolds that are more directly topological, and don't refer to metric structures.  The generalized Poincare conjecture (that a closed manifold that is homotopic to a sphere is homeomorphic to one) is an example; another simpler example is Ehresmann's theorem, stating that a submersion between closed manifolds (or more generally, any &lt;em&gt;proper&lt;/em&gt; submersion) is a fibre bundle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, these distinctions can be subtle, and may not always be well-defined, but a typical distinction bewteen geometry and topology in general (and which is borne out in the&#xA;preceding discussion) is that geometry studies metric properties of spaces, while topology studies questions which don't involve metric notions (it is the study of &lt;em&gt;pure shape&lt;/em&gt;, if you like; the old name &lt;em&gt;analysis situs&lt;/em&gt; also sheds some light on the meaning of topology).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One caveat is that classically, Euclidean geometry branched not only into other geometries&#xA;such as hyperbolic geometry (this branching being a precursor to Riemann's introduction of the general notions of Riemannian manifold and curvature), but gave rise to another branch known as projective geometry.  In projective geometry, metric notions of distance and angle aren't studied (because they are not preserved by projective transformations), but notions such as being a straight line, or being a conic section, are.  &lt;em&gt;Algebraic geometry&lt;/em&gt; is the modern subject which developed out of projective geometry (among other sources; see &lt;a href=&quot;http://math.stackexchange.com/questions/5167/group-law-for-an-elliptic-curve/5176#5176&quot;&gt;this answer&lt;/a&gt; for a discussion of a quite different problem --- computing elliptic integrals --- which was another historical precursor to algebraic geometry).   In algebraic geometry one studies &lt;em&gt;varieties&lt;/em&gt;, which are solution sets to polynomial equations; thus in its elementary form it feels a lot like what is called &lt;em&gt;analytic geomery&lt;/em&gt; in high-school,&#xA;namely studying figure in the plane, or in space, cut out by equations in the coordinates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is this &lt;em&gt;geometry&lt;/em&gt; (as opposed to &lt;em&gt;topology&lt;/em&gt;, say)?  Because it turns out that when the functions one is using to cut out figures, or describe maps between figures, are restricted to be polynomial, the objects one obtains are quite rigid, in a way very similar to the way more traditional Euclidean geometry figures are rigid.  So one has the sensation of doing geometry, rather than topology.  (In topology, by contrast, things feel rather fluid, since one is allowed to deform objects in fairly extreme ways without changing their essential topological nature.)  And in fact it turns out that there are deeper connections between algebraic and metric geometry: for example, for a compact orientable surface of genus at least 2, it turns out that the possible ways of realizing this surface as an algebraic variety over the complex numbers are in a natural bijection with the possible choices of a constant curvature -1 metric on the surface.  A more recent example is given by the Calabi conjecture proved by Yau.&lt;/p&gt;&#xA;" OwnerUserId="221" LastEditorUserId="221" LastEditorDisplayName="" LastEditDate="2010-10-03T05:12:38.343" LastActivityDate="2010-10-03T05:12:38.343" CommentCount="2" />
  <row Id="5913" PostTypeId="2" ParentId="5901" CreationDate="2010-10-03T04:51:48.407" Score="1" ViewCount="" Body="&lt;p&gt;So let $G$ be the space of all co-dimension d subspaces. ($G$ is&#xA;naturally a manifold). I claim that if $W$ is any dimension $d$ subspace,&#xA;the set of $X$ s.t. $X \cap W = 0$ is an open, dense subset of $G$.&#xA;Open is easy. If you have $X_i$ each with non-trivial intersection with&#xA;W and $X_i \to X$ then you can pick lines $L_i$ in $W$ intersect $X_i$ and find&#xA;some limit (perhaps of a subsequence) so that $L_i \to L$. Then $L$ is in&#xA;$W \cap X$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To show dense is also not hard. You need to show that if $X$ intersects&#xA;$W$, then you can change $X$ by a little so that it no longer does. This&#xA;really isn't hard but it is annoying to do without ever picking a&#xA;basis of anything. How about this. You can think of $X$ as the image&#xA;of a map $Y \to Z$ ($Y$ is a space of dim $n-d$, $Z$ is your big space). $X$&#xA;intersects $W$ trivially iff the map $Y \to Z/W$ is an injection. Pick&#xA;some $U$ so that $Z = U + V$. Then we can think of our map as $Y \to U+V$,&#xA;and we want the map $Y \to U$ to be injective. But $Y$ and  $U$ have the same&#xA;dimension and it is easy to modify a map between such spaces by&#xA;epsilon to make it a bijection (for example add a small multiple of&#xA;some particular bijection).&lt;/p&gt;&#xA;" LastActivityDate="2010-10-03T04:51:48.407" CommentCount="1" />
  <row Id="5914" PostTypeId="1" AcceptedAnswerId="5988" CreationDate="2010-10-03T05:18:39.663" Score="4" ViewCount="253" Body="&lt;p&gt;How to prove that the only integral solutions to the equation $$y^{2}=x^{3}-1$$ is $x=1, y=0$. &#xA;I rewrote the equation as $y^{2}+1=x^{3}$ and then we can factorize $y^{2}+1$ as $$y^{2}+1 = (y+i) \cdot (y-i)$$ in $\mathbb{Z}[i]$. Next i claim that the factor's $y+i$ and $y-i$ are co-prime. But i am not able to show this. Any help would be useful. Moreover, i would also like to see different proofs of this question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Extending&lt;/strong&gt; Consider the equation $$y^{a}=x^{b}-1$$ where $a,b \in \mathbb{Z}$ and $(a,b)=1$ and $a &amp;lt; b$. Then is there any result which states about the nature of the solution to this equation.&lt;/p&gt;&#xA;" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-10-03T05:30:58.677" LastActivityDate="2010-10-04T07:07:14.327" Title="Integral solutions to $y^{2}=x^{3}-1$." Tags="&lt;number-theory&gt;&lt;algebraic-number-theory&gt;&lt;diophantine-equations&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="2" />
  <row Id="5915" PostTypeId="2" ParentId="5914" CreationDate="2010-10-03T05:38:24.663" Score="3" ViewCount="" Body="&lt;p&gt;If $x$ is even then $x^3$ is divisible by 4, so $x^3-1 \equiv 3 \pmod 4$ and this cannot be a square. Thus $x$ is odd and $y$ is even.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if a prime $p$ divides both $y+i$ and $y-i$ then it divides their difference $2i$. Thus $p=2$ (up to units), but then $p$ divides $y$. That's impossible since it divides $y+i$ as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: I was too hasty in writing this, as pointed out by Moron. The prime 2 ramifies in $\mathbb{Z}[i]$, and is (up to units) the square of $1+i$. Sorry for the confusion.&lt;/p&gt;&#xA;" OwnerUserId="308" LastEditorUserId="308" LastEditorDisplayName="" LastEditDate="2010-10-04T07:07:14.327" LastActivityDate="2010-10-04T07:07:14.327" />
  <row Id="5916" PostTypeId="2" ParentId="5904" CreationDate="2010-10-03T05:38:36.380" Score="2" ViewCount="" Body="&lt;p&gt;Orientation does not play any role in Euclidean geometry. So it is most natural to define a Euclidean space as a finite-dimensional vector space with an inner product. Lengths, distances and angles are then defined in terms of this inner product. We are interested in transformations which preserve lengths and angles, which are precisely the orthogonal transformations &lt;em&gt;including reflections&lt;/em&gt;. There is no need to explicitly choose a basis at any point.&lt;/p&gt;&#xA;" OwnerUserId="1195" LastActivityDate="2010-10-03T05:38:36.380" />
  <row Id="5917" PostTypeId="2" ParentId="5909" CreationDate="2010-10-03T06:06:45.550" Score="2" ViewCount="" Body="&lt;p&gt;First, think of each edge as a unit-length path connecting two vertices.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, imagine that instead of a random walk on the vertices, we consider a continuous random walk that moves along edges at unit speed.  Every time we arrive at a vertex, we randomly decide which edge coming out of the vertex to traverse next.  Note that the sequence of vertices and edges that we traverse will be the same as in the discrete-time random walk.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now imagine that we fill the graph with a very large number of particles, each of which is performing a continuous walk.  We start the particles at random times in the interval $[0,1]$, so that each vertex continuously has particles going through it, and then wait until we are in the steady state.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At least for me, it seems obvious at this point that the graph will be uniformly filled with particles, with each edge having the same density.  Every vertex will have particles coming in from all of its edges at the same rate, and these particles will randomly distribute themselves among the outgoing edges.  This situation is clearly stable, so it must be the steady state.&lt;/p&gt;&#xA;" OwnerUserId="1726" LastActivityDate="2010-10-03T06:06:45.550" CommentCount="3" />
  <row Id="5918" PostTypeId="2" ParentId="5791" CreationDate="2010-10-03T06:17:21.190" Score="1" ViewCount="" Body="&lt;p&gt;It's not very clear what you're trying to ask, but what you're describing looks like a region &lt;a href=&quot;http://en.wikipedia.org/wiki/Quadtree&quot; rel=&quot;nofollow&quot;&gt;quadtree&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each step in your IFS corresponds to a choice of child node (or subquadrant), and the tree path of any given point is the same as the list of numeric input steps in your question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It does not really matter whether you store the point as an arbitrary-precision coordinate pair or tree traversal: the binary digits required to store its coordinates correspond exactly to its quadtree traversal path.&lt;/p&gt;&#xA;" OwnerUserId="2273" LastEditorUserId="2273" LastEditorDisplayName="" LastEditDate="2010-10-03T06:37:14.780" LastActivityDate="2010-10-03T06:37:14.780" />
  <row Id="5919" PostTypeId="1" AcceptedAnswerId="5976" CreationDate="2010-10-03T06:30:08.717" Score="8" ViewCount="184" Body="&lt;p&gt;So if I'm given $H$, when I can conclude there is a group $G$ such that $H\cong G/Z(G)$? It's easy to show that non-trivial cyclic groups are &lt;em&gt;not&lt;/em&gt; of this form. More generally, any group with the property that all finitely generated subgroups are cyclic (e.g. $\mathbb{Q},\mathbb{Q}/\mathbb{Z}$) cannot be of this form. Are there are any others?&lt;/p&gt;&#xA;" OwnerUserId="546" LastActivityDate="2010-10-04T01:43:30.420" Title="When can a (finite) group be written as the quotient of some other group by its center?" Tags="&lt;group-theory&gt;" AnswerCount="1" FavoriteCount="2" />
  <row Id="5920" PostTypeId="2" ParentId="5901" CreationDate="2010-10-03T06:34:25.707" Score="3" ViewCount="" Body="&lt;p&gt;Suppose first that $V+ V' = X$.  Then the natural map $X/(V\cap V') \to X/V \times X/V'$&#xA;is an isomorphism.  Choose a subspace of the target of this isomorphism&#xA;that projects isomorphically onto each factor (i.e. the graph of an isomorphism between the two factors; such an isomorphism exists since the two factors have the same dimension).&#xA;Its preimage under the natural map is a subspace of $X/(V\cap V')$ which&#xA;meets each of $V$ and $V'$ trivially.  Now choose any  subspace $W$&#xA;of $X$ that projects isomorphically onto this preimage; this is then a subspace of&#xA;$X$ that maps isomorphically onto each of $X/V$ and $X/V'$, and hence is a codimension $d$&#xA;subspace with the desired property.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general (i.e. if $V + V' \neq X$) then the above gives a codimension $d$ subspace $W'$ of &#xA;$V + V'$ meeting $V$ and $V'$ trivially.   Choose $W''$ to be any subspace of $X$ which&#xA;maps isomorphically onto $X/(V + V')$.  The sum $W' + W''$ is then a codimension $d$ subspace&#xA;of $X$ meeting $V$ and $V'$ trivially.&lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2010-10-03T06:34:25.707" CommentCount="1" />
  <row Id="5921" PostTypeId="1" CreationDate="2010-10-03T06:58:10.770" Score="3" ViewCount="190" Body="&lt;p&gt;F &lt;strong&gt;y&lt;/strong&gt;(w) &gt;= F &lt;strong&gt;x&lt;/strong&gt;(w), for all w.  Prove P{&lt;strong&gt;y&lt;/strong&gt;&amp;lt;&lt;strong&gt;x&lt;/strong&gt;} &gt;= 0.5&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I had this question on the exam and boy did it stump me.  It was simple to understand why it was true, but I tried several different paths and got nowhere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any ideas?&lt;/p&gt;&#xA;" OwnerUserId="377" LastEditorUserId="377" LastEditorDisplayName="" LastEditDate="2010-10-03T17:36:25.637" LastActivityDate="2011-02-23T13:39:31.353" Title="For all w, CDF of y at w is greater than the CDF of x at w. Prove that the probability ( y &lt; x ) &gt;= 0.5" Tags="&lt;probability-theory&gt;&lt;proof-strategy&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="1" />
  <row Id="5922" PostTypeId="1" CreationDate="2010-10-03T07:16:08.217" Score="1" ViewCount="164" Body="&lt;p&gt;So I wanted to figure out a meaningful probability for the card deck order problem.  I chose arbitrary values in an attempt to show the magnitude of permutations:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If you and everyone on the Earth right&#xA;  now shuffled a million randomized&#xA;  decks every millisecond that universe&#xA;  has existed, the probability that any&#xA;  two decks at one moment were the same would be:&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What is the probability?&lt;/p&gt;&#xA;" OwnerUserId="377" LastActivityDate="2010-10-04T05:40:19.807" Title="Probability of card decks being in the same order for N shufflers over X amount of time?" Tags="&lt;probability&gt;" AnswerCount="2" />
  <row Id="5923" PostTypeId="2" ParentId="5922" CreationDate="2010-10-03T07:18:02.393" Score="0" ViewCount="" Body="&lt;p&gt;Here is what I came up with and my assumptions.  Please point out any errors. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, I attempted to find the probability of 2 people having the same deck order of N people shuffling decks in one trial.  I could not figure out how to model it in a more standard way, so my leap of faith was that for N people, there were (N+N-1+N-2+...)=Nt.  For 6.7e9 people on Earth, that's 2.2445*^19 pairs of people.  The probability that one pair had the same deck order was (1/52!) = 1/8e67 = p1.  Since the events are not independent, I need to multiply by (52!)!/(52!-N).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next, I wanted the probability that any of them had a match.  So that would be 1 minus the probability that none of them had any matches:&lt;br&gt;&#xA;&lt;code&gt;(1 - (52!)!/(52!-N)*(p1)^0*(1 - p1)^(Nt)) = (1 - (52!)!/(52!-N)*(1 - p1)^(Nt)) = p2.&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, I wanted to do a million randomized trials every millisecond that the universe existed.  So that's 1.4e10 years * 365 days/year * 24 hours/day * 3600e3 ms / hour * 1e9 trials/ms = T = 4.5e29.  Again, I wanted at least one match, that would bring it to:&lt;br&gt;&#xA;&lt;code&gt;(1 - (1 - p2)^(T))&lt;/code&gt; = &#xA;&lt;code&gt;(1 - (1 - (1 - (52!)!/(52!-N)*(1 - p1)^(Nt)))^(T))&lt;/code&gt; =&lt;br&gt;&#xA;&lt;code&gt;(1-(1-(1-(52!)!/(52!-6.7e9)*(1-1/8e67)^(2.2445*^19)))^(4.5e29))&lt;/code&gt; = &lt;code&gt;6.31 × 10^-20&lt;/code&gt;&lt;/p&gt;&#xA;" OwnerUserId="377" LastEditorUserId="377" LastEditorDisplayName="" LastEditDate="2010-10-03T17:57:43.763" LastActivityDate="2010-10-03T17:57:43.763" CommentCount="5" />
  <row Id="5924" PostTypeId="2" ParentId="5904" CreationDate="2010-10-03T08:19:32.470" Score="3" ViewCount="" Body="&lt;h3&gt;On bases&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Bases do not have to be ordered. I respectfully present the following as a counterpoint to Agustí Roig's response.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First of all, a basis is simply a &lt;em&gt;set&lt;/em&gt; of vectors, by definition; and sets are unordered. If you wish to enumerate the vectors in a basis, you will of course do so in &lt;em&gt;some&lt;/em&gt; order, but this order is arbitrary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This arbitrary order of enumeration is the reason why we describe vectors as tuples. The order of the coefficients in a tuple matters &lt;em&gt;only in as much as it must be in agreement with the arbitrary order which was selected for the basis vectors&lt;/em&gt;. To wit: let &lt;strong&gt;x&lt;/strong&gt;, &lt;strong&gt;y&lt;/strong&gt; be two arbitrary linearly independent vectors. The tuple [&amp;thinsp;7&amp;nbsp;&amp;nbsp;5&amp;thinsp;] with respect to the (enumeration of the) basis &lt;strong&gt;v&lt;/strong&gt;&lt;sub&gt;1&lt;/sub&gt;&amp;nbsp;=&amp;nbsp;&lt;strong&gt;x&lt;/strong&gt;, &lt;strong&gt;v&lt;/strong&gt;&lt;sub&gt;2&lt;/sub&gt;&amp;nbsp;=&amp;nbsp;&lt;strong&gt;y&lt;/strong&gt; represents the same vector as the tuple [&amp;thinsp;5&amp;nbsp;&amp;nbsp;7&amp;thinsp;] with respect to the (&lt;em&gt;different&lt;/em&gt; enumeration of the &lt;em&gt;same&lt;/em&gt;) basis &lt;strong&gt;v&lt;/strong&gt;&lt;sub&gt;2&lt;/sub&gt;&amp;nbsp;=&amp;nbsp;&lt;strong&gt;x&lt;/strong&gt;, &lt;strong&gt;v&lt;/strong&gt;&lt;sub&gt;1&lt;/sub&gt;&amp;nbsp;=&amp;nbsp;&lt;strong&gt;y&lt;/strong&gt;. The tuple is just a &lt;em&gt;representation&lt;/em&gt; of a vector, relative to an &lt;em&gt;arbitrarily chosen&lt;/em&gt; order for the basis. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even more foundationally: &quot;tuples&quot; can be regarded as functions from the integers (&lt;em&gt;e.g.&lt;/em&gt; the indices '1', '2', etc.) to the reals, complex numbers, or whichever set you draw your coefficients from. So [&amp;thinsp;7&amp;nbsp;&amp;nbsp;5&amp;thinsp;] can be thought of as 'really being' the function mapping&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;'1'&amp;nbsp;$\mapsto$&amp;nbsp;7, &lt;br&gt;&#xA;  '2'&amp;nbsp;$\mapsto$&amp;nbsp;5.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Our enumeration of the basis has a similar role: choosing the enumeration &lt;strong&gt;v&lt;/strong&gt;&lt;sub&gt;1&lt;/sub&gt;&amp;nbsp;=&amp;nbsp;&lt;strong&gt;x&lt;/strong&gt;, &lt;strong&gt;v&lt;/strong&gt;&lt;sub&gt;2&lt;/sub&gt;&amp;nbsp;=&amp;nbsp;&lt;strong&gt;y&lt;/strong&gt; is equivalent to defining a mapping&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;'1'&amp;nbsp;$\mapsto$&amp;nbsp;&lt;strong&gt;x&lt;/strong&gt;, &lt;br&gt;&#xA;  '2'&amp;nbsp;$\mapsto$&amp;nbsp;&lt;strong&gt;y&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Why do I have '1' and '2' in quotes? Because they're just labels; any other labels would do just as well. For instance, I could replace '1' and '2' with the vectors &lt;strong&gt;x&lt;/strong&gt; and &lt;strong&gt;y&lt;/strong&gt; themselves. Then we could define vectors by coefficient functions such as&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;x&lt;/strong&gt;&amp;nbsp;$\mapsto$&amp;nbsp;7, &lt;br&gt;&#xA;  &lt;strong&gt;y&lt;/strong&gt;&amp;nbsp;$\mapsto$&amp;nbsp;5;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;that is, the coefficient 7 is associated to the vector &lt;strong&gt;x&lt;/strong&gt;, and the coefficient 5 is associated to &lt;strong&gt;y&lt;/strong&gt;, to represent the vector 7&lt;strong&gt;x&lt;/strong&gt;&amp;nbsp;+&amp;nbsp;5&lt;strong&gt;y&lt;/strong&gt;. This is what we really mean anyway; never mind any ordering of the vectors in your basis. With this, the arbitrary ordering of the basis disappears, leaving nothing but what it is we really mean by it all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So: when we order our bases, its only to make it easier to present things &amp;mdash; it is not part of the definition, or part of the structure of the objects we really care about.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;On morphisms&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;A morphism on a Euclidean space ought to preserve all of the structures of Euclidean geometry. In particular, it should map circles to other circles, to preserve the structures guaranteed by the &lt;a href=&quot;http://en.wikipedia.org/wiki/Euclidean_geometry#Axioms&quot; rel=&quot;nofollow&quot;&gt;third postulate&lt;/a&gt;; so it must be an &lt;em&gt;isothety&lt;/em&gt; (a rigid transformation up to scaling). This includes, but is not restricted to, the rigid transformations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you make any argument why the morphisms should consist &lt;em&gt;only&lt;/em&gt; of rigid transformations (excluding isothetic maps such as &lt;strong&gt;v&lt;/strong&gt;&amp;nbsp;$\mapsto$&amp;nbsp;2&lt;strong&gt;v&lt;/strong&gt;)?&lt;/p&gt;&#xA;" OwnerUserId="439" LastEditorUserId="439" LastEditorDisplayName="" LastEditDate="2010-10-03T12:06:21.923" LastActivityDate="2010-10-03T12:06:21.923" />
  <row Id="5926" PostTypeId="2" ParentId="5904" CreationDate="2010-10-03T10:23:33.440" Score="2" ViewCount="" Body="&lt;p&gt;This began as a reply to Jyotirmoy's comment on Agustí's answer, but it became way too long. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As I stated in my comment on Agustí's answer, some textbooks define a basis as a set, and I think this is wrong. Obviously, some people disagree (just read some of the other answers and comments). I'll try to explain why a basis should never be thought of as a set.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do the columns of the following matrix form a basis for $\mathbb{R}^2$? $$\begin{bmatrix}1 &amp;amp; 0 &amp;amp; 0 \\ 0 &amp;amp; 1 &amp;amp; 1\end{bmatrix}$$ The answer is &quot;yes&quot; if a basis is a set, but obviously the right answer is &quot;no.&quot; This makes it clear that whatever sort of collection to which we apply concepts like &quot;linear [in]dependence&quot; or &quot;basis&quot;, it must allow repeated vectors; otherwise, a lot of basic theorems from linear algebra would require extra clauses to deal with matrices with repeated rows or columns. This rules out sets as our collections of choice. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To find out what our collection of choice should be, let's revisit the definition of a basic concept like linear dependence (in somewhat imprecise &quot;collection-agnostic&quot; language (whose meaning should nevertheless be clear)): &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The vectors $v_1,\ldots,v_n$ are linearly dependent if there are scalars $c_1,\ldots,c_n$, not all zero, such that $c_1v_1+\cdots+c_nv_n=0$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It's obvious here that the natural object to apply such a concept to is an indexed tuple of vectors (indexed by $\{1,\ldots,n\}$ above, but we could formulate the concept equally well for an arbitrary finite index set): we begin with an indexed tuple of vectors and then consider a tuple of scalars indexed by the same set, and then we apply a function to these two tuples that is now completely specified (taking the linear combination). We could have started with a completely unindexed, unordered collection of vectors with repetition allowed (i. e., a &quot;multiset&quot;), but then to consider a linear combination, we would have had to choose some indexing of the multiset first and then to consider some scalars indexed by our arbitrary index set in order to form a linear combination. The core point is that the natural input to the &quot;linear combination&quot; operation is a tuple of vectors and a tuple of scalars indexed by the same finite set. This is the fundamental operation on a vector space, so to me, it makes sense to formulate everything in terms of indexed tuples of vectors whenever possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I talked about indexing lists of vectors in the previous paragraph, but indexing isn't the same as ordering. When I say &quot;order,&quot; though, I'm actually being a bit imprecise. It's usually not the order relation itself that matters, but the choice of index set does matter. For example, if you want to write down the coordinate matrix of a vector or the matrix of a linear transformation relative to a basis, then your index set pretty much has to be $\{1,...,n\}$ (or $\{0,...,n-1\}$ if you've ever programmed in a language other than Matlab or Fortran) because whatever you're using to index the elements of the basis, you're also using to index the rows or columns of your matrices. Since one of the main uses of bases is to reduce things to matrices, the sets $\{1,...,n\}$ are very convenient for indexing. In this very common context, all the basic concepts like &quot;basis,&quot; &quot;linear independence,&quot; &quot;spanning,&quot; etc. would apply to finite &lt;em&gt;sequences&lt;/em&gt; of vectors. In no context, however, do they naturally apply to finite &lt;em&gt;sets&lt;/em&gt; of vectors. &lt;/p&gt;&#xA;" OwnerUserId="2032" LastActivityDate="2010-10-03T10:23:33.440" CommentCount="4" />
  <row Id="5927" PostTypeId="1" AcceptedAnswerId="5930" CreationDate="2010-10-03T10:31:41.490" Score="1" ViewCount="237" Body="&lt;p&gt;Hi there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a problem with the binomial coefficient $\binom{5}{7}$. I know that the solution is zero, but I have problems to reproduce that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;${\displaystyle \binom{5}{7}=\frac{5!}{7!\times(5-7)!}=\frac{5!}{7!\times(-2)!}=\frac{120}{5040\times-2}=\frac{120}{-10080}=-\frac{1}{84}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where is my mistake?&lt;/p&gt;&#xA;" OwnerUserId="2280" LastActivityDate="2010-10-08T04:08:42.413" Title="Problem with binomial coefficients." Tags="&lt;binomial-coefficients&gt;" AnswerCount="4" CommentCount="4" />
  <row Id="5928" PostTypeId="1" AcceptedAnswerId="5950" CreationDate="2010-10-03T10:33:17.680" Score="4" ViewCount="151" Body="&lt;p&gt;The definition of the Mandelbrot set that most enthusiasts first encounter is that of the set of all complex numbers $c$ for which the sequence $z_{n+1} = z_n^2 + c$ starting from $z_0 = 0$ does not diverge. For convenience, let us name this familiar quadratic map $P_c(z) = z^2 + c$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have read that an equivalent definition of the Mandelbrot set is as the &quot;connectedness locus&quot; of the Julia sets of $P_c$. That is, $c$ is in the Mandelbrot set if and only if its corresponding filled Julia set $K_c = \{z : P_c^n(z) \not\rightarrow \infty\}$ is connected.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why are these definitions equivalent?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand that the equivalence boils down to the statement that $K_c$ is connected if and only it contains $0$, but I don't know how this is proved.&lt;/p&gt;&#xA;" OwnerUserId="856" LastActivityDate="2010-10-03T17:37:41.680" Title="Why are these two definitions of the Mandelbrot set equivalent?" Tags="&lt;complex-analysis&gt;&lt;fractals&gt;" AnswerCount="1" FavoriteCount="2" />
  <row Id="5929" PostTypeId="1" CreationDate="2010-10-03T11:17:55.953" Score="1" ViewCount="174" Body="&lt;p&gt;What is the usefulness in mathematics, I know it can provide topological invariants, any other things?&lt;/p&gt;&#xA;" LastEditorUserId="8246" LastEditorDisplayName="" LastEditDate="2011-04-26T19:51:48.870" LastActivityDate="2011-04-26T19:51:48.870" Title="Why do we need to study mathematical aspects of quantum field theory, especially axiomatic picture" Tags="&lt;physics&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="5930" PostTypeId="2" ParentId="5927" CreationDate="2010-10-03T11:30:20.510" Score="3" ViewCount="" Body="&lt;p&gt;Yes, now I see the problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, (-2)! really isn't defined. And I can't use the factorial method if $n\notin\mathbb{N}$. So I have to go these way:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;${\displaystyle \binom{5}{7}=\frac{5\times4\times3\times2\times1\times0\times-1}{7!}=\frac{0}{7!}=0}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, if $k&gt;n$ the solution will always be zero, because the numerator has always the factor zero.&lt;/p&gt;&#xA;" OwnerUserId="2280" LastActivityDate="2010-10-03T11:30:20.510" />
  <row Id="5931" PostTypeId="2" ParentId="5921" CreationDate="2010-10-03T11:51:57.280" Score="1" ViewCount="" Body="&lt;p&gt;For definiteness, let us assume that $X$ and $Y$ are positive random variables taking values &#xA;in $[0, \infty)$,and $F_Y(w)\le F_X(w)$, where $F_X(x), F_Y(y)$ are the CDFs and the equalities&#xA;take place only at zero and in the limit $w \rightarrow \infty$. For simplicity, let us assume further that &#xA;both CDFs are strictly increasing. The latter condition implies that we can invert the x-CDF and substitute in the y-CDF&#xA;to obtain the function $F_Y(F_X)$. In this function, both independent ($F_X$) and dependent ($F_Y$)&#xA;variables range in $[0, 1]$ and the value of F_Y is always less than F_X (except at 0 and 1).&#xA;(It is easy to see that, if one plots the x-CDF and y-CDF on the same axes, and see that at every value of the x-CDF, the y-CDF lies to the left of the x-CDF.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, $P(Y\lt X)$ is equal to the area under the line of $F_Y(F_X)$ , which is less than the area of a right isosceles rectangle of unit sides which is equal to 0.5.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit:&#xA;Here is the proof (as asked by Eruditass) that the required probability can be computed from the area of under the $F_Y$ graph as a function of $F_X$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the convolution formula to compute the probability density function of $Y-X$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f_{Y-X}(w) = \int_{0}^{\infty}f_X(x) f_Y(w+x) dx$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In terms of which, the required probability is given by:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$P(Y &amp;lt; X) = P(Y -X&amp;lt;0) = \int_{-\infty}^{0} f_{Y-X}(w)  dw$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Substituting the convolution formula:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$P(Y &amp;lt; X)  = \int_{-\infty}^{0} dw  \int_{0}^{\infty} dx f_X(x) f_Y(w+x) $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Changing the order of integration (the integration is over bounded functions) and performing a change of variables $ u = w +x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$P(Y &amp;lt; X)  = \int_{0}^{\infty} dx f_X(x)  \int_{-\infty}^{x} du   f_Y(u) $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the definition of the cumulative distribution function, the second integral is just the Y-CDF&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$P(Y &amp;lt; X)  = \int_{0}^{\infty} dx f_X(x)  F_Y(x) $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again using the definition of the CDF:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f_X(x) dx = dF_X(x) $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Changing the integration variable from $x$ to $F_X$, we obtain the final result:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$P(Y &amp;lt; X)  = \int_{0}^{1} dF_X  F_Y$&lt;/p&gt;&#xA;" OwnerUserId="1430" LastEditorUserId="1430" LastEditorDisplayName="" LastEditDate="2010-10-04T16:26:46.673" LastActivityDate="2010-10-04T16:26:46.673" CommentCount="3" />
  <row Id="5932" PostTypeId="1" AcceptedAnswerId="5967" CreationDate="2010-10-03T12:04:43.417" Score="7" ViewCount="214" Body="&lt;p&gt;In this &lt;a href=&quot;http://math.stackexchange.com/questions/5896/an-interesting-series/5903#5903&quot;&gt;answer&lt;/a&gt; to a question about a series, a theorem was stated:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;if $A= \{a_i \}$ is a set such that $\sum_{i = 1}^{\infty} \frac{1}{a_i}$ converges, then $d(A) = 0$, where $d(A)$ is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Natural_density&quot; rel=&quot;nofollow&quot;&gt;natural density&lt;/a&gt; of the set.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My background in number theory is basically zero and all my attempts to prove this have been utterly unseccessful; would anyone please help me, or at least provide me with a hint to prove it? &lt;/p&gt;&#xA;" OwnerUserId="1415" LastActivityDate="2010-10-04T19:58:36.130" Title="Theorem on natural density" Tags="&lt;number-theory&gt;" AnswerCount="3" FavoriteCount="2" />
  <row Id="5933" PostTypeId="2" ParentId="5932" CreationDate="2010-10-03T12:30:31.807" Score="1" ViewCount="" Body="&lt;p&gt;EDIT nr 2: Please forget about this answer; as pointed out in the comments, it wasn't correct! (I've removed the text it here; see the edit history for the embarrasing details. Next time, I'll try to think properly before I post...)&lt;/p&gt;&#xA;" OwnerUserId="1242" LastEditorUserId="1242" LastEditorDisplayName="" LastEditDate="2010-10-04T03:31:11.477" LastActivityDate="2010-10-04T03:31:11.477" CommentCount="5" />
  <row Id="5934" PostTypeId="1" CreationDate="2010-10-03T13:10:45.330" Score="0" ViewCount="153" Body="&lt;p&gt;Suppose&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;P(Z|X) = A&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;P(Z|Y) = B&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;where X and Y are independent. How does one determine P(Z|X,Y)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I haven't studied maths in years, so what I post below may be entirely off base, but I don't like to ask a question without some attempt, so:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From what I read, Baye's theorem is used to calculate update a probability based on new information. If the prior probability is P(Z|X), and the posterior P(Z|X,Y):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;P(Z|X,Y) = (P(Y|Z,X)P(Z|X))/P(Y)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(Some of the above was worked out on intuition, and I'd love for someone to give me an explanation of how to properly apply Baye's theorem to update an already dependent result, every discussion I found dealt with the prior probability as a simple probability, not a conditional). Since Y is independent of X, this is equal to:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;P(Z|X,Y) = (P(Y|Z)P(Z|X))/P(Y)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To digress:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;P(Y|Z) = P(Z|Y)P(Y)/P(Z)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(By Baye's theorem, again), so:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;P(Z|X,Y) = ((P(Z|Y)P(Y)/P(Z))P(Z|X))/P(Y)&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;P(Z|X,Y) = (P(Z|Y)P(Y)P(Z/X))/P(Y)P(Z)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Ignoring the degenerate case P(Y) = 0:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;P(Z|X,Y) = (P(Z|X)P(Z|Y))/P(Z)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Could someone first confirm whether or not this is correct? Secondly, can this be extended to more than two dependencies? It seems to work, but honestly my brain started melting when I tried to figure it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, if possible, explain simply why it works? I always like to know intuitively to reason for a result. I assume it's dependent on P(Z) because its necessary to find the overlap between P(X) and P(Y), but aside from that it's a bit hazy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, if my guess is correct and the term P(Z) is necessary to determine the overlap between P(X) and P(Y), what extra information would I need and what equation would I have to solve in order to calculate P(Z|X,Y) without knowledge of P(Z)? This one I haven't been able to make any progress on, because I can't see where to start.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I realise this is an awfully large question, and thank you all very much in advance for any contributions. If any parts of the question are too large (the last part, particularly, seems a candidate), please feel free to ignore them. This is all out of curiousity, so I don't specifically need an answer to any individual part.&lt;/p&gt;&#xA;" OwnerUserId="2058" LastActivityDate="2010-10-03T15:26:16.140" Title="Probability of consequent if dependent on two independent antecedents" Tags="&lt;probability&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="5935" PostTypeId="1" CreationDate="2010-10-03T13:29:24.257" Score="1" ViewCount="96" Body="&lt;p&gt;I have some discrete points, with time $t_n$ and value $u(t_n)$. I perform discrete fourier transormation using values $u(t_n)$. Now I have some complex values, as I understood absolute value is amplitude and argument is frequency, right? well my teacher said i need to compute &lt;strong&gt;something&lt;/strong&gt; using $t_n$'s, $u(t_n)$'s, $A_n$'s and $\omega_n$'s (assuming $A_n$ is absolute value, $\omega_n$ is argument). It's somehow related to statistics variance and uncertainity principle. As you see, I don't know about Fourier transformations :D. Now I've made a program that performs DFT and builds a plot using $\omega_n$'s as $Ox$ axis and $A_n$'s as $Oy$ axis. It looks like this: &lt;a href=&quot;http://yfrog.com/muscreenshot2otp&quot; rel=&quot;nofollow&quot;&gt;screenshot&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Value i need to compute is $\Delta T \Delta \omega$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$T_1=\frac{\sum_{n=1}^N u(t_n)*t_n}  {\sum_{n=1}^N u(t_n)}$&#xA;$T_2=\frac{\sum_{n=1}^N u(t_n)*t_n^2}  {\sum_{n=1}^N u(t_n)}$&#xA;$\Delta T = \sigma^2 = 2\sqrt{T_2-T_1^2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and same for $\Delta \omega$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$W_1=\frac{\sum_{n=1}^N A_n*\omega_n}  {\sum_{n=1}^N A_n}$&#xA;$W_2=\frac{\sum_{n=1}^N A_n*\omega_n^2}  {\sum_{n=1}^N A_n}$&#xA;$\Delta \omega = \sigma^2 = 2\sqrt{W_2-W_1^2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What my teacher wants me to compute and which values should I expect as corresponding to theory? Can you give me step by step tutorial from source values to result? May be there's something wrong with the task.&lt;/p&gt;&#xA;" OwnerUserId="2281" LastEditorUserId="2281" LastEditorDisplayName="" LastEditDate="2010-10-03T15:33:02.183" LastActivityDate="2010-10-03T15:33:02.183" Title="Link between time and complex amplitude argument, DFT." Tags="&lt;fourier-analysis&gt;" CommentCount="2" FavoriteCount="1" />
  <row Id="5937" PostTypeId="1" AcceptedAnswerId="5939" CreationDate="2010-10-03T15:17:00.147" Score="2" ViewCount="322" Body="&lt;p&gt;&lt;code&gt;(4 / (3 - sqrt(5))) ^ 2 - ((6 - 5 * sqrt(6)) / (5 - sqrt(6))) ^ 2 = 2 * sqrt(61 + 24*sqrt(5))&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left(\frac{4}{3-\sqrt{5}}\right)^2 - \left(\frac{6 - 5 \sqrt{6}}{5 - \sqrt{6}}\right)^2 = 2\sqrt{61+24\sqrt{5}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How to prove it is right equality?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I come up with $\dfrac{16}{14-6\sqrt{5}}-6=2\sqrt{61+24\sqrt{5}}$, but still can't get it to the obvious equality.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any ideas?&lt;/p&gt;&#xA;" OwnerUserId="2284" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-10-04T03:53:52.017" LastActivityDate="2010-10-04T11:13:36.483" Title="How to prove an identity in radicals?" Tags="&lt;algebra-precalculus&gt;" AnswerCount="5" CommentCount="6" FavoriteCount="2" />
  <row Id="5938" PostTypeId="2" ParentId="5934" CreationDate="2010-10-03T15:26:16.140" Score="1" ViewCount="" Body="&lt;p&gt;Unfortunately your result is not correct. Consider two coin tosses (with a fair coin). Let X be: &quot;You get heads in the first toss&quot; and Y be &quot;You get heads in the second toss&quot;. Then X and Y are independent events. Let Z be &quot;You get at least one head&quot;. Then P(Z)=3/4, but P(Z|X,Y)=P(Z|X)=P(Z|Y)=1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem in your argument is that P(Y|Z,X) is not necessary equal to P(Y|Z), or in other words, when constrained to Z, X and Y need no longer be independent.&lt;/p&gt;&#xA;" OwnerUserId="1609" LastActivityDate="2010-10-03T15:26:16.140" CommentCount="1" />
  <row Id="5939" PostTypeId="2" ParentId="5937" CreationDate="2010-10-03T15:31:00.583" Score="2" ViewCount="" Body="&lt;p&gt;Simplify both sides to $8+6\sqrt{5}$, as they are both equal to this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just square the RHS to see this and rationalise the denominators on the LHS.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After the rationalisation of the denominators on the LHS (which is very quick) you obtain $(3+\sqrt{5})^2-6,$ and hence the $8+6\sqrt{5}.$&lt;/p&gt;&#xA;" OwnerUserId="1301" LastEditorUserId="1301" LastEditorDisplayName="" LastEditDate="2010-10-03T16:06:49.753" LastActivityDate="2010-10-03T16:06:49.753" />
  <row Id="5940" PostTypeId="2" ParentId="5906" CreationDate="2010-10-03T15:37:08.807" Score="8" ViewCount="" Body="&lt;p&gt;Here's a variant of my comment: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A triangle is a topological manifold with boundary (and so can be made abstractly into a smooth manifold with boundary). But as a subspace of Euclidean space it is not a smooth manifold with boundary. On the other hand, a triangle is a &quot;smooth manifold with corners&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are a variety of stratified enhancements on the manifold concept. Manifold with no boundary is the &quot;base&quot; manifold concept. You can add boundary or various other stratifications and at some point you can let your space degenerate to the point that anything is more or less a &quot;manifold with enough degeneracies...&quot;  Common terms for highly-degenerate manifold types are things like &quot;stratified spaces&quot; and &quot;manifolds with singularities&quot; or &quot;pseudomanifolds&quot;.  Orbifolds are another variation on this thread of ideas. &lt;/p&gt;&#xA;" OwnerUserId="642" LastActivityDate="2010-10-03T15:37:08.807" />
  <row Id="5941" PostTypeId="2" ParentId="4737" CreationDate="2010-10-03T15:40:19.380" Score="1" ViewCount="" Body="&lt;p&gt;The following is the rigorous construction of desired packing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the following picture:&lt;br&gt;&#xA;&lt;img src=&quot;http://i.imgur.com/vm68m.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;br&gt;&#xA;On this picture discs with curvatures 2,2,3,3,6,6,6,6,11,11,11,11 are packed inside a unit disc. One can prove correctness of this image by solving quadratic equations. I want to cut circles with curvatures 2,3,4,5,6,... form these: use already obtained circles with curvatures 2 and 3. Use the second circle with curvature 2 for cutting 4,6,8,10,16,... (all even curvatures, starting with 4) using repeating the same procedure scaled to the circle with curvature 2. Use the second 3 for cutting 5 and 9. Use 6 for cutting 7. Use 11 for 11.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we have circles with radii 6, 6 and 6. Use the same procedure to obtain circles with curvatures 6*2, 6*3, 6*4, ... = 12, 18, 24, ... from them, each repeating 3 times. Use 12, 12 and 12 for 13, 15, 17; 18 for 19, 21, 23; 24 for 25, 27, 29 and so on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One can check, that words &quot;use the same procedure&quot; are ok, because if we will use the scheme, described above to cut circles one by one (2, then 3, then 4, then 5, then 6, then 7 in this order) immediately repeating the steps in described smaller circles, we will never use result of the step before the step itself.&lt;/p&gt;&#xA;" OwnerUserId="695" LastEditorUserId="695" LastEditorDisplayName="" LastEditDate="2010-10-03T15:46:37.793" LastActivityDate="2010-10-03T15:46:37.793" />
  <row Id="5942" PostTypeId="2" ParentId="5937" CreationDate="2010-10-03T15:40:51.907" Score="1" ViewCount="" Body="&lt;p&gt;You want to show:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left(\frac{4}{3-\sqrt{5}}\right)^2 - \left(\frac{6 - 5 \sqrt{6}}{5 - \sqrt{6}}\right)^2 = 2\sqrt{61+24\sqrt{5}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First multiply out the squares&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{16}{14-6\sqrt{5}} - \frac{186-60\sqrt{6}}{31-10\sqrt{6}} = 2\sqrt{61+24\sqrt{5}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then simplify it&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{16}{14-6\sqrt{5}} - 6 = 2\sqrt{61+24\sqrt{5}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;now we can start getting rid of the square roots on the right hand side by squaring both sides&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{256}{376-168\sqrt{5}} - \frac{192}{14-6\sqrt{5}} + 36 = 244 + 96 \sqrt{5}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;at this point it makes sense to rationalize the denominators of the fractions&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left(\frac{256}{376-168\sqrt{5}}\right)\left(\frac{376+168\sqrt{5}}{376+168\sqrt{5}}\right) - \left(\frac{192}{14-6\sqrt{5}}\right)\left(\frac{14+6\sqrt{5}}{14+6\sqrt{5}}\right) + 36 = 244 + 96 \sqrt{5}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which simplifies to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$376 + 168\sqrt{5} - 168 - 72\sqrt{5} + 36 = 244 + 96 \sqrt{5}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;collecting like terms now gives&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(376 - 168 + 36) + (168 - 72)\sqrt{5} = 244 + 96 \sqrt{5}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is easily seem to be equal.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Another approach is, using the fact that algebraic numbers cannot be too close together - compute the first few digits of both sides of the original identity and compare them.&lt;/p&gt;&#xA;" LastActivityDate="2010-10-03T15:40:51.907" CommentCount="4" />
  <row Id="5943" PostTypeId="2" ParentId="5937" CreationDate="2010-10-03T15:55:48.177" Score="1" ViewCount="" Body="&lt;p&gt;Since&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\dfrac{4^{2}}{\left( 3-\sqrt{5}\right) ^{2}}-\dfrac{\left( 6-5\sqrt{6}%&#xA;\right) ^{2}}{\left( 5-\sqrt{6}\right) ^{2}}=\dfrac{4^{2}\left( 5-\sqrt{6}%&#xA;\right) ^{2}-\left( 6-5\sqrt{6}\right) ^{2}\left( 3-\sqrt{5}\right) ^{2}}{%&#xA;\left( 3-\sqrt{5}\right) ^{2}\left( 5-\sqrt{6}\right) ^{2}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\dfrac{1}{\left( 3-\sqrt{5}\right) ^{2}\left( 5-\sqrt{6}\right) ^{2}}=\dfrac{%&#xA;\left( 3+\sqrt{5}\right) ^{2}\left( 5+\sqrt{6}\right) ^{2}}{5776},$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;to prove&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left( \dfrac{4}{3-\sqrt{5}}\right) ^{2}-\dfrac{\left( 6-5\sqrt{6}\right)&#xA;^{2}}{\left( 5-\sqrt{6}\right) ^{2}}=2\sqrt{61+24\sqrt{5}}\quad (1)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;it is enough to show that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\left( 16\left( 5-\sqrt{6}\right) ^{2}-\left( 6-5\sqrt{6}\right) ^{2}\left(&#xA;3-\sqrt{5}\right) ^{2}\right) \left( 3+\sqrt{5}\right) ^{2}\left( 5+\sqrt{6}%&#xA;\right) ^{2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$=11552\sqrt{61+24\sqrt{5}}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But (see &lt;em&gt;addendum&lt;/em&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left( 16\left( 5-\sqrt{6}\right) ^{2}-\left( 6-5\sqrt{6}\right) ^{2}\left(&#xA;3-\sqrt{5}\right) ^{2}\right) \left( 3+\sqrt{5}\right) ^{2}\left( 5+\sqrt{6}%&#xA;\right) ^{2}=3656\sqrt{5}+46208.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It remains to prove&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left( 34656\sqrt{5}+46208\right) ^{2}=11552^{2}\left( 61+24\sqrt{5}%&#xA;\right) \quad (2).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The left hand side is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left( 3656\sqrt{5}+46208\right) ^{2}=3202768896\sqrt{5}+8140370944$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and the right hand side is equal:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$11552^{2}\left( 61+24\sqrt{5}\right) =3202768896\sqrt{5}+8140370944,$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which proves your equality.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Addendum&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left( 16\left( 5-\sqrt{6}\right) ^{2}-\left( 6-5\sqrt{6}\right) ^{2}\left(&#xA;3-\sqrt{5}\right) ^{2}\right) \left( 3+\sqrt{5}\right) ^{2}\left( 5+\sqrt{6}\right) ^{2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$=16\left( 3+\sqrt{5}\right) ^{2}\left( 5+\sqrt{6}\right) ^{2}\left( 5-\sqrt{%&#xA;6}\right) ^{2}-\left( 3+\sqrt{5}\right) ^{2}\left( 5+\sqrt{6}\right)&#xA;^{2}\left( 6-5\sqrt{6}\right) ^{2}\left( 3-\sqrt{5}\right) ^{2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$=5776\left( 3+\sqrt{5}\right) ^{2}-34656=34656\sqrt{5}+46208$$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Addendum&lt;/strong&gt; 2: The second equality stated in the question&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\dfrac{16}{14-6\sqrt{5}}-6=2\sqrt{61+24\sqrt{5}}\quad (3)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is equivalent to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\dfrac{8}{14-6\sqrt{5}}-3=\sqrt{61+24\sqrt{5}}=\sqrt{61+\sqrt{2880}}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The LHS can be transformed into&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\dfrac{8}{14-6\sqrt{5}}-3=\dfrac{18\sqrt{5}-34}{14-6\sqrt{5}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$=\dfrac{18\sqrt{5}-34}{14-6\sqrt{5}}\times \dfrac{14+6\sqrt{5}}{14+6\sqrt{5}%&#xA;}=\dfrac{64+48\sqrt{5}}{16}=4+3\sqrt{5}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then we have to prove&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sqrt{61+\sqrt{2880}}=4+3\sqrt{5}.\quad (4)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we apply to the LHS the following &lt;strong&gt;general transformation involving radicals&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sqrt{A+\sqrt{B}}=\sqrt{\dfrac{A+\sqrt{A^{2}-B}}{2}}+\sqrt{\dfrac{A-\sqrt{%&#xA;A^{2}-B}}{2}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $A=61,B=2880$, then $\sqrt{A^{2}-B}=\sqrt{61^{2}-2880}=29$ and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sqrt{\dfrac{A+\sqrt{A^{2}-B}}{2}}=3\sqrt{5},$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sqrt{\dfrac{A-\sqrt{A^{2}-B}}{2}}=4,$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which completes the proof.&lt;/p&gt;&#xA;" OwnerUserId="752" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2010-10-04T11:13:36.483" LastActivityDate="2010-10-04T11:13:36.483" CommentCount="6" />
  <row Id="5944" PostTypeId="1" CreationDate="2010-10-03T16:11:05.990" Score="1" ViewCount="161" Body="&lt;p&gt;I need to find the exact solution for the lateral surface area of the solid generated by revolving the region bounded by $y=x^2$, $y=0$, $x=0$, and $x=\sqrt 2$ about the X axis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have come up with my own solution, but I'm not sure if it's right.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Before substituting values back in, I got&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \frac\pi{16} \left( (1+4x^2)^{3/2} 2x - \frac32 \ln\sqrt{1+4x^2+2x} + x\sqrt{1+4x^2} \right)  \text{ from }0\text{ to }\sqrt2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can someone please verify this? For my integration by parts, I set $u = \sec^3\theta - \sec\theta$ and $dv = \sec^2\theta d\theta$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks so much in advance.&lt;/p&gt;&#xA;" LastEditorUserId="171" LastEditorDisplayName="" LastEditDate="2010-10-03T17:15:22.853" LastActivityDate="2010-12-08T06:31:59.413" Title="Lateral surface area by hand" Tags="&lt;calculus&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="5945" PostTypeId="2" ParentId="5937" CreationDate="2010-10-03T16:24:53.097" Score="-1" ViewCount="" Body="&lt;p&gt;Trivial. Lets simplify the left part:&#xA;multiply first fraction to the sum of expressions in its denominator, and get whole part in second fraction and do with second fraction the same that we did with first:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(\frac{4(3+\sqrt{5})}{9-5})^2 - (\frac{25-19 - \sqrt{6}}{5-\sqrt{6}}) = (3+\sqrt{5})^2 - (5-\frac{19(5+\sqrt{6})}{25-6})^2 = (3+\sqrt{5})^2 - 6 = 8 + 6\sqrt{5}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now lets prove this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$8+6\sqrt{5}=2\sqrt{61+24\sqrt{5}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;divide by two and square:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$4+3\sqrt{5} = \sqrt{61+24\sqrt{5}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$16+45+24\sqrt{5}=61+24\sqrt{5}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thats all.&lt;/p&gt;&#xA;" OwnerUserId="2281" LastActivityDate="2010-10-03T16:24:53.097" />
  <row Id="5946" PostTypeId="1" AcceptedAnswerId="5948" CreationDate="2010-10-03T16:41:48.443" Score="-4" ViewCount="148" Body="&lt;p&gt;sqrt(3 - sqrt(5)) * (3 + sqrt(5)) * (sqrt(10) - sqrt(2))&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you please solve this? I've been trying to do that two hours! The result is 8.&lt;/p&gt;&#xA;" OwnerUserId="2284" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-10-04T11:01:42.187" LastActivityDate="2010-10-04T11:01:42.187" Title="Simplifying a radical expression" Tags="&lt;algebra-precalculus&gt;" AnswerCount="2" CommentCount="5" />
  <row Id="5947" PostTypeId="2" ParentId="5889" CreationDate="2010-10-03T16:54:27.373" Score="3" ViewCount="" Body="&lt;p&gt;Let &lt;code&gt;$\{g_i\}$&lt;/code&gt; be a set of generators for $G$ and &lt;code&gt;$\{c_j\}$&lt;/code&gt; a set of generators for $C$.  For each $i$, write&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$g_i= a_{i,1}b_{i,1}\ldots a_{i,n}b_{i,n}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;as usual, where $a_k\in A$, $b_k\in B$.  Clearly, the finite set &lt;code&gt;$\{a_{i,k},b_{i,k}\}$&lt;/code&gt; generates $G$.  Now, we shall see that &lt;code&gt;$\{a_{i,k}\}\cup\{c_j\}$&lt;/code&gt; generates $A$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any $a\in A$ can be written as a product&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a=\alpha_1\beta_1\ldots\alpha_m\beta_m$   (1)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\alpha_l$ is a product of $a_{i,k}^{\pm1}$'s, hence is in $A$, and $\beta_l$ is a product of $b_{i,k}^{\pm1}$'s, hence is in $B$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An expression like the right hand side of (1) is called &lt;em&gt;reduced&lt;/em&gt; if every $\alpha_i\notin C$ for $i&gt;1$ and every $\beta_i\notin C$ for $i&amp;lt; m$.  In other words, there's no 'obvious' way to reduce $m$ by combining one of the terms with its neighbouring terms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Normal Form Theorem for amalgamated products (as Lilly says below, this is Theorem 1 is Serre's &lt;em&gt;Trees&lt;/em&gt;---this is where we use that the maps $C\to A$ and $C\to B$ are injective) asserts that every element has a unique reduced expression (up to the obvious ambiguity that we can insert elements of $C$ in the $A$-terms and cancel them in neighbouring $B$-terms).  But (1) gives us two expressions for $a$, and the left hand side is clearly reduced.  It follows that the right hand side is not reduced, unless $m=1$ and $\beta_1\in C$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if $m&gt;1$ we know that the right hand side is not reduced, which means that some $\alpha_l\in C$ for $l&gt;1$ or some $\beta_l\in C$ for $l&amp;lt; m$.  Let's assume that $\beta_l\in C$---the case of $\alpha_l\in C$ is identical.  Now we can write $\beta_l$ as a product of $c_j^{\pm1}$'s. We now know that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\alpha'_l=\alpha_l\beta_l\alpha_{l+1}$ (2)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is a product of $a_{i,k}^{\pm1}$'s and $c_j^{\pm1}$'s.  Use (2) to simplify the right hand side of (1), so it becomes&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;$a=\alpha_1\beta_1\ldots\beta_{l-1}\alpha'_l \beta_{l+1} \ldots \beta_m$&lt;/code&gt; . (3)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The product on the right hand side of (3) has fewer terms than the product on the right hand side of (1), so we can continue by induction until the right hand side is reduced, at which point we have an expression of the form&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a=\alpha_1\beta_1$ (4)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with $\beta_1\in C$ and $\alpha_1$ a product of $a_{i,k}^{\pm1}$'s and $c_j$'s.  So we are done.&lt;/p&gt;&#xA;" OwnerUserId="2287" LastEditorUserId="2287" LastEditorDisplayName="" LastEditDate="2010-10-06T19:24:47.760" LastActivityDate="2010-10-06T19:24:47.760" CommentCount="7" />
  <row Id="5948" PostTypeId="2" ParentId="5946" CreationDate="2010-10-03T17:04:48.507" Score="3" ViewCount="" Body="&lt;p&gt;You want to evaluate&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(\sqrt{3 - \sqrt{5}}) (3 + \sqrt{5}) (\sqrt{10} - \sqrt{2})$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;it is equal to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sqrt{3 - \sqrt{5}} \sqrt{(3 + \sqrt{5})^2} \sqrt{(\sqrt{10} - \sqrt{2})^2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;merging the top level square roots gives&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sqrt{(3 - \sqrt{5}) (3 + \sqrt{5})^2 (\sqrt{10} - \sqrt{2})^2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;multiplying this out gives&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sqrt{144 - 24 \sqrt{20} + 48 \sqrt{5} - 8 \sqrt{100}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;pulling a factor of 48 into the radical to get like terms gives&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sqrt{64 - 24 \sqrt{20} + 24 \sqrt{20}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is $8$.&lt;/p&gt;&#xA;" LastActivityDate="2010-10-03T17:04:48.507" CommentCount="2" />
  <row Id="5949" PostTypeId="1" CreationDate="2010-10-03T17:28:13.697" Score="2" ViewCount="55" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe that if we are given two modules $M,N$ over a ring $R$, and a pairing between them&#xA;$M \otimes_R N \to R$, we can construct a perfect pairing $M'\otimes_R N' \to R$ by taking kernels.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a name for this construction?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="1007" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-10-04T00:17:22.407" LastActivityDate="2010-10-04T00:17:22.407" Title="is there a name for this construction? (taking a pairing and obtaining a perfect one)" Tags="&lt;linear-algebra&gt;&lt;abstract-algebra&gt;&lt;terminology&gt;" CommentCount="1" />
  <row Id="5950" PostTypeId="2" ParentId="5928" CreationDate="2010-10-03T17:37:41.680" Score="5" ViewCount="" Body="&lt;p&gt;Let $S_1$ be a circle of radius $100$ in the complex plane centered at the origin.  Clearly everything outside of $S_1$ diverges to $\infty$ under iteration of $P_c$, so the filled Julia set lies entirely inside of $S_1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now take the preimage $S_2$ of $S_1$ under $P_c$.  This will be a smaller curve (close to a circle, with a radius of approximately $\sqrt{100} = 10$), which again must contain the entire filled Julia set.  Iterating this process, we obtain a sequence $S_1,S_2,\ldots$ of closed curves, each of which contains the filled Julia set in its interior.  In fact, since any point outside of the filled Julia set goes to $\infty$ under iteration of $P_c$, the intersection of the interiors of the curves $S_n$ is precisely the filled Julia set.  (See &lt;a href=&quot;http://upload.wikimedia.org/wikipedia/commons/8/8e/Time_escape_Julia_set_from_coordinate_%28phi-2%2C_0%29.jpg&quot; rel=&quot;nofollow&quot;&gt;this picture&lt;/a&gt; for an example.  The curves separating the different shades of orange are the iterated preimages of some large circle.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, this reasoning is not quite correct, because the preimage of a closed curve under $P_c$ is not always a single closed curve.  Sometimes it is one closed curve, and sometimes it is two closed curves with disjoint interiors.  If we repeatedly take the preimages of $S_1$, we may find that $S_n$ is a union of a very large number of closed curves!  Note, however, that these curves still contain the filled Julia set entirely inside of them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now here is the key bit: &lt;strong&gt;the preimage of a curve will have one component if and only if $0$ lies in the the interior of the preimage&lt;/strong&gt;.  This is because $0$ is the critical point of the map $P_c$.  Thus the preimage of a curve is either a single curve that surrounds $0$, or two curves, neither of which surrounds $0$.  (In the latter case, the two curves are actually negatives of one another, i.e. symmetric across the origin.)  Therefore, there are exactly two cases:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The point $0$ lies in the filled Julia set.  In this case, each preimage $S_n$ must be a single curve, so the intersection of the interiors is connected.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The point $0$ lies outside the filled Julia set.  In this case, some preimage $S_n$ does not encircle $0$, so it must have two components.  Then each successive preimage will have twice as many curves as the last, and the resulting filled Julia set is homeomorphic to the Cantor set. (See &lt;a href=&quot;http://upload.wikimedia.org/wikipedia/en/6/65/Julia_0.45_0.1428.png&quot; rel=&quot;nofollow&quot;&gt;this picture&lt;/a&gt; for an example.  The curves separating the different shades of blue are the iterated preimages of some large circle.  You ought to be able to see the first step at which the curve separates into two components.)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="1726" LastActivityDate="2010-10-03T17:37:41.680" />
  <row Id="5951" PostTypeId="2" ParentId="5922" CreationDate="2010-10-03T17:43:48.393" Score="2" ViewCount="" Body="&lt;p&gt;Let's make some assumptions for the purposes of calculation. Suppose 6.7 billion people have been alive for 13.5 billion years each shuffling one deck a thousand times a second.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The total number of rounds of shuffles $N=1000 \times 3600 \times 24 \times 365 \times1.35 \times 10^{10}.$&#xA;So $N = 4.25736 \times 10^{20}.$&#xA;Let $n=52! \approx 8.0658 \times 10^{67}$ and let $m=6.7 \times 10^9.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When everyone shuffles the cards once in the first millisecond the probability that they are all different is&#xA;$$ \prod_{i=1}^{m-1} \left( 1- \frac{i}{n} \right) \approx 1 - \frac{m(m-1)}{2n},$$&#xA;ignoring the cross products since $n$ is much bigger than $m$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each shuffle is independent, so after $N$ steps the probabilty that we have had no match is&#xA;$$\left( 1 - \frac{m(m-1)}{2n} \right)^N \approx 1 - \frac{m(m-1)N}{2n}, \qquad (1)$$&#xA;since $m(m-1)/2n$ is very small.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence the probability that there has been a match is 1 minus RHS of (1), that is&#xA;$$\frac{m(m-1)N}{2n}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I make this about $1.18 \times 10^{-28}$ with the figures I've used.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;REMARK: Note that if you drop the &quot;at one moment&quot; requirement. That is, if you now ask if two shuffles have ever been the same, the problem is slightly quicker. The total number of shuffles that have ever taken place is $mN$ and so the probability that they were all different (as in the birthday problem) is $\prod_{i=0}^{mN-1} \left( 1- \frac{i}{n} \right),$ where as before $n=52!,$ So the probability that we've had a repeat is  $1- \prod_{i=0}^{mN-1} \left( 1- \frac{i}{n} \right),$ which, similar to before, is approximately $mN(mN-1)/2n,$ or $5 \times 10^{-8}.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that as the index $i$ in the product gets larger, better approximations are obtained by passing to logarithms (as on the wikipedia page).&lt;/p&gt;&#xA;" OwnerUserId="1301" LastEditorUserId="1301" LastEditorDisplayName="" LastEditDate="2010-10-04T05:40:19.807" LastActivityDate="2010-10-04T05:40:19.807" CommentCount="3" />
  <row Id="5952" PostTypeId="1" AcceptedAnswerId="5958" CreationDate="2010-10-03T18:37:00.610" Score="3" ViewCount="95" Body="&lt;p&gt;Are there any positive integer solutions (n,x) to the equation (x)(x+1)=n! except (2,1) and (3,2)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If not (as I suspect is the case), how do you prove that?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, is there a way to approach the Diophantine equation n!=P(x) where P is a polynomial?&lt;/p&gt;&#xA;" OwnerUserId="796" LastEditorUserId="1778" LastEditorDisplayName="" LastEditDate="2010-11-15T01:03:22.963" LastActivityDate="2010-11-15T01:03:22.963" Title="Factorial equaling a polynomial" Tags="&lt;number-theory&gt;&lt;diophantine-equations&gt;" AnswerCount="1" />
  <row Id="5953" PostTypeId="2" ParentId="5944" CreationDate="2010-10-03T18:44:43.120" Score="2" ViewCount="" Body="&lt;p&gt;Not a complete answer, but maybe a bit of help:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The surface area is $\int_{x=0}^{\sqrt{2}} (2\pi y) ds = \int_{0}^{\sqrt{2}} 2 \pi x^2 \sqrt{1+(2x)^2} dx$, and the antiderivate is, according to Mathematica,&#xA;$$\frac{\pi}{32} \left(2 \sqrt{4 x^2+1} \left(8 x^3+x\right)-\sinh^{-1}(2x)\right).$$&#xA;The inverse of $\sinh$ can be expressed using $\ln$, so you should be able to compare this to your expression.&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2010-10-03T18:44:43.120" CommentCount="0" />
  <row Id="5954" PostTypeId="2" ParentId="5937" CreationDate="2010-10-03T18:54:39.820" Score="0" ViewCount="" Body="&lt;p&gt;Specialize $\rm\ \ \ b=3,\ \ c = 5,\ \ d = 6 \ \Rightarrow\ a = 4\ \ $ in this simple derivation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\quad\quad\quad\displaystyle \bigg(\frac{b^2-c}{b-\sqrt{c}}\bigg)^2 - \bigg(\frac{d -5\sqrt{d}}{5-\sqrt d}\bigg)^2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\quad =\quad \:(\: b \ \: + \:   \sqrt{c}\ )^{\:2} \ \ \:-\ \ \ (\:-\:\sqrt{d}\:)^{2} $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\quad =\ \  2\ (\:a + b \sqrt{c}\ )\:, \quad 2\ a\ =\ b^2+c-d $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\quad =\ \ 2\:\sqrt{a^2+b^2\:c+2\:a\:b\sqrt{c} } $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;NOTE&lt;/strong&gt; $\:$ Replacing numbers by functions makes  the proof both simpler and more general - similar to your &lt;a href=&quot;http://math.stackexchange.com/questions/5972&quot;&gt;recently asked question&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-10-04T04:54:34.390" LastActivityDate="2010-10-04T04:54:34.390" />
  <row Id="5955" PostTypeId="1" AcceptedAnswerId="5957" CreationDate="2010-10-03T19:01:25.853" Score="1" ViewCount="51" Body="&lt;p&gt;If $\phi:X\rightarrow Y$ is a rational map of varieties, why is the graph of $\phi$ well-defined? That is, if $\langle \rho,U\rangle$ and $\langle \psi,V\rangle$ are two representatives of $\phi$ (so that $U,V\subseteq X$ open, and $\rho\vert_{U\cap V}=\psi\vert_{U\cap V}$), why is the closure of the image of $\rho$ equal to the closure of the image of $\psi$?&lt;/p&gt;&#xA;" OwnerUserId="264" LastActivityDate="2010-10-03T19:12:39.090" Title="Why is the graph of a rational map independent of the choice of representative?" Tags="&lt;algebraic-geometry&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="5956" PostTypeId="2" ParentId="5929" CreationDate="2010-10-03T19:01:43.920" Score="4" ViewCount="" Body="&lt;p&gt;One can prove many things (enumerative results in algebraic geometry, results in the geometric Langlands program, mirror symmetry) by making physics-style computations in various quantum field theories.  Currently, one is not able to make such arguments rigorous in general, and hence they become a method for producing conjectures in mathematics, not theorems.  One would like to upgrade their status to being proved theorems (!), and one approach would be to find&#xA;a framework in which the field-theoretic computations become valid mathematical deductions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One current framework in which one can do something like this is that supplied by the &lt;em&gt;bordism hypothesis&lt;/em&gt; of Baez and Dolan, recently proved by Jacob Lurie.    This provides a rigorous means of constructing certain topological field theories, which is already beginning to find applications.  But (for example) this framework is not encompassing enough to explain mathematically the results of Kapustin and Witten in the direction of deducing geometric Langlands from mirror symmetry.&lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2010-10-03T19:01:43.920" />
  <row Id="5957" PostTypeId="2" ParentId="5955" CreationDate="2010-10-03T19:12:39.090" Score="3" ViewCount="" Body="&lt;p&gt;Because $U\cap V$ is Zariski dense in both $U$ and $V$ (by irreducibility of $X$), and so both graphs&#xA;are obtained by closing up the graph of $\rho_{| U\cap V}$ (which of course coincides with the graph of $\phi_{| U \cap V}$).&lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2010-10-03T19:12:39.090" CommentCount="1" />
  <row Id="5958" PostTypeId="2" ParentId="5952" CreationDate="2010-10-03T19:13:54.860" Score="5" ViewCount="" Body="&lt;p&gt;This question was discussed on &lt;a href=&quot;http://mathoverflow.net/questions/39210/solve-in-positive-integers-nmm1&quot; rel=&quot;nofollow&quot;&gt;mathoverflow&lt;/a&gt;, and the question is open.&lt;/p&gt;&#xA;" OwnerUserId="425" LastActivityDate="2010-10-03T19:13:54.860" />
  <row Id="5959" PostTypeId="1" AcceptedAnswerId="6001" CreationDate="2010-10-03T19:15:53.807" Score="1" ViewCount="132" Body="&lt;p&gt;I suspect this is easy, but that I'm missing something obvious: If a ring homomorphism $f:R\rightarrow S$ is such that a maximal ideal $m$ in $S$ does not have $f^{-1}(m)$ maximal in $R$, must there be some non-unit element $x\in R$ whose image under $f$ is a unit in $S$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My intuition is just that some &quot;localization&quot; must have occurred. In the classic example of $\mathbb{Z}\hookrightarrow\mathbb{Q}$, the zero ideal becomes maximal because 2, 3, 5, ... all become units.&lt;/p&gt;&#xA;" OwnerUserId="264" LastEditorUserId="264" LastEditorDisplayName="" LastEditDate="2010-10-03T19:39:06.897" LastActivityDate="2011-05-04T06:19:04.367" Title="Is the failure of MaxSpec to be functorial due to homomorphisms which take non-units to units?" Tags="&lt;commutative-algebra&gt;" AnswerCount="2" />
  <row Id="5961" PostTypeId="2" ParentId="5231" CreationDate="2010-10-03T19:16:56.350" Score="2" ViewCount="" Body="&lt;p&gt;From your comments, what I understand of your problem is that you have the coordinates of an arbitrarily oriented rectangle centred on the origin, and you want to find the rotation that will bring it to an axis-aligned rectangle on the $xy$ plane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $u$ and $v$ be the unit vectors that should be mapped to the axis-oriented unit vectors $e_x$ and $e_y$ respectively. You can get these by subtracting adjacent points of the rectangle and normalizing. Then you want a rotation $R$ which satisfies $Ru = e_x$, $Rv = e_y$, and $Rn = e_z$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can express this as $R[u\;v\;n] = [e_x\;e_y\;e_z] = I$. Then $R$ equals $[u\;v\;n]^{-1}$, which is simply $[u\;v\;n]^T$ since $u$, $v$, and $n$ are an orthonormal set. To be more explicit, the rotation matrix you want is:&#xA;$$R = \begin{bmatrix}u_x &amp;amp; u_y &amp;amp; u_z \\ v_x &amp;amp; v_y &amp;amp; v_z \\ n_x &amp;amp; n_y &amp;amp; n_z\end{bmatrix}.$$&#xA;If you really like Euler angles (&lt;code&gt;rotateX&lt;/code&gt;, &lt;code&gt;rotateY&lt;/code&gt;, &lt;code&gt;rotateZ&lt;/code&gt;), there are ways to convert a rotation matrix like above to Euler angles, but they're ugly. You're best off using the rotation matrix explicitly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the way, if your rectangle is &lt;em&gt;not&lt;/em&gt; centred on the origin, and you want to perform the rotation keeping its centre (say $c$) fixed, you'll have to get the rotated coordinates of a point $p$ not simply as $Rp$ but as $R(p-c) + c$.&lt;/p&gt;&#xA;" OwnerUserId="856" LastActivityDate="2010-10-03T19:16:56.350" CommentCount="6" />
  <row Id="5962" PostTypeId="1" CreationDate="2010-10-03T19:39:03.560" Score="3" ViewCount="348" Body="&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;If $A$ is a differential one form then $A\wedge A .. (more\text{ }than\text{ }2\text{ }times) = 0$ Then how does the $A\wedge A \wedge A$ make sense in the Chern-Simon's form, $Tr(A\wedge dA + \frac{2}{3} A \wedge A \wedge A)$ ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I guess this anti-commutative nature of wedge product does not work for Lie algebra valued one-forms since tensoring two vectors is not commutative. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If the vector space in which the vector valued differential form is taking values is $V$ with a chosen basis ${v_i}$ then books use the notation of $A = \Sigma _{i}  A_i v_i$ where the sum is over the dimension of the vector space and $A_i$ are ordinary forms of the same rank. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I would like to know whether this $A_i v_i$ is just a notation for $A_i \otimes v_i$ ?&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Similarly say $B$ is a vector valued differential form taking values in $W$ with a chosen basis ${w_i}$ and in the same notation, $B = \Sigma _j B_j w_j$. Then the notation used is that, $A \wedge B = \Sigma _{i,j} A_i \wedge B_j v_i \otimes w_j$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wonder if in the above $A_i \wedge B_j v_i \otimes w_j$ is just a notation for $A_i \wedge B_j \otimes v_i \otimes w_j$ ?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$A$ and $B$ are vector bundle valued differential forms (like say the connection-1-form $\omega$ or the curvature-2-form $\omega$) then how is $Tr(A)$ defined and why is $d(Tr A) = Tr( d A)$ and $Tr(A \wedge B) = Tr(B \wedge A)$ ? &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Is $A \wedge A \wedge A \wedge A = 0$ ? for $A$ being a vector bundle valued $1$-form or is only $Tr(A \wedge A \wedge A \wedge A) = 0$ ? &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If A and B are two vector bundle valued $k$ and $l$ form respectively then one defines $[A,B]$ as , $[A,B] (X_1,..,X_{k+l}) = \frac{1}{(k+l)!} \Sigma _{\sigma \in S_n} (sgn \sigma) [A (X_{\sigma(1)},X_{\sigma(2)},..,X_{\sigma(k)}) , B (X_{\sigma(k+1)},X_{\sigma(k+2)},..,X_{\sigma(k+l)})]$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means that if say $k=1$ then $[A,A] (X,Y) = [A(X),A(Y)]$ and $[A,A] = 2A \wedge A$.&#xA;The Cartan structure equation states that, $d\Omega = \Omega \wedge \omega - \omega \wedge \Omega$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But some people write this as, $d\Omega = [\Omega,\omega]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is not clear to me. Because if the above were to be taken as a definition of the $[,]$ then clearly $[A,A]=0$ contradicting what was earlier derived. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="965" LastEditorUserId="965" LastEditorDisplayName="" LastEditDate="2010-10-03T20:19:48.943" LastActivityDate="2010-10-19T21:12:29.033" Title="Some questions about differential forms" Tags="&lt;differential-geometry&gt;" AnswerCount="2" CommentCount="8" FavoriteCount="4" />
  <row Id="5963" PostTypeId="2" ParentId="5959" CreationDate="2010-10-03T19:43:38.580" Score="3" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; Consider the image of a maximal ideal containing $f^{-1}(m)$.&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-10-03T19:43:38.580" CommentCount="2" />
  <row Id="5964" PostTypeId="1" CreationDate="2010-10-03T19:43:41.963" Score="2" ViewCount="80" Body="&lt;p&gt;I play a game known as Bejeweled (I'm sure some of you have heard of it) which has the following properties:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Given a planar graph with every vertex having degree 4, any path of 3 (or more) contiguous colors is considered &quot;complete.&quot;  Additionally, any two adjacent vertices may be &quot;swapped&quot; to produce said path.  If such a swap exists the board is also said to be &quot;complete.&quot;  You cannot swap if it does not produce a path of 3 or more contiguous colors.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/AALIM.jpg&quot; alt=&quot;bejeweled&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the moment I'm ignoring the property that this path must be in a straight line.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I'd like to do is to limit the number of colors I have so that the board will always be complete (having a path, or a single swap produces the path).  Now obviously this is trivially true in the 2 color case, but I want the maximal number of colors.  Or more precisely,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What is the minimum number of colors that I need to color a planar graph, with every vertex having degree 4, such that there is no contiguous path of 3 identical colors or such a path cannot be produced by a single swapping of two nodes?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2291" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-26T00:46:59.717" LastActivityDate="2010-11-26T00:46:59.717" Title="What is the minimal coloring of a planar graph with swaps?" Tags="&lt;graph-theory&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" />
  <row Id="5965" PostTypeId="1" CreationDate="2010-10-03T19:48:40.820" Score="2" ViewCount="109" Body="&lt;p&gt;Can anyone explain how to do these types of questions for me? I am having trouble wrapping my head around it. Thanks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A spherical soap bubble is seen to undergo oscillations when slightly deformed. Assuming&#xA;that the frequency of these oscillations is a function of the surface tension S of the soap bubble,&#xA;the size of the soap bubble as given by the radius r of the original spherical bubble and the density&#xA;of soap, say p. Use dimensional analysis to construct a possible dependence for f in terms of S, r&#xA;and p.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A spherical soap bubble of radius 1cm oscillates at 1 Hz. What would be the anticipated frequency*&#xA;of oscillation of a soap bubble of radius 2cm?&lt;/p&gt;&#xA;" OwnerUserId="2292" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-10-04T00:08:11.667" LastActivityDate="2010-10-04T00:08:11.667" Title="Dimensional Analysis" Tags="&lt;physics&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="5966" PostTypeId="2" ParentId="5927" CreationDate="2010-10-03T19:51:42.583" Score="2" ViewCount="" Body="&lt;p&gt;By definition $\rm\binom{n}k$ is the coefficient of $\rm x^k$ in $\rm (1+x)^n$ so it is $0$ for $\rm k &gt; n\:$.&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-10-03T19:51:42.583" />
  <row Id="5967" PostTypeId="2" ParentId="5932" CreationDate="2010-10-03T19:55:46.123" Score="3" ViewCount="" Body="&lt;p&gt;If the natural density of $A = \{a_i\}$ exists, then we can show that it must be zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $\displaystyle S_{n} = \frac{|A \cup [1,n]|}{n}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now $\displaystyle \{\frac{n}{a_n}\}$ is a subsequence of $S_{n}$ and so if the limit is $\displaystyle 2\delta &gt; 0 $ then we have that for all $\displaystyle n &gt; N_0$, $\displaystyle \frac{n}{a_n} &gt; \delta$ and so $\displaystyle \frac{1}{a_n} &gt; \frac{\delta}{n}$ for all $\displaystyle n &gt; N_0$ and so $\displaystyle \sum \frac{1}{a_n}$ diverges.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The main problem is actually showing that the limit exists.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is easy to show that $\liminf$ is zero: If the limit was $\displaystyle 2\delta &gt; 0$ then for all $n &gt; N_{0}$, $S_{n} &gt; \delta$ and an argument similar to above works.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now suppose $\displaystyle  \limsup S_n = 2\delta &gt; 0$. Then there is a subsequence $\displaystyle S_{N_1}, S_{N_2}, ..., S_{N_k}, \dots $ which converges to $\displaystyle 2\delta$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we can choose the subsequence so that $\displaystyle S_{N_i} &gt; \delta$ and $\displaystyle  N_{k+1} &gt; \frac{2N_{k}}{\delta}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the number of elements of $\displaystyle  A$ in the interval $\displaystyle (N_{k}, N_{k+1}]$ is atleast $\displaystyle  \delta N_{k+1} - N_k \ge \delta N_{k+1} - \frac{\delta N_{k+1}}{2} \ge \frac{\delta N_{k+1}}{2}$ and so the sum of reciprocals in that interval is atleast $\displaystyle \frac{\delta N_{k+1}}{2} \frac{1}{N_{k+1}} =  \displaystyle \frac{\delta}{2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And so the sum of reciprocals must diverge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence $\displaystyle \limsup S_n = 0 = \liminf S_n$ and thus $\displaystyle \lim S_n = 0$ and thus the natural density is zero.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-10-03T20:04:07.450" LastActivityDate="2010-10-03T20:04:07.450" />
  <row Id="5968" PostTypeId="1" CreationDate="2010-10-03T20:23:29.270" Score="3" ViewCount="137" Body="&lt;p&gt;My friend is taking a course on probabilities and come across a problem that has a solution we both have trouble understanding:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A single printer prints, on average, 22 jobs an hour. Let's use a simplification where no job takes more than 2 minutes to complete. Using these assumptions, calculate the upper limit for the probability that, when a job arrives to the printer, the printer is processing another job.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The solution was to use the Poisson distribution: 1 - ((22/30)^0)*e^(-22/30)/1! = 1 - e^(-22/30) = ~0.52&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm having trouble believing this: wouldn't simply 22/30 make more sense? Even though I can see how enqueueing the printing jobs is Poisson distributed, how come that goes for the waiting times, too? Or have they messed up with the solutions?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: after sending feedback the assistant said the problem was, indeed, ambiguous. Thanks for the help.&lt;/p&gt;&#xA;" OwnerUserId="2076" LastEditorUserId="2076" LastEditorDisplayName="" LastEditDate="2010-10-15T12:15:20.937" LastActivityDate="2010-10-15T12:15:20.937" Title="Using Poisson distribution to estimate if a printer is available? (I have the solution but don't understand it) [SOLVED]" Tags="&lt;probability&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" />
  <row Id="5969" PostTypeId="1" AcceptedAnswerId="5980" CreationDate="2010-10-03T20:55:47.123" Score="2" ViewCount="295" Body="&lt;p&gt;Say you have two groups $G = \langle g \rangle$ with order $n$ and $H = \langle h \rangle$ with order $m$. Then the product $G \times H$ is a cyclic group if and only if $gcd(n,m)=1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can't seem to figure out how to start proving this. I have tried with some examples, where I pick $(g,h)$ as a candidate generator of $G \times H$. I see that what we want is for the cycles of $g$ and $h$, as we take powers of $(g,h)$, to interleave such that we do not get $(1,1)$ until the $(mn)$-th power. However, I am having a hard time formalizing this and relating it to the greatest common divisor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any hints are much appreciated!&lt;/p&gt;&#xA;" OwnerUserId="2174" LastActivityDate="2010-10-03T22:44:35.837" Title="Product of two cyclic groups is cyclic iff their orders are co-prime" Tags="&lt;group-theory&gt;" AnswerCount="4" CommentCount="1" />
  <row Id="5970" PostTypeId="2" ParentId="5965" CreationDate="2010-10-03T21:00:05.173" Score="3" ViewCount="" Body="&lt;p&gt;Dimensional Analysis says ignore the variables you have and manipulate what you've got in terms of dimensions. For example, let any measure of $mass=M$, let any measure of $length=L$, let any measure of $time=T$. $Force=F=MLT^-2$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then surface tension is $\frac{MT^2}{L}$. Radius is simple, it is L whilst density is mass divided by area. Area in this case is $\pi*L^2$ and mass $M$ so density is $\frac{M}{\pi L^2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Frequency is period, or $\frac{1}{T}$, correct? Then, you simply need to attempt to equate both sides.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, in your case, $Frequency = f(S, r, p)$. Well, $\frac{1}{T} = \sqrt{\frac{L}{MT^2}}*\sqrt{\frac{M}{\pi L^2}} * \sqrt{L} = \sqrt{\frac{L^2M}{\pi L^2T^2M}}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hopefully you can see these expressions are equivalent except for a constant $\pi$. Constants don't have dimension, so they don't matter. However, you might also see we have three expressions we've already defined, as such, $F = \sqrt{S*r*p}$. We know this to be true because the units work as expected and you can check it based on the values being what you expect.&lt;/p&gt;&#xA;" OwnerUserId="892" LastActivityDate="2010-10-03T21:00:05.173" />
  <row Id="5971" PostTypeId="2" ParentId="5927" CreationDate="2010-10-03T21:05:11.313" Score="2" ViewCount="" Body="&lt;p&gt;$(-2)!$ is actually infinite. A more palatable way to phrase that, perhaps, is in terms of the reciprocal factorial: $1/(-2)! = 0$. We only need the recurrence relation $n! = n(n-1)!$, or in terms of reciprocal factorials: $$\frac{1}{(n-1)!} = n\cdot\frac{1}{n!}.$$ That means $\frac{1}{(-2)!} = \frac{0\cdot (-1)}{0!} = 0$. Then $\binom{5}{7} = \frac{5!}{7!}\cdot \frac{1}{(-2)!} = 0$, QED. &lt;/p&gt;&#xA;" OwnerUserId="2032" LastActivityDate="2010-10-03T21:05:11.313" />
  <row Id="5972" PostTypeId="2" ParentId="5946" CreationDate="2010-10-03T21:15:00.697" Score="2" ViewCount="" Body="&lt;p&gt;Square it, then put $\rm\ \: a = 3,\ b = 2\:a-1 = 5,\ c = 2\ \ $ in:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\quad\quad\quad (a-\sqrt b)\ (a+\sqrt b)^2\ ((\sqrt b - 1)\sqrt c)^2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\quad =\ (a^2 \: -\: b)\ \:(a+\sqrt b)\ (b+1-2\sqrt b)\ c$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\quad =\ (a^2\:-\: b)^2\ 2\:c\ \ \ \: $ via $\rm\:\ \ b+1 \ =\ 2\:a$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;NOTE&lt;/strong&gt; $\:$ Replacing numbers by functions makes it both simpler and more general, as in your &lt;a href=&quot;http://math.stackexchange.com/questions/5954&quot;&gt;prior question&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-10-03T21:15:00.697" CommentCount="1" />
  <row Id="5973" PostTypeId="2" ParentId="5969" CreationDate="2010-10-03T21:18:01.600" Score="0" ViewCount="" Body="&lt;p&gt;Use these facts:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Cyclic groups of the same order are isomorphic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) &lt;a href=&quot;http://mathforum.org/library/drmath/view/73205.html&quot; rel=&quot;nofollow&quot;&gt;http://mathforum.org/library/drmath/view/73205.html&lt;/a&gt;&lt;/p&gt;&#xA;" LastActivityDate="2010-10-03T21:18:01.600" />
  <row Id="5975" PostTypeId="1" CreationDate="2010-10-03T21:18:29.243" Score="0" ViewCount="85" Body="&lt;p&gt;Here is the question for my homework:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A mother wants to invest 7000 USD for her son's future education. She invests a portion of the money in a bank certificate of deposit (CD account) which earns 4% and the remainder in a savings bond that earns 7%. If the total interest earned after one year is $420.00, how much money was invested in the CD account?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, here is my steps:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Amount invested in the savings bond can be represented as 7000 - x&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The interest earned in one year on the CD account is 0.04x&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the interest earned on the savings bond in one year can be represented as: 0.07(7000-x)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The total interested earned after one year is 420.00 USD&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The equation is now 0.04x+0.07(7000-x) = 420.00&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now to SIMPLFY it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have this so far&lt;/p&gt;&#xA;&#xA;&lt;p&gt;n - 420.00 = 0.03x&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do I get the number N?&lt;/p&gt;&#xA;" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-10-04T01:48:18.633" LastActivityDate="2010-10-04T01:48:18.633" Title="Getting a number from interest equation" Tags="&lt;homework&gt;&lt;algebra&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="5976" PostTypeId="2" ParentId="5919" CreationDate="2010-10-03T21:24:33.677" Score="12" ViewCount="" Body="&lt;p&gt;The groups you are looking for are called &lt;em&gt;capable groups&lt;/em&gt;. The determination of all capable group (even all finite capable groups) is very far from done. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The earliest result is due to Baer, who characterized exactly which groups that are direct sums of cyclic groups are capable (Baer, Reinhold. &lt;em&gt;Groups with preassigned central and central quotient groups&lt;/em&gt;, Trans. Amer. Math. Soc. &lt;strong&gt;44&lt;/strong&gt; (1938), 387-412; in fact Baer considered the question of when a group has a specified center and central quotient, each of which is a direct sum of cyclic groups. You obtain the characterization of capable groups as a corollary). For finitely generated abelian groups, this becomes:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Theorem.&lt;/strong&gt; Let $G$ be a finitely generated abelian group, written as a direct sum of cyclic groups&#xA;  $$G = C_{a_1}\oplus\cdots\oplus C_{a_n}$$&#xA;  with $a_1|a_2|\cdots|a_n$ ($C_0$ denotes the infinite cyclic group). Then $G$ is capable if and only if $n\gt 1$ and $a_{n-1}=a_n$. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There is a generalization of this result, replacing abelian groups with $p$-groups and replacing the direct sum with the nilpotent product (&lt;em&gt;Capability of nilpotent products of cyclic groups&lt;/em&gt; and &lt;em&gt;Capability of nilpotent products of cyclic groups II&lt;/em&gt;, J. Group Theory &lt;strong&gt;8&lt;/strong&gt;, no. 4 (2005), 431-452; and J. Group Theory &lt;strong&gt;10&lt;/strong&gt; no. 4 (2007), 441-451):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Theorem.&lt;/strong&gt; Let $p$ be a prime and let $c$ be a positive integer, $c\leq p$. If $G$ is the $c$-nilpotent product of cyclic groups&#xA;  $$G = C_{p^{a_1}}\coprod^{c}\cdots\coprod^{c} C_{p^{a_n}}$$&#xA;  where $\coprod^{c}$ represents the $c$-nilpotent product, and $1\leq a_1\leq\cdots\leq a_n$. Then $G$ is capable if and only if $n\gt 1$ and $a_n\leq a_{n-1}+\lfloor\frac{c-1}{p-1}\rfloor$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The last condition is in fact necessary in general for $p$-groups:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Theorem.&lt;/strong&gt;(2005) Let $G$ be a nilpotent $p$-group of class $c\gt 0$, and let $x_1,\ldots,x_n$ be a generating set for $G$ with $x_i$ of order $p^{a_i}$, $a_1\leq a_2\leq\cdots\leq a_n$. If $G$ is capable, then $n\gt 1$ and $a_n\leq a_{n-1}+\lfloor\frac{c-1}{p-1}\rfloor$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The extra-special capable $p$-groups were not classified until 40 years after Baer (F.R. Beyl, U. Felgner, and P. Schmid. &lt;em&gt;On groups ocurring as central factor groups&lt;/em&gt;, J Algebra &lt;strong&gt;61&lt;/strong&gt; (1979), 161-177). The only extra-special capable $p$-groups are the nonabelian groups of order $p^3$ and exponent $p$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Aside from these classes, there is a full classification of the capable $2$-generated $p$-groups of class two. You can find it the paper (with Robert F. Morse) &lt;em&gt;Certain homological functors for $2$-generated $p$-groups of class 2&lt;/em&gt;, in Computational Group Theory and the Theory of Groups II, Contemporary Mathematics &lt;strong&gt;511&lt;/strong&gt; (2010), pp 127-166, American Mathematical Society. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are lots of known necessary conditions, but in general sufficient conditions are harder to find. Phillip Hall commented 70 years ago (&lt;em&gt;The classification of prime-power groups&lt;/em&gt;, J. Reine Angew. Math. &lt;strong&gt;182&lt;/strong&gt; (1940) 130-141) that:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The question of what conditions a group $G$ must fulfil in order that it may be the central quotient of another group $H$, $G\cong H/Z(H)$, is an interesting one. But while it is easy to write down a number of necessary conditions, it is not so easy to be sure that they are sufficient.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-10-04T01:43:30.420" LastActivityDate="2010-10-04T01:43:30.420" CommentCount="1" />
  <row Id="5977" PostTypeId="2" ParentId="5975" CreationDate="2010-10-03T21:28:45.323" Score="1" ViewCount="" Body="&lt;p&gt;I'm not sure what n is supposed to represent but instead of it you should still have that 7000*.07. Now you are left with just a number on the left and you can proceed to solve for x. $$.07(7000)-420.00 = 0.03x$$&lt;/p&gt;&#xA;" OwnerUserId="1352" LastActivityDate="2010-10-03T21:28:45.323" CommentCount="6" />
  <row Id="5978" PostTypeId="2" ParentId="5944" CreationDate="2010-10-03T21:31:53.327" Score="2" ViewCount="" Body="&lt;p&gt;I assume you have in your earlier steps:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\[ \sigma = 2\pi \int_0^\sqrt{2} x^2 \sqrt{1+4x^2} dx \]&lt;br&gt;&#xA;\[ \tan \theta = 2x \; \mathrm{and} \; \sec \theta = \sqrt{1+4x^2} \]&lt;br&gt;&#xA;\[ \sigma = \frac{\pi}2 \int_{x=0}^{x=\sqrt{2}} \sec^3\theta \tan^2\theta d\theta. \]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you have a messy antiderivative you worked out, and want to check whether it's right?  Why not see what happens if you take the derivative of your expression above?  If everything is fine, you should get back to $2\pi x^2 \sqrt{1+4x^2}$.  I get:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\[ d \sigma = \frac{\pi}{16} \left[ 24x^2\sqrt{1+4x^2}+2(1+4x^2)^{\frac 32} - \frac{3(8x+2)}{4(1+4x^2+2x)} + \sqrt{1+4x^2} + 4x^2(1+4x^2)^{-\frac 12} \right] dx. \]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yes, there are a few problems here.  More things should be cancelling out.  The middle term (from $\ln$) is conspicuous for involving $\sqrt{1+4x^2+2x}$ instead of $\sqrt{1+4x^2}$.  I'm pretty sure the $2x$ snuck under the radical by accident, and the term should really be $\ln | \sqrt{1+4x^2}+2x |$.  If I fix that in your expression and derive again, I get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\[ d \sigma = \frac{\pi}{16} \left[ 24x\sqrt{1+4x^2}+2(1+4x^2)^{\frac 32} - 3(1+4x^2)^{-\frac 12} + \sqrt{1+4x^2} + 4x^2(1+4x^2)^{-\frac 12} \right] dx. \]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now there are some like terms and related terms, but they won't cancel out far enough, because some of your coefficients are wrong.  Maybe from here you can find your mistakes?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other hints:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The correct antiderivative in terms of $x$ never involves a factor of 3, unless you count the term $(1+4x^2)^{\frac 32}$.&lt;/li&gt;&#xA;&lt;li&gt;If you have a good antiderivative in terms of $\theta$, you might want to plug in to that equation instead of converting back to $x$, by finding values for $\tan \theta$ and $\sec \theta$ at $x=0$ and $x=\sqrt{2}$.  (At $x=\sqrt{2}$, $\sec \theta$ is rational!)&lt;/li&gt;&#xA;&lt;li&gt;When you have a final answer, compare to my result $\frac{\pi}{32}\left[102\sqrt{2} - \ln\left(3+2\sqrt{2}\right)\right]$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="2236" LastActivityDate="2010-10-03T21:31:53.327" />
  <row Id="5979" PostTypeId="2" ParentId="5969" CreationDate="2010-10-03T21:35:14.430" Score="1" ViewCount="" Body="&lt;p&gt;The order of $G\times H$ is $n.m$. Thus, $G\times H$ is cyclic iff it has an element with order $n.m$. Suppose $gcd(n.m)=1$. This implies that $g^m$ has order $n$, and analogously $h^n$ has order $m$. That is, $g\times h$ has order $n.m$, and therefore $G\times H$ is cyclic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose now that $gcd(n.m) &gt;1$. Let $g^k$ be an element of $G$ and $h^j$ be an element of $H$. Since the lowest common multiple of $n$ and $m$ is lower than the product $n.m$, that is, $lcm(n,m) &amp;lt; n.m$, and since $(g^k)^{lcm(n,m)} = e_{G}$, $(h^j)^{lcm(n,m)} = e_{H}$, we have $(g^k\times h^j)^{lcm(n,m)} = e_{G\times H}$. It follows that every element of $G\times H$ has order lower than $n.m$, and therefore $G\times H$ is not cyclic.&lt;/p&gt;&#xA;" OwnerUserId="1178" LastActivityDate="2010-10-03T21:35:14.430" />
  <row Id="5980" PostTypeId="2" ParentId="5969" CreationDate="2010-10-03T21:39:48.537" Score="1" ViewCount="" Body="&lt;p&gt;Note that $|G\times H|=|G||H|=nm$; so $G\times H$ is cyclic if and only if there is an element of order $nm$ in $G\times H$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In any group $A$, if $a,b\in A$ commute with one another, $a$ has order $k$, and $b$ has order $\ell$ then the order of $ab$ will divide lcm$(k,\ell)$ (prove it).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now take an element of $G\times H$, written as $(g^a,h^b)$, where $G=\langle g\rangle$, $H=\langle h\rangle$, $0\leq a\lt n$, $0\leq b\lt m$. Then $(g^a,h^b)=(g^a,1)(1,h^b)$. In this case, what is the order? Under what conditions can you get an element of order exactly $nm$, which is what you need?&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-10-03T21:39:48.537" CommentCount="7" />
  <row Id="5981" PostTypeId="2" ParentId="5968" CreationDate="2010-10-03T21:48:55.033" Score="2" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Update&lt;/strong&gt;&#xA;This was my first answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Your answer 22/30 ignores the fact that the job preceding yours may itself have been held up by a previous job.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As Charles and tsiki point out, this is wrong: it's the Poisson-derived answer that ignores this. So now I'll try to answer the question as posed:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;By symmetry, the probability of a job having been started less than 2 minutes prior to yours is the same that a job will be started less than 2 minutes after yours. This is a Poisson distribution.&lt;/p&gt;&#xA;" OwnerUserId="1508" LastEditorUserId="1508" LastEditorDisplayName="" LastEditDate="2010-10-04T08:46:07.137" LastActivityDate="2010-10-04T08:46:07.137" CommentCount="1" />
  <row Id="5982" PostTypeId="1" AcceptedAnswerId="5994" CreationDate="2010-10-03T22:24:35.550" Score="0" ViewCount="416" Body="&lt;p&gt;Say I have the square root of 2/9. How would I convert this to a mixed radical where the radicand is a &lt;strong&gt;whole number&lt;/strong&gt;? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can convert the square root of 8 to a mixed radical easily (2 sqrt 2) but this fraction is tripping me up.&lt;/p&gt;&#xA;" OwnerUserId="2296" LastEditorUserId="2296" LastEditorDisplayName="" LastEditDate="2010-10-03T23:00:26.617" LastActivityDate="2010-10-04T01:45:39.430" Title="Converting a radical to a mixed radical" Tags="&lt;roots&gt;" AnswerCount="2" CommentCount="7" />
  <row Id="5983" PostTypeId="1" AcceptedAnswerId="6011" CreationDate="2010-10-03T22:41:26.817" Score="3" ViewCount="224" Body="&lt;p&gt;Let n = p^k be a power of a prime. Let G be a transitive permutation group on n letters. Show that there exists a ﬁxed-point free element of order p.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can show that there must be a fixed point:&#xA;The transitivity of the group implies that there is only 1 orbit. Burnside's lemma then implies that the average number of fixed points for an element in the group is 1. Since the identity fixes everything, there is a fixed point free element.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can show that there exists an element of order p:&#xA;Again we use that the group is transitive and that there is therefore only one orbit. Since there is only one orbit, the order of the orbit of x is p^k. By the orbit-stabilizer theorem, the order of the stabilizer of some element x is equal to the order of the group divided by p^k. Since the order of a stabilizer is an integer, p^k divides the order of the group. By Cauchy's theorem, there is an element of order p in the group.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I cannot, however, show that there is an element of order p with no fixed points. Does anyone have any hints?&lt;/p&gt;&#xA;" OwnerUserId="2298" LastActivityDate="2010-10-04T09:22:46.490" Title="fixed-point free elements in transitive permutation groups" Tags="&lt;group-theory&gt;" AnswerCount="1" />
  <row Id="5984" PostTypeId="2" ParentId="5969" CreationDate="2010-10-03T22:44:35.837" Score="0" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\rm\quad\ \ gcd(m,n) &gt; 1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\iff\ lcm(m,n) &amp;lt; mn$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\iff\ \mathbb Z_m \times \mathbb Z_n\ $ has all elts of order $\rm &amp;lt; mn$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\iff\ \mathbb Z_m \times \mathbb Z_n\ $ is noncyclic&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-10-03T22:44:35.837" />
  <row Id="5985" PostTypeId="2" ParentId="5982" CreationDate="2010-10-03T22:48:16.673" Score="0" ViewCount="" Body="&lt;p&gt;What do you mean by the radicand is a &quot;whole number&quot;? In your example $\sqrt{8}=2\sqrt{2}$ in both sides of the equation the radicand is a whole number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In any case, it doesn't really matter because the problem as you stated it doesn't have a radical at all when you simplify: $\sqrt{\frac{2}{8}}=\sqrt{\frac{1}{4}}=\frac{\sqrt{1}}{\sqrt{4}}=\frac{1}{2}$.&lt;/p&gt;&#xA;" OwnerUserId="2000" LastActivityDate="2010-10-03T22:48:16.673" CommentCount="1" />
  <row Id="5986" PostTypeId="1" CreationDate="2010-10-03T23:26:39.807" Score="3" ViewCount="194" Body="&lt;p&gt;My question: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A cashier has a total of 114 bills, made up of tens and hundreds. The total value of the money is 6450 USD. How much of each kind does he have?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;He has [BLANK] tens and [BLANK] hundreds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm pretty sure there is an algebra equation for this. There is the hard way of doing it on paper (that takes long) or the short way of putting it equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Help?&lt;/p&gt;&#xA;" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2010-11-15T03:22:53.767" LastActivityDate="2010-11-15T03:22:53.767" Title="$\$10$ and $\$100$ bill equation" Tags="&lt;homework&gt;&lt;algebra-precalculus&gt;" AnswerCount="3" />
  <row Id="5987" PostTypeId="2" ParentId="5986" CreationDate="2010-10-03T23:29:49.070" Score="4" ViewCount="" Body="&lt;p&gt;Let $T$ be the number of tens, and let $H$ be the number of hundreds. You want to find &lt;em&gt;integer&lt;/em&gt; solutions to the equation&#xA;$$ 10T + 100H = 6450.$$&#xA;This is what is called a &lt;a href=&quot;http://en.wikipedia.org/wiki/Diophantine_equation#Linear_Diophantine_equations&quot; rel=&quot;nofollow&quot;&gt;linear Diophantine equation&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Added&lt;/em&gt;: And in fact, you have another piece of information which I missed on my first read-through, as noted by jericson:&#xA;$$H+T = 114.$$&#xA;Once you have the two equations, this is no longer a Diophantine equation problem, but a simple system of two linear equations in two unknowns.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-10-03T23:41:17.263" LastActivityDate="2010-10-03T23:41:17.263" CommentCount="5" />
  <row Id="5988" PostTypeId="2" ParentId="5914" CreationDate="2010-10-03T23:51:06.067" Score="6" ViewCount="" Body="&lt;p&gt;Alon Amit's answer is not right, I believe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$2$ is not prime in the Gaussian integers! $(1+i)$ divides $2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The argument upto the fact that the prime $p$ divides $y+i$, $y-i$ and $2i$ is correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $p|2i$ we consider $p = 1+i$. Now any multiple of $1+i$ is of the form $2x + i2y$ or $2x+1 + i(2y+1)$. Since $y$ is even, $y+i$ cannot be divisible by $1+i$ and thus $y+i$ and $y-i$ are co-prime.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Setting $y+i$ to be a perfect cube (upto units) easily gives us $y=0$ and so $x=1, y=0$ is the only solution.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-10-03T23:51:06.067" />
  <row Id="5989" PostTypeId="2" ParentId="5986" CreationDate="2010-10-04T00:18:18.083" Score="6" ViewCount="" Body="&lt;p&gt;If all 114 were 10's then their sum is 1140, which is 5310 too low. So we need to swap some 10's for 100's. Each swap adds 90, so we need 5310/90 = 59 swaps, yielding 59 100's, $\:$ hence 114 - 59 = 55 10's.&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-10-04T00:18:18.083" />
  <row Id="5991" PostTypeId="2" ParentId="5986" CreationDate="2010-10-04T00:55:57.307" Score="2" ViewCount="" Body="&lt;p&gt;Dan,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you have $$H+T=114$$ and $$10T+100H=6450$$ as above, write $$H=114-T$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then substitute this into the second equation above&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$10T+100(114-T)=6450$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Expand, and simplify. &lt;/p&gt;&#xA;" OwnerUserId="2219" LastActivityDate="2010-10-04T00:55:57.307" CommentCount="3" />
  <row Id="5992" PostTypeId="2" ParentId="5932" CreationDate="2010-10-04T01:31:46.113" Score="2" ViewCount="" Body="&lt;p&gt;Given a set $A = (\cup a_i) \subset N$ with positive lim-sup density, it is possible to partition $N$ into intervals on which $A$ has density $\geq d$.  If the intervals grow in size at a suitable rate (for this problem, exponential growth is enough), there will be a positive number $C&gt;0$ such that on each interval, $\Sigma_{a \in A} 1/a \geq C$, and this makes the sum over all intervals infinite.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In more detail,&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;There is a partition of the positive integers into intervals $[1,n_1], [n_1 + 1, n_2], [n_2 + 1,n_3], \dots$ such that the density of $A$ is at least $d &gt; 0$ on each interval.  This can be done for any $d &amp;lt; L$ where $L$ is the lim sup density of $A$.   &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;By joining together adjacent intervals, one can also require that the sequence $n_k$ grows fast enough that density in $[1,n_i]$ is not strongly affected by what happens in preceding intervals $[1,n_j]$ with $j &amp;lt; i$.  In particular, one can keep or remove the subinterval $[1,n_{i-1}]$ with small effect on density questions in $[1,n_i]$.    For this problem, fast enough growth will mean strictly exponential growth, i.e., $n_{i+1} / n_i &gt; p $ for constant $p &gt; 1$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The most efficient packing of a given number of integers into an interval $[m,n]$ (where efficiency means minimim sum of $1/a$ with $a$ ranging over the subset) is to place the whole subset as close as possible to $n$.  If the density is at least $d$, the number of integers to be placed is at least $d(n-m+1)$, and then $\Sigma 1/a$ is at least $d(n-m)/n = d(1-m/n)$.  &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;For the exponentially growing intervals with $m/n &amp;lt; 1/p$ the sum in each interval of the partition of $N$ is bounded below by $C=d(1-1/p) &gt; 0$, so the sum of all $1/a_i$ is infinite.  &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="467" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-10-04T19:58:36.130" LastActivityDate="2010-10-04T19:58:36.130" />
  <row Id="5993" PostTypeId="1" CreationDate="2010-10-04T01:44:39.080" Score="0" ViewCount="235" Body="&lt;p&gt;I got:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After a 6% sales tax, a shirt cost 15.37 USD. Find the original cost of the shirt before sales tax.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Exactly how do I approach this? Thanks&lt;/p&gt;&#xA;" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-10-04T01:48:56.840" LastActivityDate="2010-10-04T03:02:03.823" Title="Sales Tax equation" Tags="&lt;homework&gt;&lt;algebra&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="5994" PostTypeId="2" ParentId="5982" CreationDate="2010-10-04T01:45:39.430" Score="1" ViewCount="" Body="&lt;p&gt;$\sqrt{\frac{2}{9}} = \frac{\sqrt{2}}{\sqrt{9}} = \frac{\sqrt{2}}{3}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this what you wanted?&lt;/p&gt;&#xA;" OwnerUserId="647" LastActivityDate="2010-10-04T01:45:39.430" CommentCount="3" />
  <row Id="5995" PostTypeId="2" ParentId="5993" CreationDate="2010-10-04T01:47:44.233" Score="0" ViewCount="" Body="&lt;p&gt;US\$15.37 is 106% of the original cost. So solve for the cost.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-10-04T01:47:44.233" />
  <row Id="5996" PostTypeId="2" ParentId="5993" CreationDate="2010-10-04T01:48:03.780" Score="1" ViewCount="" Body="&lt;p&gt;Let x equal the original cost of the shirt. Then x plus the sales tax (.06 of x) equals 15.37. Solve for x.&lt;/p&gt;&#xA;" OwnerUserId="1352" LastActivityDate="2010-10-04T01:48:03.780" />
  <row Id="5997" PostTypeId="1" CreationDate="2010-10-04T02:06:48.470" Score="0" ViewCount="215" Body="&lt;p&gt;I have been tasked with creating a C++ program (with GDI+ for graphics) that takes a set of user defined points and creates polynomial curve through them. For extra credit, I have to support a parametric polynomial for any number of points. The thing is.. he didn't explain what a parametric polynomial is! So I need some help understanding how to figure one out based on a set of points.. in layman's terms, please. I've taken Calc 1, so I'm not that up-to-par on my calculus skills, just so you know what level I'm at.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any help would be appreciated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-10-04T10:44:28.620" LastActivityDate="2010-10-04T10:44:28.620" Title="Graphing a Parametric Polynomial based on a given set of points" Tags="&lt;geometry&gt;&lt;polynomials&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="5998" PostTypeId="2" ParentId="5993" CreationDate="2010-10-04T02:16:44.237" Score="0" ViewCount="" Body="&lt;pre&gt;&lt;code&gt;15.37 = 1.06x&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Where &lt;em&gt;x&lt;/em&gt; is the original cost of the t-shirt.&#xA;The 6% sales tax would make the t-shirt cost 106% of the original price, so now just do some simple algebra to isolate &lt;em&gt;x&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2296" LastEditorUserId="2296" LastEditorDisplayName="" LastEditDate="2010-10-04T03:02:03.823" LastActivityDate="2010-10-04T03:02:03.823" CommentCount="4" />
  <row Id="5999" PostTypeId="2" ParentId="5997" CreationDate="2010-10-04T02:27:32.550" Score="0" ViewCount="" Body="&lt;p&gt;I suspect that by 'parameter' your professor means that your code should accept a number supplied by the user specifying how many data points they would like to input.  Then your code should write a polynomial to fit that data.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is fairly simple to write a unique $(n-1)$-degree polynomial given $n$ data points.  &lt;a href=&quot;http://en.wikipedia.org/wiki/Curve_fitting&quot; rel=&quot;nofollow&quot;&gt;Start here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="121" LastActivityDate="2010-10-04T02:27:32.550" CommentCount="2" />
  <row Id="6000" PostTypeId="1" AcceptedAnswerId="6012" CreationDate="2010-10-04T04:21:23.283" Score="4" ViewCount="200" Body="&lt;p&gt;The essence of my question is, if I have a Hermitian matrix that is linearly dependent on a set of parameters and I have an estimate of its eigenvalues, is there a &quot;simple&quot; way to determine the values of the parameters?  Ideally, I would also like to have some measure of the goodness of the fit and the degree of variation within the parameters.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;As a materials physicist, I often have to create a simple quantum mechanical model from either experimental data or a more complex calculation.  For the smaller problems (8x8, with 10 params), the parameters can be found by painstakingly working through the various relationships among the parameters, due to symmetry, etc.  But, this method is specific to each problem and does not scale well to larger problems.  For instance, one system I'm looking at would require a 20x20 matrix with 21 parameters, and that is without including spin!  Alternatively, there is the brute force method of simulated annealing, which involves taking a random walk through the parameter space, and slowly decreasing the step size in the hopes that the calculation will get stuck in the global minimum.  Neither of these methods is particularly appealing, so I'd like some ideas on how to approach this in a consistent manner. &lt;/p&gt;&#xA;" OwnerUserId="2283" LastActivityDate="2010-10-04T09:44:32.807" Title="Fitting a parameter dependent matrix to its eigenvalues" Tags="&lt;linear-algebra&gt;&lt;matrices&gt;&lt;physics&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" />
  <row Id="6001" PostTypeId="2" ParentId="5959" CreationDate="2010-10-04T05:28:28.583" Score="6" ViewCount="" Body="&lt;p&gt;I'm not sure that I properly understand the question, especially in light of your acceptance of Bill Dubuque's answer and your accompanying comment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Still, maybe these remarks will be useful:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The map $R/f^{-1}(m) \to S/m$ is a map from an integral domain to a field, and if the&#xA;source is not also a field (i.e. if $f^{-1}(m)$ is not itself maximal), then there are certainly&#xA;non-units in $R/f^{-1}(m)$ that map to units in $S/m$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One interpretation of your question is whether, in this context, there must necessarily be a non-unit of $R$ that maps to a unit in $S$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer is &quot;no&quot; in general.  Here is a counterexample:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $R$ be a domain which is not a field, and let $I$ by any subset of $R\setminus \{0\}$ which generates $R\setminus \{0\}$ as a monoid under multiplication.  Let $S$ be the polynomial ring $R[\{x_i\}_{i \in I}].$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $Q$ denotes the fraction field of $R$, then there is a natural surjection&#xA;$S \to Q$ given by mapping $x_i$ to $1/i$.  (This is where we use the assumption that&#xA;$I$ generates $R\setminus \{0\}$, so as to get a surjection.)&#xA;The kernel of this surjection is a maximal ideal $m$ of $S$, whose preimage in $R$ is the&#xA;zero ideal (hence not maximal).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, since $S$ is a polynomial ring over $R$, any non-unit in $R$ remains a non-unit in $S$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some additional comments: if $R$ is Jacobson (e.g. $\mathbb Z$, or a polynomial ring over a field), then any map between finite type $R$-algebras preserves the corresponding MaxSpecs.&#xA;On the other hand, for such domains, the set $I$ is necessarily infinite, and so $S$ is necessarily infintely generated as an $R$-algebra (and hence no contradiction ensues!).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But we can get finite type examples by taking $R$ to be e.g. a DVR.  Then $I$ can be taken&#xA;to consist of a single element (the uniformizer), and so $S = R[x]$ is a polynomial ring&#xA;in a single variable.  For example, if $R = \mathbb Z_p$ (the $p$-adic integers) then&#xA;we have &#xA;$$\mathbb Z_p \hookrightarrow \mathbb Z_p[x] \mapsto \mathbb Z_p[x]/(p x - 1) = \mathbb Q_p,$$&#xA;and so $p x - 1$ generates a maximal ideal in $\mathbb Z_p[x]$ whose preimage is not maximal, but every non-unit in $\mathbb Z_p$ remains a non-unit in $\mathbb Z_p[x]$.&lt;/p&gt;&#xA;" OwnerUserId="221" LastEditorUserId="221" LastEditorDisplayName="" LastEditDate="2011-05-04T06:19:04.367" LastActivityDate="2011-05-04T06:19:04.367" CommentCount="1" />
  <row Id="6002" PostTypeId="1" AcceptedAnswerId="6004" CreationDate="2010-10-04T06:20:03.537" Score="0" ViewCount="63" Body="&lt;p&gt;I was reading through some notes on closure operations, and the example was given that for a poset $(S,\preceq)$, the operation on $(\mathcal{P}(S),\subseteq)$ given by $E\mapsto LU(E)$ is a closure operation. Here $LU(E)$ is the set of lower bounds of the upper bounds of $E$ for some $E\subseteq S$. I verified that indeed $E\subseteq LU(E)$, and if $E\subseteq E'$, then $LU(E)\subseteq LU(E')$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is, if I take a family of sets $\xi\subseteq \mathcal{P}(S)$, such that $\xi$ consists only of $LU$-closed subsets of $S$, is it then true that the union is again $LU$-closed? That is,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\bigcup_{E\in\xi}E=\bigcup_{E\in\xi}LU(E)=LU(\bigcup_{E\in\xi}E)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where the first equality comes from the fact that $E=LU(E)$. I see that for $e\in\bigcup_{E\in\xi}E$, then $e\in LU(\bigcup_{E\in\xi}E)$, however, the other containment has eluded me. I know that the Kuratowski closure axioms require a closure operator to preserve unions in topology, but is such an equality even the case for partially ordered sets? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt; The reason I ask is that there is the following exercise in the notes, and here $\mathcal{L}(S)\subseteq\mathcal{P}(S)$ is the set of all $LU$-closed subsets of $S$. I want to show for any family $\xi\subseteq \mathcal{L}(S)$ of $LU$-closed subsets of $S$, then $\sup\xi$ exists in $\mathcal{L}(S)$ and &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sup\xi=LU(\bigcup_{E\in\xi}E).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that $\sup\xi$ in $\mathcal{P}(S)$ is the union of all sets $E\in\xi$. My strategy was if one could show that $\cup_{E\in\xi}E=LU(\cup_{E\in\xi}E)$, then it follows that $\cup_{E\in\xi}E$ does indeed exist in $\mathcal{L}(S)$, and thus $\sup\xi=\cup_{E\in\xi}E=LU(\cup_{E\in\xi}E)$. Perhaps I have misinterpreted the question?&lt;/p&gt;&#xA;" OwnerUserId="1583" LastEditorUserId="1583" LastEditorDisplayName="" LastEditDate="2010-10-04T07:52:35.483" LastActivityDate="2010-10-04T08:18:16.697" Title="Is the union of a family of $LU$-closed sets also $LU$-closed?" Tags="&lt;set-theory&gt;&lt;topology&gt;" AnswerCount="1" />
  <row Id="6003" PostTypeId="2" ParentId="5962" CreationDate="2010-10-04T06:40:28.743" Score="1" ViewCount="" Body="&lt;p&gt;I've seen some of this stuff before, but I'm definitely not an expert.  Here's my take (although I'm likely wrong on at least some of it, hopefully not too much):&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;You can think of this as a matrix of 1-forms.  Then, $A \wedge A$ means &quot;matrix multiplication, where multiplication is wedge product&quot;.&lt;/li&gt;&#xA;&lt;li&gt;Yes, I think $A_i v_i$ really means $A_i \otimes v_i$.&lt;/li&gt;&#xA;&lt;li&gt;I think also $A_i \wedge B_j v_i \otimes w_j$ means $A_i \wedge B_j \otimes v_i \otimes w_j$, although there might be some issue of skew-symmetrization.&lt;/li&gt;&#xA;&lt;li&gt;A vector-bundle-valued differential k-form locally looks like $\omega \otimes s$, where $\omega$ is a usual k-form and $s$ is a section.  Then $d$ should act by the Leibniz rule (once you've fixed a connection).  I'd imagine that the trace is just trace in the usual sense, and then it's obvious that this commutes with $d$ and that $Tr$ admits cyclic permutations of the factors in its argument.&lt;/li&gt;&#xA;&lt;li&gt;I think that whether $A\wedge A\wedge A\wedge A$ must be zero depends on the dimension of the vector bundle where it takes its values.  If for example it's the constant 1-dimensional bundle, then this is no different from usual 1-forms.&lt;/li&gt;&#xA;&lt;li&gt;Your last question is a little unclear, sorry.  Could you clarify exactly what you're referring to?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll respond to your comments in order.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;If $A$ takes a vector field and spits out a matrix, we can consider each of the entries of the matrix (in a fixed basis, of course) as an ordinary 1-form.  If you want to wedge two matrix-valued forms, you just write out the matrices, act like you're multiplying matrices like you learned a long time ago (i.e. take the dot product of rows with columns), but then when you want to simplify the expression, you consider the product to be the wedge.  So, e.g. if&#xA;$$ A = (a,b;c,d) $$&#xA;is a 2x2 matrix-valued 1-form &lt;em&gt;(sorry, for whatever reason my matrices aren't working out)&lt;/em&gt;, this means that $a$, $b$, $c$, and $d$ are just ordinary 1-forms.  So evaluating on a vector $v$ gives the 2x2 matrix (of numbers)&#xA;$$ A(v) = ( a(v) , b(v) ;c(v) , d(v) ). $$&#xA;If you want to calculate $A\wedge A$, this is the 2x2 matrix-valued 2-form&#xA;$$ A\wedge A = ( a, b ;c,d) \wedge ( a , b ;c ,d ) $$&#xA;$$ = ( a\wedge a + b\wedge c , a\wedge b + b\wedge d ; c\wedge a + d\wedge c , c\wedge b+d\wedge d) $$&#xA;$$ = ( b\wedge c , b\wedge (d-a) ;c\wedge (a-d) , c\wedge b ). $$&lt;/li&gt;&#xA;&lt;li&gt;This should be more or less answered by (1).  The decomposition of the form $\omega \otimes s$ is just saying that a vector-bundle-valued form can be thought of as an ordinary form (that's $\omega$), except that you need to decide which direction in the fiber it takes you (that's $s$).  This specializes to the case where the bundle is the trivial $\mathbb{R}$-bundle.  Then $s$ is actually just a function, and we're recovering the fact that $n$-forms are a module over $C^\infty(M,\mathbb{R})$.&lt;/li&gt;&#xA;&lt;li&gt;Like you said in your first comment, a connection takes in tangent vectors to the total space.  From here, this is exactly what I said in (2).&lt;/li&gt;&#xA;&lt;li&gt;Well, any form is nilpotent -- if you wedge it with itself enough times, you eventually get 0 for dimensional reasons.  (This assumes that the (co)tangent space is finite-dimensional, of course.  I have no idea what happens in the case that it isn't.)  The only obvious reason (to me) that $A\wedge A\wedge A\wedge A$ would be traceless is if it's 0.  But then again, I don't know anything about Chern-Simons theory.  Or maybe it's a trick along the lines of the proof that the commutator of two square matrices is always traceless (? I'm pretty sure that's true, and follows from the fact that $Tr(AB)=Tr(BA)$, which you should note does &lt;em&gt;not&lt;/em&gt; imply that $Tr(ABC)=Tr(ACB)$, only that you can &lt;em&gt;cyclically&lt;/em&gt; permute the matrices so e.g. $Tr(ABC)=Tr(CAB)=Tr(BCA)$).&lt;/li&gt;&#xA;&lt;li&gt;All I know is that back in linear algebra, you learn that the trace of a matrix is invariant under basis change.  In other words, we can have a well-defined notion of the trace of a linear map.  Of course, this requires that the dimensions of the domain and codomain agree, which doesn't seem like it'll be true in the example you gave (of a vector-valued differential form).  However, if they do agree, then (in a basis) you can consider a form to be matrix-valued, and then the trace is in all likelihood just the sum of the diagonal entries, which are themselves ordinary differential forms.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="2159" LastEditorUserId="2159" LastEditorDisplayName="" LastEditDate="2010-10-06T08:01:23.200" LastActivityDate="2010-10-06T08:01:23.200" CommentCount="7" />
  <row Id="6004" PostTypeId="2" ParentId="6002" CreationDate="2010-10-04T07:35:51.490" Score="2" ViewCount="" Body="&lt;p&gt;This isn't even true in finite posets.&#xA;Let $S=\{a,b,c\}$ where the partial order is&#xA;defined by $a &amp;lt; c$, $ b &amp;lt; c $ and $a$ and $b$ are incomparable.&#xA;Then $\{a\}$ and $\{b\}$ are LU-closed but&#xA;$\{a,b\}$ isn't.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added&lt;/strong&gt;&#xA;The supremum you seek is not&#xA;$\bigcup_{E\in\xi}E$ but rather $LU(\bigcup_{E\in\xi}E)$.&lt;/p&gt;&#xA;" OwnerUserId="226" LastEditorUserId="226" LastEditorDisplayName="" LastEditDate="2010-10-04T08:18:16.697" LastActivityDate="2010-10-04T08:18:16.697" CommentCount="1" />
  <row Id="6005" PostTypeId="1" AcceptedAnswerId="6006" CreationDate="2010-10-04T07:43:46.317" Score="1" ViewCount="104" Body="&lt;p&gt;Anybody knows whether there is a list, say first 1000 sphenic numbers? It would be much helpful to know.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance&lt;/p&gt;&#xA;" OwnerUserId="1869" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-10-04T08:25:53.513" LastActivityDate="2010-10-04T08:25:53.513" Title="Where can I find a list of sphenic numbers?" Tags="&lt;number-theory&gt;&lt;prime-numbers&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="6006" PostTypeId="2" ParentId="6005" CreationDate="2010-10-04T07:52:11.817" Score="6" ViewCount="" Body="&lt;p&gt;The first 10000 are listed &lt;a href=&quot;http://oeis.org/classic/b007304.txt&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;. (Found via &lt;a href=&quot;http://en.wikipedia.org/wiki/Sphenic_number&quot; rel=&quot;nofollow&quot;&gt;Wikipedia: Sphenic number&lt;/a&gt; and &lt;a href=&quot;http://oeis.org/classic/A007304&quot; rel=&quot;nofollow&quot;&gt;OEIS&lt;/a&gt;.)&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2010-10-04T07:52:11.817" />
  <row Id="6007" PostTypeId="2" ParentId="5997" CreationDate="2010-10-04T08:03:35.490" Score="1" ViewCount="" Body="&lt;p&gt;It's not clear to me whether this is just an exercise for its own sake, or meant to serve some real purpose. If is is meant to be like a drawing program where you want a nice curve that &quot;follows&quot; the points that the user selects, the interpolating polynomial curve is probably not a very good idea, since it will oscillate wildly when you have many points (and most likely go way outside of the screen). In such a situation, &lt;a href=&quot;http://en.wikipedia.org/wiki/B%C3%A9zier_curve&quot; rel=&quot;nofollow&quot;&gt;Bézier curves&lt;/a&gt; should be more useful.&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2010-10-04T08:03:35.490" CommentCount="2" />
  <row Id="6008" PostTypeId="1" CreationDate="2010-10-04T08:49:46.183" Score="4" ViewCount="166" Body="&lt;p&gt;Hi, folks!&lt;br&gt;&#xA;I have the following problem:&lt;br&gt;&#xA;Suppose you have a coin that has chance p of landing heads. Suppose you flip the coin N times and let&lt;br&gt;&#xA;X denote the number of &quot;head runs&quot; in N flips. A &quot;head run&quot; is defined as any sequence of heads.&lt;br&gt;&#xA;For example the sequence HHTHHHHHTTTTHHTHT contains 4 head runs. I want to compute E(X) and Var(X).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a difficulty especially with variance.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks. &lt;/p&gt;&#xA;" LastActivityDate="2010-10-05T10:24:50.807" Title="The Expectation and the Variance of the runs" Tags="&lt;probability-theory&gt;" AnswerCount="4" CommentCount="0" FavoriteCount="1" />
  <row Id="6009" PostTypeId="1" AcceptedAnswerId="6016" CreationDate="2010-10-04T08:52:16.870" Score="1" ViewCount="216" Body="&lt;p&gt;1.Find the number of ways in which a mixed double tennis game can be arranged between 10 players consisting of 6 men and 4 women.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My approach : Since,if every pair of selection of men and women gives rise to 4 matches so &lt;strong&gt;C(6,2) * C(4,2) * 4.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2.Find the number of ways in which a mixed double game can be arranged from amongst nine married couples, if no husband and wife play in the same game.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My approach : &lt;strong&gt;C(9,2) * C(7,2) * 4.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, this two questions comes from my test paper I would be grateful if somebody check whether this is correct or not since my answers doesn't matched to the given solution in the paper.&lt;/p&gt;&#xA;" OwnerUserId="2109" LastActivityDate="2010-12-19T07:45:59.363" Title="Two question on Permutation and Combination" Tags="&lt;combinatorics&gt;" AnswerCount="2" CommentCount="5" />
  <row Id="6010" PostTypeId="2" ParentId="6008" CreationDate="2010-10-04T08:59:26.910" Score="0" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Monte Carlo method&lt;/strong&gt; (actually flipping the coin)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming $x_i$ is the number of &quot;head runs&quot; in trial $i$, you can compute the sample expectation (or rather, the mean) after $n$ trials:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$E[X]=\sum\limits_{i=0}^n x_i$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The variance is calculated in a similar manner:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$V[X]=E[X^2]-E[X]^2=\sum\limits_{i=0}^n x_i^2 - \left(\sum\limits_{i=0}^n x_i\right)^2$$&lt;/p&gt;&#xA;" OwnerUserId="708" LastEditorUserId="708" LastEditorDisplayName="" LastEditDate="2010-10-04T14:17:26.210" LastActivityDate="2010-10-04T14:17:26.210" CommentCount="3" />
  <row Id="6011" PostTypeId="2" ParentId="5983" CreationDate="2010-10-04T09:22:46.490" Score="2" ViewCount="" Body="&lt;p&gt;As you ask only for hints, here are two:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Step 1: Show that some/any $p$-Sylow subgroup $P$ of $G$ acts transitively on the $n$ letters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Step 2: Then consider the action of a central element of $P$ of order $p$.&lt;/p&gt;&#xA;" OwnerUserId="669" LastActivityDate="2010-10-04T09:22:46.490" CommentCount="5" />
  <row Id="6012" PostTypeId="2" ParentId="6000" CreationDate="2010-10-04T09:44:32.807" Score="2" ViewCount="" Body="&lt;p&gt;Given:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ M = \Sigma_{i=1}^{n} t_i M_i$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $t_i$ are the unknown parameters and $M_i$ are known matrices and $n$ is the number of parameters, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $N$ be the size of the matrix $M$ and $\lambda_j$, $j = 1, . . . , N$ its eigenvalues, then one can use the identities:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ tr(M^k) = \Sigma_{j=1}^{N} \lambda_j^k$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;to obtain N polynomial equations of the parameters, for example for $k=2$, the equation has the form:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ \Sigma_{i=1}^{n} \Sigma_{j=1}^{n} t_i t_j tr(M_i M_j) = \Sigma_{j=1}^{N} \lambda_j^2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the righthand side is known and the polynomial coefficients of the left hand side are traces of products of the known submatrices.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the parameters can be found in principle from a numerical solution of the polynomial equation system.&lt;/p&gt;&#xA;" OwnerUserId="1430" LastActivityDate="2010-10-04T09:44:32.807" CommentCount="3" />
  <row Id="6013" PostTypeId="1" CreationDate="2010-10-04T10:05:55.423" Score="1" ViewCount="283" Body="&lt;p&gt;I have used logic for specification of security policies in a security model. One of my reviewers asked a question that &quot;why did you used logic and not used algebra for this purpose, and what is your justification?&quot;.&#xA;The first answer receives in my mind is that algebra is more suitable for using in modeling the behaviors (dynamic aspects). For example for specification of security protocols. However, logic is more suitable for specification of properties (static aspects). For example for specification of the security properties we require to satisfy in a computing environment, or the security properties we desire to be satisfied in a security protocol. In other words, algebra is suitable to say HOW, but logic is suitable to say WHAT.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My questions regarding this issue are:&#xA;1- Is my initial answer to this question correct or not?&#xA;2- What are the other answers for this question?&#xA;3- What are the main references describing where we should use logic, and where we should use algebra in their applications in computer science?&lt;/p&gt;&#xA;" OwnerUserId="1831" LastActivityDate="2010-10-04T17:32:40.470" Title="Applications of Logic and Algebra in Computer Science" Tags="&lt;logic&gt;&lt;algebra&gt;" AnswerCount="3" />
  <row Id="6014" PostTypeId="2" ParentId="5997" CreationDate="2010-10-04T10:33:39.387" Score="0" ViewCount="" Body="&lt;p&gt;As I mentioned in &lt;a href=&quot;http://math.stackexchange.com/questions/5166&quot;&gt;the answer to this question&lt;/a&gt;, &lt;em&gt;parametric (cubic) splines&lt;/em&gt; which are represented as &lt;em&gt;piecewise polynomials&lt;/em&gt; in each coordinate $(x,y)$ are the most useful choice, when one wants to draw a curve passing through points that do not necessarily represent a function. (The problem with Bézier is that it treats the input points as a convex hull; i.e. the curve is within the set of points but does not pass through them.)&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/hS0XJ.png&quot; alt=&quot;spline vs. Bezier&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Comparison of a parametric spline (red) and a Bézier curve (blue).&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;An interpolating polynomial is bad form as well for arbitrary point sets; the Runge phenomenon is one vivid example.&lt;/p&gt;&#xA;" OwnerUserId="498" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-10-04T10:40:22.963" LastActivityDate="2010-10-04T10:40:22.963" CommentCount="5" />
  <row Id="6015" PostTypeId="2" ParentId="6013" CreationDate="2010-10-04T11:00:50.930" Score="2" ViewCount="" Body="&lt;p&gt;At some level, algebra and logic are equivalent.  See &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0883853272&quot; rel=&quot;nofollow&quot;&gt;Logic as Algebra&lt;/a&gt; by Paul Halmos and Steven Givant.&lt;/p&gt;&#xA;" OwnerUserId="247" LastActivityDate="2010-10-04T11:00:50.930" CommentCount="1" />
  <row Id="6016" PostTypeId="2" ParentId="6009" CreationDate="2010-10-04T11:09:56.230" Score="2" ViewCount="" Body="&lt;p&gt;The answer to (1) is 2*C(6,2)*C(4,2) (and not with a 4), since M1, W1 playing M2, W2 is the same as M2, W2 playing M1, W1. I don't think the question wants you to take account of which end they're playing from. You've made a similar error in part(2).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To make it more clear you've included a factor 2 for swapping the women over, say, in our set of four people, and then included another factor of 2 for swapping the men over, but have of course ended up with the same match when both swaps have taken place.&lt;/p&gt;&#xA;" OwnerUserId="1301" LastEditorUserId="1301" LastEditorDisplayName="" LastEditDate="2010-10-04T11:21:47.560" LastActivityDate="2010-10-04T11:21:47.560" CommentCount="2" />
  <row Id="6017" PostTypeId="1" AcceptedAnswerId="6021" CreationDate="2010-10-04T12:02:46.100" Score="2" ViewCount="173" Body="&lt;p&gt;How does one find the solution of&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\dfrac{dy}{dx}\left( 1-\left( 1-t\right) x-x^{2}\right) -\left( 1+h\left( 1+t\right) +x\right) y=0\quad ?$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $h$ is an integer constant and $t$ is a real constant between $0$ and $1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$($ In Roger Apéry, &lt;em&gt;Interpolations de Fractions Continues et Irrationalité de certaines Constantes&lt;/em&gt;, Bull. section des sciences du C.T.H.S., n.º3, p.37-53, the solution is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$y=(1-x)^{-1-h}(1+tx)^{h}.)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: The sequence $(v_{h,n})$ in $y=f_{h}(x)=\displaystyle\sum_{n\ge 0}v_{h,n}x^n$  satisfies a recurrence related to $\log (1+t)$.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added&lt;/strong&gt;: Copy of the original with the equation and solution&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/DGW2b.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Addendum&lt;/strong&gt; 2: I transcribe the comment in the 1st answer: &quot;the corrected differential equation above agrees with the recurrence in your excerpt so there is clearly a typo in the printed differential equation.&quot;&lt;/p&gt;&#xA;" OwnerUserId="752" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2010-10-04T17:48:26.850" LastActivityDate="2010-10-04T17:48:26.850" Title="Solving a differential equation related to $\log (1+t)$" Tags="&lt;calculus&gt;&lt;differential-equations&gt;" AnswerCount="1" CommentCount="9" />
  <row Id="6018" PostTypeId="2" ParentId="6013" CreationDate="2010-10-04T12:49:32.743" Score="3" ViewCount="" Body="&lt;p&gt;I don't like this kind of question &amp;mdash;not yours, your referee's&amp;mdash; it's enough for authors to solve problems, the existence of other routes that other people might prefer is besides the point, unless there is some relative disadvantage that can the referee can point out.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;em&gt;Is my initial answer to this question correct or not?&lt;/em&gt; &amp;mdash; It's not what I would say.  There are plenty of approaches to handle mutability in logic, of which dynamic logic is probably the most widely applied.&lt;/li&gt;&#xA;&lt;li&gt;&lt;em&gt;What are the other answers for this question?&lt;/em&gt; &amp;mdash; I should think that ion your shoes I would say something like: &quot;The choice of logic vs. algebra is much less important than the design decisions made in formalising the specification.  This is what seemed to me to be most well suited at the time.&quot;&lt;/li&gt;&#xA;&lt;li&gt;&lt;em&gt;What are the main references describing where we should use logic, and where we should use algebra in their applications in computer science?&lt;/em&gt; &amp;mdash; I don't know of anything along the lines of what you are asking.  Blackburn, de Rijke &amp;amp; Venema (2001) &lt;em&gt;Modal Logic&lt;/em&gt; talks about the similarities and differences between algebraic and model-theoretic techniques - I left with the impression that the biggest issue was whether you preferred to work with completeness theorems or representation theorems.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;To say something in addition to John's answer, here was a thread on Math Overflow, &lt;a href=&quot;http://mathoverflow.net/questions/11974/is-there-a-relationship-between-model-theory-and-category-theory&quot; rel=&quot;nofollow&quot;&gt;Is there a relationship between model theory and category theory&lt;/a&gt;, that I think gives some impression on how little distance there is seen to be between algebra and logic.&lt;/p&gt;&#xA;" OwnerUserId="100" LastActivityDate="2010-10-04T12:49:32.743" CommentCount="3" />
  <row Id="6019" PostTypeId="2" ParentId="5774" CreationDate="2010-10-04T13:26:21.830" Score="2" ViewCount="" Body="&lt;p&gt;I think you're trying to ask:  how many such grids admit a non-trivial symmetry (in the dihedral group)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;For fixed m&lt;/em&gt;&amp;ge;1, asymptotically almost all (0,1,-1)-matrices with exactly m 1's and m -1's do not admit a non-trivial symmetry (rotation/reflection).  For these matrices, we can consider the X's as the 1's and the (letter) O's as the -1's.  The (number) 0's represent the empty cells.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since m is fixed, there are ${{N^2} \choose {m,m,N^2-2m}}=N^{4m}/m!^2+o(N^{4m})$ (0,1,-1)-matrices in total with exactly m 1's and m -1's.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The number of such matrices that are preserved under transposition is \[\sum_{i,j} {N \choose {i,j,n-i-j}} {{N \choose 2} \choose (m-i)/2} {{{N \choose 2}-(m-i)/2} \choose (m-j)/2}\] where i is the number of 1's on the main diagonal, j is the number of -1's on the main diagonal.  Therefore we require 0&amp;le;i+j&amp;le;m and m-i and m-j even.  A crude upper bound to the above summand is $\mathrm{const} \cdot N^{i+j+2(m-i)/2+2(m-j)/2} \leq N^{3m}=o(N^{4m})$.  Since m is fixed, there is only a finite number of pairs i,j which is summed over, so the overall result is $o(N^{4m})$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The number of such matrices that admit any of the other non-trivial symmetries is bounded above by the above formula also.  So asymptotically almost all (0,1,-1)-matrices with exactly m 1's and m -1's do not admit a non-trivial symmetry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Actually, I suspect that this result would be true without the fixed m condition, but I can't think of how to prove it off-hand.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in answer to the question, unless you're dealing with a very small case, and unless you're deliberately trying to create a symmetry, you're unlikely to create a matrix that has a non-trivial symmetry.  For the search tree, most of the time you will be able to identify 8 nodes corresponding to equivalent games of naughts and crosses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[PS: it would be messy (although possible) to find an exact formula along these lines for the number of such matrices that admit a non-trivial symmetry.]&lt;/p&gt;&#xA;" OwnerUserId="139" LastActivityDate="2010-10-04T13:26:21.830" CommentCount="1" />
  <row Id="6020" PostTypeId="1" AcceptedAnswerId="6022" CreationDate="2010-10-04T13:58:04.677" Score="0" ViewCount="119" Body="&lt;p&gt;Let $\mathcal{F}$ be an open cover of a metric space $(X,d)$. I have a question concerning some algebraic aspects of $\mathcal{F}$ and $X$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First of all, is it always possible to consider $(X,d)$ a subspace of some bigger space $(X^{\prime},d)$, where $d$ is the inherited metric from $X^{\prime}$? If this is possible, can we say that $\mathcal{F}$ is a subset of the power set of $X^{\prime}$? What are the conditions that we must state for this to be true?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moreover, if all of that is true, can we say that $\mathcal{F}$ is an algebra of sets? Of course, for it to be true, $\mathcal{F}$ should contain both $X^{\prime}$ and $\emptyset$, but, whatever the cover is, we can always add those two sets and not change anything about the cover, right? So, finally, whatever $\mathcal{F}$ is, can we say that $\mathcal{F} \cup X^{\prime} \cup \emptyset$ is an algebra of sets over $X^{\prime}$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some motivation: in one of my exams I could prove something just saying &quot;It is enough to notice that the cover is an algebra of sets&quot;, but I was unsure if that was always true.&lt;/p&gt;&#xA;" OwnerUserId="1415" LastActivityDate="2010-10-04T15:59:49.390" Title="A question about open covers" Tags="&lt;topology&gt;&lt;metric-spaces&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="6021" PostTypeId="2" ParentId="6017" CreationDate="2010-10-04T14:31:45.970" Score="6" ViewCount="" Body="&lt;p&gt;If the given solution is correct then the posted differential equation is wrong. Instead it should be as follows, with the corrected terms underlined:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$y^{\prime }\ \left( 1-\left( 1-t\right) x - \underline{tx^{2}}\right) -\left( 1+h\left( 1+t\right) + \underline{tx}\right)\ y\ =\ 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which of course is trivially integrable since&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \frac{y'}y\ =\ \frac{1+h}{1-x}\ +\: \frac{ht}{1+tx} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Update: the corrected differential equation above agrees with the recurrence in your excerpt so there is clearly a typo in the printed differential equation.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-10-04T17:24:12.233" LastActivityDate="2010-10-04T17:24:12.233" CommentCount="4" />
  <row Id="6022" PostTypeId="2" ParentId="6020" CreationDate="2010-10-04T14:42:05.470" Score="3" ViewCount="" Body="&lt;p&gt;Your first question is rather vague: you can always take $X'=X$, $d'=d$, and then $(X,d)$ is a subspace of $(X',d')$ with the inherited metric. Presumably you first of all want $X$ &lt;em&gt;properly contained&lt;/em&gt; in $X'$ (&quot;bigger&quot; is a too vague, sorry!). But - do you also want some other property for $X'$, besides merely containing $X$? Perhaps &quot;complete&quot;, or something along those lines?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If all you want is a metric space that properly contains $X$ and with induced metric, then the answer is yes. For example, if $X$ has only one element, then you can let $X'=X\cup\{p\}$, where $p\notin X$, and take the discrete metric. If $X$ has more than one element, you can take $X'=X\times X$, and define a metric by $d'((a,b),(x,y)) = d(a,x) + d(b,y)$ (the &quot;taxicab norm&quot;). To verify that this is indeed a metric, note that since $d(r,s)\geq 0$ for all $r,s\in X$, we have $d'((a,b),(x,y))\geq 0$; and $d'((a,b),(x,y))=0$ if and only if $d(a,x)=d(b,y)=0$, if and only if $a=x$ and $b=y$, if and only if $(a,b)=(x,y)$. Also, $d'((a,b),(x,y)) = d(a,x)+d(b,y) = d(x,a)+d(y,b) = d'((x,y),(a,b))$. Finally, for the triangle inequality, we have that &#xA;$$\begin{array}{rcl}&#xA;d'((a,b),(r,s))+d'((r,s),(x,y)) &amp;amp;=&amp;amp; d(a,r)+d(b,s) + d(r,x) + d(s,y)\\&#xA;&amp;amp; \geq &amp;amp; d(a,x) + d(b,y) = d'((a,b),(x,y)).&#xA;\end{array}$$&#xA;So $(X',d')$ is a metric space. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now fix $x_0\in X$, and embed $X$ into $X'$ by mapping $a\in X$ to $(a,x_0)$. This is an embedding, and for all $a,b\in X$, $d'((a,x_0),(b,x_0)) = d(a,b)+d(x_0,x_0)=d(a,b)$, so if we identify $X$ with its image in $X'$ you get that $d$ is the restriction of $d'$ to $X$. Since we are assuming that $X$ has more than one element, (the image of) $X$ is properly contained in $X'$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you say &quot;$\mathcal{F}$ is an open cover for a metric space $(X,d)$&quot;, I interpret this to mean that each element of $\mathcal{F}$ is an open subset of $X$ and that $\cup\mathcal{F}=X$. You seem to be thinking that the sets may be strictly larger than $X$, but in the absence of a given overset, this makes no sense. Thus, we can talk about open covers of $[0,1]$ &lt;em&gt;in $\mathbb{R}$&lt;/em&gt; (open subset of $\mathbb{R}$ whose union contains $[0,1]$); but if we just say &quot;an open cover of $\mathbb{R}$&quot;, we assume the sets are subsets  of $\mathbb{R}$ absent any other context.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if $X$ is a set, $X'$ contains $X$, and $\mathcal{F}$ is a family of subsets of $X$, then $\mathcal{F}$ is also a family of subsets of $X'$, hence a subset of the power set of $X'$; so the answer to your second question (&quot;can we say $\mathcal{F}$ is a subset of the power set of $X'$?&quot;) is an absolute yes. The fact that we are dealing with a family of open sets or any such thing is completely immaterial; this is simply a consequence of the meaning of &quot;subset&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for being an algebra: an algebra of sets must be closed under unions and complements. If you have an open cover, it is not necessarily closed under unions, and it will &lt;em&gt;usually&lt;/em&gt; not be closed under complements (the complements will be closed; if they are also part of the open cover, then each element of the open cover except for $X$ and $\emptyset$ (if they are there) would disconnect the space). So &quot;most of the time&quot; an open cover will not be closed under complementation and hence not be an algebra. (&quot;Closing&quot; the cover under unions is fairly easy; just throw in the unions of any two elements until you are closed under unions; all of them will be open, and enlarging an open cover by adding open sets will not change the fact that it is an open cover).&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-10-04T15:59:49.390" LastActivityDate="2010-10-04T15:59:49.390" CommentCount="3" />
  <row Id="6023" PostTypeId="2" ParentId="5826" CreationDate="2010-10-04T15:22:16.763" Score="0" ViewCount="" Body="&lt;p&gt;Now, I think I've got a possible real answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once connection assumed, I'm going to develop my previous comment in order to prove that, at least for compact connected surfaces, connected sums do NOT depend on anything at all. Feel free to point out if I'm missing something or any mistakes in my reasonings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem of the independence of the disks and &quot;gluing&quot; homeomorphisms can be stated as follows: Let $X$ and $Y$ be two topological surfaces, $D$ a disc in $X$ and $D'$ a disc in $Y$. Let $h: \partial D \longrightarrow  \partial D'$ be a homeomorphism. Is it necessarily true that $X \oplus_h Y $ does not depend on $h$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To see this, you could proceed as follows. First of all, you need the theorem of classification of surfaces, which says:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Classification theorem.&lt;/strong&gt; Let $X$ be a connected compact surface. Then $X$ is homeomorphic to one, and only one, of the following &lt;em&gt;standard surfaces&lt;/em&gt;:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The sphere $\mathbb{S}^2$.&lt;/li&gt;&#xA;&lt;li&gt;A connected sum of $g$ toruses $g\mathbb{T}^2 = \mathbb{T}^2 \oplus   \dots \oplus \mathbb{T}^2 $.&lt;/li&gt;&#xA;&lt;li&gt;A connected sum of $g$ projective planes $g\mathbb{RP}^2 = \mathbb{RP}^2 \oplus   \dots \oplus \mathbb{RP}^2  $.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Warning.&lt;/strong&gt; There is no circular argument here when talking about &quot;the&quot; connected sum of toruses or projective planes, provided you make the following convention. For the time being, this connected sum just means that we are representing spheres, connected sums of toruses, or projective planes by means of the standard polygonal presentations&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;(a)&lt;/strong&gt; $\mathbb{S}^2 = aa^{-1}$ .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;(b)&lt;/strong&gt; $g \mathbb{T}^2 =a_1b_1a_1^{-1}b_1^{-1} \dots a_gb_ga_g^{-1}b_g^{-1}$ .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;(c)&lt;/strong&gt; $g \mathbb{RP}^2 = a_1a_1 \dots a_ga_g$ .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, &lt;em&gt;define&lt;/em&gt; the connected sum of the standard surfaces by choosing the same &quot;standard&quot; small (such that its boundary doesn't touch the polygon) $D = D'$ disk centered at the barycenters of the polygons and with $h = \mathrm{id} : \partial D \longrightarrow \partial D'$ as homeormorphism.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course there could be still a little problem here with orientations: you could &quot;glue&quot;, for instance, two spheres with this convention in two apparently different ways: &quot;side by side&quot;, or one inside the other. Fortunately, for surfaces at least, this doesn't matter (you can &quot;extract&quot; the inner sphere without changing the homeomorphism type).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, with this definition, you can prove the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proposition 1.&lt;/strong&gt; For the standard surfaces, you have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;(1)&lt;/strong&gt; $\mathbb{S}^2 \oplus X \cong X$, for any standard surface $X$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;(2)&lt;/strong&gt; $g\mathbb{T}^2 \oplus g'\mathbb{T}^2 \cong (g+g')\mathbb{T}^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;(3)&lt;/strong&gt; $g\mathbb{RP}^2 \oplus g'\mathbb{RP}^2 \cong (g+g')\mathbb{RP}^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;(4)&lt;/strong&gt; $g\mathbb{T}^2 \oplus g'\mathbb{RP}^2 \cong (2g +g')\mathbb{RP}^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, we already know what the connected sum of the standard surfaces (with the aforementioned convention) is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;In fact, we can say more.&lt;/strong&gt; For standard surfaces, keeping the &quot;standard&quot; disk fixed once for all we can choose &lt;em&gt;any&lt;/em&gt; homeomorphism $h: \partial D \longrightarrow \partial D'$ we please. To see this we must rely on the aforementioned classification theorem and the computation of the homology of the standard surfaces, which says, among other things, that the knowledge of $H_1(X)$ suffices in order to classify our surface.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, when we try to compute, for instance, the homology of $X = g\mathbb{T}^2 \oplus g'\mathbb{T}^2$, we may use Mayer-Vietoris, choosing $g\mathbb{T}^2 \backslash V$  and $g'\mathbb{T}^2\backslash V'$ as a cover for $g\mathbb{T}^2 \oplus g'\mathbb{T}^2 $, where $V$ and $V'$ are the interior of $D$ and $D'$, respectively. We have $\mathbb{S}^1 = (g\mathbb{T}^2 \backslash V) \cap (g'\mathbb{T}^2 \backslash V')$ and the interesting part of the Mayer-Vietoris sequence is just&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;0 \longrightarrow H_2(X) \longrightarrow H_1(S^1) \stackrel{f}{\longrightarrow} H_1( g\mathbb{T}^2 \backslash V  )\oplus H_1 ( g'\mathbb{T}^2\backslash V' ) \longrightarrow H_1(X) \longrightarrow 0 \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here, $f$ is the morphism induced by the inclusions $\mathbb{S}^1 \hookrightarrow g\mathbb{T}^2 \backslash V$ and $\mathbb{S}^1 \hookrightarrow g'\mathbb{T}^2 \backslash V' $. Now, the first morphism is just the identity on $\mathbb{S}^1 \longrightarrow \mathbb{S}^1 \subset  g\mathbb{T}^2 \backslash V$ and the second one is $h:\mathbb{S}^1 \longrightarrow \mathbb{S}^1 \subset  g'\mathbb{T}^2 \backslash V' $. But any homeomorphism $\mathbb{S}^1 \longrightarrow \mathbb{S}^1$ is homotopic to $\pm \mathrm{id} : \mathbb{S}^1 \longrightarrow \mathbb{S}^1$. So, an easy computation shows that $f = 0$ and the result follows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;VERY IMPORTANT REMARK.&lt;/strong&gt; Here is the point where, for standard surfaces at least, the &quot;gluing&quot; homeomorphism $h: \partial D = \mathbb{S}^1 \longrightarrow \mathbb{S}^1 = \partial D'$ &quot;disappears&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The next result we need is that we can &quot;move&quot; our disks wherever we please on the surface.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proposition 2.&lt;/strong&gt; Let $X$ be a connected surface and $D, D'$ two disks in $X$. Then there exists a homeomorphism $H: X \longrightarrow X$ such that $H(D) = D'$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The proof of this result is more involved than the previous ones and needs &lt;a href=&quot;http://en.wikipedia.org/wiki/Sch%C3%B6nflies_problem&quot; rel=&quot;nofollow&quot;&gt;Schönflies theorem&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, we are prepared to prove the independence of the connected sum. Let $X$ and $Y$ be two connected compact surfaces, $D\subset X$ and $D'\subset Y$ two disks inside of them, $h: \partial D \longrightarrow \partial D'$ a homeomorphism. Let's show that $X \oplus_h Y$ doesn't depend on $h$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;(1)&lt;/strong&gt; Because of the classification theorem $X$ and $Y$ are homeomorphic to two standard surfaces: $S_X$ and $S_Y$, respectively. Let $\phi : X \longrightarrow S_X$ and $\psi : Y \longrightarrow S_Y$ denote the homeomorphisms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;(2)&lt;/strong&gt; These homeomorphisms $X \cong S_X$ and $Y \cong S_Y$ don't need to identify $D$ and $D'$ with our standard disks on the standard surfaces $S_X$ and $S_Y$, but for this we have proposition 2, which tells that we can &quot;move&quot; $D$, $D'$ to those standard disks using two homeomorphisms $H_X : X \longrightarrow X$ and $H_Y: Y \longrightarrow Y$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;(3)&lt;/strong&gt; Now we glue our standard surfaces along their standard disks using the homeomorphism induced by $h$ on them: $h' = H_Y\psi h  \phi^{-1} H_X^{-1}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;(4)&lt;/strong&gt; Because our &quot;we can say more&quot;, the connected sum $S_X \oplus_{h'} S_Y$ does not depend on $h'$. Hence, neither $X\oplus_h Y \cong S_X \oplus_{h'} S_Y$ depends on $h$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;References:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;V. Navarro, P. Pascual, &quot;Topologia Algebraica&quot; (Algebraic Topology), Edicions UB (1999).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;J.M. Lee, &quot;Introduction to topological manifolds&quot;, Springer (2000).&lt;/p&gt;&#xA;" OwnerUserId="664" LastEditorUserId="664" LastEditorDisplayName="" LastEditDate="2010-10-04T16:24:59.357" LastActivityDate="2010-10-04T16:24:59.357" />
  <row Id="6024" PostTypeId="1" AcceptedAnswerId="6025" CreationDate="2010-10-04T15:55:01.473" Score="0" ViewCount="89" Body="&lt;p&gt;I recently started with 2d transformations in my class. I was just working with a program when I realized the formula I am using rotates the object spirally inwards. I have no idea whats over about this. &#xA;I expected my figures to be rotated about a point but rather they all deform with every degree of rotation. I used a single point &amp;amp; realized that the point was moving spirally inwards with every degree I moved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can someone help me &amp;amp; let me know why it falls in rather than following circle or if that is what is expected of 2d rotation. I somehow felt its something related to how we view in 2d plane (or the imaginary z index here) but simply can't get around it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am following &lt;a href=&quot;http://en.wikipedia.org/wiki/Transformation_matrix&quot; rel=&quot;nofollow&quot;&gt;wiki&lt;/a&gt; &amp;amp; talking about this :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/ldY8x.png&quot; alt=&quot;alt text&quot;&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;x' = xcosθ − ysinθ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;y' = xsinθ + ycosθ&lt;/p&gt;&#xA;" OwnerUserId="2305" LastActivityDate="2010-10-04T16:16:08.507" Title="2d rotation falls spirally inwards?" Tags="&lt;transformation&gt;&lt;computational-geometry&gt;&lt;rigid-transformation&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="6025" PostTypeId="2" ParentId="6024" CreationDate="2010-10-04T16:16:08.507" Score="6" ViewCount="" Body="&lt;p&gt;My guess is that this is a &lt;em&gt;very&lt;/em&gt; classic error in computer graphics implementation; you have to make sure that you don't accidentally use the new value of X when computing the new value of y!  For instance, this code:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;x = x*cos(theta) - y*sin(theta);&#xA;&lt;br/&gt;&#xA;y = x*sin(theta) + y*cos(theta);&#xA;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;will actually perform the following operation:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;x'  =x cos &amp;theta; - y sin &amp;theta;&#xA;&lt;br/&gt;&#xA;y' = x cos &amp;theta; sin &amp;theta; + y (cos &amp;theta; - sin&lt;sup&gt;2&lt;/sup&gt;&amp;theta;)&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and for small values of &amp;theta; (i.e., incremental rotation) will generally cause the point &amp;lt;x,y&amp;gt; to spiral in towards &amp;lt;0,0&amp;gt;.&lt;/p&gt;&#xA;" OwnerUserId="785" LastActivityDate="2010-10-04T16:16:08.507" CommentCount="3" />
  <row Id="6026" PostTypeId="1" AcceptedAnswerId="6027" CreationDate="2010-10-04T16:16:14.843" Score="1" ViewCount="146" Body="&lt;p&gt;In the $d$ variable polynomial ring $R=k[x_{1},\cdots,x_{d}]$ show that&#xA;$0, x_{1}R, (x_{1},x_{2})R, \cdots , (x_{1},x_{2},...,x_{d})R$ is a strictly increasing sequence of prime ideals and there is no longer such chain.  how do i prove this claim.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, i am thinking that for this problem this result may be helpful: for a ring, $R$, to be Noetherian, one formulation dictates that any ascending chain of ideals in R terminates. But i just can't get going with it.&lt;/p&gt;&#xA;" LastActivityDate="2010-10-04T18:09:11.197" Title="Sequence of Prime ideals in a Polynomial Ring" Tags="&lt;abstract-algebra&gt;&lt;algebraic-geometry&gt;&lt;homework&gt;&lt;commutative-algebra&gt;" AnswerCount="2" />
  <row Id="6027" PostTypeId="2" ParentId="6026" CreationDate="2010-10-04T16:43:31.457" Score="3" ViewCount="" Body="&lt;p&gt;A standard proof uses Noether normalization. For a superb introduction to (Krull) dimension of rings see Chapter 8, &quot;Introduction to dimension theory&quot; in Eisenbud's &quot;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0387942696&quot; rel=&quot;nofollow&quot;&gt;Commutative algebra, with a view toward algebraic geometry&lt;/a&gt;&quot;, esp. Section 8.2.1, and for the proof see Section 13.1.&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-10-04T16:43:31.457" CommentCount="2" />
  <row Id="6028" PostTypeId="1" AcceptedAnswerId="6073" CreationDate="2010-10-04T17:15:35.760" Score="-1" ViewCount="251" Body="&lt;p&gt;I know how to count derivatives, but I actually don't understand what are those? What do they show? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wikipedia explanation: &lt;code&gt;Loosely speaking, a derivative can be thought of as how much one quantity is changing in response to changes in some other quantity&lt;/code&gt;. I understand nothing what is written here?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please, if you can, explain me as simple as you can it. Thank you very much.&lt;/p&gt;&#xA;" OwnerUserId="2284" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-10-04T18:33:05.843" LastActivityDate="2010-10-05T01:48:17.147" Title="Simple way to understand what derivative is" Tags="&lt;calculus&gt;&lt;functions&gt;" AnswerCount="5" CommentCount="3" />
  <row Id="6029" PostTypeId="2" ParentId="6008" CreationDate="2010-10-04T17:19:07.890" Score="1" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://math.stackexchange.com/questions/2763/what-is-the-expected-number-of-runs-of-same-color-in-a-standard-deck-of-cards&quot;&gt;This question&lt;/a&gt; is very similar and should give some pointers.&lt;/p&gt;&#xA;" OwnerUserId="1910" LastActivityDate="2010-10-04T17:19:07.890" />
  <row Id="6030" PostTypeId="2" ParentId="6028" CreationDate="2010-10-04T17:28:00.150" Score="2" ViewCount="" Body="&lt;p&gt;If one thing, $A$, changes when you change something else, $B$, then the &lt;em&gt;derivative of $A$ with respect to $B$&lt;/em&gt; is the rate of change of $A$ when you make very small changes in $B$.&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2010-10-04T17:28:00.150" CommentCount="1" />
  <row Id="6031" PostTypeId="2" ParentId="6013" CreationDate="2010-10-04T17:32:40.470" Score="0" ViewCount="" Body="&lt;p&gt;You may want to read about the &lt;a href=&quot;http://en.wikipedia.org/wiki/Curry%E2%80%93Howard_correspondence&quot; rel=&quot;nofollow&quot;&gt;Curry-Howard isomorphism&lt;/a&gt; which, for example, tells us that types in typed lambda calculus (thus in programming languages like OCaml, SML, Haskell, among others) actually are propositions in some logic formalism, and that values of these types are proofs of the corresponding propositions. It is e.g used to develop proof assistants that actually also are functional programming languages (Agda, Coq, Epigram, ...). Regarding algebra, you may want to read &quot;Elements of Programming&quot;, by Alexander Stepanov. &lt;/p&gt;&#xA;" OwnerUserId="1865" LastActivityDate="2010-10-04T17:32:40.470" />
  <row Id="6032" PostTypeId="2" ParentId="6008" CreationDate="2010-10-04T17:38:12.360" Score="2" ViewCount="" Body="&lt;p&gt;You can consider the random variable $Y$=number of alternations (from head to tail or viceversa) in the sequence of $N$ trials. It's clear that $Y$ follows a Binomial distribution, with values in $[0,N-1]$ and $p=1/2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$E(Y) = (N - 1) /2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$Var(Y) = (N - 1) /4$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The relation with $X$ takes slightly different forms depending on whether the first ocurrence was a head, and whether $N$ is even or odd.&#xA;This makes the problem a little cumbersone, but solvable nonetheless.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a quick approximate  (presumably quite good with N moderately large) solution:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $Y$ is odd, then: $X=(Y+1)/2$. We assume that this holds always, and then compute expectation and variance by linearity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$E(X) \approx (N-1)/4 + 1/2 \approx N/4$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$Var(X) \approx (N-1)/16  \approx N/16$&lt;/p&gt;&#xA;" OwnerUserId="312" LastActivityDate="2010-10-04T17:38:12.360" CommentCount="1" />
  <row Id="6033" PostTypeId="2" ParentId="6028" CreationDate="2010-10-04T17:45:10.163" Score="1" ViewCount="" Body="&lt;p&gt;Graphically, when you have the x axis, and the y axis with the graph of a (derivable) function f, then the derivative (the &lt;strong&gt;number&lt;/strong&gt;) of f at x0 basically tells you how much the value of the function changes if you move to x0+1. However, this won't be very accurate, since the graph may be very different at x0+1. This is why we have the derivative of f (the &lt;strong&gt;function&lt;/strong&gt;) which gives the above number for any &quot;x0&quot; you give it. &#xA;So it gives the coefficient of the tangent line at each point. It's only about how your function grows for an infinitesimal &quot;move&quot; around some x0.&lt;/p&gt;&#xA;" OwnerUserId="1865" LastActivityDate="2010-10-04T17:45:10.163" />
  <row Id="6034" PostTypeId="2" ParentId="6026" CreationDate="2010-10-04T18:09:11.197" Score="1" ViewCount="" Body="&lt;p&gt;Bill is right that the most standard proof uses Noether normalization.  Another proof exploiting the fact that $R$ is a Hilbert-Jacobson ring can be found in Section 8.2 of&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://math.uga.edu/~pete/integral.pdf&quot; rel=&quot;nofollow&quot;&gt;http://math.uga.edu/~pete/integral.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The exposition here follows Kaplansky's &lt;em&gt;Commutative Rings&lt;/em&gt;.  &lt;/p&gt;&#xA;" OwnerUserId="299" LastActivityDate="2010-10-04T18:09:11.197" CommentCount="1" />
  <row Id="6035" PostTypeId="2" ParentId="6028" CreationDate="2010-10-04T18:12:37.163" Score="0" ViewCount="" Body="&lt;p&gt;You can see this post as well:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;http://www.math.pitt.edu/~sparling/061/23561/23561notes/node5.html&quot; rel=&quot;nofollow&quot;&gt;http://www.math.pitt.edu/~sparling/061/23561/23561notes/node5.html&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;For a geometric visualization please see &lt;a href=&quot;http://www.splung.com/content/sid/1/page/differentiation&quot; rel=&quot;nofollow&quot;&gt;http://www.splung.com/content/sid/1/page/differentiation&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" LastActivityDate="2010-10-04T18:12:37.163" CommentCount="0" />
  <row Id="6036" PostTypeId="1" AcceptedAnswerId="6037" CreationDate="2010-10-04T18:18:16.423" Score="1" ViewCount="109" Body="&lt;p&gt;Let $S$ be a string of length $n$. Each character of $S$ has probability $p$ of being 'A' and probability $1-p=q$ of being 'B'. $R$ is the number of occurrences of the substring 'AB' in $S$. I'd like to determine $\mathbf{E}[R]$ and $\mathbf{var}[R]$ . I can determine the expectation via a recursive definition and total expectation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\mathbf{E}[R] = pq\left(1+\mathbf{E}\left[R_{n-1}\right]\right)+\left(1-pq\right)\mathbf{E}\left[R_{n-1}\right]$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I am not sure how I can approach the problem for determining the variance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried defining a Probability Mass Function but it looks like it will get messy very soon.&#xA;$\Pr(R=x) = \binom{n-x}{x}\cdot\left(pq\right)^{x} \cdot \text{probability of no other ABs}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And that &quot;no other&quot; probability ends up being, for example for $x=1$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sum_{i=0}^{u}p^i q^{u-1} \cdot \sum_{i=0}^{v}p^i q^{u-1}$ summed over all $u+v+2=n$&#xA;but I think there might be a better way.&lt;/p&gt;&#xA;" OwnerUserId="2256" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2011-06-03T20:08:00.037" LastActivityDate="2011-06-03T20:08:00.037" Title="Moments on number of occurrences of substring" Tags="&lt;probability&gt;&lt;combinatorics&gt;&lt;recurrence-relations&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="6037" PostTypeId="2" ParentId="6036" CreationDate="2010-10-04T18:33:45.897" Score="4" ViewCount="" Body="&lt;p&gt;I presume all the characters are independent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $X_1,\ldots,X_{n-1}$ be the random variables defined as follows:&#xA;$X_i=1$ if there is $AB$ starting at the $i$-th position, otherwise $X_i=0$.&#xA;Then $R=X_1+\cdots+X_{n-1}$. By linearity of expectation $E(R)=\sum_i E(X_i)$&#xA;and $E(R^2)=\sum_{i,j} E(X_i X_j)$. Now $E(X_i)=pq$ so $E(R)=(n-1)pq$.&#xA;IF $i=j$ then $X_i X_j=X_i^2=X_i$; if $|i-j|=1$ then $X_i X_j=0$ and&#xA;if $|i-j|\ge2$ then $X_i$ and $X_j$ are independent so that $E(X_i X_j)=E(X_i)^2&#xA;=p^2 q^2$. Now it's just book-keeping to find $E(R^2)$ and so also $Var(R)$.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-10-04T18:33:45.897" CommentCount="2" />
  <row Id="6038" PostTypeId="1" CreationDate="2010-10-04T18:52:45.167" Score="3" ViewCount="163" Body="&lt;p&gt;According to &lt;a href=&quot;http://en.wikipedia.org/wiki/Dynamical_billiards&quot; rel=&quot;nofollow&quot;&gt;this article in Wikipedia&lt;/a&gt;: A billiard is a dynamical system in which a particle alternates between motion in a straight line and specular reflections from a boundary. When the particle hits the boundary it reflects from it without loss of speed. Billiard dynamical systems are Hamiltonian idealizations of the game of billiards, but where the region contained by the boundary can have shapes other than rectangular and even be multidimensional. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is motivated by a videogame I've been playing lately, which can be seen at &lt;a href=&quot;http://www.youtube.com/watch?v=LLLmfwxNJYU&quot; rel=&quot;nofollow&quot;&gt;http://www.youtube.com/watch?v=LLLmfwxNJYU&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Essentially, the &quot;physics&quot; of the game involves several billiards in a polygon, and the player has to slash off pieces of the polygon while avoiding the billiards.  Also, the removed piece has to be void of billiards.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've been assuming that the distribution of billiards is in the long run uniform, in some hand waving sense.  That's assuming that initial distributions and velocities are random.  Is that true, or can the polygon be shaped in various ways to make that distribution non-uniform?  In other words, are certain regions of a polygon are more likely to be void than other regions?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I believe that a term like &quot;ergodic&quot; applies to this, but I'm not confident using it). &lt;/p&gt;&#xA;" OwnerUserId="1257" LastActivityDate="2010-10-04T18:52:45.167" Title="Polygonal billiards and uniform distribution" Tags="&lt;random&gt;&lt;dynamical-systems&gt;" CommentCount="2" FavoriteCount="1" />
  <row Id="6039" PostTypeId="1" AcceptedAnswerId="6205" CreationDate="2010-10-04T19:34:47.573" Score="4" ViewCount="640" Body="&lt;p&gt;Let $p(x)$ be a polynomial with integer coefficients, which is not constant. Then is this condition possible: $$p(x)=3^{x}$$ whenever $x \in \mathbb{N}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Motivation: &lt;a href=&quot;http://mathoverflow.net/questions/34059/if-f-is-infinitely-differentiable-then-f-coincides-with-a-polynomial&quot; rel=&quot;nofollow&quot;&gt;http://mathoverflow.net/questions/34059/if-f-is-infinitely-differentiable-then-f-coincides-with-a-polynomial&lt;/a&gt;&lt;/p&gt;&#xA;" LastEditorUserId="1778" LastEditorDisplayName="" LastEditDate="2010-10-05T01:52:21.237" LastActivityDate="2010-10-06T15:47:50.797" Title="Polynomial satisfying $p(x)=3^{x}$ for $ x \in \mathbb{N}$" Tags="&lt;calculus&gt;&lt;algebra&gt;&lt;polynomials&gt;&lt;diophantine-equations&gt;" AnswerCount="10" CommentCount="13" />
  <row Id="6040" PostTypeId="1" AcceptedAnswerId="6048" CreationDate="2010-10-04T19:36:27.490" Score="0" ViewCount="89" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;The strength of glue is known to follow a normal distribution with a mean of 59 and a standard deviation of 3.4.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Find the mean strength needed such that the probability  that the glue is weaker than 55 is less than 5%.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To solve this, I just guessed and checked in Excel:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;=NORMDIST(55, 50.1, 3.4, 1)&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;=NORMDIST(55, 50.2, 3.4, 1)&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;=NORMDIST(55, 50.3, 3.4, 1)&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;until the cell was close to .05. The answer I got was  60.6, which appears to work. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That probably isn't the &lt;em&gt;proper&lt;/em&gt; way to solve the problem. I searched my book for a problem like it but didn't find any examples. I did find this &quot;Percentiles of an Arbitrary Normal Distribution&quot; formula:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;(100p)th percentile for normal(mu, signma) = mu + [(100p)th for standard normal] * sigma&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But I'm not sure how to apply it (or maybe that formula doesn't help at all). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what's the smart way to solve the problem?&lt;/p&gt;&#xA;" OwnerUserId="1977" LastActivityDate="2010-10-05T06:02:25.053" Title="Find mean of normal distrubution such that it fits a condition" Tags="&lt;probability&gt;&lt;normal-distribution&gt;" AnswerCount="1" />
  <row Id="6041" PostTypeId="2" ParentId="6039" CreationDate="2010-10-04T19:45:24.350" Score="13" ViewCount="" Body="&lt;p&gt;No, a polynomial can't grow exponentially as $x\to\infty$.&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2010-10-04T19:45:24.350" CommentCount="3" />
  <row Id="6042" PostTypeId="2" ParentId="6028" CreationDate="2010-10-04T19:52:51.887" Score="0" ViewCount="" Body="&lt;p&gt;You can find some nice introductory lectures on derivatives in various &lt;a href=&quot;http://www.khanacademy.org/search?page_search_query=derivative&quot; rel=&quot;nofollow&quot;&gt;videos  from Khan's Academy&lt;/a&gt;. The site has a free online collection of a couple thousand videos on mathematics, science, history, and economics. It's worth exploring when you're looking for introductory lectures on such topics.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-10-05T00:31:12.803" LastActivityDate="2010-10-05T00:31:12.803" />
  <row Id="6043" PostTypeId="2" ParentId="6039" CreationDate="2010-10-04T19:59:23.297" Score="2" ViewCount="" Body="&lt;p&gt;Suppose $a_n x^n + a_{n-1} x^{n-1} + \cdots + a_0 = 3^x$ for $x \in \mathbb N$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then $a_n (x+1)^n + a_{n-1} (x+1)^{n-1} + \cdots + a_0 = 3 a_n x^n + 3 a_{n-1} x^{n-1} + \cdots + 3 a_0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the coefficient of $x^n$ on the LHS is $a_n$ and on the RHS it is $3 a_n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This contradicts the assumption.&lt;/p&gt;&#xA;" LastActivityDate="2010-10-04T19:59:23.297" CommentCount="4" />
  <row Id="6044" PostTypeId="1" AcceptedAnswerId="6047" CreationDate="2010-10-04T20:06:51.853" Score="2" ViewCount="463" Body="&lt;p&gt;As a part of my studies in ring theory, I've encountered the concept of an idempotent element, i.e., an element $e$ such that $e^2=e$.&#xA;Are there some interesting examples of rings with idempotent elements?&lt;/p&gt;&#xA;" OwnerUserId="766" LastActivityDate="2010-10-05T07:35:22.180" Title="Examples of rings with idempotent elements" Tags="&lt;abstract-algebra&gt;&lt;ring-theory&gt;&lt;examples&gt;" AnswerCount="6" CommentCount="0" FavoriteCount="1" />
  <row Id="6045" PostTypeId="2" ParentId="6039" CreationDate="2010-10-04T20:09:31.880" Score="1" ViewCount="" Body="&lt;p&gt;Suppose that $p(x)$ exists and has degree $n$.  $\lim_{x\to\infty}\frac{p(x)}{x^{n+1}}=0$ and $\lim_{x\to\infty}\frac{3^x}{x^{n+1}}\to\infty$, but $\frac{p(x)}{x^{n+1}}=\frac{3^x}{x^{n+1}}$ for all $x\in\mathbb{N}$, so it is impossible for the first limit to be finite and the second limit to be unbounded.&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-10-04T20:09:31.880" />
  <row Id="6047" PostTypeId="2" ParentId="6044" CreationDate="2010-10-04T20:18:06.507" Score="5" ViewCount="" Body="&lt;p&gt;As you realize, all rings have the idempotents $0$ and $1$, so the&#xA;question is whether they have any others.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If a commutative ring has a non-trivial idempotent, then it is&#xA;isomorphic to a product of two non-trivial rings. The same&#xA;is true for a non-commutative ring, as long as the idempotent&#xA;lies in its centre.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $n$ is a positive integer which is not a prime power then&#xA;$\mathbb{Z}/n\mathbb{Z}$ has nontrivial idempotents.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Matrix rings tend to have lots of idempotents, but not usually in their centres.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A group algebra $KG$ for a finite group $G$ over a characteristic zero&#xA;field $K$ has the&#xA;central idempotent $|G|^{-1}\sum_{g\in G}g$ and usually others.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-10-04T20:18:06.507" CommentCount="0" />
  <row Id="6048" PostTypeId="2" ParentId="6040" CreationDate="2010-10-04T20:18:28.393" Score="1" ViewCount="" Body="&lt;p&gt;Ok. I will show you an example.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$x$ is distributed normally with mean $a=34$ and standard deviation $\sigma=12$. Find $x_0$ such that $x&gt;x_0$ with probability $0.95$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Suppose $x_0=a-\delta$.  We need to find $\delta$, such that $P(x&gt;a-\delta)=0.95$ or, equivalently, $P(x-a\leq-\delta)=0.05$. First note, that normal distribution is symmetric. Therefore $P(x-a\geq\delta)=0.05$ and $P(|x-a|\geq\delta)=0.1$ or $P(|x-a|&amp;lt;\delta)=0.9$. &lt;a href=&quot;http://en.wikipedia.org/wiki/Normal_distribution#Standard_deviation_and_confidence_intervals&quot; rel=&quot;nofollow&quot;&gt;Here&lt;/a&gt; you can find, that $\delta=n\sigma$, where $n=1.644853626951$ (just search for this number on &lt;a href=&quot;http://en.wikipedia.org/wiki/Normal_distribution#Standard_deviation_and_confidence_intervals&quot; rel=&quot;nofollow&quot;&gt;the page&lt;/a&gt; to see the table, I've used to find it). Therefore the answer is&#xA;$$x_0=a-\delta=a-n\sigma=14.3$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; Your task is actually the same. The difference is that in my task we know $a$ and we are asked to find $a-\delta$ and in your case we know $a-\delta$ and we are asked to find $a$. That's simple. Just write $a= (a-\delta)+\delta$. I don't know Excel, but if your use of it is right, you can find the answer of your task by &lt;code&gt;= NORMINV(0.95, 55, 12)&lt;/code&gt;. Actually, the answers in the following tasks are th same (here $a,\sigma,p$ are given; you can replace $a$, $\sigma$ by any positive numbers and $p$ by any number in $[0,1]$):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;$x$ is distributed normally with mean $a$ and standard deviation $\sigma$. Find $b$ such that $P(x\leq b)=p$.  &lt;/li&gt;&#xA;  &lt;li&gt;$y$ is distributed normally with standard deviation $\sigma$. It is known, that $P(y\geq a)=p$. Find $b$ --- the mean of the distribution of $y$.&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That's because if $x$ is distributed normally with parameters $a$, $\sigma$, then distribution of $z=x-a$ depends only on $\sigma$ (it is independent of $a$) and symmetric (for each $\delta$ we have $p(z&amp;lt;\delta)=p(z&gt;-\delta)$). You can see, that tasks are the same by denoting $x-a$ in the first task by $z$ (and replacing $x$ by $z+a$) and denoting $y-b$ in the second by $z$ (and replacing $y$ by $z+b$). Then both tasks have the following form&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;$z$ is distributed normally with mean 0 and standard deviation $\sigma$. Find $b$ such that $P(z\geq b-a)=p$.&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="695" LastEditorUserId="695" LastEditorDisplayName="" LastEditDate="2010-10-05T06:02:25.053" LastActivityDate="2010-10-05T06:02:25.053" CommentCount="3" />
  <row Id="6049" PostTypeId="2" ParentId="6039" CreationDate="2010-10-04T20:23:02.060" Score="2" ViewCount="" Body="&lt;p&gt;If so then then the values of $\rm\: q(x) = p(x^2)\:$ are all powers of 3, contradicting the fact that infinitely many primes must occur among the values of a nonconstant $\rm\: p(x) \in \mathbb Z[x]\:$, by way of a simple Euclid-like proof. &lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-10-05T00:16:24.960" LastActivityDate="2010-10-05T00:16:24.960" />
  <row Id="6050" PostTypeId="2" ParentId="6039" CreationDate="2010-10-04T20:25:00.047" Score="2" ViewCount="" Body="&lt;p&gt;Here's another argument (at the other end of the real line!):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Such a polynomial would have to be non-constant, which means that it tends to $\infty$ or $-\infty$ as $x\to-\infty$. But this is impossible, since $3^x \to 0$ as $x\to-\infty$.&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2010-10-04T20:25:00.047" CommentCount="1" />
  <row Id="6051" PostTypeId="2" ParentId="6039" CreationDate="2010-10-04T20:25:57.300" Score="11" ViewCount="" Body="&lt;p&gt;Yet another proof: if $p$ is a nonzero polynomial with real coefficients&#xA;then&#xA;$$\lim_{n\to\infty}\frac{p(n+1)}{p(n)}=1$$&#xA;which certainly doesn't hold for $p(x)=3^x$.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-10-04T20:25:57.300" />
  <row Id="6052" PostTypeId="2" ParentId="6044" CreationDate="2010-10-04T20:46:40.070" Score="5" ViewCount="" Body="&lt;p&gt;Nontrivial idempotents are intimately connected to direct product decompositions. Generally, any idempotent  $\rm e^2 = e$ yields a decomposition $\rm\  R  =  e R + (1-e) R\ $  known as the &lt;a href=&quot;http://myyn.org/m/article/pierce-decomposition/&quot; rel=&quot;nofollow&quot;&gt;Pierce Decomposition&lt;/a&gt;, and vice versa. This extends to any finite set of idempotents with sum 1 which are mutually orthogonal: $\rm\ e_j\: e_k = 0\ $ if $\rm\ j \ne k\ $. For example, the Chinese Remainder Theorem (&lt;strong&gt;CRT&lt;/strong&gt;) for rings has this form.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-10-04T20:57:43.867" LastActivityDate="2010-10-04T20:57:43.867" />
  <row Id="6053" PostTypeId="2" ParentId="6039" CreationDate="2010-10-04T20:54:01.407" Score="11" ViewCount="" Body="&lt;p&gt;A polynomial of degree $n$ has the property that $p(x+1) - p(x)$ is a polynomial of degree $n-1$.  After taking $n$ finite differences you reach a contradiction.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-10-04T20:54:01.407" CommentCount="2" />
  <row Id="6054" PostTypeId="2" ParentId="6039" CreationDate="2010-10-04T21:00:41.690" Score="7" ViewCount="" Body="&lt;p&gt;The answer depends on the ring (which the question did not specify) where $x$ and $p(x)$ take their values. Polynomials with integer coefficients, and non-negative powers of integers such as $3^n$, can be evaluated in any ring, so the question makes sense in this generality.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a ring where 2=0 (such as the integers modulo 2), $3^n = 1^n$, and any polynomial with $p(0)=p(1)=1$ is an example.  It should be an interesting problem to determine all rings where there is such a $p(x)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;ADDED: Let's consider the situation over an arbitrary commutative ring $R$ with unit.  $Q(x) = p(x+1)- 3p(x)$ is zero (as a function) for all $x$ in the image of $\mathbb{Z}$ in $R$.  If $R$ is of characteristic different from $2$, $\deg Q = \deg P$ (as polynomials).  In characteristic 0 having zeros at all integers is enough to force $Q = 0$ as a polynomial, and there are no examples.  In finite characteristic $n &gt; 1$, we also have $p(n) = 3^n = p(0) = 1$ so that $n | (3^n - 1)$. If $n$ is prime then $n | (3^n - 3)$ so that $n | 2$ and thus $n=2$.  If $n$ is not prime, and $q$ is a prime dividing $n$, then an example in $R$ is an example in $R/q$ which has prime characteristic $q$ and therefore $q=2$.  So the only remaining possibility is a ring where $2^k = 0$, for $k &gt; 1$.  For such rings there are examples because in the binomial expansion of $(1+2)^n = 1 + 2n + 2n(n-1) + 4n(n-1)(n-2)/3 + 2n(n-1)(n-2)(n-3)/3 + \dots$ all terms are congruent modulo $2^k$ to polynomials with integer coefficients, and all terms of degree higher than $2^k - 1$ vanish on the integers modulo $2^k$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ADDED-2:  one can see that $2^k=0$ more directly by writing $p(n+1)-p(n) = 2p(n)$ (for images of integer $n$ in our ring).  This means that $\Delta^k p(n) = 2^k p(n)$ and for $k &gt; deg(P)$ this will be zero, then set $n=0$ to get $0 = 2^k p(0)= 2^k$.  [Note that this argument applies to $p(n)$ for integer $n$ only, not $p(x)$ as an abstract polynomial or a polynomial evaluated at non-integer values of $x$.]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conclusion: there are polynomials $p(x)$ such that $p(n)=3^n$ for all $n \geq 0$ (with the polynomial considered as a function on the commutative ring-with-unit $R$) if and only if in this ring, $2^k = 0$ for some positive integer $k$.  &lt;/p&gt;&#xA;" OwnerUserId="467" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-10-05T19:01:11.673" LastActivityDate="2010-10-05T19:01:11.673" CommentCount="0" />
  <row Id="6055" PostTypeId="1" AcceptedAnswerId="6076" CreationDate="2010-10-04T21:03:11.163" Score="2" ViewCount="159" Body="&lt;p&gt;I'm learning category theory by self-study. I have a couple of texts, and they both talk about how we ought to try not to think so much about the equality between objects in categories. Rather, the important relation is isomorphism. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Okay, sure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then they go on to talk about &quot;Cat&quot;, the category of (small) categories. One of them even segues into it by considering &quot;categories themselves as structured objects. The `morphisms' between them that preserve their structure are called functors.&quot; But then it becomes clear that equality of functors depends on equality of objects, which isn't part of the structure of categories! (Or at least, a part we're not supposed to think about.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are natural transformations, which inherit equality from the underlying equality of morphisms in the codomain category. They induce a natural isomorphism on functors. Treating functors, modulo natural isomorphism, as morphisms in a category of categories, seems to me to be the obvious &quot;Cat&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, not only is this not &quot;Cat&quot;, but I couldn't find it among the loads of exotic examples of categories I'm given. Apparently it wasn't even deemed worthy of mention. This is a bit disappointing. Doesn't it at least have a standard name and snappy acronym? Can someone point me somewhere where I can learn about its basic properties somewhere? Or is there something defective about it that I've missed? &lt;/p&gt;&#xA;" LastActivityDate="2010-10-05T03:19:58.107" Title="&quot;Cat&quot; modulo natural isomorphism?" Tags="&lt;category-theory&gt;" AnswerCount="3" FavoriteCount="2" />
  <row Id="6056" PostTypeId="1" AcceptedAnswerId="6069" CreationDate="2010-10-04T21:33:24.197" Score="1" ViewCount="165" Body="&lt;p&gt;Are there any good places for me to sell off my mathematics books online especially Springer and Dover books?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I thought perhaps this was off topic, but then I thought everybody in math probably has the problem of having bought pricey books that they would prefer to get some money back for, at one point or another.)&lt;/p&gt;&#xA;" OwnerUserId="1004" LastEditorUserId="1004" LastEditorDisplayName="" LastEditDate="2010-10-04T21:39:12.933" LastActivityDate="2010-10-05T00:03:34.907" Title="Reselling Books" Tags="&lt;books&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="1" />
  <row Id="6057" PostTypeId="1" CreationDate="2010-10-04T22:02:40.107" Score="2" ViewCount="374" Body="&lt;p&gt;I have an image encoded in the form of a list of points, like so:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;335.0 2743.0 &#xA;335.0 2754.0 &#xA;325.0 2754.0 &#xA;304.0 2743.0 &#xA;304.0 2733.0 &#xA;283.0 2723.0 &#xA;273.0 2702.0 &#xA;273.0 2691.0 &#xA;273.0 2681.0 &#xA;273.0 2670.0 &#xA;283.0 2670.0 &#xA;294.0 2670.0 &#xA;304.0 2670.0 &#xA;325.0 2670.0 &#xA;335.0 2681.0 &#xA;346.0 2702.0 &#xA;346.0 2712.0 &#xA;356.0 2723.0 &#xA;346.0 2733.0 &#xA;346.0 2743.0 &#xA;346.0 2733.0 &#xA;346.0 2723.0 &#xA;356.0 2702.0 &#xA;356.0 2670.0 &#xA;356.0 2660.0 &#xA;367.0 2660.0 &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;There is a line drawn between each point to make the image - if you sketch the points above (I'd advise doing it programatically) you'll see a lower case 'a'. This is part of a project regarding online character recognition. I am currently investigating the effects of several pre-processing techniques that can be applied to such a recognition system, and one technique that many of the papers that I have read apply is to 'smooth' the image, usually via a Gaussian blur. Unfortunately, they all proclaim it as &quot;easy&quot;, and therefore they neglect to mention &lt;em&gt;how&lt;/em&gt; one goes about doing so.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have been looking for quite some time now, but I find myself still unable to understand how I would take the idea and apply it to an image made up of a set of points like the one above. From the &lt;a href=&quot;http://en.wikipedia.org/wiki/Gaussian_blur&quot; rel=&quot;nofollow&quot;&gt;wikipedia&lt;/a&gt; article I have the function:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$G(X) = \frac{1}{\sqrt{2\pi\sigma^2}}e^{-\frac{x^2}{2\sigma^2}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I have no idea how I actually apply that to my list of points to get a smoothed image. All of the examples online are both incredibly complicated and also for pixel-images, where every pixel in your image matters. Here, I only actually have the 20 or so points. So, what I am looking for is some explanation and advice regarding how I would apply a Gaussian blur in my way of representing the image - what do I apply $G(X)$ to (each co-ordinate?), how I calculate $\sigma^2$, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The disclaimer&lt;/strong&gt;: As stated, this is part of an assignment. However, this is not breaking any rules or 'cheating' in any form. The assignment is to investigate many features of online character recognition, not to magically know how to do one of the possible techniques for pre-processing the characters. Should I gain any knowledge of how to do the smoothing from this site, the site (and the individual, should they wish) will of course be credited in full.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I thank you for any help. I hope this question is placed correctly (I felt it was more mathematical than program-based, since it's more about the concept of Gaussian blurs rather than the implementation - I'm fairly sure I'll be able to handle that part) and that it is clear. If it is not, please just ask and I will clarify any points I can.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I also apologise for the tag. I'm fairly sure it's inaccurate. But there's no 'Gaussian' tag. Tis odd.)&lt;/p&gt;&#xA;" OwnerUserId="693" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-10-05T08:57:34.567" LastActivityDate="2010-10-05T08:57:34.567" Title="How do I apply a Gaussian Blur (low-pass filter) to an image made up from a set of points?" Tags="&lt;signal-processing&gt;&lt;image-processing&gt;" AnswerCount="2" CommentCount="5" FavoriteCount="1" />
  <row Id="6058" PostTypeId="2" ParentId="6056" CreationDate="2010-10-04T22:24:49.457" Score="0" ViewCount="" Body="&lt;p&gt;Amazon will buy back a lot of textbooks at pretty fair rates. Check &lt;a href=&quot;http://www.amazon.com/Textbooks-Trade-In/b/ref=bhp_9p2_A_txbq410_05?ie=UTF8&amp;amp;node=2205237011&amp;amp;pf_rd_m=ATVPDKIKX0DER&amp;amp;pf_rd_s=center-6&amp;amp;pf_rd_r=0RH8PF985VSEMWFKG1SV&amp;amp;pf_rd_t=101&amp;amp;pf_rd_p=1275995182&amp;amp;pf_rd_i=283155&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; out.&lt;/p&gt;&#xA;" OwnerUserId="2302" LastActivityDate="2010-10-04T22:24:49.457" />
  <row Id="6059" PostTypeId="2" ParentId="6044" CreationDate="2010-10-04T23:04:37.253" Score="4" ViewCount="" Body="&lt;p&gt;Let $X$ be a disconnected space. Then the ring $C(X)$ of complex continuous functions has an idempotent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a sense, this is a motivating example, because if $R$ has a nontrivial idempotent, then $Spec(R)$ is disconnected. Since elements of $R$ can be viewed (loosely) as functions, we can construct an element of $R$ which is &quot;one&quot; on one half of $Spec(R)$ and &quot;zero&quot; on the other. (To make this precise, one thinks in terms of the sheaf of regular functions on the scheme $Spec(R)$.)&lt;/p&gt;&#xA;" OwnerUserId="536" LastActivityDate="2010-10-04T23:04:37.253" />
  <row Id="6060" PostTypeId="1" AcceptedAnswerId="6115" CreationDate="2010-10-04T23:11:29.707" Score="5" ViewCount="196" Body="&lt;p&gt;Suppose there are a finite number of disjoint unit-radii disks in the &#xA;plane, each spinning clockwise or counterclockwise at the same&#xA;angular velocity.&#xA;The plane is filled with a thin fluid layer,&#xA;and the disks can be viewed as spinning fan blades&#xA;determining vectors of fluid motion tangent to the disks.&#xA;Is the resulting flow and vector field throughout the plane known?&#xA;My initial intuition is that there should be something like a&#xA;Voronoi diagram demarcating boundaries of regions of influence.&#xA;But in exploring a bit I find it may even be nontrivial to&#xA;determine the flow between just two counter-rotating vortices.&#xA;For example, the following image was&#xA;computed by &lt;a href=&quot;http://www.bugman123.com/&quot; rel=&quot;nofollow&quot;&gt;Paul Nylander&lt;/a&gt;&#xA;based on a paper by&#xA;O.S. Kerr and J.W. Dold,&#xA;&quot;Periodic Steady Vortices in a Stagnation Point Flow,&quot;&#xA;&lt;a href=&quot;http://www.city.ac.uk/sems/mathematics/research/fluids/vortices.html&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;J. Fluid Mech.&lt;/em&gt;, 276, 307-325 (1994)&lt;/a&gt;.&#xA;&lt;br /&gt;&#xA;&lt;img src=&quot;http://i.imgur.com/QK72h.jpg&quot; alt=&quot;alt text&quot;&gt;&#xA;&lt;br /&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As I am quite unschooled in this topic, pointers to&#xA;relevant literature might suffice.  Thanks!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Edit1&lt;/b&gt;. I've now asked a revised version of &lt;a href=&quot;http://mathoverflow.net/questions/41132/vortex-voronoi-diagram&quot; rel=&quot;nofollow&quot;&gt;this question on Math Overflow&lt;/a&gt;,&#xA;incorporating the clarifying suggestions of Rahul.  I might hit a fluid dynamics expert there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Edit2&lt;/b&gt;. Thanks to Rahul and David Bar Moshe here, and Willie Wong and Bob Terrell&#xA;on MO, I have a much broader understanding of the problem, and could likely compute a numerical &#xA;solution if needed.  I appreciate the help!&lt;/p&gt;&#xA;" OwnerUserId="237" LastEditorUserId="237" LastEditorDisplayName="" LastEditDate="2010-10-06T01:04:05.323" LastActivityDate="2010-10-06T04:08:22.573" Title="Vortex Voronoi diagram?" Tags="&lt;geometry&gt;&lt;reference-request&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="1" />
  <row Id="6061" PostTypeId="2" ParentId="6055" CreationDate="2010-10-04T23:26:52.950" Score="0" ViewCount="" Body="&lt;p&gt;I'm not sure to understand what you mean, but maybe the operation for which you are asking for a name is &lt;em&gt;skeleton&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Quoting Mac Lane's &quot;Categories for the working mathematician&quot; (chapter IV, section 4), &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In any category ${\cal C}$ [not just $\mathbf{Cat}$], a &lt;em&gt;skeleton&lt;/em&gt; of ${\cal C}$ is any full subcategory ${\cal A}$ such that any object of ${\cal C}$ is isomorphic to exactly one object of ${\cal A}$. Then ${\cal A}$ is equivalent to ${\cal C}$ and the inclusion $K: {\cal A} \longrightarrow {\cal C}$ is an equivalence of categories. (...) &lt;em&gt;A category is equivalent to any of its skeletons.&lt;/em&gt; For instance, the category of all finite sets has as skeleton the full subcategory with objects all finite ordinal numbers $0, 1, 2, \dots , n , \dots$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Finally, Mac Lane adds&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A category is called &lt;em&gt;skeletal&lt;/em&gt; when any two isomorphic objects are identical; i.e., when the category is its own skeleton.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For instance, the category with objects all finite ordinal numbers is skeletal: there you can forget about the picky distinction between &quot;equal&quot; and &quot;isomorphic&quot; objects. In real life, there is another circumstance in which isomorphic objects are treated as being &quot;equal&quot; for all purposes, namely, when you have a &lt;em&gt;canonical&lt;/em&gt; isomorphism, for instance one which is derived from a universal property such as a the one verified by coproducts.&lt;/p&gt;&#xA;" OwnerUserId="664" LastActivityDate="2010-10-04T23:26:52.950" CommentCount="3" />
  <row Id="6062" PostTypeId="2" ParentId="6044" CreationDate="2010-10-04T23:30:02.463" Score="3" ViewCount="" Body="&lt;p&gt;The endomorphism ring $\mathrm{End}(V)$ of a given vector space $V$ is riddled with idempotents. For instance, all projections $\pi=\pi_{L_1,L_2}$ of $V$ onto $L_1$ along $L_2$ where $V=L_1 \oplus L_2$ is a direct decompositon of $V$ are idempotents ($\pi(x)=\pi(x_1+x_2)=x_1$ where $x \in V$ and $x_k \in L_k,$ $k=1,2$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, there are many hard/nice questions concering idempotents in group rings.  &lt;/p&gt;&#xA;" OwnerUserId="1100" LastActivityDate="2010-10-04T23:30:02.463" />
  <row Id="6063" PostTypeId="2" ParentId="6044" CreationDate="2010-10-04T23:37:04.083" Score="3" ViewCount="" Body="&lt;p&gt;Actually there are rings in which every element is idempotent. They are called boolean rings. Stone duality tells us that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$X \mapsto C_0(X,\mathbb{F}_2)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is a duality between locally compact totally disconnected hausdorff spaces and boolean rings. Compactness here corresponds to unitality. For example, every finite boolean ring is isomorphic to $(\mathbb{F}_2)^n$ for some $n$.&lt;/p&gt;&#xA;" OwnerUserId="1650" LastActivityDate="2010-10-04T23:37:04.083" />
  <row Id="6064" PostTypeId="2" ParentId="6055" CreationDate="2010-10-04T23:43:09.907" Score="0" ViewCount="" Body="&lt;p&gt;If $C$ is a $2$-category, every hom-set is a category and isomorphy yields a congruence relation on them. The quotient $\tilde{C}$ is then a $1$-category. It's objects are the objects of $C$, but the morphisms are the $1$-morphisms of $C$ modulo $2$-isomorphy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you want to compare $C$ with $2$-categories, it does not make any sense to consider $\tilde{C}$. However, this construction becomes useful when you want to compare with $1$-categories, that is, usual categories. For example, the functor&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$Ring \to \tilde{Cat}, R \mapsto Mod(R), \phi \mapsto \phi^*$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that $\phi^* \psi^* \cong (\phi \psi)^*$ holds in $Cat$, thus these morphisms are equal in $\tilde{Cat}$ and we get, indeed, a functor.&lt;/p&gt;&#xA;" OwnerUserId="1650" LastActivityDate="2010-10-04T23:43:09.907" CommentCount="2" />
  <row Id="6065" PostTypeId="2" ParentId="6057" CreationDate="2010-10-04T23:53:34.983" Score="2" ViewCount="" Body="&lt;p&gt;The operation is better described in the Wikipedia article &lt;a href=&quot;http://en.wikipedia.org/wiki/Gaussian_filter&quot; rel=&quot;nofollow&quot;&gt;Gaussian Filter&lt;/a&gt;.  First you need to convert your list of points to an image, in essence filling in the line between the points.  If you use 8 bits, those cells will have darkness 255 while all others will be 0.  Or perhaps you give the lines a thickness and calculate how much of each pixel is covered.  Now the image is a large array of numbers.  The value of sigma is yours to choose.  From sigma you choose will calculate a small array that represents G(X).  If sigma is one pixel, you can represent G(X) only over +-3 pixels, so it is 7x7.  To calculate the blurred darkness of a pixel, you take the original darknesses of the 7x7 neighborhood of the pixel, multiply them by the G(X) array and add.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-10-05T08:54:56.597" LastActivityDate="2010-10-05T08:54:56.597" CommentCount="1" />
  <row Id="6066" PostTypeId="2" ParentId="6044" CreationDate="2010-10-04T23:54:44.730" Score="1" ViewCount="" Body="&lt;p&gt;You can compute idempotents in any square matrix ring $M_n(K)$, $K$ a field, as follows:  $A \in M_n(K)$ is idempotent if and only if $A^2 = A$. But this means that the polynomial $p(t) = t(t-1)$ annihilates $A$. So the &lt;a href=&quot;http://en.wikipedia.org/wiki/Minimal_polynomial_%28linear_algebra%29&quot; rel=&quot;nofollow&quot;&gt;minimal polynomial&lt;/a&gt; $m_A(t)$ of $A$ must divide $p(t)$. This leaves only four possibilities for $m_A(t)$:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$m_A(t) = 1$, in which case the unit matrix equals the zero matrix $I=0$ and $K = 0$.&lt;/li&gt;&#xA;&lt;li&gt;$m_A(t) = t$, in which case $A = 0$.&lt;/li&gt;&#xA;&lt;li&gt;$m_A(t) = t-1$, in which case $A=I$.&lt;/li&gt;&#xA;&lt;li&gt;$m_A(t) = t(t-1)$ is the only non trivial one. It means that your matrix $A$ has only two eigenvalues: $0$ and $1$. For instance, &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;$&#xA;A=&#xA;\begin{pmatrix}&#xA;0 &amp;amp; 0 \\&#xA;0 &amp;amp; 1&#xA;\end{pmatrix} \ .&#xA;$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since all of prime factors in the minimal polynomial have multiplicity $1$, all diagonalize, so it's fairly easy to get a picture of them: all are of the form $A = S^{-1}D S$, with $D$ a diagonal matrix with just $0$ and $1$ on the main diagonal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Matrices of Olod's non-trivial projections belong to this case.&lt;/p&gt;&#xA;" OwnerUserId="664" LastEditorUserId="664" LastEditorDisplayName="" LastEditDate="2010-10-05T07:35:22.180" LastActivityDate="2010-10-05T07:35:22.180" />
  <row Id="6067" PostTypeId="2" ParentId="6056" CreationDate="2010-10-04T23:54:44.730" Score="2" ViewCount="" Body="&lt;p&gt;I find it's pretty easy to sell books on e-bay.  Local used bookstores rarely go for good math books, unfortunately. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might also want to try the big technical book resellers -- like Powell's Books, they're based in Portland Oregon.  Most on-line searches for books (like Google Books) will search Powell's stock.  So that might be one of the best options. &lt;/p&gt;&#xA;" OwnerUserId="642" LastActivityDate="2010-10-04T23:54:44.730" />
  <row Id="6068" PostTypeId="1" AcceptedAnswerId="6074" CreationDate="2010-10-04T23:56:15.170" Score="2" ViewCount="141" Body="&lt;p&gt;I know that a function of bounded variation on a closed interval can be written as a difference of increasing functions. The same is true is the function is of bounded variation on all of R, but I can't come up with a proof. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any hints?&lt;/p&gt;&#xA;" OwnerUserId="2313" LastActivityDate="2010-10-05T02:07:06.240" Title="Bounded variation: Jordan's theorem on all of R?" Tags="&lt;real-analysis&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="6069" PostTypeId="2" ParentId="6056" CreationDate="2010-10-05T00:03:34.907" Score="2" ViewCount="" Body="&lt;p&gt;You can also list your books on amazon to sell rather than have amazon buy them from you&lt;/p&gt;&#xA;" OwnerUserId="1352" LastActivityDate="2010-10-05T00:03:34.907" />
  <row Id="6070" PostTypeId="2" ParentId="6039" CreationDate="2010-10-05T00:16:36.270" Score="10" ViewCount="" Body="&lt;p&gt;$\rm\ p(x) = 3^x\:$ on $\rm\:\mathbb N\ \Rightarrow\ p(2x) = p(x)^2\:$  on $\rm\:\mathbb N \ \Rightarrow\ p(2x) = p(x)^2\:$ contra degree comparison.&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-10-05T00:16:36.270" CommentCount="4" />
  <row Id="6071" PostTypeId="1" AcceptedAnswerId="6072" CreationDate="2010-10-05T00:44:22.240" Score="1" ViewCount="151" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;A school play ran for two nights, with&#xA;  audiences totaling 1390 adults and&#xA;  students. They paid 4285 dollars for&#xA;  admission. One adult ticket cost 4 dollars&#xA;  and one student ticket cost 2.50 dollars. The&#xA;  ratio of adults to students was 3:5&#xA;  for the first night and 2:3 on the&#xA;  second night. How many students&#xA;  attended each night?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have determined that there was 850 students and 540 adults on &lt;em&gt;both&lt;/em&gt; nights, but for the life of me couldn't figure out how to incoporate those ratios into finding how much students were &lt;em&gt;per day&lt;/em&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is marked as a homework question so please don't give me the full answer but how to go about it.&#xA;I also have the answers if that is any help: for the first night, there were 400 students on the first night, and 450 on the second.&lt;/p&gt;&#xA;" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2011-01-17T01:44:14.510" LastActivityDate="2011-01-17T01:44:14.510" Title="Linear Algebra Word Question (school play audience numbers and ticket prices)" Tags="&lt;homework&gt;&lt;algebra-precalculus&gt;" AnswerCount="1" />
  <row Id="6072" PostTypeId="2" ParentId="6071" CreationDate="2010-10-05T01:19:12.573" Score="0" ViewCount="" Body="&lt;p&gt;I have verified that the adult/student totals you got is correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the second part:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose a total of $x$ people (i.e. including students and adults) attended the first day and total $y$ people attended the second day.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the total number of students in terms of $x$ and $y$?&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-10-05T01:19:12.573" CommentCount="24" />
  <row Id="6073" PostTypeId="2" ParentId="6028" CreationDate="2010-10-05T01:48:17.147" Score="1" ViewCount="" Body="&lt;p&gt;I find the visual way of thinking about it to be the easiest: if you look at the graph of $f$ and zoom in to the point $(x,f(x))$, the graph will eventually start looking very much like a line.  That line is the &quot;tangent line&quot; to $f$ at $x$, and its slope is the derivative of $f$ at $x$.  (Some functions won't ever start looking like a line, no matter how far you zoom in.  One example is $f(x)=\left\vert x\right\vert$, at $(0,0)$.  We say that this function isn't differentiable there.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The formal definition of the derivative, as&#xA;$$ f^\prime(x)=\lim_{a\rightarrow x}\frac{f(x)-f(a)}{x-a},$$&#xA;is really just another, more mathematical, way to describe &quot;zooming in&quot; and the construction of a tangent line.  If you think about it, the expression inside the limit is just the slope formula for a line going through $(x,f(x))$ and $(a,f(a))$.  This line is called a &quot;secant line.&quot;  If we let $a=x$, then we only have one point and so we don't have a unique line anymore.  But if we instead ensure that $a\ne x$ but that $a$ gets closer and closer to $x$, the secant lines approach the tangent line that we saw above.  This is just the same &quot;zooming in&quot; I was talking about above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're less of a visual person, it's often helpful to think of a physical quantity, like velocity.  Imagine driving a car or riding a bike in a straight line.  At any instant, you have a pretty good idea of how fast you're going &quot;right now,&quot; even if your speed is in the middle of changing.  Ryan Budney mentioned the example of a car with a speedometer above.  The speedometer can tell you your speed at any specific time.  This is just the derivative of your position: if you let the line be the $y$-axis and time be the $x$-axis, and graph your journey, the slope of a tangent line at a point will be exactly the speedometer reading at the that point.  On the other hand, you can also measure how much time it takes for you to get from $a$ to $b$: this is giving you the slope of a secant line.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So &lt;i&gt;instantaneous&lt;/i&gt; velocity = slope of tangent line or derivative&lt;/p&gt;&#xA;&#xA;&lt;p&gt;while &lt;i&gt;average&lt;/i&gt; velocity = slope of secant line.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These are all derivatives &quot;with respect to time,&quot; but you can easily take the derivative with respect to other things, as long as you have a function relating them.  I'm not sure what your science background is, but this is the kind of thing that pops up often in school science experiments: the rate of change of the volume of a gas with respect to pressure, etc. &lt;/p&gt;&#xA;" OwnerUserId="2215" LastActivityDate="2010-10-05T01:48:17.147" />
  <row Id="6074" PostTypeId="2" ParentId="6068" CreationDate="2010-10-05T02:07:06.240" Score="1" ViewCount="" Body="&lt;p&gt;You can use the positive and negative variation functions as usual.  The only difference is that you don't have a left endpoint to use.  Assuming BV on all of $\mathbb{R}$, a finite limit exists for each of these functions as you approach $-\infty$, and you could use this in place of the left endpoint.  More generally, it is enough to assume that the function is BV on all bounded intervals, but for this you will want to fix a finite base point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose $f$ is BV on all bounded intervals.  We want to find nondecreasing functions $g$ and $h$ such that $f=g-h$.  For nonnegative $x$, define $g(x)$ to be $f(0)$ plus the positive variation of $f$ on $[0,x]$, and define $h(x)$ to be the negative variation of $f$ on $[0,x]$.  For negative $x$, define $g(x)$ to be $f(0)$ minus the positive variation of $f$ on $[x,0]$, and define $h(x)$ to be minus the negative variation of $f$ on $[x,0]$.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-10-05T02:07:06.240" CommentCount="1" />
  <row Id="6075" PostTypeId="1" CreationDate="2010-10-05T03:02:30.793" Score="2" ViewCount="84" Body="&lt;p&gt;I have the following situation: let $m,n$ be integers such that $m|n$ and let $\zeta_m$, $\zeta_n$ denote  primitive $m$ and $n$th roots of unity. Then we have the inclusion of  fields&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathbb{Q}\subset \mathbb{Q}(\zeta_m) \subset \mathbb{Q}(\zeta_n)$$&#xA;Now suppose we also have primes (where $(p,n)=1$)&#xA;$$(p)\subset \mathbb{Z}$$&#xA; and then &#xA;$$\mathfrak{p}\subset \mathbb{Z}(\zeta_m)$$&#xA; lying over $(p)$ and &#xA;$$\mathfrak{P}\subset \mathbb{Z}(\zeta_n)$$&#xA; lying over $\mathfrak{p}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a congruence in $\mathbb{Q}(\zeta_n)$ of the form $a\equiv b \pmod{\mathfrak{P}}$, where $a,b$ are actually elements of $\mathbb{Q}(\zeta_m)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What can I say about the congruence properties of $a,b$ in $\mathbb{Q}(\zeta_m)$? Also, if I take the trace or the  norm down to $\mathbb{Q}$, can I say anything about their congruence properties there?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" LastActivityDate="2010-11-04T07:44:18.083" Title="Congruences mod primes in Galois extensions" Tags="&lt;algebraic-number-theory&gt;&lt;galois-theory&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="6076" PostTypeId="2" ParentId="6055" CreationDate="2010-10-05T03:19:58.107" Score="3" ViewCount="" Body="&lt;p&gt;You are correct that in practice one does not care so much whether two functors are equal, but rather whether they are naturally isomorphic.  The difficulty is that typically one does not want to completely forget about the natural isomorphism either!  (Which is what happens if one quotients out by natural isomorphisms as you suggest.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A typical example (which I hope will make sense to you) is to consider for each topological space the category of vector bundles on $X$. So we have a category $Vect_X$. If $f: X \to Y$ is a continuous map of spaces,&#xA;and $\mathcal V$ is a vector bundle on $Y$, then one can pull-back $\mathcal V$ to form&#xA;a vector-bundle $f^* \mathcal V$ on $Y$.  So we get a functor $f^*: Vect_Y \to Vect_X$.&#xA;If now $g: Y \to Z$ as well, then one sees that $(g f)^* $ is naturally isomorphic to&#xA;$f^* g^*$,  say by some natural isomorphism $c_{f,g}: (g f)^* \cong f^* g^*.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Morally, one would like to say that $X \mapsto Vect_X$ and $f \mapsto f^* $ gives a contravariant functor from the category of topological spaces to the category of categories Cat.  In practice, because we don't have equality between $(f g)^* $ and $f^* g^* $, we don't get such a functor, although we do get a functor into your suggested category &quot;Cat&quot;.&#xA;The problem is that in practice, one wants to remember the natural isomorphisms $c_{f , g}$,&#xA;which satisfy some important properties: for example, if $h: Z \to W$ is a third map,&#xA;then $f^* c_{g,h} \circ c_{f, h g} = c_{f,g}h^* \circ c_{gf, h}.$  (If you write this out, it is a commutative square that relates the various ways to pull back $\mathcal V$ by&#xA;$h$, $g$, and $f$, taking into account the associative law $(hg)f = h(g f)$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So really, one wants to work in a more sophisticated structure then either Cat or &quot;Cat&quot;,&#xA;namely a structure in which the objects are categories, the morphisms are functors, &#xA;and in which we add an explicit extra layer of structure, so-called 2-morphisms, which are natural isomorphisms between functors.  One then develops a theory in which one morally&#xA;regards functors as equal if they coincide up to natural isomorphism (i.e. up to a diagram involving a 2-morphism), but one also keeps track of the 2-morphisms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The resulting structure is called a 2-category, and is part of the study of higher category theory.  (Everything written above should provide some background for understanding Martin Brandenburg's answer.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This theory has a lot in common with homotopy theory: in topology, one can pass to a category in which objects are spaces and maps are continuos maps modulo homotopy, but in lots of applications one wants to remember not just that maps are homotopic, but one actually wants to remember the homotopy; often then there are homotopies between homotopies,&#xA;homotopies between homotopies between homotopies, and so on.  Similarly in category theory one can introduce not just the notion of 2-category, but notions of n-categories, in which&#xA;there are 3-morphisms beweeen the 2-morphisms, etc., up to n-morphisms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your structure &quot;Cat&quot; is analogous to passing to the homotopy category in topology; it is interesting, but forgets information one often wants to remember.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you search for &quot;higher category theory&quot;, you will find an enormous amount of material.&#xA;One good reference is the n-category cafe and the n-Lab.  Another place to look is at the&#xA;various manuscripts on Jacob Lurie's web-page (at Harvard).  What you will find is a rather intricate, and rapidbly evolving theory, blending category theory and homotopy theory in a fascinating (although sometimes daunting!) way.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In summary, your idea and your question are far from misguided, but in fact are pointing at one of the most active areas of modern research in category theory and related areas!&lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2010-10-05T03:19:58.107" CommentCount="3" />
  <row Id="6077" PostTypeId="1" AcceptedAnswerId="6086" CreationDate="2010-10-05T03:53:12.637" Score="1" ViewCount="177" Body="&lt;p&gt;So I have to find the maximum possible error $dR$ in calculating equivalent resistance for three resistors, $\displaystyle\frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}$ if the maximum error in each is 0.5%; $R_1=25\ \Omega$, $R_2=45\ \Omega$, $R_3=50\ \Omega$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, originally I did $dR_1=dR_2=dR_3=0.005$, and then did $\displaystyle\frac{dR}{R^2}=\frac{dR_1}{R_1^2}+\frac{dR_2}{R_2^2}+\frac{dR_3}{R_3^2}$ and solved for $dR$... now I realize now that that doesn't make any sense.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I recall when doing an example problem we took the derivative like normal except when doing the chain rule, replacing it with the partial; for example, when $A=\ell w$, $dA = \frac{\partial A}{\partial \ell} d\ell+\frac{\partial A}{\partial w} dw$ (finding error in area of rectangle) and plugged in what I know. How would that work here? Was I close in my original attempt? I feel like I'm not sure where to put all the partials now that there's a bunch of reciprocals everywhere.&lt;/p&gt;&#xA;" OwnerUserId="1562" LastActivityDate="2011-03-05T03:21:29.507" Title="Calculating total error based on error of variables" Tags="&lt;calculus&gt;&lt;homework&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="6078" PostTypeId="1" CreationDate="2010-10-05T04:46:56.493" Score="0" ViewCount="151" Body="&lt;p&gt;Let $G$ be a group of order $pqr$, where $p$, $q$ and $r$ be three distinct primes. By Cauchy's theorem there exist three elements, $a$, $b$ and $c$, whose orders are $p$, $q$ and $r$, respectively. If the subgroup generated by $a$ and $b$ is the whole group, then I wonder if it is possible that there exists a proper normal subgroup of $G$.&lt;/p&gt;&#xA;" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-10-05T04:55:03.043" LastActivityDate="2010-10-05T08:05:10.820" Title="A question about finite group" Tags="&lt;group-theory&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="6079" PostTypeId="2" ParentId="6078" CreationDate="2010-10-05T05:10:53.010" Score="4" ViewCount="" Body="&lt;p&gt;No, a group of order $pqr$ cannot be simple, regardless of whether you can find elements $a$ and $b$ of orders $p$ and $q$ that generate $G$ or not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose $p\lt q\lt r$, and let $\ell$, $m$, and $n$ be the number of $p$-Sylow, $q$-Sylow, and $r$-Sylow subgroups of $G$, respectively. Any two distinct Sylow subgroups will intersect trivially, because they are all cyclic of prime order.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $G$ were simple, then $\ell,m,n\gt 1$. Also, $n|pq$, $m|pr$, and $\ell|qr$. Since $n$ must be congruent to $1$ modulo $r$ and $p,q\lt r$, then we must have $n=pq$. Since $m$ must be congruent to $1$ modulo $q$ and $p\lt q$, the smallest value possible for $m$ is $r$. So the smallest values they can have are $n=pq$, $m=r$, and $\ell=q$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the distinct $p$-Sylow subgroups give at least $q(p-1)$ nonidentity elements. The distinct $q$-Sylow subgroups give at least $r(q-1)$ nonidentity elements; and the distinct $r$-Sylow subgroups give at least $pq(r-1)$ nonidentity elements. So $G$ will have at least&#xA;$$\begin{array}{rcl} q(p-1) + r(q-1) + pq(r-1) + 1 &amp;amp;=&amp;amp; &#xA;pqr + pq + rq - q - r - pq + 1\\&#xA;&amp;amp; = &amp;amp;pqr + rq +1 - q-r&#xA;\end{array}$$&#xA;distinct elements. But this is strictly larger than $pqr$, since $r$ and $q$ are greater than 1, so $qr\geq q+r$. This is impossible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The contradiction arises from the assumption that $G$ is simple; hence $G$ is not simple.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-10-05T05:22:00.113" LastActivityDate="2010-10-05T05:22:00.113" />
  <row Id="6080" PostTypeId="1" AcceptedAnswerId="6081" CreationDate="2010-10-05T05:12:18.323" Score="7" ViewCount="252" Body="&lt;p&gt;I was wondering if there is a nice way to see that $\mathbb{RP}^{3}$ is orientable without using  tools of algebraic topology, like homology.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only think I could think of was to argue that $\mathbb{RP}^{3}=\mathbb{R}^3 \cup \mathbb{RP}^{2}$ and perhaps you could argue that to get back to any starting position you have to cross the $\mathbb{RP}^{2}$ boundary but I'm pretty sure that what I'm thinking is nonsense.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This was a question on the homework for one of my topics courses and I plan on asking the professor about it tomorrow, but I was curious to see if anyone had any interesting ways of thinking about or picturing this space.&lt;/p&gt;&#xA;" OwnerUserId="249" LastActivityDate="2010-10-06T02:23:09.410" Title="Orientability of $\mathbb{RP}^3$" Tags="&lt;geometry&gt;&lt;intuition&gt;&lt;homework&gt;&lt;projective-space&gt;" AnswerCount="4" CommentCount="3" FavoriteCount="2" />
  <row Id="6081" PostTypeId="2" ParentId="6080" CreationDate="2010-10-05T05:43:07.597" Score="10" ViewCount="" Body="&lt;p&gt;The orientation on the universal cover, $S^3$, descends. It takes some checking of details, but you can do it for $\mathbb{R}P^n$ in exactly the same way for $n$ odd. Actually, much more generally if you have a covering $\pi: X\to B$, and $Aut_\pi (X)$ is cyclic and generated by some orientation preserving diffeo $f: X\to X$, then if $X$ is orientable, the orientation will descend to $B$. In this case the diffeo is the antipodal map which is orientation preserving on $S^n$ if and only if $n$ is odd.&lt;/p&gt;&#xA;" OwnerUserId="2000" LastEditorUserId="2000" LastEditorDisplayName="" LastEditDate="2010-10-06T02:23:09.410" LastActivityDate="2010-10-06T02:23:09.410" />
  <row Id="6082" PostTypeId="1" AcceptedAnswerId="6123" CreationDate="2010-10-05T05:53:04.420" Score="2" ViewCount="98" Body="&lt;p&gt;I thought I would post (as a puzzle) one of my favourite results in combinatorics.  I actually use variants of this result in research quite often.  It's not impossible that someone will post an unexpected proof that could be used in future research.  But even if that does not occur, I hope it will be a fun exercise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An &lt;em&gt;orthomorphism&lt;/em&gt; of the cyclic group $\mathbb{Z}_n$ is a permutation $\sigma:\mathbb{Z}_n \rightarrow \mathbb{Z}_n$ such that $i \mapsto \sigma(i)-i$ is also a permutation.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Prove that orthomorphisms of $\mathbb{Z}_n$ exist if and only if $n$ is odd.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Actually, one direction is easy:  $\sigma(i)=2i$ is an orthomorphism of $\mathbb{Z}_n$ when $n$ is odd.  So what's left is to prove the non-existence of orthomorphisms of $\mathbb{Z}_n$ when $n$ is even.&lt;/p&gt;&#xA;" OwnerUserId="139" LastActivityDate="2010-10-05T14:35:44.220" Title="When does an orthomorphism of the cyclic group exist?" Tags="&lt;combinatorics&gt;&lt;group-theory&gt;&lt;puzzle&gt;&lt;permutations&gt;" AnswerCount="1" />
  <row Id="6083" PostTypeId="1" AcceptedAnswerId="6085" CreationDate="2010-10-05T05:54:06.960" Score="1" ViewCount="323" Body="&lt;p&gt;If $(r, \theta)$ are polar coordinates on $\mathbb{R}^2\backslash \left\{ (0,0)\right\}$, then how do I show/prove that &#xA;$d\theta =\dfrac{x dy - y dx}{x^2 + y^2}$?&lt;/p&gt;&#xA;" OwnerUserId="2319" LastEditorUserId="171" LastEditorDisplayName="" LastEditDate="2010-10-05T09:11:55.267" LastActivityDate="2010-10-05T16:23:06.303" Title="How do you show that $d\theta = \frac{x dy - y dx }{x^2 + y^2}$?" Tags="&lt;calculus&gt;&lt;differential-geometry&gt;" AnswerCount="3" CommentCount="4" FavoriteCount="2" />
  <row Id="6084" PostTypeId="2" ParentId="6083" CreationDate="2010-10-05T06:15:59.670" Score="4" ViewCount="" Body="&lt;p&gt;On $R^2\setminus\{0\}$ we have $\theta=\Im(\ln(x+iy))$ (here $\Im$ stands for &quot;the imaginary part of&quot;). Therefore $$d\theta = \Im\left(\frac1{x+iy}\right)dx+\Im\left(\frac{i}{x+iy}\right)dy=\Im\left(\frac{x-iy}{x^2+y^2}\right)dx+\Im\left(\frac{i(x-iy)}{x^2+y^2}\right)dy=$$$$\frac{-y}{x^2+y^2}dx+\frac{x}{x^2+y^2}dy=\frac{xdy-ydx}{x^2+y^2}.$$&lt;/p&gt;&#xA;" OwnerUserId="695" LastEditorUserId="695" LastEditorDisplayName="" LastEditDate="2010-10-05T07:51:14.650" LastActivityDate="2010-10-05T07:51:14.650" />
  <row Id="6085" PostTypeId="2" ParentId="6083" CreationDate="2010-10-05T06:24:31.193" Score="11" ViewCount="" Body="&lt;p&gt;$$x=r\cos\theta$$&#xA;$$y=r\sin\theta$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So,&#xA;$$dx=\cos\theta\thinspace dr -r\sin\theta\thinspace d\theta$$&#xA;$$dy=\sin\theta\thinspace dr +r\cos\theta\thinspace d\theta$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solving for $d\theta$ and using $\sin^2\theta+\cos^2\theta=1$&#xA;$$d\theta={{dy\thinspace\cos\theta-dx\thinspace\sin\theta}\over {r}}$$&#xA;Multiplying and dividing by $r$&#xA;$$={{dy\thinspace (r\cos\theta)-dx\thinspace (r\sin\theta)}\over {r^2}}$$&#xA;Using the definition of $x$ and $y$ in terms of polar coordinates&#xA;$$={{x\thinspace dy-y\thinspace dx}\over {x^2+y^2}}$$&lt;/p&gt;&#xA;" OwnerUserId="1195" LastActivityDate="2010-10-05T06:24:31.193" CommentCount="1" />
  <row Id="6086" PostTypeId="2" ParentId="6077" CreationDate="2010-10-05T06:28:26.800" Score="3" ViewCount="" Body="&lt;p&gt;The &lt;em&gt;dR&lt;sub&gt;i&lt;/sub&gt;&lt;/em&gt; denotes the absolute error, but that 0.5% is a relative error. Therefore, you should use&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;dR_1 &amp;amp;= 0.5\% \times R_1 = 0.125\, \Omega \\&#xA;dR_2 &amp;amp;= 0.5\% \times R_2 = 0.225\, \Omega \\&#xA;dR_2 &amp;amp;= 0.5\% \times R_3 = 0.25\, \Omega&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rest is correct for obtaining a conservative value of &lt;em&gt;dR&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(However, if you can ensure the quantities are distributed normally and uncorrelated, the errors are added together quadratically:&#xA;$$ \Delta(A+B) = \sqrt{\Delta A^2 + \Delta B^2}, $$&#xA;i.e. &#xA;$$ \left(\frac{\Delta R}{R^2}\right)^2 =&#xA;\left(\frac{\Delta R_1}{R_1^2}\right)^2 &#xA;+ \left(\frac{\Delta R_2}{R_2^2}\right)^2&#xA;+ \left(\frac{\Delta R_3}{R_3^2}\right)^2 $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This may be too advanced for your course.)&lt;/p&gt;&#xA;" OwnerUserId="171" LastEditorUserId="171" LastEditorDisplayName="" LastEditDate="2010-10-05T13:45:06.927" LastActivityDate="2010-10-05T13:45:06.927" />
  <row Id="6087" PostTypeId="1" AcceptedAnswerId="6091" CreationDate="2010-10-05T06:50:02.733" Score="1" ViewCount="220" Body="&lt;p&gt;This is an Exercise from Apostol's book. Many of them seem tough and here is one of them which i am struggling with. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $n \in \mathbb{N}$, prove that this identity is true: $$\Bigl\lfloor{\sqrt{n} + \sqrt{n+1}\Bigl\rfloor} = \Bigl\lfloor{\sqrt{4n+2}\Bigl\rfloor}$$&lt;/p&gt;&#xA;" LastActivityDate="2010-10-07T01:44:05.467" Title="For $n \in \mathbb{N}$ $\lfloor{\sqrt{n} + \sqrt{n+1}\rfloor} = \lfloor{\sqrt{4n+2}\rfloor}$" Tags="&lt;number-theory&gt;&lt;functions&gt;" AnswerCount="3" FavoriteCount="1" />
  <row Id="6088" PostTypeId="2" ParentId="6057" CreationDate="2010-10-05T07:15:06.643" Score="2" ViewCount="" Body="&lt;p&gt;The industry standard way of doing this is is to calculate a &quot;mask&quot; of sorts, and apply it to each one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By &quot;mask&quot;, I mean, say, a function for the current point based on the points before and after.  For example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f(x) = \frac{1}{9}[x-2] + \frac{2}{9}[x-1] + \frac{3}{9}[x] + \frac{2}{9}[x+1] + \frac{1}{9}[x+2] $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $[p]$ is the value of the pixel at $p$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So to find the new pixel value at, say, pixel $4$, you'd use $ \frac{1}{9}[2] + \frac{2}{9}[3] + \frac{3}{9}[4] + \frac{2}{9}[5] + \frac{1}{9}[6] $ (remembering that, say, $[3]$ is the value of pixel 3)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All that's left to do is then apply your mask to every pixel.  (What should you use for pixels that are close to the edge?  That's up to you.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that to be a true blurring mask, your coefficients must add up to 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A &quot;Gaussian Blur&quot; is just applying a special mask based off of the Gaussian Curve to your pixels.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is, you make the &quot;coefficient&quot; of each term a number based off of the Gaussian Curve at that point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, &lt;a href=&quot;http://en.wikipedia.org/wiki/Gaussian_Blur#Sample_Gaussian_matrix&quot; rel=&quot;nofollow&quot;&gt;wikipedia lists&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f(x) = 0.00038771[x-3] + 0.01330373[x-2] +&#xA;0.11098164[x-1] + 0.22508352[x] + 0.11098164[x+1] +&#xA;0.01330373[x+2] + 0.00038771[x+2]$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;as a mask with σ = 0.84089642 and a &quot;cut-off&quot; radius of 3, although you can pick your own standard deviation and cut-off radius to your liking.&lt;/p&gt;&#xA;" OwnerUserId="51" LastActivityDate="2010-10-05T07:15:06.643" />
  <row Id="6089" PostTypeId="2" ParentId="6075" CreationDate="2010-10-05T07:18:08.693" Score="2" ViewCount="" Body="&lt;p&gt;If $\mathcal{A}_m$ is the ring of integers of $\mathbb{Q}(\zeta_m)$, then $\mathfrak{B} \cap \mathcal{A}_m = \mathfrak{p}$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You say $a \equiv b \pmod{\mathfrak{B}}$ where $a,b \in \mathbb{Q}(\zeta_m)$.  Does this mean $a,b$ are algebraic integers (in $\mathcal{A}_m$)?  If not, it must mean that there is some denominator $d \in \mathcal{A}_m$, $d \notin \mathfrak{B}$ such that $d(a-b) \in \mathfrak{B}$.  (Just like we could say $\frac{2}{3} \equiv \frac{7}{3} \pmod{5}$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Either way, the above fact gives us $a \equiv b \pmod{\mathfrak{p}}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't think you can relate Norm(a) to Norm(b) or Trace(a) to Trace(b) modulo (p).  Consider $1 \equiv 2i \pmod{(1-2i)}$ in the Gaussian integers $\mathbb{Z}[i]$.  Taking norm or trace gives us nothing modulo (5).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can, however, use $\mathfrak{p} \cap \mathbb{Z} = (p)$ to say that ${\rm Norm}(a-b) \in (p)$, if $a,b$ are algebraic integers.  If, however, we have some denominator $d \in \mathfrak{p}'$ where $\mathfrak{p}'$ is a conjugate ideal to $\mathfrak{p}$, then the norm will give us powers of $p$ in the numerator and denominator, so we have to be more careful.&lt;/p&gt;&#xA;" OwnerUserId="1461" LastActivityDate="2010-10-05T07:18:08.693" />
  <row Id="6090" PostTypeId="1" AcceptedAnswerId="6160" CreationDate="2010-10-05T07:26:11.997" Score="2" ViewCount="326" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have been reading about Riemann Zeta function and have been thinking about it for some time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Has anything been published regarding upper bound for the real part of zeta function zeros as the imaginary part of the zeros tend to infinity?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="2320" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-10-05T08:31:23.300" LastActivityDate="2010-12-24T23:30:36.527" Title="On Zeta function zeros in the critical strip" Tags="&lt;number-theory&gt;&lt;riemann-zeta&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="1" />
  <row Id="6091" PostTypeId="2" ParentId="6087" CreationDate="2010-10-05T07:43:15.927" Score="6" ViewCount="" Body="&lt;p&gt;$\begin{align*}(\sqrt{n} + \sqrt{n+1})^2 &amp;amp;= 4n+2 - 2\left(n+1/2 - \sqrt{n(n+1)}\right) \\ &amp;amp;= 4n+2 - 2(AM(n,n+1) - GM(n,n+1)) \\ &amp;amp;\in (4n+1, 4n+2).\end{align*}$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(The first line is just algebra. In the second line, $AM$ and $GM$ are respectively the arithmetic and geometric means. To get the third line: $n &amp;lt; GM(n,n+1) &amp;lt; AM(n,n+1)=n+1/2$ by the AM-GM inequality, so $0 &amp;lt; AM(n,n+1)-GM(n,n+1) &amp;lt; 1/2$, and the third line follows.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But there are no perfect squares between $4n+1$ and $4n+2$, and thus no integers between $\sqrt{n} + \sqrt{n+1}$ and $\sqrt{4n+2}$, QED.&lt;/p&gt;&#xA;" OwnerUserId="2032" LastEditorUserId="2032" LastEditorDisplayName="" LastEditDate="2010-10-05T08:11:42.177" LastActivityDate="2010-10-05T08:11:42.177" CommentCount="3" />
  <row Id="6092" PostTypeId="2" ParentId="6077" CreationDate="2010-10-05T08:02:29.957" Score="3" ViewCount="" Body="&lt;p&gt;This is easy, since $R$ is monotonic as a function of $R_1,R_2,R_3$ (which is obvious from physical considerations: increasing one of the resistances can only increase the total resistance). Thus, to compute error bounds for $R$, it suffices to consider the minimum and maximum possible values for your three resistors: the minimum corrected value of $R$ is $f(0.995R_1, 0.995R_2, 0.995R_3)$, and the maximum corrected value is $f(1.005R_1, 1.005R_2, 1.005R_3)$, where $f(x,y,z) = (x^{-1} + y^{-1} + z^{-1})^{-1}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But a further simplification is that $f$ is homogeneous of degree 1, i. e., $f(cx, cy, cz) = cf(x,y,z)$ (again, this is obvious intuitively: choosing different units for your resistances multiplies both the individual and total resistances by the same proportionality constant). Thus, $f(0.995R_1, 0.995R_2, 0.995R_3) = 0.995R$, and $f(1.005R_1, 1.005R_2, 1.005R_3) = 1.005R$. Thus, the maximum relative error in the total resistance is also 0.5%. &lt;/p&gt;&#xA;" OwnerUserId="2032" LastActivityDate="2010-10-05T08:02:29.957" CommentCount="2" />
  <row Id="6093" PostTypeId="2" ParentId="6078" CreationDate="2010-10-05T08:05:10.820" Score="1" ViewCount="" Body="&lt;p&gt;More generally, let $G$ be a finite group with order $p_1p_2\cdots p_n$, where $p_1&amp;lt; p_2&amp;lt; \cdots &amp;lt; p_n$ are distinct primes.  Then by the normalizer-centralizer theorem, the Sylow $p_1$-subgroup is central in its normalizer, so there is a normal $p_1$-complement in $G$ by Burnside's transfer theorem.  Inductively, we can then show the Sylow $p_n$-subgroup is normal in $G$.&lt;/p&gt;&#xA;" OwnerUserId="641" LastActivityDate="2010-10-05T08:05:10.820" />
  <row Id="6094" PostTypeId="1" AcceptedAnswerId="6095" CreationDate="2010-10-05T08:13:21.650" Score="1" ViewCount="88" Body="&lt;p&gt;Given an affine variety, we can forget it into a variety. The other way round, given a variety $X$ we can look at &#xA;$$spec(\mathcal O_X(X)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this necessarily an affine variety (a priori its just an affien scheme)?&#xA;If so, am I right that this implys that taking spec of the global functions is left adjoint to the forgetful functor on the level of varieties?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, what does this functor intuitively? If you feed in an affine variety, it spits out the same variety again. If you apply it to a complete irreducible variety (for example $\mathbb P^n$) it gets transformed into a point. &lt;/p&gt;&#xA;" OwnerUserId="667" LastActivityDate="2010-10-05T08:43:19.683" Title="Turning varieties into affine varieties?" Tags="&lt;algebraic-geometry&gt;" AnswerCount="1" />
  <row Id="6095" PostTypeId="2" ParentId="6094" CreationDate="2010-10-05T08:43:19.683" Score="4" ViewCount="" Body="&lt;p&gt;The functor $Sch \to Ring^{op}, X \mapsto \mathcal{O}_X(X)$ is left adjoint to the functor $\text{Spec} : Ring^{op}\to Sch$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By a variety you mean a separated integral scheme $X$ of finite type over a field $k$? Then $O_X(X)$ does not have to be of finite type over $k$. Namely, there are integral $k$-subalgebras $A,B$ of a $k$-algebra $C$ such that $A,B$ are of finite type, but $A \cap B$ is not. Then glue $Spec(A)$ and $Spec(B)$ along $Spec(C)$ to get a counterexample.&lt;/p&gt;&#xA;" OwnerUserId="1650" LastActivityDate="2010-10-05T08:43:19.683" CommentCount="3" />
  <row Id="6096" PostTypeId="2" ParentId="6087" CreationDate="2010-10-05T08:54:25.330" Score="4" ViewCount="" Body="&lt;p&gt;Note that $$\frac{1}{2} &gt; \sqrt{n+\frac{1}{2}}-\sqrt{n} &gt; \sqrt{n+1}- \sqrt{n+\frac{1}{2}} &gt; 0,$$&#xA;since $f(x)=\sqrt{x+1/2}-\sqrt{x}$ is a decreasing function. The inequalities and the fact that $f(x)$ is decreasing follow from noting that $\sqrt{x+1/2} - \sqrt{x} = 1/2(\sqrt{x+1/2} + \sqrt{x}).$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we can write&#xA;$$ 1&gt; \left( \sqrt{n+\frac{1}{2}}-\sqrt{n} \right) - \left( \sqrt{n+1}- \sqrt{n+\frac{1}{2}} \right) &gt; 0.$$&#xA;Hence $$1 &gt; 2\sqrt{n+\frac{1}{2}} - \left( \sqrt{n+1}+\sqrt{n} \right) &gt; 0.$$&#xA;From which the result follows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For clarity we've shown that we can write&#xA;$$\sqrt{n+1}+\sqrt{n} + r = 2\sqrt{n+\frac{1}{2}} \quad \textrm{ for } 0 &amp;lt; r &amp;lt; 1,$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where we note that we do not straddle an integer since $\sqrt{4n+1} &amp;lt; \sqrt{n} + \sqrt{n+1},$ and there are no integers between $\sqrt{4n+1}$ and $\sqrt{4n+2}.$&lt;/p&gt;&#xA;" OwnerUserId="1301" LastEditorUserId="1301" LastEditorDisplayName="" LastEditDate="2010-10-05T10:53:51.137" LastActivityDate="2010-10-05T10:53:51.137" CommentCount="1" />
  <row Id="6097" PostTypeId="2" ParentId="5964" CreationDate="2010-10-05T09:12:19.540" Score="2" ViewCount="" Body="&lt;p&gt;The answer is three.  As for a proof:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As stated above two colours is not enough.  Every square grid colouring must have either alternating colours uniformly across the grid, or somewhere there are two adjacent vertices of the same colour.  In the former case then any swap of vertices will create a three-in-a-row.  If there are two in a row of the same colour and we assume it is not possible to get to a three in a row configuration, then all the vertices marked X below must not be colour 1.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;.X..X.&#xA;XX11XX&#xA;.X..X.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;With only two colours this clearly creates a problem as all the X vertices must be colour 2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To show that 3 colours suffices consider the colouring below:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;123123&#xA;231231&#xA;312312&#xA;123123&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The pattern can be extended to cover an infinitely large grid and it's not possible to swap to get three-in-a-row or even to get a coloured path of length three.&lt;/p&gt;&#xA;" OwnerUserId="1750" LastActivityDate="2010-10-05T09:12:19.540" />
  <row Id="6098" PostTypeId="2" ParentId="6008" CreationDate="2010-10-05T10:24:50.807" Score="0" ViewCount="" Body="&lt;p&gt;Ok, i managed with &lt;a href=&quot;http://math.stackexchange.com/questions/6036/moments-on-number-of-occurrences-of-substring&quot;&gt;Robin Chapman's help&lt;/a&gt;  compute the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let X1,X2,...,XN be the random variables defined as follows:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Xi=1 if there is a run of heads starting at the position i, otherwise Xi=0.&lt;br&gt;&#xA;Then X=X1+X2+...+XN.&lt;br&gt;&#xA;By linearity of expectation E(X)=E(X1)+...+E(XN).&lt;br&gt;&#xA;Now, a run of heads can start at the position i only by the condition &quot;a tail at the position i-1&quot;. That means P(Xi=1)=(1-p)*p=q*p.&lt;br&gt;&#xA;But at i=1 there is no i=0.&lt;br&gt;&#xA;Hence P(X1=1)=p.&#xA;Last step: E(Xi)=1*P(Xi=1)+0*P(Xi=0)=P(Xi=1)&lt;br&gt;&#xA;Finally:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;E(X)=p+(N-1)*p*q&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In case of p=1/2 E(X)=(N+1)/4&#xA;and if N=&gt;oo then E(X)=&gt;N/4  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Congrats leonbloy! Your approach was very fruitful. Now, to get exact variance seems to be rather tricky. But in any case, many thanks anyway.&lt;/p&gt;&#xA;" OwnerUserId="2323" LastActivityDate="2010-10-05T10:24:50.807" />
  <row Id="6099" PostTypeId="1" CreationDate="2010-10-05T10:49:44.183" Score="6" ViewCount="189" Body="&lt;p&gt;If $M$ is a differentiable manifold, De Rham's theorem gives for each positive integer $k$ an isomorphism&#xA;$Rh^k : H^k_{DR}(M,\mathbb R) \to H^k_{singular}(M,\mathbb R)$. On the other hand, we have a canonical map $H^k_{sing}(M,\mathbb Z) \to H^k_{singular}(M,\mathbb R)$ . Allow me to denote (this is not standard)  its image by $\tilde {H}^k_{singular }(M,\mathbb Z)$. My question is : how do you recognize if, given a closed differental $k$- form $\omega$ on $M$, its image &#xA;$Rh^k([\omega]) \in H^k_{singular}(M,\mathbb R) $ is actually in &#xA;$\tilde {H}^k_{singular}(M,\mathbb Z)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would very much appreciate a concrete answer, ideally backed up by one or more explicit calculations. Thank you for your attention.&lt;/p&gt;&#xA;" LastActivityDate="2010-10-06T09:55:55.863" Title="Integral classes in de Rham cohomology" Tags="&lt;topology&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="2" />
  <row Id="6100" PostTypeId="1" AcceptedAnswerId="42229" CreationDate="2010-10-05T11:11:43.373" Score="3" ViewCount="246" Body="&lt;p&gt;Please list both the problem/area and justify why it is important philosophically. This question doesn't cover questions that are only important within the philosophy of mathematics itself.&lt;/p&gt;&#xA;" OwnerUserId="123" LastEditorUserId="96" LastEditorDisplayName="anon" LastEditDate="2011-05-30T23:12:15.093" LastActivityDate="2011-05-30T23:12:15.093" CommunityOwnedDate="2010-10-05T11:11:43.373" Title="What mathematical questions or areas have philosophical implications outside of mathematics?" Tags="&lt;soft-question&gt;&lt;big-list&gt;&lt;philosophy&gt;" AnswerCount="3" CommentCount="13" FavoriteCount="4" />
  <row Id="6101" PostTypeId="1" CreationDate="2010-10-05T11:35:36.583" Score="2" ViewCount="323" Body="&lt;p&gt;This question is intended to complement &lt;a href=&quot;http://math.stackexchange.com/questions/6100/what-mathematical-questions-or-areas-have-philosophical-implications-outside-of-m&quot;&gt;What mathematical questions or areas have philosophical implications outside of mathematics?&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="123" LastActivityDate="2010-10-13T18:02:36.500" CommunityOwnedDate="2010-10-05T11:35:36.583" Title="What are the most important questions or areas of study in the philosophy of mathematics?" Tags="&lt;philosophy&gt;" AnswerCount="4" CommentCount="5" FavoriteCount="2" />
  <row Id="6102" PostTypeId="2" ParentId="6099" CreationDate="2010-10-05T11:49:37.413" Score="6" ViewCount="" Body="&lt;p&gt;By Poincare duality a $k$-form will be integral iff its integral over all&#xA;(smooth) singular $k$-cycles is an integer.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-10-05T11:49:37.413" />
  <row Id="6103" PostTypeId="2" ParentId="6080" CreationDate="2010-10-05T11:57:19.307" Score="7" ViewCount="" Body="&lt;p&gt;One can also see that $\mathbb{R}P^3$ is a Lie group, and that all Lie groups are orientable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To see that $\mathbb{R}P^3$ is a Lie group, first think of $S^3$ as the collection of unit quaternions.  The subset $\{1,-1\}$ is not only a subgroup, but it is a normal subgroup of $S^3$.  Hence, $S^3/\{1,-1\}$ is a Lie group.  In fact, the elmement -1 acts as the antipodal map.  (It turns out, that as a Lie group, $\mathbb{R}P^3$ is isomorphic to $SO(3)$, the 3x3 real matrices satisfying $AA^t = A^tA = Id$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is every (connected component of a) Lie group orientable?  Well, suppose $G$ is a (connected) Lie group with identity $e$.  At the identity, pick any ordered basis $e_i$ for $T_e G$ (i.e., pick an orientation on the vector space $T_e G$.  To figure out what what orientation at $g\in G$ should be, notice the map $L_g:G\rightarrow G$ given by $L_g(h) = gh$ maps $e$ to $g$.  Hence its differential $d_e L_g$ maps $T_e G$ to $T_g G$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The map $L_g$ is actually a diffeomorphism (with inverse $L_{g^{-1}}$), so the map $d_e L_g$ must also be an isomorphism.  Now, simply take as an ordered basis $d_e L_g e_i$.&lt;/p&gt;&#xA;" OwnerUserId="331" LastActivityDate="2010-10-05T11:57:19.307" />
  <row Id="6104" PostTypeId="1" CreationDate="2010-10-05T12:15:21.090" Score="0" ViewCount="77" Body="&lt;p&gt;Is it possible to make up a &lt;em&gt;distinct&lt;/em&gt; number that is made up of four tuples. One of which indicates the direction of the arrangement of the other three.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Say, 9*4*7 = 252. 252 &lt;em&gt;is distinct&lt;/em&gt; in the sense that you cannot get it as a multiplication of 3 distinct single digit numbers other than 9, 4, 7. Now, the problem is 252 could mean, any of the three orders. Say, 947, 497, 974. Is there any way to add a sense of direction/arrangement to it?&lt;/p&gt;&#xA;" OwnerUserId="1869" LastActivityDate="2010-10-05T13:01:08.910" Title="Numbers that represent four tuples" Tags="&lt;number-theory&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="6105" PostTypeId="2" ParentId="6101" CreationDate="2010-10-05T12:16:47.673" Score="4" ViewCount="" Body="&lt;p&gt;Here is a rather brief summary of the major positions on a couple of important philosophy of mathematics questions. First a list of questions:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;Are we creating or discovering when we do mathematics?&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;What are mathematical entities?&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;How do you explain &lt;a href=&quot;http://en.wikipedia.org/wiki/The_Unreasonable_Effectiveness_of_Mathematics_in_the_Natural_Sciences&quot; rel=&quot;nofollow&quot;&gt;the unreasonable effectiveness of mathematics in natural science&lt;/a&gt;?&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;An early answer to these question was Plato's suggestion that mathematical entities (triangles, squares, numbers...) exist in the world of forms. So numbers really do exist, just in some kind of rarified universe &lt;a href=&quot;http://en.wikipedia.org/wiki/Allegory_of_the_Cave&quot; rel=&quot;nofollow&quot;&gt;that we only have partial access to&lt;/a&gt;. The idea of a &quot;Platonic heaven&quot; of numbers is not a particularly popular position any more.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Immanuel Kant thought that geometry and arithmetic were somehow innate (indeed, they were preconditions to experience of space and the passage of time). That is, without having an &lt;em&gt;a priori&lt;/em&gt; understanding of space and number, we wouldn't be able to perceive the world around us as we do. Mathematical entities, for Kant, were real.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Logicism is a position that grew out of Russell and Whitehead's work on mathematical logic. The idea was that logic is somehow universal, and that all of mathematics is built up from these basic axioms, which are uncontroversially, universally, absolutely True. (With a capital &quot;T&quot;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Formalism started at roughly the same time with Hilbert. Here the idea is that mathematics is a game we play, with certain rules (axioms) laid down somewhat arbitrarily. This shares a lot with logicisim, but is importantly different in its attitude to the status of the axioms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Structuralism is a much more recent position (although there are echoes of the ideas in Hilbert, Poincare and others). The idea here is that mathematics is about &lt;em&gt;structures.&lt;/em&gt; We learn about structures through abstracting away incidental details to recognise patterns. What exactly structures are depends on what kind of structuralist you ask. Some (like Stewart Shapiro) think that structures really are out there in the world, somehow. That is, structures exist over and above the things that satisfy the structural relations. Others (like Michael Resnik) think they are not quite this real. Resnik's brand of structuralism looks a lot like &lt;em&gt;nominalism:&lt;/em&gt; Nominalism about mathematical entities grows out of a more &lt;a href=&quot;http://en.wikipedia.org/wiki/Nominalism&quot; rel=&quot;nofollow&quot;&gt;general position&lt;/a&gt; towards abstract entities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a commenter said above, philosophy isn't about results. I think the most important contribution of philosophy is to ask the big questions, like the ones above, and to map out the logical space of possible responses to them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have restricted myself to philosophy of &lt;em&gt;mathematics&lt;/em&gt; and deliberately avoided philosophy of logic, which is its own cottage industry. Also, this summary reflects my own interests and biases and should not be taken as a representative summary of all the debates and positions taken in philosophy of maths. A more balanced summary can be found at the excellent &lt;em&gt;Stanford Encyclopedia of Philosophy&lt;/em&gt; &lt;a href=&quot;http://plato.stanford.edu/entries/philosophy-mathematics/&quot; rel=&quot;nofollow&quot;&gt;article&lt;/a&gt; by Leon Horsten.&lt;/p&gt;&#xA;" OwnerUserId="172" LastActivityDate="2010-10-05T12:16:47.673" CommunityOwnedDate="2010-10-05T12:16:47.673" CommentCount="2" />
  <row Id="6106" PostTypeId="2" ParentId="6100" CreationDate="2010-10-05T12:24:26.470" Score="3" ViewCount="" Body="&lt;p&gt;Goedel's incompleteness theorem is used by some (e.g. Roger Penrose) as part of a justification for why computers will never achieve consciousness.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fact that all infinite dimensional separable Hilbert spaces are isomorphic has philosophical implications for the metaphysics of quantum mechanics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Various results in dynamical systems theory related to chaotic systems limit what can be said about predictability and about what it means for a system to be deterministic. For example &lt;a href=&quot;http://www.ams.org/journals/bull/1991-24-01/S0273-0979-1991-15953-7/S0273-0979-1991-15953-7.pdf&quot; rel=&quot;nofollow&quot;&gt;this paper&lt;/a&gt; by Ornstein and Weiss (warning: it's huge and will take a long time to download on slow connections) has been used to suggest that the distinction between deterministic and stochastic systems is flawed.&lt;/p&gt;&#xA;" OwnerUserId="172" LastActivityDate="2010-10-05T12:24:26.470" CommunityOwnedDate="2010-10-05T12:24:26.470" CommentCount="6" />
  <row Id="6107" PostTypeId="1" CreationDate="2010-10-05T12:51:02.377" Score="1" ViewCount="62" Body="&lt;p&gt;i have lineA with points are(1,1) and  (8,8) and another lineB with points(2,2) and (4,4) i like to prove lineB lies on LineA ,how can do this?&lt;/p&gt;&#xA;" LastActivityDate="2010-10-05T13:13:18.930" Title="how do find the line lies on another line" Tags="&lt;geometry&gt;" AnswerCount="3" FavoriteCount="1" />
  <row Id="6109" PostTypeId="2" ParentId="6107" CreationDate="2010-10-05T12:55:11.633" Score="2" ViewCount="" Body="&lt;p&gt;The points are all on the line $y=x$.&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2010-10-05T12:55:11.633" />
  <row Id="6110" PostTypeId="2" ParentId="6107" CreationDate="2010-10-05T12:57:35.057" Score="0" ViewCount="" Body="&lt;p&gt;Hint: &lt;strike&gt; Take the perpendicular distance and prove that the distance is 0.&lt;/strike&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Use the 2 point slope form.Equation of the line passing through $(x_{1},y_{1})$ and $(x_{2},y_{2})$ is given bye $$\frac{x-x_{1}}{x_{2}-x_{1}}=\frac{y-y_{1}}{y_{2}-y_{1}}$$ and observe that they are same.&lt;/p&gt;&#xA;" LastActivityDate="2010-10-05T13:13:18.930" CommentCount="0" />
  <row Id="6111" PostTypeId="2" ParentId="6104" CreationDate="2010-10-05T13:01:08.910" Score="1" ViewCount="" Body="&lt;p&gt;Yes. Try creating your number like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose your 3-tuple has values (a,b,c). Then generate your distinct number as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2^a * 3^b * 5^c&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since these are prime bases, this number will be unique for any 3 values and will be recoverable, in that you can factor your number and get back all 3 numbers and their orders. To extend this to any n-tuple just use more prime number bases. &lt;/p&gt;&#xA;" OwnerUserId="2003" LastActivityDate="2010-10-05T13:01:08.910" />
  <row Id="6112" PostTypeId="2" ParentId="6107" CreationDate="2010-10-05T13:01:23.460" Score="0" ViewCount="" Body="&lt;p&gt;In general you can use a determinant expression&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{vmatrix}x_1&amp;amp;y_1&amp;amp;1\\x_2&amp;amp;y_2&amp;amp;1\\x_3&amp;amp;y_3&amp;amp;1\end{vmatrix}=0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;to verify that three points $(x_i,y_i)$, $i=1\dots 3$ are collinear. As Hans points out though, that the coordinates of all four points that you have have their abscissas and ordinates identical makes them obviously collinear.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-10-05T13:01:23.460" />
  <row Id="6113" PostTypeId="1" AcceptedAnswerId="6114" CreationDate="2010-10-05T13:03:01.973" Score="1" ViewCount="103" Body="&lt;p&gt;A quick question-Something that I learned that I forgot throughout the years. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If a question asked you for a multiple of 9 and 12, which would be less than 40, would &quot;36&quot; be an answer, as both 9 and 12 can multiply to get 36, yet it fits under the criteria for being under 40?&lt;/p&gt;&#xA;" OwnerUserId="2062" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-10-05T14:13:02.427" LastActivityDate="2010-10-05T14:13:02.427" Title="What's a Multiple?" Tags="&lt;arithmetic&gt;" AnswerCount="1" />
  <row Id="6114" PostTypeId="2" ParentId="6113" CreationDate="2010-10-05T13:06:41.687" Score="4" ViewCount="" Body="&lt;p&gt;Yes you are correct with your reasoning. Multiples of $9$ are of the form $9m$ where $m \in \mathbb{Z}$, the set of integers.&lt;/p&gt;&#xA;" LastActivityDate="2010-10-05T13:06:41.687" />
  <row Id="6115" PostTypeId="2" ParentId="6060" CreationDate="2010-10-05T13:10:23.830" Score="1" ViewCount="" Body="&lt;p&gt;Within the framework of &lt;a href=&quot;http://en.wikipedia.org/wiki/Potential_flow&quot; rel=&quot;nofollow&quot;&gt;Potential flow&lt;/a&gt;,the velocity field of a two dimensional potential flow is the gradient of the real part of a complex function $w(z)$ which is analytical outside regions of singularity. The imaginary part of the complex potential is the stream function, whose constant loci are the streamlines, i.e., the integral curves of the velocity field.&#xA;For example for a single vortex loacted at $z_0$, the potential function is given by: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$w(z) = -\frac{i \Gamma}{2 \pi} log(z-z_0)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;($\Gamma$ is (the constant) vorticity). It is easy to verify that this poential function&#xA;gives rize to an angular flow around $z_0$ of velocity inversly proportional to the distance from $z_0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The real part of the complex potential saisfies the two dimensional Laplace equation and by linearity, new solutions can be obtained by superposition. In the case of the multiple vortices, the complex potential function takes the form&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$w(z) = \sum_n -\frac{(-1)^n i\Gamma}{2 \pi} log(z-z_n)$, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $z_n$ are the vortex centers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, I can think of a possible candidate of a Voronoi diagram for the vortices by defining the cell walls as the loci of stagnation points in which the fluid velocity is zero. These points are characterized by the condition: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{dw(z)}{dz}=0$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This suggestion will result a simple solution in the case of vortices rotating in the same direction, but not in the case of counter-rotating vortices.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remark: From a superficial reading of the referred article, the vortex solution is a perturbation to a stagnation point flow. This is a flow towards a wall or a corner.&#xA;To take into account the existence of this wall, one must add vortices from the other side of the wall according to the &lt;a href=&quot;http://en.wikipedia.org/wiki/Method_of_images&quot; rel=&quot;nofollow&quot;&gt;method of images&lt;/a&gt;. May be the stagnation point loci of the complete solution (including the images) will have the required Voronoi cell structure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks to the correct comment by Rahul, the stagnation points are certainly discrete&#xA;and do not have one dimensional loci. I think that that the suggestion can be salvaged if we define&#xA;the Voronoi cell walls as passing through the stagnation points and having a normal in the direction of the line&#xA;connecting the two vortices.&lt;/p&gt;&#xA;" OwnerUserId="1430" LastEditorUserId="1430" LastEditorDisplayName="" LastEditDate="2010-10-05T16:29:29.913" LastActivityDate="2010-10-05T16:29:29.913" CommentCount="2" />
  <row Id="6116" PostTypeId="1" AcceptedAnswerId="6119" CreationDate="2010-10-05T13:37:55.930" Score="1" ViewCount="84" Body="&lt;p&gt;I've been doing some maths work using the rate of flow of liquids. I've used various models for the flow and various methods to integrate these models. The one thing that is confusing me is the difference between constants of proportionality, and constants of integration?&lt;/p&gt;&#xA;" OwnerUserId="1699" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-10-05T14:08:37.167" LastActivityDate="2010-10-05T14:08:37.167" Title="What is the difference between constants of proportionality and constants of integration?" Tags="&lt;terminology&gt;&lt;integral&gt;" AnswerCount="1" />
  <row Id="6117" PostTypeId="1" AcceptedAnswerId="6127" CreationDate="2010-10-05T13:40:56.157" Score="2" ViewCount="92" Body="&lt;p&gt;According to Wikipedia:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Let G be a covering group of H. The&#xA;  kernel K of the covering homomorphism&#xA;  is just the fiber over the identity in&#xA;  H and is a discrete normal subgroup of&#xA;  G.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It is easy to show that the kernel is a normal subgroup, but why is it discrete? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know this would be true if the identity of H was open, but I cannot show this (and I don't even know if it is true/the right way to prove that K is discrete). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT:&#xA;If we assume that the definition of &quot;cover space&quot; does not require the fibres to be discrete and we assume that H is connected and locally path-connected, does it still follow that the kernel is discrete? &lt;/p&gt;&#xA;" OwnerUserId="1907" LastEditorUserId="1907" LastEditorDisplayName="" LastEditDate="2010-10-05T14:55:33.127" LastActivityDate="2010-10-05T15:25:47.943" Title="Why is the kernel of a covering group discrete?" Tags="&lt;topology&gt;" AnswerCount="1" CommentCount="7" />
  <row Id="6118" PostTypeId="1" CreationDate="2010-10-05T13:46:12.883" Score="-1" ViewCount="129" Body="&lt;p&gt;I was just wondering if there is any single English word for that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We call things quantifiable, measurable etc. but none of these exactly fit the the question posted in subject.&lt;/p&gt;&#xA;" OwnerUserId="2327" LastEditorUserId="1301" LastEditorDisplayName="" LastEditDate="2010-10-05T14:00:27.930" LastActivityDate="2010-10-05T14:00:27.930" ClosedDate="2010-10-05T14:23:19.657" Title="What is the single word to describe a situation that can not be expressed as a mathematical formula" Tags="&lt;terminology&gt;" AnswerCount="2" CommentCount="5" />
  <row Id="6119" PostTypeId="2" ParentId="6116" CreationDate="2010-10-05T13:54:26.843" Score="4" ViewCount="" Body="&lt;p&gt;They are &lt;em&gt;completely different animals&lt;/em&gt;. About the only thing they have in common is the words &quot;constant of&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When two quantities, $p$ and $q$, are &quot;proportional&quot;, that means that growth in one will be &quot;mirrored&quot; by growth in the other, and vice-versa. The growth need not be exactly the same (for example, your weight is [roughly] proportional to your height, but an increase in weight of 10 pounds does not correspond to an increase in height of 10 inches). There is a certain &quot;scaling&quot; that goes on. The &quot;constant of proportionality&quot; is what represents that scaling. We will have something of the form $p=kq$, with $k$ a constant; that means that for every unit that $q$ increases, $p$ will increase by $k$ units. That's the constant of proportionality. It is a constant, that depends on the quantities being considered.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you do an indefinite integral, on the other hand, you are trying to find &lt;em&gt;all&lt;/em&gt; antiderivatives of a function. That is, $\int f(x)dx$ represents a &lt;em&gt;family&lt;/em&gt; of functions, namely, all functions $F(x)$ such that $F'(x)=f(x)$. Functions will generally have a &lt;em&gt;lot&lt;/em&gt; of antiderivatives. But one can prove that if $F(x)$ and $G(x)$ are two functions such that $F'(x)=G'(x)=f(x)$, then $F$ and $G$ will just be vertical translates of each other; that is, there will exist a constant $c_0$ such that $G(x)=F(x)+c_0$. In order to represent &lt;em&gt;all&lt;/em&gt; antiderivatives of a function $f(x)$, then, it is enough to find a single antiderivative $F(x)$, because then every other antiderivative will look like &quot;$F(x)+c$&quot; for some constant $c$. So we write &#xA;$$\int f(x)dx = F(x)+C$$&#xA;to represent the entire family of antiderivatives; this means: &quot;the antiderivatives of $f(x)$ are all the functions of the form $F(x)+C$, where $C$ is a constant.&quot; Pick a constant, you get an antiderivative. Pick a different constant, you get a different antiderivative. If $G$ is any antiderivative, then there will be some constant $C$ such that $G(x)=F(x)+C$. We call $C$ the &quot;constant of integration.&quot; So $C$ is not a specific constant, but rather corresponds to the fact that antiderivatives are families, not specific functions. &lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-10-05T13:54:26.843" />
  <row Id="6120" PostTypeId="2" ParentId="6118" CreationDate="2010-10-05T13:55:12.747" Score="0" ViewCount="" Body="&lt;p&gt;For a mathematical function which cannot be expressed as a formula we say that it has no closed-form expression. &lt;/p&gt;&#xA;" OwnerUserId="1195" LastActivityDate="2010-10-05T13:55:12.747" CommentCount="3" />
  <row Id="6121" PostTypeId="2" ParentId="6118" CreationDate="2010-10-05T13:59:44.290" Score="0" ViewCount="" Body="&lt;p&gt;I can think of a few that might fit the bill: unpredictable, incalculable, incomputable. But maybe you'd be better off asking on: &lt;a href=&quot;http://english.stackexchange.com/questions&quot;&gt;http://english.stackexchange.com/questions&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1301" LastActivityDate="2010-10-05T13:59:44.290" />
  <row Id="6122" PostTypeId="2" ParentId="6099" CreationDate="2010-10-05T14:06:05.467" Score="2" ViewCount="" Body="&lt;p&gt;I think Robin's result is also clear if we look at the definition of the isomorphism between De Rham and (${\cal C}^\infty$) singular cohomologies. (Ok, we are both saying the same thing, but I like to say it this way.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remember (see Dupont, Curvature and characteristic classes, Springer LNM 640) the definition of the isomorphism at the cochain level:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;I : \Omega^n (M) \longrightarrow C^n (M) \ , &#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here $\Omega^n(M)$ are the degree $n$-forms on $M$, $C^n(M)$ the ${\cal C}^\infty$ singular $n$-cochains. Then, for $\omega \in \Omega^n (M)$, the value of $I(\omega )$ on a ${\cal C}^\infty $ singular $n$-simplex $\sigma$ is defined as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;I (w)_{\sigma}  = \int \sigma^*  \omega \ ,&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where the integral is taken over the standard $n$-simplex $\Delta^n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, for $[I(w)]$ to be an integral class it suffices that all values of $I(w)$ on cycles $\tau = \sum_i \lambda_i\sigma_i$ are integer numbers; that is, if and only if $I(w)_\tau = \sum_i \lambda_i\int \sigma_i^*\omega \in \mathbb{Z}$ for all cycles $\tau$.&lt;/p&gt;&#xA;" OwnerUserId="664" LastEditorUserId="664" LastEditorDisplayName="" LastEditDate="2010-10-06T00:29:14.807" LastActivityDate="2010-10-06T00:29:14.807" CommentCount="1" />
  <row Id="6123" PostTypeId="2" ParentId="6082" CreationDate="2010-10-05T14:35:44.220" Score="4" ViewCount="" Body="&lt;p&gt;I suppose you are talking about the additive group $\mathbb{Z}_n$ (i.e. the numbers modulo $n$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe I am missing something, but...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $n$ is even then $\sum_{i=1}^{n} i \ne 0 \mod n$. So if $\sigma(i) -i$ were a permutation we would have that $\sum_{i=1}^{n} (\sigma(i) - i) \ne 0 \mod n$, but it is $0$ as $\sigma(i)$ is a permutation too.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-10-05T14:35:44.220" CommentCount="1" />
  <row Id="6124" PostTypeId="2" ParentId="6080" CreationDate="2010-10-05T14:40:58.833" Score="4" ViewCount="" Body="&lt;p&gt;$RP3$ can be represented as a three dimensional ball $D3$ with the antipodal points of the bounding sphere identified. One can draw a small loop on the surface of the bounding sphere and compute the direction of the normal according to the &lt;a href=&quot;http://en.wikipedia.org/wiki/Right-hand_rule&quot; rel=&quot;nofollow&quot;&gt;right hand rule&lt;/a&gt;. Now, performing the same computation for the antipodal loop, the normal will have the same direction. Thus there is a consistent choice of the normal. Notice that the same argument will fail in the case of the nonorientable $RP^2$.&lt;/p&gt;&#xA;" OwnerUserId="1430" LastActivityDate="2010-10-05T14:40:58.833" CommentCount="0" />
  <row Id="6125" PostTypeId="1" AcceptedAnswerId="6126" CreationDate="2010-10-05T14:54:51.797" Score="2" ViewCount="152" Body="&lt;p&gt;Let's say you have a set of numbers S and you want to obtain subsets of S, s&lt;sub&gt;1&lt;/sub&gt;, s&lt;sub&gt;2&lt;/sub&gt;,...s&lt;sub&gt;i&lt;/sub&gt;, where i&amp;lt;N.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a particular operation that will group the numbers that are &quot;close to each other&quot;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I will give an example to clarify the question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's say you have a set S which contains {1.4, 2, 2.5, 2.7, 14, 16, 49, 57, 58}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And let's say that you can have maximum of N=4 subsets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, you might end up with a result that looks like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;s&lt;sub&gt;1&lt;/sub&gt; = {1.4, 2, 2.5, 2.7}&lt;br/&gt;&#xA;s&lt;sub&gt;2&lt;/sub&gt; = {14, 16}&lt;br/&gt;&#xA;s&lt;sub&gt;3&lt;/sub&gt; = {49}&lt;br/&gt;&#xA;s&lt;sub&gt;4&lt;/sub&gt; = {57, 58}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am looking for either the name what such a problem would be called (which I can then use to research and write an algorithm).  If you can provide a simple solution, even better.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you for any help with this.&lt;/p&gt;&#xA;" OwnerUserId="2329" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-10-05T14:59:05.887" LastActivityDate="2010-10-06T06:22:51.217" Title="Grouping numbers that are &quot;close to each other&quot;" Tags="&lt;algorithms&gt;" AnswerCount="2" CommentCount="10" />
  <row Id="6126" PostTypeId="2" ParentId="6125" CreationDate="2010-10-05T15:19:13.367" Score="5" ViewCount="" Body="&lt;p&gt;There are many solutions to this problem extant, each rediscovered many times.  Search &quot;statistical clustering methods&quot; or &lt;a href=&quot;http://en.wikipedia.org/wiki/Cluster_analysis&quot; rel=&quot;nofollow&quot;&gt;&quot;cluster analysis.&quot;&lt;/a&gt;  A particularly nice one for your problem is called &lt;a href=&quot;http://en.wikipedia.org/wiki/K-means_clustering&quot; rel=&quot;nofollow&quot;&gt;&quot;K-means&quot;&lt;/a&gt;.  This was rediscovered by cartographers (!) where it is called &lt;a href=&quot;http://en.wikipedia.org/wiki/Jenks_Natural_Breaks_Optimization&quot; rel=&quot;nofollow&quot;&gt;&quot;Jenks' method&quot;&lt;/a&gt; (and is worth mentioning because it's used by some popular mapping software, so perhaps more people have heard of this than of any other method).  The idea behind K-means is to minimize the population-weighted sum of variances of the subsets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;BTW, the challenge in most clustering algorithms lies in determining the number of clusters; usually that needs to be specified (as $N = 4$ in this example) or at least hinted at by the user.  For instance, there always exists a minimum K-means solution: partition $S$ into singletons.  If you require $i &amp;lt; n$, merge the two nearest neighbors but leave all the other singletons as they are: the sum of variances is then just one-quarter the square of the smallest difference among the $x_i$.&lt;/p&gt;&#xA;" OwnerUserId="1489" LastEditorUserId="1489" LastEditorDisplayName="" LastEditDate="2010-10-05T15:58:08.887" LastActivityDate="2010-10-05T15:58:08.887" CommentCount="2" />
  <row Id="6127" PostTypeId="2" ParentId="6117" CreationDate="2010-10-05T15:25:47.943" Score="4" ViewCount="" Body="&lt;p&gt;By definition, if $f:Y \to X$ is a covering space and $x \in X$,&#xA;then there is some neighbourhood $U$ of $x$ such that $f^{-1}(U)$ is a union of open sets&#xA;$V_i$ such that $f$ restricted to each $V_i$ is a homeomorphism.  In particular, &#xA;$f^{-1}(x) \cap V_i$ consists of a single point, and so each point of $f^{-1}(x)$ is open&#xA;in the induced topology on $f^{-1}(x)$.  Thus, as has already been pointed out, the fibres&#xA;of a covering map are discrete.  (This is not part of the standard definition of covering&#xA;space, but is a consequence of it.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given this, you probably should explain in more detail what you mean by &quot;If we assume ...&quot;.&#xA;What kind of map do you actually want to consider?&lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2010-10-05T15:25:47.943" />
  <row Id="6128" PostTypeId="1" CreationDate="2010-10-05T15:32:56.283" Score="3" ViewCount="143" Body="&lt;p&gt;I am trying to understand, in as simple terms as possible:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;How to define integration for non-orientable manifolds, and&lt;/li&gt;&#xA;&lt;li&gt;why it is impossible to do so using only differential forms.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;In particular, I've seen some discussion of using &quot;densities&quot; instead of $n$-forms for integration, but am not really clear on why densities are required.  In other words, is it really impossible to define integration on nonorientable manifolds using forms alone?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am of course aware that any $n$-form must vanish somewhere on a nonorientable manifold, so we cannot find a volume form, hence cannot use the standard definition of integration.  I think the reason I'm not finding this answer satisfying is that it is a bit tautological: we can't define integration with respect to volume forms because there are no volume forms. But why must we define integration with respect to a (global) volume form in the first place? Is there really no other way to do it using locally-defined forms? Thinking of a manifold as a collection of local charts is common in geometry, and I'm having trouble understanding why this approach doesn't work in the case of integration.&lt;/p&gt;&#xA;" OwnerUserId="2330" LastEditorUserId="1543" LastEditorDisplayName="" LastEditDate="2011-04-21T22:11:30.083" LastActivityDate="2011-04-21T23:11:01.047" Title="Is the notion of density really needed to define integration on nonorientable manifolds?" Tags="&lt;calculus&gt;&lt;differential-geometry&gt;&lt;differential-topology&gt;" AnswerCount="2" CommentCount="18" FavoriteCount="1" />
  <row Id="6129" PostTypeId="2" ParentId="6080" CreationDate="2010-10-05T15:45:10.037" Score="4" ViewCount="" Body="&lt;p&gt;$RP^{2k-1}$ is a quotient of a codimension 1 submanifold with induced orientation ($S^{2k-1}$)in $R^{2k}$, by an orientation-preserving transformation of $R^{2k}$.  Because there is a consistent notion of normal vector (i.e., of &quot;inside&quot; and &quot;outside&quot;) on the submanifold, preserving orientation of the surrounding manifold also preserves orientation on the submanifold.  Transporting the orientation along a path between two points of the submanifold that are identified in the quotient can be viewed as transporting the orientation in the surrounding manifold, and this has to be consistent (have positive determinant on the local frames) because of the orientability of the surrounding manifold.  &lt;/p&gt;&#xA;" OwnerUserId="467" LastActivityDate="2010-10-05T15:45:10.037" />
  <row Id="6132" PostTypeId="2" ParentId="6083" CreationDate="2010-10-05T16:17:52.473" Score="1" ViewCount="" Body="&lt;p&gt;First you need a definition of $\theta$ (and/or $d \theta$) to compare with the form involving $dx$ and $dy$.    In coordinates any definition is only &quot;local&quot; and limited to regions of $(x,y)$ that exclude loops around $0$ (which would modify $\theta$ by $2\pi$).  In practice any local definition will be as a function of $x/y$ or of $y/x$ according to which line or half-line from $0$ is excluded as a &quot;branch cut&quot;.  e.g., $\theta = \arctan(y/x)$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $\theta$ is defined using trigonometric functions one then needs a definition of those functions not using angles, or the whole exercise is logically vacuous, defining $\theta$ in terms of $\theta$ (using polar coordinates is an example of this).  An angle-free definition is possible but it is ad hoc if there is no clear relation to geometry, such as defining inverse sine or tangent functions as integrals.  A more &quot;correct&quot; definition is to use arc length on the projection of $(x,y)$ to the unit circle.  Here if $u(x,y)=(x,y)/\sqrt{x^2+y^2}$ is the projection one would have $d\theta = |u(x+dx,y+dy) - u(x,y)|$ up to second order terms in $dx$ and $dy$.&lt;/p&gt;&#xA;" OwnerUserId="467" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-10-05T16:23:06.303" LastActivityDate="2010-10-05T16:23:06.303" />
  <row Id="6133" PostTypeId="2" ParentId="566" CreationDate="2010-10-05T16:35:27.453" Score="1" ViewCount="" Body="&lt;p&gt;From New Scientist some years ago:  20 teams play a round robin tournament, each gets 1 point for a win, 0 for a loss, and there are no ties.  Each team's score is a square number.  How many upsets occurred? An upset is defined as team A defeating team B where B scored more total points than A.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-10-05T16:35:27.453" CommunityOwnedDate="2010-10-05T16:35:27.453" CommentCount="1" />
  <row Id="6134" PostTypeId="2" ParentId="2273" CreationDate="2010-10-05T17:25:21.780" Score="0" ViewCount="" Body="&lt;p&gt;I recommend the book &quot;How to prove it&quot; by Daniel Velleman. The associated (free) package for learning formal aspects of proof, called Proof Designer, gives the user some practice in learning how complicated finding such paths can be if one goes completely formal! &lt;/p&gt;&#xA;" OwnerUserId="941" LastActivityDate="2010-10-05T17:25:21.780" />
  <row Id="6135" PostTypeId="2" ParentId="6101" CreationDate="2010-10-05T17:45:48.213" Score="3" ViewCount="" Body="&lt;ul&gt;&#xA;&lt;li&gt;Formal Undecidability and related topics (Tarski truth definability, Godel incompleteness, The hydra problem, Matiyasevich's negative solution of the 10th problem)&lt;/li&gt;&#xA;&lt;li&gt;Proof Theory (Gentzen's consistency proof, Girard's linear logic, Barendregt's Lambda Cube)&lt;/li&gt;&#xA;&lt;li&gt;The Continuum (Cantor's diagonalization, Cohen and Godel's independence proof, Robinson's analysis)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" LastActivityDate="2010-10-05T17:45:48.213" CommunityOwnedDate="2010-10-05T17:45:48.213" CommentCount="6" />
  <row Id="6136" PostTypeId="1" CreationDate="2010-10-05T17:56:27.483" Score="3" ViewCount="211" Body="&lt;p&gt;Any curve can be considered to be made of sufficiently small straight lines. What is the name of the theorem which states this fact?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I asked the same question at &lt;a href=&quot;http://news.ycombinator.com/item?id=1760209&quot; rel=&quot;nofollow&quot;&gt;Hacker News&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;[Edit: in the Hacker News question, the OP included the following link, which I think clarifies his/her intent:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://musr.physics.ubc.ca/~jess/hr/skept/Math/node10.html&quot; rel=&quot;nofollow&quot;&gt;http://musr.physics.ubc.ca/~jess/hr/skept/Math/node10.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;--PLC.]&lt;/p&gt;&#xA;" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2010-10-05T21:52:42.087" LastActivityDate="2010-10-06T01:35:18.803" Title="Is there a mathematical theorem that states the equivalence of curved and straight lines" Tags="&lt;calculus&gt;" AnswerCount="3" CommentCount="6" />
  <row Id="6137" PostTypeId="2" ParentId="6136" CreationDate="2010-10-05T18:37:04.533" Score="1" ViewCount="" Body="&lt;p&gt;Maybe what you need is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Stone%E2%80%93Weierstrass_theorem&quot; rel=&quot;nofollow&quot;&gt;Stone-Weierstrass Theorem&lt;/a&gt;, which you can use to show (roughly) that a continuous function on a compact set can be approximated arbitrarily well by continuous piecewise-linear functions. The boolean ring version of the theorem is directly applicable here.&lt;/p&gt;&#xA;" OwnerUserId="1195" LastActivityDate="2010-10-05T18:37:04.533" CommentCount="10" />
  <row Id="6138" PostTypeId="2" ParentId="566" CreationDate="2010-10-05T18:40:30.243" Score="0" ViewCount="" Body="&lt;p&gt;A personal favorite (albeit not ideally-worded): Jamie has a windowbox where he plants a row of iris flowers.  He plants in just two colors - blue and yellow - but never plants two yellow irises next to each other (the result is just too garish).  Assuming that he wants to keep the same number of flowers in his box every day, how many flowers will he need if he wants a different arrangement of blue and yellow every day for a year?&lt;/p&gt;&#xA;" OwnerUserId="785" LastActivityDate="2010-10-05T18:40:30.243" CommunityOwnedDate="2010-10-05T18:40:30.243" />
  <row Id="6139" PostTypeId="2" ParentId="6136" CreationDate="2010-10-05T19:03:37.177" Score="5" ViewCount="" Body="&lt;p&gt;For many calculations that one might want to perform on a &quot;reasonable&quot; curve (e.g., one coordinatized by functions that are differentiable, or piecewise differentiable/smooth/analytic/nice), taking a finely spaced mesh of points on the curve and computing the result on the polygon joining those points, instead of the curve, will produce an answer close to the one for the curve.  The smaller the spacing, the closer the result will be to the result for the curve.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Quantities approximable in this way include area enclosed by a curve, arc length, integrals of given functions along the curve, winding number around a point, splines, parametrizations, and others.  Quantities not approximable in this way include curvature, which will be zero on the sides of any polygon used as a substitute for the curve, and integer-valued &quot;global&quot; quantities such as number of tangent lines.&lt;/p&gt;&#xA;" OwnerUserId="467" LastActivityDate="2010-10-05T19:03:37.177" />
  <row Id="6140" PostTypeId="1" AcceptedAnswerId="6144" CreationDate="2010-10-05T19:05:12.547" Score="4" ViewCount="358" Body="&lt;p&gt;Allow me to preface this by saying this is not a homework problem. If I had had this thought four years ago when I was taking calculus, I probably could do it...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm trying to calculate the limit as  $n \to \infty$ of $1-\left(\frac{n-1}{n}\right)^n$ -&#xA;- it's a constant I'm inclined to call &quot;natural chance of success&quot;. I have estimated this value to be ~0.632121 but would very much like to see how it could be calculated aside from the brute-force method i employed earlier.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The background of this would be... consider n = 2 (a coin). You are given two flips of the coin to get what you pick - what is the chance you'll get your chosen outcome, assuming of course it's a fair coin. The best way to go about this would be to say there's a 1/2 chance of failing, and you have 2 flips. This means you have $(1/2)^2$ chance of failure, being 1/4. 1-1/4 is 3/4, so your chance of success here is 3/4.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now consider n = 6 (standard die). You get six rolls to get the number of your choice (assuming a fair die again). Again, you have a 5/6 chance to not get your choice, and 6 rolls at 5/6 chance would be $(5/6)^6$, or ~.334, giving you a ~.665 chance of success.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And I'm curious as n increases to infinity, what is your chance of success? Now again, I've estimated this with a double precision float (in Java) to be 0.63212 (actually, this was the point at which it could simply gain no more precision on the value, n = 296536) but this doesn't really give insight to the derivation of the number, merely its value.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I'm hoping someone a little fresher on their integrals than I can help me out here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="2333" LastEditorUserId="664" LastEditorDisplayName="" LastEditDate="2010-10-06T00:18:41.290" LastActivityDate="2010-10-06T02:06:35.027" Title="Help with a specific limit $\left( \dfrac{n-1}{n} \right)^n$ as $n \rightarrow \infty$" Tags="&lt;probability&gt;&lt;integral&gt;" AnswerCount="4" CommentCount="5" />
  <row Id="6141" PostTypeId="1" AcceptedAnswerId="6147" CreationDate="2010-10-05T19:05:23.197" Score="0" ViewCount="57" Body="&lt;p&gt;Hmmm... this question might be a bit low-brow. I'm no mathematician. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's say that I have an unordered sequence of fuzzy values $T = (t_1 \ldots t_n)$, $T \in (0,1]^n$. Each value is a positive indication towards some observation (hence, the values are not probabilities in that a value of 0.1 does not indicate a value of 0.9 against, just that we have a weak positive affirmation of the observation).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To take an example, let's say that the observation is &quot;whether or not there has been a hurricane in the US today&quot;: we have evidence from a weather forecast two days ago saying that there would be one today (0.1), there's been a high level of traffic on twitter with the keyword hurricane (0.2) and there's been breaking news reports about a hurricane in Florida (0.7).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm trying to formalise a means of aggregating these readings into a final score (the absolute value of the score makes no difference, but the relative value compared to other aggregations does). For what I'm doing, I want each fuzzy value to reduce the cumulative &quot;uncertainty&quot; about the observation. I also need to set an external $max\in(0,1]$ value on the aggregation result:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$F_0(T,max) = 0$;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$F_k(T,max) = (max - F_{k-1})t_k + F_{k-1}$; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$F(T,max) = F_n(T,max)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Taking the hurricane example, let $max = 1$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$F_0 = 0$;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$F_1 = (1 - 0) * 0.1 + 0$;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$F_2 = (1 - 0.1) * 0.2 + 0.1 = 0.28$;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$F_3 = (1 - 0.28) * 0.7 + 0.28 = 0.784$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This aggregation is nice (for my scenario) because it rewards high fuzzy values better than &#xA;multiple medium or low values, it's order indepedent, and the answer is in $(0,max]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm having a lot of trouble formalising this fairly simple idea... don't have the necessary know-how or lingo (probably clear by now). So I have the following questions:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Has anyone seen a similar such aggregation somewhere?&lt;/li&gt;&#xA;&lt;li&gt;How should I call this sequence of values, where the order is not important, and duplicates should be maintained? An unordered sequence? A set with duplicates? An unordered n-tuple?&lt;/li&gt;&#xA;&lt;li&gt;How should I formulate this function? Should I formulate it for an unordered pair first, and then for an unordered tuple? Is there a neat way of demonstrating commutativity?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Any help or nudges in the right direction would be greatly appreciated.&lt;/p&gt;&#xA;" OwnerUserId="2331" LastActivityDate="2010-10-05T20:33:41.057" Title="Aggregating multiple fuzzy values for a given observation" Tags="&lt;set-theory&gt;&lt;functions&gt;" AnswerCount="1" />
  <row Id="6142" PostTypeId="1" AcceptedAnswerId="6184" CreationDate="2010-10-05T19:16:53.577" Score="1" ViewCount="75" Body="&lt;p&gt;Let $P$ be some partial order.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We say that $x$ and $y$ are &lt;em&gt;compatible&lt;/em&gt; if $\exists r\in P (r\le x \wedge r\le y)$, we denote this by $x \perp y$. Otherwise, we say that $x$ and $y$ are &lt;em&gt;incompatible&lt;/em&gt;.&lt;br&gt;&#xA;An order $P$ is called &lt;em&gt;separative&lt;/em&gt; if whenever $\neg (x \le y)$ then there exists some $r\le x$ which incompatible with $y$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In Jech's &quot;Set Theory&quot; he proves a lemma (14.11) which construct a separative quotient, $Q$, and a mapping $h$ for a (general) partially ordered set, $P$, that satisfies the following:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$x\le y \Rightarrow h(x) \preceq h(y)$&lt;/li&gt;&#xA;&lt;li&gt;$x \perp y \iff h(x) \perp h(y)$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;And $h$ is onto $Q$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The construction is by taking a quotient of $P$ over the equivalence relation $x \sim y \iff \forall z(z \perp x \leftrightarrow z \perp y)$, and $[x] \preceq [y] \iff \forall z \le x(z \perp y)$. (And obviously enough, $h(x) = [x]$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Today I was trying to prove the following lemma (which does not appear on Jech's book):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $P$ be some partial order, and $Q$ its separative quotient. If $[x] \preceq [y]$ then there exist $x' \in [x], y' \in [y]$ such that $x' \le y'$.&lt;br&gt;&#xA;(That is to say that the separative mapping is &quot;almost&quot; order-preserving.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Intuitively it seems right, but it might very well be a not-common misbelief. Any hints, partial proofs or complete proofs (granted that I haven't proved it myself yet - in which case I'll rush to the nearest computer and update) will be most welcomed.&lt;/p&gt;&#xA;" OwnerUserId="622" LastEditorUserId="622" LastEditorDisplayName="" LastEditDate="2010-10-05T19:46:33.620" LastActivityDate="2010-10-06T07:53:22.927" Title="Separative Quotients, and the Induced Order" Tags="&lt;set-theory&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="6143" PostTypeId="2" ParentId="6140" CreationDate="2010-10-05T19:26:30.593" Score="3" ViewCount="" Body="&lt;p&gt;Your question is motivated by probability, but in itself it's pure and elemental calculus:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\lim_{x\to \infty} (1 - \frac{1}{x})^x  = e ^{-1}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From that, your number is 0.63212...&lt;/p&gt;&#xA;" OwnerUserId="312" LastActivityDate="2010-10-05T19:26:30.593" />
  <row Id="6144" PostTypeId="2" ParentId="6140" CreationDate="2010-10-05T19:29:15.100" Score="10" ViewCount="" Body="&lt;p&gt;You are essentially trying to find the limit of $\left(1-\frac{1}{n}\right)^n$ as $n\to\infty$. Rewrite this as $e^{n\ln(1 - \frac{1}{n})}$. Since the exponential function is continuous, you have&#xA;$$\lim_{n\to\infty} e^{n\ln(1-\frac{1}{n})} = e^{\lim_{n\to\infty}(n\ln(1-\frac{1}{n})}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To compute that limit, rewrite and use L'Hopital's:&#xA;$$\begin{array}{rcl}&#xA;\lim_{n\to\infty}n\ln\left(1 - \frac{1}{n}\right) &amp;amp; = &amp;amp; \lim_{n\to\infty}\frac{\ln(1-\frac{1}{n})}{\frac{1}{n}}\\&#xA;&amp;amp; = &amp;amp; \lim_{n\to\infty} \frac{\left(\frac{1}{1-\frac{1}{n}}\right)\left(1-\frac{1}{n}\right)'}{(n^{-1})'}\\&#xA;&amp;amp; = &amp;amp; \lim_{n\to\infty}\frac{\quad\frac{n^{-2}}{1-\frac{1}{n}}\quad}{-n^{-2}}\\&#xA;&amp;amp; = &amp;amp; \lim_{n\to\infty}\frac{-1}{1-\frac{1}{n}} = -1.&#xA;\end{array}$$&#xA;So, since you wanted $1$ minus this limit, your limit equals $1-e^{-1}$. &lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-10-05T19:29:15.100" CommentCount="4" />
  <row Id="6145" PostTypeId="1" AcceptedAnswerId="6154" CreationDate="2010-10-05T20:10:40.073" Score="1" ViewCount="113" Body="&lt;p&gt;Suppose we have an atom at every point with integer coordinates in $\mathbb{R}^d$. Take a ($d-1$)-dimensional hyperplane going through $\mathbf{0}$ and orthogonal to $(1,1,1,\ldots)$. What is the name of the lattice formed by atoms in that plane?&lt;/p&gt;&#xA;" OwnerUserId="998" LastActivityDate="2010-10-05T21:57:46.430" Title="What is the name of this lattice?" Tags="&lt;geometry&gt;&lt;probability-theory&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="6146" PostTypeId="2" ParentId="6101" CreationDate="2010-10-05T20:31:42.037" Score="2" ViewCount="" Body="&lt;p&gt;Paul &lt;strong&gt;Benacerraf's identification problem&lt;/strong&gt; is an argument against Platonism: Many different models satisfy Peano's arithmetic axioms. Which of these are really the Natural numbers? I.e., the uniquely intended set? We are unable to pick one account to the exclusion of all the others. Adding axioms won't help. Any such axiomatic system will be satisfied by multiple models. For example, is the number Three given by $3 = \{ \emptyset, \{ \emptyset \}, \{ \emptyset, \{ \emptyset \} \} \}$ or by $3 = \{ \{ \{ \emptyset \} \} \}$? Benacerraf concludes that numbers are not sets, as most mathematicians believe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A recent trend for dealing with Benacerraf's identification problem is &lt;strong&gt;fictionalism&lt;/strong&gt;, a form of structuralism: &quot;Fictionalism holds that mathematical theories are like fiction stories such as fairy tales and novels. Mathematical theories describe fictional entities, in the same way that literary fiction describes fictional characters.&quot; See &lt;a href=&quot;http://plato.stanford.edu/entries/philosophy-mathematics/#Fic&quot; rel=&quot;nofollow&quot;&gt;http://plato.stanford.edu/entries/philosophy-mathematics/#Fic&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fictionalists include Hartry Fields, John Burgess, and Charles Chihara.&lt;/p&gt;&#xA;" LastActivityDate="2010-10-05T20:31:42.037" CommunityOwnedDate="2010-10-05T20:31:42.037" CommentCount="2" />
  <row Id="6147" PostTypeId="2" ParentId="6141" CreationDate="2010-10-05T20:33:41.057" Score="0" ViewCount="" Body="&lt;p&gt;I'm at least acquainted with the idea of fuzzy logic, as distinct from probabilities. Having said that: while you are not looking for a probabilistic sort of formula, you have stumbled upon one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Taking &lt;em&gt;max&lt;/em&gt;&amp;nbsp;=&amp;nbsp;1, your formula can be finessed to the form&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;F&lt;sub&gt;k+1&lt;/sub&gt;&lt;br&gt;&amp;nbsp;&amp;nbsp;=&amp;nbsp;(1&amp;nbsp;&amp;minus;&amp;nbsp;F&lt;sub&gt;k&lt;/sub&gt;)&amp;thinsp;t&lt;sub&gt;k&lt;/sub&gt;&amp;nbsp;+&amp;nbsp;F&lt;sub&gt;k&lt;/sub&gt;&lt;br&gt;&amp;nbsp;&amp;nbsp;=&amp;nbsp;F&lt;sub&gt;k&lt;/sub&gt;&amp;nbsp;+&amp;nbsp;t&lt;sub&gt;k&lt;/sub&gt;&amp;nbsp;&amp;minus;&amp;nbsp;F&lt;sub&gt;k&lt;/sub&gt;t&lt;sub&gt;k&lt;/sub&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;which is the same thing as the &quot;probability&quot; of at least one of event A or event B happening, where A and B are independent random events, occurring with probability F&lt;sub&gt;k&lt;/sub&gt; and t&lt;sub&gt;k&lt;/sub&gt; respectively. &amp;mdash; This is why your construction is order independent (and in fact can be proven so using this observation).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To say the same thing in different terms: if you take the &quot;fuzzy AND&quot; (logical conjunction) of two qualities with fuzzy strengths &lt;em&gt;x&lt;/em&gt; and &lt;em&gt;y&lt;/em&gt; to be the product &lt;em&gt;xy&lt;/em&gt;, then the formula &lt;em&gt;x&lt;/em&gt;&amp;nbsp;+&amp;nbsp;&lt;em&gt;y&lt;/em&gt;&amp;nbsp;&amp;minus;&amp;nbsp;&lt;em&gt;xy&lt;/em&gt; is the corresponding &quot;fuzzy OR&quot; which will satisfy de Morgan's laws, &lt;em&gt;i.e.&lt;/em&gt; NOT(&lt;em&gt;x&lt;/em&gt;&amp;nbsp;AND&amp;nbsp;&lt;em&gt;y&lt;/em&gt;) = (NOT&amp;nbsp;&lt;em&gt;x&lt;/em&gt;)&amp;nbsp;OR&amp;nbsp;(NOT&amp;nbsp;&lt;em&gt;y&lt;/em&gt;), if you take NOT&amp;nbsp;&lt;em&gt;x&lt;/em&gt;&amp;nbsp;=&amp;nbsp;(1&amp;nbsp;&amp;minus;&amp;nbsp;&lt;em&gt;x&lt;/em&gt;). Of course, you may not want this if you don't want a small positive fuzzy value to represent a large fuzzy value for its negation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems to me that whatever you choose &lt;em&gt;max&lt;/em&gt; to be, you can re-scale it to get identical behaviour to the case &lt;em&gt;max&lt;/em&gt;&amp;nbsp;=&amp;nbsp;1. So the utility of this interpretation is unlikely to hinge on that particular parameter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For your other questions:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;A collection in which number is important, but order is not, is a &lt;em&gt;bag&lt;/em&gt; or &lt;em&gt;multi-set&lt;/em&gt; (these are synonyms). Furthermore, that's all that it is; they're basically &quot;set-like collections&quot; which are equivalent to a histogram.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I would call the function &quot;fuzzy OR&quot; or &quot;fuzzy disjunction&quot;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If you want to make the fuzzy OR of a collection (a &quot;bag&quot;) well-defined, it suffices to show commutativity (which makes it well-defined for bags of size at most two), and then show that it is associative &amp;mdash; that is, that OR(&lt;em&gt;x&lt;/em&gt;,&amp;nbsp;OR(&lt;em&gt;y&lt;/em&gt;,&lt;em&gt;z&lt;/em&gt;))&amp;nbsp;=&amp;nbsp;OR(OR(&lt;em&gt;x&lt;/em&gt;,&lt;em&gt;y&lt;/em&gt;),&amp;nbsp;&lt;em&gt;z&lt;/em&gt;) &amp;mdash; which ensures that if it is well-defined for bags of size &lt;em&gt;k&lt;/em&gt;, it is also well-defined for bags of size &lt;em&gt;k&lt;/em&gt;&amp;nbsp;+&amp;nbsp;1.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="439" LastActivityDate="2010-10-05T20:33:41.057" CommentCount="3" />
  <row Id="6148" PostTypeId="1" AcceptedAnswerId="6149" CreationDate="2010-10-05T20:44:21.713" Score="4" ViewCount="134" Body="&lt;p&gt;There's a theorem that every planar graph can be colored with 4 colors in such a way that no 2 adjacent vertices have the same color. Is the opposite true as well?&lt;/p&gt;&#xA;" OwnerUserId="2337" LastActivityDate="2010-10-05T20:46:37.347" Title="If a graph can be colored with max 4 colors, is it planar?" Tags="&lt;graph-theory&gt;" AnswerCount="1" />
  <row Id="6149" PostTypeId="2" ParentId="6148" CreationDate="2010-10-05T20:46:37.347" Score="14" ViewCount="" Body="&lt;p&gt;No. Consider $K_{3,3}$, the graph with two sets of 3 vertices each such that every vertex in one set is connected to every vertex in the other. It's not planar but can be colored with just 2 colors. More generally, take any dense bipartite graph - it's still 2-colorable, but far from planar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A picture of $K_{3,3}$ (along with $K_5$):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/fowUR.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="308" LastActivityDate="2010-10-05T20:46:37.347" CommentCount="2" />
  <row Id="6150" PostTypeId="2" ParentId="6140" CreationDate="2010-10-05T20:47:37.300" Score="7" ViewCount="" Body="&lt;p&gt;If you know one of the definitions for the number $e$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;e=\lim_{n\to\pm\infty}\Big(1+\frac{1}{n}\Big)^n,&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;then you could calculate your limit in the following way:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\lim_{n\to\infty}\Big(1-\frac{1}{n}\Big)^n =&#xA;\lim_{n\to\infty}\left[\Big(1+\frac{1}{-n}\Big)^{-n}\right]^{-1} =&#xA;\frac{1}{e}.&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="354" LastActivityDate="2010-10-05T20:47:37.300" />
  <row Id="6151" PostTypeId="2" ParentId="6101" CreationDate="2010-10-05T21:07:35.083" Score="1" ViewCount="" Body="&lt;p&gt;George Lakoff and Rafael Núñez started studying &lt;i&gt;embodied mathematics&lt;/i&gt; pretty recently.  The idea is that our ideas of mathematics are inextricable from our humanity as opposed to being platonic truths, and can all be understood in terms of metaphors for real-world concepts and our learning/acquisition of these metaphors.  To take a simple example, whenever we add numbers, our brains will always essentially be adding things to a pile, because that's what addition is to us.  If I've understood it correctly.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Their theories really haven't caught on among mathematicians, in part because they're not mathematicians themselves, which is honestly a pretty big weakness among philosophers of math.  In my opinion, a lot of what they say is really silly (my brain has nothing to do with $\mathbb{R}$ being the unique completion of $\mathbb{Q}$ as well as the largest Archimedean field), but there is a kernel of truth there: we don't study math by taking logical step after logical step, we study it by thinking about things we have ways of comprehending.  For instance, we live in three-dimensional space, and so it's hard to study higher-dimensional things that we therefore can't visualize.  I think it would be a useful philosophical project to understand how the mathematicians who study such objects conceptualize them, and how &lt;i&gt;that&lt;/i&gt; relates to our human nature, but that would obviously require a larger mathematical background.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another important question is the origin of mathematical taste and beauty.  This is probably more important to mathematicians than to outsiders who don't have as much of a sense of it.  (Yet I'm convinced that a cool proof can be appreciated by anyone, if it's explained properly!)  I don't know who, if anyone, has written about this, but it definitely exists and is worthy of study.&lt;/p&gt;&#xA;" OwnerUserId="2215" LastActivityDate="2010-10-05T21:07:35.083" CommunityOwnedDate="2010-10-05T21:07:35.083" />
  <row Id="6152" PostTypeId="1" AcceptedAnswerId="6158" CreationDate="2010-10-05T21:46:15.077" Score="0" ViewCount="153" Body="&lt;p&gt;Suppose we have a linear ODE of the form $\dfrac{dy}{dt}= Ay$. We know that $y= e^{tA}  y(0)$ is a solution to the ODE, where $e^{tA}$ represents the matrix exponential. Suppose we also knew that this is stable in the sense of Lyapunov (i.e for any $\varepsilon &gt;0$, there exists a  $\delta &gt; 0$  such that if $\vert y(0) \vert &amp;lt; \delta$, then $\vert y(t) \vert &amp;lt; \varepsilon$ for all time $t &gt; 0$). How does that affect the norm of the matrix $e^{tA}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;OK, so the actual problem is as follows. If a linear ODE $y'= Ay$ is Lyapunov stable and if $B= G^{-1}AG$ where $G$ is an invertible matrix, is $B$ also Lyapunov stable? &lt;/p&gt;&#xA;" OwnerUserId="1127" LastEditorUserId="664" LastEditorDisplayName="" LastEditDate="2010-10-06T00:15:41.610" LastActivityDate="2010-10-06T00:15:41.610" Title="Matrix exponential and Stability of Linear ODE" Tags="&lt;differential-equations&gt;" AnswerCount="1" />
  <row Id="6153" PostTypeId="1" CreationDate="2010-10-05T21:55:34.610" Score="1" ViewCount="197" Body="&lt;p&gt;Hey, sorry for the non so mathematical question but that's the only place where I'd thought someone would answer this so here we go.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm doing a chemistry homework and I have to answer this very question:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If 50 grams of a particular mineral containing copper nitrate (II) gave, after appropriate treatment 1.6933 grams of pure copper. For this ore, determine the percentage content by weight of copper and the percentage content by weight of copper nitrate (II) in ore.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Could someone help with this please?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks, any help will be very much appreciated :)&lt;/p&gt;&#xA;" OwnerUserId="2338" LastEditorUserId="16" LastEditorDisplayName="" LastEditDate="2010-10-05T22:48:48.817" LastActivityDate="2010-10-05T23:58:20.343" Title="Chemistry: Gravimetric determination of copper in ore" Tags="&lt;homework&gt;&lt;chemistry&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="6154" PostTypeId="2" ParentId="6145" CreationDate="2010-10-05T21:57:46.430" Score="4" ViewCount="" Body="&lt;p&gt;It's called $A_{d-1}$.  See &lt;a href=&quot;http://en.wikipedia.org/wiki/Root_system#An&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Root_system#An&lt;/a&gt; .&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-10-05T21:57:46.430" CommentCount="4" />
  <row Id="6155" PostTypeId="1" AcceptedAnswerId="6156" CreationDate="2010-10-05T22:25:59.347" Score="2" ViewCount="320" Body="&lt;p&gt;Hi. I'm having a little trouble with the following problem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Calculate $F'(x)$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$F(x)=\int_{1}^{x^{2}}(t-\sin^{2}t) dt$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It says we have to use substitution but I don't see why the answer can't just be:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x-\sin^{2}x$&lt;/p&gt;&#xA;" OwnerUserId="1958" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-10-05T22:43:07.457" LastActivityDate="2010-12-01T21:17:49.570" Title="Derivative of Integral" Tags="&lt;calculus&gt;&lt;integral&gt;" AnswerCount="4" CommentCount="1" FavoriteCount="1" />
  <row Id="6156" PostTypeId="2" ParentId="6155" CreationDate="2010-10-05T22:42:17.653" Score="11" ViewCount="" Body="&lt;p&gt;If you had $G(x)=\int_{1}^{x}(t-\sin^{2}t) dt$ (note that it's just $x$ at the upper limit of the integral), then $G'(x)=x-\sin^2x$.  But, you have $F(x)=\int_{1}^{x^{2}}(t-\sin^{2}t) dt=G(x^2)$, so $F'(x)=G'(x^2)\cdot2x=(x^2-\sin^2x^2)2x$ (using the chain rule).&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-10-05T22:42:17.653" CommentCount="1" />
  <row Id="6157" PostTypeId="2" ParentId="6155" CreationDate="2010-10-05T22:44:27.940" Score="1" ViewCount="" Body="&lt;p&gt;Because the upper limit of the integral is x^2, not x.  &lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-10-05T22:44:27.940" />
  <row Id="6158" PostTypeId="2" ParentId="6152" CreationDate="2010-10-05T22:45:36.090" Score="2" ViewCount="" Body="&lt;p&gt;&quot;OK, so the actual problem is as follows. If a linear ODE $y'= Ay$ is Lyapunov stable and if $B= G^{-1}AG$ where $G$ is an invertible matrix, is $B$ also Lyapunov stable?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is, is the system $z'=Bz$ Lyapunov stable around the origin? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the two systems are related by a change of basis. If $y=Gz$, then the system $y'= Ay$ can be written as $(Gz)' = A(Gz)$, that is, $z'=(G^{-1}AG)z = Bz$. And we know that $|y| = |Gz| \leq ||G||. |z|$, where $||G|| := sup_{|x|=1}|G.x|$ is the norm of the matrix $G$ (prove it!). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $G$ is invertible, $z=G^{-1}y$, an analogous argument leads to $|z| \leq ||G^{-1}||.|y|$. I think these inequalities can help you solve the problem.&lt;/p&gt;&#xA;" OwnerUserId="1178" LastActivityDate="2010-10-05T22:45:36.090" />
  <row Id="6159" PostTypeId="2" ParentId="5877" CreationDate="2010-10-05T22:55:01.363" Score="2" ViewCount="" Body="&lt;p&gt;Below is an algorithm to compute a representation $\rm\ p = a^2 + b^2\ $ for an integer $\rm\ p = 1\ \: (mod\ 4)\:.\ \ $ The algorithm is coded in $\ $ &lt;a href=&quot;http://en.wikipedia.org/wiki/Macsyma&quot; rel=&quot;nofollow&quot;&gt;$\rm Macsyma$&lt;/a&gt; $\ $ but it can easily be translated into almost any another language. The only Macsyma specific subroutines that you might need to implement in another language are exponentiation $\rm (mod\ p)$ (by repeated squaring) and the floor function in $\rm \mathbb Q[\:i\:]\:$. While this algorithm is reasonably efficient, the point is not to optimize efficiency but, rather, to emphasize the simplicity of the solution employing the Euclidean algorithm in $\rm\ \mathbb Z[\:i\:]\:.\ $ For the underlying theory see my &lt;a href=&quot;http://math.stackexchange.com/questions/5895&quot;&gt;companion post&lt;/a&gt;. For a faster algorithm see e.g. Collins, &lt;a href=&quot;http://www.csie.nuk.edu.tw/~cychen/gcd/A%20Fast%20Euclidean%20Algorithm%20for%20Gaussian%20Integers.pdf&quot; rel=&quot;nofollow&quot;&gt;A fast Euclidean algorithm for Gaussian integers&lt;/a&gt;, and see also Agarwal and Frandsen, &lt;a href=&quot;http://www.springerlink.com/content/j4up5k6v8n7ujh6d/&quot; rel=&quot;nofollow&quot;&gt;Binary GCD Like Algorithms for Some Complex Quadratic Rings.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;/* $\:$ For $\rm\ p = 1\ \: (mod\ 4)\ $ compute $\rm\ \alpha = a + b\:i\ $ with $\rm \ p = \alpha\:\alpha' = a^2 + b^2\ \ $*/  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\text {psplit(p) := block([}\ \:\text{j : 0,}\ \ \text{e : floor(p/4),}\ \ \text{algebraic : true ], }$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\text{block([ modulus : p ],}\quad\quad\quad\quad\ $ /* search for $\rm\ j = \sqrt{-1}\ \ (mod\ p)\ $ */&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\quad\ \text{for}\ \: \text{ a : 2}\ \ \text{while}\ \ \text{j^2 # -1}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\quad\quad\quad\quad\quad\quad\quad\quad\ \ \text{do}\ \: \text{ j : rat(a)^e ),}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad \text{x : p,}\ \ \: \text{y : j - %i,} \quad\quad\quad\quad\quad\quad\quad $ /* find $\rm\  gcd(p,\:j - i)\ \in\ \mathbb Z[\:i\:]\ $ by Euclidean algorithm */&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad \text{do ( if}\ \text{  y = 0}\ \ \text{then}\ \text{ return(x), }$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\quad\quad\ \ \text{j : rat(x - y * floor(rat(x/y))), }$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\quad\quad\ \ \text{x : y,}\ \text{ y : j )) }$&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-10-06T13:33:13.530" LastActivityDate="2010-10-06T13:33:13.530" />
  <row Id="6160" PostTypeId="2" ParentId="6090" CreationDate="2010-10-05T22:57:30.667" Score="7" ViewCount="" Body="&lt;p&gt;The term in analytic number theory is &quot;zero-free regions&quot;.  Any proof of the prime number theorem will produce such a region, and the region is equivalent to the error term in bounds for $\pi(x)$ and to the lower bounds in nonvanishing theorems $|\zeta (1+it)|&gt; 0$. At present, the known zero-free regions are asymptotic to the line $Re(s)=1$: at height $h$ all zeros are at least at distance $d(h)$ from the line with $\lim_{|h| \to \infty} d(h) = 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Added: on the subject of later improvements, if any region not asymptotic to $Re(s)=1$ were demonstrated it would be a giant advance in number theory, comparable to Wiles' proof of the modularity conjecture and Fermat's Last Theorem. In the analogous function field case there is an algebraic technique for boosting Beta &amp;lt; 1 to Beta &amp;lt; (1/2)+epsilon, and the latter is the Riemann hypothesis.  You can be sure that if a zero free region were proven that had finite distance from the boundary of the critical strip, we would all have heard about it.  Some known bounds are listed at &lt;a href=&quot;http://www.math.uiuc.edu/~ford/wwwpapers/zeros.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.math.uiuc.edu/~ford/wwwpapers/zeros.pdf&lt;/a&gt; )&lt;/p&gt;&#xA;" OwnerUserId="467" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-10-08T19:09:29.930" LastActivityDate="2010-10-08T19:09:29.930" CommentCount="1" />
  <row Id="6161" PostTypeId="1" CreationDate="2010-10-05T23:19:19.427" Score="1" ViewCount="188" Body="&lt;p&gt;&lt;B&gt;Ahem. I can only now laugh at how blind I am to not see such a simple answer, LMAO. Thanks to Isaac ♦, Paul VanKoughnett, and Bill Dubuque! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Case closed!&lt;/b&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How would you factor  $6x^2 + 18x - 60$  over a set of complex numbers?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I did it myself, I had  $6(x - 1.5 + 3.25i)(x - 1.5 - 3.25i)$ .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not 100% sure about my answer, so I am asking anyone who knows how to do this to help me confirm the answer! (:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;- EDIT -&lt;/b&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, first thing that I did was to factor out 6.&#xA;$6(x^2 + 3x - 10)$ , then I changed the expression into a perfect square, so : &#xA;$6(x^2 - 3x + 2.25) - 12.25$ . &lt;/p&gt;&#xA;&#xA;&lt;p&gt;After that:  $6(x - 1.5)^2 + (-12.25)$ .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then I took the square roots of that, so I got:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$6(1 - 1.5)^2 + 3.5i)$  .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus my answer:  $6(x - 1.5 + 3.25i)(x - 1.5 - 3.25i)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And I see that it is wrongggggg~&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My teacher did not go over it much; she was rushing throughout the whole lesson and her accent does not make it any easier. The textbook does not have anything on it. I looked around the index and there is nothing. -&lt;strong&gt;&lt;em&gt;_&lt;/em&gt;__&lt;em&gt;_&lt;/em&gt;_&lt;/strong&gt;-&lt;/p&gt;&#xA;" LastActivityDate="2010-10-06T00:11:51.377" Title="Factoring over a set of complex numbers ." Tags="&lt;algebra-precalculus&gt;" AnswerCount="3" />
  <row Id="6162" PostTypeId="2" ParentId="6155" CreationDate="2010-10-05T23:23:20.867" Score="3" ViewCount="" Body="&lt;p&gt;You could apply the following more general rule (&lt;a href=&quot;http://en.wikipedia.org/wiki/Leibniz_integral_rule&quot; rel=&quot;nofollow&quot;&gt;Leibnitz rule&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/Chain_rule&quot; rel=&quot;nofollow&quot;&gt;chain rule&lt;/a&gt;) -- although Isaac's answer is more straightforward for this derivative: If &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$I(x)=J(u,v,x)=\displaystyle\int_{u(x)}^{v(x)}f(t,x)dt,$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;then, under suitable conditions, we have &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$I^{\prime }(x)=\displaystyle\int_{u(x)}^{v(x)}\dfrac{\partial f(t,x)}{\partial x}dt+f(v(x),x)v^{\prime }(x)-f(u(x),x)u^{\prime }(x).$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$F(x)=I(x)=\displaystyle\int_{1}^{x^{2}}(t-\sin ^{2}t)dt,$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$u(x)=1,v(x)=x^{2},f(t,x)=(t-\sin ^{2}t),$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\dfrac{\partial f(t,x)}{\partial x}=0,v^{\prime }(x)=2x,u^{\prime }(x)=0.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$F^{\prime }(x)=I^{\prime }(x)=(x^{2}-\sin ^{2}x^{2})2x=2x^{3}-2x\sin&#xD;&#xA;^{2}x^{2}.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The conditions are: $u(x),v(x)$ are differential functions, $f(t,x)$ is a&#xA;continuous function and $\partial f/\partial x$ exists and is continuous.&lt;/p&gt;&#xA;" OwnerUserId="752" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2010-12-01T21:17:49.570" LastActivityDate="2010-12-01T21:17:49.570" />
  <row Id="6163" PostTypeId="2" ParentId="6161" CreationDate="2010-10-05T23:26:38.523" Score="0" ViewCount="" Body="&lt;p&gt;$$\begin{align}&#xA;6x^2+18x-60&amp;amp;=6(x^2+3x-10)&#xA;\\&#xA;&amp;amp;=6(x+5)(x-2)&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I expand yours:&#xA;$$6(x - 1.5 + 3.25i)(x - 1.5 - 3.25i)=6x^2-18x+76.875$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, I'm not entirely sure what went wrong.  How did you arrive at your answer?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&lt;strong&gt;edit&lt;/em&gt;&lt;/strong&gt;: Given your work, it's worth having another look.  You factored out the 6, as I did.  Then, you went to complete the square, which is reasonable, but you somehow went from $+3x$ to $-3x$ and there are some missing/misplaced parentheses:&#xA;$$\begin{align}&#xA;6(x^2+3x-10)&amp;amp;=6(x^2+3x+2.25-12.25)&#xA;\\&#xA;&amp;amp;=6((x+1.5)^2-12.25)&#xA;\\&#xA;&amp;amp;=6((x+1.5)^2-(3.5)^2)&#xA;\\&#xA;&amp;amp;=6(x+1.5+3.5)(x+1.5-3.5)&#xA;\\&#xA;&amp;amp;=6(x+5)(x-2)&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, to summarize, the ideas you tried to use would have worked, but there were one or more instances where something went wrong with +/-, and you should be careful with parentheses and the factored-out leading coefficient (the 6).&lt;/p&gt;&#xA;" OwnerUserId="72" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-10-06T00:11:51.377" LastActivityDate="2010-10-06T00:11:51.377" CommentCount="3" />
  <row Id="6164" PostTypeId="2" ParentId="6161" CreationDate="2010-10-05T23:27:32.063" Score="0" ViewCount="" Body="&lt;p&gt;Polynomials over the complex numbers still have the unique factorization property, so if a polynomial has a real factorization, that will also be its complex factorization.  This also means that you can go about factorizing in the same way you would over the reals.  The ways you already know for quadratics (quadratic formula, completing the square, etc.) still work fine, except now you keep complex answers instead of throwing them out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To check your answer, just multiply the factors back together, using complex multiplication.  In this case, your answer isn't right, since $(1.5-3.25i)(1.5+3.25i)=1.25^2+3.25^2=12.8125$.  In fact, this polynomial has real roots, which I'm sure you can find if you look at $6(x^2+3x-10)$ for a while.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(As a reminder, a quadratic has real roots if and only if its discriminant is nonnegative.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;: appears somebody else got to the answer first.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT 2&lt;/strong&gt; in light of your most recent edit: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't think you really understand completing the square.  That's actually okay, since you can do everything with the quadratic formula, but at your level in school you'll probably get tested on that specific technique, so you might as well learn it.  Also, it's how you derive the quadratic formula in the first place.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as I can see, you made three mistakes:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, you had $x^2-3x-10$ instead of $x^2+3x-10$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second, you forgot some parentheses when you completed the square.&#xA;$$6(x^2-3x-10)=6((x^2-3x+2.25)-12.25)=6(x-1.5)^2-73.5.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Third, &quot;taking the square root&quot; didn't really work out.  Here, it's better to work with the equation $f(x)=0$ than just the expression $f(x)$.  Then, you can move the $12.25$ to the other side of the equation and take the square root, giving $\pm 3.5$, not $\pm 3.5i$.  (Also, you changed $3.5i$ to $3.25i$ on the way back.)  When you add it back in, this gives you the expected answer, $6(x-5)(x+2)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That isn't the answer to the equation the first way you wrote it, but I can't tell which is the right equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I guess the moral is that you can check your work at nearly any step of these types of problems.  If you do this frequently, you can avoid little mistakes like these.  Good luck!&lt;/p&gt;&#xA;" OwnerUserId="2215" LastEditorUserId="2215" LastEditorDisplayName="" LastEditDate="2010-10-06T00:08:09.243" LastActivityDate="2010-10-06T00:08:09.243" CommentCount="2" />
  <row Id="6165" PostTypeId="2" ParentId="6161" CreationDate="2010-10-05T23:29:09.157" Score="1" ViewCount="" Body="&lt;p&gt;The answer is wrong since $\rm\ 3.25^2 \ne 10\:\ $ It appears that you made the error of computing the discriminant as $\rm\ 9^2 - 24\cdot 60\ = 39^2\ $ instead of $\rm\ 18^2 -24\cdot 60 = 42^2\:$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Simpler: apply the &lt;a href=&quot;http://en.wikipedia.org/wiki/Quadratic_equation&quot; rel=&quot;nofollow&quot;&gt;quadratic formula&lt;/a&gt; to $\rm\ p(x)/6 = x^2 + 3x -10\:$, or apply the &lt;a href=&quot;http://en.wikipedia.org/wiki/Rational_root_theorem&quot; rel=&quot;nofollow&quot;&gt;Rational Root test&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-10-05T23:51:59.420" LastActivityDate="2010-10-05T23:51:59.420" />
  <row Id="6167" PostTypeId="2" ParentId="6153" CreationDate="2010-10-05T23:58:20.343" Score="1" ViewCount="" Body="&lt;p&gt;To determine the percentage of copper (II) nitrate in the ore, you need to use a so-called &quot;gravimetric factor&quot;, which converts &quot;% Cu&quot; into &quot;% Cu(NO&lt;sub&gt;3&lt;/sub&gt;)&lt;sub&gt;2&lt;/sub&gt;&quot;. Since the two have the same amount of copper on a molar basis, multiplying the percentage Américo got with an appropriate gravimetric factor will give you the percent by weight of copper nitrate in the ore.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-10-05T23:58:20.343" CommentCount="1" />
  <row Id="6168" PostTypeId="1" AcceptedAnswerId="6169" CreationDate="2010-10-06T00:56:01.987" Score="3" ViewCount="187" Body="&lt;p&gt;I was wondering what are the general spaces that the concepts limsup and liminf can apply to?&#xA;Is complete lattice one of them? Also How about metric space?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are limsup and liminf specified with respect to? A subset? A sequence/net/filter base?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How many kinds of definitions for limsup and liminf in these various cases? Are they equivalent? If not, what are the conditions for them to be equivalent?&lt;/p&gt;&#xA;" OwnerUserId="1281" LastEditorUserId="171" LastEditorDisplayName="" LastEditDate="2010-10-06T05:55:26.863" LastActivityDate="2010-10-06T05:55:26.863" Title="Definitions for limsup and liminf" Tags="&lt;definition&gt;" AnswerCount="1" />
  <row Id="6169" PostTypeId="2" ParentId="6168" CreationDate="2010-10-06T01:01:23.167" Score="1" ViewCount="" Body="&lt;p&gt;The notion of a $\limsup$ of a filtered directed set makes sense. Namely, let $A$ be a filtered directed set and $x_\alpha, \alpha \in A$ be an $A$-indexed family in $\mathbb{R}$. Then one can define the $\limsup$ as the infimum of $\sup_{\beta &gt; \alpha} x_{\beta}$ over all $\alpha$. This makes sense for the $\liminf$ as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One needs the ordering of the range set to define the limsup, though.&lt;/p&gt;&#xA;" OwnerUserId="536" LastActivityDate="2010-10-06T01:01:23.167" />
  <row Id="6170" PostTypeId="2" ParentId="6136" CreationDate="2010-10-06T01:35:18.803" Score="3" ViewCount="" Body="&lt;p&gt;&quot;Calculus is the acknowledgment that anything but a linear function is far too complicated for us to handle.  Differentiation provides the techniques to tame functions by making them locally linear, and &lt;strong&gt;integration comprises the rules for sticking the local pieces together again&lt;/strong&gt;.&quot; -- Alf van der Poorten, &lt;em&gt;Notes on Fermat's Last Theorem&lt;/em&gt; (Lecture IX).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From this point of view, one might nominate the &lt;em&gt;Fundamental Theorem of Calculus&lt;/em&gt; (to which van der Poorten is alluding) as an answer to this question.&lt;/p&gt;&#xA;" OwnerUserId="1489" LastActivityDate="2010-10-06T01:35:18.803" CommentCount="2" />
  <row Id="6171" PostTypeId="1" CreationDate="2010-10-06T01:41:23.887" Score="2" ViewCount="128" Body="&lt;p&gt;I was under the impression that all 2-forms are the wedge (^) of two 1-forms. Is it possible to have a 2-form that you can't write as A^B with A,B 1-forms?&lt;/p&gt;&#xA;" OwnerUserId="2319" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2010-10-06T02:45:09.253" LastActivityDate="2010-10-06T02:45:09.253" Title="Can you find a 2-form not written as the wedge of two 1-forms?" Tags="&lt;exterior-algebra&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" />
  <row Id="6172" PostTypeId="2" ParentId="6171" CreationDate="2010-10-06T01:44:26.640" Score="9" ViewCount="" Body="&lt;p&gt;Yes, it is possible. (And you should find an example yourself: I will not deprive you of the joy of finding it :) )&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2010-10-06T01:44:26.640" CommentCount="11" />
  <row Id="6173" PostTypeId="2" ParentId="6140" CreationDate="2010-10-06T01:49:52.157" Score="5" ViewCount="" Body="&lt;p&gt;If you expand $\lim_{n\to\infty}\Big(1-\frac{1}{n}\Big)^n,$ by using the binomial formula, you get $1 - \frac{n}{n} +  \frac{n*(n-1)}{2*n^2} -  \frac{n*(n-1)*(n-2)}{6*n^3} + ...$&#xA;As n goes to infinity, this approaches $1 - \frac{n}{n} +  \frac{n*n}{2*n^2} -  \frac{n*n*n}{6*n^3} + ... $, which equals $1 - 1 +  \frac{1}{2} -  \frac{1}{6} + ...$ This is exactly the Taylor expansion of $e^x$ when $x=-1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Surprisingly, wolfram alpha can actually match constants to approximations, like so:&#xA;&lt;a href=&quot;http://www.wolframalpha.com/input/?i=0.632121&quot; rel=&quot;nofollow&quot;&gt;http://www.wolframalpha.com/input/?i=0.632121&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That may be a resource that you could find use for in the future.&lt;/p&gt;&#xA;" OwnerUserId="796" LastEditorUserId="796" LastEditorDisplayName="" LastEditDate="2010-10-06T02:06:35.027" LastActivityDate="2010-10-06T02:06:35.027" />
  <row Id="6174" PostTypeId="1" AcceptedAnswerId="6175" CreationDate="2010-10-06T01:52:40.920" Score="1" ViewCount="107" Body="&lt;p&gt;I'm studying for an exam an I came across a problem that I am having difficultly solving.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $\mathcal{F}$ is a family of analytic functions on the closed unit disc, $D$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\int_{D} |f|^{2} dA \le 1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for all $f \in \mathcal{F}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can I conclude that $\mathcal{F}$ is locally bounded?&lt;/p&gt;&#xA;" OwnerUserId="832" LastActivityDate="2010-10-06T02:50:04.013" Title="Locally bounded Family" Tags="&lt;complex-analysis&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="6175" PostTypeId="2" ParentId="6174" CreationDate="2010-10-06T02:50:04.013" Score="3" ViewCount="" Body="&lt;p&gt;Yes. Cauchy's formula states that $$f(z) = \frac{1}{2\pi} \int_{\theta} f(z + re^{i \theta}) d\theta$$&#xA;which gives an expression for $f(z)$ in terms of the average around a circle. &#xA;By integration it follows that &#xA;$$ f(z) = \frac{1}{r_0^2 \pi} \int_{r \leq r_0, \theta} f(z+re^{i \theta}) r dr d\theta$$&#xA;which implies that if the square integral of $f$ is bounded, then (by Cauchy-Schwarz) $f$ is locally bounded by a bound depending on the square integral of $f$ and the circle in question. From this your claim follows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically, the point is that $f$ is the average of its values in a neighborhood. Note that this implies that the space of holomorphic functions in an open set $U$ such that $\int_{U} |f|^2$ is finite is actually a Hilbert space (i.e., complete) under the usual inner product. &lt;/p&gt;&#xA;" OwnerUserId="536" LastActivityDate="2010-10-06T02:50:04.013" CommentCount="2" />
  <row Id="6176" PostTypeId="2" ParentId="6090" CreationDate="2010-10-06T03:53:13.073" Score="2" ViewCount="" Body="&lt;p&gt;I would suggest you read the book &quot;Riemann's Zeta Function&quot; by H.M.Edwards.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As to your question, this is one theorem (from the above book)&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;De La Vallee Poussin's Theorem (1899)&lt;/strong&gt; : There exist constants $c &gt; 0, K &gt; 1$ such that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \beta &amp;lt; 1 - \frac{c}{\log \gamma}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for all roots $\rho = \beta + i \gamma$, in the range $\gamma &gt; K$.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I am pretty sure there will be more such theorems (the book itself mentions that the above theorem has been improved upon). The book also mentions that (as of 1974) that the bound $\beta &amp;lt; 1$ (real part of the root) has not been improved!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope that helps.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-10-06T03:53:13.073" CommentCount="2" />
  <row Id="6177" PostTypeId="2" ParentId="6060" CreationDate="2010-10-06T03:55:20.780" Score="1" ViewCount="" Body="&lt;p&gt;&lt;em&gt;Edit: It appears an identical idea has, with far greater detail, &lt;a href=&quot;http://mathoverflow.net/questions/41132/vortex-voronoi-diagram/41225#41225&quot; rel=&quot;nofollow&quot;&gt; already been given to you by jvkersch&lt;/a&gt;. I am humbled. I should also point out that my example below, which was only meant as an illustration, would not be a steady-state solution in a physical fluid, because the interaction between the vortices themselves would cause them to move.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;David Bar Moshe's idea reminded me of some work in vector field visualization which does indeed use stagnation points to divide the fluid domain into something like &quot;regions of influence&quot;. I believe the initial paper which introduced the idea was Helman and Hesselink's &quot;Visualizing Vector Field Topology in Fluid Flows&quot; (&lt;a href=&quot;http://graphics.ethz.ch/teaching/scivis_common/Literature/HelmanHesselink91.pdf&quot; rel=&quot;nofollow&quot;&gt;PDF copy&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In our case, because we assumed that the flow is incompressible and irrotational, in a generic configuration the velocity can only be zero at saddle points, where the flow points inward along two directions and outward along two directions. Streamlines along these directions are called the &lt;em&gt;separatrices&lt;/em&gt; of the saddle point. If you place two particles close together on different sides of a separatrix and let them following the flow, they will diverge at the saddle point and follow disparate long-term trajectories. So these separatrices divide space into regions where the global topology of the streamlines is different.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's an example I cooked up in Matlab because I thought it would look pretty. There are four point vortices in a square, whose circulations are -1, -1, -2 and 1 going clockwise from top left. Here's what the direction of the velocity field looks like:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/kuHf3.png&quot; alt=&quot;quiver plot&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the diagram below, the separatrices divide space into regions around vortices (and clusters thereof). In each distinct region, the streamlines wind around a particular set of vortices. You can see the saddle points as the points where four arcs come together. I believe this sort of diagram is what you were hoping for when you asked for something like a Voronoi diagram around the vortices.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/qg78W.png&quot; alt=&quot;separatrices&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(For a complete picture, there ought to be arrows on the separatrices to indicate the direction of flow, but I couldn't figure out how to do that in Matlab.)&lt;/p&gt;&#xA;" OwnerUserId="856" LastEditorUserId="856" LastEditorDisplayName="" LastEditDate="2010-10-06T04:08:22.573" LastActivityDate="2010-10-06T04:08:22.573" />
  <row Id="6178" PostTypeId="2" ParentId="5889" CreationDate="2010-10-06T03:56:13.773" Score="0" ViewCount="" Body="&lt;p&gt;Suppose f_1:C-&gt;A, f_2:C-&gt;B and C is generated by a finite set Q. Then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;G = &amp;lt; ger(A) , ger(B) | rel(A) , rel(B) , {f_1(q)=f_2(q), q in Q} &gt; .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this presentation we can eliminate at most finitely many generators of A, namely those coming from C. Therefore if A is not finitely generated then G is not finitely generated.&lt;/p&gt;&#xA;" OwnerUserId="2342" LastActivityDate="2010-10-06T03:56:13.773" CommentCount="4" />
  <row Id="6179" PostTypeId="1" CreationDate="2010-10-06T04:09:42.187" Score="14" ViewCount="552" Body="&lt;p&gt;I'm a private tutor of High school &amp;amp; pre-engineering level mathematics. Obviously, only those people who have some trouble with maths take tuitions. Most of my students are actually good. I mean they are not dumb. Its just that they are &quot;don't like&quot;, &quot;hate&quot;, &quot;scared of&quot;, &quot;bored&quot; (these are their own words) of mathematics. They feel maths is very &lt;em&gt;abstract&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't blame them. Its their teachers and the books they follow. Most of the mathematics books just tell the mathematical concepts either bluntly or in an abstract way. I'm looking for some books which teach mathematics in &quot;fun&quot;, &quot;interesting&quot;, &quot;&lt;strong&gt;intutive&lt;/strong&gt;&quot;, &quot;&lt;strong&gt;thought provoking&lt;/strong&gt;&quot; way. Of course, as a tutor I'm doing my best to make maths simple &amp;amp; intuitive but books also help me in getting new ideas. I really like &lt;a href=&quot;http://betterexplained.com&quot; rel=&quot;nofollow&quot;&gt;http://betterexplained.com&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;These are the topics covered in my tutions&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Set theory (Sets, Relations &amp;amp; Functions)&#xA;Trigonometry&#xA;Complex Numbers&#xA;Linear Algebra&#xA;Number theory&#xA;Geometry (st.lines, triangles, circles, parabola, hyperbola, 3D: spheres, cubes, etc.)&#xA;Calculus (Limits, Differential &amp;amp; Integral calculus)&#xA;Probability (permutations &amp;amp; combinations)&#xA;Statistics&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I'm looking for any kind of resources books, papers, websites, videos, softwares, simulations or even suggestions. I need to cultivate interest at the same time make them good at mathematics.&lt;/p&gt;&#xA;" OwnerUserId="2264" LastActivityDate="2011-01-25T13:49:36.993" CommunityOwnedDate="2010-10-06T04:31:09.617" Title="Teaching mathematics in intuitive and thought provoking way" Tags="&lt;books&gt;&lt;learning&gt;" AnswerCount="5" CommentCount="4" />
  <row Id="6180" PostTypeId="2" ParentId="6179" CreationDate="2010-10-06T05:05:59.923" Score="3" ViewCount="" Body="&lt;p&gt;It depends upon the student's motivation.  When you are hired for high school, is it by the parents when the student has a problem?  And in pre-engineering is it the student?  This is a big difference.  If the student isn't motivated, I have no ideas.  If s/he is motivated, showing what math is good for, as problems in the area of interest is a good approach.  I've done a bit of this and what worked there was just to take their text and work exercises.  The one-on-one work seemed to help understanding.  For the general high school situation, again assuming motivation, there are many problem books that require creativity.  Martin Gardner and Peter Winkler are the names I would first think of, but there are many more.  Also the first problems at projecteuler.org give you a lot to work with.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-10-06T05:05:59.923" CommunityOwnedDate="2010-10-06T05:05:59.923" />
  <row Id="6181" PostTypeId="1" CreationDate="2010-10-06T05:57:19.060" Score="0" ViewCount="54" Body="&lt;p&gt;Has anyone had experience diagramming extensive form games in ditaa? I haven't quite figured out how to make diagonal lines in ditaa that scale well for a tree diagram (well, at least top-down oriented trees).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm hoping this application is what gives me an excuse to play with ditaa, but I wanted to consult others before diving in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This might belong on another stack site, but I wasn't sure.&lt;/p&gt;&#xA;" OwnerUserId="1325" LastActivityDate="2010-10-06T05:57:19.060" Title="Is it feasible to diagram extensive form games using ditaa?" Tags="&lt;game-theory&gt;" CommentCount="3" />
  <row Id="6182" PostTypeId="2" ParentId="6125" CreationDate="2010-10-06T06:22:51.217" Score="0" ViewCount="" Body="&lt;p&gt;There is a somewhat related operation, &lt;i&gt;condensation&lt;/i&gt;, in the theory of linear orderings. For example, you can take a linear ordering and identify any two elements $x,y$ which are connected by some finite chain, i.e. there are elements $x = v_1,\ldots,v_n = y$ such that $v_1 &amp;lt; \cdots &amp;lt; v_n$ and there are no elements 'in between', i.e. no $z$ such that $v_i &amp;lt; z &amp;lt; v_{i+1}$. This will compress all copies of $\mathbb{Z}$ and $\mathbb{N}$ (or its opposite) to a single point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This particular condensation can be used to (de)construct all countable linear orders, if I remember correctly. There are also some other condensations, take a look at Rosenstein's &lt;i&gt;Linear orderings&lt;/i&gt;.  &lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-10-06T06:22:51.217" />
  <row Id="6183" PostTypeId="2" ParentId="6155" CreationDate="2010-10-06T07:37:55.533" Score="0" ViewCount="" Body="&lt;p&gt;Well according to me the answer is $$F'(x) = \frac{d}{dx}(x^{2}) \cdot [x^{2}-\sin^{2}(x^{2})] - \frac{d}{dx}(1) \times \text{something} = 2x \cdot \Bigl[x^{2} -\sin^{2}(x^{2})\Bigr] - 0$$ would be the answer.&lt;/p&gt;&#xA;" LastActivityDate="2010-10-06T07:37:55.533" />
  <row Id="6184" PostTypeId="2" ParentId="6142" CreationDate="2010-10-06T07:40:16.590" Score="2" ViewCount="" Body="&lt;p&gt;My friend came up with a good counter-example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/ZSjRv.png&quot; alt=&quot;Counter example&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then it is easy to verify that $[x] \prec [y]$ but $x \not{&amp;lt;} y$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if anyone has any further insights into the lemma or this whole idea, I'd very like to hear.&lt;/p&gt;&#xA;" OwnerUserId="622" LastEditorUserId="622" LastEditorDisplayName="" LastEditDate="2010-10-06T07:53:22.927" LastActivityDate="2010-10-06T07:53:22.927" />
  <row Id="6185" PostTypeId="2" ParentId="6179" CreationDate="2010-10-06T09:36:37.373" Score="4" ViewCount="" Body="&lt;p&gt;There is no silver bullet - the hardest part of the project is getting the student curious about maths again. Once you do that, teaching them the skills to pass the exams is comparatively a breeze.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, one is &lt;em&gt;usually&lt;/em&gt; hired by parents when the student has a problem, and not before, so motivating the student is a key part of initial sessions. It helps to remember that good students who have gone off maths were once good students who enjoyed maths; you need to try and find out what they liked about maths when they enjoyed it, and when they stopped enjoying it and why.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Find out why the student is there - often students need to pass maths to do something in the future that they are interested in (to get in to college perhaps); focussing on the goal can motivate some students.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I agree with J.M. and Ross that puzzles can be an effective way to engage good students whose interest in maths has been diverted.&#xA;For example, a favourite of mine for kindling interest in trig is &lt;a href=&quot;http://rec-puzzles.org/index.php/Table%20in%20Corner&quot; rel=&quot;nofollow&quot;&gt;Table in Corner&lt;/a&gt; where elementary trig is used to solve a problem that appears to have too little information given (I drop the quarter circle condition when I state the problem - it provides an opportunity to discuss the reasonableness of the two possible solutions).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Elementary number theory is another rich source of good puzzles, and even tricks like divisibility tests can come in handy. I remember getting one student interested in practising the long division algorithm by coming up with exact multiples of horrible small primes like 13 and 17 for her to divide. She knew I couldn't possibly know the 17 times table to several thousand places, so where was I getting these numbers from? We discussed the plausibility of her suggestions and I kept her guessing for most of the session before I explained. She got the algorithm, and she got to understand essentially how it works in one go.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also remember that abstraction means different things to different people. Students often say maths is abstract because they only see it employed in textbooks, where every problem is clearly contrived, absolutely divorced from the real world, and one of a collection of standard types (puzzles usually avoid the same problem by being contrived but non-standard).  Since textbooks never give plausible real-world applications you can talk about applications often, in fact it is helpful to have a bank of non-contrived applications for most of the topics you want to teach. Books by the likes of Ian Stewart and Keith Devlin can help you here. Usually I prefer to say stuff like &quot;who cares, it's fun&quot; (and then give an application a bit later on), but then that's me - YMMV.&lt;/p&gt;&#xA;" OwnerUserId="1676" LastActivityDate="2010-10-06T09:36:37.373" CommunityOwnedDate="2010-10-06T09:36:37.373" />
  <row Id="6186" PostTypeId="2" ParentId="6099" CreationDate="2010-10-06T09:55:55.863" Score="1" ViewCount="" Body="&lt;p&gt;Here is an explicit procedure based on the isomorphism between the de-Rham and Cech cohomologies &#xA;for smooth manifolds based on R. Bott and L.W. Tu's &lt;a href=&quot;http://books.google.com/books?id=S6Ve0KXyDj8C&amp;amp;printsec=frontcover&amp;amp;dq=Differential+forms+in+algebraic+topology+By+Raoul+Bott,+Loring+W.+Tu&amp;amp;source=bl&amp;amp;ots=_kbe5djU5m&amp;amp;sig=DEVBdAp8SnxkgKWJkrWMHq-BurE&amp;amp;hl=en&amp;amp;ei=qT-sTK2uIcvHswbTwPzEBA&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=1&amp;amp;sqi=2&amp;amp;ved=0CBIQ6AEwAA#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;book&lt;/a&gt;: Differential forms in algebraic topology.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The description will be given for a three form but it can be generalized &#xA;along the same lines to forms of any degree. We suppose that the manifold has a finite good cover.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The data needed is the transition functions between the coordinate charts (which will be denoted by: $U_\alpha$, $U_\beta$, etc.) &#xA;and of course, the coordinate expression of the given form on each chart.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given a three form F, then by the Poincare lemma, on  $U_\alpha,  F = dB_\alpha$, where $B_\alpha$ are two forms on $U_\alpha$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus by the Poincare lemma on $U_\alpha \cap U_\beta$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$B_\alpha-B_\beta = dA_{\alpha\beta}$ , where $A_{\alpha\beta}$ are one forms on $U_\alpha \cap U_\beta$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since on $U_\alpha \cap U_\beta \cap U_\gamma$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$d(A_{\alpha\beta}+A_{\beta\gamma}+A_{\gamma\alpha})=0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then by the Poincare lemma&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$A_{\alpha\beta}+A_{\beta\gamma}+A_{\gamma\alpha} = d\phi_{\alpha\beta\gamma}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where: $\phi_{\alpha\beta\gamma}$ are zero forms on $U_\alpha \cap U_\beta \cap U_\gamma$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again, Since on: $U_\alpha \cap U_\beta \cap U_\gamma \cap U_\delta$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$d(\phi_{\alpha\beta\gamma}-\phi_{\beta\gamma\delta}+\phi_{\gamma\delta\alpha}-\phi_{\delta\alpha\beta})=0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\phi_{\alpha\beta\gamma}-\phi_{\beta\gamma\delta}+\phi_{\gamma\delta\alpha}-\phi_{\delta\alpha\beta} = C_{\alpha\beta\gamma\delta}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where: $C_{\alpha\beta\gamma\delta}$ are constants.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The differential form F is integral iff:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$C_{\alpha\beta\gamma\delta}= 2 \pi n_{\alpha\beta\gamma\delta}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $n_{\alpha\beta\gamma\delta}$ are integers on all quadruple intersections.&lt;/p&gt;&#xA;" OwnerUserId="1430" LastActivityDate="2010-10-06T09:55:55.863" CommentCount="1" />
  <row Id="6187" PostTypeId="1" AcceptedAnswerId="6194" CreationDate="2010-10-06T10:08:59.800" Score="3" ViewCount="102" Body="&lt;p&gt;Hi everybody!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a usual category $\mathcal{C}$ one can pull the colim out of the Hom like this: $Hom_\mathcal{C}(colim A_i,B)=lim Hom_\mathcal{C}(A_i,B)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am looking for a corresponding statement for hocolims - lets say in simplicial sets, but if there are more general statements, that's even better.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;E.g. I could imagine&#xA;$$Hom_{\mathcal{C}}(hocolim A_i,B)=lim Hom_{Ho(\mathcal{C})}(A_i,B)$$&#xA;- maybe one needs to have $B$ fibrant and the A_i cofibrant here, i.e. that the Homs on the right are $\mathbb{R}Homs$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the internal Hom in simplicial sets I could also imagine versions like this:&#xA;$$Hom(colim A_i,B)=holim Hom(A_i,B)$$&#xA;$$Hom(colim A_i,B)=holim \mathbb{R}Hom(A_i,B)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the right statement and what is the place to learn this hocolim-yoga?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&#xA;N.B.&lt;/p&gt;&#xA;" OwnerUserId="2345" LastEditorUserId="664" LastEditorDisplayName="" LastEditDate="2010-10-07T07:19:31.883" LastActivityDate="2010-10-07T07:19:31.883" Title="How does hocolim relate to Hom?" Tags="&lt;category-theory&gt;&lt;algebraic-topology&gt;&lt;homotopy-theory&gt;" AnswerCount="1" FavoriteCount="3" />
  <row Id="6188" PostTypeId="1" CreationDate="2010-10-06T10:33:50.197" Score="1" ViewCount="225" Body="&lt;p&gt;i have arc A ,i know startangle,endangle,startPoint,Endpoint,centre,radius of arc and i have point B, i like to find point B lies or not in arc A , i need formula or algorithum for this&lt;/p&gt;&#xA;" OwnerUserId="1861" LastActivityDate="2010-11-05T13:44:17.937" Title="how to find point lie on the arc" Tags="&lt;geometry&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="6189" PostTypeId="2" ParentId="6188" CreationDate="2010-10-06T10:36:29.527" Score="1" ViewCount="" Body="&lt;p&gt;Did you take a look here: &lt;a href=&quot;http://mathforum.org/library/drmath/view/54365.html&quot; rel=&quot;nofollow&quot;&gt;http://mathforum.org/library/drmath/view/54365.html&lt;/a&gt;&lt;/p&gt;&#xA;" LastActivityDate="2010-10-06T12:58:25.220" CommunityOwnedDate="2010-10-06T12:58:25.220" CommentCount="2" />
  <row Id="6190" PostTypeId="1" AcceptedAnswerId="6219" CreationDate="2010-10-06T11:51:32.463" Score="6" ViewCount="183" Body="&lt;p&gt;The card game &lt;a href=&quot;http://en.wikipedia.org/wiki/Set_%2528game%2529&quot; rel=&quot;nofollow&quot;&gt;Set&lt;/a&gt; lead to the following question. Lets call a subset $A$ of $(\mathbb{Z}/3)^n$ dependent, if there is $\{x,y,z\}\subset A$ with $x+y+z=0$. (So unlike the case of linear dependence we are not allowing any coefficients here).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $f(n)$ denote the maximal size of a independent subset of $(\mathbb{Z}/3)^n$. &#xA;Is there an explicit expression / a recursion for $f(n)$? Can anything be said about its asymptotic behaviour (like $f\in O(c^n)$ for some minimal $c$)? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As $\{0;1\}^n$ is independent, we know that $f(n)\ge 2^n$. And so $c\ge 2$.&#xA;The card game set deals with the case $n=4$ and $f(4)=20$ if i remember correctly.&lt;/p&gt;&#xA;" OwnerUserId="707" LastEditorUserId="707" LastEditorDisplayName="" LastEditDate="2010-10-06T14:33:48.323" LastActivityDate="2010-10-07T00:12:18.137" Title="A question related to the card game &quot;Set&quot;" Tags="&lt;combinatorics&gt;" AnswerCount="2" CommentCount="5" />
  <row Id="6191" PostTypeId="2" ParentId="6188" CreationDate="2010-10-06T11:59:51.823" Score="0" ViewCount="" Body="&lt;p&gt;Assuming you're talking about an arc of a circle, the first question to ask is whether the point lies on the circle.  This just means that the distance from the center of the circle to your point is equal to the radius.  Then find out the angle between the radius to your point and, say, a horizontal line, and figure out whether the point is within the bounds of the arc.&lt;/p&gt;&#xA;" OwnerUserId="2215" LastActivityDate="2010-10-06T11:59:51.823" CommentCount="1" />
  <row Id="6194" PostTypeId="2" ParentId="6187" CreationDate="2010-10-06T12:41:32.687" Score="1" ViewCount="" Body="&lt;p&gt;A formula like the one you are asking for could be the following (Bousfield-Kan, &quot;Homotopy limits, completions and localizations&quot;, chapter XII, proposition 4.1):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\mathrm{hom}_* (\mathrm{hocolim}\ \mathbf{A}, B) \cong \mathrm{holim}\ \mathrm{hom}_* (\mathbf{A}, B) \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here $B$ is a pointed simplicial set, $\mathbf{A} : I \longrightarrow \Delta^{\mathrm{o}}\mathbf{Set}_*$ a functor from a small category $I$ to the category of pointed simplicial sets, and for pointed simplicial sets $A, B$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\mathrm{hom}_* (A,B) \in \Delta^{\mathrm{o}}\mathbf{Set}_*&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is the &lt;em&gt;pointed simplicial function space&lt;/em&gt; which $n$-simplices are maps in $\Delta^{\mathrm{o}}\mathbf{Set}_* $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\left( \Delta [n] \times A \right) / \left( \Delta [n] \times * \right) \longrightarrow B&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(op.cit., chapter VIII, 4.8).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I didn't go through the details, but, if it annoys you, it seems to me that you can drop the &quot;pointed&quot; thing everywhere just deleting &quot;pointed&quot;, $*$, and $\Delta [n] \times * $ in what Bousfield-Kan say.&lt;/p&gt;&#xA;" OwnerUserId="664" LastEditorUserId="664" LastEditorDisplayName="" LastEditDate="2010-10-06T13:01:08.490" LastActivityDate="2010-10-06T13:01:08.490" CommentCount="1" />
  <row Id="6195" PostTypeId="1" CreationDate="2010-10-06T12:43:37.593" Score="15" ViewCount="657" Body="&lt;p&gt;You play a game flipping a fair coin.  You may stop after any trial, at which point you are paid in dollars &#xA;the percentage of heads flipped.  So if on the first trial you flip a head, you should stop and earn \$100&#xA;because you have 100% heads.  If you flip a tail then a head, you could either stop and earn \$50,&#xA;or continue on, hoping the ratio will exceed 1/2.  This second strategy is superior.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A paper by Medina and Zeilberger (&lt;a href=&quot;http://arxiv.org/abs/0907.0032&quot; rel=&quot;nofollow&quot;&gt;arXiv:0907.0032v2 [math.PR]&lt;/a&gt;) says that it is an unsolved&#xA;problem to determine if it is better to continue or stop after you have flipped 5 heads in 8 trials: accept \$62.50 or hope for more.  It is easy to simulate this problem and it is clear&#xA;from even limited experimental data that it is better to continue (perhaps more than 70% chance you'll improve over \$62.50):&#xA;&lt;br /&gt;&#xA;&lt;img src=&quot;http://i.imgur.com/ZVHVw.jpg&quot; alt=&quot;alt text&quot;&gt;&#xA;&lt;br /&gt;&#xA;My question is basically: Why is this difficult to prove?  Presumably it is not that difficult&#xA;to write out an expression for the expectation of exceeding 5/8 in terms of the cumulative binomial distribution.&lt;/p&gt;&#xA;" OwnerUserId="237" LastActivityDate="2010-10-07T17:58:58.357" Title="Why is this coin-flipping probability problem unsolved?" Tags="&lt;probability&gt;&lt;probability-theory&gt;" AnswerCount="3" CommentCount="10" FavoriteCount="7" />
  <row Id="6196" PostTypeId="1" AcceptedAnswerId="6198" CreationDate="2010-10-06T12:59:01.127" Score="6" ViewCount="279" Body="&lt;p&gt;Is there a reasonable statistic test one can do to standard random number generators (say, one of those that come built in in Python libs) which shows they are not really random?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(By reasonable I mean to exclude silly things like «generate a looong list and see that it is periodic»)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Later:&lt;/strong&gt; Notice I am looking for a test that &lt;em&gt;fails&lt;/em&gt; on standard generators.&lt;/p&gt;&#xA;" OwnerUserId="274" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-10-06T17:29:45.927" LastActivityDate="2010-10-06T17:29:45.927" Title="The pseudoness of pseudorandom number generators" Tags="&lt;statistics&gt;&lt;random&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="6197" PostTypeId="2" ParentId="6196" CreationDate="2010-10-06T13:06:36.777" Score="2" ViewCount="" Body="&lt;p&gt;Searches for Marsaglia (George), works by him or about his PRNG tests, would uncover a lot of information.  e.g.,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Diehard_tests&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Diehard_tests&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;was found by looking for &quot;marsaglia test&quot;.&lt;/p&gt;&#xA;" OwnerUserId="467" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-10-06T13:18:05.897" LastActivityDate="2010-10-06T13:18:05.897" CommentCount="2" />
  <row Id="6198" PostTypeId="2" ParentId="6196" CreationDate="2010-10-06T14:09:03.790" Score="7" ViewCount="" Body="&lt;p&gt;The standard reference is Knuth, The Art of Computer Programming, Volume 2, &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0201896842&quot; rel=&quot;nofollow&quot;&gt;Seminumerical&#xA;Algorithms&lt;/a&gt;. Anyone with serious interest in this topic should read Knuth's extensive exposition here. In addition to theoretical discussion, he presents tests including frequency, serial, gap, poker, coupon collector's, permutation, run, maximum-of-t,&#xA;collision, birthday spacings, and serial correlation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Marsaglia's &lt;a href=&quot;http://en.wikipedia.org/wiki/Diehard_tests&quot; rel=&quot;nofollow&quot;&gt;DIEHARD&lt;/a&gt; suite of statistical test includes birthday spacings, overlapping permutations, ranks of 31x31 and 32x32 matrices, ranks of 6x8 matrices, monkey tests on 20-bit Words,monkey tests OPSO, OQSO, DNA, count the 1's in a stream of bytes, count the 1's in specific bytes, parking lot, minimum distance, random spheres, squeeze, overlapping&#xA;sums, runs, and craps.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://www.itl.nist.gov/div898/pubs/ar/ar2000/node9.html&quot; rel=&quot;nofollow&quot;&gt;NIST Statistical Test Suite&lt;/a&gt; includes frequency, block frequency, cumulative sums, runs, long runs,Marsaglia's rank, spectral (based on the Discrete Fourier Transform), nonoverlapping template matchings, overlapping template matchings, Maurer's universal statistical, approximate entropy (based on the work of Pincus, Singer and Kalman), random&#xA;excursions (due to Baron and Rukhin), Lempel-Ziv complexity, linear complexity, and serial.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for tests that fail, as Knuth mentions when discussing a typical generator in section 3.6 p. 188 &quot;Caution: The numbers generated by ran_array fail the birthday spacings test of Section 3.3.2J, and they have other deficiencies that sometimes show up in high-resolution simulations (see exercises 3.3.2-31 and 3.3.2-35)&quot;.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-10-06T14:46:04.913" LastActivityDate="2010-10-06T14:46:04.913" />
  <row Id="6199" PostTypeId="1" AcceptedAnswerId="6592" CreationDate="2010-10-06T14:17:51.730" Score="4" ViewCount="306" Body="&lt;p&gt;I have been stumped for long by this exercise (3.12(d)) from Stokey and Lucas's &lt;em&gt;Recursive Methods in Economic Dynamics&lt;/em&gt;. Would greatly appreciate any hints.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $\phi: X \to Y$ and $\psi: X \to Y$  be lower hemicontinuous correspondences (set-valued functions), and suppose that for all $x \in X$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Gamma(x)=\{y \in Y: y \in \phi(x) \cap \psi(x)\}\neq \emptyset$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Show that if $\phi$ and $\psi$ are both convex valued, and if $\mathrm{int} \phi(x) \cap \mathrm{int} \psi(x) \neq \emptyset$, then $\Gamma(x)$ is lower hemicontinuous at $x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[A correspondence $\Gamma: X \to Y$ is said to be &lt;em&gt;lower hemicontinuous&lt;/em&gt; at $x \in X$ if $\Gamma(x)$ is nonempty and if, for every $y \in \Gamma(x)$ and every sequence $x_n \to x$, there exists $N \geq 1$ and a sequence $\{y_n\}_{n=N}^\infty$ such that $y_n \to y$ and $y_n \in \Gamma(x_n)$, all $n \geq N$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Intuitively this means that the graph of $\Gamma(x)$ cannot suddenly broaden out.]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;: We can assume that $X$ and $Y$ are subsets of $\mathbf{R}^n$.&lt;/p&gt;&#xA;" OwnerUserId="1195" LastEditorUserId="1195" LastEditorDisplayName="" LastEditDate="2010-10-08T15:12:02.357" LastActivityDate="2010-10-13T13:29:04.093" Title="Lower hemicontinuity of the intersection of lower hemicontinuous correspondences" Tags="&lt;real-analysis&gt;&lt;homework&gt;&lt;convex-analysis&gt;&lt;economics&gt;" AnswerCount="1" CommentCount="7" FavoriteCount="1" />
  <row Id="6200" PostTypeId="1" CreationDate="2010-10-06T14:56:09.460" Score="2" ViewCount="124" Body="&lt;p&gt;I've been trying to solve this problem from Elements of Applied&#xA;Bifurcation Theory, but even with the hints given I haven't found out&#xA;how to proceed yet. I would appreciate any further hints or insights&#xA;you could give me in order to make some progress.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's exercise 2.6.(2):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The following system of partial differential equations is the FitzHugh-&#xA;Nagumo caricature of the Hodgkin-Huxley equations modeling the nerve&#xA;impulse propagation along an axon:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ \frac{\partial u}{\partial t} = \frac{\partial^2 u}{\partial x^2}-f_a(u)-v$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ \frac{\partial v}{\partial t} = bu $,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $u(x,t)$ represents the membrane potential, $v=v(x,t)$ is a&#xA;&quot;recovery&quot; variable, $f_a(u)=u(u-a)(u-1)$, $1&gt;a&gt;0,b&gt;0,-\infty &amp;lt; x &amp;lt;&#xA;\infty$, and $t&gt;0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Traveling waves are solutions to these equations of the form&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$u(x,t)=U(\xi)$, $v(x,t)=V(\xi)$, $\xi = x+ct$, where $c$ is an a&#xA;priori unknown wave propagation speed. The functions $U(\xi)$,&#xA;$V(\xi)$ are the wave profiles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(a) Derive a system of three ordinary differential equations for the&#xA;profiles with &quot;time&quot; $\xi$. (Hint: introduce an extra variable:&#xA;$W=\dot{U}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(b) Check that for all $c&gt;0$ the system for the profiles (the wave&#xA;system) has a unique equilibrium with one positive eigenvalue and two&#xA;eigenvalues with negative real parts. (Hint: First, verify this&#xA;assuming that eigenvalues are real. The, show that the characteristic&#xA;equation cannot have roots on the imaginary axis, and finally, use the&#xA;continuous dependence of the eigenvalues on the parameters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(c) Conclude that the equilibrium can be either a saddle or a saddle-&#xA;focus with a one-dimensional unstable and a two-dimensional stable in-&#xA;variant manifold, and find a condition on the system parameters that de-&#xA;fines a boundary between these two cases. Plot several boundaries in the&#xA;$(a,c)$-plane for different values of $b$ and specify the region&#xA;corresponding to saddle-foci. (Hint: at the boundary the&#xA;characteristic polynomial $h(\lambda)$ has a double root&#xA;$\lambda_0:h(\lambda_0)=h'(\lambda_0)=0.$).&lt;/p&gt;&#xA;" OwnerUserId="2348" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-10-06T17:36:58.000" LastActivityDate="2010-10-06T17:59:17.500" Title="FitzHugh- Nagumo caricature of the Hodgkin-Huxley equations" Tags="&lt;differential-equations&gt;&lt;dynamical-systems&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="6203" PostTypeId="1" AcceptedAnswerId="6257" CreationDate="2010-10-06T15:38:12.927" Score="1" ViewCount="233" Body="&lt;p&gt;If I give you the following definition of the set $A$, how could you prove it is equal the set of the natural numbers without an explicit definiton for the latter?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The set $A$ is inductively defined as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i) $0 \in A$; and &lt;br/&gt;&#xA;ii) $\forall n$, a natural number, if $n \in A$, then $n+1 \in A$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can easily prove that $A$ is contained in the natural numbers, but I'm failing to see how to prove the converse without a similar definition for the natural numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for taking the time to read me.&lt;/p&gt;&#xA;" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-12-27T20:20:35.383" LastActivityDate="2010-12-27T20:20:35.383" Title="Prove that the natural numbers are present on an inductive definition of another set" Tags="&lt;homework&gt;&lt;elementary-set-theory&gt;&lt;induction&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="6204" PostTypeId="2" ParentId="6128" CreationDate="2010-10-06T15:43:21.860" Score="2" ViewCount="" Body="&lt;p&gt;On an orientable manifold, we define &lt;em&gt;integration of functions with respect to a volume form&lt;/em&gt;. On a non orientable manifold, there are no volume forms, so we have to do something else!&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2010-10-06T15:43:21.860" CommentCount="2" />
  <row Id="6205" PostTypeId="2" ParentId="6039" CreationDate="2010-10-06T15:47:50.797" Score="3" ViewCount="" Body="&lt;p&gt;$\rm\ p(x+1) = 3\:p(x)\:$ on $\rm\:\mathbb N\ \Rightarrow\ p(x+1) = 3\:p(x)\ \Rightarrow\ p(-1) = 1/3\ $ contra $\rm\ p(x)\in \mathbb Z[x]\:$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or, continuing,  $\rm\ p(-n) = 3^{-n}\: \Rightarrow \ p(x)\:p(-x) = 1\:$ on $\rm\:\mathbb N\ \Rightarrow\ p(-x)\:p(x) = 1\ $ contra degree comparison.  This proof works over much more general coefficient rings, e.g. $\mathbb Q\:$.&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-10-06T15:47:50.797" CommentCount="0" />
  <row Id="6206" PostTypeId="2" ParentId="6203" CreationDate="2010-10-06T15:55:05.287" Score="2" ViewCount="" Body="&lt;p&gt;It seems to me that your 'proof' that the set $A \subseteq \mathbb{N}$ must have a mistake because basically you're just defining what Tom Apostol calls an inductive set in his Calculus book. I mean, $A$ can just be a set of real numbers and could perfectly well satisfy your two propierties. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance you can take $A = [0, \infty[$ and then $A$ satisfies both properties, but nonetheless $A$ is not contained in $\mathbb{N}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact Apostol's book defines $\mathbb{N}$ as the set of all numbers that belong to every inductive set.&lt;/p&gt;&#xA;" OwnerUserId="900" LastActivityDate="2010-10-06T15:55:05.287" CommentCount="11" />
  <row Id="6207" PostTypeId="2" ParentId="6203" CreationDate="2010-10-06T16:11:10.790" Score="1" ViewCount="" Body="&lt;p&gt;That's not an inductive definition of $A\:$. Rather, it's an inductive proof that $\ A \supseteq\mathbb N\:$. If you know additionally that $\: A \subseteq \mathbb N\:$ then you can conclude that $\:A = \mathbb N\:$. That's standard mathematical induction.&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-10-06T16:11:10.790" CommentCount="2" />
  <row Id="6208" PostTypeId="1" AcceptedAnswerId="6210" CreationDate="2010-10-06T16:22:25.843" Score="5" ViewCount="420" Body="&lt;p&gt;Let $f$ be an analytic function on the open unit disk domain $D$.  Suppose also that $f$ is bounded.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $f$ is bounded I beleive that $f$ can be continuously extended to the closed unit disk.&#xA;I know that the zeros of $f$ in the open disk $D$ are isolated.   Are the zeros of $f$ in the closed unit disk also necessarily isolated?&lt;/p&gt;&#xA;" OwnerUserId="832" LastActivityDate="2011-03-06T00:15:18.320" Title="Isolated zeros on closure of a domain" Tags="&lt;complex-analysis&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="6209" PostTypeId="2" ParentId="566" CreationDate="2010-10-06T16:23:21.533" Score="2" ViewCount="" Body="&lt;p&gt;If you have 32 2x1 dominoes then you can cover an 8x8 board easily enough; if you throw away a domino and cut a 1x1 square from each end of a diagonal of the board, can you cover the remaining shape with the remaining dominoes?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I heard this decades ago at university.&lt;/p&gt;&#xA;" LastEditorUserId="1995" LastEditorDisplayName="" LastEditDate="2010-10-07T00:11:03.083" LastActivityDate="2010-10-07T00:11:03.083" CommunityOwnedDate="2010-10-06T16:23:21.533" CommentCount="1" />
  <row Id="6210" PostTypeId="2" ParentId="6208" CreationDate="2010-10-06T17:02:19.013" Score="10" ViewCount="" Body="&lt;p&gt;Edit (March 5, 2011): My original answer gave as justification for claims about zero sets a result of Rudin and of Carleson that not only came a half a century after Fatou's result now mentioned below, but also didn't directly address the problem.  I've edited to correct the deficiencies.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;It is not true in general that $f$ can be extended continuously to the closed disk.  The set of bounded analytic functions on the open disk is called the Hardy algebra, denoted $H^\infty$, and the set of analytic functions on the disk that have continuous extensions to the closed disk is called the disk algebra, often denoted $A(\mathbb{D})$, and $A(\mathbb D)$ is properly contained in $H^\infty$.  Elements of the Hardy algebra do have radial limits (or even nontangential limits) a.e. on the boundary, allowing the Hardy algebra to be embedded in $L^\infty$ of the circle as the subspace of functions whose negatively indexed Fourier coefficients vanish.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Elements of the Hardy algebra must be nonzero a.e. on the circle (unless they are identically zero).  In fact, this is even true for elements of $H^1$, i.e., for integrable functions on the circle whose negatively indexed Fourier coefficients vanish.  (See Theorem 17.17 of Rudin's &lt;em&gt;Real and complex analysis&lt;/em&gt; 3rd Edition for a much stronger statement.)  So for elements of $A(\mathbb D)$, the zero sets on the boundary must be closed sets of Lebesgue measure zero.  &lt;a href=&quot;http://en.wikipedia.org/wiki/Pierre_Fatou&quot; rel=&quot;nofollow&quot;&gt;Fatou&lt;/a&gt; &lt;a href=&quot;http://www.springerlink.com/content/p6710mj17w441656/&quot; rel=&quot;nofollow&quot;&gt;proved in 1906&lt;/a&gt; that for every closed subset $E$ of the boundary with Lebesgue measure zero, there is an element of $A(\mathbb D)$ whose zero set is exactly $E$.  In particular, this shows that the zeros need not be isolated.  For example, you could consider Cantor-like subsets of the circle to obtain zero sets with very many limit points.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A related theorem of &lt;a href=&quot;http://en.wikipedia.org/wiki/Walter_Rudin&quot; rel=&quot;nofollow&quot;&gt;Rudin&lt;/a&gt; says that given any closed subset $E$ of the unit circle with (one-dimensional) Lebesgue measure 0, and given any complex valued continuous function $g$ on $E$, there is an element of the disk algebra whose restriction to $E$ is $g$.  &lt;a href=&quot;http://www.jstor.org/stable/2033541&quot; rel=&quot;nofollow&quot;&gt;Here&lt;/a&gt;'s the original 1956 article of Rudin on JSTOR.  &lt;a href=&quot;http://en.wikipedia.org/wiki/Lennart_Carleson&quot; rel=&quot;nofollow&quot;&gt;Carleson&lt;/a&gt; independently proved the theorem; &lt;a href=&quot;http://www.springerlink.com/content/p4t340x045250147/&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; is a link (not open access) to his paper, published the following year.  Rudin points out in his book &lt;em&gt;Functional analysis&lt;/em&gt; on page 124 of the second edition that the result can be obtained as a corollary of a more general theorem of &lt;a href=&quot;http://en.wikipedia.org/wiki/Errett_Bishop&quot; rel=&quot;nofollow&quot;&gt;Bishop&lt;/a&gt; (Theorem 5.9 in the book) and the &lt;a href=&quot;http://en.wikipedia.org/wiki/Frigyes_Riesz&quot; rel=&quot;nofollow&quot;&gt;F.&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/Marcel_Riesz&quot; rel=&quot;nofollow&quot;&gt;M.&lt;/a&gt; Riesz &lt;a href=&quot;http://en.wikipedia.org/wiki/F._and_M._Riesz_theorem&quot; rel=&quot;nofollow&quot;&gt;theorem&lt;/a&gt;.  Bishop's result is from the 1962 paper &quot;&lt;a href=&quot;http://www.ams.org/journals/proc/1962-013-01/S0002-9939-1962-0133462-4/home.html&quot; rel=&quot;nofollow&quot;&gt;A general Rudin-Carleson theorem&lt;/a&gt;.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A good reference for everything mentioned above is Hoffman's &lt;a href=&quot;http://books.google.com/books?id=QxgdAAAACAAJ&amp;amp;dq&quot; rel=&quot;nofollow&quot;&gt;Banach spaces of analytic functions&lt;/a&gt;.  In particular, Fatou's and Rudin's theorems appear in Chapter 6, followed by a characterization of the closed ideals of $A(\mathbb D)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For some explicit examples of bounded analytic functions on the open disk with no continuous extension to the closed disk (in other words, elements of $H^\infty\setminus A(\mathbb{D})$), consider &lt;a href=&quot;http://eom.springer.de/b/b016630.htm&quot; rel=&quot;nofollow&quot;&gt;Blaschke products&lt;/a&gt; with infinitely many zeros.  If $f$ is such a function, then $f$ has boundary values of modulus 1 a.e., and since $f$ has infinitely many zeros, the set of zeros has a limit point $x$ on the circle.  If $f$ had a continuous extension to the closed disk, this would imply respectively that $f$ has modulus 1 everywhere on the circle, while at the same time $f(x)=0$, a contradiction.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2011-03-06T00:15:18.320" LastActivityDate="2011-03-06T00:15:18.320" CommentCount="2" />
  <row Id="6211" PostTypeId="2" ParentId="6179" CreationDate="2010-10-06T17:15:03.520" Score="11" ViewCount="" Body="&lt;p&gt;Provided your student has a strength, you can play to that.  In some sense that's what finding &quot;applications&quot; is about -- appealing to someone's immediate values.  But there's also a danger in that, if all you do is appeal to someone's immediate values you may be forced to build up a rather contrived and limited notion of what mathematics is. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in teaching it's also very useful to take students to foreign territory where they're weak, and build up their appreciation for what they may have found useless or &quot;abstract&quot;.  Finding applications isn't the only way to do this -- simply providing re-interpretations of things they know that seem complicated can be useful.  If something they found hard becomes easier in a different setting, this will frequently be perceived as useful. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm teaching an introductory calculus class this semester.  In the last few classes we've been differentiating trig functions. Recently students have been asking questions like &quot;can we have a formula sheet on the exams?&quot;  The bulk of the number of things to remember is starting to appear heavy.  Their concern is that things like the double-angle formula or the formula for the derivative of $\csc(x)$ are difficult to remember.   So I spent a little bit of time on how you can get by, not remembering much at all about trig functions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's the pencil-sketch of the idea that you can flesh-out: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I take it as an axiom people know what the graphs of $\sin(x)$ and $\cos(x)$ look like -- and students should be able to deduce these things from &quot;$(\cos(x),\sin(x))$ is the point on the unit circle that you get to if you lay-out a string of length $x$ along the unit circle, starting from $(1,0)$ and going in a counter-clockwise direction&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So how do you remember what $\sin'(x)$ is?  You sketch the graph of $\sin(x)$, lay down several tangent lines, and plot the slopes of the tangent lines.  Observe that the slopes vary continuously, and fill in the gaps appropriately with your knowledge of the local maxima and minimal of $\sin(x)$.  Then you observe the graph looks like $\cos(x)$, so you conclude $\sin'(x)$ must be $\cos(x)$.  This isn't a formal conclusion, more of a seat-of-the-pants &quot;well, how complicated can $\sin'(x)$ be?&quot; type argument. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, the double-angle formula for $\cos(2x)$.  How do you remember that?  The idea is that we &lt;em&gt;know&lt;/em&gt; there is a formula that relates $\cos(2x)$ to either $\cos(x)$ or $\sin(x)$ but people tend not to have the exact form on immediate recall.  I sketch the graph of $\cos(x)$, then sketch the graph of $\cos(2x)$ as the previous graph with double the frequency.  $\cos(2x)$ has twice the frequency -- you can also get twice the frequency by looking at $\cos^2(x)$ or $\sin^2(x)$ -- so spend some time sketching the graph of those functions.  This amounts to knowing the graph of $\sin(x)$ and $\cos(x)$, plus the graph of $x^2$ -- to know that $\cos^2(x)$ for example not only has twice the frequency but that when $\cos^2(x)=0$ the function looks quadratic -- like $x^2$ around those points.  From there it's just a scaling argument to say $2\cos^2(x)$ looks like $\cos(2x)+1$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the point of all this is that, with enough patience and geometric insight, all these formulas that students encounter you can cast them as things that, &lt;em&gt;by their very nature&lt;/em&gt; they must be easily re-discoverable with very little probing.  In particular, provided you're comfortable enough graphing functions it's less energy (and really far more informative) to remember the forms of these identities by intuiting them from your rough expectations.  In that way you can re-cast identities like&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sin(a+b) = \sin(a)\cos(b) + \cos(a)\sin(b)$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;in a different light.  The fact that this identity has such a simple form is somewhat remarkable -- but once you know it has a simple form, re-discovering it via non-rigorous arguments and heuristics is rather immediate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Certainly, this line of reasoning isn't great for all students as it assumes a certain familiarity and an intuition with graphing that many students may not have.  But once students have the knowledge that this is something they can think about, and that other people think this way -- that the discovery process is full of intuition and guesses before anything rigorous -- this sets them on a path. &lt;/p&gt;&#xA;" OwnerUserId="642" LastEditorUserId="642" LastEditorDisplayName="" LastEditDate="2010-10-06T17:21:13.743" LastActivityDate="2010-10-06T17:21:13.743" CommunityOwnedDate="2010-10-06T17:15:03.520" CommentCount="5" />
  <row Id="6212" PostTypeId="2" ParentId="6190" CreationDate="2010-10-06T17:41:30.893" Score="1" ViewCount="" Body="&lt;p&gt;Based on &lt;a href=&quot;http://www2.truman.edu/~erickson/presentation7.ppt&quot; rel=&quot;nofollow&quot;&gt;this presentation&lt;/a&gt; (PPT; &lt;a href=&quot;http://webcache.googleusercontent.com/search?q=cache%3a-7EZGJG-DWEJ%3awww2.truman.edu/~erickson/presentation7.ppt+proof+of+minimum+number+of+%22set+cards%22+to+guarantee+a+set&amp;amp;cd=1&amp;amp;hl=en&amp;amp;ct=clnk&amp;amp;gl=us&quot; rel=&quot;nofollow&quot;&gt;Google-based HTML conversion&lt;/a&gt;), I think a formula for your $f(n)$ is:&#xA;$$f(n)=\begin{cases}&#xA;\frac{2^{n+2}-4}{3} &amp;amp; \text{ if }n\text{ is even}&#xA;\\&#xA;\frac{2^{n+2}-5}{3} &amp;amp; \text{ if }n&gt;1\text{ is odd}&#xA;\end{cases}$$&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-10-06T17:41:30.893" />
  <row Id="6213" PostTypeId="1" AcceptedAnswerId="6215" CreationDate="2010-10-06T17:46:03.613" Score="2" ViewCount="152" Body="&lt;p&gt;So if you have a distribution &lt;code&gt;fx(x)&lt;/code&gt; and you observed x=a, how do you calculate &lt;code&gt;E{x | x=a} = integral (x*fx(x|x=a) from -inf to inf&lt;/code&gt;, or more specifically, &lt;code&gt;fx(x|x=a)&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am having trouble when using Bayes or finding &lt;code&gt;Fx(x|x=a)&lt;/code&gt; because I end up with &lt;code&gt;P(x=a)&lt;/code&gt; which is just a constant.  Is there a simple example somewhere? Searching only leads me to bivariate distributions.&lt;/p&gt;&#xA;" OwnerUserId="377" LastEditorUserId="377" LastEditorDisplayName="" LastEditDate="2010-10-06T17:51:16.300" LastActivityDate="2010-10-06T18:25:41.597" Title="How do you calculate conditional expectation for a single continuous r.v.?" Tags="&lt;probability&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="6214" PostTypeId="2" ParentId="6200" CreationDate="2010-10-06T17:59:17.500" Score="2" ViewCount="" Body="&lt;p&gt;Here's something to get you started: $u_t=c U'(\xi)$, $u_{xx}=U''(\xi)$, and $v_t=c V'(\xi)$. Plugging this into the system of PDEs gives you a system of two ODEs for $U(\xi)$ and $V(\xi)$. The first of these ODEs is second order in $U$, but you can turn it into a first order equation by replacing $U'(\xi)$ by an auxiliary function $W(\xi)$ and replacing $U''(\xi)$ by $W'(\xi)$, and then augmenting the system with the extra equation $U'=W$. Now you have a system of three first order ODEs (i.e., a dynamical system), and can proceed to part (b).&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2010-10-06T17:59:17.500" />
  <row Id="6215" PostTypeId="2" ParentId="6213" CreationDate="2010-10-06T18:04:58.960" Score="2" ViewCount="" Body="&lt;p&gt;Unless I misunderstood your problem: your conditional density is just a Dirac delta function in $x=a$:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f_x(x|x=a) =\delta(x-a)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, $E(x | x=a) = a$ &#xA;... as intuition says.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to apply the defintion, you can use a limiting procedure (because, for a continuous variable the event $X=a$ has actually zero probability), considering instead the event that $X$ takes a value around $a$, in a neighborhood of length $dx$.&#xA;But that's not necessary. You already know that the join probability -the one that goes in the numerator- is zero if $x\ne a$. Hence, as the conditional density must integrate to one, it must be a dirac delta function.  &lt;/p&gt;&#xA;" OwnerUserId="312" LastEditorUserId="312" LastEditorDisplayName="" LastEditDate="2010-10-06T18:25:41.597" LastActivityDate="2010-10-06T18:25:41.597" CommentCount="2" />
  <row Id="6216" PostTypeId="1" AcceptedAnswerId="6217" CreationDate="2010-10-06T18:10:30.797" Score="6" ViewCount="150" Body="&lt;p&gt;Hi All,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My first post here...not really a math expert, but certainly enjoy the challenge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I working writing a random string generator and would like to know how to calculate how many possible combinations there are for a particular combination.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am generating a string of 2numbers followed by 2 letters (lowercase) e.g. 12ab&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the calculation would be (breaking it down) &#xA;number combinations 10*10=100&#xA;letter combinations 26*26=676&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the number of possible combinations is 100*676=67600, but this seems a lot to me so I'm thinking I am off on my calculations!!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could someone please point me in the right direction?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thx&lt;/p&gt;&#xA;" OwnerUserId="2351" LastActivityDate="2010-10-06T18:36:17.827" Title="Calculate combinations of characters." Tags="&lt;combinatorics&gt;" AnswerCount="2" CommentCount="5" />
  <row Id="6217" PostTypeId="2" ParentId="6216" CreationDate="2010-10-06T18:29:25.730" Score="2" ViewCount="" Body="&lt;p&gt;You are right. That is the most basic/fundamental procedure for counting in combinatorics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's sometimes called the &lt;a href=&quot;http://en.wikipedia.org/wiki/Rule_of_product&quot; rel=&quot;nofollow&quot;&gt;Rule of product&lt;/a&gt;, or &lt;strong&gt;multiplication principle&lt;/strong&gt; or &lt;a href=&quot;http://people.richland.edu/james/lecture/m170/ch04-not.html&quot; rel=&quot;nofollow&quot;&gt;fundamental counting principle&lt;/a&gt;, and it can be visualized as a &lt;a href=&quot;http://www.ltcconline.net/greenl/courses/201/probdist/trees.htm&quot; rel=&quot;nofollow&quot;&gt;tree&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="312" LastEditorUserId="312" LastEditorDisplayName="" LastEditDate="2010-10-06T18:36:17.827" LastActivityDate="2010-10-06T18:36:17.827" CommentCount="1" />
  <row Id="6218" PostTypeId="2" ParentId="6216" CreationDate="2010-10-06T18:29:48.393" Score="0" ViewCount="" Body="&lt;p&gt;Some nomenclature: when you say &quot;2 number&quot;, you really mean &quot;2 &lt;strong&gt;digits&lt;/strong&gt;&quot;. Also, you need to specify if the digits can be anything or not (for example, do you allow leading zeroes?). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If each of the two digits can be anything, 0 through 9, and each of the letters can be anything, a through z, then your computation is correct. If you think about it, you can see why the number is not off: any particular pair of letters have 100 numbers that can go before them to make the string. Each particular letter going first has 26 possible &quot;second letters&quot;, and each of them has 100 possible pairs of digits to go. So there are already 2600 possible strings of the form xxxa. Another batch for xxxb, etc. They add up very quickly. &lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-10-06T18:29:48.393" />
  <row Id="6219" PostTypeId="2" ParentId="6190" CreationDate="2010-10-06T18:50:06.283" Score="2" ViewCount="" Body="&lt;p&gt;Mathematicians call these &quot;independent sets&quot; &lt;em&gt;capsets&lt;/em&gt;, and they're fairly widely studied. But surprisingly, we don't know much about their asymptotic behavior! We have a lower bound of order c^n for some c &gt; 2; and we have an upper bound of order about 3^n/n (although this is not in the least trivial to prove!) Most mathematicians seem to suspect that the upper bound can't be improved to c^n for c strictly less than 3, but no one's sure how to prove this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For more information (perhaps a little out-of-date, but I don't think much has changed) see Terry Tao's &lt;a href=&quot;http://terrytao.wordpress.com/2007/02/23/open-question-best-bounds-for-cap-sets/&quot; rel=&quot;nofollow&quot;&gt;discussion&lt;/a&gt;.&lt;/p&gt;&#xA;" LastActivityDate="2010-10-06T18:50:06.283" />
  <row Id="6220" PostTypeId="1" AcceptedAnswerId="6222" CreationDate="2010-10-06T18:55:11.880" Score="6" ViewCount="227" Body="&lt;p&gt;Can this expression be further simplified : $ \sum_{k=0}^{n-1} { n -1 \choose k } (-2)^{k} (2n - k)! $? This is the coefficient of $x^{2n}$ in the formal power series expansion of $(1-2x)^{n-1} \times \sum_{ k \geq 0} k! x^k$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Motivation: I came across this when trying to solve a problem using inclusion-exclusion principle, I am not mentioning the original problem because I am interested in this sum as an independent problem.&lt;/p&gt;&#xA;" OwnerUserId="404" LastEditorUserId="404" LastEditorDisplayName="" LastEditDate="2010-10-06T19:00:36.970" LastActivityDate="2010-10-07T02:02:29.680" Title="Can this sum be simplified?" Tags="&lt;combinatorics&gt;" AnswerCount="2" CommentCount="5" FavoriteCount="1" />
  <row Id="6221" PostTypeId="1" CreationDate="2010-10-06T19:18:09.600" Score="3" ViewCount="172" Body="&lt;p&gt;I'm currently teaching myself calculus and I'm probably trying to run before I can walk, but I've been working on this problem..&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I managed to find the correct result for:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_{0}^{\infty }(2e^{-3x}+4e^{-7x})^2dx$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;by expanding it to:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_{0}^{\infty}4e^{-6x}+16e^{-10x}+16e^{-14x}dx$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and then working from there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a better/more general approach I could have taken? I've attempted to solve it using substitution but haven't had any success...&lt;/p&gt;&#xA;" OwnerUserId="2352" LastActivityDate="2010-11-14T04:44:19.153" Title="Evaluating $\int_{0}^{\infty }(2e^{-3x}+4e^{-7x})^2dx$" Tags="&lt;calculus&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="6222" PostTypeId="2" ParentId="6220" CreationDate="2010-10-06T19:18:59.850" Score="5" ViewCount="" Body="&lt;p&gt;It's a confluent hypergeometric function $\rm\ (2n)! \phantom{f}_1 F_1(1-n;\: -2\:n;\: -2)\ $ which can be expressed in closed form in terms of modified Bessel functions of the first kind, namely&#xA;&lt;img src=&quot;http://i.imgur.com/jb2KZ.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It may well have a simpler closed form, since the various hypergeometric simplifiers in CAS are not always optimal (above is via Mma). It is not known by the OEIS Superseeker so you may wish to submit it there.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-10-06T19:27:20.237" LastActivityDate="2010-10-06T19:27:20.237" />
  <row Id="6223" PostTypeId="2" ParentId="6221" CreationDate="2010-10-06T19:37:44.793" Score="1" ViewCount="" Body="&lt;p&gt;Using substitution you can actually solve this, but generally it's more or less the same thing. Put $t=e^{-x}$ therefore you have $dx = -\frac{1}{t} \ dt$. Then your integral reduces to $$-\int\limits_{1}^{0} \Bigl[2t^{3}+4t^{7}\Bigr]^{2} \cdot \frac{dt}{t}$$&lt;/p&gt;&#xA;" LastActivityDate="2010-10-06T19:37:44.793" CommentCount="2" />
  <row Id="6224" PostTypeId="1" AcceptedAnswerId="6233" CreationDate="2010-10-06T20:31:01.510" Score="3" ViewCount="78" Body="&lt;p&gt;Theorem 2 in a &lt;a href=&quot;http://www.ams.org/journals/jams/2000-13-03/S0894-0347-00-00337-4/&quot; rel=&quot;nofollow&quot;&gt;paper&lt;/a&gt; by Borodin, Okounkov, and Olshanski  states that the discrete Bessel kernel $J(x,y,\theta)$ is given by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sqrt{\theta} \frac{J_x J_{y+1} - J_{x+1} J_y}{x-y}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $J_x = J_x(2\sqrt{\theta})$ is the Bessel function of order $x$ and argument $2\sqrt{\theta}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anybody have any ideas on how to determine the asymptotics of this kernel in the case that $x = \alpha \sqrt{n}, y = \beta \sqrt{n}$ where $\alpha,\beta$ are distinct real numbers that differ from $\pm 2$?&lt;/p&gt;&#xA;" OwnerUserId="251" LastActivityDate="2010-10-06T22:44:47.977" Title="The discrete Bessel kernel" Tags="&lt;analysis&gt;" AnswerCount="1" />
  <row Id="6225" PostTypeId="2" ParentId="4462" CreationDate="2010-10-06T21:12:40.120" Score="2" ViewCount="" Body="&lt;p&gt;I think you might find it helpful to first comprehend the essence of the matter in a slightly simpler context, e.g. see the &lt;a href=&quot;http://groups.google.com/groups?selm=y8zfxg12ckl.fsf%40nestle.csail.mit.edu&quot; rel=&quot;nofollow&quot;&gt;proof that I presented here&lt;/a&gt;. The key ideas are already there. Generally, I recommend Kaplansky, Commutative Rings, for the circle of ideas  around the generalized Nullstellensatz (Goldman, Krull, Zariski).&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-10-06T21:12:40.120" />
  <row Id="6226" PostTypeId="1" AcceptedAnswerId="6229" CreationDate="2010-10-06T21:15:28.747" Score="3" ViewCount="97" Body="&lt;p&gt;Why does the following integral converge?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;\int_{0\leq |x_1|\leq|x_2|\leq\dots\leq |x_n|\leq 1}\frac{(1-\sum_{i=1}^n\log |x_i|)^r}{|x_1x_2\dots x_n|^{1/2}}dx_1dx_2\dots dx_n&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here, $r$ is any positive real number.&lt;/p&gt;&#xA;" OwnerUserId="1101" LastActivityDate="2010-10-06T21:57:08.857" Title="Convergence of integral" Tags="&lt;calculus&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="6227" PostTypeId="1" CreationDate="2010-10-06T21:24:50.520" Score="1" ViewCount="263" Body="&lt;p&gt;$&#xA;3= \sum_{n=1}^{t} \frac{1}{1.08^n}&#xA;$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I see that it is $3 = 1.08^{-t}(12.5 \times 1.08^t{-12.5})$ (from Wolfram Alpha, but I'm not sure how to get it. I tried solving as a geometric series, I had problems and didn't get the correct answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I see that $ t\approx3.56592$, which seems like it's correct, but I have no idea where the 12.5 and all that stuff came from. Unfortunately my calculus book doesn't help much, as it is mainly focus on infinite series.&lt;/p&gt;&#xA;" LastActivityDate="2010-10-07T12:00:29.380" Title="Sum of Series with negative exponents" Tags="&lt;sequences-and-series&gt;" AnswerCount="2" CommentCount="7" FavoriteCount="1" />
  <row Id="6228" PostTypeId="1" AcceptedAnswerId="8227" CreationDate="2010-10-06T21:42:57.857" Score="0" ViewCount="149" Body="&lt;p&gt;I'd like to understand why $ \int{\rho}dV = \frac{1}{c} \int{j^0}dV = \frac{1}{c} \int{j^i}dS_i $ (the second equality), where &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$j^i = \rho \frac{dx^i}{dt} $ is the current density 4-vector&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\mathbf{j} = \rho \mathbf{v}$ is the current density 3-vector&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ j^i = (c\rho, \mathbf{j}) $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rho$ is the charge density&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$dS_i$ is the element $-dx-dy-dz+cdt$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are you able to explain me this equality?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you very much!&lt;/p&gt;&#xA;" OwnerUserId="2354" LastActivityDate="2010-10-29T01:14:36.893" Title="Why $ \int{\rho}dV = \frac{1}{c} \int{j^0}dV = \frac{1}{c} \int{j^i}dS_i $ ???" Tags="&lt;integral&gt;&lt;physics&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="6229" PostTypeId="2" ParentId="6226" CreationDate="2010-10-06T21:47:35.627" Score="4" ViewCount="" Body="&lt;p&gt;You can show (for example, using l'Hôpital's rule) that $\frac{(1-\log(t))^r}{t^{1/2}}$ is dominated by $\frac{1}{t^{2/3}}$ for sufficiently small $t$.  If you use this along with the fact that that $\sum_{i=1}^n\log|x_i|=\log|x_1\cdots x_n|$, you have your integrand dominated by $|x_1\cdots x_n|^{-2/3}$ in a neighborhood of the set where it is undefined, and bounded on the complement of this neighborhood.  Does that get you where you want to be?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, to keep things simple, you may as well just consider $0\leq x_1\leq x_2\cdots\leq x_n\leq1$, because the whole integral can be broken up into $2^n$ equal pieces depending on the various choices of sign.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2010-10-06T21:57:08.857" LastActivityDate="2010-10-06T21:57:08.857" CommentCount="2" />
  <row Id="6231" PostTypeId="2" ParentId="6227" CreationDate="2010-10-06T21:52:05.617" Score="1" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt;$\ $ Put $\rm\ a = 1.08\:$. Multiply both sides by $\rm\ a^t\ $ to get$\:$ RHS $\rm\: = \ \sum_{k=0}^{t-1}\ a^k$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for &quot;where the $12.5$ came from&quot;, $\rm\ 1/(a-1) = 1/0.08 = 100/8 = 12.5$&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-10-06T22:35:34.510" LastActivityDate="2010-10-06T22:35:34.510" />
  <row Id="6232" PostTypeId="2" ParentId="4844" CreationDate="2010-10-06T22:02:46.347" Score="1" ViewCount="" Body="&lt;p&gt;The most comprehensive webpage I have found is &lt;a href=&quot;http://www.ogre.nu/sphere.htm&quot; rel=&quot;nofollow&quot; title=&quot;Evenly distributing N points on a sphere&quot;&gt;about evenly distributing N points on a sphere&lt;/a&gt;. To be more general, this is known as the seventh of Stephen Smale's problems: the &quot;optimal&quot; distribution of points on the 2-sphere. It is still unsolved.&lt;/p&gt;&#xA;" OwnerUserId="2380" LastActivityDate="2010-10-06T22:02:46.347" />
  <row Id="6233" PostTypeId="2" ParentId="6224" CreationDate="2010-10-06T22:44:47.977" Score="2" ViewCount="" Body="&lt;p&gt;You could start with the asymptotics for $J_x$ and apply trig identities to your kernel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/8tWuG.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="247" LastActivityDate="2010-10-06T22:44:47.977" CommentCount="1" />
  <row Id="6234" PostTypeId="1" CreationDate="2010-10-06T22:47:03.750" Score="36" ViewCount="1168" Body="&lt;p&gt;One way of looking at the Vieta product&#xA;$${2\over\pi} = {\sqrt{2}\over 2}{\sqrt{2+\sqrt{2}}\over 2}{\sqrt{2+\sqrt{2+\sqrt{2}}}\over 2}\dots$$&#xA;is as the infinite product of a series of successive 'approximations' to 2, defined by $a_0 = \sqrt{2}$, $a_{i+1} = \sqrt{2+a_i}$ (or more accurately, their ratio to their limit 2).  This allows one to see that the product converges; if $|a_i-2|=\epsilon$, then $|a_{i+1}-2|\approx\epsilon/2$ and so the terms of the product go as roughly $(1+2^{-i})$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the sequence of infinite radicals $a_0=1$, $a_{i+1} = \sqrt{1+a_i}$ converges exponentially to the golden ratio $\phi$, and so the same sort of infinite product can be formed:&#xA;$$\Phi = {\sqrt{1}\over\phi}{\sqrt{1+\sqrt{1}}\over\phi}{\sqrt{1+\sqrt{1+\sqrt{1}}}\over\phi}\dots$$&#xA;and an equivalent proof of convergence goes through.  The question is, what's the value of $\Phi$?  The usual proof of Vieta's product by way of the double-angle formula for sin doesn't translate over, and from what I know of the logistic map it seems immensely unlikely that there's any function conjugate to the iteration map here in the same way that the trig functions are suitably conjugate to the version in the Vieta product.  Is there any other approach that's likely to work, or is $\Phi$ just unlikely to have any formula more explicit than its infinite product?&lt;/p&gt;&#xA;" OwnerUserId="785" LastActivityDate="2011-03-28T11:38:27.317" Title="What's the value of this Vieta-style product involving the golden ratio?" Tags="&lt;number-theory&gt;&lt;sequences-and-series&gt;" AnswerCount="1" CommentCount="7" FavoriteCount="15" />
  <row Id="6235" PostTypeId="1" AcceptedAnswerId="6236" CreationDate="2010-10-06T22:51:21.207" Score="1" ViewCount="125" Body="&lt;p&gt;and in the other way, orthogonal nullspace and range require symmetry?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;R&lt;/p&gt;&#xA;" OwnerUserId="2357" LastActivityDate="2010-10-06T23:24:27.297" Title="What is the best way to show an idempotent matrix whose nullspace and range are orthogonal must be symmetric?" Tags="&lt;linear-algebra&gt;&lt;matrices&gt;" AnswerCount="2" FavoriteCount="0" />
  <row Id="6236" PostTypeId="2" ParentId="6235" CreationDate="2010-10-06T23:13:14.167" Score="3" ViewCount="" Body="&lt;p&gt;Hint: any $v\in V$ can be written &lt;em&gt;uniquely&lt;/em&gt; as $v_1+v_2$, where $v_1$ is in the nullspace of $A$ and $v_2$ in the range space. Thus written, $Av=v_2$.&lt;/p&gt;&#xA;" OwnerUserId="1313" LastActivityDate="2010-10-06T23:13:14.167" CommentCount="3" />
  <row Id="6237" PostTypeId="2" ParentId="6235" CreationDate="2010-10-06T23:19:02.847" Score="2" ViewCount="" Body="&lt;p&gt;Let $f^2=f$. Then $f$ is self-adjoint iff $f = f^* f$ iff $\langle fx,y \rangle=\langle f^* f,y \rangle $ iff $\langle fx,y \rangle =\langle fx,fy \rangle$ iff $\langle fx,fy-y \rangle =0$ for all $x,y$ iff $ran(f)$ and $ker(f)$ are orthogonal.&lt;/p&gt;&#xA;" OwnerUserId="1650" LastEditorUserId="1650" LastEditorDisplayName="" LastEditDate="2010-10-06T23:24:27.297" LastActivityDate="2010-10-06T23:24:27.297" CommentCount="2" />
  <row Id="6238" PostTypeId="2" ParentId="6195" CreationDate="2010-10-06T23:27:29.217" Score="9" ViewCount="" Body="&lt;p&gt;I accept Qiaochu's answer &quot;Have you tried actually doing that?&quot; I did try now, and now I can appreciate the challenge. :-) The paper I cited refers to another by Chow and Robbins from 1965&#xA;that has a beautiful formulation, much clearer than the cummulative binomials with which I struggled.  Let me explain it, because it is really cool.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the natural strategy I mentioned in the comments (and echoed by Raynos),&#xA;let $f(n,h,t)$ be the expected payoff&#xA;if you start with $h$ heads and $t$ tails, and let the game continue no more than $n$ further trials.  Then there is an easy recursive formulation for $f$:&#xA;$$f(n,h,t) = \max \left( \frac{1}{2} f(n,h+1,t) + \frac{1}{2} f(n,h,t+1) , h/(h+t) \right) $$&#xA;because you have an equal chance of increasing to $h+1$ heads or to $t+1$ tails on the next&#xA;flip if you continue, and you get the current ratio if you stop.&#xA;Now, when $h+t=n$, you need to make a &quot;boundary&quot; assumption.  Assuming the law of&#xA;large numbers for large $n$ leads to the reasonable value $\max ( 1/2, h/n )$ in this case.&#xA;So now all you need to do is fill out the table for all $h$ and $t$ values up to $n=h+t$.&#xA;The real answer is the limit when &#xA;$n \rightarrow \infty$, but using large $n$ approximates this limit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After the Medina and Zeilberger paper was released, in fact just about three weeks ago,&#xA;a very careful calculation using the above recursive formulation was made by Julian Wiseman and reported on &lt;a href=&quot;http://www.jdawiseman.com/papers/easymath/chow_robbins.html&quot; rel=&quot;nofollow&quot;&gt;this web page&lt;/a&gt;.  The conclusion is to me remarkable: &quot;Choosing to continue lowers the expected payoff [from 0.625] to 0.62361957757.&quot;&#xA;This is still not a proof, but the &quot;answer&quot; is now known. &#xA;So my &quot;it is clear from even limited experimental data that&quot; was completely wrong! &#xA;I am delighted to learn from my mistake.  &lt;/p&gt;&#xA;" OwnerUserId="237" LastEditorUserId="237" LastEditorDisplayName="" LastEditDate="2010-10-06T23:33:39.617" LastActivityDate="2010-10-06T23:33:39.617" />
  <row Id="6239" PostTypeId="2" ParentId="6227" CreationDate="2010-10-06T23:29:26.880" Score="4" ViewCount="" Body="&lt;p&gt;I would solve for $t$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$S(t)=\displaystyle\sum_{n=1}^{t}\dfrac{1}{1.08^{n}}=3.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sum of a geometric progression with first term $u_{1}$ and ratio $r$ is&#xA;given $(\ast)$ by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$S(t)=u_{1}\times \dfrac{1-r^{t}}{1-r}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\dfrac{1}{1.08}\times \dfrac{1-\left( \dfrac{1}{1.08}\right) ^{t}}{1-\dfrac{1}{%&#xA;1.08}}=3$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\dfrac{1-\left( \dfrac{1}{1.08}\right) ^{t}}{0.08}=3\iff \left( \dfrac{1}{1.08}%&#xA;\right) ^{t}=0.76.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Applying logarithms, we obtain the &quot;time&quot; $t$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$t\log \left( \dfrac{1}{1.08}\right) =\log 0.76,$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$t=\dfrac{\log 0.76}{-\log 1.08}\approx 3.5659,$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which means that we need more than $3$ periods and less than $4$, at an interest rate of $8\%$, coumpounded per period, to get a total of $3$ currency units.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(\ast)$ Derivation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$S=u_{1}+u_{2}+u_{3}+\ldots +u_{t}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$rS=ru_{1}+ru_{2}+ru_{3}+\ldots +ru_{t-1}+ru_{t}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$u_{k}=ru_{k-1}=u_{1}r^{k-1}$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$S-rS=\left( u_{1}+u_{2}+u_{3}+\ldots +u_{t}\right) -\left(&#xA;ru_{1}+ru_{2}+ru_{3}+\ldots +ru_{t-1}+ru_{t}\right) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(1-r)S=u_{1}-ru_{t}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$S=\dfrac{u_{1}-ru_{t}}{1-r}=\dfrac{u_{1}-u_{1}r^{t}}{1-r}$$&lt;/p&gt;&#xA;" OwnerUserId="752" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2010-10-07T12:00:29.380" LastActivityDate="2010-10-07T12:00:29.380" />
  <row Id="6241" PostTypeId="2" ParentId="6179" CreationDate="2010-10-07T00:54:04.920" Score="2" ViewCount="" Body="&lt;p&gt;This is not a broad answer like those informative ones already posted, but just a single resource that may not be well known: the book &lt;a href=&quot;http://www.howround.com/&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;How Round is Your Circle&lt;/em&gt;&lt;/a&gt;. It emphasizes the connections between the physical and mathematics, as its subtitle indicates: &lt;em&gt;Where Engineering and Mathematics Meet&lt;/em&gt;.  It is likely too advanced in many spots for your clientele, but it has a wealth of material that can be adapted.  I've had 8th and 9th-grade children build pantograph linkages (covered in Chapter 3) and learn that vector addition is commutative in a very physical way.  You won't get that directly from this book, but this book can give you such ideas.&lt;/p&gt;&#xA;" OwnerUserId="237" LastActivityDate="2010-10-07T00:54:04.920" CommunityOwnedDate="2010-10-07T00:54:04.920" />
  <row Id="6242" PostTypeId="2" ParentId="6087" CreationDate="2010-10-07T01:05:22.640" Score="2" ViewCount="" Body="&lt;p&gt;The proof follows immediately from the  following variant of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Inequality_of_arithmetic_and_geometric_means&quot; rel=&quot;nofollow&quot;&gt;AM-GM inequality&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\quad\quad\ \ 0 &amp;lt; n &amp;lt; m \ \Rightarrow\ 3\:n+m\ &amp;lt;\ (\sqrt{n}\ \ +\: \ \sqrt{m}\ )^2\ &amp;lt;\  2\:n+2\:m  $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence $\rm\:\ m = n+1 \ \:\Rightarrow\  4\:n+1\ \ &amp;lt;\ (\sqrt{n} + \sqrt{n+1})^2\ &amp;lt;\ 4\:n+2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first inequality has an easy proof: expand the middle term, then subtract $\rm\ n+m\ $ from all the terms. Then it reduces on the left to  $\rm\ n\ &amp;lt; \sqrt{nm}\ $ via $\rm\ n^2 &amp;lt; nm\ $, and to the AM-GM inequality on the right. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Such minor variations on the $\:$ AM-GM $\:$ arise not too infrequently in practice (e.g. in competition problems).$\:$  Thus it is worthwhile to point them out in their full generality to help aid in recognizing them &quot;in the wild&quot;.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-10-07T01:44:05.467" LastActivityDate="2010-10-07T01:44:05.467" CommentCount="1" />
  <row Id="6243" PostTypeId="2" ParentId="6220" CreationDate="2010-10-07T02:02:29.680" Score="6" ViewCount="" Body="&lt;p&gt;As a way of restating Bill's answer, what you have can in fact be expressed in terms of the so-called &lt;a href=&quot;http://dlmf.nist.gov/18.34.E1&quot; rel=&quot;nofollow&quot;&gt;(generalized) Bessel polynomial&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$y_n(x;a)=(n+a-1)_n \left(\frac{x}{2}\right)^n {}_1 F_1 \left(-n;-2n-a+2;\frac{2}{x}\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(the Kummer hypergeometric series degenerates to a polynomial here because both the numerator and denominator parameters are negative integers)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your original expression, then, in terms of the (generalized) Bessel polynomial, is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(-2)^{n-1}(n+1)!y_{n-1}(-1;4)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The references in the DLMF can point you to papers where the (generalized) Bessel polynomials have been studied; for an elementary treatment, see Chihara's &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0677041500&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;An Introduction to Orthogonal Polynomials&lt;/em&gt;&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-10-07T02:02:29.680" />
  <row Id="6244" PostTypeId="1" AcceptedAnswerId="6247" CreationDate="2010-10-07T02:10:17.703" Score="11" ViewCount="1147" Body="&lt;p&gt;It would seem that one way of proving this would be to show the existence of non-algebraic numbers. Is there a simpler way to show this?&lt;/p&gt;&#xA;" OwnerUserId="647" LastActivityDate="2011-04-21T16:52:40.080" Title="Is there a quick proof as to why the vector space of $\mathbb{R}$ over $\mathbb{Q}$ is infinite-dimensional?" Tags="&lt;linear-algebra&gt;" AnswerCount="4" CommentCount="7" FavoriteCount="6" />
  <row Id="6245" PostTypeId="1" AcceptedAnswerId="6246" CreationDate="2010-10-07T02:23:42.610" Score="3" ViewCount="206" Body="&lt;p&gt;I'm currently studying basic Combinatorics for a college course and my professor is awful (and that is being generous).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore I'm looking for good resources to learn basic Combinatorics so that I can prepare for his exams and ace the course.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for any help&lt;/p&gt;&#xA;" OwnerUserId="2359" LastActivityDate="2010-10-07T02:42:36.493" Title="Good resources (book or otherwise) to learn/study basic Combinatorics" Tags="&lt;combinatorics&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="6246" PostTypeId="2" ParentId="6245" CreationDate="2010-10-07T02:42:36.493" Score="4" ViewCount="" Body="&lt;p&gt;IMO there are two great references at two different levels. The first is &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0131001191&quot; rel=&quot;nofollow&quot;&gt;Brualdi&lt;/a&gt;. This book is awesome. It is super clear, super straightfoward, and very readable by yourself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second is much more advanced, but even better! It is &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/3642072534&quot; rel=&quot;nofollow&quot;&gt;Aigner&lt;/a&gt;'s GTM on the subject. This book is fairly advanced, but rarely have I seen such a beautiful exposition of any mathematical subject. His proofs are impressively elegant, and his exercises are very interesting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope this helps.&lt;/p&gt;&#xA;" OwnerUserId="16" LastActivityDate="2010-10-07T02:42:36.493" CommentCount="2" />
  <row Id="6247" PostTypeId="2" ParentId="6244" CreationDate="2010-10-07T03:14:48.747" Score="17" ViewCount="" Body="&lt;p&gt;As Steve D. noted, a finite dimensional vector space over a countable field is necessarily countable: if $v_1,\ldots,v_n$ is a basis, then every vector in $V$ can be written uniquely as $\alpha_1 v_1+\cdots+\alpha_n v_n$ for some scalars $\alpha_1,\ldots,\alpha_n\in F$, so the cardinality of the set of all vectors is exactly $|F|^n$. If $F$ is countable, then this is countable. Since $\mathbb{R}$ is uncountable and $\mathbb{Q}$ is countable, $\mathbb{R}$ cannot be finite dimensional over $\mathbb{Q}$. (Whether it has a basis or not depends on your set theory).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your further question in the comments, whether a vector space over $\mathbb{Q}$ is finite dimensional &lt;em&gt;if and only if&lt;/em&gt; the set of vectors is countable, has a negative answer. If the vector space is finite dimensional, then it is a countable set; but there are infinite-dimensional vector spaces over $\mathbb{Q}$ that are countable as sets. The simplest example is $\mathbb{Q}[x]$, the vector space of all polynomials with coefficients in $\mathbb{Q}$, which is a countable set, and has dimension $\aleph_0$, with basis $\{1,x,x^2,\ldots,x^n,\ldots\}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added:&lt;/strong&gt; Of course, if $V$ is a vector space over $\mathbb{Q}$, then it has &lt;em&gt;countable dimension&lt;/em&gt; (finite or denumerable infinite) if and only if $V$ is countable as a set. So the counting argument in fact shows that not only is $\mathbb{R}$ infinite dimensional over $\mathbb{Q}$, but that (if you are working in an appropriate set theory) it is &lt;em&gt;uncountably&lt;/em&gt;-dimensional over $\mathbb{Q}$. &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-10-07T04:09:02.470" LastActivityDate="2010-10-07T04:09:02.470" CommentCount="4" />
  <row Id="6248" PostTypeId="2" ParentId="5119" CreationDate="2010-10-07T03:38:17.343" Score="0" ViewCount="" Body="&lt;p&gt;By continuity, the set of $\mathcal{F}_{i}$'s converge to the set $\cup_{i=1}^{\infty}\mathcal{F}_i$. The limiting set is a $\sigma$-algebra by definition of the problem. What is invalid about this argument?&lt;/p&gt;&#xA;" OwnerUserId="1527" LastActivityDate="2010-10-07T03:38:17.343" CommentCount="6" />
  <row Id="6250" PostTypeId="2" ParentId="6244" CreationDate="2010-10-07T04:01:41.873" Score="49" ViewCount="" Body="&lt;p&gt;The cardinality argument mentioned by Arturo is probably the simplest. Here is an alternative: an explicit example of an infinite $\rm\: \mathbb Q$-independent set of reals. Consider the set consisting of the logs of all primes $\rm\: p_i\:$. Then if $\rm\ \: c_1\ log\ p_1 +\:\cdots\: + c_n\ log \ p_n \: =\ 0\:,\ \: c_i\in\mathbb Q\:,\: $ multipyling by a common denominator we can assume that all $\rm\ c_i \in \mathbb Z\:,\:$ and, exponentiating, we obtain $\rm\: p_1^{\:c_1}\cdots p_n^{\:c_n} = 1\ \Rightarrow\ c_i = 0,\:$ for all $\rm\:i\:$.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-10-07T04:31:18.470" LastActivityDate="2010-10-07T04:31:18.470" CommentCount="5" />
  <row Id="6251" PostTypeId="1" AcceptedAnswerId="6252" CreationDate="2010-10-07T04:13:49.453" Score="4" ViewCount="123" Body="&lt;p&gt;Suppose $f$ is an analytic function defined on the unit disk, $D$.   I want to evaluate&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\int_{D} f(\omega) dA(\omega)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;using a change of variable.   Suppose $\phi$ is a conformal map of the $D$ onto itself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\int_{D} f(\omega) dA(\omega) = \int_{D} f(\phi(z)) |\phi^{'}(z)|^{2}  dA(z) $?,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\phi^{'}$ is the derivative of $\phi$. &lt;/p&gt;&#xA;" OwnerUserId="832" LastActivityDate="2010-10-07T04:30:05.703" Title="Change of Variable (conformal map)" Tags="&lt;complex-analysis&gt;" AnswerCount="1" FavoriteCount="2" />
  <row Id="6252" PostTypeId="2" ParentId="6251" CreationDate="2010-10-07T04:30:05.703" Score="2" ViewCount="" Body="&lt;p&gt;Because $|\phi'(z)|^2$ is the determinant of the Jacobian of $\phi$, this follows from the substitution formula for open sets in $\mathbb{R}^n$, as seen for example in &lt;a href=&quot;http://en.wikipedia.org/wiki/Integration_by_substitution#Substitution_for_multiple_variables&quot; rel=&quot;nofollow&quot;&gt;this Wikipedia article&lt;/a&gt;, which contains further references for the more general result.  Technically, to apply the result directly you would break up each side into real and imaginary parts.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-10-07T04:30:05.703" />
  <row Id="6253" PostTypeId="2" ParentId="2949" CreationDate="2010-10-07T04:47:31.343" Score="3" ViewCount="" Body="&lt;p&gt;For me it would be the &lt;a href=&quot;http://en.wikipedia.org/wiki/Green%E2%80%93Tao_theorem&quot; rel=&quot;nofollow&quot;&gt;Green-Tao&lt;/a&gt; theorem, which states: For any natural number $k$, there exist $k$-term arithmetic progressions of primes.&lt;/p&gt;&#xA;" LastActivityDate="2010-10-07T04:47:31.343" CommunityOwnedDate="2010-10-07T04:47:31.343" CommentCount="0" />
  <row Id="6254" PostTypeId="1" CreationDate="2010-10-07T04:50:55.197" Score="3" ViewCount="174" Body="&lt;p&gt;To give you some background, I have a grasp on the basics of statistics and probability theory and even remember touching Bayes theorem at the university data mining course. But being a few years away from the university made my math got &lt;em&gt;extremely&lt;/em&gt; rusty (so much for last-minute pre-exam cramming). While I remember various random basic concepts, there are a lot of gaps in my understanding of them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What would be a good material (a book, a site, or otherwise equally accessible medium) to revise the fundamentals and go beyond basics? I'd like a book that can be actually read &lt;em&gt;as a book&lt;/em&gt; (most statistics books are really dry and are close to being reference material, rather than a book).&lt;/p&gt;&#xA;" OwnerUserId="2360" LastActivityDate="2010-10-07T06:01:44.723" CommunityOwnedDate="2010-10-07T04:53:02.180" Title="Recommend a statistics fundamentals book" Tags="&lt;statistics&gt;&lt;books&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="1" />
  <row Id="6255" PostTypeId="2" ParentId="6254" CreationDate="2010-10-07T04:55:41.140" Score="1" ViewCount="" Body="&lt;p&gt;Alex i recommend you these two books:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Introduction to Probability by Sheldon Ross&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;An Introduction to Probability Theory and Its Applications: William Feller&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Both are very good books for an introductory level and the second one is a classic. It's referred by many people.&lt;/p&gt;&#xA;" LastActivityDate="2010-10-07T04:55:41.140" CommunityOwnedDate="2010-10-07T04:55:41.140" />
  <row Id="6256" PostTypeId="2" ParentId="6254" CreationDate="2010-10-07T06:01:44.723" Score="3" ViewCount="" Body="&lt;p&gt;Freedman, Pisani and Purves, &lt;em&gt;Statistics&lt;/em&gt;, followed by Freedman, &lt;em&gt;Statistical Models&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1195" LastActivityDate="2010-10-07T06:01:44.723" CommunityOwnedDate="2010-10-07T06:01:44.723" CommentCount="2" />
  <row Id="6257" PostTypeId="2" ParentId="6203" CreationDate="2010-10-07T08:00:48.810" Score="0" ViewCount="" Body="&lt;p&gt;An inductive set $A$ is any set with the following properties:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\left\{\right\}\in A$&lt;/li&gt;&#xA;&lt;li&gt;$\forall x\left(x\in A\rightarrow S\left(x\right)\in A\right)$, where $S\left(x\right)$ is the successor of $x$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;You can prove that $\mathbb{N}$ is an inductive set, and that if $A$ is any inductive set, then $\mathbb{N}\subseteq A$, but the other way around is false, that is, you can't prove that $A\subseteq\mathbb{N}$. Take for example $A=\mathbb{N}\cup\left\{a\right\}$, where $a$ is anything except a natural number. $A$ is inductive, but it is obvious that $A\subseteq\mathbb{N}$ is false.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, as far as I know, the best definition of the set of natural numbers is the following one: $\forall x\left(x\in\mathbb{N}\leftrightarrow\forall A\left(x\in A\right)\right)$, where $A$ is any inductive set.&lt;/p&gt;&#xA;" OwnerUserId="2128" LastActivityDate="2010-10-07T08:00:48.810" CommentCount="2" />
  <row Id="6258" PostTypeId="1" AcceptedAnswerId="6260" CreationDate="2010-10-07T08:16:07.097" Score="4" ViewCount="783" Body="&lt;p&gt;I've come across a paper that mentions this as a fact...where can I find the proof?&lt;/p&gt;&#xA;" OwnerUserId="998" LastActivityDate="2010-10-07T19:25:11.063" Title="matrices commute=common set of eigenvectors?" Tags="&lt;linear-algebra&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" />
  <row Id="6260" PostTypeId="2" ParentId="6258" CreationDate="2010-10-07T08:27:04.560" Score="12" ViewCount="" Body="&lt;p&gt;Suppose that $A$ and $B$ are $n\times n$ matrices, with complex entries say, that commute.&lt;br&gt;&#xA;Then we decompose $\mathbb C^n$ as a direct sum of eigenspaces of $A$, say&#xA;$\mathbb C^n = E_{\lambda_1} \oplus \cdots \oplus E_{\lambda_m}$, where $\lambda_1,\ldots,&#xA;\lambda_m$ are the eigenvalues of $A$, and $E_{\lambda_i}$ is the eigenspace for $\lambda_i$.&#xA;(Here $m \leq n$, but some eigenspaces could be of dimension bigger than one, so we needn't&#xA; have $m = n$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now one sees that since $B$ commutes with $A$, $B$ preserves each of the $E_{\lambda_i}$:&#xA;if $A v = \lambda_i v, $ then $A (B v) = (AB)v = (BA)v = B(Av) = B(\lambda_i v) = \lambda_i Bv.$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we considered $B$ restricted to each $E_{\lambda_i}$ separately, and decompose&#xA;each $E_{\lambda_i}$ into a sum of eigenspaces for $B$.   Putting all these decompositions together, we get a decomposition of $\mathbb C^n$ into a direct sum of spaces, each&#xA;of which is a simultaneous eigenspace for $A$ and $B$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;NB: I am cheating here, in that $A$ and $B$ may not be diagonalizable (and then the statement of your question is not literally true), but in this case, if you replace &quot;eigenspace&quot; by &quot;generalized eigenspace&quot;, the above argument goes through just as well.&lt;/p&gt;&#xA;" OwnerUserId="221" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-10-07T19:25:11.063" LastActivityDate="2010-10-07T19:25:11.063" CommentCount="4" />
  <row Id="6261" PostTypeId="2" ParentId="6258" CreationDate="2010-10-07T08:30:05.263" Score="8" ViewCount="" Body="&lt;p&gt;This is false in a sort of trivial way.  The identity matrix $I$ commutes with every matrix and has eigenvector set all of the underlying vector space $V$, but no non-identity matrix has this property.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is true is that two matrices which commute and are also &lt;em&gt;diagonalizable&lt;/em&gt; are &lt;a href=&quot;http://en.wikipedia.org/wiki/Diagonalizable_matrix#Simultaneous_diagonalization&quot; rel=&quot;nofollow&quot;&gt;simultaneously diagonalizable&lt;/a&gt;.  The proof is particularly simple if at least one of the two matrices has distinct eigenvalues.&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-10-07T08:43:11.853" LastActivityDate="2010-10-07T08:43:11.853" CommentCount="3" />
  <row Id="6262" PostTypeId="1" AcceptedAnswerId="6264" CreationDate="2010-10-07T09:43:08.133" Score="0" ViewCount="194" Body="&lt;p&gt;I'm trying to figure out how many ways are there to stack N cubes on top of each other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you get a particular stack of 4 cubes let's say, the order of the cubes doesn't matter, just which sides are being shown does.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm generally finding 2 numbers: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1.) 6 * 24^(n-1)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2.) 3 * 24^(n-1)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which is the correct answer, and why?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To understand the question better, you can refer to the Instant Insanity puzzle:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Instant_Insanity&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Instant_Insanity&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1630" LastEditorUserId="1630" LastEditorDisplayName="" LastEditDate="2010-10-07T10:01:23.413" LastActivityDate="2011-03-09T11:30:18.687" Title="Number of ways to stack N cubes" Tags="&lt;combinatorics&gt;" AnswerCount="3" CommentCount="4" />
  <row Id="6263" PostTypeId="1" AcceptedAnswerId="6271" CreationDate="2010-10-07T10:18:55.953" Score="3" ViewCount="247" Body="&lt;p&gt;This question is mainly to understand the meaning of my professor's correction to a proof of a theorem I gave during an oral examination.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question was to show that $\text{diam}A = \text{diam}\bar{A}$ where $\text{diam}$ is the diameter of a subspace of a metric space and $\bar{A}$ is the closure of the subset $A$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I proceded the following way:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $\bar{A} \supseteq A$, then we have $\text{diam}\bar{A}\geq \text{diam}A$. To prove the other inequality, we procede this way: $\forall x,y \in \bar{A}$ there exist $a,b \in A$ such that $d(x,a) &amp;lt; \epsilon$ and $d(y,b) &amp;lt; \epsilon$ with $\epsilon &gt; 0$. Then we know that $$d(x,y) \leq d(x,a) + d(y,b) + d(a,b) &amp;lt; 2 \epsilon + d(a,b) \leq 2 \epsilon + \text{diam}A$$ and by the arbitrarity of $\epsilon$ the inequality follows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My professor said that I coudn't do it this way and that, instead, I should do this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\forall x,y \in \bar{A}$ there exist $a_n,b_n \in A$ such that $d(x,a_n) &amp;lt; \frac{1}{n}$ and $d(y,b_n) &amp;lt; \frac{1}{n}$, then we would have $$d(x,y) \leq d(x,a_n) + d(y,b_n) + d(a_n,b_n) &amp;lt; \frac{2}{n} + d(a_n,b_n)$$, the next step would be to take the sup of it all this way: $$\sup_{x,y \in \bar{A}} d(x,y) \leq \frac{2}{n}+ \sup_{n \in \mathbb{N}} d(a_n,b_n)$$ from which the inequality follows because of the archimedean property. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is, how are these two proofs different? Why is mine wrong?&lt;/p&gt;&#xA;" OwnerUserId="1415" LastActivityDate="2010-10-23T17:55:17.097" Title="On two different proofs" Tags="&lt;proof-strategy&gt;&lt;metric-spaces&gt;" AnswerCount="1" CommentCount="8" />
  <row Id="6264" PostTypeId="2" ParentId="6262" CreationDate="2010-10-07T11:00:36.487" Score="1" ViewCount="" Body="&lt;p&gt;Your solutions presuppose that the cubes are identical; this is not the&#xA;case in Instant Insanity. If they are all distinguishable you are&#xA;missing a factor of $N!$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The smaller possible solution is valid under the assumption&#xA;that a stack is regarded as identical with that stack &#xA;produced by turning the original stack upside down. The larger&#xA;solution regards these two stacks as different.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-10-07T11:00:36.487" CommentCount="1" />
  <row Id="6265" PostTypeId="1" AcceptedAnswerId="6266" CreationDate="2010-10-07T13:27:16.890" Score="4" ViewCount="199" Body="&lt;p&gt;If $\sum a_n$ and $\sum b_n$ both converge and one of them absolutely then the Cauchy product $\sum c_n$ converges to $\sum a_n \sum b_n$. ($c_n = \sum_{k = 0}^n a_k b_{n - k}$), by Mertens Theorem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if both converge conditionally then the product does not have to converge as $a_n = b_n = (-1)^n/n$ shows.&#xA;My question now is: What if $\sum a_n$ and $\sum b_n$ both converge conditionally and $\sum c_n$ converges, then is it always true that $\sum c_n$ converges to the product?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the way, this is not homework, I'm already past the real analysis part.&lt;/p&gt;&#xA;" OwnerUserId="1120" LastActivityDate="2010-10-07T13:42:58.830" Title="Conditional convergence, Mertens theorem." Tags="&lt;real-analysis&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="6266" PostTypeId="2" ParentId="6265" CreationDate="2010-10-07T13:42:58.830" Score="5" ViewCount="" Body="&lt;p&gt;This follows readily from Abel's convergence theorem: if $\sum_0^\infty a_n$&#xA;converges then&#xA;$$\sum_0^\infty a_n=\lim_{x\to1^-}\sum_0^\infty a_n x^n.$$&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-10-07T13:42:58.830" CommentCount="1" />
  <row Id="6267" PostTypeId="2" ParentId="6195" CreationDate="2010-10-07T14:13:16.657" Score="1" ViewCount="" Body="&lt;p&gt;@joseph: on the rhs of your recursion relation for $f(n,h,t)$, shouldn't the $n$ be $n-1$?&lt;/p&gt;&#xA;" OwnerUserId="1547" LastActivityDate="2010-10-07T14:13:16.657" CommentCount="1" />
  <row Id="6268" PostTypeId="1" AcceptedAnswerId="6269" CreationDate="2010-10-07T14:23:08.193" Score="2" ViewCount="285" Body="&lt;p&gt;Suppose $f: \mathbb{R} \to \mathbb{R}$ is a continuous function which is 1-1, then does $f$ map Borel sets onto Borel sets?&lt;/p&gt;&#xA;" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-08T07:33:58.040" LastActivityDate="2010-12-08T07:33:58.040" Title="Continuous and 1-1 $\Longrightarrow$ Borel Sets $\longrightarrow$ Borel sets?" Tags="&lt;real-analysis&gt;&lt;analysis&gt;&lt;measure-theory&gt;" AnswerCount="2" FavoriteCount="2" />
  <row Id="6269" PostTypeId="2" ParentId="6268" CreationDate="2010-10-07T14:56:17.320" Score="6" ViewCount="" Body="&lt;p&gt;The continuous image of a Borel set in a Polish space is analytic. So it is Lebesgue measurable. If the complement of that analytic set is analytic then it is Borel (a theorem by Suslin). Now we know that for injective functions we have $f(A \setminus B) = f(A) \setminus f(B)$ so the complement is analytic too, hence the continuous one-to-one image of a Borel set is Borel in a Polish space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe this is unnecessarly cluttered but it is the thing that pops up in my mind.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that $f$ maps open sets to open sets. This is because injective functions $f:\mathbb{R} \to \mathbb{R}$ are strictly monotonic, so it maps open intervals to open intervals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Define $D := \{Q \in P(\mathbb{R}) : f(Q) \textrm{ is Borel}\}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now $\mathbb{R}$ is in $D$, it is closed under complements and unions (because it is injective), thus $D$ is a $\sigma$-algebra. So let $B$ be the Borel $\sigma$-algebra and $O$ the open sets of $\mathbb{R}$. Then $B \subset D$ if $O \subset D$. So we get that for all Borel sets the image is Borel.&lt;/p&gt;&#xA;" OwnerUserId="1120" LastEditorUserId="1120" LastEditorDisplayName="" LastEditDate="2010-10-07T15:31:14.710" LastActivityDate="2010-10-07T15:31:14.710" CommentCount="5" />
  <row Id="6270" PostTypeId="2" ParentId="2949" CreationDate="2010-10-07T15:16:21.143" Score="1" ViewCount="" Body="&lt;p&gt;Aside from some results I found amazing that have already been mentionned, Lagrange's Theorem in group theory is one that amazed me for some time. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For those who don't know about it, it tells us that the order of any subgroup of a group $G$ divides the order of $G$.&lt;/p&gt;&#xA;" OwnerUserId="1865" LastEditorUserId="1865" LastEditorDisplayName="" LastEditDate="2010-10-07T15:21:45.950" LastActivityDate="2010-10-07T15:21:45.950" CommunityOwnedDate="2010-10-07T15:16:21.143" />
  <row Id="6271" PostTypeId="2" ParentId="6263" CreationDate="2010-10-07T15:36:50.567" Score="7" ViewCount="" Body="&lt;p&gt;Just to elaborate on some of the comments: looking at your argument, you can see that&#xA;there is an implicit step which it might be good to make explicit:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;your last displayed equation reads&#xA;$$d(x,y) \leq \cdots \leq 2\epsilon + diam(A).$$&#xA;Now the implict step is that you want to take a sup over all pairs $x,y$ of elements&#xA;of $\overline{A}$ to conclude that&#xA;$$diam(\overline{A}) \leq 2\epsilon + diam(A).$$&#xA;Writing out this step, we see that $x$ and $y$ have disappeared, but that the quantity&#xA;$\epsilon$ persists.  And now, for this to be valid, we see that it is important to note&#xA;that $\epsilon$ can be fixed independenty of $x$ and $y$.  Luckily, this is possible: if we&#xA;fix $\epsilon &gt; 0$ in advance, then for any pair $x,y$ of points in $\overline{A}$, we &lt;em&gt;can&lt;/em&gt; find $a$ and $b$ in $A$ within distance $\epsilon$ of them, as you note.  But it is important to state that $\epsilon$ is fixed in advance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The main reason for being pedantic about this is that in other contexts, one may try to make similar arguments in which it turns out that $\epsilon$ can't be fixed in advance.  This kind of issue (whether a certain quantity (in your case, the quantity $\epsilon$) can be chosen &lt;em&gt;uniformly&lt;/em&gt;  in terms of some other quantities (in your case, $x$ and $y$) &#xA;comes up constantly in analysis, which is why people make a big deal about it in introductory courses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Final remark: the last step of your professor's argument is misstated (as was also noted).&#xA;One should add one more step to the second-to-last line, to conclude that&#xA;$$d(x,y) \leq \ldots \leq \frac{2}{n} + d(a_n,b_n) \leq \frac{2}{n} + diam(A)$$&#xA;(this is the same as the implicit step in your argument which I mentioned above),&#xA;and &lt;em&gt;then&lt;/em&gt; pass to sups, to find that&#xA;$$\sup_{x,y} d(x,y) \leq \frac{2}{n} + diam(A).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the end, the only differences between this (corrected) argument and yours are:&#xA;(a) using $1/n$ instead of $\epsilon$, which is just a matter of taste and style;&#xA;(b) making the implicit step explicit; and&#xA;(c) making it explicit that $n$ is fixed in advance of considering any particular&#xA;pair $x$ and $y$ (which is pedantic, but as I already remarked, important).&lt;/p&gt;&#xA;" OwnerUserId="221" LastEditorUserId="221" LastEditorDisplayName="" LastEditDate="2010-10-23T17:55:17.097" LastActivityDate="2010-10-23T17:55:17.097" />
  <row Id="6272" PostTypeId="1" CreationDate="2010-10-07T16:08:23.040" Score="1" ViewCount="98" Body="&lt;p&gt;I am building an android application that records accelerometer data during sleep, so as to analyze sleep trends and optionally wake the user near a desired time during light sleep. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have already built the component that collects and stores data, as well as the alarm. I still need to tackle the beast of displaying sleep data in a really meaningful way, one that preferably also lends itself to analysis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A couple of pictures say two thousand words: (I can only post one link due to low rep)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://code.google.com/p/electricsleep/downloads/list&quot; rel=&quot;nofollow&quot;&gt;Unfiltered accelerometer activity summed @ 30 second intervals with n data points AND The same data, filtered by averaging n/100 parts&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The translucent white line indicates movements that would set off the alarm, if enabled.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, neither of these are optimal solutions- the first is a little hard to understand for the average user, and the second, which is easier to read, hides a lot of what is really going on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clearly there is a better way to filter the data- I just don't have the mathematical background to know where to start looking (just look at my tags, lolz.) I'm looking for suggestions, so I can at least have some reference point to start researching a better solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-10-07T16:30:42.143" LastActivityDate="2010-10-07T16:30:42.143" Title="Meaningful way to reinterpret a sleep-movement-graph/set" Tags="&lt;statistics&gt;" CommentCount="6" FavoriteCount="0" />
  <row Id="6273" PostTypeId="1" AcceptedAnswerId="6275" CreationDate="2010-10-07T16:18:04.363" Score="3" ViewCount="589" Body="&lt;p&gt;I am not a mathematician(IANAM), however I wish I could be. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question: I often find this at the bottom of pages. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Page generated in 0.00013899803161621&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sometimes, I come across &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Page generated in 9.0122222900391E-5&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What does that time mean. &#xA;I tried searching wikipedia for E and maths but found the &lt;a href=&quot;http://en.wikipedia.org/wiki/E_%28mathematical_constant%29&quot; rel=&quot;nofollow&quot;&gt;e mathematical constant&lt;/a&gt; with a scary looking graph. My guess is E stand for Exponential or something and -5 is the power it is raised to. And the time is a really small number. But that doesn't make sense when compared to the other time in the question is 0.00013899803161621 bigger than 9.0122222900391E-5?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it means x10^-5, then 9.0122222900391E-5 will be 0.000090122222900391 which is smaller than 0.00013899803161621 &lt;/p&gt;&#xA;" OwnerUserId="2366" LastEditorUserId="2366" LastEditorDisplayName="" LastEditDate="2010-10-07T16:32:56.190" LastActivityDate="2010-10-07T16:32:56.190" Title="What doe E mean in 9.0122222900391E-5?" Tags="&lt;notation&gt;" AnswerCount="2" CommentCount="0" />
  <row Id="6274" PostTypeId="1" CreationDate="2010-10-07T16:22:25.387" Score="3" ViewCount="413" Body="&lt;p&gt;In a book about radar signal processing, they are estimating the squared distance between two objects as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$d^2 = \frac{(R_p-R_o)^2}{\sigma^2_R} + \frac{(\theta_p-\theta_o)^2}{\sigma^2_\theta}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where R is a radius, $\theta$ is an angle and $\sigma^2$ is the variance of the radius and angle respectivly. p is the predicted target and o is an observation. The book is Blackman, S. - Design and Analysis of Modern Tracking Systems&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it just me or is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Dimensional_analysis&quot; rel=&quot;nofollow&quot;&gt;dimentional analysis&lt;/a&gt; for this equation a bit off?&lt;/p&gt;&#xA;" OwnerUserId="2367" LastEditorUserId="2367" LastEditorDisplayName="" LastEditDate="2010-10-07T20:43:39.217" LastActivityDate="2011-05-07T17:50:22.133" Title="Weird distance estimate?" Tags="&lt;calculus&gt;" AnswerCount="1" CommentCount="8" FavoriteCount="1" />
  <row Id="6275" PostTypeId="2" ParentId="6273" CreationDate="2010-10-07T16:22:36.450" Score="7" ViewCount="" Body="&lt;p&gt;It means 9.0122222900391 times $10^{-5}$. It's known as &quot;&lt;a href=&quot;http://en.wikipedia.org/wiki/Scientific_notation&quot; rel=&quot;nofollow&quot;&gt;scientific notation&lt;/a&gt;&quot;.&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2010-10-07T16:22:36.450" CommentCount="8" />
  <row Id="6276" PostTypeId="2" ParentId="6273" CreationDate="2010-10-07T16:23:01.863" Score="4" ViewCount="" Body="&lt;p&gt;E-5 just means x 10^(-5) i.e. divided by 100000&lt;/p&gt;&#xA;" OwnerUserId="1470" LastActivityDate="2010-10-07T16:23:01.863" />
  <row Id="6277" PostTypeId="1" CreationDate="2010-10-07T16:53:35.450" Score="1" ViewCount="134" Body="&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; question simplified to remove confusion&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming a sorted list of items with indexes from 1 to &lt;code&gt;N&lt;/code&gt;, and given &lt;em&gt;only&lt;/em&gt; an index number &lt;code&gt;i&lt;/code&gt; and the maximum index &lt;code&gt;N&lt;/code&gt;, is there a simple function which will return the two possible indexes that would be reached next in a binary search of the list?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;i = 6&#xA;N = 8&#xA;LeftSearchIndex(i, N) = 5&#xA;RightSearchIndex(i, N) = 7&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="59" LastEditorUserId="59" LastEditorDisplayName="" LastEditDate="2010-10-07T19:07:32.607" LastActivityDate="2011-01-05T19:44:39.360" Title="Is there a simple function which can be used to determine the two &quot;next&quot; search indexes in a binary search?" Tags="&lt;algorithms&gt;&lt;discrete-mathematics&gt;&lt;searching&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="6278" PostTypeId="2" ParentId="6277" CreationDate="2010-10-07T17:05:24.320" Score="1" ViewCount="" Body="&lt;p&gt;No longer relevant, keeping answer around for comments.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-10-07T19:25:46.210" LastActivityDate="2010-10-07T19:25:46.210" CommentCount="16" />
  <row Id="6279" PostTypeId="2" ParentId="6195" CreationDate="2010-10-07T17:58:58.357" Score="3" ViewCount="" Body="&lt;p&gt;This seems to be related to Gittins Indices. Gittins Indices are a way of solving these kind of optimal stopping problems for some classes of problems, and basically give you a way of balancing how much you are expected to gain given your current knowledge and how much more you &lt;em&gt;could&lt;/em&gt; gain by risking obtaining more information about the process (or probability of flipping heads, etc).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bruno&lt;/p&gt;&#xA;" OwnerUserId="2368" LastActivityDate="2010-10-07T17:58:58.357" CommentCount="1" />
  <row Id="6280" PostTypeId="1" AcceptedAnswerId="6283" CreationDate="2010-10-07T18:05:01.580" Score="16" ViewCount="508" Body="&lt;p&gt;Let $\pi(x)$ be the number of primes less than or equal to $x$.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;em&gt;One observes that, $\pi(6) \mid 6$, $\pi(8) \mid 8$. So does $\pi(x) \mid x$ for only finitely many $x$, or is this fact true for infinitely many $x$?&lt;/em&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="9413" LastEditorUserId="9413" LastEditorDisplayName="anonymous" LastEditDate="2011-03-22T23:39:08.537" LastActivityDate="2011-03-22T23:39:08.537" Title="$\pi(x) \mid x$ for how many $x$?" Tags="&lt;number-theory&gt;&lt;prime-numbers&gt;&lt;analytic-number-theory&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="4" />
  <row Id="6282" PostTypeId="1" AcceptedAnswerId="6284" CreationDate="2010-10-07T18:45:19.383" Score="1" ViewCount="188" Body="&lt;p&gt;Let us consider that there are &lt;strong&gt;G girl students&lt;/strong&gt; and &lt;strong&gt;B boy students&lt;/strong&gt; in a class, we need to arrange them in a single row, but arrangement of students should be in order to minimize the &lt;strong&gt;gender regularity*&lt;/strong&gt;. Find the minimum gender regularity among all possible arrangements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;*The gender regularity of an arrangement is the maximum number of students of the same gender (all girls or all boys) that appear consecutively.&lt;/p&gt;&#xA;" OwnerUserId="2109" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-10-16T17:49:12.317" LastActivityDate="2010-10-16T17:49:12.317" Title="Minimizing Gender Regularity in a linear arrangement of boys and girls." Tags="&lt;combinatorics&gt;" AnswerCount="2" />
  <row Id="6283" PostTypeId="2" ParentId="6280" CreationDate="2010-10-07T18:47:19.847" Score="23" ViewCount="" Body="&lt;p&gt;Well, it's not hard to show that for every natural $k&gt;2$ equation $x=k\pi(x)$ has a positive solution.&#xA;Proof by contradiction: Imagine, that $x\ne k\pi(x)$ for every natural $x$. But then - for $x=2$:$x-k\pi(x)=2-k&amp;lt;0$ and for very large $x$: $x-k\pi(x)\sim x(1-\frac{k}{\ln x})&gt;0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So there should be such $t$, that $t-k\pi(t)&amp;lt;0$ and $(t+1)-k\pi(t+1)&gt;0$. But then from one hand:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;t+1-k\pi(t+1)-(t-k\pi(t))\ge 2&#xA;$$&#xA;as a difference of positive integer and negative integer, but from the other hand:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;t+1-k\pi(t+1)-(t-k\pi(t))=1-k(\pi(t+1)-\pi(t))\le 1&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Contradiction.&lt;/p&gt;&#xA;" OwnerUserId="425" LastEditorUserId="425" LastEditorDisplayName="" LastEditDate="2010-10-07T19:00:19.800" LastActivityDate="2010-10-07T19:00:19.800" CommentCount="4" />
  <row Id="6284" PostTypeId="2" ParentId="6282" CreationDate="2010-10-07T19:36:38.393" Score="3" ViewCount="" Body="&lt;p&gt;I will try to show the solution is ceil(M / N+1) where M is the larger of B and G, N is the lesser.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Best case is M = N, where we need merely arrange the genders alternately, and regularity = ceil(M / M+1) = 1.  Worst case is a single-gender class, i.e N=0 in which regularity = ceil(M/1) = M, i.e. the whole class size.  In all other cases, clearly 1 &amp;lt; regularity &amp;lt; M+N.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now work quasi-inductively.  If N = 1, then we place the singleton in the center of the line, with half of the others to each side.  If the larger group is even, there are M/2 to each side (the ceil of a whole number is itself.)  If the larger group is odd, there will be one more on one side of the single, and it is this group that gives regularity = ceil(M/2).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If N = 2, do a similar process, divide the M into three groups, placing the two in between them (picture ---&lt;em&gt;----&lt;/em&gt;--- where the * are the small group, - the larger).  If the larger group is divisible by 3, there are M/3 in each division.  If not, distribute the 1 or 2 remainder over distinct divisions, and the largest will contain ceil(M/3) students.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As N increases toward M, repeat the concept, dividing the larger group into M/N+1 divisions, place the N between them, and distribute the remainder.&lt;/p&gt;&#xA;" OwnerUserId="575" LastActivityDate="2010-10-07T19:36:38.393" />
  <row Id="6285" PostTypeId="2" ParentId="3085" CreationDate="2010-10-07T19:39:11.610" Score="2" ViewCount="" Body="&lt;p&gt;The most comprehensive list I know of is H.W. Gould's &lt;em&gt;Combinatorial Identities.&lt;/em&gt;  It is available directly from him if you contact him.  He also has some pdf documents available for download from his &lt;a href=&quot;http://www.math.wvu.edu/~gould/&quot; rel=&quot;nofollow&quot;&gt;web site&lt;/a&gt;.  Although he says they do &quot;NOT replace [&lt;em&gt;Combinatorial Identities&lt;/em&gt;] which remains in print with supplements,&quot; they still contain many more binomial identities even than in &lt;em&gt;Concrete Mathematics&lt;/em&gt;.  In general, Gould's work is a great resource for this sort of thing; he has spent much of his career collecting and proving combinatorial identities. &lt;/p&gt;&#xA;" OwnerUserId="2370" LastActivityDate="2010-10-07T19:39:11.610" CommentCount="1" />
  <row Id="6286" PostTypeId="2" ParentId="6280" CreationDate="2010-10-07T20:28:34.083" Score="4" ViewCount="" Body="&lt;p&gt;This is Sloane's &lt;a href=&quot;http://oeis.org/classic/A057809&quot; rel=&quot;nofollow&quot;&gt;A057809&lt;/a&gt;, but unfortunately there's not much information there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Heuristically, the chance that $\pi(x)|x$ is $\log x/x$ which suggests that there are $0.5\log^2 x$ such numbers up to x, that is, infinitely many.&lt;/p&gt;&#xA;" OwnerUserId="1778" LastActivityDate="2010-10-07T20:28:34.083" CommentCount="2" />
  <row Id="6287" PostTypeId="1" AcceptedAnswerId="6290" CreationDate="2010-10-07T20:29:39.683" Score="2" ViewCount="78" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm trying to simplify the following booleans:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Y = ¬A¬B¬C¬D + ¬A¬B¬CD + ¬A¬BC¬D + ¬A¬BCD + A¬B¬C¬D + A¬BC¬D + ABC¬D&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The solution is:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Y = ¬A¬B + ¬B¬D + AC¬D&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;My solution is:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Y = ¬A¬B (¬C¬D + ¬CD + C¬D + CD) + A¬B¬C¬D + A¬BC¬D + ABC¬D&#xA;  = ¬A¬B (¬C (¬D + D) + C (¬D + D)) + A¬B¬C¬D + A¬BC¬D + ABC¬D&#xA;  = ¬A¬B (¬C + C) + A¬B¬C¬D + A¬BC¬D + ABC¬D&#xA;  = ¬A¬B + A¬D (¬B¬C + ¬BC + BC) &#xA;  = ¬A¬B + A¬D (¬B¬C + (¬BC + ¬BC) + BC) // law of idempotency&#xA;  = ¬A¬B + A¬D (¬B (¬C + C) + C (¬B + B))&#xA;  = ¬A¬B + A¬D (¬B + C)&#xA;  = ¬A¬B + A¬B¬D + AC¬D&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;As you can see.. my answer varies with the solution by the 2nd term. Any ideas? Thanks.&lt;/p&gt;&#xA;" LastActivityDate="2010-10-07T21:22:40.763" Title="Boolean Simplification" Tags="&lt;logic&gt;&lt;algebra&gt;" AnswerCount="2" />
  <row Id="6289" PostTypeId="2" ParentId="6287" CreationDate="2010-10-07T21:07:38.527" Score="0" ViewCount="" Body="&lt;p&gt;I agree with you.  The last three terms all have A required, so anything that comes out of them (your last two terms) need an A.  Are you sure the given solution is right?&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-10-07T21:07:38.527" CommentCount="1" />
  <row Id="6290" PostTypeId="2" ParentId="6287" CreationDate="2010-10-07T21:20:35.123" Score="2" ViewCount="" Body="&lt;p&gt;All your work is right.  The first two terms of your answer simplify to:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&#xA;Y = ¬A¬B + A¬B¬D&#xA;  = ¬B(¬A + A¬D)&#xA;  = ¬B(¬AD + ¬A¬D + A¬D) // split the first term&#xA;  = ¬B(¬AD + ¬A¬D + ¬A¬D + A¬D) // idempotence&#xA;  = ¬B(¬A + ¬D) // combine the first two and last two terms&#xA;  = ¬A¬B + ¬B¬D.&#xA;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I wouldn't have believed it myself at first, but idempotence lets the &lt;code&gt;¬A&lt;/code&gt; &quot;spit out&quot; a &lt;code&gt;¬A¬D&lt;/code&gt;, which is, after all, just a subset.&lt;/p&gt;&#xA;" OwnerUserId="2215" LastActivityDate="2010-10-07T21:20:35.123" CommentCount="3" />
  <row Id="6292" PostTypeId="1" AcceptedAnswerId="6298" CreationDate="2010-10-07T23:15:20.493" Score="1" ViewCount="151" Body="&lt;p&gt;I am trying to calculate the average distance between all vertices of a complete $k$-ary tree. A complete $k$-ary tree is a tree such that all vertices have $k$ children except for the leafs of the tree. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose the tree has $r$ levels, where level 0 is the root. So the height of the tree is $r-1$. First i have to know how many vertices there are in the tree. For this i would sum up the vertices on each level of the tree which would give me&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sum_{i=1}^{r}k^i = k^1 + k^2 + ...+ k^r$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(is there an easier way of representing this?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now how many ways are there to choose paths in this tree that are of length $1, 2, 3, ..., r-1, ..., 2(r-1)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$2(r-1)$ because we can start from on leaf and go all the way up to the root and all the way down again to a vertex that is on a different branch that the initial vertex.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not sure how to count all these. For length 1 i can easily count those (i think):&#xA;$r-1 * \sum_{i=i}^{r}k^i$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But level two is a little trickier because you can go up two parents or go up one parent and down to the child on the same level as the initial vertex.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One i figure out the total distance between all vertices i have to divide it by the total number of vertices to get the average, correct?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any help would be appreciated.&lt;/p&gt;&#xA;" OwnerUserId="43" LastActivityDate="2010-10-08T21:27:08.200" Title="complete $k$-ary tree: average distance between all vertices" Tags="&lt;graph-theory&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="6293" PostTypeId="2" ParentId="6292" CreationDate="2010-10-08T00:38:38.123" Score="1" ViewCount="" Body="&lt;p&gt;Your sum $\sum_{i=1}^{r}k^i = k^1 + k^2 + ...+ k^r$ should run from $0$ to $r-1$.  This is a geometric series which can be summed in the usual way to $(k^r-1)/(k-1)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried counting the average distance from a node on each level but it became a mess.  A slicker way is to count how many times each edge gets used when you count all the paths.  It will get used if one of the vertices is above it and the other is not.  For example, an edge attached to the root has a tree of height r-1 above it, so will get used $(k^{(r-1)}-1)/(k-1) * ((k^r-1)-(k^{(r-1)}-1)/(k-1)$ times.  An edge at the top of the tree will get used if and only if its leaf is one of the vertices so will get counted $(k^r-1)/(k-1)-1$ times.  An edge at level $i$ will get used $(k^{(r-i-1)}-1)/(k-1) * ((k^r-1)-(k^{(r-i-1)}-1)/(k-1)$ times and there are $k^{(i+1)}$ of them where i goes from 0 to r-2.  Then divide by the number of pairs of vertices.  So you want $\sum_{i=0}^{r-2} k^{(i+1)}*(k^{(r-i-1)}-1)/(k-1) * ((k^r-1)-(k^{(r-i-1)}-1)/(k-1)$ divided by $((k^r-1)/(k-1)-1)*(k^r-1)/(k-1)/2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's a mess, but I'm sure Mathematica can do it.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2010-10-08T21:27:08.200" LastActivityDate="2010-10-08T21:27:08.200" CommentCount="1" />
  <row Id="6294" PostTypeId="2" ParentId="6262" CreationDate="2010-10-08T00:45:45.987" Score="0" ViewCount="" Body="&lt;p&gt;I would say 6*24(n-1).  The first cube can be placed in 6 ways, just choosing the top face.  Then you select one side as the front.  The second cube can then be placed in 24 ways as you choose the top and then which one will be front.  Each successive cube has 24 ways to be placed.  So you multiply all these numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another way to look at it is that each cube has 24 possibilities, but they fall into groups of four by rotating the whole stack.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-10-08T00:45:45.987" />
  <row Id="6295" PostTypeId="1" AcceptedAnswerId="6296" CreationDate="2010-10-08T00:47:13.110" Score="0" ViewCount="242" Body="&lt;p&gt;Suppose you have two distinct large primes $p$ and $q$.&#xA;Explain how you can find an integer $x$ such that $x^2 \equiv 49$&#xA;(mod $pq$), $x \not\equiv\pm 7$ (mod $pq$).&lt;/p&gt;&#xA;" OwnerUserId="2375" LastEditorUserId="226" LastEditorDisplayName="" LastEditDate="2010-10-09T16:01:24.587" LastActivityDate="2010-10-09T16:03:14.070" Title="Find x such that $x^2 \equiv 49$ (mod $pq$), $x \not\equiv\pm 7$ (mod $pq$)" Tags="&lt;number-theory&gt;&lt;homework&gt;&lt;prime-numbers&gt;&lt;modular-arithmetic&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="6296" PostTypeId="2" ParentId="6295" CreationDate="2010-10-08T01:10:11.323" Score="2" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\rm\ pq\:|\:(x-7)(x+7)\:,\ $ not $\rm\ pq\:|\:x\pm7\ \Rightarrow\ p\:|\:x- 7,\ q\:|\:x+7\ $ or vice versa.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, you need $\rm\: -7\not\equiv 7\ $ so $\rm\ p,\ q\ $ coprime to $\ \ldots$&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-10-08T02:03:36.410" LastActivityDate="2010-10-08T02:03:36.410" />
  <row Id="6297" PostTypeId="2" ParentId="2949" CreationDate="2010-10-08T01:26:42.237" Score="23" ViewCount="" Body="&lt;p&gt;I  found the simplicity of &lt;A HREF=&quot;http://en.wikipedia.org/wiki/Pick%27s_theorem&quot; rel=&quot;nofollow&quot;&gt;Pick's Theorem&lt;/A&gt; pretty surprising when I first stumbled across it.&lt;/p&gt;&#xA;" OwnerUserId="6916" LastActivityDate="2010-10-08T01:26:42.237" CommunityOwnedDate="2010-10-08T01:26:42.237" CommentCount="1" />
  <row Id="6298" PostTypeId="2" ParentId="6292" CreationDate="2010-10-08T01:27:31.840" Score="1" ViewCount="" Body="&lt;p&gt;Here's another approach to trying to find the total distance: assume that we have some 'initial' node on level $m$, and we're trying to find the distance to the nodes at equal or lower levels.  Start with the equal level: there are $(k-1)$ siblings, at distance 2; then $k(k-1)$ 'first cousins' (as each of our parent's $k-1$ siblings has $k$ children, at distance 4; $k^2(k-1)$ 'second cousins', at distance , ..., up to $k^{m-1}(k-1)$ 'root cousins' (i.e., children going through the root node).  Now, consider e.g. our first cousins; each of those will bring $k$ children at distance 4+1=5, $k^2$ grandchildren at distance 4+2=6, ..., $k^{n-m}$ great-great-...grandchildren, each of which contributes $4+(n-m)$ distance.  A bit of careful working through the algebra should be enough to sum these correctly (distribute out that factor of 4 first!), and that'll be enough to get the total distance from a given node on level $m$ to all nodes on deeper levels; then just sum that over all depths.  The one little 'gotcha' is that you want to divide the distance-sum from a node on level $m$ to all nodes on the same level by 2, otherwise you'll double-count it, but that's easily accounted for.  I can try and work out the algebra later this evening, but I think it'd make a pretty good exercise.&lt;/p&gt;&#xA;" OwnerUserId="785" LastActivityDate="2010-10-08T01:27:31.840" CommentCount="1" />
  <row Id="6299" PostTypeId="1" CreationDate="2010-10-08T01:35:30.443" Score="0" ViewCount="137" Body="&lt;p&gt;Suppose I have two machines, A and B. A encrypts a message m and outputs the ciphertext m^e (mod n). B outputs c such that c = m^e (mod p) and c = m^e + 1 (mod q). How can I use A and B to find p and q? I am allowed to choose m and n.&lt;/p&gt;&#xA;" OwnerUserId="2375" LastActivityDate="2010-10-08T01:53:48.823" Title="Break RSA given a correct and faulty implementation" Tags="&lt;number-theory&gt;&lt;homework&gt;&lt;prime-numbers&gt;&lt;modular-arithmetic&gt;" AnswerCount="1" />
  <row Id="6300" PostTypeId="2" ParentId="6299" CreationDate="2010-10-08T01:48:05.870" Score="2" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\ $ Consider $\rm\ gcd(A(m)-B(m),\:pq)$&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-10-08T01:53:48.823" LastActivityDate="2010-10-08T01:53:48.823" CommentCount="2" />
  <row Id="6302" PostTypeId="2" ParentId="6295" CreationDate="2010-10-08T02:01:48.180" Score="3" ViewCount="" Body="&lt;p&gt;It's helpful to know the multiplicative structure of $(\mathbb{Z}/n\mathbb{Z})^*$.  It is always a product of cyclic subgroups.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For powers of 2, we have $(\mathbb{Z}/2^k\mathbb{Z})^* \equiv C_2 \times C_{2^{k-2}}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For odd primes &lt;em&gt;p&lt;/em&gt;, we have $(\mathbb{Z}/p^k\mathbb{Z})^* \equiv C_{(p-1)p^{k-1}}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Knowing the structure for the prime powers is enough; we can use the Chinese Remainder Theorem to see that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(\mathbb{Z}/p_1^{i_1}\cdots p_r^{i_r}\mathbb{Z})^* = (\mathbb{Z}/p_1^{i_1}\mathbb{Z})^* \times \cdots \times (\mathbb{Z}/p_r^{i_r}\mathbb{Z})^*$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then $(\mathbb{Z}/pq\mathbb{Z})^* = C_{p-1} \times C_{q-1}$ has four solutions to $x^2=7^2$, and Bill's answer shows very nicely how to find them.&lt;/p&gt;&#xA;" OwnerUserId="1461" LastActivityDate="2010-10-08T02:01:48.180" />
  <row Id="6303" PostTypeId="1" AcceptedAnswerId="6304" CreationDate="2010-10-08T02:33:05.510" Score="2" ViewCount="421" Body="&lt;p&gt;Geometric multiplicity of an eigenvalue of a matrix is the dimension of the corresponding eigenspace.  The algebraic multiplicity is its multiplicity as a root of the characteristic polynomial. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is known that &lt;em&gt;the geometric multiplicity of an eigenvalue cannot be greater than the algebraic multiplicity&lt;/em&gt;.  This fact can be shown easily using the &lt;a href=&quot;http://en.wikipedia.org/wiki/Jordan_normal_form&quot; rel=&quot;nofollow&quot;&gt;Jordan normal form&lt;/a&gt; of a matrix.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was wondering if there is a more elementary way to prove this fact, possibly longer but without using the Jordan normal form?  (This is an exercise in &lt;a href=&quot;http://books.google.com/books?id=nZmpQgAACAAJ&amp;amp;dq=inauthor%3a%22Erwin+Kreyszig%22&amp;amp;hl=en&amp;amp;ei=JYKuTMe1G4KC8gbmzcTABA&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=4&amp;amp;ved=0CDsQ6AEwAw&quot; rel=&quot;nofollow&quot;&gt;Kreyszig's book on functional analysis&lt;/a&gt;, and given the author's style, I suspect that he did not intend the solution to use Jordan form, because otherwise I guess he would have given a hint about that. But I might be wrong.)&lt;/p&gt;&#xA;" OwnerUserId="145" LastActivityDate="2010-10-08T02:41:52.527" Title="Geometric multiplicity of an eigenvalue" Tags="&lt;linear-algebra&gt;" AnswerCount="1" />
  <row Id="6304" PostTypeId="2" ParentId="6303" CreationDate="2010-10-08T02:41:52.527" Score="8" ViewCount="" Body="&lt;p&gt;You don't need the Jordan form: suppose the geometric multiplicity of $\lambda$ is $k$, and let $\gamma=\{\mathbf{v}_1,\ldots,\mathbf{v}_k\}$ be a basis for the corresponding eigenspace. Extend the basis $\gamma$ to a basis $\beta$ for $F^n$, and let $Q$ be the change-of-basis matrix. Then the characteristic polynomials of $A$ and $Q^{-1}AQ$ are the same. The upper left $k\times k$ block of $Q^{-1}AQ$ is simply $\lambda I_k$, and, the $(n-k)\times k$ block under it is all zeroes. So the characteristic polynomial of $Q^{-1}AQ$ is a multiple of $(\lambda - t)^k$ hence the algebraic multiplicity of $\lambda$ is at least $k$.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-10-08T02:41:52.527" CommentCount="1" />
  <row Id="6305" PostTypeId="2" ParentId="2949" CreationDate="2010-10-08T02:42:16.150" Score="23" ViewCount="" Body="&lt;p&gt;I think one of my favorites would be Gödel's incompleteness theorem, which tells us that a formal system cannot be demonstrated to be consistent.&lt;/p&gt;&#xA;" OwnerUserId="796" LastActivityDate="2010-10-08T02:42:16.150" CommunityOwnedDate="2010-10-08T02:42:16.150" />
  <row Id="6306" PostTypeId="2" ParentId="29" CreationDate="2010-10-08T02:42:51.233" Score="2" ViewCount="" Body="&lt;p&gt;I just discovered a software called &lt;a href=&quot;http://www.spacetime.us/&quot; rel=&quot;nofollow&quot;&gt;Spacetime&lt;/a&gt;. Its free, available for windows/mac/iphone and very powerful. You can:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Graph 2D, 3D and 4D function,&#xA;  parametric, polar, vector fields,&#xA;  implicit, contour, and fractal plots&#xA;  with the tap of a stylus or click of a&#xA;  mouse.  Move, zoom, rotate and trace&#xA;  graphs in real-time.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Furthermore, it supports writing scripts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The UI is good, and the software comes with a great tutorial.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/qI8Jr.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="19" LastActivityDate="2010-10-08T02:42:51.233" CommunityOwnedDate="2010-10-08T02:42:51.233" />
  <row Id="6307" PostTypeId="2" ParentId="6282" CreationDate="2010-10-08T02:48:29.493" Score="0" ViewCount="" Body="&lt;p&gt;Assume that there are B boys and G girls. Let us say (Without loss of generality) that B is no smaller than G. Let us make the arrangement by lining up the girls and then lining up the boys among them. In this arrangement, there are G+1 places the boys can be (before girl 1, between girl 1 and 2, ... , between girl G-1 and G, and after girl G). By the pigeonhole principal, you will have at least ceil(B / G+1) boys in one of these locations, and since there are no other conditions, this minimum is achievable.&lt;/p&gt;&#xA;" OwnerUserId="796" LastActivityDate="2010-10-08T02:48:29.493" />
  <row Id="6308" PostTypeId="2" ParentId="5927" CreationDate="2010-10-08T04:08:42.413" Score="0" ViewCount="" Body="&lt;p&gt;As Bill's answer implies, this really is a problem of definition.  The definition of $\binom{n}{k}$ is not $\frac{n!}{k!(n-k)!}$.  Depending on your point of view, it's either &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) Bill's answer&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) the number of distinct sets of $k$ elements that can be formed from a set containing $n$ elements&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or even (a bit more obscurely) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(3) the solution to the recurrence relation $R^n_k = R^{n-1}_k + R^{n-1} _{k-1}$ with initial conditions $R^0_0 = 1$ and $R^0_k = 0$ for $k \neq 0$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In any of cases (1), (2), and (3) it is not hard to prove that you get $\binom{n}{k} = \frac{n!}{k!(n-k)!}$ when $n \geq k$ and $\binom{n}{k} = 0$ when $n &amp;lt; k$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Point is: The factorial formula for $\binom{n}{k}$ does not apply when $n &amp;lt; k$.&lt;/p&gt;&#xA;" OwnerUserId="2370" LastActivityDate="2010-10-08T04:08:42.413" />
  <row Id="6309" PostTypeId="1" AcceptedAnswerId="6319" CreationDate="2010-10-08T04:59:45.483" Score="2" ViewCount="108" Body="&lt;p&gt;I am trying to derive the following equation for an analytic function, $f$, defined on the unit disc, $D$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f(z)=\frac{1}{\pi}\int_{D} \frac{f(w)}{(1- \bar{w}z)^{2}}dA(w)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;{\bf Question}: Anyone have any ideas on how to derive this formula?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've been able to verify that the above is true for $z =0$.   In attempting to derive for the general case, I used conformal maps to arrive at&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f(z)=\frac{1}{\pi}\int_{D} f(w) \left| \frac{-1+|z|^{2}}{(-1+w\bar{z})^{2}} \right|^{2} dA(w)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunatly this is not the formula I was trying to derive,&lt;/p&gt;&#xA;" OwnerUserId="832" LastEditorUserId="832" LastEditorDisplayName="" LastEditDate="2010-10-08T18:48:40.690" LastActivityDate="2010-10-08T18:48:40.690" Title="Derivation of a &quot;Weighted&quot; Mean Value Thm" Tags="&lt;complex-analysis&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="6310" PostTypeId="1" AcceptedAnswerId="6312" CreationDate="2010-10-08T05:08:12.630" Score="2" ViewCount="540" Body="&lt;p&gt;How does one prove that if $n \mid (a^{n}-b^{n}) \ \Longrightarrow$ $ \displaystyle n \mid \frac{a^{n}-b^{n}}{a-b}$ where $a,b, n \in \mathbb{N}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What i thought of is to consider $$(a-b)^{n} \equiv a^{n} + (-1)^{n}b^{n} \ (\text{mod} \ n)$$ and if we suppose that $n$ is odd then we have, $$(a-b)^{n} \equiv a^{n} -b^{n} \ (\text{mod} \ n)$$ and since $n \mid (a^{n} - b^{n})$ we have $$(a-b)^{n} \equiv 0 \ (\text{mod} \ n) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think i am far away from the conclusion of the problem, but this is what i could work on regarding the problem. &lt;/p&gt;&#xA;" LastActivityDate="2011-02-06T22:29:19.163" Title="$n \mid (a^{n}-b^{n}) \ \Longrightarrow$ $n \mid \frac{a^{n}-b^{n}}{a-b}$" Tags="&lt;number-theory&gt;&lt;modular-arithmetic&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="2" />
  <row Id="6311" PostTypeId="2" ParentId="6274" CreationDate="2010-10-08T05:26:09.893" Score="0" ViewCount="" Body="&lt;p&gt;The formula is not even close.  You are right that d should have dimensions of length and the right hand side is unitless.  Worse, the formula doesn't reduce properly if the errors are zero.  Are you sure it isn't $d^2/\sigma(d)^2$ on the left?  That would fix the units.  The cosine formula gives the distance estimate you want: $d^2=R_p^2+R_o^2-2 R_p R_o \cos(\theta_p-\theta_o)$.  When you mix in sigmas you are trying to estimate the error in d based on the errors in the measurements.  So you should use the formulas you have for combining measurement errors to give the error in the measured quantity.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="1208" LastEditorDisplayName="" LastEditDate="2011-03-08T17:24:58.270" LastActivityDate="2011-03-08T17:24:58.270" CommentCount="2" />
  <row Id="6312" PostTypeId="2" ParentId="6310" CreationDate="2010-10-08T05:44:42.973" Score="5" ViewCount="" Body="&lt;p&gt;Use the fact that $x \equiv y \pmod{p^{\ell}}$ (with $\ell &gt; 0$) implies $x^p \equiv y^p \pmod{p^{\ell+1}}$ to treat the case where $n$ is a prime power.  But note that to prove the statement for $n$, it suffices to prove the statement separately for each prime power dividing $n$.  [I am of course leaving out details, but perhaps the OP might enjoy trying to fill in the sketch.]&lt;/p&gt;&#xA;" OwnerUserId="631" LastEditorUserId="631" LastEditorDisplayName="" LastEditDate="2010-10-08T05:52:25.810" LastActivityDate="2010-10-08T05:52:25.810" />
  <row Id="6313" PostTypeId="2" ParentId="2949" CreationDate="2010-10-08T06:17:04.913" Score="15" ViewCount="" Body="&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Gauss-Bonnet_Theorem&quot; rel=&quot;nofollow&quot;&gt;Gauss-Bonnet theorem&lt;/a&gt;.  The integral of the curvature of a manifold, a totally geometric concept and one that looks dependent on the embedding, is equal to $2\pi$ times its Euler characteristic, an algebraic homotopy invariant.&lt;/p&gt;&#xA;" OwnerUserId="2215" LastActivityDate="2010-10-08T06:17:04.913" CommunityOwnedDate="2010-10-08T06:17:04.913" />
  <row Id="6314" PostTypeId="1" AcceptedAnswerId="6338" CreationDate="2010-10-08T06:21:42.363" Score="16" ViewCount="1187" Body="&lt;p&gt;Can you express $[0,1]$ as a countable disjoint union of closed sets, other than the trivial way of doing this?&lt;/p&gt;&#xA;" OwnerUserId="546" LastActivityDate="2010-11-15T17:24:41.013" Title="Is $[0,1]$ a countable disjoint union of closed sets?" Tags="&lt;real-analysis&gt;&lt;analysis&gt;&lt;general-topology&gt;" AnswerCount="3" CommentCount="5" FavoriteCount="3" />
  <row Id="6315" PostTypeId="2" ParentId="6314" CreationDate="2010-10-08T06:31:06.180" Score="1" ViewCount="" Body="&lt;p&gt;No. $[0,1]$ cannot be written as the union of countable disjoint closed intervals . Try to use the Baire Category theorem. You may also refer this post &lt;a href=&quot;http://terrytao.wordpress.com/2010/10/04/covering-a-non-closed-interval-by-disjoint-closed-intervals/&quot; rel=&quot;nofollow&quot;&gt;http://terrytao.wordpress.com/2010/10/04/covering-a-non-closed-interval-by-disjoint-closed-intervals/&lt;/a&gt;&lt;/p&gt;&#xA;" LastActivityDate="2010-10-08T06:31:06.180" CommentCount="8" />
  <row Id="6316" PostTypeId="2" ParentId="2949" CreationDate="2010-10-08T09:01:43.973" Score="3" ViewCount="" Body="&lt;p&gt;The fact that the axiom of foundations (also known as regularity in some places) is independent of the rest of the axioms, and not only you can have an infinitely descending chain $x_0 \ni x_1 \ni \ldots$ but you can have $x = \{ x \}$ and even more than that! $P(a) \in a$ is possible as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Crazy set theoretic voodoo, that's what this is!&#xA;(And I'm loving every single bit of it :))&lt;/p&gt;&#xA;" OwnerUserId="622" LastActivityDate="2010-10-08T09:01:43.973" CommunityOwnedDate="2010-10-08T09:01:43.973" CommentCount="1" />
  <row Id="6317" PostTypeId="1" AcceptedAnswerId="6318" CreationDate="2010-10-08T10:36:52.747" Score="1" ViewCount="79" Body="&lt;p&gt;It seems to me not, since this would seem to imply that for all functors &lt;em&gt;F&lt;/em&gt; and all objects &lt;em&gt;A&lt;/em&gt; in &lt;strong&gt;C&lt;/strong&gt; there exists a morphism &lt;em&gt;F(A) -&gt; A&lt;/em&gt; (making all functors co-pointed?). However, intuitively it seems like the identity functor acts like a terminal object; a monad &lt;em&gt;M&lt;/em&gt; on &lt;strong&gt;C&lt;/strong&gt; is a monoid on &lt;strong&gt;[C, C]&lt;/strong&gt; where the &quot;unit&quot; is a natural transformation &lt;em&gt;η : I -&gt; M&lt;/em&gt;, while for a monoidal set &lt;em&gt;S&lt;/em&gt; in &lt;strong&gt;Set&lt;/strong&gt; the unit is a function &lt;em&gt;e : 1 -&gt; S&lt;/em&gt;. So am I misunderstanding something, or are my intuitions leading me astray?&lt;/p&gt;&#xA;" OwnerUserId="2378" LastActivityDate="2010-10-08T10:54:33.210" Title="Is the identity functor the terminal object of the category of endofunctors on C?" Tags="&lt;category-theory&gt;" AnswerCount="1" CommentCount="0" />
  <row Id="6318" PostTypeId="2" ParentId="6317" CreationDate="2010-10-08T10:48:01.960" Score="1" ViewCount="" Body="&lt;p&gt;The difference between those two examples is that in $\textbf{Set}$ the monoidal operation is the categorical product (so the identity object is the terminal object), whereas this is not true in the category of endofunctors on $\mathbf{C}$.  (I believe the latter has a product if and only if $\mathbf{C}$ does, and then it is the pointwise product.  It follows that the terminal object, if it exists, is the functor which sends all objects to $\mathbf{1}$ and all morphisms to the unique morphism $\mathbf{1} \to \mathbf{1}$.  In particular, it's not the identity functor.)&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-10-08T10:54:33.210" LastActivityDate="2010-10-08T10:54:33.210" CommentCount="4" />
  <row Id="6319" PostTypeId="2" ParentId="6309" CreationDate="2010-10-08T12:06:10.470" Score="5" ViewCount="" Body="&lt;p&gt;First, you're missing a factor of $1/\pi$ on the right-hand side, as can be seen by taking $f=1$ and $z=0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To prove the identity, you can use Green's formula, which translated into complex variables takes the form $$\int_C F(w,\bar{w}) dw = 2i \iint \frac{\partial F}{\partial \bar{w}} dA(w).$$&#xA;This transforms your area integral into a line integral around the unit circle $C$. To evaluate this line integral, use that $\bar{w}=1/w$ when $|w|=1$; this gives you an integral in $w$ alone (no $\bar{w}$) so that you can use Cauchy's formula to finish the job.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've checked that it works, but I omit the details so that I don't spoil your pleasure too much. :-)&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2010-10-08T12:06:10.470" CommentCount="1" />
  <row Id="6321" PostTypeId="1" AcceptedAnswerId="6565" CreationDate="2010-10-08T13:56:07.787" Score="8" ViewCount="296" Body="&lt;p&gt;Any number less than 1 can be expressed in base g as $\sum _{k=1}^\infty {\frac {D_k}{g^k}}$, where $D_k$ is the value of the $k^{th}$ digit. If we were interested in only the non-zero digits of this number, we could equivilantly express it as $\sum _{k=1}^\infty {\frac {C_k}{g^{Z(k)}}}$, where $Z(k)$ is the position of the $k^{th}$ non-zero digit base $g$ and $C_k$ is the value of that digit (i.e. $C_k = D_{Z(k)}$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, consider all the numbers of this form $(\sum _{k=1}^\infty {\frac {C_k}{g^{Z(k)}}})$ where the function $Z(k)$ eventually dominates any polynomial. Is there a proof that any number of this form is transcendental?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So far, I have found a paper demostrating this result for the case $g=2$; it can be found here: &lt;a href=&quot;http://www.escholarship.org/uc/item/44t5s388?display=all&quot; rel=&quot;nofollow&quot;&gt;http://www.escholarship.org/uc/item/44t5s388?display=all&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="796" LastEditorUserId="796" LastEditorDisplayName="" LastEditDate="2010-10-09T06:29:16.160" LastActivityDate="2010-10-11T21:26:28.743" Title="Have all numbers with &quot;sufficiently many zeros&quot; been proven transcendental?" Tags="&lt;number-theory&gt;&lt;algebra&gt;&lt;number&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="6322" PostTypeId="1" CreationDate="2010-10-08T14:13:40.920" Score="1" ViewCount="135" Body="&lt;p&gt;I've been self studying abstract algebra from these &lt;a href=&quot;http://jmilne.org/math/CourseNotes/gt.html&quot; rel=&quot;nofollow&quot;&gt;notes&lt;/a&gt;. I've ran across the following Lemma:&#xA;&lt;img src=&quot;http://i.imgur.com/HZM3A.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The confusing part is that $X$ is a &quot;any subset&quot; not any &quot;subgroup&quot;, and that the term &quot;normal set&quot; appeared in place where I expected &quot;normal subgroup&quot;. The definition of a &quot;normal set&quot; has not been provided and I couldn't find it with a quick Googling.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the question is &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What is a &quot;normal set&quot; and why is the proof of that Lemma obvious ?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="558" LastActivityDate="2010-10-08T14:13:40.920" Title="The smallest normal set containing a subset $X$, what is a &quot;normal set&quot; ?" Tags="&lt;abstract-algebra&gt;&lt;group-theory&gt;" CommentCount="4" />
  <row Id="6323" PostTypeId="1" CreationDate="2010-10-08T15:29:39.190" Score="2" ViewCount="125" Body="&lt;p&gt;I understand in mathematics there are many &quot;quotienting &quot; proceduce, is this the only reason that we consider equivariant theory for different &quot;unequivariant&quot; theory? Are there any more applications for equivariant theory?Thanks!&lt;/p&gt;&#xA;" OwnerUserId="2381" LastActivityDate="2010-12-08T15:45:21.453" Title="How equivariant theory (like equivariant cohomology) arise" Tags="&lt;algebraic-topology&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="1" />
  <row Id="6324" PostTypeId="1" AcceptedAnswerId="6326" CreationDate="2010-10-08T16:11:25.480" Score="11" ViewCount="606" Body="&lt;p&gt;This came up in proving non-regularity of a certain language (powers of 2 over the ternary alphabet). Any clue to the above equation could help me move forward.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit:&#xA;Of course, $x = 1, y = 1$ is a solution. I am looking for non-trivial solutions.&lt;/p&gt;&#xA;" OwnerUserId="813" LastEditorUserId="813" LastEditorDisplayName="" LastEditDate="2010-10-08T16:20:33.547" LastActivityDate="2010-10-09T04:37:24.277" Title="Are there integer solutions to $9^x - 8^y = 1$?" Tags="&lt;number-theory&gt;&lt;diophantine-equations&gt;" AnswerCount="3" CommentCount="6" />
  <row Id="6325" PostTypeId="2" ParentId="6324" CreationDate="2010-10-08T16:17:48.403" Score="13" ViewCount="" Body="&lt;p&gt;The only solution is $x=y=1$. This is a special case of &lt;a href=&quot;http://en.wikipedia.org/wiki/Catalan%27s_conjecture&quot; rel=&quot;nofollow&quot;&gt;Catalan's conjecture&lt;/a&gt; which was proven by Mihailescu in 2002: the only solution in natural numbers of $x^a - y^b = 1$ with $a,b\gt 1$ is $x=3$, $a=2$, $y=2$, and $b=3$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But as Moron points out, it can be deduced much more elementarily.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-10-08T16:33:11.307" LastActivityDate="2010-10-08T16:33:11.307" />
  <row Id="6326" PostTypeId="2" ParentId="6324" CreationDate="2010-10-08T16:26:15.627" Score="33" ViewCount="" Body="&lt;p&gt;Except $x=1$ and $y=1$ there aren't any.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$3^{2x} - 1 = 8^y$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i.e&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ (3^x + 1)(3^x - 1)  = 8^y$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus we must have that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$3^x + 1 = 2^m, 3^x - 1 = 2^n$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus $$2^m - 2^n = 2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i.e&#xA;$$ 2^n(2^{m-n} - 1) = 2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus $n=1$ and $m=2$.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-10-08T17:08:03.653" LastActivityDate="2010-10-08T17:08:03.653" CommentCount="6" />
  <row Id="6327" PostTypeId="2" ParentId="6323" CreationDate="2010-10-08T16:33:31.103" Score="3" ViewCount="" Body="&lt;p&gt;When you pass to the quotient by an action you may disastrously break some (or all!) interesting structure, so you look for ways of doing things in the quotient without actually constructing it. Doing things equivariantly upstairs is one of the ways to do that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively, the quotient, when it is a sensible object, may not contain all the information you want: for example, there are situations in which you get the same quotient by dividing a space by the action of two different groups, yet you &lt;em&gt;want&lt;/em&gt; to have different results. So you work equivariantly and are happy.&lt;/p&gt;&#xA;" OwnerUserId="274" LastEditorUserId="274" LastEditorDisplayName="" LastEditDate="2010-10-08T16:39:04.413" LastActivityDate="2010-10-08T16:39:04.413" />
  <row Id="6328" PostTypeId="1" AcceptedAnswerId="6334" CreationDate="2010-10-08T16:53:01.353" Score="2" ViewCount="146" Body="&lt;p&gt;Suppose $P(x,y)$ gives transition probabilities of a random walk. I've seen $Pf=f$ being called the discrete version of Laplace's equation. In what sense are they analogous?&lt;/p&gt;&#xA;" OwnerUserId="998" LastActivityDate="2010-10-09T01:44:36.650" Title="discrete version of Laplacian" Tags="&lt;probability-theory&gt;" AnswerCount="3" FavoriteCount="1" />
  <row Id="6329" PostTypeId="1" CreationDate="2010-10-08T17:00:37.243" Score="2" ViewCount="125" Body="&lt;p&gt;Convex polyhedron $P$ is a subset of $\mathbb{R}^n$ that satisfies system of linear inequalities&#xA;\begin{align}&#xA;a_{11}x_1 + \cdots + a_{1n}x_n &amp;amp; \sim_1\, c_1 \\&#xA;&amp;amp; \vdots \\&#xA;a_{p1}x_1 + \cdots + a_{pn}x_n &amp;amp; \sim_p\, c_p,&#xA;\end{align}&#xA;where $\sim_i \in \{\leq,\geq\}$. It can be alternatively represented by two finite sets of generators $V, W \subseteq \mathbb{R}^n$: $$P = \text{conv}(V) + \text{cone}(W),$$ where conv(V) denotes all convex combinations of points in V and cone(W) all nonnegative linear combinations of points in W.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, what if we allow $\sim_i$ to be from $\{\geq,&gt;,\leq,&amp;lt;\}$. Is there some similar representation in terms of generating points for such sets?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I possess no knowledge of this area of mathematics, so I apologize if I got the terminology wrong or if this question is just plain stupid.)&lt;/p&gt;&#xA;" LastActivityDate="2010-10-13T08:21:28.233" Title="Convex polyhedron with open faces" Tags="&lt;geometry&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="6330" PostTypeId="2" ParentId="6329" CreationDate="2010-10-08T17:23:52.060" Score="2" ViewCount="" Body="&lt;p&gt;No.  Unfortunately, it can't even be done in the 1D case.  For instance, consider the inequalities $x &gt; 0$ and $x &amp;lt; 1$.  Then $P$ is the open interval $(0,1)$.  You can't take $V = \{0,1\}$ because then conv$(V)$ is the closed interval $[0,1]$.  Taking $V$ to be any other two points between 0 and 1 would generate a conv$(V)$ that is a strict subset of $P$.&lt;/p&gt;&#xA;" OwnerUserId="2370" LastActivityDate="2010-10-08T17:23:52.060" CommentCount="2" />
  <row Id="6331" PostTypeId="2" ParentId="6328" CreationDate="2010-10-08T17:35:00.133" Score="3" ViewCount="" Body="&lt;p&gt;The infinitesimal generator for Brownian motion in $\mathbb R^n$ (and, in general, in a Riemannian manifold) is $\tfrac12\Delta$. Your matrix $P$ plays exactly the same rôle in the discrete version.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the equation: the equation $\tfrac12\Delta f=0$ is, in the case of Brownian motion in $\mathbb R^n$, the equation for stationary states. Your equation is the same thing in the discrete version.&lt;/p&gt;&#xA;" OwnerUserId="274" LastEditorUserId="274" LastEditorDisplayName="" LastEditDate="2010-10-08T17:46:53.200" LastActivityDate="2010-10-08T17:46:53.200" />
  <row Id="6332" PostTypeId="1" CreationDate="2010-10-08T17:44:49.213" Score="2" ViewCount="102" Body="&lt;p&gt;&quot;Gromov Witten potential&quot;, when does &quot;potential&quot; mean here? What does the whole thing mean in physics? Thanks!&lt;/p&gt;&#xA;" OwnerUserId="2381" LastActivityDate="2010-10-08T17:44:49.213" Title="What does &quot;Gromov Witten potential&quot; the &quot;potential&quot; mean" Tags="&lt;algebraic-geometry&gt;&lt;physics&gt;" CommentCount="0" />
  <row Id="6333" PostTypeId="2" ParentId="6328" CreationDate="2010-10-08T18:05:33.403" Score="4" ViewCount="" Body="&lt;p&gt;Here is a low-brow answer.  If you accept that the discrete version of the second derivative of a function $f$ is $f(x+1) - 2f(x) + f(x-1)$, then the discrete Laplacian, say in two dimensions, is $\Delta f(x, y) = \frac{f(x+1, y) + f(x-1, y) + f(x,y+1) + f(x,y-1)}{4} - f(x,y)$.  But the first term is just the transition probabilities of a random walk on $\mathbb{Z}^2$ where one moves to each of the four horizontally or vertically adjacent neighbors with equal probability.  A similar statement is true in $n$ dimensions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More generally one can define a &lt;a href=&quot;http://en.wikipedia.org/wiki/Discrete_Laplace_operator#On_Graphs&quot; rel=&quot;nofollow&quot;&gt;discrete Laplacian on any graph&lt;/a&gt; which mimics the usual Laplacian.  In fact there is &lt;a href=&quot;http://front.math.ucdavis.edu/math.PR/0001057&quot; rel=&quot;nofollow&quot;&gt;an entire textbook&lt;/a&gt; by Doyle and Snell dedicated to working out potential theory on graphs.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-10-08T18:05:33.403" />
  <row Id="6334" PostTypeId="2" ParentId="6328" CreationDate="2010-10-08T18:20:21.707" Score="4" ViewCount="" Body="&lt;p&gt;Some analogies between discrete and continuous Laplacians were discussed in MO:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://mathoverflow.net/questions/33602/what-is-a-reasonable-finitary-analogue-of-the-statement-that-harmonic-functions-a/&quot; rel=&quot;nofollow&quot;&gt;http://mathoverflow.net/questions/33602/what-is-a-reasonable-finitary-analogue-of-the-statement-that-harmonic-functions-a/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you search for &quot;discrete harmonic&quot; or &quot;discrete Laplacian&quot; there may be more there. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(added: the analogy is not as simple as the fact that the lattice Laplace operator converges to the continuous Laplacian as the spacing shrinks to zero.  There is a rotational symmetry of the continuous Laplacian which is an essential part of the geometric theory. For the analogies to be meaningful, at least some of the geometry of the continuum story should be be visible in the lattice or graph Laplacians, and for this eigenvalue inequalities, Hodge theory, vector bundles, zeta functions and other constructs are considered for graphs, in addition to the more easily perceived analogies with harmonic theory and Brownian motion.)&lt;/p&gt;&#xA;" OwnerUserId="467" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-10-09T01:44:36.650" LastActivityDate="2010-10-09T01:44:36.650" />
  <row Id="6335" PostTypeId="1" AcceptedAnswerId="6348" CreationDate="2010-10-08T18:23:03.220" Score="6" ViewCount="250" Body="&lt;p&gt;Using generating functions one can see that the $n^{th}$ Bell number, i.e., the number of all possible partitions of a set of $n$ elements, is equal to $E(X^n)$ where $X$ is a Poisson random variable with mean 1. Is there a way to explain this connection intuitively?&lt;/p&gt;&#xA;" OwnerUserId="404" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-10-08T22:57:39.597" LastActivityDate="2010-10-08T22:57:39.597" Title="Bell numbers and moments of the Poisson distribution" Tags="&lt;combinatorics&gt;&lt;probability&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="6336" PostTypeId="1" AcceptedAnswerId="6342" CreationDate="2010-10-08T18:32:36.663" Score="3" ViewCount="282" Body="&lt;p&gt;If not, is there only a limited rangle of polynomials for which the root can be found?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, if $u=x^{\frac{3}{2}}+x$, is $x$ expressible in terms of $u$?&lt;/p&gt;&#xA;" OwnerUserId="1958" LastActivityDate="2010-10-08T21:32:55.553" Title="Are all polynomials solvable?" Tags="&lt;polynomials&gt;" AnswerCount="3" CommentCount="4" FavoriteCount="1" />
  <row Id="6337" PostTypeId="2" ParentId="6336" CreationDate="2010-10-08T19:06:36.403" Score="3" ViewCount="" Body="&lt;p&gt;Yes, there is just a limited amount of polynomials for which we can find the exact, i.e. algebraic roots by some general formula (like it's possible for quadratic polynomials). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Particularly, the &lt;a href=&quot;http://en.wikipedia.org/wiki/Abel-Ruffini_theorem&quot; rel=&quot;nofollow&quot;&gt;Abel-Ruffini-theorem&lt;/a&gt; states that there is no such general solution for polynomials of degree five or higher in terms of radicals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This doesn't mean they don't have roots (in fact they &lt;a href=&quot;http://en.wikipedia.org/wiki/Fundamental_theorem_of_algebra&quot; rel=&quot;nofollow&quot;&gt;always&lt;/a&gt; have), but one has to write them down or approximate them in other ways.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;As to your equation (note that this is no longer a polynomial equation, which would require it's exponents to be integers, but a general power sum)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ u = x ^ {\frac{3}{2}} + x$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We substitute $k = x^{\frac{1}{2}}$ and now have to solve a cubic (polynomial) equation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ k^3 + k^2 - u = 0 $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For degree three, we can use a well-known but kinda unwieldy &lt;a href=&quot;http://en.wikipedia.org/wiki/Cardano_formulae#Roots_of_a_cubic_function&quot; rel=&quot;nofollow&quot;&gt;general solution formula&lt;/a&gt; and finally obtain zero to three results.&lt;/p&gt;&#xA;" OwnerUserId="917" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-10-08T20:26:59.870" LastActivityDate="2010-10-08T20:26:59.870" CommentCount="9" />
  <row Id="6338" PostTypeId="2" ParentId="6314" CreationDate="2010-10-08T19:07:30.290" Score="16" ViewCount="" Body="&lt;p&gt;The answer is no. In fact, as Steve D said, we have a theorem that holds for a wide class of spaces, which includes closed intervals, circles, balls and cubes. It was proved by Sierpinski in $1918$ $[1]$. You can find the proof in the book &quot;General Topology&quot; by Ryszard Engelking, but I'll post here since it's not easy to find it online. First a definition: a topological space is called a continuum if it is a compact connected Hausdorff space. The precise statement is the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Theorem (Sierpinski).&lt;/strong&gt; If a continuum $X$ has a countable cover $\{X_i\}_{i=1}^{\infty}$ by pairwise disjoint closed subsets, then at most one of the sets $X_i$ is non-empty.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to prove this we'll need the following lemmas:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Lemma $1$.&lt;/strong&gt; Let $X$ be a continuum. If $F$ is a non-trivial closed subset of $X$, then for every component $C$ of $F$ we have that $\text{Bd}(F) \cap C$ is non-empty.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Proof&lt;/em&gt;. Let $x_0$ be in $C$. Since $X$ is Hausdorff compact, quasicomponents coincide with components, so $C$ is the intersection of all open-closed sets in $F$ which contain $x_0$. Suppose that $C$ is disjoint from $\text{Bd}(F)$. Then, by compactness of $\text{Bd}(F)$, there is one open-closed set $A$ in $F$ containing $x_0$ and disjoint from $\text{Bd}(F)$. Take an open set $U$ such that $A = U \cap F$. Thus the equality $A \cap \text{Bd}(F) = \emptyset$ implies that $A = U \cap \text{Int}(F)$, so $A$ is open in $X$. But $A$ is also closed in $X$, and contains $x_0$, so $A=X$. But then $\text{Bd}(F) = \emptyset$, which is not possible since $F$ would be non-trivial open-closed in $X$. $\bullet$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Lemma $2$.&lt;/strong&gt; If a continuum $X$ is covered by pairwise disjoint closed sets $X_1, X_2, \ldots$ of which at least two are non-empty, then for every $i$ there exists a continuum $C \subseteq X$ such that $ C \cap X_i = \emptyset$ and at least two sets in the sequence $C \cap X_1, C \cap X_2, \ldots$ are non-empty.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Proof&lt;/em&gt;. If $X_i$ is empty then we can take $C = X$; thus we can assume that $X_i$ is non-empty. Take $j \ne i$ such that $X_j \ne \emptyset$. Since $X$ is Hausdorff compact, there are disjoint open sets $U,V \subseteq X$ satisfying $X_i \subseteq U$ and $X_j \subseteq V$. Let $x$ be a point of $X_j$ and $C$ the component of $x$ in the subspace $\overline{V}$. Clearly, $C$ is a continuum, $ C \cap X_i = \emptyset$ and $ C \cap X_j \ne \emptyset$. By the previous lemma, $C \cap \text{Bd}( \overline{V}) \ne \emptyset$ and since $X_j \subseteq \text{Int}(\overline{V})$, there exist a $k \ne j$ such that $C \cap X_k \ne \emptyset$. $\bullet$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we can prove the theorem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Proof&lt;/em&gt;. Assume that at least two of the sets $X_i$ are non-empty. From lemma $2$ it follows that there exists a decreasing sequence $C_1 \supseteq C_2 \ \supseteq \ldots$ of continua contained in $X$ such that $C_i \cap X_i = \emptyset$ and $C_i \ne \emptyset$ for $i=1,2, \ldots$ The first part implies that $\bigcap_{i=1}^{\infty} C_i = \emptyset$ and from the second part and compactness of $X$ it follows that $\bigcap_{i=1}^{\infty} C_i \ne \emptyset$. $\bullet$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Hausdorff hypothesis is fundamental. For example, consider $X$ a countable infinite set with the cofinite topology. Then $X$ is compact, connected and a $T_1$-space. However, we can write $X$ as a disjoint union of countable singletons, which are closed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$[1]$ Sierpinski, W: Un théorème sur les continus, Tohoku Math. J. $13$ ($1918$), $300-305$.&lt;/p&gt;&#xA;" OwnerUserId="329" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2010-11-15T17:07:32.963" LastActivityDate="2010-11-15T17:07:32.963" />
  <row Id="6339" PostTypeId="1" AcceptedAnswerId="6343" CreationDate="2010-10-08T19:24:14.447" Score="1" ViewCount="134" Body="&lt;p&gt;Suppose that $A \in {\cal S}_+^n$ is a symmetric positive semidefinite matrix. Let $B = {\rm sign}(A)$, where the sign is taken elementwise. Is the resulting matrix $B$ always positive semidefinite? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If not, under what conditions can we say that $B \in {\cal S}_+^n$ ?&lt;/p&gt;&#xA;" OwnerUserId="2384" LastActivityDate="2010-11-03T21:48:52.257" Title="Is a &quot;sign matrix&quot; obtained from a symmetric positive-semidefinite matrix itself symmetic positive-semidefinite?" Tags="&lt;linear-algebra&gt;&lt;matrices&gt;" AnswerCount="2" />
  <row Id="6340" PostTypeId="1" CreationDate="2010-10-08T19:38:41.347" Score="1" ViewCount="141" Body="&lt;p&gt;Just to clarify, I want to show that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If f is entire and $\int_{\mathbb{C}} |f|^p dxdy &amp;lt;\infty$, then f=0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think I can show that this is the case for p=2, but I'm not sure about other values of p...&lt;/p&gt;&#xA;" LastActivityDate="2010-10-14T07:13:28.553" Title="How do you show that an L^p entire (holomorphic on the complex plane) function is 0?" Tags="&lt;complex-analysis&gt;" AnswerCount="3" FavoriteCount="2" />
  <row Id="6341" PostTypeId="2" ParentId="6340" CreationDate="2010-10-08T19:46:34.637" Score="3" ViewCount="" Body="&lt;p&gt;Use Hölder's inequality and Cauchy's integral formula to show the function and its derivatives all vanishes at zero.&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2010-10-08T19:46:34.637" CommentCount="1" />
  <row Id="6342" PostTypeId="2" ParentId="6336" CreationDate="2010-10-08T20:03:04.637" Score="11" ViewCount="" Body="&lt;p&gt;The answer to your question depends entirely on what &quot;can be found&quot; means. I will note, before we start, that the expression you have with fractional exponents is not generally considered a polynomial: a polynomial in the variable $x$, with real coefficients, is an expression of the form&#xA;$$a_0 + a_1x + a_2x^2+ \cdots + a_nx^n$$&#xA;where $a_0,a_1,\ldots,a_n$ are real numbers, and $n$ is a nonnegative integer. Note that all the powers of the variable are &lt;em&gt;integral&lt;/em&gt; powers. If $a_n\neq 0$, we say the polynomial has degree $n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When $n\leq 4$, then there are formulas that express all the roots of the polynomial in terms of the coefficients ($a_0$, $a_1,\ldots,a_n$). For $n=1$, that is a polynomial $a_0+a_1x$ with $a_1\neq 0$, the solution is simply $x=-\frac{a_0}{a_1}$. For $n=2$, that is a polynomial $a_0 + a_1x + a_2x^2$ with $a_2\neq 0$, you get the well-known quadratic formula that gives the two roots:&#xA;$$ r_1 = \frac{-a_1 + \sqrt{a_1^2 - 4a_0a_2}}{2a_2}\quad\text{and}\quad r_2 = \frac{-a_1-\sqrt{a_1^2-4a_0a_2}}{2a_2}.$$&#xA;When $n=3$ and when $n=4$ there are also formulas to express all the roots of the polynomial in terms of the coefficients and using only the operations of addition, subtractions, multiplication, division, and root extraction. For $n=3$, these are the &lt;a href=&quot;http://en.wikipedia.org/wiki/Cardano_formulae&quot; rel=&quot;nofollow&quot;&gt;Cardano formulae&lt;/a&gt;, and for $n=4$ &lt;a href=&quot;http://en.wikipedia.org/wiki/Quartic_equation&quot; rel=&quot;nofollow&quot;&gt;the solution&lt;/a&gt; is due to Ferrari.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are no similar formulas when $n\geq 5$; this is the celebrated &lt;a href=&quot;http://en.wikipedia.org/wiki/Abel%E2%80%93Ruffini_theorem&quot; rel=&quot;nofollow&quot;&gt;Abel-Ruffini&lt;/a&gt; theorem. This does not mean we have no way of finding the roots, just that there is no formula that applies to all polynomials that gives the roots in terms of the coefficients, using only certain kinds of operations. For example, quintic equations (degree 5 polynomials) can be solved using other more complicated kinds of functions and operations (theta functions). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, there are plenty of methods for finding &lt;em&gt;approximate&lt;/em&gt; values for roots of polynomials, or in fact values that are &quot;as close as you want&quot; to the roots of the polynomial. For example, &lt;a href=&quot;http://en.wikipedia.org/wiki/Sturm%27s_theorem&quot; rel=&quot;nofollow&quot;&gt;Sturm's theorem&lt;/a&gt; can help you locate the roots approximately, and then you can combine it with &lt;a href=&quot;http://en.wikipedia.org/wiki/Newton%27s_method&quot; rel=&quot;nofollow&quot;&gt;Newton's method&lt;/a&gt; to find very good approximations to the roots of the polynomial. See for example the &lt;a href=&quot;http://en.wikipedia.org/wiki/Root-finding_algorithm&quot; rel=&quot;nofollow&quot;&gt;Wikipedia page on root-finding algorithms&lt;/a&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But of course, this turns on whether &quot;finding an approximation&quot; qualifies as &quot;can be found&quot; (or even if &quot;can be found, theoretically, given enough time to work&quot; qualifies as &quot;can be found&quot;). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For your &quot;also&quot; question, you want to express $x$ in terms of $u$ if $u=x^{3/2}+x$. This is slightly different from &quot;solving a polynomial&quot;; you are really trying to find a formula for the inverse of $x^{3/2}+x$ (assuming it has one, which it does because the function is increasing). You can turn it into a problem of solving a polynomial because if you let $z=x^{1/2}$, then you have $u=z^3+z^2$, which is equivalent to $z^3+z^2-u=0$; this is a cubic polynomial, so the roots of the polynomial can be expressed in terms of the coefficients (in this case, $1$, $1$, $0$, and $-u$) using Cardano's formulae. Then you would replace $x^{1/2}$ for $z$ in the case in which $z$ is nonnegative and real, and squaring gives you the answer. So, yes, it can be done.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-10-08T20:03:04.637" />
  <row Id="6343" PostTypeId="2" ParentId="6339" CreationDate="2010-10-08T20:20:42.663" Score="2" ViewCount="" Body="&lt;p&gt;A small symmetric perturbation of the identity matrix is positive definite, but the corresponding sign matrix need not be positive semidefinite.  For example, let $A$ be the 3-by-3 matrix with 1 on the diagonal, and, say, -1/100 in the other 6 entries.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I have nothing to say about your second question.)&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-10-08T20:20:42.663" />
  <row Id="6344" PostTypeId="1" AcceptedAnswerId="6347" CreationDate="2010-10-08T20:37:39.853" Score="3" ViewCount="129" Body="&lt;p&gt;I am looking for a good reference that could clearly explain finite field arithemtic.  Specifically I want to understand the importance and utility of the field generator polynomial and why it is important in multiplication.&lt;/p&gt;&#xA;" OwnerUserId="2390" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-10-08T22:55:58.883" LastActivityDate="2010-10-08T22:55:58.883" Title="Good Resources for Understanding Finite Field Arithemtic" Tags="&lt;number-theory&gt;&lt;reference-request&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="2" />
  <row Id="6345" PostTypeId="1" CreationDate="2010-10-08T20:59:52.720" Score="2" ViewCount="87" Body="&lt;p&gt;I know this question is easy, but for the life of me, I cannot remember what we call this thing. Googling for this has offered no help.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider an object $A$ and a second object $B$(let them be groups if you so choose). We wish to consider and action of $A$ on $B$. Moreover there is a subobject $C \hookrightarrow B$(subgroup) which is annihilated by the action of $A$, i.e. the restriction of the action of $A$ on $B$ to $C$ sends $C$ to the zero object(the zero in $B$ which corresponds to the trivial group).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I thought it would be the kernel of the action, but this term is reserved for something else(in particular those objects which fix everything).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think that this should be referred to as Torsion, and in particular, in the back of my mind, I keep thinking it is called the $A$-Torsion of $B$. But I am not sure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anyone know what this has been called in the past?&lt;/p&gt;&#xA;" OwnerUserId="16" LastEditorUserId="16" LastEditorDisplayName="" LastEditDate="2010-10-08T21:07:27.253" LastActivityDate="2010-11-07T21:44:17.970" Title="The name for a subobject(subgroup) which is annihilated by action." Tags="&lt;group-theory&gt;&lt;terminology&gt;&lt;category-theory&gt;" AnswerCount="1" CommentCount="6" />
  <row Id="6346" PostTypeId="2" ParentId="6344" CreationDate="2010-10-08T21:02:54.527" Score="0" ViewCount="" Body="&lt;p&gt;When I was an undergrad, I was trying to learn this material for the purpose of finite geometry. I found that even with little background in Abstract Algebra, that &lt;a href=&quot;http://www-ma4.upc.es/~simeon/IFG.pdf&quot; rel=&quot;nofollow&quot;&gt;this book&lt;/a&gt; by Simeon Ball and Zsuzsa Weiner, was very readable. The beginning in particular focuses on what you asked about.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Not to mention, it's free!&lt;/p&gt;&#xA;" OwnerUserId="16" LastActivityDate="2010-10-08T21:02:54.527" />
  <row Id="6347" PostTypeId="2" ParentId="6344" CreationDate="2010-10-08T21:09:44.537" Score="0" ViewCount="" Body="&lt;p&gt;One area where finite fields is widely applied is in the theory of error-correcting codes. A very nice book that treats these two subjects jointly is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Error-Correcting Codes and Finite Fields by Oliver Pretzel; Clarendon Press, Oxford, 1992.&lt;/p&gt;&#xA;" OwnerUserId="1369" LastActivityDate="2010-10-08T21:09:44.537" />
  <row Id="6348" PostTypeId="2" ParentId="6335" CreationDate="2010-10-08T21:14:34.140" Score="7" ViewCount="" Body="&lt;p&gt;One way may be to use these facts.  You can decide if this is intuitive enough or not. :)&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;$B_n = \sum_{k=0}^n S(n,k)$, where $S(n,k)$ is a Stirling number of the second kind. ($S(n,k)$ counts the number of ways to partition a set of $n$ elements into sets of size $k$.)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Stirling numbers of the second kind are used to convert ordinary powers to falling powers via $x^n = \sum_{k=0}^n x^{\underline{k}} S(n,k)$, where $x^{\underline n} = x(x-1)(x-2) \cdots (x-n+1)$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The factorial moments of a Poisson(1) distribution are all 1; i.e., $E[X^{\underline{n}}] = 1$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Putting them together yields &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$E[X^n] = \sum_{k=0}^n E[X^{\underline{k}}] S(n,k) = \sum_{k=0}^n  S(n,k) = B_n.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Facts 1 and 2 are well-known properties of the Bell and Stirling numbers.  Here is a quick proof of #3.  The second step is the definition of expected value, using the Poisson probability mass function.  The second-to-last step is the Maclaurin series expansion for $e^x$ evaluated at 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$E[X^{\underline{n}}] = E[X(X-1)(X-2) \cdots (X-n+1)] = \sum_{x=0}^{\infty} (x)(x-1) \cdots (x-n+1) \frac{e^{-1}}{x!}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$= \sum_{x=n}^{\infty} (x)(x-1) \cdots (x-n+1) \frac{e^{-1}}{x!} = \sum_{x=n}^{\infty} \frac{x!}{(x-n)!} \frac{e^{-1}}{x!} = \sum_{y=0}^{\infty} \frac{e^{-1}}{y!} = e/e = 1.$$&lt;/p&gt;&#xA;" OwnerUserId="2370" LastActivityDate="2010-10-08T21:14:34.140" CommentCount="1" />
  <row Id="6349" PostTypeId="2" ParentId="6323" CreationDate="2010-10-08T21:15:03.657" Score="0" ViewCount="" Body="&lt;p&gt;i'm not sure what the context is for your first question, but in reply to the second one, i can mention applications in statistics that come with the word 'equivariant' attached. [they involve 'homogeneous spaces', topological spaces on which act a group of isomorphisms of interest.]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a simple example might suffice to illustrate the idea. [the ensuing is motivated by changing scales of measurement, such as celsius to kelvin or celsius to fahrenheit.]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;suppose ${\mathbf X} = (X_1,\cdots,X_n)$, where the coordinates are independent observations from the same normal distribution $N(\mu, \sigma^2)$ and we want to estimate $\sigma$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;if the {$X_j$} are transformed in an affine way, it does not change the model of normality for the data, but it could change $\sigma$ [or $\mu$].&lt;/p&gt;&#xA;&#xA;&lt;p&gt;consider first a location shift $X\to X+a$, where $a$ is some constant. [the notation is meant to suggest that the shift transformation on $\mathbb R$ is extended to $\mathbb R^n$ by applying it to each coordinate of $\mathbf X$.]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;there is a corresponding transformation induced on the parameters, taking ($\mu, \sigma)\to (\mu+a, \sigma)$. as this transformation of the data does not change $\sigma$, it is reasonable to require of an estimator $\hat\sigma (X_1,\cdots,X_n)$ of $\sigma$ that it be invariant under shifts; that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\kern-57pt (1)\kern 57pt\hat\sigma (X_1+a,\cdots,X_n+a) = \hat\sigma (X_1,\cdots,X_n)\ \forall a\in\mathbb R.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;since (1) holds for all $a$, we can set $a = -\bar X$ in (1), where $\bar X$ is the sample mean. then (1) implies&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\kern-70pt (2)\kern 70pt\hat\sigma (X_1,\cdots,X_n) = \hat\sigma (X_1-\bar X,\cdots,X_n-\bar X).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so any estimator of $\sigma$ satisfying (1) can depend on the data only thru ${\mathbf D} := (X_1-\bar X,\cdots,X_n-\bar X)$, the [vector of] deviations from the sample mean. [${\mathbf D}$ is a &lt;em&gt;maximal&lt;/em&gt;$\kern2pt$ invariant for shift transformations in that it indexes the orbits in $\mathbb R^n$ for the shifts $(x_1,\cdots,x_n)\to (x_1+a,\cdots,x_n+a)$.]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;another group of affine transformations acting on the data is scale changes: $X\to X+b,\ b &gt; 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;in turn, we get the following induced transformations: $(\mu,\sigma)\to (b\mu,b\sigma)$ and&#xA;${\mathbf D}\to b{\mathbf D} = (b(X_1-\bar X),\cdots,b(X_n-\bar X))$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;since applying $b$ to the data rescales $\sigma$, this suggests [using (2)] that $\hat\sigma$ should be required to satisfy&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\kern-20pt (3)\kern 20pt\hat\sigma(b(X_1-\bar X),\cdots,b(X_n-\bar X)) = b\hat\sigma (X_1-\bar X,\cdots,X_n-\bar X), \forall b&gt;0.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(3) requires that $\hat\sigma$ be equivariant for scale changes of the data.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;letting $D^2 = \sum_1^n (X_j - \bar X)^2 = |{\mathbf D}|^2$ denote the sum of the squared deviations from $\bar X$, putting $b = 1/D$ in (3) gives&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\kern-55pt (4)\kern 55pt \hat\sigma (X_1,\cdots,X_n) = D\hat\sigma \big( \frac{X_1-\bar X}{D},\cdots,\frac{X_n-\bar X}{D}\big).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the vector ${\mathbf S} := \frac{\mathbf D}{|\mathbf D|} = \big( \frac{X_1-\bar X}{D},\cdots,\frac{X_n-\bar X}{D}\big)$ in (4) is a maximal invariant for the affine group on $\mathbb R$ extended [coordinatewise] to $\mathbb R^n$. note that its distribution does not depend on $(\mu,\sigma)$ [and it is thus an example of what is called an ancillary statistic]. in fact, $\mathbf S$ is also independent of $(\bar X, D)$. the multiplier of $D$ [$\hat\sigma ({\mathbf S})$] in (4) is usually chosen to be a constant [else $\hat\sigma$ is a randomized estimator].&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[it is also well-known that $\bar X$ and $D$ are independent in the normal case, so that&#xA;$\bar X, D$ and $\mathbf S$ are mutually independent. this is due to the fact that $\bar X$ and ${\mathbf D}$ are obtained from projections of $\mathbf X$ into two orthogonal subspaces,  and $\bf 1^\perp$, where  ${\bf 1}$ = (1,$\cdots$,1) $\in \mathbb R^n$. this implies that $\bar X$ and $\mathbf D$ are independent - and that $\mathbf D$ is spherically distributed in $\mathbf 1^\perp$, so that its length $D$ and orientation $\mathbf S$ are independent as well.]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;this argument extends to jointly estimating $(\mu,\sigma)$ and shows that the nonrandomized equivariant estimators are of the form $(\hat\mu, \hat\sigma) = (\bar X +bD), cD)$, where $b$ and $c$ are constants. the common choices $b=0$ and $c = \frac{1}{\sqrt{n-1}}$ give $\hat\mu = \bar X$ and &#xA;$\hat\sigma^2 := s^2 = \frac{\sum (X_j-\bar X)^2}{n-1}$, which are unbiased for $\mu$ and $\sigma^2$, respectively, [$s^2$ being the sample variance].&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the transformation group involved here is perhaps not so exciting. the situation becomes more interesting in the multivariate normal case. see, &lt;em&gt;e.g.&lt;/em&gt; &lt;a href=&quot;http://projecteuclid.org/DPubS/Repository/1.0/Disseminate?view=body&amp;amp;id=pdf_1&amp;amp;handle=euclid.aos/1176346733&quot; rel=&quot;nofollow&quot;&gt;eaton (1983) or muirhead (1982)&lt;/a&gt;  or &lt;a href=&quot;http://projecteuclid.org/DPubS?service=UI&amp;amp;version=1.0&amp;amp;verb=Display&amp;amp;handle=euclid.lnms/1196285102&quot; rel=&quot;nofollow&quot;&gt;eaton (2007)&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1547" LastActivityDate="2010-10-08T21:15:03.657" CommentCount="0" />
  <row Id="6350" PostTypeId="2" ParentId="6345" CreationDate="2010-10-08T21:20:12.277" Score="1" ViewCount="" Body="&lt;p&gt;in linear algebra, the subspace annihilated by a linear mapping $A$ is the nullity of $A$.&lt;/p&gt;&#xA;" OwnerUserId="1547" LastActivityDate="2010-10-08T21:20:12.277" CommentCount="3" />
  <row Id="6351" PostTypeId="2" ParentId="6344" CreationDate="2010-10-08T21:29:15.933" Score="0" ViewCount="" Body="&lt;p&gt;Try the book &lt;a href=&quot;http://www.ams.org/bookstore-getitem/isbn=0-8218-4418-0&quot; rel=&quot;nofollow&quot;&gt;Finite Fields and Applications&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="589" LastActivityDate="2010-10-08T21:29:15.933" />
  <row Id="6352" PostTypeId="2" ParentId="6336" CreationDate="2010-10-08T21:32:55.553" Score="2" ViewCount="" Body="&lt;p&gt;Your equation can be solved: &lt;a href=&quot;http://www.wolframalpha.com/input/?i=solve+u%3Dx%5E%283%2F2%29%2Bx+for+x&quot; rel=&quot;nofollow&quot;&gt;http://www.wolframalpha.com/input/?i=solve+u%3Dx%5E%283%2F2%29%2Bx+for+x&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The mathematics that decides when a polynomial is solvable by radicals is called &lt;a href=&quot;http://en.wikipedia.org/wiki/Galois_theory&quot; rel=&quot;nofollow&quot;&gt;Galois Theory&lt;/a&gt; and is one of the hallmarks of modern algebra.&lt;/p&gt;&#xA;" OwnerUserId="589" LastActivityDate="2010-10-08T21:32:55.553" />
  <row Id="6353" PostTypeId="1" CreationDate="2010-10-08T23:42:19.380" Score="2" ViewCount="163" Body="&lt;p&gt;Suppose $U$ and $V$ are 2 dimensional subspaces of $\mathbb{R}^4$. How do you determine the dimension of $U \cap V$? I know that $\text{dim}(U + V) = \text{dim}(U)+ \text{dim}(V) - \text{dim}(U \cap V)$. So it seems that $\text{dim}(U \cap V) = 0$.&lt;/p&gt;&#xA;" LastEditorUserId="664" LastEditorDisplayName="" LastEditDate="2010-10-09T03:10:39.053" LastActivityDate="2010-10-09T03:10:39.053" Title="How do you determine dimension" Tags="&lt;linear-algebra&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="6354" PostTypeId="2" ParentId="6353" CreationDate="2010-10-08T23:48:36.053" Score="1" ViewCount="" Body="&lt;p&gt;If you have B and B', basis of U and V respectively, you can check if the vectors of B are linearly independent with those of B'. If so, then $\dim(U\cap V) = 0$. If not, you may be able to write a new basis of $U+V$ eliminating the linearly dependent vectors. $\dim(U\cap V)$ is the amount of vectors you have eliminated from $U+V$'s basis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, $\dim(U+V) = \dim(U) + \dim(V) - \dim(U\cap V)$.&lt;/p&gt;&#xA;" OwnerUserId="1523" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-10-09T02:36:33.083" LastActivityDate="2010-10-09T02:36:33.083" CommentCount="3" />
  <row Id="6355" PostTypeId="1" CreationDate="2010-10-08T23:54:21.337" Score="0" ViewCount="50" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm looking for an algorithm, that can process an amount of (GPS-) positions and determine, if the actual position gets closer to a way point. &#xA;What is the &quot;best&quot; way to filter variability in the positions? So far, I've implemented a moving average over the last positions and compare that to the new one. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does it make sense to use a low-pass filter? Any other ideas?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" LastEditorUserId="1489" LastEditorDisplayName="" LastEditDate="2010-10-09T00:53:52.733" LastActivityDate="2010-10-09T00:53:52.733" Title="Algorithm for approaching a position" Tags="&lt;geometry&gt;&lt;statistics&gt;" CommentCount="2" />
  <row Id="6357" PostTypeId="1" AcceptedAnswerId="6362" CreationDate="2010-10-09T00:32:48.293" Score="4" ViewCount="309" Body="&lt;p&gt;Hello,&#xA;I'm currently studying the notion of a quotient topology. The one thing I'm having trouble with understanding is what we're actually doing to the points as we're identifying them. Say we have a [0,1] x [0,1] in E^2 (euclidean 2-space) with the subspace topology and we partition X into:&#xA;1) The set {(0,0),(1,0),(0,1),(1,1)} of four corner points&#xA;2) sets of pairs of points (x,0),(x,1) where 0&#xA;&#xA;&lt;p&gt;I wonder whether there's a any natural way to see how to &quot;place&quot; the points? This example is of course kinda easy, but just given this set, is there any way to figure out how it looks geometrically? If we'd have say the set consisting of all points from a fixed distance from a fixed point, we'd identify it as a circle, is there anything similar here?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And now, for the last question here:&#xA;Let B^n denote the unit ball in n-dimensiona euclidean space, and let S^(n-1) denote its boundary. Consider the partition of B^n which has as members:&#xA;1) the set S^(n-1);&#xA;b) the individual points of B^n-S^(n-1). &#xA;OK, when we now identify S^(n-1) it collapses into one point, right? Exactly what does it collapse into and where do we &quot;put it&quot;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope you can shed some light on this.&lt;/p&gt;&#xA;" OwnerUserId="2394" LastActivityDate="2010-10-10T02:35:15.503" Title="Quotient topologies and equivalence classes." Tags="&lt;topology&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="2" />
  <row Id="6358" PostTypeId="2" ParentId="6353" CreationDate="2010-10-09T02:26:58.037" Score="3" ViewCount="" Body="&lt;p&gt;First: it is incorrect, in general, to even write $\dim(U\cup V)$, because $U\cup V$ is &lt;em&gt;almost never&lt;/em&gt; a subspace: it is a subspace if and only if $U\subseteq V$ or $V\subseteq U$. Not being a subspace, it doesn't even make sense to talk about its dimension.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rather, what you probably meant is the correct equation&#xA;$$\dim(U+V) = \dim(U) + \dim(V) - \dim(U\cap V)$$&#xA;(though I prefer to put the intersection on the left hand side, because in that form it is valid even in the infinite dimensional case). Here, $U+V$ is the smallest subspace that contains $U$ and $V$, and it happens to equal the set of all vectors of the form $u+v$ with $u\in U$ and $v\in V$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The equation does not give you the full answer, because there are several situations that can occur: you could have $U$ and $V$ intersect trivially: this is what happens when $U+V=\mathbb{R}^4$. For an explicit example, you could have $U=\{(a,b,0,0):a,b\in\mathbb{R}\}$, and $V=\{(0,0,c,d) : c,d\in\mathbb{R}\}$. Here, $\dim(U\cap V)=0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or you could have that $U$ and $V$ intersect in a one-dimensional subspace (for example, take $U$ as above, but take $V=\{(0,b,c,0) : b,c\in\mathbb{R}\}$). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or you could have $U=V$, in which case the intersection has dimension $2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you can say is: if the spaces are distinct, then the intersection will either have dimension 1 or dimension 0. But that is all you can say with the given information.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-10-09T02:38:30.113" LastActivityDate="2010-10-09T02:38:30.113" />
  <row Id="6359" PostTypeId="2" ParentId="6357" CreationDate="2010-10-09T02:38:20.660" Score="10" ViewCount="" Body="&lt;p&gt;There are two things you can do with Topology: (1) try to imagine what's going on intuitively, and (2) write it down rigorously. Both are useful, necessary steps.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, if I understand your first example, you have the unit square $[0,1]^2 \subset \mathbb{R}^2$ with the subspace topology. Let me write the coordinates of its points $(\theta, z) \in [0,1]^2$, for psychological reasons that will become apparent in a minute.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the intuitive side, what you are doing is bending your square $[0,1]^2$ in such a way that  sides $\theta = 0$ and $\theta = 1$ become closer and closer... You put some glue on them, stick them... Ok: and what do you &quot;see&quot; now? -A cylinder, right?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Except for the fact that in real life you cannot stick just two segments -or it's really difficult-, you can try to do this experiment with a piece of paper and some glue.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the rigorous side, you may do the following: you recover that parametrization of the cylinder you have probably seen in some geometry or integral calculus course&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\varphi: [0,1]^2 \longrightarrow \mathbb{R}^3 \ , \qquad \varphi (\theta , z) = (\cos (2\pi\theta ), \sin (2\pi\theta ), z) \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This $\varphi$ is obviously continuous (&quot;obviously&quot; means: &quot;someone taught you this in some calculus course a year ago&quot;) and surjective, if we restrict the codomain to the cylinder&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;C = \left\{ (x,y,z) \in \mathbb{R}^3 \ \vert \ x^2 + y^2 = 1 \ , \ z \in [0,1]      \right\}  \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is, we are looking to $\varphi$ as a map $\varphi : [0,1]^2 \longrightarrow C$. Notice that also this way it is continuous with the subspace topologies on both sides.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now $\varphi $ is &lt;em&gt;almost&lt;/em&gt; injective too, except for those points $(0,z)$ and $(1,z)$ that have the same image &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\varphi (0,z) = (1,0,z) = \varphi (1,z) \ , \qquad \text{for all}\qquad z \in [0,1] \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to fix this lack of injectivity, you say: &quot;Ok, let's identify (&quot;identify&quot; is the picky, correct, way to say &quot;glue&quot; in Topology) each $(0,z)$ with its corresponding $(1,z)$ for all $z$.&quot; So, you are defining an equivalence relation $\sim$ among the points of the square $[0,1]^2$, the one generated by:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;(0,z) \sim (1,z) \qquad \text{for all} \qquad z \in [0,1] \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You call $X = [0,1]^2/\sim$ the space obtained from the square $[0,1]^2$ after gluing (sorry, &quot;identifying&quot;) each $(0,z)$ with its corresponding $(1,z)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's write $\widetilde{(\theta ,z)} \in X$ the equivalence classes produced by this equivalence relation. Notice that, if $\theta \neq 0, 1$, you just have $\widetilde{(\theta , z)} = \left\{ (\theta , z) \right\}$, but $\widetilde{(0,z)} = \widetilde{(1,z)} = \left\{ (0,z), (1,z)   \right\}$. That is, points of $X$ are &quot;the same&quot; as the points of $[0,1]^2$, except for those $(0,z)$ and $(1,z)$ that are now glued togehter (&quot;identified&quot;, yes).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In doing this, we also get a natural map (&quot;projection&quot;, &quot;identification&quot;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\pi : [0,1]^2 \longrightarrow X \ , \qquad \pi (\theta , z) = \widetilde{(\theta , z)} \ ,&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is continuous by definition of the quotient topology on $X$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Back to our $\varphi$: since it identifies the same points as $\sim$ does, $\varphi$ induces a well-defined map&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\widetilde{\varphi} : X \longrightarrow C \ , \qquad \widetilde{\varphi}\widetilde{(\theta , z)} = \varphi (\theta ,z) \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, since we have glued the points with the same image by $\varphi$,  this new $\widetilde{\varphi}$ is bijective, but it's also continuous, because of the universal property of the quotient topology and the fact that $\widetilde{\varphi} \circ \pi = \varphi$. Right?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Last step. A continuous bijective map doesn't need to be a homeomorphism, as you probably know: the inverse $\widetilde{\varphi}^{-1}$ is not necessarily continuous.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What could we do? Well, you can try to write down an explicit formula for $\widetilde{\varphi}^{-1}$ and check its continuity directly, but this is hard and painful, so nobody does it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instead, everybody resorts to the following marvelous, great, second to none trick (the best one you can buy in elementary Topology; btw, &quot;trick&quot; means &quot;proposition&quot;: you can prove it, of course): &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;$X$ is compact, because it is a quotient of a compact space (namely, $[0,1]^2$). $C$ is Hausdorff, because it is a subspace of a Hausdorff space (namely, $\mathbb{R}^3$). &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;The GTET (Greatest Trick in Elementary Topology)&lt;/strong&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Now, I have a continuous, bijective map $\widetilde{\varphi} : X \longrightarrow C$ between a compact space and a Hausdorff one. &lt;strong&gt;Hence&lt;/strong&gt; $\widetilde{\varphi}$ is a homeomorphism.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Isn't that fantastic?   :-)  (Don't forget it the next time you need to prove that some map is a homeomorphism: it will save your life.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, you can also prove rigorously your first intuition: your space is a cylinder, $X \cong C$.&lt;/p&gt;&#xA;" OwnerUserId="664" LastEditorUserId="664" LastEditorDisplayName="" LastEditDate="2010-10-09T23:12:28.000" LastActivityDate="2010-10-09T23:12:28.000" CommentCount="2" />
  <row Id="6360" PostTypeId="2" ParentId="6357" CreationDate="2010-10-09T02:38:51.353" Score="1" ViewCount="" Body="&lt;p&gt;Trying to place points in topology is risky.  The geometric intuition it gives can be very useful, but can also be very misleading.  Formally, a topology is defined by the open sets regardless of any capability to imagine them.  See &quot;Counterexamples in Topology&quot; for examples of the problems you face here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Collapsing points can change the shape of a space considerably.  I think your specific example of collapsing the boundary of $B^n$ gives you $S^n$ as you have a compact n-dimensional manifold without boundary in which every loop is contractible to a point, but I am willing to be corrected. &lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-10-09T02:38:51.353" CommentCount="4" />
  <row Id="6361" PostTypeId="2" ParentId="6324" CreationDate="2010-10-09T04:37:24.277" Score="3" ViewCount="" Body="&lt;p&gt;Equation $\rm\ 3^{2x}-2^{3y}=1\ $ is an instance of various special cases of &lt;a href=&quot;http://en.wikipedia.org/wiki/Catalans_conjecture&quot; rel=&quot;nofollow&quot;&gt;Catalan's Conjecture&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First,$\ $ making the specialization $\rm\ \ \: z,\:p^n = 3^x,2^{3y}\ $ below yields $\rm\ x = 1 = y\ $ as desired.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;LEMMA&lt;/strong&gt;$\ \ $ $\rm z^2 - p^n = 1\ \ \Rightarrow\ \ z,\:p^n = \:3\ ,\:2^3\ $  or $\ 2,\:3\ $ for $\rm\ \ z,\:p\:,n\in \mathbb N,\ \ p\: $ prime  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof&lt;/strong&gt; $\rm\ \ \ (z+1)\:(z-1)\: =\: p^n\ \  \Rightarrow\ \ z+1 = p^{\:j},\ \ z-1 = p^k\ $ for some $\rm\ j,\:k\in \mathbb N$   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad \:\Rightarrow\ \ \ \ 2\ =\ p^{\:j} - p^k\ =\ p^k\: (p^{\:j-k}-1) \ \Rightarrow\ p^k=2\ $ or $\rm\ p^k = 1 \ \Rightarrow\ \ldots$   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second, it's simply the special case $\rm\: X = 3^x,\ Y = 2^y\: $ of $\rm\ X^2 - Y^3 = 1\:,\: $ solved by Euler in 1738. Nowadays one can present this solution quite easily using elementary properties of $\rm\ \mathbb Z[\sqrt[3]{2}]\:$, e.g see p.44 of Metsankyla: &lt;a href=&quot;http://www.ams.org/journals/bull/2004-41-01/S0273-0979-03-00993-5/S0273-0979-03-00993-5.pdf&quot; rel=&quot;nofollow&quot;&gt;Catalan's Conjecture: another old diophantine problem solved&lt;/a&gt;. See also &lt;a href=&quot;http://mathoverflow.net/questions/39561/is-there-an-elementary-way-to-find-the-integer-solutions-to-x2-y31&quot; rel=&quot;nofollow&quot;&gt;this MO thread&lt;/a&gt; and &lt;a href=&quot;http://mathoverflow.net/questions/25674/integer-points-of-an-elliptic-curve&quot; rel=&quot;nofollow&quot;&gt;this MO thread&lt;/a&gt; and Schoof: &lt;a href=&quot;http://www.maths.mq.edu.au/~alf/Humid%20Summer/catalan.pdf&quot; rel=&quot;nofollow&quot;&gt;Catalan's Conjecture&lt;/a&gt;. Note also that Catalan equations are a special case of the theory of generalized Fermat (FLT) equations, e.g. see &lt;a href=&quot;http://www.math.mcgill.ca/~darmon/pub/Articles/Expository/09.Ribenboim/paper.pdf&quot; rel=&quot;nofollow&quot;&gt;Darmon's exposition&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-10-09T04:37:24.277" />
  <row Id="6362" PostTypeId="2" ParentId="6357" CreationDate="2010-10-09T08:30:59.550" Score="5" ViewCount="" Body="&lt;p&gt;As for your second example, Ross Milikan's intuition is right. Let's see how we can arrive to the conclusion that $B^n/S^{n-1} \cong S^n$ examining what happens in low dimensions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;For $n=1$&lt;/strong&gt;, the (closed) unit ball is the interval $B^1 = [-1,1]$ and its boundary the &quot;sphere&quot; $S^0 = \left\{ -1, 1\right\}$. Now, if we try to imagine what happens to $B^1$ when we glue (&quot;identify&quot;) the points $-1$ and $1$, we see the segment $[-1,1]$ bending, points $-1$ and $1$ approaching one to each other... Our glue provider has already put some of it on them, so when they meet, they stick together and we get... a circumference! So $B^1 / S^0 \cong S^1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ok, before going to higher dimensions: how can we prove this? Well, it turns out that, with some adaptations, we already know how to do it. First, notice that $B^1 \cong [0,1]$ and here $S^0 = \left\{ 0,1 \right\}$. So we resort again to a parametrization; this time, of the circumference&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\varphi : [0,1] \longrightarrow \mathbb{R}^2 \ ,\qquad \varphi (\theta ) = (\cos (2\pi\theta) , \sin (2\pi \theta) )&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and proceed along the lines of the former cylinder example: $\varphi$ identifies the points of the &quot;sphere&quot; $S^0$, $\varphi (0) = (1,0) = \varphi (1)$, so $\varphi$ induces a well-defined map&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\widetilde{\varphi} : B^1/S^0 \longrightarrow  S^1 \ , \qquad \widetilde{\varphi}(\widetilde{\theta}) = \varphi (\theta) \ ,&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is continuous and bijective. Now use the GTET (Greatest Trick of Elementary Topology) you already know to conclude that $\widetilde{\varphi} $ is a homeomorphism  $B^1/S^0 \cong S^1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So far, so good. What's next? &lt;strong&gt;Let's take a look at $n=2$.&lt;/strong&gt; The unit ball $B^2$ is now a disk and its boundary the circumference $S^1$. We put some glue all over $S^1$ and we bend the disk and at the same time shrink the circumference making it become smaller and smaller till it's just one point. What do we get? A sphere!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we may suspect that $B^2 / S^1 \cong S^2$. In order to prove it, we may try to reproduce our &quot;visual&quot; homeomorphism with formulae. Let's put our disk $B^2$ in $\mathbb{R}^3$ in the $xy$-plane, centered at $(0,0,0)$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;B^2 = \left\{ (x,y,z) \in \mathbb{R}^3 \ \vert \ x^2 + y^2 \leq 1 \ , \ z= 0  \right\} \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And we put our sphere centered at $(0,0, 1)$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;S^2 = \left\{ (x,y,z) \in \mathbb{R}^3 \ \vert \ x^2 + y^2 + (z-1)^2 = 1 \right\}  \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the South Pole is $(0,0,0)$ -the center of the disk $B^2$-, the North Pole is $(0,0,2)$ and the equator lies in the plane $z= 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Our &quot;visual&quot; homeomorphism bends the disk over the sphere and in doing this shrinks its boundary till it is identified with the North Pole of the sphere. So, let's try to do the same with a map like&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\varphi : B^2 \longrightarrow S^2 \ , \qquad \varphi (x,y) = (f(r)x , f(r)y, 2r) \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here $r= + \sqrt{x^2+y^2}$. Let's analyse what this kind of map is doing:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;When going from $r = 0$ to $r=1$, we want the points $\varphi (x,y)$ go up the sphere, from height $z=0$ (South Pole) to $z=2$ (North Pole). So $z$ has to climb up at double speed as $r$ goes from the center of the disk to its boundary. Hence we put that $2$.&lt;/li&gt;&#xA;&lt;li&gt;To make things easier, we want the projections of the points $\varphi (x,y)$ onto the $xy$-plane to lie in the same direction as the point of the disk $(x,y)$ they come from. So this projection should be something like $(\lambda x, \lambda y)$. At the same time this lambda will vary with the distance $r$ to the center of the disk in order that $\varphi (x,y)$ remains on the sphere. So we try this $(f(r)x , f(r)y)$.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;To find which $f(r)$ we need, we impose the condition that $\varphi (x,y) $ lies on the sphere:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;1 = x^2 + y^2 + (z-1)^2 = f(r)^2 r^2 + (2r-1)^2&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And we get $f(r) = 2 \sqrt{\frac{1-r}{r}}$. So our map is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\varphi (x,y) = &#xA;\begin{cases}&#xA;\left( 2 \sqrt{\frac{1-r}{r}}x , 2 \sqrt{\frac{1-r}{r}}y, 2r  \right) &amp;amp; \text{if}\ (x,y) \neq (0,0) \\&#xA;(0,0,0) &amp;amp; \text{if}\ (x,y) = (0,0) \ .&#xA;\end{cases}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Exercise.&lt;/strong&gt; Prove that $\varphi$ is continuous.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, we proceed much in the same way as in the cylinder example. By construction, $\varphi$ is clearly surjective. But not injective: all the points in the boundary $r=1$ of the disk  go to the North Pole. So we quotient out this boundary and, as before, $\varphi$ induces a bijective continuous map&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\widetilde{\varphi} : B^2/S^1 \longrightarrow S^2 \ , \qquad \widetilde{\varphi}\widetilde{(x,y)} = \varphi (x,y) \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We apply again our nuclear weapon GTET and we are done.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All right, &lt;strong&gt;what about higher dimensions?&lt;/strong&gt; -We've got no geometric or visual intuition there (at least, I've got none; if you think you can &quot;see&quot; what happens in, say, $\mathbb{R}^4$, congratulations, but just in case do a visit to your psychiatrist and tell him/her). Anyway, formulae allow us to go beyond our visual intuition. We put our ball and sphere inside $\mathbb{R}^{n+1}$ as before:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;B^n = \left\{ (x_1, \dots , x_n, x_{n+1}) \ \vert \ x_1^2 + \cdots + x_n^2 \leq 1 \ , \ x_{n+1} = 0 \right\}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;S^n = \left\{ (x_1, \dots , x_n, x_{n+1}) \ \vert \ x_1^2 + \cdots + x_n^2   + (x_{n+1}-1)^2 = 1 \right\} \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And we try some map of the same kind:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\varphi : B^n \longrightarrow S^n \ , \qquad \varphi (x_1, \dots , x_n ) = (f(r)x_1, \dots , f(r)x_n, 2r) \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Exercises.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Complete the details.&lt;/li&gt;&#xA;&lt;li&gt;This is not the same homeomorphism we previously found for $n=1$. Write this general formula for $n=1$.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="664" LastEditorUserId="664" LastEditorDisplayName="" LastEditDate="2010-10-10T02:35:15.503" LastActivityDate="2010-10-10T02:35:15.503" />
  <row Id="6363" PostTypeId="2" ParentId="6321" CreationDate="2010-10-09T08:40:23.810" Score="2" ViewCount="" Body="&lt;p&gt;Unless I am confused (and I hope someone will correct me!) you are describing a subset of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Liouville_number&quot; rel=&quot;nofollow&quot;&gt;Liouville numbers&lt;/a&gt; and they are all transcendental.  The point is - the numbers you describe are not rational, but are very very well approximated by rational numbers.  I think, also, that it suffices for $Z(k)$ to be super-quadratic (so its difference $Z(k+1) - Z(k)$ is super-linear).  The canonical example is discussed on the same webpage.&lt;/p&gt;&#xA;" OwnerUserId="1307" LastEditorUserId="1307" LastEditorDisplayName="" LastEditDate="2010-10-09T09:04:51.113" LastActivityDate="2010-10-09T09:04:51.113" CommentCount="5" />
  <row Id="6364" PostTypeId="1" CreationDate="2010-10-09T08:58:46.173" Score="6" ViewCount="296" Body="&lt;p&gt;In particular for two real numbers $a$ and $b$, I'd like to know if there are formulas for $\zeta (a+b)$ and $\zeta (a-b)$ as a function of $\zeta (a)$ and $\zeta (b)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The closest I could find online is a paper by Harry Yosh &quot;General Addition Formula for Meromorphic Functions Derived from Residue Theorem&quot; in some little known journal, but unfortunately I have no access to it and don't know if it would answer my question. Maybe this is well-known and I didn't search correctly...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any help appreciated, thanks!&lt;/p&gt;&#xA;" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-10-09T11:33:57.857" LastActivityDate="2010-12-01T18:55:59.120" Title="Are there addition formulas for the Riemann Zeta function?" Tags="&lt;number-theory&gt;&lt;special-functions&gt;&lt;riemann-zeta&gt;" AnswerCount="4" FavoriteCount="1" />
  <row Id="6365" PostTypeId="1" AcceptedAnswerId="6438" CreationDate="2010-10-09T09:08:01.233" Score="0" ViewCount="377" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can't seem to solve this one. I've tried Dirichlet's test but to no avail.&#xA;Any help is greatly appreciated.&lt;/p&gt;&#xA;" OwnerUserId="2399" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-10-09T11:36:27.880" LastActivityDate="2010-10-10T17:55:56.297" Title="Convergence of $\sum_{n=1}^{\infty} \frac{\sin (n x)}{2^n}$" Tags="&lt;sequences-and-series&gt;&lt;fourier-series&gt;" AnswerCount="2" />
  <row Id="6366" PostTypeId="2" ParentId="6365" CreationDate="2010-10-09T09:15:24.220" Score="6" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT:&lt;/strong&gt; To show absolute convergence use that $\sin$ is bounded by 1 (I assume that you want $x$ to take real values).&lt;/p&gt;&#xA;" OwnerUserId="367" LastActivityDate="2010-10-09T09:15:24.220" CommentCount="1" />
  <row Id="6367" PostTypeId="2" ParentId="6364" CreationDate="2010-10-09T09:35:57.320" Score="2" ViewCount="" Body="&lt;p&gt;I've never seen any such formulas. If there were any, the would most likely be listed in &lt;a href=&quot;http://functions.wolfram.com/ZetaFunctionsandPolylogarithms/Zeta/&quot; rel=&quot;nofollow&quot;&gt;the Zeta section at functions.wolfram.com&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2010-10-09T09:35:57.320" />
  <row Id="6368" PostTypeId="2" ParentId="6364" CreationDate="2010-10-09T09:46:50.070" Score="9" ViewCount="" Body="&lt;p&gt;Almost certainly, the answer is no. Consider $a$ with real part in the interval (1/2,1). The Riemann hypothesis states that $\zeta(a)\not=0$. If $\zeta(a)$ were a function of $\zeta(2+a)$ and $\zeta(2)$ then this would convert the RH into a statement about the values of $\zeta(s)$ on $5/2&amp;lt;\Re[s]&amp;lt;3$, which would be just too easy.&lt;/p&gt;&#xA;" OwnerUserId="1321" LastEditorUserId="1321" LastEditorDisplayName="" LastEditDate="2010-10-09T09:54:25.557" LastActivityDate="2010-10-09T09:54:25.557" />
  <row Id="6369" PostTypeId="1" AcceptedAnswerId="6373" CreationDate="2010-10-09T11:57:44.643" Score="1" ViewCount="123" Body="&lt;p&gt;How does one &lt;strike&gt;find the no of positive integers such that&lt;/strike&gt; find all &lt;em&gt;possible&lt;/em&gt; numbers such that $$\lfloor{\sqrt{n}\rfloor} \mid n$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What i did was to subsitute $n=t^{2}$ so that the equation becomes $\lfloor{t\rfloor} \mid t^{2}$ But this means that we want $t^{2} = k \lfloor{t\rfloor}$, where $k \in \mathbb{N}$. I don't really know what to do from here. By the way, this problem is in Apostol. &lt;/p&gt;&#xA;" LastActivityDate="2010-10-09T15:36:34.200" Title="Finding the Number of Positive integers such that $\lfloor{\sqrt{n}\rfloor} \mid n$" Tags="&lt;number-theory&gt;&lt;functions&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="6370" PostTypeId="1" AcceptedAnswerId="6374" CreationDate="2010-10-09T13:35:19.900" Score="5" ViewCount="351" Body="&lt;p&gt;A map $s : \mathbb{N} \to X$ is a &lt;em&gt;computable sequence&lt;/em&gt; in $(X,\nu_X)$ when there exists a computable map $f : \mathbb{N} \to \mathbb{N}$ such that $s(n) = \nu_X(f(n))$ for all $n \in \mathrm{dom}(\nu_X)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My best guess would be, &quot;A map s taking N onto X is a computable sequence in the ??? (X, nu??) when there exists a computable map f taking N onto N such that s at n equals ??? of f ??? for all n elements of the domain of nu ???.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am searching for a way to read it aloud that encodes all the elements of the sentence into speech.&lt;/p&gt;&#xA;" OwnerUserId="2163" LastActivityDate="2010-10-10T12:15:35.987" Title="How can I read this mathematical sentence aloud in English?" Tags="&lt;terminology&gt;" AnswerCount="4" CommentCount="3" FavoriteCount="1" />
  <row Id="6371" PostTypeId="1" AcceptedAnswerId="6375" CreationDate="2010-10-09T13:51:42.060" Score="2" ViewCount="211" Body="&lt;p&gt;In the UK the lottery uses numbers 1 to 49 and a total of six numbers are picked. It has been said may times that there is as much chance of numbers 1 2 3 4 5 6 to be picked as any other random combination.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's say that the first 3 numbers to come out are 1,2 and 3. What are the chances of a number between 1 and 10 coming out next Vs a number between 11 and 20? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are obviously less numbers between 1 and 10 now that we already lost 1 to 3, so surely the probability is that a number between 11 and 20 is more likely? In which case, the chances of a lottery selection of 1,2,3,4,5,6 is less likely than 2,12,21,28,32,47 for example...&lt;/p&gt;&#xA;" OwnerUserId="2400" LastActivityDate="2010-10-09T15:43:46.663" Title="Lottery probability" Tags="&lt;probability&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="6372" PostTypeId="2" ParentId="6370" CreationDate="2010-10-09T14:05:08.647" Score="4" ViewCount="" Body="&lt;p&gt;I would say &quot;A map s is a computable sequence when there exists a computable map f satisfying certain properties&quot; while writing down the property.&lt;/p&gt;&#xA;" LastActivityDate="2010-10-09T14:05:08.647" CommentCount="2" />
  <row Id="6373" PostTypeId="2" ParentId="6369" CreationDate="2010-10-09T14:18:30.247" Score="8" ViewCount="" Body="&lt;p&gt;I suppose you want to find all &lt;em&gt;possible&lt;/em&gt; numbers such that $\displaystyle [\sqrt{n}] \mid n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume $\displaystyle n$ is not a perfect square, then there must be some $\displaystyle k$ such that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle  k^2 &amp;lt; n &amp;lt; (k+1)^2$. We have that $\displaystyle k = [\sqrt{n}]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only numbers in the range $\displaystyle  k^2 &amp;lt; n &amp;lt; (k+1)^2$ which are divisible by $k$ are $\displaystyle k^2+k$ and $\displaystyle k^2+2k$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus the numbers $\displaystyle n$ such that $\displaystyle [\sqrt{n}] \mid n$ are of the form&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle k^2, k^2+k, k^2+2k$&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-10-09T14:18:30.247" CommentCount="1" />
  <row Id="6374" PostTypeId="2" ParentId="6370" CreationDate="2010-10-09T15:05:08.833" Score="11" ViewCount="" Body="&lt;p&gt;&quot;A map ess from en to ex is a computable sequence in ex nu-ex when there exists a computable map eff from en to en such that ess of en equals nu-ex of eff of en for all en in the domain of nu-ex.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The hyphens are meant to indicate that the pause between these syllables is shorter than between two separated words.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yes, I am not verbalizing that the X is subscripted, the ordered pair, or the difference between $n$ and $\mathbb{N}$. I assume you are talking about a situation where you are writing this statement on the board as you speak. Communicating any significant amount of math in a purely oral manner is incredibly hard because mathematical notation is much denser than ordinary spoken language; I find that I have to leave out some detail in order for the sentence to fit in my listener's head.&lt;/p&gt;&#xA;" OwnerUserId="448" LastActivityDate="2010-10-09T15:05:08.833" CommentCount="4" />
  <row Id="6375" PostTypeId="2" ParentId="6371" CreationDate="2010-10-09T15:16:21.393" Score="2" ViewCount="" Body="&lt;p&gt;The chances of the next number being between 1 and 10 is $\frac{7}{49}$, as opposed to the probability of it being between 11 and 20 being $\frac{10}{49}$. So, among other things, it is less likely that a lottery ticket will have only numbers between 1 and 10, as opposed to numbers between 1 and 20. However, that does not mean that a given ticket with numbers between 1 and 10 is less likely then a given ticket with numbers between 1 and 20. The fact that there are more tickets with numbers between 1 and 20 exactly cancels with the higher chance of getting such a ticket.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example:&#xA;You have a bag of pens; a red pen, and 99 blue pens numbered 1-99. The chance of you getting a red pen is only .01 whereas the chance of oyu getting a blue pen is .99, but the chance of getting any given blue pen (say, #42) is .99 / 99 = .01, which is the same as the chance of getting the red pen.  &lt;/p&gt;&#xA;" OwnerUserId="796" LastActivityDate="2010-10-09T15:16:21.393" />
  <row Id="6376" PostTypeId="1" CreationDate="2010-10-09T15:30:28.190" Score="3" ViewCount="150" Body="&lt;p&gt;For instance, say $G = \langle x , y \ | \ x^{12}y=yx^{18} \rangle$. I want to know what is the &lt;a href=&quot;http://groupprops.subwiki.org/wiki/Normal_subgroup_generated_by_a_subset&quot; rel=&quot;nofollow&quot;&gt;normal closure&lt;/a&gt; of $y$ in $G$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, what are the standard approaches to compute the normal closure of a subset of a finitely presented group? Are there algorithms?  &lt;/p&gt;&#xA;" OwnerUserId="2342" LastActivityDate="2010-10-13T13:23:04.123" Title="Normal closure in groups" Tags="&lt;group-theory&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="1" />
  <row Id="6377" PostTypeId="2" ParentId="6376" CreationDate="2010-10-09T15:42:48.880" Score="8" ViewCount="" Body="&lt;p&gt;You can compute the normal closure by computing the quotient, and then considering the kernel of the quotient homomorphism.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the example you gave, let $N$ be the normal closure of $y$ in $G$.  Then $G/N$ has presentation&#xA;$$&#xA;\langle x,y \mid x^{12}y = yx^{18},y=1\rangle&#xA;$$&#xA;This presentation reduces to $\langle x \mid x^{12} = x^{18}\rangle$, which is the same as $\langle x \mid x^6 = 1\rangle$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus $G/N$ is a cyclic group of order 6, and $N$ is the kernel of the homomorphism $G\to G/N$. In particular, the normal closure of $y$ consists of all words for which the total power of $x$ is a multiple of 6.&lt;/p&gt;&#xA;" OwnerUserId="1726" LastActivityDate="2010-10-09T15:42:48.880" CommentCount="1" />
  <row Id="6378" PostTypeId="2" ParentId="6371" CreationDate="2010-10-09T15:43:46.663" Score="2" ViewCount="" Body="&lt;p&gt;Having chosen 1,2,3 all possibilities (there are 46*45*44/6) that include 1,2,3 are equally likely and all possibilities that do not include 1,2,3 have probability zero.  The psychological error comes from grouping &quot;numbers less than 10&quot; together.  As Moron commented you could also group &quot;numbers containing a digit 2&quot;.  Each of these are legal groupings and the probability of selecting all numbers from within them can be calculated.  But it is still true that any particular combination is equally likely at the outset.  A draw of 13,18,29,33,36,45 looks &quot;random&quot; to our eye and we group it with all other &quot;random looking&quot; combinations.  From this point of view, a &quot;random looking&quot; combination is highly likely.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-10-09T15:43:46.663" />
  <row Id="6379" PostTypeId="2" ParentId="6295" CreationDate="2010-10-09T16:03:14.070" Score="0" ViewCount="" Body="&lt;p&gt;You could solve $y^2\equiv1$ (mod $pq$) with $y\not\equiv\pm1$ (mod $pq$)&#xA;and then take $x=7y$. :-)&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-10-09T16:03:14.070" CommunityOwnedDate="2010-10-09T16:03:14.070" />
  <row Id="6380" PostTypeId="1" AcceptedAnswerId="6439" CreationDate="2010-10-09T16:05:28.807" Score="7" ViewCount="130" Body="&lt;p&gt;Let $X$ be a Hausdorff space and $A$ a closed subspace. Suppose the inclusion $A \hookrightarrow X$ is a cofibration. Let $f, g: X \to Y$ be maps that agree on $A$ and which are homotopic. Are they homotopic relative to $A$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My motivation for asking this question comes from the following result:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Let $i: A \to X, j: A \to Y$ be cofibrations. Suppose $f: X \to Y$ is a map which makes the natural triangle commutative. Suppose $f$ is a homotopy equivalence. Then $f$ is a cofiber homotopy equivalence.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;On the other hand, I'm having trouble adapting the proof in Peter May's book of this to the question I asked. Nonetheless, the standard examples of pairs of maps which are homotopic but not with respect to which some subset on which they agree (say, the identity map of a comb space and its collapsing to a suitably chosen point), don't seem to involve NDR pairs. &lt;/p&gt;&#xA;" OwnerUserId="536" LastActivityDate="2010-10-10T17:25:49.837" Title="Are homotopic maps over a cofibration homotopic relative to the cofibration?" Tags="&lt;algebraic-topology&gt;" AnswerCount="1" FavoriteCount="2" />
  <row Id="6381" PostTypeId="1" CreationDate="2010-10-09T16:30:03.147" Score="2" ViewCount="169" Body="&lt;p&gt;Is there any relation between the homology of a space with local coefficients (in $\mathbb Q$ vector space) and the homology with coefficients in $\mathbb Q$? Thanks!&lt;/p&gt;&#xA;" OwnerUserId="1944" LastEditorUserId="8246" LastEditorDisplayName="" LastEditDate="2011-04-26T22:38:45.140" LastActivityDate="2011-04-26T22:38:45.140" Title="Homology with local coefficients" Tags="&lt;algebraic-topology&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" />
  <row Id="6382" PostTypeId="1" CreationDate="2010-10-09T17:00:17.160" Score="0" ViewCount="221" Body="&lt;p&gt;(Hint: notice that (n+1)(n!+1) = (n+1)!+n+1)&lt;/p&gt;&#xA;" OwnerUserId="2401" LastActivityDate="2010-10-09T21:17:49.833" Title="Let n be a positive integer, n! denotes the factorial of n. Let d = gcd(n! + 1, (n + 1)! + 1). Show that d divides n." Tags="&lt;linear-algebra&gt;&lt;homework&gt;" AnswerCount="2" CommentCount="8" />
  <row Id="6383" PostTypeId="1" AcceptedAnswerId="6386" CreationDate="2010-10-09T17:38:15.293" Score="1" ViewCount="116" Body="&lt;p&gt;Let's say I am distributing a bunch of encryption keys, where if two users have the same key, they can communicate.  I have 20 different keys, and I am giving each user 2.  Since encryption keys don't disappear when I hand them out, it can be considered 'with replacement', but I'm not giving the same user the same key twice.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to know what the probability that two users can communicate under these circumstances.  Doing a bit of the math from what I know, each user has 20 * 19 = 380 permutations of keys.  The order that they get the keys in doesn't matter, though, so it's actually 190 different combinations of keys.  So, two users have 190 possible combinations of four keys, and what is the probability that one of their two keys is in common.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That reasoning is about as far as I get.  How does one move forward from there?  Thanks!&lt;/p&gt;&#xA;" LastActivityDate="2010-10-10T00:53:05.037" Title="Basic probability... probability that two users share a key out of a keypool of 20?" Tags="&lt;probability&gt;&lt;statistics&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="6384" PostTypeId="2" ParentId="6382" CreationDate="2010-10-09T17:43:17.953" Score="2" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT:&lt;/strong&gt; Use the definition of GCD and the fact that $$(n+1)!+1 = n! \times (n+1) +1 = n! \cdot n + (n!+1)$$ We know that $d \mid (n!+1)$ since $d$ is the &lt;strong&gt;GCD&lt;/strong&gt;&lt;/p&gt;&#xA;" LastActivityDate="2010-10-09T17:43:17.953" CommentCount="3" />
  <row Id="6385" PostTypeId="2" ParentId="6382" CreationDate="2010-10-09T17:59:07.643" Score="4" ViewCount="" Body="&lt;p&gt;The given hint shows that $\rm\ n\ $ is an integral linear combination of $\rm\ n!+1\ $ and $\rm\ (n+1)! + 1\:,\: $ so $\rm\ n\ $ is divisible by all common divisors, including the GCD. In fact we can go further and show that the GCD = 1. Namely, since the GCD divides the coprime numbers $\rm\:n\:$ and $\rm\ n!+1\ $ it must be 1. Below is an alternate derivation using explicit gcd laws, along with an explicit Bezout linear representation of the GCD.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Putting $\rm\ k = n!+1\ $ below shows that the GCD equals $\rm\ gcd(n!+1,n) = 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\quad\ \  gcd(k,(n+1)k-n)\ =\ gcd(k,n)\ \ $ via $\rm\ \ n = (n+1)k - ((n+1)k-n)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\quad\quad$ recalling $\rm\quad\quad\ \  gcd(k,m)\ =\ gcd(k,\:jk\pm m)\ $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\quad\quad$ since if $\rm\ \ d|k\ $ then $\rm\ \: d|m\ \ \iff\ \: d\ |\ jk\pm m\quad\ \ $ Above $\rm \ j = n+1 $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact we can unwind the above to obtain an explicit Bezout linear representation of the GCD:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\quad\quad\quad\quad\quad \rm n\ =\ n!\ ((n+1)!+1) + (n-(n+1)!)\ (n!+1) $ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dividing the above through by $\rm\:n\:$ shows that the gcd is 1! $\ $ But, alas, I fear I've exclaimed too much ...&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-10-09T21:17:49.833" LastActivityDate="2010-10-09T21:17:49.833" CommentCount="5" />
  <row Id="6386" PostTypeId="2" ParentId="6383" CreationDate="2010-10-09T17:59:09.550" Score="1" ViewCount="" Body="&lt;p&gt;In this case, the probability is just $1$ minus the probability that they &lt;em&gt;can't&lt;/em&gt; communicate, i. e. that they've been given $4$ different keys.  The number of possible arrangements of keys among the two users is $190^2=36100$, since, as you noted, they each have $190$ possible pairs.  The number of ways they can be given four different keys in total is $\frac{20\times 19\times 18\times 17}{4}=29070$.  The numerator should be obvious, and we are dividing by $4$ to account for swaps in either Alice's or Bob's keys.  So the probability is $1-(29070/36100)=37/190$.&lt;/p&gt;&#xA;" OwnerUserId="2215" LastActivityDate="2010-10-09T17:59:09.550" CommentCount="5" />
  <row Id="6387" PostTypeId="1" AcceptedAnswerId="6388" CreationDate="2010-10-09T19:24:40.357" Score="3" ViewCount="332" Body="&lt;p&gt;So I was bored and decided to figure out the indefinite integral of the absolute value function, $|x|$. Using integration by parts ($u=|x|, dv=dx$, $dx = \text{sgn}(x)=\frac{|x|}{x}$), it can be shown that $\displaystyle\int |x| dx = \frac{x |x|}{2}+C$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I decided to take the integral again, finding that $\displaystyle\int\left(\int |x| dx \right) dx=\frac{x^2 |x|}{3}+C$. Continuing, I found the pattern in the title, that the $n$th indefinite integral of $|x|$ is $\displaystyle\frac{x^n |x|}{n+1}+C$. Is there a way to prove this general result?&lt;/p&gt;&#xA;" OwnerUserId="1562" LastEditorUserId="1562" LastEditorDisplayName="" LastEditDate="2010-10-09T19:30:07.987" LastActivityDate="2010-11-11T20:08:36.847" Title="How can I prove $\displaystyle\underbrace{\int \ldots \int}_{n} |x| dx = \frac{x^n |x|}{n+1}+C$?" Tags="&lt;calculus&gt;&lt;algebra&gt;" AnswerCount="3" CommentCount="4" FavoriteCount="3" />
  <row Id="6388" PostTypeId="2" ParentId="6387" CreationDate="2010-10-09T19:33:44.530" Score="4" ViewCount="" Body="&lt;p&gt;Yes, there is a way: Use &lt;a href=&quot;http://en.wikipedia.org/wiki/Mathematical_induction&quot; rel=&quot;nofollow&quot;&gt;Mathematical induction&lt;/a&gt;. I don't add more details because I think it's elementary, isn't it?&lt;/p&gt;&#xA;" OwnerUserId="367" LastActivityDate="2010-10-09T19:33:44.530" CommentCount="7" />
  <row Id="6389" PostTypeId="2" ParentId="6387" CreationDate="2010-10-09T19:57:06.080" Score="2" ViewCount="" Body="&lt;p&gt;The $C$ in the title of this question should be replaced by an arbitrary polynomial $p(x)$ of degree $\le n-1$. After all, we are talking of an $n$-fold integration here.&lt;/p&gt;&#xA;" OwnerUserId="1303" LastActivityDate="2010-10-09T19:57:06.080" CommentCount="2" />
  <row Id="6390" PostTypeId="2" ParentId="6364" CreationDate="2010-10-09T20:01:49.957" Score="6" ViewCount="" Body="&lt;p&gt;There is a functional equation relating values of $\zeta$ at $s$ and $(1-s)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there were formulas relating values along a continuous family of 1-dimensional curves, such as $x + y = C$, one would get a differential equation for $\zeta$, or some comparably strong constraint.  It is known that $\zeta$ does not satisfy any ODE's with algebraic functions as coefficients. Of course there could be gamma functions or other more complicated coefficients but prospects for this kind of additional structure in $\zeta$ seem dim.  There are no extra functional equations for finite field zetas, for example.&lt;/p&gt;&#xA;" OwnerUserId="467" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-12-01T18:55:59.120" LastActivityDate="2010-12-01T18:55:59.120" CommentCount="2" />
  <row Id="6391" PostTypeId="1" CreationDate="2010-10-09T20:09:05.127" Score="1" ViewCount="225" Body="&lt;p&gt;I'm looking for a fast algorithm to perform division of large numbers (by hand). Traditional long division just isn't fast enough for my needs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In most of these cases, I'm only looking for the modulus/remainder of the division, so if there is a fast algorithm specifically for this purpose, it would suffice. (But, if there is a way to get the quotient as well, it would also help).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;: Now that I think about it, the algorithm for synthetic division/synthetic substitution involves no division (and addition/multiplication is typically easier and faster than division). Is there perhaps a clever way to adapt this algorithm for fast numerical division?&lt;/p&gt;&#xA;" OwnerUserId="2403" LastEditorUserId="2403" LastEditorDisplayName="" LastEditDate="2010-10-10T01:46:25.713" LastActivityDate="2010-10-10T01:46:25.713" Title="Fast algorithm for modular division (residue)" Tags="&lt;number-theory&gt;&lt;algorithms&gt;&lt;modular-arithmetic&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="6392" PostTypeId="2" ParentId="6391" CreationDate="2010-10-09T20:22:37.470" Score="2" ViewCount="" Body="&lt;p&gt;The key idea is to make a good estimate of the quotient based on the most significant digits of the dividend and divisor. This is described in detail in the division algorithm presented in section 4.3.1 of Knuth, The art of computer programming, Volume 2, Seminumerical algorithms - the standard reference. Similar ideas (Lehmer at al.) are used in optimizing the computation of the remainder sequence in the Euclidean algorithm for the GCD, which is also described in Knuth's book, viz. p. 345, section 4.5.2.&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-10-09T20:22:37.470" />
  <row Id="6393" PostTypeId="1" AcceptedAnswerId="6396" CreationDate="2010-10-09T20:52:44.067" Score="3" ViewCount="486" Body="&lt;p&gt;I am trying to understand left/right cosets in group theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is the example in my text:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $G = \lbrace 1, a, b, c, d ,e \rbrace$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lets define the group operation $.$ by the following table, where the entry at row $x$ and column $y$ gives $x.y$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ex. $d.e = b$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;&amp;nbsp;&lt;b&gt;1&lt;/b&gt;&amp;nbsp;&lt;b&gt;a&lt;/b&gt;&amp;nbsp;&lt;b&gt;b&lt;/b&gt;&amp;nbsp;&lt;b&gt;c&lt;/b&gt;&amp;nbsp;&lt;b&gt;d&lt;/b&gt;&amp;nbsp;&lt;b&gt;e&lt;/b&gt;&lt;br /&gt;&#xA;&lt;b&gt;1&lt;/b&gt;&amp;nbsp;1&amp;nbsp;a&amp;nbsp;b&amp;nbsp;c&amp;nbsp;d&amp;nbsp;e&lt;br /&gt;&#xA;&lt;b&gt;a&lt;/b&gt;&amp;nbsp;a&amp;nbsp;b&amp;nbsp;1&amp;nbsp;d&amp;nbsp;e&amp;nbsp;c&lt;br /&gt;&#xA;&lt;b&gt;b&lt;/b&gt;&amp;nbsp;b&amp;nbsp;1&amp;nbsp;a&amp;nbsp;e&amp;nbsp;c&amp;nbsp;d&lt;br /&gt;&#xA;&lt;b&gt;c&lt;/b&gt;&amp;nbsp;c&amp;nbsp;e&amp;nbsp;d&amp;nbsp;1&amp;nbsp;b&amp;nbsp;a&lt;br /&gt;&#xA;&lt;b&gt;d&lt;/b&gt;&amp;nbsp;d&amp;nbsp;c&amp;nbsp;e&amp;nbsp;a&amp;nbsp;1&amp;nbsp;b&lt;br /&gt;&#xA;&lt;b&gt;e&lt;/b&gt;&amp;nbsp;e&amp;nbsp;d&amp;nbsp;c&amp;nbsp;b&amp;nbsp;a&amp;nbsp;1&#xA;&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is no problem i understand this. But then we get the left and right cosets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $G$ be a group and let $H \leq G$. A left coset of $H$ in $G$  ($G / H$) is a set of the form $gH = \lbrace gh : h \in H \rbrace$ for some $g \in G$. A right coset of $H$ in $G$ ($H$ \ $G$) is a set of the form $Hg = \lbrace hg : h \in H \rbrace$ for some $g \in G$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are some examples that i am trying to figure out how they are generated. I guess i just do not understand the theory fully. I would like a little help explaining and possible a few more examples.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\lbrace 1, a, b, c, d, e \rbrace / \lbrace 1, a, b \rbrace = \lbrace \lbrace1, a, b\rbrace , \lbrace c, d, e \rbrace\rbrace$&lt;br /&gt;&#xA;$\lbrace 1, a, b \rbrace$ \ $\lbrace 1, a, b, c, d, e\rbrace = \lbrace \lbrace 1, a, b \rbrace, \lbrace c, d, e \rbrace \rbrace$&lt;br /&gt;&#xA;$\lbrace 1, a, b, c, d, e \rbrace / \lbrace 1, c \rbrace = \lbrace \lbrace 1, c \rbrace, \lbrace a, d \rbrace, \lbrace b, e \rbrace \rbrace$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A few more examples which i want to figure out are:&lt;br /&gt;&#xA;$\lbrace 1, a, b, c, d, e \rbrace / \lbrace 1, e \rbrace = ?$&lt;br /&gt;&#xA;$\lbrace 1, d \rbrace$ \ $\lbrace 1, a, b, c, d, e \rbrace = ?$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="43" LastEditorUserId="43" LastEditorDisplayName="" LastEditDate="2010-10-09T21:39:27.093" LastActivityDate="2010-10-10T00:32:59.547" Title="Left/Right Cosets" Tags="&lt;group-theory&gt;" AnswerCount="5" CommentCount="2" />
  <row Id="6394" PostTypeId="2" ParentId="6370" CreationDate="2010-10-09T20:57:34.077" Score="0" ViewCount="" Body="&lt;p&gt;I would try to keep it really simple. Rather than verbalizing all the function and set symbols and subscripts, I would just say, &lt;strong&gt;&quot;A &lt;em&gt;computable sequence&lt;/em&gt; is the result of composing a &lt;em&gt;notation&lt;/em&gt; after a &lt;em&gt;computable function&lt;/em&gt;.&quot;&lt;/strong&gt; I assume you're talking Computable Analysis theory and by $\nu_X$ you intend a &lt;em&gt;notation&lt;/em&gt; $\nu_X:\subseteq \Sigma^* \to X$. Of course, this reading assumes your audience already knows what you mean by a &lt;em&gt;notation&lt;/em&gt; and a &lt;em&gt;computable function&lt;/em&gt;.&lt;/p&gt;&#xA;" LastActivityDate="2010-10-09T20:57:34.077" />
  <row Id="6395" PostTypeId="2" ParentId="6393" CreationDate="2010-10-09T21:30:19.507" Score="1" ViewCount="" Body="&lt;p&gt;gprime, thinking about cosets in the following context may help you out. Leg $G$ be $\mathbb{R}^3$, the group operation will be vector addition. Suppose you want to know the left cosets of $H$, where $H$ is a straight line through the origin, which as you know, is a subspace. You look at the sets &quot;$x+H$&quot;. Fix such an $x$, and see what happens when you add $x+h_1$, $x+h_2$, etc. where $h_1, h_2, ...$ are elements of $H$. You get a straight line that is parallel to $H$, but no longer passes through the origin. Which point does it surely pass through? For one, $x$, because, being $H$ a subspace of $\mathbb{R}^3$, $x+0$ is one of the elements of $x+H$. If you change the $x$ for $y$, another point not in the same straight line that is $x+H$, you'll obtain another straight line parallel to $H$. The left cosets (which in this case are the same as the right cosets, because vector addition is commutative) are thus all straight lines parallel to $H$. Since in this case the left and right cosets are the same, all these parallel straight lines form a group themselves, with the group operation defined simply as this: to add two straight lines, just take one point from on one of them, another point on the other, and add them, and all points that you obtain doing this will be a new straight line parallel to $H$. This new group is called the quotient group.&#xA;As your textbook says, however not all left cosets will be right cosets, in which case there's no defining a quotient group.&#xA;Well, I hope I didn't make it worse, but that particular example was helpful for me.&lt;/p&gt;&#xA;" OwnerUserId="1313" LastActivityDate="2010-10-09T21:30:19.507" />
  <row Id="6396" PostTypeId="2" ParentId="6393" CreationDate="2010-10-09T21:30:36.543" Score="7" ViewCount="" Body="&lt;p&gt;If your table is correct, it should be the symmetric group in three letters, with &lt;strong&gt;a&lt;/strong&gt; and &lt;strong&gt;b&lt;/strong&gt; the 3-cycles, and &lt;strong&gt;c&lt;/strong&gt;, &lt;strong&gt;d&lt;/strong&gt;, and &lt;strong&gt;e&lt;/strong&gt; the transpositions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strike&gt;Now. Your first equations is, alas, nonsense. The set that contains $1$, $a$, $b$, $c$, $d$, and $e$ is certainly not equal to the set whose elements are the sets $\{1,a,b\}$ and $\{c,d,e\}$. You are writing nonsense.  The rest are a bit better because you are looking at cosets.&lt;/strike&gt; Fixed in problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$G$ has six different subgroups: the trivial subgroup $\{1\}$; the whole subgroup $G$; one subgroup of order three, $H=\{1,a,b\}$, and three subgroups of order two: $K_1=\{1,c\}$, $K_2=\{1,d\}$, and $K_3=\{1,e\}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are the left cosets of $H$ in $G$? They are the sets $1H$, $aH$, $bH$, $cH$, $dH$, and $eH$. As it happens, $1H=aH=bH = H$, and $cH=dH=eH=\{c,d,e\}$. You can verify it exlicitly; for instance, &#xA;$$aH = \{ ah : h \in H\} = \{a1, aa, ab\} = \{a, b, 1\} = H$$&#xA;and&#xA;$$cH = \{ ch: h \in H\} = \{c1, ca, cb\} = \{c, e, d\}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The right costs of $H$ are the same as the left cosets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, what are the left cosets of $K_3=\{1,e\}$ in $G$? They are the sets $1K_3$, $aK_3$, $bK_3$, $cK_3$, $dK_3$, and $eK_3$. They are:&#xA;$$\begin{array}{rcl}&#xA;1K_3 &amp;amp; = &amp;amp; \{1k : k \in K_3\} = \{11, 1e\} = \{1,e\} = K_3\\&#xA;aK_3 &amp;amp; = &amp;amp; \{ak : k \in K_3\} = \{a1, ae\} = \{a, c\}.\\&#xA;bK_3 &amp;amp; = &amp;amp; \{bk : k \in K_3\} = \{b1, be\} = \{b, d\}.\\&#xA;cK_3 &amp;amp; = &amp;amp; \{ck : k \in K_3\} = \{c1, ce\} = \{c, a\} = aK_3.\\&#xA;dK_3 &amp;amp; = &amp;amp; \{dk : k \in K_3\} = \{d1, de\} = \{d, b\} = bK_3.\\&#xA;eK_3 &amp;amp; = &amp;amp; \{ek : k \in K_3\} = \{ei, ee\} = \{e, 1\} = K_3.&#xA;\end{array}$$&#xA;So there are three distinct left cosets, and they are $\{1,e\}$, $\{a,c\}$, and $\{b,d\}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are the right cosets of $\{1,d\}$ in $G$? They are $K_21$, $K_2a$, $K_2b$, $K_2c$, $K_2d$, and $K_2e$. Let us be a bit smarter this time, rather than computing them directly: we know that any two &lt;em&gt;distinct&lt;/em&gt; cosets are disjoint, and that the right coset of $x$ contains $x$. So $K_21$ contains both $1$ and $d$, hence must be equal to the right coset of $d$, $K_2d$; that is, $K_21=K_2d = \{1,d\}$. The right coset of $a$ contains $a$ and contains $da = c$, hence must be equal to the right coset of $c$, $K_2c$; indeed it is, as $dc = a$; so $K_2a=K_2c=\{a,c\}$. And the right coset of $b$, $K_2b$, contains $b$ and $db=e$, hence must equal the right coset of $e$, $K_2e$, which it does (since $de=b$ and $1e=1$). So $K_2b=K_2e=\{b,e\}$. So the three right cosets of $K_2$ in $G$ are $\{1,d\}$, $\{a,c\}$, and $\{b,e\}$. &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-10-09T22:06:13.987" LastActivityDate="2010-10-09T22:06:13.987" CommentCount="2" />
  <row Id="6397" PostTypeId="2" ParentId="6393" CreationDate="2010-10-09T21:40:12.660" Score="3" ViewCount="" Body="&lt;p&gt;The fundamental idea of a coset is that of taking a subgroup and &quot;translating&quot; it to fill up $G$.  A coset is just one such &quot;translate.&quot;  No coset but $H$ itself is a subgroup of $G$, since they don't contain $1$, but there is a lot of ground that can be covered by treating them like elements of a group or set instead.  Since the cardinality of the cosets is constant, they represent ways of dividing up $G$, and you get nice formulas like $|G|=|H|[H:G]$, where $[H:G]$ is called the &lt;strong&gt;index&lt;/strong&gt; of $H$ in $G$ and is equal to the number of cosets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In my opinion, the biggest use of cosets is to set up the idea of a &lt;strong&gt;quotient group&lt;/strong&gt;.  You may have seen the general idea -- formally dividing one mathematical structure by a substructure -- in other fields, like topology.  What you do in almost every case is add, to the laws defining your mathematical structure, another law that sets everything in the substructure equal to a point, or the identity, or zero.  Then you try to make everything else work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In group theory, in particular, the set $G/H$ (or $H\backslash G$) always has an induced group structure when $H$ is a normal subgroup.  And in fact, it's easy most of the time to just think of its elements as abstract and indivisible group elements.  But sometimes, especially when you're proving that this works, it's useful to think of them as equivalence classes instead, where the equivalence relation is of the form $a\sim b$ when $a=bh$ for some $h\in H$.  Of course, such an equivalence class is just the coset $bH$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll give one nice example with abelian groups before moving on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The quotient group $\mathbb{R}/\mathbb{Z}$ is the set of cosets of the form $x+\mathbb{Z}$, with $(x+\mathbb{Z})+(y+\mathbb{Z})=(x+y)+\mathbb{Z}$, which is just the induced addition.  But you can also think of the elements of the group as &lt;em&gt;numbers&lt;/em&gt; of the form $0\le x&amp;lt;1$, with addition being done &quot;modulo $\mathbb{Z}$&quot;: that is, $x+y$ in $\mathbb{R}/\mathbb{Z}$ is equal to the &lt;em&gt;fractional part&lt;/em&gt; of $x+y$ in $\mathbb{R}$.  So all we really have is a circle with an addition operation.  One place where this pops up is the group formed by numbers $e^{i\theta}$, the unit circle in $\mathbb{C}$, under multiplication.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Another, weirder example is $\mathbb{R}/\mathbb{Q}$.  I don't know much about it in terms of group theory, but it gives rise to a standard example of a non-measurable set in analysis.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another use of cosets will pop up when you study group actions.  I think this is just called the &lt;strong&gt;coset action&lt;/strong&gt;.  Basically, given a subgroup $H$, $G$ acts on $G/H$ by multiplication: $(g,kH)\mapsto gkH$.  Surprisingly, this action works differently than $G$'s normal multiplication on itself.  I don't know any good examples of this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let's look at the coset questions you mentioned.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, the line $\lbrace 1,a,b,c,d,e\rbrace=\lbrace\lbrace 1,a,b\rbrace,\lbrace c,d,e\rbrace\rbrace$ looks like a typo or something.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let's look at the &lt;em&gt;right&lt;/em&gt; cosets of $\lbrace 1,a,b\rbrace$ in $G$.  If you look at the top left corner of the multiplication table, you see that right multiplication by $a$ or $b$ just gives you the same set $\lbrace 1,a,b\rbrace$, though its elements are permuted.  Multiplying by $c,d,$ or $e$ gives something different.  We don't have to check that these all give the same coset, since the sizes of the cosets are identical.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the second example, you can look at the two columns labeled $1$ and $c$.  Then just scan down the rows: $a\mapsto\lbrace a,d\rbrace,b\mapsto\lbrace b,e\rbrace,c\mapsto\lbrace c,1\rbrace$, and so on.  These are all cosets, and identifying three distinct ones is easy.  They also show you something important: if, for example, $aH=\lbrace a,d\rbrace$, you know for a fact that $dH$ is the same set, simply because $H$ contains $1$.  Thus, the elements of a coset always give the same coset when multiplied by the subgroup.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let's do the same thing for the left cosets of $\lbrace 1,e\rbrace$.  Again, ignore everything but the $1$ and $e$ columns, and scan down.  We get $aH=\lbrace a,c\rbrace =cH,bH=\lbrace b,d\rbrace =dH$.  And, of course, $eH=H$ since $e\in H$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The right cosets of $\lbrace 1,d\rbrace$ will be similar, but you'll be using the rows $1$ and $d$ instead of columns.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Good luck, and I hope this was at least somewhat helpful!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, what book are you using?  I learned all this from Artin, but it's possible that you might be getting a different perspective if you have a different book.&lt;/p&gt;&#xA;" OwnerUserId="2215" LastActivityDate="2010-10-09T21:40:12.660" CommentCount="1" />
  <row Id="6398" PostTypeId="2" ParentId="6393" CreationDate="2010-10-09T21:45:19.707" Score="0" ViewCount="" Body="&lt;p&gt;Some general notes: the notation G/H is used to denote the set of left cosets of H in G. Note that this is a &lt;em&gt;set&lt;/em&gt; whose elements are also sets (in particular, they are the left cosets of H). Similarly, the notation H\G is used to denote the set of right cosets of H in G; again, this is a &lt;em&gt;set&lt;/em&gt; whose elements are sets. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, let's look at the first three equations. Unfortunately, the first one makes no sense (OP has since edited the first equation), so let's start with the second one. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notationally, as per what I wrote above, {1,a,b} \ {1,a,b,c,d,e} denotes the set of right cosets of {1,a,b}. To actually compute this, you take each element of H={1,a,b} and multiply on the right by each element of G={1,a,b,c,d,e}. So for example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;H*1 = {1,a,b}*1 = {1*1, a*1, b*1} = {1,a,b} = H&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But that's too easy. Try the element a:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;H*a = {1,a,b}*a = {1*a, a*a, b*a} = {a,b,1} = H, where we're using the multiplication table given to figure out each of these products. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proceed similarly for each of the remaining elements of G (which are b,c,d,e). That is, compute the cosets H*b, H*c, H*d, H*e in exactly the same way we did above. You should find that every coset is either H or {c,d,e}. Thus, H\G = { H , {c,d,e} }.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the third equation, you want to compute the &lt;em&gt;left&lt;/em&gt; cosets of the subset {1,c}. You do this exactly as I've done above, except remember to multiply on the &lt;em&gt;left&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the two examples you'd like to see worked out, we again proceed exactly as above. The only difference is that the first example is &lt;em&gt;left&lt;/em&gt; cosets of {1,e} while the second example is &lt;em&gt;right&lt;/em&gt; cosets of {1,d}.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I highly recommend working through these examples on your own. Working with cosets can be a little goofy at first, but doing the legwork on your own is really invaluable. &lt;/p&gt;&#xA;" OwnerUserId="2313" LastEditorUserId="2313" LastEditorDisplayName="" LastEditDate="2010-10-09T21:53:36.893" LastActivityDate="2010-10-09T21:53:36.893" CommentCount="1" />
  <row Id="6401" PostTypeId="1" CreationDate="2010-10-09T23:52:00.067" Score="0" ViewCount="51" Body="&lt;p&gt;The fist &lt;a href=&quot;http://en.wikipedia.org/wiki/Weyl_algebra&quot; rel=&quot;nofollow&quot;&gt;Weyl algebra&lt;/a&gt; $A_{1}$ can be represented by the following matrices:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$P=\left(\begin{array}{ccccc}&#xA;0 &amp;amp; 1 &amp;amp; 0 &amp;amp; 0 &amp;amp; \cdots\\&#xA;0 &amp;amp; 0 &amp;amp; 2 &amp;amp; 0 &amp;amp; \cdots\\&#xA;0 &amp;amp; 0 &amp;amp; 0 &amp;amp; 3 &amp;amp; \cdots\\&#xA;0 &amp;amp; 0 &amp;amp; 0 &amp;amp; 0 &amp;amp; \cdots\\&#xA;\vdots &amp;amp; \vdots &amp;amp; \vdots &amp;amp; \vdots &amp;amp; \ddots\end{array}\right)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$Q=\left(\begin{array}{ccccc}&#xA;0 &amp;amp; 0 &amp;amp; 0 &amp;amp; 0 &amp;amp; \cdots\\&#xA;1 &amp;amp; 0 &amp;amp; 0 &amp;amp; 0 &amp;amp; \cdots\\&#xA;0 &amp;amp; 1 &amp;amp; 0 &amp;amp; 0 &amp;amp; \cdots\\&#xA;0 &amp;amp; 0 &amp;amp; 1 &amp;amp; 0 &amp;amp; \cdots\\&#xA;\vdots &amp;amp; \vdots &amp;amp; \vdots &amp;amp; \vdots &amp;amp; \ddots\end{array}\right)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where P corresponds to the differential operator and Q corresponds to multiplication by x.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am interested in how this correspondence generalizes for $A_{n}$ because it seems like the language of matrices is in adequate for higher n but I'm afraid I don't know enough algebra to know what the appropriate tool to use would be.&lt;/p&gt;&#xA;" OwnerUserId="249" LastActivityDate="2010-10-09T23:52:00.067" Title="Generalizing the matrix formulation of the Weyl algebra" Tags="&lt;abstract-algebra&gt;&lt;modules&gt;" CommentCount="2" />
  <row Id="6402" PostTypeId="2" ParentId="6370" CreationDate="2010-10-10T00:28:35.237" Score="4" ViewCount="" Body="&lt;p&gt;David Speyer wrote how I would say it in practise, in a context where I was writing it on a black/whiteboard. Here's how I would say it in a pub or walking down the street:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Let's define a 'representation map for X' [&lt;em&gt;or your own preferred jargon&lt;/em&gt;] to be just some partial function nu, from the natural numbers to X. Then we can define &lt;em&gt;a computable sequence for that representation map nu&lt;/em&gt; to be any function s, from the natural numbers to X, which [is consistent with&amp;nbsp;/&amp;nbsp;agrees with&amp;nbsp;/&amp;nbsp;extends] the composition of nu with a computable function f on the natural numbers.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When using natural language, choose your nouns wisely and characterize them. Do you care about the ordered pair $(X,\nu_X)$, or really just the map $\nu_X$ (for which $X$ is just the background against which the idea is presented)? What is the &lt;em&gt;role&lt;/em&gt; of the partial map $\nu_X$ in the idea you are communicating? Do you care about the integers $f(n) \in \mathop{\mathrm{dom}}(\nu_X)$ over which you quantify, or really just the domain of the composite function $\nu_X \circ f$? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Identify the main characters in the synopsis of your play, and their roles: you will have a better chance of transporting the objects and morphisms of your idea faithfully to your interlocutors.&lt;/p&gt;&#xA;" OwnerUserId="439" LastEditorUserId="439" LastEditorDisplayName="" LastEditDate="2010-10-10T12:15:35.987" LastActivityDate="2010-10-10T12:15:35.987" CommentCount="2" />
  <row Id="6403" PostTypeId="2" ParentId="6393" CreationDate="2010-10-10T00:32:59.547" Score="1" ViewCount="" Body="&lt;p&gt;Cosets arise when you want to model the idea that certain elements of a group are &lt;em&gt;effectively equal&lt;/em&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To see this, rather that looking at one coset at a time it is best to think of all possible cosets of a subgroup. Then you will find that cosets partition a group $G$ into equivalence classes such that two elements of a class differ by an element of the subgroup $H$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example sometimes in number theory we do not want to distinguish two numbers that differ by a multiple of a given number $n$. Say $n=4$. Then let $G$ be the set of integers under addition and $H$ be the set of multiples of $4$. The cosets of $H$ are&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$H\cdot 0=\{\ldots,-4,0,4,8,\ldots\}$$&#xA;$$H\cdot 1=\{\ldots,-3,1,5,9,\ldots\}$$&#xA;$$H\cdot 2=\{\ldots,-2,2,6,10,\ldots\}$$&#xA;$$H\cdot 3=\{\ldots,-1,3,7,11,\ldots\}$$&#xA;$$H\cdot 4=\{\ldots,0,4,8,12,\ldots\}$$&#xA;$$H\cdot 5=\{\ldots,1,5,9,13,\ldots\}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that $H\cdot 0 =H\cdot 4$ and $H\cdot 1 =H\cdot 5$. In fact there are only $4$ distinct cosets, each corresponding to one congruence class of integers modulo $4$. If we care only about the remainder of an integer after division by $4$ then all the elements in a coset are equivalent and we can think of them as a single entity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To take another example, let $G$ be the group possible rotations of a point on a unit circle and we care only about where a point ends up on the circle after the rotation. Then let $H$ be rotations that are a multiple of $2\pi$. The cosets of $H$ will now be rotations which differ by multiples of $2\pi$ and which therefore have the same effect on the final position of the point rotated. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, two elements $g_1$ and $g_2$ of $G$ are defined to be equivalent ($g_1 \equiv g_2$) if $g_1 \cdot g_2^{-1} \in H$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can prove the following&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;For all $g$ in $G$, $g \equiv g$ since the identity element belongs to $H$ as it is a subgroup.&lt;/li&gt;&#xA;&lt;li&gt;If $g_1 \equiv g_2$ the $g_2 \equiv g_1$ since if an element belongs to $H$ then so does its inverse.&lt;/li&gt;&#xA;&lt;li&gt;If $g_1 \equiv g_2$ and $g_2 \equiv g_3$ then $g_1 \equiv g_3$ since $H$ is closed under multiplication.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;This show that $\equiv$ is a genuine equivalence relation. We define the equivalence class of an element $x$ of $G$ as&#xA;$$[x]=\{y \in G\mid x \equiv y\}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can show that&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$x \in [x]$&lt;/li&gt;&#xA;&lt;li&gt;For any $x,y \in G$, either $[x] \cap [y] =\emptyset$ or $[x]=[y]$.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;So the equivalence classes are either equal or disjoint and they cover $G$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, back to cosets. The equivalence classes we have defined above are the same as the right cosets. If $y \equiv x$ then $yx^{-1}=h$ for some $h \in H$ or $y=hx$. So $[x]=Hx$. If we had instead defined our equivalence relation by the condition $x^{-1}y \in H$ then we would have got the left cosets.&lt;/p&gt;&#xA;" OwnerUserId="1195" LastActivityDate="2010-10-10T00:32:59.547" />
  <row Id="6404" PostTypeId="2" ParentId="6383" CreationDate="2010-10-10T00:53:05.037" Score="3" ViewCount="" Body="&lt;p&gt;A solution which doesn't involve quite so large numbers:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;without loss of generality, say Alice gets keys 1 and 2. (Renumber the keys to make this the case.) Then there are ${20 \choose 2} = (20\times 19)/2 = 190$ possible pairs of keys that Bob can get. Of these, ${18 \choose 2} = (18 \times 17)/2 = 153$ include just keys chosen from ${ 3, \ldots, 20 }$. So the probability that Alice and Bob can't communicate is $153/190$, and the probability that they can is one minus this, or $37/190$.&lt;/p&gt;&#xA;" OwnerUserId="173" LastActivityDate="2010-10-10T00:53:05.037" />
  <row Id="6406" PostTypeId="1" CreationDate="2010-10-10T02:08:18.003" Score="4" ViewCount="193" Body="&lt;p&gt;Suppose we have two schemes $X, Y$ and a map $f: X\to Y$. Then we know that $Hom_X(f^*\mathcal{G}, \mathcal{F})\simeq Hom_Y(\mathcal{G}, f_*\mathcal{F})$, where $\mathcal{F}$ is an $\mathcal{O}_X$-module and $\mathcal{G}$ an $\mathcal{O}_Y$-module (and the Homs are in the category of $\mathcal{O}_X$-modules etc). This gives a natural map $f^* f_* \mathcal{F}\to \mathcal{F}$, just by setting $\mathcal{G}=f_* \mathcal{F}$ and looking at where the identity map goes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there any well-known conditions on the map or sheaves that give this is an isomorphism? For instance, I was looking through a book and saw that the map is surjective if $\mathcal{F}$ is a very ample invertible sheaf (and maybe some more hypothesis on the map and $X$ and $Y$ were assumed as well).&lt;/p&gt;&#xA;" OwnerUserId="2000" LastActivityDate="2010-10-10T02:08:18.003" Title="Pullback and Pushforward Isomorphism of Sheaves" Tags="&lt;algebraic-geometry&gt;" CommentCount="2" FavoriteCount="1" />
  <row Id="6407" PostTypeId="1" AcceptedAnswerId="6408" CreationDate="2010-10-10T02:18:25.580" Score="3" ViewCount="246" Body="&lt;p&gt;Suppose you would like to find the two tangent lines that support two given disks in the plane&#xA;to the same side.  Parameterizing the circles using $( \cos \theta, \sin \theta )$,&#xA;I find that ultimately I am computing the roots of a 4th-degree polynomial in $x = \cos \theta$&#xA;(two of whose roots are imaginary, and two real).&#xA;But I may not have formalized this in the most perspicacious manner.&#xA;Rather than detail my approach, let me just pose the general question of whether&#xA;there is a way to compute the tangent lines without effectively finding the roots of&#xA;a 4th-degree polynomial. With two solutions one might hope for a 2nd-degree polynomial.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Addendum&lt;/b&gt;.&#xA;I should have made clear that what I need are the coordinates of the two points of tangency for&#xA;two circles arbitrarily placed in the plane.  Nevertheless, both Ross's and Moron's solutions&#xA;can answer this version of the question as well.  With appropriate translation, scaling, and rotation, I can place the two circles so that one is a unit-radius circle centered on the origin,&#xA;and the other has $r \ge 1$ with its center on the $+x$ axis.  Then I am in the situation Moron drew.&#xA;Knowing the side lengths of the triangles yields $\sin \theta$ where $\theta$ is the&#xA;angle of the triangle on the $x$-axis, and from that I can compute the coordinates&#xA;of the points of tangencies&#xA;in terms of $\sin \theta$ and $\cos \theta$.  Much simpler than my brute-force calculation.&#xA;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="237" LastEditorUserId="237" LastEditorDisplayName="" LastEditDate="2010-10-10T18:41:14.210" LastActivityDate="2010-10-11T19:53:49.743" Title="Tangent to two disks: Roots of a 4th-degree polynomial?" Tags="&lt;geometry&gt;" AnswerCount="3" CommentCount="5" />
  <row Id="6408" PostTypeId="2" ParentId="6407" CreationDate="2010-10-10T02:59:52.143" Score="2" ViewCount="" Body="&lt;p&gt;I hope I understood the question correctly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the following (sorry for the crude image):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/XtwA4.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using similar triangles we get that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{D'+D}{D'} = \frac{R}{r}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can easily solve for $D'$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$r,R$ are the radii of the circles and and $D$ is the distance between the centres of the circles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$D'$ is the distance between the common point of the tangents to the center of the smaller circle.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-10-10T02:59:52.143" CommentCount="2" />
  <row Id="6409" PostTypeId="2" ParentId="6407" CreationDate="2010-10-10T03:00:09.543" Score="2" ViewCount="" Body="&lt;p&gt;In the wikipedia article&lt;a href=&quot;http://en.wikipedia.org/wiki/Tangent_lines_to_circles&quot; rel=&quot;nofollow&quot;&gt; tangent lines to circles&lt;/a&gt; there is a construction that appears to be second order.  You construct the tangents from the center of the smaller circle to a circle, centered on the center of the larger, with radius the difference of the radii.  Your tangents are parallel to these, offset by the smaller radius.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sorry for the broken link-an extra / got in the way&lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2010-10-11T19:53:49.743" LastActivityDate="2010-10-11T19:53:49.743" CommentCount="1" />
  <row Id="6410" PostTypeId="1" CreationDate="2010-10-10T03:27:52.747" Score="1" ViewCount="374" Body="&lt;p&gt;I know that a universal quantifier can be distributed over conjunction and not disjunction, but I'm having a hard time wrap my head around it. Why is this the case? Is there an example of a statement that would demonstrate this principle?&lt;/p&gt;&#xA;" OwnerUserId="2403" LastActivityDate="2010-10-11T01:16:04.503" Title="Distribution of Universal Quantifiers" Tags="&lt;logic&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="6411" PostTypeId="1" AcceptedAnswerId="6455" CreationDate="2010-10-10T04:06:38.390" Score="1" ViewCount="191" Body="&lt;p&gt;Let $(X,||\cdot||_X)$, $(Y,||\cdot||_Y)$ be a pair of normed linear spaces, and $(X \times Y, ||\cdot||_{X \times Y})$ the induced product space and norm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $(x,y)$ is an element in $X \times Y$, is it true that $||(x,y)||_{X \times Y} \lt \delta$ implies that $||x||_{X \times Y} \lt \delta$, where $(||x||_{X\times Y}=||(x,0)||_{X\times Y}$)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This has been true with every product norm that I have encountered, but I am not sure if it is true in general and/or if there is an easy way to show it. &lt;/p&gt;&#xA;" OwnerUserId="1736" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-10-10T19:50:45.393" LastActivityDate="2010-10-10T20:05:49.500" Title="Relation between the norms on X and Y and the induced product norm on X x Y" Tags="&lt;functional-analysis&gt;" AnswerCount="2" CommentCount="5" />
  <row Id="6412" PostTypeId="1" AcceptedAnswerId="6425" CreationDate="2010-10-10T04:25:00.083" Score="2" ViewCount="126" Body="&lt;p&gt;I'm guessing that this is a standard enough problem, but can't seem to find an intuitive solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have an undirected weighted graph $G = (V,E,f)$ with vertices $V$, edges $E$ and $f$ the weighting function for an edge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a set $\mathbf{I} = \{ I_1, ..., I_n \}$ such that for $I \in \mathbf{I}$, $I \subset V$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to perform a minimum-cut type operation on $G$ to produce $G'$ such that no pair of vertices in a set $I \in \mathbf{I}$ are reachable from each other, and such that the cut minimises the summation of edge weights removed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you can tell, I'm not a maths wizard, but I need to implement this for small graphs. (Preferably in the next two hours :/). Should I be looking at something like linear programming?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any pointers would be greatly appreciated. Thanks!&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt; Not a homework problem (unfortunately)... if I had had the type of education where this was a homework problem, I wouldn't need to ask now. :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The background is that I have a set of equivalence relations which hold between a set of things. I can perform the transitive closure of these equivalence relations to determine equivalence classes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, these relations are fallible. Afterwards, I can look at an equivalence class $\{A, B, C, D, E\}$ and say hey, crap!: thing $A$ and thing $B$ are different, and thing $B$ and $C$ are different. I need to break up this equivalence class.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, I look at the original non-transitive graph of equivalences. I want to do a minimum cut, but generally the number of vertices in the graph are small, and ties frequently occur. I want to avoid trivial decisions on what to cut. Turns out I can weight the original equivalence relations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given the original non-transitive weighted graph of equivalences, and the set of sets of things I know to be different, I want to perform some minimal-cut of the graph, and then reapply transitive closure over the sub-graphs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any help appreciated for a sleep-deprived Java hacker (even if you hate Java hackers).&lt;/p&gt;&#xA;" OwnerUserId="2331" LastEditorUserId="2331" LastEditorDisplayName="" LastEditDate="2010-10-10T04:51:32.270" LastActivityDate="2010-10-12T14:59:13.333" Title="Minimum cuts for weighted graphs" Tags="&lt;graph-theory&gt;" AnswerCount="2" CommentCount="5" />
  <row Id="6413" PostTypeId="2" ParentId="6411" CreationDate="2010-10-10T04:29:32.173" Score="0" ViewCount="" Body="&lt;p&gt;Your question doesn't show up well, but the wikipedia entry &lt;a href=&quot;http://en.wikipedia.org/wiki/Normed_vector_space&quot; rel=&quot;nofollow&quot;&gt;Normed vector space&lt;/a&gt; claims that if you have the product of a finite number of normed vector spaces, the sum of the norms in the individual spaces is a norm in the product.  &lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-10-10T04:29:32.173" CommentCount="5" />
  <row Id="6414" PostTypeId="1" CreationDate="2010-10-10T04:47:06.387" Score="10" ViewCount="452" Body="&lt;p&gt;...subject to the conditions that (i) the projection be smooth and that (ii) smooth sections correspond to smooth vector fields.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This homework problem is really bugging the hell out of me.  Of course it can be checked locally, so we'll look at an open neighborhood $U\in \mathbb{R}^n$ and check that $TU\cong U\times \mathbb{R}^n$ (where the isomorphism is as topological spaces with sheaves of functions that we declare to be smooth).  On the one hand, it's easy to see that $U \times \mathbb{R}^n$ satisfies these two conditions, since the projection is literally just a projection (in coordinates) and the correspondence is&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$X$ a vector field --&gt; $s = (x_1,\ldots, x_n, X(\partial/\partial x_1), \ldots, X(\partial/\partial x_n))$&lt;/li&gt;&#xA;&lt;li&gt;$s=(s_1,\ldots, s_n)$ a smooth section --&gt; $X = \sum_i s_i\cdot\partial/\partial x_i $.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;On the other hand, I'm having a lot of trouble showing that the obvious sheaf (which I'll denote $O_{TU}$) is the unique one with properties (i) and (ii).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, suppose $\mathcal{F}$ is another sheaf of functions on $U\times \mathbb{R}^n$.  Suppose that $f \in \mathcal{F}(V) \backslash \mathcal{O}_{TU}$ for some open set $V\subseteq U$.  There really doesn't seem to be any way to show that this violates property (i), because what it really means is that for any $g\in O_U$, $g \circ \pi \in O_{TU}$, and the hypothesis of this statement begins with a function on $U$, not $TU$.  On the other hand, I don't see any way to make a vector field on $U$ out of $f$, and (according to &lt;a href=&quot;http://mathoverflow.net/questions/37423/is-there-an-easy-way-to-describe-the-sheaf-of-smooth-functions-on-a-product-manif&quot; rel=&quot;nofollow&quot;&gt;this question&lt;/a&gt;) it feels likely that I can't necessarily get a section that detects the failure of $f$ to be $O_{TU}$-smooth.  Nor am I having much luck deriving a contradiction from $f \in O_{TU}(V) \backslash \mathcal{F}(V)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, I'd welcome any suggestions as to how I should proceed.&lt;/p&gt;&#xA;" OwnerUserId="2159" LastActivityDate="2011-03-19T10:33:41.717" Title="Why is the manifold structure on the tangent bundle unique?" Tags="&lt;differential-geometry&gt;" AnswerCount="2" CommentCount="5" FavoriteCount="4" />
  <row Id="6415" PostTypeId="1" CreationDate="2010-10-10T04:58:32.847" Score="0" ViewCount="245" Body="&lt;p&gt;First question: Let $a$, $b$, $c$ be distinct primes larger than $3$, and let  $x = abc$. Show that if $p$ is a positive integer and $p^2 \equiv 9 \pmod {x}$, then $p \equiv \pm 3 \pmod {a}$, $p \equiv \pm 3 \pmod{b}$, and $p \equiv \pm 3 \pmod{c}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second question: Using the fact that $1001 = (7)(11)(13)$ find the eight solutions to $y^2 \equiv 9 \pmod{1001}$ in the set $\{ 1, 2, 3,\ldots, 1000 \}$.&lt;/p&gt;&#xA;" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-28T14:58:41.230" LastActivityDate="2011-03-28T14:58:41.230" Title="Two little number theory questions" Tags="&lt;elementary-number-theory&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" />
  <row Id="6416" PostTypeId="2" ParentId="6410" CreationDate="2010-10-10T05:18:57.067" Score="-1" ViewCount="" Body="&lt;p&gt;Matt E, in his comment, gives two statements. He shows that a universal quantifier cannot be distributed over disjunction, because if it could, a false statement would follow logically from a true one. The statement &quot;A universal quantifier can be distributed over disjunction&quot; is itself a universal statement, and requires only one counterexample to &lt;strong&gt;disprove&lt;/strong&gt; it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;A universal quantifier can be distributed over conjunction&quot; is also a universal statement. It cannot be &lt;strong&gt;proved&lt;/strong&gt; by giving examples. There might be some counterexample we didn't think of. So how do we know it's true? -- Good question!&lt;/p&gt;&#xA;" OwnerUserId="1813" LastActivityDate="2010-10-10T05:18:57.067" />
  <row Id="6418" PostTypeId="2" ParentId="6364" CreationDate="2010-10-10T07:04:15.037" Score="1" ViewCount="" Body="&lt;p&gt;Many thanks for the answers, all three are interesting in their own right (I'm not registered so can't upvote nor comment, apologies).&lt;/p&gt;&#xA;" LastActivityDate="2010-10-10T07:04:15.037" />
  <row Id="6419" PostTypeId="2" ParentId="6415" CreationDate="2010-10-10T07:12:34.167" Score="1" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt;$\rm\ \ a|x|(p-3)(p+3)\ \Rightarrow \ a\:|\:p-3\ \: or\ \: a\:|\:p+3\:,\:$ since $\rm\:a\:$ is prime. Combine this with the Chinese remainder theorem to solve the second question. See also my &lt;a href=&quot;http://math.stackexchange.com/questions/6296&quot;&gt;answer here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-10-10T07:12:34.167" />
  <row Id="6421" PostTypeId="1" AcceptedAnswerId="6422" CreationDate="2010-10-10T09:06:47.943" Score="3" ViewCount="179" Body="&lt;p&gt;As I'm trying to improve my maths I've now reached the subject of differentiation of more general exponential functions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've already studied the so called &quot;chain rule&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, my book tells me:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$y=a^x \rightarrow y^{\prime}=a^x\ln(a)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, there's the following function:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f(x) = 10^{-x}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Applying the above rule, this would lead me to believe that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f^{\prime}(x) = 10^{-x} \ln(10)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the book tells me it's&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f^{\prime}(x) = -10^{-x} \ln(10)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why's that? They say they also used the chain rule, but if I use the chain rule I get something totally different:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f(x) = 10^{-x}$$&#xA;$$f^{\prime}(x) = -x(10)^{-x-1}\cdot(-1*1)$$&#xA;$$f^{\prime}(x) = x(10)^{-x-1}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What am I doing wrong in all these cases?&lt;/p&gt;&#xA;" OwnerUserId="1771" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-10-10T17:49:04.870" LastActivityDate="2010-10-10T22:15:16.057" Title="How does differentiation of &quot;more general exponential functions&quot; work" Tags="&lt;calculus&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="6422" PostTypeId="2" ParentId="6421" CreationDate="2010-10-10T09:28:26.897" Score="5" ViewCount="" Body="&lt;p&gt;Let $N: \mathbb{R} \to \mathbb{R}$ be the negation function, i.e. $N(x) = -x$ and let $g(x) = 10^x$. Then it should be clear that your $f = g \circ N$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f'(x) = (g \circ N)'(x) = g'(N(x))N'(x) = \log 10 \cdot 10^{-x} \cdot (-1)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS: Learn to use $\LaTeX$ instead of code. It takes up less space and looks prettier.&lt;/p&gt;&#xA;" OwnerUserId="1269" LastActivityDate="2010-10-10T09:28:26.897" CommentCount="4" />
  <row Id="6424" PostTypeId="1" CreationDate="2010-10-10T14:10:30.063" Score="2" ViewCount="107" Body="&lt;p&gt;The space of connection has no measure right?&lt;/p&gt;&#xA;" OwnerUserId="2381" LastActivityDate="2010-10-10T14:10:30.063" Title="How does physicist compute path integral in Chern-Simons theory" Tags="&lt;topology&gt;&lt;physics&gt;" FavoriteCount="1" />
  <row Id="6425" PostTypeId="2" ParentId="6412" CreationDate="2010-10-10T14:32:05.887" Score="1" ViewCount="" Body="&lt;p&gt;So you want to find a minimum cost cut so that for each $I \in \mathbf{I}$, $I$ is 'cut' by the set of edges?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general this is NP-Hard, even in the case when each $|I| = 2$ (i.e each set has exactly two elements). (I believe this is called as the multi-cut problem).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When $\mathbf{I} = \{ \{s, t\}\}$, this becomes the max-flow min-cut problem which is solvable in polynomial time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope that helps.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-10-10T14:32:05.887" CommentCount="1" />
  <row Id="6426" PostTypeId="2" ParentId="6407" CreationDate="2010-10-10T15:11:31.620" Score="2" ViewCount="" Body="&lt;p&gt;The degree 4 polynomial appears because the circles have 4, not 2, tangent lines algebraically: two internal and two external.  In the question, the phrase &quot;tangent lines that &lt;em&gt;support&lt;/em&gt;&quot; singles out the external tangents, which are lines of support, but this is not a distinction that can be made algebraically given a coordinatization of the problem.  The subtlety is that the equations of the circles contain only $r^2$ and $R^2$ while a choice of which pair of tangents lines counts as &quot;internal&quot; and which pair is &quot;external&quot; requires a choice of sign for $r/R$ to determine the relative orientation of the circles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The internal and external pairs of tangent lines to two circles can be constructed by ruler and compass, using one intersection of circles in each case, which in coordinates is solution of a quadratic equation.  The diagram posted in the other answer, with $(R-r)$, is the same as the one for the Euclidean construction.  The quartic with roots corresponding to the tangent lines is a product of two quadratics $Q_{int}(x)Q_{ext}(x)$ for the internal and external construction problems. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Among the 4 tangent lines the difference between the internal and external pairs can be seen algebraically, as choices of sign for different square roots.  This naturally partitions the set of 4 roots into two pairs.  However, a further labelling of the pairs as &quot;internal&quot; and &quot;external&quot; requires a further choice beyond a coordinatization of the plane.   This is because the coefficient field of the Cartesian equations defining the tangent lines is $k(d_x,d_y,R^2,r^2)$ where $d_x$ and $d_y$ are the $x,y$ distance between centers and $k$ is the underlying field of coordinates, but the selection of which pair of tangents is external relies on a value of $r/R$ or $(R-r)^2$ and this requires some sign choices for the radii, or equivalently, an extension of the coefficient field by some square roots. &lt;/p&gt;&#xA;" OwnerUserId="467" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-10-10T19:27:13.790" LastActivityDate="2010-10-10T19:27:13.790" />
  <row Id="6427" PostTypeId="1" AcceptedAnswerId="6429" CreationDate="2010-10-10T15:22:08.143" Score="2" ViewCount="127" Body="&lt;p&gt;Let's say I have a bag of coins, which contains 1 quarter, 2 dimes, 3 nickles and 4 pennies.  If I were to randomly pull out 3 coins, on average, how much money would I get?&lt;/p&gt;&#xA;" LastActivityDate="2010-10-10T17:51:32.187" Title="Average amount of money received from pulling three random coins" Tags="&lt;probability&gt;&lt;statistics&gt;&lt;average&gt;" AnswerCount="3" CommentCount="20" />
  <row Id="6428" PostTypeId="1" CreationDate="2010-10-10T15:34:36.010" Score="3" ViewCount="116" Body="&lt;p&gt;What I currently meet are all fiber bundles.&lt;/p&gt;&#xA;" OwnerUserId="2381" LastActivityDate="2010-11-09T17:59:53.627" Title="Is there anywhere we use a fibration which is not a fiber bundle" Tags="&lt;topology&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="6429" PostTypeId="2" ParentId="6427" CreationDate="2010-10-10T15:53:53.970" Score="5" ViewCount="" Body="&lt;p&gt;Each coin is pulled 30% of the time.  So on average you will get 30% of 64 cents=19.2 cents&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-10-10T15:53:53.970" CommentCount="2" />
  <row Id="6430" PostTypeId="1" AcceptedAnswerId="6434" CreationDate="2010-10-10T16:01:22.007" Score="0" ViewCount="117" Body="&lt;p&gt;The vector space $V(n,d)$ has dimension $n$ and the vector coordinates are from set &#xA;$\Sigma$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$d:=|\Sigma|$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How many vectors can the vector space contain? Why?&lt;/p&gt;&#xA;" OwnerUserId="2411" LastEditorUserId="2411" LastEditorDisplayName="" LastEditDate="2010-10-10T16:14:12.500" LastActivityDate="2010-10-10T17:01:28.767" Title="Maximal number of vectors in vector space" Tags="&lt;linear-algebra&gt;&lt;vector-space&gt;" AnswerCount="2" CommentCount="5" />
  <row Id="6431" PostTypeId="1" AcceptedAnswerId="6432" CreationDate="2010-10-10T16:11:14.810" Score="2" ViewCount="195" Body="&lt;p&gt;While searching a question about fibre bundles, which was asked here, i got directed to Vector bundles. I noticed this word &quot;Hairy Ball&quot; which sounded eccentric and made a search at Wikipedia. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How is the hairy ball theorem related to this statement: You can't comb the hair on a coconut.&lt;/p&gt;&#xA;" LastActivityDate="2010-10-10T17:02:03.897" Title="Hairy Ball theorem and its applications." Tags="&lt;soft-question&gt;&lt;topology&gt;&lt;algebraic-topology&gt;" AnswerCount="2" />
  <row Id="6432" PostTypeId="2" ParentId="6431" CreationDate="2010-10-10T16:30:25.430" Score="3" ViewCount="" Body="&lt;p&gt;I think the picture at the top right of the wikipedia page does a pretty good job at visualization of this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The idea is that if you think of each hair on a coconut as a vector with base at a point on the sphere, then we have a notion of what it means for these to change continuously. So combing the hairs is essentially the analogy for making all the hairs lie tangent to the sphere and for them to vary continuously. This cannot be done, since if you have a continuous vector field on the sphere, it must be zero somewhere. The way the picture shows this is that you get hairs that point outward or &quot;normal&quot; to the sphere (which I think is not a really good way to think about the theorem) and hence have not been combed. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice that you can't just comb that one part down, or else it won't be continuous. There will be some &quot;overlapping&quot;.&lt;/p&gt;&#xA;" OwnerUserId="2000" LastActivityDate="2010-10-10T16:30:25.430" />
  <row Id="6433" PostTypeId="2" ParentId="6430" CreationDate="2010-10-10T16:39:50.107" Score="1" ViewCount="" Body="&lt;p&gt;V(n,d) contains d&lt;sup&gt;n&lt;/sup&gt; vectors. As you know, each vector in V(n,d) can be described as an n-tuple (v&lt;sub&gt;1&lt;/sub&gt;,&amp;nbsp;...,&amp;nbsp;v&lt;sub&gt;n&lt;/sub&gt;) where each coefficient v&lt;sub&gt;j&lt;/sub&gt; is drawn from the coefficient-set &amp;Sigma;. It is straightforward to show that the number of such tuples is |&amp;Sigma;|&lt;sup&gt;n&lt;/sup&gt;&amp;nbsp;=&amp;nbsp;d&lt;sup&gt;n&lt;/sup&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular: if &amp;Sigma; is a finite field (such as the integers modulo 2), then the vector space has finitely many elements. Otherwise, it has infinitely many elements.&lt;/p&gt;&#xA;" OwnerUserId="439" LastActivityDate="2010-10-10T16:39:50.107" />
  <row Id="6434" PostTypeId="2" ParentId="6430" CreationDate="2010-10-10T16:55:04.357" Score="2" ViewCount="" Body="&lt;p&gt;If $\rm V$ is an $\rm n$-dimensional vector space over a field $\rm F$ then it has cardinality $\rm |V| = |F|^n$ if both $\rm |F|$ and $\rm n$ are finite; otherwise $\rm |V| = n\:|F| = max(n,|F|)\:$, as follows from basic properties of cardinal arithmetic. For example, this implies that the dimension of $\mathbb R$ over $\mathbb Q\:$ is $\rm |\mathbb R| &gt; |\mathbb Q| = |\mathbb N|\:$.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-10-10T17:01:28.767" LastActivityDate="2010-10-10T17:01:28.767" />
  <row Id="6435" PostTypeId="2" ParentId="6431" CreationDate="2010-10-10T17:02:03.897" Score="3" ViewCount="" Body="&lt;p&gt;More of a comment than an answer, but it didn't fit:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, the hairy ball theorem works on any even-dimensional sphere.  However, odd-dimensional spheres &lt;em&gt;do&lt;/em&gt; admit nonvanishing vector fields: if you use the usual embedding of $S^{2k-1}$ in $\mathbb{R}^{2k}$, then at the point $(x_1,y_1,\ldots,x_k,y_k)$ you can put the (nonzero) tangent vector $(y_1,-x_1,\ldots,y_k,-x_k)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The natural generalization is: How many linearly independent tangent vector fields can exist on $S^{n-1}$?  As it turns out, the answer only depends on how many factors of 2 there are in $n$.  So e.g. $S^3$ and $S^{35}$ both admit exactly 3 linearly independent tangent vector fields, because $3+1=2^2$ and $35+1=2^2\cdot 9$.  This is crazy!&lt;/p&gt;&#xA;" OwnerUserId="2159" LastActivityDate="2010-10-10T17:02:03.897" CommentCount="2" />
  <row Id="6436" PostTypeId="2" ParentId="6428" CreationDate="2010-10-10T17:05:05.237" Score="0" ViewCount="" Body="&lt;p&gt;I think that most examples of fibrations that you can think of off the top of your head are going to be fiber bundles (although this is probably only because of my limited experience).  On the other hand, it's much easier to just prove that things are fibrations, because the definition is much weaker.  As far as homotopy theory is concerned, this is going to be the same thing.&lt;/p&gt;&#xA;" OwnerUserId="2159" LastActivityDate="2010-10-10T17:05:05.237" />
  <row Id="6437" PostTypeId="1" CreationDate="2010-10-10T17:22:56.677" Score="1" ViewCount="301" Body="&lt;p&gt;Five microcomputers are to be connected to three printers.How many connections are&#xA;necessary between computers and printers in order to ensure that whenever any three&#xA;computers require a printer the printers are available?&lt;/p&gt;&#xA;" OwnerUserId="1337" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-11-09T21:14:25.380" LastActivityDate="2010-11-09T21:14:25.380" Title="Pigeon Hole Principle" Tags="&lt;discrete-mathematics&gt;" AnswerCount="2" CommentCount="5" />
  <row Id="6438" PostTypeId="2" ParentId="6365" CreationDate="2010-10-10T17:24:51.910" Score="8" ViewCount="" Body="&lt;p&gt;If you want, you can actually calculate what it converges to by appealing to complex numbers.  The sum is the imaginary part of &#xA;$$&#xA;\sum \frac{e^{nix}}{2^n} = \sum \left(\frac{e^{ix}}{2}\right)^n = \frac{2}{2-e^{ix}},&#xA;$$&#xA;so your sum converges to $\frac{2 \sin x}{5 - 4\cos x}$.&lt;/p&gt;&#xA;" OwnerUserId="848" LastEditorUserId="848" LastEditorDisplayName="" LastEditDate="2010-10-10T17:55:56.297" LastActivityDate="2010-10-10T17:55:56.297" />
  <row Id="6439" PostTypeId="2" ParentId="6380" CreationDate="2010-10-10T17:25:49.837" Score="2" ViewCount="" Body="&lt;p&gt;I think the machinery of obstruction theory deals with the special case where the spaces are  skeleta of a CW-complex.  Here's the setup (I'm basically just copying from pp. 6-7 of Mosher &amp;amp; Tangora here):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $Y$ be simply-connected for simplicity.  First, let $B$ be a complex and $A$ be a subcomplex.  Let $f:A\cup B^n\rightarrow Y$.  Then we get an &lt;em&gt;obstruction cochain&lt;/em&gt; $c(f)\in C^{n+1}(B,A;\pi_n(Y))$ (i.e. a function on relative $(n+1)$-cells with values in $\pi_n(Y)$).  Similarly, let $K$ be a complex; then for any two maps $f,g:K\rightarrow Y$ that agree on $K^{n-1}$, we similarly get a &lt;em&gt;difference cochain&lt;/em&gt; $d(f,g)\in C^n(K;\pi_n(Y))$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are the two results.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Theorem: There is a map $g:A\cup B^{n+1}\rightarrow Y$ agreeing with $f$ on $A\cup B^{n-1}$ iff $[c(f)]=0 \in H^{n+1}(B,A;\pi_n(Y))$.&lt;/li&gt;&#xA;&lt;li&gt;Theorem: The restrictions of $f$ and $g$ to $X^n$ are homotopic rel $K^{n-1}$ iff $d(f,g)=0 \in C^n(K;\pi_n(Y))$.  They are homotopic rel $K^{n-2}$ iff $[d(f,g)]=0 \in H^n(K;\pi_n(Y))$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="2159" LastActivityDate="2010-10-10T17:25:49.837" CommentCount="2" />
  <row Id="6440" PostTypeId="2" ParentId="6427" CreationDate="2010-10-10T17:26:09.710" Score="5" ViewCount="" Body="&lt;p&gt;Ross Millikan's answer is entirely correct, but perhaps it deserves slightly more elaboration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The expected amount of money you receive is the sum of the expected amounts of money that each coin gives you, which is the value of the coin times the probability that you draw it.  (Linearity of expectation!)  The probability that you draw each coin is the same: it is one minus the probability that you do not draw it, or $1 - \frac{ {9 \choose 3} }{ {10 \choose 3} } = \frac{3}{10}$.  So the expected amount of money you receive is $\frac{3}{10}$ of the total value of the coins.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that I am basically treating coins with the same value as distinguishable (pretend they have the same value but look different; it can't affect the answer to the question).  The question of distinguishability is a red herring.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-10-10T17:26:09.710" />
  <row Id="6441" PostTypeId="1" AcceptedAnswerId="6443" CreationDate="2010-10-10T17:51:22.747" Score="1" ViewCount="155" Body="&lt;p&gt;&lt;strong&gt;What number of positive integers satisfying the given inequality :&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;              C(n + 1,n – 2) – C(n + 1,n – 1) &amp;lt;= 100 &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;What I did so far :&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;C(n + 1,n – 2) – C(n + 1,n – 1) &amp;lt;= 100 &#xA;=&amp;gt; C(n + 1,3)  – C(n + 1,2) &amp;lt;= 100 &#xA;=&amp;gt; n(n+1)(n-4) &amp;lt;= 600&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Now to derive after this ? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS: This is not my homework, It originated my test paper without proper explanation of answers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT:The time constraint is only a mint so please suggest shortest possible method.&lt;/p&gt;&#xA;" OwnerUserId="2109" LastEditorUserId="2109" LastEditorDisplayName="" LastEditDate="2010-10-10T18:58:03.017" LastActivityDate="2010-10-10T19:49:30.447" Title="What number of positive integers satisfying the given inequality ?" Tags="&lt;combinatorics&gt;&lt;algebra-precalculus&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="6442" PostTypeId="2" ParentId="6427" CreationDate="2010-10-10T17:51:32.187" Score="4" ViewCount="" Body="&lt;p&gt;Ross Millikan has already given a very elegant solution, but if you're still in doubt, this problem is small enough that you can easily verify it by letting a computer enumerate all the possibilities. Here's a Python program to do it:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;coins = [25, 10, 10, 5, 5, 5, 1, 1, 1, 1]&#xA;total = count = 0&#xA;for k in range(10):&#xA;    for l in range(10):&#xA;        for m in range(10):&#xA;            if k!=l and k!=m and l!=m:&#xA;                total += coins[k] + coins[l] + coins[m]&#xA;                count += 1&#xA;print &quot;Got&quot;, total, &quot;cents in&quot;, count, &quot;draws; on average&quot;, total*1.0/count, &quot;cents.&quot;&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The answer comes instantly: &quot;Got 13824 cents in 720 draws; on average 19.2 cents.&quot;&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2010-10-10T17:51:32.187" CommentCount="3" />
  <row Id="6443" PostTypeId="2" ParentId="6441" CreationDate="2010-10-10T17:56:12.567" Score="2" ViewCount="" Body="&lt;p&gt;To solve $n(n+1)(n-4)\le 600$ algebraically, expand the left side and subtract 600 from both sides, so that you're comparing a polynomial to 0, find the zeros of the polynomial&amp;mdash;which divide the number line into regions&amp;mdash;and test a value in each region to see which regions satisfy the inequality.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Simpler, especially since you're only looking for positive integer solutions, would be to graph $y=x(x+1)(x-4)-600$ and look for where it is below the x-axis.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&lt;strong&gt;edit&lt;/em&gt;&lt;/strong&gt;: Under very tight time constraints and without a calculator/computer, I'd have approached the original problem differently.  The original statement  ${n+1\choose n–2}–{n+1\choose n–1}\le 100$  (as well as the restatement ${n+1\choose 3}–{n+1\choose 2}\le 100$) can be interpreted as a statement about the (n+1)st row of Pascal's triangle: that the difference between the numbers 2 and 3 positions in from the edge is less than or equal to 100.  It should be clear that the difference between adjacent entries in a row of Pascal's triangle is increasing, so it is sufficient to write successive rows until the statement fails:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/DTQS9.png&quot; alt=&quot;Pascal's triangle, through the 11th row&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The bold/darkest numbers are the relevant ones:&#xA;n=1: $0-1\le 100$;&#xA;n=2: $1-3\le 100$;&#xA;n=3: $4-6\le 100$;&#xA;n=4: $10-10\le 100$;&#xA;n=5: $20-15\le 100$;&#xA;n=6: $35-21\le 100$;&#xA;n=7: $56-28\le 100$;&#xA;n=8: $84-36\le 100$;&#xA;n=9: $120-45\le 100$;&#xA;n=10: $165-55&gt;100$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, note that the lightest numbers, below and to the left of the relevant numbers, are not used at all, so you don't actually need to compute them.&lt;/p&gt;&#xA;" OwnerUserId="72" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-10-10T19:49:30.447" LastActivityDate="2010-10-10T19:49:30.447" CommentCount="6" />
  <row Id="6444" PostTypeId="2" ParentId="6441" CreationDate="2010-10-10T18:11:20.713" Score="2" ViewCount="" Body="&lt;p&gt;$n(n+1)(n-4) \le 600$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is just a matter of trial and error, as $n(n+1)(n-4)$ is increasing for $n \ge 4$. For $ 0 &amp;lt; n &amp;lt; 4$ it is negative.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $n=10$, the value is $660 &gt; 600$ and for $n=9$ the value is $450$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus the number is $9$ got by $1 \le n \le 9$.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-10-10T18:11:20.713" CommentCount="1" />
  <row Id="6445" PostTypeId="1" AcceptedAnswerId="6449" CreationDate="2010-10-10T18:20:44.087" Score="1" ViewCount="174" Body="&lt;p&gt;Given finite-dimensional vector spaces $V,W$, there is an isomorphism $\text{Hom}(V,W) \rightarrow V^* \otimes W$. In particular, any linear map $\phi : V \rightarrow W$ has a tensor expansion $\sum v^*_i \otimes w_i$ where $v^*_i \in V^*, w_i \in W$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, if one chooses dual bases $\{x_i\}, \{e_i\}$ of $V^*$ and $V$, then $\sum x_i \otimes e_i \in V^* \otimes V$ is a tensor expansion for the identity map on $V$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What's the best way to intuitively understand tensor expansions of linear maps?&lt;/p&gt;&#xA;" OwnerUserId="251" LastEditorUserId="171" LastEditorDisplayName="" LastEditDate="2010-10-10T19:40:11.677" LastActivityDate="2010-10-10T19:40:11.677" Title="Intuition behind tensor expansions of linear maps" Tags="&lt;linear-algebra&gt;&lt;abstract-algebra&gt;&lt;intuition&gt;" AnswerCount="3" />
  <row Id="6446" PostTypeId="2" ParentId="6437" CreationDate="2010-10-10T18:22:37.657" Score="1" ViewCount="" Body="&lt;p&gt;mcAfee, here's a hint: call P1, P2, P3 the printers and C1, ..., C5 the computers. Suppose P1 is connected to C1, C4, C5; P2 is connected to C2, C4, C5, and P3 is connected to C3, C4, C5. That is 9 connections (and this is not the only way to connect them; you should try others). Check out whether this works, and if it does, try and see whether it's possible to do it with 8 connections. &lt;/p&gt;&#xA;" OwnerUserId="1313" LastActivityDate="2010-10-10T18:22:37.657" />
  <row Id="6447" PostTypeId="1" AcceptedAnswerId="6450" CreationDate="2010-10-10T18:34:23.107" Score="2" ViewCount="112" Body="&lt;p&gt;Let $T_f g = f \cdot g$ where $f, g, f \cdot g$ are in $H^2(\mathbb{D})$ (where $H^2$ is the Hardy space on the open unit disk). Now $T_f$ is a bounded operator.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to show this by showing that $f \in H^\infty$. So I try to write $f = G_1 h_1$ and $g = G_2 h_2$ where $G_i$ are outer functions and $h_i$ inner functions. So, what I need to do is if $G_1 G_2$ is in $H^2$ for all $G_2$ outer, then $G_1$ is in $H^\infty$. Does someone have a hint how I could obtain this?&lt;/p&gt;&#xA;" OwnerUserId="1120" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2010-10-11T00:39:41.807" LastActivityDate="2010-10-11T00:39:41.807" Title="Bounded operator Hardy space." Tags="&lt;complex-analysis&gt;&lt;functional-analysis&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="6448" PostTypeId="2" ParentId="6310" CreationDate="2010-10-10T18:56:32.583" Score="4" ViewCount="" Body="&lt;p&gt;At the heart this is really trivial, a  consequence of $\rm\ f(x) = x^n\ \Rightarrow\ f{\:&amp;#39;}(x)\ =\ n\: x^{n-1}\ \equiv\ 0\:\ (mod\ n)\:$. First I give the simple proof, then I explain the viewpoint in terms of derivatives.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\ n\:|\:(a-b)\frac{a^n-b^n}{a-b}\ \Rightarrow\ n = m\:k,\ \ m\:|\:a-b,\ \ k\:\big|\frac{a^n-b^n}{a-b}\ $ by unique factorization. Thus it suffices&lt;/p&gt;&#xA;&#xA;&lt;p&gt;to show $\rm\ \ m\:|\:n,\:a-b\ \Rightarrow\ m\:\big|\frac{a^n-b^n}{a-b}\ =\ a^{n-1}+a^{n-2}\:b+\:\cdots\:+a\:b^{n-2}+b^{n-1}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, $\rm\ mod\ \:m\:,\ \ a\equiv b\ \ \ \Rightarrow\ \ \ \frac{a^n-b^n}{a-b}\ \equiv\ a^{n-1}+\cdots+a^{n-1} \equiv n\: a^{n-1}\equiv 0\ $ via $\rm\ m|n\quad\quad $ &lt;strong&gt;QED&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The prior line is the special case $\rm\ f = x^n,\ x = b\ $ in this polynomial Taylor series approximant:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\displaystyle\quad\quad\quad\quad\quad  \frac{f(x)-f(a)}{x-a} \: \equiv\ f\:&amp;#39;(a)\ \ \ (mod\ \:x-a)\quad$ for $\rm\ f(x)\in \mathbb Z[x]$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\quad\ \ $ i.e. $\rm\quad\ f(x)\ =\ f(a)  +\: f\:&amp;#39;(a)\ (x-a) \:+\: (x-a)^2\: g(x)\quad$ for some $\rm\ g(x) \in \mathbb Z[x]$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore this result about &lt;em&gt;numbers&lt;/em&gt; is just a special case of the following well-known result about &lt;em&gt;functions&lt;/em&gt; (here polynomials):  a root $\rm\ x = a\ $ of $\rm\ f(x)\ $ has multiplicity $\rm &amp;gt; 1\ \iff\ f\:&amp;#39;(a) \:=\: 0\:.\:$ In fact, like above, many results about numbers are actually specializations of results about functions. Moreover, because functions have richer structure than numbers - e.g. having derivatives available - we can exploit this structure in the function realm before specializing to numbers. A powerful example of this is Mason's ABC theorem - which has a trivial high-school level proof for polynomials but is a deep conjecture for numbers. It yields as a consequence a trivial proof of FLT for polynomials. The moral is: to prove a result about numbers, try to interpret it as special case of a result about functions.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2011-02-06T22:29:19.163" LastActivityDate="2011-02-06T22:29:19.163" />
  <row Id="6449" PostTypeId="2" ParentId="6445" CreationDate="2010-10-10T19:03:28.097" Score="1" ViewCount="" Body="&lt;p&gt;Ok, let's try to incorporate your question in a more gneral setting: at least one of the vector spaces in the following is finite-dimensional.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $f\colon U\to U'$ and $g\colon V\to V'$ be linear maps. We easily define the &quot;tensor product&quot; of f and g to be the map $f\otimes g$ such that&#xA;$$ (f\otimes g)(u\otimes v)=f(u)\otimes g(v) $$&#xA;for all $u\in U$ and $v\in V$. So, we obtain a linear map&#xA;$$ \lambda\colon\hom(U,U')\otimes \hom(V,V')\cong \hom(V\otimes U, V'\otimes U')$$&#xA;which is easily checked to be an isomorphism, provided at least one of the pairs $(U,U')$, $(V,V')$, $(U,V)$ consists of finite-dimensional vector spaces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With an intelligent choice of the spaces, the isomorphism $\lambda$ allows us to prove every sort of useful identity relating tensor product of spaces and their duals: Try putting $U'=V'=\mathbb{K}$ (the underlying field), you'll obtain $(V\otimes U)^\star\cong U^\star\otimes V^\star $ (don't worry for the twist). Try putting $U=V'=\mathbb{K}$, you'll obtain... $\lambda_{UV}\colon V\otimes U^\star\cong \hom(U,V)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, let $f\colon U\to V$ be a linear map. Using bases for $U$ and $V$, we have&#xA;$$f(u_j) = \sum_i f_i^j v_i$$&#xA;for a family $(f_i^j)$ of scalars. Try proving that&#xA;$$  f = \lambda_{UV}\Big(\sum_{ij} f_i^j v_i\otimes u^j\Big) $$&lt;/p&gt;&#xA;" OwnerUserId="685" LastActivityDate="2010-10-10T19:03:28.097" CommentCount="1" />
  <row Id="6450" PostTypeId="2" ParentId="6447" CreationDate="2010-10-10T19:05:05.367" Score="3" ViewCount="" Body="&lt;p&gt;This isn't the same approach you had in mind, but you can show that $T_f$ is bounded using the closed graph theorem and the fact that evaluation at a point in the open disk is bounded on $H^2$.  You can then show that $f$ is in $H^\infty$ by showing that the complex conjugates of elements of its image on the disk are eigenvalues for the adjoint of $T_f$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is an elaboration on the last sentence.  For each $w\in\mathbb{D}$, define $k_w:\mathbb{D}\to\mathbb{C}$ by $k_w(z)=\frac{1}{1-\overline{w}z}=\sum_{k=0}^\infty \overline{w}^k z^k$.  Each $k_w$ is in $H^\infty$ and thus in $H^2$.  Using the second expression for $k_w$ and the characterization of the inner product on $H^2$ in terms of the $\ell^2$ sequences of Maclaurin coefficients, notice that $\langle g,k_w\rangle=g(w)$ for all $w\in\mathbb{D}$ and all $g\in H^2$.  It then follows that for all $w$ and $z$ in $\mathbb{D}$, &#xA;$$(T_f^*k_w)(z)=\langle T_f^* k_w,k_z\rangle=\overline{\langle T_f k_z,k_w\rangle}=\overline{f(w)k_z(w)}$$ $$=\overline{f(w)}\overline{\langle k_z,k_w\rangle}=\overline{f(w)}\langle k_w,k_z\rangle=\overline{f(w)}k_w(z).$$  Since $z$ was arbitrary, this shows that $T_f^*k_w=\overline{f(w)}k_w$, so $\overline{f(w)}$ is an eigenvalue for $T_f^*$ with eigenvector $k_w$.  Thus, $\|f\|_\infty\leq \|T_f^*\|=\|T_f\|&amp;lt;\infty$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a standard fact about reproducing kernel Hilbert spaces, and only the particular form of the function $k_w$ is special to the Hardy space. The way I have presented this, it might seem that $k_w$ was summoned by magic, but in fact one could rediscover them without too much work.  The important point is that there exist elements of $H^2$ whose corresponding inner product functionals are point evaluations.  These exist by Riesz's lemma using continuity of the point evaluations, which can be shown by other means.  You don't need to know what these elements are for the argument to carry through.  However, if you did want to discover them, then &quot;working backwards&quot; and considering Maclaurin series would lead you to the second expression for $k_w$ given above.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2010-10-10T21:12:58.727" LastActivityDate="2010-10-10T21:12:58.727" CommentCount="5" />
  <row Id="6451" PostTypeId="1" CreationDate="2010-10-10T19:18:37.097" Score="4" ViewCount="217" Body="&lt;p&gt;Hello,&#xA;I'm trying to show that the Möbius strip with boundary circle identified to a point is homeomorphic to P^2 (real projective space). I get geometrically why this is so, but, how does one generally construct maps between these spaces to show that they're homeomorphic? &lt;/p&gt;&#xA;" LastActivityDate="2010-10-11T21:12:28.947" Title="Möbius strip with edge identified - constructing map?" Tags="&lt;topology&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="1" />
  <row Id="6452" PostTypeId="2" ParentId="6445" CreationDate="2010-10-10T19:31:06.850" Score="1" ViewCount="" Body="&lt;p&gt;Not entirely sure what you're looking for so I'll just throw out some thoughts:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Recall that $V^{*}$ is the set of linear maps from $V$ to $\mathbb{R}$.  Tensoring with $W$ effectively replaces the $\mathbb{R}$ with $W$.  So the star on $V$ represents that the maps are coming from $V$; the lack of one on $W$ represents that you're getting elements of $W$.  Likewise, I'd expect that elements of $\text{Hom}(V,W)$ would covary with a change of basis of $V$ and contravary with one of $W$ (if you're unfamiliar with these terms, here's &lt;a href=&quot;http://en.wikipedia.org/wiki/Covariance_and_contravariance_of_vectors&quot; rel=&quot;nofollow&quot;&gt;the Wikipedia article&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a bit more rigor, recall that you have a natural pairing between $V$ and $V^{*}$, which is just a map $$\langle -,- \rangle: V\times V^*\rightarrow\mathbb{R}$$ defined by $\langle v,\alpha\rangle=\alpha(v)$.  As you can check, this is invariant under change of basis (hence the &quot;natural&quot;) -- this actually follows from $V$ and $V$ varying oppositely.  This also has the property that if $v_i,\alpha^i$ are dual bases for $V$ and $V^{*}$, then $\langle v_i,\alpha^i\rangle=1$, and $\langle v_i,\alpha^j\rangle=0$ when $i\ne j$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So now choose bases $v_i,w_j$ for $V$ and $W$ and a dual basis $\alpha^i$ for $V^{*}$.  Given an element $v=\sum b^iv_i$ of $V$ and an element $f=\sum a_i^jv^i\otimes w_j$ of $V^*\otimes W$, we get&#xA;$$f(v)=\sum_{i,j,k}(a_i^j\alpha^i\otimes w_j)(b^kv_k)=\sum_{i,j,k}(a_i^jb^k\langle v_k,\alpha^i\rangle\otimes w_j).$$&#xA;Since the pairing is only nonzero for $k=i$, this reduces to&#xA;$$f(v)=\sum_{i,j}a_i^jb^i\langle v_i,\alpha^i\rangle\otimes w_j=\sum_{i,j}a_i^jb^iw_j$$&#xA;where the tensor product has disappeared because the output of the pairing is just a real number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But this is just a vector in $W$!  And conversely, given a linear map $V\rightarrow W$, we can pick bases and write the map as a matrix $(a_i^j)$, which then transforms to the above tensor product.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found &lt;a href=&quot;http://www.math.binghamton.edu/paul/601-S06/AS5.pdf&quot; rel=&quot;nofollow&quot;&gt;these notes&lt;/a&gt; very helpful when I was learning this stuff.  You can change the 5 in the URL to other numbers to read all the notes, though I believe they stop at 9.&lt;/p&gt;&#xA;" OwnerUserId="2215" LastActivityDate="2010-10-10T19:31:06.850" />
  <row Id="6453" PostTypeId="2" ParentId="6445" CreationDate="2010-10-10T19:31:58.333" Score="1" ViewCount="" Body="&lt;p&gt;Here's a very &quot;low-tech&quot; answer, in terms of coordinates. With respect to the dual bases (I take $\dim V=\dim W=2$ for simplicity), the linear transformation with matrix $$\begin{pmatrix} a&amp;amp;b \\ c&amp;amp;d \end{pmatrix}$$ can be expanded as&#xA;$$a \begin{pmatrix} 1 \\ 0 \end{pmatrix} \begin{pmatrix} 1 &amp;amp; 0 \end{pmatrix}&#xA;+ b \begin{pmatrix} 1 \\ 0 \end{pmatrix} \begin{pmatrix} 0 &amp;amp; 1 \end{pmatrix}&#xA;+ c \begin{pmatrix} 0 \\ 1 \end{pmatrix} \begin{pmatrix} 1 &amp;amp; 0 \end{pmatrix}&#xA;+ d \begin{pmatrix} 0 \\ 1 \end{pmatrix} \begin{pmatrix} 0 &amp;amp; 1 \end{pmatrix}.$$&#xA;Here the column vectors are coordinate vectors for elements in $W$, while the row vectors are coordinate vectors for elements in $V^*$.&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2010-10-10T19:31:58.333" />
  <row Id="6454" PostTypeId="2" ParentId="6451" CreationDate="2010-10-10T19:47:33.887" Score="2" ViewCount="" Body="&lt;p&gt;I would contruct a homeomorphism from $A/\sim$ to the Mobius strip&#xA;where $A$ is the region on the sphere bounded by two antipodal lines&#xA;of latitude (let's think of them as the arctic and antarctic circles)&#xA;and $\sim$ is the equivelance relation identifying antipodal points.&#xA;Now deforming the artic/antartic regions into the north and south poles,&#xA;induces a homeomorphism from the Mobius strip with edge squeezed to a point&#xA;and ${S^2/\sim}\ = P^2$.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-10-10T19:47:33.887" />
  <row Id="6455" PostTypeId="2" ParentId="6411" CreationDate="2010-10-10T19:49:17.530" Score="2" ViewCount="" Body="&lt;p&gt;No, this need not be true, unless I am missing some assumption implicit in your use of the phrase &quot;&lt;em&gt;the&lt;/em&gt; induced norm&quot; (emphasis added).  Here is an example where $X=Y=\mathbb{R}$.  All norms on $X\times Y=\mathbb{R}^2$ are equivalent.  The norm $\|(x,y)\|^2=2x^2-2xy+y^2=x^2+(x-y)^2$ does not satisfy your property because, for example $\|(x,x)\|=|x|$ while $\|(x,0)\|=\sqrt{2}|x|$.  (This norm actually comes from an inner product on $\mathbb{R}^2$.)&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The following was a result of misreading the question.  I'll leave it here, for now at least, if for no other reason than leaving Arturo's correction comprehensible:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is true is that the projection map $(x,y)\mapsto x$ is continuous.  This implies as a special case the following (which turns out to actually be equivalent to continuity):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For all $\epsilon&gt;0$, there is a $\delta&gt;0$ such that for all $x\in X$ and $y\in Y$, $\|(x,y)\|&amp;lt;\delta$ implies $\|x\|&amp;lt;\epsilon$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, the projection is Lipschitz continuous, and $\delta$ can be taken to be $\epsilon$ divided by the Lipschitz constant.  For the reason Qiaochu Yuan gave in a comment on Ross Millikan's answer, this is the best you could hope for.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2010-10-10T20:05:49.500" LastActivityDate="2010-10-10T20:05:49.500" CommentCount="4" />
  <row Id="6456" PostTypeId="1" CreationDate="2010-10-10T20:58:11.020" Score="4" ViewCount="118" Body="&lt;p&gt;Given the binary alphabet {a,b}, I'm trying to find the generating function that distinguishes, for all words of fixed length $n$, the count of blocks of a's and the number of a's. Let $x^p$ count the number of blocks of size $p$ and $y^q$ count the number of a's in a word. As a first example, take the word:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;abbaaabaabaabba&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This has nine a's broken up into a two blocks of length one, two blocks on length two and a block of length three. This would contribute to the generating function a term of $y^9(2x + 2x^2 + x^3)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a second example, with $n=3$ there are eight words and the term associated with each of them is:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;bbb = 1&#xA;bba = xy&#xA;bab = xy &#xA;baa = (xy)^2&#xA;abb = xy&#xA;aba = 2xy^2&#xA;aab = (xy)^2&#xA;aaa = (xy)^3&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Note that the term &lt;code&gt;aba = 2xy^2&lt;/code&gt; is not a typo, there are two blocks of length one. With $A$ as the generating function, this would give: $A_3(x,y) = 1 + 3xy + (2x + 2x^2)y^2 + y^3x^3$. I've been able to come up with a special case of this, when we don't care about the counts of the a's (equivalent to setting all $y=1$). This restricted form $B_n(x) = A_n(x,y=1)$ is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ B_n = B_{n-1} + x^n + \sum_{j=0}^{n-2} ( 2^j x^{n-j-1} -1 + B_j ) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with $B_0 = 1$. Is there such a formula for $A_n$, or even better, a non-recursive closed form of $A_{npq}$?&lt;/p&gt;&#xA;" OwnerUserId="196" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2011-06-03T20:07:35.417" LastActivityDate="2011-06-03T20:07:35.417" Title="Generating function of words in a binary alphabet counting blocks and appearances" Tags="&lt;generating-functions&gt;&lt;recurrence-relations&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="6457" PostTypeId="2" ParentId="6414" CreationDate="2010-10-10T21:04:34.010" Score="0" ViewCount="" Body="&lt;p&gt;First, I am pretty rusty on this stuff, so hopefully somebody&#xA;with more confidence will chime in. I started writing this as another&#xA;comment, but it didn't fit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, I still think what I said makes sense. If&#xA;$\mathcal{E}_{\mathbb{R}^n}$ is the standard differentiable structure&#xA;on $\mathbb{R}^n$, and $\mathcal{C}_M$ is the sheaf of continuous&#xA;functions on a topological manifold $M$, then a subsheaf $\mathcal{F}$&#xA;of $\mathcal{C}_M$ is a differentiable structure on $M$ if (and only if)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) at each point $p\in M$ there exists a neighbourhood $V$, and&#xA;  $x_1,...x_n \in \mathcal{F}(V)$ such that $q \mapsto (x_1, ...,&#xD;&#xA;  x_n)$ is a homeomorphism onto an open $U \subset \mathbb{R}^n$, and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) $f \in \mathcal{F}(V)$ iff there exists $F \in&#xD;&#xA;  \mathcal{E}_{\mathbb{R}^n}(U)$, such that $f = F(x_1,...,x_n)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think this definition is derived from Warner (&quot;Foundations on&#xA;differentiable manifolds and Lie groups&quot;), but it has paraphrased by&#xA;me and possibly been distorted through the lens of an absurdly long&#xA;time. In particular, that second condition should probably be reworked&#xA;to be more obviously a purely sheaf theoretic statement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In any event, a proper definition of a differentiable structure is&#xA;going to resemble something like this (likewise for holomorphic&#xA;structure, or analytic structure or whatever). And the point is, that&#xA;the \emph{definition} will require that your sheaf of differentiable&#xA;functions on $M$ will be locally isomorphic to the \emph{standard}&#xA;sheaf of differentiable functions on $\mathbb{R}^n$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This should apply even if the topological manifold $M$ is&#xA;$\mathbb{R}^n$. This is why I say that you can't solve the problem&#xA;locally: Locally all differentiable structures are isomorphic to the&#xA;standard one on $\mathbb{R}^n$ by definition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As to my second statement, about an &quot;appropriate&quot; differentiable&#xA;structure on $TM$ defining a unique differentiable structure on $M$,&#xA;this is mostly a guess. However, the qualifier &quot;appropriate&quot; is&#xA;important here. In particular, whatever differentiable structure on&#xA;$TM$ you choose, it has to restrict to the standard structure on&#xA;$\mathbb{R}^n$ on the fibres. In this way, I don't think there is any&#xA;choice left once you've defined the restriction on the zero section.&lt;/p&gt;&#xA;" OwnerUserId="1728" LastEditorUserId="6490" LastEditorDisplayName="" LastEditDate="2011-02-17T10:05:14.657" LastActivityDate="2011-02-17T10:05:14.657" CommunityOwnedDate="2010-10-11T02:38:01.230" />
  <row Id="6458" PostTypeId="2" ParentId="6415" CreationDate="2010-10-10T21:12:41.350" Score="0" ViewCount="" Body="&lt;p&gt;okay ive figured out the first part, now ive been told the chinese remainder theorem for the second part but i dont see how it will give me the extra 2 answers since it only allows the set 1-1000&#xA;I assume the first 6 answers are x ≡ +-3(mod y) where y=7,11,13 yet what about the other 2 answers?&lt;/p&gt;&#xA;" LastActivityDate="2010-10-10T21:12:41.350" CommentCount="2" />
  <row Id="6459" PostTypeId="1" AcceptedAnswerId="6480" CreationDate="2010-10-10T21:28:26.737" Score="1" ViewCount="223" Body="&lt;p&gt;I need to write a piece of software that analyzes packet switched computer networks and calculates certain characteristics based on packet loss (the application is Voice over IP quality monitoring).  One approach is to use the following 4-state Markov model (as described &lt;a href=&quot;http://www.voiptroubleshooter.com/indepth/burstloss.html&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/CXMxu.jpg&quot; alt=&quot;4-state markov model&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;State 1 – packet received successfully&lt;br&gt;&#xA;State 2 – packet received within a burst&lt;br&gt;&#xA;State 3 – packet lost within a burst&lt;br&gt;&#xA;State 4 – isolated packet lost within a gap&lt;/p&gt;&#xA;&#xA;&lt;p&gt;State transitions are bound to certain criteria.  This is how it is expected to work for a stream of network packets:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Loss pattern 000001100101010110110000000000000000000000001000000000&#xA;State        111113322323232332331111111111111111111111114111111111&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;An example algorithm I found stores the state changes in counter variables (e.g. state change 1-&gt;3 increments counter c13).  So far I get it and I end up with these counters: c14, c11, c13, c22, c23, c33.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, for the actual analysis I need to calculate the state transition probabilities from these counters.  This is where the algorithm chokes.  Besides the issue with zero divisions, I don't see the total number of packets incorporated into the probability calculation (this is from the pseudo code of the algorithm definition, p14 is not included because they added c14 to c11):&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;p11 = c11/(c11 + c13)&#xA;p13 = 1 - p11&#xA;p31 = c31/(c31 + c32 + c33)&#xA;p32 = c32/(c31 + c32 + c33)&#xA;p33 = 1 - p31 - p32&#xA;p22 = c22/(c22 + c23)&#xA;p23 = 1 - p22&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Is this even correct?  I don't quite get what the author is doing here.  Am I missing something?  What's the correct way to calculate probabilities p14, p11, p13, p22, p23, p33, p31 and p32?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope you understand why I asked this here instead of on &lt;a href=&quot;http://stackoverflow.com&quot;&gt;Stackoverflow&lt;/a&gt;: this is basically just about the calculation of the probabilities, the rest was provided for context.&lt;/p&gt;&#xA;" OwnerUserId="2415" LastEditorUserId="2415" LastEditorDisplayName="" LastEditDate="2010-10-10T23:57:19.910" LastActivityDate="2010-10-11T02:13:20.227" Title="Calculating probabilities for a 4-state Markov model" Tags="&lt;probability&gt;&lt;algorithms&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="6460" PostTypeId="1" AcceptedAnswerId="6465" CreationDate="2010-10-10T21:32:17.260" Score="5" ViewCount="210" Body="&lt;p&gt;I'm studiying for an exam I have on combinatorics next week and I'm stuck with the solution to this question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine you have a pentagon and you want to color the corners of the pentagon so that no two adjacent corners have the same color. If you have 'q' colors, how many ways of coloring the pentagon are possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I say that the answer is $q^2*(q-1)^2*(q-2) + q*(q-1)*(q-2)^2*(q-3)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My friend says the answer is $MyAnswer + q*(q-1)*(q-2)*(q-3)*(q-4)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Who is right? (or are we both wrong).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Further if you could point out a resource that explains how to calculate this question for other numbers of corners (say an hexagon) we would really appreciate it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Answering to some comments in the answer, rotations of the pentagon should not be counted. What I mean is that if we have 5 colors (a,b,c,d,e) the pentagon with corners {(1,a),(2,b),(3,c),(4,d),(5,e)} is exactly the same as {(1,c),(2,d),(3,e),(4,a),(5,b)}&lt;/p&gt;&#xA;" OwnerUserId="2359" LastEditorUserId="2359" LastEditorDisplayName="" LastEditDate="2010-10-11T01:40:53.827" LastActivityDate="2010-10-11T01:40:53.827" Title="Calculate how many ways can you paint the corners of a Pentagon" Tags="&lt;combinatorics&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" />
  <row Id="6461" PostTypeId="1" AcceptedAnswerId="6469" CreationDate="2010-10-10T21:48:05.170" Score="2" ViewCount="86" Body="&lt;p&gt;Let $A$ and $B$ be compact subsets of a metric space. Denote by $\mathcal{P}(A)$ the set of (Borel) probability measures in $A$, and similarly define $\mathcal{P}(A \cup B)$. They are compact metric. Define $C$ as the set of all elements of $\mathcal{P}(A \cup B)$ which assign probability 1 to $A$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question&lt;/strong&gt;: Is there an embedding between $C$ and $\mathcal{P}(A)$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as I can see, when $A$ and $B$ are finite (and assuming without loss they are also disjoint), the answer is yes because one can identify $\mathcal{P}(A)$ with a closed subset of $\mathcal{P}(A \cup B)$. But I am a little lost on how to (try to) show it in general.&lt;/p&gt;&#xA;" LastActivityDate="2010-10-11T00:16:28.717" Title="Existence of an embedding" Tags="&lt;real-analysis&gt;&lt;probability-theory&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="6462" PostTypeId="2" ParentId="6421" CreationDate="2010-10-10T22:15:16.057" Score="2" ViewCount="" Body="&lt;p&gt;The biggest problem here is that you seem to be confusing the chain rule with the power rule.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The power rule is as follows: Suppose you have some real number $n$. Then the function $f(x) =x^n$ has derivative $f'(x)=n\cdot x^{n-1}$. Notice that $n$ is fixed, but $x$ is variable. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The chain rule is as follows: Given functions $f$ and $g$, the function $h(x)=(f\circ g)(x) = f(g(x))$ has derivative $h'(x)= (f'\circ g)(x)\cdot g'(x)= f'(g(x))\cdot g'(x)$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the function you have is $f(x)=10^{-x}$. In this case, your &lt;em&gt;exponent&lt;/em&gt; is variable, but $10$, obviously, is fixed. So the power rule does not apply; that is, we cannot simply bring down the exponent, multiply, and subtract one from the exponent, which is part of what you did in your work. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, given an exponential function $g(x)=a^x$, the derivative is $g'(x)=a^x\cdot\ln{(a)}$. So why does this rule not apply directly to the function $f(x)=10^{-x}$? Well, notice that the exponent in your $f$ is actually itself a function, namely, $-x$. The exponent in the &lt;em&gt;rule&lt;/em&gt; from your book is merely $x$. That's why we have to apply the chain rule; we technically have a function &quot;inside&quot; another function. From here, follow kahen's solution above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I again want to emphasize that the power rule is not applied to your $f$ because $f$ is exponential: it has a fixed base of $10$ and a variable exponent $-x$ (which is also a function).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lastly, I should note that the power rule and chain rule &lt;em&gt;as I have written them&lt;/em&gt; are neither rigorous nor precise, but I sacrificed details for readability. &lt;/p&gt;&#xA;" OwnerUserId="2313" LastActivityDate="2010-10-10T22:15:16.057" CommentCount="2" />
  <row Id="6463" PostTypeId="2" ParentId="6460" CreationDate="2010-10-10T22:21:49.817" Score="3" ViewCount="" Body="&lt;p&gt;You need to specify the problem better.  Do you count rotations and reflections as different?  For example, imagine a regular pentagon drawn on a sheet of paper with a corner up.  Starting from the top corner and going clockwise, maybe we have 0,1,2,3,4 (using numbers instead of colors).  Is this different from 1,2,3,4,0?  How about from 0,4,3,2,1?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The easiest situation is if you say these are all different.  Then the naive approach would be to say you have $q$ choices for the top corner, $q-1$ choices for each of the next three corners (as you can't have neighboring corners that match) and $q-2$ choices for the last, giving $q*(q-1)^3*(q-2)$, but this ignores the fact that the first and fourth corners could be the same color, giving $q-1$ choices for the last.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One way to deal with this situation is to separate into classes based on which corners match the top one.  If we designate the top corner 0 and count clockwise, you can match the top color only in no corner, corner 2, or corner 3.  So matching none, we have $q*(q-1)*(q-2)^3 $ as at corner 1 you can use any color but the color at corner 0, but for each other you have two eliminated.  Matching corner 2, we have $q*(q-1)*1*(q-1)*(q-2)$ and matching corner 3 we have $q*(q-1)*(q-1)*1*(q-2)$.  Add them all up and you have your answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, the answer has to be a fifth degree polynomial.  For large $q$ the constraint of not matching colors won't reduce the count very much.  So if you count the solutions for six different q (I suggest 0 through 5), you can fit a fifth degree polynomial through them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to count rotations and reflections as the same there are two possibilites.  One way is to define a standard position and only count those cases in standard position.  In this case this is easy.  You would say that the smallest number has to be at the top, and that corner 2 is less than corner 3.  A pitfall would be to say the smallest is at the top and corner 1 is less than corner 4.  It could be that corner 1 and 4 are the same, and it could be that one of corner 2 and 3 is the same as corner 0.  But counting how many configurations satisfy this constraint may not be easy.  It is again a fifth degree polynomial.  The other case is to list the configurations of matching corners and see how many times they each get counted.  So if no corners match there are $q*(q-1)*(q-2)*(q-3)*(q-4)$ possibilities, but you have counted each one 10 times, so divide these by 10.  Then you can have one pair or two pair of corners matching.  These are probably easier to count by the standard position approach.  If one pair matches, say they have to be positions 0 and 2.  So we have $q*(q-1)*1*(q-1)*(q-2)$ choices for this.  And so on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sorry for taking out the asterisks from my expressions, but it rendered in italics and run over itself when they were there.  I think they made it easier to read.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2010-10-10T22:29:16.207" LastActivityDate="2010-10-10T22:29:16.207" CommentCount="2" />
  <row Id="6464" PostTypeId="1" AcceptedAnswerId="6476" CreationDate="2010-10-10T22:22:01.990" Score="5" ViewCount="312" Body="&lt;p&gt;In physics there are lots of identities like:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\nabla \times (\nabla \times A) = \nabla (\nabla \cdot A) - (\nabla \cdot \nabla) A$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm wondering if there is an algorithmic algebraic method to prove and/or derive these identities (something like using $e^{i\theta}$ to prove trigonometric identities)?&lt;/p&gt;&#xA;" LastActivityDate="2010-10-11T03:13:47.490" Title="Identities with Div, Grad, Curl" Tags="&lt;calculus&gt;&lt;algebra&gt;" AnswerCount="1" FavoriteCount="2" />
  <row Id="6465" PostTypeId="2" ParentId="6460" CreationDate="2010-10-10T22:33:08.617" Score="5" ViewCount="" Body="&lt;p&gt;Here I am going to assume that we are not looking for distinct configurations, that is, we count a rotation or reflection of the pentagon to be a distinct possibility.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You both have incorrect terms. Consider having only $3$ colors, ie $q = 3$. In this condition there must be a vertex which lies between two points which have the same color. There are $3 \times 2 = 6$ ways to color these three points, leaving $2$ ways to color the remaining two points for a total of $12$ possibilities. But $3^2 \times 2^2 \times 1 + 0 + (0) = 36$, which is too many. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint: If you are coloring a pentagon, there are three possible cases:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The pentagon can be colored with $k = 5$ distinct colors chosen from the $q$ possibilities.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The pentagon can be colored with $k = 4$ distinct colors and one repeat.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The pentagon can be colored with $k = 3$ distinct colors where two colors appear twice. (We also have that one color could appear three times, but then it would be impossible to color a pentagon without the same color being adjacent.)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;It is easy to see that it is impossible to color the vertices of a pentagon with only 2 colors such that the same color is not adjacent anywhere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your question, your friend is accounting for the $q(q-1)(q-2)(q-3)(q-4)$ ways to color a pentagon using $5$ distinct colors. To find the final solution, we need to count how many ways we can color a pentagon in each of the given cases and then we sum all the possibilities together.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, for each of the three cases, $q$ choose $k$ to fix the k elements you are working with, then consider how many ways you can color a pentagon using those $k$ colors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any more assistance will be moving towards an outright solution so I believe I should stop here.&lt;/p&gt;&#xA;" OwnerUserId="557" LastEditorUserId="557" LastEditorDisplayName="" LastEditDate="2010-10-10T22:42:07.160" LastActivityDate="2010-10-10T22:42:07.160" CommentCount="1" />
  <row Id="6466" PostTypeId="1" AcceptedAnswerId="6479" CreationDate="2010-10-10T22:50:19.357" Score="5" ViewCount="256" Body="&lt;p&gt;Hello!  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $G$ be the fundamental group of the Klein bottle,  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$G = \langle x,y \ ; \ yxy^{-1}=x^{-1} \rangle = {\mathbb Z} \rtimes {\mathbb Z} \ .$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are the nilpotent subgroups of $G$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was only able to find a normal series of abelian subgroups with cyclic quotients in $G$, namely  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$1\leq \langle y^2 \ ; \ \ \rangle\leq \langle x,y^2 \ ; \ xy^2=y^2x \rangle\leq G \ .$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since I'm not an algebraist, I'm sorry if this a silly question. Thanks!&lt;/p&gt;&#xA;" OwnerUserId="2412" LastActivityDate="2010-11-28T06:21:27.400" Title="Subgroups - Klein bottle" Tags="&lt;group-theory&gt;&lt;algebraic-topology&gt;" AnswerCount="3" CommentCount="3" />
  <row Id="6467" PostTypeId="1" AcceptedAnswerId="6468" CreationDate="2010-10-10T23:16:36.207" Score="2" ViewCount="107" Body="&lt;p&gt;I am wondering about the quickest way to prove the following from scratch: the lower shift is not similar to a direct sum of other linear transformations. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are the definitions of the terms used in the question. The lower shift is a mapping from $R^n$ to $R^n$ which maps $(x_1, x_2, \ldots, x_{n-1}, x_n)$ to $(0, x_1, \ldots, x_{n-1})$. Given two linear mappings $T_1: U \rightarrow U, ~T_2: V \rightarrow V$ their direct sum is the map from $U \times V~$ to $~U \times V$ which maps $(u,v)$ to $(T_1 u, T_2 v)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My current proof of this appeals to the Cayley-Hamilton theorem. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you!&lt;/p&gt;&#xA;" OwnerUserId="2120" LastActivityDate="2010-10-10T23:49:49.980" Title="The impossibility of decomposing the lower shift operator" Tags="&lt;linear-algebra&gt;" AnswerCount="2" />
  <row Id="6468" PostTypeId="2" ParentId="6467" CreationDate="2010-10-10T23:29:18.627" Score="4" ViewCount="" Body="&lt;p&gt;Tensor by $\mathbb{C}$.  Then the direct sum of two linear transformations has at least two eigenvectors, but the lower shift has only one.  (If you don't want to tensor by $\mathbb{C}$, the corresponding statement over $\mathbb{R}$ is that the direct sum of two linear transformations has at least two one- or two-dimensional invariant subspaces, but the lower shift has only one.)&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-10-10T23:29:18.627" CommentCount="1" />
  <row Id="6469" PostTypeId="2" ParentId="6461" CreationDate="2010-10-10T23:35:17.190" Score="1" ViewCount="" Body="&lt;p&gt;Disclaimer: I am new to this metric on spaces of probability measures, so hopefully there are not silly errors in what follows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $\mu\in C$, define $\mu'\in\mathcal{P}(A)$ by $\mu'(E)=\mu(E)$.  The map $\mu\mapsto\mu'$ is surjective, because if $\nu$ is in $\mathcal{P}(A)$, then $\nu=\tilde{\nu}'$, where $\tilde{\nu}\in C$ is defined by $\tilde{\nu}(E)=\nu(E\cap A)$.  The map is also isometric with respect to the &lt;a href=&quot;http://en.wikipedia.org/wiki/L%C3%A9vy%E2%80%93Prokhorov_metric&quot; rel=&quot;nofollow&quot;&gt;Lévy–Prokhorov metric&lt;/a&gt;.  To see this, note that although the $\epsilon$-neighborhoods of a Borel subset $E$ of $A$ will be different in $A$ versus in $A\cup B$, the $\mu$-measures of these neighborhoods are the same for all $\mu\in C$, due to the fact that $\mu(B\setminus A)=0$ for such $\mu$.  Hence the sets appearing in the definition of the metric (and therefore also their infima) will be identical whether considering the distance between $\mu$ and $\nu$ in $C$ or between $\mu'$ and $\nu'$ in $\mathcal{P}(A)$.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2010-10-11T00:16:28.717" LastActivityDate="2010-10-11T00:16:28.717" />
  <row Id="6470" PostTypeId="1" AcceptedAnswerId="6493" CreationDate="2010-10-10T23:37:27.447" Score="1" ViewCount="125" Body="&lt;p&gt;Given representations $\rho_1 : G \to \mathrm{GL}(V)$ and $\rho_2 : G \to \mathrm{GL}(W)$, how can we define explicitly the representation of $G$ over $\mathrm{Hom}(V, W) \cong V^* \otimes W$?&lt;/p&gt;&#xA;" OwnerUserId="314" LastActivityDate="2010-10-11T06:41:50.533" Title="Representation over Hom(V, W)" Tags="&lt;representation-theory&gt;" AnswerCount="1" />
  <row Id="6472" PostTypeId="2" ParentId="6467" CreationDate="2010-10-10T23:49:49.980" Score="3" ViewCount="" Body="&lt;p&gt;The shift is nilpotent.  If it were nontrivially a direct sum, then each of the summands would be nilpotent on spaces of positive dimension.  The kernel of a nilpotent on a nontrivial space is nontrivial, and the kernel of a direct sum is the direct sum of the kernels, so this would imply that the kernel of the shift has dimension at least 2.  But its kernel is 1-dimensional.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-10-10T23:49:49.980" CommentCount="1" />
  <row Id="6473" PostTypeId="1" CreationDate="2010-10-11T01:15:21.443" Score="0" ViewCount="196" Body="&lt;p&gt;suppose p,q are two prime number, how to quickly calculate how many numbers x are there such that gcd(x, (q-1)*(p-1)) = 1, without using brute force!&lt;/p&gt;&#xA;" OwnerUserId="2419" LastActivityDate="2010-10-11T02:09:09.537" Title="how to find co-prime numbers" Tags="&lt;cryptography&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="6474" PostTypeId="2" ParentId="6410" CreationDate="2010-10-11T01:16:04.503" Score="5" ViewCount="" Body="&lt;p&gt;Intuitively, the reason that universal quantification can be distributed over conjunction is that universal quantification can already be viewed as conjunction: $(\forall x)\Phi(x)$ can be viwed as the conjunction of $\Phi(d)$ taken over every element $d$ of the domain.  Viewed this way, $(\forall x)(\Phi(x) \land \Psi(x))$ and $(\forall x)\Phi(x) \land (\forall x)\Psi(x)$ are equivalent because they both represent a giant conjunction of every instance of $\Phi(d)$ along with every instance of $\Psi(d)$. This can be made more precise by looking at Tarski's schema for truth in a structure. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the same reason, existential quantification can be distributed over disjunction, because $(\exists x)\Phi(x)$ can be viewed as the disjunction of every possible substitution instance of $\Phi$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In very old literature, people actually used $\bigwedge_x$ for $\forall$ and  $\bigvee_x$ for $\exists$, for this reason. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, continuing this informal viewpoint, the reason that $\forall$ cannot be distributed over disjunction is that a certain distributive rule does not hold:&#xA;$$&#xA;\bigwedge_x\left(\Phi(x) \lor \Psi(x)\right)&#xA;$$&#xA;is not in general the same as &#xA;$$&#xA;(\bigwedge_x \Phi(x)) \lor (\bigwedge_x \Psi(x))&#xA;$$&#xA;Indeed, this rule already fails when there are just two elements in the domain, because the propositional formula&#xA;$(P \lor Q) \land (R \lor S)$ is not equivalent to $(P \land R) \lor (Q \land S)$. &lt;/p&gt;&#xA;" OwnerUserId="630" LastActivityDate="2010-10-11T01:16:04.503" />
  <row Id="6475" PostTypeId="1" AcceptedAnswerId="6477" CreationDate="2010-10-11T01:27:20.067" Score="1" ViewCount="122" Body="&lt;p&gt;Dear all,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wish to prove the following inequality. I have done a plot and it looks ok. $\mu \geq 1$ is an integer and so is $h$, where $0 \leq h \leq \mu-1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I started like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{e^{-\mu}\mu^{(\mu+h)}}{(\mu+h)!} \geq \frac{e^{-\mu}\mu^{(\mu-h-1)}}{(\mu-h-1)!}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have rewritten this to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(\mu-h-1)!\mu^{(\mu+h)} \geq (\mu+h)!\mu^{(\mu-h-1)}$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;however I am not able to go further from there. Can you help me out?&lt;/p&gt;&#xA;" OwnerUserId="2174" LastActivityDate="2010-10-11T01:48:07.920" Title="An inequality involving factorials and powers" Tags="&lt;calculus&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="6476" PostTypeId="2" ParentId="6464" CreationDate="2010-10-11T01:35:09.047" Score="4" ViewCount="" Body="&lt;p&gt;The most straightforward route I know of is through &lt;a href=&quot;http://en.wikipedia.org/wiki/Einstein_notation&quot; rel=&quot;nofollow&quot;&gt;Einstein tensor notation&lt;/a&gt;, which renders the derivation of such identities largely mechanical by virtue of suppressing distracting notational noise. (I should warn you that I'm largely self-taught on this subject, so the following may contain errors of rigour, but it &quot;works&quot; as long as you're working in Cartesian coordinates with the Euclidean metric.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me work through your example to illustrate. In tensor notation, the $i$th component of the curl of a vector field $v_i$ is given by $\varepsilon_{ijk}\partial_j v_k$, where $\varepsilon_{ijk}$ is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Levi-Civita_symbol&quot; rel=&quot;nofollow&quot;&gt;Levi-Civita symbol&lt;/a&gt;, that takes the values $1$, $-1$ and $0$ depending on the order in which the coordinates appear in the subscripts $ijk$. You can think of this as a notational shorthand for the plus and minus signs in the curl formula. So the left-hand side of the desired identity is&#xA;$$\varepsilon_{ijk}\partial_j (\varepsilon_{k\ell m}\partial_\ell A_m)$$&#xA;$$= \varepsilon_{ijk}\varepsilon_{k\ell m}\partial_j\partial_\ell A_m$$&#xA;because derivatives commute with constants and with each other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At this point, to simplify simplify $\varepsilon_{ijk}\varepsilon_{k\ell m}$, I just looked up the relevant identity of the Levi-Civita symbol, but it should be possible to derive it simply by algebraic manipulation from &lt;a href=&quot;http://en.wikipedia.org/wiki/Levi-Civita_symbol#Definition&quot; rel=&quot;nofollow&quot;&gt;the purely combinatorial definition of the symbol&lt;/a&gt;. It turns out that $\varepsilon_{ijk}\varepsilon_{k\ell m} = \delta_{i\ell}\delta_{jm} - \delta_{im}\delta_{j\ell}$, where $\delta_{ij}$ is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Kronecker_delta&quot; rel=&quot;nofollow&quot;&gt;Kronecker delta&lt;/a&gt; which is $1$ if and only if $i = j$, and $0$ otherwise. This symbol acts like a substitution operator: $\delta_{ij}v_j = v_i$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we have&#xA;$$\varepsilon_{ijk}\varepsilon_{k\ell m}\partial_j\partial_\ell A_m$$&#xA;$$= (\delta_{i\ell}\delta_{jm} - \delta_{im}\delta_{j\ell})\partial_j\partial_\ell A_m$$&#xA;$$= \partial_m\partial_i A_m - \partial_\ell\partial_\ell A_i$$&#xA;$$= \partial_i (\partial_m A_m) - (\partial_\ell\partial_\ell) A_i.$$&#xA;The first term is $\nabla(\nabla\cdot A)$, while the second is $-(\nabla\cdot\nabla)A$.&lt;/p&gt;&#xA;" OwnerUserId="856" LastEditorUserId="856" LastEditorDisplayName="" LastEditDate="2010-10-11T03:13:47.490" LastActivityDate="2010-10-11T03:13:47.490" />
  <row Id="6477" PostTypeId="2" ParentId="6475" CreationDate="2010-10-11T01:48:07.920" Score="5" ViewCount="" Body="&lt;p&gt;Cancel a factor of $\mu^\mu$ on both sides, move the powers of $\mu$ to the left, and move the factorials to the right to obtain the equivalent statement&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mu^{2h+1}\geq (\mu+h)(\mu+h-1)(\mu+h-2)\cdots(\mu-h+1)(\mu-h).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Break up the right hand side into the middle term, $\mu$, along with $h$ pairs of the form $(\mu+k)(\mu-k)$.  Since $(\mu+k)(\mu-k)=\mu^2-k^2\lt\mu^2$ for $k&gt;0$, it should now be clear.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-10-11T01:48:07.920" CommentCount="4" />
  <row Id="6478" PostTypeId="2" ParentId="6473" CreationDate="2010-10-11T02:09:09.537" Score="3" ViewCount="" Body="&lt;p&gt;The answer is $\phi((q-1)(p-1))$, but to compute that you probably need to factor $p-1$ and $q-1$.&lt;/p&gt;&#xA;" OwnerUserId="589" LastActivityDate="2010-10-11T02:09:09.537" />
  <row Id="6479" PostTypeId="2" ParentId="6466" CreationDate="2010-10-11T02:10:55.320" Score="6" ViewCount="" Body="&lt;p&gt;Every cyclic subgroup is abelian, hence nilpotent. Every such subgroup is generated by an element of the form $x^ay^b$ with $a$ and $b$ integers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In addition, $y^2$ commutes with $x$. Since $y^2$ is central, to determine the commutator of two elements we only need to consider the parity of the exponents of $y$ in their normal forms. Elements of the form $x^ay^{2b}$ and $x^ry^{2s}$ commute; elements of the form $x^ay^{2k+1}$ and $x^r$ commute if and only if $r=0$; and elements of the form $x^ay^{2k+1}$ and $x^ry^{2s+1}$ commute if and only if $a=r$. That gives you lots of abelian subgroups that are isomorphic to $\mathbb{Z}\times\mathbb{Z}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What about higher nilpotency? Suppose you have two noncommuting elements in your subgroup. It is not hard to check that $[x^ay^{2b+1},x^r] = x^{2r}$ and  $[x^ay^{2b+1},x^r y^{2s+1}] = x^{2a-2r}$. But no nontrivial power of $x$ commutes with an element of the form $x^a y^{2b+1}$; and further commutators will just yield further nontrivial powers of $x$ that still do not commute with an element of the form $x^ay^{2b+1}$. So if your subgroup $H$ has an element of the form $h_1=x^a y^{2b+1}$, and $h$ is any nontrivial element of $H$ that is not a power of $h_1$, then $h_1$ and $h$ do not commute, and $h_1$ does not commute with any of $[h_1,h]$, $[h_1,h,h_1]$, $[h_1,h,h_1,h_1],\ldots,[h_1,h,h_1,\ldots,h_1]$, etc. But if $H$ is nilpotent of class $c$, then any commutator of weight $c$ would be central in $H$. Thus, $H$ cannot be nilpotent. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the only  nilpotent subgroups of $H$ are abelian, and they are given as above.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-28T06:21:27.400" LastActivityDate="2010-11-28T06:21:27.400" CommentCount="1" />
  <row Id="6480" PostTypeId="2" ParentId="6459" CreationDate="2010-10-11T02:13:20.227" Score="2" ViewCount="" Body="&lt;p&gt;There is one important rule for assigning probabilities to edges in a Markov model:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Let $i$ be a node, and $out(i)$ be the set of nodes to which $i$ is connected (via an outgoing edge). Then $\sum_{j \in out(i)}{P_{ij}} = 1$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Consider your own model: As an instance, $out(4)=\{1\}$. Therefore, $P_{41}=1$. In other words, if we are in state 4, we will certainly go to state 1 at the next transition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now consider node 1. We have $out(1)= \{1,3,4\}$. Hence $P_{11}+P_{13}+P_{14}=1$. Using the counts:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$P_{11} = \frac{C_{11}}{C_{11}+C_{13}+C_{14}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$P_{13} = \frac{C_{13}}{C_{11}+C_{13}+C_{14}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$P_{14} = \frac{C_{14}}{C_{11}+C_{13}+C_{14}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm sure you can do the math for other nodes as well.&lt;/p&gt;&#xA;" OwnerUserId="2420" LastActivityDate="2010-10-11T02:13:20.227" CommentCount="1" />
  <row Id="6481" PostTypeId="2" ParentId="6466" CreationDate="2010-10-11T03:02:22.747" Score="3" ViewCount="" Body="&lt;p&gt;Here's a topological version of Arturo's argument.  (I'd leave this as a comment, but I don't have enough rep.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Every covering space of the Klein bottle is a surface, and it's fairly easy to convince yourself that the infinite-degree ones have cyclic, hence nilpotent, fundamental group.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It remains to consider the finite-degree covering spaces.  Well, these are all closed surfaces of Euler characteristic zero (because Euler characteristic is &lt;strike&gt;nilpotent&lt;/strike&gt; multiplicative), so by the classification of surfaces they are either tori or Klein bottles.  As already noted, the Klein-bottle group itself is not nilpotent, so only the abelian subgroups are left.&lt;/p&gt;&#xA;" OwnerUserId="2287" LastEditorUserId="2287" LastEditorDisplayName="" LastEditDate="2010-10-11T11:42:54.530" LastActivityDate="2010-10-11T11:42:54.530" CommentCount="2" />
  <row Id="6482" PostTypeId="1" CreationDate="2010-10-11T03:12:27.370" Score="1" ViewCount="294" Body="&lt;p&gt;How would I start proving&#xA;If&#xA;$$\lim_{(x,y)\to(a,b)} f(x,y) = L$$&#xA;and both the $x$ and $y$ limits exist&#xA;that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\lim_{x\to a}\left(\lim_{y\to b} f(x,y)\right)  = \lim_{y\to b}\left(\lim_{x\to a} f(x,y)\right) = L$$&lt;/p&gt;&#xA;" OwnerUserId="2319" LastEditorUserId="2319" LastEditorDisplayName="" LastEditDate="2010-10-11T03:40:19.317" LastActivityDate="2010-10-16T17:07:53.413" Title="Limits of 2 variable functions" Tags="&lt;real-analysis&gt;&lt;algebra&gt;&lt;limit&gt;" AnswerCount="2" />
  <row Id="6483" PostTypeId="2" ParentId="6482" CreationDate="2010-10-11T03:23:32.117" Score="6" ViewCount="" Body="&lt;p&gt;You cannot prove that the two-variable limit equals the iterated limits even if they both exist, since the two-variable limit may fail to exist even if both iterated limits exists and are equal. For example, take $f(x,y) = \frac{xy}{x^2+y^2}$, with $a=b=0$. The iterated limits both exist:&#xA;$$\lim_{x\to 0}\left(\lim_{y\to 0}\frac{xy}{x^2+y^2}\right) = \lim_{x\to 0}\frac{0}{x^2} = 0,$$&#xA;$$\lim_{y\to 0}\left(\lim_{x\to 0}\frac{xy}{x^2+y^2}\right) = \lim_{y\to 0}\frac{0}{y^2} = 0.$$&#xA;But the limit does not exist as $(x,y)\to(0,0)$. If you approach $(0,0)$ along the line $y=x$ you get the limit as $x\to 0$ of $\frac{x^2}{2x^2}$, which goes to $\frac{1}{2}$, not $0$. So the two-variable limit does not exist, even though the two iterated limits both exist and are equal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you assume that the two-sided limit exists and is equal to $L$, and then you want to prove that the iterated limits will therefore both exist and be equal, then you might have a better shot. Start by thinking about the definitions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added later:&lt;/strong&gt; And then look at Robin Chapman's example to see that it does not quite work out. So you don't get implications either way.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-10-16T17:07:53.413" LastActivityDate="2010-10-16T17:07:53.413" CommentCount="2" />
  <row Id="6484" PostTypeId="1" CreationDate="2010-10-11T03:35:03.267" Score="1" ViewCount="71" Body="&lt;p&gt;why in homological mirror symmetry, we restrict us to a projective variety (calabi-yau)? Because in physics we don't need this condition. What's the general picture for general calabi-yau?&lt;/p&gt;&#xA;" OwnerUserId="2381" LastActivityDate="2010-10-11T03:35:03.267" Title="about homological mirror symmetry" Tags="&lt;algebraic-geometry&gt;&lt;physics&gt;" CommentCount="1" />
  <row Id="6485" PostTypeId="1" CreationDate="2010-10-11T03:48:49.973" Score="2" ViewCount="234" Body="&lt;p&gt;Does Slater condition hold trivially (because there are no inequality constraints) for the problem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\min_{x,y} \:\: cx+dy$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;s.t.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$e^x + e^y = 1.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can I conclude there is a zero duality gap here?&lt;/p&gt;&#xA;" OwnerUserId="7076" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-12-10T19:41:44.140" LastActivityDate="2011-03-01T04:33:42.893" Title="Does Slater's condition hold for the following problem?" Tags="&lt;optimization&gt;&lt;nonlinear-optimization&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="6486" PostTypeId="2" ParentId="6485" CreationDate="2010-10-11T04:21:56.487" Score="-1" ViewCount="" Body="&lt;p&gt;You can rewrite the equality constraint as $\exp{x} + \exp{y} \leq 1$ and $-\exp{x} -\exp{y} \leq -1$. If slater conditions hold for this new formulation, (since the problem is convex), strong duality (gap zero) would follow.&lt;/p&gt;&#xA;" OwnerUserId="1527" LastActivityDate="2010-10-11T04:21:56.487" CommentCount="1" />
  <row Id="6487" PostTypeId="2" ParentId="6485" CreationDate="2010-10-11T04:28:25.087" Score="1" ViewCount="" Body="&lt;p&gt;A Naga says, you can rewrite the equality constraint $h(x)=0$ as a pair of inequality constraints&#xA;$$h(x) \leq 0$$&#xA;$$-h(x) \leq 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Slater condition would require&#xA;$$h(x)&amp;lt;0$$&#xA;$$-h(x)&amp;lt;0$$&#xA;which is an impossibility. So the Slater condition cannot be true of a problem with equality constraints. Other regularity conditions may be usable though.&lt;/p&gt;&#xA;" OwnerUserId="1195" LastActivityDate="2010-10-11T04:28:25.087" CommentCount="1" />
  <row Id="6488" PostTypeId="2" ParentId="6466" CreationDate="2010-10-11T04:32:52.427" Score="0" ViewCount="" Body="&lt;p&gt;Is there a shorter (yet elementary) argument showing that $G$ is not nilpotent?&lt;/p&gt;&#xA;" OwnerUserId="2424" LastActivityDate="2010-10-11T04:32:52.427" CommentCount="3" />
  <row Id="6489" PostTypeId="1" AcceptedAnswerId="6490" CreationDate="2010-10-11T04:45:45.797" Score="20" ViewCount="1259" Body="&lt;p&gt;As I'm sure many of you do, I read the &lt;a href=&quot;http://xkcd.com/&quot; rel=&quot;nofollow&quot;&gt;XKCD&lt;/a&gt; webcomic regularly. The &lt;a href=&quot;http://xkcd.com/804/&quot; rel=&quot;nofollow&quot;&gt;most recent one&lt;/a&gt; involves a joke about the Axiom of Choice, which I didn't get. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/ff9dS.png&quot; alt=&quot;The Banach-Tarski theorem was actually first developed by King Solomon, but his gruesome attempts to apply it set back set theory for centuries.&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I went to Wikipedia to see what the Axiom of Choice is, but as often happens with things like this, the Wikipedia entry is not in plain, simple, understandable language. Can someone give me a nice simple explanation of what this axiom is, and perhaps explain the XKCD joke as well?&lt;/p&gt;&#xA;" OwnerUserId="496" LastEditorUserId="496" LastEditorDisplayName="" LastEditDate="2010-10-11T06:10:32.383" LastActivityDate="2010-10-15T10:53:51.693" Title="Can you explain the &quot;Axiom of choice&quot; in simple terms?" Tags="&lt;set-theory&gt;&lt;axioms&gt;" AnswerCount="3" CommentCount="5" FavoriteCount="5" />
  <row Id="6490" PostTypeId="2" ParentId="6489" CreationDate="2010-10-11T05:11:47.357" Score="21" ViewCount="" Body="&lt;p&gt;The joke is really about the Banach-Tarski theorem, which says that you can cut up a sphere into a finite number of pieces which when reassembled give you two spheres of the same size as the original sphere. This theorem is extremely counterintuitive since we seem to be doubling volume without adding any material or stretching the material that we have.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The proof the the theorem requires the Axiom of Choice (AC), which says that if you have a collection of sets then there is a way to select one element from each set. It has been proved that AC cannot be derived from the rest of set theory but must be introduced as an additional axiom. Since AC can be used to derive counterintuitive results such as the Banach-Tarski theorem, some mathematicians are very careful to specify when their arguments depend on AC.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a formal statement of AC. Suppose we have a set $W$ and a rule associating a set $S_w$ to each $w \in W$. Then AC says that there is a function &#xA;$$f:W \to \bigcup_{w \in W} S_w$$&#xA;such that&#xA;$$f(w) \in S_w$$&lt;/p&gt;&#xA;" OwnerUserId="1195" LastEditorUserId="1195" LastEditorDisplayName="" LastEditDate="2010-10-15T01:00:54.307" LastActivityDate="2010-10-15T01:00:54.307" CommentCount="2" />
  <row Id="6491" PostTypeId="1" AcceptedAnswerId="6520" CreationDate="2010-10-11T05:22:46.290" Score="2" ViewCount="281" Body="&lt;p&gt;Reffered here by &lt;a href=&quot;http://mathoverflow.net/questions/41750/determining-n-in-sigma-x0n&quot; rel=&quot;nofollow&quot;&gt;http://mathoverflow.net/questions/41750/determining-n-in-sigma-x0n&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not entirely sure if this question falls under MathOverflow but neither of my Calculus AP  teachers in high school could help me with this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given $\Sigma_{x=0}^n {f(x)\over2}$ and the output of the summation, how would you find $n$? I've learned how to determine the $n$ given an arithmetic or geometric sequence, but not for an arbitrary function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Specifically, when $f(x) = 40 + 6\sqrt{x}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;12 Oct 2010. Edit:&#xA;It seems like I need to explain the entire situation for finding $n$, the number of trapezoids, for trapezoidal rule. It started on a simple review question for Calc AP and a TI-83 program that my calc teacher gave to me to solve the definite integral with trapezoidal rules. Aiming to major in Computer Science, I took it a bit further and completely took apart the program resulting in my original question on StackOverflow: &lt;a href=&quot;http://stackoverflow.com/questions/3886899/determining-the-input-of-a-function-given-an-output-calculus-involved&quot;&gt;http://stackoverflow.com/questions/3886899/determining-the-input-of-a-function-given-an-output-calculus-involved&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since there were tumbleweeds for a response, I took it as a personal challenge to reverse engineer the trapezoidal program into an algebraic form with my notes found on my forum: &lt;a href=&quot;http://www.zerozaku.com/viewtopic.php?f=19&amp;amp;t=6041&quot; rel=&quot;nofollow&quot;&gt;http://www.zerozaku.com/viewtopic.php?f=19&amp;amp;t=6041&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After reverse engineering the code into some algebra, I derived the formula: &#xA;$$TrapRule(A, B, N) = {(B-A)\over N}({F(A)\over2}+\sum_{k=1}^NF(k)+{F(B)\over2})$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given the values of A and B are constant for the definite integral, I should be able to isolate and solve for $N$. The problem, however, was determining $N$ in $sum_{k=1}^N$ and I came to the conclusion that it was an issue that I called &lt;em&gt;recursive complexity&lt;/em&gt; because it was impossible to determine without recursively adding for the summation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Eventually, I found MathOverflow and they referred me here. I was hoping only to get help on the issue for a summation because its beyond my skill as a high school student. Now that others have proposed other solutions for my dilemma, I guess I can throw out my thesis Dx&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for the help though, I'll definitely be returning for more.&lt;/p&gt;&#xA;" OwnerUserId="2426" LastEditorUserId="2426" LastEditorDisplayName="" LastEditDate="2010-10-13T01:46:16.893" LastActivityDate="2010-10-13T01:46:16.893" Title="Determining n in sigma ($\Sigma_{x=0}^n$)" Tags="&lt;calculus&gt;&lt;sequences-and-series&gt;" AnswerCount="7" />
  <row Id="6492" PostTypeId="2" ParentId="6491" CreationDate="2010-10-11T05:39:36.853" Score="3" ViewCount="" Body="&lt;p&gt;You are trying to calculate the inverse of the function&#xA;$$S(n)=\sum_{x=0}^n f(x)/2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general the inverse would not exist (think $f(x)=0$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the inverse exists, a trivial way to solve your problem would be to start calculating the sum for $n=0,1,2,\ldots$ and stop when the result matches the number provided.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are looking for a formula for $S^{-1}(n)$, there is no reason for there to be such a formula for every $f(x)$. The arithmetic and geometric progression examples you gave were lucky exceptions.&lt;/p&gt;&#xA;" OwnerUserId="1195" LastEditorUserId="1195" LastEditorDisplayName="" LastEditDate="2010-10-11T07:57:32.533" LastActivityDate="2010-10-11T07:57:32.533" />
  <row Id="6493" PostTypeId="2" ParentId="6470" CreationDate="2010-10-11T06:41:50.533" Score="4" ViewCount="" Body="&lt;p&gt;The isomorphism of $\mathrm{Hom}(V,W)$&#xA;with $V^*\otimes W$ only holds when $V$ is finite-dimensional. I'll assume&#xA;both $V$ and $W$ are finite-dimensional. If we write our group actions&#xA;on the right, if $\phi\in\mathrm{Hom}(V,W)$&#xA;then&#xA;$$\phi g:v\mapsto\phi(vg^{-1})g.$$&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-10-11T06:41:50.533" CommentCount="1" />
  <row Id="6494" PostTypeId="1" AcceptedAnswerId="6496" CreationDate="2010-10-11T06:50:55.073" Score="1" ViewCount="613" Body="&lt;p&gt;Can we have a measurable function $f$, whose inverse is not measurable?&lt;/p&gt;&#xA;" LastActivityDate="2010-10-11T18:46:38.530" Title="Example of a non measurable function!" Tags="&lt;analysis&gt;&lt;measure-theory&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="0" />
  <row Id="6495" PostTypeId="1" AcceptedAnswerId="6792" CreationDate="2010-10-11T06:57:35.303" Score="2" ViewCount="328" Body="&lt;p&gt;Think of $t$ and $r$ as two independent variables. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Suppose $E$ be a function of $r$ and $V$ be a function of $(t,r)$ such that both go to $0$ at $r=0$. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;There exists a positive function $M(r)$ such that $M(0)=0$ and $V(t,r) = -\frac{M(r)}{R(t,r)}$ where $R$ is another positive function such that $R(0,r)=r$. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Let $p(r) = \frac{E(r)}{V(0,r)}$ be a function regular at $r=0$ such that $p(0) \in (-\infty,1)$. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Also define a function $a$ of $r$ such that, $a(r) = \frac{M(r)}{\frac{4}{3}\pi r^3}$. &#xA;Then $a$ is also a positive definite function with a well-defined value at $r=0$.  &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Define $\alpha = a(0)$ &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Now look at this differential equation,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\dot{R}^2}{2} + V(t,r) = E(r)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Apparently this differential equation has a solution of the form,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{t}{t_0} = \sqrt{\frac{\alpha}{a(r)}}\frac{F(p(r))}{F(p(0))} \left [1 - \left ( \frac{R(t,r)}{r} \right)^{\frac{3}{2}}\frac{F\left( \frac{p(r)R(t,r)}{r} \right) }{F(p(r))} \right ] $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $t_0 = \sqrt {\frac{3}{8\pi \alpha}} F(p(0))$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;and the function $F$ is defined over the interval $(-\infty,1)$ as,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$F(x) = \left\{ \begin{array}{c c}&lt;br&gt;&#xA;                  -\frac{\sqrt{1-x}}{x} - \frac{1}{(-x)^{\frac{3}{2}}} tanh^{-1} \left [ \sqrt{\frac{x}{x-1}} \right ] &amp;amp; x&amp;lt;0 \\&lt;br&gt;&#xA;         \frac{2}{3}  &amp;amp;  x =0 \\&#xA;         \frac{1}{x^{\frac{3}{2}}}tan^{-1} \left [ \sqrt{\frac{x}{1-x}} \right ] - \frac{\sqrt{1-x}}{x}  &amp;amp; 0&amp;lt;x&amp;lt;1 &#xA;             \end {array} \right. $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How does one get the above solution? &lt;/p&gt;&#xA;" OwnerUserId="965" LastEditorUserId="274" LastEditorDisplayName="" LastEditDate="2010-10-13T20:25:42.500" LastActivityDate="2010-10-14T16:20:19.650" Title="A differential equation" Tags="&lt;differential-equations&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="6496" PostTypeId="2" ParentId="6494" CreationDate="2010-10-11T07:13:22.073" Score="1" ViewCount="" Body="&lt;p&gt;I think I have an answer that works now.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously the problem is trivial if we allow $f$ to be any function between measure spaces (just take the domain to have its powerset as $\sigma$-algebra and/or the trivial $\sigma$-algebra on the codomain), so this must be a question about real functions and the Lebesgue measure. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $X$ be the Cantor set, $Y$ a nonmeasureable set (and hence uncountable) contained in $[0,1]$ and $g: X \to Y$ be any bijective function. Then define &#xA;$$f(x) = \begin{cases}&#xA;  g(x) &amp;amp; \text{if } x \in X \\&#xA;  e^x + 2 &amp;amp; \text{otherwise.}&#xA;\end{cases}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let $U$ be any measurable set and let us show that $f^{-1}(U)$ is measurable. If $U \cap Y = \emptyset$ then $f^{-1}(U) \subset \mathbb{R} \setminus X$ on which $f$ is continuous. In particular $f^{-1}(U)$ must be measurable. If $U \cap Y \neq \emptyset$ then we split into two more cases. First consider the possibility that $U \cap (2,\infty) = \emptyset$. Then $f^{-1}(U) \subset X$, i.e. it is a null set and hence Lebesgue measurable. And if $U \cap (2,\infty) \neq \emptyset$ then by a similar argument we see that $f^{-1}(U)$ is the disjoint union of a null set contained in $X$ and a measurable subset of $\mathbb{R}\setminus X$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus $f$ is measurable, but its inverse is obviously not measurable since $(f^{-1})^{-1}(X) = f(X) = Y$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now if we also require $f$ to be bijective rather than just injective, then I don't know the answer. It seems much harder to construct an example in that case.&lt;/p&gt;&#xA;" OwnerUserId="1269" LastEditorUserId="1269" LastEditorDisplayName="" LastEditDate="2010-10-11T09:00:50.990" LastActivityDate="2010-10-11T09:00:50.990" />
  <row Id="6498" PostTypeId="2" ParentId="6489" CreationDate="2010-10-11T09:28:42.423" Score="15" ViewCount="" Body="&lt;p&gt;It might be a good idea to say something about the connection between the axiom of choice and the Banach-Tarski paradox.  The reason the Banach-Tarski paradox is counterintuitive is that we expect that if you split up a sphere into finitely many pieces, the total &quot;mass&quot; of all the pieces is still the same - so you shouldn't be able to put those pieces together again into something of twice the total &quot;mass.&quot;  The reason this reasoning doesn't apply is that the pieces in question &lt;strong&gt;don't have mass&lt;/strong&gt; at all!  This is not the same as saying that they have zero mass.  It's something much more terrifying: the notion of &quot;mass&quot; (which to a mathematician generally means something precise called &lt;a href=&quot;http://en.wikipedia.org/wiki/Measure_(mathematics)&quot; rel=&quot;nofollow&quot;&gt;measure&lt;/a&gt;) can't be defined for these pieces (in a way that still preserves all the reasonable properties we want of our notion of mass).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What does this have to do with the axiom of choice?  Well, it turns out that the axiom of choice is one way we can construct these bizarre pieces (&lt;a href=&quot;http://en.wikipedia.org/wiki/Non-measurable_set&quot; rel=&quot;nofollow&quot;&gt;non-measurable sets&lt;/a&gt;).  Without the axiom of choice, it's not possible to prove that such pieces exist.  With the axiom of choice, we can construct things like the &lt;a href=&quot;http://en.wikipedia.org/wiki/Vitali_set&quot; rel=&quot;nofollow&quot;&gt;Vitali set&lt;/a&gt; and like the pieces that occur in the Banach-Tarski paradox because AC greatly increases our ability to write down weird sets.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-10-11T09:28:42.423" CommentCount="5" />
  <row Id="6500" PostTypeId="2" ParentId="6491" CreationDate="2010-10-11T10:26:36.103" Score="0" ViewCount="" Body="&lt;p&gt;Further to Jyotirmoy's answer, for the $f(x)$ you have given&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$S(n)=\sum_{x=0}^{n}\frac{f(x)}{2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$=\sum_{x=0}^{n}\frac{(40+6\sqrt{x})}{2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$=\sum_{x=0}^{n}(20+3\sqrt{x})$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$=\sum_{x=0}^{n}20+3\sum_{x=0}^{n}\sqrt{x}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second sum is a generalised harmonic number, which doesn't have a nice closed form expression&lt;/p&gt;&#xA;" OwnerUserId="2219" LastEditorUserId="2219" LastEditorDisplayName="" LastEditDate="2010-10-11T10:40:03.420" LastActivityDate="2010-10-11T10:40:03.420" CommentCount="1" />
  <row Id="6501" PostTypeId="1" AcceptedAnswerId="6504" CreationDate="2010-10-11T10:46:33.987" Score="10" ViewCount="383" Body="&lt;p&gt;So, from what I understand, the axiom of choice is equivalent to the claim that every set can be &lt;a href=&quot;http://en.wikipedia.org/wiki/Well-order&quot; rel=&quot;nofollow&quot;&gt;well ordered&lt;/a&gt;. A set is well ordered by a relation, $R$ , if every subset has a least element. My question is: as anyone constructed a well ordering on the reals?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, I was going to ask this question about the rationals, but then I realised that if you pick your favourite bijection between rationals and the integers, this determines a well ordering on the rationals through the natural well order on $\mathbb{Z}$ . So it's not the denseness of the reals that makes it hard to well order them. So is it just the &lt;em&gt;size&lt;/em&gt; of $\mathbb{R}$ that makes it difficult to find a well order for it? Why should that be?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To reiterate:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Is there a known well order on the Reals?&lt;/li&gt;&#xA;&lt;li&gt;If there is, does a similar construction work for larger cardinalities?&lt;/li&gt;&#xA;&lt;li&gt;Is there a largest cardinality for which the construction works?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="172" LastEditorUserId="622" LastEditorDisplayName="" LastEditDate="2011-05-18T18:15:20.750" LastActivityDate="2011-05-18T18:15:20.750" Title="Is there a known well ordering of the reals?" Tags="&lt;set-theory&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="2" />
  <row Id="6502" PostTypeId="1" CreationDate="2010-10-11T11:06:28.470" Score="2" ViewCount="91" Body="&lt;p&gt;One of the definitions of stochastic continuity is: $\forall a &gt; 0$ and  $\forall s \geq 0$ $$\lim_{t\rightarrow s}\;P(|X(t)-X(s)|&gt;a) = 0.$$ What does it mean intuitively? I know that it implies the jumps but am not sure how and why. &lt;/p&gt;&#xA;" OwnerUserId="626" LastEditorUserId="1489" LastEditorDisplayName="" LastEditDate="2010-10-11T15:17:09.860" LastActivityDate="2010-10-11T15:35:24.707" Title="About stochastic continuity" Tags="&lt;stochastic-processes&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" />
  <row Id="6503" PostTypeId="2" ParentId="6501" CreationDate="2010-10-11T11:11:52.417" Score="7" ViewCount="" Body="&lt;p&gt;No, it's not just the size. One can &lt;a href=&quot;http://en.wikipedia.org/wiki/Hartogs_number&quot; rel=&quot;nofollow&quot;&gt;constructively prove&lt;/a&gt; the existence of large well-ordered sets, but for example even when one has the &lt;a href=&quot;http://en.wikipedia.org/wiki/First_uncountable_ordinal&quot; rel=&quot;nofollow&quot;&gt;first uncountable ordinal&lt;/a&gt; in hand, one can't show that it is in bijection with $\mathbb{R}$ without the continuum hypothesis. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;All the difficulty in the problem has to do with what you mean by &quot;constructed.&quot;  If one has a well-ordering on $\mathbb{R}$ then it is possible to carry out the construction of a &lt;a href=&quot;http://en.wikipedia.org/wiki/Vitali_set&quot; rel=&quot;nofollow&quot;&gt;Vitali set&lt;/a&gt;, which is a non-measurable subset of $[0, 1]$.  And &lt;a href=&quot;http://en.wikipedia.org/wiki/Solovays_model&quot; rel=&quot;nofollow&quot;&gt;it is known&lt;/a&gt; that the existence of non-measurable subsets of $\mathbb{R}$ is independent of ZF.  In other words, it is impossible to write down a well-ordering of $\mathbb{R}$ in ZF.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand given AC one can obviously write down a well-ordering in a non-constructive way (choose the first element, then the second element, then...).  This is probably not what you meant by &quot;construct,&quot; though.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-10-11T11:11:52.417" CommentCount="1" />
  <row Id="6504" PostTypeId="2" ParentId="6501" CreationDate="2010-10-11T11:37:37.577" Score="20" ViewCount="" Body="&lt;p&gt;I assume you know the general theorem that, using the axiom of choice, every set can be well ordered. Given that, I think you're asking how hard it is to actually define the well ordering. This is a natural question but it turns out that the answer may be unsatisfying.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, of course, without the axiom of choice it's consistent with ZF set theory that there is &lt;em&gt;no&lt;/em&gt; well ordering of the reals. So you can't just write down a formula of set theory akin to the quadratic formula that will &quot;obviously&quot; define a well ordering. Any formula that does define a well-ordering of the reals is going to require a nontrivial proof to verify that it's correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, there is not even a formula that unequivocally defines a well ordering of the reals in ZFC.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;The theorem of &quot;Borel determinacy&quot; implies that there is no well ordering of the reals whose graph is a Borel set. This is provable in ZFC.  The stronger hypothesis of &quot;projective determinacy&quot; implies there is no well ordering of the reals definable by a formula in the projective hierarchy. This is consistent with ZFC but not provable in ZFC. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Worse, it's even consistent with ZFC that &lt;em&gt;no&lt;/em&gt; formula in the language of set theory defines a well ordering of the reals (even though one exists). That is, there is a model of ZFC in which no formula defines a well ordering of the reals. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;A set theorist could tell you more about these results.  They are in the set theoretic literature but not in the undergraduate literature. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a positive result. If you work in $L$ (that is, you assume the axiom of constructibility) then a specific formula is known that defines a well ordering of the reals in that context.  However, the axiom of constructibility is not provable in ZFC (although it is consistent with ZFC), and the formula in question does not define a well ordering of the reals in arbitrary models of ZFC. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A second positive result, for relative definability. By looking at the standard proof of the well ordering principle (Zermelo's proof), we see that there is a single formula $\phi(x,y,z)$ in the language of set theory such that if we have any choice function $F$ on the powerset of the reals then the formula $\psi(x,y) = \phi(x,y,F)$ defines a well ordering of the reals, in any model of ZF that happens to have such a choice function. Informally, this says that the reason the usual proof can't explicitly construct a well ordering is because we can't explicitly construct the choice function that the proof takes as an input. &lt;/p&gt;&#xA;" OwnerUserId="630" LastEditorUserId="630" LastEditorDisplayName="" LastEditDate="2010-10-11T11:46:23.770" LastActivityDate="2010-10-11T11:46:23.770" />
  <row Id="6505" PostTypeId="1" AcceptedAnswerId="6524" CreationDate="2010-10-11T11:42:59.813" Score="4" ViewCount="264" Body="&lt;p&gt;Online, I have read contradicting opinions on whether &lt;code&gt;intersect&lt;/code&gt; should take precedence over &lt;code&gt;union&lt;/code&gt; (by analogy to logical &lt;code&gt;and&lt;/code&gt; and &lt;code&gt;or&lt;/code&gt;), or whether all set operators should have equal precedence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which way makes more sense and why?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And where does &lt;code&gt;difference&lt;/code&gt; fit in?  (I'd say it should be equal precedence to &lt;code&gt;intersect&lt;/code&gt; because &lt;code&gt;A - B = A intersect B'&lt;/code&gt;.)&lt;/p&gt;&#xA;" OwnerUserId="2430" LastEditorUserId="329" LastEditorDisplayName="anon" LastEditDate="2010-11-23T04:05:06.540" LastActivityDate="2010-11-23T04:05:06.540" Title="Precedence of set union, intersect, and difference?" Tags="&lt;notation&gt;&lt;elementary-set-theory&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="6506" PostTypeId="2" ParentId="6451" CreationDate="2010-10-11T12:36:42.130" Score="1" ViewCount="" Body="&lt;p&gt;Starting from a flat rectangular Mobius strip $ABA'B'$ with the identification $AB'\cong A'B$. The side $AB$ is to be shrunk to a point and similarly the side $A'B'$ to obtain the identification of the boundary circle to a point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To construct a transformation, we take the side $AB$ is along the x axis and $BA'$ along the y-axis of a Cartesian system. We take the origin of the system at the center of the rectangle and the side  lengths 2a (along the x-axis) and 2b along the (y-axis). ($AB$ is along the line $y=-b$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then the transformation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$r = \sqrt {a^2b^2-(a^2-x^2)(b^2-y^2)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\theta = \arcsin{\frac{y}{b}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;($r$ and $\theta$ are polar coordinates) maps the Mobius strip to a disc of radius $ab$ with the boundary antipodal points identified (which is homeomorphic to $RP^2$. The identified lines $AB$ $A'B'$ are mapped to  south and the north poles respectively. The lines $BA'$ and $B'A$ are mapped to the right and left boundary semicircles.&lt;/p&gt;&#xA;" OwnerUserId="1430" LastActivityDate="2010-10-11T12:36:42.130" />
  <row Id="6507" PostTypeId="1" AcceptedAnswerId="6642" CreationDate="2010-10-11T12:40:41.460" Score="1" ViewCount="270" Body="&lt;p&gt;I am of, and I would like to retain a, mindset that mathematics does not have to have numbers as the central object of interest. With that in mind, I have done a fair amount of self-study on topics in undergrad modern algebra, looking for examples to show this is the case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For objects like groups, it is often stated explicitly, at some point, that group elements can be though of as automorphisms on a particular set. I find this pretty interesting, that numbers do not come into play at all here. This makes groups pretty useful for problems that have symmetries.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, for objects like fields and rings, it seems that their application often ends up dealing with numbers somehow. I'm aware there are probably a handful of cases where rings and fields do not show up as numbers, but those tend to be side-cases that are not of main study. I'm only vaguely aware of Galois theory and it's uses for fields as extensions of groups, so perhaps that is the direction I should follow there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But as for rings, aside from the example I just gave, I still come up a bit short on any powerful or broad theorems that make use of those two operations but do not involve numbers in some fundamental way. Perhaps math.stackexchange can prove me wrong?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S. Answers with really good links to books or other resources will at least get an upvote.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;An update for those who pointed out my lack of precision.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So admittedly, the choice of words was bad. I did use the phrase &quot;involve numbers in some fundamental way&quot;, which, in my mind, means &quot;does not use facts about numbers&quot;. Perhaps I can  rephrase it better.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose I have a group (G, *). There are many examples of groups if I want to let G be the integers, rationals, reals, complex, etc, where the group operation (*) can be addition, multiplication, or some other operation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, each of these are &quot;numerical&quot; examples. There are groups which do not behave like numbers, if we let G be the set of all bijections on a set S, and let (*) be the composition, we may not be able to find a group where the set S is a collection of numbers and (*) is a &quot;numerical&quot; operation on them. Our group operation (*) may end just being some random rearrangement of numbers that does not take advantage of the properties of numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this way, and as others often point out, all groups are just specific cases of bijections &amp;amp; composition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now suppose I have a ring (R, +, *). The ring of functions on integers, rationals, reals, etc. takes advantage of the properties of numbers when defining (+) and (*). As would the ring of polynomials, or the ring of matrices. Each of these examples of rings, while the set R is not a set of numbers, ends up being an abstraction of numbers, and the operations (+) and (*) make use of this fact.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;An example of what I mean&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To give a example (counterexample?) of what I mean, try Boolean rings. The operations (+) and (*) are isomorphic to the logical operations &quot;xor&quot; and &quot;and&quot;. There is also examples of lattices with join and meet. One could try to squeeze in natural numbers into this with lcd and gcd, but these properties are not unique to numbers so they are not exactly &quot;numerical&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to add more structure, say an additive and multiplicative identity, you can use  Stone's Representation Theorem to show it is isomorphic to a collection of sets with symmetric difference and intersection.&lt;/p&gt;&#xA;" OwnerUserId="1892" LastEditorUserId="1892" LastEditorDisplayName="" LastEditDate="2010-10-12T20:31:23.423" LastActivityDate="2010-10-18T02:32:36.800" CommunityOwnedDate="2010-10-12T20:31:23.423" Title="Are there broad or powerful theorems of rings that do not involve the familiar numerical operations (+) and (*) in some fundamental way?" Tags="&lt;abstract-algebra&gt;&lt;soft-question&gt;&lt;group-theory&gt;&lt;ring-theory&gt;&lt;field-theory&gt;" AnswerCount="6" CommentCount="12" FavoriteCount="1" />
  <row Id="6508" PostTypeId="1" AcceptedAnswerId="6514" CreationDate="2010-10-11T13:01:03.740" Score="0" ViewCount="123" Body="&lt;p&gt;I want to prove that $P(|A+B|&gt;c)\leq P(|A|&gt;c/2)+P(|B|&gt;c/2)$, where $A$ and $B$ are random variables and $c&gt;0$.&lt;/p&gt;&#xA;" OwnerUserId="626" LastActivityDate="2010-10-13T04:20:12.140" Title="how to do i prove this simple probabilty result" Tags="&lt;probability&gt;" AnswerCount="2" />
  <row Id="6509" PostTypeId="2" ParentId="6491" CreationDate="2010-10-11T13:05:33.530" Score="0" ViewCount="" Body="&lt;p&gt;Perhaps you can also work it out by approximating (lower and upper bounding) the sum by an integral:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \int_0^N \sqrt{x} \; dx \le \sum_0^N \sqrt x \le \int_0^N  \sqrt{x+1} \; dx$$&lt;/p&gt;&#xA;" OwnerUserId="312" LastActivityDate="2010-10-11T13:05:33.530" />
  <row Id="6510" PostTypeId="2" ParentId="6507" CreationDate="2010-10-11T13:07:48.627" Score="1" ViewCount="" Body="&lt;p&gt;If $F$ is a field, then $F[X]$ is a PID.&#xA;If $D$ is a UFD, then so is $D[X]$.&#xA;These are broad and powerful theorems.&lt;/p&gt;&#xA;" OwnerUserId="589" LastActivityDate="2010-10-11T13:07:48.627" />
  <row Id="6511" PostTypeId="2" ParentId="6507" CreationDate="2010-10-11T13:21:44.340" Score="0" ViewCount="" Body="&lt;p&gt;The question is not very well defined. For example, the nullstellensatz shows that the points in $\mathbb{C}^n$ (or any algebraically closed field) are in correspondence with the maximal ideals in $\mathbb{C}[x_1,...,x_n]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Indeed this correspondence between geometry and algebra (notice that we can talk about generalized points: prime ideals in $\mathbb{C}[x_1,...,x_n]$) is what's in the &quot;core&quot; of modern algebra. It's at the very least the main reason that we look at rings (arguably the primary reason we look at fields is Galois Theory; and Group Theory has intrinsic value).&lt;/p&gt;&#xA;" LastActivityDate="2010-10-11T13:21:44.340" />
  <row Id="6512" PostTypeId="2" ParentId="6491" CreationDate="2010-10-11T13:26:18.930" Score="0" ViewCount="" Body="&lt;p&gt;Following on from leonbloy's answer, one can use the &lt;a href=&quot;http://en.wikipedia.org/wiki/Euler-Maclaurin_formula&quot; rel=&quot;nofollow&quot;&gt;Euler--Maclaurin summation formula&lt;/a&gt; to obtain&#xA;a rather precise approximation of the sum via an integral.  In particular, the error&#xA;term in that formula involves higher derivatives of the function $f(x)$.  Since &#xA;the higher derivatives of your function tend to zero as $x \to \infty$, you should be&#xA;able to obtain reasonably precise approximations by playing around a bit.&lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2010-10-11T13:26:18.930" CommentCount="1" />
  <row Id="6513" PostTypeId="2" ParentId="6507" CreationDate="2010-10-11T13:30:33.143" Score="0" ViewCount="" Body="&lt;p&gt;Well,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the &lt;a href=&quot;http://en.wikipedia.org/wiki/Wedderburn%27s_little_theorem&quot; rel=&quot;nofollow&quot;&gt;Wedderburn's theorem&lt;/a&gt;, &lt;a href=&quot;http://en.wikipedia.org/wiki/Gauss%27s_lemma_%28polynomial%29&quot; rel=&quot;nofollow&quot;&gt;Gauss Lemma,&lt;/a&gt; and Eisentein's Criteria are good ones.&lt;/p&gt;&#xA;" LastActivityDate="2010-10-11T13:37:04.897" />
  <row Id="6514" PostTypeId="2" ParentId="6508" CreationDate="2010-10-11T13:31:31.367" Score="1" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Hint:&lt;/strong&gt; Use the following: If $|A|\leq c/2$ and $|B|\leq c/2$, then $|A+B|\leq c$. It might be useful to negate this statement and then use the subadditivity of the probability measure.&lt;/p&gt;&#xA;" OwnerUserId="367" LastActivityDate="2010-10-11T13:31:31.367" />
  <row Id="6516" PostTypeId="2" ParentId="6507" CreationDate="2010-10-11T13:39:38.853" Score="1" ViewCount="" Body="&lt;p&gt;Rings of &quot;functions&quot; (vs. numbers) are heavily studied - whether formally or truly functionally. Even in a first algebra course you'll find significant theorems on polynomials and rational functions, rings of linear transformations on vector spaces (matrices)  and, perhaps, some nonlinear analogs in commutative algebra (e.g. via Grobner bases). Combinatorics makes heavy use of generating functions (formal power series). Algebraicized analysis employs differential/difference algebra to model elementary functions and their asymptotics (&lt;a href=&quot;http://arxiv.org/abs/0801.4877&quot; rel=&quot;nofollow&quot;&gt;transseries&lt;/a&gt;), e.g. Liouville-Risch theory of integration/summation in finite terms, Rosenlicht's Hardy fields. Differential Galois theory studies solutions of differential equations and similaly for the difference analog (recurrences).  Entire books have been written on rings of continuous functions (e.g. Gillman and Jerison). Abe Robinson's NSA (nonstandard analysis) is another algebraic approach to analysis.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-10-18T02:32:36.800" LastActivityDate="2010-10-18T02:32:36.800" />
  <row Id="6517" PostTypeId="2" ParentId="6244" CreationDate="2010-10-11T13:48:07.590" Score="9" ViewCount="" Body="&lt;p&gt;No transcendental numbers are needed for this question.  Any set of algebraic numbers of unbounded degree spans a vector space of infinite dimension.  Explicit examples of linearly independent sets of algebraic numbers are also relatively easy to write down. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The set $\sqrt{2}, \sqrt{\sqrt{2}}, \dots, = \cup_{n&gt;0} 2^{2^{-n}} $ is linearly independent over $Q$. (Proof: Any expression of the $n$th iterated square root $a_n$ as a linear combination of earlier terms $a_i, i &amp;lt; n$ of the sequence could also be read as a rational polynomial of degree dividing $2^{n-1}$ with $a_n$ as a root and this contradicts the irreducibility of $X^m - 2$, here with $m=2^n$). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The square roots of the prime numbers are linearly independent over $Q$.  (Proof: this is immediate given the ability to extend the function &quot;number of powers of $p$ dividing $x$&quot; from the rational numbers to algebraic numbers. $\sqrt{p}$ is &quot;divisible by $p^{1/2}$&quot; while any finite linear combination of square roots of other primes is divisible by an integer power of $p$, i.e., is contained in an extension of $Q$ unramified at $p$). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Generally any infinite set of algebraic numbers that you can easily write down and is not dependent for trivial reasons usually is independent. This because the only algebraic numbers for which we have a simple notation are fractional powers, and valuation (order of divisibility) arguments work well in this case.  Any set of algebraic numbers where, of the ones ramified at any prime $p$, the amount of ramification is different for different elements of the set, will be linearly independent.  Proof: take the most ramified element in a given linear combination, express it in terms of the others, and compare valuations.&lt;/p&gt;&#xA;" OwnerUserId="467" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-10-11T14:09:27.680" LastActivityDate="2010-10-11T14:09:27.680" CommentCount="16" />
  <row Id="6518" PostTypeId="2" ParentId="2037" CreationDate="2010-10-11T14:03:30.083" Score="1" ViewCount="" Body="&lt;p&gt;This isn't a complete answer to your question, but at least it is a closed form. For 5 normal (6-sided), dice, the exact probability of achieving a yahtzee, following the optimum strategy, given $n$ rolls is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$1+\frac{53}{13} \left(\frac{5}{6}\right)^{2n+1}+\frac{11 \cdot 5^n}{13 \cdot 2^{n+5} \cdot 3^{3n+1}}  -\frac{5^n}{8 \cdot 3^{2n-2}}  -\frac{7 \cdot 5^{n+1}}{ 2^{n+5} 3^{n-1}} $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, setting $n=3$ gives $\frac{347897}{7558272}$.&lt;/p&gt;&#xA;" OwnerUserId="130" LastActivityDate="2010-10-11T14:03:30.083" />
  <row Id="6519" PostTypeId="1" CreationDate="2010-10-11T14:08:35.300" Score="13" ViewCount="241" Body="&lt;p&gt;Let $M$ be an $n \times n$ matrix whose elements are random reals in [0,1].&#xA;Two questions.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt; What is the growth rate of the magnitude of the elements of $M^k$ as a function of $k$?  It is definitely&#xA;exponential, but maybe the exponent is known?&lt;/li&gt;&#xA;&lt;li&gt; Is it the case that eventually one element of $M^k$ dominates, as $k \rightarrow \infty$?&#xA;I have some ambiguous experimental evidence that this is the case, but because&#xA;of the exponential growth, exact computation is difficult, rendering my &quot;evidence&quot; tenuous&#xA;at best and perhaps worthless.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;One can ask the same question for matrices whose elements are random reals in [-1,1],&#xA;or random 0's and 1's, or random choices among $\lbrace -1, 0, 1\rbrace$, ...&#xA;These question have likely been studied.  Thanks for pointers and/or ideas! &lt;/p&gt;&#xA;" OwnerUserId="237" LastActivityDate="2010-10-11T19:51:33.727" Title="Powers of random matrices" Tags="&lt;probability&gt;&lt;reference-request&gt;&lt;matrices&gt;&lt;random&gt;" AnswerCount="4" FavoriteCount="2" />
  <row Id="6520" PostTypeId="2" ParentId="6491" CreationDate="2010-10-11T14:35:37.783" Score="3" ViewCount="" Body="&lt;p&gt;An approximation to $\displaystyle \sum_{k=1}^{n} \sqrt{k}$ can be found here (on this very site): &lt;a href=&quot;http://math.stackexchange.com/questions/5676/how-closely-can-we-estimate-sum-i0n-sqrti/5728#5728&quot;&gt;http://math.stackexchange.com/questions/5676/how-closely-can-we-estimate-sum-i0n-sqrti/5728#5728&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It was shown that $\displaystyle \frac{2n\sqrt{n}}{3} + \frac{\sqrt{n}}{2} -\frac{2}{3} &amp;lt; \sum_{k=1}^{n} \sqrt{k} &amp;lt; \frac{2n\sqrt{n}}{3} + \frac{\sqrt{n}}{2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus if $\displaystyle S = \sum_{k=0}^{n} f(x)/2$, then &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{k=0}^{n} \sqrt{n} = \frac{S - 10n(n+1)}{3}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the approximation above&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{2n\sqrt{n}}{3} + \frac{\sqrt{n}}{2} = \frac{S - 10n(n+1)}{3}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a fourth degree equation in $\sqrt{n}$ which can be solved exactly (closed formula) in terms of $S$ and would give you a value of $n$ which is close to $n$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My guess is that taking the integer part of the square of an appropriate root will be sufficient to give $n$ or $n-1$ (and so a formula might exist, after all!)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are not looking for a formula, but a procedure, you can always try binary search.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope that helps.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-10-11T14:59:40.443" LastActivityDate="2010-10-11T14:59:40.443" CommentCount="1" />
  <row Id="6521" PostTypeId="2" ParentId="6502" CreationDate="2010-10-11T14:57:15.267" Score="2" ViewCount="" Body="&lt;p&gt;$a$ is a kind of 'tolerance' parameter on $X$. Imagine an adversary of yours setting this parameter at some level, as small as they wish. They also set an upper bound, say $\epsilon$, on the proportion of possible sample paths that may violate the tolerance condition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You on the other hand are allowed to choose a time window around the time $s$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Continuity then requires that regardless of how tight the two tolerances are set by your adversary, you can always choose a time window narrow enough that within that time window the values of $X$ remain within a distance $a$ of $X(s)$ for no less than $(1-\epsilon)$ proportion of the possible sample paths.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The tricky part here is that you are not required to commit to the sample paths which will violate the tolerance. Depending on the choice of $a$ and $\epsilon$ you may allow different sample paths to 'fail'. All that is required is that the proportion of paths that fail be less than $\epsilon$. This makes your task easier. Therefore this type of continuity is know as &lt;em&gt;weak&lt;/em&gt; continuity. On the other hand we could have required that almost all sample paths be continuous in the ordinary deterministic sense:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$P\left(\lim_{t\to s} X(t)=X(s)\right)=1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which would have been a stronger condition to fulfil.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As Rasmus points out, these different notions of continuity are a direct counterpart of the more fundamental notions of convergence of random variables.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;['Proportion' above should be interpreted in terms of probability mass.]&lt;/p&gt;&#xA;" OwnerUserId="1195" LastEditorUserId="1195" LastEditorDisplayName="" LastEditDate="2010-10-11T15:35:24.707" LastActivityDate="2010-10-11T15:35:24.707" CommentCount="1" />
  <row Id="6522" PostTypeId="2" ParentId="6519" CreationDate="2010-10-11T15:14:10.343" Score="2" ViewCount="" Body="&lt;p&gt;Powers of a matrix are more easily calculated by first diagonalizing it. Let $P, D$ be matrizes, $D$ being a diagonal matrix with $M = PDP^{-1}$, then $$M^k = PD^{k}P^{-1}.$$ The entries in D are the eigenvalues of $M$, so the entries in $M^k$ are growing exponentially with the rate of the logarithm of the largest eigenvalue of $M$, each entry being a linear combination of these eigenvalues.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So to point 2: No entry should dominate the others significantly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is only a partial answer.. maybe you can find something about the distribution of eigenvalues of random matrizes. =)&lt;/p&gt;&#xA;" OwnerUserId="505" LastEditorUserId="505" LastEditorDisplayName="" LastEditDate="2010-10-11T15:37:00.253" LastActivityDate="2010-10-11T15:37:00.253" CommentCount="2" />
  <row Id="6523" PostTypeId="2" ParentId="6505" CreationDate="2010-10-11T15:39:34.557" Score="6" ViewCount="" Body="&lt;p&gt;Since there are contradicting opinions, I recommend assuming that whoever is reading your work assumes a different precedence order than you do, and using lots of parentheses. &lt;/p&gt;&#xA;" OwnerUserId="173" LastActivityDate="2010-10-11T15:39:34.557" CommentCount="1" />
  <row Id="6524" PostTypeId="2" ParentId="6505" CreationDate="2010-10-11T15:43:33.797" Score="6" ViewCount="" Body="&lt;p&gt;There is no sensible way of preferring one of intersection and union more than the other, since complement switches them.  I don't think you should assume an order at all and you should always use parentheses.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-10-11T15:43:33.797" />
  <row Id="6526" PostTypeId="1" AcceptedAnswerId="6528" CreationDate="2010-10-11T16:08:48.173" Score="3" ViewCount="169" Body="&lt;p&gt;Given an $n &gt;0$ is it possible to partition the set $\mathcal{P} = \{1,2, \cdots, 2n\}$ into $n$ pairs $(a_{i},b_{i})$ such that $a_{i} + b_{i}$ is a prime?&lt;/p&gt;&#xA;" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-11-23T06:45:12.477" LastActivityDate="2010-11-23T06:45:12.477" Title="Partitioning sets such that the sum of 2 elements is Prime" Tags="&lt;number-theory&gt;&lt;combinatorics&gt;&lt;prime-numbers&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="6528" PostTypeId="2" ParentId="6526" CreationDate="2010-10-11T16:18:48.790" Score="10" ViewCount="" Body="&lt;p&gt;Yes.  The proof is by strong induction.  The base case is obvious.  By &lt;a href=&quot;http://en.wikipedia.org/wiki/Bertrands_postulate&quot; rel=&quot;nofollow&quot;&gt;Bertrand's postulate&lt;/a&gt; there exists a prime $p$ between $2n+1$ and $4n$, so pick the pairs $\{ p-2n, 2n \}, \{ p-2n+1, 2n-1 \}, ...$ and so forth.  Now it remains to pair up the numbers $\{ 1, 2, ... p-2n-1 \}$, which is possible by the inductive hypothesis.  &lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-10-11T16:25:14.707" LastActivityDate="2010-10-11T16:25:14.707" />
  <row Id="6529" PostTypeId="2" ParentId="6519" CreationDate="2010-10-11T16:33:36.780" Score="2" ViewCount="" Body="&lt;p&gt;The following analysis treats only the particular case of symmetric matrices $M$, which can be diagonalized by&#xA;an orthogonal transformation. The key observation is that the diagonalizing&#xA;matrix $P$ and the eigenvalue matrix $D$ are statistically independent. The diagonalizing matrix&#xA;elements are $O(n)$ Haar distributed. Now, it is easy to see that every matrix element is a scalar product of&#xA;two $O(n)$ Haar distributed n-vectors weighted by the k-th power of the eigenvalues. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;At large values of k, the maximal eigenvalue dominates all eigenvalues and the contribution from the diagonalizing matrix is fixed. &#xA;Thus the growth rate of all matrix elements is the natural logarithm of the maximal eigenvalue.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moreover, the contribution of the maximal eigenvalue to the $(i,j)$ element is given by $P_{im} D_{mm}^k  P_{jm}$&#xA;($D_{mm}$ is the maximal eigenvalue). Thus the matrix element for which the product $P_{im} P_{jm}$ is maximal dominates all matrix elements of the power matrix.&#xA;This element will be most likely on the diagonal because it is multiplied by the square of a Haar distributed element having a non-zero mean.&#xA;The off-diagonal elements are multiplied by the product of two different Haar distributed elements thus having a zero mean.&lt;/p&gt;&#xA;" OwnerUserId="1430" LastActivityDate="2010-10-11T16:33:36.780" CommentCount="2" />
  <row Id="6530" PostTypeId="2" ParentId="6519" CreationDate="2010-10-11T16:35:03.340" Score="5" ViewCount="" Body="&lt;p&gt;The best results have been obtained for random matrices with Normally distributed entries, but some of them--like &lt;a href=&quot;http://mathworld.wolfram.com/WignersSemicircleLaw.html&quot; rel=&quot;nofollow&quot;&gt;Wigner's Circular Law&lt;/a&gt;--extend to uniformly distributed entries.  (Independence is crucial.)  Wigner's law applies only to symmetric random matrices.  &lt;a href=&quot;http://mathworld.wolfram.com/GirkosCircularLaw.html&quot; rel=&quot;nofollow&quot;&gt;Girko's Circular Law&lt;/a&gt; holds without the symmetry condition, but afaik requires Normally distributed entries.  It says that for largish $n$ the eigenvalues are approximately uniformly distributed on a disk.  For smaller $n$, before these asymptotics are reached, there is a preference for real eigenvalues extending beyond the disk.  At any rate, these asymptotics will immediately give you the distribution of eigenvalues of positive integral powers of such matrices, especially when you consider that the probability all eigenvalues are distinct (and therefore the matrix is diagonalizable over $\mathbb{C}$) is one.  For example, the largest eigenvalue of $M^k$ will be approximately $n^{k/2} \left( 1 - \frac{1}{2} (3 \pi / (2 n) ) ^ {2/3} \right) ^ k$.&lt;/p&gt;&#xA;" OwnerUserId="1489" LastActivityDate="2010-10-11T16:35:03.340" CommentCount="2" />
  <row Id="6531" PostTypeId="1" CreationDate="2010-10-11T16:39:19.573" Score="1" ViewCount="217" Body="&lt;p&gt;Given n segments of line (into the X axis) with coordinates [li; ri]. You are to choose the&#xA;minimum number of segments that cover the segment [0;M]. All segment are within [0,M].&lt;/p&gt;&#xA;" OwnerUserId="1997" LastActivityDate="2010-10-12T21:09:38.420" Title="line segment covering problem" Tags="&lt;algorithms&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="6532" PostTypeId="1" CreationDate="2010-10-11T16:44:21.370" Score="3" ViewCount="92" Body="&lt;p&gt;Let $k$ be a field, $V$ a finite-dimensional $k$-vectorspace and $M \in End(V)$. How can I determine $Z$, the centralizer of $M \otimes M$ in $End(V) \otimes End(V)$? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, if &#xA;$$M=[[1,0],[0,2]],$$ then $M$ is 6-dimensional, consisting of block matrices of shape 1,2,1. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was confused at first, because this seems to be a contradiction to the fact that the centralizer of a subalgebra of the form $A \otimes B$ is just the tensor product of the centralizers of $A$ and $B$; but here we are considering only the element $M \otimes M$, not $A \otimes A$, where $A$ is the subalgebra generated by $M$.&lt;/p&gt;&#xA;" OwnerUserId="2436" LastActivityDate="2010-10-11T18:18:45.940" Title="Centralizer of $M \otimes M$" Tags="&lt;algebra&gt;&lt;tensors&gt;" AnswerCount="1" />
  <row Id="6533" PostTypeId="2" ParentId="4783" CreationDate="2010-10-11T16:45:57.253" Score="6" ViewCount="" Body="&lt;p&gt;This does not exactly answer the question (which was answered very nicely by Qiaochu), but I am adding this as a curiosity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can write $\displaystyle \sum_{k=0}^{n} {n \choose k}^{-1}$ as an integral (which can be generalized to other expressions involving the reciprocals of the binomial coefficients).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have the following beta integral identity&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$${n \choose k}^{-1} = (n+1)\int_{0}^{1} t^{n-k}(1-t)^k \ dt$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(For a nice application of Beta integrals, see this answer by Robin Chapman: &lt;a href=&quot;http://math.stackexchange.com/questions/5058/formula-for-the-harmonic-series-h-n-sum-k1n-1-k-due-to-gregorio-fontana/5234#5234&quot;&gt;Formula for the harmonic series due to Gregorio Fontana.&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{k=0}^{n} {n \choose k}^{-1} = (n+1) \sum_{k=0}^{n} \int_{0}^{1} t^{n-k}(1-t)^k \ dt$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ = (n+1)\int_{0}^{1} \sum_{k=0}^{n} t^{n-k}(1-t)^k \ dt$$&#xA;$$ = (n+1)\int_{0}^{1} \frac{t^{n+1} - (1-t)^{n+1}}{2t-1} \ dt$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We should be able to use the analytical tools available to investigate the properties of this integral.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, at least two different proofs that the limit is $2$ can be found here: &lt;a href=&quot;http://math.stackexchange.com/questions/6544/different-proofs-of-lim-n-rightarrow-infty-n-int-01-xn-1-xn&quot;&gt;http://math.stackexchange.com/questions/6544/different-proofs-of-lim-n-rightarrow-infty-n-int-01-xn-1-xn&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For an example of a different use of this integral formula:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Setting $\displaystyle 2t-1 = y$ we get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{k=0}^{n} {n \choose k}^{-1} = \frac{n+1}{2^{n+2}} \int_{-1}^{1} \frac{(1+y)^{n+1} - (1-y)^{n+1}}{y} \ dy$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Expanding out the right side gives us this nice looking formula&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{k=0}^{n} {n \choose k}^{-1} = \frac{n+1}{2^n}(\frac{{n+1 \choose 1}}{1} + \frac{{n+1 \choose 3}}{3} + \frac{{n+1 \choose 5}}{5} + \dots)$$&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-10-12T16:13:56.180" LastActivityDate="2010-10-12T16:13:56.180" />
  <row Id="6534" PostTypeId="2" ParentId="6531" CreationDate="2010-10-11T16:50:36.197" Score="1" ViewCount="" Body="&lt;p&gt;I think the greedy algorithm will work.  You have to cover 0, so search all segments with a left end of zero and take the longest.  Now search all segments with a left end that you have already covered and take the one with the greatest right end.  Continue until you hit M.  At each stage you have a minimal solution for the range so far.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If overlap (see the comment below) is not allowed, you need a tree structure.  Make a list of all the segments that have left end zero.  For each one, attach all segments you can, that have the same left end as the existing right end.  If you want, you can prune out paths that have as many or more segments than some other path with the same right endpoint.  Continue to the end.  Then search over all the paths for ones that reach M and take the shortest.  I wouldn't call this a greedy algorithm (as called out in the Math Overflow version) but if overlap is not allowed we can't be sure that any given path will reach M.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2010-10-12T21:09:38.420" LastActivityDate="2010-10-12T21:09:38.420" CommentCount="4" />
  <row Id="6535" PostTypeId="1" AcceptedAnswerId="6547" CreationDate="2010-10-11T16:52:18.427" Score="1" ViewCount="89" Body="&lt;p&gt;So i am reading up on some interesting network topologies for a CS course and what i am reading is $r$-dimensional meshes of tree's. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;An $r$-dimensional mesh of trees is formed by adding complete binary tree's to an $N$-sided $r$-dimensional grid where the grid vertices act as the leafs of each binary tree.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Something like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/s2p0C.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This book says this about the graph $G = (V, E)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$|V| = (r+1)N^r - rN^{r-1}$&lt;br /&gt;&#xA;$|E| = 2rN^r - 2rN^{r-1}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And the diameter of this graph (the longest path) is $2rlogN$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How did they come about getting this result? The book does not explain it at all and i do not see how they calculated it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The book i am reading is: Introduction to Parallel Algorithms and Architectures: Arrays, Trees, Hypercubes is any of you are interested.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt;&lt;br /&gt;&#xA;I have tried to derive this myself by first finding $|V|$ when $r=2$ which would make it a 2-dimensional mesh of trees.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For this i know there are $2n$ complete binary tree's. Each binary tree have $2^{n-1}-1$ vertices. So this means $2n * 2 ^{n-1} -1$ vetices but we are double counting some because there is a binary tree for every row and column. Therefore we can minus out all the leaves of the column binary trees and they will be accounted in the row binary tree's. So that gives us $n * 2^{n-2}$ vertices to minus from the original.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$2n * (2^{n-1} - 1) - n2^{n-2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From this can i just times it by $n$ for 3 dimensions and $n^2$ for 4 dimensions? Then up to $n^{r-2}$ to find the total in $r$-dimensions? Was i calculating this correct for $r=2$? I read somewhere else that when $r=2$ $|V| = 3n^2 - 2n$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Second Edit&lt;/strong&gt;&lt;br /&gt;&#xA;Maybe i should use the fact that i know the leafs of each complete binary tree have $n$ vertices which means each binary tree is made out of $n + (n-1)$ vertices.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are $2n$ complete binary tree's which gives me: $2n(n + (n - 1))$ total vertices which contains a few we counted twice. A total of $n^2$ we counted twice. So &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$2n(n + (n-1)) - n^2 = 2n^2 + 2n^2 - 2n - n^2 = 3n^2 - 2n$&lt;br /&gt;&#xA;That looks right, doesn't it? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So now how would I apply this for 3 dimensions, 4, 5 ect.. all the way to $r$-dimensions?&lt;/p&gt;&#xA;" OwnerUserId="43" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-10-11T21:02:01.173" LastActivityDate="2010-11-10T21:44:17.957" Title="$r$-dimensional Meshes of Trees" Tags="&lt;graph-theory&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="6536" PostTypeId="2" ParentId="4406" CreationDate="2010-10-11T17:16:40.700" Score="1" ViewCount="" Body="&lt;p&gt;Morteza, I'd guess you'll get different answers from everyone you ask.  Here is the way I see it (which is, by necessity, somewhat circular):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Inference&lt;/strong&gt;, in the narrowest sense, is a single step in a deductive chain.  If I know $P\rightarrow Q$ and $\neg Q$, I can infer $\neg P$ from modus tollens (which itself is sometimes called a &quot;rule of inference&quot;).  More loosely, we can call any conclusion from premises an inference, even if it's not properly deductive (i.e. I look outside, see a clear sky, and infer that it's not raining).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Reasoning&lt;/strong&gt; is the mental process of logic -- what goes on inside my head when I use deduction.  Though a lot of people would probably use it as a synonym of &quot;deduction,&quot; I'd say the differences are far more important.  When you reason, you skip steps, explore multiple pathways, and use your intuition, which are all things unavailable to, say, a computer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Deduction&lt;/strong&gt; is the formal process of logic, and an inference is deductive when it follows from an axiom or logical rule.  This is the makeup of most mathematical proofs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Induction&lt;/strong&gt; has two meanings.  The first is some sort of mathematical induction (strong, weak, or transfinite), all of which are based on the idea of an infinite deductive chain that can be collapsed into a few steps.  For instance, in normal (weak) induction, you prove the proposition $P(0)$, and you prove that $\forall n:P(n)\rightarrow P(n+1)$.  Modus ponens then gives you $P(1)$, then $P(2)$ using $P(1)$, then $P(3)$ using $P(2)$, and so forth.  In all but the most intuitionist systems of arithmetic/set theory, induction is either provable or taken as an axiom, so its use is properly a deductive inference!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, there is another, more abstract meaning of induction that contrasts properly with deduction.  The logician Charles Sanders Peirce had an analogy with bags of beans that I find compelling.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Deduction&lt;/strong&gt; has you applying a generalization to a specific case to get a result.  So you know that all the beans in my bag are white (the generalization), and you take a bean from my bag (the case): then that bean must be white (the result).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Induction&lt;/strong&gt; has you taking a case and a result and generalizing them.  You take a bean from my bag (the case), and you see it is white (the result): therefore, all the beans in my bag are white (generalization).  In this rudimentary form, induction looks like guesswork, and nobody would trust this logic.  &lt;em&gt;But&lt;/em&gt; if you took five or ten beans, and they were all white, your confidence in the generalization would increase.  In fact, everybody uses this form of induction many times a day: you expect the sun to rise, your bed to be in the same place, your food to cook properly, etc. just because they've all happened many times before.  This kind of inference is not deductive at all, yet we still trust it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Abduction&lt;/strong&gt; was the third of Peirce's forms of inference.  It is the most suspect of the three, and is best thought of as a sort of &quot;reverse implication.&quot;  The idea is to take a generalization and a result and observe that a case is probably true because it implies the result.  So you have a white bean (the result), and you know that all the beans in my bag are white (the generalization): thus, this bean must be from my bag, for if it were, it would have to be white.  This seems like the sort of reasoning you see a lot on detective shows.&lt;/p&gt;&#xA;" OwnerUserId="2215" LastActivityDate="2010-10-11T17:16:40.700" />
  <row Id="6537" PostTypeId="1" AcceptedAnswerId="6542" CreationDate="2010-10-11T17:50:17.910" Score="0" ViewCount="153" Body="&lt;p&gt;I need help understanding how to derivate this function:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f(x) = x^{\arctan(x)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any suggestions?&lt;/p&gt;&#xA;" OwnerUserId="713" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-10-11T17:52:38.287" LastActivityDate="2010-10-11T18:14:32.160" CommunityOwnedDate="2010-10-11T17:50:17.910" Title="Finding derivative of f(x) = x^(arctan x)" Tags="&lt;calculus&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="1" />
  <row Id="6538" PostTypeId="1" CreationDate="2010-10-11T17:52:51.900" Score="0" ViewCount="228" Body="&lt;p&gt;It is clear that a finte group of order 105 is not simple since it contains a normal Sylow 7-subgroup or a normal Sylow 5-subgroup. I wonder if it is possible that there exists a Sylow 7-subgroup and a Sylow 5-subgroup ?&lt;/p&gt;&#xA;" LastActivityDate="2010-10-18T02:06:19.707" Title="Group of order 105" Tags="&lt;group-theory&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="6540" PostTypeId="2" ParentId="6494" CreationDate="2010-10-11T17:59:18.483" Score="2" ViewCount="" Body="&lt;p&gt;My previous answer is wrong.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take $f(x) = x$ ($f: \mathbb{R} \to \mathbb{R}$) where the domain has the Borel $\sigma$-algebra and the range the trivial $\sigma$-algebra. Then $f$ is clearly measurable but $f$ maps for example $[0,1]$ to $[0,1]$ which is not measurable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or do you want both $\sigma$-algebra's to be Borel? In that case I should modify my first answer (which was wrong because not every subset of a measure zero set is Borel measurable).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another idea would be the following (based on the one below me): &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $g:[0,1] \to \mathbb{R}$ be the Cantor function. Extend it to all of $\mathbb{R}$ by defining $g(x) = 0$ for $x &amp;lt; 0$ and $g(x) = 1$. Now define $h(x) = x + g(x)$ (this is standard), this function has a lot of nice properties like: $h$ is a bijection and $h(C) = 1$ where $C$ is the Cantor set.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we know that a set of positive measure contains a non-measurable subset. So let $A$ be an non-measurable subset of $h(C)$. So $M = h^{-1}(A)$ is a subset of the Cantor set and has thus zero Lebesgue measure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So define $u := h^{-1}$. This function is measurable (Lebesgue/Borel). But $u^{-1}(M) = A$ is non-measurable.&lt;/p&gt;&#xA;" OwnerUserId="1120" LastEditorUserId="1120" LastEditorDisplayName="" LastEditDate="2010-10-11T18:46:38.530" LastActivityDate="2010-10-11T18:46:38.530" CommentCount="2" />
  <row Id="6541" PostTypeId="2" ParentId="6537" CreationDate="2010-10-11T18:00:53.413" Score="2" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\ \  g^{\:h}\ =\ e^{h\: \log(g)}\:.\ $ Or, take logs, cf. &lt;a href=&quot;http://en.wikipedia.org/wiki/Logarithmic_derivative&quot; rel=&quot;nofollow&quot;&gt;logarithmic derivative,&lt;/a&gt; and my &lt;a href=&quot;http://math.stackexchange.com/questions/3773&quot;&gt;post here.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-10-11T18:07:16.773" LastActivityDate="2010-10-11T18:07:16.773" CommunityOwnedDate="2010-10-11T18:00:53.413" CommentCount="2" />
  <row Id="6542" PostTypeId="2" ParentId="6537" CreationDate="2010-10-11T18:03:28.083" Score="3" ViewCount="" Body="&lt;p&gt;Let $f(x) = x^{\arctan{x}}$ then $\log{f(x)} = \arctan{x} \cdot \log x$. Therefore $$\frac{1}{f(x)} \times f'(x) = \frac{d}{dx} \Bigl[ \arctan{x} \cdot \log x \Bigr] \Longrightarrow f'(x) = f(x) \times  \frac{d}{dx} \Bigl[ \arctan{x} \cdot \log x \Bigr]$$&lt;/p&gt;&#xA;" LastEditorUserId="1242" LastEditorDisplayName="" LastEditDate="2010-10-11T18:14:32.160" LastActivityDate="2010-10-11T18:14:32.160" CommunityOwnedDate="2010-10-11T18:03:28.083" />
  <row Id="6543" PostTypeId="2" ParentId="6482" CreationDate="2010-10-11T18:07:06.367" Score="3" ViewCount="" Body="&lt;p&gt;This is a rejoinder to Arturo's comment that if the two-variable limit&#xA;exists then both iterated limits exist.&#xA;Define&#xA;$$f(x,y)=\left\{\begin{array}{cl}&#xA;(x^2+y^2)\sin(1/x+1/y)&amp;amp;\textrm{if $xy\ne0$,}\\&#xA;0&amp;amp;\textrm{if $xy=0$.}&#xA;\end{array}\right.$$&#xA;Then certainly $f$ is continuous at $(0,0)$; that is&#xA;$$\lim_{(x,y)\to(0,0)}f(x,y)=0.$$&#xA;But for any nonzero $x$,&#xA;$$\lim_{y\to0}f(x,y)$$&#xA;doesn't exist (the function is wildly oscillatory for $y$ near zero). So&#xA;$$\lim_{x\to0}\left(\lim_{y\to0}f(x,y)\right)$$&#xA;is meaningless.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-10-11T18:07:06.367" CommentCount="1" />
  <row Id="6544" PostTypeId="1" AcceptedAnswerId="6603" CreationDate="2010-10-11T18:17:18.727" Score="13" ViewCount="430" Body="&lt;p&gt;It can be shown that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ n \int_{0}^{1} \frac{x^n - (1-x)^n}{2x-1} \ dx = \sum_{k=0}^{n-1} {n-1 \choose k}^{-1}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(For instance see my answer here: &lt;a href=&quot;http://math.stackexchange.com/questions/4783/finding-displaystyle-lim-n-to-infty-sum-limits-k0n-n-choose-k/6533#6533&quot;&gt;http://math.stackexchange.com/questions/4783/finding-displaystyle-lim-n-to-infty-sum-limits-k0n-n-choose-k/6533#6533&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It can also be shown that $$\lim_{n \rightarrow \infty} \  \sum_{k=0}^{n-1} {n-1 \choose k}^{-1} = 2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(For instance see Qiaochu's answer: &lt;a href=&quot;http://math.stackexchange.com/questions/4783/finding-displaystyle-lim-n-to-infty-sum-limits-k0n-n-choose-k/4784#4784&quot;&gt;http://math.stackexchange.com/questions/4783/finding-displaystyle-lim-n-to-infty-sum-limits-k0n-n-choose-k/4784#4784&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Combining those two shows that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \lim_{n \rightarrow \infty} \ n \int_{0}^{1} \frac{x^n - (1-x)^n}{2x-1} \ dx  = 2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a different, (preferably analytic) proof of this fact? Please do feel free to add a proof which is not analytic.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-12-05T08:39:55.590" Title="Different proofs of $\lim_{n \rightarrow \infty} n \int_{0}^{1} (x^n - (1-x)^n)/(2x-1) = 2$" Tags="&lt;analysis&gt;&lt;sequences-and-series&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="1" />
  <row Id="6545" PostTypeId="2" ParentId="6532" CreationDate="2010-10-11T18:18:45.940" Score="1" ViewCount="" Body="&lt;p&gt;This looks like it can get complicated. Generically, at least over an algebraically&#xA;closed field, a matrix $M$ will have distinct eigenvalues $m_1,\ldots,m_n$,&#xA;and generically $M\otimes M$ will have distinct eigenvalues $m_1^2,\ldots,m_n^2$&#xA;with multiplicity one, and $m_1m_2,m_1m_3,\ldots,m_{n-1}m_n$ with&#xA;multiplicity two. Thus the centralizer will have dimension $n+4{n\choose 2}=2n^2-n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But there are many degenerate cases: for instance if $M$ has eigenvalues&#xA;$1,a,\ldots,a^{n-1}$ then $M\otimes M$ will have eigenvalues&#xA;$1,a,\ldots,a^{2n-2}$ with multiplicities $1,2,\ldots n-1,n,n-1,\ldots,1$.&#xA;Things can get more complicated still.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then $M$ might have non-trivial Jordan blocks, and then the real fun starts!&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-10-11T18:18:45.940" />
  <row Id="6546" PostTypeId="1" AcceptedAnswerId="11869" CreationDate="2010-10-11T18:34:51.803" Score="1" ViewCount="157" Body="&lt;p&gt;Motivating factor behind this question is the comments which are given for this question asked some hours ago &lt;a href=&quot;http://math.stackexchange.com/questions/6538/group-of-order-105&quot;&gt;http://math.stackexchange.com/questions/6538/group-of-order-105&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So given a group $G$ of order $n$ what are the different methods for  constructing a non abelian group of order $n$. Well, i have seen a method which Herstein uses in his book. He takes a cyclic group of given order and defines an Automorphism on the cyclic group and then places some restrictions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to know, whether there are anymore methods for obtaining a Non Abelian group of a given order.&lt;/p&gt;&#xA;" LastActivityDate="2010-11-25T19:24:01.080" Title="Constructing Non Abelian Groups of Given order" Tags="&lt;abstract-algebra&gt;&lt;group-theory&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="6547" PostTypeId="2" ParentId="6535" CreationDate="2010-10-11T18:38:46.660" Score="1" ViewCount="" Body="&lt;p&gt;You seem to have mixed up two different formulas for the number of nodes in a complete binary tree. If the &lt;em&gt;height&lt;/em&gt; of the tree is $h$, it has $2^h-1$ nodes. But in this case, $N$ is not the height but the number of &lt;em&gt;leaves&lt;/em&gt;, so the tree has $N-1$ internal nodes, and the total number of nodes is $2n-1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From here you get, for $r = 2$, a total of $2N\cdot(2N-1)$ vertices in each tree, less the overcounting of shared leaves, $N\cdot N$, equalling $|V| = 2N(2N-1) - N^2 = 3N^2 - 2N$, which also agrees with the formula $|V| = (r+1)N^r - rN^{r-1}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A better way to do it would be to notice that the shared nodes are simply the nodes in the grid, and count them once: $N^r$. The internal nodes of the trees are not shared, there are $N-1$ of them for each tree, and there are $rN^{r-1}$ trees (see below). Add this up and you'll get your solution without worrying about overcounting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The number of trees is $rN^{r-1}$ because along any direction, a tree connects $N$ nodes, so you need $N^{r-1}$ of them to cover the grid. And the graph contains $r$ copies of this arrangement, one each for the $r$ directions along the grid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A similar approach should work for counting the number of edges in the graph -- count the edges in the trees and the edges in the grid separately.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the diameter, first think about what the shortest path between two nodes is in a traditional mesh (without trees), then consider how the additional tree structure shortens this path. Then it should be clear which pair of nodes remains the farthest apart; their distance is the diameter of the graph.&lt;/p&gt;&#xA;" OwnerUserId="856" LastEditorUserId="856" LastEditorDisplayName="" LastEditDate="2010-10-11T19:11:16.643" LastActivityDate="2010-10-11T19:11:16.643" />
  <row Id="6548" PostTypeId="2" ParentId="6538" CreationDate="2010-10-11T18:43:37.097" Score="2" ViewCount="" Body="&lt;p&gt;Of &lt;em&gt;course&lt;/em&gt; there exist Sylow $7$-subgroups and Sylow $5$-subgroups: Sylow's Theorems tell you that the group has Sylow $p$-subgroups for every $p$ (with the Sylow $p$-subgroup being trivial if $p$ does not divide $105$). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you meant to ask &quot;Can the group have &lt;em&gt;normal&lt;/em&gt; Sylow $5$- &lt;em&gt;and&lt;/em&gt; Sylow $7$ subgroup?&quot; then the answer is &quot;yes&quot;: the (unique) abelian group of order $105$ will do (since $105=3\times 5\times 7$ is square free, the only abelian group of order $105$ is the cyclic one). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, you can &lt;em&gt;always&lt;/em&gt; find a finite group of order $n$ in which all Sylow subgroups are normal: any abelian group of order $n$, and there is always at least one, will do. If any of the primes in the factorization of $n$ is raised to a degree greater than $2$, then you can also find a &lt;em&gt;nonabelian&lt;/em&gt; group in which all the Sylow subgroups are normal, just by taking the direct product of $p$-groups of the correct size, and picking a nonabelian one when the prime is raised to a sufficiently large degree.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And in this case, you can get a nonabelian group in which the Sylow $7$- and the Sylow $5$-subgroups are normal by first constructing a nonabelian group of order $21$ (there are no nonabelian groups of orders $15$ or $35$) which has a normal subgroup of order $7$, and then taking the direct product with with $C_5$.  &lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-10-11T18:43:37.097" />
  <row Id="6549" PostTypeId="1" AcceptedAnswerId="6550" CreationDate="2010-10-11T19:04:38.237" Score="0" ViewCount="75" Body="&lt;p&gt;I have multiple objects in my lab which are fully described in two coordinate systems. Both systems are Cartesian and in meters. If they had a common origin but were rotated relative to each other I could easily reconcile the two. Additionally, if it was just a displacement it would be even easier. However the two are offset in rotation and displacement. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I want to do is find how to convert between the two of them. I would think I would just need the coordinates of the same three objects in the two systems. I know how to do this for two coordinate systems that are rotated OR displaced, but I can't figure out how to do it in a scenario with both rotation and displacement. Anyone got a formula for me or (even better) a link describing how it is done?&lt;/p&gt;&#xA;" OwnerUserId="2440" LastActivityDate="2010-10-11T19:46:07.670" Title="Coordinate substitution between two cartesian systems" Tags="&lt;geometry&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="6550" PostTypeId="2" ParentId="6549" CreationDate="2010-10-11T19:46:07.670" Score="0" ViewCount="" Body="&lt;p&gt;Look at &lt;a href=&quot;http://en.wikipedia.org/wiki/Affine_transformation&quot; rel=&quot;nofollow&quot;&gt;Affine transformation&lt;/a&gt;&#xA;Basically if you know how to do rotations and translations you can chain them together.  If you have the coordinates of a few objects you can solve the simultaneous equations to get the matrix.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-10-11T19:46:07.670" />
  <row Id="6551" PostTypeId="1" AcceptedAnswerId="6624" CreationDate="2010-10-11T19:48:45.823" Score="3" ViewCount="138" Body="&lt;p&gt;I'm looking to brush up on SAT math.  (That's basic algebra, probability and trigonometry.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm interested in books and websites.&lt;/p&gt;&#xA;" OwnerUserId="2414" LastEditorUserId="2414" LastEditorDisplayName="" LastEditDate="2010-10-11T23:25:54.640" LastActivityDate="2010-10-19T04:25:12.650" CommunityOwnedDate="2010-10-11T21:14:15.947" Title="Good SAT learning resources?" Tags="&lt;probability&gt;&lt;algebra&gt;&lt;trigonometry&gt;&lt;learning&gt;&lt;self-learning&gt;" AnswerCount="2" CommentCount="6" FavoriteCount="0" />
  <row Id="6552" PostTypeId="1" AcceptedAnswerId="6561" CreationDate="2010-10-11T19:50:36.893" Score="2" ViewCount="74" Body="&lt;p&gt;I've been trying to find a way to do this: given a point $P(\alpha,\beta)$, a function $f(x)$, and an angle $\theta$, find the area of the sector determined by extending a horizontal line from $P$ to $f$, and then another one from $P$ to $f$ at the angle $\theta$ from the first line. If you look at this picture:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://imgur.com/ekTqY.png&quot; alt=&quot;diagram&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;it's a little more clear what I'm talking about. I want to find the area of that sector $S$, which is determined by that triangle $R_1$ and the other region $R_2$. Let the value of $x$ where the horizontal line from $P$ hits $f$ be $q$ (that is, $q=f^{-1}(\beta)$). Now, the line $\overline{PA}$ has the equation $y=\tan\theta(x-\alpha)+\beta$. The intersection of that and $f$, the point $A$ (the first intersection point), is the beginning of the interval determining $R_2$; I'll call the root $p$ of $f(x)=\tan\theta(x-\alpha)+\beta$ that value of $x$, so $p=f^{-1}(\tan\theta(x-\alpha)+\beta)$. The area of $R_1$ is simply the area of the triangle: $R_1=\frac{1}{2}(p-\alpha)(f(p)-\beta)$. The area of $R_2$ is the area under the curve from $p$ to $q$, minus the initial height $\beta$: $R_2=\displaystyle\int_p^q (f(x)-\beta) dx$. Therefore, the total area is $$S=\frac{1}{2}\tan\theta(p-\alpha)^2-\beta(q-p)+\displaystyle\int_p^q f(x) dx$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the way I did it, but there has to be a simpler way; is there one involving polar equations? Is there a general formula when $f$ is non-invertible?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; If the function $f$ is translated so that $P$ is now the origin, the area is just $\displaystyle\frac{1}{2}\int_0^\theta r^2 d\phi$ where $r(\phi)$ is $f(x)$ in polar form. However, translating $f$ over to where $P$ is on the origin turns it into $f(x+\alpha)-\beta$. Is there a nice way of putting that into polar form?&lt;/p&gt;&#xA;" OwnerUserId="1562" LastEditorUserId="1562" LastEditorDisplayName="" LastEditDate="2010-10-11T20:58:20.433" LastActivityDate="2010-10-11T20:58:20.433" Title="Area of a sector between a point and a function determined by an angle" Tags="&lt;geometry&gt;&lt;calculus&gt;&lt;algebra&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="6553" PostTypeId="2" ParentId="6519" CreationDate="2010-10-11T19:51:33.727" Score="1" ViewCount="" Body="&lt;p&gt;I don't have enough reputation to leave a comment, so here goes....&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As other have observed, the growth rate of $M^k$ is determined by the largest eigenvalue of $M$. I just want to note that for a nonnegative matrix, the largest eigenvalue is always &#xA;between the largest and smallest column sum. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That, in principle, could allow you to obtain upper and lower bounds by trying to bound how big or small the column sums usually get for the random matrix you described. &lt;/p&gt;&#xA;" OwnerUserId="2120" LastActivityDate="2010-10-11T19:51:33.727" CommentCount="1" />
  <row Id="6554" PostTypeId="2" ParentId="6544" CreationDate="2010-10-11T19:55:02.453" Score="2" ViewCount="" Body="&lt;p&gt;Here's a sketch of an elementary analytic demonstration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) The integrals all exceed $2/n$ when $x$ exceeds 1: subtract $2/n = \int_0^1{(t^{n-1} + (1-t)^{n-1}) \ dt}$ to obtain a positive integrand (using elementary inequalities).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) The integrals are bounded by a sequence decreasing to the desired limit.  By looking at the Taylor series of the log of the integrand near 1, for example, and assuming $n \ge 3$, you can easily establish an upper bound of $\exp((n-2)(x-1))$ in the interval $(1/2, 1]$, with a similar bound in the interval $[0, 1/2)$ by symmetry.  The estimate (integrating over one of those two intervals) equals $(1 - \exp(2-n)) / (n-2)$ which, when doubled and multiplied by $n$, converges to $2$.&lt;/p&gt;&#xA;" OwnerUserId="1489" LastActivityDate="2010-10-11T19:55:02.453" CommentCount="1" />
  <row Id="6555" PostTypeId="2" ParentId="6451" CreationDate="2010-10-11T20:15:16.837" Score="1" ViewCount="" Body="&lt;p&gt;Maybe you're going to tell me that if all you've got is a hammer, all problems look like nails, but here goes my proposal -following the same strategy as in my answers to &lt;a href=&quot;http://math.stackexchange.com/questions/6357/quotient-topologies-and-equivalence-classes/6362#6362&quot;&gt;this question&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First of all, take as a model for the projective plane the disk&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;D^2 = \left\{ (x,y) \in \mathbb{R}^2\ \vert \ x^2 + y^2 \leq 1   \right\}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and quotient by the equivalence relation among antipodal points on $S^1$: $(x,y) \sim -(x,y) $ if $x^2 + y^2 = 1$.  So&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\mathbb{RP}^2 = D^2/\sim \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Put the Moebius strip inside $\mathbb{RP}^2$ as the vertical strip&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;M = \left\{\widetilde{(x,y)} \in \mathbb{RP}^2 \ \vert \ -\frac{1}{2} \leq x \leq \frac{1}{2}   \right\}  \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's denote by $M'$ the same set of points inside $D^2$, without any quotient at all. That is, ordinary points of $\mathbb{R}^2$. Thus&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;M = M'/\sim \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, you want to prove that, when you quotient out the boundary of the Moebius band&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\partial M = \left\{\widetilde{(x,y)} \in \mathbb{RP}^2 \ \vert \ x = -\frac{1}{2} \ \quad \text{or} \quad  x = \frac{1}{2}   \right\}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;you get the projective plane:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;M/\partial M \ \cong \ \mathbb{RP}^2 \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ok, the homeomorphism is easily described in some kind of &quot;eliptical&quot; coordinates: for every point $(x,y) \in D^2$ there is one and only one elipse of the form&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;x^2 + \frac{y^2}{b^2} = 1  \qquad \text{for} \qquad -1 \leq b \leq 1 \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All these elipses have a common major axis, namely the segment $-1\leq x \leq 1, y = 0$. We make the abuse of considering $b=0$ too as the degenerate &quot;elipse&quot; $y=0$ and in order to distingish the &quot;positive&quot; $y\geq 0$ and &quot;negative&quot; $y\leq 0$ branches of the elipse, we  put a sign in $b$: the same as $y$. If all this doesn't sound too rigorous to you, it's ok: look at it as just a motivation. Wait for the resulting formulae.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So a point inside the disk is determined once you know its $x$-coordinate and the elipse to which it belongs; that is, its $b$-coordinate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Our homeomorphism is going to do the following: points inside the Moebius strip  are going to &quot;travel&quot; along the elipse to which they belong. How far? Well, we want all the points in $\partial M$, those with $x=-1/2$ or $x= 1/2$, to reach points $(-1,0)$ or $(1,0)$, depending on the sign of $x$. Hence, we try something like&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\widetilde{\varphi }: M \longrightarrow \mathbb{RP}^2 \ , \qquad \widetilde{\varphi} \widetilde{(x,b)}  = \widetilde{(2x, b)} \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, let's write a decent $\varphi$ in Cartesian coordinates. We want our $(2x, b)$ to belong to the $b$-elipse all the time, hence we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;(2x)^2 + \frac{y^2}{b^2} = 1 \qquad \Longrightarrow \qquad b = \frac{y}{\sqrt{1-(2x)^2}} \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence we begin with a map&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\varphi : M' \longrightarrow D^2 &#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;defined as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{cases}&#xA;\varphi (x,y) = \left( 2x , \frac{y}{\sqrt{1-(2x)^2}}\right) &amp;amp; \text{if} \ x \neq \frac{\pm 1}{2}  \\&#xA;\varphi (\pm 1/2, y)  = (\pm 1,0) &amp;amp; {}&#xA;\end{cases}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Exercise.&lt;/strong&gt; Check that $\varphi$ is continuous, surjective and compatible with the antipodal equivalence relation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, $\varphi$ induces a well-defined map&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;M \longrightarrow \mathbb{RP}^2 \ ,&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is continuous, because of the universal property of the quotient topology, and surjective. It is &quot;almost&quot; injective too, except for all the points in the boundary of the Moebius band $x=-1/2$ and $x=1/2$ that go to $\widetilde{(-1,0)} = \widetilde{(1,0)}$. So, if we quotient out them, we get a homeomorphism&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\widetilde{\varphi} : M/\partial M \longrightarrow \mathbb{RP}^2 \ ,&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;whit the help, as usual, of the GTET (see the link above, taking into account that, despite being a quotient, $\mathbb{RP}^2$ &lt;strong&gt;is&lt;/strong&gt; a Hausdorff space -see Rotman, &quot;An introduction to Algebraic Topology&quot;, theorem 8.4, &lt;a href=&quot;http://books.google.es/books?id=bApVT-6A43QC&amp;amp;pg=PA180&amp;amp;lpg=PA180&amp;amp;dq=hausdorff+%22projective+plane%22&amp;amp;source=bl&amp;amp;ots=Ge6niTS6G2&amp;amp;sig=SKjgSQKf15psgqzerEDH0OdpXhg&amp;amp;hl=ca&amp;amp;ei=wGOzTJaeIo674gbtrPHDBg&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=7&amp;amp;ved=0CDkQ6AEwBg#v=onepage&amp;amp;q=hausdorff%20%22projective%20plane%22&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;Hausdorff&lt;/a&gt;).&lt;/p&gt;&#xA;" OwnerUserId="664" LastEditorUserId="664" LastEditorDisplayName="" LastEditDate="2010-10-11T21:12:28.947" LastActivityDate="2010-10-11T21:12:28.947" />
  <row Id="6556" PostTypeId="1" AcceptedAnswerId="6560" CreationDate="2010-10-11T20:20:10.803" Score="1" ViewCount="167" Body="&lt;p&gt;Can you help me understand how&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$g^{-1}(x) = f^{-1}(x + 2)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is the inverse of&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$g(x) = f(x) - 2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I cannot get my head around how the 2 ends up inside $f$ :/&lt;/p&gt;&#xA;&#xA;&lt;p&gt;thanks&lt;/p&gt;&#xA;" OwnerUserId="713" LastActivityDate="2010-10-11T21:30:02.710" CommunityOwnedDate="2010-10-11T20:27:09.447" Title="Find the inverse of g(x) = f(x) - 2" Tags="&lt;calculus&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="6557" PostTypeId="1" AcceptedAnswerId="6558" CreationDate="2010-10-11T20:23:15.133" Score="1" ViewCount="218" Body="&lt;p&gt;Consider I have two sets that neither one is a subset of the other (for example, the set of prime numbers and the set of odd integers). Is there a specific notation that combines the meanings of both ⊄ and ⊅ ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, is there a notation to mean that two sets are completely distinct and share no elements in common (for example, the sets of odd and even integers)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;≠ seems very inappropriate, and I'd like to avoid it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Context:&lt;/strong&gt; My area is actually Computer Science. I'm using this refine my vocabulary in order to teach my students about &lt;a href=&quot;http://en.wikipedia.org/wiki/Subnetwork&quot; rel=&quot;nofollow&quot;&gt;subnetting&lt;/a&gt; (which is basically about sets and subsets of IP addresses), and to write some exercises for them. I'll probably use the words &lt;em&gt;incomparable&lt;/em&gt; and &lt;em&gt;disjoint&lt;/em&gt; as suggested.&lt;/p&gt;&#xA;" OwnerUserId="2442" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2010-11-23T04:01:37.307" LastActivityDate="2010-11-23T04:01:37.307" Title="Proper notation for distinct sets" Tags="&lt;notation&gt;&lt;elementary-set-theory&gt;" AnswerCount="2" />
  <row Id="6558" PostTypeId="2" ParentId="6557" CreationDate="2010-10-11T20:29:06.830" Score="3" ViewCount="" Body="&lt;p&gt;No, there is no special symbol to denote the situation in which neither set is contained in the other. There &lt;em&gt;is&lt;/em&gt; a special word, though: we say the two sets are &lt;strong&gt;incomparable.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is also no special symbol to denote the fact that two sets are disjoint; we simply write $A\cap B=\emptyset$. &lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-10-11T20:29:06.830" />
  <row Id="6559" PostTypeId="2" ParentId="6556" CreationDate="2010-10-11T20:29:32.800" Score="3" ViewCount="" Body="&lt;p&gt;It's simply $\rm\ g = h\: f\ \Rightarrow\ g^{-1} = (h\:f)^{-1} = f^{-1} h^{-1},\ \ h(x) = x-2\:.\ $   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or you can explicitly check that they're inverses, namely&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\ g^{-1}(g(x)) = f^{-1}(g(x)+2) = f^{-1}(f(x)-2 + 2) = x$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\ g(g^{-1}(x)) = f(g^{-1}(x))-2 = f(f^{-1}(x+2))-2  = x$&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-10-11T20:34:44.497" LastActivityDate="2010-10-11T20:34:44.497" CommunityOwnedDate="2010-10-11T20:29:32.800" />
  <row Id="6560" PostTypeId="2" ParentId="6556" CreationDate="2010-10-11T20:34:57.757" Score="3" ViewCount="" Body="&lt;p&gt;If you want to complement Bill Dubuque's answer with trying to figure out &lt;em&gt;how&lt;/em&gt; you might have come up with the formula, think about what $g(x)=f(x)-2$ is telling you to do: in order to compute $g(x)$, you first compute $f(x)$, and then you subtract $2$. The inverse of this should be something that &quot;unravels&quot; the process you just went to. So the first thing you need to unravel is the &lt;em&gt;last&lt;/em&gt; thing you did when you computed $g(x)$, namely subtracting $2$. So the first thing you are going to do is add $2$ from whatever they give you. What did you did just prior to subtracting $2$ when you computed $g$? You computed $f$. So the next thing you need to do in the &quot;unraveling&quot; process is to undo the computation of $f$, which is achieved by computing $f^{-1}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In summary: to invert (unravel) the process of (i) first compute $f$, then (ii) subtract $2$; you will (I) first add $2$; then (II) compute $f^{-1}$. Symbolically: if you are given $x$ and you want to unravel, first you compute $x+2$, then you compute $f^{-1}$ of that. So if $g(x) = f(x)-2$, then $g^{-1}(x) = f^{-1}(x+2)$.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-10-11T20:34:57.757" CommunityOwnedDate="2010-10-11T20:34:57.757" />
  <row Id="6561" PostTypeId="2" ParentId="6552" CreationDate="2010-10-11T20:45:59.213" Score="2" ViewCount="" Body="&lt;p&gt;Yes, it is easier using polar coordinates. If you have a region in the plane determined by two straight lines $OA$ and $OB$ starting in a point $O$ -like those in your picture- and forming angles $\theta_1$ and $\theta_2$ with respect the $x$-axis ($OB$ doesn't need to be horizontal) and you close your region with a curve which polar equation is $r = r(\theta )$, then you apply the &lt;a href=&quot;http://en.wikipedia.org/wiki/Change_of_variables_theorem&quot; rel=&quot;nofollow&quot;&gt;change of variables&lt;/a&gt; theorem, and you have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;A(D) = \int_D dxdy = \int_{\theta_1}^{\theta_2} \int_0^{r(\theta)} rdrd\theta = \frac{1}{2} \int_{\theta_1}^{\theta_2} r(\theta)^2 d\theta \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here $D$ is your sector and $A(D)$ its area.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, if you want to compute the area of a circle of radius $R$ (you already know, but just in case...), $r(\theta ) = R$. Hence&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;A(D) = \frac{1}{2}\int_0^{2\pi} R^2d\theta = \pi R^2 \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Amazing, isn't it?   :-)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the only problem could be that your data necessarily start with the equation of the curve given in Cartesain coordinates $y = f(x)$ and you have to translate it into polar coordinates.&lt;/p&gt;&#xA;" OwnerUserId="664" LastActivityDate="2010-10-11T20:45:59.213" CommentCount="3" />
  <row Id="6562" PostTypeId="2" ParentId="6557" CreationDate="2010-10-11T20:54:33.600" Score="2" ViewCount="" Body="&lt;p&gt;Let $\rm\:P\: $ be a poset = &lt;a href=&quot;http://en.wikipedia.org/wiki/Partially_ordered_set&quot; rel=&quot;nofollow&quot;&gt;partially-ordered set&lt;/a&gt;, i.e. a set equipped wth a relation $\rm\le$ that is reflexive, antisymmetric and transitive. Then one says that $\rm\: x\: $ and $\rm\: y\: $ are &lt;em&gt;comparable&lt;/em&gt;$\ $ if $\rm\ x \le y\ $ or $\rm\ y \le x\:;\: $ otherwise they are &lt;em&gt;incomparable&lt;/em&gt;, sometimes written $\rm\ x||y\:.\:$ A &lt;em&gt;chain&lt;/em&gt;$\ $ of  $\rm\: P\:$ is a subset in which every two elements are comparable, and dually, an &lt;em&gt;antichain&lt;/em&gt;$\ $ is a subset in which every two elements are incomparable.&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-10-11T20:54:33.600" CommentCount="1" />
  <row Id="6563" PostTypeId="1" AcceptedAnswerId="6564" CreationDate="2010-10-11T21:10:56.173" Score="1" ViewCount="70" Body="&lt;p&gt;&lt;em&gt;This is part of a homework question&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm trying to figure out the perimeter of the black-filled object below (sorry, it's crudely drawn in paint, but I think you get the idea). To find the perimeter, I'm assuming that I should first find the perimeter of the rectangle (12*2+3*2 in this case) then subtract each little triangle (which is suppose to be all the same size by the way). The question mentions only one fact other than the size of each side- that it's trisected. I'm pretty sure the answer to this question lies in here...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/2sLxC.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-10-11T22:54:04.540" LastActivityDate="2010-10-11T22:54:04.540" Title="Properties of a trisected rectangle?" Tags="&lt;geometry&gt;&lt;homework&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="6564" PostTypeId="2" ParentId="6563" CreationDate="2010-10-11T21:13:26.667" Score="3" ViewCount="" Body="&lt;p&gt;Well, if each side is trisected, that means it's split into three identical parts. The base of each triangle is one of the parts of the 12 side, and the height of each triangle is one of the parts of the 3 side. You can figure it out from there.&lt;/p&gt;&#xA;" OwnerUserId="1562" LastActivityDate="2010-10-11T21:13:26.667" CommentCount="3" />
  <row Id="6565" PostTypeId="2" ParentId="6321" CreationDate="2010-10-11T21:26:28.743" Score="4" ViewCount="" Body="&lt;p&gt;The answer to your question is yes. All numbers of the form $x=\sum_{n\ge0}\frac{C_k}{g^{Z(k)}}$ for Z(k) eventually dominating any polynomial are indeed transcendental. As in the question, g and C&lt;sub&gt;k&lt;/sub&gt; are integers with 1&amp;nbsp;&amp;le;&amp;nbsp;C&lt;sub&gt;k&lt;/sub&gt;&amp;nbsp;&amp;le;&amp;nbsp;g-1. In fact, the methods used by the paper linked in the question generalize in a quite straightforward way to handle this situation. I don't know of this result appearing in any published paper, but note the following points.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;We can say straight-away that x is irrational. This follows from Z(n) eventually dominating any linear function of n, so its base-g expansion is not eventually periodic. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If Z(n+1)/Z(n) is unbounded then, as noted in the comments, x will be a &lt;a href=&quot;http://en.wikipedia.org/wiki/Liouville_number&quot; rel=&quot;nofollow&quot;&gt;Liouville number&lt;/a&gt; so, by Liouville's theorem, it is transcendental. For any N&amp;nbsp;&amp;gt;&amp;nbsp;0, Z(n+1)&amp;nbsp;&amp;ge;&amp;nbsp;NZ(n) for infinitely many n. The fact that it is a Liouville number follows from taking $p=\sum_{k=1}^nC_kg^{Z(n)-Z(k)}$ and $q=g^{Z(n)}$, giving the rational approximation $|x-p/q|&amp;lt; g^{1+Z(n)-Z(n+1)}\le gq^{-N}$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;By the &lt;a href=&quot;http://en.wikipedia.org/w/index.php?title=Thue%E2%80%93Siegel%E2%80%93Roth_theorem&amp;amp;oldid=385480595&quot; rel=&quot;nofollow&quot;&gt;Thue–Siegel–Roth theorem&lt;/a&gt;, if Z(n+1)/Z(n)&amp;nbsp;&amp;ge;&amp;nbsp;2+&amp;epsilon; infinitely often (any &amp;epsilon;&amp;nbsp;&amp;gt;&amp;nbsp;0) then x will be transcendental. The theorem says that an irrational algebraic number has only finitely many rational approximations $\vert x-p/q\vert\le cq^{-2-\epsilon}$ for any fixed c,&amp;epsilon;&amp;nbsp;&amp;gt;&amp;nbsp;0. That x has infinitely many such rational approximations follows in the same way as for point 2 above. Every Z(n+1)&amp;nbsp;&amp;ge;&amp;nbsp;(2+&amp;epsilon;)Z(n) gives a rational approximation $\vert x-p/q\vert&amp;lt; gq^{-2-\epsilon}$, so x cannot be algebraic. This covers the case where Z(n) grows exponentially of rate a&lt;sup&gt;n&lt;/sup&gt; for any a&amp;nbsp;&amp;gt;&amp;nbsp;2, but is not strong enough to cover cases such as Z(n)&amp;nbsp;=&amp;nbsp;2&lt;sup&gt;n&lt;/sup&gt;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If Z(n+1)/Z(n)&amp;nbsp;&amp;gt;&amp;nbsp;1+&amp;epsilon; infinitely often (any &amp;epsilon;&amp;nbsp;&amp;gt;&amp;nbsp;0) then x will be transcendental. This is a consequence of the Roth-Ridout theorem, from the 1957 paper &lt;a href=&quot;http://dx.doi.org/10.1112/S0025579300001182&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;Rational approximations to algebraic numbers&lt;/em&gt;&lt;/a&gt; (not free access, but is also quoted in the freely available paper &lt;a href=&quot;http://seminariomatematico.dm.unito.it/rendiconti/cartaceo/53-3.html&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;An explicit version of the theorem of Roth-Ridout&lt;/em&gt;&lt;/a&gt;, Theorem 2). The Roth-Ridout theorem strengthens the Thue-Siegel-Roth theorem implying, in particular, for irrational and algebraic x, there are only finitely many rational approximations $\vert x-p/q\vert\le cq^{-1-\epsilon}$ when the prime factors of q all belong to some fixed finite set &lt;em&gt;P&lt;/em&gt;. In our case, we can let &lt;em&gt;P&lt;/em&gt; be the set of prime factors of g and the result follows in the same way as for point 3 above. This shows that x is transcendental if Z(n) grows exponentially. (thanks to Mike Bennett over at mathoverflow for pointing out the Roth-Ridout theorem).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;A paper by Bugeaud, &lt;a href=&quot;http://rendiconti.math.unipd.it/volumes/vol118.php?lan=english&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;On the b-ary expansion of an algebraic number&lt;/em&gt;&lt;/a&gt; shows that, if x is irrational and algebraic then for large enough n, there are at least (log n)&lt;sup&gt;1+1/(&amp;omega;+4)&lt;/sup&gt; (loglog n)&lt;sup&gt;-1/4&lt;/sup&gt; nonzero digits among the first n digits of the base g expansion. Here, &amp;omega; is the the number of prime divisors of g. This shows that, if Z(n)&amp;nbsp;&amp;ge;&amp;nbsp;exp(cn&lt;sup&gt;&amp;alpha;&lt;/sup&gt;) for large n and any fixed c&amp;nbsp;&amp;gt;&amp;nbsp;0, &amp;alpha;&amp;nbsp;&amp;gt;&amp;nbsp;1/(1+1/(&amp;omega;+4)) then x is transcendental.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;After reading through the details of the paper linked in the original question, I note that they do generalize to the base g&amp;nbsp;&amp;ge;&amp;nbsp;2 case. So x is transcendental as long as Z(n) eventually dominates any polynomial. I don't know of any published paper proving this, but posted my proof on &lt;a href=&quot;http://mathoverflow.net/questions/41609/have-all-numbers-with-sufficiently-many-zeros-been-proven-transcendental/41760#41760&quot; rel=&quot;nofollow&quot;&gt;mathoverflow&lt;/a&gt; where this question was also asked. I have re-read through this proof a few times to be sure, and am now confident that it is correct (modulo small typos, etc). Also, Bugeaud posted an answer to the question agreeing that the method generalizes. Using #(x,n) to denote the number of non-zero digits in the first n digits of the base g expansion of x, the precise statement is as follows.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If x is irrational and satisfies a rational polynomial of degree D then #(x,n)&amp;nbsp;&amp;ge;&amp;nbsp;cn&lt;sup&gt;1/D&lt;/sup&gt; for a positive constant c and all large enough n.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;In fact, you can easily remove the &quot;large enough&quot; from this statement, although I find it convenient stated in this way. The proof I wrote out is a generalization of the methods used in the paper linked in the question. There is one change worth noting though. Whereas the paper made use of the Thue-Siegel-Roth theorem at one point (Theorem 3.1), I used &lt;a href=&quot;http://en.wikipedia.org/wiki/Liouville_number&quot; rel=&quot;nofollow&quot;&gt;Liouville's theorem&lt;/a&gt;. This means that the constant c appearing in the statement above is not quite as good (if you go through the proof and work it out explicitly) although, in any case, the paper linked in the question could have obtained a better value by using the Roth-Ridout theorem instead. Using Liouville's theorem does have two advantages though. Firstly, it is elementary. A proof of Liouville's theorem is given in the linked Wikipedia article. Secondly, it is &lt;em&gt;effective&lt;/em&gt;. That is, not only can the constant c be calculated but you can also work out exactly what &quot;large enough&quot; means for n in the statement above (which will depend on the polynomial satisfied by x). The strengthened versions of Liouville's theorem such as Thue-Siegel-Roth and Roth-Ridout are not effective.&lt;/p&gt;&#xA;" OwnerUserId="1321" LastActivityDate="2010-10-11T21:26:28.743" />
  <row Id="6566" PostTypeId="2" ParentId="6556" CreationDate="2010-10-11T21:30:02.710" Score="0" ViewCount="" Body="&lt;p&gt;$$g(x) = f(x) - 2 = (h \circ f)(x)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $h(x) = x - 2$ and of course $h^\circ(x) = x + 2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then the inverse $g^{\circ}(x) = (h \circ f)^{\circ}(x) = (f^{\circ}\circ h^{\circ})(x) = f^\circ(x + 2)$ by &lt;a href=&quot;http://math.stackexchange.com/questions/2349/how-to-prove-f-circ-g-1-g-1-circ-f-1/2426#2426&quot;&gt;the usual rules&lt;/a&gt;.&lt;/p&gt;&#xA;" LastActivityDate="2010-10-11T21:30:02.710" CommunityOwnedDate="2010-10-11T21:30:02.710" />
  <row Id="6567" PostTypeId="1" AcceptedAnswerId="6600" CreationDate="2010-10-11T21:33:01.970" Score="8" ViewCount="146" Body="&lt;p&gt;The other day I &lt;a href=&quot;http://caicedoteaching.wordpress.com/2008/11/11/275-positive-polynomials/&quot; rel=&quot;nofollow&quot;&gt;came across&lt;/a&gt; the following statement:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt; A polynomial $f(x,y)$ of degree at most $3$ that vanishes at $8$ of the $9$ points $(x,y)$ with $x, y \in \{-1,0,1\}$ must also vanish at the $9$th point. &lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am wondering about how this statement generalizes. Specifically, I am looking for a theorem of the form&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt; Suppose a polynomial $f(x_1, \ldots, x_n)$ of degree $d$ vanishes on a some discrete set $S \subset R^n$, which satisfies _______. Then, defining a discrete set $U$ as ______, the polynomial $f$ must also vanish on $U$.  &lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Can anyone fill in the blanks? &lt;/p&gt;&#xA;" OwnerUserId="2120" LastActivityDate="2010-10-12T08:50:01.127" Title="The zeros of a multivariable polynomial" Tags="&lt;polynomials&gt;" AnswerCount="1" FavoriteCount="0" />
  <row Id="6568" PostTypeId="1" AcceptedAnswerId="6569" CreationDate="2010-10-11T21:37:06.213" Score="2" ViewCount="191" Body="&lt;p&gt;I was wondering about the relation between complement of a subset, different between two subsets, union and intersection of subsets. Can we reduce the set of the above operations in a minimal way so that other operations can be represented by these &quot;independent&quot; operations? And how many ways to do this? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, the intersection (or union) can be represented by complement and union (or intersection) in the same way as in De Morgan law.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But can we represent complement or difference in terms of union and intersection?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="1281" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2010-11-22T14:36:24.820" LastActivityDate="2010-11-22T18:34:27.377" Title="relation between set operations" Tags="&lt;elementary-set-theory&gt;" AnswerCount="3" />
  <row Id="6569" PostTypeId="2" ParentId="6568" CreationDate="2010-10-11T21:46:16.300" Score="2" ViewCount="" Body="&lt;p&gt;Well, set difference is defined in terms of intersection and complement: $A\setminus B = A\cap B^C$. (Personally, I prefer the notation $A-B$ to $A\setminus B$, but the latter is relatively standard in my experience.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you note, De Morgan's laws allow us to translate between union and intersection via complements. Beyond this, we must have at least union and complement (or, equivalently, intersection and complement). That is, we cannot define complement solely in terms union and intersection.&lt;/p&gt;&#xA;" OwnerUserId="2313" LastActivityDate="2010-10-11T21:46:16.300" CommentCount="4" />
  <row Id="6570" PostTypeId="1" AcceptedAnswerId="6571" CreationDate="2010-10-11T21:52:42.047" Score="2" ViewCount="249" Body="&lt;p&gt;(1). According to Wikipedia, a &lt;a href=&quot;http://en.wikipedia.org/wiki/Field_of_sets&quot; rel=&quot;nofollow&quot;&gt;field of subsets&lt;/a&gt; of X is defined to be a non-empty subset of the power set of X closed under the intersection and union of pairs of sets and under complements of individual sets. I was wondering if there is any redundancy in this definition?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can it be defined to be closed only under pairwise intersection and pairwise union? Or only closed under complement and pairwise intersection (or pairwise union)? If yes in each case, how can we show it also closed under the other operation?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2). A sigma algebra of subsets of X is defined to be a nonempty subset of the power set of X closed under complementation and under countable union.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can it be defined to be closed under countable union and countable intersection? If yes, how can we show it is subsequently closed under complement?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks and regards!&lt;/p&gt;&#xA;" OwnerUserId="1281" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2010-11-22T14:37:08.733" LastActivityDate="2010-11-22T14:37:08.733" Title="Field of sets and Sigma algebra of sets" Tags="&lt;elementary-set-theory&gt;&lt;field-theory&gt;&lt;sigma-algebra&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" />
  <row Id="6571" PostTypeId="2" ParentId="6570" CreationDate="2010-10-11T21:59:31.920" Score="5" ViewCount="" Body="&lt;p&gt;No, you can't get complements from unions and intersections.  For example, let $X$ be a nonempty set.  Then $\{\emptyset\}$ is nonempty, closed under (arbitrary) intersection and union, but not closed under complements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can get intersections from unions and complements using &lt;a href=&quot;http://en.wikipedia.org/wiki/De_Morgan%27s_laws&quot; rel=&quot;nofollow&quot;&gt;De Morgan's laws&lt;/a&gt;.  To get an intersection, just take the complement of the union of the complements.  Similarly, you get unions from intersections and complements.  So yes, the definition of field of subsets was redundant.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-10-11T21:59:31.920" CommentCount="2" />
  <row Id="6572" PostTypeId="1" CreationDate="2010-10-11T22:11:22.543" Score="0" ViewCount="193" Body="&lt;p&gt;If G is a finite group containing an irreducible character of degree 5, deduce its order.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that the order of G is divisble by 5&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also know that the order of G is equal to the sum of the squares of the degrees of irreducible characters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However I am unable to deduce the order of G with just this information.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could someone please please please help me determine the exact order of G.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many thanks&#xA;Sarah&lt;/p&gt;&#xA;" OwnerUserId="2444" LastActivityDate="2010-10-12T18:29:24.150" Title="Finding the order of a group" Tags="&lt;group-theory&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="6573" PostTypeId="1" AcceptedAnswerId="6574" CreationDate="2010-10-11T23:22:48.207" Score="0" ViewCount="447" Body="&lt;p&gt;Since union of sets is defined to be independent of the number of sets as in &lt;a href=&quot;http://en.wikipedia.org/wiki/Union_%28set_theory%29#Forms&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If M is a set whose elements are themselves sets, then x is an element of the union of M if and only if for at least one element A of M, x is an element of A.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I was wondering why for a class of subsets of a set, it being closed under finite union and it being closed under countable union are not same?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similar question for intersection.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks and regards!&lt;/p&gt;&#xA;" OwnerUserId="1281" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2010-11-23T04:03:19.957" LastActivityDate="2010-11-23T04:03:19.957" Title="Closed under finite union and Closed under countable union" Tags="&lt;elementary-set-theory&gt;" AnswerCount="2" CommentCount="0" />
  <row Id="6574" PostTypeId="2" ParentId="6573" CreationDate="2010-10-11T23:45:32.610" Score="2" ViewCount="" Body="&lt;p&gt;Starting from a initial collection of sets being allowed to take countable unions and intersections lets you create many more sets that being allowed to take only finite unions and intersections. Therefore it seems plausible to me that the former can take you out of your starting collection even if the latter does not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can find examples in any standard measure theory text. Here is a very simple one. Let&#xA;$$A_n=\{1,2,3,\ldots,n\}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $\Omega$ be the collection of the sets $A_n$ for $n \in \mathbb{N}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then for a finite set of indices, $n_1,n_2,\ldots,n_k$ say, the union&#xA;$$A_{n_1} \cup A_{n_2} \cup \cdots \cup A_{n_k} = A_{\max\{n_1,\ldots,n_k\}}$$&#xA;and hence belongs to $\Omega$. So $\Omega$ is closed under finite unions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the countably infinite union&#xA;$$\cup_{n=1}^{\infty} A_n = \{1,2,3,\ldots\}$$&#xA;and hence does not belong to $\Omega$. So $\Omega$ is not closed under countable unions.&lt;/p&gt;&#xA;" OwnerUserId="1195" LastActivityDate="2010-10-11T23:45:32.610" CommentCount="0" />
  <row Id="6575" PostTypeId="1" CreationDate="2010-10-11T23:48:35.367" Score="0" ViewCount="141" Body="&lt;p&gt;prove by structural induction that in any tree T, the number of leaves is 1 more than the number of nodes that have right siblings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My proof so far:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;s(n).  in any tree T, the number of leaves(L) is 1 more than the number of nodes(N) that have right siblings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;s(1) for a tree of one node, there are no right siblings so n=1. The root itself is a leave so L =1. Thus L=n+1 is true, so the basis is true.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;s(n+1)...Im stuck at the induction part here.&lt;/p&gt;&#xA;" LastActivityDate="2010-11-25T05:16:57.633" Title="structural induction" Tags="&lt;induction&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="6576" PostTypeId="1" AcceptedAnswerId="6579" CreationDate="2010-10-11T23:56:39.193" Score="1" ViewCount="101" Body="&lt;p&gt;This is a contest problem. It asks for all the real solutions of&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sqrt{x^2-p}+2\sqrt{x^2-1}=x$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for arbitrary real $p$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the eq. it's clear that $p$ cannot be negative. By squaring once, splitting the radicals from the other terms and squaring again, I found a formula for $x^2$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x^2 = 1+\frac{p^2}{16(1-p/2)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From here we have $p&amp;lt;2$ for the second radical to be defined, and by comparing this expression with $p$ we get that $p&amp;lt;2$ makes the first radical be defined as well. Yet solutions of the original equation can only be found when $p\leq 4/3$, and $4/3$ is the (double) root of $9p^2-24p+16=0$ (obtained from comparing $1+\frac{p^2}{16(1-p/2)}$ with $p$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't understand why this happens. Perhaps I've done something wrong. If I haven't, I'd appreciate any help in proving that there are no solutions if $p$ is greater than $4/3$.&lt;/p&gt;&#xA;" OwnerUserId="1313" LastActivityDate="2010-10-12T00:51:37.040" Title="Equation with radicals." Tags="&lt;algebra-precalculus&gt;" AnswerCount="1" CommentCount="0" />
  <row Id="6577" PostTypeId="1" CreationDate="2010-10-12T00:02:28.387" Score="1" ViewCount="148" Body="&lt;p&gt;Given a certain coordinate frame, I can compute a new one by applying a set of rotations in a given order (what I call Euler Z-Y-X). So I yaw, then pitch then roll.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now imagine that I want to do exactly the opposite: given two coordinate frames (same origin to simplify), how do I find out roll, pitch and yaw angles that were used to transform from one to the other?&lt;/p&gt;&#xA;" OwnerUserId="2449" LastActivityDate="2010-10-12T00:02:28.387" Title="How to extract Euler angles from a a point in a plane?" Tags="&lt;geometry&gt;" CommentCount="4" />
  <row Id="6578" PostTypeId="2" ParentId="6573" CreationDate="2010-10-12T00:13:57.693" Score="2" ViewCount="" Body="&lt;p&gt;Hmm.  There is a somewhat sneaky and quite false conclusion being made here.  It's true that the Wikipedia definition doesn't mention the number of sets.  However, that doesn't mean that union and intersection are &quot;independent of&quot; the number of sets.  I can't even figure out what this would mean.  To state an obvious counterexample, adding another set to your union will give a set with at least as many elements as before, and adding another set to your intersection will give a set with at most as many elements as before.  (Not even in the sense of cardinality, but just the simple fact that&#xA;$$\left(\bigcup_\alpha A_\alpha\right)\cup B\supset \bigcup_\alpha A_\alpha$$&#xA;and the dual statement for intersections.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Jyotirmoy's example shows that the countable union of finite sets isn't finite.  Another one comes from topology.  A topology on a set $X$ is &lt;em&gt;defined&lt;/em&gt; to be a set of subsets of $X$ that is closed under unions and finite intersections.  We call such sets &quot;open,&quot; and they're roughly a way of saying what points are close to each other.  A typical example is the Euclidean topology on $\mathbb{R}$, which consists of the open intervals $(a,b)$ and their unions.  This is well-defined because any finite intersection of open intervals is either the empty set or another, smaller open interval.  On the other hand, the set $\bigcap_n (-1/n,1/n)$ is just the point $\lbrace 0\rbrace$, which is definitely not a union of open intervals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, countable intersections are still kind of important in the same general situation.  A $\sigma$-algebra is a set of subsets of $X$ that is closed under complementation and countable intersections (and thus also countable unions).  You can use such a thing to define a &quot;measure,&quot; which is an assignment of a sort of &quot;size&quot; to each set in the $\sigma$-algebra.  The algebra of Lebesgue-measurable on $\mathbb{R}^n$ is probably the best example, but you can turn any topology into a $\sigma$-algebra, called its Borel algebra, just by iterating countable intersections and countable unions.  A set that's a countable intersection of open sets is called a $G_\delta$, from the German words &quot;gebiet Durchschnitt&quot; meaning &quot;open intersection.&quot;  So we know, for example, that every $G_\delta$ in $\mathbb{R}$, though not necessarily open, has a Lebesgue measure.&lt;/p&gt;&#xA;" OwnerUserId="2215" LastActivityDate="2010-10-12T00:13:57.693" />
  <row Id="6579" PostTypeId="2" ParentId="6576" CreationDate="2010-10-12T00:29:52.030" Score="4" ViewCount="" Body="&lt;p&gt;When solving equations, by doing algebra like squaring etc, one could potentially include more solutions than the original equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x = 1$ implies $x^2 = 1$, by squaring.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second equation also has $x = -1$ as a solution, which is not a solution of the first one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your case, when you square etc, you might lose constraints like $x \ge 1$ and $x^2 \ge p$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suppose when you squared, you had the term $x - 2\sqrt{x^2 -1}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The constraint that $x \ge 2 \sqrt{x^2-1}$ (i.e. $x^2 \le 4/3$) gives us the constraint that $p \le 4/3$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you noticed,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$1 + \frac{p^2}{16(1-p/2)} -p \ge 0$ for all $p$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$1 + \frac{p^2}{16(1-p/2)} \le 4/3$ implies that $0 \le 1 + \frac{p^2}{16(1-p/2)} -p \le 4/3 - p$ which implies $4/3 \ge p$.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-10-12T00:51:37.040" LastActivityDate="2010-10-12T00:51:37.040" CommentCount="5" />
  <row Id="6580" PostTypeId="1" AcceptedAnswerId="6585" CreationDate="2010-10-12T01:35:40.797" Score="2" ViewCount="306" Body="&lt;p&gt;A field is an abelian group under addition and a group under multiplication. But for the power set of a set S, under union and intersection operations and with S and the empty set being their identities, the inverse of any set is not unique for either union or intersection. So is the power set a field of sets?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Must the inverse of a set for either union or intersection be restricted to the complement of the set?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="1281" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2010-11-22T14:37:18.190" LastActivityDate="2010-11-22T14:37:18.190" Title="inverse element in a field of sets" Tags="&lt;elementary-set-theory&gt;&lt;field-theory&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="6581" PostTypeId="1" CreationDate="2010-10-12T01:40:22.923" Score="3" ViewCount="392" Body="&lt;p&gt;Hey guys, I have a couple things I want to prove. I'm pretty sure a proof by induction is the best route for these.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;First, I need to show that $5^n &amp;lt; n!$ from some $n_{0} &gt; 0$. I'm choosing $n_{0} = 12$ since that's the smallest positive integer where $n! &gt; 5^n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So: $P(k) : 5^k &amp;lt; k!$&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Show $P(12)$: $5^{12} &amp;lt; 12!$&lt;/li&gt;&#xA;&lt;li&gt;Assume $P(k)$, show $P(k+1)$:&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;$P(k+1):$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$5^{(k+1)} \leq (k+1)!$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$5*5^k \leq (k+1)*k!$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is where I'm stuck. Since our previous assumption was that $5^k \leq k!$, couldn't we conclude the proof by showing:&#xA;$5 \leq (k+1)$ for $n \geq 12$?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The second one is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$P(k): n! \leq n^n$&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Show $P(1)$: $1 \leq 1$&lt;/li&gt;&#xA;&lt;li&gt;Assume $P(k)$, show $P(k+1)$:&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;$P (k+1):$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(n+1)! \leq {n+1}^{n+1}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(n+1)*n! \leq {(n+1)} *{(n+1)}^{n}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we cancel out the common factor (we can do that, right?) $(n+1)$ we get:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$n! \leq (n+1)^n$ which is true for $n \geq1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, $n! \leq n^n$ for $n \geq 1 $&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;So in conclusion, I'm wondering if these proofs are sufficient (and correct). I kinda feel like I'm missing something in the last few steps - that it's &quot;turtles all the way down&quot;&lt;/p&gt;&#xA;" OwnerUserId="2451" LastEditorUserId="664" LastEditorDisplayName="" LastEditDate="2010-10-12T17:15:16.863" LastActivityDate="2010-10-18T00:44:56.747" Title="Proof of inequalities by induction" Tags="&lt;homework&gt;&lt;induction&gt;" AnswerCount="2" />
  <row Id="6582" PostTypeId="2" ParentId="6580" CreationDate="2010-10-12T01:42:24.527" Score="2" ViewCount="" Body="&lt;p&gt;One word, two meanings. A 'field of sets' is not a field in the abstract algebra sense.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt; While a 'field of sets' is not a field, a Boolean Algebra (BA) of sets (of which a field of sets is a special case) is indeed an algebra. One can check that a BA is a ring, with symmetric difference $(A \cap B^c) \cup (A^c \cap B)$ as addition and intersection as multiplication. It is then trivially an algebra over the two-element field $\mathbb{F}_2$ by defining $1\cdot A=A$ and $0\cdot A=\emptyset$ and checking that $A+A=\emptyset$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[Arturo's answer discussed this before me in a more general setting. The edit above is to set right my incorrect assertion in the comments that a BA is not an algebra.]&lt;/p&gt;&#xA;" OwnerUserId="1195" LastEditorUserId="1195" LastEditorDisplayName="" LastEditDate="2010-10-12T03:03:29.713" LastActivityDate="2010-10-12T03:03:29.713" CommentCount="10" />
  <row Id="6583" PostTypeId="2" ParentId="6581" CreationDate="2010-10-12T02:29:50.853" Score="4" ViewCount="" Body="&lt;p&gt;Your first proof is correct.  (If $a\le c$ and $b\le d$ and they're all positive, then $ab\le cd$.)  Your second one needs just a tiny bit more.  How do we know that $n!\le (n+1)^n$?  It follows from your inductive hypothesis, but it's not completely trivial.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a general tip, try not to work your proofs backwards.  When you're trying to prove that $P(k)\Rightarrow P(k+1)$, you start out by stating $P(k+1)$ and then deriving a true statement.  It works here because pretty much everything you're doing is reversible (dividing by $k+1$ vs. multiplying), but in later life, this might no longer be true and so it's a good practice to start now.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Otherwise, good work!&lt;/p&gt;&#xA;" OwnerUserId="2215" LastActivityDate="2010-10-12T02:29:50.853" CommentCount="1" />
  <row Id="6584" PostTypeId="1" AcceptedAnswerId="6589" CreationDate="2010-10-12T02:57:44.090" Score="1" ViewCount="258" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am an amateur mathematician (but I do have degree's in computer science (with mathematics)). Anyways, I have written this paper, where I have proved that for $\zeta(\rho) = 0$ if $\Im(\rho) \to \infty$ then $\Re(\rho) \leq \log_2(3) - 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now this is a big claim, and I am a cranky fella ;) who has a number of withdrawn papers on arxiv. So, chances of me being taken seriously $\to 0$ even though I do put my sweat in for checking any paper I write, before releasing it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the problem is how do I get my paper discussed and verified, before I put it on arxiv (cause I do not want to withdraw it again) and also I do not know any local mathematician working on analytic number theory.&#xA;So here's my question:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Before posting on arxiv are there any&#xA;  other avenues through which I can get&#xA;  my paper checked (on the internet)?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2320" LastActivityDate="2010-10-12T04:12:07.087" Title="Alternatives to arxiv" Tags="&lt;education&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" />
  <row Id="6585" PostTypeId="2" ParentId="6580" CreationDate="2010-10-12T02:58:50.503" Score="4" ViewCount="" Body="&lt;p&gt;A &quot;field of sets&quot; is a collection $\mathcal{F}$ of subsets of a given set $X$ which is closed under binary unions, intersections, and under complementation. Alas, it is not an &quot;field&quot; in the sense of abstract algebra.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It &lt;em&gt;is&lt;/em&gt; a &quot;subalgebra&quot; of the Boolean algebra of the power set of $X$. Alas, here as well, &quot;Boolean algebra&quot; and &quot;subalgebra&quot; do not have the meaning they have in abstract algebra/ring theory: a &quot;Boolean algebra&quot; is not really an algebra: rather, it is a special kind of &lt;a href=&quot;http://en.wikipedia.org/wiki/Lattice_%28order%29&quot; rel=&quot;nofollow&quot;&gt;lattice&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Namely, a Boolean algebra is a set $A$ together with two binary operations, $\wedge$ (meet) and $\vee$ (join), one unary operation ${}^c$ (complementation), and two nulary operations (distinguished elements) $0$ and $1$. We require $\vee$ and $\wedge$ to be associative, commutative, and to distribute over each other (this is already different from the case of &quot;algebra&quot; in the ring-theoretic sense, in which only multiplication distributes over sums, and not the other way around); we also have two rules of &quot;absorption&quot; that further describe how $\wedge$ and $\vee$ interact: $x\wedge(x\vee y) = x$ and $x\vee(x\wedge y) = x$. Finally, we require $x\vee x^{c}=1$ and $x\wedge x^{c}=0$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The typical example of a Boolean algebra/Boolean lattice is the lattice of subsets of a given set $X$, with $\wedge$ corresponding to intersection, $\vee$ to union, ${}^c$ to complementation, $0$ to $\emptyset$, and $1$ to $X$. Another example is propositional calculus, with $\wedge$ corresponding to conjunction, $\vee$ to disjunction, ${}^{c}$ to negation, $1$ to the class of tautologies, and $0$ to the class of contradictions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason for calling it &lt;em&gt;algebra&lt;/em&gt;, in defiance of the meaning of &quot;algebra&quot; from ring theory, is historical: Boole talked about &quot;algebra of thought&quot; or &quot;algebra of logic&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, any Boolean algebra gives rise to a ring. We define $+$ by &quot;symmetric difference&quot;: $x+y = (x\wedge y^{c})\vee(x^{c}\wedge y)$. If you do this, then you get an abelian group, with identity element being $0$, and with every element being its own inverse (that is, every nontrivial element is of additive order $2$). We define $*$ by $x*y=x\wedge y$. The unity of this ring is the $1$ from the Boolean lattice. This ring is an algebra over the field of $2$ elements, since it has characteristic $2$ and a unity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You may ask: will this ring be a field? The only time it is a field is if the original Boolean algebra was the trivial algebra, with $0$ and $1$ the only elements. For if $x$ is any other element, $0\neq x$, $x\neq 1$, then if you look at the order induced by the lattice structure ($x\preceq y$ if and only if $x\wedge y = x$) then you have $0\prec x \prec 1$, and $x\wedge y \preceq x$ for all $y$. Therefore, since $x$ is strictly smaller than $1$, then so will $x*y$ for any $y$, so $x$ cannot have a multiplicative inverse in the corresponding ring.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-10-12T03:07:56.660" LastActivityDate="2010-10-12T03:07:56.660" CommentCount="1" />
  <row Id="6586" PostTypeId="2" ParentId="6581" CreationDate="2010-10-12T03:35:24.027" Score="4" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\ $ Rewrite them as products so to make the induction completely trivial.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Namely $\quad\quad\displaystyle  \frac{n^n}{n\:!}\ =\ \ \: \frac{n}1\ \ \ \frac{n}2\: \ \ \frac{n}3\ \:\cdots\ \frac{n}n\ \ge\ 1\ \ $ since each factor is $\:\ge 1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\quad$ and $\rm\quad\quad\displaystyle\ \frac{n\:!}{5^n}\ =\ \frac{12\:!}{5^{12}}\ \frac{13}{5}\ \frac{14}5\ \cdots\ \frac{n}5\ &gt;\ 1\ \ $ since each factor is $\:  &gt; 1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now you need only inductively prove the lemma that a product is $\ge 1\:$ or $&gt; 1\:$ if each factor is - which is completely trivial. Many inductive proofs can be similarly drastically simplified by conceptual preprocessing. Here we've effectively employed &lt;strong&gt;multiplicative telescopy&lt;/strong&gt; to reduce generally intractable exponential inequalities to tractable polynomial inequalities. $\: $ This is a powerful technique with widespread applications. See also the analogous case of &lt;a href=&quot;http://math.stackexchange.com/questions/6814&quot;&gt;&lt;strong&gt;additive telescopy&lt;/strong&gt;&lt;/a&gt; and the fundamental theorem of difference calculus.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-10-18T00:44:56.747" LastActivityDate="2010-10-18T00:44:56.747" />
  <row Id="6587" PostTypeId="1" AcceptedAnswerId="6590" CreationDate="2010-10-12T03:46:43.520" Score="2" ViewCount="145" Body="&lt;p&gt;If f is in $L^1(\mathbb{R})$, is it true that $\lim_{h \to 0^+} \frac1{h} \int_x^{x+h} f(t)\mathrm{d}t$ exists and is finite for every x in $\mathbb{R}$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would it be possible to use something along the lines of the following argument: The Lebesgue Differentiation Theorem says that this integral is equal to f(x) a.e., which is finite a.e. if f is in $L^1(\mathbb{R})$. And since the integral doesn't change on a set of measure 0, then the limit itself must be finite a.e. &lt;/p&gt;&#xA;" OwnerUserId="1736" LastEditorUserId="1736" LastEditorDisplayName="" LastEditDate="2010-10-12T06:20:19.313" LastActivityDate="2010-10-12T06:20:19.313" Title="Finiteness of $\lim_{h \to 0^+} \frac1{h} \int_x^{x+h} f(t)\mathrm{d}t$" Tags="&lt;real-analysis&gt;" AnswerCount="1" />
  <row Id="6588" PostTypeId="2" ParentId="6584" CreationDate="2010-10-12T04:11:05.663" Score="6" ViewCount="" Body="&lt;p&gt;No one on the internet (or elsewhere) is obliged to check your paper. So you have to catch people's interest and minimize the amount of investment they have to make. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are some suggestion based not on personal experience but from being an spectator round the Net:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Make your work easy to understand: Present your work in a theorem-lemma format, clearly specifying links to existing literature and pinpointing your own specific contributions.&lt;/li&gt;&#xA;&lt;li&gt;Make your work publicly accessible: Put it up on you website. Or on Google Documents. Or on &lt;a href=&quot;http://vixra.org&quot; rel=&quot;nofollow&quot;&gt;http://vixra.org&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;Publicise the key points: Make a newsgroup posting, or create a question here or on Mathoverflow that explains your main insight or innovation and link to your paper. &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I was a bit hesitant in writing (3) since if you do it wrong you are just creating spam. No one is obliged to read on if you say &quot;I have proved X. Can you please check?&quot;. Much better to say &quot;So far people did not succeed in approach A to problem B because of C. But I think things can be made to work if you do D. Here's an attempt.&quot;&lt;/p&gt;&#xA;" OwnerUserId="1195" LastActivityDate="2010-10-12T04:11:05.663" CommunityOwnedDate="2010-10-12T04:11:05.663" />
  <row Id="6589" PostTypeId="2" ParentId="6584" CreationDate="2010-10-12T04:12:07.087" Score="7" ViewCount="" Body="&lt;p&gt;There are other places to post papers, but most are either harder to get into (most online journals) or are less reputable than the arXiv (like viXra).  Finding math forums and posting there may be your best bet.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://mymathforum.com/&quot; rel=&quot;nofollow&quot;&gt;http://mymathforum.com/&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://mersenneforum.org/&quot; rel=&quot;nofollow&quot;&gt;http://mersenneforum.org/&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://physicsforums.com/&quot; rel=&quot;nofollow&quot;&gt;http://physicsforums.com/&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;to name just a few.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;On this specific point:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your result seems wrong.  Aren't there vertical asymptotes and, in particular, values unbounded above for $\Re(\zeta(x+iy))$ for any fixed $y$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have (or think you have) an &lt;em&gt;effective&lt;/em&gt; proof, can you give bounds Y and Z &gt; 0 where, for any $y&gt;Y$, $\Re(\zeta(x+iy)) &amp;lt; Z$?&lt;/p&gt;&#xA;" OwnerUserId="1778" LastActivityDate="2010-10-12T04:12:07.087" CommentCount="1" />
  <row Id="6590" PostTypeId="2" ParentId="6587" CreationDate="2010-10-12T04:15:41.437" Score="5" ViewCount="" Body="&lt;p&gt;No.  As you said, it is true almost everywhere, but for example if $f(x)=x^{-1/2}$ on $(0,1)$ (and whatever you like elsewhere), then for $0\lt h\lt 1$, $1/h\int_0^h f(t)dt=2/\sqrt{h}$.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-10-12T04:15:41.437" CommentCount="0" />
  <row Id="6591" PostTypeId="2" ParentId="6568" CreationDate="2010-10-12T04:51:25.983" Score="4" ViewCount="" Body="&lt;p&gt;There is a &quot;minimal&quot; definition of set operations, though it's not really a restriction of the classical ones, and is more a fun trick than a clever insight.  It was discovered by Charles Sanders Peirce, for general Boolean algebras.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $A\uparrow B=(A\cap B)^c$.  Then $A\uparrow A=A^c$.  So $(A\uparrow A)\uparrow(B\uparrow B)=(A^c\cap B^c)^c=A\cup B$.  Similarly, $(A\uparrow B)\uparrow(A\uparrow B)=((A\cap B)^c)^c=A\cap B$.  And then you can get difference or whatever else, as expected.  You can do the same thing with $(A\cup B)^c$.&lt;/p&gt;&#xA;" OwnerUserId="2215" LastActivityDate="2010-10-12T04:51:25.983" CommentCount="2" />
  <row Id="6592" PostTypeId="2" ParentId="6199" CreationDate="2010-10-12T04:51:32.547" Score="2" ViewCount="" Body="&lt;p&gt;Here is a somewhat detailed outline of an argument that I think works.  As this is a homework problem, some of the pieces of the argument do need to be filled in.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume we're in $\mathbb{R}^m$.  For fixed $y \in \Gamma(x)$, the fact that $\Gamma(x)$ has a nonempty interior means that there are $m$ points $z_1, z_2, \ldots, z_m$ in the interior of $\Gamma(x)$ such that these $m$ points and $y$ together are affinely independent.  Thus you can take sufficiently small balls around each of these $m+1$ points such that the balls do not intersect and that any set consisting of one point from each ball is also affinely independent.  Let $z_0 = y$.  Since $\phi$ is lower hemicontinuous, for each $z_i$ there exists a sequence $z_{i_n} \to z_i$ and an $N_i$ such that $z_{i_n} \in \phi(x_n)$ and $z_{i_n}$ is inside that small ball around $z_i$ for all $n \geq N_i$.  For each $n \geq \max \{N_i\}$, construct the convex hull $C_n$ of $\{z_{0_n}, z_{1_n}$, $z_{2_n}, \ldots, z_{i_n}\}$.  Since $\phi$ is convex-valued, $C_n$ is a subset of $\phi(x_n)$.  Do the same thing for each $n$ for the $\psi$ function to obtain sets $D_n$.  Let $S_n = C_n \cap D_n$.  The intersection of the convex hulls of two sets of $m+1$ affinely independent points in $\mathbb{R}^m$ that are pairwise close to each other must be nonempty.  (Consider the supporting hyperplanes.)  Let $y_n$ be the point in $S_n$ closest to $y$.  Since the extreme points of $S_n$ converge to the extreme points of the convex hull of $\{z_0, z_1, z_2, \ldots, z_m\}$, $S_n$ converges as a set to the convex hull of $\{z_0, z_1, z_2, \ldots, z_m\}$.  Thus the point in $S_n$ closest to $y$ $(= z_0)$ must converge to $y$; i.e., $y_n \to y$. &lt;/p&gt;&#xA;" OwnerUserId="2370" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-10-13T13:29:04.093" LastActivityDate="2010-10-13T13:29:04.093" CommentCount="2" />
  <row Id="6593" PostTypeId="1" AcceptedAnswerId="6693" CreationDate="2010-10-12T05:49:46.957" Score="1" ViewCount="156" Body="&lt;p&gt;If $c$ is a singular 1-cube in $\mathbb{R}^2 \backslash \left\{ (0,0)\right\}$ with $c(0)=c(1)$, show that there is an integer $n$ such that: $c-c_{(1,n)} = \partial (c^2)$ for some $2$-chain $c^2$.&lt;/p&gt;&#xA;" OwnerUserId="2319" LastEditorUserId="664" LastEditorDisplayName="" LastEditDate="2010-10-13T04:39:00.107" LastActivityDate="2010-10-13T17:56:39.443" Title="Boundary of a 2-chain" Tags="&lt;differential-geometry&gt;" AnswerCount="1" CommentCount="7" FavoriteCount="1" />
  <row Id="6594" PostTypeId="1" CreationDate="2010-10-12T06:26:30.657" Score="1" ViewCount="124" Body="&lt;p&gt;I am trying to work through an example from Jordan and Smiths book Nonlinear Ordinary Differential Equations&#xA;Its example 6.1 on page 195&#xA;The question reads: obtain an approximate solution of the family of equations x&quot;+x=ex^3&#xA;with x(e,0)=1  x'(e,0)=0 and error O(e&lt;sup&gt;3&lt;/sup&gt;) uniformly on t&gt;=0, by the method of coordinate pertubation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This what I have so far:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;x&quot;&lt;sub&gt;0&lt;/sub&gt;+x&lt;sub&gt;0&lt;/sub&gt;=0&#xA;So we get: &#xA;x&lt;sub&gt;0&lt;/sub&gt;=A&lt;sub&gt;0&lt;/sub&gt;cost + B&lt;sub&gt;0&lt;/sub&gt;sint&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the conditions given: x&lt;sub&gt;0&lt;/sub&gt;(0)=1 and x'&lt;sub&gt;0&lt;/sub&gt;(0)=0&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have x&lt;sub&gt;0&lt;/sub&gt;=cost&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second equation is x&quot;&lt;sub&gt;1&lt;/sub&gt;+x&lt;sub&gt;1&lt;/sub&gt;=x&lt;sub&gt;0&lt;/sub&gt;&lt;sup&gt;3&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which becomes x&lt;sub&gt;1&lt;/sub&gt; + x&lt;sub&gt;1&lt;/sub&gt;=cos&lt;sup&gt;3&lt;/sup&gt;t&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So x&quot;&lt;sub&gt;1&lt;/sub&gt;+x&lt;sub&gt;1&lt;/sub&gt; = 3/4cost+ 1/4cos3t&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I assume x&lt;sub&gt;1&lt;/sub&gt;=Acost+Bcos3t&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So x'&lt;sub&gt;1&lt;/sub&gt;=-Asint-3Bsin3t&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And&#xA;ẍ&lt;sub&gt;1&lt;/sub&gt;=-Acost-9Bcos3t&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore ẍ&lt;sub&gt;1&lt;/sub&gt; + x&lt;sub&gt;1&lt;/sub&gt; = -8Bcos3t = 3/4cost+ 1/4cos3t&lt;/p&gt;&#xA;&#xA;&lt;p&gt;therefore -8B= 1/4&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so B = -1/32&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which gives me x&lt;sub&gt;1&lt;/sub&gt;=Acost - 1/32cos3t&lt;/p&gt;&#xA;&#xA;&lt;p&gt;using this and the condition that x&lt;sub&gt;0&lt;/sub&gt;(0)=0 gives me A = 1/32&lt;/p&gt;&#xA;&#xA;&lt;p&gt;however the solution the book gives contains another term : 3/8tsint&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i dont know how to get this term.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;could someone please help&lt;/p&gt;&#xA;" OwnerUserId="2444" LastEditorUserId="2444" LastEditorDisplayName="" LastEditDate="2010-10-12T13:37:52.307" LastActivityDate="2010-10-12T13:37:52.307" Title="differential equations" Tags="&lt;differential-equations&gt;" CommentCount="1" />
  <row Id="6595" PostTypeId="1" CreationDate="2010-10-12T06:51:50.620" Score="1" ViewCount="283" Body="&lt;p&gt;$\displaystyle\lim_{\epsilon \rightarrow 0} \int \frac{1}{x^{1+\epsilon}} \mathrm{d}x $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How should I go about evaluating this integral?&#xA;Does this integral converge to $log_e x $ or to something else?&lt;/p&gt;&#xA;" OwnerUserId="1798" LastEditorUserId="367" LastEditorDisplayName="" LastEditDate="2010-10-12T15:03:31.177" LastActivityDate="2010-10-12T15:03:31.177" Title="How do I evaluate this integral?" Tags="&lt;integral&gt;&lt;limit&gt;" AnswerCount="4" CommentCount="2" />
  <row Id="6596" PostTypeId="2" ParentId="6595" CreationDate="2010-10-12T07:03:03.027" Score="0" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;edit&lt;/strong&gt; The integral asked for was $\displaystyle\lim_{\epsilon\to0}\int\frac1{x^{1+\epsilon}}dx$, so it's $=\lim\int x^{-(1+\epsilon)}dx = \lim \frac1{-\epsilon}x^{-\epsilon} = -\mathrm{sign}(\epsilon)\cdot\infty$ (just use $\int x^a dx = \frac1{a+1} x^{a+1}$ for $a\neq -1$).&lt;/p&gt;&#xA;" OwnerUserId="163" LastEditorUserId="163" LastEditorDisplayName="" LastEditDate="2010-10-12T08:57:19.577" LastActivityDate="2010-10-12T08:57:19.577" CommentCount="9" />
  <row Id="6597" PostTypeId="2" ParentId="6595" CreationDate="2010-10-12T07:04:21.230" Score="0" ViewCount="" Body="&lt;p&gt;Yes, the limit quantity whether its outside the integral or inside the integral doesn't make any sense. So the answer is $$\lim_{\epsilon \to 0} \frac{1}{1+ \epsilon} \cdot \int \frac{1}{x} \ dx = \log{x}$$&lt;/p&gt;&#xA;" LastActivityDate="2010-10-12T07:04:21.230" CommentCount="2" />
  <row Id="6598" PostTypeId="2" ParentId="6595" CreationDate="2010-10-12T07:06:25.783" Score="0" ViewCount="" Body="&lt;p&gt;Evaluate the integral first using u-substitution and you'll end up with (1/(1+epsilon))ln(u) and the the limit of this as epsilon approaches 0 is indeed ln(u).&lt;/p&gt;&#xA;" OwnerUserId="2302" LastActivityDate="2010-10-12T07:06:25.783" CommentCount="1" />
  <row Id="6599" PostTypeId="1" AcceptedAnswerId="6602" CreationDate="2010-10-12T08:43:11.790" Score="1" ViewCount="134" Body="&lt;p&gt;If $f(x) = \sum_{j=1}^n c_j 1_{A_j}(x)$ ie a simple function. Why is $$\prod_{j=1}^n \exp(t[exp(i(u,c_j))-1]\mu(A_j)) =\exp[t\int_A[\exp(i(u,f(x)))-1]\mu(dx)].$$ The notation (u,c) is dot product, $1_{A}(x)$ is the indicator function and $i=\sqrt{-1}$.&lt;/p&gt;&#xA;" OwnerUserId="626" LastEditorUserId="626" LastEditorDisplayName="" LastEditDate="2010-10-12T18:16:53.847" LastActivityDate="2010-10-12T18:16:53.847" Title="integrating a simple functions" Tags="&lt;integral&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="6600" PostTypeId="2" ParentId="6567" CreationDate="2010-10-12T08:50:01.127" Score="6" ViewCount="" Body="&lt;p&gt;The statement you mention is a special case of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Cayley%E2%80%93Bacharach_theorem&quot; rel=&quot;nofollow&quot;&gt;Cayley-Bacharach theorem&lt;/a&gt;. &#xA;For one way of generalizing the theorem, see the beautiful article &lt;a href=&quot;http://www.ams.org/bull/1996-33-03/S0273-0979-96-00666-0/S0273-0979-96-00666-0.pdf&quot; rel=&quot;nofollow&quot;&gt;&quot;Cayley-Bacharach theorems and conjectures&quot;&lt;/a&gt; by Eisenbud-Green-Harris.&lt;/p&gt;&#xA;" OwnerUserId="2436" LastActivityDate="2010-10-12T08:50:01.127" />
  <row Id="6601" PostTypeId="2" ParentId="6491" CreationDate="2010-10-12T09:19:02.737" Score="1" ViewCount="" Body="&lt;p&gt;This was supposed to be a comment to reply to Gio's comment in Matt's answer, but it got too long.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gio, you should have mentioned that little detail about your application to determining the number of panels needed for the trapezoidal rule in your question, because it now looks to me that you're going about it the wrong way. what you actually need here is an estimate of the truncation error of the trapezoidal rule.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One expression you can use for the error estimate is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{h^2}{12}\left(f^{\prime}(b)-f^{\prime}(a)\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $a$ and $b$ are the upper and lower limits, $h=\frac{b-a}{n}$ is the panel size, and $f^{\prime}$ is the derivative of your integrand. To use this, suppose you want the error to be around $10^{-11}$. Equate the error estimate to your desired tolerance, and solve for $n$. This error estimate assumes that $f$ is continuous up to the second derivative ($C^2$); otherwise, the estimate can fail badly. Apart from this, the other possible way the error estimate can fail is if $f^{\prime}$ vanishes at the endpoints (e.g. when integrating a periodic function over one period); for this particular case, the trapezoidal rule is actually more accurate than usual.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-10-12T09:19:02.737" CommentCount="2" />
  <row Id="6602" PostTypeId="2" ParentId="6599" CreationDate="2010-10-12T09:41:43.043" Score="0" ViewCount="" Body="&lt;p&gt;Because $e^a \cdot e^b = e^{a+b}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only thing $t(e^{i\langle u,c_j\rangle}-1)$ does (for the purpose of evaluating the integral) is just to make the manipulations a bit more ugly.&lt;/p&gt;&#xA;" OwnerUserId="1269" LastActivityDate="2010-10-12T09:41:43.043" />
  <row Id="6603" PostTypeId="2" ParentId="6544" CreationDate="2010-10-12T10:10:43.350" Score="11" ViewCount="" Body="&lt;p&gt;Let $x=(1+s)/2$, so that the expression becomes&#xA;$$ \frac{n}{2^{n+1}} \int_{-1}^1 \frac{(1+s)^n-(1-s)^n}{s} ds =&#xA;\frac{n}{2^{n}} \int_{0}^1 \frac{(1+s)^n-(1-s)^n}{s} ds. $$&#xA;(The integrand is an even function.)&#xA;Fix some $c$ between 0 and 1, say $c=1-\epsilon$.&#xA;Then the integral from 0 to $c$ will be small in comparison to $2^n$,&#xA;since the integrand is bounded by a constant times $(1+c)^n$,&#xA;so as far as the limit is concerned it is enough to look at the&#xA;integral from $c$ to 1,&#xA;and in that integral we can neglect the term $(1-s)^n$ since it will&#xA;also contribute something much smaller than $2^n$.&#xA;The surviving contribution therefore comes from&#xA;$$ \int_c^1 \frac{(1+s)^n}{s} ds, $$&#xA;which lies between&#xA;$$ \int_c^1 \frac{(1+s)^n}{1} ds $$&#xA;and&#xA;$$ \int_c^1 \frac{(1+s)^n}{c} ds, $$&#xA;that is,&#xA;$$ \frac{2^{n+1} - (1+c)^{n+1}}{n+1} &amp;lt; \int_c^1 \frac{(1+s)^n}{s} ds &amp;lt; \frac{1}{c} \frac{2^{n+1} - (1+c)^{n+1}}{n+1} .$$&#xA;Multiplying by $n/2^n$ and letting $n\to\infty$ shows that the liminf&#xA;is at least 2 and the limsup is at most $2/c$.&#xA;But since this holds for any $c$ between 0 and 1,&#xA;it follows that liminf=limsup=2, hence the limit is 2.&lt;/p&gt;&#xA;" OwnerUserId="1242" LastEditorUserId="1242" LastEditorDisplayName="" LastEditDate="2010-10-13T06:41:54.610" LastActivityDate="2010-10-13T06:41:54.610" CommentCount="5" />
  <row Id="6604" PostTypeId="2" ParentId="6595" CreationDate="2010-10-12T10:34:51.367" Score="4" ViewCount="" Body="&lt;p&gt;As it stands, the limit doesn't exist (see Tobias's answer).&#xA;But if you consider the &lt;em&gt;definite&lt;/em&gt; integral from 1 to $x$,&#xA;then the limit is $\ln x$:&#xA;$$ \int_1^x \frac{dt}{t^{1+\epsilon}} =&#xA;\left[ \frac{t^{-\epsilon}}{-\epsilon} \right]_1^x =&#xA;\frac{1-x^{-\epsilon}}{\epsilon} =&#xA;\frac{1-e^{-\epsilon \ln x}}{\epsilon} =&#xA;\frac{1-(1-\epsilon \ln x+O(\epsilon^2))}{\epsilon}&#xA;\to \ln x$$&#xA;as $\epsilon\to 0$.&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2010-10-12T10:34:51.367" CommentCount="3" />
  <row Id="6605" PostTypeId="1" AcceptedAnswerId="6658" CreationDate="2010-10-12T11:56:57.037" Score="2" ViewCount="237" Body="&lt;p&gt;We know that the Mandelbrot set is derived from the iterations of z^2 + c.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do anyone know something about magnet Mandelbrot? I found it in the software UltraFractal, and it is much more beautiful than the original Mandelbrot, in my opinion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do you know anything about it? What is it iterating?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Here's a picture of a magnet Julia set&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/yoZGK.jpg&quot; alt=&quot;Magnet Julia&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/tZVDM.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="2354" LastEditorUserId="2354" LastEditorDisplayName="" LastEditDate="2010-10-24T10:55:17.007" LastActivityDate="2010-10-24T10:55:17.007" Title="Magnet Mandelbrot Set" Tags="&lt;fractals&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="1" />
  <row Id="6606" PostTypeId="1" AcceptedAnswerId="6607" CreationDate="2010-10-12T13:34:19.990" Score="1" ViewCount="203" Body="&lt;p&gt;In a boolean algebra, 0 (the lattice's bottom) is the identity element for the join operation $\lor$, and 1 (the lattice's top) is the identity element for the meet operation $\land$. For an element in the boolean algebra, its inverse/complement element for $\lor$ is wrt 1 and its inverse/complement element for $\land$ is wrt 0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A Boolean algebra can be defined to be a complemented lattice that is also distributive. For a distributive lattice, the complement of x, when it exists, is unique. See Wikipedia (http://en.wikipedia.org/wiki/Lattice_(order)#Complements_and_pseudo-complements).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The power set of a set $S$ is an example of Boolean algebra. $S$ is the identity for union and $\emptyset$ is the identity for intersection. However, for union, the complement of a set wrt $S$ is not unique; For intersection, the complement of a set wrt $\emptyset$ is not unique either. So is this a contradiction?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks and regards!&lt;/p&gt;&#xA;" OwnerUserId="1281" LastActivityDate="2010-10-12T13:51:07.550" Title="identity and inverse/complement elements in a boolean algebra" Tags="&lt;set-theory&gt;&lt;boolean-algebra&gt;" AnswerCount="1" />
  <row Id="6607" PostTypeId="2" ParentId="6606" CreationDate="2010-10-12T13:51:07.550" Score="5" ViewCount="" Body="&lt;p&gt;By definition, $\rm b\:$ is a complement of $\rm a\:$  if $\rm\ a\vee b = 1,\ a\wedge b = 0\:$. So a unique complement must be a unique solution to &lt;em&gt;both&lt;/em&gt;$\ $ equations (involving &lt;em&gt;both&lt;/em&gt;$\ $ operations), not just a single operation - as you consider above. So there is no contradiction.&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-10-12T13:51:07.550" />
  <row Id="6608" PostTypeId="2" ParentId="6412" CreationDate="2010-10-12T14:59:13.333" Score="0" ViewCount="" Body="&lt;p&gt;Not an answer to my question, as opposed to a workaround I applied for my case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Firstly, in my particular case, all $\lvert I \rvert = 2$ for all $I \in \mathbf{I}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sketching, I applied the following:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;model the weighted graph of non-transitive, symmetric (directionless) equivalences;&lt;/li&gt;&#xA;&lt;li&gt;apply iterations until fixpoint as follows:&#xA;&lt;ul&gt;&#xA;&lt;li&gt;order the edges based on edge weight (assumes a total ordering, and granular ordering such that you don't have to make trivial decisions between equals weights);&lt;/li&gt;&#xA;&lt;li&gt;iterate over the edges in decreasing order, and derive a &quot;strong, transitive equivalence&quot; for the highest weighted edge:&#xA;&lt;ul&gt;&#xA;&lt;li&gt;merge those nodes in the graph, and their edges and edge weights;&lt;/li&gt;&#xA;&lt;li&gt;do not merge nodes which conflict with $\mathbf{I}$;&lt;/li&gt;&#xA;&lt;li&gt;do not merge nodes which have already been involved in a merge in that iteration.&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Again, this is not precisely an answer to my question, but the core idea is to rebuild the equivalences in decreasing order of edge weights, avoiding any equivalences that would equate two identifiers known to be different. You might not end up with a &quot;minimal-cut&quot;, but it's a fairly intuitive, tractable solution (which I hope makes sense).&lt;/p&gt;&#xA;" OwnerUserId="2331" LastActivityDate="2010-10-12T14:59:13.333" />
  <row Id="6609" PostTypeId="1" AcceptedAnswerId="6621" CreationDate="2010-10-12T15:32:18.920" Score="2" ViewCount="543" Body="&lt;p&gt;These operators are written in different forms in Cartesian,&#xA;cylindrical and spherical coordinates. For instance, in spherical&#xA;coordinate system, one has&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\nabla \cdot \overrightarrow{F}=\frac{1}{r^{2}}\frac{\partial }{\partial r}&#xA;\left( r^{2}F_{r}\right) +\frac{1}{r\sin \theta }\frac{\partial }{\partial&#xA;\theta }\left( \sin \theta \cdot F_{\theta }\right) +\frac{1}{r\sin \theta }&#xA;\frac{\partial F_{\varphi }}{\partial \varphi }.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question&lt;/strong&gt;: Do identities such as &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\nabla \cdot \left( \overrightarrow{A}\times \overrightarrow{B}\right) =%&#xA;\overrightarrow{B}\cdot \nabla \times \overrightarrow{A}-\overrightarrow{A}%&#xA;\cdot \nabla \times \overrightarrow{B}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;hold in general when cylindrical and spherical coordinate systems are used or do they have to be adapted?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added&lt;/strong&gt;: After having read the comments it occured to me that the invariance of&#xA;these identities with regard to the coordinate system is a consequence of&#xA;the definitions of the mentioned operators in terms of integrals, e.g.:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\nabla \cdot \overrightarrow{F}=\underset{V\rightarrow 0}{\lim }\frac{1}{V}%&#xA;\underset{S}{\int \int }\overrightarrow{F}\cdot \overrightarrow{n}\;dA$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $V$ is the volume of a bounded closed region $T$, $S$ is the surface&#xA;of $T$, and $\overrightarrow{n}$ the unit outer normal vector to $S$.&lt;/p&gt;&#xA;" OwnerUserId="752" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2010-12-26T12:15:21.323" LastActivityDate="2010-12-26T12:15:21.323" Title="Do Vector Calculus Cartesian coordinates identities with Div, Grad, Curl hold in cylindrical and spherical coordinates?" Tags="&lt;calculus&gt;&lt;physics&gt;&lt;vector-calculus&gt;&lt;multivariable-calculus&gt;" AnswerCount="2" CommentCount="5" />
  <row Id="6610" PostTypeId="1" AcceptedAnswerId="6611" CreationDate="2010-10-12T15:57:31.830" Score="1" ViewCount="78" Body="&lt;p&gt;Suppose I have a binary vector $v$ that is copied twice by two separate machines $M$ and $N$, resulting in two new vectors $x$ and $y$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both machines are faulty in the sense that they both might fail to copy each character correctly. More formally, $M$ has an associated vector $e_M$ of length $|v|$, where $(e_M)_i$ is the probability that bit $i$ in $M$'s input will be inverted in $M$'s output (and $N$ has a similarly defined, but not necessarily identical, associated vector $e_N$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Questions:&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;if the only information I receive is $x$, $y$, and two real numbers $p$ and $q$ describing the probabilities that a mistake has been made in producing $x$ and $y$ (i.e. $p$ is the probability that $M$ made an error when producing $x$ from $v$, and $q$ is the probability that $N$ made an error when producing $y$ from $v$), is there a way to reconstruct, or approximate $e_M$ and $e_N$? &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;does the answer to question 1 depend on $|v|$ (which is of course assumed to be at least $2$ for this question to make sense), and if so, how?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;(all apologies if the question is trivial, and if I'm not using the right terms)&lt;/p&gt;&#xA;" OwnerUserId="689" LastActivityDate="2010-10-12T16:38:58.290" Title="Separate probabilities from &quot;joint&quot; probability" Tags="&lt;probability&gt;" AnswerCount="1" CommentCount="0" />
  <row Id="6611" PostTypeId="2" ParentId="6610" CreationDate="2010-10-12T16:38:58.290" Score="0" ViewCount="" Body="&lt;p&gt;$1-p$ gives you the product of $(1-(e_M)_i)$ over $i$ and similarly for $1-q$.  If you only have one set of $x$ and $y$ and don't know anything about the $(e_M)_i$ and $(e_N)_i$ there is no hope.  Even if you get a lot of $x,y$ pairs but don't get any information of what is the truth all you can tell is the combination of error rates per bit.  The chance of disagreement at a given bit is $e_M+e_N-e_M*e_N$ but you can't sort out any more than that.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-10-12T16:38:58.290" CommentCount="1" />
  <row Id="6613" PostTypeId="1" AcceptedAnswerId="6614" CreationDate="2010-10-12T16:51:30.630" Score="5" ViewCount="185" Body="&lt;p&gt;In MK (Morse-Kelley) set theory life is easy: $\forall X\forall y\left(y\in\bigcap X\leftrightarrow\forall x\left(x\in X\rightarrow y\in x\right)\right)$. If $X=\left\{\right\}$ then $\bigcap X=U$, where $U$ is the universal class. So the (unary) intersection of the empty set is the class that contains all sets as elements. In ZF (Zermelo-Fraenkel) set theory, instead, proper classes are not allowed. So, how can I define $\bigcap X$ in ZF? I tried with the following definitions:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$\forall X\left(X\not=\left\{\right\}\rightarrow\forall y\left(y\in\bigcap X\leftrightarrow\forall x\left(x\in X\rightarrow y\in x\right)\right)\right)$. This means that $\bigcap\left\{\right\}$ is undefined, which is not that good.&lt;/li&gt;&#xA;&lt;li&gt;$\forall X\forall y\left(y\in\bigcap X\leftrightarrow X\not=\left\{\right\}\land\forall x\left(x\in X\rightarrow y\in x\right)\right)$. This means that $\bigcap\left\{\right\}=\left\{\right\}$, which is the opposite of MK.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I couldn't find any other valuable definition. Any ideas? Thank you.&lt;/p&gt;&#xA;" OwnerUserId="2128" LastActivityDate="2010-10-12T17:48:01.470" Title="Unary intersection of the empty set" Tags="&lt;set-theory&gt;" AnswerCount="3" />
  <row Id="6614" PostTypeId="2" ParentId="6613" CreationDate="2010-10-12T17:02:59.853" Score="3" ViewCount="" Body="&lt;p&gt;The fact that proper classes are handled differently in ZF than in MK doesn't really affect the definition of the unary intersection or other operations that can produce proper classes. So in ZF, $\bigcap z$ is defined using the same definition as in MK: &#xA;$$ &#xA;y \in \bigcap z \Leftrightarrow (\forall w \in z) ( y \in w).&#xA;$$&#xA;Just like in MK, $\bigcap \emptyset$ is a proper class in ZF. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are treatments of ZF in which the authors don't handle proper classes at all, but there are other treatments in which proper classes are &quot;allowed&quot;. The simplest way to allow them is to restrict to definable proper classes, and silently replace each such class with its definition whenever the class is used. In this way, for any class or set $z$ we have a (possibly proper) class $\bigcap z$ defined as above. We can prove as a theorem in ZF (just as in MK) is that if $z$ is a nonempty set then there is a set $\hat{z}$ such that $\hat{z} = \bigcap z$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Addendum: I want to emphasize that this does not conflict with the answer given by Arturo Magidin. The definition I stated here (in which $\bigcap \emptyset$ is a proper class) is the one used by Levy, &lt;em&gt;Basic Set Theory&lt;/em&gt;). The definition in the other answer, in which $\bigcap \emptyset = \emptyset$, is used by other authors (I believe). Kunen, Jech, and Halmos all leave $\bigcap \emptyset$ undefined in their well-known texts. &lt;/p&gt;&#xA;" OwnerUserId="630" LastEditorUserId="630" LastEditorDisplayName="" LastEditDate="2010-10-12T17:48:01.470" LastActivityDate="2010-10-12T17:48:01.470" CommentCount="4" />
  <row Id="6615" PostTypeId="2" ParentId="6613" CreationDate="2010-10-12T17:07:53.237" Score="0" ViewCount="" Body="&lt;p&gt;The only other way I can think of is to use specification, which is really the only way you know that such a set exists.  Given a universe $U$, define $\bigcap X$ to be the unique $B$ satisfying the following formula:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\forall x(x\in B\Leftrightarrow(x\in U\wedge \exists y(y\in X\wedge x\in y)))$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then $\bigcap\emptyset=U$.  Is this any more or less desirable than having $\bigcap\emptyset$ be a proper class or $\emptyset$ itself?  I don't know.  My topology instructor used it when defining a topology to ensure that the universe was always open, but this is just cosmetic -- there's nothing to be lost by just including this as another axiom.  Ultimately, I suppose everyone but the most hardcore set theorists will just use what works and ignore the axiomatic background.&lt;/p&gt;&#xA;" OwnerUserId="2215" LastActivityDate="2010-10-12T17:07:53.237" />
  <row Id="6616" PostTypeId="1" AcceptedAnswerId="6618" CreationDate="2010-10-12T17:16:48.733" Score="6" ViewCount="536" Body="&lt;p&gt;While proving that some normed spaces were complete, two questions came to my mind. They relate the topological and the metric structures induced by a norm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Is it possible to find two equivalent norms $\|\cdot\|_1$ and $\|\cdot\|_2$ on a vector space $V \ $ such that $(V \ ,\|\cdot\|_1)$ is complete and $(V \ ,\|\cdot\|_2)$ is not?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Is there a vector space $V \ $ and two non-equivalent norms such that $V \ $ is complete relative to both?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here I'm assuming $V \ $ a vector space over a subfield of $\mathbb{C}$. Also I know that the answer is no if we only consider finite-dimensional vector spaces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[Edit: I'm considering two norms equivalent if they define the same topology. I think it's the usual notion Jonas referred in his comment.]&lt;/p&gt;&#xA;" OwnerUserId="329" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2010-10-12T19:44:58.100" LastActivityDate="2010-10-12T22:55:23.273" Title="Metric and Topological structures induced by a norm" Tags="&lt;analysis&gt;&lt;functional-analysis&gt;" AnswerCount="1" CommentCount="10" FavoriteCount="2" />
  <row Id="6618" PostTypeId="2" ParentId="6616" CreationDate="2010-10-12T17:24:10.453" Score="10" ViewCount="" Body="&lt;p&gt;1) No.  It is straightforward to show that equivalent norms yield both the same convergent sequences and the same Cauchy sequences. (Written before Rasmus's answer was posted, but posted afterward.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Yes.  One way to see this is to note that isomorphism classes of vector spaces depend only on linear dimension, so the question amounts to finding 2 nonisomorphic Banach spaces of the same linear dimension.  There are lots of examples of these.  &lt;a href=&quot;http://www.jstor.org/stable/2318458&quot; rel=&quot;nofollow&quot;&gt;Every infinite dimensional separable Banach space has linear dimension $2^{\aleph_0}$&lt;/a&gt;.  However, for example, $\ell^1$ and $c_0$ are separable Banach spaces that are not isomorphic (as Banach spaces).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Actually, &quot;amounts to&quot; wasn't quite accurate.  It is certainly sufficient that the 2 Banach spaces are not isomorphic, but it is not necessary because you are only asking that one particular map (the identity in the original formulation) is not an isomorphism.  So what I gave above is actually stronger.  To just answer 2), you could just take any infinite dimensional Banach space and induce a new norm via an unbounded linear isomorphism with itself.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The answer above was assuming that equivalent norms are defined as in &lt;a href=&quot;http://planetmath.org/encyclopedia/EquivalentNorms.html&quot; rel=&quot;nofollow&quot;&gt;this PlanetMath article&lt;/a&gt;.  If instead you meant only that the spaces are homeomorphic in the norm topologies, as Jyotirmoy Bhattacharya suspected, then the examples alluded to above won't work.  However, there are also examples of pairs of Banach spaces that have the same linear dimension but are not homeomorphic, and this will work in either case.  For example, $\ell^\infty$ and $c_0$ are not homeomorphic because $\ell^\infty$ is nonseparable.  Both spaces have linear dimension $2^{\aleph_0}$.  This was already mentioned for $c_0$, and for $\ell^\infty$ it follows because $c_0$ embeds in $l^\infty$ (which gives the lower bound on dimension) and because the cardinality of $\ell^\infty$ is $2^{\aleph_0}$ (which gives the upper bound).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I'm now pretty sure this isn't what you want, based on your edit, but this still gives another example for the actual question as well as an answer to Jyotirmoy's comment.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Incidentally, another way to see that $2^{\aleph_0}$ is a lower bound for the linear dimensions of $\ell^1$ and friends is to consider the linearly independent set $\{(1,t,t^2,t^3,\ldots):0\lt t\lt 1\}$.  Cardinality of the spaces gives an upper bound.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2010-10-12T22:55:23.273" LastActivityDate="2010-10-12T22:55:23.273" CommentCount="6" />
  <row Id="6619" PostTypeId="2" ParentId="6613" CreationDate="2010-10-12T17:31:27.520" Score="2" ViewCount="" Body="&lt;p&gt;The way I learned it, in ZF, we define the unary union by&#xA;$$\forall y \left(y\in\cup X \Leftrightarrow \exists z(z\in X\wedge y\in z)\right).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The unary intersection is defined using the unary union and the Axiom of Separation:&#xA;$$\cap X = \left\{ y\in\cup X\,|\, \forall z(z\in X\rightarrow y\in z)\right\}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using this definition, since $\cup\emptyset = \emptyset$, then $\cap\emptyset=\emptyset$ as well. &lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-10-12T17:31:27.520" CommentCount="1" />
  <row Id="6620" PostTypeId="2" ParentId="6491" CreationDate="2010-10-12T18:01:07.540" Score="-1" ViewCount="" Body="&lt;p&gt;If you don't require trapezoids of the same width, you can get a closed form by taking a partition of $[0,n]$ into intervals with endpoints  $n,nq,nq^2,\dots nq^k$ for $q&amp;lt;1$ and $k$ large (or infinite), or by using $(i/n)^2, i=0 \dots n$ as the endpoints.  The first approach leads to the sum of a geometric series and the second is the sum of values of a polynomial.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Better answers are possible if you specify what is the problem you are ultimately trying to solve.  Is it finding the number of equal-width trapezoids needed to achieve a given accuracy in the integral $\int_0^n f(x) dx$? &lt;/p&gt;&#xA;" OwnerUserId="467" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-10-12T18:06:24.847" LastActivityDate="2010-10-12T18:06:24.847" CommentCount="1" />
  <row Id="6621" PostTypeId="2" ParentId="6609" CreationDate="2010-10-12T18:24:58.650" Score="5" ViewCount="" Body="&lt;p&gt;[This answer doesn't add much that is new, but it became too painful trying to post is as a comment.]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One doesn't need to talk about differential forms, or integral identities, in order to prove the validity of vector calculus formulas in spherical coordinates (although both those points of view are very nice!).  If $\vec{F}(x,y,z)$ is a vector field expressed in Cartesian coordinates, say, and $\vec{F}(r,\theta,\varphi)$ is the same vector field but now expressed in spherical coordinates, then the formula for $\nabla$ in spherical coordinates is &lt;em&gt;determined&lt;/em&gt; by the requirement that $\nabla\cdot \vec{F}(x,y,z)$ and $\nabla\cdot \vec{F}(r,\theta,\varphi)$ correspond to one another as vector fields when you change from Cartesian to shperical coordinates.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, if $\vec{A}(x,y,z)$ and $\vec{B}(x,y,z)$ are a pair of vector fields expressed in Cartesian coordinates, with $\vec{A}(r,\theta,\varphi)$ and $\vec{B}(r,\theta,\varphi)$ being the same vector fields but now expressed in spherical coordiantes, then $\vec{A}\times \vec{B}$ (expressed in Cartesian coordinates) will correspond to $\vec{A}\times&#xA;\vec{B}$ (expressed in spherical coordinates).  So the left-hand side of the identity to be checked, when computed in Cartesian coordinates, will match (under the change from spherical to Cartesian coordinates) with the same &#xA;expression when computed in spherical coordiantes.  Similarly for the right-hand side of the identity.  Thus the identity is equally valid in Cartesian or spherical coordinates.  (Again, the key point is that the formulas for grad, div, and curl under a coordinate change are &lt;em&gt;defined&lt;/em&gt; so as to make the above argument valid.)&lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2010-10-12T18:24:58.650" CommentCount="1" />
  <row Id="6622" PostTypeId="1" AcceptedAnswerId="6632" CreationDate="2010-10-12T18:27:30.800" Score="1" ViewCount="87" Body="&lt;p&gt;I have an inequality:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$n \ge C\log^k n \times (A\log(n^r\log^l n) + B)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to turn it into an inquality&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$n &gt; F(A,B,C,k,l,r)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;that implies the first inequality, while making it $F$ as tight as possible.&#xA;How can I do that?&lt;/p&gt;&#xA;" OwnerUserId="7076" LastActivityDate="2010-10-12T23:22:14.050" Title="how do I transform an inequality on $n$?" Tags="&lt;arithmetic&gt;&lt;natural-number&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="6623" PostTypeId="2" ParentId="6572" CreationDate="2010-10-12T18:29:24.150" Score="5" ViewCount="" Body="&lt;p&gt;I am going to assume that you meant to state that this was a faithful representation, meaning that the map $G \to GL_n(\mathbb{C})$ has no kernel. Otherwise, as Akhil says, the problem is dramatically underconstrained.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An infinite class of examples is the semidriect product of $S_5$ acting on $(\mathbb{Z}/N)^4$, for $N$ relatively prime to $30$. In other words, look at all permutation matrices whose nonzero entries are $N$-th roots of unity, with product $1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is another infinite family of examples: Let $G$ be any subgroup of $GL_5(\mathbb{C})$. Let $A$ be an abelian cyclic group, embedded in $GL_5(\mathbb{C})$ as a subgroup of the scalar matrices. Then $A$ is central in $GL_5(\mathbb{C})$, so $GA$ is a finite subgroup of $GL_5(\mathbb{C})$. If $\mathbb{C}^5$ is an irreducible representation of $G$, it is also an irreducible representation of $GA$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Roughly speaking, the rest of my answer says that every example looks like one of these.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;There is a theorem of Jordan which states that there is a bound $J(n)$ such that any subgroup $G$ of $GL_n(\mathbb{C})$ has a normal abelian subgroup $A$ with $|G/A| \leq J(n)$. See &lt;a href=&quot;http://terrytao.wordpress.com/2010/02/18/a-proof-of-gromovs-theorem/#jordan&quot; rel=&quot;nofollow&quot;&gt;Terry Tao's blog&lt;/a&gt; for a nice exposition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using modern improvements on Jordan's result, we can get some more precise bounds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Specifically, every $G$ as in your question is either:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) An &lt;a href=&quot;http://en.wikipedia.org/wiki/Group_extension&quot; rel=&quot;nofollow&quot;&gt;extension&lt;/a&gt; $0 \to A \to G \to H \to 0$, where $A$ is abelian and $H$ is a subgroup of $S_5$ or&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) A &lt;a href=&quot;http://en.wikipedia.org/wiki/Central_extension_%28mathematics%29#Central_extension&quot; rel=&quot;nofollow&quot;&gt;central extension&lt;/a&gt; $0 \to A \to G \to H \to 0$ where $|H| \leq 25920$ and $A$ is cyclic. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think if I knew more about group cohomology I could, in principle, turn this into an infinite  but comprehensible, list of all groups that may occur.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;A representation $V$ of $G$ is called primitive it is irreducible and not induced from a proper subgroup $H$ of $G$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Case 1:&lt;/strong&gt; $\mathbb{C}^5$ is an irreducible, but imprimitive representation, of $G$. Let $\mathbb{C}^5$ be induced from a representation $L$ of $B \subset G$. Then $(\dim L) |G/B| = 5$, so $L$ is one dimensional and $H$ has index $5$ in $G$. Let $A = \bigcap_{g \in G/B} g B g^{-1}$. So $A$ is normal and $G/A$ acts faithfully on the $5$ element set $G/B$, showing that $G/A$ is a subgroup of $S_5$. We must show that $A$ is abelian. Suppose not, and let $C$ be the commutator of $A$. Then $C$ is killed by the map to $GL(L)$, as $L$ is one dimensional. But $C$ is normal in $G$, so it is also killed by the map to $GL(\mathbb{C}^5)$. This contradicts that we are supposed to be dealing with a faithful representation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Case 2:&lt;/strong&gt; $\mathbb{C}^5$ is primitive.&#xA;According to a &lt;a href=&quot;http://www.ams.org/mathscinet-getitem?mr=2334748&quot; rel=&quot;nofollow&quot;&gt;recent paper of Collins&lt;/a&gt;, if $G$ has a a primitive representation of dimension $5$, then $|G/Z(G)| \leq 25920$. (Here $Z(G)$ is the center of $G$.) We take $A=Z(G)$. Since $\mathbb{C}^5$ is irreducible, we know that $Z(G)$ acts by scalars. So $Z(G)$ is a subgroup of $\mathbb{C}^*$, and must be cyclic. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I couldn't find a list of all groups that could occur as $G/Z(G)$, but perhaps you can extract it from Collins' work.&lt;/p&gt;&#xA;" OwnerUserId="448" LastActivityDate="2010-10-12T18:29:24.150" CommentCount="1" />
  <row Id="6624" PostTypeId="2" ParentId="6551" CreationDate="2010-10-12T19:06:52.260" Score="0" ViewCount="" Body="&lt;p&gt;I highly reccomend the Princeton Review series of prep books, they have worked wonders for me. The College Board website also has a lot of useful tools, such as a complete free test that you can take online, and sample questions.&#xA;&lt;a href=&quot;http://sat.collegeboard.com/practice&quot; rel=&quot;nofollow&quot;&gt;http://sat.collegeboard.com/practice&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2254" LastActivityDate="2010-10-12T19:06:52.260" CommunityOwnedDate="2010-10-12T19:06:52.260" CommentCount="2" />
  <row Id="6625" PostTypeId="1" AcceptedAnswerId="7030" CreationDate="2010-10-12T20:08:22.757" Score="3" ViewCount="192" Body="&lt;p&gt;Suppose I'm given a very slowly converging sequence $\sum a_k$. In the literature the Levin's U- transfomation is mentioned as a good universal technique for convergence acceleration. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm have difficult understanding how this method actually works and what the transformation actually is. Can anyone help? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;After understand how the method works I would like to try it out on a computer. Does anyone know any software packages which have black-boxed this process. I have access to Mathematica, maple, matlab and C++. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you very much in advance. &lt;/p&gt;&#xA;" OwnerUserId="1794" LastActivityDate="2010-10-17T14:29:54.280" Title="Levin's u-transformation" Tags="&lt;sequences-and-series&gt;&lt;numerical-methods&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="6626" PostTypeId="1" AcceptedAnswerId="6627" CreationDate="2010-10-12T20:28:12.213" Score="3" ViewCount="141" Body="&lt;p&gt;Consider the function $Pt(n)$. It tells us how many primitive Pythagorean Triples there are (below $n$) when any argument $n \in \mathbb{N}$ is plugged in. Is there an 'exact formula'; i.e. an elementary function of even a combination of known special functions like the Gamma and Error Function, that describes $Pt(n)$ ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Max&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: I'm also interested in the exact value of the limit of $Pt(n)/n$ when $n$ tends to infinity. &lt;/p&gt;&#xA;" OwnerUserId="1055" LastEditorUserId="1778" LastEditorDisplayName="" LastEditDate="2010-10-12T21:35:50.827" LastActivityDate="2010-10-12T21:35:50.827" Title="Number of Pythagorean Triples under a given Quantity" Tags="&lt;number-theory&gt;&lt;analysis&gt;&lt;diophantine-equations&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="6627" PostTypeId="2" ParentId="6626" CreationDate="2010-10-12T20:49:17.923" Score="2" ViewCount="" Body="&lt;p&gt;To answer the sharpened version of the question I suggested (the number of primitive Pythagorean triples with largest element ${\lt}n$ ): by the parametrization of pythagorean triples as sums of two squares, this is (essentially) equal to the question of how many ways there are of expressing all the odd numbers ${\lt}n$ as a sum of two squares.  Mathworld's page on the sum-of-two-squares function at &lt;a href=&quot;http://mathworld.wolfram.com/SumofSquaresFunction.html&quot; rel=&quot;nofollow&quot;&gt;http://mathworld.wolfram.com/SumofSquaresFunction.html&lt;/a&gt; indicates that this is proportional to $n$ (though it might take some work to explicitly work out the constant of proportionality for the odd case), and so in fact your intuition is wrong; the limit you suggest tends to a finite positive value.&lt;/p&gt;&#xA;" OwnerUserId="785" LastActivityDate="2010-10-12T20:49:17.923" CommentCount="1" />
  <row Id="6628" PostTypeId="2" ParentId="20" CreationDate="2010-10-12T20:53:23.093" Score="2" ViewCount="" Body="&lt;p&gt;Going closer to the foundations of mathematics than most ever need or are comfortable with, we can also write down the classic model of $\mathbb{N_0}$ in ZF. This is of course going to be &lt;em&gt;very&lt;/em&gt; bird's eye view and rather devoid of details.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;There exists a set, $I$, such that $\emptyset \in I$ and $x \in I \Rightarrow x \cup \{x\} \in I$.&quot; This is the axiom of infinity. We've the chosen the letter &quot;I&quot; to represent it since it is infinite and we can do induction over $I$ (thus sets satisfying this axiom are also called &quot;&lt;a href=&quot;http://en.wikipedia.org/wiki/Inductive_set_(axiom_of_infinity)&quot; rel=&quot;nofollow&quot;&gt;inductive sets&lt;/a&gt;&quot;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then we construct the &lt;a href=&quot;http://en.wikipedia.org/wiki/Ordinal_number&quot; rel=&quot;nofollow&quot;&gt;ordinals&lt;/a&gt; starting with $0 = \emptyset$, $1 = 0 \cup \{0\} = \{\emptyset, \{\emptyset\}\}, 2 = 1 \cup \{1\} \ldots$ Let us not discuss limit ordinals or why &quot;the ordinals&quot; is not a set - it is all rather complicated. Instead let me point out that from the existense of $I$ it's possible to show the existense of the smallest infinite ordinal, $\omega$, the first ordinal with infitely many preceding ordinals. Then we define $\mathbb{N}_0$ to be $\omega$ along with this &quot;simple&quot; definition of sum: $x+0 = x$, $x + 1 = x \cup \{x\}$ and $x + (y \cup \{y\}) = x+y \cup \{x+y\}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The next couple of steps are easy if you're familiar with &lt;a href=&quot;http://en.wikipedia.org/wiki/Equivalence_class&quot; rel=&quot;nofollow&quot;&gt;equivalence classes&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\mathbb{Z} = \mathbb{N_0}\times\mathbb{N_0}/\sim$, where $(n,m) \sim (n',m')$ iff $(n+m') = (m+n')$. E.g. the integer -2 is the equivalence class consisting of $\{(0,2),(1,3),(2,4),\ldots\}$. Then you show that there is a natural inclusion $\mathbb{N_0} \hookrightarrow \mathbb{Z}$ so it makes sense to write things like $0 \in \mathbb{Z}$ even though we technically defined 0 to be $\emptyset$ and $\emptyset \notin \mathbb{Z}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\mathbb{Q} = \mathbb{Z}\times\mathbb{Z}\setminus\{0\}/\sim$, where $(a,b) \sim (c,d)$ iff $ad=bc$. Again we have natural inclusions $\mathbb{N}_0 \hookrightarrow \mathbb{Q}$ and $\mathbb{Z} \hookrightarrow \mathbb{Q}$ and each equivalence class is one rational number, i.e. a fraction. E.g. $\tfrac{2}{3}$ is the set $\{(2,3),(4,6),(6,9),\ldots\}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When we want the real numbers it gets hard. Typically in a first course in analysis you would give the students the reals in the form of an extra axiom:&#xA;There exists a field, $\mathbb{R}$, together with an order (&quot;inequality sign&quot;) $\leq$ such that:&#xA;1) $\mathbb{R}$ is an &lt;a href=&quot;http://en.wikipedia.org/wiki/Ordered_field&quot; rel=&quot;nofollow&quot;&gt;ordered field&lt;/a&gt;, and&#xA;2) Every subset of $\mathbb{R}$ that has an upper bound has a &lt;a href=&quot;http://en.wikipedia.org/wiki/Least_upper_bound_property&quot; rel=&quot;nofollow&quot;&gt;least upper bound&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why don't you start from scratch? Well... first of all there's the issue of time, but also that of mathematical maturity. Any construction of $\mathbb{R}$ is going to require quite a bit of machinery. Some algebra, some topology and being able to manipulate things like &quot;sequences of equivalence classes of sequences of rational numbers&quot; without getting lost. Even formulating the notion of uniqueness of the real numbers can be a bit tricky. We want not just that any two fields satisfying 1) and 2) to be isomorphic (both as fields and w.r.t. order), but also that the isomorphism should be unique. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what different constructions are there? Quite a lot actually. My 3 personal favourites are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A) Dedekind cuts. Based on the &quot;obvious&quot; observation that a real number is uniquely determined by which rational numbers it is smaller than and which it is larger than. Pros: Easy to prove that this construction has the least upper bound property. Difficulties: It has to be a field, so we need to define multiplication and division. How do you divide subsets of rational numbers? Especially when the divisor contains $0$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;B) Equivalence classes of rational cauchy sequences. Based on the observation that the only way a rational cauchy sequence can &lt;em&gt;not&lt;/em&gt; converge is if the limit is somehow &quot;not there&quot;. So we just set the limit to be the sequence itself! This is of course nonsense, but it can be made precise. Pros: How easy it is to show it's a field: The set of rational cauchy sequences under coordinate-wise addition and multiplication is a ring and the rational sequences that converge to $0$ are a maximal ideal! Difficulties: Defining the order (note that $(\tfrac{1}{n})$ belongs to the same equivalence class as $0$) and then showing that the order is in fact total. Keeping track of sequences of equivalence classes of rational cauchy sequences when you're trying to prove that your field is complete can be a bit tricky.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;C) &quot;Almost-homomorphisms&quot; of $\mathbb{Z}$. Based on the observation (at least since we got pixel-based displays) that a real number $\alpha$ is in some sense the same as choosing a rule for how to draw a line with slope $\alpha$ on an infinite ($\mathbb{Z}\times\mathbb{Z}$) computer display. Pros and cons? Not sure actually. It could be a fun little project to work out the details I guess.&lt;/p&gt;&#xA;" OwnerUserId="1269" LastEditorUserId="1269" LastEditorDisplayName="" LastEditDate="2010-10-13T06:01:26.353" LastActivityDate="2010-10-13T06:01:26.353" CommentCount="2" />
  <row Id="6629" PostTypeId="1" CreationDate="2010-10-12T21:11:27.143" Score="2" ViewCount="256" Body="&lt;p&gt;Suppose I have a variable $s$ that has a geometric distribution with success parameter x. So the probability of success on  trial $s$ is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$p_s = (1 - x)^{s - 1} x$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the following function of $s$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$V(s) = \delta^s$,  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\delta &amp;lt; 1 $.  Then the expected value of this function is given by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sum_ 1^{\infty} V(s) p_s = \frac{\delta x} {1 - \delta (1 - x)}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider now the deviation of V(s) from its mean :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(V(s) - \frac{\delta x} {1 - \delta (1 - x)}) $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What distribution does this have?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I am really interested in is the behavior of the follow ordinary stochastic differential equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\dot{V} =  \delta^s  - V$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;stochastic because the law of motion depends on the random variable $s$.  The stochastic approximation technique allows me to focus on the mean dynamics, which are given by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\dot{V} =  \frac{\delta x} {1 - \delta (1 - x)}  - V$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;whose equilibrium is simply $\hat{V} = \frac{\delta x} {1 - \delta (1 - x)}$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to characterize the large deviation properties of the original ODE, IE calculate the exponential likelihood $V$ crosses a particular boundary $c$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, if $\delta = .95$ and $x= .01$, $\hat{V} = 0.159664$.  If we set $V_0 = 0.159664$, and let $V_t = V_{t-1} + .2 (\delta^s - V_{t-1})$, and $s$ has the above distribution, how do I calculate the expected time to $V_t$ crossing, say, $c=.4$?  What is the associated rate function?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Editing to give the background to the problem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am interested in the following stochastic dynamic system:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$V_t = V_{t-1} + \gamma (\delta^{S_{t-1}} - V_{t-1})$,&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;where $\delta &amp;lt;1$, $\gamma&amp;lt;1$, both positive, and each $S_t$ is an i.i.d. geometric random variable with success parameter $x$.  What this models is agents who have to wait a geometric length of time to get a reward of value 1, and they have time discount factor $\delta$, and they are learning  the expected value of their reward using a constant gain adaptive  learning procedure, with gain $\gamma$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So considering $S$ as following a Poisson, rather than Geometric, makes little difference to the sense of the problem.  The goal is&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Find the equilibria of the learning dynamics&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Characterize the mean time to escape from this equilibrium, the large deviation properties.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;So, for large deviation properties, I should be able to use something like Cramer's theorem, I think.  As Mike says below, iterating the dynamics gives&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$V_t = (1−γ)^tV_0+γ\sum_{k=0}^{t−1}(1−γ)^{t−1−k}δ^{S_k}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So it all depends on $\sum_{k=0}^{t−1}(1−γ)^{t−1−k}δ^{S_k}$, the discounted sum of some random variables.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Probability that $V_t &gt;c$ is then given by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$Pr(V_t &gt; c ) = Pr(\sum_{k=0}^{t−1}(1−γ)^{t−1−k}δ^{S_k} &gt; \frac{c-(1−γ)^tV_0}{\gamma})$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $Z:=\frac{c-(1−γ)^tV_0}{\gamma}$.  so we need the distribution of $\sum_{k=0}^{t−1}(1−γ)^{t−1−k}δ^{S_k}$, which as Mike says is a discounted sum of independent random variables, and so we have that this is approximately normal, for large $t$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Turning now to the question of large deviations, if things were not discounted by powers of $(1-\gamma)$, we could appeal directly to Cramer's Theorem; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ Pr(\sum \delta^{S_k} &gt; n a) \leq Exp[-n(r^* a - Log(E[Exp(r^* \delta^S)])] $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $r^* $ is chosen to maximize $r a - Log(E[Exp(r \delta^S)]$.  The problem would be to simply calculate the moment generating function of  $\delta^S$.  But, I don't exactly want $Pr(\sum \delta^{S_k} &gt; n a)$, I need $ Pr(\sum (1-\gamma)^{t-1-k} \delta^{S_k} &gt; a)$; so not an average of $\delta^{S_k}$s, but a discounted sum.  so it is not quite a direct application.&lt;/p&gt;&#xA;" OwnerUserId="2464" LastEditorUserId="2464" LastEditorDisplayName="" LastEditDate="2010-10-19T19:48:58.187" LastActivityDate="2010-10-19T19:48:58.187" Title="Large Deviation Properties of a function of a geometric random variable" Tags="&lt;probability&gt;&lt;probability-theory&gt;&lt;differential-equations&gt;&lt;approximation&gt;&lt;dynamical-systems&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="6630" PostTypeId="2" ParentId="6609" CreationDate="2010-10-12T21:13:00.187" Score="1" ViewCount="" Body="&lt;p&gt;[I post this answer as suggested in a comment]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Vector Calculus identities with Div, Grad, Curl are independent of the  coordinate system. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A possible explanation is that they may be expressed in terms of integrals whose values do not depend on the particular coordinate system one uses. As an example, one has the following limit for the divergence of the vector field $\overrightarrow{F}$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\nabla \cdot \overrightarrow{F}=\underset{V\rightarrow 0}{\lim }\frac{1}{V}\underset{S}{\int \int }\overrightarrow{F}\cdot \overrightarrow{n}\;dA$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $V$ is the volume of a bounded closed region $T$, $S$ is the surface&#xA;of $T$, and $\overrightarrow{n}$ the unit outer normal vector to $S$.&lt;/p&gt;&#xA;" OwnerUserId="752" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2010-10-13T21:00:28.560" LastActivityDate="2010-10-13T21:00:28.560" />
  <row Id="6631" PostTypeId="1" CreationDate="2010-10-12T21:49:20.877" Score="1" ViewCount="130" Body="&lt;p&gt;I have a homework assignment to find the two smallest probable primes with 12 digits, where a probable prime is defined as a number such that a^p-1 = 1 (mod p), where a = 2,3,5,7. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm obviously not asking for this to be done for me, but I'm having trouble figuring out a way to do anything other than to brute-force it by checking each of the four congruences for each (odd) 12 digit number n until i find two of them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can someone offer me some hint, relevant theorem, or something that I might be able to use to figure out how to do find these numbers in a more efficient manner? Thanks for any help. &lt;/p&gt;&#xA;" OwnerUserId="2466" LastEditorUserId="1778" LastEditorDisplayName="" LastEditDate="2010-10-13T03:23:38.213" LastActivityDate="2010-10-13T03:23:38.213" Title="Finding probable primes with large numbers of digits" Tags="&lt;homework&gt;&lt;prime-numbers&gt;" CommentCount="14" />
  <row Id="6632" PostTypeId="2" ParentId="6622" CreationDate="2010-10-12T22:04:26.930" Score="0" ViewCount="" Body="&lt;p&gt;I don't think you will get a satisfying bound on n without iteration.  You can expand the right side to $n \ge AC*log^{(k+r)}n+AC*log^kn*log  log^ln+BClog^kn$.  If n is very large, you can even ignore all but the first term as it will dominate.  Using the fact that $log n$ varies slowly with $n$, start with some guess for $log n$, such as 1 or 2.  Plug it in to the right hand side and calculate a new value of n.  Then calculate $logn$.  Continue to convergence, which should be quick.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2010-10-12T23:22:14.050" LastActivityDate="2010-10-12T23:22:14.050" CommentCount="2" />
  <row Id="6633" PostTypeId="2" ParentId="6629" CreationDate="2010-10-12T22:09:10.370" Score="1" ViewCount="" Body="&lt;p&gt;If $\delta &gt; 0$, then the transformations you are applying to the random variable $S$ to get the quantity $T = \delta^S - \frac{\delta x}{1 - \delta(1-x)}$ are all one-to-one.  Since $S$ is discrete, this means you can construct the probability distribution of $T$ by inverting those transformations.  This would yield&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$P(T = t) = P\left(\delta^S - \frac{\delta x}{1 - \delta(1-x)} = t\right) = P\left(S = \frac{\ln\left(\frac{\delta x}{1 - \delta(1-x)} + t\right)}{\ln \delta}\right) = (1 - x)^{y-1} x,$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $$y = \frac{\ln\left(\frac{\delta x}{1 - \delta(1-x)} + t\right)}{\ln \delta},$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for any value of $t$ such that $y$ is in the set $\{1, 2, \ldots\}$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe someone else can help you with the other questions.&lt;/p&gt;&#xA;" OwnerUserId="2370" LastActivityDate="2010-10-12T22:09:10.370" CommentCount="6" />
  <row Id="6634" PostTypeId="1" AcceptedAnswerId="6637" CreationDate="2010-10-12T23:01:23.867" Score="2" ViewCount="91" Body="&lt;p&gt;I'm trying to find a factorization for $f(z) = z - c$ in inner and outer functions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, first I try to find for what $c$, $f$ is cyclic for the unilateral shift $M_z$. There is a theorem that states that if $A$ is a bounded operator on a Hilbert space $H$ then $x \in H$ is cyclic for $A$ iff zero is the only element that is orthogonal to all $A^n x$ for $n = 0,1,2,\ldots$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $M_z$ be the multiplication operator by $z$ (&quot;the unilateral shift on $H^2$&quot;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, if $g_n(z) = M_z^n (z - c) = -cz^n + z^{n + 1}$ I find that the power series coefficients of $g_n$ (which is given by $g_n(z) = \sum_m a_{n,m} z^m$) are $a_{n,n} = -c$ and $a_{n,n+1} = 1$. If I insert this in the inner product (where $h(z) = \sum \overline{b_n} z^n$), $(g_n, h) = 0$ for all $n = 0, 1, 2, \ldots$. I obtain that $b_{n + 1} - c b_n = 0$ for $n \geq 0$. This gives that $b_n = C \cdot c^n$. But $\sum c^n z^n = 0$ only if $c = 0$. This would imply that $f(z) = z$ is cyclic but a cyclic vector cannot have a zero in the unit disk! What is wrong here?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I would find this $c$, then I would find a factorization if the $c$ makes $f$ cyclic because then the inner function is $1$ and the outer function $f$. How to do it if $f$ is not cyclic?&lt;/p&gt;&#xA;" OwnerUserId="1120" LastEditorUserId="1120" LastEditorDisplayName="" LastEditDate="2010-10-12T23:37:49.313" LastActivityDate="2010-10-12T23:44:40.983" Title="Factorization inner and outer functions." Tags="&lt;complex-analysis&gt;&lt;functional-analysis&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="6635" PostTypeId="1" AcceptedAnswerId="6703" CreationDate="2010-10-12T23:14:03.337" Score="2" ViewCount="82" Body="&lt;p&gt;Let $f : {\cal S}_+^n \mapsto \mathbb{R}$ be a function defined as &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f(Q) := {\rm tr} WQ^{-1}W + {\rm tr} Q$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $W \in {\cal S}_{+}^{n}$ is a symmetric positive definite matrix.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How to compute a subdifferential of $f$ at a point $Q \in {\cal S}_+^n$ for which $f(Q) &amp;lt; \infty$? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that $f(Q) = \infty$ for $Q$ such that range of $W$ is not contained within the range of Q. Furthermore, we know that $\min_{Q \in {\cal S}_+^n} f(Q) = 2{\rm tr}W$ which is obtained at $Q = W$.&lt;/p&gt;&#xA;" OwnerUserId="2384" LastActivityDate="2010-10-13T15:55:11.740" Title="How to compute the subdifferential of a variational represention of the trace norm?" Tags="&lt;matrices&gt;&lt;convex-analysis&gt;&lt;differential&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="6636" PostTypeId="2" ParentId="6544" CreationDate="2010-10-12T23:15:05.130" Score="5" ViewCount="" Body="&lt;p&gt;Here's another idea.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $P_n(x)$ be the polynomial defined by&#xA;$$P_n(x) = \frac{x^n – (1-x)^n}{2x-1},$$&#xD;&#xA;and note that $P_n(x)=(1-x)P_{n-1}(x) + x^{n-1} \qquad (1), $ where $P_1(x)=1.$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Define $I_n = \int_0^{1} P_n(x) dx.$ Integrating (1) between 0 and 1 we obtain&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$I_n = \int_0^{1} (1-x)P_{n-1}(x) dx + \frac{1}{n},$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and using the symmetry of $P_n(x)$ about 1/2 we note that&#xA;$\int_0^{1} (1-x)P_{n-1}(x) dx = \frac{1}{2}I_{n-1}.$ Hence we obtain the recurrence relation for $n &amp;gt; 1$&#xD;&#xA;$$I_n = \frac{1}{2}I_{n-1} + \frac{1}{n}, \textrm{ where } I_1=1. \qquad (2)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From which it follows that if $\lim_{n \rightarrow \infty} n I_n$ exists then it must equal 2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To show that the limit exists we can solve (2) to obtain&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$I_n = \sum_{k=1}^n \frac {1}{k2^{n-k}},$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and straightforward algebra shows that $(n-1)I_{n-1} – nI_n &amp;gt; 0$ for sufficiently large $n$ (in fact, $n&gt;6$), and since $nI_n$ is bounded below by 0 the result follows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just for completeness:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(n-1)I_{n-1} – nI_n = \left\lbrace \sum_{k=1}^{n-2} \frac{k}{2^k(n-k)(n-k-1)} \right\rbrace - \frac{n}{2^{n-1}} \quad \textrm{ for } n&amp;gt;2.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Comparing the first term in the summation with the negative term we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \frac{1}{2(n-1)(n-2)} &amp;gt; \frac{n}{2^{n-1}} \quad \textrm{ for } n \ge 13.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence $\lbrace nI_n \rbrace$ eventually forms a strictly decreasing sequence.&lt;/p&gt;&#xA;" OwnerUserId="1301" LastEditorUserId="1301" LastEditorDisplayName="" LastEditDate="2010-12-05T08:39:55.590" LastActivityDate="2010-12-05T08:39:55.590" CommentCount="1" />
  <row Id="6637" PostTypeId="2" ParentId="6634" CreationDate="2010-10-12T23:38:48.157" Score="1" ViewCount="" Body="&lt;p&gt;What is wrong is the reasoning where you wrote &quot;But $\sum c^n z^n = 0$ only if $c=0$.&quot;  What you need to know is for which choices of $c$ does the condition $b_{n+1}=c b_n$ for all $n$, with $b_0,b_1,\ldots$ being a square summable sequence, force all of the $b_n$'s to be zero.  If $c=0$, this is not the case; just take $h(z)=1$.  The coefficients are not forced to be zero when $|c|&amp;lt;1$, where you can take $h(z)=1 + cz +c^2z^2 +c^3z^3 +\cdots$.  But the condition $b_{n+1}=c b_n$ is impossible for $|c|\geq 1$ unless $h=0$, so your criterion is $|c|\geq1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When $f$ is not cyclic, for inner part you can take the holomorphic automorphism of the unit disk that swaps $c$ and 0, and what remains will be an invertible element of $H^\infty$, hence cyclic.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2010-10-12T23:44:40.983" LastActivityDate="2010-10-12T23:44:40.983" CommentCount="1" />
  <row Id="6638" PostTypeId="1" AcceptedAnswerId="6697" CreationDate="2010-10-13T00:22:36.790" Score="2" ViewCount="132" Body="&lt;p&gt;Let $p(x)$ be a fixed distribution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $A, C &gt; 0$ be constants.&#xA;Let $\epsilon &gt; 0$. Can we find an example of a distribution $q_{\epsilon}$ such that $\mathrm{KL}(p||q_{\epsilon}) &amp;lt; \epsilon$, but $E[(\log (p(x)/q(x)))^2] \ge A * \epsilon^C$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $p$ and $q_{\epsilon}$ would be a coin toss or a geometric distribution, that would be even better.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope I am making myself clear. This is a rather hard question.&#xA;Perhaps I could also get a clue if someone could tell me whether there is a name for the quantity $E[(\log (p(x)/q(x)))^2]$. (KL divergence is defined as $E[\log (p(x)/q(x))]$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All expectations are taken with respect to $p$.&#xA;So, for example, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$E[(\log (p(x)/q(x)))^2] = \sum_x p(x) \left(\log (p(x)/q(x))\right)^2$&lt;/p&gt;&#xA;" OwnerUserId="7076" LastEditorUserId="7076" LastEditorDisplayName="" LastEditDate="2010-10-13T11:47:30.240" LastActivityDate="2010-10-13T14:38:54.583" Title="can KL divergence converge to 0, but variance of log odds not converge to 0?" Tags="&lt;probability&gt;&lt;statistics&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="1" />
  <row Id="6639" PostTypeId="1" AcceptedAnswerId="6654" CreationDate="2010-10-13T00:37:38.827" Score="2" ViewCount="123" Body="&lt;p&gt;If $x$ is an element in a standard convex linear optimization set constrained by $Ax = b, x \geq 0$, then how can I prove $d$ is a feasible direction only if $Ad=0$ and $di \geq 0$ for every $i$ where $xi=0$?&lt;/p&gt;&#xA;" OwnerUserId="2014" LastEditorUserId="664" LastEditorDisplayName="" LastEditDate="2010-10-13T01:01:38.007" LastActivityDate="2010-10-13T15:41:02.193" Title="Prove feasible direction" Tags="&lt;linear-programming&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="6640" PostTypeId="1" AcceptedAnswerId="6641" CreationDate="2010-10-13T00:48:43.073" Score="2" ViewCount="203" Body="&lt;p&gt;I am looking for algorithms that compute the inverse of Euler's totient function.&lt;/p&gt;&#xA;" OwnerUserId="2468" LastEditorUserId="1352" LastEditorDisplayName="" LastEditDate="2010-10-13T03:36:00.213" LastActivityDate="2010-10-13T03:36:00.213" Title="Algorithms for computing inverse of Euler's phi" Tags="&lt;number-theory&gt;&lt;algorithms&gt;" AnswerCount="1" />
  <row Id="6641" PostTypeId="2" ParentId="6640" CreationDate="2010-10-13T01:23:43.747" Score="3" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://library.wolfram.com/infocenter/MathSource/696/&quot; rel=&quot;nofollow&quot;&gt;Here&lt;/a&gt; is a nice article off of wolfram that gives an efficient algorithm for computing the inverse of the Euler totient function.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;-Download the &lt;strong&gt;invphi.nb&lt;/strong&gt; file at the bottom &lt;/p&gt;&#xA;" OwnerUserId="1352" LastEditorUserId="1352" LastEditorDisplayName="" LastEditDate="2010-10-13T01:30:57.920" LastActivityDate="2010-10-13T01:30:57.920" CommentCount="3" />
  <row Id="6642" PostTypeId="2" ParentId="6507" CreationDate="2010-10-13T01:30:57.120" Score="1" ViewCount="" Body="&lt;p&gt;You can think of a group (e.g., via Cayley's Theorem) as a collection of bijections that contains the identity and is closed under composition and inverses. So groups are objects that &quot;act on&quot; arbitrary sets (as bijections). That seems to be what you are talking about when you talk about group elements being &quot;automorphisms of a particular set&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As Jack Schmidt pointed out in &lt;a href=&quot;http://mathoverflow.net/questions/27971/why-is-there-no-cayleys-theorem-for-rings&quot; rel=&quot;nofollow&quot;&gt;MathOverflow&lt;/a&gt;, every associative unital ring is a subring of the endomorphism ring of its underlying additive group. That is, you can think of any ring as being a collection of group endomorphisms of some group, with the ring addition corresponding to the usual addition of endomorphisms (inherited from the group), and ring multiplication corresponding to composition of functions. This closely parallels the view of &quot;groups as collections of bijections on a set&quot;: rings are &quot;collections of endomorphisms of a group&quot;. If the group has nothing to do with numbers, then the corresponding ring will also have &quot;nothing to do with numbers&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you think of groups as objects that act on sets as bijections, then you can think of (associative unital) rings as objects that act on abelian groups as endomorphisms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know if this answer is satisfying, but it seems to me to be along the lines you are inquiring.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-10-13T01:30:57.120" CommunityOwnedDate="2010-10-13T01:30:57.120" CommentCount="1" />
  <row Id="6643" PostTypeId="1" AcceptedAnswerId="6644" CreationDate="2010-10-13T01:38:01.660" Score="2" ViewCount="126" Body="&lt;p&gt;Construct a subset of [0,1] in the same manner as the Cantor set by removing from each remaining interval a subinterval of relative length $\theta$, $0&amp;lt;\theta &amp;lt;1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the first statement in a homework exercise. I don't know if my English parsing skills are lacking today or what, but it's unclear to me what this means. I just need help understanding what this construction &quot;looks like.&quot; &lt;/p&gt;&#xA;" OwnerUserId="2313" LastActivityDate="2010-10-13T16:57:01.107" Title="Cantor-like construction" Tags="&lt;real-analysis&gt;&lt;homework&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="6644" PostTypeId="2" ParentId="6643" CreationDate="2010-10-13T01:47:14.100" Score="3" ViewCount="" Body="&lt;p&gt;The construction of the Cantor set begins with the removal of the interval $[1/3,2/3]$, an interval of length $1/3$.  The remaining intervals have length $1/3$; you remove intervals of length $1/9$ from each of them.  At stage $k$, you're removing $2^{k-1}$ intervals of length $1/3^k$.  The total length removed is then $$\sum_{k=1}^\infty\frac{2^{k-1}}{3^k}=\frac{1}{3}\frac{1}{1-\frac{2}{3}}=1$$&#xA;so the Cantor set has measure zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let $\theta\ne 1/3$.  You can repeat the same construction with $\theta$ instead of $1/3$: at step $k$, you remove $2^{k-1}$ intervals of length $\theta^k$.  This is going to look very much like the Cantor set -- in particular, it has empty interior, which you can prove.  If your homework problem is going where I'm guessing it's going, you should calculate the measure of the set you get for different $\theta$.  Actually, you should do that no matter what the homework says.  The results are quite surprising.&lt;/p&gt;&#xA;" OwnerUserId="2215" LastEditorUserId="2215" LastEditorDisplayName="" LastEditDate="2010-10-13T16:57:01.107" LastActivityDate="2010-10-13T16:57:01.107" CommentCount="3" />
  <row Id="6645" PostTypeId="1" AcceptedAnswerId="6651" CreationDate="2010-10-13T02:17:44.817" Score="3" ViewCount="167" Body="&lt;p&gt;How do you prove $|\sqrt[3]{x} - \sqrt[3]{y}|$ $\leq$ $\sqrt[3]{|x-y|}$?&lt;/p&gt;&#xA;" OwnerUserId="2467" LastEditorUserId="274" LastEditorDisplayName="" LastEditDate="2010-10-13T02:39:32.917" LastActivityDate="2010-10-13T02:52:57.243" Title="Cube Root Inequality" Tags="&lt;real-analysis&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="6646" PostTypeId="1" AcceptedAnswerId="6656" CreationDate="2010-10-13T02:18:04.863" Score="2" ViewCount="184" Body="&lt;p&gt;I can't seem to understand Generalized Set Operations. Can someone explain it for me? For example, in Example 1.12, suppose &lt;code&gt;k = 3&lt;/code&gt; then set &lt;code&gt;A(of k) = {3, 4, 5, 6 ...}&lt;/code&gt;, then how is &lt;code&gt;{3,4,5 ...} = {1,2,3 ...}&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/NvXFE.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="2006" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2010-11-21T02:02:31.353" LastActivityDate="2010-11-21T02:02:31.353" Title="Generalized Set Operation" Tags="&lt;elementary-set-theory&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="6647" PostTypeId="1" AcceptedAnswerId="6650" CreationDate="2010-10-13T02:23:20.670" Score="5" ViewCount="412" Body="&lt;p&gt;I'm facing a practical problem where I've calculated a formula that, with the help of some programming, can bring me to my final answer. However, the numbers involved are so big that it takes ages to compute, and I think it might not be necessary. I have the following formula, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sum\limits^{k}_{i=m}(N-i)^{k-i}(\frac{1}{N})^k\frac{k!}{(k-i)!i!} \leq a$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;from which I need to calculate N. The rest of the values are constants, but in the range of the 100,000s. The factorial there is giving me a headache, since the values involved are too large; what simplifications could I make that will loosen the bounds slightly and thereby simplify the calculation? Are there any standard tricks? Or perhaps a way to calculate this in matlab / octave?&lt;/p&gt;&#xA;" OwnerUserId="2471" LastEditorUserId="1489" LastEditorDisplayName="" LastEditDate="2010-10-13T03:04:34.030" LastActivityDate="2010-11-25T18:48:17.030" Title="How to simplify or calculate a formula with very big factorials" Tags="&lt;approximation&gt;&lt;binomial-coefficients&gt;&lt;big-numbers&gt;" AnswerCount="3" />
  <row Id="6648" PostTypeId="1" AcceptedAnswerId="6772" CreationDate="2010-10-13T02:28:30.037" Score="4" ViewCount="201" Body="&lt;p&gt;California issues license plates in numeric order (if we turn the letters into numbers).  I have fun noticing the latest plate I have seen.  I am interested in what you can derive from a series of these observations.  I understand that sampling from $\{1,2,3...n\}$ the only useful data is the highest value you have seen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's oversimplify the problem.  Assume the highest plate issued is $N_0+n*t$, $n$ in plates/day and $t$ in days.  Assume a similar number of low valued plates come off the road each day.  I don't observe a consistent number of plates each day, but it averages out.  Over a long time, the increase in highest plate seen should give a measure of $n$.  The only other data I have is how frequently I see a new highest plate.  Does that give some measure of how far my highest plate is from the highest issued?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As we are asked to cite the source of a question, I made it up.  You probably guessed.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-10-14T11:42:35.893" LastActivityDate="2010-10-14T11:45:52.643" Title="License Plate Statistics" Tags="&lt;statistics&gt;&lt;puzzle&gt;&lt;cryptography&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="6649" PostTypeId="1" AcceptedAnswerId="6674" CreationDate="2010-10-13T02:45:48.660" Score="6" ViewCount="274" Body="&lt;p&gt;I've been trying to work out the best way to understand why Fourier series converge, and it's a little embarrassing but I don't even know a rigorous proof.  Can someone please help put me on the right track to thinking about these issue's in the proper way?  I am especially interested any geometric ways to think about the convergence issue (something I suppose which takes advantage of the fact that each component $e^{in\theta}$ corresponds to some point along the unit circle).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="1417" LastEditorUserId="664" LastEditorDisplayName="" LastEditDate="2010-10-13T09:48:38.927" LastActivityDate="2010-10-24T13:20:30.903" Title="Geometric intuition behind convergence of Fourier series." Tags="&lt;fourier-series&gt;" AnswerCount="4" CommentCount="6" FavoriteCount="1" />
  <row Id="6650" PostTypeId="2" ParentId="6647" CreationDate="2010-10-13T02:46:31.170" Score="7" ViewCount="" Body="&lt;p&gt;You need &lt;a href=&quot;http://en.wikipedia.org/wiki/Stirling%27s_approximation&quot; rel=&quot;nofollow&quot;&gt;Stirling's approximation&lt;/a&gt;.  It is very accurate for large factorials.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-10-13T02:46:31.170" />
  <row Id="6651" PostTypeId="2" ParentId="6645" CreationDate="2010-10-13T02:52:57.243" Score="7" ViewCount="" Body="&lt;p&gt;It's not true. Try $x=1$ and $y=-1$.  If you assume $x$ and $y$ have the same sign, and you might as well assume $x\gt y\gt 0$, then it reduces to showing $(x-y)^3\leq x^3-y^3$ (WLOG replacing variables in the original inequality with cubes and cubing both sides).  This is true because $x\gt y\gt 0$ implies $3xy^2&amp;lt;3x^2y$.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-10-13T02:52:57.243" CommentCount="1" />
  <row Id="6653" PostTypeId="1" CreationDate="2010-10-13T03:08:47.753" Score="3" ViewCount="229" Body="&lt;p&gt;I'm trying to prove &quot;the following generalization of Theorem 5 [ Th.5: if $a|bc$ and $(a,b)=1$ then $a | c$ ], which uses the same argument for its proof&quot; (Sierpinski, The Theory of Numbers): if $a$, $b$, and $c$ are integers such that $b | ac$, then $b | (a,b)(b,c)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I haven't been able to prove it without any reference to prime numbers (which the author introduces way later), using only divisibility and facts like $(a,b)[a,b]=ab$. Here's what I've done so far:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $(a,b)=d_a$, $(b,c)=d_c$, $a=a'd_a$, $c=c'd_c$. Since $b | ac$ and $a | ac$ $\Rightarrow [b,a] | ac$ (this is the argument used in Th5's proof) $\Rightarrow ab/(a,b) | ac \Rightarrow a'b | ac \Rightarrow b | d_a c$. Doing the same for $c$, we get $b | ad_c$. From this we also have $b | (ad_c,d_a c)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you for your help.            &lt;/p&gt;&#xA;" OwnerUserId="1694" LastEditorUserId="1313" LastEditorDisplayName="" LastEditDate="2010-10-13T03:12:26.200" LastActivityDate="2010-10-19T01:50:34.103" Title="Basic divisibility fact" Tags="&lt;number-theory&gt;" AnswerCount="2" />
  <row Id="6654" PostTypeId="2" ParentId="6639" CreationDate="2010-10-13T03:23:00.823" Score="2" ViewCount="" Body="&lt;p&gt;[&lt;strong&gt;EDITED&lt;/strong&gt; after OP's clarification that $i$ is a subscript.]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I get the idea of feasible direction right then it means that $x+\lambda d$ should be in the constraint set for some $\lambda&gt;0$. &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Suppose that $Ad \neq 0$. Then $A(x+\lambda d)=b+\lambda Ad \neq b$, violating the equality constraint.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Suppose that there is an $i$ such that $x_i=0$ but $d_i&amp;lt;0$. Then $x_i+\lambda d_i&amp;lt;0$ for all $\lambda&gt;0$ and hence the non-negativity constraint would be violated.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="1195" LastEditorUserId="1195" LastEditorDisplayName="" LastEditDate="2010-10-13T15:41:02.193" LastActivityDate="2010-10-13T15:41:02.193" />
  <row Id="6655" PostTypeId="2" ParentId="6653" CreationDate="2010-10-13T03:26:53.747" Score="1" ViewCount="" Body="&lt;p&gt;Hint: write $(a,b)$ as a linear combination of $a$ and $b$, $(b,c)$ as a linear combination of $b$ and $c$. &lt;/p&gt;&#xA;" OwnerUserId="1195" LastActivityDate="2010-10-13T03:26:53.747" CommentCount="2" />
  <row Id="6656" PostTypeId="2" ParentId="6646" CreationDate="2010-10-13T03:31:30.127" Score="7" ViewCount="" Body="&lt;p&gt;When you're learning set theory, the definitions of union and intersection you're first given are binary: they're defined on a pair of sets.&#xA;\begin{align}A\cup B&amp;amp;=\lbrace x:x\in A\text{ or }x\in B\rbrace\\&#xA;A\cap B&amp;amp;=\lbrace x:x\in A\text{ and }x\in B\rbrace\end{align}&#xA;Since they're also associative (the order you do a chain of consecutive unions or intersections in doesn't matter), you can define the union or intersection of any finite collection of sets.  This is the first pair of equations your book gives above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, it's always good to generalize your definitions!  In particular, those definitions don't allow us to take a union or intersection of an &lt;em&gt;infinite&lt;/em&gt; collection of sets -- even though the operation makes intuitive sense to us!  (The intersection of sets is just the elements that are in all the sets, for example, and that should make just as much sense with an infinite number of sets as with a finite number of sets.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what we do is &lt;em&gt;redefine&lt;/em&gt; union and intersection to apply to a &lt;em&gt;single&lt;/em&gt; set -- this time, a set of sets.  This is $\mathcal{A}$ above, and the second pair of equations your book gives.  There are a couple different notations for this:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&lt;li&gt;$\bigcup\mathcal{A}$&lt;/li&gt;&#xA;&lt;li&gt;$\bigcup(A|A\in\mathcal{A})$ (the one your book uses)&lt;/li&gt;&#xA;&lt;li&gt;$\bigcup_{A\in\mathcal{A}}A$&lt;/li&gt;&#xA;&lt;li&gt;$\bigcup_{i\in I}A_i$, where $I$ is what's called an _indexing set_.  (For example, if $I=\mathbb{N}$, this is equal to $A_1\cup A_2\cup A_3\cup\dotsb$.&lt;/li&gt;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;For me, the last notation is the most revealing.  I don't know if you're familiar with cardinality yet.  If not, it's a measure of the size of a set, and one of the most important theorems in set theory is that there is no greatest cardinality -- that is to say, there are multiple sizes of infinity. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, in the indexing set notation, $I$ itself doesn't really matter as much as its cardinality.  If it is $\mathbb{N}$, you're taking the union of $A_1,A_2,\dotsc$, a countable number of sets.  But it could also be $\mathbb{R}$, in which case you're taking the union of the sets $A_x$ for every real number $x$.  Because $\mathbb{R}$ has a higher cardinality than $\mathbb{N}$, there's no way to write this out in a list!  So thinking of this as $$A_{x_1}\cup A_{x_2}\cup\dotsb$$ is just wrong.  &lt;em&gt;You can't get this union from a generalization of the binary union.&lt;/em&gt;  Hopefully this shows how powerful this generalization is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To conclude, I'll explain the example at the bottom of the page.  In this case, the set $A_n=\lbrace n,n+1,n+2,\dotsc\rbrace$: that is, all integers greater than or equal to $n$.  The union $\bigcup_{n\in\mathbb{N}}A_n$ is the set of integers that are in at least one set $A_n$.  But every $n\ge 1$ is in such a set -- for example, $n\in A_n$.  So this union is equal to $\mathbb{N}$.  On the other hand, &lt;em&gt;no integer&lt;/em&gt; is in all sets $A_n$ -- for example, $n\not\in A_{n+1}$.  So $\bigcap_{n\in\mathbb{N}}A_n=\emptyset$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope this helps.  Ask if you're still confused!&lt;/p&gt;&#xA;" OwnerUserId="2215" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-10-13T04:51:01.303" LastActivityDate="2010-10-13T04:51:01.303" CommentCount="6" />
  <row Id="6657" PostTypeId="2" ParentId="6653" CreationDate="2010-10-13T03:45:40.113" Score="0" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\ \ \ \rm bd = ac\ \:\Rightarrow\:\ (a,b)\:(b,c)\: =\ b\: (a,b,c,d)\ $ via GCD distributive, associative laws.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively, if you know $\rm\ [a,b]\: (a,b)\: =\: ab\ $ then using this LCM*GCD law&lt;/p&gt;&#xA;&#xA;&lt;p&gt;we have $\rm\ \ \: a,b\:|\:ac\ \:\Rightarrow\ \: [a,b]\:|\:ac\ \ \Rightarrow\:\ ab\:|(a,b)\:ac\ \Rightarrow\ b\:|\:(a,b)\:c$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Doing the same to $\rm\:c\:$ yields the result. This appears to be the proof that Sierpinski has in mind since his prior proof is merely the special case where $\rm\ \ (a,b)= 1\:,\: $ and it employs the consequent specialization of the above $\ $ LCM*GCD $\ $ law, explicitly that $\rm\ (a,b) = 1\ \Rightarrow\ [a,b] = ab\:$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a proof of said LCM*GCD law simpler than Sierpinski's see this &lt;a href=&quot;http://math.stackexchange.com/questions/3149&quot;&gt;one line universal proof  $\:$ [THEOREM here]&lt;/a&gt;. Not only is this proof much simpler but it is also more general - it works in any domain. Note also that the result that you seek is a special case of the powerful $\:$ &lt;strong&gt;Euler four number theorem (Vierzahlensatz)&lt;/strong&gt;,$\:$ or $\:$ &lt;strong&gt;Riesz interpolation&lt;/strong&gt;, or $\:$ &lt;strong&gt;Schreier refinement&lt;/strong&gt;,$\:$ see &lt;a href=&quot;http://groups.google.com/groups?selm=y8z7k9leoku.fsf%40nestle.ai.mit.edu&quot; rel=&quot;nofollow&quot;&gt;my post here&lt;/a&gt;. For another example of the simplicity of proofs employing the fundamental GCD laws (associative, commutative, distributive, absorptive), see my &lt;a href=&quot;http://groups.google.com/groups?selm=y8zej18x733.fsf%40nestle.csail.mit.edu&quot; rel=&quot;nofollow&quot;&gt;post on the Freshman's Dream&lt;/a&gt; $\rm\: (A+B)^n =\: A^n + B^n\ $ for GCDs / Ideals, $\:$ if $\rm\: A+B\ $ is cancellative. It's advantageous to present gcd proofs using these basic laws (vs. the Bezout linear form) since such proofs will generalize better (e.g. to ideal arithmetic) and, moreover, since these laws are so similar to integer arithmetic, we can reuse are well-honed expertise manipulating expressions obeying such laws (associative, commutative, distributive). For examples see said Freshman's Dream post.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-10-19T01:50:34.103" LastActivityDate="2010-10-19T01:50:34.103" />
  <row Id="6658" PostTypeId="2" ParentId="6605" CreationDate="2010-10-13T04:06:19.407" Score="4" ViewCount="" Body="&lt;p&gt;I searched Google for &quot;magnetic fractal&quot;, and found the answer on &lt;a href=&quot;http://www.icd.com/tsd/fractals/beginner4.htm&quot; rel=&quot;nofollow&quot;&gt;the first hit&lt;/a&gt;. It quotes &lt;a href=&quot;http://spanky.triumf.ca/www/fractint/magnet_type.html&quot; rel=&quot;nofollow&quot;&gt;the Fractint documentation&lt;/a&gt; (I can't resist mentioning that Fractint is the grand-daddy of freeware fractal-generating software for personal computers -- it had its first release in 1988, and is still being maintained!):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;These fractals use formulae derived from the study of hierarchical lattices, in the context of magnetic renormalisation transformations. This kinda stuff is useful in an area of theoretical physics that deals with magnetic phase-transitions (predicting at which temperatures a given substance will be magnetic, or non-magnetic). In an attempt to clarify the results obtained for Real temperatures (the kind that you and I can feel), the study moved into the realm of Complex Numbers, aiming to spot Real phase-transitions by finding the intersections of lines representing Complex phase-transitions with the Real Axis. The first people to try this were two physicists called Yang and Lee, who found the situation a bit more complex than first expected, as the phase boundaries for Complex temperatures are (surprise!) fractals.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The formulas for the two fractals are also given there. They are&#xA;$$z \mapsto \left(\frac{z^2 + (c-1)}{2z + (c-2)}\right)^2$$&#xA;for magnet 1, and&#xA;$$z \mapsto \left(\frac{z^3 + 3(c-1)z + (c-1)(c-2)}{3z^2 + 3(c-2)z + (c-1)(c-2) + 1}\right)^2$$&#xA;for magnet 2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm by no means knowledgeable on this subject, but I've been looking at some of &lt;a href=&quot;http://math.bu.edu/people/bob/papers.html&quot; rel=&quot;nofollow&quot;&gt;Robert Devaney's papers&lt;/a&gt;, which I came across &lt;a href=&quot;http://www.tetration.org/Fractals/PPEM/index.html&quot; rel=&quot;nofollow&quot;&gt;via tetration.org&lt;/a&gt;. Looking at Devaney's images, I'd guess that the reason why these fractals have the beautiful Sierpinski-gasket-like structures, while the standard quadratic Julia and Mandelbrot sets don't, is that each of the formulas defining these fractals is a &lt;em&gt;rational function&lt;/em&gt;, that is, a ratio of two polynomials, rather than a single polynomial. I believe the field that studies these things is called complex dynamics. My knowledge doesn't extend to &lt;em&gt;how&lt;/em&gt; rational functions with poles give rise to Julia sets with gaskets, but you could try looking for the answer to that in some of Devaney's papers, or in the book &lt;em&gt;Iteration of Rational Functions&lt;/em&gt; by Alan F. Beardon which is cited in &lt;a href=&quot;http://en.wikipedia.org/wiki/Julia_set&quot; rel=&quot;nofollow&quot;&gt;the Wikipedia article on Julia sets&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="856" LastActivityDate="2010-10-13T04:06:19.407" />
  <row Id="6659" PostTypeId="2" ParentId="6508" CreationDate="2010-10-13T04:20:12.140" Score="0" ViewCount="" Body="&lt;p&gt;Yes, it directly follows from the union bound.&lt;/p&gt;&#xA;" OwnerUserId="1527" LastActivityDate="2010-10-13T04:20:12.140" />
  <row Id="6660" PostTypeId="2" ParentId="6507" CreationDate="2010-10-13T04:39:33.907" Score="2" ViewCount="" Body="&lt;p&gt;I think this is a legitimate observation -- it is possible to do lots of group theory without knowing much about integers or real numbers, but these number systems seem to come up immediately whenever one is dealing with rings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems to me that the main reason is that every ring has a characteristic, and therefore a ring with unity necessarily includes a copy of either the integers or some $\mathbb{Z}_n$.    Furthermore, any element of a finitely-generated ring can be written as a (possibly noncommutative) polynomial in the generators, where the coefficients of the polynomial are integers (or perhaps some elements of $\mathbb{Z}_n$).  Adding and multiplying these polynomials inherently requires addition and multiplication of integers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you get past this basic &quot;number-ness&quot;, there are many rings that are otherwise unrelated to either integers or real numbers.  For example, it's possible to develop the theory of polynomial rings quite a bit without using numbers for anything other than coefficients.  The same goes for group rings, whose elements are hardly more number-like than typical group elements.  For a more sophisticated example, the elements of the (integer) cohomology ring of a topological space are essentially geometric objects, not numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fields are the same way.  Every field inherently contains either the rational numbers $\mathbb{Q}$ or a prime field $\mathbb{Z}_p$, but aside from that one could argue that many fields have very little to do with numbers.&lt;/p&gt;&#xA;" OwnerUserId="1726" LastActivityDate="2010-10-13T04:39:33.907" CommunityOwnedDate="2010-10-13T04:39:33.907" />
  <row Id="6661" PostTypeId="1" AcceptedAnswerId="6662" CreationDate="2010-10-13T04:49:14.667" Score="11" ViewCount="530" Body="&lt;p&gt;Is there a simple explanation of what the Laplace transformation do exactly and how they work?  Reading my math book has left me in a foggy haze of proofs that I don't completely understand. I'm looking for an explanation on lamans terms so that I understand what it is doing as I make these seemingly &quot;magical&quot; transformations. I searched the site and closest to an answer was &lt;a href=&quot;http://math.stackexchange.com/questions/954/inverse-of-laplace-transform&quot;&gt;this&lt;/a&gt; however it doesn't explain things for my simplistic mathematical mind. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Note: please edit tags as I don't really know what to put down for this topic&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="1602" LastEditorUserId="1543" LastEditorDisplayName="" LastEditDate="2011-04-17T20:27:35.347" LastActivityDate="2011-04-17T20:27:35.347" Title="Laplace transformations for dummies" Tags="&lt;integral-transforms&gt;" AnswerCount="3" CommentCount="4" FavoriteCount="5" />
  <row Id="6662" PostTypeId="2" ParentId="6661" CreationDate="2010-10-13T04:55:38.210" Score="5" ViewCount="" Body="&lt;p&gt;There are beautiful video lessons at &lt;a href=&quot;http://ocw.mit.edu/index.htm&quot; rel=&quot;nofollow&quot;&gt;MIT Opencourseware&lt;/a&gt;. I'm particularly in love with &lt;a href=&quot;http://ocw.mit.edu/courses/mathematics/18-03-differential-equations-spring-2006/video-lectures/embed19/&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; presentation of the Laplace transform.&lt;/p&gt;&#xA;" OwnerUserId="664" LastActivityDate="2010-10-13T04:55:38.210" CommentCount="2" />
  <row Id="6663" PostTypeId="1" AcceptedAnswerId="6664" CreationDate="2010-10-13T05:10:08.627" Score="2" ViewCount="135" Body="&lt;p&gt;This is probably a silly question, but a couple of people that I have talked to have had different responses. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does $C_0$ denote the set of continuous functions with compact support or the set of continuous functions which vanish at infinity?  &lt;/p&gt;&#xA;" OwnerUserId="1736" LastActivityDate="2010-10-13T06:57:33.737" Title="Definition of $C_0$" Tags="&lt;notation&gt;" AnswerCount="2" />
  <row Id="6664" PostTypeId="2" ParentId="6663" CreationDate="2010-10-13T05:20:10.217" Score="2" ViewCount="" Body="&lt;p&gt;I have always seen $C_0(X)$ denoting the continuous functions vanishing at infinity, and $C_c(X)$ or $C_{00}(X)$ denoting the continuous functions with compact support, where $X$ is usually a locally compact Hausdorff space.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A special case is $c_0$, which is shorthand for $C_0(\mathbb{N})$, and $c_{00}$ means $C_{00}(\mathbb{N})$.  In this case $\mathbb{N}$ has the discrete topology, and &quot;continuous&quot; is redundant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By analogy, sometimes the compact operators on a Hilbert space are denoted by $B_0(H)$, and the finite rank operators by $B_{00}(H)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See this &lt;a href=&quot;http://eom.springer.de/f/f130270.htm&quot; rel=&quot;nofollow&quot;&gt;Springer Online Reference Works article&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2010-10-13T05:31:11.717" LastActivityDate="2010-10-13T05:31:11.717" />
  <row Id="6665" PostTypeId="1" AcceptedAnswerId="6679" CreationDate="2010-10-13T06:41:52.317" Score="5" ViewCount="170" Body="&lt;p&gt;Consider a candy bag that contains &lt;strong&gt;N=100&lt;/strong&gt; candies. There are only two types of candy in the bag. Say the caramel candy and the chocolate candy. Nothing more is known about the contents of the bag.&lt;br&gt;&#xA;Now, you are going to draw (randomly) one candy at a time from the bag until the first caramel appear. Suppose that the first caramel appeared at &lt;strong&gt;k=7th&lt;/strong&gt; drawing. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;At this moment, what can we say about  the number of caramel candies in the bag?&lt;/p&gt;&#xA;" OwnerUserId="2323" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-10-13T07:25:54.997" LastActivityDate="2010-10-13T09:48:48.437" Title="The probability theory around a candy bag" Tags="&lt;probability&gt;" AnswerCount="3" FavoriteCount="1" />
  <row Id="6666" PostTypeId="2" ParentId="6663" CreationDate="2010-10-13T06:57:33.737" Score="2" ViewCount="" Body="&lt;p&gt;The reason people have different responses is that the notation is not completely standardized. For example, &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0125850506&quot; rel=&quot;nofollow&quot;&gt;Reed &amp;amp; Simon&lt;/a&gt; use $C_0^{\infty}(X)$ for smooth functions with compact support in a space $X$, and $C_{\infty}(X)$ for continuous functions vanishing at infinity. (But instead of $C_0(X)$ for continuous functions with compact support, they write $\kappa(X)$ for some reason...)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you just have to check in every case which convention the text that you're reading uses.&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2010-10-13T06:57:33.737" CommentCount="1" />
  <row Id="6667" PostTypeId="2" ParentId="6665" CreationDate="2010-10-13T07:23:08.670" Score="1" ViewCount="" Body="&lt;p&gt;If the number of caramel candies is $a$, then the probability that the first 6 drawn will not be caramel and the 7th drawn will be caramel (assuming that we do not put back the drawn candies) is&#xA;$P(\text{7th}|a)=\frac{93!(100-a)(99-a)(98-a)(97-a)(96-a)(95-a)a}{100!}$.  Now, given that this has occurred, the probability $P(a|\text{7th})$ of any particular value of $a$ given that the first caramel is the 7th drawn should be $P(\text{7th}|a)$ for that particular $a$ divided by the sum of all possible $P(\text{7th}|a)$. $\sum_{a=0}^{100}P(\text{7th}|a)=\frac{101}{56}$, so&#xA;$$P(a|\text{7th})=\frac{P(\text{7th}|a)}{\frac{101}{56}}=\frac{56\cdot 93!(100-a)(99-a)(98-a)(97-a)(96-a)(95-a)a}{101!}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bashing out some values and adding things up, the probability that $a\le 19$ is slightly less than 50% (49.673%) and the expected value of $a$ is $\frac{65}{3}=21\frac{2}{3}$.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&lt;strong&gt;edit&lt;/em&gt;&lt;/strong&gt;:  (I've slightly altered my original answer above, mostly in the notation, to better accommodate the work below; I believe that the work above assumed that, without knowing how long it took to draw the first caramel, each possible number of caramels was equally likely.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose that $P(a)$ is the probability that there are $a$ caramels.  As above, for any particular value of $a$, the probability $P(\text{7th}|a)$ that the first caramel drawn is the 7th candy drawn is&#xA;$$P(\text{7th}|a)=\frac{93!(100-a)(99-a)(98-a)(97-a)(96-a)(95-a)a}{100!}.$$&#xA;So, the probability that there are $a$ caramels and that the first caramel drawn is the 7th candy drawn is $P(a\text{ and 7th})=P(a)\cdot P(\text{7th}|a)$.  By Bayes's Theorem: &#xA;$$\begin{align}&#xA;P(a|\text{7th})&amp;amp;=\frac{P(a\text{ and 7th})}{P(\text{7th})}=\frac{P(a\text{ and 7th})}{\sum_{k=0}^{100}P(k\text{ and 7th})}&#xA;\\&#xA;&amp;amp;=\frac{P(a)P(\text{7th}|a)}{\sum_{k=0}^{100}P(k)P(\text{7th}|k)}&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if $P(a)=\frac{1}{100}$ for all $a$, this yields the results in my original answer.  If $P(a)={100 \choose a}\frac{1}{2^{100}}$ (a binomial distribution with caramel and not equally likely for each individual candy when the bag is originally filled), the expected value of $a$ is 47.5.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $P(a)={100\choose a}p^a(1-p)^{100-a}$ (a binomial distribution where the probability of each single candy being caramel is $p$ when the bag is originally filled), the expected value of $a$ is $1+93p$.  If this expected value of $a$ given that the first caramel drawn was the 7th candy drawn is to equal the expected value of $a$ without having drawn any candies, which is $100p$, then $p=\frac{1}{7}$, so the expected value of $a$ is $\frac{100}{7}=14\frac{2}{7}$.&lt;/p&gt;&#xA;" OwnerUserId="72" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-10-13T09:48:48.437" LastActivityDate="2010-10-13T09:48:48.437" CommentCount="6" />
  <row Id="6668" PostTypeId="1" AcceptedAnswerId="6672" CreationDate="2010-10-13T07:36:45.530" Score="5" ViewCount="109" Body="&lt;p&gt;The field of real numbers has a nice property - for a fixed $n\in \mathbb{N}$, every $r\in \mathbb{R}$ is either an n-th power (if n is odd or if r&gt;0) or -r is an n-th power.&#xA;I want to generalize this - let K be a field extension of F, then $a\in K$ is almost an n-th power if there is $b\in F$ and $c\in K$ such that $ab=c^n$. In other words, if $A=\{ x^n \mid x\in K^\times \}$ then $A F^\times = K^\times$ as groups (In the example above the smaller field is $\mathbb{Q}$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My first question is how to find such field extensions $F\leq K$ where every element in K is almost an n-th power for a fixed n (or even better - for all n). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Two such examples are when K is algebraically closed field (so every element has an n-th root), or if it is a formally real closed field. Another such example is when K is finite and n is prime to $|K|-1$ (since $K^\times$ is cyclic of order $|K|-1$), though I'm more interested with fields of char 0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since my first two examples are fields that are algebraically closed or &quot;almost&quot; closed, I thought that I should start with such fields F where $F(\alpha)$, for $\alpha$ algebraic over F, is algebraically closed (or is formally real closed in the case where F can be ordered). The problem is that besides the real numbers I never encounter such fields. I mean, it is &quot;easy&quot; to start with a field F and ask what are the properties of $F(\alpha)$ but I never saw someone do it in the opposite direction. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So my second question is what are the properties of fields that have finite algebraic extensions which are algebraically closed or formally real closed, and specially properties that can help with my first question.&lt;/p&gt;&#xA;" OwnerUserId="2125" LastEditorUserId="2125" LastEditorDisplayName="" LastEditDate="2010-10-13T07:42:49.650" LastActivityDate="2010-10-13T09:42:29.293" Title="almost n-th power in a field" Tags="&lt;field-theory&gt;" AnswerCount="3" />
  <row Id="6669" PostTypeId="2" ParentId="6647" CreationDate="2010-10-13T07:48:19.200" Score="4" ViewCount="" Body="&lt;p&gt;You don't need to compute the individual factorials in order to compute $k!/(k-i)!i!$, since that's the binomial coefficient $\binom{k}{i}$. A simple algorithm for computing binomial coefficients can be found on &lt;a href=&quot;http://en.wikipedia.org/wiki/Binomial_coefficient#Binomial_coefficient_in_programming_languages&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;. A more sophisticated algorithm is due to Goetgheluck (&lt;a href=&quot;http://www.jstor.org/pss/2323099&quot; rel=&quot;nofollow&quot;&gt;JSTOR&lt;/a&gt;); implementations can be found &lt;a href=&quot;http://www.numbertheory.org/php/binomial.html&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; and &lt;a href=&quot;http://www.luschny.de/math/factorial/FastBinomialFunction.html&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, with numbers of the size that you have, this might still not be feasible, and in this case I also recommend Stirling's formula.&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2010-10-13T07:48:19.200" />
  <row Id="6670" PostTypeId="2" ParentId="6649" CreationDate="2010-10-13T07:50:24.017" Score="2" ViewCount="" Body="&lt;p&gt;You can write the partial sum $S_n(x)$ as an integral $${1\over 2\pi}\int_{-\pi}^\pi D_n(t) f(x-t)dt,$$&#xA;where the weight function or &quot;kernel&quot; $D_n(t)$ can be easily computed and graphed once and for all. One obtains $$D_n(t)={\sin((n+1/2)t)\over \sin(t/2)}.$$ So $S_n(x)$ is an &quot;average&quot; of $f$-values from the neighborhood of $x$. The essential point is that $D_n(t)$ is heavily concentrated around $t=0$ and oscillates quickly far away from $0$.&lt;/p&gt;&#xA;" OwnerUserId="1303" LastEditorUserId="1303" LastEditorDisplayName="" LastEditDate="2010-10-13T10:04:01.300" LastActivityDate="2010-10-13T10:04:01.300" />
  <row Id="6671" PostTypeId="2" ParentId="6649" CreationDate="2010-10-13T07:59:31.700" Score="2" ViewCount="" Body="&lt;p&gt;The way I see Fourier series (especially the trigonometric expansions) you simply draw the initial sine and cosine lines at the macro level, and then you start dealing with higher frequencies that correct the smaller details.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in the case of an infinite sum, you always go about correcting a bit more on a smaller scale, and at the limit point you have your original function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I'm pretty sure I wasn't clear about it, and that I need to wave my arms around, so I've set this CW so if anyone gets the idea and thinks they can clarify it will be easier to do so.)&lt;/p&gt;&#xA;" OwnerUserId="622" LastActivityDate="2010-10-13T07:59:31.700" CommunityOwnedDate="2010-10-13T07:59:31.700" CommentCount="4" />
  <row Id="6672" PostTypeId="2" ParentId="6668" CreationDate="2010-10-13T08:12:24.727" Score="3" ViewCount="" Body="&lt;p&gt;Your first question is interesting but I don't know any answers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer to your second question is well-known. If a field $F$ has&#xA;a finite extension which is real-closed then it has a finite extension&#xA;which is algebraically closed (by adjoining $\sqrt{-1}$ to the real-closed&#xA;field). But the only fields which have finite extensions $K$ which are&#xA;algebraically closed are either real-closed or algebraically closed&#xA;already, and the degree of $K/F$ is $2$ or $1$. This is a theorem of&#xA;Artin and Schreier and can be found in the more comprehensive standard&#xA;texts, for instance Lang's &lt;em&gt;Algebra&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-10-13T08:12:24.727" CommentCount="1" />
  <row Id="6673" PostTypeId="2" ParentId="6329" CreationDate="2010-10-13T08:21:28.233" Score="2" ViewCount="" Body="&lt;p&gt;Seems like such polyhedra are called &lt;em&gt;not necessarily closed&lt;/em&gt; (NNC) and are usually represented as closed polyhedra with additional dimension $\varepsilon$: every strict inequality $a_{i1}x_1 + \ldots + a_{in}x_n &gt; c_i$ is replaced by $a_{i1}x_1 + \ldots + a_{in}x_n - \varepsilon \geq c_i$ and two additional inequalities $0 \leq \varepsilon \leq 1$ are added to the system. If we call this polyhedron $P'$, the desired polyhedron is the set $\{ (x_1,\ldots,x_n) \mid (x_1,\ldots,x_n,\varepsilon) \in P', \varepsilon &gt; 0 \}$. Such a system of constraints can be converted to representation by generators.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively, they can be characterized directly by three sets of generators $R,P,C \in \mathbb{R}^n$, i.e. every point can be obtained as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\alpha_1r_1 + \ldots + \alpha_kr_k + \beta_1p_1 + \ldots + \beta_lp_l + \gamma_1c_1 + \ldots + \gamma_mp_m,$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $r_i \in R, p_i \in P, c_i \in C$ and $\alpha_i, \beta_i, \gamma_i \in \mathbb{R}^+$ and $\sum_{i=1}^l \beta_i + \sum_{i=1}^m \gamma_i = 1$ and there is $1 \leq i \leq l$ such that $\beta_i \neq 0$. The trick here is that points in $C$ don't have to lie within the NNC polyhedron, but its closure, and whenever they appear in the sum, there must also be point from $P$ with nonzero coefficient. This representation can easily be converted to the one mentioned above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;References:&lt;br&gt;&#xA;R. Bagnara, P. M. Hill, E. Zaffanella: &lt;em&gt;A New Encoding of Not Necessarily Closed Convex Polyhedra&lt;/em&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.12.5503&quot; rel=&quot;nofollow&quot;&gt;R. Bagnara, E. Ricci, E. Zaffanella, P. M. Hill: &lt;em&gt;Possibly Not Closed Convex Polyhedra and the Parma Polyhedra Library&lt;/em&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Seems like this stuff is used mostly in static analysis/verification and is not interesting to mathematicians, maybe I should have asked on cstheory instead? Any feedback regarding the question welcome, as well as corrections regarding my misuse of terminology/notation.&lt;/p&gt;&#xA;" LastActivityDate="2010-10-13T08:21:28.233" CommentCount="1" />
  <row Id="6674" PostTypeId="2" ParentId="6649" CreationDate="2010-10-13T08:38:41.670" Score="5" ViewCount="" Body="&lt;p&gt;I don't know about a geometric interpretation, but here is a brief sketch of a proof.  First we need to be precise about what we mean by &quot;convergence.&quot;  In the naive sense, Fourier series &lt;em&gt;don't&lt;/em&gt; always converge - that is, pointwise.  (If you change the value of a function at a single point, the Fourier series remains unchanged.)  The sense in which they do always converge is in the &lt;a href=&quot;http://en.wikipedia.org/wiki/Hilbert_space&quot; rel=&quot;nofollow&quot;&gt;Hilbert space&lt;/a&gt; $L^2([0, 1])$, which has inner product defined by $\langle f, g \rangle = \int_0^1 \overline{g(x)} f(x) dx$ inducing a norm, which induces a metric.  In $L^2([0, 1])$ let $X$ be the subspace spanned by the functions $e^{2\pi i nx}, n \in \mathbb{Z}$.  It is fairly straightforward to verify that the functions $e^{2\pi i nx}$ are orthogonal and have norm $1$; generally I think about this in a representation-theoretic way, as a special case of the orthogonality relations for characters.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then the statement that Fourier series converge is equivalent to the statement that $X$ is dense in $L^2([0, 1])$.  Why?  Given a sequence in $X$ converging to an element of $L^2([0, 1])$ we can compute the Fourier coefficients, which depend continuously on the sequence and hence which converge to a limit.  That these coefficients actually represent the element of $L^2([0, 1])$ is a standard Hilbert space argument and you should take a course in functional analysis if you want to learn this kind of stuff thoroughly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, something else you need to know about $L^2([0, 1])$ is that the subspace $Y$ consisting of all step functions is dense in it.  (If you have trouble believing this, first convince yourself that $Y$ is dense in the continuous functions on $[0, 1]$ and then believe me that the continuous functions are dense in $L^2([0, 1])$.  In fact, $L^2([0, 1])$ can be defined as the completion of $C([0, 1])$ with respect to the $L^2$ norm.)  So to show that $X$ is dense, it suffices to show that the closure of $X$ contains $Y$.  In fact, it suffices to show that $X$ has as a limit point a step function with a single bump, say&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$a(x) = \begin{cases} 0 \text{ if } 0 \le x \le \frac{1}{3}, \frac{2}{3} \le x \le 1 \\ 1 \text{ otherwise} \end{cases}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and to take linear combinations, translations, and dilations of this.  In other words, it suffices to prove convergence for square waves.  But one can do the computations directly here.  There is a &lt;a href=&quot;http://cnx.org/content/m0041/latest/fourier4.png&quot; rel=&quot;nofollow&quot;&gt;standard picture&lt;/a&gt; to stare at, and of course if you have ever actually heard a square wave you should believe that audio engineers, at least, are perfectly capable of approximating square waves by sines and cosines.  &lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-10-13T08:38:41.670" CommentCount="1" />
  <row Id="6675" PostTypeId="1" AcceptedAnswerId="6699" CreationDate="2010-10-13T08:43:46.943" Score="2" ViewCount="199" Body="&lt;p&gt;Let G be a group with generators and relations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that in general it is difficult to determine what a group is from its generators and relations. I am interested in learning about techniques for figuring out the order of a group from the given information.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, I know that if the number of generators exceeds the number of relations then the group has infinite order. If the number of generators equals the number of relations then the group is cyclic or has infinite order.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let G= [x, y|x^2 = y^3 = (xy)^4 = 1]. My hunch is that G has finite order because (xy)^4 is somehow independent of x^2 and y^3. But if the exponent on xy were bigger, say (xy)^6=1 that relation becomes redundant. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is: is this sort of thinking correct? Furthermore: my method will only tell me if G, or its modification, is finite (or infinite). If G is finite how can I figure out the order of the group? I know that the orders divide the order of the group, but I am looking for a specific number.&lt;/p&gt;&#xA;" OwnerUserId="2298" LastActivityDate="2010-10-13T14:54:56.393" Title="Order of a Group from its Presentation" Tags="&lt;group-theory&gt;&lt;presentation&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="6676" PostTypeId="2" ParentId="6675" CreationDate="2010-10-13T08:53:35.660" Score="3" ViewCount="" Body="&lt;p&gt;Intuition in these cases is treacherous. Your example $G$ is an&#xA;von Dyck group, a subgroup of index 2 in the more well-known&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Triangle_group&quot; rel=&quot;nofollow&quot;&gt;triangle groups&lt;/a&gt;. The general von Dyck group has presentation&#xA;$$\langle x, y\mid x^m=y^n=(xy)^p=1\rangle$$&#xA;and is finite iff $1/m+1/n+1/p&gt;1$.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-10-13T08:53:35.660" CommentCount="2" />
  <row Id="6677" PostTypeId="2" ParentId="6665" CreationDate="2010-10-13T08:58:59.540" Score="0" ViewCount="" Body="&lt;p&gt;Since 100 is quite a bit larger than 7 we will assume for the purposes of calculation that the probability of drawing a sweet is not affected by the previous draw (of course it is altered a little, but not too much).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let the probability of drawing a caramel be $p$ and the probability of drawing a chocolate be $q,$ where $p+q=1.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then the expected number of draws to draw a caramel is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$E= p+2pq+3pq^2+4pq^3+ \cdots  = \frac{1}{p}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since we drew a caramel at the seventh draw set $7=1/p,$ so $p = 1/7.$ Thus $100/7 \approx 14$ of the sweets are caramel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we used the correct figures the probability of drawing a caramel at the next draw would go up with each chocolate drawn, which will push the estimated number of caramels down, perhaps to 13, but I don't expect it will alter much since we started with 100 sweets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The smaller the number of the first draw of a caramel, the less that can be reliably said. Imagine if you drew a caramel on the first draw, that could imply they are all caramels which could be wildly wrong.&lt;/p&gt;&#xA;" OwnerUserId="1301" LastEditorUserId="1301" LastEditorDisplayName="" LastEditDate="2010-10-13T09:11:12.183" LastActivityDate="2010-10-13T09:11:12.183" CommentCount="1" />
  <row Id="6678" PostTypeId="2" ParentId="6668" CreationDate="2010-10-13T09:11:49.673" Score="1" ViewCount="" Body="&lt;p&gt;There exist lots of fields $K$ that are closed under taking $n$-th roots for fixed $n$.&#xA;They can be contructed using valuation theory: take a field $K$ which is henselian&#xA;with respect to some valuation $v$ and has the following additional properties:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$n$ is not divisible by the characteristic of the residue field $\overline{K}$.&lt;/li&gt;&#xA;&lt;li&gt;The residue field is algebraically closed.&lt;/li&gt;&#xA;&lt;li&gt;The value group $v(K)$ is divisible by $n$.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Then $K$ is closed under taking $n$-th roots by Hensel's lemma.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;H&lt;/p&gt;&#xA;" OwnerUserId="2479" LastActivityDate="2010-10-13T09:11:49.673" />
  <row Id="6679" PostTypeId="2" ParentId="6665" CreationDate="2010-10-13T09:21:00.353" Score="4" ViewCount="" Body="&lt;p&gt;The question you are asking here is the classical question of &lt;a href=&quot;http://en.wikipedia.org/wiki/Inferential_statistics&quot; rel=&quot;nofollow&quot;&gt;inferential statistics&lt;/a&gt;: &quot;Given the outcome of an experiment, what can be said about the underlying probability distribution?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could, for example, give an &lt;a href=&quot;http://en.wikipedia.org/wiki/Estimator&quot; rel=&quot;nofollow&quot;&gt;estimator&lt;/a&gt; for the unknown quantity &quot;number of caramels&quot; (called $a$ from here on). The one most often used (since its easy to calculate) would be the maximum likelyhood estimator, where you estimate $a$ to the the value that maximizes the probability of the outcome.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case, you'd choose $a$ to maximize $P_a(7)$ (the probability of drawing the first caramel in the seventh draw, assuming there are $a$  of them). A little Excel calculation, along with Isaac's way to calculate $P_a(7)$ results in $a$ to be estimated as 14.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To judge, what this result is worth, you'd need to calculate the mean squared error of this estimator, which is not as easily done.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you already had a hypothesis about $a$ (say $a$ &amp;lt; 20), you could use your experimental result to test it, using &lt;a href=&quot;http://en.wikipedia.org/wiki/Hypothesis_testing&quot; rel=&quot;nofollow&quot;&gt;statistical hypothesis testing&lt;/a&gt;, too.&lt;/p&gt;&#xA;" OwnerUserId="505" LastEditorUserId="505" LastEditorDisplayName="" LastEditDate="2010-10-13T09:28:16.283" LastActivityDate="2010-10-13T09:28:16.283" CommentCount="3" />
  <row Id="6680" PostTypeId="2" ParentId="5248" CreationDate="2010-10-13T09:36:08.967" Score="3" ViewCount="" Body="&lt;p&gt;Maybe I'm flogging a dead horse, but nobody has mentioned the &lt;strike&gt;standard&lt;/strike&gt; &lt;strong&gt;suspiciously circular (see the comments)&lt;/strong&gt; Fourier analytic proof yet:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $f(t)=1$ for $|t|&amp;lt;1$ and 0 otherwise. Then the Fourier transform is&#xA;$$ F(\omega) = \int_{-\infty}^{\infty} f(t) e^{-i\omega t} dt&#xA;= \int_{-1}^{1} e^{-i\omega t} dt&#xA;= \frac{e^{-i\omega} - e^{i\omega}}{-i\omega}&#xA;= \frac{2\sin\omega}{\omega}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fourier's inversion formula states that&#xA;$$ f(t) = \frac{1}{2\pi} \int_{-\infty}^{\infty} F(\omega) e^{i\omega t} d\omega $$&#xA;if $f$ is (say) differentiable at $t$.&#xA;In our case, we get in particular that&#xA;$$ 1 = f(0)&#xA;= \frac{1}{2\pi} \int_{-\infty}^{\infty} F(\omega) d\omega&#xA;= \frac{1}{2\pi} \int_{-\infty}^{\infty} \frac{2\sin\omega}{\omega} d\omega&#xA;= \frac{2}{\pi} \int_{0}^{\infty} \frac{\sin\omega}{\omega} d\omega.&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(EDIT: Even if this is not really a proof, it's still a good thing to be aware of, since one can use similar ideas to integrate powers of $\sin\omega/\omega$, or integrals &lt;a href=&quot;http://mathoverflow.net/questions/11517/computer-algebra-errors/11607#11607&quot; rel=&quot;nofollow&quot;&gt;like these&lt;/a&gt;.)&lt;/p&gt;&#xA;" OwnerUserId="1242" LastEditorUserId="1242" LastEditorDisplayName="" LastEditDate="2010-10-13T17:42:30.527" LastActivityDate="2010-10-13T17:42:30.527" CommentCount="4" />
  <row Id="6681" PostTypeId="2" ParentId="6668" CreationDate="2010-10-13T09:42:29.293" Score="0" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://mathoverflow.net/questions/18374/if-l-is-a-field-extension-of-k-how-big-is-l-k&quot; rel=&quot;nofollow&quot;&gt;This question&lt;/a&gt; and its answer is relevant here.&lt;/p&gt;&#xA;" OwnerUserId="2436" LastActivityDate="2010-10-13T09:42:29.293" />
  <row Id="6682" PostTypeId="1" CreationDate="2010-10-13T10:25:17.270" Score="2" ViewCount="394" Body="&lt;p&gt;Show that for any integer n&gt;1 all the numbers n!+2, n!+3, ..., n!+n are composite (i.e. not prime).&lt;/p&gt;&#xA;" OwnerUserId="2401" LastEditorUserId="1778" LastEditorDisplayName="" LastEditDate="2010-10-14T20:07:51.007" LastActivityDate="2010-10-15T18:03:49.170" Title="Show that a number is not prime?" Tags="&lt;number-theory&gt;&lt;homework&gt;&lt;prime-numbers&gt;" AnswerCount="4" CommentCount="12" />
  <row Id="6684" PostTypeId="2" ParentId="6682" CreationDate="2010-10-13T10:43:40.570" Score="3" ViewCount="" Body="&lt;p&gt;Hint:&#xA;Try to show, that if for two numbers $a$ and $b$, $a$ is divisible by $d$ and $b$ is divisible by $d$, then so is their sum. Then go looking for such a common divisor in your sums.&lt;/p&gt;&#xA;" OwnerUserId="505" LastActivityDate="2010-10-13T10:43:40.570" CommentCount="3" />
  <row Id="6685" PostTypeId="1" AcceptedAnswerId="6688" CreationDate="2010-10-13T11:49:46.960" Score="3" ViewCount="102" Body="&lt;p&gt;Given $n$ smooth real functions $f_1, f_2, \dots, f_n$, define a composite function like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f(x) = \max(f_1(x), f_2(x), \dots, f_n(x)) - \min(f_1(x), f_2(x), \dots, f_n(x))$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it possible to say anything useful in general about the shape of this function?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Intuitively, it seems like $f$ will be at least $C^0$ continuous but $f^\prime$ may have arbitrarily many discontinuities. How much would we have to know about the individual $f_k$ to be more specific?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, if we know that each $f_k$ has $m_k$ extrema, it seems like we should be able to place bounds on both the number of extrema in $f$ and number of discontinuities in $f^\prime$, but I'm having trouble visualising all the possible interactions as $n$ increases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I've tried to put this in general terms, but for context my particular interest is somewhat related to my &lt;a href=&quot;http://math.stackexchange.com/questions/1632/analysing-an-optics-model-in-discrete-and-continuous-forms&quot;&gt;earlier optics question&lt;/a&gt;. A different but similar imaging process produces something like the above function with constituent $f_k$ of this form:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f_k(x) = \sum_i e_i \sin(x_i + \frac{2k\pi}{n}) P(x - x_i)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once again, simulations suggest that this process can noticeably improve lateral resolution because of the corners introduced between maxima, but it would be nice to have a more formal characterisation.)&lt;/p&gt;&#xA;" OwnerUserId="144" LastActivityDate="2010-10-13T14:40:56.973" Title="Characterising a function involving the max and min of other functions" Tags="&lt;real-analysis&gt;&lt;physics&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" />
  <row Id="6688" PostTypeId="2" ParentId="6685" CreationDate="2010-10-13T13:12:19.057" Score="2" ViewCount="" Body="&lt;p&gt;Everything you need to know can be deduced from the identity $\text{max}(a, b) = \frac{a+b}{2} + \frac{|a-b|}{2}$ and similarly for $\text{min}$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit:  Okay, maybe this is less useful than I thought.  You want to know about the extrema of $f$ and the discontinuities of $f'$.  If each $f_k$ has finitely many extrema, it may still be the case that $f$ has infinitely many discontinuities; consider two functions which are increasing in a &quot;race&quot; in which the victor changes infinitely often.  As a deleted answer says, you want in addition that $f_i - f_j$ has finitely many extrema for every pair $i, j$.  Then outside a bounded range of values of $x$ the relative order of the $f_i$ won't change and everything will be hunky-dorky.&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-10-13T14:40:56.973" LastActivityDate="2010-10-13T14:40:56.973" CommentCount="7" />
  <row Id="6689" PostTypeId="2" ParentId="6661" CreationDate="2010-10-13T13:13:57.487" Score="3" ViewCount="" Body="&lt;p&gt;I'm going to come at this one from left-field.  In quantum mechanics, we deal with infinite dimensional vector spaces (Hilbert spaces), so I tend to think of integral transforms in those terms.  For instance, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_{-\infty}^{\infty} K(x,y) f(y) dy = F(x)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;can be thought of as &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \mathbf{K} f = F $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and $x$ and $y$ from the first equation are the indices of the infinite dimensional vectors and matrix (kernel) $f$, $F$, and $\mathbf{K}$.  Using that interpretation, if $\mathbf{K}$ is unitary then the integral is just a changing the bases of the function (Hilbert) space.  In other words, the integral can be viewed as the decomposition of original vector, $f$, in terms the new basis.  For Fourier transforms the kernel is unitary, and while not true of Laplace transforms, the idea of it being a change of basis still holds.  It should be noted that unlike in the finite case, in the infinite dimensional case care must be taken to ensure that the transform actually converges, but that is another problem entirely.&lt;/p&gt;&#xA;" OwnerUserId="2283" LastEditorUserId="2283" LastEditorDisplayName="" LastEditDate="2010-10-13T13:53:52.283" LastActivityDate="2010-10-13T13:53:52.283" CommentCount="2" />
  <row Id="6691" PostTypeId="2" ParentId="6682" CreationDate="2010-10-13T13:23:13.620" Score="-4" ViewCount="" Body="&lt;p&gt;Take an arbitrary number $n! + k$, where $n &gt; 1, 1 &amp;lt; k \leq n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The following is the definition of the factorial function:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$n! = \prod_{i=1}^n i$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence every i where $1 \leq i \leq n$ is a factor of $n!$. Since the range k covers is within the range of i, this is clearly true for all k.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know that k is a factor of $n!$, so let us say $n! = qk, k \in \mathbb{Z}$. Hence $n! + k = qk + k = (q + 1)k$, which implies q + 1 and k are integral factors of $n! + k$. QED&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a reasonably rigorous proof. When you become more familiar with the concepts of divisibility and factorials the result should be fairly apparent from first sight.&lt;/p&gt;&#xA;" OwnerUserId="56" LastEditorUserId="56" LastEditorDisplayName="" LastEditDate="2010-10-15T18:03:49.170" LastActivityDate="2010-10-15T18:03:49.170" CommentCount="19" />
  <row Id="6692" PostTypeId="2" ParentId="6682" CreationDate="2010-10-13T13:24:50.577" Score="0" ViewCount="" Body="&lt;p&gt;$8! = 1\cdot 2\cdot 3\cdot 4\cdot 5\cdot 6\cdot 7\cdot 8$ so combine $k|8!$ and $k|k$ to get $k|8!+k$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Definition&lt;/strong&gt;: $a|b :\iff \exists k, a k = b.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Theorem&lt;/strong&gt;: $a|m$ and $a|n$ imply that $a|m+n$.&#xA;&lt;strong&gt;Proof&lt;/strong&gt;: Our hypothesis says that $a k = m$ and $a h = n$, add this and use distributivity to get $a (k + h) = n + m$ which proves that $a|n+m$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that if $k|b$ with ($k$ not $1$ or $b$) then $b$ has a divisor, so it is not prime.&lt;/p&gt;&#xA;" LastActivityDate="2010-10-13T13:47:13.823" CommentCount="3" />
  <row Id="6693" PostTypeId="2" ParentId="6593" CreationDate="2010-10-13T13:38:29.940" Score="1" ViewCount="" Body="&lt;p&gt;Ok, so, as Jesse says, this is problem 4-24 from Little Spivak. Maybe you should look at problem 4-23, where he explains what $c_{(1,n)}$ is: a parametrization of the circumference of radius $1$, centered at the origin, that goes round the origin $n$ times&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;c_{(1,n)} (t) = (\cos 2\pi nt, \sin 2\pi nt) \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the problem says that, given any closed parametrized curve $c(t)$, there is an integer $n$ such that this $c$ together with $c_{(1,n)}$ are the boundary of some $2$-chain $c^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I'm not wrong, this $n$ is the &lt;em&gt;winding number&lt;/em&gt; of $c$; that is, the number of times $c$ goes round the origin. So, if $c$ goes round the origin one time, you need the circumference $c_{(1,1)}$ that goes round the origin once too; if $c$ goes round twice, you need $c_{(1,2)}$...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe you should try first problem 4-23, where Spivak asks a similar question, but where $c$ is also a circumference (of different radius). There you see too that you need both circumferences to go round the origin the same number of times in order they are the boundary of some $2$-chain. For instance, if $n=1$ (just once round the origin), these circumferences are the boundary of a regular anulus.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, you may try to take a look at &lt;a href=&quot;http://www.ms.uky.edu/~ken/ma570/&quot; rel=&quot;nofollow&quot;&gt;Little Spivak solutions&lt;/a&gt;. They are quite right. For instance, in the solution to 4-24, Ken Kubota (the author of these solutions), begins computing $L = \int_0^1 c^*(d\theta)$, which is the winding number of $c$ multiplied by $2\pi$.&lt;/p&gt;&#xA;" OwnerUserId="664" LastEditorUserId="664" LastEditorDisplayName="" LastEditDate="2010-10-13T17:56:39.443" LastActivityDate="2010-10-13T17:56:39.443" />
  <row Id="6694" PostTypeId="1" CreationDate="2010-10-13T13:43:46.387" Score="3" ViewCount="255" Body="&lt;p&gt;In &lt;a href=&quot;http://math.stackexchange.com/q/6647/2471&quot;&gt;previous question&lt;/a&gt;, I asked how one would simplify the following equation for the case where the variables are very big:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sum\limits^{k}_{i=m}(N-i)^{k-i}(\frac{1}{N})^k\frac{k!}{(k-i)!i!} \leq a$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This answer was basically to use an approximation like Stirling's formula. Having implemented this with some code, it still takes too long to find the maximum value of N so that the inequality holds true. Therefore, I need a direct solution for N. So the new question is, how would you go about solving this equation for N?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Some simplifications are acceptable, but I would like to have it as accurate as possible. &#xA;The values this equation will be used for are all in the 100,000-1,000,000 range, except $m$, which is in the 100s range.)&lt;/p&gt;&#xA;" OwnerUserId="2471" LastEditorUserId="2471" LastEditorDisplayName="" LastEditDate="2010-10-13T14:03:37.717" LastActivityDate="2010-12-12T14:44:25.123" Title="How to solve an inequality containing the sum of factorials and powers" Tags="&lt;binomial-coefficients&gt;&lt;big-numbers&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="6695" PostTypeId="1" AcceptedAnswerId="6700" CreationDate="2010-10-13T14:00:43.433" Score="18" ViewCount="887" Body="&lt;p&gt;The standard approach for showing $\int \sec \theta  d \theta = \ln |\sec \theta + \tan \theta| + C$ is to multiply by $\frac{\sec \theta + \tan \theta}{\sec \theta + \tan \theta}$ and then do a substitution with $u = \sec \theta + \tan \theta$.  I like the fact that this trick leads to a fast and clean derivation, but I also find it unsatisfying: It's not very intuitive, nor does it seem to have applicability to any integration problem other than $\int \csc \theta  d \theta$.  Does anyone know of another way to evaluate $\int \sec \theta  d \theta$?&lt;/p&gt;&#xA;" OwnerUserId="2370" LastActivityDate="2011-03-31T19:01:48.440" Title="Ways to evaluate $\int \sec \theta d \theta$" Tags="&lt;calculus&gt;&lt;trigonometry&gt;" AnswerCount="4" CommentCount="4" FavoriteCount="4" />
  <row Id="6696" PostTypeId="2" ParentId="6695" CreationDate="2010-10-13T14:12:00.983" Score="16" ViewCount="" Body="&lt;p&gt;A useful technique is to use the half angle formulas in terms of $\tan (\theta/2)$ in order to convert trigonometric (rational) functions into rational functions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example if $t = \tan(\theta/2)$ we have that $\sec \theta = \frac{1+t^2}{1-t^2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have $2dt = (1 + tan^2(\theta/2))\ d\theta$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And so&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int \sec \theta \ d\theta = \int \frac{2dt}{1-t^2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which can easily be evaluated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly we get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int \csc \theta \ d\theta = \int \frac{dt}{t}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;using $\csc \theta = \frac{1+t^2}{2t}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Check this page out: &lt;a href=&quot;http://en.wikipedia.org/wiki/Tangent_half-angle_formula&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Tangent_half-angle_formula&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-10-13T16:10:15.950" LastActivityDate="2010-10-13T16:10:15.950" CommentCount="4" />
  <row Id="6697" PostTypeId="2" ParentId="6638" CreationDate="2010-10-13T14:38:54.583" Score="0" ViewCount="" Body="&lt;p&gt;The idea is to modify just one term of a distribution $p$ in a way that changes the distribution very little but alters the variance of the log ratio profoundly.  In order to make this work, the term that gets modified changes at each step, taking us farther and farther into the tail of $p$, in the hope that this will assure the convergence of the modified distributions to $p$ in the sense of the KL divergence between them going to $0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $p$ be the geometric distribution&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Pr(x) = p(x) = 2^{-x}, x = 1, 2, \ldots$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and for each $n = 1, 2, \ldots$ let &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$q_n(x) = C_n p(x) \quad \text{if}\quad x \ne n$$&#xA;$$= 2^{-f_n} \quad \text{if} \quad x = n$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $f_n$ is a sequence of positive real numbers to be determined and $C_n$ is a sequence of normalizing constants.  The requirement that $q_n$ be a probability function determines $C_n$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$1 = \sum_x {q(x)} = C_n \sum_{x \ne n} {p(x)} + 2^{-f_n} = C_n (\sum_x {p(x)} - p(n)) + 2^{-f_n} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$= C_n (1 - p(n)) + 2^{-f_n} = C_n(1 - 2^{-n}) + 2^{-f_n},$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;whence&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$C_n = (1 - 2^{-f_n}) / (1 - 2^{-n}).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can now compute the KL divergence:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathrm{KL}(p||q_n) = \sum_x {p(x) \log( p(x) / q_n(x))}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Breaking this into a sum over $x \ne n$ and the remaining term as before gives&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$= \sum_x {p(x) \log(C_n)} - p(n) \log(C_n) + p(n) \log( 2^{-n} / 2^{-f_n} )$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$= \log(C_n) \left(1 - p(n) \right) + p(n) \left( f_n - n \right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$= \log(C_n) \left(1 - 2^{-n} \right) + 2^{-n} \left( f_n - n \right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(using the logarithm base 2 for ease of computation).  Notice that the sequence $\log(C_n)$ converges to $0$ provided $f_n$ diverges, so the first summand goes to $0$.  To get the whole thing to converge to zero we therefore require that $f_n - n = o(2^n)$; &lt;em&gt;i.e.&lt;/em&gt;, $f_n$ should not diverge too rapidly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A similar computation produces the expectation of the squared log ratio:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \sum_x {p(x) ( \log( p(x) / q_n(x)) )^2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$= \sum_x {p(x) (\log(C_n))^2} - p(n) (\log(C_n))^2 + p(n) \left(\log( 2^{-n} / 2^{-f_n} ) \right)^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$= (\log(C_n))^2 \left(1 - 2^{-n} \right) + 2^{-n} \left( f_n - n \right)^2 .$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again the first summand converges to zero.  However, we can choose $f_n$ (subject to the previous restriction not to diverge too quickly) to make the right summand behave almost as we wish.  Three regimes are exhibited by these choices:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;$f_n = n^2$.  The right summand converges to $0$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$f_n = 2^{n/2} + n$.  The right summand equals a constant $1$, converging to $1$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$f_n = (2^{n/2} + n)n$.  The right summand (equal to $n^2$) diverges.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;In all cases $f_n - n = o(2^n)$ as required, ensuring that $\mathrm{KL}(p||q_n) \to 0$.  Finally, the variance of $\log( p / q )$ behaves like the expectation of $(\log( p/q ))^2$ because the square of the expectation is converging to $0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This shows that the asymptotic behavior of the variance of the log ratio is essentially independent of the convergence of $q$ to $p$ in the sense of the KL divergence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$q$, by the way, is the probability distribution for a simple truncated experiment provided $f_n \ge n$ (which includes all three cases): let $X$ be the number of flips of a fair coin needed to reach heads the first time.  If the number of flips is not $n$, accept $X$.  Otherwise, only in the case $X = n$, generate a uniform random number $U$.  If $U \lt 2^{n - f_n}$, accept the result $X$.  Otherwise, re-run the same experiment recursively until eventually some value is accepted.&lt;/p&gt;&#xA;" OwnerUserId="1489" LastActivityDate="2010-10-13T14:38:54.583" CommentCount="1" />
  <row Id="6698" PostTypeId="2" ParentId="6694" CreationDate="2010-10-13T14:39:18.403" Score="1" ViewCount="" Body="&lt;p&gt;Now you are basically into root finding.  I love &lt;a href=&quot;http://www.nr.com/&quot; rel=&quot;nofollow&quot;&gt;Numerical Recipes&lt;/a&gt;.  I have had the book over 30 years.  Good discussion of the techniques and code supplied in various languages.   Your formula should be an easy one.  Other numerical analysis books will work, too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A small thought:  if m is as small as you say, there might be some algebraic simplification in making it 1.  I haven't found it, but others are better at these combinatoric identities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are actually doing the sum every time, you might be able to identify the range of i that is the big contribution.  It looks like it should be somewhat below k/2.  You can probably reduce the terms in the sum by a large factor, at least to get close to N.  Then use the full formula for final refinement.  And take the N^(-k) out of the sum.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-10-13T14:39:18.403" CommentCount="4" />
  <row Id="6699" PostTypeId="2" ParentId="6675" CreationDate="2010-10-13T14:46:11.667" Score="4" ViewCount="" Body="&lt;p&gt;The group generated by two elements a,b subject only to the relations aa=1, aba=bab has the same number of generators and relations, but is neither cyclic nor infinite (it is the non-abelian group of order 6).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Be careful that a relation like a^6 = 1 does not imply a has order 6, only that it has order dividing 6.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is no practical and general method of determining the order of a group given by a presentation.  You may want to look into &quot;quotient methods&quot;, where you can determine the order of nice quotients of your group.  If that order is small, then you can find a finite presentation for the kernel, and try again.  For instance, finding the abelian invariants of a finitely presented group is easy, and if those invariants are small, then finding a presentation of the derived subgroup is easy.  Repeat, until you hit the perfect residuum, the largest perfect subgroup.  Now apply simple group recognition methods to determine the simple quotients of your group.  Find the entire top layer, and then hope beyond hope you can find the kernel.  Now repeat the derived subgroup method on the kernel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can often use &lt;a href=&quot;http://www.gap-system.org/&quot; rel=&quot;nofollow&quot;&gt;GAP&lt;/a&gt; to compute abelian invariants and derived subgroups fairly easily, as long as the index is kept under a thousand or so.&lt;/p&gt;&#xA;" OwnerUserId="583" LastActivityDate="2010-10-13T14:46:11.667" />
  <row Id="6700" PostTypeId="2" ParentId="6695" CreationDate="2010-10-13T14:49:34.913" Score="19" ViewCount="" Body="&lt;p&gt;Another way is: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int \sec x dx = \int \frac{\cos x}{1-\sin^2 x} dx =&#xA;\frac{1}{2} \int \left( \frac{1}{1-\sin x} + \frac{1}{1+\sin x} \right) \cos x dx $$ &#xA;$$= \frac{1}{2} \log \left| \frac{1+\sin x}{1-\sin x} \right| + C.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's worth noting that the answer can appear in many disguises. Another is&#xA;$$\log \left| \tan \left( \frac{\pi}{4} + \frac{x}{2} \right) \right| $$&lt;/p&gt;&#xA;" OwnerUserId="1301" LastActivityDate="2010-10-13T14:49:34.913" CommentCount="2" />
  <row Id="6701" PostTypeId="2" ParentId="6675" CreationDate="2010-10-13T14:54:56.393" Score="0" ViewCount="" Body="&lt;p&gt;While somewhat &quot;dated&quot; at this point you might want to look at the book of Coxeter and Moser, Generators and Relations, for work in this area.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www-history.mcs.st-and.ac.uk/Extras/Coxeter_Moser.html&quot; rel=&quot;nofollow&quot;&gt;http://www-history.mcs.st-and.ac.uk/Extras/Coxeter_Moser.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1369" LastActivityDate="2010-10-13T14:54:56.393" />
  <row Id="6702" PostTypeId="2" ParentId="6695" CreationDate="2010-10-13T15:09:45.863" Score="13" ViewCount="" Body="&lt;p&gt;Using the definitions $$\sec \theta = 1/\cos \theta \quad \text{and} \quad \cos \theta = (\exp(i \theta) + \exp(-i \theta))/2$$ gives  $$\int \sec \theta  d \theta = \int \frac{2 d \theta}{\exp(i \theta) + \exp(-i \theta)}.$$  The only insight needed is to find the substitution $u = \exp( i \theta )$ (what else is there to try?), leading to a multiple of $\int \frac{du}{1+u^2}$, the inverse tangent.  Thus, in an essentially mechanical fashion you obtain the generic solution $$-2 i \arctan(\exp(i \theta)).$$  Unwinding this via the usual algebraic identities between exponential and trig functions not only shows it is equal to the usual solutions, but also reveals why half angles might be involved and where an offset of $\pi /4$ might come from (as in @Derek Jennings' answer): it's a constant of integration, of course.&lt;/p&gt;&#xA;" OwnerUserId="1489" LastEditorUserId="1489" LastEditorDisplayName="" LastEditDate="2010-10-14T01:51:09.333" LastActivityDate="2010-10-14T01:51:09.333" CommentCount="5" />
  <row Id="6703" PostTypeId="2" ParentId="6635" CreationDate="2010-10-13T15:55:11.740" Score="1" ViewCount="" Body="&lt;p&gt;I use the following definition of the  matrix derivative of $f(Q)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(\delta f)(Q) = \lim_{t \to 0} f(Q+t \delta Q)-f(Q)$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;where, $ \delta Q $ is a symmetric not necessarily positive definite matrix.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is important to notice that the matrix derivative operator $\delta$ obeys the Leibniz rule.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore, in the following,  the cyclicity of the trace, and the identity&#xA;$\delta Q^{-1} = - Q^{-1} \delta Q Q^{-1}$ obtained from the matrix differentiation of $Q Q^{-1} = 1$ are used. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first matrix derivative of $f(Q)$ is given by:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\delta f(Q) = tr(-W Q^{-1} \delta Q Q^{-1} W + \delta Q )= tr((-Q^{-1} W^2 Q^{-1}+1) \delta Q  )$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, the condition $\delta f(Q) = 0$, implies:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$-Q^{-1} W^2 Q^{-1} = 1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is equivalent to:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$Q^2 = W^2 $.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, $Q = W $ is the only solution, because both $Q$ and $W$ are defined to be positive definite. To verify that this solution is a minimum,&#xA;we check the second matrix derivative is positive for an arbitrary $\delta Q$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\delta^2 f(Q) = tr((Q^{-1} \delta Q Q^{-1} W^2 Q^{-1}+Q^{-1}  W^2 Q^{-1}\delta Q Q^{-1}) \delta Q  ) = 2 tr(Q^{-1} (\delta Q)^2)$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where, the second equality is obtained after the substitution of the solution. &lt;/p&gt;&#xA;" OwnerUserId="1430" LastActivityDate="2010-10-13T15:55:11.740" CommentCount="1" />
  <row Id="6704" PostTypeId="1" AcceptedAnswerId="6711" CreationDate="2010-10-13T16:36:57.360" Score="1" ViewCount="125" Body="&lt;p&gt;I have the following sequences:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;a = [0, 1, 2, 3, 4, 5]&#xA;b = [1, 2, 4, 7, 13, 26]&#xA;c = [1, 1, 2, 4, 7, 13]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I'm trying to reverse-engineer an undocumented API and believe that the values in &lt;code&gt;a&lt;/code&gt; can be used to determine the corresponding values in &lt;code&gt;b&lt;/code&gt; and &lt;code&gt;c&lt;/code&gt;. For instance, we could say that third element in &lt;code&gt;b&lt;/code&gt; (&lt;code&gt;4&lt;/code&gt;) is two times the value in &lt;code&gt;a&lt;/code&gt; (&lt;code&gt;2&lt;/code&gt;), while the corresponding value in &lt;code&gt;c&lt;/code&gt; is the same as &lt;code&gt;a&lt;/code&gt;'s. Unfortunately that algorithm doesn't work across all the values f the sequences. Does this make sense? If so, can anyone think of any common series that would give the values of &lt;code&gt;b&lt;/code&gt; and &lt;code&gt;c&lt;/code&gt;?&lt;/p&gt;&#xA;" OwnerUserId="2487" LastActivityDate="2010-10-13T18:58:08.187" Title="Help needed to figure out relationships between a few sequences" Tags="&lt;sequences-and-series&gt;" AnswerCount="2" CommentCount="9" />
  <row Id="6705" PostTypeId="2" ParentId="6704" CreationDate="2010-10-13T16:46:58.463" Score="2" ViewCount="" Body="&lt;p&gt;The sequence $b$ can be obtained from $a$ by the function $$t \mapsto \frac{t^4}{12}-\frac{t^3}{2}+\frac{17 t^2}{12}+1$$ and the sequence $c$ using $$t \mapsto \frac{t^5}{60}-\frac{t^4}{6}+\frac{7 t^3}{12}-\frac{t^2}{3}-\frac{t}{10}+1.$$ I am pretty sure this is not what you want!&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2010-10-13T16:46:58.463" />
  <row Id="6706" PostTypeId="1" CreationDate="2010-10-13T16:52:37.850" Score="1" ViewCount="71" Body="&lt;p&gt;Using stereoscopic cameras, I track a certain object through it's path in space. For every frame, I compute it's pose in 3D. I can represent it's pose by either a translation vector + rotation matrix or a 6 element parameter vector (X, Y, Z, roll, pitch, yaw).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is that the output generated by the cameras is noisy, and I'd like to filter some of that noise off by smoothing the poses in 3D.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any pointers on that? What fitting works better for multidimensional problems?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some more information: in my case, the XYZ path can probably be fitted with a 2nd order curve, and roll, pitch and yaw also demonstrate smooth change over samples. &lt;/p&gt;&#xA;" OwnerUserId="2449" LastActivityDate="2011-05-22T06:00:25.380" Title="Fitting object poses in 3D space" Tags="&lt;analysis&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="6707" PostTypeId="1" AcceptedAnswerId="6727" CreationDate="2010-10-13T17:27:22.590" Score="5" ViewCount="121" Body="&lt;p&gt;Let $G = \langle x , y \ | \ x^{12}y=yx^{18} \rangle$. I want to continue a discussion &lt;a href=&quot;http://math.stackexchange.com/questions/6376/normal-closure-in-groups&quot;&gt;here&lt;/a&gt; on the normal closure of $y$ in $G$.&lt;br&gt;&#xA;How can we determine a group presentation for $N$?&lt;/p&gt;&#xA;" OwnerUserId="2424" LastActivityDate="2010-10-14T00:05:35.377" Title="Normal closure in groups II" Tags="&lt;group-theory&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="1" />
  <row Id="6708" PostTypeId="2" ParentId="6682" CreationDate="2010-10-13T17:32:00.737" Score="3" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\rm\quad  k\: $ divides $\rm\: k\:m + k\:,\ $ and, of course, $\rm\ k\:$ divides $\rm\: n!\ $ for $\rm\:k =2,3,\:\ldots\:,n\:$.&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-10-13T17:32:00.737" />
  <row Id="6709" PostTypeId="2" ParentId="2234" CreationDate="2010-10-13T17:53:47.030" Score="0" ViewCount="" Body="&lt;p&gt;Hey guys.... &#xA;There is no correct simple general formula for sum to n terms of the series &#xA;1+1/2+1/3+1/4+ ............. + 1/n &lt;/p&gt;&#xA;&#xA;&lt;p&gt;but the following formula will be a good approximation for sum to n-terms of the above series when n&gt;5 &lt;/p&gt;&#xA;&#xA;&lt;p&gt;S = log(n+0.5) + 0.5772 + 0.04021/(n*n + 0.8848)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Deviation from the actual value fluctuates but remains relatively low.. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;so i guess this may be a good approximation&lt;/p&gt;&#xA;" OwnerUserId="2488" LastEditorUserId="2488" LastEditorDisplayName="" LastEditDate="2010-10-27T21:17:37.567" LastActivityDate="2010-10-27T21:17:37.567" CommentCount="8" />
  <row Id="6710" PostTypeId="2" ParentId="2234" CreationDate="2010-10-13T18:02:20.463" Score="4" ViewCount="" Body="&lt;p&gt;A good quick approximation: $\log(n + 0.5) + \gamma$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A yet more accurate approximation: $\log(n)+\gamma+1/2z+1/12z^2-1/120z^4+1/252z^6-1/240z^8+1/132z^{10}-691/32760z^{12}+\cdots$, taking as many terms as is convenient.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Error analysis:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first formula above has maximum error of $1/24n^2$.  uday's first answer has error of about 0.0024 ln n.  Mine is a better approximation for all $n\neq6$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first formula is also better than uday's revised formula for all $n\ge46$. The expected error in uday's answer is about 0.000016.&lt;/p&gt;&#xA;" OwnerUserId="1778" LastEditorUserId="1778" LastEditorDisplayName="" LastEditDate="2010-10-24T23:58:31.310" LastActivityDate="2010-10-24T23:58:31.310" />
  <row Id="6711" PostTypeId="2" ParentId="6704" CreationDate="2010-10-13T18:31:31.877" Score="1" ViewCount="" Body="&lt;p&gt;Unfortunately, we don't have enough information here to reverse-engineer your problem. For instance, the Generalized Catalan Numbers have been referenced in the comments (by Qiaochu Yuan) because they appear in your sequence $b.$ This sequence starts $1,1,1,2,4,7,13,26,52,104,\ldots$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I could define a “random” sequence by $u_1=1$ and for $n&gt;1$&#xA;$$U_n=2U_{n-1} \quad \textrm{ for } n \equiv 0,1 \textrm{ mod } 4$$&#xA;and&#xA;$$U_n=2U_{n-1}-1 \quad \textrm{ for } n \equiv 2,3 \textrm{ mod } 4.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This follows the stated Catalan sequence exactly, and so contains your sequence $b,$ up until we get 103 instead of 104. I give this as an illustration that with the current information this problem has no definitive answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: To be fair to the question asker, I'd like to point out that the fact that he/she cannot do what he/she wants to do is often not so clear to people who've been brought up on a diet of school mathematics where one is frequently asked to fill in the missing number in a sequence or find the next number.&lt;/p&gt;&#xA;" OwnerUserId="1301" LastEditorUserId="1301" LastEditorDisplayName="" LastEditDate="2010-10-13T18:58:08.187" LastActivityDate="2010-10-13T18:58:08.187" CommentCount="4" />
  <row Id="6712" PostTypeId="2" ParentId="5303" CreationDate="2010-10-13T19:16:20.467" Score="3" ViewCount="" Body="&lt;p&gt;One essential piece of intuition from theoretical computer science: since P $\neq$ NP (we assume!), there are many statements which are easy to state but hard to prove.&lt;/p&gt;&#xA;" OwnerUserId="1778" LastActivityDate="2010-10-13T19:16:20.467" />
  <row Id="6713" PostTypeId="1" AcceptedAnswerId="6973" CreationDate="2010-10-13T19:59:06.930" Score="3" ViewCount="121" Body="&lt;p&gt;Rectangles will all be the same size, circle may vary.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was discussing this with a friend and he said to just start placing them in the middle and moving outwards but it seems that you could free up more space for rectangles around the edges if you moved that center of mass to one side and maybe put some space between the rectangles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me know if any of this is confusing or if you have heard of a good method for doing this.&lt;/p&gt;&#xA;" OwnerUserId="2490" LastActivityDate="2010-10-16T19:54:20.263" Title="Is there an algorithm out there that will show you how to fit the most rectangles into a circle?" Tags="&lt;geometry&gt;&lt;calculus&gt;" AnswerCount="1" CommentCount="6" />
  <row Id="6714" PostTypeId="1" AcceptedAnswerId="6725" CreationDate="2010-10-13T20:37:00.130" Score="0" ViewCount="93" Body="&lt;p&gt;Suppose x is the solution to a standard linear programming problem ($Ax=b$, $x&gt;=0$) and the set $S$ is every $i$ where $x_{i} = 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I show this is optimal only where&lt;/p&gt;&#xA;&#xA;&lt;p&gt;minimize $c'f$ subject to $Af=0, f_{i} \ge 0$, $i$ is in $S$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;has a cost of zero.&lt;/p&gt;&#xA;" OwnerUserId="2014" LastEditorUserId="1195" LastEditorDisplayName="" LastEditDate="2010-10-14T14:33:26.177" LastActivityDate="2010-10-14T14:33:26.177" Title="Convert linear programming problem" Tags="&lt;linear-programming&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="6716" PostTypeId="1" AcceptedAnswerId="10476" CreationDate="2010-10-13T20:49:44.203" Score="4" ViewCount="64" Body="&lt;p&gt;Hi all,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose that $\mathcal{E}$ is a well-pointed elementary topos, that $X$ and $Y$ are objects of $\mathcal{E}$, and that $F$ is a function which maps global elements $p: 1 \to X$ to global elements $F(p): 1 \to Y$ (here $1$ is the terminal object of $\mathcal{E}$). Does there exist a (necessarily unique) arrow $f: X \to Y$ in $\mathcal{E}$ such that $fp = F(p)$ for all $p$? Equivalently, is any object in a well-pointed topos the coproduct over its global elements of $1$? It's easy to show that the answer is &quot;yes&quot; if the coproduct exists since the induced map $\coprod_{p \in \Gamma X} 1 \to X$ is iso. But I don't know whether the coproduct exists in general.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Could somebody with enough reputation create a &quot;topos-theory&quot; tag and add it to this? Thanks)&lt;/p&gt;&#xA;" OwnerUserId="2491" LastEditorUserId="2215" LastEditorDisplayName="" LastEditDate="2010-10-13T21:13:00.280" LastActivityDate="2010-11-15T21:31:26.457" Title="Do functions defined on global elements give rise to arrows in a well-pointed topos?" Tags="&lt;category-theory&gt;&lt;topos-theory&gt;" AnswerCount="1" FavoriteCount="2" />
  <row Id="6717" PostTypeId="2" ParentId="6695" CreationDate="2010-10-13T20:55:55.750" Score="5" ViewCount="" Body="&lt;p&gt;Instead of presenting another way of evaluating this integral, I justify a more general case in an approach which uses partial fractions and trigonometric identities, at the level of a Calculus class, I think:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int \dfrac{1}{a+b\cos x}dx=\dfrac{1}{\sqrt{b^{2}-a^{2}}}\ln \left\vert &#xA;\dfrac{\sqrt{a+b}+\sqrt{b-a}\tan x/2}{\sqrt{a+b}-\sqrt{b-a}\tan x/2}\right\vert \quad a\lt b.\quad (\ast)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$a+b\cos x=(a-b)+2b\cos ^{2}x/2,$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\dfrac{1}{a+b\cos x}=\dfrac{\sec ^{2}x/2}{(a-b)\sec ^{2}x/2+2b}=\dfrac{\sec&#xA;^{2}x/2}{(a-b)\sec ^{2}x/2+2b}=\dfrac{\sec ^{2}x/2}{a+b-(b-a)\tan ^{2}x/2}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\dfrac{1}{a+b-(b-a)\tan ^{2}x/2}=$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$=\dfrac{1}{2\sqrt{a+b}}\left( \dfrac{1}{%&#xA;\sqrt{a+b}-\sqrt{b-a}\tan x/2}+\dfrac{1}{\sqrt{a+b}+\sqrt{b-a}\tan x/2}%&#xA;\right) .$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int \dfrac{1}{a+b\cos x}dx=$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$=\dfrac{1}{2\sqrt{a+b}}\int \left( \dfrac{\sec&#xA;^{2}x/2}{\sqrt{a+b}-\sqrt{b-a}\tan x/2}+\dfrac{\sec ^{2}x/2}{\sqrt{a+b}+%&#xA;\sqrt{b-a}\tan x/2}\right) dx$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$=\dfrac{1}{\sqrt{b^{2}-a^{2}}}\ln \left\vert \dfrac{\sqrt{a+b}+\sqrt{b-a}%&#xA;\tan x/2}{\sqrt{a+b}-\sqrt{b-a}\tan x/2}\right\vert .$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, we have your particular case&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int \dfrac{1}{\cos x}dx=\int \dfrac{1}{0+1\cos x}dx=\ln \left\vert \dfrac{%&#xA;1+\tan x/2}{1-\tan x/2}\right\vert . \qquad (\ast\ast)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From $\tan \dfrac{x}{2}=\dfrac{\sin x}{1+\cos x}$ and $\sec x+\tan x=\dfrac{1+\sec x+\tan x}{1+\sec x-\tan x}$ it follows that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\dfrac{1+\tan x/2}{1-\tan x/2}=\dfrac{1+\dfrac{\sin x}{1+\cos x}}{1-\dfrac{%&#xA;\sin x}{1+\cos x}}=\dfrac{1+\cos x+\sin x}{1+\cos x-\sin x}=\sec x+\tan x$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and, finally&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int \sec x\; dx=\ln \left\vert \sec x+\tan x\right\vert .$$&lt;/p&gt;&#xA;" OwnerUserId="752" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2010-10-15T22:51:37.493" LastActivityDate="2010-10-15T22:51:37.493" CommentCount="2" />
  <row Id="6718" PostTypeId="1" AcceptedAnswerId="6719" CreationDate="2010-10-13T21:02:11.520" Score="1" ViewCount="123" Body="&lt;p&gt;I'm playing around with a collection of subsets of $\mathbb{R}^n$ let's call them $X_i$. What I want to know is, is the following condition sufficient for some $X_i$ being measurable?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Almost all elements of $X_i$ are surrounded by a neighbourhood contained in $X_i$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is motivated by the idea that non measurable sets are &quot;fractally wiggly&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming standard topology and measure on $\mathbb{R}^n$ but I'm also interested in minimal assumptions regarding topology and measure that keep this a sufficient condition for being measurable.&lt;/p&gt;&#xA;" OwnerUserId="172" LastEditorUserId="172" LastEditorDisplayName="" LastEditDate="2010-10-14T10:10:34.170" LastActivityDate="2010-10-14T10:10:34.170" Title="Sufficient condition for being measurable?" Tags="&lt;topology&gt;&lt;measure-theory&gt;" AnswerCount="1" />
  <row Id="6719" PostTypeId="2" ParentId="6718" CreationDate="2010-10-13T21:08:48.463" Score="4" ViewCount="" Body="&lt;p&gt;I believe this is sufficient, though not necessary.  The set of $x\in X_i$ that are contained in a neighborhood in $X_i$ is the interior of $X_i$, which is open.  If $X_i$ satisfies this condition, then it's an open set union a set of measure zero, both of which are measurable.  This will still be true for any measure defined at least on the Borel $\sigma$-algebra for your space.&lt;/p&gt;&#xA;" OwnerUserId="2215" LastActivityDate="2010-10-13T21:08:48.463" CommentCount="2" />
  <row Id="6720" PostTypeId="1" AcceptedAnswerId="28583" CreationDate="2010-10-13T21:16:41.097" Score="1" ViewCount="245" Body="&lt;p&gt;This may be a really basic question but the typical atan2 method isn't working when I write it in a program.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need a camera object to face a certain target in 3D space.&lt;br&gt;&#xA;I need to calculate the 3 Euler angles (yaw, pitch, roll) for the camera.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/q9SAq.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I do this?&lt;/p&gt;&#xA;" OwnerUserId="2493" LastActivityDate="2011-03-23T00:26:50.683" Title="Calculate Euler yaw pitch and roll for a 3d point to face another" Tags="&lt;geometry&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="6721" PostTypeId="2" ParentId="6707" CreationDate="2010-10-13T22:01:37.720" Score="0" ViewCount="" Body="&lt;p&gt;I don't know the answer myself but you can take a look on Chapter 9 (Presentation of Subgroups) in Johnson's &quot;Presentations of Groups&quot; - I just discovered this very nice book. I'll do that when I get to university.&lt;br&gt;&#xA;Unfortunately not many people participated in the first question, so thanks for continuing the discussion... I'll be voting you up when I can.&lt;/p&gt;&#xA;" OwnerUserId="2342" LastActivityDate="2010-10-13T22:01:37.720" CommentCount="1" />
  <row Id="6722" PostTypeId="1" CreationDate="2010-10-13T22:43:26.157" Score="-1" ViewCount="272" Body="&lt;p&gt;&lt;strike&gt;Tell me everything you know about Euklid, Pythagoras etcetera.&lt;/strike&gt;&#xA;I'm about to begin reading about greek algebra used in calculating geometric figures. After reading I'm going to write a 15-20 pages report.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So any kind of introduction would be awesome, I'm primarily interested in Euklid and how the theories at that time are important in math today.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ps. sorry for the bad english, I'm tired, made a 300 words spanish assignment and I'm not a native. &lt;em&gt;Be nice, it's my first time.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="2495" LastEditorUserId="2495" LastEditorDisplayName="" LastEditDate="2010-10-17T21:39:26.530" LastActivityDate="2010-10-22T03:57:54.393" ClosedDate="2011-04-23T16:16:45.313" Title="Greek Geometry Algebra" Tags="&lt;geometry&gt;&lt;reference-request&gt;&lt;euclidean-geometry&gt;" AnswerCount="1" CommentCount="9" />
  <row Id="6723" PostTypeId="1" CreationDate="2010-10-13T22:44:03.253" Score="1" ViewCount="168" Body="&lt;p&gt;So let's say you have a curved mirror, $y=y(x)$ with this property:&#xA;Whenever a ray of light emanates from the origin, it reflects parallel to the x axis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Find the equation of the mirror.&lt;/p&gt;&#xA;" OwnerUserId="2193" LastEditorUserId="1543" LastEditorDisplayName="" LastEditDate="2011-04-18T12:14:55.327" LastActivityDate="2011-04-18T12:14:55.327" Title="Equation of a mirror" Tags="&lt;physics&gt;&lt;analytic-geometry&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="1" />
  <row Id="6724" PostTypeId="2" ParentId="6723" CreationDate="2010-10-13T22:48:41.120" Score="1" ViewCount="" Body="&lt;p&gt;This sounds like homework.  Can you translate the words into an equation?  Think about what happens when the light ray strikes the mirror.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-10-13T22:48:41.120" />
  <row Id="6725" PostTypeId="2" ParentId="6714" CreationDate="2010-10-13T23:57:36.873" Score="1" ViewCount="" Body="&lt;p&gt;$f=0$, yielding a cost of $0$, is a feasible point for your second problem. So if the optimal cost is non-zero then it must be negative. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But then $y=x+\lambda f$ would be an improvement over $x$ in the original problem for all $\lambda&gt;0$. Because of the nature of constraints in the second problem $y$ satisfies the equality and binding non-negativity constraints in the first problem. We can choose $\lambda$ small enough so that it also satisfies the non-binding non-negativity constraints. But then $y$ would be superior to $x$ in the original problem, proving that the latter was not an optimal point.&lt;/p&gt;&#xA;" OwnerUserId="1195" LastActivityDate="2010-10-13T23:57:36.873" />
  <row Id="6726" PostTypeId="1" AcceptedAnswerId="8595" CreationDate="2010-10-13T23:59:52.813" Score="1" ViewCount="100" Body="&lt;p&gt;I read that Smale recast classical mechanics in terms of symplectic geometry. I know a bit about classical mechanics but nothing about symplectic geometry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there any writings from Smale on this which I could read?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And also what other books or texts are there which cover this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you!&lt;/p&gt;&#xA;" LastEditorUserId="8575" LastEditorDisplayName="" LastEditDate="2011-05-23T17:00:01.660" LastActivityDate="2011-05-23T17:00:01.660" Title="References for Smale's Geometry" Tags="&lt;reference-request&gt;&lt;symplectic-geometry&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="0" />
  <row Id="6727" PostTypeId="2" ParentId="6707" CreationDate="2010-10-14T00:05:35.377" Score="4" ViewCount="" Body="&lt;p&gt;This is a much harder question than your last one, and involves two separate parts: finding generators for $N$ and finding relations for $N$.  I'll try to sketch the answers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For context, recall from the previous thread that every element of $G$ can be written as a word in $x$ and $y$, and $N$ is the subgroup consisting of all words for which the total power of $x$ is a multiple of $6$.  In particular, $N$ is a subgroup of $G$ with index $6$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Generators&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finding generators for $N$ involves something called a &lt;strong&gt;&lt;em&gt;Schreier graph&lt;/em&gt;&lt;/strong&gt;.  This is a directed graph defined as follows:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;It has one vertex for each coset of $N$ in $G$ (so six vertices total)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;There is a directed edge labeled $x$ from each coset $aN$ to the coset $xaN$, and a directed edge labeled $y$ from each coset $aN$ to the coset $yaN$.  (Thus there will be $12$ edges total, $6$ labeled $x$ and $6$ labeled $y$.)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;For the case you are looking at, the six vertices are $N,xN,x^2N,\ldots,x^5N$.  These six vertices are arranged in a directed hexagon of $x$ edges), and each vertex has a loop labeled $y$:&#xA;&lt;img src=&quot;http://i.imgur.com/TO7fZ.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that each path in this graph corresponds to a word in $x$ and $y$, with backwards travel along directed edges corresponding to inverse generators.  In particular, the elements of $N$ are the paths that start and end at the vertex $N$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the statement that is true is that generators for $N$ are the same as generators for the fundamental group of this graph, using $N$ as the basepoint.  In this case, what that means is that $N$ is generated by the following elements:&#xA;$$&#xA;A=x^6,\quad B_0 = y,\quad B_1 = xyx^{-1},\quad B_2 = x^2yx^{-2},\quad\ldots,\quad B_5 = x^5yx^{-5}&#xA;$$&#xA;(In general, one can obtain generators for the fundamental group of a graph by first choosing a spanning tree for the graph, and then considering all closed paths starting at the basepoint that pass through exactly one edge not in the spanning tree.  In this case, the spanning tree consists of the first five edges of the hexagon.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Relations&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next, the relations for $N$ come from the relations for $G$.  There is one relation for $G$, namely:&#xA;$$&#xA;x^{12}yx^{-18}y^{-1} = 1.&#xA;$$&#xA;This will lead to six different relations for $N$, obtained by conjugating this relation by representatives for the six cosets of $N$.  Thus, the relations in $N$ can be written as&#xA;$$&#xA;x^{12}yx^{-18}y^{-1} = 1, \quad x(x^{12}yx^{-18}y^{-1})x^{-1} = 1, \quad\ldots,\quad x^5(x^{12}yx^{-18}y^{-1})x^{-5}=1.&#xA;$$&#xA;Of course, these are expressed in terms of $x$ and $y$, and we must rewrite them in terms of the generators $A,B_0,\ldots,B_5$:&#xA;$$&#xA;A^2B_0 A^{-3} B_0^{-1} = 1,\quad A^2B_1 A^{-3} B_1^{-1} = 1,\quad\ldots,\quad A^2B_5A^{-3}B_5^{-1} = 1&#xA;$$&#xA;(The justification for this comes from algebraic topology.  Each group $G$ has an associated 2-complex, which has a single vertex, one loop at the vertex for each generator, and one 2-cell for each relation.  The fundamental group of this 2-complex is equal to $G$.  The finite-index subgroup $N$ corresponds to a finite-sheeted cover of this 2-complex.  This cover has a Schreier graph as its 1-skeleton, and the six 2-cells of the cover are attached along the six words listed above.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Final Result&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In any case, we have now obtained a presentation of the group $N$:&#xA;$$&#xA;N = \langle A,B_0,\ldots,B_5 \mid A^2B_k = B_kA^3\text{ for each $k$}\rangle&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the way, the group $G$ that you seem to be interested in is called a &lt;a href=&quot;http://en.wikipedia.org/wiki/Baumslag-Solitar_group&quot; rel=&quot;nofollow&quot;&gt;Baumslag-Solitar group&lt;/a&gt;, and is usually denoted $B(12,18)$.&lt;/p&gt;&#xA;" OwnerUserId="1726" LastActivityDate="2010-10-14T00:05:35.377" CommentCount="4" />
  <row Id="6728" PostTypeId="1" CreationDate="2010-10-14T00:09:05.827" Score="0" ViewCount="52" Body="&lt;p&gt;what are some &lt;em&gt;lower&lt;/em&gt; bounds (if they exist) on $w^t A w$ in terms of $||w||_2^2$ for $A$ an matrix with all positive values? The lower bound can depend on $A$ and $||w||_2^2$. $w$ is $k \times 1$. Note when $A = I$ then we exactly get the squared norm of $w$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there are no such values, what in the case that $A$ is positive semi-definite? (i.e. $A$ has a $B$ such that $A = BB^T$ and then we are talking $w^tA w = ||Bw||^2_2$ and we want it to be larger than something that depends on $||w||^2_2$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="7076" LastActivityDate="2010-10-14T07:42:12.417" Title="what are simple lower bounds on w^t A w in terms of ||w||_2^2?" Tags="&lt;linear-algebra&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="6729" PostTypeId="2" ParentId="6728" CreationDate="2010-10-14T00:22:55.537" Score="2" ViewCount="" Body="&lt;p&gt;The bounds on $w^TAw$ are&#xA;$$\lambda_{\min} \|w\|_2^2 \le w^TAw \le \lambda_{\max}\|w\|_2^2,$$&#xA;where $\lambda_{\min}$ and $\lambda_{\max}$ are the smallest and largest eigenvalues, respectively, of $\frac{1}{2}\left(A + A^T\right)$. Each bound is attained when $w$ is parallel to the corresponding eigenvector.&lt;/p&gt;&#xA;" OwnerUserId="856" LastEditorUserId="1794" LastEditorDisplayName="" LastEditDate="2010-10-14T07:42:12.417" LastActivityDate="2010-10-14T07:42:12.417" />
  <row Id="6730" PostTypeId="2" ParentId="5303" CreationDate="2010-10-14T01:40:33.340" Score="1" ViewCount="" Body="&lt;p&gt;The question may look almost self-evident, yet when translated into math, it sometimes looses that self-evident feature and so it requires (often fairly complicated) proof. My favorite example is the Jordan Theorem. A closed non-intersecting curve separates the plane into &quot;inner&quot; and &quot;outer&quot; regions. Looks like there isn't anything to prove here. But translate the notion of &quot;closed non-intersecting curve&quot; into the formula, and we no longer have any intuitive feel of &quot;inner region&quot;, &quot;outer region&quot;, and what is that region anyway. The simplest proof requires the machinery of algebraic topology and still quite long.&lt;/p&gt;&#xA;" OwnerUserId="2450" LastActivityDate="2010-10-14T01:40:33.340" CommentCount="2" />
  <row Id="6731" PostTypeId="2" ParentId="6340" CreationDate="2010-10-14T02:03:27.597" Score="0" ViewCount="" Body="&lt;p&gt;Right! Thank you, I didn't think about using H\&quot;older's.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In short, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;|f(0)| = |\int_{|z|=R} \frac{f(z)dz}{z} | &amp;lt;  (\int_{|z|=R} |f(z)|^p |dz|)^{1/p} (\int_{|z|=R} |z|^{-q} |dz|)^{1/q}&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And $(\int_{|z|=R} |z|^{-q} |dz|)^{1/q} = R^{-1+1/q} (2\pi)^{1/q} = R^{-1/p} (2\pi)^{1/q}$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;\int_0^\infty ((2\pi)^{1/q} |f(0)| R^{1/p})^p dR &amp;lt; || f ||_p^p &amp;lt;\infty&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And the only way for this to fly is for $f(0)=0$, for $p\le 1$&lt;/p&gt;&#xA;" OwnerUserId="2499" LastActivityDate="2010-10-14T02:03:27.597" CommentCount="1" />
  <row Id="6732" PostTypeId="1" AcceptedAnswerId="6762" CreationDate="2010-10-14T02:27:53.990" Score="1" ViewCount="123" Body="&lt;p&gt;If I have a bounded set F in N dimensional space and another set G where every element g in G has h'g=c and also must exist in F. H is a vector in the N dimensional space and c is any constant 1x1 matrix (scalar). h is a vector of appropriate dimension.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I prove every extreme point of G lies on the boundary of F?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is to say if x and y are extreme points in F then xλ + (1-λ)y = g&lt;/p&gt;&#xA;" OwnerUserId="2014" LastEditorUserId="1195" LastEditorDisplayName="" LastEditDate="2010-10-14T14:34:47.490" LastActivityDate="2010-10-14T14:34:47.490" Title="Show convex combination" Tags="&lt;linear-algebra&gt;&lt;convex-analysis&gt;" AnswerCount="1" CommentCount="8" />
  <row Id="6733" PostTypeId="1" AcceptedAnswerId="6736" CreationDate="2010-10-14T03:11:59.973" Score="1" ViewCount="44" Body="&lt;p&gt;Hello,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've been trying to simplify this equation:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Y = (¬AB + ¬C)(¬A + ¬D)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;.. into this equation.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Y  = ¬A¬C + ¬C¬D + ¬AB&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Unfortunately I keep going in circles with expanding and minimizing the booleans. Any tips or advice? Thanks.&lt;/p&gt;&#xA;" LastActivityDate="2010-10-14T03:31:43.533" Title="Boolean Simplification" Tags="&lt;boolean-algebra&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="6734" PostTypeId="1" AcceptedAnswerId="6735" CreationDate="2010-10-14T03:17:52.503" Score="2" ViewCount="212" Body="&lt;p&gt;The following is a question from the sample GRE Mathematics Subject Test&#xA;found on the ETS website:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $M$ be a $5\times 5$ real matrix.&#xA;Exactly four of the following five conditions on $M$ are equivalent&#xA;to each other.&#xA;Which of the five conditions is equivalent to NONE of the other four?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(A) For any two distinct column vectors $u$ and $v$ of $M$,&#xA;the set {$u,v$} is linearly independent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(B) The homogeneous system $Mx=0$ has only the trivial solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(C) The system of equations $Mx=b$ has a unique solution for each real $5\times 1$&#xA;column vector $b$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(D) The determinant of $M$ is nonzero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(E) There exists a $5\times 5$ real matrix $N$ such that $NM$ is the $5\times 5$ identity matrix.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Apparently, the correct answer is (A), but I can't figure out why this is true. If $M$ is&#xA;nonsingular, as is implied by statements (B)-(E), then isn't that equivalent to its column vectors being linearly&#xA;independent? And if the 5 column vectors are independent, then I can easily show that&#xA;each pair of vectors are independent. What am I missing?&lt;/p&gt;&#xA;" OwnerUserId="2029" LastActivityDate="2011-01-15T23:25:07.550" Title="GRE linear algebra question" Tags="&lt;linear-algebra&gt;&lt;matrices&gt;" AnswerCount="3" />
  <row Id="6735" PostTypeId="2" ParentId="6734" CreationDate="2010-10-14T03:25:43.047" Score="4" ViewCount="" Body="&lt;p&gt;Linear independence of $n$ vectors, for $n&gt;2$, is &lt;em&gt;not&lt;/em&gt; equivalent to &quot;pairwise&quot; independence. Take three different vectors in the plane, for an example.&lt;/p&gt;&#xA;" OwnerUserId="1313" LastActivityDate="2010-10-14T03:25:43.047" CommentCount="2" />
  <row Id="6736" PostTypeId="2" ParentId="6733" CreationDate="2010-10-14T03:31:43.533" Score="1" ViewCount="" Body="&lt;p&gt;Use the following simplification:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\neg AB(\neg A + \neg D) = \neg AB\neg A + \neg AB\neg D = \neg AB + \neg AB\neg D = \neg AB$.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-10-14T03:31:43.533" CommentCount="1" />
  <row Id="6737" PostTypeId="1" AcceptedAnswerId="6741" CreationDate="2010-10-14T03:42:15.737" Score="9" ViewCount="368" Body="&lt;p&gt;&lt;strong&gt;Note:&lt;/strong&gt; &lt;a href=&quot;http://stackoverflow.com/q/3924602/33686&quot;&gt;This question has been posted on StackOverflow&lt;/a&gt;. I have moved it here because:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;I am curious about the answer&lt;/li&gt;&#xA;&lt;li&gt;The OP has not shown any interest in moving it himself&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;In the Communications of the ACM, &lt;a href=&quot;http://delivery.acm.org/10.1145/1790000/1787260/p128-winkler.html?key1=1787260&amp;amp;key2=3588796821&amp;amp;coll=portal&amp;amp;dl=ACM&amp;amp;CFID=108548737&amp;amp;CFTOKEN=59330921&quot; rel=&quot;nofollow&quot;&gt;August 2008 &quot;Puzzled&quot; column&lt;/a&gt;, Peter Winkler asked the following question:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;On the table before us are 10 dots,&#xA;  and in our pocket are 10 $1 coins.&#xA;  Prove the coins can be placed on the&#xA;  table (no two overlapping) in such a&#xA;  way that all dots are covered. Figure&#xA;  2 shows a valid placement of the coins&#xA;  for this particular set of dots; they&#xA;  are transparent so we can see them.&#xA;  The three coins at the bottom are not&#xA;  needed.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In the &lt;a href=&quot;http://delivery.acm.org/10.1145/1820000/1810917/p110-winkler.html?key1=1810917&amp;amp;key2=2298796821&amp;amp;coll=portal&amp;amp;dl=ACM&amp;amp;CFID=108548737&amp;amp;CFTOKEN=59330921&quot; rel=&quot;nofollow&quot;&gt;following issue&lt;/a&gt;, he presented his proof:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;We had to show that any 10 dots on a&#xA;  table can be covered by&#xA;  non-overlapping $1 coins, in a problem&#xA;  devised by Naoki Inaba and sent to me&#xA;  by his friend, Hirokazu Iwasawa, both&#xA;  puzzle mavens in Japan.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The key is to note that packing disks&#xA;  arranged in a honeycomb pattern cover&#xA;  more than 90% of the plane. But how do&#xA;  we know they do? A disk of radius one&#xA;  fits inside a regular hexagon made up&#xA;  of six equilateral triangles of&#xA;  altitude one. Since each such triangle&#xA;  has area &lt;code&gt;sqrt(3)/3&lt;/code&gt;, the hexagon&#xA;  itself has area &lt;code&gt;2*sqrt(3)&lt;/code&gt;; since the&#xA;  hexagons tile the plane in a honeycomb&#xA;  pattern, the disks, each with area π,&#xA;  cover &lt;code&gt;π /(2*sqrt(3))&lt;/code&gt; ~ .9069 of the&#xA;  plane's surface.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;It follows that if the disks are&#xA;  placed randomly on the plane, the&#xA;  probability that any particular point&#xA;  is covered is .9069. Therefore, if we&#xA;  randomly place lots of $1 coins&#xA;  (borrowed) on the table in a hexagonal&#xA;  pattern, on average, 9.069 of our 10&#xA;  points will be covered, meaning at&#xA;  least some of the time all 10 will be&#xA;  covered. (We need at most only 10&#xA;  coins so give back the rest.)&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;What does it mean that the disks cover&#xA;  90.69% of the infinite plane? The easiest way to answer is to say,&#xA;  perhaps, that the percentage of any&#xA;  large square covered by the disks&#xA;  approaches this value as the square&#xA;  expands. What is &quot;random&quot; about the&#xA;  placement of the disks? One way to&#xA;  think it through is to fix any packing&#xA;  and any disk within it, then pick a&#xA;  point uniformly at random from the&#xA;  honeycomb hexagon containing the disk&#xA;  and move the disk so its center is at&#xA;  the chosen point.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I don't understand. Doesn't the probabilistic nature of this proof simply mean that in the &lt;strong&gt;majority&lt;/strong&gt; of configurations, all 10 dots can be covered. Can't we still come up with a configuration involving 10 (or&amp;nbsp;less) dots where one of the dots can't be covered?&lt;/p&gt;&#xA;" OwnerUserId="59" LastEditorUserId="856" LastEditorDisplayName="" LastEditDate="2010-10-14T06:00:22.397" LastActivityDate="2010-10-14T06:00:22.397" Title="Covering ten dots on a table with ten equal-sized coins: explanation of proof" Tags="&lt;geometry&gt;&lt;probability&gt;&lt;puzzle&gt;" AnswerCount="3" CommentCount="4" FavoriteCount="6" />
  <row Id="6738" PostTypeId="1" AcceptedAnswerId="6742" CreationDate="2010-10-14T03:50:17.947" Score="2" ViewCount="281" Body="&lt;p&gt;Originally I was just going to ask the problem on my practice math contest, which is asking how many ways there are to write a nine-digit number containing each digit 1-9 so that the first digit is twice as big as the second and the last two digits are odd. (e.g. 846527913).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understood that in this case, the number would be in the form of EE(EEEOO)OO or EO(EEOOO)OO where E is an even, O is an odd, the numbers in the parentheses can be ordered in any way, and the only four cases for the first two are 21, 42, 63, and 84. Calculating this gave me 240, which is definitely wrong (I made a program to check, and it says 7680).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So here are the questions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) How do I solve this particular problem correctly? Am I on the right track?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) How are these solved in general? Given that it's a 9 digit number and each digit 1-9 is used once, how does one handle constraints such as these? This has always confused me (I'm not very strong in combinatorics)&lt;/p&gt;&#xA;" OwnerUserId="1562" LastActivityDate="2010-10-14T04:10:15.780" Title="Different ways to write a number given digit constraints" Tags="&lt;combinatorics&gt;&lt;probability&gt;&lt;homework&gt;" AnswerCount="1" />
  <row Id="6739" PostTypeId="2" ParentId="6737" CreationDate="2010-10-14T04:02:11.663" Score="8" ViewCount="" Body="&lt;p&gt;The key point is that the 90.69% probability is with respect to &quot;the &lt;em&gt;disks&lt;/em&gt; [being] placed randomly on the plane&quot;, not the &lt;em&gt;points&lt;/em&gt; being placed randomly on the plane. That is, the set of points on the plane is &lt;em&gt;fixed&lt;/em&gt;, but the honeycomb arrangement of the disks is placed over it at a random displacement. Since the probability that any such placement covers a given point is 0.9069, a random placement of the honeycomb will cover, on average, 9.069 points (this follows from linearity of expectation; I can expand on this if you like). Now the only way random placements can cover 9.069 points &lt;em&gt;on average&lt;/em&gt; is if &lt;em&gt;some&lt;/em&gt; of these placements cover 10 points -- if all placements covered 9 points or less, the average number of points covered would be at most 9. Therefore, &lt;em&gt;there exists&lt;/em&gt; a placement of the honeycomb arrangement that covers 10 points (though this proof doesn't tell you what it is, or how to find it).&lt;/p&gt;&#xA;" OwnerUserId="856" LastActivityDate="2010-10-14T04:02:11.663" CommentCount="1" />
  <row Id="6740" PostTypeId="2" ParentId="6737" CreationDate="2010-10-14T04:05:05.393" Score="1" ViewCount="" Body="&lt;p&gt;If you read carefully, this proof is for an &lt;em&gt;arbitrary&lt;/em&gt; placement of dots. So given &lt;em&gt;any&lt;/em&gt; dot arrangement, if we just place the coins randomly (in the honeycomb arrangement,) then on average we will cover slightly more than 9 of the dots. But since we can't cover &quot;part&quot; of a dot (in this problem) then that means that there &lt;strong&gt;exits&lt;/strong&gt; a random placement of the coins that cover all 10 dots. So no matter the configuration of dots, we know that there is always a way to cover the dots with at most 10 coins :) &lt;/p&gt;&#xA;" OwnerUserId="1352" LastEditorUserId="1352" LastEditorDisplayName="" LastEditDate="2010-10-14T04:11:41.460" LastActivityDate="2010-10-14T04:11:41.460" />
  <row Id="6741" PostTypeId="2" ParentId="6737" CreationDate="2010-10-14T04:07:38.023" Score="12" ViewCount="" Body="&lt;p&gt;Nice! The above proof proves that &lt;em&gt;any&lt;/em&gt; configuration of 10 dots can be covered. What you have here is an example of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Probabilistic_method&quot; rel=&quot;nofollow&quot;&gt;probabilistic method&lt;/a&gt;, which uses probability but gives a certain (not a probabilistic) conclusion (an example of &lt;a href=&quot;http://en.wikipedia.org/wiki/Probabilistic_proofs_of_non-probabilistic_theorems&quot; rel=&quot;nofollow&quot;&gt;probabilistic proofs of non-probabilistic theorems&lt;/a&gt;). This proof also implicitly uses the linearity of expectation, a fact that seem counter-intuitive in some cases until you get used to it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To clarify the proof: given &lt;em&gt;any&lt;/em&gt; configuration of 10 dots, &lt;strong&gt;fix&lt;/strong&gt; the configuration, and consider placing honeycomb-pattern disks randomly. Now, what is the expected number $X$ of dots covered? Let $X_i$ be 1 if dot $i$ is covered, and $0$ otherwise. We know that $E[X] = E[X_1] + \dots + E[X_{10}]$, and also that $E[X_i] = \Pr(X_i = 1) \approx 0.9069$ as explained above, for all $i$. So $E[X] = 9.069$. (Note that we have obtained this result using linearity of expectation, even though it would be hard to argue about the events of covering the dots being independent.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, since the average over placements of the disks (for the fixed configuration of points!) is 9.069, not &lt;em&gt;all&lt;/em&gt; placements can cover ≤9 dots — at least one placement must cover all 10 dots.&lt;/p&gt;&#xA;" OwnerUserId="205" LastActivityDate="2010-10-14T04:07:38.023" CommentCount="1" />
  <row Id="6742" PostTypeId="2" ParentId="6738" CreationDate="2010-10-14T04:10:15.780" Score="2" ViewCount="" Body="&lt;p&gt;You're on the right track.  I'd break the problem up into four cases: where the number starts with 21, 42, 63 or 84.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case where you start with 21, you must have 21(EEEOO)OO in your notation. (There are four even digits, so three of them must be in the middle part of the number.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we can ask the questions: &#xA;- in how many different places can 4 go? &#xA;- in how many different places can  6 go (given that we've already placed 4?)&#xA;- in how many different places can  8 go (given that we've already placed 4 and 6)? &#xA;The answers are 5, 4, and 3 respectively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then we need to place the four remaining odd digits among the four remaining places; there are of course $4! = 24$ ways to do this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So there are a total of $5 \times 4 \times 3 \times 24 = 1440$ such numbers beginning with 21.  The same is true if we start with 63 instead of 21. When we start with 42 or 84 the result is different but the method is much the same, and adding up the results from the four cases gives 7680.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for (2), there aren't really &quot;general&quot; methods. But one thing that is good to keep in mind is trying to place the digits one at a time (as I've done above) instead of trying to do everything in one fell swoop.&lt;/p&gt;&#xA;" OwnerUserId="173" LastActivityDate="2010-10-14T04:10:15.780" CommentCount="1" />
  <row Id="6743" PostTypeId="2" ParentId="4175" CreationDate="2010-10-14T04:12:32.343" Score="2" ViewCount="" Body="&lt;p&gt;You can also use the following finite calculus formula for alternating binomial transforms.  Let $B(f(k),n)$ denote the alternating binomial transform; i.e., $$B(f(k),n) = \sum_{k=0}^n (-1)^k \binom{n}{k} f(k).$$ &#xA;Then (the formula)&#xA;$$B(f(k),n) = -B(\Delta f(k),n-1) + f(0)[n=0].$$&#xA;(Here, $\Delta f(k)$ is the finite difference $f(k+1) - f(k)$, and the expression $[n=0]$ evaluates to $1$ if $n = 0$ and $0$ otherwise.  Also, note that the first argument to $B$ is a function, while the second is a number.)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Starting with $f(k) = 1$, we have $\Delta f(k) = 0$.  Since $B(0,n-1)$ is clearly $0$, $B(1,n) = [n=0]$.  The latter is just the known formula &#xA;$$\sum_{k=0}^n (-1)^k \binom{n}{k} = [n=0].$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, continuing to take antidifferences, and using the notation $k^{\underline{m}}$ for the falling factorial $k(k-1)\cdots (k-m+1)$ as well as the power rule for finite differences $\Delta k^{\underline{m}} = m k^{\underline{m-1}}$, we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{k=0}^n (-1)^k \binom{n}{k}k = -B(1,n-1) = - [n-1=0] = -[n=1],$$&#xA;$$\sum_{k=0}^n (-1)^k \binom{n}{k} k^{\underline{2}} = -2B(k,n-1) = 2[n-1 = 1] = 2[n=2],$$&#xA;$$\sum_{k=0}^n (-1)^k \binom{n}{k} k^{\underline{3}} = -3B(k^{\underline{2}},n-1) = -6[n-1 = 2] = -6[n=3],$$&#xA;and so forth, until we get to &#xA;$$\sum_{k=0}^n (-1)^k \binom{n}{k} k^{\underline{m}} = -mB(k^{\underline{m-1}},n-1) = (-1)^m m![n=m].$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $k(k-1)\cdots (k-m+1) = \frac{k!}{(k-m)!}$, dividing both sides of this last identity by $(-1)^m m!$ proves the OP's identity.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(The finite calculus formula for alternating binomial transforms and this argument are excerpted from my paper &quot;Combinatorial sums and finite differences,&quot; &lt;em&gt;Discrete Mathematics&lt;/em&gt; 307 (24): 3130-3146, 2007.) &lt;/p&gt;&#xA;" OwnerUserId="2370" LastActivityDate="2010-10-14T04:12:32.343" />
  <row Id="6744" PostTypeId="2" ParentId="6734" CreationDate="2010-10-14T04:46:10.127" Score="1" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\rm\ \ u,v\ $ linear independent $\rm\ \Rightarrow\:\ u + v,\:u+2v,\: u+3v,\:\ldots\: $   pairwise independent&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-10-14T04:46:10.127" />
  <row Id="6745" PostTypeId="1" AcceptedAnswerId="6761" CreationDate="2010-10-14T04:59:21.177" Score="2" ViewCount="65" Body="&lt;p&gt;It's a well known fact in probability theory that if $\lbrace X_i\rbrace$ is a family of random variables, $\phi : [0,\infty) \to [0,\infty)$ satisfies $$\lim_{x \to \infty} \frac{\phi(x)}{x} = \infty \quad (*) $$ and $\sup_i E\phi(|X_i|) &amp;lt; \infty$, then the family $\lbrace X_i\rbrace$ is uniformly integrable.  I want to know whether there is a single function $\phi$ satisfying (*) such that $\sup_i E\phi(|X_i|) &amp;lt; \infty$ iff $\lbrace X_i\rbrace$ is uniformly integrable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suspect the answer is no, and that even a family of one absolutely continuous random variable should suffice.  So I conjecture the following: given any $\phi$ satisfying (*), there exists a nonnegative function $f$ such that&#xA;$$\int_0^\infty f(x)dx=1$$&#xA;$$\int_0^\infty x f(x)dx &amp;lt; \infty$$&#xA;$$\int_0^\infty \phi(x)f(x)dx = \infty.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I can't see offhand what to use for $f$.  Ideas?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nb.  This is not homework.  It was going to be, but I thought I should be able to solve it myself before assigning it to my students :)&lt;/p&gt;&#xA;" OwnerUserId="822" LastActivityDate="2010-10-14T08:13:38.290" Title="No perfect function for uniform integrability" Tags="&lt;calculus&gt;&lt;real-analysis&gt;&lt;probability-theory&gt;" AnswerCount="1" />
  <row Id="6746" PostTypeId="1" AcceptedAnswerId="6747" CreationDate="2010-10-14T05:07:39.977" Score="3" ViewCount="112" Body="&lt;p&gt;Let $T:X \times X \to \mathbb{R}$ be a continuous bilinear operator defined on a normed linear space $X$ s.t. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$T(\alpha x + \beta y,z) = \alpha T(x,z) + \beta T(y,z)$) and $T(x,y) = T(y,x)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does there exist a constant $C$ s.t. $||T(x,y)|| \leq C$ $||x||$ $||y|| \forall x,y$? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that the result is true if $X$ and $Y$ are complete spaces, by using the uniform boundedness principle on $T$ as a continuous function of x for fixed y (and/or the other way around). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I'm not sure if completeness is necessary, since it is true that a continuous linear operator $T: X \to \mathbb{R}$ has the property $||T(x)|| \leq C ||x|| \forall x$ on any normed linear space $X$ (although linear and bilinear operators are not exactly the same). &lt;/p&gt;&#xA;" OwnerUserId="1736" LastEditorUserId="1736" LastEditorDisplayName="" LastEditDate="2010-10-14T13:41:57.733" LastActivityDate="2010-10-14T13:41:57.733" Title="Boundedness of Continuous Bilinear Operators" Tags="&lt;functional-analysis&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" />
  <row Id="6747" PostTypeId="2" ParentId="6746" CreationDate="2010-10-14T06:09:29.650" Score="2" ViewCount="" Body="&lt;p&gt;I think you can show this using the same argument as in the continuous linear operators.&#xA;Since T is continuous, then $U=T^{-1}( (-1,1) )$ is open and contains $(0,0)$. find a c&gt;0 small enough such that if $|x|,|y|\leq c$ then $(x,y)\in U$ and then for a general point (x,y) you have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$|T(x,y)| = |T(\frac{c x}{|x|} \frac {|x|}{c}, \frac{c y}{|y|} \frac {|y|}{c} )| = \frac {|x||y|}{c^2} |T(\frac{c x}{|x|} , \frac{c y}{|y|})| \leq \frac {|x||y|}{c^2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;if x=0 or y=0 then T(x,y)=0 so you can use the argument above for $x,y \neq 0$. &lt;/p&gt;&#xA;" OwnerUserId="2125" LastActivityDate="2010-10-14T06:09:29.650" CommentCount="1" />
  <row Id="6748" PostTypeId="1" AcceptedAnswerId="6783" CreationDate="2010-10-14T06:12:06.323" Score="2" ViewCount="237" Body="&lt;p&gt;I'm trying to wrap my brain around adjoint functors. One of the examples I've seen is the categories $\bf IntLE \bf = (\mathbb{Z}, ≤)$ and $\bf RealLE \bf = (\mathbb{R}, ≤)$, where the ceiling functor $ceil : \bf RealLE \rightarrow IntLE$ is left adjoint to the inclusion functor $incl : \bf IntLE \rightarrow RealLE$. I want to check that the following are true, as they seem to be:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$floor : \bf RealLE \rightarrow IntLE$ would be right adjoint to $incl$&lt;/li&gt;&#xA;&lt;li&gt;Between the dual categories of $\bf IntGE \bf = (\mathbb{Z}, ≥)$ and $\bf RealGE \bf = (\mathbb{R}, ≥)$, $ceil$ would be right adjoint to $incl$&lt;/li&gt;&#xA;&lt;li&gt;Between $\bf RealGE$ and $\bf IntGE$, $floor$ would be left adjoint to $incl$&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Is my understanding correct on these points?&lt;/p&gt;&#xA;" OwnerUserId="2378" LastEditorUserId="2378" LastEditorDisplayName="" LastEditDate="2010-10-14T06:25:13.353" LastActivityDate="2011-06-03T07:28:24.343" Title="Adjoint functors" Tags="&lt;category-theory&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="1" />
  <row Id="6749" PostTypeId="1" AcceptedAnswerId="6754" CreationDate="2010-10-14T06:18:15.003" Score="2" ViewCount="236" Body="&lt;p&gt;First of all, i'll confess i'm no math geek. I'm from Stackoverflow, but this question seemed more apt here, so i decided to ask you guys :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, i know noone has discovered (or ever will) a Polynomial that generates Prime Numbers. But i've read about Curve Fitting (or Polynomial Fitting) so i was wondering if there was a way, we could have a simple n-degree Polynomial that could generate the first 1000 (or X) primes accurately.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't need it to generate all the primes, maybe just upto 1 million, since we already have the data, can we deduce that polynomial?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How big will be the polynomial for it to be accurate? Could you give an example for the first 100 primes? Am i just plain naive? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in Advance. :)&lt;/p&gt;&#xA;" OwnerUserId="2214" LastActivityDate="2010-10-14T13:02:22.380" Title="The Prime Polynomial : Generating Prime Numbers." Tags="&lt;prime-numbers&gt;&lt;polynomials&gt;" AnswerCount="2" CommentCount="9" />
  <row Id="6750" PostTypeId="1" AcceptedAnswerId="6756" CreationDate="2010-10-14T06:18:16.373" Score="4" ViewCount="346" Body="&lt;p&gt;What is the difference complete metric space and a closed set?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can a set be closed but not complete?&lt;/p&gt;&#xA;" OwnerUserId="2504" LastActivityDate="2010-10-14T07:13:14.613" Title="Difference between complete and closed set." Tags="&lt;real-analysis&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="6751" PostTypeId="1" CreationDate="2010-10-14T06:21:55.907" Score="0" ViewCount="65" Body="&lt;p&gt;Let $g_{ij}$ be the components of a symmetric rank-2 positive definite tensor (metric on a Riemannian manifold). Let ${C^i}$ and ${ \beta ^i }$ be components of a vector field on it, the former of which is a functions of a variable $t$ and let $N$ and $M$ be two positive constants. Now let the following inequality hold,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sqrt {g_{ij} \left( \frac{dC^i}{dt} + \beta ^i \right) \left( \frac{dC^j}{dt} + \beta ^j  \right)} &amp;lt; N &amp;lt; M    $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And there exists a constant $B$ such that $\sqrt{g_{ij} \beta ^i  \beta ^j}$ is uniformly bounded by $B$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then apparently the following is true,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sqrt {g_{ij} \frac{dC^i}{dt} \frac{dC^j}{dt} } \leq N + \sqrt{g_{ij} \beta ^i  \beta ^j} \leq  M + B $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How does this follow?&lt;/p&gt;&#xA;" OwnerUserId="965" LastEditorUserId="965" LastEditorDisplayName="" LastEditDate="2010-10-14T06:35:43.323" LastActivityDate="2010-11-13T08:44:19.097" Title="An inequality about norms of vector fields on Riemannian manifolds" Tags="&lt;analysis&gt;&lt;differential-geometry&gt;" AnswerCount="1" />
  <row Id="6754" PostTypeId="2" ParentId="6749" CreationDate="2010-10-14T06:51:09.463" Score="5" ViewCount="" Body="&lt;p&gt;While Robin was commenting, I cooked up a Mathematica one-liner which gives you the Lagrange interpolation polynomial for the first n primes:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;f[n_] := Sum[Prime[k] Product[x-m,{m,Join[Range[1,k-1], Range[k+1,n]]}] / Product[k-m,{m,Join[Range[1,k-1], Range[k+1,n]]}] ,{k,1,n}] // Expand&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;And I don't think anyone would call the results &quot;simple&quot;. For n=100 it's too big to paste here, but n=15 gives&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4748 - (1752758563*x)/120120 + (71043957851*x^2)/3783780 - &#xA; (4320411427*x^3)/316800 + (378496362427*x^4)/59875200 - &#xA; (7239131749*x^5)/3628800 + (11528263*x^6)/25920 - &#xA; (2075348983*x^7)/29030400 + (5090997277*x^8)/609638400 - &#xA; (977071*x^9)/1382400 + (743507*x^10)/17418240 - (568871*x^11)/319334400 + &#xA; (2207*x^12)/45619200 - (3151*x^13)/4151347200 + (89*x^14)/17435658240&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2010-10-14T06:51:09.463" CommentCount="3" />
  <row Id="6755" PostTypeId="2" ParentId="6340" CreationDate="2010-10-14T07:01:25.227" Score="0" ViewCount="" Body="&lt;p&gt;$\newcommand\RR{\mathbb{R}}$&#xA;$\newcommand\NN{\mathbb{N}}$&#xA;$\newcommand\e{\varepsilon}$&#xA;$\newcommand\lbg{\lambda}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $f: \RR \to \RR$ is an analytic unbounded Lebesgue integrable function, then it must in particular satisfy this property (because it is continuous):&#xA;\[&#xA;   \forall \e,M &gt; 0 \; \exists x_0 : |x| &gt; |x_0| \Rightarrow&#xA;   \lbg(\{x \in \RR : |f(x)| &gt; M\}) &amp;lt; \e,&#xA;\]&#xA;where $\lbg$ is the usual Lebesgue measure. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then you just use that $f$ is analytic iff $\forall \text{ compact } K \subset \RR \; \exists C &gt; 0 : x \in K, n \in \NN \Rightarrow \left\vert\frac{\partial^n f}{\partial x^n}(x)\right\vert &#xA;\leq C^{n+1}n!$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But it is easy to see that this is impossible since $|f'|$ will have to be arbitrarily large - or in other words that you can go far enough out to the right on the x-axis that $|f|$ has to go from being larger than some big $M$ to smaller than some small $m$ on a very small interval.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is missing quite a few details, but it should be pretty easy to fill out.&lt;/p&gt;&#xA;" OwnerUserId="1269" LastEditorUserId="1269" LastEditorDisplayName="" LastEditDate="2010-10-14T07:13:28.553" LastActivityDate="2010-10-14T07:13:28.553" />
  <row Id="6756" PostTypeId="2" ParentId="6750" CreationDate="2010-10-14T07:13:14.613" Score="6" ViewCount="" Body="&lt;p&gt;A metric space is complete if every Cauchy sequence converges.  A subset $F$ of a metric space $X$ is closed if $F$ contains all of its limit points; this can be characterized by saying that if a sequence in $F$ converges to a point $x$ in $X$, then $x$ must be in $F$.  It also makes sense to ask whether a subset of $X$ is complete, because every subset of a metric space is a metric space with the restricted metric.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It turns out that a complete subspace must be closed, which essentially results from the fact that convergent sequences are Cauchy sequences.  However, closed subspaces need not be complete.  For a trivial example, start with any incomplete metric space, like the rational numbers $\mathbb{Q}$ with the usual absolute value distance.  Like every metric space, $\mathbb{Q}$ is closed in itself, so there you have a subset that is closed but not complete.  If taking the whole space seems like cheating, just take the rationals in $[0,1]$, which will be closed in $\mathbb{Q}$ but not complete.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $X$ is a complete metric space, then a subset of $X$ is closed if and only if it is complete.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-10-14T07:13:14.613" CommentCount="2" />
  <row Id="6757" PostTypeId="1" AcceptedAnswerId="6798" CreationDate="2010-10-14T07:45:32.363" Score="3" ViewCount="234" Body="&lt;p&gt;What is the continued fraction for $\displaystyle\sum_{i=1}^n\frac{1}{2^{2^i}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems to be &quot;almost&quot; periodic, but I can't figure out the exact way to express it.&lt;/p&gt;&#xA;" OwnerUserId="2505" LastActivityDate="2010-10-16T15:56:23.260" Title="Continued Fraction of an Infinite Sum" Tags="&lt;number-theory&gt;&lt;continued-fractions&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="3" />
  <row Id="6758" PostTypeId="1" AcceptedAnswerId="6765" CreationDate="2010-10-14T07:55:24.650" Score="3" ViewCount="235" Body="&lt;p&gt;I'm still making my way along in Niven's Intro to Number Theory, and the title problem is giving me a little trouble near the end, and I was hoping someone could help get me through it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now $x^8\equiv 16\pmod{2}$ is solvable with $x\equiv 0\pmod{2}$, so I assume $p$ is an odd prime. From a theorem earlier in the text, &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If $p$ is a prime and $(a,p)=1$, then the congruence $x^n\equiv a\pmod{p}$ has $(n,p-1)$ solutions or no solution according as $a^{(p-1)/(n,p-1)}\equiv 1\pmod{p}$ or not.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So since $(16,p)=1$, the problem reduces to showing that $16^{(p-1)/(8,p-1)}\equiv 1\pmod{p}$ holds for all $p$. I note that $(8,p-1)$ can only take values $2,4,8$. For $2$, the above equivalence is then $4^{p-1}\equiv 1\pmod{p}$, which is true by Fermat's little Theorem. For $4$, it is then $2^{p-1}\equiv 1\pmod{p}$, which again holds by FlT. However, the case where $(8,p-1)=8$ is throwing me off. At best I see that $16^{(p-1)/8}\equiv 2^{(p-1)/2}\pmod{p}$, but I'm not sure how to show this is congruent to $1$ modulo $p$. Maybe there's a more elegant way to do it without looking at cases. Thanks for any insight.&lt;/p&gt;&#xA;" OwnerUserId="1583" LastActivityDate="2010-10-14T15:13:46.610" Title="Showing $x^8\equiv 16 \pmod{p}$ is solvable for all primes $p$." Tags="&lt;number-theory&gt;&lt;modular-arithmetic&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="6759" PostTypeId="2" ParentId="98" CreationDate="2010-10-14T08:01:40.487" Score="2" ViewCount="" Body="&lt;p&gt;The book is Coffee in Memphis by Bollobas. It's the first problem, and there are loads more :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0521693950&quot; rel=&quot;nofollow&quot;&gt;http://www.amazon.com/Art-Mathematics-Coffee-Time-Memphis/dp/0521693950&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2506" LastActivityDate="2010-10-14T08:01:40.487" />
  <row Id="6760" PostTypeId="2" ParentId="6751" CreationDate="2010-10-14T08:11:13.687" Score="2" ViewCount="" Body="&lt;p&gt;This is simply the triangle inequality:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \lVert\frac{dC}{dt}\rVert = \lVert\frac{dC}{dt} + \beta - \beta\rVert \le \lVert\frac{dC}{dt} + \beta \rVert + \lVert-\beta\rVert &amp;lt; N + \lVert\beta\rVert &amp;lt; M + B$$&lt;/p&gt;&#xA;" OwnerUserId="434" LastActivityDate="2010-10-14T08:11:13.687" CommentCount="1" />
  <row Id="6761" PostTypeId="2" ParentId="6745" CreationDate="2010-10-14T08:13:38.290" Score="2" ViewCount="" Body="&lt;p&gt;By hypothesis, for any positive $k$, there is an $N_k$ such that $\phi(x) &gt; kx$ if $x &gt; N_k$.&#xA;We can of course assume that $(N_k)$ is an increasing sequence, and so it splits the real line into intervals, which we furthermore can assume are of length at least one (say).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now take $f$ to be piecewise constant: zero if $0 &amp;lt; x &amp;lt; N_1$, and chosen on the other intervals such&#xA;that $\int_{N_k}^{N_{k+1}} x f(x) dx = \frac{c}{k^2}$, where $c$ is just a&#xA;normalization constant (since you require $\int f(x) dx=1$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(That is, choose &#xA;$$ f(x) = \frac{c/k^2}{\int_{N_k}^{N_{k+1}} x dx}&#xA;= \frac{2c}{k^2 (N_{k+1}^2-N_k^2)} $$&#xA;for $N_k \le x &amp;lt; N_{k+1}$.&#xA;Then, because we arranged for the intervals to be sufficiently long,&#xA;$\int_0^{\infty} f(x) dx$ reduces to a &lt;em&gt;convergent&lt;/em&gt; sum.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then&#xA;$$ \int_0^{\infty} \phi(x) f(x) dx \ge \sum_{k\ge 1} \int_{N_k}^{N_{k+1}} kx f(x) dx = \sum_{k\ge 1} k \frac{c}{k^2} = \infty.$$&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2010-10-14T08:13:38.290" CommentCount="1" />
  <row Id="6762" PostTypeId="2" ParentId="6732" CreationDate="2010-10-14T08:25:56.257" Score="1" ViewCount="" Body="&lt;p&gt;I'll prove that every point of $G$ in the interior of $F$ is not an extreme point of $G$. I'll assume that $N&gt;1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;LEMMA.&lt;/em&gt; There is a vector $v\neq 0$ such that $h'v=0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Proof.&lt;/em&gt; Since $N&gt;1$ there is a vector $u$ which is not a multiple of $h$. Let&#xA;$$v=u-\left({{h'u}\over{h'h}}\right)h$$&#xA;Then $h'v=0$. Since $u$ is not a multiple of $h$, $v\neq 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Answer&lt;/em&gt;. Suppose $x \in G$ and $x$ is also in the interior of $F$. Let $v$ be as in the lemma above. Then there is a small enough $\lambda$ such that $(x \pm \lambda v) \in F$. But $h'(x \pm \lambda v)=h'x$ so $(x \pm \lambda v) \in G$. But,&#xA;$$x= {(x + \lambda v) \over 2} + {(x - \lambda v) \over 2}$$&#xA;So $x$ is not an extreme point of $G$.&lt;/p&gt;&#xA;" OwnerUserId="1195" LastActivityDate="2010-10-14T08:25:56.257" />
  <row Id="6763" PostTypeId="1" AcceptedAnswerId="6773" CreationDate="2010-10-14T09:07:02.870" Score="1" ViewCount="416" Body="&lt;p&gt;In measure theory there are three fundamentally related theorems about exchanging limits and integrals: Fatou's lemma, Lebesgue's Dominated Convergence Theorem, and Monotone Convergence Theorem. It is difficult to prove any of these from scratch, but once you have one, the others are easier.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is, for those who have learned these theorems: which one do you prefer to prove first? Difficulty, length, and, perhaps most importantly, how enlightening each path is are the key considerations. I suppose you could also phrase the question: if you were teaching a class in what order would you prove these theorems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've read through all of the proofs and there doesn't seem to be a big difference, but perhaps someone can shed some new light on this question.&lt;/p&gt;&#xA;" OwnerUserId="2298" LastActivityDate="2010-10-14T13:00:41.277" Title="Theorems in Measure Theory: Fatou's Lemma, Lebesgue DCT, Monotone CT" Tags="&lt;soft-question&gt;&lt;measure-theory&gt;" AnswerCount="4" CommentCount="1" FavoriteCount="1" />
  <row Id="6764" PostTypeId="2" ParentId="457" CreationDate="2010-10-14T09:47:58.157" Score="6" ViewCount="" Body="&lt;p&gt;There's a whole room devoted to physical demonstrations of mathematical concepts in the Museum of Science in Boston. I'll list the ones I can remember, and others who have been there please feel free to add to the list.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;A little train that goes around on a track that's on a mobius strip.&lt;/li&gt;&#xA;&lt;li&gt;A &quot;Pachinko&quot; type machine where the balls fall into different slots and form a bell curve as the slots fill up.&lt;/li&gt;&#xA;&lt;li&gt;A pendulum that is free to swing forward and backward as well as left and right. At the bottom is a container of sand with a small hole in it, so as the sand leaks out, it forms &lt;a href=&quot;http://en.wikipedia.org/wiki/Lissajous_curve&quot; rel=&quot;nofollow&quot;&gt;Lissajous curves&lt;/a&gt;. &lt;/li&gt;&#xA;&lt;li&gt;Something about the Riemann Zeta function that I didn't really understand.&lt;/li&gt;&#xA;&lt;li&gt;A straight, solid bar, rotated around an axis that is skew to the bar itself, that fits right through a hyperbola-shaped slot. I'm pretty sure the Exploratorium in San Francisco has this too, and here's a &lt;a href=&quot;http://youtube.com/watch?v=OJGkdO9gcdY&quot; rel=&quot;nofollow&quot;&gt;video of a similar exhibit&lt;/a&gt; thanks to commenter Rahul.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="496" LastEditorUserId="496" LastEditorDisplayName="" LastEditDate="2010-10-18T05:13:16.207" LastActivityDate="2010-10-18T05:13:16.207" CommunityOwnedDate="2010-10-14T09:47:58.157" CommentCount="1" />
  <row Id="6765" PostTypeId="2" ParentId="6758" CreationDate="2010-10-14T09:57:43.250" Score="2" ViewCount="" Body="&lt;p&gt;One way is to use the &lt;a href=&quot;http://en.wikipedia.org/wiki/Legendre_symbol&quot; rel=&quot;nofollow&quot;&gt;Legendre symbol&lt;/a&gt; identity $2^{(p-1)/2} \equiv (\frac{2}{p}) \equiv (-1)^{(p^2-1)/8} \pmod p$ (for odd primes p), keeping in mind that if (8,p-1)=8 then $p \equiv 1 \pmod 8$.&lt;/p&gt;&#xA;" OwnerUserId="139" LastActivityDate="2010-10-14T09:57:43.250" CommentCount="2" />
  <row Id="6766" PostTypeId="2" ParentId="6763" CreationDate="2010-10-14T10:06:21.260" Score="0" ViewCount="" Body="&lt;p&gt;I prefer the direction: Dominated convergence theorem -&gt; Beppo-Levi (monotone convergence) -&gt; Fatou.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This direction requires more machinery (like Egoroff and the absolute continuity of the Lebesgue integral) but it is in my opinion more tidy. A book that uses this approach is &lt;a href=&quot;http://books.google.com/books?id=CoSIe7h5mTsC&amp;amp;printsec=frontcover&amp;amp;dq=Bogachev&amp;amp;hl=nl&amp;amp;ei=ENW2TIH0IsqZOozSxOwJ&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=1&amp;amp;ved=0CCcQ6AEwAA#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;Bogachev - Measure Theory&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: This way the later theorems are just &quot;corollaries&quot; of DCT, the &quot;biggest&quot; one!&lt;/p&gt;&#xA;" OwnerUserId="1120" LastEditorUserId="1120" LastEditorDisplayName="" LastEditDate="2010-10-14T12:14:34.530" LastActivityDate="2010-10-14T12:14:34.530" />
  <row Id="6767" PostTypeId="2" ParentId="6758" CreationDate="2010-10-14T10:36:53.263" Score="0" ViewCount="" Body="&lt;p&gt;I usually set this as an exercise when teaching Number Theory.&#xA;My hint is to ask the students: what are the solutions of&#xA;$z^8=16$ in the complex numbers?&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-10-14T10:36:53.263" />
  <row Id="6768" PostTypeId="2" ParentId="6763" CreationDate="2010-10-14T10:41:22.527" Score="0" ViewCount="" Body="&lt;p&gt;I believe I learned the monotone convergence theorem first, then dominated convergence, then Fatou.  You could argue that this sequence orders the theorems in increasing order of hypothesis complexity.  &lt;/p&gt;&#xA;" OwnerUserId="247" LastActivityDate="2010-10-14T10:41:22.527" />
  <row Id="6769" PostTypeId="2" ParentId="6763" CreationDate="2010-10-14T10:45:08.260" Score="0" ViewCount="" Body="&lt;p&gt;In fact i would add something more. For me it would be the Bounded Convergence theorem, then the Fatou's lemma and then the Monotone convergence theorem followed by Dominated convergence theorem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the book by Royden has a very good way proving these theorems&lt;/p&gt;&#xA;" LastActivityDate="2010-10-14T10:45:08.260" />
  <row Id="6770" PostTypeId="1" CreationDate="2010-10-14T10:49:41.527" Score="1" ViewCount="63" Body="&lt;p&gt;Let $u$ be a fixed vector of length $K$ and $A$ be a matrix of all positive numbers of size $K \times K$. Let $V$ be a set of vectors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $V(\epsilon) \subseteq V$ be a set of vectors of length $K$ such that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$0 \le \sup_{v \in V(\epsilon)} u \cdot v &amp;lt; \epsilon$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For an arbitrary vector $w \in V$, define $L(w) = \sum_{i=1,j=1}^K (u_i u_j / A_{ij}) \times w_i w_j$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Claim: $\sup_{v \in V(\epsilon)} L(v) &amp;lt; C*\epsilon^r$ for some $C,r &gt; 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proof: $L(v)$ can also be rewritten as:&#xA;$L(v) = A \colon (u \otimes v) (u \otimes v)^T$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\otimes$ is Hadamard product and $\colon$ is the Frobenius inner product.&#xA;We also know that the Frobenis inner product $A:B = tr(A^t B)$, therefore we have:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$L(v) = tr(A^T (u \otimes v) (u \otimes v)^T)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which equals (because the reordering of the matrices is a cyclic permutation):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$L(v) = tr((u \otimes v)^T A^T (u \otimes v))$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here I am stuck. Not sure how to deduce the rest, if it is at all possible.&#xA;Basically, if I show the following, then the proof is done:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Claim??: let $A$ be a matrix of positive numbers. let $u$ be a vector of all positive numbers. There is a $C$ and an $r$ such that: let $v \in V$ be a vector. let $w = u \otimes v$ be a vector such that $0 &amp;lt; \sum_i w_i &amp;lt; \epsilon$. then $0 \le tr(w A^T w) \le C\epsilon^r$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I don't think the above claim is true.&#xA;Yet, I have more structure to my problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$V$ in fact has the form $V = ${$ (\log \frac{x}{x^\ast}, \log \frac{(1-x)}{(1-x^\ast)}) \mid x \in [0,1] $} for an $x^\ast \in [0,1]$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(i.e. $K=2$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or for $K = 4$ we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$V = ${$ (\log \frac{x}{x^\ast}, \log \frac{(1-x)}{(1-x^\ast)},\log \frac{y}{y^\ast}, \log \frac{(1-y)}{(1-y^\ast)}) \mid y,x \in [0,1] $} for an $x^\ast,y^\ast \in [0,1]$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;etc. (we can have any even $K$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;could it be true for this $V$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for any insight.&lt;/p&gt;&#xA;" OwnerUserId="7076" LastActivityDate="2010-10-14T10:49:41.527" Title="can we prove a connection between sum of numbers and $L_2$-like norm?" Tags="&lt;linear-algebra&gt;" />
  <row Id="6771" PostTypeId="2" ParentId="2427" CreationDate="2010-10-14T11:32:01.850" Score="6" ViewCount="" Body="&lt;p&gt;What do we mean by  'sporadic' ?  I suggest that we may find some extension of the&#xA;Chevalley programme (Tohoku J 1950+)  that brings in the sporadics. This is about&#xA;construction - not the classification as we know it.&lt;/p&gt;&#xA;" LastActivityDate="2010-10-14T11:32:01.850" CommentCount="2" />
  <row Id="6772" PostTypeId="2" ParentId="6648" CreationDate="2010-10-14T11:39:27.663" Score="1" ViewCount="" Body="&lt;p&gt;Joseph Gallian has decrypted many of the US state license plate and driver's license codes.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://books.google.com/books?id=PD0clAlF8O4C&amp;amp;pg=PA27&quot; rel=&quot;nofollow&quot;&gt;http://books.google.com/books?id=PD0clAlF8O4C&amp;amp;pg=PA27&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think he used Markov chain models.  As whuber mentioned your problem is similar to the German tanks for which the subject reference is &quot;extreme value statistics&quot;.&lt;/p&gt;&#xA;" OwnerUserId="467" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-10-14T11:45:52.643" LastActivityDate="2010-10-14T11:45:52.643" />
  <row Id="6773" PostTypeId="2" ParentId="6763" CreationDate="2010-10-14T13:00:41.277" Score="2" ViewCount="" Body="&lt;p&gt;I've generally seen MCT -&gt; Fatou -&gt; DCT.  MCT is nice if the integral is defined as the supremum of the integrals of all simple functions less than $f$.  Fatou points out that you can lose mass when passing to the limit, but cannot gain it.  And DCT is nice to prove with two applications of Fatou, since turning your head upside down shows that you cannot gain mass either positively or negatively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I disagree with Jonas's idea that DCT is the &quot;biggest&quot; one, since it doesn't speak about functions not in $L^1$, which the others do; this is often very important.  Also, I see the hypothesis of the DCT as somewhat ad hoc.  To my mind, the &quot;biggest&quot; one is the Vitali convergence theorem, whose hypothesis is uniform integrability, which is necessary and sufficient.  But since it is more complicated it is often skipped.&lt;/p&gt;&#xA;" OwnerUserId="822" LastActivityDate="2010-10-14T13:00:41.277" CommentCount="1" />
  <row Id="6774" PostTypeId="1" AcceptedAnswerId="6776" CreationDate="2010-10-14T13:01:13.723" Score="1" ViewCount="109" Body="&lt;p&gt;Let $f: X \rightarrow Y$ be a local homeomorphism with X, Y connected, locally path connected, Hausdorff and with X also compact. Then f is also a covering with finite fibers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know how to show that the fibers are finite.&#xA;Given that f is a surjection, I know how to show that f is a covering map.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;How do I show that f is surjective?&lt;/strong&gt; &lt;/p&gt;&#xA;" OwnerUserId="1907" LastActivityDate="2010-10-14T17:05:11.953" Title="Local homeomorphism and coverings" Tags="&lt;topology&gt;" AnswerCount="1" />
  <row Id="6775" PostTypeId="2" ParentId="6749" CreationDate="2010-10-14T13:02:22.380" Score="3" ViewCount="" Body="&lt;p&gt;From &lt;a href=&quot;http://mathworld.wolfram.com/Prime-GeneratingPolynomial.html&quot; rel=&quot;nofollow&quot;&gt;Mathworld&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, there exists a polynomial  in 10 variables with integer coefficients such that the set of primes equals the set of positive  values of this polynomial obtained as the variables run through all nonnegative integers, although it is really a set of Diophantine equations in disguise (Ribenboim 1991). Jones, Sato, Wada, and Wiens have also found a polynomial of degree 25 in 26 variables whose positive values are exactly the prime numbers (Flannery and Flannery 2000, p. 51).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, the primes do not come out in order, so this will not help for what you want.  But it is interesting.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-10-14T13:02:22.380" />
  <row Id="6776" PostTypeId="2" ParentId="6774" CreationDate="2010-10-14T13:07:55.590" Score="3" ViewCount="" Body="&lt;p&gt;I think I have it. The image of X under f is open (use &quot;local homeomorphism&quot;) and closed (use &quot;compact&quot; and &quot;Hausdorff&quot;) in Y and since Y is connected this shows that the image is the whole of Y. &lt;/p&gt;&#xA;" OwnerUserId="1907" LastEditorUserId="1907" LastEditorDisplayName="" LastEditDate="2010-10-14T17:05:11.953" LastActivityDate="2010-10-14T17:05:11.953" />
  <row Id="6777" PostTypeId="1" CreationDate="2010-10-14T13:19:33.000" Score="4" ViewCount="496" Body="&lt;p&gt;What is motivation behind the definition of a complete metric space?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Intuitively,a complete metric is complete if they are no points missing from it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How does the definition of completeness (in terms of convergence of cauchy sequences) show that?&lt;/p&gt;&#xA;" OwnerUserId="2504" LastEditorUserId="274" LastEditorDisplayName="" LastEditDate="2010-10-14T16:04:20.357" LastActivityDate="2010-10-15T11:08:07.987" Title="Motivation behind the definition of complete metric space." Tags="&lt;metric-spaces&gt;" AnswerCount="5" CommentCount="1" />
  <row Id="6778" PostTypeId="2" ParentId="6777" CreationDate="2010-10-14T13:21:31.397" Score="2" ViewCount="" Body="&lt;p&gt;When working with metric spaces, we tend to deal with sequences &lt;em&gt;a lot&lt;/em&gt;. Completeness is just the condition that those that &lt;em&gt;should&lt;/em&gt; converge actually converge.&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2010-10-14T13:21:31.397" CommentCount="6" />
  <row Id="6779" PostTypeId="1" AcceptedAnswerId="6781" CreationDate="2010-10-14T13:21:41.100" Score="0" ViewCount="78" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Sum of a set of sphenic numbers can't be equal to the sum of &lt;em&gt;any other&lt;/em&gt; set of sphenic numbers.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;By that I meant, Say S is the set of sphenic numbers. Let, S$_1$ $\subset$ S.&#xA;Then, there is no such S$_2$ $\subset$ S so that,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;S$_1$ $\neq$ S$_2$ AND $\sum S_1$ $=$ $\sum S_2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Question 1 : Is this statement correct?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Question 2 : Is this formulation mathematically right? I mean even if the statement is wrong, is the way that I expressed it conveys mathematical notations/rules etc. Or, how a mathematician would write it if s/he intended to convey the same message?&lt;/p&gt;&#xA;" OwnerUserId="1869" LastActivityDate="2010-10-14T16:18:46.650" Title="Conjecture about the set of Sphenic numbers" Tags="&lt;number-theory&gt;" AnswerCount="2" />
  <row Id="6780" PostTypeId="2" ParentId="6777" CreationDate="2010-10-14T13:23:36.843" Score="3" ViewCount="" Body="&lt;p&gt;I tend to think of it this way: A Cauchy sequence in any metric space is one such that for all $\varepsilon &gt; 0$ the tail of the sequence is &lt;em&gt;eventually&lt;/em&gt; in some $\varepsilon$-ball (not necessarily around the same point for all $\varepsilon$). In other words the only way a Cauchy sequence can fail to converge is if the limit is somehow &quot;not there&quot;. Hence the choice of the word &quot;complete&quot; because all of the limits that should be there, are in fact there.&lt;/p&gt;&#xA;" OwnerUserId="1269" LastEditorUserId="1269" LastEditorDisplayName="" LastEditDate="2010-10-14T16:03:06.003" LastActivityDate="2010-10-14T16:03:06.003" CommentCount="2" />
  <row Id="6781" PostTypeId="2" ParentId="6779" CreationDate="2010-10-14T13:30:08.787" Score="3" ViewCount="" Body="&lt;p&gt;1) The statement is not correct.  230+1310=231+1309.  &lt;a href=&quot;http://en.wikipedia.org/wiki/Sphenic_number&quot; rel=&quot;nofollow&quot;&gt;Sphenic numbers&lt;/a&gt; gives the first sequential pair (and implies there are more) and the first run of three.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) I'm not a mathematician, but the formulation seems fine to me.  You might define Sphenic numbers so people don't have to look it up&lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2010-10-14T16:18:46.650" LastActivityDate="2010-10-14T16:18:46.650" CommentCount="6" />
  <row Id="6782" PostTypeId="2" ParentId="6777" CreationDate="2010-10-14T14:11:13.283" Score="5" ViewCount="" Body="&lt;p&gt;To &quot;fill the holes&quot; or &quot;add the missing points&quot; would presumably mean embedding the metric space as a subspace of a larger metric space.  To avoid trivialities like placing a line inside the plane, it is required (and it appears to be the only sensible interpretation) that the given space is &lt;em&gt;dense&lt;/em&gt; in the larger space: every neighborhood of a point in the larger space contains points of the smaller space.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A complete metric space is one to which nothing new can be added in this way.  The &quot;no holes&quot; definition of completeness is then equivalent to completeness defined using Cauchy sequences.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A more precise term than holes would &quot;punctures&quot;.  Holes also have a topological meaning such as the hole surrounded by an annulus in the plane, the hole in a torus, or the keyhole in a lock.&lt;/p&gt;&#xA;" OwnerUserId="467" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-10-14T17:47:17.680" LastActivityDate="2010-10-14T17:47:17.680" CommentCount="3" />
  <row Id="6783" PostTypeId="2" ParentId="6748" CreationDate="2010-10-14T14:39:47.683" Score="2" ViewCount="" Body="&lt;p&gt;Given categories $\mathcal{C}$ and $\mathcal{D}$, and functors $\mathbf{F}\colon\mathcal{C}\to\mathcal{D}$ and $\mathbf{U}\colon\mathcal{D}\to\mathcal{C}$, $\mathbf{F}$ is the left adjoint of $\mathbf{U}$ if and only if for every objects $C\in\mathcal{C}$ and $D\in\mathcal{D}$ there is a natural bijection between $\mathcal{C}(C,\mathbf{U}(D))$ and $\mathcal{D}(\mathbf{F}(C),D)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me use $\leq$ and $\geq$ for the relation among reals, and $\preceq, \succeq$ for the relation among integers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $ceil$ to be the left adjoint of the inclusion functor, you would need that for all real numbers $r$ and all integers $z$, $\lceil r\rceil \preceq z$ if and only if $r\leq z$. This holds, so you do have an ajunction (this works because in these categories, the morphism set &lt;strong&gt;IntLE&lt;/strong&gt;$(a,b)$ is empty if $a\not\preceq b$, and contains a unique arrow if $a\preceq b$; and similarly with &lt;strong&gt;RealLE&lt;/strong&gt;; so you get a natural bijection between the sets of arrows if and only if they are either both empty or both are singletons at the same time).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $floor$ to be a right adjoint to $incl$, you would need that for all real numbers $r$ and all integers $z$, $z\preceq \lfloor r\rfloor$ if and only if $z\leq r$, which again is true; so $floor$ is a right adjoint to the inclusion functor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $ceil$ to be the right adjoint to the inclusion functor in the dual categories, you would need $z\succeq \lceil r \rceil$ if and only if $z\geq r$; and for $floor$ to be the left adjoint, you would need $\lfloor r\rfloor \succeq z$ if and only if $r\geq z$. Both hold, so your assertions 1 through 3 are correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S. Let me second Mariano's suggestion in the comments to keep in mind the case of the underlying set functor and the free group functor for thinking about right and left adjoints. I find myself going back to those two every time I need to remind myself of how things work with adjoints, what adjoints respect or don't respect, and especially when thinking about some of the other equivalent definitions, in particular the one in terms of the unit and co-unit of the adjunction (which are natural transformations between the identity functors and the functors $\mathbf{FU}$ and $\mathbf{UF}$). &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="308" LastEditorDisplayName="" LastEditDate="2011-06-03T07:28:24.343" LastActivityDate="2011-06-03T07:28:24.343" CommentCount="2" />
  <row Id="6784" PostTypeId="1" CreationDate="2010-10-14T14:50:01.800" Score="2" ViewCount="324" Body="&lt;p&gt;I need to calculate all the numbers in a certain row of Pascal's triangle. Obviously, this is easy to do with combinatorics. However, what do you do when you need to estimate all the numbers in, say, the 100000th row of Pascal's triangle? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any way to estimate the number so that the costly multiplications and divisions of binomials can be avoided? I'm already estimating factorials with Stirling's formula, but it still takes a number of seconds to calculate only one number - and I need about 100000/2 (since a row is symmetric).&lt;/p&gt;&#xA;" OwnerUserId="2471" LastActivityDate="2010-10-14T17:35:52.343" Title="Estimating a certain row of Pascal's triangle" Tags="&lt;binomial-coefficients&gt;&lt;big-numbers&gt;" AnswerCount="4" CommentCount="5" />
  <row Id="6785" PostTypeId="2" ParentId="6758" CreationDate="2010-10-14T15:01:07.857" Score="2" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\rm\ \ \ x^8 - 16\ =\ (x^2 - 2)\: (x^2 + 2)\: (x^4 + 4)\:.\ \:$ If the first two factors have no roots in $\rm\ \mathbb Z/p\ $ then $\:2, -2\:$ are nonsquares so their product $-4\: $ is a square, so $\rm\: i = \sqrt{-1} \in \mathbb Z/p\:$. Thus the third factor has a root since $\rm\ x^4 + 4\ $ has roots $\rm\: \pm 1\pm i\:$.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-10-14T15:13:46.610" LastActivityDate="2010-10-14T15:13:46.610" CommentCount="2" />
  <row Id="6786" PostTypeId="2" ParentId="6784" CreationDate="2010-10-14T15:14:06.207" Score="4" ViewCount="" Body="&lt;p&gt;Are you computing the factorials and &lt;em&gt;then&lt;/em&gt; dividing?  Don't do that.  You can combine the estimates you get from Stirling's formula into &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$${n+m \choose n} \approx \sqrt{ \frac{m+n}{2 \pi mn} } \frac{(m+n)^{m+n}}{m^m n^n}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and even then you can optimize, for example rewriting the second factor as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left( 1 + \frac{n}{m} \right)^m \left( 1 + \frac{m}{n} \right)^n$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and, depending on the relative sizes of $m$ and $n$, applying the approximation $\left( 1 + \frac{x}{n} \right)^n \approx e^x$.  This will work if one of $m$ and $n$ are large compared to the other and in the intermediate case the above powers should be straightforward to compute.  See also Exercises 1 and 2 at &lt;a href=&quot;http://terrytao.wordpress.com/2010/01/02/254a-notes-0a-stirlings-formula/&quot; rel=&quot;nofollow&quot;&gt;this blog post by Terence Tao&lt;/a&gt; on the subject.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Depending on what kind of precision you need you should consider just working with the logarithms and not with the numbers directly.&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-10-14T16:12:27.050" LastActivityDate="2010-10-14T16:12:27.050" CommentCount="1" />
  <row Id="6788" PostTypeId="2" ParentId="6784" CreationDate="2010-10-14T15:35:37.850" Score="0" ViewCount="" Body="&lt;p&gt;Depending on the accuracy you need, this bound might help:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left(\frac{n}{k}\right)^k \leq \binom{n}{k} \leq \left(\frac{ne}{k}\right)^k.$$&lt;/p&gt;&#xA;" OwnerUserId="2370" LastActivityDate="2010-10-14T15:35:37.850" CommentCount="1" />
  <row Id="6789" PostTypeId="2" ParentId="6784" CreationDate="2010-10-14T15:43:43.043" Score="3" ViewCount="" Body="&lt;p&gt;If you want all the numbers in one particular row, you can get each from the last with one multiply and one divide.  ${n \choose m}={n \choose m-1}*(n-m+1)/m$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For individual components, you can use the fact that the distribution is approximately normal with standard deviation $\sqrt n$ and the central coefficient is about $4^n/\sqrt n\pi$.  I'm not sure that is any faster than Qiaochu's suggestion&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-10-14T15:43:43.043" CommentCount="1" />
  <row Id="6790" PostTypeId="2" ParentId="6779" CreationDate="2010-10-14T15:44:39.123" Score="2" ViewCount="" Body="&lt;p&gt;Question 2: Instead of $\sum S_1$ you can write $$\sum_{k \in S_1} k.$$&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2010-10-14T15:44:39.123" CommentCount="3" />
  <row Id="6791" PostTypeId="1" CreationDate="2010-10-14T15:49:08.723" Score="26" ViewCount="506" Body="&lt;p&gt;A high-school student named Erna Fekete made a conjecture to me via email three years ago, &#xA;which I could not answer.  I've since lost touch with her.&#xA;I repeat her interesting conjecture here, in case anyone can provide updated&#xA;information on it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is how she phrased it.  Let $b(0) = 1$ and $b(n)= \tan( b(n-1) )$.&#xA;In other words, $b(n)$ is the repeated application of $\tan(\;)$ to 1:&#xA;$$\tan(1) = 1.56, \; \tan(\tan(1)) = 74.7, \; \tan^3(1) = -0.9, \; \ldots $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $a(n) = \lfloor b(n) \rfloor$.&#xA;Her conjecture is: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Every integer eventually appears in the $a(n)$ sequence.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This sequence is not unknown; it&#xA;is &lt;a href=&quot;http://akpublic.research.att.com/~njas/sequences/A000319&quot; rel=&quot;nofollow&quot;&gt;A000319&lt;/a&gt; in Sloane's integer sequences.&#xA;Essentially hers is a question about the orbit of 1 under repeated $\tan(\;)$-applications.&#xA;Her and my investigations at the time led us to believe it was an open problem.&lt;/p&gt;&#xA;" OwnerUserId="237" LastActivityDate="2011-01-06T15:11:25.920" Title="Fekete's conjecture on repeated applications of the tangent function" Tags="&lt;reference-request&gt;&lt;trigonometry&gt;&lt;dynamical-systems&gt;" AnswerCount="2" CommentCount="7" FavoriteCount="6" />
  <row Id="6792" PostTypeId="2" ParentId="6495" CreationDate="2010-10-14T16:20:19.650" Score="3" ViewCount="" Body="&lt;p&gt;I'm sort of just guessing here (partly based on the solution already found). By explicitly plugging in $V(t,r) = - \frac{M(r)}{R(t,r)}$, you arrive at the ordinary differential equation (for each fixed $r$) for $R$ as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \dot{R}^2 - \frac{2M}{R} = 2 E $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $M$ and $E$ are constants in time. Now, re-scale the original equation by $t = \lambda s$ and $R = \mu \rho$. Then $\partial_t R = \frac{\mu}{\lambda} \partial_s \rho$. Then you can solve $(\frac{\mu}{\lambda})^2 = \frac{2M}{\mu} = 2E$ to reduce the equation to &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \dot{\rho}^2 - \frac{1}{\rho} = 1 $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(the weights $\mu$ and $\lambda$ will, roughly speaking, give you the weights $p(r)$ and $a(r)$ in your question). Now note that this scaling degenerates if $E = 0$. In the case that $E = 0$, the equation can be solved by quadrature:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \dot{x}^2 = x^{-1} \Rightarrow \sqrt{x} dx = dt \Rightarrow x^{3/2} \sim t $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This gives the solution to the homogeneous case. In the inhomogeneous case, you take that as a sort of integrating factor: assume that $\rho^{3/2} f(\rho) \sim t$, this implies that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \dot{\rho} \left( \rho^{3/2} f(\rho) \right)' = 1 $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;we plug this into the equation, which we first re-arrange as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \dot{\rho} = \sqrt{\frac{1}{\frac{\rho}{1+\rho}}} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and we conclude that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \sqrt{\frac{\rho}{1+\rho}} = \frac{d}{d\rho}( \rho^{3/2} f(\rho)) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and you solve this by directly integrating it. I think this $f$ you find should be exactly the $F$ you wrote down above (I didn't check it myself). Note that due to the singular weight $\rho$ which degenerates as $\rho \to 0$, you will have to separately integrate in the regime where $\rho &gt; 0$ and $\rho &amp;lt; 0$. The existence and uniqueness of solution is guaranteed by the theory of Fuchsian ODEs. &lt;/p&gt;&#xA;" OwnerUserId="1543" LastActivityDate="2010-10-14T16:20:19.650" CommentCount="2" />
  <row Id="6793" PostTypeId="2" ParentId="6748" CreationDate="2010-10-14T16:23:43.427" Score="3" ViewCount="" Body="&lt;p&gt;Arturo has already posted a nice answer. I'd merely like to emphasize that such universal definitions often enable slick proofs, e.g. see below. For a much more striking example see the theorem in &lt;a href=&quot;http://math.stackexchange.com/questions/3149&quot;&gt;my post here&lt;/a&gt;, which presents a slick one-line proof of the  LCM * GCD law via their universal definitions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;LEMMA&lt;/strong&gt; $\rm\: \  \lfloor x/(mn)\rfloor\ =\ \lfloor{\lfloor x/m\rfloor}/n\rfloor\ \  $&#xA;for $\rm\ \ n &gt; 0$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof&lt;/strong&gt; $\rm\quad\quad\quad\quad\quad\quad\quad k\ \le \lfloor{\lfloor x/m\rfloor}/n\rfloor$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\quad\quad\quad\quad\iff\quad\ \ k\ \le\ \:{\lfloor x/m\rfloor}/n$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\quad\quad\quad\quad\iff\ \ nk\ \le\ \ \lfloor x/m\rfloor$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\quad\quad\quad\quad\iff\ \ nk\ \le\:\ \ \ x/m$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\quad\quad\quad\quad\iff\ \ \ \ k\ \le\:\ \ \ x/(mn)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\quad\quad\quad\quad\iff\ \ \ \  k\ \le\ \ \lfloor x/(mn)\rfloor $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Compare the above trivial proof to more traditional proofs, e.g. the &lt;a href=&quot;http://chandrumath.wordpress.com/2010/10/07/more-on-the-floor-function/&quot; rel=&quot;nofollow&quot;&gt;special case $\rm\ m = 1\ $ here.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-10-14T20:37:58.503" LastActivityDate="2010-10-14T20:37:58.503" CommentCount="1" />
  <row Id="6794" PostTypeId="1" AcceptedAnswerId="6796" CreationDate="2010-10-14T16:41:23.873" Score="10" ViewCount="586" Body="&lt;p&gt;The Cantor set is uncountable so I expect countably many of them to be able to cover $\mathbb R$, but the set has measure $0$ so countably many of them also has set of measure $0$ and thus can't cover the real line.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why/where is my intuition broken?&lt;/p&gt;&#xA;" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-10-14T16:42:27.290" LastActivityDate="2010-10-25T11:23:19.250" Title="Why can't Cantor sets cover $\mathbb{ R}$?" Tags="&lt;soft-question&gt;&lt;real-analysis&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="2" />
  <row Id="6795" PostTypeId="1" CreationDate="2010-10-14T16:41:50.467" Score="1" ViewCount="241" Body="&lt;p&gt;A &lt;a href=&quot;http://en.wikipedia.org/wiki/Perfect_number&quot; rel=&quot;nofollow&quot;&gt;perfect number&lt;/a&gt; is a number that is the sum of its proper divisors:&#xA;&lt;code&gt;28 = 1 + 2 + 4 + 7 + 14&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a name for numbers whose only proper divisors are 1, 2, and N/2? For example&lt;/p&gt;&#xA;&#xA;&lt;p&gt;6 is divisible by 1, 2, and 3&lt;/p&gt;&#xA;&#xA;&lt;p&gt;8 is divisible by 1, 2, and 4&lt;/p&gt;&#xA;&#xA;&lt;p&gt;34 is divisible by 1, 2, and 17&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a name for these numbers?&lt;/p&gt;&#xA;" OwnerUserId="2517" LastEditorUserId="1778" LastEditorDisplayName="" LastEditDate="2010-10-14T23:53:23.747" LastActivityDate="2010-10-15T18:30:03.327" Title="Is there a name for this kind of number?" Tags="&lt;number-theory&gt;&lt;terminology&gt;" AnswerCount="1" CommentCount="8" />
  <row Id="6796" PostTypeId="2" ParentId="6794" CreationDate="2010-10-14T16:57:23.117" Score="23" ViewCount="" Body="&lt;p&gt;You can't cover a complete metric space with a countable union of closed nowhere dense subsets, by Baire's theorem. Cardinality tells us very little.  Even infinite dimensional but separable Banach spaces have the same cardinality as lines, but perhaps here it is more intuitive that you will never cover such a space with countably many lines.  Similarly, you can't cover the plane with countably many lines, polygons, conic sections, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To see how Baire's theorem is more fundamental than measure here, note that the same is true for &quot;fat&quot; Cantor sets.  That is, you take out smaller intervals to obtain Cantor-like sets with positive measure.  These sets are still closed and nowhere dense, so by Baire's theorem the line is not a countable union of such sets, even though comparing measures wouldn't tell you this.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-10-14T16:57:23.117" CommentCount="1" />
  <row Id="6797" PostTypeId="2" ParentId="6794" CreationDate="2010-10-14T16:57:54.597" Score="11" ViewCount="" Body="&lt;p&gt;It's not just that the Cantor set has measure 0, it's also that the reals are a Baire space and the Cantor set is nowhere dense. E.g. take a fat Cantor set, $C$, and look at $C + \mathbb{Z}$ which has infinite measure but is still be nowhere dense.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;UPDATE: There is a &lt;a href=&quot;http://mathoverflow.net/questions/43478/is-there-a-measure-zero-set-which-isnt-meagre&quot; rel=&quot;nofollow&quot;&gt;related discussion on MO&lt;/a&gt; that could be of interest.&lt;/p&gt;&#xA;" OwnerUserId="1269" LastEditorUserId="1269" LastEditorDisplayName="" LastEditDate="2010-10-25T11:23:19.250" LastActivityDate="2010-10-25T11:23:19.250" CommentCount="1" />
  <row Id="6798" PostTypeId="2" ParentId="6757" CreationDate="2010-10-14T17:02:03.437" Score="4" ViewCount="" Body="&lt;p&gt;We can apply the following general transformation formula of a series into a&#xA;continued fraction, which one can justify (see Addendum 1 and Addendum 2) by comparing the continued fraction fundamental recurrence relations with the series partial sums recurrence:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{n=1}^{N}\dfrac{u_{n}}{v_{n}}=\dfrac{u_{1}}{v_{1}+\underset{n=1}{%&#xA;\overset{N-1}{\mathbb{K}}}\left(\left( -\dfrac{u_{n+1}}{u_{n}}v_{n}^{2}\right) /\left( v_{n+1}+\dfrac{u_{n+1}}{u_{n}}v_{n}\right)\right) }.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case, we have $u_{n}=1$, $v_{n}=2^{\left( 2^{n}\right) }$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{n=1}^{N}\dfrac{1}{v_{n}}=\dfrac{1}{4+\underset{n=1}{\overset{N-1}{%&#xA;\mathbb{K}}}\left( \left( -v_{n}^{2}\right) /\left( v_{n+1}+v_{n}\right)\right) }$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{n=1}^{N}\dfrac{1}{2^{2^{n}}}=\dfrac{1}{4+\underset{n=1}{\overset{N-1}{\mathbb{K}}}\left(\left( -2^{2^{n+1}}\right)/\left( 2^{2^{n+1}}+2^{2^{n}}\right)\right) }$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$=\dfrac{1}{4+}\dfrac{-16}{20+}\cdots \dfrac{-2^{2^{n+1}}}{%&#xA;2^{2^{n+1}}+2^{2^{n}}+}{\cdots }\dfrac{-2^{2^{N}}}{%&#xA;2^{2^{N}}+2^{2^{N-1}}}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The transformation of the series into a continued fraction is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{n=1}^{\infty}\dfrac{1}{2^{2^{n}}}=\dfrac{1}{4+\underset{n=1}{\overset{\infty}{\mathbb{K}}}\left(\left( -2^{2^{n+1}}\right)/\left( 2^{2^{n+1}}+2^{2^{n}}\right) \right) }.$$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Addendum 1&lt;/strong&gt;: The series partial sums&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$s_{n}=\sum_{k=1}^{n}\frac{u_{k}}{v_{k}}=\frac{A_{n}}{B_{n}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;verify, for $n\geq 2$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$s_{n}=s_{n-1}+\frac{u_{n}}{v_{n}}=\frac{A_{n-1}}{B_{n-1}}+\frac{u_{n}}{v_{n}%&#xA;}=\frac{v_{n}A_{n-1}+u_{n}B_{n-1}}{v^{n}B_{n-1}}=\frac{A_{n}}{B_{n}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which means that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$A_{n}=v_{n}\;A_{n-1}+u_{n}\;B_{n-1}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$B_{n}=v_{n}\;B_{n-1}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The truncated continued fraction&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\underset{k=1}{\overset{n}{\mathbb{K}}}\left( u_{k}/v_{k}\right) =\frac{%&#xA;A_{n}}{B_{n}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;verifies:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$A_{n}=b_{n}\;A_{n-1}+a_{n}\;A_{n-2}\qquad A_{0}=0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$B_{n}=b_{n}\;B_{n-1}+a_{n}\;B_{n-2}\qquad B_{0}=1.$$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Addendum 2&lt;/strong&gt;: Detailed algebraic computation. For $n=1$ we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{u_{1}}{v_{1}}=\frac{a_{1}}{b_{1}}=\frac{A_{1}}{B_{1}}\qquad u_{1}=a_{1}\qquad v_{1}=b_{1}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Replacing $n-1$ for $n$ in the first recurrence we get for $n\geq 3$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$A_{n-1}=v_{n-1}\;A_{n-2}+u_{n-1}\;B_{n-2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$B_{n-1}=v_{n-1}\;B_{n-2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which in turn gives:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$A_{n}=v_{n}\;A_{n-1}+u_{n}\;B_{n-1}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$=v_{n}\;\left( v_{n-1}\;A_{n-2}+u_{n-1}\;B_{n-2}\right) +u_{n}\;\left( v_{n-1}\;B_{n-2}\right) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$=v_{n}\;v_{n-1}\;A_{n-2}+\left( v_{n}\;u_{n-1}+u_{n}\;v_{n-1}\right)\;B_{n-2}$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$B_{n}=v_{n}\;B_{n-1}=v_{n}\;v_{n-1}\;B_{n-2}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same substitution in the second recurrence yields (for $n\geq 3$):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$A_{n-1}=b_{n-1}\;A_{n-2}+a_{n-1}\;A_{n-3}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$B_{n-1}=b_{n-1}\;B_{n-2}+a_{n-1}\;B_{n-3}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Combining everything we obtain:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$A_{n}=b_{n}\;A_{n-1}+a_{n}\;A_{n-2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$=b_{n}\;\left( v_{n-1}\;A_{n-2}+u_{n-1}\;B_{n-2}\right) +a_{n}\;A_{n-2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$=\left( b_{n}\;v_{n-1}+a_{n}\;\right) \;A_{n-2}+b_{n}\;u_{n-1}\;B_{n-2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$B_{n}=b_{n}\;B_{n-1}+a_{n}\;B_{n-2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$=b_{n}\;\left( v_{n-1}\;B_{n-2}\right) +a_{n}\;B_{n-2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$=\left( b_{n}\;v_{n-1}+a_{n}\right) \;B_{n-2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Comparing both $A_{n}$ and $B_{n}$ formulae&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$A_{n}=v_{n}\;v_{n-1}\;A_{n-2}+\left( v_{n}\;u_{n-1}+u_{n}\;v_{n-1}\right)\;B_{n-2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$A_{n}=\left( b_{n}\;v_{n-1}+a_{n}\;\right) \;A_{n-2}+b_{n}\;u_{n-1}\;B_{n-2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$B_{n}=v_{n}\;v_{n-1}\;B_{n-2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$B_{n}=\left( b_{n}\;v_{n-1}+a_{n}\right) \;B_{n-2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;one concludes that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$v_{n}\;u_{n-1}+u_{n}\;v_{n-1}=b_{n}\;u_{n-1}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$v_{n}\;v_{n-1}=b_{n}\;v_{n-1}+a_{n}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$a_{n}=v_{n}\;v_{n-1}-b_{n}\;v_{n-1}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$=v_{n}\;v_{n-1}-\left( v_{n}\;u_{n-1}+u_{n}\;v_{n-1}\right)\;v_{n-1}/u_{n-1}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$=v_{n}\;v_{n-1}-v_{n}\;v_{n-1}-u_{n}\;v_{n-1}\;v_{n-1}/u_{n-1}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$=-\frac{u_{n}}{u_{n-1}}v_{n-1}^{2},$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$b_{n}\;u_{n-1}=v_{n}\;u_{n-1}+u_{n}\;v_{n-1}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$b_{n}=v_{n}+\frac{u_{n}}{u_{n-1}}v_{n-1}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus for $n\geq 2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$a_{n}=-\frac{u_{n}}{u_{n-1}}v_{n-1}^{2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$b_{n}=v_{n}+\frac{u_{n}}{u_{n-1}}v_{n-1}.$$&lt;/p&gt;&#xA;" OwnerUserId="752" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2010-10-16T15:56:23.260" LastActivityDate="2010-10-16T15:56:23.260" CommentCount="2" />
  <row Id="6799" PostTypeId="2" ParentId="6784" CreationDate="2010-10-14T17:35:52.343" Score="0" ViewCount="" Body="&lt;p&gt;The entries in the $n$th row of Pascal's triangle (for large $n$) follow a Gaussian curve quite closely, as a consequence of the central limit theorem. See &lt;a href=&quot;http://en.wikipedia.org/wiki/Binomial_distribution#Normal_approximation&quot; rel=&quot;nofollow&quot;&gt;the Wikipedia article about the binomial distribution&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2010-10-14T17:35:52.343" CommentCount="2" />
  <row Id="6800" PostTypeId="2" ParentId="6777" CreationDate="2010-10-14T17:47:13.323" Score="4" ViewCount="" Body="&lt;p&gt;I'm not going to add nothing directly related to your question and previous answers, but make some propaganda of a theorem I like since I was student and which, I believe, says something stronger than comparing some intuitive notion of completness with its definition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A somewhat related notion of completeness is the geodesical one. The definition may not be too much appealing unless you're interested in differential geometry, but one of its consequences is easy to explain: if a Riemann manifold is &lt;strong&gt;geodesically complete&lt;/strong&gt;, you can join any two points by a length minimizing geodesic. (But geodesic already implies that it minimizes length, doesn't it? Not quite: just locally. So, for instance, the meridian joining the North Pole with London, but going &quot;backward&quot;, through the Bering Strait and the Pacific Ocean, then the South Pole, Africa and finally London, is a geodesic, but not a length minimizing one blatantly.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, $\mathbb{R^2} \backslash \left\{ (0,0)\right\} $ is &lt;strong&gt;not&lt;/strong&gt; geodesically complete, since there is no length minimizing geodesic joining, say, $(-1,0)$ and $(1,0)$, due to the &quot;hole&quot; $(0,0)$. At the same time, as a metric space, $\mathbb{R^2} \backslash \left\{ (0,0)\right\}$ is &lt;strong&gt;not&lt;/strong&gt; complete: the Cauchy sequence $(\frac{1}{n}, 0)$ converges to $(0,0)$, but since $(0,0)$ is not in $\mathbb{R^2} \backslash \left\{ (0,0)\right\}$ it doesn't have a limit there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, the &lt;a href=&quot;http://en.wikipedia.org/wiki/Hopf-Rinow_theorem&quot; rel=&quot;nofollow&quot;&gt;Hopf-Rinow theorem&lt;/a&gt; tells us that this kind of things always happen together: a &quot;hole&quot; for geodesics is the same as a &quot;hole&quot; for Cauchy sequences, since for a (finite-dimensional) Riemann manifold $M$, both notions agree: $M$ is complete as a metric space if and only if it is geodesically complete.&lt;/p&gt;&#xA;" OwnerUserId="664" LastEditorUserId="664" LastEditorDisplayName="" LastEditDate="2010-10-15T11:08:07.987" LastActivityDate="2010-10-15T11:08:07.987" CommentCount="3" />
  <row Id="6801" PostTypeId="1" CreationDate="2010-10-14T17:54:40.847" Score="1" ViewCount="227" Body="&lt;p&gt;I'm not quite sure what the mathematical term for what I'm asking is, so let me just describe what I'm trying to figure out.  Let's say that I have two ordered sets of numbers {1, 2} and {3, 4}.  I'm trying to figure out the number of possible ways to combine these two sets into one without breaking the ordering of the two sets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So for instance, {1, 2, 3, 4}, {3, 4, 1, 2}, and {1, 3, 2, 4} are valid combinations, but {2, 1, 4, 3} isn't.  How do I figure out the number of valid combinations?  This feels like something I should remember from college, but I'm drawing a blank.  It feels somewhere in between a combination and a permutation.  Maybe I'm looking for a partially-ordered permutation (which seems to be a somewhat difficult concept if google is to be believed)?&lt;/p&gt;&#xA;" OwnerUserId="2519" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-10-14T18:59:56.220" LastActivityDate="2010-10-14T18:59:56.220" Title="How do I determine the possible number of combinations of two ordered sets?" Tags="&lt;combinatorics&gt;&lt;binomial-coefficients&gt;&lt;permutations&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="6802" PostTypeId="2" ParentId="6801" CreationDate="2010-10-14T17:59:21.073" Score="3" ViewCount="" Body="&lt;p&gt;You are talking about (resulting) tuples I presume (and not sets).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the sets are $S$ with $s$ elements and $T$ with $t$ elements, then the total possible tuples is $\displaystyle {s+t \choose s}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically, you have $s+t$ slots and you pick the slots (say $s$) for one of the sets in $\displaystyle {s+t \choose s}$ ways. Once the slots are chosen, all the $s+t$ numbers can now be filled in only one way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the total is $\displaystyle {s+t \choose s}$.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-10-14T17:59:21.073" />
  <row Id="6803" PostTypeId="1" CreationDate="2010-10-14T18:07:34.813" Score="3" ViewCount="126" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do you think the following limits are correct? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle\lim_{d\to\infty}\frac{\sum\limits_{k=1}^{d} {\varphi(N) \choose  k} {d-1 \choose k-1}}{\varphi(N)^d}=0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle\lim_{N\to\infty}\frac{\sum\limits_{k=1}^{d} {\varphi(N) \choose  k} {d-1 \choose k-1}}{\varphi(N)^d}=c$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I plotted the equations and guessed the results according to the graphs but I could not prove them mathematically by myself. Any hints would be appreciated. Graphs are as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://deniz.cs.utsa.edu/plots/&quot; rel=&quot;nofollow&quot;&gt;http://deniz.cs.utsa.edu/plots/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks,&lt;/p&gt;&#xA;" OwnerUserId="2518" LastActivityDate="2011-06-03T14:35:52.430" Title="Limit involving the totient function and combination" Tags="&lt;number-theory&gt;&lt;combinatorics&gt;&lt;limit&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="1" />
  <row Id="6805" PostTypeId="1" CreationDate="2010-10-14T18:09:41.087" Score="1" ViewCount="482" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm trying to prove using induction that 9 divides $n^3 + (n+1)^3 + (n+2)^3$ whenever $n$ is a non-negative integer. So far, I have:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Base case: P(1) = (1) + (8) + (27) = 36, 36 can be divided by 9 so the base case is valid&lt;/li&gt;&#xA;&lt;li&gt;Inductive step: let p(n) be the statement 9 divides $n^3 + (n+1)^3 + (n+2)^3$ Assume p(k) is true, so 9 divides $k^3 + (k+1)^3 + (k+2)^3$&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;And this is where I'm stuck. I'm not sure how to demonstrate that 9 divides P(n) when n = k+1. If someone could step me in the right direction that would be awesome. Thanks!&lt;/p&gt;&#xA;" OwnerUserId="2520" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-10-16T17:58:19.803" LastActivityDate="2010-10-16T19:21:03.143" Title="Simple Proof by induction: &quot;9 divides $n^3 + (n+1)^3 + (n+2)^3$&quot;" Tags="&lt;algebra-precalculus&gt;&lt;induction&gt;" AnswerCount="4" CommentCount="7" />
  <row Id="6806" PostTypeId="2" ParentId="6805" CreationDate="2010-10-14T18:12:25.247" Score="9" ViewCount="" Body="&lt;p&gt;You want to prove that if $9\mid[n^3+(n+1)^3+(n+2)^3]$&#xA;then $9\mid[(n+1)^3+(n+2)^3+(n+3)^3]$. This boils down to proving&#xA;that $9\mid[(n+3)^3-n^3]$.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-10-14T18:12:25.247" />
  <row Id="6808" PostTypeId="1" CreationDate="2010-10-14T18:23:19.223" Score="1" ViewCount="80" Body="&lt;p&gt;Let's call $S$ the (infinite) string that is made by concatenating the consecutive positive integers (starting from 1) written down in base 10.&#xA;Thus, $S = 1234567891011121314151617181920212223242\ldots$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any number in $S$ occurs multiple times. The first occurence of 3 is in position 3 of the series, 2nd occurence is in position 17.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do I find the position of 3 in 100th occurence? Is there a pattern?&lt;/p&gt;&#xA;" LastEditorUserId="1154" LastEditorDisplayName="" LastEditDate="2010-10-14T20:23:24.217" LastActivityDate="2010-10-14T20:55:15.457" Title="Finding n-th position of occurence of a number in an infinite series of numbers" Tags="&lt;combinatorics&gt;&lt;sequences-and-series&gt;" AnswerCount="3" CommentCount="5" />
  <row Id="6809" PostTypeId="2" ParentId="6777" CreationDate="2010-10-14T18:30:19.243" Score="2" ViewCount="" Body="&lt;p&gt;This answer only applies to the order version of completeness rather than the metric version, but I've found it quite a nice way to think about what completeness means intuitively: consider the real numbers. There the completeness property is what guarantees that the space is connected. The rationals can be split into disjoint non-empty open subsets, for example the set of all positive rationals whose squares are greater than two, and its complement, and the reason this works is because, roughly speaking, there is a &quot;hole&quot; in between the two sets which lets you pull them apart. In the reals this is not possible; there are always points at the ends of intervals, so whenever you partition the reals into two non-empty subsets, one of them will always fail to be open.&lt;/p&gt;&#xA;" OwnerUserId="2491" LastActivityDate="2010-10-14T18:30:19.243" CommentCount="1" />
  <row Id="6810" PostTypeId="2" ParentId="6801" CreationDate="2010-10-14T18:33:01.463" Score="0" ViewCount="" Body="&lt;p&gt;Here is another way of visualising Moron's answer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $S$ be the first set, and $T$ be the second set. Now, imagine the elements of $S$ are listed, horizontally, in their order; we will not disturb their order, but will only &quot;place&quot; the elements of $T$ among them. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The possible locations for any element of $T$ are (i) before the first elements of $S$; (ii) between two elements of $S$; and (iii) after the last element of $S$. If $S$ has $s$ elements, then this gives $s+1$ possible locations. We can put more than one element of $T$ in each location, though. For example, if $S=\{1,2,3\}$ and $T=\{a,b,c,d\}$, then we can place elements of $T$ either before $1$, between $1$ and $2$, between $2$ and $3$, or after $3$ (four locations); and we could place one element before $1$, none between $1$ and $2$, two between $2$ and $3$, and one after $3$. But note that once we decide &lt;em&gt;where&lt;/em&gt; to place the elements of $T$, the &lt;em&gt;order&lt;/em&gt; they will appear on is completely determined. If we go the way I just described, you would end up with $a,1,2,b, c, 3, d$, and that is the only way to place elements of $T$ as described while preserving the order.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we are placing $t$ elements, then, we just need to &quot;select&quot; $t$ locations from $s+1$ possibilities. The order in which we pick them doesn't matter, because in the end we will just put the elements of $T$ in their appropriate order in those locations. And we can select the same location more than once. So we need to compute &quot;combinations with repetitions&quot; (order does not matter, repetitions allowed). The formula for making $n$ selections, with repetitions allowed but where order does not matter, from among $m$ possibilities is $\binom{n+m-1}{n}$ (see for example &lt;a href=&quot;http://en.wikipedia.org/wiki/Combination#Number_of_combinations_with_repetition&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;) so here we have $n=t$ and $m=s+1$, giving $\binom{t+s}{t} = \binom{t+s}{s}$ possibilities. &lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-10-14T18:33:01.463" />
  <row Id="6811" PostTypeId="1" AcceptedAnswerId="6816" CreationDate="2010-10-14T19:03:34.217" Score="3" ViewCount="112" Body="&lt;p&gt;This from an exercise in &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/3540390324&quot; rel=&quot;nofollow&quot;&gt;Aigner's book&lt;/a&gt; where one has to evaluate $\sum_{k\ge 0} (-1)^k \binom{n}{k}^2$ using sign reversing involutions. When $n$ is odd, the problem is trivial : let $[n] = \{1,2,\dots,n\}$ consider all pairs of subsets $\{ (A,B) \in 2^{[n]} \times 2^{[n]} : |A| = |B| \}$, and $(A,B)$ has sign $(-1)^{|A|}$ and the sign-reversing involution is $(A,B) \to ([n]-A,[n]-B)$. Any hints on how to approach this problem for even $n$ will be appreciated.&lt;/p&gt;&#xA;" OwnerUserId="404" LastEditorUserId="404" LastEditorDisplayName="" LastEditDate="2010-10-14T20:07:26.823" LastActivityDate="2010-10-14T20:07:26.823" Title="sign reversing involution proof of a combinatorial identity" Tags="&lt;combinatorics&gt;" AnswerCount="1" />
  <row Id="6812" PostTypeId="2" ParentId="6808" CreationDate="2010-10-14T19:05:09.293" Score="1" ViewCount="" Body="&lt;p&gt;You could try some careful counting. If mine is careful enough, I make the 100th occurence of 3 to be in the 889th position.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's the first digit of 333.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I counted like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In 1 to 99 there are 10 threes in the unit's digit and 10 in the ten's digit, so that's makes 20, and hence there are $3 \times 20=60$ in 1 to 299.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now in 300 to 329 there are 30 in the hundred's digit and 3 in the unit's digit, so that's another 33, making 93 so far. Now just look at 330331332333 and you will see that the 100th three is the first digit of 333 which, by some straightforward calculations, is in the 889th position.&lt;/p&gt;&#xA;" OwnerUserId="1301" LastEditorUserId="1301" LastEditorDisplayName="" LastEditDate="2010-10-14T20:55:15.457" LastActivityDate="2010-10-14T20:55:15.457" />
  <row Id="6813" PostTypeId="1" AcceptedAnswerId="6817" CreationDate="2010-10-14T19:20:04.697" Score="3" ViewCount="82" Body="&lt;p&gt;I note that the ordinals of L are the same as V, so I guess that it has no $\Pi_1^1$ consequences. On the other hand Wikipedia tells me that it asserts the existance of a $\Delta_2^1$ non-measurable set of reals. &quot;Measurable&quot; involves third-order concepts but I know that there's often a &quot;coding trick&quot; that gets around that sort of thing, so I guess it has some analytic consequences. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course I am guessing -- I am not very good at this stuff. But I am curious. What's the least point in the analytic hierarchy where V=L matters (if any)?&lt;/p&gt;&#xA;" LastActivityDate="2010-10-14T20:49:06.120" Title="Where in the analytic hierarchy does V=L start having consequences?" Tags="&lt;real-analysis&gt;&lt;set-theory&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="6814" PostTypeId="2" ParentId="6805" CreationDate="2010-10-14T19:24:04.923" Score="3" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\ \ $ If $\rm\ 9\:|\:P(n)\ $ then $\rm \ 9\:|\:P(n+1)\ \iff\ 9\ | \ P(n+1)-P(n)\  =\ (n+3)^3 - n^3$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;REMARK&lt;/strong&gt; $\ $ This reduction can be viewed as arising from rewriting $\rm\: P(n)\: $ as a &lt;strong&gt;telescoping&lt;/strong&gt; series&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\rm P(n) = (P(n)-P(n-1)) + (P(n-1)-P(n-2)) +\: \cdots\: + (P(1)-P(0))+\ P(0) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus $\ 9\ $ divides $\rm\ P(n)\ $ if it divides all RHS terms,$\ $ i.e.$\ $ all the differences $\rm \ P(i+1)-P(i)\ $ and $\rm\ P(0)$. Since $\rm\: P\:$ is a polynomial, its difference has lower degree than $\rm\: P\:$, so the problem reduces to a simpler one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Such &lt;strong&gt;telescopic&lt;/strong&gt; transformations often simplify inductive proofs. See &lt;a href=&quot;http://math.stackexchange.com/questions/6586&quot;&gt;here for similar multiplicative telescopy.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: if you continue to iterate the above reduction then it amounts to showing that the coefficients $\rm \Delta^k P(0)$ in &lt;a href=&quot;http://en.wikipedia.org/wiki/Finite_difference&quot; rel=&quot;nofollow&quot;&gt;Newton's forward difference formula&lt;/a&gt; are all divisible by 9. Indeed one has&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\rm  P(n)\ =\ 9 + 27\ n + 36\ \binom{n}2 + 18\ \binom{n}3 $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Equivalently, one could simply check that four consecutive values of $\rm\:P\:$ are multiples of 9. Analogous remarks hold true for any $\rm\:P(n)\:$ that satisfies a monic linear recurrence with polynomial coefficients (or, much more generally, for proving identities of &lt;a href=&quot;http://www.mathematik.uni-kassel.de/~koepf/Publikationen/Algebra.pdf&quot; rel=&quot;nofollow&quot;&gt;multivariate holonomic functions&lt;/a&gt;). While this viewpoint is overkill here, it is essential when working with nontrivial examples.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-10-16T17:54:46.930" LastActivityDate="2010-10-16T17:54:46.930" CommentCount="1" />
  <row Id="6815" PostTypeId="2" ParentId="6808" CreationDate="2010-10-14T19:31:06.750" Score="2" ViewCount="" Body="&lt;p&gt;You can solve this by thinking of some lists:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How many 3's is there in 1-10? -- 1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How many 3's is there in 11-20? -- 1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How many 3's is there in 21-30? -- 2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How many 3's is there in 31-40? -- 10&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How many 3's is there in 41-50? -- 1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How many 3's is there in 91-100? -- 1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, how many 3's is there in 1-100?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How many 3's is there in 101-200?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How many 3's is there in 201-300?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How many 3's is there in 301-400? (obviously sufficient)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, when you know the what number comes at the 100:th place you just need count the place - for example by &quot;counting lists&quot; again.&lt;/p&gt;&#xA;" OwnerUserId="1154" LastEditorUserId="1154" LastEditorDisplayName="" LastEditDate="2010-10-14T20:22:28.447" LastActivityDate="2010-10-14T20:22:28.447" CommentCount="2" />
  <row Id="6816" PostTypeId="2" ParentId="6811" CreationDate="2010-10-14T19:33:00.507" Score="4" ViewCount="" Body="&lt;p&gt;Consider the set of pairs $(A,B)$ of subsets of $\{1,\ldots,n\}$&#xA;such that $|A|+|B|=n$ with weight $(-1)^{|A|}$.&#xA;The exceptional pairs where the involution is not defined are those&#xA;where $A=B$. For each other pair $(A,B)$ &#xA;take the smallest element of the symmetric difference of $A$ and $B$&#xA;and move it from one set to the other. The symmetric difference of&#xA;the two new sets is the same, so this is an involution, and it is&#xA;weight-reversing.&lt;/p&gt;&#xA;" OwnerUserId="226" LastEditorUserId="226" LastEditorDisplayName="" LastEditDate="2010-10-14T19:38:46.673" LastActivityDate="2010-10-14T19:38:46.673" CommentCount="1" />
  <row Id="6817" PostTypeId="2" ParentId="6813" CreationDate="2010-10-14T20:03:14.847" Score="3" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Absoluteness#Shoenfield.27s_absoluteness_theorem&quot; rel=&quot;nofollow&quot;&gt;Shoenfield's absoluteness theorem&lt;/a&gt; ensures that any $\Sigma^1_2$ statement (and therefore, any $\Pi^1_2$ one) has the same truth value in $L$ as in $V$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, $\Sigma^1_3$ statements differ in general whether they are in $V$ or in $L$. For a silly example, $\exists x\in{\mathbb R}\,(x\notin L)$ can be rewritten as a $\Sigma^1_3$ statement, which is false in $L$ but true in general. Given a $\Sigma^1_2$ formula $\phi(x,y)$, the statement ``$\phi$ defines a well-ordering of the reals'' is $\Pi^1_3$. Again, there is a specific such $\phi$ for which this is true in $L$ and false in general. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(&quot;In general&quot; means here that it can be made false by forcing. There are deeper discrepancies if one allows large cardinals into the picture.)&lt;/p&gt;&#xA;" OwnerUserId="462" LastEditorUserId="462" LastEditorDisplayName="" LastEditDate="2010-10-14T20:49:06.120" LastActivityDate="2010-10-14T20:49:06.120" CommentCount="1" />
  <row Id="6818" PostTypeId="1" AcceptedAnswerId="6822" CreationDate="2010-10-14T20:09:30.890" Score="7" ViewCount="423" Body="&lt;p&gt;We know that the &quot;Harmonic Series&quot; $$ \sum \frac{1}{n}$$ diverges. And for $p &gt;1$ we have the result that the series converges $$\sum \frac{1}{n^{p}}$$ converges. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One can then ask the question of testing the convergence the following 2 Series:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum\limits_{n=1}^{\infty} \frac{1}{n^{k + \cos{n}}}, \quad \sum\limits_{n=1}^{\infty} \frac{1}{n^{k + \sin{n}}}$$ where $ k \in (0,2)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Only thing which i have as tool for this problem is the inequality $| \sin{n} | \leq 1$, which i am not sure whether would applicable or not. &lt;/p&gt;&#xA;" OwnerUserId="9413" LastEditorUserId="9413" LastEditorDisplayName="anonymous" LastEditDate="2010-10-15T07:38:03.877" LastActivityDate="2010-10-15T07:38:03.877" Title="Testing the series $\sum\limits_{n=1}^{\infty} \frac{1}{n^{k + \cos{n}}}$" Tags="&lt;real-analysis&gt;&lt;analysis&gt;&lt;sequences-and-series&gt;" AnswerCount="3" FavoriteCount="2" />
  <row Id="6819" PostTypeId="2" ParentId="6808" CreationDate="2010-10-14T20:33:40.640" Score="1" ViewCount="" Body="&lt;p&gt;Brute force:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;[mariano@godot ~]$ ghci&#xA;GHCi, version 6.12.1: http://www.haskell.org/ghc/  :? for help  &#xA;Prelude&amp;gt; [i | (d,i) &amp;lt;- concat (map show [1..]) `zip` [1..], d == '3'] !! 100 - 1&#xA;889&#xA;Prelude&amp;gt;&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="274" LastActivityDate="2010-10-14T20:33:40.640" CommentCount="3" />
  <row Id="6820" PostTypeId="2" ParentId="6818" CreationDate="2010-10-14T20:42:53.603" Score="5" ViewCount="" Body="&lt;p&gt;Here is a handwaving argument that the series diverges for all $k\in(0,2)$:&#xA;Given $k$, there will be a certain density of $n$ for which $k+cos n &amp;lt;1$.  A &quot;random sample&quot; of the harmonic series will diverge by the same argument that shows the harmonic series diverges.  Our series will be greater, term by term, than the harmonic series that has numerator 0 if $k+cos n &gt;1$ and numerator 1 if $k+cos n &amp;lt;1$&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-10-14T20:42:53.603" CommentCount="4" />
  <row Id="6822" PostTypeId="2" ParentId="6818" CreationDate="2010-10-14T20:58:30.637" Score="3" ViewCount="" Body="&lt;p&gt;Ross' handwaving can be justified.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We use Weyl's theorem on the equidistribution of polynomials (mod 1) with at least one irrational coefficient (see here: &lt;a href=&quot;http://math.stackexchange.com/questions/128/how-do-you-prove-that-pn-xi-for-xi-irrational-and-p-a-polynomial-is-uni/1163#1163&quot;&gt;http://math.stackexchange.com/questions/128/how-do-you-prove-that-pn-xi-for-xi-irrational-and-p-a-polynomial-is-uni/1163#1163&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Choose $\displaystyle p(x) = 2\pi x + \pi$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $\displaystyle k \in (0,2)$. Let $\displaystyle \delta = k$ if $\displaystyle k \in (0,1)$ and $\displaystyle \delta = k-1$ otherwise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By Weyl's theorem, density of $n$ such that the fractional part of $\displaystyle p(n) &amp;lt; 1- \delta$ is non zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus the density (among the set of natural numbers) of the odd numbers $\displaystyle 2n+1$ for which the fractional part of $\displaystyle (2n+1)\pi &amp;lt; 1 - \delta$ is non-zero. Thus the density of $\displaystyle [(2n+1)\pi]$ is non-zero too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For such an $\displaystyle n$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $\displaystyle N = [(2n+1)\pi]$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have that $\displaystyle (2n+1)\pi - N &amp;lt; 1 - \delta$ and so $\displaystyle \cos(N) - \cos((2n+1)\pi) &amp;lt; 1 - \delta$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And thus $\displaystyle \cos(N) &amp;lt; -\delta$ and so $\displaystyle \cos(N) + k &amp;lt; k - \delta \leq 1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus for we have a positive density subset $\displaystyle S$ of $\displaystyle \mathbb{N}$ for which $\displaystyle \cos(N) + k &amp;lt; 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \sum_{N \in S} \frac{1}{N^{k+\cos(N)}} &gt; \sum_{N \in S} \frac{1}{N}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $\displaystyle S$ is of positive density, the sum of reciprocals is divergent, for a proof see here: &lt;a href=&quot;http://math.stackexchange.com/questions/5932/theorem-on-natural-density/&quot;&gt;http://math.stackexchange.com/questions/5932/theorem-on-natural-density/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I will leave my earlier attempt here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $k \in (0,1)$ consider the below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Lemma:&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given any $\displaystyle k \in (0,1)$, we can find an infinite number of odd numbers $\displaystyle M$ such that $\displaystyle M\pi - [M\pi] &amp;lt; 1 - k$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This easily follows from Vinogradov's theorem that $\displaystyle \{p_{n}\alpha\}$ is equidistributed for irrational $\displaystyle \alpha$ where $\displaystyle p_{n}$ is the $\displaystyle n^{th}$ prime. (See here: &lt;a href=&quot;http://en.wikipedia.org/wiki/Equidistributed_sequence&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Equidistributed_sequence&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle \circ$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $\displaystyle n = [M\pi]$ we have &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle \cos(n) - \cos(M\pi) = \cos(n)  + 1 &amp;lt; 1- k$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(using $\displaystyle |\cos x - \cos y| &amp;lt; |x-y|$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And so $\displaystyle \cos(n) + k &amp;lt; 0$ for an infinite number of $\displaystyle n$ and so the series $\displaystyle \sum \frac{1}{n^{k+\cos n}}$ must be divergent. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the $\displaystyle \sin n$ case, we need the sequence $\displaystyle \{(4k+3)\pi/2\}$ which is an infinite subsequence of $\displaystyle \{p_{n}\pi/2\}$ and so the series  $\displaystyle \sum \frac{1}{n^{k+\sin n}}$ is divergent.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-10-14T22:55:01.647" LastActivityDate="2010-10-14T22:55:01.647" />
  <row Id="6823" PostTypeId="1" AcceptedAnswerId="6837" CreationDate="2010-10-14T21:38:16.150" Score="0" ViewCount="108" Body="&lt;p&gt;Let $X_1,....,X_n$ be some positive random variables.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is $A_{ij} = \frac{E[X_i*X_j]}{E[X_i]E[X_j]}$ a positive (semi) definite matrix?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(note that $B_{ij} = Cov(X_i,X_j) = E[X_i*X_j] - E[X_i]E[X_j]$ is PSD because it is a covariance matrix.)&lt;/p&gt;&#xA;" OwnerUserId="7076" LastActivityDate="2010-10-14T23:52:48.807" Title="is the following a positive definite matrix?" Tags="&lt;linear-algebra&gt;&lt;probability&gt;&lt;statistics&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="6824" PostTypeId="2" ParentId="6818" CreationDate="2010-10-14T21:48:15.917" Score="4" ViewCount="" Body="&lt;p&gt;If $k$ is a real number less than 2 then $\sum_{n=1}^\infty n^{-k-\sin n}$ diverges.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is problem 11162 from the &lt;em&gt;American Mathematical Monthly&lt;/em&gt;, the solution by the Microsoft Research Problems Group appears in the February 2007 issue.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's the solution (if I've transcribed it correctly).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $k\leq 0$, then the series diverges by comparison with the harmonic series.&#xA;If $0&amp;lt; k &amp;lt;2$, then let $\alpha,\beta$ be such that $\sin\alpha=\sin\beta=1-k$ and&#xA;$\pi/2 &amp;lt; \alpha &amp;lt; \beta &amp;lt; 5\pi/2$. Note that $\sin x &amp;lt; 1-k$ for all $x$ with $\alpha &amp;lt; x&amp;lt; \beta$. Define&#xA;$$ S=(n\in{\mathbb N}:\sin n&amp;lt;1-k )={\mathbb N}\cap\bigcup_{j\in {\mathbb Z}}(2\pi j+\alpha,2\pi j+\beta).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $\pi$ is irrational, the sequence $n/(2\pi)$ mod 1 is dense in [0,1], so $S\not=\emptyset$&#xA;and there exist positive integers $a,A,b,B$ with&#xA;$$0 &amp;lt; a-2\pi A &amp;lt; {\beta-\alpha\over2}\quad\mbox{and}\quad -{\beta-\alpha\over2} &amp;lt; b-2\pi B &amp;lt; 0.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $C=\max(a,b)$. Given $s\in S$, we claim that there exists $s^\prime\in S$ with $s &amp;lt; s^\prime\leq s+C$.&#xA;Indeed, there exists $j\in{\mathbb Z}$ with $\alpha &amp;lt; s-2\pi j&amp;lt;\beta$. Compare $s-2\pi j$ to $(\alpha+\beta)/2$.&#xA;If $\alpha &amp;lt; s-2\pi j\leq (\alpha+\beta)/2$, then $\alpha&amp;lt; (s+a)-2\pi(j+A)&amp;lt;\beta$, so we can use&#xA;$s^\prime=s+a$. On the other hand,  if $(\alpha+\beta)/2 &amp;lt; s-2\pi j &amp;lt; \beta$, then&#xA;$a&amp;lt;(s+b)-2\pi(j+B) &amp;lt; \beta$, so we can use $s^\prime=s+b$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus $S$ is infinite. Index it in increasing order, $s_1 &amp;lt; s_2 &amp;lt; \cdots.$ Now $s_j\leq s_1+C(j-1)$, so &#xA;$$\sum_{n=1}^{s_m}n^{-k-\sin n}\geq \sum_{1\leq n\leq s_m, n\in S} n^{-k-\sin n}&#xA;\geq \sum_{1\leq n\leq s_m , n\in S} {1\over n} \geq \sum_{j=1}^m {1\over s_j}\geq&#xA;\sum_{j=1}^m {1\over s_1+C(j-1)},$$ which diverges as $m\to\infty$ by the integral test.&lt;/p&gt;&#xA;" OwnerUserId="940" LastEditorUserId="940" LastEditorDisplayName="" LastEditDate="2010-10-14T22:31:34.943" LastActivityDate="2010-10-14T22:31:34.943" CommunityOwnedDate="2010-10-14T21:48:15.917" />
  <row Id="6825" PostTypeId="1" AcceptedAnswerId="6827" CreationDate="2010-10-14T22:02:17.110" Score="0" ViewCount="33" Body="&lt;p&gt;Here's an horrible drawing that tries to explain what I'm asking:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://dl.dropbox.com/u/1164414/SO/Four%20hectagons%20an%20hectagon%20do%20not%20make.png&quot; alt=&quot;Okay, so if you're reading or hearing this you can't see images, so I will take the effort of explaining it with words. Consider four $n$-sided polygons arranged so that their centers are the vertices of a square. The square is exactly as large as the diameter of the smallest circle enclosing each polygon. The polygons do not overlap. Polygons are oriented so that each one has either exactly two sides in common or exactly one side and one vertex in common with the other shapes.&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Trying this with small numbers gives me $f: 4 \to 0, 6 \to 4, 8 \to 4, 10 \to 8, 12 \to 8.$ This suggests that $$f(2n) = 4 \times \left( \lceil \frac{2n}{4} \rceil - 1 \right), 2n &gt; 4.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can this result be extended for $n \to \infty$? Bonus: What about odd values of $n$?&lt;/p&gt;&#xA;" OwnerUserId="92" LastEditorUserId="92" LastEditorDisplayName="" LastEditDate="2010-10-14T22:19:55.137" LastActivityDate="2010-10-14T22:19:55.137" Title="What is the number of sides of the hollow area generated by juxtaposition of four $2n$-sided regular polygons?" Tags="&lt;geometry&gt;&lt;discrete-mathematics&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="6826" PostTypeId="2" ParentId="6825" CreationDate="2010-10-14T22:14:33.437" Score="1" ViewCount="" Body="&lt;p&gt;Yes, and your answer is correct.  Following your figure, assume the polygon has a side at the top and bottom.  If n is even, so the polygons have a multiple of 4 sides, there will be a pair of sides at the left and right.  There will be $n/2-1$ sides between them and four sets of that bounding the central region.  If n is odd, there will be points at the left and right.  From the point to the flat on top is $(n-1)/2$ sides, again multiplied by four to make the boundary of the central region.  This supports your formula, using the ceiling function to combine the cases.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-10-14T22:14:33.437" />
  <row Id="6827" PostTypeId="2" ParentId="6825" CreationDate="2010-10-14T22:16:29.590" Score="1" ViewCount="" Body="&lt;p&gt;There are two clear cases.  When $2n$ is divisible by $4$, each polygon will have a horizontal side and a vertical side touching other polygons.  So the number of sides each polygon contributes to the interior is just the number of sides between a horizontal and a vertical side.  (Each side of the &quot;interior&quot; polygon belongs to exactly one of the exterior ones.)  This is just $(2n-4)/4$, which is then multiplied by $4$ again to give $2n-4$ as the answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the other case, you have the points of tangency being a side and a vertex.  We can reduce this to the other case by &quot;stretching&quot; the polygons so the vertices become edges.  So the number of sides in this case is $(2n+2)-4=2n-2$.  This is your result.&lt;/p&gt;&#xA;" OwnerUserId="2215" LastActivityDate="2010-10-14T22:16:29.590" />
  <row Id="6828" PostTypeId="1" AcceptedAnswerId="6835" CreationDate="2010-10-14T22:43:46.393" Score="10" ViewCount="276" Body="&lt;p&gt;Can anyone see a nice way to prove the following for $0\le x \le 1$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sqrt{1-x^2}\ge \text{erf}(\sqrt{-\log x})$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$erf$ is defined as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$erf(z) = \frac{2}{\sqrt{\pi}}\int_{0}^{z} e^{-t^2} \  dt$$&lt;/p&gt;&#xA;" OwnerUserId="998" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-10-15T01:38:43.720" LastActivityDate="2010-10-15T01:38:43.720" Title="Proving $\sqrt{1-x^2}\ge \text{erf}(\sqrt{-\log x})$" Tags="&lt;inequality&gt;" AnswerCount="1" FavoriteCount="2" />
  <row Id="6829" PostTypeId="2" ParentId="6805" CreationDate="2010-10-14T23:01:09.670" Score="5" ViewCount="" Body="&lt;p&gt;Here's an all-purpose technique for proving your kind of induction problems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are trying to prove using induction that 9 divides $n^3 + (n+1)^3 + (n+2)^3$ whenever n is a non-negative integer. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have already demonstrated the base case, so here's the inductive step.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming it is true for n, is it also true for n+1?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That means to say: is $(n+1)^3 + (n+2)^3 + (n+3)^3$ divisible by 9?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, if $[(n+1)^3 + (n+2)^3 + (n+3)^3] -[n^3 + (n+1)^3 + (n+2)^3]$ (the difference of the inductive step and the formula) is divisible by 9, so is $(n+1)^3 + (n+2)^3 + (n+3)^3$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why? A bit of modular arithmetic here.&#xA;If (x mod 9) - (0 mod 9) = (0 mod 9) then x = (0 mod 9). This means that if the difference is divisible by 9, and one of the numbers in our subtraction is divisible by 9, the other number must be too. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not going to do the algebra on here, but you can verify that $[(n+1)^3 + (n+2)^3 + (n+3)^3] -[n^3 + (n+1)^3 + (n+2)^3]$ simplifies into $9n^2+27n+27$ which you can rewrite as $9(n^2+3n+3)$. Hence the difference is divisible by 9, and so is our inductive step.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Q.E.D.&lt;/p&gt;&#xA;" OwnerUserId="647" LastActivityDate="2010-10-14T23:01:09.670" />
  <row Id="6830" PostTypeId="1" CreationDate="2010-10-14T23:25:42.310" Score="1" ViewCount="118" Body="&lt;p&gt;I approached this exam question the wrong way apparently, help please?&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Consider the word &quot;mathematics&quot;. In how&#xA;  many ways can you rearrange all the&#xA;  letters so that the vowels are paired&#xA;  and always apart?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" LastEditorUserId="6724" LastEditorDisplayName="" LastEditDate="2011-05-08T17:03:01.850" LastActivityDate="2011-05-08T17:03:01.850" Title="Permutations and Combination rearrangement" Tags="&lt;permutations&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="6831" PostTypeId="2" ParentId="6830" CreationDate="2010-10-14T23:30:48.743" Score="1" ViewCount="" Body="&lt;p&gt;Given there are four vowels, list all the ways to split them into pairs.  Now for each set of pairs think of each pair as a single character.  How many ways can you permute the resulting collection?  How many of those have the vowel pairs next to each other?&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-10-14T23:30:48.743" CommentCount="3" />
  <row Id="6832" PostTypeId="1" AcceptedAnswerId="6833" CreationDate="2010-10-14T23:46:04.107" Score="1" ViewCount="429" Body="&lt;p&gt;Problably a simple question and possibly not asked very well. What I want to know is..&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In binary, a decimal value of 1 is also 1.&#xA;It can be expressed as  x = 1 X 2º&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(probably the wrong use of a character symbol above, I was trying to represent &quot;to the power of zero&quot;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Question:&#xA;Why is two to the power of zero equal to one?&#xA;I get that two to the power of one is equal to two, or binary 10, but why is to the power of zero equal to one, is this a math convention? is there a link I could read?&lt;/p&gt;&#xA;" OwnerUserId="2496" LastActivityDate="2010-10-18T00:07:15.683" Title="Why is two to the power of zero equal to binary one?" Tags="&lt;powers&gt;" AnswerCount="5" CommentCount="3" />
  <row Id="6833" PostTypeId="2" ParentId="6832" CreationDate="2010-10-14T23:48:20.563" Score="11" ViewCount="" Body="&lt;p&gt;Because we want $2^{m+n} = 2^m \cdot 2^n$, and if $n = 0$ this requires that $2^0 = 1$.  More combinatorially, $a^b$ is the number of functions from a set with $b$ elements to a set with $a$ elements, and there is exactly one function from the empty set to any other set (the empty function).  This is the same reason that $0! = 1$.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-10-14T23:48:20.563" CommentCount="1" />
  <row Id="6834" PostTypeId="2" ParentId="6832" CreationDate="2010-10-14T23:48:46.597" Score="5" ViewCount="" Body="&lt;p&gt;Any number (except zero) to the zero power equals one.  There are many ways to justify it.  Easiest (to me) is to make the laws of exponents work.  We know that $a^b/a^c=a^{b-c}$.  Now put b=c&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-10-14T23:48:46.597" CommentCount="1" />
  <row Id="6835" PostTypeId="2" ParentId="6828" CreationDate="2010-10-14T23:50:47.453" Score="16" ViewCount="" Body="&lt;p&gt;Let $y = \sqrt{-\log x}$. Then the inequality reduces to $\text{erf}(y) \leq \sqrt{(1-e^{-2y^2})}$ or equivalently $\text{erf}^2(y) + e^{-2y^2} \leq 1$. Now $\text{erf}^2(y)$ can be written as a double integral $\text{erf}^2(y) = \frac{4}{\pi} \int_{0}^y \int_{0}^{y} e^{- (a^2+b^2)} da db$ (As Qiaochu Yuan points out, the functions involved are well behaved and the double integral is well defined). Replace the area of integration from the square of side $y$ in the first quadrant to a quarter-circle of radius $y\sqrt{2}$ in the first quadrant and switch to polar co-ordinates. This would give the inequality $\text{erf}^2(y) \leq 1-e^{-2y^2}$ which is what we wanted.&lt;/p&gt;&#xA;" OwnerUserId="618" LastEditorUserId="618" LastEditorDisplayName="" LastEditDate="2010-10-15T01:00:18.827" LastActivityDate="2010-10-15T01:00:18.827" CommentCount="5" />
  <row Id="6836" PostTypeId="2" ParentId="6795" CreationDate="2010-10-14T23:52:03.487" Score="5" ViewCount="" Body="&lt;p&gt;Aside from 8 and arguably 2, these are the even semiprimes, Sloane's &lt;a href=&quot;http://oeis.org/classic&quot; rel=&quot;nofollow&quot;&gt;A100484&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1778" LastEditorUserId="1778" LastEditorDisplayName="" LastEditDate="2010-10-15T18:30:03.327" LastActivityDate="2010-10-15T18:30:03.327" />
  <row Id="6837" PostTypeId="2" ParentId="6823" CreationDate="2010-10-14T23:52:48.807" Score="3" ViewCount="" Body="&lt;p&gt;Yes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose we choose arbitrary $\lambda_1,\ldots,\lambda_n$ and form a new random variable&#xA;$$Z={{\lambda_1} \over {E[X_1]}}X_1+\cdots+{{\lambda_n} \over {E[X_n]}}X_n$$&#xA;Then&#xA;$$E[Z^2]=\sum_{i=1}^n\sum_{j=1}^n A_{ij} \lambda_i \lambda_j$$&#xA;But being the expectation of a non-negative variable, $E[Z^2]$ is non-negative for any choice of $\lambda_1,\ldots,\lambda_n$. Which proves that your matrix $A$ is non-negative semi-definite.&lt;/p&gt;&#xA;" OwnerUserId="1195" LastActivityDate="2010-10-14T23:52:48.807" CommentCount="1" />
  <row Id="6838" PostTypeId="2" ParentId="6832" CreationDate="2010-10-15T00:29:24.427" Score="2" ViewCount="" Body="&lt;p&gt;The definition $\ 2^0 = 1\ $ is &quot;natural&quot; since it makes the arithmetic of exponents have the same structure as $\mathbb N$ (or $\mathbb Z\:$ if you extend to negative exponents). In more algebraic language: the definition is the canonical extension of the powering homomorphism from $\rm\ \mathbb N_+\: $ to $\rm \mathbb N\ $ (or $\rm\: \mathbb Z\:$),$\ $ viz. $\rm\ 2^n\ =\ 2^{n+0}\ =\ 2^n\ 2^0\ $ $\rm\Rightarrow\ 2^0 = 1\:$. It's just a special case of the fact that the identity element must be preserved by structure preserving maps of certain multiplicative structures (e.g. commutative cancellative monoids).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It may be viewed as a special case of adjoining an identity element to a commutative semigroup. And it proves very convenient to do so, for the same reason it proves convenient to adjoin the identity element 0 to the positive natural numbers, e.g. it allows every element to be viewed as a sum, so one can write general formulas for sums that work even in extremal cases where an element is indecomposable (e.g. by writing $ 1 = 1 + 0 $ vs. having to separate a special case for the sum-indecomposable element 1 or,$\ $ for $\ 2^\mathbb N\:$, $ 2 = 2\: *\: 1 $). Empty sums and products prove quite handy for naturally terminating recursive definitions.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-10-18T00:07:15.683" LastActivityDate="2010-10-18T00:07:15.683" CommentCount="1" />
  <row Id="6839" PostTypeId="1" AcceptedAnswerId="6841" CreationDate="2010-10-15T02:05:43.957" Score="0" ViewCount="86" Body="&lt;p&gt;12, 13, 11, 14, 10, 15, 9, 16... 0, maxNum&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Start with x then alternate on each side of x until you reach 0 on one side and maxNum on the other. Need to do this in some programming code but can't nail it down. How would you write out the formula for that sequence?&lt;/p&gt;&#xA;" LastActivityDate="2010-10-15T02:54:33.663" Title="formula for number sequence" Tags="&lt;number-theory&gt;&lt;sequences-and-series&gt;" AnswerCount="3" CommentCount="3" />
  <row Id="6840" PostTypeId="2" ParentId="6839" CreationDate="2010-10-15T02:18:07.610" Score="2" ViewCount="" Body="&lt;p&gt;let x denote the start number, the the value of the i'th term can be written:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;f_i = x + (-1)^(i+1) * floor((i+1)/2) , if (i &amp;lt;= 2x)&#xA;undefined, otherwise&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="2526" LastEditorUserId="2526" LastEditorDisplayName="" LastEditDate="2010-10-15T02:45:28.777" LastActivityDate="2010-10-15T02:45:28.777" CommunityOwnedDate="2010-10-15T02:45:28.777" CommentCount="0" />
  <row Id="6841" PostTypeId="2" ParentId="6839" CreationDate="2010-10-15T02:22:44.080" Score="2" ViewCount="" Body="&lt;p&gt;Let n start at 0 and x be the initial value for the sequence then take &#xA;$a_{n}(x)=\begin{cases}&#xA;x-\frac{n}{2} &amp;amp; ,n\equiv0\mbox{mod}(2)\\&#xA;x+\frac{n+1}{2} &amp;amp; ,n\equiv1\mbox{mod}(2)\end{cases}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You'll just need to add some kind of check statement to notice when you get 0 and some variable to store the maxNum and spit it out at the end.&lt;/p&gt;&#xA;" OwnerUserId="249" LastEditorUserId="249" LastEditorDisplayName="" LastEditDate="2010-10-15T02:28:47.997" LastActivityDate="2010-10-15T02:28:47.997" CommentCount="5" />
  <row Id="6842" PostTypeId="2" ParentId="709" CreationDate="2010-10-15T02:25:24.300" Score="5" ViewCount="" Body="&lt;p&gt;Already so many answers, but I haven't seen my favorite one posted, so here's another.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The vertex occurs on the vertical line of symmetry, which is not affected by shifting up or down.  So subtract $c$ to obtain the parabola $y=ax^2+bx$ having the same axis of symmetry.  Factoring $y=x(ax+b)$, we see that the $x$-intercepts of this parabola occur at $x=0$ and $x=-\frac{b}{a}$, and hence the axis of symmetry lies halfway between, at $x=-\frac{b}{2a}$.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-10-15T02:25:24.300" CommentCount="2" />
  <row Id="6843" PostTypeId="2" ParentId="2324" CreationDate="2010-10-15T02:35:44.463" Score="0" ViewCount="" Body="&lt;p&gt;As n0vakovic mentioned, it depends on the type of input signal, what you expect the classification to depend on, and the type of noise, if any in the signal. Also, remember that expanding the input to a higher dimensional space also helps a lot, if your classification surface is non complicated (highly nonlinear). This idea is used in nonlinear SVMs and Deep Learning (Geoff Hinton). Now, as to the possible encodings:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Raw input at each data point&lt;/li&gt;&#xA;&lt;li&gt;Scaled input to [0.0,1.0]&lt;/li&gt;&#xA;&lt;li&gt;Take FFT or DCT and use the coefficients as input&lt;/li&gt;&#xA;&lt;li&gt;Wavelet transform and use the coefficients. You have the added advantage of selecting different wavelet bases.&lt;/li&gt;&#xA;&lt;li&gt;Expand into pairwise products of raw input along with the inputs. Now you have approximately 100 + 100*50 inputs&lt;/li&gt;&#xA;&lt;li&gt;Do some denoising (for example in wavelet domain) if you have an idea of what kind of noise it is.&lt;/li&gt;&#xA;&lt;li&gt;Expand the input space by taking moving averages in addition to the raw inputs&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;.... you have many other choices. Also, you can mix and match from the above. i.e., do denoising with wavelets if you think you have some intermittent noise in certain frequencies, and then expand into inputs and pairwise products. You could take pairwise products locally instead of all pairs. Basically, the answer depends, though hopefully the above will give you something to start on.&lt;/p&gt;&#xA;" OwnerUserId="2532" LastEditorUserId="2532" LastEditorDisplayName="" LastEditDate="2010-10-15T18:17:26.450" LastActivityDate="2010-10-15T18:17:26.450" />
  <row Id="6844" PostTypeId="2" ParentId="6839" CreationDate="2010-10-15T02:54:33.663" Score="1" ViewCount="" Body="&lt;p&gt;WWright and anonymous_21321 have given nice formulas for your sequence.  (Note that x is the zero-th term in those formulas-- you plug in n=0 or i=0 to get x.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if you're trying to understand how to write a program for the sequence, a formula may not be the easiest way to go.  What you wrote in your question is an algorithm for the sequence.  To make it a program, you need to make it completely precise and clear.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In SAGE, I would write:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;print x&#xA;i=1&#xA;while x-i &amp;gt;= 0:&#xA;     print x+i&#xA;     print x-i&#xA;     i = i+1&#xA;print x+i&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This is a direct translation of your description, written precisely for SAGE.  With a little insight into the sequence, you could write a better program or even a formula for the n-th term, as the other answers do.  If you're just interested in maxNum, there's a simple formula-- you can skip all the other terms in the sequence once you understand what's going on.&lt;/p&gt;&#xA;" OwnerUserId="1461" LastActivityDate="2010-10-15T02:54:33.663" CommentCount="1" />
  <row Id="6845" PostTypeId="1" CreationDate="2010-10-15T03:04:44.453" Score="12" ViewCount="233" Body="&lt;p&gt;What is the current state of the art in summing (where by 'summing', I mean 'representing in terms of already known constants and whatnot') series such as these:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{n=1}^{\infty} \frac{1}{\sqrt[3]{1+7^{n}}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{n=1}^{\infty} \frac{1}{\sqrt{n!}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{n=1}^{\infty} e^{-\sqrt{n}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{n=1}^{\infty} \frac{1}{n^{3}+\sqrt[7]{n}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a copy of &lt;a href=&quot;http://en.wikipedia.org/wiki/Konrad_Knopp&quot; rel=&quot;nofollow&quot;&gt;Konrad Knopp&lt;/a&gt;'s book /Theory and Application of Infinite Series/ , but that's fifty years old, and I've been hoping that there have been improvements in the techniques since then. &lt;/p&gt;&#xA;" OwnerUserId="890" LastEditorUserId="890" LastEditorDisplayName="" LastEditDate="2010-10-15T04:42:27.857" LastActivityDate="2011-03-01T15:25:30.147" Title="what is the current state of the art in methods of summing &quot;exotic&quot; series?" Tags="&lt;sequences-and-series&gt;" AnswerCount="2" FavoriteCount="5" />
  <row Id="6846" PostTypeId="1" CreationDate="2010-10-15T03:38:53.953" Score="0" ViewCount="88" Body="&lt;p&gt;Consider that I have 3 parametric equations as function of time and describe the motion of a body in space:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;x = f(t)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;y = g(t)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;z = h(t)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These curves are pretty simple and can be modeled within a certain interval by a 2nd order polynomial.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it as simple as performing individual fitting on each function or that would break the time relationship between the functions?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any pointers appreciated.&lt;/p&gt;&#xA;" OwnerUserId="2449" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-10-17T14:37:57.727" LastActivityDate="2010-10-17T14:37:57.727" Title="How to fit a 3-D parametric equation to datapoints" Tags="&lt;approximation&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="6847" PostTypeId="2" ParentId="6846" CreationDate="2010-10-15T03:48:44.827" Score="3" ViewCount="" Body="&lt;p&gt;It should be appropriate to treat them independently since each is a function of time. The case where you have to worry is if f,g,h involve x,y,z. For example, if y=g(t,z,x) you could not treat them independently. Also if x=f(t,x) y=g(t,y) and z=h(t,z) you could still treat the x,y,z independently.&lt;/p&gt;&#xA;" OwnerUserId="249" LastActivityDate="2010-10-15T03:48:44.827" />
  <row Id="6848" PostTypeId="1" AcceptedAnswerId="6850" CreationDate="2010-10-15T03:54:53.137" Score="10" ViewCount="270" Body="&lt;p&gt;The problem is to find:&lt;br&gt;&#xA;$\lim_{x \to 0}\ \dfrac{\sin(\cos(x))}{\sec(x)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I rewrite the equation as follows:&lt;br&gt;&#xA;$\lim_{x \to 0}\ \dfrac{\sin(\cos(x))}{\dfrac{1}{\cos(x)}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And multiply by $\dfrac{\cos(x)}{\cos(x)}$, producing:&lt;br&gt;&#xA;$\lim_{x \to 0}\ \dfrac{\cos(x)*\sin(\cos(x))}{\dfrac{\cos(x)}{\cos(x)}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And rewrite as:&lt;br&gt;&#xA;$\lim_{x \to 0}\ \cos^2(x)\ \dfrac{\sin(\cos(x))}{\cos(x)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which then becomes:&lt;br&gt;&#xA;$\lim_{x \to 0}\ \cos^2(x) * 1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which becomes 1. However, the answer is apparently $\sin(1)$. What am I doing wrong?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; I found a different way to solve this, but I'm still not sure what I did wrong originally.&lt;/p&gt;&#xA;" OwnerUserId="451" LastEditorUserId="451" LastEditorDisplayName="" LastEditDate="2010-10-15T04:20:03.167" LastActivityDate="2011-01-05T17:51:43.657" Title="Finding $\lim_{x \to 0}\ \frac{\sin(\cos(x))}{\sec(x)}$" Tags="&lt;homework&gt;&lt;limit&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="6849" PostTypeId="2" ParentId="5625" CreationDate="2010-10-15T03:58:15.300" Score="1" ViewCount="" Body="&lt;p&gt;What is n here?&#xA;Is it no .of digits or the value of the number?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it is the value of the no,, then just convert each digit of base-9 no. in to its base-3 equivalent. &lt;/p&gt;&#xA;" OwnerUserId="2504" LastActivityDate="2010-10-15T03:58:15.300" />
  <row Id="6850" PostTypeId="2" ParentId="6848" CreationDate="2010-10-15T04:07:21.890" Score="9" ViewCount="" Body="&lt;p&gt;As $\displaystyle x \to 0$, $\displaystyle \cos x \to 1$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you cannot use the limit $\displaystyle \lim_{h \to 0} \frac{\sin h}{h} = 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The given answer is $\displaystyle \sin(1)$ I presume and not $\displaystyle \sin(0)$ (which is $0$)...&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-10-15T04:07:21.890" />
  <row Id="6851" PostTypeId="1" AcceptedAnswerId="6969" CreationDate="2010-10-15T04:08:09.287" Score="2" ViewCount="284" Body="&lt;p&gt;I am learning to work with inequalities.  For an example, let's say I want to find the set of $(x, y) \in \mathbb{R}^{2}$ that satisfy: $|x \cdot y| \leq 2$.&lt;/p&gt;&#xA;" OwnerUserId="2533" LastEditorUserId="742" LastEditorDisplayName="anonymous" LastEditDate="2011-03-30T18:56:18.387" LastActivityDate="2011-03-30T18:56:18.387" Title="Determine and sketch the pairs $(x, y)$ in $\mathbb{R} \times \mathbb{R}$ that satisfy: $|x\cdot y| \leq 2$" Tags="&lt;homework&gt;&lt;real-analysis&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="1" />
  <row Id="6852" PostTypeId="1" AcceptedAnswerId="6856" CreationDate="2010-10-15T04:37:03.577" Score="4" ViewCount="163" Body="&lt;p&gt;The following is a lemma I read online, but I don't understand part of the proof. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $d$ be the maximum possible order among integers $a$ prime to $p$. Then for any integer $a$ not divisible by $p$, the order of $a$ divides $d$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proof: Let $b$ be an element of order $d$, let $k$ be the order of $a$, and let $g=gcd(k,d)$. Then $a^g$ has order $k/(k,g)=k/g$, and $k/g$ is relatively prime to $d$. Therefore $ba^g$ has order $d\cdot k/g$, which by maximality of $d$ implies that $k/g=1$. Hence $k|d$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The one part I don't understand is how $k/g$ and $d$ are relatively prime. If I take $d=6$, $k=4$, $g=2$, then $k/g$ and $d$ are not relatively prime. However, I've gone through a few cases for small primes, and such a situation as never arisen, so there must be some constraints on which values $k$ and $d$ can take. Could someone please explain this?&lt;/p&gt;&#xA;" OwnerUserId="1583" LastActivityDate="2010-10-15T20:52:42.890" Title="Maximum order of integers coprime to a prime $p$." Tags="&lt;number-theory&gt;" AnswerCount="1" CommentCount="11" FavoriteCount="1" />
  <row Id="6855" PostTypeId="2" ParentId="6832" CreationDate="2010-10-15T05:49:19.240" Score="3" ViewCount="" Body="&lt;p&gt;The easy explanation is:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$2^0 = 1_2$&lt;/li&gt;&#xA;&lt;li&gt;$2^1 = 10_2$ &lt;/li&gt;&#xA;&lt;li&gt;$2^2 = 100_2$&lt;/li&gt;&#xA;&lt;li&gt;$2^3 = 1000_2$&lt;/li&gt;&#xA;&lt;li&gt;$2^4 = 10000_2$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;If we multiply $2^1$ by $2$, we get $2^2$; naturally if we divide $2^1$ by $2$, we should get $2^0$. And it happens that &#xA;${2^1 \over 2} = {2 \over 2} = 1$.&lt;/p&gt;&#xA;" OwnerUserId="2166" LastActivityDate="2010-10-15T05:49:19.240" CommentCount="1" />
  <row Id="6856" PostTypeId="2" ParentId="6852" CreationDate="2010-10-15T05:51:23.997" Score="0" ViewCount="" Body="&lt;p&gt;Below is the key theorem as it applies to arbitrary finite abelian groups. See my &lt;a href=&quot;http://groups.google.com/groups?selm=y8zr7a0j6us.fsf@nestle.csail.mit.edu&quot; rel=&quot;nofollow&quot;&gt;post here&lt;/a&gt; for an example of how it is applied to deduce the more general result that  a finite multiplicative subgroup of a domain is cyclic. The lemma is famous as &quot;Herstein's hardest problem&quot; - see the note below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;THEOREM&lt;/strong&gt;  $\rm\quad\quad\: maxord(G)\ =\ expt(G)\ $ for a finite abelian group $\rm\: G\:,\ $ i.e.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\ \ max\ \{ ord(g) : \: g \in G\}\ =\  min\ \{ n&gt;0 : \: g^n = 1\ \:\forall\ g \in G\}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof&lt;/strong&gt; $\ \:$ By the lemma below, $\rm\: S\: =\: \{ ord(g) : \:g \in G \}$  is a finite &#xA;set of naturals closed under$\rm\ lcm\:$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence every elt $\rm\ s \in S\:$  is a &#xA;divisor of the max elt $\rm\: m\: $  [else  $\rm\: lcm(s,m) &gt; m\:$],$\ $ so $\rm\ m = expt(G)\:$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;LEMMA&lt;/strong&gt; $\ $ A finite abelian group $\rm\:G\:$ has an  lcm-closed  order set, i.e. with $\rm\: o(X) = $ order of $\rm\: X$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\quad\quad\quad\ \ X,Y \in G\ \Rightarrow\ \exists\ Z \in G:\  o(Z) = lcm(o(X),o(Y))$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof&lt;/strong&gt;$\ \ $  By induction on  $\rm o(X)\: o(Y)\:.\ $ If it's $\:1\:$ then trivially $\rm\:Z = 1\:$. $\ $ Otherwise&lt;/p&gt;&#xA;&#xA;&lt;p&gt;write  $\rm\ o(X) =\: AP,\: \ o(Y) = BP',\ \ P'|P = p^m &gt; 1\:,\ $  prime $\rm\: p\:$ coprime to $\rm\: A,B$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then  $\rm\: o(X^P) = A,\ o(Y^{P'}) = B\:.\ $  By induction there's a $\rm\: Z\:$ with $\rm \: o(Z) = lcm(A,B)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so  $\rm\ o(X^A\: Z)\: =\: P\ lcm(A,B)\: =\: lcm(AP,BP')\: =\: lcm(o(X),o(Y))\:$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;NOTE&lt;/strong&gt; $\ \ $ This lemma was presented as problem 2.5.11, p. 41 in the first edition of Herstein's popular textbook &quot;Topics in Algebra&quot;. In the 2nd edition Herstein added the following note (problem 2.5.26, p.48) &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Don't be discouraged if you don't get this problem with what you know of group theory up to this stage. I don't know anybody, including myself, who has done it subject to the restriction of using material developed so far in the text.  But it is fun to try. I've had more correspondence about this problem than about any other point in the whole book.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-10-15T20:52:42.890" LastActivityDate="2010-10-15T20:52:42.890" CommentCount="2" />
  <row Id="6857" PostTypeId="2" ParentId="6832" CreationDate="2010-10-15T05:52:08.487" Score="5" ViewCount="" Body="&lt;p&gt;I think it's helpful to think about sums for a bit, and then come back to products.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With sums… we’re all happy with the idea that you can add up a list of 2 things, or 3 things, or more (so e.g. the sum of the list $(x,y,z)$ is $x+y+z$).  And after moment’s thought, we’re happy with the idea that you can add up a list of 1 thing, eg $(x)$, and get just $x$ itself.  And it takes a moments thought, but it's still pretty intuitive, to decide that you can add up the empty list $()$, and that its sum is $0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(“I went shopping today, with nothing on my shopping list.  How much did it cost in total?”)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, play exactly the same game with products!  It's just as easy until you get to the empty list.  For some reason, it's much less intuitive at first (at least it was for me, and I think it is for most people) that one can make sense of the product of the empty list, and that it should be $1$.  But every argument I know to justify why the sum of the empty list should be $0$ also works as an argument that its product should be $1$!  (See the arguments about making exponents work nicely, etc, in the earlier comments and answers.)  And, of course, once the lightbulb flashes, it suddenly becomes completely clear and natural that the product is 1, and that that's not just a convention, and how could anyone possibly think otherwise!? :-P&lt;/p&gt;&#xA;&#xA;&lt;p&gt;…unfortunately I don't know an example for products that’s quite as intuitively convincing as the shopping list example is for sums.  The best I can do: at the checkout, they apply a bunch of taxes and/or discounts.  Each one of those is a multiplier: e.g. 7% sales tax multiplies my total by 1.07, a 20% discount is a multiplier of 0.8, and so on.  To get the overall multiplier from all the taxes/discounts currently in effect, you just multiply the list of them together (of course)!  But if there are no taxes or discounts at the store today, what's the overall multiplier on my shopping?&lt;/p&gt;&#xA;" OwnerUserId="2439" LastActivityDate="2010-10-15T05:52:08.487" CommentCount="1" />
  <row Id="6858" PostTypeId="1" AcceptedAnswerId="6859" CreationDate="2010-10-15T07:14:12.730" Score="-2" ViewCount="297" Body="&lt;p&gt;I've recently volunteered as a tutor for &quot;higher-level mathematics&quot; as a Calculus student because I'm required by my school to achieve at least twenty volunteer hours for service learning. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When applying to the NPO as a volunteer, their only position available was for a &quot;higher-level mathematics tutor&quot;. I've been a Calculus AB student and T/A for my Calculus AB teacher since the start of the year (we're on our third chapter) and I've only recently touched upon Calculus so I'm unsure whether my skills are on par with &quot;higher-level mathematics&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Soon after I left their building, the connection between &quot;higher-level education&quot; and &quot;higher-level mathematics&quot; hit me. This NPO is from the elementary to the high school level so I became confused.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could someone define &quot;higher-level mathematics&quot;?&lt;/p&gt;&#xA;" OwnerUserId="2426" LastEditorUserId="2215" LastEditorDisplayName="" LastEditDate="2010-10-15T07:43:37.310" LastActivityDate="2010-10-15T07:54:55.917" ClosedDate="2010-10-15T18:53:42.897" Title="Define &quot;higher-level mathematics&quot;" Tags="&lt;calculus&gt;&lt;soft-question&gt;&lt;education&gt;" AnswerCount="1" CommentCount="6" />
  <row Id="6859" PostTypeId="2" ParentId="6858" CreationDate="2010-10-15T07:54:55.917" Score="0" ViewCount="" Body="&lt;p&gt;I think this is a soft question (and have tagged it as such; you can roll it back if you disagree) but I'll try to answer.  First, since the NPO only works up to high school, their definition of &quot;higher-level&quot; probably doesn't correspond to the typical one.  Just think of it in terms of your school.  If it has AP classes, Calc BC is probably the highest math class available, and it's possible a few people are taking basic college-level classes (say, multivariable calc or linear algebra), though they probably won't need tutors.  As a blind guess, I'd say that you'd probably have to tutor kids in calculus, which could be a problem if you're still learning it.  The best solution is to tell them where you are now and ask them or previous tutors what level is required.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(For me, what &quot;higher-level mathematics&quot; really is is the things a math major starts doing around the middle of his or her career -- not linear algebra, multivariable calculus, or basic group theory, but topology, commutative algebra, real/complex analysis, etc. would all be there.  Presumably, mathematicians would find some of this elementary, but there is an objective change in the way the math is done around this time.  You move from calculations and applications of theorems to proving the theorems yourself and learning to work rigorously, axiomatically, intuitively, etc.)&lt;/p&gt;&#xA;" OwnerUserId="2215" LastActivityDate="2010-10-15T07:54:55.917" CommentCount="2" />
  <row Id="6860" PostTypeId="1" CreationDate="2010-10-15T08:49:38.467" Score="-2" ViewCount="250" Body="&lt;p&gt;This is a weird logic problem that was posed to me today. This question isn't about the actual logic problem (or the trivial paradox) but in its implications, so read to the end!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A man has been sentenced to death. The executioner comes to his cell on May 31st and tells him that he will die before June 30th. He also tells him that he will not be able to know when he is killed, and that if he is not killed before 9a.m. he will not be killed on that day.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On what day(s) can he be killed? The answer can be shown through induction. If the man isn't killed by June 29th, then he knows he will be killed on June 30th, hence He cannot be killed on the 30th because that would violate one of the logical axioms (he cannot know when he will be killed). The inductive step: The 29th now becomes the &quot;last&quot; day of the month, and using similiar logic, he cannot be killed on the 29th. Therefore the man CANNOT be killed without him knowing when it happens.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Logically we have a contradiction in the following statement. $A \land B \Rightarrow \sim A$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, if we assume we live in a logical world, such a system or situation should not realizable because if we have both $A$ and $\sim A$ true, then we can really prove anything to be true (this is a logic theorem).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question then becomes, how can you reconcile the fact that we can easily do this in the physical world? Find a friend, and tell him that at some point in the next thirty days you will send him an email that says &quot;you're dead&quot;, and that he will not be able to know when it will happen. It works out just fine...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So how can this be? How can we reconcile the physical realizability of the problem, with its obvious logical paradox.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And I don't know the answer, so don't expect one :P&lt;/p&gt;&#xA;" OwnerUserId="1906" LastActivityDate="2010-10-15T13:34:41.840" ClosedDate="2010-10-15T14:07:03.403" Title="The Hangman's Noose - A Logical Paradox" Tags="&lt;logic&gt;" AnswerCount="1" CommentCount="6" />
  <row Id="6861" PostTypeId="1" AcceptedAnswerId="6930" CreationDate="2010-10-15T09:39:55.187" Score="2" ViewCount="120" Body="&lt;p&gt;Given two complex varieties over a common base, I can take their fiber product in the category of varieties, or I can take their fiber product in the category of schemes and then take the reduced subscheme. I have heard, that these two operations yield the same.&#xA;Has someone a reference? &lt;/p&gt;&#xA;" OwnerUserId="667" LastActivityDate="2010-10-16T04:41:42.113" Title="Fiber product of varieties vs schemes reference" Tags="&lt;algebraic-geometry&gt;&lt;schemes&gt;" AnswerCount="1" />
  <row Id="6862" PostTypeId="2" ParentId="6845" CreationDate="2010-10-15T10:17:15.290" Score="2" ViewCount="" Body="&lt;p&gt;Well, much depends on the individual series but one interesting development in the last 50 years is on the acceleration of convergence, which enables us to express some slowly convergent series in terms of one(s) with much faster convergence, or a sum of known constants plus faster converging series. For example, I really like this transformation due to Simon Plouffe&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\zeta(7) = \frac{19}{56700}\pi^7 – 2\sum_{n=1}^{\infty} \frac{1}{n^7(e^{2\pi n}-1)},$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\zeta(7) = \sum_ {n=1}^{\infty} 1/n^7,$ which is given on &lt;a href=&quot;http://en.wikipedia.org/wiki/Zeta_constant&quot; rel=&quot;nofollow&quot;&gt;this wikipedia page.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Andrei Markov's 1890 method, which was the basis for Apéry's proof of the irrationality of $\zeta(3),$ has been pushed further by &lt;a href=&quot;http://www.springerlink.com/content/q87617rp46135qwm/&quot; rel=&quot;nofollow&quot;&gt;Mohammed and Zeilbeger&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can see some other nice transformations of $\zeta(3)$ &lt;a href=&quot;http://numbers.computation.free.fr/Constants/Zeta3/zeta3.html&quot; rel=&quot;nofollow&quot;&gt;here.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1301" LastActivityDate="2010-10-15T10:17:15.290" CommentCount="4" />
  <row Id="6863" PostTypeId="1" CreationDate="2010-10-15T10:41:19.527" Score="1" ViewCount="177" Body="&lt;p&gt;I'm interested, how is the dual input in a sensor fusioning setup in a Kalman filter modeled? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Say for instance that you have an accelerometer and a gyro and want to present the &quot;horizon level&quot;, like in an airplane, a good demo of something like this &lt;a href=&quot;http://www.youtube.com/watch?v=AWAFFZ7rPDc&amp;amp;feature=related&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do you actually harvest the two sensors positive properties and minimize the negative?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this modeled in the Observation Model matrix (&lt;a href=&quot;http://en.wikipedia.org/wiki/Kalman_filter#Underlying_dynamic_system_model&quot; rel=&quot;nofollow&quot;&gt;usually&lt;/a&gt; symbolized by capital H)?&lt;/p&gt;&#xA;" OwnerUserId="2367" LastEditorUserId="2367" LastEditorDisplayName="" LastEditDate="2010-10-15T10:49:17.733" LastActivityDate="2011-05-23T18:00:53.227" Title="Sensor fusioning in Kalman filter" Tags="&lt;stochastic-processes&gt;&lt;signal-processing&gt;" AnswerCount="2" />
  <row Id="6864" PostTypeId="2" ParentId="6489" CreationDate="2010-10-15T10:53:51.693" Score="4" ViewCount="" Body="&lt;p&gt;The annotation to &lt;a href=&quot;http://www.irregularwebcomic.net/2339.html&quot; rel=&quot;nofollow&quot;&gt;http://www.irregularwebcomic.net/2339.html&lt;/a&gt; has an explanation of the Banach-Tarski paradox in easy to read language. It does ignore some of the technical details, but it covers most of the idea of the construction nicely.&lt;/p&gt;&#xA;" OwnerUserId="2538" LastActivityDate="2010-10-15T10:53:51.693" />
  <row Id="6865" PostTypeId="1" AcceptedAnswerId="6866" CreationDate="2010-10-15T11:57:33.463" Score="1" ViewCount="120" Body="&lt;p&gt;Hi, I'm taking a course on machine learning and came across a model answer which I can't wrap my head around.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The whole model answer is the 2nd exercise here: &lt;a href=&quot;https://noppa.tkk.fi/noppa/kurssi/t-61.3050/viikkoharjoitukset/T-61_3050_proposed_solutions.pdf&quot; rel=&quot;nofollow&quot;&gt;https://noppa.tkk.fi/noppa/kurssi/t-61.3050/viikkoharjoitukset/T-61_3050_proposed_solutions.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The part I'm specifically having trouble understanding is the last 2 lines of the 2nd page, ie this: &lt;a href=&quot;http://i.imgur.com/SZEPC.png&quot; rel=&quot;nofollow&quot;&gt;http://i.imgur.com/SZEPC.png&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How does the left side of the upper equation become just the sum of indicator variables? How does the right side simplify to just lambda*p_i?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any help is appreciated.&lt;/p&gt;&#xA;" OwnerUserId="2076" LastActivityDate="2010-10-15T12:30:18.190" Title="Weird differentation of probabilities, please explain?" Tags="&lt;probability&gt;&lt;differential-equations&gt;" AnswerCount="1" />
  <row Id="6866" PostTypeId="2" ParentId="6865" CreationDate="2010-10-15T12:30:18.190" Score="2" ViewCount="" Body="&lt;p&gt;I say there be typos. Particular the suddenly appearing subscript $k$ does not make sense.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Straightforward differentiation gives:&#xA;$${\partial \over {\partial p_i}} L(p|X)={1 \over p_i}\sum_{t=1}^N x^t_i + \lambda$$&#xA;Setting this equal to zero gives us:&#xA;$$\sum_{t=1}^N x^t_i + \lambda p_i=0$$&#xA;Proceeding from which and reading the subscript $k$ in your text as $i$ leads to the same ultimate answer.&lt;/p&gt;&#xA;" OwnerUserId="1195" LastActivityDate="2010-10-15T12:30:18.190" CommentCount="3" />
  <row Id="6867" PostTypeId="2" ParentId="6860" CreationDate="2010-10-15T13:34:41.840" Score="1" ViewCount="" Body="&lt;p&gt;While I agree there is no academic consensus on the matter, to me it seems there is a pretty simple resolution. The man goes through all the logical steps to work out that he cannot possibly be hanged, because he would know it was going to happen, so he settles down to wait for the end of June. Then on the day he is actually killed his last thought will be &quot;I didn't know that was going to happen&quot;.&lt;/p&gt;&#xA;" OwnerUserId="397" LastActivityDate="2010-10-15T13:34:41.840" />
  <row Id="6868" PostTypeId="1" AcceptedAnswerId="6888" CreationDate="2010-10-15T13:36:36.477" Score="3" ViewCount="264" Body="&lt;p&gt;Hi all,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I encounter a problem recently stated as below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How many pieces of cheese we can obtain from a single thick piece by making five straight slices?(we can't move the cheese when slicing)If we wanna maximize the number of pieces which is denoted by P(n), is there any recurrence relation for P(n)?(n is number of slices)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any hints will be highly appreciated and thank you all in advance!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Best regards.&lt;/p&gt;&#xA;" OwnerUserId="1294" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2011-06-03T20:07:01.643" LastActivityDate="2011-06-03T20:07:01.643" Title="How to slice the cheese" Tags="&lt;combinatorics&gt;&lt;recurrence-relations&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="6869" PostTypeId="2" ParentId="6868" CreationDate="2010-10-15T13:54:37.077" Score="12" ViewCount="" Body="&lt;p&gt;This is a special case of the problem of counting the number of&#xA;regions $\mathbb{R}^n$ is divided into by $k$ hyperplanes &lt;em&gt;in&#xA;general position&lt;/em&gt;. The answer is&#xA;$$\sum_{j=0}^n {k\choose j}.$$&#xA;This is mentioned in Richard Stanley's notes on&#xA;&lt;a href=&quot;http://www-math.mit.edu/~rstan/arrangements/arr.html&quot; rel=&quot;nofollow&quot;&gt;hyperplane arrangements&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-10-15T13:54:37.077" CommentCount="1" />
  <row Id="6870" PostTypeId="1" CreationDate="2010-10-15T14:56:17.993" Score="6" ViewCount="474" Body="&lt;p&gt;I often see notation/wording like &quot;let f(x) be a continuous function&quot; or &quot;let $f(x) \in C^0(\mathbb{R})$&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would say that sin and $x \mapsto \sin(x)$ are functions, while sin(x) is a real number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there any correct or best practices in this regard?&lt;/p&gt;&#xA;" OwnerUserId="2043" LastEditorUserId="2043" LastEditorDisplayName="" LastEditDate="2010-10-15T15:26:43.113" LastActivityDate="2010-10-15T17:07:18.370" Title="In written mathematics, is f(x) a function or a number?" Tags="&lt;functions&gt;&lt;notation&gt;" AnswerCount="6" CommentCount="8" />
  <row Id="6872" PostTypeId="2" ParentId="6870" CreationDate="2010-10-15T15:02:50.963" Score="1" ViewCount="" Body="&lt;p&gt;It all depends on what you have said $x$ is. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $x$ is previously defined as a number (e.g. $x  = 3$) then $f(x)$ is also a number, since you can evaluate $f$ at a certain point $x$ to get the number. (A more common way to notate that would be to say $x_1$ instead of $x$...)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If, on the other hand, $x$ is defined as part of some set (e.g. $x\in[0,1]$) then $f$ is a function defined for input values in that interval. $\sin x$ is defined as a function for $-\infty &amp;lt; x &amp;lt; \infty$, but as a number for $x=\pi/2$.&lt;/p&gt;&#xA;" OwnerUserId="161" LastActivityDate="2010-10-15T15:02:50.963" />
  <row Id="6873" PostTypeId="1" AcceptedAnswerId="6880" CreationDate="2010-10-15T15:04:10.787" Score="3" ViewCount="184" Body="&lt;p&gt;This problem was asked in a test, couple of years ago. Looked interesting!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a chess tournament, each pair of players plays exactly one game. No game is drawn. Suppose the $i^{th}$ player wins $a_{i}$ games and loses $b_{i}$ games. Show that $$\sum a_{i}^{2} = \sum b_{i}^{2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can't think of a better title. It would be nice in case someone comes up with a nice title.&lt;/p&gt;&#xA;" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-10-15T16:30:52.090" LastActivityDate="2010-10-15T16:30:52.090" Title="Sum of squares of in-degrees vs out-degrees in a Tournament Graph." Tags="&lt;combinatorics&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="6874" PostTypeId="2" ParentId="6870" CreationDate="2010-10-15T15:06:01.383" Score="9" ViewCount="" Body="&lt;p&gt;You are right. Properly speaking, the function is called just $f$, and its value at the point $x$ is denoted by $f(x)$. Speaking of &quot;the function $f(x)$&quot; is what mathematicians call &quot;abuse of notation&quot;, but it is of course often very practical.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(By the way, one usually writes $x \mapsto \sin x$, not $x \to \sin x$; it's &quot;\mapsto&quot; in TeX.)&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2010-10-15T15:06:01.383" CommentCount="1" />
  <row Id="6875" PostTypeId="2" ParentId="6870" CreationDate="2010-10-15T15:06:55.637" Score="1" ViewCount="" Body="&lt;p&gt;The wording &quot;let $f(x) \in C^0(\mathbb{R})$...&quot; is really sloppy. Of course it doesn't lead to confusion, but there is something to be said for correctness. Ideally whenever you see $f(x)$ it should be because we care about what $f$'s values are rather than $f$ in and of itself as say a member of some space of functions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Oh, and your example of &quot;$x \to sin(x)$&quot; is IMO better written as &quot;$\mathbb{R} \owns x \mapsto \sin x \in \mathbb{R}$. Obviously not so useful for this particular function, but very smart for things like &quot;$V \times V \owns (x,y) \mapsto \langle x,y \rangle \in \mathbb{C}$&quot;.&lt;/p&gt;&#xA;" OwnerUserId="1269" LastActivityDate="2010-10-15T15:06:55.637" />
  <row Id="6876" PostTypeId="2" ParentId="6870" CreationDate="2010-10-15T15:07:28.490" Score="14" ViewCount="" Body="&lt;p&gt;Saying that $f$ or $x\mapsto f(x)$ is the function rather than $f(x)$ is the more common modern perspective in higher mathematics.  I have even seen a precalculus book that stresses that $f$ is a function, whereas $f(x)$ is a value of the function.  The notation $f(x)$ to denote a function remains because it is often more convenient, and it is especially prominent from high school mathematics up to calculus because it is psychologically easier to become accustomed to.  This may lead to ambiguity, because in any case $f(x)$ sometimes &lt;em&gt;does&lt;/em&gt; refer to a value of the function, but the strictest logical correctness of notation isn't for everyone, or for all situations. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I find it off-putting when I read mathematics that was written many decades ago and come across something like, &quot;Let $\varphi(x)$ be a bounded linear functional on $X$....&quot;.  I think to myself, &quot;No, $\varphi$ is the functional!&quot;  Then I take a deep breath and relax, and find that this notation did not hinder the authors from doing and writing great mathematics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ahlfors's complex analysis text has a footnote on page 21 of the 2nd edition, 1966:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Modern students are well aware that $f$ stands for the function and $f(z)$ &#xA;  for a value of the function. However, analysts are traditionally minded and continue to speak of &quot;the function $f(z)$.&quot; &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2010-10-15T15:19:21.327" LastActivityDate="2010-10-15T15:19:21.327" />
  <row Id="6877" PostTypeId="1" CreationDate="2010-10-15T15:16:15.147" Score="5" ViewCount="232" Body="&lt;p&gt;When I first saw the title of &lt;a href=&quot;http://meta.math.stackexchange.com/questions/909/write-an-elevator-pitch-tagline&quot;&gt;this question&lt;/a&gt;, I forgot for a moment I was on meta, and thought it was asking about quick, catchy, attractive, informative one-or-two-liner summaries of various fields of mathematics.  It turned out not to be… so here’s the question I thought it was!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are lots of times when one wants a quick way to explain to someone (family, students, teachers, colleagues) what some field is about, and (hopefully) catch their interest as well — perhaps to come and learn about it themselves, perhaps just to understand why you find it interesting, perhaps to convince them why they should give you funding to work on it…&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question linked nicely describes how a good such pitch should work: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;This isn't as easy as it sounds. Imagine the user who will never read your FAQ and you have two seconds to grab their attention. It should be catchy but descriptive. It should be thoroughly clear but painfully concise. &lt;strong&gt;Make every... word... count.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A couple of extra thoughts: the level of technicality of the pitch should probably depend on how specialised the field is.  You don't need to describe “algebra” or “calculus” to a professional mathematician; dually, you probably aren't trying to explain “homotopy theory of CDGA’s” to your friend who never took any maths courses (at least, not in one sentence).  This isn’t about any level of maths in particular — whether in recreational maths or cutting-edge research, everyone can benefit from a good slogan sometimes.  Posting a few examples myself, to get the ball rolling&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I’m not sure to what extent guidelines/conventions for big-list community-wiki questions have solidified on this site, but a couple which work well on MO and SE are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;just one example (tagline) per posted answer, so that they can be voted up/down individually;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;don't be shy of making near-duplicates if you think an idea could have been executed a bit better, nor of suggesting improvements to answers in their comments.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Related questions: &lt;a href=&quot;http://math.stackexchange.com/questions/2601/best-intuitive-metaphors-for-math-concepts-of-any-level&quot;&gt;Best intuitive metaphors for math concepts&lt;/a&gt;; &lt;a href=&quot;http://mathoverflow.net/questions/5450/cocktail-party-math&quot; rel=&quot;nofollow&quot;&gt;Cocktail party math&lt;/a&gt; (at MO).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&lt;strong&gt;p.s. I can't see how to make this question community wiki; anyone with enough rep to do so, feel free to do so &amp;amp; delete this request!&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="2439" LastEditorUserId="2439" LastEditorDisplayName="" LastEditDate="2010-10-15T16:05:38.630" LastActivityDate="2010-10-21T16:51:04.640" Title="Elevator pitch for a (sub)field of maths?" Tags="&lt;soft-question&gt;&lt;big-list&gt;&lt;intuition&gt;&lt;education&gt;" AnswerCount="8" CommentCount="6" FavoriteCount="3" />
  <row Id="6878" PostTypeId="2" ParentId="6877" CreationDate="2010-10-15T15:20:52.687" Score="2" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Calculus&lt;/strong&gt;: the mathematics of motion and change.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Not original, of course!)&lt;/p&gt;&#xA;" OwnerUserId="2439" LastActivityDate="2010-10-15T15:20:52.687" />
  <row Id="6879" PostTypeId="2" ParentId="6877" CreationDate="2010-10-15T15:25:48.790" Score="2" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Category Theory&lt;/strong&gt;:  A birds-eye view of mathematics; a framework for investigating how different areas relate to one another.&lt;/p&gt;&#xA;" OwnerUserId="2439" LastActivityDate="2010-10-15T15:25:48.790" />
  <row Id="6880" PostTypeId="2" ParentId="6873" CreationDate="2010-10-15T15:27:13.670" Score="6" ViewCount="" Body="&lt;p&gt;$$\sum (a_i^2 - b_i^2) =  \sum (a_i+b_i)(a_i-b_i) = (n-1) \sum (a_i-b_i) $$&#xA;$$= (n-1) \left(\sum a_i - \sum b_i\right) = (n-1)\left(\binom{n}{2} - \binom{n}{2}\right) = 0,$$&#xA;where $n$ is the total number of players.  (Each player plays $n-1$ games, so $a_i+b_i = n-1$, and there are a total of $\binom{n}{2}$ games played, so $\sum a_i = \sum b_i = \binom{n}{2}$.)&lt;/p&gt;&#xA;" OwnerUserId="2370" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-10-15T15:41:28.333" LastActivityDate="2010-10-15T15:41:28.333" CommentCount="0" />
  <row Id="6881" PostTypeId="2" ParentId="6873" CreationDate="2010-10-15T15:33:14.407" Score="1" ViewCount="" Body="&lt;p&gt;Let the number of wins of a random player be A, and let the number of losses of a random player be B. Then A and B have the same mean (half the number of games) and the same variance (since $B = n-A$, where n is the number of players). So $A^2$ and $B^2$ have the same expectation, which is what you want.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Of course, if you write this out more explicitly it'll resemble Mike Spivey's answer quite closely. But this is what went through my head.)&lt;/p&gt;&#xA;" OwnerUserId="173" LastActivityDate="2010-10-15T15:33:14.407" />
  <row Id="6883" PostTypeId="1" CreationDate="2010-10-15T15:46:25.823" Score="0" ViewCount="170" Body="&lt;p&gt;If a quantity is successively increased by &lt;strong&gt;20%&lt;/strong&gt; for two times and another quantity is successively decreased by &lt;strong&gt;10%&lt;/strong&gt; for three times, the first quantity is &lt;strong&gt;80%&lt;/strong&gt; of the second. What percentage is the original first quantity as compared to the original second quantity?&lt;/p&gt;&#xA;" LastActivityDate="2010-10-15T16:28:10.617" Title="Average--Average---Average" Tags="&lt;algebra&gt;" AnswerCount="1" CommentCount="14" />
  <row Id="6884" PostTypeId="2" ParentId="6873" CreationDate="2010-10-15T16:13:11.720" Score="3" ViewCount="" Body="&lt;p&gt;Consider flipping the result of one game, say between player 1 and 2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then $\sum (a_i)^2 - \sum (b_i)^2$ changes by $2(a_1 + b_1) - 2(a_2 + b_2) = 0$ as $a_i + b_i = n-1$, where $n$ is the number of players.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus $\sum (a_i)^2 - \sum (b_i)^2$ is the same no matter what the results and thus is same as the extreme case when the winner wins all matches, the runner up all but one etc, for which the sum is 0.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-10-15T16:13:11.720" />
  <row Id="6885" PostTypeId="2" ParentId="6877" CreationDate="2010-10-15T16:18:22.660" Score="3" ViewCount="" Body="&lt;p&gt;There's the old &quot;&lt;strong&gt;Topology:&lt;/strong&gt; Geometry on a rubber sheet.&quot; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For both Category Theory and Universal Algebra, I also like &lt;em&gt;The Esperanto of Mathematics/Algebra&lt;/em&gt; (except for the negative connotation of &quot;nobody actually uses it&quot;...).&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-10-15T16:18:22.660" CommunityOwnedDate="2010-10-15T16:18:22.660" CommentCount="2" />
  <row Id="6886" PostTypeId="2" ParentId="6870" CreationDate="2010-10-15T16:28:03.537" Score="0" ViewCount="" Body="&lt;p&gt;(I don't have the points yet to make comment)&#xA;Are you a computer scientist? Because I think your view is more about typed languages.&lt;/p&gt;&#xA;" OwnerUserId="2468" LastActivityDate="2010-10-15T16:28:03.537" />
  <row Id="6887" PostTypeId="2" ParentId="6883" CreationDate="2010-10-15T16:28:10.617" Score="1" ViewCount="" Body="&lt;p&gt;Say your quantity is $Q$. To get 80% of that, you multiply by $0.8$, so $0.8Q$. How do you get 80% of &lt;em&gt;that&lt;/em&gt;? Well, you multiply by $0.8$ again, so you get $0.8(0.8Q) = (0.8)^2Q$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do you take 80% successively three times? Four times? Five times? $n$ times?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you first take $k$% ($k$ between $0$ and $100$), then  you multiply by $\frac{k}{100}$. So if you started with $Q$, you now have $\frac{k}{100}Q$. If you then take $\ell$% (again, $\ell$ between $0$ and $100$), then you need to multiply &lt;em&gt;that&lt;/em&gt; by $\frac{\ell}{100}$. So you would have $\left(\frac{\ell}{100}\right)\left(\frac{k}{100}\right)Q$. Etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, say that instead of your problem, I wanted to first decrease by 10%, then the result by 20%, then the result of &lt;em&gt;that&lt;/em&gt; by 45%, and finally the result of that by 15%. How would I do it? I would get $(.85)(.55)(.8)(.9)Q$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are &lt;em&gt;increasing&lt;/em&gt; by 20%, then instead of multiplying by 0.8, you multiply by $1.2$ (since adding 20% means taking 120% of the original). Same idea.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, you have two quantities, $Q_1$ and $Q_2$; you are doing different things to them. At the end you will have some multiple of $Q_1$ and some multiple of $Q_2$, say $aQ_1$ and $bQ_2$. But you are told that &quot;the first quantity is 80% of the second&quot;. That means that you can write $aQ_1$ as a multiple of $bQ_2$, from which you can &quot;solve for $Q_1$&quot; in terms of $Q_2$. That should tell you exactly what you want to know.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-10-15T16:28:10.617" />
  <row Id="6888" PostTypeId="2" ParentId="6868" CreationDate="2010-10-15T16:31:19.577" Score="5" ViewCount="" Body="&lt;p&gt;This was a very old Monthly problem - see below. For an excellent introduction to the general topic see Richard Stanley's paper &lt;a href=&quot;http://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.118.5996&quot; rel=&quot;nofollow&quot;&gt;An Introduction to Hyperplane Arrangements (2004)&lt;/a&gt; and also Renteln's lecture slides &lt;a href=&quot;http://physics.csusb.edu/~prenteln/papers/hyperintro.pdf&quot; rel=&quot;nofollow&quot;&gt;It All Depends on How You Slice It: An Introduction to Hyperplane Arrangements, 2008&lt;/a&gt;&#xA;&lt;img src=&quot;http://i.imgur.com/U07We.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-10-15T16:39:17.810" LastActivityDate="2010-10-15T16:39:17.810" />
  <row Id="6889" PostTypeId="2" ParentId="6870" CreationDate="2010-10-15T17:07:18.370" Score="7" ViewCount="" Body="&lt;p&gt;You are quite right: $f$ by itself should denote a function, $f(x)$ by&#xA;itself should denote the element in the codomain of $f$ (in this case,&#xA;the real number) that results when you evaluate $f$ on the element $x$&#xA;of its domain, where $x$ should previously have been defined.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, this rule is honored as much in the breach as the observance:&#xA;there are many situations where it is convenient to break it.  When&#xA;defining a function by a formula, it's hard to avoid a dummy variable,&#xA;and so one likes to say &quot;let $f(x)=e^{-x}$&quot; instead of the more&#xA;correct but awkward &quot;let $f$ be the function $x \mapsto e^{-x}$&quot;.  In&#xA;particular, with functions that have multi-letter symbols like $\sin$,&#xA;I find that people generally prefer to avoid writing them without an&#xA;argument like $\sin x$.  One does not like to talk about $\sin$ as a&#xA;function in itself, so instead of writing something like $\sin'' =&#xA;-\sin$, one would rather say &quot;if $f(x)=\sin x$, then $f'' = -f$&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An alternative to a dummy variable that's sometimes used is a dot:&#xA;$\cdot$.  People sometimes write &quot;let $f=g(\cdot + 5)$&quot; to avoid the&#xA;less correct &quot;let $f(x)=g(x+5)$&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When working with functions of several variables, using dummy&#xA;variables often helps keep track of which variable is which.  One&#xA;often writes something like: &quot;let $u(x,t)$ be a solution of the heat&#xA;equation $\frac{\partial u}{\partial t} - \frac{\partial^2 u}{\partial&#xA;x^2} = 0$&quot;.  Of course, one is really talking about the function $u$&#xA;and not any particular real number of the form $u(x,t)$, but it would&#xA;be much more awkward to write otherwise.  It also reminds you that the&#xA;first argument of $u$ should be interepreted as space and the second&#xA;one as time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In short: mathematicians are not compilers.  Written mathematics has&#xA;some syntax rules, but they are not quite hard-and-fast, and need not&#xA;be followed at the expense of clarity.&lt;/p&gt;&#xA;" OwnerUserId="822" LastActivityDate="2010-10-15T17:07:18.370" CommentCount="2" />
  <row Id="6890" PostTypeId="1" AcceptedAnswerId="7276" CreationDate="2010-10-15T17:19:51.373" Score="1" ViewCount="278" Body="&lt;p&gt;Hi everybody!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd like to understand the physical meaning of the Feynman's vector potential definition:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ A_{m}^{(b)}(x) = e_b \int \delta (xb_{\mu}xb^{\mu})db_m(b), \qquad m=0,1,2,3 $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(component m of the vector potential of the particle b at the point x)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;$\delta$ is the Dirac delta function&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$ xb_{\mu}xb^{\mu} $ is the 4-norm of $x - b$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;So the $\delta$ function is non-zer $\iff x$ is in the boundary of the light cone of b.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Can you give me any clues?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="2354" LastEditorUserId="2354" LastEditorDisplayName="" LastEditDate="2010-10-15T18:08:44.357" LastActivityDate="2010-10-21T07:04:42.917" Title="Feynman's vector potential" Tags="&lt;physics&gt;" AnswerCount="2" CommentCount="6" FavoriteCount="1" />
  <row Id="6891" PostTypeId="1" CreationDate="2010-10-15T19:36:40.820" Score="4" ViewCount="91" Body="&lt;p&gt;A phase change is only possible in a physical system which obeys the laws of statistical mechanics if the infinite series for the partition function of that system converges non-uniformly (i.e. converges point-wise).  This is because in order for a phase change to occur, the partition function must converge to two different continuous functions in different regions of the phase plane; and this can only happen if the number of terms in the partition function is infinite.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It might be possible, using the analogy with phase changes, to make some general statements about partition functions which converge non-uniformly to two different continuous functions.  This is interesting to me because point-wise convergence is a very weak condition, and the analogy with phase changes seems to suggest a path towards a deeper understanding of this form of convergence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of the most interesting subjects in the statistical mechanics of phase transitions is the so-called &quot;renormalization group&quot; (which is not a group) and the critical exponents it predicts (these are non integer numbers which are associated with a &quot;critical point&quot; in the phase plane).  Another powerful result from renormalization group theory is the concept of universality, in which the thermodynamic variables of the system are symmetric across all length scales at a certain critical point.  Universality is directly related to the critical exponents and critical point behavior, but as is fairly standard for theoretical physics, the mathematics used is non-rigorous (I recommend the book &quot;scaling and renormalization in statistical physics&quot; by Cardy)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the point-wise convergence of a partition function, can we put the foundations of critical exponents on a rigorous mathematical basis; in the sense that we can somehow predict their values from properties (such as the partial derivatives) of the partition function?&lt;/p&gt;&#xA;" OwnerUserId="1417" LastEditorUserId="1417" LastEditorDisplayName="" LastEditDate="2010-10-15T19:50:51.087" LastActivityDate="2010-10-15T19:50:51.087" Title="Critical exponents and point-wise convergence." Tags="&lt;analysis&gt;&lt;statistics&gt;" FavoriteCount="1" />
  <row Id="6892" PostTypeId="2" ParentId="6877" CreationDate="2010-10-15T19:55:37.850" Score="3" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Algebra:&lt;/strong&gt;&#xA;  The mathematics of structures.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Set Theory:&lt;/strong&gt; The main foundation of mathematics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Logic:&lt;/strong&gt; The main foundation of set theory.&lt;/p&gt;&#xA;" OwnerUserId="1154" LastActivityDate="2010-10-15T19:55:37.850" />
  <row Id="6893" PostTypeId="1" AcceptedAnswerId="6901" CreationDate="2010-10-15T20:01:39.893" Score="2" ViewCount="138" Body="&lt;p&gt;A loan for $8,000$ must be repaid with 6 year end payments at an annual rate of $11 \%$. What is the annual payment? I know that the present value of an annuity with end payments is $\frac{1-v^n}{i}$ where $v = 1/(1+i)$. Likewise, the future value is $(1+i)^{n}\frac{1-v^n}{i}$. How do I use this to solve this problem?&lt;/p&gt;&#xA;" OwnerUserId="2548" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2010-10-18T10:14:29.473" LastActivityDate="2010-10-18T10:14:29.473" Title="Annuities and Loans" Tags="&lt;homework&gt;&lt;finance&gt;" AnswerCount="1" CommentCount="6" FavoriteCount="1" />
  <row Id="6894" PostTypeId="1" AcceptedAnswerId="6957" CreationDate="2010-10-15T20:03:10.483" Score="2" ViewCount="141" Body="&lt;p&gt;I wanted to prove that $\varphi$ is multiplicative (that for $(a,b)=1$, $\varphi(ab)=\varphi(a)\varphi(b)$) using the following idea:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Define $\varphi'$ by $n = \varphi'(n) + \varphi(n)$.&lt;/li&gt;&#xA;&lt;li&gt;Multiply out $\varphi(a)\varphi(b) = (a - \varphi'(a))(b - \varphi'(b)) = ab - \varphi'(a)b - a \varphi'(b) + \varphi'(a)\varphi'(b)$&lt;/li&gt;&#xA;&lt;li&gt;Use the principle of inclusion exclusion ($|A \cup B| = |A| + |B| - |A \cap B|$).&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;but I could not get it to work out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this approach possible at all? How can it be saved?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;So I just need to construct sets $A(a,b)$ and $B(a,b)$ that satisfy the following&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{matrix}&#xA;\varphi'(ab) &amp;amp;=&amp;amp; \varphi'(a) b &amp;amp;+&amp;amp; a \varphi'(b) &amp;amp;-&amp;amp; \varphi'(a)\varphi'(b)\\&#xA;|| &amp;amp; &amp;amp; || &amp;amp; &amp;amp; || &amp;amp; &amp;amp; || \\&#xA;|A \cup B| &amp;amp;=&amp;amp; |A| &amp;amp;+&amp;amp; |B| &amp;amp;-&amp;amp; |A \cap B|&#xA;\end{matrix}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;then the proof follows from simple algebra. I just cannot find any way to construct such sets - since the identity is true I imagine they do exist but I am not be completely sure especially since I couldn't construct them.&lt;/p&gt;&#xA;" LastActivityDate="2010-10-16T15:38:21.377" Title="Can this broken proof (of multiplicative of $\varphi$) attempt be fixed?" Tags="&lt;number-theory&gt;" AnswerCount="1" CommentCount="7" />
  <row Id="6895" PostTypeId="2" ParentId="6868" CreationDate="2010-10-15T20:08:30.097" Score="1" ViewCount="" Body="&lt;p&gt;This is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Lazy_caterer%27s_sequence&quot; rel=&quot;nofollow&quot;&gt;lazy caterer's sequence&lt;/a&gt;. As others have mentioned, arbitrary dimensional analogues are called hyperplane arrangements.&lt;/p&gt;&#xA;" OwnerUserId="2549" LastActivityDate="2010-10-15T20:08:30.097" />
  <row Id="6896" PostTypeId="2" ParentId="6851" CreationDate="2010-10-15T20:19:28.447" Score="0" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT:&lt;/strong&gt; When you look at the answer of Chandru1 you must think twice. Do you know any algebraic properties of the absolute value? How do you solve an inequality?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might also like to think of different points $(x,y)=(5,5)$, $(x,y)=(5,-5)$, $(x,y)=(-5,5)$, $(x,y)=(-5,-5)$ -- are they in you set? $(x,y)=(1/5,5)$, $(x,y)=(1/5,-5)$, $(x,y)=(-1/5,5)$, $(x,y)=(-1/5,-5)$ -- are they in you set?&lt;/p&gt;&#xA;" OwnerUserId="1154" LastActivityDate="2010-10-15T20:19:28.447" />
  <row Id="6897" PostTypeId="1" AcceptedAnswerId="6900" CreationDate="2010-10-15T20:26:48.423" Score="6" ViewCount="89" Body="&lt;h2&gt;The Setting&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Let $K$ be an &lt;a href=&quot;http://en.wikipedia.org/wiki/Archimedean_field&quot; rel=&quot;nofollow&quot;&gt;Archimedean field&lt;/a&gt;. TFAE:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$K$ has the least upper bound property.&lt;/li&gt;&#xA;&lt;li&gt;Every &lt;a href=&quot;http://en.wikipedia.org/wiki/Cauchy_sequence#In_topological_groups&quot; rel=&quot;nofollow&quot;&gt;Cauchy sequence&lt;/a&gt; in $K$'s additive group converges.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Now proving that 1 implies 2 is easy, but the other direction is slightly harder. Not that that's a problem. Rather the problem is that I can't see a route that doesn't invoke at least &lt;a href=&quot;http://en.wikipedia.org/wiki/Dependent_choice&quot; rel=&quot;nofollow&quot;&gt;dependent choice&lt;/a&gt; at some point.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Strategy 1&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Starting with a nonempty set $A$ that's bounded above, you could construct a monotonely non-decreasing Cauchy sequence of upper bounds that has the supremum as its limit. Here's a short sketch: Pick an upper bound $B_0$ of $A$. Pick an $a_0 \in A$. Recursively define&#xA;$$ B_{i+1} = \begin{cases}&#xA;\frac{B_i+a_i}{2}, &amp;amp; \text{ if that's an upper bound for } A \\&#xA;B_i,                &amp;amp; \text{ otherwise}&#xA;\end{cases} $$&#xA;and&#xA;$$ a_{i+1} = \begin{cases}&#xA;a_i,  &amp;amp; \text{ if $\tfrac{a_i+B_i}{2}$ is an upper bound for $A$}\\&#xA;\text{choose any } a \in A \text{ s.t. } \frac{a_i+B_i}{2} &amp;lt; a,  &amp;amp; \text{ otherwise.}&#xA;\end{cases} $$&#xA;I can't see a way to get rid of the choice because you really want the $a_i$ to be in $A$ for the argument to go through.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Strategy 2&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Okay, let's go the long way instead! First we show that $K$ complete implies $[a,b]$ compact. Then we show that &lt;em&gt;that&lt;/em&gt; implies that closed and bounded subsets are compact (&quot;Heine-Borel property&quot;). And finally we show that not(Heine-Borel property) implies not(least upper bound property). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I already get stumped on the first part. Clearly it's easy to show that $K$ complete implies $[a,b]$ is sequentially compact. And from here it'd be nice to use that $K$ is 2nd countable (the intervals with rational endpoints are a basis) to get that $[a,b]$ is in fact compact. So you start with an open covering $U_\alpha$ of $[a,b]$. 2nd countable spaces are Lindelöf... wait... let's make sure and prove that. Let $\{B_i\}$ be a countable basis. Then for each $B_i$ you choose a $U_\alpha$... oh. Choice crept up again.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So my question is this: Does this really require (an admittedly weak form of) choice? Or is there a way to do without?&lt;/p&gt;&#xA;" OwnerUserId="1269" LastActivityDate="2010-10-15T20:53:31.847" Title="Archimedian field $K$ has LUB property iff it's complete requires DC?" Tags="&lt;real-analysis&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="6899" PostTypeId="1" AcceptedAnswerId="6904" CreationDate="2010-10-15T20:46:09.007" Score="3" ViewCount="239" Body="&lt;p&gt;Can we find all positive integers $a,b$ such that $a^{n}+b^{n}$ is an $(n+1)^{th}$ power? I think this question equivalent to solving the statement $$a^{n} + b^{n} = c^{n+1}$$ for $a,b,c \in \mathbb{N}$. But i don't know as to how i can solve this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I attempted by subsituting $n$ as $n+1$ so that we get $$a^{n+1} + b^{n+1} = c^{n+2} = c \times c^{n+1}=(a^{n}+b^{n}) \cdot c$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but it seems that this doesn't help!&lt;/p&gt;&#xA;" LastEditorUserId="1778" LastEditorDisplayName="" LastEditDate="2010-10-16T17:31:37.543" LastActivityDate="2010-10-16T17:31:37.543" Title="Finding $a,b$ such that $a^{n} + b^{n} $ is $(n+1)^{th}$ power" Tags="&lt;number-theory&gt;&lt;diophantine-equations&gt;" AnswerCount="4" CommentCount="1" />
  <row Id="6900" PostTypeId="2" ParentId="6897" CreationDate="2010-10-15T20:53:31.847" Score="4" ViewCount="" Body="&lt;p&gt;It seems to me that you can modify strategy 1 as follows.  Given a nonempty set $A$ that's bounded above, let $x_0$ be a non-upper-bound for $A$ and let $y_0$ be an upper bound for $A$.  Recursively define&#xA;$$&#xA;y_{n+1} \;=\; \begin{cases}(x_n+y_n)/2 &amp;amp; \text{if this is an upper bound for $A$} \\&#xA; y_n &amp;amp; \text{otherwise.}  \end{cases}&#xA;$$&#xA;and&#xA;$$&#xA;x_{n+1} \;=\; \begin{cases}(x_n+y_n)/2 &amp;amp; \text{if this is a non-upper-bound for $A$} \\&#xA; x_n &amp;amp; \text{otherwise.}  \end{cases}&#xA;$$&#xA;Note that $x_1 \leq x_2 \leq \cdots \leq y_2 \leq y_1$. Furthermore, each $y_n$ is an upper bound for $A$, and each $x_n$ is a non-upper-bound for $A$.  It is easy to prove that both sequences are Cauchy sequences, and that they converge to the same limit $L$.  We claim that $L$ is the least upper bound for $A$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both directions are fairly easy.  If $a \in A$, then $a \leq y_n$ for all $n$, which proves that $a \leq L$.  Thus $L$ is in fact an upper bound for $A$.  Next, if $u$ is any upper bound for $A$, then $u \geq x_n$ for all $n$, and therefore $u \geq L$.  Thus $L$ is the least upper bound for $A$.&lt;/p&gt;&#xA;" OwnerUserId="1726" LastActivityDate="2010-10-15T20:53:31.847" CommentCount="1" />
  <row Id="6901" PostTypeId="2" ParentId="6893" CreationDate="2010-10-15T21:11:00.633" Score="2" ViewCount="" Body="&lt;p&gt;We have to find the value of &lt;strong&gt;constant&lt;/strong&gt; (I assume, since nothing is specified on the contrary) payments $A$ given the principal $P$ during $n$ yearly periods and the interest rate $i$. The value of $A$ in the period $k$ is equivalent to the present value $A/\left( 1+i\right) ^{k}$ monetary units, where $i$ is the interest rate in each capitalization period. Summing in $k$, from 1 to $n$, we get the sum&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle\sum_{k=1}^{n}\dfrac{A}{\left( 1+i\right) ^{k}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a geometric progression with ratio $r=1/(1+i)$ and first term $u_{1}=A/\left( 1+i\right)$ whose sum is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\dfrac{A}{1+i}\dfrac{\left( \dfrac{1}{1+i}\right)^{n}-1}{\dfrac{1}{1+i}-1}=A\dfrac{\left( 1+i\right) ^{n}-1}{i\left( 1+i\right) ^{n}}=P.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$A=P\dfrac{i\left( 1+i\right) ^{n}}{\left( 1+i\right) ^{n}-1}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the given problem, the payments will be made during $n=6$ years, with $i=11\%=\dfrac{11}{100}$ and $P=8000$: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$A=8000\times \dfrac{0.11(1.11)^{6}}{(1.11)^{6}-1}\approx 1891$$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Sum of a geometric progression:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$S=u_{1}+u_{2}+u_{3}+\ldots +u_{n}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$rS=ru_{1}+ru_{2}+ru_{3}+\ldots +ru_{n-1}+ru_{n}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$u_{k}=ru_{k-1}=u_{1}r^{k-1}$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$S-rS=\left( u_{1}+u_{2}+u_{3}+\ldots +u_{n}\right) -\left(&#xA;ru_{1}+ru_{2}+ru_{3}+\ldots +ru_{n-1}+ru_{n}\right) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(1-r)S=u_{1}-ru_{n}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$S=\dfrac{u_{1}-ru_{n}}{1-r}=\dfrac{u_{1}-u_{1}r^{n}}{1-r}=u_1\times\dfrac{1-r^{n}}{1-r}=u_1\times\dfrac{r^{n}-1}{r-1}.$$&lt;/p&gt;&#xA;" OwnerUserId="752" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2010-10-16T18:45:51.023" LastActivityDate="2010-10-16T18:45:51.023" CommentCount="3" />
  <row Id="6902" PostTypeId="2" ParentId="6899" CreationDate="2010-10-15T21:11:30.927" Score="1" ViewCount="" Body="&lt;p&gt;A few to start you off:  for n=1 there are many solutions.  a=b=c=2 works for all n.  There is &lt;a href=&quot;http://en.wikipedia.org/wiki/Fermat%27s_theorem_on_sums_of_two_squares&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Fermat%27s_theorem_on_sums_of_two_squares&lt;/a&gt;.  You can apply that to c^3 to see if all factors of c which are 3 mod 4 come with even exponents you can solve it.  I suspect that is it as there just aren't that many powers to choose from, but don't have a proof.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-10-15T21:11:30.927" />
  <row Id="6903" PostTypeId="2" ParentId="6877" CreationDate="2010-10-15T21:19:06.357" Score="2" ViewCount="" Body="&lt;p&gt;&lt;b&gt;Geometry&lt;/b&gt;: The ticket to Plato's Academy. :-)&lt;/p&gt;&#xA;" OwnerUserId="237" LastActivityDate="2010-10-15T21:19:06.357" />
  <row Id="6904" PostTypeId="2" ParentId="6899" CreationDate="2010-10-15T21:30:49.183" Score="7" ViewCount="" Body="&lt;p&gt;When $n = 2$ it's straightforward to give a more-or-less complete solution: $c^3$ is a sum of squares if and only if $c$ is, so write $c = d^2 + e^2$ and the rest is factorization over the Gaussian integers.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For all $n$ there is the family of solutions $a = x(x^n + y^n), b = y(x^n + y^n), c = x^n + y^n$ for integers $x, y$.  But I think finding all solutions is likely to be incredibly hard.  &lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-10-15T21:30:49.183" CommentCount="1" />
  <row Id="6905" PostTypeId="2" ParentId="6877" CreationDate="2010-10-15T21:41:44.977" Score="2" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Statistics&lt;/strong&gt;: Making decisions under uncertainty&lt;/p&gt;&#xA;" OwnerUserId="247" LastActivityDate="2010-10-15T21:41:44.977" />
  <row Id="6906" PostTypeId="1" AcceptedAnswerId="7044" CreationDate="2010-10-15T21:47:23.527" Score="5" ViewCount="431" Body="&lt;p&gt;By &quot;Sylvester's Law of Inertia,&quot; I mean:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Sylvester%27s_law_of_inertia&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Sylvester%27s_law_of_inertia&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can guess that Sylvester is the name of the dude that proved this theorem, but I have no idea as to why it's called &quot;Law of Inertia.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where is this name coming from? I guess it's from physics, but... I just don't see the connection.&lt;/p&gt;&#xA;" OwnerUserId="2499" LastEditorUserId="1120" LastEditorDisplayName="" LastEditDate="2010-10-15T21:51:08.800" LastActivityDate="2010-10-17T17:53:28.147" Title="Why is it called Sylvester's Law of Inertia?" Tags="&lt;soft-question&gt;&lt;math-history&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="6907" PostTypeId="2" ParentId="6906" CreationDate="2010-10-15T21:59:38.250" Score="10" ViewCount="" Body="&lt;p&gt;From Sylvester's &lt;em&gt;On the Theory of the Syzygetic Relations&lt;/em&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;This constant number of positive signs which attaches to a quadratic function under all its transformations [...] may be termed conveniently its &lt;em&gt;inertia&lt;/em&gt;, until a better word is found.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="274" LastEditorUserId="274" LastEditorDisplayName="" LastEditDate="2010-10-15T22:53:01.527" LastActivityDate="2010-10-15T22:53:01.527" CommentCount="5" />
  <row Id="6908" PostTypeId="1" AcceptedAnswerId="6925" CreationDate="2010-10-15T22:00:28.357" Score="6" ViewCount="215" Body="&lt;p&gt;This is a follow-up to &lt;a href=&quot;http://math.stackexchange.com/questions/6828/proving-sqrt1-x2-ge-texterf-sqrt-log-x&quot;&gt;this&lt;/a&gt; question. I came across a tighter potential bound and checked it numerically for $0\le x\le 5$. I think it holds for all positive $x$, can anyone see a proof?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$1-\exp(-4x^2/ \pi) \ge \text{erf}(x)^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: using analysis for previous &lt;a href=&quot;http://math.stackexchange.com/questions/6828/proving-sqrt1-x2-ge-texterf-sqrt-log-x&quot;&gt;question&lt;/a&gt; you can show that $1-\exp(-k x^2)$ is an upper bound on $\text{erf}(x)^2$ for $k=2$ and a lower bound when $k=1$. The factor $k=4/\pi$ comes out when numerically searching for tightest upper bound. Not only does it seem to give an upper bound, but it also tracks $\text{erf}(x)^2$ very closely.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dashed graph below is $\text{erf}(x)^2$, red is $1-\exp(-k x^2)$ for $k=4/\pi$, other two graphs are for $k=1$ and $k=2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://yaroslavvb.com/upload/bounds.png&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="998" LastEditorUserId="998" LastEditorDisplayName="" LastEditDate="2010-10-16T00:04:42.547" LastActivityDate="2010-10-16T02:26:03.177" Title="Proving $1-\exp(-4x^2/ \pi) \ge \text{erf}(x)^2$" Tags="&lt;inequality&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="6909" PostTypeId="1" CreationDate="2010-10-15T22:57:35.767" Score="0" ViewCount="129" Body="&lt;p&gt;1.&lt;strong&gt;How many  number of three digit even numbers than can be formed out of the digits 0 to 9 ?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question seems confusing since there is no mention of whether repetition is allowed or not ?! since if repetition is allowed then the answer would be &lt;strong&gt;450&lt;/strong&gt; (&lt;code&gt;5*9*10&lt;/code&gt;) where as if not then the answer would be &lt;strong&gt;328&lt;/strong&gt; (&lt;code&gt;9p2+ 4*(8*8) = 328&lt;/code&gt;). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Am I correct about the question or am missing some point ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;2.How many calendars that can be prepared for the month of February ?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;3.A golf player wants to put the ball in the hole in 5 shots. He says that he will put the ball in the hole at most by 3 shots and will qualify for the next round. how many number of possibilities of shots he had played ?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't have any clue for this two.&lt;/p&gt;&#xA;" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-10-16T03:06:42.757" LastActivityDate="2010-10-16T03:06:42.757" Title="Three question on Permutation and combination" Tags="&lt;combinatorics&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="6911" PostTypeId="2" ParentId="6909" CreationDate="2010-10-15T23:14:37.593" Score="1" ViewCount="" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;I get the same answers you do.  I also agree the question is unclear on repetition.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Think of how many ways the pattern of days in February can be different, starting day of week and length&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I don't understand this question&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2010-10-16T00:07:37.430" LastActivityDate="2010-10-16T00:07:37.430" CommentCount="1" />
  <row Id="6913" PostTypeId="2" ParentId="6899" CreationDate="2010-10-15T23:46:35.703" Score="3" ViewCount="" Body="&lt;p&gt;For $n \ge 3$, the Fermat-Catalan conjecture (http://en.wikipedia.org/wiki/Fermat–Catalan_conjecture) implies that there are only finitely many solutions to your equation. (But the Fermat-Catalan conjecture is just a conjecture!)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: as pointed out by yrudoy, this doesn't actually follow. I still think there should be only finitely many solutions, though, based on some probabilistic considerations. See for example &lt;a href=&quot;http://www.thehcmr.org/issue1_1/elkies.pdf&quot; rel=&quot;nofollow&quot;&gt;Noam Elkies, The ABCs of Number Theory&lt;/a&gt;, in vol. 1, issue 1 of the &lt;a href=&quot;http://www.thehcmr.org/&quot; rel=&quot;nofollow&quot;&gt;Harvard College Mathematics Review&lt;/a&gt;, p. 61. (This gives a standard heuristic argument for Fermat's last theorem only having finitely many solutions for each $n \ge 4$, which adapts to this case.&lt;/p&gt;&#xA;" OwnerUserId="173" LastEditorUserId="173" LastEditorDisplayName="" LastEditDate="2010-10-16T14:39:01.830" LastActivityDate="2010-10-16T14:39:01.830" CommentCount="1" />
  <row Id="6914" PostTypeId="1" CreationDate="2010-10-15T23:49:12.827" Score="7" ViewCount="357" Body="&lt;p&gt;I'm curious about the rights of ownership regarding problems presented in math textbooks. Can a math problem in general be considered intellectual property of its &quot;inventor&quot;? I'm assuming textual word problems can be &quot;owned&quot; in the same way other literature is by publishers/authors but it becomes more gray with purely numerical expressions. Is it merely a contextual ownership, for example, say no other math author can include such a problem in another textbook or is it more inclusive regarding other mediums like software or blogs etc where money is being made? An &quot;elucidation&quot; here would be much appreciated, thanks. &lt;/p&gt;&#xA;" OwnerUserId="2054" LastActivityDate="2010-10-16T05:30:33.967" Title="Can someone copyright/own a math problem?" Tags="&lt;books&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="1" />
  <row Id="6916" PostTypeId="1" AcceptedAnswerId="6918" CreationDate="2010-10-16T00:59:14.553" Score="1" ViewCount="193" Body="&lt;p&gt;I've always been taught that one way to look at complex numbers is as a cartesian space, where the &quot;real&quot; part is the x component and the &quot;imaginary&quot; part is the y component.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this sense, these complex numbers are like vectors, and they can be added geometrically like normal vectors can.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, is there a geometric interpretation for the multiplication of two complex numbers?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried out two test ones, $3+i$ and $-2+3i$, which multiply to $-9+7i$.  But no geometrical significance seems to be found.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a geometric significance for the multiplication of complex numbers?&lt;/p&gt;&#xA;" OwnerUserId="51" LastActivityDate="2010-10-16T01:22:26.853" Title="Geometric interpretation of the multiplication of complex numbers?" Tags="&lt;ring-theory&gt;&lt;complex-numbers&gt;&lt;vector-space&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="6918" PostTypeId="2" ParentId="6916" CreationDate="2010-10-16T01:16:02.960" Score="6" ViewCount="" Body="&lt;p&gt;Suppose we multiply the complex numbers $z_1$ and $z_2$. If these numbers are written in the polar form as $r_1 e^{i \theta_1}$ and $r_2 e^{i \theta_2}$, the product will be $r_1 r_2 e^{i (\theta_1 + \theta_2)}$. Equivalently, we are stretching the first complex number $z_1$ by a factor equal to the magnitude of the second complex number $z_2$ and then rotating the stretched $z_1$ counter-clockwise by an angle $\theta_2$ to arrive at the product. There are several websites that expand upon this intuition with graphics and more explanation. See this site for example - &lt;a href=&quot;http://www.suitcaseofdreams.net/Geometric_multiplication.htm&quot; rel=&quot;nofollow&quot;&gt;http://www.suitcaseofdreams.net/Geometric_multiplication.htm&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="618" LastActivityDate="2010-10-16T01:16:02.960" CommentCount="1" />
  <row Id="6919" PostTypeId="2" ParentId="6916" CreationDate="2010-10-16T01:20:16.810" Score="4" ViewCount="" Body="&lt;p&gt;Yes, there is a simple geometric meaning, but you need to convert to the polar form of the complex numbers to see it clearly.  $3+i$ has magnitude $\sqrt{10}$ and angle about $18^\circ$;&#xA;$-2+3i$ has magnitude $\sqrt{13}$ and angle about $124^\circ$.&#xA;Multiplication of the complex numbers multiplies the two magnitudes, resulting in $\sqrt{130}$,&#xA;and adds the two angles, $142^\circ$.&#xA;In other words, you can view the second number as scaling and rotating the first (or the first scaling and rotating the second).&lt;/p&gt;&#xA;" OwnerUserId="237" LastActivityDate="2010-10-16T01:20:16.810" />
  <row Id="6920" PostTypeId="1" CreationDate="2010-10-16T01:21:09.657" Score="2" ViewCount="143" Body="&lt;p&gt;I'm having difficulty justifying the change of limits in the derivation of the Riemann-Liouville derivative at &lt;a href=&quot;http://www.xuru.org/fc/RiemannLiouville.asp&quot; rel=&quot;nofollow&quot;&gt;xuru.org&lt;/a&gt;. What I don't undestand is how $\int_0^{t_2}$ becomes $\int_{t_1}^x$ in the following statement,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\int_0^x \int_0^{t_2} f(t_1) dt_1 dt_2 = \int_0^x \int_{t_1}^x f(t_1) dt_2 dt_1$&lt;/p&gt;&#xA;" OwnerUserId="1752" LastEditorUserId="1752" LastEditorDisplayName="" LastEditDate="2010-10-16T02:18:32.623" LastActivityDate="2011-03-17T23:24:27.627" Title="Change of limits in derivation of Riemann-Liouville (Fractional) Derivative" Tags="&lt;calculus&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="6921" PostTypeId="2" ParentId="6916" CreationDate="2010-10-16T01:22:26.853" Score="6" ViewCount="" Body="&lt;p&gt;Add the angles and multiply the lengths.&lt;/p&gt;&#xA;" OwnerUserId="130" LastActivityDate="2010-10-16T01:22:26.853" />
  <row Id="6922" PostTypeId="2" ParentId="6920" CreationDate="2010-10-16T01:42:58.927" Score="5" ViewCount="" Body="&lt;p&gt;Notice that $0\leq t_2\leq x$ and $0\leq t_1\leq t_2$ is equivalent to $0\leq t_1\leq x$ and $t_1\leq t_2\leq x$.  You could shorten the description to $0\leq t_1\leq t_2\leq x$.  Whether you believe me or not, I recommend that you draw a picture.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The point is that both integrals are over the same triangular region.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2010-10-16T02:28:04.307" LastActivityDate="2010-10-16T02:28:04.307" CommentCount="1" />
  <row Id="6923" PostTypeId="1" AcceptedAnswerId="6929" CreationDate="2010-10-16T02:10:25.573" Score="2" ViewCount="93" Body="&lt;p&gt;What is, in the language of Schemes, a G-galois branched cover?&lt;/p&gt;&#xA;" OwnerUserId="1971" LastEditorUserId="1971" LastEditorDisplayName="" LastEditDate="2010-10-16T02:21:28.887" LastActivityDate="2010-10-17T13:35:29.860" Title="What is a G-Galois Branched Cover" Tags="&lt;algebraic-geometry&gt;&lt;schemes&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="6924" PostTypeId="2" ParentId="6899" CreationDate="2010-10-16T02:10:29.363" Score="4" ViewCount="" Body="&lt;p&gt;The hard case is relatively prime solutions. For $n &gt; 2$ it is conjectured that there are no solutions.  This follows from the more precise conjecture that there are no relatively prime solutions to $x^a + y^b = z^c$ in the hyperbolic case $1/a + 1/b + 1/c &amp;lt; 1$ other than the ten or so examples found by Beukers and Zagier in a computer search (there are large solutions with $(a,b,c)=(2,3,7)$ and a few others) shortly after Wiles announced the proof of Fermat's Last Theorem.   It was proven by Darmon and Granville that the number of relatively prime solutions is finite for any given $n &gt; 2$, and this required the Mordell conjecture (Faltings' theorem) on finiteness of rational points on curves of genus $g \geq 2$.  There is no known elementary approach to these questions, and resolution of the problem for particular values of $(a,b,c)$ uses the full machinery of modern algebraic geometry and number theory. &lt;/p&gt;&#xA;" OwnerUserId="467" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-10-16T14:19:38.887" LastActivityDate="2010-10-16T14:19:38.887" />
  <row Id="6925" PostTypeId="2" ParentId="6908" CreationDate="2010-10-16T02:20:52.313" Score="7" ViewCount="" Body="&lt;p&gt;As before we consider &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$erf(x)^2={4\over \pi}\int_0^x\int_0^x \exp{-(s^2+t^2)}\  ds \ dt$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now compare this with the same over the area which is given by the quarter of a circle of radius $\displaystyle \frac{2x}{\sqrt{\pi}}$. The area of this is same as the area of the square of side $x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $\displaystyle e^{-(s^2 + t^2)}$ decreases as $\displaystyle s^2 + t^2$ increases, we are done!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The integral over the non-common area for the quarter circle is greater than the integral over the non-common area of the square (which lies outside the circle and so $\displaystyle s^2 + t^2$ is higher in that region).&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-10-16T02:26:03.177" LastActivityDate="2010-10-16T02:26:03.177" CommentCount="9" />
  <row Id="6926" PostTypeId="1" CreationDate="2010-10-16T02:53:42.427" Score="0" ViewCount="93" Body="&lt;p&gt;$r$ is a primitive root for the prime $p$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x_1\equiv r^a \pmod{p}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x_2\equiv r^b \pmod{p}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $gcd(b, p-1)=1$, how can I determine $r$ if $p$, $x_2$, and $b$ are known?&lt;/p&gt;&#xA;" OwnerUserId="2375" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-10-16T03:08:52.533" LastActivityDate="2010-10-16T03:51:35.970" Title="Find $r$ from $x_1\equiv r^a \pmod{p}, x_2\equiv r^b \pmod{p}$ and gcd$(b, p-1)=1$" Tags="&lt;number-theory&gt;&lt;homework&gt;&lt;modular-arithmetic&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="6927" PostTypeId="2" ParentId="6926" CreationDate="2010-10-16T02:59:29.793" Score="5" ViewCount="" Body="&lt;p&gt;HINT: If gcd$(b,p-1)=1$, then we can write $1 = tb + s(p-1)$. So $tb\equiv 1 \pmod{p-1}$. &lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-10-16T02:59:29.793" CommentCount="4" />
  <row Id="6928" PostTypeId="2" ParentId="6926" CreationDate="2010-10-16T03:33:58.087" Score="2" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\ $The isomorphism $\rm n\to r^n\:$ from the additive group $\rm \mathbb Z/(p-1)$ to the multiplicative group $\rm \mathbb Z/p^*\:$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;maps $\rm\  b\to x,\ 2b\to x^2,\: \cdots\:,\: nb\to x^n$. You seek $\rm 1\to r\ $  so you need $\rm\ nb\equiv 1\ $ so $\rm\ n\equiv \:\ldots$&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-10-16T03:51:35.970" LastActivityDate="2010-10-16T03:51:35.970" CommentCount="2" />
  <row Id="6929" PostTypeId="2" ParentId="6923" CreationDate="2010-10-16T04:14:09.557" Score="4" ViewCount="" Body="&lt;p&gt;I don't think this phrase has a single precise translation into the general language of schemes, but it will mean something like the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a finite surjective morphism $X \to Y$ of integral schemes such that the corresponding extension of function fields $K(Y) \subset K(X)$ is Galois with Galois group $G$.&#xA;It would also be reasonable to require that $X$ and $Y$ be normal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To understand what it means geometrically, one should imagine that $X$ and $Y$ are projective varieties over some field.  What this will mean then is that $X \to Y$ is surjection with finite fibres, that $G$ acts as a group of automorphisms of $X$ over $Y$,&#xA;and that if we remove the branch locus (i.e. the closed subset of $Y$ along&#xA;which the fibres contain points with multiplicity $&gt; 1$) then each fibre is acted on&#xA;faithfully and transitively by $G$ (so, away from the ramification locus, there&#xA;are $|G|$ sheets of $X$ over $Y$, which are permuted by $G$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A concrete example is given by (the projectivization of) the map $(x,y) \mapsto x$&#xA;from the elliptic curve $E$ defined by $y^2 = x^3 - x$ to $\mathbb P^1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The group $G$ is cyclic of order two, acting by $(x,y) \mapsto (x,-y)$, and &#xA;the branch points are precisely the points where $y = 0$ (four of them; three&#xA;three finite points $(0,0), (\pm 1,0)$, and also one at infinity).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this context, an important result is the Zariski--Nagata purity theorem, which says &#xA;(under mild hypotheses, e.g. that $Y$ and $X$ are smooth, or more generally,&#xA;that $Y$ is smooth and $X$ is normal) that the set of branch points has pure codimension one,&#xA;i.e. is a divisor. &lt;/p&gt;&#xA;" OwnerUserId="221" LastEditorUserId="221" LastEditorDisplayName="" LastEditDate="2010-10-17T13:35:29.860" LastActivityDate="2010-10-17T13:35:29.860" CommentCount="2" />
  <row Id="6930" PostTypeId="2" ParentId="6861" CreationDate="2010-10-16T04:41:42.113" Score="2" ViewCount="" Body="&lt;p&gt;Here is a proof of this fact:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $\mathcal C$ be the category of (not necessarily irreducible) complex varieties.&#xA;Then $\mathcal C$ can be identified with the category of reduced finite type $\mathbb C$-schemes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $\mathcal D$ be the category of all finite type $\mathbb C$-schemes.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then obviously $\mathcal C$ is a full subcategory of $\mathcal D$, and the inclusion&#xA;$\mathcal C \subset \mathcal D$ has a right adjoint, namely passage to the underlying reduced subscheme.  General nonsense (i.e. an easy categorical argument) then shows that if $X\to S$ and $Y \to S$ in $\mathcal C$ are two morphisms, the fibre product in $\mathcal C$&#xA;can be computed by first computing the fibre product in the bigger category $\mathcal D$,&#xA;and then applying the right adjoint to the inclusion.  That is,&#xA;the fibre product in the category of varieties over $\mathcal C$ is equal to&#xA;the reduced subscheme of the fibre product in $\mathcal D$ (which coincides with&#xA;the fibre product in the category of all schemes, just because the fibre product&#xA;of morphisms of finite type $\mathbb C$-schemes is again finite type over $\mathbb C$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure of a reference.  Because the proof is easy when you have the right&#xA;categorical framework, it is the kind of thing that is well-known to experts but whose proof&#xA;is not necessarily written down explicitly.&lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2010-10-16T04:41:42.113" CommentCount="1" />
  <row Id="6931" PostTypeId="1" CreationDate="2010-10-16T04:49:51.737" Score="6" ViewCount="111" Body="&lt;p&gt;I understand that the only free groups that are abelian are 1 and Z, hence a difference between 'free abelian groups' and 'abelian free groups'.&lt;br&gt;&#xA;Can someone please tell me what are the solvable free groups?&lt;br&gt;&#xA;How can I construct 'free solvable groups'?&lt;br&gt;&#xA;What is known about free groups in other classes of groups like polycyclic, nilpotent ...?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="2554" LastActivityDate="2010-10-16T13:38:11.163" Title="Free groups in some classes?" Tags="&lt;group-theory&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="2" />
  <row Id="6932" PostTypeId="2" ParentId="6931" CreationDate="2010-10-16T05:16:01.410" Score="4" ViewCount="" Body="&lt;p&gt;I think that you might have to bound the derived length to get precisely the notion that you want.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notation: For any group $G$, let $G^{(i)}$ denote the $i$th iterated derived subgroup of $G$, so $G^{(0)} = G$ and $G^{(i+1)} = [G^{(i)},G^{(i)}]$).  A group is solvable if &#xA;$G^{(i)}$ is trivial for some $i$, and has derived length $d$ if $G^{(d)}$ is trivial,&#xA;but $G^{(i)}$ is non-trivial for $i &amp;lt; d$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let $F(X)$ be the free group on a non-empty set $X$ of generators, and consider $F(X)/F(X)^{(d)}$. On the one hand, this group is clearly solvable of derived length $d$. On the other hand, if $G$ is any solvable group of derived length $\leq d$, then giving a map of sets $X \to G$ will be equivalent to giving a map of groups $F(X)/F(X)^{(d)} \to G$.&#xA;Thus $F(X)/F(X)^{(d)}$ is the free object on $X$ in the category of solvable groups of&#xA;derived length $\leq d$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One can make a similar construction if you consider e.g. the category of nilpotent groups&#xA;of nilpotency class $\leq c$ (i.e. whose central series is of length $\leq c$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you really want to consider all solvable groups at once, then you can consider the projective limit $$\lim_{\leftarrow} F(X)/F(X)^{(d)},$$ the so-called &lt;em&gt;pro-solvable&lt;/em&gt; completion of $F(X)$.  Similarly, in the nilpotent setting, you can form the &lt;em&gt;pro-nilpotent&lt;/em&gt; completion of $F(X)$.  But now the &quot;free object&quot; is not literally in the category you want, but is instead is a projective limit of groups from that category.  (This is often okay, though; is you search for &quot;pro-nilpotent completion&quot;, you will see that pro-nilpotent completions of free groups come up a lot, e.g. in arithmetic geometry.)&lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2010-10-16T05:16:01.410" CommentCount="0" />
  <row Id="6933" PostTypeId="2" ParentId="6914" CreationDate="2010-10-16T05:24:13.493" Score="3" ViewCount="" Body="&lt;p&gt;I asked this very question to a copywriter some years ago and he said that you can not copyright protect a formula but he also said that there are copyright protected algorithms. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, where is the borderline? I do not know, but I guess this has to be tested in each case. Here is a &lt;a href=&quot;http://imranontech.com/2006/12/04/are-algorithms-copyrightable/&quot; rel=&quot;nofollow&quot;&gt;discussion&lt;/a&gt; on the question, and here is a link to &lt;a href=&quot;http://www.rsa.com/glossary/default.asp?id=1045&quot; rel=&quot;nofollow&quot;&gt;RSA&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, here is a link about &lt;a href=&quot;http://www.math.columbia.edu/~woit/wordpress/?p=590&quot; rel=&quot;nofollow&quot;&gt;plagiarism&lt;/a&gt; in science.  &lt;/p&gt;&#xA;" OwnerUserId="1154" LastEditorUserId="1154" LastEditorDisplayName="" LastEditDate="2010-10-16T05:30:33.967" LastActivityDate="2010-10-16T05:30:33.967" CommentCount="5" />
  <row Id="6934" PostTypeId="2" ParentId="6931" CreationDate="2010-10-16T05:28:39.150" Score="5" ViewCount="" Body="&lt;p&gt;There is no &quot;free solvable group&quot;, but there &lt;em&gt;are&lt;/em&gt; &quot;free solvable groups of length $n$&quot; for any $n$. There are also free nilpotent groups of class $n$, free Burnside groups of exponent $k$; all groups that are abelian-by-exponent $k$; and more.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More precisely: a &lt;strong&gt;variety of groups&lt;/strong&gt; is a class of groups that is closed under taking isomorphisms, subgroups, quotients, and arbitrary direct products (that is, if $G\in\mathcal{C}$ and $K\cong G$, then $K\in\mathcal{C}$; if $G\in\mathcal{C}$ and $H\lt G$, then $H\in \mathcal{C}$; if $G\in\mathcal{C}$ and $N\triangleleft G$, then $G/N\in\mathcal{C}$; and if $\{G_i\}$ is an arbitrary family of groups with $G_i\in\mathcal{C}$ for every $i$, then $\prod G_i\in\mathcal{C}$). Examples of varieties include &quot;all abelian groups&quot;, &quot;all groups of solvability length at most $n$&quot;; &quot;all nilpotent groups of class at most $c$&quot;; &quot;all groups such that every element is of exponent $k$&quot;; and more.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $\mathcal{C}$ is a variety of groups, and $G$ is any group, then there is a smallest normal subgroup $\mathcal{C}(G)$ of $G$ such that $G/\mathcal{C}(G)\in\mathcal{C}$ (this normal subgroup is, in fact, not merely normal, but fully invariant).  For example, when $\mathcal{C}$ is the class of all abelian groups, $\mathcal{C}(G)$ is none other than the commutator subgroup of $G$. &lt;em&gt;Proof of the existence of $\mathcal{C}(G)$:&lt;/em&gt; Let $\{N_i\}_{i\in I}$ be the collection of all normal subgroups of $G$ such that $G/N_i\in\mathcal{C}$; the class is nonempty, since $G/G$ is always in any variety. Then consider the obvious map from $G$ to $\prod G/N_i$; the product must be in $\mathcal{C}$ (it is a product of groups in $\mathcal{C}$), so the image of $G$ is in $\mathcal{C}$; the kernel of this map is the group $\mathcal{C}(G)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if $X$ is any set, and $F(X)$ is the free group on $X$, then there is a smallest subgroup $\mathcal{C}(F(X))$ such that $F(X)/\mathcal{C}(F(X))\in\mathcal{C}$. The resulting quotient can be seen to have the same universal property &lt;em&gt;relative to groups in $\mathcal{C}$&lt;/em&gt; as $F(X)$ does relative to all groups. So we say that this quotient is the &quot;relatively free $\mathcal{C}$-group on $X$.&quot;  We call the original group $F(X)$ the &quot;absolutely free group on $X$&quot; to distinguish it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added:&lt;/strong&gt; Just in case, remember that if $\mathcal{D}$ is a category in which the objects are sets and the arrows are maps of sets, then given a set $X$, the &lt;strong&gt;free $\mathcal{D}$-object on $X$&lt;/strong&gt; is an object $F(X)$ in $\mathcal{D}$, together with a set-theoretic inclusion $i\colon X\to F(X)$, such that for every object $D\in\mathcal{D}$ and every set-theoretic map $f\colon X\to D$ there exists a unique $\mathcal{D}$-morphism $\varphi\colon F(X)\to D$ such that $f=\varphi\circ i$. For the &quot;absolutely free groups&quot;, the category is the category of all groups; for &quot;free abelian groups&quot;, the category is the category of abelian groups. For &quot;relatively free $\mathcal{C}$-group&quot;, the category is the category of all groups in $\mathcal{C}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason there is no &quot;free solvable group&quot; is that the class of all solvable groups is &lt;strong&gt;not&lt;/strong&gt; a variety: it is not closed under arbitrary direct products. But if you put a bound on the solvability length then you &lt;em&gt;do&lt;/em&gt; get a variety, and you can construct the corresponding &quot;free solvable group of length $n$&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In addition you can take the pro-$\mathcal{C}$ approach that Matt E mentions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are interested in learning more about varieties, these lie at the intersection of Universal Algebra and Group Theory. I would recommend Hanna Neumann's book &lt;strong&gt;Varieties of Groups&lt;/strong&gt; (a bit old, but still the standard reference); also George Bergman's Universal Algebra notes (available &lt;a href=&quot;http://math.berkeley.edu/~gbergman/245/&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;).&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-10-16T13:38:11.163" LastActivityDate="2010-10-16T13:38:11.163" />
  <row Id="6935" PostTypeId="1" AcceptedAnswerId="6937" CreationDate="2010-10-16T05:48:20.043" Score="1" ViewCount="46" Body="&lt;p&gt;I'm trying to understand the approach to calculating an answer to the following question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need to do some processing on a list of n items.&#xA;I know that the total processing time takes log2(n) milliseconds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Problem: What is the maximum number of items that can be processed in 10 seconds (i.e., 10,000 milliseconds) ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the right general approach / technique? I should mention that I am a math neophyte.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's clear that if I have a list of 1,000,000 items it will take 13.8 milliseconds, as log2(1000000) = 13.8.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What's not clear is how to compute the maximum n that can be processed in 10,0000 milliseconds. I know it's a gigantic number, but I am not sure how to calculate it. If the number was small I could probably find it by trial and error, but:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A) It's a giant number such that trial and error seems impractical.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;B) Even if it was not so huge, I'd really like to understand how to compute it determinsitically rather than by experiment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks very much for any pointers.&lt;/p&gt;&#xA;" OwnerUserId="2558" LastEditorUserId="1269" LastEditorDisplayName="" LastEditDate="2010-10-16T06:25:25.380" LastActivityDate="2010-10-16T06:26:31.430" Title="Finding the maximum n items that can be processed in x time using a function that gives the time for some n." Tags="&lt;algebra-precalculus&gt;" AnswerCount="2" />
  <row Id="6936" PostTypeId="2" ParentId="6935" CreationDate="2010-10-16T06:24:25.153" Score="0" ViewCount="" Body="&lt;p&gt;When I read the headline I thought you were asking about the &lt;a href=&quot;http://en.wikipedia.org/wiki/Knapsack_problem&quot; rel=&quot;nofollow&quot;&gt;knapsack problem&lt;/a&gt;, but then I read your question and it turns out you are just asking for an inverse to binary logarithm. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem asks &quot;find $n$ such that $\log_2 n \geq 10^4$.&quot;&#xA;Note that $\log_2 x \geq y \Leftrightarrow 2^{\log_2 x} = x \geq 2^y$ because $x \mapsto 2^x$ and $\log_2$ are monotonely increasing. This should be enough for you to figure out the answer (and yes, it is huge).&lt;/p&gt;&#xA;" OwnerUserId="1269" LastActivityDate="2010-10-16T06:24:25.153" CommentCount="1" />
  <row Id="6937" PostTypeId="2" ParentId="6935" CreationDate="2010-10-16T06:26:31.430" Score="2" ViewCount="" Body="&lt;p&gt;If I understand your question you want to solve an equation of the form: $\log_{2}(x)=100,000$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The key to this is &lt;a href=&quot;http://en.wikipedia.org/wiki/Exponentiation#Real_powers&quot; rel=&quot;nofollow&quot;&gt;exponentiation&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your case we want to do the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$2^{\log_{2}(x)}=2^{100,000}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which simplifies to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x=2^{100,000}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is quite a large number as you correctly predicted :)&lt;/p&gt;&#xA;" OwnerUserId="249" LastActivityDate="2010-10-16T06:26:31.430" CommentCount="7" />
  <row Id="6938" PostTypeId="1" CreationDate="2010-10-16T08:12:02.700" Score="3" ViewCount="476" Body="&lt;p&gt;Okay, I have this not so pretty 2nd order non-linear ODE I should be able to solve numerically. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f''(R) + \frac{2}{R} f'(R)=\frac{0.7}{R} \left( \frac{1}{\sqrt{f(R)}} - \frac{0.3}{\sqrt{1-f(R)}} \right),$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f(1)=1.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The function around the origin is behaving very wildly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was thinking of breaking this guy up into a system of two first order ODE's and then solve, but I have no idea how to set this up. What method should I use to set up the system of ODE's?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there is some other method rather than numerically solving a system of differential equations, please feel welcome to share. Thanks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/32jhJ.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="2559" LastEditorUserId="2559" LastEditorDisplayName="" LastEditDate="2010-10-19T15:31:56.740" LastActivityDate="2010-10-19T15:31:56.740" Title="Numerically Solving a Second Order Nonlinear ODE" Tags="&lt;analysis&gt;&lt;differential-equations&gt;&lt;math-software&gt;&lt;numerical-methods&gt;&lt;matlab&gt;" AnswerCount="1" CommentCount="7" FavoriteCount="0" />
  <row Id="6939" PostTypeId="1" AcceptedAnswerId="6946" CreationDate="2010-10-16T08:42:17.640" Score="12" ViewCount="418" Body="&lt;p&gt;In the course of playing around with $\sum_{n=1}^{\infty} \frac{1}{\sqrt{1+n^{n}}}$, I used w|α to obtain the power series for $f(x)=\frac{1}{\sqrt{1+x^{x}}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align*}&#xA;\frac{1}{\sqrt{1+x^{x}}} =&amp;amp; \frac{1}{\sqrt{2}} - \frac{x\log(x)}{4\sqrt{2}} -\frac{x^{2}\log^{2}(x)}{32\sqrt{2}}+ \frac{5x^{3}\log^{3}(x)}{384\sqrt{2}}\\&#xA; \\&#xA;&amp;amp;+ \frac{17x^{4}\log^{4}(x)}{6144\sqrt{2}} - \frac{121x^{5}\log^{5}(x)}{122880\sqrt{2}} - \frac{721x^{6}\log^{6}(x)}{2949120\sqrt{2}} \ldots&#xA;\end{align*}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Before I realized that I couldn't really use this to help me with the sum, I found that the denominators (ignoring the $\sqrt{2}$, because all of them have it in common) correspond to &lt;a href=&quot;http://www.research.att.com/~njas/sequences/A047053&quot; rel=&quot;nofollow&quot;&gt;$4^{n}n!$&lt;/a&gt;, what is baffling is that the numerators appear to correspond to the coefficients in the exponential generating function for &lt;a href=&quot;http://www.research.att.com/~njas/sequences/A012259&quot; rel=&quot;nofollow&quot;&gt;$f(x)=e^{\tanh^{-1}(\tan(x))}$&lt;/a&gt; (I believe that the 7th entry should be 1369 and not 6845), and I'm curious what the explanation is, because $f(x)=e^{\tanh^{-1}(\tan(x))}$ is a mighty weird looking function.&lt;/p&gt;&#xA;" OwnerUserId="890" LastEditorUserId="7676" LastEditorDisplayName="" LastEditDate="2011-03-15T06:04:22.320" LastActivityDate="2011-03-15T06:04:22.320" Title="mystery regarding power series of $\frac{1}{\sqrt{1+x^{x}}}$" Tags="&lt;real-analysis&gt;&lt;sequences-and-series&gt;" AnswerCount="2" CommentCount="9" />
  <row Id="6940" PostTypeId="2" ParentId="6938" CreationDate="2010-10-16T09:48:25.527" Score="1" ViewCount="" Body="&lt;p&gt;The general method to reinterpret a higher-order ODE as a system of first order ODEs is to regard the derivatives of the unknown function as additional unknown functions. In your case, regard $f'$ as a new function $g$. Then the system of first order ODEs consists of two equations, the first being the original equation with $f'$ replaced by $g$ and $f''$ replaced by $g'$, and the second equation being $f'=g$.&lt;/p&gt;&#xA;" OwnerUserId="1609" LastActivityDate="2010-10-16T09:48:25.527" CommentCount="3" />
  <row Id="6941" PostTypeId="1" AcceptedAnswerId="7113" CreationDate="2010-10-16T10:48:00.313" Score="1" ViewCount="114" Body="&lt;p&gt;Hi, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've got a 2D quadratic Bézier curve which, by construction, is a f(x) function : no loops, a single solution for each defined x.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a common mean to convert this curve to a 3rd degree polynomial ? 3 should be enough since there can only be two &quot;bumps&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="2561" LastActivityDate="2010-10-18T11:50:51.630" Title="bézier to f(x) polynomial function" Tags="&lt;polynomials&gt;&lt;bezier-curve&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="6942" PostTypeId="1" CreationDate="2010-10-16T11:52:03.363" Score="5" ViewCount="351" Body="&lt;p&gt;The following is a geometry puzzle from a math school book. Even though it has been a long time since I finished school, I remember this puzzle quite well, and I don't have a nice solution to it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So here is the puzzle:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/QGBoB.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The triangle ABC is known to be isosceles, that is, AC=BC. The labelled angles are known to be $\alpha=\gamma=20°$, $\beta=30°$. The task is to find the angle labelled &quot;?&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only solution that I know of is to use the sine formula and cosine formula several times. From this one can obtain a numerical solution. Moreover this number can be algebraically shown to be correct (all sines and cosines are contained in the real subfield of the 36th cyclotomic field). So in this sense I solved the problem, but the solution is kind of a brute force attack (for example, some of the polynomials that show up in the computation have coefficients &gt; 1000000). Since the puzzle originates from a book that deals only with elemetary geometry (and not even trigonometry if I remember correctly) there has to be a more elegant solution.&lt;/p&gt;&#xA;" OwnerUserId="1609" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-10-16T17:45:40.083" LastActivityDate="2010-10-16T17:45:40.083" Title="Finding an angle within an 80-80-20 isosceles triangle." Tags="&lt;geometry&gt;" AnswerCount="2" CommentCount="5" FavoriteCount="1" />
  <row Id="6943" PostTypeId="2" ParentId="6851" CreationDate="2010-10-16T12:09:18.733" Score="1" ViewCount="" Body="&lt;p&gt;My suggestion:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(A) Start with the definition of absolute value $\left\vert z\right\vert $, with $z=xy,x,y\in\mathbb{R}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left\vert xy\right\vert =xy\qquad \text{if}\ xy\geq 0\iff \left( x\geq 0\wedge y\geq 0\right) \vee \left( x\leq 0\wedge y\leq 0\right) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left\vert xy\right\vert =-xy\qquad \text{if}\ xy&amp;lt;0\iff \left( x\geq 0\wedge y\leq 0\right) \vee \left( x\leq 0\wedge y\geq 0\right) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(B) Add the condition  $\left\vert xy\right\vert \leq 2\iff -2\le xy\leq 2$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Why?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(C) Observe that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(i) $xy\gt 0$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;in the 1st and 3rd quadrants; and &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(ii)  $xy\lt 0$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;in the 2nd and 4th quadrants. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Why?)&lt;/p&gt;&#xA;" OwnerUserId="752" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2010-10-16T14:26:51.683" LastActivityDate="2010-10-16T14:26:51.683" CommentCount="3" />
  <row Id="6944" PostTypeId="2" ParentId="6890" CreationDate="2010-10-16T12:15:02.180" Score="5" ViewCount="" Body="&lt;p&gt;Okay, I'm still not very sure about the fact that there's no distance function in the integrand, but if you assume it is somehow absorbed in the definition of $db_m(b)$, you can make some sort of sense out of this expression. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;First you should recall that the classical electromagnetic potentials and their relations to the position and motion of a point particle. The electric potential generated by a charge density $\rho(x)$ is usually written as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ A_0(x) = V(x) = C_1 \int \frac{\rho(y)}{|x - y|} dy$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and the magnetic vector potential generated by a current is usually written as (Biot-Savart law form)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \vec{A}(x) = C_2 \int \frac{\vec{J}(y)}{|x-y|} dy$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $C_1,C_2$ are constants which depends on the unit chosen. Then in this case, in units where the speed of light $c = 1$,  $(\rho,\vec{J})$ is a four-vector representing the movement of the particles. (Recall that an object with velocity $\vec{v}$ is traversing $\vec{v}$ spatial units in 1 time unit, so the four-vector corresponding to the tangent vector of its world line will be $(1,\vec{v})$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, you also realize that the signal (electromagnetic potential) should propagate at the speed of light. So while the above classical notion is valid for small object velocities, they are only ultimately valid in the electrostatics/magnetostatics cases, in which case there is a preferred time direction (the one in which everything stays constant) and we don't have to worry about this propagation); in other words, in those cases the particles on the past light cone can be transported forward using the stationarity to particles on your instantaneous surface and the computations can be done there. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;By direct analogy then, the proper way to compute the electromagnetic vector four potential is to set $j$ the four-current of your charge distribution (so that conservation of charge demands that the space-time divergence of $j$ vanishes, and we also write $\hat{j}$ for the unit time vector in the $j$ direction) is  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ A_\nu(x) = k\int \frac{j_\nu(y)}{(x-y)_\mu \hat{j}^\mu} \delta((x-y)_\mu(x-y)^\mu) dy $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $x,y$ are space-time events. ($k$ is just some normalizing constant, usually set to $\frac{1}{4\pi\epsilon_0}$). Basically the idea is not to compute $A$'s dependence on $j$, but rather $j$'s contribution to $A$. The influence of $j$ originating from space-time event $y$ will propagate out along the light cone (the $\delta$ function term). In the inertial frame corresponding to the instantaneous velocity $\hat{j}(y)$, the propagation is uniform in directions (because instantaneous at $y$, the distribution is at rest), and decays as $1/r$ where $r$ is the spatial distance, in $\hat{j}(y)$'s rest frame. A direct computation tells you that $r = (x-y)_\mu \hat{j}^\mu$. So the integrand tells you the infinitesmial contribution by $j(y)$ at space-time event $x$ to the four-potential. Then you just add them all up. If you look carefully at the integrand, you'd notice that it is the convolution of the Green's function for the linear wave equation against $j$! So in fact from this expression you recover the differential form of the law for the four-potential&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \nabla_\mu\nabla^\mu A = 4\pi k j $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is just equation (25.22) in volume 2 of Feynman's lectures. If you just take the formula above, and restrict $j$ to be the tangent vector to the world line of a charged particle, you will recover the equation you wrote down, if you assume $dB_m(b)$ is given by $dy$ times the fraction that appears in the formula above. &lt;/p&gt;&#xA;" OwnerUserId="1543" LastActivityDate="2010-10-16T12:15:02.180" CommentCount="6" />
  <row Id="6945" PostTypeId="2" ParentId="6942" CreationDate="2010-10-16T12:20:45.313" Score="7" ViewCount="" Body="&lt;p&gt;The solutions are not as trivial as one would expect from the statement. It's called Langley's problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Check out &lt;a href=&quot;http://mathcircle.berkeley.edu/BMC4/Handouts/geoprob.pdf&quot; rel=&quot;nofollow&quot;&gt;http://mathcircle.berkeley.edu/BMC4/Handouts/geoprob.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1313" LastActivityDate="2010-10-16T12:20:45.313" CommentCount="2" />
  <row Id="6946" PostTypeId="2" ParentId="6939" CreationDate="2010-10-16T13:14:52.330" Score="19" ViewCount="" Body="&lt;p&gt;Well,&#xA;$$\frac1{\sqrt{1+x^x}}=\frac1{\sqrt{1+e^y}}$$&#xA;where $y=x\log x$ so it's unsurprising you get a series in terms of $y=x\log x$.&#xA;Then&#xA;$$\frac1{\sqrt{1+e^y}}=\frac1{\sqrt2}\frac1{\sqrt{1+u}}$$&#xA;where $u=\frac12(e^y-1)=o(y)$. This explains the fact that the coefficients&#xA;are rationals over $\sqrt2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now consider $f(x)=\exp(\tanh^{-1}(\tan x))$. Now&#xA;$$\tanh z=\frac{e^{2z}-1}{e^{2z}+1}$$&#xA;so that&#xA;$$e^z=\sqrt{\frac{1+\tanh z}{1-\tanh z}}.$$&#xA;Hence,&#xA;$$\exp(\tanh^{-1} t)=\sqrt{\frac{1+t}{1-t}}.$$&#xA;Putting $t=\tan x$ gives&#xA;$$f(x)=\sqrt{\frac{1+\tan x}{1-\tan x}}=\sum_{n=0}^\infty a_n x^n.$$&#xA;Then&#xA;\begin{eqnarray*}&#xA;&amp;amp;&amp;amp;(1+i)f(ix)+(1-i)f(-ix)\\&#xA;&amp;amp;=&amp;amp;&#xA;2\sum_{m=0}^\infty&#xA;(a_{4m} x^{4m}-a_{4m+1} x^{4m+1}-a_{4m+2} x^{4m+2}+a_{4m+3} x^{4m+3}).&#xA;\end{eqnarray*}&#xA;But&#xA;$$f(ix)=\sqrt{\frac{1+i\tanh x}{1-i\tanh x}}&#xA;=\frac{1+i\tanh x}{\sqrt{1+\tanh^2x}}$$&#xA;and so&#xA;\begin{eqnarray*}&#xA;&amp;amp;&amp;amp;(1+i)f(ix)+(1-i)f(-ix)\\&#xA;&amp;amp;=&amp;amp;&#xA;\frac{2-2\tanh x}{\sqrt{1+\tanh^2 x}}&#xA;=\frac{2\cosh x-2\sinh x}{\sqrt{\cosh^2x+\sinh^2x}}\\&#xA;&amp;amp;=&amp;amp;\frac{2e^{-x}}{\sqrt{(e^{2x}+e^{-2x})/2}}=\frac{2\sqrt2}{\sqrt{1+e^{4x}}}&#xA;\end{eqnarray*}&#xA;which explains why the coefficients in the two series are the same&#xA;up to signs and powers of 4.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-10-16T13:14:52.330" CommentCount="1" />
  <row Id="6947" PostTypeId="1" AcceptedAnswerId="6974" CreationDate="2010-10-16T13:39:02.690" Score="2" ViewCount="176" Body="&lt;p&gt;A smooth surface $S$ embedded in $\mathbb{R}^3$ whose metric is inherited from $\mathbb{R}^3$&#xA;(i.e., distance measured by shortest paths on $S$) is a Riemannian 2-manifold: differentiable&#xA;because smooth and with the metric just described.  Two questions:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt; Are such surfaces a subset of all Riemannian 2-manifolds? Are there Riemannian 2-manifolds&#xA;that are not &quot;realized&quot; by any surface embedded in $\mathbb{R}^3$?  I assume _Yes_.&lt;/li&gt;&#xA;&lt;li&gt; If so, is there any characterization of which Riemannian 2-manifolds are realized by&#xA;such surfaces? In the absence of a characterization, examples would help.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Edit.&lt;/b&gt; In light of the useful responses, a sharpening of my question occurs to me:&#xA;3. Is the &lt;em&gt;only&lt;/em&gt; impediment embedding vs. immersion? Is every Riemannian 2-manifold realized by&#xA;a surface &lt;em&gt;immersed&lt;/em&gt; in $\mathbb{R}^3$?&lt;/p&gt;&#xA;" OwnerUserId="237" LastEditorUserId="237" LastEditorDisplayName="" LastEditDate="2010-10-16T16:30:09.707" LastActivityDate="2010-10-16T21:39:19.510" Title="Riemannian 2-manifolds not realized by surfaces in $\mathbb{R}^3$?" Tags="&lt;geometry&gt;&lt;differential-geometry&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="2" />
  <row Id="6948" PostTypeId="2" ParentId="6629" CreationDate="2010-10-16T13:50:58.183" Score="0" ViewCount="" Body="&lt;p&gt;I've thought some more about your problem.  The recursion in your dynamical system yields the expression&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$V_t = (1- \gamma)^t V_0 + \gamma \sum_{k=0}^{t-1} (1 - \gamma)^{t-1-k} \delta^{S_k}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you have a sum of discounted random variables.  If they were not discounted the distribution would be approximately normal for large $t$.  It turns out, though, that there is an analogous result for the distribution of a sum of discounted random variables -- a discounted central limit theorem, as it were.  You might have some success using this discounted central limit theorem as an approximation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are the specifics.  Let $Y_v = \sum_{k=0}^{\infty} v^k X_k$, where the $X_k$'s are independent and have a common distribution function.  Let &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$Z_v = \frac{\sqrt{1-v}}{\sigma} \left(Y_v - \frac{\mu}{1-v}\right).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then $Z_v$ is asymptotically normal with mean 0 and variance $\frac{1}{1+v}$, for $v \to 1$, where $\mu$ and $\sigma$ are the common mean and standard deviation of the $X_k$'s.  (The expected value of $|X_k - \mu|^3$ must exist, too.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moreover, there is a bound on the error in this approximation.  Let $F_v(x)$ and $N_v(x)$ be the cdf's of the $Z_v$ random variable and of a normal with mean 0 and variance $\frac{1}{1+v}$, respectively.  Then &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$|F_v(x) - N_v(x)| \leq \frac{C \rho \sqrt{1-v}}{\sigma^3},$$&#xA;where $\rho = E[|X_k - \mu|^3]$ and $C = 5.4$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reference is Hans Gerber, &quot;The Discounted Central Limit Theorem and Its Berry-Esseen Analogue,&quot; &lt;em&gt;Annals of Mathematical Statistics&lt;/em&gt; 42(1), pp. 389-392, 1971.&lt;/p&gt;&#xA;" OwnerUserId="2370" LastActivityDate="2010-10-16T13:50:58.183" />
  <row Id="6949" PostTypeId="2" ParentId="6947" CreationDate="2010-10-16T14:14:43.590" Score="4" ViewCount="" Body="&lt;p&gt;The hyperbolic plane cannot be smoothly isometrically embedded in $\mathbb{R}^3$. It can be so in $\mathbb{R}^5$. It is open (as far as I know) if it can be embedded in $\mathbb{R}^4$. I believe this is mentioned in Do Carmo's book on curves and surfaces. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Edit:&lt;/em&gt; Not a complete characterization, but Amsler has shown (see below for reference) that any Riemannian surface with constant negative curvature, if attempted to be imbedded in $\mathbb{R}^3$, must have singularities. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Amsler, M.H., Des surfaces a courbure negative constante dans l'espace a trois dimensions et de leurs singularites, Math. Ann. 130, 1955, 234-256&lt;/p&gt;&#xA;" OwnerUserId="1543" LastEditorUserId="1543" LastEditorDisplayName="" LastEditDate="2010-10-16T14:20:06.060" LastActivityDate="2010-10-16T14:20:06.060" CommentCount="2" />
  <row Id="6950" PostTypeId="1" CreationDate="2010-10-16T14:29:32.303" Score="1" ViewCount="162" Body="&lt;p&gt;How to prove that for all positive a,b,c this formula holds true:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{a}{\sqrt{a+b}} + \frac{b}{\sqrt{b+c}} + \frac{c}{\sqrt{c+a}} &gt; \sqrt{a+b+c}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any help will be invaluable.&lt;/p&gt;&#xA;" OwnerUserId="2562" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-10-16T17:51:44.213" LastActivityDate="2010-10-16T17:51:44.213" Title="$\frac{a}{\sqrt{a+b}} + \frac{b}{\sqrt{b+c}} + \frac{c}{\sqrt{c+a}} &gt; \sqrt{a+b+c}$ is true for positive a,b,c" Tags="&lt;algebra-precalculus&gt;&lt;inequality&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" />
  <row Id="6951" PostTypeId="1" CreationDate="2010-10-16T14:49:45.137" Score="1" ViewCount="222" Body="&lt;p&gt;Are there any odd positive numbers that satisfy the equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a^2 - b^3 = 4$&#xA;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am certain that there are none but can't prove it. How would you prove that?&lt;/p&gt;&#xA;" OwnerUserId="2562" LastEditorUserId="2562" LastEditorDisplayName="" LastEditDate="2010-10-18T13:28:23.330" LastActivityDate="2010-10-18T13:28:23.330" Title="Odd positive integers that satisfy $a^2 - b^3 = 4$" Tags="&lt;number-theory&gt;&lt;homework&gt;&lt;diophantine-equations&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="0" />
  <row Id="6952" PostTypeId="2" ParentId="6951" CreationDate="2010-10-16T14:53:48.037" Score="8" ViewCount="" Body="&lt;p&gt;Start by rewriting it as $a^2-4=b^3$, and do what comes naturally.&lt;/p&gt;&#xA;" OwnerUserId="625" LastActivityDate="2010-10-16T14:53:48.037" CommentCount="14" />
  <row Id="6954" PostTypeId="2" ParentId="6950" CreationDate="2010-10-16T15:02:05.303" Score="4" ViewCount="" Body="&lt;p&gt;Since&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$a+b+c=\sqrt{(a+b+c)^2}=\sqrt{a+b+c}\ \sqrt{a+b+c}\iff \dfrac{a+b+c}{\sqrt{a+b+c}}=\sqrt{a+b+c}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\dfrac{a}{\sqrt{a+b}}&gt;\dfrac{a}{\sqrt{a+b+c}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\dfrac{b}{\sqrt{b+c}}&gt;\dfrac{b}{\sqrt{a+b+c}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\dfrac{c}{\sqrt{c+a}}&gt;\dfrac{c}{\sqrt{a+b+c}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\dfrac{a}{\sqrt{a+b}}+\dfrac{b}{\sqrt{b+c}}+\dfrac{c}{\sqrt{c+a}}&gt;\dfrac{a+b+c}{%&#xA;\sqrt{a+b+c}}=\sqrt{a+b+c}.$$&lt;/p&gt;&#xA;" OwnerUserId="752" LastActivityDate="2010-10-16T15:02:05.303" CommentCount="2" />
  <row Id="6956" PostTypeId="2" ParentId="6942" CreationDate="2010-10-16T15:32:33.430" Score="3" ViewCount="" Body="&lt;p&gt;Here is the solution I had for this (I had the writeup lying around in an old email I had sent regarding this, so, no tex, also, A and B are interchanged):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/9rhDL.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can see that an 80-80-20 triangle is nothing but a part of the triangulation of an 18 sided regular polygon, whose 6 triangles are shown in the the bottom part of the circle above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now consider an 80-80-20 triangle (ABC on top part of figure) and shoot a light ray from one of the base vertices  (B in the figure) at an angle of 50 degrees to the base (or 30 from one of the equal sides). (See the triangle in the top part of the circle and the red arrows)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can show that the ray will reflect twice (first at D then E)  and strike at a 90 degree angle the third time (F) i.e. after  5 refections, the light ray will return back to the vertex! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This reflection process can be pictured in another way, by reflecting the triangle each time instead of reflecting the ray (see the red arrows in the 6 triangles in the bottom part)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the point of the third reflection F (i.e. the 90 degree incidence point) is exactly the midpoint of the side on which the ray is incident on. This can be seen by considering the bottom part:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the right most B and corresponding triangle CBF. This is a 90-60-30 triangle. Thus CF is half CB which is half CA.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Back to triangle at top) i.e. F is midpoint of AC. Thus triangle ACE is isosceles, thus angle CAE = 20.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, we see that angle BDE must be the angle x in the problem, which must be 180-(50+50) = 80. (as DE is BD reflected on AC).  That angle y is 30, follows…&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: To get more context about what x and y are, this was the figure when the problem was shown to me:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/2bfkT.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-10-16T15:40:54.427" LastActivityDate="2010-10-16T15:40:54.427" />
  <row Id="6957" PostTypeId="2" ParentId="6894" CreationDate="2010-10-16T15:38:21.377" Score="4" ViewCount="" Body="&lt;p&gt;Oh, I see.  In $\mathbb{Z}/(ab)\mathbb{Z}$ let $A$ be the set of elements not relatively prime to $a$ and let $B$ be the set of elements not relatively prime to $b$.  From here what you want follows by the Chinese Remainder Theorem.  (Of course, you could just use CRT from the get-go.)&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-10-16T15:38:21.377" />
  <row Id="6958" PostTypeId="1" AcceptedAnswerId="6960" CreationDate="2010-10-16T15:39:08.657" Score="3" ViewCount="211" Body="&lt;p&gt;How does the factoring of polynomials over Galois fields work? I cannot seem to understand the basic concept.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;For example:&lt;/strong&gt; How do I factorize x&lt;sup&gt;6&lt;/sup&gt; - 1 over GF(3)? I know that the result is (x+1)&lt;sup&gt;3&lt;/sup&gt; (x+2)&lt;sup&gt;3&lt;/sup&gt;, but I'm unable to compute it myself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've studied the articles on Wikipedia:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Galois_field&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Galois_field&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Factorization_of_polynomials&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Factorization_of_polynomials&lt;/a&gt; &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;but I found it very difficult to understand. Is there some algorithm that would help me factorize polynomials like x&lt;sup&gt;n&lt;/sup&gt; - 1 over GF(k)?&lt;/p&gt;&#xA;" OwnerUserId="2411" LastActivityDate="2010-10-16T15:50:02.347" Title="How do I factorize polynomial over Galois field?" Tags="&lt;abstract-algebra&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" />
  <row Id="6959" PostTypeId="2" ParentId="6958" CreationDate="2010-10-16T15:42:23.690" Score="3" ViewCount="" Body="&lt;p&gt;Factoring the polynomials $x^n - 1$ is very different from factoring general polynomials, since you already know what the roots are; they are, in a suitably large finite extension, precisely the elements of order dividing $n$.  Since you know that the multiplicative group of a finite field is cyclic, the conclusion follows from here.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-10-16T15:42:23.690" />
  <row Id="6960" PostTypeId="2" ParentId="6958" CreationDate="2010-10-16T15:50:02.347" Score="5" ViewCount="" Body="&lt;p&gt;If $n=pm$ is a multiple of $p$ then over $GF(p)$ one has&#xA;$$x^n-1=(x^m-1)^p$$&#xA;so the problem reduces swiftly to the case where $n$ is coprime to $p$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $p$ is not a factor of $n$ then over an algebraic closure of&#xA;$GF(p)$&#xA;$$x^n-1=\prod_{k=0}^{n-1}(x-\zeta^j)$$&#xA;where $\zeta$ is a primitive $n$-th root of unity.&#xA;One makes this into a factorization over $GF(p)$ by combining conjugate&#xA;factors together. For each $k$, the polynomial&#xA;$$(x-\zeta^k)(x-\zeta^{pk})(x-\zeta^{p^2k})\cdots(x-\zeta^{p^{r-1}k})$$&#xA;has coefficients in, and is irreducible over, $GF(p)$ where $r$&#xA;is the least positive integer with $p^r k\equiv k$ (mod $n$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using this, it's easy to work out the degrees of the irreducible&#xA;factors of $x^n-1$, but to find the factors themselves needs a bit more&#xA;work, using for instance &lt;a href=&quot;http://en.wikipedia.org/wiki/Berlekamp%27s_algorithm&quot; rel=&quot;nofollow&quot;&gt;Berlekamp's algorithm&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-10-16T15:50:02.347" CommentCount="1" />
  <row Id="6961" PostTypeId="2" ParentId="6947" CreationDate="2010-10-16T17:12:35.663" Score="3" ViewCount="" Body="&lt;p&gt;The Whitney embedding theorem says you can always embed a smooth $n$-manifold in $\mathbb{R}^{2n}$, and immerse it in $\mathbb{R}^{2n-1}$.  Nonorientable Riemann surfaces, for example, don't embed in $\mathbb{R}^3$, but there are some pretty good immersions (the typical picture of the Klein bottle is a good example).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a Riemannian manifold, Nash and Kuiper proved that there's a $C^1$ globally isometric embedding into $\mathbb{R}^{2n+1}$ (and, in fact, that you can arbitrarily closely approximate any metric $C^\infty$ embedding into at least $\mathbb{R}^{n+1}$ by a global isometric $C^1$ embedding).  For a global isometric $C^\infty$ embedding, it looks like the current lower bound is max$(n(n+1)/2+2n,n(n+1)/2+n+5)$.  For a local one, you can do it into $n(n+1)/2+n$-space.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means that for a globally isometric and analytic embedding of a surface, you might have to go up to $\mathbb{R}^{10}$.  Ew.&lt;/p&gt;&#xA;" OwnerUserId="2215" LastEditorUserId="1489" LastEditorDisplayName="" LastEditDate="2010-10-16T21:39:19.510" LastActivityDate="2010-10-16T21:39:19.510" CommentCount="1" />
  <row Id="6962" PostTypeId="1" CreationDate="2010-10-16T17:15:39.810" Score="0" ViewCount="80" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Possible Duplicate:&lt;/strong&gt;&lt;br&gt;&#xA;  &lt;a href=&quot;http://math.stackexchange.com/questions/6282/minimizing-gender-regularity-in-a-linear-arrangement-of-boys-and-girls&quot;&gt;Minimizing Gender Regularity in a linear arrangement of boys and girls.&lt;/a&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;p&gt;There are G girl students and B boy students in a class that is about to graduate. You&#xA;need to arrange them in a single row for the graduation. To give a better impression of&#xA;diversity, you want to avoid having too many girls or too many boys seating consecutively.&#xA;You decided to arrange the students in order to minimize the gender regularity. The&#xA;gender regularity of an arrangement is the maximum number of students of the same&#xA;gender (all girls or all boys) that appear consecutively.&#xA;Given G and B, calculate the minimum gender regularity among all possible arrange-&#xA;ments.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example&#xA; g=?,b=?.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1- 10 10&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2- 5 1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3- 0 1000&lt;/p&gt;&#xA;&#xA;&lt;p&gt;result &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1- 1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2- 3&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3- 1000&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i want a formula to solve this problem&lt;/p&gt;&#xA;" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-10-16T17:48:37.073" LastActivityDate="2010-10-16T17:48:37.073" ClosedDate="2010-10-16T19:16:51.747" Title="Minimizing Gender regularity" Tags="&lt;combinatorics&gt;" CommentCount="0" />
  <row Id="6963" PostTypeId="2" ParentId="6939" CreationDate="2010-10-16T17:20:09.453" Score="3" ViewCount="" Body="&lt;p&gt;It's simple: $\rm\quad\quad  f(x)\ =\  \cos x - \sin x \ =\ \frac{1+i}2 e^{ix} + \frac{1-i}2 e^{-ix}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\displaystyle \Rightarrow\quad\quad\quad\quad\ f(\tan^{-1} x)\ =\ \frac{1-x}{\sqrt{x^2+1}}\quad $ via $\rm\displaystyle\quad e^{\:i\:tan^{-1} x}\ =\ \frac{1+ x\: i}{\sqrt{x^2 + 1}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\displaystyle \Rightarrow\quad f(\tan^{-1}\tanh x)\ =\ \ \frac{\sqrt 2}{e^{4x}+1}\quad\ $ via $\rm\displaystyle\quad\ \tanh{x}\ =\ 1 - \frac{2}{e^{2x}+1}$&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-10-18T14:34:06.757" LastActivityDate="2010-10-18T14:34:06.757" CommentCount="1" />
  <row Id="6964" PostTypeId="1" AcceptedAnswerId="7150" CreationDate="2010-10-16T18:06:31.270" Score="2" ViewCount="116" Body="&lt;p&gt;Let $T_\varphi$ be an analytic Toeplitz operator (meaning $\varphi \in H^\infty$). Further let $K$ be a compact operator that commutes with $T_\varphi$. Now I want to show that the spectrum of $K$ only consists of $0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm trying the following things: $T_\varphi K$ is compact so we can apply the spectral theorem for compact operators. So we know that at least $0$ is in the spectrum of this operator (not of $K$) and that the spectrum is countable with the only limit point $0$. Can I somehow use this? I also know that if $C$ is a compact operator $\|T_\varphi - C\| \geq \|T_\varphi \|$ so $\|T_\varphi(1 - K)\| = \|(1 - K)T_\varphi\| \geq \|T_\varphi \|$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any suggestions how I can continue?&lt;/p&gt;&#xA;" OwnerUserId="1120" LastActivityDate="2010-10-18T20:41:05.187" Title="Toeplitz operator commuting with a compact operator." Tags="&lt;complex-analysis&gt;&lt;functional-analysis&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="6965" PostTypeId="1" AcceptedAnswerId="6990" CreationDate="2010-10-16T18:33:41.290" Score="2" ViewCount="186" Body="&lt;p&gt;$$u + v = 2$$&#xA; $$ux + vy = 1$$&#xA; $$ux^2 + vy^2 = - 1$$&#xA; $$ux^3 + vy^3 = -5$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone give me a hint for solving this ? I'm kinda stuck. &lt;/p&gt;&#xA;" OwnerUserId="2227" LastActivityDate="2010-10-17T21:37:27.267" Title="system of equations" Tags="&lt;polynomials&gt;" AnswerCount="4" CommentCount="2" FavoriteCount="2" />
  <row Id="6966" PostTypeId="2" ParentId="83" CreationDate="2010-10-16T18:45:23.647" Score="2" ViewCount="" Body="&lt;p&gt;Some of my favorite ways of engaging students from jump street are:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;the Chinese magic square&lt;/li&gt;&#xA;&lt;li&gt;Erostophenes’ way of calculating the circumference of the Earth&lt;/li&gt;&#xA;&lt;li&gt;the “Why are manhole covers always round?” question&lt;/li&gt;&#xA;&lt;li&gt;the Twin-Sisters paradox&lt;/li&gt;&#xA;&lt;li&gt;Morley’s Miracle&lt;/li&gt;&#xA;&lt;li&gt;Pythagorean Triples&lt;/li&gt;&#xA;&lt;li&gt;Euler’s formula for connected planar graphs&lt;/li&gt;&#xA;&lt;li&gt;the reason for the dimensions of A4 paper&lt;/li&gt;&#xA;&lt;li&gt;Kaprekar’s Constant&lt;/li&gt;&#xA;&lt;li&gt;Casting out nines&lt;/li&gt;&#xA;&lt;li&gt;the “Which part of a train is always moving backwards?” question&lt;/li&gt;&#xA;&lt;li&gt;the question about how far the bee flying between two trains goes&lt;/li&gt;&#xA;&lt;li&gt;the red-herring question about “...met a man with seven wives...”&lt;/li&gt;&#xA;&lt;li&gt;the red-herring question about the trains leaving Chicago and Dallas (“When they meet, which one is closer to Dallas?”)&lt;/li&gt;&#xA;&lt;li&gt;the red-herring question about the rowing team that dropped a bottle into the river and row back downstream to retrieve it&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Note that the curriculum need not be improved or reformed in order to drop one or two of these in at an odd moment or two during class. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regards,&#xA;- Mike Jones&lt;/p&gt;&#xA;" OwnerUserId="2344" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-10-16T20:14:28.650" LastActivityDate="2010-10-16T20:14:28.650" CommunityOwnedDate="2010-10-16T18:45:23.647" />
  <row Id="6967" PostTypeId="2" ParentId="6965" CreationDate="2010-10-16T18:54:12.107" Score="5" ViewCount="" Body="&lt;p&gt;One way to do this is using Groebner basis (which is implemented in most serious computer algebra systems, and which you can learn to do by hand in many places, like the amazing book by Cox &lt;em&gt;et al.&lt;/em&gt; called &lt;em&gt;Ideals, Varieties and Algorithms&lt;/em&gt;). For example, using &lt;em&gt;Mathematica&lt;/em&gt; you can compute:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;In[1]:= GroebnerBasis[{&#xA;  u + v - 2,&#xA;  u x + v y - 1,&#xA;  u x^2 + v y^2 + 1,&#xA;  u x^3 + v y^3 + 5&#xA;  }, {u, v, x, y}&#xA; ]                                                &#xA;                     2&#xA;Out[1]= {2 - 3 y + y , -3 + x + y, -7 + v + 4 y, 5 + u - 4 y}&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The result is an equivalent system of equations. You'll notice that the first equation&#xA;depends only on $y$, and that the other three allow you to determine $x$, $v$ and $u$, respectively, in terms of $y$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The book I mentioned explains why this works.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively, you can use the (freely available, great) Macaulay2:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Macaulay2, version 1.3.1&#xA;with packages: ConwayPolynomials, Elimination, IntegralClosure, LLLBases, PrimaryDecomposition,&#xA;               ReesAlgebra, SchurRings, TangentCone&#xA;&#xA;i1 : R = QQ[u,v,x,y,z];&#xA;&#xA;i2 : I = ideal (u + v - 2, u*x + v*y-1, u*x^2+v*y^2+1, u*x^3+v*y^3+5);&#xA;&#xA;o2 : Ideal of R&#xA;&#xA;i3 : gens gb I&#xA;&#xA;o3 = | x+y-3 v+4y-7 u-4y+5 y2-3y+2 |&#xA;&#xA;             1       4&#xA;o3 : Matrix R  &amp;lt;--- R&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The polynomials listed in the &lt;code&gt;o3&lt;/code&gt; line are a Groebner basis, &lt;em&gt;i.e.&lt;/em&gt;, an equivalent system of equations, and it allows you to solve it at once.&lt;/p&gt;&#xA;" OwnerUserId="274" LastEditorUserId="274" LastEditorDisplayName="" LastEditDate="2010-10-16T19:01:33.860" LastActivityDate="2010-10-16T19:01:33.860" CommentCount="1" />
  <row Id="6968" PostTypeId="2" ParentId="6965" CreationDate="2010-10-16T19:03:27.880" Score="5" ViewCount="" Body="&lt;p&gt;An elementary way would be:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose $\displaystyle x$ and $\displaystyle y$ are roots of $\displaystyle z^2 + az + b = 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus we have that $\displaystyle x^2 + ax + b = 0$ and $\displaystyle y^2 + ay + b= 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Multiply the first by $\displaystyle u$, the second by $\displaystyle v$ and adding we get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$-1 +a + 2b = 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mutiplying the first $ux$ and second by $vy$ and adding we get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$-5 - a + b = 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can easily find $\displaystyle a$ and $\displaystyle b$ now and so can find the possibilities for $\displaystyle x$ and $\displaystyle y$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It should be easy to solve the whole system now.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-10-16T19:03:27.880" CommentCount="3" />
  <row Id="6969" PostTypeId="2" ParentId="6851" CreationDate="2010-10-16T19:10:38.727" Score="1" ViewCount="" Body="&lt;p&gt;By taking the seperate cases where $x = 0$ or $y = 0$ and showing that they are in the set, then we can assume $x,y \ne 0$  Therefore we can take $-2 \le xy \le 2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$-2/y \le x \le 2/y$&lt;/li&gt;&#xA;&lt;li&gt;$-2/x \le y \le 2/x$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So for all x,y real numbers satisfying the above, and (0, y real number) (x real number, 0), x,y are in R x R.&lt;/p&gt;&#xA;" OwnerUserId="2533" LastEditorUserId="2533" LastEditorDisplayName="" LastEditDate="2010-10-18T04:27:28.247" LastActivityDate="2010-10-18T04:27:28.247" CommentCount="2" />
  <row Id="6970" PostTypeId="2" ParentId="6805" CreationDate="2010-10-16T19:21:03.143" Score="1" ViewCount="" Body="&lt;p&gt;It's clearer if you write it instead in a symmetric fashion: $(n-1)^3+n^3+(n+1)^3$, which simplifies to $3n(n^2+2)$. If $n$ is a multiple of $3$ then this is a multiple of $9$. Otherwise, $n$ leaves a remainder of $\pm1$ when divided by $3$ and so $n^2+2$ is multiple of $3$, which implies that $3n(n^2+2)$ is a multiple of $9$.&#xA;It's not a proof by induction, though.&lt;/p&gt;&#xA;" OwnerUserId="589" LastActivityDate="2010-10-16T19:21:03.143" CommentCount="2" />
  <row Id="6971" PostTypeId="1" AcceptedAnswerId="6995" CreationDate="2010-10-16T19:46:06.293" Score="1" ViewCount="152" Body="&lt;p&gt;Suppose you want to accumulate $12,000$ in a $5 \%$ account by making a level deposit at the beginning of each of the next 9 years. Find the required level payment. So this seems to be an annuity due problem. I know the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ \displaystyle \sum_{k=1}^{n} \frac{A}{(1+i)^{k}} = \frac{A}{1+i} \left[\frac{1- \left(\frac{1}{1+i} \right)^{n}}{1- \left(\frac{1}{1+i} \right)} \right] = P$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in this problem, we are trying to solve for $P$? Just plug in the numbers? Or do we need to calculate the discount rate $d = i/(i+1)$ since the annuity is being payed at the beginning of the year? &lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2010-10-18T10:16:50.500" Title="Principal and Annuities" Tags="&lt;homework&gt;&lt;finance&gt;" AnswerCount="2" CommentCount="5" />
  <row Id="6972" PostTypeId="2" ParentId="6971" CreationDate="2010-10-16T19:53:45.100" Score="1" ViewCount="" Body="&lt;p&gt;The problem statement is missing the time when you want to have the 12,000.  If it is at the end of the ninth year, the value of the deposit at the beginning of year n will have increased by 1.05^(10-n).  So if A is the deposit you have $\displaystyle \sum_{k=1}^{n} A*1.05^{(10-k)}=12000$.  Solve for A&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-10-16T19:53:45.100" CommentCount="2" />
  <row Id="6973" PostTypeId="2" ParentId="6713" CreationDate="2010-10-16T19:54:20.263" Score="2" ViewCount="" Body="&lt;p&gt;As Rahul's comment indicates, there is no algorithm known to achieve an optimal packing.&#xA;However, some effective heuristics have been developed. For example, there is a 2010 paper by &#xA;Andrea Cassioli and Marco Locatelli entitled&#xA;&quot;A heuristic approach for packing rectangles in convex regions&quot;&#xA;(&lt;a href=&quot;http://www.optimization-online.org/DB_FILE/2010/06/2637.pdf&quot; rel=&quot;nofollow&quot;&gt;PDF download&lt;/a&gt;).&#xA;Here is their abstract:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In this paper we propose a heuristic approach for the problem of packing equal rectangles within a convex region. The approach is based on an Iterated Local Search scheme (or, using a terminology employed for continuous problems, a Monotonic Basin Hopping), in which the key step is the perturbation move. Different perturbation moves, both combinatorial and continuous ones, are pro- posed and compared through extensive computational experiments on a set of test instances. The overall results are quite encouraging.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This recent paper has 38 references, so this should give you much to pursue and ponder.&lt;/p&gt;&#xA;" OwnerUserId="237" LastActivityDate="2010-10-16T19:54:20.263" />
  <row Id="6974" PostTypeId="2" ParentId="6947" CreationDate="2010-10-16T20:13:28.120" Score="2" ViewCount="" Body="&lt;p&gt;There are pointers to a wealth of information on this question in the responses to the &lt;a href=&quot;http://mathoverflow.net/questions/37708/&quot; rel=&quot;nofollow&quot;&gt;Math Overflow question&lt;/a&gt;, which you mentioned in your comment to Paul VanKoughnett's response.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, Deane Yang's response gives a nice summary of the situation, and Bill Thurston's response seems to give a good perspective on the problem of trying to find a characterization of Riemannian manifolds that admit such an embedding.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regarding the third question you mention in an edit. This is essentially a local problem. All this from the same MO question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From BS's response:&#xA;there is not even a local isometric embedding in general:&#xA;&lt;a href=&quot;http://www.springerlink.com/content/m775p64w64351260/&quot; rel=&quot;nofollow&quot;&gt;http://www.springerlink.com/content/m775p64w64351260/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;from Will Jaggy's response (and Deane Yang's comments on it):&#xA;If the metric is analytic, then you can construct a local isometric embedding.&#xA;Some recent progress on characterizing the requirements when the degree of smoothness is relaxed:&#xA;&lt;a href=&quot;http://arxiv.org/abs/1009.6214&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/1009.6214&lt;/a&gt;&#xA;The bibliography for that last one has no shortage of other relevant sounding titles.&lt;/p&gt;&#xA;" OwnerUserId="1728" LastActivityDate="2010-10-16T20:13:28.120" CommentCount="1" />
  <row Id="6975" PostTypeId="1" AcceptedAnswerId="6982" CreationDate="2010-10-16T20:34:46.983" Score="3" ViewCount="175" Body="&lt;p&gt;I'm reading &lt;em&gt;A Computational Introduction to Number Theory and Algebra&lt;/em&gt;, which can be found &lt;a href=&quot;http://www.shoup.net/ntb/&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; as a free download. From the book's exercises, I'm stuck with a proof to show that $\gcd(a,b)=|a| \iff a | b$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My current reasoning goes as follows: since $\gcd(a,b) = |a|$, we see that $a\mathbb{Z} + b\mathbb{Z}= |a|\mathbb{Z}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $a|b \iff az = b$, for some $z \in \mathbb{Z}$, we see that both $az, b \in |a|\mathbb{Z}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So there exist $s, t \in \mathbb{Z}$ such that $as + bt = az = b$, which is true for $s = 0$ and $t = 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Q.E.D.&lt;/em&gt;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I guess that in the end my reasoning is wrong, but I can't explain it.&lt;/p&gt;&#xA;" OwnerUserId="2566" LastEditorUserId="622" LastEditorDisplayName="" LastEditDate="2011-05-07T21:15:45.593" LastActivityDate="2011-05-07T21:15:45.593" Title="Show that $\gcd(a,b)=|a| \iff a | b$" Tags="&lt;number-theory&gt;&lt;abstract-algebra&gt;" AnswerCount="3" FavoriteCount="1" />
  <row Id="6976" PostTypeId="1" AcceptedAnswerId="6980" CreationDate="2010-10-16T20:39:53.287" Score="2" ViewCount="242" Body="&lt;p&gt;I have not been able to solve this problem. Any insights would be appreciated!&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Let $x, n &gt; 1$ be integers. Suppose that for each $k &gt; 1$ there exists an integer $a_{k}$&#xA;such that $x − a_k^n$ is divisible by $k$. Prove that $x = A^{n}$ for some integer $A$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" LastEditorUserId="1778" LastEditorDisplayName="" LastEditDate="2010-11-28T07:00:38.503" LastActivityDate="2010-11-28T07:00:38.503" Title="Proving that an integer is the $n$ th power" Tags="&lt;number-theory&gt;&lt;diophantine-equations&gt;&lt;arithmetic&gt;" AnswerCount="3" CommentCount="4" />
  <row Id="6977" PostTypeId="2" ParentId="6975" CreationDate="2010-10-16T20:44:38.970" Score="2" ViewCount="" Body="&lt;p&gt;It is true that the final assertion holds if $s=0$ and $t=1$, but it is not true that  that $as+bt = az = b$ can only hold if $s=0$ and $t=1$ is false. Take, for example, $a=2$ and $b=4$. Then $(-5)a + 3(b) = b$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remember that, by definition, gcd$(x,y)|x$ and gcd$(x,y)|y$. That gives one implication.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the converse, remember that $d|x$ and $d|y$ implies $d|$gcd$(x,y)$.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-10-16T20:44:38.970" />
  <row Id="6979" PostTypeId="1" AcceptedAnswerId="6984" CreationDate="2010-10-16T20:55:54.397" Score="11" ViewCount="218" Body="&lt;p&gt;Suppose we rotate the graph of $y = f(x)$ about the $x$-axis from $a$ to $b$.  Then (using the disk method) the volume is $$\int_a^b \pi f(x)^2 dx$$ since we approximate a little piece as a cylinder.  However, if we want to find the surface area, then we approximate it as part of a cone and the formula is $$\int_a^b 2\pi f(x)\sqrt{1+f'(x)} dx.$$  But if approximated it by a circle with thickness $dx$ we would get $$\int_a^b 2\pi f(x) dx.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So my question is how come for volume we can make the cruder approximation of a disk but for surface area we can't.&lt;/p&gt;&#xA;" OwnerUserId="9" LastActivityDate="2010-10-17T01:24:29.310" Title="Areas versus volumes of revolution." Tags="&lt;calculus&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="1" />
  <row Id="6980" PostTypeId="2" ParentId="6976" CreationDate="2010-10-16T20:58:51.893" Score="6" ViewCount="" Body="&lt;p&gt;This is an old chestnut: an integer which is an $n$-th power modulo&#xA;all primes is an $n$-th power.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a sledgehammer proof via the Chebotarev density theorem.&#xA;Suppose for the moment that $n$ is prime.&#xA;Consider $K=\mathbb{Q}(x^{1/n})$ and its Galois closure&#xA;$L=\mathbb{Q}(x^{1/n},\exp(2\pi i/n))$. Then each prime $p$ that is unramified&#xA;in $L$ splits in $K$ into various prime ideals at least one of which has norm $p$.&#xA;So its Frobenius has a fixed point on the permutation representation on the&#xA;$n$-th roots of $x$. By Chebotarev, the Galois group $G$ of $L/\mathbb{Q}$&#xA;has no element of degree $n$ and so must have a fixed point; that is one&#xA;of the $n$-th roots of $x$ must lie in $\mathbb{Q}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm sure something similar works for any positive integer $n$, but it's&#xA;too late tonight for me to work out the details :-)&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-10-16T20:58:51.893" CommentCount="2" />
  <row Id="6981" PostTypeId="1" CreationDate="2010-10-16T21:06:19.600" Score="4" ViewCount="292" Body="&lt;p&gt;I'm having trouble wrapping my head around the following issue. My book solves a problem without using complex exponential solution like $C_1  e^{it}$ and using either $A \cos(t) + B \sin(t)$ or $A \cos(t+\phi)$ to a second-order &quot;undamped&quot; linear ODE. I understand that the solution can be purely real or complex. The trouble arises when I use the solution to this ODE inside another function which cubes this solution. Cubing $C_1  e^{it}$ results in another complex sinusoid with 3 times the frequency, which is a very different effect from cubing something like $A \cos(t+\phi)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is my work. I've highlighted where I think things have gone awry.&#xA;&lt;img src=&quot;http://i.imgur.com/ORCn9.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/tKTRS.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="2256" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-12-08T21:06:27.980" LastActivityDate="2011-01-07T21:44:39.953" Title="Using complex exponentials as solution of ODE" Tags="&lt;complex-numbers&gt;&lt;fourier-series&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="6982" PostTypeId="2" ParentId="6975" CreationDate="2010-10-16T21:13:29.813" Score="3" ViewCount="" Body="&lt;p&gt;Assume that $a,b\in\mathbb{N}$. I use the following definition of $D=\gcd (a,b)$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(i) $D$ is a common divisor of $a$ and $b$;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(ii) Every integer $d\in\mathbb{N}$ which is a common divisor of $a$ and $b$ divides $D$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proposition 1: If $\gcd (a,b)=a$, then $a|b$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proposition 2: If $a|b$, then $\gcd (a,b)=a$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proof: Since $a|a$ and $a|b$, then any integer $d$ such that $d|a$ and $d|b$&#xA;satisfies also the condition $d|a$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From $a|a$, $a|b$ and $d|a$ we conclude that $\gcd (a,b)=a$.&lt;/p&gt;&#xA;" OwnerUserId="752" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2010-10-16T22:45:02.767" LastActivityDate="2010-10-16T22:45:02.767" CommentCount="4" />
  <row Id="6983" PostTypeId="2" ParentId="6976" CreationDate="2010-10-16T21:20:37.067" Score="4" ViewCount="" Body="&lt;p&gt;To be a little more concrete, here's the proof for $n = 2$.  We proceed by proving the contrapositive.  Suppose $x$ is not a square, so write $x = k^2 p_1 p_2 ... p_n$ where the $p_i$ are distinct primes, one of which may be $-1$.  By quadratic reciprocity, together with the CRT, there exists an arithmetic progression such that any prime $q$ in that arithmetic progression has the property that $\left( \frac{p_i}{q} \right) = 1$ for $1 \le i \le n-1$ and $\left( \frac{p_i}{q} \right) = -1$ for $i = n$.  A prime $q$ exists in this arithmetic progression by Dirichlet's theorem (a special case of Chebotarev's density theorem!), and $x$ is not a square $\bmod q$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly for $n = 3$ one can give a proof using cubic reciprocity, and so on.  In general it should be possible to give a proof along these lines using Artin reciprocity, but if you're using Artin reciprocity and Dirichlet's theorem you might as well use the full strength of Chebotarev.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-10-16T21:20:37.067" CommentCount="1" />
  <row Id="6984" PostTypeId="2" ParentId="6979" CreationDate="2010-10-16T21:21:29.950" Score="7" ViewCount="" Body="&lt;p&gt;In the case of volume, the error is second order (it is basically revolving a triangle with sides $dx$ and $f'(x)dx$ so is negligible compared to the cylinder which is first order.  This is the same as the 1 dimensional case, where to measure area (the standard integral) you use rectangles but to measure arc length you need to integrate $\sqrt{1+f'(x)^2}dx$, which is the hypotenuse of the triangle.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2010-10-17T01:24:29.310" LastActivityDate="2010-10-17T01:24:29.310" CommentCount="1" />
  <row Id="6985" PostTypeId="2" ParentId="6981" CreationDate="2010-10-16T21:25:13.100" Score="1" ViewCount="" Body="&lt;p&gt;If you cube $cos(t)$ you get $(3cos(t) + cos(3t))/4$ so the triple frequency shows up here, too.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-10-16T21:25:13.100" CommentCount="5" />
  <row Id="6986" PostTypeId="2" ParentId="6965" CreationDate="2010-10-16T21:37:11.830" Score="2" ViewCount="" Body="&lt;p&gt;Looking for linear terms suggests we write the system as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{pmatrix} 1 &amp;amp; 1 \cr x &amp;amp; y  \cr x^2 &amp;amp; y^2 \cr x^3 &amp;amp; y^3\end{pmatrix}&#xA;\begin{pmatrix} u \cr v\end{pmatrix}&#xA;=\begin{pmatrix} 2 \cr 1 \cr -1 \cr -5 \end{pmatrix}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Row-reduction has to give two entirely zero rows of coefficients,yielding two equations for $x$ and $y$, one of which is linear in $y$.  (It will help to recognize that $x-y$ is a factor of $x^2 - y^2$ and of $x^3 - y^3$.)  That makes it straightforward to find $x$, from which you obtain $y$.  With $x$ and $y$ in hand the coefficients are completely determined, allowing you to solve for $(u, v)$ in the usual ways.&lt;/p&gt;&#xA;" OwnerUserId="1489" LastActivityDate="2010-10-16T21:37:11.830" />
  <row Id="6987" PostTypeId="2" ParentId="6979" CreationDate="2010-10-16T21:56:53.440" Score="10" ViewCount="" Body="&lt;p&gt;The problem here is that volume behaves nicely under small deformations of 3D regions in 3D, but surface area does not.  Similarly, area behaves nicely under small deformations of 2D regions in 2D, but circumference / arc length does not.  You can see the essence of the 2D problem, hence the essence of the 3D problem, in the following: the length of the diagonal from $(0, 0)$ to $(1, 1)$ is $\sqrt{2}$, but if we approximate the diagonal by a &quot;staircase&quot; of horizontal and vertical lines of length $\frac{1}{n}$ and let $n \to \infty$ we get a length of $2$ instead.  &lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-10-16T21:56:53.440" />
  <row Id="6988" PostTypeId="1" CreationDate="2010-10-16T22:04:55.940" Score="4" ViewCount="189" Body="&lt;p&gt;Given two Riemannian Manifolds of dimension 2, and a point on each. If the scalar curvatures are isomorphic (as functions) in some neighbourhoods of these points, are then the manifolds necessarily locally isometric?&lt;/p&gt;&#xA;" LastEditorUserId="8246" LastEditorDisplayName="" LastEditDate="2011-05-10T18:41:27.247" LastActivityDate="2011-05-10T18:41:27.247" Title="Is the scalar curvature the only isometric invariant of a Riemannian 2-manifold?" Tags="&lt;differential-geometry&gt;" AnswerCount="3" FavoriteCount="2" />
  <row Id="6989" PostTypeId="2" ParentId="6988" CreationDate="2010-10-16T22:20:30.820" Score="4" ViewCount="" Body="&lt;p&gt;This is &lt;a href=&quot;http://en.wikipedia.org/wiki/Gaussian_curvature#Surfaces_of_constant_curvature&quot; rel=&quot;nofollow&quot;&gt;true if the curvatures are constant&lt;/a&gt; (Minding's theorem), but not in general.  &lt;strike&gt;I actually don't know a counterexample, but (edit: I think that)&lt;/strike&gt; Stillwell &lt;a href=&quot;http://books.google.com/books?id=V7mxZqjs5yUC&amp;amp;lpg=PP1&amp;amp;pg=PA344#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;points out on page 344&lt;/a&gt; of &lt;em&gt;Mathematics and its history&lt;/em&gt; that Strubecker's &lt;em&gt;Differentialgeometrie&lt;/em&gt; gives &lt;strike&gt;such an&lt;/strike&gt; a related example in Volume III on page 124, but I'm not 100% sure if that is a counterexample of what you're asking, and I don't currently have access to it.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Update: Okay, here's an example I can access.  In Audin's &lt;em&gt;Geometry&lt;/em&gt;, after the &lt;em&gt;Theorema egregium&lt;/em&gt;, there is a remark that the &quot;converse is wrong; a classical counter-example will be found in Exercise VIII.24.&quot; Thankfully &lt;a href=&quot;http://books.google.com/books?id=U_cTJMCIzdUC&amp;amp;lpg=PP1&amp;amp;pg=PA300#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;Exercise VIII.24&lt;/a&gt; is available on Google Books, at least to me.  There is a section with hints and solutions, too, but I can't see whether it includes that exercise.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2010-10-16T22:56:53.140" LastActivityDate="2010-10-16T22:56:53.140" CommentCount="2" />
  <row Id="6990" PostTypeId="2" ParentId="6965" CreationDate="2010-10-16T22:23:59.137" Score="6" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\ \ $ Exploit the innate symmetry: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\quad\quad\quad  u x^2 + v y^2\ =\ (x+y)\ (ux\ +\ vy)\  -\  xy\ (u+v)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\quad\quad\quad  u x^3 + v y^3\ =\ (x+y)\ (ux^2+vy^2) -\: xy\ (ux+vy)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;yields  $\rm\quad\quad\quad\quad\:\ \ -1 \ =\ \ x+y\ -\ 2\ xy $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\quad\quad\quad\quad\quad\quad\quad\quad\  5\ =\ \ x+y\ \:\ +\:\ \ xy$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solving yields $\rm\quad\quad\ \  x+y = 3,\ \ \ xy = 2\ \ \Rightarrow\ \ \{x,\:y\}\ =\ \{1,\:2\}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;REMARK&lt;/strong&gt; $\ $ Readers familiar with the  theory of &lt;strong&gt;Lucas-Lehmer sequences&lt;/strong&gt; may recognize this recursion as one of many useful identities satisfied by sums of powers.  Such identities arise in many diverse contexts. Here is another example: Capdegelle's work on &lt;strong&gt;FLT&lt;/strong&gt; (Fermat's Last Theorem) employed the recursion $\rm\ F_{n+3} + S_2\: F_{n+2} + S_1\: F_{n+1} + S_0\: F_n = 0\ $ where $\rm\ F_n = X^n + Y^n - Z^n\ $ is the Fermat polynomial and the polynomials $\rm\ S_k\ $ are the elementary symmetric polynomials $\rm\ S_0 = -XYZ,$ $\rm\ S_1 = XY + XZ + YZ,\ $ $\rm S_2 = -(X+Y+Z)\:.\ $ &lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-10-17T21:37:27.267" LastActivityDate="2010-10-17T21:37:27.267" CommentCount="10" />
  <row Id="6991" PostTypeId="2" ParentId="6975" CreationDate="2010-10-16T22:41:51.320" Score="0" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\rm\ \ $ If $\rm\ a\:|\:b\ $ then $\rm\ c\:|\:a,b\ \iff\ c\:|\:a\ \ \ $ [and conversely via $\rm\ c = a\:$]  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus $\rm\ a,\:b\ $  and $\rm\:a\:$ have the same common divisors $\rm\:c\:$, hence the same &lt;em&gt;greatest&lt;/em&gt;$\ $ common divisor.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-10-18T13:59:03.193" LastActivityDate="2010-10-18T13:59:03.193" />
  <row Id="6992" PostTypeId="2" ParentId="6988" CreationDate="2010-10-16T22:44:06.717" Score="1" ViewCount="" Body="&lt;p&gt;I'm a little rusty on this stuff, but here's a sketch of a general answer.  Basically the idea is that for any differential-geometric structure on a manifold (Riemannian metric, almost-complex structure, Finsler structure, symplectic structure, jet structure, web structure, etc.) you define the &quot;frame bundle&quot; to be the space of ways of identifying the structure with the canonical one on $\mathbb{R}^n$.  So in the case of a Riemannian metric, point by point you choose an identification of the tangent space with $\mathbb{R}^n$ so that the inner product agrees with the usual one on $\mathbb{R}^n$, and then you allow yourself to post-compose with any element of the orthogonal group $O(n)$ in order to allow for differences in the original choices.  (This is because the orthogonal group is exactly the space of linear maps that preserves the inner product.)  This gives you what is known as a &lt;em&gt;G-structure&lt;/em&gt; on the manifold (here $G=O(n)$).  This is a bundle over the manifold where every fiber is a $G$-torsor, i.e. a space which is topologically the same as $G$ but there's no choice of which point is the identity because there was no canonical choice of how to identify $T_pM$ with $\mathbb{R}^n$.  (So e.g. a real line is a torsor for the group $\mathbb{R}$.)  The next step is to try and make a connection for this bundle, and if I remember correctly, this is the place where we tease out geometric invariants.  There's a Lie-algebra-valued 1-form (with values in $Lie(G)$) that measures the failure of the connection to be trivial(izable), I think, and then you calculate something called the &quot;Spencer cohomology&quot;.  The main point is that this gives &lt;em&gt;all&lt;/em&gt; the invariants, so for example it falls out as an immediate consequence that two manifolds are locally isomorphic if and only if there's a local diffeomorphism that lines up their Gaussian curvatures.  (Something else really cool is that this perspective tells you exactly why we can have the Riemann mapping theorem in dimension 2, but then in all higher dimensions the only conformal maps are Mobius transformations.  Also this can be used to prove Darboux's theorem, and the same method gives the Nijenhaus tensor.)&lt;/p&gt;&#xA;" OwnerUserId="2159" LastEditorUserId="2159" LastEditorDisplayName="" LastEditDate="2010-10-17T03:50:26.637" LastActivityDate="2010-10-17T03:50:26.637" CommentCount="3" />
  <row Id="6993" PostTypeId="1" CreationDate="2010-10-16T22:45:59.347" Score="3" ViewCount="93" Body="&lt;p&gt;During the summer, I did an REU where we focused primarily on one-dimensional dynamics and more specifically kneading theory. One thing that I was always confused about is why the Schwarzian derivatives always seem to pop up in discussions of iterated dynamics on the real line. I understand what a Schwarzian derivative is, but I don't see any intuitive reason that it should show up in this area.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was wondering if anyone could explain or provide me with a reference that makes the appearance of Schwarzian derivatives in one-dimensional dynamics on the real line seem natural.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another question I have, is there an intuitive motivation for the Schwarzian derivative itself?&lt;/p&gt;&#xA;" OwnerUserId="249" LastEditorUserId="249" LastEditorDisplayName="" LastEditDate="2010-10-17T00:58:35.490" LastActivityDate="2010-10-17T00:58:35.490" Title="Schwarzian Derivative and One-Dimensional Dynamics - how are they connected?" Tags="&lt;intuition&gt;&lt;reference-request&gt;&lt;dynamical-systems&gt;" CommentCount="4" FavoriteCount="1" />
  <row Id="6994" PostTypeId="1" AcceptedAnswerId="7106" CreationDate="2010-10-16T23:46:40.097" Score="10" ViewCount="195" Body="&lt;p&gt;Krull's Hauptidealsatz (principal ideal theorem) says that for a Noetherian ring $R$ and any $r\in R$ which is not a unit or zero-divisor, all primes minimal over $(r)$ are of height 1. How badly can this fail if $R$ is a non-Noetherian ring? For example, if $R$ is non-Noetherian, is it possible for there to be a minimal prime over $(r)$ of infinite height?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: The answer is yes. See &lt;a href=&quot;http://mathoverflow.net/questions/42510/how-badly-can-krulls-hauptidealsatz-fail-for-non-noetherian-rings&quot; rel=&quot;nofollow&quot;&gt;http://mathoverflow.net/questions/42510/how-badly-can-krulls-hauptidealsatz-fail-for-non-noetherian-rings&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="264" LastEditorUserId="264" LastEditorDisplayName="" LastEditDate="2010-10-17T23:17:16.380" LastActivityDate="2010-10-18T07:23:06.007" Title="How badly can Krull's Hauptidealsatz fail for non-Noetherian rings?" Tags="&lt;commutative-algebra&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="6995" PostTypeId="2" ParentId="6971" CreationDate="2010-10-17T00:15:35.477" Score="0" ViewCount="" Body="&lt;p&gt;I get $F = A \frac{(1+i)^{2n}-(1+i)^n}{i(1+i)^n}$. But when I solve for $A$, I get the level payment as being $1088.28$. But the correct answer is $1036.46$. Why is there a discrepancy?&lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2010-10-17T00:15:35.477" CommentCount="4" />
  <row Id="6996" PostTypeId="1" AcceptedAnswerId="6997" CreationDate="2010-10-17T02:30:41.267" Score="1" ViewCount="87" Body="&lt;p&gt;According to an &lt;a href=&quot;http://www.cs.bu.edu/faculty/crovella/paper-archive/ss-lect-notes.ps&quot; rel=&quot;nofollow&quot;&gt;article&lt;/a&gt; I'm studying (&quot;Time series, self-similarity&#xA;and network traffic by Mark Crovella) the expectation of the square of&#xA;a time-scaled Brownian motion process $E[ B(ct)^2 ]$ where $c$ is the time&#xA;scaling is equal to $ct$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd appreciate help proving this; i.e.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$E[  B(ct)^2 ] = c t$&lt;/p&gt;&#xA;" OwnerUserId="1752" LastActivityDate="2010-10-17T02:49:47.730" Title="Expectation of squared time-scaled Brownian process" Tags="&lt;probability&gt;&lt;statistics&gt;" AnswerCount="1" />
  <row Id="6997" PostTypeId="2" ParentId="6996" CreationDate="2010-10-17T02:44:08.583" Score="3" ViewCount="" Body="&lt;p&gt;This follows from the fact that $B(t)$ has a Gaussian$(0,t)$ distribution.  Therefore, $B(ct)$ has a Gaussian$(0,ct)$ distribution.  Thus $E[B(ct)^2] = Var[B(ct)] + (E[B(ct)])^2 = ct + 0 = ct$.&lt;/p&gt;&#xA;" OwnerUserId="2370" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-10-17T02:49:47.730" LastActivityDate="2010-10-17T02:49:47.730" CommentCount="4" />
  <row Id="6998" PostTypeId="1" AcceptedAnswerId="7009" CreationDate="2010-10-17T03:47:07.083" Score="10" ViewCount="1109" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm teaching a group theory course now, and I wanted to give my students a proof that every subgroup of a cyclic group is cyclic. The easiest way I could think to do this is to say that any cyclic group corresponds to a $\mathbb{Z}/n\mathbb{Z}$ and then (implicitly) use the first isomorphism theorem (which they haven't learned yet) to &quot;reduce&quot; it to the theorem that every subgroup of $\mathbb{Z}$ is of the form $m\mathbb{Z}$. The problem is that this latest statement is not simple at all (every subgroup is an ideal, and $\mathbb{Z}$ is a PID because $\mathbb{Z}$ is Euclidean).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This all seems way too convoluted for such an easy statement. What's the simplest most elementary proof that I can give?&lt;/p&gt;&#xA;" LastActivityDate="2010-10-17T22:34:22.857" Title="How to show every subgroup of a cyclic group is cyclic?" Tags="&lt;group-theory&gt;&lt;education&gt;" AnswerCount="5" CommentCount="2" FavoriteCount="3" />
  <row Id="6999" PostTypeId="2" ParentId="6998" CreationDate="2010-10-17T03:51:25.963" Score="7" ViewCount="" Body="&lt;p&gt;If the original group is generated by $a$, and $d$ is the least positive integer such that $a^d$ is in the subgroup, then show using the division algorithm that $a^d$ generates the subgroup.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;To elaborate, the division algorithm says:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If $m$ and $n$ are integers and $n$ is positive, then there are unique integers $q$ and $r$ such that $m=nq+r$ and $0\leq r\lt n$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You could prove this by taking $r$ to be the least nonnegative integer expressible in the form $m-np$ as $p$ ranges over the integers, which then determines $q$.  (There are details here which I am omitting.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose $G$ is a cyclic group (finite or infinite) with generator $a$, i.e., $G=\langle a\rangle$.  Let $H$ be a subgroup of $G$.  Let $d$ be the smallest element of the set of positive integers $n$ such that $a^n$ is in $H$.  (Unless $G$ is infinite and $H$ is trivial, this exists.)  If $b$ is an arbitrary element of $H$, then because $b$ is in $G$, $b=a^m$ for some integer $m$.  Apply the division algorithm to write $m=dq+r$ for integers $q$ and $r$ with $0\leq r\lt d$.  Then $b=a^m=(a^d)^q\cdot a^r$.  Since $b$ is in $H$ and $a^d$ is in $H$, it follows that $a^r=a^m\cdot(a^d)^{-q}$ is in $H$.  If $r$ were positive, this would contradict the choice of $d$, so $r=0$, and $b=a^m=(a^d)^q$.  Since $b$ was arbitrary, this shows that $H\subseteq \langle a^d\rangle$.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2010-10-17T04:32:42.360" LastActivityDate="2010-10-17T04:32:42.360" CommentCount="9" />
  <row Id="7000" PostTypeId="2" ParentId="6976" CreationDate="2010-10-17T04:12:50.847" Score="6" ViewCount="" Body="&lt;p&gt;If I am not mistaken, the question has a more elementary answer than those provided so far.  I will use the functions $\operatorname{ord}_p: \mathbb{Q}^{\times} \rightarrow \mathbb{Z}$, defined to be the largest power of $p$ appearing in the numerator minus the largest power of $p$ appearing in the denominator.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Step 1: A positive rational number $x$ is a rational $n$th power iff for all primes $p$, $\operatorname{ord}_p(x)$ is divisible by $n$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is an easy consequence of unique factorization in $\mathbb{Z}$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Step 2: A positive integer $x$ is an integral $n$th power iff for all primes $p$, &#xA;$\operatorname{ord}_p(x)$ is divisible by $n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This follows easily from Step 1 using the fact that since $\mathbb{Z}$ is a UFD, it is integrally closed.  (Alternately, apply the rational roots theorem to the polynomial $t^n - x$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Step 3: As in lhf's comment above, I claim that the given condition on $x$ implies that $x$ is an nth power in $\mathbb{Q}_p$ for all primes $p$.  Indeed, taking $k$ to be a prime power, it implies that for all positive integers $a$, the congruence $t^n -x \equiv 0 \pmod{p^a}$ has a solution, and by a routine application of Hensel's Lemma, this implies that $x$ is an $n$th power in $\mathbb{Q}_p$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Step 4: Since $x$ is an $n$th power in $\mathbb{Q}_p$, the $p$-adic valuation $v_p(x)$ is divisible by $n$.  For a rational number $x$, this means that $\operatorname{ord}_p(x)$ is divisible by $n$.  And we are done.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As regards the fancy stuff, perhaps people are thinking of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Grunwald%E2%80%93Wang_theorem&quot; rel=&quot;nofollow&quot;&gt;Grunwald-Wang Theorem&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This says that if $x$ is an element of a global field $K$ which is an $n$th power in all but finitely many completions at finite places $v$.  Actually, this is &quot;Grunwald's theorem&quot;, i.e., it isn't quite true!  Wang showed that there are counterexamples to this statement, even over $\mathbb{Q}$ (if one uses all but finitely many places, rather than all places): see the wikipedia article for an explanation.  The easy proof that I give above works in any global field which is the fraction field of a PID $R$, with the conclusion that $x$ comes out to be an $n$th power up to a unit of $R$.  (When $R = \mathbb{Z}$, requiring $x$ to be positive fixes the unit ambiguity.)  &lt;/p&gt;&#xA;" OwnerUserId="299" LastActivityDate="2010-10-17T04:12:50.847" CommentCount="4" />
  <row Id="7001" PostTypeId="2" ParentId="6998" CreationDate="2010-10-17T04:27:56.593" Score="2" ViewCount="" Body="&lt;p&gt;See Proposition 5 in &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.math.uga.edu/~pete/4400algebra2point5.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.math.uga.edu/~pete/4400algebra2point5.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that part of this argument is left as an exercise for the reader.  It &lt;em&gt;is&lt;/em&gt; an incarnation of the same argument that shows that a Euclidean domain is a PID, but one certainly doesn't need this terminology or any prior results.  Namely, one is supposed to show that -- upon identifying the elements of a cyclic group $Z_n$ with the integers $0,1,\ldots,n-1$ as usual -- if $H$ is a nontrivial subgroup of $Z_n$, then it is generated by its smallest nonzero element, say $k$.  Why?  Well, if not, there would exist an element $\ell$ of the subgroup which is not a multiple of $k$, and then by subtracting off a suitable multiple of $k$ -- to be completely explicit, let $a$ be the largest non-negative integer $a$ such that $\ell - ak$ is positive, and subtract $ak$ -- we get an element of the subgroup which is positive but smaller than $k$, contradiction.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the more explicitly ring-theoretic proof you have in mind is the one alluded to in the Remark at the bottom of page 2.&lt;/p&gt;&#xA;" OwnerUserId="299" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2010-10-17T04:38:53.113" LastActivityDate="2010-10-17T04:38:53.113" />
  <row Id="7002" PostTypeId="1" AcceptedAnswerId="7007" CreationDate="2010-10-17T04:37:46.953" Score="2" ViewCount="228" Body="&lt;p&gt;Prove the following: If $H$ is a subgroup of finite index in a group $G$, and $K$ is a subgroup of $G$ containing $H$, then $K$ is of finite index in $G$ and $[G:H] = [G:K][K:H]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So this is basically a bijective proof? The number of cosets of $H$ in $G$ equals the number of cosets of $K$ in $G$ times the number of cosets of $H$ in $K$ by the multiplication principle? &lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2011-02-17T22:27:04.477" Title="Subgroups and finite index" Tags="&lt;group-theory&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="1" />
  <row Id="7004" PostTypeId="1" AcceptedAnswerId="7017" CreationDate="2010-10-17T05:01:18.760" Score="1" ViewCount="215" Body="&lt;p&gt;Suppose $f(x) = \begin{cases} 0 \ \ \text{if} \ x \in \mathbb{R}- \mathbb{Q} \newline \frac{1}{q} \ \ \text{if} \ x \in \mathbb{Q} \ \text{and} \ x = \frac{p}{q} \ \text{in lowest terms} \end{cases}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(i) Is $f$ continuous on the irrationals?&#xA;(ii) Is $f$ continuous on the rationals?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For (i) you could use the sequence definition of continuity? Maybe try $a_n = \frac{\sqrt{2}}{n}$ and show that $a_n \to 0$ but $f(a_n) \not \to 0$? So its discontinuous on the irrationals?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For (ii) I don't see why there are $1+2+ \cdots + (q-1)$ rational numbers? I know that we need to use this fact to choose an appropriate $\delta$ (e.g  $0&amp;lt; |x-a| &amp;lt; \delta \Rightarrow |f(x)-L| &amp;lt; \epsilon$). &lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2010-10-17T11:51:47.543" Title="Continuity of function" Tags="&lt;real-analysis&gt;" AnswerCount="3" CommentCount="6" />
  <row Id="7005" PostTypeId="2" ParentId="7004" CreationDate="2010-10-17T05:11:30.873" Score="1" ViewCount="" Body="&lt;p&gt;Your $a_n$'s are irrational, so $f(a_n)=0$.  The function is continuous on the irrationals.  You can show this using the $\epsilon$-$\delta$ definition of continuity.  Once you're given an irrational $x$ and an $\epsilon&gt;0$, there is an integer $n&gt;1/\epsilon$, and there are only finitely many rational numbers in, say, $(x-1,x+1)$ having denominator smaller than $n$ is lowest terms.  Thus there is a closest one to $x$, and you can use this to find your $\delta$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using similar reasoning, you can show that the limit at each rational is 0, but, as AD. points out, without knowing this there is a quick way to see that the function is discontinuous at each rational number. &lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2010-10-17T05:35:31.370" LastActivityDate="2010-10-17T05:35:31.370" CommentCount="2" />
  <row Id="7006" PostTypeId="2" ParentId="7004" CreationDate="2010-10-17T05:17:08.753" Score="0" ViewCount="" Body="&lt;p&gt;Here's a picture that might help you see things more clearly: &lt;a href=&quot;http://en.wikipedia.org/wiki/Thomae%27s_function&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Thomae%27s_function&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1736" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-10-17T06:23:34.850" LastActivityDate="2010-10-17T06:23:34.850" />
  <row Id="7007" PostTypeId="2" ParentId="7002" CreationDate="2010-10-17T06:21:49.177" Score="2" ViewCount="" Body="&lt;p&gt;In fact, the result holds even if you do not assume that we are dealing with finite index, provided you take the equality to be an equality of cardinalities. Just follow the idea Pete Clark suggests, but without assuming that the number of cosets is finite in either case.  And remember that the number of distinct cosets of a subgroup $B$ in a group $A$ is equal, by definition, to the index of $B$ in $A$, $[A:B]$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As to whether the $x_i$ are &quot;coset representatives for $H$&quot;, remember that for a collection of elements to be &quot;a set of coset representatives&quot; we require that if $x_iH = x_jH$, then $i=j$, and that for every $g\in G$ there exists $i$ such that $gH = x_iH$ (thinking left cosets; right cosets work the same way). So $x_1,\ldots,x_n$ will not generally be &quot;a coset representative&quot; nor a set of coset representatives, but that is immaterial. Simply show that if $\{x_i\}_{i\in I}$ is a set of coset representatives for $K$ in $G$ (meaning, (i) for every $g\in G$ there exists $i\in I$ such that $gK = x_iK$; and (ii) for all $i,i&amp;#39;\in I$, if $x_iK = x_{i&amp;#39;}K$ then $i=i'$), and $\{y_j\}_{j\in J}$ is a set of coset representatives for $H$ in $K$ (meaning, (1) for every $k\in K$ there exists $j\in J$ such that $kH = y_jH$; and (2) for all $j,j&amp;#39;\in J$, if $y_jH = y_{j&amp;#39;}H$, then $j=j'$); then it follows that $\{x_iy_j\}_{(i,j)\in I\times J}$ is a set of coset representatives for $H$ in $G$ (meaning, you need to prove that (I) for all $g\in G$ there exists $(i,j)\in I\times J$ such that $gH = (x_iy_j)H$; and (II) for all $(i,j),(i&amp;#39;,j&amp;#39;)\in I\times J$, if $(x_iy_j)H = (x_{i&amp;#39;}y_{j&amp;#39;})H$, then $(i,j)=(i',j')$). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A fact that will no doubt be useful is to remember that for any group $A$ and any subgroup $B$ of $A$, $cB = dB$ if and only if $cB\cap dB\neq\emptyset$.  &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-07T05:36:58.663" LastActivityDate="2010-12-07T05:36:58.663" />
  <row Id="7008" PostTypeId="2" ParentId="6998" CreationDate="2010-10-17T07:37:07.680" Score="3" ViewCount="" Body="&lt;p&gt;It is enough to show that every maximal proper subgroup $H$ of $G$ is cyclic, by induction;&#xA;to start the induction, notice that if $G$ is simple, then there is nothing to prove.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So let $H\subset G$ be a maximal proper subgroup, so that $G/H$ is simple. There exists then a prime $p$ such that $G/H$ is isomorphic to $\mathbb Z/p\mathbb Z$ and then $pG\subset H$. From the exact sequence of abelian groups $$0\to H\to G\to G/H\to 0$$ we get the exact sequence $$0\to H/pG\to G/pG\to G/H\to 0.$$ Now this is an exact sequence of $\mathbb F_p$-vector spaces in an obvious way, and $G/H$ and $G/pG$ are vector spaces of dimension $1$ (for the first, this is obvious; for the second, it is enough to notice that $G/pG$ is cyclic as an abelian group and non-zero because it surjects onto $G/H$) It follows that $H/pG=0$, &lt;em&gt;i.e.&lt;/em&gt;, that $H=pG$. Since $G$ is cyclic, then $H$ is cyclic too.&lt;/p&gt;&#xA;" OwnerUserId="274" LastEditorUserId="274" LastEditorDisplayName="" LastEditDate="2010-10-17T14:59:25.293" LastActivityDate="2010-10-17T14:59:25.293" CommentCount="4" />
  <row Id="7009" PostTypeId="2" ParentId="6998" CreationDate="2010-10-17T07:42:23.057" Score="4" ViewCount="" Body="&lt;p&gt;If you accept Lagrange's theorem and&#xA;the existence of quotient groups, there's another&#xA;argument in the finite case.&#xA;Let $G=\langle a\rangle$ be a cyclic group of order $n$.&#xA;By Lagrange's theorem if $H$ is a subgroup of $G$ then $m=|H|$&#xA;is a factor of $n$, and is $m\mid n$ then $H_m=\langle a^{n/m}\rangle$&#xA;is a subgroup of order $m$. I claim $G$ has no other order $m$ subgroup.&#xA;The group $G/H$ has order $n/m$ so $x^{n/m}=e$ for all $x\in G/H$.&#xA;This translates to $x^{n/m}\in H$ for all $x\in G$ which means $H_m\le H$.&#xA;These have the same order so $H=H_m$.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-10-17T07:42:23.057" CommentCount="3" />
  <row Id="7010" PostTypeId="2" ParentId="6723" CreationDate="2010-10-17T08:05:11.033" Score="0" ViewCount="" Body="&lt;p&gt;Making Ross's answer more explicit: remember that in the vicinity of the point where the light ray from the origin strikes your mirror, you can replace the curved mirror with the corresponding tangent line. Try deriving a differential equation for your $f(x)$ (hint: you may have to use the formula for the tangent of the difference of two angles) such that the reflected ray has zero slope (i.e., is horizontal).&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-10-17T08:05:11.033" />
  <row Id="7011" PostTypeId="1" AcceptedAnswerId="7037" CreationDate="2010-10-17T09:59:57.057" Score="5" ViewCount="196" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm in the process of writing my Master's Thesis on automata theory. The writing must be in English which isn't my mother tongue. &#xA;  So the question is, given that this is my first time long (hundred pages) writing and since English is a second language for me, what is the book you recommend for academic writing style? Ideally the book:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;must address things like how to structure the writing and/or align it with the research &lt;/li&gt;&#xA;&lt;li&gt;address the technical nature of the writing&lt;/li&gt;&#xA;&lt;li&gt;recommend general styling and language rules&lt;/li&gt;&#xA;&lt;li&gt;and as a plus, some hints to those of us for whom English is a second language&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Best regards.&lt;/p&gt;&#xA;" OwnerUserId="982" LastActivityDate="2010-10-17T16:36:44.010" Title="Book about technical and academical writting" Tags="&lt;books&gt;" AnswerCount="6" CommentCount="1" FavoriteCount="2" />
  <row Id="7012" PostTypeId="1" AcceptedAnswerId="7015" CreationDate="2010-10-17T10:08:47.150" Score="1" ViewCount="206" Body="&lt;p&gt;very sorry for the noobish post, I'm studying math on my own and lot of times im very confused about little things.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is my question, the original problem starts out like this:  Solve the inequality $\vert x -2 \vert \geq \vert 2x -3\vert $ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is the solved problem from my book &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$\vert x -2 \vert \geq \vert 2x -3\vert $  &lt;/li&gt;&#xA;&lt;li&gt;$ (x-2)^2 \geq (2x-3)^2 $  &lt;/li&gt;&#xA;&lt;li&gt;$ 3x^2 - 8x + 5 \leq 0 $&lt;/li&gt;&#xA;&lt;li&gt;$ (x-1)(3x-5) \leq 0$&lt;/li&gt;&#xA;&lt;li&gt;$1 \leq x \leq 5/3$.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;How does: $(x-1)(3x-5) \leq 0$. (Step 4)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Become solved as : $1 \leq x \leq 5/3$. (Step 5)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it were separate wouldn't it be:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;x \leq 1 \qquad  \text{and} \qquad   x \leq 5/3&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So how does &lt;strong&gt;$x\leq 1$&lt;/strong&gt; become &lt;strong&gt;$1\leq  x$&lt;/strong&gt;??&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks so much!! =)&lt;/p&gt;&#xA;" OwnerUserId="2576" LastEditorUserId="940" LastEditorDisplayName="" LastEditDate="2010-10-17T15:26:20.360" LastActivityDate="2010-10-17T16:17:19.180" Title="simple inequality problem with modulus" Tags="&lt;algebra&gt;&lt;inequality&gt;" AnswerCount="3" CommentCount="5" />
  <row Id="7013" PostTypeId="1" AcceptedAnswerId="7135" CreationDate="2010-10-17T10:08:50.980" Score="3" ViewCount="322" Body="&lt;p&gt;How does one evaluate the integral $$\int\limits_{0}^{\infty} \frac{x^{a-1} - x^{b-1}}{1-x} \ dx \quad \text{for} \ a,b \in (0,1)$$&lt;/p&gt;&#xA;" LastEditorUserId="1301" LastEditorDisplayName="" LastEditDate="2011-02-22T14:43:10.340" LastActivityDate="2011-02-22T14:43:10.340" Title="Evaluating the improper integral $\int\limits_{0}^{\infty} \frac{x^{a-1} - x^{b-1}}{1-x} \ dx $" Tags="&lt;calculus&gt;&lt;analysis&gt;" AnswerCount="4" FavoriteCount="1" />
  <row Id="7014" PostTypeId="2" ParentId="7012" CreationDate="2010-10-17T10:19:17.273" Score="4" ViewCount="" Body="&lt;p&gt;Hint: In order for a product to be negative, one of the factors must be positive and the other one negative.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;UPDATE: It's often helpful to draw a table such as this:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;         x  |    1    5/3   &#xA;------------+---------------&#xA;       x-1  | -  0  +  +  +&#xA;      3x-5  | -  -  -  0  +&#xA;------------+---------------&#xA;(x-1)(3x-5) | +  0  -  0  + &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;On each line you keep track of the signs of the individual factors in the intervals of interest. Then you construct the bottom line by using &quot;minus times minus equals plus&quot;, &quot;zero times anything equals zero&quot;, etc.&#xA;In the bottom line you can then read off that $(x-1)(3x-5) \le 0$ is true exactly when $-1 \le x \le 5/3$.&lt;/p&gt;&#xA;" OwnerUserId="1242" LastEditorUserId="1242" LastEditorDisplayName="" LastEditDate="2010-10-17T12:01:57.230" LastActivityDate="2010-10-17T12:01:57.230" />
  <row Id="7015" PostTypeId="2" ParentId="7012" CreationDate="2010-10-17T10:19:35.953" Score="1" ViewCount="" Body="&lt;p&gt;So you want $(x-1) (3x-5) \leq 0$. So either one of the terms has to be positive and the other has to be negative. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;So if $x \geq 5/3$, then both the terms are positive and the product is positive.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Similarly if $x \leq 1$, then both the terms are negative and their product is positive. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So observe that you don't want either of these things to happen, that is $x \notin (- \infty,1) \cup (5/3,\infty)$&lt;/p&gt;&#xA;" LastActivityDate="2010-10-17T10:19:35.953" CommentCount="4" />
  <row Id="7016" PostTypeId="2" ParentId="7013" CreationDate="2010-10-17T11:45:35.917" Score="2" ViewCount="" Body="&lt;p&gt;I think it might be like the proof of $B(x,1-x)=\pi\csc(\pi x)$.  Let $x=\exp(y)$ then evaluate the integral $\int_{-\infty}^\infty \frac{\exp(ay)-\exp(by)}{\exp(y)-1}=\pi(\cot(a\pi)-\cot(b\pi))$ using contour integration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll leave the details as an exercise (that I can't be bothered completing!)&lt;/p&gt;&#xA;" OwnerUserId="954" LastActivityDate="2010-10-17T11:45:35.917" CommentCount="0" />
  <row Id="7017" PostTypeId="2" ParentId="7004" CreationDate="2010-10-17T11:51:47.543" Score="0" ViewCount="" Body="&lt;p&gt;It's an well known function, continuous at all irrationals and discontinuous at all rationals. I will give sketch of the proof below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First let's see why it's continuous at irrationals. To be continuous at an irrational point $x_0$, we need to show that $\forall\delta&gt;0\exists\epsilon&gt;0:|x-x_0|&amp;lt;\epsilon\Rightarrow |f(x)-f(x_0)|&amp;lt;\delta$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice that within any range of length 1, in particular within $[x_0-1/2,x_0+1/2)$, $f(x)$ takes the value 1/2 exactly once, 1/3 twice, 1/4 thrice, and so on, taking 1/k exactly k+1 times. So if we remove these points upto say $\lceil1/\delta\rceil$, we will be removing only finitely many points, and we will be left with points on which $f(x)&amp;lt;1/\lceil1/\delta\rceil\le\delta$. Notice that $x_0$ being irrational was not one of the points that was removed. So find the closest point $a$ to $x_0$ that was removed, and call the distance $|x_0-a|$ as $\epsilon$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the same way, one can argue that the function is discontinuous for rationals. If $x_1$ is a rational point, notice that if we remove all the points of the form $p/q$ upto $0\lt q\le N$, any interval around $x_1$ will contain all points with $f(x)\lt 1/N$. So we can form intervals around $x_1$ with arbitrarily small $f$. Hence $f$ is discontinuous at $x_1$ since $f(x_1)&gt;0$.&lt;/p&gt;&#xA;" OwnerUserId="1310" LastActivityDate="2010-10-17T11:51:47.543" />
  <row Id="7018" PostTypeId="2" ParentId="7011" CreationDate="2010-10-17T11:52:02.687" Score="1" ViewCount="" Body="&lt;p&gt;I found Krantz's &lt;em&gt;A Primer of Mathematical Writing&lt;/em&gt; helpful, though not deep.&lt;/p&gt;&#xA;" OwnerUserId="1195" LastEditorUserId="1195" LastEditorDisplayName="" LastEditDate="2010-10-17T11:57:55.433" LastActivityDate="2010-10-17T11:57:55.433" CommunityOwnedDate="2010-10-17T11:52:02.687" />
  <row Id="7019" PostTypeId="2" ParentId="7011" CreationDate="2010-10-17T11:54:45.907" Score="0" ViewCount="" Body="&lt;p&gt;&lt;em&gt;The Chicago Manual of Style&lt;/em&gt; is one of the standard style guides for English and has lot of useful advice for academic writing. &lt;/p&gt;&#xA;" OwnerUserId="1195" LastActivityDate="2010-10-17T11:54:45.907" CommunityOwnedDate="2010-10-17T11:54:45.907" />
  <row Id="7020" PostTypeId="2" ParentId="7011" CreationDate="2010-10-17T11:57:20.970" Score="3" ViewCount="" Body="&lt;p&gt;A set of lecture notes from a course on mathematical writing run by Don Knuth&#xA;&lt;a href=&quot;http://tex.loria.fr/typographie/mathwriting.pdf&quot; rel=&quot;nofollow&quot;&gt;http://tex.loria.fr/typographie/mathwriting.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1195" LastEditorUserId="1195" LastEditorDisplayName="" LastEditDate="2010-10-17T12:05:31.033" LastActivityDate="2010-10-17T12:05:31.033" CommunityOwnedDate="2010-10-17T11:57:20.970" />
  <row Id="7021" PostTypeId="2" ParentId="7013" CreationDate="2010-10-17T12:20:13.063" Score="1" ViewCount="" Body="&lt;p&gt;To finish off Simon's answer, see Example 4.3.3 (p. 244–245) in &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0521534291&quot; rel=&quot;nofollow&quot;&gt;Complex Variables: Introduction and Applications&lt;/a&gt; by Ablowitz &amp;amp; Fokas.&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2010-10-17T12:20:13.063" />
  <row Id="7022" PostTypeId="1" AcceptedAnswerId="7023" CreationDate="2010-10-17T12:46:11.870" Score="2" ViewCount="156" Body="&lt;p&gt;Young's inequality for convolutions states that if $1 \leq p, q, r \leq \infty$ satisfy&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{1}{q} + 1 = \frac{1}{p} + \frac{1}{r}$$&#xA;for all $f \in L^p(G)$ and all $g \in L^r(G)$ where $g$ and $g'$ have the same $L^r$-norm  and $g'(x) = g(x^{-1})$ ($G$ is a topological group) we have that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\|f * g\|_q \leq \|g\|_r \|f\|_p.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now Grafakos claims we can use this to prove the following inequality due to Hardy:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left ( \int_0^\infty \left ( \frac{1}{x} \int_0^x |f(t)| \, dt \right )^p \, dx \right  )^{1/p} \leq \frac{p}{p - 1} \|f\|_{L^p(0, \infty)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The hint is to consider on the multiplicative group $(\mathbb{R}^+, \frac{dt}{t})$ the convolution of $|f(x)| x^{1/p}$ and ${x^{-1/p'}} 1_{[1, \infty)}$. So if we use this, the RHS is no problem, it is just a direct computation (I can add it if someone wants it for future reference). However, if I compute the convolution I get:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_0^{x - 1} |f(t)| (y(t - y))^{1/p'} \, dt$$ but I don't see how this is larger (or equal) to the inner integral on the LHS of the inequality. Any suggestions?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: As Willie Wong points out below, the convolution is wrong. It is an multiplicative group, not an additive one.&lt;/p&gt;&#xA;" OwnerUserId="1120" LastEditorUserId="1120" LastEditorDisplayName="" LastEditDate="2010-10-17T13:52:17.007" LastActivityDate="2010-10-18T11:13:01.237" Title="An inequality by Hardy." Tags="&lt;fourier-analysis&gt;" AnswerCount="1" CommentCount="6" />
  <row Id="7023" PostTypeId="2" ParentId="7022" CreationDate="2010-10-17T13:45:55.163" Score="4" ViewCount="" Body="&lt;p&gt;You are doing the convolution wrong. On the multiplicative group $(\mathbb{R}_+, dt/t)$, the convolution is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f * g(x) = \int_0^\infty f(y) g(x/y) dy/y$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(for harmonic analysis on an abelian group, you need to re-interpret the $+$ and $-$ signs in formulae to be the group binary operator). If you plug in, as $g$, the weight function Grafakos suggested, you should get exactly the LHS. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just to be more general: let $(G,\mu)$ be an Abelian group with an invariant measure $\mu$, where $\cdot$ denotes the group binary operator, then the convolution of two functions $f,g: G\to \mathbb{R}$ is defined as the function $G\to \mathbb{R}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ f*g(y) = \int_G f(x) g(y\cdot x^{-1})\mu(dx) $$&lt;/p&gt;&#xA;" OwnerUserId="1543" LastEditorUserId="1543" LastEditorDisplayName="" LastEditDate="2010-10-18T11:13:01.237" LastActivityDate="2010-10-18T11:13:01.237" CommentCount="10" />
  <row Id="7024" PostTypeId="1" CreationDate="2010-10-17T13:49:03.350" Score="1" ViewCount="42" Body="&lt;p&gt;Let $M$ be the mean matrix of a multitype branching process.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am trying to figure out whether $A = (I-M)^{-1}$ has any signifiance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think $A_{ij}$ would stand for the number of times that type $j$ would be generated from type $j$, but I can't find any material on that in the web. Anyone knows anything about that?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="7076" LastActivityDate="2010-11-16T16:44:18.893" Title="what is the connection between the mean matrix and properties of a branching process?" Tags="&lt;probability&gt;&lt;statistics&gt;" AnswerCount="1" />
  <row Id="7025" PostTypeId="1" AcceptedAnswerId="7060" CreationDate="2010-10-17T13:51:21.977" Score="3" ViewCount="183" Body="&lt;p&gt;We have two functions:&#xA;$F(x)$ and $G(x)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose the improper integral has the value 1.&#xA;$$\int_{1}^{\infty} F(x) G(x) \mathrm{d}x = 1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can we find the value of the integral, where c is a constant?&#xA;$$\int_{1}^{\infty} F(x+c) G(x) \mathrm{d}x$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am interested in generalized answers and not specific examples.&#xA;Or, if any links/books' info are given where I can read similar problems, that would be highly appreciated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Due to the highly generalized nature of this question, I am giving the examples I am working upon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$F(x) = frac(x)$$&#xA;Where frac(x) denotes the fractional part of x, and&#xA;$$G(x) = \frac{sin(xlog(x))}{x^{h+1}}$$&#xA;where constants are $0&amp;lt; h &amp;lt; 1$ and $c = 1/2$&lt;/p&gt;&#xA;" OwnerUserId="2320" LastEditorUserId="2320" LastEditorDisplayName="" LastEditDate="2010-10-17T14:29:50.863" LastActivityDate="2010-10-18T02:55:15.560" Title="A question on integrals" Tags="&lt;improper-integrals&gt;" AnswerCount="5" />
  <row Id="7027" PostTypeId="2" ParentId="7025" CreationDate="2010-10-17T14:10:52.173" Score="1" ViewCount="" Body="&lt;p&gt;You asked for references for problems similar to yours.  One such is the &lt;em&gt;convolution&lt;/em&gt; of two functions $f$ and $g$, one form of which is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(f*g)(t) = \int_{-\infty}^{\infty} f(\tau) g(t - \tau) d \tau.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Convolutions have lots of interesting properties and interpretations.  See MathWorld's article on &lt;a href=&quot;http://mathworld.wolfram.com/Convolution.html&quot; rel=&quot;nofollow&quot;&gt;convolutions&lt;/a&gt; for more information.&lt;/p&gt;&#xA;" OwnerUserId="2370" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-10-18T02:55:15.560" LastActivityDate="2010-10-18T02:55:15.560" CommentCount="1" />
  <row Id="7028" PostTypeId="2" ParentId="7025" CreationDate="2010-10-17T14:13:32.807" Score="0" ViewCount="" Body="&lt;p&gt;If $c$ is negative, the second integral involves values $F(t)$ for $t&amp;lt;1$, which can be changed arbitrarily without affecting the first integral! But even if you restrict yourself to positive $c$, I guess there's nothing you can say in general without additional information.&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2010-10-17T14:13:32.807" />
  <row Id="7029" PostTypeId="2" ParentId="7025" CreationDate="2010-10-17T14:21:31.963" Score="3" ViewCount="" Body="&lt;p&gt;Do you know anything about $F$ and $G$? If not, then certainly no. Let $G(x) = 1$ if $x \leq 2$ and $0$ otherwise. Let $F(x) = 1$ if $x&amp;lt; 2$; we leave $F(x)$ undefined elsewhere for now. Then $\int_1^{\infty} F(x)G(x) = 1$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the function $h(c) = \int_1^\infty F(x+c)G(x)dx$ defined on $0 &amp;lt; c$. I claim that for any continuously differentiable function $H(c)$ such that $H(0) = 1$, I can find an extension to $F(x)$ (i.e. filling in the undefined spots) such that $h(c) = H(c)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proof: Notice that by definition $h(c) = \int_{1+c}^{c+2} F(x) dx$. Then $h'(c) = F(c+2) - F(c+1)$.  So start with $F(x) = 1$ for $1\leq x &amp;lt; 2$, then define $F(x) = H'(x-2) + H'(x - 3) + \ldots + H'(x - k) + F(x - k + 1)$, where $k$ is the integer where $1 \leq x - k + 1 &amp;lt; 2$. The by construction, $h'(c) = H'(c)$ and $h(0) = H(0)$, so the two functions agree. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;From the above we can conclude that just knowing $\int_1^\infty F(x) G(x) dx = 1$ tells us absolutely nothing about $\int_1^\infty F(x+c) G(x) dx$; it can be pretty much anything you want it to be. &lt;/p&gt;&#xA;" OwnerUserId="1543" LastActivityDate="2010-10-17T14:21:31.963" CommentCount="1" />
  <row Id="7030" PostTypeId="2" ParentId="6625" CreationDate="2010-10-17T14:29:54.280" Score="3" ViewCount="" Body="&lt;p&gt;The article you should be looking at, apart from &lt;a href=&quot;http://dx.doi.org/10.1080/00207167308803075&quot; rel=&quot;nofollow&quot;&gt;David Levin's original paper&lt;/a&gt; is E.J. Weniger's &lt;a href=&quot;http://arxiv.org/abs/math/0306302&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;Nonlinear sequence transformations for the acceleration of convergence and the summation of divergent series&lt;/em&gt;&lt;/a&gt;; in there he gives a short FORTRAN routine for implementing Levin's transformations. A more elaborate &lt;a href=&quot;http://doi.acm.org/10.1145/356044.356051&quot; rel=&quot;nofollow&quot;&gt;implementation by Fessler, Ford, and Smith&lt;/a&gt; is &lt;a href=&quot;http://www.netlib.org/toms/602&quot; rel=&quot;nofollow&quot;&gt;available at Netlib&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just to show how easy it is to implement the Levin transformation, here's a demonstration program I wrote for the &lt;a href=&quot;http://education.ti.com/educationportal/sites/US/productDetail/us_ti83p.html&quot; rel=&quot;nofollow&quot;&gt;TI-83 Plus calculator&lt;/a&gt; for summing the alternating harmonic series, based on Weniger's short FORTRAN routine (comments are delimited by a backslash):&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;PROGRAM:XTRPOL&#xA;Prompt N \\ number of terms of the series to use&#xA;1→P:0→S \\ P: alternating sign S: stores partial sums&#xA;For(K,1,N)&#xA;P/K→U \\ K-th term of the series, change to sum a different series&#xA;S+U→S&#xA;1→B \\ adjustable parameter for the Levin transformation&#xA;(B+K-1)U→T \\ required for the Levin u-transform; for t version, remove the (B+K-1)&#xA;prgmLEVINT&#xA;Pause Y&#xA;-P→P&#xA;End&#xA;Y&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;which uses the subroutine&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;PROGRAM:LEVINT&#xA;(B+K-1)ֿ¹→W&#xA;W/T→U&#xA;If K=1&#xA;ClrList ∟DL,∟NL&#xA;U→∟DL(K)&#xA;SU→∟NL(K)&#xA;1-W→V&#xA;For(J,K-1,1,-1)&#xA;(B+J-1)W→U&#xA;∟NL(J+1)-U∟NL(J)→∟NL(J)&#xA;∟DL(J+1)-U∟DL(J)→∟DL(J)&#xA;WV→W&#xA;End&#xA;10^(99)→Y&#xA;If abs(∟DL(1))≥10^-99&#xA;∟NL(1)/∟DL(1)→Y&#xA;Y&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I haven't bothered to implement the stopping rules described by Fessler, Ford, and Smith in that demo program, but it's doable. Translating that short routine to your language of choice should be straightforward.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-10-17T14:29:54.280" CommentCount="5" />
  <row Id="7031" PostTypeId="2" ParentId="7025" CreationDate="2010-10-17T14:36:56.637" Score="1" ViewCount="" Body="&lt;p&gt;In reasonably general situations (using, e.g, distributions if needed) the second integral can be differentiated with respect to $c$ and the derivatives will be $\int F^{(n)}(x)G(x) dx$.  Just as a function is not determined by its value at a point, these derivatives are not determined by the value at $c=0$, so even the local behavior of perturbations of the original integral (the asymptotics of the second integral for small $c&gt;0$) is not determined by the integral with $c=0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's hard to think of examples where the second integral is determined, no matter how constrained $F$ and $G$ are.  Consider the situation where $F(x)$ and $1/G(x)$ are polynomials of degrees 2 and 4 with positive coefficients.  This is a very well behaved problem but the variation with respect to $c$ is undetermined by a single piece of information.  If you require integer coefficients or a similar discretization then there is some hope of recovering $F$ and $G$ from the value of the first integral, but this is a Diophantine problem and no longer a question of real analysis.&lt;/p&gt;&#xA;" OwnerUserId="467" LastActivityDate="2010-10-17T14:36:56.637" />
  <row Id="7032" PostTypeId="2" ParentId="7011" CreationDate="2010-10-17T14:49:23.220" Score="1" ViewCount="" Body="&lt;p&gt;Here is another: &lt;a href=&quot;http://books.google.com/books?id=9gQd2fJA7Y4C&amp;amp;printsec=frontcover&amp;amp;dq=Nicholas+J.+Higham+-+Handbook+of+Writing+for+the+Mathematical+Sciences&amp;amp;hl=pt-PT&amp;amp;ei=IQy7TJXNJM7rsgaq-PHIDQ&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=1&amp;amp;ved=0CCcQ6AEwAA#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;Handbook of Writing for the Mathematical Sciences - Nicholas J. Higham&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1747" LastActivityDate="2010-10-17T14:49:23.220" />
  <row Id="7033" PostTypeId="2" ParentId="7002" CreationDate="2010-10-17T15:09:07.667" Score="2" ViewCount="" Body="&lt;p&gt;The canonical map $G/H \to G/K$ is surjective. The fiber of $gK$ is $\{gkH : k \in K\}$, which can be identified with $K/H$.&lt;/p&gt;&#xA;" OwnerUserId="1650" LastActivityDate="2010-10-17T15:09:07.667" />
  <row Id="7034" PostTypeId="1" AcceptedAnswerId="7045" CreationDate="2010-10-17T15:31:15.787" Score="11" ViewCount="298" Body="&lt;p&gt;Prove that the sequence $c_{1} = 1$, $c_{(n+1)}= 4/(1 + 5c_{n})$     , $n \geq 1$  is convergent and find its limit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ok so up to now I've worked out a couple of things.&lt;br&gt;&#xA;$c_1 = 1$&lt;br&gt;&#xA;$c_2 = 2/3$&lt;br&gt;&#xA;$c_3 = 12/13$&lt;br&gt;&#xA;$c_4 = 52/73$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the odd $c_n$ are decreasing and the even $c_n$ are increasing.  Intuitively, it's clear the the two sequences for odd and even $c_n$ are decreasing/increasing less and less.&#xA;Therefore it seems like the sequence may converge to some limit $L$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the sequence has a limit, let $L=\underset{n\rightarrow \infty }{\lim }a_{n}.$   Then $L = 1/(1+5L).$&#xA;So we yield $L = 4/5$ and $L = -1$.  But since the even sequence is increasing and &gt;0, then $L$ must be $4/5$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ok, here I am stuck.  I'm not sure how to go ahead and show that the sequence converges to this limit (I tried using the definition of the limit but I didn't manage) and and not sure about the separate sequences how I would go about showing their limits.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A few notes :&#xA;I am in 2nd year calculus.&#xA;This is a bonus question, but I enjoy the challenge and would love the extra marks.&#xA;Note : Once again I apologize I don't know how to use the HTML code to make it nice.&lt;/p&gt;&#xA;" OwnerUserId="2533" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2010-10-17T16:32:33.857" LastActivityDate="2010-10-17T23:00:44.417" Title="Prove that the sequence$ c_1 = 1$, $c_{n+1} = 4/(1 + 5c_n) $ , $ n \geq 1$ is convergent and find its limit." Tags="&lt;calculus&gt;&lt;homework&gt;" AnswerCount="5" CommentCount="8" FavoriteCount="2" />
  <row Id="7035" PostTypeId="2" ParentId="7024" CreationDate="2010-10-17T15:56:10.150" Score="1" ViewCount="" Body="&lt;p&gt;The matrix $M$ has $(i,j)$th entry $M_{ij}=E(Z^j_1 | Z_0=e_i)$, that is, $M_{ij}$ is the mean number of type $j$ offspring starting with one type $i$ individual. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, for the $n$th power of $M$ we get $M_{ij}^n=E(Z^j_n | Z_0=e_i)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Expanding $A=(I-M)^{-1}=\sum_{n\geq 0} M^n$, we get (ignoring issues of convergence), that  $A_{ij}=E(\sum_{n\geq 0} Z^j_n | Z_0=e_i)$,  the mean number of type $j$ descendants starting with a single type $i$ individual. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A good reference is Harris's book &lt;em&gt;The Theory of Branching Processes&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="940" LastActivityDate="2010-10-17T15:56:10.150" />
  <row Id="7036" PostTypeId="2" ParentId="7013" CreationDate="2010-10-17T15:57:02.653" Score="1" ViewCount="" Body="&lt;p&gt;This looks reminiscent of &lt;em&gt;Frullani's Integrals&lt;/em&gt;; see for example the &lt;a href=&quot;http://arxiv.org/abs/1005.2940&quot; rel=&quot;nofollow&quot;&gt;article:&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1709" LastActivityDate="2010-10-17T15:57:02.653" CommentCount="3" />
  <row Id="7037" PostTypeId="2" ParentId="7011" CreationDate="2010-10-17T16:02:06.463" Score="2" ViewCount="" Body="&lt;p&gt;Nicholas Higham's book &lt;em&gt;Handbook of writing for the mathematical sciences&lt;/em&gt; is detailed, clear, and overall very useful. He covers all the four bullet points that you have raised. Moreover, he gives an annotated reference to useful style guides, technical writing guides, dictionaries, and so on. I am sure you won't be disappointed!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; I guess you might already be familiar with the classic: &lt;em&gt;The Elements of Style&lt;/em&gt; by Strunk &amp;amp; White --- this book is a gem on writing, and is worth reading again and again every few months.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, here is a paper with very practical advice. This paper really underscores the importance of structure, and offers concrete advice on how to attain good structure: &lt;em&gt;The Science of Scientific Writing&lt;/em&gt;, by G. D. Gopen and J. A. Swan. The&#xA;fundamental thesis of the paper is: Writing to meet the reader’s expectation.&lt;/p&gt;&#xA;" OwnerUserId="1709" LastEditorUserId="1709" LastEditorDisplayName="" LastEditDate="2010-10-17T16:36:44.010" LastActivityDate="2010-10-17T16:36:44.010" CommentCount="1" />
  <row Id="7038" PostTypeId="2" ParentId="7011" CreationDate="2010-10-17T16:05:41.107" Score="2" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://retro.seals.ch/digbib/view?rid=ensmat-001:1970:16%3a:59&amp;amp;id=browse&amp;amp;id2=browse2&quot; rel=&quot;nofollow&quot;&gt;How to write mathematics&lt;/a&gt; is a free essay by Paul Halmos which fulfills your first three criteria. It is short, but has several good points.&lt;/p&gt;&#xA;" OwnerUserId="1823" LastActivityDate="2010-10-17T16:05:41.107" CommentCount="2" />
  <row Id="7039" PostTypeId="2" ParentId="6998" CreationDate="2010-10-17T16:09:41.617" Score="1" ViewCount="" Body="&lt;p&gt;A simple way to present this and related results is to employ concrete fractional models of cyclic groups, i.e. $\rm\ C_n\: =\: \frac{1}n\mathbb Z\ \:(mod\ \mathbb Z)\ =\ \{ 0,\: 1/n,\: 2/n,\: \ldots,\: (n-1)/n\}\:.\ $ Suppose $\rm G$ is a nontrivial subgroup. Since $\rm G\:$ is closed under subtraction, every element of $\rm G\:$ must be an integer multiple of its least element $\rm\:g\:$ (else if $\rm\: h\:$ were the least nonmultiple, then $\rm\: h - g\:$ would be a smaller one). Thus $\rm\:G = \langle g \rangle\:$ is cyclic. $\ $ &lt;strong&gt;QED&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-10-17T22:34:22.857" LastActivityDate="2010-10-17T22:34:22.857" />
  <row Id="7040" PostTypeId="1" AcceptedAnswerId="7048" CreationDate="2010-10-17T16:15:03.007" Score="4" ViewCount="163" Body="&lt;p&gt;Why does $ \text{Var}(Y) = E(\text{Var}(Y|X))+ \text{Var}(E(Y|X))$? What is the intuitive explanation for this? In laymen's terms it seems to say that the variance of $Y$ equals the expected value of the conditional variance plus the variance of the conditional expectation. &lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2010-10-24T09:45:05.033" Title="Conditional and Total Variance" Tags="&lt;probability&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="7041" PostTypeId="1" CreationDate="2010-10-17T16:15:13.983" Score="6" ViewCount="216" Body="&lt;p&gt;Hi, I'm a math PhD who has been asked to interview a high school student and determine what he/she is interested in and how strong the student is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Usually I would want them to talk as much as possible about what classes they've taken and what parts of the classes they like the best. Generally I'd like to allow them to get comfortable talking about what they like rather than what they think I'd like. However, that might not work this time as the student is a non-native english speaker and may get less comfortable as I try to get him/her to talk more.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus far I'm thinking about watching how he/she solves different problems in different areas of math, but I would like to know if anyone has done this before or has any suggestions for evaluating a student who might not be comfortable talking much.&lt;/p&gt;&#xA;" OwnerUserId="2580" LastEditorUserId="2580" LastEditorDisplayName="" LastEditDate="2010-10-17T17:35:42.207" LastActivityDate="2010-10-17T20:21:17.217" Title="How to assess a non-natively english speaking high-schooler's mathematical ability?" Tags="&lt;soft-question&gt;&lt;education&gt;" AnswerCount="2" CommentCount="7" FavoriteCount="1" />
  <row Id="7042" PostTypeId="1" CreationDate="2010-10-17T16:17:05.293" Score="0" ViewCount="53" Body="&lt;p&gt;Brownian motion is a self-similar process with a Hurst index $H = \frac{1}{2}$. Is it a coincidence that the probability of a positive (forward) or negative (backward) step of the random walk that generates Brownian motion is also $\frac{1}{2}$.&lt;/p&gt;&#xA;" OwnerUserId="1752" LastActivityDate="2010-10-17T16:17:05.293" Title="Is there a connection between the Hurst index and the probabiliy of a random walk that produces Brownian motion?" Tags="&lt;probability&gt;&lt;statistics&gt;" CommentCount="1" />
  <row Id="7043" PostTypeId="2" ParentId="7012" CreationDate="2010-10-17T16:17:19.180" Score="0" ViewCount="" Body="&lt;p&gt;In the end here when you solve (x-1)(3x-5) &amp;lt;= 0   You are really finding the bounds of what x can be.  In this case, the smallest number is 1, and the greatest number is 5/3.  Therefore x has to be between 1 and 5/3.&#xA;Which is where 1 &amp;lt;= x &amp;lt;= 5/3 comes from.&lt;/p&gt;&#xA;" OwnerUserId="2533" LastActivityDate="2010-10-17T16:17:19.180" CommentCount="1" />
  <row Id="7044" PostTypeId="2" ParentId="6906" CreationDate="2010-10-17T16:32:32.047" Score="2" ViewCount="" Body="&lt;p&gt;The quote in Mariano's answer is from the introduction to &lt;a href=&quot;http://www.jstor.org/stable/108572&quot; rel=&quot;nofollow&quot;&gt;Sylvester's paper.&lt;/a&gt; Typical of Sylvester's mathematical papers, he used so many nonstandard terms in that paper that he appended a five-page &quot;Glossary of new or unusual Terms, or of Terms used in a new or unusual sense in the preceding Memoir&quot;. There he lists:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Inertia. -- The unchangeable number of integers in the excess of positive over negative signs which adheres to a quadratic form expressed as the sum of positive and negative squares, notwithstanding any real linear transformations impressed upon such form.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sylvester did similarly for many mathematical terms, i.e. coined them or used them in a &quot;new or unusual ways&quot; mathematically. You can find many such examples in Jeff Miller's &lt;a href=&quot;http://jeff560.tripod.com/mathword.html&quot; rel=&quot;nofollow&quot;&gt;Earliest Known Uses of Some of the Words of mathematics,&lt;/a&gt; including: allotrious factor, anallagmatic, Bezoutiant, catalecticant, combinant covariant cumulant cyclotomy, cyclotomic, dialytic, discriminant, Hessian, invariant, isomorphic, Jacobian, latent, law of intertia of quadratic forms, matrix, minor, nullity, plagiograph, quintic, Schur complement, sequence, syzygy, totient, tree, umbral calculus, umbral notation, universal algebra, x/y/z-coordinate, zero matrix, zetaic multiplication. Please see each entry for  Sylvester's role - some are major, others are minor. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Apparently Sylvester's penchant for colorfully naming mathematical objects arose from his love of language and poetry. Indeed, &lt;a href=&quot;http://www-groups.dcs.st-and.ac.uk/~history/Biographies/Sylvester.html&quot; rel=&quot;nofollow&quot;&gt;Karen Parshall&lt;/a&gt; wrote:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Sylvester's love of poetry and language manifested itself in notable ways even in his mathematical writings. His mastery of French, German, Italian, and Greek was often reflected in the mathematical neologisms - like &quot;meicatecticizant&quot; and &quot;tamisage&quot; - for which he gained a certain notoriety. Moreover, literary illusions, poetic quotations, and unfettered hyperbole spiced his published papers and lectures.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sylvester wrote about such:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Perhaps I may without immodesty lay claim to the appellation of Mathematical Adam, as I believe that I have given more names (passed into general circulation) of the creatures of mathematical reason than all the other mathematicians of the age combined.-- James Joseph Sylvester, Nature 37 (1888), p. 152.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You can find a short &lt;a href=&quot;http://www-groups.dcs.st-and.ac.uk/~history/Biographies/Sylvester.html&quot; rel=&quot;nofollow&quot;&gt;Sylvester biography here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-10-17T17:53:28.147" LastActivityDate="2010-10-17T17:53:28.147" CommentCount="5" />
  <row Id="7045" PostTypeId="2" ParentId="7034" CreationDate="2010-10-17T16:47:59.260" Score="3" ViewCount="" Body="&lt;p&gt;Well done! Your observations are correct and can be completed to give a proof that the sequence is convergent to $\frac{4}{5}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The tough part in proving your observations it to prove that the odd/even sub-sequences are monotonic and appropriately bounded (what do I mean by this?).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a hint:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Show that $$ c_{2n-1} \ge \frac{4}{5} \ge c_{2n} \ \ \forall n \ge 1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint2:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Try writing $\displaystyle c_{n+2}$ in terms of $\displaystyle c_{n}$ and see if that helps you prove the above bounds (and as a next step, the monotonicity).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(For a simpler proof of the above bound, you can also try showing that if $c_{n} \ge 4/5$ then $c_{n+1} \le 4/5$ and similarly if $c_n \le 4/5$ then $c_{n+1} \ge 4/5$).&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-10-17T18:52:25.757" LastActivityDate="2010-10-17T18:52:25.757" CommentCount="6" />
  <row Id="7046" PostTypeId="2" ParentId="7034" CreationDate="2010-10-17T16:52:04.490" Score="2" ViewCount="" Body="&lt;p&gt;Here is a more explicit version of my hint.  I said it would be easier to show that $d_n = c_n - \frac{4}{5}$ tends to zero as $n \to \infty$.  This gives $d_1 = \frac{1}{5}$ and the recurrence&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$d_{n+1} + \frac{4}{5} = \frac{4}{1 + 5(d_n + \frac{4}{5})} = \frac{4}{5} \frac{1}{1 + d_n}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;hence&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$d_{n+1} = - \frac{4}{5} \frac{d_n}{1 + d_n}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you see what to do from here?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt;  Some more hints.  If the $1 + d_n$ in the denominator were just a $1$, we would be done because then $|d_n|$ would decrease exponentially at least as fast as $\left( \frac{4}{5} \right)^n$.  But it's not.  However, $d_n$ is small, so the denominator should be close enough to $1$ that this argument should still go through.  More precisely, you just need to find a constant $0 &amp;lt; c &amp;lt; \frac{1}{5}$ such that you can prove that $|d_n| \le c$, say for $n \ge 2$.  (Maybe by induction.)  From there it will follow that $|d_n|$ actually decays exponentially at least as fast as $\left( \frac{4}{5(1 - c)} \right)^n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is an example of a general technique called &quot;bootstrapping,&quot; where you use weak estimates together with relations that a sequence satisfies to get stronger estimates.&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-10-17T23:00:44.417" LastActivityDate="2010-10-17T23:00:44.417" CommentCount="9" />
  <row Id="7047" PostTypeId="2" ParentId="7034" CreationDate="2010-10-17T16:59:15.437" Score="0" ViewCount="" Body="&lt;p&gt;Here's one way to prove it: let $f(x) = 4/(1+5x)$. Say $|x-4/5| \le C$ for some constant $C$. Can you find $C$ and some constant $0 \le k &amp;lt; 1$ so that if $|x-4/5| \le C$, then $|f(x)-4/5| \le k|x-4/5|$? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you do this, then you can iterate to get $|f^j(x)-4/5| \le k^j |x-4/5|$, for all $j$, and so if you make $j$ large enough then you can get $f^j(x)$ as close to $4/5$ as you like.&lt;/p&gt;&#xA;" OwnerUserId="173" LastActivityDate="2010-10-17T16:59:15.437" />
  <row Id="7048" PostTypeId="2" ParentId="7040" CreationDate="2010-10-17T17:25:35.100" Score="6" ViewCount="" Body="&lt;p&gt;A rigorous proof is &lt;a href=&quot;http://en.wikipedia.org/wiki/Law_of_total_variance&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;; it relies on the law of total expectation, which says that $E(E(X|Y))=E(X)$.  The intuitive explanation of &lt;em&gt;that&lt;/em&gt; is that $E(X|Y)$ is the expected value of $X$ given a particular value of $Y$, and that $E(E(X|Y))$ is the expected value of that over all values of $Y$.  So $Y$ no longer matters, and we're just looking at $E(X)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The variance law is a bit more difficult to parse, but this is what it says to me.  &quot;How much does $Y$ vary?  We &lt;em&gt;expect&lt;/em&gt; it to vary by the average value of the variances we get by fixing $X$.  But even when we fix $X$, there is some swing in $Y$, and thus swing in $E(Y|X)$.  So we add on the variance of $E(Y|X)$.  The first term is the expected variance &lt;em&gt;from the mean&lt;/em&gt; of $Y|X$; the second is the variance &lt;em&gt;of that mean&lt;/em&gt;.&quot;&lt;/p&gt;&#xA;" OwnerUserId="2215" LastActivityDate="2010-10-17T17:25:35.100" />
  <row Id="7049" PostTypeId="1" CreationDate="2010-10-17T18:28:33.240" Score="2" ViewCount="97" Body="&lt;p&gt;Need to use Fermat's Little Theorem (Let &lt;strong&gt;p&lt;/strong&gt; be a prime number and let &lt;strong&gt;a&lt;/strong&gt; be an integer. Then &lt;strong&gt;a&lt;/strong&gt; ^ &lt;strong&gt;p&lt;/strong&gt; = &lt;strong&gt;a&lt;/strong&gt; mod &lt;strong&gt;p&lt;/strong&gt;. If &lt;strong&gt;p&lt;/strong&gt; does not divide &lt;strong&gt;a&lt;/strong&gt; then &lt;strong&gt;a&lt;/strong&gt; ^ &lt;strong&gt;p&lt;/strong&gt;-1 = 1 mod &lt;strong&gt;p&lt;/strong&gt;.)&#xA;154 is not prime, but 154 = 22*7 and 23 is prime, so a^22 = 1 mod 23.&#xA;a^154 = a^22 * a^7 = 1 mod 23 * 6 mod 23 = 6 mod 23.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Not sure how to proceed further. Would be grateful for any help.&lt;/p&gt;&#xA;" OwnerUserId="2401" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-10-17T18:32:42.517" LastActivityDate="2010-10-17T18:38:26.607" Title="Prove that for any integer a &gt; 1 s.t. gcd(a, 23) = 1, 23 divides a^154 - 1." Tags="&lt;number-theory&gt;&lt;homework&gt;" AnswerCount="2" />
  <row Id="7050" PostTypeId="2" ParentId="7049" CreationDate="2010-10-17T18:32:09.647" Score="3" ViewCount="" Body="&lt;p&gt;You solved it! apart from an arithmetic error.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle a^{mn} = (a^{m})^n$, not $\displaystyle a^{m}* a^{n}$ (which is actually $\displaystyle a^{m+n}$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, not sure how you got $a^7 = 6 \mod 23$.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-10-17T18:32:09.647" CommentCount="6" />
  <row Id="7051" PostTypeId="2" ParentId="7049" CreationDate="2010-10-17T18:38:26.607" Score="2" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\rm\ \ a^{22}\ \equiv 1\ \Rightarrow\ a^{22\: N}\ \equiv\ 1^N\ \equiv\ 1\:.\ $ Now put $\rm\ N = 7\:$.&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-10-17T18:38:26.607" CommentCount="1" />
  <row Id="7052" PostTypeId="2" ParentId="7041" CreationDate="2010-10-17T18:46:26.497" Score="1" ViewCount="" Body="&lt;p&gt;Well, I'm a non-native English speaker, so I can provide some insight. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think that one of the first problems would be names and notation in general. As far as I can see, there is no easy way to solve this unless you already know the customs of the country the student is coming from, but your education will probably make it easier to guess what means what. For example in my country we use $log_{base}number$ for general case logarithms, $lg$ for $log_{10}$ and $ln$ for $log_e$. Another example would be popular sets. In some countries 0 is a natural number and in some other countries it isn't. Another example would be popular trigonometric functions. In my country for example secant and cosecant are considered deprecated and aren't used. Same thing with grads as angle measurement units.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Problems with this part can be especially tricky because they can stay undetected for a long time. I've seen some math problems whose solutions depend on whether zero is a natural number or not. Another example here can be integrals. As far as I can see the most popular name in English for operation opposite of finding a derivative is antiderivation. In my area that's called indefinite integration and the indefinite part if often omitted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Still I think that most of the notational problems can be solved with Wikipedia.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next part would be names of fields of mathematics and popular theorems. How big problems this is going to be is closely related to which language terms come from. Sometimes terms are going to be shared between languages and sometimes there will be no direct translation. For example in my language, there is no term for calculus. Instead we use mathematical analysis, which as far as I know is superset of calculus.&lt;br&gt;&#xA;Also in some cultures some theorems are named after famous mathematicians and in some other cultures they are named after what the theorem describes. I can't find any examples right now, but I know I encountered such problems when asking few questions here. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another problem could be non-existence of a concept. For example I was unable to find any mention of &lt;a href=&quot;http://translate.google.com/translate?js=n&amp;amp;prev=_t&amp;amp;hl=sr&amp;amp;ie=UTF-8&amp;amp;layout=2&amp;amp;eotf=1&amp;amp;sl=auto&amp;amp;tl=en&amp;amp;u=http%3A%2F%2Fru.wikipedia.org%2Fwiki%2F%25D0%25A3%25D1%2580%25D0%25B0%25D0%25B2%25D0%25BD%25D0%25B5%25D0%25BD%25D0%25B8%25D0%25B5_%25D0%259C%25D0%25B5%25D1%2589%25D0%25B5%25D1%2580%25D1%2581%25D0%25BA%25D0%25BE%25D0%25B3%25D0%25BE&quot; rel=&quot;nofollow&quot;&gt;Ivan Vsevolodovich Meshcherskiy's equation&lt;/a&gt; &lt;a href=&quot;http://ru.wikipedia.org/wiki/%D0%A3%D1%80%D0%B0%D0%B2%D0%BD%D0%B5%D0%BD%D0%B8%D0%B5_%D0%9C%D0%B5%D1%89%D0%B5%D1%80%D1%81%D0%BA%D0%BE%D0%B3%D0%BE&quot; rel=&quot;nofollow&quot;&gt;(original article here&lt;/a&gt;) for bodies with variable mass in any English language web-sites. Here it's considered one of the fundamental rocketry equations, right next to Tsiolkovsky's formula. While the example is related more to physics than mathematics, be prepared to find similar examples in mathematics if the student is coming from a different culture.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since you said that student knows some English, here's my recommendation: Try starting a conversation about some topic in English and see if student is familiar with it. If it doesn't ring a bell, be ready to provide simple examples related to the area. I sometimes had problems finding exact name for some field of mathematics but immediately recognized it as soon as I saw a few simple problems. Another good idea would be to make a list of alternative names for every operation and see if some of them are familiar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yet another important thing is to have at least some familiarity with mathematics program which the student followed. Sometimes focus of what's studied in classes can be a bit different than what's written on paper. I remember now back when I was in secondary school we had formal mathematical proofs in books and they were supposed to be a part of the curriculum, but teachers in schools just decided that proofs are not important and did not teach them. Also in some schools less attention was payed to proper use of terminology. I know some of my friends who didn't know what was derivative and it's graphical representations, but knew that $x^2{'}=2x$ and same thing for integrals. Another point can be some obvious areas which may have been skipped. When  was in secondary school, we didn't learn almost any probability and statistics and did very little work with percentages. On the other hand I see that in some other countries it's expected that someone who has graduated from primary school can without any problems calculate his interest rate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can't think of anything else to add at the moment...&lt;/p&gt;&#xA;" OwnerUserId="348" LastEditorUserId="348" LastEditorDisplayName="" LastEditDate="2010-10-17T19:14:41.027" LastActivityDate="2010-10-17T19:14:41.027" />
  <row Id="7053" PostTypeId="2" ParentId="7034" CreationDate="2010-10-17T18:48:47.777" Score="2" ViewCount="" Body="&lt;p&gt;As shown by the other answers, there are a few nice ways to approach this problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could concentrate only on $C_{2n-1},$ say, since if you establish that $C_{2n-1}$ tends to a limit you automatically nail $C_{2n}$ as well, because&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$C_{2n} = \frac{4}{1+5C_{2n-1}} \textrm { so }  \lim C_{2n}  = \lim  \frac{4}{1+5C_{2n-1}}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now it's easy to show $C_{2n-1} &gt; 4/5$ and so expand&#xA;$$(5C_{2n-1}-4)(C_{2n-1}+1)&gt;0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and manipulate (add $20C_{2n-1}$ to both sides and take the 4 to the RHS) to obtain&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ C_{2n-1} &gt; \frac{4+20C_{2n-1}}{21+5C_{2n-1}} = C_{2n+1},$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and since $C_{2n-1}$ is bounded below by 4/5 the result follows immediately.&lt;/p&gt;&#xA;" OwnerUserId="1301" LastActivityDate="2010-10-17T18:48:47.777" CommentCount="2" />
  <row Id="7054" PostTypeId="1" AcceptedAnswerId="7057" CreationDate="2010-10-17T19:00:13.097" Score="2" ViewCount="214" Body="&lt;p&gt;So my brain is frazzled which is probably why this seems like a big deal to me right now, but I just can't get over this reasoning:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose you have F = {All 1-1, increasing functions from N to N}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1-1: means that every value of the domain maps to some unique value of the range and every value of the range is equal to f(n) for some n.  Hence, since f is 1-1 AND increasing, the only function that exists is the trivial function, f(n) = n.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please tell me why that is wrong :-p&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(ftr, this isn't the homework question.  I am proving uncountability of F, which is why my brain-fart is bothering me more)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Answer: @Prometheus left this as a comment and then deleted it, probably because he didn't wish to be associated with stupidity as great as mine.  The error is that I'm assuming one-to-one =&gt; onto, which it clearly doesn't. &lt;em&gt;huddles in a ball and cries&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="355" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2010-11-21T02:02:50.073" LastActivityDate="2010-11-21T02:02:50.073" Title="Simple Question on 1-1, Increasing integer functions" Tags="&lt;elementary-set-theory&gt;" AnswerCount="2" CommentCount="5" />
  <row Id="7055" PostTypeId="2" ParentId="7034" CreationDate="2010-10-17T19:07:03.847" Score="0" ViewCount="" Body="&lt;p&gt;So $c_{1} = 1$ and $c_{n+1} = 4/(1+5c_{n})$ for $n \geq 1$. Let $C(x) = \sum_{n \geq 1} c_{n}x^n$. Then maybe try to express $\sum_{n \geq 1} c_{n+1}x^n$ and $\sum_{n \geq 1} \frac{4x^n}{1+5c_{n}}$ in terms of $C(x)$ to get a closed form. &lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2010-10-17T19:07:03.847" CommentCount="1" />
  <row Id="7056" PostTypeId="2" ParentId="7054" CreationDate="2010-10-17T19:11:52.810" Score="4" ViewCount="" Body="&lt;p&gt;First, a theorem: $|P(\mathbb{N})| &gt; \aleph_0$ (this is Cantor's theorem applied to the natural numbers. I will not go through the proof here.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now a second theorem: Let $A$ be the set of all finite or co-finite (that is - the complement is finite) subsets of the natural numbers, then $A$ is countable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proof: The set of all subsets of size $n$ can be mapped to a subset of $\mathbb{N}^n$ (as images of functions from $n \to \mathbb{N}$), since this is a countable set itself, and we have countably many $n$ then we have countably many finite subsets of the natural numbers, and since every one of those can be mapped to its complement - we have the needed proof.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Corollary: There are uncountably many subsets of $\mathbb{N}$ which are infinite and their complement is also infinite.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For each of these subsets we know that we can order it, and thus have a 1-1 function from $\mathbb{N}$ onto it, which is also increasing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, the set of all functions 1-1 and increasing from $\mathbb{N}$ to itself is uncountable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I hope this is clear, if not - let me know which points needs further expansion.)&lt;/p&gt;&#xA;" OwnerUserId="622" LastActivityDate="2010-10-17T19:11:52.810" CommentCount="1" />
  <row Id="7057" PostTypeId="2" ParentId="7054" CreationDate="2010-10-17T19:16:12.550" Score="3" ViewCount="" Body="&lt;p&gt;The question has been answered in the comments.  The functions are not assumed to be onto.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a hint for an approach that is different from Asaf's: You could consider the sequences of differences $f(n+1)-f(n)$.  It is enough to restrict to functions that only increase by 1 or 2 at each integer.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-10-17T19:16:12.550" CommentCount="1" />
  <row Id="7058" PostTypeId="2" ParentId="7041" CreationDate="2010-10-17T20:15:36.753" Score="1" ViewCount="" Body="&lt;p&gt;Seeing textbooks or other material, in any language, from which this person has learned mathematics, would be useful and relatively easy to interpret for an Anglophone mathematician.   If these are not available then equivalents may exist on the internet.  Pointing to material in books in English and seeing if it is familiar may also reveal something. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the high end, the International Mathematical Olympiad problems are available online in dozens of languages, probably including that of the student's native country.  This could at least tell you whether the student has heard of some concept used in a problem, and you would have a translation into English so you know you are discussing the same thing.  &lt;/p&gt;&#xA;" OwnerUserId="467" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-10-17T20:21:17.217" LastActivityDate="2010-10-17T20:21:17.217" />
  <row Id="7060" PostTypeId="2" ParentId="7025" CreationDate="2010-10-17T20:59:05.267" Score="1" ViewCount="" Body="&lt;p&gt;Now that you have updated your question with the fact that $F(x) = \{x\}$ (the standard notation for the fractional part of $x$), there is quite a bit more that can be said.  In particular, you can give an interpretation to $\int_1^{\infty} \{x + c\} G(x) dx$.  Let $0 &amp;lt; c &amp;lt; 1$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then $$\{x + c\} = \{x\} + c, 0 \leq \{x\} &amp;lt; 1 - c;$$ &#xA;$$\{x + c\}= \{x\} + c - 1, 1-c \leq \{x\} &amp;lt; 1.$$&#xA;(Sorry; I can't seem to make the &lt;em&gt;cases&lt;/em&gt; command work right.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have &#xA;$$\int_1^{\infty} \{x + c\} G(x) dx = \int_{1 \leq x &amp;lt; \infty, 0 \leq \{x\} &amp;lt; 1 - c} (\{x \} + c) G(x) dx + \int_{1 \leq x &amp;lt; \infty, 1-c \leq \{x\} &amp;lt; 1 } (\{x \} + c - 1) G(x) dx$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ = \int_1^{\infty} \{x \} G(x) dx + c \int_{1 \leq x &amp;lt; \infty, 0 \leq \{x\} &amp;lt; 1 - c} G(x) dx + (c-1) \int_{1 \leq x &amp;lt; \infty, 1-c \leq \{x\} &amp;lt; 1 } G(x) dx$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ = 1 + c \int_{1 \leq x &amp;lt; \infty, 0 \leq \{x\} &amp;lt; 1 - c} G(x) dx - (1-c) \int_{1 \leq x &amp;lt; \infty, 1-c \leq \{x\} &amp;lt; 1 } G(x) dx.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The two remaining integrals constitute an average of sorts, weighted to account for the fact that they are being taken over different percentages of the interval $[1,\infty)$.  The first integral gets weighted by $c$ but includes $1-c$ of the interval $[1,\infty)$, as it is being taken over the set $\cup_{i=1}^{\infty} [i,i+1-c)$.  (Remember that $c$ is a fraction between 0 and 1.)  The second integral gets weighted by $1-c$ but includes $c$ of the interval $[1,\infty)$, as it is being taken over the set $\cup_{i=2}^{\infty} [i-c,i)$.  So $\int_1^{\infty} \{x + c\} G(x) dx$ just shifts the weights on the values of $G(x)$ in $\int_1^{\infty} \{x \} G(x) dx$ in the manner I just described.  The resulting value for $\int_1^{\infty} \{x + c\} G(x) dx$ will be either greater or smaller than 1, depending on whether the larger values of $G(x)$ over $[1, \infty)$ tend to clump just above each integer value of $x$ or just below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other than this, I think Willie Wong's answer still applies.  In particular, you still can't get an exact answer for $\int_1^{\infty} \{x + c\} G(x) dx$ -- just an interpretation of it.&lt;/p&gt;&#xA;" OwnerUserId="2370" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-10-17T21:04:21.277" LastActivityDate="2010-10-17T21:04:21.277" CommentCount="2" />
  <row Id="7061" PostTypeId="1" AcceptedAnswerId="7063" CreationDate="2010-10-17T22:42:08.123" Score="2" ViewCount="232" Body="&lt;p&gt;Prove: If X is NOT divisible by 3 and Y is NOT divisible by 3 Then X^2 - y^2 is Divisible by 3&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I cannot find a good way to prove this.. Tried contrapositive and contradiction but cannot see the answer. Anyone have any ideas?&lt;/p&gt;&#xA;" OwnerUserId="2587" LastActivityDate="2010-10-18T17:46:14.870" Title="Prove: If X is NOT divisible by 3 and Y is NOT divisible by 3 Then X^2 - y^2 is Divisible by 3" Tags="&lt;divisibility&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="7062" PostTypeId="2" ParentId="7061" CreationDate="2010-10-17T22:54:57.743" Score="2" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\ \ $ mod 3: $\rm\ x\not\equiv 0\ \Rightarrow\ x \equiv \pm 1\ \Rightarrow\ x^2 \equiv 1\:.\ $ Alternatively, without modular arithmetic,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\quad\quad\quad\ $ 3 divides $\rm\: (x-1)\:x\:(x+1)\ $ but not $\:x\ \Rightarrow\ $ 3 divides $(x-1)(x+1) = x^2 - 1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, if both $\rm\ x^2-1\ $ and $\rm\ y^2-1\ $ are multiples of 3 then so too is their difference $\rm\ x^2 - y^2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;REMARK&lt;/strong&gt; $\: $ This proof easily generalizes to show that $\:24\:$ divides $\rm x^2-y^2\:$ if $\rm\:x\:$  and $\rm\:y\:$ are coprime to $6\:$. See &lt;a href=&quot;http://math.stackexchange.com/questions/910&quot;&gt;here for proof&lt;/a&gt;, and for generalizations using Carmichael's function (universal exponent of a group).&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-10-18T17:46:14.870" LastActivityDate="2010-10-18T17:46:14.870" CommentCount="3" />
  <row Id="7063" PostTypeId="2" ParentId="7061" CreationDate="2010-10-17T23:00:10.957" Score="4" ViewCount="" Body="&lt;p&gt;If $x$ and $y$ are not divisible by 3 then they must be of the form $x = 3k \pm 1$ and $y = 3t \pm 1$. Therefore $x^2 - y^2 = 9k^2 \pm 6k + 1 -9t^2 \mp 6t -1 = 3(3k^2 - 3t^2 \pm 2k \mp 2t).$ So $x^2 - y^2$ is divisible by 3.&lt;/p&gt;&#xA;" OwnerUserId="900" LastEditorUserId="900" LastEditorDisplayName="" LastEditDate="2010-10-18T01:29:02.183" LastActivityDate="2010-10-18T01:29:02.183" CommentCount="2" />
  <row Id="7064" PostTypeId="1" CreationDate="2010-10-17T23:33:06.067" Score="5" ViewCount="255" Body="&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Long_division&quot; rel=&quot;nofollow&quot;&gt;Wikipedia article&lt;/a&gt; on long division explains the different notations. I still use the European notation I learned in elementary school in Colombia. I had difficulty adapting to the US/UK notation when I moved to the US. However, I did enjoy seeing my classmates' puzzled faces in college whenever we had a professor that preferred the European notation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What long division notation do you use and where did you learn it?&lt;/p&gt;&#xA;" OwnerUserId="2586" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-11-17T00:14:30.427" LastActivityDate="2010-11-18T21:22:15.223" CommunityOwnedDate="2010-11-18T06:46:45.763" Title="Long division notation (census of nations)" Tags="&lt;big-list&gt;&lt;notation&gt;&lt;arithmetic&gt;" AnswerCount="4" CommentCount="2" />
  <row Id="7065" PostTypeId="1" AcceptedAnswerId="7074" CreationDate="2010-10-17T23:40:29.537" Score="6" ViewCount="186" Body="&lt;p&gt;My question is essentially this: Why does the principle of transfinite induction not suffice to show the axiom of choice when the sets to be chosen from are indexed by a well ordered set?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have read that one can prove the axiom of finite choice from simple induction. You induct on the size of the system of sets you are choosing from and pick an element from each set. I understand this. However, my grasp of the details is sketchy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Why does standard induction alone not suffice to show the axiom of choice for systems of countable sets? Doesn't induction show the truth of the statement for all natural numbers, and therefore for any system of sets that can be indexed by the natural numbers (countable sets)? I know this to be false, but I do not know why.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Why can't the above &quot;proof&quot; that induction implies the AoC for countable sets not be repaired by using transfinite induction? Isn't this the purpose of transfinite induction, to allow one to induct on sets of infinite size? Shouldn't transfinite induction suffice to prove the axiom of choice for any system of sets indexed by a well-ordered set? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am reading Jech right now, but my knowledge of ordinals and transfinite induction is very, very poor, so I would greatly prefer answers with a great amount of explanation and hand-holding.&lt;/p&gt;&#xA;" OwnerUserId="2591" LastEditorUserId="2591" LastEditorDisplayName="" LastEditDate="2010-10-18T00:50:18.523" LastActivityDate="2010-10-18T01:02:24.780" Title="Transfinite Induction and the Axiom of Choice" Tags="&lt;set-theory&gt;" AnswerCount="1" FavoriteCount="2" />
  <row Id="7066" PostTypeId="2" ParentId="7064" CreationDate="2010-10-17T23:41:07.507" Score="1" ViewCount="" Body="&lt;p&gt;I use the same one that is used in the U.S.; it is the one that is (or at least, was up until the 80s) taught in Mexico. I had some classmates in high school who had emigrated form Argentina: they used the one described in the Wikipedia article as &quot;European&quot;, not the one described as &quot;Latin America&quot;. (I'll also note that beginners in Mexico would write the full computation as described in &quot;US notation&quot;; doing the computation mentally was more an 'advanced shortcut' than part of the notation, as I remember).&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-10-17T23:41:07.507" CommunityOwnedDate="2010-10-17T23:41:07.507" />
  <row Id="7067" PostTypeId="1" AcceptedAnswerId="7076" CreationDate="2010-10-18T00:01:20.017" Score="4" ViewCount="187" Body="&lt;p&gt;Can someone give me an explanation targeted to a high school student as to why finding thegcd of two numbers is faster using the euclidean algorithm compared to using factorization, there should be no algorithm efficiency involved, just a general explanation, something my brother in grade 9 can understand.&lt;/p&gt;&#xA;" OwnerUserId="2588" LastActivityDate="2010-10-18T06:45:19.593" Title="Euclidean Algorithm vs Factorization" Tags="&lt;number-theory&gt;" AnswerCount="3" CommentCount="6" />
  <row Id="7068" PostTypeId="2" ParentId="4990" CreationDate="2010-10-18T00:01:51.407" Score="3" ViewCount="" Body="&lt;p&gt;The approved answer has caused some risibility at mathoverflow, and I'll elaborate on Robin's&#xA;more reasonable comment (but I'm inclined to attribute the descent argument in this case to&#xA;Euler--at least he wrote it down). The version I give in an undergrad  number theory class is this:  First one develops the standard facts about Z[w] where w^2+w+1=0. (It has unique&#xA;factorization, 2 is prime, the units are 1,-1,w,-w,w^2 and -w^2, any element not 0 or a unit&#xA;has absolute value &gt;1, and each congruence class mod 2 is represented by 0 or a unit). Then one notes that it's enough to prove:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Theorem--There are no a,b,c in R with a+b+c=0, abc a non-zero cube and a=b=1 mod 2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The proof of the theorem is a reductio. Let H be max(/a/,/b/,/c/) and choose a solution&#xA;a,b,c with minimal H. (H^2 is an integer). a, b and c are evidently pairwise prime. Since&#xA;their product is a non-zero cube, each is (unit)(cube). Since a=b=1 mod 2, a=A^3 and&#xA;b=B^3, and we may assume that A=B=1 mod 2. Since abc is a cube, c=C^3 for some C in R. Since&#xA;2 divides c, 2 divides C and H is at least 8.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let S=Aw+Bw^2, T=Aw^2+Bw, and U=A+B. Then S+T+U=0 while STU is A^3+B^3=-C^3. Also S=T=1 mod 2, while max(/S/,/T/,/U/) is at most 2(H^(1/3)). This contradicts the minimality assumption.&lt;/p&gt;&#xA;" OwnerUserId="2594" LastActivityDate="2010-10-18T00:01:51.407" />
  <row Id="7069" PostTypeId="2" ParentId="7067" CreationDate="2010-10-18T00:24:41.913" Score="3" ViewCount="" Body="&lt;p&gt;It boils down to the fact that currently known factorization algorithms are much slower than the fastest known gcd algorithms. It would be difficult to say much more than that that would be comprehensible to a 9'th grade student (in fact one might argue that not much more than that is even known to experts).&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-10-18T00:24:41.913" />
  <row Id="7070" PostTypeId="1" AcceptedAnswerId="7072" CreationDate="2010-10-18T00:27:05.137" Score="4" ViewCount="96" Body="&lt;p&gt;I am trying to calculate the average distance between any two nodes in an $r$-dimensional mesh.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a $3$-dimensional mesh with $n=3$&#xA;&lt;img src=&quot;http://i.imgur.com/uulgk.gif&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To choose any two points in the mesh there are $\left(n^r \left(n^r - 1\right)\right)/2$ ways of doing this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if we choose 2 points in the mesh $p\lbrace 1, 2, ..., r \rbrace$ and $q\lbrace 1, 2, ..., r \rbrace$ The distance is calculated as the &lt;a href=&quot;http://en.wikipedia.org/wiki/Taxicab_geometry&quot; rel=&quot;nofollow&quot;&gt;Manhattan Distance&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sum_{i=1}^{r} \left| {q_i - p_i} \right|$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now i have to find the total distance between all nodes and divide that by the the number of ways of choosing two nodes (given above).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know how to do this by looking at this 3-dimensional case which should work on the r-dimensional case also but i am having trouble expressing it in some sort of sum notation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is how i think i can get the total distance between all nodes, i hope this is clear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this 3-d mesh let's assume the nodes are labelled $\left( 1, 1, 1 \right)$ to $\left( 3, 3, 3\right)$. Lets start at $\left( 1, 1, 1 \right)$ and sum up the distance between this node and all other nodes. Then we move to $\left( 1, 1, 2 \right)$ and sum up the distances between this node and all other nodes except for $\left( 1, 1, 1 \right)$ because we have already counted the distance between $\left( 1, 1, 1 \right)$ and $\left( 1, 1, 2 \right)$. Then we move onto $\left( 1, 1, 3 \right)$ and sum up all except for $\left( 1, 1, 1 \right)$ and $\left( 1, 1, 2 \right)$. We continue this until we permute through in order $\left( 1, 1, 1 \right)$, $\left( 1, 1, 2 \right)$, $\left( 1, 1, 3 \right)$, $\left( 1, 2, 1 \right)$, $\left( 1, 2, 2 \right)$, ect... Does that make sense?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That should give me the total distance without counting anything twice. Then i divide that by the number of ways of choosing 2 nodes and i will have the average distance. Does this sound correct to you? Any help would be appreciated.&lt;/p&gt;&#xA;" OwnerUserId="43" LastActivityDate="2010-11-17T01:44:19.080" Title="r-dimensional mesh question" Tags="&lt;graph-theory&gt;" AnswerCount="1" FavoriteCount="0" />
  <row Id="7071" PostTypeId="1" CreationDate="2010-10-18T00:54:16.490" Score="-2" ViewCount="291" Body="&lt;p&gt;Is the sum of primes up to p a multiple of p?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i.e Is 1+2+...+p divisible by p and how would you prove it?&lt;/p&gt;&#xA;" OwnerUserId="2588" LastActivityDate="2010-11-10T16:56:26.647" Title="Sum of primes up to p is multiple of p?" Tags="&lt;prime-numbers&gt;" AnswerCount="3" />
  <row Id="7072" PostTypeId="2" ParentId="7070" CreationDate="2010-10-18T00:59:27.587" Score="1" ViewCount="" Body="&lt;p&gt;This approach is correct, but more work than required.  Using the taxicab metric each dimension is independent.  Taking a cube of side n you can count the vertical segments used as follows:  from the bottom layer to the next layer up, we count $n^2$ (points on the bottom layer) times $n^2*(n-1)$ (points above the bottom layer) = $n^4*(n-1)$.  For the next layer you have $2*n^2*n^2*(n-2)$ and so on.  The $n^4$s distribute out and you can just think about the total distance along a line of length n.  This is $\sum_{i=1}^{i=n}i*(n-i)$.  Then multiply by $n^4$ and you have the total usage in the vertical direction.  Then multiply by 3 for the total usage in each direction.  And divide by the number of pairs of points, which you have calculated already.  The extension to different number of dimensions or non-cubical meshes should be clear.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-10-18T00:59:27.587" CommentCount="2" />
  <row Id="7073" PostTypeId="2" ParentId="7071" CreationDate="2010-10-18T01:00:17.543" Score="5" ViewCount="" Body="&lt;p&gt;What you wrote is not the sum of the primes up to $p$.  1 is not prime, and the meaning of your ellipsis is unclear.  Based on one possible interpretation, a counterexample is that 2 does not divide $1+2$.  Based on another, a counterexample is that 3 does not divide $2+3$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if $p$ is an odd number, and in particular if it is a prime other than 2, then $p$ divides the sum of the first $p$ positive integers, because this sum is $p\cdot\frac{p+1}{2}$.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-10-18T01:00:17.543" CommentCount="2" />
  <row Id="7074" PostTypeId="2" ParentId="7065" CreationDate="2010-10-18T01:02:24.780" Score="4" ViewCount="" Body="&lt;p&gt;The problem is that you also do a &quot;meta&quot; induction when proving choice for a finite number of sets and this &quot;meta&quot; induction fails in the transfinite or countable case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To be more clear, the proof of choice for a number of sets sets usually goes something like this (I'm assuming choice is &quot;the product of nonempty sets is nonempty&quot;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Base case is vacuously true (you're given one nonempty set and want to prove it's nonempty).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next, assume inductively that for any nonempty sets $X_1,...,X_n$, the product $X_1\times...\times X_n$ is nonempty.  Now, given $n+1$ sets, $X_1,...,X_{n+1}$, you want to show their product is nonempty.  By the induction hypothesis, there is an element $y\in X_1\times....\times X_n$.  Also, there is an element $x_{n+1}\in X_{n+1}$ by assumption.  Then $\{\{y,x_{n+1}\}, x_{n+1}\} = (y, x_{n+1})$ is an element of the product.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the meta level, what you're saying by unraveling the induction is &quot;I have $n+1$ sets $X_1,...,X_{n+1}$, and I know there is an element (which is itself a set because we're doing set theory!) in each $X_i$ which I'll call $x_i$.  By using the axiom of pairing once, I can form the set $\{x_1,x_2\}$.  By using it again, I can form the set $\{\{x_1,x_2\}, x_1\}$, which is the definition of $(x_1,x_2)$.  Using the axiom of pairing two more times, I can form $(x_1, x_2, x_3)$.  In general (i.e., by induction), by using the axiom of pairing $2(n+1)$ times, I can form the set $(x_1,x_2,...,x_{n+1})$.  This set (element) is a member of $X_1\times...\times X_{n+1}$, and hence this product is nonempty!&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What goes wrong as soon as you try to adapt this to the countable or larger setting is on the meta level.  Essentially, you'd have to apply the pairing axiom countably (or more) times to create the set $(x_1,...)$ which &quot;should&quot; be in $X_1\times...\times $.  But naive set theory has taught us that just because we think something &quot;should&quot; be a set, doesn't mean it should (see, for example, Russel's Paradox).  Thus, we at least cannot trust this proof to work in countable or larger settings.  Of course, the failure of this approach doesn't mean there is NO way of proving choice from $ZF$, but it's known from other methods that you cannot prove choice from $ZF$ alone.&lt;/p&gt;&#xA;" OwnerUserId="331" LastActivityDate="2010-10-18T01:02:24.780" CommentCount="3" />
  <row Id="7075" PostTypeId="2" ParentId="7071" CreationDate="2010-10-18T01:03:24.933" Score="2" ViewCount="" Body="&lt;p&gt;Or if by the ellipsis you mean the sum of naturals up to p, it is true and not just for primes.  The sum of naturals up to n is $n*(n+1)/2$, which can be proved by induction, and is divisible by n. (as commented, only if n is odd)&lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2010-11-10T16:56:26.647" LastActivityDate="2010-11-10T16:56:26.647" CommentCount="1" />
  <row Id="7076" PostTypeId="2" ParentId="7067" CreationDate="2010-10-18T01:16:37.730" Score="7" ViewCount="" Body="&lt;p&gt;The Euclidean algorithm is a definite recipe that tell you exactly what to do at any given step; there is no guessing, no trial and error involved. Trying to factor a number will (even with some of the best methods currently known) involve guesses and trial-and-error; trial division is of course the classic example, but even some of our best methods (elliptic curve factorization and number field sieve, to name two) all involve some &quot;random guessing&quot; and trials to try to find factors. Sure, they are more clever ways of testing than simply trying everything out there, but you still usually end up doing a lot of grunt work along the way that leads nowhere (dividing by a number that is not a factor in trial division; not getting good relations in the number field sieve; performing all computations on an elliptic curve modulo $n$ and not finding any non-invertible elements), or performing good-looking computations that end up with a trivial factor ($1$ or $n$). In essence, this is &quot;wasted effort&quot;, waste that simply does not occur with the Euclidean algorithm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added:&lt;/strong&gt; Note that this is a reflection of our current known factoring methods, and not necessarily (as Bill Dubuque points out, we just don't know either way) an inherent difficulty in factoring. You don't want to compare &quot;factoring&quot; with &quot;Euclidean algorithm&quot;, you want to compare &lt;em&gt;specific ways&lt;/em&gt; of factoring with the Euclidean algorithm. And the ways we know (and the ways the high school students know, which are likely to be trial division plus a handful or two of divisibility tests to make the former simpler) have these drawbacks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps an analogy would be that the Euclidean algorithm is like having a full recipe to prepare a dish, and all the ingredients laid out ready to be used; factorization involves starting to prepare the dish, rummaging through your supplies for ingredients, and possibly realizing part-way through that you don't have the right ingredients, forcing you to start over from scratch with a new dish for which you hope you do have the ingredients. Even if the former situation involves a complex dish while the latter is a series of attempts at very simple and quick dishes, chances are you will spend less total time with the full-recipe-and-all-ingredients-laid-out-method than the let's-try-this-and-hope-we-have-all-the-stuff method.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-10-18T03:32:58.980" LastActivityDate="2010-10-18T03:32:58.980" CommentCount="2" />
  <row Id="7077" PostTypeId="2" ParentId="7071" CreationDate="2010-10-18T01:18:09.147" Score="2" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\rm\quad\ \ S\ =\ 1 \ \ \ +\ \ \: 2\ \ \ \ +\ \:\cdots\  +\ p-1\ +\ p $&lt;br&gt;&#xA;$\rm\quad\quad\quad\quad\quad\quad\ S\ =\ p \ +\  p-1\  +\ \cdots\ +\ \  \ \ 2\ \ \ +\ \ \:1 $  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Adding $\rm\ \ 2\: S\ =\ p\ (p+1)$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A &lt;a href=&quot;http://www.americanscientist.org/issues/pub/gausss-day-of-reckoning&quot; rel=&quot;nofollow&quot;&gt;famous legend&lt;/a&gt; says Gauss used this trick to quickly compute $ 1+2+\:\cdots\:+100\ $ in grade school.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This trick of pairing up reflections around the average value is a special case of exploiting innate symmetry - here a reflection or involution. It's a ubiquitous powerful technique, e.g. see my post on &lt;a href=&quot;http://math.stackexchange.com/questions/1865&quot;&gt;Wilson's Theorem&lt;/a&gt; and it's &lt;a href=&quot;http://math.stackexchange.com/questions/9311/product-of-all-elements-in-an-odd-finite-abelian-group-is-1/9345#9345&quot;&gt;group theoretic generalization&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-11-08T22:01:57.580" LastActivityDate="2010-11-08T22:01:57.580" CommentCount="4" />
  <row Id="7078" PostTypeId="2" ParentId="7061" CreationDate="2010-10-18T01:20:40.107" Score="4" ViewCount="" Body="&lt;p&gt;$x^2-y^2 = (x-y)(x+y)$. If neither $x$ nor $y$ are multiples of $3$, then either they have the same remainder and $x-y$ is a multiple of $3$, or they have different remainders ($1$ and $2$) and $x+y$ is a multiple of $3$.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-10-18T01:20:40.107" />
  <row Id="7079" PostTypeId="1" AcceptedAnswerId="7080" CreationDate="2010-10-18T01:37:00.417" Score="-1" ViewCount="140" Body="&lt;p&gt;Let n be a natural number with d(n) = 33 (number of factors are 33)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How would I prove that n and n + 1 don't have a common factor &gt; 1?&lt;/p&gt;&#xA;" OwnerUserId="2588" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-10-18T16:07:32.277" LastActivityDate="2010-10-18T16:07:32.277" Title="Number of factors proof" Tags="&lt;number-theory&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="7080" PostTypeId="2" ParentId="7079" CreationDate="2010-10-18T01:38:35.260" Score="6" ViewCount="" Body="&lt;p&gt;The first thing you would do is ignore the information that $d(n)=33$, because it is completely irrelevant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;HINT: Any common factor of $a$ and $b$ divides $a-b$. &lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-10-18T01:38:35.260" CommentCount="3" />
  <row Id="7081" PostTypeId="1" AcceptedAnswerId="7110" CreationDate="2010-10-18T01:41:44.460" Score="3" ViewCount="191" Body="&lt;p&gt;Let G be the group defined by generators a,b and relations $a^4=e$, $a^2b^{-2}=e$, $abab^{-1}=e$. Since the quaternion group of order 8 is generated by elements $a,b$ satisfying the previous relations, there is an epimorphism from $G$ onto $Q_8$. Let $F$ be the free group on $\{a,b\}$ and $N$ the normal subgroup generated by $\{ a^4, a^2b^{-2}, abab^{-1} \}$. How to write down the normal subgroup N and express the group F/N ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="2177" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-10-18T01:56:16.293" LastActivityDate="2010-10-18T10:14:07.423" Title="presentation of quaternion group of order 8" Tags="&lt;group-theory&gt;" AnswerCount="2" />
  <row Id="7082" PostTypeId="2" ParentId="7081" CreationDate="2010-10-18T01:53:31.450" Score="2" ViewCount="" Body="&lt;p&gt;To write down the group $F/N$, the best way in this case is to find a normal form. Note that the last relation $abab^{-1}=e$ means that $ba=a^{-1}b = a^3b$ (that is, the element $b^{-1}a^{-3}ba$ is in $N$); so any coset representative that contains the string $ba$ can be replaced with a coset representative in which the string $ba$ is replaced with the string $a^3b$. The fact that $a^2b^{-2}\in N$ means that any coset representative that in which you have a $b^2$ can be replaced with one that has an $a^2$ instead. Proceeding in this way, you can conclude that every coset can be represented by an element of the form $a^i b^j$ with $0\leq i\leq 3$ and $0\leq j\leq 1$ (you can erase any $a^4$ since $a^4\in N$). Thus, the group $F/N$ has at most $8$ elements. You already know it has at least $8$ elements, so that means that each of these elements represent distinct cosets of $N$, so that gives you a way to express $F/N$ using the representatives.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Writing down&quot; the group $N$ is a bit more difficult, since it is an infinite subgroup in $F$. What exactly do you mean by &quot;writing down $N$&quot;? A way to determine if a given element of $F$ lies in $N$? This can be achieved by finding its &quot;coset representative&quot; from among the special set identified above. If you get $a^0b^0$, then the element was in $N$; if not, then it was not.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-10-18T02:27:32.537" LastActivityDate="2010-10-18T02:27:32.537" CommentCount="2" />
  <row Id="7083" PostTypeId="2" ParentId="6538" CreationDate="2010-10-18T02:06:19.707" Score="0" ViewCount="" Body="&lt;p&gt;For any group of order 105, there always exists a normal sylow 7-subgroup and a normal sylow 5-group. &lt;/p&gt;&#xA;" OwnerUserId="2177" LastActivityDate="2010-10-18T02:06:19.707" />
  <row Id="7084" PostTypeId="1" AcceptedAnswerId="7087" CreationDate="2010-10-18T02:06:54.557" Score="-2" ViewCount="243" Body="&lt;p&gt;How would I prove that gcd(ab,n) = gcd(a,n)*gcd(b,n)?&lt;/p&gt;&#xA;" OwnerUserId="2588" LastEditorUserId="1543" LastEditorDisplayName="" LastEditDate="2011-05-03T09:59:16.773" LastActivityDate="2011-05-03T09:59:16.773" Title="GCD Proof Question" Tags="&lt;algebra-precalculus&gt;&lt;elementary-number-theory&gt;" AnswerCount="3" CommentCount="9" />
  <row Id="7085" PostTypeId="2" ParentId="7084" CreationDate="2010-10-18T02:13:10.710" Score="1" ViewCount="" Body="&lt;p&gt;It would be difficult, as it is not true.  Take a=b=n=2&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-10-18T02:13:10.710" />
  <row Id="7086" PostTypeId="2" ParentId="7084" CreationDate="2010-10-18T02:17:47.787" Score="2" ViewCount="" Body="&lt;p&gt;See this recently-asked &lt;a href=&quot;http://math.stackexchange.com/questions/6653/basic-divisibility-fact/6657&quot;&gt;equivalent question&lt;/a&gt;, after imposing the correct hypotheses, e.g. $\rm \gcd(a,b,n) = 1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a proof using said hypothesis and gcd laws (associative, commutative, distributive)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\quad\quad (a,n)(b,n)\ =\ (ab, n\:(a,b,n))\ =\ (ab,n)$&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-10-18T13:49:02.273" LastActivityDate="2010-10-18T13:49:02.273" />
  <row Id="7087" PostTypeId="2" ParentId="7084" CreationDate="2010-10-18T02:21:08.857" Score="3" ViewCount="" Body="&lt;p&gt;fmunshi, it isn't true, as it was pointed out. But you can try and see what happens there when $\gcd(a,b)=1$.&lt;/p&gt;&#xA;" OwnerUserId="1313" LastActivityDate="2010-10-18T02:21:08.857" />
  <row Id="7089" PostTypeId="2" ParentId="4844" CreationDate="2010-10-18T02:40:32.613" Score="2" ViewCount="" Body="&lt;p&gt;The canonical thing to do for a question like this is to look at Neal Sloane's home page. Sure enough, there is a table giving some good arrangements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www2.research.att.com/~njas/electrons/index.html&quot; rel=&quot;nofollow&quot;&gt;http://www2.research.att.com/~njas/electrons/index.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This was indeed one of the links on the page in wok's answer, but it may be the most complete resource.&lt;/p&gt;&#xA;" OwnerUserId="2501" LastActivityDate="2010-10-18T02:40:32.613" />
  <row Id="7090" PostTypeId="1" AcceptedAnswerId="11796" CreationDate="2010-10-18T02:55:42.190" Score="1" ViewCount="382" Body="&lt;p&gt;I was given a proof of this in class, but it does not make sense to me, can someone provide a better proof, or explain the one I have provided, thanks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose $\gcd(a,n) = 1$, where $a$ and $n$ are natural numbers. Prove that $0$, $a$, $2a$, $3a,\ldots, (n-1)a$, is complete residue system modulo $n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proof given in class:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align*}&#xA;ka &amp;amp;\equiv ra \pmod{n}\\&#xA;&amp;amp;\Rightarrow  k \equiv r \pmod{n}\\&#xA;&amp;amp;\Rightarrow  \text{$k-r$ divides $n$}\\&#xA;&amp;amp;\Rightarrow \text{$k=r$ because $0\leq k\lt n$, $0\leq r\lt n$}&#xA;\end{align*}&lt;/p&gt;&#xA;" OwnerUserId="2588" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-25T04:59:12.177" LastActivityDate="2010-12-03T22:45:09.197" Title="Complete residue system proof" Tags="&lt;number-theory&gt;" AnswerCount="2" CommentCount="7" />
  <row Id="7091" PostTypeId="2" ParentId="4894" CreationDate="2010-10-18T03:35:50.797" Score="4" ViewCount="" Body="&lt;p&gt;Any two of (a) symplectic structure, (b) almost-complex structure, and (c) Riemannian structure determine the third.  This reflects the fact that $$ U(n) = O(2n) \cap GL(n,\mathbb{C}) \cap Sp(2n),$$ but the intersection of any two is already $U(n)$.  This gives rise to the theory of Kahler manifolds (cf. wikipedia), which are central to mirror symmetry (and the mirror symmetry conjecture).&lt;/p&gt;&#xA;" OwnerUserId="2159" LastEditorUserId="2159" LastEditorDisplayName="" LastEditDate="2010-10-18T20:19:27.963" LastActivityDate="2010-10-18T20:19:27.963" />
  <row Id="7092" PostTypeId="1" AcceptedAnswerId="7093" CreationDate="2010-10-18T03:57:46.073" Score="0" ViewCount="450" Body="&lt;p&gt;What is the highest power of 2 dividing 100!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is what I have so far:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;50 multiples of 2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;25 multiples of 4&lt;/p&gt;&#xA;&#xA;&lt;p&gt;12 multiples of 8&lt;/p&gt;&#xA;&#xA;&lt;p&gt;6 multiples of 16&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3 multiples of 32&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1 multiple of 64&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: giving a highest power of &lt;strong&gt;2^97&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;am I missing anything here?&lt;/p&gt;&#xA;" OwnerUserId="2588" LastEditorUserId="2588" LastEditorDisplayName="" LastEditDate="2010-10-18T04:09:04.563" LastActivityDate="2010-10-18T20:02:42.620" Title="What is the highest power of 2 dividing 100!" Tags="&lt;number-theory&gt;" AnswerCount="3" />
  <row Id="7093" PostTypeId="2" ParentId="7092" CreationDate="2010-10-18T04:06:35.200" Score="2" ViewCount="" Body="&lt;p&gt;Just look at this page &lt;a href=&quot;http://planetmath.org/encyclopedia/PrimePowerDividingAFactorial.html&quot; rel=&quot;nofollow&quot;&gt;http://planetmath.org/encyclopedia/PrimePowerDividingAFactorial.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="900" LastActivityDate="2010-10-18T04:06:35.200" CommentCount="1" />
  <row Id="7094" PostTypeId="2" ParentId="7092" CreationDate="2010-10-18T04:07:34.017" Score="1" ViewCount="" Body="&lt;p&gt;Your list of multiples is correct.  But then the total is 97, not 2^97.  But good to see some work.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-10-18T04:07:34.017" CommentCount="2" />
  <row Id="7095" PostTypeId="2" ParentId="7092" CreationDate="2010-10-18T04:11:55.470" Score="1" ViewCount="" Body="&lt;p&gt;You are not missing anything. Using WolframAlpha to factor 100! into primes, we&#xA;get the answer:&#xA;2^97×3^48×5^24×7^16×11^9×13^7×17^5×19^5×23^4×29^3×31^3×37^2×41^2×43^2×47^2×53×59×61×67×71×73×79×83×89×97(239 factors, 25 distinct)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;therefore, the highest power of 2 that divides 100! is 2^97.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;100!/2^97 = 588971222367687651371627846346807888288472382883312574253249804256440585603406374176100610302040933304083276457607746124267578125&lt;/p&gt;&#xA;" OwnerUserId="1605" LastEditorUserId="1605" LastEditorDisplayName="" LastEditDate="2010-10-18T20:02:42.620" LastActivityDate="2010-10-18T20:02:42.620" CommentCount="8" />
  <row Id="7096" PostTypeId="1" CreationDate="2010-10-18T04:31:00.267" Score="6" ViewCount="172" Body="&lt;p&gt;Kodaira defines a complex analytic family of compact complex manifolds as the data $(E,B,\pi)$, where $E$ and $B$ are complex manifolds, and $\pi$ is a surjective holomorphic submersion such that the preimage $\pi^{-1}(x)$ of any point $x \in B$ is a compact complex submanifold of $E$. (Here complex manifolds are required to be connected.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A key property of this definition is that this specifies a differentiable fibre bundle. This fact can almost be obtained from the Ehresmann fibration theorem, which might be stated as follows:&#xA;&quot;Let $X$, $Y$ be differentiable manifolds, and let $f: X \rightarrow Y$ be a proper surjective submersion. Then $f$ is the projection of a differentiable fibre bundle.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, the map $\pi$ defining a family lacks the properness needed to apply Ehresmann's theorem. It is, however, a corollary of the fact that a family gives a fibre bundle that $\pi$ is indeed proper.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a more elementary way of seeing that $\pi$ must be proper?&lt;/p&gt;&#xA;" OwnerUserId="2598" LastActivityDate="2010-12-30T19:44:30.230" Title="Proper maps and families of compact complex manifolds" Tags="&lt;differential-topology&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="7100" PostTypeId="1" AcceptedAnswerId="7103" CreationDate="2010-10-18T04:58:29.037" Score="0" ViewCount="190" Body="&lt;p&gt;Calculate 1^30 + 2^30 + 3^30 + ... + 17^30 mod 31&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using Fermat's Theorem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1^30 = 1 (mod 31) (since 31 is prime)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This implies the above is congruent to 17 (mod 31)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is correct, right?&lt;/p&gt;&#xA;" OwnerUserId="2588" LastActivityDate="2010-10-18T13:25:55.490" Title="Calculate 1^30 + 2^30 + 3^30 + ... + 17^30 mod 31" Tags="&lt;number-theory&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="7103" PostTypeId="2" ParentId="7100" CreationDate="2010-10-18T05:16:46.200" Score="5" ViewCount="" Body="&lt;p&gt;True.  But 1^n=1 mod whatever.  Fermat's Little Theorem says that n^30=1 for all n prime to 31.  So your answer of 17 is correct.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-10-18T05:16:46.200" />
  <row Id="7105" PostTypeId="2" ParentId="7067" CreationDate="2010-10-18T06:45:19.593" Score="1" ViewCount="" Body="&lt;p&gt;For very small numbers then factorization is quicker than the Euclidean algorithm.&#xA;But the Euclidea algorithm gives more; it allows one to compute reciprocals&#xA;in modular arithmetic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For larger numbers, finding factorizations may be slow or even practically&#xA;impossible, but the Eulcidean algorithm still works well. Also modern&#xA;factorization methods from Pollard rho to the number field sieve, make&#xA;essential use of the Euclidean algorithm.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-10-18T06:45:19.593" />
  <row Id="7106" PostTypeId="2" ParentId="6994" CreationDate="2010-10-18T07:23:06.007" Score="2" ViewCount="" Body="&lt;p&gt;Valuation rings demonstrate quite clearly the failure of Krull's principal ideal theorem:&#xA;take a valuation ring O of finite dimension. The prime ideals then form a chain &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$p_0:=0\subset p_1\subset\ldots\subset p_d$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so that for every $i\in{1,\ldots ,d}$ there exists $r_i\in p_i\setminus p_{i-1}$. Obviously $p_i$ is a minimal prime over $r_iO$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For valuation domains of infinite dimension one has to consider the so-called limit-primes: a prime ideal $p$ of a commutative ring $R$ is called limit-prime if&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$p=\bigcup\limits_{q\in\mathrm{Spec} (R): q\subset p}q$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There exist valuation domains $O$ of infinite Krull dimension such that the maximal ideal $m$ of $O$ is no limit-prime. For example take a valuation ring such that the corresponding value group is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\mathbb{Z}\times\mathbb{Z}\times\ldots$ (countably many factors ordered lexigraphically).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then one can find $r\in m$ such that $m$ is minimal over $rO$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;H&lt;/p&gt;&#xA;" OwnerUserId="2479" LastActivityDate="2010-10-18T07:23:06.007" />
  <row Id="7107" PostTypeId="1" AcceptedAnswerId="7128" CreationDate="2010-10-18T08:09:32.607" Score="1" ViewCount="172" Body="&lt;p&gt;As part of my solution to a problem, I come to a point where I need to find the solutions to&#xA;$-2\partial_{T}B\left(T\right)+\frac{3}{4}B\left(T\right)\left(A\left(T\right)^{2}+B\left(T\right)^{2}\right)=0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$2\partial_{T}A\left(T\right)+\frac{3}{4}A\left(T\right)\left(B\left(T\right)^{2}+A\left(T\right)^{2}\right)=0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\partial_{T}(f)$ is the derivative with respect to $T$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is possible that I made a mistake in the steps leading to this because I am supposed to be able to get a not-so-ugly solution for $A(T)$ and $B(T)$. Is there one that exists and I don't see it? I've tried the following:&#xA;&lt;img src=&quot;http://i.imgur.com/yi0U1.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="2256" LastEditorUserId="2256" LastEditorDisplayName="" LastEditDate="2010-10-18T14:45:35.177" LastActivityDate="2010-10-18T14:53:48.223" Title="System of Non-linear ODEs -- Analytic Solution" Tags="&lt;differential-equations&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="7108" PostTypeId="1" CreationDate="2010-10-18T08:39:17.287" Score="3" ViewCount="230" Body="&lt;p&gt;What are the conditions for torsion to be zero other than having a plane curve?&#xA;The only thing I can thing of is an equation that have the torsion that cancels out each other.&lt;/p&gt;&#xA;" OwnerUserId="2603" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-10-19T04:28:33.667" LastActivityDate="2010-10-19T04:28:33.667" Title="Conditions that torsion is zero in a space curve" Tags="&lt;differential-geometry&gt;" AnswerCount="3" CommentCount="4" />
  <row Id="7109" PostTypeId="1" AcceptedAnswerId="7116" CreationDate="2010-10-18T09:19:05.287" Score="4" ViewCount="197" Body="&lt;p&gt;$$S = \sum_{k=1}^{\infty} \frac{cos(\theta\log(k))}{k^a}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do I go about finding the value of S, given that $\theta \to \infty$ and $0 &amp;lt; a &amp;lt; 1$.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Any special techniques that might be&#xA;  helpful in calculating this sum?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;EDIT: &#xA;Just to give some background,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was actually trying to figure out&#xA;$$\sum_{k=1}^{\infty} \frac{cos(\theta\log(k))}{k^a} - \sum_{k=1}^{\infty} \frac{cos(\theta\log(k + 0.5))}{(k+0.5)^a}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since that expression was a bit complicated, I decided to write the common version...&lt;/p&gt;&#xA;" OwnerUserId="2320" LastEditorUserId="2320" LastEditorDisplayName="" LastEditDate="2010-10-18T14:56:34.490" LastActivityDate="2010-10-18T14:56:34.490" Title="How to find the sum of this cos series" Tags="&lt;sequences-and-series&gt;&lt;trigonometry&gt;" AnswerCount="1" />
  <row Id="7110" PostTypeId="2" ParentId="7081" CreationDate="2010-10-18T10:14:07.423" Score="7" ViewCount="" Body="&lt;p&gt;Two points about this. As a subgroup of a free group, N is itself a free group. There is an algorithmic procedure for finding a free set of generators of N, known as the Schreier generators. An example of this was presented in detail in a recent item &quot;Normal Closure in groups II&quot;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://math.stackexchange.com/questions/6707/normal-closure-in-groups-ii&quot;&gt;http://math.stackexchange.com/questions/6707/normal-closure-in-groups-ii&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using a computer program (MAGMA), I got the following set of 9 generators of N:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;N.1 = b * a * b * a^-1&#xA;N.2 = b^2 * a^-2&#xA;N.3 = a^-1 * b^2 * a^-1&#xA;N.4 = b^-1 * a * b^-1 * a^-1&#xA;N.5 = a^4&#xA;N.6 = a^2 * b^2&#xA;N.7 = a * b * a * b^-1&#xA;N.8 = a * b^2 * a&#xA;N.9 = a * b^-1 * a * b&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The second point is that the first relation $a^4=e$ in your presentation is actually redundant. The quaternion group is defined by the presentation with two generators $a,b$ and just the two relations $a^2=b^2$ and $abab^{-1} = e$. It is a nice exercise to prove that!&lt;/p&gt;&#xA;" OwnerUserId="2604" LastActivityDate="2010-10-18T10:14:07.423" />
  <row Id="7111" PostTypeId="1" AcceptedAnswerId="7134" CreationDate="2010-10-18T10:28:33.173" Score="1" ViewCount="262" Body="&lt;p&gt;I came across this question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $d(n)$ denote the number of divisors of $n$. Let $$\nu(z) = \sum\limits_{n=1}^{\infty} d(n) z^{n}$$ Whats the radius of convergence of this power series. We also have to show that $$\nu(z) = \sum\limits_{n=1}^{\infty} \frac{z^{n}}{1-z^{n}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strike&gt;Regarding the divisor function, we have Dirichlet's formula in hand. But will that help!&lt;/strike&gt;&lt;/p&gt;&#xA;" LastActivityDate="2010-10-18T16:26:45.870" Title="Radius of Convergence of a given Power series" Tags="&lt;number-theory&gt;&lt;complex-analysis&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="7112" PostTypeId="2" ParentId="7111" CreationDate="2010-10-18T11:35:33.597" Score="2" ViewCount="" Body="&lt;p&gt;Write your series as&#xA;$$\sum_{n=1}^\infty \sum_{m=1}^\infty z^{mn},$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and consider $mn=k,$ say. Does this help you?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you get stuck look &lt;a href=&quot;http://en.wikipedia.org/wiki/Lambert_series&quot; rel=&quot;nofollow&quot;&gt;here.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1301" LastEditorUserId="1301" LastEditorDisplayName="" LastEditDate="2010-10-18T11:41:51.807" LastActivityDate="2010-10-18T11:41:51.807" />
  <row Id="7113" PostTypeId="2" ParentId="6941" CreationDate="2010-10-18T11:50:51.630" Score="0" ViewCount="" Body="&lt;p&gt;Well, it turns out it's not really possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The following Processing code shows my results for a 4th order polynomial. The equations of the parameters are from Maple.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;float x1 = 10;&#xA;float y1 = 20;&#xA;float x3 = 90;&#xA;float y3 = 50;&#xA;float d1 = 3;&#xA;float d3 = 1;&#xA;float f = 40;&#xA;&#xA;float x2 = 30;&#xA;float y2 = mouseY;&#xA;&#xA;void draw(){&#xA;background(100);&#xA;   x2 = mouseX;&#xA;   y2 = mouseY;&#xA;&#xA;  noFill();&#xA;  stroke(0, 0, 0);&#xA;  bezier(x1, y1,  x1+f, y1+f*d1, x3-f, y3-f*d3,    x3,y3);&#xA;&#xA;  stroke(255, 102, 0);&#xA;&#xA;  for (float i=x1; i&amp;lt;x3;i++){&#xA;    point(i, approx(i));&#xA;  }&#xA;}&#xA;&#xA;float approx(float time){&#xA;  float a,b,c,d,e;&#xA;&#xA;a = -(8*x1*pow(x3,2)*x2*y1-2*pow(x1,4)*y3+2*pow(x3,4)*y1-pow(x1,2)*d3*pow(x3,3)+2*pow(x1,4)*d3*x3-pow(x1,3)*d3*pow(x3,2)+pow(x1,2)*d1*pow(x3,3)+pow(x1,3)*d1*pow(x3,2)-2*x1*pow(x3,4)*d1-8*x1*pow(x3,2)*x2*y3-8*x3*pow(x1,2)*x2*y3+8*x3*pow(x1,2)*x2*y1-2*pow(x1,3)*x3*x2*d1-2*pow(x3,2)*pow(x1,2)*x2*d1+x3*pow(x1,2)*pow(x2,2)*d1-x3*pow(x1,2)*pow(x2,2)*d3-2*x3*pow(x1,3)*x2*d3+2*x1*pow(x3,3)*x2*d1+x1*pow(x3,2)*pow(x2,2)*d1-x1*pow(x3,2)*pow(x2,2)*d3+2*pow(x1,2)*pow(x3,2)*x2*d3+2*pow(x3,3)*x2*x1*d3+3*pow(x1,3)*pow(x2,2)*d3-2*pow(x1,4)*x2*d3+pow(x1,3)*pow(x2,2)*d1+4*pow(x1,3)*x3*y3+2*pow(x1,4)*y2-2*pow(x3,4)*y2-2*pow(x2,4)*y3+2*pow(x2,4)*y1+4*x1*y3*x3*pow(x2,2)-4*x1*y1*x3*pow(x2,2)+2*pow(x3,4)*x2*d1-3*pow(x3,3)*pow(x2,2)*d1-pow(x3,3)*pow(x2,2)*d3+pow(x2,4)*x3*d3-pow(x2,4)*x1*d3-pow(x2,4)*x1*d1+pow(x2,4)*x3*d1-4*pow(x3,3)*x1*y1+4*pow(x3,3)*x1*y2-4*pow(x3,2)*pow(x1,2)*y1-4*x3*pow(x1,3)*y2-4*pow(x1,2)*pow(x2,2)*y1+4*pow(x1,2)*pow(x2,2)*y3-4*pow(x3,2)*pow(x2,2)*y1+4*pow(x3,2)*pow(x2,2)*y3+4*pow(x3,2)*pow(x1,2)*y3)/((-pow(x3,3)-3*x3*pow(x1,2)+pow(x1,3)+3*x1*pow(x3,2))*(pow(x3,2)*pow(x1,2)-2*x3*pow(x1,2)*x2+pow(x1,2)*pow(x2,2)+4*x1*x3*pow(x2,2)-2*x1*pow(x3,2)*x2-2*x1*pow(x2,3)+pow(x2,4)-2*x3*pow(x2,3)+pow(x3,2)*pow(x2,2)));&#xA;b = (y1*pow(x3,5)-pow(x1,2)*pow(x3,4)*d1+2*pow(x1,3)*pow(x3,3)*d1-2*pow(x1,3)*pow(x3,3)*d3+pow(x1,4)*pow(x3,2)*d3+pow(x1,5)*x3*d3-pow(x1,4)*x3*y3-x1*pow(x3,5)*d1+x1*pow(x3,4)*y1-pow(x1,5)*y3-pow(x3,3)*d3*pow(x2,3)+pow(x3,5)*x2*d1-pow(x3,5)*y2-3*pow(x3,3)*d1*pow(x2,3)+pow(x3,2)*pow(x2,4)*d3-3*y3*x3*pow(x2,4)-3*pow(x1,3)*pow(x3,2)*x2*d1-pow(x1,3)*pow(x3,2)*x2*d3-pow(x1,4)*x3*x2*d3+pow(x1,2)*x3*d1*pow(x2,3)+3*pow(x1,2)*pow(x3,3)*x2*d3+pow(x1,2)*pow(x3,3)*x2*d1-pow(x1,2)*x3*d3*pow(x2,3)+x1*pow(x2,4)*x3*d3-x1*pow(x2,4)*x3*d1+2*pow(x2,4)*pow(x3,2)*d1+pow(x1,5)*y2+3*x3*pow(x2,4)*y1-pow(x1,5)*x2*d3+pow(x1,4)*x3*y2+3*pow(x1,3)*d3*pow(x2,3)+pow(x1,3)*d1*pow(x2,3)-2*pow(x1,2)*pow(x2,4)*d3-pow(x1,2)*pow(x2,4)*d1-3*x1*pow(x2,4)*y3-x1*pow(x3,4)*y2+3*x1*pow(x2,4)*y1+x1*pow(x3,4)*x2*d1-x1*pow(x3,2)*d3*pow(x2,3)+x1*pow(x3,2)*d1*pow(x2,3)+4*pow(x3,2)*y3*pow(x2,3)-12*pow(x1,2)*y3*pow(x3,2)*x2+12*pow(x1,2)*y1*pow(x3,2)*x2+4*x1*x3*pow(x2,3)*y3-4*x1*x3*pow(x2,3)*y1+8*pow(x1,3)*y3*pow(x3,2)-8*pow(x1,3)*pow(x3,2)*y2+4*pow(x1,2)*y3*pow(x2,3)-4*pow(x1,2)*y1*pow(x2,3)-8*pow(x1,2)*pow(x3,3)*y1+8*pow(x1,2)*pow(x3,3)*y2-4*pow(x3,2)*pow(x2,3)*y1)/((-x3+x1)*(pow(x3,2)*pow(x1,4)-2*pow(x1,4)*x3*x2+pow(x1,4)*pow(x2,2)-2*pow(x3,3)*pow(x1,3)+2*pow(x1,3)*x3*pow(x2,2)+2*pow(x1,3)*pow(x3,2)*x2-2*pow(x1,3)*pow(x2,3)+pow(x1,2)*pow(x2,4)+2*pow(x1,2)*x3*pow(x2,3)+pow(x3,4)*pow(x1,2)-6*pow(x1,2)*pow(x2,2)*pow(x3,2)+2*pow(x1,2)*pow(x3,3)*x2+2*x1*pow(x3,3)*pow(x2,2)-2*x1*x3*pow(x2,4)-2*x1*pow(x3,4)*x2+2*x1*pow(x3,2)*pow(x2,3)+pow(x2,4)*pow(x3,2)-2*pow(x3,3)*pow(x2,3)+pow(x3,4)*pow(x2,2)));&#xA;c = -(pow(x1,3)*pow(x3,4)*d1-pow(x1,4)*pow(x3,3)*d3+pow(x1,5)*pow(x3,2)*d3-2*pow(x1,5)*x3*y3-pow(x1,2)*pow(x3,5)*d1+2*pow(x1,4)*d3*pow(x2,3)+2*y1*pow(x3,5)*x1-2*x1*pow(x3,5)*y2+pow(x3,5)*pow(x2,2)*d1-2*d1*pow(x3,4)*pow(x2,3)-2*pow(x1,2)*pow(x2,4)*x3*d1-pow(x1,2)*pow(x2,4)*x3*d3-pow(x1,4)*x3*pow(x2,2)*d3+x1*pow(x3,4)*pow(x2,2)*d1+pow(x3,3)*pow(x1,2)*pow(x2,2)*d1+3*pow(x3,3)*pow(x1,2)*pow(x2,2)*d3-pow(x3,2)*pow(x1,3)*pow(x2,2)*d3-3*pow(x3,2)*pow(x1,3)*pow(x2,2)*d1+2*pow(x3,2)*pow(x2,4)*x1*d3+pow(x3,2)*pow(x2,4)*x1*d1+pow(x3,3)*pow(x2,4)*d1-pow(x1,3)*pow(x2,4)*d3-pow(x1,5)*pow(x2,2)*d3+2*pow(x1,5)*x3*y2-6*x3*x1*pow(x2,4)*y3+6*x3*x1*pow(x2,4)*y1+2*pow(x1,3)*x3*d1*pow(x2,3)+2*pow(x1,3)*x3*d3*pow(x2,3)+2*pow(x1,2)*pow(x3,2)*d1*pow(x2,3)-2*pow(x1,2)*pow(x3,2)*d3*pow(x2,3)-2*x1*pow(x3,3)*d3*pow(x2,3)-2*x1*pow(x3,3)*d1*pow(x2,3)+8*y3*pow(x2,3)*x3*pow(x1,2)+8*y3*pow(x2,3)*x1*pow(x3,2)-8*y1*pow(x2,3)*x3*pow(x1,2)-8*y1*pow(x2,3)*x1*pow(x3,2)-12*pow(x1,2)*pow(x3,2)*pow(x2,2)*y3+12*pow(x1,2)*pow(x3,2)*pow(x2,2)*y1+4*pow(x1,4)*pow(x3,2)*y3-4*pow(x1,4)*pow(x3,2)*y2-4*pow(x1,2)*pow(x3,4)*y1+4*pow(x1,2)*pow(x3,4)*y2)/((-x3+x1)*(pow(x3,2)*pow(x1,4)-2*pow(x1,4)*x3*x2+pow(x1,4)*pow(x2,2)-2*pow(x3,3)*pow(x1,3)+2*pow(x1,3)*x3*pow(x2,2)+2*pow(x1,3)*pow(x3,2)*x2-2*pow(x1,3)*pow(x2,3)+pow(x1,2)*pow(x2,4)+2*pow(x1,2)*x3*pow(x2,3)+pow(x3,4)*pow(x1,2)-6*pow(x1,2)*pow(x2,2)*pow(x3,2)+2*pow(x1,2)*pow(x3,3)*x2+2*x1*pow(x3,3)*pow(x2,2)-2*x1*x3*pow(x2,4)-2*x1*pow(x3,4)*x2+2*x1*pow(x3,2)*pow(x2,3)+pow(x2,4)*pow(x3,2)-2*pow(x3,3)*pow(x2,3)+pow(x3,4)*pow(x2,2)));&#xA;d = (3*x1*pow(x3,2)*pow(x2,4)*y1+2*x3*d3*pow(x1,4)*pow(x2,3)-pow(x3,5)*pow(x1,2)*x2*d1+pow(x3,4)*pow(x1,3)*x2*d1+pow(x3,4)*pow(x1,2)*pow(x2,2)*d1+pow(x3,5)*x1*pow(x2,2)*d1+2*pow(x3,3)*pow(x1,3)*pow(x2,2)*d3-pow(x3,3)*pow(x1,4)*x2*d3-2*pow(x3,3)*pow(x1,3)*pow(x2,2)*d1+pow(x3,3)*pow(x2,4)*x1*d1-pow(x3,2)*pow(x1,4)*pow(x2,2)*d3+pow(x3,2)*pow(x1,5)*x2*d3-x3*pow(x1,5)*pow(x2,2)*d3-3*x3*pow(x1,2)*pow(x2,4)*y3+pow(x3,2)*pow(x1,2)*pow(x2,4)*d3-x3*pow(x1,3)*pow(x2,4)*d3-pow(x3,2)*pow(x1,2)*pow(x2,4)*d1-3*pow(x3,3)*pow(x1,4)*y2-pow(x3,3)*pow(x2,4)*y1-pow(x3,5)*pow(x1,2)*y2+3*pow(x1,3)*pow(x3,4)*y2+pow(x1,3)*pow(x2,4)*y3+pow(x1,5)*pow(x3,2)*y2-2*y3*pow(x1,4)*pow(x2,3)+y3*pow(x1,5)*pow(x2,2)+2*y1*pow(x3,4)*pow(x2,3)-y1*pow(x3,5)*pow(x2,2)-2*y3*pow(x1,5)*x3*x2+y3*x3*pow(x1,4)*pow(x2,2)-2*x1*d1*pow(x3,4)*pow(x2,3)-y1*x1*pow(x3,4)*pow(x2,2)+2*y1*pow(x3,5)*x2*x1+pow(x1,3)*pow(x3,2)*d1*pow(x2,3)-pow(x1,3)*pow(x3,2)*d3*pow(x2,3)+pow(x1,2)*pow(x3,3)*d1*pow(x2,3)-pow(x1,2)*pow(x3,3)*d3*pow(x2,3)+4*pow(x1,4)*pow(x3,2)*y3*x2-4*pow(x1,2)*pow(x3,4)*y1*x2-4*pow(x1,2)*pow(x3,2)*pow(x2,3)*y1+4*pow(x1,2)*pow(x3,2)*y3*pow(x2,3)-4*pow(x3,3)*x1*pow(x2,3)*y1+8*pow(x3,3)*pow(x1,2)*pow(x2,2)*y1+4*pow(x1,3)*x3*pow(x2,3)*y3-8*pow(x1,3)*pow(x3,2)*pow(x2,2)*y3)/((-x3+x1)*(pow(x3,2)*pow(x1,4)-2*pow(x1,4)*x3*x2+pow(x1,4)*pow(x2,2)-2*pow(x3,3)*pow(x1,3)+2*pow(x1,3)*x3*pow(x2,2)+2*pow(x1,3)*pow(x3,2)*x2-2*pow(x1,3)*pow(x2,3)+pow(x1,2)*pow(x2,4)+2*pow(x1,2)*x3*pow(x2,3)+pow(x3,4)*pow(x1,2)-6*pow(x1,2)*pow(x2,2)*pow(x3,2)+2*pow(x1,2)*pow(x3,3)*x2+2*x1*pow(x3,3)*pow(x2,2)-2*x1*x3*pow(x2,4)-2*x1*pow(x3,4)*x2+2*x1*pow(x3,2)*pow(x2,3)+pow(x2,4)*pow(x3,2)-2*pow(x3,3)*pow(x2,3)+pow(x3,4)*pow(x2,2)));&#xA;e = (pow(x3,3)*y1-pow(x1,3)*y3-x1*pow(x3,3)*d1+pow(x1,2)*pow(x3,2)*d1-3*pow(x3,2)*y1*x1-pow(x3,2)*d3*pow(x1,2)+pow(x1,3)*x3*d3+3*x3*y3*pow(x1,2)+x3*d3*pow(x2,3)+pow(x3,3)*x2*d1-2*pow(x3,2)*pow(x2,2)*d1-pow(x3,2)*pow(x2,2)*d3+3*y3*x3*pow(x2,2)-3*y1*x3*pow(x2,2)+x3*d1*pow(x2,3)-pow(x3,3)*y2-2*y3*pow(x2,3)+2*y1*pow(x2,3)-2*pow(x1,2)*x3*x2*d1-pow(x1,2)*x3*x2*d3+x1*pow(x3,2)*x2*d1+2*x1*pow(x3,2)*x2*d3-x1*x3*pow(x2,2)*d3+6*x1*y1*x3*x2-6*x1*y3*x3*x2+x1*x3*pow(x2,2)*d1-x1*d1*pow(x2,3)+3*x1*y3*pow(x2,2)-3*x1*y1*pow(x2,2)+3*x1*pow(x3,2)*y2-pow(x1,3)*x2*d3+2*pow(x1,2)*pow(x2,2)*d3+pow(x1,2)*pow(x2,2)*d1-3*pow(x1,2)*x3*y2-x1*d3*pow(x2,3)+pow(x1,3)*y2)/(-pow(x2,4)*pow(x3,3)+pow(x1,5)*pow(x3,2)-3*pow(x1,4)*pow(x3,3)-pow(x3,5)*pow(x2,2)-pow(x1,2)*pow(x3,5)+3*pow(x1,3)*pow(x3,4)+pow(x2,4)*pow(x1,3)+pow(x1,5)*pow(x2,2)+2*pow(x3,4)*pow(x2,3)-2*pow(x1,4)*pow(x2,3)+3*pow(x2,4)*x1*pow(x3,2)-8*pow(x1,3)*pow(x3,2)*pow(x2,2)+2*pow(x3,5)*x2*x1-3*pow(x2,4)*x3*pow(x1,2)-x1*pow(x3,4)*pow(x2,2)+x3*pow(x1,4)*pow(x2,2)+4*pow(x1,4)*pow(x3,2)*x2-4*pow(x1,2)*pow(x3,4)*x2+8*pow(x3,3)*pow(x1,2)*pow(x2,2)-2*pow(x1,5)*x3*x2+4*pow(x1,3)*x3*pow(x2,3)-4*pow(x3,3)*x1*pow(x2,3));&#xA;  return &#xA;  e*time*time*time*time+&#xA;  a*time*time*time+&#xA;  b*time*time+&#xA;  c*time+&#xA;  d;   &#xA;}&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I'll try with adding the derivative of x2 and adding another point in the middle (with a 5rth order polynomial) but I'm not sure it'll appropriate for my use.&lt;/p&gt;&#xA;" OwnerUserId="2561" LastActivityDate="2010-10-18T11:50:51.630" CommentCount="2" />
  <row Id="7114" PostTypeId="2" ParentId="2777" CreationDate="2010-10-18T12:25:21.527" Score="1" ViewCount="" Body="&lt;p&gt;There isn't any simple approximation of the sort you seek. Simply defining an angle $\theta_{k\ell}$ such that $\cos(\theta_{k\ell}) = \tfrac12 \bigl( \cos(\tfrac{2\pi k}n) + \cos(\tfrac{2\pi \ell}n) \bigr)$ is not sufficient. There are infinitely many different angles $\theta_{k\ell}$ which satisfy this equation, after all, which are very different from one another. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what you want is to obtain &lt;em&gt;some single&lt;/em&gt; value $\theta_{k\ell}$ which you can approximate well using rational multiples of a full rotation, as you describe. This will require mor than just a definition in terms of a periodic function such as cosine.&lt;/p&gt;&#xA;" OwnerUserId="439" LastEditorUserId="439" LastEditorDisplayName="" LastEditDate="2010-10-18T14:36:40.240" LastActivityDate="2010-10-18T14:36:40.240" />
  <row Id="7115" PostTypeId="1" AcceptedAnswerId="7117" CreationDate="2010-10-18T12:32:45.877" Score="3" ViewCount="128" Body="&lt;p&gt;Am I correct to say:&#xA;A difference between an integral and a function is: that an integral can describe an area on a graph, while a function can't?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or am I completely off course here?&lt;/p&gt;&#xA;" OwnerUserId="2605" LastEditorUserId="2605" LastEditorDisplayName="" LastEditDate="2010-10-18T12:52:21.843" LastActivityDate="2010-10-18T13:04:57.277" Title="Can a function describe an area on a graph like an integral?" Tags="&lt;calculus&gt;&lt;intuition&gt;" AnswerCount="2" CommentCount="0" />
  <row Id="7116" PostTypeId="2" ParentId="7109" CreationDate="2010-10-18T12:53:13.837" Score="4" ViewCount="" Body="&lt;p&gt;Your value is basically the value of the Riemann zeta function:&#xA;$$&#xA;S = \sum_{k=1}^{\infty} \frac{Re (e^{i \theta \log(k)})}{k^a} = Re( \sum_{k=1}^{\infty} k^{i \theta - a} ) = Re( \zeta ( a - i \theta ))&#xA;$$&#xA;You want to evaluate this on the critical strip $0 &amp;lt; a &amp;lt; 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The good news is that there is an enormous amount of literatue on the Riemann zeta. The bad news is that this function is nasty on the critical strip.&lt;/p&gt;&#xA;" OwnerUserId="960" LastActivityDate="2010-10-18T12:53:13.837" CommentCount="4" />
  <row Id="7117" PostTypeId="2" ParentId="7115" CreationDate="2010-10-18T13:00:21.640" Score="2" ViewCount="" Body="&lt;p&gt;You can view the integral as a function too, assuming it is defined. For instance, if you fix an interval $I=[a,b]$, and consider all functions that are integrable on $I$, you can view the integral as a function that maps each of these functions to a real number, that represents the 'area under the curve'.&#xA;Or, if you fix a function $f$ on the same interval, you can view the integral from $a$ to $x$ (where $a\leq x\leq b$) as a new function of $x$, defined on the interval, and the value at each $x$ would represent the area under the curve up to $x$.&lt;/p&gt;&#xA;" OwnerUserId="1313" LastActivityDate="2010-10-18T13:00:21.640" />
  <row Id="7118" PostTypeId="2" ParentId="7115" CreationDate="2010-10-18T13:04:57.277" Score="5" ViewCount="" Body="&lt;p&gt;A function $f : A \rightarrow B$ is just an object which, given any $a \in A$ then $f(a) \in B$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The graph of a function $\{(a,f(a))|a\in A\}$ in the case of $f : \mathbb R \rightarrow \mathbb R$ can be viewed as a curve but that's just an illustration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An anti-derivative, say $F(t) = \int_0^{t} f(x) \mathrm{d}x$ is just another function $F : \mathbb R \rightarrow \mathbb R$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While it is true that say, $F(3)$ gives the area under the curve $f$ defines between $0$ and $3$ on the x-axis.. but this idea of 'area' is just an intuitive guide to help motivate the definitions of integral and similar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The integral $\int_{a}^{b}$ could be thought of as a map that takes $(\mathbb R \rightarrow \mathbb R) \rightarrow \mathbb R$ but that is not usual and it can also just been seen as a notation which only has meaning when completed (given an integrand).&lt;/p&gt;&#xA;" LastActivityDate="2010-10-18T13:04:57.277" />
  <row Id="7120" PostTypeId="2" ParentId="7108" CreationDate="2010-10-18T13:31:13.553" Score="1" ViewCount="" Body="&lt;p&gt;Wikipedia &lt;a href=&quot;http://en.wikipedia.org/wiki/Torsion_of_a_curve#Properties&quot; rel=&quot;nofollow&quot;&gt;states&lt;/a&gt; that &quot;if the torsion of a regular curve is identically zero then this curve belongs to a fixed plane.&quot; By &quot;regular curve&quot; I expect they mean that the curve's first and second derivatives are never zero. I imagine that a planar curve connected via a linear segment to another planar curve lying in a different plane would still have zero torsion everywhere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the limit of my knowledge; I'm posting it as an answer so others can see it and point out if there are any errors.&lt;/p&gt;&#xA;" OwnerUserId="856" LastActivityDate="2010-10-18T13:31:13.553" CommentCount="3" />
  <row Id="7121" PostTypeId="2" ParentId="7108" CreationDate="2010-10-18T13:42:06.937" Score="3" ViewCount="" Body="&lt;p&gt;As has been said, the curve is planar iff the torsion is zero. This is clear if you look at the &lt;a href=&quot;http://en.wikipedia.org/wiki/Frenet%E2%80%93Serret_formulas&quot; rel=&quot;nofollow&quot;&gt;Frenet formulas&lt;/a&gt;. For fun, you can get from those a formula that the curve $\alpha: \mathbb{R} \rightarrow \mathbb{R}^3$ must satisfy. We need&#xA;$$&#xA;\frac{d\mathbf{N}}{ds} + \kappa\mathbf{T} = 0.&#xA;$$&#xA;Assuming $\alpha$ is parameterized by arc length, this could be written&#xA;$$&#xA;\frac{d}{ds}\left(\frac{1}{\kappa}\frac{d^2\alpha}{ds^2}\right) + \kappa\frac{d\alpha}{ds} = 0&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="1728" LastActivityDate="2010-10-18T13:42:06.937" CommentCount="2" />
  <row Id="7122" PostTypeId="1" AcceptedAnswerId="7124" CreationDate="2010-10-18T13:54:15.243" Score="2" ViewCount="188" Body="&lt;p&gt;Is there a more efficient algorithm besides &lt;a href=&quot;http://en.wikipedia.org/wiki/Gram-Schmidt_process#Algorithm&quot; rel=&quot;nofollow&quot;&gt;Gram-Schmidt&lt;/a&gt; that would produce an orthonormal matrix of rank N, with first row equal to [1 1 1 1 1 ... 1] / sqrt(N)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;e.g. for N = 3, the matrix $\begin{align}&#xA;\mathsf A_3 &amp;amp;= \begin{bmatrix}&#xA;  1/\sqrt 3 &amp;amp; 1/\sqrt 3 &amp;amp; 1/\sqrt 3 \\&#xA;  2/\sqrt 6 &amp;amp; -1/\sqrt 6 &amp;amp; -1/\sqrt 6 \\&#xA;  0 &amp;amp; 1/\sqrt 2 &amp;amp; -1/\sqrt 2&#xA;\end{bmatrix}\end{align}$ suffices, but I'm not sure how to generalize.&lt;/p&gt;&#xA;" OwnerUserId="120" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-10-18T14:09:23.647" LastActivityDate="2010-10-18T21:26:16.480" Title="determining orthonormal matrix of rank N with special first row" Tags="&lt;matrices&gt;&lt;algorithms&gt;&lt;orthonormal&gt;" AnswerCount="5" CommentCount="4" />
  <row Id="7123" PostTypeId="2" ParentId="7122" CreationDate="2010-10-18T14:04:52.967" Score="1" ViewCount="" Body="&lt;p&gt;Given the special form, you can probably construct a rotation matrix that would turn one of the standard basis vectors there.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-10-18T14:04:52.967" CommentCount="2" />
  <row Id="7124" PostTypeId="2" ParentId="7122" CreationDate="2010-10-18T14:26:37.320" Score="6" ViewCount="" Body="&lt;p&gt;There's a reflection swapping $v=(\sqrt{n},0,\ldots,0)$&#xA;with $w=(1,1,\ldots,1)$ which will do what you want.&#xA;It will fix $v+w$ and all vectors orthogonal to $v$ and $w$.&#xA;It will negate $v-w$.&#xA;It will be a symmetric matrix; its first row and column&#xA;is all $1/\sqrt{n}$. The remaining entries $a_{i,j}$ for $i$, $j\ge2$&#xA;will depend only on whether $i=j$ or not. This should be enough&#xA;information to reconstruct it.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-10-18T14:26:37.320" CommentCount="5" />
  <row Id="7125" PostTypeId="1" AcceptedAnswerId="7129" CreationDate="2010-10-18T14:41:52.890" Score="2" ViewCount="118" Body="&lt;p&gt;Hello, I am a new user!&lt;br&gt;&#xA;I don't understand what is &lt;em&gt;polynomial fitting&lt;/em&gt;.&lt;br&gt;&#xA;Can anyone explain me how to fit a curve to given points?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDITED&lt;/p&gt;&#xA;" OwnerUserId="2582" LastEditorUserId="2582" LastEditorDisplayName="" LastEditDate="2010-10-18T16:22:44.563" LastActivityDate="2010-10-18T16:22:44.563" Title="Polynomial fitting - how to fit and what is _polynomial fitting_" Tags="&lt;linear-algebra&gt;&lt;polynomials&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="7126" PostTypeId="1" AcceptedAnswerId="7133" CreationDate="2010-10-18T14:48:14.133" Score="2" ViewCount="215" Body="&lt;p&gt;Here's the Karnaugh map:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://imgur.com/NpaAB.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer I should be getting from the Karnaugh should be:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;T = R ∙ (CGM)'&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I'm really not seeing how this was arrived at through any simplification methods I've learned thus far.  I can see the answers that are intended are correct (I think), though.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From what I know, the best answer I can come up with to simplify (only) the Kargnaugh map is:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;T=RCG'+RCM'+RC'&#xA;  to:&#xA;T=R∙(CG'+CM'+C')&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Help appreciated!&lt;/p&gt;&#xA;" OwnerUserId="2609" LastEditorUserId="2609" LastEditorDisplayName="" LastEditDate="2010-10-19T22:09:58.167" LastActivityDate="2010-10-19T22:09:58.167" Title="Boolean Algebra, Simplification: Don't know the method used" Tags="&lt;boolean-algebra&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="7127" PostTypeId="2" ParentId="7122" CreationDate="2010-10-18T14:51:09.650" Score="1" ViewCount="" Body="&lt;p&gt;You should be able to use K-1 rows of Normalized &lt;a href=&quot;http://en.wikipedia.org/wiki/Hadamard_matrix&quot; rel=&quot;nofollow&quot;&gt;Hadamard&lt;/a&gt; matrices of size K, where K is a power of 2 as building blocks. (How to normalize see this: &lt;a href=&quot;http://math.stackexchange.com/questions/4797/what-is-sum-of-rows-of-hadamard-matrix/4798#4798&quot;&gt;http://math.stackexchange.com/questions/4797/what-is-sum-of-rows-of-hadamard-matrix/4798#4798&lt;/a&gt;) You should be able to construct these rows recursively as K is a power of 2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Something like&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;[1 1 ..       1 ]&#xA;[ H_1 | 0  |0 0 ]&#xA;[ 0   | H_2|0 0 ]&#xA;[ 0   | 0  | H_4]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;etc&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the end O(LogN) rows will remain to be filled, which I suppose you can use Gram-Schmidt on.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-10-18T14:51:09.650" />
  <row Id="7128" PostTypeId="2" ParentId="7107" CreationDate="2010-10-18T14:53:48.223" Score="3" ViewCount="" Body="&lt;p&gt;You can make the second terms in both equations vanish by multiplying the first by $A(T)$, the second by $B(T)$, and subtracting.  The resulting equation is readily solved for the product $A(T)B(T)$, reducing the system to a single ODE which is directly integrable.&lt;/p&gt;&#xA;" OwnerUserId="1489" LastActivityDate="2010-10-18T14:53:48.223" CommentCount="2" />
  <row Id="7129" PostTypeId="2" ParentId="7125" CreationDate="2010-10-18T14:59:21.157" Score="8" ViewCount="" Body="&lt;p&gt;There are two concepts frequently conflated: interpolation and fitting. I'll discuss both since you don't seem to know what you really want.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's say you have a set of $n$ points&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(x_i,y_i)\qquad i=1\dots n$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Interpolation is the problem of finding the (unique) polynomial $p(x)$ that passes through all your given points (under the assumption that no two points have the same abscissa), i.e. $p(x_i)=y_i \qquad i=1\dots n$. Your resulting polynomial is of degree $n-1$. The usual techniques for finding the interpolating polynomial are the methods of Lagrange, Newton, and Neville-Aitken.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fitting on the other hand assumes your data is contaminated with error, and you want the polynomial that is the &quot;best approximation&quot; to your data. Here polynomial interpolation does not make much sense since you do not want your function to be reproducing the inherent errors in your data as well. Least-squares is a common technique: it finds the polynomial $f(x)$ such that the quantity&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{j=1}^{n}\left(f(x_i)-y_i\right)^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which measures the departure of your polynomial from the ordinates is minimized (here the assumption is that your abscissas are error-free, and the error in your ordinates is normally distributed). The degree of $f(x)$ can be (and is often) less than $n$. A number of techniques for this are used as well: there's the normal equations, and then there are special matrix decompositions that can be used to efficiently solve this problem.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-10-18T14:59:21.157" CommentCount="3" />
  <row Id="7130" PostTypeId="1" AcceptedAnswerId="7132" CreationDate="2010-10-18T15:05:38.490" Score="2" ViewCount="472" Body="&lt;p&gt;I'm looking for an explanation on how reducing the Hamiltonian cycle problem to the Hamiltonian path's one (to proof that also the latter is NP-complete). I couldn't find any on the web, can someone help me here? (linking a source is also good).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you.&lt;/p&gt;&#xA;" OwnerUserId="2610" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-10-18T15:37:41.550" LastActivityDate="2010-10-19T03:46:36.883" Title="Reduction from Hamiltonian cycle to Hamiltonian path" Tags="&lt;graph-theory&gt;&lt;computer-science&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="7131" PostTypeId="2" ParentId="7122" CreationDate="2010-10-18T15:08:37.963" Score="0" ViewCount="" Body="&lt;p&gt;hmm, I think one possibility for orthonormal vectors for N odd is&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;[ 1 1 1 1 1 ... 1 ]&#xA;[ 1, cos phi, cos 2*phi, cos 3*phi, ... cos (N-1)*phi ]&#xA;[ 0, sin phi, sin 2*phi, sin 3*phi, ... sin (N-1)*phi ]&#xA;[ 1, cos 2*phi, cos 4*phi, cos 6*phi, ... cos 2*(N-1)*phi ]&#xA;[ 0, sin 2*phi, sin 4*phi, sin 6*phi, ... sin 2*(N-1)*phi ]&#xA;   ...&#xA;[ 1, cos k*phi, cos 2*k*phi, cos 3*k*phi, ... cos k*(N-1)*phi ]&#xA;[ 0, sin k*phi, sin 2*k*phi, sin 3*k*phi, ... sin k*(N-1)*phi ]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;where k = (N-1)/2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but I'm not sure how to extend to N even.&lt;/p&gt;&#xA;" OwnerUserId="120" LastActivityDate="2010-10-18T15:08:37.963" />
  <row Id="7132" PostTypeId="2" ParentId="7130" CreationDate="2010-10-18T15:37:17.523" Score="1" ViewCount="" Body="&lt;p&gt;From Wikipedia's article on &lt;a href=&quot;http://en.wikipedia.org/wiki/Hamiltonian_path_problem&quot; rel=&quot;nofollow&quot;&gt;the Hamiltonian path problem&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;The Hamiltonian path problem for graph G is equivalent to the Hamiltonian cycle problem in a graph H obtained from G by adding a new vertex and connecting it to all vertices of G.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(EDIT: This answer has been accepted but is incorrect.  Please see Moron's answer instead.)&lt;/p&gt;&#xA;" OwnerUserId="2370" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-10-19T03:46:36.883" LastActivityDate="2010-10-19T03:46:36.883" CommentCount="4" />
  <row Id="7133" PostTypeId="2" ParentId="7126" CreationDate="2010-10-18T16:19:43.280" Score="2" ViewCount="" Body="&lt;p&gt;Hi Hamster. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using sum of products you should be able to derive:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;RC' + RCG' + RCGM' &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Substitute out the R:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;R(C' + CG' + CGM')&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Use the identity A + A'B = A + B&lt;/p&gt;&#xA;&#xA;&lt;p&gt;R(C' + G' + CGM')&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That same identity works as a multivariable expression&lt;/p&gt;&#xA;&#xA;&lt;p&gt;R(C' + G' + M')&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then apply DeMorgans&lt;/p&gt;&#xA;&#xA;&lt;p&gt;R(CGM)'&lt;/p&gt;&#xA;" OwnerUserId="592" LastActivityDate="2010-10-18T16:19:43.280" CommentCount="8" />
  <row Id="7134" PostTypeId="2" ParentId="7111" CreationDate="2010-10-18T16:20:34.863" Score="1" ViewCount="" Body="&lt;p&gt;Elaborating upon Robin's hint about the radius of convergence, we have that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle 1 \le d(n)^{1/n} \le n^{1/n}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $\displaystyle n^{1/n} \to 1$, the radius of convergence of the power series is $\displaystyle 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason I posted this is because I wanted to mention this quick proof of $\displaystyle  n^{1/n} \to 1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $\displaystyle n &gt; 3$ then we have that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle \frac{1 + 1 + \dots + 1 + \sqrt{n} + \sqrt{n}}{n} \ge \sqrt[n]{n}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;using $AM \ge GM$ on $\displaystyle n-2 \ 1$s and two $\sqrt{n}$s.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus&#xA;$\displaystyle \frac{n-2 + 2\sqrt{n}}{n} \ge n^{1/n} \ge 1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i.e&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle 1- \frac{2}{n} + \frac{2}{\sqrt{n}} \ge n^{1/n} \ge 1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and so $\displaystyle n^{1/n} \to 1$.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-10-18T16:26:45.870" LastActivityDate="2010-10-18T16:26:45.870" CommentCount="2" />
  <row Id="7135" PostTypeId="2" ParentId="7013" CreationDate="2010-10-18T16:56:06.660" Score="9" ViewCount="" Body="&lt;p&gt;Here's a sketch of how we can evaluate the integral without complex analysis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\text{Let} \quad I(a) = \int_0^\infty \frac{x^{a-1}}{1-x} dx.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Split the range into two intervals $(0,1)$ and $(1,\infty)$ then use the substitution $x=1/t$ in the latter part to obtain&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$I(a) = \int_0^1 \frac{x^{a-1}-x^{-a}}{1-x} dx.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Expand the integrand as a power series using $(1-x)^{-1} = \sum_{n=0}^\infty x^n$ and integrate to obtain&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$I(a) = \frac{1}{a} + \sum_{n=1}^\infty \left( \frac{1}{a+n} + \frac{1}{a-n} \right).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, by differentiating logarithmically the product formula for $\sin x,$&#xA;$$\sin \pi x = \pi x \prod_{n=1}^{\infty} \left( 1 - \frac{x^2}{n^2} \right),$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;we note that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\pi \cot \pi x = \frac{1}{x} + \sum_{n=1}^\infty \left( \frac{1}{x+n} + \frac{1}{x-n} \right).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus $$I(a) = \pi \cot(\pi a)$$ and the result follows since the integral in question is $I(a)-I(b).$&lt;/p&gt;&#xA;" OwnerUserId="1301" LastEditorUserId="1301" LastEditorDisplayName="" LastEditDate="2010-10-18T17:05:47.713" LastActivityDate="2010-10-18T17:05:47.713" CommentCount="2" />
  <row Id="7136" PostTypeId="1" AcceptedAnswerId="7139" CreationDate="2010-10-18T17:16:10.253" Score="2" ViewCount="287" Body="&lt;p&gt;An Exercise from Apostol which i am not able to solve.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $\mathsf{S_{n}}$ denote the $n$-th partial sum of the series: $$\sum\limits_{r=1}^{\infty} \frac{1}{r(r+1)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Prove that for every integer $k&gt;1$, there exists integers $m$ and $n$, such that $$s_{m}-s_{n} =\frac{1}{k}$$&lt;/p&gt;&#xA;" LastActivityDate="2010-10-19T19:57:40.087" Title="Partial sum of a given series" Tags="&lt;real-analysis&gt;&lt;analysis&gt;&lt;sequences-and-series&gt;" AnswerCount="4" CommentCount="5" />
  <row Id="7137" PostTypeId="2" ParentId="7136" CreationDate="2010-10-18T17:24:07.467" Score="1" ViewCount="" Body="&lt;p&gt;First prove that $S_n=1-\frac{1}{n+1}$ by induction.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2010-10-18T17:30:59.753" LastActivityDate="2010-10-18T17:30:59.753" />
  <row Id="7139" PostTypeId="2" ParentId="7136" CreationDate="2010-10-18T17:40:10.307" Score="2" ViewCount="" Body="&lt;p&gt;When you've worked out your formula for $s_m-s_n,$ consider $m=(n+1)^2+n.$&lt;/p&gt;&#xA;" OwnerUserId="1301" LastActivityDate="2010-10-18T17:40:10.307" />
  <row Id="7140" PostTypeId="2" ParentId="7136" CreationDate="2010-10-18T17:52:36.030" Score="2" ViewCount="" Body="&lt;p&gt;(EDIT: My original post took $p$ to be the smallest prime that divides $k$.  This is unnecessary.  You can take $p$ to be &lt;em&gt;any&lt;/em&gt; number other than 1 that divides $k$.  Doing so gives you not just one solution but $d(k)-1$ solutions, one for each of the divisors of $k$ other than 1.)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(SECOND EDIT: See the answer to &lt;a href=&quot;http://math.stackexchange.com/questions/7173/&quot;&gt;this question&lt;/a&gt; for all solutions in $m$ and $n$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First,&#xA;$$\frac{1}{r(r+1)} = \frac{1}{r} - \frac{1}{r+1}.$$&#xA;Then $$s_n = \sum_{r=1}^n \left(\frac{1}{r} - \frac{1}{r+1}\right) = 1 - \frac{1}{n+1}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, for any $m,n \geq 1$, $$s_m - s_n = \frac{1}{n+1} - \frac{1}{m+1} = \frac{m-n}{(m+1)(n+1)}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We want to find $m,n$ such that this last expression is equal to $\frac{1}{k}$.  Let $p$ be any positive integer other than 1 that divides $k$.  Take $m = (p-1)k-1$.  Clearly $m$ is a positive integer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then we want to show that the resulting $n$ that solves $$\frac{m-n}{(m+1)(n+1)} = \frac{1}{k}$$ is also an integer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{m-n}{(m+1)(n+1)} = \frac{1}{k} \Rightarrow (m-n)k = (m+1)(n+1) $$&#xA;$$\Rightarrow ((p-1)k-1 - n)k = (p-1)k(n+1) \Rightarrow (p-1)k-1 - n = (p-1)n + p-1 $$&#xA;$$\Rightarrow pn = (p-1)k - p \Rightarrow n = \frac{(p-1)k}{p} - 1,$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which means that $n$ is an integer because $p|k$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus we have a family of solutions $$m = (p-1)k-1, n = \frac{(p-1)k}{p} - 1,$$&#xA;where $p$ is any positive integer other than 1 that divides $k$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This works even in the case $k = 2$ because then we just get $m = 1$, $n = 0$.&lt;/p&gt;&#xA;" OwnerUserId="2370" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-10-19T19:57:40.087" LastActivityDate="2010-10-19T19:57:40.087" CommentCount="1" />
  <row Id="7141" PostTypeId="2" ParentId="7130" CreationDate="2010-10-18T18:16:19.550" Score="5" ViewCount="" Body="&lt;p&gt;I don't think Mike's answer is correct. That gives the reduction from Hamiltonian Path &lt;em&gt;to&lt;/em&gt; cycle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a reduction from Hamiltonian Cycle to Path.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given a graph $G$ of which we need to find Hamiltonian Cycle, for a single edge $e = \{u,v\}$ add new vertices $u'$ and $v'$ such that $u'$ is connected only to $u$ and $v'$ is connected only to $v$ to give a new graph $G_e$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$G_e$ has a hamiltonian path if and only if $G$ has a hamiltonian cycle with the edge $e=\{u,v\}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Run the Hamiltonian path algorithm on each $G_e$ for each edge $e \in G$. If all graphs have no hamiltonian path, then $G$ has no hamiltonian cycle. If at least one $G_e$ has a hamiltonian path, then $G$ has a hamiltonian cycle which contains the edge $e$.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-10-18T18:16:19.550" CommentCount="0" />
  <row Id="7142" PostTypeId="1" CreationDate="2010-10-18T18:57:40.590" Score="8" ViewCount="202" Body="&lt;p&gt;I have a problem of the following form:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;minimize $\|Dx\|_2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;subject to  $\|x*x\|_2 = 1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $x\in\mathbb R^n$, $D$ is a given diagonal matrix of positive entries, and $*$ represents convolution, i.e. $(x*x)_n = \sum_{i+j=n}x_ix_j$ and $x*x\in\mathbb R^{2n-1}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What approach could be used in dealing with this problem numerically? Could this problem be converted to one of the known problem classes that have available solvers?&lt;/p&gt;&#xA;" OwnerUserId="145" LastActivityDate="2011-06-02T13:09:09.627" Title="A numerical optimization problem with a convolution in the constraint" Tags="&lt;optimization&gt;&lt;nonlinear-optimization&gt;&lt;numerical-optimization&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="1" />
  <row Id="7143" PostTypeId="1" AcceptedAnswerId="7147" CreationDate="2010-10-18T19:01:30.843" Score="13" ViewCount="478" Body="&lt;p&gt;The scatter graph at the &lt;a href=&quot;http://en.wikipedia.org/wiki/Pythagorean_triple#Distribution_of_triples&quot; rel=&quot;nofollow&quot;&gt;Wikipedia article&lt;/a&gt; seems to suggest so. Has anyone attacked this before? Is there a known proof?&lt;/p&gt;&#xA;" OwnerUserId="2612" LastActivityDate="2011-04-05T06:57:58.150" Title="Is it possible to get arbitrarily near any acute angle with Pythagorean triangles?" Tags="&lt;number-theory&gt;" AnswerCount="4" FavoriteCount="2" />
  <row Id="7144" PostTypeId="1" AcceptedAnswerId="7149" CreationDate="2010-10-18T19:05:22.520" Score="3" ViewCount="193" Body="&lt;p&gt;Let $G=\{x,y|x^2=y^3=(xy)^3=1\}$&#xA;I would like to show that G is isomorphic to A_4.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $f:\mathbf{F}_{2} \to G$ be a surjective homomorphism from the free group on two elements to $G$. Let $f$ map $x \to (12)(34)$ and $y \mapsto (123)$. I'm not sure how to show that these elements generate the kernel of $f$. If they do generate the kernel, how do I conclude that the order of $G$ is 12?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once I have that the order of the group is of order 12 then I can show that $G$ contains $V$ (the Klein four group) as a subgroup, or that $A_4$ is generated by the image of $x$ and $y$.&lt;/p&gt;&#xA;" OwnerUserId="2298" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-10-18T19:55:54.920" LastActivityDate="2010-10-19T04:37:55.797" Title="Generators and Relations for A_4" Tags="&lt;group-theory&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="7146" PostTypeId="2" ParentId="7136" CreationDate="2010-10-18T19:14:46.713" Score="0" ViewCount="" Body="&lt;p&gt;By partial fractions it &lt;a href=&quot;http://en.wikipedia.org/wiki/Telescoping_series&quot; rel=&quot;nofollow&quot;&gt;&lt;strong&gt;telescopes&lt;/strong&gt;&lt;/a&gt; to a sum matching the RHS of this well-known &lt;a href=&quot;http://en.wikipedia.org/wiki/Egyptian_fraction&quot; rel=&quot;nofollow&quot;&gt;&lt;strong&gt;Egyptian fraction&lt;/strong&gt;&lt;/a&gt; sum&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\rm \frac{1}k\ =\ \frac{1}{k-1}\ -\ \frac{1}{k\:(k-1)}$$&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-10-18T19:14:46.713" />
  <row Id="7147" PostTypeId="2" ParentId="7143" CreationDate="2010-10-18T19:16:37.730" Score="17" ViewCount="" Body="&lt;p&gt;If you mean a pythagorean triangle having one of the angles arbitrary close to a given acute angle then the answer is yes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice that $\displaystyle (m^2-n^2, 2mn, m^2+n^2)$ is always a pythagorean triangle for any positive integers $\displaystyle m \gt n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have that $\displaystyle \tan \theta = \frac{2mn}{m^2-n^2} = \frac{2t}{1-t^2}$ where $\displaystyle t = \frac{n}{m}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By continuity (and surjectivity) of $\displaystyle \frac{2t}{1-t^2}$ we can find an interval of $\displaystyle t$ (which is infact a subinterval of $\displaystyle (0,1)$)  for which $\displaystyle \frac{2t}{1-t^2}$ is arbitrarily close to any positive number we want.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the rationals are dense, we can find $\displaystyle m,n$ which make the tangent of one the angles arbitrarily close to the tangent of a given acute angle, and hence the angles can be made arbitrarily close.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-12-08T08:05:14.350" LastActivityDate="2010-12-08T08:05:14.350" CommentCount="4" />
  <row Id="7149" PostTypeId="2" ParentId="7144" CreationDate="2010-10-18T20:18:49.573" Score="3" ViewCount="" Body="&lt;p&gt;Abelsh, your definition of $f$ does not make sense: you cannot define the image of $x$ to &lt;strong&gt;be&lt;/strong&gt; $(12)(34)$ unless you already know that $G$ &lt;strong&gt;is&lt;/strong&gt; a group of permutations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instead, what you want to do is define $f$ from $\mathbf{F}_2$ to $A_4$ by defining it as you want: $f(x) = (12)(34)$ and $f(y)=(123)$. If you do this, then it is straightfoward to check that the images satisfy the given relations, $(f(x))^2 = 1$, $(f(y))^3 = 1$, and since $f(x)f(y) = (12)(34)(123) = (243)$ (I'm composing my permutations right to left), then $(f(x)f(y))^3 = 1$. That means that the kernel of $f$ &lt;em&gt;contains&lt;/em&gt; $x^2$, $y^3$, and $(xy)^3$, and therefore, by van Dyck's Theorem, the homomorphism factors through $G$. That means that $G$ has $A_4$ as a quotient, and thus has at least 12 elements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But unless you find a precise description of the kernel, it would be difficult to prove that the kernel is exactly the subgroup generated by $x^2$, $y^3$, and $(xy)^3$. Instead, you might want to find a normal form for elements of $G$ to show that $G$ has &lt;em&gt;at most&lt;/em&gt; 12 elements; if you can do that, you get that $G$ has exactly $12$ elements, and has $A_4$ as a quotient, which gives you what you want.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So... do we have a sufficiently nice normal form for elements of $G$? Every element of $F$ can be written as a word in $x$ and $y$; using the relations in $G$ we can replace any power of $x$ with either $x^0$ or $x$, and any power of $y$ with either $y^0$, $y^1$, or $y^2$. So every element of $G$ can be written as a string that alternates $x$s and either $y$'s or $y^2$'s. And $(xy)^3 = 1$ so $(y^2x)^3=1$. That means that you can replace any string of the form $xyx$ with $y^2xy^2$, and replace $xy^2x$ with $yxy$. Note that in each of these replacements, you go from a string with two $x's$ to a string with just one $x$. That means that you can take any string that alternates $xs$ with $y$s or $y^2$s and use those replacements to get it down to one that has at most one $x$ in it. How many such words are there? That gives an upper bound for the number of distinct elements that $G$ can have. &lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-10-18T20:18:49.573" />
  <row Id="7150" PostTypeId="2" ParentId="6964" CreationDate="2010-10-18T20:41:05.187" Score="1" ViewCount="" Body="&lt;p&gt;I think I got the idea:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lemma: If $T_\varphi$ is an nonconstant analytic Toeplitz operator then the only invariant finite-dimensional subspace for $T_\varphi$ is $\{0\}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proof: If $M$ is a finite dimensional subspace them $T_\varphi$ restricted to $M$ has an non-zero eigenvector (because it is compact). But this is not possible for an the Toeplitz operator on the whole space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So now let $K$ be the compact operator that commutes with $T_\varphi$ and let $\lambda \neq 0$. Define $K_\lambda := \textrm{ker}(K - \lambda)$. Now we see that $K_\lambda$ is an invariant subspace for $T_\varphi$ as follows: $f \in T_\varphi K_\lambda$ then $f = T_\varphi  g$ for some $g \in K_\lambda$, that is $(K - \lambda)g = 0$. Because they commute we have $(K - \lambda) T_\varphi g = (K - \lambda)f = 0$, so $f \in K_\lambda$. Also note that $K_\lambda$ is finite dimensional by the spectral theorem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, since $\varphi$ is nonconstant $K_\lambda = 0$. So $\lambda$ is not in the spectrum of $K$. $\{0\}$ is in the spectrum of $K$ by the spectral theorem for compact operators.&lt;/p&gt;&#xA;" OwnerUserId="1120" LastActivityDate="2010-10-18T20:41:05.187" CommentCount="4" />
  <row Id="7151" PostTypeId="1" AcceptedAnswerId="7156" CreationDate="2010-10-18T20:47:21.373" Score="10" ViewCount="311" Body="&lt;p&gt;What is the least number of comparisons we need to find the median of 6 distinct numbers?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am able to find the answer to the median of 5 distinct numbers to be 6 comparisons, and it makes sense, however in the case of 6 numbers I can't find an answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The best I was able to do it in by hand was 9 comparisons.  Can that be minimized further?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt;  Median in this case, we are assuming to be the lower median.&lt;/p&gt;&#xA;" OwnerUserId="1630" LastEditorUserId="1630" LastEditorDisplayName="" LastEditDate="2010-10-19T08:39:31.533" LastActivityDate="2010-10-19T08:39:31.533" Title="Median of distinct numbers" Tags="&lt;algorithms&gt;&lt;sorting&gt;" AnswerCount="2" FavoriteCount="3" />
  <row Id="7152" PostTypeId="1" AcceptedAnswerId="7157" CreationDate="2010-10-18T20:57:31.250" Score="3" ViewCount="111" Body="&lt;p&gt;I'm familiar with &lt;a href=&quot;http://en.wikipedia.org/wiki/Householder_transformation&quot; rel=&quot;nofollow&quot;&gt;Householder reflections&lt;/a&gt;; they are a simple transformation that, given a normal vector, describes reflection in the hyperplane perpendicular to that vector.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But do Householder reflections describe the complete space of matrices describing reflection about the origin, or are there other reflections that cannot be described by Householder reflections?&lt;/p&gt;&#xA;" OwnerUserId="120" LastEditorUserId="120" LastEditorDisplayName="" LastEditDate="2010-10-18T21:05:09.327" LastActivityDate="2010-10-19T04:21:30.230" Title="do Householder reflections describe all reflections?" Tags="&lt;linear-algebra&gt;&lt;matrices&gt;&lt;reflection&gt;" AnswerCount="2" FavoriteCount="2" />
  <row Id="7153" PostTypeId="2" ParentId="7152" CreationDate="2010-10-18T20:59:47.440" Score="0" ViewCount="" Body="&lt;p&gt;Yes, they are all reflections.&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2010-10-18T20:59:47.440" />
  <row Id="7154" PostTypeId="2" ParentId="7122" CreationDate="2010-10-18T21:26:16.480" Score="1" ViewCount="" Body="&lt;p&gt;I took @Robin Chapman's answer and ran with it:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The matrix A that is the Householder reflection of v-w where v = [sqrt(N), 0, 0, 0, ... ] and w = [1 1 1 1 1 ... 1] can be defined by:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;A(1,i) = A(i,1) = 1/sqrt(N)&#xA;A(i,i) = 1 - K   for i &amp;gt;= 2 &#xA;A(i,j) = -K      for i,j &amp;gt;= 2, i != j&#xA;K = 1/sqrt(N)/(sqrt(N)-1)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="120" LastActivityDate="2010-10-18T21:26:16.480" />
  <row Id="7155" PostTypeId="2" ParentId="7108" CreationDate="2010-10-18T21:31:52.360" Score="1" ViewCount="" Body="&lt;p&gt;Here's an example of an infinitely differentiable regular curve with identically zero torsion but not contained in any plane.  (By &lt;em&gt;regular&lt;/em&gt;, I mean that the velocity never vanishes.)&#xA;$$&#xA;\alpha(t) = &#xA;\left\{&#xA;\begin{aligned}&#xA;&amp;amp;(t,e^{-1/t},0), &amp;amp; t&gt;0,\\&#xA;&amp;amp;(0,0,0), &amp;amp;t=0,\\&#xA;&amp;amp;(t,0,e^{1/t}), &amp;amp;t&amp;lt;0.&#xA;\end{aligned}&#xA;\right.&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: As Mariano Suárez-Alvarez points out in his comment, the torsion is only defined at points where the curvature is nonzero.  Since the curvature of this curve is zero at the origin, the torsion is not defined there.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus the argument given by yasmar shows that if the curve is regular, its curvature is nowhere zero, and its torsion is everywhere zero, then it's a plane curve.&lt;/p&gt;&#xA;" OwnerUserId="1421" LastEditorUserId="1421" LastEditorDisplayName="" LastEditDate="2010-10-18T21:56:49.867" LastActivityDate="2010-10-18T21:56:49.867" CommentCount="6" />
  <row Id="7156" PostTypeId="2" ParentId="7151" CreationDate="2010-10-18T21:53:54.173" Score="8" ViewCount="" Body="&lt;p&gt;It takes at least 8 and can be done in exactly 8.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Lower Bound&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A lower bound to find all the $k$ smallest elements is given here: &lt;a href=&quot;http://en.wikipedia.org/wiki/Selection_algorithm#Lower_bounds&quot; rel=&quot;nofollow&quot;&gt;Lower bounds for Selection&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which says we need at least&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ n - k + \sum_{n+1-k &amp;lt; j \le n} \lceil \log_{2} j \rceil $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $n=6, k=3$, this gives $9$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ 6 - 3 + \lceil \log_{2} 5 \rceil + \lceil \log_{2} 6 \rceil = 9$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus to find the smallest, second smallest and the third smallest, we need at least $9$ compares.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now any algorithm which finds the $3^{rd}$ smallest element (say $z$) will have to find the set $S = \{x, y \}$ such that $x &amp;lt; z$ and $y &amp;lt; z$ (otherwise how does it know that $z$ is the third largest?).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An extra compare between $x$ and $y$ will now give us the smallest, second smallest and third smallest numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus any algorithm which finds the third smallest must use at least 8 comparisons.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Upper Bound&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As I.J. Kennedy notes, there is an algorithm to find the 3rd smallest element of 6 elements in 8 comparisons and can be found in Knuth's Art of Computer Programming Vol 3, Page 212.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to Knuth, this is a result obtained by A. Hadian and M. Sobel [Univ of Minnesota, Dept of Statistics Report 121 (169)] and they show that the $k^{th}$ largest element can be found in&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ n - k + (k-1) \lceil \log_{2} (n+2-k) \rceil $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;comparisons. For $n=6$ and $k=4$ we get the upper bound of $8$.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Algorithm&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The way they do it (Knuth's book has all this information) is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First create a knockout tournament on $n-k+2$ items, which takes $n-k+1$ compares.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The largest item (say $L$) of this cannot be a candidate. So of the remaining $k-2$ pick one and follow the path of $L$ up the tree, which gives the the new largest in at most $\lceil \log_{2} n+2-k \rceil$ compares. Remove this new largest, replacing it with one of the remaining and following up the path of this new largest.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Continue till the rest of the $k-2$ are exhausted. Finally replace the largest with $-\infty$ and determine the largest of the remaining $n+1-k$ which will be the $k^{th}$ largest as we have removed the largest $k-1$ so far.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Related: &lt;a href=&quot;http://compgeom.cs.uiuc.edu/~jeffe/teaching/497/02-selection.pdf&quot; rel=&quot;nofollow&quot;&gt;http://compgeom.cs.uiuc.edu/~jeffe/teaching/497/02-selection.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-10-19T03:35:59.740" LastActivityDate="2010-10-19T03:35:59.740" CommentCount="15" />
  <row Id="7157" PostTypeId="2" ParentId="7152" CreationDate="2010-10-18T22:32:39.720" Score="2" ViewCount="" Body="&lt;p&gt;Mariano's answer should be enough, if you read your link to Wikipedia carefully, but maybe I could add some hints.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to do a reflection about a hyperplane containing the origin, you pick a unitary orthogonal vector to that hyperplane, $v$, and then write the vector $x$ you want to reflect about $[v]^\bot$ as a sum&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;x = \lambda v + u&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with $u \in [v]^\bot$. Now, you want to determine $\lambda \in \mathbb{R}$ in order this to be true:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;v \cdot x = \lambda v\cdot v + v\cdot u = \lambda  \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, necessarily&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;x = (v\cdot x) v + u \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, if you want to reflect $x$ about $[v]^\bot$, you just substract two times its $v$-component:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;P_v x = x - 2 (v\cdot x) v = x - 2 v (v^t x) = (I - 2vv^t ) x \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So any (orthogonal) reflexion about any hyperplane $[v]^\bot$ is of the form&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;P_v = I - 2vv^t \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is, a Housholder reflection.&lt;/p&gt;&#xA;" OwnerUserId="664" LastEditorUserId="664" LastEditorDisplayName="" LastEditDate="2010-10-19T04:21:30.230" LastActivityDate="2010-10-19T04:21:30.230" />
  <row Id="7158" PostTypeId="2" ParentId="7151" CreationDate="2010-10-19T00:06:01.760" Score="5" ViewCount="" Body="&lt;p&gt;It looks like the answer is 8.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My Knuth Volume Three (1970s&amp;mdash;not as dusty as you think) reports an &lt;strong&gt;upper bound&lt;/strong&gt; of 8, which, paired with Moron's &lt;strong&gt;lower bound&lt;/strong&gt; of 8, ...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The general form of this question is called the Selection Problem. If you google that phrase you will get lots of useful results.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;: Knuth doesn't give an explicit algorithm for finding the median of 6 elements in at most 8 steps (at least in the first edition). However, in exercise 12 of section 5.3.3, he does give the explicit method for finding the median of 7 elements using at most 10 comparisons, which may be of some help.&lt;/p&gt;&#xA;" OwnerUserId="130" LastEditorUserId="130" LastEditorDisplayName="" LastEditDate="2010-10-19T03:44:48.620" LastActivityDate="2010-10-19T03:44:48.620" CommentCount="4" />
  <row Id="7159" PostTypeId="1" AcceptedAnswerId="7161" CreationDate="2010-10-19T01:04:35.897" Score="4" ViewCount="163" Body="&lt;p&gt;I'm trying to come up with some math problems (word or otherwise) that get to the meaning of adding zero, but I'm getting stuck because it seems just too simple to me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have come up with questions like &quot;John is having a birthday party and he invites five friends over but nobody shows up. How many people are at the party?&quot;. While this question gets to the problem at hand, it might be too simple to understand the meaning of adding zero. This problem also suffers from the possibility that first graders will answer like so, &quot;Well there would be three people at the party because John's mom and dad would be there too.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any ideas?&lt;/p&gt;&#xA;" OwnerUserId="2616" LastEditorUserId="2215" LastEditorDisplayName="" LastEditDate="2010-10-19T02:52:29.003" LastActivityDate="2010-10-19T02:52:29.003" Title="What are some deep questions that are applicable to first graders in regards to adding zero?" Tags="&lt;teaching&gt;" AnswerCount="2" FavoriteCount="2" />
  <row Id="7160" PostTypeId="2" ParentId="7143" CreationDate="2010-10-19T01:31:49.140" Score="9" ViewCount="" Body="&lt;p&gt;In fact, more is true -- consider the triples $(3,4,5), (-7,24,25), (-117,44,125), (-527,-336,625), \ldots$, where the $n$th triple is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ (a_n, b_n, c_n) := (Re((3+4i)^n), Im((3+4i)^n), 5^n). $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $\tan^{-1} 4/3$ is irrational, the sequence of pairs $(a_n/c_n,b_n/c_n)$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ (3/5, 4/5), (-7/25, 24/25), (-117/125, 44/125), (-527/625, -336/625), \ldots $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is &lt;a href=&quot;http://en.wikipedia.org/wiki/Equidistribution_theorem&quot; rel=&quot;nofollow&quot;&gt;equidistributed&lt;/a&gt; on the unit circle. Therefore the sequence $(|a_1/c_1|, |b_1/c_1|), (|a_2/c_2|, |b_2/c_2|)$ obtained by taking absolute values is equidistributed on the arc of the unit circle in the first quadrant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure if we can say that the Pythagorean triples (as opposed to some subsequence of them) are equidistributed in the first quadrant in this sense.  Anyone?&lt;/p&gt;&#xA;" OwnerUserId="173" LastActivityDate="2010-10-19T01:31:49.140" CommentCount="8" />
  <row Id="7161" PostTypeId="2" ParentId="7159" CreationDate="2010-10-19T01:39:33.367" Score="3" ViewCount="" Body="&lt;p&gt;I suggest you modify your birthday party idea to include different types of people, such as friends from school, friends from soccer, etc.  Then in general you need to add together the number of attendees from the various categories, and sometimes that will entail adding zero.&lt;/p&gt;&#xA;" OwnerUserId="1392" LastActivityDate="2010-10-19T01:39:33.367" />
  <row Id="7162" PostTypeId="2" ParentId="7159" CreationDate="2010-10-19T01:44:20.093" Score="0" ViewCount="" Body="&lt;p&gt;Assuming this is a two-student group.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Give student 1 three pieces. Give student 2 two pieces. Tell student 1 to take all of their pieces in the pot. Ask the students how much are in the pot. Now ask student 2 to put all their pieces in the pot. Ask the students how much are in the pot. Now ask student 1 to take all their pieces and put them in the pot. Now ask the students how much is in the pot.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the last addition to the pot, student 1 didn't have any more pieces, therefore student 1 added zero and the count did not increase.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This solution can scale to larger groups of students and the students addition order could be rearranged to show that adding zero at the beginning or the end will not change the outcome.&lt;/p&gt;&#xA;" OwnerUserId="2616" LastActivityDate="2010-10-19T01:44:20.093" />
  <row Id="7163" PostTypeId="1" AcceptedAnswerId="7165" CreationDate="2010-10-19T02:21:45.443" Score="2" ViewCount="60" Body="&lt;p&gt;Is it possible to have 2 different points (non-degenerate) in n-dimensions for any value of n where n&gt;1, share the same cost and both be visited by the simplex method?&lt;/p&gt;&#xA;" OwnerUserId="2014" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-10-19T04:26:01.903" LastActivityDate="2010-10-19T04:26:01.903" Title="Simplex - 2 different points with the same cost?" Tags="&lt;optimization&gt;&lt;linear-programming&gt;" AnswerCount="1" />
  <row Id="7164" PostTypeId="1" AcceptedAnswerId="7166" CreationDate="2010-10-19T03:25:21.050" Score="2" ViewCount="186" Body="&lt;p&gt;How can I compute the pmf of a random sum(1 to N) of iid bernoulli random variables, where N is distributed geometrically?&lt;/p&gt;&#xA;" OwnerUserId="2617" LastActivityDate="2010-10-19T04:53:31.520" Title="How can I compute the pmf of a random sum?" Tags="&lt;probability&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="7165" PostTypeId="2" ParentId="7163" CreationDate="2010-10-19T03:28:08.130" Score="1" ViewCount="" Body="&lt;p&gt;No, I think.  At first I thought &quot;Yes&quot; but then convinced myself otherwise by this argument:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At each iteration the simplex method looks for an improving feasible direction.  If there is one, then the next basic feasible solution has a lower cost than the current one (except in the case of degeneracy, which you've already ruled out).  Since the costs of the basic feasible solutions are thus strictly decreasing (again, except for degeneracy) there cannot be two nonoptimal points with the same cost that are visited by the simplex method.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there isn't an improving feasible direction, then the simplex method stops, because it has found an optimal solution.  So it wouldn't visit two optimal points with the same cost, either.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since two such points would both have to be optimal or both be nonoptimal, it appears they cannot exist.&lt;/p&gt;&#xA;" OwnerUserId="2370" LastActivityDate="2010-10-19T03:28:08.130" />
  <row Id="7166" PostTypeId="2" ParentId="7164" CreationDate="2010-10-19T04:07:00.543" Score="2" ViewCount="" Body="&lt;p&gt;Let $S = \sum_{i=1}^{N} X_i$ be the random sum that you are interested in. $X_i$ are the iid Bernoulli random variables. To compute the probability that the random sum $S$ equals an integer $s$, use the law of total probability as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$P(S=s) = \sum_{n=1}^{\infty} P(S=s \mid N=n) P(N=n)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once we condition on $N = n$, the sum of $n$ iid Bernoulli random variables is easy to compute (the answer will be another familiar distribution). Since $P(N=n)$ follows the geometric law, we can compute both the terms of the product and then compute the infinite sum.&lt;/p&gt;&#xA;" OwnerUserId="618" LastActivityDate="2010-10-19T04:07:00.543" CommentCount="2" />
  <row Id="7167" PostTypeId="2" ParentId="6551" CreationDate="2010-10-19T04:25:12.650" Score="3" ViewCount="" Body="&lt;p&gt;I'm not familiar with SAT, as I'm not from USA, but you might find this useful: &lt;a href=&quot;http://www.khanacademy.org/&quot; rel=&quot;nofollow&quot;&gt;http://www.khanacademy.org/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My guess is that this guy explains all of the subjects you might need. It might lack exercises though, but it might be worth checking it out if you need to review a subject.&lt;/p&gt;&#xA;" OwnerUserId="1605" LastActivityDate="2010-10-19T04:25:12.650" CommunityOwnedDate="2010-10-19T04:25:12.650" />
  <row Id="7168" PostTypeId="2" ParentId="7144" CreationDate="2010-10-19T04:37:55.797" Score="3" ViewCount="" Body="&lt;p&gt;Perhaps this answer will use too much technology.  Still, I think it's pretty.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider $A_4$ as the group of orientation-preserving symmetries of a tetrahedron $S$.  The quotient $X=S/A_4$ is a 2-dimensional &lt;a href=&quot;http://en.wikipedia.org/wiki/Orbifold&quot; rel=&quot;nofollow&quot;&gt;orbifold&lt;/a&gt;.  Let's try to analyse it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Two-dimensional orbifolds have three different sorts of singularities that set them apart from surfaces: cone points, reflector lines and corners where reflector lines meet.  Because $A_4$ acts preserving orientation, all the singularities of $X$ are cone points, and we can write them down: they're precisely the images of the points of $S$ that are fixed by non-trivial elements of $A_4$, and to give $X$ its orbifold structure you just have to label them with their stabilisers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what are these points?  There are the vertices of $S$, which are fixed by a rotation of order 3; there are the midpoints of the edges of $S$, which are fixed by a rotation of order 2; and finally, the midpoints of faces, which are fixed by a rotation of order 3.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A fundamental domain for the action of $A_4$ is given by a third of one of the faces, and if you're careful about which sides get identified you can check that $X$ is a sphere with three cone points, one labelled with the cyclic group $C_2$ and the other two labelled with the cyclic group $C_3$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, we can compute a presentation for $A_4$ by thinking of it as the &lt;em&gt;orbifold fundamental group&lt;/em&gt; of $X$ and applying van Kampen's Theorem.  This works just as well for orbifolds, as long as you remember to consider each cone point as a space with fundamental group equal to its label.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The complement of the cone points is a 3-punctured sphere, whose fundamental group is free on $x,y$.  The boundary loops correspond to the elements $x$, $y$ and $xy$.  Next, we take account of each cone point labelled $C_n$ by inserting a relation that makes the $n$th power of the appropriate boundary loop equal to $1$.  So we get the presentation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\langle x,y\mid x^2=y^3=(xy)^3=1\rangle$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;as required.&lt;/p&gt;&#xA;" OwnerUserId="2287" LastActivityDate="2010-10-19T04:37:55.797" CommentCount="2" />
  <row Id="7169" PostTypeId="1" AcceptedAnswerId="7181" CreationDate="2010-10-19T04:47:10.533" Score="2" ViewCount="189" Body="&lt;p&gt;I was asked to find a solution to&#xA;$$\frac{\sin^2(nx)}{n^2\sin^2(x)}=2^{-1/2}$$&#xA;where $n$ is a fixed integer greater than 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Numerically, there's a solution just above 1/n so I decided to find this one (rather than one of the other roots).  I hoped to find an obvious pattern that I could use to find some sort of closed-form solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It turns out that the power series of the root jumps out, numerically.  Since x is only used inside a sine, the power series has only odd terms.  The coefficients look like&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1.0019063576966069771401037205331682344703361047262779848987578751140847/n +&lt;/p&gt;&#xA;&#xA;&lt;p&gt;0.466422280109844953629560577622677335783739044717/n^3 +&lt;/p&gt;&#xA;&#xA;&lt;p&gt;0.3247444758314010835749/n^5 + ...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which are annoyingly close to 1, 1/2, 1/3, ... but are clearly distinct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any idea how to solve this?  This is, I'm sure, well beyond anything the original question-asker cares about, but now my interest is piqued.&lt;/p&gt;&#xA;" OwnerUserId="1778" LastEditorUserId="1778" LastEditorDisplayName="" LastEditDate="2010-10-20T13:59:09.993" LastActivityDate="2010-10-20T13:59:09.993" Title="Identify this power series / solve this trig equation" Tags="&lt;real-analysis&gt;&lt;trigonometry&gt;&lt;closed-form&gt;&lt;power-series&gt;" AnswerCount="2" CommentCount="5" />
  <row Id="7170" PostTypeId="2" ParentId="7164" CreationDate="2010-10-19T04:53:31.520" Score="2" ViewCount="" Body="&lt;p&gt;if $N$ is independent of the bernoulli sequence, you have what is called a compound distribution. the pmf can be obtained by conditioning: if $X$ is the [random] sum,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$P(X = k | N = n) = {\rm bin}(k,n,p) = {\binom{n}{k}}p^k(1-p)^{n-k}.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;if $N$ is geometric-$t, P(N = n) = (1- t)t^n, n = 0,1,\dots .$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$P(X=k, N=n) = \binom{n}{k}p^k(1-p)^{n-k}(1-t)t^n.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;now sum out $n$ for $n \ge k$ to get &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$P(X=k) = \sum_{n=k}^\infty \binom{n}{k}p^k(1-p)^{n-k}(1-t)t^n$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$= p^kt^k(1-t)\sum_k^\infty \binom{n}{k}[(1-p)t)]^{n-k}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$= (pt)^k(1-t)\sum_{m=0}^\infty \binom{m+k}{k}[(1-p)t)]^m$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$=(pt)^k(1-t)(1- (1-p)t)^{-(k+1)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$=\frac{1-t}{1-t+pt}[\frac{pt}{1-t+pt}]^k.$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;so $  X \sim {\rm geom}-\frac{pt}{1-t+pt}.$&lt;/p&gt;&#xA;" OwnerUserId="1547" LastActivityDate="2010-10-19T04:53:31.520" CommentCount="2" />
  <row Id="7171" PostTypeId="2" ParentId="7169" CreationDate="2010-10-19T05:02:15.533" Score="2" ViewCount="" Body="&lt;p&gt;Let $x=\arccos y$, replace the $\sin^2$ in the numerator by $1-\cos^2$, and use the trigonometric definition of &lt;a href=&quot;http://en.wikipedia.org/wiki/Chebyshev_polynomials&quot; rel=&quot;nofollow&quot;&gt;Chebyshev's polynomials&lt;/a&gt; to transform the equation into a polynomial equation on $y$. Then solve that using any of the usual ways to solve it (numerically!)&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2010-10-19T05:02:15.533" />
  <row Id="7172" PostTypeId="1" CreationDate="2010-10-19T05:10:38.887" Score="8" ViewCount="440" Body="&lt;p&gt;A function $f: \mathbb{R} \to \mathbb{R}$ is said to have the intermediate-value property if for any $a$,$b$ and $\lambda \in [f(a),f(b)]$ there is a $x \in [a,b]$ such that $f(x)=\lambda$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A function $f$ is injective if $f(x)=f(y) \Rightarrow x=y$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now it is the case that every injective function with the intermediate-value property is continuous. I can prove this using the following steps:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;An injective function with the intermediate-value property must be monotonic.&lt;/li&gt;&#xA;&lt;li&gt;A monotonic function possesses left- and right-handed limits at each point.&lt;/li&gt;&#xA;&lt;li&gt;For a function with the intermediate-value property the left- and right-handed limits at $x$, if they exist, equal $f(x)$.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I am not really happy with this proof. Particularly I don't like having to invoke the intermediate-value property twice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can there be a shorter or more elegant proof?&lt;/p&gt;&#xA;" OwnerUserId="1195" LastEditorUserId="1195" LastEditorDisplayName="" LastEditDate="2010-10-20T03:18:18.297" LastActivityDate="2010-10-20T03:18:18.297" Title="Injective functions with intermediate value property are continuous. Better proof?" Tags="&lt;real-analysis&gt;" AnswerCount="3" CommentCount="4" FavoriteCount="1" />
  <row Id="7173" PostTypeId="1" AcceptedAnswerId="7174" CreationDate="2010-10-19T05:24:33.910" Score="3" ViewCount="118" Body="&lt;p&gt;This is a follow-up to a &lt;a href=&quot;http://math.stackexchange.com/questions/7136/&quot;&gt;question posted recently&lt;/a&gt;.&#xA;Let&#xA;$$s_n = \sum_{r=1}^{n} \frac{1}{r(r+1)},$$&#xA;where we take $s_0 = 0$.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The problem I am interested in is this: For fixed $k \geq 2$, find all solutions $(m,n)$ in nonnegative integers to the Diophantine equation $$s_m - s_n = \frac{1}{k}.$$   &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Current state of knowledge (see the original question):&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;$s_m - s_n$ can be expressed as &#xA;$$s_m - s_n = \frac{m-n}{(m+1)(n+1)},$$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;At least some of the solutions are given by taking each divisor $a$ of $k$ such that $a &gt; 1$ and setting $$m = (a-1)k-1,$$ $$n = \frac{(a-1)k}{a} - 1.$$ &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="2370" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-10-20T19:44:35.290" LastActivityDate="2010-10-20T19:44:35.290" Title="Finding all solutions to a Diophantine equation involving partial sums of a given series" Tags="&lt;number-theory&gt;&lt;diophantine-equations&gt;" AnswerCount="1" />
  <row Id="7174" PostTypeId="2" ParentId="7173" CreationDate="2010-10-19T05:41:18.707" Score="3" ViewCount="" Body="&lt;p&gt;We are to solve $$\displaystyle \frac{1}{x} - \frac{1}{y} = \frac{1}{k}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This can be rearranged to &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(k+y)(k-x) = k^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So each divisor $\displaystyle a &gt; k$ of $\displaystyle k^2$ will give a solution and all solutions are gotten by that. So just by taking divisors of $\displaystyle k$, you might  be missing some solutions for many values of $\displaystyle k$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance for $\displaystyle k=15$ you will be missing the solution $\displaystyle m = 9, n = 5$&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-10-19T06:15:55.217" LastActivityDate="2010-10-19T06:15:55.217" CommentCount="1" />
  <row Id="7175" PostTypeId="1" AcceptedAnswerId="7199" CreationDate="2010-10-19T05:49:31.627" Score="4" ViewCount="242" Body="&lt;p&gt;I had already posted this on mathoverflow and was advised to post the same here. So here it goes:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;X={x,y,z|$x^2+y^2+z^2≤1$ and $ z≥0$} i.e. X is the top half of a 3-Disk.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Z=X/E, where E is the equivalence relation on the the plane z = 0 which is as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(x,y,0)∼(−x,−y,0).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was told that this space is equivalent to a cone of RP2 (Real Projective Plane).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to know the following facts about &quot;Z&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Is this a manifold with a boundary?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) If it is a manifold with a boundary, what are the points of Z that make the boundary?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) Is it simply connected?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4) What is the minimum Euclidean dimension in which Z can be embedded in?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you very much for your help. I am new to topology and this problem came up as a part of my project. Any help is appreciated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you. Will.&lt;/p&gt;&#xA;" LastActivityDate="2010-10-19T21:56:13.423" Title="Topology of a cone of RP^2." Tags="&lt;algebraic-topology&gt;" AnswerCount="3" FavoriteCount="2" />
  <row Id="7176" PostTypeId="2" ParentId="7175" CreationDate="2010-10-19T06:15:23.160" Score="2" ViewCount="" Body="&lt;p&gt;First consider just the hemisphere $X'\subseteq X$, and define $Z'= X'/E$.    This is the real projective plane, in one of its many incarnations.  You may have seen $\mathbb{RP}^2$ presented as the space of lines through the origin in $\mathbb{R}^3$.  To get from that presentation to this one, note that each such line intersects $S^2 \subseteq \mathbb{R}^3$ in exactly two points, and for those lines where one point is above the xy-plane and the other is below, we've just chosen the $z&gt;0$ representative.  This is a manifold without boundary, and it is not simply-connected.  Indeed, its universal cover is $S^2$, which is a two-fold cover.  Hence its fundamental group has two elements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, note that in $Z$, we're literally just taking the cone over $Z'$ -- those rays from the origin that end at points in $X'$ that were identified in $Z'$ are themselves identified.  It is a non-trivial (?) fact that $\mathbb{RP}^2$ is not the boundary of any 3-manifold.  I'm pretty sure that if the cone over a space is a manifold, then the boundary is the manifold itself, although I can't instantly tell you why this should be true.  (In fact, $\mathbb{RP}^2$ generates the unoriented cobordism ring in dimension 3.)  Note that the cone over &lt;em&gt;any&lt;/em&gt; space is simply-connected, since it is contractible (to the cone point).  I don't know the answer to 4, though I'd suspect it's either 4 or 5, since 4 is the minimum dimension you need in order to embed $\mathbb{RP}^2$.  In general, the question of minimum dimension for embedding a particular manifold is not so easy, beyond the a priori bounds given by the Whitney embedding theorem.  And this isn't even a manifold...&lt;/p&gt;&#xA;" OwnerUserId="2159" LastEditorUserId="2159" LastEditorDisplayName="" LastEditDate="2010-10-19T06:28:03.430" LastActivityDate="2010-10-19T06:28:03.430" CommentCount="8" />
  <row Id="7177" PostTypeId="2" ParentId="7172" CreationDate="2010-10-19T06:27:15.127" Score="3" ViewCount="" Body="&lt;p&gt;I don't know if this is shorter or elegant, and in particular it does use the hypotheses more than once, but here's something.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given your 1., assume WLOG that $f$ is increasing.  Then for all $a\lt b$, $f(a,b)\subseteq (f(a),f(b))$, and thus $(a,b)\subseteq f^{-1}(f(a),f(b))$.  Since each element of $(f(a),f(b))$ has a preimage point in $(a,b)$ and $f$ is injective, $f^{-1}(f(a),f(b))\subseteq (a,b)$.  Combining, $f^{-1}(f(a),f(b))=(a,b)$.  I'll show that this implies that the inverse image of each open interval is open, which implies continuity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $x\lt y$ and suppose that $t$ is in $f^{-1}(x,y)$.  Let $b&gt;t$.  Since $f^{-1}(f(t),f(b))=(t,b)$ and $f(t)\lt y$, there is a $b'$ with $t\lt b' \lt b$ and $f(b')\lt y$.  Thus $f^{-1}(f(t),f(b'))=(t,b')$ is contained in $f^{-1}(x,y)$.  Applying a similar argument on the other side, there is an $a'\lt t$ such that $(a',t)$ is contained in $f^{-1}(x,y)$.  Together this implies that $t$ is in the interior of $f^{-1}(x,y)$.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2010-10-19T06:41:17.417" LastActivityDate="2010-10-19T06:41:17.417" />
  <row Id="7178" PostTypeId="1" AcceptedAnswerId="7179" CreationDate="2010-10-19T07:23:21.860" Score="3" ViewCount="115" Body="&lt;p&gt;Although this is not actually homework, it's an exercise from the book;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to rewrite the variance,  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$var[f]=E[(f(x)-E[f(x)])^2]$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;as  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$E[f(x)^2]-E[f(x)]^2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure how it reduces, and it has had me stumped for a while.  The answer would be nice. Any &lt;em&gt;insight&lt;/em&gt; [a brief, yet, interesting explanation] would be great.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance&lt;/p&gt;&#xA;" OwnerUserId="1774" LastEditorUserId="1774" LastEditorDisplayName="" LastEditDate="2010-10-19T07:37:33.253" LastActivityDate="2010-10-19T07:59:49.867" Title="Expected value and Variance" Tags="&lt;homework&gt;&lt;statistics&gt;&lt;probability&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="7179" PostTypeId="2" ParentId="7178" CreationDate="2010-10-19T07:59:49.867" Score="5" ViewCount="" Body="&lt;p&gt;Let $X = f(x)$ denote the random variable in question. Let $\mu = E(f(x))$. Then, you are looking to prove the following&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$E((X-\mu)^2) = E(X^2)-\mu^2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me outline the steps needed to prove this and you can try to fill in the gaps.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Expand the LHS algebraically and use linearity of expectation to split it into $3$ different terms.&lt;/li&gt;&#xA;&lt;li&gt;Note that $\mu$ is a constant and $E(aX)$ is related in a simple way to $E(X)$ when $a$ is a constant.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;These two steps combined should get you to the right hand side. &lt;/p&gt;&#xA;" OwnerUserId="618" LastActivityDate="2010-10-19T07:59:49.867" CommentCount="2" />
  <row Id="7180" PostTypeId="2" ParentId="7172" CreationDate="2010-10-19T08:14:41.090" Score="2" ViewCount="" Body="&lt;p&gt;As you note, &lt;em&gt;f&lt;/em&gt; is injective and has the intermediate value property =&gt; &lt;em&gt;f&lt;/em&gt; is monotonic.  We may assume &lt;em&gt;f&lt;/em&gt; is increasing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now for any &lt;em&gt;x&lt;/em&gt; and any small* &amp;epsilon; &gt; 0, we have by the IVP &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;c&lt;/em&gt; in [&lt;em&gt;x-1, x&lt;/em&gt;] such that &lt;em&gt;f(c) =  f(x)&lt;/em&gt;-&amp;epsilon; and&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;d&lt;/em&gt; in [&lt;em&gt;x, x+1&lt;/em&gt;] such that &lt;em&gt;f(d)=f(x)+&lt;/em&gt;&amp;epsilon;.  &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Then we choose &amp;delta;=min(&lt;em&gt;x-c,d-x&lt;/em&gt;).  Since &lt;em&gt;f&lt;/em&gt; is increasing, any point within &amp;delta; of &lt;em&gt;x&lt;/em&gt; maps to a point within &amp;epsilon; of &lt;em&gt;f(x)&lt;/em&gt;; thus &lt;em&gt;f&lt;/em&gt; is continuous.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;*By &quot;small&quot; &amp;epsilon;, I mean &amp;epsilon; smaller than both &lt;em&gt;f(x)-f(x-1)&lt;/em&gt; and &lt;em&gt;f(x+1)-f(x)&lt;/em&gt;, so that the condition of the IVP is satisfied.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also invoke the IVP more than once.  Perhaps the other Jonas will supply a proof with just one use of the IVP!&lt;/p&gt;&#xA;" OwnerUserId="1461" LastActivityDate="2010-10-19T08:14:41.090" />
  <row Id="7181" PostTypeId="2" ParentId="7169" CreationDate="2010-10-19T08:32:04.590" Score="2" ViewCount="" Body="&lt;p&gt;Just the $x=\arccos\;y$ substitution suggested by Mariano suffices; what you will then have will be the equation $\left(\frac{U_{n-1}(y)}{n}\right)^2=\frac1{\sqrt{2}}$ where $U_n(y)$ is the &lt;a href=&quot;http://dlmf.nist.gov/18.3&quot; rel=&quot;nofollow&quot;&gt;Chebyshev polynomial of the second kind&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-10-19T08:32:04.590" CommentCount="0" />
  <row Id="7182" PostTypeId="2" ParentId="7172" CreationDate="2010-10-19T09:18:21.980" Score="3" ViewCount="" Body="&lt;p&gt;[I thought of another proof that uses the IVP and injectiveness once. Putting it as a community wiki answer.]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume on the contrary that $f$ is not continuous at $x$. Then there is a sequence $x_n$ converging to $x$ such that $f(x_n)$ does not converge to $f(x)$. Then there is $\epsilon&gt;0$  and a subsequence $x_{n_k}$ such that $f(x_{n_k}) \notin (f(x)-\epsilon,f(x)+\epsilon)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There must either be a further subsequence $x_{n_j}$ such that $f(x_{n_j}) \leq f(x)-\epsilon$ or a subsequence $x_{n_q}$ such that $f(x_{n_q}) \geq f(x)+\epsilon$ (or both). Assume without loss of generality the former. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since  $f(x_{n_j}) \leq f(x)-\epsilon &amp;lt; f(x)$, by the IVP for every $j$ there is a $y_j$ such that $x_{n_j}\leq y_j &amp;lt; x$ and $f(y_j)=f(x)-\epsilon$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because $f$ is injective, all the $y_j$ must be the same, say $y$. Because $x_{n_j}$ converges to $x$, $y=x$ by the sandwich theorem. But $f(y)\neq f(x)$. Hence a contradiction.&lt;/p&gt;&#xA;" OwnerUserId="1195" LastEditorUserId="1195" LastEditorDisplayName="" LastEditDate="2010-10-19T14:55:15.543" LastActivityDate="2010-10-19T14:55:15.543" CommunityOwnedDate="2010-10-19T09:18:43.603" />
  <row Id="7183" PostTypeId="1" CreationDate="2010-10-19T11:41:43.753" Score="2" ViewCount="156" Body="&lt;p&gt;There is an intricate connection between Hurwitz Zeta and the (traditional) polygamma function: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\psi_n(z)=(-1)^{n+1}n!\zeta(n+1,z)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If to use a generalization for Bernoulli numbers, this can be considered a formula, connecting polygamma and Bernoulli polynomials of negative order:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\psi_n(z)=(-1)^{n+1}n!\frac{B_{-n}(x)}n$$ (1)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As much as this equality is impressing, it is limited as it only holds for natural n.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After a couple of unsuccessful attempts to find a more general formula, I encountered a more natural, &quot;balanced&quot; generalization of polygamma function explained in &lt;a href=&quot;http://www.math.tulane.edu/~vhm/papers_html/genoff.pdf&quot; rel=&quot;nofollow&quot;&gt;this paper&lt;/a&gt;. It turned out that while the old formula still holds for natural n (since the old polygamma and balanced polygamma coincide in integer positive orders), a completely new formula connecting this balanced polygamma with Zeta and Bernoulli numbers can be derived which holds for any z:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\zeta(z,q)=\frac{\Gamma (1-z) \left(2^{-z} \left(\psi \left(z-1,\frac{q}{2}+\frac{1}{2}\right)+\psi \left(z-1,\frac{q}{2}\right)\right)-\psi(z-1,q)\right)}{\ln(2)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$B_z(q) = -\frac{\Gamma (z+1) \left(2^{z-1} \left(\psi\left(-z,\frac{q}{2}+\frac{1}{2}\right)+\psi\left(-z,\frac{q}{2}\right)\right)-\psi(-z,q)\right)}{\ln (2)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both of them can be expressed completely in terms of balanced polygamma and elementary functions if to notice that $\Gamma(x)=e^{\psi(-1,x)+\frac 12 \ln(2\pi)}$, which allows to get rid of the Gamma function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While the target was reached, these expressions still leave a bad impression. I cannot simplify it as no CAS system is capable of operations with the balanced polygamma. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence I am asking for help on how to simplify the expressions so they could be easier to manage and use. It is also not evident how the letter formulas become the former ones at positive real z.&lt;/p&gt;&#xA;" OwnerUserId="2513" LastEditorUserId="247" LastEditorDisplayName="" LastEditDate="2010-10-19T14:09:46.450" LastActivityDate="2010-10-19T14:09:46.450" Title="Connection between Bernoulli polynomials and polygamma function" Tags="&lt;gamma-function&gt;&lt;zeta-function&gt;" CommentCount="2" />
  <row Id="7184" PostTypeId="1" CreationDate="2010-10-19T11:50:55.590" Score="3" ViewCount="126" Body="&lt;p&gt;s and t are two nodes in a graph.&#xA;How to prove that the maximum number of edge-disjoint cut sets which divide s and t equals the length of the shortest path from s to t?&lt;/p&gt;&#xA;" OwnerUserId="2621" LastEditorUserId="2621" LastEditorDisplayName="" LastEditDate="2010-10-19T15:54:08.210" LastActivityDate="2010-10-19T16:58:07.590" Title="Number of cut sets and the length of the shortest path?" Tags="&lt;homework&gt;&lt;graph-theory&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" />
  <row Id="7185" PostTypeId="1" AcceptedAnswerId="7190" CreationDate="2010-10-19T14:19:14.600" Score="1" ViewCount="178" Body="&lt;p&gt;A very simple solution to the &lt;a href=&quot;http://en.wikipedia.org/wiki/Diffusion_equation&quot; rel=&quot;nofollow&quot;&gt;diffusion equation&lt;/a&gt; is $u(x,t)=x^2+2 t$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My question:&lt;/strong&gt;&lt;br&gt;&#xA;How can this be a solution to the diffusion equation when nothing really diffuses, but just stays the same - see plot: &lt;a href=&quot;http://www.wolframalpha.com/input/?i=plot+x+x%2B2+t&quot; rel=&quot;nofollow&quot;&gt;Here&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can see that it is a solution but can't believe my eyes... Is there some kind of intuition other than: &quot;it satisfies the PDE&quot;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there other examples of solutions to well-known equations that behave completely un-intuitive in some situations?&lt;/p&gt;&#xA;" OwnerUserId="346" LastActivityDate="2010-10-19T15:13:07.763" Title="Simple Solution to Diffusion equation" Tags="&lt;intuition&gt;&lt;differential-equations&gt;" AnswerCount="3" FavoriteCount="2" />
  <row Id="7186" PostTypeId="1" AcceptedAnswerId="7193" CreationDate="2010-10-19T14:52:09.437" Score="2" ViewCount="79" Body="&lt;p&gt;I'm currently reading Hilton &amp;amp; Stammbach's A First Course in Homological Algebra, and the following point has stumped me:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In section 1.8, they construct co-free modules (&quot;left moodule&quot; over some ring) as essentially coming from the right adjoint to the forgetful functor from $\Lambda$-Modules to Abelian Groups. On the other hand, the free module is constructed as the left adjoint to the forgetful functor from $\Lambda$-modules to Sets. This turns out to be equivalent to requiring free modules to be direct sums of copies of $\Lambda$ considered as a module over itself, and to requiring co-free modules as direct products of the $\Lambda^*=Hom_\mathbb{Z}(\Lambda, \mathbb{Q}/\mathbb{Z})$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I guess my question is: what does the right adjoint to the forgetful functor to Set look like, and why is the right adjoint to the forgetful functor to Abelian Groups more useful?&lt;/p&gt;&#xA;" OwnerUserId="400" LastActivityDate="2010-10-19T15:30:19.787" Title="Why is the co-free module defined as the right adjoint to the forgetful functor to Ab rather than Set?" Tags="&lt;category-theory&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" />
  <row Id="7187" PostTypeId="1" AcceptedAnswerId="7198" CreationDate="2010-10-19T14:57:14.870" Score="10" ViewCount="438" Body="&lt;p&gt;Quaternions came up while I was interning not too long ago and it seemed like no one really know how they worked. While eventually certain people were tracked down and were able to help with the issue, it piqued my interest in quaternions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After reading many articles and a couple books on them, I began to know the formulas associated with them, but still have no clue &lt;em&gt;how&lt;/em&gt; they work (why they allow rotations in 3D space to be specific). I back-tracked a little bit and looked at normal complex numbers with just one imaginary component and asked myself if I even understood how they allow rotations in 2D space. After a couple awesome moments of understanding, I understood it for imaginary numbers, but I'm still having trouble extending the thoughts to quaternions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can someone intuitively think about quaternions and how they allow for rotations in 3D space?&lt;/p&gt;&#xA;" OwnerUserId="2623" LastEditorUserId="2623" LastEditorDisplayName="" LastEditDate="2010-10-19T20:58:23.217" LastActivityDate="2010-10-20T12:52:51.393" Title="How can one intuitively think about quaternions?" Tags="&lt;linear-algebra&gt;&lt;complex-numbers&gt;&lt;quaternions&gt;" AnswerCount="6" CommentCount="6" FavoriteCount="8" />
  <row Id="7188" PostTypeId="2" ParentId="7184" CreationDate="2010-10-19T14:58:08.393" Score="1" ViewCount="" Body="&lt;p&gt;Perhaps this paper might be of use:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.iis.sinica.edu.tw/page/jise/2001/200103_02.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.iis.sinica.edu.tw/page/jise/2001/200103_02.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1369" LastActivityDate="2010-10-19T14:58:08.393" CommentCount="1" />
  <row Id="7189" PostTypeId="2" ParentId="7185" CreationDate="2010-10-19T14:59:19.577" Score="1" ViewCount="" Body="&lt;p&gt;It is difficult to interpret a PDE without knowing its boundary conditions and/or initial values.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, at first guess, I'd say you should look at individual time slices of the solution on the same graph and see if that helps your interpretation.  The 3D image is nice for showing everything at once, but that can obscure an interpretation.&lt;/p&gt;&#xA;" OwnerUserId="2283" LastActivityDate="2010-10-19T14:59:19.577" />
  <row Id="7190" PostTypeId="2" ParentId="7185" CreationDate="2010-10-19T15:12:14.120" Score="3" ViewCount="" Body="&lt;p&gt;Here are a few pointers, from the more obtuse to the more direct.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;&quot;Meeting a friend in a corridor, Wittgenstein said: 'Tell me, why do people always say it was &lt;em&gt;natural&lt;/em&gt; for men to assume that the sun went round the earth, rather than that the earth was rotating?'  His friend said, 'Well, obviously, because it &lt;em&gt;looks&lt;/em&gt; as if the sun is going round the earth.'  To which the philosopher replied, 'Well, what would it have looked like if it had looked as if the earth was rotating?'&quot; Apparently from Tom Stoppard's play &lt;em&gt;Jumpers&lt;/em&gt;. More to the point: what does your intuition think the solution to the diffusion equation with initial conditions $u(x,0) = x^2$ &lt;em&gt;should&lt;/em&gt; have looked like?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$u_1(x,t) = \text{const}$, and $u_2(x,t) = x$ are also solutions of the diffusion equation. Do you find them more or less surprising than the $x^2 + t$ solution?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The standard intuitive interpretation of the diffusion equation is that it &quot;smooths out&quot; variations in the solution. The way it finds these variations is by looking at the second derivative. But the second derivative of $x^2$ is constant everywhere, so what can the diffusion equation do?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Locally, every point &quot;wants to be&quot; equal to the average of the value in its neighbourhood. If the second derivative at a point is positive, the function looks like a valley, so the point increases its value over time; if the second derivative is negative, the point is on a peak, and tries to decrease. With $x^2$, all of $\mathbb{R}$ is a valley.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="856" LastActivityDate="2010-10-19T15:12:14.120" CommentCount="2" />
  <row Id="7191" PostTypeId="2" ParentId="7185" CreationDate="2010-10-19T15:13:07.763" Score="5" ViewCount="" Body="&lt;p&gt;But there is diffusion! Heat – or whatever quantity $u$ represents – flows from the far left and right ($x\to\pm\infty$, where there's lots of it!) towards the middle ($x=0$). For every fixed point $x$ there is always more heat flowing towards it from the outside than what it emits inwards, and the result is that the heat keeps increasing at each point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now this might seem unphysical, and that's because it is. For this particular solution, the total amount of heat, $\int_{-\infty}^{\infty} u(x,t) dx$, is infinite for each $t$.&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2010-10-19T15:13:07.763" CommentCount="1" />
  <row Id="7192" PostTypeId="1" CreationDate="2010-10-19T15:27:00.493" Score="2" ViewCount="67" Body="&lt;p&gt;Can any of you guys think of a topological space $(X,\tau)$ and a family of subsets {A&lt;sub&gt;s&lt;/sub&gt;}&lt;sub&gt;${s \in S}$&lt;/sub&gt; of $X$ such that for a certain $x \in X$ you can find a subset $V$ such that $x \in V$ and {$s \in S: V \cap A_{s} \neq \emptyset$} is finite, whereas for every $W \subseteq X$, with $x \in W$, we have that {$s \in S: W \cap \mathrm{cl}(A_{s}) \neq \emptyset$} is never finite?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I thank you in advance for your replies.&lt;/p&gt;&#xA;" OwnerUserId="2624" LastEditorUserId="2624" LastEditorDisplayName="" LastEditDate="2010-10-19T15:47:17.323" LastActivityDate="2010-10-19T15:47:17.323" Title="A finiteness condition" Tags="&lt;general-topology&gt;" AnswerCount="1" />
  <row Id="7193" PostTypeId="2" ParentId="7186" CreationDate="2010-10-19T15:30:19.787" Score="2" ViewCount="" Body="&lt;p&gt;If a functor has a right adjoint, it preserves colimits; but the&#xA;forgetful functor from $R$-Mod to Set doesn't. For instance it&#xA;doesn't preserve binary coproducts. So there isn't a right adjoint.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-10-19T15:30:19.787" />
  <row Id="7194" PostTypeId="2" ParentId="7187" CreationDate="2010-10-19T15:37:48.827" Score="3" ViewCount="" Body="&lt;p&gt;I highly recommend that your read the presentation in Conway and Smith: On quaternions and octonions: Their geometry, arithmetic, and symmetry. Here's an excerpt from John Baez's &lt;a href=&quot;http://www.ams.org/journals/bull/2005-42-02/S0273-0979-05-01043-8/S0273-0979-05-01043-8.pdf&quot; rel=&quot;nofollow&quot;&gt;very informative review:&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;It follows that the quaternions of norm 1 form a group under multiplication. This group is usually called SU(2), because people think of its elements as 2 × 2 unitary matrices with determinant 1. However, the quaternionic viewpoint is better adapted to seeing how this group describes rotations in 3 and 4 dimensions. The unit quaternions act via conjugation as rotations of the 3d space of &quot;pure imaginary&quot; quaternions, namely those with Re(q) = 0. This gives a homomorphism from SU(2) onto the 3d rotation group SO(3). The kernel of this homomorphism is {±1}, so we see SU(2) is a double cover of SO(3). The unit quaternions also act via left and right multiplication as rotations of the 4d space of all quaternions. This gives a homomorphism from SU(2) × SU(2) onto the 4d rotation group SO(4). The kernel of this homomorphism is {±(1, 1)}, so we see SU(2) × SU(2) is a double cover of SO(4).&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;These facts are incredibly important throughout mathematics and physics. With their help, Conway and Smith classify the finite subgroups of the 3d rotation group SO(3), its double cover SU(2), the 3d rotation/reflection group O(3), and the 4d rotation group SO(4). These classifications are all in principle &quot;well known'. However, they seem hard to find in one place, so Conway and Smith's elegant treatment is very helpful&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;See also the Wikipedia article &lt;a href=&quot;http://en.wikipedia.org/wiki/Quaternions_and_spatial_rotation&quot; rel=&quot;nofollow&quot;&gt;Quaternions and spatial rotation.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-10-19T15:46:24.207" LastActivityDate="2010-10-19T15:46:24.207" />
  <row Id="7195" PostTypeId="1" CreationDate="2010-10-19T15:39:53.380" Score="6" ViewCount="593" Body="&lt;p&gt;In my attempts to derive the closed form for the arclength of the hyperbola, I wound up with the following integral:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int\frac{\sqrt{1-m\;\sin^2 u}}{\sin^2 u}\mathrm{d}u$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am aware that such integrals are expressible as combinations of elliptic integrals and trigonometric functions, and &lt;em&gt;Mathematica&lt;/em&gt; does return a result containing elliptic integrals. However, the result I got from &lt;em&gt;Mathematica&lt;/em&gt; seems to suggest that the original integrand was split like so:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\sqrt{1-m\;\sin^2 u}}{\sin^2 u}=\frac{1-m}{\sqrt{1-m\;\sin^2 u}}-\sqrt{1-m\;\sin^2 u}+\frac{\csc^2u-m\;\sin^2 u}{\sqrt{1-m\;\sin^2 u}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and then integrated to yield the expression containing the elliptic integrals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I'm stuck with is how this splitting was thought of so that the final integral is expressible in terms of the Legendre elliptic integrals. A possible clue lay in the fact that through an appropriate substitution, the original integral can be expressed as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int\mathrm{ds}^2(v|m)\mathrm{d}v$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\mathrm{ds}(v|m)=\frac{\mathrm{dn}(v|m)}{\mathrm{sn}(v|m)}$ is a Jacobian elliptic function, from which this &lt;a href=&quot;http://dlmf.nist.gov/22.16.E25&quot; rel=&quot;nofollow&quot;&gt;formula&lt;/a&gt; applies. However, I don't know how the formula listed in the DLMF was derived, and I suspect that the answer to my question will also explain that formula.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would appreciate any help in understanding why the integrand was split in that manner.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Trivially,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathrm{ds}^2(v|m)=\mathrm{ns}^2(v|m)-m=1-m+\mathrm{cs}^2(v|m)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The elliptic integral of the second kind is supposed to arise due to the fact that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int\mathrm{dn}^2(v|m)\mathrm{d}v$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$=\int\mathrm{dn}(v|m)\mathrm{d}(\mathrm{am}(v|m))$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$=\int\sqrt{1-m\;\mathrm{sn}^2(v|m)}\mathrm{d}(\mathrm{am}(v|m))$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$=\int\sqrt{1-m\;\sin^2(\mathrm{am}(v|m))}\mathrm{d}(\mathrm{am}(v|m))$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$=E(\mathrm{am}(v|m)|m)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so I suppose figuring out how to integrate the squares of any of $\mathrm{cs}$, $\mathrm{ns}$, or $\mathrm{ds}$ would do the trick. I suspect something like integration by parts or tricks similar to those used for integrating ratios of trigonometric functions would work here, but I can't seem to find the proper strategy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yet another way would be to show that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$-\int\left(\mathrm{cs}^2(v|m)+\mathrm{dn}^2(v|m)\right)\mathrm{d}v=\mathrm{cs}(v|m)\mathrm{dn}(v|m)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Differentiating the RHS gives the integrand, but supposing I did not know this expression, how would this integral be evaluated?&lt;/p&gt;&#xA;" OwnerUserId="498" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-10T17:00:38.860" LastActivityDate="2011-03-27T05:34:02.060" Title="On deriving the arclength of a hyperbola" Tags="&lt;calculus&gt;&lt;special-functions&gt;&lt;plane-curves&gt;&lt;conic-sections&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="1" />
  <row Id="7196" PostTypeId="2" ParentId="7192" CreationDate="2010-10-19T15:46:58.207" Score="4" ViewCount="" Body="&lt;p&gt;If I read it correctly you want $V$ to meet only finitely many of the $A_s$, but every set containing $x$ to meet infinitely many of the closures of the $A_s$.  Without some restriction on $V$ and the $A_s$ this is easy.  Just have all the $A_s$ have x as a limit point but not a member.  Then $V$ is just the point $x$ and any $W$ containing $x$ will meet the closure of all the $A_s$.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-10-19T15:46:58.207" />
  <row Id="7197" PostTypeId="2" ParentId="7187" CreationDate="2010-10-19T15:47:28.533" Score="1" ViewCount="" Body="&lt;p&gt;I find the conversion between quaternions and the &lt;a href=&quot;http://en.wikipedia.org/wiki/Axis_angle&quot; rel=&quot;nofollow&quot;&gt;axis-angle representation&lt;/a&gt; quite instructive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the axis-angle representation, you describe a rotation by specifying the axis of rotation as a unit vector $\omega$ and an angle $\theta$ about which to rotate around this axis. An interesting fact is that any possible rotation can be described in this way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The corresponding quaternion is given simply by $\left(\cos(\theta/2), \omega\sin(\theta/2)\right)$. Here the notation $(a, v)$, where $a$ is a scalar and $v$ a real vector, denotes the quaternion $a + v_xi + v_yj + v_zk$, or $(a,v_x,v_y,v_z)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is the intuitive interpretation of this. Given a particular rotation axis $\omega$, if you restrict the 4D quaternion space to the 2D plane containing $(1,0,0,0)$ and $(0,\omega_x,\omega_y,\omega_z)$, the unit quaternions representing all possible rotations about the axis $\omega$ form the unit circle in that plane. A rotation of $\theta$ about the axis $\omega$ is the point at an angle $\theta/2$ from $(1,0,0,0)$ on that circle. For example, not rotating at all is $(1,0,0,0)$, rotating 180° is $(0,\omega_x,\omega_y,\omega_z)$, and rotating 360° is $(-1,0,0,0)$, which is the same as not rotating at all (see final paragraph).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Multiplying&lt;/em&gt; two quaternions is unintuitive, but I'm not bothered by that, because the composition of two rotations in real life is quite unintuitive in the first place.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Note that the quaternions are a &quot;double cover&quot; of the space of rotations, in that any rotation actually has &lt;em&gt;two&lt;/em&gt; quaternions, say $q$ and $-q$, that represent it: $\theta$ and $\theta + 2\pi$ are the same angle, but $\theta/2$ and $(\theta+2\pi)/2$ are not. This is the only &quot;glitch&quot; in the quaternion representation of rotations.)&lt;/p&gt;&#xA;" OwnerUserId="856" LastActivityDate="2010-10-19T15:47:28.533" CommentCount="1" />
  <row Id="7198" PostTypeId="2" ParentId="7187" CreationDate="2010-10-19T15:55:33.217" Score="9" ViewCount="" Body="&lt;p&gt;Here's one way.  The group of unit quaternions is isomorphic to the &lt;a href=&quot;http://en.wikipedia.org/wiki/Special_unitary_group&quot; rel=&quot;nofollow&quot;&gt;special unitary group&lt;/a&gt; $\text{SU}(2)$, the group of $2 \times 2$ unitary complex matrices with determinant $1$.  This group acts on $\mathbb{C}^2$ in the obvious way, and so it also acts on &lt;em&gt;lines&lt;/em&gt; in $\mathbb{C}^2$.  (These are complex lines, so they have real dimension $2$.)  The space of lines in $\mathbb{C}^2$ is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Complex_projective_space&quot; rel=&quot;nofollow&quot;&gt;complex projective line&lt;/a&gt; $\mathbb{CP}^1$, and it turns out there is a natural way to think about this space as a sphere - namely, the &lt;a href=&quot;http://en.wikipedia.org/wiki/Riemann_sphere&quot; rel=&quot;nofollow&quot;&gt;Riemann sphere&lt;/a&gt;.  There is a beautiful projection which is pictured at the Wikipedia article which shows this; essentially one thinks of $\mathbb{CP}^1$ as $\mathbb{C}$ plus a &quot;point at infinity&quot; and then projects the latter onto the former in a way which misses one point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So $\text{SU}(2)$ naturally acts on a sphere, and as it turns out it naturally acts by rotations.  This describes the famous 2-to-1 map $\text{SU}(2) \to \text{SO}(3)$ which allows quaternions to describe 3D rotations.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-10-19T15:55:33.217" CommentCount="3" />
  <row Id="7199" PostTypeId="2" ParentId="7175" CreationDate="2010-10-19T16:02:19.677" Score="3" ViewCount="" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;It's not a manifold either with or without boundary.  For the proof, look at the cone point c.  If U is a small open set around c, then $U-c$ is homeomorphic to $\mathbb{R}P^2 \times (0,1]$.  However, this is not homeomorphic to either $\mathbb{R}^3-pt$ nor a half space minus a point.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;In fact, it's contractible (the cone over anything is contracitble).  Thus, it's simply connected.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I'm not sure, but it can definitely be topological embedded in R^5.  To see this, let $f$ be an embedding of $\mathbb{R}P^2$ into $\mathbb{R}^4$ (which exists by Whitney's embedding theorem).  Define $G:Z\rightarrow\mathbb{R}^5$ by $G(p, t) = (tf(p),t)$.  This will be a topological embedding.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The minimal possible answer is 4 since that's the smallest in which $\mathbb{R}P^2$ can be embedded into.&lt;/p&gt;&#xA;" OwnerUserId="331" LastEditorUserId="331" LastEditorDisplayName="" LastEditDate="2010-10-19T18:46:53.087" LastActivityDate="2010-10-19T18:46:53.087" CommentCount="8" />
  <row Id="7200" PostTypeId="1" CreationDate="2010-10-19T16:21:35.037" Score="1" ViewCount="128" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Possible Duplicate:&lt;/strong&gt;&lt;br&gt;&#xA;  &lt;a href=&quot;http://math.stackexchange.com/questions/7126/boolean-algebra-simplification-dont-know-the-method-used&quot;&gt;Boolean Algebra, Simplification: Don&amp;#39;t know the method used&lt;/a&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;p&gt;Based on a &lt;a href=&quot;http://math.stackexchange.com/questions/7126/boolean-algebra-simplification-dont-know-the-method-used&quot;&gt;previous question&lt;/a&gt;, the answer I got for the &quot;sum of products&quot; method was:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;RC' + RCG' + RCGM'&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Just for completeness, can someone help walk me through to how this was arrived at?  Thanks again in advance!&lt;/p&gt;&#xA;" OwnerUserId="2609" LastActivityDate="2010-10-19T16:21:35.037" ClosedDate="2010-10-20T05:41:33.297" Title="Boolean Algebra, Simplification: Sum of products" Tags="&lt;boolean-algebra&gt;" CommentCount="2" />
  <row Id="7201" PostTypeId="1" AcceptedAnswerId="10947" CreationDate="2010-10-19T16:41:23.990" Score="1" ViewCount="141" Body="&lt;p&gt;Can someone help direct me towards information on calculating the polynomial remainder?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know how to do the basics, but what if I don't have access to the polynomial in expanded form?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, $\frac{1}{1-x}$ represents the polynomial $\sum_{i=0}^\infty{x^i}$.  I want to be able to find the remainder using an expression like the fraction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please help explain where I can find more resources on this.  I'm not a math expert, so something very readable would be great.  However, I'm willing to learn the math if I need to.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could someone help me find resources with the equivalent and generating functions?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just to show that this operation is well defined, Dr. Herwig Hauser has written a paper concerning the operation &lt;a href=&quot;http://homepage.univie.ac.at/herwig.hauser/Publications/linear_recurrences_and_powerseries_division.pdf&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2626" LastEditorUserId="2626" LastEditorDisplayName="" LastEditDate="2010-10-19T18:43:43.030" LastActivityDate="2010-11-19T09:59:39.423" Title="Polynomial Remainder / Generating Function Remainder" Tags="&lt;calculus&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="7202" PostTypeId="2" ParentId="7184" CreationDate="2010-10-19T16:58:07.590" Score="1" ViewCount="" Body="&lt;p&gt;Hint:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First show that the size of any collection of edge-disjoint cut-sets cannot exceed the length of the shortest path.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-10-19T16:58:07.590" CommentCount="1" />
  <row Id="7204" PostTypeId="1" AcceptedAnswerId="7216" CreationDate="2010-10-19T17:06:09.683" Score="5" ViewCount="243" Body="&lt;p&gt;How does one evaluate show that this limit: $$\lim_{n \to \infty}\sqrt{1 + 2 \sqrt{1+3 \sqrt{1+ \cdots \sqrt{1+(n-1) \sqrt{1+n}}}}}=3$$&lt;/p&gt;&#xA;" LastEditorUserId="367" LastEditorDisplayName="" LastEditDate="2010-10-19T17:31:53.287" LastActivityDate="2010-10-19T19:53:57.927" Title="Limit of Nested Radical" Tags="&lt;real-analysis&gt;&lt;analysis&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="7205" PostTypeId="2" ParentId="7204" CreationDate="2010-10-19T17:13:33.360" Score="11" ViewCount="" Body="&lt;p&gt;This is Ramanujan's famous nested radical.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More information can be found here: &lt;a href=&quot;http://www.isibang.ac.in/~sury/ramanujanday.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.isibang.ac.in/~sury/ramanujanday.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See Also: &lt;a href=&quot;http://mathworld.wolfram.com/NestedRadical.html&quot; rel=&quot;nofollow&quot;&gt;http://mathworld.wolfram.com/NestedRadical.html&lt;/a&gt; (number 26).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Apparently, this is how he came up with it (sorry, no reference for this claim).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Start with&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$3 = \sqrt{9} = \sqrt{1 + 8} = \sqrt{1 + 2 \cdot 4}$$&#xA;$$ = \sqrt{1 + 2\sqrt{16}} = \sqrt{1 + 2\sqrt{1 + 3 \cdot 5}}$$&#xA;$$ = \sqrt{1 + 2\sqrt{1 + 3 \sqrt{25}}} = \sqrt{1 + 2\sqrt{1 + 3 \sqrt{1 + 4 \cdot 6}}}$$&#xA;etc.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-10-19T17:20:31.400" LastActivityDate="2010-10-19T17:20:31.400" CommentCount="1" />
  <row Id="7206" PostTypeId="2" ParentId="7187" CreationDate="2010-10-19T17:25:04.893" Score="0" ViewCount="" Body="&lt;p&gt;One way to see the analogy between the transition from real to complex numbers and from complex numbers to quaternions&#xA;is by means of the fact that the spaces of unit complex numbers and the unit quaternions, namely,&#xA;the unit circle $S^1$ and the unit 3-sphere $S^3$ respectively, are Lie groups.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The analogy can be seen in the group laws, which can be casted in a very similar form as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A point on $S^1$ can be represented by the complex number $z = (x, y)$ , $x^2+y^2= 1$.&#xA;A point on $S^3$ can be represented by a quaternion $q = (z, w)$ , where $z$ and $w$ are complex numbers and&#xA;$|z|^2+|w|^2= 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The $S^1$ multiplication law has the form:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$z_1 z_2 = (x_1x_2 - y_1 y_2, x_1y_2 + y_1 x_2)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The $S^3$ multiplication law is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$q_1 q_2 = (z_1z_2 - w_1 \bar{w_2}, z_1w_2 + w_1 \bar{z_2})$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It easy to verify that these multiplication laws satisfy the group laws. &#xA;Of course the main difference is that the quaternion group law is no longer commutative: since&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$q_2 q_1 = (z_1z_2 - \bar{w_1} w_2, z_1w_2 + \bar{w_1} z_2)$&lt;/p&gt;&#xA;" OwnerUserId="1430" LastActivityDate="2010-10-19T17:25:04.893" />
  <row Id="7207" PostTypeId="1" AcceptedAnswerId="7208" CreationDate="2010-10-19T17:31:20.327" Score="7" ViewCount="109" Body="&lt;p&gt;Given two values $a$ and $b$, how should one go about solving the following inequality for $r$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a \le b^{r}-r$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Applying $log_{b}$ on both sides of the inequality doesn't help me much since that yields the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$log_{b}{a} \le log_{b}{(b^{r}-r)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\log_{z}{x} - \log_{z}{y} = \frac{\log_{z}{x}}{\log_{z}{y}} $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But that doesn't help me to eliminate the exponentiation and solve for $r$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's been ages that I've done algebra, and now I'm back in grad school, and I'm finding this equation in a homework. Can't remember ever seeing a rule of logarithms involving such expressions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The actual homework (after simplification) involves $9 \le 2^{r} - r$ which is trivially solvable by just eyeballing it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But is there a methodical way to solve $a \le b^{r}-r$ for $r$ given any arbitrary numbers $a$ and $b$?&lt;/p&gt;&#xA;" OwnerUserId="2522" LastEditorUserId="367" LastEditorDisplayName="" LastEditDate="2010-10-19T17:40:21.310" LastActivityDate="2010-10-19T18:05:09.120" Title="How do we solve $a \le b^{r}-r$ for $r$?" Tags="&lt;algebra&gt;" AnswerCount="3" />
  <row Id="7208" PostTypeId="2" ParentId="7207" CreationDate="2010-10-19T17:41:59.800" Score="6" ViewCount="" Body="&lt;p&gt;The case of equality can be solved in terms of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Lambert_W_function#Applications&quot; rel=&quot;nofollow&quot;&gt;Lambert W&lt;/a&gt; function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It can be used to solve equations of the form $$p^{ax+b} = cx + d$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(quoted from the wiki page).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$b^{r} -r$ is 'mostly' monotonic, so I suppose solving the equality will be enough to find the solutions to your inequality.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-10-19T17:41:59.800" CommentCount="1" />
  <row Id="7209" PostTypeId="2" ParentId="7207" CreationDate="2010-10-19T17:45:58.910" Score="2" ViewCount="" Body="&lt;p&gt;If you don't have the W function, you need an iterative root finding type of solution.  Note that for your example r is somewhere between 3 and 4.  This equation is nicely behaved, so any of them will work.  See &lt;a href=&quot;http://en.wikipedia.org/wiki/Root-finding_algorithm&quot; rel=&quot;nofollow&quot;&gt;Root finding&lt;/a&gt; for a start&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-10-19T17:45:58.910" CommentCount="1" />
  <row Id="7210" PostTypeId="1" AcceptedAnswerId="7229" CreationDate="2010-10-19T17:47:33.297" Score="6" ViewCount="499" Body="&lt;p&gt;A subset $S$ of a metric space $X$ is totally bounded if for any $r&gt;0$, $S$ can be covered by a finite number of $X$-balls of radius $r$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A metric space $X$ is totally bounded if it is a totally bounded subset of itself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, bounded subsets of $\mathbb{R}^n$ are totally bounded.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there any interesting necessary and/or sufficient conditions for a metric space or its subsets to be totally bounded?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[Background: I was trying to generalize problem 4.8 of baby Rudin which asks you to prove that a real uniformly continuous function on a bounded subset $E$ of the real line is bounded. It seems after a little googling that a more general true statement would require $E$ to be a totally bounded subset of some metric space. But where might we meet such subsets?]&lt;/p&gt;&#xA;" OwnerUserId="1195" LastEditorUserId="1195" LastEditorDisplayName="" LastEditDate="2010-10-21T01:32:44.120" LastActivityDate="2010-11-04T12:41:50.667" Title="Which metric spaces are totally bounded?" Tags="&lt;analysis&gt;&lt;metric-spaces&gt;" AnswerCount="3" CommentCount="0" />
  <row Id="7211" PostTypeId="2" ParentId="7207" CreationDate="2010-10-19T18:05:09.120" Score="2" ViewCount="" Body="&lt;p&gt;You can also get a quick-and-dirty approximation for the equality case with the following approach, which should hold in most situations when $a, b &gt; 1$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rearrange the problem and take logarithms to rewrite it as $$\ln\left(1+ \frac{r}{a}\right) = r \ln b - \ln a.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if $\frac{r}{a}$ is less than 1 (and it should be in most cases where $a, b &gt; 1$), &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ln\left(1+ \frac{r}{a}\right) \approx \frac{r}{a},$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which gives you $$r \approx \frac{- \ln a}{1/a - \ln b}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For your specific problem, this method yields $r \approx 3.78$, whereas the actual answer is closer to $3.66$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can improve this by using the quadratic approximation &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ln\left(1+ \frac{r}{a}\right) \approx \frac{r}{a} - \frac{r^2}{2a^2},$$&#xA;which requires solving a quadratic equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(These approximations come from the Taylor expansion of $\ln (1+x)$.)&lt;/p&gt;&#xA;" OwnerUserId="2370" LastActivityDate="2010-10-19T18:05:09.120" />
  <row Id="7212" PostTypeId="2" ParentId="7210" CreationDate="2010-10-19T18:14:30.323" Score="5" ViewCount="" Body="&lt;p&gt;A metric space is totally bounded if and only if its completion is compact.  (However, a little googling reveals that totally bounded doesn't necessarily imply compact completion without the axiom of choice.)  This gives one explanation of the result in your background, because uniformly continuous functions extend to completions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For subsets of a complete metric space, total boundedness is equivalent to having compact closure, which is what we see for bounded subsets of $\mathbb{R}^n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know, is this interesting?&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-10-19T18:14:30.323" CommentCount="5" />
  <row Id="7213" PostTypeId="1" AcceptedAnswerId="7215" CreationDate="2010-10-19T18:14:37.700" Score="3" ViewCount="329" Body="&lt;p&gt;Use the Lagrange remainder to determine how many terms of the Taylor series of cos expanded&#xA;at 0 are needed to estimate $cos(\frac{21\pi}{5})$ . Accuracy to 3 decimal places&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I have a lot of rough work on paper. I know math.stackexchange dosn't really answer assignment questions but I am looking for a head start on even where to start. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you guys want, i can type up my notes here. Bascically the f'(c) term from the lagrange formula can be 1. I know my inequality has something to do with &amp;lt; 0.001. I need taylor's expansion (obviously to find the n I need). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I just can't put it together. This is my first real-analysis course.&lt;/p&gt;&#xA;" OwnerUserId="287" LastEditorUserId="287" LastEditorDisplayName="" LastEditDate="2010-10-19T18:23:59.620" LastActivityDate="2010-10-19T20:40:57.847" Title="Lagrange remainder and Taylor series" Tags="&lt;real-analysis&gt;&lt;homework&gt;" AnswerCount="3" />
  <row Id="7214" PostTypeId="2" ParentId="7213" CreationDate="2010-10-19T18:21:49.270" Score="1" ViewCount="" Body="&lt;p&gt;You are right that the derivative term in the error is bounded by 1, so you just need the $\frac{(x-a)^{(n+1)}}{(n+1)!}$ less than .001.  As $21\pi/5$ is rather large, it will take a lot of terms.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2010-10-19T18:28:39.420" LastActivityDate="2010-10-19T18:28:39.420" CommentCount="2" />
  <row Id="7215" PostTypeId="2" ParentId="7213" CreationDate="2010-10-19T19:10:43.370" Score="1" ViewCount="" Body="&lt;p&gt;To cut down the number of terms it seems reasonable to use&#xA;$$\cos \left( \frac {21\pi}{5} \right) = \cos \left( \frac{\pi}{5} \right),$$&#xA;so we look for the smallest $n$ such that&#xA;$$ \frac{1}{(n+1)!} \left( \frac{\pi}{5} \right)^{n+1} &amp;lt; \frac{1}{1000}.$$&lt;/p&gt;&#xA;" OwnerUserId="1301" LastActivityDate="2010-10-19T19:10:43.370" CommentCount="3" />
  <row Id="7216" PostTypeId="2" ParentId="7204" CreationDate="2010-10-19T19:12:24.983" Score="5" ViewCount="" Body="&lt;p&gt;This is the special case $\rm\ x,\:n,\:a = 2,\:1,\:0\ $ in Ramanujan's second notebook, chapter XII, entry 4:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\rm x + n + a\ =\ \sqrt{ax + (n+a)^2 + x\sqrt{a(x+n) + (n+a)^2 + (x+n) \sqrt{\cdots}}} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Below is Ramanujan's solution of the given special case - which was submitted to a journal in April 1911. Note that his solution is incomplete (exercise: why?). For further discussion see this 1935 Monthly article, &lt;a href=&quot;http://www.jstor.org/stable/2301294&quot; rel=&quot;nofollow&quot;&gt;Herschfeld: On infinite radicals.&lt;/a&gt;  It also appeared as Problem A6 on the 27th Putnam competition, 1966. Vijayaraghavan proved that a sufficient criterion for the convergence of the following sequence $\ \sqrt{a_1 + \sqrt{a_2 +\:\cdots\: +\sqrt{a_n}}}\ \ $ is that $\rm\displaystyle\ \ {\overline \lim}_{n\to\infty}\frac{\log{a_n}}{2^n}\ &amp;lt; \infty\:.\ $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/UBCbs.jpg&quot; alt=&quot;alt text&quot;&gt; &lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-10-19T19:53:57.927" LastActivityDate="2010-10-19T19:53:57.927" />
  <row Id="7217" PostTypeId="1" AcceptedAnswerId="7286" CreationDate="2010-10-19T19:32:11.200" Score="7" ViewCount="230" Body="&lt;p&gt;Everything that follows takes place in the Borel $\sigma$-algebra with Lebesgue measure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Lebesgue mean of $f$ at $x$ is defined as $\displaystyle \lim_{\epsilon\to 0} \int_{x-\epsilon}^{x+\epsilon} \frac{f}{2\epsilon}$. The function defined by $[f](x) =$ &quot;Lebesgue mean of $f$ at $x$&quot; is equal to $f$ almost everywhere so integrals of it will be equal to integrals of $f$, call this property $P$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Starting with the interval $[0,1]$ and removing the middle $1/4$ then the two middle $1/8$-ths then the four middle $1/16$-ths and so on produces the fat Cantor set which has measure $1/2$ but does not contain any interval (Since measures are countably additive this informs us that the set contains uncountably many points). Let $I$ be the indicator function for the fat Cantor set - it is equal to $1$ when applied on a point of the set and $0$ otherwise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Intuitively I would have thought that the Lebesgue mean $[I]$ of the indicator for the fat Cantor set would be the zero function. Since that contradicts $P$, it seems more plausible that $[I] = I$ but I cannot prove this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can we construct the function $[I]$?&lt;/p&gt;&#xA;" LastActivityDate="2010-10-20T12:28:17.247" Title="What is the Lebesgue mean of the fat Cantor set?" Tags="&lt;analysis&gt;&lt;real-analysis&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="2" />
  <row Id="7218" PostTypeId="1" CreationDate="2010-10-19T19:35:39.167" Score="1" ViewCount="116" Body="&lt;p&gt;What is the name of the following matrix:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{pmatrix} a &amp; b &amp; 0 \\ &#xA;c &amp; d &amp; 0 \\ &#xA;0 &amp; a &amp; b\\ &#xA;0&amp; c&amp; d&amp; \\ &#xA;b &amp; 0 &amp; a \\&#xA;d &amp; 0 &amp; c\end{pmatrix}\ &#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It looks like a &lt;a href=&quot;http://en.wikipedia.org/wiki/Block_matrix#Block_Toeplitz_matrices&quot; rel=&quot;nofollow&quot;&gt;Block Toeplitz&lt;/a&gt; matrix, but usually one defines those by full shifts by (in this case) 2x2 matrices. In particular i'm interested in solving linear equations of this form. Any reference would be appreciated. &lt;/p&gt;&#xA;" LastActivityDate="2010-10-20T17:33:22.947" Title="Block Toeplitz Matrix" Tags="&lt;linear-algebra&gt;&lt;numerical-methods&gt;" AnswerCount="0" CommentCount="3" />
  <row Id="7219" PostTypeId="1" CreationDate="2010-10-19T20:11:04.437" Score="1" ViewCount="67" Body="&lt;p&gt;Let $\Sigma$ be a $n-1$ dimensional space-like submanifold of a $n+1$ dimensional space-time $(V,g)$ and let $x \in \Sigma$. Then $(T_x \Sigma)^\perp$ is of dimension $2$ and is time-like. Such a $2$-plane as $(T_x \Sigma)^\perp$ cuts the future lightcone at $x$ along two linearly independent future null directions. Let $l ^+$ and $l^-$ be two future null vectors along these directions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $e_a$ with $a \in {2,3,4,5,...}$ be a reference frame on $\Sigma$ and hence orthogonal to $l^+$ and $l^-$. Then the following sequence of identities hold,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$K^+_{ab} = K^+(e_a,e_b) = g(l^+,\nabla _{e_a} e_b) = g(e_b,\nabla _ {e_a} l^+) = \nabla _b l^+_a$&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;By metric compatibility and the fact that $l^+$ and $e_b$ are orthogonal I would have expected a minus between the last but two steps. Why is there no negative sign? &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;and similarly,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$K^-_{ab} = \nabla _b l^-_a$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If ${h_{ab}}$ is the induced metric on the submanifold $\Sigma$ then the {\bf Null Mean Curvature} of $\Sigma$ is defined as,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\chi ^+ = h^{ab}K^+_{ab}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\chi ^- = h^{ab}K^-_{ab}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now suppose that $\Sigma$ is an oriented $n-1$ submanifold of a space-like n-submanifold $M$ of $V$. Because of the orientability assumption one can make a continuous choice of unit normal $\nu$ to $\Sigma$ (and tangent to $M$) distinguishing the interior and exterior of $\Sigma$ in $M$. Also define $n$ as the future unit normal to $M$. Choose $\nu$ to be pointing outwards with respect to $\Sigma$ and $n$ to be pointing outwards with respect to $M$. Then one can choose $l^+$ and $l^-$ such that $l^+ = n + \nu$ and $l^- = n - \nu$ ensuring $l^+_\mu l^{-\mu} = -2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The metric ${h_{ab}}$ defined on $\Sigma$ can be lifted to a metric $h$ (calling it the same by abuse of notation) on $V$ by extending it trivially along the two orthogonal directions.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Then apparently the following equation is true which I do not understand, &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;$h^{\mu \nu} = g^{\mu \nu} + \frac{1}{2}(l^{+^\mu} l^{-\nu} + l^{-^\mu} l^{+^\nu})$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(where $\mu$ and $\nu$ run over coordinates of $V$ unlike $a$ and $b$ which run over coordinates of $\Sigma$) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How does one see that the lift to $V$ ($h_{\mu \nu}$) of the pulled back metric $h_{ab}$ and the original metric on $V$ ($g_{\mu \nu}$) are related thus?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now by parallel translation extend $l^+$ and $l^-$ to be tangent field to geodesic congruences. Then the following equation will be true for any $\nu$, &lt;code&gt;$l^{+-\mu}\nabla _{\mu} l^{+-}_\nu = 0$&lt;/code&gt;. Further since these vector fields are also of constant norm one has for any $\nu$, $l^{+-\mu}\nabla &lt;em&gt;{\nu} l^{+-}&lt;/em&gt;\mu = 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the above one gets the following identity for the expansion $\theta = \nabla _\mu l^{+\mu}$ of the geodesic congruence to which $l^+$ is tangent,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\nabla_\mu l^{+\mu} = g^{\mu \nu}\nabla_\mu l^{+}&lt;em&gt;\nu = h^{\mu \nu} \nabla&lt;/em&gt;\mu l^{+}_\nu$&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Now I don't understand why $h^{ab}\nabla_a l^+&lt;em&gt;b = h^{\mu \nu} \nabla&lt;/em&gt;\mu l^+_\nu$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="965" LastEditorUserId="965" LastEditorDisplayName="" LastEditDate="2010-10-19T20:29:35.247" LastActivityDate="2010-10-19T20:29:35.247" Title="A particular pulling back and lifting of metric" Tags="&lt;differential-geometry&gt;" />
  <row Id="7220" PostTypeId="2" ParentId="7213" CreationDate="2010-10-19T20:40:57.847" Score="1" ViewCount="" Body="&lt;p&gt;The Wikipedia article &lt;a href=&quot;http://en.wikipedia.org/wiki/Lagrange_remainder&quot; rel=&quot;nofollow&quot;&gt;Lagrange remainder&lt;/a&gt; has a good discussion.  The Taylor series is a polynomial that approximates a function near a point, calculated by taking derivatives at the point.  The error term is just that-the difference between the function and the polynomial.  There are various forms, given there, for the error of the Taylor series approximation.  All of them are right, but generally not very useful except as bounds, like you have in this problem.  The difficulty is that you don't know the value of $\xi$, just that it is between $a$ and $x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think one of the points of the problem is to see that when $(x-a)$ gets large you need a lot of terms for accuracy.  If you calculate the number of terms at $21\pi/5$ and then again at $\pi/5$ it will be evident.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-10-19T20:40:57.847" />
  <row Id="7221" PostTypeId="1" CreationDate="2010-10-19T20:41:50.637" Score="2" ViewCount="109" Body="&lt;p&gt;If a space has curvature, then the curvature can be seen intrinsically by finding sums of angles in triangles made of geodesics. Under general relativity, space-time is curved on local scales. On global scales, experiments have determined that space-time is almost totally flat.&#xA;If one were to, for each of three points in a gravitational gradient (like near a black hole), find the angle between the sight lines to the other two points and then add up all of the angles, would it add up to 180 (degrees) or perhaps it would show intrinsic curvature? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is the way of thinking about this good, because it is quite general and conceptual. I don't actually know the equations for general relativity, but this is what I thought of when I was thinking about intrinsic geometric properties.&lt;/p&gt;&#xA;" OwnerUserId="2630" LastActivityDate="2011-01-18T14:00:49.563" Title="Effect of curvature of spacetime on intrinsic geometric properties (under general relativity)" Tags="&lt;differential-topology&gt;&lt;curvature&gt;&lt;geodesy&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="7222" PostTypeId="2" ParentId="5962" CreationDate="2010-10-19T21:02:07.453" Score="1" ViewCount="" Body="&lt;p&gt;Let me note that &lt;a href=&quot;http://en.wikipedia.org/wiki/Vector-valued_form&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Vector-valued_form&lt;/a&gt; seems to be highly relevant. They essentially define the bracket. Note that in the case of a 1-form A valued in $\mathfrak{g}$ you get $[A,A]=[A\wedge A (v \otimes w)]= A(v)\otimes A(w)- A(w)\otimes A(v)$, which has value in  $\mathfrak{g} \otimes \mathfrak{g}$. To combine them you use the bracket in $\mathfrak{g}$, so you get $2 [A(x), A(y)]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your definition for $[A, B]$ has a factor of $\frac{1}{(k+l)!}$, which in the case of your vector bundle being over a point does not reproduce the bracket on $\mathfrak{g}$. Maybe it should be $\frac{1}{(k!l!)}$, as in Wikipedia?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: Mmm, this was nonsense. Over a point you'd be doing k=l=0, so it would not matter. Anyhow, Wikipedia definition seems to give constants more consistent with the rest. But I can't do arithmetic, so no guarantees.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, as suggested by Aaron, you can view A as a matrix of 1-forms, and wedge-multiply the matrices. Then $A\wedge A (x, y) = A(x)A(y)-A(y)A(x)= [A(x), A(y)]$ and indeed, $2 A\wedge A= [A,A]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am a bit confused by this wedge notation you give in the question. $A_i$ and $B_j$ differential forms - components of A and B. We then tensor things up, using wedge instead of product, getting a vector in forms valued in the tensor product. We would then need a map $V \otimes V=\mathfrak{g} \otimes \mathfrak{g}$ to $V=\mathfrak{g}$ to get a form with values in $V$ again. Presumably this is the Lie bracket on V, and the result is the 'wedge multiplication of matrices' as above.  The 'issue with skew symmetrization' is that this wedge operation is not the bracket operation on forms, but it's less skewsymmetrized version.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Cartan equation should work out fine. We have 2-form $\Omega$ and 1-form $\omega$. Their bracket on the triple $a,b,c$ is a sum of 6 terms, with a factor of 1/2 in front. These 6 terms pair up into 3 pairs of 2 equal terms each(as $\Omega$ is antisymmetric). So you get 3 terms with coefficient 1, which we then bracket. On the other hand, wedge multiplying the matrices for $\Omega$ and $\omega$ in different order results in same 3 bracket terms (wedging 2 form and 1 form produces 3 distinct terms). One just needs to write down the formulas. (I'm sure there is a general formula for bracket in terms of antisymmetrization of the wedge).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe another point is that $[A,A]=0$ for 'ordinary' $\mathbb{R}$ valued forms is a consequence of the Lie bracket on $\mathbb{R}$ being trivial.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, trace is evidently defined as before - viewing A as a matrix of forms, trace is the summ of the diagonal forms. It is then clear why it commutes with the differential. However, $Tr(A\wedge B) = (-1)^{deg A \times deg B} Tr(B\wedge A)$. This follows from the usual formulas for trace and commuting the forms $A_{i,j}$ past $B_{j,k}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, the trace of $A\wedge A\wedge A\wedge A$. It is $\Sigma_i A^4_{ii}= \Sigma_i \Sigma_j A^2_{ji} \wedge A^2_{ij}$. $A^2_{kl}=\Sigma_m A_{km} \wedge A_{ml}$ is in general non-zero (just take A  a 3 by 3 matrix with 0 in $A_{12}$ entry and  $A_{2,3}\wedge A_{3,1}  \wedge A_{1,3} \wedge A_{3,2}$ non 0). BUT, you are on a 3-manifold (or you'd be doing a differnt C-S form), so there is no 4-form at all!&lt;/p&gt;&#xA;" OwnerUserId="2633" LastEditorUserId="2633" LastEditorDisplayName="" LastEditDate="2010-10-19T21:12:29.033" LastActivityDate="2010-10-19T21:12:29.033" CommentCount="7" />
  <row Id="7223" PostTypeId="1" AcceptedAnswerId="7225" CreationDate="2010-10-19T21:12:18.287" Score="2" ViewCount="98" Body="&lt;p&gt;More specifically, assume d taken from [1, 2^n] and m is taken from [1, n]. What is an upper bound on the probability that d is a multiple of m?&lt;/p&gt;&#xA;" OwnerUserId="2634" LastActivityDate="2010-10-19T22:59:31.537" Title="For an arbitrary positive integer d and random modulus m, what is the probability that d mod m = 0?" Tags="&lt;number-theory&gt;&lt;probability&gt;" AnswerCount="1" />
  <row Id="7224" PostTypeId="1" CreationDate="2010-10-19T21:15:39.150" Score="1" ViewCount="223" Body="&lt;p&gt;The Complex plane, set of all z=x+iy where x and y are real, surface area equals cross product of x and y equals aleph-something (that's not the question). Projecting the plane onto a sphere, and adding the complex infinity to the set (without a value for arg(z)), gives the riemann sphere, which, being a sphere, is compact. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does the addition of infinity, as some sort of bounder, make the complex plane compact? is the riemann sphere without infinity compact, since it has an open boundary? also, if the hyperbolic plane can be shown in a disk, what does that mean? The answer to this question might primarily clarify the definitions in my new studies of topology of manifolds, etc. &#xA;thanks.&lt;/p&gt;&#xA;" OwnerUserId="2630" LastActivityDate="2010-10-20T13:36:24.847" Title="Is the riemann sphere compact even though the complex plane isn't?" Tags="&lt;topology&gt;" AnswerCount="3" FavoriteCount="2" />
  <row Id="7225" PostTypeId="2" ParentId="7223" CreationDate="2010-10-19T21:21:28.030" Score="6" ViewCount="" Body="&lt;p&gt;Fixing $m$, the probability that $d$ is a multiple of $m$ is approximately $1/m$. Now, if you choose $m$ randomly from $[1,n]$, and average $1/m$, you get $\frac{1}{n} \sum_{m=1}^n \frac{1}{m}$, which is approximately $\frac{\log n}{n}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: I didn't read the question carefully enough. Sorry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The least common multiple of the first $n$ numbers grows like $e^n$. So if you take the lcm of the numbers $1 \ldots n \cdot ln 2$, you get a number somewhere around $2^n$ which is going to have as factors all the numbers between $1$ and $n \cdot \ln 2 $. This shows the probability you're interested in is at least $\ln 2$. In fact, this lcm is going to contain as a factor all numbers except prime powers larger than $n \cdot \ln 2$.  Since any prime has at most one power in this interval, then by the prime number theorem, you then get a lower bound for the probability of around $1-1/\ln n$. I expect the upper bound matches, but offhand I don't see how to prove it.&lt;/p&gt;&#xA;" OwnerUserId="2501" LastEditorUserId="2501" LastEditorDisplayName="" LastEditDate="2010-10-19T22:59:31.537" LastActivityDate="2010-10-19T22:59:31.537" CommentCount="1" />
  <row Id="7226" PostTypeId="2" ParentId="7224" CreationDate="2010-10-19T21:28:22.693" Score="4" ViewCount="" Body="&lt;p&gt;Riemann sphere without infinity is just the complex plane again, which is not compact.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason that $\mathbb C$ is not compact is because a sequence like $\{1,2,3,\cdots\}$ which tends to infinity does not have a convergent subsequence, whereas any sequence on the Riemann sphere does.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The addition of infinity of called the one point compactification.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(N.B. what you said about surface area equals cross product of x and y equals aleph-something is probably wrong)&lt;/p&gt;&#xA;" LastActivityDate="2010-10-19T21:28:22.693" />
  <row Id="7227" PostTypeId="2" ParentId="7175" CreationDate="2010-10-19T21:33:20.090" Score="0" ViewCount="" Body="&lt;p&gt;&lt;em&gt;This should be a comment somehwere, but got to long.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Will, the cone on $P^2_\mathbb R$ is &lt;em&gt;not&lt;/em&gt; a manifold. Consider the long exact sequence for integral reduced homology of the pair $(C,C')$ where $C=C(P^2_\mathbb R)$ is the cone over $P^2_\mathbb R$ and $C'=C\setminus\{a\}$ is the complement of the apex of the cone. Since $C$ is contractible and $C'$ deformation-retracts onto $P^2_\mathbb R$, you get isomorphisms $H^\sharp_2(C,C')\cong H^\sharp_{1}(P^2_\mathbb R)\cong\mathbb Z/2\mathbb Z$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If follows that $C$ is not a manifold: in a manifold $M$, for every point $p\in M$ we have that the integral reduced homology $H^\#_\bullet(M,M\setminus\{p\})$ is that of a sphere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Generalizing this reasoning, you get a rather strong condition on a manifold for its cone to be also a manifold. I'm sure the topologists among our fellow M.SEers know of a precise characterization.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the same way, we see that $C$ is not a manifold with boundary, because in such a space $H^\#_\bullet(M,M\setminus\{p\})$ is, for every $p$, either identically zero or that of a sphere.&lt;/p&gt;&#xA;" OwnerUserId="274" LastEditorUserId="274" LastEditorDisplayName="" LastEditDate="2010-10-19T21:56:13.423" LastActivityDate="2010-10-19T21:56:13.423" CommentCount="3" />
  <row Id="7228" PostTypeId="1" AcceptedAnswerId="7233" CreationDate="2010-10-19T21:35:15.473" Score="2" ViewCount="151" Body="&lt;p&gt;True or False? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $\displaystyle \sum_{n=1}^{\infty}a_{n}^2$ is convergent, and$\displaystyle \sum_{n=1}^{\infty} (\frac{\pi}{2} - b_{n})$ is divergent, then $\displaystyle {a_{n}(\sin(b_{n})-1)}$ is a convergent sequence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we know that $(a_{n})^2$ is convergent.&#xA;Therefore, $\lim_{n \to \infty} a_{n}^2 = 0$&#xA;I'm not sure what this tells us about or how to relate this to $(a_{n})$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, we can see that since $-1 \ge \sin(b_{n}) \le 1$&#xA;And therefore, $-2 \ge \sin(b_{n}) - 1 \le 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, again since I don't know what I can conclude about $(a_{n})$, I'm not sure how I can use this fact.&lt;/p&gt;&#xA;" OwnerUserId="2533" LastEditorUserId="2533" LastEditorDisplayName="" LastEditDate="2010-10-19T21:59:47.303" LastActivityDate="2010-10-19T22:18:07.867" Title="Given a convergent $\{(a_{n})^2\}$, divergent $\{b_{n}\}$, prove $\{a_{n}(\sin(b_{n})-1)\}$ converges." Tags="&lt;calculus&gt;&lt;homework&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="7229" PostTypeId="2" ParentId="7210" CreationDate="2010-10-19T21:35:17.313" Score="6" ViewCount="" Body="&lt;p&gt;A metric space is totally bounded if and only if every sequence has a Cauchy subsequence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Try and prove this!)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you might suspect, this is basically equivalent to what Jonas has said.  The key between these two is provided by:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A metric space is compact if and only if it is totally bounded and complete.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, every sequence has a convergent subsequence (compact) if and only if every sequence has a Cauchy sequence (Totally bounded) and every Cauchy sequence converges (complete).&lt;/p&gt;&#xA;" OwnerUserId="640" LastEditorUserId="640" LastEditorDisplayName="" LastEditDate="2010-11-04T12:41:50.667" LastActivityDate="2010-11-04T12:41:50.667" CommentCount="2" />
  <row Id="7230" PostTypeId="2" ParentId="7224" CreationDate="2010-10-19T21:39:54.003" Score="9" ViewCount="" Body="&lt;p&gt;For the Riemann sphere, it is the &lt;a href=&quot;http://en.wikipedia.org/wiki/One-point_compactification&quot; rel=&quot;nofollow&quot;&gt;one point compactification&lt;/a&gt; of the plane. So yes, rather by definition it is compact. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Generally, a non-compact topological space &lt;a href=&quot;http://en.wikipedia.org/wiki/Compactification&quot; rel=&quot;nofollow&quot;&gt;can be compactified by suitable addition of &quot;points at infinity&quot;&lt;/a&gt;. Particularly useful ones (besides the one-point compactification above) include the Stone-Cech compactification, which in some sense adds the &quot;most points&quot;, and whose existence for any topological space follows from axiom of choice. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another compactification that often arises is the Penrose/conformal compactification of space-time in general relativity. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A propos your question on hyperbolic plane. Topologically the hyperbolic plane is the same as the normal plane. By considering the Poincare disk model and &quot;adding&quot; the boundary of the disk, you compactify the hyperbolic plane. As mentioned above, any topological space admits some compactification. What's more useful is to consider special compactifications: for example the one-point compactification of the complex plane into the Riemann sphere is conformal relative to the additional &lt;em&gt;geometrical structure&lt;/em&gt; on the two surfaces. Similarly the Penrose compactifications respect the geometrical structure on space-times. &lt;/p&gt;&#xA;" OwnerUserId="1543" LastActivityDate="2010-10-19T21:39:54.003" />
  <row Id="7231" PostTypeId="1" AcceptedAnswerId="7236" CreationDate="2010-10-19T22:14:10.950" Score="0" ViewCount="275" Body="&lt;p&gt;Based on a &lt;a href=&quot;http://math.stackexchange.com/questions/7126/boolean-algebra-simplification-dont-know-the-method-used&quot;&gt;previous question&lt;/a&gt; (and this ought to be the last of this sequence I promise), I just need to know one more thing.  According to this truth table:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/9T1x4.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm told the answer I should be getting by using the &lt;strong&gt;sum of products&lt;/strong&gt; method on the truth table should be:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;T = R ∙ (RCGM)'&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Unlike before with the Karnaugh map, I can't see how this was arrived at.  All I can get is:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;T = RC'G'M'+RC'G'M+RC'GM'+RC'GM+RCG'M'+RCG'M+RCGM'&#xA;T = R(C'G'M'+C'G'M+C'GM'+C'GM+CG'M'+CG'M+CGM')&#xA;T = R(C'(G'M'+G'M+GM'+GM)+C(G'M'+G'M+GM'))&#xA;T = R(C'(G'(M'+M)+G(M'+M))+C(G'(M'+M)+GM'))&#xA;T = R(C'(G'+G)+C(G'+GM'))&#xA;T = R(C'+C(G'+M'))&#xA;T = R(C'+G'+M')&#xA;T = R(CGM)'&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This is the right answer I believe, but it's not the same as what I should supposedly have found.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks again in advance!&lt;/p&gt;&#xA;" OwnerUserId="2609" LastEditorUserId="2609" LastEditorDisplayName="" LastEditDate="2010-10-19T22:34:56.920" LastActivityDate="2010-10-20T19:54:48.807" Title="Boolean Algebra, Simplification: Sum of products from truth table" Tags="&lt;boolean-algebra&gt;" AnswerCount="2" />
  <row Id="7232" PostTypeId="1" CreationDate="2010-10-19T22:14:37.643" Score="0" ViewCount="289" Body="&lt;p&gt;I have a text book which gives two diagrams of beads suspended from strings with different angles, and worked examples to calculate the tension in the strings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first bead is suspended by 2 strings attached to a rod, both strings at angles to the vertical (the bead is at the bottom, the rod is across the top). The text book tells me to label the tensions T1 and T2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the next diagram, one of the threads is now vertical, while the other is at an angle.  This time the text book tells me the tensions are equal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why are they different in one layout, but equal in another?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Tim&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Agh, sorry - missed some vital info!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the second diagram, there is a horizontal force applied to the bead to keep it in equilibrium!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I still don't understand why the tensions in the strings would be equal though!)&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;And some more vital info. I think I'm almost there now.  In the second example, the bead is suspended not by two pieces of string, but by 1! The bead is somewhere in the middle of the string.  The diagrams are so similar, I thought there were two strings in each diagram! (That'll teach me to read the question!)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suspect the tensions in the two parts of the string have to be equal because otherwise the string wouldn't be in equilibrium. Or something.  Can anyone confirm that for me?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(For the interested, the text book is the Edexcel AS and A Level Modular Mathematics. The page is 98.)&lt;/p&gt;&#xA;" OwnerUserId="2635" LastEditorUserId="2635" LastEditorDisplayName="" LastEditDate="2010-10-22T06:31:09.973" LastActivityDate="2011-01-20T06:46:40.000" Title="If a bead is suspended by 2 strings, one vertical, the other diagonally, why are the tensions equal?" Tags="&lt;physics&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="7233" PostTypeId="2" ParentId="7228" CreationDate="2010-10-19T22:18:07.867" Score="3" ViewCount="" Body="&lt;p&gt;In the current form of the question, the condition on $b_n$ is irrelevant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you said, $a_n^2\to0$.  This implies that $a_n\to0$.  Then because $|a_n(\sin(b_n)-1)|=|a_n||\sin(b_n)-1|$ and, for the reason you indicated, $|\sin(b_n)-1|\leq2$ for all $n$, the $n^{th}$ term in the sequence has absolute value less than or equal to $2|a_n|$, which converges to 0.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-10-19T22:18:07.867" CommentCount="5" />
  <row Id="7234" PostTypeId="2" ParentId="6414" CreationDate="2010-10-19T22:35:34.120" Score="1" ViewCount="" Body="&lt;p&gt;This is actually easier if you use charts. Suppose your smooth structure is given by a maximal atlas C.&#xA;Near any point $p$ in $TU$ you have a neighborhood $V=\pi^{-1} W$ for $W$ - a chart neighborhood of $\pi(p)$, so that on $V$ which there are functions $x_i \cdot\pi$ which are smooth by property (1) and functions $\frac{\partial}{\partial x_i}$, which are smooth by property (2).  But they also define a chart - a map $V$ to $R^{2n}$. So this chart has to be smooth. So we have a cover by charts smooth in both the standard atlas and the maximal atlas C. Hence the two smooth structures coincide.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Trying to do it in sheave seems a bit strange. If your structure sheaf is F, you get $2n$ functions in $F(V)$ and the fact that they define a chart should mean that you can pull back (restriction of) any function f in F to a function on part of $R^{2n}$, so that restriction is in the standard structure sheaf, and hence, (by sheaf axiom) f is too. This seems slightly fishy, even though can probably be fixed, in which case it will be the sheaf translation of the above argument.&lt;/p&gt;&#xA;" OwnerUserId="2633" LastActivityDate="2010-10-19T22:35:34.120" CommentCount="3" />
  <row Id="7235" PostTypeId="1" AcceptedAnswerId="7239" CreationDate="2010-10-19T22:37:05.047" Score="2" ViewCount="61" Body="&lt;p&gt;Suppose we look at $k$-digit strings with digits between $1$ and $n$. How many distinct classes of strings are there when digit identity and order doesn't matter?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More formally, what is the number of equivalence classes over strings $d_1,\ldots,d_k$ with $d_i\in \{1,\ldots, n\}$ where strings $d,\hat{d}$ are considered equivalent iff there exist bijections $\sigma$ over $\{1,\ldots, k\} $ and $\delta$ over $\{1,\ldots, n\}$ such that the following holds for all $j \in \{1,\ldots,k\}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$d_{\sigma(j)}=\delta(\hat{d}_j)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Edit&lt;/b&gt;: this seems to correspond to the number of ways to partition integer $k$ into at most $n$ parts&lt;/p&gt;&#xA;" OwnerUserId="998" LastEditorUserId="998" LastEditorDisplayName="" LastEditDate="2010-10-20T02:31:26.853" LastActivityDate="2010-10-20T02:31:26.853" Title="Number of classes of k-digit strings when digit order and identity doesn't matter" Tags="&lt;combinatorics&gt;" AnswerCount="1" />
  <row Id="7236" PostTypeId="2" ParentId="7231" CreationDate="2010-10-19T23:17:24.630" Score="0" ViewCount="" Body="&lt;p&gt;The two results are the same.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;R(RCGM)'=RR'+RC'+RG'+RM'=RC'+RG'+RM'=R(CGM)'&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;On the other hand, this means the second &lt;code&gt;R&lt;/code&gt; is kind of redundant, so I don't know why you'd want to write the solution this way rather than your way.&lt;/p&gt;&#xA;" OwnerUserId="2215" LastActivityDate="2010-10-19T23:17:24.630" />
  <row Id="7237" PostTypeId="2" ParentId="7143" CreationDate="2010-10-19T23:18:34.767" Score="5" ViewCount="" Body="&lt;p&gt;Although you already have a couple of answers proving the result you ask for, I think it is worth adding a further answer using a completely different approach. There is a simple way of parameterizing all points on an algebraic curve of degree two which, once you know, the answer to your question just drops out really easily.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[Edit: Actually, I didn't realize how close this was to Moron's answer when I posted. Still, my parameterization is a bit more general, so I'll leave it up.]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, the parameterization: For any Pythagorean triple &lt;em&gt;a&lt;/em&gt;&lt;sup&gt;2&lt;/sup&gt;&amp;nbsp;+&amp;nbsp;&lt;em&gt;b&lt;/em&gt;&lt;sup&gt;2&lt;/sup&gt;&amp;nbsp;=&amp;nbsp;&lt;em&gt;c&lt;/em&gt;&lt;sup&gt;2&lt;/sup&gt; you can divide through by &lt;em&gt;c&lt;/em&gt; to get the equation &lt;em&gt;a&lt;/em&gt;&lt;sup&gt;2&lt;/sup&gt;&amp;nbsp;+&amp;nbsp;&lt;em&gt;b&lt;/em&gt;&lt;sup&gt;2&lt;/sup&gt;&amp;nbsp;=&amp;nbsp;1 in rational pairs (&lt;em&gt;a&lt;/em&gt;,&lt;i&gt;b&lt;/i&gt;)&amp;nbsp;&amp;isin;&amp;nbsp;Q&lt;sup&gt;2&lt;/sup&gt;. The set of all such rational (&lt;em&gt;a&lt;/em&gt;,&lt;i&gt;b&lt;/i&gt;) is nothing other than the rational points in the unit circle S&lt;sup&gt;1&lt;/sup&gt;. Conversely, if you have rational points (&lt;em&gt;a&lt;/em&gt;,&lt;i&gt;b&lt;/i&gt;)&amp;nbsp;&amp;isin;&amp;nbsp;S&lt;sup&gt;1&lt;/sup&gt; then you obtain a Pythagorean triple by multiplying through by a common denominator &lt;em&gt;c&lt;/em&gt; (which does not affect the angle).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now pick any rational point on the unit circle (&lt;em&gt;a&lt;/em&gt;&lt;sub&gt;0&lt;/sub&gt;,&lt;em&gt;b&lt;/em&gt;&lt;sub&gt;0&lt;/sub&gt;) and a nonzero vector (&lt;em&gt;u&lt;/em&gt;,&lt;i&gt;v&lt;/i&gt;), and consider the line t&amp;nbsp;&amp;rarr;&amp;nbsp;(&lt;em&gt;a&lt;/em&gt;&lt;sub&gt;0&lt;/sub&gt;+&lt;em&gt;tu&lt;/em&gt;,&lt;i&gt;b&lt;/i&gt;&lt;sub&gt;0&lt;/sub&gt;+&lt;em&gt;tv&lt;/em&gt;). This must intersect the circle at exactly one other point (&lt;em&gt;a&lt;/em&gt;&lt;sub&gt;1&lt;/sub&gt;,&lt;em&gt;b&lt;/em&gt;&lt;sub&gt;1&lt;/sub&gt;) -- unless (&lt;em&gt;u&lt;/em&gt;,&lt;i&gt;v&lt;/i&gt;) is tangent to.the circle (in which case we can say that it intersects it twice at (&lt;em&gt;a&lt;/em&gt;&lt;sub&gt;0&lt;/sub&gt;,&lt;em&gt;b&lt;/em&gt;&lt;sub&gt;0&lt;/sub&gt;) ). In fact, the equation &lt;em&gt;a&lt;/em&gt;&lt;sup&gt;2&lt;/sup&gt;&amp;nbsp;+&amp;nbsp;&lt;em&gt;b&lt;/em&gt;&lt;sup&gt;2&lt;/sup&gt;&amp;nbsp;-&amp;nbsp;1&amp;nbsp;=&amp;nbsp;0 is a quadratic in t with zero constant term (as it is solved by &lt;em&gt;t&lt;/em&gt;&amp;nbsp;=&amp;nbsp;0), so extracting out the factor of &lt;em&gt;t&lt;/em&gt; gives a linear expression. This means that it intersects the curve at a rational point. To be explicit, the equation you get for &lt;em&gt;t&lt;/em&gt; is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;(u^2+v^2)t^2+2(a_0u+b_0v)t=0\ \Rightarrow\ t=-2(a_0u+b_0v)/(u^2+v^2).&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conversely, given any rational point (&lt;em&gt;a&lt;/em&gt;&lt;sub&gt;1&lt;/sub&gt;,&lt;em&gt;b&lt;/em&gt;&lt;sub&gt;1&lt;/sub&gt;) on the circle distinct from (&lt;em&gt;a&lt;/em&gt;&lt;sub&gt;0&lt;/sub&gt;,&lt;em&gt;b&lt;/em&gt;&lt;sub&gt;0&lt;/sub&gt;) you can always take (&lt;em&gt;u&lt;/em&gt;,&lt;i&gt;v&lt;/i&gt;)&amp;nbsp;=&amp;nbsp;(&lt;em&gt;a&lt;/em&gt;&lt;sub&gt;1&lt;/sub&gt;&amp;nbsp;-&amp;nbsp;&lt;em&gt;a&lt;/em&gt;&lt;sub&gt;0&lt;/sub&gt;,&lt;em&gt;b&lt;/em&gt;&lt;sub&gt;1&lt;/sub&gt;&amp;nbsp;-&amp;nbsp;&lt;em&gt;b&lt;/em&gt;&lt;sub&gt;1&lt;/sub&gt;) to see that every such point is reached in this way.&#xA;This parameterizes the rational points on the circle by the nonzero rational vectors (&lt;em&gt;u&lt;/em&gt;,&lt;i&gt;v&lt;/i&gt;)&amp;nbsp;&amp;isin;&amp;nbsp;Q&lt;sup&gt;2&lt;/sup&gt; (up to recaling). In fact, dividing through by &lt;em&gt;u&lt;/em&gt; (if it is nonzero), this parameterizes the rational points on the circle by &lt;em&gt;v&lt;/em&gt;&amp;nbsp;&amp;isin;&amp;nbsp;Q and the additional point parameterized by (0,1) curresponding to &lt;em&gt;u&lt;/em&gt;&amp;nbsp;=&amp;nbsp;0. This can be seen to be nothing other than &lt;a href=&quot;http://en.wikipedia.org/wiki/Stereographic_projection&quot; rel=&quot;nofollow&quot;&gt;stereographic projection&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now to your question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pick any point (&lt;em&gt;x&lt;/em&gt;,&lt;i&gt;y&lt;/i&gt;) on the unit circle and approximate the vector (&lt;em&gt;x&lt;/em&gt;&amp;nbsp;-&amp;nbsp;a&lt;sub&gt;0&lt;/sub&gt;,&lt;em&gt;y&lt;/em&gt;&amp;nbsp;-&amp;nbsp;b&lt;sub&gt;0&lt;/sub&gt;) as closely as you like by rational (&lt;em&gt;u&lt;/em&gt;,&lt;i&gt;v&lt;/i&gt;). The curve t&amp;nbsp;&amp;rarr;&amp;nbsp;(&lt;em&gt;a&lt;/em&gt;&lt;sub&gt;0&lt;/sub&gt;+&lt;em&gt;tu&lt;/em&gt;,&lt;i&gt;b&lt;/i&gt;&lt;sub&gt;0&lt;/sub&gt;+&lt;em&gt;tv&lt;/em&gt;) intersects the circle at a rational point, which can be made as close as you like to (&lt;em&gt;x&lt;/em&gt;,&lt;i&gt;y&lt;/i&gt;). That is, &lt;i&gt;the rational points are dense in the unit circle&lt;/i&gt;. So, the angles of pythagorean triples will also be dense in the first quadrant.&lt;/p&gt;&#xA;" OwnerUserId="1321" LastEditorUserId="1321" LastEditorDisplayName="" LastEditDate="2010-10-20T00:57:03.840" LastActivityDate="2010-10-20T00:57:03.840" />
  <row Id="7238" PostTypeId="1" CreationDate="2010-10-20T00:18:02.650" Score="1" ViewCount="441" Body="&lt;p&gt;So for a writing assignment in one of my classes we are asked to discuss and prove some basic results about compact sets in general topological spaces. I like proving these things, but they dont help me understand what a compact( locally compact, paracompact,...) set in a topological space &quot;looks like.&quot; That said I'm asking for some examples of compact (locally compact, ...) sets in a variety of topological spaces. I'm also interested in some explanation of what extra &quot;benefits&quot; are brought about by singling out these compact sets. For example the p-adic numbers are locally compact and locally compact things (abelian groups to be precise) are a good setting in which to carry out fourier analyis.&lt;/p&gt;&#xA;" OwnerUserId="2636" LastActivityDate="2010-10-20T09:23:13.720" Title="What does a compact set look like?" Tags="&lt;real-analysis&gt;&lt;topology&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="2" />
  <row Id="7239" PostTypeId="2" ParentId="7235" CreationDate="2010-10-20T00:19:00.453" Score="3" ViewCount="" Body="&lt;p&gt;If I've read this right, you're looking to count k-tuples of integers in {1,..,n} that are inequivalent under the operations (a) permute the coordinates and (b) permute the integers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, for example, when k=7 and n=2, 1122212 would be equivalent to 2211121, 1112222 and 1111222 (and many others).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe it would be useful to construct a &lt;em&gt;canonical form&lt;/em&gt; for each class:   In this case, we can take the lexicographic first element in the equivalence class.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So these would be in canonical form:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;1111222&#xA;1111122&#xA;1111112&#xA;1111111&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;whereas 1112222 would not be since it belongs to the same class as 1111222.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These canonical forms are equivalent to ordered partitions of k into n parts (some of which can be zero).  For example 1111222 &amp;lt;-&gt; 4+3 since there are four 1's and three 2's.  [If instead n=3, then 1111222 &amp;lt;-&gt; 4+3+0 since there are no 3's in the string.]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The number of these is the number $p_n(k)$ of partitions of k into at most n parts (afterwards, we can append the zeroes to achieve n parts).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can compute $p_n(k)$ using the recurrence relation $p_n(k)=p_n(k-n)+p_{n-1}(k-1)$ (with some appropriate boundary conditions).  [see: Enumerative combinatorics, Volume 1 By Richard P. Stanley p. 28]&lt;/p&gt;&#xA;" OwnerUserId="139" LastActivityDate="2010-10-20T00:19:00.453" CommentCount="1" />
  <row Id="7240" PostTypeId="2" ParentId="7238" CreationDate="2010-10-20T01:18:42.860" Score="2" ViewCount="" Body="&lt;p&gt;I don't know if I understand your question, but here are some elementary examples of compact and non compact spaces. You can find more examples in &lt;a href=&quot;http://en.wikipedia.org/wiki/Compactness&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$[0,1]$ is compact with the subspace topology of the usual, Euclidean topology on $\mathbb{R}$.&lt;/li&gt;&#xA;&lt;li&gt;Neither $(0,1)$, nor $\mathbb{R}$ are compact with the same topology.&lt;/li&gt;&#xA;&lt;li&gt;A set with a finite number of points $A \subset \mathbb{R}$ is always compact. In fact, it is always compact, with no matter which topology.&lt;/li&gt;&#xA;&lt;li&gt;With the usual topology on $\mathbb{R}^n$, $A \subset \mathbb{R}^n$ is compact if and only if it is bounded and closed. Beware: this characterization of compactness is no longer true for arbitrary metric spaces: for instance, with the bounded metric $d(x,y) = \mathrm{min} \left\{ \| x-y\|,1 \right\}$ (which is equivalent to the Euclidean one), every subset $A \subset \mathbb{R}^n$, even the whole $\mathbb{R}^n$, is bounded. So, if the former characterization was still true with this metric, every closed subset of $\mathbb{R}^n$ ($\mathbb{R}^n$ itself included) would be compact.&lt;/li&gt;&#xA;&lt;li&gt;This countable set $\left\{\frac{1}{n}\ \vert \ n \in \mathbb{N} \right\} \cup \left\{ 0\right\} \subset \mathbb{R}$ is compact (with the usual topology).&lt;/li&gt;&#xA;&lt;li&gt;This countable set $\left\{\frac{1}{n}\ \vert \ n \in \mathbb{N} \right\} \subset \mathbb{R}$ is not compact (with the usual topology).&lt;/li&gt;&#xA;&lt;li&gt;This countable set $\mathbb{N} \subset \mathbb{R}$ is not compact (with the usual topology).&lt;/li&gt;&#xA;&lt;li&gt;With the &lt;a href=&quot;http://en.wikipedia.org/wiki/Finite_complement_topology&quot; rel=&quot;nofollow&quot;&gt;finite complement topology&lt;/a&gt;, $\mathbb{R}$ is compact. In fact, with this topology on the real line, every subset $A \subset \mathbb{R}$ is compact. In fact, every topological space with the finite complement topology is compact.&lt;/li&gt;&#xA;&lt;li&gt;$\mathbb{Q}$ is not compact with the subspace topology of the usual, Euclidean topology on $\mathbb{R}$. &lt;/li&gt;&#xA;&lt;li&gt;Btw, the empty set is always compact.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="664" LastActivityDate="2010-10-20T01:18:42.860" />
  <row Id="7241" PostTypeId="2" ParentId="7143" CreationDate="2010-10-20T01:30:53.873" Score="4" ViewCount="" Body="&lt;p&gt;The other answers here explain that it is possible to find arbitrarily good approximations to any angle using Pythagorean triples.  Here's a footnote on how to turn this into a constructive procedure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moron above says to pick a rational number $n/m$ near $t$ where $t$ satisfies $\tan \theta = 2t/(1 - t^2)$.  You could pick $n$ and $m$ using a &lt;a href=&quot;http://www.johndcook.com/blog/2010/10/20/best-rational-approximation/&quot; rel=&quot;nofollow&quot;&gt;Farey algorithm&lt;/a&gt; to find the best rational approximation to $t = (1 - \cos \theta)/\sin\theta$ with a given denominator.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Start with $a = 0$, $b = c = d = 1$.  The number $t$ is between $a/b$ and $c/d$.  Next compute the &quot;mediant&quot; $(a+c)/(b+d)$.  If $t$ is bigger than the mediant, update $a$ and $b$ with $a+c$ and $b+d$.  Otherwise update $c$ with $a+c$ and $d$ with $b+d$. Each iteration of this process produces the best rational approximation to $t$ for a given denominator size.&lt;/p&gt;&#xA;" OwnerUserId="247" LastEditorUserId="247" LastEditorDisplayName="" LastEditDate="2010-10-21T00:10:27.133" LastActivityDate="2010-10-21T00:10:27.133" />
  <row Id="7242" PostTypeId="1" AcceptedAnswerId="7244" CreationDate="2010-10-20T01:31:06.883" Score="0" ViewCount="290" Body="&lt;p&gt;For example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the limit of $e^t$ as $t$ approaches infinity is simply infinity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but the limit of $e^{t^2}$ cannot be found because it explodes?  Is this due to the fact that there are differing types of infinity and the second limit produces a non-&quot;countable&quot; infinity?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://math.stackexchange.com/questions/1/different-kinds-of-infinities&quot;&gt;see referenced question&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1602" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-10-20T02:30:38.910" LastActivityDate="2010-10-21T16:22:34.890" Title="Why do some limits &quot;explode&quot; to infinity?" Tags="&lt;limit&gt;&lt;asymptotics&gt;" AnswerCount="3" CommentCount="5" />
  <row Id="7243" PostTypeId="2" ParentId="7224" CreationDate="2010-10-20T01:47:41.023" Score="3" ViewCount="" Body="&lt;p&gt;The complex plane $\mathbb{C}$ is not compact. For instance, here is an open cover that does not admit any finite subcover:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;{\mathcal U} = \left\{ U_n\right\} \qquad \text{with} \qquad U_n = \left\{ z \in \mathbb{C} \ \vert \ \vert z \vert &amp;lt; n \right\} \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason why ${\mathcal U}$ does not admit any finite subcover is easy: if you pick just a finite number of $U_n$'s, then their union would be equal to the biggest one of them, which is not the whole $\mathbb{C}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, thanks to the &lt;a href=&quot;http://en.wikipedia.org/wiki/Stereographic_projection&quot; rel=&quot;nofollow&quot;&gt;stereographic projection&lt;/a&gt;, think of these $U_n$'s inside the Riemann sphere $S^2$. They do not form a cover of $S^2$, since $\infty$ (= the North Pole) is not included in any of them. So, we need to add an open neighborhood of the infinity. For instance, the image of &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;V = \left\{ z \in \mathbb{C} \ \vert \ \vert z \vert &gt; 394 \right\}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;by the stereographic projection. Then, we would have an open cover of $S^2$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;{\mathcal U}' = \left\{ U_n \ \vert \ n \in \mathbb{N}\right\} \cup \left\{ V\right\} \ ,&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which admits and open subcover. For instance,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;U_{395} \ , \quad V \ .&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="664" LastEditorUserId="664" LastEditorDisplayName="" LastEditDate="2010-10-20T13:36:24.847" LastActivityDate="2010-10-20T13:36:24.847" />
  <row Id="7244" PostTypeId="2" ParentId="7242" CreationDate="2010-10-20T01:53:28.493" Score="8" ViewCount="" Body="&lt;p&gt;The limit can be found. What are we saying when we assert that $lim_{t\to\infty} f(t) = \infty$, with $f:\mathbb{R} \to \mathbb{R}$? What does it mean?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It means, &lt;em&gt;by definition&lt;/em&gt;, that given $M &gt;0$ there exists $\varepsilon &gt;0$ such that if $t&gt;\varepsilon$ then $f(t) &gt; M$. Simply this. In (not so rigorous) words, for arbitrary large values of the parameter $t$ the function assumes arbitrary large values. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is easy to see that the limit you want is indeed $\infty$ (in the sense just described). This can be seen noticing that $e^t &amp;lt; e^{t^2}$ for $t&gt;1$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think that this question is not related to the linked one. For me, they are different things. &lt;/p&gt;&#xA;" OwnerUserId="1178" LastEditorUserId="1178" LastEditorDisplayName="" LastEditDate="2010-10-21T16:22:34.890" LastActivityDate="2010-10-21T16:22:34.890" />
  <row Id="7245" PostTypeId="2" ParentId="7242" CreationDate="2010-10-20T02:13:56.213" Score="1" ViewCount="" Body="&lt;p&gt;Given your mentioning of the word &quot;countable&quot; (and looking at the link), I think that the &quot;differing types of infinity&quot; that you are thinking of are different infinite cardinalities (i.e. different possible infinite sizes of sets).  This notion of infinity is not the same as the notion of infinity that is used when one says that a limit approaches infinity.  (Of course, they are related at a basic level, in that both allude to the idea of being larger than any finite number, but at a more technical level, they are different notions.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ronaldo in his answer has given an explanation of what it means for &#xA;$\displaystyle \lim_{t\to\infty} f(t)$ to equal $\infty$, and this is just as valid for $e^{t^2}$ as it is for $e^t$.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As Dan Ramras notes in one of his comments, the &lt;em&gt;rate&lt;/em&gt; at which $e^{t^2}$ approaches infinity&#xA;is much greater than the rate at which $e^t$ approaches infinty (which is turn much&#xA;greater than the rate at which $t$ approaches infinity, for example).  In terms of Ronaldo's answer, this means that for a given $M$, the $\epsilon$ that you have to choose&#xA;to be sure that $e^{t^2} &gt; M$ when $t &gt; \epsilon$ is smaller than in the corresponding situation for $e^t$.  But both quantities approach infinity as $t$ does.&lt;/p&gt;&#xA;" OwnerUserId="221" LastEditorUserId="221" LastEditorDisplayName="" LastEditDate="2010-10-20T14:57:44.253" LastActivityDate="2010-10-20T14:57:44.253" CommentCount="2" />
  <row Id="7246" PostTypeId="2" ParentId="7242" CreationDate="2010-10-20T02:37:07.833" Score="5" ViewCount="" Body="&lt;p&gt;To take a different perspective on it (if this is what you're looking for), there &lt;em&gt;are&lt;/em&gt; limits that can't be found; the limit of $1/x$ as $x\rightarrow 0$ is one example.  The limit doesn't exist because the limit coming from the right and from the left are different: one is $\infty$ and one is $-\infty$.  I guess you could say that $1/x$ &quot;explodes&quot; at 0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's important to distinguish the &lt;em&gt;type&lt;/em&gt; of infinity you're talking about, too.  The &quot;countable and uncountable&quot; infinities you mention are &lt;em&gt;cardinalities&lt;/em&gt;: they measure the number of elements in sets.  The infinities you get with limits of functions like these are just limits of real numbers, and don't describe sets at all.  You could call them &quot;uncountable&quot; since $\mathbb{R}$ is uncountable, I suppose, but then there wouldn't be a &quot;countable&quot; infinity in between $\infty$ and the real numbers.  Basically, &quot;infinity&quot; can mean a lot of different things.&lt;/p&gt;&#xA;" OwnerUserId="2215" LastActivityDate="2010-10-20T02:37:07.833" />
  <row Id="7247" PostTypeId="1" AcceptedAnswerId="7279" CreationDate="2010-10-20T02:38:38.527" Score="5" ViewCount="90" Body="&lt;p&gt;Let $f(x) \in L^p(\mathbb{R})$ and $K \in C^m(\mathbb{R})$. Can I then say that $(f \ast K) (x) = \int_{\mathbb{R}} f(t) K(x-t) dt$ is in $C^m$?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that this is true if $K$ has compact support, but I was wondering if it is possible to have a stronger result (perhaps $K$ vanishing at $\infty$?). &lt;/p&gt;&#xA;" OwnerUserId="1736" LastActivityDate="2010-10-20T10:01:11.697" Title="Differentiability of Convolutions" Tags="&lt;real-analysis&gt;" AnswerCount="1" />
  <row Id="7248" PostTypeId="1" CreationDate="2010-10-20T02:43:11.423" Score="1" ViewCount="112" Body="&lt;p&gt;I'm attempting to come up with a first order language $L$ that is able to describe vector spaces over fields. I came up with a few sets of nonlogical symbols.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$Rs_L=\{Scal,Vec\}$ where $Scal,Vec$ are unary relation symbols which I interpret to be true or false depending on whether $x$ is a scalar or vector.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, $Fs_L=\{+,\cdot,\}$ which are the usual addition and multiplication for both vectors and scalars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For constant symbols, $Cs_L=\{0,1,0_V\}$, where $0$ and $1$ are regular field scalars for the additive and multiplicative identities, respectively, and $0_V$ is the additive identity for vectors. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using these, I was able to formulate axioms for scalar multiplication on a given structure, and I define addition between a scalar and a vector to simply always give the 0 vector and define multiplication between two vectors to always be the 0 vector. This gives a theory such that a structure models it if and only if it is a vector space in the regular algebraic sense. It is easy to see that all linear transformations in the algebraic sense are included in the class of homomorphisms (in the logical sense), but I am having difficulty proving they are the only ones in this class. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suppose I want to show that for any $L$-homomorphism $T$, $T(av+w)=aTv+Tw$ for $a$ a scalar and $v,w$ vectors. Since $T$ is a homomorphism, I see that $T(av+w)=Tav+Tw=TaTv+Tw$, so it would be enough to show that all homomorphisms act as the identity on all scalars? Is there a way to show this, or an extra axiom to include?&lt;/p&gt;&#xA;" LastActivityDate="2010-10-20T11:00:03.710" Title="First Order Language for vector spaces over fields" Tags="&lt;linear-algebra&gt;&lt;logic&gt;&lt;vector-space&gt;" AnswerCount="2" />
  <row Id="7249" PostTypeId="1" CreationDate="2010-10-20T03:57:21.363" Score="7" ViewCount="141" Body="&lt;p&gt;It's kind of an infamous problem in differential equations to find the correct road surface so that a car with square wheels (and an axle located in the center) keeps its axle level as it drives along.  I hope I won't offend anybody by saying that one smooth piece of the solution (for a wheel with sides of length 2) is $y = -\cosh(x)$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you actually take this solution and describe the position of the axle at any given point, unless I have calculated incorrectly you find that the axle is always positioned directly over the point where the wheel makes contact with the road.  I've been unable to come up with a physical justification of this phenomenon and it seems fairly non-obvious to me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a straightforward reason why this must be true?  Is it specific to this wheel shape?&lt;/p&gt;&#xA;" OwnerUserId="1925" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-10-20T04:38:35.830" LastActivityDate="2010-10-21T04:15:25.823" Title="Question about square-wheeled cars." Tags="&lt;calculus&gt;&lt;differential-equations&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="1" />
  <row Id="7250" PostTypeId="2" ParentId="7248" CreationDate="2010-10-20T04:01:43.637" Score="0" ViewCount="" Body="&lt;p&gt;It's not clear what your motivations are but, in any case, I thought it would be worthwhile to point out that usually the most convenient way to specify a vector space is to view it as a &quot;group with operators&quot;, i.e. for each scalar $c\:$ introduce a unary operation $\ c(v) = c\:v\:$. &lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-10-20T04:01:43.637" CommentCount="1" />
  <row Id="7252" PostTypeId="1" CreationDate="2010-10-20T04:10:38.073" Score="0" ViewCount="150" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Possible Duplicate:&lt;/strong&gt;&lt;br&gt;&#xA;  &lt;a href=&quot;http://math.stackexchange.com/questions/4764/sine-function-dense-in-1-1&quot;&gt;Sine function dense in $[-1,1]$&lt;/a&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;p&gt;How do i prove this is dense??&lt;/p&gt;&#xA;" LastActivityDate="2010-10-20T04:16:43.613" ClosedDate="2010-10-20T04:47:15.867" Title="Is sin(n) for all n in the intergers dense?" Tags="&lt;real-analysis&gt;" AnswerCount="1" CommentCount="0" />
  <row Id="7253" PostTypeId="2" ParentId="7249" CreationDate="2010-10-20T04:14:55.080" Score="1" ViewCount="" Body="&lt;p&gt;Quite an exposition is in &lt;a href=&quot;http://poncelet.math.nthu.edu.tw/disk5/js/cardioid/1.pdf&quot; rel=&quot;nofollow&quot;&gt;Stan Wagon's paper&lt;/a&gt;  If you look at Fig 13, it appears that the axle of the ellipse is not over the contact point.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-10-20T04:14:55.080" CommentCount="4" />
  <row Id="7254" PostTypeId="2" ParentId="7252" CreationDate="2010-10-20T04:16:43.613" Score="0" ViewCount="" Body="&lt;p&gt;This was already covered in &lt;a href=&quot;http://math.stackexchange.com/questions/4764/sine-function-dense-in-1-1&quot;&gt;http://math.stackexchange.com/questions/4764/sine-function-dense-in-1-1&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would vote to close as identical but don't have the rep&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-10-20T04:16:43.613" CommentCount="2" />
  <row Id="7255" PostTypeId="1" AcceptedAnswerId="7257" CreationDate="2010-10-20T04:56:40.297" Score="2" ViewCount="168" Body="&lt;p&gt;Trying to remember my high school formulas, and coming up dry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Say I have two choices: A and B. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;P(A) = 0.25; P(B) = 0.75&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are no conditional probabilities or anything. Each choice is independent of the last.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do I go about calculating the probability that I will choose A exactly twice, having chosen ten items?&lt;/p&gt;&#xA;" OwnerUserId="1885" LastActivityDate="2010-10-20T05:36:27.487" Title="Probability of picking an item r times out of n attempts" Tags="&lt;combinatorics&gt;&lt;probability&gt;" AnswerCount="2" />
  <row Id="7256" PostTypeId="2" ParentId="7255" CreationDate="2010-10-20T05:06:18.760" Score="3" ViewCount="" Body="&lt;p&gt;You want the binomial distribution with $p = 0.25$ and $n = 10$ and $k = 2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See Wikipedia's article on the &lt;a href=&quot;http://en.wikipedia.org/wiki/Binomial_distribution&quot; rel=&quot;nofollow&quot;&gt;binomial distribution&lt;/a&gt;, which says,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;The probability of getting exactly $k$ successes in $n$ trials [when each trial has success probability $p$] is given by ...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \binom{n}{k} p^k (1-p)^{n-k}.&quot;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Paul VanKoughnett's answer gives you an explanation of why this formula is correct.&lt;/p&gt;&#xA;" OwnerUserId="2370" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-10-20T05:24:27.557" LastActivityDate="2010-10-20T05:24:27.557" CommentCount="1" />
  <row Id="7257" PostTypeId="2" ParentId="7255" CreationDate="2010-10-20T05:10:33.640" Score="3" ViewCount="" Body="&lt;p&gt;You want to pick A twice and B eight times, so the probability of doing this in a certain order is $P(A)^2P(B)^8$.  But for you, the order doesn't matter, so you have to divide by the number of ways to choose two things from 10.  This is $\binom{10}{2}=\frac{10!}{8!2!}=45$.  So all in all you get $\frac{(45)3^8}{4^{10}}$, which you can calculate yourself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general the formula for $\binom{n}{r}=\frac{n!}{(n-r)!r!}$, where the exclamation point means &quot;factorial,&quot; $n!=n\times(n-1)\times\dotsb\times 2\times 1$.  Since $n!$ represents the number of arrangements of $n$ things, you can interpret this formula as giving you the number of arrangements of your $n$ things, and then cancelling out the ways you can arrange the $r$ things you want, and the $n-r$ things you don't want.&lt;/p&gt;&#xA;" OwnerUserId="2215" LastEditorUserId="2215" LastEditorDisplayName="" LastEditDate="2010-10-20T05:36:27.487" LastActivityDate="2010-10-20T05:36:27.487" CommentCount="3" />
  <row Id="7258" PostTypeId="1" AcceptedAnswerId="7271" CreationDate="2010-10-20T06:29:46.767" Score="-1" ViewCount="102" Body="&lt;p&gt;In $\mathbb{R}^2$ specify a linearly independent set that has many elements as possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What do such sets have in common?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In $\mathbb{R}^3$ do the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I have an answer in mind, but I would like to know if this wording is clear... and if I have really answered my own question fully and correctly. Thank you.)&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I removed the 2nd question since people are responding as if it is related or as if I don't know the definition of linear independence.  &lt;/p&gt;&#xA;" OwnerUserId="2563" LastEditorUserId="2563" LastEditorDisplayName="" LastEditDate="2010-10-20T13:31:05.700" LastActivityDate="2010-10-20T13:31:05.700" Title="linearly independent set that has many elements as possible" Tags="&lt;linear-algebra&gt;&lt;matrices&gt;" AnswerCount="3" />
  <row Id="7259" PostTypeId="1" AcceptedAnswerId="7260" CreationDate="2010-10-20T06:34:48.980" Score="7" ViewCount="144" Body="&lt;p&gt;Is there an expansion for the following summation?&#xA;$$ (a_1 + a_2 + \cdots + a_k)^n $$&lt;/p&gt;&#xA;" OwnerUserId="2468" LastActivityDate="2010-10-20T06:44:30.667" Title="Expansion of $ (a_1 + a_2 + \cdots + a_k)^n $" Tags="&lt;binomial-coefficients&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="7260" PostTypeId="2" ParentId="7259" CreationDate="2010-10-20T06:37:28.610" Score="10" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Multinomial_theorem&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Multinomial_theorem&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is what you seek.&lt;/p&gt;&#xA;" OwnerUserId="2563" LastActivityDate="2010-10-20T06:37:28.610" CommentCount="3" />
  <row Id="7261" PostTypeId="2" ParentId="7258" CreationDate="2010-10-20T06:37:42.853" Score="0" ViewCount="" Body="&lt;p&gt;think about what linear independence means, and make sure you understand the definition of &quot;the subspace generated (or spanned) by the set {x,y,...}.&quot; Once you have done that, it should be clear. If you still have trouble try looking at the vector space $\mathbb{R}$ and do the same. &lt;/p&gt;&#xA;" LastActivityDate="2010-10-20T06:37:42.853" CommentCount="1" />
  <row Id="7262" PostTypeId="2" ParentId="7259" CreationDate="2010-10-20T06:44:30.667" Score="1" ViewCount="" Body="&lt;p&gt;Since you know what $(a+b)^{n}$ is you can take $A = a_{1} + a_{2} + \cdots + a_{k-1}$ and $B=a_{k}$ and try to simplify the big expression by the binomial theorem.&lt;/p&gt;&#xA;" LastActivityDate="2010-10-20T06:44:30.667" CommentCount="2" />
  <row Id="7263" PostTypeId="2" ParentId="7187" CreationDate="2010-10-20T06:53:17.330" Score="0" ViewCount="" Body="&lt;p&gt;I don't know if the following counts as &quot;intuition&quot;, but perhaps it&#xA;can put things in some perspective.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's a theorem that any rotation in $R^n$ can be written as the&#xA;composition of an even number of reflections through hyperplanes.&#xA;To construct a way of doing rotions by algebra (not matrix algebra,&#xA;but rather by multiplying vectors and similar geometric objects), it&#xA;is therefore enough to figure out two things:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) Given a vector $f$, how do we represent the reflection $R_f$&#xA;through the hyperplane orthogonal to $f$? We want $R_f(x)$ to be some&#xA;kind of product involving the vectors $x$ and $f$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) The correspondence constructed in (1) should be such that for two&#xA;vectors $f$ and $g$, the algebraic object $fg$ (whatever that is)&#xA;should represent the composition $R_f \circ R_g$. Then any rotation&#xA;can be written as&#xA;$R_{f_1} \circ R_{f_2} \circ \cdots \circ R_{f_{2m}}$ and therefore represented by the object&#xA;$f_1 f_2 \cdots f_{2m}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For this to work, we have to give up the idea that the product of two&#xA;vectors must be a vector (as Hamilton famously struggled with for a&#xA;long time). Instead we let the vectors sit inside some bigger space of&#xA;&quot;complex&quot; (compound) objects that can be multiplied with each other&#xA;according to suitable rules (associative, but not commutative).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of quaternions, these complex objects consist of a &quot;scalar&#xA;part&quot; plus a &quot;vector part&quot;, and the construction (1) which makes&#xA;everything work is $R_f(x)=-fxf^{-1}$. To see that this is a&#xA;reflection, note that it's linear in $x$, it maps $f$ to $-f$, and it&#xA;maps vectors orthogonal to $f$ to themselves (since they satisfy&#xA;$fx=-xf$ because of the way quaternion multiplication is defined).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To do rotations in $R^n$ one constructs a so-called Clifford algebra,&#xA;consisting of complex objects with a &quot;scalar part&quot;, a &quot;vector part&quot;, a&#xA;&quot;bivector part&quot;, a &quot;trivector part&quot;, etc., up to an &quot;n-vector part&quot;,&#xA;and the rules to multiply these objects are generated by the&#xA;requirement that for vectors $x$ and $y$, one wants $xy+yx=2 x\cdot y$&#xA;(or the negative of that, depending on convention). This makes&#xA;orthonal vectors anticommute, so that $R_f(x)=-fxf^{-1}$ gives a&#xA;hyperplane reflection, and everything works as desired.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I won't go into what a &quot;k-vector&quot; is, but the space of k-vectors&#xA;is $\binom{n}{k}$-dimensional, so that the whole Clifford algebra is&#xA;$2^n$-dimensional.&#xA;The Clifford product of two pure vectors is a scalar plus a bivector,&#xA;which is a little bit different from quaternions where vector times&#xA;vector equals a scalar plus a vector.&#xA;The full Clifford algebra for $R^3$ is 8-dimensional, and it's a bit&#xA;of a &quot;coincidence&quot; that one can actually manage just as well with the&#xA;4-dimensional quaternion algebra if one only wants to work with&#xA;rotations in $R^3$.&lt;/p&gt;&#xA;" OwnerUserId="1242" LastEditorUserId="1242" LastEditorDisplayName="" LastEditDate="2010-10-20T12:41:03.220" LastActivityDate="2010-10-20T12:41:03.220" />
  <row Id="7264" PostTypeId="2" ParentId="7258" CreationDate="2010-10-20T06:54:17.547" Score="0" ViewCount="" Body="&lt;p&gt;Just to clarify about the three dimensional case, it is not enough to ask that it won't be a multiple of another vector in the set, you must request for something stronger:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose you have $\{v_1,\ldots ,v_k\}$ which are linear independent, then $v_{k+1}$ can be added to this set while retaining the independence if $v_{k+1}$ cannot be written as the sum: $c_1 v_1 + c_2 v_2 + \cdots + c_k v_k$, where $c_i$ are scalars (in this case real numbers).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(The said sum is called &quot;linear combination&quot;, so using this term one can restate that $v_{k+1}$ is linearly independent of $\{v_1,\ldots, v_k\}$ if it cannot be written as a linear combination of $v_1,\ldots ,v_k$.)&lt;/p&gt;&#xA;" OwnerUserId="622" LastActivityDate="2010-10-20T06:54:17.547" CommentCount="6" />
  <row Id="7265" PostTypeId="1" AcceptedAnswerId="7267" CreationDate="2010-10-20T06:54:31.660" Score="3" ViewCount="96" Body="&lt;p&gt;Suppose that a rational generating function is given representing the function&#xA;$f(x) = \sum_{i=0}^{2n}{c_i x^i}$ where $c_i \in \mathbb{N}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The goal is to determine if the coefficient in the middle, $c_n$, is nonzero or not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, we could be given $f(x) = \frac{1 - 10x^9 + 9x^{10}}{(x - 1)^2}$, which represents $1 + 2x + 3x^2 + \cdots + 9x^8$.  We are interested, in this case, in the middle coefficient, which happens to be 5.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Original Formulation&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The actual value can be determined by evaluating a contour integral.  This is accomplished by first multiplying $f(x)$ by $x^{-n}$.  Then, rewriting the variable $x$ as $x\cdot e^{i t}$ and finally evaluating the integral $\int_C{f(x)dt}$.  The result of all of this is the coefficient in question.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The Goal&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I seek to find alternative formulations to this problem.  Let me give some examples.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Example 1 : &quot;Tiling&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Start with the original formula $f(x)$.  Make a copy, for example $f(y)$, by rewriting $x$ as $y$.  Then we can subtract the second function from the first, eliminating the first coefficient.  We repeat additions and subtractions (possibly infinitely many times) until all coefficients except for the middle one have been eliminated.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Example 2 : &quot;Division&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We multiply $f(x)$ by a function $g(y)$, where $g(y) = \frac{1}{1-x^{2n+1}}$, essentially making the series repeat forever.  We then &lt;a href=&quot;http://homepage.univie.ac.at/herwig.hauser/Publications/linear_recurrences_and_powerseries_division.pdf&quot; rel=&quot;nofollow&quot;&gt;divide&lt;/a&gt; this by $x^n\cdot g(y)$, and get the remainder.  This remainder is the first half of the original series, and $c_n$ can be easily obtained from this.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What is desired / Motivations&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'd simply like to have alternative formulations for the solution.  I'm using this for an algorithm, and I'd like to publish this with anyone who gives a formulation that provides a quick result.&lt;/p&gt;&#xA;" OwnerUserId="2626" LastActivityDate="2010-10-21T00:01:42.607" Title="What is an alternative formulation to a contour integral?" Tags="&lt;soft-question&gt;&lt;sequences-and-series&gt;&lt;big-list&gt;&lt;complex-analysis&gt;" AnswerCount="2" />
  <row Id="7266" PostTypeId="1" AcceptedAnswerId="7275" CreationDate="2010-10-20T07:04:52.720" Score="8" ViewCount="286" Body="&lt;ol&gt;&#xA;&lt;li&gt;Take any number $x$ (edit: x should be positive, heh)&lt;/li&gt;&#xA;&lt;li&gt;Add 1 to it $x+1$&lt;/li&gt;&#xA;&lt;li&gt;Find its reciprocal $1/(x+1)$&lt;/li&gt;&#xA;&lt;li&gt;Repeat from 2&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;So, taking $x = 1$ to start:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;1&lt;/li&gt;&#xA;&lt;li&gt;2 (the + 1)&lt;/li&gt;&#xA;&lt;li&gt;0.5 (the reciprocal)&lt;/li&gt;&#xA;&lt;li&gt;1.5 (the + 1)&lt;/li&gt;&#xA;&lt;li&gt;0.666... (the reciprocal)&lt;/li&gt;&#xA;&lt;li&gt;1.666... (the + 1)&lt;/li&gt;&#xA;&lt;li&gt;0.6 (the reciprocal)&lt;/li&gt;&#xA;&lt;li&gt;1.6 (the + 1)&lt;/li&gt;&#xA;&lt;li&gt;0.625&lt;/li&gt;&#xA;&lt;li&gt;1.625&lt;/li&gt;&#xA;&lt;li&gt;0.61584...&lt;/li&gt;&#xA;&lt;li&gt;1.61584...&lt;/li&gt;&#xA;&lt;li&gt;0.619047...&lt;/li&gt;&#xA;&lt;li&gt;1.619047...&lt;/li&gt;&#xA;&lt;li&gt;0.617647058823..&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we look at just the &quot;step 3&quot;'s (the reciprocals), we get:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;1&lt;/li&gt;&#xA;&lt;li&gt;0.5&lt;/li&gt;&#xA;&lt;li&gt;0.666...&lt;/li&gt;&#xA;&lt;li&gt;0.6&lt;/li&gt;&#xA;&lt;li&gt;0.625&lt;/li&gt;&#xA;&lt;li&gt;0.61584...&lt;/li&gt;&#xA;&lt;li&gt;0.619047...&lt;/li&gt;&#xA;&lt;li&gt;0.617647058823..&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This appears to always converge to 0.61803399... no matter where you start from.  I looked up this number and it is often called &quot;The golden ratio&quot; - 1, or $\frac{1+\sqrt{5}}{2}-1$.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Is there any &quot;mathematical&quot; way to represent the above procedure (or the terms of the second series, of &quot;only reciprocals&quot;) as a limit or series?&lt;/li&gt;&#xA;&lt;li&gt;Why does this converge to what it does for every starting point $x$?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;edit: darn, I just realized that the golden ratio is actually 1.618... and not 0.618...; I edited my answer to change what the result is apparently (golden ratio - 1).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I think I could easily make it the golden ratio by taking the +1 &quot;steps&quot; of the original series, instead of the reciprocation steps of the original series:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;2&lt;/li&gt;&#xA;&lt;li&gt;1.5&lt;/li&gt;&#xA;&lt;li&gt;1.666...&lt;/li&gt;&#xA;&lt;li&gt;1.6&lt;/li&gt;&#xA;&lt;li&gt;1.625&lt;/li&gt;&#xA;&lt;li&gt;1.61584...&lt;/li&gt;&#xA;&lt;li&gt;1.619047...&lt;/li&gt;&#xA;&lt;li&gt;1.617647058823..&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;which does converge to $\frac{1+\sqrt{5}}{2}-1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Explaining either of these series is adequate as I believe that explaining one also explains the other.&lt;/p&gt;&#xA;" OwnerUserId="51" LastEditorUserId="51" LastEditorDisplayName="" LastEditDate="2010-10-20T07:41:47.950" LastActivityDate="2011-05-13T11:35:45.973" Title="Why does this process, when iterated, tend towards a certain number? (the golden ratio?)" Tags="&lt;limit&gt;&lt;convergence&gt;" AnswerCount="5" CommentCount="1" FavoriteCount="1" />
  <row Id="7267" PostTypeId="2" ParentId="7265" CreationDate="2010-10-20T07:10:02.337" Score="1" ViewCount="" Body="&lt;p&gt;I'm not sure I understand exactly what you want, but if you differentiate the function $n$ times and then evaluate the result at $x=0$ you get $n! c_n$.&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2010-10-20T07:10:02.337" CommentCount="4" />
  <row Id="7268" PostTypeId="2" ParentId="7266" CreationDate="2010-10-20T07:30:21.483" Score="17" ViewCount="" Body="&lt;p&gt;You are iterating the operation&#xA;$$x\mapsto\frac1{x+1}.$$&#xA;We can represent this in matrix terms. Set&#xA;$$A=\begin{pmatrix}&#xA;0&amp;amp;1\\&#xA;1&amp;amp;1\end{pmatrix}.$$&#xA;If&#xA;$$A&#xA;\begin{pmatrix}&#xA;x\\&#xA;1\end{pmatrix}&#xA;=\begin{pmatrix}&#xA;y_1\\&#xA;z_1\end{pmatrix}$$&#xA;then $1/(x+1)=y_1/z_1$. After $n$ iterations one gets&#xA;$$A^n&#xA;\begin{pmatrix}&#xA;x\\&#xA;1\end{pmatrix}&#xA;=\begin{pmatrix}&#xA;y_n\\&#xA;z_n\end{pmatrix}$$&#xA;and $x_n=y_n/z_n$ is got by applying the map $x\mapsto 1/(x+1)$ $n$ times to $x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Naturally you won't be surprised to find that the eigenvalues of&#xA;$A$ are $\tau=\frac12(1+\sqrt5)$ and $\tau=\frac12(1-\sqrt5)$.&#xA;The eigenvectors are $v=(1\ \tau)^t$ and $w=(1\ \tau')^t$. We can write&#xA;$$\begin{pmatrix}&#xA;x\\&#xA;1\end{pmatrix}=av+bw$$&#xA;and so&#xA;$$\begin{pmatrix}&#xA;y_n\\&#xA;z_n\end{pmatrix}=a\tau^n v+b\tau'^nw.$$&#xA;Now $\tau&gt;1&gt;|\tau'|$&#xA;so that for large $n$, $y_n$ and $z_n$ are very close to $a\tau^n$and $a\tau^{n+1}$&#xA;so that $x_n\to\infty$ if $a=0$. The only exception is when $a=0$&#xA;which only arises for $x=-\frac12(1+\sqrt5)$. (I bet you didn't test that one!)&lt;/p&gt;&#xA;" OwnerUserId="226" LastEditorUserId="226" LastEditorDisplayName="" LastEditDate="2010-10-20T08:10:56.497" LastActivityDate="2010-10-20T08:10:56.497" CommentCount="4" />
  <row Id="7269" PostTypeId="2" ParentId="7266" CreationDate="2010-10-20T07:30:41.717" Score="0" ViewCount="" Body="&lt;p&gt;Define the function $f(x) = \frac1{1+x} $ so the process you defined is just going from x to f(x).&#xA;let $x_0\in\mathbb{R}$ and $x_{i+1} = f(x_i)$ then your claim is that $\lim\; x_i = \frac {1+\sqrt{5}}{2} $.&#xA;if this series converges then&#xA;$x = \lim\; x_i = \lim\; f(x_{i-1}) = f( \lim\; x_{i-1} ) = f(x)$ so the limit is $x=\frac{1}{1+x}$ or in other words you have $x^2+x-1=0$. the solutions are $\frac{-1\pm \sqrt{5}}{2}$ &#xA;this isn't exactly the golden ratio. if you take $\frac1{x-1}$ instead then you'll have it. you still need to find out for which $x_0$ this sequence converges - this could be done using numerical analysis (sorry, I'm a bit rusty on this, so I'll leave this part to someone else).&lt;/p&gt;&#xA;" OwnerUserId="2125" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2011-05-13T11:35:45.973" LastActivityDate="2011-05-13T11:35:45.973" />
  <row Id="7270" PostTypeId="1" AcceptedAnswerId="7278" CreationDate="2010-10-20T07:41:00.310" Score="3" ViewCount="146" Body="&lt;p&gt;Thats it! Thats what I want to know. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If θ is the angle subtended by an arc of length L at its center with radius R.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know, θ = L/R.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How did we get this? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please don't say we got it from generalizing the idea of perimeter of circle.&lt;/p&gt;&#xA;" OwnerUserId="2264" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-10-20T10:25:00.597" LastActivityDate="2010-10-20T10:26:21.283" Title="θ = (length of arc)/(angle subtended by it). How?" Tags="&lt;geometry&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="0" />
  <row Id="7271" PostTypeId="2" ParentId="7258" CreationDate="2010-10-20T07:49:26.873" Score="1" ViewCount="" Body="&lt;p&gt;The first question is clear.  It appears to be a basic question designed to make &lt;em&gt;you&lt;/em&gt; apply the definition of linear independence to familiar objects, the vector spaces $\mathbb{R}^2$ and $\mathbb{R}^3$.  (That's why the responses so far give hints rather than solutions.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You should note that linear independence is a property of sets of vectors-- so the &quot;sets&quot; in the first question are implied to be sets of vectors.  So both questions are really about sets of vectors.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Asaf Karagila is pointing out that the property in question 2, that &quot;no vector is a constant multiple of another vector in the set,&quot; is a different property than linear independence.  In fact, question 2 has a very different answer than question 1.  If you aren't sure of the answers, explain your thinking and your uncertainty, and I'm sure you'll get a more direct hint.&lt;/p&gt;&#xA;" OwnerUserId="1461" LastActivityDate="2010-10-20T07:49:26.873" CommentCount="1" />
  <row Id="7272" PostTypeId="2" ParentId="7270" CreationDate="2010-10-20T08:03:56.297" Score="1" ViewCount="" Body="&lt;p&gt;This is because using &lt;a href=&quot;http://mathworld.wolfram.com/Radian.html&quot; rel=&quot;nofollow&quot;&gt;radian measure&lt;/a&gt; (as opposed to using degrees) in working with things related to circles is mathematically very convenient. In more advanced mathematical work you rarely see people use degrees since the formulae that have the angles in degrees require having to keep around inconvenient factors of $\frac{\pi}{180}$ or its reciprocal.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-10-20T08:03:56.297" CommentCount="4" />
  <row Id="7273" PostTypeId="2" ParentId="7238" CreationDate="2010-10-20T08:23:22.880" Score="1" ViewCount="" Body="&lt;p&gt;You could try reading &lt;a href=&quot;http://terrytao.wordpress.com/2009/02/09/245b-notes-10-compactness-in-topological-spaces/&quot; rel=&quot;nofollow&quot;&gt;Terence Tao's notes&lt;/a&gt; on the subject; I found them very informative.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for examples, here is an example and a non-example which I think are informative.  The example is that, by Tychonoff's theorem, $[0, 1]^I$ is compact for any index set $I$.  The non-example is that the closed unit ball in any infinite-dimensional Banach space, say $\ell_1(\mathbb{Z})$, is &lt;em&gt;not&lt;/em&gt; compact.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit:  Here is another perspective from which to think about compact &lt;em&gt;Hausdorff&lt;/em&gt; spaces (and LCH spaces).  A compact Hausdorff space $X$ is completely determined by the ring of continuous functions $C(X, \mathbb{R})$; this is a standard exercise which is worked out, for example, &lt;a href=&quot;http://qchu.wordpress.com/2009/11/24/spectra-of-rings-of-continuous-functions/&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.  One gets the points of $X$ as the maximal ideals of $C(X, \mathbb{R})$ and the topology as the initial topology making every function in $C(X, \mathbb{R})$ continuous.  Slightly generalized, this is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Gelfand_representation#Statement_of_the_commutative_Gelfand-Naimark_theorem&quot; rel=&quot;nofollow&quot;&gt;commutative Gelfand-Naimark theorem&lt;/a&gt;, and it says that studying commutative C*-algebras with unit is the same thing as studing compact Hausdorff spaces.  Removing the requirement that we have a unit is the same thing as studying locally compact Hausdorff spaces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So one can write down compact Hausdorff spaces by writing down commutative C*-algebras with unit.  One choice is to take the space $C_b(X, \mathbb{R})$, the space of bounded continuous functions on an arbitrary topological space $X$, with the sup norm.  If $X$ is completely regular Hausdorff, this gives the &lt;a href=&quot;http://en.wikipedia.org/wiki/Stone%E2%80%93%C4%8Cech_compactification#Construction_using_C.2A-algebras&quot; rel=&quot;nofollow&quot;&gt;Stone-Cech compactification&lt;/a&gt; of $X$.  (Which is hopeless to think about in general; even for $X = \mathbb{N}$ it's very weird, as the Wikipedia article describes.)&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-10-20T09:23:13.720" LastActivityDate="2010-10-20T09:23:13.720" CommentCount="1" />
  <row Id="7274" PostTypeId="2" ParentId="7249" CreationDate="2010-10-20T08:29:49.090" Score="1" ViewCount="" Body="&lt;p&gt;This is supposed to be in reply to Rahul's second comment, but it got too long.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would suppose that the requirement of Hall and Wagon for the axle to have the same horizontal displacement as the contact point of the road and wheel is a generalization of the concept of a circle rolling on a line. In that situation it is obvious that the axle is indeed directly above the point of contact.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remember that rolling without slip can be decomposed as a translation, followed by a rotation, and then another translation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the case of the wheel parametrically defined as $(x\qquad y)^T=(\sin\;t\qquad 1-\cos\;t)^T$, the axle is initially at $(0\;1)^T$. If we look at the expression&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\bigl(\begin{smallmatrix}t\\0\end{smallmatrix}\bigr)+\bigl(\begin{smallmatrix}\cos\;t&amp;amp;\sin\;t\\-\sin\;t&amp;amp;\cos\;t\end{smallmatrix}\bigr)\cdot\left(\bigl(\begin{smallmatrix}0\\1\end{smallmatrix}\bigr)-\bigl(\begin{smallmatrix}\sin\;t\\1-\cos\;t\end{smallmatrix}\bigr)\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(in English: translate axle to the point on the circle that would be the future point of contact of the circle with the road, rotate by some angle, and then translate by an amount equal to the length of the arc subtended by said angle horizontally)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;we find that this whole mess simplifies to the vector $(t\qquad 1)^T$ which indeed has the same x-coordinate as the current point of contact $(t\qquad 0)$.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-10-20T08:29:49.090" CommentCount="2" />
  <row Id="7275" PostTypeId="2" ParentId="7266" CreationDate="2010-10-20T08:46:54.120" Score="12" ViewCount="" Body="&lt;p&gt;We want to show that the function $f(x) = \frac{1}{1+x}$ has a unique fixed point to which it converges when iterated on positive $x$.  If $x$ is positive, then $f(x)$ is between $1$ and $0$, so $f(f(x))$ is between $1$ and $\frac{1}{2}$.  It is not hard to see that in fact $f$ fixes the interval $\left[ \frac{1}{2}, 1 \right]$.  Now, for $x, y$ in this interval,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left| \frac{1}{x+1} - \frac{1}{y+1} \right| = \left| \frac{y-x}{(1+x)(1+y)} \right| \le \frac{4}{9} |x - y|$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so on this interval $f$ satisfies the conditions of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Banach_fixed_point_theorem&quot; rel=&quot;nofollow&quot;&gt;Banach fixed point theorem&lt;/a&gt;.  The unique fixed point to which everything always converges is the unique solution to $f(x) = x$, which you have already found.&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-10-20T15:10:31.087" LastActivityDate="2010-10-20T15:10:31.087" CommentCount="5" />
  <row Id="7276" PostTypeId="2" ParentId="6890" CreationDate="2010-10-20T09:07:31.323" Score="1" ViewCount="" Body="&lt;p&gt;The 4-vector density current is&#xA;\begin{equation*}&#xA;j^{\mu} := \rho \frac{\partial x^{\mu}}{\partial \tau} = \rho {\dot{x}}^{\mu} = \rho (c, \mathbf{\dot{x}}) = (\rho c, \mathbf{j})&#xA;\end{equation*}&#xA;where $\tau$ is the proper time.\&#xA;We can ask ${\dot{x}}^{\mu}$ to be a 4-versor, or in other words to be of unit Minkowski norm.&#xA;Usually we define the vector potential $\mathbf{A}$ in terms of $\mathbf{j}$, and the scalar potential $\phi$ in terms of the current density $\rho$&#xA;\begin{equation*}&#xA; \mathbf{A}(\mathbf{x}) = C_1 \int_{space} \frac{\mathbf{j}(\mathbf{y})}{r}d\mathbf{y}, \qquad &#xA; \phi(\mathbf{x}) = C_2 \int_{space} \frac{\rho(\mathbf{y})}{r}d\mathbf{y}&#xA;\end{equation*}&#xA;where $r$ is the spatial distance between $\mathbf{x}$ and $\mathbf{y}$, where $C_1$ and $C_2$ are costants. \&#xA;Let's put all together in the 4-vector $A$&#xA;\begin{equation*}&#xA; A_{\mu}(x) = k \int_{spacetime} \frac{j_{\mu}(y)}{r} dy&#xA;\end{equation*}&#xA;But we want $A$ to be non-zero at the point $\mathbf{x}$ only when the light comes from $\mathbf{y}$ to $\mathbf{x}$, or when the event $x$ os on the boundary of the light cone centered in $y$. To impose this is sufficient to multiply the integrand with $\delta((x-y)^{\nu} (x-y)_{\nu})$&#xA;\begin{equation*}&#xA; A_{\mu}(x) = k \int_{spacetime} \frac{j_{\mu}(y)}{r} \delta((x-y)^{\nu} (x-y)_{\nu}) dy&#xA;\end{equation*}&#xA;We can simplify the expression of $A$, posing&#xA;\begin{equation*}&#xA;\int \delta(f(\alpha,\beta))g(\alpha,\beta) d\beta := \sum_i \frac{g(\alpha,\beta^i)}{| \frac{\partial f}{\partial \beta}(\alpha, \beta^i) |}&#xA;\end{equation*}&#xA;where $\alpha$ is fixed, and the sum is extended to all the $\beta^i$ for which $f(\alpha,\beta^i) = 0$.&#xA;Doing like this we have, speaking about a single particle $b$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation*} \int \delta((x-b)^{\nu} (x-b){\nu})j{\mu}(\beta) d\beta = e_b \int \delta((x-b)^{\nu} (x-b){\nu}) \dot{b}{\mu}(\beta) d\beta = \frac{e_b}{2} \sum_i \frac{j_{\mu}(\beta^i)}{| (x-b)^{\nu} \dot{b}_{\nu} |} \end{equation*}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we only consider the events $x$ that $(x-b)^{\nu} (x-b)_{\nu} = 0$, we have&#xA;\begin{equation*}&#xA; r = \lvert \mathbf{x-b} \rvert = (x-b)^0&#xA;\end{equation*}&#xA;In the rest frame of the current flux $\dot{b} = (1,0,0,0)$. So we find&#xA;\begin{equation*}&#xA; r = (x-b)^0 = (x-b)^{\nu}{\dot{b}}_{\nu}&#xA;\end{equation*}&#xA;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation*} \int \delta((x-b)^{\nu} (x-b){\nu})j{\mu}(\beta) d\beta = \frac{e_b}{2} \sum_i \frac{j_{\mu}(\beta^i)}{r} \end{equation*}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally we notice $j_{\mu} d\beta = e_b (db_{\mu} / d\beta) d\beta = e_b db_{\mu}$, and so we define the 4-vector potential as follows&#xA;The 4 dimensional vector potential of particle $b$ at the point $x$, with charge $e_b$ is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}\label{A} A_{\mu}^{(b)}(x) := e_b \int \delta((x-b)^{\nu} (x-b){\nu}) db{\mu} \end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where we integrate over the whole world line of $b$.&lt;/p&gt;&#xA;" OwnerUserId="2354" LastEditorUserId="2354" LastEditorDisplayName="" LastEditDate="2010-10-21T07:04:42.917" LastActivityDate="2010-10-21T07:04:42.917" CommentCount="2" />
  <row Id="7277" PostTypeId="2" ParentId="7266" CreationDate="2010-10-20T09:08:28.263" Score="4" ViewCount="" Body="&lt;p&gt;This was supposed to be an addendum to Hans's comment, but it got too long.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The iteration Justin considered formally generates a continued fraction:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\cfrac{1}{1+\cfrac{1}{1+\dots}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and as I mentioned in &lt;a href=&quot;http://math.stackexchange.com/questions/3382/continued-fraction-for-frac1e-2/3387#3387&quot;&gt;this answer&lt;/a&gt;, the numerators and denominators of the nth convergent of a continued fraction can be computed recursively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we apply the formula in that answer to this situation, we get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{bmatrix}C_n\\D_n\end{bmatrix}=\begin{bmatrix}C_{n-1}\\D_{n-1}\end{bmatrix}+\begin{bmatrix}C_{n-2}\\D_{n-2}\end{bmatrix}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with initial conditions&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{bmatrix}C_{-1}\\D_{-1}\end{bmatrix}=\begin{bmatrix}1\\0\end{bmatrix},\qquad \begin{bmatrix}C_{0}\\D_{0}\end{bmatrix}=\begin{bmatrix}0\\1\end{bmatrix}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From here, the setup is now similar to Robin's answer, since the two recursions in fact generate the Fibonacci numbers: $C_n=F_n$ and $D_n=F_{n+1}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your continued fraction then is the limit&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\lim_{n\to\infty}\frac{F_n}{F_{n+1}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and the equivalence to Robin's answer is due to the &lt;a href=&quot;http://mathworld.wolfram.com/BinetsFibonacciNumberFormula.html&quot; rel=&quot;nofollow&quot;&gt;Binet formula&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$F_n=\frac{\varphi^n-\left(-\varphi^{-1}\right)^n}{\sqrt{5}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(which in fact can be derived from Robin's answer).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Substituting that into the limit and evaluating gets you the answer.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-10-20T09:08:28.263" />
  <row Id="7278" PostTypeId="2" ParentId="7270" CreationDate="2010-10-20T09:21:28.653" Score="1" ViewCount="" Body="&lt;p&gt;What is your definition of an angle? The formula you quoted follows directly from the definition of an angle as the length of the arc on the unit circle, combined with the intercept theorems (for “scaling the circle to a unit circle,” if you want).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using angles defined this way is, as J.M. said, much more convenient than degrees. E.g., it connects triangles with trigonometric functions with nice derivatives – from $\sin' = \cos$, $\cos' = -\sin$, $\sin(0) = 0$, and $\cos(0) = 1$, you directly find that $\cos(\pi/2)=0$, i.e., these formulas only hold if the arguments to $\sin$ and $\cos$, traditionally angles, are given in radians.&lt;/p&gt;&#xA;" OwnerUserId="1354" LastActivityDate="2010-10-20T09:21:28.653" />
  <row Id="7279" PostTypeId="2" ParentId="7247" CreationDate="2010-10-20T10:01:11.697" Score="3" ViewCount="" Body="&lt;p&gt;No. If $K$ does not have compact support, then $f*K$ need not even be bounded. Consider the function $K = 1 / \log( 3 + |x|)$, and $f(x) = (1 + |x|)^{-1/p - \epsilon}$. For $\epsilon &gt; 0$. Then $f\in L^p$, but the convolution integral does not converge anywhere, even when $K$ vanishes at $\infty$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you can do is to require $K$ decays sufficiently rapidly. What you need is also &lt;a href=&quot;http://en.wikipedia.org/wiki/Young%27s_inequality&quot; rel=&quot;nofollow&quot;&gt;Young's inequality for convolutions&lt;/a&gt;, which implies that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \sup |f * K| \leq \|f \|_p \|K\|_q $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;if $1/p + 1/q = 1$. So in particular if $K\in C^m(\mathbb{R}) \cap W^{m,q}(\mathbb{R})$, where $W^{m,q}$ is the Sobolev space of $m$-times weakly differentiable, $q$-integrable functions with $q = p / (p-1)$, you can conclude that the convolution is $C^m$. &lt;/p&gt;&#xA;" OwnerUserId="1543" LastActivityDate="2010-10-20T10:01:11.697" />
  <row Id="7280" PostTypeId="2" ParentId="6661" CreationDate="2010-10-20T10:03:06.003" Score="0" ViewCount="" Body="&lt;p&gt;Have also a look here - many great resources for the Laplace transform:&lt;br&gt;&#xA;&lt;a href=&quot;http://mathoverflow.net/questions/383/motivating-the-laplace-transform-definition/2141#2141&quot; rel=&quot;nofollow&quot;&gt;http://mathoverflow.net/questions/383/motivating-the-laplace-transform-definition/2141#2141&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="346" LastActivityDate="2010-10-20T10:03:06.003" />
  <row Id="7281" PostTypeId="2" ParentId="7270" CreationDate="2010-10-20T10:12:24.727" Score="1" ViewCount="" Body="&lt;p&gt;The length of the arc is proportional to the angle by proposition 6.33 in Euclid's &lt;em&gt;&lt;a href=&quot;http://farside.ph.utexas.edu/euclid/Elements.pdf&quot; rel=&quot;nofollow&quot;&gt;Elements&lt;/a&gt;&lt;/em&gt;. We can choose units to make the constant of proportionality equal to $1$ for a circle of radius $1$. Those units are called radians.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[I give the Euclid reference since it uses the notion of angle that might be familiar to you. Nowadays we find it more economical to define the trignometric functions without any appeal to geometry (for eg. via the complex exponential), then define angles by inverting one of these functions, and then derive your result as an exercise in calculus.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a simple hand-waving argument along those lines. Let OA and OB be two radii of a circle of radius $r$. Then by simple trignometry the length of the line segment AB is $2r\sin(\theta/2)$. For small enough $\theta$, $\sin(\theta/2) \approx \theta/2$ so the length of AB is approximately $r\theta$. In this case the length of AB is also close to the length of the arc AB hence that is also approximately $r\theta$. We can make the approximation as good as we like by making $\theta$ small enough. We can handle large arcs by approximating it by a large number of small segments of this sort.]&lt;/p&gt;&#xA;" OwnerUserId="1195" LastEditorUserId="1195" LastEditorDisplayName="" LastEditDate="2010-10-20T10:26:21.283" LastActivityDate="2010-10-20T10:26:21.283" CommentCount="1" />
  <row Id="7282" PostTypeId="1" AcceptedAnswerId="7283" CreationDate="2010-10-20T10:23:20.667" Score="3" ViewCount="102" Body="&lt;p&gt;For a probability distribution, its quantile function is defined in terms of its distribution function as &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$Q(p)=F^{-1}(p) = \inf \{ x\in R : p \le F(x) \}$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I was wondering if, conversely, a quantile function can uniquely determine a distribution and therefore fully describe the probability distribution just as a distribution function does?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks and regards!&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;UPDATE:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please let me be more specific. Because a CDF is nondecreasing, right-continous and limit is 0 when $x \rightarrow -\infty$ and 1 when $x \rightarrow \infty$, its quantile function is nondecreasing, left-continous and a map from (0,1) into $R$. If a function is nondecreasing, left-continous and a map from (0,1) into $R$, can it become a quantile function of some CDF? When it can, is there a way to represent the CDF in terms of the quantile function using infimum or supremum similar as quantile function in terms of CDF?&lt;/p&gt;&#xA;" OwnerUserId="1281" LastEditorUserId="1281" LastEditorDisplayName="" LastEditDate="2010-10-20T10:58:43.133" LastActivityDate="2011-04-15T19:12:56.240" Title="Does Quantile Function Uniquely Determine Distribution Function?" Tags="&lt;probability-theory&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="7283" PostTypeId="2" ParentId="7282" CreationDate="2010-10-20T10:34:19.707" Score="3" ViewCount="" Body="&lt;p&gt;Well, if $Q(p)$ is well-defined and monotonic in the interval $(0,1)$, then certainly.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-10-20T10:34:19.707" CommentCount="8" />
  <row Id="7284" PostTypeId="2" ParentId="7248" CreationDate="2010-10-20T11:00:03.710" Score="2" ViewCount="" Body="&lt;p&gt;If you work in two-sorted first-order logic, it is possible to formalize vector spaces by using one sort for the field of scalars and another sort for the vectors, just as you describe. But there is no way, within the theory itself, to force an embedding of models to act as the identity on the scalars. For example, if you take a structure for your theory, every automorphism of the field will give a homomorphism from the structure to itself.  And every proper subfield of the field of scalars gives rise to a proper substructure in which the vectors remain the same and the scalars are cut down.  This is the motivation for the usual method in which the field is fixed through the addition of a lot of constant symbols to the language. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One response is to simply refuse to consider such things when you work on your proofs. That is, you could simply limit your attention to those homomorphisms that do preserve the field, and those substructures that do not change the field.  It depends on what you are trying to get out of the formalization. &lt;/p&gt;&#xA;" OwnerUserId="630" LastActivityDate="2010-10-20T11:00:03.710" />
  <row Id="7285" PostTypeId="2" ParentId="2234" CreationDate="2010-10-20T12:18:57.773" Score="0" ViewCount="" Body="&lt;p&gt;as i have already told you the expression is a theoretical expressions obtained using the curve fitting tool in matlab. &lt;/p&gt;&#xA;" OwnerUserId="2488" LastEditorUserId="2488" LastEditorDisplayName="" LastEditDate="2010-10-20T17:20:28.497" LastActivityDate="2010-10-20T17:20:28.497" CommentCount="1" />
  <row Id="7286" PostTypeId="2" ParentId="7217" CreationDate="2010-10-20T12:28:17.247" Score="1" ViewCount="" Body="&lt;p&gt;The Lebesgue mean does not exist for &quot;boundary points&quot;. Let $A_n$ be the union of the intervals of step $n$, that is, $A_1=[0,3/8]\cup [5/8,1]$, $A_2=[0,9/64]\cup [15/64,3/8]\cup [5/8,49/64]\cup [55/64,1]$ etc, $A=\cap_n A_n$ is the fat Cantor set. Consider a boundary point $x$ of a set $A_n$. Consider the sequence $\varepsilon_k:=(3/8)^{n+k}$. Then by symmetry we have $I_k=|(x-\varepsilon_k,x+\varepsilon_k)\cap A|=1/4\cdot(1/2)^{n+k}$ for $k$ large enough, thus $I_k/\varepsilon_k \sim (4/3)^{n+k} \to \infty$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can think of elements of $A$ as infinite sequences of {Left,Right}. Those sequences which are eventually constant correspond to &quot;boundary points&quot;. My guess is that the points for which the mean is 1 correspond to sequences that are normal in some sense (for example, all patterns of {Left,Right} appear with the same asymptotic frequency).&lt;/p&gt;&#xA;" OwnerUserId="1609" LastActivityDate="2010-10-20T12:28:17.247" />
  <row Id="7287" PostTypeId="1" AcceptedAnswerId="7299" CreationDate="2010-10-20T12:44:46.447" Score="3" ViewCount="74" Body="&lt;p&gt;I'm trying to show that if $f$ and $1/f$ are in $H^2$ then $f$ is an outer function. I think I somewhat have a proof (if it is correct) but it doesn't feel like it is the easiest one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know that if $f$ is in $H^2$ and non-zero and $B$ is the Blanschke product of its zeros, then there exists a constant $K$ of modulus $1$ and a singular measure $\mu$ on $S^1$ such that $f(z)$ has the form:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f(z) = K B(z) \text{exp} \left (-\frac{1}{2\pi} \int_0^{2\pi} \frac{e^{i\theta} + z}{e^{i\theta} - z} \, d\mu(\theta) \right ) \text{exp} \left (\frac{1}{2\pi} \int_0^{2\pi} \frac{e^{i\theta} + z}{e^{i\theta} - z} \log |\tilde{f}(e^{i \theta}) | \, d\theta \right )$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where the first exponential is the singular inner function part. So by uniqueness of this form (up to constants I believe?) and the fact that $1/f$ is in $H^2$ we can write:&#xA;$$\frac{1}{f(z)} = \frac{1}{KB(z)} \text{exp} \left (\frac{1}{2\pi} \int_0^{2\pi} \frac{e^{i\theta} + z}{e^{i\theta} - z} \, d\mu(\theta) \right ) \text{exp} \left (-\frac{1}{2\pi} \int_0^{2\pi} \frac{e^{i\theta} + z}{e^{i\theta} - z} \log |\tilde{f}(e^{i \theta}) | \, d\theta \right )$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, we see that $B(z) = 1$. Further we can see that the singular inner function-part has modulus $&gt;1$ a.e. in the open unit disk if $\mu \neq 0$, and so the limit $r \to 1^-$ is not $1$ because of the maximum modulus principle (maybe this is flawed since $f$ is not analytic on the boundary). So we are left with:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f(z) = K \text{exp} \left (\frac{1}{2\pi} \int_0^{2\pi} \frac{e^{i\theta} + z}{e^{i\theta} - z} \log |\tilde{f}(e^{i \theta}) | \, d\theta \right )$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which is an outer function. Are there any flaws in this? If not, is there a less &quot;thick&quot; way to show this (I mean, not invoke all the theorems about the form these functions must have and so on) since I'm not really satisfied with this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: Oh, a nice corollary of this result I found is that if $f$ in $H^\infty$ and $\|f\|_\infty &amp;lt; 1$, then $1 + f$ is outer. (By noting that if $f$ is in $H^\infty$, then $f$ in $H^2$.)&lt;/p&gt;&#xA;" OwnerUserId="1120" LastEditorUserId="1120" LastEditorDisplayName="" LastEditDate="2010-10-20T14:38:36.513" LastActivityDate="2010-10-20T16:03:23.447" Title="$f$, $1/f$ in $H^2$, then $f$ outer." Tags="&lt;complex-analysis&gt;&lt;functional-analysis&gt;" AnswerCount="1" />
  <row Id="7288" PostTypeId="2" ParentId="7187" CreationDate="2010-10-20T12:52:51.393" Score="0" ViewCount="" Body="&lt;p&gt;Re the last answer, in Geometric Algebra (a Clifford Algebra interpreted geometrically), this is very intuitive, you can even visualize it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See &lt;a href=&quot;http://www.geometricalgebra.net/quaternions.html&quot; rel=&quot;nofollow&quot;&gt;http://www.geometricalgebra.net/quaternions.html&lt;/a&gt;&lt;/p&gt;&#xA;" LastActivityDate="2010-10-20T12:52:51.393" />
  <row Id="7289" PostTypeId="1" AcceptedAnswerId="7292" CreationDate="2010-10-20T13:28:17.123" Score="1" ViewCount="120" Body="&lt;p&gt;In $\mathbb{R}^2$ specify a set of vectors, that has many elements as possible, so that no vector is a constant multiple of another vector in the set.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What do such sets have in common?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In $\mathbb{R}^3$ do the same&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I asked both of these earlier, but people are acting like it is a HW question (giving me &quot;hints&quot; etc.) but I'm not looking for HW help. I made these questions up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answers are the equivalence classes of fractions for $\mathbb{R}^2$ and for $\mathbb{R}^3$ the equivalence classes of 3-part &lt;em&gt;ordered&lt;/em&gt; ratios.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But now that you have heard that answer are there any more interesting answers?&lt;/p&gt;&#xA;" OwnerUserId="2563" LastEditorUserId="2563" LastEditorDisplayName="" LastEditDate="2010-10-20T13:42:24.903" LastActivityDate="2010-10-21T01:46:58.110" Title="What do these sets have in common?" Tags="&lt;linear-algebra&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="7291" PostTypeId="1" CreationDate="2010-10-20T13:42:38.267" Score="5" ViewCount="173" Body="&lt;p&gt;It is well known that the differential calculus has a nice algebraization in terms of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Differential_algebra#Differential_ring&quot; rel=&quot;nofollow&quot;&gt;differential rings&lt;/a&gt; but what about integral calculus? Of course, one sometimes defines an integral in a differential ring $R$ with a derivation $\partial$ as a projection $\pi: R\rightarrow \tilde R$, where $\tilde R$ is a quotient of $R$ w.r.t. the following equivalence relation: $f\sim g$ iff $f-g$ is in the image of $\partial$, but this is not very intuitive and apparently corresponds to the idea of &lt;em&gt;definite&lt;/em&gt; integral over a fixed domain rather than to that of an indefinite one. So my question is: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Are there algebraic counterparts for the concept of an &lt;em&gt;indefinite&lt;/em&gt; integral?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1399" LastEditorUserId="1399" LastEditorDisplayName="" LastEditDate="2010-10-20T14:19:36.293" LastActivityDate="2010-10-20T22:52:44.423" Title="Algebraization of integral calculus" Tags="&lt;calculus&gt;&lt;abstract-algebra&gt;&lt;integral&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="4" />
  <row Id="7292" PostTypeId="2" ParentId="7289" CreationDate="2010-10-20T13:58:21.900" Score="5" ViewCount="" Body="&lt;p&gt;An easy way to do this is to start with a circle resp. a sphere and delete one of each pair of antipodal vectors.  Any answer can be obtained from this answer by replacing any point with a scalar multiple of itself.  I don't know what else there is to say.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-10-20T13:58:21.900" CommentCount="1" />
  <row Id="7293" PostTypeId="2" ParentId="7289" CreationDate="2010-10-20T14:03:46.210" Score="2" ViewCount="" Body="&lt;p&gt;The intuitive way to answer this, for me, is the following.  Let's start with $\mathbb{R}^2$.  We're going to imagine the vectors as points in $\mathbb{R}^2-{(0,0)}$.  Once you choose one vector, you have basically blocked off the line through the origin that contains it.  So you can choose one vector for every line through the origin.  The same goes for $\mathbb{R}^3$.  There isn't much more to say about this, but if you require the set to be &lt;em&gt;connected&lt;/em&gt;, so that each vector you choose is &quot;close&quot; to other vectors in your set, you've got an embedding of the &quot;real projective plane&quot;, $\mathbb{RP}^{n-1}$, in $\mathbb{R}^n$.  The real projective plane is (equivalently), the space of lines through the origin, the sphere modulo the map sending $x$ to $-x$, or $\mathbb{R}^n$ modulo scalar multiplication.  The last definition is just what you said about equivalence classes of ordered pairs/triples.  Oh, and $\mathbb{RP}^2$ is what you get if you glue a circle to the single edge of a Moebius strip!&lt;/p&gt;&#xA;" OwnerUserId="2215" LastActivityDate="2010-10-20T14:03:46.210" />
  <row Id="7294" PostTypeId="1" AcceptedAnswerId="7349" CreationDate="2010-10-20T14:18:42.723" Score="0" ViewCount="64" Body="&lt;p&gt;I have some geometry in 3d (like road surface or terrain surface) and there could be several not connected parts of a road. &#xA;So I'd like to know if I may unite this two not connected pieces of road in a term &quot;surface&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And if I couldn't, then what is the best mathematical term for this?&lt;/p&gt;&#xA;" OwnerUserId="2647" LastEditorUserId="2647" LastEditorDisplayName="" LastEditDate="2010-10-20T15:10:46.030" LastActivityDate="2010-10-20T20:24:32.917" Title="is surface always locally connected area in euklidian 3d ?" Tags="&lt;geometry&gt;&lt;topology&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="7295" PostTypeId="1" AcceptedAnswerId="16344" CreationDate="2010-10-20T14:27:28.727" Score="4" ViewCount="112" Body="&lt;p&gt;Is there some place (on the internet or elsewhere) where I can find the number and preferably a list of all (isomorphism classes of) finite connected $T_0$-spaces with, say, 5 points?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In know that a $T_0$-topology on a finite set is equivalent to a partial ordering, and &lt;a href=&quot;http://en.wikipedia.org/wiki/Partial_order&quot; rel=&quot;nofollow&quot;&gt;wikipedia&lt;/a&gt; tells me that there are, up to isomorphism, 63 partially ordered sets with precisely 5 elements. However, I am only interested in connected spaces, and I'd love to have a list (most preferably in terms of Hasse diagrams).&lt;/p&gt;&#xA;" OwnerUserId="367" LastActivityDate="2011-01-04T18:33:19.823" Title="Is there a list of all connected $T_0$-spaces with 5 points?" Tags="&lt;topology&gt;&lt;reference-request&gt;" AnswerCount="1" CommentCount="0" />
  <row Id="7296" PostTypeId="1" AcceptedAnswerId="7298" CreationDate="2010-10-20T14:35:14.883" Score="7" ViewCount="176" Body="&lt;p&gt;I've seen recently for the first time in &lt;em&gt;Special Functions&lt;/em&gt; (by G. Andrews, R. Askey and R. Roy) the definitions of fractional integral &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(I_{\alpha }f)(x)=\frac{1}{\Gamma (\alpha )}\int_{a}^{x}(x-t)^{\alpha -1}f(t)dt\qquad \text{Re}\alpha &gt;0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and fractional derivative &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{d^{\nu }w^{\mu }}{dw^{\nu }}=\frac{\Gamma (\mu +1)}{\Gamma (\mu -\nu +1)}w^{\mu -\nu },$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;in &lt;em&gt;The Hypergeometric Functions&lt;/em&gt; Chapter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to know some applications for Fractional Calculus and/or which results can only be obtained by it, if any.&lt;/p&gt;&#xA;" OwnerUserId="752" LastEditorUserId="2548" LastEditorDisplayName="" LastEditDate="2010-12-23T05:27:56.867" LastActivityDate="2010-12-23T05:27:56.867" Title="Applications of Fractional Calculus" Tags="&lt;calculus&gt;&lt;analysis&gt;" AnswerCount="4" FavoriteCount="3" />
  <row Id="7297" PostTypeId="2" ParentId="7266" CreationDate="2010-10-20T14:45:57.773" Score="1" ViewCount="" Body="&lt;p&gt;Here is another way of looking at it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First consider the special case of starting with $1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider what happens when  $\displaystyle x = \frac{f_n}{f_{n+1}}$ where $f_n$ is the $n^{th}$ fibonacci number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You get $$\frac{1}{\frac{f_n}{f_{n+1}} + 1} = \frac{f_{n+1}}{f_n + f_{n+1}} = \frac{f_{n+1}}{f_{n+2}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $\displaystyle 1 = \frac{f_1}{f_2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;we see that after $n$ iterations, $\displaystyle x = \frac{f_{n+1}}{f_{n+2}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This can be generalized to any other starting value, by using a Fibonacci like sequence, which satisfies the recurrence $\displaystyle a_{n+2} = a_{n+1} + a_{n}$ and choosing appropriate $a_{2}$ and $a_{1}$ so that $\displaystyle \frac{a_1}{a_2}$ is the initial guess for $x$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The $n^{th}$ value for $x$ will be given by $\displaystyle \frac{a_n}{a_{n+1}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The general formula for such sequences is given by $a_{n} = A\alpha^n + B\beta^n$ where $\alpha,\beta$ are roots of $\displaystyle z^2 = z + 1$ and thus the limit of $\displaystyle \frac{a_n}{a_{n+1}}$ can be easily found, which will be one of $1/\alpha$ or $1/\beta$ (which you can also see, by assuming there is a limit $1/L$ and setting $\displaystyle 1/L = \frac{1}{1+1/L}$).&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-10-20T14:51:36.180" LastActivityDate="2010-10-20T14:51:36.180" />
  <row Id="7298" PostTypeId="2" ParentId="7296" CreationDate="2010-10-20T14:51:44.417" Score="2" ViewCount="" Body="&lt;p&gt;Fractional derivatives can be used to establish connections between various special functions.  The book &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0387488065&quot; rel=&quot;nofollow&quot;&gt;An Atlas of Functions&lt;/a&gt; makes heavy use of this, especially derivatives of order 1/2 and -1/2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, the &lt;em&gt;existence&lt;/em&gt; of fractional derivatives is related to the convergence of Fourier transforms.  For example, if a function has a 1/2 a derivative that means you can multiply its Fourier transform by $x^{1/2}$ and it is still in $L^2$. But I haven't seen much use in actually computing fractional derivatives, only knowing that they exist.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On a somewhat related note, see my answer to a &lt;a href=&quot;http://mathoverflow.net/questions/28147/why-do-we-care-about-lp-spaces-besides-p-1-p-2-and-p-infinity/28149#28149&quot; rel=&quot;nofollow&quot;&gt;question on Math Overflow&lt;/a&gt; related to Sobolev spaces.&lt;/p&gt;&#xA;" OwnerUserId="247" LastEditorUserId="247" LastEditorDisplayName="" LastEditDate="2010-10-20T14:58:25.197" LastActivityDate="2010-10-20T14:58:25.197" />
  <row Id="7299" PostTypeId="2" ParentId="7287" CreationDate="2010-10-20T14:53:34.950" Score="1" ViewCount="" Body="&lt;p&gt;I think your proof i fine. Here is pretty much the same thing as you did, but with less formulas. Write $f(z)= B(z)S(Z)O(z)$, where $B$ is the Blaschke factor, $S$ is the singular inner function of $f$ and where $O$ is the outer part of $f$. &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;First note that $1/O$ is outer (because $\log\tilde{f}\in L^1$ in your notation, then $\log(1/\tilde{f})=-\log\tilde{f}\in L^1$ too)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Since $1/f$ is analytic in the disc, we get $B(z)=1$. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$S/f=1/O$, is outer so $1/f=(1/S)(1/O)$ is the unique inner outer factorization.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$|S|\le 1$ and $|1/S|\le 1$ means $|S|=1$, so by open mapping $S$ is a unimodular constant.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="1154" LastEditorUserId="1154" LastEditorDisplayName="" LastEditDate="2010-10-20T16:03:23.447" LastActivityDate="2010-10-20T16:03:23.447" CommentCount="3" />
  <row Id="7300" PostTypeId="2" ParentId="7296" CreationDate="2010-10-20T15:00:43.870" Score="1" ViewCount="" Body="&lt;p&gt;I wouldn't say there are results that can only be obtained through differintegration. It only happens that there are problems whose solutions look neater when we bring in the machinery of differintegrals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.sciencedirect.com/science/bookseries/00765392&quot; rel=&quot;nofollow&quot;&gt;Spanier and Oldham&lt;/a&gt; and &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0471588849&quot; rel=&quot;nofollow&quot;&gt;Miller and Ross&lt;/a&gt; remain useful references on the applications of differintegration. The first reference has a chapter on how certain diffusion problems have a neater formulation when differintegrals are used. For the second reference, the application that jumped out at me was Abel's solution to the so-called &lt;a href=&quot;http://mathworld.wolfram.com/TautochroneProblem.html&quot; rel=&quot;nofollow&quot;&gt;tautochrone problem&lt;/a&gt;: finding the curve such that the time needed for a particle to descend from a given position to the bottom of the curve (assuming there is no friction) is &lt;em&gt;independent&lt;/em&gt; of position.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Though Huygens and other mathematicians have already obtained this solution long before Abel, he decided to use an integral equation formulation that can then be solved with the help of differintegration. In particular, he arrived at the equation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sqrt{2g}T=\int_0^y\frac{s^{\prime}(\eta)}{\sqrt{y-\eta}}\mathrm{d}\eta$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which when reformulated as a differintegral is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sqrt{\frac{2g}{\pi}}T=\frac{\mathrm{d}^{-\frac12}}{\mathrm{d}y^{-\frac12}}s^{\prime}(y)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I won't spoil the rest of the solution; I'd suggest that you read Miller and Ross if you're interested.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-10-20T15:00:43.870" />
  <row Id="7301" PostTypeId="1" AcceptedAnswerId="7312" CreationDate="2010-10-20T15:29:00.573" Score="12" ViewCount="782" Body="&lt;p&gt;Both &lt;a href=&quot;http://en.wikipedia.org/wiki/Fourier_transform&quot; rel=&quot;nofollow&quot;&gt;fourier transform&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/Taylor_series&quot; rel=&quot;nofollow&quot;&gt;taylor series&lt;/a&gt; are means to represent functions in a different form. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My question:&lt;/strong&gt;&lt;br&gt;&#xA;What is the connection between these two? Is there a way to get from one to the other (and back again)? Is there an overall, connceting (geometric?) intuition?&lt;/p&gt;&#xA;" OwnerUserId="346" LastActivityDate="2011-03-19T08:49:08.250" Title="Connection between fourier transform and taylor series" Tags="&lt;intuition&gt;&lt;fourier-analysis&gt;&lt;taylor-expansion&gt;&lt;integral-transforms&gt;" AnswerCount="4" CommentCount="1" FavoriteCount="14" />
  <row Id="7302" PostTypeId="1" AcceptedAnswerId="7306" CreationDate="2010-10-20T15:30:58.647" Score="4" ViewCount="262" Body="&lt;p&gt;The Wronskian lets us determine if a set of functions (possibly the solutions to a differential equation) are linearly dependent or not. But, for every example in the book, it is very obvious if one of the functions is a linear combination of the others. The examples in the book use 3-5 functions. What would be an example of a small number of functions where this isn't obvious?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or is the application of the Wronskian mostly to deal with large sets of functions... where the sheer number makes it hard to tell if they are dependent or not?&lt;/p&gt;&#xA;" OwnerUserId="2563" LastActivityDate="2010-10-20T18:53:03.183" Title="Example of functions where linear dependence isn't obvious." Tags="&lt;linear-algebra&gt;" AnswerCount="3" FavoriteCount="2" />
  <row Id="7303" PostTypeId="1" CreationDate="2010-10-20T15:32:26.977" Score="2" ViewCount="329" Body="&lt;p&gt;After some searching I found:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Triangulation&quot; rel=&quot;nofollow&quot;&gt;Triangulation&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;Using theta and &quot;b&quot; to find out the height of a building.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance.&lt;/p&gt;&#xA;" OwnerUserId="1112" LastActivityDate="2010-10-20T15:36:05.963" Title="What are the uses of trigonometry in architecture?" Tags="&lt;trigonometry&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="7304" PostTypeId="2" ParentId="7303" CreationDate="2010-10-20T15:36:05.963" Score="3" ViewCount="" Body="&lt;p&gt;Too many to name. Anything that involves the study of the measure of triangles is trigonometry. The Egyptians used Pythagorean triplets to produce right angles using the &quot;rope stretching method.&quot; (a circular rope with 12 knots. stretch so that you have 3, 4, 5)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, it has played a role as long as architecture has been architecture.&lt;/p&gt;&#xA;" OwnerUserId="2563" LastActivityDate="2010-10-20T15:36:05.963" />
  <row Id="7305" PostTypeId="2" ParentId="7301" CreationDate="2010-10-20T15:43:04.507" Score="12" ViewCount="" Body="&lt;p&gt;A holomorphic function in an annulus containing the unit circle has a &lt;a href=&quot;http://en.wikipedia.org/wiki/Laurent_series&quot; rel=&quot;nofollow&quot;&gt;Laurent series&lt;/a&gt; about zero which generalizes the Taylor series of a holomorphic function in a neighborhood of zero.  When restricted to the unit circle, this Laurent series gives a Fourier series of the corresponding periodic function.  (This explains the connection between the Cauchy integral formula and the integral defining the coefficients of a Fourier series.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But it's worth mentioning that the Fourier transform is much more general than this and applies in a broad range of contexts.  I don't know that there's a short, simple answer to this question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit:  I guess it's also worth talking about intuition.  One intuition for the Taylor series of a function $f(x)$ at at a point is that its coefficients describe the displacement, velocity, acceleration, &lt;a href=&quot;http://en.wikipedia.org/wiki/Jerk_(physics)&quot; rel=&quot;nofollow&quot;&gt;jerk&lt;/a&gt;, and so forth of a particle which is at location $f(t)$ at time $t$.  And one intuition for the Fourier series of a periodic function $f(x)$ is that it describes the decomposition of $f(x)$ into pure tones of various frequencies.  In other words, a periodic function is like a chord, and its Fourier series describes the notes in that chord.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(The connection between the two provided by the Cauchy integral formula is therefore quite remarkable; one takes an integral of $f$ over the unit circle and it tells you information about the behavior of $f$ at the &lt;em&gt;origin&lt;/em&gt;.  But this is more a magic property of holomorphic functions than anything else.  One intuition to have &lt;em&gt;here&lt;/em&gt; is that a holomorphic function describes, for example, the flow of some ideal fluid, and integrating over the circle gives you information about &quot;sources&quot; and &quot;sinks&quot; of that flow within the circle.)&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-10-20T21:07:13.843" LastActivityDate="2010-10-20T21:07:13.843" CommentCount="2" />
  <row Id="7306" PostTypeId="2" ParentId="7302" CreationDate="2010-10-20T15:43:10.893" Score="5" ViewCount="" Body="&lt;p&gt;It can be concealed in various ways.  If the functions are expressed as trig functions the dependence gets hidden easily.  Think of $sin^2(x)$ and $cos(2x)$.  As the functions get messier it gets easier to hide.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem with the Wronskian is that the functions must be sufficiently differentiable and you need to be able to calculate it.  Think of the indicator function on the rationals and the indicator function on the irrationals.  These are dependent, but the Wronskian won't help.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-10-20T15:43:10.893" CommentCount="6" />
  <row Id="7307" PostTypeId="1" AcceptedAnswerId="7311" CreationDate="2010-10-20T15:45:14.610" Score="2" ViewCount="183" Body="&lt;p&gt;Perhaps this has a simple answer, but I don't know (I wouldn't be asking if I did). Every set of outer measure zero is Lebesgue measurable with Lebesgue measure zero. Is the converse true? That is, if a set has Lebesgue measure zero, does it necessarily have outer measure zero?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And I suppose I should specify that I'm thinking in $\mathbb{R}^n$. &lt;/p&gt;&#xA;" OwnerUserId="2313" LastActivityDate="2010-10-20T16:26:50.297" Title="Concerning sets of (Lebesgue) measure zero." Tags="&lt;real-analysis&gt;&lt;measure-theory&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="7308" PostTypeId="2" ParentId="7302" CreationDate="2010-10-20T15:49:04.777" Score="5" ViewCount="" Body="&lt;p&gt;The cute application of the Wronskian I like to keep in mind is that the invertibility of the Vandermonde matrix implies that the set $\{ e^{\lambda z} : \lambda \in \mathbb{C} \}$ is linearly independent.  The Wronskian also shows up, for example, in the method of &lt;a href=&quot;http://en.wikipedia.org/wiki/Variation_of_parameters&quot; rel=&quot;nofollow&quot;&gt;variation of parameters&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-10-20T15:49:04.777" CommentCount="2" />
  <row Id="7309" PostTypeId="2" ParentId="7301" CreationDate="2010-10-20T15:49:30.457" Score="12" ViewCount="" Body="&lt;p&gt;There is a big difference between the Taylor series and Fourier transform-the Taylor series is a local approximation, while the Fourier transform uses information over a range of the variable.  The theorem Qiaochu mentions is very important in complex analysis and is one indication of how restrictive having a derivative in the complex plane is on functions.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-10-20T15:49:30.457" />
  <row Id="7310" PostTypeId="2" ParentId="7302" CreationDate="2010-10-20T15:50:05.827" Score="2" ViewCount="" Body="&lt;p&gt;Have you tried to prove by hand (i.e., only using the definition of linear independence)  that $\sin \theta$ and $ \cos \theta$ are linearly independent? Of course, this can be done with the help of the Wronskian.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And what about $e^{i\theta}$ and $ e^{i (\theta + \frac{\pi}{2})}$? This is geometrically clear, but: can you see the difference between linear independence over the real and complex numbers?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT. Just to add still a more elementary example: what about $\sin^2\theta $ and $\cos^2\theta$?&lt;/p&gt;&#xA;" OwnerUserId="664" LastEditorUserId="664" LastEditorDisplayName="" LastEditDate="2010-10-20T18:53:03.183" LastActivityDate="2010-10-20T18:53:03.183" CommentCount="5" />
  <row Id="7311" PostTypeId="2" ParentId="7307" CreationDate="2010-10-20T15:51:49.773" Score="6" ViewCount="" Body="&lt;p&gt;If $\mu^{*}$ is an outer measure, then we define the measurable sets in terms of $\mu^{*}$; the measure is then &lt;em&gt;defined&lt;/em&gt; to be the restriction of the outer measure to the measurable sets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To be more explicit: if you have a $\sigma$-algebra and an outer measure $\mu^{*}$ on the algebra, then we say that a set $E$ is $\mu^{*}$-measurable if and only if for every $A$ in the $\sigma$-algebra,&#xA;$$\mu^{*}(A) = \mu^{*}(A\cap E) + \mu^{*}(A\cap E').$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As Halmos says in his book &lt;strong&gt;Measure Theory&lt;/strong&gt;,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;&lt;em&gt;It is rather difficutl to get an intuitive understanding of the meaning of $\mu^{*}$-measurability except through familiarity with its implications.&lt;/em&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Once you have the definiiton of $\mu^{*}$-measurable, then let $S$ be the set of all measurable sets, and you &lt;em&gt;define&lt;/em&gt; the measure $\mu$ on $S$ by $\mu(E) = \mu^{*}(E)$ for all $E\in S$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, this holds for the Lebesgue measure: if $E$ is Lebesgue measurable, then the Lebesgue measure of $E$ is equal to the outer measure of $E$, because the Lebesgue measure of $E$ is &lt;em&gt;defined&lt;/em&gt; to be the outer measure of $E$. This holds for any Lebesgue measure, not just for measure $0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The point of the theorem you state earlier is that having outer measure zero &lt;em&gt;implies&lt;/em&gt; that the set is measurable; that's the nontrivial part of the statement (not that the Lebesgue measure of the set will then be zero). &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-10-20T16:26:50.297" LastActivityDate="2010-10-20T16:26:50.297" CommentCount="3" />
  <row Id="7312" PostTypeId="2" ParentId="7301" CreationDate="2010-10-20T15:52:19.800" Score="14" ViewCount="" Body="&lt;p&gt;Assume that the Taylor expansion $f(x)=\sum_{k=0}^\infty a_k x^k$ is convergent for some $|x|&amp;gt;1$. Then $f$ can be extended in a natural way into the complex domain by writing $f(z)=\sum_{k=0}^\infty a_k z^k$ with $z$ complex and $|z|≤1$. So we may look at $f$ on the unit circle $|z|=1$. Consider $f$ as a function of the polar angle $\phi$ there, i.e., look at the function $F(\phi):=f(e^{i\phi})$. This function $F$ is $2\pi$-periodic, and its Fourier expansion is nothing else but $F(\phi)=\sum_{k=0}^\infty a_k e^{ik\phi}$ where the $a_k$ are the Taylor coefficients of the &quot;real&quot; function $x\mapsto f(x)$ we started with.&lt;/p&gt;&#xA;" OwnerUserId="1303" LastEditorUserId="1303" LastEditorDisplayName="" LastEditDate="2011-03-19T08:49:08.250" LastActivityDate="2011-03-19T08:49:08.250" />
  <row Id="7313" PostTypeId="2" ParentId="6791" CreationDate="2010-10-20T16:45:52.127" Score="3" ViewCount="" Body="&lt;p&gt;This isn't a proof, but's too long for a comment, and may just be a &#xA;restatement of the problem. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For contradiction, let k be any integer such that b(n) = k never holds. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means k &amp;lt;= a(n) &amp;lt; k+1 never holds. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since a(n) can't be between k and k+1, the arctan of a(n) can't be either. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, there's an interval between -Pi/2 and -Pi/2 that a(n) may not &#xA;touch. Let's call in [c,d). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since tan is periodic, a(n) must also avoid m*Pi+[c,d) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since Pi is irrational, m*Pi+[c,d) must contain an infinite number of &#xA;integers (pretty sure this is true, but I could be wrong). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, there are an infinite number of intervals (approaching &#xA;Pi/2) that a(n) must avoid. Further, a(n) must avoid the arctans of &#xA;these intervals, and the arctans of those intervals, etc. The repeated &#xA;arctan intervals approach 0. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, a(n) also has to avoid those intervals plus any multiple of Pi.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This non-proof actually applies to any interval a(n) misses, so, if true, shows that a(n) is dense in R.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope that helps.  &lt;/p&gt;&#xA;" OwnerUserId="2469" LastActivityDate="2010-10-20T16:45:52.127" />
  <row Id="7314" PostTypeId="2" ParentId="7291" CreationDate="2010-10-20T16:52:06.263" Score="3" ViewCount="" Body="&lt;p&gt;You might be interested in &lt;a href=&quot;http://en.wikipedia.org/wiki/Rota%E2%80%93Baxter_algebra&quot; rel=&quot;nofollow&quot;&gt;Rota-Baxter algebras&lt;/a&gt;.  They are, among other things, an abstraction of how indefinite integration by parts works.  If you are interested, &lt;a href=&quot;http://www.ams.org/bull/1969-75-02/S0002-9904-1969-12156-7/S0002-9904-1969-12156-7.pdf&quot; rel=&quot;nofollow&quot;&gt;Rota's original paper&lt;/a&gt; on the subject is a good read.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-10-20T16:52:06.263" CommentCount="1" />
  <row Id="7315" PostTypeId="1" AcceptedAnswerId="7316" CreationDate="2010-10-20T16:57:00.573" Score="6" ViewCount="122" Body="&lt;p&gt;So here is the problem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solved for a in terms of x:&lt;br&gt;&#xA;$$a^{x} = 10^{2x + 1}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried:&lt;br&gt;&#xA;$\displaystyle x \cdot \log(a) = (2x+1) \cdot \log\;10 $ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle \frac{x}{2x + 1} = \frac{\log\;10} {\log\;a} $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But this is not going in the right direction, the answer according to the book is:&lt;br&gt;&#xA;$$ \frac{1} {\log\;a - 2} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Excuse the 'power' tag for this question, there is no logarithm tag&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="2576" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-10-20T17:17:41.097" LastActivityDate="2010-10-20T17:55:13.893" Title="Solving $a^{x} = 10^{2x + 1}$" Tags="&lt;algebra-precalculus&gt;&lt;homework&gt;" AnswerCount="3" CommentCount="3" />
  <row Id="7316" PostTypeId="2" ParentId="7315" CreationDate="2010-10-20T17:00:40.740" Score="2" ViewCount="" Body="&lt;p&gt;Hint: The answer is using $\log_{10}$.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-10-20T17:00:40.740" CommunityOwnedDate="2010-10-20T17:01:01.110" CommentCount="14" />
  <row Id="7317" PostTypeId="1" AcceptedAnswerId="7327" CreationDate="2010-10-20T17:11:33.780" Score="3" ViewCount="1437" Body="&lt;p&gt;Assume I have only basic math knowledge, what specific areas of math would I need to learn in order to understand the following webpage:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Black%E2%80%93Scholes&quot; rel=&quot;nofollow&quot;&gt;Black-Scholes&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many thanks.&lt;/p&gt;&#xA;" OwnerUserId="2648" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-10-21T00:07:02.507" LastActivityDate="2010-10-21T15:20:10.657" Title="Understanding Black-Scholes" Tags="&lt;probability&gt;&lt;applications&gt;&lt;stochastic-processes&gt;&lt;finance&gt;" AnswerCount="5" CommentCount="2" FavoriteCount="4" />
  <row Id="7318" PostTypeId="2" ParentId="7315" CreationDate="2010-10-20T17:24:06.590" Score="0" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt;$\ $  Putting $\rm\ a = 10^{\:b}\ $ yields $\rm\ x = 1/(b - 2)$&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-10-20T17:54:35.253" LastActivityDate="2010-10-20T17:54:35.253" CommentCount="2" />
  <row Id="7320" PostTypeId="1" AcceptedAnswerId="7333" CreationDate="2010-10-20T17:39:56.413" Score="2" ViewCount="108" Body="&lt;p&gt;How to show that  $$\text{g.c.d} \Biggl( {m \choose k} , {m+1 \choose k } , {m+2 \choose k}, \cdots, {m+k \choose k} \Biggr)=1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $m,k \in \mathbb{N}$ and $m \geq k $.&lt;/p&gt;&#xA;" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-10-20T17:58:09.693" LastActivityDate="2010-10-20T18:48:43.627" Title="To show that $\text{g.c.d} \Bigl( {m \choose k} , {m+1 \choose k } , {m+2 \choose k}, \cdots, {m+k \choose k} \Bigr)=1$" Tags="&lt;number-theory&gt;&lt;binomial-coefficients&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" />
  <row Id="7321" PostTypeId="2" ParentId="7210" CreationDate="2010-10-20T17:49:49.660" Score="3" ViewCount="" Body="&lt;p&gt;The question asked &quot;where we might meet&quot; total boundedness.   One situation where you do &lt;em&gt;not&lt;/em&gt; meet it is classical analysis or geometry in finite-dimensional spaces; in $R^n$, bounded and totally bounded are equivalent concepts.   So the intuition and conditions for appearance are necessarily more subtle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Geometrically, a space fails to be totally bounded if and only if it contains an infinite set of points with all pairwise distances at least $d$ for some $d &gt; 0$.  This is just the negation of the definition of totally bounded: if for some $\epsilon &gt; 0$ there is no finite set of points whose $\epsilon$-neighborhood covers the space, then we can build an infinite set of mutually $\epsilon$-separated points by placing in the set  any point $p_1$ and for $n&gt;1$ inductively adding to the set any point $p_n$ not inside the $\epsilon$-neighborhood of the preceding points.  For finite-dimensional spaces this can happen only for a sequence of points that escapes to infinity.  But in infinite dimensions an infinite set of $d$-separated points can be packed into a ball of finite radius.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Boundedness and total boundedness are both properties of the completion of the space.  A metric space has either boundedness property if and only if its completion does. Considering complete metric spaces is convenient because total boundedness (for a complete metric space) is equivalent to the space being compact.  So one might want a topological characterization of the difference between bounded and totally bounded, for complete spaces.  However, any metric space can be converted into a bounded one (without affecting total boundedness, completeness, or the topology) by replacing the metric $d$ with $d/(1+d)$ or some other bounded monotonic function that preserves the property of being a metric.  So topological characterization suitable for understanding the difference between bounded and totally bounded (complete) spaces is more elusive. Maybe one should consider uniform spaces instead. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Returning to the question of situations where total boundedness appears, there are at least two:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;As far as total boundedness requires infinite-dimensionality and a metric, it comes up naturally in functional analysis.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;It also appears in so-called &quot;constructive analysis&quot;, i.e., analysis using direct and generally negation-free arguments, as in Errett Bishop's book.   The concept of total boundedness first arose from a (classical) logical analysis of the Heine-Borel theorem and what was required to extend it to general metric spaces.   In a situation where nonconstructive arguments, such as determining whether a given sequence has infinitely many positive terms, are not present, spaces like the real numbers are not all that different from general metric spaces, and an analysis of which hypotheses were &quot;really at work&quot; in the classical proofs is useful for building the theory in the more general and direct fashion that Bishop advocated.  So where boundedness might be used in a finite-dimensional classical argument, one might use total boundedness in the constructive one.  This is similar in flavor to the question above in the comments, of what characterizations are true without the axiom of choice.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="467" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-10-21T21:34:00.283" LastActivityDate="2010-10-21T21:34:00.283" CommentCount="4" />
  <row Id="7322" PostTypeId="1" AcceptedAnswerId="7375" CreationDate="2010-10-20T17:50:07.480" Score="2" ViewCount="80" Body="&lt;p&gt;We can easily see that if a sequence of inner functions converges then certainly the limit is still inner (using the fact that a countable union of measure zero sets still has measure zero). This implies that the set of inner functions is closed in $H^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, is the set of the &lt;em&gt;singular&lt;/em&gt; inner functions closed?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know that a singular inner function has the form:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$S(z) = K \text{exp} \left ( -\frac{1}{2\pi} \int_0^{2\pi} \frac{e^{i \theta} + z}{e^{i  \theta} - z} \, d\mu(\theta) \right )$$&#xA;where $\mu$ is a positive Borel measure singular with respect to the Lebesgue measure and $K$ is a constant of modulus $1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus if we take $S_n \to S$ in $H^2$ then $S_n \to S$ uniformly on compact sets. Clearly $S$ is inner because $S_n$ is inner, now we need to show it is singular. My question now is, does anyone have a hint how I show this? I know that if $(\mu_n)$ is a sequence of positive finite Borel measures, then by Riesz and Alaoglu we have that a subsequence converge to a finite positive Borel measure $\mu$. Can I use this?&lt;/p&gt;&#xA;" OwnerUserId="1120" LastEditorUserId="1120" LastEditorDisplayName="" LastEditDate="2010-10-20T17:59:51.000" LastActivityDate="2010-10-21T01:08:58.330" Title="Singular inner functions." Tags="&lt;complex-analysis&gt;&lt;functional-analysis&gt;" AnswerCount="1" />
  <row Id="7323" PostTypeId="2" ParentId="7315" CreationDate="2010-10-20T17:55:13.893" Score="0" ViewCount="" Body="&lt;p&gt;HINT: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe you can find useful to look at &lt;a href=&quot;http://en.wikipedia.org/wiki/Logarithm#Change_of_base&quot; rel=&quot;nofollow&quot;&gt;Logarithm - Change of base&lt;/a&gt;, after solving your equation $\displaystyle \frac{x}{2x+1}=\frac{\text{log} 10}{\text{log}\thinspace a}$. You should finish with something like $x = \displaystyle \frac{1}{\frac{\displaystyle \text{log} \thinspace a}{\displaystyle \text{log} 10}-2}$&lt;/p&gt;&#xA;" OwnerUserId="2569" LastActivityDate="2010-10-20T17:55:13.893" />
  <row Id="7324" PostTypeId="1" AcceptedAnswerId="7326" CreationDate="2010-10-20T17:58:43.577" Score="2" ViewCount="177" Body="&lt;p&gt;Suppose $f_{X}(x) = xe^{-x^2/2}$ for $x&gt;0$ and $Y = \ln X$. Find the density function for $Y$. So we want to find $P(Y \leq y)$. This is the same thing as $P(\ln X \leq y)$ or $P(X \leq e^{y})$. Thus $f_{Y}(y) = f_{X}(e^y)$? Or does $f_{Y}(y) = F_{X}(e^y)$ since $P(X \leq x) = F_{X}(x)$? &lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2010-10-20T18:05:53.110" Title="Transformation of Random Variables" Tags="&lt;probability&gt;" AnswerCount="1" />
  <row Id="7325" PostTypeId="2" ParentId="7317" CreationDate="2010-10-20T18:01:28.877" Score="0" ViewCount="" Body="&lt;p&gt;I found &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0521497892&quot; rel=&quot;nofollow&quot;&gt;this book&lt;/a&gt; quite useful.  But it does take the background listed.  The basic result that the equation is the same as one dimensional heat flow and the graphs of roughly how the value flows are not too tough.  There are also free calculators on the web that let you play with it.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-10-20T18:01:28.877" />
  <row Id="7326" PostTypeId="2" ParentId="7324" CreationDate="2010-10-20T18:05:53.110" Score="2" ViewCount="" Body="&lt;p&gt;You're correct up to the point where you have $P(Y \leq y) = P(\ln X \leq y) = P(X \leq e^y)$.  The correct next step, though, is that $F_Y(y) = F_X(e^y)$.  To obtain $f_Y(y)$, differentiate both sides of this equation with respect to $y$ (and don't forget to use Leibniz's rule on the right-hand side). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(In case you don't remember, Leibniz's rule is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Fundamental_theorem_of_calculus#First_part&quot; rel=&quot;nofollow&quot;&gt;first part of the Fundamental Theorem of Calculus&lt;/a&gt; combined with the chain rule.)&lt;/p&gt;&#xA;" OwnerUserId="2370" LastActivityDate="2010-10-20T18:05:53.110" CommentCount="5" />
  <row Id="7327" PostTypeId="2" ParentId="7317" CreationDate="2010-10-20T18:15:02.630" Score="14" ViewCount="" Body="&lt;p&gt;The standard low technology argument for Black-Scholes (the famous &quot;binomial tree&quot;) requires only basic material, though there is also a standard medium technology approach using stochastic calculus (informally) and an advanced approach using the rigorous mathematical apparatus of stochastic processes, Brownian motion, and diffusion equations.  I think summation of a finite geometric series, very basic probability, and knowledge of how to take a limit of the binomial distribution to get the Gaussian normal distribution, is enough for the binary tree approach.   This is in almost every elementary text on options pricing, such as the ones used in business schools or quant basic training, where by elementary I mean not starting immediately from stochastic calculus.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(This relatively simple derivation applies to the Black-Scholes &lt;em&gt;formula&lt;/em&gt; for pricing the simplest types of options and, in some developments of the theory, also to the Black-Scholes &lt;em&gt;model&lt;/em&gt; giving a more general formalism for pricing arbitrary European options as expected destinations of a random walk.  The binomial tree is not ordinarily presented as a method for producing the Black-Scholes &lt;em&gt;partial differential equation&lt;/em&gt; satisfied by prices in their model, although in theory it could do that.  The differential equation is usually arrived at by a simple heuristic argument using stochastic calculus (given in all finance books that introduce stochastic calculus).  A fully rigorous derivation is quite complicated and does require a lot of mathematical background, but this is not necessary for understanding the formula and its use in finance.  An elementary mathematical derivation of the differential equation not explicitly using the binomial tree was posted at Terry Tao's webpage, though it does not dwell on financial intuitions or interpretations.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem can be discretized, as pricing of an option to buy a security whose value goes up and down in an (exponentiated) random walk, where there are a finite number of time steps and at each step the price is multiplied by $r$ or $1/r$ with a constant probability of up- or down- motion in the price at each step.  Taking a limit you get the Black-Scholes formula.  The key point is to consider a single time step and show that the price is determined by the relationship between the payoffs for the security and for the option.   The rest is a matter of algebraically propagating this known answer back up through the binary tree (the $2^n$ possible paths for the price of the security between time 0 and time $n$) until you reach the root.  This gives the answer for the finite problem, then taking the continuous limit of the problem &quot;derives&quot; Black-Scholes for the continuous time case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To show that the continuous time theory makes sense mathematically involves more than just suggesting that such a theory might exist as a limit of a well-defined discrete theory.  For that you need to construct the limit theory directly, which requires the machinery of stochastic analysis. To understand the limitations of the finite or continuous theory requires knowledge of finance in addition to the mathematics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[References added:  low-tech approach (Cox-Ross-Rubinstein binomial trees): &lt;a href=&quot;http://en.wikipedia.org/wiki/Binomial_options_pricing_model&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Binomial_options_pricing_model&lt;/a&gt;  ; medium-tech is in most introductory books on derivatives pricing that cover stochastic calculus -- Hull and its competitors;  high-tech is in books on stochastic PDE such as Oksendal or advanced finance books like Shreve.  ]&lt;/p&gt;&#xA;" OwnerUserId="467" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-10-21T01:22:46.387" LastActivityDate="2010-10-21T01:22:46.387" CommentCount="6" />
  <row Id="7328" PostTypeId="1" AcceptedAnswerId="7330" CreationDate="2010-10-20T18:15:11.600" Score="2" ViewCount="105" Body="&lt;p&gt;Let R be a ring with identity. An R-module M is Artinian if it satisfies the descending chain condition on submodules. What is an example of an Artinian module with a proper submodule that is not finitely generated?&lt;/p&gt;&#xA;" OwnerUserId="941" LastEditorUserId="941" LastEditorDisplayName="" LastEditDate="2010-10-20T19:06:33.980" LastActivityDate="2010-10-20T19:06:33.980" Title="Is there an Artinian module with infinitely generated proper submodule?" Tags="&lt;abstract-algebra&gt;&lt;ring-theory&gt;" AnswerCount="1" />
  <row Id="7329" PostTypeId="1" AcceptedAnswerId="7474" CreationDate="2010-10-20T18:24:00.413" Score="0" ViewCount="234" Body="&lt;p&gt;Two players put a dollar in a pot. They decide to throw a pair of dice alternatively. The first one who throws a total of $ 5$ on both dice wins the pot. How much should the player who starts add to the pot to make this a fair game? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So my interpretation of this problem is that I first throw a dice and then the other person throws a dice afterwards. E.g. if I throw a 1, and if the other person throws a 4 wins the pot. So we are trying to find the expected payout?&lt;/p&gt;&#xA;" OwnerUserId="2548" LastEditorUserId="2548" LastEditorDisplayName="" LastEditDate="2010-10-20T18:39:54.987" LastActivityDate="2010-10-22T00:53:42.100" Title="Fair Game and Dice" Tags="&lt;probability&gt;" AnswerCount="3" CommentCount="7" />
  <row Id="7330" PostTypeId="2" ParentId="7328" CreationDate="2010-10-20T18:24:36.707" Score="5" ViewCount="" Body="&lt;p&gt;Take a look at &lt;a href=&quot;http://en.wikipedia.org/wiki/Artinian_module#Relation_to_the_Noetherian_condition&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Artinian_module#Relation_to_the_Noetherian_condition&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There you can find a concrete example of an Artinian module over the integers which is non-Noetherian and hence has a non-finitely generated submodule.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the example at hand, the module itself is not finitely generated, and there are also lots of non-finitely generated proper submodules.&lt;/p&gt;&#xA;" OwnerUserId="367" LastEditorUserId="367" LastEditorDisplayName="" LastEditDate="2010-10-20T18:49:35.843" LastActivityDate="2010-10-20T18:49:35.843" CommentCount="9" />
  <row Id="7331" PostTypeId="1" AcceptedAnswerId="7336" CreationDate="2010-10-20T18:24:40.443" Score="1" ViewCount="217" Body="&lt;p&gt;Maple tells me that $\lim_{n \to \infty} n^2 \log^n(1 - \frac{c \log n}{n}) = 0$ for any constant $c$, but I can't find a way to prove it. Any suggestions?&lt;/p&gt;&#xA;" OwnerUserId="2649" LastEditorUserId="2649" LastEditorDisplayName="" LastEditDate="2010-10-20T19:15:43.400" LastActivityDate="2010-10-20T19:15:43.400" Title="The limit of $n^2 \log^n(1 - \frac{c \log n}{n})$" Tags="&lt;limit&gt;" AnswerCount="2" CommentCount="5" />
  <row Id="7332" PostTypeId="2" ParentId="7296" CreationDate="2010-10-20T18:25:06.823" Score="1" ViewCount="" Body="&lt;p&gt;Miller and Ross looks very nice indeed. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as applications are concerned: &lt;a href=&quot;http://docs.google.com/viewer?a=v&amp;amp;q=cache:aWfxS6fj7WgJ:www.m-hikari.com/ams/forth/rahimyAMS49-52-2010.pdf+An+Introduction+to+the+Fractional+Calculus+and+Fractional+Differential+Equation&amp;amp;hl=en&amp;amp;gl=mx&amp;amp;pid=bl&amp;amp;srcid=ADGEESg0ul80vQSr4JNRvdX8z_LgojseL92CgSyOs19upoGguvZPFjwzTYwtXe-baJIxbksAo2AFdc9hZ2NNOzyS8nmDrkUQJTNZ5a3sCiZml0Cjj3cVCYJPOJKzsI8mQI7XXQSRUNa4&amp;amp;sig=AHIEtbRNtVslJk9HBm6E0JtV-1mAytq9BQ&quot; rel=&quot;nofollow&quot;&gt;Applications of Fractional Diﬀerential Equations&lt;/a&gt;. For some strange reason, the original link is not available. Instead, you can look at it by using Google Docs viewer.&lt;/p&gt;&#xA;" OwnerUserId="2569" LastActivityDate="2010-10-20T18:25:06.823" />
  <row Id="7333" PostTypeId="2" ParentId="7320" CreationDate="2010-10-20T18:48:43.627" Score="1" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\ \ $ if $\rm\ \ f_{k}(m+1)\ \ mod \ \ f_{k}(m)\ =\ f_{k-1}(m)\ \ $ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;then $\rm\ \ \gcd(f_k(m),\ f_k(m+1),\:\cdots,\:f_k(m+k)) $  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad =\ \ \gcd(f_k(m),\ \ f_{k-1}(m),\ \cdots,\ f_0(m))\ \ \ $ [$\rm\ = 1\ $ if $\rm\ f_0(m) = 1\:$ ]&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-10-20T18:48:43.627" />
  <row Id="7334" PostTypeId="1" AcceptedAnswerId="7339" CreationDate="2010-10-20T18:59:24.933" Score="1" ViewCount="542" Body="&lt;p&gt;I know that in general, $H_{n}(X)$ counts the number of $n$-cycles that are not $n$-boundaries of a simplicial complex $X$. So for the sphere, $H_{0}(X) \cong \mathbb{Z}$ since it is connected. Also $H_{n}(X) = 0$ for all $n&gt;0$ (e.g. all $1$-cycles are $1$-boundaries, etc..). How do you use this geometric interpretation to deduce that $H_{1}(X) = \mathbb{Z} \times \mathbb{Z}_2$ where $X$ is the Klein bottle? This doesn't seem to correspond to the Betti number. &lt;/p&gt;&#xA;" OwnerUserId="2548" LastEditorUserId="367" LastEditorDisplayName="" LastEditDate="2010-10-20T19:17:42.607" LastActivityDate="2010-10-20T19:42:28.680" Title="Homology of the Klein Bottle" Tags="&lt;algebraic-topology&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="7335" PostTypeId="2" ParentId="7331" CreationDate="2010-10-20T19:00:40.203" Score="3" ViewCount="" Body="&lt;p&gt;The Taylor series expansion $- \log (1-x) = \sum_{k \ge 1} \frac{x^k}{k}$ shows that $\log(1 - x) = O(x)$ for small $x$.  So $\log \left( 1 - \frac{c \log n}{n} \right) = O \left( \frac{\log n}{n} \right)$, hence the absolute value of the expression in question is bounded by a constant times $\frac{(\log n)^n}{n^{n-2}}$ which rather clearly goes to zero.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a good example of why it's more flexible to use big-O notation than to deal directly with the definition of a limit.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-10-20T19:00:40.203" CommentCount="1" />
  <row Id="7336" PostTypeId="2" ParentId="7331" CreationDate="2010-10-20T19:05:07.463" Score="5" ViewCount="" Body="&lt;p&gt;Well&#xA;$$\log\left(1-c\frac{\log n}{n}\right)=-c\frac{\log n}{n}&#xA;+O\left(\frac{(\log n)^2}{n^2}\right)$$&#xA;and so&#xA;$$n^2\log\left(1-c\frac{\log n}{n}\right)^n=(-c)^n\frac{(\log n)^n}{n^{n-2}}&#xA;\left(1+O\left(\frac{\log n}{n}\right)\right)^n.$$&#xA;Now&#xA;$$n\log\left(1+O\left(\frac{\log n}{n}\right)\right)=O(\log n)$$&#xA;so that&#xA;$$\left(1+O\left(\frac{\log n}{n}\right)\right)^n$$&#xA;grows as at most a polynomial in $n$. The $n^n$ in the denominator&#xA;will swamp everything else...&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-10-20T19:05:07.463" />
  <row Id="7337" PostTypeId="2" ParentId="7329" CreationDate="2010-10-20T19:14:21.890" Score="1" ViewCount="" Body="&lt;p&gt;First, find the probability &lt;em&gt;p&lt;/em&gt; that a 5 is thrown and define &lt;em&gt;q&lt;/em&gt; = 1 - &lt;em&gt;p&lt;/em&gt;.  The chance That the first player wins is C=p+pq^2+pq^4+\cdots; calculate this value.  Now set C=(1-C)(1+x) and solve for the extra contribution &lt;em&gt;x&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1778" LastActivityDate="2010-10-20T19:14:21.890" CommentCount="2" />
  <row Id="7338" PostTypeId="2" ParentId="7334" CreationDate="2010-10-20T19:23:02.437" Score="4" ViewCount="" Body="&lt;p&gt;Think of the bottle as two Möbius strips glued along their edges. A closed loop that circles one of the bands (along its center, say) is not a boundary, but if you follow it twice it becomes one: you can think of it as a loop that follows precisely the edge of one of the original bands, so it's just the boundary of that band. This gives you the &quot;torsion&quot; piece of the homology, and it shows you that homology is a slightly more refined invariant than just counting the number of closed &quot;empty&quot; loops. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: perhaps worth to point out that the group-theoretic term &quot;torsion&quot;, which corresponds to elements of finite order, is derived precisely from this geometric picture: the torsion element result from the &quot;twist&quot; in the band.&lt;/p&gt;&#xA;" OwnerUserId="308" LastActivityDate="2010-10-20T19:23:02.437" />
  <row Id="7339" PostTypeId="2" ParentId="7334" CreationDate="2010-10-20T19:42:28.680" Score="5" ViewCount="" Body="&lt;p&gt;For starters, the Betti numbers are not the homology.  The Betti number is what does the &quot;counting&quot; of cycles you mention, but there's more to the story.  The classification theorem for finitely generated abelian groups says that you can express any such group as a direct sum (equivalently, a product) of copies of $\mathbb{Z}$ and of $\mathbb{Z}_n$ for some values of $n$.  The Betti number is the number of copies of $\mathbb{Z}$ you get, i. e. the largest possible free subgroup (the rest is called the &lt;em&gt;torsion&lt;/em&gt; of the group).  If you've done homology with coefficients yet, a good way to get this is by taking coefficients in $\mathbb{Q}$, which annihilates all the elements of finite order and leaves you with a $\mathbb{Q}$-vector space whose dimension is the Betti number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But back to the main story.  Rather than finding torsion and Betti numbers individually, for simplicial complexes especially, I find it easier to just compute the homology via $H_n=\text{ker}(d_n)/\text{im}(d_{n+1})$.  Let's use the following picture:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://www.sagemath.org/doc/reference/_images/klein.png&quot; alt=&quot;Klein bottle Delta-complex&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have a single $0$-simplex, which I'll call $v$; three $1$-simplices, of which the horizontal one will be $a$, the vertical one $b$, and the diagonal one $c$; and two $2$-simplices, of which the upper is $U$ and the lower $L$.  I'm orienting the edges in the direction of their arrows; the faces are oriented so that their boundaries are in the direction of two edges, rather than one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $H_1$, you want the $1$-cycles mod those that bound a $2$-cell.  Since each edge is a cycle, the group of $1$-cycles is the free abelian group on $a,b,c$.  The boundary of $U$ is $a+b-c$ and that of $L$ is $c+a-b$.  So we're looking at $\langle a,b,c\rangle/\langle a+b-c,a-b+c\rangle$.  Let's simplify this to $\langle a+b-c,b,c\rangle/\langle a+b-c,2b-2c\rangle=\langle b,c\rangle/\langle 2b-2c\rangle$... and again to $\langle b-c,c\rangle/\langle 2b-2c\rangle$.  Setting $d=b-c$, this is just $\langle d\rangle/\langle 2d\rangle\oplus\langle c\rangle$, which is $\mathbb{Z}\oplus\mathbb{Z}_2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're familiar at all with the theory of the fundamental group, there's a cool theorem that says that $H_1(X)$ is the abelianization of $\pi_1(X,x_0)$ for path-connected $X$.  This is another good way to visualize torsion in this specific instance.  I don't know about the Klein bottle, but $\mathbb{RP}^2$ has an element of order $2$ in its fundamental group given by (visualizing it as a disk with its boundary quotiented by the antipodal map) a path from a point on the boundary to the &quot;opposite&quot; point on the boundary.  In the case of $H_1$, you can imagine torsion as the &quot;homology version&quot; of this phenomenon.&lt;/p&gt;&#xA;" OwnerUserId="2215" LastActivityDate="2010-10-20T19:42:28.680" />
  <row Id="7340" PostTypeId="1" CreationDate="2010-10-20T19:47:51.307" Score="2" ViewCount="138" Body="&lt;p&gt;Each person from a group of 3 people can choose his dish from a menu of 5 options. Knowing that each person eats only 1 dish what are the number of different orders the waiter can ask the chef?&lt;/p&gt;&#xA;" OwnerUserId="2651" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-10-20T19:49:42.700" LastActivityDate="2010-10-21T06:51:58.133" Title="basic combinatorics question" Tags="&lt;combinatorics&gt;" AnswerCount="3" />
  <row Id="7341" PostTypeId="1" CreationDate="2010-10-20T19:48:53.037" Score="1" ViewCount="135" Body="&lt;p&gt;I would like to solve:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\frac{d^2y}{dx^2} - \frac{2}{y^2}=0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with $y(0)=a$ and $y'(0)=0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $a$ is a known constant. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance. &lt;/p&gt;&#xA;" OwnerUserId="2148" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-10-20T19:52:58.610" LastActivityDate="2010-10-20T20:04:02.470" Title="Nonlinear Ordinary differential equation" Tags="&lt;differential-equations&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" />
  <row Id="7342" PostTypeId="1" CreationDate="2010-10-20T19:51:12.653" Score="1" ViewCount="239" Body="&lt;p&gt;i couldn't do the following question for hours&lt;/p&gt;&#xA;&#xA;&lt;p&gt;minimize $\sum_{i=1}^{n}x_{i}^{3}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;s.t.  $\sum_{i=1}^{n}x_{i}=0$&#xA;      and $\sum_{i=1}^{n}x_{i}^{2}=n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;by Lagrange multiplier rule&lt;/p&gt;&#xA;&#xA;&lt;p&gt;?&lt;/p&gt;&#xA;" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2010-10-22T19:20:55.463" LastActivityDate="2011-02-27T15:54:37.110" Title="lagrange multiplier for more than 2 equality constraints" Tags="&lt;nonlinear-optimization&gt;" AnswerCount="2" CommentCount="6" FavoriteCount="1" />
  <row Id="7343" PostTypeId="2" ParentId="7340" CreationDate="2010-10-20T19:52:08.720" Score="3" ViewCount="" Body="&lt;p&gt;This is a problem of &lt;em&gt;combinations with repetitions&lt;/em&gt;. See for example &lt;a href=&quot;http://en.wikipedia.org/wiki/Combination#Number_of_combinations_with_repetition&quot; rel=&quot;nofollow&quot;&gt;the discussion on Wikipedia&lt;/a&gt; (which lacks a proof right now). &lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-10-20T19:52:08.720" CommentCount="1" />
  <row Id="7344" PostTypeId="2" ParentId="7231" CreationDate="2010-10-20T19:54:48.807" Score="0" ViewCount="" Body="&lt;p&gt;Rather I think the suggestion of $R(RCGM)'$ is a product of sums which -can- be read off directly:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$R$ for -all- the rows where R is negative&lt;/li&gt;&#xA;&lt;li&gt;$(RCGM)' = R'+C'+G'+M'$ for the single row where all are true &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1919" LastActivityDate="2010-10-20T19:54:48.807" />
  <row Id="7345" PostTypeId="2" ParentId="7341" CreationDate="2010-10-20T20:04:02.470" Score="8" ViewCount="" Body="&lt;p&gt;Change variables so that the independent variable is $y$ and the dependent on $p=y'$. Then $p\dot p=y''$ (dots stand for derivatives w.r.t. $y$, primes w.r.t. $x$) and the equation becomes $$p\dot p-\frac2{y^2}=0$$ or $$\frac{d}{dy}(\frac{p^2}2)=\frac2{y^2}.$$ This can be solved by integrating into a first order equation for $y$ as a function of $x$.&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2010-10-20T20:04:02.470" CommentCount="4" />
  <row Id="7346" PostTypeId="1" CreationDate="2010-10-20T20:04:34.370" Score="1" ViewCount="143" Body="&lt;p&gt;Reading &quot;Random number generation and Monte Carlo methods&quot; by James E. Gentle, I encountered an example concerning the Bays-Durham Shuffling algorithm whose result I can't reproduce.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically, such an example is written in &lt;a href=&quot;http://en.wikibooks.org/wiki/Statistics/Numerical_Methods/Random_Number_Generation#Bays-Durham_Shuffling_of_Uniform_Deviates&quot; rel=&quot;nofollow&quot;&gt;Random Number Generation - Durham Shuffling of Uniform Deviates&lt;/a&gt;. For the full text you can go to &lt;a href=&quot;http://books.google.com/books?id=ERSSDBDcYOIC&quot; rel=&quot;nofollow&quot;&gt;Random number generation and Monte Carlo methods&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, I'm trying to reproduce &lt;a href=&quot;http://books.google.com/books?id=ERSSDBDcYOIC&amp;amp;pg=PA23&amp;amp;dq=bays%20durham%20shuffling%20of%20uniform%20deviates&quot; rel=&quot;nofollow&quot;&gt;Figure 1.7&lt;/a&gt; for the linear congruential generator of $x_{i} \equiv 3x_{i-1} \text{mod} 31$. In order to do so, I wrote the following piece of code in Mathematica: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Defines initial variables for the generator:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;a = 3; mod = 31; x = 9; n = 0; m = {}; &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Produce a list with the initial set of numbers (In this case: 27, 19, 26, ... &amp;lt;&amp;lt;27&gt;&gt; ... 1, 3, 9):&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;While[n &amp;lt; 30, AppendTo[m, Mod[a x, mod]]; x = Mod[a x, mod]; n++]; &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The algorithm generates the first sequence of numbers in a slightly different way using the input value k, so in order to account for this, I had to use this While for the first sequence:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;k = 8; i = 1; y = m[[k + 1]];&#xA;&#xA;While[i == 1, choose = Mod[y, 8] + 1; y = l[[choose]]; &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Then, for the following sequences:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;i = i + 1; m[[choose]] = k; k = k + 2];&#xA;&#xA;While[i &amp;lt; 10000, choose = Mod[y, 8] + 1; y = m[[choose]];&#xA;&#xA;i = i + 1; k++; &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Given that the congruence relation is congruence modulo 31, the sequence will repeat itself after 30 digits. To avoid looking for the non-existent element m[[31]], I added the conditional:&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;  If[k &amp;lt; 30, m[[choose]] = m[[k]], k = 1; m[[choose]] = m[[k]]]];&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The book says:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;'By continuing in this manner to yield 10,000 deviates and plotting the successive pairs, we get Figure 1.7'. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Well, as you can see, the procedure takes places 10,000 times. The output is:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;{4, 4, 21, 18, 2, 4, 21, 3, 13, 8, 24, 10, 30, 28, 22, 4, 12, 5, 15, 14, 11, 2, 6, 18, 23, 7, 21, 1, 3, 9}.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Plotting the successive pairs, I get this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://miql.zxq.net/modulo.jpg&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, even if some of the points look like the ones in Figure 1.7, obviously those images are not the same. Actually, Figure 1.7 contains 122 points (assuming I didn't miscount), however, given the generator, I don't know how you can produce more than 30 points. Actually, the author writes:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;... Remember that there are only 30 different values, however. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Am I missing something? Any ideas?&lt;/p&gt;&#xA;" OwnerUserId="2569" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-10-21T02:05:07.187" LastActivityDate="2010-10-21T02:05:07.187" Title="Bays-Durham Shuffling" Tags="&lt;random&gt;" CommentCount="2" />
  <row Id="7349" PostTypeId="2" ParentId="7294" CreationDate="2010-10-20T20:24:32.917" Score="1" ViewCount="" Body="&lt;p&gt;The only requirement for a surface is that it be two dimensional. It can be connected or not connected. So it's OK mathematically to call your multi-component terrain object a &quot;surface&quot;, even though in everyday language people might think of it as several surfaces.&lt;/p&gt;&#xA;" OwnerUserId="1257" LastActivityDate="2010-10-20T20:24:32.917" />
  <row Id="7350" PostTypeId="1" CreationDate="2010-10-20T20:25:06.457" Score="1" ViewCount="118" Body="&lt;p&gt;What is a good introduction in gradient flows in metric spaces? I know the book Gradient flows: in metric spaces and in the space of probability measures by Luigi Ambrosio, Nicola Gigli and Giuseppe Savaré, but is too hard for an introduction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can someone make this a community wiki? I cannot find how you have to do that...&lt;/p&gt;&#xA;" OwnerUserId="1120" LastEditorUserId="1543" LastEditorDisplayName="" LastEditDate="2011-04-09T15:03:32.453" LastActivityDate="2011-04-09T15:03:32.453" Title="Gradient flows in metric spaces." Tags="&lt;soft-question&gt;&lt;reference-request&gt;&lt;metric-spaces&gt;&lt;gradient-flows&gt;" CommentCount="6" />
  <row Id="7351" PostTypeId="1" AcceptedAnswerId="7352" CreationDate="2010-10-20T20:25:58.937" Score="0" ViewCount="103" Body="&lt;p&gt;So, I thought of something nerdy...I was making an algebra problem to disguise it so that to see the joke, you'd have to a bit of work, but I think that I might be doing something wrong, because neither &lt;a href=&quot;http://www.wolframalpha.com/&quot; rel=&quot;nofollow&quot;&gt;Wolfram|Alpha&lt;/a&gt; nor &lt;a href=&quot;http://myalgebra.com/&quot; rel=&quot;nofollow&quot;&gt;MyAlgebra&lt;/a&gt; can solve it...I fixed a few things that I had wrong initially, but I don't see anything wrong now...you're supposed to solve for i.&#xA;$$\begin{array}{rcl}&#xA;    (5u)^2-(i+2u)^2 &amp;amp;\gt&amp;amp; 20u^2+4u(i+2u)\\&#xA;    (5u+(i+2u))(5u-(i+2u))&amp;amp;\gt&amp;amp; 4u(5u+(i+2u))\\&#xA;    5u-(i+2u)&amp;amp;\gt&amp;amp; 4u\\&#xA;    3u-i&amp;amp;\gt&amp;amp; 0\\ &#xA;    i&amp;amp;\lt&amp;amp; 3u&#xA;\end{array}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I figured that you could factor the left side and then divide &lt;code&gt;(5u+(i+2u))&lt;/code&gt; out of both sides, but maybe not?&lt;/p&gt;&#xA;" OwnerUserId="2654" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-10-20T20:39:32.727" LastActivityDate="2010-10-20T21:02:40.933" Title="Am I doing something wrong in making this inequality or do algebra tools just usually not factor?" Tags="&lt;algebra&gt;&lt;inequality&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="7352" PostTypeId="2" ParentId="7351" CreationDate="2010-10-20T20:33:26.010" Score="6" ViewCount="" Body="&lt;p&gt;Your mistake lies in trying to cancel $5u +(i+2u)$ without being careful about whether it is positive, negative, or $0$. &lt;strong&gt;Added:&lt;/strong&gt; There is also a miscalculation in the third line.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remember that &quot;cancelling factors&quot; is &lt;em&gt;really&lt;/em&gt; multiplying by the inverse. But when you multiply an inequality by a number, you have to be careful about whether the number is positive or negative: if you have $a\gt b$ and $c\gt 0$, then $ac\gt bc$; but if $c\lt 0$, then you have $ac\lt bc$ (multiplying by negative numbers reverses the inequality).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you try cancelling $5u+(i+2u) = 7u+i$, by not reversing the inequality you are implicitly assuming that $7u+i\gt 0$; that is, that $-7u\lt i$. So in fact, your conclusion is that &lt;strong&gt;if&lt;/strong&gt; $-7u\lt i$ &lt;em&gt;and&lt;/em&gt; the inequality holds, then $i\lt 3u$. But what if $7u+i\lt 0$? Then when you divide both sides by $7u+i$ in order to cancel the factor, you must reverse the inequality, so you actually get $5u -(i+2u) \lt 4u$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added:&lt;/strong&gt; At this point you made a slight mistake with the signs. The left hand side is $5u - (i+2u) = 3u-i$; you had $5u - i + 2u$ (missing the parenthesis) so you incorrectly got $7u - i$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, if $7u+i \gt 0$, then you get $3u-i\gt 4u$, or $i\lt -u$. Thus, if $i\gt -7u$, then you must also have $i\lt -u$, or $-7u\lt i \lt -u$. Note that this can only occur if $u\gt 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $7u+i \lt 0$, then when you cancel that factor you get $3u-i \lt 4u$, or $i\gt -u$; so you have that both $i\lt -7u$ and $i\gt -u$ must hold. That is, $-u\lt i\lt -7u$. Note that this can only occur if $u\lt 0$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in fact you have two parts of the solution: you either have $u\gt 0$ and $-7u \lt i\lt -u$, or else you have $u\lt 0$ and $-u\lt i\lt-7u$. &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-10-20T21:02:40.933" LastActivityDate="2010-10-20T21:02:40.933" CommentCount="2" />
  <row Id="7353" PostTypeId="1" AcceptedAnswerId="7357" CreationDate="2010-10-20T21:18:50.490" Score="1" ViewCount="140" Body="&lt;p&gt;I believe I'm right in saying that an isometry of a convex figure will be convex. There are other automorphisms that have this property, for example &quot;stretching along one axis&quot; (e.g. $f(x,y) = (2x,y)$) won't transform a convex shape into a non-convex one or vice versa.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What other maps have this property?&lt;/p&gt;&#xA;" OwnerUserId="172" LastActivityDate="2010-11-01T02:20:47.147" Title="What automorphisms of R^n preserve convexity?" Tags="&lt;geometry&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="7354" PostTypeId="1" CreationDate="2010-10-20T21:21:07.340" Score="1" ViewCount="546" Body="&lt;p&gt;Determine whether this series converges or diverges: $$\sum\limits_{n=0}^{\infty} \frac{1}{(2n+1)!}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thought about using the limit theorem or by comparison but am so stuck.  any pointers would be appreciated guys&lt;/p&gt;&#xA;" OwnerUserId="2657" LastActivityDate="2010-10-21T14:37:58.867" Title="Determine whether this series converges or diverges: $\sum\limits_{n=0}^{\infty} \frac{1}{(2n+1)!}$" Tags="&lt;calculus&gt;&lt;sequences-and-series&gt;" AnswerCount="6" CommentCount="2" />
  <row Id="7355" PostTypeId="2" ParentId="7354" CreationDate="2010-10-20T21:22:03.353" Score="2" ViewCount="" Body="&lt;p&gt;Can you bound the series from above by one that you know converges?  The factorials grow very fast, so you should be able to.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-10-20T21:22:03.353" />
  <row Id="7356" PostTypeId="1" CreationDate="2010-10-20T21:23:09.400" Score="1" ViewCount="367" Body="&lt;p&gt;Given the coordinates of two rectangles on a coordinate plane, what would be the easiest way to find the coordinates of the intersecting rectangle of the two?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am trying to do this programatically.&lt;/p&gt;&#xA;" OwnerUserId="2658" LastEditorUserId="2658" LastEditorDisplayName="" LastEditDate="2010-10-24T14:39:37.287" LastActivityDate="2010-10-24T14:39:37.287" Title="How to find rectangle intersection on a coordinate plane" Tags="&lt;geometry&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="7357" PostTypeId="2" ParentId="7353" CreationDate="2010-10-20T21:24:05.770" Score="10" ViewCount="" Body="&lt;p&gt;Any transformation that sends lines to lines will do, since convexity is defined by the existence of certain lines in a figure.  In particular, any transformation of the form $x \mapsto Ax + b$ where $A$ is a matrix and $b$ is a vector works.  (It's not clear to me what you mean by &quot;automorphism&quot; here - in what category? - so depending on your meaning these are the only ones, by which I mean that as an affine space these are the only automorphisms of $\mathbb{R}^n$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you don't trust that explanation, one way to see this is to prove the result for translations (obvious) and then prove it for elementary matrices.  It should be geometrically obvious for all of them, as long as you have a reasonably good handle on &lt;a href=&quot;http://en.wikipedia.org/wiki/Shear_mapping&quot; rel=&quot;nofollow&quot;&gt;shearing&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; In fact, I think this condition is necessary.  If a line $L$ is not sent to a line, draw a circle which is split in half by $L$.  Then the image of either one half or the other will not be convex.  &lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-10-20T21:30:42.027" LastActivityDate="2010-10-20T21:30:42.027" CommentCount="4" />
  <row Id="7358" PostTypeId="2" ParentId="7354" CreationDate="2010-10-20T21:47:48.310" Score="4" ViewCount="" Body="&lt;p&gt;The series you have is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$1 + \frac{1}{3!} + \frac{1}{5!} \ldots $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you add the even factorial terms, you get an upper bound i.e.,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$1 + \frac{1}{3!} + \frac{1}{5!} \ldots &amp;lt; \frac{1}{0!} + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \frac{1}{4!}+ \frac{1}{5!} \ldots$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This can be written more compactly as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sum_{n=0}^\infty \frac{1}{(2n+1)!} &amp;lt; \sum_{n=0}^\infty \frac{1}{n!} = e^1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore the series converges.&lt;/p&gt;&#xA;" OwnerUserId="2641" LastActivityDate="2010-10-20T21:47:48.310" CommentCount="3" />
  <row Id="7359" PostTypeId="2" ParentId="7353" CreationDate="2010-10-20T22:07:45.043" Score="3" ViewCount="" Body="&lt;p&gt;Search google books for &quot;fundamental theorem of geometry&quot; (it's in a book by M. Berger).&#xA;Using the fact that the only field automorphism of $\mathbb{R}$ is the identity, you get your result (for $n \geq 2$): the affine bijections.&lt;/p&gt;&#xA;" OwnerUserId="2660" LastEditorUserId="2660" LastEditorDisplayName="" LastEditDate="2010-11-01T02:20:47.147" LastActivityDate="2010-11-01T02:20:47.147" />
  <row Id="7360" PostTypeId="2" ParentId="7356" CreationDate="2010-10-20T22:21:07.120" Score="0" ViewCount="" Body="&lt;p&gt;In order to simplify the problem, I will assume that the rectangles are orientated &quot;straight up and down.&quot; By this I mean that each of the rectangles has an edge which is parallel to the x (or y) axis. Judging by the context, this should be a fair assumption. Let me know if this is not the case (in other words, what class is this for?) It might help to first graph the two rectangles whose coordinates are given. In the &quot;simplest&quot; case they will intersect in exactly two places.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The coordinates for some intersection will just be the point where &lt;em&gt;&quot;edge a&quot;&lt;/em&gt; of &lt;em&gt;&quot;rectangle 1&quot;&lt;/em&gt; intersects &lt;em&gt;&quot;edge x&quot;&lt;/em&gt; of &lt;em&gt;&quot;rectangle 2.&quot;&lt;/em&gt; That is, the intersection point will have the x value of (edge a or x), and the y value of the other edge (since these rectangles are parallel to the axis the x &lt;strong&gt;or&lt;/strong&gt; y value of an edge will be constant across that edge.) You know the values of these edges by the coordinates given for the corners of your original rectangles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After drawing the rectangles this should be a fairly easy and straight-forward process to work through. If you are trying to write a computer program to do this for you it is slightly more complicated but follows the same logic. Hope this helps!&lt;/p&gt;&#xA;" OwnerUserId="1352" LastActivityDate="2010-10-20T22:21:07.120" CommentCount="1" />
  <row Id="7361" PostTypeId="1" AcceptedAnswerId="7365" CreationDate="2010-10-20T22:27:07.853" Score="5" ViewCount="138" Body="&lt;p&gt;Let $I$ be a finitely generated ideal of a commutative ring $R$. Assume every element of $I$ is a zero divisor. Does then exist a $x \neq 0$ in $R$ with $xI=0$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is true if $0$ is a decomposable ideal, for example if $R$ is noetherian. I wonder if we actually need this. Doesn't it sound plausible? The problem is that we cannot just multiply the elements which kill the generators of $I$, the product can vanish.&lt;/p&gt;&#xA;" OwnerUserId="1650" LastActivityDate="2010-10-20T23:15:08.443" Title="Ideal consisting of zero divisors" Tags="&lt;commutative-algebra&gt;" AnswerCount="1" />
  <row Id="7362" PostTypeId="2" ParentId="7356" CreationDate="2010-10-20T22:30:04.757" Score="1" ViewCount="" Body="&lt;p&gt;You can think of an axis-aligned rectangle as the set of points $(x,y) \in \mathbb{R}^2$ such that $x_1 \le x \le x_2$ and $y_1 \le y \le y_2$. What can you say about the intersection of two such sets?&lt;/p&gt;&#xA;" OwnerUserId="856" LastActivityDate="2010-10-20T22:30:04.757" />
  <row Id="7363" PostTypeId="2" ParentId="7317" CreationDate="2010-10-20T22:34:05.227" Score="1" ViewCount="" Body="&lt;p&gt;Take a look at the syllabi for the first three &lt;a href=&quot;http://www.beanactuary.org/exams/exam_info.cfm&quot; rel=&quot;nofollow&quot;&gt;actuarial preliminary examinations&lt;/a&gt;. If you study for these tests, in order, you will learn everything you need in order to understand and apply the Black-Scholes formula.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each exam typically requires three hundred hours of study to do well. If you pace yourself, and you take one exam every six months, you will be studying 11 to 12 hours per week for one and a half years to &lt;i&gt;master&lt;/i&gt; this material. Of course, you could cover this material much faster, if you're more interested in breadth than depth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first three exams are 2.5 to 3 hours each. They are&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt; &lt;a href=&quot;http://www.beanactuary.org/exams/exam1p.cfm&quot; rel=&quot;nofollow&quot;&gt;1/P - The Probability Exam&lt;/a&gt; - A thorough command of calculus and probability topics is assumed.&#xA;&lt;li&gt; &lt;a href=&quot;http://www.beanactuary.org/exams/exam2fm.cfm&quot; rel=&quot;nofollow&quot;&gt;2/FM - Financial Mathematics Exam&lt;/a&gt; - Interest theory (discrete and continuous) and an introduction to derivative securities.&#xA;&lt;li&gt; &lt;a href=&quot;http://www.beanactuary.org/exams/examam.cfm&quot; rel=&quot;nofollow&quot;&gt;3F/MFE - Financial Economics Exam&lt;/a&gt; - Interest rate models, rational valuation of derivative securities, simulation, and financial risk management techniques.&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The syllabi list the textbooks and papers you should study. Additionally, there are &lt;a href=&quot;http://www.actexmadriver.com/&quot; rel=&quot;nofollow&quot;&gt;study guides&lt;/a&gt; that cover each test.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Exam 3F/MFE covers &lt;b&gt;Black-Scholes&lt;/b&gt;. Specifically, you must be able to&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Calculate the value of European and American options using the Black-Scholes option-pricing model.&#xA;&lt;li&gt;Interpret the option &lt;a href=&quot;http://en.wikipedia.org/wiki/Greeks_%28finance%29&quot; rel=&quot;nofollow&quot;&gt;Greeks&lt;/a&gt;.&#xA;&lt;li&gt;Explain the properties of a lognormal distribution and explain the Black-Scholes formula as a limited expected value for a &lt;a href=&quot;http://en.wikipedia.org/wiki/Lognormal&quot; rel=&quot;nofollow&quot;&gt;lognormal distribution&lt;/a&gt;.&#xA;&lt;/ul&gt;&#xA;" LastActivityDate="2010-10-20T22:34:05.227" />
  <row Id="7364" PostTypeId="2" ParentId="7291" CreationDate="2010-10-20T22:52:44.423" Score="4" ViewCount="" Body="&lt;p&gt;If you find the Rota-Baxter algebra viewpoint of interest then an excellent entry point into the literature on differential-algebraic aspects is the the recent work of Guo, e.g. &lt;a href=&quot;http://arxiv.org/abs/math.RA/0703780&quot; rel=&quot;nofollow&quot;&gt;On differential Rota-Baxter algebras&lt;/a&gt; and &lt;a href=&quot;http://andromeda.rutgers.edu/~liguo/psfiles/guo6.pdf&quot; rel=&quot;nofollow&quot;&gt;Baxter algebras and differential algebras&lt;/a&gt;. See also other papers listed on his &lt;a href=&quot;http://andromeda.rutgers.edu/~liguo/&quot; rel=&quot;nofollow&quot;&gt;home page.&lt;/a&gt; The early papers don't focus so much on these aspects so I would not recommend reading them initially.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also you may find of interest this paper on classification of related operator identities.&lt;br&gt;&#xA;Freeman, J. M. On the classification of operator identities. Studies in Appl. Math. 51 (1972), 73-84.&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-10-20T22:52:44.423" CommentCount="1" />
  <row Id="7365" PostTypeId="2" ParentId="7361" CreationDate="2010-10-20T23:09:31.460" Score="3" ViewCount="" Body="&lt;p&gt;No, for a counterexample see Exercises 2-2-6,7 pp. 62-63 in Kaplansky: Commutative Rings (excerpted below). See also the discussion following Theorem 82 p.56.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/MMRqu.jpg&quot; alt=&quot;alt text&quot;&gt;&#xA;&lt;img src=&quot;http://i.imgur.com/mFafo.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-10-20T23:15:08.443" LastActivityDate="2010-10-20T23:15:08.443" CommentCount="2" />
  <row Id="7366" PostTypeId="2" ParentId="6791" CreationDate="2010-10-20T23:44:39.037" Score="12" ViewCount="" Body="&lt;p&gt;I had made the same conjecture as Fekete, apparently around the same time -- mid-2007.  In 2008 I verified that the first twenty million terms do not include 319.  (I actually pushed the verification further, but I can't find the more recent records at the moment.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because tan(x) - x = x^3/3 + O(x^5), the function spends a lot of its time in a small neighborhood around 0.  It escapes when it nears $\pi/2$ and quickly returns for many iterations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A mostly-unexplained phenomenon presumably related to the above: there are long spans of small numbers followed by short, 'productive' spans with large numbers.  $\tan^k(1)$ is &quot;below 20 or so&quot; (according to a 2008 email I sent) for $360110\le k\le1392490$ but in the next 2000 numbers there are five which are above 20.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More theory is needed!&lt;/p&gt;&#xA;" OwnerUserId="1778" LastEditorUserId="1778" LastEditorDisplayName="" LastEditDate="2011-01-06T15:11:25.920" LastActivityDate="2011-01-06T15:11:25.920" CommentCount="9" />
  <row Id="7367" PostTypeId="1" CreationDate="2010-10-20T23:58:24.050" Score="4" ViewCount="424" Body="&lt;p&gt;This question and the described solution are copied from a test-paper :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the equation $x^2$ + |x| - 6 = 0 analyze the four statements below for correctness.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;there is only one root&lt;/li&gt;&#xA;&lt;li&gt;sum of the roots is + 1&lt;/li&gt;&#xA;&lt;li&gt;sum of the roots is zero&lt;/li&gt;&#xA;&lt;li&gt;the product of the roots is +4&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Answer : (3)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Answer Explanation :&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If x &gt; 0 |x| = x.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given equation will be $x^2 + x - 6 = 0$⇒ x = 2,- 3 ⇒ x = 2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;If x &amp;lt; 0 |x| = - x.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given equation will b e  $x^2$ - x - 6 = 0 ⇒ x = -2, 3 ⇒x = - 2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sum of roots is 2 - 2 = 0&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I have a doubt on the statment &quot;&lt;strong&gt;If x &amp;lt; 0 |x| = - x.&lt;/strong&gt;&quot; I think modulus means that |x| is always positive ?! Also I can see that (2) seems to be the correct option isn't ?!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please post your views.&lt;/p&gt;&#xA;" LastEditorUserId="367" LastEditorDisplayName="" LastEditDate="2010-10-21T16:39:57.530" LastActivityDate="2010-10-21T16:39:57.530" Title="Exploring the quadratic equation $x^2 + \lvert x\rvert - 6 = 0$" Tags="&lt;algebra-precalculus&gt;" AnswerCount="4" CommentCount="7" />
  <row Id="7368" PostTypeId="2" ParentId="7367" CreationDate="2010-10-21T00:01:12.203" Score="2" ViewCount="" Body="&lt;p&gt;If x&amp;lt;0, then -x&gt;0, which means that the modulus is indeed positive.&lt;/p&gt;&#xA;" OwnerUserId="2029" LastActivityDate="2010-10-21T00:01:12.203" CommentCount="3" />
  <row Id="7369" PostTypeId="2" ParentId="7265" CreationDate="2010-10-21T00:01:42.607" Score="1" ViewCount="" Body="&lt;p&gt;If the coefficients are naturals, there is no problem with rounding.  So just divide the polynomials and look at the middle term.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rather equivalently, you can evaluate it at a number of points and do Lagrange interpolation.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-10-21T00:01:42.607" CommentCount="1" />
  <row Id="7370" PostTypeId="2" ParentId="7354" CreationDate="2010-10-21T00:13:42.417" Score="5" ViewCount="" Body="&lt;p&gt;I think svenkatr's response is correct. He is using the comparison test, in particular, comparing with the exponential function for $x=1$, that is  obviously a number, so he doesn't have to prove that the series for e converges.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe you can prove the same by using the ratio test $\lim_{n \rightarrow \infty} \displaystyle |\frac{a_{n+1}}{a_{n}}|$. For example, you have $a_{n}=\displaystyle \frac{1}{(2n+1)!}$ and $a_{n+1}=\displaystyle \frac{(2n+1)!}{(2n+3)!}$, then using the definition for the factorial you have $\lim_{n \rightarrow \infty} \displaystyle \frac{1}{(2n+3)(2n+2)}$ which is 0. According to the &lt;a href=&quot;http://en.wikipedia.org/wiki/Convergence_tests&quot; rel=&quot;nofollow&quot;&gt;ratio test&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If r &amp;lt; 1, then the series converges. If r &gt; 1, then the series diverges. If r = 1, the ratio test is inconclusive, and the series may converge or diverge.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Therefore, the series converges.&lt;/p&gt;&#xA;" OwnerUserId="2569" LastEditorUserId="2569" LastEditorDisplayName="" LastEditDate="2010-10-21T01:01:30.610" LastActivityDate="2010-10-21T01:01:30.610" CommentCount="2" />
  <row Id="7371" PostTypeId="2" ParentId="7367" CreationDate="2010-10-21T00:15:11.463" Score="1" ViewCount="" Body="&lt;p&gt;The statement is saying:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$|x| = -x$ (for $x &amp;lt; 0$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and divide both sides by $x$ to give us:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{|x|}{x} = -1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let us test a few values to make sure this holds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x = -3$ and $|x| = 3$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{|-3|}{-3} = -1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{3}{-3} = -1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$-1 = -1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;True.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x = -5$ and $|x| = 5$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{5}{-5} = -1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$-1 = -1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;True.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can verify this for all $x &amp;lt; 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you don't believe me, here is a plot of $y = \frac{|x|}{x}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://img716.imageshack.us/img716/4149/wolframalpha20101020192.gif&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Alternative:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we define&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;-a is the number such that a + (-a) = 0&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;ie,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;if a = 3, -a = -3, because $3 + (-3) = 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;if a = -2, -a = 2, because $-2 + 2 = 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can call 3 and -3 &quot;opposites&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The opposite of -2 is 2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The opposite of 5 is -5.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then we can translate the original statement then as:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;If x &amp;lt; 0, |x| is the opposite of x&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Let's test this out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x = -4$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$|x| = 4$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$-4 + 4 = 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$|-4|$ (or, 4) is indeed the opposite of $-4$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More generally, the statement says that, if $x &amp;lt; 0$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x + |x| = 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which we can verify as true for all values $x &amp;lt; 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you don't believe me, you can look at the plot of $y = x + |x|$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://img409.imageshack.us/img409/6834/wolframalpha20101020191.gif&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="51" LastEditorUserId="51" LastEditorDisplayName="" LastEditDate="2010-10-21T00:28:12.173" LastActivityDate="2010-10-21T00:28:12.173" CommentCount="3" />
  <row Id="7372" PostTypeId="2" ParentId="7367" CreationDate="2010-10-21T00:33:15.223" Score="23" ViewCount="" Body="&lt;p&gt;$\rm\ 0\ =\ x^2 + |x| -6\ =\ (|x| - 2)\:(|x| + 3)\ \Rightarrow\ |x| = 2\ \Rightarrow\ x = \pm2 $&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-10-21T00:33:15.223" CommentCount="1" />
  <row Id="7375" PostTypeId="2" ParentId="7322" CreationDate="2010-10-21T01:08:58.330" Score="1" ViewCount="" Body="&lt;p&gt;You could use &lt;a href=&quot;http://en.wikipedia.org/wiki/Hurwitz%27s_theorem_%28complex_analysis%29&quot; rel=&quot;nofollow&quot;&gt;Hurwitz's theorem&lt;/a&gt;.  For some $R$ with $0\lt R\lt1$, $S$ has no zeros on $\partial B(0,R)$, and $R$ can be taken arbitrarily close to 1.  By Hurwitz's theorem, $S$ has the same number of zeros in $B(0,R)$ as $S_n$ for sufficiently large $n$.  Each $S_n$ is never zero in the open disk, so this shows that $S$ is never zero on $B(0,R)$.  Thus $S$ is never zero on the disk.  This means that $S$ has no Blaschke factor, hence it is singular.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I do not know whether $(\mu_n)$ converges weak-$*$ to the singular measure of $S$.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-10-21T01:08:58.330" />
  <row Id="7376" PostTypeId="1" AcceptedAnswerId="7509" CreationDate="2010-10-21T01:15:09.850" Score="10" ViewCount="505" Body="&lt;p&gt;Let $C$ be a convex subset of $\mathbb{R}^n$. I've been trying for hours to prove that $\dot{\overline{C}}=\dot{C}$. Somehow my intuition completely fails me. I found a proof in a textbook, but just got stuck on another statement the author considered obvious. Could someone please give a proof that uses little more than elementary linear algebra, topology, and the definition of a convex set?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; &#xA;The proof mentioned above is from &lt;a href=&quot;http://books.google.de/books?id=1F2B-8ap4wwC&amp;amp;pg=PA32&amp;amp;dq=convex+closure+%22same+boundary%22&amp;amp;hl=de&amp;amp;ei=K_i_TISsCsO54AaN-qS6DA&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=5&amp;amp;ved=0CDkQ6AEwBA#v=onepage&amp;amp;q=convex%20closure%20%22same%20boundary%22&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;Blackwell and Girshick&lt;/a&gt;:&#xA;Let $y\in\dot{\overline{C}}$ and $T$ be a ball around $y$ contained in $\overline{C}$. Then $C\cap T$ has an inner point, as otherwise $C\cap T$ would be contained in a hyperplane and $\overline{C\cap T}=\overline{T}$ would be contained in the same hyperplane. The problematic statement is &quot;as otherwise $C\cap T$ would be contained in a hyperplane&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another thing: I would be interested in a proof that doesn't use the theorem about separating a convex set from a point by a hyperplane, as I came across this problem in a proof of that very theorem (in the appendix of &lt;a href=&quot;http://books.google.de/books?id=cL-bZSOrqWoC&amp;amp;printsec=frontcover#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;Stochastic Finance by Föllmer and Schied)&lt;/a&gt;. To be more precise, it occurs in the case of the point in question being in the boundary of $C$, when it is tacitly assumed, that it is also in the boundary of $\overline{C}$. I know this isn't strictly necessary, as I could use another proof, e.g. the one referred to by Mike, but now I'm curious.  &lt;/p&gt;&#xA;" OwnerUserId="1878" LastEditorUserId="1878" LastEditorDisplayName="" LastEditDate="2010-10-21T09:25:40.213" LastActivityDate="2011-02-04T23:12:14.703" Title="Why does a convex set have the same interior points as its closure?" Tags="&lt;topology&gt;&lt;convex-analysis&gt;" AnswerCount="4" CommentCount="1" FavoriteCount="3" />
  <row Id="7377" PostTypeId="1" AcceptedAnswerId="7381" CreationDate="2010-10-21T01:31:07.137" Score="13" ViewCount="796" Body="&lt;p&gt;Dear Math.so,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Firstly, you guys are awesome, and I learn quite a bit just from reading the questions of others.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Secondly, a friend asked me recently why large primes are important for data security, and I was unable to give him an answer with which I myself was satisfied.  Various wikipedia articles have mostly pointed out an embarrassing paucity in mathematical knowledge on my part, and since this happens to be a very math-related question (and not a programming-related question) I was hoping someone could shed some light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;tl;dr: question reads as title.&lt;/p&gt;&#xA;" OwnerUserId="1774" LastActivityDate="2010-10-22T18:21:23.480" Title="Why are very large prime numbers important in cryptography?" Tags="&lt;prime-numbers&gt;&lt;cryptography&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="6" />
  <row Id="7378" PostTypeId="2" ParentId="7377" CreationDate="2010-10-21T01:37:45.000" Score="5" ViewCount="" Body="&lt;p&gt;Some cryptographic algorithms use 2 very large primes (such as 128 bit long) and multiply them together.  The only way we know how to crack that is to try and find the only 2 factors that are available for that number (the 2 large primes).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, it turns out, it takes A LOT of computer power to be able to find those 2 factors.  Therefore, even if one tries, it'll take so long that the encryption is considered secure enough.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most numbers have many factors, but a product of 2 primes &lt;em&gt;only&lt;/em&gt; has the 2 primes as factors.&lt;/p&gt;&#xA;" OwnerUserId="1630" LastEditorUserId="1630" LastEditorDisplayName="" LastEditDate="2010-10-21T07:08:37.677" LastActivityDate="2010-10-21T07:08:37.677" CommentCount="5" />
  <row Id="7379" PostTypeId="2" ParentId="7289" CreationDate="2010-10-21T01:41:21.367" Score="1" ViewCount="" Body="&lt;p&gt;To gratuitously generalize the question, in $\mathbb{R}^n$ with $n\geq2$ there are continuum-sized subsets each of whose $n$-element subsets is linearly independent.  You can use &lt;a href=&quot;http://en.wikipedia.org/wiki/Vandermonde_matrix&quot; rel=&quot;nofollow&quot;&gt;Vandermonde matrices&lt;/a&gt; to see that $\{(1,x,x^2,\ldots,x^{n-1}):x\in\mathbb{R}\}$ is an example of this.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2010-10-21T01:46:58.110" LastActivityDate="2010-10-21T01:46:58.110" />
  <row Id="7380" PostTypeId="2" ParentId="7301" CreationDate="2010-10-21T02:07:10.163" Score="5" ViewCount="" Body="&lt;p&gt;There is an analogy, more direct for fourier series. Both Fourier series and Taylor series are decompositions of a function $f(x)$, which is represented as a linear combination of a (countable) set of functions.  The function is then fully specified by a sequence of coefficients, instead of by its values $f(x)$ for each $x$.  In this sense, both can be called a &lt;strong&gt;transform&lt;/strong&gt; ($f(x) \leftrightarrow \{ a_0, a_1, ...\}$). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the Taylor series (around 0, for simplicity), the set of functions is $\{1, x , x^2, x^3...\}$. For the Fourier series is $\{1, \sin(\omega x), \cos(\omega x), \sin(2 \omega x), \cos(2 \omega x) ...\}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Actually the Fourier series is one the many transformations that uses an orthonomal basis of functions. It is shown that, in that case, the coefficients are obtained by &quot;projecting&quot; $f(x)$ onto each basis function, which amounts to an inner product, which (in the real scalar case) amounts to an integral. This implies that the coefficients depends on a &lt;strong&gt;global&lt;/strong&gt; property of the function (over the full &quot;period&quot; of the function).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Taylor series (which does &lt;strong&gt;not&lt;/strong&gt; use a orthonormal basis) is conceptually very different, in that the coeffients depends only in &lt;strong&gt;local&lt;/strong&gt; properties of the function, i.e., its behaviour in a neighbourhood  (it's derivatives).&lt;/p&gt;&#xA;" OwnerUserId="312" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2011-03-19T00:33:50.017" LastActivityDate="2011-03-19T00:33:50.017" />
  <row Id="7381" PostTypeId="2" ParentId="7377" CreationDate="2010-10-21T02:52:53.937" Score="24" ViewCount="" Body="&lt;p&gt;There is a whole class of cryptographic/security systems which rely on what are called &quot;trap-door functions&quot;. The idea is that they are functions which are generally easy to compute, but for which finding the &lt;em&gt;inverse&lt;/em&gt; is very hard (here, &quot;easy&quot; and &quot;hard&quot; refer to how quickly we know how to do it), but such that if you have an extra piece of information, then finding the inverse is easy as well. Primes play a very important role in many such systems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One such example is the function that takes two integers and multiplies them together (something we can do very easily), versus the &quot;inverse&quot;, which is a function that takes an integer and gives you proper factors (given $n$, two numbers $p$ and $q$ such that $pq=n$ and $1\lt p,q\lt n$).  If $n$ is the product of two primes, then there is one and only one such pair.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another example is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Discrete_logarithm&quot; rel=&quot;nofollow&quot;&gt;discrete logarithm&lt;/a&gt;. To consider a simple example, look at the integers modulo, say, $7$. The integers between $1$ and $6$, inclusively, form a group under multiplication, and in fact every number between $1$ and $6$ is a power of $3$. The &quot;discrete logarithm problem&quot; would be, given a number $x$ between $1$ and $6$, to find a number $a$ such that $3^a$ equals $x$ modulo $7$. In this case, you can just try powers of $3$ until you hit the right answer. But if the modulo is very large, then this would take too much time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One method for exchanging information over an open channel relies on the fact that we do not have very good methods of finding discrete logarithms in general, but we &lt;em&gt;do&lt;/em&gt; have very good methods for computing modular powers. The idea is: suppose you and I need to exchange information. We want to use some very secure cryptographic system that relies on a complicated key. But, how can we agree on a key? If we have some secure way of communicating so that when we agree on the key nobody will overhear us, then why bother with the entire exercise? We should just communicate using that secure way. So instead we need to communicate at a place where we can be overhead. How can we agree on a &lt;em&gt;secret&lt;/em&gt; key if everyone can hear us? Well, &lt;a href=&quot;http://en.wikipedia.org/wiki/Diffie-Hellman_key_exchange&quot; rel=&quot;nofollow&quot;&gt;Diffie and Hellman&lt;/a&gt; proposed the following method:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pick a very large prime $p$, and a number $r$ such that every number between $1$ and $p-1$ is a power of $r$ modulo $p$ (such numbers $r$ are known to exist for &lt;em&gt;every&lt;/em&gt; prime; they are called &lt;a href=&quot;http://en.wikipedia.org/wiki/Primitive_root_modulo_n&quot; rel=&quot;nofollow&quot;&gt;primitive roots&lt;/a&gt;). Everyone knows $p$ and everyone knows $r$. Then I pick a secret number $a$, and you pick a secret number $b$. I cannot tell you my secret number (it's &lt;em&gt;secret&lt;/em&gt;). But I tell you what $r^a \mod p$ is. Because computing modular powers is easy, I can do this computation easy enough; but because we don't know how to do discrete logarithms easily, we are hoping that nobody will be able to figure out $a$ just from knowing $r^a$... at least, not very quickly. Likewise, you tell me $r^b \mod p$. Now, you know $r^a$, &lt;em&gt;and&lt;/em&gt; you know what $b$ is, so you compute $(r^a)^b \mod p$. By the laws of exponent, you now know (secretly!) the number $r^{ab} \mod p$. I, on the other hand, know $r^b$ (because you told me that number) and I know what $a$ is. So &lt;em&gt;I&lt;/em&gt; compute $(r^b)^a\mod p$. But this is the same as $r^{ab} \mod p$. So now we both have a piece of information, namely the number $r^{ab}\mod p$.  This is going to be our &quot;secret key&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if someone can figure out either $a$ or $b$, then since they also know $r^a$ and $r^b$, they'll be able to figure out our secret key. We hope this is hard, but we certainly need $p$ to be very big: otherwise, they can just try all powers of $r$ until they hit the right one. We need the &quot;search space&quot; to be very big, so we need $p$ to be very big. &lt;strong&gt;Added:&lt;/strong&gt; As jug points out, having $p$ big is not sufficient. There are algorithms for computing discrete logarithms that are particularly good with certain kinds of primes, so we generally also require that $p$ satisfy some additional &quot;good&quot; properties relative to the cryptographic application. You generally want $p$ and $(p-1)/2$ to be both primes, for example. On the other hand, in practice one does not &lt;em&gt;really&lt;/em&gt; need $r$ to be a primitive root. Instead, it is enough that it generate a &quot;large&quot; subgroup of the multiplicative group, which one generally wants to be of prime order.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Note: figuring out $a$ or $b$ is just &lt;em&gt;one&lt;/em&gt; way in which they could figure out our secret key $r^{ab}$, since everyone knows $p$, $r$, $r^a$, and $r^b$. It is &lt;em&gt;not&lt;/em&gt; known whether this is essentially the only way to break this &quot;key exchange&quot; method; the method really relies on whether one can figure out $r^{ab}$ from knowing $r$, $p$, $r^a$, and $r^b$; this is called the &lt;strong&gt;Diffie-Hellman problem&lt;/strong&gt;; the Diffie-Hellman problem is at most as hard as the Discrete Logarithm Problem, but we do not know if it is &lt;em&gt;just&lt;/em&gt; as hard (it could be easier); and we don't know just how hard the Discrete Logarithm Problem is, we just know that we don't have any easy ways of doing it yet).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So key exchange is one place where big primes are very important. (Diffie-Hellman is not the only way to do key exchanges). Another place where big primes play a big role is in &lt;a href=&quot;http://en.wikipedia.org/wiki/RSA&quot; rel=&quot;nofollow&quot;&gt;RSA&lt;/a&gt; which is a cryptosystem that also relies on big primes (this time, two big primes $p$ and $q$, and we do arithmetic modulo $n=pq$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added:&lt;/strong&gt; Might as well add a quick overview of RSA and how the primes come into play. Here, once again modular exponentiation is part of the process. This is an &quot;public key&quot; system: I will tell everyone how to send me secret messages, which hopefully only I can decode. (In Diffie-Hellman, we did not exchange a message; we agreed on a secret key that we will use with a separate system that requires a secret key; for example, &lt;a href=&quot;http://en.wikipedia.org/wiki/Advanced_Encryption_Standard&quot; rel=&quot;nofollow&quot;&gt;AES&lt;/a&gt;). I pick two large primes $p$ and $q$, and compute $n=pq$. I also pick a number $e$ that is relatively prime to $(p-1)(q-1)$ (I can do that because I know $p$ and $q$). Then I use the Euclidean algorithm, which is pretty quick, to find a $d$ such that $ed\equiv 1 \pmod{(p-1)(q-1)}$. Finally, I tell everyone what $n$ and $e$ are. If you want to send me a message, you first convert it to a number $M$ using some standard mechanism. Then you compute $M^e \mod n$, and you tell me what $M^e\mod n$ is. I will take $M^e$ and compute $(M^e)^d = M^{ed}\mod n$. Because $ed\equiv 1 \pmod{(p-1)(q-1)}$, then $M^{ed}\equiv M\pmod{n}$, so that is how I recover $M$. The security of the system relies in hoping that from knowing $n$ and $e$, it is difficult to figure out $d$ (it is easy if I know $p$ and $q$; this is why this is believed to be a &quot;trap-door function&quot; as described in the first paragraph). The problem is at most as hard as factoring $n$, because if you can factor $n$ then you can find $d$ the same way I did; it is not known if the problem of finding $M$ from $n$, $M^e$, and $e$  is at least as hard as factoring (it has been shown that some variants &lt;em&gt;are&lt;/em&gt; at least as hard as factoring), and again we don't know just how hard factoring &lt;em&gt;is&lt;/em&gt;.  But: because we know that if you can factor $n$ then you can read the message, then we want to make $n$ difficult to factor. It only has two factors, but you don't want them to be easy to find, so you want $p$ and $q$ to be large for sure. (Again, there are other conditions one usually puts on $e$, $p$, and $q$ to make sure that certain special attacks do not succeed easily, but at &lt;em&gt;least&lt;/em&gt; we need $p$ and $q$ to be very big).&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-10-22T18:21:23.480" LastActivityDate="2010-10-22T18:21:23.480" CommentCount="15" />
  <row Id="7382" PostTypeId="1" AcceptedAnswerId="7387" CreationDate="2010-10-21T03:44:32.410" Score="2" ViewCount="93" Body="&lt;p&gt;Page 13 of &lt;a href=&quot;http://books.google.com/books?id=4JP7ofjHVh8C&amp;amp;lpg=PP1&amp;amp;ots=tWhNW0T4T4&amp;amp;dq=mikhalkin%20tropical%20algebraic%20geometry&amp;amp;pg=PA13#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;Tropical Algebraic Geometry&lt;/a&gt; by Itenberg, Mikhalkin, and Shustin mentions 1-valent vertices, but I haven't been able to find a source that actually defines this term or managed to guess the definition myself.  Either a definition or reference would be greatly appreciated. Thanks!&lt;/p&gt;&#xA;" OwnerUserId="249" LastActivityDate="2010-10-21T04:36:16.827" Title="What are valent vertices?" Tags="&lt;terminology&gt;&lt;reference-request&gt;&lt;graph-theory&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="7383" PostTypeId="2" ParentId="7249" CreationDate="2010-10-21T04:15:25.823" Score="2" ViewCount="" Body="&lt;p&gt;The physical justification is quite simple. The point of contact between the wheel and the road is instantaneously stationary. Since the wheel moves rigidly, the velocity of any point on the wheel is as though the wheel were rotating about the point of contact. In particular, the velocity of the axle is perpendicular to the line joining it to the point of contact. Since we require that the axle stays level, it velocity must be purely horizontal, so it has to be directly above the contact point.&lt;/p&gt;&#xA;" OwnerUserId="856" LastActivityDate="2010-10-21T04:15:25.823" CommentCount="4" />
  <row Id="7384" PostTypeId="2" ParentId="7367" CreationDate="2010-10-21T04:19:33.227" Score="3" ViewCount="" Body="&lt;p&gt;$f(x)=x^2+|x|-6$ is an even function—that is, $f(x)=f(-x)$ for all $x$, or the graph of $y=f(x)$ is symmetric over the y-axis—so if $f(c)=0$ then $f(-c)=0$, so the sum of the zeros of $f$ must be 0.&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-10-21T04:19:33.227" CommentCount="5" />
  <row Id="7385" PostTypeId="2" ParentId="7376" CreationDate="2010-10-21T04:30:23.893" Score="2" ViewCount="" Body="&lt;p&gt;One direction is straightforward, so I'll only give the proof in the other direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $x \in \dot{\bar{C}}$.  Then there exists an open ball $B_x$ containing $x$ such that $B_x \subset \bar{C}$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose $x \not\in \dot{C}$.  But $x \in \bar{C}$, which means that $x$ is on the boundary of $C$.   Therefore, $B_x$ contains at least one point $z$ not in $C$.  Since $C$ is convex, there is a hyperplane that separates $z$ and $C$.  But $z \in \dot{\bar{C}}$.  Thus there exists another open ball $B_z$ containing $z$ such that $B_z \subset \bar{C}$.  Now, intersect $B_z$ with the open half-space (from the separating hyperplane) that does not contain $C$ to obtain set $D$.  We have $D \subset \bar{C}$ but $D \cap C = \emptyset$.  Thus $D$ is a subset of the boundary of $C$.  However, $D$ is open because it is the intersection of two open sets.  $D$ thus has the property that it is an open set containing boundary points of $C$ yet fails to contain any points of $C$.  Such a set cannot exist, and we have our contradiction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This proof does bring in convexity via the separating hyperplane theorem - the version with a convex set and a point not in the set - rather than via the definition of a convex set.  I'm not sure how to prove this without the separating hyperplane theorem (which is one of the fundamental properties of a convex set).  If you want a proof of the separating hyperplane theorem it can be found on page 47 of &lt;a href=&quot;http://books.google.com/books?id=VdMcrxUfeMUC&amp;amp;pg=PA46&amp;amp;lpg=PA46&amp;amp;dq=separation+theorems+for+convex+sets&amp;amp;source=bl&amp;amp;ots=IQIbpQtnOj&amp;amp;sig=jQLbA7sZol8R6FplrmVJ1Jji0Bc&amp;amp;hl=en&amp;amp;ei=Mba_TP_uGo7CsAOCyfizCQ&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=8&amp;amp;ved=0CDwQ6AEwBw#v=onepage&amp;amp;q=separation%20theorems%20for%20convex%20sets&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;Mangasarian&lt;/a&gt;.  I'm also assuming that you're using the usual topology on $\mathbb{R}^n$.&lt;/p&gt;&#xA;" OwnerUserId="2370" LastActivityDate="2010-10-21T04:30:23.893" CommentCount="1" />
  <row Id="7386" PostTypeId="2" ParentId="7342" CreationDate="2010-10-21T04:33:31.637" Score="2" ViewCount="" Body="&lt;p&gt;Using Lagrange multipliers we get $x_i^2 = \lambda x + \mu$, so that $x_i$ can take only two values, say $X &gt; 0$ and $Y &amp;lt; 0$ (we can't have $X=Y=0$ for obvious reasons). Suppose that $X$ is taken $a$ times, and $Y$ is taken $b = n-a$ times. We have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$aX + bY = 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$aX^2 + bY^2 = n$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can check that the solution is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$X^2 = b/a, Y^2 = a/b$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The objective function that we want to minimize is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$aX^3 + bY^3 = b\sqrt{b/a} - a\sqrt{a/b}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We want to make the first summand small and the second big. Note that as we increase $a$, the first summand becomes smaller, and the second one becomes bigger, so we might as well choose $a=n-1$ and $b=1$. So $X = 1/\sqrt{(n-1)}$ and $Y = -\sqrt{n-1}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Concluding, the optimum is obtained with the sequence $-\sqrt{n-1}, 1/\sqrt{(n-1)} \times (n-1)$, and the optimal value is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ -(n-1)\sqrt{n-1} + 1/\sqrt{n-1} $&lt;/p&gt;&#xA;" OwnerUserId="1277" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2010-10-23T03:30:01.450" LastActivityDate="2010-10-23T03:30:01.450" CommentCount="2" />
  <row Id="7387" PostTypeId="2" ParentId="7382" CreationDate="2010-10-21T04:36:16.827" Score="4" ViewCount="" Body="&lt;p&gt;In graph theory, the &lt;em&gt;valency&lt;/em&gt; of a vertex is sometimes used to mean its &lt;em&gt;degree&lt;/em&gt;, i.e. the number of edges incident on it. Both &lt;a href=&quot;http://en.wikipedia.org/wiki/Degree_%28graph_theory%29&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt; and &lt;a href=&quot;http://mathworld.wolfram.com/VertexDegree.html&quot; rel=&quot;nofollow&quot;&gt;MathWorld&lt;/a&gt; mention both terms as synonyms.&lt;/p&gt;&#xA;" OwnerUserId="856" LastActivityDate="2010-10-21T04:36:16.827" />
  <row Id="7388" PostTypeId="2" ParentId="7340" CreationDate="2010-10-21T05:52:41.477" Score="3" ViewCount="" Body="&lt;p&gt;(5+2 choose 3)=(7 choose 3)=35 not (5 choose 3). The 35 different orders orders are {{1, 1, 1}, {1, 1, 2}, {1, 1, 3}, {1, 1, 4}, {1, 1, 5}, {1, 2, 2}, {1, 2, 3}, {1, 2, 4}, {1, 2, 5}, {1, 3, 3}, {1, 3, 4}, {1, 3, 5}, {1, 4, 4}, {1, 4, 5}, {1, 5, 5}, {2, 2, 2}, {2, 2, 3}, {2, 2, 4}, {2, 2, 5}, {2, 3, 3}, {2, 3, 4}, {2, 3, 5}, {2, 4, 4}, {2, 4, 5}, {2, 5, 5}, {3, 3, 3}, {3, 3, 4}, {3, 3, 5}, {3, 4, 4}, {3, 4, 5}, {3, 5, 5}, {4, 4, 4}, {4, 4, 5}, {4, 5, 5}, {5, 5, 5}}.  The notation and formula were from &lt;a href=&quot;http://www.research.att.com/~njas/sequences/A000292&quot; rel=&quot;nofollow&quot;&gt;http://www.research.att.com/~njas/sequences/A000292&lt;/a&gt;&lt;/p&gt;&#xA;" LastActivityDate="2010-10-21T05:52:41.477" />
  <row Id="7389" PostTypeId="2" ParentId="7340" CreationDate="2010-10-21T06:51:58.133" Score="3" ViewCount="" Body="&lt;p&gt;If we assume that the chef doesn't care which person ordered which dish, just how many of each dish to make, the problem is equivalent to placing 3 balls (each representing a dish the chef needs to make) in 5 urns (representing the 5 possible dishes).  This type of balls/urns problem (or pieces of identical candy to children, donuts of different types, etc.) can be solved with a technique sometimes called &quot;stars and bars&quot;: represent the balls by stars in a line:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;* * *&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Now, to divide these balls among 5 urns, add 4 bars, denoting the breaking points between different urns, e.g.:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;* * * | | | |   all the balls in the first urn&#xA;* | | * * | |   1 balls in the first urn, 2 in the third urn, none in the rest&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The number of possible ways to do this is the number of ways to rearrange the line of 3 *s and 4 |s, which is the number of ways to choose which 3 of the 7 symbols in the line should be *s, so ${7\choose 3}$.  In general, for $k$ balls into $n$ urns, there are $k$ *s, $n-1$ |s, $k+n-1$ total symbols in the line, so ${k+n-1\choose k}$.&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-10-21T06:51:58.133" />
  <row Id="7390" PostTypeId="2" ParentId="7354" CreationDate="2010-10-21T08:42:12.293" Score="2" ViewCount="" Body="&lt;p&gt;We have $$e^{1} = 1 + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \cdots$$ and $$e^{-1} = 1 - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!} + \cdots $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Subtracting these two we get $$e - e^{-1} = 2 \cdot \Bigl( 1 + \frac{1}{3!} + \frac{1}{5!} + \cdots \Bigr)$$ Therefore the series converges to $$\frac{e-e^{-1}}{2} = \sum\limits_{n=0}^{\infty} \frac{1}{(2n+1)!}$$&lt;/p&gt;&#xA;" LastActivityDate="2010-10-21T08:42:12.293" CommentCount="1" />
  <row Id="7391" PostTypeId="1" AcceptedAnswerId="8881" CreationDate="2010-10-21T08:48:57.103" Score="5" ViewCount="617" Body="&lt;p&gt;How to solve $$\int_0^{\pi} \sin{2x}\sin{x} dx$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; Sorry! I should have described more. This is not a homework. Recently, Out of the blue I got interest in physics and started reading and solving problems. This is part of a physics problem where I got stuck (because I forgot all high school formulae.). Thanks all of you guys for wonderful solutions. &lt;/p&gt;&#xA;" OwnerUserId="2264" LastEditorUserId="2264" LastEditorDisplayName="" LastEditDate="2010-10-27T18:55:18.683" LastActivityDate="2011-04-17T06:38:17.240" Title="How to solve $ \int_0^{\pi} \sin{2x}\sin{x} dx $?" Tags="&lt;calculus&gt;&lt;integral&gt;" AnswerCount="10" CommentCount="12" FavoriteCount="0" />
  <row Id="7392" PostTypeId="1" AcceptedAnswerId="7409" CreationDate="2010-10-21T08:56:38.103" Score="5" ViewCount="267" Body="&lt;p&gt;&lt;strong&gt;Question 1:&lt;/strong&gt; Does such a ring can be found? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: The definition of a noetherian &lt;em&gt;topological space&lt;/em&gt; is similar to that in rings or sets. Every descending chain of closed subsets stops after a finite number of steps &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(&lt;strong&gt;Question 2:&lt;/strong&gt; is this equivalent to saying that every &lt;em&gt;descending&lt;/em&gt; chain of &lt;em&gt;opens&lt;/em&gt; stops?).&lt;/p&gt;&#xA;" OwnerUserId="685" LastActivityDate="2011-02-15T14:14:47.020" Title="A non-noetherian ring with $\text{Spec}(R)$ noetherian" Tags="&lt;commutative-algebra&gt;" AnswerCount="4" CommentCount="1" FavoriteCount="3" />
  <row Id="7393" PostTypeId="2" ParentId="7391" CreationDate="2010-10-21T09:01:55.463" Score="8" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT:&lt;/strong&gt; Use this formula  $$\cos C - \cos D = 2 \sin\frac{(C+D)}{2} \cdot \sin\frac{(D - C)}{2}$$&lt;/p&gt;&#xA;" OwnerUserId="9413" LastEditorUserId="9413" LastEditorDisplayName="" LastActivityDate="2010-10-21T09:01:55.463" CommentCount="9" />
  <row Id="7394" PostTypeId="2" ParentId="7391" CreationDate="2010-10-21T09:03:51.727" Score="8" ViewCount="" Body="&lt;p&gt;I'd suggest using an identity for $\sin 2x$ that rewrites it in terms of $\sin x$ and $\cos x$, then using substitution ($u=\sin x$) on the result.&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-10-21T09:03:51.727" CommentCount="1" />
  <row Id="7395" PostTypeId="2" ParentId="7391" CreationDate="2010-10-21T09:06:22.147" Score="8" ViewCount="" Body="&lt;p&gt;Two more ways: use Euler's formula $\sin ax=(e^{iax}-e^{-iax})/2i$, or integrate by parts twice to get an equality where your sought integral appears twice and can be solved for.&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2010-10-21T09:06:22.147" CommentCount="3" />
  <row Id="7396" PostTypeId="2" ParentId="7354" CreationDate="2010-10-21T09:31:45.847" Score="1" ViewCount="" Body="&lt;p&gt;To elaborate on the first answer given to this question by Ross Millikan.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{n=0}^\infty \frac{1}{(2n+1)!} = 1 + \sum_{n=1}^\infty \frac{1}{(2n+1)!}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&amp;lt; 1 + \sum_{n=1}^\infty \frac{1}{4^n} = \frac{4}{3}, \quad \textrm{ as } &#xA;\frac{1}{(2n+1)!} &amp;lt; \frac{1}{4^n} \textrm{ for } n \ge1.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence by &lt;a href=&quot;http://www.maths.abdn.ac.uk/~igc/tch/ma2001/notes/node50.html&quot; rel=&quot;nofollow&quot;&gt;the comparison test&lt;/a&gt; the series converges.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Comparing with another more manageable series could be useful in this case for possible follow-on questions as, with this approach, it's not much extra work to prove that it converges to an irrational number. Such a proof might include: Let $S$ be the series and $S_N$ the $N$th partial sum and $R_N$ the remainder then $S=S_N + R_N,$ where we note that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$R_N &amp;lt; \frac{1}{(2n+3)!} &#xA;\left( 1 +  \frac{1}{(2n+3)^2} + \frac{1}{(2n+3)^4} + \cdots \right).$$&lt;/p&gt;&#xA;" OwnerUserId="1301" LastActivityDate="2010-10-21T09:31:45.847" />
  <row Id="7397" PostTypeId="2" ParentId="7376" CreationDate="2010-10-21T09:58:15.540" Score="6" ViewCount="" Body="&lt;p&gt;I understand the problematic statement to mean that if a convex set $X$ has empty interior, then it is contained in a hyperplane. Let me know if I've misunderstood your problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a sketch of a proof by contrapositive. If $X$ is not contained in a hyperplane, then you can find $n+1$ affinely independent points in $X$. Their convex hull is a simplex contained in $X$, and a nondegenerate simplex has nonempty interior. (I'm pretty sure that last statement is true, but I'd be hard pressed to prove it.)&lt;/p&gt;&#xA;" OwnerUserId="856" LastActivityDate="2010-10-21T09:58:15.540" CommentCount="2" />
  <row Id="7398" PostTypeId="2" ParentId="7391" CreationDate="2010-10-21T10:46:04.467" Score="14" ViewCount="" Body="&lt;p&gt;Plotting $ \sin{2x}\sin{x} $ in $[0,\pi]$ shows that it's symmetrical with respect to $x=\pi/2$ and so the integral is zero.&lt;/p&gt;&#xA;" OwnerUserId="589" LastActivityDate="2010-10-21T10:46:04.467" CommentCount="0" />
  <row Id="7399" PostTypeId="2" ParentId="7391" CreationDate="2010-10-21T10:53:13.260" Score="8" ViewCount="" Body="&lt;p&gt;Have a look here:&lt;br&gt;&#xA;&lt;a href=&quot;http://www.wolframalpha.com/input/?i=integrate+sin%282x%29+sin%28x%29+from+0+to+Pi&quot; rel=&quot;nofollow&quot;&gt;http://www.wolframalpha.com/input/?i=integrate+sin%282x%29+sin%28x%29+from+0+to+Pi&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;...and for all the steps of the derivation (click on &quot;show steps&quot;):&lt;br&gt;&#xA;&lt;a href=&quot;http://www.wolframalpha.com/input/?i=integrate+sin%282x%29+sin%28x%29&quot; rel=&quot;nofollow&quot;&gt;http://www.wolframalpha.com/input/?i=integrate+sin%282x%29+sin%28x%29&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="346" LastActivityDate="2010-10-21T10:53:13.260" />
  <row Id="7400" PostTypeId="2" ParentId="7391" CreationDate="2010-10-21T11:05:10.310" Score="5" ViewCount="" Body="&lt;p&gt;Somewhat inspired by lhf's answer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_0^\pi\sin\;2u\;\sin\;u\;\mathrm{d}u$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_{0-\frac{\pi}{2}}^{\pi-\frac{\pi}{2}}\sin\left(2\left(u+\frac{\pi}{2}\right)\right)\;\sin\left(u+\frac{\pi}{2}\right)\mathrm{d}u$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$-\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\sin\;2u\;\cos\;u\;\mathrm{d}u$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$-\left(\int_{-\frac{\pi}{2}}^0\sin\;2u\;\cos\;u\;\mathrm{d}u+\int_0^{\frac{\pi}{2}}\sin\;2u\;\cos\;u\;\mathrm{d}u\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$-\left(-\int_0^{\frac{\pi}{2}}\sin\;2u\;\cos\;u\;\mathrm{d}u+\int_0^{\frac{\pi}{2}}\sin\;2u\;\cos\;u\;\mathrm{d}u\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and we see that the integral is zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is convenient here that $\sin\;2u\;\cos\;u$ is an odd function.&lt;/p&gt;&#xA;" OwnerUserId="498" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-10-21T11:10:48.713" LastActivityDate="2010-10-21T11:10:48.713" CommentCount="4" />
  <row Id="7401" PostTypeId="1" CreationDate="2010-10-21T11:23:12.523" Score="3" ViewCount="107" Body="&lt;p&gt;Suppose there are 9 events, that have a probability of 10%, 20%, 30%, ..., 90% of being a success.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How would I find the probability of exactly n number of these events succeeding?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For n = 1, I'm thinking it is something like&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;(first succeeds) -&amp;gt; (1-0.9) * 0.8 * 0.7 * 0.6 ... 0.1&#xA;(second succeeds) -&amp;gt; 0.9 * (1-0.8) * 0.7 * 0.6 ... 0.1&#xA;(...)&#xA;(last)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;and then adding them up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The probabilities might not always be an an Arithmetic Progression, hoping to find a solution that doesn't depend on that.&lt;/p&gt;&#xA;" LastActivityDate="2010-10-21T15:14:38.010" Title="Probability Distribution with different probabilities" Tags="&lt;combinatorics&gt;&lt;probability&gt;" AnswerCount="3" />
  <row Id="7402" PostTypeId="2" ParentId="7317" CreationDate="2010-10-21T11:27:57.567" Score="1" ViewCount="" Body="&lt;p&gt;Well, it depends if you want to understand esp. &lt;em&gt;this&lt;/em&gt; derivation or &lt;em&gt;an&lt;/em&gt; derivation (so what is going on there).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The easiest and most intuitive derivation I have seen, ever - is this one:&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Intuitive Proof of Black-Scholes Formula Based on Arbitrage and Properties of Lognormal Distribution by Alexei Krouglov&lt;br&gt;&#xA;Link: &lt;a href=&quot;http://arxiv.org/abs/physics/0612022&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/physics/0612022&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="346" LastActivityDate="2010-10-21T11:27:57.567" CommentCount="1" />
  <row Id="7403" PostTypeId="2" ParentId="7392" CreationDate="2010-10-21T12:27:41.513" Score="4" ViewCount="" Body="&lt;p&gt;Answer to question 1: Yes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take a non-noetherian valuation domain $R$ of finite Krull dimension. The spectrum of $R$ is totally ordered under inclusion hence finite.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;H&lt;/p&gt;&#xA;" OwnerUserId="2479" LastActivityDate="2010-10-21T12:27:41.513" />
  <row Id="7404" PostTypeId="2" ParentId="7401" CreationDate="2010-10-21T12:49:18.407" Score="1" ViewCount="" Body="&lt;p&gt;Take the events in order.  After the first, you have 0.1 chance of 1 success and 0.9 chance of 0 successes.  Then add on the second.  The chance of one success after two events is 0.2*the chance of failure on the first + 0.8*the chance of success on the first.  A spreadsheet is your friend for this.  After all the events, you will have a column with the chance of each possible number of successes.  Adding to see if you still get 1 is a good check that your equations are right.  If you use the absolute/relative reference codes correctly, you can copy down and right most of the equations.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-10-21T12:49:18.407" />
  <row Id="7405" PostTypeId="1" AcceptedAnswerId="7430" CreationDate="2010-10-21T12:56:48.643" Score="1" ViewCount="76" Body="&lt;p&gt;A quantile function Q is defined in terms of its distribution function F as:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$Q(p) = \inf \{ x\in R : p \le F(x) \}, p \in (0,1).$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I was wondering:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) if the definition of Q can be extended to $[0,1]$ by letting $ x \in R \cup \{\infty, -\infty \}$? if not, why?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) if it is correct that a distribution function can always be represented in terms of a quantile function defined on $(0,1)$ as:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$F(x) = \sup \{ p\in (0,1) : Q(p) \le x\}, x \in R-\{x|F(x)=0\text{ or }1\}.$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(3) if it is correct that a distribution function can always be represented in terms of a quantile function defined on $[0,1]$ (if (1) is correct) as:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$F(x) = \sup \{ p\in [0,1] : Q(p) \le x\}, x \in R \cup \{\infty, -\infty\}.$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Thanks and regards!&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;UPDATE:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One more question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(4) if the followings are correct:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$F$ is discontiunous at $x_0$ with $F(x_0)=p_0$ and $lim_{x \rightarrow x_0-}F(x)= p_1$ $\Longleftrightarrow$ $Q$ is constant over $[p_1, p_0]$, taking value $x_0$.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$F$ is constant over $[x_1, x_0]$ taking value $p_0$ $\Longleftrightarrow$ $Q$ is discontiunous at $p_0$ with $Q(p_0)=x_1$ and $lim_{x \rightarrow x_1+}F(x)= x_0$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="1281" LastEditorUserId="1281" LastEditorDisplayName="" LastEditDate="2010-10-21T13:18:47.350" LastActivityDate="2010-10-21T17:01:45.637" Title="quantile function and distribution function" Tags="&lt;probability-theory&gt;" AnswerCount="1" />
  <row Id="7406" PostTypeId="2" ParentId="7392" CreationDate="2010-10-21T14:03:49.080" Score="8" ViewCount="" Body="&lt;p&gt;If the non-noetherianness of the ring is hidden inside the nilradical, then $\mathrm{Spec}$ won't see it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $k$ be any ring, and let $V$ be a free $k$-module of infinite rank. Consider $R=k\oplus V$, and turn it into a ring by defining $$(a,v)\cdot(b,w)=(ab,aw+bv).$$ (Representation-people call this a &lt;em&gt;trivial extension&lt;/em&gt;) Then $R$ is not-noetherian, because every $k$-submodule of $V$ is an ideal in $R$. Yet $V$ is contained in the nilradical of $R$: if you look at $\mathrm{Spec}\;R$ and at $\mathrm{Spec}\;k$, you'll see that they are very similar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for your second question: no. If $k$ is an infinite field, then $\mathrm{Spec}\;k[X]$ is noetherian, yet you'll easily find a decreasing chain of open sets init which does not stop. (What examples did you consider before asking the question? :) )&lt;/p&gt;&#xA;" OwnerUserId="274" LastEditorUserId="274" LastEditorDisplayName="" LastEditDate="2010-10-21T17:00:35.807" LastActivityDate="2010-10-21T17:00:35.807" CommentCount="1" />
  <row Id="7407" PostTypeId="2" ParentId="7329" CreationDate="2010-10-21T14:05:22.860" Score="0" ViewCount="" Body="&lt;p&gt;Building on your interpretation, let us define the rules as follows:  A and B alternately throw a single die.  If the sum of the last two throws is 5, the one who just threw wins.  Note that as stated, an initial 5 could win, or a series of throws of a different length than 2.  The analysis would change, but follow the same route.  The below requires precisely two throws to add to 5.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let p be the probability that the first player wins.  Let q be the probability that the next player to throw wins given that he has received a chance of winning, that is that the last throw is less than 5.  Then if you receive a throw of 5 or 6 your chance of winning is p.  So q=1/6 (that you win on this throw) + (1-p)/3 (that you throw 5 or 6 and then win) +(1-q)/2 (that you throw &amp;lt;5, don't win this throw, but finally win.  p=(1-p)/3 (that you throw 5 or 6 and win)+2(1-q)/3 (that you throw &amp;lt;5 and win).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$q=\frac{1}{6}+\frac{1-p}{3}+\frac{1-q}{2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$p=\frac{1-p}{3}+\frac{2(1-q)}{3}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I have the algebra right, p=15/32 and q=9/16, so the first player should withdraw $2/17. &lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2010-10-21T17:45:25.887" LastActivityDate="2010-10-21T17:45:25.887" />
  <row Id="7408" PostTypeId="2" ParentId="7391" CreationDate="2010-10-21T14:25:15.453" Score="8" ViewCount="" Body="&lt;p&gt;If you use the identity $\sin{2x} = 2 \cos{x}\sin{x}$ then you get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int \sin{2x} \sin{x} \, dx = 2 \int \sin^2{x} \cos{x} \, dx  $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and now you can make the substitution $u = \sin{x}$ to get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$2 \int u^2 \, du  = \frac{2}{3} u^3 + C = \frac{2}{3} \sin^3{x} + C$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore $$\int_{0}^{\pi} \sin{2x} \sin{x} \, dx = \left ( \frac{2}{3} \sin^3{x} \right )_{0}^{\pi} = 0 $$&lt;/p&gt;&#xA;" OwnerUserId="900" LastEditorUserId="900" LastEditorDisplayName="" LastEditDate="2010-10-21T14:35:48.983" LastActivityDate="2010-10-21T14:35:48.983" CommentCount="2" />
  <row Id="7409" PostTypeId="2" ParentId="7392" CreationDate="2010-10-21T14:30:48.357" Score="8" ViewCount="" Body="&lt;p&gt;It is easy to come up with examples when you keep in mind that $X$ and $X_{red}$ are homeomorphic: The topology is not changed when you divide out (all) nilpotent sections.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, $A = k[x_1,x_2,...] / (x_1^2,x_2^2,...)$ satisfies $A/rad(A)=k$, so that $Spec(A)$ consists of exactly one point, namely the maximal ideal $(x_1,x_2,...)$, which is not finitely generated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, $Spec(A)$ is a noetherian topological space iff $A$ satisfies the ascending chain condition for &lt;em&gt;radical&lt;/em&gt; ideals.&lt;/p&gt;&#xA;" OwnerUserId="1650" LastActivityDate="2010-10-21T14:30:48.357" CommentCount="1" />
  <row Id="7410" PostTypeId="2" ParentId="7354" CreationDate="2010-10-21T14:37:58.867" Score="2" ViewCount="" Body="&lt;p&gt;Another way is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $\displaystyle S_n = 1 + \frac{1}{3!} + \dots + \frac{1}{(2n+1)!}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle S_n \le 1 + \frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \dots + \frac{1}{n(n+1)}$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle = 1 + (1 - \frac{1}{2}) + (\frac{1}{2} - \frac{1}{3}) + \dots + (\frac{1}{n} - \frac{1}{n+1}) = 2 - \frac{1}{n+1} &amp;lt; 2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus $S_n &amp;lt; 2$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;thus we have the $\displaystyle S_n$ is monotonically increasing and bounded above and so is convergent.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-10-21T14:37:58.867" />
  <row Id="7411" PostTypeId="2" ParentId="7401" CreationDate="2010-10-21T14:41:28.570" Score="1" ViewCount="" Body="&lt;p&gt;Let the probability that the $i$th event occurs be $p_i = i/10$ and let the probability that exactly $n$ events occur be $P(n).$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the probability that no events occur is $P(0) = \prod_{i=1}^9 (1-p_i) = 3.6288 \times 10^{-4}.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $S= \lbrace 1,2,3,\ldots,9 \rbrace $ then for $n \ge 1$ we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$P(n) = P(0) \times \sum_{i_1,i_2,\ldots,i_n \in S, \quad i_k \textrm { distinct} }&#xA;\frac{p_{i_1} p_{i_2} \ldots p_{i_n} } {(1-p_{i_1})(1-p_{i_2}) \ldots (1-p_{i_n}) },$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where there are ${9 \choose n}$ terms in the summation.&lt;/p&gt;&#xA;" OwnerUserId="1301" LastActivityDate="2010-10-21T14:41:28.570" />
  <row Id="7412" PostTypeId="1" CreationDate="2010-10-21T14:52:00.710" Score="6" ViewCount="260" Body="&lt;p&gt;Hi all,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was playing around with &lt;a href=&quot;http://en.wikipedia.org/wiki/Sierpinski_carpet&quot; rel=&quot;nofollow&quot;&gt;Sierpinski carpets&lt;/a&gt; (see &lt;a href=&quot;http://stackoverflow.com/questions/3972902/is-it-possible-to-perform-floating-point-operations-on-gpu-when-using-opengl/3973433#3973433&quot;&gt;pretty GPU-produced picture here&lt;/a&gt;), and came up with a variation that I have been unable to find mentioned elsewhere. I'm wondering if anyone can tell me an existing name for it, and help me determine where it fits into existing classifications of fractals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you probably know, &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The construction of the &lt;strong&gt;Sierpinski&#xA;  carpet&lt;/strong&gt; begins with a square. The&#xA;  square is cut into 9 congruent&#xA;  subsquares in a 3-by-3 grid, and the&#xA;  [open] central subsquare is removed. The same&#xA;  procedure is then applied recursively&#xA;  to the remaining 8 subsquares, ad&#xA;  infinitum.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In this variation, &lt;strong&gt;instead of merely removing the central subsquare, we recursively apply the same procedure to it &quot;inversely.&quot;&lt;/strong&gt; (Is there a better term for that?) That is, we treat the central hole as if we want to replace it with the complement of [this variant of] a Sierpinski carpet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So when we recur on the central hole, we put an &quot;anti-hole&quot; (like a &quot;donut hole&quot; without a donut) in the middle of it. We will then recur on the anti-hole using the normal (positive) procedure, and recur on the 8 subsquares surrounding the anti-hole using the inverse procedure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It looks something like this, in its 5th iteration:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/vVMDk.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you think I should post the code for this, I'm happy to, but I didn't think it was very relevant to the math question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've been calling it &quot;Sierpinski tablecloth&quot; since I couldn't find another name for it, and because it reminds me of a lace or crocheted tablecloth design.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Correction:&lt;/strong&gt; @Rahul pointed out that &quot;the limiting set of the recursive process does not exist.&quot; This is because the central point of the unit square must be part of both the tablecloth and its complement, or neither - which is impossible. (Of course we can still do finite iterations of the constructive process without hitting a contradiction.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To fix this problem, I tried proposing that we define the tablecloth set T (arbitrarily) to include the central point. The central subsquare is replaced with the &lt;em&gt;union of the central point&lt;/em&gt; with the scaled-down complement of T. As before, the subsquares around the center are replaced with scaled-down copies of T (and therefore the central point of each of the subsquares is contained in T).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Another update:&lt;/strong&gt; Unfortunately the same problem is true of any point with an irrational coordinate, and of many rational ones. If we express the coordinates in base 3, as with the Cantor set, a point is a member of T iff the number of 1's in the base-3 expansion of each of its coordinates is even. But this doesn't work for numbers with repeating 1's.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we tried to fix this by making all &quot;problem points&quot; belong to T, we'd be left with &quot;almost&quot; a solid unit square, minus a &quot;dust&quot; of rational points with an even number of 1's in their base-3 coordinate expansions. Then T would cease to generate interesting fractal pictures (in the limit). Its area would be 1, like the irrationals$^2$. It's not clear to me that T would be different from the set of the irrationals$^2$ in any interesting way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mind you, we can still draw pictures of the nth iteration of the tablecloth that are fun to look at, but it loses a good deal of its bite if it doesn't exist as a fractal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you think of a way to save the tablecloth from extinction as a fractal?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rest of this question was written under the misapprehension that the tablecloth exists (is a well-defined set) in the limit.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Fractal dimension&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The traditional carpet has &lt;a href=&quot;http://en.wikipedia.org/wiki/List_of_fractals_by_Hausdorff_dimension#Deterministic_fractals&quot; rel=&quot;nofollow&quot;&gt;Hausdorff dimension&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$log(8)/log(3) = ~1.8928$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;derived from the fact that it takes 8 self-similar copies at reduction factor 1/3 to cover itself (see &lt;a href=&quot;http://en.wikipedia.org/wiki/Fractal_dimension#Specific_definitions&quot; rel=&quot;nofollow&quot;&gt;example for Sierpinski triangle&lt;/a&gt;). In this vein I figure that the tablecloth takes 9 self-similar copies at reduction factor 1/3 to cover itself (no smaller copy will cover the middle mini-square), so it would have Hausdorff dimension&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$log(9)/log(3) = 2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This would put it in the same company (in terms of Hausdorff dimension) as space-filling curves like the Sierpinski curve, Hilbert, Peano, etc. Does this sound reasonable so far?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not quite comfortable with that, though, because a 1/3-size copy of the tablecloth will not actually cover the whole middle subsquare. In fact I can't see how any finite number of reduced-size copies of the tablecloth would cover the whole middle subsquare. And if we say we need an infinite number, the Hausdorff dimension would be infinite. :-(&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an alternative, I considered &lt;a href=&quot;http://en.wikipedia.org/wiki/Multifractal&quot; rel=&quot;nofollow&quot;&gt;multifractals&lt;/a&gt;, since we seem to have different scaling behavior occurring in different parts of the attractor. But that's deeper water than I'm prepared to wade into.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Can anyone point me to literature on&#xA;fractals that are defined using&#xA;&lt;em&gt;complements&lt;/em&gt; of reduced-size self-similar sets (as well as the&#xA;usual reduced-size self-similar&#xA;sets)? Is that a class of fractals&#xA;that has been studied? &lt;strong&gt;(Update:)&lt;/strong&gt; Or are such &quot;fractals&quot; doomed to&#xA;be ill-defined (self-contradictory) by their nature?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Similarly to Hausdorff dimension, the &lt;strong&gt;box-counting dimension&lt;/strong&gt; (using the criterion &quot;if the space were divided up into a grid of boxes of size ε, how many boxes of that scale would contain part of the attractor?&quot;) also seems to end up at 2, for much the same reasons. And I'm more comfortable with it because a &quot;box&quot; (as opposed to a reduced copy of the tablecloth) will indeed cover the whole middle subsquare. It seems clear that in the limit, the box-counting dimension converges to 2, because there are no gaps, so every box, at whatever size ε, will always contain part of the fractal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In terms of &lt;strong&gt;area&lt;/strong&gt; (standard &lt;a href=&quot;http://en.wikipedia.org/wiki/Lebesgue_measure&quot; rel=&quot;nofollow&quot;&gt;Lebesgue measure&lt;/a&gt;), I'm pretty sure that the tablecloth has zero, like the carpet, because it has no intervals (rectangles) that are entirely in the set. Right? (&lt;strong&gt;Update:&lt;/strong&gt; I stand corrected: the tablecloth seems to have area 1/2.)&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Summary&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;There are several questions above, so to tie them all together: I'm wondering if this tablecloth variant is something new and interesting, or something well-known, or an unknown but trivial variant that can be classified under existing categories. E.g. if there is a variant of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Cantor_set&quot; rel=&quot;nofollow&quot;&gt;Cantor set&lt;/a&gt; that is defined similarly to the tablecloth (for one dimension), I would appreciate a link to info about that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;(Update:)&lt;/strong&gt; One thing that seems interesting to me about it is that it has points in every interval of the unit square, but it also has holes in every interval. I guess it's like rational numbers in [0,1] in that regard, or like space-filling curves/trees, yet is not a curve or tree (not connected). I can't think of other fractals off-hand that are like that, can you? It seems that this property is connected with the fact that we recur on the inverse instead of leaving holes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;P.S.&lt;/strong&gt; I just read about the term &lt;em&gt;mutually dense&lt;/em&gt; (&lt;a href=&quot;http://www.mathpages.com/home/kmath172.htm&quot; rel=&quot;nofollow&quot;&gt;Interleaving Ad Infinitum&lt;/a&gt;): 'If a set S with linear ordering O can be partitioned into two subsets&#xA;A and B such that between every two distinct elements of A is an &#xA;element of B, and vice versa, then the sets A and B are said to be &#xA;&quot;mutually dense&quot;.' So the tablecloth and its complement are mutually dense, assuming that this term can be generalized to the unit square. At least in any horizontal or vertical linear slice of the unit square, they are mutually dense. &lt;/p&gt;&#xA;" OwnerUserId="2656" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2011-06-03T20:06:43.967" LastActivityDate="2011-06-03T20:06:43.967" Title="variant on Sierpinski carpet: rescue the tablecloth!" Tags="&lt;topology&gt;&lt;fractals&gt;" AnswerCount="3" CommentCount="9" />
  <row Id="7413" PostTypeId="1" AcceptedAnswerId="7416" CreationDate="2010-10-21T14:52:44.483" Score="7" ViewCount="467" Body="&lt;p&gt;I am currently learning about the concept of convolution  between two functions in my university course. The course notes are vague about what convolution is, so I was wondering if anyone could give me a good explanation. I can't seem to grasp other than the fact that it is just a particular integral of two functions. What is the physical meaning of convolution and why is it useful? Thanks a lot. &lt;/p&gt;&#xA;" OwnerUserId="2673" LastEditorUserId="900" LastEditorDisplayName="" LastEditDate="2010-10-26T01:06:54.320" LastActivityDate="2011-03-31T14:42:41.693" Title="Meaning of convolution?" Tags="&lt;real-analysis&gt;&lt;intuition&gt;&lt;convolution&gt;" AnswerCount="4" CommentCount="1" FavoriteCount="2" />
  <row Id="7415" PostTypeId="2" ParentId="7413" CreationDate="2010-10-21T14:59:01.713" Score="0" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Convolution&quot; rel=&quot;nofollow&quot;&gt;The Wikipedia&lt;/a&gt; has some nice graphical explanations.&lt;/p&gt;&#xA;" OwnerUserId="505" LastActivityDate="2010-10-21T14:59:01.713" CommentCount="1" />
  <row Id="7416" PostTypeId="2" ParentId="7413" CreationDate="2010-10-21T15:02:15.240" Score="4" ViewCount="" Body="&lt;p&gt;Have a look here:&lt;br&gt;&#xA;&lt;a href=&quot;http://answers.yahoo.com/question/index?qid=20070125163821AA5hyRX&quot; rel=&quot;nofollow&quot;&gt;http://answers.yahoo.com/question/index?qid=20070125163821AA5hyRX&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;...and lots of good answers here:&lt;br&gt;&#xA;&lt;a href=&quot;http://mathoverflow.net/questions/5892/what-is-convolution-intuitively&quot; rel=&quot;nofollow&quot;&gt;http://mathoverflow.net/questions/5892/what-is-convolution-intuitively&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="346" LastActivityDate="2010-10-21T15:02:15.240" CommentCount="1" />
  <row Id="7417" PostTypeId="2" ParentId="7401" CreationDate="2010-10-21T15:14:38.010" Score="1" ViewCount="" Body="&lt;p&gt;The generating function for the probabilities $a_n$ for $n$ successes is&#xA;$$\sum_{n=0}^9 a_n t^n=\prod_{k=1}^9(p_k t+(1-p_k))$$&#xA;where $p_k$ is the probability of the $k$-th event occuring, in this&#xA;example $p_k=k/10$. Maybe in this case the product won't simplify&#xA;so nicely, but the method is quite general.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-10-21T15:14:38.010" CommentCount="1" />
  <row Id="7418" PostTypeId="2" ParentId="7317" CreationDate="2010-10-21T15:20:10.657" Score="1" ViewCount="" Body="&lt;p&gt;I found this concise paper one of the most accessible ones giving an overview of the traditional approach:&lt;br&gt;&#xA;&lt;a href=&quot;http://www.williams.edu/Mathematics/fmorgan/MorganBlackScholes.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.williams.edu/Mathematics/fmorgan/MorganBlackScholes.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="346" LastActivityDate="2010-10-21T15:20:10.657" />
  <row Id="7419" PostTypeId="1" AcceptedAnswerId="7420" CreationDate="2010-10-21T15:20:18.527" Score="1" ViewCount="156" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm trying to solve this non-linear differential equation using substitution $\dfrac{y}{x}$ to $t$. However, I can't solve this equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\text{xy$\prime $} = \left(\frac{y}{x}\right)^3+y&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How to solve this equation and what's the general solution?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="2674" LastEditorUserId="2674" LastEditorDisplayName="" LastEditDate="2010-10-21T16:51:10.150" LastActivityDate="2010-10-22T00:14:37.857" Title="How to solve this non-linear differential equation?" Tags="&lt;differential-equations&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="7420" PostTypeId="2" ParentId="7419" CreationDate="2010-10-21T15:24:49.827" Score="4" ViewCount="" Body="&lt;p&gt;if you change the dependent variable to $u=y/x$, as you said, the equation becomes $x^2u'=u^3$, which can be immediately be solved by separating variables.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Show us what you tried and we may help you with it.&lt;/p&gt;&#xA;" OwnerUserId="274" LastEditorUserId="274" LastEditorDisplayName="" LastEditDate="2010-10-21T15:38:02.317" LastActivityDate="2010-10-21T15:38:02.317" CommentCount="2" />
  <row Id="7421" PostTypeId="2" ParentId="7412" CreationDate="2010-10-21T15:29:19.673" Score="2" ViewCount="" Body="&lt;p&gt;The area, if it exists, is 1/2.  Let the area inside the unit square be A.  Then it is made up of 8 copies at 1/3 linear size, plus the complement of a 1/3 linear size copy (the center square).  So $A=\frac{8}{9}*A+\frac{1-A}{9}$.  There is no problem having an area without a rectangle.  See the &lt;a href=&quot;http://en.wikipedia.org/wiki/Fat_Cantor_set&quot; rel=&quot;nofollow&quot;&gt;Fat Cantor Set&lt;/a&gt;.  I believe having positive area also supports dimension 2&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-10-21T15:29:19.673" CommentCount="1" />
  <row Id="7422" PostTypeId="2" ParentId="7419" CreationDate="2010-10-21T15:41:14.077" Score="1" ViewCount="" Body="&lt;p&gt;The substitution $u = \frac{y}{x}$ works. What you have is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$y = ux$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i.e., $ y' = x u' + u$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or $x y' = x^2 u' + y$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Substituting this into your equation, we get &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x^2 u' = u^3$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which can be solved using standard methods.&lt;/p&gt;&#xA;" OwnerUserId="2641" LastActivityDate="2010-10-21T15:41:14.077" CommentCount="2" />
  <row Id="7423" PostTypeId="1" AcceptedAnswerId="7424" CreationDate="2010-10-21T16:04:53.337" Score="2" ViewCount="101" Body="&lt;p&gt;Choose numbers from 1 to 2n uniformly at random.  How many numbers must be chosen, on average, before at least n have been picked?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is similar to the coupon-collector problem, but looking for only partial completion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: Choosing an appropriate meaning of 'random' is part of the question.&lt;/p&gt;&#xA;" OwnerUserId="1778" LastActivityDate="2010-10-21T16:18:01.997" Title="How long to choose n out of 2n numbers?" Tags="&lt;combinatorics&gt;&lt;probability&gt;" AnswerCount="1" />
  <row Id="7424" PostTypeId="2" ParentId="7423" CreationDate="2010-10-21T16:15:44.600" Score="4" ViewCount="" Body="&lt;p&gt;If you stop the sum from the coupon collector problem half-way, you get your answer.  It takes 1 draw on average to get the first different ticket, then $\frac{2n}{2n-1}$ draws for the second, and so on until $\frac{2n}{n+1}$ for the $n^{th}$.  So this is $2n*(H_{2n}-H_n)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, if you choose without replacement, the answer is n draws.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-10-21T16:15:44.600" CommentCount="1" />
  <row Id="7426" PostTypeId="2" ParentId="7419" CreationDate="2010-10-21T16:23:59.233" Score="1" ViewCount="" Body="&lt;p&gt;Here is my solution. Is this right? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;\text{xy$\prime $} = \left(\frac{y}{x}\right)^3+y \\ \\&#xA;\frac{y}{x}=u, \\&#xA;y = \text{ux}, \\&#xA;\text{y$\prime $} = \text{u$\prime $x} + u \\ \\&#xA;x^2\text{u$\prime $} = u^3 \\&#xA;\int \frac{1}{u^3} \, du =\int \frac{1}{x^2} \, dx \\&#xA;-\frac{1}{2u^2} = -\frac{1}{x} + c \\&#xA;u^{2 }= \frac{1}{\frac{2}{x}+c} \\&#xA;y^{2 }= \frac{x^2}{\frac{2}{x}+c} \\&#xA;y = x\left(\frac{2}{x}+c\right)^{-\frac{1}{2}}&#xA;\end{align}&lt;/p&gt;&#xA;" OwnerUserId="2674" LastEditorUserId="2674" LastEditorDisplayName="" LastEditDate="2010-10-22T00:14:37.857" LastActivityDate="2010-10-22T00:14:37.857" CommentCount="5" />
  <row Id="7427" PostTypeId="2" ParentId="7412" CreationDate="2010-10-21T16:36:45.437" Score="1" ViewCount="" Body="&lt;p&gt;Regarding your update:  Using &lt;a href=&quot;http://en.wikipedia.org/wiki/Conway_base_13_function&quot; rel=&quot;nofollow&quot;&gt;Conway's base 13 function&lt;/a&gt; or other constructions (I have one I prefer) you can find a function $f: (0,1) \rightarrow (0,1)$ such that for any interval $(a,b) \subset (0,1)$ and $y \in (0,1)$ there is an $x \in (a,b)$ such that $f(x)=y$.  Then for given $y$ the points such that $f(x)=y$ are dense in $(0,1)$.  So you have continuum many disjoint sets, each dense in $(0,1)$.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-10-21T16:36:45.437" CommentCount="2" />
  <row Id="7428" PostTypeId="2" ParentId="6877" CreationDate="2010-10-21T16:47:03.940" Score="4" ViewCount="" Body="&lt;p&gt;Combinatorics: Counting without your fingers&lt;/p&gt;&#xA;" OwnerUserId="2677" LastActivityDate="2010-10-21T16:47:03.940" />
  <row Id="7429" PostTypeId="2" ParentId="6877" CreationDate="2010-10-21T16:51:04.640" Score="2" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Numerical Analysis&lt;/strong&gt;: The search for fast, accurate, and stable algorithms for finding numerical solutions.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-10-21T16:51:04.640" />
  <row Id="7430" PostTypeId="2" ParentId="7405" CreationDate="2010-10-21T16:55:27.533" Score="1" ViewCount="" Body="&lt;p&gt;I think your last shoud be $Q(p_0)=x_1$ and $lim_{p\rightarrow p_0+} Q(p) =x_0$ with the rest OK&lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2010-10-21T17:01:45.637" LastActivityDate="2010-10-21T17:01:45.637" />
  <row Id="7431" PostTypeId="2" ParentId="7412" CreationDate="2010-10-21T17:16:17.320" Score="2" ViewCount="" Body="&lt;p&gt;I wrote a bunch of stuff about the properties of this set assuming it existed, then I realized that it wasn't even obvious that the set was well-defined.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's the problem: Does the central point belong to the set? If it does, then it does not belong to the middle block, and vice versa, leading to a contradiction. So the limiting set of the recursive process does not exist.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, looking at the picture, it feels as though the limit &quot;should&quot; exist in some expanded sense. This could possibly be related to the way that some sequences of functions which do not converge to functions do actually converge to distributions. Or I could just be wrong and the limit simply cannot exist in any sense. &lt;em&gt;Edit:&lt;/em&gt; Well, it turns out that the limiting measure contained in any rectangular region is just half the area of the rectangle. So I suppose it does have a limit in the sense of a measure on $[0,1]^2$, but it's not a very interesting measure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The obvious corresponding version of the Cantor set would be to replace every middle third with the complement of the set, rather than leaving it empty. However, the Cantor set has a very nice interpretation in terms of ternary expansions: a real number $x \in [0,1]$ is in the Cantor set if and only if its ternary expansion contains no digit 1 (allowing, for example, $1/3 = 0.1_3$ to be $0.0222\ldots_3$). For this variant, the corresponding condition would have to be that the ternary expansion contains an &lt;em&gt;even&lt;/em&gt; number of ones. This &lt;em&gt;is&lt;/em&gt; well-defined for fractions of the form $a/3^b$ whose ternary expansion terminates, but like above, it is not clear how to deal with rationals like $1/2$, or all the irrationals.&lt;/p&gt;&#xA;" OwnerUserId="856" LastEditorUserId="856" LastEditorDisplayName="" LastEditDate="2010-10-21T18:03:46.790" LastActivityDate="2010-10-21T18:03:46.790" CommentCount="7" />
  <row Id="7432" PostTypeId="1" CreationDate="2010-10-21T17:30:04.557" Score="0" ViewCount="73" Body="&lt;p&gt;I have a probability space $\mathcal{X} \times \mathcal{Y}$. $\mathcal{X}$ and $\mathcal{Y}$ could be infinite, but they are at worst case countable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a matrix $A_{xy}$ ranging over $\mathcal{X}$ for rows $\mathcal{Y}$ for columns, which is either 0 or 1 for each $x,y$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have the information that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$P(${$ (x,y) \mid A_{xy} = 1$}$) \le \epsilon$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it possible to upper bound the probability $P(${$x \mid \exists y A_{xy} = 1$}$)$, where the bound depends on $\epsilon$ and other parameters given in the question? (say, structure of $A$ perhaps... but I'd rather it be $A$-independent.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is that we can always create an $A$ such that it has $1$ in row (for each $x$ somewhere, therefore the probability that I want to bound would be 1), but still the probability of total 1s would be $\epsilon$. So I am pretty sure I am not approaching the whole problem formulation the right way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So maybe you have an idea what assumptions I can make to get a better bound.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One idea that I had is doing the following:&#xA;assume that we have $\alpha = \sup_{x,y} \frac{P(x)}{P(x,y)}$.&#xA;Then, we know that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$P(${$x \mid \exists y A_{xy} = 1$}$)  = \sum_x P(x) I(\exists y A_{xy} = 1)\le \sum_{x,y \mathrm{s.t.}\, A_{xy}=1} \alpha P(x,y) \le \alpha \epsilon $ (where $I$ is a 0 1 indicator.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The thing is that in my case, $\alpha$ could get pretty big if I really define it that way.&lt;/p&gt;&#xA;" OwnerUserId="7076" LastActivityDate="2010-11-20T18:44:19.987" Title="can I bound the following probability?" Tags="&lt;probability&gt;&lt;statistics&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="7433" PostTypeId="1" AcceptedAnswerId="7448" CreationDate="2010-10-21T17:51:54.947" Score="0" ViewCount="113" Body="&lt;p&gt;The pdf of the geomtric distribution is the following: $f(x) = (1-p)^{x}p$. Also $E[X] = \frac{1-p}{p}$ and $\text{Var}[X] = \frac{1-p}{p^2}$. The pdf of the negative binomial distribution is: $f(x) = \binom{r+x-1}{x}p^{r}(1-p)^{x}$. Also $E[X] = \frac{r(1-p)}{p}$ and $\text{Var}[X] = \frac{r(1-p)}{p^2}$. So is the geomtric distribution really a special case of the negative binomial? In each case, $X$ is the number of failures until a success occurs?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, does the geometric distribution have a different version as well? The $\binom{r+x-1}{x}$ seems analogous to $\binom{n+k-1}{k}$. &lt;/p&gt;&#xA;" OwnerUserId="2548" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-10-21T19:31:20.567" LastActivityDate="2010-10-21T19:31:20.567" Title="Geometric Distribution versus Negative Binomial Distribution" Tags="&lt;probability-theory&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="7434" PostTypeId="2" ParentId="7432" CreationDate="2010-10-21T17:59:13.067" Score="1" ViewCount="" Body="&lt;p&gt;You have a two sets of nodes, $X$ and $Y$, with $A_{i,j}=1$ when there is an edge from $i\in X$ to $j \in Y$.  You are given that only a fraction $\epsilon$ of the possible edges are occupied and want the chance that a given $i\in X$ has at least one edge occupied.  If $|Y|$ is large the chance can be quite high.  Your bound would be to recognize that the worst case is if each node in $X$ has only one edge occupied.  Then the bound is $\epsilon * |Y|$ (or 1 if smaller).  The size of $X$ doesn't matter.  Essentially the elements of $X$ are all independent.  If the sets are infinite you need to worry about how you take limits.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-10-21T17:59:13.067" />
  <row Id="7435" PostTypeId="1" CreationDate="2010-10-21T18:10:04.110" Score="2" ViewCount="106" Body="&lt;p&gt;In the past couple years, I've read many words pertaining to D-branes without feeling I have really comprehended them. In broad terms, I think I get what they're about: They're supposed to serve as habitats for the ends of open strings and can be viewed as submanifolds (of the target manifold in a sigma model), possibly augmented with a vector bundle, or a sheaf of [somethings], or maybe some other kind of label. (Please correct me if I got that wrong.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the hopes of tightening my grasp on the concept, here are some of the questions that have been nagging me during my reading.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;What &lt;em&gt;specifically&lt;/em&gt; is the definition of a D-brane, say in the context of a topological field theory? (Or what are the most promising provisional definitions?) What references are most accessible to a mathematical audience?&lt;/li&gt;&#xA;&lt;li&gt;What picture should I have in my head when an author talks about &quot;the moduli space of D-branes&quot;?&lt;/li&gt;&#xA;&lt;li&gt;What is the idea behind the &quot;dynamics of D-branes&quot; that researchers sometimes talk about? (Perhaps when I understand better how to think about these gadgets, it will be easier to conceive of how they should change over time.)&lt;/li&gt;&#xA;&lt;li&gt;What goes into verifying (or at least asserting/conjecturing) that the elements of twisted K-theory classify D-brane charges?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;(I considered putting this on MO, but it feels too basic relative to what TFT people talk about there for it to be appropriate.)&lt;/p&gt;&#xA;" OwnerUserId="1347" LastActivityDate="2010-10-21T18:10:04.110" Title="What are D-branes (in a topological field theory)?" Tags="&lt;reference-request&gt;&lt;physics&gt;" CommentCount="2" />
  <row Id="7436" PostTypeId="1" AcceptedAnswerId="7444" CreationDate="2010-10-21T18:11:13.387" Score="2" ViewCount="660" Body="&lt;p&gt;I'm having trouble finding a good explanation of the Lebesgue integral. As per the definition, it is the expectation of a random variable. Then how does it model the area under the curve? Let's take for example a function $f(x) = x^2$.  How do we find the integral of $f(x)$ under $[0,1]$ using the Lebesgue integral?&lt;/p&gt;&#xA;" OwnerUserId="957" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2011-02-23T06:33:35.173" LastActivityDate="2011-02-23T07:43:47.827" Title="Lebesgue integral basics" Tags="&lt;analysis&gt;&lt;integral&gt;&lt;probability-theory&gt;&lt;probability&gt;" AnswerCount="7" CommentCount="4" FavoriteCount="9" />
  <row Id="7437" PostTypeId="1" AcceptedAnswerId="7439" CreationDate="2010-10-21T18:20:25.740" Score="0" ViewCount="137" Body="&lt;p&gt;Prove(A∩B)’=A’∪B’.&lt;br&gt;&#xA;let x ∈ (A ∩ B)’&lt;br&gt;&#xA;∴ (x ∈ A ∩ B)’&lt;br&gt;&#xA;∴ (x ∈ A ∧ x ∈ B)’&lt;br&gt;&#xA;∴ (x ∈ A)’ ∨ (x ∈ B)’&lt;br&gt;&#xA;∴ x ∈ A’ ∨ x ∈ B’&lt;br&gt;&#xA;∴ x ∈ A’ ∪ B’&lt;br&gt;&#xA;∴ (A ∩ B)’ ⊆ A’ ∪ B’&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The above is the solution provided. Sorry if it seems trivial but I don't understand how&lt;br&gt;&#xA;∴ (x ∈ A ∧ x ∈ B)’&lt;br&gt;&#xA;leads to&lt;br&gt;&#xA;∴ (x ∈ A)’ ∨ (x ∈ B)’&lt;br&gt;&#xA;and not&lt;br&gt;&#xA;∴ (x ∈ A)’ ∧ (x ∈ B)’&lt;/p&gt;&#xA;" OwnerUserId="2679" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2010-11-20T17:54:58.943" LastActivityDate="2010-11-20T17:54:58.943" Title="Proving two sets are equal using bi-directional set inclusion" Tags="&lt;homework&gt;&lt;discrete-mathematics&gt;&lt;elementary-set-theory&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="7438" PostTypeId="2" ParentId="7436" CreationDate="2010-10-21T18:26:39.677" Score="1" ViewCount="" Body="&lt;p&gt;The definition &lt;a href=&quot;http://en.wikipedia.org/wiki/Lebesgue_measure&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; doesn't mention probability or expectation or random variable.  Intuitively, it just says the measure (in $\mathbb R^2$) is the area of the smallest set of rectangles that will cover the set.  Then for an area the Lebesgue integral is just the integral of 1 over the set.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-10-21T18:26:39.677" />
  <row Id="7439" PostTypeId="2" ParentId="7437" CreationDate="2010-10-21T18:29:37.233" Score="2" ViewCount="" Body="&lt;p&gt;This is just one of DeMorgan's Laws.  Not (A and B) is equivalent to (Not A) or (Not B), that is that (A and B) is false if either one is false.  If you go to (Not A) and (Not B) you need both to be false.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-10-21T18:29:37.233" CommentCount="1" />
  <row Id="7440" PostTypeId="2" ParentId="7436" CreationDate="2010-10-21T18:45:45.153" Score="4" ViewCount="" Body="&lt;p&gt;The Lebesgue integral is a generalization of the usual Riemann integral taught in basic calculus. If the Riemann integral of a function over a set exists then it equals the Lebesgue integral. So the Lebesgue integral of $x^2$ over $[0,1]$ is just the old $(1/3) 1^3-(1/3)0^3$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Lebesgue integral has the benefit of being defined for many more functions than the Riemann integral. Even more importantly the Lebesgue integral has useful limit properties:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Dominated_convergence_theorem&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Dominated_convergence_theorem&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Monotone_convergence_theorem&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Monotone_convergence_theorem&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The expectation of a random variable is a particular application of the Lebesgue integral where the function to be integrated is the random variable (viewed as a function on the sample space) and the integration is with respect to a probability measure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You need to look at one of the many probability and measure books for the details. My own favourites are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Pollard, &lt;em&gt;A User's Guide to Measure-Theoretic Probability&lt;/em&gt;&lt;/li&gt;&#xA;&lt;li&gt;Dudley, &lt;em&gt;Real Analysis and Probability&lt;/em&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Terence Tao has some online lecture notes:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://terrytao.wordpress.com/2010/09/09/245a-notes-1-lebesgue-measure/&quot; rel=&quot;nofollow&quot;&gt;http://terrytao.wordpress.com/2010/09/09/245a-notes-1-lebesgue-measure/&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://terrytao.wordpress.com/2010/09/19/245a-notes-2-the-lebesgue-integral/&quot; rel=&quot;nofollow&quot;&gt;http://terrytao.wordpress.com/2010/09/19/245a-notes-2-the-lebesgue-integral/&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://terrytao.wordpress.com/2010/01/01/254a-notes-0-a-review-of-probability-theory/&quot; rel=&quot;nofollow&quot;&gt;http://terrytao.wordpress.com/2010/01/01/254a-notes-0-a-review-of-probability-theory/&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1195" LastActivityDate="2010-10-21T18:45:45.153" />
  <row Id="7441" PostTypeId="1" AcceptedAnswerId="7463" CreationDate="2010-10-21T18:57:37.303" Score="5" ViewCount="287" Body="&lt;p&gt;For $x&gt;0$ and $y&gt;x+1$, how do we prove that $$\sum\limits_{n=1}^{\infty} \frac{x(x+1) \cdots (x+n-1)}{y(y+1) \cdots (y+n-1)} = \frac{x}{y-x-1}$$&lt;/p&gt;&#xA;" LastActivityDate="2010-10-21T22:50:42.490" Title="Summation of $\sum\limits_{n=1}^{\infty} \frac{x(x+1) \cdots (x+n-1)}{y(y+1) \cdots (y+n-1)}$" Tags="&lt;real-analysis&gt;&lt;analysis&gt;&lt;sequences-and-series&gt;" AnswerCount="4" />
  <row Id="7442" PostTypeId="2" ParentId="7436" CreationDate="2010-10-21T19:04:35.317" Score="3" ViewCount="" Body="&lt;p&gt;One of my graduate school professors, Erhan Cinlar, used to give the following analogy to explain the intuitive difference between the Lebesgue integral and the Riemann integral.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose you have a pile of coins of different denominations, and you want to know how much money you have.  The Riemann integral is like picking up the coins, one-by-one, and adding the denomination of each to a running total.  The Lebesgue integral is like sorting the coins by denomination first, and then getting the total by multiplying each denomination by how many you have of that denomination and then adding up those numbers.  The methods are different, but you obtain the same result by either method.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the same way, when both the Riemann integral and the Lebesgue integral are defined, they give the same value.  As others have said, though, there are functions for which the Lebesgue integral is defined but the Riemann integral is not, and so in that sense the Lebesgue integral is more general than the Riemann.&lt;/p&gt;&#xA;" OwnerUserId="2370" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-10-22T04:12:51.830" LastActivityDate="2010-10-22T04:12:51.830" CommentCount="2" />
  <row Id="7443" PostTypeId="2" ParentId="7436" CreationDate="2010-10-21T19:05:34.983" Score="0" ViewCount="" Body="&lt;p&gt;You may also like to refer, these two books:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;A radical approach to Lebesgue Theory of Integration: Bressoud&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Real Analysis by G.B. Folland.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;As Jyotirmoy pointed out, Lebesgue integral is the generalization of Riemann Integral. There are shortcomings of the Riemann Integral, due to which the Lebesgue integral, was discovered. A rigorous definition of the Lebesgue integral needs, you to know what a Simple Function is, and you can read more on this at &lt;a href=&quot;http://en.wikipedia.org/wiki/Lebesgue_integration&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Lebesgue_integration&lt;/a&gt;&lt;/p&gt;&#xA;" LastActivityDate="2010-10-21T19:05:34.983" />
  <row Id="7444" PostTypeId="2" ParentId="7436" CreationDate="2010-10-21T19:09:34.727" Score="18" ViewCount="" Body="&lt;p&gt;As has been noted, the usual definition of the Lebesgue integral has little to do with probability or random variables (though the notions of measure theory and the integral can then be applied to the setting of probability, where under suitable interpretations it will turn out that the (Lebesgue) integral of (a certain) functions corresponds to the expectation of (a certain) random variable).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But this is not the &lt;em&gt;origin&lt;/em&gt; of the Lebesgue integral. Here is an intuitive idea of what the Lebesgue integral is, as compared to the Riemann integral.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Recall from Calculus the idea behind the Riemann integral: the integral $\int_a^b f(x)\,dx$ is meant to represent the net signed area between the $x$-axis, the graph of $y=f(x)$, and the lines $x=a$ and $x=b$. The way we attempt to do this is by breaking up the domain, $[a,b]$, into subintervals $[a=x_0,x_1]$, $[x_1,x_2],\ldots,[x_{n-1},x_n=b]$. Then, on each subinterval $[x_i,x_{i+1}]$ we pick a point $x_i^*$, and we estimate the area under the graph of the function with the rectangle of height $f(x_i^*)$ and base $[x_i,x_{i+1}]$. This leads to the Riemann sums&#xA;$$ \sum_{i=0}^{n-1} f(x_i^*)(x_{i+1}-x_i)$$&#xA;as estimates of the area under the graph. We then consider finer and finer partitions of $[a,b]$ and take limits to estimate the area.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lebesgue's idea was that instead of partitioning the domain, we will partition the &lt;em&gt;range&lt;/em&gt;; if the function takes values between $c$ and $d$, we can divide the range $[c,d]$ into subintervals $[c=y_0,y_1]$, $[y_1,y_2],\ldots,[y_{m-1},y_m=d]$. Then, we let $E_i$ be the set of all points in $[a,b]$ whose value under $f$ lies between $y_i$ and $y_{i+1}$. That is,&#xA;$$ E_i = f^{-1}([y_i,y_{i+1}]) = \{ x\in[a,b]\,|\, y_i \leq f(x) \leq y_{i+1}\}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we have a way of assigning a &quot;size&quot; to $E_i$, call it its &quot;measure&quot; $\mu(E_i)$, then the portion of the graph of $y=f(x)$ that lies between the horizontal lines $y=y_i$ and $y=y_{i+1}$ will be $A$, where,&#xA;$$ y_i\mu(E_i) \leq A \leq y_{i+1}\mu(E_i).$$&#xA;So Lebesgue suggests to approximate the the area by picking a number $y_i^*$ between $y_i$ and $y_{i+1}$, and considering the sums&#xA;$$ \sum_{i=0}^{n-1} \mu(E_i)y_i^*.$$&#xA;Then consider finer and finer partitions of $[c,d]$, and this gives finer and finer approximations of of the area by these sums. The Lebesgue integral will be the limit of these sums. (The analogy given by Mike Spivey is very apt for the distinction between partitioning the domain and partitioning the range to find the sum.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But in order for this to make sense, we need to develop a way of measuring fairly intricate subsets of the line, so that we can compute $\mu(E_i)$. So we first develop a way of doing this; turns out that if you accept the Axiom of Choice, then it is impossible to come up with a way of measuring that will (i) assign to an interval $[a,b]$ the &quot;measure&quot; $b-a$; (ii) will be invariant under translation, so so that if $F=E+c = \{e+c | e\in E\}$ then $\mu(F)=\mu(E)$; (iii) will be countably additive: if $E = \cup_{i=1}^{\infty}E_i$ and the $E_i$ are pairwise disjoint, then $\mu(E) = \sum\mu(E_i)$; and (iv) every subset of the line will have a well-defined (possibly infinite) measure. (If you don't accept the Axiom of Choice, then there are models of the reals where we &lt;em&gt;can&lt;/em&gt; achieve this). So one drops the restriction (iv), and constructs a measure for which some sets will be &quot;too weird&quot; to have a measure. We then restrict attention to certain kinds of functions (called the measurable functions), which are the ones for which the sets we get in the process described above are all measurable sets. And then we define the Lebesgue integral for those functions, following the idea described above (but one does not define it exactly that way; instead the usual way is to describe $f$ as a limit of functions for which the integral is easy, and then compute the integral of $f$ as a limit of the integrals that are easy).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For your function, $f(x)=x^2$, this is fairly easy: the value all lie between $0$ and $1$, so say that we break up the range into subintervals of length $1/n$, so $y_i = i/n$, $i=0,\ldots,n$. Then &#xA;$$f^{-1}([y_i,y_{i+1}]) = f^{-1}([i/n, (i+1)/n]) = [\sqrt{i/n},\sqrt{(i+1)/n}],$$&#xA;so the $n$th estimate, picking $y_i^* = y_i = i/n$ is just&#xA;$$\sum_{i=0}^n (i/n)\left(\sqrt{(i+1)/n} - \sqrt{i/n}\right).$$&#xA;Take the limit as $n\to\infty$, and you will get that the limit is $\frac{1}{3}$, as expected. (I will spare you the details; see the end of this answer for a high-power way of getting the answer similar to the way you do it with the Riemann integral).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It turns out that not every function is Lebesgue-integrable, just like not every function is Riemann-integrable. But every function that is Riemann-integrable will &lt;em&gt;also&lt;/em&gt; be Lebesgue integrable, &lt;em&gt;and&lt;/em&gt; the value of its Lebesgue integral will be the same as the value of its Riemann integral. But there are functions that are &lt;em&gt;not&lt;/em&gt; Riemann-integrable but &lt;em&gt;are&lt;/em&gt; Lebesgue-integrable (for example, the characteristic function of the rationals is Lebesgue-integrable, with integral $0$ over any interval, but is not Riemann-integrable). We also have a &quot;Fundamental Theorem of Calculus&quot; for the Lebesgue Integral:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&lt;strong&gt;Theorem.&lt;/strong&gt; If $F$ is a differentiable function, and the derivative $F'$ is bounded on the interval $[a,b]$, then $F'$ is Lebesgue integrable on $[a,b]$ and&#xA;$$\int_a^x F'\,d\mu = F(x) - F(a).$$&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here, the integral is the Lebesgue integral. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, to &lt;em&gt;finally&lt;/em&gt; answer the question you ask about your example, since $F(x)=\frac{x^3}{3}$ is a differentiable function whose derivative is bounded over any finite interval, in particular over $[0,1]$, then from this theorem you can deduce that the integral over the interval $[0,1]$ of the derivative $F'(x)=x^2$ is equal to $F(1)-F(0)$; that is,&#xA;$$\int_0^1 x^2\,d\mu = \int_0^1 \left(\frac{x^3}{3}\right)'\,d\mu = \frac{1}{3} - \frac{0}{3} =\frac{1}{3}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I recommend the book &lt;em&gt;A Garden of Integrals&lt;/em&gt; by Frank E. Burk (Dolciani Mathematical Expositions 31, MAA, 2007, ISBN 9-780883-853375); it discusses and compares the Cauchy integral, the Riemann integral, the Riemann-Stieltjes integral, the Lebesgue integral, the Lebesgue-Stieltjes integral, and the Henstock-Kurzweil integral; it also discusses the Wiener and Feynman integral. I just finished reading it recently.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-10-22T03:19:27.760" LastActivityDate="2010-10-22T03:19:27.760" CommentCount="5" />
  <row Id="7446" PostTypeId="1" AcceptedAnswerId="7447" CreationDate="2010-10-21T19:11:38.707" Score="7" ViewCount="361" Body="&lt;p&gt;In &lt;a href=&quot;http://meta.math.stackexchange.com/questions/901/request-for-reopening-equivalence-of-curved-and-straight-lines/903#903&quot;&gt;this answer on meta&lt;/a&gt;, Pete L. Clark said:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I think the question concerns the idea that a differentiable curve becomes more and more like a straight line segment the closer one zooms in on its graph. (And I must say that I regard part of this confusion as an artifact of badly written recent calculus books who describe this phenomenon as &quot;local linearity&quot;. Ugh!)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So, what's wrong with calling it &quot;local linearity&quot;?  (Examples of the specific language from some relatively recent books follow.)&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;From Finney, Demana, Waits, and Kennedy's &lt;em&gt;Calculus: Graphical, Numerical, Algebraic&lt;/em&gt;, 1st ed, p107:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A good way to think of differentiable functions is that they are &lt;strong&gt;locally linear;&lt;/strong&gt; that is, a function that is differentiable at &lt;em&gt;a&lt;/em&gt; closely resembles its own tangent line very close to &lt;em&gt;a&lt;/em&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;From Hughes-Hallett, Gleason, et al's &lt;em&gt;Calculus: Single Variable&lt;/em&gt;, 2nd ed, pp138-9:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;When we zoom in on the graph of a differentiable function, it looks like a straight line.  In fact, the graph is not exactly a straight line when we zoom in; however, its deviation from straightness is so small that it can't be detected by the naked eye.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Following that, there is discussion of the tangent line approximation, then a theorem titled &quot;Differentiability and Local Linearity&quot; (the first time &quot;local linearity&quot;/&quot;locally linear&quot; appears) stating that if a function &lt;em&gt;f&lt;/em&gt; is differentiable at &lt;em&gt;a&lt;/em&gt;, then the limit as &lt;em&gt;x&lt;/em&gt; goes to &lt;em&gt;a&lt;/em&gt; of the quotient of the error in the tangent line approximation and the difference between &lt;em&gt;x&lt;/em&gt; and &lt;em&gt;a&lt;/em&gt; goes to 0.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Ostebee and Zorn's &lt;em&gt;Calculus from Graphical, Numerical, and Symbolic Points of View&lt;/em&gt;, 1st ed, p110:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Remarkably, the just-illustrated strategy of zooming in to estimate slope almost always works.  Zooming in on the graph of almost any calculus function $f$, at almost any point $(a,f(a))$, eventually produces what looks like a straight line with slope $f'(a)$.  A function with this property is sometimes called &lt;strong&gt;locally linear&lt;/strong&gt; (or &lt;strong&gt;locally straight&lt;/strong&gt;) at $x=a$. [Margin note: &lt;em&gt;These aren't formal definitions, just descriptive phrases.&lt;/em&gt;]  Local linearity says, in effect, that $f$ &quot;looks like a line&quot; near $x=a$ and therefore has a well-defined slope at $x=a$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;(I did not find the term &quot;local linearity&quot; or &quot;locally linear&quot; at a quick glance in Stewart's &lt;em&gt;Calculus: Concepts and Contexts&lt;/em&gt;, 2nd ed, or Leithold's &lt;em&gt;The Calculus 7&lt;/em&gt;; the rest of the calculus books I have on hand predate the inclusion of graphing calculators/software in textbooks, so are not suitable for comparison.)&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-10-22T04:30:33.770" Title="Is &quot;locally linear&quot; an appropriate description of a differentiable function?" Tags="&lt;calculus&gt;&lt;education&gt;&lt;teaching&gt;" AnswerCount="3" CommentCount="8" FavoriteCount="2" />
  <row Id="7447" PostTypeId="2" ParentId="7446" CreationDate="2010-10-21T19:25:54.090" Score="8" ViewCount="" Body="&lt;p&gt;None of the books you quoted actually called differentiability &quot;local linearity&quot;; they just used as a good analogy.  It is a good analogy, but it is not a good definition.  A lot of mathematical terminology, especially from topology, uses the word &quot;local,&quot; but it is almost always used with the same purpose.  A locally compact set is one where points have arbitrarily small compact neighborhoods.  A locally connected set is one where points have arbitrarily small connected neighborhoods.  On the other hand, a non-linear differentiable function is &lt;em&gt;not linear on any neighborhood&lt;/em&gt; of a point about which it is differentiable.  It just looks linear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, such functions do have tangent lines, which is an equivalent definition of differentiability.  I don't think that a tangent counts as local linearity, though.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're asking whether people learning calculus should be taught the words &quot;local linearity&quot; instead of &quot;differentiability,&quot; I mean, that isn't really much more helpful than using the analogy but keeping the terminology.  If you're asking whether we should use the analogy at all, I don't see why not, as long as it's clear that linearity is just an approximation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;: &quot;Locally linear&quot; only describes differentiable functions $\mathbb{R}\rightarrow\mathbb{R}$.  &quot;Differentiable&quot; can be extended to functions between arbitrary differential manifolds.  I think it's better to keep the extensible definition.&lt;/p&gt;&#xA;" OwnerUserId="2215" LastEditorUserId="2215" LastEditorDisplayName="" LastEditDate="2010-10-21T19:55:22.090" LastActivityDate="2010-10-21T19:55:22.090" CommentCount="7" />
  <row Id="7448" PostTypeId="2" ParentId="7433" CreationDate="2010-10-21T19:29:33.753" Score="1" ViewCount="" Body="&lt;p&gt;The geometric distribution describes the probability of &quot;x trials are made before a success&quot;, and the negative binomial distribution describes that of &quot;x trials are made before $r$ successes are obtained&quot;, where $r$ is fixed. So you see that the latter is a particular case of the former, namely, when $r=1$.&#xA;As to your last question, it doesn't matter whether you call the variable $x$ or $k$, as long as you keep in mind that it takes on the values $0,1,2...$.&lt;/p&gt;&#xA;" OwnerUserId="1313" LastActivityDate="2010-10-21T19:29:33.753" />
  <row Id="7449" PostTypeId="1" AcceptedAnswerId="7452" CreationDate="2010-10-21T19:30:57.073" Score="1" ViewCount="30" Body="&lt;p&gt;Given that a population (eg. of bacteria) grows exponentially with the function&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$p 2^{\frac{t}{10}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $p$ is the initial population and $t$ is the time in minutes. Given that the population consumes (eg. food) at a steady rate, how would you calculate how much the population consumes?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume that each bacterium consumes $1µg$ of food per minute, how much food would an initial population of &lt;strong&gt;100&lt;/strong&gt; bacteria have consumed after 10 minutes? The solution lies somewhere between the consumption of 100 and 200 bacteria over 10 minutes, thus between $1000µg$ and $2000µg$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I figured I needed an integral to solve this problem. If I'm not mistaken, the integral of the function above is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_0^{10} p 2^{\frac{t}{10}} dt = \frac{10 p 2^{\frac{t}{10}}}{\ln{2}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $p=100$ the result is $2885.39µg$, which is higher than the expected maximum of $2000µg$. Where did I go wrong?&lt;/p&gt;&#xA;" OwnerUserId="2680" LastActivityDate="2010-10-21T19:43:22.280" Title="Consumption of an exponentially growing population" Tags="&lt;integral&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="7451" PostTypeId="2" ParentId="7436" CreationDate="2010-10-21T19:33:38.973" Score="1" ViewCount="" Body="&lt;p&gt;The Riemann integral is pretty good and very intuitive, however the main reason to consider other types of integrals is that &quot;the space of functions that are Riemann integrable&quot;, say $R(I)$ where $I\subset\mathbb{R}^n$ is compact, is to small (it is a linear space in the sense you can add them and multiply by constants). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you just look at a piecewise continuous function that vanish outside a bounded region and then you can go on with the Riemann integral. In mathematical analysis we look at various kinds of limits of functions and we would like the limit functions to stay in &quot;the space&quot; (we want the space to be complete). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;About the best we can do in the Riemann case is to look at uniformly convergent sequences $f_n$ on a compact interval $I\subset\mathbb{R}^n$ - in that case the limit $\lim f_n\in R(I)$  and $\lim\int f_n =\int \lim f_n$. However, uniform convergent is very rare! (Many Fourier series are not continuous even though there partial sums are, etc..).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Lebesgue integral can be constructed in several ways (ending up with the same space though). A first try might be to start with norming $R(I)$, $\|f\|=\int|f|$ and then we would get a distance between $f,g\in R(I)$ by $\|f-g\|$, thus turning up with a metric space which we may complete by adding all possible limits - this will not work however because even though $R(I)$ is small it is to large (there are unbounded functions such that $\|f\|=\infty$). A better start would be to look at $C(I)$ = the space of continuous functions on (the compact set) $I$, (certainly each $f\in R(I)$ is a point-wise limit  of $C(I)$ functions) if we norm $C(I)$ in the same manner we would be a normed space and the completion of that space is $L^1(I)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In $L^1$ you can sure take limits in norm and moreover, as has already been pointed out in other answers, you have many other better limit theorems such as Lebesgue dominated theorem or the monotone convergence theorem. Also, bounded functions of $R(I)$ do belong to $L^1(I)$.&lt;/p&gt;&#xA;" OwnerUserId="1154" LastActivityDate="2010-10-21T19:33:38.973" />
  <row Id="7452" PostTypeId="2" ParentId="7449" CreationDate="2010-10-21T19:43:22.280" Score="1" ViewCount="" Body="&lt;p&gt;You have a definite integral&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\int_{0}^{10} p 2^{\frac{t}{10}} dt = \left[ \frac{10p2^{\frac{t}{10}}}{ln 2} \right]_{0}^{10} = \frac{10p}{ln2} = 1442.69 \mu g $ of food consumed.&lt;/p&gt;&#xA;" OwnerUserId="2641" LastActivityDate="2010-10-21T19:43:22.280" CommentCount="3" />
  <row Id="7453" PostTypeId="2" ParentId="7441" CreationDate="2010-10-21T19:57:10.563" Score="7" ViewCount="" Body="&lt;p&gt;The sum telescopes since the summand $\rm\ f_n = g_{n+1} - g_n\:$ where $\rm\displaystyle\ g_n = \frac{1-n-y}{y-x-1}\ f_n$&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-10-21T20:21:59.007" LastActivityDate="2010-10-21T20:21:59.007" CommentCount="4" />
  <row Id="7454" PostTypeId="1" CreationDate="2010-10-21T20:04:33.807" Score="14" ViewCount="460" Body="&lt;p&gt;A cohort in a school consists of 75 students who study for 6 years.  Each year, the students are randomly distributed into 3 classrooms of 25 students each.  What is the probability that, after 6 years, each student has at some point been in a classroom with every other student?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More generally:  Starting with an edgeless (undirected) graph on &lt;em&gt;cn&lt;/em&gt; vertices, a &lt;strong&gt;round&lt;/strong&gt; consists of first randomly partitioning the vertices into &lt;em&gt;c&lt;/em&gt; disjoint sets of &lt;em&gt;n&lt;/em&gt; vertices each, then adding an edge between every pair of not-yet-joined vertices that lie in the same set.  What is the probability that, after &lt;em&gt;y&lt;/em&gt; rounds, the result is a complete graph on &lt;em&gt;cn&lt;/em&gt; vertices?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have estimates and solutions to special cases, and it's straightforward to find&#xA;the probability that a single given student sees all the others, but I don't know how to tackle the question in general.  (I do have a very pretty but completely useless expression for the exact answer, which I can supply if there's interest.)  In the case &lt;em&gt;c&lt;/em&gt;=3, &lt;em&gt;n&lt;/em&gt;=25, &lt;em&gt;y&lt;/em&gt;=6 it's clear that the answer is &quot;so close to zero that nobody can tell the difference&quot; but I was hoping for a more precise result.  Any guidance appreciated.&lt;/p&gt;&#xA;" OwnerUserId="2682" LastActivityDate="2010-11-06T13:36:21.480" Title="What is the probability that every pair of students studies together at some point?" Tags="&lt;combinatorics&gt;&lt;probability&gt;&lt;graph-theory&gt;" AnswerCount="2" CommentCount="5" FavoriteCount="2" />
  <row Id="7455" PostTypeId="2" ParentId="7436" CreationDate="2010-10-21T20:18:54.660" Score="1" ViewCount="" Body="&lt;p&gt;You may want to consider the following sources:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Lebesgue's theory of integration: its origins and development - Thomas Hawkins&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://books.google.com/books?id=QnXSqvTiEjYC&amp;amp;printsec=frontcover&amp;amp;dq=The+Calculus+Gallery+Masterpieces+from+Newton+to+Lebesgue&amp;amp;hl=en&amp;amp;ei=Jp3ATPT0I4voOfG25YwM&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=1&amp;amp;ved=0CCsQ6AEwAA#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;The calculus gallery: masterpieces from Newton to Lebesgue - William Dunham&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://books.google.com/books?id=qgiLag9dPpMC&amp;amp;pg=PR3&amp;amp;dq=Carter+M.,+van+Brunt+B.+The+Lebesgue-Stieltjes+integral+-++A+practical+introduction&amp;amp;hl=en&amp;amp;ei=Mp7ATPz6G4SYOruT7ZcM&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=1&amp;amp;ved=0CCkQ6AEwAA#v=onepage&amp;amp;q=Carter%20M.%2C%20van%20Brunt%20B.%20The%20Lebesgue-Stieltjes%20integral%20-%20%20A%20practical%20introduction&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;The Lebesgue-Stieltjes integral: a practical introduction - Michael Carter, Bruce Van Brunt&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1747" LastActivityDate="2010-10-21T20:18:54.660" />
  <row Id="7456" PostTypeId="2" ParentId="7441" CreationDate="2010-10-21T20:33:41.847" Score="2" ViewCount="" Body="&lt;p&gt;This is not a stand-alone proof like the very nice one provided by Bill, but one can note that the sum is a special case of &lt;a href=&quot;http://en.wikipedia.org/wiki/Hypergeometric_function#Special_values_at_z.C2.A0.3D.C2.A01&quot; rel=&quot;nofollow&quot;&gt;this identity for the hypergeometric function ${}_2F_1$&lt;/a&gt;. Let $a=x$, $b=1$, $c=y$, use $\Gamma(z+1)/\Gamma(z)=z$, and subtract 1 to compensate for the fact that the hypergeometric series starts with $n=0$.&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2010-10-21T20:33:41.847" />
  <row Id="7457" PostTypeId="1" AcceptedAnswerId="7458" CreationDate="2010-10-21T20:54:14.590" Score="0" ViewCount="98" Body="&lt;p&gt;I'm trying to remember a particular theorem of ZF but unfortunately my memory is quite incomplete.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The theorem is of the form (some set operation) is either (expected answer) or the empty set.&lt;/li&gt;&#xA;&lt;li&gt;If the Axiom of Choice holds, (some set operation) is (expected answer).&lt;/li&gt;&#xA;&lt;li&gt;I believe that if (some set operation) is (expected answer) for all infinite sets, then the Axiom of Choice holds -- that is, the statement is equivalent to AC.&lt;/li&gt;&#xA;&lt;li&gt;The theorem has a formalized proof at &lt;a href=&quot;http://us.metamath.org/&quot; rel=&quot;nofollow&quot;&gt;Metamath&lt;/a&gt;.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I think it was something like $|S\times S|=|S|$ for infinite sets $S$, or $\bigcup S$ is... something... I'm just not sure.  Does this ring a bell?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My apologies for the lack of well-defined information, and all the more for the possibility that some part of this is misleading.  I figured that, short of MathOverflow, this is just about the best place I could come to find my answer.&lt;/p&gt;&#xA;" OwnerUserId="1778" LastActivityDate="2010-10-21T21:02:31.000" Title="Specific equivalent to the Axiom of Choice involving the empty set" Tags="&lt;set-theory&gt;&lt;axiom-of-choice&gt;" AnswerCount="1" />
  <row Id="7458" PostTypeId="2" ParentId="7457" CreationDate="2010-10-21T20:56:04.380" Score="3" ViewCount="" Body="&lt;p&gt;I believe what you are looking for is that the Cartesian product of non-empty sets is non-empty.  It is equivalent to AC.  Without AC you cannot be sure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Tarski's theorem, which you quote in the form of cardinals:  for infinite $S$ there is a bijective map between $S×S$ and $S$ is also an equivalent.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2010-10-21T21:02:31.000" LastActivityDate="2010-10-21T21:02:31.000" CommentCount="4" />
  <row Id="7459" PostTypeId="1" AcceptedAnswerId="7507" CreationDate="2010-10-21T21:03:45.427" Score="2" ViewCount="106" Body="&lt;p&gt;I have a polygon P, that may or may not be convex. Is there an algorithm that will enable me to find the collection of points A that are at a distance less than d from P?&#xA;Is A in turn always a polygon?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does the solution change materially if we try to solve the problem on the surface of a sphere instead of on a Euclidean plane?&lt;/p&gt;&#xA;" OwnerUserId="2684" LastActivityDate="2011-02-07T19:24:07.557" Title="Find all points with a distance less than d to a (potentially not convex) polygon" Tags="&lt;geometry&gt;&lt;euclidean-geometry&gt;" AnswerCount="3" />
  <row Id="7460" PostTypeId="2" ParentId="7459" CreationDate="2010-10-21T21:16:10.443" Score="3" ViewCount="" Body="&lt;p&gt;It will not be a polygon.  If you think about the original polygon being a square of side s, the set A is a square of side s+2d, but with the corners rounded.  The corners become quarter circles with radius d and centered on the original corners of the square.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a general polygon the situation is much the same.  Draw a parallel to the sides offset by d.  Then round the outer corners with a circular arc of radius d centered on the original corners and tangent to the new parallels.  The meeting points will be the intersection of the new line parallel to one side and the extension of the other side of the corner.  The inner corners will stay corners but get less deep and eventually disappear if d is large enough..&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-10-21T21:16:10.443" />
  <row Id="7461" PostTypeId="2" ParentId="7459" CreationDate="2010-10-21T21:41:37.660" Score="3" ViewCount="" Body="&lt;p&gt;I heard Ravi Vakil give a series of plenary talks at an MAA conference on this question.  The talks were entitled &quot;The Mathematics of Doodling.&quot;  He generalized the question, though, by asking what happens in the limit when you start with some set in the plane and then iterate the process of finding the set of points at a distance $r$ from the current set.  Some interesting mathematics came out as he showed that the resulting sequence of points gets more and more circular.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's a good picture of him doing this &lt;a href=&quot;http://www.maa.org/news/121407vakil.html&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; on the MAA website.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;He also has a series of related problems that he developed for the Stanford Math Circle available &lt;a href=&quot;http://math.stanford.edu/circle/notes06f/doodling.pdf&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also according to his website, he has an article, &quot;The Mathematics of Doodling,&quot; that will be appearing in the February 2011 issue of the &lt;em&gt;American Mathematical Monthly&lt;/em&gt;.  (Update: The article has just appeared in print.  The rest of the reference is Vol. 118(2), pp. 116-129.)&lt;/p&gt;&#xA;" OwnerUserId="2370" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2011-02-07T19:24:07.557" LastActivityDate="2011-02-07T19:24:07.557" />
  <row Id="7462" PostTypeId="2" ParentId="7441" CreationDate="2010-10-21T21:52:07.323" Score="0" ViewCount="" Body="&lt;p&gt;Nice observation Hans. I wanted to derive a very similar sum:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle \sum\limits_{n=1}^{\infty} \frac{x(x+1) \cdots (x+n-1)}{y(y+1) \cdots (y+n-1)}=\sum \limits_{n=1}^{\infty}\displaystyle  \frac{x^{(n)}}{y^{(n)}}$ where $x^{(n)}$ is the rising factorial which can be written as $x^{(n)} = \displaystyle  \frac{\Gamma(x+n)}{\Gamma(x)}$. Then we have: $\displaystyle  \frac{\Gamma(y)}{\Gamma(x)}\sum\limits_{n=1}^{\infty}\frac{\Gamma(x+n)}{\Gamma(y+n)}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Apparently, this sum converges to $\displaystyle \frac{\Gamma(x+1)\Gamma(y-x-1)}{\Gamma(y)\Gamma(y-x)}$. After simplication we obtain $\displaystyle  \frac{x}{y-x-1}$. However, I haven't been able to derive the result for  $\displaystyle \sum\limits_{n=1}^{\infty}\frac{\Gamma(x+n)}{\Gamma(y+n)}$&lt;/p&gt;&#xA;" OwnerUserId="2569" LastActivityDate="2010-10-21T21:52:07.323" />
  <row Id="7463" PostTypeId="2" ParentId="7441" CreationDate="2010-10-21T22:43:55.433" Score="6" ViewCount="" Body="&lt;p&gt;This is based on Robert Smith's observation above and Robin Chapman's beta trick in some previous problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{\Gamma{(x+n)}\Gamma{(y-x)}}{\Gamma{(y+n)}} = \int_{0}^{1} t^{x+n-1} (1-t)^{y-x-1} dt$, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;summing over $n$  we get,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Gamma{(y-x)}\sum_{n \geq 1}\frac{\Gamma{(x+n)}}{\Gamma{(y+n)}} = \int_{0}^{1} \sum_{n \geq 1} t^{x+n-1} (1-t)^{y-x-1} dt = \int_{0}^{1} t^x (1-t)^{y-x-2}dt, $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sum_{n \geq 1}\frac{\Gamma{(x+n)}}{\Gamma{(y+n)}} = \frac{1}{\Gamma(y-x)}\int_{0}^{1} t^{x+1-1}(1-t)^{y-x-1-1}dt = \frac{\Gamma{(x+1)}\Gamma{(y-x-1)}}{\Gamma{(y-x)}\Gamma(y)} = \frac{\Gamma{(x+1)}}{(y-x-1)\Gamma(y)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and hence,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{\Gamma(y)}{\Gamma(x)} \sum_{n \geq 1}\frac{\Gamma{(x+n)}}{\Gamma{(y+n)}} = \frac{\Gamma(x+1)}{(y-x-1)\Gamma(x)} = \frac{x}{y-x-1}.$&lt;/p&gt;&#xA;" OwnerUserId="404" LastEditorUserId="404" LastEditorDisplayName="" LastEditDate="2010-10-21T22:50:42.490" LastActivityDate="2010-10-21T22:50:42.490" CommentCount="2" />
  <row Id="7464" PostTypeId="1" AcceptedAnswerId="7530" CreationDate="2010-10-21T22:48:55.670" Score="2" ViewCount="292" Body="&lt;p&gt;The question &lt;a href=&quot;http://math.stackexchange.com/questions/571/what-is-the-optimum-angle-of-projection-when-throwing-a-stone-off-a-cliff&quot;&gt;what-is-the-optimum-angle-of-projection-when-throwing-a-stone-off-a-cliff&lt;/a&gt; was asked and answered a while back.  This one has a much cleaner answer.  Now you are on a uniform slope, a line through the origin that is not horizontal and you want to throw a stone as far as possible.  As a function of the angle of the slope, what angle should you throw at?  Usual assumptions:  uniform gravity, no friction.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-10-22T01:30:58.460" LastActivityDate="2010-10-23T03:09:06.060" Title="What is the optimal angle when throwing a stone on a slope?" Tags="&lt;calculus&gt;&lt;trigonometry&gt;&lt;physics&gt;&lt;optimization&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" />
  <row Id="7465" PostTypeId="1" CreationDate="2010-10-21T23:49:03.207" Score="1" ViewCount="177" Body="&lt;p&gt;List all the elements of the alternating group A_3 (A sub 3) written in cyclic notation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I come up with &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Identity (1)&#xA;Obviously (123)&lt;/p&gt;&#xA;" LastActivityDate="2010-10-22T00:08:07.760" Title="Elements of alternating group A_3 (A sub 3)" Tags="&lt;abstract-algebra&gt;" AnswerCount="2" />
  <row Id="7466" PostTypeId="2" ParentId="7465" CreationDate="2010-10-21T23:59:28.463" Score="0" ViewCount="" Body="&lt;p&gt;Hint: If a group is cyclic, then find an element that is not the identity and find the group that is generated by that element. That is a subset of your group (and even more so a subgroup, albeit unrelated here). And if the group generated is the size of your desired group (you should know the size of $A_3$), then you have all the elements within your group. If you are unfamiliar with cyclic notation please reference the &lt;a href=&quot;http://en.wikipedia.org/wiki/Cycle_notation&quot; rel=&quot;nofollow&quot;&gt;Wikipedia article&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1016" LastActivityDate="2010-10-21T23:59:28.463" CommentCount="1" />
  <row Id="7467" PostTypeId="2" ParentId="7465" CreationDate="2010-10-22T00:08:07.760" Score="1" ViewCount="" Body="&lt;p&gt;(1) (123) (132)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;got it&lt;/p&gt;&#xA;" LastActivityDate="2010-10-22T00:08:07.760" />
  <row Id="7468" PostTypeId="1" AcceptedAnswerId="7472" CreationDate="2010-10-22T00:21:51.960" Score="1" ViewCount="60" Body="&lt;p&gt;Denoted as $\zeta(s,a)$ for a &gt; 0&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Where do I find topics on the Hurwitz zeta function for a &amp;lt; 0? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Any links or resources would be appreciated.&#xA;(Please dont mention wiki or mathworld)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="2320" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-10-22T00:47:31.233" LastActivityDate="2010-10-22T01:19:54.827" Title="On functions similar to Hurwitz zeta function" Tags="&lt;special-functions&gt;&lt;zeta-function&gt;&lt;analytic-number-theory&gt;" AnswerCount="1" />
  <row Id="7469" PostTypeId="1" CreationDate="2010-10-22T00:23:03.087" Score="1" ViewCount="81" Body="&lt;p&gt;Consider the group of permutation matrices G ={I3, P1, P2, P3, P4, P5}&#xA;For n = 3 the permutation matrices are I3 and the five matrices are:&#xA;P1 = [1,0,0;0,0,1;0,1,0]&#xA;P2 = [0,1,0;1,0,0;0,0,1]&#xA;P3 = [0,1,0;0,0,1;1,0,0]&#xA;P4 = [0,0,1;0,1,0;1,0,0]&#xA;P5 = [0,0,1;1,0,0;0,1,0]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Write out the elements of a group of permutations that is isomorphic to G, and exhibit an isomorphism from G to this group!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think it has to do with Cayley's Theorem.&#xA;With f_a:G-&gt;G defined by f_a(x) = ax for each a that exists in G...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I thought about making a table, but realize I don't know how to since I am dealing with matrices. &lt;/p&gt;&#xA;" LastActivityDate="2010-10-22T00:27:36.747" Title="Consider group of permutation matrices and write out elements isomorphic to the group and exhibit it." Tags="&lt;abstract-algebra&gt;" AnswerCount="1" />
  <row Id="7470" PostTypeId="2" ParentId="7469" CreationDate="2010-10-22T00:27:36.747" Score="1" ViewCount="" Body="&lt;p&gt;HINT: Look at the result of multiplying each of those matrices by the vector $(1,2,3)^T$.&lt;/p&gt;&#xA;" OwnerUserId="1313" LastActivityDate="2010-10-22T00:27:36.747" CommentCount="6" />
  <row Id="7471" PostTypeId="1" CreationDate="2010-10-22T00:44:53.897" Score="1" ViewCount="110" Body="&lt;p&gt;I have a homework question &quot;Show the following is true using theorems. State which theorem you use at each step.&quot; This is just one of many problems I have! So, if you can help me with this one problem than I can apply what I learn to finish the rest. I don't want a handout; I just don't know where to start or what to do. I have looked through the different theorems in the book and I don't see how any would apply to this!&#xA;Am i suppose to prove that the first part  = the second?&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;A'BD' + BCD + ABC' + AB'D = A'BD' + BCD + ABC' + AB'D + BC'D' + A'BC + ABD&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Should I group two or more and work from there? Since, we are only working with '+' then I should only need these types of theorems?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a book that shows the different theorems. But it still doesn't help me tackle this problem. Are we to prove how the first part equals the second part? To me that's expanding and not reducing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can work either side i guess whichever is easier..&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a list of theorems the book provides&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;X + 0 = X&#xA;X + 1 = 1&#xA;X + X = X&#xA;X + X' = 1&#xA;X + y = Y + x&#xA;(X + Y) + Z = X +(Y + Z)&#xA;x times (X + Y) = x&#xA;X times Y + X times Y' = x&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;these are most of them. there are a few others..&#xA;The book i am using is digital logic - principles and practices.&lt;/p&gt;&#xA;" OwnerUserId="2688" LastEditorUserId="2688" LastEditorDisplayName="" LastEditDate="2010-10-22T02:20:37.667" LastActivityDate="2010-10-22T02:20:37.667" Title="boolean algebra - theorems" Tags="&lt;boolean-algebra&gt;&lt;duality-theorems&gt;" CommentCount="13" />
  <row Id="7472" PostTypeId="2" ParentId="7468" CreationDate="2010-10-22T00:47:00.593" Score="2" ViewCount="" Body="&lt;p&gt;Well, there's the &lt;a href=&quot;http://dlmf.nist.gov/25.11&quot; rel=&quot;nofollow&quot;&gt;DLMF&lt;/a&gt; and the &lt;a href=&quot;http://functions.wolfram.com/ZetaFunctionsandPolylogarithms/Zeta2/&quot; rel=&quot;nofollow&quot;&gt;Wolfram Functions&lt;/a&gt; site... which people really should be checking out first when they encounter an unfamiliar &quot;special function&quot;.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-10-22T00:47:00.593" CommentCount="3" />
  <row Id="7473" PostTypeId="1" CreationDate="2010-10-22T00:47:26.863" Score="5" ViewCount="673" Body="&lt;p&gt;For all $a, m, n \in \mathbb{Z}^+$, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\gcd(a^n - 1, a^m - 1) = a^{\gcd(n, m)} - 1$$&lt;/p&gt;&#xA;" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-04T13:20:43.707" LastActivityDate="2011-03-24T03:22:28.687" Title="Prove that $\gcd(a^n - 1, a^m - 1) = a^{\gcd(n, m)} - 1$" Tags="&lt;number-theory&gt;&lt;homework&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="7474" PostTypeId="2" ParentId="7329" CreationDate="2010-10-22T00:53:42.100" Score="0" ViewCount="" Body="&lt;p&gt;Player 1 throws on odd throws and player 2 throws on even throws. This is a geometric distribution. So his distribution is $\left(\frac{8}{9} \right)^{2k} \cdot \frac{1}{9}$ and player 2's distribution is $\left(\frac{8}{9} \right)^{2k-1} \cdot \frac{1}{9}$. So player 1 must give $1/8$. &lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2010-10-22T00:53:42.100" />
  <row Id="7475" PostTypeId="1" CreationDate="2010-10-22T00:59:09.117" Score="0" ViewCount="214" Body="&lt;p&gt;Write out the elements of group permutation that is isomorphic to U(14)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;G ={I3, P1, P2, P3, P4, P5} For n = 3 the permutation matrices are I3 and the five matrices are:&#xA;P1 = [1,0,0;0,0,1;0,1,0]&#xA;P2 = [0,1,0;1,0,0;0,0,1]&#xA;P3 = [0,1,0;0,0,1;1,0,0]&#xA;P4 = [0,0,1;0,1,0;1,0,0]&#xA;P5 = [0,0,1;1,0,0;0,1,0]&lt;/p&gt;&#xA;" LastActivityDate="2010-12-26T20:44:27.977" Title="Find elements of group permutation that is isomorphic to U(14)" Tags="&lt;abstract-algebra&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="7476" PostTypeId="1" CreationDate="2010-10-22T01:04:23.040" Score="5" ViewCount="103" Body="&lt;p&gt;Let $\mathbb{T}^n = { (z_1,\ldots,z_n) \in \mathbb{C}^n : |z_l| = 1, \; 1 \leq l \leq n }$ denote the $n$-torus, and let $t_1, \ldots, t_n$ be arbitrary real numbers. Then it can be shown that the topological closure $H$ in $\mathbb{T}^n$ of the one-parameter subgroup&#xA;$$H' = {(e^{2\pi i t_1 y}, \ldots, e^{2\pi i t_n y}) \in \mathbb{T}^n : y \in \mathbb{R}}$$&#xA;is an $r$-dimensional subtorus of $\mathbb{T}^n$, where $0 \leq r \leq n$ is the dimension over $\mathbb{Q}$ of the span of $t_1, \ldots, t_n$, and furthermore that&#xA;$$\lim_{Y \to \infty} \frac{1}{Y} \int^{Y}_{0}{g(e^{2\pi i t_1 y}, \ldots, e^{2\pi i t_n y}) \: dy} = \int_{H}{g(z) \: d\mu_H(z)}$$&#xA;for any continuous function $g : \mathbb{T}^n \to \mathbb{C}$, where $\mu_H$ is the normalised Haar measure on $H$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is, for which sets $B \subset \mathbb{T}^n$ do we have that&#xA;$$\lim_{Y \to \infty} \frac{1}{Y} \int\limits_{{y \in [0,Y] : (e^{2\pi i t_1 y}, \ldots, e^{2\pi i t_n y}) \in B}}{dy} = \mu_H(B).&#xA;$$&#xA;I have a feeling that some sort of approximation by continuous functions should tell you, but I can't seem to make it work for some reason.&lt;/p&gt;&#xA;" OwnerUserId="2689" LastActivityDate="2010-10-22T01:04:23.040" Title="Ergodic flow in tori" Tags="&lt;measure-theory&gt;&lt;fourier-analysis&gt;" FavoriteCount="2" />
  <row Id="7478" PostTypeId="1" CreationDate="2010-10-22T01:30:16.043" Score="1" ViewCount="108" Body="&lt;p&gt;Find an isomorphism from octic group G to group G'&lt;/p&gt;&#xA;&#xA;&lt;p&gt;G = {e, s, s^2, s^3, b, g, d, t}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;e =  (1)  identity &#xA;s =  (1234)&#xA;s^2= (13)(24)&#xA;s^3= (1432)&#xA;b =  (14)(23)&#xA;g =  (24)&#xA;d =  (12)(34)&#xA;t =  (13)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;G' = {I2, R, R^2, R^3, H, D, V, T}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I2 = [1,0;0,1]  identity matrix&#xA;R =  [0,-1;1,0]&#xA;R^2= [-1,0;0,-1]&#xA;R^3= [0,1;-1,0]&#xA;H =  [1,0;0,-1]&#xA;D =  [0,1;1,0]&#xA;T =  [0,-1;-1,0]&lt;/p&gt;&#xA;" LastActivityDate="2010-10-22T13:31:48.637" Title="Find an isomorphism from octic group G to group G'" Tags="&lt;abstract-algebra&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="7479" PostTypeId="2" ParentId="7473" CreationDate="2010-10-22T01:51:26.423" Score="1" ViewCount="" Body="&lt;p&gt;Below is a proof which has the neat feature that it immediately specializes&#xA;to a proof of the integer Bezout identity for $\rm\:x = 1\:$. It allows us to interpret the result as a &lt;strong&gt;q-analog of the integer Bezout identity&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;E.g. for $\rm\ m,\:n\ =\ 15,\:21$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\displaystyle\quad\quad\quad\quad\quad\quad\quad\quad\ \frac{x^3-1}{x-1}\ =\ (x^{15} + x^9 + 1)\ \frac{x^{15}-1}{x-1}\ -\ (x^9+x^3)\ \frac{x^{21}-1}{x-1}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for  $\rm\ x = 1\ $  specializes to  $\ 3\ =\ \ 3\ (15)\ -\ 2\ (21)\:,\ $ i.e. $\ (3)\ =\ (15,21)$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;DEFINITION&lt;/strong&gt; $\rm\displaystyle \quad n' \: :=\ \frac{x^n - 1}{x-1}\:$. $\quad$ &lt;strong&gt;NOTE&lt;/strong&gt; $\rm\quad  n' = n\ $  for  $\rm\ x = 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;THEOREM&lt;/strong&gt; $\rm\quad (m',n')\ =\ ((m,n)')\ $ for naturals $\rm\:m,\:n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;PROOF&lt;/strong&gt; $\ $ It is trivially true if $\rm\ m = n\ $  or  if  $\rm\ m = 0\ $  or  $\rm\ n = 0\:$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence suppose  $\rm\:n &gt; m &gt; 0\:$. We proceed by induction on  $\rm\:n + m\:$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since  $\rm\:\ x^n - 1\  =\  x^r\ (x^m - 1)\  +\  x^r - 1\ \ $  for  $\rm\ \ r = n - m$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;we obtain  $\rm\quad\: n'\  =\  x^r\ m'\ +\ r' \quad $   via  dividing the above by  $\rm\: x-1 $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So $\rm\ \ \ (m', n')\ =\ (m', r')\quad\ \ \ $   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\quad\quad\quad\quad\quad\quad\quad = \  ((m,r)') \quad\ $   via  induction: $\rm\ m+r = n &amp;lt; m+n$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\quad\quad\quad\quad\quad\quad\quad = \  ((m,n)')\quad\:$   via  $\rm\ r = n\ \:(mod\ m)\quad\quad\quad\quad\quad$ &lt;strong&gt;QED&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;COROLLARY&lt;/strong&gt; $\ $ Integer Bezout Theorem $\ $ Proof: set $\rm\ x = 1\ $ above: i.e. erase primes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A deeper understanding comes when one studies Divisibility Sequences&#xA;and Divisor Theory.&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-10-22T01:51:26.423" />
  <row Id="7480" PostTypeId="2" ParentId="7232" CreationDate="2010-10-22T02:20:34.060" Score="1" ViewCount="" Body="&lt;p&gt;Simply put, the tensions are equal because that's the way the problem is set up--it's a given, and you're supposed to use this information to solve for something.&lt;/p&gt;&#xA;" OwnerUserId="2527" LastActivityDate="2010-10-22T02:20:34.060" CommentCount="1" />
  <row Id="7481" PostTypeId="1" CreationDate="2010-10-22T02:23:39.170" Score="4" ViewCount="315" Body="&lt;p&gt;The question is:&#xA;Give an example of a function $f$, continuous on $S=(0,1) \times (0,1)$, such that $f(S)=\mathbb{R}^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm getting stuck on $f(x)=\tan\left(\pi \left(x-\frac{1}{2}\right)\right)$,&#xA;which covers all of $\mathbb{R}$ but not all of $\mathbb{R}^2$.&lt;/p&gt;&#xA;" OwnerUserId="2319" LastEditorUserId="856" LastEditorDisplayName="" LastEditDate="2010-10-22T02:45:23.580" LastActivityDate="2010-11-08T12:49:21.850" Title="Example of a continuous function from $(0,1) \times (0,1)$ to $\mathbb{R}^2$" Tags="&lt;real-analysis&gt;" AnswerCount="1" CommentCount="9" FavoriteCount="1" />
  <row Id="7482" PostTypeId="2" ParentId="7446" CreationDate="2010-10-22T02:35:32.310" Score="3" ViewCount="" Body="&lt;p&gt;@Paul and @Issac's comments on a wrong answer I had initially posted made me see why &quot;linear on zooming in&quot; is the right picture of differentiability. Suppose $f$ is some function and &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$g(x)=f(a)+m(x-a)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is a proposed linear approximation to it at a point $a$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose I zoom in on the graph of $f$ and $g$ by a multiple $\lambda&gt;0$ and look at a point $h$ distance apart from $a$ in my zoomed coordinates. Then the vertical error in the approximation in my zoomed coordinates is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$e(\lambda)=\lambda[f(a+h/\lambda)-g(a+h/\lambda)]&#xA;=\lambda[f(a+h/\lambda)-f(a)]-mh$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I keep zooming the approximation will look better and better only if $\lim_{\lambda \to \infty} e(\lambda)=0$. This will be the case only if $f$ is differentiable at $a$ and $f'(a)=m$.&lt;/p&gt;&#xA;" OwnerUserId="1195" LastEditorUserId="1195" LastEditorDisplayName="" LastEditDate="2010-10-22T02:52:26.653" LastActivityDate="2010-10-22T02:52:26.653" CommunityOwnedDate="2010-10-22T02:52:26.653" />
  <row Id="7484" PostTypeId="2" ParentId="7475" CreationDate="2010-10-22T03:37:02.520" Score="1" ViewCount="" Body="&lt;p&gt;If U(14) means the group of units modulo 14 under multiplication, then the units are {1,3,5,9,11,13}.  (as on page 2 of this document:  &lt;a href=&quot;http://math.bu.edu/people/rpollack/Teach/541fall09/HW3_Solutions.pdf&quot; rel=&quot;nofollow&quot;&gt;http://math.bu.edu/people/rpollack/Teach/541fall09/HW3_Solutions.pdf&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint:  3 generates U(14), and therefore has order |U(14)|=6.  Since the group you're seeking is isomorphic, it is also generated by an element of order 6.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS: there are no elements of order 6 in the symmetric group of order 3 (so 3 by 3 permutation matrices won't work).&lt;/p&gt;&#xA;" OwnerUserId="139" LastActivityDate="2010-10-22T03:37:02.520" CommentCount="1" />
  <row Id="7486" PostTypeId="1" CreationDate="2010-10-22T03:48:22.513" Score="0" ViewCount="656" Body="&lt;p&gt;I am trying to make a truth table from an SOP boolean algebra expression. I understand AND, OR, NOT truth tables. I just dont understand these types of tables and there outputs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the expression&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;A'BD' + BCD + ABC' + AB'D = A'BD' + BCD + ABC' + AB'D  + BC'D' + A'BC + ABD&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I can use either side whichever is easier.. just let me know what side.&#xA;Would A' be a 1 and the others be a zero? I am also not sure how they get the output?&#xA;I understand the outputs of a AND, OR truth tables. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But i cant figure out these outputs. would this be considered an OR table since the expression is &quot;+&quot;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;would i just construct A, B, D with nots = 1 or zero?&#xA;Than, how do i determine the output?&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; -----------------------&#xA; A  | B  | D  |  output&#xA;-----------------------&#xA;| 1 | 0  | 1  |   1?    |  A'BD'&#xA;------------------------&#xA;| 0 | 0  | 0  |   0?    |  BCD&#xA;------------------------&#xA;| 0 | 0  | 1  |   1?    |  ABC'&#xA;-------------------------&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;something like that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What i am trying to achieve is how the below expression is true using theorems.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;A'BD' + BCD + ABC' + AB'D = A'BD' + BCD + ABC' + AB'D  + BC'D' + A'BC + ABD&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="2688" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-10-22T15:06:37.087" LastActivityDate="2010-10-23T03:24:38.780" Title="how to make a truth table from an boolean expression" Tags="&lt;boolean-algebra&gt;" AnswerCount="4" />
  <row Id="7487" PostTypeId="2" ParentId="6722" CreationDate="2010-10-22T03:57:54.393" Score="1" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://aleph0.clarku.edu/~djoyce/java/elements/elements.html&quot; rel=&quot;nofollow&quot;&gt;link text&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would first really grasp his elements to add math into your report. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just for fun check out how many postage stamps were done in his name around the world.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://jeff560.tripod.com/stamps.html&quot; rel=&quot;nofollow&quot;&gt;link text&lt;/a&gt;&lt;/p&gt;&#xA;" LastActivityDate="2010-10-22T03:57:54.393" CommentCount="3" />
  <row Id="7488" PostTypeId="2" ParentId="7486" CreationDate="2010-10-22T04:02:39.280" Score="1" ViewCount="" Body="&lt;p&gt;Your truth table will have 2^4=16 lines.  Start with four columns labeled A,B,C,D.  On each line put a different combination of truth values for A,B,C, and D.  Then add more columns for the pieces of your expression.  So to build up the left hand side, make a column A'B and put in the truth values of that combination.  The prime on A is not and the multiply is AND.  So this column would show (NOT A) AND B for the values of A and B on that line.  Continue adding columns for other terms until you have built up the entire left hand side.  Then start on the terms that make up right hand side.  When you have built up the right hand side, compare the left hand side column and the right hand side column.  If they agree on all 16 lines, your equation is correct.  For a simple example, say we want to prove (A+B)'=A'B'  The table would look like this&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/LIcli.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the (A+B)' column and the A'B' column agree everywhere, the equation is correct.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-10-22T04:02:39.280" CommentCount="1" />
  <row Id="7489" PostTypeId="2" ParentId="7486" CreationDate="2010-10-22T04:02:42.633" Score="0" ViewCount="" Body="&lt;p&gt;I think the &quot;using theorems&quot; part of the question is intended to prevent you from using the exhaustive &quot;try every possibility&quot; technique (so don't use this technique if it's a homework question).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, this technique will work fine, you can just run through the $2^4=16$ possible values of (A,B,C,D) and check that both sides of the equation match.  This will be a rigorous proof of the identity -- just not a particularly intelligent one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming I understand the notation correctly:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;concatenation (i.e. writing the letters next to each other) is short for &quot;and&quot;, for example ABC is &quot;A and B and C&quot;,&lt;/li&gt;&#xA;&lt;li&gt;addition (i.e. +) is short for &quot;or&quot;, for example AB+CD is &quot;(A and B) or (C and D)&quot;&lt;/li&gt;&#xA;&lt;li&gt;adding a dash ' is the notation for negation, i.e. &quot;not&quot;, for example AB' is &quot;A and (not B)&quot;.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So A'BD' is the same as (not A) and B and (not D).  In your table, the first row is wrong since (not A) is false (because A is true=1).  The second row is correct since B is false=0 (regardless of what value C has).  The third row is wrong since A is false=0 (again, regardless of what values C has).&lt;/p&gt;&#xA;" OwnerUserId="139" LastActivityDate="2010-10-22T04:02:42.633" CommentCount="7" />
  <row Id="7490" PostTypeId="2" ParentId="7446" CreationDate="2010-10-22T04:30:33.770" Score="4" ViewCount="" Body="&lt;p&gt;The &quot;microscope&quot; or &quot;zooming in&quot; explanation of differentiation is basically nonsense. It contradicts the notion of zooming in actually used in calculus (asymptotic expansion of a function near a point, revealing finer and finer information).  Magnification matches the mathematics only at order 0 and 1, that is, it captures linear information but also destroys the higher order picture.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose that $f(0)=0$. The $N$-fold magnification of the graph near 0 is the plot of $Nf(x/N)$.  This is the straight line $xf'(0)$ to within an error that shrinks to an invisible size as the picture is &quot;zoomed&quot; by increasing $N$.   This will, literally, linearize any differentiable function.  The higher order approximations -- whose effect is beautifully visible in a &lt;em&gt;fixed scale&lt;/em&gt; picture of the first several Taylor polynomials for $f(x)$ near 0 all superimposed on the same coordinate system -- are made indistinguishable from a straight line by flattening everything with the magnification.  This is a step backward considering that the difference between a tangent line and an osculating circle is easily illustrated with any curve, and many people see this picture in some form before learning calculus.    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What one actually wants to display are the effects &lt;em&gt;away&lt;/em&gt; from the point, with the higher approximations being visibly closer to graph over a longer range.   The &quot;zooming&quot; picture may be defensible if presented together with these more informative images, but by itself as an explanation of the derivative as a local linear approximation, it is using diagrams that tell a story opposite to the mathematical concepts involved.&lt;/p&gt;&#xA;" OwnerUserId="467" LastActivityDate="2010-10-22T04:30:33.770" />
  <row Id="7491" PostTypeId="1" CreationDate="2010-10-22T07:40:53.127" Score="1" ViewCount="100" Body="&lt;p&gt;If $X\sim \Gamma(a,\sigma_x^2)$ and $Y\sim \Gamma(b,\sigma_y^2)$. What will be the probability density function of R? Where $R=\frac{X+C}{X+Y}$, here $C$ is a positive constant, $\Gamma(.,.)$ denotes standard gamma probability density function and '$\sim$' represents 'distributed as'. X and Y are independent random variables.   &lt;/p&gt;&#xA;" OwnerUserId="1273" LastEditorUserId="1273" LastEditorDisplayName="" LastEditDate="2010-10-22T08:09:24.967" LastActivityDate="2010-10-26T00:43:34.470" Title="what will be the distribution of ratio of correlated gamma distributed random variables?" Tags="&lt;probability-theory&gt;&lt;statistics&gt;&lt;probability&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="7492" PostTypeId="1" CreationDate="2010-10-22T07:57:31.737" Score="0" ViewCount="193" Body="&lt;p&gt;Let $S \subset L^2(\mathbb{R})$, and let us define $S^{\perp} =\{f \in L^2(\mathbb R) | \left&amp;lt; f,g\right&gt; =0, g \in S\}$. Show that $S^{\perp}$ is a vector subspace of $L^2$.&lt;/p&gt;&#xA;" OwnerUserId="2723" LastActivityDate="2010-10-22T15:50:37.417" Title="Show that $S^{\perp}$ is a vector subspace of $L^2$." Tags="&lt;linear-algebra&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="7493" PostTypeId="1" CreationDate="2010-10-22T08:03:37.243" Score="1" ViewCount="248" Body="&lt;p&gt;Can you solve rudin problem #19, I'm having a lot of difficult with it &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've proven the first party namely lim N-&gt;infinity Sum n=1 to N exp(ik(x+nalpha)=1/(2pi)Integral[-pi,pi]… = { 1 if k=0 and 0 otherwise&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I want to prove if $f$ is continuous in $\mathbb{R}$ and $f(x+2\pi)=f(x)$ for all $x$ then &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\lim_{N-&gt;\infty} \sum\limits_{n=1}^{N} f(x+n\alpha)=\frac{1}{2\pi} \int\limits_{-\pi}^{\pi}f(t) dt$$ for any $x$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\alpha/\pi$ is irrational&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've tried writing it as lim n-&gt;infinity 1/N Sum n=1 to N Sum k=0 to N 1/(2pi)integral[-pi,pi]e^(ikt)f(x+nalpha) but that was no help&lt;/p&gt;&#xA;" OwnerUserId="2143" LastActivityDate="2010-10-22T08:09:52.783" Title="Fejer's Theorem (Problem in Rudin)" Tags="&lt;real-analysis&gt;" CommentCount="6" />
  <row Id="7494" PostTypeId="1" CreationDate="2010-10-22T08:18:48.517" Score="0" ViewCount="387" Body="&lt;p&gt;compute the fourier transform of f(t) = sin(t). Does that converge?&lt;/p&gt;&#xA;" OwnerUserId="2723" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-10-22T08:20:53.283" LastActivityDate="2010-10-22T12:52:50.863" Title="Compute the fourier transform of f(t) = sin(t)?" Tags="&lt;complex-analysis&gt;" AnswerCount="2" CommentCount="6" FavoriteCount="1" />
  <row Id="7496" PostTypeId="2" ParentId="7494" CreationDate="2010-10-22T09:21:13.797" Score="-1" ViewCount="" Body="&lt;p&gt;$sin(t)=\frac{e^{it}-e^{-it}}{2i}$. Since, $e^{it}$ and $e^{-it}$ are basis function of the fourier transform. From this easily can be computed using duality property and there will be no convergence issue. &lt;/p&gt;&#xA;" OwnerUserId="1273" LastEditorUserId="1273" LastEditorDisplayName="" LastEditDate="2010-10-22T10:24:50.793" LastActivityDate="2010-10-22T10:24:50.793" CommentCount="2" />
  <row Id="7497" PostTypeId="1" AcceptedAnswerId="7501" CreationDate="2010-10-22T09:28:06.820" Score="3" ViewCount="449" Body="&lt;p&gt;Hi I have just found the proof that 90 equals 95 and was wondering if I have made some mistake. If so, which step in my proof is not true?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Definitions:&lt;/strong&gt;&lt;br&gt;&#xA; 1. $\angle ABC=90^{\circ}$&lt;br&gt;&#xA; 2. $\angle BCD=95^{\circ}$&lt;br&gt;&#xA; 3. $|AB|=|CD|$&lt;br&gt;&#xA; 4. $M:=$ the center of $BC$&lt;br&gt;&#xA; 5. $N:=$ the center of $AD$&lt;br&gt;&#xA; 6. $l:=$ a line perpendicular to $BC$ passing through $M$&lt;br&gt;&#xA; 7. $m:=$ a line perpendicular to $AD$ passing through $N$&lt;br&gt;&#xA; 8. $S:=$ is the cross-section of $l$ and $m$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Based on definitions 1 through 8 we can draw the following image:&lt;/strong&gt;&#xA;&lt;img src=&quot;http://i.imgur.com/CBKzz.jpg&quot; alt=&quot;90 = 95&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Based on the definitions we can derive the following:&lt;/strong&gt;&lt;br&gt;&#xA; 9. $\triangle BSC$ is isosceles (follows from 4, 6 and 8)&lt;br&gt;&#xA; 10. $\triangle ASD$ is isosceles (follows from 5, 7 and 8)&lt;br&gt;&#xA; 11. $|BS|=|CS|$ (follows from 9)&lt;br&gt;&#xA; 12. $|AS|=|DS|$ (follows from 10)&lt;br&gt;&#xA; 13. $\triangle ABS\cong\triangle DCS$ (follows from 3, 11 and 12)&lt;br&gt;&#xA; 14. $\angle ABS=\angle DCS$ (follows from 13)&lt;br&gt;&#xA; 15. $\angle CBS=\angle BCS$ (follows from 9)&lt;br&gt;&#xA; 16. $\angle ABC=\angle ABS - \angle CBS=\angle DCS-\angle BCS=\angle BCD$ (follows from 14 and 15)&lt;br&gt;&#xA; 17. $90^{\circ}=95^{\circ}$ (follows from 1, 2 and 16)   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: point $S$ is indeed lying above $BC$. If however it would be below $BC$, then the 'minus' in step 16 would simply have to be changed into a 'plus'. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also note: The image is not drawn to scale. It only serves as to provide the reader with a intuitive view of the proof.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also also note: before posting, I have first investigated if this type of question would be appropriate. Based on &lt;a href=&quot;http://math.stackexchange.com/questions/287/how-come-32-5-31-5&quot;&gt;How come 32.5 = 31.5?&lt;/a&gt; and the following meta &lt;a href=&quot;http://meta.math.stackexchange.com/questions/68/questions-about-math-jokes/70#70&quot;&gt;Questions about math jokes&lt;/a&gt; I have decided to post this question.&lt;/p&gt;&#xA;" OwnerUserId="2262" LastEditorUserId="2262" LastEditorDisplayName="" LastEditDate="2010-10-25T07:21:44.120" LastActivityDate="2010-10-25T07:21:44.120" Title="What is wrong in my proof that 90 = 95? Or is it correct?" Tags="&lt;geometry&gt;&lt;euclidean-geometry&gt;&lt;puzzle&gt;" AnswerCount="1" CommentCount="11" FavoriteCount="1" />
  <row Id="7498" PostTypeId="2" ParentId="7494" CreationDate="2010-10-22T10:19:22.380" Score="2" ViewCount="" Body="&lt;p&gt;If you consider $f:=\sin$ as a function on $\mathbb R$ then it has a Fourier transform only in the distribution sense, as remarked in H.L.'s comment. If, however, you consider $\sin$ as a function on $\mathbb R/(2\pi)$ then its Fourier transform $\hat f$ is a function on $\mathbb Z$, namely the doubly infinite sequence $(c_k)_{k\in\mathbb Z}$ obtained by integration of $\sin$ against the functions $t\mapsto e^{-ikt}$: $$c_k={1\over 2\pi}\int_{-\pi}^\pi \sin(t)e^{-ikt}dt\qquad (k\in\mathbb Z).\qquad(*)$$ As $\sin$ is a nice function Fourier theory guarantees that we can get $\sin (t)$ back using the $c_k$ as follows:  $$\sin (t)=\sum_{k=-\infty}^\infty  c_k e^{ikt}.$$ Now, by Euler's formula, $\sin(t) =(e^{it}-e^{-it})/(2i)$, so we can immediately read off $\hat f(1)=c_1=-i/2$, $\hat f(-1)=c_{-1}= i/2$ and $\hat f(k)=c_k=0$ otherwise, without doing the integration (*).&lt;/p&gt;&#xA;" OwnerUserId="1303" LastEditorUserId="1303" LastEditorDisplayName="" LastEditDate="2010-10-22T12:52:50.863" LastActivityDate="2010-10-22T12:52:50.863" CommentCount="2" />
  <row Id="7499" PostTypeId="1" AcceptedAnswerId="9509" CreationDate="2010-10-22T10:40:11.833" Score="2" ViewCount="144" Body="&lt;p&gt;I have a question pertaining to linear separability with hyperplanes in a support vector machine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to &lt;a href=&quot;http://en.wikipedia.org/wiki/Support_vector_machineclassifier.&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;...formally, a support vector machine&#xA;  constructs a hyperplane or set of&#xA;  hyperplanes in a high or infinite&#xA;  dimensional space, which can be used&#xA;  for classification, regression or&#xA;  other tasks. Intuitively, a good&#xA;  separation is achieved by the&#xA;  hyperplane that has the largest&#xA;  distance to the nearest training data&#xA;  points of any class (so-called&#xA;  functional margin), since in general&#xA;  the larger the margin the lower the&#xA;  generalization error of the&#xA;  classifier.classifier.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The linear separation of classes by hyperplanes intuitively makes sense to me. And I think I understand linear separability for two-dimensional geometry. However, I'm implementing an SVM using a popular SVM library (libSVM) and when messing around with the numbers, I fail to understand how an SVM can create a curve between classes, or enclose central points in category 1 within a circular curve when surrounded by points in category 2 if a hyperplane in an n-dimensional space V is a &quot;flat&quot; subset of dimension n − 1, or for two-dimensional space - a 1D line.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Here is what I mean:&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/wqZMd.png&quot; alt=&quot;circularly enclosed class separation for a 2D binary SVM&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's not a hyperplane. That's circular. How does this work? Or are there more dimensions inside the SVM than the two-dimensional 2D input features?&lt;/p&gt;&#xA;" OwnerUserId="1832" LastActivityDate="2010-11-09T00:36:34.167" Title="Help me understand linearly separability in a binary SVM" Tags="&lt;algorithms&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="7500" PostTypeId="2" ParentId="7492" CreationDate="2010-10-22T10:47:51.643" Score="3" ViewCount="" Body="&lt;p&gt;I think you mean, show that  $S^\bot = \lbrace f \in L^2| \int_\mathbb{R} fg dx = 0 , \forall g \in S \rbrace$ is a vector subspace of $L^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $f, g  \in S^\bot$ and $\alpha, \beta \in \mathbb{R}$, then for $h \in S$,  $\int_\mathbb{R} (\alpha f + \beta g)h dx = \alpha\int_\mathbb{R}fh dx + \beta\int_\mathbb{R}gh dx = \alpha 0 + \beta 0 = 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore $\alpha f + \beta g \in S^\top$, and so $S^\bot$ is a vector subspace.&lt;/p&gt;&#xA;" OwnerUserId="780" LastActivityDate="2010-10-22T10:47:51.643" />
  <row Id="7501" PostTypeId="2" ParentId="7497" CreationDate="2010-10-22T11:00:35.287" Score="4" ViewCount="" Body="&lt;p&gt;Your statement 16 depends on the way in which the diagram is drawn to know that ∠DCS = ∠DCB + ∠BCS.&lt;/p&gt;&#xA;" OwnerUserId="72" LastEditorUserId="856" LastEditorDisplayName="" LastEditDate="2010-10-22T16:00:55.150" LastActivityDate="2010-10-22T16:00:55.150" CommentCount="5" />
  <row Id="7503" PostTypeId="1" CreationDate="2010-10-22T11:56:16.970" Score="-4" ViewCount="250" Body="&lt;p&gt;Use fourier transform and convolution to show that  $$\int_{-\infty}^{\infty}\frac{(\sin(t))^3}{t^3}dt=\frac{3\pi}{4}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;convolution is defined by $$(f * g)(t) = \int_{-\infty}^{\infty}(f(s)*g(t-s)ds $$. Fourier transform defined by $$\int_{-\infty}^{\infty}f(t)e^{-i\omega t}dt$$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="2723" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2010-11-21T06:26:42.560" LastActivityDate="2010-11-21T06:26:42.560" Title="show that $$\int_{-\infty}^{\infty}\frac{(\sin(t))^3}{t^3}dt=\frac{3\pi}{4}$$" Tags="&lt;calculus&gt;&lt;fourier-analysis&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="7505" PostTypeId="2" ParentId="7486" CreationDate="2010-10-22T13:12:27.797" Score="3" ViewCount="" Body="&lt;p&gt;A, B, C, and D are boolean variables, meaning that each takes the value&#xA;&quot;true&quot; or &quot;false&quot;.  More complex expressions have value &quot;true&quot; or &quot;false&quot;&#xA;depending on the values of these variables, so for example A'BD' is true if&#xA;A is false, B is true, and D is false, and C is either true or false.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To say that F = G, where F and G are complex expressions, means that, no&#xA;matter what values the boolean variables have, the value of F is the same&#xA;as the value of G (that is, F and G are either both true or both false).&#xA;So, for example, we have&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;     AB + AC = A(B+C)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;because if A is true and either B or C is true, then both sides are true,&#xA;and in any other case both sides are false---there's no way to assign&#xA;values to A, B, C such that the two side come out differently.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As previous posters have mentioned, you can always prove (or disprove!)  an&#xA;equality by going through all possible assignments of &quot;true&quot; and &quot;false&quot; to&#xA;the variables.  The goal seems to be to prove &quot;by theorems&quot;, that is, using&#xA;operations previously proven true.  As you say, we can either manipulate&#xA;one side of the equation until it has the form of the other side, or we&#xA;could manipulate both sides and get them into a common form.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case, the first thing to notice is that the right-hand side (RHS)&#xA;is a copy of the LHS with some extra stuff tacked onto the end.  For this&#xA;reason it's simplest to manipulate just the RHS to get rid of the&#xA;surplussage!  We can start with this basic theorem:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;  if Q is true whenever P is true, then Q = Q + P&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;We can prove this theorem by systematically considering all possibilities&#xA;for P and Q.  Or look at it this way:  If Q is true, then both sides of the&#xA;equation are true.  And if Q is false, then P must be false (since, by&#xA;assumption, if P were true Q would be true) hence both sides are false.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given this theorem we prove:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;  XY + X'Z = XY + X'Z + YZ     &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Proof:  Let P be YZ and let Q be XY+X'Z.  Suppose P is true, that is, Y and&#xA;Z are both true.  But if Y and Z are both true, then XY+X'Z must be true.&#xA;(Reason:  If X is true then, since Y is true, XY is true.  And if X is&#xA;false then, since Z is true, X'Z is true.  So in either case XY+X'Z is&#xA;true.)  So we've shown that Q is true whenever P is true, hence by the&#xA;previous theorem Q = Q + P, which in this case is what we wanted to prove.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we can prove&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; A'D' + AC' = A'D' + AC' + C'D'&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This is just the same as the previous theorem, putting A for X, D' for Y,&#xA;and C' for Z.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This last theorem implies&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; B(A'D' + AC') = B(A'D' + AC' + C'D')&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;which is the same as&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; A'BD' + ABC' = A'BD' + ABC' + BC'D'&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Given this, we can take the RHS of the original and substitute A'BD' + ABC'&#xA;for A'BD' + ABC' + BC'D', which is to say, we can drop the term BC'D'.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With steps similar to the above we can prove these two theorems:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; A'BD' + BCD = A'BD' + BCD + A'BC&#xA; BCD + ABC' = BCD + ABC' + ABD&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;which permit us to drop the final two terms of the RHS of the original,&#xA;completing the proof.&lt;/p&gt;&#xA;" OwnerUserId="2682" LastActivityDate="2010-10-22T13:12:27.797" CommentCount="2" />
  <row Id="7506" PostTypeId="1" AcceptedAnswerId="8192" CreationDate="2010-10-22T13:14:28.230" Score="2" ViewCount="95" Body="&lt;p&gt;Let $A$ be an abelian category. There are various types of full subcategories. I often wonder if it is assumed that these are nonempty, since in most proofs this is used implicitely, but also the empty case then works as a special case. To be sure, I ask you:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;$S \subseteq A$ is called thick iff $S$ is closed under subquotients and extensions. Note that $\emptyset$ is thick according to this definition. Is this allowed in the literature you know? Remark that $S$ is nonempty iff $0 \in S$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$S \subseteq A$ is called topologizing iff $S$ is closed under subquotients and direct sums. The same as above.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$S \subseteq A$ is called Serre iff $S = S^{-}$. Here $S^-$ consists of those objects, such that every nonzero subquotient of it contains a nonzero subobject isomorphic to an object of $S$. Since $0 \in S^{-}$, this always implies that $S$ is nonempty, ok.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$S \subseteq A$ is called localizing iff $S$ is thick and $A \to A/S$ admits a right adjoint. I think that the construction $A/S$ makes only sense if $S$ is nonempty. Thus we should add that $S$ is nonempty (i.e. $0 \in S$)?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1650" LastActivityDate="2010-11-28T07:44:24.003" Title="Is the empty subcategory thick, localizing, topologizing, etc." Tags="&lt;category-theory&gt;&lt;homological-algebra&gt;" AnswerCount="1" />
  <row Id="7507" PostTypeId="2" ParentId="7459" CreationDate="2010-10-22T13:18:27.520" Score="5" ViewCount="" Body="&lt;p&gt;What you seek is known as the &lt;em&gt;offset curve&lt;/em&gt; of a polygon.  &lt;a href=&quot;http://en.wikipedia.org/wiki/Offset_curve&quot; rel=&quot;nofollow&quot;&gt;Here&lt;/a&gt; is the Wikipedia article.&#xA;This is closely related to the &lt;a href=&quot;http://en.wikipedia.org/wiki/Minkowski_sum&quot; rel=&quot;nofollow&quot;&gt;Minkowski sum construction&lt;/a&gt;, in this case the Minkowski sum of a polygon and a disk.  As others already indicated, the offset of a polygon is not in general&#xA;a polygon.  There is a substantial literature on the topic, which can be reached with these&#xA;key phrases.  There are algorithms; none are easy.  Here is one:&#xA;&quot;An offset algorithm for polyline curves,&quot; &lt;em&gt;Computers in Industry&lt;/em&gt;,&#xA;&lt;a href=&quot;http://portal.acm.org/citation.cfm?id=1228072&quot; rel=&quot;nofollow&quot;&gt;Volume 58,Issue 3 (April 2007)&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="237" LastActivityDate="2010-10-22T13:18:27.520" CommentCount="3" />
  <row Id="7508" PostTypeId="2" ParentId="7478" CreationDate="2010-10-22T13:31:48.637" Score="1" ViewCount="" Body="&lt;p&gt;Jessica: Preserving the order of all elements isn't sufficient; you have to show that it preserves the group operation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a hint though.  $G^\prime$ is the set of symmetries of a square, right?  Try labelling the square's corners.  Then each symmetry is just a permutation of its corners...&lt;/p&gt;&#xA;" OwnerUserId="2215" LastActivityDate="2010-10-22T13:31:48.637" />
  <row Id="7509" PostTypeId="2" ParentId="7376" CreationDate="2010-10-22T14:18:26.187" Score="3" ViewCount="" Body="&lt;p&gt;Interior points of a convex set $K$ in $R^n$ are the points in what could be called the &quot;convex interior&quot; of the set: they are inside (the part with positive barycentric coordinates of) at least one nondegenerate &lt;em&gt;n&lt;/em&gt;-dimensional simplex with vertices in $K$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given a point in the interior of the convex set $\overline{C}$, surround it by a nondegenerate $n$-dimensional simplex with vertices in $\overline{C}$.  Because $C$ is dense in its closure, we can perturb the simplex very slightly into one with vertices in $C$, and this simplex will continue to contain the given interior point.  (This is because, for example, the barycentric coordinates are continuous functions of the simplex vertices, as long as the simplex does not degenerate, so a small perturbation will keep the coordinates positive and the point strictly inside the simplex).&lt;/p&gt;&#xA;" OwnerUserId="467" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-10-23T03:47:20.380" LastActivityDate="2010-10-23T03:47:20.380" CommentCount="2" />
  <row Id="7510" PostTypeId="1" CreationDate="2010-10-22T15:54:20.227" Score="0" ViewCount="71" Body="&lt;p&gt;I need to show that :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;||(ATA)^-1||2=(||A+||2)^2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;hint:use singular decomposition.I need to proof this equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you for your help.&lt;/p&gt;&#xA;" LastActivityDate="2010-10-22T15:54:20.227" Title="proof with singular value decomposition help me?" Tags="&lt;linear-algebra&gt;" CommentCount="3" />
  <row Id="7511" PostTypeId="1" AcceptedAnswerId="7513" CreationDate="2010-10-22T16:14:47.553" Score="7" ViewCount="410" Body="&lt;p&gt;Consider (non-trivial) functions that are their own nth derivatives.  For instance&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{\mathrm{d}}{\mathrm{d}x} e^x = e^x$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{\mathrm{d}^2}{\mathrm{d}x^2} e^{-x} = e^{-x}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{\mathrm{d}^3}{\mathrm{d}x^3} e^{\frac{-x}{2}}\sin(\frac{\sqrt{3}x}{2}) = e^{\frac{-x}{2}}\sin(\frac{\sqrt{3}x}{2})$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{\mathrm{d}^4}{\mathrm{d}x^4} \sin x = \sin x$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\cdots$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $f_n(x)$ be the function that is it's own nth derivative.  I believe (but I'm not sure) for nonnegative integer n, this function can be written as the following infinite polynomial:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f_n(x) = 1 + \cos(\frac{2\pi}{n})x + \cos(\frac{4\pi}{n})\frac{x^2}{2!} + \cos(\frac{6\pi}{n})\frac{x^3}{3!} + \cdots + \cos(\frac{2t\pi}{n})\frac{x^t}{t!} + \cdots$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there some sense in which this function can be extended to real n using fractional derivatives?  Would it then be possible to graph $z(n, x) = f_n(x)$, and would this function be smooth and continuous on both n and x axes?  Or would it have many discontinuities?&lt;/p&gt;&#xA;" OwnerUserId="2704" LastActivityDate="2010-10-23T18:28:39.117" Title="Functions that are their Own nth Derivatives for Real n" Tags="&lt;calculus&gt;&lt;differential-equations&gt;" AnswerCount="5" CommentCount="2" FavoriteCount="1" />
  <row Id="7512" PostTypeId="1" CreationDate="2010-10-22T16:14:54.200" Score="5" ViewCount="659" Body="&lt;p&gt;Hello all,&#xA;  I am a EE grad. student who has had one undergraduate course in real analysis (which was pretty much the only pure math course that I have ever done). I would like to do a self study of some basic functional analysis so that I can be better prepared to take a graduate course in that material in my university. I plan to do that next fall so I do have some time to work through a book fully. Could some one recommend some good books to start working on this? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance&lt;/p&gt;&#xA;" OwnerUserId="2707" LastActivityDate="2011-03-17T18:55:39.297" Title="Good book for self study of functional analysis" Tags="&lt;functional-analysis&gt;" AnswerCount="9" CommentCount="1" FavoriteCount="5" />
  <row Id="7513" PostTypeId="2" ParentId="7511" CreationDate="2010-10-22T16:20:50.443" Score="19" ViewCount="" Body="&lt;p&gt;The set of functions $f$ such that $f^{(n)}-f=0$ is a vector space of dimension $n$, spanned by the functions $e^{\lambda t}$ with $\lambda$ an $n$th root of unity. In particular, there are many such functions, not just one: the general such function is of the form $$f(t)=\sum_{k=0}^{n-1}a_ke^{\frac{2\pi i k}{n}t}.$$&#xA;This is explained in every text on ordinary diffential equations; I remember fondly, for example, &lt;em&gt;Theory of Ordinary Differential Equations&lt;/em&gt; by Earl A. Coddington and Norman Levinson, but I am sure you can find more modern expositions in every library.&lt;/p&gt;&#xA;" OwnerUserId="274" LastEditorUserId="274" LastEditorDisplayName="" LastEditDate="2010-10-22T16:29:26.920" LastActivityDate="2010-10-22T16:29:26.920" CommentCount="2" />
  <row Id="7514" PostTypeId="1" AcceptedAnswerId="17510" CreationDate="2010-10-22T16:27:49.493" Score="5" ViewCount="610" Body="&lt;p&gt;I sincerely apologize if this question isn't appropriate here. Originally, I wanted to post this question on Overflow because I wanted to hear the opinions of mathematicians who are doing research, but it probably really is not appropriate to post there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Long story short, I was looking for alternative ways to carry my batch of papers aside from the laptop, and Kindle came into my mind. I know they recently came out with a new one, and I was wondering if anyone has tried using it for storing, carrying, and reading math article or books, and what their experience was.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Aside from the Kindle, are there any other electronic tablet or pads, or other devices that you'd recommend for this type of purpose?&lt;/p&gt;&#xA;" OwnerUserId="2499" LastActivityDate="2011-01-14T18:34:34.470" Title="Kindle as a Tool for Mathematicians?" Tags="&lt;soft-question&gt;" AnswerCount="3" CommentCount="8" />
  <row Id="7515" PostTypeId="2" ParentId="7511" CreationDate="2010-10-22T16:28:41.713" Score="3" ViewCount="" Body="&lt;p&gt;The solutions to a homogeneous linear equation with constant coefficients&#xA;$$ a_{n}y^{(n)} + a_{n-1}y^{(n-1)} + \cdots + a_1y' + a_0y = 0$$&#xA;correspond to roots of the &quot;auxiliary polynomial&quot; $a_nt^n + \cdots + a_0$. If the polynomial has roots $r_1,\ldots,r_k$ (in complex numbers), with multiplicities $a_1,\ldots,a_k$, then a basis for the solutions is given by&#xA;$${ e^{r_1x}, xe^{r_1x},\ldots, x^{a_1-1}e^{r_1x},e^{r_2x},\ldots,x^{a_k-1}e^{r_kx}}.$$&#xA;Here, the complex exponential is used, so that if $a$ and $b$ are real numbers and $i$ is the square root of $-1$, we have&#xA;$$e^{a+bi} = e^a(\cos(b) + i \sin(b)).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your case, you are looking at the polynomial $t^n - 1 = 0$, whose roots are the $n$th roots of unity. They are all distinct; so you can either take the complex exponentials, or the real and complex parts. So if $\lambda$ is a primitive $n$th root of $1$, then a basis for the space of solutions is&#xA;$$ e^x, e^{\lambda x}, e^{\lambda^2 x},\ldots,e^{\lambda^{n-1}x}.$$&#xA;The general solution is a linear combination of these with complex coefficients:&#xA;$$f(x) = a_0e^x + a_1e^{\lambda x} + a_2e^{\lambda^2x}+\cdots+a_{n-1}e^{\lambda^{n-1}x},\qquad a_0,\ldots,a_{n-1}\in\mathbb{C}.$$&#xA;If you don't want complex values, you can take a general form as above, and take the real and complex parts separately.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-10-22T16:28:41.713" CommentCount="3" />
  <row Id="7516" PostTypeId="1" CreationDate="2010-10-22T16:36:35.627" Score="1" ViewCount="121" Body="&lt;p&gt;For example, I have heard of a topological one wherein negation means the interior of the complement (but still would like a reference).&lt;/p&gt;&#xA;" OwnerUserId="1575" LastActivityDate="2010-12-02T23:07:35.427" Title="What are good elementary examples for teaching/introducing/learning about Intuitionistic Logic or Heyting Algebras?" Tags="&lt;logic&gt;&lt;teaching&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="7517" PostTypeId="2" ParentId="7511" CreationDate="2010-10-22T16:50:53.760" Score="1" ViewCount="" Body="&lt;p&gt;When you say non-trivial it looks like you mean such that the $m^{th}$ derivative of $f$ is not equal to $f$ for every $m &amp;lt; n$.  Unfortunately, this is still not enough to define $f_n$ uniquely; for any positive integer $1 \le k \le n$ such that $\gcd(k, n) = 1$, any nontrivial linear combination of the $\phi(n)$ functions $e^{ \frac{2\pi i k}{n} x}$ works.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One way to make a consistent choice across all values of $n$ is to take $k = 1$; in that case, you might want to choose&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f_n(x) = e^{ \frac{2\pi i}{n} x } = \cos \frac{2\pi x}{n} + i \sin \frac{2\pi x}{n}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is well-defined for any complex number $n \neq 0$ and all complex numbers $x$, and it is probably true that the $n^{th}$ fractional derivative of $f_n$ is equal to $f_n$.  There is an &lt;a href=&quot;http://en.wikipedia.org/wiki/Essential_singularity&quot; rel=&quot;nofollow&quot;&gt;essential singularity&lt;/a&gt; at $n = 0$ for fixed nonzero $x$.&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-10-22T19:32:51.900" LastActivityDate="2010-10-22T19:32:51.900" CommentCount="1" />
  <row Id="7518" PostTypeId="2" ParentId="7514" CreationDate="2010-10-22T16:52:39.887" Score="-1" ViewCount="" Body="&lt;p&gt;I am also very interested in this question. I am particularly interested in how .pdf files display - are they all good, or do some look wonky, under different magnification levels, etc.&lt;/p&gt;&#xA;" OwnerUserId="2710" LastActivityDate="2010-10-22T16:52:39.887" CommentCount="2" />
  <row Id="7519" PostTypeId="2" ParentId="7512" CreationDate="2010-10-22T17:06:28.217" Score="0" ViewCount="" Body="&lt;p&gt;Not to scare you, but list of requirements for a first course in functional analysis is rather long:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Basic theorems of metric spaces including, but not limited to:&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The Baire category theorem&lt;/li&gt;&#xA;&lt;li&gt;$\ell^p$ is complete&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Arzel%C3%A0%E2%80%93Ascoli_theorem&quot; rel=&quot;nofollow&quot;&gt;Arzelà-Ascoli&lt;/a&gt; (how else will you show that an operator is compact?)&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;Measure theory --- or at least be ready to accept that you have to learn some while reading functional analysis. Because the &lt;a href=&quot;http://en.wikipedia.org/wiki/Riesz_representation_theorem&quot; rel=&quot;nofollow&quot;&gt;Riesz representation theorem&lt;/a&gt; essentially says that for a big class of &quot;reasonable&quot; spaces, continuous linear functionals and measures are the same. In other words a lot of the theory will make no sense without at least knowing some measure theory.&lt;/li&gt;&#xA;&lt;li&gt;Topology. If you want to go beyond Banach spaces and study Fréchet spaces. The continuous dual of a Fréchet space that is not a Banach space is not necessarily metrisable --- and you get to work with multiple different topologies on your spaces (weak, strong, weak-*)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;If that doesn't scare you off, I can recommend the information-dense &quot;Introduction to Functional Analysis&quot; by Reinhold Meise and Dietmar Vogt. ISBN 0-19-851485-9. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And when I say dense i mean &lt;em&gt;very&lt;/em&gt; dense. It clocks in at a modest 437 pages, yet in a late undergraduate course in functional analysis we covered less than a third of that book (plus some notes on convexity) in a semester. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for Rudin's Real &amp;amp; Complex Analysis: it's a great book, but I don't know if I'd really call it a book on functional analysis. I'd say it's on analysis in general --- hence the title.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;UPDATE: If you find that you need to brush up on real analysis, Terence Tao has notes for 3 courses on his webpage: Real Analysis &lt;a href=&quot;http://en.wordpress.com/tag/245a-real-analysis/&quot; rel=&quot;nofollow&quot;&gt;245A&lt;/a&gt;  (in progress at the time of writing), &lt;a href=&quot;http://en.wordpress.com/tag/245b-real-analysis/&quot; rel=&quot;nofollow&quot;&gt;245B&lt;/a&gt; and &lt;a href=&quot;http://en.wordpress.com/tag/245c-real-analysis/&quot; rel=&quot;nofollow&quot;&gt;245C&lt;/a&gt;. Actually I think I can highly recommend the entirety of his webpage.&lt;/p&gt;&#xA;" OwnerUserId="1269" LastEditorUserId="1269" LastEditorDisplayName="" LastEditDate="2010-10-23T09:56:25.497" LastActivityDate="2010-10-23T09:56:25.497" CommentCount="6" />
  <row Id="7520" PostTypeId="1" CreationDate="2010-10-22T17:17:05.830" Score="2" ViewCount="663" Body="&lt;p&gt;I am a graduate student of math right now but i was not able to get a topology subject in my undergrad... i just would like to know if you guys know the best one..&lt;/p&gt;&#xA;" OwnerUserId="2231" LastActivityDate="2010-10-22T23:54:22.627" Title="best book for topology?" Tags="&lt;topology&gt;" AnswerCount="6" CommentCount="5" FavoriteCount="3" />
  <row Id="7521" PostTypeId="2" ParentId="7520" CreationDate="2010-10-22T17:24:11.843" Score="5" ViewCount="" Body="&lt;p&gt;I would suggest the following options:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Topology by James Munkres&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;General Topology by Stephen Willard&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Basic Topology by M.A. Armstrong&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Perhaps you can take a look at Allen hatcher's webpage for more books on introductory topology. He has a .pdf file containing some very good books. &lt;/p&gt;&#xA;" LastActivityDate="2010-10-22T17:24:11.843" CommunityOwnedDate="2010-10-22T17:24:11.843" CommentCount="5" />
  <row Id="7522" PostTypeId="2" ParentId="7514" CreationDate="2010-10-22T17:24:21.373" Score="1" ViewCount="" Body="&lt;p&gt;I have the previous-generation Kindle, though I think the software and screen are essentially the same.  I think PDF would probably be the only convenient format and letter-size PDFs are a bit small to read without having to zoom and pan.  Personally, I wouldn't want to deal with a textbook on it—in fact, I wouldn't want to deal with textbooks on any tablet device I've seen thus far because you lose physical features like looking at two books at once and flipping through the book—but I'd certainly consider using it for articles or other shorter works.&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-10-22T17:24:21.373" CommentCount="4" />
  <row Id="7523" PostTypeId="2" ParentId="7516" CreationDate="2010-10-22T17:30:03.603" Score="2" ViewCount="" Body="&lt;p&gt;The example that you mention in you question is explained in Maclane and Moerdijk  's book &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/3540977104&quot; rel=&quot;nofollow&quot;&gt;Sheaves in Geom and Logic&lt;/a&gt; very near the beginning. They discuss this in relation to Heyting algebras for the purpose of Topos Theory. This book is very very nice, I highly recommend it.&lt;/p&gt;&#xA;" OwnerUserId="16" LastActivityDate="2010-10-22T17:30:03.603" CommentCount="2" />
  <row Id="7524" PostTypeId="1" AcceptedAnswerId="7546" CreationDate="2010-10-22T17:31:30.777" Score="3" ViewCount="84" Body="&lt;p&gt;I am working with $G_{n,p}$ random graphs. Let $p$ (the probability of an edge) be $c/n$, so $c$ is the expected number of neighbors for a vertex when $n$ is large.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let $N$ be the number of components with more edges than vertices in the graph. I am trying to find the largest $c$ such that $\mathbb{E}(N)$ tends to zero as $n$ grows large.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\mathbb{E}(N) = \sum_{k=0}^n {n \choose k} \mathbb{E}(X_k)$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $X_k$ is the indicator that we have a component with $k$ vertices and $k+1$ edges or more.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\mathbb{E}(X_k) = \mathbb{P}(\text{no outgoing edges}) \mathbb{P}(k+1 \text{ edges or more})$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\mathbb{E}(X_k) = (1-p)^{k(n-k)} \mathbb{P}(\text{Binomial}({k \choose 2}, p) \ge k+1)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here I am stuck when trying to find an upper bound for this expression. The inequalities I've tried either diverge or cause trouble in the summation. Hoeffding's inequality looked promising but gave me $k^2$ in the exponent which gave me a hard time with the summation. Any tips?&lt;/p&gt;&#xA;" OwnerUserId="2649" LastEditorUserId="2649" LastEditorDisplayName="" LastEditDate="2010-10-22T18:38:09.800" LastActivityDate="2010-10-22T20:40:36.427" Title="A component with more edges than vertices in a random graph" Tags="&lt;probability&gt;&lt;graph-theory&gt;&lt;inequality&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="7525" PostTypeId="2" ParentId="7499" CreationDate="2010-10-22T17:37:00.163" Score="0" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://stackoverflow.com/questions/3997454/help-me-understand-linear-separability-in-a-binary-svm&quot;&gt;Cross-posted&lt;/a&gt; my question at StackOverflow. More responses there.&lt;/p&gt;&#xA;" OwnerUserId="1832" LastActivityDate="2010-10-22T17:37:00.163" />
  <row Id="7526" PostTypeId="2" ParentId="7512" CreationDate="2010-10-22T17:38:49.133" Score="1" ViewCount="" Body="&lt;p&gt;Considering the fact that you have only had one undergraduate course in analysis and will be taking an actual functional analysis class, I don't think you actually want to self-study functional analysis. It would be much more useful for you to &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;bulk up on your linear algebra &lt;/li&gt;&#xA;&lt;li&gt;review your real analysis&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Functional analysis is, for a large part, linear algebra on a infinite dimensional vector space over the real or complex numbers. Having a good intuition from linear algebra is essential: you'll know what is reasonable to expect when the dimensional infinities can be controlled (by some sort of compactness), and when they cannot be controlled, what parts of the argument cannot possibly go wrong. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A bit of real analysis is also helpful because a lot of topological notions are introduced in those books, and familiarity with them is necessary. Furthermore, notions involved in the normed/metric spaces, basic notions of convergence and compactness, and many such are used all the time in functional analysis. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore I think you will be better off reviewing the notes for your undergraduate analysis course (or going through Rudin's &lt;em&gt;Principles of Mathematical Analysis&lt;/em&gt;) and studying some linear algebra (unfortunately I can't think of any good book to recommend there). &lt;/p&gt;&#xA;" OwnerUserId="1543" LastActivityDate="2010-10-22T17:38:49.133" CommentCount="1" />
  <row Id="7527" PostTypeId="1" AcceptedAnswerId="7529" CreationDate="2010-10-22T17:42:20.920" Score="4" ViewCount="251" Body="&lt;p&gt;I have attempting to solve this using the infinite ramsey theorem, with colouring based on whether the sum of two vertices has an even or odd number of distinct prime factors.&#xA;This is leading to an infinite recursion.&#xA;Is this ok? At the end of all time I will be done.&lt;/p&gt;&#xA;" OwnerUserId="2712" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-10-22T19:21:59.113" LastActivityDate="2010-11-06T19:15:47.177" Title="Find an infinite set of positive integers such that the sum of any two distinct elements has an even number of distinct prime factors." Tags="&lt;number-theory&gt;&lt;combinatorics&gt;&lt;homework&gt;&lt;graph-theory&gt;" AnswerCount="2" CommentCount="9" FavoriteCount="0" />
  <row Id="7528" PostTypeId="1" CreationDate="2010-10-22T17:56:40.600" Score="1" ViewCount="90" Body="&lt;p&gt;Given a continuous function $f$ of period $2\pi$ and $\alpha$ in $\mathbb{R}$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;can you show for any real $x$ that $\lim_{ N\to \infty} \sum_{n=1}^N f(x+n\alpha) = \lim_{ N\to\infty} \sum_{n=1}^N f(x_n)$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;that is, the 'average value' of $f$ diluted by $N$ is the same as the average value of the sequence $x+n\alpha$ since $\alpha/\pi$ is never irrational&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\alpha/\pi$ is irrational&lt;/p&gt;&#xA;" OwnerUserId="2143" LastEditorUserId="1543" LastEditorDisplayName="" LastEditDate="2010-12-21T23:08:43.270" LastActivityDate="2010-12-21T23:08:43.270" Title="Proving a certain sequence is dense in the image of a periodic function" Tags="&lt;real-analysis&gt;" AnswerCount="1" />
  <row Id="7529" PostTypeId="2" ParentId="7527" CreationDate="2010-10-22T18:44:06.123" Score="1" ViewCount="" Body="&lt;p&gt;Start with $\{5n+1\}$ and construct an infinite complete graph, colouring the edge between $x$ and $y$ with the parity of the number of distinct prime factors of $x+y$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the infinite Ramsey theorem, there is an infinite clique of one colour.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If that corresponds to the number of distinct prime factors being even, we are done. Otherwise consider each number multiplied by $5$. Since the sum of any (original) two is not divisible by $5$, after multiplying by $5$, we have that the number of distinct prime factors for each sum must be even.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, I am guessing there are an infinite number of such sets, such that any two such sets have finite intersection.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-11-06T19:15:47.177" LastActivityDate="2010-11-06T19:15:47.177" CommentCount="1" />
  <row Id="7530" PostTypeId="2" ParentId="7464" CreationDate="2010-10-22T18:45:57.660" Score="6" ViewCount="" Body="&lt;p&gt;I am surprised no one seems to have answered this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I get $$\pi/4 + \alpha/2$$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the line is given by $y = x \tan \alpha$, with $\alpha$ acute and we throw from the origin at an angle $\theta$ from the x-axis, at velocity $1$, then we have that the projectile satisfies, assuming gravity $g=2$ (in appropriate units)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle y = t\sin\theta - t^2$, $\displaystyle x = t\cos \theta$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The time at which it intersects the line again is given by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle t\sin\theta - t^2 = t \tan\alpha \cos \theta$ and so&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle t = \sin \theta - \tan \alpha \cos \theta$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is enough to maximize the horizontal distance travelled by the projectile, which is given by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle \cos \theta (\sin \theta - \tan \alpha \cos \theta)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With little manipulation, we need to maximize&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle  \sin(2\theta - \alpha)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which gives the result.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-10-22T21:20:26.790" LastActivityDate="2010-10-22T21:20:26.790" CommentCount="1" />
  <row Id="7531" PostTypeId="1" AcceptedAnswerId="8398" CreationDate="2010-10-22T18:50:55.893" Score="1" ViewCount="152" Body="&lt;p&gt;Actually, i had posted this long ago in MO, and i didn't get a reply to this question as it was unfit. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, this is an exercise, in some textbook ( i think Apostol) and i would be happy if i can receive some answers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $P(n)$ be the product of positive integers which are $\leq n$ and relatively prime $n$. Prove that $$ \displaystyle P( n) =n^{\phi(n)} \prod\limits_{d \mid n} \biggl(\frac{d!}{d^d} \biggr)^{\mu(n/d)}$$&lt;/p&gt;&#xA;" LastActivityDate="2010-10-30T20:32:14.653" Title="Proving $P( n) =n^{\phi(n)} \prod\limits_{d \mid n} \biggl(\frac{d!}{d^d} \biggr)^{\mu(n/d)}$" Tags="&lt;number-theory&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="1" />
  <row Id="7532" PostTypeId="2" ParentId="7520" CreationDate="2010-10-22T18:59:18.843" Score="1" ViewCount="" Body="&lt;p&gt;You might look at the answers to this previous MSE question, which had a slightly different&#xA;slant: &lt;a href=&quot;http://math.stackexchange.com/questions/903/choosing-a-topology-text&quot;&gt;&quot;choosing a topology text&quot;&lt;/a&gt;.  Apparently the poser was also interested in self-learning,&#xA;but with less preparation than you.&lt;/p&gt;&#xA;" OwnerUserId="237" LastActivityDate="2010-10-22T18:59:18.843" />
  <row Id="7533" PostTypeId="2" ParentId="7520" CreationDate="2010-10-22T19:06:17.430" Score="1" ViewCount="" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;I also like Bourbaki's treatise, but some times it is a bit too logical. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Simmons General topology&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Also, &quot;A topological picture book&quot; (I don't remember the author though). &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;K Jänich Topology.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;J. Kelly General topology.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="1154" LastActivityDate="2010-10-22T19:06:17.430" />
  <row Id="7534" PostTypeId="1" AcceptedAnswerId="7540" CreationDate="2010-10-22T19:13:03.593" Score="1" ViewCount="85" Body="&lt;p&gt;Suppose one is given an arbitrary moment generating function $M_{X}(t)$. How would you determine $P(X=k)$ from this? We know that $M_{X}(t) = E[e^{tX}]$ and $M_{X}(0) = 1$. &lt;/p&gt;&#xA;" OwnerUserId="2548" LastEditorUserId="2548" LastEditorDisplayName="" LastEditDate="2010-10-22T19:30:16.523" LastActivityDate="2010-10-22T19:46:43.510" Title="Moment Generating Functions" Tags="&lt;probability&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" />
  <row Id="7535" PostTypeId="1" AcceptedAnswerId="7543" CreationDate="2010-10-22T19:13:27.703" Score="0" ViewCount="89" Body="&lt;p&gt;Suppose $E$ is a set of edges on graph $G$ over $n$ vertices, $\mathbf{x} \in \{-1,1\}^n$. The following seems to hold if G is not &lt;a href=&quot;http://mathworld.wolfram.com/EulerianGraph.html&quot; rel=&quot;nofollow&quot;&gt;Eulerian&lt;/a&gt;, can anyone see a proof?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_\mathbf{x} \prod_{ij\in E} x_i x_j=0$$&lt;/p&gt;&#xA;" OwnerUserId="998" LastEditorUserId="998" LastEditorDisplayName="" LastEditDate="2010-10-22T22:03:00.193" LastActivityDate="2010-10-22T22:03:00.193" Title="Sum that is 0 for Eulerian graphs" Tags="&lt;graph-theory&gt;" AnswerCount="2" />
  <row Id="7536" PostTypeId="2" ParentId="7512" CreationDate="2010-10-22T19:19:12.377" Score="0" ViewCount="" Body="&lt;p&gt;There's a book that could fit your actual level perfectly. The book is &lt;a href=&quot;http://books.google.co.cr/books?id=QALoZC64ea0C&amp;amp;pg=PP1&amp;amp;dq=beginning+functional+analysis&amp;amp;hl=es&amp;amp;ei=S-HBTLT6PIbGlQeX1ojTCg&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=1&amp;amp;ved=0CCcQ6AEwAA#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;Beginning Functional Analysis&lt;/a&gt; by Karen Saxe. It is aimed at undergraduates whose background is a basic course in linear algebra and real analysis. It is pretty well suited for self study since it is very readable (I've done it myself), and the author claims that one of its aims is that the book can be used for self study. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It does not require any measure theory because it develops the basics of Lebesgue measure in chapter 3.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I highly recommend it if you want to read a little bit about functional analysis without having to master more things in order to read more advanced books.&lt;/p&gt;&#xA;" OwnerUserId="900" LastActivityDate="2010-10-22T19:19:12.377" CommentCount="1" />
  <row Id="7537" PostTypeId="2" ParentId="7512" CreationDate="2010-10-22T19:19:15.437" Score="2" ViewCount="" Body="&lt;p&gt;Check out &quot;Introductory Functional Analysis with Applications&quot; by Erwin Kreyszig. I have not read it myself, but I have heard great things. Also, in the preface, he writes that Calculus and a familiarity with Linear Algebra are all that's needed as prerequisites.&lt;/p&gt;&#xA;" OwnerUserId="2029" LastActivityDate="2010-10-22T19:19:15.437" />
  <row Id="7538" PostTypeId="1" CreationDate="2010-10-22T19:38:49.860" Score="3" ViewCount="65" Body="&lt;p&gt;Hi everyone, first time poster.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm trying to determine the length of a repeating sequence, with noise.&#xA;I.e. given the sequence 4, 12, 3, 4, 12, 3, 4, 12, 3, ...  the period is apparently 3.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, what if the sequence was more complicated. What if each value had added noise?&#xA;I was thinking of something with autocorrelation, or Fourier transform (my intuition tells me the repeating signal has a major frequency component, right? How can I determine which?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It feels like this should be easy, but I'm having trouble implementing a solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="2714" LastEditorUserId="2714" LastEditorDisplayName="" LastEditDate="2010-10-22T19:45:27.040" LastActivityDate="2010-10-23T01:30:16.613" Title="Determining the length of a repeating sequence, with noise." Tags="&lt;sequences-and-series&gt;&lt;complex-analysis&gt;" AnswerCount="2" />
  <row Id="7539" PostTypeId="1" AcceptedAnswerId="7551" CreationDate="2010-10-22T19:40:44.117" Score="4" ViewCount="384" Body="&lt;p&gt;Given a large univariate polynomial, say of degree 200 or more, is there a procedural way of finding all the complex roots? By &quot;roots&quot;, I mean complex decimal approximations to the roots, though the multiplicity of the root is important. I have access to MAPLE and the closest function I've seen is:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;with(RootFinding):&#xA;Analytic(Z,x,-(2+2*I)..2+2*I);&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;but this chokes if Z is of high degree (in fact it fails to complete even if deg(Z)&gt;15). &lt;/p&gt;&#xA;" OwnerUserId="196" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-10-22T23:06:07.193" LastActivityDate="2010-10-23T11:11:19.287" Title="Finding all complex zeros of a high-degree polynomial" Tags="&lt;complex-analysis&gt;&lt;numerical-methods&gt;&lt;roots&gt;" AnswerCount="4" CommentCount="6" FavoriteCount="1" />
  <row Id="7540" PostTypeId="2" ParentId="7534" CreationDate="2010-10-22T19:46:43.510" Score="2" ViewCount="" Body="&lt;p&gt;$M_X(\log(t))$ is the probability generating function. Differentiate $k$ times and set $t=0$. Divide the result by $k!$.&lt;/p&gt;&#xA;" OwnerUserId="940" LastActivityDate="2010-10-22T19:46:43.510" CommentCount="2" />
  <row Id="7541" PostTypeId="2" ParentId="7520" CreationDate="2010-10-22T19:50:45.017" Score="2" ViewCount="" Body="&lt;p&gt;As an introductory book, &quot;Topology without tears&quot; by S. Morris. You can download PDF for free, but you might need to obtain a key to read the file from the author. (He wants to make sure it will be used for self-studying.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, another great introductory book is Munkres, Topology. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On graduate level (non-introductory books) are Kelley and Dugunji (or Dugundji?).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Munkres said when he started writing his Topology, there wasn't anything accessible on undergrad level, and both Kelley and Dugunji wasn't really undergrad books. He wanted to write something any undergrad student with an appropriate background (like the first 6-7 chapters of Rudin's Principles of Analysis) can read. He also wanted to focus on Topological spaces and deal with metric spaces mostly from the perspective &quot;whether topological space is metrizable&quot;. That's the first half of the book. The second part is a nice introduction to Algebraic Topology. Again, quoting Munkres, at the time he was writing the book he knew very little of Algebraic Topology, his speciality was General (point-set) topology. So, he was writing that second half as he was learning some basics of algebraic topology. So, as he said, &quot;think of this second half as an attempt by someone with general topology background, to explore the Algebraic Topology.&lt;/p&gt;&#xA;" OwnerUserId="2450" LastActivityDate="2010-10-22T19:50:45.017" CommentCount="1" />
  <row Id="7542" PostTypeId="2" ParentId="7520" CreationDate="2010-10-22T20:02:29.207" Score="1" ViewCount="" Body="&lt;p&gt;You might consider &lt;em&gt;Topology Now!&lt;/em&gt; by Messer and Straffin. Their idea is to introduce the intuitive ideas of continuity, convergence, and connectedness so that students can quickly delve into knot theory, the topology of surfaces and three dimensional manifolds, fixed points, and elementary homotopy theory. I wish this book had been around when I was a student!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://people.albion.edu/ram/TopologyNow!/&quot; rel=&quot;nofollow&quot;&gt;http://people.albion.edu/ram/TopologyNow!/&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="940" LastActivityDate="2010-10-22T20:02:29.207" CommunityOwnedDate="2010-10-22T20:02:29.207" CommentCount="1" />
  <row Id="7543" PostTypeId="2" ParentId="7535" CreationDate="2010-10-22T20:06:43.183" Score="2" ViewCount="" Body="&lt;p&gt;if G is Eulerian - as in you have one cyclic route that goes over all the edges, then the degree of each vertex is even so $\prod _{ij\in E} x_i x_j = \prod_{i\in [n]} x_i ^{deg(i)} = 1$ and so the total sum is $2^n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;if by eulerian you refer to a route that is not a circle then you have exactly two vertices with odd degree so  $\prod _{ij\in E} x_i x_j = x_s x_t$ and it is easy to see that $\sum_x x_s x_t = 0$ &lt;/p&gt;&#xA;" OwnerUserId="2125" LastActivityDate="2010-10-22T20:06:43.183" CommentCount="1" />
  <row Id="7544" PostTypeId="2" ParentId="7539" CreationDate="2010-10-22T20:12:04.597" Score="1" ViewCount="" Body="&lt;p&gt;I think this might help. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Rouche's Theorem:&lt;/strong&gt; Let $D$ be a bounded domain, with piecewise smooth boundary, $\partial{D}$. Let $f(z)$ and $h(z)$ be analytic on $D \cup \partial{D}$. If $|h(z)| &amp;lt; |f(z)$ for $z \in \partial{D}$, then $f(z)$ and $f(z)+h(z)$ have the same number of zero's in $D$, counting multiplicities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Example:&lt;/strong&gt; We find the zeros of the function $p(z)=z^{6}+9z^{4}+z^{3}+2z+4$, inside the unit circle. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For using Rouche's theorem, let $p(z)= f(z) + h(z)$, where $f(z)$ dominates, $h(z)$ inside the unit circle. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider $f(z) = 9z^{4}$, which has four zeros inside the unit circle, all at the origin. $h(z) =z^{6}+z^{3}+2z+4$, which satisfies, $|h(z)| &amp;lt; |f(z)|$ for $|z|=1$. Therefore by Rouche's theorem $p(z)$ also has 4 zeros inside the unit circle. &lt;/p&gt;&#xA;" LastActivityDate="2010-10-22T20:12:04.597" CommentCount="8" />
  <row Id="7545" PostTypeId="2" ParentId="7538" CreationDate="2010-10-22T20:27:33.573" Score="2" ViewCount="" Body="&lt;p&gt;If the noise is additive, autocorrelation or Fourier transform seem the way to go. In general, time-domain and frequency-domain approaches are the most basic and standard tools in this area. Beware that the spectra of such sequence will have extra peaks corresponding to the &quot;harmonics&quot; of your signal. You can read about &lt;a href=&quot;http://en.wikipedia.org/wiki/Pitch_detection_algorithm&quot; rel=&quot;nofollow&quot;&gt;pitch detection&lt;/a&gt;, which is very related to your problem.&lt;/p&gt;&#xA;" OwnerUserId="312" LastActivityDate="2010-10-22T20:27:33.573" />
  <row Id="7546" PostTypeId="2" ParentId="7524" CreationDate="2010-10-22T20:40:36.427" Score="2" ViewCount="" Body="&lt;p&gt;The threshold for cycles in $G(n,p)$ is $p=1/n$. The same threshold should work in your case, so the solution is basically $c = 1$.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-10-22T20:40:36.427" CommentCount="0" />
  <row Id="7547" PostTypeId="2" ParentId="7528" CreationDate="2010-10-22T20:45:20.473" Score="2" ViewCount="" Body="&lt;p&gt;If $\alpha/\pi$ is rational, then your claim is false (in general, unless we got lucky), since you're sampling the same points over and over again.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conversely, if $\alpha/\pi$ is irrational, then you're sampling random points of the function (by Weyl's equidistribution theorem), and then with some work you can probably prove that in the limit you do get the mean value of $f$.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-10-22T20:45:20.473" />
  <row Id="7548" PostTypeId="2" ParentId="7535" CreationDate="2010-10-22T20:46:11.753" Score="0" ViewCount="" Body="&lt;p&gt;I don't think you need the graph to be Eulerian.  You have a bunch of terms $x_i x_j$ for $i,j \in \{1,2,\dots,n\}$ and want the terms to add to zero as the various $x_i$ are $\pm1$.  It is true for n=2 as there are no pairs of $x_i x_j$.  If it is true for n, it will be true for n+1 because all the added terms are of the form $x_i x_{n+1}$ and the ones where $x_{n+1}=+1$ cancel the ones where $x_{n+1}=-1$.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-10-22T20:46:11.753" CommentCount="3" />
  <row Id="7549" PostTypeId="2" ParentId="7539" CreationDate="2010-10-22T20:54:03.177" Score="0" ViewCount="" Body="&lt;p&gt;I think the following page summarizes the root-finding well, by showing how to reduce it to an appropriate Eigenvalue computation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://mathworld.wolfram.com/PolynomialRoots.html&quot; rel=&quot;nofollow&quot;&gt;http://mathworld.wolfram.com/PolynomialRoots.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1709" LastActivityDate="2010-10-22T20:54:03.177" />
  <row Id="7550" PostTypeId="1" CreationDate="2010-10-22T22:51:22.107" Score="1" ViewCount="287" Body="&lt;p&gt;Let’s define the codicil of the Geometric Mean – Arithmetic Mean Inequality to be the statement that if the means are equal, then all the terms are equal. Then: I conjecture that most of the GM-AM Inequality is actually in its codicil, in the sense that if you can use the codicil (that is, assume the codicil as part of the hypothesis), then it is a hundred times easier to prove the AM-GM Inequality.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt; Rephrased by Y.F. &lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Is a proof along the following lines possible? Show that as long as not all variables are equal, there is a way to make them &quot;more equal&quot; which decreases the difference $\mathrm{AM}-\mathrm{GM}$. Deduce somehow that &quot;the worst case&quot; is when all variables are equal, in which case we know that $\mathrm{AM}=\mathrm{GM}$.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt; Addendum by OP Mike Jones &lt;/h3&gt;&#xA;&#xA;&lt;p&gt;I thought it was obvious that although the setting of this question is real analysis (specifically, inequalities), the germane issue is proof theory. I want to explore the interrelationships between the various parts of the hypothesis and conclusion of this celebrated theorem. I would have tagged it as “proof theory” in the first place, but didn’t see “proof theory” in my cursory glance at the tags. Now, I have found it buried under the tag “logic”. I was attempting to add the “logic” tag when I got thrown into the comment box, so, could someone who knows how please add this tag to the question? Also, if the answer to the question is affirmative, then it should also be tagged with “education” and “teaching”. About 80% of learning consists of simply acquiring (deep) familiarity, and so the teacher could pose the problem to the students, or to an especially talented/eager younger student: “Prove the GM-AM Inequality, but use its codicil as part of the hypothesis this first time around.”. Indeed, that is the original motivation for my question: how to make this celebrated result much more accessible to students. (I teach mathematics in high school.) It occurred to me that perhaps a great deal, even most, of the difficulty is tied up in the codicil, precisely as user “Moron” has described:&#xA;(GM = AM implies values equal) implies GM &amp;lt;= AM.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found that this is too long for a comment, so I bailed out of the comment box and am putting this in as an addendum to the original question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Oh, I see now how to do addtional tags:-)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regards,&#xA;Mike Jones&#xA;American expatriate in Beijing&#xA;5.May.2011&lt;/p&gt;&#xA;" OwnerUserId="2344" LastEditorUserId="2344" LastEditorDisplayName="" LastEditDate="2011-05-04T19:57:50.933" LastActivityDate="2011-05-06T21:54:18.853" Title="Is most of the GM-AM Inequality in its codicil?" Tags="&lt;logic&gt;&lt;inequality&gt;&lt;education&gt;&lt;teaching&gt;" AnswerCount="2" CommentCount="9" />
  <row Id="7551" PostTypeId="2" ParentId="7539" CreationDate="2010-10-22T23:04:35.823" Score="4" ViewCount="" Body="&lt;p&gt;Everyone's first starting point when dealing with the polynomial rootfinding problem should be a peer at &lt;a href=&quot;http://www.yorku.ca/mcnamee/&quot; rel=&quot;nofollow&quot;&gt;J.M. McNamee's excellent bibliography&lt;/a&gt; and &lt;a href=&quot;http://books.google.com/books?id=4PMqxwG-eqQC&quot; rel=&quot;nofollow&quot;&gt;book&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, it is a fact that polynomials of very high degree tend to make most polynomial rootfinders choke. Even the standard blackbox, the Jenkins-Traub algorithm, can choke if not properly safeguarded. Eigenmethods, while they can have nice accuracy, can be very demanding of space and time (O(n²) space and O(n³) operations for a problem with only O(n) inputs!)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My point is that unless you are prepared to devote some time and extra precision, this is an insoluble problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Having been pessimistic in those last few sentences, one family of methods you might wish to peer at (and I have had personal success with) are the so-called &quot;simultaneous iteration&quot; methods. The simplest of them, (Weierstrass-)Durand-Kerner, is essentially an application of Newton's method to the Vieta formulae, treated as n equations in n unknowns (the assumption taken by (W)DK is that your polynomial is monic, but that is easily arranged).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you wish for more details and references, the &lt;a href=&quot;http://books.google.com/books?id=4PMqxwG-eqQC&amp;amp;pg=PA67&quot; rel=&quot;nofollow&quot;&gt;book by McNamee&lt;/a&gt; I mentioned earlier is a good start.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-10-22T23:04:35.823" />
  <row Id="7552" PostTypeId="2" ParentId="7511" CreationDate="2010-10-22T23:38:12.077" Score="2" ViewCount="" Body="&lt;p&gt;What you are interested here are what &lt;a href=&quot;http://www.sciencedirect.com/science/bookseries/00765392&quot; rel=&quot;nofollow&quot;&gt;Spanier and Oldham&lt;/a&gt; term &quot;cyclodifferential functions&quot;, functions that are regenerated after being differintegrated to the appropriate order (or to put it another way, functions that are eigenfunctions of the differintegration operator).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the cyclodifferential equation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$${}_0 D_x^{\alpha}y=y$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(or in more familiar notation, $$\frac{\mathrm{d}^{\alpha}y}{\mathrm{d}x^{\alpha}}=y$$, but the problem with this notation in the setting of differintegrals is that it neglects to take the lower limit that is present in both the Caputo and Riemann-Liouville definitions into account).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;as you might have seen, $\exp(x)$ is a cyclodifferential function for ${}_0 D_x^1 y$ (i.e. $\exp(x)$ is its own derivative), $\cosh(x)$ and $\sinh(x)$ are cyclodifferentials for ${}_0 D_x^2 y$ (differentiating those two functions twice gives you the originals);&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac1{\sqrt{\pi x}}+\exp(x)\mathrm{erfc}(-\sqrt{x})$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;($\mathrm{erfc}(x)$ is the complementary error function)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is a cyclodifferential for ${}_0 D_x^{\frac12} y$ (it is its own &lt;em&gt;semiderivative&lt;/em&gt;), and in general&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x^{\alpha-1}\sum_{j=0}^\infty \frac{C^j x^{\alpha j}}{\Gamma(\alpha(j+1))}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for $\alpha &gt; 0$ and $C$ an appropriate eigenvalue, is a cyclodifferential for ${}_0 D_x^{\alpha}$.&lt;/p&gt;&#xA;" OwnerUserId="498" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-10-23T08:07:53.183" LastActivityDate="2010-10-23T08:07:53.183" CommentCount="1" />
  <row Id="7553" PostTypeId="2" ParentId="7520" CreationDate="2010-10-22T23:54:22.627" Score="1" ViewCount="" Body="&lt;p&gt;Singer and Thorpe, &lt;em&gt;Lecture Notes on Elementary Topology and Geometry&lt;/em&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A slim book that gives an intro to point-set, algebraic and differential topology and differential geometry. It does not have any exercises and is very tersely written, so it is not a substitute for a standard text like Munkres, but as a beginner I liked this book because it gave me the big picture in one place without many prerequisites.&lt;/p&gt;&#xA;" OwnerUserId="1195" LastActivityDate="2010-10-22T23:54:22.627" CommentCount="1" />
  <row Id="7554" PostTypeId="1" AcceptedAnswerId="7557" CreationDate="2010-10-23T00:23:42.890" Score="4" ViewCount="197" Body="&lt;p&gt;Is there an algorithm that will allow me to numerically compute the limit of a function f(x) in a principled way?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most naive algorithm would be to continue to compute the function for larger values of x. The first problem is how to figure out the 'large' values for x to compute the function for.  How do I know when to stop?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can we construct some error bars for this calculation perhaps based on some kind of statistical rationale?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any books for further reading will be much appreciated.&lt;/p&gt;&#xA;" OwnerUserId="1004" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-10-23T01:02:56.900" LastActivityDate="2010-12-24T07:10:56.393" Title="Numerically estimate the limit of a function" Tags="&lt;analysis&gt;&lt;numerical-methods&gt;&lt;asymptotics&gt;" AnswerCount="3" CommentCount="0" />
  <row Id="7555" PostTypeId="2" ParentId="7342" CreationDate="2010-10-23T00:42:24.193" Score="3" ViewCount="" Body="&lt;p&gt;This is a development of my comment, changing the restraints from two to one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $x_{n}=-\sum_{i=1}^{n-1}x_{i}$, the question &quot;find &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\min \sum_{i=1}^{n}x_{i}^{3}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;s.t. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{i=1}^{n}x_{i}=0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{i=1}^{n}x_{i}^{2}=n\ &amp;quot;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is equivalent to find &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\min f(x_{1},x_{2},\ldots ,x_{n-1}),$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $f(x_{1},x_{2},\ldots ,x_{n-1})=\left(&#xD;&#xA;\sum_{i=1}^{n-1}x_{i}^{3}\right) -\left( \sum_{i=1}^{n-1}x_{i}\right) ^{3}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;s.t. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$c(x_{1},x_{2},\ldots ,x_{n-1})=0,$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$c(x_{1},x_{2},\ldots ,x_{n-1})=\left( \sum_{i=1}^{n-1}x_{i}\right)&#xD;&#xA;^{2}-n+\sum_{i=1}^{n-1}x_{i}^{2}=0.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The conditions regarding first derivatives are&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\nabla f(x_{1},x_{2},\ldots ,x_{n-1})-\nabla c(x_{1},x_{2},\ldots&#xD;&#xA;,x_{n-1})\lambda =0\qquad (\ast ),$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\nabla f(x)=\begin{pmatrix}\frac{\partial f}{\partial x_{1}} &amp;amp; \ldots  &amp;amp; \frac{\partial f}{\partial x_{n-1}}\end{pmatrix}^{T}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$=%&#xD;&#xA;\begin{pmatrix}&#xD;&#xA;3x_{1}^{2}-3\left( \sum_{i=1}^{n-1}x_{i}\right) ^{2} &amp;amp; \ldots  &amp;amp; &#xD;&#xA;3x_{n-1}^{2}-3\left( \sum_{i=1}^{n-1}x_{i}\right) ^{2}%&#xD;&#xA;\end{pmatrix}%&#xD;&#xA;^{T}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\nabla c(x)=%&#xD;&#xA;\begin{pmatrix}&#xD;&#xA;\frac{\partial c}{\partial x_{1}} &amp;amp; \ldots  &amp;amp; \frac{\partial c}{\partial&#xD;&#xA;x_{n-1}}%&#xD;&#xA;\end{pmatrix}%&#xD;&#xA;^{T}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$=%&#xD;&#xA;\begin{pmatrix}&#xD;&#xA;2\left( \sum_{i=1}^{n-1}x_{i}\right) +2x_{1} &amp;amp; \ldots  &amp;amp; 2\left(&#xD;&#xA;\sum_{i=1}^{n-1}x_{i}\right) +2x_{n-1}%&#xD;&#xA;\end{pmatrix}%&#xD;&#xA;^{T}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Condition $(\ast )$ is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$3x_{1}^{2}-3\left( \sum_{i=1}^{n-1}x_{i}\right) ^{2}-2\lambda \left(&#xD;&#xA;\sum_{i=1}^{n-1}x_{i}\right) -\lambda 2x_{1}=0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ldots $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$3x_{n-1}^{2}-3\left( \sum_{i=1}^{n-1}x_{i}\right) ^{2}-2\lambda \left(&#xD;&#xA;\sum_{i=1}^{n-1}x_{i}\right) -\lambda 2x_{n-1}=0.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All these equations have the same form (the coefficients of $x_k$, $1\lt k\lt n-1$, are independent of $k$). Thus the solution is atained when all the $n-1$ variables are equal&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x_{1}=\ldots =x_{n-1}=x^{\ast }$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, we have to solve the new system of two equations&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$3(x^{\ast })^{2}-3\left( \sum_{i=1}^{n-1}x^{\ast }\right) ^{2}-2\lambda&#xD;&#xA;\left( \sum_{i=1}^{n-1}x^{\ast }\right) -2\lambda x^{\ast }=0\qquad (\ast&#xD;&#xA;\ast )$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left( \sum_{i=1}^{n-1}x^{\ast }\right)&#xD;&#xA;^{2}-n+\sum_{i=1}^{n-1}(x^{\ast })^{2}=0.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We obtain, successively&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$3(x^{\ast })^{2}-3n+3\sum_{i=1}^{n}(x^{\ast })^{2}-2\lambda \left(&#xD;&#xA;\sum_{i=1}^{n-1}x^{\ast }\right) -2\lambda x^{\ast }=0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$3(x^{\ast })^{2}-3n+3(x^{\ast })^{2}(n-1)-2\lambda x^{\ast }(n-1)-2\lambda&#xD;&#xA;x^{\ast }=0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(x^{\ast })^{2}(n-1)-1=0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x^{\ast }=x_{1}=\ldots =x_{n-1}=\pm \frac{1}{\sqrt{n-1}}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x_{n}=-\sum_{i=1}^{n-1}x_{i}=\mp \frac{n-1}{\sqrt{n-1}}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since for $n&amp;gt;2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left( \frac{1}{\sqrt{n-1}}\right) ^{3}-\left( \frac{n-1}{\sqrt{n-1}}%&#xD;&#xA;\right) ^{3}&amp;lt;-\left( \frac{1}{\sqrt{n-1}}\right) ^{3}+\left( \frac{n-1}{%&#xD;&#xA;\sqrt{n-1}}\right) ^{3},$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the minimum is obtained at&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x^{\ast }=x_{1}=\ldots =x_{n-1}=\frac{1}{\sqrt{n-1}},x_{n}=-\frac{n-1}{%&#xD;&#xA;\sqrt{n-1}}$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;and its value is equal to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left( \frac{1}{\sqrt{n-1}}\right) ^{3}-\left( \frac{n-1}{\sqrt{n-1}}%&#xD;&#xA;\right) ^{3}.$$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Added: The original problem in $n$ variables is the solution of $n$ equations&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\nabla f(x_{1},x_{2},\ldots ,x_{n})-\nabla c(x_{1},x_{2},\ldots&#xD;&#xA;,x_{n})\lambda =0\qquad (\ast\ast\ast )$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;plus the two conditions&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$c_{1}(x_{1},x_{2},\ldots ,x_{n})=0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$c_{2}(x_{1},x_{2},\ldots ,x_{n})=0,$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\nabla f=%&#xD;&#xA;\begin{pmatrix}&#xD;&#xA;\frac{\partial f}{\partial x_{1}} &amp;amp; \ldots  &amp;amp; \frac{\partial f}{\partial&#xD;&#xA;x_{n}}%&#xD;&#xA;\end{pmatrix}%&#xD;&#xA;^{T}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\nabla c=%&#xD;&#xA;\begin{pmatrix}&#xD;&#xA;\left( \nabla c_{1}\right) ^{T} &amp;amp; \left( \nabla c_{2}\right) ^{T}%&#xD;&#xA;\end{pmatrix}%&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\lambda =\begin{pmatrix}&#xD;&#xA;\lambda _{1} &amp;amp; \lambda _{2}\end{pmatrix}^{T}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: The equations would be &lt;em&gt;similar&lt;/em&gt; for $m$ constraints $c_1,\dots ,c_m.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remark: Since I don't know how to write here matrices with more than one row, I wrote them in the transposed form.  &lt;/p&gt;&#xA;" OwnerUserId="752" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2011-02-27T15:54:37.110" LastActivityDate="2011-02-27T15:54:37.110" />
  <row Id="7556" PostTypeId="2" ParentId="7554" CreationDate="2010-10-23T00:47:15.590" Score="4" ViewCount="" Body="&lt;p&gt;You have to assume something about f(x).  For the case where the approach to the limit is expected to have an asymptotic expansion:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;John Stalker, &lt;em&gt;A convergence speeding algorithm with applications to numerical integration&lt;/em&gt;, Advances in Applied Mathematics Volume 22, Issue 1, January 1999, Pages 119-153 &lt;/p&gt;&#xA;" OwnerUserId="467" LastActivityDate="2010-10-23T00:47:15.590" />
  <row Id="7557" PostTypeId="2" ParentId="7554" CreationDate="2010-10-23T01:07:18.670" Score="4" ViewCount="" Body="&lt;p&gt;To expand on T's answer, &lt;em&gt;convergence acceleration&lt;/em&gt; methods assume that the error terms of the sequence you are interested in are of a certain form.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of the general-purpose methods for accelerating the convergence of a sequence is the Shanks transformation. For the case of a sequence of partial sums of a series, the Shanks transformation amounts to constructing a sequence of Padé approximants, which hopefully converge faster to the limit of the sequence of partial sums. (The justly famous Wynn ε algorithm is an efficient realization of the Shanks transformation)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another popular technique is Richardson extrapolation, which assumes that the error can be expressed as a power series of some form. This is in effect the application of the usual algorithms for polynomial interpolation to estimate the limit of a sequence. (Richardson extrapolation is the machinery behind the Romberg algorithm for accelerating the convergence of the trapezoidal rule, since the error of the trapezoidal rule is expressible as a series in powers of the panel size.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have been deliberately vague here since you have given absolutely no information on the nature of your sequence. There are many convergence acceleration methods to choose from (a good reference is Brezinski and Redivo-Zaglia's &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0444888144&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;Extrapolation Methods: Theory and Practice&lt;/em&gt;&lt;/a&gt;), and the best sequence transformation to use depends a lot on the provenance of your sequence.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-10-23T01:07:18.670" CommentCount="7" />
  <row Id="7558" PostTypeId="2" ParentId="7538" CreationDate="2010-10-23T01:30:16.613" Score="0" ViewCount="" Body="&lt;p&gt;In connection with certain algorithms for integer factorization, there are methods due to &lt;a href=&quot;http://doi.acm.org/10.1145/800135.804400&quot; rel=&quot;nofollow&quot;&gt;Floyd&lt;/a&gt; and &lt;a href=&quot;http://dx.doi.org/10.1007/BF01933190&quot; rel=&quot;nofollow&quot;&gt;Brent&lt;/a&gt; for the so-called &quot;cycle finding&quot; problem. It would look like you can adapt these to your situation, but in the noisy case you will have to use a &quot;fuzzy equality&quot; comparison function, e.g. &lt;code&gt;|x - y| &amp;lt;= tol&lt;/code&gt; where &lt;code&gt;x&lt;/code&gt; and &lt;code&gt;y&lt;/code&gt; are the two entities being compared, and &lt;code&gt;tol&lt;/code&gt; is a small quantity.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-10-23T01:30:16.613" />
  <row Id="7559" PostTypeId="2" ParentId="7512" CreationDate="2010-10-23T01:37:46.290" Score="0" ViewCount="" Body="&lt;p&gt;I am also an EE.   And I think Kreyszig's Funcational Analysis is a good book for your background.&lt;/p&gt;&#xA;" OwnerUserId="832" LastActivityDate="2010-10-23T01:37:46.290" CommentCount="1" />
  <row Id="7561" PostTypeId="2" ParentId="7473" CreationDate="2010-10-23T02:21:16.483" Score="3" ViewCount="" Body="&lt;p&gt;Note $\rm\ f_n\: :=\ a^n-1\ =\ a^{n-m} \: (a^m-1) + a^{n-m}-1\:$,$\ $ i.e. $\rm\ f_n = k\ f_m + f_{n-m}\:$.$\ $ Now apply&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;THEOREM&lt;/strong&gt; $\: $  Let $\rm\ f_n\: $ be an integer sequence with $\rm\ f_{\:0} =\: 0\:\ $ and $\rm\ \: f_n\equiv f_{n-m}\ (mod\ f_m)\ $ for $\rm\: n &amp;gt; m\:$.  Then $\rm\:\ (f_n,f_m)\ =\ f_{(n,\:m)}\ \: $ where $\rm\ (i,\:j)\ $ denotes $\rm\ gcd(i,\:j)\:$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof&lt;/strong&gt; $\ $ By induction on  $\rm\:n + m\:$. The theorem is trivially true if $\rm\ n = m\ $ or $\rm\ n = 0\ $ or $\rm\: m = 0\:$.&lt;br&gt;&#xA;So we may assume $\rm\:n &amp;gt; m &amp;gt; 0\:$.$\ $  Note $\rm\ (f_n,f_m)\: =\ (f_{n-m},f_m)\ $ follows from the hypothesis.&lt;br&gt;&#xA;Since $\rm\:\ (n-m)+m \ &amp;lt;\ n+m\: ,\ $ induction yields $\rm\ \ (f_{n-m},f_m)\ =\ f_{(n-m,\:m)}\ =\ f_{(n,\:m)}\ $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See also &lt;a href=&quot;http://math.stackexchange.com/questions/11567/gcdbx-1-by-1-b-z-1-b-gcdx-y-z-1/11636#11636&quot;&gt;this post&lt;/a&gt; for a conceptual proof exploiting the innate structure - an &lt;em&gt;order ideal&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2011-03-24T03:22:28.687" LastActivityDate="2011-03-24T03:22:28.687" />
  <row Id="7562" PostTypeId="2" ParentId="7464" CreationDate="2010-10-23T03:09:06.060" Score="4" ViewCount="" Body="&lt;p&gt;When the solution is so neat, I feel like it should have a high-level explanation &quot;from the book&quot; that makes it immediately obvious. Sadly, I was not able to find one, but I'm posting what I did get in hopes that it might inspire someone else.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine firing projectiles at all angles from the origin. Assume unit speed and gravity $g = 2$, as in Moron's answer. Consider a frame of reference which starts at the origin and falls freely. There is no gravity in this frame of reference, so all projectiles expand radially outward forming a cone,&#xA;$$x^2 + y^2 = t^{2}.$$&#xA;However, the ground is no longer given by $y = x \tan\alpha$ but rather by&#xA;$$y = x\tan\alpha + t^{2},$$&#xA;which is a parabolic cylinder. The projectiles hit the ground at the points where the cone and the parabolic cylinder intersect. Subtracting the two equations, we find that this intersection satisfies&#xA;$$\left(x^2 + x\tan\alpha\right) + \left(y^2 - y\right) = 0,$$&#xA;&lt;em&gt;i.e.&lt;/em&gt; on the $xy$ plane it projects to a circle. On physical grounds, one can see that the circle passes through the origin, and is tangent to the line $y = x\tan\alpha$ there. The projectile that reaches the farthest corresponds to the point with the largest $x$-coordinate, and the angle of the projectile is simply the angle of the line joining that point to the origin (because projectiles just move radially outward in this frame of reference). I encourage you to draw a little diagram and see that this line bisects the angle between the vertical, at $\pi/2$, and the ground, at $\alpha$.&lt;/p&gt;&#xA;" OwnerUserId="856" LastActivityDate="2010-10-23T03:09:06.060" CommentCount="1" />
  <row Id="7563" PostTypeId="2" ParentId="7486" CreationDate="2010-10-23T03:24:38.780" Score="1" ViewCount="" Body="&lt;p&gt;This is more of an extended remark out of academic curiosity.  I thought I would try to use &lt;a href=&quot;http://www.cs.unm.edu/~mccune/mace4/&quot; rel=&quot;nofollow&quot;&gt;Prover9&lt;/a&gt; to prove the given identity.  [here v denotes &quot;or&quot; and ^ denotes &quot;and&quot;]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To test this, I inputted the assumptions into Prover9 (this is the OP's list of axioms from &lt;a href=&quot;http://math.stackexchange.com/questions/7471/boolean-algebra-theorems&quot;&gt;here&lt;/a&gt;):&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;x v x' = y v y'.  % false&#xA;x ^ x' = y ^ y'.  % true&#xA;&#xA;y v (x ^ x') = y.&#xA;y v (x v x') = y.&#xA;x v x = x.&#xA;x v y = y v x.&#xA;(x v y) v z = x v (y v z).&#xA;x ^ (x v y) = x.&#xA;(x ^ y) v (x ^ y') = x.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;from which it proved the OP's identity (and if I understand correctly, it was by contradiction):&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;((((w' ^ x) ^ z') v ((x ^ y) ^ z)) v ((w ^ x) ^ y')) v ((w ^ x') ^ z) = (((((((w' ^ x) ^ z') v ((x ^ y) ^ z)) v ((w ^ x) ^ y')) v ((w ^ x') ^ z)) v ((x ^ y') ^ z')) v ((w' ^ x) ^ y)) v ((w ^ x) ^ z).&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="139" LastActivityDate="2010-10-23T03:24:38.780" CommentCount="1" />
  <row Id="7564" PostTypeId="1" CreationDate="2010-10-23T03:59:37.630" Score="3" ViewCount="480" Body="&lt;p&gt;It's been a while since my last statistics class...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have 404 files that went through some automated generation process. I would like to manually verify some of them to make sure that their data is indeed correct. I want to use probability to help me out so that I don't need to check every single file.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How would I calculate what sample size I should use to reach a certain confidence level?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, if I would like to say with 95% confidence that the files are correct, how many of them do I have to check?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found an &lt;a href=&quot;http://www.surveysystem.com/sscalc.htm&quot; rel=&quot;nofollow&quot;&gt;online calculator&lt;/a&gt;, but I'm not entirely sure what I should put for the confidence interval. Say I put 20% and leave the confidence factor at 95%. I get a sample size of 23. Let's say now that I tested 23 random files and all of them were fine. Does that mean that &quot;I can be 95% confident that 80% to 100% of the files are correct&quot;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does this mean, then, that for my original question, I would need to use a 99% confidence level with a 4% confidence interval, then I would need to verify that the 291 files (the sample size it gave me) are all correct. And only then I can say with 95% confidence that the files are correct? (99% +- 4% = 95% to 100%)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It also mentions something about percentages which I'm not quite clear on... does the fact that most (i.e. 100%) of the files I test are valid (since if I found an invalid one, I would stop the whole process and examine my generation process for errors) mean that I can use a smaller sample to get the same confidence factor? If so, how would I calculate it?&lt;/p&gt;&#xA;" OwnerUserId="2718" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-10-23T04:17:52.553" LastActivityDate="2010-10-23T08:27:42.933" Title="Calculating a sample size based on a confidence level" Tags="&lt;probability&gt;&lt;statistics&gt;" AnswerCount="2" CommentCount="5" FavoriteCount="2" />
  <row Id="7566" PostTypeId="1" CreationDate="2010-10-23T05:02:36.593" Score="0" ViewCount="74" Body="&lt;p&gt;Is there a way to determine the probability of winning a particular sweepstakes with only the following information:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-the estimated odds of winning are 1 in 13,000&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-contestants must be owners (property holders) within the corporation running the sweepstakes and the total number of owners is 222,600&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-each owner can enter the sweepstakes up to 6 times&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-there is one grand prize and three runner-up prizes&lt;/p&gt;&#xA;" LastActivityDate="2010-10-23T05:44:59.540" Title="Sweepstakes Probability" Tags="&lt;probability&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="7567" PostTypeId="2" ParentId="7564" CreationDate="2010-10-23T05:03:33.567" Score="3" ViewCount="" Body="&lt;p&gt;It's not surprising you're a bit confused; understanding what's really going on with confidence intervals can be tricky.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The short version: If you don't want to check all the files you have to choose two different percentages: the confidence level (95% in your example), and how far off you're willing to be at that level (20% in your example).  These percentages refer to two different quantities, and so it doesn't make sense to add or subtract them from each other.  Once you've made these choices then I think it is fine to use the online calculator to get a sample size.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want more detail on what's going on, here's the explanation: You're trying to estimate the true percentage of files that have correct data.  Let's call that percentage $p$.  Since you don't want to calculate $p$ exactly, you have to choose how far off you are willing to be with your estimate, say 20%.  Unfortunately, you can't even be certain that your estimate of $p$ will be within 20%, so you have to choose a level of confidence that that estimate will be within 20% of $p$.  You have chosen 95%.  Then the online calculator gives you the sample size of 23 you need to estimate $p$ to within 20% at 95% confidence. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But what does that 95% really mean?  Basically, it means that if you were to choose lots and lots of samples of size 23 and calculate a confidence interval from each one, 95% of the resulting confidence intervals would actually contain the unknown value of $p$.  The other 5% would give an interval of some kind that does not include $p$.  (Some would be too large, others would be too small.)  Another way to look at it is that choosing a 95% confidence interval means that you're choosing a method that gives correct results (i.e., produces a confidence interval that actually contains the value of $p$) 95% of the time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To answer your specific questions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Does that mean that 'I can be 95% confident that 80% to 100% of the files are correct'?&quot;  Not precisely.  It really means that you can be 95% confident that the true percentage of correct files is between 80% and 100%.  That's a subtle distinction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;And only then I can say with 95% confidence that the files are correct? (99% +- 4% = 95% to 100%)?&quot;  No, this is confusing the two kinds of percentages.  The 99% refers to 99% of all confidence intervals constructed if you were to construct lots of them.  The 4% refers to an error margin of $\pm$ 4% for the files.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One other thing to remember is that the sample size estimator assumes that the population you're drawing from is much, much larger than the size of the sample you end up going with.  Since your population is fairly small you can get away with a smaller-sized sample with the same level of confidence.  The determination of exactly how small, though, is a much more difficult calculation.  It's beyond what you would have seen in a basic statistics class.  I'm not sure how to do it; maybe someone else on the site does.  (EDIT: Even better: take Jyotirmoy Bhattacharya's suggestion and ask on &lt;a href=&quot;http://stats.stackexchange.com/&quot;&gt;Stats Stack Exchange&lt;/a&gt;.)  But this is the only justification for being able to use a smaller sample size than 23 - not the fact that you would abort the confidence interval calculation if you found anything other than 100% for your sample's estimate of the true value of $p$.&lt;/p&gt;&#xA;" OwnerUserId="2370" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-10-23T05:21:47.637" LastActivityDate="2010-10-23T05:21:47.637" />
  <row Id="7568" PostTypeId="2" ParentId="7566" CreationDate="2010-10-23T05:31:49.187" Score="1" ViewCount="" Body="&lt;p&gt;If the estimated odds of winning are $1$ in $13,000$ per entrance, then if you enter once, your odds of winning are $1:13000$, of course.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, entering more than once, it's a bit more complicated.  It's easier to calculate the odds of you &lt;em&gt;not&lt;/em&gt; winning, and subtract it from $1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The odds of you &lt;em&gt;not&lt;/em&gt; winning once is $12999/13000$.  The odds of you &lt;em&gt;not&lt;/em&gt; winning twice is $(12999/13000)^2$, and three times is $(12999/13000)^3$, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We could then say we can calculate your odds of not winning $n(x)$, where $x$ is the number of entries, as $n(x) = (12999/13000)^x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, you want to calculate the odds of winning $p(x)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know that $n(x) + p(x) = 1$ (Think about it.  The sum of the odds of not winning and the odds of winning equal up to 1, because there is a 100% chance that one or the other will happen).  So, $p(x) = 1 - n(x)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So to calculate your odds of winning, you would use&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$p(x) = 1 - (12999/13000)^x$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $x$ is the amount of times you've entered.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you enter 6 times, then your total odds are $\frac{2227329628937465077999}{4826809000000000000000000}$, which is about $0.0004614497$.&lt;/p&gt;&#xA;" OwnerUserId="51" LastEditorUserId="51" LastEditorDisplayName="" LastEditDate="2010-10-23T05:44:59.540" LastActivityDate="2010-10-23T05:44:59.540" CommentCount="1" />
  <row Id="7569" PostTypeId="1" AcceptedAnswerId="8682" CreationDate="2010-10-23T05:55:16.840" Score="3" ViewCount="173" Body="&lt;p&gt;Are there known sharp upper bounds for sums of the form $\sum_{p \mid k} \frac{1}{p+1}$ for $k &gt; 1$ subject to the constraint $\sum_{p \mid k} \frac{1}{p+1} &amp;lt; 1$? (The factor of +1 in the denominator and the constraint of the sum of inverse &quot;shifted prime divisors&quot; is bounded by 1 are both necessary.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A related question: Suppose $(p_i)$ is a set of $n$ consecutive primes which minimizes $1 - \sum_{i = 1}^{n} \frac{1}{p_i+1} &gt; 0$ for a given $n &gt; 1$. Are there known bounds for $1 - \sum_{i = 1}^{n} \frac{1}{p_i + 1}$ from below in terms of $n$, e.g., $n^{-\delta n}$ for some fixed $\delta &gt; 0$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="2720" LastEditorUserId="2720" LastEditorDisplayName="" LastEditDate="2010-10-26T14:49:48.390" LastActivityDate="2010-11-02T23:12:17.947" Title="Sharp upper bounds for sums of the form $\sum_{p \mid k} \frac{1}{p+1}$" Tags="&lt;number-theory&gt;&lt;real-analysis&gt;&lt;analysis&gt;&lt;asymptotics&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="2" />
  <row Id="7570" PostTypeId="1" CreationDate="2010-10-23T06:00:07.927" Score="2" ViewCount="317" Body="&lt;p&gt;$$\lim_{n\to\infty} \lim_{N\to\infty} a(N,n)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the reason I want to know is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a sequence of functions $\{f_n\}$ which converges uniformly to a function $f$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and I want to say that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\lim_{n\to\infty} \lim_{N\to\infty} \sum_{k=1}^N \frac{f_k(x_n)}N = \lim_{N\to\infty} \lim_{n\to\infty} \sum_{k=1}^N \frac{f_k(x_n)}N $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;here my $a(N,n)$ is that sum ($\sum_{k=1}^N f_k(x_n)/N$) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;and somehow I want to use that $f_n$ converges uniformly to do this &lt;/p&gt;&#xA;&#xA;&lt;p&gt;sorry.edited&lt;/p&gt;&#xA;" OwnerUserId="2143" LastEditorUserId="171" LastEditorDisplayName="" LastEditDate="2010-10-23T13:55:42.027" LastActivityDate="2010-12-24T17:23:18.287" Title="When can you interchange limits of a sequence of 2 variables?" Tags="&lt;real-analysis&gt;&lt;limit&gt;" AnswerCount="3" CommentCount="4" />
  <row Id="7571" PostTypeId="1" CreationDate="2010-10-23T06:18:31.647" Score="1" ViewCount="75" Body="&lt;p&gt;Given a smooth vector field on a closed smooth manifold, does the flow of that vector field exists for all time $t \in R$? This may be a well-known and elementary conclusion, but I just wnat to confirm it to me. &lt;/p&gt;&#xA;" OwnerUserId="2381" LastActivityDate="2010-10-23T21:26:12.643" Title="flow on a closed manifold" Tags="&lt;differential-equations&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="7572" PostTypeId="2" ParentId="7564" CreationDate="2010-10-23T06:39:05.473" Score="1" ViewCount="" Body="&lt;p&gt;Let's forget about confidence intervals for a while and try to think of your sample size problem from first principles. This might also answer @Mike's point about finite samples.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose among your $N$ files $M$ are good and $N-M$ are bad. If you pick a random sample of $n$ files the probability that all of them are good is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$p(n,M) = {{M \choose n} \over {N \choose n}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[This is a special case of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Hypergeometric_distribution&quot; rel=&quot;nofollow&quot;&gt;hypergeometric&lt;/a&gt; distribution. You can calculate it in Excel as &lt;code&gt;HYPGEOMDIST(n,n,M,N)&lt;/code&gt; or in &lt;a href=&quot;http://www.r-project.org/&quot; rel=&quot;nofollow&quot;&gt;R&lt;/a&gt; as &lt;code&gt;dhyper(n,M,N-M,n)&lt;/code&gt;]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$p(n,M)$ will be smaller for larger $n$. A large sample is more likely to uncover bad files even if they are rare.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$p(n,M)$ will be larger for larger $M$. Rarer bad files are, the less likely are they to be caught by a sample of a given size.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How large an $n$ should you choose? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you knew $M$ then you could plot $p(n,M)$ and choose a $n$ large enough to put $p(n,M)$ below the threshold of error you are willing to tolerate.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the whole point of the exercise is that you don't know $M$. Now it becomes a question of how optimistic or pessimistic you are. If you are an optimist then you will think that your process will either succeed or fail very badly. So if $M$ is not $N$ then it is a very small number. In this case choosing a small $n$ will let you reach your error tolerance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are a pessimist you will think that if your process fails it will produce a bad file only occasionally, i.e. $M$ will be large but not equal to $N$. In this case you will be forced to take a large sample to reach any given error tolerance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to be formal you could try to codify you pre-sampling beliefs about the value of $M$ in terms of a probability distribution. [I have not seen anyone really do it, though I have seen people sometimes pretend to do so.] Let the probability mass function be $f(m)$. Then the probability of a sample of size $n$ having all good files even if some files are bad is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\pi(n)={1 \over {1-f(N)}}\sum_{m=0}^{N-1} p(n,m)f(m)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now you can choose a large enough $n$ to keep $\pi(n)$ below your error tolerance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some additional points:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;I assumed that the 404 files are all you care about. On the other hand if you care about the process that generated the files, then you should model the process as generating bad files with a probability $q$ and instead of $p(n,M)$ have $p(n,q)=(1-q)^n$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I assumed from your question that you are interested only in the two possibilities $M=N$ and $M \neq N$. The extension of the approach above when you care about how many files are bad is left as an exercise.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The &quot;error tolerance&quot; mentioned about should not be a number picked from the air or chosen by convention. It should be the result of considering the costs of inspections, the damages that would result from missing a bad file, and your attitude towards risk.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Finally, what about confidence intervals? They are a barbaric relic of the past. In most situations they don't answer any interesting questions.&lt;/p&gt;&#xA;" OwnerUserId="1195" LastEditorUserId="1195" LastEditorDisplayName="" LastEditDate="2010-10-23T08:27:42.933" LastActivityDate="2010-10-23T08:27:42.933" CommentCount="1" />
  <row Id="7573" PostTypeId="2" ParentId="7569" CreationDate="2010-10-23T06:40:42.727" Score="0" ViewCount="" Body="&lt;p&gt;Actually, there is a bound for $\displaystyle \frac{1}{p}$. Perhaps, this should help. You can find this result in a standard Analytic Number theory textbook.(I have taken this from Apostol's &lt;em&gt;Analytic Number Theory&lt;/em&gt; textbook!)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proposition:&lt;/strong&gt; There is a constant $A$ such that $$\sum\limits_{p \leq x} \frac{1}{p}= \log\log{x} + A + \mathcal{O}\Bigl(\frac{1}{\log{x}}\Bigr)$$ for all $x \geq 2$, where the constant $A$ is given by $$A = 1 -\log\log{2} + \int\limits_{2}^{\infty} \frac{ R(t)}{t \log^{2}{t}} \ dt$$&lt;/p&gt;&#xA;" LastActivityDate="2010-10-23T06:40:42.727" CommentCount="1" />
  <row Id="7574" PostTypeId="2" ParentId="7512" CreationDate="2010-10-23T07:33:26.610" Score="0" ViewCount="" Body="&lt;p&gt;I like Zeidler's &lt;em&gt;Applied Functional Analysis&lt;/em&gt; (&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0387944427&quot; rel=&quot;nofollow&quot;&gt;Volume 1&lt;/a&gt; and &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0387944222&quot; rel=&quot;nofollow&quot;&gt;Volume 2&lt;/a&gt;).&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2010-10-23T07:33:26.610" />
  <row Id="7575" PostTypeId="2" ParentId="7512" CreationDate="2010-10-23T08:15:49.810" Score="0" ViewCount="" Body="&lt;p&gt;One unconventional book is &lt;em&gt;Infinite Dimensional Analysis: A Hitchiker's Guide&lt;/em&gt; by Aliprantis and Border. It's fully rigorous but written for &quot;practical people, such as engineers and economists&quot; rather than math students.&lt;/p&gt;&#xA;" OwnerUserId="1195" LastActivityDate="2010-10-23T08:15:49.810" />
  <row Id="7577" PostTypeId="2" ParentId="7539" CreationDate="2010-10-23T11:11:19.287" Score="2" ViewCount="" Body="&lt;p&gt;I think one of the biggest problems is approximating multiple roots. The approach described in&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;L.Brugnano, D.Trigiante. &quot;Polynomial Roots: the Ultimate Answer?&quot;, Linear Algebra and its Applications 225 (1995) 207-219&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;relies on the approximation of eigenvalues of a tridiagonal matrix, obtained via the application of Euclid's GCD algorithm to the original polynomial, and seems to work pretty well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I couldn't find the pdf for the article though, sorry.&lt;/p&gt;&#xA;" OwnerUserId="2691" LastActivityDate="2010-10-23T11:11:19.287" CommentCount="1" />
  <row Id="7578" PostTypeId="1" AcceptedAnswerId="7581" CreationDate="2010-10-23T11:15:36.210" Score="12" ViewCount="607" Body="&lt;p&gt;I'm looking for an example of two metrics that induce the same topology, but so that one metric is complete and the other is not (Since it is known that completeness isn't a topological invariant).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance for any hints or ideas.&lt;/p&gt;&#xA;" OwnerUserId="766" LastActivityDate="2010-10-24T03:48:33.183" Title="Two metrics induce the same topology, but one is complete and the other isn't." Tags="&lt;topology&gt;&lt;homework&gt;&lt;metric-spaces&gt;" AnswerCount="6" CommentCount="2" FavoriteCount="3" />
  <row Id="7580" PostTypeId="2" ParentId="7578" CreationDate="2010-10-23T11:43:36.807" Score="4" ViewCount="" Body="&lt;p&gt;Look for a homeomorphism $f: \mathbb{R} \to \mathbb{R}^+$ (you should know a good one). Then take the usual metric on $\mathbb{R}$ and use it to define a metric $d$ on $\mathbb{R}^+$ by $d(x,y) = |f(x)-f(y)|$ (*). Now what can you say about the relationship between $d$ and the usual metric on $\mathbb{R}^+$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Made the question here correct and more illustrative.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the identity $\iota: (\mathbb{R}^+,d) \to (\mathbb{R}^+,|\cdot|)$ is a homeomorphism. Now just ask yourself which property, that continuous functions between metric spaces can have, $\iota$ &lt;em&gt;doesn't&lt;/em&gt; have? Can you formulate a theorem about what property a homeomorphism must have to preserve completeness?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(*): &lt;a href=&quot;http://en.wikipedia.org/wiki/Pullback&quot; rel=&quot;nofollow&quot;&gt;Pullback&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/Pushforward&quot; rel=&quot;nofollow&quot;&gt;pushforward&lt;/a&gt; might be of interest.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS: &lt;code&gt;\mathbb{R}_+&lt;/code&gt; makes the math renderer choke.&lt;/p&gt;&#xA;" OwnerUserId="1269" LastEditorUserId="1269" LastEditorDisplayName="" LastEditDate="2010-10-23T12:13:36.320" LastActivityDate="2010-10-23T12:13:36.320" CommentCount="0" />
  <row Id="7581" PostTypeId="2" ParentId="7578" CreationDate="2010-10-23T11:56:34.643" Score="25" ViewCount="" Body="&lt;p&gt;The metric space $\{\frac {1}{n} \mid n\in \mathbb{N} \}$ with the usual metric is incomplete (since we don't have zero), and it has the discrete topology.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the same space with the metric $d(x,y)=1 \iff x\neq y$ also has a discrete topology but is complete, since any cauchy sequence will eventually be constant.&lt;/p&gt;&#xA;" OwnerUserId="2125" LastActivityDate="2010-10-23T11:56:34.643" CommentCount="2" />
  <row Id="7583" PostTypeId="1" AcceptedAnswerId="7591" CreationDate="2010-10-23T13:00:44.417" Score="3" ViewCount="201" Body="&lt;p&gt;Let $\zeta$ be the Riemann- Zeta function. For any integer, $n \geq 2$, how to prove $$\zeta(2) \zeta(2n-2) + \zeta(4)\zeta(2n-4) + \cdots + \zeta(2n-2)\zeta(2) = \Bigl(n + \frac{1}{2}\Bigr)\zeta(2n)$$&lt;/p&gt;&#xA;" LastActivityDate="2010-10-24T16:45:49.443" Title="An Identity Concerning the Riemann Zeta Function" Tags="&lt;number-theory&gt;&lt;analysis&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="3" />
  <row Id="7586" PostTypeId="1" AcceptedAnswerId="7588" CreationDate="2010-10-23T13:42:13.490" Score="4" ViewCount="168" Body="&lt;p&gt;hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need to solve this :$$ z^2 -6z + 25 = 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My book says 'complete the square' so :&lt;br&gt;&#xA;1.$$ (z - 6/2)^2 -36/4 + 25 $$&lt;br&gt;&#xA;2.$$ (z - 3)^2 -9 + 25 $$&lt;br&gt;&#xA;3.$$ (z - 3)^2 + 16 $$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now how exactly does the above turn into this:&lt;br&gt;&#xA;$$ 3\pm 4\imath $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks so Much&lt;br&gt;&#xA;Gideon&lt;/p&gt;&#xA;" OwnerUserId="2576" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-10-23T23:43:23.267" LastActivityDate="2010-10-24T15:51:50.350" Title="solve $ z^2 -6z + 25 $ into complex conjugate" Tags="&lt;algebra-precalculus&gt;&lt;homework&gt;&lt;complex-numbers&gt;" AnswerCount="4" CommentCount="4" />
  <row Id="7587" PostTypeId="2" ParentId="7586" CreationDate="2010-10-23T13:44:49.430" Score="3" ViewCount="" Body="&lt;p&gt;The square roots of -16 are $4i$ and $-4i$.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;As a further hint to anybody who might encounter this sort of thing in the future, a polynomial with complex conjugate roots $\alpha$ and $\alpha^\ast$ will be of the form&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x^2-2(\Re\alpha)x+|\alpha|^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that $3^2+4^2=5^2=25$ and that $6=2\times 3$. Thus, once you verify that your quadratic has a negative discriminant, you can almost instantly write down the complex conjugate roots.&lt;/p&gt;&#xA;" OwnerUserId="498" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-10-24T05:24:45.643" LastActivityDate="2010-10-24T05:24:45.643" CommentCount="4" />
  <row Id="7588" PostTypeId="2" ParentId="7586" CreationDate="2010-10-23T13:49:12.540" Score="6" ViewCount="" Body="&lt;p&gt;I think your confusion here stems from the fact you're working with an expression, not an equation. You can only solve equations; it makes no sense to 'solve' an expression!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; You seem to have just edited your post so the first line is an equation. Best to keep it in equation form throughout however.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm presuming you have:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$z^2 - 6z + 25 = 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, completing the square:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(z - 6/2)^2 -36/4 + 25 = 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(z - 3)^2 - 9 + 25 = 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(z - 3)^2 = -16$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$z - 3 = \pm 4i$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$z = 3 \pm 4i$$&lt;/p&gt;&#xA;" OwnerUserId="56" LastActivityDate="2010-10-23T13:49:12.540" CommentCount="3" />
  <row Id="7589" PostTypeId="1" CreationDate="2010-10-23T14:20:46.227" Score="7" ViewCount="311" Body="&lt;p&gt;During my study of plane algebraic curves, I got curious if there is a nontrivial example of a plane algebraic curve that has a &lt;a href=&quot;http://mathworld.wolfram.com/Crunode.html&quot; rel=&quot;nofollow&quot;&gt;node&lt;/a&gt;, a &lt;a href=&quot;http://mathworld.wolfram.com/Cusp.html&quot; rel=&quot;nofollow&quot;&gt;cusp&lt;/a&gt; (for my purposes I do not care which of the two kinds of cusps would the example exhibit), a &lt;a href=&quot;http://mathworld.wolfram.com/Tacnode.html&quot; rel=&quot;nofollow&quot;&gt;tacnode&lt;/a&gt;, and an &lt;a href=&quot;http://mathworld.wolfram.com/Acnode.html&quot; rel=&quot;nofollow&quot;&gt;isolated point&lt;/a&gt;. By &quot;nontrivial&quot; I mean a curve that was not constructed as a chimera of two or more simpler curves, e.g. $(x-y)(x^2+y^2-1)=0$. Of course, it would be a quintic at the very least (i.e. the algebraic degree should be 5 at the minimum).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Apart from an explicit example, I would also be interested in a general procedure for constructing algebraic curves with a prescribed number and type of singular points.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;After trying out Qiaochu's and T..'s suggestions, I have a follow-up question: does the problem become more difficult if the requirement that the curve be bounded (i.e. one can draw a circle such that the whole curve, including the isolated point, is within the circle) is imposed?&lt;/p&gt;&#xA;" OwnerUserId="498" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-10-24T00:28:56.103" LastActivityDate="2010-10-25T05:50:12.987" Title="A plane algebraic curve with all four kinds of double points" Tags="&lt;algebraic-geometry&gt;&lt;plane-curves&gt;" AnswerCount="3" CommentCount="5" FavoriteCount="2" />
  <row Id="7590" PostTypeId="2" ParentId="2110" CreationDate="2010-10-23T14:23:05.567" Score="1" ViewCount="" Body="&lt;p&gt;You can try the &lt;a href=&quot;http://en.wikipedia.org/wiki/Kalman_filter&quot; rel=&quot;nofollow&quot;&gt;Kalman filter&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://calypso.inesc-id.pt/FCUL/psm/docs/kalman-dan-simon.pdf&quot; rel=&quot;nofollow&quot;&gt;http://calypso.inesc-id.pt/FCUL/psm/docs/kalman-dan-simon.pdf&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1195" LastActivityDate="2010-10-23T14:23:05.567" CommentCount="2" />
  <row Id="7591" PostTypeId="2" ParentId="7583" CreationDate="2010-10-23T14:29:59.800" Score="8" ViewCount="" Body="&lt;p&gt;This is too nice an exercise to give away. Your starting point should be:&#xA;$$\frac{t}{e^t-1} = 1 - t/2 + \frac{2 \zeta(2)}{(2 \pi)^2} t^2 - \frac{2 \zeta(4)}{(2 \pi)^4} t^4 + \frac{2 \zeta(6)}{(2 \pi)^6} t^6 - \cdots.$$&lt;/p&gt;&#xA;" OwnerUserId="448" LastActivityDate="2010-10-23T14:29:59.800" CommentCount="1" />
  <row Id="7592" PostTypeId="2" ParentId="7571" CreationDate="2010-10-23T14:49:51.163" Score="4" ViewCount="" Body="&lt;p&gt;The local existence theorem for ODEs gives a lower bound for the domain of existence of solutions in terms of a Lipschitz constant. On a compact manifold $M$, you can find a &lt;em&gt;global&lt;/em&gt; Lipschitz constant, and then the theorem gives you an $\epsilon&gt;0$ such that for every point $x\in M$, the solution that starts at $x$ is defined at least in $(-\epsilon,\epsilon)$. With this you immediately get existence for all values of $t$.&lt;/p&gt;&#xA;" OwnerUserId="274" LastEditorUserId="274" LastEditorDisplayName="" LastEditDate="2010-10-23T21:26:12.643" LastActivityDate="2010-10-23T21:26:12.643" />
  <row Id="7593" PostTypeId="1" AcceptedAnswerId="7998" CreationDate="2010-10-23T15:36:04.023" Score="9" ViewCount="214" Body="&lt;p&gt;A recent &lt;a href=&quot;http://people.missouristate.edu/lesreid/POW12_0910.html&quot; rel=&quot;nofollow&quot;&gt;Missouri State problem&lt;/a&gt; stated that it is easy to decompose the plane into half-open intervals and asked us to do so with intervals pointing in every direction.  That got me trying to decompose the plane into closed or open intervals.  The best I could do was to make a square with two sides missing (which you can do out of either type) and form a checkerboard with the white squares missing the top and bottom and the black squares missing the left and right.  That gets the whole plane except the lattice points.  This seems like it must be a standard problem, but I couldn't find it on the web.  So can the plane be decomposed into unit open intervals?  closed intervals?&lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-10-23T19:19:43.510" LastActivityDate="2010-10-27T04:25:29.060" Title="Decomposing the plane into intervals" Tags="&lt;geometry&gt;&lt;euclidean-geometry&gt;&lt;tiling&gt;" AnswerCount="2" CommentCount="8" FavoriteCount="1" />
  <row Id="7594" PostTypeId="1" CreationDate="2010-10-23T15:48:47.543" Score="2" ViewCount="163" Body="&lt;p&gt;I am interested in the sum of a Hadamard product of generating functions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we are given $n$ functions, where $0 &amp;lt; i \leq n$, of the form:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f_i(x_i) = \sum_{j=0}^m{c_{i,j} x_i^j}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The hadamard product is defined as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$g(x) = \sum_{j=0}^m{( \prod_{i=0}^n {c_{i,j}})x^j}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's essentially the same as going through all the generating functions simultaneously and multiplying all the $j$th coefficients together to obtain a new coefficient.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Some given information&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have polynomials; I think of them as finite length generating functions.  I know that all of the coefficients are natural numbers.  I'd like the sum of the resulting coefficients.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's a trick.  Calculating the Hadamard product generates an extremely complicated expression.  I would like to avoid dealing with this expression explicitly, if possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to know ways that I could do this.&lt;/p&gt;&#xA;" OwnerUserId="2626" LastEditorUserId="1269" LastEditorDisplayName="" LastEditDate="2010-10-23T18:11:13.023" LastActivityDate="2010-10-23T18:11:13.023" Title="How can I get this tricky sum?" Tags="&lt;analysis&gt;&lt;sequences-and-series&gt;&lt;functions&gt;" CommentCount="5" />
  <row Id="7595" PostTypeId="2" ParentId="7589" CreationDate="2010-10-23T15:55:21.460" Score="9" ViewCount="" Body="&lt;p&gt;Here is the idea.  Everything depends on what the defining polynomial $f(x, y) = 0$ looks like in local coordinates.  To simplify things we'll take $f(x, y) = y^2 - p(x)$ where $p$ is some polynomial in $x$.  Now, I claim that&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If $y^2 = x^2 + \text{higher terms}$ then there is a crunode at the origin,&lt;/li&gt;&#xA;&lt;li&gt;If $y^2 = x^3 + \text{higher terms}$ then there is a cusp at the origin,&lt;/li&gt;&#xA;&lt;li&gt;If $y^2 = x^4 + \text{higher terms}$ then there is a tacnode at the origin, and&lt;/li&gt;&#xA;&lt;li&gt;If $y^2 = -x^2 + \text{higher terms}$ then there is an acnode at the origin.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The key point is that in each of these cases we get something locally diffeomorphic to the same guy but without the higher-order terms.  In other words, to get a bunch of double points with arbitrary behavior it suffices to pick a bunch of points $(x_i, 0)$ and place various conditions on the values of $p$ and its derivatives at $x_i$.  In particular:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If $p(0) = p'(0) = 0, p''(0) = 1$ then there is a crunode at $(0, 0)$.&lt;/li&gt;&#xA;&lt;li&gt;If $p(1) = p'(1) = p''(1) = 0, p^{(3)}(1) = 1$ then there is a cusp at $(1, 0)$.&lt;/li&gt;&#xA;&lt;li&gt;If $p(2) = p'(2) = p''(2) = p^{(3)}(2) = 0, p^{(4)}(2) = 1$ then there is a tacnode at $(2, 0)$.&lt;/li&gt;&#xA;&lt;li&gt;If $p(3) = p'(3) = 0, p''(3) = -1$ then there is an acnode at $(3, 0)$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I claim that this kind of polynomial interpolation is always possible.  In fact, it follows by the Chinese Remainder Theorem!  The conditions are equivalent to&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$p(x) \equiv x^2 \bmod x^3$&lt;/li&gt;&#xA;&lt;li&gt;$p(x) \equiv (x-1)^3 \bmod (x-1)^4$&lt;/li&gt;&#xA;&lt;li&gt;$p(x) \equiv (x-2)^4 \bmod (x-2)^5$&lt;/li&gt;&#xA;&lt;li&gt;$p(x) \equiv -(x-3)^2 \bmod (x-3)^3.$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Of course this discussion generalizes.  And it shouldn't be hard for you to convince yourself that you can choose $p$ to be a non-square.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt;  And if you also want to be able to choose the $y$-coordinates arbitrarily, it suffices to replace $y^2$ by a polynomial $q(y)$ which is equal to $y^2 + \text{higher terms}$ in local coordinates at each of the points you're interested in, which can be done by the same method.  &lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-10-23T19:29:24.753" LastActivityDate="2010-10-23T19:29:24.753" CommentCount="5" />
  <row Id="7596" PostTypeId="2" ParentId="7578" CreationDate="2010-10-23T16:24:34.997" Score="3" ViewCount="" Body="&lt;p&gt;The interval $M:=]-1,1[$ with the usual line element $ds:=|dx|$ is an incomplete metric space: The sequence $x_n:=1-{1\over n} \ (n\to\infty)$ converges in $\mathbb R$, so it is a Cauchy sequence, but it diverges in $M$. On the other hand, the &quot;hyperbolic metric&quot; defined by $ds:=|dx|/(1-x^2)$ induces the same topology on $M$ but is complete. The latter statement needs of course a proof. Suffice it here to say that now the endpoints $\pm1$ are &quot;infinitely far away&quot;.&lt;/p&gt;&#xA;" OwnerUserId="1303" LastActivityDate="2010-10-23T16:24:34.997" />
  <row Id="7597" PostTypeId="2" ParentId="7589" CreationDate="2010-10-23T16:43:22.560" Score="5" ViewCount="" Body="&lt;p&gt;Everything but the isolated point is a local algebraic requirement that can be satisfied by one component of a rational curve parametrized by polynomials, $(x,y)=(f(z),g(z))$.  For each singularity that requires $k$ segments of the curve to meet in specified ways, write down values of $f$ and $g$ and their first few derivatives at $k$ (or fewer) different values of $z$ that would provide the desired picture.   So this part of the problem reduces to finding polynomial $f$ and $g$ whose coefficients satisfy some linear equations.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, to get two branches to meet we could require $f(1)=f(2)$ and $g(1)=g(2)$; to have the intersection be transversal (resp., a tangency) we need $(f'(1),g'(1))$ different from (resp., equal to) $(f'(2),g'(2))$ and higher tangency conditions can be expressed in the same form.  You can actually draw the curve freely (but unicursally, not lifting the pencil) on paper to define what the pattern of intersections and cusps should be and then choose the values and derivatives at the intersection points to create that behavior in the parametrization.  It can be arbitrarily complicated, such as a point where two branches meet to order $5$ while two other segments meet to order $6$ and are tangent to the first two branches to order $2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To add an isolated point at $(0,0)$, perform the above construction so that all singularities occur for $z \neq 0$ and the curve does not pass through 0, then replace the parametrization by $x = (1+z^2)f(z), \quad y= (1+z^2)g(z)$.  The algebraic curve you want is the Zariski closure of the parametrized thing.  This is the equation between $x$ and $y$ obtained by elimination of $z$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To get a bounded locus, replace any rational parametrization $(A(t),B(t))$ by $(A,B)/(1+(PAB)^2)$ where $P$ is a polynomial vanishing to high order at all the singular points arranged for $A$ and $B$. &lt;/p&gt;&#xA;" OwnerUserId="467" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-10-25T05:50:12.987" LastActivityDate="2010-10-25T05:50:12.987" CommentCount="5" />
  <row Id="7598" PostTypeId="1" AcceptedAnswerId="7599" CreationDate="2010-10-23T16:54:36.140" Score="0" ViewCount="462" Body="&lt;p&gt;I want to know what is the Taylor series of $\sqrt{1+x^2}$  while $x \to 0$. &lt;/p&gt;&#xA;" OwnerUserId="2728" LastEditorUserId="171" LastEditorDisplayName="" LastEditDate="2010-10-23T17:32:51.983" LastActivityDate="2010-10-23T22:37:14.237" Title="Taylor series of $\sqrt{1 + x^2}$" Tags="&lt;sequences-and-series&gt;&lt;homework&gt;&lt;taylor-expansion&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="7599" PostTypeId="2" ParentId="7598" CreationDate="2010-10-23T16:58:51.567" Score="3" ViewCount="" Body="&lt;p&gt;If you just want the answer, you can find it at &lt;a href=&quot;http://www.wolframalpha.com/input/?i=series%28sqrt%281%2Bx%5e2%29%29&quot; rel=&quot;nofollow&quot;&gt;WolframAlpha&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2010-10-23T17:40:08.100" LastActivityDate="2010-10-23T17:40:08.100" CommentCount="4" />
  <row Id="7600" PostTypeId="1" CreationDate="2010-10-23T17:25:31.133" Score="0" ViewCount="96" Body="&lt;p&gt;Is there a way to calculate a f test using a beta distribution?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, if I have&#xA;(x1,x2 ..., xn)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and &#xA;(y1, y2 ..., yk)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can easily compute the variance of x and y.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, how does it relate to a beta distribution?&lt;/p&gt;&#xA;" OwnerUserId="2729" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-10-23T17:35:46.550" LastActivityDate="2010-10-23T17:35:46.550" Title="What is the relationship between f test and the beta distribution" Tags="&lt;probability&gt;&lt;statistics&gt;" CommentCount="1" />
  <row Id="7601" PostTypeId="1" CreationDate="2010-10-23T17:41:33.230" Score="0" ViewCount="67" Body="&lt;p&gt;how do you go from (πr²/r)²s² to (2πr)²s² ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;its supposed to be part of propagation of error in radius to circle area&#xA;should be fairly simple but i am newbie :-)&lt;/p&gt;&#xA;" LastActivityDate="2010-10-23T17:41:33.230" ClosedDate="2010-10-24T00:48:35.780" Title="how do you go from (πr²/r)²s² to (2πr)²s² ?" Tags="&lt;geometry&gt;" CommentCount="1" />
  <row Id="7602" PostTypeId="2" ParentId="7511" CreationDate="2010-10-23T18:28:39.117" Score="1" ViewCount="" Body="&lt;p&gt;Sorry to give yet another answer that does not address the issue of fractional $n$ [it seems that fractional derivatives are not such a familiar topic to many research mathematicians; certainly they're not to me], but:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a little issue here which has not been addressed.  By the context of the OP's question, I gather s/he is looking for &lt;strong&gt;real-valued&lt;/strong&gt; functions which are equal to their $n$th derivative (and not their $k$th derivative for $k &amp;lt; n$).  Several answerers have mentioned that the set of solutions to $f^{n} = 0$ forms an $n$-dimensional vector space.  But over what field?  It is easier to identify the space of such &lt;strong&gt;complex-valued&lt;/strong&gt; functions, i.e., $f: \mathbb{R} \rightarrow \mathbb{C}$: namely, a $\mathbb{C}$-basis is given by $f(x) = e^{2 \pi i k/n}$ for $0 \leq k &amp;lt; n$.  But what does this tell us about the $\mathbb{R}$-vector space of real-valued solutions to this differential equation?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer is that it is $n$-dimensional as a $\mathbb{R}$-vector space, though it does not have such an immediately obvious and nice basis.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $W$ be the $\mathbb{R}$-vector space of real-valued functions $f$ with $f^{(n)} = 0$ and $V$ the $\mathbb{C}$-vector space of $\mathbb{C}$-valued functions $f$ with $f^{(n)} = 0$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a natural inclusion map $W \mapsto V$.  Phrased algebraically, the question is whether the induced map $L: W \otimes_{\mathbb{R}} \mathbb{C} \rightarrow V$ is an isomorphism of $\mathbb{C}$-vector spaces.  In other words, this means that any given $\mathbb{R}$-basis of $W$ is also a $\mathbb{C}$-basis of $V$.  This is certainly not automatic.  For instance, viewing the Euclidean plane as first $\mathbb{R}^2$ and second as $\mathbb{C}$ gives a map &#xA;$\mathbb{R}^2 \rightarrow \mathbb{C}$ which certainly does not induce an isomorphism upon tensoring with $\mathbb{C}$, since the first space has (real) dimension $2$ but the second space has (complex) dimension $1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For more on this, see Theorem 1.6 of&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.math.uconn.edu/~kconrad/blurbs/galoistheory/galoisdescent.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.math.uconn.edu/~kconrad/blurbs/galoistheory/galoisdescent.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It turns out that this is actually a problem in &lt;strong&gt;Galois descent&lt;/strong&gt;: according to Theorem 2.14 of the notes of Keith Conrad already cited above, the map $L$ is an isomorphism iff there exists a conjugate-linear involution $r: V \rightarrow V$, i.e., i.e., a map which is self-inverse and satisfies, for any $z \in \mathbb{C}$ and $v \in V$, $r(zc) = \overline{z} r(c)$.&lt;br&gt;&#xA;But indeed we have such a thing: an element of $V$ is just a complex-valued function $f$, &#xA;so we put $r(f) = \overline{f}$.  Note that this stabilizes $V$ since the differential equation $f^{(n)}) = 0$ &quot;is defined over $\mathbb{R}$&quot;: or more simply, the complex conjugate of the $n$th derivative is the $n$th derivative of the complex conjugate.  Thus we have &quot;descent data&quot; (or, in Keith Conrad's terminology, a &lt;strong&gt;G-structure&lt;/strong&gt;) and the real solution space has the same dimension as the complex solution space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is a nice exercise to use these ideas to construct an explicit real basis of $W$.  &lt;/p&gt;&#xA;" OwnerUserId="299" LastActivityDate="2010-10-23T18:28:39.117" CommentCount="2" />
  <row Id="7603" PostTypeId="2" ParentId="7593" CreationDate="2010-10-23T18:46:39.030" Score="0" ViewCount="" Body="&lt;p&gt;Seems pretty easy. We're going to cover the complex plane instead since that's obviously equivalent to covering $\mathbb{R}^2$. Start with the family of unit length line segments $I_\theta = \{a\cdot e^{i\theta} \mid a \in (0,1] \}$, $\theta \in (0,2\pi)$. Then define $I_{\theta,n}$, $n \in \mathbb{N}_0$ to be $I_\theta$ translated a length of $n$ in the direction of $\theta$. This collection covers the entire complex plane except the ray $[0,\infty)$ which we cover with the half-open intervals $[k,k+1)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Armed with this idea, it should be easy to see how to for any $k$ fill $\mathbb{R}^k$ with intervals of unit length that &quot;point in all directions&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;FYI: &lt;code&gt;\mathbb{N}_0&lt;/code&gt; breaks stuff. You have to escape the underscore: &lt;code&gt;\mathbb{N}\_0&lt;/code&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1269" LastEditorUserId="1269" LastEditorDisplayName="" LastEditDate="2010-10-23T18:53:09.080" LastActivityDate="2010-10-23T18:53:09.080" CommentCount="3" />
  <row Id="7604" PostTypeId="1" CreationDate="2010-10-23T19:08:28.073" Score="1" ViewCount="73" Body="&lt;p&gt;I remember hearing somewhere that if $p^n$ is the largest power of a prime $p$ dividing the class number $h_K$ of a number field $K$, then there is unique factorization of the $p^n$-th powers of elements in $\mathcal{O}_K$ into $p^n$-th powers of irreducibles. I can see that there can't be any $p$-torsion for a $p^n$-th power of an ideal class (because $p^n$ is the largest power dividing $h_K$), but I'm confused as to how to translate this into a statement about factorization of elements, and why $p$ is arbitrary. Am I remembering this result correctly (if not, what is the right formulation)? Can someone provide a proof or reference?&lt;/p&gt;&#xA;" OwnerUserId="264" LastEditorUserId="264" LastEditorDisplayName="" LastEditDate="2010-10-23T20:26:16.397" LastActivityDate="2010-10-23T20:26:16.397" Title="Recovering unique factorization of elements by only looking at $p^n$-th powers?" Tags="&lt;algebraic-number-theory&gt;" CommentCount="5" />
  <row Id="7605" PostTypeId="1" AcceptedAnswerId="7613" CreationDate="2010-10-23T19:28:34.320" Score="3" ViewCount="235" Body="&lt;p&gt;In general, how would one find the distribution of $f(X)$ where $X$ is a random variable? Or consider the inverse problem of finding the distribution of $X$ given the distribution of $f(X)$. For example, what is the distribution of $\max(X_1, X_2, X_3)$ if $X_1, X_2$ and $X_3$ have the same distribution? Likewise, if one is given the distribution of $ Y = \log X$, then the distribution of $X$ is deduced by looking at $\text{exp}(Y)$?&lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2011-06-05T07:27:46.540" Title="Distribution of Functions of Random Variables" Tags="&lt;probability&gt;" AnswerCount="4" CommentCount="6" />
  <row Id="7606" PostTypeId="1" AcceptedAnswerId="7611" CreationDate="2010-10-23T19:30:28.760" Score="1" ViewCount="244" Body="&lt;p&gt;Suppose $p$ is prime, $q|(p-1)$. Prove the congruence $1+x+\cdots+x^{q-1}\equiv 0 \pmod{p}$ has exactly $q$ solutions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have this as a homework assignment and I've having trouble proving it. I'm trying to use Lagrange's theorem:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;If $p$ is prime, an $n$th degree polynomial has at most $n$ solutions mod $p$.&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My attempt so far:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$q|(p-1)$ implies $p-1 = qr$ for some $r$ in $\mathbb{N}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$1+x+\cdots+x^{p-1} = (1+x+\cdots+x^{q-1})  h(x)$&#xA;where $h(x)$ is a polynomial of degree $q(r-1)-1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(since $x^{q(r-1)-1}x^{q-1}=x^{qr-q-1+q-1}=x^{qr-2}=x^{p-1}$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but here I'm stuck. I'm not quite sure where to go from here. I think I can say that all the solutions of $1+x+\cdots+x^{p-1} \mod p$ are either solutions of $1+x+\cdots+x^{q-1}$ or $h(x)$ modilo $p$, but I'm not quite sure if that's true, or if it is, where to go from there. I'd appreciate any help or hints, thanks! &lt;/p&gt;&#xA;" OwnerUserId="2466" LastEditorUserId="1242" LastEditorDisplayName="" LastEditDate="2010-10-23T20:32:16.937" LastActivityDate="2010-10-23T21:55:43.290" Title="solutions of polynomials using lagrange's theorem" Tags="&lt;number-theory&gt;&lt;homework&gt;" AnswerCount="2" CommentCount="5" />
  <row Id="7607" PostTypeId="2" ParentId="7578" CreationDate="2010-10-23T19:32:56.383" Score="2" ViewCount="" Body="&lt;p&gt;If $(X,d)$ is a connected, complete metric space and $G$ is a nonempty proper open subset of $X$, then $G$ is not complete with the restricted metric (because it is not closed), but $G$ is homeomorphic to a complete metric space, because the map $x\mapsto(x,1/d(x,X\setminus G))$ embeds $G$ onto a closed subspace of $X\times \mathbb{R}$.  (This is more or less &lt;a href=&quot;http://books.google.com/books?id=zRep59fhoGkC&amp;amp;lpg=PP1&amp;amp;dq=arveson%20invitation&amp;amp;pg=PA61#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;Lemma 3.1.1&lt;/a&gt; of Arveson's &lt;em&gt;Invitation&lt;/em&gt;.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A topological space that is homeomorphic to a separable complete metric space is called a &lt;a href=&quot;http://books.google.com/books?id=bQwhdmL6IjUC&amp;amp;lpg=PP1&amp;amp;pg=PA195#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;Polish space&lt;/a&gt;.  A subspace of a Polish space is Polish if and only if it is a $G_\delta$, while $G_\delta$s in complete metric spaces are not typically complete with the restricted metric.  For example, there is a complete metric inducing the usual topology on the set of irrational numbers.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-10-23T19:32:56.383" />
  <row Id="7608" PostTypeId="2" ParentId="7578" CreationDate="2010-10-23T19:57:33.527" Score="1" ViewCount="" Body="&lt;p&gt;Consider one complete normed vector space $(V,\|\cdot\|)$, that is, a Banach space, and let $$B = \{ x \in V : \|x\| &amp;lt; 1 \}$$ be the unit ball centered at the origin. Define $f: V \rightarrow B$ by $x \mapsto x /(1+ \|x\|)$. Then f is continuous and is easy to show that the inverse of $f$ is also continuous, so f is a homeomorphism. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now $B$ isn't complete, since if $x \in V$ and $\| x \| = 1$, then $x_n = (1-1/n)x$ is a cauchy sequence in $B$ that doesn't converge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As Kahen referred, if we have two metric spaces $X$ and $Y$ with the latter complete and a  uniformly continuous homeomorphism $g : X \rightarrow Y$, then $X$ is complete. To prove this you'll need to observe that since $g$ is uniformly continuous then it sends Cauchy sequences to  Cauchy sequences.&lt;/p&gt;&#xA;" OwnerUserId="329" LastActivityDate="2010-10-23T19:57:33.527" />
  <row Id="7610" PostTypeId="1" AcceptedAnswerId="7617" CreationDate="2010-10-23T20:01:25.327" Score="-1" ViewCount="169" Body="&lt;p&gt;Hello&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which mathematical subjects I have to learn in order to create my own formula about a topic something. For example, I am curious how Einstein could find his final formulas. It is all about logic or he had to know deeply some math rules and subjects in order to get his first formula?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was also wondering how a equation can appear from nothing, and what would be its start.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What subjects should I study to create my own formula and test it to see if if its results are plausible?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;thanks in advance&lt;/p&gt;&#xA;" OwnerUserId="2732" LastActivityDate="2010-10-23T21:56:59.247" ClosedDate="2010-10-23T21:29:49.093" Title="Which subjects do I have to learn to create a formula from scratch?" Tags="&lt;soft-question&gt;" AnswerCount="2" CommentCount="8" />
  <row Id="7611" PostTypeId="2" ParentId="7606" CreationDate="2010-10-23T20:21:25.070" Score="2" ViewCount="" Body="&lt;p&gt;I'm assuming that when you say &quot;congruence&quot;, you mean that you want to show that the congruence &#xA;$$x^{q-1} + \cdots + 1 \equiv 0 \pmod{p}$$&#xA;has exactly $q$ solutions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are incorrect in saying that $1+x+\cdots + x^{p-1}$ is a multiple of $1+x+\cdots+x^{q-1}$. For example, if $p=7$ and $q=3$ (which divides $7-1=6$), you would need $1+x+x^2$ to divide $1+x+x^2+x^3+x^4+x^5+x^6$. But doing long division, you get&#xA;$$x^6 + x^5 + x^4 + x^3 + x^2 + x + 1 = (x^2+x+1)(x^4+x) + 1,$$&#xA;so there is a remainder. Your assertion is false, which is probably why you are stuck.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(The reason I knew you were incorrect in the assertion is: notice that $x^q-1 = (x^{q-1}+\cdots+1)(x-1)$ and $x^p-1 = (x^{p-1}+\cdots+1)(x-1)$. The only solution modulo $p$ to $x^{p}-1$ is $x=1$, because by Fermat's Little Theorem you have $x^p \equiv x \pmod{p}$ for all $x$; if $x^{q-1}+\cdots+1$ divides $x^{p-1}+\cdots + 1$, then $(x^{q-1}+\cdots+1)(x-1)$ divides $(x^{p-1}+\cdots+1)(x-1)$; so every root of $x^{q-1}+\cdots+1$ has to be a root of $x^p-1$. But $x^{q-1}+\cdots+1$ has a lot of roots, and that contradicts the fact that $x^p-1$ only has one).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As Qiaochu suggests, consider the polynomial $x^q - 1 = (x-1)(x^{q-1}+\cdots+1)$. Since $p$ is prime, $x^q-1\equiv 0\pmod{p}$ if and only if at least one of $x-1$ and $x^{q-1}+\cdots+1$ are congruent to $0$ modulo $p$; and since $q|p-1$, then $1^{q-1}+1^{q-2}+\cdots+1 = q\not\equiv 0 \pmod{p}$. So $a$ is a solution to the original congruence if and only if $a$ is a solution to $x^q-1 \equiv 0 \pmod{p}$ &lt;strong&gt;and&lt;/strong&gt; $a\not\equiv 1 \pmod{p}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now look at the multiplicative group modulo $p$, which has order $p-1$, and look at a &lt;a href=&quot;http://en.wikipedia.org/wiki/Primitive_root_modulo_n&quot; rel=&quot;nofollow&quot;&gt;primitive root modulo $p$.&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you do not know about primitive roots, you can still more or less push through your attempt if you are a little more careful. Instead of claiming that $x^{q-1}+\cdots+1$ divides $x^{p-1}+\cdots + 1$ (which is &lt;em&gt;false&lt;/em&gt;), try to see if it divides $x^{p-1}-1$. By Fermat's Little Theorem, you know exactly how many roots modulo $p$ $x^{p-1}-1$ has. If you can find the $h(x)$ such that $(x^{q-1}+\cdots+1)h(x) = x^{p-1}-1$, then you know that every root of $x^{p-1}-1$ is either a root of $h(x)$ or a root of the polynomial you want. Then show that the two polynomials have no root in common, and use Lagrange's Theorem as you want to give an upper bound for the number of roots of $h(x)$, and hence a lower bound to the number of roots of $x^{q-1}+\cdots+x+1$. Then use Lagrange's Theorem again to get an upper bound.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-10-23T20:31:48.040" LastActivityDate="2010-10-23T20:31:48.040" CommentCount="9" />
  <row Id="7612" PostTypeId="2" ParentId="7610" CreationDate="2010-10-23T20:22:53.213" Score="0" ViewCount="" Body="&lt;p&gt;If there had been a simple rule/algorithm that would allow us to derive new things out of the air, everybody would have been an Einstein now.&lt;br /&gt;&lt;br /&gt;&#xA;The background of such formulas is usually a simple idea that the scientist plays with in his/her head and expands it to a more and more complex theory. I personally think that the formulas are the last thing - just a formal formulation of the ideas behind.&lt;/p&gt;&#xA;" OwnerUserId="2733" LastActivityDate="2010-10-23T20:22:53.213" />
  <row Id="7613" PostTypeId="2" ParentId="7605" CreationDate="2010-10-23T20:32:38.323" Score="1" ViewCount="" Body="&lt;p&gt;Qiaochu is right.  There isn't a magic wand.  That said, there is a set of common procedures that can be applied to certain kinds of transformations.  One of the most important is the cdf (cumulative distribution function) method that you are already aware of.  (It's the one used in your &lt;a href=&quot;http://math.stackexchange.com/questions/7324/&quot;&gt;previous question&lt;/a&gt;.)  Another is to do a change of variables, which is like the method of substitution for evaluating integrals.  You can see that procedure and others for handling some of the more common types of transformations at this &lt;a href=&quot;http://www.math.uah.edu/stat/dist/Transformations.xhtml&quot; rel=&quot;nofollow&quot;&gt;web site&lt;/a&gt;.  (Some of the other examples there include finding maxes and mins, sums, convolutions, and linear transformations.)&lt;/p&gt;&#xA;" OwnerUserId="2370" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-10-23T20:42:52.840" LastActivityDate="2010-10-23T20:42:52.840" />
  <row Id="7614" PostTypeId="1" AcceptedAnswerId="7615" CreationDate="2010-10-23T20:42:25.010" Score="5" ViewCount="173" Body="&lt;p&gt;This is a chug-plug formula given in my book :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1.The number of ways in which &lt;strong&gt;mn&lt;/strong&gt; different items can be divided equally int &lt;strong&gt;m groups&lt;/strong&gt;, each containing &lt;strong&gt;n objects and the order of the groups is not important&lt;/strong&gt;,is : $\frac {(mn)!}{(n!)^m} \frac {1}{m!} $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2.The number of ways in which &lt;strong&gt;mn&lt;/strong&gt; different items can be divided equally int &lt;strong&gt;m groups&lt;/strong&gt;, each containing &lt;strong&gt;n objects and the order of the groups is important&lt;/strong&gt;,is : $\frac {(mn)!}{(n!)^m} \frac {1}{m!} * (m!) $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am having some troubles in understanding the above two formulas and hence making mistakes while solving problems based on the same, so I am very inquisitive to know the proof.&lt;/p&gt;&#xA;" OwnerUserId="2109" LastEditorUserId="2109" LastEditorDisplayName="" LastEditDate="2010-11-11T15:17:23.533" LastActivityDate="2010-11-11T15:17:23.533" Title="Grouping items into groups" Tags="&lt;combinatorics&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="7615" PostTypeId="2" ParentId="7614" CreationDate="2010-10-23T20:51:16.010" Score="4" ViewCount="" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Imagine first that both the order of the groups &lt;em&gt;and&lt;/em&gt; the order within the groups is important. You have $mn$ items; one way to divide them into $m$ groups of $n$ objects each is to simply list the items. Items 1 through $n$ will go into the first group; items $n+1$ through $2n$ will go into the second group; etc. There are $(mn)!$ ways of listing the items. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But in fact, you don't care about the order in which the groups appear on the list: if you take the same list you have, and now list items $n+1$ through $2n$ first, then, say, times $4n+1$ through $5n$, then items $1$ through $n$, etc., you will still get the same division of objects into the same groupings. So any shuffling of the blocks in your list will yield the same distribution of items into the groups. Each distribution into groups can then be described in $m!$ ways, one for each ordering of the $m$ groups. So you have counted each distribution $m!$ times, and you need to divide $(mn)!$ by $m!$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are still not done, though, because you &lt;em&gt;also&lt;/em&gt; don't care about the order within each group. Again, say you have a listing: if you take the first $n$ items in the list, and you list them in a different order (say, from last to first) while keeping them in the first $n$ slots, then in the end you will have the same elements separated into the same groups: you don't care who goes into group $1$ first or last, just that they end up in that group. So you have also overcounted by as many ways as there are to list the members of each group in order. In each group, you have $n$ objects, which can be listed in $n!$ different ways. There are $m$ such groups, so each listing of the elements in the groups can be done in $(n!)^m$ ways; so you have &lt;em&gt;also&lt;/em&gt;, separately, overcounted the lists by $(n!)^m$, forcing you to also divide by $(n!)^m$. Putting these two overcount factors together gives you the first formula.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Try to adapt the explanation above to see how to obtain this formula; note that a simpler way of writing it is just as $\frac{(mn)!}{(n!)^m}$, since the two $m!$ will cancel out.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-10-23T21:07:52.750" LastActivityDate="2010-10-23T21:07:52.750" CommentCount="4" />
  <row Id="7617" PostTypeId="2" ParentId="7610" CreationDate="2010-10-23T20:54:59.410" Score="1" ViewCount="" Body="&lt;p&gt;Just consider doing an experiment: Let's take a bowling ball up a scaffold, the higher we drop it off the building the harder a dent it will put into the soft earth below. Suppose we did a whole series of experiments and measured each case - when plotting the results we might get a graph that curves up quickly - since graphs are and functions are essentially the same thing - we have a formula too. This is how &lt;a href=&quot;http://en.wikipedia.org/wiki/Willem_%27s_Gravesande&quot; rel=&quot;nofollow&quot;&gt;Willem Gravesande&lt;/a&gt; found the famous formula for kinetic energy $\frac{1}{2}m v^2$. Another fun example of this is &lt;a href=&quot;http://en.wikipedia.org/wiki/Rydberg_formula#Rydberg_formula_for_hydrogen&quot; rel=&quot;nofollow&quot;&gt;Rydberg formula&lt;/a&gt;, data had been measured and he just tried out lots of different equations until finding one which fit the data nicely. This is the experimental derivation of a physical formula.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is not always necessary to do experiments to discover physical formulae though. Newtons theory of mechanics gives the formula $F = ma$. Using calculus (i.e. purely mathematical reasoning) we can derive the formula $E_k= \frac{1}{2}m v^2$. &lt;a href=&quot;http://en.wikipedia.org/wiki/Kinetic_energy#Derivation&quot; rel=&quot;nofollow&quot;&gt;Here&lt;/a&gt; is the explicit derivation but the point is just that it is possible to derive new formula as consequences of the mathematical model. This is the mathematical derivation of a physical formula.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another approach (you can follow this approach to derive the equations for relativity) is to define some physical principles that you want to model - define a physical theory as a set of postulates. For example &quot;The speed of light is constant for all observers&quot;, etc.. then construct mathematical models of these ideas. Once you have the model you can derive consequences from it and build experiments to check them. Another example: Niels Bohr devised a model of the atom from which he was able to derive the Rydberg formula. I think this is called Mathematical Physics, or maybe just Physics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, to do this sort of work you need to be a very strong mathematician - but also have a good intuition about reality. If you just study basic physics (like watching the Caltech lectures) you can practice the skills needed.&lt;/p&gt;&#xA;" LastEditorUserId="664" LastEditorDisplayName="" LastEditDate="2010-10-23T21:56:59.247" LastActivityDate="2010-10-23T21:56:59.247" />
  <row Id="7618" PostTypeId="1" AcceptedAnswerId="7620" CreationDate="2010-10-23T21:38:18.373" Score="5" ViewCount="98" Body="&lt;p&gt;I am trying to self-read through Ian Stewart's Galois theory and am stuck at this paragraph. I quote - &quot;Lagrange observed that all methods for solving polynomial equations by radicals involve constructing rational functions of the roots that take a small number of values when the roots $\alpha_j$ are permuted. Prominent among these is the expression&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\delta = \prod_{j&amp;lt; k} (\alpha_j - \alpha_k)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which takes just two values, $\pm \delta$, plus for even permutations and minus for the odd ones. Therefore, $\Delta = \delta^{2}$ is a rational function of the coefficients. This gets us started and it yields a complete solution for the quadratic.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is as follows: Why should an expression that is unchanged by permuting the roots of a polynomial be expressible as a rational function of the polynomial's coefficients? Is this statement even true or have I misunderstood the book? &lt;/p&gt;&#xA;" OwnerUserId="618" LastActivityDate="2010-11-02T20:16:12.967" Title="Expressions unchanged by permuting the roots of a polynomial" Tags="&lt;group-theory&gt;&lt;polynomials&gt;" AnswerCount="1" FavoriteCount="2" />
  <row Id="7619" PostTypeId="1" CreationDate="2010-10-23T21:42:13.557" Score="2" ViewCount="155" Body="&lt;p&gt;I am trying to solve some problems based on this &lt;a href=&quot;http://math.stackexchange.com/questions/7614/how-to-prove-the-following&quot;&gt;formula&lt;/a&gt;,but am facing some issues in determine whether or not consider ordering as important.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For Example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;In how many ways 15 different books can be divided five heaps of equal number of books ?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So here according to the answer the ordering of groups is not considered important,hence the answer given is $\frac {15!} {5!(3!)^5}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and the same thing goes for this one :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;In how many ways 3 piles can be formed out of 18 different books&lt;/strong&gt;, which will be $\frac {18!} {3!(6!)^3}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But for the following problem &lt;strong&gt;&quot;There are three copies, each of four different books. In how many ways they can be arranged on a self ?&quot;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This this is similar to divide 12 books into 4 sets of 3 books each, the answer given here considers the ordering important,why this change ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why in the first two the ordering not and on the third one it is important ? I guess (if the solution is correct) it is something related to copies but am not confident enough.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A good example of a problem where ordering should be considered important is say &quot;&lt;strong&gt;In how many ways can 18 different books be divided equally among three students ?&quot;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The solution should be $\frac {18!} {(6!)^3}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I am not much confident while in these problems so I will be grateful if somebody helps me to understand when to consider ordering important and when not to.&lt;/p&gt;&#xA;" OwnerUserId="2109" LastEditorUserId="2109" LastEditorDisplayName="" LastEditDate="2010-11-11T15:18:50.770" LastActivityDate="2010-11-11T15:18:50.770" Title="Dealing with grouping problem in combinatorics" Tags="&lt;combinatorics&gt;" AnswerCount="2" />
  <row Id="7620" PostTypeId="2" ParentId="7618" CreationDate="2010-10-23T21:42:57.873" Score="8" ViewCount="" Body="&lt;p&gt;This is a straightforward consequence of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Elementary_symmetric_polynomial#The_fundamental_theorem_of_symmetric_polynomials&quot; rel=&quot;nofollow&quot;&gt;Fundamental Theorem on Symmetric Functions&lt;/a&gt;. Indeed, the coefficients of the polynomial are the evaluation of the so called elementary symmetric functions at the roots.&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2010-10-23T21:42:57.873" CommentCount="3" />
  <row Id="7621" PostTypeId="1" AcceptedAnswerId="7658" CreationDate="2010-10-23T21:46:33.733" Score="7" ViewCount="340" Body="&lt;p&gt;This thing came up in a combinatorics course I am taking.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Choose a fixed set of primes $p_1,p_2,\dots,p_k$ and let $A_n$ be number of integers in $\{1,2,\dots,n\}$ which are not divisible by any of the $p_i$'s. $A_n$ is given by $ n - \sum_{1\leq i_1 \leq k} \lfloor \frac{n}{p_{i_1}} \rfloor + \sum_{1\leq i_1 &amp;lt; i_2 &amp;lt; k} \lfloor \frac{n}{p_{i_1}p_{i_2}}\rfloor - \dots $. Now, if $n$ is divisible by each of the $p_i$'s then we have the simpler expression : $A_n = n \prod_{i=1}^{k}(1-\frac{1}{p_i})$(added later : note I am not assuming $n = \prod_{i=1}^{k}p_i$ here, this holds true whenever $n = c \prod_{i=1}^{k} p_i $ for some integer c, since each $\lfloor \frac{n}{p_{i_1} \dots p_{i_j}} \rfloor = \frac{n}{p_{i_1} \dots p_{i_j}}\ )$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another student pointed out that even if $n$ does not have $\prod_{i=1}^{k} p_i$ as a factor, the approximation $B_n = n \prod_{i=1}^{k}(1-\frac{1}{p_i})$ to $A_n$ is quite close in some specific cases. It is easy to see that $\lim_{n\to\infty}\frac{A_n - B_n}{n}=0$ as $\lim_{n\to\infty}\frac{1}{n}\lfloor \frac{n}{p_{i_1}p_{i_2}\dots p_{i_j}} \rfloor \to \frac{1}{p_{i_1}p_{i_2} \dots p_{i_j}},$ however that is not strong enough to imply that $A_n - B_n$ will be always close.&#xA;Is there any way to analyze how well this approximation will perform in general? I am interested in the worst case for small to moderately sized $n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just to give a feel of how close $A_n$ and $B_n$ can get, assuming the set of primes is $\{2,3,7\}$ we have (assuming my program is correct):&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;n     A(n)     B(n)&#xA;17    5    4.86  &#xA;27    8    7.71  &#xA;37    11      10.57  &#xA;107   31      30.57  &#xA;1111  318     317.43  &#xA;3001  858     857.43  &#xA;4007  1145    1144.86  &#xA;5000  1429    1428.57 .&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="404" LastEditorUserId="404" LastEditorDisplayName="" LastEditDate="2010-10-25T18:00:58.520" LastActivityDate="2010-10-26T12:19:54.940" Title="Accuracy of approximation to inclusion-exclusion formula in prime sieve" Tags="&lt;number-theory&gt;&lt;combinatorics&gt;&lt;analytic-number-theory&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="2" />
  <row Id="7622" PostTypeId="2" ParentId="7606" CreationDate="2010-10-23T21:50:32.350" Score="2" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\: $ Since $\rm\ x^q - 1\ $ divides the prime product $\rm\ x^{p-1} - 1\ =\ (x-1)\:(x-2)\:\cdots\:(x-p+1)\ $ by uniqueness it is uniquely a product of $\rm\:q\:$ of them, so it has precisely $\rm\:q\:$ roots.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;REMARK&lt;/strong&gt; $\ $ Products of primes factor uniquely over any integral domain, i.e. any other factorization into irreducibles is the same up to order and unit factors. Such &lt;strong&gt;uniqueness&lt;/strong&gt; of prime factorizations is the trivial part of being a UFD - the nontrivial part is the &lt;strong&gt;existence&lt;/strong&gt; of prime factorizations for all nonzero nonunits.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-10-23T21:55:43.290" LastActivityDate="2010-10-23T21:55:43.290" />
  <row Id="7623" PostTypeId="2" ParentId="7619" CreationDate="2010-10-23T21:54:13.900" Score="0" ViewCount="" Body="&lt;p&gt;In the first one, the problem only states that you need to group them, as in, let's say, put the books in a bag, so there could be no order.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second one is very similar. It would be different if it had stated it wanted a stack of books, which would have been something with an order, where you can say: book A is in the bottom, book B is right on top of A, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The third, however, asks for an arrangement of books, which would be like putting them all on a shelf. Since you can't distinguish between 2 copies of the same book, swapping them around won't change the order of the books, so that is why you need to divide by the number of copies of the same book, so you don't count the very same arrangement twice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Overall, I think that in these kind of problems, grouping stuff together means that order is not important (picture putting books in several big bags) and arranging things implies that there is order (imagine putting books in a shelf).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your last question, ordering is not important, as you are not asking in which order the books can be given, but which different groups of books can be given to each student. Say we give books A B C D E F to student 1. He would end up with the same books if we had given him B C A E F D, so order is not important.&lt;/p&gt;&#xA;" OwnerUserId="1605" LastEditorUserId="1605" LastEditorDisplayName="" LastEditDate="2010-10-23T21:59:48.140" LastActivityDate="2010-10-23T21:59:48.140" />
  <row Id="7625" PostTypeId="2" ParentId="5921" CreationDate="2010-10-23T22:42:55.797" Score="3" ViewCount="" Body="&lt;p&gt;Here's a rough sketch of one approach:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;\mathbb{P}[Y\lt X] &#xA;&amp;amp;= \int_{-\infty}^{\infty} \mathbb{P}[Y\lt X|X=x] \cdot f_X(x) \,dx&#xA;\\&#xA;&amp;amp;= \int_{-\infty}^{\infty} F_Y(x) \cdot f_X(x) \,dx&#xA;\\&#xA;&amp;amp;\geq \int_{-\infty}^{\infty} F_X(x) \cdot f_X(x) \,dx&#xA;\\&#xA;&amp;amp;= \left[ \frac{1}{2} \left(F_X(x)\right)^2 \right]_{-\infty}^{\infty}&#xA;\\&#xA;&amp;amp;= \frac{1}{2}.&#xA;\end{align}&lt;/p&gt;&#xA;" OwnerUserId="1023" LastActivityDate="2010-10-23T22:42:55.797" />
  <row Id="7626" PostTypeId="2" ParentId="7619" CreationDate="2010-10-23T22:45:58.127" Score="1" ViewCount="" Body="&lt;p&gt;There are different notions of &quot;ordered&quot; and &quot;unordered&quot; that can arise in relation to set partitions.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;An unordered set of unordered sets,&lt;/li&gt;&#xA;&lt;li&gt;An unordered set of ordered sets,&lt;/li&gt;&#xA;&lt;li&gt;An ordered set of unordered sets,&lt;/li&gt;&#xA;&lt;li&gt;An ordered set of ordered sets,&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;all of which are counted differently.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In how many ways 15 different books can be divided five heaps of equal number of books ?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;An example of what is counted here is:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;{{1,3,8},{2,14,15},{4,5,7},{6,12,13},{9,10,11}}&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;is an unordered set of unordered sets.  This is what the answer to the first question suggests it's counting.  [Although, the question itself does not resolve the question of order within the heaps.]&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;There are three copies, each of four different books. In how many ways they can be arranged on a self ?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;An example of what is counted here is:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;1 2 3 1 4 1 3 3 2 2 4 4&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;which we can interpret as a set partition counting problem -- place the i-th element in the s(i)-th set, where s(i) denotes the element in the i-th position.  So the above example gives rise to:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;({1,4,6},{2,9,10},{3,7,8},{5,11,12})&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;which is an ordered set of unordered sets.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;To highlight the difference between these two situations, for the first question:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;{{1,3,8},{2,14,15},{4,5,7},{6,12,13},{9,10,11}}&#xA;{{2,14,15},{1,3,8},{4,5,7},{6,12,13},{9,10,11}}&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;should be considered the same (this is the same as swapping the positions of the first two heaps), whereas in the second question:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;({1,4,6},{2,9,10},{3,7,8},{5,11,12})&#xA;({2,9,10},{1,4,6},{3,7,8},{5,11,12})&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;should be considered different (this is the same as swapping all the books labelled 1 with all the books labelled 2).&lt;/p&gt;&#xA;" OwnerUserId="139" LastActivityDate="2010-10-23T22:45:58.127" />
  <row Id="7627" PostTypeId="1" AcceptedAnswerId="7631" CreationDate="2010-10-23T22:46:41.163" Score="6" ViewCount="207" Body="&lt;p&gt;I want to graph an equilateral triangle. It would be ideal if I had a set of three points: $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ with $x_i, y_i \in \mathbb{Z}$ as the vertices. However, this is impossible. I am willing to &quot;settle&quot; for a triangle that is off by the width of a pencil lead. (That is, if we draw circles of radius $\delta$ around all of the lattice points, the true location of the vertices of the equilateral triangle are inside of these error circles.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;For a given $\delta$ what is the smallest such triangle?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My instinct is to let the computer guess, is there a more elegant way?&lt;/p&gt;&#xA;" OwnerUserId="2563" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-10-23T23:52:51.653" LastActivityDate="2010-10-23T23:52:51.653" Title="I want to graph an equilateral triangle on graph paper..." Tags="&lt;number-theory&gt;&lt;approximation&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" />
  <row Id="7628" PostTypeId="1" AcceptedAnswerId="7847" CreationDate="2010-10-23T23:00:35.083" Score="7" ViewCount="242" Body="&lt;p&gt;It is a theorem that every locally free coherent sheaf on $\mathbb{P}^1$ over an algebraically closed field is isomorphic to a unique sum of sheaves $\mathcal{O}(n)$ for various integers $n$. In particular, the K-ring of locally free coherent sheaves (or all coherent sheaves, $\mathbb{P}^1$ being nonsingular) is isomorphic to $\mathbb{Z}[t, t^{-1}]$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The topological K-ring of vector bundles on $S^2$ is, by Bott periodicity, isomorphic to $\mathbb{Z}[H]/(H-1)^2$, where $H$ is the canonical bundle. But $S^2$ is homeomorphic to $\mathbb{P}^1_{\mathbb{C}}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Every locally free sheaf corresponds to a vector bundle on $S^2$. It follows that the map on the K-groups from locally free sheaves to vector bundles is surjective but not injective. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Questions:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;What goes wrong? &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Is there a version of Bott periodicity for algebraic varieties (or schemes)? (I.e., relating K-groups of $X$ and $X \times \mathbb{P}^1$.) I understand that there is one for the Picard groups. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="536" LastActivityDate="2010-10-25T21:00:23.740" Title="Bott periodicity and algebraic geometry" Tags="&lt;algebraic-geometry&gt;&lt;algebraic-topology&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="7629" PostTypeId="1" CreationDate="2010-10-23T23:09:42.143" Score="1" ViewCount="71" Body="&lt;p&gt;I'm wondering if there is other terminology that describes row-completeness, outside the context of a latin square, or if row-complete is actually a general term.&lt;/p&gt;&#xA;" LastEditorUserId="139" LastEditorDisplayName="" LastEditDate="2010-11-19T10:36:40.500" LastActivityDate="2011-01-18T11:46:34.280" Title="Does the term Row-Complete have any synonyms?" Tags="&lt;combinatorics&gt;&lt;terminology&gt;&lt;latin-square&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="7630" PostTypeId="2" ParentId="7627" CreationDate="2010-10-23T23:10:16.000" Score="3" ViewCount="" Body="&lt;p&gt;If you want one of the sides to be along the axis, you can use (0,0),(2x,0),and (x,y).  Then you want $sqrt(y^2+x^2)=2x$ or $y^2=3x^2$.  &lt;a href=&quot;http://en.wikipedia.org/wiki/Pell%27s_equation&quot; rel=&quot;nofollow&quot;&gt;Pell's equation&lt;/a&gt; will help you find solutions in the integers to $y^2=3*x^2\pm1$.  $x=30, y=26$ is within .017 in y.  Once you have one solution (say (a,b)=(1,0)), you can get another one as (2a+3b,a+2b).  These get very close to equilateral.  I don't know if you can do better if you don't put one side along an axis.  &lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-10-23T23:10:16.000" />
  <row Id="7631" PostTypeId="2" ParentId="7627" CreationDate="2010-10-23T23:17:38.953" Score="5" ViewCount="" Body="&lt;p&gt;Great question!  Here is an upper bound: let $\frac{p_n}{q_n}$ be the sequence of convergents to the &lt;a href=&quot;http://en.wikipedia.org/wiki/Continued_fraction&quot; rel=&quot;nofollow&quot;&gt;continued fraction&lt;/a&gt; of $\sqrt{3}$.  It is known that $| \sqrt{3} - \frac{p_n}{q_n} | &amp;lt; \frac{1}{q_n q_{n+1}}$.  Then the vertices&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(0, 0), (2q_n, 0), (q_n, p_n)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;are within $|\sqrt{3} q_n - p_n| &amp;lt; \frac{1}{q_{n+1}}$ of an equilateral triangle.  Asymptotically I believe that $q_n \sim C \cdot \sqrt{3}^n$ for some constant $C$, which means that for small $\delta$ we can always find a triangle with side length something like $\frac{2}{\delta}$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a way to get a lower bound: suppose $(x_i, y_i)$ is a collection of points which are $\delta$-close to an equilateral triangle, and suppose WLOG that $(x_1, y_1) = (0, 0)$.  Then $\frac{x_3 + iy_3}{x_2 + iy_2}$ is a rational approximation to $\frac{1}{2} + \frac{\sqrt{3}}{2} i$, hence taking imaginary parts, $2 \frac{x_2 y_3 - x_3 y_2}{x_2^2 + y_2^2}$ is a rational approximation to $\sqrt{3}$.  It is known that the convergents $\frac{p_n}{q_n}$ give the best rational approximations to $\sqrt{3}$ (in the sense that better ones must have larger denominators), so one should be able to write down how good this approximation is in terms of $\delta$ and use this to give a lower bound on how large $x_2^2 + y_2^2$ must be.  But the details look messy.  In any case, I would be very surprised if one didn't end up with a lower bound of the form $\frac{C}{\delta}$.  &lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-10-23T23:27:46.337" LastActivityDate="2010-10-23T23:27:46.337" CommentCount="3" />
  <row Id="7632" PostTypeId="2" ParentId="7586" CreationDate="2010-10-23T23:48:40.123" Score="1" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://math.stackexchange.com/questions/7586/solve-z2-6z-25-into-complex-conjugate/7588#7588&quot;&gt;Noldorin's answer&lt;/a&gt; pretty much covers what I'd say, but I would note that you can proceed in working with the expression until it is factored into linear terms and read off the zeros (the solutions to the equation $(z - 3)^2 + 16=0$) from there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Continuing from $(z - 3)^2 + 16$:&#xA;$$\begin{align}&#xA;(z - 3)^2 + 16 &amp;amp;= (z - 3)^2 - (-16)&#xA;\\&#xA;&amp;amp;= (z - 3)^2 - (4i)^2&#xA;\\&#xA;&amp;amp;= ((z-3)-4i)((z-3)+4i)&#xA;\\&#xA;&amp;amp;= (z-(3+4i))(z-(3-4i))&#xA;\end{align}$$&#xA;So, the zeros of the expression are $3+4i$ and $3-4i$.&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-10-23T23:48:40.123" CommentCount="1" />
  <row Id="7633" PostTypeId="2" ParentId="7629" CreationDate="2010-10-23T23:55:25.800" Score="2" ViewCount="" Body="&lt;p&gt;For those who are unfamiliar, a &lt;em&gt;Latin square&lt;/em&gt; $L=(l_{ij})$ is an n by n matrix in which each symbol occurs exactly once in each row an each column, such as the following:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;0 1 5 2 4 3&#xA;1 2 0 3 5 4&#xA;5 0 4 1 3 2&#xA;2 3 1 4 0 5&#xA;4 5 3 0 2 1&#xA;3 4 2 5 1 0&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;A Latin square L is called &lt;em&gt;row-complete&lt;/em&gt; if there are n(n-1) distinct tuples of the form $(l_{ij},l_{i(j+1)})$ in L.  The above Latin square is a row-complete Latin square (sourced from &lt;a href=&quot;http://designtheory.org/library/encyc/exs/clsex.html&quot; rel=&quot;nofollow&quot;&gt;http://designtheory.org/library/encyc/exs/clsex.html&lt;/a&gt;) since if we look through it we find each ordered pair (0,1), (0,2), ..., (5,4) exactly once.  These Latin squares (and Latin squares in general) are sometimes used in statistical experiments (or so I'm led to believe).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For which orders n does there exist a row-complete Latin square is still not completely resolved (although some partial results are known).  In the abstract of J. T. Higham's PhD thesis Construction methods for row-complete Latin squares, row-complete Latin squares are also known as &lt;em&gt;roman squares&lt;/em&gt;.  I haven't heard this term previously myself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know of any cases where &quot;row-complete&quot; has been used in other contexts, and searching through Google and MathSciNet didn't yield anything.  There are probably equivalent notions used in graph theory but under a different name.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you really need an answer to the question, my feeling is you're going to need to ask a bunch of experts in the field.  E.g. you could present your work at a seminar or conference, and ask the audience if they know of an alternative name or usage of &quot;row-complete&quot;.&lt;/p&gt;&#xA;" OwnerUserId="139" LastActivityDate="2010-10-23T23:55:25.800" />
  <row Id="7634" PostTypeId="1" AcceptedAnswerId="7637" CreationDate="2010-10-23T23:57:36.480" Score="16" ViewCount="646" Body="&lt;p&gt;I use stories like these to develop intuition... or perhaps to destroy it. I have my own answers in mind, but I want to see if I have made any mistakes...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are standing at the origin of an &quot;infinite forest&quot; holding an &quot;infinite bb-gun.&quot; The &quot;trees&quot; in this forest are at the lattice points all around you. (The lattice points are like those on graph paper and they align with the cardinal directions: N, S, E, W.) The &quot;forest&quot; is Euclidean in the sense that the trees have no width. To hit a tree with your bb-gun you must aim perfectly at it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You would, for example, hit a tree if you fired the gun due north, south east or west. (Your bullets also have no width.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A. You fire the gun in an arbitrary direction without bothering to aim. What happens?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;B. You get a new bb-gun and the bullets have a little width to them. ($\delta$?) You fire the gun in an arbitrary direction without bothering to aim. What happens?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;C. All of the trees are removed that have coordinates whose absolute values are not perfect squares. (So, only points such as $(25, 100)$ and $(4,-1600)$ remain.)  Again you use width-less bullets. You fire the gun in an arbitrary direction without bothering to aim. What happens?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;D. Again, only with perfect squares, but now the bullets have width. What happens?&lt;/p&gt;&#xA;" OwnerUserId="2563" LastEditorUserId="2563" LastEditorDisplayName="" LastEditDate="2010-10-24T05:11:15.807" LastActivityDate="2010-10-24T17:55:29.573" Title="You are standing at the origin of an &quot;infinite forest&quot; holding an &quot;infinite bb-gun&quot;" Tags="&lt;number-theory&gt;&lt;intuition&gt;&lt;infinity&gt;" AnswerCount="4" CommentCount="3" FavoriteCount="2" />
  <row Id="7635" PostTypeId="2" ParentId="7629" CreationDate="2010-10-24T00:12:25.630" Score="0" ViewCount="" Body="&lt;p&gt;Alright, thanks for all the help. ;)  I don't really need to exhaust all possibilities, just wondering (but like you mentioned, it isn't the easiest thing to search for).&lt;/p&gt;&#xA;" LastActivityDate="2010-10-24T00:12:25.630" CommentCount="1" />
  <row Id="7636" PostTypeId="1" CreationDate="2010-10-24T00:30:09.093" Score="2" ViewCount="192" Body="&lt;p&gt;Or has it remained a terminal node at the frontier of mathematics?&lt;/p&gt;&#xA;" OwnerUserId="2344" LastActivityDate="2011-04-20T23:24:15.660" Title="Has anything further been done with Morley’s Miracle?" Tags="&lt;euclidean-geometry&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="7637" PostTypeId="2" ParentId="7634" CreationDate="2010-10-24T00:31:36.020" Score="18" ViewCount="" Body="&lt;p&gt;A., C.  The probability that you hit a tree is $0$.  This is equivalent to the statement that the set of angles at which you can see a tree is of measure zero in $[0, 2\pi)$ with the Lebesgue measure, which follows because it is countable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;B.  The probability that you hit a tree is $1$.  By &lt;a href=&quot;http://en.wikipedia.org/wiki/Dirichlets_approximation_theorem&quot; rel=&quot;nofollow&quot;&gt;Dirichlet's approximation theorem&lt;/a&gt;, for any real $\alpha$ there exist infinitely many pairs $p_n, q_n$ of integers such that $|\alpha - \frac{p_n}{q_n}| &amp;lt; \frac{1}{q_n^2}$.  It follows that the distance between the points $(q_n, q_n \alpha)$ and $(q_n, p_n)$ is at most $\frac{1}{q_n}$, and since the sequence $q_n$ tends to $\infty$ it follows by letting $\alpha = \tan \theta$ where $\theta$ is the angle at which you fired that for any bullet size $\delta &gt; 0$ we will hit a tree at $(q_n, p_n)$ where $q_n &gt; \frac{1}{\delta}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;D.  The probability that you hit a tree is &lt;strong&gt;still&lt;/strong&gt; $1$.  This is a consequence of the following theorem, which I just learned about &lt;a href=&quot;http://mathoverflow.net/questions/43381/what-numbers-can-be-approximated-pretty-well-by-rationals&quot; rel=&quot;nofollow&quot;&gt;by asking this MO question&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://eom.springer.de/d/d032580.htm&quot; rel=&quot;nofollow&quot;&gt;Theorem (Khinchin):&lt;/a&gt;  Let $\phi(q) : \mathbb{N} \to \mathbb{R}$ be a monotonically decreasing function.  For almost all real numbers $\alpha$, the number of pairs of positive integers $(q, p)$ satisfying&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left| p - q\alpha \right| &amp;lt; \phi(q)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is infinite if $\sum \phi(q)$ diverges, and finite if $\sum \phi(q)$ converges.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, taking $\phi(q) = \frac{1}{q \ln q}$ (the sum of which diverges, for example by the integral test), we get that for almost all real $\alpha$ there are infinitely many solutions $(q, p)$ to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left| \frac{p}{q} - \alpha \right| &amp;lt; \frac{1}{q^2 \ln q}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let $\alpha = \sqrt{\tan \theta}$.  With probability $1$ there will be infinitely many $(q, p)$ satisfying the above condition.  Then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left|p^2 - q^2 \tan \theta \right| &amp;lt; \frac{\left| \frac{p}{q} + \sqrt{\tan \theta} \right|}{\ln q}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By taking $q$ large enough so that the RHS is less than $\frac{\delta}{2}$ it follows that a bullet shot at angle $\theta$ will hit the tree at $(q^2, p^2)$.  (I'm only working in the positive quadrant but the generalization to the other quadrants should be clear.)  &lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-10-24T16:56:05.127" LastActivityDate="2010-10-24T16:56:05.127" CommentCount="6" />
  <row Id="7638" PostTypeId="2" ParentId="7634" CreationDate="2010-10-24T00:44:38.157" Score="6" ViewCount="" Body="&lt;p&gt;Interestingly, the &lt;em&gt;fraction&lt;/em&gt; of the lattice points that you could hit (in question A&amp;mdash;not what you asked) from&#xA;the origin is &#xA;$$ 6 / {\pi}^2 $$&#xA;or about 61%.&#xA;See &lt;em&gt;Research Problems in Discrete Geometry&lt;/em&gt;,&#xA;Peter Brass, W. O. J. Moser, János Pach, p.430, &quot;Visibility problems for lattice points.&quot;&lt;/p&gt;&#xA;" OwnerUserId="237" LastActivityDate="2010-10-24T00:44:38.157" CommentCount="8" />
  <row Id="7639" PostTypeId="2" ParentId="7634" CreationDate="2010-10-24T00:58:29.737" Score="1" ViewCount="" Body="&lt;p&gt;B. in the annulus from $r$ to $r+dr$ there are $2*\pi *r*dr$ trees.  If we move the diameter from the bullet to the trees, they subtend an angle $\delta*2*\pi*r*dr/r$.  This reaches $2\pi$ at $r=1/\delta$, so that is about where we would expect to hit a tree.  As the integral diverges, we are guaranteed to hit a tree.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;D. If we make the tree size δ it subtends an angle $\delta /\sqrt {i^4+j^4}$ Does the sum of this diverge? We need it to exceed $\pi /2$ as I have just covered the first quadrant. &lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-10-24T00:58:29.737" CommentCount="6" />
  <row Id="7640" PostTypeId="1" AcceptedAnswerId="7642" CreationDate="2010-10-24T01:03:37.797" Score="1" ViewCount="108" Body="&lt;p&gt;You have 3 bins.  You can toss balls one at a time to the bins until any one of them has 2 balls in it, and then you stop.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The tosses are independent, and each bin is equally likely to be hit.  It's impossible to not make it in a bin on any given toss.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the expected number of tosses so that 1 bin has 2 balls in it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;--&#xA;Not sure if it's important, but by the pigeon hole principle we know that the maximum number of balls tosses required to do this would be 4.&lt;/p&gt;&#xA;" OwnerUserId="1630" LastActivityDate="2010-10-24T01:14:47.203" Title="Expected number of tosses so that 1 out of 3 bins has 2 balls in it" Tags="&lt;probability&gt;" AnswerCount="2" />
  <row Id="7641" PostTypeId="2" ParentId="7640" CreationDate="2010-10-24T01:07:08.070" Score="0" ViewCount="" Body="&lt;p&gt;Hint:  So what is the chance that you get success in two throws?  Call it x.  In exactly three throws?  Call it y.  And that tells you the chance that you will need four is 1-x-y.  Then what is the expected number of throws?&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-10-24T01:07:08.070" />
  <row Id="7642" PostTypeId="2" ParentId="7640" CreationDate="2010-10-24T01:14:47.203" Score="1" ViewCount="" Body="&lt;p&gt;It can't happen in 1 toss.  In 2 tosses, the first toss doesn't matter (probability 1), but the second must match the first (probability $\frac{1}{3}$), so it happens in 2 tosses with probability $1\cdot\frac{1}{3}=\frac{1}{3}$.  In 3 tosses, the first toss doesn't matter, the second toss must not match the first toss ($\frac{2}{3}$), and the third toss must match one of the first two ($\frac{2}{3}$), so it happens in exactly 3 tosses with probability $1\cdot\frac{2}{3}\cdot\frac{2}{3}=\frac{4}{9}$.  In 4 tosses, the first toss doesn't matter, the second toss must not match the first toss, the third toss must not match either of the first two tosses ($\frac{1}{3}$), and the fourth toss will match one of the first three no matter what (1), so it happens in exactly 4 tosses with probability $1\cdot\frac{2}{3}\cdot\frac{1}{3}\cdot 1=\frac{2}{9}$.  Note that the sum of these three probabilities is 1, which confirms that there is no need to look past 4 tosses.  The expected number of tosses is $2\cdot\frac{1}{3}+3\cdot\frac{4}{9}+4\cdot\frac{2}{9}=\frac{26}{9}$.&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-10-24T01:14:47.203" />
  <row Id="7643" PostTypeId="1" AcceptedAnswerId="7667" CreationDate="2010-10-24T03:04:24.913" Score="5" ViewCount="326" Body="&lt;p&gt;I remember my professor in college challenging me with this question, which I failed to answer satisfactorily: I know there exists a bijection between the rational numbers and the natural numbers, but can anyone produce an explicit formula for such a bijection?&lt;/p&gt;&#xA;" OwnerUserId="506" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2010-11-21T14:21:39.320" LastActivityDate="2010-11-21T14:21:39.320" Title="Produce an explicit bijection between rationals and naturals?" Tags="&lt;elementary-set-theory&gt;" AnswerCount="1" CommentCount="7" FavoriteCount="1" />
  <row Id="7644" PostTypeId="1" CreationDate="2010-10-24T03:34:07.280" Score="2" ViewCount="149" Body="&lt;p&gt;The following comes from questions comes from a recent combinatorics paper I attended :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;1.27 people are to travel by a bus which can carry 12 inside and 15 outside. In how many ways can the party be distributed between inside and outside if 5 people refuse to go outside and 6 will not go inside?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The solution given C(16,7), I have no clue how they got it ?!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;2.The number of functions f from the set A = {0, 1, 2} into the set B = {1, 2, 3, 4, 5, 6, 7} such that $f(i) \le f(j) $ for $i \lt j $ and $i,j$ belongs to A is&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The solution given is C(8,3). I didn't really understood this one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;3.The number of ordered pairs $(m, n) m, n $ is in {1 , 2, … , 100} such that $7^m + 7^n$ is divisible by 5 is&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The solution given is 2500, but how ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;4.The coefficient of $x^{20}$in the expansion of $(1 + 3x + 3x^2 + x^3)^{20}$, is ?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How to solve this one elegantly ? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;5.An eight digit number divisible by 9 is to be formed by using 8 digits out of the digits 0, 1, …, 9 without replacement. The number of ways in which this can be done is:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now this one seems impossible for me to solve in 1 mint,or is it ? Given soln is 36(7!)&lt;/p&gt;&#xA;" LastActivityDate="2010-10-24T05:30:08.113" Title="Problems on combinatorics" Tags="&lt;combinatorics&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="7645" PostTypeId="2" ParentId="7644" CreationDate="2010-10-24T03:48:05.667" Score="0" ViewCount="" Body="&lt;p&gt;For number 4 you would use the multinomial theorem.&lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2010-10-24T03:48:05.667" />
  <row Id="7646" PostTypeId="2" ParentId="7578" CreationDate="2010-10-24T03:48:33.183" Score="1" ViewCount="" Body="&lt;p&gt;As many others have pointed out, the way to do this is to think about pairs of metric spaces that are homeomorphic but not isometric.  In many cases, the metric will be complete on one space but not the other.  For example:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;The set $\{1/n \mid n \in \mathbb{N}\}$ is incomplete, but the natural numbers $\mathbb{N}$ are complete.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The ray $(0,\infty)$ and the open interval $(0,1)$ are incomplete, but the real line is complete.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;An open disc in the plane is incomplete, but the entire plane is complete.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The punctured plane $\mathbb{R}-\{0\}$ is incomplete, but the infinite cylinder $\mathbb{R} \times S^1$ (where $S^1$ is the circle) is complete.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The real line minus the $x$-axis is incomplete, but a disjoint pair of planes is complete.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The twice-punctured plane $\mathbb{R}^2 - \{(-1,0),(1,0)\}$ is incomplete, but the graph of  $$&#xA;z = \frac{1}{(x^2-1)^2 + y^2}&#xA;$$&#xA;in $\mathbb{R}^3$ is complete.  (This graph has asymptotic &quot;cusps&quot; at $(-1,0)$ and $(1,0)$.)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In each case, the homeomorphism between the two spaces can be used to define a nonstandard metric on the latter space that makes it incomplete.&lt;/p&gt;&#xA;" OwnerUserId="1726" LastActivityDate="2010-10-24T03:48:33.183" />
  <row Id="7647" PostTypeId="1" AcceptedAnswerId="7648" CreationDate="2010-10-24T04:06:57.383" Score="3" ViewCount="170" Body="&lt;p&gt;Does anyone know if there's a client or web app that generates Penrose patterns which can then be converted to a tileable rectangular background image for web site?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found this &lt;a href=&quot;http://stephencollins.net/penrose&quot; rel=&quot;nofollow&quot;&gt;http://stephencollins.net/penrose&lt;/a&gt; but I can't seem to get a rectangular shape out of it for web use. Thanks.&lt;/p&gt;&#xA;" OwnerUserId="5" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-10-24T04:12:53.247" LastActivityDate="2010-10-24T04:12:53.247" Title="Penrose Tile generator" Tags="&lt;geometry&gt;&lt;math-software&gt;&lt;tiling&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="7648" PostTypeId="2" ParentId="7647" CreationDate="2010-10-24T04:08:36.793" Score="8" ViewCount="" Body="&lt;p&gt;I believe, since Penrose tilings are aperiodic (lacking translational symmetry), there isn't such a rectangular shape.&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-10-24T04:08:36.793" CommentCount="1" />
  <row Id="7649" PostTypeId="2" ParentId="7644" CreationDate="2010-10-24T04:29:43.297" Score="5" ViewCount="" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Five people refuse to go outside, and therefore will go inside. Six people refuse to go inside, so will go outside. That means that you still have $27 - (5+6)=16$ people to accommodate. There are 12 spots for people inside, but five are already &quot;taken&quot; by those who refuse to be outside. That leaves 7 seats inside to assign. So you need to choose which seven people, out of the 16 that are left, will go inside. The number of ways of doing this is precisely $\binom{16}{7}$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; I misread the question as saying that $f(i)\lt f(j)$ if $i\lt j$. That answer follows: since you require the values of the function to increase (the condition just says that $f(0)\lt f(1)\lt f(2)$), if you know the three values of $f(0)$, $f(1)$, and $f(2)$ you know which one corresponds to which ($f(0)$ is the smallest value, $f(1)$ is the middle value, and $f(2)$ is the largest value). So all you need to do in order to determine a function is to pick three values from $B$. There are $7$ possibilities, you need to pick $3$, so there are $\binom{7}{3}$ ways of doing it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, it seems I misread the question. It actually says that $f(i)\leq f(j)$ if $i\lt j$. I gave the number of functions in which all inequalities are strict. There are $\binom{7}{1}$ functions in which all inequalities are actually equalities (just one value). Now, to count the number of functions in which $f(0)=f(1)\lt f(2)$, you just need to pick two values from the set $B$, which there are $\binom{7}{2}$ ways of doing; the same holds for the case in which you have $f(0)\lt f(1)=f(2)$. So the total is $\binom{7}{3}+2\binom{7}{2} + \binom{7}{1}$. Since $\binom{n}{k}+\binom{n}{k-1} = \binom{n+1}{k}$,  this total is equal to &#xA;$$\left(\binom{7}{3} + \binom{7}{2}\right) + \left(\binom{7}{2}+\binom{7}{1}\right) = \binom{8}{3}+\binom{8}{2} = \binom{9}{3}.$$ It seems to me, then, that the answer you give is incorrect, or perhaps you mistyped it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(If you know the formula for combinations with repetitions, then there is a more direct way of getting this result: simply pick $3$ out of the $7$ possible images, with repetitions allowed; smallest value is $f(0)$, middle value is $f(1)$, largest values $f(3)$ (equality allowed). The total for this is $\binom{7+3-1}{3} = \binom{9}{3}$, as above).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Assume $m\leq n$; then $7^m+7^n = 7^m(1 + 7^{n-m})$. The product is divisible by $5$ if and only if $1 + 7^{n-m}$ is divisible by $5$ (since $5$ is prime and never divides a power of $7$). For $1+7^{n-m}$ to be divisible by $5$, you need $7^{n-m}$ to have a remainder of $4$ when divided by $5$. If you run over the powers of $7$ and see the remainder when divided by $5$, you will notice that they go $2$, $4$, $3$, $1$, and repeat. So basically, you need $n-m$ to be a multiple of $4$ plus $1$. That is, you need $n-m = 4k+1$. Note in particular that if the pair has $n=m$, then it does not satisfy the condition. So count how many pairs there are where the two differ by a multiple of four plus 1.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strike&gt;One possibility is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Multinomial_theorem&quot; rel=&quot;nofollow&quot;&gt;Multinomial theorem&lt;/a&gt; You would need to figure out all the ways in which you can obtain $x^{20}$ as products of powers of $x$, $x^2$, and $x^3$, and add the appropriate coefficients.&lt;/strike&gt; &lt;strong&gt;Edit:&lt;/strong&gt; But the intended answer is almost certainly the one given by Larry Denenberg.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;In order for the number to be divisible by $9$, the digits must add up to a multiple of $9$. The digits $0$ through $9$ add up to $45$, which is a multiple of $9$. So if you omit two of them, they must add up to $9$: thus, if you omit $0$, then you must also omit $9$; if you omit $1$, then you must also omit $8$; etc. So you only have five possible pairs of numbers that you can omit. So pick which of the five pairs you will omit. If you omit $0$ and $9$, then the remaining $8$ digits can be arranged in any order, giving $8!$ possibilities. In all other cases, you cannot place $0$ in the first position, but otherwise can place the rest in any order. That gives $7(7!)$ possible ways of ordering the numbers. Thus, you have one choice that leads to $8!$ numbers, and four choices that lead each to $7(7!)$ numbers. Adding them up gives $8!+(4\times 7)(7!) = 8(7!)+28(7!) = 36(7!)$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-10-24T05:30:08.113" LastActivityDate="2010-10-24T05:30:08.113" CommentCount="2" />
  <row Id="7650" PostTypeId="1" CreationDate="2010-10-24T04:51:02.483" Score="1" ViewCount="153" Body="&lt;p&gt;I am just beginning to read about the use of &quot;Concave Programming&quot; methods and use of the Karush-Kuhn-Tucker conditions to identify the maximum value of a non-linear objective function subject to inequality constraints.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The examples I have seen in the text I have at hand, all involve only linear constraints. Is this method equally applicable to situations where not only are there multiple constraints, but where one or more of those constraints are non-linear ?&lt;/p&gt;&#xA;" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-10-24T19:29:35.807" LastActivityDate="2010-10-24T19:29:35.807" Title="Non-linear programming" Tags="&lt;nonlinear-optimization&gt;&lt;economics&gt;" CommentCount="5" />
  <row Id="7651" PostTypeId="2" ParentId="7644" CreationDate="2010-10-24T04:57:53.557" Score="3" ViewCount="" Body="&lt;p&gt;For number 4, the multinomial theorem is overkill.  The given expression is just $((x+1)^3)^{20} = (x+1)^{60}$. The coefficient of $x^{20}$ is $60\choose 20$.&lt;/p&gt;&#xA;" OwnerUserId="2682" LastEditorUserId="2682" LastEditorDisplayName="" LastEditDate="2010-10-24T05:03:20.980" LastActivityDate="2010-10-24T05:03:20.980" />
  <row Id="7652" PostTypeId="1" AcceptedAnswerId="7653" CreationDate="2010-10-24T05:51:11.260" Score="0" ViewCount="94" Body="&lt;p&gt;Let $X$ be a Poisson random variable with mean $\lambda$. How do I show that $P[X \geq \lambda] = 1/2$? Also, I was wondering what distributions have this property that the density is concentrated equally around the mean. I see that it's true for Gaussian and Uniform distributions.&lt;/p&gt;&#xA;" OwnerUserId="1527" LastActivityDate="2010-10-24T06:52:50.887" Title="Concentration of poisson distribution" Tags="&lt;probability&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="7653" PostTypeId="2" ParentId="7652" CreationDate="2010-10-24T06:24:03.337" Score="2" ViewCount="" Body="&lt;p&gt;P[X ≥ &amp;mu;] = 1/2 if and only if the mean (&amp;mu;) and median &quot;coincide&quot;. This is true for Gaussian and uniform distribution, but not always true for Poisson.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is exactly one &amp;mu; between every unit interval [n, n+1) such that P[X &amp;ge; &amp;mu;] = 1/2. Some of them are: ln 2, 1.6784…, 2.674…, 3.672…, ...&lt;/p&gt;&#xA;" OwnerUserId="171" LastEditorUserId="171" LastEditorDisplayName="" LastEditDate="2010-10-24T06:52:50.887" LastActivityDate="2010-10-24T06:52:50.887" CommentCount="3" />
  <row Id="7654" PostTypeId="2" ParentId="734" CreationDate="2010-10-24T06:30:05.757" Score="2" ViewCount="" Body="&lt;p&gt;For my experience, the &lt;a href=&quot;http://www.open.ac.uk/&quot; rel=&quot;nofollow&quot;&gt;Open University&lt;/a&gt; is well-respected internationally.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[Note this is different to Open Universities, such as &lt;a href=&quot;http://www.open.edu.au/&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; in Australia, which is a collaboration of several universities.]&lt;/p&gt;&#xA;" OwnerUserId="139" LastActivityDate="2010-10-24T06:30:05.757" CommunityOwnedDate="2010-10-24T06:30:05.757" CommentCount="1" />
  <row Id="7655" PostTypeId="1" AcceptedAnswerId="7661" CreationDate="2010-10-24T06:40:56.027" Score="0" ViewCount="130" Body="&lt;p&gt;If $S$ is a surface which is the complement of finitely many points in a compact surface, and the metric in $S$ is complete, then is Gauss-Bonnet theorem still valid for $S$?&lt;/p&gt;&#xA;" LastEditorUserId="367" LastEditorDisplayName="" LastEditDate="2010-10-24T12:35:04.030" LastActivityDate="2010-10-24T14:21:11.990" Title="A question about Gauss Bonnet theorem" Tags="&lt;differential-geometry&gt;" AnswerCount="3" />
  <row Id="7656" PostTypeId="2" ParentId="7454" CreationDate="2010-10-24T07:05:57.753" Score="4" ViewCount="" Body="&lt;p&gt;As promised in my response to Byron, here is my very partial progress to&#xA;date on this problem.  Though long-winded, it doesn't amount to much.&#xA;I'm still looking for an answer to the general case, or to other special&#xA;cases (like $n=2$ or $c=2$) or even just better estimates.  I'm also&#xA;interested to know about any applicable combinatorial tools even without&#xA;a full answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, a caution:  The graphs I'm about to use are not related to the&#xA;graphs in the general formulation of the original question.  Ignore&#xA;those graphs and just think of students and classrooms!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $G$ be a graph with a vertex for each student, $cn$ vertices in all.&#xA;Say that a given year's assignment of students to classrooms&#xA;&lt;strong&gt;respects&lt;/strong&gt; $G$ if every edge in $G$ joins two students that are&#xA;assigned to different classrooms.  That is, $G$ encodes constraints on&#xA;assignments; each edge in $G$ represents two students that must be kept&#xA;apart.  Let $R(G)$ be the probability that a random assignment of&#xA;students to classrooms respects $G$.  Note that $R(G)$ does not depend&#xA;on $y$ since the definition of &quot;respects&quot; refers to a single year's&#xA;assignment.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let $e(G)$ be the number of edges in $G$.  Then the probability that&#xA;after $y$ years every pair of students has at some point shared a&#xA;classroom is exactly&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum (-1)^{e(G)} (R(G))^y$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where the sum is taken over all graphs $G$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This result follows from a straightforward application of&#xA;Inclusion/Exclusion.  It's very pretty (at least I think so) but useless&#xA;if you want a real answer.  The problem isn't finding $R(G)$; this is tedious&#xA;but can be automated.  The real difficulty is that you have to handle&#xA;$2^{cn\choose 2}$ graphs, which is far too many.  The graphs come in&#xA;isomorphic bunches (e.g. there are $cn\choose 2$ one-edged graphs that all &#xA;have the probability of being respected) and you don't have to handle most graphs with lots of edges since $R(G) = 0$ for any graph $G$ that is not $c$-partite.  But this doesn't help enough.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We get some use out of this wretched formula by generalizing the&#xA;problem.  Suppose we ask, given the same conditions, for the probability&#xA;that each student &lt;em&gt;in a given group of $p$ students&lt;/em&gt; at some point sees&#xA;each of the other students &lt;em&gt;in that group&lt;/em&gt;.  The formula above gives the&#xA;exact answer if we sum over all graphs with $p$ vertices, each labelled&#xA;with a student in the group.  For example, with $p = 2$ the formula&#xA;gives the probability that two given students at some point see each&#xA;other:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$1 - \left(\frac{n(c-1)}{cn-1}\right)^y$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Of course we can get this result by a simpler route.)  For $p=3$ the&#xA;formula gives&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$1 - \left(\frac{N-n+1}{N(N-1)}\right)^y(3(N-1)^y - 3(N-n)^y + (N-2n+1)^y)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $N = cn-1$.  I've fought through the formula for $p$ up to 6, but&#xA;it becomes impossible quickly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now some approximations.  Define $U_m$ to be the probability that a&#xA;given student $S$ is at some point in a classroom with each student in a&#xA;given set of $m$ students not containing $S$.  Another straightforward&#xA;application of inclusion/exclusion gives&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$U_m = \sum_{j=0}^m(-1)^j{m\choose j}{N-j\choose n-1}^y \bigg/ {N\choose n-1}^y$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $N = cn-1$ as before.  $U_N$ is then the probability that a single&#xA;given student will see all the others.  This is a very weak upper bound&#xA;on the probability that &lt;em&gt;all&lt;/em&gt; the students see all the others, though in&#xA;trivial cases one student seeing all the others does imply that they all&#xA;do!  $U_m$ is rather easy to calculate; with the original parameters,&#xA;for example, $U_{74}$ turns out to be a bit less than 1/7792.  Note, BTW,&#xA;that $U_1$ correctly gives the same result as the $p=2$ answer above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If &quot;student $A$ sees all the others&quot; were independent of &quot;student $B$&#xA;sees all the others&quot;, then we'd have the answer we seek:  $(U_N)^{cn}$.&#xA;But of course these events are unlikely to be independent.  If many&#xA;students have seen all the others, then surely the probability increases&#xA;that all have seen the others.  &lt;strike&gt;So $(U_N)^{cn}$ may be a lower bound on&#xA;the true answer, if a weak one.&lt;/strike&gt; &lt;strong&gt;Edit:&lt;/strong&gt; Actually, I don't think it can be a lower bound, since I'm pretty sure one could find a case where &lt;em&gt;one&lt;/em&gt; student can see all the others, but it's impossible for &lt;em&gt;all&lt;/em&gt; students to do so.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's another way of estimating the answer:  A group of $p$ students&#xA;all see each other if and only if both (a) a particular one of these&#xA;students sees all the other $p-1$ students, and (b) the other $p-1$&#xA;students all see each other.  The probability of (a) is exactly&#xA;$U_{p-1}$.  Let $C_{p-1}$ denote the probability of (b).  If events (a)&#xA;and (b) were independent, then the exact answer to the original question&#xA;would be $U_{p-1}C_{p-1}$.  But by identical reasoning, $C_{p-1} =&#xA;U_{p-2}C_{p-2}$.  Continuing in this way, the exact answer to the original&#xA;question would be&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$U_NU_{N-1}U_{N-2}\ldots U_2U_1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As above, the events in question are not likely to be independent.  But&#xA;most of the time they're close, right?  With the original parameters,&#xA;the fact that a group of five students all see each other has very&#xA;little impact on whether a sixth student sees each of the five, though&#xA;it should increase that probability slightly.  This would make the&#xA;expression above a lower bound on the actual answer, though I don't know&#xA;how good it is.&lt;/p&gt;&#xA;" OwnerUserId="2682" LastEditorUserId="2682" LastEditorDisplayName="" LastEditDate="2010-11-06T13:36:21.480" LastActivityDate="2010-11-06T13:36:21.480" CommentCount="3" />
  <row Id="7657" PostTypeId="2" ParentId="7655" CreationDate="2010-10-24T07:51:09.733" Score="0" ViewCount="" Body="&lt;p&gt;No.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you remove a finite set of points from a compact surface, it will no longer be complete. If you change the metric so that you do have a complete surface, you cannot apply the Gauss Bonnet theorem anyway, because the surface isn't compact.&lt;/p&gt;&#xA;" OwnerUserId="1728" LastActivityDate="2010-10-24T07:51:09.733" CommentCount="2" />
  <row Id="7658" PostTypeId="2" ParentId="7621" CreationDate="2010-10-24T08:10:25.360" Score="3" ViewCount="" Body="&lt;p&gt;To help with the notation I will change your $k$ to an $m$, so we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ A_n = n - \sum \lfloor \frac{n}{p_i} \rfloor + \sum \lfloor \frac{n}{p_i p_j} \rfloor - &#xA;\sum \lfloor \frac{n}{p_i p_j p_k} \rfloor + \cdots, $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where the summations are over distinct primes in our set $ \lbrace p_1,p_2,\ldots,p_m \rbrace ,$ and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ B_n = n \prod_{i=1}^m \left( 1- \frac{1}{p_i} \right). $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using $ x \ge \lfloor x \rfloor &gt; x – 1 $ we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ A_n \ge n - \sum \frac{n}{p_i} + \sum \frac{n}{p_i p_j} - \binom{m}{2} -&lt;br&gt;&#xA;\sum \frac{n}{p_i p_j p_k} + \sum \frac{n}{p_i p_j p_k p_l} - \binom{m}{4} + \cdots.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Noting that $$\binom{m}{0}+\binom{m}{2}+\binom{m}{4} \cdots = 2^{m-1},$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;we obtain $A_n \ge B_n – 2^{m-1}.$ Similarly, we have $A_n \le B_n + 2^{m-1}$ and hence we can bound&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \left| \frac{A_n – B_n}{n} \right| \le \frac{2^{m-1}}{n} = O(1/n).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if we take $n \equiv -1 \textrm{ mod } \prod_{i=1}^m p_i$ then we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$A_n – B_n = \prod_{i=1}^m \left( 1- \frac{1}{p_i} \right) = \text{a constant.} \quad (1)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the difference is a constant infinitely often.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT:&#xA;Since the differences between the $ \lfloor \frac{n}{p_i p_j \ldots} \rfloor $ and the&#xA;$ \frac{n}{p_i p_j \ldots} $ are a maximum when $n \equiv -1 \textrm{ mod } \prod_{i=1}^m p_i$&#xA;it appears we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$A_n – B_n \le \prod_{i=1}^m \left( 1- \frac{1}{p_i} \right), \quad (2)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with equality when $n \equiv -1 \textrm{ mod } \prod_{i=1}^m p_i.$ I do not have a proof of this latter claim, although I have a proof for (1), which I hope to be able to modify to prove (2).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT2: I could not find a proof of (2) because, sadly, it's not true! However, (1) is exact for the given $n.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT3:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a vast improvement on my previous comments, which I believe solves your problem completely.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose $n \equiv -r \textrm{ mod } \prod_{i=1}^m p_i, $ where $ 0 \le r &amp;lt; \prod_{i=1}^m p_i$ then we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$A_n – B_n \le r \prod_{i=1}^m \left( 1- \frac{1}{p_i} \right).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I proved this simply by expanding $A_n – B_n$ and doing the summations. Also, I believe the exact formula to be&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$A_n – B_n = \delta +  r \prod_{i=1}^m \left( 1- \frac{1}{p_i} \right) - \Phi_P(r),$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\delta = 1$ when $r$ is coprime to all the $p_i$ and $0$ otherwise, and&#xA;$\Phi_P(r)$ is the number of positive integers $\le r$ which are coprime to the $p_i.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that when $n=711,$ $r=9,$ $\delta = 0$ and $\Phi_P(9) = 2,$&#xA;$P=\lbrace 2,3,5 \rbrace,$&#xA;so the above formula gives&#xA;$$A_{711}-B_{711} = 9 \left( 1 - \frac{1}{2} \right) \left( 1 - \frac{1}{3} \right)&#xA;\left( 1 - \frac{1}{5} \right) – 2 = 0.4$$&#xA;which agrees with your $190-189.6 = 0.4$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When $n=107,$ $r=19,$ $\delta = 1$ and $\Phi_P(19) = 6,$&#xA;$P=\lbrace 2,3,7 \rbrace,$&#xA;so we have&#xA;$$A_{107}-B_{107} = 1 + 19 \left( 1 - \frac{1}{2} \right) \left( 1 - \frac{1}{3} \right)&#xA;\left( 1 - \frac{1}{7} \right) – 6 = 0.42857\ldots,$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is also in agreement with the rounded figures you've given in the question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sorry, I do not have time to type up the full proof.&lt;/p&gt;&#xA;" OwnerUserId="1301" LastEditorUserId="1301" LastEditorDisplayName="" LastEditDate="2010-10-26T12:19:54.940" LastActivityDate="2010-10-26T12:19:54.940" CommentCount="6" />
  <row Id="7659" PostTypeId="1" AcceptedAnswerId="7662" CreationDate="2010-10-24T08:51:40.203" Score="5" ViewCount="181" Body="&lt;p&gt;I've started a little reading on quadratic reciprocity, and a reason for this has eluded me. Here's a little of what I came up with so far. I decided I want to show that for all primes $p$, if $p|x^2-2$, then $p$ does not divide $2y^2+3$. Then, by way of contradiction, if $(x^2-2)/(2y^2+3)$ is an integer, then any $p$ such that $p|2y^2+3$ would have to divide $x^2-2$, a contradiction. I see this is true for $p=2$. I want to find all $p$ such that $x^2\equiv 2\pmod{p}$, and since for any odd $p$, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left(\frac{2}{p}\right)=(-1)^{(p^2-1)/8}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I see $(2|p)=1$ iff $p\equiv 1,7\pmod{8}$. So only primes of the form $8k+1$ or $8k+7$ divide $x^2-2$. However, I don't see a way to show that primes of the form $8k+1$ or $8k+7$ do not divide $2y^2+3$, so maybe I'm completely off the mark. Does anyone know how to resolve this, or have a better idea of what to do? Thanks!&lt;/p&gt;&#xA;" OwnerUserId="1583" LastEditorUserId="1583" LastEditorDisplayName="" LastEditDate="2010-10-24T09:37:35.283" LastActivityDate="2010-10-24T09:37:35.283" Title="Why is $(x^2-2)/(2y^2+3)$ never an integer for any integers $x$ and $y$?" Tags="&lt;number-theory&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="7660" PostTypeId="1" AcceptedAnswerId="7665" CreationDate="2010-10-24T08:57:57.737" Score="3" ViewCount="233" Body="&lt;p&gt;Assume you want to prove an identity such as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{k=m+1}^{n}A(k,m)-B(k,m)=S(m)+T(n,m)\qquad\text{for } n,m\in &#xA;\mathbb{Z},n,m\geq 0.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Added&lt;/em&gt;: I applied mathematical induction on $m,n$ to prove it. I am unsure because up to now I have seen it applied to properties depending on a single variable only. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Question: does application of two inductive arguments, one on $m$ and the&#xA;other on $n$, guarantee the validity of such a proof? &lt;/p&gt;&#xA;" OwnerUserId="752" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2010-10-24T10:16:19.987" LastActivityDate="2010-10-24T12:04:54.287" Title="Induction on two integer variables" Tags="&lt;induction&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="0" />
  <row Id="7661" PostTypeId="2" ParentId="7655" CreationDate="2010-10-24T09:12:22.727" Score="1" ViewCount="" Body="&lt;p&gt;Sort of.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.ams.org/proc/1982-086-01/S0002-9939-1982-0663893-8/S0002-9939-1982-0663893-8.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.ams.org/proc/1982-086-01/S0002-9939-1982-0663893-8/S0002-9939-1982-0663893-8.pdf&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="2633" LastEditorUserId="2633" LastEditorDisplayName="" LastEditDate="2010-10-24T09:18:06.187" LastActivityDate="2010-10-24T09:18:06.187" CommentCount="1" />
  <row Id="7662" PostTypeId="2" ParentId="7659" CreationDate="2010-10-24T09:31:54.707" Score="3" ViewCount="" Body="&lt;p&gt;Try to show that $2y^2 + 3$ must have at least one prime divisor which is not of the form $8k+1$ or $8k+7$.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-10-24T09:31:54.707" CommentCount="1" />
  <row Id="7663" PostTypeId="2" ParentId="7040" CreationDate="2010-10-24T09:45:05.033" Score="0" ViewCount="" Body="&lt;p&gt;Geometrically it's just the Pythagorean theorem. We may measure the &quot;length&quot; of random variables by standard deviation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We start with a random variable Y. E(Y|X) is the projection of this Y to the set of random variables wich may be expressed as a deterministic function of X. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have a hypotenuse Y with squared length Var(Y).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first leg is E(Y|X) with squared length Var(E(Y|X)).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second leg is Y-E(Y|X) with squared length Var(Y-E(Y|X))=...=E(Var(Y|X)).&lt;/p&gt;&#xA;" OwnerUserId="2481" LastActivityDate="2010-10-24T09:45:05.033" />
  <row Id="7664" PostTypeId="1" CreationDate="2010-10-24T10:48:45.657" Score="1" ViewCount="255" Body="&lt;p&gt;Do somebody knows anything about the Dirac's identity?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation} \label{Dirac}&#xA;    \frac{\partial^2}{\partial x_{\mu}\partial x^{\mu}} \delta(xb_{\mu}xb^{\mu}) =&#xA;    -4\pi  \delta(xb_0)\delta(xb_1)\delta(xb_2)\delta(xb_3)&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;here&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;$xb$, is the 4-vector $x-b$ in Minkowsky spacetime&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$\delta$ is the Dirac delta function&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$x_0 = -x^0, \quad x_1 = x^1, \quad x_2 = x^2, \quad x_3, = x^3$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Do you know where can i find some material about it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;UPDATE:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Following Willie's &lt;a href=&quot;http://farside.ph.utexas.edu/teaching/jk1/lectures/node19.html&quot; rel=&quot;nofollow&quot;&gt;link&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i've understood that a solution for the linear wave equation&#xA;$$ \square \psi(\mathbf{r},t) = g(\mathbf{r},t) $$&#xA;for a given $g(\mathbf{r},t)$ is&#xA;$$ \psi = \int \int g(r',t')G(r,r',t,t')dV'dt' $$&#xA;where&#xA;$$ G(r,r',t,t') = AG^+(r,r',t,t') + BG^-(r,r',t,t') , \qquad A + B = 1 $$&#xA;and&#xA;$$ G^{\pm}(r,r',t,t') = \frac{\delta(t' - (t \mp | \mathbf{r} - \mathbf{r'} | / c))}{4\pi | \mathbf{r} - \mathbf{r'} | } $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think Dirac's follow from the solution of&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \square \psi(r,t) = \delta(\mathbf{r},t) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But i'm not sure of the details. Can you Willie help me?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="2354" LastEditorUserId="2354" LastEditorDisplayName="" LastEditDate="2010-10-27T12:50:40.817" LastActivityDate="2010-10-27T12:50:40.817" Title="Dirac's identity" Tags="&lt;physics&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="7665" PostTypeId="2" ParentId="7660" CreationDate="2010-10-24T10:48:56.790" Score="4" ViewCount="" Body="&lt;p&gt;Here are some induction principles for two variables:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$P(0,0)$&lt;/li&gt;&#xA;&lt;li&gt;$\forall x,y. P(x,y) \Rightarrow P(x+1,y)$&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$\forall x,y. P(x,y) \Rightarrow P(x,y+1)$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$\forall x,y. P(x,y)$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$P(0,0)$&lt;/li&gt;&#xA;&lt;li&gt;$\forall x,y. P(x,0) \Rightarrow P(x+1,0)$&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$\forall x,y. P(x+1,y) \Rightarrow P(x,y+1)$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$\forall x,y. P(x,y)$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" LastActivityDate="2010-10-24T10:48:56.790" CommentCount="4" />
  <row Id="7666" PostTypeId="1" CreationDate="2010-10-24T11:00:50.780" Score="1" ViewCount="118" Body="&lt;p&gt;Let L^2(R)={f:R-&gt;C| \int_{\infty}^{\infty} \abs(f(t))^2 &amp;lt; \infty } and L^1(R)={f:R-&gt;C| \int_{\infty}^{\infty} \abs(f(t)) &amp;lt; \infty }. Give and example of a function such that f \in L^2(R) and f \notin L^1(R).&lt;/p&gt;&#xA;" OwnerUserId="2723" LastEditorUserId="630" LastEditorDisplayName="" LastEditDate="2010-10-24T12:07:34.937" LastActivityDate="2010-10-24T12:07:34.937" ClosedDate="2010-10-24T12:33:21.173" Title="Give and example of a function such that f \in L^2(R) and f \notin L^1(R)." Tags="&lt;real-analysis&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" />
  <row Id="7667" PostTypeId="2" ParentId="7643" CreationDate="2010-10-24T11:05:22.580" Score="8" ViewCount="" Body="&lt;p&gt;Wikipedia contains several explicit examples:&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Cantor_pairing_function#Cantor_pairing_function&quot; rel=&quot;nofollow&quot;&gt;Cantor pairing function&lt;/a&gt;, &lt;a href=&quot;http://en.wikipedia.org/wiki/Stern-Brocot_tree&quot; rel=&quot;nofollow&quot;&gt;Stern–Brocot tree&lt;/a&gt;, &lt;a href=&quot;http://en.wikipedia.org/wiki/Calkin-Wilf_tree&quot; rel=&quot;nofollow&quot;&gt;Calkin–Wilf tree&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="589" LastActivityDate="2010-10-24T11:05:22.580" />
  <row Id="7668" PostTypeId="2" ParentId="7666" CreationDate="2010-10-24T11:50:27.353" Score="0" ViewCount="" Body="&lt;p&gt;Try to look at a function of the form $f(x) = x^{-a}$ for some $a &gt; 0$ on an appropriate domain.&lt;/p&gt;&#xA;" OwnerUserId="1120" LastEditorUserId="1120" LastEditorDisplayName="" LastEditDate="2010-10-24T12:03:24.737" LastActivityDate="2010-10-24T12:03:24.737" CommentCount="3" />
  <row Id="7669" PostTypeId="2" ParentId="7660" CreationDate="2010-10-24T12:04:54.287" Score="3" ViewCount="" Body="&lt;p&gt;Suppose you are trying to prove a family of statements $P(x, y)$.  This is the same as proving the family of statements $F(x)$, where $F(x) = \forall y : P(x, y)$.  Each statement $F(x)$ can be proven by induction on $y$ (for fixed $x$), and then you can prove $P(x, y)$ by induction on $x$.  You might want to try proving&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$${n+1 \choose k+1} = {n \choose k+1} + {n \choose k}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;this way.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But actually you can be much trickier than this.  Sometimes it suffices to induct on $x + y$, for example.  &lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-10-24T12:04:54.287" CommentCount="1" />
  <row Id="7670" PostTypeId="1" AcceptedAnswerId="7671" CreationDate="2010-10-24T12:14:31.650" Score="2" ViewCount="154" Body="&lt;p&gt;How many solutions does the following equation have?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$k+l+m+n=30,$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $k,l,m,n\in\mathbb{Z},0\leq k,l,m,n\leq 10.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: this is an exercise in the &lt;em&gt;inclusion-exclusion principle&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="752" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2010-10-24T13:51:39.303" LastActivityDate="2010-10-24T13:51:39.303" Title="Number of integer solutions in [0,10]" Tags="&lt;combinatorics&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" />
  <row Id="7671" PostTypeId="2" ParentId="7670" CreationDate="2010-10-24T12:36:35.373" Score="7" ViewCount="" Body="&lt;p&gt;This is the coefficient of $x^{30}$ in $(1 + x + ... + x^{10})^4$.  (To see this, think about what happens when you choose a term from each factor.)  But this is $\frac{(1 - x^{11})^4}{(1 - x)^4}$.  Now,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{1}{(1 - x)^4} = \sum_{n \ge 0} {n+3 \choose 3} x^n$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(which you'll recognize as the generating function giving the answer to the problem without the range restriction.)  On the other hand, $(1 - x^{11})^4 = 1 - {4 \choose 1} x^{11} + {4 \choose 2} x^{22} - {4 \choose 3} x^{33} + x^{44}$.  Of these terms only the first three contribute to the coefficient of $x^{30}$, making the final answer&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$${33 \choose 3} - {4 \choose 1} {22 \choose 3} + {4 \choose 2} {11 \choose 3} = 286.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is equivalent to the following inclusion-exclusion argument: start with the set of all solutions with no restriction.  Remove all the solutions where one of the four numbers is greater than $10$.  Add back in all the solutions where two of the four numbers are greater than $10$.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-10-24T12:36:35.373" CommentCount="3" />
  <row Id="7672" PostTypeId="1" AcceptedAnswerId="7750" CreationDate="2010-10-24T12:48:35.310" Score="10" ViewCount="260" Body="&lt;p&gt;Consider commutative semigroup S and its Grothendieck completion group &#xA;G(S).Suppose I insist on defining G(S) as  free abelian group on  basis $[a]$ (with $a\in S$) divided out by the relations $[a+b]-[a]-[b]$. How do I show &lt;em&gt;with that definition&lt;/em&gt; that if the images of $a,b\in S$ become equal in G(S), then necessarily there existed $c\in S$ with a+c=b+c ? I know it true because I can prove it with other construction of  grothendieck group, but I should like  direct proof with above free abelian group construction.&lt;/p&gt;&#xA;" OwnerUserId="2872" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2010-11-06T03:32:18.320" LastActivityDate="2010-11-06T03:32:18.320" Title="How to show directly that two elements become equal in Grothendieck group?" Tags="&lt;group-theory&gt;&lt;algebraic-k-theory&gt;" AnswerCount="1" CommentCount="0" />
  <row Id="7673" PostTypeId="2" ParentId="7670" CreationDate="2010-10-24T12:49:10.173" Score="5" ViewCount="" Body="&lt;p&gt;A geometric way of seeing the answer is this: this is the number lattice points in the cube $[0,10]^4$ that intersects the plane $k+l+m+n = 30$. Note that the corners $(0,10,10,10)$ are points on this plane. Since the weights on $k,l,m,n$ are equal, the solutions are in bijection (it would be clear once you draw a picture) with the lattice points in the $k,l,m$ plane that is inside the tetrahedron formed by the vertices $(0,10,10)$, $(10,0,10)$, $(10,10,0)$ and $(0,0,0)$, the total number of which would be the 11th &lt;a href=&quot;http://en.wikipedia.org/wiki/Tetrahedral_number&quot; rel=&quot;nofollow&quot;&gt;tetrahedral number&lt;/a&gt; which is 286. &lt;/p&gt;&#xA;" OwnerUserId="1543" LastEditorUserId="1543" LastEditorDisplayName="" LastEditDate="2010-10-24T13:08:43.517" LastActivityDate="2010-10-24T13:08:43.517" CommentCount="1" />
  <row Id="7674" PostTypeId="1" AcceptedAnswerId="7679" CreationDate="2010-10-24T12:55:55.613" Score="0" ViewCount="133" Body="&lt;p&gt;I didn't get this method.&#xA;$3a^2-9a+6 = 3(a^2-3a+2) = 3(a^2-1a*3+3^2-3^2+2)=$ $= 3[(a-3)^2-7]$, then?&lt;/p&gt;&#xA;" OwnerUserId="1795" LastEditorUserId="367" LastEditorDisplayName="" LastEditDate="2010-10-24T17:21:42.417" LastActivityDate="2010-10-24T17:21:42.417" Title="Perfect square method" Tags="&lt;algebra&gt;" AnswerCount="4" CommentCount="3" />
  <row Id="7675" PostTypeId="2" ParentId="7664" CreationDate="2010-10-24T13:00:39.173" Score="1" ViewCount="" Body="&lt;p&gt;Your identity is in fact the expression for the fundamental solution of the linear wave equation in (1+3) dimensions. This should be in most textbooks on electrodynamics or intro to quantum field theory. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Google also tells me:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.math.ucsd.edu/~lindblad/110b/l18.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.math.ucsd.edu/~lindblad/110b/l18.pdf&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.math.ucla.edu/~ralston/266b.1.08w/Funwave.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.math.ucla.edu/~ralston/266b.1.08w/Funwave.pdf&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;page 15 of &lt;a href=&quot;http://www.math.princeton.edu/~seri/homepage/courses/Analysis2008.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.math.princeton.edu/~seri/homepage/courses/Analysis2008.pdf&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1543" LastActivityDate="2010-10-24T13:00:39.173" CommentCount="15" />
  <row Id="7676" PostTypeId="1" CreationDate="2010-10-24T13:06:17.733" Score="1" ViewCount="162" Body="&lt;p&gt;Is there a specific way to attack congruences with two variables? The reason I ask is because up until now, I had never encountered such a problem. While browsing tonight, I came across this challenge problem in one of my books:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose that $(ab,p)=1$ and that $p&gt;2$. Show that the number of solutions $(x,y)$ of the congruence $ax^2+by^2\equiv 1\pmod{p}$ is $p-\left(\frac{-ab}{p}\right)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;After experimenting with it for a few hours, I've barely gotten anywhere. So far, I've only managed to break it into some cases. $(ab,p)=1$ implies $(a,p)=(b,p)=1$ and $\left(\frac{-ab}{p}\right)=\left(\frac{-1}{p}\right)\left(\frac{a}{p}\right)\left(\frac{b}{p}\right)$, so if&#xA;$\left(\frac{a}{p}\right)=\left(\frac{b}{p}\right)=1$ or $-1$, then $\left(\frac{ab}{p}\right)=1$, so I want to show there are $p-(-1)^{(p-1)/2}$ solutions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, if $\left(\frac{a}{p}\right)=1$ and $\left(\frac{b}{p}\right)=-1$, I need to show there are $p+(-1)^{(p-1)/2}$ solutions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the first case, $a$ and $b$ would both be quadratic residues modulo $p$, and hence $c^2\equiv a\pmod{p}$ and $d^2\equiv b\pmod{p}$ for some $c,d$. Then the original problem would come to $(cx)^2+(dy)^2\equiv 1\pmod{p}$, but I'm not sure a sum of squares is any better. Is there a slick solution to a problem like this?, I'm very puzzled over it.&lt;/p&gt;&#xA;" OwnerUserId="1583" LastActivityDate="2010-10-24T15:50:44.710" Title="Solving congruences in two variables" Tags="&lt;number-theory&gt;&lt;modular-arithmetic&gt;" AnswerCount="2" />
  <row Id="7677" PostTypeId="2" ParentId="6649" CreationDate="2010-10-24T13:20:30.903" Score="0" ViewCount="" Body="&lt;p&gt;please lisen this vedio lecture &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.youtube.com/watch?v=3lS5ZMsfUdQ&quot; rel=&quot;nofollow&quot;&gt;http://www.youtube.com/watch?v=3lS5ZMsfUdQ&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You will understand the geometry behind the convergence of fourier series &lt;/p&gt;&#xA;" LastActivityDate="2010-10-24T13:20:30.903" />
  <row Id="7678" PostTypeId="2" ParentId="7676" CreationDate="2010-10-24T13:35:47.353" Score="5" ViewCount="" Body="&lt;p&gt;The answer to your first question is somewhat long.  Counting solutions to congruences in two variables is equivalent to counting points on (affine) curves over finite fields, which happens to be quite hard in general and is an active field of study.  Even studying cubic polynomials rapidly leads one to deep waters: elliptic curves, modular forms, ...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The quadratic case is much simpler.  When working in projective coordinates the underlying curve has genus $0$, hence either has no points or is isomorphic to $\mathbb{P}^1$ (and has $p+1$ points).  So to count points on an affine curve it suffices to determine if any solutions exist and, if so, to find an explicit map from $\mathbb{P}^1$ and see what points it misses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course this might just be gibberish to you, so let me illustrate with the example case $a = b = 1$.  It's clear that $x^2 + y^2 = 1$ has nontrivial solutions such as $(1, 0)$.  The existence of these nontrivial solutions allows one to deduce the parameterization&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x = \frac{1 - t^2}{1 + t^2}, y = \frac{2t}{1 + t^2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is valid for any $t$ such that $t^2 \not \equiv -1 \bmod p$.  (The argument is roughly as follows: if $(x, y)$ is any other rational solution, then the line from $(x, y)$ to $(1, 0)$ has rational slope, and conversely given any such line it intersects the curve in at most one other point.)  This parameterization misses the solution $(1, 0)$ which corresponds to sending $t$ to $\infty$, so in total there are $p+1$ solutions if $p \equiv 3 \bmod 4$ (so $t$ can be anything) and $p-1$ solutions if $p \equiv 1 \bmod 4$) (so $t$ cannot be the two nontrivial square roots of $-1$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For general $a, b$ you get cases depending on whether $a$ and $b$ are quadratic residues.  &lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-10-24T13:35:47.353" CommentCount="4" />
  <row Id="7679" PostTypeId="2" ParentId="7674" CreationDate="2010-10-24T13:47:21.723" Score="7" ViewCount="" Body="&lt;p&gt;Ah, so you want to factor the polynomial. Here's how:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$3(a^2-2 \frac{3}{2} a + 2) = 3\left(\left(a-\frac{3}{2}\right)^2 - \frac{9}{4} + 2\right) = 3\left(z^2 - \frac{1}{4}\right)$$&#xA;$$= 3\left(z-\frac{1}{2}\right)\left(z+\frac{1}{2}\right) = 3(a-2)(a-1),$$&#xA;where I temporarily used the substitution $z=a-\frac{3}{2}$.&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2010-10-24T13:47:21.723" CommentCount="1" />
  <row Id="7680" PostTypeId="1" AcceptedAnswerId="7682" CreationDate="2010-10-24T14:00:15.597" Score="0" ViewCount="61" Body="&lt;p&gt;Show that the only regular, self-complementary, planar graph is the 5-cycle. Pay special attention to the 4-regular case.&lt;/p&gt;&#xA;" OwnerUserId="2738" LastEditorUserId="2738" LastEditorDisplayName="" LastEditDate="2010-10-24T15:01:06.620" LastActivityDate="2010-10-24T15:01:06.620" Title="Strange Case in Self-Complementary Planarity" Tags="&lt;homework&gt;&lt;graph-theory&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="7681" PostTypeId="2" ParentId="7655" CreationDate="2010-10-24T14:14:41.123" Score="0" ViewCount="" Body="&lt;p&gt;No.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take the sphere $S^2$ and remove one point, giving you a surface $S$ diffeomorphic to $\mathbb{R}^2$.  So we can equip $S$ with a metric making it isometric to $\mathbb{R}^2$, which is certainly complete.  $\mathbb{R}^2$ has zero curvature but its Euler characteristic is $1$, so&#xA;$$ 0 = \int_{\mathbb{R}^2} K dVol \ne \chi(\mathbb{R}^2) = 1.$$&lt;/p&gt;&#xA;" OwnerUserId="822" LastEditorUserId="822" LastEditorDisplayName="" LastEditDate="2010-10-24T14:21:11.990" LastActivityDate="2010-10-24T14:21:11.990" CommentCount="4" />
  <row Id="7682" PostTypeId="2" ParentId="7680" CreationDate="2010-10-24T14:48:47.757" Score="1" ViewCount="" Body="&lt;p&gt;Here's something to get you started. Let G be such a graph, with n vertices. If G is k-regular, then its complement is n-1-k-regular, so to be self-complementary, we must have k = n-1-k, so k = (n-1)/2. Now try finding a reason why the graph is nonplanar if k &gt; 2.&lt;/p&gt;&#xA;" OwnerUserId="759" LastActivityDate="2010-10-24T14:48:47.757" CommentCount="1" />
  <row Id="7683" PostTypeId="2" ParentId="7674" CreationDate="2010-10-24T15:00:02.070" Score="1" ViewCount="" Body="&lt;p&gt;You can guess and then confirm that $a=1$ is a zero of $3a^{2}-9a+6$. Thus&#xA;you can factor it as &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$3a^{2}-9a+6=3(a-1)(a-x)=3x-3a-3ax+3a^{2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and solve for $x$. Comparing the coefficients of $a^{2},a,a^{0}$, you get &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$3=3,3x=6,-3-3x=-9$, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is equivalent to $x=2$ and $1+x=3$. Hence $x=2$ and &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$3a^{2}-9a+6=3(a-1)(a-2)$.&lt;/p&gt;&#xA;" OwnerUserId="752" LastActivityDate="2010-10-24T15:00:02.070" />
  <row Id="7684" PostTypeId="2" ParentId="7674" CreationDate="2010-10-24T15:10:24.017" Score="7" ViewCount="" Body="&lt;p&gt;There is also a simple way that involves no fraction calculation,it goes like this :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ 3a^{2}-9a+6 $ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$  = 3a^{2}-6a-3a+6 $ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$  = a(3a-6)-(3a-6) $ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$  = (3a-6)(a-1) $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$  = 3(a-1)(a-2) $&lt;/p&gt;&#xA;" OwnerUserId="2109" LastActivityDate="2010-10-24T15:10:24.017" CommentCount="3" />
  <row Id="7685" PostTypeId="2" ParentId="7676" CreationDate="2010-10-24T15:36:21.123" Score="3" ViewCount="" Body="&lt;p&gt;What you have is a quadratic form, which makes the problem significantly easier than the general case, and can be split into the following steps.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Find at least one solution.&lt;/li&gt;&#xA;&lt;li&gt;Given one solution, use this to parameterize all solutions.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;This is roughly what Qiaochu did in his answer and this method is quite general, applying to quadratic forms in any number of variables and for finding solutions mod p, in the rationals or, in fact, in any field. I also used this method recently in answer to a question about &lt;a href=&quot;http://math.stackexchange.com/questions/7143/is-it-possible-to-get-arbitrarily-near-any-acute-angle-with-pythagorean-triangles/7237#7237&quot;&gt;Pythagorean triples&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me start with (2) first, so we suppose that you already have one solution (x&lt;sub&gt;0&lt;/sub&gt;,y&lt;sub&gt;0&lt;/sub&gt;). The idea is to find other solutions of the form (x,y)&amp;nbsp;=&amp;nbsp;(x&lt;sub&gt;0&lt;/sub&gt;,y&lt;sub&gt;0&lt;/sub&gt;)&amp;nbsp;+&amp;nbsp;(u,v)t. Where you choose nonzero (u,v)&amp;nbsp;&amp;isin;&amp;nbsp;(&amp;#x02124;/(p))&lt;sup&gt;2&lt;/sup&gt; and then solve for t. This will give a quadratic in t (as long as the t&lt;sup&gt;2&lt;/sup&gt; term doesn't cancel) with t&amp;nbsp;=&amp;nbsp;0 as one solution. Extracting out a factor of t gives a linear form with a unique solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;ax^2+by^2-1=(au^2+bv^2)t^2+2(ux_0+vy_0)t=0\ \Rightarrow\ t=-2\frac{ux_0+vy_0}{au^2+bv^2}\ (*)&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This works as long as au&lt;sup&gt;2&lt;/sup&gt;+bv&lt;sup&gt;2&lt;/sup&gt;&amp;nbsp;&amp;ne;&amp;nbsp;0, and gives us a parameterization of the solutions to the quadratic form in terms of nonzero (u,v)&amp;nbsp;&amp;isin;&amp;nbsp;(&amp;#x02124;/(p))&lt;sup&gt;2&lt;/sup&gt;. The solution we get does not change if (u,v) are rescaled (the projective space &amp;#x2119;&lt;sup&gt;1&lt;/sup&gt; mentioned by Qiaochu is precisely the set of nonzero (u,v) up to rescaling). Rescaling by 1/u, we only look at solutions corresponding to (u,v)&amp;nbsp;=&amp;nbsp;(1,v) and (in case u&amp;nbsp;=&amp;nbsp;0), (u,v)&amp;nbsp;=&amp;nbsp;(0,1). That is p&amp;nbsp;+&amp;nbsp;1 possibilities. The parameterization in terms of (1,v) only fails when a&amp;nbsp;+bv&lt;sup&gt;2&lt;/sup&gt;&amp;nbsp;=&amp;nbsp;0 so that (bv)&lt;sup&gt;2&lt;/sup&gt;&amp;nbsp;=&amp;nbsp;-ab. This can only happen if -ab is a quadratic residue, and occurs for bv one of the 2 square roots of -ab. This gives a total number of solutions to the quadratic form as p&amp;nbsp;+&amp;nbsp;1 if -ab is not a quadratic residue and (p&amp;nbsp;+&amp;nbsp;1)&amp;nbsp;-&amp;nbsp;2 if it is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Looking at step (1) now. You could try finding a solution by looking at cases depending on whether a,b, ab are quadratic residues, etc. However, there is a quick counting method. For any right hand side c, ax&lt;sup&gt;2&lt;/sup&gt;&amp;nbsp;+&amp;nbsp;by&lt;sup&gt;2&lt;/sup&gt;&amp;nbsp;=&amp;nbsp;c can have at most p&amp;nbsp;+&amp;nbsp;1 solutions in x,y (mod p) -- using the above argument. As there are p&lt;sup&gt;2&lt;/sup&gt; possibilities for (x,y), this gives at least p&lt;sup&gt;2&lt;/sup&gt;/(p&amp;nbsp;+&amp;nbsp;1) values of c which occur. As p&lt;sup&gt;2&lt;/sup&gt;/(p&amp;nbsp;+&amp;nbsp;1)&amp;nbsp;&amp;gt;&amp;nbsp;p&amp;nbsp;&amp;minus;&amp;nbsp;1, this shows that you can solve ax&lt;sup&gt;2&lt;/sup&gt;&amp;nbsp;+&amp;nbsp;by&lt;sup&gt;2&lt;/sup&gt;&amp;nbsp;=&amp;nbsp;c (mod p) for &lt;em&gt;any&lt;/em&gt; value of c.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You should be able to see that this whole argument carries through almost word-for-word to &lt;em&gt;any&lt;/em&gt; finite field of characteristic &amp;ne;&amp;nbsp;2 (characteristic 2 is a bit different because (*) gives t&amp;nbsp;=&amp;nbsp;0).&lt;/p&gt;&#xA;" OwnerUserId="1321" LastEditorUserId="1321" LastEditorDisplayName="" LastEditDate="2010-10-24T15:50:44.710" LastActivityDate="2010-10-24T15:50:44.710" />
  <row Id="7686" PostTypeId="2" ParentId="7586" CreationDate="2010-10-24T15:37:11.300" Score="1" ViewCount="" Body="&lt;p&gt;You can also use the &lt;a href=&quot;http://en.wikipedia.org/wiki/Quadratic_equation#Quadratic_formula&quot; rel=&quot;nofollow&quot;&gt;quadratic formula&lt;/a&gt; to compute the result elegantly, it goes like this :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$z^2 - 6z + 25 = 0 $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ z = \frac { 6 \pm \sqrt{36 -100} } {2} $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ = \frac{ 6 \pm 8i}{2} $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$=  3 \pm 4i $&lt;/p&gt;&#xA;" OwnerUserId="2109" LastEditorUserId="2109" LastEditorDisplayName="" LastEditDate="2010-10-24T15:51:50.350" LastActivityDate="2010-10-24T15:51:50.350" />
  <row Id="7688" PostTypeId="1" AcceptedAnswerId="7722" CreationDate="2010-10-24T16:05:32.087" Score="2" ViewCount="170" Body="&lt;p&gt;&lt;strong&gt;Question 1:&lt;/strong&gt; The volume of a parallelepiped in $\mathbb{R}^n$ with n sides given by the vectors $(x_{1_1}, x_{1_2} ... x_{1_n}), (x_{2_1}, x_{2_2} ... x_{2_n}) ... (x_{n_1}, x_{n_2} ... x_{n_n})$ and diagonal $(x_{1_1} + x_{2_1}+ \cdots + x_{n_1}, x_{1_2} + x_{2_2} + \cdots + x_{n_2}, ..., x_{1_n} + x_{2_n} + \cdots + x_{n_n})$  Is the determinant of the matrix $A = (x_{j, i})$ where i and J are the double sub-scripts on each $x_{i_j}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The case where n= 2 has a nice geometric proof. Shown in this diagram:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/ybk6O.gif&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$u= (a,c)^T$ and $v= (b, d)^T$ then the area of the parallelogram is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ det \left| [ u | v ] \right| = | ad - bc| $ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Is there a &quot;nice&quot; geometric proof for n=3?&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The idea behind the n=2 proof is you take the area of the rectangle and subtract the triangles. So, I'm thinking: take the volume of the rectangular solid and subtract tetrahedrons. How many will I need to subtract? Is it possible to draw this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question 2:&lt;/strong&gt; Suppose that (a,b) and (b, a) are points on the unit circle. Find parallelogram with vertices (0,0), (a,b) (a+b, a+b), (b,a) is where $\theta$ is between the x-axis and (a,b).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know the answer is $|2cos^2 \theta -1|$. I want to get a sense of how difficult this question is. Could an undergrad taking linear for the first time do this? Is it easier if one is asked to prove that the area is $|2cos^2 \theta -1|$ rather than finding it from scratch? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The graphic is from: &#xA;&lt;a href=&quot;http://algebra.math.ust.hk/determinant/01_geometry/lecture1.shtml&quot; rel=&quot;nofollow&quot;&gt;http://algebra.math.ust.hk/determinant/01_geometry/lecture1.shtml&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2563" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-10-24T21:21:04.080" LastActivityDate="2010-10-24T21:21:04.080" Title="Nice geometric parallelepiped proof?" Tags="&lt;geometry&gt;&lt;linear-algebra&gt;&lt;analytic-geometry&gt;" AnswerCount="3" CommentCount="3" />
  <row Id="7689" PostTypeId="2" ParentId="734" CreationDate="2010-10-24T16:24:38.453" Score="1" ViewCount="" Body="&lt;p&gt;The Rochester Institute of Technology offers a MS in applied statistics that is completely online (I think, I'm not enrolled yet.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.rit.edu/emcs/ptgrad/_program_detail.php?id=441&quot; rel=&quot;nofollow&quot;&gt;http://www.rit.edu/emcs/ptgrad/_program_detail.php?id=441&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's pretty expensive, since it combines the increased out-of-state tuition with the extra online cost. However an online degree such as this is good for working professionals who &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Can't take classes during the day and&#xA;2) Can get their employer to pay for most of the degree.&lt;/p&gt;&#xA;" OwnerUserId="592" LastActivityDate="2010-10-24T16:24:38.453" CommunityOwnedDate="2010-10-24T16:24:38.453" />
  <row Id="7691" PostTypeId="2" ParentId="7674" CreationDate="2010-10-24T16:40:35.117" Score="2" ViewCount="" Body="&lt;p&gt;If you desire to factor it by way of completing the square then it is simpler to first multiply by 4, namely  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\rm\ \ 4\:(a^2-3\: a+2)\ =\ (2\:a-3)^2 - 1\ =\ (2\:a-2)\:(2\:a-4)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, divide both sides through by $\rm\ \: 4\ =\ 2\cdot 2$&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-10-24T16:40:35.117" />
  <row Id="7692" PostTypeId="2" ParentId="7688" CreationDate="2010-10-24T16:43:15.617" Score="1" ViewCount="" Body="&lt;p&gt;For question 2, I think it is easy to find the area if you know a bit of trig, not using the linear algebra.  But maybe the students are too limited by the class at hand to use another approach.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-10-24T16:43:15.617" CommentCount="2" />
  <row Id="7693" PostTypeId="1" AcceptedAnswerId="7702" CreationDate="2010-10-24T16:53:33.320" Score="3" ViewCount="105" Body="&lt;p&gt;I'm trying to work out a basic example where formal smoothness should fail.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm considering $\mathbb{R} \to \mathbb{R}[x,y]/(x^2-y^2)$.&lt;br&gt;&#xA;The idea is that not every $\mathbb{R}$-homomorphism $\mathbb{R}[x,y]/(x^2-y^2) \to \mathbb{R}$ should lift to a homomorphism $\mathbb{R}[x,y]/(x^2-y^2) \to \mathbb{R}[\varepsilon]/(\varepsilon^2)$. But I can't see that ever being possible: after all, I can just take the exact same homomorphism, with image $\mathbb{R} \subset \mathbb{R}[\varepsilon]/(\varepsilon^2)$; this gives a valid lift.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do I need to consider something else than $R = \mathbb{R}[\varepsilon]/(\varepsilon^2)$ with $I = (\varepsilon)$ in order to witness the failure of formal smoothness? I would think that was enough, given that it should fully explain lifting of points to tangent vectors.&lt;/p&gt;&#xA;" OwnerUserId="8729" LastActivityDate="2010-10-24T17:39:41.923" Title="Have I got the right definition of formal smoothness?" Tags="&lt;algebraic-geometry&gt;&lt;schemes&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="7694" PostTypeId="2" ParentId="7621" CreationDate="2010-10-24T16:57:10.303" Score="3" ViewCount="" Body="&lt;p&gt;This might be &lt;em&gt;slightly&lt;/em&gt; off tangent, but it's definitely in the spirit of your post, and is found at the beginning of most texts in sieve theory. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When your fixed set of primes is the first $k$ primes and $n = p_{k}^2$ then $B_{n}$ is the main term of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Legendre_sieve&quot; rel=&quot;nofollow&quot;&gt;sieve of Eratosthenes-Legendre&lt;/a&gt;, which gives an approximation for the number of primes in the interval $[p_{k}, p_{k}^2]$. In this setting &lt;a href=&quot;http://en.wikipedia.org/wiki/Mertens%27_theorems&quot; rel=&quot;nofollow&quot;&gt;Merten's theorem&lt;/a&gt; gives an asymptotic for $B_{n}$. (It's called Merten's Third theorem in the Wikipedia page).&#xA;However, an appeal to the prime number theorem shows that this is a reasonably bad approximation, and that $B_{n}$ overestimates the number of primes by a factor of $2e^{-\gamma}$ where $\gamma$ is the Euler-Mascheroni constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Terry Tao has a good &lt;a href=&quot;http://terrytao.wordpress.com/2007/06/05/open-question-the-parity-problem-in-sieve-theory/&quot; rel=&quot;nofollow&quot;&gt;blog posting&lt;/a&gt; on why repeated use of the inclusion-exclusion principle yields non-optimal results.&lt;/p&gt;&#xA;" OwnerUserId="592" LastEditorUserId="592" LastEditorDisplayName="" LastEditDate="2010-10-24T17:06:15.017" LastActivityDate="2010-10-24T17:06:15.017" CommentCount="1" />
  <row Id="7695" PostTypeId="1" AcceptedAnswerId="7700" CreationDate="2010-10-24T16:57:13.780" Score="8" ViewCount="523" Body="&lt;p&gt;I would like to find the apothem of a regular pentagon. It follows from &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\cos \dfrac{2\pi }{5}=\dfrac{-1+\sqrt{5}}{4}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But how can this be proved (geometrically or trigonometrically)? &lt;/p&gt;&#xA;" OwnerUserId="752" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2010-10-25T12:38:31.337" LastActivityDate="2010-10-25T12:38:31.337" Title="How to prove $\cos \frac{2\pi }{5}=\frac{-1+\sqrt{5}}{4}$?" Tags="&lt;geometry&gt;&lt;trigonometry&gt;&lt;euclidean-geometry&gt;" AnswerCount="5" FavoriteCount="3" />
  <row Id="7696" PostTypeId="2" ParentId="7695" CreationDate="2010-10-24T17:00:15.313" Score="6" ViewCount="" Body="&lt;p&gt;Look up the &quot;construction of a regular pentagon&quot; using the straight edge and compass. If you keep track each step in this construction you will find that the angle 72 degrees comes in a few places, and this expression follows from it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's a fun excercise-- you should do it.&lt;/p&gt;&#xA;" OwnerUserId="2563" LastEditorUserId="2563" LastEditorDisplayName="" LastEditDate="2010-10-24T17:05:45.467" LastActivityDate="2010-10-24T17:05:45.467" CommentCount="3" />
  <row Id="7697" PostTypeId="1" AcceptedAnswerId="7703" CreationDate="2010-10-24T17:04:35.490" Score="3" ViewCount="135" Body="&lt;p&gt;This question is taken from a contest in India. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Prove that we can construct an infinite set of positive integers of the form $2^{n}-3$, where $n \in \mathbb{N}$, such that any two numbers from the set are relatively prime&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I would like to have an answer for this question, and i would also like to know why $2^{n}-3$ has importance here. Can this question be generalized. That is:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Can we construct an infinite set of positive integers of the form $k^{n} -(k+1)$ such that any two numbers from the set are relatively prime?.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" LastActivityDate="2010-10-24T20:54:35.507" Title="Construction of an infinite set such that any two number from the set are relatively prime" Tags="&lt;number-theory&gt;&lt;combinatorics&gt;" AnswerCount="2" />
  <row Id="7698" PostTypeId="2" ParentId="7695" CreationDate="2010-10-24T17:06:14.283" Score="2" ViewCount="" Body="&lt;p&gt;Or you can go to &lt;a href=&quot;http://mathworld.wolfram.com/TrigonometryAnglesPi5.html&quot; rel=&quot;nofollow&quot;&gt;Mathworld&lt;/a&gt; for $\pi/5$ and use the multiple angle formula&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-10-24T17:06:14.283" />
  <row Id="7699" PostTypeId="1" AcceptedAnswerId="7701" CreationDate="2010-10-24T17:08:06.213" Score="0" ViewCount="83" Body="&lt;p&gt;Hello! I have &lt;strong&gt;differential calculus&lt;/strong&gt; problem where I must find the coordinates of a point C on a parabola. There is a point given which is D(8,4) and D is on a line that is intersecting with the parabola at C. Also, the line where there is the point D is perpendicular to the tangent of C.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;This is the equation of the parabola:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;f(x) = -x^2 + 4x + 2&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This is coordinates of the single point given:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;D (8,4) &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;What are the coordinates of C?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I am having trouble with the fact that there is only point given for the intersecting line.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you for helping out!&lt;/p&gt;&#xA;" OwnerUserId="2751" LastEditorUserId="2751" LastEditorDisplayName="" LastEditDate="2010-10-24T17:24:14.060" LastActivityDate="2010-10-24T17:24:14.060" Title="Finding the coordinates of point on a parabola with a single point on an intersecting line" Tags="&lt;calculus&gt;&lt;differential-equations&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="7700" PostTypeId="2" ParentId="7695" CreationDate="2010-10-24T17:12:18.737" Score="16" ViewCount="" Body="&lt;p&gt;Since $x := \cos \frac{2 \pi}{5} = \frac{z + z^{-1}}{2}$ where $z:=e^{\frac{2 i \pi}{5}}$, and $1+z+z^2+z^3+z^4=0$ (for $z^5=1$ and $z \neq 1$), $x^2+\frac{x}{2}-\frac{1}{4}=0$, and voilà.&lt;/p&gt;&#xA;" OwnerUserId="2660" LastEditorUserId="1242" LastEditorDisplayName="" LastEditDate="2010-10-24T17:45:57.673" LastActivityDate="2010-10-24T17:45:57.673" CommentCount="5" />
  <row Id="7701" PostTypeId="2" ParentId="7699" CreationDate="2010-10-24T17:20:55.863" Score="0" ViewCount="" Body="&lt;p&gt;Using your differential calculus, you should be able to write the equation for the line tangent to the parabola at a given point on the parabola.  So if $(x,y)$ satisfies $y=-x^2+4x+2$ what is the equation of the tangent line?  Then you should be able to find the equation for the line perpendicular to the tangent at the point on the parabola.  One of these perpendiculars will pass through $(8,4)$.  &lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-10-24T17:20:55.863" CommentCount="8" />
  <row Id="7702" PostTypeId="2" ParentId="7693" CreationDate="2010-10-24T17:39:41.923" Score="3" ViewCount="" Body="&lt;p&gt;Indeed, you need to consider a different artinian ring to witness the failure of formal smoothness.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Look instead at $\mathbb{R}[\epsilon]/\epsilon^3 \to \mathbb{R}[\epsilon]/\epsilon^2$ and map $\mathbb{R}[x,y]/(x^2-y^2)$ to  $\mathbb{R}[\epsilon]/\epsilon^2$ by $x \mapsto \epsilon$, $y \mapsto 0$. If you wanted to lift this to a map to $\mathbb{R}[\epsilon]/\epsilon^3$, then you would have $x \mapsto \epsilon + a \epsilon^2$ and $y \to b \epsilon^2$, but then $x^2 = \epsilon^2$ and $y^2=0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &quot;geometric&quot; intuition here is that the Zariski tangent space to $x^2=y^2$ at $0$ is two dimensional but those tangent vectors which have slope other than $\pm 1$ can not be extended to higher order jets.&lt;/p&gt;&#xA;" OwnerUserId="448" LastActivityDate="2010-10-24T17:39:41.923" CommentCount="1" />
  <row Id="7703" PostTypeId="2" ParentId="7697" CreationDate="2010-10-24T17:49:36.740" Score="4" ViewCount="" Body="&lt;p&gt;Given a set of odd primes $\displaystyle S = \{p_1, p_2, \dots, p_m \}$, we can find an $n$ such that $\displaystyle 2^n - 3 = 1 \mod p_i$, by choosing $n = 2 + \prod (p_i-1)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus starting with &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\{2^2 -3,2^3 - 3\}$ we can add a new member which is relatively prime to all the previous members.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe a similar argument will hold for other $k$ (when considering $k^n -(k+1)$), since we will have $gcd(p_i,k) =1$.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-10-24T17:49:36.740" CommentCount="0" />
  <row Id="7704" PostTypeId="2" ParentId="7688" CreationDate="2010-10-24T17:54:12.657" Score="1" ViewCount="" Body="&lt;p&gt;Question 2: Since $a=\cos\theta$ and $b=\sin\theta$, it follows from the determinant formula for the area:&#xA;$$ A = \left| \det\begin{pmatrix} a &amp;amp; b \\ b &amp;amp; a \end{pmatrix} \right| = | \cos^2\theta - \sin^2 \theta |.$$&#xA;So I would say this is an easy problem for a linear algebra student.&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2010-10-24T17:54:12.657" CommentCount="2" />
  <row Id="7705" PostTypeId="2" ParentId="7634" CreationDate="2010-10-24T17:55:29.573" Score="3" ViewCount="" Body="&lt;p&gt;The problem with trees of finite width at all lattice points is called &quot;Polya's Orchard Problem&quot;.  Other variants such as higher dimensions and different shapes of the orchard are in the literature under the same name, or variations on &lt;em&gt;visibility&lt;/em&gt;, &lt;em&gt;lattice point&lt;/em&gt;, and &lt;em&gt;orchard&lt;/em&gt;.  Sample search result:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.rose-hulman.edu/mathjournal/archives/2006/vol7-n2/paper9/v7n2-9pd.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.rose-hulman.edu/mathjournal/archives/2006/vol7-n2/paper9/v7n2-9pd.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the problem with zero-width trees and visibility from the origin (ie., primitive lattice points) I think the $6/\pi^2$ density is attributed to Chebyshev in the 1800's, but it seems like the type of result that could have been known considerably earlier and frequently rediscovered. &lt;/p&gt;&#xA;" OwnerUserId="467" LastActivityDate="2010-10-24T17:55:29.573" />
  <row Id="7706" PostTypeId="1" AcceptedAnswerId="7707" CreationDate="2010-10-24T18:09:23.027" Score="0" ViewCount="102" Body="&lt;p&gt;How can you prove that $\Bigl\lfloor{x + \dfrac{1}{2}}\Bigr\rfloor$ is a $\theta(x)$ function?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm just practicing questions that could come up on an exam and this one was giving me a tough time trying to formally prove it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f(x)$ is $\theta(g(x)) \leftrightarrow f(x)$ is $O(g(x))$ and $f(x)$ is $\Omega(g(x))$&lt;/p&gt;&#xA;" OwnerUserId="1880" LastActivityDate="2010-10-24T18:36:34.350" Title="Proving theta complexity for floor function" Tags="&lt;functions&gt;&lt;proof-strategy&gt;" AnswerCount="1" CommentCount="7" />
  <row Id="7707" PostTypeId="2" ParentId="7706" CreationDate="2010-10-24T18:32:53.533" Score="1" ViewCount="" Body="&lt;p&gt;You can write&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x - \frac{1}{2} \leq \Bigl\lfloor x+\frac{1}{2}\Bigr\rfloor \leq x+\frac{1}{2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Doesn't this give you the result?&lt;/p&gt;&#xA;" OwnerUserId="2641" LastActivityDate="2010-10-24T18:36:34.350" CommentCount="1" />
  <row Id="7708" PostTypeId="1" CreationDate="2010-10-24T18:39:06.733" Score="2" ViewCount="128" Body="&lt;p&gt;If $q$ and $r$ are quaternions, and $p$ is a point, applying $q$ then $r$ to $p$ is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;(qr)p\dfrac{1}{qr}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What if I want to go the other way? Instead of concatenating rotations, I want to remove them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, I have $q$ and $m$, and I know that $qx =m$, but I don't know what $x$ is. How can I discover $x$? (How can I do quaternion divison?)&lt;/p&gt;&#xA;" OwnerUserId="2755" LastEditorUserId="664" LastEditorDisplayName="" LastEditDate="2010-10-24T19:50:58.193" LastActivityDate="2010-10-25T00:35:03.980" Title="Quaternion Division" Tags="&lt;quaternions&gt;" AnswerCount="3" CommentCount="4" FavoriteCount="1" />
  <row Id="7709" PostTypeId="2" ParentId="7708" CreationDate="2010-10-24T18:43:33.770" Score="0" ViewCount="" Body="&lt;p&gt;This is described in &lt;a href=&quot;http://en.wikipedia.org/wiki/Quaternion&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt; under the section Conjugation, the norm, and reciprocal.  You just need the reciprocal of q.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-10-24T18:43:33.770" />
  <row Id="7710" PostTypeId="2" ParentId="7708" CreationDate="2010-10-24T18:43:52.100" Score="4" ViewCount="" Body="&lt;p&gt;In algebra, multiplication is denoted by juxtaposition, not by an asterisk.&#xA;The inverse of the operation $p\mapsto qpq^{-1}$ is $p\mapsto q^{-1}pq$.&#xA;The solution of $qx=m$ is $x=q^{-1}m$. Note that&#xA;$$(a+bi+cj+dk)^{-1}=(a^2+b^2+c^2+d^2)^{-1}(a-bi-cj-dk).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All this can be found in &lt;a href=&quot;http://en.wikipedia.org/wiki/Quaternion&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-10-24T18:43:52.100" />
  <row Id="7711" PostTypeId="1" AcceptedAnswerId="7725" CreationDate="2010-10-24T19:20:02.490" Score="1" ViewCount="107" Body="&lt;p&gt;Suppose we have two sets of discrete events, $A$ and $B$. Then I think it is true that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$2\sum_{i \in A, j \in B}\Pr[i\ \textrm{AND}\ j] \leq \sum_{i \in A}\Pr[i]+ \sum_{j \in B}\Pr[j] +\sum_{i, j \in A}\Pr[i\ \textrm{AND}\ j] + \sum_{i, j \in B}\Pr[i\ \textrm{AND}\ j]$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My intuition for why this should be true is just by analogy to the simple inequality for $a, b \in \mathbb{R}:$&#xA;$$2ab \leq a + b + a(a-1) + b(b-1)$$&#xA;To see why this identity is true, note:&#xA;$$a + b + a(a-1) + b(b-1) = a^2 + b^2 = 2ab + (a-b)^2 \geq 2ab$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't have a proof for the more complex identity though. Is it true? Why does it hold?&lt;/p&gt;&#xA;" LastActivityDate="2010-10-24T20:57:56.667" Title="Why does this inequality in probability theory hold?" Tags="&lt;probability&gt;&lt;probability-theory&gt;&lt;inequality&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="7712" PostTypeId="1" AcceptedAnswerId="7714" CreationDate="2010-10-24T19:28:58.043" Score="0" ViewCount="72" Body="&lt;p&gt;Hi, if A = { Market research predicates strong demand } and B = { Market demand is strong }, can we reasonably assume that P(A or B) = P(A) * P(B)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is that I know &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;P(B|A) = 0.8 &lt;/li&gt;&#xA;&lt;li&gt;P(not B | not A) = 0.9, &lt;/li&gt;&#xA;&lt;li&gt;P(B) = 0.2&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I need to calculate P(A) and P(not A).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For me, it seems that if P(A or B) != P(A) * P(B), it's impossible to know the answer.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Actually, I'm drawing a decision tree to determine whether a market research is worth trying.  All the information is listed above.  Without P(A) and P(not A), I can not determine calculate the expected value of the branch of taking the research and can not draw the tree.&lt;/p&gt;&#xA;" OwnerUserId="1618" LastEditorUserId="1618" LastEditorDisplayName="" LastEditDate="2010-10-24T19:50:15.680" LastActivityDate="2010-10-24T19:50:15.680" Title="A problem of probability." Tags="&lt;probability&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="7713" PostTypeId="1" CreationDate="2010-10-24T19:34:22.303" Score="0" ViewCount="113" Body="&lt;p&gt;Let f(t)=H(t)-H(t-2) and \psi(t)=t*exp(-abs(t)). Compute the wavelet transform of f. Analyse the behavior of the transform when the scaling parameter a approaches zero. H(t) = piecewise( t&gt;0, 1, t&amp;lt;0, 0), W_f(a,b)== \int_{-\infty}^{\infty} f(t)*\psi_{a,b}(t) dt and \psi_{a,b}(t)=a^{-1/2} \psi((t-b)/a)&lt;/p&gt;&#xA;" OwnerUserId="2723" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-10-25T09:06:50.430" LastActivityDate="2010-10-25T09:06:50.430" ClosedDate="2010-10-25T01:57:50.297" Title="Let f(t)=H(t)-H(t-2) and \psi(t)=t*exp(-abs(t)). Compute the wavelet transform W_f(a,b) of f." Tags="&lt;wavelet-transform&gt;" CommentCount="2" />
  <row Id="7714" PostTypeId="2" ParentId="7712" CreationDate="2010-10-24T19:41:20.303" Score="1" ViewCount="" Body="&lt;p&gt;Your first question is not a mathematical question, but I'd&#xA;say it was reasonable to assume the events are not independent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Drawing a Venn diagram of events gives four regions, but&#xA;you only have three equations for the probabilities of each, so&#xA;not enough information to solve.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added&lt;/strong&gt;&#xA;With your edit, you introduce a new piece of information.&#xA;I would assign probabilities to the four regions in the Venn&#xA;diagram, say $p_1=P(A \textrm{ and }B)$,&#xA;$p_2=P(A \textrm{ and not }B)$,&#xA;$p_3=P(\textrm{not }A \textrm{ and }B)$ and&#xA;$p_4=P(\textrm{not }A \textrm{ and not }B)$.&#xA;Then you know $p_1+p_2+p_3+p_4=1$, $p_1=0.8(p_1+p_2)$,&#xA;$p_4=0.9(p_3+p_4)$ and $p_1+p_3=0.2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Four linear equations in four unknowns!&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-10-24T19:41:20.303" CommentCount="1" />
  <row Id="7715" PostTypeId="2" ParentId="7527" CreationDate="2010-10-24T20:06:17.533" Score="3" ViewCount="" Body="&lt;p&gt;For what it's worth, such a sequence constructed by a greedy algorithm begins:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2, 4, 8, 10, 16, 18, 36, 199, 208, 1131, 1347, 3984, 5751, 7310, 27315, 129313, 134101, 169400, 589570&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is, we start with $A_1 = 2$, and then each $A_n$ is the smallest number greater than $A_{n-1}$ with $A_i + A_n$ having an even number of distinct prime factors for each $n \le 1$. (So, for example, $A_3 = 8$ because $5+4, 6+2, 7+2$ each have an odd number of distinct prime factors, but $8+2$ and $8+4$ both have even numbers.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another such sequence, starting with 1, begins&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1, 5, 9, 13, 35, 39, 286, 290, 381, 385, 866, 4376&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and one starting with 3 begins&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3, 7, 11, 15, 33, 41, 47, 65, 101, 203, 4102, 6392, 8507, 18608.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From these it &lt;i&gt;seems&lt;/i&gt; that these sequences grow roughly exponentially; that is, $A_n \approx k^n$ for some constant $k$. This makes sense. Since approximately half of all integers have an even number of distinct prime factors, once you have $n$ numbers in such a sequence it should take about $2^n$ tries to find the next one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course this isn't a proof, but it's at least an argument why such sequences should exist. And judging from the irregularity of the greedily constructed sequences, a greedy method probably isn't the best way to go here even if you want to explicitly construct the sequence.&lt;/p&gt;&#xA;" OwnerUserId="173" LastActivityDate="2010-10-24T20:06:17.533" CommentCount="1" />
  <row Id="7716" PostTypeId="1" AcceptedAnswerId="7718" CreationDate="2010-10-24T20:12:57.467" Score="4" ViewCount="180" Body="&lt;p&gt;I was just wondering where the y'/(dy/dx) in implicit differentiation comes from.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;x^2 + y^2 = 25&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(d/dx) x^2 + (d/dy) y^2 &lt;strong&gt;(dy/dx)&lt;/strong&gt; = 25 (d/dx)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2x + 2y (dy/dx) = 0&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(dy/dx) = -x/y&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where does the bold part come from? Wikipedia says it's a byproduct of the chain rule, but it's just not clicking for me.&lt;/p&gt;&#xA;" OwnerUserId="1389" LastEditorUserId="1543" LastEditorDisplayName="" LastEditDate="2011-04-18T12:07:55.457" LastActivityDate="2011-04-18T12:07:55.457" Title="Implicit Differentiation" Tags="&lt;calculus&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="7717" PostTypeId="1" CreationDate="2010-10-24T20:15:21.210" Score="0" ViewCount="827" Body="&lt;p&gt;I've heard all these terms thrown about in proofs and in geometry, but what are the differences and relationships between them? Examples would be awesome! :)&lt;/p&gt;&#xA;" OwnerUserId="104" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-10-25T00:19:54.450" LastActivityDate="2010-10-25T02:56:15.230" Title="Difference between axioms, theorems, postulates, corollaries, and hypothesis" Tags="&lt;terminology&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="7718" PostTypeId="2" ParentId="7716" CreationDate="2010-10-24T20:26:06.123" Score="5" ViewCount="" Body="&lt;p&gt;When you implicitly differentiate $x^2+y^2=25$, you are differentiating with respect to a particular variable—in this case, $x$, so:&#xA;$$\begin{align}&#xA;\frac{d}{dx}(x^2+y^2)&amp;amp;=\frac{d}{dx}25&#xA;\\&#xA;\frac{d}{dx}(x^2)+\frac{d}{dx}(y^2)&amp;amp;=0&#xA;\\&#xA;2x+2y\frac{dy}{dx}&amp;amp;=0&#xA;\\&#xA;2y\frac{dy}{dx}&amp;amp;=-2x&#xA;\\&#xA;\frac{dy}{dx}&amp;amp;=-\frac{x}{y}&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the 3rd line to the 4th line, $\frac{d}{dx}(y^2)$ is the derivative with respect to $x$ of $y^2$, in which (as in Ryan Budney's comment) we assume that $y$ is some function of $x$, so we apply the chain rule, differentiating $y^2$ with respect to $y$ and multiplying by the derivative of $y$ with respect to $x$ to get $2y\frac{dy}{dx}$.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;edit&lt;/em&gt;: Based on the comments below, I think it might be useful if I introduced a slightly different notation: Let $D_x$ be the &lt;a href=&quot;http://en.wikipedia.org/wiki/Differential_operator&quot; rel=&quot;nofollow&quot;&gt;differential operator&lt;/a&gt; with respect to $x$, which you have previously written as $\frac{d}{dx}$ (and, similarly, $D_y$ is the differential operator with respect to $y$).  When we apply the differential operator to something, we read and write it like a function: $D_x(x^2)=2x$ is &quot;the derivative with respect to $x$ of $x^2$ is $2x$.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, rewriting the work above in this notation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xA;D_x(x^2+y^2)&amp;amp;=D_x(25)&#xA;\\&#xA;D_x(x^2)+D_x(y^2)&amp;amp;=0&#xA;\\&#xA;2x+D_y(y^2)D_x(y)&amp;amp;=0&#xA;\\&#xA;2x+2yD_x(y)&amp;amp;=0&#xA;\\&#xA;2yD_x(y)&amp;amp;=-2x&#xA;\\&#xA;D_x(y)=\frac{dy}{dx}&amp;amp;=-\frac{x}{y}&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And, to your question of finding $\frac{dx}{dy}$:&#xA;$$\begin{align}&#xA;D_y(x^2+y^2)&amp;amp;=D_y(25)&#xA;\\&#xA;D_y(x^2)+D_y(y^2)&amp;amp;=0&#xA;\\&#xA;D_x(x^2)D_y(x)+2y&amp;amp;=0&#xA;\\&#xA;2xD_y(x)+2y&amp;amp;=0&#xA;\\&#xA;2xD_y(x)&amp;amp;=-2y&#xA;\\&#xA;D_y(x)=\frac{dx}{dy}&amp;amp;=-\frac{y}{x}&#xA;\end{align}$$&lt;/p&gt;&#xA;" OwnerUserId="72" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-10-25T04:32:46.083" LastActivityDate="2010-10-25T04:32:46.083" CommentCount="9" />
  <row Id="7719" PostTypeId="2" ParentId="7717" CreationDate="2010-10-24T20:27:58.157" Score="5" ViewCount="" Body="&lt;p&gt;In Geometry, &quot;Axiom&quot; and &quot;Postulate&quot; are essentially interchangeable. In antiquity, they referred to propositions that were &quot;obviously true&quot; and only had to be stated, and not proven. In modern mathematics there is no longer an assumption that axioms are &quot;obviously true&quot;. Axioms are merely 'background' assumptions we make. The best analogy I know is that axioms are the &quot;rules of the game&quot;. In Euclid's Geometry, the main axioms/postulates are: (I) Given any two distinct points, there is a line that contains them. (II) Any line segment can be extended to an infinite line. (III) Given a point and a radius, there is a circle with center in that point and that radius. (IV) All right angles are equal to one another. (V) If a straight line falling on two straight lines makes the interior angles on the same side less than two right angles, the two straight lines, if produced indefinitely, meet on that side on which are the angles less than the two right angles. (The &lt;em&gt;parallel postulate&lt;/em&gt;). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A theorem is a logical consequence of the axioms. In Geometry, the &quot;propositions&quot; are all theorems: they are derived using the axioms and the valid rules. A &quot;Corollary&quot; is a theorem that is usually considered an &quot;easy consequence&quot; of another theorem. What is or is not a corollary is entirely subjective. Sometimes what an author thinks is a 'corollary' is deemed more important than the corresponding theorem. (The same goes for &quot;Lemma&quot;s, which are theorems that are considered auxiliary to proving some other, more important in the view of the author, theorem). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A &quot;hypothesis&quot; is an assumption made. For example, &quot;If $x$ is an even integer, then $x^2$ is an even integer&quot; I am not asserting that $x^2$ is even or odd; I am asserting that if &lt;em&gt;something&lt;/em&gt; happens (namely, if $x$ happens to be an even integer) then &lt;em&gt;something else&lt;/em&gt; will also happen. Here, &quot;$x$ is an even integer&quot; is the hypothesis being made to prove it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See the Wikipedia pages on &lt;a href=&quot;http://en.wikipedia.org/wiki/Axiom&quot; rel=&quot;nofollow&quot;&gt;axiom&lt;/a&gt;, &lt;a href=&quot;http://en.wikipedia.org/wiki/Theorem&quot; rel=&quot;nofollow&quot;&gt;theorem&lt;/a&gt;, and &lt;a href=&quot;http://en.wikipedia.org/wiki/Corollary&quot; rel=&quot;nofollow&quot;&gt;corollary&lt;/a&gt;. The first two have many examples.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-10-25T00:33:48.633" LastActivityDate="2010-10-25T00:33:48.633" CommentCount="2" />
  <row Id="7720" PostTypeId="1" AcceptedAnswerId="7721" CreationDate="2010-10-24T20:29:47.700" Score="1" ViewCount="96" Body="&lt;p&gt;I know it is possible for a group $G$ to have normal subgroups $H, K$, such that $H\cong K$ but $G/H\not\cong G/K$, but I couldn't think of any examples with $G$ finite. What is an illustrative example? &lt;/p&gt;&#xA;" OwnerUserId="264" LastActivityDate="2010-10-24T20:35:27.730" Title="Finite group with isomorphic normal subgroups and non-isomorphic quotients?" Tags="&lt;group-theory&gt;" AnswerCount="1" CommentCount="0" />
  <row Id="7721" PostTypeId="2" ParentId="7720" CreationDate="2010-10-24T20:35:27.730" Score="6" ViewCount="" Body="&lt;p&gt;Take $G = Z_4 \times Z_2$, $H$ generated by $(0,1)$, $K$ generated by $(2,0)$.  Then $H \cong K \cong Z_2$ but $G/H \cong Z_4$ while $G/K \cong Z_2 \times Z_2$.&lt;/p&gt;&#xA;" OwnerUserId="822" LastActivityDate="2010-10-24T20:35:27.730" CommentCount="2" />
  <row Id="7722" PostTypeId="2" ParentId="7688" CreationDate="2010-10-24T20:53:48.067" Score="1" ViewCount="" Body="&lt;p&gt;It is a fact of linear algebra that any nonsingular matrix $A$ is a product of &quot;elementary matrices&quot;. An elementary matrix $E$ is either (a) an identity matrix with one diagonal entry replaced by an arbitrary $a\ne 0$ or (b) an identity matrix with one off-diagonal zero replaced by some $b\ne 0$. In case (a) the corresponding linear map multiplies volumes by $|a|$, which coincides with $|\det(E)|$, and in case (b) volumes are unchanged by $E$, which can be verified by looking at a two-dimensional figure. It follows that in case (b) the volume factor is $1=|\det(E)|$ as well. By the multiplicativity of determinants it follows that any nonsingular $A:\mathbb R^n\to\mathbb R^n$ multiplies volumes by the factor $|\det(A)|$. The parallelepiped $P$ considered in the question is the linear image of the unit cube; therefore vol$(P)=|\det(A)|$ where the matrix $A$ has the spanning vectors of $P$ in its columns.&lt;/p&gt;&#xA;" OwnerUserId="1303" LastActivityDate="2010-10-24T20:53:48.067" />
  <row Id="7723" PostTypeId="2" ParentId="7695" CreationDate="2010-10-24T20:53:49.530" Score="10" ViewCount="" Body="&lt;p&gt;How about &lt;em&gt;combinatorially&lt;/em&gt;?  This follows from the following two facts.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;The eigenvalues of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Adjacency_matrix&quot; rel=&quot;nofollow&quot;&gt;adjacency matrix&lt;/a&gt; of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Path_graph&quot; rel=&quot;nofollow&quot;&gt;path graph&lt;/a&gt; on $n$ vertices are $2 \cos \frac{k \pi}{n+1}, k = 1, 2, ... n$.  &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The number of closed walks from one end of the path graph on $4$ vertices to itself of length $2n$ is the Fibonacci number $F_{2n}$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The first can be proven by direct computation (although it also somehow falls out of the theory of quantum groups) and the second is a nice combinatorial argument which I will leave as an exercise.  I discuss some of the surrounding issues in &lt;a href=&quot;http://qchu.wordpress.com/2009/06/07/the-catalan-numbers-regular-languages-and-orthogonal-polynomials/&quot; rel=&quot;nofollow&quot;&gt;this blog post&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-10-24T20:53:49.530" CommentCount="2" />
  <row Id="7724" PostTypeId="2" ParentId="7697" CreationDate="2010-10-24T20:54:35.507" Score="3" ViewCount="" Body="&lt;p&gt;Further to Moron's answer, consider any sequence of the form x&lt;sub&gt;n&lt;/sub&gt;&amp;nbsp;=&amp;nbsp;a&lt;sup&gt;n&lt;/sup&gt;&amp;nbsp;+&amp;nbsp;b for fixed integers a,b. Is there an infinite set of the x&lt;sub&gt;n&lt;/sub&gt; for which any two are coprime?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, any common factor of a and b will divide every x&lt;sub&gt;n&lt;/sub&gt;. Furthermore, for any n,&#xA;$$&#xA;x_n=(a^{n-1}+a^{n-2}+\cdots+a+1)(a-1)+(b+1).&#xA;$$&#xA;so any common factor of a&amp;nbsp;&amp;minus;&amp;nbsp;1 and b&amp;nbsp;+&amp;nbsp;1 will divide every x&lt;sub&gt;n&lt;/sub&gt;. The answer to the question is no if either a, b or a&amp;nbsp;&amp;minus;&amp;nbsp;1, b&amp;nbsp;+&amp;nbsp;1 have a common prime factor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If a, b are coprime and every factor of b&amp;nbsp;+&amp;nbsp;1 divides a then the answer is yes.&#xA;Let S&amp;nbsp;=&amp;nbsp;{p&lt;sub&gt;1&lt;/sub&gt;,...,p&lt;sub&gt;m&lt;/sub&gt;} be a finite set of primes not dividing a. Taking n to be a multiple of &amp;prod;&lt;sub&gt;k&lt;/sub&gt;(p&lt;sub&gt;k&lt;/sub&gt;&amp;nbsp;&amp;minus;&amp;nbsp;1) gives x&lt;sub&gt;n&lt;/sub&gt;&amp;nbsp;=&amp;nbsp;1&amp;nbsp;+&amp;nbsp;b&amp;nbsp;&amp;ne;&amp;nbsp;0 (mod p&lt;sub&gt;k&lt;/sub&gt;), so is not a multiple of p&lt;sub&gt;k&lt;/sub&gt;. If S is the set of prime factors of any finite collection of the x&lt;sub&gt;n&lt;/sub&gt;, this shows that we can find a new x&lt;sub&gt;n&lt;/sub&gt; coprime to each of the ones so far and, by induction, we can find an infinite set.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This shows that the answer is yes in the cases you mention a&amp;nbsp;=&amp;nbsp;2, b&amp;nbsp;=&amp;nbsp;-3 and a&amp;nbsp;=&amp;nbsp;k, b&amp;nbsp;=&amp;nbsp;-(k&amp;nbsp;+&amp;nbsp;1). It does, however, leave open the cases where a,b are coprime and b&amp;nbsp;+&amp;nbsp;1 has prime factors not dividing a or a&amp;nbsp;&amp;minus;&amp;nbsp;1.&lt;/p&gt;&#xA;" OwnerUserId="1321" LastActivityDate="2010-10-24T20:54:35.507" />
  <row Id="7725" PostTypeId="2" ParentId="7711" CreationDate="2010-10-24T20:57:56.667" Score="1" ViewCount="" Body="&lt;p&gt;If I've understood your notation, $A$ and $B$ are index sets of events denoted $i,j$ etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The inequality $2ab\leq a^2+b^2$ applied to the sums $a:=\sum_{i\in A}1_i$ and &#xA;$b:=\sum_{j\in B}1_j$ of indicator functions, followed by integration, gives &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$2\sum_{i \in A, j \in B}\Pr[i\ \textrm{AND}\ j] \leq  \sum_{i, j \in A}\Pr[i\ \textrm{AND}\ j] + \sum_{i, j \in B}\Pr[i\ \textrm{AND}\ j],$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is stronger than what you asked for. &lt;/p&gt;&#xA;" OwnerUserId="940" LastActivityDate="2010-10-24T20:57:56.667" />
  <row Id="7726" PostTypeId="1" AcceptedAnswerId="7732" CreationDate="2010-10-24T21:03:07.057" Score="0" ViewCount="403" Body="&lt;p&gt;Consider the probability space $([0,1]; B[0,1], L)$, where $B[0,1]$ contains the Borel sets&#xA;intersecting $[0,1]$ and $L$ is the Lebesgue measure. How do I find the sigma algebra generated by a random variable defined on this space, $X = 1_{[0,1/2]}$? Secondly, how do I determine whether random variables defined on this space are independent or not, e.g., are $X = 1_{[0,1/2}]$ and $Y = 1_{[1/4,3/4}]$ independent?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For finding sigma algebra generated by $X$, I find $X^{-1}(1_{[0,1/2]})$ but what will be this inverse in borel sets? For finding independence, it should be sufficient to show the sigma algebras generated by $X$ and $Y$ are independent right?&lt;/p&gt;&#xA;" OwnerUserId="957" LastActivityDate="2011-02-06T17:38:43.710" Title="sigma algebra generated by random variable" Tags="&lt;probability-theory&gt;&lt;measure-theory&gt;&lt;sigma-algebra&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="7727" PostTypeId="2" ParentId="7695" CreationDate="2010-10-24T21:04:36.103" Score="14" ViewCount="" Body="&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/IvfAH.png&quot; alt=&quot;diagram&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider a $\triangle ABC$ with $AB=1$, $\mathrm{m}\angle A=\frac{\pi}{5}$ and $\mathrm{m}\angle B=\mathrm{m}\angle C=\frac{2\pi}{5}$, and point $D$ on $\overline{AC}$ such that $\overline{BD}$ bisects $\angle ABC$.  Now, $\mathrm{m}\angle CBD=\frac{\pi}{5}$ and $\mathrm{m}\angle BDC=\frac{2\pi}{5}$, so $\triangle ABC\sim\triangle BCD$.  Also note that $\triangle ABD$ is isosceles so that $BC=BD=AD$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $x=BC=BD=AD$.  From the similar triangles, $\frac{AB}{BC}=\frac{BC}{CD}$ or $\frac{1}{x}=\frac{x}{1-x}$, so $1-x=x^2$ and $x=\frac{\sqrt{5}-1}{2}$ (the other solution is negative and lengths cannot be negative).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, apply the Law of Cosines to $\triangle ABC$:&#xA;$$\begin{align}&#xA;\cos\frac{2\pi}{5}=\cos C&amp;amp;=\frac{a^2+b^2-c^2}{2ab}&#xA;\\&#xA;&amp;amp;=\frac{(\frac{\sqrt{5}-1}{2})^2+1^2-1^2}{2\cdot\frac{\sqrt{5}-1}{2}\cdot 1}&#xA;\\&#xA;&amp;amp;=\frac{\frac{3-\sqrt{5}}{2}}{\sqrt{5}-1}&#xA;\\&#xA;&amp;amp;=\frac{-1+\sqrt{5}}{4}.&#xA;\end{align}$$&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-10-24T21:04:36.103" CommentCount="3" />
  <row Id="7729" PostTypeId="1" AcceptedAnswerId="7730" CreationDate="2010-10-24T21:17:37.490" Score="2" ViewCount="105" Body="&lt;p&gt;If $X$ is a random variable with finite mean $\mu$ and variance $\sigma^2$, how do I show that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$P[\mu − d\sigma &amp;lt; X &amp;lt; \mu + d\sigma] ≥ 1 − 1/d^2~~~~~~\forall d&gt;1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found this in a book but unable to see the proof. Note that $X$ may not be normal.&lt;/p&gt;&#xA;" OwnerUserId="957" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-10-24T22:15:56.037" LastActivityDate="2010-10-24T22:15:56.037" Title="lower bound for probability distribution of a random variable" Tags="&lt;probability&gt;&lt;probability-theory&gt;&lt;measure-theory&gt;" AnswerCount="1" />
  <row Id="7730" PostTypeId="2" ParentId="7729" CreationDate="2010-10-24T21:24:31.817" Score="3" ViewCount="" Body="&lt;p&gt;This is &lt;a href=&quot;http://en.wikipedia.org/wiki/Chebyshev%27s_inequality&quot; rel=&quot;nofollow&quot;&gt;Chebyshev's inequality&lt;/a&gt;, which holds for any probability distribution.  There are two proofs given on the linked Wikipedia page - a measure-theoretic one, and one that uses &lt;a href=&quot;http://en.wikipedia.org/wiki/Markov%27s_inequality&quot; rel=&quot;nofollow&quot;&gt;Markov's inequality&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your expression is in a different form, though, than the one on the Wikipedia page.  To see how they are the same, observe that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$P[\mu - d \sigma &amp;lt; X &amp;lt; \mu + d \sigma] \geq 1 - 1/d^2$$&#xA;is equivalent to&#xA;$$P[|X - \mu| &amp;lt; d \sigma] \geq 1 - 1/d^2,$$&#xA;which is equivalent to&#xA;$$P[|X - \mu| &gt; d \sigma] \leq 1/d^2,$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is the one on the Wikipedia page.&lt;/p&gt;&#xA;" OwnerUserId="2370" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-10-24T21:42:26.383" LastActivityDate="2010-10-24T21:42:26.383" />
  <row Id="7731" PostTypeId="1" CreationDate="2010-10-24T21:38:35.420" Score="1" ViewCount="83" Body="&lt;p&gt;This is one of the descriptions I've seen online:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For any even n, say n = 2m, a row complete latin square of order n can be formed by writing down&lt;/p&gt;&#xA;&#xA;&lt;p&gt;0, 1, 2m - 1, 2, 2m - 2, 3,..., m + 1, m&lt;/p&gt;&#xA;&#xA;&lt;p&gt;as the first row and then developing subsequent rows by adding 1 modulo n.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not quite clear on how that goes, such as what the difference is between 2m and m.  2m stands for n, does m stand for modulo?  I'd like to see and example with an even number or two, such as 4, and 6. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2m = 4&lt;/p&gt;&#xA;&#xA;&lt;p&gt;0, 1, 3, 2, 2... wait, that can't be right.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It should turn out with non repeating sequences, and non recurring pairs between rows.&#xA;I can outline the other methods I've found (but don't understand) if this one is not the best.&lt;/p&gt;&#xA;" LastEditorUserId="139" LastEditorDisplayName="" LastEditDate="2010-11-19T10:38:00.530" LastActivityDate="2010-11-19T10:38:00.530" Title="How to Form a Row-Complete Latin Square" Tags="&lt;combinatorics&gt;&lt;latin-square&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="7732" PostTypeId="2" ParentId="7726" CreationDate="2010-10-24T21:41:53.210" Score="2" ViewCount="" Body="&lt;p&gt;The sigma-algebra generated by $1_{[0,1/2]}$ is simply&#xA;$$&#xA;\bigl\{\emptyset,[0,1],[0,1/2],(1/2,1]\bigr\}.&#xA;$$&#xA;It consists of the preimages under the function $1_{[0,1/2]}$ of all Borel sets in the codomaion of the function $1_{[0,1/2]}$, namely, $(\mathbb R,B(\mathbb R))$. (Notice that the preimage $1_{[0,1/2]}^{-1}(M)$ is completely determined by the information of whether 0 and 1 do or do not belong to $M$ respectively.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The situation for $1_{[1/4,3/4]}$ is similar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The random variables $1_{[0,1/2]}$ and $1_{[1/4,3/4]}$ on $([0,1],B[0,1],L)$ are indeed independent: For this you have to check that $L(A\cap B)=L(A)\cdot L(B)$ for all $A\in 1_{[0,1/2]}^{-1}(B(\mathbb R))$ and $B\in 1_{[1/4,3/4]}^{-1}(B(\mathbb R))$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most interesting case is $L([0,1/2]\cap [1/4,3/4])=L([0,1/2])\cdot L([1/4,3/4])$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Check that both sides are equal!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also think about the following question: Are the random variables $1_{[0,1/2]}$ and $1_{[1/4,1]}$ on $([0,1],B[0,1],L)$ also independent?&lt;/p&gt;&#xA;" OwnerUserId="367" LastEditorUserId="367" LastEditorDisplayName="" LastEditDate="2010-10-24T21:48:03.480" LastActivityDate="2010-10-24T21:48:03.480" CommentCount="7" />
  <row Id="7733" PostTypeId="2" ParentId="7731" CreationDate="2010-10-24T21:42:53.803" Score="2" ViewCount="" Body="&lt;p&gt;If I understand your description correctly, for 2m=4:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;0  1  3  2&#xA;1  2  0  3&#xA;2  3  1  0&#xA;3  0  2  1&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;For 2m=6:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;0  1  5  2  4  3&#xA;1  2  0  3  5  4&#xA;2  3  1  4  0  5&#xA;3  4  2  5  1  0&#xA;4  5  3  0  2  1&#xA;5  0  4  1  3  2&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;edit&lt;/em&gt;: In each case, the first row follows the pattern you gave; in subsequent rows, each number is one more than the number above it, but wrapping around at the size of the square (e.g. in the case where n=2m=6, after 5 comes 0).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's look at n=2m=12 (m=6) a bit more closely.  The pattern that you gave said:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;0, 1, 2m - 1, 2, 2m - 2, 3,..., m + 1, m&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So, the first row is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;0, 1, 12 - 1, 2, 12 - 2, 3,..., 6 + 1, 6&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;which is&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;0, 1, 12-1=11, 2, 12-2=10, 3, 12-3=9, 4, 12-4=8, 5, 12-5=6+1=7, 6&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now, the second row is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;0+1=1, 1+1=2, 11+1=12 becomes 0, 2+1=3, 9+1=10, 4+1=5, 8+1=9, 5+1=6, 7+1=8, 6+1=7&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Continuing gives:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;0  1 11  2 10  3  9  4  8  5  7  6&#xA;1  2  0  3 11  4 10  5  9  6  8  7&#xA;2  3  1  4  0  5 11  6 10  7  9  8&#xA;3  4  2  5  1  6  0  7 11  8 10  9&#xA;               .&#xA;               .&#xA;               .&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="72" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-10-24T22:10:56.053" LastActivityDate="2010-10-24T22:10:56.053" CommentCount="2" />
  <row Id="7735" PostTypeId="2" ParentId="7716" CreationDate="2010-10-24T22:12:40.720" Score="6" ViewCount="" Body="&lt;p&gt;Isaac and Ryan have already answered your question in words.  Now, in symbols, the chain rule gives:&#xA;$$\frac{d(y^2)}{dx} = \frac{d(y^2)}{dy}\frac{dy}{dx} = 2y\frac{dy}{dx}$$&lt;/p&gt;&#xA;" OwnerUserId="640" LastActivityDate="2010-10-24T22:12:40.720" CommentCount="1" />
  <row Id="7737" PostTypeId="1" AcceptedAnswerId="7747" CreationDate="2010-10-24T22:48:52.347" Score="0" ViewCount="331" Body="&lt;p&gt;How do I use Lebesgue Dominated Convergence Theorem to evaluate &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\lim_{n \to \infty}\int_{[0,1]}\frac{n\sin(x)}{1+n^2\sqrt(x)}dx$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;what dominating function to use here?&lt;/p&gt;&#xA;" OwnerUserId="957" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-10-24T22:51:43.560" LastActivityDate="2010-10-25T00:50:01.333" Title="Lebesgue Dominated Convergence example" Tags="&lt;probability-theory&gt;&lt;integral&gt;&lt;measure-theory&gt;&lt;convergence&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="7738" PostTypeId="1" CreationDate="2010-10-24T22:51:06.243" Score="-1" ViewCount="186" Body="&lt;p&gt;I have trouble understanding Radon-Nikodym theorem. Here's an example.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the measurable space $(\Omega,F)$, where $\Omega = R$, $F~$ is the $\sigma$-algebra of Borel sets. Let $P[dw] = \frac{1}{\sqrt(2\pi)}e^{-w^2/2}dw$ and $\tilde{P}[dw] = e^{-w}1_{[0,\infty)(w)}dw$ be two probability measures on $\omega$. How to test whether $\tilde{P} &amp;lt;&amp;lt; P$ and $P &amp;lt;&amp;lt; \tilde{P}$. How do we compute Radon-Nikodym derivative $\frac{d\tilde{P}}{dP}$?&lt;/p&gt;&#xA;" OwnerUserId="957" LastActivityDate="2010-10-25T12:20:09.047" Title="Radon-Nikodym example" Tags="&lt;probability-theory&gt;&lt;integral&gt;&lt;measure-theory&gt;&lt;convergence&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" />
  <row Id="7740" PostTypeId="1" CreationDate="2010-10-24T23:00:08.350" Score="2" ViewCount="105" Body="&lt;p&gt;I want to refer to functions of the form $f(x) = \sum_{i=1}^n a_i x^{\alpha_i}$ where $\alpha_i &amp;lt; 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is not a polynomial, because $\alpha_i$ could be just real arbitrary numbers (though positive ones).&#xA;Is there a name for this class of functions that I can refer to? Wikipedia does claim these are not polynomials.&lt;/p&gt;&#xA;" OwnerUserId="7076" LastEditorUserId="30" LastEditorDisplayName="" LastEditDate="2010-10-25T00:45:25.417" LastActivityDate="2010-10-25T00:45:25.417" Title="polynomial of $x$?" Tags="&lt;terminology&gt;&lt;algebra&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="7741" PostTypeId="2" ParentId="7740" CreationDate="2010-10-24T23:07:54.570" Score="3" ViewCount="" Body="&lt;p&gt;In 1983 they still had no official name, see for example &lt;a href=&quot;http://www.cecm.sfu.ca/personal/pborwein/PAPERS/P16.pdf&quot; rel=&quot;nofollow&quot;&gt;this paper&lt;/a&gt; by Peter Borwein.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-10-24T23:07:54.570" CommentCount="2" />
  <row Id="7742" PostTypeId="2" ParentId="7737" CreationDate="2010-10-24T23:26:11.120" Score="4" ViewCount="" Body="&lt;p&gt;Consider this for a solution (it does not use dominated convergence). In $[0,1]$, we have the following inequality&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;0 \leq \frac{n \sin(x)}{1+ n^2 \sqrt{x}} &amp;lt; \frac{n}{n^2 \sqrt{x}} = \frac{1}{n \sqrt{x}}&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Taking integral over $[0,1]$ on both sides, we get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;\int_{[0,1]} \frac{n \sin(x)}{1+ n^2 \sqrt{x}} &amp;lt; \int_{[0,1]} \frac{1}{n \sqrt{x}} = \left[ \frac{2 \sqrt{x}}{n} \right]_0^1&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;taking limit $n \to \infty$ on both sides, you can show that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;\lim_{n \to \infty} \int_{[0,1]} \frac{n \sin(x)}{1+ n^2 \sqrt{x}} = 0&#xA;\end{equation}&lt;/p&gt;&#xA;" OwnerUserId="2641" LastActivityDate="2010-10-24T23:26:11.120" CommentCount="1" />
  <row Id="7743" PostTypeId="1" AcceptedAnswerId="8203" CreationDate="2010-10-25T00:07:12.920" Score="19" ViewCount="872" Body="&lt;p&gt;So, I don't like proofs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To me building a proof feels like constructing a steel trap out of arguments to &lt;strong&gt;make true&lt;/strong&gt; what you're trying to assert.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Oftentimes the proof in the book is something that I get if I study, but hard to come up with on my own.  In other words I can't make steel traps, but I feel fine buying them from others.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How does one acquire the ability to create steel traps with fluency and ease?  Are there any particular reference books that you found helped you really get how to construct a proof fluently?  Or is it just practice?&lt;/p&gt;&#xA;" OwnerUserId="1069" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-10-28T22:55:12.823" LastActivityDate="2010-10-28T23:25:21.243" Title="Getting better at proofs" Tags="&lt;soft-question&gt;&lt;proof-strategy&gt;" AnswerCount="5" CommentCount="7" FavoriteCount="11" />
  <row Id="7744" PostTypeId="2" ParentId="7731" CreationDate="2010-10-25T00:10:05.193" Score="0" ViewCount="" Body="&lt;p&gt;That seems clear.  I guess it was the switch from n to m and modulo that I was confused about.  Thank you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(By the way, I meant for this to be a comment, not an answer)&lt;/p&gt;&#xA;" LastActivityDate="2010-10-25T00:20:48.570" />
  <row Id="7745" PostTypeId="1" AcceptedAnswerId="7749" CreationDate="2010-10-25T00:28:24.627" Score="2" ViewCount="192" Body="&lt;p&gt;Suppose $B_t$ is a Brownian motion. As I understand, $B_2-B_1$ is independent of $B_3-B_2$ from properties of Brownian motion. Does it also mean that $B_1$ and $B_2$ are also independent? Can I use this independence to find the joint density of $B_1+B_2+B_3$ as each Brownian process is a normal process of mean 0 and variance t, it should be trivial.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've another related question. To find the expectation over a Brownian process, can I integrate my stochastic process over the normal density function for Brownian motion (mean 0 and variance t)? I hope this makes sense.&lt;/p&gt;&#xA;" OwnerUserId="957" LastEditorUserId="1195" LastEditorDisplayName="" LastEditDate="2010-10-25T01:31:31.923" LastActivityDate="2010-10-28T16:07:06.293" Title="combination of brownian motion" Tags="&lt;probability&gt;&lt;probability-theory&gt;&lt;measure-theory&gt;&lt;stochastic-processes&gt;" AnswerCount="3" />
  <row Id="7746" PostTypeId="2" ParentId="7743" CreationDate="2010-10-25T00:32:03.377" Score="28" ViewCount="" Body="&lt;p&gt;You get better at proofs the same way you get better at basketball or carpentry: lots and lots of practice.  (In particular, like in basketball and carpentry, you can only get so far by reading books.)  Of course, there's good practice and bad practice.  For general experience writing and coming up with proofs I think the same kind of material that people use to prepare for Olympiad-style mathematics is very helpful.  To that end, here are a few references you might find helpful:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0387982191&quot; rel=&quot;nofollow&quot;&gt;Problem-Solving Strategies&lt;/a&gt;, Engel (somewhat high-level)&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/How_to_Solve_It&quot; rel=&quot;nofollow&quot;&gt;How to Solve It&lt;/a&gt;, Polya&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0471135712&quot; rel=&quot;nofollow&quot;&gt;The Art and Craft of Problem Solving&lt;/a&gt;, Zeitz&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;For writing proofs specific to a course (e.g. real analysis) you'll generally find that the same basic ideas are being used over and over again, and once you learn to recognize when these ideas will be useful your life will be much easier.  But I think this is a hard skill to teach.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It doesn't help that many proofs in textbooks are written in a style that makes it nearly impossible to see how someone could have come up with the proof from first principles.  This is an unfortunate tendency, and you should find a different textbook if this is too much of an issue.  (Alternately, you should see if there are better proofs online, for example on someone's blog.  &lt;a href=&quot;http://gowers.wordpress.com/&quot; rel=&quot;nofollow&quot;&gt;Tim Gowers&lt;/a&gt; and &lt;a href=&quot;http://terrytao.wordpress.com/&quot; rel=&quot;nofollow&quot;&gt;Terence Tao&lt;/a&gt; are fond of writing up conceptual proofs of things, and more generally their blogs are a great source of insight into how mathematicians think.)  Once you build up a little problem-solving skill, another way to fix this is to reprove things yourself.  It helps if you can't remember what the proof in the textbook is.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I'm serious about the practice.  $10000$ hours and all that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt;  I also find this &quot;steel trap&quot; analogy very depressing.  While it might seem that way in some textbooks, a properly presented proof should feel more like a poem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit #2:&lt;/strong&gt;  I'd also like to mention that the general outline of Polya's advice is &lt;a href=&quot;http://en.wikipedia.org/wiki/How_to_Solve_It#Four_principles&quot; rel=&quot;nofollow&quot;&gt;on Wikipedia&lt;/a&gt;.  &lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-10-28T23:25:21.243" LastActivityDate="2010-10-28T23:25:21.243" />
  <row Id="7747" PostTypeId="2" ParentId="7737" CreationDate="2010-10-25T00:32:21.473" Score="3" ViewCount="" Body="&lt;p&gt;Note that for $x\in(0,1]$ and $n\geq 1$, we have:&#xA;$$\frac{n\sin(x)}{1+n^2\sqrt{x}} \leq \frac{n}{1+n^2\sqrt{x}} \leq \frac{n}{n^2\sqrt{x}} \leq \frac{1}{n\sqrt{x}} \leq \frac{1}{\sqrt{x}}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, if $g(x) = \frac{1}{\sqrt{x}}$ for $0\lt x\leq 1$ and $g(x)=0$ if $x=0$, then $0\leq f_n(x)\leq g(x)$ almost everywhere in $[0,1]$, where $f_n(x) = \frac{n\sin(x)}{1+n^2\sqrt{x}}$. Since&#xA;$$\int g(x)\,d\mu = \int_{[0,1]}\frac{1}{\sqrt{x}}\,d\mu= 2\sqrt{x}\Biggm|_0^1 = 2\lt \infty$$&#xA;by Lebesgue's Dominated Convergence Theorem you know that if $f(x)$ equals the pointwise limit of the $f_n(x)$ almost everywhere on $[0,1]$, then &#xA;$$\lim_{n\to\infty}\int_{[0,1]}\frac{n\sin(x)}{1+n^2\sqrt{x}}\,d\mu = \int_{[0,1]}f(x)\,d\mu.$$&#xA;So to use Dominated Convergence you now need to figure out an $f(x)$. &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-10-25T00:50:01.333" LastActivityDate="2010-10-25T00:50:01.333" CommentCount="1" />
  <row Id="7748" PostTypeId="2" ParentId="7708" CreationDate="2010-10-25T00:35:03.980" Score="2" ViewCount="" Body="&lt;p&gt;I would note that the situation is similar to matrix multiplication: since multiplication here is noncommutative, you have to distinguish between &quot;premultiplying&quot; with the inverse $p^{-1}q$ and &quot;postmultiplying&quot; with the inverse $qp^{-1}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note also that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(q\cdot r)^{-1}=r^{-1}\cdot q^{-1}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only other thing I can say is that what you're essentially doing is the quaternion analog of a &quot;similarity transformation&quot; of a matrix.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-10-25T00:35:03.980" />
  <row Id="7749" PostTypeId="2" ParentId="7745" CreationDate="2010-10-25T01:02:09.157" Score="3" ViewCount="" Body="&lt;p&gt;$B_1$ and $B_2$ are not independent. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $B_1$ and $B_2-B_1$ are independent,&#xA;$$0=\mathrm{Cov}(B_1,B_2-B_1)=E[B_1 (B_2-B_1)]=E[B_1 B_2]-E[B_1^2]=E[B_1 B_2] - t_1 \sigma^2$$&#xA;So,&#xA;$$\mathrm{Cov}(B_1,B_2)=E[B_1 B_2]=t_1 \sigma^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can use this result and the fact that linear combinations of normal variables are normal to calculate the distribution of $B_1+B_2+B_3$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure what exactly you meant by &quot;expectation over a Brownian process&quot;. Can you please give us a little background of the problem you are trying to solve? Your method of taking an expectation with respect to $N(0,\sigma\sqrt{t})$ is fine when you want the expected value of some function of $B_t$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is also a theory of &lt;a href=&quot;http://en.wikipedia.org/wiki/It%C5%8D_calculus&quot; rel=&quot;nofollow&quot;&gt;stochastic integration&lt;/a&gt; &lt;em&gt;with respect to&lt;/em&gt; stochastic processes which might be of use to you.&lt;/p&gt;&#xA;" OwnerUserId="1195" LastEditorUserId="1195" LastEditorDisplayName="" LastEditDate="2010-10-25T01:09:49.757" LastActivityDate="2010-10-25T01:09:49.757" CommentCount="3" />
  <row Id="7750" PostTypeId="2" ParentId="7672" CreationDate="2010-10-25T01:03:23.510" Score="8" ViewCount="" Body="&lt;p&gt;Well your notation is a little bit confusing. Since we are considering free abelian groups, we should avoid additive notations for the commutative semigroup S. Let's talk of $(S, \ast )$. If $F(S)$ is the free abelian group with basis $S$, we can assume $S \subseteq F(S)$. So the set of relations is $$R = \{ (a \ast b) -a - b : a,b \in S \}.$$ Then, the Grothendieck completion is the abelian group $$G(S) = F(S)/ \langle R \rangle.$$ If $a \in S$, I'll denote the coset $a + \langle R \rangle$ by $[a]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The proof isn't hard. You'll just need to remember that in free abelian groups the expression in terms of a basis is unique. You can find one proof, with an ugly notation, in Rotman's 'Advanced Modern Algebra'. I'll reproduce below the proof on Magum's book 'An Algebraic Introduction to K-theory'.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose that $[a]=[b]$. Then, $a-b \in \langle R \rangle$. That is, $$a-b = \sum_{i=1}^n n_i((a_i \ast b_i) - a_i - b_i),$$ where $n_i = 1$ or $-1$ and $a_i, b_i \in S$. Bringing terms with negative coefficients to the other side, $$ a + \sum_{n_i = -1}(a_i \ast b_i) + \sum_{n_i = 1}(a_i + b_i) = b + \sum_{n_i = 1}(a_i \ast b_i) + \sum_{n_i = -1}(a_i + b_i).$$ Since $S$ is a basis of $F(S)$, the terms on one side of the equation are a permutation of those on the other side. Since $(S, \ast )$ is commutative, it follow that, in S, $$ a \ast \prod_{n_i = -1}(a_i \ast b_i) \ast \prod_{n_i = 1}(a_i \ast b_i) = b \ast \prod_{n_i = 1}(a_i \ast b_i) \ast \prod_{n_i = -1}(a_i \ast b_i). $$ So $a \ast c = b \ast c$, where $$ c = \prod_{i = 1}^n (a_i \ast b_i).$$&lt;/p&gt;&#xA;" OwnerUserId="329" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2010-10-25T01:30:25.030" LastActivityDate="2010-10-25T01:30:25.030" CommentCount="2" />
  <row Id="7751" PostTypeId="2" ParentId="7743" CreationDate="2010-10-25T01:05:59.813" Score="3" ViewCount="" Body="&lt;p&gt;As Arturo and Qiaochu have pointed out you get better at proving things by practicing a lot, solving exercises and by seeing how other people have proved things. There's a big part in learning how to prove statements by repeating certain strategies that may have worked for particular types of problems. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, a particular argument that is used countless times in analysis is the &quot;technique&quot; of adding and substracting a quantity in a way that let's you group things in a useful way. And there are lots and lots of tricks such as this one that you will learn during the way by seeing them being used several times (and also by using them yourself).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But there's certainly something about proving things that I must say. I don't know how common this is for people studying at &quot;top universities&quot; around the world were only really talented and smart people are admitted, but in my country it is relatively easy to get accepted in a mathematics program. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a consequence, I've seen lots of people fail several times in the first mathematics course in my university for mathematics students (where actually students are introduced to proof techniques and basic logic and set theory, basically what in some American universities is referred to as a transition course from the usual calculus courses to proof oriented courses). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The main reason for most of them failing such a course from what I've seen, is that some of them just don't seem to get used to the idea of having to prove something rigorously (as a mathematician will expect). I remember very well when one my classmates argued with the professor because he was asked in the test to prove an equality of sets, and my classmate draw a Venn diagram which showed the sets in question. The professor told him repeatedly that as an aid for intuition the drawing was perfectly fine, but that the drawing didn't constitute a proof by itself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the end (as you'll imagine) my classmate didn't win the argument (or extra points for that matter) and he ended up going home frustrated and finally he gave up on the course. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My point is that for some reason some people are just better than others at proving things, and I'm not saying that they're just smarter, it's only that the thought processes involved just seem to come more naturally to some persons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That being said, there are some books that specifically address the topic of introducing students to the task of proving things. What they usually do is to begin by explaining some basic logic and them they build up some easy facts from set theory, binary relations and functions and in the way they introduce some proof techniques, such as &quot;proof by contradiction&quot;, &quot;proof by contrapositive&quot;, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, one book that helped me a lot when I was starting is &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0817641114&quot; rel=&quot;nofollow&quot;&gt;Proofs and Fundamentals: A First Course in Abstract Mathematics&lt;/a&gt;, but there a lot of other books that I'm pretty sure work really well when making your first steps in proving mathematical statements, such as &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0471680583&quot; rel=&quot;nofollow&quot;&gt;How To Read and Do Proofs&lt;/a&gt; by Daniel Solow or &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0521675995&quot; rel=&quot;nofollow&quot;&gt;How To Prove It&lt;/a&gt; by Daniel Velleman.&lt;/p&gt;&#xA;" OwnerUserId="900" LastActivityDate="2010-10-25T01:05:59.813" />
  <row Id="7752" PostTypeId="2" ParentId="7745" CreationDate="2010-10-25T01:06:21.477" Score="1" ViewCount="" Body="&lt;p&gt;Expressing $B_1+B_2+B_3=3B_1+2(B_2-B_1)+(B_3-B_2)$ as a sum of independent normal random variables will help you find its density. I'm not sure why you call it a joint density.&lt;/p&gt;&#xA;" OwnerUserId="940" LastActivityDate="2010-10-25T01:06:21.477" />
  <row Id="7753" PostTypeId="1" CreationDate="2010-10-25T01:20:57.087" Score="1" ViewCount="54" Body="&lt;p&gt;The usual action of $fg$  on $u⊗v$, where $f,g$  are elements in the Universal Enveloping Algebra $U(G)$ of a Lie algebra $G$ and $u,v$ are elements of a representation $V$  of $G$, is given by $fg(u⊗v)=fgu⊗v+fu⊗gv+gu⊗fv+u⊗fgv$,  right? How to state this fact for $V^{\otimes n}$, i.e. $fg$  acting on $u⊗v$, where $u=⊗_{i=1}^{n-k} u_i$ and $v=\otimes_{i=1}^k v_i$, for each $k=1,...,n−1$? &#xA;Thanks,&lt;/p&gt;&#xA;" OwnerUserId="2764" LastActivityDate="2010-10-25T03:26:05.367" Title="How to define the action of U(G) in this situation?" Tags="&lt;universal-algebra&gt;&lt;representation-theory&gt;" AnswerCount="1" />
  <row Id="7754" PostTypeId="1" CreationDate="2010-10-25T02:12:35.540" Score="0" ViewCount="127" Body="&lt;p&gt;we have a truck in our game we have a rod against the wall of that truck, and we know the acceleration of truck when the rod start slipping... we just want to know what is acceleration of rod when it touch surface of truck.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/l2wJa.jpg&quot; alt=&quot;diagram for problem&quot;&gt;&lt;/p&gt;&#xA;" LastEditorUserId="723" LastEditorDisplayName="" LastEditDate="2010-10-27T05:03:11.427" LastActivityDate="2010-10-27T05:03:11.427" Title="slipping rod on moving truck." Tags="&lt;physics&gt;&lt;classical-mechanics&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="7755" PostTypeId="2" ParentId="7754" CreationDate="2010-10-25T02:35:50.510" Score="1" ViewCount="" Body="&lt;p&gt;To make the question more precise: there is a max static friction coefficient at points A and B.  Calculating the perpendicular force at A and B is not hard, so the acceleration where it breaks loose is known.  The acceleration increases exceeding this.  So the stick starts to slide aft.  Now you have a sliding friction, which would usually be modeled as a (lower) coefficient of the force on the surface.  You want an equation for the height of A as a function of time. As the acceleration depends upon user input, it seems you want something for short timesteps.  If we could give you delta A=f(acceleration,A,dA/dt previous step), would this meet your needs?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the rod hits A=0.  This is hard.  One approach would be that it doesn't bounce, but has a certain (different?) sliding friction and continues to move aft.  Another would be that it bounces with a certain recovery fraction, so dA/dt is -(some number times the previous) dA/dt, but there is no more friction at point A.  The sliding friction at B continues unchanged.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It might even lose contact at A before hitting the deck.  This shouldn't be too hard to detect.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2010-10-25T02:49:39.057" LastActivityDate="2010-10-25T02:49:39.057" CommentCount="3" />
  <row Id="7756" PostTypeId="2" ParentId="7743" CreationDate="2010-10-25T02:51:06.327" Score="1" ViewCount="" Body="&lt;p&gt;I recommend the book by Lay, &quot;Analysis with introduction to proofs&quot;. The first chapter focuses on logic and proofs and personally I found it rather helpful. At least it helped me. Of course to master the art of proofs you have to keep practicing, but knowledge of some basics such that instead of proving the statement &quot;A implies B you can prove contrapositive &quot;not B implies not A&quot;, and how it differs from &quot;Proof by contradiction&quot;, and the fact that negation of &quot;A implies B&quot; is &quot;A and not B&quot;, and how to deal with existential and universal quantifiers, and so forth and so forth. Those are useful techniques one should master. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Incidentally, as an introduction to analysis, the aforementioned book is quite mediocre. I had a first hand experience with that book, since it was a textbook when I took the &quot;low level introduction to analysis (wasn't sure if I was ready for Rudin's Principles of Analysis). So, this is only the chapter on logic and proof which is worth reading. Of course, purchasing the whole book just for a single chapter is funny, but may be you can find the used one or get one in the library. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But that chapter on proofs did help me. When subsequently I took a course on analysis and we used Baby Rudin as a textbook, I was in a very good shape and in fact did quite well. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And, of course, another suggestion: learn by example. See how other people write the proofs. Rudin's style, for instance, is very terse, you literally have to dig through his proofs, but after a while you start getting used to it, and still after a while you start writing similar proofs. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Good luck.&lt;/p&gt;&#xA;" OwnerUserId="2450" LastActivityDate="2010-10-25T02:51:06.327" />
  <row Id="7757" PostTypeId="1" AcceptedAnswerId="7762" CreationDate="2010-10-25T02:53:55.737" Score="4" ViewCount="733" Body="&lt;p&gt;I am trying to prove this binomial identity $\sum_{r=0}^n {r {n \choose r}} = n2^{n-1}$ but am not able to think something except induction,which is of-course not necessary (I think) here, so I am inquisitive to prove this in a more general way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The left side of an identity occurs while solving another problem (concerning binomial theorem) so I am more interested in deriving the right side from the left side, else I have to remember it now onward.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt; I am more interested in an algebraic proof rather than combinatorial argument or something involving calculus (however I liked &lt;a href=&quot;http://math.stackexchange.com/users/2641/svenkatr&quot;&gt;svenkatr&lt;/a&gt; and &lt;a href=&quot;http://math.stackexchange.com/users/242/bill-dubuque&quot;&gt;Bill Dubuque&lt;/a&gt; soution),hence I am removing the combinatorics tag.&lt;/p&gt;&#xA;" OwnerUserId="2109" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-10-25T14:45:45.310" LastActivityDate="2010-12-06T18:07:50.067" Title="How to prove this binomial identity $\sum_{r=0}^n {r {n \choose r}} = n2^{n-1}$ ?" Tags="&lt;algebra-precalculus&gt;&lt;binomial-coefficients&gt;" AnswerCount="10" CommentCount="1" FavoriteCount="1" />
  <row Id="7758" PostTypeId="2" ParentId="7717" CreationDate="2010-10-25T02:56:15.230" Score="0" ViewCount="" Body="&lt;p&gt;Add &quot;lemmas&quot;, &quot;conjectures&quot; and &quot;principles&quot; to your question  and you have the full gamut!&lt;/p&gt;&#xA;" OwnerUserId="2450" LastActivityDate="2010-10-25T02:56:15.230" CommentCount="1" />
  <row Id="7761" PostTypeId="2" ParentId="7757" CreationDate="2010-10-25T03:01:05.980" Score="13" ViewCount="" Body="&lt;p&gt;Here's a hint:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$r\binom{n}{r} = n\binom{n-1}{r-1}, \quad r\geq 1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here be spoilers:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;\sum_{r=0}^{n} r\binom{n}{r}&#xA;&amp;amp;= \sum_{r=1}^{n} r\binom{n}{r} &amp;amp; &amp;amp;\text{drop the zero term}&#xA;\\&#xA;&amp;amp;= \sum_{r=1}^{n} n\binom{n-1}{r-1} &amp;amp; &amp;amp;\text{apply the above identity}&#xA;\\&#xA;&amp;amp;= n \sum_{k=0}^{n-1} \binom{n-1}{k} &amp;amp; &amp;amp;\text{replace index $r$ with $k=r-1$} &#xA;\\&#xA;&amp;amp;= n 2^{n-1} &amp;amp; &amp;amp;\text{the sum is just the total number of subsets of $n-1$ elements} &#xA;\end{align}&lt;/p&gt;&#xA;" OwnerUserId="1023" LastEditorUserId="1023" LastEditorDisplayName="" LastEditDate="2010-10-26T06:05:05.743" LastActivityDate="2010-10-26T06:05:05.743" CommentCount="3" />
  <row Id="7762" PostTypeId="2" ParentId="7757" CreationDate="2010-10-25T03:02:53.513" Score="19" ViewCount="" Body="&lt;p&gt;Take the binomial expansion of $(1+x)^n$. Differentiate both sides with respect to $x$, then substitute $x=1$ and that will give you the identity.&lt;/p&gt;&#xA;" OwnerUserId="2641" LastActivityDate="2010-10-25T03:02:53.513" CommentCount="6" />
  <row Id="7763" PostTypeId="2" ParentId="7757" CreationDate="2010-10-25T03:04:01.563" Score="3" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\ $ Differentiate $\rm (1+x)^n\:$, use the binomial theorem, then set $\rm\ x = 1\:$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;NOTE&lt;/strong&gt; $\ $ Using derivatives, we can pull out of a sum any polynomial function of the index variable, namely&lt;/p&gt;&#xA;&#xA;&lt;p&gt;since we have $\rm\:\ k^i\ x^k\ =\ (xD)^i \ x^k\ \ $ for $\rm\ \ D = \frac{d}{dx},\ \ k &gt; 0\ $  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;it follows that $\rm\ \sum a_k\: f(k)\: x^k\ =\ f(x\:D) \sum a_k\: x^k\ \ $&#xA; for polynomial $\rm\:f\:$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &quot;trick&quot; of representing a numerical series as the value f(1) of a generating function f(x) goes back at least to Euler, who employed it to sum divergent series (among other applications). The power arises from the fact that at the function level one has available much more powerful tools, esp. derivatives, e.g. above. Such techniques come in quite handy when shifting between differential and difference (recurrence) viewpoints, e.g. for hypergeometric functions and their special values.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-10-25T13:07:53.400" LastActivityDate="2010-10-25T13:07:53.400" />
  <row Id="7764" PostTypeId="2" ParentId="7757" CreationDate="2010-10-25T03:07:29.823" Score="1" ViewCount="" Body="&lt;p&gt;We have $x^{n} = (1+(x-1))^{n}$ and $$[1+(x-1)]^{n} = {n \choose 0 } + {n \choose 1} \cdot (x-1) + {n \choose 2} \cdot (x-1)^{2} + \cdots + {n \choose n} \cdot (x-1)^{n}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Differentiate on both sides w.r.t x and substitute the value $x=2$ in your equation.&lt;/p&gt;&#xA;" LastActivityDate="2010-10-25T03:07:29.823" />
  <row Id="7765" PostTypeId="2" ParentId="7753" CreationDate="2010-10-25T03:26:05.367" Score="1" ViewCount="" Body="&lt;p&gt;If $V_1,V_2\dots,V_n$ are $U(\mathfrak g)$-modules, then the $U(\mathfrak g)$-module structure on the tensor product $V_1\otimes V_2\otimes\cdots\otimes V_n$ is the unique one such that whenever $x\in\mathfrak g\subset U(\mathfrak g)$ we have $$\begin{aligned}g\cdot v_1\otimes\cdots\otimes v_n&amp;amp;=(x\cdot v_1)\otimes v_2\otimes\cdots\otimes v_n\\&amp;amp;+v_1\otimes(x\cdot v_2)\otimes v_3\otimes\cdots\otimes v_n\\{}&amp;amp;+v_1\otimes v_2\otimes(x\cdot v_3)\otimes v_4\otimes\cdots\otimes v_n\\{}&amp;amp;+\cdots\\{}&amp;amp;+v_1\otimes v_2\otimes\cdots\otimes v_{n-1}\otimes(x\cdot v_n).\end{aligned}$$ Notice that this formula is only valid for elements of $\mathfrak g$, not all elements of $U(\mathfrak g)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to know how an arbitrary element of $U(\mathfrak g)$ acts on the tensor product, write it as a sum of products of elements of $\mathfrak g$ and make use of the relation $$xy\cdot v = x\cdot(y\cdot v)$$ valid for all $x$, $y\in U(\mathfrak g)$ and all $v\in V_1\otimes V_2\otimes\cdots\otimes V_n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You should really pick up a book on Hopf algebras. My favorite is Sweedler's &lt;em&gt;Hopf Algebras&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2010-10-25T03:26:05.367" />
  <row Id="7766" PostTypeId="2" ParentId="7743" CreationDate="2010-10-25T03:36:52.323" Score="3" ViewCount="" Body="&lt;p&gt;It seems you are asking two questions.  The first is how to get better at doing proofs, and the previous answers are better than I can do.  The second is why to do them, which has not been addressed.  From your questions on this site, you seem more an engineer than a mathematician.  This is not a negative-I am educated as a physicist and practicing as an engineer(ing manager), but I enjoy proof-type math.  It just is a different view.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In response to your question &quot;How can I characterize the type of solution vector that comes out of a matrix?&quot; Greg Graviton not only gave an answer of what matrices met your criterion, he proved that matrices that did not met his answer did not work.  This is a major improvement as you know there aren't others out there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proofs are designed to make sure you covered all the loose ends.  Ideally, each statement in a proof would be derived from the previous statements through an accepted rule of logic.  But aside from the automated theorem provers, this is unachievable because it would make the proofs too long.  So you only need to display enough steps to convince the audience it can be done.  How big a leap is acceptable varies with the audience.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As physicists/engineers our numbers and functions are better behaved than the ones the mathematicians worry about.  The mathematical world is not uniformly continuous, but ours is.  As such, we interchange limits, integrals, derivatives, and sums freely.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-10-25T03:36:52.323" />
  <row Id="7767" PostTypeId="2" ParentId="7757" CreationDate="2010-10-25T04:43:15.057" Score="6" ViewCount="" Body="&lt;p&gt;I can't resist giving the standard bijective proof of this fact. Let $[n]$ be the set of integers from $1$ to $n$. Then $r {n \choose r}$ is the number of elements in all the subsets of $[n]$ which have $r$ elements. So your sum is the number of elements in all the subsets of $[n]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But we can pair the sets up into $2^{n-1}$ pairs of the form $S, [n] \setminus S$. Each pair contains $n$ elements, so all the sets together contain $n 2^{n-1}$ elements.&lt;/p&gt;&#xA;" OwnerUserId="173" LastActivityDate="2010-10-25T04:43:15.057" />
  <row Id="7768" PostTypeId="2" ParentId="7757" CreationDate="2010-10-25T04:45:37.833" Score="16" ViewCount="" Body="&lt;p&gt;Use &lt;a href=&quot;http://www.jimloy.com/algebra/gauss.htm&quot; rel=&quot;nofollow&quot;&gt;Gauss' trick&lt;/a&gt; for summing a sequence: Combine the sum term-by-term with its reversed-order self, noting that the binomial coefficients are symmetric.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$S = \sum_{r=0}^n r { n \choose r } = \sum_{r=0}^n (n-r) {n \choose n-r }$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{eqnarray}&#xA;2 S &amp;amp;=&amp;amp; \sum_{r=0}^n \left( r {n\choose r} + (n-r) {n \choose n-r} \right)\\&#xA; &amp;amp;=&amp;amp; \sum_{r=0}^n \left( r {n\choose r} + (n-r) {n \choose r} \right)\\&#xA;&amp;amp;=&amp;amp; \sum_{r=0}^n \left( n {n\choose r} \right)\\&#xA;&amp;amp;=&amp;amp; n \sum_{r=0}^n {n\choose r}\\&#xA;&amp;amp;=&amp;amp; n 2^n \hspace{0.1in} \text{, by familiar identity}&#xA;\end{eqnarray}$$&lt;/p&gt;&#xA;" OwnerUserId="409" LastActivityDate="2010-10-25T04:45:37.833" CommentCount="1" />
  <row Id="7769" PostTypeId="1" CreationDate="2010-10-25T05:08:08.093" Score="0" ViewCount="233" Body="&lt;p&gt;How can I find the critical points of a square root function?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example: What are the critical points of $f(x) = \sqrt{7x^2 + 2x - 2}$?&lt;/p&gt;&#xA;" OwnerUserId="2751" LastEditorUserId="51" LastEditorDisplayName="" LastEditDate="2010-10-25T05:59:24.223" LastActivityDate="2010-10-25T14:33:25.540" Title="Critical Points of a Square Root Function" Tags="&lt;calculus&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="7770" PostTypeId="2" ParentId="7589" CreationDate="2010-10-25T05:35:48.277" Score="2" ViewCount="" Body="&lt;p&gt;I'm setting this answer community wiki so other people can attach their own examples of curves constructed with the methods in this thread.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, Qiaochu's suggestion for building a curve $q(y)=p(x)$ with collinear singular points on the horizontal axis amounts to the construction of an appropriate &lt;a href=&quot;http://mathworld.wolfram.com/HermitesInterpolatingPolynomial.html&quot; rel=&quot;nofollow&quot;&gt;Hermite interpolation&lt;/a&gt; problem. More explicitly, one wants to find a polynomial (or rational function) whose first few derivatives at preset points vanish.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In &lt;em&gt;Mathematica&lt;/em&gt; for instance, the function &lt;code&gt;InterpolatingPolynomial&lt;/code&gt; can be used to generate a Hermite interpolant. (For systems that do not have such a function handy, the Hermite interpolation problem is solved through either an appropriate &lt;a href=&quot;http://books.google.com/books?id=Mp8-z5mHptcC&amp;amp;pg=PA189&quot; rel=&quot;nofollow&quot;&gt;modification of the Newton divided differences scheme&lt;/a&gt; or by solving an associated &lt;a href=&quot;http://dx.doi.org/10.1007/BF01436299&quot; rel=&quot;nofollow&quot;&gt;confluent Vandermonde system&lt;/a&gt;.) The rational interpolant case is a bit tougher, and I am still experimenting with algorithms for the rational Hermite problem so I won't be considering them for now (but might include them in a later edit).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, taking the conditions in Qiaochu's answer, here is how one builds a curve with a node at $(-1,0)$, a tacnode at $(0,0)$, and a cusp at $(1,0)$:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Expand[InterpolatingPolynomial[{{-1, {0, 0, 1}}, {0, {0, 0, 0, 0, 1}}, {1, {0, 0, 0, 1}}}, x]]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The result has fractional coefficients, but you can multiply with an appropriate factor so that all the coefficients are integers, resulting in the polynomial&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$3x^{11}+2x^{10}-15x^9+21x^7-6x^6-9x^5+4x^4$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here for instance is a plot of $$y^2=3x^{11}+2x^{10}-15x^9+21x^7-6x^6-9x^5+4x^4$$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/n0bhG.png&quot; alt=&quot;monster #1&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and a more complicated curve, $$y^2-y^3=(x^2+xy+y^2)(3x^{11}+2x^{10}-15x^9+21x^7-6x^6-9x^5+4x^4)$$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://imgur.com/hBmL8.png&quot; alt=&quot;monster #2&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have yet to make the prescription for getting an isolated point to work, since the curves generated only manage to have separate branches passing through the desired point, but no isolated points at all.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;As for T..'s suggestion, the &lt;em&gt;Mathematica&lt;/em&gt; code I have requires some serious cleanup, so I shall be editing this answer later to include his parametric construction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Currently I am trying to find a curve with four-fold symmetry that has four of each of the types of double points. If I manage to find it, I shall be naming it after Qiaochu and T.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-10-25T05:35:48.277" CommunityOwnedDate="2010-10-25T05:35:48.277" CommentCount="1" />
  <row Id="7771" PostTypeId="2" ParentId="7769" CreationDate="2010-10-25T05:51:52.290" Score="6" ViewCount="" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Take the derivative. This is straightforward here using the chain rule:&#xA;$$\frac{14x + 2}{2\sqrt{7x^2 + 2x - 2}} = \frac{7x+1}{\sqrt{7x^2 + 2x - 2}}.$$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Find the points in the domain of $f(x)$ where the derivative does not exist or is equal to zero.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint the first&lt;/strong&gt;: There are two points that are in the domain of $f(x)$ where the derivative does not exist.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint the second&lt;/strong&gt;: A fraction is equal to zero if and only if the numerator is $0$ and the denominator is not zero.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-10-25T14:33:25.540" LastActivityDate="2010-10-25T14:33:25.540" />
  <row Id="7772" PostTypeId="1" AcceptedAnswerId="7784" CreationDate="2010-10-25T06:33:36.910" Score="-2" ViewCount="103" Body="&lt;p&gt;A friend of mine who is quite an aggressive Nominalist told me the other day:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Mathematics and numbers are arbitrary; they can accurately predict physical systems in real life only because they are consistent, but any consistent framework will do.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm curious as to what other examples of &quot;consistent frameworks&quot; he is talking about.  Perhaps there are alternative definitions of addition?  Of subtraction, or multiplication?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Different things you can do to numbers that have a &quot;mapping&quot; (isomorphic) to real-world, physical phenomenon?  (Such as mathematical addition isomorphically mapping to, say, putting together two groups of apples, or differentiation of position mapping to velocity)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there any examples of this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There might not be; this is somewhat of a vague question and it might end up being very stupid.  But I'm interested in what other completely different frameworks of &quot;math&quot; might exist that are meaningful in describing/modeling physical events.&lt;/p&gt;&#xA;" OwnerUserId="51" LastEditorUserId="51" LastEditorDisplayName="" LastEditDate="2010-10-25T08:24:18.020" LastActivityDate="2010-10-25T08:58:21.333" Title="Alternative, consistent frameworks of mathematics with isomorphic mappings to physical phenomenon" Tags="&lt;abstract-algebra&gt;&lt;math-history&gt;&lt;philosophy&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="7773" PostTypeId="2" ParentId="7757" CreationDate="2010-10-25T06:35:08.780" Score="5" ViewCount="" Body="&lt;p&gt;Another way is to consider the following arrangement with $n$ rows and $n+1$ columns&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\begin{pmatrix}&#xA;{n \choose 0} &amp;amp; {n \choose 1} &amp;amp; \dots &amp;amp; {n \choose n} \\&#xA;{n \choose 0} &amp;amp; {n \choose 1} &amp;amp; \dots &amp;amp; {n \choose n} \\&#xA;\vdots &amp;amp; \vdots &amp;amp; \dots &amp;amp; \vdots \\&#xA;{n \choose 0} &amp;amp; {n \choose 1} &amp;amp; \dots &amp;amp; {n \choose n} \\&#xA;\end{pmatrix}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What we want is the sum of the last $n$ elements of row 1, plus the sum of last $n-1$ elements of row 2 etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which is same as the sum of the first $n$ elements of row $n$, plus the sum of the first $n-1$ elements of row $n-1$ etc, as ${n \choose r} = {n \choose n-r}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus the required sum is half the sum of all elements, which is $n2^{n-1}$, as the sum of each row is $2^n$ and there are $n$ rows.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-10-25T06:35:08.780" />
  <row Id="7774" PostTypeId="1" AcceptedAnswerId="22442" CreationDate="2010-10-25T06:40:06.330" Score="8" ViewCount="413" Body="&lt;p&gt;Exercise 1.14 of &lt;a href=&quot;http://books.google.ru/books?id=SMiB8VIB5UIC&amp;amp;printsec=frontcover&amp;amp;dq=Rordam+Larsen+Laustsen&amp;amp;source=bl&amp;amp;ots=o_tWQGf66z&amp;amp;sig=IfsFtbHbCz3d-wZc3q7O2ZXQulo&amp;amp;hl=en&amp;amp;ei=CyDFTPPlLMPrOceP0e8L&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=1&amp;amp;ved=0CBIQ6AEwAA#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;the book&lt;/a&gt; Rordam, Larsen and Laustsen &quot;An introduction to K-theory for C*-algebras&quot; asks to prove, that upper triangular matrix with elements from some C*-algebra $A$ is invertible in $M_n(A)$ iff all diagonal entries are invertible in $A$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Trying to solve this I've found that if $a$ is invertible and $\delta$ is such that $(a^{-1}\delta)^n=0$ then $a+\delta$ is invertible too and its inverse is given by $(a+\delta)^{-1}=\sum_{k=0}^{n} (-a^{-1}\delta)^ka^{-1}$. Using this fact I can show that if diagonal is invertible, then upper-triangular matrix with this diagonal is invertible too, and also that if upper-triangular matrix has upper-triangular inverse, then its diagonal is invertible. So all I need to prove is that if upper-triangular matrix invertible, then its inverse is upper-triangular. I've failed to prove this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also there is a hint for this exercise: &quot;Solve the equation $ab=1$ where $a$ is as above [i.e. upper-triangular matrix] and where $b$ is unknown upper triangular matrix&quot;. Solution of this equation follows from my reasoning above, but this doesn't help.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update (counterexample attempt):&lt;/strong&gt; I've made one more attempt and it looks for me like I have found a counterexample. However I think there is a mistake in it (because otherwise there is a mistake in &lt;a href=&quot;http://books.google.ru/books?id=SMiB8VIB5UIC&amp;amp;printsec=frontcover&amp;amp;dq=Rordam+Larsen+Laustsen&amp;amp;source=bl&amp;amp;ots=o_tWQGf66z&amp;amp;sig=IfsFtbHbCz3d-wZc3q7O2ZXQulo&amp;amp;hl=en&amp;amp;ei=CyDFTPPlLMPrOceP0e8L&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=1&amp;amp;ved=0CBIQ6AEwAA#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;the book&lt;/a&gt;). Here it is. Let $A=B(l^2(\mathbb{N}))$ --- algebra of bounded operators on sequences $x=\{x_i\}_ {i=1}^ \infty:\|x\|^2=\sum_{x=1}^{\infty}|x_i|^2&amp;lt;\infty$. Let $z\in A$ be defined by $(zx)_ {2n-1}=0$, $(zx)_{2n}=x_n$, and $t\in A$ be defined by $(tx)_{2n-1}=x_n$, $(tx)_ {2n}=0$. Then we have $t^*t=z^ * z=tt^* +zz^* =1$, $t^* z=z^* t=0$. From these we have that&#xA;$$\begin{pmatrix}z&amp;amp;tz^* \\0&amp;amp;t^* \end{pmatrix}\begin{pmatrix}z^* &amp;amp;0\\zt^* &amp;amp;t\end{pmatrix}=\begin{pmatrix}1&amp;amp;0\\0&amp;amp;1\end{pmatrix}$$ and&#xA;$$\begin{pmatrix}z^* &amp;amp;0\\zt^* &amp;amp;t\end{pmatrix}\begin{pmatrix}z&amp;amp;tz^* \\0&amp;amp;t^* \end{pmatrix}=\begin{pmatrix}1&amp;amp;0\\0&amp;amp;1\end{pmatrix}.$$ &#xA;So now my question should say &quot;Where am I wrong?&quot;.&lt;/p&gt;&#xA;" OwnerUserId="695" LastEditorUserId="695" LastEditorDisplayName="" LastEditDate="2010-10-25T09:28:31.007" LastActivityDate="2011-02-17T05:02:44.560" Title="Upper-triangular matrix is invertible iff its diagonal is invertible: C*-algebra case." Tags="&lt;linear-algebra&gt;&lt;abstract-algebra&gt;&lt;algebra&gt;&lt;matrices&gt;&lt;c-star-algebras&gt;" AnswerCount="1" CommentCount="7" FavoriteCount="1" />
  <row Id="7775" PostTypeId="1" AcceptedAnswerId="7803" CreationDate="2010-10-25T06:49:27.110" Score="7" ViewCount="179" Body="&lt;p&gt;Consider the scheme of random placing balls into N=1000 cells. We continue the procedure of placing balls as long as a last cell remains empty. The process terminates when a ball is placed into this cell. At this moment several cells (or a certain cell) contain(s) a maximum number of balls among all cells. What is the expectation of this maximum?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an application of this scheme consider N people who enter a lottery game. Each raffle is equivalent to the random placing of a ball into N cells. We take a look at this process as long as each person has won at least once.&lt;/p&gt;&#xA;" OwnerUserId="2323" LastActivityDate="2010-10-25T16:13:26.320" Title="An occupancy problem" Tags="&lt;probability&gt;" AnswerCount="2" CommentCount="5" />
  <row Id="7776" PostTypeId="1" CreationDate="2010-10-25T06:59:06.740" Score="1" ViewCount="113" Body="&lt;p&gt;I've noticed a type of sum for the number of solutions of a type of quadratic form. For example, the number of solutions to $X^2-Y^2\equiv a\pmod{p}$ for $p$ an odd prime is given by the sum&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{Y=1}^p(1+\left(\frac{Y^2+a}{p}\right))$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\left(\frac{Y^2+a}{p}\right)$ is the Legendre symbol. It seems that the trick is to isolate the $X^2$, add 1 to the RHS, and then let $Y$ range over a complete residue system. It seems that for each value $Y$ takes, there will either be 2 or 0 solutions added to the running total. Is there a simple explanation why this is true?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suspect that for each $y$ that is a solution, we could take both $x$ and $-x$ to get two solutions $(x,y)$ and $(-x,y)$? My only concern is when $x\equiv -x\pmod{p}$, in which case we would have $x\equiv 0\pmod{p}$. Why does the formula still work in that case as well? Thanks for any explanation.&lt;/p&gt;&#xA;" LastActivityDate="2010-10-25T14:42:39.507" Title="Formula for counting solutions to a quadratic form?" Tags="&lt;number-theory&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="7777" PostTypeId="2" ParentId="7776" CreationDate="2010-10-25T07:06:47.200" Score="5" ViewCount="" Body="&lt;p&gt;This is a general (and well-known) trick. The number of solutions to the&#xA;congruence&#xA;$$x^2\equiv f(y)\pmod{p}$$&#xA;for a polynomial $f$ is&#xA;$$\sum_{y=0}^{p-1}\left[1+\left(\frac{f(y)}{p}\right)\right].$$&#xA;The reason for this is that the congruence $x^2\equiv a$ (mod $p$)&#xA;always has $1+(x/p)$ solutions. Note that when $a\equiv0$ (mod $p$)&#xA;the only solution to $x^2\equiv0$ (mod $p$) is $x\equiv0$ (mod $p$)&#xA;but this is in accord with $(0/p)=0$.&lt;/p&gt;&#xA;" OwnerUserId="226" LastEditorUserId="226" LastEditorDisplayName="" LastEditDate="2010-10-25T08:36:40.903" LastActivityDate="2010-10-25T08:36:40.903" CommentCount="2" />
  <row Id="7778" PostTypeId="1" AcceptedAnswerId="7782" CreationDate="2010-10-25T08:25:48.333" Score="3" ViewCount="508" Body="&lt;p&gt;How to prove this  binomial identity  :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ { 2n \choose n } = 2^n \frac{ 1 \cdot 3 \cdot 5 \cdots (2n-1)}{n!} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The left hand side arises while solving a standard binomial problem the right hand side is given in the solution , I checked using induction that this is true but I am inquisitive to prove it in a rather general way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt; I have gone through all of the answers posted here,I particularly liked &lt;a href=&quot;http://math.stackexchange.com/users/72/isaac&quot;&gt;Isaac♦&lt;/a&gt;&#xA;answers after which it was not much difficult for me to figure out something  i would rather say an easy and straight algebraic proof, I am posting it here if somebody needs in future :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ { 2n \choose n } = \frac{(2n)!}{n! \cdot n!} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ = \frac{ 1 \cdot 2 \cdot 3 \cdots n \cdot (n+1) \cdots (2n-1)\cdot (2n) }{n! \cdot n!}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ = \frac{ [1 \cdot 3 \cdot 5 \cdots (2n-1)] \cdot [2 \cdot 4 \cdot 6 \cdots (2n)]}{n! \cdot n!}  $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ = \frac{ [1 \cdot 3 \cdot 5 \cdots (2n-1)] \cdot [(2.1) \cdot (2.2) \cdot (2.3) \cdots (2.n)]}{n! \cdot n!}  $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$= \frac{ [1 \cdot 3 \cdot 5 \cdots (2n-1)] \cdot [ 2^n \cdot (n)! ]}{n! \cdot n!}  $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ = 2^n \frac{ 1 \cdot 3 \cdot 5 \cdots (2n-1)}{n!} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I do always welcome your comments :-)&lt;/p&gt;&#xA;" OwnerUserId="2109" LastEditorUserId="2109" LastEditorDisplayName="" LastEditDate="2010-10-25T19:16:53.647" LastActivityDate="2010-10-26T04:55:39.623" Title="Proving $ { 2n \choose n } = 2^n \frac{ 1 \cdot 3 \cdot 5 \cdots (2n-1)}{n!} $" Tags="&lt;algebra-precalculus&gt;&lt;binomial-coefficients&gt;" AnswerCount="5" FavoriteCount="1" />
  <row Id="7779" PostTypeId="2" ParentId="7778" CreationDate="2010-10-25T08:34:52.840" Score="7" ViewCount="" Body="&lt;p&gt;If you have proved it via induction, the you have &lt;strong&gt;proved&lt;/strong&gt; it.&#xA;I don't know what you mean by &quot;in a rather general way&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But note that&#xA;$$1\cdot3\cdot5\cdot\cdots\cdot(2n-1)$$&#xA;is just crying out to be rewritten as&#xA;$$\frac{1\cdot2\cdot3\cdot\cdots\cdot(2n)}&#xA;{2\cdot4\cdot6\cdot\cdots\cdot(2n)}.$$&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-10-25T08:34:52.840" CommentCount="2" />
  <row Id="7780" PostTypeId="1" AcceptedAnswerId="7785" CreationDate="2010-10-25T08:36:06.483" Score="0" ViewCount="147" Body="&lt;p&gt;i have an object and a 3d direction vector and position for it . I would like to know how do i determine if a certain point X is in the space below the plan determined by my direction ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is an image that i have drawn to make it more clear . In this image i've made the vector 2d &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/bgAYJ.png&quot; alt=&quot;diagram&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please Help!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regards, &#xA;Aleandru Badescu&lt;/p&gt;&#xA;" OwnerUserId="2777" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-10-25T08:41:47.130" LastActivityDate="2010-10-25T09:55:49.553" Title="vector question" Tags="&lt;vector-calculus&gt;" AnswerCount="1" CommentCount="11" FavoriteCount="1" />
  <row Id="7781" PostTypeId="2" ParentId="7778" CreationDate="2010-10-25T08:36:51.850" Score="1" ViewCount="" Body="&lt;p&gt;Multiply right side by $\frac{n!}{n!}$, and distribute the $2^n$.&lt;/p&gt;&#xA;" OwnerUserId="2778" LastActivityDate="2010-10-25T08:36:51.850" />
  <row Id="7782" PostTypeId="2" ParentId="7778" CreationDate="2010-10-25T08:39:09.597" Score="4" ViewCount="" Body="&lt;p&gt;Your original identity, ${ 2n \choose n } = 2^n \frac{ 1 \cdot 3 \cdot 5 \cdots (2n-1)}{n!}$, can be rewritten (by multiplying both sides by $(n!)^2$) as $(2n)!=2^n\cdot 1\cdot 3\cdot 5\cdots (2n-1)\cdot n!$.  Now, $2^n\cdot 1\cdot 3\cdot 5\cdots (2n-1)\cdot n!=$ $1\cdot 3\cdot 5\cdots (2n-1)\cdot 2\cdot 4\cdot 6\cdots 2n=$ $1\cdot 2\cdot 3\cdots (2n-1)(2n)=(2n)!$.&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-10-25T08:39:09.597" CommentCount="1" />
  <row Id="7783" PostTypeId="1" AcceptedAnswerId="7852" CreationDate="2010-10-25T08:53:56.213" Score="10" ViewCount="537" Body="&lt;p&gt;Much effort has been expended on a famous unsolved problem about the [Riemann Zeta function $ \zeta(s) $][1].  Not surprisingly, it&amp;#39;s called the Riemann hypothesis, which asserts:&#xD;&#xA;&#xD;&#xA;$$ \zeta(s) = 0 \Rightarrow \text{Re } s = 1/2 \text{ or Im } s = 0. $$&#xD;&#xA;&#xD;&#xA;Now there are numerical methods for approximating $ \zeta(s) $, but as I understand, no one knows any exact values except at even integers, which include the trivial zeroes for which $ \text{Im } s = 0 $.  So I&amp;#39;ve always wondered:  *For the rest, how does one prove $ \text{Re } s = 1/2 $ &lt;strong&gt;exactly&lt;/strong&gt;?*&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(All I know that seems vaguely useful is the argument principle, which I'm not sure helps here, but I'd be happy to learn about other techniques that aren't too far advanced.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;:  Looking over this again, I found out I missed the closed-form values at &lt;a href=&quot;http://en.wikipedia.org/wiki/Riemann_zeta_function#Specific_values&quot; rel=&quot;nofollow&quot;&gt;odd negative integers&lt;/a&gt;.  This doesn't affect the question but seemed worth correcting.  Thanks to the people who contributed.&lt;/p&gt;&#xA;" OwnerUserId="336" LastEditorUserId="336" LastEditorDisplayName="" LastEditDate="2011-01-03T00:36:50.637" LastActivityDate="2011-01-03T01:30:23.843" Title="Proving a known zero of the Riemann Zeta has real part exactly 1/2" Tags="&lt;complex-analysis&gt;&lt;special-functions&gt;&lt;riemann-zeta&gt;&lt;riemann-hypothesis&gt;" AnswerCount="3" CommentCount="4" FavoriteCount="3" />
  <row Id="7784" PostTypeId="2" ParentId="7772" CreationDate="2010-10-25T08:58:21.333" Score="4" ViewCount="" Body="&lt;p&gt;Philosophers of a nominalistic/constructivist bent often come out with&#xA;stuff like this, asserting that maths, physics or whatever is some&#xA;sort of arbitrary, contingent system which could have been totally&#xA;different if history had turned out differently. What these&#xA;philosophers are not very good at doing is proposing credible alternative&#xA;systems. Of course, the burden of proposing alternative&#xA;&quot;constructive frameworks&quot; should be on your chum. Only he can say&#xA;what he has in mind (if anything), not us.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-10-25T08:58:21.333" CommunityOwnedDate="2010-10-25T08:58:21.333" />
  <row Id="7785" PostTypeId="2" ParentId="7780" CreationDate="2010-10-25T09:55:49.553" Score="1" ViewCount="" Body="&lt;p&gt;You have a point $x$, vector $v$, and also some vertical vector, pointing down. Lets call it $g$. If I have understood your problem correctly, your plane contains $(0,0,0)$ (or otherwise you can write $x$ in the coordinate system with origin on the plane), vector $v$, and a vector $w=v\times g$ --- horizontal vector, orthogonal to $v$. I want to interpret $x$ as a vector (from the origin to the point $x$). Then for some $\alpha\in\mathbb{R}$ vector $x-\alpha g$ is on the plane and $x$ is below the plane if and only if $\alpha&gt;0$. We have&#xA;$$(x-\alpha g, v\times w)=0,$$&#xA;$$(x-\alpha g, v\times[v\times g])=0,$$&#xA;$$(x-\alpha g, v(v,g)-g(v,v))=0,$$&#xA;$$(x, v)(v,g)-(x,g)(v,v)=\alpha\bigl((g,v)(g,v)-(g,g)(v,v)\bigr),$$&#xA;$$\alpha=\bigl((x, v)(v,g)-(x,g)(v,v)\bigr)/\bigl((g,v)(g,v)-(g,g)(v,v)\bigr).$$&lt;/p&gt;&#xA;" OwnerUserId="695" LastActivityDate="2010-10-25T09:55:49.553" CommentCount="9" />
  <row Id="7786" PostTypeId="1" CreationDate="2010-10-25T10:14:26.367" Score="0" ViewCount="89" Body="&lt;p&gt;When dealing with Dirichlet Processes, according to [Teh, 2007], a DP is defined as by a base Probability H and a scale factor &quot;alpha&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to the Stick Breaking Construction, the random draws G from a DP:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;G~DP(alpha,H)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;are given by:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;G=sum(pi_k*delta_theta_k) over k from 1 to infinity&lt;/p&gt;&#xA;&#xA;&lt;p&gt;pi_k are ordered draws from a Beta Distribution given the length of an unitary stick&lt;/p&gt;&#xA;&#xA;&lt;p&gt;delta_theta_k is a point mass centered in &quot;theta_k&quot; (theta_k are random draws from the base distribution)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have pretty much a clear understanding of all the variables, but I do not know what do they mean by &quot;mass point&quot; is it the probability density of that draw, or is it something else.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It would be great if you could point me in any direction, only a reference would be amazing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="2780" LastActivityDate="2010-10-25T10:14:26.367" Title="Mass Point, Dirac Delta in Dirichlet Processes" Tags="&lt;statistics&gt;" CommentCount="1" />
  <row Id="7787" PostTypeId="2" ParentId="7783" CreationDate="2010-10-25T10:32:27.827" Score="12" ViewCount="" Body="&lt;p&gt;From what I've found, it's mostly presented as a &lt;em&gt;fait accompli&lt;/em&gt;, for instance, the &lt;a href=&quot;http://mathworld.wolfram.com/Riemann-SiegelFunctions.html&quot; rel=&quot;nofollow&quot;&gt;Riemann-Siegel function $Z(t)$&lt;/a&gt;, a function frequently used in the zero-finding, is defined with a $\zeta\left(\frac12+it\right)$ factor. What now one looks for when using constructs like these is if the function ever &quot;turns away&quot; from the horizontal axis without crossing it (if this does happen, the &lt;strong&gt;hypothesis&lt;/strong&gt; is false).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is why things like &lt;a href=&quot;http://mathworld.wolfram.com/LehmersPhenomenon.html&quot; rel=&quot;nofollow&quot;&gt;Lehmer's phenomenon&lt;/a&gt; are interesting behavior for the ζ function. Put simply, these are sections of the critical strip where there is (very!) nearly no crossing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are two traditional ways of visualizing the Lehmer phenomenon: one can look at the graph of the Riemann-Siegel function:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/l1dHQ.png&quot; alt=&quot;Riemann-Siegel Z(t)&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(&lt;em&gt;Mathematica&lt;/em&gt; code: &lt;code&gt;Plot[RiemannSiegelZ[z], {z, 0, 100}, AspectRatio -&amp;gt; 1/5, Frame -&amp;gt; True]&lt;/code&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or the so-called &quot;zeta spiral&quot; $z(t)=\zeta\left(\frac12+it\right)$ in the complex plane:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/tOoxm.png&quot; alt=&quot;zeta spiral&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(&lt;em&gt;Mathematica&lt;/em&gt; code: &lt;code&gt;ParametricPlot[Through[{Re, Im}[Zeta[1/2 + I t]]], {t, 0, 40}, AspectRatio -&amp;gt; Automatic, Frame -&amp;gt; True]&lt;/code&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, here is the first instance of the Lehmer phenomenon, seen using both viewpoints:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/n7rFh.png&quot; alt=&quot;Lehmer for Riemann-Siegel&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(&lt;em&gt;Mathematica&lt;/em&gt; code: &lt;code&gt;Plot[RiemannSiegelZ[z], {z, 7004, 7006}, Frame -&amp;gt; True, PlotRange -&amp;gt; {-2, 2}]&lt;/code&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/JUxe8.png&quot; alt=&quot;Lehmer for zeta spiral&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(&lt;em&gt;Mathematica&lt;/em&gt; code: &lt;code&gt;ParametricPlot[Through[{Re, Im}[Zeta[1/2 + I t]]], {t, 7004 + 1/2, 7005 + 1/2}, AspectRatio -&amp;gt; Automatic, Frame -&amp;gt; True]&lt;/code&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;strong&gt;hypothesis&lt;/strong&gt; would have been false if for the Riemann-Siegel function, the function displayed a local extremum without crossing the x-axis, or for the zeta spiral, the apparent &quot;cusp&quot; was actually a cusp or did not even pass through the origin. (I won't spoil the fun by posting zoomed-in versions of those last two images, you can do it yourself with &lt;em&gt;Mathematica&lt;/em&gt; or some other computing environment that can evaluate ζ(s) for complex values).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The two zeroes are in fact quite close: &lt;code&gt;FindRoot[RiemannSiegelZ[x], {x, ##}, WorkingPrecision -&amp;gt; 20] &amp;amp; @@@ {{7005 + 1/50, 7005 + 7/100}, {7005 + 9/10, 7005 + 11/100}}&lt;/code&gt; returns the two approximate zeroes &lt;code&gt;7005.0628661749205932&lt;/code&gt; and &lt;code&gt;7005.1005646726748389&lt;/code&gt;.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-10-25T10:32:27.827" CommentCount="2" />
  <row Id="7788" PostTypeId="1" AcceptedAnswerId="7790" CreationDate="2010-10-25T10:43:00.373" Score="4" ViewCount="170" Body="&lt;p&gt;As far as i know LU decomposition allow you to calculate matrix determinant following easy and cheap formula: Det[A] = Det[L] Det[U] = Det[U]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Trying this out in Mathematica 7 gives me correct result by absolute value, i.e. it ignores negative determinants and transforms them to positive ones.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sample code:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;matrixA = {{-1, 2, 3, 5}, {-7, -4, 5, 4}, {-89, 7, 8, -6}, {8, 6, -1, 4}};&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;Det[matrixA]&lt;/code&gt; gives out -2067    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;but&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;{lu, p, c} = LUDecomposition[matrixA]&#xA;&#xA;u = lu SparseArray[{i_, j_} /; j &amp;gt;= i -&amp;gt; 1, {4, 4}]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;Det[u]&lt;/code&gt; is 2067&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, the question is obvious - how to get correct determinant in Mathematica using LU decomposition?&lt;/p&gt;&#xA;" OwnerUserId="2781" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-10-25T11:12:47.777" LastActivityDate="2010-10-25T11:12:47.777" Title="Calculating determinant from LUDecomposition" Tags="&lt;linear-algebra&gt;&lt;matrices&gt;&lt;mathematica&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="7789" PostTypeId="2" ParentId="7788" CreationDate="2010-10-25T11:00:20.960" Score="3" ViewCount="" Body="&lt;p&gt;From the documentation: &quot;Therefore multiplying L and U recreates the original matrix permuted with the row pivot vector.&quot; So you need to multiply by &lt;code&gt;Signature[p]&lt;/code&gt; to get the correct sign.&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2010-10-25T11:00:20.960" CommentCount="2" />
  <row Id="7790" PostTypeId="2" ParentId="7788" CreationDate="2010-10-25T11:01:03.643" Score="8" ViewCount="" Body="&lt;p&gt;Well, this is because you forgot to take into account the &lt;em&gt;permutation matrix&lt;/em&gt; that is output by the LU decomposition routine. You have to remember two facts: 1.) Gaussian elimination generally performs row interchanges for numerical stability reasons, and 2.) the determinant of a matrix changes sign if you interchange rows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In &lt;em&gt;Mathematica&lt;/em&gt; at least, the function that will rescue you is &lt;code&gt;Signature[]&lt;/code&gt;, which gives the signature of the permutation required to turn the permutation matrix (which &lt;em&gt;Mathematica&lt;/em&gt; outputs as a scrambled list of the numbers from 1 to n, where n is the size of your matrix) into the identity matrix.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, to use your example in &lt;em&gt;Mathematica&lt;/em&gt;:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;m={{-1, 2, 3, 5}, {-7, -4, 5, 4}, {-89, 7, 8, -6}, {8, 6, -1, 4}};&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;we compare&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Det[m]&#xA;-2067&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;with the following:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;{lu, p, c} = LUDecomposition[m]&#xA;{{{-1, 2, 3, 5}, {7, -18, -16, -31}, {-8, -11/9, 31/9, 55/9}, {89, 19/2, -963/31, 2067/62}}, {1, 2, 4, 3}, 1}&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Now, &lt;code&gt;LUDecomposition[]&lt;/code&gt; here outputs three things: the merged $\mathbf{L}$ and $\mathbf{U}$ matrices, the permutation, and the condition number. We can get the tentative determinant by multiplying together the diagonal elements of &lt;code&gt;lu&lt;/code&gt;, thus:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Tr[lu, Times]&#xA;2067&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Here is where &lt;code&gt;Signature&lt;/code&gt; comes in:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Signature[{1, 2, 4, 3}]&#xA;-1&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Note that to turn &lt;code&gt;{1, 2, 4, 3}&lt;/code&gt; into &lt;code&gt;{1, 2, 3, 4}&lt;/code&gt;, one needs one swap; namely, to swap the third and fourth elements. Since 1 is odd, the signature is -1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus,&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Signature[p]*Tr[lu, Times]&#xA;-2067&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;gives the correct answer.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-10-25T11:01:03.643" CommentCount="1" />
  <row Id="7792" PostTypeId="1" AcceptedAnswerId="7794" CreationDate="2010-10-25T11:12:11.570" Score="2" ViewCount="121" Body="&lt;p&gt;One of the most famous 2D-curves are &lt;a href=&quot;http://en.wikipedia.org/wiki/Logarithmic_spiral&quot; rel=&quot;nofollow&quot;&gt;logarithmic spirals&lt;/a&gt; (or Spira mirabilis). They can be constructed by using a machinery that ensures a constant angle between the tangent and the radial lines all the time while plotting it.&lt;br&gt;&lt;br&gt; &#xA;&lt;img src=&quot;http://i.imgur.com/yP6mK.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My question&lt;/strong&gt;&lt;br&gt;&#xA;I can see it in the picture that the spiral arms are getting bigger each turn and I see the math. What I don't understand intuitively is where the logarithm comes from - or put differently: Why does a geometric progression arise just by holding an angle constant. Could you give me some hints? Thank you,&lt;/p&gt;&#xA;" OwnerUserId="346" LastActivityDate="2010-10-25T15:39:13.173" Title="Why does an equiangular spiral become logarithmic (intuitively)?" Tags="&lt;intuition&gt;&lt;plane-curves&gt;&lt;logarithms&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="7793" PostTypeId="1" AcceptedAnswerId="7804" CreationDate="2010-10-25T11:43:26.380" Score="1" ViewCount="167" Body="&lt;p&gt;If $x \geq 2$, then how do we prove that $$\int\limits_{2}^{x} \frac{dt}{\log^{n}{t}} = O\Bigl(\frac{x}{\log^{n}{x}}\Bigr)$$&lt;/p&gt;&#xA;" LastActivityDate="2010-10-25T14:08:00.517" Title="Dominating the integral $\int\limits_{2}^{x} \frac{dt}{\log^{n}{t}}$" Tags="&lt;number-theory&gt;&lt;analysis&gt;&lt;analytic-number-theory&gt;" AnswerCount="2" />
  <row Id="7794" PostTypeId="2" ParentId="7792" CreationDate="2010-10-25T11:44:03.493" Score="3" ViewCount="" Body="&lt;p&gt;Much is explained by looking at the polar equation of the spiral:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$r=\exp(\theta\cot\alpha)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here, $\alpha$ is the constant angle any tangent to the curve makes with the radius vector (a line segment joining the origin and the point of tangency). This explains the adjective &lt;em&gt;equiangular&lt;/em&gt; (the verification of this property from the defining equation is left as an exercise). As an aside, insects flying towards a point light source like a candle or a light bulb follow the path of an equiangular spiral, since the usual strategy of an insect flying at the daytime to get their bearing is to fly at a constant angle from the sun's rays, and this strategy works against them when encountering man-made light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, suppose we have an arithmetic progression of angles $\theta,\theta+\Delta\theta,\theta+2\Delta\theta,\dots$; if we get the corresponding values of the radius vector using the defining equation for the logarithmic spiral (geometrically speaking, this corresponds to a clockwise rotation by $0,\Delta\theta,2\Delta\theta,\dots$ radians), we get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\exp(\theta\cot\alpha),\exp((\theta+\Delta\theta)\cot\alpha),\exp((\theta+2\Delta\theta)\cot\alpha),\dots$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which can be re-expressed as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\exp(\theta\cot\alpha),\exp(\theta\cot\alpha)\cdot\exp(\Delta\theta\cot\alpha),\exp(\theta\cot\alpha)\cdot\exp(\Delta\theta\cot\alpha)^2,\dots$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which as you can see is a &lt;em&gt;geometric&lt;/em&gt; progression; that is to say, the &lt;em&gt;logarithms&lt;/em&gt; of the members of this sequence form an arithmetic progression. This is where the &lt;em&gt;logarithmic&lt;/em&gt; adjective arises from.&lt;/p&gt;&#xA;" OwnerUserId="498" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-10-25T15:39:13.173" LastActivityDate="2010-10-25T15:39:13.173" CommentCount="2" />
  <row Id="7795" PostTypeId="1" CreationDate="2010-10-25T11:52:04.000" Score="0" ViewCount="332" Body="&lt;p&gt;$$\frac{57^{46}}{17}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is best and quickest way to evaluate an approximate answer to this by hand?&lt;/p&gt;&#xA;" OwnerUserId="2784" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-04-16T03:25:50.813" LastActivityDate="2011-04-16T03:25:50.813" Title="What is $57^{46}$ divided by 17?" Tags="&lt;arithmetic&gt;" AnswerCount="5" CommentCount="5" />
  <row Id="7796" PostTypeId="2" ParentId="7795" CreationDate="2010-10-25T11:55:47.700" Score="2" ViewCount="" Body="&lt;p&gt;Euler's theorem gives you the remainder.&lt;/p&gt;&#xA;" OwnerUserId="589" LastActivityDate="2010-10-25T11:55:47.700" />
  <row Id="7797" PostTypeId="2" ParentId="7795" CreationDate="2010-10-25T12:04:03.937" Score="2" ViewCount="" Body="&lt;p&gt;You can do it by Fermats little theorem as well.&lt;/p&gt;&#xA;" LastActivityDate="2010-10-25T12:04:03.937" />
  <row Id="7798" PostTypeId="2" ParentId="7795" CreationDate="2010-10-25T12:17:40.397" Score="5" ViewCount="" Body="&lt;p&gt;It really depends what you mean by &quot;approximate answer&quot;. Since you tagged arithmetic...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice 57 = 3 . 19, you can say that $57^{46} / 17 = 3 \cdot 57^{45} \cdot \frac{19}{17}$. So if you know what $57^{45}$ is, you can approximate to within 2% if you multiply that by 3.3. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, compute by hand $57^5 = 601692057$, which is less than 0.3% off from $6\cdot 10^8$. Observe that $57^{45} = (57^5)^9$. Using the binomial theorem you see that you can approximate that by $(6\cdot 10^8)^9 = 10 077 696 \cdot 10^{72}$ to within 3%. Rounding off the lower digits won't matter much in the error, so you have $57^{45} \sim 10^{79}$ to 3%. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So putting it all together you have that to within 5%, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ 57^{46} \sim 3.3 \cdot 10^{79} $$&lt;/p&gt;&#xA;" OwnerUserId="1543" LastActivityDate="2010-10-25T12:17:40.397" CommentCount="2" />
  <row Id="7799" PostTypeId="2" ParentId="7738" CreationDate="2010-10-25T12:20:09.047" Score="1" ViewCount="" Body="&lt;p&gt;$\tilde P &amp;lt;&amp;lt; P$ is true but $P &amp;lt;&amp;lt; \tilde P$ is false. We have&#xA;$$\frac{d\tilde P}{dP}(w)=\sqrt{2\pi}\frac{e^{-w} 1_{[0,\infty)}}{e^{-w^2/2}}$$&#xA;Proving these facts is an exercise that you should really do on your own, as it is not that hard and you would not learn anything if you were just told the proof.&lt;/p&gt;&#xA;" OwnerUserId="1609" LastActivityDate="2010-10-25T12:20:09.047" CommentCount="0" />
  <row Id="7800" PostTypeId="2" ParentId="7793" CreationDate="2010-10-25T12:33:10.047" Score="5" ViewCount="" Body="&lt;p&gt;Consider&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \frac{d}{dx} \frac{x}{\log^n(x)} = \frac{1}{\log^n(x)} - \frac{n}{\log^{n+1}(x)} = \frac{1}{\log^n(x)}\left( 1 - \frac{n}{\log x}\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For sufficiently large $x$, the RHS is bigger than $\frac12 \frac{1}{\log^n(x)}$, and hence we get the asymptotic bound we need. The usual arguments then gives you a sufficiently large constant for the Big O. &lt;/p&gt;&#xA;" OwnerUserId="1543" LastActivityDate="2010-10-25T12:33:10.047" />
  <row Id="7801" PostTypeId="2" ParentId="7795" CreationDate="2010-10-25T12:44:31.223" Score="1" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\rm \mod\ 17:\ \ 57\equiv 6,\ \ 6^{46}\equiv 1/6^2,\ \ 1/6\equiv 3\ \ $ which immediately yields the result&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-10-25T13:10:19.407" LastActivityDate="2010-10-25T13:10:19.407" />
  <row Id="7802" PostTypeId="2" ParentId="7778" CreationDate="2010-10-25T13:51:13.647" Score="4" ViewCount="" Body="&lt;p&gt;Here's the more general way that you seek. The LHS and RHS are both products of rational functions of $\rm\:n\:$, i.e. they satisfy $\rm\ f(n+1) = r(n)\ f(n)\ $ for some rational function $\rm r(n)\:$. To test whether two such functions are equal you need only test that they have the same value of $\rm\ r(n) = f(n+1)/f(n)\ $ and the same initial condition $\rm\: f(0)$.$\ $ For the above it's trivial: $\rm\ f(0) = 1\:$ and the common value of $\rm\ r(n)\ $ is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\quad\quad\displaystyle f(n)\ \ =\  \ {2n \choose n}\ \: =\: \ \frac{(2n)!}{n!^2}\ \ \Rightarrow\ \frac{f(n+1)}{f(n)}\ =\ \frac{2\:(n+1)\:(2n+1)}{\ (n+1)\ (n+1)}$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\quad\quad\displaystyle f(n)\ =\ 2^n \frac{ 1 \cdot 3  \cdots (2n-1)}{n!} \ \Rightarrow\ \frac{f(n+1)}{f(n)}\ =\ \frac{2\ (2n+1)}{n+1}$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice that this method is completely algorithmic - it requires no insight whatsoever. In essence one is employing the &lt;strong&gt;uniqueness theorem&lt;/strong&gt; for first order difference equations  (recurrences) to prove an equality by simply checking that both sides satisfy the same difference / differential equation and initial conditions. Such techniques generalize to much wider classes of multivariate mixed differential - difference equations, e.g.  &lt;strong&gt;holonomic&lt;/strong&gt; functions. For some further examples of this powerful technique see my &lt;a href=&quot;http://math.stackexchange.com/questions/3514&quot;&gt;post here.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-10-26T00:52:42.443" LastActivityDate="2010-10-26T00:52:42.443" CommentCount="7" />
  <row Id="7803" PostTypeId="2" ParentId="7775" CreationDate="2010-10-25T13:56:13.540" Score="3" ViewCount="" Body="&lt;p&gt;You can get an approximate answer.  The &lt;a href=&quot;http://en.wikipedia.org/wiki/Coupon_collector%27s_problem&quot; rel=&quot;nofollow&quot;&gt;coupon collector's problem&lt;/a&gt; tells you that the expected number of balls is $n*H_n$ or about 7485 for n=1000.  So the average bin will have 7.4 balls.  If you look at the Poisson distribution for 7.485 and find the number where it falls to 1/1000.  I make that 17 or 18.  The motivation is that you have 999 bins with an average occupancy of 7.485.  The highest will be at a probability of about 1/999&lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2010-10-25T15:13:22.670" LastActivityDate="2010-10-25T15:13:22.670" CommentCount="2" />
  <row Id="7804" PostTypeId="2" ParentId="7793" CreationDate="2010-10-25T14:08:00.517" Score="6" ViewCount="" Body="&lt;p&gt;Here is a way to prove it without knowing the answer before hand.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $c &gt; 2$ be a number such that $c &gt; e^{n+1}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is enough to consider&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_{c}^{x} \frac{dt}{\log^n t}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Integrating by parts, we get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_{c}^{x} \frac{dt}{\log^n t} =  \frac{x}{\log^n x} - K + n \int_{c}^{x} \frac{dt}{\log^{n+1} t}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now $$n\int_{c}^{x} \frac{dt}{\log^{n+1} t} \le \frac{n}{n+1} \int_{c}^{x} \frac{dt}{\log^n t} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;as $\log t \ge n+1$ for $t \ge c$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus $$\frac{1}{n+1}\int_{c}^{x} \frac{dt}{\log^n t} &amp;lt; \frac{x}{\log^n x}  $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, we have that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_{2}^{x} \frac{dt}{\log^n t} = \theta(\frac{x}{\log^n x})$$&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-10-25T14:08:00.517" CommentCount="9" />
  <row Id="7805" PostTypeId="2" ParentId="7795" CreationDate="2010-10-25T14:28:50.060" Score="2" ViewCount="" Body="&lt;p&gt;Why not simply do the computation? The most time-consuming part is to compute the $46$th power of $57$, but that can be done quite fast using &lt;a href=&quot;http://en.wikipedia.org/wiki/Exponentiation_by_squaring&quot; rel=&quot;nofollow&quot;&gt;repeated squaring&lt;/a&gt; in 6 multiplications. That results in a number with 81 digits. Dividing by 17 using long division is faster, and it takes about 80 easy steps.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you are done, you can check your result with &#xA;&lt;a href=&quot;http://www.google.com/search?q=57%5E46+divided+by+17&quot; rel=&quot;nofollow&quot;&gt;Google&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="274" LastEditorUserId="274" LastEditorDisplayName="" LastEditDate="2010-10-25T15:01:22.637" LastActivityDate="2010-10-25T15:01:22.637" CommentCount="3" />
  <row Id="7806" PostTypeId="2" ParentId="7776" CreationDate="2010-10-25T14:42:39.507" Score="0" ViewCount="" Body="&lt;p&gt;It's a composition of two obvious facts. First, over a finite ring $\rm\:R\:$, any formula that counts the roots of a univariate polynomial extends to bivariate polynomials since the number of roots of $\rm\:f(X,Y)\:$ is simply the sum of the number of roots of $\rm\:f(X,r)\:$ for all $\rm\ r\in R\:$. Second, for quadratic polynomials over some finite field $\rm\:\mathbb F_p\:$ with $\rm\:p\:$ odd, the problem of root counting reduces to testing if the discriminant is a square, hence there exists an elementary closed-form root counting formula expressed in terms of the Legendre symbol. Apply these two observations immediately yields the formula at hand.&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-10-25T14:42:39.507" />
  <row Id="7807" PostTypeId="1" AcceptedAnswerId="7820" CreationDate="2010-10-25T14:44:25.123" Score="5" ViewCount="278" Body="&lt;p&gt;The coordinate transformation (due to Beukers, Calabi and Kolk)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x=\frac{\sin u}{\cos v}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$y=\frac{\sin v}{\cos u}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;transforms the square domain $0\lt x\lt 1$ and $0\lt y\lt 1$ into the triangle domain $u,v&amp;gt;0,u+v&amp;lt;\pi /2$ (in &lt;em&gt;&lt;a href=&quot;http://books.google.pt/books?id=CuTpgZaLahkC&amp;amp;printsec=frontcover&amp;amp;dq=proofs+from+the+book&amp;amp;source=bl&amp;amp;ots=swepaSdSxh&amp;amp;sig=j1gB1zxabJE8qxEgbFh1ig8RA3Q&amp;amp;hl=pt-PT&amp;amp;ei=Pnj2TKKcCZOKhQfb67yABg&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=3&amp;amp;ved=0CDIQ6AEwAg#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;Proofs from the BOOK&lt;/a&gt;&lt;/em&gt; by M. Aigner and G. Ziegler).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the invert transformation is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$u=\arccos \sqrt{\dfrac{1-x^{2}}{1-x^{2}y^{2}}}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$v=\arccos \sqrt{\dfrac{1-y^{2}}{1-x^{2}y^{2}}}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;it is easy to see that three of the vertices (although not belonging to the domain) are transformed as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(x,y)=(0,0)\mapsto (0,0)=(u,v),$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(x,y)=(1,0)\mapsto (\pi /2,0)=(u,v),$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(x,y)=(0,1)\mapsto (0,\pi /2)=(u,v).$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question 1&lt;/strong&gt; - But how is the fourth vertex $(x,y)=(1,1)$ transformed? In the plot below of $\dfrac{1-x^{2}}{1-x^{2}y^{2}}$ seems that the following limit does not exist&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\underset{(x,y)\rightarrow (1,1)}{\lim }\sqrt{\dfrac{1-x^{2}}{1-x^{2}y^{2}}}.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/0Um5U.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question 2&lt;/strong&gt; - As a second question I would like to know how can one &quot;discover&quot; a&#xA;transformation of a square into a triangle such as this. &lt;em&gt;Is there any systematic study of this kind of transformations?&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="752" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2010-12-23T23:46:34.417" LastActivityDate="2010-12-23T23:46:34.417" Title="A limit and a coordinate trigonometric transformation of the interior points of a square into the interior points of a triangle" Tags="&lt;calculus&gt;&lt;trigonometry&gt;&lt;limit&gt;&lt;transformation&gt;&lt;multivariable-calculus&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="7808" PostTypeId="2" ParentId="7775" CreationDate="2010-10-25T14:49:46.030" Score="0" ViewCount="" Body="&lt;p&gt;To refine Ross' answer(+1), at the moment when the last (say, 7400) ball fills the last cell, the distribution of the remaining cells should approach that of 999 iid ZERO-TRUNCATED Poisson variables with media u=7399/999. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Updated: With media = 7.484, this gives $\lambda \approx 7.480 $&lt;/p&gt;&#xA;" OwnerUserId="312" LastEditorUserId="312" LastEditorDisplayName="" LastEditDate="2010-10-25T16:13:26.320" LastActivityDate="2010-10-25T16:13:26.320" CommentCount="1" />
  <row Id="7809" PostTypeId="2" ParentId="7757" CreationDate="2010-10-25T14:53:59.160" Score="1" ViewCount="" Body="&lt;p&gt;Count all pairs $(x,S)$ where $S \subseteq \{1,2,\dots,n\}$ and $x \in S$ in two ways. Having chosen a $S$ to be an r-element subset, we have $r$ choices for $x$, giving us the left hand side, $\sum_{r \geq 1} r {n \choose r}$. Next having chosen an arbitrary element $x \in \{1,2,\dots,n\}$ we can choose $S$ in $2^{n-1}$ ways, and $x$ can be chosen in $n$ ways giving us the right hand side.&lt;/p&gt;&#xA;" OwnerUserId="404" LastActivityDate="2010-10-25T14:53:59.160" CommentCount="1" />
  <row Id="7810" PostTypeId="2" ParentId="7757" CreationDate="2010-10-25T14:59:28.327" Score="5" ViewCount="" Body="&lt;p&gt;There are already some good answers here, but you did say that you wanted something general, so I'll add the following.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose you are interested, for some function $f(k)$, in the binomial sum &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$B(n) = \sum_{k=0}^n \binom{n}{k} f(k).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Your problem has $f(k) = k$.)  Then, taking $\Delta f(k) = f(k+1) - f(k)$, denote $A(n)$ by &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$A(n) = \sum_{k=0}^n \binom{n}{k} \Delta f(k).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A few years ago I proved the following relationship between $B(n)$ and $A(n)$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$B(n) = 2^n \left(f(0) + \sum_{k=1}^n \frac{A(k-1)}{2^k}\right).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For your problem, $f(k) = k$.  So $\Delta f(k) = k+1 - k = 1$.  If you're willing to take &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$A(n) = \sum_{k=0}^n \binom{n}{k} = 2^n,$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;then the formula gives &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{k=0}^n \binom{n}{k} k = 2^n \sum_{k=1}^n \frac{2^{k-1}}{2^k} = 2^n \sum_{k=1}^n \frac{1}{2} = n 2^{n-1}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Even if you're not willing to take $\sum_{k=0}^n \binom{n}{k} = 2^n,$ you can derive that immediately from the expression for $B(n)$ by letting $f(k) = 1$, as $\Delta f(k) = 0$ in that case.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The expression for $B(n)$ is Theorem 4 in my paper &quot;&lt;a href=&quot;http://math.pugetsound.edu/~mspivey/CombSum.pdf&quot; rel=&quot;nofollow&quot;&gt;Combinatorial Sums and Finite Differences&lt;/a&gt;,&quot; &lt;em&gt;Discrete Mathematics&lt;/em&gt;, 307 (24): 3130-3146, 2007.  This identity is in the paper as well, as an illustration of Theorem 4.&lt;/p&gt;&#xA;" OwnerUserId="2370" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-10-25T15:56:34.397" LastActivityDate="2010-10-25T15:56:34.397" />
  <row Id="7811" PostTypeId="1" AcceptedAnswerId="7813" CreationDate="2010-10-25T15:13:52.557" Score="8" ViewCount="391" Body="&lt;p&gt;Let $G$ be a group and let $G'$ be the derived subgroup, defined as the subgroup generated by the commutators of $G$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there an example of a finite group $G$ where not every element of $G'$ is a commutator? $G'$ is only generated by commutators, but with all of the properties of commutators (ie: what happens under conjugation, exponentation, etc) I can't think of an example.&lt;/p&gt;&#xA;" OwnerUserId="2298" LastActivityDate="2010-10-26T08:16:17.717" Title="Derived Subgroups and Commutators" Tags="&lt;group-theory&gt;" AnswerCount="4" />
  <row Id="7812" PostTypeId="1" AcceptedAnswerId="7817" CreationDate="2010-10-25T15:16:47.880" Score="1" ViewCount="192" Body="&lt;p&gt;I am trying to understand a simple calculation in &lt;a href=&quot;http://en.wikipedia.org/wiki/Polar_coordinate_system&quot; rel=&quot;nofollow&quot;&gt;polar coordinates&lt;/a&gt; - and I am getting totally discombobulated (the original source can be found: &lt;a href=&quot;http://online.redwoods.cc.ca.us/instruct/darnold/CalcProj/Fall07/KableKoch/EquiangularPresentation.pdf&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;).&lt;br&gt;&lt;br&gt;Please have a look at the following picture:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/EulN4.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My questions&lt;/strong&gt;&lt;br&gt;&lt;br&gt;(1) I don't understand how the length labeled $dr \over d \theta$ can possibly be the change of $r$ with respect to a change in $\theta$. &lt;br&gt;(2) It is then said that $tan \alpha = {r \over {dr \over d \theta}}$. But how can you calculate the tan-function if you don't even have a right angled triangle - and anyway, the formular would suggest that you calculate the tan of the angle that is not labeled in the pic (because $tan={opposite \over adjecent}$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sorry, if this is too elementary - but it really bothers me right now... Thank you!&lt;/p&gt;&#xA;" OwnerUserId="346" LastActivityDate="2010-10-25T16:16:09.133" Title="Difficulty with differentiation in polar coordinates" Tags="&lt;calculus&gt;&lt;polar-coordinates&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" />
  <row Id="7813" PostTypeId="2" ParentId="7811" CreationDate="2010-10-25T15:23:23.380" Score="4" ViewCount="" Body="&lt;p&gt;For a nice source of examples, see &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I. M. Isaacs: Commutators and the Commutator Subgroup. The American Mathematical Monthly. Vol. 84, No. 9 (Nov., 1977), pp. 720-722 &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;which you can get from &lt;a href=&quot;http://www.jstor.org/stable/2321253&quot; rel=&quot;nofollow&quot;&gt;JSTOR&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using GAP I find that the two examples of order 96, which is the minimal possible order, are the groups generated by the following two lists of permutations on $32$ elements.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;[ (3,8,6)(4,7,5)(9,27,17)(10,28,18)(11,30,22)(12,29,21)(13,26,23)(14,25,24)(15,31,20)(16,32,19), &#xA;    (1,17,7,23)(2,18,8,24)(3,19,5,21)(4,20,6,22)(9,26,15,32)(10,25,16,31)(11,28,13,30)(12,27,14,29), &#xA;    (1,9,5,13)(2,10,6,14)(3,11,7,15)(4,12,8,16)(17,25,21,29)(18,26,22,30)(19,27,23,31)(20,28,24,32), &#xA;    (1,5)(2,6)(3,7)(4,8)(9,13)(10,14)(11,15)(12,16)(17,21)(18,22)(19,23)(20,24)(25,29)(26,30)(27,31)(28,32), &#xA;    (1,3)(2,4)(5,7)(6,8)(9,11)(10,12)(13,15)(14,16)(17,19)(18,20)(21,23)(22,24)(25,27)(26,28)(29,31)(30,32), &#xA;    (1,2)(3,4)(5,6)(7,8)(9,10)(11,12)(13,14)(15,16)(17,18)(19,20)(21,22)(23,24)(25,26)(27,28)(29,30)(31,32) ]&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;[ (3,7,5)(4,8,6)(9,25,17)(10,26,18)(11,31,21)(12,32,22)(13,27,23)(14,28,24)(15,29,19)(16,30,20), &#xA;    (1,17)(2,18)(3,19)(4,20)(5,21)(6,22)(7,23)(8,24)(9,25)(10,26)(11,27)(12,28)(13,29)(14,30)(15,31)(16,32), &#xA;    (1,9)(2,10)(3,11)(4,12)(5,13)(6,14)(7,15)(8,16)(17,25)(18,26)(19,27)(20,28)(21,29)(22,30)(23,31)(24,32), &#xA;    (1,5,2,6)(3,8,4,7)(9,13,10,14)(11,16,12,15)(17,21,18,22)(19,24,20,23)(25,29,26,30)(27,32,28,31), &#xA;    (1,3,2,4)(5,7,6,8)(9,11,10,12)(13,15,14,16)(17,19,18,20)(21,23,22,24)(25,27,26,28)(29,31,30,32), &#xA;    (1,2)(3,4)(5,6)(7,8)(9,10)(11,12)(13,14)(15,16)(17,18)(19,20)(21,22)(23,24)(25,26)(27,28)(29,30)(31,32) ]&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Later:&lt;/strong&gt; I used the following very simple-minded code to find the groups&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;chk := function (g)&#xA;  local comms, sub;&#xA;  comms := Set(List(Cartesian(g, g), p -&amp;gt; p[1]*p[2]*p[1]^-1*p[2]^-1));&#xA;  sub := Set(DerivedSubgroup(g));&#xA;  return Size(comms) = Size(sub);    &#xA;end;;&#xA;examples := AllSmallGroups(96, chk, false);&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This works because I remembered there were examples of order 96. To get the permutation representations I did, for example,&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;GeneratorsOfGroup(Image(SmallerDegreePermutationRepresentation(Image(IsomorphismPermGroup(examples[1])))));&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="274" LastEditorUserId="274" LastEditorDisplayName="" LastEditDate="2010-10-25T16:43:36.557" LastActivityDate="2010-10-25T16:43:36.557" CommentCount="1" />
  <row Id="7814" PostTypeId="2" ParentId="7811" CreationDate="2010-10-25T15:29:34.433" Score="2" ViewCount="" Body="&lt;p&gt;Keith Dennis has also done some work on this problem.  In the paper:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dennis, R. K.(1-CRNL); Vaserstein, L. N.(1-PAS)&#xA;Commutators in linear groups.&#xA;K-Theory 2 (1989), no. 6, 761–767. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;they consider conditions under which all elements of the commutator subgroup are products of at most 2 commutators.  I seem to recall that Dennis did computer calculations to find the smallest group for which the commutator subgroup contains elements that are not commutators (and my recollection is that the size of the group was around 100 elements) but I can't see anything on-line about that.&lt;/p&gt;&#xA;" OwnerUserId="1392" LastActivityDate="2010-10-25T15:29:34.433" CommentCount="2" />
  <row Id="7815" PostTypeId="1" AcceptedAnswerId="7816" CreationDate="2010-10-25T15:41:22.503" Score="2" ViewCount="97" Body="&lt;p&gt;Let $Z$ be a countable set. Let $f_1,....,f_n$ be a collection of real functions over $Z$. Let $z_1,...,z_m,...$ be an enumeration of elements of $Z$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Define $V_1 = \{ (f_1(z),...,f_n(z)) | z \in Z\}$ a set of $n$ dimensional vectors.&#xA;Define $V_2 =\{ (f_i(z_1),...,f_i(z_m),..., )) | i \in\{ 1,...,n\} \}$ a set of infinite dimensional vectors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $span(V_i)$ be the span (linear span) of $V_i$. Is the linear dimension of $span(V_i)$ the same for $i=1,2$?&lt;/p&gt;&#xA;" OwnerUserId="7076" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-10-25T16:15:24.713" LastActivityDate="2010-10-25T16:15:24.713" Title="do two vector spaces have the same linear dimension?" Tags="&lt;linear-algebra&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="7816" PostTypeId="2" ParentId="7815" CreationDate="2010-10-25T15:53:31.117" Score="1" ViewCount="" Body="&lt;p&gt;Let me rephrase your question in a way that's less confusing.  Define a matrix $A$ such that $A_{ij} = f_i(z_j)$.  This matrix has finitely many rows but infinitely many columns.  Then the dimension of the span of $V_1$ is the row rank of this matrix, and the dimension of the span of $V_2$ is the column rank of this matrix.  And you want to know if they are the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer is yes.  Here's one proof: define $A_m$ to be $A$, but with all columns after the $m^{th}$ column chopped off, and let $a_m$ denote the column rank of $A_m$.  Then it's not hard to see that there exists $m$ such that $a_m$ is the column rank of $A$.  Now apply the theorem that row rank equals column rank for ordinary matrices to $A_m$.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-10-25T15:53:31.117" />
  <row Id="7817" PostTypeId="2" ParentId="7812" CreationDate="2010-10-25T16:09:05.663" Score="2" ViewCount="" Body="&lt;p&gt;Something is screwy with that picture. I think &lt;a href=&quot;http://beares.net/r/astro/galaxies/galaxies_docs/pitch_angle.htm&quot; rel=&quot;nofollow&quot;&gt;http://beares.net/r/astro/galaxies/galaxies_docs/pitch_angle.htm&lt;/a&gt; has a better picture. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$dr$ should be represented by swinging $r_0$ over to the arm that is denoted by $r$, and looking at the length difference. The $\tan$ is just an approximation: in the infinitesimal limit, the arc that &quot;swings&quot; $r_0$ over to the arm denoted by $r$ approaches a straight line that is perpendicular to the arm at $r$. So $dr/(r\dtheta)$ is the slope, which is just $\tan \alpha$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope this is sort of understandable. I don't have the software to draw a picture right now, if you need one to illustrate what I wrote in the previous paragraph, I can probably upload a picture in a few hours. &lt;/p&gt;&#xA;" OwnerUserId="1543" LastActivityDate="2010-10-25T16:09:05.663" CommentCount="3" />
  <row Id="7818" PostTypeId="2" ParentId="7812" CreationDate="2010-10-25T16:10:44.507" Score="2" ViewCount="" Body="&lt;p&gt;(Willie beat me by a few seconds, but this is what I had written.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;No wonder you're confused. That's a horrible picture! You should draw an arc of a circle ($\approx$ a line) at right angles to the side labeled $r_0$; this will have length $r d\theta$ and split the line labeled $r$  into two pieces of lenghts $r_0$ and $dr$ (approximately, for small $d\theta$). Then you have a small right triangle such that $\tan\alpha \approx (r d\theta)/dr$.&lt;/p&gt;&#xA;" OwnerUserId="1242" LastEditorUserId="1242" LastEditorDisplayName="" LastEditDate="2010-10-25T16:16:09.133" LastActivityDate="2010-10-25T16:16:09.133" />
  <row Id="7819" PostTypeId="2" ParentId="7811" CreationDate="2010-10-25T16:12:59.783" Score="1" ViewCount="" Body="&lt;p&gt;The &quot;simplest&quot; example is the free group on two generators, but is perhaps not the easiest to play with unless you are comfortable with free groups.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A nice survey of the problem of commutators vs. commutators subgroups in finite groups is:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Kappe, Luise-Charlotte, and Morse, Robert Fitzgerald. &lt;strong&gt;On commutators in groups.&lt;/strong&gt; &lt;em&gt;Groups St. Andrews 2005, Vol. 2&lt;/em&gt;, pp. 531-558. London Math. Soc. Lecture Notes Ser., &lt;strong&gt;340&lt;/strong&gt;, Cambridge University Press, 2007. MR 2331612 (2008g:20068)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The smallest groups in which the commutator subgroup does not equal the set of commutators have order 96; in fact, there are two non-isomorphic groups of order 96 in which the set of commutators does not equal the commutator subgroup. (This result also appeared in Robert Guralnick's 1977 thesis, &quot;Expressing group elements as products of commutators&quot;, UCLA). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Kappe also has an upcoming joint paper in which she discusses the same question, restricted to $p$-groups. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another interesting item is the Ore Conjecture, which stated that if $G$ is a finite nonabelian simple group, then the set of commutators equals the commutator subgroup. Its proof was recently completed, and it appears in:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Liebeck, Martin W.; O'Brien, E. A.; Shalev, Aner; Tiep, Pham Huu. &lt;strong&gt;The Ore conjecture.&lt;/strong&gt; &lt;em&gt;J. Eur. Math. Soc. (JEMS)&lt;/em&gt; 12 (2010), no. 4, 939–1008. MR MR2654085. A PDF preprint can be found at &lt;a href=&quot;http://www.math.auckland.ac.nz/~obrien/research/ore.pdf&quot; rel=&quot;nofollow&quot;&gt;O'Brien's website&lt;/a&gt;.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-10-25T17:23:51.830" LastActivityDate="2010-10-25T17:23:51.830" CommentCount="2" />
  <row Id="7820" PostTypeId="2" ParentId="7807" CreationDate="2010-10-25T16:27:52.537" Score="4" ViewCount="" Body="&lt;p&gt;Question 1: Setting $u=\pi/2-v$ in your first set of formulas gives $(x,y)=(1,1)$, so that point corresponds to the whole hypotenuse $u+v=\pi/2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Question 2: I have no idea. (Luck? Inspiration? Trial and error?)&lt;/p&gt;&#xA;" OwnerUserId="1242" LastEditorUserId="1242" LastEditorDisplayName="" LastEditDate="2010-10-25T16:37:27.703" LastActivityDate="2010-10-25T16:37:27.703" CommentCount="1" />
  <row Id="7821" PostTypeId="1" AcceptedAnswerId="7825" CreationDate="2010-10-25T16:45:03.437" Score="2" ViewCount="399" Body="&lt;p&gt;Ok. This question may sound very easy, but actually i am in great need of it. I have been facing trouble in constructing functions, which are only continuous at some particular sets. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For e.g, the standard example of a function which is only continuous at one point, is the function,  $f(x) = x, \ x \in \mathbb{Q}$ and $f(x) = -x, x  \in \mathbb{R} \setminus \mathbb{Q}$. Similarly, i would like to know as to how to construct a function which is &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Continuous at exactly $2,3,4$ points.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Continuous exactly at integers&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Continuous exactly at Natural numbers&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Continuous exactly at Rationals. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I would like to see many examples (with proof!), so that i can don't struggle when somebody asks me to construct such functions. &lt;/p&gt;&#xA;" LastActivityDate="2010-10-26T03:26:54.163" Title="Constructing Continuous functions at given points." Tags="&lt;real-analysis&gt;&lt;analysis&gt;&lt;functions&gt;" AnswerCount="2" CommentCount="11" FavoriteCount="2" />
  <row Id="7822" PostTypeId="1" AcceptedAnswerId="7826" CreationDate="2010-10-25T16:51:22.803" Score="4" ViewCount="191" Body="&lt;p&gt;So this is some I partially understand, I'm not sure what I dont and do understand because most of my understanding is based on assumtions... sorry if I sound a little stupid!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The equation $6 \cos x - \sin x = 5 $ needs to be turned into the form $ \Re  \cos (x + \alpha) $ then solved for &lt;strong&gt;in the interval&lt;/strong&gt; $ -1/2\pi &amp;lt; x &amp;lt; 1/2\pi $.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I turned it into this: $ \sqrt{37} \cos( x + 0.165) $ then&lt;br&gt;&#xA; $ \sqrt{37} \cos( x + 0.165)  = 5$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ \cos( x + 0.165)  = \frac{5}{\sqrt{37}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ x + 0.165= \arccos( \frac{5}{\sqrt{37}} )$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ x  = \arccos( \frac{5}{\sqrt{37}} ) - 0.165$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ x = 0.44$ YAY! but... the answer is 0.44 &lt;strong&gt;and&lt;/strong&gt; -0.771  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Im thinking its asking what other value would of the above equation would end up in 5 also? Correct? How do I do this? &lt;strong&gt;Could someone explain&lt;/strong&gt; what is meant by &lt;em&gt;&quot;solve this equation for that interval&quot;&lt;/em&gt;, and how does one go about it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A problem I think might be related that I JUST cannot get my head around is this one:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;em&gt;The angle made by a wasps wings horizontally is given by the equation $ \theta = 0.4 \sin600t $, where t is time is seconds. How many times a second does its wing oscillate?&lt;/em&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I tried solving this, honest but I do not know where to begin!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks so much&lt;/p&gt;&#xA;" OwnerUserId="2576" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2010-10-25T17:28:18.903" LastActivityDate="2010-10-25T17:28:18.903" Title="understanding intervals in trigonometry" Tags="&lt;algebra-precalculus&gt;&lt;homework&gt;&lt;trigonometry&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="7824" PostTypeId="2" ParentId="7821" CreationDate="2010-10-25T17:04:35.487" Score="6" ViewCount="" Body="&lt;p&gt;Continuous at 2, 3, 4: $f(x)=(x-2)(x-3)(x-4)$ if $x$ is rational, $f(x)=0$ if $x$ is irrational.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Continuous at the integers: $f(x)=\sin(\pi x)$ if $x$ is rational, 0 if $x$ is irrational.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Continuous at the natural numbers: $f(x)=\sin(\pi x)$ if $x$ is rational and not a nonpositive integer, 0 if $x$ is irrational, 1 if $x$ is a nonpositive integer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Continuous exactly at the rationals: Impossible, because the set of rational numbers is not a $G_\delta$.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-10-25T17:04:35.487" CommentCount="10" />
  <row Id="7825" PostTypeId="2" ParentId="7821" CreationDate="2010-10-25T17:07:30.687" Score="3" ViewCount="" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;One simple way of constructing a function which is continuous only at a finite number of points, $x=a_1,\ldots,a_n$, is to do a slight modification to the function you give: take a polynomial $p(x)$ that has roots exactly at $x=a_1,\ldots,a_n$ (e.g., $p(x) = (x-a_1)\cdots(x-a_n)$) , and then define&#xA;$$ g(x) = \left\{\begin{array}{ll}&#xA;p(x) &amp;amp; \text{if $x\in\mathbb{Q}$;}\\&#xA;0 &amp;amp; \text{if $x\notin\mathbb{Q}$.}&#xA;\end{array}\right.$$&#xA;The function is continuous at $a_1,\ldots,a_n$, and since $p(x)\neq 0$ for any $x\notin\{a_1,\ldots,a_n\}$ then $g(x)$ is not continuous at any point &lt;em&gt;other than&lt;/em&gt; $a_1,\ldots,a_n$. Other possibilities should suggest themselves easily enough.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;A function that is continuous exactly at the integers: a similar idea will work: find a function that has zeros exactly at the integers, for example $f(x)=\sin(\pi x)$, and then take&#xA;$$g(x) = \left\{\begin{array}{ll}&#xA;\sin(\pi x) &amp;amp; \text{if $x\in\mathbb{Q}$;}\\&#xA;0 &amp;amp; \text{if $x\notin\mathbb{Q}$.}&#xA;\end{array}\right.$$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;A function continuous exactly in the natural numbers: take a function that is continuous at the integers, and redefine it as the characteristic function of the rationals in appropriate places(what happens at $0$ depends on whether you believe $0$ is in the natural numbers or not). Assuming that $0\in\mathbb{N}$, one possibility is:&#xA;$$g(x) = \left\{\begin{array}{ll}&#xA;\sin(\pi x)&amp;amp;\text{if $x\in\mathbb{Q}$ and $x\geq 0$;}\\&#xA;x &amp;amp; \text{if $x\in\mathbb{Q}$ and $-\frac{1}{2}\lt x\leq 0$;}\\&#xA;1 &amp;amp; \text{if $x\in\mathbb{Q}$ and $x\leq -\frac{1}{2}$;}\\&#xA;0 &amp;amp; \text{if $x\notin\mathbb{Q}$.}&#xA;\end{array}\right.$$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;A function continuous exactly on the rationals. This one is a bit trickier. There is no such function. This follows because the set of discontinuities of a real valued function must be a countable union of closed sets. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps then, we might anticipate the next question:&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;A function that is continuous exactly on the irrationals. An example is the following: let $s\colon\mathbb{N}\to\mathbb{Q}$ be an enumeration of the rationals (that is, a bijection from $\mathbb{N}$ to $\mathbb{Q}$. Define $f(x)$ as follows:&#xA;$$f(x) = \sum_{\stackrel{n\in\mathbb{N}}{s_n\leq x}} \frac{1}{2^n}.$$&#xA;The function has a jump at every rational, so it is not continuous at any rational. However, if $x$ is irrational, let $\epsilon\gt 0$. Then there exists $N$ such that $\sum_{k\geq N}\frac{1}{2^k}\lt \epsilon$. Find a neighborhood of $x$ which excludes every $q_m$ with $m\leq N$, and conclude that the difference between the value of $f$ at $x$ and at any point in the neighborhood is at most $\sum_{k\geq N}\frac{1}{2^k}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; As I was reminded in the comments by jake, in fact the &quot;standard example&quot; of a function that is continuous at every rational and discontinuous at every rational is &lt;a href=&quot;http://en.wikipedia.org/wiki/Thomae%27s_function&quot; rel=&quot;nofollow&quot;&gt;Thomae's function&lt;/a&gt;. The example I give is a monotone function, and although it is discontinuous at every rational, it is continuous &lt;em&gt;from the right&lt;/em&gt; at every number.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-10-26T03:26:54.163" LastActivityDate="2010-10-26T03:26:54.163" CommentCount="14" />
  <row Id="7826" PostTypeId="2" ParentId="7822" CreationDate="2010-10-25T17:13:20.227" Score="2" ViewCount="" Body="&lt;p&gt;As the cosine is periodic, there are many values of $\theta$ which have the same $\cos(\theta)$.  So they are just asking for all the values between $-\pi/2$ and $\pi/2$ that solve the equation.  Your solution got one of them, $\cos(0.605)$ does equal $5/\sqrt(37)$.  But so does $\cos(-.605)$  You were supposed to find that one, too. It leads to the solution -.771 when the .165 is subtracted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For your second problem I presume the &gt; sign is supposed to be *.  How much does the argument of the sine function (the thing you take the sine of) have to increase to go through one cycle?  How much does t have to increase to go through one cycle.  This gives you the period.  The frequency is one divided by this.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2010-10-25T17:26:02.823" LastActivityDate="2010-10-25T17:26:02.823" CommentCount="9" />
  <row Id="7827" PostTypeId="1" AcceptedAnswerId="7828" CreationDate="2010-10-25T17:15:10.713" Score="0" ViewCount="176" Body="&lt;p&gt;ok so I did all the revision problems and noted the ones I couldn't do today and Im posting them together, hope thats not a problem with the power that be?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have exhibit A:&lt;br&gt;&#xA;$e^{-x} -x + 2 $&lt;br&gt;&#xA;So I differentiate to find where the derivative hits 0:&lt;br&gt;&#xA;$-e^{-x} -1 = 0 $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now HOW do I figure when this &lt;em&gt;hits&lt;/em&gt; zero!?&lt;br&gt;&#xA;$-1 = e^{-x} $&lt;br&gt;&#xA;$\ln(-1) = \ln(e^{-x})$ ???&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More to come ... as one day rests between me and my final exam/attempt at math!&lt;/p&gt;&#xA;" OwnerUserId="2576" LastActivityDate="2010-10-25T18:07:31.470" Title="finding the minima and maxima of some tough functions" Tags="&lt;calculus&gt;&lt;homework&gt;" AnswerCount="3" CommentCount="4" />
  <row Id="7828" PostTypeId="2" ParentId="7827" CreationDate="2010-10-25T17:20:56.540" Score="3" ViewCount="" Body="&lt;p&gt;$e^{-x}&gt;0$ for all real $x$. Hence $-e^{-x}&amp;lt;0$ for all real $x$, whence $-e^{-x} - 1 &amp;lt;-1$ for all real $x$. So it nevers &quot;hits&quot; zero. Look at a graph of your function.&lt;/p&gt;&#xA;" OwnerUserId="1313" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-10-25T18:07:31.470" LastActivityDate="2010-10-25T18:07:31.470" CommentCount="6" />
  <row Id="7829" PostTypeId="1" CreationDate="2010-10-25T17:48:39.607" Score="1" ViewCount="211" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need calculate ray (line) intersection with torus for my ray-tracing program (I know, its to graphics, but i need math behind it).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can solve equation of order x^4, but thats too way slow (Cardano's method). So is there better way, how to calculate this ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My torus has center [0,0,0] and R, r diameters&#xA;Line (Ray) is point and its direction vector&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-10-26T02:46:53.520" LastActivityDate="2010-10-26T02:46:53.520" Title="Annuloid (Torus)-line intersection" Tags="&lt;geometry&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" />
  <row Id="7830" PostTypeId="2" ParentId="7827" CreationDate="2010-10-25T17:49:25.830" Score="0" ViewCount="" Body="&lt;p&gt;I don't understand why the book wants you to find the max and min.  You have correctly deduced that the derivative is never zero, which says there isn't a max or min.  Looking for the root of this function is not hard.  If you graph it over a reasonable range, say -5 to 5, you will find it close enough to get N correctly&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-10-25T17:49:25.830" CommentCount="2" />
  <row Id="7831" PostTypeId="2" ParentId="7827" CreationDate="2010-10-25T17:50:03.477" Score="1" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\rm\ e^{-x}\:$ and $\rm\: -x\: $ are both strictly descreasing on $\:\mathbb R\:$, hence so is their sum + 2.&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-10-25T17:50:03.477" />
  <row Id="7832" PostTypeId="2" ParentId="7829" CreationDate="2010-10-25T18:18:08.710" Score="1" ViewCount="" Body="&lt;p&gt;What kind of speed are you looking for?  In general, smart interval-based solvers work well for this sort of thing: the core idea is that by using interval arithmetic you can bound the values of the 'potential function' (the quartic polynomial whose roots you're trying to find) within any given interval, and then fast-reject if those bounds don't contain zero - that is, if you can confirm that the function is always-positive or always-negative on the interval.  Find starting values of t&lt;sub&gt;min&lt;/sub&gt; and t&lt;sub&gt;max&lt;/sub&gt; by intersecting your ray with the bounding box of the torus, use the interval arithmetic mentioned above to determine whether it's possible for an intersection to appear within that range, and then use binary subdivision (always exploring the 'earlier' - closer to the camera - branch of the subdivision first) to narrow in on the first zero to whatever accuracy is desired.  In practice this isn't much slower than explicitly solving the quartic, and it tends to be more robust, particularly around tangent points/double zeroes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One question I think it's worth asking, though: why are you trying to explicitly handle ray-torus intersection?  Tori don't, in general, make very good primitives for composing a scene; they're &lt;i&gt;too&lt;/i&gt; geometric and not particularly organic enough.  IMHO you're much better off going the pile-of-triangles route and giving yourself more flexibility, rather than tying yourself too closely to complicated geometric primitives.&lt;/p&gt;&#xA;" OwnerUserId="785" LastActivityDate="2010-10-25T18:18:08.710" CommentCount="7" />
  <row Id="7833" PostTypeId="2" ParentId="7413" CreationDate="2010-10-25T18:48:08.273" Score="0" ViewCount="" Body="&lt;p&gt;If $X$ and $Y$ are random variables, then their sum $X+Y$ is the convolution of their distributions. &lt;/p&gt;&#xA;" LastActivityDate="2010-10-25T18:48:08.273" CommentCount="1" />
  <row Id="7835" PostTypeId="1" AcceptedAnswerId="7841" CreationDate="2010-10-25T19:32:00.323" Score="1" ViewCount="297" Body="&lt;p&gt;I am stuck while proving this identity I verified it using induction but like the other two &lt;a href=&quot;http://math.stackexchange.com/questions/7757/how-to-prove-this-binomial-identity-sum-r0n-r-n-choose-r-n2n-1&quot;&gt;(1)&lt;/a&gt; &lt;a href=&quot;http://math.stackexchange.com/questions/7778/proving-2n-choose-n-2n-frac-1-cdot-3-cdot-5-cdots-2n-1n&quot;&gt;(2)&lt;/a&gt;, I am seeking a more of a general way (algebraic will be much appreciated)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac {1}{(n-1)!} + \frac {1}{3!(n-3)!} + \frac {1}{5!(n-5)!} +\frac {1}{7!(n-7)!} + \cdots = \frac {2^{n-1}}{(n)!}  $$&lt;/p&gt;&#xA;" OwnerUserId="2109" LastActivityDate="2010-10-26T05:11:42.643" Title="Showing $\frac {1}{(n-1)!} + \frac {1}{3!(n-3)!} + \frac {1}{5!(n-5)!} +\frac {1}{7!(n-7)!} + \cdots = \frac {2^{n-1}}{(n)!} $" Tags="&lt;algebra-precalculus&gt;&lt;binomial-coefficients&gt;" AnswerCount="4" CommentCount="12" />
  <row Id="7837" PostTypeId="1" AcceptedAnswerId="7844" CreationDate="2010-10-25T20:03:30.947" Score="4" ViewCount="293" Body="&lt;p&gt;Let $r$, $s$ be positive real numbers and $\theta$, $\phi$ real numbers with $|\theta -\phi|&amp;lt;\pi$. Then an argument of $re^{i\theta}+se^{i\phi}$ lies between $\theta$ and $\phi$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can someone give a short, clean proof of the statement above that doesn't rely on geometric intuition? Of course the proof may use the not so trivial fact that every nonzero complex number can be written in polar form as well as trigonometrical functions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have tried, but I have no idea how to start, probably because it seems so obvious.&lt;/p&gt;&#xA;" OwnerUserId="1878" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-10-26T03:08:52.697" LastActivityDate="2010-10-26T03:08:52.697" Title="Argument of the sum of two complex numbers" Tags="&lt;complex-numbers&gt;&lt;analytic-geometry&gt;" AnswerCount="2" />
  <row Id="7839" PostTypeId="2" ParentId="7835" CreationDate="2010-10-25T20:07:05.583" Score="0" ViewCount="" Body="&lt;p&gt;Seems easy is $n$ is odd. Multiplying both sides by $n!$ we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ {n \choose 1} + {n \choose 3} + ...  {n \choose n} = \frac{1}{2} \left[  {n \choose 0} + {n \choose 1} + ... {n \choose n}  \right]  = 2^{n-1}$$&lt;/p&gt;&#xA;" OwnerUserId="312" LastActivityDate="2010-10-25T20:07:05.583" />
  <row Id="7840" PostTypeId="1" AcceptedAnswerId="7851" CreationDate="2010-10-25T20:10:21.757" Score="3" ViewCount="246" Body="&lt;p&gt;If I have a doubly stochastic matrix, how can I find the set of all BFS?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Doubly_stochastic_matrix&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Doubly_stochastic_matrix&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2014" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-10-26T03:06:45.383" LastActivityDate="2011-05-10T19:52:52.927" Title="Finding all n×n permutation matrices" Tags="&lt;algorithms&gt;&lt;matrices&gt;&lt;linear-programming&gt;&lt;permutations&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="7841" PostTypeId="2" ParentId="7835" CreationDate="2010-10-25T20:15:22.850" Score="5" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(1+x)^{n} - (1-x)^{n} = 2 \Biggl[ {n \choose 1} x + {n \choose 3} x^{3} + \cdots \Biggr]$$&lt;/p&gt;&#xA;" LastActivityDate="2010-10-26T04:02:26.553" CommentCount="1" />
  <row Id="7842" PostTypeId="2" ParentId="7835" CreationDate="2010-10-25T20:18:02.787" Score="1" ViewCount="" Body="&lt;p&gt;For even $n$, look at the coefficient of $x^n$ in the expansion of $\sinh^2 x = \frac12(\cosh 2x-1)$. For odd $n$, modify this idea suitably.&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2010-10-25T20:18:02.787" />
  <row Id="7843" PostTypeId="2" ParentId="7837" CreationDate="2010-10-25T20:29:11.647" Score="4" ViewCount="" Body="&lt;p&gt;It's enough to do the case $\theta=0$ (just factor out $e^{i\theta}$).  Assume $0&amp;lt;\phi&amp;lt;\pi/2$. Then&#xA;$$0 &amp;lt; \tan\arg(r+s e^{i\phi}) = \frac{s\sin\phi}{r+s\cos\phi} &amp;lt; \frac{s\sin\phi}{s\cos\phi} = \tan\phi,$$&#xA;so the argument of the sum is between $0$ and $\phi$ as desired. Something similar should work when $\phi$ is obtuse.&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2010-10-25T20:29:11.647" />
  <row Id="7844" PostTypeId="2" ParentId="7837" CreationDate="2010-10-25T20:31:50.457" Score="5" ViewCount="" Body="&lt;p&gt;You want to show that if you write $re^{i\theta}  + se^{i\phi}$ as $te^{i\chi}$, then $\chi$ lies between $\theta$ and $\phi$. Without loss of generality, assume $\theta\leq\phi$. Factoring out $e^{i\theta}$ you get $re^{i\theta}+se^{i\phi} = e^{i\theta}(r + se^{i(\phi-\theta)})$. Since multiplying by $e^{i\theta}$ is just a rotation by an angle $\theta$, it is enough to consider the case where $\theta=0$ and $0\lt \phi\lt \pi$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In that case, you have $r + se^{i\phi} = r + s(\cos(\phi)+i\sin(\phi))$, and you want to express it in the form $t(\cos(\chi) + i \sin(\chi))$. Looking at real and complex parts, you see that $t\sin(\chi)=s\sin(\phi)$, and $r+s\cos(\phi) = t\cos(\chi)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume first that $0\lt \phi\lt \frac{\pi}{2}$. Then $\chi$ must also lie in the first quadrant, since we need both $\sin(\chi)$ and $\cos(\chi)$ to be positive (since $t$, $s\sin(\phi)$, and $r+s\cos(\phi)$ are all positive).  If, on the other hand, $\frac{\pi}{2}\lt \phi\lt \pi$, then $\cos(\phi)$ is negative. If $r+s\cos(\phi)\gt 0$, then we need $\cos(\chi)\gt 0$, so $\chi$ is in the first quadrant and automatically smaller than $\phi$ and we are done.  If $r+s\cos(\phi)$ is negative then we need $\chi$ in the second quadrant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the former case, $0\lt \chi,\phi\lt \frac{\pi}{2}$, then from $\sin(\chi)=\frac{s}{t}\sin(\phi)$ and since $\sin(x)$ is increasing on $0\leq x\leq\frac{\pi}{2}$, then $0\lt\chi\lt\phi$ if and only if $\frac{s}{t}\lt 1$, if and only if $s\lt t$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now note that $t^2 = ||r+se^{i\phi}||^2 = r^2 + s^2 + 2rs\cos(\phi) \gt s^2$, since all of $r$, $s$, and $\cos(\phi)$ are positive. Since $s$ and $t$ are both positive, $s\lt t$, which shows that $0\lt\chi\lt\phi$, as desired.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the other case, where $\frac{\pi}{2}\lt\phi\lt \pi$ and $r+s\cos(\phi)\lt 0$, then we know $\chi$ is also in the second quadrant where $\sin(x)$ is decreasing, so from $\sin(\chi)=\frac{s}{t}\sin(\phi)$ we get that $\frac{\pi}{2}\lt \chi\lt \phi$ if and only if $\frac{s}{t}\gt 1$, if and only if $s\gt t$. Here, since $\cos(\theta)\lt 0$, then you have again $t^2 = ||r+se^{i\phi}||^2 = s^2 + r(r + 2s\cos(\phi)) \lt s^2$ (since $r+s\cos(\phi)\lt 0$ in this situation), so you get $t\lt s$ and hence $\frac{\pi}{2}\lt\chi\lt\phi$, as desired&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-10-25T20:45:40.010" LastActivityDate="2010-10-25T20:45:40.010" CommentCount="2" />
  <row Id="7845" PostTypeId="1" CreationDate="2010-10-25T20:52:06.173" Score="1" ViewCount="173" Body="&lt;p&gt;Suppose we observe 84 alchoholics with cirrhosis of the liver, of whom 29 have hepatomas, that is, a liver-cell carcinoma. Suppose we know, based on a large sample, that the risk of hepatoma among alcoholics without cirrhosis of the liver is 24%.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the probability that we observe exactly 29 alcoholics with cirrhosis of the liver who have hepatomas if the true rate of hepatoma among alcoholics (with or without cirrhosis of the liver) is .24?&lt;/p&gt;&#xA;" LastActivityDate="2010-10-25T21:24:42.780" Title="Question from Fundamentals of Biostatistics" Tags="&lt;homework&gt;&lt;statistics&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="7846" PostTypeId="1" AcceptedAnswerId="7855" CreationDate="2010-10-25T20:54:12.523" Score="1" ViewCount="198" Body="&lt;p&gt;UPDATE: I feel like my original question was too vague and didn't provide enough information (as others have mentioned). So I'm going to restate it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Purpose:&lt;/strong&gt; I want to find an equation in order to provide a smooth slope for an animation that I am creating. My animation consists of a lever and a set of numbers on a horizontal line. As you move the lever up and down the numbers will slide down and up the line (opposite of the lever). I want to create a smooth curve so that the user can &quot;feel&quot; the speed increasing the further they get away from the center. Most of the time, the user will stay within the main part of the screen, therefore I want the sliding numbers to go slowly for that space. But once they get closer to the top or bottom I want the increase in speed to be relatively quick.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Requirements:&lt;/strong&gt; The variables are the distance from center that the lever is, $x$, and the distance that the numbers should move, $y$. I need $x=y=0$ so there is no movement at the center. $y$ should be positive when $x$ is positive, but $y$ can be positive or negative when $x$ is negative. The graph just needs to be symmetric for $x=0$ or $x=y$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Right now I'm using a simple parabola $y = (0.03*x)^2$ but I would like the middle space to be &quot;stretched out.&quot; Therefore, is there a function that I can use that would have some constants (maybe some points) to define what kind of slope is found for, lets say $x=0$ to $x=100$ and then it curving to another point at $x=160$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't want to use piecewise because unless I use a lot of pieces, the user will be able to &quot;feel&quot; the positions where the equation changes. I think Quadratic Bézier curve might do the trick, but if it can, I would need help in understanding how to use it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; is the Quadratic Bézier curve what I should be looking into, if not, what do you suggest? If it is, do you have a suggestion of what equation I should use or at least a reference were I can relearn it.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Old question:&#xA;&lt;strike&gt;I need to come up with an equation for a curve but I have forgotten most of my calculus by this point. What I need very close to a parabola that opens up. I'm currently using y = (0.03*x)^2 but this does not have a fast enough slope on the ends.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need the slope in the middle to be very gradual, and then near the ends to increase faster.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm guessing this won't be a standard parabola, but I'm not just sure what I should be looking for.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;UPDATE: Attempting to be more detailed with my &quot;requirements.&quot; This is for a computer animation and a user is sliding a &quot;lever&quot; up and down. As they do so a set of numbers are scrolling the opposite direction on the side. The further away the lever is from center, the faster the numbers should go. Thus, I want the edges (top and bottom) to make the numbers go much faster. But for most of the main section of the screen, they shouldn't go so fast. But I want the transition from slow to fast to be noticeable so that it feels intuitive to the user.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, I need a function that has a high value for y as x gets larger, but I want the edges to have a greater slope. Bézier curve looks like it might work? But it has been a long time since I've used it. Anyone have suggestions for how to use it in this context?&lt;/strike&gt;&lt;/p&gt;&#xA;" OwnerUserId="2798" LastEditorUserId="2798" LastEditorDisplayName="" LastEditDate="2010-10-26T16:51:11.110" LastActivityDate="2010-10-26T21:48:28.837" Title="Parabola attributes - custom curve?" Tags="&lt;calculus&gt;&lt;polynomials&gt;&lt;bezier-curve&gt;&lt;conic-sections&gt;" AnswerCount="1" CommentCount="7" />
  <row Id="7847" PostTypeId="2" ParentId="7628" CreationDate="2010-10-25T21:00:23.740" Score="7" ViewCount="" Body="&lt;p&gt;It is &lt;strong&gt;not&lt;/strong&gt; true that the Grothendieck ring of coherent sheaves on $\mathbb{P}^1$ is isomorphic to $\mathbb{Z}[t, t^{-1}]$. Although $\mathcal{O} \oplus \mathcal{O}(2)$ is not isomorphic to $\mathcal{O}(1) \oplus \mathcal{O}(1)$, they do have the same class in $K^0$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The definition of the Grothendieck group of coherent sheaves on a scheme $X$ is that it is generated by isomorphism classes of coherent sheaves, modulo the relation that $[A] + [C] = [B]$ whenever there is a short exact sequence&#xA;$$0 \to A \to B \to C \to 0.$$&#xA;In particular, we have the short exact sequence&#xA;$$0 \to \mathcal{O} \to \mathcal{O}(1)^2 \to \mathcal{O}(2) \to 0,$$&#xA;where the maps are given by $(x \ y)$ and $\binom{-y}{x}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This makes $K^0$ into $\mathbb{Z}[t, t^{-1}]/(t^2 - 2t +1) \cong \mathbb{Z}[u]/u^2$, just like you wanted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When working in the categories of smooth or of topological vector bundles, all short exact sequences split, so you can get away with defining $K$-theory with direct sums. You can't do that in the coherent or the algebraic categories.&lt;/p&gt;&#xA;" OwnerUserId="448" LastActivityDate="2010-10-25T21:00:23.740" CommentCount="1" />
  <row Id="7848" PostTypeId="2" ParentId="5159" CreationDate="2010-10-25T21:07:05.927" Score="1" ViewCount="" Body="&lt;p&gt;As you have thought you should solve&#xA;$$\left\{\begin{aligned}3y^2-1=0,\\x^3+y^3-y=0\end{aligned}\right.&#xA;$$&#xA;which gives you 6 ramification points $y=\pm 1/\sqrt{3}, x=\pm e^{2\pi i k/3}\sqrt[3]{2}/\sqrt{3}$ where $k=0,1,2$.&lt;/p&gt;&#xA;" OwnerUserId="695" LastActivityDate="2010-10-25T21:07:05.927" />
  <row Id="7849" PostTypeId="2" ParentId="7783" CreationDate="2010-10-25T21:16:30.090" Score="9" ViewCount="" Body="&lt;p&gt;First, one uses the argument principle from complex analysis to compute the number of zeroes with real parts between 0 and 1 and with imaginary parts between 0 and some postive number $T$.  Since the answer is &lt;em&gt;a priori&lt;/em&gt; an integer, one can get a precise answer even though one can't compute the $\zeta$-function exactly; it is &quot;just&quot; a matter of bounding the error in all approximations carefully enough.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, one has to show that this number of zeroes actually lie on the line where $\Re s = 1/2$.&#xA;For this, one factors $\zeta(1/2 + i t)$ as a product of a certain nowhere zero function,&#xA;and a real valued function $Z(t)$ (the function appearing in J.M.'s answer).  (That this can&#xA;be done in a useful way is part of the theory of the $\zeta$-function.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To count zeroes of $\zeta$ with $\Re s = 1/2$ is now the same as counting zeroes of $Z(t)$,&#xA;which can be determined by sign changes in $Z(t)$ (provided one estimates $Z(t)$ accurately enough).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming that all works out (which it will if RH is true!) one gets a rigorous proof that all zeroes with imaginary part $\leq T$ lie on the line $\Re s = 1/2$.  (In fact, for this to work out, one also needs the zeroes &#xA;of $\zeta$ to be simple, which they are conjectured to be; see T..'s answer&#xA;for an elaboration on this.)&lt;/p&gt;&#xA;" OwnerUserId="221" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2011-01-03T01:30:23.843" LastActivityDate="2011-01-03T01:30:23.843" CommentCount="3" />
  <row Id="7850" PostTypeId="2" ParentId="7845" CreationDate="2010-10-25T21:24:42.780" Score="4" ViewCount="" Body="&lt;p&gt;When you have $n$ people, with probability $p$ for each of them to have hepatoma, and you want to know the probability that exactly $k$ of them have hepatoma, then the formula is given by:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$${n \choose k}p^k(1-p)^{n-k}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your case, $n$ = 84, $k$ = 29, $p$ = 0.24. This is called the Binomial Distribution. You can learn more about it  here: &lt;a href=&quot;http://en.wikipedia.org/wiki/Binomial_distribution&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Binomial_distribution&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2778" LastActivityDate="2010-10-25T21:24:42.780" />
  <row Id="7851" PostTypeId="2" ParentId="7840" CreationDate="2010-10-25T22:07:13.637" Score="6" ViewCount="" Body="&lt;p&gt;Don Knuth's Volume 4, Fascicle 2, of &lt;em&gt;The Art of Computer Programming&lt;/em&gt; has a long section on generating all permutations, including algorithms for doing so.  I found a draft &lt;a href=&quot;http://www.google.com/url?sa=t&amp;amp;source=web&amp;amp;cd=1&amp;amp;ved=0CBMQFjAA&amp;amp;url=http%3A%2F%2Fwww-cs-faculty.stanford.edu%2F~uno%2Ffasc2b.ps.gz&amp;amp;rct=j&amp;amp;q=generating%20all%20permutations%20knuth&amp;amp;ei=AvfFTMm_LouksQPf8-DLDQ&amp;amp;usg=AFQjCNEw5tfBK7QyvtTni9utWTNbPAJbIQ&amp;amp;sig2=rwj1e_RcSsJgDYcSpYNI-w&amp;amp;cad=rja&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; online.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, going from a permutation to a permutation matrix is fairly straightforward.  For example, suppose you have the permutation 1342 of the numbers 1, 2, 3, and 4.  That can be represented in two-line form as &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{matrix}1&amp;amp;2&amp;amp;3&amp;amp;4\\1&amp;amp;4&amp;amp;2&amp;amp;3\end{matrix}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;because the permutation sends 1 to the first position, 2 to the fourth position, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then the permutation matrix is the matrix with 1's in entries (1,1), (2,4), (3,2), (4,3), and 0's elsewhere; i.e.,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{pmatrix}1&amp;amp;0&amp;amp;0&amp;amp;0\\0&amp;amp;0&amp;amp;0&amp;amp;1\\0&amp;amp;1&amp;amp;0&amp;amp;0\\0&amp;amp;0&amp;amp;1&amp;amp;0\end{pmatrix}$$&lt;/p&gt;&#xA;" OwnerUserId="2370" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2011-05-10T19:52:52.927" LastActivityDate="2011-05-10T19:52:52.927" CommentCount="4" />
  <row Id="7852" PostTypeId="2" ParentId="7783" CreationDate="2010-10-25T22:58:01.737" Score="13" ViewCount="" Body="&lt;p&gt;As long as there are no multiple zeros on the line up to height  $T$  you can determine, rigorously and using a finite amount of computation:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;the number of zeros (counted with multiplicity) off the critical line, i.e., the number of counterexamples to the Riemann hypothesis, of height less than $T$.   This is done by contour integrals of $d \log \zeta$ -- the &quot;argument principle&quot; in complex analysis.  For each zero one can locate it to any specified accuracy, and determine its multiplicity.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;the number of zeros on the critical line, up to height $T$.  For each zero, its ordinate on the line can be calculated to any desired accuracy.  This is done by counting sign changes of real-analytic functions with a $\zeta (1/2 + it)$ factor, on the critical line.  As Matt E mentioned, the ability to write down a real-valued function capturing the behavior of a complex function on a line is a special phenomenon particular to zeta and L-functions, coming from the functional equation relating $\zeta(s)$ to $\zeta(1-s)$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;If there is a multiple zero of order $n$ at some point on the line, all you can determine by computations of the above type is that a very small neighborhood of that point contains $n$ zeros (counted with multiplicity).  This could mean zeros on the line, counterexamples to the Riemann conjecture, or both.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is part of the package of conjectures surrounding the Riemann hypothesis, that there are no multiple zeros, and that the zeros in fact &quot;repel&quot; each other in a quantifiable sense. There is a lot of numerical and theoretical support for the zero repulsion, especially from random matrix theory.  If the Riemann hypothesis is true but there are multiple zeros, that would be almost as surprising as finding a zero off the line.&lt;/p&gt;&#xA;" OwnerUserId="467" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-10-25T23:17:49.193" LastActivityDate="2010-10-25T23:17:49.193" />
  <row Id="7853" PostTypeId="1" CreationDate="2010-10-25T23:17:41.890" Score="4" ViewCount="179" Body="&lt;p&gt;I'm searching for an algorithm that forms a balanced (or quasi-complete) latin square, in which every element is a horizontal neighbor to every other element exactly twice, and a vertical neighbor to every other element exactly twice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've found one example (for &lt;strong&gt;n = 5&lt;/strong&gt;), but am not clear about a couple of steps:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Step 1&lt;/strong&gt;: Write &lt;strong&gt;1 2 ... n&lt;/strong&gt; as the first row&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;12345&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Step 2&lt;/strong&gt;: For &lt;strong&gt;i&lt;/strong&gt; even in the first row, fill in the diagonal from upper left to lower right starting with &lt;strong&gt;i&lt;/strong&gt; and alternating with &lt;strong&gt;i - 1&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;12345&#xA;  1 3&#xA;   2&#xA;    1&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Step 3&lt;/strong&gt;: For &lt;strong&gt;i&lt;/strong&gt; odd and less than &lt;strong&gt;n&lt;/strong&gt; in the first row fill in the diagonal from upper right to lower left beginning with &lt;strong&gt;i&lt;/strong&gt; and alternating with &lt;strong&gt;i + 1&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;12345&#xA; 41 3&#xA;3  2&#xA;    1&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Step 4&lt;/strong&gt;: Fill in &lt;strong&gt;n&lt;/strong&gt; for the main off diagonal&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;12345&#xA; 4153&#xA;3 52&#xA; 5  1&#xA;5&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Step 5&lt;/strong&gt;: For the last column write &lt;strong&gt;n n-2 n-1 n-3 ... 1 2&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;12345&#xA; 4153&#xA;3 524&#xA; 5  1&#xA;5   2&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Step 6&lt;/strong&gt;: For the even entries of the last column, fill in the diagonal from upper left to lower right beginning with &lt;strong&gt;i&lt;/strong&gt; and alternating with &lt;strong&gt;i - 1&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;?&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Step 7&lt;/strong&gt;: Complete by symmetry about the main diagonal&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;?&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;For &lt;strong&gt;n = 5&lt;/strong&gt;, the following is obtained:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;12345&#xA;24153&#xA;31524&#xA;45231&#xA;53412&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Questions...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[RE: &lt;strong&gt;Step 5&lt;/strong&gt;] I was able to fill in that portion of the square, but am not sure what would come after &lt;strong&gt;n-3&lt;/strong&gt; in a larger sequence, as it would go &lt;strong&gt;n n-2 n-1 n-3 ??? ... 1 2&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[RE: &lt;strong&gt;Step 6&lt;/strong&gt;] Are even entries even numbers, or does that refer to even positions in the column? Is &quot;i&quot; in this case the number at the upper left or is it the even entry?  I don't see how the completed square corresponds to the instructions in this step.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[RE: &lt;strong&gt;Step 7&lt;/strong&gt;] What does &quot;complete by symmetry about the main diagonal&quot; mean?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Otherwise if there's a less involved algorithm for balancing a latin square, I'd like to know how it goes.&lt;/p&gt;&#xA;" OwnerUserId="2768" LastEditorUserId="139" LastEditorDisplayName="" LastEditDate="2010-11-19T10:37:50.607" LastActivityDate="2010-12-19T10:44:26.240" Title="Balancing a Latin Square" Tags="&lt;combinatorics&gt;&lt;algorithms&gt;&lt;latin-square&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="7854" PostTypeId="1" CreationDate="2010-10-26T00:09:48.490" Score="1" ViewCount="113" Body="&lt;p&gt;When pricing a european option by monte carlo over 30 days for instance, what's the difference between one big 30 day jump vs 30 one day steps?&lt;/p&gt;&#xA;" OwnerUserId="2809" LastActivityDate="2010-10-26T03:11:49.900" Title="Stochastic/finance monte carlo question" Tags="&lt;finance&gt;&lt;monte-carlo&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="7855" PostTypeId="2" ParentId="7846" CreationDate="2010-10-26T00:41:37.683" Score="2" ViewCount="" Body="&lt;p&gt;I think the easiest way may be piecewise—not piecewise-linear, but a piecewise-defined function that is relatively smooth, so there isn't a point where the speed suddenly changes.  Let's say that you want slope $m$ for $-a\le x\le a$ (and $y=0$ when $x=0$), then curving up to the point $(b,c)$.&#xA;$$y=\begin{cases}&#xA;\frac{-a^2 c+a^2 b m}{(a-b)^2}-\frac{\left(2 a c-a^2 m-b^2 m\right) x}{(a-b)^2}+\frac{(-c+b m) x^2}{(a-b)^2} &amp;amp; \text{ if } x&amp;lt;-a&#xA;\\ &#xA;mx &amp;amp; \text{ if } -a\le x\le a&#xA;\\ &#xA;-\frac{-a^2 c+a^2 b m}{(a-b)^2}-\frac{\left(2 a c-a^2 m-b^2 m\right) x}{(a-b)^2}-\frac{(-c+b m) x^2}{(a-b)^2} &amp;amp; \text{ if } x&gt;a&#xA;\end{cases}$$&#xA;(I arrived at this formula by having Mathematica solve for the coefficients of the quadratic polynomial that would go through the point at the end of the linear segment and the other given point and have the same slope as the linear segment at the splice point: &lt;code&gt;a1 x^2 + a2 x + a3 /. &#xA; Solve[a1 a^2 + a2 a + a3 == a m &amp;amp;&amp;amp; 2 a1 a + a2 == m &amp;amp;&amp;amp; &#xA;   a1 b^2 + a2 b + a3 == c, {a1, a2, a3}]&lt;/code&gt;.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, if you wanted a slope of $\frac{1}{4}$ from -100 to 100, then curving up to (180,150), $a=100$, $m=\frac{1}{4}$, $b=180$, $c=150$, and $$y=\begin{cases}&#xA;-\frac{2625}{16}-\frac{97 x}{32}-\frac{21 x^2}{1280} &amp;amp; \text{ if } x&amp;lt;-100&#xA;\\ &#xA;\frac{1}{4}x &amp;amp; \text{ if } -100\le x\le 100&#xA;\\ &#xA;\frac{2625}{16}-\frac{97 x}{32}+\frac{21 x^2}{1280} &amp;amp; \text{ if } x&gt;100&#xA;\end{cases}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/FW2qS.png&quot; alt=&quot;graph of example&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;edit&lt;/em&gt;: even smoother would be to use a cubic polynomial for the end parts and force it to match the first and second derivatives at the splice points (&lt;code&gt;a1 x^3 + a2 x^2 + a3 x + a4 /. &#xA; Solve[a1 a^3 + a2 a^2 + a3 a + a4 == a m &amp;amp;&amp;amp; &#xA;   3 a1 a^2 + 2 a2 a + a3 == m &amp;amp;&amp;amp; 6 a1 a + 2 a2 == 0 &amp;amp;&amp;amp; &#xA;   a1 b^3 + a2 b^2 + a3 b + a4 == c, {a1, a2, a3, a4}]&lt;/code&gt;):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\begin{cases}&#xA; \frac{-a^3 c+a^3 b m}{(a-b)^3}-\frac{\left(3 a^2 c-a^3 m-3 a b^2 m+b^3 m\right) x}{(a-b)^3}+\frac{3 a (-c+b m) x^2}{(a-b)^3}-\frac{(c-b m) x^3}{(a-b)^3} &amp;amp; x&amp;lt;-a \\&#xA; m x &amp;amp; -a\leq x\leq a \\&#xA; -\frac{-a^3 c+a^3 b m}{(a-b)^3}-\frac{\left(3 a^2 c-a^3 m-3 a b^2 m+b^3 m\right) x}{(a-b)^3}-\frac{3 a (-c+b m) x^2}{(a-b)^3}-\frac{(c-b m) x^3}{(a-b)^3} &amp;amp; x&gt;a&#xA;\end{cases}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, if you wanted a slope of $\frac{1}{4}$ from -100 to 100, then curving up to (180,150), as above, $a=100$, $m=\frac{1}{4}$, $b=180$, $c=150$, and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$y=\begin{cases}&#xA; \frac{13125}{64}+\frac{1639 x}{256}+\frac{63 x^2}{1024}+\frac{21 x^3}{102400} &amp;amp; x&amp;lt;-100 \\&#xA; \frac{x}{4} &amp;amp; -100\leq x\leq 100 \\&#xA; -\frac{13125}{64}+\frac{1639 x}{256}-\frac{63 x^2}{1024}+\frac{21 x^3}{102400} &amp;amp; x&gt;100 &#xA;\end{cases}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/Mfp1H.png&quot; alt=&quot;graph of example&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;original answer&lt;/em&gt;:&#xA;Would $y=\cot(3x+0.07)$ for $0\le x\le 1$ work?&#xA;Or $y=-\frac{x-0.5}{(x+0.1)(x-1.1)}$?&lt;/p&gt;&#xA;" OwnerUserId="72" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-10-26T21:48:28.837" LastActivityDate="2010-10-26T21:48:28.837" CommentCount="8" />
  <row Id="7856" PostTypeId="2" ParentId="7491" CreationDate="2010-10-26T00:43:34.470" Score="1" ViewCount="" Body="&lt;p&gt;Please see the answer in &lt;a href=&quot;http://mathoverflow.net/questions/43181/what-will-be-the-distribution-of-ratio-of-correlated-gamma-distributed-random-var&quot; rel=&quot;nofollow&quot;&gt;link text&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2810" LastActivityDate="2010-10-26T00:43:34.470" />
  <row Id="7857" PostTypeId="2" ParentId="7853" CreationDate="2010-10-26T01:29:18.693" Score="2" ViewCount="" Body="&lt;p&gt;The Latin square L:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;12345&#xA;24153&#xA;31524&#xA;45231&#xA;53412&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;is isomorphic to the Cayley table of the cyclic group of order 5 via the isomorphism (3,5,4).  That is, if you permute the rows, columns and symbols of L according to the permutation (3,5,4) you will generate&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;12345&#xA;23451&#xA;34512&#xA;45123&#xA;51234&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The n=7 case gives rise to the Latin square&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;1234567&#xA;2416375&#xA;3152746&#xA;4627153&#xA;5371624&#xA;6745231&#xA;7563412&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;which is isomorphic to the Cayley table of the cyclic group of order 7 via the isomorphism (3,7,5,6,4).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With this in mind, my suspicion is that these Latin squares are special cases of those generated in the constructions described by Rosemary Bailey in &lt;a href=&quot;http://www.jstor.org/pss/2345518&quot; rel=&quot;nofollow&quot;&gt;Quasi-complete Latin squares: construction and randomization&lt;/a&gt;, although I haven't gone into the details.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the smaller questions:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;In papers, by an &lt;em&gt;entry&lt;/em&gt; I exclusively refer to a triple $(i,j,l_{ij})$, where $i$ is the row index, $j$ is the column index and $l_{ij}$ is the symbol in the $(i,j)$-th cell.  For example (2,3,1) would be an entry of the first Latin square.  Importantly, in an entry, the position in the Latin square is important, whereas in a symbol, there is no notion of position (aside from being somewhere in the Latin square).  Consequently, I would use the term &quot;even symbol&quot; (although note that the symbols you use make no difference to whether or not you have a quasi-complete Latin square).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Latin squares L such that $L^T=L$ are called symmetric Latin squares (just as for symmetric matrices).  When giving details of a construction of these types of Latin squares, I find myself writing &quot;...then L admits a unique completion to a symmetric Latin square&quot; or similar.  There's no need to identify every entry since some entries are determined by other entries.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="139" LastActivityDate="2010-10-26T01:29:18.693" CommentCount="3" />
  <row Id="7858" PostTypeId="2" ParentId="7854" CreationDate="2010-10-26T02:24:28.757" Score="1" ViewCount="" Body="&lt;p&gt;You are in effect replacing many detailed paths by their average effect. This changes the answer since pricing options involves applying the non-linear $\max$ or $\min$ operators. If you are trying to compute the answer corresponding to a continuous time model, then you have to take a large enough number of steps.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It may help your intuition to draw two binomial tree models, one with one period and one with two periods and solve them by hand.&lt;/p&gt;&#xA;" OwnerUserId="1195" LastActivityDate="2010-10-26T02:24:28.757" CommentCount="1" />
  <row Id="7859" PostTypeId="2" ParentId="7854" CreationDate="2010-10-26T03:11:49.900" Score="2" ViewCount="" Body="&lt;p&gt;European options have an efficiently computable deterministic answer in both the continuous time and discrete lattice (binomial tree) settings.  Unless you are outside the Black-Scholes framework, random walk simulations are needed only for path-dependent options.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are asking about the accuracy of the discretization, it is not only a function of the number of steps.  Both the &quot;time&quot; and &quot;money&quot; coordinates are discretized, and the random walk answer may not converge at all unless the time and price steps are sized correctly relative to each other.&lt;/p&gt;&#xA;" OwnerUserId="467" LastActivityDate="2010-10-26T03:11:49.900" />
  <row Id="7860" PostTypeId="1" AcceptedAnswerId="7861" CreationDate="2010-10-26T03:20:40.857" Score="2" ViewCount="87" Body="&lt;p&gt;I am given a lemma and it states:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Lemma: Let A be a real n x n matrix.  Then given any e &gt; 0, there is a norm such that &lt;code&gt;norm(A) &amp;lt;= p(A) + e&lt;/code&gt;&lt;/strong&gt;  Where p(A) is the spectral radius of A.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They then go on to state that based on this lemma, &lt;code&gt;if p(A) &amp;lt; 1, then norm(A) &amp;lt; 1&lt;/code&gt; for the correct choice of the norm.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I'm looking for help on is understanding how they came to that conclusion based only on that lemma.  For example, given &lt;code&gt;e = .001&lt;/code&gt;, then the lemma states there is a norm such that &lt;code&gt;norm(A) &amp;lt;= p(A) + .001&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since it is a &quot;or equal&quot; then this means there is the possibility that &lt;code&gt;norm(A) = p(A) + .001&lt;/code&gt;, and in such a case then &lt;code&gt;norm(A) &amp;gt; p(A)&lt;/code&gt;.  So it seems that based on the lemma, then &lt;code&gt;p(A) &amp;lt; 1&lt;/code&gt; does not necessarily always imply that &lt;code&gt;norm(A) &amp;lt; 1&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I even tried to do a proof of this but still came up with the same result:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;p(A) &amp;lt; 1 -&amp;gt; &#xA;p(A) + e &amp;lt; 1 + e -&amp;gt; &#xA;norm(A) &amp;lt;= p(A) +e &amp;lt; 1 + e&#xA;norm(A) - e &amp;lt;= p(A) &amp;lt; 1&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This pretty much says the same thing, if norm(A) - e &amp;lt;= p(A), then p(A) &amp;lt; 1, but if p(A) close to 1 by a difference less than e, then norm(A) &gt; 1 in the &quot;equal to&quot; case where norm(A) - e = p(A)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now of course you could pick a smaller e, but the way the lemma reads it says that for a given e, you can find a norm s.t. A &amp;lt;= p(A) + e, so the &quot;or equal&quot; part will still get you using a counter example I gave similar to the e = .001 above&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I have seen a different proof on wikipedia that shows if p(A) &amp;lt; 1 then norm(A) &amp;lt; 1, so I am not disputing that fact, but that proof gets into looking at entries of the Jordan Normal Form.  So my point is, yes p(A) &amp;lt; 1 then norm(A) &amp;lt; 1, &lt;strong&gt;BUT&lt;/strong&gt; I don't understand how you can conclude that based only on the Lemma I gave at the beginning.  It's these big jumps in rational that confuse me.&lt;/p&gt;&#xA;" OwnerUserId="2812" LastActivityDate="2010-10-26T03:26:41.473" Title="How does norm(A) &lt;= p(A) + e imply norm(A) &lt; 1 if p(A) &lt; 1 ?" Tags="&lt;matrices&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="7861" PostTypeId="2" ParentId="7860" CreationDate="2010-10-26T03:26:41.473" Score="3" ViewCount="" Body="&lt;p&gt;Apply the lemma with $e=\frac{1}{2}(1-p(A))$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, the fact that it is not strict inequality is insignificant.  Given $e$, you could apply the lemma with $e/2$ to get strict inequality with $e$.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-10-26T03:26:41.473" />
  <row Id="7862" PostTypeId="1" AcceptedAnswerId="7863" CreationDate="2010-10-26T03:51:54.073" Score="3" ViewCount="117" Body="&lt;p&gt;Two questions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1.) Construct a subset of $[0,1]$ in the same manner as the Cantor set, except that at the $k$-th stage, each interval removed has length $\delta 3^{-k}$, $0&amp;lt;\delta &amp;lt;1$. Show that the resulting set is perfect, has measure $1-\delta$, and contains no intervals. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Showing that it's perfect is not difficult. The resulting set is an intersection of closed intervals, so is closed. Any point in the set is a limit point of endpoints of intervals (obviously I have more details in my written solution, but that's the idea.) But I don't see how the measure is $1-\delta$... At stage $k$, we remove $2^{k-1}$ intervals of length $\delta 3^{-k}$, so the measure of the resulting set is $$1-\sum\limits_{k=0}^{\infty}{2^{k}\cdot\dfrac{\delta}{3^{k+1}}} = 1-\dfrac{\delta}{3}\cdot\sum\limits_{k=0}^{\infty}{\dfrac{2^k}{3^k}} = 1-\dfrac{2\delta}{3} .$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thoughts?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2.) Construct a Cantor-type set subset of $[0,1]$ by removing from each interval remaining at the $k$-th stage a subinterval of relative length $\theta_k$, $0&amp;lt;\theta_k&amp;lt;1$. Show that the remainder has measure zero if and only if $\sum{\theta_k}=\infty$. (Use the fact that for $a_k&gt;0$, $\prod\limits_{k=1}^{\infty}{a_k}$ converges, in the sense that $\lim\limits_{N\to\infty}{\prod\limits_{k=1}^N{a_k}}$ exists and is not zero, if only if $\sum\limits_{k=1}^{\infty}{\log{a_k}}$ converges. )&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't really understand the construction. Suppose we have our $\theta_1$ and we remove an interval of that length. So then we choose $\theta_2$ so that $\theta_2 &amp;lt; \dfrac{1-\theta_1}{2}$, or something like that? Or... ? The phrasing is weird to me. Secondly, the hint does not help at all; it is complete opaque to me, so any insight you can offer would be great. &lt;/p&gt;&#xA;" OwnerUserId="2313" LastActivityDate="2010-10-26T04:43:44.787" Title="More Cantor-like constructions..." Tags="&lt;real-analysis&gt;&lt;homework&gt;&lt;measure-theory&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="7863" PostTypeId="2" ParentId="7862" CreationDate="2010-10-26T04:07:55.533" Score="3" ViewCount="" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Your last step in the computation is incorrect. The sum $\sum_{n=0}^{\infty}\frac{2^k}{3^k}$ is:&#xA;$$\sum_{n=0}^{\infty}\frac{2^k}{3^k} = \sum_{n=0}^{\infty}\left(\frac{2}{3}\right)^k = \frac{1}{1-\frac{2}{3}} = 3$$&#xA;so when you multiply it by $\frac{\delta}{3}$ you get $\delta$, exactly what you expect it to be. I'm not sure why you thought you get $1- \frac{2\delta}{3}$. (Recall that if $|x|&amp;lt;1$, then $\sum_{n=0}^{\infty}x^n = \frac{1}{1-x}$). &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Notice that it says &quot;relative length&quot;, not &quot;length&quot;. So at each step, you are taking out an interval which is $\theta_k\cdot\ell$, where $\ell$ is the length of the interval you are looking at. In the Cantor set, you take $\theta_k=\frac{1}{3}$ for every $k$; in the example in (1), you are setting $\theta_k = \frac{\delta}{3}$ for every $k$. If you take out the central $\theta_k\ell$ portion of an interval of length $\ell$, then you are left with two intervals of length $\frac{\ell}{2}(1-\theta_k)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Added:&lt;/em&gt; As per your question in the comments, in &lt;a href=&quot;http://math.stackexchange.com/questions/6643/cantor-like-construction&quot;&gt;this question&lt;/a&gt;, all the $\theta_k$ are constant and equal to $\theta$, much like in the regular Cantor set all $\theta_k$ are equal to $\frac{1}{3}$. To see an example, suppose that you set $\theta_k = \frac{1}{2^{k+1}}$, $k=0,1,\ldots$. First you remove $\theta_0 = \frac{1}{2}$ of the interval, leaving you with $[0,\frac{1}{4}]\cup[\frac{3}{4},1]$. Then you remove $\theta_1=\frac{1}{4}$th of each remaining interval; each interval is of length $\frac{1}{4}$, so you are removing the central $\frac{1}{16}$th part of the interval; this leaves you with $[0,\frac{3}{32}]\cup[\frac{5}{32},\frac{1}{4}]\cup[\frac{3}{4},\frac{27}{32}]\cup[\frac{29}{32},1]$. At the next step, you are removing the central $\theta_2=\frac{1}{8}$th of each interval; the intervals are of length $\frac{3}{32}$, so you are removing the central $\frac{3}{256}$th portion. Etc. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-10-26T04:43:44.787" LastActivityDate="2010-10-26T04:43:44.787" CommentCount="2" />
  <row Id="7864" PostTypeId="1" CreationDate="2010-10-26T04:24:18.303" Score="5" ViewCount="677" Body="&lt;p&gt;Over on &lt;a href=&quot;http://mathoverflow.net/&quot; rel=&quot;nofollow&quot;&gt;mathoverflow&lt;/a&gt;, there is a popular CW question titled: &lt;a href=&quot;http://mathoverflow.net/questions/23478/examples-of-common-false-beliefs-in-mathematics&quot; rel=&quot;nofollow&quot;&gt;Examples of common false beliefs in mathematics&lt;/a&gt;. I thought it would be nice to have a parallel question on this site to serve as a reference for false beliefs within less obscure mathematics. &#xA;That said, it would be good not get bogged down with misconceptions that are generally assumed to be elementary such as: $(x + y)^{2} = x^{2} + y^{2}$.&lt;/p&gt;&#xA;" OwnerUserId="19" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-10-26T15:47:19.153" LastActivityDate="2010-10-28T13:03:18.580" CommunityOwnedDate="2010-10-26T20:30:14.583" Title="False beliefs in mathematics (conceptual errors made despite, or because of, mathematical education)." Tags="&lt;big-list&gt;&lt;education&gt;&lt;teaching&gt;&lt;learning&gt;" AnswerCount="9" CommentCount="9" FavoriteCount="3" />
  <row Id="7865" PostTypeId="1" AcceptedAnswerId="7869" CreationDate="2010-10-26T04:29:06.783" Score="3" ViewCount="102" Body="&lt;p&gt;Define a function $F(A, B, C)$ as the number of ways you can roll $B$ $C$-sided dice to sum up to $A$, counting different orderings (rolling a $2$, $2$, and $3$ with three dice is different from rolling a $2$, $3$, and $2$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With three $5$-sided dice, the list of $F(A, B, C)$ values in the domain of the possible values of $A$ for $B = 3$ and $C = 5$ is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$F(3, 3, 5), F(4, 3, 5), F(5, 3, 5), F(6, 3, 5), ... , F(15, 3, 5)$$ is evaluated to:&#xA;$$1, 3, 6, 10, 15, 18, 19, 18, 15, 10, 6, 3, 1$$&#xA;Call this list $L_1$.&lt;br/&gt;&lt;br/&gt;&#xA;Let $s$ be the number of sides on each die, let $n$ be the number of dice, and let $v$ be the total value to roll from the $n$ dice.&#xA;Let $L_2$ be the list of ${v - 1}\choose{v - n}$ in the domain of $v$ values for $n = 3$.&#xA;Then $L_2$ is:&#xA;$${{3 - 1}\choose{3 - 3}}, {{4 - 1}\choose{4 - 3}}, {{5 - 1}\choose{5 - 3}}, {{6 - 1}\choose{6 - 3}}, ... , {{15 - 1}\choose{15 - 3}}$$&#xA;Which is evaluated to:&#xA;$$1, 3, 6, 10, 15, 21, 28, 36, 45, 55, 66, 78, 91$$&#xA;Comparing $L_1$ with $L_2$, we see that only the first $s$ values of the lists are equal:&#xA;$$1, 3, 6, 10, 15$$&#xA;I have observed that this property holds with other values of $s$, $v$, and $n$, and $A$, $B$, and $C$.&#xA;&lt;br/&gt;&lt;br/&gt;Can someone please explain why $L_1$ and $L_2$ share the first $s$ values?&lt;/p&gt;&#xA;" OwnerUserId="2771" LastEditorUserId="2771" LastEditorDisplayName="" LastEditDate="2010-10-26T04:39:41.280" LastActivityDate="2010-10-26T04:51:26.470" Title="Combinatorics and Rolling Dice Similarity?" Tags="&lt;combinatorics&gt;" AnswerCount="1" />
  <row Id="7866" PostTypeId="2" ParentId="7864" CreationDate="2010-10-26T04:34:21.457" Score="4" ViewCount="" Body="&lt;p&gt;These are 2 instances which i have seen to happen with my friends. If $A$ and $B$ are 2 matrices, then they believe that $(A+B)^{2}=A^{2}+ 2 \cdot A \cdot B +B^{2}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another mistake is if one i asked to solve this equation, $ \displaystyle\frac{\sqrt{x}}{2}=-1$, people generally square both the sides and do get $x$ as $4$. &lt;/p&gt;&#xA;" LastActivityDate="2010-10-26T04:34:21.457" CommunityOwnedDate="2010-10-26T20:30:52.290" CommentCount="2" />
  <row Id="7867" PostTypeId="1" CreationDate="2010-10-26T04:44:28.173" Score="0" ViewCount="200" Body="&lt;p&gt;I have an expected value say 5(minimum) or 6(minimum).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then I have 4 values [3, 10, 6, 8]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to compute the percentage on how far the 4 values are from the expected value in terms of percentage. Given 100% means they are all within the minimum and maximum and the farther they are from the expected the lower the percentage would be.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was looking for a mathematical way of computing this. One way is to assign true or false score for each. Meaning either they match or not the minimum or maximum value.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[3(0.0), 10(0.0), 6(1.0), 8(0.0)] would give 25%&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But this solution is too crude because I want to give more fine grained computation for how far I am from the expected value.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[3(0.0), 10(0.0), 6(1.0), 8(0.0)] would give X%&#xA;[1(0.0), 10(0.0), 6(1.0), 9(0.0)] would give Y% &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Y% should be lower than X% since their value deviates farther from the mean than X%.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I do have a boundary of expected values, 1 to 30. I tried to read on statistics like standard deviation but it is the deviation from the mean. Even If i change it to deviation from expected value, How do I get percentage of how far is it from expected value?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Updated:&#xA;I guess I already got an idea on this one. Let say there are N=20 available number to be taken. Then there are X=4 who would share this 20 slots. What I will do is get the average which is A=N/X=5. So I would compute for the summation of the difference of each item from the average. S=sum(abs(Ni-A)) I get the sum of the difference from the maximum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The percentage would be 1-(S/N)&lt;/p&gt;&#xA;" OwnerUserId="2813" LastEditorUserId="2813" LastEditorDisplayName="" LastEditDate="2010-10-27T04:22:21.003" LastActivityDate="2010-12-27T13:09:44.917" Title="How do you get a Percentage of how far the value is from expected Value" Tags="&lt;statistics&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="7868" PostTypeId="1" AcceptedAnswerId="7874" CreationDate="2010-10-26T04:50:46.317" Score="2" ViewCount="88" Body="&lt;p&gt;What is the degree of the polynomial of the following expression ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ [x + (x^3 – 1)^{1/2}]^5 + [x – (x^3 – 1)^{1/2}]^5 $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I am not very wrong the highest power of x is  $\frac {15}{2} $ ?! So the degree is floor(15/2) = 7 ?! The answer is 7 but I am not sure of this approach.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please comment.&lt;/p&gt;&#xA;" LastActivityDate="2010-10-26T05:03:30.440" Title="Degree of polynomial :$ [x + (x^3 – 1)^{1/2}]^5 + [x – (x^3 – 1)^{1/2}]^5 $" Tags="&lt;algebra-precalculus&gt;&lt;polynomials&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="7869" PostTypeId="2" ParentId="7865" CreationDate="2010-10-26T04:51:26.470" Score="5" ViewCount="" Body="&lt;p&gt;The first $s$ terms of $L_1$ are the compositions of $A$.  They stop being the compositions of $A$ at that point because you hit the limit.  In your example, you miss the composition of 8 as 6+1+1 because the dice only have 5 sides.  The compositions of $n$ into exactly $k$ parts are given by ${{n - 1}\choose{k - 1}}$ as shown &lt;a href=&quot;http://en.wikipedia.org/wiki/Composition_%28number_theory%29&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-10-26T04:51:26.470" />
  <row Id="7870" PostTypeId="2" ParentId="7778" CreationDate="2010-10-26T04:55:39.623" Score="0" ViewCount="" Body="&lt;p&gt;Here's an enumerative proof. Consider the following procedure for dividing a set $S$ of size $2n$ into two sets $A,B$ of size $n$. The procedure proceeds in $n$ rounds. We start with $A,B$ empty. In each round, let $m = \min S$ and pick some $x \in S \setminus m$. Now decide whether to put $m$ in $A$ and $x$ in $B$, or to put $m$ in $B$ and $x$ in $A$. Remove both $m,x$ from $S$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clearly, the procedure ends with the required partition, and requires $(2n-1)(2)(2n-3)(2)\cdots(1)(2)$ choices. Furthermore, I claim that each partition can be generated in $n!$ ways. Indeed, at each round, either $m \in A$ or $m \in B$; suppose wlog that $m \in A$. This forces our second choice, and for our force choice, we can choose an arbitrary member of $B$. So our total degrees of freedom are $(n)(n-1)\cdots(1)$.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-10-26T04:55:39.623" />
  <row Id="7871" PostTypeId="2" ParentId="7867" CreationDate="2010-10-26T04:57:48.873" Score="2" ViewCount="" Body="&lt;p&gt;Do you mean 5 (minimum) to 6 (maximum)?  You could calculate how many ranges away each item is.  So 3 would get a score of 2 (from being 2 away from 5).  10 would get a score of 4 (10-6).  6 would get a score of 0 for being in range and 8 would get a score of 2.  This doesn't tell you what percentage are in range, but it does tell you how far out the others are.  Then if your range was 5 to 7 the scores would be 1 for 3, 3/2 for 10, and 1/2 for 8.  Maybe you need both figures.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-10-26T04:57:48.873" CommentCount="3" />
  <row Id="7872" PostTypeId="2" ParentId="7868" CreationDate="2010-10-26T05:00:46.660" Score="1" ViewCount="" Body="&lt;p&gt;If you expand the 5th powers, the terms that have square roots in them will cancel.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-10-26T05:00:46.660" />
  <row Id="7873" PostTypeId="1" CreationDate="2010-10-26T05:02:16.667" Score="0" ViewCount="169" Body="&lt;p&gt;I'm solving this question, but I don't know what it is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;============================================================&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider a plane $x - y + z = 0$, and a point $b = (1, 2, 0)^T$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a) Find a basis of this plane, and a basis of orthogonal subspace to this plane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;b) Find the closest point $\hat{b}$ on the plane to b.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;c) What is the error between b and $\hat{b}$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;============================================================&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this a linear algebra problem? What's a basis of orthogonal subspace and closest point $\hat{b}$? What's the error between b and $\hat{b}$??&lt;/p&gt;&#xA;" OwnerUserId="2674" LastActivityDate="2010-10-26T08:47:25.067" Title="Is this a linear algebra problem?" Tags="&lt;linear-algebra&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="7874" PostTypeId="2" ParentId="7868" CreationDate="2010-10-26T05:03:30.440" Score="4" ViewCount="" Body="&lt;p&gt;Usually, a polynomial is understood to have only non-negative powers of $x$ (i.e. $x^0=1,x^1=x,x^2$ etc.), so a-priori it's not even clear that this is a polynomial. However, if we open up the binomial theorem, we see that the odd powers cancel:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(x+y)^n + (x-y)^n = \sum_{k=0}^n \binom{n}{k} x^{n-k} [y^k + (-y)^{k}]$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When $k$ is odd, $y^k + (-y)^k$ cancels. In your case, the term corresponding to $k = 2l$ has degree $n-2l+3l = n+l$, and so the maximum is achieved for $l=2$ and $n+l = 7$.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-10-26T05:03:30.440" CommentCount="1" />
  <row Id="7875" PostTypeId="2" ParentId="7835" CreationDate="2010-10-26T05:11:42.643" Score="1" ViewCount="" Body="&lt;p&gt;Multiplying by $n!$, the left-hand side it the number of subsets of $[n]$ of odd size, and the right-hand side is $2^{n-1}$. This is because for each set $S \subseteq [n-1]$, exactly one of $S,S \cup {n}$ has odd size.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-10-26T05:11:42.643" />
  <row Id="7876" PostTypeId="1" CreationDate="2010-10-26T05:42:22.483" Score="2" ViewCount="79" Body="&lt;p&gt;If in the expansion of $(1 + x)^m \cdot (1 – x)^n $, the coefficients of $ x $ and $ x^2 $are 3 and -6 respectively, then m is ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I solved it in the following way :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Expanding we get, the coefficient of $ x $ as $ (m-n) = 3 \cdots (1)$ and coefficient of $ x^2 $ as $ \frac{n(n-1)}{2} + \frac{m(m-1)}{2} +- m \cdot n = -6 \cdots (2)$ after substituting and some algebraic treatment I got m = 12 and n = 9. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now this is correct but I am interested if there exists any short procedure such that the entire problem could be done under a mint.&lt;/p&gt;&#xA;" LastActivityDate="2010-10-26T06:17:51.643" Title="How to calculate this efficiently ?" Tags="&lt;algebra-precalculus&gt;&lt;algebra&gt;&lt;polynomials&gt;&lt;binomial-coefficients&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="7877" PostTypeId="2" ParentId="7876" CreationDate="2010-10-26T06:02:51.717" Score="3" ViewCount="" Body="&lt;p&gt;For the coefficient of $x$ to be 3, $m-n=3$ as you said, so $m&gt;n$ and $m-n&gt;0$ and we can factor the original product:&#xA;$$\begin{align}&#xA;(1+x)^m(1-x)^n&amp;amp;=(1+x)^{m-n+n}(1-x)^n&#xA;\\&#xA;&amp;amp;=(1+x)^{m-n}(1+x)^n(1-x)^n&#xA;\\&#xA;&amp;amp;=(1+x)^{m-n}\left((1+x)(1-x)\right)^n&#xA;\\&#xA;&amp;amp;=(1+x)^{m-n}(1-x^2)^n&#xA;\\&#xA;&amp;amp;=(1+x)^3(1-x^2)^n&#xA;\end{align}$$&#xA;Now, the coefficient of $x^2$ is the sum of the coefficients of $x^2$ in $(1-x^2)^n=1-nx^2+\cdots$ and $(1+x)^3=1+3x+3x^2+x^3$, so $-6=-n+3$ and $n=9$.  Since $m-n=3$, $m=12$.&lt;/p&gt;&#xA;" OwnerUserId="72" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-10-26T06:17:51.643" LastActivityDate="2010-10-26T06:17:51.643" CommentCount="9" />
  <row Id="7878" PostTypeId="2" ParentId="7873" CreationDate="2010-10-26T06:30:07.107" Score="2" ViewCount="" Body="&lt;p&gt;Yes, this is a linear algebra problem. Looks like a homework.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint: You are working in $\mathbb{R}^3$. Your plane is given by 1 equation, which strongly suggest that it has dimension 2. Hence you should try to find 2 vectors linear independant and you will have your first basis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, 2 vector spaces are orthogonal if any vector of the first one is orthogonal to any vector of the second one (if $v_1 = (x_1, y_1, z_1)^T$ and $v_2 = (x_2, y_2, z_2)^T$, then they are orthogonal if and only if $x_1x_2 + y_1y_2 + z_1z_2 = 0$). When we are talking about the orthogonal subspace in this context, you are supposed to find one with maximal dimension. For example $V = {(0,0,0)^T}$ is always orthogonal to any other subspace, therefore try to find one with dimension &gt; 0. Hint: the dimension of a subspace + dimension of it's orthogonal complement should equals the dimension of the space you are working in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you have solved a), then we might help you to solve b) and c).&lt;/p&gt;&#xA;" OwnerUserId="2778" LastEditorUserId="2778" LastEditorDisplayName="" LastEditDate="2010-10-26T06:39:53.180" LastActivityDate="2010-10-26T06:39:53.180" />
  <row Id="7879" PostTypeId="1" AcceptedAnswerId="7880" CreationDate="2010-10-26T06:47:25.473" Score="1" ViewCount="70" Body="&lt;p&gt;Let $f(x,y)=1$ if $(x,y)$ is in the unit disk and $f=0$ otherwise. I would like to approximate $f$ by a sequence of smooth functions. The functions need to be evaluated quickly so the results I'm getting from mollification are to awkward to deal with... &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance..&lt;/p&gt;&#xA;" OwnerUserId="1794" LastEditorUserId="1794" LastEditorDisplayName="" LastEditDate="2010-10-26T07:03:26.790" LastActivityDate="2010-10-26T07:03:26.790" Title="Sequence of smooth functions approximating a 2d cylinder step function" Tags="&lt;approximation&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="7880" PostTypeId="2" ParentId="7879" CreationDate="2010-10-26T06:53:31.550" Score="3" ViewCount="" Body="&lt;p&gt;You might want to try any of formulae 17 to 27 in the &lt;a href=&quot;http://mathworld.wolfram.com/HeavisideStepFunction.html&quot; rel=&quot;nofollow&quot;&gt;MathWorld page for the unit step function&lt;/a&gt;, since the function you're interested in is expressible as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$1-H(x^2+y^2-1)$$&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-10-26T06:53:31.550" />
  <row Id="7881" PostTypeId="1" AcceptedAnswerId="7903" CreationDate="2010-10-26T07:14:43.230" Score="3" ViewCount="238" Body="&lt;p&gt;For some collection of sets $A$, let $\sigma(A)$ denote the $\sigma$-algebra generated by $A$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $C$ be some collection of subsets of a set $Y$, and let $f$ be a function from some set $X$ to $Y$. I want to prove:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f^{-1}(\sigma(C))=\sigma(f^{-1}(C))$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I could prove that &#xA;$$\sigma(f^{-1}(C)) \subset f^{-1}(\sigma(C))$$ &#xA;since complements and unions are 'preserved' by function inverse. But how do I go the other way? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: One way to go the other way would be to argue that any set in $\sigma(C)$ must be built by &lt;em&gt;repeatedly&lt;/em&gt; applying the complement, union and intersection operations to elements of $C$ and all these operations are preserved when taking the inverse. The problem I am facing with the approach is formalizing the word &quot;repeatedly&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;[not-homework]&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="1195" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2010-11-21T14:21:12.830" LastActivityDate="2010-11-21T14:21:12.830" Title="Preimage of generated $\sigma$-algebra" Tags="&lt;elementary-set-theory&gt;&lt;sigma-algebra&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="7883" PostTypeId="1" AcceptedAnswerId="7891" CreationDate="2010-10-26T07:27:56.463" Score="1" ViewCount="113" Body="&lt;p&gt;This is a problem from Mathematical Analysis by Zorich, from the chapter on continuous functions.&#xA;This is also a homework problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Definition:&#xA;Let $P_n$ be a polynomial of degree $n$. For a function $f:[a,b]\to\mathbb{R}$, Let $\Delta(P_n) = \sup_{x\in[a,b]} |f(x)-P_n(x)|$. and $E_n(f) = \inf_{P_n} \Delta(P_n)$. A polynomial $P_n$ is the best approximation of degree $n$ of $f$ is $\Delta(P_n) = E_n(f)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have already proved the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There exist a polynomial $P_0(x) = a$ of best approximation of degree 0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $Q_\lambda(x) = \lambda P_n(x)$ for some fixed polynomial $P_n$. Then there exist a polynomial $Q_{\lambda_0}$ such that $\Delta(Q_{\lambda_0}) = \min_{\lambda\in \mathbb{R}} \Delta(Q_\lambda)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm stuck on proving the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there exists a polynomial of best approximation of degree $n$, there also exists a polynomial of best approximation of degree $n+1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My intuition is to prove $\lambda_0 x^{n+1}+P_n$ is the $n+1$ best approximation for $f$. Where $\Delta(\lambda_0 x^{n+1}) =  \min_{\lambda\in \mathbb{R}} \Delta(\lambda x^{n+1})$ and $\Delta(P_n) = E(f(x) - \lambda_0 x^{n+1})$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know if this approach is right. I'm stuck on how to proceed. &lt;/p&gt;&#xA;" OwnerUserId="96" LastActivityDate="2010-10-26T09:24:03.243" Title="There exist a degree n+1 polynomial of best approximation if there exist a degree n polynomial of best approximation" Tags="&lt;analysis&gt;&lt;polynomials&gt;" AnswerCount="1" />
  <row Id="7884" PostTypeId="2" ParentId="7881" CreationDate="2010-10-26T07:56:13.927" Score="-2" ViewCount="" Body="&lt;p&gt;Looks like a homework.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I didn't have time to check the details (I am at work and my boss is watching), however, it seems to me that the following approach should work:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Step 1: prove that $f^{-1}(\sigma(C))$ is a $\sigma$-algebra.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Step 2: you should be able to conclude, if you understand what $\sigma(f^{-1}(C))$ means.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sorry, you are right, I read your question too quickly. My steps only help you to get the inclusion you already had.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Not sure if this new approach is working, but I would:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1/ consider $f(\sigma(f^{-1}(C)))$ and try to prove it is a $\sigma$-algebra.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2/ when 1/ is done, prove that it is a $\sigma$-algebra that contains $C$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3/ If 1/ and 2/ are ok, then that means we have $\sigma(C) \subset f(\sigma(f^{-1}(C)))$ and you can then conclude.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems to me that 1 is true, I don't have time to do the details however.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT 2:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ok, me second approach doesn't seem to be working either. I decided to do a little bit of research, and you can actually find the proof online. Here it is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $S = ${ $B \in \mathcal{P}(Y); f^{-1}(B) \in \sigma(f^{-1}(C)) $}.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then $S$ is a $\sigma$-algebra that contains $C$. Therefore $\sigma(C) \subset S$ and $f^{-1}(\sigma(C)) \subset f^{-1}(S)$. But, by definition, $ f^{-1}(S) \subset \sigma(f^{-1}(C))$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sorry for my &quot;hints&quot;, they were not really useful.&lt;/p&gt;&#xA;" OwnerUserId="2778" LastEditorUserId="2778" LastEditorDisplayName="" LastEditDate="2010-10-26T11:34:29.793" LastActivityDate="2010-10-26T11:34:29.793" CommentCount="3" />
  <row Id="7885" PostTypeId="2" ParentId="7811" CreationDate="2010-10-26T08:16:17.717" Score="5" ViewCount="" Body="&lt;p&gt;For any prime $p$ and $n&gt;1$, there are nilpotent groups $G$ of class 2 and order $p^{n(n+1)/2}$ with generators $a_i$ $(1 \le i \le n)$, $b_{ij}$ $(1 \le i &amp;lt; j \le n)$, such that $[a_i,a_j] = b_{ij}$, the $b_{ij}$ are all central in the group, and all generators have order $p$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then $G'$ is the group of order $p^{n(n-1)/2}$ generated by the $b_{ij}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In any group, we have $[ax,by] = [a,b]$ when $x,y$ are central in the group, so $G$ has at most $p^{2n}$ distinct elements that are commutators.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, for any fixed $k&gt;0$, by choosing $n$ sufficiently large we can find $G$ such that not all element of $G'$ are products of at most $k$ commutators.&lt;/p&gt;&#xA;" OwnerUserId="2820" LastActivityDate="2010-10-26T08:16:17.717" CommentCount="1" />
  <row Id="7886" PostTypeId="1" AcceptedAnswerId="7887" CreationDate="2010-10-26T08:23:13.410" Score="3" ViewCount="201" Body="&lt;p&gt;How does one evaluate the limit: $$ \lim_{n \to \infty} \frac{1}{n}\sum\limits_{k=1}^{\lfloor{\frac{n}{2}\rfloor}} \cos\Bigl(\frac{k\pi}{n}\Bigr)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yes, i recognize this as soon as i saw the problem: $$\int\limits_{0}^{1}f(x) \ dx = \lim_{n \to \infty} \frac{1}{n} \sum\limits_{k=1}^{n} f\Bigl(\frac{r}{n}\Bigr)$$ but the problem is there is $\lfloor{\frac{n}{2}\rfloor}$.&lt;/p&gt;&#xA;" LastActivityDate="2010-10-26T14:49:56.700" Title="Evaluating limit of Summation" Tags="&lt;calculus&gt;&lt;real-analysis&gt;&lt;analysis&gt;&lt;sequences-and-series&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="7887" PostTypeId="2" ParentId="7886" CreationDate="2010-10-26T08:35:47.903" Score="0" ViewCount="" Body="&lt;p&gt;Hint: Consider two separate limits.  The limit of n even going to infinity, and the limit of n odd going to infinity.  If they both converge to the same limit you are done.  Both problems should be easy to solve using your observation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first one converges to $1/2\int_0^1 cos(\pi x/2) dx$ and the second converges to $\int_0^{1/2}cos(\pi x) dx$&lt;/p&gt;&#xA;" OwnerUserId="780" LastEditorUserId="780" LastEditorDisplayName="" LastEditDate="2010-10-26T09:27:48.477" LastActivityDate="2010-10-26T09:27:48.477" CommentCount="8" />
  <row Id="7888" PostTypeId="2" ParentId="7873" CreationDate="2010-10-26T08:47:25.067" Score="1" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;(a)&lt;/strong&gt; To find a basis for the plane $x-y+z = 0$, you could solve this equation in terms of $x$ and $z$: $y= x+z$. Then the set of vectors of your plane could be described as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;V = \left\{ (x, x+z, z) \ \vert \ x,z \in \mathbb{R}   \right\} \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From this description it's easy to find a basis for your plane $V$: it will have two vectors, say $u,v \in \mathbb{R}^3$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;V = [u,v] \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, the orthonormal subspace to your plane is a straight line, generated by the &lt;a href=&quot;http://en.wikipedia.org/wiki/Vector_product&quot; rel=&quot;nofollow&quot;&gt;cross product&lt;/a&gt; of $u$ and $v$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;V^\bot = [u \times v] \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;(b)&lt;/strong&gt; The closest point $\hat{b}$ to $b$ in the plane $V$ is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Orthogonal_projection&quot; rel=&quot;nofollow&quot;&gt;orthogonal projection&lt;/a&gt; of $b$ onto $V$. In this case, you don't really need the formulae you'll see in Wikipedia, because it's just the intersection of $V$ with the straight line with direction $u\times v$ that passes through $b$. That is, you have to solve the system of linear equations&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{align}&#xA;x - y +z &amp;amp;= 0    \\&#xA;(x,y,z) &amp;amp;= (1,2,0) + \lambda u\times v  \ .&#xA;\end{align}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;(c)&lt;/strong&gt; The &quot;error&quot; between $b$ and $\hat{b}$ is the same as the distance from $b$ to $\hat{b}$; that is, $\| b - \hat{b} \|$.&lt;/p&gt;&#xA;" OwnerUserId="664" LastActivityDate="2010-10-26T08:47:25.067" />
  <row Id="7889" PostTypeId="1" CreationDate="2010-10-26T09:20:17.003" Score="0" ViewCount="104" Body="&lt;p&gt;Hello,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a way to find the class of all bounded functions $f(w,y)$ which satisfies the following inequality:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\int\limits_{-\infty}^{+\infty} f(w,y) ~\mathcal{N}(y,w,1) dy &gt;0 ~~~~\forall w&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;it is assumed that the function $f(w,y)$ is integrable w.r.t. to the Gaussina pdf $\mathcal{N}(y,w,1) $. For instance, the function $I_A(y)-0.5$, where $I_A$ is the indicator function over the set $A={(w,y): w-a\leq y&amp;lt;\infty}$ satisfies the inequality for each scalar $a&gt;0$.&lt;/p&gt;&#xA;" OwnerUserId="2821" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-10-26T18:59:13.530" LastActivityDate="2010-10-26T18:59:13.530" Title="Functions with constant expected values" Tags="&lt;probability-theory&gt;&lt;integral&gt;&lt;probability&gt;" />
  <row Id="7890" PostTypeId="1" CreationDate="2010-10-26T09:23:41.137" Score="2" ViewCount="136" Body="&lt;p&gt;According to Mathworld, hyperbolic geometry is well understood in 2 dimensions but not in 3 dimensions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://mathworld.wolfram.com/HyperbolicGeometry.html&quot; rel=&quot;nofollow&quot;&gt;http://mathworld.wolfram.com/HyperbolicGeometry.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What isn't well understood about hyperbolic geometry in 3D, that is in 2D or that is in 3D-Euclidean geometry?&lt;/p&gt;&#xA;" OwnerUserId="2822" LastActivityDate="2010-10-26T10:44:17.423" Title="Hyperbolic geometry. 3 dimensions. What is not well understood ?" Tags="&lt;geometry&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" />
  <row Id="7891" PostTypeId="2" ParentId="7883" CreationDate="2010-10-26T09:24:03.243" Score="2" ViewCount="" Body="&lt;p&gt;Determining the best $n$th degree approximation explicitely is a very difficult problem and cannot be done by a simple induction (see the chapter on Tschebyscheff approximation in your favorite textbook of numerical analysis). But existence is easy, it follows from general principles: You have to set up a certain continuous function $\Phi$ on a certain compact set $K$ in a high-dimensional space.&lt;/p&gt;&#xA;" OwnerUserId="1303" LastActivityDate="2010-10-26T09:24:03.243" CommentCount="2" />
  <row Id="7892" PostTypeId="1" AcceptedAnswerId="9459" CreationDate="2010-10-26T09:29:48.480" Score="5" ViewCount="439" Body="&lt;p&gt;How to calculate without calculator or something like this the values of   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\pi^{e}$    and    $e^{\pi}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;in oder to compare them?&lt;/p&gt;&#xA;" OwnerUserId="2779" LastEditorUserId="723" LastEditorDisplayName="" LastEditDate="2010-11-08T16:13:04.513" LastActivityDate="2010-11-08T17:31:15.737" Title="Comparing $\pi^{e}$ and $e^{\pi}$" Tags="&lt;real-analysis&gt;&lt;inequality&gt;" AnswerCount="5" FavoriteCount="1" />
  <row Id="7893" PostTypeId="1" CreationDate="2010-10-26T09:34:14.160" Score="3" ViewCount="158" Body="&lt;p&gt;Let $\psi$ be a wavelet. Can its Fourier transform $\hat{\psi}$ be also wavelet? produce an example or prove that it is not possible. A wavelet is a function $\psi:\mathbb R\to\mathbb R$ such that (i) $\psi \in L^1(R) \cap L^2(R)$, (ii) $\int_{-\infty}^{\infty} \psi(t) dt = 0$, fourier transform is $\hat f(\omega) =\int_{-\infty}^{\infty} f(t)e^{-i\omega t} dt$.&lt;/p&gt;&#xA;" OwnerUserId="2723" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-10-26T11:10:24.810" LastActivityDate="2010-11-22T21:42:37.057" Title="Let $\psi$ be a wavelet. Can its Fourier transform $\hat{\psi}$ be also wavelet?" Tags="&lt;fourier-analysis&gt;" AnswerCount="1" CommentCount="9" FavoriteCount="1" />
  <row Id="7894" PostTypeId="2" ParentId="7892" CreationDate="2010-10-26T09:47:53.660" Score="15" ViewCount="" Body="&lt;p&gt;This is an old chestnut. As a hint, it's easier to consider the more&#xA;general problem: for which positive $x$ is $e^x&gt;x^e$?&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-10-26T09:47:53.660" CommentCount="8" />
  <row Id="7895" PostTypeId="2" ParentId="7892" CreationDate="2010-10-26T09:52:54.003" Score="2" ViewCount="" Body="&lt;p&gt;Elaborating Robin's answer take $f(x) = \log{x} - \frac{x}{e}$.  We have $$f'(x)= \frac{e-x}{xe}$$ Thus $f'(x)&gt;0$ for $0 &amp;lt; x &amp;lt; e$ and $f'(x) &amp;lt;0$ if $x &gt; e$. Consequently, we have $f(x) &amp;lt; f(e)$ if $x \neq e$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Exercise:&lt;/strong&gt; Try to prove this using the same methods: $2^{\sqrt{2}} &amp;lt; e$.&lt;/p&gt;&#xA;" LastActivityDate="2010-10-26T10:01:45.413" />
  <row Id="7896" PostTypeId="1" AcceptedAnswerId="7898" CreationDate="2010-10-26T10:13:24.387" Score="2" ViewCount="273" Body="&lt;p&gt;I was wondering this today, and my algebra professor didn't know the answer. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are subgroups of finitely generated groups finitely generated? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suppose it is necessarily true for finitely generated abelian groups, but is it true in general?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And if not, is there a simple example of a finitely generated group with a non-finitely generated subgroup?&lt;/p&gt;&#xA;" OwnerUserId="1906" LastActivityDate="2011-02-15T07:12:48.673" Title="Subgroups of finitely generated groups" Tags="&lt;abstract-algebra&gt;&lt;group-theory&gt;" AnswerCount="4" />
  <row Id="7897" PostTypeId="2" ParentId="7896" CreationDate="2010-10-26T10:18:21.240" Score="3" ViewCount="" Body="&lt;p&gt;No.  The example given on &lt;a href=&quot;http://en.wikipedia.org/wiki/Generating_set_of_a_group#Finitely_generated_group&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt; is that the free group $F_2$ contains a subgroup generated by $y^n x y^{-n}, n \ge 1$, which is free on countably many generators.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-10-26T10:18:21.240" />
  <row Id="7898" PostTypeId="2" ParentId="7896" CreationDate="2010-10-26T10:20:26.220" Score="4" ViewCount="" Body="&lt;p&gt;It is well-known that the free group $F_2$ on two generators&#xA;has as a subgroup a group isomorphic to a free group on a countably&#xA;infinite set of generators. See Qiaochu's example.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However a &lt;strong&gt;finite index&lt;/strong&gt; subgroup of a finitely generated&#xA;group is finitely generated.&lt;/p&gt;&#xA;" OwnerUserId="226" LastEditorUserId="226" LastEditorDisplayName="" LastEditDate="2010-10-26T10:27:26.653" LastActivityDate="2010-10-26T10:27:26.653" />
  <row Id="7899" PostTypeId="2" ParentId="7896" CreationDate="2010-10-26T10:21:18.087" Score="0" ViewCount="" Body="&lt;p&gt;See this for more details: &lt;a href=&quot;http://planetmath.org/encyclopedia/FinitelyGeneratedSubgroup.html&quot; rel=&quot;nofollow&quot;&gt;http://planetmath.org/encyclopedia/FinitelyGeneratedSubgroup.html&lt;/a&gt;&lt;/p&gt;&#xA;" LastActivityDate="2010-10-26T10:21:18.087" />
  <row Id="7900" PostTypeId="1" AcceptedAnswerId="7946" CreationDate="2010-10-26T10:36:25.470" Score="2" ViewCount="119" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Diablo 3 has &lt;em&gt;97 billion&lt;/em&gt; possible skill/trait builds. Per class.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;sub&gt;&lt;a href=&quot;http://chat.gaming.stackexchange.com/rooms/3/conversation/97-billion&quot;&gt;LessPop_MoreFizz&lt;/a&gt;, emphasis is mine.&lt;/sub&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I used base two logarithm to claim &quot;97 billion&quot; configurations only are roughly 37 binary choices because $log_2(97×10^9) \simeq 36.5$, but I would like to know if there are more refined techniques to remove the marketing out of these big numbers and see their true meaning.&lt;/p&gt;&#xA;" OwnerUserId="92" LastActivityDate="2010-10-26T18:15:25.560" Title="Making sense of combinatorics-based marketing hyperboles" Tags="&lt;combinatorics&gt;&lt;big-numbers&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="7901" PostTypeId="2" ParentId="7890" CreationDate="2010-10-26T10:44:17.423" Score="2" ViewCount="" Body="&lt;p&gt;I don't know what &quot;well-understood&quot; means in this context.  It is true that hyperbolic geometry is a very rich area in both dimensions two and three.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In dimension two every closed surface admitting one hyperbolic structure admits uncountably many, and these fit together to form moduli space.  Moduli space is an object of intense study from the point of view of algebraic geometry, number theory, dynamical systems, etc.  The &quot;fundamental group&quot; of moduli space is the mapping class group (aka the modular group) and this is again studied by many different groups.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In dimension three we have Mostow rigidity saying that a closed three-manifold admits either zero or essentially one hyperbolic structure.  It follows that the geometric invariants of the hyperbolic structure give topological invariants of the three-manifold, which was a wonderful shock to topologists in the 70's.  (I was recently told a story about Thurston wandering around a conference at Warwick, in the 70's, with a programmable calculator on his belt, saying that he was computing volumes of knots in the three-sphere.  Nobody had any idea what he was saying, apparently.)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It follows that the moduli spaces of hyperbolic three-manifolds are much more simple.  To get an interesting theory one allows hyperbolic structures that are not complete (or discrete, etc) and then the moduli spaces become very wild and are not well understood.&lt;/p&gt;&#xA;" OwnerUserId="1307" LastActivityDate="2010-10-26T10:44:17.423" CommentCount="2" />
  <row Id="7902" PostTypeId="2" ParentId="7881" CreationDate="2010-10-26T11:05:26.667" Score="3" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt;  The original answer that was here is wrong; Carl Mummert's answer contains the correct way to do what I was trying to do.  So instead here is the solution by transfinite induction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Define a sequence of functions $\sigma_i$ where $i$ is an &lt;a href=&quot;http://en.wikipedia.org/wiki/Ordinal_number&quot; rel=&quot;nofollow&quot;&gt;ordinal&lt;/a&gt; as follows.  For any subset $C$ of the base set $E$ we take $\sigma_0(C) = C$.  If $i$ is a &lt;a href=&quot;http://en.wikipedia.org/wiki/Successor_ordinal&quot; rel=&quot;nofollow&quot;&gt;successor ordinal&lt;/a&gt;, take $\sigma_i(C)$ to be the set of all countable unions or complements of elements of $\sigma_{i-1}(C)$; otherwise $i$ is a &lt;a href=&quot;http://en.wikipedia.org/wiki/Limit_ordinal&quot; rel=&quot;nofollow&quot;&gt;limit ordinal&lt;/a&gt; and we take $\sigma_i(C) = \bigcup_{j &amp;lt; i} \sigma_j(C)$.  For example, $\sigma_{\omega}(C)$ is the set of all sets that can be obtained from $C$ by performing countable union or complement countably many times.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We would like to say that $\sigma(C)$ is the union of the $\sigma_i(C)$ over all ordinals $i$, but the ordinals don't form a set so we can't actually do this.  What I believe is true is that we only need to take ordinals of cardinality at most the cardinality of the powerset of the underlying set.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If that's true, then the proof is as follows.  It's enough to show that $f^{-1}(\sigma_i(C)) = \sigma_i(f^{-1}(C))$ for all $i$.  This is obvious for $i = 0$.  If it's true for all ordinals $j &amp;lt; i$, then it's true for $i$ by the obvious argument.  And now we are done by the &lt;a href=&quot;http://en.wikipedia.org/wiki/Transfinite_induction&quot; rel=&quot;nofollow&quot;&gt;principle of transfinite induction&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me remark that, on the one hand, this is more complicated than Carl Mummert's proof because it requires knowledge of ordinals.  On the other hand, getting into the nitty-gritty like this really shows how complicated $\sigma$-algebras can be.  &lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-10-26T14:17:45.693" LastActivityDate="2010-10-26T14:17:45.693" CommentCount="4" />
  <row Id="7903" PostTypeId="2" ParentId="7881" CreationDate="2010-10-26T11:16:37.420" Score="7" ViewCount="" Body="&lt;p&gt;Yes, we &lt;em&gt;can&lt;/em&gt; use transfinite induction to prove this (formalizing the word &quot;repeatedly&quot;). That would be the bottom-up approach. There is also a top-down approach, using the characterization of $\sigma(C)$ as the smallest $\sigma$-algebra containing $C$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A key fact here is the the preimage operation commutes with all the set algebra operations: if $f \colon X \to Y$ then&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$f^{-1}(A \cup B) = f^{-1}(A) \cup f^{-1}(B)$ &lt;/li&gt;&#xA;&lt;li&gt;$f^{-1}(A \cap B) = f^{-1}(A) \cap f^{-1}(B)$ &lt;/li&gt;&#xA;&lt;li&gt;$f^{-1}(A - B) = f^{-1}(A) - f^{-1}(B)$. In particular $f^{-1}(Y - A) = X - f^{-1}(A)$. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;and so forth. So $f^{-1}\colon P(Y) \to P(X)$ is a lattice homomorphism on the powerset lattices, to use that terminology.  More importantly for us, the preimage operation even commutes with infinite unions and intersections. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To show that $f^{−1}(\sigma(C)) \subseteq \sigma(f^{−1}(C))$, you could follow this strategy:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Show that $f^{-1}(C) \in \sigma(f^{-1}(C))$&lt;/li&gt;&#xA;&lt;li&gt;Show that if $f^{-1}(A_i) \in \sigma(f^{-1}(C))$ for $i \in \omega$ then  $f^{-1}(\bigcup A_i) \in \sigma(f^{-1}(C))$&lt;/li&gt;&#xA;&lt;li&gt;Show that if $f^{-1}(A) \in \sigma(f^{-1}(C))$ then $f^{-1}(Y - A) \in \sigma(f^{-1}(C))$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The point here is that, if we let $D$ be the collection of sets $A$ such that $f^{-1}(A) \in \sigma(f^{-1}(C))$, then $D$ is itself a $\sigma$ algebra containing $C$, which means $\sigma(C) \subseteq D$. But by the definition of $D$ this means $f^{-1}(\sigma(C)) \subseteq \sigma(f^{-1}(C))$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;None of the three bullets will require taking forward images under $f$. For example, for the third one, let $A$ be as stated. This means $f^{-1}(A)$ is in $\sigma(f^{-1}(C))$, which means that $X - f^{-1}(A)$ is also in $\sigma(f^{-1}(C))$. But $f^{-1}(Y - A)$ is exactly $X - f^{-1}(A)$, so we see that $f^{-1}(Y - A)$ is indeed in $\sigma(f^{-1}(C))$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The underlying point here is that the entire proof is algebraic and that a more general theorem is true: you can replace $f^{-1}$ with any other homomorphism of the powerset lattices that preserves countable unions. &lt;/p&gt;&#xA;" OwnerUserId="630" LastEditorUserId="630" LastEditorDisplayName="" LastEditDate="2010-10-26T11:37:48.353" LastActivityDate="2010-10-26T11:37:48.353" CommentCount="5" />
  <row Id="7904" PostTypeId="2" ParentId="7893" CreationDate="2010-10-26T12:01:27.243" Score="5" ViewCount="" Body="&lt;p&gt;$f(x)=\sin(x)\cdot\exp(-x^2)$ should do, because:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The decay of $f$ ensure $f,\hat{f}\in L^1\cap L^\infty$. &lt;/li&gt;&#xA;&lt;li&gt;$f$ is odd $f(-x)=-f(x)$.&lt;/li&gt;&#xA;&lt;li&gt;$\hat{f}(-\xi)=\int_{-\infty}^\infty e^{ix\xi}f(x)dx=\int_{-\infty}^\infty e^{-i(-x)\xi}f(x)dx=\int_{\infty}^{\infty} e^{-it\xi}f(-t)(-dt)=-\hat{f}(\xi)$&#xD;&#xA;where we used that $f$ is odd.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt;&#xA;If $g$ is integrable and odd then $$\int_{-\infty}^0g(x)dx =\int_{-\infty}^0-g(-x)dx=\int_{+\infty}^0g(t)dt=-\int_0^{+\infty}g(t)dt$$&#xD;&#xA;hence &#xA;$$\int_{-\infty}^\infty g(t)dt = 0.$$&#xD;&#xA;This imply that $$\int_{-\infty}^\infty f dx= \int_{-\infty}^\infty\hat{f}d\xi=0$$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="1154" LastEditorUserId="1154" LastEditorDisplayName="" LastEditDate="2010-11-22T21:42:37.057" LastActivityDate="2010-11-22T21:42:37.057" CommentCount="2" />
  <row Id="7905" PostTypeId="2" ParentId="5645" CreationDate="2010-10-26T12:26:52.290" Score="4" ViewCount="" Body="&lt;p&gt;Suppose that $M$ satisfies all the easy things like extensionality, pairing, etc., plus $\Delta_0$ separation and suppose also that $M$ thinks for every ordinal $\alpha$ that $V_\alpha$ exists (that is, $V_\alpha^M$ exists in $M$). All this is expressible by a single axiom, but when $M$ is a proper class, it ensures that $M$ satisfies the entire Collection scheme, because for any formula $\varphi(x,y)$ and set $A\in M$ for which we want to collect, we may apply Collection in $V$ to find an ordinal $\alpha$ such that whenever $a\in A$ has $\exists y\varphi(a,y)$ in $M$, then such a $y$ may be found in $V_\alpha$ and hence in $V_\alpha^M$. Thus, $V_\alpha^M$ serves as a collection set inside $M$. Similarly, one can get full Separation in $M$ from $\Delta_0$ Separation (or much less, if you care to optimize it), by applying the Reflection theorem, which allows you to bound all the quantifiers by a suitably large $V_\alpha^M$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The argument works only when $M$ is a proper class, however, because it appeals to Collection in $V$, and uses that $M$ grows taller than the resulting collection set. This method would break down completely when $M$ is a set, since the collection set in $V$ could be unbounded in $M$. &lt;/p&gt;&#xA;" OwnerUserId="413" LastActivityDate="2010-10-26T12:26:52.290" CommentCount="1" />
  <row Id="7906" PostTypeId="2" ParentId="7900" CreationDate="2010-10-26T12:48:45.213" Score="2" ViewCount="" Body="&lt;p&gt;I think this is a fine approach.  Some of the choices could be three-way, but to get a general idea of how many choices it takes to get that many total possibilities you are spot on.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-10-26T12:48:45.213" />
  <row Id="7907" PostTypeId="2" ParentId="7864" CreationDate="2010-10-26T13:04:45.623" Score="2" ViewCount="" Body="&lt;p&gt;I have seen this one time too many&#xA;$$\frac{a}{b}+\frac{c}{d}=\frac{a+c}{b+d}$$&lt;/p&gt;&#xA;" OwnerUserId="1154" LastActivityDate="2010-10-26T13:04:45.623" CommunityOwnedDate="2010-10-26T20:31:08.403" CommentCount="2" />
  <row Id="7908" PostTypeId="1" AcceptedAnswerId="7911" CreationDate="2010-10-26T13:40:24.583" Score="0" ViewCount="194" Body="&lt;p&gt;This is NOT a homework, I am over homework stage ages ago :-)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What function draws semicircle?&#xA;$$y=f(x) = \text{something}$$ &#xA;and I end up with a unit circle.&lt;/p&gt;&#xA;" OwnerUserId="2743" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-10-26T14:15:20.433" LastActivityDate="2011-02-15T20:43:53.610" Title="Function in Y semi-circle" Tags="&lt;functions&gt;" AnswerCount="2" CommentCount="0" />
  <row Id="7909" PostTypeId="2" ParentId="7908" CreationDate="2010-10-26T13:44:09.027" Score="5" ViewCount="" Body="&lt;p&gt;On the interval $[-1,1]$ you can use the function $\sqrt{1-x^2}$. Using the &lt;a href=&quot;http://en.wikipedia.org/wiki/Pythagorean_theorem&quot; rel=&quot;nofollow&quot;&gt;Pythagorean theorem&lt;/a&gt; you can see that the graph of this function is the closed upper semicircle with radius 1 and centre (0,0).&lt;/p&gt;&#xA;" OwnerUserId="367" LastActivityDate="2010-10-26T13:44:09.027" />
  <row Id="7910" PostTypeId="2" ParentId="7864" CreationDate="2010-10-26T13:55:05.007" Score="5" ViewCount="" Body="&lt;p&gt;&lt;em&gt;Every torsion-free Abelian group is free.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(This only holds for finitely generated Abelian groups.)&lt;/p&gt;&#xA;" OwnerUserId="367" LastActivityDate="2010-10-26T13:55:05.007" CommunityOwnedDate="2010-10-26T20:31:25.143" CommentCount="2" />
  <row Id="7911" PostTypeId="2" ParentId="7908" CreationDate="2010-10-26T13:58:18.760" Score="6" ViewCount="" Body="&lt;p&gt;The only &quot;unit semicircles&quot; that can be expressed as $y=f(x)$ are the upper half of the unit circle and the lower half (any other half circle fails the &quot;vertical line test&quot;, so it cannot be expressed as an explicit function of $x$). Since the unit circle consists of all points $(x,y)$ that satisfy $x^2+y^2=1$, solving for $y$ gives you that $y=\sqrt{1-x^2}$ is the upper unit semicircle, and $y=-\sqrt{1-x^2}$ is the lower unit semicircle. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are willing to use &lt;a href=&quot;http://en.wikipedia.org/wiki/Parametric_equation&quot; rel=&quot;nofollow&quot;&gt;parametric equations&lt;/a&gt;, you can cover more possibilities. The unit circle can be parametrized as&#xA;$$\begin{array}{rcl}&#xA;x &amp;amp; = &amp;amp; \cos t,\\&#xA;y &amp;amp; = &amp;amp; \sin t;&#xA;\end{array}\qquad 0\leq t\leq 2\pi;$$&#xA;this &quot;traces&quot; the unit circle exactly once, counterclockwise, starting at $0$ radians (the point $(1,0)$). If you want the unit semicircle that starts at $\theta$ radians from the positive $x$-axis (and ends at $\theta+\pi$ radians), then you can obtain it parametrically by $x=\cos(t+\theta)$ and $y=\sin(t+\theta)$, $0\leq t\leq \pi$ (or simply $x=\cos t$, $y=\sin t$ with $\theta\leq t\leq \theta+\pi$). &lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-10-26T13:58:18.760" CommentCount="0" />
  <row Id="7912" PostTypeId="1" AcceptedAnswerId="7913" CreationDate="2010-10-26T13:58:44.757" Score="1" ViewCount="193" Body="&lt;p&gt;I have to prove this question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$2^{k} \,\left|\, {2k \choose 0} \cdot 3^{0} + {2k \choose 2} \cdot 3^{1} + \cdots + {2k \choose 2i} \cdot 3^{i} + \cdots + {2k \choose 2k} \cdot 3^{k}\right.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What i did was to get a closed form for this summation, and i haven't been able to get one. Any ideas, or is there any other method for proving this.&lt;/p&gt;&#xA;" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-10-26T14:01:57.187" LastActivityDate="2010-10-26T19:46:06.853" Title="$2^{k} \mid {2k \choose 0} \cdot 3^{0} + {2k \choose 2} \cdot 3^{1} + \cdots + {2k \choose 2i} \cdot 3^{i} + \cdots + {2k \choose 2k} \cdot 3^{k}$" Tags="&lt;combinatorics&gt;&lt;arithmetic&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="7913" PostTypeId="2" ParentId="7912" CreationDate="2010-10-26T14:13:28.157" Score="10" ViewCount="" Body="&lt;p&gt;The expression on the right just screams &quot;&lt;em&gt;Binomial Theorem! Binomial Theorem!&lt;/em&gt;&quot;. It's a quick matter to figure out &lt;em&gt;what&lt;/em&gt; to apply the Binomial Theorem to...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added:&lt;/strong&gt; So, how can we find out to what to apply the Binomial Theorem? Well, I can tell you what &lt;em&gt;my&lt;/em&gt; thought process was; I'm sure there are better or more direct ones. I'm going to take $(a+b)^{2k}$; $2k$ because the binomial coefficients are $2k$-choose-something. The $a$ should be equal to $1$, because we only have the $3$'s in the expansion. Oh, wait... the power of $3$ that shows up is &lt;em&gt;half&lt;/em&gt; the &quot;choose-something&quot;, and in the binomial theorem, if you take $(1+b)^{2k}$, then the $2i$th term is $\binom{2k}{2i}b^{2i}$... oh, well, let's take $b=3^{1/2}$ then! The exponents will work out. Of course, then now I only have the even powers... but that's okay, because the odd powers are going to have that pesky $\sqrt{3}$ in them. So, I'm going to look at $(1+\sqrt{3})^{2k}$. Let's take it out for a spin and see if that works... Well, what do you know? It does!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider $(1+\sqrt{3})^{2k}$; using the binomial theorem, you get&#xA;$$(1+\sqrt{3})^{2k} = \binom{2k}{0}1^{2k} + \binom{2k}{1}1^{2k-1}3^{1/2}+\cdots + \binom{2k}{2k}3^{k}.$$&#xA;The left hand side is $(4 + 2\sqrt{3})^{k} = 2^k(2+\sqrt{3})^{k}$. This can be written uniquely as $x + y\sqrt{3}$ with $x,y\in\mathbb{Z}$, so $x$ is divisible by $2^k$. The right hand side can likewise be written uniquely as $a + b\sqrt{3}$, with $a,b\in\mathbb{Z}$, and $a$ is exactly the sum you have. So $a = x$, and $2^k$ divides $x$, giving the result you want.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-10-26T17:32:36.923" LastActivityDate="2010-10-26T17:32:36.923" CommentCount="3" />
  <row Id="7914" PostTypeId="2" ParentId="7892" CreationDate="2010-10-26T14:23:31.827" Score="5" ViewCount="" Body="&lt;p&gt;Alternatively, we can compare $e^{1/e}$ and $\pi^{1/\pi}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $f(x) = x^{1/x}$. Then $f'(x) = x^{1/x} (1 - \log(x))/x^2$. Since $\log(x) &gt; 1$ for $x &gt; e$, we see that $f'(x) &amp;lt; 0$ for $e &amp;lt; x &amp;lt; \pi$. We conclude that $\pi^{1/\pi} &amp;lt; e^{1/e}$, and so $\pi^e &amp;lt; e^\pi$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same calculation shows that $f(x)$ reaches its maximum at $e^{1/e}$, and so in general $x^e &amp;lt; e^x$.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2010-10-26T19:03:40.620" LastActivityDate="2010-10-26T19:03:40.620" CommentCount="1" />
  <row Id="7915" PostTypeId="1" AcceptedAnswerId="7918" CreationDate="2010-10-26T14:26:18.923" Score="1" ViewCount="78" Body="&lt;p&gt;Hello! i have 2 perpendicular vectors in space . How can i determine the plane determined by the 2 vectors?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regards,&#xA;Alexandru Badescu&lt;/p&gt;&#xA;" OwnerUserId="2777" LastActivityDate="2010-10-26T14:32:05.057" Title="plane determined by 2 vectors" Tags="&lt;vector-space&gt;" AnswerCount="2" />
  <row Id="7916" PostTypeId="2" ParentId="7864" CreationDate="2010-10-26T14:28:30.360" Score="8" ViewCount="" Body="&lt;p&gt;I recently caught myself thinking that the formula for the determinant of a 2-by-2 matrix also works for a block matrix, i.e. $det (A B; C D) = det(A)det(D) - det(B)det(C)$.&lt;/p&gt;&#xA;" OwnerUserId="360" LastActivityDate="2010-10-26T14:28:30.360" CommunityOwnedDate="2010-10-26T20:30:36.610" CommentCount="1" />
  <row Id="7917" PostTypeId="2" ParentId="7915" CreationDate="2010-10-26T14:30:45.400" Score="1" ViewCount="" Body="&lt;p&gt;Take their cross product.  This will be perpendicular to the plane they determine.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-10-26T14:30:45.400" CommentCount="2" />
  <row Id="7918" PostTypeId="2" ParentId="7915" CreationDate="2010-10-26T14:32:05.057" Score="2" ViewCount="" Body="&lt;p&gt;The plane determined by two noncollinear vectors $\mathbf{v}_1$ and $\mathbf{v}_2$ is the collection of all vectors of the form $\alpha\mathbf{v}_1 + \beta\mathbf{v}_2$, with $\alpha$ and $\beta$ scalars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If by &quot;space&quot; you happen to mean $\mathbb{R}^3$, $\mathbf{v}_1=(a,b,c)$ and $\mathbf{v}_2=(r,s,t)$, then $(a,b,c)\times(r,s,t)$ (the cross product) is perpendicular to both $(a,b,c)$ and $(r,s,t)$, hence perpendicular to the plane they determine, so it will be the normal to the plane. One you have the normal, presumably you know how to get the usual equation of the plane. &lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-10-26T14:32:05.057" />
  <row Id="7919" PostTypeId="1" AcceptedAnswerId="7920" CreationDate="2010-10-26T14:40:16.993" Score="2" ViewCount="112" Body="&lt;p&gt;Venn diagrams captured my attention a few years ago. One night I decided to see if I could come up with some new Venn Diagrams. After a few hours, this is what I came up with. I'm a web developer not a mathematician, so I'm quite proud of this little creation. I've searched the internet and several books to confirm that this construction is unique. Lastly, I wanted to post it here because I know that the this crowd has more resources and knowledge than are available to me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/nzk3P.png&quot; alt=&quot;http://i55.tinypic.com/303jksh.jpg&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="2828" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-10-26T14:43:44.030" LastActivityDate="2010-10-26T14:48:13.783" Title="Is this 2d five-fold Venn Diagram original and valid?" Tags="&lt;logic&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="7920" PostTypeId="2" ParentId="7919" CreationDate="2010-10-26T14:48:13.783" Score="6" ViewCount="" Body="&lt;p&gt;It seems okay to me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might want to take a look at a very nice paper that appeard in the &lt;em&gt;Notices of the American Mathematical Society&lt;/em&gt; in 2006 about Venn diagrams:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Ruskey, Frank; Savage, Carla D.; Wagon, Stan.&#xA;&lt;strong&gt;The search for simple symmetric Venn diagrams.&lt;/strong&gt;&#xA;&lt;em&gt;Notices Amer. Math. Soc.&lt;/em&gt; &lt;strong&gt;53&lt;/strong&gt; (2006), no. 11, 1304–1312.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;At least I (an AMS member) can access it &lt;a href=&quot;http://www.ams.org/notices/200611/ea-wagon.pdf&quot; rel=&quot;nofollow&quot;&gt;in PDF format on the web&lt;/a&gt;. I don't know if the link will work for someone who is not a member, though. &lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-10-26T14:48:13.783" CommentCount="2" />
  <row Id="7921" PostTypeId="2" ParentId="7886" CreationDate="2010-10-26T14:49:56.700" Score="1" ViewCount="" Body="&lt;p&gt;You can probably also use the formula&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{k=0}^{m} \cos (\phi + k \alpha) = \frac{\sin((m+1)\alpha/2)\cos(\phi + m\alpha/2)}{\sin \alpha/2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which can be derived using complex numbers, or in a more elementary fashion, using the fact that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle 2\cos(\phi + k \alpha) \sin(\alpha/2)$  $\displaystyle = \sin (\phi + \alpha(k+1/2)) - \sin(\phi + \alpha(k-1/2))$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and telescoping the sum.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-10-26T14:49:56.700" />
  <row Id="7922" PostTypeId="2" ParentId="7757" CreationDate="2010-10-26T14:50:59.743" Score="2" ViewCount="" Body="&lt;p&gt;Here's a dirty trick.  $\frac{1}{2^n} \sum_{r \ge 0} r {n \choose r}$ is the expected size of a random subset of an $n$-element set.  But by linearity of expectation, this is $n$ times the probability that any given element is in a subset, which is $\frac{1}{2}$.  So&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{1}{2^n} \sum_{r \ge 0} r {n \choose r} = \frac{n}{2}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt;  And as long as I have this written up somewhere, I might as well use it.  In &lt;a href=&quot;http://math.stackexchange.com/questions/7912/2k-mid-2k-choose-0-cdot-30-2k-choose-2-cdot-31-cdots/7955#7955&quot;&gt;this math.SE answer&lt;/a&gt; I prove the following.  If $a_n, b_n$ are sequences satisfying $b_n = \sum_{k=0}^n {n \choose k} a_k$, and if $A(x) = \sum_{n \ge 0} a_n x^n, B(x) = \sum_{n \ge 0} b_n x^n$, then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$B(x) = \frac{1}{1 - x} A \left( \frac{x}{1 - x} \right).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this example $a_n = n$.  One can prove by various arguments that in this case $A(x) = \frac{x}{(1 - x)^2}$; this is a special case of an identity I use in the above answer.  It follows that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$B(x) = \frac{1}{1 - x} \left( \frac{ \frac{x}{1-x} }{ \left( 1 - \frac{x}{1-x} \right)^2 } \right) = \frac{x}{(1 - 2x)^2} = \frac{1}{2} \frac{2x}{(1 - 2x)^2} = \frac{1}{2} \sum_{n \ge 0} n \cdot 2^n x^n$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;as desired.  &lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-10-26T20:22:36.833" LastActivityDate="2010-10-26T20:22:36.833" CommentCount="1" />
  <row Id="7923" PostTypeId="1" CreationDate="2010-10-26T15:10:31.907" Score="7" ViewCount="589" Body="&lt;p&gt;are continuous functions always differentiable?&lt;/p&gt;&#xA;" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-10-26T15:45:31.473" LastActivityDate="2010-10-27T06:21:23.473" Title="Are Continuous Functions Always Differentiable?" Tags="&lt;calculus&gt;&lt;real-analysis&gt;" AnswerCount="5" CommentCount="3" />
  <row Id="7924" PostTypeId="1" AcceptedAnswerId="7929" CreationDate="2010-10-26T15:18:07.307" Score="5" ViewCount="216" Body="&lt;p&gt;I am learning about various signal processing methods in my university course, and I can't seem to grasp what 'energy' in signals represent. I mean, I know that it is the integral of the absolute value of energy squared (side question: why is absolute value needed?), but I don't know the meaning of what this energy is. Is there an intuitive way to think about these energies? Thanks a lot.&lt;/p&gt;&#xA;" OwnerUserId="2673" LastActivityDate="2010-10-27T12:53:48.000" Title="How can I interpret &quot;energy&quot; in signals?" Tags="&lt;signal-processing&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="1" />
  <row Id="7925" PostTypeId="2" ParentId="7923" CreationDate="2010-10-26T15:21:01.950" Score="15" ViewCount="" Body="&lt;p&gt;No. &lt;a href=&quot;http://en.wikipedia.org/wiki/Karl_Weierstrass&quot; rel=&quot;nofollow&quot;&gt;Weierstraß&lt;/a&gt; gave in 1872 the first published example of a &lt;a href=&quot;http://en.wikipedia.org/wiki/Weierstrass_function&quot; rel=&quot;nofollow&quot;&gt;contiuous function that's nowhere differentiable&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1269" LastActivityDate="2010-10-26T15:21:01.950" CommentCount="2" />
  <row Id="7926" PostTypeId="1" AcceptedAnswerId="7928" CreationDate="2010-10-26T15:33:03.453" Score="1" ViewCount="169" Body="&lt;p&gt;Notation question (I believe the notes that I'm reading uses pretty common notation):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $V $ be a vector space over a field $K$. What is $V^\mathbb{N}$? Is it a vector space of infinite dimensions? What are the elements of $V^\mathbb{N}$? Are they infinite length tuples?&lt;/p&gt;&#xA;" OwnerUserId="2468" LastActivityDate="2010-10-26T16:02:57.067" Title="If $V$ is a vector space, that what is $V^\mathbb{N}$?" Tags="&lt;notation&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="7927" PostTypeId="2" ParentId="7912" CreationDate="2010-10-26T15:36:28.990" Score="0" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt;$\ \ $  Specialize $\rm\ a = 1,\ b = 3 \ $ below and then apply the binomial theorem to the RHS&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\quad\quad\quad\ \ \ 2\:|\:a-b\ \Rightarrow\ 2\:|\:(a\pm\sqrt b)^2\ \:in\ \: \mathbb Z[\sqrt b]\ \Rightarrow\ 2^k\ |\ (\:a+\sqrt b)^{2k} + (\:a-\sqrt b)^{2k}\:\ in\ \:\mathbb Z$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;i.e. the sum is the trace of the $\rm\:$k'th power of an even quadratic integer so it is divisible by $\rm\:2^k\:$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As in your similar recently posted problems the key is to recognize the sum as a slight perturbation of a well-known sum. Here one notices that the given sum is the even bisection of the binomial expansion of power of a quadratic integer, which immediately yields the result as above.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-10-26T16:21:43.467" LastActivityDate="2010-10-26T16:21:43.467" CommentCount="1" />
  <row Id="7928" PostTypeId="2" ParentId="7926" CreationDate="2010-10-26T15:40:31.183" Score="7" ViewCount="" Body="&lt;p&gt;In my opinion, this notation is ambiguous (edit: but maybe I'm wrong).  It could mean two things: it could mean the space of functions $\mathbb{N} \to V$, which is a vector space in the obvious way.  This is the countable product in the category of vector spaces.  This vector space is very large; in particular, it has uncountable dimension and it is &lt;em&gt;not&lt;/em&gt; spanned by the functions $e_i$ which are equal to $1$ at $i$ and equal to $0$ elsewhere.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or it could mean the space of functions $\mathbb{N} \to V$ which are zero except at finitely many places, which is also a vector space in the obvious way, but countable-dimensional, and this is the vector space spanned by the $e_i$.  This is the countable &lt;em&gt;coproduct&lt;/em&gt; in the category of vector spaces.  (Note that it is not necessary to distinguish between finite products and finite coproducts because they are the same.)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I had to guess, I would guess that it means the former.  Can you give a link to the notes?  It might be possible to figure out which one is meant from context.&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-10-26T16:02:57.067" LastActivityDate="2010-10-26T16:02:57.067" CommentCount="9" />
  <row Id="7929" PostTypeId="2" ParentId="7924" CreationDate="2010-10-26T15:41:34.473" Score="8" ViewCount="" Body="&lt;p&gt;The energy is the energy carried in the signal. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The absolute value is probably because signal amplitudes are often represented by &lt;a href=&quot;http://en.wikipedia.org/wiki/Phasor&quot; rel=&quot;nofollow&quot;&gt;phasors&lt;/a&gt;, especially in the context of electromagnetic radiation or in the context of processing via Fourier analysis. And since it uses complex numbers, and the square of a complex number $z^2$ is not the same as its 'absolute value squared' $|z|^2$ (simplest case, take $z = -i$, the imaginary unit, then the former is -1 and the latter 1$), in the formla it is customary to write an absolute value. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That we associate energy with the square of the wave amplitude is due to the mechanical model of small amplitude waves of d'Alembert. Essentially the wave medium is thought of as a bunch of springs coupled together. For springs, with Hooke's law we have that the mechanical energy is proportional to the square of the amplitude of displacement. The analogy that the displacement of the media due to the wave should be well approximated by a simple spring leads us to take the infinitesimal mechanical energy associated to an infinitesimal chunck of the medium to be the square of the wave amplitude. So adding them all up you get an integral. &lt;/p&gt;&#xA;" OwnerUserId="1543" LastActivityDate="2010-10-26T15:41:34.473" CommentCount="5" />
  <row Id="7930" PostTypeId="2" ParentId="7926" CreationDate="2010-10-26T15:46:36.267" Score="2" ViewCount="" Body="&lt;p&gt;It would be the vector space consisting of an infinite product V&amp;nbsp;&amp;times;&amp;nbsp;V&amp;nbsp;&amp;times;&amp;nbsp;... of copies of V.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More generally: for two sets A and B, A&lt;sup&gt;B&lt;/sup&gt; often denotes the set of functions &lt;em&gt;f&lt;/em&gt;&amp;thinsp;:&amp;thinsp;B&amp;nbsp;&amp;rarr;&amp;nbsp;A. (The only exception I'm aware of is when the sets A and B happen to be ordinals, in set theory; then a different notion of exponentiation is often used.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How are these two concepts related? Well: as a vector space, V is also a set of vectors. And so a function &lt;em&gt;f&lt;/em&gt;&amp;thinsp;:&amp;thinsp;&amp;#8469;&amp;nbsp;&amp;rarr;&amp;nbsp;V is essentially a block-vector description of a vector &lt;strong&gt;v&lt;/strong&gt;&amp;nbsp;=&amp;nbsp;(&lt;em&gt;f&lt;/em&gt;(0),&amp;nbsp;&lt;em&gt;f&lt;/em&gt;(1),&amp;nbsp;...)&amp;nbsp;&amp;isin;&amp;nbsp;V&amp;nbsp;&amp;times;&amp;nbsp;V&amp;nbsp;&amp;times;&amp;nbsp;...&amp;nbsp;.&lt;/p&gt;&#xA;" OwnerUserId="439" LastActivityDate="2010-10-26T15:46:36.267" CommentCount="2" />
  <row Id="7931" PostTypeId="1" AcceptedAnswerId="7933" CreationDate="2010-10-26T15:52:16.840" Score="3" ViewCount="74" Body="&lt;p&gt;in R3 (3d) , having a vector perpendicular with a plane ( so we know where is 'up') , how do we determine if a certain point is below our plane ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regards,&#xA;Alexandru Badescu&lt;/p&gt;&#xA;" OwnerUserId="2777" LastActivityDate="2010-10-26T16:03:04.443" Title="point below a plane" Tags="&lt;vector-space&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="7932" PostTypeId="2" ParentId="7926" CreationDate="2010-10-26T15:55:53.427" Score="1" ViewCount="" Body="&lt;p&gt;I'd think of it as the set of sequences of elements of $V$, i.e.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ V^\mathbb{N} = \prod_{n=1}^\infty V.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I don't think there is a really nice description. In the case where $V$ is finite-dimensional, it'd just be $K^\mathbb{N}$ in a clever (and non-canonically isomorphic) disguise, but if $V$ were to be infinite-dimensional, $V^\mathbb{N}$ could be rather ugly. E.g. $\ell_1^\mathbb{N}$: sequences of absolutely summable sequences. Then a sequence of sequences would converge to a sequence iff it converged coordinatewise. Complicated and very confusing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Following Munkres' &quot;Topology&quot;, I prefer to use $V^\infty$ for set of sequences in $V$ for which only finitely many elements are nonzero, i.e. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$V^\infty = \bigoplus_{n=1}^\infty V = \left\lbrace (v_n) \in V^\mathbb{N} \;{\large\mid}\; \text{$v_n = 0$ from some index onwards}\right\rbrace. $$&lt;/p&gt;&#xA;" OwnerUserId="1269" LastActivityDate="2010-10-26T15:55:53.427" />
  <row Id="7933" PostTypeId="2" ParentId="7931" CreationDate="2010-10-26T16:01:26.490" Score="1" ViewCount="" Body="&lt;p&gt;Isn't this the same as your question &lt;a href=&quot;http://math.stackexchange.com/questions/7780/vector-question&quot;&gt;vector question&lt;/a&gt;?  Take the dot product of the perpendicular and the coordinates of the point.  If it is less than zero, the point is below the plane.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-10-26T16:01:26.490" CommentCount="1" />
  <row Id="7934" PostTypeId="2" ParentId="7931" CreationDate="2010-10-26T16:03:04.443" Score="5" ViewCount="" Body="&lt;p&gt;If $v$ is the vector that points 'up' and $p_0$ is some point on your plane, and&#xA;finally $p$ is the point that might be below the plane, compute the dot product&#xA;$v \cdot (p-p_0)$.  This projects the vector to $p$ on the up-direction.&#xA;This product is $\lbrace -, 0, + \rbrace$ if $p$ is below, on, above the plane, respectively.&lt;/p&gt;&#xA;" OwnerUserId="237" LastActivityDate="2010-10-26T16:03:04.443" CommentCount="1" />
  <row Id="7938" PostTypeId="1" AcceptedAnswerId="7940" CreationDate="2010-10-26T16:32:56.330" Score="6" ViewCount="325" Body="&lt;p&gt;One observes that $4!+1 =25=5^{2}$, $5!+1=121=11^{2}$ is a perfect square. Similarly for $n=7$ also we see that $n!+1$ is a perfect square. So one can ask the truth of this question:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Is $n!+1$ a perfect square for infinitely many $n$? If yes, then how to prove.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2010-10-27T05:05:45.630" LastActivityDate="2010-10-27T05:05:45.630" Title="$n!+1$ being a perfect square" Tags="&lt;number-theory&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="7939" PostTypeId="2" ParentId="7938" CreationDate="2010-10-26T16:41:33.707" Score="3" ViewCount="" Body="&lt;p&gt;The sequence of factorials $n!+1$ which are also perfect squares is &lt;a href=&quot;http://www.research.att.com/~njas/sequences/A085692&quot; rel=&quot;nofollow&quot;&gt;here in Sloane&lt;/a&gt;. It contains three terms, and notes that there are no more terms below $(10^9)!+1$, but as far as I  know there's no proof.&lt;/p&gt;&#xA;" OwnerUserId="173" LastActivityDate="2010-10-26T16:41:33.707" />
  <row Id="7940" PostTypeId="2" ParentId="7938" CreationDate="2010-10-26T16:43:33.457" Score="10" ViewCount="" Body="&lt;p&gt;This is Brocard's problem, and it is still open.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Brocard%27s_problem&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Brocard%27s_problem&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="940" LastActivityDate="2010-10-26T16:43:33.457" CommunityOwnedDate="2010-10-26T16:43:33.457" CommentCount="1" />
  <row Id="7941" PostTypeId="2" ParentId="7938" CreationDate="2010-10-26T16:44:15.780" Score="3" ViewCount="" Body="&lt;p&gt;My intuition would be that there are very few.  There are just not many squares and even fewer factorials.  &lt;a href=&quot;http://www.research.att.com/~njas/sequences/A025494&quot; rel=&quot;nofollow&quot;&gt;OEIS A025494&lt;/a&gt; lists the squares which are a sum of distinct factorials, which is less restrictive than what you ask and says the list is probably finite.  In particular, there are no more below 31!&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-10-26T16:44:15.780" />
  <row Id="7942" PostTypeId="2" ParentId="2601" CreationDate="2010-10-26T17:03:34.217" Score="0" ViewCount="" Body="&lt;p&gt;Using parallel, vertical &quot;gates&quot; to visualize high-dimensional or even infinite cartesian products of intervals. It comes up in topological and measurable spaces (specifically in probability theory), and is nearly essential to intuiting convergence in these spaces.&lt;/p&gt;&#xA;" OwnerUserId="2122" LastActivityDate="2010-10-26T17:03:34.217" CommunityOwnedDate="2010-10-26T17:03:34.217" />
  <row Id="7943" PostTypeId="2" ParentId="7924" CreationDate="2010-10-26T17:11:25.087" Score="3" ViewCount="" Body="&lt;p&gt;I try to give an explanation in the time and frequency domains, as normaly&#xA;used in Telecomunication Systems.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have a physical process described in the time by an oscillating&#xA;signal $s(t)$ such as an electrical voltage or current, the description&#xA;in the frequency domain is given by its spectrum&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\overrightarrow{S}(\omega )=\frac{1}{%&#xA;2\pi }\int_{-\infty }^{\infty }s(t)e^{-i\omega t}dt$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The signal is $$%&#xA;s(t)=\int_{-\infty }^{\infty }\overrightarrow{S}(\omega )e^{-i\omega&#xA;t}d\omega ,$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\omega =2\pi f=\frac{2\pi }{T}$ is the angular frequency. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The function $\overrightarrow{S}(\omega )$ is the &lt;em&gt;Fourier Transform&lt;/em&gt; of $s(t)$ and $s(t)$ is the inverse transform. Note: a &lt;a href=&quot;http://en.wikipedia.org/wiki/Fourier_transform&quot; rel=&quot;nofollow&quot;&gt;definition&lt;/a&gt; of the pair $(s(t),\overrightarrow{S}(\omega ))$  with different constant factors is possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &quot;energetic activity&quot; during the infinitesimal time interval $%&#xA;\delta t$ is $\delta E=s^{2}(t)\delta t$ (&lt;strong&gt;excluding a constant with physical&#xA;dimensions&lt;/strong&gt;, see note below). The instantaneous power is $$p(t)=\underset{\delta t\rightarrow 0%&#xA;}{\lim }\frac{\delta E}{\delta t}=s^{2}(t).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore the &quot;energy&quot; of the full process is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$E=\int_{-\infty }^{\infty }s^{2}(t)dt=2\pi \int_{-\infty }^{\infty&#xA;}S^{2}(\omega )d\omega ,$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $S(\omega )=\left\vert \overrightarrow{S}(\omega )\right\vert $ and $\overrightarrow{S}(\omega )=S(\omega )e^{i\psi (\omega )}$ is the&#xA;distribution or spectrum with the properties $S(-\omega )=S(\omega )$, $\psi&#xA;(-\omega )=\psi (\omega ).$ The equality between both integrals is known as the &lt;a href=&quot;http://en.wikipedia.org/wiki/Parseval%27s_theorem&quot; rel=&quot;nofollow&quot;&gt;Parseval's Theorem&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Note on the  physical dimensions&lt;/strong&gt;. e.g. in &lt;a href=&quot;http://en.wikipedia.org/wiki/Electrical_network&quot; rel=&quot;nofollow&quot;&gt;electrical circuits&lt;/a&gt; the product of a voltage $V$ by a current $I$ is a power  (energy/time).  By the Ohm's law, in d.c. circuits, $V=RI$ and $VI=RI^2=\dfrac{1}{R}V^2$; in a.c. circuits $| \overrightarrow{V}|| \overrightarrow{I}|=R |\overrightarrow{I}|^2$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Spectral_density&quot; rel=&quot;nofollow&quot;&gt;From the Wikipedia&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;The power spectral density (PSD)&#xA;  (...) describes how the power of a&#xA;  signal or time series is distributed&#xA;  with frequency. Here power can be the&#xA;  actual physical power, or more often,&#xA;  for convenience with abstract signals,&#xA;  can be defined as the squared value of&#xA;  the signal, that is, as the actual&#xA;  power dissipated in a load if the&#xA;  signal were a voltage applied to it.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="752" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2010-10-27T12:53:48.000" LastActivityDate="2010-10-27T12:53:48.000" CommentCount="3" />
  <row Id="7944" PostTypeId="1" AcceptedAnswerId="7948" CreationDate="2010-10-26T18:05:54.607" Score="5" ViewCount="145" Body="&lt;p&gt;I was trying to teach myself some basic Lie theory, and I came across this statement &lt;a href=&quot;http://mathworld.wolfram.com/Commutator.html&quot; rel=&quot;nofollow&quot;&gt;on Mathworld&lt;/a&gt;, relating the commutator of a group, $\alpha\beta\alpha^{-1}\beta^{-1}$, to the commutator of its Lie algebra, $[A,B] = AB-BA$:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;For instance, let $A$ and $B$ be square matrices, and let $\alpha(s)$ and $\beta(t)$ be paths in the Lie group of nonsingular matrices which satisfy&#xA;  $$\begin{align}&#xA;  \alpha(0)=\beta(0)    &amp;amp;=    I    \\&#xA;  \left.\frac{\partial\alpha}{\partial s}\right|_{s=0}    &amp;amp;=    A    \\&#xA;  \left.\frac{\partial\beta}{\partial s}\right|_{s=0}    &amp;amp;=    B,&lt;br&gt;&#xA;  \end{align}$$&#xA;  then&#xA;  $$\left.\frac{\partial}{\partial s}\frac{\partial}{\partial t}\alpha(s)\beta(t)\alpha^{-1}(s)\beta^{-1}(t)\right|_{(s=0,t=0)}=2[A,B].$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When I tried to derive this for myself, using the fact that&#xA;$$\left.\frac{\partial\alpha^{-1}}{\partial s}\right|_{s=0} = \left.-\alpha^{-1}\frac{\partial\alpha}{\partial s}\alpha^{-1}\right|_{s=0} = -A,$$&#xA;I expanded the expression to get&#xA;$$\left.\frac{\partial}{\partial s}\frac{\partial}{\partial t}\alpha\beta\alpha^{-1}\beta^{-1}\right|_{(s=0,t=0)}$$&#xA;$$=\left.\left( \frac{\partial\alpha}{\partial s}\frac{\partial\beta}{\partial t}\alpha^{-1}\beta^{-1} + \alpha\frac{\partial\beta}{\partial t}\frac{\partial\alpha^{-1}}{\partial s}\beta^{-1} + \frac{\partial\alpha}{\partial s}\beta\alpha^{-1}\frac{\partial\beta^{-1}}{\partial t} + \alpha\beta\frac{\partial\alpha^{-1}}{\partial s}\frac{\partial\beta^{-1}}{\partial t} \right)\right|_{(s=0,t=0)}$$&#xA;$$=AB - BA - AB + AB$$&#xA;$$=[A,B].$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The difference is that the factor of 2 is missing. This seems to agree with &lt;a href=&quot;http://ocw.mit.edu/courses/mathematics/18-755-introduction-to-lie-groups-fall-2004/lecture-notes/&quot; rel=&quot;nofollow&quot;&gt;the lecture notes I found on MIT OCW&lt;/a&gt;, which state (in Ch. 2, PDF 1) that if $X, Y \in \mathfrak{g}$,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$\exp(-tX)\exp(-tY)\exp(tX)\exp(tY) = \exp\{t^2[X,Y]+O(t^3)\}$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Since this is not my area of expertise, I wanted to make sure I got things right before I contacted MathWorld about a typo. Have I done something wrong somewhere, or is the MathWorld statement actually an error?&lt;/p&gt;&#xA;" OwnerUserId="856" LastActivityDate="2010-10-26T18:35:23.250" Title="On the relationship between the commutators of a Lie group and its Lie algebra" Tags="&lt;abstract-algebra&gt;&lt;group-theory&gt;&lt;lie-algebra&gt;" AnswerCount="1" CommentCount="8" FavoriteCount="1" />
  <row Id="7945" PostTypeId="2" ParentId="7923" CreationDate="2010-10-26T18:10:19.370" Score="-4" ViewCount="" Body="&lt;p&gt;differentiability implies continuity... the other way is not always true.&lt;/p&gt;&#xA;" OwnerUserId="2231" LastActivityDate="2010-10-26T18:10:19.370" CommentCount="1" />
  <row Id="7946" PostTypeId="2" ParentId="7900" CreationDate="2010-10-26T18:15:25.560" Score="1" ViewCount="" Body="&lt;p&gt;It's right and wrong.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a certain raw number of possibilities and let's assume from the earlier remarks that 97 billion is the correct number.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What the hype does not mention is that there is also some statistical distribution to these possibilities. Some are much likelier than others, and some selections with different &quot;genotypes&quot; have similar &quot;phenotypes&quot;: they are almost the same in effect.  The amount of information in one selection from the 97 billion is, on average, a lot less than the number of bits needed to specify a unique integer between 1 and 97000000000.&lt;/p&gt;&#xA;" OwnerUserId="467" LastActivityDate="2010-10-26T18:15:25.560" />
  <row Id="7947" PostTypeId="2" ParentId="7924" CreationDate="2010-10-26T18:18:00.633" Score="1" ViewCount="" Body="&lt;p&gt;The electrical power in a resistive circuit is given by $P = I V$. Frequently one has a fixed resistance (or more in general, impedance), so that voltage and current are variable; taking voltage as independent variable, we have $P = V^2/R$. This gives the instantaneous power (energy over time unit), and hence also gives the average power (or energy over some finite time interval) if the voltage is constant (DC current). When the voltage varies peridocally in time (for example, AC current), the above equation applies to each instant of time ($ P(t) = V(t)^2/R $) , so that, for getting the average power over a period one must integrate (&lt;a href=&quot;http://en.wikipedia.org/wiki/Root_mean_square#Average_electrical_power&quot; rel=&quot;nofollow&quot;&gt;see here&lt;/a&gt;). That's why AC energy is typically measured in RMS units: a 120V power outlet gives a sinuosid of approx. 170V peak voltage: that sinusoid &lt;em&gt;has the same energy&lt;/em&gt; as  a constant 120V voltage.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the case of electric signals, but the analogy applies in other fields.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another (related) motivation comes from the &lt;a href=&quot;http://en.wikipedia.org/wiki/Parseval%27s_theorem&quot; rel=&quot;nofollow&quot;&gt;Parseval's theorem&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="312" LastActivityDate="2010-10-26T18:18:00.633" />
  <row Id="7948" PostTypeId="2" ParentId="7944" CreationDate="2010-10-26T18:35:23.250" Score="1" ViewCount="" Body="&lt;p&gt;Ignoring terms of order 3 and higher, and inverting 1+X using the power series,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1+A)(1+B)(1-A+A^2)(1-B+B^2)&#xA;= (1 + (A+B) + AB) (1 - (A+B) + (A^2 + AB + B^2))&#xA;= 1 - (A+B)^2 + AB + A^2 + AB + B^2 &#xA;= 1 - (A^2 + AB + BA + B^2) + 2AB + A^2 + B^2&#xA;= 1 + AB - BA&lt;br&gt;&#xA;= 1 + [A,B]&lt;/p&gt;&#xA;" OwnerUserId="467" LastActivityDate="2010-10-26T18:35:23.250" />
  <row Id="7949" PostTypeId="1" AcceptedAnswerId="13637" CreationDate="2010-10-26T18:42:51.067" Score="10" ViewCount="187" Body="&lt;p&gt;Let $\mathbf C$ be an abelian category admitting projective limits. Let's consider the category whose objects are those of the form&#xA;$$&#xA;0\to A\to B\to C\to 0&#xA;$$&#xA;and whose morphisms are triples of morphisms of $\mathbf C$ such that the diagram&#xA;$$&#xA;\begin{array}{ccccc}&#xA;A&amp;amp;\hookrightarrow &amp;amp; B &amp;amp; \to &amp;amp; C &amp;amp; \newline&#xA;\downarrow &amp;amp;&amp;amp;\downarrow &amp;amp;&amp;amp; \downarrow \newline&#xA;A' &amp;amp;\hookrightarrow &amp;amp; B' &amp;amp;\to &amp;amp;  C'&#xA;\end{array}&#xA;$$&#xA;commutes in all its parts. Call this category $\boldsymbol\Sigma(\mathbf C)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How could one characterize the limits in $\boldsymbol\Sigma(\mathbf C)$? A little meditation shows that inverse systems are objects of the form&#xA;$$&#xA;\mathcal E_i\colon 0\rightarrow A_i\rightarrow B_i\rightarrow C_i\to 0&#xA;$$&#xA;(every object in the sequence can be thought as an element in a separate inverse system), so the universal property of $\varprojlim_\mathbf J \mathcal E_i$, whatever it turns out to be, must be enjoyed by&#xA;$$&#xA;\textstyle \varprojlim_\mathbf J \mathcal E_i : 0\to \varprojlim_\mathbf J A_i\to \varprojlim_\mathbf J B_i \to \varprojlim_\mathbf J C_i\to 0&#xA;$$&#xA;as soon as one looks to  ${A_i},{B_i},{C_i}$ as three inverse systems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What condition(s?) has(ve?) to be imposed on them to make sure the sequence is exact?&lt;/p&gt;&#xA;" OwnerUserId="685" LastActivityDate="2010-12-09T10:22:23.303" Title="Limits in the category of exact sequences" Tags="&lt;category-theory&gt;&lt;abelian-categories&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="4" />
  <row Id="7950" PostTypeId="2" ParentId="7923" CreationDate="2010-10-26T18:43:13.230" Score="2" ViewCount="" Body="&lt;p&gt;For a nice simple example of an everywhere continuous, nowhere differentiable function it's hard to beat &lt;a href=&quot;http://www-formal.stanford.edu/jmc/weierstrass.pdf&quot; rel=&quot;nofollow&quot;&gt;this example&lt;/a&gt; of John McCarthy.&lt;/p&gt;&#xA;" OwnerUserId="1301" LastActivityDate="2010-10-26T18:43:13.230" CommentCount="1" />
  <row Id="7951" PostTypeId="1" AcceptedAnswerId="7960" CreationDate="2010-10-26T18:50:35.453" Score="1" ViewCount="444" Body="&lt;p&gt;If I have 5 students (A-E) that score 80%, 70%, 70%, 60% and 50% on a test what percentiles do they fall in?&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;A - 20th percentile (80%)&lt;/li&gt;&#xA;&lt;li&gt;B - 40th percentile (70%)&lt;/li&gt;&#xA;&lt;li&gt;C - 40th percentile (70%)&lt;/li&gt;&#xA;&lt;li&gt;D - 80th percentile (60%)&lt;/li&gt;&#xA;&lt;li&gt;E - 100th percentile (50%)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Is this correct?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me expand the question as I wasn't clear earlier: If A is in the top 20th percentile, are B and C in the top 40th percentile or top 60th percentile?&lt;/p&gt;&#xA;" OwnerUserId="2835" LastEditorUserId="2835" LastEditorDisplayName="" LastEditDate="2010-10-27T00:43:23.840" LastActivityDate="2010-10-27T02:45:55.840" Title="How do percentiles work?" Tags="&lt;statistics&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" />
  <row Id="7952" PostTypeId="1" AcceptedAnswerId="7957" CreationDate="2010-10-26T19:22:39.827" Score="2" ViewCount="102" Body="&lt;p&gt;This is a question on the homework for my finite fields class. The beginning of the assignment defines the following notation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $i\geq1 $, define the following elements of $A=\mathbb{F}_{q}[T]$ : &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$[i]=T^{q^{i}}-T$, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$L_{i}=[i][i-1]\cdots[1]$, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$D_{i}=[i][i-1]^{q}\cdots[1]^{q^{i-1}}$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;and put $D_{0}=L_{0}=1$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $n\geq1$.&#xA;Put $A(n):=\{a \in A\mid \text{deg} (a)&amp;lt; n \} $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and $e_{n}(x):=\prod_{a\in A(n)}\left(x+a\right)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am trying to prove: $\sum\frac{1}{b}=\frac{(-1)^{n}}{L_{n}}$ where the LHS ranges over all monic $b$ in $A$ of degree equal to $n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Earlier on the assignment, we showed $e_{n}(x)+D_{n}=\prod(x+b)$ where $b$ is monic with degree $n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I thought I would be able to take the Carlitz analog of the logarithm $\log_{c}(x)$ and then take the formal derivative and evaluate it at zero. But I've realized it is not true that $\log_{c}(xy)=\log_{c}(x)+\log_{c}(y)$. I've been stuck on this problem for some time, and can't think of any other way to approach it. I would appreciate a nudge in the right direction. &lt;/p&gt;&#xA;" OwnerUserId="249" LastActivityDate="2010-10-27T19:11:09.583" Title="Using Carlitz's exponential formula to prove an identity" Tags="&lt;homework&gt;&lt;finite-fields&gt;" AnswerCount="2" />
  <row Id="7954" PostTypeId="1" AcceptedAnswerId="7968" CreationDate="2010-10-26T19:33:27.977" Score="4" ViewCount="295" Body="&lt;p&gt;Let's assume we have a commutative ring with identity. Can the Nil-Radical and the Jacobson Radical be equal in a non-trivial case (i.e. not every nonzero prime ideal in said ring is maximal)?&#xA;Are there any interesting examples of this case?&lt;/p&gt;&#xA;" OwnerUserId="766" LastEditorUserId="766" LastEditorDisplayName="" LastEditDate="2010-10-26T20:03:18.730" LastActivityDate="2010-10-26T21:29:40.077" Title="Nil-Radical equals Jacobson Radical even though not every prime ideal is maximal?" Tags="&lt;abstract-algebra&gt;&lt;commutative-algebra&gt;&lt;ring-theory&gt;" AnswerCount="2" CommentCount="6" FavoriteCount="1" />
  <row Id="7955" PostTypeId="2" ParentId="7912" CreationDate="2010-10-26T19:46:06.853" Score="4" ViewCount="" Body="&lt;p&gt;Let's talk about some general methods.  This is a special case of the following problem: given that you understand a sequence $a_n$, how can you understand the sequence $b_n = \sum_{k=0}^n {n \choose k} a_k$?  Generating function techniques will come to your rescue.  If you let $A(x) = \sum_{n \ge 0} a_n x^n$, then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$B(x) = \sum_{n \ge 0} b_n x^n = \sum_{n \ge 0} \sum_{k=0}^{n} {n \choose k} a_k x^n.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At this point we employ one of the simplest but most effective tools in &lt;a href=&quot;http://www.math.upenn.edu/~wilf/DownldGF.html&quot; rel=&quot;nofollow&quot;&gt;generatingfunctionology&lt;/a&gt;, namely we exchange the order of summation.  This gives&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$B(x) = \sum_{k \ge 0} \sum_{n \ge k} {n \choose k} a_k x^n.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since we want an answer in terms of $A$, it seems natural to factor $a_k x^k$ out of the inner sum.  This gives&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$B(x) = \sum_{k \ge 0} a_k x^k \sum_{n \ge k} {n \choose k} x^{n-k}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now a generatingfunctionologist will immediately recognize the inner sum as $\sum_{m \ge 0} {m+k \choose k} x^m = \frac{1}{(1 - x)^{k+1}}$.  This gives&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$B(x) = \sum_{k \ge 0} a^k \frac{x^k}{(1 - x)^{k+1}} = \frac{1}{1 - x} A \left( \frac{x}{1 - x} \right).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Beautiful.  Now, in this problem we are dealing with the sequence $a_{2k} = 3^k, a_{2k+1} = 0$.  This has generating function $A(x) = \frac{1}{1 - 3x^2}$, hence&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$B(x) = \frac{1}{1 - x} \left( \frac{1}{1 - \frac{3x^2}{(1 - x)^2} } \right) = \frac{1 - x}{(1 - x)^2 - 3x^2} = \frac{1 - x}{1 - 2x - 2x^2}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, what we want to prove is that the even coefficients of $B \left( \frac{x}{\sqrt{2}} \right)$ are integral.  To extract the even coefficients we appeal to another general tool in generatingfunctionology: if $B(x) = \sum_{n \ge 0} b_n x^n$ is a generating function you want to extract the even coefficients from, then these are given by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{n \ge 0} b_{2n} x^{2n} = \frac{B(x) + B(-x)}{2}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a special case of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Discrete_Fourier_transform&quot; rel=&quot;nofollow&quot;&gt;discrete Fourier transform&lt;/a&gt;.  Once you know this, it is completely mechanical to verify that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{B \left( \frac{x}{\sqrt{2}} \right) + B \left( \frac{-x}{\sqrt{2}} \right)}{2} = \frac{1 - x^2}{1 - 4x^2 + x^4}$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;which has integer coefficients (since the constant term of the denominator is $1$).  This is a bit tedious, but my point here is that once one has picked up a few generating function techniques most of the work in solving problems like these is automatic.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-10-26T19:46:06.853" CommentCount="1" />
  <row Id="7956" PostTypeId="2" ParentId="7954" CreationDate="2010-10-26T19:49:30.177" Score="4" ViewCount="" Body="&lt;p&gt;Let $k$ be a field, and let $R=k[x_1,\ldots,x_n]/\mathfrak{I}$, where $\mathfrak{I}$ is an ideal of $k[x_1,\ldots,x_n]$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The nilradical of $R$ is $\sqrt{\mathfrak{I}}/\mathfrak{I}$, and the Jacobson radical of $R$ is $\mathfrak{M}/\mathfrak{I}$, where $\mathfrak{M}$ is the intersection of all maximal ideals of $k[x_1,\ldots,x_n]$ that contain $\mathfrak{I}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But by Hilbert's Nullstellensatz, $\sqrt{\mathfrak{I}}$, the radical of $\mathfrak{I}$, &lt;em&gt;is&lt;/em&gt; the intersection of all maximal ideals of $k[x_1,\ldots,x_n]$ that contain $\mathfrak{I}$, so the nilradical of $R$ equals the Jacobson radical of $R$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, in these rings there are generally prime ideals that are not maximal. For example, if $n\geq 3$ and you take $\mathfrak{I}=(x_1)$, then $(x_1,x_2)+\mathfrak{I}$ is prime but not maximal. &lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-10-26T19:49:30.177" />
  <row Id="7957" PostTypeId="2" ParentId="7952" CreationDate="2010-10-26T19:49:59.490" Score="3" ViewCount="" Body="&lt;p&gt;From $e_n(x)+D_n=\prod(x+b)$&#xA;it follows that $\prod b=D_n$ and the $x^1$ coefficient of $e_n(x)$&#xA;is $D_n\sum 1/b$. But the $x^1$ coefficient is equal to $\prod'a$&#xA;where the product is over all nonzero polynomials $a$ of degree $ &amp;lt; n$.&#xA;All you have to do is to prove this product equals $(-1)^nD_n/L_n$.&#xA;The leading coefficient should be easy enough. For the rest you need&#xA;to do a census of the degrees of the irreducible factors of polynomials&#xA;of degree $ &amp;lt; n$. Possibly some earlier results in your course might&#xA;cover this?&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-10-26T19:49:59.490" CommentCount="1" />
  <row Id="7960" PostTypeId="2" ParentId="7951" CreationDate="2010-10-26T20:02:55.093" Score="5" ViewCount="" Body="&lt;p&gt;You've got them backwards. The 20th percentile would be the level below which only 20% of the observations fall. Since 80% of the scores fall below A's score (four out of 5), then A is in the 80th percentile. Since 20% of the scores fall below D's score, then D is in the 20th percentile, not the 80th. See &lt;a href=&quot;http://en.wikipedia.org/wiki/Percentile&quot; rel=&quot;nofollow&quot;&gt;Wikipedia's page&lt;/a&gt; on percentiles (it's rather rough, but the lead will do it). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; You edited the question, so here goes the answer to that: &#xA;It seems that you are not really asking about &lt;em&gt;percentiles&lt;/em&gt;, but about &lt;strong&gt;percents&lt;/strong&gt;. They are not the same thing. A is in the top 20 percent, but not &quot;the top 20 percentile&quot;. By &lt;em&gt;definition&lt;/em&gt;, &lt;strong&gt;percentile&lt;/strong&gt; refers to how many quantities are &lt;em&gt;below&lt;/em&gt;. It makes no sense to talk about &quot;top 20 percentile&quot;.  Your intended question is about percents, not 'percentiles'.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To answer that: You have 5 people; A is the top, B and C are tied in second, D is in fourth, and E in fifth. A is certainly in the top 20% of the group, E in the bottom 20%; D and E are in the bottom 40%. A, B, and C are in the top 60% (not the top 40%); B, C, D, and E are in the bottom 80%. But you are using &quot;percentile&quot; wrong. If X is in the 40th percentile, then that means that 40% of the scores are below him; that would mean he is at the bottom of the top 60%. A is in the 80 percentile, B and C are in the 40th percentile, D is in the 20th percentile. &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-10-27T02:45:55.840" LastActivityDate="2010-10-27T02:45:55.840" />
  <row Id="7961" PostTypeId="1" AcceptedAnswerId="7962" CreationDate="2010-10-26T20:04:54.647" Score="1" ViewCount="86" Body="&lt;p&gt;Suppose for company A, there is a $60 \%$ chance that no claim is made in the coming year. If one or more claims are made, then the total amount is normally distributed with mean $1000$ and standard deviation $2000$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose for company B, there is a $70 \%$ chance that no claim is made during the coming year. If one or more claims are made, then the total amount is normally distributed with mean $9000$ and standard deviation $2000$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume the total claim amounts are independent. &lt;strong&gt;What is the probability that in the coming year, company B's total claim amount will exceed company A's total claim amount?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $X_A$ be company A's total claim amount and $X_B$ likewise. We want to find $P(X_B - X_A &gt;0)$. Now $E[X_B-X_A] = 8000$ and $\text{Var}[X_B-X_A] = 2000^2+2000^2$. So what we want to find is:   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ \displaystyle P \left[Z \geq \frac{-8000}{\sqrt{2000^2+2000^2}} \right]$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I feel that we are given two conditional distributions: $P(X_A|I_A)$ and $P(X_B|I_B)$ where $I_A$ and $I_B$ indicate that more than one claim is made for company A and B respectively.&lt;/p&gt;&#xA;" OwnerUserId="2548" LastEditorUserId="2548" LastEditorDisplayName="" LastEditDate="2010-10-26T20:31:54.517" LastActivityDate="2010-10-26T20:46:47.243" Title="Normal Distributions and Claim Amounts" Tags="&lt;probability&gt;" AnswerCount="1" />
  <row Id="7962" PostTypeId="2" ParentId="7961" CreationDate="2010-10-26T20:19:05.690" Score="1" ViewCount="" Body="&lt;p&gt;As the expected difference in claims is 2.8 standard deviations, there is little (how little?) chance that A's claims will exceed B's claims assuming they each get one.  If we ignore it, and assume that the chance of a claim at A is independent of the chance of a claim at B (can we?)  We have probability of no claim either place, so a tie, =60%*70%=42%.  Probability of claim at A, none at B =40%*70%=28% chance of higher claim at A.  Chance of higher claim at B=30%, as if there is a claim it will be higher than A's.  These add to 100%, a good check.  I'll let you figure the chance that they both get a claim and A's is higher than B's.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is also a little hard to fathom a claim being made on A which is less than zero, but if you are only 1/2 standard deviation low, that will happen.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2010-10-26T20:46:47.243" LastActivityDate="2010-10-26T20:46:47.243" CommentCount="4" />
  <row Id="7963" PostTypeId="1" AcceptedAnswerId="7964" CreationDate="2010-10-26T20:43:21.010" Score="1" ViewCount="156" Body="&lt;p&gt;I'm looking for a function that will replicate logical operators. In particular consider the AND operator. Then I am looking for a function z = f(x,y) such that&#xA;$$x,y,z \in \{0,1\}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;f(0,0)=0&#xA;f(1,0)=0&#xA;f(0,1)=0&#xA;f(1,1)=1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More importantly I need to generalize over a space where Xi,Yi ∈ {0,1} and I need to do this for other sorts of operators too. (X XOR Y AND Z for example.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So given a vector Xi and a vector Yi I need to be able to generate Zi w&lt;/p&gt;&#xA;" OwnerUserId="2121" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-10-27T02:37:29.687" LastActivityDate="2011-05-31T16:48:30.330" Title="How to represent logical operators as functions?" Tags="&lt;functions&gt;" AnswerCount="3" CommentCount="4" FavoriteCount="2" />
  <row Id="7964" PostTypeId="2" ParentId="7963" CreationDate="2010-10-26T20:47:08.930" Score="6" ViewCount="" Body="&lt;p&gt;If $x, y \in [0, 1]$ you can take $f(x, y) = xy$.  This comes from replacing &quot;true&quot; and &quot;false&quot; with probabilities and from assuming that $x, y$ are probabilities of independent events.  Similarly for NOT take $f(x) = 1 - x$, and for OR take $f(x, y) = 1 - (1 - x)(1 - y)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If this isn't what you're looking for then you're going to have to be more precise about what you want.  If you want to do something with vectors with entries in $\{ 0, 1 \}$, what's wrong with applying the operations pointwise?&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-10-26T20:47:08.930" CommentCount="6" />
  <row Id="7965" PostTypeId="1" CreationDate="2010-10-26T20:48:16.740" Score="7" ViewCount="141" Body="&lt;p&gt;I'm aware that one can use a Fast Fourier Transform (FFT) to take the cost of multiplication of two polynomials of degree N from O$(N^2)$ to O$(N \ln N)$ (which is an amazing reduction when dealing with large polynomials!). Does a similar transformation procedure exist for multinomials?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm interested in the special case where the number of independent variables is only two, ie. $h(x,y) = f(x,y)g(x,y)$, but I'd love to read up on the general procedure. &lt;/p&gt;&#xA;" OwnerUserId="196" LastEditorUserId="196" LastEditorDisplayName="" LastEditDate="2010-10-26T20:56:42.473" LastActivityDate="2010-10-26T20:56:42.473" Title="Use of FFT in the multiplication of multinomials" Tags="&lt;algorithms&gt;&lt;polynomials&gt;" CommentCount="8" FavoriteCount="4" />
  <row Id="7966" PostTypeId="1" AcceptedAnswerId="9539" CreationDate="2010-10-26T21:16:44.867" Score="9" ViewCount="349" Body="&lt;p&gt;What are good expositions of spectral sequences, which include a thorough introduction to the topic as well as the most important examples of applications - maybe with an emphasis an topological and/or operator algebraic uses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that the wikipedia article is quite extensive, but I'd like to have something... nicer.&lt;/p&gt;&#xA;" OwnerUserId="367" LastEditorUserId="367" LastEditorDisplayName="" LastEditDate="2011-02-08T14:27:39.057" LastActivityDate="2011-02-08T14:27:39.057" Title="Reference for spectral sequences" Tags="&lt;reference-request&gt;&lt;algebraic-topology&gt;&lt;homological-algebra&gt;&lt;operator-algebras&gt;" AnswerCount="7" CommentCount="8" FavoriteCount="4" />
  <row Id="7967" PostTypeId="2" ParentId="7966" CreationDate="2010-10-26T21:29:15.183" Score="4" ViewCount="" Body="&lt;p&gt;I really like the last chapter of Aluffi's book &quot;Algebra: Chapter 0.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps one of the canonical references (especially to see tons of applications) is John McCleary's &quot;A User's Guide to Spectral Sequences.&quot;&lt;/p&gt;&#xA;" OwnerUserId="1007" LastActivityDate="2010-10-26T21:29:15.183" />
  <row Id="7968" PostTypeId="2" ParentId="7954" CreationDate="2010-10-26T21:29:40.077" Score="4" ViewCount="" Body="&lt;p&gt;There are indeed very many rings in which the nilradical equals the Jacobson radical.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the following property of a ring: every prime ideal is the intersection of the maximal ideals containing it.  These rings are called &lt;strong&gt;Hilbert-Jacobson&lt;/strong&gt; rings.  In a Hilbert-Jacobson ring, the intersection of all the prime ideals is therefore also the intersection of all maximal ideals, i.e., the nilradical and Jacobson radical coincide. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And indeed there are very many Hilbert-Jacobson rings.  One useful result in this direction is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Theorem: Let $R$ be a Hilbert Jacobson ring, and let $S$ be a ring which is finitely generated as an $R$-algebra.  Then $S$ is also a Hilbert-Jacobson ring.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since any field is trivially a Hilbert-Jacobson ring, it follows that all algebras which are finitely generated over a field are Hilbert-Jacobson, as Arturo Magidin pointed out in his answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For some information on this subject, including a proof of the theorem, see Section 8 of&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.math.uga.edu/~pete/integral.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.math.uga.edu/~pete/integral.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and in particular Theorem 128 on p. 82.&lt;/p&gt;&#xA;" OwnerUserId="299" LastActivityDate="2010-10-26T21:29:40.077" />
  <row Id="7973" PostTypeId="2" ParentId="7923" CreationDate="2010-10-26T22:23:47.313" Score="6" ViewCount="" Body="&lt;p&gt;No, consider the example of $f(x) = |x|$. This function is continuous but not differentiable at $x = 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are even more bizare functions that are not differentiable everywhere, yet still continuous. This class of functions lead to the development of the study of fractals.&lt;/p&gt;&#xA;" OwnerUserId="1906" LastActivityDate="2010-10-26T22:23:47.313" />
  <row Id="7974" PostTypeId="1" AcceptedAnswerId="8043" CreationDate="2010-10-26T22:43:53.690" Score="1" ViewCount="75" Body="&lt;p&gt;I'm trying to sample a lot of points efficiently.  I'm wondering if the following method is possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I sample points of a function (evaluate the function) mod $n$. I.e. I calculate f(element one), f(element 2)...f(element n).  I do this for all of the points in a ring.  Now I take a second ring (mod $m$, for instance) and calculate the points of a function for all elements of this ring.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So my idea is this: I've sampled only $n+m$ points, but the combinations of these sample points number $n \cdot m$.  I'd like to calculate the sum of (evaluations of) this larger set of points efficiently.  I'm wondering if this is possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd like to know as much about this idea as possible.  I'd like to know related ideas, what's possible, or just general ideas on how to do this.  Any pointers would constitute an acceptable answer to me, and an actual method would be absolutely wonderful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I realize that this may be more of a numerical method than mathematics, but I'm hoping that everyone will accept this question, since it may allow new methods of integration as a result.&lt;/p&gt;&#xA;" OwnerUserId="2626" LastActivityDate="2010-10-27T13:51:14.493" Title="Efficient Sampling" Tags="&lt;analysis&gt;&lt;intuition&gt;&lt;statistics&gt;&lt;numerical-methods&gt;&lt;ring-theory&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="0" />
  <row Id="7976" PostTypeId="2" ParentId="7512" CreationDate="2010-10-26T23:44:06.313" Score="1" ViewCount="" Body="&lt;p&gt;We used Conway (about 3/4th) of the book for functional analysis. It was very dense and I felt like I wanted to stab myself in the guts, but I made it! ;-).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I agree with the other posts that you need to brush up some material before you do this.&lt;/p&gt;&#xA;" OwnerUserId="1120" LastActivityDate="2010-10-26T23:44:06.313" />
  <row Id="7978" PostTypeId="1" AcceptedAnswerId="7985" CreationDate="2010-10-27T00:58:55.023" Score="3" ViewCount="243" Body="&lt;p&gt;Is there a succinct proof for the fact that the rank of a non-zero skew-symmetric matrix ($A = -A^T$)  is at least 2? I can think of a proof by contradiction: Assume rank is 1. Then you express all other rows as multiple of the first row. Using skew-symmetric property, this matrix has to be a zero matrix. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why does such a matrix have at least 2 non-zero eigenvalues?&lt;/p&gt;&#xA;" OwnerUserId="1527" LastActivityDate="2010-10-27T02:45:27.853" Title="Rank of skew-symmetric matrix" Tags="&lt;linear-algebra&gt;&lt;matrices&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="7979" PostTypeId="2" ParentId="7966" CreationDate="2010-10-27T01:02:18.833" Score="3" ViewCount="" Body="&lt;p&gt;Allen Hatcher has some textbooks available for free download on his website, including the first 100 pages of an unfinished text on &lt;a href=&quot;http://www.math.cornell.edu/~hatcher/SSAT/SSATpage.html&quot; rel=&quot;nofollow&quot;&gt;Spectral Sequences in Algebraic Topology&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It should qualify as a thorough first introduction to the topic, but I don't think all of the important examples have made it into the text.  I haven't read it myself, but I remember his more basic text on algebraic topology being very friendly and a good introduction.&lt;/p&gt;&#xA;" OwnerUserId="1461" LastActivityDate="2010-10-27T01:02:18.833" />
  <row Id="7980" PostTypeId="2" ParentId="7864" CreationDate="2010-10-27T01:03:59.343" Score="2" ViewCount="" Body="&lt;p&gt;The question I've heard on many levels (including the grad level): what is the square root of $a^2$? And everyone says: it's $a$! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact it is $|a|$. &lt;/p&gt;&#xA;" OwnerUserId="2450" LastActivityDate="2010-10-27T01:03:59.343" CommunityOwnedDate="2010-10-27T01:03:59.343" CommentCount="2" />
  <row Id="7981" PostTypeId="1" AcceptedAnswerId="7987" CreationDate="2010-10-27T01:31:23.443" Score="13" ViewCount="1300" Body="&lt;p&gt;I've read so much about it but none of it makes a lot of sense. Also, what's so unsolvable about it?&lt;/p&gt;&#xA;" OwnerUserId="1141" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-10-27T05:00:33.927" LastActivityDate="2010-10-28T12:58:38.593" Title="Can someone please explain the Riemann Hypothesis to me... in English?" Tags="&lt;number-theory&gt;&lt;soft-question&gt;&lt;complex-analysis&gt;&lt;riemann-hypothesis&gt;" AnswerCount="4" CommentCount="7" FavoriteCount="4" />
  <row Id="7982" PostTypeId="2" ParentId="7978" CreationDate="2010-10-27T01:45:49.933" Score="2" ViewCount="" Body="&lt;p&gt;There exists an invertible matrix $P$ such that $^t P A P$ is diagonal with blocks equal to $\begin{pmatrix} 0 &amp;amp; 1 \\ -1 &amp;amp; 0 \end{pmatrix}$ or $0$ (it is a simple exercise in bilinear forms), so that the rank of $A$ is necessarily even.&lt;/p&gt;&#xA;" OwnerUserId="2660" LastActivityDate="2010-10-27T01:45:49.933" />
  <row Id="7983" PostTypeId="1" AcceptedAnswerId="8137" CreationDate="2010-10-27T02:10:11.333" Score="2" ViewCount="162" Body="&lt;p&gt;Suppose I have an optimization problem of the form:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\min \mathbf{d}^{T}\mathbf{y}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;subject to &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathbf{M}\mathbf{y} \geq  \mathbf{d}, \qquad \mathbf{y} \geq 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If a solution $\mathbf{s}$ which satisfies $\mathbf{M}\mathbf{s} = \mathbf{d}$ and $\mathbf{s} \geq 0$, how can it be shown $\mathbf{s}$ is optimal for the above linear programming problem?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Initial thought was to form the dual:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;max p'd&lt;/p&gt;&#xA;&#xA;&lt;p&gt;p&gt;=0&lt;/p&gt;&#xA;&#xA;&lt;p&gt;p'M&amp;lt;=d&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus weak duality gives&#xA;d'y &amp;lt;= p'd&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And the constraint of the dual gives p'M&amp;lt;=d&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so : p'My&amp;lt;=dy&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and thus p'd&amp;lt;=dy&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Combining the two equalities shows that p'd must equal dy, and thus with strong duality this means that p'd is the optimal solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where did I go wrong, and how does being symmetric fit into the problem?&lt;/p&gt;&#xA;" OwnerUserId="2014" LastEditorUserId="2014" LastEditorDisplayName="" LastEditDate="2010-10-28T02:37:45.800" LastActivityDate="2010-10-28T12:43:41.203" Title="Linear optimization constrained by cost function" Tags="&lt;optimization&gt;&lt;linear-programming&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="7984" PostTypeId="1" AcceptedAnswerId="8000" CreationDate="2010-10-27T02:31:44.707" Score="7" ViewCount="190" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;It is a profoundly erroneous truism, repeated by copybooks and by eminent people when they are making speeches, that we should cultivate the habit of thinking of what we are doing. The precise opposite is the case. Civilization advances by extending the number of important operations which we can perform without thinking about them.&quot; ---Alfred North Whitehead&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&quot;&lt;em&gt;All&lt;/em&gt; math instruction problems are primarily insufficient &lt;em&gt;speed&lt;/em&gt; at a previous step.  If you can't run multiplication in your head, you can't do long division and have it make sense.  If you can't factor numbers in your head, you can't do decent fraction problems.  If math facts aren't instant, basic algebraic manipulations (2x + 3 = 17) take too long to do ... and you never learn the patterns. Algebraic fractions, quadratic factoring, the fundamental theorem of calculus, integration over the complex half-plane ... it's all the same thing.  How fast are you at the prior step?  Can you do it in your head?&quot; ---&lt;a href=&quot;http://aretae.blogspot.com/2010/06/homeschool-curricula.html&quot; rel=&quot;nofollow&quot;&gt;Aretae&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What specific exercises are helpful in internalizing and automatizing computational skills?&lt;/strong&gt; I realize that the obvious answer is &quot;drill and repetition,&quot; but I was hoping the community would have some something more specific to add, so I have a number of related subquestions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Say, are &lt;em&gt;timed&lt;/em&gt; drills helpful? Should one try to do simpler exercises in one's head? Does anyone have tips on how to develop the mental toughness to persist through computations that might be boring or painful in the moment even as one understands that they are ultimately necessary for building understanding? Is there some esoteric metacognitive art of debugging one's own mental algorithms: figuring out the &lt;em&gt;most&lt;/em&gt; efficient way of doing something, and training oneself to do that thing without thinking?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Motivation&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After two years of pure autodidacticism in mathematics, I'm taking a differential equations course at the local community college, and it's been more difficult than I expected. Humiliating as it is to admit, I don't seem to yet have the patience or mental toughness for large problem sets, and I worry that I may have developed a few bad study habits. I have a tendency to leisurely examine proofs and casually attempt a few exercises, without (I fear) taking care to establish the firm base of &lt;em&gt;quick and reliable&lt;/em&gt; skills needed for higher understanding. To take one example, I can &lt;em&gt;do&lt;/em&gt; integration by parts, but only after a noticeable hesitation; the operation is not introspectively obvious to me in the way that (say) the distributivity of multiplication over addition is introspectively obvious and hardly even feels like a step. Integration by parts still &lt;em&gt;feels like a step&lt;/em&gt;---and I wonder if perhaps it &lt;em&gt;shouldn't&lt;/em&gt;, supposing one really wants a deep understanding. So while I'm proud of myself for (say) having picked up a lot of cool complex analysis insights from my few months with Mathews and Howell and my quick skim of Tristan Needham, I also feel as if I am missing some fundamental skills that are needed to achieve deeper insight, and I was hoping maybe this community would have some ideas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously this is a soft question (perhaps too soft?---I dearly apologize), and should therefore be community wiki; however, I don't have the reputation to enable that. Perhaps a moderator could be so kind? I thank you-all in advance for any advice you have to offer, and remain yours.&lt;/p&gt;&#xA;" OwnerUserId="2773" LastActivityDate="2010-10-27T04:42:40.877" Title="Automatizing computational skills" Tags="&lt;soft-question&gt;&lt;learning&gt;&lt;self-learning&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="2" />
  <row Id="7985" PostTypeId="2" ParentId="7978" CreationDate="2010-10-27T02:45:27.853" Score="6" ViewCount="" Body="&lt;p&gt;For a skew symmetric (real) matrix, the eigenvalues are all purely imaginary.  This is because if $Av = \lambda v$, then we have $\lambda \langle v,v\rangle = \langle \lambda v, v\rangle = \langle Av,v\rangle = \langle v, -Av \rangle = \langle v, -\lambda v\rangle = -\overline{\lambda} \langle v,v\rangle$, so we conclude that $\lambda = -\overline{\lambda}$, i.e., that $\lambda$ is purely imaginary.  Here, we're using an Hermitian inner product.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a real matrix, complex eigenvalues come in conjugate pairs, so the rank must be even.&lt;/p&gt;&#xA;" OwnerUserId="331" LastActivityDate="2010-10-27T02:45:27.853" CommentCount="2" />
  <row Id="7986" PostTypeId="1" AcceptedAnswerId="7997" CreationDate="2010-10-27T03:15:00.440" Score="10" ViewCount="376" Body="&lt;p&gt;So I feel a bit strange asking a Calculus question, but this came up today while teaching.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One can check that if you start with some integral, which can be see as an &quot;obvious u-substitution problem&quot;, that you can instead use integration by parts, and wind up with the scenario where you have have the original integral on both sides of your equation so you solve for the integral.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Example:&lt;/strong&gt; Given $I=\int g^n(x)g'(x)dx$ we can clearly use u-substitution, but if we use integration by parts we get the equation $I=-nI+g^{n+1}$. This is nothing exciting or surprising, but it yields the observation that u-sub leads to one of these int by parts equations.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Question&lt;/strong&gt; Is the opposite true?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What I mean to say is, if you do integration by parts and you wind up with an equation of this type, does it mean that you could of used some very clever u-substitution?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I feel like I should know this, but I have thought about it today, and asked a friend or two, and we don't see an immediate proof of this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="16" LastActivityDate="2010-10-27T20:51:21.483" Title="Integration by Parts implies U-substitution?" Tags="&lt;calculus&gt;" AnswerCount="2" CommentCount="11" FavoriteCount="1" />
  <row Id="7987" PostTypeId="2" ParentId="7981" CreationDate="2010-10-27T03:20:56.997" Score="25" ViewCount="" Body="&lt;p&gt;The prime number theorem states that the number of primes less than or equal to&#xA;$x$ is approximately equal to $\int_2^x \dfrac{dt}{\log t}.$    The Riemann hypothesis gives a precise answer to how good this approximation is; namely, it states that the difference between the exact number of primes below $x$, and the given integral, is (essentially) $\sqrt{x} \log x$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Here &quot;essentially&quot; means that one should actually take the absolute value of the difference, and also that one might have to multiply $\sqrt{x} \log x$ by some positive constant.  Also, I should note that the Riemann hypothesis is more usually stated in terms of the location of the zeroes&#xA;of the Riemann zeta function; the previous paragraph is giving an equivalent form, which may be easier to understand, and also may help to explain the interest of the statement.  See &lt;a href=&quot;http://en.wikipedia.org/wiki/Riemann_hypothesis#Distribution_of_prime_numbers&quot; rel=&quot;nofollow&quot;&gt;the wikipedia entry&lt;/a&gt; for the formulation in terms of counting primes, as well as various other formlations.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The difficulty of the problem is (it seems to me) as follows: there is no approach currently known to understanding the distribution of prime numbers well enough to establish the desired approximation, other than by studying&#xA;the Riemann zeta function and its zeroes.  (The information about the primes&#xA;comes from information about the zeta function via a kind of Fourier transform.)  On the other hand, the zeta function is not easy to understand; there is no straightforward formula for it that allows one to study its zeroes, and because of this any such study ends up being somewhat indirect.&#xA;So far, among the various possible such indirect approaches, no-one has found&#xA;one that is powerful enough to control all the zeroes.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A very naive comment, that nevertheless might give some flavour of the problem, is that there are an infinite number of zeroes that one must contend with, so there is no obvious finite computation that one can make to solve&#xA;the problem; ingenuity of some kind is necessarily required.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, one can remark that the Riemann hypothesis, when phrased in terms of the location of the zeroes, is very simple (to state!) and very beautiful: it says that all the non-trivial zeros have real part $1/2$.  This suggests that perhaps there is some secret symmetry underlying the Riemann zeta function that would &quot;explain&quot; the Riemann hypothesis.  Mathematicians have had, and continue to have, various ideas about what this secret symmetery might be (in this they are inspired by an analogy with what is called &quot;the function field case&quot; and the&#xA;deep and beautiful theory of &lt;a href=&quot;http://en.wikipedia.org/wiki/Weil_conjectures&quot; rel=&quot;nofollow&quot;&gt;the Weil conjectures&lt;/a&gt;), but so far they haven't managed to establish any underlying phenonemon which implies the Riemann hypothesis.&lt;/p&gt;&#xA;" OwnerUserId="221" LastEditorUserId="221" LastEditorDisplayName="" LastEditDate="2010-10-27T04:40:47.893" LastActivityDate="2010-10-27T04:40:47.893" CommentCount="1" />
  <row Id="7988" PostTypeId="1" AcceptedAnswerId="7991" CreationDate="2010-10-27T03:24:48.497" Score="3" ViewCount="265" Body="&lt;p&gt;I'm trying to describe a counterexample for a theorem which includes the figure eight or &quot;infinity&quot; symbol, but I'm having trouble finding a good piecewise function to draw it. I need it to be the symbol, except at the &quot;crossing point&quot; the function jumps (not continuous) so that we still have a manifold.&lt;/p&gt;&#xA;" OwnerUserId="2319" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-10-27T03:44:18.000" LastActivityDate="2010-10-27T13:56:20.420" Title="The function that draws a figure eight" Tags="&lt;differential-geometry&gt;&lt;plane-curves&gt;" AnswerCount="3" CommentCount="3" />
  <row Id="7989" PostTypeId="2" ParentId="7988" CreationDate="2010-10-27T03:30:20.093" Score="3" ViewCount="" Body="&lt;p&gt;There are no functions that describe the &quot;lemniscate&quot;, but there are parametric and polar equations for the &lt;a href=&quot;http://mathworld.wolfram.com/Lemniscate.html&quot; rel=&quot;nofollow&quot;&gt;lemniscate of Bernoulli&lt;/a&gt; and the &lt;a href=&quot;http://mathworld.wolfram.com/EightCurve.html&quot; rel=&quot;nofollow&quot;&gt;lemniscate of Gerono&lt;/a&gt;, to name two of the more famous lemniscates.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Might as well share this, since the question already has an accepted answer. Here is my favorite way of generating the lemniscate of Bernoulli, as an &lt;a href=&quot;http://mathworld.wolfram.com/Envelope.html&quot; rel=&quot;nofollow&quot;&gt;envelope&lt;/a&gt; of circles centered on a rectangular hyperbola, and passing through the origin:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/HMyER.png&quot; alt=&quot;lemniscate envelope&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="498" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-10-27T13:56:20.420" LastActivityDate="2010-10-27T13:56:20.420" CommentCount="2" />
  <row Id="7990" PostTypeId="1" AcceptedAnswerId="7994" CreationDate="2010-10-27T03:30:28.533" Score="0" ViewCount="147" Body="&lt;p&gt;Let $R$ be a ring and $f(X) \in R[x]$ be a non-constant polynomial. We know that the number of roots, of $f(X)$ in $R$ has no relation, to its degree if $R$ is not &lt;em&gt;commutative&lt;/em&gt;, or &lt;em&gt;commutative but not a domain&lt;/em&gt;. But,&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;em&gt;The number of roots, of a non-zero polynomial over commutative integral domain, is at most its degree.&lt;/em&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;How does one prove above result?&lt;/p&gt;&#xA;" LastActivityDate="2010-10-27T03:54:16.347" Title="Roots of a polynomial in an integral domain." Tags="&lt;abstract-algebra&gt;&lt;algebra&gt;&lt;polynomials&gt;&lt;ring-theory&gt;" AnswerCount="2" CommentCount="5" />
  <row Id="7991" PostTypeId="2" ParentId="7988" CreationDate="2010-10-27T03:34:01.600" Score="4" ViewCount="" Body="&lt;p&gt;You can use the function $$t\in(-\tfrac12\pi,\tfrac32\pi)\mapsto(\cos t,\sin t\cos t)\in\mathbb R^2.$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The resulting curve is (I'm omitting a little bit on the left and a little bit on the right from the domain in this picture):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/CyNo6.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="274" LastEditorUserId="274" LastEditorDisplayName="" LastEditDate="2010-10-27T03:47:15.260" LastActivityDate="2010-10-27T03:47:15.260" CommentCount="5" />
  <row Id="7992" PostTypeId="2" ParentId="7988" CreationDate="2010-10-27T03:34:08.873" Score="3" ViewCount="" Body="&lt;p&gt;Is the &lt;a href=&quot;http://mathworld.wolfram.com/Lemniscate.html&quot; rel=&quot;nofollow&quot;&gt;lemniscate&lt;/a&gt; what you want?  I don't know what you need as a jump at the crossing point, but maybe you can get that with a trigonometric parameterization.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-10-27T03:34:08.873" />
  <row Id="7994" PostTypeId="2" ParentId="7990" CreationDate="2010-10-27T03:43:15.973" Score="2" ViewCount="" Body="&lt;p&gt;This is a consequence of the fact that over a commutative ring with identity $A$, an element $a \in A$ is a zero of a polynomial $f \in A[x]$ if and only if $f(x) = (x - a)q(x)$ for some polynomial $q(x) \in A[x]$.&lt;/p&gt;&#xA;" OwnerUserId="900" LastActivityDate="2010-10-27T03:43:15.973" />
  <row Id="7995" PostTypeId="2" ParentId="7990" CreationDate="2010-10-27T03:54:16.347" Score="0" ViewCount="" Body="&lt;p&gt;It follows from the division algorithm, the fact that the evaluation maps gives a homomorphism from $R[x]$ to $R^R$ (functions from $R$ to $R$ with pointwise operations), and that $R$ is a domain. It is the exact same argument as for fields, with the division algorithm suitably restricted to certain kinds of polynomials over $R$ as divisors. &lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-10-27T03:54:16.347" />
  <row Id="7997" PostTypeId="2" ParentId="7986" CreationDate="2010-10-27T04:12:51.760" Score="6" ViewCount="" Body="&lt;p&gt;&lt;em&gt;Added:&lt;/em&gt; I interpreted the question as follows: suppose that we use the integration by parts formula,&#xA;$$\int u(x)v'(x)\,dx = u(x)v(x) - \int v(x)u'(x)\,dx$$&#xA;and $\int v(x)u'(x)\,dx = \alpha \int u(x)v'(x)\,dx$ for some constant $\alpha$, so that we can &quot;solve&quot; for the original integral as&#xA;$$\int u(x)v'(x)\,dx = \frac{1}{1+\alpha}u(x)v(x) + C.$$&#xA;Is it the case that we can solve the original integral by doing a direct substitution instead of integration by parts?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think this works: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose that $v(x)u'(x) = \alpha u(x)v'(x)$ for some $\alpha\neq -1$. (I think this is essentially what you have, since the integral on the left hand side is $\int u(x)v'(x)\,dx$, and the integral on the right is $\int v(x)u'(x)\,dx$). I restrict to $\alpha\neq -1$, because if $\alpha=-1$, then you cannot &quot;solve&quot; for the original integral. But in fact, from the work below we will get that $u=\frac{B}{|v|}$, so that integration by parts will simply result in the true (but useless) $\int u\,dv = B + \int u\,dv$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $u$ or $v$ are zero, then the original integral was the integral of $0$, so we may discard that case. So we get that $\frac{u'}{u} = \alpha \frac{v'}{v}$, and integrating we get $\ln|u|=\alpha\ln|v|+C$, or $|u| = A|v|^{\alpha}$ for some $A\gt 0$; hence $u=B|v|^{\alpha}$ for some $B\neq 0$. So the original integral was&#xA;$$\int u(x)v'(x)\,dx = \int B|v(x)|^{\alpha}v'(x)\,dx$$&#xA;which suggests the substitution $g=v(x)$ to get $B\int |g|^{\alpha}\,dg$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added:&lt;/strong&gt; Note that if $\alpha = -1$, we still get $u=B|v|^{-1}$, and&#xA;so again the substitution $g=|v(x)|$ will do the trick. &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-10-27T16:51:12.063" LastActivityDate="2010-10-27T16:51:12.063" CommentCount="8" />
  <row Id="7998" PostTypeId="2" ParentId="7593" CreationDate="2010-10-27T04:25:29.060" Score="5" ViewCount="" Body="&lt;p&gt;I posted this to &lt;a href=&quot;http://mathoverflow.net/questions/43611/decomposing-the-plane-into-intervals&quot; rel=&quot;nofollow&quot;&gt;Math Overflow&lt;/a&gt; and Jeff Strom gave the following answer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conway and Croft show it can be done for closed intervals and cannot be done for open intervals in the paper:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Covering a sphere with congruent great-circle arcs. Proc. Cambridge Philos. Soc. 60 1964 787–800.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-10-27T04:25:29.060" />
  <row Id="7999" PostTypeId="2" ParentId="4961" CreationDate="2010-10-27T04:38:14.523" Score="5" ViewCount="" Body="&lt;p&gt;I came across this slick proof of &lt;a href=&quot;http://en.wikipedia.org/wiki/Heron%27s_formula&quot; rel=&quot;nofollow&quot;&gt;Heron's formula&lt;/a&gt; on artofproblemsolving.com the other day. Heron's formula yields the area of a triangle given the lengths of its three sides:&#xA;$$A = \sqrt{s(s-a)(s-b)(s-c)}$$ where $ s = \frac{1}{2}(a+b+c)$.&#xA;The &lt;a href=&quot;http://www.artofproblemsolving.com/Resources/Papers/Heron.pdf&quot; rel=&quot;nofollow&quot;&gt;entire proof&lt;/a&gt;, by high schooler Miles Dillon Edwards, is reproduced here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/0PpRN.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $I$ be the center of the incircle of $\triangle ABC$. Let $a = y + z$, $b = x + z$, and&#xA;$c = x + y$ be the lengths of the sides opposite $A$, $B$, and $C$, respectively, and let&#xA;$s = x + y + z$ be the semiperimeter of the triangle. Clearly $2 \alpha + 2 \beta + 2 \gamma = 2\pi$, so&#xA;$\alpha + \beta + \gamma = \pi$. Now notice that&#xA;$$(r + ix)(r + iy)(r + iz) = (u e^{i \alpha})(v e^{i \beta})(w e^{i \gamma}) = u v w e^{i(\alpha+\beta+\gamma)} = u v w e^{\pi i} = −uvw.$$&#xA;Therefore&#xA;$$0 = \text{Im}[(r + ix)(r + iy)(r + iz)] = r^2(x + y + z) − xyz,$$&#xA;so&#xA;$$r = \sqrt{\frac{x y z}{x + y + z}} = \sqrt{\frac{(s-a)(s-b)(s-c)}{s}}$$.&#xA;Thus the area of $\triangle ABC$ is&#xA;$$\frac{r a}{2} + \frac{r b}{2} + \frac{r c}{2} = r s =&#xA;\sqrt{s(s-a)(s-b)(s-c)}$$&lt;/p&gt;&#xA;" OwnerUserId="130" LastActivityDate="2010-10-27T04:38:14.523" CommunityOwnedDate="2010-10-27T04:38:14.523" />
  <row Id="8000" PostTypeId="2" ParentId="7984" CreationDate="2010-10-27T04:42:40.877" Score="1" ViewCount="" Body="&lt;p&gt;I too am a self-learner of mathematics working in a field where I have to use math professionally. In my experience speed or automatism has not been important. The trouble in using or learning math comes from not being able to figure out what line of reasoning to follow or how to adequately model a real-world phenomena. Once you have that right being slow or fast only makes a marginal difference.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Practice is important but for a different reason. It is not possible to grasp the relation between different ideas without thinking your way through many concrete instances. Also, every area of mathematics has certain tricks or idioms that are repeated again and again. You get a hang for them only through practice and exposure to good work done by others (good textbooks for beginners like me). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unless you are being too modest and underplaying your effort, it is a very bad idea to &quot;leisurely examine proofs and casually attempt a few exercises&quot;. Use pen and paper. Know the strategy of the proof well enough to be able to reproduce it on your own. Do at least 90% of the exercises.&lt;/p&gt;&#xA;" OwnerUserId="1195" LastActivityDate="2010-10-27T04:42:40.877" CommentCount="1" />
  <row Id="8001" PostTypeId="1" CreationDate="2010-10-27T05:01:34.877" Score="1" ViewCount="111" Body="&lt;p&gt;I manually arranged two complete latin squares from  ordered pairs of 4 and 6 like so:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;1 2 3 4&#xA;2 4 1 3&#xA;3 1 4 2&#xA;4 3 2 1&#xA;&#xA;1 2 3 4 5 6&#xA;2 4 1 6 3 5&#xA;3 1 5 2 6 4&#xA;4 6 2 5 1 3&#xA;5 3 6 1 4 2&#xA;6 5 4 3 2 1&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I like this kind of sequence, but can tell it is going to be more difficult to manually arrange as the number of pairs increase.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a related question about calculating row-complete latin squares, the following method generated a slightly different arrangement:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For any even n, say n = 2m, a row complete latin square of order n can be formed by writing down&lt;/p&gt;&#xA;&#xA;&lt;p&gt;0, 1, 2m - 1, 2, 2m - 2, 3,..., m + 1, m&lt;/p&gt;&#xA;&#xA;&lt;p&gt;as the first row and then developing subsequent rows by adding 1 modulo n.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;0 1 3 2&#xA;1 2 0 3&#xA;2 3 1 0&#xA;3 0 2 1&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;which can be made column-complete by rearranging the rows (incremented for comparison with the first example)...&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;1 2 4 3&#xA;2 3 1 4&#xA;4 1 3 2&#xA;3 4 2 1&#xA;&#xA;1 2 6 3 5 4&#xA;2 3 1 4 6 5&#xA;6 1 5 2 4 3&#xA;3 4 2 5 1 6&#xA;5 6 4 1 3 2&#xA;4 5 3 6 2 1&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;What I'm trying to figure out is how to generate the first pattern with a method similar to that row-complete calculation, or a way to convert the above squares into the first pattern by swapping some values.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It looks like they both have their values reversed in rows and columns in the same way, yet their sequences are different.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That makes it seem like a simple matter, but I'm not finding a simple (enough) pattern between them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example putting one of each square together and looking at it:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;11 22 63 34 55 46&#xA;22 34 11 46 63 55&#xA;63 11 55 22 46 34&#xA;34 46 22 55 11 63&#xA;55 63 46 11 34 22&#xA;46 55 34 63 22 11&#xA;&#xA;      63 34    46&#xA;   34    46 63   &#xA;63          46 34&#xA;34 46          63&#xA;   63 46    34   &#xA;46    34 63 &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Does anyone see a way of transposing one to the other, or modifiying the row-complete calculation to match the other row sequence?&lt;/p&gt;&#xA;" OwnerUserId="2768" LastEditorUserId="139" LastEditorDisplayName="" LastEditDate="2010-11-19T10:37:01.763" LastActivityDate="2010-11-19T10:50:15.237" Title="Finding a Pattern Between Latin Squares" Tags="&lt;combinatorics&gt;&lt;algorithms&gt;&lt;latin-square&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" />
  <row Id="8002" PostTypeId="1" AcceptedAnswerId="8005" CreationDate="2010-10-27T05:11:37.780" Score="7" ViewCount="528" Body="&lt;p&gt;Let $\varphi(n)$ be Euler's totient function, the number of positive integers less than or equal to $n$ and relatively prime to $n$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Challenge: Prove&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{k=1}^n \left\lfloor \frac{n}{k} \right\rfloor \varphi(k) = \frac{n(n+1)}{2}.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have two proofs, one of which is partially combinatorial.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm posing this problem partly because I think some folks on this site would be interested in working on it and partly because I would like to see a purely combinatorial proof.  (But please post any proofs; I would be interested in noncombinatorial ones, too.  I've learned a lot on this site by reading alternative proofs of results I already know.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll wait a few days to give others a chance to respond before posting my proofs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: The two proofs in full are now given among the answers.&lt;/p&gt;&#xA;" OwnerUserId="2370" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-11-22T05:25:23.223" LastActivityDate="2010-11-22T05:25:23.223" Title="Identity involving Euler's totient function: $\sum_{k=1}^n \left\lfloor \frac{n}{k} \right\rfloor \varphi(k) = \frac{n(n+1)}{2}$" Tags="&lt;number-theory&gt;&lt;combinatorics&gt;" AnswerCount="4" CommentCount="3" />
  <row Id="8003" PostTypeId="2" ParentId="7981" CreationDate="2010-10-27T05:12:37.653" Score="8" ViewCount="" Body="&lt;p&gt;A direct translation of RH (Riemann Hypothesis) would be very baffling in layman's terms. But, there are many problems that are equivalent to RH and hence, defining them would be actually indirectly stating RH. Some of the equivalent forms of RH are much easier to understand than RH itself. I give what I think is the most easy equivalent form that I have encountered:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The Riemann hypothesis is equivalent&#xA;  to the statement that an integer has&#xA;  an equal probability of having an odd&#xA;  number or an even number of distinct&#xA;  prime factors. (Borwein page. 46)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2320" LastActivityDate="2010-10-27T05:12:37.653" CommentCount="3" />
  <row Id="8004" PostTypeId="1" AcceptedAnswerId="8006" CreationDate="2010-10-27T05:42:01.813" Score="6" ViewCount="209" Body="&lt;p&gt;Hi, I'm just starting to learn some basic topology, and I've mostly encountered definitions. For instance, I read that for a topological space $X$, for any $E\subseteq X$, define its closure $\overline{E}$ as the set of points $p\in X$ such that each neighborhood $N$ of $p$ has a nonempty intersection with $E$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wanted to verify that this is indeed a closure operation, and here's what I have:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) Suppose $p\in E$, then for any $N$, $p\in N$, so $N\cap E\neq\emptyset$, so $p\in\overline{E}$, and thus $E\subseteq\overline{E}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) Suppose $p\in\overline{E\cup F}$. Then for any $N$, $N\cap(E\cup F)\neq\emptyset$, so $(N\cap E)\cup(N\cap F)\neq\emptyset$, and thus either $N\cap E\neq\emptyset$ or $N\cap F\neq\emptyset$, so $p\in\overline{E}$ or $p\in\overline{F}$, and thus $\overline{E\cup F}\subseteq\overline{E}\cup\overline{F}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(3) For any $p\in X$, the intersection of any $N$ and $\emptyset$ is empty, so there are no points such that every neighborhood has nonempty intersection with $\emptyset$. Hence $\overline{\emptyset}=\emptyset$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(4) &lt;strong&gt;I'm stuck showing that $\overline{\overline{E}}=\overline{E}$&lt;/strong&gt;. From (1), I know that $\overline{E}\subseteq\overline{\overline{E}}$, but I can't show the other containment. I took $p\in\overline{\overline{E}}$, and so every $N\cap\overline{E}\neq\emptyset$. I want to show $p\in N\cap\overline{E}$, and I think this would be easy if $X$ contains isolated points. But then I found out there are such things as perfect sets or dense-in-itself sets, so that can't work. Is there some trick to show this containment?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks, I know this is probably very simple, but I'm going a little mad staring at it.&lt;/p&gt;&#xA;" OwnerUserId="1583" LastActivityDate="2010-12-04T12:09:59.680" Title="Verifying a Closure Operation" Tags="&lt;topology&gt;" AnswerCount="2" />
  <row Id="8005" PostTypeId="2" ParentId="8002" CreationDate="2010-10-27T05:42:28.257" Score="11" ViewCount="" Body="&lt;p&gt;One approach is to use the formula $\displaystyle \sum_{d \mid k} \varphi(d) = k$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we have that $\displaystyle \sum_{k=1}^{n} \sum_{d \mid k} \varphi(d) = n(n+1)/2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Exchanging the order of summation we see that the $\displaystyle \varphi(d)$ term appears $\displaystyle \left\lfloor \frac{n}{d} \right\rfloor$ times&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and thus&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle \sum_{d=1}^{n} \left\lfloor \frac{n}{d} \right\rfloor \varphi(d) = n(n+1)/2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or in other words, if we have the $n \times n$ matrix $A$ such that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle A[i,j] = \varphi(j)$ if $j \mid i$ and $0$ otherwise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sum of elements in row $i$ is $i$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sum of elements in column $j$ is $\displaystyle \left\lfloor \frac{n}{j} \right\rfloor \varphi(j)$ and the identity just says the total sum by summing the rows is same as the total sum by summing the columns.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-10-27T05:58:46.663" LastActivityDate="2010-10-27T05:58:46.663" CommentCount="1" />
  <row Id="8006" PostTypeId="2" ParentId="8004" CreationDate="2010-10-27T05:49:13.517" Score="7" ViewCount="" Body="&lt;p&gt;From $N\cap\overline{E}\neq\emptyset$ you know that $N$ is a neighborhood of some element of $\overline{E}$, and thus $N\cap E\neq\emptyset$ by your definition of $\overline{E}$.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-10-27T05:49:13.517" CommentCount="5" />
  <row Id="8008" PostTypeId="1" AcceptedAnswerId="8106" CreationDate="2010-10-27T06:00:17.800" Score="2" ViewCount="326" Body="&lt;p&gt;Just wondering if equations used as props in SG1, atlantis and most recently Universe are just random and cool looking, or real/meaningful and cool looking.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most recent episode of Stargate Universe S02E04 has a &quot;corridor of equations&quot; where dr nick Rush is trying to decipher something. Has anyone familiar with the show got an opinion on this?&lt;/p&gt;&#xA;" OwnerUserId="2496" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-10-27T06:02:51.353" LastActivityDate="2010-10-27T21:49:07.530" Title="Do the equations used in Stargate make sense or are they gibberish?" Tags="&lt;popular-math&gt;" AnswerCount="1" CommentCount="8" FavoriteCount="1" />
  <row Id="8009" PostTypeId="2" ParentId="7981" CreationDate="2010-10-27T06:16:21.590" Score="1" ViewCount="" Body="&lt;p&gt;In very layman's terms it states that there is some order in the distribution of the primes (which seem to occur totally chaotic at first sight). Or to say it like Shakespeare: &quot;Though this be madness, yet there is method in 't.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to know more there is a new trilogy about that topic where the first volume has just arrived:&#xA;&lt;a href=&quot;http://www.secretsofcreation.com/volume1.html&quot; rel=&quot;nofollow&quot;&gt;http://www.secretsofcreation.com/volume1.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is a marvelous and easy to understand book from a number theorist who knows his stuff! &lt;/p&gt;&#xA;" OwnerUserId="346" LastActivityDate="2010-10-27T06:16:21.590" />
  <row Id="8010" PostTypeId="2" ParentId="7923" CreationDate="2010-10-27T06:21:23.473" Score="1" ViewCount="" Body="&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Wiener_process&quot; rel=&quot;nofollow&quot;&gt;Wiener process&lt;/a&gt; is a continuous everwhere, but differentiable nowhere function (quite an impressive beast by the way...) &lt;/p&gt;&#xA;" OwnerUserId="346" LastActivityDate="2010-10-27T06:21:23.473" />
  <row Id="8011" PostTypeId="2" ParentId="8002" CreationDate="2010-10-27T06:25:49.213" Score="5" ViewCount="" Body="&lt;p&gt;Look at the problem of counting all the triples (d,m+1,k) where $d\leq m\leq n$ and $gcd(d,m)=\frac{m}{k}$.&#xA;For each (d,m+1) we have exactly one k such that $gcd(d,m)=\frac{m}{k}$, and so we are counting exactly all the pairs (i,j)  where $ i &amp;lt; j \leq n+1 $ which is $\binom{n+1}{2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, for each k , if $ gcd(d,m)=\frac{m}{k} $, then $ k \mid m $ and there are $ \left\lfloor \frac{n}{k}\right\rfloor $ such m's, and for each m we have $\varphi(k)$ d's that satisfies this equality, so there are $ \left\lfloor \frac{n}{k}\right\rfloor \varphi(k) $ triples that end with k.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The idea is counting the number of ways to choose two numbers from 1,2,...,n+1 (which is the right side of the equation). Once you choose the larger number m+1, you have m options which is exactly $\sum_{d\mid m} \varphi(d)$ (this is basically what Moron wrote).&#xA;What I wanted to do is to change the counting according to the &quot;relation&quot; between d and m, and the one that worked out is that $\frac{m}{gcd(d,m)}=k$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, going over triples (d,m+1,k) where $\frac{m}{gcd(d,m)}=k$ and $d\leq m\leq n$ just mean that (d,m+1) is an ordered pair from {1,...,n} and k is decided as above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, for specific k, if $\frac{m}{gcd(d,m)}=k$ then $k \mid m$ and there are $ \left\lfloor \frac{n}{k}\right\rfloor$ such m's. For each m, if $gcd(d,m)=m/k$ then $gcd(\frac{d}{m/k},\frac{m}{m/k})=gcd(\frac{d}{m}k,k)=1$ and there are $\varphi(k)$ such d's (because $d\leq m$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, or each k there are $ \left\lfloor \frac{n}{k}\right\rfloor \varphi(k) $ options , and summing over k will give the left side of the equation.&lt;/p&gt;&#xA;" OwnerUserId="2125" LastEditorUserId="2125" LastEditorDisplayName="" LastEditDate="2010-10-28T06:22:09.847" LastActivityDate="2010-10-28T06:22:09.847" CommentCount="2" />
  <row Id="8013" PostTypeId="1" AcceptedAnswerId="8014" CreationDate="2010-10-27T07:10:58.670" Score="8" ViewCount="318" Body="&lt;p&gt;Given a set $S$, one can easily find a set with greater cardinality -- just take the power set of $S$. In this way, one can construct a sequence of sets, each with greater cardinality than the last. Hence there are at least countably infinite many orders of infinity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;But do there exist uncountably infinite orders of infinity?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To be precise, does there exist an uncountable set of sets whose elements all have distinct cardinalities?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first answer to &lt;a href=&quot;http://math.stackexchange.com/questions/5378/types-of-infinity&quot;&gt;http://math.stackexchange.com/questions/5378/types-of-infinity&lt;/a&gt; suggests the answer is &quot;yes&quot;, but only establishes a countable number of cardinalities (which, to be fair, was what the question was asking about).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've been exposed to enough mathematical logic to realize that I'm walking in a minefield; let me know if I've already mis-stepped.&lt;/p&gt;&#xA;" OwnerUserId="1126" LastActivityDate="2010-10-27T16:21:52.773" Title="Are there uncountably infinite orders of infinity?" Tags="&lt;set-theory&gt;&lt;infinity&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="1" />
  <row Id="8014" PostTypeId="2" ParentId="8013" CreationDate="2010-10-27T07:15:14.833" Score="9" ViewCount="" Body="&lt;p&gt;The answer is yes.  See &quot;Fact 20&quot; at the end of the following handout:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://math.uga.edu/~pete/settheorypart1.pdf&quot; rel=&quot;nofollow&quot;&gt;http://math.uga.edu/~pete/settheorypart1.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="299" LastActivityDate="2010-10-27T07:15:14.833" CommentCount="1" />
  <row Id="8015" PostTypeId="1" CreationDate="2010-10-27T07:37:21.070" Score="4" ViewCount="130" Body="&lt;p&gt;I'm looking for lightweight, free, Windows, Math software.&lt;br&gt;&#xA;Something I can put an expression and get an answer, or graph it.&lt;br&gt;&#xA;I tried Euler, but it is quiet complicated and HUGE.&lt;br&gt;&#xA;Basic needs:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Expression Based&lt;/li&gt;&#xA;&lt;li&gt;Supports Variables&lt;/li&gt;&#xA;&lt;li&gt;Support Functions, User defined and auto loaded.&lt;/li&gt;&#xA;&lt;li&gt;Supports graphs, 2D.  Not really needing 3D.&lt;/li&gt;&#xA;&lt;li&gt;Supports History.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;What do you use?  What do you recommend?&lt;/p&gt;&#xA;" OwnerUserId="2743" LastActivityDate="2010-10-27T09:56:10.793" CommunityOwnedDate="2010-10-27T07:37:21.070" Title="Windows lightweight Math Software" Tags="&lt;math-software&gt;" AnswerCount="6" CommentCount="2" />
  <row Id="8016" PostTypeId="2" ParentId="8015" CreationDate="2010-10-27T07:41:39.840" Score="1" ViewCount="" Body="&lt;p&gt;I'm also interested in answers to this questions!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a &lt;a href=&quot;http://www.microsoft.com/windowsxp/downloads/powertoys/xppowertoys.mspx&quot; rel=&quot;nofollow&quot;&gt;Power Calculator&lt;/a&gt; from MS, but has a lot to be desired.&lt;/p&gt;&#xA;" OwnerUserId="1099" LastActivityDate="2010-10-27T07:41:39.840" CommentCount="2" />
  <row Id="8017" PostTypeId="2" ParentId="8015" CreationDate="2010-10-27T07:45:06.137" Score="1" ViewCount="" Body="&lt;p&gt;I haven't personally used &lt;a href=&quot;http://www.wmich.edu/cpmp/CPMP-Tools/index.html&quot; rel=&quot;nofollow&quot;&gt;CPMP-Tools&lt;/a&gt;, but I've heard it referred to as a free CAS.&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-10-27T07:45:06.137" />
  <row Id="8018" PostTypeId="2" ParentId="8015" CreationDate="2010-10-27T07:52:30.353" Score="1" ViewCount="" Body="&lt;p&gt;Graphmatica does a fairly decent job at plotting functions. Not very powerful, but it can graph a lot of functions and their derivatives and such.&lt;/p&gt;&#xA;" OwnerUserId="1906" LastEditorUserId="1906" LastEditorDisplayName="" LastEditDate="2010-10-27T09:00:06.583" LastActivityDate="2010-10-27T09:00:06.583" />
  <row Id="8019" PostTypeId="2" ParentId="8015" CreationDate="2010-10-27T07:58:25.013" Score="1" ViewCount="" Body="&lt;p&gt;In your position, I'd just use an interactive &lt;a href=&quot;http://www.python.org/&quot; rel=&quot;nofollow&quot;&gt;Python&lt;/a&gt; shell with &lt;a href=&quot;http://matplotlib.sourceforge.net/&quot; rel=&quot;nofollow&quot;&gt;Matplotlib&lt;/a&gt;. (But then, I'm a computer science guy first and foremost.)&lt;/p&gt;&#xA;" OwnerUserId="856" LastActivityDate="2010-10-27T07:58:25.013" CommunityOwnedDate="2010-10-27T07:58:25.013" CommentCount="2" />
  <row Id="8020" PostTypeId="1" AcceptedAnswerId="8063" CreationDate="2010-10-27T08:00:22.360" Score="4" ViewCount="149" Body="&lt;p&gt;Well, in a previous post regarding the explanation of &lt;a href=&quot;http://math.stackexchange.com/questions/7981/can-someone-please-explain-the-riemann-hypothesis-to-me-in-english/7987#7987&quot;&gt;&lt;em&gt;Riemann Hypothesis&lt;/em&gt;&lt;/a&gt; Matt answered that:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The prime number theorem states that the number of primes less than or equal to $x$ is approximately equal to $\int_2^x \dfrac{dt}{\log t}.$ The Riemann hypothesis gives a precise answer to how good this approximation is; namely, it states that the difference between the exact number of primes below $x$, and the given integral, is (essentially) $\sqrt{x} \log x$. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;What i have heard about RH is:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The non-trivial Zeros of the &lt;em&gt;Riemann- $\zeta$ -function&lt;/em&gt; have real part as $\frac{1}{2}$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Can anyone tell me how these two statements are related?&lt;/p&gt;&#xA;" LastActivityDate="2010-10-27T15:58:35.210" Title="A Question on RH relating to Prime Number theorem" Tags="&lt;number-theory&gt;&lt;soft-question&gt;&lt;prime-numbers&gt;&lt;analytic-number-theory&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" />
  <row Id="8021" PostTypeId="2" ParentId="8020" CreationDate="2010-10-27T08:10:42.117" Score="4" ViewCount="" Body="&lt;p&gt;A brief explanation is sufficient I think, since the question is a little vague, because there are many ways to relate the two. Regarding relation between RH and PNT one will have,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$| \pi(x) - Li(x) | = O({x^\theta}\log{x})$ if and only if $\zeta(s)$ has no non-trivial zeros $\rho$ with $\Re(\rho) &gt; \theta$. Naturally, since the non-trivial zeros of the Riemann Zeta function are symmetric about $\frac{1}{2}$. Hence $| \pi(x) - Li(x) | = O(\sqrt{x}\log{x})$ implies RH.&lt;/p&gt;&#xA;" OwnerUserId="2320" LastActivityDate="2010-10-27T08:10:42.117" CommentCount="1" />
  <row Id="8022" PostTypeId="2" ParentId="8002" CreationDate="2010-10-27T08:17:18.613" Score="6" ViewCount="" Body="&lt;p&gt;We can use induction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have $ \lfloor (n+1)/k \rfloor - \lfloor n/k \rfloor = 1 $ when $n \equiv -1 \textrm{ mod } k,$ (for $k&gt;1$) and $0$ otherwise. Hence&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \phi(1) + \sum_{k=2}^n \left( \left\lfloor \frac{n+1}{k} \right\rfloor -&lt;br&gt;&#xA;\left\lfloor \frac{n}{k} \right\rfloor \right) \phi(k) =&#xA;\sum_{k=1, k \textrm{ s.t. } n \equiv -1 \textrm{ mod } k}^n \phi(k)$$&#xA;$$= \sum_{ k | (n+1), k \ne n+1} \phi(k) = (n+1) - \phi(n+1),$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;using $\sum_{d|k} \phi(d) = k.$&#xA;So assuming the result is true for $n,$ we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{k=1}^{n+1} \left\lfloor \frac{n+1}{k} \right\rfloor \phi(k)&#xA;= \phi(n+1) + \sum_{k=1}^n \left( \left\lfloor \frac{n+1}{k} \right\rfloor -&lt;br&gt;&#xA;\left\lfloor \frac{n}{k} \right\rfloor \right) \phi(k) + \frac{n(n+1)}{2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$= \phi(n+1) + \phi(1) + \sum_{k=2}^n \left( \left\lfloor \frac{n+1}{k} \right\rfloor -&lt;br&gt;&#xA;\left\lfloor \frac{n}{k} \right\rfloor \right) \phi(k) + \frac{n(n+1)}{2},$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which, using the previous result to substitute for the summation,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$= (n+1) +  \frac{n(n+1)}{2} =  \frac{(n+1)(n+2)}{2}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Noting that the result is true when $n=1 \textrm{ and } 2$ completes the proof.&lt;/p&gt;&#xA;" OwnerUserId="1301" LastEditorUserId="1301" LastEditorDisplayName="" LastEditDate="2010-10-27T09:38:06.373" LastActivityDate="2010-10-27T09:38:06.373" CommentCount="1" />
  <row Id="8023" PostTypeId="2" ParentId="8015" CreationDate="2010-10-27T08:37:22.050" Score="2" ViewCount="" Body="&lt;p&gt;As calculator &lt;a href=&quot;http://wxmaxima.sourceforge.net/wiki/index.php/Main_Page&quot; rel=&quot;nofollow&quot;&gt;wxMaxima&lt;/a&gt; (&lt;a href=&quot;http://wxmaxima.sourceforge.net/wiki/index.php/Screenshots&quot; rel=&quot;nofollow&quot;&gt;screenshots&lt;/a&gt;) with export to png and LaTeX syntax.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For playing with graphs I use &lt;a href=&quot;http://www.geogebra.org/cms/&quot; rel=&quot;nofollow&quot;&gt;GeoGebra&lt;/a&gt; with export to png, eps, pdf...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For some special cases &lt;a href=&quot;http://www.wolframalpha.com/&quot; rel=&quot;nofollow&quot;&gt;Wolfram|Alpha&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And also MS Excel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And also a normal calculator - is often faster than PC programs ;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(And for hard differential/integral/calculus &lt;a href=&quot;http://wood.mendelu.cz/math/maw-html/index.php?lang=en&quot; rel=&quot;nofollow&quot;&gt;Czech Mathematical Assistant on Web&lt;/a&gt;).&lt;/p&gt;&#xA;" OwnerUserId="2839" LastActivityDate="2010-10-27T08:37:22.050" CommunityOwnedDate="2010-11-13T18:39:23.673" />
  <row Id="8024" PostTypeId="1" AcceptedAnswerId="8026" CreationDate="2010-10-27T08:50:11.893" Score="5" ViewCount="231" Body="&lt;p&gt;Lots of people (including myself) face lot of problems in tackling Mathematical Problems, which appear as if we can solve it, but then writing out a solution becomes difficult. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let us consider some examples:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;I was asked this question, some time back in an exam. Give an example of a continuous function on $(a,b)$ which is not uniformly continuous. Well, one's obvious choice is $$f(x) = \frac{1}{x-a} \quad \text{or} \ \frac{1}{x-b}$$ I knew this as soon as i saw the problem, and started proving it. One actually has to make an observation that as $x \to a$ then, $\frac{1}{x-a}$ will be larger. But i found that i couldn't actually formally prove it.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Similarly, to prove that $f(x)=x^{2}$ is not uniformly continuous on $\mathbb{R}$, one again has to play with the quantifiers, to get the contradiction part. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So, these are two instances, where i have found the problem, which to me appeared that i could solve it, but writing out a formal solution became difficult. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can students improve upon this? Are there any instances, which happened to you like this!&lt;/p&gt;&#xA;" LastActivityDate="2010-10-27T15:27:39.633" Title="Difficulty in Mathematical Writing" Tags="&lt;soft-question&gt;" AnswerCount="4" CommentCount="3" FavoriteCount="3" />
  <row Id="8025" PostTypeId="1" CreationDate="2010-10-27T09:11:16.853" Score="6" ViewCount="181" Body="&lt;p&gt;See &lt;a href=&quot;http://stackoverflow.com/questions/4014090/is-it-safe-to-ignore-the-possibility-of-sha-collisions-in-practice&quot;&gt;this question&lt;/a&gt; on applying probability theory principles in software design.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is generally the following: you design some system (say software) and rely on some well-known mathematical concept (say hash function). You know that when this concept is used without caution your system can sometimes fail, however the probability of such failure is extremely low.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You need to evaluate whether you want to alter the design or can just ignore that drawback. Consequences of a failure are usually taken into account when such evaluations are done. For example if a failure leads to a person being mildly offended then it is not that of a problem but if a failure leads to a nuclear power explosion that it is a serious problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now &lt;a href=&quot;http://stackoverflow.com/questions/4014090/is-it-safe-to-ignore-the-possibility-of-sha-collisions-in-practice/4014407#4014407&quot;&gt;the accepted answer&lt;/a&gt; goes like this (all numbers here are exaggerated for better perception): probability of Earth colliding with a space rock is 10E-50 and probability of that drawback causing a problem is 10E-100. You see - Earth colliding with a space rock is a gazillion times more likely. So relax, that design is good enough.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is that reasoning correct? Can it be accepted at all times?&lt;/p&gt;&#xA;" OwnerUserId="2844" LastActivityDate="2011-04-14T20:59:28.760" Title="How safe is it to ignore low probability events?" Tags="&lt;probability&gt;&lt;probability-theory&gt;" AnswerCount="3" CommentCount="5" />
  <row Id="8026" PostTypeId="2" ParentId="8024" CreationDate="2010-10-27T09:45:31.357" Score="5" ViewCount="" Body="&lt;p&gt;I have a (joke) template file for writing papers, which contains in it&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Lemma&lt;/strong&gt; &lt;em&gt;(Main technical lemma)&lt;/em&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Let $D$ be a domain in (INSERT SPACE HERE) such that the following properties hold: &lt;/p&gt;&#xA;  &#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;Technical condition 1&lt;/li&gt;&#xA;  &lt;li&gt;Technical condition 2&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;  &#xA;  &lt;p&gt;Then $D$ is both opened and closed in (INSERT SPACE AGAIN).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;What is the point of the above? At some point in your mathematical career you will come to the realisation that the proofs &lt;em&gt;you&lt;/em&gt; personally are going to write are all based on the few small set of technical arguments. For what I do the main tool happens to be the Method of Continuity. By the time you have this realisation, it will also be completely obvious to you how to formulate a given proof to fit the template. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But how to you come to this realisation? My only suggestion is to read more papers/books/proofs and write more of them yourselfs. Just like a foreign language, the only way to get better and converting your intuitive ideas into formal arguments is through practice and immersion. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an aside, from the examples you gave, it is not quite clear whether your difficulty is with implementing $\epsilon$ - $\delta$ s, or with setting up the proof by contradiction. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, besides the usual proofs in textbooks, a good place to read up on proof techniques is &lt;a href=&quot;http://en.wikipedia.org/wiki/Proofs_from_THE_BOOK&quot; rel=&quot;nofollow&quot;&gt;Proofs from the BOOK&lt;/a&gt; by Aigner and Ziegler. Try to really figure out the details of each proof so you can explain the idea of it, a few days later, without having the book open. Another good resource for problem solving techniques is &lt;a href=&quot;http://www.tricki.org/&quot; rel=&quot;nofollow&quot;&gt;the Tricki&lt;/a&gt;. Of course, reading up on answers (and providing them) on this website would also help. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;A bit on the quantifier issue. First you need to mentally nest the various implications. For example, diagrammatically I think (the following is not formal logic notation) uniform continuity to be something like&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \forall \epsilon \to \left( \exists \delta \to ( \forall |x-y| &amp;lt; \delta \to |f(x) - f(y)| &amp;lt; \epsilon ) \right) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So to contradict it by example, you want something that satisfies the hypothesis $\forall \epsilon$ but not the conclusion $\exists \delta \ldots$. Which means that for each $\epsilon$ there &lt;em&gt;cannot exist&lt;/em&gt; a $\delta$ with the requisite property. Which means that for each $\epsilon$ and for each $\delta$ the requisite property must be false. So after step 1 we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \forall \epsilon, \delta \to \mbox{ not } ( \forall |x-y| &amp;lt; \delta \to |f(x) - f(y)| &amp;lt; \epsilon ) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the statement inside the parentheses were to be false, since it is a $\forall$ statement, you just need one example. So you can convert the negation to &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \forall \epsilon, \delta \to \left( \exists |x-y| &amp;lt; \delta \mbox{ and } |f(x) - f(y)| &gt; \epsilon \right) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So there, we have converted a &quot;negative statement&quot; which we want to contradict, to a &quot;positive statement&quot; of a property we want our function to have. The rest is picking the mapping $(\epsilon, \delta) \to (x,y)$ for your chosen function $f$ so that the above implication is true. &lt;/p&gt;&#xA;" OwnerUserId="1543" LastEditorUserId="1543" LastEditorDisplayName="" LastEditDate="2010-10-27T10:06:46.010" LastActivityDate="2010-10-27T10:06:46.010" CommentCount="4" />
  <row Id="8027" PostTypeId="2" ParentId="8015" CreationDate="2010-10-27T09:56:10.793" Score="0" ViewCount="" Body="&lt;p&gt;I have also found SpeQ Mathematics.  &lt;a href=&quot;http://www.speqmath.com&quot; rel=&quot;nofollow&quot;&gt;http://www.speqmath.com&lt;/a&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is very lightweight, starts quickly and has some good functions.&lt;/p&gt;&#xA;" OwnerUserId="2743" LastActivityDate="2010-10-27T09:56:10.793" />
  <row Id="8028" PostTypeId="2" ParentId="8025" CreationDate="2010-10-27T09:58:13.523" Score="2" ViewCount="" Body="&lt;p&gt;Maybe do it like the industry does (e.g. with hard drives) and estimate the mean time between failures. If the unsafe operation is used at most once each time interval, and the probability of a failure is $p$, you can expect a time between failures of $\frac{1}{p}$ time intervals. If this time is significantly longer than you expect your system to be used, you are of the safe side.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1e-100, is ridiculously low, by any standards. If such an error might occure once every nanosecond, you have a mean time between failures of 3e83 years. Compare that to the estimated age of the universe.... =)&lt;/p&gt;&#xA;" OwnerUserId="505" LastActivityDate="2010-10-27T09:58:13.523" CommentCount="3" />
  <row Id="8029" PostTypeId="2" ParentId="8024" CreationDate="2010-10-27T11:00:13.150" Score="1" ViewCount="" Body="&lt;p&gt;The problem isn't really about writing. It's about being able to do math.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of the big problems in math is the P = NP problem. Most mathematicians believe that the know that the two aren't equal.&#xA;There's however no accepted proof. As a result a lot of smart people work on the problem. Writing proofs is the main activity of professional mathematicians.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you 'or'-sign instead of the 'and'-sign because you don't remember with is turned with way, then you have a problem with writing stuff down in a formal matter. &#xA;If you however have no idea about what theorems you need to use to prove the solution then you misstate your problem by saying it's about writing stuff down in a formal matter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Math centers around finding a way to prove your result. Faith isn't enough.&lt;/p&gt;&#xA;" OwnerUserId="2263" LastEditorUserId="2263" LastEditorDisplayName="" LastEditDate="2010-10-27T11:16:33.130" LastActivityDate="2010-10-27T11:16:33.130" CommentCount="4" />
  <row Id="8030" PostTypeId="1" AcceptedAnswerId="8032" CreationDate="2010-10-27T11:15:31.853" Score="0" ViewCount="185" Body="&lt;p&gt;Problem with this question&lt;/p&gt;&#xA;&#xA;&lt;p&gt;f(x)=3sin(x)cos(x/2) a=1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have been breaking up f(x) into g(x) = 3sin(x) and h(x)cos(x/2), and then multiplying them back (g(x))(h(x)) this was suggested to my by the lecturer. but for the life of me I cant get the answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Answer is: y = 0.817x+1.398&lt;/p&gt;&#xA;&#xA;&lt;p&gt;any help would be greatly appreciated.&lt;/p&gt;&#xA;" OwnerUserId="2874" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-10-27T11:22:16.233" LastActivityDate="2010-11-26T11:44:23.947" Title="Taylor Series Expansions" Tags="&lt;calculus&gt;&lt;taylor-expansion&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="8031" PostTypeId="2" ParentId="8024" CreationDate="2010-10-27T11:17:57.167" Score="1" ViewCount="" Body="&lt;p&gt;What makes writing mathematics difficult is lack of confidence.  Someone confident who happens to be wrong will not face difficulties writing up their misapprehensions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Willie's argument, getting familiar with lots of proof techniques that together solve most problems in some area, builds confidence: when you see such a problem, your reaction is &quot;I know just what to do with you&quot;.  And if you've seen this technique applied to many different problems, you'll have a good understanding of the difficulties it faced.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even then, some problems will be tricky.  What happens when your favourite technique comes apart at the seams, and you say to yourself &quot;I'm lost&quot;?  My favourite trick is to look at whatever sublemma I'm trying to prove and to try to doubt it, and once I'm not quite sure it's true, to go looking for a counterexample.  This hopeless search often informs you of the contours of the proof you can then reattempt.  To put into slogan form: &lt;em&gt;It's often hard to provbe what you can't doubt.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Polya (1945)'s &lt;em&gt;How to Solve It&lt;/em&gt; has many ideas about techniques to help you keep going with proving things.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To sum up, being &lt;em&gt;knowledgeable&lt;/em&gt;, as Willie suggests, is one key to being confident, and could call having the resources to deal with problem-solving setbacks being &lt;em&gt;indefatigability&lt;/em&gt;, another key.  Self-help books are full of &quot;you can do it!&quot; advice that will help you write out what you think is a proof, but unfortunately, unlike the above, they reduce the chances that the proof is valid.&lt;/p&gt;&#xA;" OwnerUserId="100" LastEditorUserId="100" LastEditorDisplayName="" LastEditDate="2010-10-27T11:23:10.197" LastActivityDate="2010-10-27T11:23:10.197" />
  <row Id="8032" PostTypeId="2" ParentId="8030" CreationDate="2010-10-27T11:28:39.247" Score="2" ViewCount="" Body="&lt;p&gt;You apparently only need the 1st-degree Taylor polynomial; note that the 1st-degree Taylor polynomial for a function $f(x)$ at $x=a$ is the same as the tangent line to the graph of $f$ at $a$, $y=f(a)+f'(a)\cdot(x-a)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given your $g$ and $h$, you can write 1st-degree Taylor polynomials for each of them, then take the product, as you said.  For $g$, $$y=g(1)+g'(1)\cdot(x-1)=3\sin(1)+3\cos(1)\cdot(x-1)\approx 1.621x+0.904.$$  I'll leave $h$ for you to do.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, you've got (for some numbers $a$ and $b$): $$(1.621x+0.904)(ax+b)=1.621ax^2+(0.904a+1.621b)x+0.904b,$$ in which you don't care about the quadratic term (it's not accurate anyway).&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Given that you only need a 1st-degree Taylor polynomial, it might be easier to just differentiate $f$ as written and compute it directly (not bother with $g$ and $h$ and multiplying).&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-10-27T11:28:39.247" CommentCount="1" />
  <row Id="8034" PostTypeId="2" ParentId="8025" CreationDate="2010-10-27T11:33:30.577" Score="7" ViewCount="" Body="&lt;p&gt;Extremely small probabilities are always suspect.  See &lt;a href=&quot;http://www.johndcook.com/blog/2010/10/18/titanic-effect/&quot; rel=&quot;nofollow&quot;&gt;The Titanic Effect&lt;/a&gt;.  If you think the probability of a software failure is 10^-50 then you're not looking at the most probable source of failure.  If you're hash algorithm, for example, has one chance in 10^50 of failing, then it's more likely that your program, operating system, or hardware has a bug.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general it's not enough to look at probabilities of failure.  You have to look at probabilities of failure and the consequences of the failures.  In some contexts, a 10% probability of failure is acceptable.  In other contexts, a one-in-a-million chance of failure is quite serious.&lt;/p&gt;&#xA;" OwnerUserId="247" LastActivityDate="2010-10-27T11:33:30.577" />
  <row Id="8035" PostTypeId="1" AcceptedAnswerId="8285" CreationDate="2010-10-27T11:40:22.107" Score="6" ViewCount="176" Body="&lt;p&gt;&lt;strong&gt;Statement:&lt;/strong&gt;&#xA;There exist a polynomial P such that |P(x) - cos(x)| &amp;lt;= 10^-6 for all (real) x.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My answer&lt;/strong&gt;: False. All polynomials of a degree n &gt;= 1 are unbound as x tends to infinity. A polynomial of degree n = 0 is bound only when it is in the form y = a (horizontal line) but this will not help because cos(x) varies between [-1,1].&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My question:&lt;/strong&gt; Is my answer reasonable? I am particularly concerned about 'unbound' polynomials as this is the term I just made up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please don't give me complete answer, I just want to know the flaws in my argument and get some hints to better solutions.&lt;/p&gt;&#xA;" OwnerUserId="2846" LastEditorUserId="2846" LastEditorDisplayName="" LastEditDate="2010-10-27T11:45:38.940" LastActivityDate="2010-10-29T14:27:29.533" Title="Statement about existence of a polynomial - true or false?" Tags="&lt;polynomials&gt;" AnswerCount="2" CommentCount="5" />
  <row Id="8036" PostTypeId="2" ParentId="7864" CreationDate="2010-10-27T11:47:16.677" Score="8" ViewCount="" Body="&lt;p&gt;Many well-educated people believe that a p-value is the probability that a study conclusion is wrong.  For example, they believe that if you get a 0.05 p-value, there's a 95% chance that your conclusion is correct.  In fact there may be less than a 50% chance that the conclusion is correct, depending on the context.  Read more &lt;a href=&quot;http://www.johndcook.com/blog/2008/11/18/five-criticisms-of-significance-testing/&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="247" LastActivityDate="2010-10-27T11:47:16.677" CommunityOwnedDate="2010-10-27T11:47:16.677" />
  <row Id="8037" PostTypeId="1" CreationDate="2010-10-27T11:54:54.593" Score="0" ViewCount="171" Body="&lt;p&gt;$$&#xA;\begin{pmatrix}&#xA;Y \\ Cb \\ Cr&#xA;\end{pmatrix}&#xA; =&#xA;\mathrm{Round}&#xA;\left\{&#xA;\begin{pmatrix}&#xA;0.2126\times\frac{219}{255} &amp;amp;&#xA;0.7152\times\frac{219}{255} &amp;amp;&#xA;0.0722\times\frac{219}{255} \\&#xA;-\frac{0.2126}{1.8556}\times\frac{224}{255} &amp;amp;&#xA;-\frac{0.7152}{1.8556}\times\frac{224}{255} &amp;amp;&#xA;0.5\times\frac{224}{255} \\&#xA;0.5\times\frac{224}{255} &amp;amp;&#xA;-\frac{0.7152}{1.5748}\times\frac{224}{255} &amp;amp;&#xA;-\frac{0.0722}{1.5748}\times\frac{224}{255} &#xA;\end{pmatrix}&#xA;\begin{pmatrix}&#xA;R \\ G \\ B&#xA;\end{pmatrix}&#xA;\right\}&#xA;+&#xA;\begin{pmatrix}&#xA;16 \\ 128 \\ 128&#xA;\end{pmatrix}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it possible to derive formula from above matrix that converts YCbCr to RGB? I am newbie to maths.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have YCbCr value already available, I want to convert it to RGB.&lt;/p&gt;&#xA;" OwnerUserId="2849" LastEditorUserId="171" LastEditorDisplayName="" LastEditDate="2010-10-27T13:54:05.133" LastActivityDate="2010-10-27T14:51:44.597" Title="YCbCr to RGB conversion with given matrix." Tags="&lt;matrices&gt;" AnswerCount="0" CommentCount="3" />
  <row Id="8039" PostTypeId="1" AcceptedAnswerId="8041" CreationDate="2010-10-27T12:45:36.033" Score="0" ViewCount="99" Body="&lt;p&gt;Let A be and nxn matrix over R or C. Suppose B is an nxn matrix over the same field as A and Bx=Ax for all x \in R^n. Prove that A=B, that is a_{ij}=b_{ij}, for alla i,j&lt;/p&gt;&#xA;" OwnerUserId="2723" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-10-27T12:52:41.870" LastActivityDate="2010-10-27T13:28:07.870" Title="Prove that A=B, that is a_{ij}=b_{ij}, for alla i,j" Tags="&lt;linear-algebra&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="8040" PostTypeId="1" AcceptedAnswerId="8047" CreationDate="2010-10-27T13:04:41.467" Score="0" ViewCount="161" Body="&lt;p&gt;I have a problem understanding this equation (I encountered it in a mechanics problem).&#xA;Given:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x = x(t)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$q = q(t)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$v = dx/dt$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$w = dq/dt$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac {\partial v}{ \partial w} = \frac{\partial x}{\partial q}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is that true? If yes, why is that true? How can I proof that?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT:&#xA;Actually it's a passage in the derivation of the Lagrange equations in Analytical Mechanics:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$v = dOP/dt$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\dot{q} = dq/dt$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\partial v / \partial \dot{q} = \partial OP/ \partial q$&lt;/p&gt;&#xA;" OwnerUserId="1228" LastEditorUserId="1228" LastEditorDisplayName="" LastEditDate="2010-10-27T13:58:29.073" LastActivityDate="2010-11-25T20:03:39.290" Title="Derivating in respect of a function that's a derivative of another function" Tags="&lt;calculus&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="8041" PostTypeId="2" ParentId="8039" CreationDate="2010-10-27T13:28:07.870" Score="1" ViewCount="" Body="&lt;p&gt;You don't even need to look at the coordinate entries of your matrices. If E and F are vector spaces, and A and B are linear maps from E to F such that for all x in E: Ax=Bx, then by linearity (A-B)x = 0 for all x. Now what can you say about the linear map A-B, and what does that imply about A and B?&lt;/p&gt;&#xA;" OwnerUserId="1269" LastActivityDate="2010-10-27T13:28:07.870" CommentCount="2" />
  <row Id="8042" PostTypeId="1" AcceptedAnswerId="8171" CreationDate="2010-10-27T13:35:46.393" Score="5" ViewCount="315" Body="&lt;p&gt;How can I prove that for every positive integer $n$ we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{n\pi}{4}−\frac{1}{\sqrt{8n}}&amp;lt;\frac{1}{2}+\sum_{k=1}^{n−1}\sqrt{1−\frac{k^2}{n^2}}$?&lt;/p&gt;&#xA;" OwnerUserId="2856" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-10-28T17:28:42.340" LastActivityDate="2010-10-28T17:28:42.340" Title="An inequality like Riemann sum involving $\sqrt{1-x^2}$" Tags="&lt;sequences-and-series&gt;&lt;inequality&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="8043" PostTypeId="2" ParentId="7974" CreationDate="2010-10-27T13:51:14.493" Score="1" ViewCount="" Body="&lt;p&gt;I recommend chapters 3 and 5 of &lt;a href=&quot;http://nr.com/&quot; rel=&quot;nofollow&quot;&gt;Numerical Recipes&lt;/a&gt;.  It sounds like you have a function f which is hard/expensive to evaluate.  You can evaluate it wherever you like, just not too many times.  You then use those values to model f in some way that is much easier to compute and use that model to get values at other points.  What do you know about f?  If you know it is a ninth-degree polynomial (just not which one) you can evaluate it at ten points, do a fit, and have exact values anywhere else (ignoring important issues of numeric accuracy).  If you know it is a finite sum of trig functions, the Nyquist theorem tells you that proper sampling can also give you perfect accuracy on extrapolation.  If you know it is multiplicative, evaluate it at a set of primes.  If you don't know, maybe the best you can do is evaluate it on a grid and use splines.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An important question is what you will use the values at other points for.  If you need an integral over some region, the formulas will tell you where to evaluate it to get high accuracy &lt;strong&gt;as long as your functional model is correct&lt;/strong&gt;.  For this, you wouldn't need the value at any other points.  If you need derivatives, numeric differentiation is a noise amplifier and dangerous.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If your f is not in the class of functions you think, things can go badly wrong.  NR has a classic example where poles off the real line can cause problems, even for fitting a real function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is important to remember that although you can get approximate values at other points, evaluating these values teaches you nothing about f.  They teach you about g, which is the function within the class you selected to approximate f.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-10-27T13:51:14.493" />
  <row Id="8044" PostTypeId="1" AcceptedAnswerId="8051" CreationDate="2010-10-27T14:09:03.573" Score="0" ViewCount="143" Body="&lt;p&gt;I have made up this &quot;fun&quot; problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first row of a 3x3 matrix is $(a_{11} a_{12} a_{13})$. The next row consists of the variables $h, g$ and $c$ in any order. The last row are distinct non-zero fixed scalars, such as $2, 1, 7$ or $-4$ etc. They could be any integer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given an expression for $det(A)$ for example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(-c -2h)a_{11} -3ca_{13} + 3ha_{12} + 2ga_{13} + ga_{12}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the probability that a matrix you construct meeting the conditions given and with this determinate will be the original matrix A?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ans: 1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the last two rows are sets of distinct non-zero integers. What is the probability that a matrix you construct meeting the conditions given will be the original matrix A?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example $det(A) = -71a_{11} +13a_{12} +11a_{13}$ how many matrices could produce this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(This one seems to be limited to the number of solutions to a Diophantine system, and it's not I type I can solve, unless there is another way...)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we let the scalars be any real number will the set of possible matrices become infinite?&lt;/p&gt;&#xA;" OwnerUserId="2563" LastEditorUserId="2563" LastEditorDisplayName="" LastEditDate="2010-10-27T14:37:54.780" LastActivityDate="2010-10-27T15:12:54.390" Title="Recovering a matrix A from det(A)" Tags="&lt;linear-algebra&gt;&lt;matrices&gt;&lt;diophantine-equations&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="8045" PostTypeId="2" ParentId="8013" CreationDate="2010-10-27T14:09:20.750" Score="7" ViewCount="" Body="&lt;p&gt;Pete's excellent notes have correctly explained that there is no set containing sets of unboundedly large size in the infinite cardinalities, because from any proposed such family, we can produce a set of strictly larger size than any in that family. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This observation by itself, however, doesn't actually prove that there are uncountably many infinities. For example, Pete's argument can be carried out in the classical Zermelo set theory (known as Z, or ZC, if you add the axiom of choice), but to prove that there are uncountably many infinities requires the axiom of Replacement. In particular, it is actually consistent with ZC that there are only countably many infinities, although this is not consistent with ZFC, and this fact was the historical reason for the switch from ZC to ZFC. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The way it happened was this. Zermelo had produced sets of size $\aleph_0$, $\aleph_1,\ldots,\aleph_n,\ldots$ for each natural number $n$, and wanted to say that therefore he had produced a set of size $\aleph_\omega=\text{sup}_n\aleph_n$. Frankael objected that none of the Zermelo axioms actually ensured that $\{\aleph_n\mid n\in\omega\}$ forms a set, and indeed, it is now known that in the least Zermelo universe, this class does not form a set, and there are in fact only countably many infinite cardinalities in that universe; they cannot be collected together there into a single set and thereby avoid contradicting Pete's observation. One can see something like this by considering the universe $V_{\omega+\omega}$, a rank initial segment of the von Neumann hierarchy, which satisfies all the Zermelo axioms but not ZFC, and in which no set has size $\beth_\omega$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;By adding the Replacement axiom, however, the Zermelo axioms are extended to the ZFC axioms, from which one can prove that $\{\aleph_n\mid n\in\omega\}$ does indeed form a set as we want, and everything works out great. In particular, in ZFC using the Replacement axiom in the form of transfinite recursion, there are huge uncountable sets of different infinite cardinalities. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The infinities $\aleph_\alpha$, for example, are defined by transfinite recursion:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\aleph_0$ is the first infinite cardinality, or $\omega$.&lt;/li&gt;&#xA;&lt;li&gt;$\aleph_{\alpha+1}$ is the next (well-ordered) cardinal after $\aleph_\alpha$. (This exists by Hartog's theorem.)&lt;/li&gt;&#xA;&lt;li&gt;$\aleph_\lambda$, for limit ordinals $\lambda$, is the supremum of the $\aleph_\beta$ for $\beta\lt\lambda$. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Now, for any ordinal $\beta$, the set $\{\aleph_\alpha\mid\alpha\lt\beta\}$ exists by the axiom of Replacement, and this is a set containing $\beta$ many infinite cardinals. In particular, for any cardinal $\beta$, including uncountable cardinals, there are at least $\beta$ many infinite cardinals, and indeed, strictly more.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The cardinal $\aleph_{\omega_1}$ is the smallest cardinal having uncountably many infinite cardinals below it.&lt;/p&gt;&#xA;" OwnerUserId="413" LastEditorUserId="413" LastEditorDisplayName="" LastEditDate="2010-10-27T16:21:52.773" LastActivityDate="2010-10-27T16:21:52.773" CommentCount="4" />
  <row Id="8046" PostTypeId="1" CreationDate="2010-10-27T14:21:18.437" Score="2" ViewCount="66" Body="&lt;p&gt;&lt;em&gt;Wondering if my following question is an application of information theory:&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lets say we have a factory and ship boxes of stuff outside. If a competitor stands outside my factory, observes the stream of boxes coming out of a factory and note their sizes, he can get valuable information about my manufacturing process. So in order to hoodwink him, I manually expand the boxes to sizes different than the original ones. In order to make sure that I am doing a good job at masking the original size distribution from the resultant distribution, I must make sure that I maximize relative entropy between output distribution q and input distribution p, given that I cannot just create a uniform output distribution. Is this premise correct? Do you find any flaw in the question?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The constraint, of course, is that we must use minimum amount of box material. In other words, sum of box sizes for a unit period of time must not exceed a threshold B.&lt;br&gt;&#xA;                 i.e. $\frac{\sum_{j=1}^{i}{BS_i}}{T} \le B$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other constraints are:            &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1] Both p and q must be probability distributions i.e. sum up to 1&lt;br&gt;&#xA; 2] $BS_i \le MS$, where MS is the maximum size of a box.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;--&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we take $BS_i$ as the size of the $i^{th}$ box, then we have an optimization function applying Lagrange multiplier $\lambda$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ \Lambda(BS_i,\lambda,p,q) = D(q||p) + \lambda (\frac{\sum_{j=1}^{i}{BS_i}}{T} - B) $&#xA;$                           = \sum_{S=1}^{n}{q(S) \log  \frac{q(S)}{p(S)}} + \lambda (\frac{\sum_{j=1}^{i}{BS_i}}{T} - B)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$                           = \sum_{S=1}^{n}{q(S) \log q(S)} -\sum_{S=1}^{n}{q(S) \log p(S)} + \lambda (\frac{\sum_{j=1}^{i}{BS_i}}{T} - B)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;To proceed:&lt;/strong&gt;&#xA;Lets say we at the $i^{th}$ iteration. We know the actual box sizes that we already sent out for 1...i-1 iterations, and the actual size of the present box. i.e. $BS_0..BS_i$ are known, and hence p(BS) can be calculated and hence is a known value. We know the output box sizes till the $i-1^{th}$ iteration. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, we have,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ \Lambda(BS_i,\lambda,q) = \sum_{S=1}^{n}{q(S) \log q(S)} -\sum_{S=1}^{n}{k*q(S) } + \lambda (\frac{\sum_{j=1}^{i}{BS_i}}{T} - B)  $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do we now solve the optimization function, in order to get the output box size for this $i^{th} iteration?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have,&#xA;$\frac{\partial\Lambda}{\partial BS_i} = 0$,&#xA;$\frac{\partial\Lambda}{\partial q(S)} = 0$, and&#xA;$\frac{\partial\Lambda}{\partial \lambda} = 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The pertial differentiation wrt $\lambda$ results in a trivial solution. Since q(S) is dependent on $BS_i$, we will get a factor of $\frac{\partial q(S)}{\partial BS_i}$ in the other two equations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do I proceed from here? What I am thinking wrong?&lt;/p&gt;&#xA;" OwnerUserId="2858" LastActivityDate="2010-10-27T14:21:18.437" Title="Optimization of relative entropy" Tags="&lt;optimization&gt;&lt;convex-optimization&gt;" />
  <row Id="8047" PostTypeId="2" ParentId="8040" CreationDate="2010-10-27T14:25:10.610" Score="5" ViewCount="" Body="&lt;p&gt;If you have a parametrized curve $(x,y)=(f(t),g(t))$, then the tangent vector is $(f'(t),g'(t))$. Provided that the tangent vector isn't pointing straight up or straight down (that is, provided that $f'(t)\neq 0$), the curve looks locally like the graph of a function $y=H(x)$, and its slope $dy/dx=H'(x)$ at some point $(x,y)=(a,H(a))$  is just $g'(t_0)/f'(t_0)$ where $t_0$ is the parameter value corresponding to that point on the curve (that is, $a=f(t_0)$ and $H(a)=g(t_0)$). To see this, think of the right triangle with the tangent vector as its hypotenuse; it has $\Delta y=g'(t_0)$ and $\Delta x=f'(t_0)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what the statement $dy/dx=(dy/dt)/(dx/dt)$ really means is $H'(a)=g'(t_0)/f'(t_0)$.&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2010-10-27T14:25:10.610" />
  <row Id="8048" PostTypeId="1" AcceptedAnswerId="8049" CreationDate="2010-10-27T14:26:35.907" Score="1" ViewCount="105" Body="&lt;p&gt;In this formula for Euler numbers:&#xA;$\begin{align*}A_n = i^{n+1}\sum _{k=1}^{n+1} \sum _{j=0}^k{k\choose{j}} \frac{(-1)^j(k-2j)^{n+1}}{2^ki^kk}\end{align*}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;what is $i$? I have to generate the $n-th$ Euler number, and I don't have $i$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Isn't it the imaginary unit, is it?&lt;/p&gt;&#xA;" OwnerUserId="2582" LastActivityDate="2010-10-27T14:39:02.243" Title="Euler numbers formula" Tags="&lt;linear-algebra&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="8049" PostTypeId="2" ParentId="8048" CreationDate="2010-10-27T14:39:02.243" Score="7" ViewCount="" Body="&lt;p&gt;Yes, it is $\sqrt{-1}$.  They will all divide out in the end.  A derivation of your formula is &lt;a href=&quot;http://www.voofie.com/content/117/an-explicit-formula-for-the-euler-zigzag-numbers-updown-numbers-from-power-series/&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.  The $i$'s come from changing the $\sin(x)$ function to its definition in terms of $\exp(x)$&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-10-27T14:39:02.243" CommentCount="3" />
  <row Id="8051" PostTypeId="2" ParentId="8044" CreationDate="2010-10-27T14:58:28.897" Score="1" ViewCount="" Body="&lt;p&gt;Letting the entries in the other two rows be $a, b, c, d, e, f$, we are solving a system of Diophantine equations&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$bf - ce = P, cd - af = Q, ae - bd = R.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;($P, Q, R$ are $-71, -13, 11$ in your example.)  This is a system of three equations in six variables so we should expect lots of solutions.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's one way to write them down.  Let $Q = \gcd(Q, R) Q', R = \gcd(Q, R) R'$ and fix $a = \gcd(Q, R), d = 2 \gcd(Q, R)$.  Then the last two equations become $f = c - Q', e = R' + 2b$ and the first equation becomes&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$b(c - Q') - c(R' + b) = - bQ' - cR' = P.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;since $\gcd(Q', R') = 1$, this always has a solution by &lt;a href=&quot;http://en.wikipedia.org/wiki/Bezouts_identity&quot; rel=&quot;nofollow&quot;&gt;Bezout's lemma&lt;/a&gt;.  It can easily be found using the &lt;a href=&quot;http://en.wikipedia.org/wiki/Extended_Euclidean_algorithm&quot; rel=&quot;nofollow&quot;&gt;extended Euclidean algorithm&lt;/a&gt;.  In fact, if it has one solution it has infinitely many, since if $(b, c)$ are a pair of solutions, so is $(b + Rt, c - Qt)$ for every $t \in \mathbb{Z}$.  In particular, by choosing $t$ large enough you can arrange for $b$ to be very positive and $c$ to be very negative, which ensures that $a, b, c, d, e, f$ are distinct (although this is not a particularly natural restriction to me).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, for any choice of $P, Q, R$ with at least two of them nonzero, you get infinitely many solutions, so this &quot;probability&quot; stuff doesn't make sense without more qualification.  If two of them are equal to zero then the vectors $(a, d), (b, e), (c, f)$ are proportional and there are either zero or infinitely many solutions depending on whether the third is zero or nonzero.&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-10-27T15:12:54.390" LastActivityDate="2010-10-27T15:12:54.390" />
  <row Id="8052" PostTypeId="1" CreationDate="2010-10-27T15:00:52.733" Score="1" ViewCount="163" Body="&lt;p&gt;I know it is a very general question, but i'm not sure how to get started and improve number theory related skills related to ACM ICPC and other programming competitions.Any good way or book recommendations ?&lt;/p&gt;&#xA;" OwnerUserId="2860" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-10-27T15:34:00.030" LastActivityDate="2010-10-27T17:40:32.573" Title="How to improve number theory related skills ?" Tags="&lt;number-theory&gt;&lt;soft-question&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="8053" PostTypeId="2" ParentId="3978" CreationDate="2010-10-27T15:01:13.520" Score="0" ViewCount="" Body="&lt;p&gt;OK, after some days of struggling hard with this problem, i think i have found one more solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By Fermats little theorem, we have $$a^{p} + p  \equiv a ( \text{mod} \ p)$$ and $$b^{p}+p \equiv b (\text{mod} \ p)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next, $$\frac{b^{p}+p}{a^{p}+p}=\frac{k_{2}p+b}{k_{1}p+a} = C(\text{an integer})$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Choose a prime $p$ such that $p \mid a$, then by the above we can see that $p \mid b$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, we choose a prime such that $p \mid a$, then we have, $$\frac{k_{2}p+b}{k_{1}p+a}=\frac{k_{4}p}{k_{1}p+a} \Longrightarrow p\mid a \ \text{or} \ p=c$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now if $p=c$, then we will have $$\frac{b+1}{a+1}=p \Longrightarrow p\mid (p-1)$$ a contradiction. So $p \mid a$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we have concluded:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Whenever $p \mid a \Longrightarrow p \mid b$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Whenever $p \mid b \Longrightarrow p \mid a$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Does this help!&lt;/p&gt;&#xA;" LastActivityDate="2010-10-27T15:01:13.520" />
  <row Id="8054" PostTypeId="1" AcceptedAnswerId="8057" CreationDate="2010-10-27T15:06:17.300" Score="2" ViewCount="118" Body="&lt;p&gt;Hello.&#xA;What meaning of &quot;()&quot; in this formula:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$C_n=\frac1{n+1}\binom{2n}{n}=\frac{(2n)!}{n!(n+1)!}\qquad\mathrm{for}\;n\geq 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(That is Catalan number formula (http://en.wikipedia.org/wiki/Catalan_number))&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it a division?&lt;/p&gt;&#xA;" OwnerUserId="2861" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-10-27T15:12:06.033" LastActivityDate="2010-10-27T15:12:06.033" Title="What meaning of () in Catalan number formula?" Tags="&lt;notation&gt;" AnswerCount="1" />
  <row Id="8055" PostTypeId="2" ParentId="8052" CreationDate="2010-10-27T15:06:24.620" Score="1" ViewCount="" Body="&lt;p&gt;Well, i am not anyone great in answering this question, but atleast on my behalf, i shall tell you some good introductory books:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Introduction to theory of numbers by David Burton&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Introduction to theory of numbers by G.H.Hardy, E.M.Wright&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Introduction to Analytic number theory by Tom Apostol&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;104 problems in Number theory by T.Andreescu&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;All the above books are getting a very good foundation on the subject. As i said, i am no one to answer, this as i myself struggle in solving problems. But Solving problems are the only way one learns a subject. There are many peers, who would like to answer this question, and you can learn a lot from them!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Happy Reading!&lt;/p&gt;&#xA;" LastActivityDate="2010-10-27T15:06:24.620" />
  <row Id="8057" PostTypeId="2" ParentId="8054" CreationDate="2010-10-27T15:10:42.927" Score="6" ViewCount="" Body="&lt;p&gt;The notation $n\choose r$ means &quot;the number of ways of choosing $r$ things out of a total of $n$ given things, where order is not important.&quot;  Its value is $n!/r!(n-r)!$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, given five coins, say a penny, a nickel, a dime, a quarter, and a dollar, there are 10 ways to choose two coins with order unimportant, because ${5\choose 2} = 5!/2!3! = 120/(2)(6) = 10$.&lt;/p&gt;&#xA;" OwnerUserId="2682" LastActivityDate="2010-10-27T15:10:42.927" CommentCount="3" />
  <row Id="8058" PostTypeId="1" AcceptedAnswerId="8059" CreationDate="2010-10-27T15:22:27.177" Score="1" ViewCount="111" Body="&lt;p&gt;If $2 x^4 + x^3 - 11 x^2 + x + 2 = 0$, then find the value of $x + \frac{1}{x}$ ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would be very grateful if somebody show me how to factor this polynomial by hand, as of now I have used to mathematica to get  $(x-2) (2x - 1) (1 + 3x + x^2)$  but I am having trouble to factor this manually.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Best regards,&lt;/p&gt;&#xA;" LastActivityDate="2010-10-27T16:48:09.800" Title="How to solve this ?" Tags="&lt;algebra-precalculus&gt;&lt;polynomials&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="8059" PostTypeId="2" ParentId="8058" CreationDate="2010-10-27T15:26:36.907" Score="6" ViewCount="" Body="&lt;p&gt;You don't need to factor the polynomial to find the value of $x + \frac{1}{x}$.  Note that the given condition implies $\left( 2x^2 + \frac{2}{x^2} \right) + \left( x + \frac{1}{x} \right) - 11 = 0$ by dividing by $x^2$ and collecting symmetric terms.  Now observe that $\left( x + \frac{1}{x} \right)^2 = x^2 + \frac{1}{x^2} + 2$...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if you want to factor it by hand anyway, your best friend is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Rational_root_theorem&quot; rel=&quot;nofollow&quot;&gt;rational root test&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-10-27T15:26:36.907" CommentCount="1" />
  <row Id="8060" PostTypeId="2" ParentId="8024" CreationDate="2010-10-27T15:27:39.633" Score="2" ViewCount="" Body="&lt;p&gt;You need practice in formulating your intutions.  This is a skill that can be learned through practice, but like all skills, you can't begin just by jumping into the deep-end: you need to work consistently at extending your comfort zone, one small step at a time; over time you will then make substantial improvements.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What form should your practice take?  Well, you have to practice writing careful and complete proofs.   (The word &quot;writing&quot; here is important; it is not enough just to think aboutit and convince yourself that you &lt;em&gt;could&lt;/em&gt; write a correct answer --- rather, sit down with pen and paper and carefully write out the full and correct answer.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a practical suggestion of where to begin (in light of your specific problem): rather than beginning directly with a concept like (failure of) uniform continuity, which involves a rather large number of quantifiers, be sure you can completely prove some simpler statements, such as &quot;$1/(x-a)$ is unbounded on $(a,b)$&quot; or &quot;$x^2$ is unbounded on $\mathbb R$&quot;.   Once you can comfortably prove such statements, you can build up to carefully proving analytic statements with more quantifiers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One more thing: When trying to prove statements (using precise defintions)&#xA;which you feel that you have a good intuitive understanding for, try to self-conciously refelect on how your intuition accords with the formal definitions.  For example, ask yourself &quot;why do I think that $x^2$ is unbounded on $\mathbb R$?&quot; Try to analyze your own intuition and extract some statement out of it.  Then compare this with the formal defintion of unbounded and see if they are the same.  If so, great (!) ; you've got good intuition.  If they dont' quite agree, try to understand why; what was missing in your intuition?  what extra aspect of the situation is the formal definition trying to capture?  Again, reflecting on and analyzing your mathematical intuition is a skill that can be practiced and which you can improve --- but, like all skills, you have to practice it if you want to get better.&lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2010-10-27T15:27:39.633" CommentCount="3" />
  <row Id="8062" PostTypeId="1" CreationDate="2010-10-27T15:45:14.707" Score="1" ViewCount="64" Body="&lt;p&gt;If a,b,c are in arithmetic progression., p,q,r in harmonic progression and ap,bq,cr are in geometric progression., then $\frac{p}{r}+\frac{r}{p} = $ ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt; I have tried to use the basic/standard properties of the respective progressions to get the desired result, but I am not yet successful.&lt;/p&gt;&#xA;" LastActivityDate="2010-10-27T18:09:38.983" Title="A problem on progression" Tags="&lt;algebra-precalculus&gt;" AnswerCount="3" />
  <row Id="8063" PostTypeId="2" ParentId="8020" CreationDate="2010-10-27T15:49:18.110" Score="3" ViewCount="" Body="&lt;p&gt;The point is that there is an explicit formula (due to Riemann) relating&#xA;$\pi(x)$ to the zeroes of the zeta-function.  (The proof is via a kind of Fourier transform.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rough shape is that&#xA;$$\pi(x) = \mathrm{Li}(x) + \sum_{\rho} \mathrm{Li}(x^{\rho}) + \text{ lower order terms},$$&#xA;where the sum is over zeroes $\rho$ of $\zeta(s)$ in the critical strip (i.e. with real&#xA;parts between $0$ and $1$).&#xA;(See &lt;a href=&quot;http://en.wikipedia.org/wiki/Explicit_formula&quot; rel=&quot;nofollow&quot;&gt;the wikipedia entry&lt;/a&gt; for a more precise statement; this is the same link as in Qiaochu's comment above.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the (simple but) key fact to remember is that&#xA;$| x^{\rho}| = x^{\Re \rho}$, for a positive real number $x$.&#xA;So to get asymptotics on $\pi(x)$ from this, one has to give &#xA;upper bounds on $\Re \rho$.  For example, to get the prime number theorem, one has to show that $\Re \rho &amp;lt; 1$ for all $\rho$ (i.e. that $\zeta(s)$&#xA;has no zeroes on the line $\Re s = 1$).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The best possible estimate comes if you assume RH.  Then $\Re \rho = 1/2$&#xA;for all $\rho$, so $| x^{\rho}| = x^{1/2}$, and (careful) estimates&#xA;give the error term $\sqrt{x} \log x$ for the difference between&#xA;$\pi(x)$ and $\mathrm{Li}(x)$.&lt;/p&gt;&#xA;" OwnerUserId="221" LastEditorUserId="221" LastEditorDisplayName="" LastEditDate="2010-10-27T15:58:35.210" LastActivityDate="2010-10-27T15:58:35.210" />
  <row Id="8064" PostTypeId="2" ParentId="8062" CreationDate="2010-10-27T15:49:57.823" Score="0" ViewCount="" Body="&lt;p&gt;If a,b,c are in arithmetic progression, then you can write them as a, a+d, a+2d.  Maybe all you need is 2b=a+c.  Similarly for geometric progression, you can write them as ap, apz, apz^2 where z is the common ration (often we use r, but that was taken).  Again, maybe (bq)^2=ap*cr is all you need.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-10-27T15:49:57.823" CommentCount="1" />
  <row Id="8065" PostTypeId="2" ParentId="8062" CreationDate="2010-10-27T16:08:50.487" Score="3" ViewCount="" Body="&lt;p&gt;Let's represent your arithmetic progression as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$a,a+h,a+2h$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $h=b-a$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and your harmonic progression as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$p,\frac{p}{l+1},\frac{p}{2l+1},$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(following &lt;a href=&quot;http://en.wikipedia.org/wiki/Harmonic_progression_%28mathematics%29&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and the condition that the products of corresponding terms of the arithmetic and harmonic sequences form a geometric sequence implies&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{a+h}{a(l+1)}=\frac{a+2h}{a+h}\frac{l+1}{2l+1}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{p}{r}+\frac{r}{p}=2l+1+\frac1{2l+1}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;we solve for $l$, which has two solutions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I leave the rest to you. ;)&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-10-27T16:08:50.487" />
  <row Id="8066" PostTypeId="1" AcceptedAnswerId="8068" CreationDate="2010-10-27T16:25:24.443" Score="0" ViewCount="196" Body="&lt;p&gt;And if not, what are the most common hypotheses (on open sets it induces, for example) under which distance is a continuous function?&lt;/p&gt;&#xA;" LastActivityDate="2010-10-27T16:36:59.843" Title="Is the function distance continuous?" Tags="&lt;analysis&gt;&lt;topology&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="8067" PostTypeId="2" ParentId="8066" CreationDate="2010-10-27T16:33:40.327" Score="2" ViewCount="" Body="&lt;p&gt;Yes.  The standard definition of the topology induced by a metric ensures this; in fact it's not hard to see that it's the coarsest topology such that $d(x, y)$ is continuous for fixed $x$.  &lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-10-27T16:33:40.327" CommentCount="1" />
  <row Id="8068" PostTypeId="2" ParentId="8066" CreationDate="2010-10-27T16:36:59.843" Score="0" ViewCount="" Body="&lt;p&gt;As Qiaochu points out $d(x,y)$ is continuous for fixed $x$. You may like to see this as well, as this is a familiar result in Topology.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If $A$ is a non empty closed subset of a metric space $(X,d)$ then show that the function f on X given by $f(x)=d(x,A)$ is continuous.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Because $$| f(x) - f(y) | = | d(x,A) - d(x,B) | \leq d(x,y)$$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means that $f$ is uniformly continuous (use $\delta = \epsilon$ in any point)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $x$ and $y$ be points in $X$, and $p$ any point of $A$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then $d(x,p) \leq d(x,y) + d(y,p)$ (triangle inequality)&#xA;and so $d(x,A) \leq d(x,y) + d(y,p)$ (as $d(x,A)$ is the infimum).&#xA;But then $d(y,p) \geq d(x,A) - d(x,y)$ (for all $p$)&#xA;so that $d(y,A) \geq d(x,A) - d(x,y)$ (as $d(y,A)$&#xA; is the infimum).&#xA;So : $d(x,A) - d(y,A) \leq d(x,y)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now reverse the roles of $x$ and $y$ to get&#xA;$d(y,A) - d(x,A) \leq d(x,y)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is taken from &lt;a href=&quot;http://at.yorku.ca/cgi-bin/bbqa?forum=homework_help_2004;task=show_msg;msg=1323.0001&quot; rel=&quot;nofollow&quot;&gt;http://at.yorku.ca/cgi-bin/bbqa?forum=homework_help_2004;task=show_msg;msg=1323.0001&lt;/a&gt;&lt;/p&gt;&#xA;" LastActivityDate="2010-10-27T16:36:59.843" />
  <row Id="8069" PostTypeId="2" ParentId="7963" CreationDate="2010-10-27T16:44:05.623" Score="1" ViewCount="" Body="&lt;p&gt;In one of the standard approaches to &lt;a href=&quot;http://en.wikipedia.org/wiki/Fuzzy_logic&quot; rel=&quot;nofollow&quot;&gt;fuzzy logic&lt;/a&gt;--used in household appliances everywhere!--the truth values are taken to be real numbers in the unit interval, where $0$ represents &lt;em&gt;false&lt;/em&gt; and $1$ represents &lt;em&gt;true&lt;/em&gt;, and the Boolean connectives are defined by the following equations:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;(not)   $\neg p = 1-p$&lt;/li&gt;&#xA;&lt;li&gt;(and)   $p\wedge q= \text{min}(p,q)$&lt;/li&gt;&#xA;&lt;li&gt;(or)    $p\vee q= \text{max}(p,q)$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;These values agree with the classical values when restricted to $0$ and $1$, and thus agree with Qiaochu's answer on the classical values, but they disagree on non-classical values. In particular, this version of fuzzy logic gives the same truth value to $p$ as $p\wedge p$, but this isn't the case with the probabilistic formulas, which (inappropriately) interprets $p\wedge p$ as a conjunction of independent events. Similarly, fuzzy logic also says $p$ is equivalent to $p\vee p$, whereas the probabilistic formula does not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is a fun exercise to show that fuzzy logic doesn't work with classical tautologies very well. For example, $p\vee \neg p$ is not uniformly value $1$, although it is always at least $\frac 12$. Indeed, a statement is a classical tuatology if and only if it gets fuzzy value at least $\frac 12$ on all input. There are numerous other such fun problems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are also numerous variations on the fuzzy theme, however, with many different fuzzy logics, and the Wikipedia page appears extensive.&lt;/p&gt;&#xA;" OwnerUserId="413" LastActivityDate="2010-10-27T16:44:05.623" />
  <row Id="8070" PostTypeId="1" CreationDate="2010-10-27T16:45:58.630" Score="0" ViewCount="119" Body="&lt;p&gt;2x ≡ 1 (mod p) where p is an odd prime&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm really stuck on this one.&lt;/p&gt;&#xA;" OwnerUserId="2868" LastActivityDate="2010-10-27T17:15:24.893" Title="Solving 2x ≡ 1 (mod p) where p is an odd prime" Tags="&lt;prime-numbers&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="8071" PostTypeId="1" AcceptedAnswerId="8076" CreationDate="2010-10-27T16:47:24.417" Score="11" ViewCount="804" Body="&lt;p&gt;At the Blizzcon 2010, StarCraft II multiplayer panel, this stuff was supposed to explain the ladder matchmaking system. I look at this and go eh? what!?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/CYV2b.png&quot; alt=&quot;StarCraft II: Ladder math&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is any of this real? or are they just messing with me, because I can't make heads or tails of this humongous equation. Can anyone break it down for me? or at least explain the principle behind this, I don't get how these integrals, derivatives, Euler-functions and multiplication sums yield anything meaningful...&lt;/p&gt;&#xA;" OwnerUserId="2867" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-10-28T02:40:37.200" LastActivityDate="2010-10-28T02:40:37.200" Title="StarCraft II: Ladder math" Tags="&lt;calculus&gt;&lt;statistics&gt;&lt;applications&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="4" />
  <row Id="8072" PostTypeId="2" ParentId="8058" CreationDate="2010-10-27T16:48:09.800" Score="3" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\rm\displaystyle\quad \bigg(a\ \bigg(x+\frac{1}x\bigg)^2 + b\ \bigg(x+\frac{1}x\bigg)+c\bigg)\ x^2\: \ =\ \ a\ (x^4 + 1) + b\ (x^3+x) + (c+2a)\ x^2 $  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The above allows you to determine the values of $\rm\:a,\:b,\:c\:$ by inspection. The same works for arbitrary even degree symmetric polynomials: $\ $ simply subtract $\rm\ a\ (x^2 +1)^k$ and recurse on what remains.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is essentially a special case of Gauss's algorithm for rewriting symmetric polynomials in terms of elementary symmetric polynomials. It's one of the earliest known occurrences of Grobner basis rewriting, e.g. see p. 314 of Cox, Little, O'Shea:$\ $ Ideals, Varieties and Algorithms. In chapter 7 they present a generalization to compute the ring of invariants of a finite matrix group $\rm\: G \subset GL(n,k)\:$. &lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-10-27T16:48:09.800" />
  <row Id="8073" PostTypeId="1" AcceptedAnswerId="8074" CreationDate="2010-10-27T17:05:25.377" Score="0" ViewCount="97" Body="&lt;p&gt;I need to do a series expansion of this forumla&#xA;$$z = \sqrt{\frac{x+1}{y+1}}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need to get Zx Zxx and so on. but I really dont know where to start with this this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;just a push in the right direction should be enough.&lt;/p&gt;&#xA;" OwnerUserId="2874" LastEditorUserId="1543" LastEditorDisplayName="" LastEditDate="2010-10-27T18:52:58.270" LastActivityDate="2010-10-27T18:52:58.270" Title="partial derivatives." Tags="&lt;calculus&gt;&lt;sequences-and-series&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="8074" PostTypeId="2" ParentId="8073" CreationDate="2010-10-27T17:10:07.260" Score="1" ViewCount="" Body="&lt;p&gt;If you know how to differentiate already, remember that partial differentiation with respect to a certain independent variable treats the other independent variables as constants.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To use $(x+y)^2$ as an example, $\partial_x (x+y)^2$ is just $2(x+y)$ (chain rule). Now try this operation on your function.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-10-27T17:10:07.260" CommentCount="1" />
  <row Id="8075" PostTypeId="2" ParentId="8070" CreationDate="2010-10-27T17:15:24.893" Score="1" ViewCount="" Body="&lt;p&gt;The trick proposed by Qiaochu in the comment is actually a very special case of &lt;a href=&quot;http://math.stackexchange.com/questions/3230&quot;&gt;Gauss's algorithm&lt;/a&gt; for computing inverses $\rm\:(mod\ p)\:$ for $\rm\:p\:$ prime. This special case of the Euclidean algorithm works simply by repeatedly scaling the fraction so to reduce the denominator until it reaches one. See my linked post for further details.&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-10-27T17:15:24.893" />
  <row Id="8076" PostTypeId="2" ParentId="8071" CreationDate="2010-10-27T17:24:16.150" Score="17" ViewCount="" Body="&lt;p&gt;It looks like an ordinary statistical calculation.  The numerator with $\Pi_{g=1}^G$ is a likelihood or probability density, presumably of some outcomes for games $1$ to $G$.  The denominator with $\int \Pi \dots$ is the integral of the numerator over all outcomes; it is a normalization constant to ensure the total probability of all results is $1$.  Everything in the formulas is a calculation of (z-scores in) independent normal distributions, so they have a fairly simple probability model for how a player's ranking parameters drive the game outcomes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The goal of the calculation might be to calculate a player's set of ranking parameters $y$ (a vector of numbers measuring strength, speed, skill, wins, or whatever interpretation the quantities have for the game) that maximize the conditional probability $P(g_j | y)$ of having observed the game outcomes $g_i$ for $i = 1$ to $G$.  In other words, Maximum Likelihood Estimation of a player's parameters from game data. I can't read everything in the formula -- can you post a larger magnification? -- but the $\theta_{1,g} - \theta_{2,g}$ look like a measure of how one side of the game performed compared relative to the other, such as a difference in number of points, or a measure of how the sides were &lt;em&gt;expected&lt;/em&gt; to perform relative to each other, given their ratings.  Alternatively, $P(g_j | y)$ could be a Bayesian &quot;posterior&quot; distribution on $y$ in light of the game outcomes, so that the formula is a rule for updating the rankings given some game results. Here $\Phi(\theta_0 + \gamma_i + \psi_{i,0})$ can be understood as implying an initial rating, where the distribution of skills in the player population is assumed to be normal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One can also infer from the formula that either they are doing the wrong calculation (after $G$ new games), or the big formula is actually a summary of what has happened after $G$ separate re-estimation steps, one after each game (so that in a single step there is no product involved, and the formula would involve only the ratings just before the game, and the game outcome).  The probabilies they are computing for the $G$ games are of the form &quot;what is the chance the player had performance at least $x$ in game 1, and at least $y$ in game 2, $z$ or better in game 3, ...&quot;.  This is not the correct way to assess the probability that the whole set of $G$ game results is, collectively, above a certain level of performance. But if the parameters are re-estimated after every game, and the older game outcomes forgotten, then &quot;chances of a victory at least $X$ big&quot; is the only thing you can do, so this would account for the shape of their formula.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, as the old joke goes:  &quot;and by the way, what's StarCraft?&quot;.&lt;/p&gt;&#xA;" OwnerUserId="467" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-10-27T20:02:45.843" LastActivityDate="2010-10-27T20:02:45.843" CommentCount="5" />
  <row Id="8077" PostTypeId="1" CreationDate="2010-10-27T17:25:44.387" Score="1" ViewCount="97" Body="&lt;p&gt;I've got a function whose general form is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f(x) = \frac{1}{x^\alpha}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $x &gt; 0$ and $\alpha &gt; 0$. I would like to show that if $0&amp;lt; \alpha &amp;lt; 1$ $f(x)$ has slow decay and if $\alpha &gt; 1$ the $f(x)$ has rapid decay. (I've already verified these properties of $f(x)$ using a graphing application.)&lt;/p&gt;&#xA;" OwnerUserId="1752" LastActivityDate="2010-10-27T22:54:54.007" Title="How to show that a function has slow or rapid decay" Tags="&lt;calculus&gt;&lt;functions&gt;" AnswerCount="3" CommentCount="5" />
  <row Id="8078" PostTypeId="1" AcceptedAnswerId="8083" CreationDate="2010-10-27T17:34:55.917" Score="2" ViewCount="125" Body="&lt;p&gt;Suppose $B \subset X$ where $X$ is a vector space. $B$ is called balanced if $\alpha B \subset B$ for every $\alpha \in \Phi$ with $|\alpha| \leq 1$. Note that $\Phi = \textbf{R}$ or $\Phi = \textbf{C}$. What is the intuition behind balanced sets? Why do we define them? Is it &quot;good&quot; for a vector space to have more balanced sets? Why do we require $| \alpha| \leq 1$?  &lt;/p&gt;&#xA;" OwnerUserId="2548" LastEditorUserId="171" LastEditorDisplayName="" LastEditDate="2010-10-27T17:39:38.470" LastActivityDate="2010-10-27T20:26:22.227" Title="Intuition Behind Balanced Sets" Tags="&lt;linear-algebra&gt;&lt;intuition&gt;&lt;functional-analysis&gt;" AnswerCount="3" CommentCount="3" />
  <row Id="8079" PostTypeId="2" ParentId="8052" CreationDate="2010-10-27T17:40:32.573" Score="1" ViewCount="" Body="&lt;p&gt;For the ACM you should probably focus on their problem sets and could also look into problems from discrete mathematics. I really don't think they ask questions that rely on a graduate level understanding of number theory so reading books could be overkill for your purposes and would only confuse you if you rush it. It should take several weeks to read several chapters in a good number theory book - assuming you're not too busy (but you're in cs, right?). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Short answer: Go and ask a math professor for some problem sets.&lt;/p&gt;&#xA;" OwnerUserId="2738" LastActivityDate="2010-10-27T17:40:32.573" />
  <row Id="8081" PostTypeId="1" AcceptedAnswerId="8082" CreationDate="2010-10-27T17:44:11.203" Score="1" ViewCount="200" Body="&lt;p&gt;Does any compact subset belong to Borel Sigma Algebra? &#xA;Why?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here Borel Sigma algebra is defined to be the sigma algebra generated by all open subsets in the topological space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks and regards!&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;UPDATE:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found that there are two kinds of definitions for a Borel sigma algebra: one generated by all open subsets, and the other by all compact subsets.&#xA;These two definitions are equivalent for most typical spaces, including any locally compact separable metrizable space (or more generally any σ-compact space), &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also in the definition for inner regular measure, is it correct that all the compact subsets are required to be Borel measurable, where the Borel sigma algebra is defined to be generated by all open sets instead of compact sets? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I saw in Wikipedia, inner regular measure is defined to be on Hausdorff space. Is a Haursdorff space a locally compact separable metrizable space or a σ-compact space? Or in a Hausdorff space, the sigma algebra generated by all compact subsets is a proper subset of the sigma algebra generated by all open subsets?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="1281" LastEditorUserId="1281" LastEditorDisplayName="" LastEditDate="2010-10-27T19:07:35.443" LastActivityDate="2010-10-27T19:07:35.443" Title="Does any compact subset belong to Borel Sigma Algebra ?" Tags="&lt;topology&gt;&lt;sigma-algebra&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="8082" PostTypeId="2" ParentId="8081" CreationDate="2010-10-27T17:48:16.183" Score="11" ViewCount="" Body="&lt;p&gt;If your space is Hausdorff then any compact subset is closed (hence the complement of something open), in which case the answer is &lt;em&gt;yes&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If your set is a finite set with more than one element and with the indiscrete topology (just the whole set and the empty set are open),&#xA;then every subset is compact, but proper subsets are not Borel.  So in complete generality the answer is &lt;em&gt;no&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2010-10-27T17:48:16.183" CommentCount="1" />
  <row Id="8083" PostTypeId="2" ParentId="8078" CreationDate="2010-10-27T17:52:48.830" Score="5" ViewCount="" Body="&lt;p&gt;[Added in response to the comments below:] Despite what is written below,&#xA;&lt;em&gt;balanced&lt;/em&gt; needn't imply convex; it simply doesn't follow from the definition.  So one should take the following discussion as rather an explanation about &lt;em&gt;convex&lt;/em&gt; balanced sets, for whatever that is worth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[Original answer follows:]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Balanced&lt;/em&gt; means that the set is symmetric around the origin (i.e. invariant under $v \mapsto -v$) and is convex.  These are nice properties, generalizing some useful features of balls around the origin in $\mathbb R^n$.  (Incidentally, if you didn't have the condition $|\alpha| = 1$, then any balanced subset with non-empty interior (in $\mathbb R^n$ say) would be all of $\mathbb R^n$, so that wouldn't be such an interesting condition.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, any vector space will have plenty of such sets; you can form them just by taking the convex hull of a collection of points invariant under $v \mapsto -v$.  What is more relevant in functional analysis is whether the origin of a topological vector space contains a &lt;em&gt;basis&lt;/em&gt; of balaced neighbourhoods.  The topological vector space is then called &lt;em&gt;locally convex&lt;/em&gt; and this has important consequences for the space (especially for&#xA;its duality theory).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that balls around the origin in a normed space are automatically balanced, so normed spaces are locally convex; so again, in the topological setting, one sees that this is a way of capturing some the geometric aspects&#xA;of normed vector spaces, without insisting on the existence of a norm.&#xA;(So, while normed spaces are not closed under certain operations, such as&#xA;forming infinite direct sums, locally convex spaces are --- any direct sum&#xA;(just to take one example) of a collection of locally convex spaces has &#xA;a natural locally convex topology of its own.)&lt;/p&gt;&#xA;" OwnerUserId="221" LastEditorUserId="221" LastEditorDisplayName="" LastEditDate="2010-10-27T20:26:22.227" LastActivityDate="2010-10-27T20:26:22.227" CommentCount="3" />
  <row Id="8084" PostTypeId="2" ParentId="8078" CreationDate="2010-10-27T17:53:49.747" Score="0" ViewCount="" Body="&lt;p&gt;In &lt;em&gt;Complex analysis in locally convex spaces&lt;/em&gt; by Seán Dineen, it states that:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;Balanced open sets arise as the natural domain of convergence of Taylor series expansion at the origin of &lt;i&gt;holomorphic functions.&lt;/i&gt;&lt;/blockquote&gt;&#xA;" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-10-27T18:42:35.153" LastActivityDate="2010-10-27T18:42:35.153" CommentCount="8" />
  <row Id="8085" PostTypeId="1" AcceptedAnswerId="8087" CreationDate="2010-10-27T18:01:11.143" Score="6" ViewCount="231" Body="&lt;p&gt;Suppose $A$ is a set of finite measure.   Is it possible that $A$ can be an &lt;strong&gt;uncountable&lt;/strong&gt; union of disjoint subsets of $A$, each of which has positive measure?  &lt;/p&gt;&#xA;" OwnerUserId="832" LastActivityDate="2010-10-27T18:04:51.227" Title="Set of Finite Measure: Uncountable disjoint subsets of non-zero measure." Tags="&lt;real-analysis&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="8086" PostTypeId="1" AcceptedAnswerId="8089" CreationDate="2010-10-27T18:02:12.203" Score="3" ViewCount="246" Body="&lt;p&gt;How is a Borel measure on Euclidean space defined? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I mean the Borel measure on the Borel sigma algebra generated by usual toplogy on an Euclidean space, which coincides with the normal definition of volume of subsets in the Euclidean space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For Lebesgue measure on an Euclidean space, it is defined in terms of the outer measure on the power set of the Euclidean space. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the Borel measure can be defined independent of the Lebesgue measure, then the Lebesgue measure can be defined in terms of the Borel measure by completion of measure. Right?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks and regards!&lt;/p&gt;&#xA;" OwnerUserId="1281" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2010-11-06T03:37:25.243" LastActivityDate="2010-11-06T03:37:25.243" Title="How is a Borel measure on Euclidean space defined?" Tags="&lt;topology&gt;&lt;measure-theory&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="0" />
  <row Id="8087" PostTypeId="2" ParentId="8085" CreationDate="2010-10-27T18:04:51.227" Score="7" ViewCount="" Body="&lt;p&gt;No.  Suppose $A$ is an uncountable disjoint union of measurable subsets $A_i, i \in I$ with positive measure.  Then $I$ is a countable union of the sets of indices $i$ such that $\mu(A_i) &gt; \frac{1}{n}, n \in \mathbb{N}$, so it follows that one of these sets must be uncountable.  In particular a countable union of some subcollection of the $A_i$ has arbitrarily large measure; contradiction.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-10-27T18:04:51.227" CommentCount="7" />
  <row Id="8088" PostTypeId="2" ParentId="8062" CreationDate="2010-10-27T18:09:38.983" Score="0" ViewCount="" Body="&lt;p&gt;Notice that $\rm\:\ \quad\displaystyle \frac{p}r+\frac{r}p\ =\ \frac{(p+r)^2}{pr} - 2$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But we have that$\rm\quad\displaystyle p\:r\ =\ \frac{(bq)^2}{ac}\ \ $ via $\rm\ ap,\:bq,\:cr\ $ geometric  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;and we have $\rm\quad\ \ \displaystyle p+r\ =\ \frac{2pr}q\quad\ \ \ $ via $\rm\ p,q,r\ $ harmonic&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-10-27T18:09:38.983" />
  <row Id="8089" PostTypeId="2" ParentId="8086" CreationDate="2010-10-27T18:15:47.440" Score="3" ViewCount="" Body="&lt;p&gt;The Borel $\sigma$-algebra on $\mathbb{R}$ is generated by the ring of finite disjoint unions of half-open intervals.  On this ring one can define a countably additive measure in the obvious way (although one should prove that it is countably additive).  By &lt;a href=&quot;http://en.wikipedia.org/wiki/Carath%C3%A9odorys_extension_theorem&quot; rel=&quot;nofollow&quot;&gt;Carathéodory's extension theorem&lt;/a&gt; this measure extends to the entire Borel $\sigma$-algebra, and by &lt;a href=&quot;http://en.wikipedia.org/wiki/Dynkin_system#Dynkin.27s_.CF.80-.CE.BB_Theorem&quot; rel=&quot;nofollow&quot;&gt;Dynkin's lemma&lt;/a&gt; this extension is unique.  Then one can take products to get a Borel measure on $\mathbb{R}^n$ and complete to get Lebesgue measure on the Lebesgue $\sigma$-algebra.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Of course, this isn't essentially different - the proof of Caratheodory that I've seen uses outer measure anyway.)&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-10-27T18:21:51.950" LastActivityDate="2010-10-27T18:21:51.950" />
  <row Id="8090" PostTypeId="2" ParentId="8077" CreationDate="2010-10-27T18:24:27.363" Score="1" ViewCount="" Body="&lt;p&gt;If you start at 1 (your sum is not defined at x=0), you can bound the sums as integrals: $ \int_{x=2}^{\infty}x^{-\alpha} dx \lt \sum_{x = 1}^\infty x^{-\alpha} \lt \int_{x=1}^{\infty}x^{-\alpha} dx$.  One will solve it for $\alpha \le 1$ and one for $\alpha \gt 1$&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-10-27T18:24:27.363" />
  <row Id="8091" PostTypeId="2" ParentId="8078" CreationDate="2010-10-27T18:25:23.080" Score="0" ViewCount="" Body="&lt;p&gt;It's like balancing a tire, you spin the set around! (A bit tongue in cheek, but that's how I've always felt about it...)&lt;/p&gt;&#xA;" OwnerUserId="941" LastActivityDate="2010-10-27T18:25:23.080" />
  <row Id="8092" PostTypeId="2" ParentId="8077" CreationDate="2010-10-27T18:30:59.417" Score="1" ViewCount="" Body="&lt;p&gt;Another approach is to use&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Cauchy_condensation_test&quot; rel=&quot;nofollow&quot;&gt;Cauchy's condensation test&lt;/a&gt;&#xA;which states that a series $\sum a_n$, where the terms $a_n$&#xA;are positive and decreasing, is convergent if and only if&#xA;$\sum 2^m a_{2^m}$ is convergent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this example the line $\mathbf{Re}(s)=1$ is the&#xA;edge of the region of ansolute convergence of the usual&#xA;formula $\sum_{n=1}^\infty 1/n^s$ for the Riemann zeta function.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-10-27T18:30:59.417" CommentCount="1" />
  <row Id="8093" PostTypeId="2" ParentId="7952" CreationDate="2010-10-27T19:11:09.583" Score="0" ViewCount="" Body="&lt;p&gt;In class today we asked the professor about this problem and he told us to look up logarithmic derivatives. This approach leads to the following solution:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the previous homework problem we know that: $e_{n}(x)+D_{n}=\prod(x+a)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider $\frac{f'}{f}$ where we are formally differentiating with respect to x. This transforms the LHS to $\frac{e\prime_{n}(x)}{e_{n}(x)+D_{n}}$ where &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ e\prime_{n}(x)=\frac{d}{dx}\sum_{i=0}^{n-1}x^{q^{i}}\frac{D_{n}}{D_{i}L_{n-i}}=(-1)^{n}\frac{D_{n}}{L_{n}}+\sum_{i=1}^{n-1}(-1)^{n-i}q^{i}\frac{D_{n}}{D_{i}L_{n-i}^{q}}x^{q^{i-1}}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;At $x=0$, $e\prime_{n}(x)=(-1)^{n}\frac{D_{n}}{L_{n}}$ and so the LHS becomes $\frac{(-1)^{n}}{L_{n}}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, the RHS gets transformed to $\sum\frac{1}{x+b}$ since $f'=\sum_{b}\frac{\prod(x+a)}{x+b}$ and $f=\prod(x+a)$. At $x=0$ this becomes $\sum\frac{1}{b}$ which is what we wanted.&lt;/p&gt;&#xA;" OwnerUserId="249" LastActivityDate="2010-10-27T19:11:09.583" />
  <row Id="8094" PostTypeId="1" CreationDate="2010-10-27T19:38:46.403" Score="1" ViewCount="149" Body="&lt;p&gt;I am not a math expert nor a native English speaker, so maybe I am using wrong terminology... In this case, help me to improve my question :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Part 1:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is my set of equations, I have to calculate values $x_n$, $y_n$ and $z_n$. I know the values  of A, B, C, Total and sums of parameters ($S_1$ = $x_1$ + $x_2$ + $x_3$, $S_2$ = $y_1$  + $y_2$ + $y_3$, $S_3$ = $z_1$ + $z_2$ + $z_3$). It is not necessary, that each equation gives us exactly the same result, but each equation should have its solution it's close to others as possible - in perfect case this would be Total/3.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x_1A + y_1B + z_1C = \text{Total}/3$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x_2A + y_2B + z_2C = \text{Total}/3$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x_3A + y_3B + z_3C = \text{Total}/3$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I figured it out, this is a set of diophantine equations. I could solve this with a genetic algorithm:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;First randomly assign values $x_n$, $y_n$ and $z_n$ (just watch for the sum of parameters)&lt;/li&gt;&#xA;&lt;li&gt;Then mutate this numbers so, that the solution of each equation is as close to others as possible. By mutating I mean to increase one parameter while decreasing another and calculate fitness function.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Part 2:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the first solution should work (have to try it out though). But it was a simplification of my original problem, which is like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$i(x_1A + y_1B + z_1C) = \text{Total}/3$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$j  (x_2A + y_2B + z_2C) = \text{Total}/3$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$k  (x_3A + y_3B + z_3C) = \text{Total}/3$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the problem is, that I cannot just randomly increase/decrease parameters, because $i$, $j$ and $k$ could be different and hence the set of these equations could yield different solution as previous mutation. Also - i, j and k are not given, I have to calculate them by my self.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the best way to solve my two problems?&lt;/p&gt;&#xA;" OwnerUserId="2873" LastEditorUserId="2873" LastEditorDisplayName="" LastEditDate="2010-10-28T07:59:17.083" LastActivityDate="2010-12-27T08:44:28.197" Title="How to calculate this set of diophantine equations" Tags="&lt;number-theory&gt;&lt;diophantine-equations&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="8095" PostTypeId="1" AcceptedAnswerId="8098" CreationDate="2010-10-27T19:49:27.543" Score="0" ViewCount="75" Body="&lt;p&gt;For each $n&gt;0$, how do we prove that $$\Gamma'(n+1)&gt; \log{n} \cdot \Gamma(n+1)$$&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;I had spent about half an hour on this question, but just could find any way of proceeding for the solution. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Wikipedia page gave me an interesting identity $$\Gamma'(n+1)= n! \cdot \Biggl( - \gamma + \sum\limits_{k=1}^{n} \frac{1}{k}\Biggr)$$ But i don't know how it can be applied here.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" LastActivityDate="2010-10-27T20:00:35.377" Title="Inequality concerning the Gamma function" Tags="&lt;real-analysis&gt;&lt;analysis&gt;&lt;special-functions&gt;&lt;inequality&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="8096" PostTypeId="1" CreationDate="2010-10-27T19:51:33.937" Score="2" ViewCount="117" Body="&lt;p&gt;Let $X$ be integral scheme and $\mathcal K$ sheaf of rational functions on $X$. For any &#xA;point $y\in X$ different of generic point we know that fiber of $\mathcal K$ (defined as usual as $\mathcal K _y / \mathcal m_y \mathcal K_y$) is zero. I'll be very gratefull if you explain intuitively why this is so, in language of restriction of $\mathcal K$ to reduced subscheme $Y=\overline{\{y \} }$. I have difficulty because many rational functions on $X$ can be restricted to nonzero rational functions on  $Y$ . How is that compatible with fiber of $\mathcal K$ equals zero at $y$?  &lt;/p&gt;&#xA;" OwnerUserId="2872" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-10-27T19:57:23.583" LastActivityDate="2010-10-27T19:57:23.583" Title="Intuition for fiber of rational function sheaf" Tags="&lt;algebraic-geometry&gt;" CommentCount="7" />
  <row Id="8097" PostTypeId="2" ParentId="7475" CreationDate="2010-10-27T19:59:31.807" Score="0" ViewCount="" Body="&lt;p&gt;I made a table using the labels {I3, p1, p2, p3, p4, p5} instead of filling the table with the matrices.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;---  I3  p1  p2  p3  p4  p5&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I3  I3  p1  p2  p3  p4  p5&lt;/p&gt;&#xA;&#xA;&lt;p&gt;p1  p1  I3  p3  p2  p5  p4&lt;/p&gt;&#xA;&#xA;&lt;p&gt;p2  p2  p5  I3  p4  p3  p1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;p3  p3  p4  p1  p5  p2  I3&lt;/p&gt;&#xA;&#xA;&lt;p&gt;p4  p4  p3  p5  p1  I3  P2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;p5  p5  p2  p4  I3  p1  p3&lt;/p&gt;&#xA;&#xA;&lt;p&gt;f_I3 (x) = I3&lt;/p&gt;&#xA;&#xA;&lt;p&gt;then looking at a matrix mapping looking at p1 line:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;G = I3 p1 p2 p3 p4 p5&#xA;    p1 I3 p3 p2 p5 p4&lt;/p&gt;&#xA;&#xA;&lt;p&gt;p1*I3 = p1  p1*p1 = I3  this brings it back to start so &#xA;(I3, p1)&#xA;p1*p2 = p3  p1*p3 = p2 brings it back to start so &#xA;(I3, p1),(p2, p3)&#xA;p1*p4 = p5  p1*p5 = p4 brings it back to start so &#xA;(I3, p1),(p2, p3),(p4, p5)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;f_p1 (x) = (I3, p1),(p2, p3),(p4, p5)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;then looking at a matrix mapping looking at p2 line:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;G = I3 p1 p2 p3 p4 p5&#xA;    p2 p5 I3 p4 p3 p1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;f_p2 (x) = (I3, p2), (p1, p5), (p3, p4)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and so on... &lt;/p&gt;&#xA;" LastActivityDate="2010-10-27T19:59:31.807" CommentCount="1" />
  <row Id="8098" PostTypeId="2" ParentId="8095" CreationDate="2010-10-27T20:00:35.377" Score="6" ViewCount="" Body="&lt;p&gt;Well, if we rearrange the inequality to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\Gamma^{\prime}(n+1)}{\Gamma(n+1)} &gt; \log n$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and then note that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\Gamma^{\prime}(n+1)}{\Gamma(n+1)}=-\gamma+\sum_{j=1}^n \frac1{j}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;you should be able to easily deduce the inequality.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-10-27T20:00:35.377" CommentCount="2" />
  <row Id="8099" PostTypeId="1" CreationDate="2010-10-27T20:07:39.767" Score="1" ViewCount="128" Body="&lt;p&gt;Given octic group G = {e, s, s^2, s^3, b, g, d, t}   (sigma, betea, delta... )&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Find subgroup of G that has order 2 and is a normal subgroup of G.&#xA;Find subgroup of G that has order 2 and is NOT a normal subgroup of G.&lt;/p&gt;&#xA;" LastActivityDate="2010-10-27T20:07:39.767" Title="Find subgroup of octic group that is normal and NOT normal" Tags="&lt;abstract-algebra&gt;&lt;homework&gt;" CommentCount="6" />
  <row Id="8100" PostTypeId="2" ParentId="58" CreationDate="2010-10-27T20:20:10.033" Score="15" ViewCount="" Body="&lt;p&gt;Some of the data generated by Math Stack Exchange itself (and, presumably, by similar sites) ought to approximately follow Benford's law.  These would include the distribution of first digits of &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Frequency of tag use,&lt;/li&gt;&#xA;&lt;li&gt;Number of votes for questions,&lt;/li&gt;&#xA;&lt;li&gt;User reputation,&lt;/li&gt;&#xA;&lt;li&gt;Number of views for questions.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;This is because Benford's law applies to exponentially growing quantities, and the total number of all of these quantities ought to be growing exponentially.  It's only approximate because of artifacts in the way that some of these quantities are determined and because you need several orders of magnitude in order to see Benford's law really kick in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, I decided to test this for the first three.  The first digits weren't actually that hard to compile because the site allows you to sort the first three from highest to lowest.  (Unfortunately, it does not allow you to sort the fourth that way, and I don't feel like wading through 2200+ questions to collect the data.) Here are the results I got on the first three.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Tag Use&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Data collected October 25, 2010.)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/njTkU.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Not a bad fit, especially when you consider that there are only three orders of magnitude represented in this measure.  There are a disproportionately large number of tags that were created but only used once, which explains the larger frequency of 1 as a first digit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Votes Per Question&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Data collected October 26, 2010.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/zC1pM.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also not a bad fit, especially since there are currently only two orders of magnitude represented in this measure.  Also, I threw out the questions with 0 and negative numbers of votes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;User Reputations&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Data collected October 25 and 26, 2010.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/kJYsl.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the worst fit of the three, as the frequency of 1 as a first digit is so much larger than the others.  However, there are a very large number of users who have never posted a question or an answer and so have a rep of 1.  And, for reasons unknown to me, there are also a large number of users who have a reputation of 101, despite never having asked or answered a question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you remove 1 as a possible first digit and then rescale the Benford law probabilities to consider only 2 through 9 as possible first digits then the picture looks like the following, which is a much better Benford fit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/ZM83s.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Admittedly, none of these data sets would pass Pearson's chi-square test for goodness-of-fit with respect to the Benford probabilities at a reasonable level of significance.  However, given some of the artifacts in the data and the fact that there are relatively few orders of magnitude represented, the fit with Benford's law is really not that bad.&lt;/p&gt;&#xA;" OwnerUserId="2370" LastActivityDate="2010-10-27T20:20:10.033" CommentCount="3" />
  <row Id="8101" PostTypeId="1" AcceptedAnswerId="8119" CreationDate="2010-10-27T20:24:36.073" Score="6" ViewCount="160" Body="&lt;p&gt;Polynomial $P$ satisfies $P(n)&gt;n$ for all positive integers $n$. Every positive integer $m$ is a factor of some number of the form $P(1),P(P(1)),P(P(P(1))),\ldots $. Prove that $P(x)=x+1$.&lt;/p&gt;&#xA;" OwnerUserId="2856" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-10-27T20:28:24.170" LastActivityDate="2010-10-28T00:03:54.260" Title="Iterated polynomial problem" Tags="&lt;number-theory&gt;&lt;polynomials&gt;&lt;functional-equations&gt;" AnswerCount="1" CommentCount="8" FavoriteCount="1" />
  <row Id="8102" PostTypeId="2" ParentId="7986" CreationDate="2010-10-27T20:51:21.483" Score="5" ViewCount="" Body="&lt;p&gt;This expands on a comment that might not have been perfectly clear (perhaps because of its obviousness, for which I apologize).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose you have in hand some magic method of integration (such as the one in the question).  By this I mean &lt;em&gt;any&lt;/em&gt; procedure that takes a functional &lt;em&gt;expression&lt;/em&gt; $f(x)$ and returns some &lt;em&gt;expression&lt;/em&gt; $F(x)$ for its indefinite integral. (I also assume you accept $F(x)$ as a valid solution to your integration; &lt;em&gt;e.g.&lt;/em&gt;, some people might not accept an elliptic function, or imaginary values are &lt;em&gt;verboten,&lt;/em&gt; or they might not like power series, &lt;em&gt;etc&lt;/em&gt;.  I leave the criterion of acceptability to you as a matter of taste.  All I require is that you know how to differentiate $F$ and are able to compare that result to $f$ in order to check the validity of your magic method.)  A trivial application of the Fundamental Theorem of Calculus asserts that the &quot;very clever&quot; substitution&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$u = F(x)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;will enable you to perform the integral.  This happens, of course, because you can calculate that $du = F'(x)dx = f(x)dx$, whence the substitution converts the original integral into $\int{du}$ with general solution $u + C = F(x) + C$ by back-substitution.  In other words, the question as posed merely asks whether the FTC is true in a special case.&lt;/p&gt;&#xA;" OwnerUserId="1489" LastActivityDate="2010-10-27T20:51:21.483" CommentCount="5" />
  <row Id="8103" PostTypeId="2" ParentId="8035" CreationDate="2010-10-27T20:54:37.780" Score="0" ViewCount="" Body="&lt;p&gt;You could write $y(x)=a$ instead of $y=a$ to point out you are meaning a constant function, as $y(a)=a$ is not constant. Part (horizontal line) is not necessary. &lt;/p&gt;&#xA;" OwnerUserId="2856" LastActivityDate="2010-10-27T20:54:37.780" CommentCount="0" />
  <row Id="8104" PostTypeId="1" AcceptedAnswerId="8129" CreationDate="2010-10-27T21:01:13.140" Score="3" ViewCount="106" Body="&lt;p&gt;If i have a standard LP problem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\min \mathbf{d}^T \mathbf{x}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;subject to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathbf{B}\mathbf{x}=\mathbf{f},\qquad \mathbf{x} \geq 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\mathbf{y}$ is the optimal solution and $\mathbf{z}$ is the optimal solution to the dual problem&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, for the same cost function $\mathbf{d}$, if $\mathbf{f}$ is replaced by $\mathbf{b}$ then $\mathbf{x}$ becomes the new optimal solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can it be shown:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathbf{z}^T (\mathbf{b}-\mathbf{f}) \leq \mathbf{d}^T (\mathbf{x}-\mathbf{y})$$&lt;/p&gt;&#xA;" OwnerUserId="2014" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-10-28T01:55:05.023" LastActivityDate="2010-10-28T01:55:05.023" Title="Prove property of dual Linear Programming problem" Tags="&lt;optimization&gt;&lt;linear-programming&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="8105" PostTypeId="1" AcceptedAnswerId="8216" CreationDate="2010-10-27T21:27:08.070" Score="1" ViewCount="97" Body="&lt;p&gt;I'm quite positive that I saw before an encyclopedia (more accurately, dictionary/glossary) on Elsevier website. I remember I looked up the definition of a hyperplane there, few months ago. But I can't find it through Google anymore.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anyone know the link?&lt;/p&gt;&#xA;" OwnerUserId="2468" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-10-28T01:50:36.107" LastActivityDate="2010-10-29T19:29:25.217" Title="Website for Elsevier Encyclopedia (or glossary)?" Tags="&lt;reference-request&gt;&lt;online-resources&gt;" AnswerCount="2" CommentCount="5" />
  <row Id="8106" PostTypeId="2" ParentId="8008" CreationDate="2010-10-27T21:41:32.567" Score="6" ViewCount="" Body="&lt;p&gt;I can't believe this question is not closed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As near as I can tell, Chloe solved the problem which has stymied Dr. Rush by successfully integrating a constant. Apparently Rush's expertise in ancient technology does not cover first year calculus. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/kv6EZ.jpg&quot; alt=&quot;Before&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/Z4lZ7.jpg&quot; alt=&quot;After&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update&lt;/strong&gt;: Actually, looking over it, I might be mistaken. I interpreted the question as $-\int_0^t g \sin \gamma dt$ and Chloe's answer as $-\ddot{r} t \sin \gamma$, which is integrating a constant and making a gratuitous substitution of $\ddot{r}$ for $g$. I can't make any sense at all of the middle line in Chloe's answer.&lt;/p&gt;&#xA;" OwnerUserId="317" LastEditorUserId="317" LastEditorDisplayName="" LastEditDate="2010-10-27T21:49:07.530" LastActivityDate="2010-10-27T21:49:07.530" CommentCount="1" />
  <row Id="8107" PostTypeId="1" AcceptedAnswerId="8198" CreationDate="2010-10-27T22:27:44.467" Score="2" ViewCount="71" Body="&lt;p&gt;Imagine I have a company that makes widgets, where each widget costs me A dollars to make. Each month I can allocate money toward research and development with the aim of finding a new process that will allow me to build widgets for a cost of A/B dollars. Presume that I know that for each C dollars I spend on research and development there's a D% chance of finding a breakthrough. Of course, spending money on research and development means that I have less to spend on building widgets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a monthly budget of E dollars. This budget is not directly tied to my profit margin, but it is safe to say that it my profit margins influence future budgets (i.e., if I make no widgets for three straight months b/c I do all research and development, it's likely that my budget will be reduced, whereas if I discover a breakthrough the first month my profits will skyrocket and I'll likely see that budget grow over time).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In case that is too abstract, here's the real world scenario I'm interested in solving (although I'd like a more general approach, as well):&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;A = 15 dollars&lt;/li&gt;&#xA;&lt;li&gt;B = 3&lt;/li&gt;&#xA;&lt;li&gt;C = 5 dollars&lt;/li&gt;&#xA;&lt;li&gt;D = 2.75%&lt;/li&gt;&#xA;&lt;li&gt;E = 30 dollars&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;That is, today widgets cost me 15 dollars to build but if I can find a breakthrough I know I can make them at 1/3 the cost (5 dollars). For each 5 dollars I spend on research and development there is a 2.75% chance I'll find the breakthrough. However, I have only 30 dollars to spend each month. If I spend it all on research and development and have no success then I have made no widgets for sale. If I spend it all on widget construction I have no chance of finding a breakthrough.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there some statistical distribution or formula that can let me plug in these variables and see some sort of breakdown that gives me an idea of whether it's a good idea to spend any money on research and development each month and, if so, how much?&lt;/p&gt;&#xA;" OwnerUserId="2876" LastEditorUserId="2876" LastEditorDisplayName="" LastEditDate="2010-10-28T15:43:09.940" LastActivityDate="2010-10-28T16:51:58.243" Title="How to analyze risk vs. reward for spending on research and development work?" Tags="&lt;statistics&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="8108" PostTypeId="2" ParentId="8077" CreationDate="2010-10-27T22:36:43.830" Score="0" ViewCount="" Body="&lt;p&gt;I'm probably talking nonsense but &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\int_1^\infty \frac{1}{x^\alpha} = \left. \frac{x^{1 - \alpha}}{1 - \alpha} \right|_1^\infty$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The integral diverges if $1 - \alpha &gt; 0$ i.e. if $\alpha &amp;lt; 1$, but converges if $1 - \alpha &amp;lt; 0$ i.e. $\alpha &gt; 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My difficulty now is in showing that the integral is a bound on the sum, which from what I've studied can only be done if the terms of the sum are taken as follows&#xA;&lt;img src=&quot;http://i.imgur.com/OniXk.gif&quot; alt=&quot;alt text&quot;&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;instead of &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/4J5KC.gif&quot; alt=&quot;alt text&quot;&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is more natural and intuitive.&lt;/p&gt;&#xA;" OwnerUserId="1752" LastEditorUserId="1752" LastEditorDisplayName="" LastEditDate="2010-10-27T22:54:54.007" LastActivityDate="2010-10-27T22:54:54.007" CommentCount="1" />
  <row Id="8109" PostTypeId="2" ParentId="7391" CreationDate="2010-10-27T22:37:16.643" Score="2" ViewCount="" Body="&lt;p&gt;To supplement lhf's observation of the symmetry of the graph, note that if $f(x)=\sin(2x)\sin(x)$, then the symmetry observed is that $f(\pi/2+x)=-f(\pi/2-x)$, which means that the integral over $[0,\pi/2]$ exactly cancels the integral over $[\pi/2,\pi]$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The symmetry can be seen algebraically by noting that $\sin(\pi/2+x)=\sin(\pi/2-x)$, and that $$\sin(2(\pi/2+x))=\sin(\pi+2x)=-\sin(2x)=\sin(-2x)=-\sin(\pi-2x)=-\sin(2(\pi/2-x)),$$ with each equality being evident from the unit circle definition of $\sin$.  Multiplying yields $f(\pi/2+x)=-f(\pi/2-x)$.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-10-27T22:37:16.643" />
  <row Id="8110" PostTypeId="1" CreationDate="2010-10-27T22:47:10.720" Score="5" ViewCount="61" Body="&lt;p&gt;&lt;code&gt;Given three containers of specified volume, how many different volumes can you measure ?&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, suppose we have cans with capacities 2, 3, 4 litres.&#xA;We can measure :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;&#xA;1 litre by filling the 3 litre can and pouring it in the 2 litre can. &lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt; 2, 3, 4 litres are trivial.&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt; 5 litres by filling both the 2 and 3 litre cans. &lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;6 litres by filling the 4 litre can with 3 litres and filling the 3 litre can fully or filling the 2 and 4 litre cans. &lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt; 7 litres by filling the 3 and 4 litre cans.&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;8 litres by getting 6 litres and filling the 2 litre can.&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;9 litres by filling all the cans. &lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer for this case is 9.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a general way to answer this for three capacities a, b, c?&lt;/p&gt;&#xA;" OwnerUserId="2865" LastActivityDate="2010-10-28T19:58:26.160" Title="Measure different volumes based on specified capacities" Tags="&lt;combinatorics&gt;" AnswerCount="2" />
  <row Id="8111" PostTypeId="1" AcceptedAnswerId="8120" CreationDate="2010-10-27T22:58:37.800" Score="3" ViewCount="245" Body="&lt;p&gt;Is there a short way to say $f(f(f(f(x))))$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know you can use recursion:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$g(x,y)=\begin{cases}&#xA;f(g(x,y-1)) &amp;amp; \text{if } y &amp;gt; 0, \&#xA;\newline x &amp;amp; \text{if } y = 0.&#xA;\end{cases}$&lt;/p&gt;&#xA;" OwnerUserId="2736" LastEditorUserId="2736" LastEditorDisplayName="" LastEditDate="2010-10-27T23:20:24.920" LastActivityDate="2010-10-27T23:34:25.377" Title="A short way to say f(f(f(f(x))))" Tags="&lt;functions&gt;&lt;notation&gt;" AnswerCount="4" CommentCount="1" FavoriteCount="1" />
  <row Id="8112" PostTypeId="2" ParentId="8094" CreationDate="2010-10-27T23:00:22.363" Score="1" ViewCount="" Body="&lt;p&gt;It looks like you have six equations, the three you list in the post and $x_1+x_2+x_3=y_1+y_2+y_3=z_1+z_2+z_3=S$, where $S$ is given to you.  If we add the three equations in your post together we find $S(A+B+C)=Total$.  As all the terms in this equation are givens, it had better be true.  This will not apply to Part 2.  So you have five equations in nine unknowns and it should not be hard to satisfy.  If you just do a solution of linear equations you should find that there is a four parameter solution set and all the terms are rational multiples of each other.  The only thing that can go wrong is for things to go negative.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For Part 2, it looks like $i, j, k$ are givens as well.  You still have more unknowns than equations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I have misunderstood something, please help me out.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-10-27T23:00:22.363" CommentCount="3" />
  <row Id="8113" PostTypeId="1" CreationDate="2010-10-27T23:05:49.600" Score="3" ViewCount="235" Body="&lt;p&gt;&lt;strong&gt;1.How many proper subset of {1,2,3,4,5,6,7} contain the numbers 1 and 7 ?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lets consider {1,7} as a single element then the number of possible subset is $2^6$ and hence the number of proper subset is &lt;strong&gt;62&lt;/strong&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2.A survey show that 63% of the Americans like cheese where as 73% like apples.If x% of Americans like both cheese and apples, then we have :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(A) $x \ge 39 $   (B) $x \le 63 $ (C) $39 \le x \le 63 $ (D) None of these&lt;/p&gt;&#xA;&#xA;&lt;p&gt;if a% and b%  like only cheese and only apples then we have, $ a + x + x + b = 100 $ ,$ a + x  = 63 $ and $ b + x  = 63 $ solving we get $x = 39%$. So (D) is my answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Am I correct ?&lt;/p&gt;&#xA;" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2010-11-20T18:04:16.467" LastActivityDate="2010-11-20T18:20:02.643" Title="Two basic question on set theory" Tags="&lt;algebra-precalculus&gt;&lt;elementary-set-theory&gt;" AnswerCount="4" CommentCount="2" FavoriteCount="1" />
  <row Id="8114" PostTypeId="2" ParentId="8111" CreationDate="2010-10-27T23:09:21.813" Score="0" ViewCount="" Body="&lt;p&gt;You should define it this way:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{eqnarray}&#xA;\text{iterate}_0(f) &amp;amp;:=&amp;amp; id \\&#xA;\text{iterate}_{n + 1}(f) &amp;amp;:=&amp;amp; \text{iterate}_{n}(f) \circ f&#xA;\end{eqnarray}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then write $\text{iterate}_4(f)(x)$.&lt;/p&gt;&#xA;" LastActivityDate="2010-10-27T23:09:21.813" CommentCount="4" />
  <row Id="8115" PostTypeId="2" ParentId="8111" CreationDate="2010-10-27T23:11:16.223" Score="2" ViewCount="" Body="&lt;p&gt;Some will say $f^4(x)$.  But it becomes confused with the fourth power  or fourth derivative of $f(x)$.  I'm not sure what you mean here by &quot;piecewise&quot;.  To me &quot;piecewise&quot; would be something like a step function:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f(x)=1$ if $x\gt 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f(x)=0$ if $x\le 0$&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-10-27T23:11:16.223" CommentCount="13" />
  <row Id="8116" PostTypeId="2" ParentId="8111" CreationDate="2010-10-27T23:11:18.800" Score="0" ViewCount="" Body="&lt;p&gt;See &quot;function powers&quot; in Wikipedia &quot;Function composition&quot;. &lt;/p&gt;&#xA;" LastActivityDate="2010-10-27T23:11:18.800" CommentCount="2" />
  <row Id="8117" PostTypeId="2" ParentId="8113" CreationDate="2010-10-27T23:15:47.827" Score="4" ViewCount="" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;No, the subsets must have 1 and 7.  The other five elements are optional, but you can't have them all.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;You have $a + b + x +  \text{those who like neither cheese nor apples} =100%$.  Don't count x twice when adding to 100%.  But you are right that $a+x=63, b+x=39$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-10-28T03:16:04.797" LastActivityDate="2010-10-28T03:16:04.797" CommentCount="6" />
  <row Id="8118" PostTypeId="2" ParentId="8110" CreationDate="2010-10-27T23:19:57.077" Score="1" ViewCount="" Body="&lt;p&gt;If gcd(a,b,c) is not 1, you can only get multiples of this gcd.  You can't get more than a+b+c.  I think if you play around you can convince yourself you can get any multiple of the gcd, but you would need to define the allowable activities carefully to be able to prove it.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-10-27T23:19:57.077" CommentCount="1" />
  <row Id="8119" PostTypeId="2" ParentId="8101" CreationDate="2010-10-27T23:34:06.347" Score="2" ViewCount="" Body="&lt;p&gt;Denote the iterates by $x_0 = 1, x_{n+1} = P(x_n)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume that the coefficients of $P$ are integral.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If at some point $P(x_n) &gt; 2x_n$, then I claim that $m = P(x_n)-x_n$ does not divide any iterate. First, $x_n &amp;lt; m$, so $x_0,\ldots,x_n$ cannot be divisible by $m$. Second, we prove by induction that for $k \geq n$, $x_k \equiv x_n \pmod{m}$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x_{k+1} = P(x_k) \equiv P(x_n) \equiv x_n \pmod{m}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $x_n &amp;lt; m$, we see that $m$ doesn't divide any of the iterates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We conclude that always $P(x_n) \leq 2x_n$. Thus $P(x) = ax+b$ with $a \leq 2$. On the one hand $P(1) &gt; 1$, and on the other hand $P(1) \leq 2$. Thus $P(1) = 2$, and therefore either $a = 1$ or $a = 2$. If $a = 2$ then $P(x) = 2x$, and we generate only powers of $2$. Thus $a = 1$ and $P(x) = x + 1$.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2010-10-28T00:03:54.260" LastActivityDate="2010-10-28T00:03:54.260" CommentCount="3" />
  <row Id="8120" PostTypeId="2" ParentId="8111" CreationDate="2010-10-27T23:34:25.377" Score="8" ViewCount="" Body="&lt;p&gt;I personally prefer &#xA;$f^{\circ n} = f \circ f^{\circ n-1} = \dotsb = \kern{-2em}\underbrace{f \circ \dotsb \circ f}_{n-1\text{ function compositions}}$&lt;/p&gt;&#xA;" OwnerUserId="1269" LastActivityDate="2010-10-27T23:34:25.377" CommentCount="4" />
  <row Id="8121" PostTypeId="1" AcceptedAnswerId="8122" CreationDate="2010-10-27T23:50:53.587" Score="6" ViewCount="300" Body="&lt;p&gt;I'm asking because numerical tests seem to give nonsensical answers, and I thought I would check if there was an analytic way of checking for divergence, but I couldn't think of one offhand.&lt;/p&gt;&#xA;" OwnerUserId="890" LastActivityDate="2010-10-28T01:17:23.090" Title="does $\sum_{n=1}^{\infty} \frac{1}{\zeta(1+\frac{1}{n})}$ diverge or converge?" Tags="&lt;sequences-and-series&gt;" AnswerCount="2" />
  <row Id="8122" PostTypeId="2" ParentId="8121" CreationDate="2010-10-28T00:11:32.677" Score="13" ViewCount="" Body="&lt;p&gt;$(s-1)\zeta(s) = 1 + a(s-1) + b(s-1)^2 + \dots$ is analytic near $1$ (in fact entire, but we don't need that for this problem).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $s=1+1/n$ this gives $\zeta(1+1/n)=n + a + b/n + \dots  =n + O(1)$, so the sum diverges.&lt;/p&gt;&#xA;" OwnerUserId="467" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-10-28T00:47:47.343" LastActivityDate="2010-10-28T00:47:47.343" CommentCount="0" />
  <row Id="8123" PostTypeId="2" ParentId="8121" CreationDate="2010-10-28T00:20:48.543" Score="13" ViewCount="" Body="&lt;p&gt;You could use an integral comparison to get a bound on $\zeta(1+1/n)$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\zeta(1+1/n)\lt 1+\int_1^\infty x^{-1-\frac{1}{n}}=1+n.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More generally, if $0\lt a\lt 1$, then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{1}{a}=\int_1^\infty x^{-1-a}\lt\zeta(1+a)\lt1+\int_1^\infty x^{-1-a}=1+\frac{1}{a}&amp;lt;\frac{2}{a}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus if $a_1,a_2,\ldots$ is a sequence of positive numbers converging to $0$, then&#xA; $\sum_{n=1}^{\infty} \frac{1}{\zeta(1+a_n)}$ converges if and only if $\sum_{n=1}^\infty a_n$ does.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2010-10-28T01:17:23.090" LastActivityDate="2010-10-28T01:17:23.090" />
  <row Id="8124" PostTypeId="2" ParentId="8105" CreationDate="2010-10-28T00:29:22.713" Score="1" ViewCount="" Body="&lt;p&gt;What you are looking for doesn't seem to exist (if you go to Elsevier site and type any combination of what you are looking for, like math glossary, you don't get any relevant answers).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might want to check Wolfram Mathworld: &lt;a href=&quot;http://mathworld.wolfram.com/&quot; rel=&quot;nofollow&quot;&gt;http://mathworld.wolfram.com/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1605" LastEditorUserId="1605" LastEditorDisplayName="" LastEditDate="2010-10-29T19:29:25.217" LastActivityDate="2010-10-29T19:29:25.217" CommentCount="1" />
  <row Id="8125" PostTypeId="1" AcceptedAnswerId="8127" CreationDate="2010-10-28T01:32:31.730" Score="3" ViewCount="189" Body="&lt;p&gt;A union of finite sets can always be converted to a union of finite disjoint sets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was wondering if a union of countable sets can always be converted to a union of countable disjoint sets? If yes how to do that?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Update with Paul's question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;I wonder about the natural extension of this question to uncountable unions. Can we always refine an arbitrary set {Aα} to {Bα} (i.e. with Bα⊂Aα for each α) such that ⋃αBα=⋃αAα? The answer below uses countability pretty blatantly and I can't think of a way to do it without. &quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="1281" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2010-11-20T17:55:22.470" LastActivityDate="2010-11-20T17:55:22.470" Title="convert union of sets into union of disjoint sets" Tags="&lt;elementary-set-theory&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" />
  <row Id="8126" PostTypeId="1" AcceptedAnswerId="8166" CreationDate="2010-10-28T01:33:29.357" Score="16" ViewCount="245" Body="&lt;p&gt;I'm trying to think of an example of a homomorphism of commutative rings $f:A\rightarrow B$ and ideals $I,J$ of $B$ such that $f^{-1}(I)+f^{-1}(J)$ is not a preimage of any ideal of $B$.  I can't seem to come up with one... anyone know one?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: To clear up some basic facts / head off some mistakes:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As Arturo points out, we can assume $f$ is an inclusion.  Perhaps I should have written the question in terms of inclusions in the first place, but, eh.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;No, $f^{-1}(I)+f^{-1}(J)$ is not equal to $f^{-1}(I+J)$ in general.  A counterexample would be the inclusion of $\mathbb{C}$ in $\mathbb{C}[x]$; consider $(x)$ and $(1-x)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To show an ideal $K\subseteq A$ is not a preimage of any ideal of $B$, it suffices to show that it's not equal to $f^{-1}(Bf(K))$.&lt;/p&gt;&#xA;" OwnerUserId="2884" LastEditorUserId="2884" LastEditorDisplayName="" LastEditDate="2010-10-28T08:02:05.840" LastActivityDate="2010-10-28T14:44:47.643" Title="Basic counterexample re: preimages of ideals" Tags="&lt;abstract-algebra&gt;&lt;algebra&gt;&lt;commutative-algebra&gt;&lt;counter-examples&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="2" />
  <row Id="8127" PostTypeId="2" ParentId="8125" CreationDate="2010-10-28T01:39:43.593" Score="12" ViewCount="" Body="&lt;p&gt;Given $A_1,A_2,\ldots$, let $B_1=A_1$, and for $n\gt1$, $B_n=A_n\setminus\cup_{k=1}^{n-1}A_k$.  Then for $m\lt n$, $B_m$ is contained in $A_m$ while $B_n$ is contained in $A_n\setminus A_m$, so $B_m$ and $B_n$ are disjoint.  You can check that $\cup_{k=1}^n B_k=\cup_{k=1}^n A_k$ for all $n$, and $\cup_{k=1}^\infty B_k=\cup_{k=1}^\infty A_k$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have a collection of sets indexed by an arbitrary set, then assuming the axiom of choice you can put a well-order on the index set.  Given $\{A_i\}_{i\in I}$, where $(I,\lt)$ is a well-ordered set, we can define $\{B_i\}_{i\in I}$ as follows.  Let $B_{i_0}=A_{i_0}$, where $i_0$ is the least element of $I$, and for $i\gt i_0$, $B_i=A_i\setminus\cup_{j\lt i}A_j$.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2010-10-28T02:22:01.230" LastActivityDate="2010-10-28T02:22:01.230" CommentCount="3" />
  <row Id="8128" PostTypeId="2" ParentId="8113" CreationDate="2010-10-28T01:48:02.377" Score="2" ViewCount="" Body="&lt;p&gt;On #1, you are over counting. You have thought of {1,7} as a single elements but your answer has included the possibility that it is not in there. Think of it as we are setting 1 and 7 to the side. Now for each remaining element, we can either include it or exclude it from a subset. If you still don't see how to obtain the answer, I would recommend the following short explanation of the &lt;a href=&quot;http://www.icoachmath.com/SiteMap/MultiplicationPrinciple.html&quot; rel=&quot;nofollow&quot;&gt;multiplication principle&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On #2, you are correct in thinking that each person must fall into 1 of four groups: people who like apples and cheese, people who just like cheese, people who like just apples, and people who like neither. If you think abut the information you are given, you should see an upper bound for x. For a lower bound on x, assume that everyone either likes apple or cheese and use the &lt;a href=&quot;http://users.csc.tntech.edu/~srini/DM/chapters/review3.6.html&quot; rel=&quot;nofollow&quot;&gt;inclusion exclusion principle&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="249" LastActivityDate="2010-10-28T01:48:02.377" CommentCount="4" />
  <row Id="8129" PostTypeId="2" ParentId="8104" CreationDate="2010-10-28T01:51:41.503" Score="2" ViewCount="" Body="&lt;p&gt;You are almost there.  The feasible region of the dual for the new problem is exactly the same as that of the dual for the original problem.  (Write down the two duals to see that.)  Since &lt;strong&gt;z&lt;/strong&gt; is feasible for the dual of the original problem, it is also feasible for the dual of the new problem.&lt;/p&gt;&#xA;" OwnerUserId="2370" LastActivityDate="2010-10-28T01:51:41.503" />
  <row Id="8131" PostTypeId="1" CreationDate="2010-10-28T01:58:04.443" Score="0" ViewCount="70" Body="&lt;p&gt;What is the easiest way to find the maximum of 9cos(t) - 8cost(3t)?&lt;/p&gt;&#xA;" LastActivityDate="2010-10-28T02:38:26.140" Title="What is the easiest way to find the maximum of 9cos(t) - 8cos(3t)?" Tags="&lt;calculus&gt;&lt;trigonometry&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="8132" PostTypeId="2" ParentId="8131" CreationDate="2010-10-28T02:06:32.160" Score="2" ViewCount="" Body="&lt;p&gt;Express $\cos(3t)$ in terms of $\cos(t)$ and differentiate with respect to $t$ or $\cos(t)$.&lt;/p&gt;&#xA;" OwnerUserId="618" LastActivityDate="2010-10-28T02:06:32.160" />
  <row Id="8133" PostTypeId="1" CreationDate="2010-10-28T02:18:05.457" Score="5" ViewCount="188" Body="&lt;p&gt;Of course, Euler proved that the Riemann zeta function can be defined as the analytic continuation of a product over all primes.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\zeta(s) = \prod_{p \in \mathbb{P}}\frac1{1-p^{-s}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is well known (but not something I understand) that the positions of zeros of the zeta function allows one to make inferences about the asymptotic behavior of primes.  Is this a general phenomenon? Does Euler's transform generalize to products over other subsets of the natural numbers $\mathbb{A}$?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\alpha(s) = \prod_{a \in (\mathbb{A} \subset \mathbb{N})}\frac1{1-a^{-s}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can one then reverse Euler's transform and derive the generating subset $\mathbb{A}$ completely from the new function's zero set?  More generally, how do properties of the derived function's zeros translate to properties of the generating subset?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And, specifically for the standard Riemann zeta function, if it was shown that exactly one single zero existed off the critical line, what would its position say about the distribution of primes?&lt;/p&gt;&#xA;" OwnerUserId="948" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-10-28T02:45:47.927" LastActivityDate="2010-10-30T00:43:37.700" Title="An inverse for Euler's zeta function product formula" Tags="&lt;number-theory&gt;&lt;prime-numbers&gt;&lt;riemann-zeta&gt;&lt;riemann-hypothesis&gt;" AnswerCount="1" FavoriteCount="3" />
  <row Id="8134" PostTypeId="1" AcceptedAnswerId="8139" CreationDate="2010-10-28T02:49:56.470" Score="11" ViewCount="294" Body="&lt;p&gt;Let $z_1, z_2 \ldots z_m$ be complex numbers, $m \in \mathbb{N}$. Can anybody tell me how to prove the following inequality?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$| z_1 z_2 \ldots z_m - 1 | \leq e^{|z_1 - 1| + \ldots + |z_m - 1|} - 1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In case you're wondering, this is asserted without proof in a paper by Von Neumann, about infinite tensor products of Hilbert spaces.&lt;/p&gt;&#xA;" OwnerUserId="2491" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-10-28T17:42:01.470" LastActivityDate="2010-10-28T17:42:01.470" Title="Complex-number inequality $| z_1 z_2 \ldots z_m - 1 | \leq e^{|z_1 - 1| + \ldots + |z_m - 1|} - 1$" Tags="&lt;complex-numbers&gt;&lt;inequality&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="2" />
  <row Id="8135" PostTypeId="1" AcceptedAnswerId="8136" CreationDate="2010-10-28T03:00:42.727" Score="0" ViewCount="96" Body="&lt;p&gt;Define the relation $\sim$ on $\mathbb{Z}$ by $a \sim b$ if and only if $5a \equiv 2b \pmod{3}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Prove that $\sim$ is an equivalence relation on $\mathbb{Z}$.&#xA;Identify the distinct equivalence classes on $\mathbb{Z}/\sim$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I prove something like this? Do I need to prove something for reflexiveness, symmetry, and transitivity?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="2587" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-10-28T03:07:52.963" LastActivityDate="2010-10-28T04:14:18.693" Title="Define the relation $\sim$on $\mathbb{Z}$ by $a \sim b$ if and only if $5a \equiv 2b \pmod{3}$" Tags="&lt;modular-arithmetic&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="8136" PostTypeId="2" ParentId="8135" CreationDate="2010-10-28T03:10:57.157" Score="6" ViewCount="" Body="&lt;p&gt;Yes, you need to &quot;prove something&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To show that $\sim$ is an equivalence relation, you need to prove that it is reflexive, symmetric, and transitive. That is, you need to prove that:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;For every $a\in\mathbb{Z}$, $a\sim a$ is true. Since &quot;$x\sim y$&quot; means &quot;$5x\equiv 2y\pmod{3}$&quot;, then that means that you need to prove that for every $a\in\mathbb{Z}$, $5a\equiv 2a\pmod{3}$ is true. &lt;/li&gt;&#xA;&lt;li&gt;For every $a,b\in\mathbb{Z}$, if $a\sim b$ is true, then $b\sim a$ is true. If you are unsure what this means, then &quot;unpack&quot; the meaning of &quot;$a\sim b$&quot; and of &quot;$b\sim a$&quot; the way I did above.&lt;/li&gt;&#xA;&lt;li&gt;For every $a,b,c\in\mathbb{Z}$, if $a\sim b$ and $b\sim c$ are both true, then $a\sim c$ is true.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;That will prove that $\sim$ is an equivalence relation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then you need to think about the equivalence classes. That is, given $a\in\mathbb{Z}$, what is $[a]=\{b\in\mathbb{Z}\mid a\sim b\}$, the set of all integers that are related to $a$? The different sets you get this way are the equivalence classes of $\mathbb{Z}/\sim$. &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-10-28T04:14:18.693" LastActivityDate="2010-10-28T04:14:18.693" />
  <row Id="8137" PostTypeId="2" ParentId="7983" CreationDate="2010-10-28T03:31:17.993" Score="3" ViewCount="" Body="&lt;p&gt;You can go this route: You know &lt;strong&gt;Ms&lt;/strong&gt; = &lt;strong&gt;d&lt;/strong&gt;.  Taking transposes, you have &lt;strong&gt;s&lt;/strong&gt;$^T$ &lt;strong&gt;M&lt;/strong&gt;$^T$ = &lt;strong&gt;d&lt;/strong&gt;$^T$.  But (the key step) since &lt;strong&gt;M&lt;/strong&gt;$^T$ = &lt;strong&gt;M&lt;/strong&gt;, this means that &lt;strong&gt;s&lt;/strong&gt; is feasible for the dual problem.  Since &lt;strong&gt;s&lt;/strong&gt; is feasible for the primal and for the dual, and since the primal and dual have the same objective function, weak duality shows you that &lt;strong&gt;s&lt;/strong&gt; must be optimal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If &lt;strong&gt;M&lt;/strong&gt;$^T \neq $ &lt;strong&gt;M&lt;/strong&gt; then &lt;strong&gt;s&lt;/strong&gt; might not be optimal.  For example, suppose we have the following linear program.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\min x + y$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;subject to &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{1}{2}x + y \geq 1,$$&#xA;$$\frac{1}{2}x + 2y \geq 1,$$&#xA;$$x,y \geq 0.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The objective and right-hand side coefficient vectors are the same, as in the problem statement.  Solving the inequalities at equality gives you $x = 2$, $y=0$.  However, this is not optimal.  The solution $x = 0$, $y = 1$ yields a smaller objective function value (and is actually the optimal solution).&lt;/p&gt;&#xA;" OwnerUserId="2370" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-10-28T12:43:41.203" LastActivityDate="2010-10-28T12:43:41.203" />
  <row Id="8138" PostTypeId="2" ParentId="8134" CreationDate="2010-10-28T03:37:03.900" Score="-1" ViewCount="" Body="&lt;p&gt;EDIT: This answer is wrong.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It all boils down to the inequality $|xy-1| \leq |x-1|+|y-1|$, which I expect to be true.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given this inequality, prove by induction that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$|z_1\cdots z_m - 1| \leq |z_1 - 1| + \cdots + |z_m - 1|$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now use $e^x \geq 1 + x$.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2010-10-28T04:20:39.117" LastActivityDate="2010-10-28T04:20:39.117" CommentCount="2" />
  <row Id="8139" PostTypeId="2" ParentId="8134" CreationDate="2010-10-28T04:01:38.487" Score="16" ViewCount="" Body="&lt;p&gt;The inequality in question bounds how far you can get from $1$ by multiplying several complex numbers that may individually not be far from $1$. So it makes sense to try to derive a bound for the product of just &lt;em&gt;two&lt;/em&gt; complex numbers, and then proceed by induction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Lemma:&lt;/strong&gt; Suppose $\lvert z_1 - 1\rvert = \alpha_1$ and $\lvert z_2 - 1\rvert = \alpha_2$. Then $\lvert z_1 z_2 - 1\rvert \le (1 + \alpha_1)(1 + \alpha_2) - 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof:&lt;/strong&gt; We know $\alpha_1\alpha_2 = \lvert z_1z_2 - z_1 - z_2 + 1\rvert$. By triangle inequality on the three points $z_1 z_2$, $z_1 + z_2 - 1$, and $1$, we have&#xA;$$\begin{align}&#xA;\lvert z_1 z_2 - 1\rvert &amp;amp;\le \lvert z_1 z_2 - z_1 - z_2 + 1\rvert + \lvert z_1 + z_2 - 2\rvert \\&#xA;&amp;amp;\le \alpha_1\alpha_2 + \alpha_1 + \alpha_2 \\&#xA;&amp;amp;= (1 + \alpha_1)(1 + \alpha_2) - 1.&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, for several numbers,&#xA;$$\begin{align}&#xA;\lvert z_1 z_2 \cdots z_m - 1\rvert &amp;amp;\le (1 + \alpha_1)(1 + \alpha_{2,\ldots,m}) - 1 \\&#xA;&amp;amp;\le (1 + \alpha_1)(1 + \alpha_2)(1 + \alpha_{3,\ldots,m}) - 1 \\&#xA;&amp;amp;\vdots \\&#xA;&amp;amp;\le (1 + \alpha_1)(1 + \alpha_2)\cdots(1 + \alpha_m) - 1,&#xA;\end{align}$$ where $\alpha_{2,\ldots,m}$, for example, is my hopefully transparent abuse of notation to denote $\lvert z_2\cdots z_m - 1\rvert.$ Finally, since $1+x \le e^x$ for real $x$, the desired inequality follows.&lt;/p&gt;&#xA;" OwnerUserId="856" LastActivityDate="2010-10-28T04:01:38.487" CommentCount="7" />
  <row Id="8140" PostTypeId="1" AcceptedAnswerId="15069" CreationDate="2010-10-28T04:12:39.473" Score="1" ViewCount="532" Body="&lt;p&gt;I have a graph $G$ with a list of edges $E$, all the edges can form cycles, as shown below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://docs.google.com/drawings/pub?id=1Ux3b0FtuIMpVGX4e-AbblH05h2gIZII9u8pACRzymPE&amp;amp;w=960&amp;amp;h=720&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All the coordinates for the vertex $V$ &lt;strong&gt;are known&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wat is the algorithm to find all the faces $F$? For the example above, there are 4 such faces, as indicated by the looping arrow. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Update: Why is it that the algorithm to find the face in the case without coordinate is different from the one with coordinates? Simply because by specifying the coordinates there is a unique set of faces, and without coordinates there could be sets of faces that are topologically equal. See the diagram below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://docs.google.com/drawings/pub?id=1wAxp6gJgN7wAkr5xVpSfBv7_8Gfd2FHIFX6CY5bdgdc&amp;amp;w=960&amp;amp;h=720&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Update 2: Clear up the terminology&lt;/p&gt;&#xA;&#xA;&lt;p&gt;UPdate 3: &lt;a href=&quot;http://math.stackexchange.com/questions/8140/find-all-cycles-faces-in-a-graph/15069#15069&quot;&gt;It is suggested&lt;/a&gt; that I construct a cyclically linked list at each vertex, sorted according to the angles made between the incoming edge and the outgoing edge. Then one would just have to continue along the edge, but I afraid this algorithm doesn't work with the so called butterfly graph. Consider this graph:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://docs.google.com/drawings/pub?id=1QF2JouvziCYaPCkERy5kld1No8NORf63O7XwKN22nXY&amp;amp;w=960&amp;amp;h=720&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume that we start at $e_1$, and the arrow denotes the direction of the movement. If we take the left turn as our orientation then we would be visiting $v_1$ twice before completing a cycle-- and the cycle is not really a cycle. &lt;/p&gt;&#xA;" OwnerUserId="550" LastEditorUserId="550" LastEditorDisplayName="" LastEditDate="2010-12-22T02:43:27.190" LastActivityDate="2010-12-24T18:49:30.593" Title="Find All Cycles (Faces) In a Graph" Tags="&lt;combinatorics&gt;&lt;graph-theory&gt;" AnswerCount="3" CommentCount="11" />
  <row Id="8141" PostTypeId="1" CreationDate="2010-10-28T04:18:00.483" Score="0" ViewCount="144" Body="&lt;p&gt;This animation from wikipedia shows basically what I want to accomplish, however - I'm hoping to have it flipped around, where it starts progressing more towards the destination and &quot;up&quot; (in this image), and then arcs more directly to the end point. However, I only have access to a starting point and ending point, what I am hoping to do is be able to determine the other points by specifying a &quot;height&quot; (or width, whatever you want to call it), to determine how high the arc actually goes.&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/File:Bezier_3_big.png&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/File:Bezier_3_big.png&lt;/a&gt; (unable to post image due to no rep)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Help or direction would be appreciated.&lt;/p&gt;&#xA;" LastActivityDate="2011-05-23T21:29:16.577" Title="How to calculate a bezier curve with only start and end points?" Tags="&lt;geometry&gt;&lt;algorithms&gt;&lt;bezier-curve&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="8142" PostTypeId="1" AcceptedAnswerId="8146" CreationDate="2010-10-28T04:18:41.040" Score="7" ViewCount="181" Body="&lt;p&gt;Let $C$ be a circle of radius $r$ with $n$ points. Prove that there is a point on the circle such that the product of the distances from this point to the other $n$ points is greater than $r^n$. So we seem to be looking at chords that are of length $\leq 2r$. To show the existence of such a point, would you use a constructive argument? Or more of an indirect argument? Likewise, to show that the inequality holds, would we invoke certain &quot;famous&quot; inequalities such as AM-GM or Cauchy-Schwartz? &lt;/p&gt;&#xA;" OwnerUserId="2548" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-10-28T17:25:16.443" LastActivityDate="2010-10-28T20:16:14.783" Title="Lower bound on product of distances from points on a circle" Tags="&lt;geometry&gt;&lt;inequality&gt;" AnswerCount="3" CommentCount="3" />
  <row Id="8143" PostTypeId="1" AcceptedAnswerId="8148" CreationDate="2010-10-28T04:50:33.747" Score="1" ViewCount="71" Body="&lt;p&gt;How many distinct integral value of $n$ satisfies the equation $2^{2n} - 3 \cdot (2^{n+2}) + 32 = 0 $ ?&lt;/p&gt;&#xA;" OwnerUserId="2109" LastActivityDate="2010-10-28T04:57:37.300" Title="How many distinct integral value of $n$ satisfies the equation $2^{2n} - 3 \cdot (2^{n+2}) + 32 = 0 $ ?" Tags="&lt;algebra-precalculus&gt;" AnswerCount="2" />
  <row Id="8144" PostTypeId="2" ParentId="8143" CreationDate="2010-10-28T04:53:58.980" Score="1" ViewCount="" Body="&lt;p&gt;You should be able to prove a fairly small limit on n.  Think about the fact that 3&amp;lt;2^2.  Then you can just try them all up to there.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-10-28T04:53:58.980" />
  <row Id="8145" PostTypeId="2" ParentId="8142" CreationDate="2010-10-28T04:55:10.010" Score="3" ViewCount="" Body="&lt;p&gt;We can take $r=1$ by scaling.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The distance between $e^{i\alpha}$ and $e^{i\beta}$ is $2|\sin\frac{\alpha-\beta}{2}|$. Denoting the given points by $e^{i\alpha_i}$ and the required point by $e^{i\beta}$, the total distance is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ \prod_i 2|\sin\frac{\alpha_i-\beta}{2}| $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let's take a random angle $\beta$. The average multiplicative value of sine on $[0,\pi]$ is $1/2$, see &lt;a href=&quot;http://en.wikipedia.org/wiki/List_of_trigonometric_identities#Identities_without_variables&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;. Therefore the multiplicative expected value of the product is $1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to be more formal, replace &quot;multiplicative&quot; by taking of the logarithm.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-10-28T04:55:10.010" />
  <row Id="8146" PostTypeId="2" ParentId="8142" CreationDate="2010-10-28T04:55:58.220" Score="10" ViewCount="" Body="&lt;p&gt;Let $r&gt;0$, $D=\{z\in\mathbb C:|z|&amp;lt;r\}$, and $S=\{z\in\mathbb C:|z|=r\}$. Pick $n\in\mathbb N$ and distinct points $z_1,\dots,z_n\in S$, and consider the polynomial $f(z)=\prod_{i=1}^n(z_i-z)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $f$ is holomorphic on $\mathbb C$ and non-constant, its absolute value attains its maximum on the closed disc $D\cup S$ at some point $p$ of $S$ and not inside $D$. Since $|f(0)|=r^n$, it follows that $|f(p)|&amp;gt; r^n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is precisely what you wanted.&lt;/p&gt;&#xA;" OwnerUserId="274" LastEditorUserId="274" LastEditorDisplayName="" LastEditDate="2010-10-28T05:19:17.437" LastActivityDate="2010-10-28T05:19:17.437" CommentCount="2" />
  <row Id="8148" PostTypeId="2" ParentId="8143" CreationDate="2010-10-28T04:57:37.300" Score="6" ViewCount="" Body="&lt;p&gt;HINT: make the substitution $x=2^{n}$ This will reduce your equation to a quadratic.&lt;/p&gt;&#xA;" OwnerUserId="249" LastActivityDate="2010-10-28T04:57:37.300" CommentCount="2" />
  <row Id="8149" PostTypeId="1" AcceptedAnswerId="8156" CreationDate="2010-10-28T05:23:59.040" Score="1" ViewCount="152" Body="&lt;p&gt;We know that the points of discontinuity of a monotone function on an interval $[a,b]$ are countable.Using this can we prov that: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Any bijection $f: \mathbb{R} \to [0,\infty)$ has infinitely many points of discontinuity.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;If yes, how or otherwise how to prove the above result?&lt;/p&gt;&#xA;" LastActivityDate="2010-10-28T05:55:33.890" Title="Points of discontinuity of a bijective function $f:\mathbb{R} \to [0,\infty)$" Tags="&lt;real-analysis&gt;&lt;analysis&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="8150" PostTypeId="1" AcceptedAnswerId="8153" CreationDate="2010-10-28T05:24:34.213" Score="2" ViewCount="82" Body="&lt;p&gt;Is the localization of a reduced ring (no nilpotents) still reduced?&lt;/p&gt;&#xA;" OwnerUserId="1491" LastActivityDate="2010-10-28T05:46:06.727" Title="Does localization preserve reducedness?" Tags="&lt;abstract-algebra&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="8152" PostTypeId="2" ParentId="8150" CreationDate="2010-10-28T05:32:15.087" Score="1" ViewCount="" Body="&lt;p&gt;EDIT: This argument is incorrect but I feel others could learn from Mariano Suárez-Alvarez's comments so I've made the post CW.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I understand correctly you are asking does reduce ring imply reduced localization.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Argue by contrapositive, assume that localization is not reduced, i.e. contains nilpotents. Since elements of the localization of our ring are of the form $\frac{r}{s}$ where s is a subset of our ring that does not contain 0. Then choose a nilpotent element in the localization $(\frac{r}{s})^{n}=0$ Since 0 is not in S it must be the case that $r^{n}=0$, i.e. r is nilpotent. Hence the original ring is nilpotent.&lt;/p&gt;&#xA;" OwnerUserId="249" LastEditorUserId="249" LastEditorDisplayName="" LastEditDate="2010-10-28T05:46:06.727" LastActivityDate="2010-10-28T05:46:06.727" CommunityOwnedDate="2010-10-28T05:46:06.727" CommentCount="4" />
  <row Id="8153" PostTypeId="2" ParentId="8150" CreationDate="2010-10-28T05:32:23.867" Score="4" ViewCount="" Body="&lt;p&gt;Let $A$ be a ring, $S\subset A$ a multiplicatively closed subset, and suppose that $0\neq a/b\in A_S$ is nilpotent. Then there exists $n$ such that $(a/b)^n=0$, &lt;em&gt;i.e.&lt;/em&gt;, such that there exists $t\in S$ with $ta^n=0$. But then $ta$ is nilpotent in $A$. If it is zero, then $a/b=0$ in $A_S$, which it isn't.&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2010-10-28T05:32:23.867" />
  <row Id="8154" PostTypeId="2" ParentId="8113" CreationDate="2010-10-28T05:45:30.033" Score="2" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;1.&lt;/strong&gt; Say {1,7} is a single element, then total number of subsets possible = $2^6$ Now, how many subsets among these &lt;strong&gt;don't have {1,7}&lt;/strong&gt; ? It will be $2^5$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence,number of subsets &lt;strong&gt;having {1,7}&lt;/strong&gt; is  $2^6$ -  $2^5$ = 32.But since you have asked for proper subset so the answer would be 31.&lt;/p&gt;&#xA;" OwnerUserId="2109" LastEditorUserId="2109" LastEditorDisplayName="" LastEditDate="2010-10-28T05:52:01.060" LastActivityDate="2010-10-28T05:52:01.060" CommentCount="2" />
  <row Id="8155" PostTypeId="1" AcceptedAnswerId="8168" CreationDate="2010-10-28T05:50:17.987" Score="0" ViewCount="142" Body="&lt;p&gt;I want to write the equation of a line in $\mathbb C^n$ passing through a point $(z_1,z_2,...,z_n)$. Actually I have a set of points and I suspect they all lie in the same line which passes through this point and I want a convenient way to check it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you&lt;/p&gt;&#xA;" OwnerUserId="2890" LastActivityDate="2010-10-29T18:40:28.157" Title="How to write the equation of a line in $\mathbb C^n$" Tags="&lt;algebra&gt;&lt;linear-algebra&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="8156" PostTypeId="2" ParentId="8149" CreationDate="2010-10-28T05:55:33.890" Score="4" ViewCount="" Body="&lt;p&gt;Suppose to the contrary that $f$ has a finite number $n$ of discontinuities, at $x_1, x_2, \ldots x_n$. Then $\mathbb{R} - ( x_1, \ldots x_n )$ is a union of open intervals $I_1, \ldots I_{n + 1}$. $f$ restricted to each $I_m$ is continuous and injective, and therefore monotone, so each image $f|_{I_m}(I_m)$ is an open (in $\mathbb{R}$) interval $J_m$. The $J_m$ are non-empty, pairwise disjoint and contained in $(0, \infty)$; suppose that $J_{m_1} &amp;lt; J_{m_2} &amp;lt; \ldots &amp;lt; J_{m_{n + 1}}$. Then between each $J_{m_p}$ and $J_{m_{p + 1}}$ there is at least one point, which must be the image of some $x_q$. But this exhausts all the $x$'s, so that $0$ is not in the image of $f$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;BTW, can anybody tell me how to get the forum to display braces in math mode?&lt;/p&gt;&#xA;" OwnerUserId="2491" LastActivityDate="2010-10-28T05:55:33.890" CommentCount="6" />
  <row Id="8157" PostTypeId="1" AcceptedAnswerId="8161" CreationDate="2010-10-28T05:59:19.727" Score="1" ViewCount="104" Body="&lt;p&gt;If a set contain $(2n+1)$ elements and if the number of subsets which contain at most n elements is 4096, then what is the value of $n$?&lt;/p&gt;&#xA;" OwnerUserId="2109" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2010-11-21T02:02:57.343" LastActivityDate="2010-11-21T02:02:57.343" Title="If a set contain $(2n+1)$ elements and if the number of subsets which contain at most n elements is 4096, then what is the value of $n$?" Tags="&lt;elementary-set-theory&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="8159" PostTypeId="2" ParentId="8157" CreationDate="2010-10-28T06:11:57.187" Score="0" ViewCount="" Body="&lt;p&gt;Use the identity $(1+1)^n= n_{C_0}+n_{C_1}+\dots+n_{C_n}$ and the fact that $n_{C_r}=n_{C_{n-r}}$.&lt;/p&gt;&#xA;" OwnerUserId="2890" LastActivityDate="2010-10-28T06:11:57.187" />
  <row Id="8160" PostTypeId="1" AcceptedAnswerId="8163" CreationDate="2010-10-28T06:14:21.163" Score="4" ViewCount="109" Body="&lt;p&gt;In a compact hyperbolic Reimann surface without boundary is length minimizing geodesic between two points $p$ and $q$ unique.&lt;/p&gt;&#xA;" OwnerUserId="2893" LastActivityDate="2010-10-28T06:49:46.053" Title="Uniqueness of length minimizing geodesic" Tags="&lt;geometry&gt;&lt;riemann-surface&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="8161" PostTypeId="2" ParentId="8157" CreationDate="2010-10-28T06:23:44.687" Score="3" ViewCount="" Body="&lt;p&gt;There are $2^{2n + 1}$ subsets in total. Now note that the function which takes a subset to its complement is a bijection, and that a subset has $n$ or fewer elements if, and only if, its complement has more than $n$ elements. Thus the number of subsets with $n$ or fewer elements is equal to the number with more than $n$; thus the number of each is $2^{2n + 1}/2 = 2^{2n}$. Now solve for $n$.&lt;/p&gt;&#xA;" OwnerUserId="2491" LastActivityDate="2010-10-28T06:23:44.687" />
  <row Id="8162" PostTypeId="2" ParentId="8155" CreationDate="2010-10-28T06:40:23.527" Score="0" ViewCount="" Body="&lt;p&gt;How much do you know about vectors?  You can take two points and find the difference between them, as a vector; this is the analogue of the slope of a line in $\mathbb{R}^2$.  If the vector is $\mathbf{v}$ and one point is $a$, then points on the line are of the form $a+\lambda\mathbf{v}$, where $\lambda$ is real.  (Unless you're talking about a &quot;complex line,&quot; which is really a plane.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, if you're just uncomfortable with $\mathbb{C}^n$, you can just rewrite all your points as points in $\mathbb{R}^{2n}$, like $(Re(z_1),Im(z_1),\dotsc,Re(z_n),Im(z_n))$.  This is an isomorphism of the real vector space structure of $\mathbb{C}^n$, so in particular it preserves lines and addition.  The only thing you lose is complex multiplication.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;: If you're looking for a complex line, just let $\lambda$ be complex in the above vector expression.&lt;/p&gt;&#xA;" OwnerUserId="2215" LastEditorUserId="2215" LastEditorDisplayName="" LastEditDate="2010-10-28T09:03:23.223" LastActivityDate="2010-10-28T09:03:23.223" CommentCount="2" />
  <row Id="8163" PostTypeId="2" ParentId="8160" CreationDate="2010-10-28T06:47:46.877" Score="6" ViewCount="" Body="&lt;p&gt;Think about a Riemann surface of genus 2 say. This is a 2-holed doughnut.&#xA;It has a curve winding about one of its holes. (Or more mathematically speaking&#xA;it has an embedded circle representing a nonzero homology class.)&#xA;We can deform this curve so that it is length-minimizing in its homotopy&#xA;class. Then the curve becomes a geodesic $C$. Then let $p$ and $q$ be &quot;opposite&quot;&#xA;points of $C$, that is the distance from $p$ to $q$ along $C$ is a half of&#xA;the length of $C$. Then $C$ splits into two equal-length geodesics between&#xA;$p$ and $q$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suppose there may be even shorter geodesics between $p$ and $q$, but&#xA;I think that choosing the homology class to minimize the length of $C$&#xA;shoudl sort that out.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-10-28T06:47:46.877" />
  <row Id="8164" PostTypeId="2" ParentId="8160" CreationDate="2010-10-28T06:49:46.053" Score="3" ViewCount="" Body="&lt;p&gt;I'm not a hyperbolic geometer, but I'm going to say &lt;strong&gt;no&lt;/strong&gt;.  Here is my reasoning:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $C$ be your compact hyperbolic Riemann surface, so it is the quotient of its universal cover, the hyperbolic plane, by a cocompact Fuchsian group $\Gamma$.  Let $\pi: \mathbb{H} \rightarrow \Gamma \backslash \mathbb{H} = C$ be the covering map.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have points $P$ and $Q$ on $C$.  Fix one lift $\tilde{P}$ of $P$ and consider all possible lifts &lt;code&gt;$\tilde{Q}_{\gamma}$&lt;/code&gt; of $Q$: this will be a full orbit of $\Gamma$.  For &#xA;each $\gamma \in \Gamma$, there is a unique geodesic in $\mathbb{H}$ from $P$ to $\tilde{Q}_{\gamma}$ and its image under $\pi$ is a geodesic from $P$ to $Q$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So certainly there are lots of geodesics in question.  The ones of minimal length are going to correspond to the choices of $\gamma$ such that the hyperbolic distance from $P$ to $\tilde{Q}_{\gamma}$ is minimized.  But it seems pretty clear that we can choose $P$ so as to be equidistant between two such points: choose any point $Q_1$, and then choose $Q_2$ to be distinct from $Q_1$, to lie in the same $\Gamma$-orbit as $Q_1$ and to be closest (not necessarily uniquely closest!) to $Q_1$ among all such points.  Let $P$ be the midpoint of the geodesic arc from $Q_1$ to $Q_2$.  Then so long as $P$ is at least as far away from any other point in the $\Gamma$-orbit of $Q$ as it is from $Q_1$ and $Q_2$, on the quotient we have two length-minimizing geodesics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am pretty sure you can arrange for this to happen with a suitable choice of $\Gamma$.  For instance, if $\Gamma$ is such that the fundamental domain is a regular hyperbolic octagon with edges identified in the usual way to get a genus $2$ quotient, then I think it works to take $Q_1$ and $Q_2$ to be adjacent vertices of the octagon.  See for instance the picture towards the bottom of this page:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.math.cornell.edu/~mec/Winter2009/Victor/part4.htm&quot; rel=&quot;nofollow&quot;&gt;http://www.math.cornell.edu/~mec/Winter2009/Victor/part4.htm&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="299" LastActivityDate="2010-10-28T06:49:46.053" />
  <row Id="8165" PostTypeId="2" ParentId="7864" CreationDate="2010-10-28T07:24:28.983" Score="2" ViewCount="" Body="&lt;p&gt;I have seen this many times:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$a^2 + a^3 = a^5$$&lt;/p&gt;&#xA;" OwnerUserId="2778" LastActivityDate="2010-10-28T07:24:28.983" CommunityOwnedDate="2010-10-28T07:24:28.983" CommentCount="1" />
  <row Id="8166" PostTypeId="2" ParentId="8126" CreationDate="2010-10-28T07:30:53.207" Score="5" ViewCount="" Body="&lt;p&gt;Unless I'm missing something, this is the preimage of I+J (I missed something... this is not correct)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $I,J$ are ideals in $B$ then $f^{-1}(I),f^{-1}(J)\subseteq f^{-1}(I+J)$ so $f^{-1}(I)+f^{-1}(J)\subseteq f^{-1}(I+J)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, if $x\in f^{-1}(I+J)$ then $f(x)=i+j,\; i\in I,j\in J$. Let $a\in f^{-1}(I)$ such that $f(a)=i$, and $b\in f^{-1}(J),\;f(b)=j$ then $f(a+b)=i+j=f(x)$ so $a+b=x+k$ where $k \in ker(f)$. $ker(f)\subseteq f^{-1}(I)+f^{-1}(J)$ and $a,b\in  f^{-1}(I)+f^{-1}(J)$ so also $x \in f^{-1}(I)+f^{-1}(J)$ and we get that $f^{-1}(I+J)\subseteq f^{-1}(I)+f^{-1}(J)$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;ok, another try&lt;/p&gt;&#xA;&#xA;&lt;p&gt;let $f:\mathbb{Z}[x]\rightarrow \mathbb{Q}[x]$ be the inclusion function. If $I=(x-3)\mathbb{Q}[x],\; J=x \mathbb{Q}[x]$ then their preimage is $(x-3)\mathbb{Z}[x], \; x\mathbb{Z}[x]$ (I used here Gauss lemma). The sum of the preimages is not all of $\mathbb{Z}[x]$ and it contains 3. If it is in itself a preimage of K then it  $3\in K$ in $\mathbb{Q}[x]$ which is invertible and so this is all of $\mathbb{Q}[x]$ and we get a contradiction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope this one is ok....&lt;/p&gt;&#xA;" OwnerUserId="2125" LastEditorUserId="2125" LastEditorDisplayName="" LastEditDate="2010-10-28T08:41:19.400" LastActivityDate="2010-10-28T08:41:19.400" CommentCount="3" />
  <row Id="8168" PostTypeId="2" ParentId="8155" CreationDate="2010-10-28T07:51:31.100" Score="1" ViewCount="" Body="&lt;p&gt;Since $\mathbb{C}$ is 2-dimensional, as a real vector space,&#xA;a one-dimensional subspace of $\mathbb{C^n}$ is actually a complex plane, and is therefore not an ordinary line.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, such subspaces/&quot;lines&quot; can be expressed as in the real case, as&#xA;a solution to $n-1$ equations of the form&#xA;$\lambda_1z_1+\lambda_2z_2+\dots+\lambda_nz_n = c$&#xA;where c is 0 if you want a subspace, any complex number if you want any set that &quot;looks&quot; like $\mathbb{C}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: Fixed error.&lt;/p&gt;&#xA;" OwnerUserId="934" LastEditorUserId="934" LastEditorDisplayName="" LastEditDate="2010-10-29T13:52:32.743" LastActivityDate="2010-10-29T13:52:32.743" CommentCount="3" />
  <row Id="8170" PostTypeId="1" AcceptedAnswerId="8197" CreationDate="2010-10-28T09:46:39.330" Score="6" ViewCount="503" Body="&lt;p&gt;I'm trying to understand basic tensor analysis. I understand the basic concept  that the valency of the tensor determines how it is transformed, but I am having trouble visualizing the difference between different valencies when it comes to higher order tensors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have this picture in my mind for the lower order tensors&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$X^i = \left(\begin{array}{x} x^1 \\ x^2 \\ x^3\end{array}\right)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$X_i = \left(\begin{array}{ccc} x_1 &amp;amp; x_2 &amp;amp; x_3\end{array}\right)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$X^i_j = \left(\begin{array}{ccc} x^1_1 &amp;amp; x^1_2 &amp;amp; x^1_3 \\&#xA; x^1_2 &amp;amp; x^2_2 &amp;amp; x^3_2 \\ x^1_3 &amp;amp; x^2_3 &amp;amp; x^3_3\end{array} \right)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for $X^{ij}$ and $X_{ij}$ they are represented in the same 2d array, but the action on a vector isn't defined in the same way as with matrices.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I am having trouble with is intuitively understanding the difference between $X^{ijk}$, $X_{k}^{ij}$, $X_{jk}^{i}$, and $X_{ijk}$ (other permutations of the valence $(2,1)$ and $(1,2)$ omitted for brevity).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;ADDED&lt;/strong&gt;&#xA;After reading the responses and their comments I came up with this new picture in my head for higher order tensors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since I am somewhat comfortable with tensor products in quantum mechanics, I can draw a parallel with the specific tensor space I'm used to.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we consider a rank-5 tensor with a valence of (2,3) then we can consider it in the braket notation as &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ \langle \psi_i \mid \otimes \ \langle \psi_j \mid \otimes \ \langle \psi_k \mid \otimes \mid \psi_l \rangle \ \otimes \mid \psi_m \rangle = X_{ijk}^{lm} $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now if we operate with this tensor on rank-3 contravariant tensor, we are-left with a constant (from the inner product) and a rank-2 contravariant tensor, unmixed tensor product &#xA;$\begin{eqnarray}(\langle \psi_i \mid \otimes \ \langle \psi_j \mid \otimes \ \langle \psi_k \mid \otimes \mid \psi_l \rangle \ \otimes \mid \psi_m \rangle)(\mid \Psi_i \rangle \ \otimes \mid \Psi_j \rangle \ \otimes \mid \Psi_k \rangle) &amp;amp;=&amp;amp;  c \mid \psi_l \rangle \ \otimes \mid \psi_m \rangle \\ &amp;amp;=&amp;amp; X_{ijk}^{lm}\Psi^{ijk} = cX'^{lm}\end{eqnarray}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we were to further operate with a rank-2 covariant tensor (from the right, per convention that a covector and vector facing each other is an implied direct product) we would simply get a number out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One thing I am confused about though, is that in one of the answer to this question there was a point made that we are taking tensor products of a Vector space with itself (and possibly it's dual), however in the quantum mechanics picture (although I didn't rely on it in this example) we often take tensor products between different, often disjoint, subspaces of the enormous Hilbert space that describes the quantum mechanical universe. Does the tensor picture change in this case? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any comments on my example would be appreciated.&lt;/p&gt;&#xA;" OwnerUserId="1906" LastEditorUserId="1906" LastEditorDisplayName="" LastEditDate="2010-10-28T20:19:31.977" LastActivityDate="2010-10-28T20:19:31.977" Title="Intuitive way to understand covariance and contravariance in Tensor Algebra" Tags="&lt;intuition&gt;&lt;tensors&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" />
  <row Id="8171" PostTypeId="2" ParentId="8042" CreationDate="2010-10-28T09:55:36.467" Score="8" ViewCount="" Body="&lt;p&gt;Write the inequality as&#xA;$$\frac{\pi}{4} &amp;lt; \frac{1}{2n} + \frac{1}{n} \sum_{k=1}^{n-1} \sqrt{1-\left(\frac{k}{n}\right)^2} + \frac{1}{2n} \sqrt{\frac{1}{2n}}.$$&#xA;The left-hand side $\pi/4$ is the area of the part of the unit circle that&#xA;lies in the first quadrant (below the curve $y=f(x)=\sqrt{1-x^2}$).&#xA;We want to interpret the right-hand side as the area of a region $D$ which&#xA;covers that quarter circle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that $f$ is concave, so that its graph lies below any tangent line.&#xA;Thus the trapezoid bounded by the lines $x=a-\epsilon$ and $x=a+\epsilon$&#xA;and by the $x$ axis and the tangent line through $(a,f(a))$ will cover the&#xA;corresponding part of the circle:&#xA;$$\int_{a-\epsilon}^{a+\epsilon} f(x) dx &amp;lt; 2\epsilon f(a).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, taking $D$ to be the union of the following pieces does the trick:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;A rectangle of height 1 between $x=0$ and $x=1/2n$.&lt;/li&gt;&#xA;&lt;li&gt;Trapezoids as above, of width $\frac{1}{n}$ and centered at $x=k/n$ for $k=1,\ldots,n-1$.&lt;/li&gt;&#xA;&lt;li&gt;A trapezoid as above, of width $\frac{1}{2n}$ and centered at $x=1-1/4n$. This last one has area&#xA;$$\frac{1}{2n} f(1-1/4n) = \frac{1}{2n} \sqrt{\frac{1}{2n} - \frac{1}{16n^2}} &amp;lt;   \frac{1}{2n} \sqrt{\frac{1}{2n}}.$$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2010-10-28T09:55:36.467" />
  <row Id="8172" PostTypeId="2" ParentId="8170" CreationDate="2010-10-28T10:03:00.327" Score="3" ViewCount="" Body="&lt;p&gt;I prefer to think of them as maps instead of matrices.  When you move to tensor bundles over manifolds, you won't have global coordinates, so it might be preferable to think this way.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So $x_i$ is a map which sends vectors to reals.  Since it's a tensor, you're only concerned with how it acts on basis elements.  It's nice to think of them in terms of dual bases: then $x_i(x^j)=\delta_{ij}$, which is defined as $1$ when $i=j$ and $0$ otherwise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, $x^i$ is a map which sends covectors to reals, and is defined by $x^i(x_j)=\delta_{ij}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have more indices, then you're dealing with a tensor product $V^*\otimes\dotsb\otimes V^*\otimes V\otimes\dotsb\otimes V$, say with $n$ copies of the vector space and $m$ copies of the dual.  An element of this vector space takes in $m$ vectors and gives you back $n$, again in a tensorial way.  So, for example, $X_{ijk}$ is a trilinear map; $X^{ijk}$ is a trivector (an ordered triple of vectors up to linearity); $X_{ij}^k$ is a bilinear map taking two vectors to one vector; and so on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's worth thinking about these in terms of the tensors you've seen already.  The dot product, for example, is your basic (0,2)-tensor.  The cross product is a (1,2)-tensor.  If you study Riemannian manifolds, it turns out you can use the metric to &quot;raise and lower indices&quot;; so the Riemannian curvature tensor, for example, is alternately defined as a (1,3)-tensor and a (0,4)-tensor, depending on the author's needs.&lt;/p&gt;&#xA;" OwnerUserId="2215" LastEditorUserId="2215" LastEditorDisplayName="" LastEditDate="2010-10-28T10:39:57.533" LastActivityDate="2010-10-28T10:39:57.533" CommentCount="10" />
  <row Id="8173" PostTypeId="1" AcceptedAnswerId="8184" CreationDate="2010-10-28T10:27:06.037" Score="3" ViewCount="217" Body="&lt;p&gt;What is an example of a lambda-system that is not a sigma algebra?&lt;/p&gt;&#xA;" OwnerUserId="774" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-10-28T19:10:48.930" LastActivityDate="2011-01-12T17:40:04.647" Title="What is an example of a lambda-system that is not a sigma algebra?" Tags="&lt;probability-theory&gt;&lt;measure-theory&gt;" AnswerCount="1" CommentCount="0" />
  <row Id="8175" PostTypeId="2" ParentId="7864" CreationDate="2010-10-28T11:13:26.270" Score="3" ViewCount="" Body="&lt;p&gt;Both my students and some of my colleagues (!) believe that the graph of a function cannot cross a horizontal asymptote.  Obviously this implies that they misunderstand the definition of an asymptote.  More worryingly (in my eyes), it also seems to imply that they don't understand &lt;em&gt;why&lt;/em&gt; we even care about asymptotes.&lt;/p&gt;&#xA;" OwnerUserId="506" LastActivityDate="2010-10-28T11:13:26.270" CommunityOwnedDate="2010-10-28T11:13:26.270" CommentCount="1" />
  <row Id="8176" PostTypeId="1" AcceptedAnswerId="8201" CreationDate="2010-10-28T12:11:21.717" Score="5" ViewCount="298" Body="&lt;p&gt;This is probably an obvious parallel that most people are aware of, but I only just noticed it the other day and it made me quite excited. The &lt;strong&gt;determinant&lt;/strong&gt;  in algebra has a lot in common with the &lt;strong&gt;integral&lt;/strong&gt; in analysis. For example:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;They are both applied to functions, the integral to integrable functions, the determinant to linear transformations $T:V \rightarrow V$.&lt;/li&gt;&#xA;&lt;li&gt;They are both &quot;sums of products.&quot;&lt;/li&gt;&#xA;&lt;li&gt;They both &lt;em&gt;can&lt;/em&gt; be used to give a scalar result. (Not always, of course, but this is how they are first developed.)&lt;/li&gt;&#xA;&lt;li&gt;They are both important major structures in algebra and analysis.&lt;/li&gt;&#xA;&lt;li&gt;They are both defined in ways that feel 'backwards'-- the formal definition isn't always useful for calculating them-- then they come to represent &lt;strong&gt;multiple&lt;/strong&gt; important concepts acting as a fulcrum in their fields.  (ie. AREA is connected to ANTI-DERIVATIVES... or that SOLUTIONS TO AX=B are connected to LINEAR TRANSFORMATIONS.)&lt;/li&gt;&#xA;&lt;li&gt;They can both be used to give area and volume. (under a curve, or of a parallelepiped)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Question: &lt;strong&gt;What mathematical structure encompasses both?&lt;/strong&gt; (If the answer is category theory, please go slowly with me, I don't understand that stuff yet.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What else could we add to this list? &lt;strong&gt;Are there any problems or proofs that bring these parallels in to the light?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Are there any other mathematical structures that follow the pattern established by these two structures&lt;/strong&gt;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Were they developed independently (what I suspect) or is the determinant in some way patterned after the integral or vice versa?&lt;/strong&gt; (I know my math history and have not come across anything about this.)&lt;/p&gt;&#xA;" OwnerUserId="2563" LastEditorUserId="2563" LastEditorDisplayName="" LastEditDate="2010-10-28T15:43:44.263" LastActivityDate="2010-10-28T17:24:52.717" Title="The determinant is the integral of algebra. The integral is the determinant of analysis." Tags="&lt;linear-algebra&gt;&lt;analysis&gt;&lt;intuition&gt;" AnswerCount="2" CommentCount="8" />
  <row Id="8177" PostTypeId="2" ParentId="7981" CreationDate="2010-10-28T12:58:38.593" Score="1" ViewCount="" Body="&lt;p&gt;Here is a very simply description of the Riemann Hypothesis that requires nothing more than a 3rd grade education to understand:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.jstor.org/pss/2323497&quot; rel=&quot;nofollow&quot;&gt;http://www.jstor.org/pss/2323497&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is also a beautiful proof linking the Farey sequence of fractions to the Riemann hypothesis by Jerome Franel.  It's only three pages long and should be able to be understood by any undergraduate mathematics major.&lt;/p&gt;&#xA;" OwnerUserId="2899" LastActivityDate="2010-10-28T12:58:38.593" />
  <row Id="8178" PostTypeId="2" ParentId="7864" CreationDate="2010-10-28T13:03:18.580" Score="3" ViewCount="" Body="&lt;p&gt;To generalize a few of the answers, for pretty much any function, someone somewhere will make the mistake of treating it as if it is linear in all of its variables. Thus we get:&#xA;$e^a + e^b = e^{a+b}$, $\sqrt{a + b} = \sqrt{a} + \sqrt{b}$, $a/(b+c) = a/b + a/c$, ...&lt;/p&gt;&#xA;" OwnerUserId="759" LastActivityDate="2010-10-28T13:03:18.580" CommunityOwnedDate="2010-10-28T13:03:18.580" />
  <row Id="8179" PostTypeId="2" ParentId="8126" CreationDate="2010-10-28T13:06:31.280" Score="2" ViewCount="" Body="&lt;p&gt;Let $k$ be field of caracteristic $\neq 2$, $A=k[x,y], B=k[x,y,x^{-1},y^{-1}]$ and take $f:A\hookrightarrow B$ be inclusion. Example for what you want is $I=(x+y)B, J=(x-y)B$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then $f^{-1}I=(x+y)A, f^{-1}J=(x-y)A$ and $f^{-1}I+f^{-1}J=(x,y)A$ . Since $(x,y)B=B$,  sum &#xA;$f^{-1}I+f^{-1}J$ can not be $f^{-1}$ of ideal in $B$&lt;/p&gt;&#xA;" OwnerUserId="2872" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-10-28T14:44:47.643" LastActivityDate="2010-10-28T14:44:47.643" CommentCount="2" />
  <row Id="8180" PostTypeId="1" CreationDate="2010-10-28T13:49:35.613" Score="1" ViewCount="70" Body="&lt;p&gt;Short question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can an element in the Kalman gain vector ever be over one?&lt;/p&gt;&#xA;" OwnerUserId="2367" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2010-11-20T07:04:59.780" LastActivityDate="2010-11-20T07:04:59.780" Title="Kalman gain over one" Tags="&lt;signal-processing&gt;" AnswerCount="1" />
  <row Id="8181" PostTypeId="1" AcceptedAnswerId="8183" CreationDate="2010-10-28T14:04:04.380" Score="2" ViewCount="130" Body="&lt;p&gt;I remember seeing something like the following problem in the past and would like to know if it has a solution (or if I can find a source for it).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Problem&lt;/strong&gt; Given a finite set of points in the plane in general position (no three are collinear), does there exist a line that intersects no points?&lt;/p&gt;&#xA;" OwnerUserId="565" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-10-28T19:02:46.863" LastActivityDate="2010-10-28T19:49:30.153" Title="Existence of lines not containing given points in general position." Tags="&lt;geometry&gt;&lt;euclidean-geometry&gt;&lt;discrete-geometry&gt;&lt;projective-geometry&gt;" AnswerCount="5" CommentCount="3" />
  <row Id="8182" PostTypeId="2" ParentId="8142" CreationDate="2010-10-28T14:09:02.360" Score="1" ViewCount="" Body="&lt;p&gt;Using complex numbers, but a more elementary (and constructive) approach than the Maximum Modulus theorem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the unit circle. It is enough if we show that the product of distances is greater than $1$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, we will show that the product of distances of some point is at least $2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let the points be $\displaystyle z_1, z_2, \dots z_n$. Rotate the circle so that $\displaystyle (-1)^n \prod_{j=1}^{n} z_j = 1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $\displaystyle P(z) = \prod_{j=1}^{n}(z-z_j)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The product of distances from $\displaystyle  z$ to $\displaystyle z_j$ is given by $\displaystyle |P(z)|$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let $\displaystyle \omega_j$ be the $\displaystyle n$ $n^{th}$ roots of unity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since for $\displaystyle 1 \le k &amp;lt; n$, we have that $\displaystyle \sum_{j=1}^{n} (\omega_j)^k = 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;we have that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle \sum_{j=1}^{n} P(w_j) = 2n$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus $\displaystyle \sum_{j=1}^{n} |P(w_j)| \ge |\sum_{j=1}^{n} P(w_j)| = 2n$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence there is some $j$ for which $|P(w_j)| \ge 2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we have shown that there is a point whose product of distances from the $n$ points is at least $2r^n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, I believe we can also show that if for all points, the product of distances is  $\leq 2r^n$, then the $n$ points must be equally spaced! (I will leave that for you :-)).&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-10-28T20:16:14.783" LastActivityDate="2010-10-28T20:16:14.783" />
  <row Id="8183" PostTypeId="2" ParentId="8181" CreationDate="2010-10-28T14:09:30.550" Score="5" ViewCount="" Body="&lt;p&gt;Yes.  Choose any direction in the plane.  There are continuum many lines parallel to that direction.  Only finitely many hit one of the points.  So continuum many of these lines do not hit any points.  This argument works as well for countably many points.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-10-28T14:09:30.550" CommentCount="1" />
  <row Id="8184" PostTypeId="2" ParentId="8173" CreationDate="2010-10-28T14:26:13.380" Score="3" ViewCount="" Body="&lt;p&gt;Here is a somewhat more natural example.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $(\Omega, \mathcal{F})$ be a measurable space, and let $P,Q$ be two probability measures on $\mathcal{F}$.  It is a good exercise to verify that $$\mathcal{L} := \{ A \in \mathcal{F} : P(A) = Q(A) \}$$ is a $\lambda$-system.  (This is a common application of the $\pi$-$\lambda$ theorem : if one can show that $P$ and $Q$ agree on a $\pi$-system that generates $\mathcal{F}$, then $P$ and $Q$ must be the same.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, $\mathcal{L}$ need not be a $\sigma$-algebra.  For instance, consider a sample space consisting of two coin flips: $$\Omega = \{ HH, HT, TH, TT \}, \quad \mathcal{F} = 2^\Omega.$$ Let $P$ be the probability measure under which the coins are independent and unbiased, and let $Q$ be the measure under which the first coin is unbiased but the second coin is stuck to the first so that they always come up the same.  Explicitly,&#xA;$$P(HH)=P(HT)=P(TH)=P(TT)=\frac{1}{4}$$&#xA;$$Q(HH)=Q(TT)=\frac{1}{2}; \quad Q(HT)=Q(TH)=0.$$&#xA;Then one can check that the events on which $P$ and $Q$ agree are those which only look at one of the coins (or none), so that&#xA;$$\mathcal{L} = \{ \{HH,HT\}, \{HH,TH\}, \{HT,TT\}, \{TH,TT\}, \emptyset, \Omega\}.$$&#xA;This is not a $\sigma$-algebra since it is not closed under intersections.&lt;/p&gt;&#xA;" OwnerUserId="822" LastActivityDate="2010-10-28T14:26:13.380" CommentCount="4" />
  <row Id="8185" PostTypeId="2" ParentId="8181" CreationDate="2010-10-28T14:35:23.490" Score="1" ViewCount="" Body="&lt;p&gt;Ross' argument with equations says that, if you choose a Cartesian coordinate system in the plane, every straight line can be written as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;y = ax + b \ , \qquad \text{or} \qquad x = 0 \ ,&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for some real numbers $a, b$. Now, choosing a direction means picking up some $a \in \mathbb{R}$. Everyone of your finite (or countable) number of points $(x_n, y_n)_{n=1,2, \dots}$ determines a straight line: you just solve the equation for $b$ in terms of $y_n, a, x_n$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;y_n = a x_n + b \qquad \Longleftrightarrow \qquad b = y_n - ax_n \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And then you just need to choose any $b \neq y_n - ax_n, \ n = 1,2,\dots$, which you can do, since there is just a finite (or countable) number of them, but an uncountable quantity of real numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively, you could fix the height $b$ instead of the slope $a$; for instance, $b=0$, and the points $(x_n, y_n)_{n=1,2,\dots}$ would give you a finite (or countable) number of slopes of straight lines from the origin $(0,0)$ to each of your points, using the same equation as above:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;a = \dfrac{y_n}{x_n} \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(In case you have $x_n = 0$, you just have the straight line $x=0$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now you pick up any $a\neq\dfrac{y_n}{x_n} $, which you can do for the same reason as before, and you are done.&lt;/p&gt;&#xA;" OwnerUserId="664" LastActivityDate="2010-10-28T14:35:23.490" />
  <row Id="8186" PostTypeId="1" CreationDate="2010-10-28T14:51:18.943" Score="6" ViewCount="354" Body="&lt;p&gt;It is well-known that given two primes &lt;em&gt;p&lt;/em&gt; and &lt;em&gt;q&lt;/em&gt;, &lt;em&gt;pZ + qZ = Z&lt;/em&gt;  where &lt;em&gt;Z&lt;/em&gt; stands for all integers.&#xA;It seems to me that the set of natural number multiples, i.e. &lt;em&gt;pN + qN&lt;/em&gt; also span all natural numbers that are large enough. That is, there exists some &lt;em&gt;K&gt;0&lt;/em&gt;, such that&#xA;&lt;em&gt;pN + qN = [K,K+1,...)&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is, given &lt;em&gt;p&lt;/em&gt; and &lt;em&gt;q&lt;/em&gt;, can we get a upper bound on &lt;em&gt;K&lt;/em&gt;?&lt;/p&gt;&#xA;" LastActivityDate="2010-11-07T00:54:29.040" Title="When do the multiples of two primes span all large enough natural numbers?" Tags="&lt;number-theory&gt;&lt;diophantine-equations&gt;&lt;cryptography&gt;&lt;number&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="8187" PostTypeId="2" ParentId="8186" CreationDate="2010-10-28T14:55:05.867" Score="14" ViewCount="" Body="&lt;p&gt;$K = pq + 1$ if $\mathbb{N} = \{ 1, 2, 3, ... \}$, and $K = pq - p - q + 1$ if $\mathbb{N} = \{ 0, 1, 2, 3, ... \}$.  This is known as the &lt;a href=&quot;http://en.wikipedia.org/wiki/Coin_problem&quot; rel=&quot;nofollow&quot;&gt;coin problem&lt;/a&gt;, or Frobenius problem (and you only need $p, q$ relatively prime).  It frequently appears on middle- and high-school math competitions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt;  I completely misremembered how hard the proof is.  Here it is.  If $n$ is at least $pq+1$, then the positive integers&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$n-p, n-2p, ... n-qp$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;have distinct residue classes $\bmod q$, so one of them must be divisible by $q$.  On the other hand, it's not hard to see that $pq$ itself cannot be written in the desired way.&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-11-07T00:01:15.903" LastActivityDate="2010-11-07T00:01:15.903" CommentCount="6" />
  <row Id="8188" PostTypeId="1" AcceptedAnswerId="8189" CreationDate="2010-10-28T14:56:53.023" Score="2" ViewCount="165" Body="&lt;p&gt;For a ring $R$, an $R$-right-module $M$, an $R$-left-module $N$, and an abelian group $P$, one can use the universal property of the tensor product to construct maps&#xA;$$&#xA;M\otimes_R N\to P.&#xA;$$&#xA;It concrete cases, it is often easy to see that the constructed map is surjective by just writing down pre-images.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems to be harder to verify that the map is injective, because then one has to consider general sums of elementary tensors in $M\otimes_R N$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there any tricks to avoid this, and to achieve injectivity more elegantly?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;More concretely, an example I have in mind is the following: I have a pre-additive category $C$ with finitely many objects. I consider modules over this category (that is, functors from $C$ to Abelian groups; or, equivalently, modules over the category ring of $C$). Now, I consider the &quot;free $C$-right-module $Q_Y$ over an object $Y$ in $C$&quot; which is the hom functor $C(\bullet,Y)$. I want to show that for an arbitrary left-module $M$:&#xA;$$&#xA;Q_Y\otimes_C M\cong M(Y)&#xA;$$&#xA;as abelian groups. Using the module structure of $M$, I can accomplish a natural epimorphism $Q_Y\otimes_C M\to M(Y)$ which should be injective.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Currently I think that the formula&#xA;$$&#xA;M(Y)\to Q_Y\otimes_C M;\quad m\mapsto\mathrm{id}_Y\otimes m&#xA;$$&#xA;gives indeed a (two-sided) inverse. Is this correct?&lt;/p&gt;&#xA;" OwnerUserId="367" LastEditorUserId="367" LastEditorDisplayName="" LastEditDate="2010-11-02T12:48:07.210" LastActivityDate="2010-11-02T12:48:07.210" Title="Is there a trick to prove injectivity for maps out of tensor products?" Tags="&lt;algebra&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="1" />
  <row Id="8189" PostTypeId="2" ParentId="8188" CreationDate="2010-10-28T15:26:34.100" Score="1" ViewCount="" Body="&lt;p&gt;If the image Q is easy to find and well understood, then you can try to construct an inverse from Q→M⊗N. If Q is well understood, then you can define homomorphisms from it (say by defining the images of some generators, and checking that all the relations are sent to 0 in M⊗N).  Then you just check that the composition is the identity (say by sending generators of Q into M⊗N and then back into P, and checking relations to check equality).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This depends heavily on the image being well understood.  It works well to prove an abstract tensor product is equal to some concretely understood module.  If the image is confusing, then checking that the homomorphism is well defined (or even just defining it) can be hard, and even checking equality of elements in P can be hard in some cases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are some other tricks using embeddings and exact sequences, but they also have limited applicability (usually when the entire problem has exact sequences floating around).  Feel free to give some more concrete examples for more concrete tricks.&lt;/p&gt;&#xA;" OwnerUserId="583" LastActivityDate="2010-10-28T15:26:34.100" CommentCount="2" />
  <row Id="8190" PostTypeId="2" ParentId="8181" CreationDate="2010-10-28T16:01:06.853" Score="6" ViewCount="" Body="&lt;p&gt;What is this general position stuff?  Take any line outside the convex hull.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe you're thinking of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Sylvester%E2%80%93Gallai_theorem&quot; rel=&quot;nofollow&quot;&gt;Sylvester-Gallai theorem&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-10-28T16:01:06.853" />
  <row Id="8191" PostTypeId="2" ParentId="7745" CreationDate="2010-10-28T16:07:06.293" Score="1" ViewCount="" Body="&lt;p&gt;I can not post comments, so I will post an answer..  Using $S_0 + B_t$ to model a stock price is not a very good idea, because $S_t$ may well turn negative. Have you considered using &lt;a href=&quot;http://en.wikipedia.org/wiki/Geometric_Brownian_motion&quot; rel=&quot;nofollow&quot;&gt;Geometric Brownian Motion&lt;/a&gt; ?&lt;/p&gt;&#xA;" OwnerUserId="731" LastActivityDate="2010-10-28T16:07:06.293" />
  <row Id="8192" PostTypeId="2" ParentId="7506" CreationDate="2010-10-28T16:21:05.780" Score="1" ViewCount="" Body="&lt;p&gt;Normally subcategories of Abelian Categories are assumed nonempty for the simple reason that they need contain the zero object. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want your subcats to be abelian, then they need this zero. For this reason, I would say no.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is at least the viewpoint in NCAG.&lt;/p&gt;&#xA;" OwnerUserId="16" LastEditorUserId="16" LastEditorDisplayName="" LastEditDate="2010-10-29T07:28:12.907" LastActivityDate="2010-10-29T07:28:12.907" CommentCount="2" />
  <row Id="8193" PostTypeId="2" ParentId="8181" CreationDate="2010-10-28T16:22:36.730" Score="2" ViewCount="" Body="&lt;p&gt;Such statements are often easier to see in the projectively dual version.&#xA;The dual statement would be that given any finite set of lines in the projective plane (which secretely we are thinking of the dual plan of lines in&#xA;the projective plane where our original points lay), and a point $P$ not on them (this is the point in the dual plane corresponding to the line at infinity&#xA;in our original plan), there is a point $Q$ distinct from $P$, and also not lying on any of the lines.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just stating it again without the parenthetical explanations:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Given any finite set of lines in the projective plane, and a point $P$ not lying on them any of them, there exists a point $Q$ distinct from $P$ and not lying on any of them.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This will (pretty clearly) be true just provided we are working over an&#xA;infinite field.  (Thus we don't need that $\mathbb R$ is uncountable;&#xA;the fact that it is infinite is enough.  This is made clear in Agusti Roig's answer, of which the above is slighty more theoretical reformulation.)&lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2010-10-28T16:22:36.730" CommentCount="1" />
  <row Id="8194" PostTypeId="1" AcceptedAnswerId="8196" CreationDate="2010-10-28T16:32:25.517" Score="3" ViewCount="115" Body="&lt;p&gt;Consider a function $f : \mathcal X \times \mathcal Y \mapsto \mathbb R$. I want to define $g_x(y) = f(x,y) : \mathcal Y \mapsto \mathbb R$. I want to say that &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$g_x$ is a &lt;strong&gt;&lt;em&gt;_&lt;/em&gt;__&lt;/strong&gt; of function $f$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What is the appropriate word for &lt;strong&gt;&lt;em&gt;_&lt;/em&gt;____&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="2904" LastActivityDate="2010-10-28T16:49:26.260" Title="Function of two variables as a function of a function" Tags="&lt;terminology&gt;&lt;functions&gt;" AnswerCount="2" FavoriteCount="2" />
  <row Id="8195" PostTypeId="2" ParentId="8194" CreationDate="2010-10-28T16:35:32.497" Score="5" ViewCount="" Body="&lt;p&gt;This is sometimes called &lt;a href=&quot;http://en.wikipedia.org/wiki/Currying&quot; rel=&quot;nofollow&quot;&gt;currying&lt;/a&gt;.  It is closely related to the notion of an &lt;a href=&quot;http://en.wikipedia.org/wiki/Exponential_object&quot; rel=&quot;nofollow&quot;&gt;exponential object&lt;/a&gt;.  But you don't really need to use either of these terms to perform this construction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt;  Ah, I was assuming you were varying $x$.  If $x$ is fixed, you might want to call $g_x$ a &lt;a href=&quot;http://en.wikipedia.org/wiki/Function_(mathematics)#Restrictions_and_extensions&quot; rel=&quot;nofollow&quot;&gt;restriction&lt;/a&gt; of $f$.&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-10-28T16:45:30.393" LastActivityDate="2010-10-28T16:45:30.393" CommentCount="2" />
  <row Id="8196" PostTypeId="2" ParentId="8194" CreationDate="2010-10-28T16:39:19.737" Score="7" ViewCount="" Body="&lt;p&gt;I've seen your $g_x$ called the $x$-&lt;strong&gt;section&lt;/strong&gt; of $f$.  E.g. Folland's &lt;em&gt;Real Analysis&lt;/em&gt;, section 2.5.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: Another notation that's often useful is to write $f(x, \cdot)$ instead of $g_x$.&lt;/p&gt;&#xA;" OwnerUserId="822" LastEditorUserId="822" LastEditorDisplayName="" LastEditDate="2010-10-28T16:49:26.260" LastActivityDate="2010-10-28T16:49:26.260" CommentCount="1" />
  <row Id="8197" PostTypeId="2" ParentId="8170" CreationDate="2010-10-28T16:47:27.600" Score="4" ViewCount="" Body="&lt;p&gt;Since you asked for an intuitive way to understand covariance and contravariance, I think this will do. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;First of all, remember that the reason of having covariant or contravariant tensors is because you want to represent the same thing in a different coordinate system. Such a new representation is achieved by a transformation using a set of partial derivatives. In tensor analysis, a good transformation is one that leaves invariant the quantity you are interested in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, we consider the transformation from one coordinate system $x^1,...,x^{n}$ to another $x^{'1},...,x^{'n}$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x^{i}=f^{i}(x^{'1},x^{'2},...,x^{'n})$ where $f^{i}$ are certain functions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take a look at a couple of specific quantities. How do we transform coordinates? The answer is: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$dx^{i}=\displaystyle \frac{\partial x^{i}}{\partial x^{'k}}dx^{k}$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Every quantity which under a transformation of coordinates, transforms like the coordinate differentials is called a contravariant tensor.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do we transform some scalar $\Phi$? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle \frac{\partial \Phi}{\partial x^{i}}=\frac{\partial \Phi}{\partial x^{'k}}\frac{\partial x^{'k}}{\partial x^{i}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Every quantity which under a coordinate transformation, transforms like the derivatives of a scalar is called a covariant tensor.&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Accordingly, a reasonable generalization is having a quantity which transforms like the product of the components of two contravariant tensors, that is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$A^{ik}=\displaystyle \frac{\partial x^{i}}{\partial x^{'l}}\frac{\partial x^{k}}{\partial x^{'m}}A^{'lm}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is called a contravariant tensor of rank two. The same applies to covariant tensors of rank n or mixed tensor of rank n. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Having in mind the analogy to coordinate differentials and derivative of a scalar, take a look at this picture, which I think will help to make it clearer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From Wikipedia:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/8ZQfK.gif&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;blockquote&gt;&#xA;    &lt;p&gt;The contravariant components of a vector are obtained by projecting onto the coordinate axes. The covariant components are obtained by projecting onto the normal lines to the coordinate hyperplanes.&lt;/p&gt;&#xA;  &lt;/blockquote&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Finally, you may want to read: &lt;a href=&quot;http://en.wikipedia.org/wiki/Orthogonal_coordinates#Basis_vectors&quot; rel=&quot;nofollow&quot;&gt;Basis vectors&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the way, I don't recommend to rely blindly in the picture given by matrices, specially when you are doing calculations.&lt;/p&gt;&#xA;" OwnerUserId="2569" LastActivityDate="2010-10-28T16:47:27.600" CommentCount="2" />
  <row Id="8198" PostTypeId="2" ParentId="8107" CreationDate="2010-10-28T16:51:58.243" Score="2" ViewCount="" Body="&lt;p&gt;Since your question is about seeing how your different options play out, and you have a small number of them (the point of my previous question), you can use a &lt;a href=&quot;http://en.wikipedia.org/wiki/Decision_tree&quot; rel=&quot;nofollow&quot;&gt;decision tree&lt;/a&gt;.  (From Wikipedia: &quot;In decision analysis, a 'decision tree'... is used as a visual and analytical decision support tool, where the expected values (or expected utility) of competing alternatives are calculated.&quot;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You will want the first level of branches of the tree to be the various decisions you have (spend 0 dollars on research, spend 5 dollars on research, etc.).  The second level of branches of the tree will be whether or not the breakthrough is found.  Based on the outcomes and probabilities for the second level, you can calculate an expected value for each option in the first level.  Presumably you want to choose the option with the highest expected value (or expected utility, if you are comfortable with utility functions).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I think you need some more information to get a really good answer.  In particular, your example doesn't quantify exactly how your budget would change over time.  That needs to be included somehow.  In addition, perhaps going with the highest expected monthly value isn't what you really want, as your description of the scenario implies that you might be willing to go three months without profit in order to increase your chance of a breakthrough.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In short: I think a decision tree is what you want (and there are lots of resources out there besides the Wikipedia page that can help you with that), but you need to quantify some of the other aspects of your problem before you can get a good answer.&lt;/p&gt;&#xA;" OwnerUserId="2370" LastActivityDate="2010-10-28T16:51:58.243" />
  <row Id="8199" PostTypeId="2" ParentId="8176" CreationDate="2010-10-28T16:52:35.823" Score="5" ViewCount="" Body="&lt;p&gt;I guess one very natural answer to the question is provided by differential forms.  On a smooth $n$-manifold, an $n$-form is a type of &quot;field&quot; whose value at each point is a determinant on the tangent space at that point.  (I say &quot;a determinant&quot; because it's only on $\mathbb R^n$ -- or, more generally, an inner product space -- that there's a natural way to define &lt;em&gt;the&lt;/em&gt; determinant.)  An $n$-form is exactly the type of field that can be integrated over an $n$-manifold in a coordinate-independent way.  More generally, a $k$-form on an $n$-manifold is a field that yields a determinant on each $k$-dimensional subspace of each tangent space, and it is the type of field that can be integrated over $k$-dimensional submanifolds.  (You can find more about this in my book &lt;em&gt;Introduction to Smooth Manifolds&lt;/em&gt;.)&lt;/p&gt;&#xA;" OwnerUserId="1421" LastActivityDate="2010-10-28T16:52:35.823" CommentCount="3" />
  <row Id="8200" PostTypeId="2" ParentId="8186" CreationDate="2010-10-28T16:56:17.387" Score="1" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\rm\ \ p\ (p^{-1}\: mod\ q) + q\ (q^{-1}\: mod\ p)\ =\ pq+1\ $ since it's $1$ mod $\rm\:p,q\:$ and $\rm &gt;2\:$ and $\rm &amp;lt; 2\:pq$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To represent largers numbers keep adding $\rm\ 1 = ap+bq\ $ while, if need be, adding $\rm\:\pm (q ,-p)\:$ to the coefficients to keep them positive.&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-10-28T16:56:17.387" />
  <row Id="8201" PostTypeId="2" ParentId="8176" CreationDate="2010-10-28T17:18:03.240" Score="4" ViewCount="" Body="&lt;p&gt;It's an interesting question. If there were any strong and formalizable analogy it probably would have been developed a long time ago and inscribed in the textbooks.  A few observations.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The linear algebra analogue of integration is a trace.   Determinants are an exponentiated trace. For example $\det \exp A = \exp Tr(A)$ for matrices.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If you view integration as solution of differential equations rather than measure, then determinant appears as the Wronskian.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Integration (as measure) and determinants are closely related in the theory behind the change of variables formula in integrals: differential forms.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The integral is a trace on an infinite-dimensional space (the commutative algebra of functions on, e.g., a closed interval or the real line) while the determinant is specifically finite-dimensional.  The Lebesgue measure used in ordinary $n$-dimensional integrals is in some sense defined using determinants (volume), which is why it does not generalize well to infinite dimensional spaces.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Thinking about integrals and determinants in terms of formal properties they satisfy leads to &quot;K-theory&quot;, specifically $K_1$, but I don't think this produces any deep or striking analogies between the concepts.  &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="467" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-10-28T17:24:52.717" LastActivityDate="2010-10-28T17:24:52.717" />
  <row Id="8202" PostTypeId="1" CreationDate="2010-10-28T18:28:01.187" Score="0" ViewCount="217" Body="&lt;p&gt;I have a function that is difficult to integrate.  So I elect to work with power series representations.  Suppose the power series representation for this function is the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f(x) = \sum_{i=M}^N{c_i x^i}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I attempt to integrate each term of the power series seperately.  For instance, looking at a few terms, we can integrate this to obtain $g(x)$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f(x) = \cdots + c_0 x^0 + c_1 x^1 + c^2 x^2 + \cdots$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$g(x) = \cdots + c_0 x^1 + \frac{1}{2}c_1 x^2 + \frac{1}{3}c_2 x^3 + \cdots$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In effect, we take $c_i x_i \mapsto \frac{1}{i+1}c_i x^{i+1}$ integrating terms individually.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Its possible to do this using generating function methods, which is fairly routine.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The Problem&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Two terms are problematic.  The term $c_{-1} x^{-1}$ should integrate to $\log(x)$.  The constant of integration is not known, so this is the second problem term.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The Question&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm looking for a way to perhaps solve this problem.  For instance, finding a way to approximate the two problem terms would be a good start.  In the end, I want to get an approximation to the integral within an error bounds.  I really want to consider this particular method, though.  I'd be interested in seeing related literature.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Some Ideas&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We may be able to use approximations to similar integrals to solve for approximations to the two problem terms.  I'm interested in trying to get an approximation (to the original integral) within a certain amount of error.  If this is possible, I'd consider the problem solved.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;An Example:  Particular integrals that I'm using this on.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here's a description of the function(s) I want to integrate.  I take:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle x^b \frac{\prod_{i=0}^m{(\pm x^{c_i} \pm 1)}}{\prod_{j=0}^n{(\pm x^{d_j} \pm 1)}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that both the quotient and the denominator consist of products of the form $\pm x^r \pm 1$.  Here, $r \in \mathbb{N}$ and $-(2^s) \leq r \leq 2^s$ for some natural $s$.  Also, $b$ is similarly restricted/defined as $r$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I take this formula and make the substitution $x \mapsto e^{i t}$ and integrate from $t=-\pi$ to $t=\pi$.&lt;/p&gt;&#xA;" OwnerUserId="2626" LastEditorUserId="2626" LastEditorDisplayName="" LastEditDate="2010-10-29T16:25:17.917" LastActivityDate="2010-10-29T16:25:17.917" Title="How effective is this alternative to integration?" Tags="&lt;complex-analysis&gt;&lt;intuition&gt;&lt;integral&gt;&lt;numerical-methods&gt;&lt;approximation&gt;" CommentCount="7" />
  <row Id="8203" PostTypeId="2" ParentId="7743" CreationDate="2010-10-28T19:16:44.633" Score="33" ViewCount="" Body="&lt;p&gt;I'd like to second one part of Qiaochu Yuan's answer: the recommendation to read Polya's book.  Unlike many other books I've seen (albeit none of the others recommended above), it actually does contain guidance on how to construct a proof &quot;out of nothing&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And that's one problem with the &quot;practise, practise, practise&quot; mantra.  Practise what?  Where are the lists of similar-but-not-quite-identical things to prove to practise on?  I can find lists of integrals to do and lists of matrices to solve, but it's hard coming up with lists of things to prove.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, practise is correct.  But just as with anything else in mathematics, there's guidelines to help get you started.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first thing to realise is that reading others proofs is not guaranteed to give you any insight as to how the proof was developed.  A proof is meant to convince someone of a result, so a proof points to the theorem (or whatever) and knowing how the proof was constructed does not (or at least, &lt;em&gt;should not&lt;/em&gt;) lend any extra weight to our confidence in the theorem.  Proofs can be written in this way, and when teaching we should make sure to present some proofs in this way, but to do it every time would be tedious.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, what are the guidelines for constructing a proof?  You'll probably get different answers from different mathematicians so these should be construed as being my opinion and not a(n attempt at a) definitive answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My recommendation is that you take the statement that you want to prove and apply the following steps to it as often as you can:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;strong&gt;Expand out unfamiliar terms.&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Replacing generic statements by statements about generic objects.&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Including implicit information.&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Once you've done all that, the hope is that the proof will be much clearer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's an example.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Original statement:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The composition of linear transformations is again linear.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Replace generic statements:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If $S$ and $T$ are two composable linear transformations then their composition, $S T$, is again linear.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It is important to be precise here.  The word &quot;composable&quot; could have been left out, as the statement only makes sense if $S$ and $T$ are composable, but until you are completely familiar with this kind of process, it is better to be overly precise than otherwise.  In this case, leaving in the word &quot;composable&quot; reminds us that there is a restriction on the domains and codomains which will be useful later.  (However, one has to draw the line somewhere: even the word &quot;composable&quot; is not quite enough since it leaves open the question as to whether it is $S T$ or $T S$!)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Include implicit information:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If $S \colon V \to W$ and $T \colon U \to V$ are linear transformations then $S T \colon U \to W$ is again linear.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here's where remembering that $S$ and $T$ are composable in the previous step helps keep things clear.  As $S$ and $T$ are composable, we only need $3$ vector spaces.  Then, since we explicitly have the vector spaces the fact that $S$ and $T$ are composable is plain, though some may prefer to keep that fact in the statement.  Also, some may like to have the fact that $U$, $V$, and $W$ are &lt;strong&gt;vector&lt;/strong&gt; spaces explicitly stated.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Expand out definitions:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If $S \colon V \to W$ and $T \colon U \to V$ are such that $S(v_1 + \lambda v_2) = S(v_1) + \lambda S(v_2)$ and $T(u_1) + \mu T(u_2)$ for all $v_1, v_2 \in V$, $u_1, u_2 \in U$, and $\lambda, \mu \in \mathbb{R}$, then $S T(x_1 + \eta x_2) = S T(x_1) + \eta S T(x_2)$ for all $x_1, x_2 \in U$ and $\eta \in \mathbb{R}$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Note that I have been careful not to repeat myself with the newly introduced symbols.  It would be technically alright to reuse $u_1$ and $u_2$ in place of $x_1$ and $x_2$ since these are &lt;em&gt;local&lt;/em&gt; declarations (restricted by the phrases &quot;for all ...&quot;).  However, humans are not good at differentiating between local and global declarations so it is best not to reuse symbols unless the scope is very clear.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Replace generic statements:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If $S \colon V \to W$ and $T \colon U \to V$ are such that $S(v_1 + \lambda v_2) = S(v_1) + \lambda S(v_2)$ and $T(u_1) + \mu T(u_2)$ for all $v_1, v_2 \in V$, $u_1, u_2 \in U$, and $\lambda, \mu \in \mathbb{R}$, then whenever  $x_1, x_2 \in U$ and $\eta \in \mathbb{R}$,  $S T(x_1 + \eta x_2) = S T(x_1) + \eta S T(x_2)$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Up to now, the rephrasing has not taken into account the fact that there is a &lt;em&gt;conclusion&lt;/em&gt; and a &lt;em&gt;hypothesis&lt;/em&gt;.  This rephrasing modifies a part of the conclusion to turn it from a generic statement &quot;$P(p)$ is true for all $p \in Q$&quot; to a statement about a generic object &quot;whenever $p \in Q$ then $P(p)$ is true&quot;.  We do not do this for the similar statements in the hypothesis.  This is because these two pieces are treated differently in the proof.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Replace generic statements, and reorganise to bring choices to the fore:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Let $S \colon V \to W$ and $T \colon U \to V$ be such that $S(v_1 + \lambda v_2) = S(v_1) + \lambda S(v_2)$ and $T(u_1) + \mu T(u_2)$ for all $v_1, v_2 \in V$, $u_1, u_2 \in U$, and $\lambda, \mu \in \mathbb{R}$.  Let  $x_1, x_2 \in U$ and $\eta \in \mathbb{R}$.  Then  $S T(x_1 + \eta x_2) = S T(x_1) + \eta S T(x_2)$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In this form, the distinction between &lt;em&gt;hypothesis&lt;/em&gt; and &lt;em&gt;conclusion&lt;/em&gt; is all the clearer.  Parts of the hypothesis use the word &quot;Let&quot;, parts of the conclusion use the word &quot;Then&quot;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;With this formulation, the proof essentially writes itself.  With all it's gory details:&lt;/p&gt;&#xA;&#xA;Proof&#xA;&#xA;&lt;p&gt;Let $S \colon V \to W$ and $T \colon U \to V$ be such that $S(v_1 + \lambda v_2) = S(v_1) + \lambda S(v_2)$ and $T(u_1) + \mu T(u_2)$ for all $v_1, v_2 \in V$, $u_1, u_2 \in U$, and $\lambda, \mu \in \mathbb{R}$.  Let  $x_1, x_2 \in U$ and $\eta \in \mathbb{R}$.[^quick]  Then:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;S T(x_1 + \eta x_2) = S \big( T(x_1) + \eta T(x_2)\big)&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;using the hypothesis on $T$ as $x_1, x_2 \in U$ and $\eta \in \mathbb{R}$.  So:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;S T(x_1 + \eta x_2) = S T(x_1) + \eta S T(x_2)&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;using the hypothesis on $S$ as $T(x_1), T(x_2) \in V$ and $\eta \in \mathbb{R}$.  Hence the conclusion is true.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notes:&#xA;1. This could be condensed, but the important thing here is how to &lt;em&gt;find&lt;/em&gt; it, not what the final form should be.&#xA;2. Notice that I wrote &quot;as $x_1, x_2 \in U$&quot; rather than &quot;with $u_1 = x_1$ and $u_2 = x_2$&quot;.  This is partly style, and partly because in the statement of linearity, $u_1$ and $u_2$ are &lt;em&gt;placeholders&lt;/em&gt; into which we put $x_1$ and $x_2$.   So saying $u_1 = x_1$ is semantically incorrect as it equates a virtual vector with an actual vector.  This is a very minor point, though.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Finally, I would like to disagree with one part of Qiaochu's answer.  I actually like the imagery of a steel trap.  A proof is a bit like a trap: we want to capture the theorem in a trap so that it can't wriggle out.  We construct the proof so that there is no possibility of escape.  Eventually, yes, we want the proof to be beautiful but when it's first constructed we just want it to do the job.  Only once the theorem is caught can we spend a little time decorating the cage to make it look pretty and set it off to its best advantage.  So build the trap because theorems can be dangerous!  An escaped theorem can do untold damage, rampaging across the countryside, laying waste like an unchecked viking.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Okay, not quite finally.  The step-by-step proof above was taking from a page I wrote for my students on the nature of proof.  The original can be found &lt;a href=&quot;http://syvert.math.ntnu.no:2500/tma4145/show/proof&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.)&lt;/p&gt;&#xA;" OwnerUserId="2907" LastActivityDate="2010-10-28T19:16:44.633" CommentCount="3" />
  <row Id="8204" PostTypeId="1" AcceptedAnswerId="8208" CreationDate="2010-10-28T19:31:33.153" Score="2" ViewCount="205" Body="&lt;p&gt;Is it true that for all proofs $\forall x \exists y : R(x, y)$, then $y = y(x)$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A while back I remember reading a book on functional programming that was leading into some questions about what exactly mathematical proofs were and how they were related to functions. In fact, I just found the book and page: here on PDF page 16 (marked page 3):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.cs.kent.ac.uk/people/staff/sjt/TTFP/ttfp.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.cs.kent.ac.uk/people/staff/sjt/TTFP/ttfp.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the author mentions that if we have a proof that  $\forall x \exists y : R(x, y)$, then, in a sense, because x is bound before y, then y is dependent on x, and therefore $y = y(x)$. There may be counterexamples. Perhaps y is a constant itself, but then y(x) can also just be a constant function. I'm trying to understand under what circumstances this is not true.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Example:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $f : \mathbb{R}^n \to \mathbb{R}$ be continous at a point $\vec{v_0}$ and $f(\vec{v_0}) = A &amp;lt; 0$. Prove that there exists a number $\delta &gt; 0$ such that if $\| \vec{v} - \vec{v_0} \| &amp;lt; \delta $ then $f(\vec{v}) &amp;lt; 0$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My argument was simply to use the statement of continuity at $\vec{v_0}$ to prove this. That is, since f is continuous, the following holds:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\forall \epsilon &gt; 0 \exists \delta &gt; 0 : \| \vec{v} - \vec{v_0} \| &amp;lt; \delta \rightarrow \| f(\vec{v}) - f(\vec{v_0}) \| &amp;lt; \epsilon$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, as referred to above, we must have $\delta = \delta(\epsilon)$. Then we can make the replacement $\delta = \delta(|A|)$, and we have the necessary $\delta$ to prove that $f(\vec{v}) &amp;lt; 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have one main question (the one that closes this), and one side question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Under what conditions can we assume that $\forall x \exists y : R(x, y)$, then $y = y(x)$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there indirect ways to prove a proposition of the form $\forall x \exists y : R(x, y)$ where y is somehow not dependent on x and also not a constant? Perhaps it varies on R, but that is given, I suppose. Certainly all $\epsilon-\delta$ proofs that I've come across have been constructive in this way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Is there a canonical way to prove the proposition above?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I actually struggled with a possible proof for a while before I came up with the technique above. I used this technique in a HW assignment and on my latest Advanced Calc test. The HW was marked as correct although with the note that it was &quot;indirect&quot;, while the test was marked as wrong. Granted, the HW was done by the TA, the test by my prof. I personally don't see a more &quot;direct&quot; way to prove this result.&lt;/p&gt;&#xA;" OwnerUserId="1892" LastEditorUserId="1892" LastEditorDisplayName="" LastEditDate="2010-10-28T19:50:28.730" LastActivityDate="2010-10-29T16:21:57.180" Title="Is it true that for all proofs of the statement, $\forall x \exists y : R(x, y)$, then we can say $y = y(x)$? (Example given)" Tags="&lt;real-analysis&gt;&lt;soft-question&gt;&lt;logic&gt;&lt;proof-strategy&gt;" AnswerCount="2" CommentCount="6" />
  <row Id="8205" PostTypeId="2" ParentId="8204" CreationDate="2010-10-28T19:47:37.667" Score="0" ViewCount="" Body="&lt;p&gt;Some texts define a relation as simply a set of ordered pairs and a function as a set of ordered pairs in which every first element is unique. So if for all x there is a unique y such that R(x,y), then you can say you have a function y(x).  You have not specified that y has to be unique.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure what you mean by &quot;prove a relation&quot; under question 1.  Your proof seems fine to me.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-10-28T19:47:37.667" />
  <row Id="8206" PostTypeId="2" ParentId="8181" CreationDate="2010-10-28T19:49:30.153" Score="1" ViewCount="" Body="&lt;p&gt;For a very nice treatment of the combinatorics of point and line configurations and the symmetry of such configurations, look at Branko Grünbaum's book: Configurations of Points and Lines, American Mathematical Society, 2009.&lt;/p&gt;&#xA;" OwnerUserId="1369" LastActivityDate="2010-10-28T19:49:30.153" CommentCount="1" />
  <row Id="8207" PostTypeId="1" AcceptedAnswerId="8209" CreationDate="2010-10-28T19:50:04.783" Score="2" ViewCount="226" Body="&lt;p&gt;Simple example in order to clarify what I mean:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sequence - {2 , -1, 4, 2, 9 }&#xA;Sum and product combination: { 16, -144 }&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Question:&#xA;Is a combination of sum and product unique for any* sequence?&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&quot;any sequence&quot; except that one which consists only of zeros.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;UPD:&lt;/strong&gt; The answer to my initial question is &quot;no&quot;. Thanks.&#xA;But what's the answer if we restrict ourselves to sequences of the same length and made of non-negative integers? Do two or more non-negative integer sequences exist, each of which has the same {sum,product} combination and the same length?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;UPD:&lt;/strong&gt; Both answers are &quot;no&quot;. Thanks a lot.&lt;/p&gt;&#xA;" OwnerUserId="1934" LastEditorUserId="630" LastEditorDisplayName="" LastEditDate="2010-10-31T01:43:11.697" LastActivityDate="2010-10-31T01:43:11.697" Title="Is a combination of sum and product of any sequence unique?" Tags="&lt;sequences-and-series&gt;&lt;algebra&gt;" AnswerCount="2" CommentCount="8" />
  <row Id="8208" PostTypeId="2" ParentId="8204" CreationDate="2010-10-28T19:50:27.060" Score="8" ViewCount="" Body="&lt;p&gt;You seem to be asking: if for every $x$ there is a $y$ for which $R(x,y)$ holds, can we then define a function $f$ (in the usual sense) such that for every $x$, $R(x,f(x))$ holds?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From a set-theoretic point of view, that statement is in fact none other than the &lt;a href=&quot;http://en.wikipedia.org/wiki/Axiom_of_Choice&quot; rel=&quot;nofollow&quot;&gt;Axiom of Choice&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Axiom of Choice has many forms, but one form is the following: given a nonempty family of nonempty sets, $S_i$ with $i\in I$ (&quot;nonempty family&quot; means that $I$ is not empty, so we have some sets; &quot;nonempty sets&quot; means that each $S_i$ is not empty), there exists a &quot;choice function&quot;: a way to select for each $i\in I$ an element of the corresponding $S_i$. Formally, there exists a function $f\colon I\to \cup S_i$ from the index set to the union of the sets, with the property that $f(i)\in S_i$ for each $i$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose that by assumption, for every $x$ in our universe $X$, the set $S_x = \{y\mid R(x,y)\}$ is nonempty, and you have a function $f$ with the desried property. The function $f$ will be a &quot;choice function&quot;: a function  $f:X \to \cup_{x\in U}S_x$ such that for each $x\in X$, $f(x)\in S_x$. Conversely, a choice function for the family $\{S_x\}_{x\in X}$ will yield and $f$ with the desired property. The existence of such functions for &lt;em&gt;every&lt;/em&gt; nonempty family of nonempty sets is precisely the Axiom of Choice. Conversely, suppose that the Axiom of Choice fails, and let $\{S_x\}_{x\in X}$ be a nonempty family of nonempty sets for which no choice function exists. Then let $R(x,y)$ be &quot;$y$ is in $S_x$&quot;. For every $x$ there exists a $y$ such that $R(x,y)$, since $S_x\neq\emptyset$. But since we are assuming that the family has no choice function, there does not exist any function $f$ such that for every $x$, $f(x)$ is in $S_x$. Thus, the question not only depends on the Axiom of Choice, but is &lt;em&gt;equivalent&lt;/em&gt; to the Axiom of Choice. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In some circumstances, the exact nature of the sets $S_x$ mean that you do not need to invoke the Axiom of Choice, because we have some &quot;natural way&quot; of defining the choice. For instance, if every set $S_x$ is a subset of the natural numbers, then you can define $f(x)$ to be &quot;the least element of $S_x$&quot;.  Another standard example is when $S_x$ is a singleton for every $x$; that is, for each $x$ there exists one &lt;strong&gt;and only one&lt;/strong&gt; $y$ for which $R(x,y)$ holds. In that case, you can define your function $f$ to be &quot;$f(x)$ is the unique $y$ for which $R(x,y)$ holds&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But in general, the Axiom of Choice is, in some sense, &quot;inherently non-constructive&quot;.  And the thing is, the Axiom of Choice is &lt;em&gt;independent&lt;/em&gt; of the usual axioms of set theory. Like the parallel postulate in geometry, you can accept it or not at your taste (the consequences of accepting and of not accepting it are different; for example, if you accept the Axiom of Choice, then that means that there is a way to slice a ball of radius 1 into a finite number of parts, and then rearrange the parts without changing their shape or size, into &lt;em&gt;two&lt;/em&gt; balls of radius of 1; this is called the &lt;a href=&quot;http://en.wikipedia.org/wiki/Banach-Tarski_paradox&quot; rel=&quot;nofollow&quot;&gt;Banach-Tarski paradox&lt;/a&gt;; on the other hand, if you want all your &lt;a href=&quot;http://en.wikipedia.org/wiki/Vector_space&quot; rel=&quot;nofollow&quot;&gt;vector spaces&lt;/a&gt; to have &lt;a href=&quot;http://en.wikipedia.org/wiki/Basis_%28linear_algebra%29&quot; rel=&quot;nofollow&quot;&gt;bases&lt;/a&gt;, then you &lt;em&gt;must&lt;/em&gt; accept the Axiom of Choice. In short, it has good consequences and weird/bad consequences). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; In the situation you have, it is possible to describe a function that will work; however, the &quot;description&quot; is a bit like cheating, in the sense that you can give an explicit rule for what the value of the function will be, but in practice it would be difficult to actually find the value. Namely, for every $\epsilon\gt 0$ you know that the set of $\delta\gt 0$ for which the implication $||v-v_0||\lt\delta\Rightarrow ||f(v)-f(v_0)||\lt\epsilon$ holds is nonempty. In particular, by the Archimedean property, the set $A_{\epsilon}=\{n\in\mathbb{N}| ||v-v_0||\lt\frac{1}{n} \Rightarrow ||f(v)-f(v_0)||\lt\epsilon\}$ is nonempty. This is a nonempty set of positive integers, so it has a least element. You can then define your function $\mathbf{f}\colon (0,\infty)\to(0,\infty)$ by the &quot;formula&quot; $\mathbf{f}(\epsilon)=1/\min(A_{\epsilon})$, the reciprocal of the smallest $n$ in the set. This definition would not require you to invoke the Axiom of Choice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As to your question 2: you almost had it, I think, but you should not invoke (implicitly) the Axiom of Choice: pick a specific $\epsilon$ (for example, $|A|/2$). You &lt;em&gt;know&lt;/em&gt; that there exists a $\delta\gt 0$ that works for &lt;em&gt;that&lt;/em&gt; epsilon, so you pick &lt;em&gt;any&lt;/em&gt; $\delta_0$ that works and use &lt;em&gt;that&lt;/em&gt;. That would be the standard way of proving the proposition at hand. If you assume you have some kind of &quot;black-box function&quot; that will hand you a $\delta$ whenever you hand it an $\epsilon$, you are invoking the Axiom of Choice and you are making your proof somewhat non-constructive. &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-10-29T16:21:57.180" LastActivityDate="2010-10-29T16:21:57.180" CommentCount="8" />
  <row Id="8209" PostTypeId="2" ParentId="8207" CreationDate="2010-10-28T19:53:04.940" Score="8" ViewCount="" Body="&lt;p&gt;Not in general.  Knowing the sum and product of a sequence $r_1, ... r_n$ only fixes two coefficients of the polynomial $(x - r_1)...(x - r_n) = x^n + a_{n-1} x^{n-1} + ... + a_0$, namely $a_0$ and $a_{n-1}$, and for $n &gt; 2$ one can choose the other coefficients arbitrarily and solve for the complex roots of the resulting polynomial.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With certain restrictions, this might sometimes be possible, for example if you restrict the sequence to be integers.  &lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-10-28T19:53:04.940" CommentCount="2" />
  <row Id="8210" PostTypeId="2" ParentId="8110" CreationDate="2010-10-28T19:58:26.160" Score="2" ViewCount="" Body="&lt;p&gt;For the case where we have cans with $2,3,4$ liters, consider the polynomial&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;(1+x^2)^3 (1+x^3)^2(1+x^4)^2\left(1+\frac{1}{x^2}\right)^2\left(1+\frac{1}{x^3}\right) \left(1+\frac{1}{x^4}\right)&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Look at the terms $x^k$ where $k\geq 1$. I believe all those powers of $x$ are the capacities you can measure using the three cans.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reasoning:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any term of the form $1+x^a$ allows for two choices - You either fill the can of volume $a$ or you don't fill it. If we multiplied three terms and wrote $(1+x^2)(1+x^3)(1+x^4)$, this will allow you to find out how many different volumes you can measure if you are allowed to fill each can at most once and if a can once filled is never emptied.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, in the case of cans of size $2,3$ and $4$, it is possible to fill the 2 liter can once, empty it into the 4 liter can, fill the 2 liter can again, pour it into the 4 liter can again and fill the 2 liter can a third time. This is why we have $(1+x^2)^3$. Similarly for the 3 litre can. The 4 litre can could be emptied into the 2 and 3 liter cans combined and the refilled, so we have the term $(1+x^4)^2$ as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We should also allow for subtracting of volumes - This implies that terms of the form $(1+\frac{1}{x^a})$ should be present. If we look at the 2 liter can, you could fill it (and empty it) from the 4 liter can twice. Therefore, we have a term $(1+1/x^2)^2$. Similarly we get the term $(1+1/x^3)$. The 4 liter can could also be emptied once after filling it from the other two cans. Combining all of this, we get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;(1+x^2)^3 (1+x^3)^2(1+x^4)^2\left(1+\frac{1}{x^2}\right)^2\left(1+\frac{1}{x^3}\right) \left(1+\frac{1}{x^4}\right)&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This idea can be extended to other values of $a,b$ and $c$. I tried it for a few other values (using Wolfram Alpha to expand the expressions) and it seems to work. &lt;/p&gt;&#xA;" OwnerUserId="2641" LastActivityDate="2010-10-28T19:58:26.160" />
  <row Id="8212" PostTypeId="2" ParentId="8207" CreationDate="2010-10-28T20:19:08.007" Score="1" ViewCount="" Body="&lt;p&gt;Let's say we have a finite tuple (not set) of positive integers, $A = (a_1,a_2,\dotsb,a_n)$, with $\sum a_i = s$, $\prod a_i = p$. Consider the simplest prime factorization $p = p_1^{n_1} \dotsb p_k^{n_k}$ and construct $B = (1,\dotsb,1,p_1,\dotsb,p_1,p_2,\dotsb,p_k,\dotsb,p_k)$ where each $p_i$ occurs $n_i$ times and we fill out with enough $1$s to make $\sum B = s$ (why can we do this?).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proving that this produces a $B$ with different elements than $A$ in &quot;most&quot; cases is left as an exercise for the reader. As is extending it to the case where $A$ may contain negative integers. In that case you want to fill up with a clever combination of $1$s and $-1$s.&lt;/p&gt;&#xA;" OwnerUserId="1269" LastActivityDate="2010-10-28T20:19:08.007" CommentCount="1" />
  <row Id="8213" PostTypeId="1" AcceptedAnswerId="8214" CreationDate="2010-10-28T22:17:39.343" Score="2" ViewCount="149" Body="&lt;p&gt;Hi all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Quick formulation of the problem:&lt;/strong&gt;&lt;br&gt;&#xA;Given two points: $(x_l, y_l)$ and $(x_u, y_u)$&lt;br&gt;&#xA;with: $x_l &amp;lt; x_u$ and $y_l &amp;lt; y_u$,&lt;br&gt;&#xA;and given lower asymptote=0 and higher asymptote=1,&#xA;what's the logistic function that passes through the two points?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Explanatory image:&lt;/strong&gt;&lt;br&gt;&#xA;&lt;img src=&quot;http://i.imgur.com/QoQix.png&quot; alt=&quot;logistic&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Other details:&lt;/strong&gt;&lt;br&gt;&#xA;I'm given two points in the form of Pareto 90/10 (green in the example above) or 80/20 (blue in the example above), and I know that the upper bound is one and the lower bound is zero.&lt;br&gt;&#xA;How do I get the formula of a sigmoid function (such as the logistic function) that has a lower asymptote on the left and higher asymptote on the right and passes via the two points?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="2916" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-10-31T08:36:38.767" LastActivityDate="2010-10-31T08:36:38.767" Title="Logistic function passing through two points?" Tags="&lt;statistics&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" />
  <row Id="8214" PostTypeId="2" ParentId="8213" CreationDate="2010-10-28T22:29:57.373" Score="7" ViewCount="" Body="&lt;p&gt;I believe you're looking for constants $a$ and $b$ so that $f(x_\ell) = y_\ell$ and $f(x_u) = y_u$ where $f(x) = \exp(a + bx) / (1 + \exp(a + bx))$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is equivalent to the linear system $a + b x_\ell = g(y_\ell)$ and $a + b x_u = g(y_u)$ where $g(y) = f^{-1}(y) = \log(y/(1-y))$.&lt;/p&gt;&#xA;" OwnerUserId="247" LastActivityDate="2010-10-28T22:29:57.373" CommentCount="1" />
  <row Id="8215" PostTypeId="2" ParentId="3590" CreationDate="2010-10-28T22:32:10.090" Score="1" ViewCount="" Body="&lt;p&gt;Since &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\zeta(z)=\frac{\Gamma (1-z) \left(2^{-z} \left(\psi \left(z-1,1\right)+\psi \left(z-1,\frac{1}{2}\right)\right)-\psi(z-1,1)\right)}{\ln(2)}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\psi(x,z)$ is the generalized polygamma following Espinosa's &lt;a href=&quot;http://www.math.tulane.edu/~vhm/papers_html/genoff.pdf&quot; rel=&quot;nofollow&quot;&gt;generalization&lt;/a&gt;, whatever we say about Zeta function we can also say about the right hand part of this identity. It consists only of Gamma function, its (fractional) derivatives and integrals.&lt;/p&gt;&#xA;" OwnerUserId="2513" LastEditorUserId="2513" LastEditorDisplayName="" LastEditDate="2010-12-05T03:25:14.210" LastActivityDate="2010-12-05T03:25:14.210" CommentCount="2" />
  <row Id="8216" PostTypeId="2" ParentId="8105" CreationDate="2010-10-28T22:33:38.210" Score="1" ViewCount="" Body="&lt;p&gt;You may be confusing Elsevier with Springer. &lt;a href=&quot;http://eom.springer.de/&quot; rel=&quot;nofollow&quot;&gt;The Springer Online Encyclopaedia of Mathematics&lt;/a&gt; is quite an impressive reference.&lt;/p&gt;&#xA;" OwnerUserId="856" LastActivityDate="2010-10-28T22:33:38.210" />
  <row Id="8217" PostTypeId="1" AcceptedAnswerId="8218" CreationDate="2010-10-28T22:53:36.287" Score="2" ViewCount="58" Body="&lt;p&gt;I am working on a game program.&#xA;I have an array of 8 points in 3d space (X,Y,Z) that are the 8 corners of a cube whose W=H=D. The 8 points are listed in no particular order. For the sake of determining this algorithm, The cube's center is at the world origin, (0,0,0) but the cube's orientation is not aligned with the Cartesian axis (it's arbitrarily rotated). I'd like to create 6 sets of 4 points each that represent the faces of the cube but I'm stumped as to how to proceed&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any idea of an algorithm that would help me? &lt;/p&gt;&#xA;" OwnerUserId="2915" LastActivityDate="2010-10-28T22:58:27.233" Title="Determine which of 8 points make up the 4 corners of a cube's face" Tags="&lt;linear-algebra&gt;&lt;linear-programming&gt;&lt;vector-space&gt;" AnswerCount="1" CommentCount="0" />
  <row Id="8218" PostTypeId="2" ParentId="8217" CreationDate="2010-10-28T22:58:27.233" Score="4" ViewCount="" Body="&lt;p&gt;Pick a point $p$, calculate the distance from that point to all the others, three will be the same length and smaller than the other lengths. Call these three points $x_1$, $x_2$, $x_3$. Do the same for $x_1$, $x_2$, $x_3$. Each pair of those will share two points, one will be $p$ the other will be $y_{12}$, $y_{13}$, $y_{23}$ respectively. You just defined three sides by points. Now, pick the only remaining point that hasn't been named. Call it $q$. Run the above algorithm. Done.&lt;/p&gt;&#xA;" OwnerUserId="16" LastActivityDate="2010-10-28T22:58:27.233" CommentCount="4" />
  <row Id="8220" PostTypeId="1" CreationDate="2010-10-28T23:44:55.900" Score="7" ViewCount="198" Body="&lt;p&gt;This was suggested by a problem on FreeNode's #math a little while ago...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Construct a directed graph $\Gamma$ with vertex set the set of compact convex sets in $\mathbb R^2$, and an arrow $A\to B$ if $A$ and $B$ are disjoint and there is a point $a$ in $A$ and a $t&gt;0$ such that $a+(t,0)$ is in $B$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I guess there are no oriented cycles in $\Gamma$, so it is a directed acyclic graph. Can you give a non-ugly proof?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Later:&lt;/strong&gt; Rahul's example below shows that this is too much to hope for. In fact, what I &lt;em&gt;really&lt;/em&gt; wanted is to know if the subgraph spanned by every finite set of disjoint convex sets has &lt;em&gt;no&lt;/em&gt; cycles.&lt;/p&gt;&#xA;" OwnerUserId="274" LastEditorUserId="274" LastEditorDisplayName="" LastEditDate="2010-10-30T14:05:49.563" LastActivityDate="2010-10-30T14:05:49.563" Title="A tree of convex sets?" Tags="&lt;convex-sets&gt;" AnswerCount="1" CommentCount="8" FavoriteCount="2" />
  <row Id="8221" PostTypeId="1" CreationDate="2010-10-29T00:02:54.290" Score="1" ViewCount="173" Body="&lt;p&gt;How does one go about solving the following quadratic congruence?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$4x^2 \equiv 2 \ (\text{mod} \ 7)$&lt;/p&gt;&#xA;" OwnerUserId="2868" LastEditorUserId="2868" LastEditorDisplayName="" LastEditDate="2010-10-29T00:24:42.700" LastActivityDate="2010-10-29T05:10:36.363" Title="Solving a quadratic congruence: $4x^2 \equiv 2 \ (\text{mod} \ 7)$" Tags="&lt;number-theory&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="8222" PostTypeId="2" ParentId="8221" CreationDate="2010-10-29T00:14:41.687" Score="1" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\ $ Multiply both sides by 2.&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-10-29T00:14:41.687" />
  <row Id="8223" PostTypeId="2" ParentId="8221" CreationDate="2010-10-29T00:28:32.990" Score="1" ViewCount="" Body="&lt;p&gt;Although not as useful in general, with a small modulus like $7$, one can let $x$ run through all possible congruence classes modulo $7$. Using Bill Dubuque's hint will make the mental calculation easier to see which $x$ actually satisfy the congruence.&lt;/p&gt;&#xA;" OwnerUserId="1583" LastActivityDate="2010-10-29T00:28:32.990" />
  <row Id="8224" PostTypeId="1" CreationDate="2010-10-29T00:36:59.367" Score="0" ViewCount="224" Body="&lt;p&gt;In a survey of political preference, 78% of those asked were in favor of at-least one of the proposals: I, II and III. 50% of those asked favored proposal I, 30% favored proposal II, and 20% favored proposal III. If 5% favored all the three proposals, what % of those asked favored more than one of the three proposals?&lt;/p&gt;&#xA;" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2010-11-28T06:58:32.987" LastActivityDate="2010-11-28T16:16:04.023" Title="A homework problem involving counting using a special case of inclusion/exclusion" Tags="&lt;combinatorics&gt;&lt;homework&gt;" AnswerCount="3" CommentCount="3" />
  <row Id="8225" PostTypeId="2" ParentId="8220" CreationDate="2010-10-29T00:39:04.527" Score="5" ViewCount="" Body="&lt;p&gt;Am I misunderstanding the conditions of the problem or is this a counterexample?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/lFHHH.png&quot; alt=&quot;Two ellipses in an X, and two circles between them&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Later:&lt;/strong&gt; I think the answer to your updated question is yes -- or rather, no, any subgraph induced by disjoint convex sets has no cycles. (However, that would make it a &lt;em&gt;directed acyclic graph&lt;/em&gt; or DAG, and not a tree, unless that's computer science terminology.) I have a sketch of what I think should be a valid proof, but filling in the details is a little ugly. Unfortunately, I think there is no completely non-ugly proof, for two reasons:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The power of convexity is a little lost because separating hyperplanes are of no help here. If $A \rightarrow B \rightarrow C$, there is not necessarily a hyperplane separating $C$ from $A \cup B$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The proposition is not valid in more than 2 dimensions, so any proof must fundamentally depend on the 2D nature of the problem.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;That said, here's the fairly ugly proof I got. Suppose for the sake of contradiction that there is a cycle of disjoint compact convex sets, $A_1 \rightarrow A_2 \rightarrow \cdots \rightarrow A_n \rightarrow A_1$. Each set $A_i$ contains a pair of points $p_i$ and $q_i$ such that $q_{i-1} \rightarrow p_i$ and $q_i \rightarrow p_{i+1}$ (where $i+1$ and $i-1$ are modulo $n$, of course). Therefore, it suffices to prove the impossibility of a cycle on the line segments $p_1 q_1 \rightarrow p_2 q_2 \rightarrow \cdots \rightarrow p_n q_n \rightarrow p_1 q_1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the rightmost &quot;fringe&quot; of the first $i$ segments, as a function of $y$; that is, let $f_{i}(y) = \sup \{x : (x,y) \in p_1 q_1 \cup p_2 q_2 \cup \cdots \cup p_i q_i\}$. While $f_i$ is not continuous, I think you can prove by induction that if $p_1 q_1 \rightarrow p_2 q_2 \rightarrow \cdots \rightarrow p_i q_i \rightarrow p_{i+1} q_{i+1}$, then the discontinuities of $f_i$ are not &quot;visible&quot; to $p_{i+1}$. What I mean is that $f_i$ jumps leftwards rather than rightwards as you move in $y$ away from $p_{i+1}$, so there is no way for $q_{i+1}$ to sneak in to the left of $f_i$ without the segment $p_{i+1}q_{i+1}$ intersecting $f_i$ somewhere. Therefore, every $p_{i+1}$ and $q_{i+1}$ is to the right of $f_i$. In particular, $q_n$ is to the right of $f_{n-1}$. But $p_1$ is on or to the left of $f_{n-1}$, so $q_n \rightarrow p_1$ is impossible.&lt;/p&gt;&#xA;" OwnerUserId="856" LastEditorUserId="856" LastEditorDisplayName="" LastEditDate="2010-10-30T07:28:44.270" LastActivityDate="2010-10-30T07:28:44.270" CommentCount="3" />
  <row Id="8226" PostTypeId="1" CreationDate="2010-10-29T00:53:43.227" Score="0" ViewCount="144" Body="&lt;p&gt;The question reads:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $\mathfrak{p}$ be a prime ideal of $A$. Show that $A_\mathfrak{p}$ has a unique maximal ideal, consisting of all elements $a/s$ with $a \in \mathfrak{p}$ and $s \notin \mathfrak{p}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If &quot;$A_\mathfrak{p}$&quot; is $A/\mathfrak{p}$, then this is equivalent to saying $A$ is entire $\Rightarrow A$ has a unique maximal ideal, which is false, no? (consider $\mathbb{Z}$).&lt;/p&gt;&#xA;" OwnerUserId="654" LastActivityDate="2010-10-29T00:53:43.227" Title="What does this notation mean, in Lang's Algebra Ch2 Problem3?" Tags="&lt;abstract-algebra&gt;&lt;notation&gt;&lt;books&gt;" CommentCount="2" />
  <row Id="8227" PostTypeId="2" ParentId="6228" CreationDate="2010-10-29T01:05:57.240" Score="1" ViewCount="" Body="&lt;p&gt;Following the advice of Weltschmerz, from 'The Classical Theory of Fields':&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;blockquote&gt;&#xA;    &lt;p&gt;The total charge present in all of space is equal to the integral $\int \rho dV$ over all space. We can write this integral in four-dimensional form:&lt;/p&gt;&#xA;  &lt;/blockquote&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Then, you have the equality you wrote: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle \int{\rho}dV = \displaystyle \frac{1}{c} \int{j^0}dV = \frac{1}{c} \int{j^i}dS_i$$&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;blockquote&gt;&#xA;    &lt;p&gt;where the integral is taken over the entire four-dimensional hyperplane perpendicular to the $x^{0}$ axis (clearly this integration means integration over the whole three dimensional space). Generally, the integral &lt;/p&gt;&#xA;  &lt;/blockquote&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle \frac{1}{c}\int j^{i}dS_{i}$$&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;blockquote&gt;&#xA;    &lt;p&gt;over an arbitrary hypersurface is the sum of the charges whose world lines pass through this surface. &lt;/p&gt;&#xA;  &lt;/blockquote&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Then, $dS_{i}$ is not a surface in the conventional sense. Instead, is a hyper-surface. &lt;/p&gt;&#xA;" OwnerUserId="2569" LastEditorUserId="2569" LastEditorDisplayName="" LastEditDate="2010-10-29T01:14:36.893" LastActivityDate="2010-10-29T01:14:36.893" />
  <row Id="8229" PostTypeId="2" ParentId="8140" CreationDate="2010-10-29T01:09:52.947" Score="3" ViewCount="" Body="&lt;p&gt;Consider your graph is planarly embedded in the plane. And think of edge unoriented edge as a pair of oriented arcs in opposite directions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pick one oriented arc $a$. There is a unique face whose boundary, when drawn in counterclockwise direction, &quot;contains&quot; $a$, and in that boundary $a$ is followed by a well-defined arc $b$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To construct the faces, it is enough to be able to decide for each $a$ what  the corresponding $b$ is. This can be done easily by sorting the edges incident to each vertex cyclically.&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2010-10-29T01:09:52.947" CommentCount="2" />
  <row Id="8230" PostTypeId="1" AcceptedAnswerId="8238" CreationDate="2010-10-29T01:17:43.783" Score="1" ViewCount="66" Body="&lt;p&gt;I'd like to get the exact value of the following sum:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sum_{i=0}^{\lceil \frac{k}{2} \rceil}{({k \choose i}\cdot i)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd also like to know the asymptotic limits of the function.&lt;/p&gt;&#xA;" OwnerUserId="2626" LastActivityDate="2010-10-29T04:13:44.660" Title="What is the sum of half of the choices (weighted)?" Tags="&lt;combinatorics&gt;" AnswerCount="1" CommentCount="8" FavoriteCount="2" />
  <row Id="8231" PostTypeId="2" ParentId="8224" CreationDate="2010-10-29T01:19:21.417" Score="0" ViewCount="" Body="&lt;p&gt;hmm, I got 17, not 27...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we count |A|+|B|+|C|, then we overcount intersections of any 2 sets twice, and intersections of all three 3 times. Then, if we subtract $|A \cup B \cup C|$, we get the intersections of any 2 sets once and 3 sets 2 times. (most easily seen by drawing a Venn diagram). Subtracting the 3-set intersection once more gives 100-78-5=17 .&lt;/p&gt;&#xA;" OwnerUserId="654" LastActivityDate="2010-10-29T01:19:21.417" />
  <row Id="8233" PostTypeId="1" CreationDate="2010-10-29T01:30:28.827" Score="1" ViewCount="68" Body="&lt;p&gt;why does this algorithm fail to factor polynomials over Z/2Z, what would I need to do to fix this? &lt;/p&gt;&#xA;" OwnerUserId="2143" LastActivityDate="2010-10-29T01:30:28.827" Title="Cantor-Zassenhaus split" Tags="&lt;number-theory&gt;&lt;number&gt;" />
  <row Id="8234" PostTypeId="1" AcceptedAnswerId="8320" CreationDate="2010-10-29T02:26:27.343" Score="1" ViewCount="170" Body="&lt;p&gt;While reading Munkres' Topology section about integers and reals (Chapter 1, Section 4), he defines the set $\mathbb{Z}_+$ as:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Definition: A subset $A$ of the real&#xA;  numbers is said to be inductive if it&#xA;  contains the number 1, and if for&#xA;  every $x$ in $A$, the number $x+1$ is&#xA;  also in $A$. Let $\cal{A}$ be the&#xA;  collection of all inductive subsets of&#xA;  $\mathbb{R}$. Then the set&#xA;  $\mathbb{Z}_+$ of positive integers is&#xA;  defined by the equation $\bigcap _{A&#xA;  \in \cal{A}} A$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Isn't it possible to define $\mathbb{Z}_+$ as the set defined this way:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$1\in B \wedge (j \in B, n-1 = j) \Rightarrow n\in B$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then $B = \mathbb{Z}_+$&lt;/p&gt;&#xA;" OwnerUserId="1523" LastActivityDate="2010-10-29T22:52:41.660" Title="Defining $\mathbb{Z}_+$" Tags="&lt;set-theory&gt;" AnswerCount="1" CommentCount="6" />
  <row Id="8235" PostTypeId="1" AcceptedAnswerId="8574" CreationDate="2010-10-29T02:29:35.393" Score="9" ViewCount="338" Body="&lt;p&gt;Inspired by &#xA;&lt;a href=&quot;http://math.stackexchange.com/questions/8186/when-do-the-multiples-of-two-primes-span-all-large-enough-natural-numbers/8187#8187&quot;&gt; this question&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $A$ be a subset of ${\mathbb Z}^d$ that generates the whole additive group ${\mathbb Z}^d$, and let $S$ be the additive semigroup generated by $A$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Prove that there is a $d$-dimensional convex cone $C\subset {\mathbb R}^d$ so that&#xA;all elements of $C \cap {\mathbb Z}^d$ with sufficiently large norm are contained in $S$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S. Feel free to retag!&lt;/p&gt;&#xA;" OwnerUserId="940" LastActivityDate="2010-11-04T11:07:01.237" Title="A multi-dimensional Frobenius problem" Tags="&lt;discrete-geometry&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="8236" PostTypeId="2" ParentId="4054" CreationDate="2010-10-29T03:11:06.173" Score="2" ViewCount="" Body="&lt;p&gt;See Needham's &lt;a href=&quot;http://usf.usfca.edu/vca/&quot; rel=&quot;nofollow&quot;&gt;Visual Complex Analysis&lt;/a&gt; for lots of geometric intuition and much more.&lt;/p&gt;&#xA;" OwnerUserId="589" LastActivityDate="2010-10-29T03:11:06.173" />
  <row Id="8237" PostTypeId="2" ParentId="8221" CreationDate="2010-10-29T03:21:15.660" Score="0" ViewCount="" Body="&lt;p&gt;Since $2 \equiv 9 \pmod 7$, you have $(2x-3)(2x+3) = 4x^2-9 \equiv 0 \pmod 7$. Now use that $7$ is prime.&lt;/p&gt;&#xA;" OwnerUserId="589" LastEditorUserId="589" LastEditorDisplayName="" LastEditDate="2010-10-29T05:10:36.363" LastActivityDate="2010-10-29T05:10:36.363" CommentCount="2" />
  <row Id="8238" PostTypeId="2" ParentId="8230" CreationDate="2010-10-29T04:13:44.660" Score="3" ViewCount="" Body="&lt;p&gt;Use $\binom{k}{i} i = k \binom{k-1}{i-1}$ to get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sum_{i=0}^{\lceil k/2 \rceil} \binom{k}{i} = k\sum_{j=0}^{\lceil k/2 \rceil - 1} \binom{k-1}{j}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $k = 2m$ is even then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$k\sum_{j=0}^{m-1} \binom{2m-1}{j} = k 2^{2m-2} = k2^{k-2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $k = 2m+1$ is odd then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$k\sum_{j=0}^m \binom{2m}{j} = k \left(2^{2m-1} +\frac{1}{2} \binom{2m}{m}\right) = k\left(2^{k-2} + \frac{1}{2} \binom{k-1}{(k-1)/2}\right)$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the asymptotics, $\frac{k}{2} \binom{k-1}{(k-1)/2} \approx 2^{k-1} \sqrt{\frac{k}{2\pi}}$.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-10-29T04:13:44.660" CommentCount="1" />
  <row Id="8239" PostTypeId="1" AcceptedAnswerId="8251" CreationDate="2010-10-29T04:20:10.690" Score="2" ViewCount="249" Body="&lt;p&gt;This is a problem I'm stuck on that our professor gave us for additional practice (not homework, but its recommended that we understand how to prove it). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know X and Y are complete metric spaces, and we need to show that $X \times Y$ is complete. I'm really lost on the proof technique. We were given an outline as follows, but I could only fully figure out (1). Part 3 is what we've been really stuck on though.  I was wondering whether someone could give an proof for say a more specific space where $X = \mathbb{R}, Y = \mathbb{R}$, so I could understand the principle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Outlined: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Show that $d_{X \times Y} ( (a_1,b_1) , (a_2,b_2)) = \max \left\{ d_X (a_1,a_2) , d_Y (b_1, b_2) \right\}$ is the product metric. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Prove that this gives the product topology on $X \times Y$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) Prove that  if  $a_n, b_n$ are Cauchy sequences, where $a_n \in X$ and $b_n \in Y$, then $(a_n,b_n )$ is Cauchy.&lt;/p&gt;&#xA;" OwnerUserId="2919" LastEditorUserId="664" LastEditorDisplayName="" LastEditDate="2010-10-29T08:42:11.853" LastActivityDate="2010-10-29T08:55:23.043" Title="Cartesian Product of Two Complete Metric Spaces is Complete" Tags="&lt;real-analysis&gt;&lt;analysis&gt;&lt;topology&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="2" />
  <row Id="8240" PostTypeId="2" ParentId="8239" CreationDate="2010-10-29T04:33:06.273" Score="2" ViewCount="" Body="&lt;p&gt;For 3) if you are given an $\epsilon$, you can find $N$ such that $|a_n-a_m|&amp;lt;\epsilon \text{ if } n,m&gt;N$ and similarly for $M$ and the $b$ series.  Doesn't $\max{(N,M)}$ give a limit that $|(a_n,b_n)-(a_m,b_m)|&amp;lt;\epsilon$?&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-10-29T04:33:06.273" />
  <row Id="8241" PostTypeId="1" AcceptedAnswerId="8246" CreationDate="2010-10-29T04:34:11.173" Score="5" ViewCount="195" Body="&lt;p&gt;Given are some positive integers $t_1, ..., t_n$ such that $\mathrm{gcd}(t_1, ..., t_n) = 1$. Now we create a linear combination of these: $k_1 t_1 + ... + k_n t_n$ where $k_1, ..., k_n$ are nonnegative integers that you can choose to generate different integers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the smallest $N$ such that all integers greater than or equal to $N$ can be generated by the linear combination?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example: $k_1 3 + k_2 4$ can generate $0, 3, 4, 6, 7, 8, 9, ...$. So here $N = 6$ and $k_1 = 2, k_2 = 0$.&lt;/p&gt;&#xA;" OwnerUserId="2649" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-10-29T15:49:17.480" LastActivityDate="2010-10-29T15:49:17.480" Title="Generating integers from a linear combination of integers" Tags="&lt;number-theory&gt;&lt;diophantine-equations&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="8242" PostTypeId="2" ParentId="8239" CreationDate="2010-10-29T04:34:37.210" Score="2" ViewCount="" Body="&lt;p&gt;If ${a_n}$, ${b_n}$ are Cauchy, then for some N, $d_X(a_n,a_m) &amp;lt; \epsilon$, $d_Y(b_n,b_m) &amp;lt; \epsilon$ when $n,m \geq N$. Because of this, $d_{X \times Y} ((a_n,b_n),(a_m,b_m)) &amp;lt; \epsilon$ for $n,m \geq N$. &lt;/p&gt;&#xA;" OwnerUserId="1736" LastActivityDate="2010-10-29T04:34:37.210" />
  <row Id="8243" PostTypeId="1" AcceptedAnswerId="8245" CreationDate="2010-10-29T04:43:20.223" Score="1" ViewCount="243" Body="&lt;p&gt;If $\gcd(a,b) = 1$, $x \equiv 0 \pmod{a}$, $x \equiv 0 \pmod{b}$ then prove that $x \equiv 0 \pmod{ab}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My attempt at answering the question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align*}&#xA;x &amp;amp;\equiv 0 \pmod{a}\\&#xA;&amp;amp;\Longrightarrow x\text{ is divisible by $a$}\\&#xA;&amp;amp;\Longrightarrow x = ma\text{ for some integer $m$}\\&#xA;\ \\&#xA;x &amp;amp;\equiv 0 \pmod{b}\\&#xA;&amp;amp;\Longrightarrow  x\text{ is divisible by $b$}\\&#xA;&amp;amp;\Longrightarrow x = mb\text{ for some integer $m$}\\&#xA;\ \\&#xA;x^2 &amp;amp;= (ma)(mb)\\&#xA;x^2 &amp;amp;= (m^2)(ab)\\&#xA;x &amp;amp;= \sqrt{m^2ab}\\&#xA;x &amp;amp;= m\sqrt{a}\sqrt{b}&#xA;\end{align*}&#xA;Let $m$ be $k\sqrt{a}\sqrt{b}$. Then&#xA;\begin{align*}&#xA;x &amp;amp;= kab\\&#xA;&amp;amp;\Longrightarrow x \equiv 0 \pmod{ab}&#xA;\end{align*}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this correct, if not can someone point me in the right direction?&lt;/p&gt;&#xA;" OwnerUserId="2588" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-04T13:12:34.463" LastActivityDate="2010-12-04T13:12:34.463" Title="If $\gcd(a,b) = 1$, $x \equiv 0 \pmod{a}$, $x \equiv 0 \pmod{b}$ then prove that $x \equiv 0 \pmod{ab}$" Tags="&lt;number-theory&gt;" AnswerCount="4" CommentCount="1" />
  <row Id="8244" PostTypeId="2" ParentId="8243" CreationDate="2010-10-29T04:52:38.480" Score="1" ViewCount="" Body="&lt;p&gt;No. This is not correct.&lt;br&gt;&#xA;1. Integers $m_1$ and $m_2$ in $x=m_1 a$ and $x=m_2 b$ can be different.&lt;br&gt;&#xA;2. In the beginning of your proof all numbers are integers. However when you write &quot;let m be k*sqrt(a)*sqrt(b)&quot;, you don't know, that k is integer too.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Right direction: use, the fact, that each number $y$ can be uniquely represented as $p_1^{\alpha_1} p_2^{\alpha_2}\dots p_n^{\alpha_n}$, where $p_i$ are primes and $\alpha_i$ are integers.&lt;/p&gt;&#xA;" OwnerUserId="695" LastActivityDate="2010-10-29T04:52:38.480" CommentCount="1" />
  <row Id="8245" PostTypeId="2" ParentId="8243" CreationDate="2010-10-29T04:57:40.680" Score="7" ViewCount="" Body="&lt;p&gt;You are given that $a$ divides $x$. Therefore, you can write $x = ma$ where m is an integer. You are also given that $b$ divides $x$. This implies that $b$ divides $ma$. But $b$ and $a$ are coprime. Therefore $b$ must divide $m$. So you can write $m = kb$ where $k$ is another integer. Therfore, you have $x = kab$.&lt;/p&gt;&#xA;" OwnerUserId="2641" LastActivityDate="2010-10-29T04:57:40.680" CommentCount="1" />
  <row Id="8246" PostTypeId="2" ParentId="8241" CreationDate="2010-10-29T05:07:16.700" Score="5" ViewCount="" Body="&lt;p&gt;This is a famous problem, sometimes called the coin problem of Frobenius. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $n=2$ the answer to your question is known to be $N = t_1 t_2 - (t_1 + t_2) + 1$.  A proof of this can be found in the answer to this &lt;a href=&quot;http://math.stackexchange.com/questions/8186/&quot;&gt;recent question&lt;/a&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For three or more integers, there is no known closed-form solution for $N$.  There are some bounds on the values of $N$ in the $n = 3$ case, as well as some algorithms for determining $N$.  For more information and references, see the &lt;a href=&quot;http://en.wikipedia.org/wiki/Coin_problem&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt; and &lt;a href=&quot;http://mathworld.wolfram.com/CoinProblem.html&quot; rel=&quot;nofollow&quot;&gt;MathWorld&lt;/a&gt; pages on the Frobenius coin problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically, the problem is considered solved when $n = 2$, partially solved (because of the bounds and algorithms) when $n = 3$, and unsolved when $n \geq 4$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Update: Guy's &lt;em&gt;Unsolved Problems in Number Theory&lt;/em&gt; says, &quot;The case $n = 3$ was first solved explicitly by Selmer and Beyer, using a continued fraction algorithm.&quot;  So I guess the $n=3$ case has been solved.  I suppose you have would have to dig up their paper (it's in the MathWorld references) to see their solution. &lt;/p&gt;&#xA;" OwnerUserId="2370" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-10-29T15:46:37.763" LastActivityDate="2010-10-29T15:46:37.763" CommentCount="4" />
  <row Id="8247" PostTypeId="1" AcceptedAnswerId="29081" CreationDate="2010-10-29T05:19:58.367" Score="1" ViewCount="193" Body="&lt;p&gt;&lt;strong&gt;Suppose n is an integer, such that the sum of the digits of n is 2, and $10^{10} \lt n \lt 10^{11} $. The number of different values for n is:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me try to list them :&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;(1) 11000000000&#xA;(2) 10100000000&#xA;(3) 10010000000&#xA;...&#xA;(10)10000000001&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;So I am getting 10 possible values but the answer is 11.What  am I missing here ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: I was missing 20000000000.&lt;/p&gt;&#xA;" OwnerUserId="2109" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2011-03-26T15:34:05.007" LastActivityDate="2011-03-26T15:34:05.007" Title="Counting eleven digit  integers with the sum of the digits 2" Tags="&lt;combinatorics&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="8248" PostTypeId="1" CreationDate="2010-10-29T07:02:12.220" Score="0" ViewCount="233" Body="&lt;p&gt;By listing the first six prime numbers: 2, 3, 5, 7, 11, and 13, we can see that the 6th prime is 13.  What is the 10001st prime number?&lt;/p&gt;&#xA;" OwnerUserId="2784" LastActivityDate="2010-10-29T09:43:42.307" Title="How to find next prime number?" Tags="&lt;prime-numbers&gt;&lt;arithmetic&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="1" />
  <row Id="8249" PostTypeId="1" CreationDate="2010-10-29T07:15:37.680" Score="2" ViewCount="76" Body="&lt;p&gt;Define $f$ on $[-1,1]$ by &#xA;$$f(x) = \left\{ \begin{array}{ll}&#xA;0 &amp;amp; \mbox{if $x\lt 0$;}\\&#xA;0 &amp;amp; \mbox{if $x=0$;}\\&#xA;1 &amp;amp; \mbox{if $x\gt 0$.}&#xA;\end{array}\right.$$&#xA;Let the integrator $a$ be defined by &#xA;$$a(x) = \left\{\begin{array}{ll}&#xA;0 &amp;amp; \mbox{if $x\lt 0$;}\\&#xA;0 &amp;amp; \mbox{if $x=0$;}\\&#xA;1 &amp;amp; \mbox{if $x\gt 0$.}&#xA;\end{array}\right.$$&#xA;Show that $f$ is Riemann-Stieltjes integrable on  $[-1,1]$ even though &#xA;$$\lim_{||P||\to 0} S(P,f,a)$$&#xA;does not exist.&lt;/p&gt;&#xA;" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-10-29T16:30:45.890" LastActivityDate="2010-10-29T18:49:51.933" Title="to show a function is RS integrable even when lim S(P,f,a) does not exist" Tags="&lt;real-analysis&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="8250" PostTypeId="1" CreationDate="2010-10-29T07:47:18.270" Score="10" ViewCount="499" Body="&lt;p&gt;Here is a statement of &lt;a href=&quot;http://en.wikipedia.org/wiki/Farkas_lemma&quot; rel=&quot;nofollow&quot;&gt;Farkas Lemma&lt;/a&gt; from the Wikipedia. Let $A$ be an $m \times n$ matrix and $b$ an $m$-dimensional vector. Then, exactly one of the following two statements is true:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;There exists an $x \in \mathbb{R}^n$ such that $Ax = b$ and $x \geq 0$.&lt;/li&gt;&#xA;&lt;li&gt;There exists a $y \in \mathbb{R}^m$ such that $A^T y \geq 0$ and $b^T y &amp;lt; 0$.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;This result has a simple geometric interpretation: either $b$ lies in the cone formed by the columns of $A$ or it is possible to find a vector $y$ such that $y$ forms an acute angle with all columns of $A$ and an obtuse angle with $b$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I was wondering if there is a way to make the result intuitive purely at the algebraic level of solving linear equations and inequalities?&lt;/em&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The lemma is important in finance and the geometric intuition is not much help there since the vectors are payoffs and prices of financial assets which have no natural geometric meaning. Gale's &lt;em&gt;Theory of Linear Economic Models&lt;/em&gt; has a purely algebraic proof, but that is an opaque induction argument.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: A little more about my application. We have $m$ assets and $n$ possible future states of nature. $A_{ij}$ is the payoff by asset $i$ in state $j$. $b_i$ is the price of asset $i$. $x_j$ is the value of one dollar in state $j$. $y_i$ is the amount of asset $i$ in a portfolio. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So Farkas' lemma tells us that either (1) there is a way of assigning a non-negative price to a dollar in each state in a way such that the price of each asset is just the sum total of the value of its payoffs, or (2) there is a portfolio whose price is negative, so you get paid for holding it, but whose payoffs are non-negative, which means that you do not have to pay anything back. I want to understand, in terms which make economic sense, why (1) and (2) should be mutually exclusive. Acute and obtuse angles are no help here.&lt;/p&gt;&#xA;" OwnerUserId="1195" LastEditorUserId="1195" LastEditorDisplayName="" LastEditDate="2010-10-29T18:12:43.890" LastActivityDate="2010-11-05T22:18:18.113" Title="Farkas’ lemma: purely algebraic intuition" Tags="&lt;convex-analysis&gt;&lt;finance&gt;" AnswerCount="2" CommentCount="7" FavoriteCount="4" />
  <row Id="8251" PostTypeId="2" ParentId="8239" CreationDate="2010-10-29T08:55:23.043" Score="2" ViewCount="" Body="&lt;p&gt;As for (2), it suffices to prove that the balls produced by the distance $d_{X\times Y}$ are a basis for the product topology. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So let's write down what it is a ball for the distance $d_{X\times Y}$ with center $(a,b)$ and radius $\varepsilon &gt; 0$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;B_{X\times Y}((a,b); \varepsilon) = \left\{ (x,y) \in X\times Y \ \vert \ d_{X\times Y} ((a,b), (x,y))= \max  \left\{ d_X(a,x) , d_Y (b,y) \right\}   &amp;lt; \varepsilon \right\} \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\max  \left\{ d_X(a,x) , d_Y (b,y) \right\}   &amp;lt; \varepsilon \quad \Longleftrightarrow \quad d_X(a,x) &amp;lt;\varepsilon \ \text{and} \ d_Y(b,y) &amp;lt; \varepsilon \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence we see that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;B_{X\times Y }((a,b); \varepsilon)\  = \ B_X (a;\varepsilon ) \times B_Y (b; \varepsilon ) \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is: a ball for the distance $d_{X \times Y}$ is the same as a product of balls for the distances $d_X$ and $d_Y$. Which means that balls for the distance $d_{X\times Y}$ form a basis for the product topology.&lt;/p&gt;&#xA;" OwnerUserId="664" LastActivityDate="2010-10-29T08:55:23.043" />
  <row Id="8252" PostTypeId="2" ParentId="8248" CreationDate="2010-10-29T08:58:40.983" Score="2" ViewCount="" Body="&lt;p&gt;It is 104743, according to &lt;a href=&quot;http://primes.utm.edu/lists/small/100000.txt&quot; rel=&quot;nofollow&quot;&gt;http://primes.utm.edu/lists/small/100000.txt&lt;/a&gt; and the related list of the first 10,000 primes ;-)&lt;/p&gt;&#xA;" OwnerUserId="689" LastActivityDate="2010-10-29T08:58:40.983" CommentCount="1" />
  <row Id="8253" PostTypeId="1" AcceptedAnswerId="8304" CreationDate="2010-10-29T09:05:35.963" Score="0" ViewCount="337" Body="&lt;p&gt;In the proof of Rouche's theorem I saw in a book, there are two points I failed to understand, or failed to prove myself. (if you aren't familiar with the theorem, please try to look at the two statements here below and explain them regardless of the theorem):&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;&quot;Consider the complex plane. It's not difficult to show that the index of a curve C with respect to a vector field v is equal to the sum of indices of singular points, i.e., those at which v(z) = 0.&quot;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&quot;If f(z) is a polynomial, and let v(z) = f(z), where v is a vector field, then the index of the singular point z0 is equal to the multiplicity of the root z0 of f.&quot;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I am having trouble understanding this, can anyone please help?&lt;/p&gt;&#xA;" OwnerUserId="1214" LastEditorUserId="1214" LastEditorDisplayName="" LastEditDate="2010-10-29T09:19:01.707" LastActivityDate="2010-10-29T18:56:59.080" Title="Polynomials, Rouche's theorem and index of vector fields" Tags="&lt;complex-analysis&gt;&lt;algebra&gt;&lt;polynomials&gt;&lt;plane-curves&gt;&lt;roots&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="8255" PostTypeId="2" ParentId="8248" CreationDate="2010-10-29T09:43:42.307" Score="1" ViewCount="" Body="&lt;p&gt;I assume you want to program this? Then the question belongs in a CS forum. However, I will try to motivate how to program this using properties of prime numbers:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) For a number you &lt;em&gt;could&lt;/em&gt; try to divide it by every single positive integer smaller than the respective number (except for 1). This is stupid, as you obviously could just search through the odd numbers (checking 2 first). But can you do even better?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Do you have to stop at n-1, or can you stop way earlier? You can. Find out where.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now count primes.&lt;/p&gt;&#xA;" OwnerUserId="2900" LastActivityDate="2010-10-29T09:43:42.307" CommentCount="3" />
  <row Id="8256" PostTypeId="1" CreationDate="2010-10-29T09:48:24.633" Score="2" ViewCount="158" Body="&lt;p&gt;I was recently asked to help someone with the following question on their first year analysis course.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Recall that  $\mathbb{N}$ is the set of all positive integers. Use the principle of induction to show that $n \ge 1$ for all $ n \in \mathbb{N}.$ [ Hint: Let $ S = \lbrace n \in \mathbb{N} | n \ge 1 \rbrace.$ ]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am concerned that the first sentence of the question assumes that we already know exactly what the positive integers are, which renders the rest of the question redundant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would prefer to see the question written:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Recall that $\mathbb{N}$ is the set of $ x \in \mathbb{R}$ such that $x$ is a member of every inductive set. [ $P$ is an inductive set if (a) $ 1 \in P $ and (b) $ x \in P \Rightarrow x+1 \in P.$ ] Then we can answer as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With $S$ as in the hint, by definition $ S \subseteq \mathbb{N}.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$1 \in S$ since $1 \in \mathbb{N},$ as it is an inductive set.&#xA;For $ n \in S, $ $n+1 &gt; n \ge 1,$ therefore $n+1 \in S.$&#xA;Hence $S$ is an inductive set, and so  $ \mathbb{N} \subseteq S.$ Therefore $ S=\mathbb{N}.$&#xA;Therefore $1$ is the least element of $\mathbb{N}.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Am I being way too picky, or is the question flawed as it stands?&lt;/p&gt;&#xA;" OwnerUserId="1301" LastEditorUserId="1301" LastEditorDisplayName="" LastEditDate="2010-10-29T13:00:46.093" LastActivityDate="2010-10-29T18:09:14.360" Title="Is this analysis problem involving induction flawed?" Tags="&lt;analysis&gt;&lt;induction&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="8257" PostTypeId="1" CreationDate="2010-10-29T10:23:16.027" Score="0" ViewCount="139" Body="&lt;p&gt;&lt;strong&gt;How many even positive integers are there that are divisors of 720 ?&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know how to compute the number of divisors but how to compute the number of &lt;strong&gt;even or odd positive divisors of a number&lt;/strong&gt; ? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we list the divisors of 720 (using mathematica) : &lt;strong&gt;{1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, 16, 18, 20, 24, 30, 36, 40, 45,48, 60, 72, 80, 90, 120, 144, 180, 240, 360, 720}&lt;/strong&gt; among these only &lt;strong&gt;24&lt;/strong&gt; are even,I am looking for some tricks that can be used in solving similar kinds of problems during exam (under a minute solution).&lt;/p&gt;&#xA;" OwnerUserId="3744" LastActivityDate="2010-10-29T10:58:37.557" Title="How many even positive integers are there that are divisors of 720 ?" Tags="&lt;number-theory&gt;" AnswerCount="3" CommentCount="4" />
  <row Id="8258" PostTypeId="2" ParentId="8257" CreationDate="2010-10-29T10:28:28.007" Score="1" ViewCount="" Body="&lt;p&gt;You have an even number given, so just cut it in half and count the divisors of this.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-10-29T10:28:28.007" CommentCount="4" />
  <row Id="8259" PostTypeId="1" AcceptedAnswerId="8276" CreationDate="2010-10-29T10:39:05.123" Score="0" ViewCount="113" Body="&lt;p&gt;Hi everybody!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please can you explain me why&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \nabla^2 G_{\omega}(R) = f(R) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is equivalent to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \frac{1}{R} \frac{d^2(RG_\omega(R))}{d R^2} = f(R) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$G_\omega(R)$ is the Fourier transform of $G$&lt;/li&gt;&#xA;&lt;li&gt;$G$ is the Green's function&lt;/li&gt;&#xA;&lt;li&gt;We suppose that $G$ and $f$ are spherically symmetric&lt;/li&gt;&#xA;&lt;li&gt;$R = \sqrt{x^2 + y^2 + z^2}$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Thanks! &lt;/p&gt;&#xA;" OwnerUserId="2354" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-10-29T10:46:56.343" LastActivityDate="2010-10-29T13:15:27.627" Title="Laplacian and spherical symmetry" Tags="&lt;differential-equations&gt;" AnswerCount="1" />
  <row Id="8260" PostTypeId="2" ParentId="8257" CreationDate="2010-10-29T10:39:52.423" Score="0" ViewCount="" Body="&lt;p&gt;1*2*2*2*2*3*3*5. Odd divisors: (1,3,3,5) =&gt; (1,3,5,9,15,45). &#xA;Let A = (1,3,5,9,15,45). Then all factors are, A, 2*A, 4*A, 8*A, 16*A. &lt;/p&gt;&#xA;" OwnerUserId="2900" LastActivityDate="2010-10-29T10:39:52.423" CommentCount="1" />
  <row Id="8261" PostTypeId="2" ParentId="4054" CreationDate="2010-10-29T10:50:27.307" Score="2" ViewCount="" Body="&lt;p&gt;I don't have enough reputation to add a comment, but this is just a minor add-on to KCd's answer. He points out that the whole thing reduces to computing integrals of the form $\int z^m dz$, where &lt;em&gt;m&lt;/em&gt; is an integer. Certainly one can compute these integrals by polar coordinates, but the following way to think about this computation is in my opinion closer to geometric intuition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Namely, when $m \neq -1$ the function $z^m$ has a single-valued antiderivative  $z^{m+1}/(m+1)$ on $\mathbb C \setminus \{0\}$, and so the integral of $z^m$ over &lt;em&gt;any&lt;/em&gt; closed loop in $\mathbb C \setminus \{0\}$ vanishes, even though the punctured plane is not simply connected. The only &lt;em&gt;m&lt;/em&gt; for which one fails to get a single-valued antiderivative is $m=-1$, in which case one gets the multi-valued function $\log z$. The value of $\log z$ for two ''adjacent'' branches always differs by exactly $2\pi i$, so the integral of $1/z$ over a closed loop in $\mathbb C \setminus \{0\}$ just gives you $2\pi i$ times the winding number of the loop around the origin, i.e. the total number of times it has ''jumped'' to another branch in the counterclockwise direction. In the assumptions of your theorem the boundary &lt;em&gt;C&lt;/em&gt; will traverse the singularity once counterclockwise, which gives the result.&lt;/p&gt;&#xA;" OwnerUserId="677" LastActivityDate="2010-10-29T10:50:27.307" />
  <row Id="8262" PostTypeId="2" ParentId="8257" CreationDate="2010-10-29T10:58:37.557" Score="2" ViewCount="" Body="&lt;p&gt;An even number $2m$ is a factor of $720$ iff $m$ is a factor of $360$.&#xA;So it's the same problem as counting divisors of $360$. For&#xA;that it helps to consider the prime factorization of $360$.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-10-29T10:58:37.557" />
  <row Id="8263" PostTypeId="1" CreationDate="2010-10-29T11:01:17.350" Score="0" ViewCount="162" Body="&lt;p&gt;1.What should be added to $15^{81}$ to make it divisible by 7 ? Suggest a general technique to counter problems like this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2.For how many two-digit numbers, the sum of the digits is greater than the product of the digits?&lt;/p&gt;&#xA;" OwnerUserId="3744" LastEditorUserId="3744" LastEditorDisplayName="" LastEditDate="2010-10-29T11:07:28.390" LastActivityDate="2010-12-22T14:44:13.733" Title="Some problems in Number theory" Tags="&lt;number-theory&gt;" AnswerCount="3" CommentCount="5" FavoriteCount="1" />
  <row Id="8264" PostTypeId="1" AcceptedAnswerId="8271" CreationDate="2010-10-29T11:06:31.703" Score="0" ViewCount="119" Body="&lt;p&gt;I'm a student and I'm struggling to understand the basics of discrete maths. If such a noob question wont offend you, please help me understand the equation and how to prepare a truth table for this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$p \rightarrow (\neg q \lor \neg r) \land \neg p \land q$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you use any substitution or such, explain how you derived them too .&lt;/p&gt;&#xA;" OwnerUserId="2925" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-10-29T11:15:26.450" LastActivityDate="2010-10-29T12:46:04.857" Title="Provide truth table" Tags="&lt;logic&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="8265" PostTypeId="1" AcceptedAnswerId="8267" CreationDate="2010-10-29T11:15:11.473" Score="1" ViewCount="122" Body="&lt;p&gt;Hi, I have the following problem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Show that no uncountable subset of $P(\omega)$ is well-ordered by the inclusion relation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think they want me to do by embedding it in a separable complete dense linear order (like $\mathbb{R}$), which I have already proved cannot have any uncountable well-ordered subset.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any hints or ideas are welcome.&lt;/p&gt;&#xA;" OwnerUserId="2926" LastEditorUserId="664" LastEditorDisplayName="" LastEditDate="2010-10-29T11:27:06.643" LastActivityDate="2010-10-29T11:53:06.993" Title="Showing that well-ordered subsets of $P(\omega)$ are countable" Tags="&lt;set-theory&gt;&lt;order-theory&gt;" AnswerCount="2" />
  <row Id="8266" PostTypeId="2" ParentId="8263" CreationDate="2010-10-29T11:20:22.480" Score="0" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT:&lt;/strong&gt; For the first one use the fact that $15^{2} \equiv 1 \ (\text{mod} \ 7)$&lt;/p&gt;&#xA;" LastActivityDate="2010-10-29T11:20:22.480" CommentCount="7" />
  <row Id="8267" PostTypeId="2" ParentId="8265" CreationDate="2010-10-29T11:36:34.403" Score="2" ViewCount="" Body="&lt;p&gt;Say $X$ is a well-ordered subset of $P(\omega)$.&#xA;For each $x \in X$ which is not the biggest element of $X$, let $y$ be the smallest amongst the elements of $X$ which are strictly bigger than $x$, and choose $f(x) \in y \setminus x$.&#xA;That way you get an injective ($x$ is the biggest element of $X$ not containing $f(x)$) map from $X$ (possibly minus one element) to $\omega$, which is countable.&lt;/p&gt;&#xA;" OwnerUserId="2660" LastActivityDate="2010-10-29T11:36:34.403" />
  <row Id="8268" PostTypeId="2" ParentId="8264" CreationDate="2010-10-29T11:40:47.053" Score="0" ViewCount="" Body="&lt;p&gt;When first doing truth tables, it's best to break a sentence into its constituent parts. Here the propositions are $p,q,r$, so have the first three columns of your truth table be possible truth values for these. Then the remaining columns should be things such as $\neg q$, $\neg r$, $\neg q \vee\neg r$, $\neg p\wedge q$, $p\to (\neg q\vee\neg r)$, finally building up to $p\to (\neg q\vee\neg r)\wedge\neg p\wedge q$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To get started, the possible triples for $(p,q,r)$ are $(T,T,T),(T,T,F),(T,F,T),(F,T,T),(T,F,F),(F,T,F),(F,F,T),(F,F,F)$, and the rest of the table should follow easily from there. and the sentence should be easier to understand in retrospect, now that you see how each part affects each other part.&lt;/p&gt;&#xA;" OwnerUserId="1583" LastEditorUserId="1583" LastEditorDisplayName="" LastEditDate="2010-10-29T12:46:04.857" LastActivityDate="2010-10-29T12:46:04.857" />
  <row Id="8269" PostTypeId="2" ParentId="8263" CreationDate="2010-10-29T11:52:30.457" Score="1" ViewCount="" Body="&lt;p&gt;A good approach to these is to reduce your base number according to the modulus. So since $15\equiv 1\pmod{7}$, you would have $15^{89}\equiv 1^{89}\equiv 1\pmod{7}$. In other words, $15^{89}$ is 1 greater than a multiple of $7$, or 6 less than a multiple of $7$. So to get a complete solution set, you could add any integer of the form $6+7k$, or $-1+7k$, and these two sets coincide. So you can take either what must be subtracted or added to get the number to a multiple of your modulus. For example, for $29^2=841$, you could either add 4 to get 845, or subtract 1 to get 840, both of which are divisible by 5.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the second, as was noted in the comments, $29^2\equiv(-1)^2\equiv 1\pmod{5}$, and you should be able to get a complete solution set by the same reasoning. I just mentioned $29\equiv -1$ since $(-1)^2$ is nicer to deal with than $4^2$, and you would have to reduce 16 modulo 5 anyway.&lt;/p&gt;&#xA;" OwnerUserId="1583" LastActivityDate="2010-10-29T11:52:30.457" CommentCount="2" />
  <row Id="8270" PostTypeId="2" ParentId="8265" CreationDate="2010-10-29T11:53:06.993" Score="1" ViewCount="" Body="&lt;p&gt;I assume by &quot;inclusion&quot; that you mean the relation A&amp;nbsp;&amp;sube;&amp;nbsp;B.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider a well-ordered subset &amp;sigma;&amp;nbsp;&amp;sube;&amp;nbsp;P(&amp;omega;). Clearly, with respect to inclusion, &amp;sigma; has some ordinal type; the question is whether this type is countable. For any a,b&amp;nbsp;&amp;isin;&amp;nbsp;&amp;sigma; such that a&amp;nbsp;&amp;sub;&amp;nbsp;b, we have |b&amp;nbsp;\&amp;nbsp;a|&amp;nbsp;&amp;ge;&amp;nbsp;1. Thus, the cardinality of V&amp;nbsp;=&amp;thinsp;[&amp;nbsp;&amp;cup;&amp;nbsp;&amp;sigma;&amp;nbsp;]&amp;thinsp;&amp;sube;&amp;nbsp;&amp;omega; is at least that of the order type of &amp;sigma;, and at most that of &amp;omega;.&lt;/p&gt;&#xA;" OwnerUserId="439" LastActivityDate="2010-10-29T11:53:06.993" />
  <row Id="8271" PostTypeId="2" ParentId="8264" CreationDate="2010-10-29T12:20:42.320" Score="4" ViewCount="" Body="&lt;p&gt;This answer is meant to elaborate on yunone's earlier answer. I also assume an order of precedence so that &quot;IF a THEN b&quot; takes lower precedence than &quot;a AND b&quot; or &quot;a OR b&quot; &amp;mdash; that is, I assume that the expression you must evaluate is p&amp;nbsp;→&amp;nbsp;((¬q&amp;nbsp;∨&amp;nbsp;¬r)&amp;nbsp;∧&amp;nbsp;¬p&amp;nbsp;∧&amp;nbsp;q).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Draw out the following table (drawn below in monospaced unicode): spaces in between columns are just to indicate separation because ASCII double-lines look ugly.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&#xA; p | q | r    ¬q | ¬r | ¬q ∨ ¬r    ¬p | ¬p ∧ q | (¬q ∨ ¬r) ∧ ¬p ∧ q    p → ((¬q ∨ ¬r) ∧ ¬p ∧ q)&#xA;===|===|===  ====|====|=========  ====|========|====================  ==========================&#xA; F | F | F       |    |               |        |&#xA; F | F | T       |    |               |        |&#xA; F | T | F       |    |               |        |&#xA; F | T | T       |    |               |        |&#xA; T | F | F       |    |               |        |&#xA; T | F | T       |    |               |        |&#xA; T | T | F       |    |               |        |&#xA; T | T | T       |    |               |        |&#xA;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The columns pre-filled with T and F describe, in each row, a different possible setting of your variables p, q, and r. From these, you can evaluate other expressions such as ¬q and ¬r, use these to evaluate other expressions such as (¬q&amp;nbsp;∨&amp;nbsp;¬r), and so forth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it helps, introduce auxiliary variables for expressions that you have to evaluate along the way. For instance, you can define A&amp;nbsp;=&amp;nbsp;¬q&amp;nbsp;∨&amp;nbsp;¬r&amp;nbsp; and &amp;nbsp;B&amp;nbsp;=&amp;thinsp;&amp;thinsp;¬p&amp;nbsp;∧&amp;nbsp;q&amp;nbsp;, which will help you to evaluate the expression A&amp;nbsp;∧&amp;nbsp;B&amp;thinsp;&amp;thinsp;=&amp;thinsp;&amp;thinsp;(¬q&amp;nbsp;∨&amp;nbsp;¬r)&amp;nbsp;∧&amp;nbsp;¬p&amp;nbsp;∧&amp;nbsp;q&amp;nbsp;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Generally, just break down the expression into pieces that you can manage, and re-assemble the pieces to get the whole. (Which is a high-level description of mathematical problem solving in general, of course)&lt;/p&gt;&#xA;" OwnerUserId="439" LastActivityDate="2010-10-29T12:20:42.320" />
  <row Id="8272" PostTypeId="1" CreationDate="2010-10-29T12:35:28.253" Score="0" ViewCount="73" Body="&lt;p&gt;Let $G$ be a topological group and $G^{[0,1]}$ the set of paths $[0,1]\rightarrow G$ with the compact-open topology. This should make $G^{[0,1]}$ a topological group. I'd like to be able to refer to a simple proof that multiplication $p\ast q(t)=p(t)q(t)$ of paths is continuous without having to give all the details. I'm not so much interested in the proof itself as I am in a reference. Any suggestions for texts that this (or a generalization) shows up in?&lt;/p&gt;&#xA;" OwnerUserId="2423" LastActivityDate="2010-10-29T12:35:28.253" Title="Continuity of multiplication in Map([0,1],G)" Tags="&lt;topology&gt;&lt;reference-request&gt;" CommentCount="2" FavoriteCount="1" />
  <row Id="8273" PostTypeId="1" AcceptedAnswerId="8282" CreationDate="2010-10-29T12:37:25.743" Score="2" ViewCount="148" Body="&lt;p&gt;Hello, I'm learning about continued fractions, and I've enjoyed them so far, but I'm unsure if I've done the following correctly. I have no real experience with analysis, so I'm not sure if my reasoning is formal enough, or correct. Any feedback would be appreciated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $\xi$ be an irrational number with continued fraction expansion $\langle a_0,a_1,a_2,a_3\dots\rangle$. Let $b_1,b_2,b_3,\cdots$ be any sequence of positive integers, either finite or infinte. Prove that $\lim_{n\to\infty}\langle a_0,a_1,a_2,\dots,a_n,b_1,b_2,b_3\dots\rangle=\xi$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I let $r_n=\langle a_0,a_1,\dots,a_n\rangle$ and $\xi'=\langle b_1,b_2,b_3,\dots\rangle$, and let $\beta_n=\langle a_0,a_1,a_2,a_3\dots,a_n,\xi'\rangle$. So* &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\beta_n-r_n=\beta_n-\frac{h_n}{k_n}=\frac{\xi'h_n+h_{n-1}}{\xi'k_n+k_{n-1}}-\frac{h_n}{k_n}$$&#xA;$$=\frac{-(h_nk_{n-1}-h_{n-1}k_n)}{k_n(\xi'k_n+k_{n-1})}=\frac{(-1)^n}{k_n(\xi'k_n+k_{n-1})}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But ${k_n}$ is a positive increasing series, $\xi'$ is a positive real number, so as $n$ approaches $\infty$, the denominator goes to $\infty$ while the numerator alternates between $-1$ and $1$, so the fraction tends to $0$. Hence we have that $\lim_{n\to\infty}(\beta_n-r_n)=0$, so $$\lim_{n\to\infty}\beta_n=\lim_{n\to\infty}r_n=\lim_{n\to\infty}\langle a_0,a_1,a_2,\dots,a_n\rangle=\langle a_0,a_1,a_2\dots\rangle=\xi.$$ Hence $\lim_{n\to\infty}\langle a_0,a_1,a_2,\dots,a_n,b_1,b_2,b_3\dots\rangle=\xi$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If this the correct route to go? As a small side question, how does it make sense to have integers $b_i$ at the end of this sequence, if the sequence is infinite? Thanks!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;*If it's not well known, $\{h_n\}$ is the sequence defined by $h_{-2}=0,h_{-1}=1,h_i=a_ih_{i-1}+h_{i-2}$ and $\{k_n\}$ is defined as $k_{-2}=1,k_{-1}=-1, k_i=a_ik_{i-1}+k_{i-2}$, and $r_n=\langle a_0,a_1,\dots,a_n\rangle$, for any sequence of integers $a_0,a_1,a_2\dots$ all positive except perhaps $a_0$. &lt;/p&gt;&#xA;" OwnerUserId="1583" LastEditorUserId="1583" LastEditorDisplayName="" LastEditDate="2010-10-30T02:23:07.807" LastActivityDate="2010-10-30T18:19:01.090" Title="Adding integers to an infinite continued fraction expansion doesn't change the value?" Tags="&lt;number-theory&gt;&lt;continued-fractions&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="8275" PostTypeId="1" AcceptedAnswerId="8281" CreationDate="2010-10-29T13:03:29.780" Score="0" ViewCount="78" Body="&lt;p&gt;Suppose $S^n$ is the n-sphere with basepoint $x$. If based map $f:S^n\rightarrow S^n$ is such that the pull-back $f^{-1}(S^{n}-x)$ is connected do we necessarily have that $deg(f)$ is either -1,0, or 1?&lt;/p&gt;&#xA;" OwnerUserId="2423" LastEditorUserId="2423" LastEditorDisplayName="" LastEditDate="2010-10-29T13:59:02.127" LastActivityDate="2010-10-29T14:02:40.330" Title="Degrees of certain maps of spheres" Tags="&lt;algebraic-topology&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="8276" PostTypeId="2" ParentId="8259" CreationDate="2010-10-29T13:15:27.627" Score="3" ViewCount="" Body="&lt;p&gt;If $H(R)$ is spherically symmetric, we can write the Laplacian of $H$ in spherical coordinates:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\nabla^2 H = \frac{1}{R^2}\frac{d}{dR}\bigg(R^2\frac{dH(R)}{dR}\bigg)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(see &lt;a href=&quot;http://en.wikipedia.org/wiki/Laplace_operator&quot; rel=&quot;nofollow&quot;&gt;wikipedia&lt;/a&gt;). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, $\frac{d}{dR}\bigg(R^2\frac{dH(R)}{dR}\bigg) = R^2\frac{d^2H}{dR^2} + 2R \frac{dH}{dR}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other way, $\frac{d}{dR}(RH(R)) = R\frac{dH}{dR} + H$&#xA;and thus&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{d^2}{dR^2}(RH(R)) = R\frac{d^2H}{dR^2} + 2\frac{dH}{dR}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Rightarrow \frac{1}{R^2}\frac{d}{dR}\bigg(R\frac{dH(R)}{dR}\bigg) = \frac{1}{R}\frac{d^2}{dR^2}(RH(R))$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Rightarrow \nabla^2 H = \frac{1}{R^2}\frac{d}{dR}\bigg(R^2\frac{dH(R)}{dR}\bigg) = \frac{1}{R}\frac{d^2}{dR^2}(RH(R))$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, apply it to $G_{\omega}$.&lt;/p&gt;&#xA;" OwnerUserId="1178" LastActivityDate="2010-10-29T13:15:27.627" CommentCount="1" />
  <row Id="8277" PostTypeId="1" AcceptedAnswerId="8278" CreationDate="2010-10-29T13:18:14.783" Score="2" ViewCount="122" Body="&lt;p&gt;How to prove that if you take any 400 consecutive Octobers then exactly 14 % of those years have five Fridays and Sundays?&lt;/p&gt;&#xA;" OwnerUserId="2856" LastEditorUserId="2856" LastEditorDisplayName="" LastEditDate="2010-10-29T14:13:12.973" LastActivityDate="2010-10-29T16:47:36.007" Title="Five Fridays and Sundays on October" Tags="&lt;recreational-mathematics&gt;" AnswerCount="2" />
  <row Id="8278" PostTypeId="2" ParentId="8277" CreationDate="2010-10-29T13:30:35.807" Score="2" ViewCount="" Body="&lt;p&gt;To have five Fridays and Sundays, October must start on Friday.  A span of 400 years has 146097 days in it-remember the years that are multiples of 100 but not 400 are not leap years.  This is a multiple of 7.  So the pattern repeats.  Count how many of the 400 start October with Thursday or Friday.  You should find 56.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Corrected-used to say start on Thursday or Friday. &lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2010-10-29T16:47:36.007" LastActivityDate="2010-10-29T16:47:36.007" CommentCount="4" />
  <row Id="8279" PostTypeId="2" ParentId="8273" CreationDate="2010-10-29T13:39:42.047" Score="3" ViewCount="" Body="&lt;p&gt;Your work looks good to me.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It does make sense to have the $b_n$ at the end of the sequence as each stage of the sequence is finite.  This is an important point.  At each stage, the list of $a$'s is finite, so it makes sense to have more numbers (of any sort) at the end.  But as you let $n \rightarrow \inf $ they get pushed farther and farther out.  The way continued fractions work, they have less and less impact, as you showed.  So, &quot;in the limit&quot;, they don't matter.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-10-30T18:19:01.090" LastActivityDate="2010-10-30T18:19:01.090" CommentCount="1" />
  <row Id="8280" PostTypeId="1" AcceptedAnswerId="8283" CreationDate="2010-10-29T13:55:44.033" Score="0" ViewCount="81" Body="&lt;p&gt;If 72% percent of new apps are reviewed within the last 7 days, then there is a 72% chance that any app is reviewed in the last week.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$72\%  = \sum\limits_{i = 0}^{i = 6} {x\left( {1 - x} \right)^i }$$&lt;br&gt;&#xA;          where $x$ represents the probability that an app is reviewed each day.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $p = \left( {1 - x} \right)$, then the following is true:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$&#xA;\eqalign{&#xA;  &amp;amp; 72\%  = x + xp + xp^2  + xp^3  + xp^4  + xp^5  + xp^6   \cr &#xA;  &amp;amp; 72\%  = x\left( {1 + p + p^2  + p^3  + p^4  + p^5  + p^6 } \right)  \cr &#xA;  &amp;amp; 72\%  = 1 - p^7   \cr &#xA;  &amp;amp; p^7  = 1 - 72\%   \cr &#xA;  &amp;amp; p^7  = 28\%   \cr &#xA;  &amp;amp; \left( {1 - x} \right)^7  = 28\%   \cr &#xA;  &amp;amp; 1 - x \approx 0.8337  \cr &#xA;  &amp;amp; x \approx 16.63\%  \cr} &#xA;$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is my work correct?&lt;/p&gt;&#xA;" OwnerUserId="2929" LastActivityDate="2010-10-29T14:09:11.293" Title="Apple App Store Review Rate" Tags="&lt;probability&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="0" />
  <row Id="8281" PostTypeId="2" ParentId="8275" CreationDate="2010-10-29T14:02:40.330" Score="3" ViewCount="" Body="&lt;p&gt;Look at $S^2$ as the result of adding $\infty$ to $\mathbb C$, and let $f:S^2\to S^2$ be the map $z\mapsto z^2$. Then the preimage of $S^2-z$ is connected for all $z\in S^2$ (because the preimage of $z$ is finite). The degree is, of course, 2.&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2010-10-29T14:02:40.330" CommentCount="1" />
  <row Id="8282" PostTypeId="2" ParentId="8273" CreationDate="2010-10-29T14:03:30.247" Score="4" ViewCount="" Body="&lt;p&gt;You are on the right track, but some statements do not make any sense at all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) &quot;Since $\xi'$ is a sequence of positive integers&quot; : This does not make any sense. $\xi'$ is a constant real number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) You are using confusing notation. $\beta = \langle a_0, a_1, \dots, a_n, \xi' \rangle$. This varies with $n$, so you need to talk about $\beta_n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) &quot;Hence we have $\lim_{n \to \infty} (\xi' -r_n) = 0$&quot; is not right. This implies that any two real numbers are equal (think about it)! You need to consider $\lim_{n \to \infty} \beta_n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As to you side question, it is not one sequence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is a sequence of sequences!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a_0, a_1, b_1, b_2, b_3, \dots$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a_0, a_1, a_2, b_1, b_2, b_3, \dots$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\vdots$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a_0, a_1, a_2, \dots, a_n, b_1, b_2, b_3 \dots$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\vdots$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice that in each sequence, we only had a finite number of the $a_i$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each sequence corresponds to a real number (the $\beta_n$ above).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus we get a new sequence&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\beta_1, \beta_2, \dots$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And the problem is asking you to prove that $\lim_{n \to \infty} \beta_n = \xi$.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-10-29T14:37:43.720" LastActivityDate="2010-10-29T14:37:43.720" CommunityOwnedDate="2010-10-29T14:18:54.353" CommentCount="2" />
  <row Id="8283" PostTypeId="2" ParentId="8280" CreationDate="2010-10-29T14:09:11.293" Score="2" ViewCount="" Body="&lt;p&gt;Looks good.  You have made the assumption, probably incorrect, that the probability of an app being reviewed is independent of the time since release and independent of whether it has already been reviewed.  But for math problems (as opposed to the real world) we often assume such.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-10-29T14:09:11.293" />
  <row Id="8284" PostTypeId="1" CreationDate="2010-10-29T14:16:09.110" Score="3" ViewCount="95" Body="&lt;p&gt;Let $(B_t)$ denote the standard Brownian motion on the interval $[0,1]$. For a given confidence level $\alpha \in (0,1)$ a confidence band on $[0,1]$ is any function $u$ with the property that &#xA;$$&#xD;&#xA;P(\omega; |B_t(\omega)| &amp;lt; u(t), \quad \forall t\in [0,1])=\alpha.&#xD;&#xA;$$&#xA;In other words, the probability that a path of the Brownian motion stays within a confidence band is $\alpha$. Additionally the boundary hitting position for those paths leaving the band must be uniformly distributed on $[0,1]$. This condition can be stated using the stopping time &#xA;$$\tau(\omega) = \inf [ t \in [0,1], |B_t(\omega)|=u(t) ].&#xD;&#xA;$$&#xA;Then $\tau $ is the time of the first hitting, and one asks that $\tau$ is uniformly distributed on $[0,1]$ conditionally on the event that $\tau$ is finite.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am interested in&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;References and links to literature or papers considering this or similar problems&lt;/li&gt;&#xA;&lt;li&gt;Thoughts, ideas, discussion&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The context of the problem is a rather boring one, so will not state it here. The problem itself seem to be non-trivial and interesting.&lt;/p&gt;&#xA;" OwnerUserId="731" LastEditorUserId="6179" LastEditorDisplayName="" LastEditDate="2011-03-20T18:04:29.797" LastActivityDate="2011-03-20T18:19:31.923" Title="Confidence band for Brownian Motion with uniformly distributed hitting position" Tags="&lt;probability&gt;&lt;probability-theory&gt;&lt;stochastic-processes&gt;" AnswerCount="0" CommentCount="8" />
  <row Id="8285" PostTypeId="2" ParentId="8035" CreationDate="2010-10-29T14:27:29.533" Score="0" ViewCount="" Body="&lt;p&gt;Thanks everyone! The model answer I was given is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;False. Roughly speaking, any polynomial can be made as large as you like by taking x to be very large (provided it is of degree greater than zero), whereas | cos x| &amp;lt;= 1 . There is obviously no polynomial of degree zero (i.e. no constant number) for which the statement holds.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is from &lt;em&gt;Advanced Problems in Core Mathematics&lt;/em&gt; by Siklos. I am not very worried about the details, I am more than happy to be broadly right which I am on this occasion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks again, everyone!&lt;/p&gt;&#xA;" OwnerUserId="2846" LastActivityDate="2010-10-29T14:27:29.533" />
  <row Id="8286" PostTypeId="2" ParentId="8243" CreationDate="2010-10-29T14:30:10.300" Score="2" ViewCount="" Body="&lt;p&gt;Since $\rm\ \gcd(a,b) = 1\:,\ $ by &lt;a href=&quot;http://en.wikipedia.org/wiki/Euclid%27s_lemma&quot; rel=&quot;nofollow&quot;&gt;Euclid's Lemma,&lt;/a&gt; $\rm\ \ a\:|\:b\:(x/b)\ \Rightarrow\ a\:|\:x/b\ \Rightarrow\ ab\:|\:x$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively $\rm\ \ b,a\:|\:x\ \Rightarrow\ ab\: |\: ax,bx\ \Rightarrow\ ab\ |\ gcd(ax,bx)\ =\ x\ gcd(a,b)\ =\ x $&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the special case $\rm\ gcd(a,b) = 1\ $ of $\rm\ gcd(a,b)\ lcm(a,b)\ =\ ab\ $ which has a &lt;a href=&quot;http://math.stackexchange.com/questions/3149&quot;&gt;similar proof.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-12-03T06:38:50.247" LastActivityDate="2010-12-03T06:38:50.247" />
  <row Id="8287" PostTypeId="2" ParentId="8277" CreationDate="2010-10-29T14:52:00.510" Score="0" ViewCount="" Body="&lt;p&gt;As people have said, the best way to do this is brute force, once you know that the calendar repeats with period 400 years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd note that to have five Fridays and five Sundays, October must &lt;i&gt;start&lt;/i&gt; on a Friday (like it does this year), and therefore run from Friday, October 1 to Sunday, October 31. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it feels like cheating to use date rules that somebody has already computed, you could (for example) note that:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;this year, October 1 was a Friday, the sixth day of the week;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;if October 1, year $n$ is the $k$th day of the week, then October 1, year $n+1$ is either the $k+1$st or $k+2$nd day of the week, depending on whether $n+1$ is an ordinary year or a leap year. (Of course the $8$th and $9$th days of the week can be converted to the $1$st and $2$nd days.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Finally, it's easy to see that the answer should be &quot;approximately 1/7&quot; because the calendar rules don't make reference to any particular days of the week, although of course this is not a proof.  This is why the claim that went around the Internet earlier this month that an October like the current one only occurs every 823 years was so laughable; see &lt;a href=&quot;http://www.geekosystem.com/5-fridays-saturdays-sundays-not-823-years/&quot; rel=&quot;nofollow&quot;&gt;here for a debunking&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="173" LastActivityDate="2010-10-29T14:52:00.510" />
  <row Id="8288" PostTypeId="1" AcceptedAnswerId="8330" CreationDate="2010-10-29T15:35:17.317" Score="3" ViewCount="154" Body="&lt;p&gt;Is it possible to divide an equilateral triangle into 5 equal (i.e., obtainable&#xA;from each other by a rigid motion) parts?&lt;/p&gt;&#xA;" OwnerUserId="2856" LastActivityDate="2010-10-30T01:54:01.897" Title="Splitting equilateral triangle to 5 parts" Tags="&lt;geometry&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" />
  <row Id="8289" PostTypeId="1" AcceptedAnswerId="8294" CreationDate="2010-10-29T15:35:34.140" Score="2" ViewCount="219" Body="&lt;p&gt;I need to find the number of natural numbers between 1 and 1000 that are divisible by 3, 5 or 7.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that given a set of numbers, 1 ... n, the number of numbers divisible by d is given by $\lfloor \frac{n}{d} \rfloor$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if we let $A$ be the set of numbers divisible by 3, $B$ be the set of numbers divisible by 5 and $C$ be the set of numbers divisible by 7, then the size of the union of those sets is going to be the number required.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, to work out $A \cup B \cup C$, I need to know $A \cap B$ (and $A \cap C$ etc).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My only thought was that to work out $A \cap B$, I could use $\lfloor \frac{1000}{3 \times 5} \rfloor$, but I am unsure if this misses out some numbers. Is that the correct way to do it or not? If no, can anyone suggest the correct way.&lt;/p&gt;&#xA;" OwnerUserId="108" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-10-29T15:47:10.620" LastActivityDate="2010-10-29T15:49:12.350" Title="How many numbers between 1 and 1000 can be divided by both $a$ and $b$" Tags="&lt;number-theory&gt;&lt;homework&gt;" AnswerCount="2" CommentCount="0" />
  <row Id="8290" PostTypeId="1" CreationDate="2010-10-29T15:39:06.807" Score="2" ViewCount="117" Body="&lt;p&gt;A triple of positive integers $(a,b,c)$ is an $abc$-triple if $a$ and $b$ are coprime and $c = a + b$. Define the &lt;em&gt;quality&lt;/em&gt; or &lt;em&gt;power&lt;/em&gt; of an $abc$-triple as $P(a,b,c) = \frac{\log c}{\log \text{rad}(abc)}$, where $\text{rad}(k)$ denotes the product of distinct prime divisors of $k$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One version of the $abc$-Conjecture is that for each $\varepsilon &gt; 0$, there are finitely many $abc$-triples such that $P(a,b,c) &gt; 1 + \varepsilon$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are finitely many known triples satisfying $P &gt; 1.4$, the so called &lt;em&gt;good triples&lt;/em&gt;,  and the largest (quality) is $P(2,3^{10} \cdot 109, 23^{5}) = 1.629911684 \dots$ (discovered by E. Reyssat).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; Is there an upper bound for $P(a,b,c)$ simply in terms of $c$ and absolute constants (sharper than $\log_{2} c$)? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; Is there an absolute upper bound for $P(a,b,c)$ so that no triple has higher quality?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Best Answer:&lt;/strong&gt; If there were such a bound, then asymptotic FLT would be in hand. (Thanks Ace of Base on MO!)&lt;/p&gt;&#xA;" OwnerUserId="2720" LastEditorUserId="1778" LastEditorDisplayName="" LastEditDate="2010-11-04T03:55:05.947" LastActivityDate="2010-11-04T03:55:05.947" Title="Upper bound for the quality of an $abc$-triple" Tags="&lt;number-theory&gt;&lt;real-analysis&gt;&lt;analysis&gt;&lt;diophantine-equations&gt;&lt;asymptotics&gt;" FavoriteCount="2" />
  <row Id="8291" PostTypeId="2" ParentId="8289" CreationDate="2010-10-29T15:40:53.577" Score="5" ViewCount="" Body="&lt;p&gt;That's right. A number is divisible by 3 and 5 if and only if it is divisible by 15.&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2010-10-29T15:40:53.577" />
  <row Id="8293" PostTypeId="1" AcceptedAnswerId="8296" CreationDate="2010-10-29T15:48:34.763" Score="2" ViewCount="113" Body="&lt;p&gt;In a problem my teacher did the following:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;5a - 2b = 3m &#xA;5b - 2c = 3n    +&#xA;-----------------&#xA;5a - 2c + 3b = 3m + 3n&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I tried solving for b in the first equation and then plugging it in to the second but could not get the same result.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When is adding full equations like this valid?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks! &lt;/p&gt;&#xA;" OwnerUserId="2587" LastActivityDate="2010-10-30T16:39:57.753" Title="When can you add full equations together?" Tags="&lt;algebra&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="8294" PostTypeId="2" ParentId="8289" CreationDate="2010-10-29T15:49:12.350" Score="5" ViewCount="" Body="&lt;p&gt;In general, you have to use $\left\lfloor\frac{1000}{[a,b]}\right\rfloor$, where $[a,b]$ is the least common multiple of $a$ and $b$.&lt;/p&gt;&#xA;" OwnerUserId="589" LastActivityDate="2010-10-29T15:49:12.350" />
  <row Id="8295" PostTypeId="1" CreationDate="2010-10-29T16:34:30.903" Score="1" ViewCount="94" Body="&lt;p&gt;I want to solve a simple stochastic problem. Imagine there are n agents who want to access a resource, with a probability p at a given time t. What ist the probability that the resource will be free and not accessed by an agent. How does the result change with 2n and 2 available resources?&#xA;How do i solve this problem?&lt;/p&gt;&#xA;" OwnerUserId="2931" LastActivityDate="2010-12-28T18:44:29.343" Title="n agents accessing a resource" Tags="&lt;probability&gt;&lt;statistics&gt;&lt;stochastic-processes&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="8296" PostTypeId="2" ParentId="8293" CreationDate="2010-10-29T16:50:57.093" Score="1" ViewCount="" Body="&lt;p&gt;If A = B and C = D then A+C = B+D.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More generally, if $f$ is any function, then $x = y$ implies $f(x) = f(y)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case we have used the function f(x) = x + C. to go from A = B to A + C = B + C as well as g(x) = B + x to go from C = D to B + C = B + D. Then we paste the two equations together (transitivity) to get A + C = B + D.&lt;/p&gt;&#xA;" LastActivityDate="2010-10-29T16:50:57.093" />
  <row Id="8297" PostTypeId="2" ParentId="8243" CreationDate="2010-10-29T17:22:33.513" Score="2" ViewCount="" Body="&lt;p&gt;gcd(a,b)=1 so there are integers s and t such that sa+tb=1, whence sax+tbx=x. But a divides x so x=ax' for some integer x'. As b divides x, x=bx&quot; for some integer x&quot;. Substituting on the left-hand side of the preceding equation yields sabx&quot;+tabx'=x. Factor out the commn ab on the left to get ab(ax&quot;+tx&quot;)=x, so that ab divides x.&lt;/p&gt;&#xA;" OwnerUserId="2933" LastActivityDate="2010-10-29T17:22:33.513" />
  <row Id="8298" PostTypeId="2" ParentId="8288" CreationDate="2010-10-29T17:42:17.143" Score="0" ViewCount="" Body="&lt;p&gt;I -think- there's a proof related to sums of angles that says this can't be done.&lt;/p&gt;&#xA;" LastActivityDate="2010-10-29T17:42:17.143" CommentCount="1" />
  <row Id="8299" PostTypeId="1" CreationDate="2010-10-29T17:53:42.420" Score="2" ViewCount="111" Body="&lt;p&gt;Is there a relation between the max of a Gaussian random walk of 10 &#xA;steps vs the max of 10 Gaussian random walks? Specifics (in &#xA;Mathematica notation): &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; &#xA;(* a Gaussian random walk with standard deviation 1 *) &#xA;a[0] := 0 &#xA;a[n_] := a[n-1] + RandomReal[NormalDistribution[0, 1]] &#xA;&#xA;(* the max of the walk over 10 steps *) &#xA;b := Max[Table[a[i],{i,1,10}]] &#xA;&#xA;(* calculate max many times to get good sample set *) &#xA;(* Mathematica &quot;magic&quot; insures we're not using the same random #s each time *) &#xA;c = Table[b,{i,1,10000}] &#xA;&#xA;(* distribution isn't necessarily normal, but we can still compute mu + SD *) &#xA;Mean[c] (* 3.66464 *) &#xA;StandardDeviation[c] (* 1.61321 *) &#xA;&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Now, consider 10 people doing a Gaussian random walk of 1 step and we &#xA;take the max of these 10 values. &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; &#xA;(* max of 10 standard-normally distributed numbers *) &#xA;d := Max[Table[RandomReal[NormalDistribution[0, 1]],{i,1,10}]] &#xA;&#xA;(* get a good sample set *) &#xA;f = Table[d,{i,10000}] &#xA;&#xA;(* and now the mean and SD *) &#xA;Mean[f] (* 1.54843 *) &#xA;StandardDeviation[f] (* 0.580593 *)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The two means/SDs are obviously different, but I sense they're related &#xA;somehow, perhaps by Sqrt[10], since the sum (not max) of 10 random &#xA;walks is normal with SD of Sqrt[10], and I sense that somehow the &#xA;cumulative sum of the first 9 somehow cancel out. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are these known distributions? &lt;/p&gt;&#xA;" OwnerUserId="2469" LastActivityDate="2011-05-28T04:03:58.737" Title="Max of multistep Gaussian walk vs max of multiple single-step Gaussian walks" Tags="&lt;random&gt;&lt;standard-deviation&gt;" AnswerCount="1" />
  <row Id="8300" PostTypeId="2" ParentId="8256" CreationDate="2010-10-29T18:09:14.360" Score="1" ViewCount="" Body="&lt;p&gt;I agree that if we define $\mathbb{N}$ as a &quot;set of positive integers&quot;, the question is redundant. However if we say that $\mathbb{N}$ is a set of natural numbers (and, after all, that is exactly what $\mathbb{N}$ stands for), defined by Peano axioms, then the question is no longer redundant. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a side note, calling $\mathbb{N}$ a &quot;set of positive integers&quot; is, IMHO, a bad style. Unfortunately, there is no standard convention for &quot;set of positive integers&quot;. Many books use $\mathbb{Z}^{+}$ for that, however at least one book I'm familiar with uses $\mathbb{Z}^{+}$ for &quot;group of integers under the regular addition&quot;.  May be $\mathbb{Z}_{+}$ is better (?)&lt;/p&gt;&#xA;" OwnerUserId="2450" LastActivityDate="2010-10-29T18:09:14.360" />
  <row Id="8301" PostTypeId="2" ParentId="8295" CreationDate="2010-10-29T18:38:12.747" Score="2" ViewCount="" Body="&lt;p&gt;The resource will be free with probability $(1-p)^n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When there are $2n$ agents and $2$ resources, one of the resources will be free with probability $(1-p)^{2n} + 2np(1-p)^{2n-1}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, you want to calculate the tail of a Binomial distribution. When $p = c/n$ and $n$ is large, the distribution is approximately Poisson.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-10-29T18:38:12.747" />
  <row Id="8302" PostTypeId="2" ParentId="8155" CreationDate="2010-10-29T18:40:28.157" Score="0" ViewCount="" Body="&lt;p&gt;It doesn't matter if you work with complex, real numbers, or elements of any field $\mathbb{K}$: if you have a point $p = (z_1, \dots , z_n) \in \mathbb{K}^n$, or any $\mathbb{K}$-vector space $V$, an equation for a straight line in $\mathbb{K}^n$ (or in $V$) passing through $p$ may always be written, for instance, as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;p + \lambda v \ ,&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with $v= (v_1, \dots , v_n) \in \mathbb{K}^n$ (or $v\in V$) and $\lambda \in \mathbb{K}$.&lt;/p&gt;&#xA;" OwnerUserId="664" LastActivityDate="2010-10-29T18:40:28.157" />
  <row Id="8303" PostTypeId="2" ParentId="8249" CreationDate="2010-10-29T18:49:51.933" Score="1" ViewCount="" Body="&lt;p&gt;HINT: Take a sequence of partitions which alternate between including $0$ and not including $0$ while having their size go to $0$ (the most natural choice will do, since $0$ is the midpoint of the interval); pick your &quot;tag&quot; appropriately in the ones that do not include $0$ so that the sum gives you one value, and see what you get when the partition does include $0$. With some appropriate choices, you'll get a sequence that does not converge, showing that the limit does not exist.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To prove the function is Riemman-Stieltjes integrable relative to $a$, break up the integral into two interals, possibly redefining $f$ and $a$ at a single point in one or both of them, so that you can use the standard formula for Riemman-Stieltjes when the functions are &quot;nice&quot;. &lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-10-29T18:49:51.933" CommentCount="2" />
  <row Id="8304" PostTypeId="2" ParentId="8253" CreationDate="2010-10-29T18:56:59.080" Score="2" ViewCount="" Body="&lt;ol&gt;&#xA;&lt;li&gt;If by index of the curve C they mean the integral:&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;$ \frac{1}{2 \pi i} \int_\gamma \frac{v'(z)}{v(z)} dz $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is what I am guessing that they mean, then this follows from the residue theorem applied to the complex valued function $v(z)$.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The index of a vector field at $z_0$ is defined to be the winding number you get for taking a small circle about $z_0$ of winding number $1$ and computing the winding number of its image about $0$. This can be computed as:&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;$ \frac{1}{2 \pi i} \int_\gamma \frac{v'(z)}{v(z)} dz $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\gamma$ is a parametrization of a sufficiently small circle which winds around $z_0$ once. For $v(z) = f(z)$ and $f(z_0) = 0$, we can write $f(z) = (z-z_0)^n g(z)$ where $n$ is maximal such that $g(z_0) \neq 0$. Then,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ f'(z) = n (z-z_0)^{n-1} g(z) + (z-z_0)^n g'(z) $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ \frac{f'(z)}{f(z)} = \frac{n}{z-z_0} + (z-z_0)^n \frac{g'(z)}{g(z)} $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second term above is analytic in this region if we make $\gamma$ a small enough circle so that $g(z)$ is non-zero in the interior. Thus, when we integrate we just get that the index is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ \frac{1}{2 \pi i} \int_\gamma \frac{n}{z - z_0} dz = n $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is the multiplicity of $z_0$.&lt;/p&gt;&#xA;" OwnerUserId="1089" LastActivityDate="2010-10-29T18:56:59.080" />
  <row Id="8305" PostTypeId="1" AcceptedAnswerId="8311" CreationDate="2010-10-29T19:09:46.660" Score="1" ViewCount="190" Body="&lt;p&gt;If we have a uniform continuous function $f: X \to Y$, then&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$f$ takes a cauchy sequence in $X$ to a cauchy sequence in $Y$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Now is this statement true: $f$ is uniformly continuous iff,  given $\epsilon &gt; 0$, there is an $N &gt; 0$, such that for every $x,y \in I$,where $I$ is an interval ($x \neq y$) we have $$ \Biggl| \frac{f(x)-f(y)}{x-y} \Biggr| &gt; N \ \Longrightarrow |f(x)-f(y)|&amp;lt; \epsilon.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If yes, how to prove it?&lt;/p&gt;&#xA;" LastActivityDate="2010-10-29T20:24:57.473" Title="Necessary and sufficient condition for uniform continuity" Tags="&lt;real-analysis&gt;&lt;analysis&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="8306" PostTypeId="2" ParentId="8299" CreationDate="2010-10-29T19:29:19.043" Score="0" ViewCount="" Body="&lt;p&gt;I don't have an answer but only some trivial observations about your question that I post below. It got too big to be left as a comment. If I understand your question correctly, you want to compare the following two problems:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Let $X_i$ be i.i.d unit normals and you are interested in $X^{*} = \max(X_i, 1 \leq i \leq 10)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Let $Y_i$ be i.i.d unit normals. Define $S_0 = 0$ and $S_i = S_{i-1} + Y_i$ for $i &gt; 0$ and $S^{*} = \max(S_i, 1 \leq i \leq 10)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is easy to see that $P(X^{*} \leq x) = P(X_i \leq x)$ for all $1 \leq i \leq 10$ and therefore the distribution of $X^{*}$ is given by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f_{X^{*}}(x) = 10 \Phi(x)^9 \phi(x)$ where $\phi$ and $\Phi$ are the pdf and cdf of the standard normal. From this, you can technically compute the mean of $X^{*}$ though I don't know if the messy integration yields a nice solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The case of determining the distribution of $S^{*}$ seems much trickier. I found some references &lt;a href=&quot;http://projecteuclid.org/DPubS?service=UI&amp;amp;version=1.0&amp;amp;verb=Display&amp;amp;handle=euclid.aoap/1174323252&quot; rel=&quot;nofollow&quot;&gt;online&lt;/a&gt; that study the asymptotic behavior for long random walks and even that seems very hard. You can argue that $P(S^{*} \leq s) = P(S_i \leq s)$ for all $1 \leq i \leq n$ but the $S_i$ are dependent random variables. Their joint distribution is easy to derive but once again, I don't know if the integration is manageable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, if you are only interested in comparing the first moment of $X^{*}$ and $S^{*}$, there might be a clever way to do it that avoids all the integrations and such. If there exists such a proof, I would love to learn it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Update: I found some links that give the asymptotics for these random variables as the number of random variables $n \rightarrow \infty$. See &lt;a href=&quot;http://www.springerlink.com/content/lt45q14201550468/&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; and &lt;a href=&quot;http://projecteuclid.org/DPubS/Repository/1.0/Disseminate?view=body&amp;amp;id=pdf_1&amp;amp;handle=euclid.pjm/1102911985&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt;. As best as I can determine, $E(X^{*})$ grows as $\sqrt{2\ln(n)}$ and $E(S^{*})$ grows as $\sqrt{\frac{n}{\pi}}$.&lt;/p&gt;&#xA;" OwnerUserId="618" LastEditorUserId="618" LastEditorDisplayName="" LastEditDate="2010-10-30T00:47:18.573" LastActivityDate="2010-10-30T00:47:18.573" CommentCount="3" />
  <row Id="8307" PostTypeId="1" AcceptedAnswerId="37411" CreationDate="2010-10-29T19:35:40.070" Score="4" ViewCount="125" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Van_der_Waerden%27s_theorem&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Van_der_Waerden%27s_theorem&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The usual approach is to induct on the length of the arithmetic progression, which is difficult to simplify directly to the case of two colors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anyone know a different approach?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you.&lt;/p&gt;&#xA;" OwnerUserId="2934" LastActivityDate="2011-05-06T13:20:01.340" Title="Is there a simpler proof of Van Der Waerden's Theorem when there are only two colors?" Tags="&lt;graph-theory&gt;" AnswerCount="1" />
  <row Id="8308" PostTypeId="1" AcceptedAnswerId="8310" CreationDate="2010-10-29T19:36:20.327" Score="0" ViewCount="157" Body="&lt;p&gt;I'm trying to understand the intuitions behind working with very large numbers.  Specifically, I'm talking about numbers of the form $a^b$ where $a &gt; 10,000$ and $b &gt; 10,000$, and in general $a$ and $b$ are small, but $a^b$ has millions of digits.  Obviously, computing the number is infeasible, but I still would like to determine things about them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am particularly interested in ideas or rules about evaluating which of two large numbers is larger.  I believe I want to use logarithms somehow, but they've always confused me a little bit, so I can't quite see where to go from there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More generally, I'd love to hear more generic discussion of how one can work with such large numbers easily, sacrificing precision, but not accuracy.&lt;/p&gt;&#xA;" OwnerUserId="1887" LastActivityDate="2010-10-29T19:58:05.103" Title="Working With Large Exponents" Tags="&lt;number-theory&gt;" AnswerCount="1" />
  <row Id="8309" PostTypeId="2" ParentId="8250" CreationDate="2010-10-29T19:56:53.153" Score="-1" ViewCount="" Body="&lt;p&gt;First suppose both conditions are true. From the first of them we know $x\geq 0$, from the second --- $y^TA\geq 0$. Therefore $y^Tb=y^TAx\geq 0$. Contradiction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now suppose that both conditions are false. Let $\{e_k\}_{k=1}^n$ be the standard basis in $\mathbb{R}^n$. Note that the set $\{Ax\mid x\geq 0\} = \{\sum_kx_k Ae_k\mid x_k\geq 0\}$ is closed and take the nearest point to $b$ in it. From minimality of $\|b-Ax\|^2$ for any $k$ we have either $x_k&gt;0$ and $\partial_{x_k}\|b-Ax\|^2=0$ either $x_k=0$ and $\partial_{x_k}\|b-Ax\|^2\geq 0$. Thus&#xA;$$\left[\begin{aligned}x_k&amp;amp;&gt;0,&amp;amp;(b-Ax,Ae_k)&amp;amp;=0,\\x_k&amp;amp;=0,&amp;amp; (b-Ax,Ae_k)&amp;amp;&amp;lt;0\end{aligned}\right.\tag{1}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take $y=Ax-b$. From (1) we have $(y,Ae_k)\geq 0$ and hence $A^Ty\geq 0$. Finally&#xA;$$b^Ty=(y,b)=(Ax-b,b)=(Ax-b,b-Ax)+(Ax-b,Ax)=$$&#xA;$$-\|b-Ax\|^2-\sum_kx_k (b-Ax,Ae_k)=-\|b-Ax\|^2&amp;lt;0.$$&#xA;So the second condition should be true. Contradiction.&lt;/p&gt;&#xA;" OwnerUserId="695" LastActivityDate="2010-10-29T19:56:53.153" />
  <row Id="8310" PostTypeId="2" ParentId="8308" CreationDate="2010-10-29T19:58:05.103" Score="7" ViewCount="" Body="&lt;p&gt;Since logarithms are monotonically increasing, you can determine if $$a^b &gt; c^d$$ by instead checking if $$\log(a^b) &gt; \log(c^d)$$ which can be rewritten (using basic properties of logarithms) as $$b\log(a) &gt; d\log(c)$$ This results in the comparison of two numbers on the order of those given in the input (that is, both numbers will be within an order of magnitude or so of $a,b,c,d$), which is definitely feasible on a computer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can also use logs to calculate other properties, as long as you remember to convert your solution back to the original numbers (by means of an exponential) when you're done. For example, to compute whether $${a_1}^{b_1}{a_2}^{b_2} &gt; {c_1}^{d_1}{c_2}^{d_2}$$, you can take the log of both sides to get &#xA;\begin{align}&#xA;\log({a_1}^{b_1}{a_2}^{b_2}) &amp;amp;&gt; \log({c_1}^{d_1}{c_2}^{d_2}) \\&#xA;\log({a_1}^{b_1}) + \log({a_2}^{b_2}) &amp;amp;&gt; \log({c_1}^{d_1}) + log({c_2}^{d_2}) \\&#xA;b_1\log(a_1) + b_2\log(a_2) &amp;amp;&gt; d_1\log(c_1) + d_2\log(c_2) \\&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is again computationally feasible.&lt;/p&gt;&#xA;" OwnerUserId="1507" LastActivityDate="2010-10-29T19:58:05.103" CommentCount="2" />
  <row Id="8311" PostTypeId="2" ParentId="8305" CreationDate="2010-10-29T20:24:57.473" Score="2" ViewCount="" Body="&lt;p&gt;Well, the standard definition of uniform continuous on an interval $I$ is given by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\forall \epsilon &gt; 0,  \exists \delta &gt; 0 \text{ such that } \forall x, y \in I,  |x-y| &amp;lt; \delta \Rightarrow |f(x)-f(y)| &amp;lt; \epsilon$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you want to prove is that this is equivalent to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\forall \epsilon &gt; 0,  \exists N &gt; 0 \text{ such that } \forall x, y \in I,  \frac{|f(x)-f(y)|}{|x-y|} &gt; N \Rightarrow |f(x)-f(y)| &amp;lt; \epsilon$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{|f(x)-f(y)|}{|x-y|} &gt; N \iff |x-y| &amp;lt; \frac{|f(x) - f(y)|}{N}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proof (of your question):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Rightarrow$ : take $N$ such that $\delta=\frac{ \sup( f(x)) - \inf(f(y))}{N}$. The supremum and infimum are well-defined because of uniform continuity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Leftarrow$ : take $\delta := \frac{\epsilon}{N}$&lt;/p&gt;&#xA;" OwnerUserId="2778" LastActivityDate="2010-10-29T20:24:57.473" />
  <row Id="8312" PostTypeId="1" AcceptedAnswerId="8313" CreationDate="2010-10-29T20:25:00.587" Score="2" ViewCount="163" Body="&lt;p&gt;Let $f\in\mathbb{Z}[x]$ be monic and irreducible, let $K=$ splitting field of $f$ over $\mathbb{Q}$. What can we say about the relationship between $disc(f)$ and $\Delta_K$? I seem to remember that one differs from the other by a multiple of a square, but I don't know which is which. On a more philosophical note: why are these quantities related at all? Is there an explanation for why they can be different, i.e. some information that one keeps track of that the other doesn't? &lt;/p&gt;&#xA;" OwnerUserId="264" LastActivityDate="2011-05-07T19:27:31.000" Title="Discriminant of a monic irreducible integer polynomial vs. discriminant of its splitting field" Tags="&lt;algebraic-number-theory&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="8313" PostTypeId="2" ParentId="8312" CreationDate="2010-10-29T20:32:34.303" Score="5" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; Every time I say &quot;splitting field&quot; in the following answer I really mean $\mathbb{Q}[x]/f(x)$. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The two are the same if the roots of $f$ form an integral basis of the ring of integers of the splitting field (e.g. if $f$ is a cyclotomic polynomial) because then, well, they're defined by the same determinant (see &lt;a href=&quot;http://en.wikipedia.org/wiki/Discriminant_of_an_algebraic_number_field#Examples&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;), but in general they don't.  In the general case $\mathbb{Z}[\alpha_1, ... \alpha_n]$ is an order in $\mathcal{O}_K$ so one can write the $\alpha_i$ as an integer linear combination of an integral basis, so the matrices whose determinants define the two discriminants should be related by the square of a matrix with integral entries, hence integral determinant.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact if I'm not totally mistaken, the quotient of the two discriminants should be precisely the index of $\mathbb{Z}[\alpha_1, ... \alpha_n]$ in $\mathcal{O}_K$ as lattices, or maybe its square...?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In any case, since the discriminant of the field is defined in terms of $\mathcal{O}_K$ it is the &quot;right&quot; choice for carrying information about, for example, ramification.  One can see this even in the quadratic case: if $d \equiv 1 \bmod 4$ then the discriminant of $x^2 - d$ is $4d$ but the discriminant of $\mathbb{Q}(\sqrt{d})$ is $d$, and the latter is the &quot;right&quot; choice because $2$ doesn't ramify in $\mathbb{Z} \left[ \frac{1 + \sqrt{d}}{2} \right]$.  &lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2011-05-07T19:27:31.000" LastActivityDate="2011-05-07T19:27:31.000" CommentCount="1" />
  <row Id="8314" PostTypeId="1" AcceptedAnswerId="8316" CreationDate="2010-10-29T20:45:59.703" Score="1" ViewCount="192" Body="&lt;p&gt;I know a few proofs over this theorem (where $\mathfrak{c}$ is the cardinality of $[0,1]$ and $\aleph_0$ is the cardinality of $\mathbb{N}$) where they construct two injections and then use Schröder-Bernstein (via the Cantor set or something like that).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I was wondering if I could do something like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Define $s: \{0,1\}^\mathbb{N} \to [0,1]$ by&#xA;$$s(x) = \sum_{i = 1}^\infty \frac{x(i)}{2^i}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now this is clearly a surjection because this is just a binary expansion of numbers in $[0,1]$, but not injective because these expansions are not unique. Is there a way to make this work? Can I use this to construct a bijection?&lt;/p&gt;&#xA;" OwnerUserId="1120" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2010-11-21T14:21:57.040" LastActivityDate="2010-11-21T14:21:57.040" Title="$\mathfrak{c} = 2^{\aleph_0}$." Tags="&lt;elementary-set-theory&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="8315" PostTypeId="1" CreationDate="2010-10-29T20:49:21.623" Score="2" ViewCount="164" Body="&lt;p&gt;Consider a function $f_1$ defined by $f_1(x)=1-x+o(x)$ and $f_1(2x)=f_1(x)^2 + 0$. It's simple to find that $f_1(x)=e^{-x}$ (for example by writing series near $x=0$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider a function $f_2$ defined by $f_2(x)=2-x^2+o(x^2)$ and $f_2(2x)=f_2(x)^2-2$. It can be proven that $f_2(x)=2 \cos(x)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any formula (probably with use of special functions) for the generalization of this, i.e. function $f_n$ defined by $f_n(x)=2^{n-1}-x^n+o(x^n)$ and $f_n(2x)=f_n(x)^2-2^{2n-2}+2^{n-1}$?&lt;/p&gt;&#xA;" OwnerUserId="695" LastEditorUserId="695" LastEditorDisplayName="" LastEditDate="2010-10-30T07:18:58.207" LastActivityDate="2010-10-30T07:18:58.207" Title="Generalization of cos: is this function known?" Tags="&lt;functions&gt;&lt;special-functions&gt;&lt;functional-equations&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="8316" PostTypeId="2" ParentId="8314" CreationDate="2010-10-29T20:53:46.353" Score="7" ViewCount="" Body="&lt;p&gt;Yes, but in my opinion, it's really not worth it.  Binary expansions are unique for all real numbers except dyadic rationals $\frac{k}{2^n}$, since they can end with a string of zeroes or a string of ones.  But there are only countably many dyadic rationals.  So define $s : \{ 0, 1 \}^{\mathbb{N}} \to [0, 1]$ to be what you said for all strings that don't end with a string of zeroes or a string of ones, and for the countably many exceptions pick any bijection you like.  For example, if a sequence $a_i \in \{ 0, 1 \}^{\mathbb{N}}$ ends with a string of ones, send it to $0.1 a_1 a_2 a_3 ...$, and if it ends with a string of zeroes, send it to $0.0 a_1 a_2 a_3 ...$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But really, there's no point in not using Schroeder-Bernstein.  Explicit bijections are highly overrated; in general it is much easier to construct an injection and a surjection.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-10-29T20:53:46.353" CommentCount="3" />
  <row Id="8318" PostTypeId="1" CreationDate="2010-10-29T21:32:04.870" Score="0" ViewCount="164" Body="&lt;p&gt;I'm reading through &lt;a href=&quot;http://en.wikipedia.org/wiki/LFSR#Fibonacci_LFSRs&quot; rel=&quot;nofollow&quot;&gt;wikipedia&lt;/a&gt; about Linear Feedback Shift Registers (Specifically Fibonacci LFSRs) and the only restrictions it mentions about the taps are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The LFSR will only be maximum-length if the number of taps is even; just 2 or 4 taps can suffice even for extremely long sequences.&lt;/li&gt;&#xA;&lt;li&gt;The set of taps must be relatively prime, and share no common divisor to all taps.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;It has two other bullet points basically saying that there can be more than one working set of taps, and if you find one set of taps, you can use that to determine another set of taps. Is this all it takes to find taps?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are these the only restrictions to finding taps in a Fibonacci LFSR?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know there are plenty of reference tables to look up taps, but I'm trying to understand the fundamentals of an LFSR better.&lt;/p&gt;&#xA;" OwnerUserId="757" LastActivityDate="2010-10-30T18:08:44.893" Title="How are the taps of an LFSR found?" Tags="&lt;number-theory&gt;" AnswerCount="1" />
  <row Id="8319" PostTypeId="1" AcceptedAnswerId="8381" CreationDate="2010-10-29T21:44:35.780" Score="7" ViewCount="281" Body="&lt;p&gt;Why $\frac{1}{\cos x}=\sum_{n=1}^\infty \frac{(-1)^n(2n-1)\pi }{x^2-\left (n-\frac{1}{2}\right )^2\pi^2}$?&lt;/p&gt;&#xA;" OwnerUserId="2856" LastActivityDate="2010-10-31T06:44:08.973" Title="Series representation for 1/cos x" Tags="&lt;sequences-and-series&gt;" AnswerCount="1" CommentCount="7" FavoriteCount="1" />
  <row Id="8320" PostTypeId="2" ParentId="8234" CreationDate="2010-10-29T22:52:41.660" Score="0" ViewCount="" Body="&lt;p&gt;Quoting Adrián Barquero:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;This doesn't determine B as the set of&#xA;  positive integers. Basically you're&#xA;  just saying that B is an inductive&#xA;  set. In fact, if you look at the&#xA;  definition you gave from Munkres's&#xA;  book, the set of positive integers&#xA;  being defined as the intersection of&#xA;  all inductive sets is just saying that&#xA;  the positive integers constitute a&#xA;  minimal inductive set.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1523" LastActivityDate="2010-10-29T22:52:41.660" CommunityOwnedDate="2010-10-29T22:52:41.660" />
  <row Id="8321" PostTypeId="1" AcceptedAnswerId="8333" CreationDate="2010-10-29T23:24:28.970" Score="2" ViewCount="185" Body="&lt;p&gt;Hello,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In case you don't know what I'm talking about:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Weierstrass%27s_elliptic_functions&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Weierstrass%27s_elliptic_functions&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm reading Farkas and Kra's &quot;Riemann Surfaces,&quot; and they talk briefly about these guys in section 0.2. Please tell me if I am misunderstanding something or saying something wrong or funny. Let $\wp$ be the Weierstrass function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From what I understand, $\wp$ maps the torus to $CP^1 \times CP^1$ in the following way:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a \mapsto (\wp(a),\wp'(a)) = (z,w)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore, the image of this map lies on the zero set of the polynomial $P(z,w) = 4(z-e_1)(z-e_2)(z-e_3) - w^2.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I don't get is the description of the inverse to this map. It says that the integral of the differential form $\frac{dz}{w}$ from $\infty$ to a point $Q$ along a path $c$ is the inverse of this map.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;I don't understand what they mean by $\infty$ here.&lt;/li&gt;&#xA;&lt;li&gt;I don't understand why this is the inverse to the map.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Heuristically, I see that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;\int_\infty^Q \frac{dz}{w} = \int_0^z \frac{\wp'(u)}{\wp'(u)} du = \int_0^a du = a&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but unfortunately, I have no clue what I just did here. I mean, I guess I did something like a pull back? (Set $z=\wp$ and $w=\wp'$. Isn't this a map into the complex plane, and not the Torus?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a related question, but why is infinity the branch point?  In general, when you have $w^2 = p(z)$, with degree of $p$ odd, why is infinity one of the branch points? Why is it when the degree of $p$ is even, it is not a branch point?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you for your time!&lt;/p&gt;&#xA;" OwnerUserId="2499" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-10-30T13:04:06.677" LastActivityDate="2010-10-30T13:04:06.677" Title="A Torus and the Weierstrass P function??" Tags="&lt;complex-analysis&gt;&lt;riemann-surface&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="8322" PostTypeId="1" CreationDate="2010-10-29T23:51:12.830" Score="4" ViewCount="144" Body="&lt;p&gt;Given an infinite dimensional manifold modeled on a Banach space, what does it mean for it to have a Riemannian metric? Does it necessarily mean that it is actually a Hilbert manifold?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My understanding is that Hilbert Spaces have inner products, whereas Banach Spaces just have norms. If a Banach manifold has a Riemannian metric, it means that at each tangent space, which is a Banach space, it has an inner product... so wouldn't that make any Banach manifold a Hilbert Manifold?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm obviously completely confused and mixing up something.&lt;/p&gt;&#xA;" OwnerUserId="2499" LastActivityDate="2010-10-30T19:13:54.393" Title="A Banach Manifold with a Riemannian Metric?" Tags="&lt;soft-question&gt;&lt;inner-product-space&gt;&lt;manifold&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" />
  <row Id="8323" PostTypeId="1" AcceptedAnswerId="8340" CreationDate="2010-10-30T00:42:41.487" Score="8" ViewCount="411" Body="&lt;p&gt;Dear friends,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is known that the number of digits of a natural number n&gt;0, which represent by d (n) is given by:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;d(n)= 1 + floor(log n)   (I)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(&lt;em&gt;log indicates logarithm base 10&lt;/em&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well .. the classical approach to the Stirling factorial natural number n&gt;1 is given by:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;n! ~ f(n) = [(2.n.pi) ^ (1 / 2)]. [(n / e) ^ n]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(The symbol ~ indicates &quot;approximately.&quot;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The number of digits n!, according to equality (I), is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;d(n!) = 1 + floor(log n!)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;It seems to me that for all natural n&gt; 1, log n! and log [f (n)] have the same floor:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;floor[log(n!)] = floor[log(f(n))]&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Here's my big question!&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, we could write:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;d (n!) = 1 + floor[log(f(n))]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope someone has some little time for the theme.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sincerely,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Paulo Sergio Argolo&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rio de Janeiro, Brazil&lt;/p&gt;&#xA;" OwnerUserId="2938" LastEditorUserId="2938" LastEditorDisplayName="" LastEditDate="2010-11-01T19:39:18.293" LastActivityDate="2010-11-02T13:50:06.387" Title="Does Stirling's formula give the correct number of digits of n!?" Tags="&lt;number-theory&gt;" AnswerCount="2" CommentCount="14" />
  <row Id="8324" PostTypeId="2" ParentId="8133" CreationDate="2010-10-30T00:43:37.700" Score="4" ViewCount="" Body="&lt;p&gt;I'll just answer your last question for now.  It is not possible for the zeta function to have a single zero off the critical line.  The functional equation for the Riemann zeta function immediately shows that the zeros in the critical strip are symmetric about the line Re z=1/2.  Hence, if there is one zero, there must be two.  Plus, $\overline{\zeta(\bar s)}=\zeta(s)$.  So, if you have one zero, you have four.&lt;/p&gt;&#xA;" OwnerUserId="2937" LastActivityDate="2010-10-30T00:43:37.700" />
  <row Id="8326" PostTypeId="1" CreationDate="2010-10-30T01:05:42.380" Score="2" ViewCount="272" Body="&lt;p&gt;Is there a good quick introduction to knot theory?  I am relatively mathematically savvy so any level is appreciated.&lt;/p&gt;&#xA;" OwnerUserId="355" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-10-30T01:18:30.303" LastActivityDate="2010-11-02T21:45:05.527" Title="A good quick introduction to Knot Theory?" Tags="&lt;recreational-mathematics&gt;&lt;knot-theory&gt;" AnswerCount="7" CommentCount="3" FavoriteCount="1" />
  <row Id="8327" PostTypeId="2" ParentId="8315" CreationDate="2010-10-30T01:25:00.610" Score="5" ViewCount="" Body="&lt;p&gt;Assuming Willie Wong's correction, by a method of undetermined coefficients it's not too hard to generate a series for  $f_3$, and it begins&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ 4-x^3+(1/56)x^6-(1/14112)x^9+(13/115379712)x^{12}-(53/629973227520)x^{15}+(17413/495415985907548160)x^{18} + \cdots$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but it's not clear what to do with this. For example, the denominators have large-ish prime factors: $495415985907548160 = 2^{14} 3^4 5^1 7^5 13^1 31^1 73^1 151^1$. This seems to suggest that there's not going to be an easy third-order differential equation, whereas your $f_1$ and $f_2$ satisfy first- and second-order differential equations.&lt;/p&gt;&#xA;" OwnerUserId="173" LastActivityDate="2010-10-30T01:25:00.610" />
  <row Id="8328" PostTypeId="2" ParentId="8326" CreationDate="2010-10-30T01:39:09.663" Score="7" ViewCount="" Body="&lt;p&gt;As Michael comments, Colin Adams has a well regarded text called &quot;The Knot Book&quot;. Adams has also written a comic book about knot theory called &quot;Why Knot?&quot;. It's very humorous but is a genuine introduction to the mathematics involved. This comic book comes with a plastic &quot;rope&quot; that can be knotted, unknotted, and twisted into different shapes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think &quot;Why Knot?&quot; qualifies well as a &quot;good, quick&quot; introduction to the topic. Well worth tracking down.  &lt;/p&gt;&#xA;" OwnerUserId="940" LastActivityDate="2010-10-30T01:39:09.663" CommunityOwnedDate="2010-10-30T01:39:09.663" CommentCount="2" />
  <row Id="8329" PostTypeId="2" ParentId="8326" CreationDate="2010-10-30T01:47:30.627" Score="1" ViewCount="" Body="&lt;p&gt;Two books I would recommend are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Knots and Surfaces, David Farmer and Theodore Standford (American Mathematical Society; 1996)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Knots and Links, Peter Cromwell (Cambridge U. Press; 2004).&lt;/p&gt;&#xA;" OwnerUserId="1369" LastActivityDate="2010-10-30T01:47:30.627" />
  <row Id="8330" PostTypeId="2" ParentId="8288" CreationDate="2010-10-30T01:54:01.897" Score="2" ViewCount="" Body="&lt;p&gt;You might want to look at:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.michaelbeeson.com/research/papers/TriangleTiling.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.michaelbeeson.com/research/papers/TriangleTiling.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and the references given there.&lt;/p&gt;&#xA;" OwnerUserId="1369" LastActivityDate="2010-10-30T01:54:01.897" CommentCount="2" />
  <row Id="8331" PostTypeId="1" AcceptedAnswerId="8335" CreationDate="2010-10-30T02:17:21.750" Score="3" ViewCount="272" Body="&lt;p&gt;I've just started learning about schemes, so maybe I'm missing something basic. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is exercise I-24(a):&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Take Z = Spec$\mathbb{C}[x]$, let $X$ be the result of identifying the two closed points (x) and (x-1) of |Z|, and let $\phi: Z \to X$ be the natural projection. Let $\mathcal{O}$ be $\phi_* \mathcal{O}_Z$, a sheaf of rings on&#xA;$X$. Show that $(X, \mathcal{O})$ satisfies condition (i) above for all elements $f \in \mathcal{O}(X) = \mathbb{C}[x]$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The condition (i) referred to: For any $f \in \mathbb{C}[x]$ define $U_f \subset X$ as the set of points $x \in X$ such that $f$ maps to a unit of the stalk $\mathcal{O}_x$. (i) means that $\mathcal{O}(U_f) = \mathbb{C}[x][f^{-1}]$ for all f. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;But how can this be? Put f = x. Then &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$U_f = X \setminus \{(x)\}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\phi^{-1}(U_f) = Z \setminus \{ (x), (x-1) \}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\mathcal{O}(U_f) = \mathcal{O}_Z(\phi^{-1}(U_f)) = \mathbb{C}[x][ ((x)(x-1))^{-1} ]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And that is not $\mathbb{C}[x][f^{-1}]$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: Regarding the answer and comments. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;evgeniamerkulova's answer reassures me that I'm not out of my mind, but obviously Matt E and Mariano know what they're talking about, so I don't know what to think. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both Mariano and Matt E imply that $\mathcal{O}(X)$ is not $\mathbb{C}[x]$, but that seems obviously wrong (and contradicts the book itself). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's my reasoning, spelled out. O(X) is C[x]. This is because $\mathcal{O}_Z(\phi^{-1}(X)) = \mathcal{O}_Z(Z) = \mathbb{C}[x]$. In order for the condition to be satisfied, we need $\mathcal{O}(U) = \mathbb{C}[x,x^{-1}]$ for some open U in X. So we need $\mathcal{O}_Z(\phi^{-1}(U)) = \mathbb{C}[x,x^{-1}]$. For that to happen we need $\phi^{-1}(U) = Z \setminus \{ (x) \}$. But all the inverse images of sets in X either include both (x) and (x-1) or neither of them, so this can never happen. &lt;/p&gt;&#xA;" OwnerUserId="566" LastEditorUserId="566" LastEditorDisplayName="" LastEditDate="2010-10-30T15:08:25.887" LastActivityDate="2010-10-30T15:08:25.887" Title="Exercise from Eisenbud &amp; Harris's The Geometry of Schemes" Tags="&lt;algebraic-geometry&gt;" AnswerCount="1" CommentCount="9" FavoriteCount="1" />
  <row Id="8332" PostTypeId="1" AcceptedAnswerId="22240" CreationDate="2010-10-30T03:02:11.513" Score="7" ViewCount="278" Body="&lt;p&gt;Can we always break an arbitrary field extension $L/K$ into an extension $F/K$ in which the only roots of unity of $F$ are those in $K$, followed by an extension $L/F$ which is of the form $L=F(\{\omega_i\})$ where the $\omega_i$ are roots of unity? What if we restrict to $L/K$ separable, or finite?&lt;/p&gt;&#xA;" OwnerUserId="264" LastEditorUserId="264" LastEditorDisplayName="" LastEditDate="2010-11-07T01:08:48.857" LastActivityDate="2011-02-15T22:57:36.960" Title="Roots of unity and field extensions" Tags="&lt;abstract-algebra&gt;&lt;field-theory&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="5" />
  <row Id="8333" PostTypeId="2" ParentId="8321" CreationDate="2010-10-30T03:17:24.580" Score="5" ViewCount="" Body="&lt;p&gt;It is better to think of mapping to $\mathbb C P^2$ rather than a product of $\mathbb C P^1$s.  The point $\infty$ is then the unique point at infinity on the cubic $y^2 = 4(x-e_1)(x-e_2)(x-e_3).$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, the integral depends on the path, not just the endpoint; the ambiguity in the value of the integral if you just specify the endpoints is given by integrating over closed loops, i.e. over representatives of the 1st homology.  These integrals span a lattice $\Lambda$ in $\mathbb C$ (the original lattice with respect to which you defined $\wp$), and so the integrals really take values in $\mathbb C/\Lambda$, not in $\mathbb C$.&lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2010-10-30T03:17:24.580" CommentCount="3" />
  <row Id="8334" PostTypeId="1" AcceptedAnswerId="8685" CreationDate="2010-10-30T04:44:38.530" Score="4" ViewCount="138" Body="&lt;p&gt;I find confusing some examples I have seen. Maybe you can help me to determine what is going on with them. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A Generalized Feedback Shift Register (GFSR) sequence defines a sequence $\{W_{i}\}$ satisfying the equation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$W_{k+p}=c_{0}W_{k}\bigoplus c_{1}W_{k+1}\bigoplus...\bigoplus c_{p-1}W_{k+p-1} \qquad \qquad (1)$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\bigoplus$ is the binary exclusive-or operation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the polynomial $f(x)=c_{0}+c_{1}x+c_{2}x^{2}+...+c_{p-1}x^{p-1}+x^{p}$ is a primitive polynomial over $GF(2)$, then the sequence $\{W_{i}\}$ will have maximal sequence $2^{p}-1$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example 1: Let's consider the trinomial $1+x+x^{4}$ and a bit sequence $1, 0, 1, 0$. For the polynomial we have $c_{0}=1$, $c_{1}=1$, $c_{2}=0$, $c_{3}=0$ and $p=4$. Therefore, the equation $(1)$ should be $W_{k+4}=W_{k}\bigoplus W_{k+1}$. According to this, we calculate the values for $W_{5}, W_{6},...$ etc (since we already know that $W_{1}=1, W_{2}=0, W_{3}=1, W_{4}=0$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This procedure generates the following sequence&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$1,0,1,0,1,1,1,1,0,0,0,1,0,0,1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then the example takes 4 bit chunks (changing to decimal representation):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$1010=10, 1111=15, 0001=1, 0011=3, 1010+1111=0101=5, 0001+0011=0010=2$ and so on. So a '4-wise decimation' using the recurrence yields the numbers&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$10, 15, 1, 3, 5, 2, ...$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this a standard way to generate a bigger sequence?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example 2: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;blockquote&gt;&#xA;    &lt;p&gt;By the using the bit stream from the trinomial $1+x+x^{4}$ and the starting sequence $1,0,1,0$, and... forming 4-bit words by putting the bits into a fixed binary position with a delay of 3 between binary positions, we have&#xA;    $$1010=10, 1110=14, 0011=3, 0101=5, 1111=15, 0001 = 1, 0010=2, 0111=7,...$$&lt;/p&gt;&#xA;  &lt;/blockquote&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Well, both examples are dealing with exactly the same problem. However, they lead to different sequences. I don't even know how the second example generates its sequence (it looks like it is taking the first bit sequence $1, 0, 1, 0$ and applying the binary exclusive-or operation for the first two terms $1 \bigoplus 0 = 1$ which is the first term of the following bit sequence, then take the second and third terms $0 \bigoplus 1=1$ which is the second term of the new sequence and so on). However, I don't know how it gets the last term. Such a pattern works for all the sequences of the Example 2, which makes me thing that I'm not seeing the full picture.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ideas?&lt;/p&gt;&#xA;" OwnerUserId="2569" LastEditorUserId="2569" LastEditorDisplayName="" LastEditDate="2010-10-30T16:06:18.307" LastActivityDate="2011-01-01T20:44:30.010" Title="Generalized Feedback Shift Registers" Tags="&lt;cryptography&gt;&lt;random&gt;" AnswerCount="1" />
  <row Id="8335" PostTypeId="2" ParentId="8331" CreationDate="2010-10-30T08:47:33.240" Score="2" ViewCount="" Body="&lt;p&gt;Everything you say is right and Eisenbud and  Harris are false. I don't inderstand Matt E ' comments either because you did no mistake:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a) That &quot;x&quot; is not function on $X$ has nothing to do with problem, and you never said it was function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;b) He writes &quot;To compute correctly on $X$, you need to figure out what $\mathcal O(X)$ is&quot; : you have computed that it is   $\mathbb C[x]$ and you are right.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For completeness stalk $\mathcal O_a$ of $\mathcal O$ at  quotient point $a\in X$  corresponding to $0,1$ [you write x, but you may not because x is already polynomial] is  ring $S \subset \mathbb C(x)$ of all fractions $f(x)/g(x)$  ($f(x), g(x) \in \mathbb C [x])$)  such that $g(0)\neq 0$ and $ g(1)\neq 0$&lt;/p&gt;&#xA;" OwnerUserId="2872" LastActivityDate="2010-10-30T08:47:33.240" CommentCount="3" />
  <row Id="8336" PostTypeId="1" AcceptedAnswerId="8370" CreationDate="2010-10-30T09:41:47.130" Score="0" ViewCount="122" Body="&lt;p&gt;There was a problem in &lt;em&gt;Apostol's&lt;/em&gt; book namely, to prove that:&#xA;$$\sum\limits_{k=1}^{n} \Biggl\lfloor{\frac{k}{2}\Biggr\rfloor} = \Biggl\lfloor{\frac{n^{2}}{4}\Biggr\rfloor}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which i could solve. The following probelm is eluding me:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If $a =1,2, \cdots, 7$, prove that there exists an integer $b$ depending on $a$ such that $$\sum\limits_{k=1}^{n} \Biggr\lfloor{\frac{k}{a}\Biggr\rfloor} =\Biggl\lfloor{\frac{(2n+b)^{2}}{8a}\Biggr\rfloor} $$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" LastActivityDate="2010-10-30T18:01:15.967" Title="proving $\sum\limits_{k=1}^{n} \Bigl\lfloor{\frac{k}{a}\Bigr\rfloor} =\Bigl\lfloor{\frac{(2n+b)^{2}}{8a}\Bigr\rfloor} $" Tags="&lt;number-theory&gt;" AnswerCount="2" CommentCount="5" />
  <row Id="8337" PostTypeId="1" CreationDate="2010-10-30T10:04:43.110" Score="23" ViewCount="1084" Body="&lt;p&gt;As I have heard people did not trust Euler when he first discovered the formula&#xA;$$\zeta(2)=\sum_{n=1}^\infty \frac{1}{n^2}=\frac{\pi^2}{6}.$$&#xA;However, Euler was Euler and he gave other proofs. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe many of you know some nice proofs of this, can you please share it with us?!&lt;/p&gt;&#xA;" OwnerUserId="1154" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-10-31T08:32:00.257" LastActivityDate="2011-04-18T14:57:34.097" Title="Different methods to compute $\sum_{n=1}^\infty \frac{1}{n^2}$." Tags="&lt;sequences-and-series&gt;&lt;complex-analysis&gt;&lt;big-list&gt;&lt;fourier-analysis&gt;" AnswerCount="8" CommentCount="10" FavoriteCount="14" />
  <row Id="8338" PostTypeId="2" ParentId="8326" CreationDate="2010-10-30T10:17:24.640" Score="5" ViewCount="" Body="&lt;p&gt;I love the book &quot;On Knots&quot; by Louis Kauffman. It's got a playful style, yet he develops a lot of deep mathematics. I read this in high school, and I got quite a lot out of it, and as my mathematical knowledge progressed, I got more and more out of it.&lt;/p&gt;&#xA;" OwnerUserId="2941" LastActivityDate="2010-10-30T10:17:24.640" />
  <row Id="8339" PostTypeId="1" AcceptedAnswerId="8407" CreationDate="2010-10-30T10:39:18.517" Score="0" ViewCount="586" Body="&lt;p&gt;Suppose A and B are similar matrices. Show that A and B have the same eigenvalues with the same geometric multiplicities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similar matrices; Suppose A and B are $n\times n$ matrices over $\mathbb R$ or $\mathbb C$. We say A and B are similar, or that A is similar to B, if there exists a matrix P such that $B = P^{-1}AP$&lt;/p&gt;&#xA;" OwnerUserId="2723" LastEditorUserId="171" LastEditorDisplayName="" LastEditDate="2010-10-30T10:42:23.983" LastActivityDate="2010-10-30T21:36:00.027" Title="Suppose A and B are similar matrices. Show that A and B have the same eigenvalues with the same geometric multiplicities." Tags="&lt;linear-algebra&gt;&lt;matrices&gt;" AnswerCount="4" CommentCount="1" />
  <row Id="8340" PostTypeId="2" ParentId="8323" CreationDate="2010-10-30T10:52:30.393" Score="4" ViewCount="" Body="&lt;p&gt;Here are some thoughts on the conjecture that may lead one to suspect that it is true. This is not a proof that it is true.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We want to know whether&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \left\lfloor \log_{10} n! \right\rfloor = \left\lfloor \log_{10} \lfloor f(n) \rfloor \right\rfloor$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is true for all $n &gt; 1.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We note that this would NOT be true if the interval&#xA;$I_n = ( \log_{10} n!,\log_{10} \lfloor f(n) \rfloor)$ contains an integer but is true otherwise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So let's look at the length of $I_n.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the &lt;a href=&quot;http://en.wikipedia.org/wiki/Stirling%27s_approximation&quot; rel=&quot;nofollow&quot;&gt;Stirling series&lt;/a&gt; we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{1}{\log_{e}10} \left( \frac{1}{12n} - \frac{1}{360n^3} \right) &amp;lt; &#xA; \log_{10} n! -  \log_{10} f(n) &amp;lt; &#xA;\frac{1}{\log_{e}10} \frac{1}{12n}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And so taking the integer part of $f(n)$ we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{1}{\log_{e}10} \left( \frac{1}{12n} - \frac{1}{360n^3} \right) &amp;lt; &#xA; \log_{10} n! -  \log_{10} \lfloor f(n) \rfloor &amp;lt; &#xA;\frac{1}{\log_{e}10} \left( \frac{1}{12n} + \frac{1}{ \lfloor f(n) \rfloor } \right),$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where to achieve the RHS we note that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \log_{10} f(n) - \log_{10} \lfloor f(n) \rfloor  &amp;lt; \frac{1}{ \lfloor f(n) \rfloor \log_e 10 }.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence&#xA;$$\text{Length}(I_n) &amp;lt; \frac{1}{\log_{e}10} \left( \frac{1}{360n^3} +&#xA;\frac{1}{ \lfloor f(n) \rfloor } \right).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can verify that the conjecture holds for $n=2,3,\ldots,10,$ so summing up the remaining lengths of the $I_n$ we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{n=11}^\infty \text{Length}(I_n) &amp;lt; \frac{1}{360 \log_e 10}&#xA;\left( \sum_{n=11}^\infty \frac{1}{n^3} +  \sum_{n=11}^\infty \frac{1}{ \lfloor f(n) \rfloor } \right).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now $ \lfloor f(n) \rfloor  \ge (n-1)! $ and so, doing some calculations (replacing the&#xA;$ \lfloor f(n) \rfloor $ on the RHS by $ (n-1)! $ , we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{n=11}^\infty \text{Length}(I_n) &amp;lt; \frac{1}{360 \log_e 10}&#xA;\left( 0.00452492 + 0.00000030 \right) &amp;lt; 5.5 \times 10^{-6}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence the probability that an integer falls in any of the intervals is very small.&lt;/p&gt;&#xA;" OwnerUserId="1301" LastEditorUserId="1301" LastEditorDisplayName="" LastEditDate="2010-10-30T16:49:09.177" LastActivityDate="2010-10-30T16:49:09.177" CommentCount="2" />
  <row Id="8341" PostTypeId="1" AcceptedAnswerId="8343" CreationDate="2010-10-30T11:09:29.820" Score="2" ViewCount="139" Body="&lt;p&gt;What is the name for I can only describe as the circumference of a sphere. I mean like the equator on a planet. What is the line called which goes around the entire sphere, it can be anywhere (so not just in the middle of the top and the bottom, if your talking about a planet or something) and the length of it equals 2πr.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another way to describe it is if you put a plane through sphere at any angle so that 2 equal halves are created, what is the name of the line where the plane intersect the sphere's surface.&lt;/p&gt;&#xA;" OwnerUserId="2075" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-10-30T13:06:54.600" LastActivityDate="2010-10-30T13:06:54.600" Title="&quot;circumference&quot; of a sphere" Tags="&lt;geometry&gt;&lt;terminology&gt;" AnswerCount="1" />
  <row Id="8342" PostTypeId="2" ParentId="8339" CreationDate="2010-10-30T11:12:53.913" Score="2" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;hint&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \det (P^{-1}P - P^{-1}A P) = \det( P^{-1}(I - A)P ) = \det P^{-1} \cdot \det (I - A) \cdot \det P $$ &lt;/p&gt;&#xA;" OwnerUserId="1543" LastActivityDate="2010-10-30T11:12:53.913" CommentCount="2" />
  <row Id="8343" PostTypeId="2" ParentId="8341" CreationDate="2010-10-30T11:14:51.287" Score="6" ViewCount="" Body="&lt;p&gt;A &quot;great circle&quot;, so called because they are the largest possible circles on a sphere. If you're on a sphere (say, Earth) and you want to go between two points using the shortest path, then a great circle is the path you should follow. This makes them like straight lines for spheres.&lt;/p&gt;&#xA;" OwnerUserId="1149" LastActivityDate="2010-10-30T11:14:51.287" CommentCount="1" />
  <row Id="8345" PostTypeId="2" ParentId="8339" CreationDate="2010-10-30T11:20:29.667" Score="0" ViewCount="" Body="&lt;p&gt;It's basically something like:  If $v_1,\dots,v_k$ is a basis for the eigenspace of A corresponding to eigenvalue $\lambda$, then $P^{-1}v_1,\dots,P^{-1}v_k$ is a basis for the eigenspace of B corresponding to the same eigenvalue.  There are some details to be filled in, like showing that any vector $w$ with $Bw=\lambda w$ can be written as a linear combination of the $P^{-1}v_i$ by using that the same holds for the matrix $A$ and the vectors $v_i$ by assumption.&lt;/p&gt;&#xA;" OwnerUserId="2942" LastActivityDate="2010-10-30T11:20:29.667" />
  <row Id="8346" PostTypeId="1" CreationDate="2010-10-30T11:48:18.783" Score="2" ViewCount="92" Body="&lt;p&gt;What requirements should a CRC polynomial of a given degree satisfy to make the CRC catch a maximum of errors?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;edit&lt;/strong&gt;&lt;br&gt;&#xA;I'm talking about &lt;strong&gt;GF(2) polynomials&lt;/strong&gt;.&lt;br&gt;&#xA;As an example of the kind of requirements I'm looking for: I can imagine (but don't know for sure) that a prime polynomial catches more errors than a composite polynomial.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not a mathematician, so please type slowly :-)&lt;/p&gt;&#xA;" OwnerUserId="923" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-11-01T12:17:04.637" LastActivityDate="2010-11-01T12:17:04.637" Title="What requirements should a CRC polynomial satisfy?" Tags="&lt;computer-science&gt;&lt;finite-fields&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="8347" PostTypeId="1" AcceptedAnswerId="8414" CreationDate="2010-10-30T12:21:58.610" Score="0" ViewCount="111" Body="&lt;p&gt;I hate min and max questions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a function&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;G(x, y) = 1/x + 4/y + 9/(4-x-y)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I need to prove that G(x,y) has a minimum value of 9&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've partialy derived them for x and y giving&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;(1) 1/x^2 + 9/(4-x-y)^2 = 0&#xA;(2) 4/y^2 + 9/(4-x-y)^2 = 0&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This is where I get stuck I'm not entirely sure how to continue. I need to find x = ? and y = ? but this simultaneous equation stumps me.&lt;/p&gt;&#xA;" OwnerUserId="2874" LastActivityDate="2010-10-31T04:39:48.330" Title="Two Var Function." Tags="&lt;calculus&gt;&lt;functions&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="0" />
  <row Id="8348" PostTypeId="1" CreationDate="2010-10-30T12:22:30.330" Score="7" ViewCount="352" Body="&lt;p&gt;It is a rather surprising fact (to me, at least) that $\displaystyle \binom{14}{4} = 1001$; $\displaystyle \binom{14}{5} = 2002$; $\displaystyle \binom{14}{6} = 3003$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Actually, this is the only instance where three consecutive binomial coefficients are in the ratio $\displaystyle 1 : 2 : 3$. I found it quite interesting to investigate under what conditions consecutive members of a row of Pascal’s triangle can be in the ratio $\displaystyle a : b : c$, where $\displaystyle a,b,c$ are positive integers with  $\displaystyle \mathrm{gcd}(a,b,c) = 1$ and $\displaystyle a &amp;lt; b &amp;lt; c$, except where otherwise stated. That is, only the left-hand side of the triangle will be considered.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\binom{n}{k} : \binom{n}{k+1} : \binom{n}{k+2} = a : b : c$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cancelling and rearranging,  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle b(k + 1) = a(n - k)$......….....[1]&lt;br&gt;&#xA;$\displaystyle c(k + 2) = b(n - k - 1)$.........[2]               &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$n = \frac{a(b + c) + c(a + b)}{(b^2 - ac)}$$                                  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$k = \frac{a(b + c)}{b^2 - ac} - 1 $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore n and k are integers iff $\displaystyle b^2-ac$ divides both $\displaystyle a(b + c)$ and $\displaystyle c(a + b)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle n &gt; 0$ implies $\displaystyle b^2 &gt; ac$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle k\ge 0$ implies $\displaystyle c \ge \frac{b(b - a)}{2a}$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence a third condition is $\displaystyle \frac{b^2}{a} &gt; c \ge \frac{b(b - a)}{2a}$        &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps the most interesting special case is $\displaystyle c = a + b$, when for &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle a,b &amp;lt; 100$ there are only five solutions. Namely,  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle (a,b,c) = (1,2,3)$ gives $\displaystyle \{n,k\} = \{14,4\}$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle (a,b,c) = (3,5,8)$ gives $\displaystyle \{n,k\} = \{103,38\} $ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle (a,b,c) = (8,13,21)$ gives $\displaystyle \{n,k\} = \{713,271\}$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle (a,b,c) = (21,34,55)$ gives $\displaystyle \{n,k\} = \{4894,1868\}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle (a,b,c) = (55,89,144)$ gives $\displaystyle \{n,k\} = \{33551,12814\}$    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is, there are solutions only when $\displaystyle (a,b,c) = (F(2m), F(2m + 1), F(2m + 2)) $ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle m = 1,2,3...,$ where $\displaystyle F(m)$ is the $\displaystyle m^{th}$ Fibonacci number.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Generally,  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle \{n,k\} = \{F(2m + 2)F(2m + 3) - 1, F(2m)F(2m + 3) - 1\} $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All solutions satisfy $\displaystyle F(2m)n = F(2m+2)k + F(2m+1) $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where possible I have been able to derive formulae for $\displaystyle \{n,k\}$ for all special cases I could think of (eg. $\displaystyle a,b,c$ in arithmetic progression) except for the case $\displaystyle c = a^2$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $\displaystyle a,b &amp;lt; 3000$ there are only three solutions:   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle (a,b,c) = (1,2,1)$ gives $\displaystyle \{n,k\} = \{2,0\}$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle (a,b,c) = (2,3,4)$ gives $\displaystyle \{n,k\} = \{34,13\}$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle (a,b,c) = (13,47,169)$ gives $\displaystyle \{n,k\} = \{1079,233\}$   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Letting $\displaystyle c = a^2$ and dividing equation [2] by [1] leads to&#xA;$\displaystyle a(k + 1)(k + 2) = (n - k)(n - k - 1)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rearranging, all solutions satisfy $\displaystyle n^2 - (2k + 1)n - (k + 1)[(a - 1)k + 2a] = 0$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The discriminant $\displaystyle D$ of the above quadratic is $\displaystyle 4a(k + 1)(k + 2) + 1$, and a necessary and sufficient condition for $\displaystyle n$ to be an integer is that this expression be a perfect square.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle a = 1, k = 0$ gives $\displaystyle D = 9 = 3^2$     &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle a = 2, k = 13$ gives $\displaystyle D = 1681 = 41^2$    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle a = 13, k = 233$ gives $\displaystyle D = 2859481 = 1691^2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And my question is: can one prove (or disprove) that there are no more solutions?&lt;/p&gt;&#xA;" OwnerUserId="1770" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-11-10T19:40:34.187" LastActivityDate="2011-02-08T20:17:53.063" Title="$\binom{n}{k} : \binom{n}{k+1} : \binom{n}{k+2} = a : b : c$" Tags="&lt;binomial-coefficients&gt;&lt;recreational-mathematics&gt;" AnswerCount="1" CommentCount="8" />
  <row Id="8349" PostTypeId="1" CreationDate="2010-10-30T13:07:06.367" Score="1" ViewCount="172" Body="&lt;p&gt;Is there any way to decipher, manually of course , whether a number (may be large enough) is a prime?&lt;/p&gt;&#xA;" OwnerUserId="2943" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-02T15:08:37.630" LastActivityDate="2010-12-02T15:08:37.630" Title="How do you determine if a number is prime or composite?" Tags="&lt;number-theory&gt;&lt;prime-numbers&gt;" AnswerCount="2" />
  <row Id="8350" PostTypeId="2" ParentId="5248" CreationDate="2010-10-30T13:50:53.383" Score="5" ViewCount="" Body="&lt;p&gt;Let's consider the integrals&lt;br&gt;&#xA;$$I_1(t)=\int_t^{\infty}\frac{\sin(x-t)}{x}dx\qquad\mbox{ and }\qquad&#xA;I_2(t)=\int_0^{\infty}\frac{e^{-tx}}{1+x^2}dx,\qquad t\geq 0.$$&#xA;A direct calculation shows that $I_1(t)$ and $I_2(t)$ satisfy the ordinary differential equation&#xA;$$y''+y=\frac{1}{t},\qquad t&gt;0.$$&#xA;Therefore, the difference $I(t)=I_1(t)-I_2(t)$ satisfy the homogeneous differential equation&#xA;$$y''+y=0,\qquad t&gt;0,$$&#xA;hence it should be of the form &#xA;$$I(t)=A\sin (t+B) $$&#xA;with some constants $A$, $B$. But  $I_1(t)$ and $I_2(t)$ both converge to $0$ as $t\to\infty$. This implies that $A=0$ and $I_1(t)=I_2(t)$ for all $t\geq 0$. Finally, we have that&#xA;$$\int_0^{\infty}\frac{\sin x}{x}dx=\int_{0}^{\infty}\frac{1}{1+x^2}dx=\arctan(\infty)-\arctan(0)=\frac{\pi}{2}.$$&lt;/p&gt;&#xA;" OwnerUserId="723" LastEditorUserId="723" LastEditorDisplayName="" LastEditDate="2010-10-30T23:35:44.327" LastActivityDate="2010-10-30T23:35:44.327" CommentCount="2" />
  <row Id="8352" PostTypeId="2" ParentId="8349" CreationDate="2010-10-30T14:29:50.953" Score="4" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Primality_test&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Primality_test&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2900" LastActivityDate="2010-10-30T14:29:50.953" CommentCount="1" />
  <row Id="8353" PostTypeId="2" ParentId="8337" CreationDate="2010-10-30T14:32:30.077" Score="22" ViewCount="" Body="&lt;p&gt;OK, here's my favorite. I thought of this after reading a proof from the book &quot;Proofs from the book&quot; by Aigner &amp;amp; Ziegler, but later I found more or less the same proof as mine in &lt;a href=&quot;http://homepage.univie.ac.at/josef.hofbauer/02amm.pdf&quot; rel=&quot;nofollow&quot;&gt;a paper&lt;/a&gt; published a few years earlier by Josef Hofbauer. On Robin's list, the proof most similar to this is number 9&#xA;(EDIT: ...which is actually the proof that I read in Aigner &amp;amp; Ziegler).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When $0 &amp;lt; x &amp;lt; \pi/2$ we have $0&amp;lt;\sin x &amp;lt; x &amp;lt; \tan x$ and thus&#xA;$$\frac{1}{\tan^2 x} &amp;lt; \frac{1}{x^2} &amp;lt; \frac{1}{\sin^2 x}.$$&#xA;Note that $1/\tan^2 x = 1/\sin^2 x - 1$.&#xA;Split the interval $(0,\pi/2)$ into $2^n$ equal parts, and sum&#xA;the inequality over the (inner) &quot;gridpoints&quot; $x_k=(\pi/2) \cdot (k/2^n)$:&#xA;$$\sum_{k=1}^{2^n-1} \frac{1}{\sin^2 x_k} - \sum_{k=1}^{2^n-1} 1 &amp;lt; \sum_{k=1}^{2^n-1} \frac{1}{x_k^2} &amp;lt; \sum_{k=1}^{2^n-1} \frac{1}{\sin^2 x_k}.$$&#xA;Denoting the sum on the right-hand side by $S_n$, we can write this as&#xA;$$S_n - (2^n - 1) &amp;lt; \sum_{k=1}^{2^n-1} \left( \frac{2 \cdot 2^n}{\pi} \right)^2 \frac{1}{k^2} &amp;lt; S_n.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Although $S_n$ looks like a complicated sum, it can actually be computed fairly easily. To begin with,&#xA;$$\frac{1}{\sin^2 x} + \frac{1}{\sin^2 (\frac{\pi}{2}-x)} = \frac{\cos^2 x + \sin^2 x}{\cos^2 x \cdot \sin^2 x} = \frac{4}{\sin^2 2x}.$$&#xA;Therefore, if we pair up the terms in the sum $S_n$ except the midpoint $\pi/4$ (take the point $x_k$ in the left half of the interval $(0,\pi/2)$ together with the point $\pi/2-x_k$ in the right half) we get 4 times a sum of the same form, but taking twice as big steps so that we only sum over every other gridpoint; that is, over those gridpoints that correspond to splitting the interval into $2^{n-1}$ parts. And the midpoint $\pi/4$ contributes with $1/\sin^2(\pi/4)=2$ to the sum. In short,&#xA;$$S_n = 4 S_{n-1} + 2.$$&#xA;Since $S_1=2$, the solution of this recurrence is&#xA;$$S_n = \frac{2(4^n-1)}{3}.$$&#xA;(For example like this: the particular (constant) solution $(S_p)_n = -2/3$ plus the general solution to the homogeneous equation $(S_h)_n = A \cdot 4^n$, with the constant $A$ determined by the initial condition $S_1=(S_p)_1+(S_h)_1=2$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We now have&#xA;$$ \frac{2(4^n-1)}{3} - (2^n-1) &amp;lt;  \frac{4^{n+1}}{\pi^2} \sum_{k=1}^{2^n-1} \frac{1}{k^2}  &amp;lt; \frac{2(4^n-1)}{3}.$$&#xA;Multiply by $\pi^2/4^{n+1}$ and let $n\to\infty$. This squeezes the partial sums between two sequences both tending to $\pi^2/6$. Voilà!&lt;/p&gt;&#xA;" OwnerUserId="1242" LastEditorUserId="1242" LastEditorDisplayName="" LastEditDate="2010-10-30T17:44:03.363" LastActivityDate="2010-10-30T17:44:03.363" CommentCount="1" />
  <row Id="8354" PostTypeId="2" ParentId="8326" CreationDate="2010-10-30T14:54:19.533" Score="4" ViewCount="" Body="&lt;p&gt;Rolfsen's textbook &quot;Knots and links&quot; is quite nice. It assumes a 1st course in algebraic topology, and is pleasant self-learner text.  Plenty of nice exercises. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the higher-end of the knots textbook world, Burde and Zieschang's &quot;Knots&quot; covers quite a lot of ground in much detail.   Kawauchi's &quot;A survey of knot theory&quot; covers much more ground but in less detail.  Hillman's &quot;Algebraic invariants of links&quot; is more specialized and tends to focus on ideas such as Alexander modules, but it goes into them in more detail than I've seen anywhere outside of Jerry Levine's papers. &lt;/p&gt;&#xA;" OwnerUserId="642" LastActivityDate="2010-10-30T14:54:19.533" />
  <row Id="8355" PostTypeId="2" ParentId="8339" CreationDate="2010-10-30T14:56:26.740" Score="5" ViewCount="" Body="&lt;p&gt;The matrices $A$ and $B$ describe the same linear transformation $L$ of some vector space $V$ with respect to different bases. The eigenspaces $E_\lambda:=\lbrace x\in V\ |\ Lx=\lambda x\rbrace$ of $L$ have an invariant geometrical meaning and a well defined dimension. So there is actually nothing to prove.&lt;/p&gt;&#xA;" OwnerUserId="1303" LastActivityDate="2010-10-30T14:56:26.740" />
  <row Id="8356" PostTypeId="1" CreationDate="2010-10-30T14:56:41.467" Score="1" ViewCount="252" Body="&lt;p&gt;If we have two Random Variables X and Y, what is the interpretation for the three cases:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a] p(X=x,Y=y) = $p_1$(X=x) * $p_2$(Y=y)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;b] p(X=x,Y=y) &amp;lt; $p_1$(X=x) * $p_2$(Y=y)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;c] p(X=x,Y=y) &gt; $p_1$(X=x) * $p_2$(Y=y)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;where p -&gt; joint pdf of X and Y,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$p_1$ -&gt; marginal pdf of X&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$p_2$ -&gt; marginal pdf of Y&lt;/p&gt;&#xA;" OwnerUserId="2858" LastEditorUserId="2858" LastEditorDisplayName="" LastEditDate="2010-10-31T04:47:57.580" LastActivityDate="2011-06-01T20:35:53.007" Title="Interpretation of joint pdf/product of marginal pdfs" Tags="&lt;probability&gt;&lt;probability-theory&gt;" AnswerCount="3" CommentCount="5" FavoriteCount="1" />
  <row Id="8357" PostTypeId="1" AcceptedAnswerId="8358" CreationDate="2010-10-30T15:29:36.377" Score="1" ViewCount="558" Body="&lt;p&gt;Give &lt;code&gt;a &amp;gt; b&lt;/code&gt;, where &lt;code&gt;a,b ∈ ℝ&lt;/code&gt;, is it always true that &lt;code&gt;a² &amp;gt; b²&lt;/code&gt;?&lt;/p&gt;&#xA;" OwnerUserId="2946" LastActivityDate="2010-11-08T16:06:32.617" Title="If a &gt; b is a² &gt; b² ?" Tags="&lt;inequality&gt;" AnswerCount="4" CommentCount="1" FavoriteCount="0" />
  <row Id="8358" PostTypeId="2" ParentId="8357" CreationDate="2010-10-30T15:30:39.303" Score="5" ViewCount="" Body="&lt;p&gt;no its not. When $a,b$ are positive it happens. Consider a=-2 and b =-3. notice that inequality reverses.&lt;/p&gt;&#xA;" LastActivityDate="2010-10-30T15:30:39.303" />
  <row Id="8359" PostTypeId="1" AcceptedAnswerId="8372" CreationDate="2010-10-30T15:37:53.723" Score="0" ViewCount="124" Body="&lt;p&gt;Could I calculate the answer to either&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum\left[ x^{2} y \right ] - \sum xy$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum \left[ \left ( x - \sum xy \right )^{2} y \right ]$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I had any of these variables&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ \sum xy $; $ \sum x $; $ \sum y $; $ \sum x^2 $; $ \sum y^2 $;&lt;br&gt;&#xA;and &lt;code&gt;n&lt;/code&gt; ( the amount of items in the list )&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;adding more info&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need to calculate the &lt;strong&gt;variance&lt;/strong&gt; of a discrete probability distribution on my calculator. There is no function $\sum \left[ \left ( x - \sum xy \right )^{2} y \right ]$ on my model, and I was wondering if I could get there using the other &lt;strong&gt;functions&lt;/strong&gt;. ( those listed above )&lt;/p&gt;&#xA;" OwnerUserId="2945" LastEditorUserId="856" LastEditorDisplayName="" LastEditDate="2010-10-30T18:28:02.420" LastActivityDate="2010-10-30T18:28:02.420" Title="Calculating $\sum\left[ x^{2} y \right ]$" Tags="&lt;sequences-and-series&gt;&lt;statistics&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="8361" PostTypeId="2" ParentId="8356" CreationDate="2010-10-30T16:13:55.797" Score="0" ViewCount="" Body="&lt;p&gt;Two random variables $X$ and $Y$ are independent if $P(X=x, Y=y) = P(X=x)P(Y=y)$, i.e. their joint pdf factorizes. E.g. $X$ = getting heads on the first toss of a coin, $Y$ = getting heads on the second toss of the coin. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, for random variables $X$ and $Y$, $P(X = x, Y = y) \leq P(X=x)$ and $P(X = x, Y=y) \leq P(Y=y)$. This implies that (c) can't hold true. There's no specific interpretation to (b).&lt;/p&gt;&#xA;" OwnerUserId="1527" LastActivityDate="2010-10-30T16:13:55.797" CommentCount="5" />
  <row Id="8362" PostTypeId="2" ParentId="8357" CreationDate="2010-10-30T16:21:47.373" Score="27" ViewCount="" Body="&lt;p&gt;If $\rm\ a &gt; b\  $ then $\rm\ a^2 - b^2 = (a-b)\:(a+b) &gt; 0\ \iff\ a+b &gt;0 $&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-10-30T16:21:47.373" CommentCount="1" />
  <row Id="8363" PostTypeId="2" ParentId="8293" CreationDate="2010-10-30T16:39:57.753" Score="0" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;LEMMA&lt;/strong&gt; $\rm\quad\ \: A = A',\ \ \: B = B'\ \Rightarrow\ A+B\  =\:\ A'+B'$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof&lt;/strong&gt; $\rm\ \ 0 = A-A' + B-B' \: =\ (A+B) - (A'+B')$&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-10-30T16:39:57.753" CommentCount="3" />
  <row Id="8364" PostTypeId="2" ParentId="123" CreationDate="2010-10-30T16:44:29.150" Score="3" ViewCount="" Body="&lt;p&gt;Many kinds of nonlinear PDE have wave solutions explicitly expressed using hyperbolic tangents and secants: shock-wave profiles, solitons, reaction-diffusion fronts, and phase-transition fronts, for starters.&lt;/p&gt;&#xA;" OwnerUserId="2947" LastActivityDate="2010-10-30T16:44:29.150" />
  <row Id="8365" PostTypeId="2" ParentId="8322" CreationDate="2010-10-30T17:02:23.970" Score="1" ViewCount="" Body="&lt;p&gt;Lang has written several (very similar) books that discuss these topics.  Basically, you're right: Riemannian metrics only make sense when the manifold is modeled on a Hilbert space.  There is, however, a notion of psuedo-Riemannian manifold, which makes sense when the manifold is modeled on a self-dual Banach space.  Take a look at Chapter 7 of Lang's book, visible through Google Books:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://books.google.com/books?id=VfxGB5nYv1MC&amp;amp;printsec=frontcover#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;http://books.google.com/books?id=VfxGB5nYv1MC&amp;amp;printsec=frontcover#v=onepage&amp;amp;q&amp;amp;f=false&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1392" LastActivityDate="2010-10-30T17:02:23.970" CommentCount="1" />
  <row Id="8366" PostTypeId="1" CreationDate="2010-10-30T17:17:52.210" Score="0" ViewCount="73" Body="&lt;p&gt;I want to count integral of f(x)=x^2 in range for example 0-10.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What should I read to do it by myself and apply it to other funcs ?&lt;/p&gt;&#xA;" OwnerUserId="1755" LastActivityDate="2010-10-30T18:37:55.423" Title="How to use integral - x^2" Tags="&lt;integral&gt;" AnswerCount="1" />
  <row Id="8367" PostTypeId="2" ParentId="8366" CreationDate="2010-10-30T17:20:52.737" Score="2" ViewCount="" Body="&lt;p&gt;First learn differential calculus, then integral calculus. The formula to be used here is :$$\int_{a}^{b} x^{n} \ dx = \bigg[\displaystyle \frac{x^{n+1}}{n+1}\bigg]^{b}_{a}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: the answer will be $\frac{b^{n+1}}{n+1} - \frac{a^{n+1}}{n+1}$ &lt;/p&gt;&#xA;" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-10-30T18:37:55.423" LastActivityDate="2010-10-30T18:37:55.423" CommentCount="4" />
  <row Id="8368" PostTypeId="2" ParentId="8336" CreationDate="2010-10-30T17:35:33.423" Score="1" ViewCount="" Body="&lt;p&gt;It seems that as of now, doing individual cases might be a good way to get a feel for things. However, the outline below is motivated by 1.2.4 from TAOCP Volume 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Recall that&#xA;$$\sum_{k=1}^n c_k = n c_n - \sum_{k=1}^{n-1} k (c_{k+1}-c_k)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried this identity for the case where $c_k = \lfloor k/2\rfloor$, and it rapidly leads to the alleged proof. I have not worked out the details, but it seems that the other cases that you mention above should also follow similarly, noting for example that when $a=3$, then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$c_{k+1}-c_k = \left\lfloor \frac{k+1}{3}\right\rfloor - \left\lfloor \frac{k}{3}\right\rfloor,$$ which is zero for all $k$, except for $k=3j-1$, as $j=1,2,\ldots$, in which case it equals $1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using these ideas, the latter sum in the identity above can be greatly simplified, which should help prove your claims.&lt;/p&gt;&#xA;" OwnerUserId="1709" LastActivityDate="2010-10-30T17:35:33.423" />
  <row Id="8369" PostTypeId="1" CreationDate="2010-10-30T17:57:58.307" Score="0" ViewCount="81" Body="&lt;p&gt;Convex optimization problem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Minimize,&#xA; $\sum{|y-x|^2}$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;such that,&lt;br&gt;&#xA;a] K-L divergence between probability distributions for x and y, is maximum i.e. D(p(y)||q(x)) is maximum    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;b] $\sum{y} &amp;lt; \alpha$, where $\alpha$ is constant.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;c] x,y &amp;lt; $\beta$&lt;/p&gt;&#xA;" OwnerUserId="2858" LastActivityDate="2010-10-30T17:57:58.307" Title="convex optimization problem" Tags="&lt;optimization&gt;&lt;convex-optimization&gt;" />
  <row Id="8370" PostTypeId="2" ParentId="8336" CreationDate="2010-10-30T18:01:15.967" Score="1" ViewCount="" Body="&lt;p&gt;You should be able to use the fact that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $\displaystyle n = aj+r$, where $ 0 \le r &amp;lt; a$ then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle \sum_{k=1}^{n} \left\lfloor \frac{k}{a} \right\rfloor = aj(j-1)/2 + rj$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Multiplying and dividing the RHS by $8a$ and using $aj = n-r$ would help, I think.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Note: I haven't worked out the remaining details myself).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope that helps.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-10-30T18:01:15.967" CommentCount="1" />
  <row Id="8371" PostTypeId="2" ParentId="8318" CreationDate="2010-10-30T18:08:44.893" Score="1" ViewCount="" Body="&lt;p&gt;The taps, including an extra &quot;zero&quot; tap, form a polynomial $P(X)$ over $GF(2)$, we should satisfy two properties:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;$P(X)$ must be irreducible. Otherwise, the sequence decomposes (in some way) as a &quot;direct sum&quot; of several different LFSRs (try an example to see what this means).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The order of $X$ modulo $P(X)$ is the period of the LFSR sequence. If the period is maximal ($2^{\deg P} - 1$), then the polynomial is called primitive. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;If the number of taps is odd then $P(X)$ will have an even number of elements, so $P(1) = 0$. This implies that $P(X)$ is divisible by $X+1$, and so is reducible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the taps have a common factor $k$, then taking $Y=X^k$ you can recover LFSR sequences of a smaller polynomial $Q(Y) = P(X)$ (look at the output at intervals of $k$), and this implies that the period of the original sequence can't be maximal. Try and see.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's not difficult to calculate (using Moebius inversion) the number of irreducible/primitive polynomials, and there turn out to be lots of them. There are also ways to test if a polynomial is irreducible (in general, to factor it) and whether it's primitive (this requires factorizing $2^{\deg P}-1$ and then becomes trivial). So in practice, you can just &quot;do it yourself&quot; - try several polynomials, and soon enough you'll hit upon a nice one.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-10-30T18:08:44.893" CommentCount="2" />
  <row Id="8372" PostTypeId="2" ParentId="8359" CreationDate="2010-10-30T18:22:24.923" Score="2" ViewCount="" Body="&lt;p&gt;No, you cannot recover the variance of the distribution from the information you have. Here is a simple counterexample:&#xA;$$A = \{(0,1), (1,0), (2,0), (3,1)\},$$&#xA;$$B = \{(0,0), (1,1), (2,1), (3,0)\}.$$&#xA;Both $A$ and $B$ have the same $n = 4$, $\sum x = 6$, $\sum y = 2$, $\sum x^2 = 14$, $\sum y^2 = 2$, and $\sum xy = 3$, but $\sum x^2 y$ is $9$ for $A$ and $5$ for $B$.&lt;/p&gt;&#xA;" OwnerUserId="856" LastActivityDate="2010-10-30T18:22:24.923" CommentCount="2" />
  <row Id="8373" PostTypeId="2" ParentId="8337" CreationDate="2010-10-30T18:26:36.583" Score="8" ViewCount="" Body="&lt;p&gt;I have two favorite proofs.  One is the last proof in Robin Chapman's collection; you really should take a look at it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other is a proof that generalizes to the evaluation of $\zeta(2n)$ for all $n$, although I'll do it &quot;Euler-style&quot; to shorten the presentation.  The basic idea is that meromorphic functions have infinite partial fraction decompositions that generalize the partial fraction decompositions of rational functions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The particular function we're interested in is $B(x) = \frac{x}{e^x - 1}$, the exponential generating function of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Bernoulli_number&quot; rel=&quot;nofollow&quot;&gt;Bernoulli numbers&lt;/a&gt; $B_n$.  $B$ is meromorphic with poles at $x = 2 \pi i n, n \in \mathbb{Z}$, and at these poles it has residue $2\pi i n$.  It follows that we can write, a la Euler,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{x}{e^x - 1} = \sum_{n \in \mathbb{Z}} \frac{2\pi i n}{x - 2 \pi i n} = \sum_{n \in \mathbb{Z}} - \left( \frac{1}{1 - \frac{x}{2\pi i n}} \right).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we can expand each of the terms on the RHS as a geometric series, again a la Euler, to obtain&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{x}{e^x - 1} = - \sum_{n \in \mathbb{Z}} \sum_{k \ge 0} \left( \frac{x}{2\pi i n} \right)^k = \sum_{k \ge 0} (-1)^{n+1} \frac{2 \zeta(2n)}{(2\pi )^n} x^{2n}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;because, after rearranging terms, the sum over odd powers cancels out and the sum over even powers doesn't.  (This is one indication of why there is no known closed form for $\zeta(2n+1)$.)  Equating terms on both sides, it follows that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$B_{2n} = (-1)^{n+1} \frac{2 \zeta(2n)}{(2\pi)^n}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\zeta(2n) = (-1)^{n+1} \frac{B_{2n} (2\pi)^n}{2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;as desired.  To compute $\zeta(2)$ it suffices to compute that $B_2 = \frac{1}{6}$, which then gives the usual answer.  &lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-10-30T18:26:36.583" CommentCount="1" />
  <row Id="8374" PostTypeId="1" CreationDate="2010-10-30T18:51:34.797" Score="1" ViewCount="64" Body="&lt;p&gt;I would like a family of functions $f_a(x)$ so that $f_a$ is $a\in\mathbb{R}$ fractionally differentiable but not $a+\epsilon$ fractionally differentiable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anyone know such functions which are easily described? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance. &lt;/p&gt;&#xA;" OwnerUserId="1794" LastActivityDate="2010-10-31T20:03:40.423" Title="Family of function with fractional derivatives" Tags="&lt;functions&gt;" AnswerCount="1" />
  <row Id="8375" PostTypeId="1" AcceptedAnswerId="8392" CreationDate="2010-10-30T18:54:52.700" Score="1" ViewCount="88" Body="&lt;p&gt;While reading a probability paper titled &lt;a href=&quot;http://www.springerlink.com/content/j83542h24q561674/&quot; rel=&quot;nofollow&quot;&gt;Gröbner bases and factorisation in discrete probability and Bayes&lt;/a&gt; (can't find free version, sorry),  I came across the explanation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The set of indicator functions $I_i$ form a Boolean algebra that is in one-to-one correspondence with the quotient of the ring $\mathbb{R}[x_1, \ldots, x_n]$ with respect to the relationships $x_i(x_i - 1)=0$ where $x_i$ represents $I_i$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$I_i$ is defined to be the indicator function $I_i(\omega) = 1 \;\textrm{if}\;  \omega \in A_i \;\textrm{else}\; 0$, $i = 1, \ldots, n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a rudimentary understanding of rings, but can you please explain the correspondence shown here?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cheers&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;: In the rest of the paper, the authors don't seem to limit themselves to $\mathbb{F}_2$. In fact, I can't see a need for the use of the minus sign in $x_i(x_i-1)$ if the field is $\mathbb{F}_2$. I can't verify if the coefficients of the polynomials are disregarded yet.&lt;/p&gt;&#xA;" OwnerUserId="2949" LastEditorUserId="2949" LastEditorDisplayName="" LastEditDate="2010-10-31T11:57:13.757" LastActivityDate="2010-10-31T11:57:13.757" Title="Quotient ring and Boolean algebra correspondence" Tags="&lt;algebra&gt;&lt;ring-theory&gt;&lt;boolean-algebra&gt;" AnswerCount="2" />
  <row Id="8376" PostTypeId="2" ParentId="8375" CreationDate="2010-10-30T18:59:57.810" Score="2" ViewCount="" Body="&lt;p&gt;I assume $i$ is an element of $\{ 1, 2, ... n \}$.  The correspondence comes from identifying $x_i$ with the indicator function of $\{ i \}$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the sentence you quote is imprecise.  If you wanted a Boolean algebra (more precisely, a Boolean ring), you would take the base field to be $\mathbb{F}_2$, not $\mathbb{R}$.  If you wanted real multiples of indicator functions, you wouldn't say &quot;the set of indicator functions.&quot;&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-10-30T18:59:57.810" />
  <row Id="8377" PostTypeId="1" AcceptedAnswerId="8441" CreationDate="2010-10-30T19:00:10.603" Score="0" ViewCount="114" Body="&lt;p&gt;Greetings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Besides Un+1 - Un ?&lt;/p&gt;&#xA;" OwnerUserId="2950" LastEditorUserId="2950" LastEditorDisplayName="" LastEditDate="2010-10-30T23:55:34.450" LastActivityDate="2010-10-31T11:03:26.620" Title="Another way to calculate monotonicity of a sequence" Tags="&lt;calculus&gt;&lt;sequences-and-series&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="8378" PostTypeId="2" ParentId="8337" CreationDate="2010-10-30T19:01:54.847" Score="8" ViewCount="" Body="&lt;p&gt;We can use the function $f(x)=x^{2}$ with $-\pi \leq x\leq \pi $ and find&#xA;its expansion into a trigonometric Fourier series&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\dfrac{a_{0}}{2}+\sum_{n=1}^{\infty }(a_{n}\cos nx+b_{n}\sin x),$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is periodic and converges to $f(x)$ in $-\pi \leq x\leq \pi $.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Observing that $f(x)$ is even, it is enough to determine the coefficients &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$a_{n}=\dfrac{1}{\pi }\int_{-\pi }^{\pi }f(x)\cos nx\;dx\qquad n=0,1,2,3,...,$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;because &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$b_{n}=\dfrac{1}{\pi }\int_{-\pi }^{\pi }f(x)\sin nx\;dx=0\qquad n=1,2,3,... .$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $n=0$ we have &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$a_{0}=\dfrac{1}{\pi }\int_{-\pi }^{\pi }x^{2}dx=\dfrac{2}{\pi }\int_{0}^{\pi&#xA;}x^{2}dx=\dfrac{2\pi ^{2}}{3}.$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And for $n=1,2,3,...$ we get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$a_{n}=\dfrac{1}{\pi }\int_{-\pi }^{\pi }x^{2}\cos nx\;dx$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$=\dfrac{2}{\pi }\int_{0}^{\pi }x^{2}\cos nx\;dx=\dfrac{2}{\pi }\times \dfrac{%&#xA;2\pi }{n^{2}}(-1)^{n}=(-1)^{n}\dfrac{4}{n^{2}},$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;because&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int x^2\cos nx\;dx=\dfrac{2x}{n^{2}}\cos nx+\left( \frac{x^{2}}{%&#xA;n}-\dfrac{2}{n^{3}}\right) \sin nx.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f(x)=\dfrac{\pi ^{2}}{3}+\sum_{n=1}^{\infty }\left( (-1)^{n}\dfrac{1}{n^{2}}%&#xA;\cos nx\right) .$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $f(\pi )=\pi ^{2}$, we obtain&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\pi ^{2}=\dfrac{\pi ^{2}}{3}+\sum_{n=1}^{\infty }\left( (-1)^{n}\dfrac{4}{%&#xA;n^{2}}\cos \left( n\pi \right) \right) $$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\pi ^{2}=\dfrac{\pi ^{2}}{3}+4\sum_{n=1}^{\infty }\left( (-1)^{n}(-1)^{n}%&#xA;\dfrac{1}{n^{2}}\right) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\pi ^{2}=\dfrac{\pi ^{2}}{3}+4\sum_{n=1}^{\infty }\dfrac{1}{n^{2}}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{n=1}^{\infty }\dfrac{1}{n^{2}}=\dfrac{\pi ^{2}}{4}-\dfrac{\pi ^{2}}{12}=%&#xA;\dfrac{\pi ^{2}}{6}$$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Second method (available on-line a few years ago) by Eric Rowland. From&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\log (1-t)=-\sum_{n=1}^{\infty}\dfrac{t^n}{n}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and making the substitution $t=e^{ix}$ one gets the series expansion&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$w=\text{Log}(1-e^{ix})=-\sum_{n=1}^{\infty }\dfrac{e^{inx}}{n}=-\sum_{n=1}^{%&#xA;\infty }\dfrac{1}{n}\cos nx-i\sum_{n=1}^{\infty }\dfrac{1}{n}\sin nx,$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;whose radius of convergence is $1$. Now if we take the imaginary part of both sides, the RHS becomes&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Im w=-\sum_{n=1}^{\infty }\dfrac{1}{n}\sin nx,$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and the LHS&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Im w=\arg \left( 1-\cos x-i\sin x\right) =\arctan \dfrac{-\sin x}{%&#xA;1-\cos x}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\arctan \dfrac{-\sin x}{1-\cos x}=-\arctan \dfrac{2\sin \dfrac{x}{2}\cdot \cos \dfrac{x}{2}}{2\sin ^{2}\dfrac{x}{2}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$=-\arctan \cot \dfrac{x}{2}=-\arctan \tan \left( \dfrac{\pi }{2}-\dfrac{x}{2}%&#xA;\right) =\dfrac{x}{2}-\dfrac{\pi }{2},$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the following expansion holds&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\dfrac{\pi }{2}-\frac{x}{2}=\sum_{n=1}^{\infty }\dfrac{1}{n}\sin nx.\qquad&#xA;(\ast )$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Integrating the identity $(\ast )$, we obtain&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\dfrac{\pi }{2}x-\dfrac{x^{2}}{4}+C=-\sum_{n=1}^{\infty }\dfrac{1}{n^{2}}\cos&#xA;nx.\qquad (\ast \ast )$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Setting $x=0$, we get the relation between $C$ and $\zeta (2)$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$C=-\sum_{n=1}^{\infty }\dfrac{1}{n^{2}}=-\zeta (2).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And for $x=\pi $, since&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\zeta (2)=2\sum_{n=1}^{\infty }\dfrac{(-1)^{n-1}}{n^{2}},$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;we deduce&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\dfrac{\pi ^{2}}{4}+C=-\sum_{n=1}^{\infty }\dfrac{1}{n^{2}}\cos n\pi&#xA;=\sum_{n=1}^{\infty }\dfrac{(-1)^{n-1}}{n^{2}}=\dfrac{1}{2}\zeta (2)=-\dfrac{1}{%&#xA;2}C.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solving for $C$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$C=-\dfrac{\pi ^{2}}{6},$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;we thus prove&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\zeta (2)=\dfrac{\pi ^{2}}{6}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Note&lt;/em&gt;: this 2nd method can generate all the zeta values $\zeta (2n)$ by integrating repeatedly $(\ast\ast )$. This is the reason why I appreciate it. Unfortunately it does not work for $\zeta (2n+1)$.&lt;/p&gt;&#xA;" OwnerUserId="752" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2010-10-31T22:19:36.730" LastActivityDate="2010-10-31T22:19:36.730" />
  <row Id="8379" PostTypeId="1" AcceptedAnswerId="8402" CreationDate="2010-10-30T19:13:45.920" Score="2" ViewCount="343" Body="&lt;p&gt;Well, this is an exercise problem from &lt;em&gt;Herstein&lt;/em&gt; which sounds difficult:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;How does one prove that if $|G|&gt;2$, then $G$ has non-trivial automorphism?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The only thing, i know which connects a group with its automorphism is the theorem, $$G/Z(G) \cong \mathcal{I}(G)$$ where $\mathcal{I}(G)$ denotes the &lt;em&gt;Inner- Automorphism&lt;/em&gt; of $G$. So for a group with $Z(G)=(e)$, we can conclude that it has a non-trivial automorphism, but what about groups with center?&lt;/p&gt;&#xA;" LastActivityDate="2010-10-30T21:07:41.440" Title="$|G|&gt;2$ implies $G$ has non trivial automorphism." Tags="&lt;abstract-algebra&gt;&lt;group-theory&gt;&lt;algebra&gt;" AnswerCount="2" CommentCount="8" />
  <row Id="8380" PostTypeId="2" ParentId="8322" CreationDate="2010-10-30T19:13:54.393" Score="3" ViewCount="" Body="&lt;p&gt;There are two notions of what it means for a(n infinite dimensional) manifold to have a Riemannian structure.  A &lt;strong&gt;strong&lt;/strong&gt; Riemannian structure means a (smooth) choice of inner product on each tangent space which induces an isomorphism of each tangent space with its corresponding cotangent space.  A &lt;strong&gt;weak&lt;/strong&gt; Riemannian structure simply means a (smooth) choice of inner product on each tangent space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Strong Riemannian structures only exist if the manifold is modelled on a Hilbert space, and even then they have to be chosen correctly (so the usual $L^2$-metric on the space of $L^{1,2}$-loops is not a strong structure, even though the manifold is Hilbertian).  Weak Riemannian structures exist much more widely.  For a weak Riemannian structure you only need to know that the manifold admits smooth partitions of unity and that the model spaces admit continuous inner products.  So, for example, continuous loops in a smooth manifold admit a weak Riemannian structure but not a strong one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Although strong Riemannian structures are very good for generalising much of ordinary differential geometry to infinite dimensions, there are occasions where the requirement of having a Hilbert manifold is too strong, and one can get away with merely having a weak Riemannian structure.  I've written an article where having a weak Riemannian structure on the space of smooth loops was an essential step and where the construction &lt;em&gt;would not have worked&lt;/em&gt; on a Hilbertian manifold (though actually it was a &lt;em&gt;co&lt;/em&gt;-Riemannian structure that I needed).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Declaration of interests: I've actually proposed a refinement of the &quot;weak/strong&quot; classification as I found it too harsh.  See my article &lt;a href=&quot;http://www.math.ntnu.no/~stacey/Research/Preprints/constructdirac.html&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; for this, and the above-mentioned result, and a load of examples of spaces with different types of Riemannian structure.)&lt;/p&gt;&#xA;" OwnerUserId="2907" LastActivityDate="2010-10-30T19:13:54.393" CommentCount="2" />
  <row Id="8381" PostTypeId="2" ParentId="8319" CreationDate="2010-10-30T19:18:07.930" Score="4" ViewCount="" Body="&lt;p&gt;You can find such partial fraction expansions given as examples in many texts on complex analysis. For example, you will find a derivation of this series in Example 1 at &lt;a href=&quot;http://books.google.com/books?id=DYCOCBCBwoIC&amp;amp;lpg=PA64&amp;amp;dq=partial%20fraction%20series%20csc%20cot%20sec%20complex&amp;amp;pg=PA59#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;this link&lt;/a&gt; to Markushevich and Silverman's book.  It uses &quot;Cauchy's theorem on partial fraction expansions&quot;, given a few pages earlier at &lt;a href=&quot;http://books.google.com/books?id=DYCOCBCBwoIC&amp;amp;lpg=PA64&amp;amp;dq=partial%20fraction%20series%20csc%20cot%20sec%20complex&amp;amp;pg=PA56#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;this link&lt;/a&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example 2 gives the series for $\cot$, although from the preview I can't see whether $\tan$ and $\csc$ are included. The analogous series for $\tan$ is derived on &lt;a href=&quot;http://en.wikipedia.org/wiki/Partial_fractions_in_complex_analysis#Example&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;.  Each of these series can be useful for evaluating numerical series.  For example, taking $x=0$ in your series you get &#xA;$$\frac{\pi}{4}=1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\cdots.$$&#xA; And although it is not directly related to your question, I thought I'd add in light of the &lt;a href=&quot;http://math.stackexchange.com/questions/8337/different-methods-to-compute-sum-n1-infty-frac1n2&quot;&gt;recent question&lt;/a&gt; on evaluating the sum of the reciprocals of the squares that if you take the partial fraction series &#xA;$$\tan(x) = \sum_{k=0}^{\infty} \frac{-2x}{x^2 - \left(k + \frac{1}{2}\right)^2\pi^2},$$&#xA;divide by $x$ and let $x$ go to $0$, rearranging yields &#xA;$$\frac{\pi^2}{8}=1+\frac{1}{9}+\frac{1}{25}+\frac{1}{49}+\cdots,$$&#xA;which in turn (because the sum over the evens is $\frac{1}{4}$ the total sum) leads to &#xA;$$\frac{\pi^2}{6}=1+\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+\cdots.$$&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2010-10-31T06:44:08.973" LastActivityDate="2010-10-31T06:44:08.973" />
  <row Id="8382" PostTypeId="2" ParentId="8326" CreationDate="2010-10-30T19:20:28.033" Score="2" ViewCount="" Body="&lt;p&gt;This answer is a little less popular of a suggestion, but it was how I first learned about knots, and I really enjoyed it. This book is primarily focused on Vassiliev Diagrams, and is currently unpublished, but available through the authors webpages. Yay! free stuff!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a link to &lt;a href=&quot;http://www.math.ohio-state.edu/~chmutov/preprints/&quot; rel=&quot;nofollow&quot;&gt;the CDbook&lt;/a&gt; by Chmutov, Duzhin, and Mostovoy. It is the first link on the page.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I personally found the focus on invariants very useful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My second favorite would be Rolfsen, as Ryan suggested.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Good luck!&lt;/p&gt;&#xA;" OwnerUserId="16" LastActivityDate="2010-10-30T19:20:28.033" />
  <row Id="8383" PostTypeId="2" ParentId="8346" CreationDate="2010-10-30T19:21:01.450" Score="3" ViewCount="" Body="&lt;p&gt;Typically one considers the Hamming distance for the possible lengths of the messages, the larger the better.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See &lt;a href=&quot;http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.5.5027&amp;amp;rep=rep1&amp;amp;type=pdf&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;Koopman &amp;amp; Chakravarty&lt;/em&gt;, Cyclic Redundancy Code (CRC) Polynomial Selection For Embedded Networks&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="210" LastActivityDate="2010-10-30T19:21:01.450" />
  <row Id="8384" PostTypeId="1" CreationDate="2010-10-30T19:26:31.030" Score="2" ViewCount="79" Body="&lt;p&gt;Given the following game,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;there is a hider and a seeker, each game consists of the hider randomly selecting one of two places to hide that is A with probability p and B with 1 - p, and the seeker randomly choses to search in A with probability q and B with 1 - q. If seeker finds hider then game ends, else both randomly pick again. This continues each time both randomly picking where to hide and search, until seeker finds hider.&#xA;I want to know how to formulate the average number of game iterations to expect using theory...that is using p and q, how can i express how long i should expect it to take for seeker to find hider?&lt;/p&gt;&#xA;" LastActivityDate="2010-10-30T20:18:45.320" Title="how do i theoretically find the expected length of the following game?" Tags="&lt;game-theory&gt;" AnswerCount="2" />
  <row Id="8385" PostTypeId="1" AcceptedAnswerId="8395" CreationDate="2010-10-30T19:30:08.840" Score="3" ViewCount="299" Body="&lt;p&gt;Using n_th root of unity&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;$(e^{\frac{2ki\pi}{n}})^{n} = 1$&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Prove that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;$\prod_{k=1}^{n-1}\sin\frac{k \pi}{n} = \frac{n}{2^{n-1}}$&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="2951" LastEditorUserId="2951" LastEditorDisplayName="" LastEditDate="2010-10-30T19:45:00.543" LastActivityDate="2010-11-01T03:47:26.787" Title="Prove that $\prod_{k=1}^{n-1}\sin\frac{k \pi}{n} = \frac{n}{2^{n-1}}$" Tags="&lt;trigonometry&gt;&lt;complex-numbers&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="3" />
  <row Id="8386" PostTypeId="2" ParentId="8348" CreationDate="2010-10-30T19:33:40.837" Score="1" ViewCount="" Body="&lt;p&gt;Sketch:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose $\binom{n}{k},\binom{n}{k+1},\binom{n}{k+2}$ are in ratio $1:2:3$. Then $\frac{n-k}{k+1}=2$ and $\frac{n-k-1}{k+2}=\frac{3}{2}$. Solving these two equations gives $n=14$ and $k=4$.&lt;/p&gt;&#xA;" OwnerUserId="2856" LastActivityDate="2010-10-30T19:33:40.837" CommentCount="2" />
  <row Id="8387" PostTypeId="2" ParentId="8384" CreationDate="2010-10-30T19:39:00.053" Score="4" ViewCount="" Body="&lt;p&gt;The probability that the game doesn't end in a particular round is $x = p(1 - q) + q(1 - p)$, so the probability that the game lasts for exactly $n$ rounds is $x^{n-1}(1 - x)$.  This means that the expected length of a game is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{n \ge 1} n x^{n-1} (1 - x).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This sum can be found by various methods, depending on what you're comfortable with; for now I'll just say that the answer works out to be $\frac{1}{1 - x}$.  In other words, it's the reciprocal of the probability that the game ends.  &lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-10-30T19:39:00.053" CommentCount="3" />
  <row Id="8388" PostTypeId="2" ParentId="3087" CreationDate="2010-10-30T19:39:05.890" Score="1" ViewCount="" Body="&lt;p&gt;In high school in the 1970s I got tremendous benefit out of a one-term elective course called Mathematical Logic, which was nothing fancy, just basic prepositional stuff.  Understanding the basic structure of logical arguments empowers one to learn anything more efficiently. If college-bound students could ditch half of their AP Calculus for a course in &lt;b&gt;Logical Reasoning&lt;/b&gt; that includes basic material from Daniel Velleman's book &lt;i&gt;How to Prove It&lt;/i&gt; and some basic facts about statistics, they would be in a far better position to pursue many subjects, not just math.&lt;/p&gt;&#xA;" OwnerUserId="2947" LastActivityDate="2010-10-30T19:39:05.890" CommunityOwnedDate="2010-10-30T19:39:05.890" CommentCount="1" />
  <row Id="8389" PostTypeId="1" CreationDate="2010-10-30T19:39:40.910" Score="1" ViewCount="177" Body="&lt;p&gt;Let G be a finite p-group and has a nontrivial normal subgroup H, then the intersection of H and its center is not trivial.&lt;/p&gt;&#xA;" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-10-31T04:24:48.123" LastActivityDate="2010-10-31T17:55:41.333" Title="A question about p-group" Tags="&lt;group-theory&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="8390" PostTypeId="2" ParentId="8389" CreationDate="2010-10-30T19:54:38.257" Score="1" ViewCount="" Body="&lt;p&gt;Let $a_{1}, . . . , a_{k}$ be representatives of the conjugacy classes of $G$, ordered such that $a_{m} \in H$ and $a_{m+1}, \cdots , a_{k} \notin H$. The conjugacy class $C(a_{i})$ have either $C(a_{i}) \subset H$ or $C(a_{i}) \cap H = {e}$. First arrange the ${a_{1}, . . . , a_{m}}$ so that the first $r$  represent conjugacy classes of size 1, (i.e. elements in $H \cap Z$) and the latter $m − r$ represent classes of size larger than 1. Then we can write the&#xA;class equation for $H \cap Z = H$ as: $$|H| = \sum\limits_{i=1}^{m} |C(a_{i} \cap H| = |H \cap Z| + \sum\limits_{i=r}^{m} |C(a_{i})| =  |H \cap Z| + \sum\limits_{i=r}^{m}\frac{|G|}{|N(a_{i})|}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As $|H| &amp;lt; p^{n}$ every term in the sum is divisible by $p$ so $|H \cap Z|$ from above is divisible by $p$.&lt;/p&gt;&#xA;" LastActivityDate="2010-10-30T19:54:38.257" />
  <row Id="8391" PostTypeId="2" ParentId="3087" CreationDate="2010-10-30T20:03:32.207" Score="4" ViewCount="" Body="&lt;p&gt;I strongly believe that even if something can be done by calculator, a high school student should understand some nuts and bolts of how it's being done. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I spent some time doing volunteer tutoring at CRLS (Cambridge Rindge and Latin School). I still remember the girl at 8th or 9th grade who was studying inequalities, yet she couldn't compute $2-1$ without the help of calculator. The most frustrating case was the kid in 9th or 10th grade who didn't know multiplication tables, he couldn't compute, say, 3 times 7 in his head. I was supposed to help him with square roots. Well, how do you explain that $\sqrt{8} = 2\sqrt{2}$? First of, you have to realize that $8 = 2 \times 2 \times 2$, and that's exactly what he couldn't do. I ended up spending some extra time with him, afterward been told that he passed the test or exam and was very happy with my help - but I wasn't! I felt we were wasting lots of time and just because he didn't know multiplication tables, he couldn't break down numbers into factors, and calculator wasn't really help. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The bottom line: I think in school students &lt;em&gt;still&lt;/em&gt; should learn basic arithmetic before moving to more advanced stuff. I think it's a pure idiocy to introduce the kids to set theory while having them relying on their calculators to compute $1 + 1$. Honestly, the advanced stuff can wait till college. &lt;/p&gt;&#xA;" OwnerUserId="2450" LastActivityDate="2010-10-30T20:03:32.207" CommunityOwnedDate="2010-10-30T20:03:32.207" />
  <row Id="8392" PostTypeId="2" ParentId="8375" CreationDate="2010-10-30T20:08:16.737" Score="0" ViewCount="" Body="&lt;p&gt;A nice exposition of Stone's cryptomorphism between Boolean algebras and Boolean rings is in Burris and Sankappanavar's textbook &lt;a href=&quot;http://www.math.uwaterloo.ca/~snburris/htdocs/UALG/univ-algebra.pdf#page=154&quot; rel=&quot;nofollow&quot;&gt;A Course in Universal Algebra&lt;/a&gt;. There you'll also find a presentation of Boolean products - a powerful tool that facilitates the transfer of Boolean algebra results to other algebraic structures - leading to very general representation theories.&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-10-30T20:08:16.737" />
  <row Id="8393" PostTypeId="1" AcceptedAnswerId="8399" CreationDate="2010-10-30T20:09:06.007" Score="1" ViewCount="297" Body="&lt;p&gt;Simple question-&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it possible for a linear programming optimization problem possible to have a degenerate optimal solution whereas the dual has a unique optimal solution? I can't find a scenario where this is possible but intuitively it seems that it is possible.&lt;/p&gt;&#xA;" OwnerUserId="2014" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-10-30T21:55:28.157" LastActivityDate="2010-10-30T21:55:28.157" Title="Primal- degenerate optimal, Dual - unique optimal" Tags="&lt;optimization&gt;&lt;linear-programming&gt;" AnswerCount="1" />
  <row Id="8395" PostTypeId="2" ParentId="8385" CreationDate="2010-10-30T20:18:15.243" Score="8" ViewCount="" Body="&lt;p&gt;$$P=\prod_{k=1}^{n-1}\sin(k\pi/n)=(2i)^{1-n}\prod_{k=1}^{n-1}(e^{ik\pi/n}-e^{-ik\pi/n})=(2i)^{1-n}e^{-i\pi n(n-1)/(2n)}\prod_{k=1}^{n-1}(e^{2ik\pi/n}-1)=$$&#xA;$$(-2)^{1-n}\prod_{k=1}^{n-1}(\xi^k-1)=2^{1-n}\prod_{k=1}^{n-1}(1-\xi^k),$$&#xA;where $\xi=e^{2i\pi/n}$. Now note, that $x^n-1=(x-1)\sum_{k=0}^{n-1}x^k$ and $x^n-1=\prod_{k=0}^{n-1} (x-\xi^k)$, thus cancelling $x-1$ we have $\prod_{k=1}^{n-1} (x-\xi^k) =\sum_{k=0}^{n-1}x^k$. Substituting $x=1$ we have $\prod_{k=1}^{n-1} (1-\xi^k)=n$. Therefore $P=n2^{1-n}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt;&lt;br&gt;&#xA;In order to note that $x^n-1=\prod_{k=0}^{n-1} (x-\xi^k)$, note that $1,\xi,\dots,\xi^{n-1}$ are roots of $x^n-1$. Therefore by &lt;a href=&quot;http://en.wikipedia.org/wiki/Polynomial_remainder_theorem&quot; rel=&quot;nofollow&quot;&gt;polynomial reminder theorem&lt;/a&gt; we have $x^n-1=Q(x) \prod_{k=0}^{n-1} (x-\xi^k)$. Comparing degrees of lhs and rhs we can find, that $Q(x)$ has degree 0. Comparing highest coefficients we can conclude $Q(x)=1$.&lt;/p&gt;&#xA;" OwnerUserId="695" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-01T03:47:26.787" LastActivityDate="2010-11-01T03:47:26.787" CommentCount="10" />
  <row Id="8396" PostTypeId="2" ParentId="8384" CreationDate="2010-10-30T20:18:45.320" Score="6" ViewCount="" Body="&lt;p&gt;You can sometimes answer such questions using the idea of &quot;conditioning&quot;. Students often prefer this method as there are no infinite sums to calculate. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Two things can happen on the first round of this game: either the seeker finds the hider (probability = $pq+(1-p)(1-q)$), or the game starts over (probability = $p+q-2pq$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $t$ is the average number of iterations that the game lasts, then averaging over  these two scenarios gives $$t=(1)(pq+(1-p)(1-q))+(1+t)(p+q-2pq).$$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solving for $t$ gives $t=1/(pq+(1-p)(1-q))$.   &lt;/p&gt;&#xA;" OwnerUserId="940" LastActivityDate="2010-10-30T20:18:45.320" />
  <row Id="8398" PostTypeId="2" ParentId="7531" CreationDate="2010-10-30T20:32:14.653" Score="1" ViewCount="" Body="&lt;p&gt;Success finally!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f( n) = \sum_{(k,n)=1;1\leq k\leq n} \log\Bigl(\frac{k}{n}\Bigr)$$ therefore we have &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{d|n}f(d) =\log\Bigl(\frac{1}{n}\Bigr)+...+\log\Bigl(\frac{n}{n}\Bigr)=\log\left(\frac{n!}{n^n}\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus by Moebius Inversion Formula: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f(n) = \sum_{d|n}\log\left(\frac{d!}{d^d}\right)\cdot \mu\left(\frac{n}{d}\right) = \log\left(\prod_{d|n}\left(\frac{d!}{d^d}\right)^{\mu\left(\frac{n}{d}\right) }\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f(n) = \sum_{(k,n)=1;1\leq k\leq n} {\log(k)} -\phi(n)\cdot \log( n) = \log(P(n))-\log(n^{\phi(n)})$$&lt;/p&gt;&#xA;" LastActivityDate="2010-10-30T20:32:14.653" />
  <row Id="8399" PostTypeId="2" ParentId="8393" CreationDate="2010-10-30T20:55:13.407" Score="2" ViewCount="" Body="&lt;p&gt;The answer is yes, but only if there are other optimal solutions than the degenerate one.  For example, suppose the primal problem is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\max x_1 + x_2$$&#xA;subject to&#xA;$$x_1 \leq 1,$$&#xA;$$x_1 + x_2 \leq 1,$$&#xA;$$x_1, x_2 \geq 0.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The solution $(1,0)$ is optimal and degenerate, but every solution $(a,1-a)$, for $0 \leq a \leq 1$ is also optimal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The dual is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\min y_1 + y_2$$&#xA;subject to&#xA;$$y_1 + y_2 \geq 1,$$&#xA;$$y_2 \geq 1,$$&#xA;$$y_1, y_2 \geq 0.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The dual has the unique (degenerate) optimal solution $(0,1)$.  So we do have a situation with a degenerate optimal solution in the primal but a unique dual optimal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if the degenerate optimal solution is unique, then there &lt;em&gt;must&lt;/em&gt; be multiple optimal solutions in the dual.  The following table is from Sierksma's &lt;em&gt;Linear and Integer Programming: Theory and Practice&lt;/em&gt;, Volume 1, page 144.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Primal Optimal Solution                     Dual Optimal Solution&#xA;(a) Multiple                     implies    Degenerate&#xA;(b) Unique and nondegenerate     implies    Unique and nondegenerate&#xA;(c) Multiple and nondegenerate   implies    Unique and degenerate&#xA;(d) Unique and degenerate        implies    Multiple&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="2370" LastActivityDate="2010-10-30T20:55:13.407" />
  <row Id="8400" PostTypeId="2" ParentId="8389" CreationDate="2010-10-30T21:03:12.257" Score="7" ViewCount="" Body="&lt;p&gt;Perhaps a slightly less computationally-intensive argument is to simply note that since $H$ is normal, it must be a union of conjugacy classes. Each conjugacy class of $G$ has $p^i$ elements for some $i$; since $H$ contains at least one conjugacy class with $p^0 = 1$  elements (the class of the identity), and $|H|\equiv 0 \pmod{p}$, it must contain other classes with just one element, which must be classes of central elements of $G$.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-10-30T21:03:12.257" CommentCount="6" />
  <row Id="8401" PostTypeId="2" ParentId="8379" CreationDate="2010-10-30T21:05:58.573" Score="10" ViewCount="" Body="&lt;p&gt;If $G$ is not abelian, then conjugation by a noncentral element will do.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $G$ is abelian, then $x\mapsto x^{-1}$ is an automorphism. It will be nontrivial unless every element of $G$ equals its inverse, that is, if every element of $G$ is of exponent $2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If every element of $G$ is of exponent $2$, then $G$ is a vector space over the field of $2$ elements, so it is isomorphic to a (possibly infinite) sum of copies of $C_2$, the cyclic group of two elements. Since $|G|\gt 2$, there are at least two copies, so the linear transformation that swaps two copies of $C_2$ is a nontrivial automorphism.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-10-30T21:05:58.573" CommentCount="16" />
  <row Id="8402" PostTypeId="2" ParentId="8379" CreationDate="2010-10-30T21:07:41.440" Score="6" ViewCount="" Body="&lt;p&gt;As you note in the question, the group of inner automorphisms is isomorphic to $G/Z(G)$. In particular, it's non-trivial if and only if $Z(G)=G$.  So there is a non-trivial (inner) automorphism unless $G=Z(G)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, notice that, by definition, $Z(G)=G$ if and only if $G$ is abelian; so we have reduced to the abelian case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $G$ is abelian then $g\mapsto -g$ is an automorphism, and it is non-trivial unless $g=-g$ for all $g\in G$.  But $g=-g$ if and only if the order of $g$ divdes two.  So we have now reduced to the case in which $2g=0$ for all $g\in G$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case, $G$ is a vector space over the field $\mathbb{Z}/2$.  As&#xA;$|G|$ is equal to 2 raised to the power of the $\mathbb{Z}/2$-dimension of $G$,&#xA;the hypothesis that $|G|&gt;2$ implies that $\mathrm{dim}_{\mathbb{Z/2}} G&gt;1$.  But now we can write down lots of linear automorphisms of $G$. For instance, you could fix any basis $g_1,g_2,\ldots$ and take the automorphism $g_1\mapsto g_2$, $g_2\mapsto g_1$ and $g_i\mapsto g_i$ for every $i&gt;2$.&lt;/p&gt;&#xA;" OwnerUserId="2287" LastActivityDate="2010-10-30T21:07:41.440" />
  <row Id="8403" PostTypeId="1" AcceptedAnswerId="8406" CreationDate="2010-10-30T21:19:09.143" Score="0" ViewCount="240" Body="&lt;p&gt;I'm sorry if the English isn't correct, I'm not native.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's say for example these two:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{3n}{2n+3}(-1)^{n+1}-\frac{(-1)^n}{2n}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$((-1)^n+1/n)^2$$&lt;/p&gt;&#xA;" OwnerUserId="2950" LastEditorUserId="2950" LastEditorDisplayName="" LastEditDate="2010-10-31T14:24:00.877" LastActivityDate="2010-10-31T14:24:00.877" Title="How to check if a sequence is bounded from above, bounded from below or bounded?" Tags="&lt;calculus&gt;&lt;sequences-and-series&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="8404" PostTypeId="2" ParentId="8002" CreationDate="2010-10-30T21:29:49.207" Score="3" ViewCount="" Body="&lt;p&gt;In case anyone is interested, here are the full versions of my two proofs.  (I constructed the combinatorial one from my original partially combinatorial one after I posted the question.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;HR&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The non-combinatorial proof&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As Derek Jennings observes, $\lfloor \frac{n+1}{k} \rfloor - \lfloor \frac{n}{k} \rfloor$ is $1$ if $k|(n+1)$ and $0$ otherwise.  Thus, if $$f(n) = \sum_{k=1}^n \left\lfloor\frac{n}{k} \right\rfloor \varphi (k),$$&#xD;&#xA;then $$\Delta f(n) = f(n+1) - f(n) = \sum_{k|(n+1)} \phi(k) = n+1,$$&#xD;&#xA;where the last equality follows from the well-known formula Moron cites.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then&#xA;$$\sum_{k=1}^n \left\lfloor\frac{n}{k} \right\rfloor \varphi (k) = f(n) = \sum_{k=0}^{n-1} \Delta f(k) = \sum_{k=0}^{n-1} (k+1) = \frac{n(n+1)}{2}.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;HR&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The combinatorial proof&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both sides count the number of fractions (reducible or irreducible) in the interval (0,1] with denominator $n$ or smaller.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the right side, the number of ways to pick a numerator and a denominator is the number of ways to choose two numbers with replacement from the set $\{1, 2, \ldots, n\}$.  This is known to be &#xA;$$\binom{n+2-1}{2} = \frac{n(n+1)}{2}.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now for the left side.  The number of irreducible fractions in $(0,1]$ with denominator $k$ is equal to the number of positive integers less than or equal to $k$ and relatively prime to $k$; i.e., $\varphi(k)$.  Then, for a given irreducible fraction $\frac{a}{k}$, there are $\left\lfloor \frac{n}{k} \right\rfloor$ total fractions with denominators $n$ or smaller in its equivalence class.  (For example, if $n = 20$ and $\frac{a}{k} = \frac{1}{6}$, then the fractions $\frac{1}{6}, \frac{2}{12}$, and $\frac{3}{18}$ are those in its equivalence class.)  Thus the sum &#xA;$$\sum_{k=1}^n \left\lfloor\frac{n}{k} \right\rfloor \varphi (k)$$&#xD;&#xA;also gives the desired quantity.&lt;/p&gt;&#xA;" OwnerUserId="2370" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-11-22T05:15:18.223" LastActivityDate="2010-11-22T05:15:18.223" />
  <row Id="8405" PostTypeId="1" CreationDate="2010-10-30T21:31:55.070" Score="2" ViewCount="84" Body="&lt;p&gt;In a 365 day year, Joe, Greg &amp;amp; Dean visit the park multiple times for a walk.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Joe visits every 3rd day&lt;/li&gt;&#xA;&lt;li&gt;Greg visits every 5th day&lt;/li&gt;&#xA;&lt;li&gt;Dean visits every 7th day&lt;/li&gt;&#xA;&lt;li&gt;All three visit the park on the first day of the year&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In the year:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;On how many days is Greg alone in the park?&lt;/li&gt;&#xA;&lt;li&gt;On how many days will Dean meet Joe?&lt;/li&gt;&#xA;&lt;li&gt;On how many days will all three meet?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I can work this out brute force by laying down the multiples of 3,5,7 from 1 to 365. I'm looking for an efficient way of formalizing the calculation step by step. What'd be the best way?&lt;/p&gt;&#xA;" OwnerUserId="2956" LastEditorUserId="1352" LastEditorDisplayName="" LastEditDate="2010-10-30T21:50:54.800" LastActivityDate="2010-10-30T21:54:23.353" Title="Simple Sequence Problem: Walk in the park" Tags="&lt;combinatorics&gt;&lt;discrete-mathematics&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="1" />
  <row Id="8406" PostTypeId="2" ParentId="8403" CreationDate="2010-10-30T21:33:41.123" Score="2" ViewCount="" Body="&lt;p&gt;If you just want the bounded property, you can simply find crude upper and lower bounds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;E.g. if $(-1)^n+1/n$ is bounded, then so is $((-1)^n+1/n)^2$ (so we can essentially disregard the power of 2).  Then we note that both $(-1)^n$ and $1/n$ are between $-1$ and $1$ for all $n \geq 1$ (I told you it was crude).  Hence $-2 \leq (-1)^n+1/n \leq 2$ and $((-1)^n+1/n)^2 \leq 4$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first sequence $\frac{3n}{2n+3}(-1)^{n+1}-\frac{(-1)^n}{2n}$ can be resolved by similar techniques.&lt;/p&gt;&#xA;" OwnerUserId="139" LastActivityDate="2010-10-30T21:33:41.123" CommentCount="2" />
  <row Id="8407" PostTypeId="2" ParentId="8339" CreationDate="2010-10-30T21:36:00.027" Score="0" ViewCount="" Body="&lt;p&gt;$B = P^{-1}AP \ \Longleftrightarrow \ PBP^{-1} = A$. If $Av = \lambda v$, then $PBP^{-1}v = \lambda v \ \Longrightarrow \ BP^{-1}v = \lambda P^{-1}v$. so, if $v$ is an eigenvector of $A$, with eigenvalue $\lambda$, then $P^{-1}v$ is an eigenvector of $B$ with the same eigenvalue. So, every eigenvalue of $A$  is an eigenvalue of $B$ and since you can interchange the roles of $A$ and $B$ in the previous calculations, every eigenvalue of $B$ is an eigenvalue of $A$ too. Hence, $A$ and $B$ have the same eigenvalues. Geometrically, in fact, also $v$ and $P^{-1}v$ are the same vector, written in different coordinate systems.&lt;/p&gt;&#xA;" OwnerUserId="664" LastActivityDate="2010-10-30T21:36:00.027" />
  <row Id="8408" PostTypeId="2" ParentId="8405" CreationDate="2010-10-30T21:54:23.353" Score="4" ViewCount="" Body="&lt;p&gt;Lets say its the $x$th day. Then, Greg is alone if and only if $x$ is a multiple of $5$ but not of $3$ or $7$. Dividing 365 by $5$ we get that Greg will go to the park 73 times.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Joe will go to the park every 3 times that Greg goes to the park. 73 = 24*3 + 1, so will be 24 occasions in which Joe is also in the park. Also, as 73 = 10*7 + 3, there will be 10 occasions when Dean is also in the park. Finally they are all in the park if $x$ is a multiple of $3*5*7=105$, which is only on 3 days. Thus, tallying it all up Greg is alone 73 - 24 - 10 + 3 = 42 days out of the year (the reason I add 3 is because in subtracting 24 and 10 I have also included days when they are both there twice, but we are only counting how many days he is alone).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, Dean will meet Joe if and only if $x$ is both a multiple of 3 and 7, or if and only if $x$ is a multiple of 21. As $365 = 17*21 + 8$, they will meet on $17$ occasions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically to attack these problems just think about dividing the 365 days and using the quotient, as it is equal to the number of days a given person goes to the park.&lt;/p&gt;&#xA;" OwnerUserId="1089" LastActivityDate="2010-10-30T21:54:23.353" CommentCount="1" />
  <row Id="8409" PostTypeId="1" AcceptedAnswerId="8411" CreationDate="2010-10-30T22:29:43.837" Score="4" ViewCount="427" Body="&lt;p&gt;I'm self-studying field extensions. I ran over an exercise which I can't completely solve. (I haven't yet started studying Galois theory, and I think this exercise isn't meant to be solved using it, just in case):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is:&#xA;a) Prove $\sqrt{2}+\sqrt[3]{5}$ is algebraic over $\mathbb{Q}$ of degree 6.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Done: I know it has degree $\leq 6$ because $\mathbb{Q}\subset \mathbb{Q}(\sqrt{2}+\sqrt[3]{5})\subset \mathbb{Q}(\sqrt{2},\sqrt[3]{5})$ which has degree 6; then I explicitly found the polynomial by solving a 6-equation linear system, and Wolfram Alpha proved it irreducible (btw: how can I prove it by hand?). The polynomial is $t^6-6t^4-10t^3+12t^2-60t+17$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;b) What's its degree over $\mathbb{Q}(\sqrt{2})$ and $\mathbb{Q}(\sqrt[3]{5})$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is this part b) which I can't solve. Of course its degree is $\leq 6$ in both cases, but I don't know what else to do.&lt;/p&gt;&#xA;" OwnerUserId="2614" LastActivityDate="2010-10-31T02:50:19.037" Title="Degree of $\sqrt{2}+\sqrt[3]{5}$ over $\mathbb{Q}(\sqrt{2})$ and $\mathbb{Q}(\sqrt[3]{5})$" Tags="&lt;abstract-algebra&gt;&lt;field-theory&gt;" AnswerCount="2" CommentCount="8" FavoriteCount="1" />
  <row Id="8410" PostTypeId="2" ParentId="8374" CreationDate="2010-10-30T22:33:04.320" Score="3" ViewCount="" Body="&lt;p&gt;x^a is fractionally differentiable at $x=0$ with $b$-fractional derivative proportional to $x^{a-b}$ for $b \leq a$. However, if $b &gt; a$, then the fractional derivative blows up at the origin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want a collection of functions which have $a$-fractional derivatives but such that the $b$-fractional derivatives for $b&gt;a$ simply do not exist, consider the $a$-fractional anti-derivatives of $|x|$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These are easy to find, because $|x| = x$ for $x&gt;0$, and $|x| = -x$ for $x \leq 0$, and so you can just use the formula for the $k$th anti-derivative of $x$.&lt;/p&gt;&#xA;" OwnerUserId="1089" LastEditorUserId="1089" LastEditorDisplayName="" LastEditDate="2010-10-31T20:03:40.423" LastActivityDate="2010-10-31T20:03:40.423" CommentCount="1" />
  <row Id="8411" PostTypeId="2" ParentId="8409" CreationDate="2010-10-30T22:39:57.727" Score="4" ViewCount="" Body="&lt;p&gt;Bruno: The degree over ${\mathbb Q}(\sqrt2)$ is 3: Consider $(x-\sqrt 2)^3-5$. Since $\root 3\of 5$ is in ${\mathbb Q}(\sqrt 2)(\sqrt 2+\root 3\of 5)$, the only other option is that it already belongs to ${\mathbb Q}(\sqrt 2)$. But this is impossible, since its minimal polynomial is $x^3-5$ over ${\mathbb Q}$ and ${\mathbb Q}(\sqrt 2)$ is an extension of degree 2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[ By the way, one can use this and the tower law, to show that ${\mathbb Q}(\sqrt 2+\root 3\of 5)$ has degree 6 over ${\mathbb Q}$. The irreducibility of your polynomial then follows for free. ]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, the degree over ${\mathbb Q}(\root 3\of 5)$ is 2.&lt;/p&gt;&#xA;" OwnerUserId="462" LastActivityDate="2010-10-30T22:39:57.727" CommentCount="8" />
  <row Id="8413" PostTypeId="2" ParentId="123" CreationDate="2010-10-30T23:05:47.493" Score="7" ViewCount="" Body="&lt;p&gt;The catenary has been mentioned a number of times, but apparently not the corresponding surface of revolution, the &lt;a href=&quot;http://mathworld.wolfram.com/Catenoid.html&quot; rel=&quot;nofollow&quot;&gt;catenoid&lt;/a&gt;. It and the plane are the only surfaces of revolution that have zero mean curvature (i.e. they are &lt;em&gt;minimal surfaces&lt;/em&gt;). This surface is the form a soap bubble (approximately) takes when it is stretched across two rings:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/8Zo0d.jpg&quot; alt=&quot;catenoid bubble&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(image from &lt;a href=&quot;http://www.soapbubble.dk/&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;)&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-10-30T23:05:47.493" />
  <row Id="8414" PostTypeId="2" ParentId="8347" CreationDate="2010-10-30T23:33:53.137" Score="4" ViewCount="" Body="&lt;p&gt;As always, to find the maximum and minimum of a function, you have to calculate first the critical points, that is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\partial G(x,y)}{\partial x}=0$$&#xA;$$\frac{\partial G(x,y)}{\partial y}=0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case, the respective equations are&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\partial G(x,y)}{\partial x}=-\frac{1}{x^{2}}+\frac{9}{(4-x-y)^{2}}$$&#xA;$$\frac{\partial G(x,y)}{\partial y}=-\frac{4}{y^{2}}+\frac{9}{(4-x-y)^{2}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Simplifying you should obtain $x=1-\displaystyle \frac{y}{4}$ and $\displaystyle \frac{5y}{2}=4-x$ which lead to the solution $(x,y)=\bigg\{\displaystyle \frac{2}{3},\displaystyle \frac{4}{3}\bigg\}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then you substitute those values in your original function $G(\frac{2}{3},\frac{4}{3})=9$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As I said, such a point is just a critical point (which means, it could be a local minimum, a local maximum or a saddle point). In order to figure out, we have to use the &lt;a href=&quot;http://en.wikipedia.org/wiki/Second_partial_derivative_test&quot; rel=&quot;nofollow&quot;&gt;Second partial derivative test&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can see the details in the Wikipedia article. As you already know what you want to prove (that $(\frac{2}{3},\frac{4}{3})$ is a local minimum), we expect that $M(\frac{2}{3},\frac{4}{3})&gt;0$ and $f_{xx}(\frac{2}{3},\frac{4}{3})&gt;0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think I'll leave it there. You can do the rest of the calculations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the way, the aforementioned procedure works to obtain only a local minimum, maximum or saddle point, not global extrema.&lt;/p&gt;&#xA;" OwnerUserId="2569" LastEditorUserId="2569" LastEditorDisplayName="" LastEditDate="2010-10-31T04:39:48.330" LastActivityDate="2010-10-31T04:39:48.330" />
  <row Id="8415" PostTypeId="1" AcceptedAnswerId="22582" CreationDate="2010-10-30T23:57:11.313" Score="3" ViewCount="157" Body="&lt;p&gt;The origin of this question is the identity&#xA;$$\sum_{k=0}^n \binom{n}{k} H_k = 2^n \left(H_n - \sum_{k=1}^n \frac{1}{k 2^k}\right),$$&#xA;where $H_n$ is the $n$th harmonic number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dividing by $2^n$, we have&#xA;$$2^{-n} \sum_{k=0}^n \binom{n}{k} H_k = H_n - \sum_{k=1}^n \frac{1}{k 2^k}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sum on the left can now be interpreted as a weighted average of the harmonic numbers through $H_n$ -- where the weights, of course, are the binomial coefficients.  Thus the difference between $H_n$ and its &quot;binomial average&quot; (I'm guessing there's no term for this) is&#xA;$$H_n - 2^{-n} \sum_{k=0}^n \binom{n}{k} H_k =  \sum_{k=1}^n \frac{1}{k 2^k}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sum on the right is known to converge to $\ln 2$ as $n \to \infty$.  (Substitute $-\frac{1}{2}$ into the Maclaurin series for $\ln (1+x)$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This leads me to my question: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Can we classify nonnegative functions $f(n)$ for which &#xA;  $$\lim_{n \to \infty} \left(f(n) - 2^{-n} \sum_{k=0}^n \binom{n}{k} f(k) \right)$$ is finite and nonzero?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It would seem that if $f$ increases sufficiently rapidly, then the limit would be $\infty$.  This is the case with both $f(n) = a^n$ and $f(n) = n$.  If $f$ decreases or is constant, then the limit is zero.  If $f$ has basically logarithmic growth, then it seems the limit would behave as $H_n$.  But can this be proved?  And what about other sublinear, increasing functions?&lt;/p&gt;&#xA;" OwnerUserId="2370" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2011-02-21T06:14:36.400" LastActivityDate="2011-02-21T06:14:36.400" Title="Asymptotic difference between a function and its binomial average" Tags="&lt;real-analysis&gt;&lt;binomial-coefficients&gt;&lt;asymptotics&gt;&lt;average&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="0" />
  <row Id="8416" PostTypeId="2" ParentId="8409" CreationDate="2010-10-31T00:31:53.580" Score="1" ViewCount="" Body="&lt;p&gt;For part b, we can find the minimal polynomial in both cases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider $a =\sqrt{2} + \sqrt[3]{5}$ over $\mathbb{Q}(\sqrt[3]{5})$. Notice that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ (x - \sqrt[3]{5})^2 -2 = 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is a degree two polynomial in $\mathbb{Q}(\sqrt[3]{5})[x]$ that $a$ is a root of. We know that $a \notin \mathbb{Q}(\sqrt[3]{5})$, so the degree of its minimal polynomial must be greater than 1, and now we have a polynomial of degree 2 that it is a root of, so its minimal polynomial must be that one above. Therefore $a$ has degree two over $\mathbb{Q}(\sqrt[3]{5})$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, consider $a$ over $\mathbb{Q}(\sqrt{2})$. Notice that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ (x - \sqrt{2})^3 - 5 = 0 $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is a degree 3 polynomial in $\mathbb{Q}(\sqrt{2})[x]$ that $a$ is a root of, and hence the minimal polynomial, which must divide this polynomial, can have degree either 2, or 3. But, if it has degree two, then it is attained from the above one by dividing by $x - a$. We get by polynomial division that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ (x - \sqrt{2})^3 - 5 = (x - \sqrt{2} - \sqrt[3]{5})(-2 - \sqrt{2} \sqrt[3]{5} + \sqrt[3]{25} + (-2\sqrt{2} + \sqrt[3]{5})x + x^2) $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and $-2 - \sqrt{2} \sqrt[3]{5} + \sqrt[3]{25} + (-2\sqrt{2} + \sqrt[3]{5})x + x^2 \notin \mathbb{Q}(\sqrt{2})[x]$ because the $x$ coefficient is not in $\mathbb{Q}(\sqrt{2})$. If it were, then as $2\sqrt{2} \in \mathbb{Q}(\sqrt{2})$, we would get that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$2 \sqrt{2} + (-2\sqrt{2} + \sqrt[3]{5}) = \sqrt[3]{5} \in \mathbb{Q}(\sqrt{2})$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is impossible. Thus that quadratic polynomial cannot be the minimal polynomial for $a$. Hence, the minimal polynomial of $a$ has degree 3.&lt;/p&gt;&#xA;" OwnerUserId="1089" LastEditorUserId="1089" LastEditorDisplayName="" LastEditDate="2010-10-31T02:50:19.037" LastActivityDate="2010-10-31T02:50:19.037" CommentCount="2" />
  <row Id="8417" PostTypeId="1" CreationDate="2010-10-31T00:54:43.520" Score="1" ViewCount="145" Body="&lt;p&gt;Hi all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I had recently asked the question:&lt;br&gt;&#xA;&lt;a href=&quot;http://math.stackexchange.com/q/8213/2916&quot;&gt;http://math.stackexchange.com/q/8213/2916&lt;/a&gt;&lt;br&gt;&#xA;which received a very helpful answer. However, I'd like to ask your help again as the formulation of the problem has slightly changed and i'm lost...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Problem:&lt;/strong&gt;&lt;br&gt;&#xA;Given two points: $(x_l, y_l)$ and $(x_u, y_u)$&#xA;with: $x_l &amp;lt; x_u$ and $y_l &amp;lt; y_u$,&lt;br&gt;&#xA;and given that the higher asymptote is one ($\lim_{x\to+\infty}f(x)=1$)&lt;br&gt;&#xA;what's the logistic function that passes through the two points &lt;strong&gt;and the origin $(0, 0)$&lt;/strong&gt;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="2916" LastEditorUserId="2916" LastEditorDisplayName="" LastEditDate="2010-10-31T10:40:40.093" LastActivityDate="2010-10-31T10:40:40.093" Title="Logistic function passing through two points AND the origin?" Tags="&lt;statistics&gt;" CommentCount="8" />
  <row Id="8418" PostTypeId="2" ParentId="8337" CreationDate="2010-10-31T01:28:49.853" Score="8" ViewCount="" Body="&lt;p&gt;Here is one more nice proof, I learned it from Grisha Mikhalkin: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Lemma:&lt;/strong&gt; Let $Z$ be a complex curve in $\mathbb{C}^2$. Let $R(Z) \subset \mathbb{R}^2$ be the projection of $Z$ onto its real parts and $I(Z)$ the projection onto its complex parts. If these projections are both one to one, then the area of $R(Z)$ is equal to the area of $I(Z)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof:&lt;/strong&gt; There is an obvious map from $R(Z)$ to $I(Z)$, given by lifting $(x_1, x_2) \in R(Z)$ to $(x_1+i y_1, x_2 + i y_2) \in Z$, and then projecting to $(y_1, y_2) \in I(Z)$. We must prove this map has Jacobian $1$. WLOG, translate $(x_1, y_1, x_2, y_2)$ to $(0,0,0,0)$ and let $Z$ obey $\partial z_2/\partial z_1 = a+bi$ near $(0,0)$. To first order, we have $x_2 = a x_1 - b y_1$ and $y_2 = a y_1 + b x_1$. So $y_1 = (a/b) x_1 - (1/b) x_2$ and $y_2 = (a^2 + b^2)/b x_1 - (a/b) x_2$. So the derivative of $(x_1, x_2) \mapsto (y_1, y_2)$ is $\left( \begin{smallmatrix} a/b &amp;amp; - 1/b \\ (a^2 + b^2)/b &amp;amp; -a/b \end{smallmatrix} \right)$ and the Jacobian is $1$. QED&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, consider the curve $e^{-z_1} + e^{-z_2} = 1$, where $z_1$ and $z_2$ obey the following inequalities: $x_1 \geq 0$, $x_2 \geq 0$, $-\pi \leq y_1 \leq 0$ and $0 \leq y_2 \leq \pi$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given a point on $e^{-z_1} + e^{-z_2} = 1$, consider the triangle with vertices at $0$, $e^{-z_1}$ and $e^{-z_1} + e^{-z_2} = 1$. The inequalities on the $y$'s states that the triangle should lie above the real axis; the inequalities on the $x$'s state the horizontal base should be the longest side.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Projecting onto the $x$ coordinates, we see that the triangle exists if and only if the triangle inequality $e^{-x_1} + e^{-x_2} \geq 1$ is obeyed. So $R(Z)$ is the region under the curve $x_2 = - \log(1-e^{-x_1})$. The area under this curve is&#xA;$$\int_{0}^{\infty} - \log(1-e^{-x}) dx = \int_{0}^{\infty} \sum \frac{e^{-kx}}{k} dx = \sum \frac{1}{k^2}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, project onto the $y$ coordinates. Set $(y_1, y_2) = (-\theta_1, \theta_2)$ for convenience, so the angles of the triangle are $(\theta_1, \theta_2, \pi - \theta_1 - \theta_2)$. The largest angle of a triangle is opposite the largest side, so we want $\theta_1$, $\theta_2 \leq \pi - \theta_1 - \theta_2$, plus the obvious inequalities $\theta_1$, $\theta_2 \geq 0$. So $I(Z)$ is the quadrilateral with vertices at $(0,0)$, $(0, \pi/2)$, $(\pi/3, \pi/3)$ and $(\pi/2, 0)$ and, by elementary geometry, this has area $\pi^2/6$.&lt;/p&gt;&#xA;" OwnerUserId="448" LastEditorUserId="448" LastEditorDisplayName="" LastEditDate="2010-10-31T12:11:28.040" LastActivityDate="2010-10-31T12:11:28.040" CommentCount="4" />
  <row Id="8419" PostTypeId="1" AcceptedAnswerId="8420" CreationDate="2010-10-31T01:38:08.503" Score="1" ViewCount="282" Body="&lt;p&gt;Please suggest a suitable approach for this problem.&lt;/p&gt;&#xA;" OwnerUserId="3744" LastEditorUserId="3744" LastEditorDisplayName="" LastEditDate="2010-10-31T01:52:36.230" LastActivityDate="2010-10-31T16:21:56.547" Title="What is the remainder when $4^{44}$ is divided by 44 ?" Tags="&lt;number-theory&gt;" AnswerCount="6" CommentCount="3" />
  <row Id="8420" PostTypeId="2" ParentId="8419" CreationDate="2010-10-31T01:58:43.080" Score="8" ViewCount="" Body="&lt;p&gt;Well... $4^{44}$ is obviously divisible by 4, that is, $4^{44} \equiv 0 \pmod 4$.  So you only need to calculate $4^{44} \pmod {11}$, which can be done using &lt;a href=&quot;http://en.wikipedia.org/wiki/Fermat%27s_little_theorem&quot; rel=&quot;nofollow&quot;&gt;Fermat's Little Theorem&lt;/a&gt; (since 4 and 11 are coprime), then use the &lt;a href=&quot;http://en.wikipedia.org/wiki/Chinese_remainder_theorem&quot; rel=&quot;nofollow&quot;&gt;Chinese Remainder Theorem&lt;/a&gt; to combine the two congruences.&lt;/p&gt;&#xA;" OwnerUserId="139" LastActivityDate="2010-10-31T01:58:43.080" CommentCount="3" />
  <row Id="8421" PostTypeId="2" ParentId="8419" CreationDate="2010-10-31T02:03:57.967" Score="4" ViewCount="" Body="&lt;p&gt;$4^{44}=4^4\times4^{40}=4^4\times(4^{10}-1+1)^4$. The second factor leaves remainder 1 one divided by 11. This is because of Fermat's little theorem, but can be seen directly: $4^{10}-1=(2^{10}-1)(2^{10}+1)=11\times93\times1025$, so $4^{10}$ is a multiple of 11 plus 1, and so is $4^{40}$. Finally, $4^{44}=4^4\times(11k+1)\equiv 4^4=256=44\times5+36\equiv 36\pmod{44}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The key is to consider the prime decomposition of 44 and treat each factor separately.&lt;/p&gt;&#xA;" OwnerUserId="462" LastActivityDate="2010-10-31T02:03:57.967" />
  <row Id="8422" PostTypeId="2" ParentId="8419" CreationDate="2010-10-31T02:13:36.383" Score="2" ViewCount="" Body="&lt;p&gt;Another method is to simply compute $4^{44} \pmod {44}$, which could be done in any self-respecting computer algebra system (or you could write your own code for it, watching out for overflow errors).  For example:  you could use &lt;a href=&quot;http://www.wolframalpha.com/&quot; rel=&quot;nofollow&quot;&gt;WolframAlpha&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Warning:  If this is a homework question, finding the answer computationally is likely to annoy the lecturer.  However, it's still a good way to check your answer.&lt;/p&gt;&#xA;" OwnerUserId="139" LastActivityDate="2010-10-31T02:13:36.383" CommentCount="2" />
  <row Id="8423" PostTypeId="2" ParentId="8419" CreationDate="2010-10-31T02:41:54.237" Score="1" ViewCount="" Body="&lt;p&gt;As previously stated, since $4^{44} \equiv 0 \pmod 4$, all you need to find is $4^{44} \mod 11$ (which makes sense if you think about the fact that if $4^{44}$ is evenly divisible by one factor of 44, it won't affect the result of modding, so you only need to consider the other factor, 11).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A faster way to do this is note that $4^{44}=2^{88}=2^{85}\cdot8=32^{17}\cdot8$, and that $32\equiv -1 \pmod{11}$, which turns it into $(-8)\mod 11$. However, it is important to note that this is still in mod 11, so the answer must be turned into $(-8)\mod 44$. (which is the same thing, as we noted earlier)&lt;/p&gt;&#xA;" OwnerUserId="1562" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-10-31T08:16:12.957" LastActivityDate="2010-10-31T08:16:12.957" CommentCount="0" />
  <row Id="8424" PostTypeId="1" AcceptedAnswerId="8428" CreationDate="2010-10-31T03:06:53.563" Score="1" ViewCount="180" Body="&lt;p&gt;I was browsing through Wikipedia for some reason or another, and it says that there is a way to do Galois theory without worrying about whether or not polynomials are irreducible.  That's a bit surprising to me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is Wikipedia wrong (as is often the case when I read something surprising)?  If not, where can I hear more about it?  &lt;/p&gt;&#xA;" OwnerUserId="126" LastActivityDate="2010-10-31T15:19:50.647" Title="Galois theory with non-irreducible polynomials" Tags="&lt;galois-theory&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="8425" PostTypeId="2" ParentId="8419" CreationDate="2010-10-31T03:08:28.627" Score="3" ViewCount="" Body="&lt;p&gt;One way to do this is to figure out that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ 4^6 \equiv 64^2 \equiv 20^2 \equiv 4 \mod 44 $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ 4^{44} \equiv (4^6)^7 * 4^2 \equiv 4^7 * 4^2 \equiv 4^4 \equiv 36 \mod 44 $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so the answer is $36$, and all you had to do is calculate a few powers of $4$ mod $44$.&lt;/p&gt;&#xA;" OwnerUserId="1089" LastActivityDate="2010-10-31T03:08:28.627" />
  <row Id="8426" PostTypeId="1" AcceptedAnswerId="8427" CreationDate="2010-10-31T03:14:07.963" Score="2" ViewCount="140" Body="&lt;p&gt;I'm attempting to implement an equation (for calculating &lt;a href=&quot;http://dx.doi.org/10.1109/TMAG.2010.2049026&quot; rel=&quot;nofollow&quot;&gt;magnetic forces between coils&lt;/a&gt;, eqs (22–24) in the linked paper) that requires the use of elliptic integrals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately these equations require the evaluation of the elliptic integrals far outside their standard parameter range of $0\le m\le 1$ and the numerical implementations I have available to evaluate them give inconsistent results.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe that Mathematica is correct in its answer:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;EllipticF[ArcSin[Sqrt[1/c5]], c5] //. c5 -&amp;gt; 817.327&#xA;=&amp;gt; 0.054961 - 1.17196*10^-17 i&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Whereas Matlab's MuPad engine gives:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;mfun('EllipticF',asin(sqrt(1/817.327)),sqrt(817.327))&#xA;=&amp;gt; 0.054961 - 0.000707i&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;(Mma's function takes parameter $m=k^2$ whereas MuPad takes modulus $k$, explaining the &lt;code&gt;sqrt&lt;/code&gt;)&#xA;While I can use Mathematica for my own work, my colleagues only have Matlab available and I'd like them to be able to use this code.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm a pretty unfamiliar with the theory behind elliptic integrals, but &lt;a href=&quot;http://www.gutenberg.org/ebooks/31076.bibrec.html&quot; rel=&quot;nofollow&quot;&gt;Baker's ‘Elliptic Functions’&lt;/a&gt; says&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;We shall see later on that the quantity $k^2$ [...] can always be considered real and less than unity.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Which leads me to ask: can the arguments to these elliptic integrals be re-stated in terms of an input of $k&gt;1$, such as I seem to require above?&lt;/p&gt;&#xA;" OwnerUserId="2961" LastEditorUserId="2961" LastEditorDisplayName="" LastEditDate="2010-11-01T00:12:18.057" LastActivityDate="2010-11-01T00:12:18.057" Title="Elliptic integrals with parameter outside 0&lt;m&lt;1" Tags="&lt;numerical-methods&gt;&lt;special-functions&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="8427" PostTypeId="2" ParentId="8426" CreationDate="2010-10-31T03:45:05.133" Score="2" ViewCount="" Body="&lt;p&gt;What you seem to require here are the so-called &lt;a href=&quot;http://dlmf.nist.gov/19.7#ii&quot; rel=&quot;nofollow&quot;&gt;&quot;reciprocal-modulus transformations&quot;&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your specific case, upon applying the reciprocal-modulus transformation, your elliptic integral can actually be simplified to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac1{\sqrt{c_5}}F\left(\frac{\pi}{2}\vert \frac1{c_5}\right)=\frac1{\sqrt{c_5}}K\left(\frac1{c_5}\right)$$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $K(m)$ is the complete elliptic integral of the first kind with parameter $m$ (though I understand MuPAD uses $k=\sqrt{m}$ as argument, in which case you seem to know the conversion formulae already).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are similar formulae for the elliptic integrals of the second and third kinds, and I direct you to the DLMF link I gave above for them.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-10-31T03:45:05.133" CommentCount="2" />
  <row Id="8428" PostTypeId="2" ParentId="8424" CreationDate="2010-10-31T05:25:45.047" Score="6" ViewCount="" Body="&lt;p&gt;There is a very big difference between &quot;you can do Galois theory without worrying about whether or not a polynomial is irreducible&quot; and &quot;one can do Galois theory while working with polynomials that are not irreducible&quot;. The former implies that the results will hold regardless of whether the polynomial is irreducible or not; the latter merely says that one can adjust the work so as to deal with reducible polynomials.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $f(x)$ is not irreducible, but factors as $f(x)=p_1(x)^{a_1}\cdots p_k(x)^{a_k}$ with $p_j(x)$ irreducible and separable, then you can certainly still talk about the Galois group of the splitting field of $f(x)$, and it will have all the usual properties: it would simply be compositum of the splitting fields of the different $p_i$, and so the Galois group of $f$ would have normal subgroups corresponding to the splitting fields of each $p_i$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the polynomials are inseparable one runs into greater difficulties; the usual technique is to separate the extension into a separable extension and a purely inseparable extension. For the separable ones, the usual techniques of Galois Theory work; for the purely inseparable ones, we have special results that apply.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That said, the quote from Wikipedia is rather poor, since separability is a property of the &lt;strong&gt;irreducible factors&lt;/strong&gt; of the polynomials. Repeated factors do not affect separability or lack thereof. But even if we were to take it at face value, it does not claim that one can do Galois theory &quot;without worrying&quot;; it would say that if you are doing Galois theory with non-irreducible polynomials (in the way that one usually does: by splitting the polynomial into irreducible factors), then you would not worry about whether the polynomial itself has repeated roots or not, because repeated roots could come from repeated &lt;em&gt;separable&lt;/em&gt; irreducible factors. &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-10-31T15:19:50.647" LastActivityDate="2010-10-31T15:19:50.647" CommentCount="4" />
  <row Id="8429" PostTypeId="1" AcceptedAnswerId="8437" CreationDate="2010-10-31T06:08:17.423" Score="0" ViewCount="62" Body="&lt;p&gt;I was always thinking about this questions when it comes to lottery. lets say you have a number range from 1 - 45. You need 5-6 numbers to win.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let says you have 2 options&#xA;Draw 3 sets of numbers with 10 numbers on each set.&#xA;or &#xA;1 set of numbers with 19 numbers for that?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which chance will have higher of winning?&lt;/p&gt;&#xA;" OwnerUserId="2962" LastActivityDate="2010-10-31T09:36:32.240" Title="More choices lesser draw or more draw lesser choices?" Tags="&lt;probability&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="8430" PostTypeId="2" ParentId="392" CreationDate="2010-10-31T06:09:01.513" Score="7" ViewCount="" Body="&lt;p&gt;This is related to &lt;a href=&quot;http://math.stackexchange.com/questions/392/intuitive-understanding-of-the-derivatives-of-sinx-and-cosx/395#395&quot;&gt;Justin L.'s answer&lt;/a&gt;, because I have basically the same interpretation, but whereas that answer (as I'm interpreting it) gives a great intuitive &lt;em&gt;check&lt;/em&gt; that the derivatives are correct, I intend to present how one might actually (somewhat intuitively) arrive at the derivatives.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By definition, $s(x)=(\cos(x),\sin(x))$ gives the point on the unit circle after traveling an arclength $x$ from the point $(1,0)$, oriented counterclockwise.  Parametrization with respect to arclength is precisely the condition that guarantees that the curve has unit speed, i.e., $|s'(x)|\equiv 1$.  Because $s$ also has constant length, the product rule can be used to show that $s'(x)$ is perpendicular to $s(x)$ for all $x$: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$s\cdot s \equiv 1 \Rightarrow s'\cdot s + s\cdot s'\equiv 0 \Rightarrow s\cdot s'\equiv 0.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus for each $x$, $s'(x)$ is a unit length vector perpendicular to $s(x)$.  This leaves 2 possibilities: either a counterclockwise or clockwise rotation by $\frac{\pi}{2}$ from $s(x)$.  But now, because $s'$ tells us how $s$ is changing, it must point in the direction of motion of $s$, namely counterclockwise.  Thus $s'(x)$ is a counterclockwise rotation by $\frac{\pi}{2}$ from $s(x)$, which means &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$s'(x)=s\left(x+\frac{\pi}{2}\right)=\left(\cos\left(x+\frac{\pi}{2}\right),\sin\left(x+\frac{\pi}{2}\right)\right)=(-\sin(x),\cos(x)).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But we also have $s'(x)=(\cos'(x),\sin'(x))$, so matching coordinates yields $\cos'=-\sin$ and $\sin'=\cos$.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-10-31T06:09:01.513" CommentCount="2" />
  <row Id="8431" PostTypeId="1" AcceptedAnswerId="8432" CreationDate="2010-10-31T06:22:47.240" Score="1" ViewCount="110" Body="&lt;p&gt;Let G=(V,E) be a connex graph. Color it's edges randomly with red/blue.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-prove that there exists an Eulerian circuit, without any two adjacent edges of the same color.. only if for any &lt;code&gt;v&lt;/code&gt; vertex of G the number of incident blue edges is equal to the number of red ones.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;-if G is complete and &lt;code&gt;x,y,z&lt;/code&gt; are three distinct vertices prove that, if there is a path without consecutive edges of the same color from &lt;code&gt;x&lt;/code&gt; to &lt;code&gt;y&lt;/code&gt; passing through &lt;code&gt;z&lt;/code&gt;,  then there is a path with the same property that has &lt;code&gt;xz&lt;/code&gt; as the first edge or &lt;code&gt;zy&lt;/code&gt; as the last edge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please help me by at least providing some suitable references that may help me solve this problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you in advance.&lt;/p&gt;&#xA;" OwnerUserId="2963" LastActivityDate="2010-10-31T06:34:32.473" Title="Eulerian graphs proof" Tags="&lt;homework&gt;&lt;graph-theory&gt;" AnswerCount="1" FavoriteCount="0" />
  <row Id="8432" PostTypeId="2" ParentId="8431" CreationDate="2010-10-31T06:34:32.473" Score="2" ViewCount="" Body="&lt;p&gt;These probably should be comments but I don't have sufficient reputation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint for problem 1: If there are an equal number of red and blue edges at each vertex, what does that tell you about the parity of the degree of each vertex?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint for problem 2: If you have a path from x to y passing through z, can you use the fact that the graph is complete to find another path that fits the alternating condition?&lt;/p&gt;&#xA;" OwnerUserId="2954" LastActivityDate="2010-10-31T06:34:32.473" CommentCount="2" />
  <row Id="8433" PostTypeId="1" AcceptedAnswerId="8438" CreationDate="2010-10-31T07:18:27.503" Score="0" ViewCount="148" Body="&lt;p&gt;pseudocode:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;void recursive('k'){ // 'k' and 'i' vertices&#xA;  sumA = 0;&#xA;  sumB = 0;&#xA;  for each non visited 'i' neighbor do{&#xA;     recursive('i');&#xA;     sumA = sumA + b['i'];&#xA;     sumB = sumB + max(a['i'], b['i']);&#xA;     }&#xA;  a['k'] = 1 + sumA;&#xA;  b['k'] = sumB;&#xA;  }&#xA;&#xA;&#xA;void main(){&#xA; a = b = 0; //initialize tables with 0 (zeros)&#xA; recursive('X');  //let 'X' be an arbitrary root&#xA; cout&amp;lt;&amp;lt;max(a['X'], b['X']);&#xA; }&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;need proof that &lt;code&gt;max(a['X'], b['X'])&lt;/code&gt; is the cardinal of the maximum independent set in the tree.&#xA;What am I missing ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you in advance.&lt;/p&gt;&#xA;" OwnerUserId="2963" LastActivityDate="2010-10-31T09:52:27.187" Title="Maximum independent set in a tree. Review algorithm, need proof." Tags="&lt;homework&gt;&lt;graph-theory&gt;" AnswerCount="1" FavoriteCount="0" />
  <row Id="8434" PostTypeId="2" ParentId="8419" CreationDate="2010-10-31T07:59:57.717" Score="0" ViewCount="" Body="&lt;p&gt;$\rm\ 4^5\equiv 2^{10}\equiv 1\ \:(mod\ 11)\ \Rightarrow\ 4^{4+5\cdot8}\equiv 4^4\equiv 4\cdot 20\equiv -8\ (mod\ 11)\:$ and $\rm (mod\ 4)\:$, so $\rm (mod\ 44)$&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-10-31T16:21:56.547" LastActivityDate="2010-10-31T16:21:56.547" CommentCount="1" />
  <row Id="8435" PostTypeId="1" AcceptedAnswerId="8471" CreationDate="2010-10-31T08:00:09.887" Score="3" ViewCount="115" Body="&lt;p&gt;Suppose n is an integer. What kind of bounds do we know for how close the closest prime p &gt; n will be?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd especially appreciate an answer that pushes me in the right direction of proving a good bound for this without showing the proof fully.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Context: I ran into this while coming up to a solution for &lt;a href=&quot;http://stackoverflow.com/questions/4058172/tricky-algorithm-interview-question&quot;&gt;http://stackoverflow.com/questions/4058172/tricky-algorithm-interview-question&lt;/a&gt;. Depending on the bound, the algorithm I come up with may or may not be deterministic.&lt;/p&gt;&#xA;" OwnerUserId="181" LastActivityDate="2010-10-31T22:07:13.360" Title="Bounds on prime number spacing" Tags="&lt;prime-numbers&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="8436" PostTypeId="2" ParentId="8429" CreationDate="2010-10-31T09:00:17.110" Score="1" ViewCount="" Body="&lt;p&gt;Hopefully I understand your question properly. To make sure, here's how I understand it:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's a lottery, where 5 distinct random numbers will be chosen from 1-45 inclusive. If on your lottery card, you have all 5, you win.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Options:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;a) draw 3 cards with 10 random numbers per card&lt;/li&gt;&#xA;&lt;li&gt;b) draw 1 card with 19 random numbers&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Which option is best?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a) The probability of winning if you drew one card with 10 numbers is 10/45 * 9/44 * 8/43 * 7/42 * 6/41 = 4/19393. Your probability of not winning is 1-4/19393. Hence your probability of winning with at least one card is 1-(1-4/19393)^3, which is approximately equal to 0.0006187.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;b) On the other hand, your probability of winning with the 1 card option is 19 / 45 * 18/44 * 17/43 * 16/42 * 15/41 = 1292/135751, which is approximately equal to 0.009517.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So clearly the second option is the best.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&#xA;&lt;strong&gt;A more intuitive (less numerical) explanation:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your chances of winning with a card with 10 numbers is very low since half of the numbers on the card half to be winning numbers. Clearly with 19 numbers the chances are greater. If you crunch the numbers, it turns out that 3 10-number cards are not sufficient to overcome the advantage held by the 19-number card.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps you can extend the calculations above to show how many 10-number cards you'd need before that becomes the better option...?&lt;/p&gt;&#xA;" OwnerUserId="181" LastActivityDate="2010-10-31T09:00:17.110" />
  <row Id="8437" PostTypeId="2" ParentId="8429" CreationDate="2010-10-31T09:36:32.240" Score="3" ViewCount="" Body="&lt;p&gt;We have $45$ numbers $1,2,\ldots,45$ of which $6$ are winners.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The choice is between (a) buying three cards, each of which contains $10$ distinct numbers chosen from the $45$  (the same number can appear on different cards) and (b) buying one card which contains $19$ distinct numbers chosen from the $45.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To win we must have at least $5$ of the winning numbers on a card.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the probability of at least $5$ winning number on a card of size $10$ is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{ \binom{6}{5} \binom{39}{5} +  \binom{6}{6} \binom{39}{4} }{ \binom{45}{10} }&#xA;= \frac{1}{902}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And so the probability of $ \le 4 $ winners is $ 1 – 1/902 = 901/902.$ Hence the probability of at least one of our three cards of size $10$ winning is ($1$ minus the probability that they all lose)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ 1 - \left( \frac{901}{902} \right)^3 \approx 0.003322.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The probability of at least five winners on a card of size $19$ is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{ \binom{6}{5} \binom{39}{14} +  \binom{6}{6} \binom{39}{13} }{ \binom{45}{19} }&#xA;= \frac{5491}{135751} \approx 0.040449.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $0.040449 &gt; 0.003322$ you have a better chance of winning by buying one card of size $19.$&lt;/p&gt;&#xA;" OwnerUserId="1301" LastActivityDate="2010-10-31T09:36:32.240" CommentCount="3" />
  <row Id="8438" PostTypeId="2" ParentId="8433" CreationDate="2010-10-31T09:52:27.187" Score="1" ViewCount="" Body="&lt;p&gt;The element $a[i]$ is the size of the maximal independent set in the subtree rooted in $i$ which contains $i$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The element $b[i]$ is the size of the maximal independent set in the subtree rooted in $i$ which doesn't contain $i$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The algorithm computes these recursively, and then choose the maximal of the two at the root.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-10-31T09:52:27.187" CommentCount="2" />
  <row Id="8439" PostTypeId="1" AcceptedAnswerId="8440" CreationDate="2010-10-31T10:06:53.540" Score="1" ViewCount="239" Body="&lt;p&gt;$$\cos A \cdot \cos 2A  \cdot \cos 2^{2}A  \cdot \cos 2^{3}A  ... \cos 2^{n-1}A = \frac { \sin 2^n A}{ 2^n \sin  A } $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am very much inquisitive to see how this trigonometrical identity can be proved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS:I am not much of interested about an inductive proof.&lt;/p&gt;&#xA;" OwnerUserId="2109" LastEditorUserId="2109" LastEditorDisplayName="" LastEditDate="2010-10-31T11:05:39.540" LastActivityDate="2010-10-31T14:54:24.227" Title="Prove: $\cos A \cdot \cos 2A \cdot \cos 2^{2}A \cdot \cos 2^{3}A ... \cos 2^{n-1}A = \frac { \sin 2^n A}{ 2^n \sin A } $" Tags="&lt;trigonometry&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="1" />
  <row Id="8440" PostTypeId="2" ParentId="8439" CreationDate="2010-10-31T10:39:33.043" Score="8" ViewCount="" Body="&lt;p&gt;The proof is just repeated application of the double-angle formula for the sine function. For example, the case $n=3$:&#xA;$$\sin 8A = 2 \sin 4A \cos 4A = 2 (2 \sin 2A \cos 2A) \cos 4A = 2 (2 (2 \sin A \cos A) \cos 2A) \cos 4A.$$&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2010-10-31T10:39:33.043" CommentCount="4" />
  <row Id="8441" PostTypeId="2" ParentId="8377" CreationDate="2010-10-31T11:03:26.620" Score="1" ViewCount="" Body="&lt;p&gt;I'll write something so this question doesn't remain unanswered.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sometimes it can be quite difficult to prove that a sequence is increasing.  For example, Smetaniuk gave a proof that the number of Latin squares $L_n$ (Sloane's &lt;a href=&quot;http://www.research.att.com/~njas/sequences/A002860&quot; rel=&quot;nofollow&quot;&gt;A002860&lt;/a&gt;) is increasing (actually, he proved $L_{n+1} \geq (n+1)!L_n$), which is one of the best results around regarding the mysterious $L_n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Smetaniuk, Bohdan&#xA;A new construction of Latin squares. II. The number of Latin squares is strictly increasing. &#xA;Ars Combin. 14 (1982), 131–145.&lt;/p&gt;&#xA;" OwnerUserId="139" LastActivityDate="2010-10-31T11:03:26.620" />
  <row Id="8442" PostTypeId="2" ParentId="5892" CreationDate="2010-10-31T11:49:06.070" Score="1" ViewCount="" Body="&lt;p&gt;this link provides some good resources ,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;check it up&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://entrance-exam.net/gre-papers/&quot; rel=&quot;nofollow&quot;&gt;http://entrance-exam.net/gre-papers/&lt;/a&gt;&lt;/p&gt;&#xA;" LastActivityDate="2010-10-31T11:49:06.070" CommunityOwnedDate="2010-10-31T11:49:06.070" />
  <row Id="8443" PostTypeId="1" CreationDate="2010-10-31T12:56:48.943" Score="3" ViewCount="108" Body="&lt;p&gt;I just read this document - &lt;a href=&quot;http://www.math.ohio-state.edu/~econrad/Jacobi/sumofsq/sumofsq.html&quot; rel=&quot;nofollow&quot;&gt;Jacobi's Four Square Theorem&lt;/a&gt;. It shows how to count the number of representations of a number as the sum of four squares.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can follow the proof but currently it just appears like a brilliant coincidence, so I am wondering what else can you count with these?&lt;/p&gt;&#xA;" LastActivityDate="2010-10-31T19:46:34.110" Title="What else can the elliptic integral count?" Tags="&lt;fourier-series&gt;&lt;elliptic-curves&gt;&lt;counting&gt;" AnswerCount="1" FavoriteCount="0" />
  <row Id="8444" PostTypeId="1" CreationDate="2010-10-31T13:49:03.427" Score="1" ViewCount="147" Body="&lt;p&gt;What could be the possible way to find the &lt;code&gt;Nth term&lt;/code&gt; of following series where the sign toggles after each triangular number?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;1 -2 -3 4 5 6 -7 -8 -9 -10 11 12 13 14 15 -16 -17 ....&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The series cannot be in a Geometric Progression because there are 4 distinct triangular numbers in the above series.&lt;/p&gt;&#xA;" LastActivityDate="2010-11-01T09:07:10.250" Title="Nth term of the series where sign toggles after a triangular number" Tags="&lt;number-theory&gt;&lt;sequences-and-series&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="8445" PostTypeId="2" ParentId="3087" CreationDate="2010-10-31T14:07:26.733" Score="1" ViewCount="" Body="&lt;p&gt;Whatever you think the high school mathematics curriculum &quot;should&quot; be, in the United States a curriculum has been put in place, known as the Common Core Standards (CCS), which will significantly - I believe - change American mathematics education. (The CCS extend to all K-12 mathematics.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.corestandards.org/the-standards/mathematics&quot; rel=&quot;nofollow&quot;&gt;http://www.corestandards.org/the-standards/mathematics&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't think these changes are for the better. A short summary of my personal views are available here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.education.umd.edu/MathEd/conference/vbook/public-perceptions_Malkevitch.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.education.umd.edu/MathEd/conference/vbook/public-perceptions_Malkevitch.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1369" LastActivityDate="2010-10-31T14:07:26.733" CommunityOwnedDate="2010-10-31T14:07:26.733" />
  <row Id="8446" PostTypeId="2" ParentId="3087" CreationDate="2010-10-31T14:43:33.380" Score="1" ViewCount="" Body="&lt;p&gt;I recall the Asimov short story about humanity that has completely lost the knowledge of how to do simple arithmetic without use of a computer, called &quot;The Feeling of Power&quot;. The point is, despite computers being able to do these things for us, it is still important and valuable to know how to do simple operations. At times, we are amazingly close to that point in this very day, certainly in the current crop of kids growing up now.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basic mathematical skills are tremendously valuable in a variety of places. Just because you CAN factor a polynomial by computer does not mean that at least understanding the tools to do so are no longer necessary for 99% of the people in the world.&lt;/p&gt;&#xA;" OwnerUserId="1608" LastActivityDate="2010-10-31T14:43:33.380" CommunityOwnedDate="2010-10-31T14:43:33.380" />
  <row Id="8447" PostTypeId="2" ParentId="1116" CreationDate="2010-10-31T14:48:27.290" Score="12" ViewCount="" Body="&lt;p&gt;I had my student, Tim Michaels, work on this. We came up with a complicated recurrence relation, &#xA;indicated below. The first few answers are 1, 2, 6, 25, 128, 758, 5014, 36194, 280433, 2303918, 19885534, 179028087, 1671644720, 16114138846, 159761516110, 1623972412726, &#xA;16880442523007, 179026930243822, 1933537655138482, 21231023519199575, &#xA;236674460790503286, 2675162663681345170, 30625903703241927542, &#xA;354767977792683552908, 4154708768196322925749, &#xA;49152046198035152483150, 587011110939295781585102, &#xA;7072674305834582713614923. Note that we are counting rotations and reflections as distinct tilings. An interesting pattern is that mod 2, there is an 8-fold periodicity which we don't understand and can't prove in general.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a picture of the cases n=1,2,3,4, with 1,2,6,25 tilings in each case.The way to systematically generate these is to &quot;push in&quot; a vertical line from the right to all previously constructed tilings in all possible ways. That's how the recurrence relation is defined.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/OftaF.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Okay, here is the recurrence:&#xA;Let $a_{\ell,j,m}$ be the number of distinct tilings with $\ell$ tiles, $j$ edges that meet the right-hand side of the square and $m$ 4-valent vertices.&#xA;$$a_{\ell,j,m}=\sum_{p=1}^\ell(-1)^{p+1}\sum_{i=0}^m\sum_{n=1}^{\ell-1}\sum_{k=0}^{n-1}\alpha_{n,k,i,\ell,j,m,p} a_{n,k,i}$$&#xA;where, letting $t=m-i, s=\ell-n-p-t$ and $r=k+s+t+p-j$,&#xA;$$\alpha_{n,k,i,l,j,m,p}=\binom{r-1}{p-1}\binom{k-r+2}{p}\binom{s+r-1}{r-1}\binom{r-p}{t}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;B&gt;Edit:&lt;/B&gt; I have posted a preprint describing the recurrence relation &lt;a href=&quot;http://www.math.utk.edu/~jconant/counttilings.pdf&quot; rel=&quot;nofollow&quot;&gt;here.&lt;/a&gt; Comments are welcome. Is this sort of thing publishable anywhere to anyone's knowledge?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;B&gt;Edit 2:&lt;/B&gt; Nathan Reading has just posted a relevant &lt;a href=&quot;http://arxiv.org/abs/1105.3093&quot; rel=&quot;nofollow&quot;&gt;preprint.&lt;/a&gt; He finds a bijection between generic tilings (no 4-valent vertices) and a set of permutations that avoid a certain pattern.&lt;/p&gt;&#xA;" OwnerUserId="2941" LastEditorUserId="2941" LastEditorDisplayName="" LastEditDate="2011-05-20T17:11:57.147" LastActivityDate="2011-05-20T17:11:57.147" CommentCount="12" />
  <row Id="8449" PostTypeId="1" AcceptedAnswerId="8451" CreationDate="2010-10-31T15:41:43.990" Score="3" ViewCount="178" Body="&lt;p&gt;Consider the set $F_n: = [n]^{[n]}$ of all functions $f:[n] \rightarrow [n]$, $[n] = \lbrace 1,2,...,n\rbrace$. It is well known that $|F_n| = n^n$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;: Let two functions $f, g$ in $F_n$ be of the same isomorphism type ($f\sim g$) iff there exists a permutation $\pi$ such that $f\pi = \pi g $&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What is the number of isomorphism types of functions $f:[n] \rightarrow [n]$, i.e. what is $|F_n/_{\sim}|$?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Examples:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$|F_2| = 4$, $|F_2/_{\sim}| = 3$ &lt;/li&gt;&#xA;&lt;li&gt;$|F_3| = 27$, $|F_3/_{\sim}| = 7$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1792" LastEditorUserId="1792" LastEditorDisplayName="" LastEditDate="2010-10-31T23:19:24.577" LastActivityDate="2010-11-01T03:52:00.047" Title="Number of isomorphism types of functions f:[n]-&gt;[n]" Tags="&lt;combinatorics&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="8450" PostTypeId="1" AcceptedAnswerId="8454" CreationDate="2010-10-31T16:00:15.657" Score="0" ViewCount="157" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Part1(Solved)&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it possible to create a function that maps x to y as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;x 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 ...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;y 0 1 2 3 4 5 4 3 2 1  0  1  2  3  4  5 ...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So far, this is what I have been able to come up with:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;|5 sin ((1/10)*pi*x)|&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which will give the following y's for x = 0 to 10:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; 0.00000&#xA; 1.5451&#xA; 2.9389&#xA; 4.0451&#xA; 4.7553&#xA; 5.0000&#xA; 4.7553&#xA; 4.0451&#xA; 2.9389&#xA; 1.5451&#xA; 0.00000&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This sequence will repeat every 10 steps, with the peak (5) at step 5. This is getting close to what I want so all that is left is to somehow adjust the function so that the required integers appear (1,2,3,4), but I don't know how.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the function is cycling through a series of numbers over and over, I figured that it might be possible to write the function in terms of a periodic function and not piece by piece.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please comment. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Part2&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How would I go about writing a function that produce the following plot:(x is an integer &gt;=0 , the pattern repeats after x = 8)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/7tXVu.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can this plot be expressed using a single formula or must it be defined piecewise?&#xA;  Thank you.&lt;/p&gt;&#xA;" OwnerUserId="2973" LastEditorUserId="2973" LastEditorDisplayName="" LastEditDate="2010-11-20T16:06:40.383" LastActivityDate="2010-11-20T16:06:40.383" Title="Creating a sine function with specific parameters" Tags="&lt;trigonometry&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="8451" PostTypeId="2" ParentId="8449" CreationDate="2010-10-31T16:08:00.060" Score="2" ViewCount="" Body="&lt;p&gt;If you want to count the number of functions up to renamings on the domain and codomain, then the number is $p_n$, the number of &lt;a href=&quot;http://en.wikipedia.org/wiki/Partition_%28number_theory%29&quot; rel=&quot;nofollow&quot;&gt;partitions&lt;/a&gt; of $n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Later:&lt;/strong&gt; now that you've made precise what you wanted... This is counted &lt;a href=&quot;http://www.research.att.com/~njas/sequences/A001372&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;, with references.&lt;/p&gt;&#xA;" OwnerUserId="274" LastEditorUserId="274" LastEditorDisplayName="" LastEditDate="2010-11-01T03:52:00.047" LastActivityDate="2010-11-01T03:52:00.047" CommentCount="2" />
  <row Id="8452" PostTypeId="1" AcceptedAnswerId="8460" CreationDate="2010-10-31T16:30:26.610" Score="2" ViewCount="280" Body="&lt;p&gt;How do I solve for $x$ from this equation?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$-1/x^2 + 9/(4-x-y)^2 = 0$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need to get this into $x=$&quot;blah&quot;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also. How do you guys get that funky formatting? &lt;/p&gt;&#xA;" OwnerUserId="2874" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2010-11-19T00:11:40.827" LastActivityDate="2010-11-19T00:11:40.827" Title="Make x subject of $-1/x^2 + 9/(4-x-y)^2 = 0$." Tags="&lt;algebra&gt;" AnswerCount="4" CommentCount="6" />
  <row Id="8453" PostTypeId="2" ParentId="8450" CreationDate="2010-10-31T16:40:54.990" Score="3" ViewCount="" Body="&lt;p&gt;You ask for a function. A function is a mapping from one set to&#xA;another set but you neglect to specify which sets. Your list of&#xA;$x$-values consists of small nonnegative integers together with&#xA;an ellipsis. From this, one might surmise that your intended&#xA;domain is $\mathbb{N}_0$, the set of nonnegative integers. Likewise&#xA;you do not specify a codomain, but let's say for argument's sake&#xA;that it is $\mathbb{N}_0$, the same as the domain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course there are uncountably many functions from $\mathbb{N}_0$&#xA;to $\mathbb{N}_0$ containing your specified values. But you mention&#xA;it has to be a &quot;sine&quot; function. The term sine refers to a particular&#xA;function on the real numbers, taking all velues between $-1$ and $1$&#xA;and having period $2\pi$. Your function cannot have all of these&#xA;properties. Perhaps by a &quot;sine&quot; function you mean a periodic function?&#xA;If so here is one defined on $\mathbb{N}_0$ which extends your table&#xA;of values:&#xA;$$f(n)=\left\{&#xA;\begin{array}{cl}&#xA;0&amp;amp;\textrm{if $n\equiv0$ (mod 10),}\\&#xA;1&amp;amp;\textrm{if $n\equiv\pm1$ (mod 10),}\\&#xA;2&amp;amp;\textrm{if $n\equiv\pm2$ (mod 10),}\\&#xA;3&amp;amp;\textrm{if $n\equiv\pm3$ (mod 10),}\\&#xA;4&amp;amp;\textrm{if $n\equiv\pm4$ (mod 10),}\\&#xA;5&amp;amp;\textrm{if $n\equiv5$ (mod 10).}&#xA;\end{array}&#xA;\right.$$&#xA;Of course, unlike the acutal sine function, this never takes negative&#xA;values.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps as in Hans's comment, you want a function on the reals&#xA;restricting to your table of values. Here is a piecewise linear&#xA;example as Hans suggests:&#xA;$$g(x)=\left\{&#xA;\begin{array}{cl}&#xA;10\{x/10\}&amp;amp;\textrm{if $\{x/10\}&amp;lt;1/2$,}\\&#xA;10-10\{x/10\}&amp;amp;\textrm{if $\{x/10\}\ge 1/2$}\\&#xA;\end{array}&#xA;\right.$$&#xA;where $\{t\}$ denotes the fractional part of a real number $t$.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-10-31T16:40:54.990" />
  <row Id="8454" PostTypeId="2" ParentId="8450" CreationDate="2010-10-31T16:41:34.857" Score="9" ViewCount="" Body="&lt;p&gt;$\frac{5}{\pi}\arccos\left(\cos\left(\frac{\pi x}{5}\right)\right)$ seems to do what you want, unless you need something smoother.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-10-31T16:41:34.857" CommentCount="6" />
  <row Id="8455" PostTypeId="2" ParentId="8452" CreationDate="2010-10-31T17:17:42.423" Score="1" ViewCount="" Body="&lt;p&gt;When you have fractions involved, the usual trick is to put everything on the same side, so you have &quot;= 0&quot; on the other side (This step is already ok in your case). Then, you put all the fractions on the same denominator and you multiply by this denominator (which vanishes since you have 0 on the other side). Usually it's then easier, since you don't have fractions anymore (well, it depends on the situation, but this is a good idea in general).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, in your case:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{-1}{x^2} + \frac{9}{(4-x-y)^2} = 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;becomes&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{-(4-x-y)^2}{x^2(4-x-y)^2} + \frac{9x^2}{x^2(4-x-y)^2} = 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$-(4-x-y)^2 + 9x^2 = 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now you can extand $(4-x-y)^2$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(4-x-y)^2 = ((4-y)-x)^2 = (4-y)^2 - 2(4-y)x + x^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And thus your equation becomes:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$-(4-y)^2 + 2(4-y)x + 8x^2 = 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which is a standard equation, the same as $ax^2 + bx + c = 0$, that can be solved for $x$ (consider $y$ as if it was any number).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the end, don't forget to check if the solution you obtain is ok with your beginning equation (no division by 0 for example).&lt;/p&gt;&#xA;" OwnerUserId="2778" LastActivityDate="2010-10-31T17:17:42.423" />
  <row Id="8456" PostTypeId="2" ParentId="3087" CreationDate="2010-10-31T17:30:22.860" Score="2" ViewCount="" Body="&lt;p&gt;Students should learn &lt;em&gt;math&lt;/em&gt;. You know, the stuff that most of us &lt;em&gt;do&lt;/em&gt;. When we use the word &quot;math&quot; we certainly don't have in our minds adding, multiplying, factoring polynomials, or even calculus. So why do we teach children that this is what math is?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This view has been in several published places like Devlin's &lt;em&gt;The Math Gene&lt;/em&gt;. He makes a good argument that the concept of a group isn't any harder than other stuff done in grade school. The problem is that it is introduced later.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The best thing I've read about math education is &lt;em&gt;A Mathematician's Lament&lt;/em&gt; by Paul Lockhart.&lt;/p&gt;&#xA;" OwnerUserId="2000" LastActivityDate="2010-10-31T17:30:22.860" CommunityOwnedDate="2010-10-31T17:30:22.860" />
  <row Id="8457" PostTypeId="2" ParentId="8452" CreationDate="2010-10-31T17:31:46.183" Score="1" ViewCount="" Body="&lt;p&gt;If you're working on the same problem as asked about in &lt;a href=&quot;http://math.stackexchange.com/q/8347/1242&quot;&gt;this question&lt;/a&gt;, the answer is that you don't need to right away, since you can first use the other equation to get a relation between $x$ and $y$ that will simplify things a little. See my comment there.&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2010-10-31T17:31:46.183" CommentCount="1" />
  <row Id="8458" PostTypeId="2" ParentId="8452" CreationDate="2010-10-31T17:45:24.073" Score="3" ViewCount="" Body="&lt;p&gt;Put $z = 4-x-y\:.\ $ Multiplying by $(xz)^2$ yields $$0 = 9x^2-z^2 = (3x-z)\:(3x+z) = (4x+y-4)(2x-y+4)$$&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-10-31T17:45:24.073" />
  <row Id="8460" PostTypeId="2" ParentId="8452" CreationDate="2010-10-31T17:56:40.003" Score="5" ViewCount="" Body="&lt;p&gt;If you add $\frac{1}{x^{2}}$ to both sides of your equation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$-\frac{1}{x^{2}}+\frac{9}{\left( 4-x-y\right) ^{2}}=0\qquad (1)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;you get this equivalent one (provided that $\frac{1}{x^{2}}$ is finite,&#xA;i.e $x\neq 0$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{1}{x^{2}}=\frac{9}{\left( 4-x-y\right) ^{2}}.\qquad (2)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is satisfied if the square root of one side is equal or&#xA;symmetric to the square root of the other side: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{1}{x}=\pm \frac{3}{4-x-y}.\qquad (3)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Equation $(3)$ is equivalent to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$3x=\pm \left( 4-x-y\right) \qquad (4)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;provided that $x\neq 0$ and $4-x-y\neq 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The equation $(4)$ represents two equations. One is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$3x=4-x-y,\qquad (5)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is equivalent to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$4x=4-y\Leftrightarrow x=1-\frac{1}{4}y\qquad (6)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and the other&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$3x=-\left( 4-x-y\right), \qquad (7)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is equivalent to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$3x=-4+x+y\Leftrightarrow x=-2+\frac{1}{2}y.\qquad (9)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus $(1)$ is equivalent to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x=1-\frac{1}{4}y\qquad (10)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x=-2+\frac{1}{2}y,\qquad (11)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;provided that $y\neq 4$ because the conditions $x\neq 0$ and $4-x-y\neq 0$&#xD;&#xA;correspond to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x\neq 0\Leftrightarrow 1-\frac{1}{4}y\neq 0\Leftrightarrow y\neq 4$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x\neq 0\Leftrightarrow -2+\frac{1}{2}y\neq 0\Leftrightarrow y\neq 4$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$4-x-y\neq 0\Leftrightarrow 4-\left( 1-\frac{1}{4}y\right) -y\neq&#xD;&#xA;0\Leftrightarrow y\neq 4$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$4-x-y\neq 0\Leftrightarrow 4-\left( -2+\frac{1}{2}y\right) -y\neq&#xD;&#xA;0\Leftrightarrow y\neq 4.$$&lt;/p&gt;&#xA;" OwnerUserId="752" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2010-11-18T21:41:35.687" LastActivityDate="2010-11-18T21:41:35.687" />
  <row Id="8461" PostTypeId="1" AcceptedAnswerId="8538" CreationDate="2010-10-31T18:25:54.540" Score="1" ViewCount="112" Body="&lt;p&gt;claims arrive in an insurance portfolio, according to a homogenous Poission process X(t), t&gt;=0. Assume that each of the 12 months in a year has exactly 30 days. Calculate the value of the following conditional probability:&#xA;P{X(2) = 3, X(3) = 5 | X(4) = 5}&#xA;where X(k) is the number of claims during the first k months of the year&#xA;can anyone help?&lt;/p&gt;&#xA;" OwnerUserId="2976" LastActivityDate="2010-11-01T17:56:54.300" Title="claims in an insurance portfolio" Tags="&lt;probability&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="8462" PostTypeId="1" CreationDate="2010-10-31T19:08:48.297" Score="1" ViewCount="112" Body="&lt;p&gt;I have a three pairs of points namely : p{1,1} p{1,2} p{2,1} p{2,2} p{3,1} p{3,2}, I want to find p such that : |p-p{i,1}| = |p-p{i,2}| for all i. I know that these represent three planes, but I am not sure whether these planes do meet in a single point. My simple question is even if the three planes are not parallel do they always meet in a single point. Intuitively I can visualize three planes which are not parallel but don't meet at a single point but how do I know that mathematically ?&lt;/p&gt;&#xA;" OwnerUserId="2978" LastActivityDate="2010-10-31T20:23:24.737" Title="How to determine a point in R3 which is which is equidistant to three pairs of point ?" Tags="&lt;linear-algebra&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="8463" PostTypeId="2" ParentId="8462" CreationDate="2010-10-31T19:26:29.970" Score="5" ViewCount="" Body="&lt;p&gt;The planes described by the equations $z=0$, $z=y$, and $z=2y$ have the $x$-axis in common, so you may have infinitely many points.  The planes described by the equations $z=0$, $z=y$, and $z=1-y$ have no points in common, but no two are parallel.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One way to determine whether the given planes will have a unique point of intersection is using linear algebra.  Given the equations $a_1x+b_1y+c_1z=d_1$, $a_2x+b_2y+c_2z=d_2$, and $a_3x+b_3y+c_3z=d_3$ describing the planes, there is a unique solution to the system of equations, corresponding to the point where the planes meet, if and only if the coefficient matrix&#xA;$$A= \begin{bmatrix}&#xA;  a_1 &amp;amp; b_1 &amp;amp; c_1 \\&#xA;  a_2 &amp;amp; b_2 &amp;amp; c_2 \\&#xA;  a_3 &amp;amp; b_3 &amp;amp; c_3&#xA;\end{bmatrix}$$&#xA;has an inverse, in which case the solution is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{bmatrix}&#xA;  x\\&#xA;  y \\&#xA;  z&#xA;\end{bmatrix}&#xA;=A^{-1}&#xA;\begin{bmatrix}&#xA;  d_1\\&#xA;  d_2 \\&#xA;  d_3&#xA;\end{bmatrix}.&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2010-10-31T20:23:24.737" LastActivityDate="2010-10-31T20:23:24.737" CommentCount="1" />
  <row Id="8464" PostTypeId="2" ParentId="8443" CreationDate="2010-10-31T19:46:34.110" Score="2" ViewCount="" Body="&lt;p&gt;&quot;What else can you count with these?&quot;: Solutions to the Fermat equation $x^n + y^n = z^n$.  (If this doesn't make immediate sense, I will add more material later, when I have more time.)&lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2010-10-31T19:46:34.110" />
  <row Id="8465" PostTypeId="1" AcceptedAnswerId="8474" CreationDate="2010-10-31T20:07:31.527" Score="1" ViewCount="213" Body="&lt;p&gt;Greetings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{i=x}^{2x} 5 - \sum_{k=2x+1}^{5x+2} 5 = -25$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I discover the value of X?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Turns out it was 5x+2, not 5x+1. My bad, apologies for the trouble.&lt;/p&gt;&#xA;" OwnerUserId="2950" LastEditorUserId="2950" LastEditorDisplayName="" LastEditDate="2010-10-31T20:40:09.020" LastActivityDate="2010-11-01T00:23:09.960" Title="Solving summation problem" Tags="&lt;calculus&gt;&lt;sequences-and-series&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="8466" PostTypeId="2" ParentId="8465" CreationDate="2010-10-31T20:13:52.160" Score="0" ViewCount="" Body="&lt;p&gt;Use the fact that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sum_{i=x}^{2x} 5 = 5 \cdot (2x - x) = 5x$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sum_{k=2x+1}^{5x+1} 5 = 5 \cdot (5x+1-2x-1) = 15x $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, the equation becomes&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$5x - 15x = -25$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is easy to solve, you just get $x = 25/10$ or in simpler terms $x = 5/2$&lt;/p&gt;&#xA;" OwnerUserId="1089" LastActivityDate="2010-10-31T20:13:52.160" CommentCount="6" />
  <row Id="8467" PostTypeId="2" ParentId="8435" CreationDate="2010-10-31T20:23:51.460" Score="2" ViewCount="" Body="&lt;p&gt;There are many things that are known about this. Most of which I believe come from something called the Prime Number Theorem, which you can read about on wikipedia.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Something that might be helpful with regards to your question are the formulas in this section:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Prime_number_theorem#Approximations_for_the_nth_prime_number&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Prime_number_theorem#Approximations_for_the_nth_prime_number&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The proofs for most of these theorems are really tricky and require a lot of knowledge of analysis though, so be warned!&lt;/p&gt;&#xA;" OwnerUserId="1089" LastActivityDate="2010-10-31T20:23:51.460" />
  <row Id="8468" PostTypeId="1" AcceptedAnswerId="8469" CreationDate="2010-10-31T21:34:46.363" Score="0" ViewCount="78" Body="&lt;p&gt;Hello.&#xA;$$\sum_{i=1}^{x} x + \sum_{i=1}^{x} 2 = 8$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Trying to discover $x$ here, so I made:&#xA;$$x(x-1) + 2(x-1) = 8$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which should give me the $x$, but instead of an integer I get:&#xA;$$x = \frac{-1\pm\sqrt{41}}{2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the answer should be $x=2$, according to the solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What am I doing wrong?&lt;/p&gt;&#xA;" OwnerUserId="2950" LastActivityDate="2010-10-31T21:49:10.947" Title="Summation problem not giving expected answer" Tags="&lt;calculus&gt;&lt;sequences-and-series&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="8469" PostTypeId="2" ParentId="8468" CreationDate="2010-10-31T21:43:10.330" Score="2" ViewCount="" Body="&lt;p&gt;You have $x$ integers between $1$ and $x$. Look: $1, 2, 3, \ldots, x-1, x$. So you are just not counting properly.&lt;/p&gt;&#xA;" OwnerUserId="329" LastActivityDate="2010-10-31T21:43:10.330" />
  <row Id="8471" PostTypeId="2" ParentId="8435" CreationDate="2010-10-31T21:54:34.807" Score="6" ViewCount="" Body="&lt;p&gt;The keyword you want is &lt;a href=&quot;http://en.wikipedia.org/wiki/Prime_gap&quot; rel=&quot;nofollow&quot;&gt;prime gap&lt;/a&gt;.  The best unconditional result I can see on that page appears to be $O(n^{3/4 + \epsilon})$ for any $\epsilon &gt; 0$.  With the Riemann hypothesis you get $O(\sqrt{n} \log n)$, but a famous conjecture asserts that you should be able to do considerably better, something like $O((\log n)^2)$.  &lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-10-31T22:07:13.360" LastActivityDate="2010-10-31T22:07:13.360" CommentCount="2" />
  <row Id="8472" PostTypeId="2" ParentId="392" CreationDate="2010-10-31T21:55:42.093" Score="1" ViewCount="" Body="&lt;p&gt;One important fact that ought to be mentioned explicitly is that &#xA;$$\frac{d \sin x}{dx} = \cos x$$&#xA;&lt;em&gt;only when $x$ is measured in radians.&lt;/em&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a general angle measure, the derivative of $\sin x$ is some scalar multiple of $\cos x$.  In fact, it could be argued that this is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Radian#Advantages_of_measuring_in_radians&quot; rel=&quot;nofollow&quot;&gt;major reason for the usefulness of radians&lt;/a&gt;: The radian is the angle measure that makes that scalar multiple equal to 1.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is something that not a lot of people realize.  For example, here's a quote from a Math Overflow answer to the question &lt;a href=&quot;http://mathoverflow.net/questions/40082/&quot; rel=&quot;nofollow&quot;&gt;&quot;Why do we teach calculus students the derivative as a limit?&quot;&lt;/a&gt;  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I would like to point out a simple question that very few calculus students and even teachers can answer correctly: Is the derivative of the sine function, where the angle is measured in degrees, the same as the derivative of the sine function, where the angle is measured in radians. In my department we audition all candidates for teaching calculus and often ask this question. So many people, including some with Ph.D.'s from good schools, couldn't answer this properly that I even tried it on a few really famous mathematicians. Again, the difficulty we all have with this question is for me a sign of how badly we ourselves learn calculus. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To see why radians are crucial, look at the slopes of the graphs of $\sin x$ at $x = 0$ when $x$ is measured in radians and when $x$ is measured in degrees.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, when $x$ is in radians:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/4AsQJ.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The slope appears to be close to 1.  (And, of course, we know that it is 1.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second, when $x$ is in degrees:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/1I74Q.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The slope is much, much smaller than 1.  So the derivative of $\sin x$ at $x = 0$ when $x$ is in degrees cannot be $\cos(0) = 1$.    The correct answer, if $x$ is in degrees, is that the derivative of $\sin x$ is $\frac{\pi}{180}\cos x$ (via the chain rule).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, all of the answers to the OP's question given here implicitly assume that $x$ is being measured in radians.  (It might be an interesting exercise for students reading this to go through each of the other arguments to see exactly where that assumption is being made.)  However, as the Math Overflow quote points out, this is something that a lot of people don't realize.  &lt;/p&gt;&#xA;" OwnerUserId="2370" LastActivityDate="2010-10-31T21:55:42.093" />
  <row Id="8473" PostTypeId="2" ParentId="8337" CreationDate="2010-10-31T22:05:39.633" Score="3" ViewCount="" Body="&lt;p&gt;Here is an other one which is more or less what Euler did in one of his proofs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The function $\sin(x)$ where $x\in\mathbb{R}$ is zero exactly at $x=n\pi$ for each integer $n$. If we factorized it as an infinite product we get&#xA;$$\sin(x)= \cdots(1+\frac{x}{3\pi})(1+\frac{x}{2\pi})(1+\frac{x}{\pi})x(1-\frac{x}{\pi})(1-\frac{x}{2\pi})(1-\frac{x}{3\pi})\cdots =$$&#xA;$$\qquad x(1-\frac{x^2}{\pi^2})(1-\frac{x^2}{2^2\pi^2})(1-\frac{x^2}{3^2\pi^2})\cdots $$&#xA;We can also represent $\sin(x)$ as a Taylor series at $x=0$&#xA;$$\sin(x) = x - \frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\cdots$$&#xA;Multiplying the product and identifying the coefficient of $x^3$ we see that&#xA;$$\frac{x^3}{3!}=x(\frac{x^2}{\pi^2} + \frac{x^2}{2^2\pi^2}+ \frac{x^2}{3^2\pi^2}+\cdots)$$&#xA;or &#xA;$$\frac{\pi^2}{3!}=\sum_{n=1}^\infty\frac{1}{n^2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are two interesting links: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;http://www.math.dartmouth.edu/~euler/date-publish.html&quot; rel=&quot;nofollow&quot;&gt;Euler's papers&lt;/a&gt; &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;&lt;a href=&quot;http://www.southernct.edu/~sandifer/Ed/History/Preprints/Talks/NYU%20Basel%20Problem%20Paper.PDF&quot; rel=&quot;nofollow&quot;&gt;Euler’s Solution of the Basel Problem – The Longer Story&lt;/a&gt;&lt;/em&gt; an essay on the subject written by Ed Sandifer. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1154" LastActivityDate="2010-10-31T22:05:39.633" />
  <row Id="8474" PostTypeId="2" ParentId="8465" CreationDate="2010-10-31T22:33:14.637" Score="1" ViewCount="" Body="&lt;p&gt;You asked a similar question, so I think you didn't understand Derek's comment. Here we need to count how many integers there are between $x$ and $2x$ and between $2x+1$ and $5x+2$. Since you are not familiar with these things, try to write all possible numbers. You'll get something like $$x,\ x+1, \ldots, \ x+ (x-1), \ 2x$$ and $$2x+1, \ldots, \ 2x + (x-1), \ 3x, \ 3x+1, \ldots, \ 3x+ (x-1), \ 4x, $$ $$4x+1, \ldots, \ 4x +(x-1), \ 5x, \ 5x+1, \ 5x+2.$$ Now it's easy to see that you have $x+1$ numbers between $x$ and $2x$ and $3x+2$ between $2x+1$ and $5x+2$.&lt;/p&gt;&#xA;" OwnerUserId="329" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2010-11-01T00:23:09.960" LastActivityDate="2010-11-01T00:23:09.960" />
  <row Id="8475" PostTypeId="1" CreationDate="2010-10-31T22:37:30.213" Score="1" ViewCount="69" Body="&lt;p&gt;I need a proof of this well known fact:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If a group  $G$  is given as a quotient of a free group $F/R$ then its pro-finite&#xA;completion is given by the quotient $\hat{F}/ \bar{R}$ where $\bar{R}$ is &#xA;the closure of $R$ in $\hat{F}$. (since $F$ embeds into $\hat{F}$ densely &#xA;we can think of $R$ inside $\hat{F}$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance.&lt;/p&gt;&#xA;" OwnerUserId="2980" LastActivityDate="2010-11-01T00:32:27.777" Title="Pro-finite presentations" Tags="&lt;group-theory&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="8476" PostTypeId="1" CreationDate="2010-10-31T22:59:18.900" Score="1" ViewCount="206" Body="&lt;p&gt;What I'm attempting to do is rearrange the formula for the sum of a geometric series so as to find the value of $r$. I've tried several different methods, all of which have failed; though I can't understand why.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This was my last attempt using logs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$S_n=\frac{a(1-r^n)}{1-r}$$&#xA;$$S_n(1-r) = a(1-r^n)$$&#xA;$$S_n-S_nr = a - ar^n$$&#xA;$$\frac{S_n}{a}-1 = \frac{S_n}{a}r-r^n$$&#xA;$$\log{\frac{S_n}{a}} = \log{\frac{S_n}{a}}+\log{r}-n\log{r}$$&#xA;$$0 = (\log{r})(n-1)$$&#xA;$$r = 10^{\frac{0}{n-1}}$$&#xA;$$r=1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can't understand where I'm going wrong. Any advice would be great. Also, if you know a formula to find $r$ using $S_n$ &amp;amp; $a$ then that would be equally fantastic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ben.&lt;/p&gt;&#xA;" LastActivityDate="2010-11-01T16:01:15.017" Title="Rearrange the formula for the sum of a geometric series to find &quot;r&quot;?" Tags="&lt;sequences-and-series&gt;&lt;algebra&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="8477" PostTypeId="2" ParentId="8476" CreationDate="2010-10-31T23:16:42.447" Score="1" ViewCount="" Body="&lt;p&gt;No need to solve the full polynomial as &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$S_{n+1} - rS_n = a$$ (by def. of geometric series)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I would not leave this as an own entry if I had enough points to comment everywhere, sorry about that)&lt;/p&gt;&#xA;" OwnerUserId="2900" LastEditorUserId="2900" LastEditorDisplayName="" LastEditDate="2010-10-31T23:48:20.410" LastActivityDate="2010-10-31T23:48:20.410" />
  <row Id="8478" PostTypeId="2" ParentId="8476" CreationDate="2010-10-31T23:21:33.190" Score="1" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\ \ $ Consider $\ S_{n+1} - r\ S_n $&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-10-31T23:21:33.190" />
  <row Id="8479" PostTypeId="1" CreationDate="2010-10-31T23:30:53.227" Score="0" ViewCount="147" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Possible Duplicate:&lt;/strong&gt;&lt;br&gt;&#xA;  &lt;a href=&quot;http://math.stackexchange.com/questions/8323/does-stirlings-formula-give-the-correct-number-of-digits-of-n&quot;&gt;Does Stirling&amp;#39;s formula give the correct number of digits of n!?&lt;/a&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;p&gt;Dear friends,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let&lt;/p&gt;&#xA;&#xA;&lt;p&gt;f(n) = [(2.n.pi) ^ (1 / 2)]. [(n / e) ^ n]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(f(n) is the classical approach of Stirling's factorial natural number n&gt;1.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;QUESTION:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;floor[log(f(n)] = floor(log n!), whatever the natural number n&gt;1?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(log indicates logarithm to base 10.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I STILL CAN NOT GET NO COUNTEREXAMPLE.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Already, many thanks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Paulo Argolo&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rio de Janeiro, Brazil&lt;/p&gt;&#xA;" OwnerUserId="2938" LastEditorUserId="2938" LastEditorDisplayName="" LastEditDate="2010-11-01T09:53:47.973" LastActivityDate="2010-11-01T09:53:47.973" ClosedDate="2010-11-01T13:15:39.687" Title="About Stirling's formula" Tags="&lt;number-theory&gt;" CommentCount="5" />
  <row Id="8480" PostTypeId="2" ParentId="8476" CreationDate="2010-10-31T23:38:27.110" Score="3" ViewCount="" Body="&lt;p&gt;Added: I assume you don't know both $S_n$ and $S_{n+1}$. Also I thought you had a geometric progression, i.e. not an infinite series. If you mean a series then you have to apply limits to $S_n$. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;From &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{S_{n}}{a}-1=\frac{S_{n}}{a}r-r^{n}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;you cannot deduce&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\log \frac{S_{n}}{a}=\log \frac{S_{n}}{a}+\log r-n\log r$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;because&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\log (x+y)\neq \log x+\log y.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The correct property of the $\log $ function is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\log (xy)=\log x+\log y.$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The algebraic identity&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x^{n}-c^{n}=(x-c)(x^{n-1}+cx^{n-2}+c^{2}x^{n-3}+\cdots +c^{n-2}x+c^{n-1})$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;gives for $c=1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{x^{n}-1}{x-1}=x^{n-1}+x^{n-2}+x^{n-3}+\cdots +x+1.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore you have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{S_{n}}{a}=\frac{1-r^{n}}{1-r}=r^{n-1}+r^{n-2}+r^{n-3}+\cdots +r+1.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You would have to find $r$ such that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$r^{n-1}+r^{n-2}+r^{n-3}+\cdots +r+1-\frac{S_{n}}{a}=0.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However this polynomial equation has no algebraic solution, in general, for $n\ge 6$. You need a different method (a numerical one).&lt;/p&gt;&#xA;" OwnerUserId="752" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2010-11-01T16:01:15.017" LastActivityDate="2010-11-01T16:01:15.017" CommentCount="2" />
  <row Id="8481" PostTypeId="2" ParentId="6234" CreationDate="2010-10-31T23:46:38.267" Score="1" ViewCount="" Body="&lt;p&gt;What you're basically looking for is a function $f(x)$ such that $f(2x)=f^2(x)-1$ and $f(0)=\phi$, from there:&#xA;\begin{align}&#xA;2f'(2x)&amp;amp;=2f(x)f'(x)\\&#xA;\frac{f'(2x)}{f'(x)}&amp;amp;=f(x)\\&#xA;\frac{f'(x)}{f'(x/2)}&amp;amp;=f(x/2)\\&#xA;\frac{f'(x)}{f'(x/2^n)}&amp;amp;=\prod_{k=1}^n f(x/2^k)&#xA;\end{align}&#xA;and, given a value $x_0$ such that $f(x_0)=1$,&#xA;\begin{align}&#xA;\Phi&amp;amp;=\prod_{k=1}^{\infty} \frac{f(x_0/2^k)}{\phi}\\&#xA;    &amp;amp;=\lim_{n\rightarrow\infty}\phi^{-n} \prod_{k=1}^n f(x_0/2^k)\\&#xA;    &amp;amp;=\lim_{n\rightarrow\infty}\phi^{-n} \frac{f'(x_0)}{f'(x_0/2^n)}\\&#xA;    &amp;amp;=\lim_{h\rightarrow0}h^\alpha \frac{f'(x_0)}{f'(hx_0)}&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\alpha=\frac{ln(\phi)}{ln(2)}$.  Unfortunately, I have no idea how to get $f(x)$, and the fact that $f(x)=1+O(x^{1+\alpha})$ does not make finding this function look easy.&lt;/p&gt;&#xA;" OwnerUserId="2957" LastActivityDate="2010-10-31T23:46:38.267" CommentCount="3" />
  <row Id="8482" PostTypeId="2" ParentId="8461" CreationDate="2010-11-01T00:27:06.320" Score="1" ViewCount="" Body="&lt;p&gt;(Major rewrite: The following is a much simpler and cleaner answer than my previous one involving binomial distributions.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Start by considering what $X(4) = 5$ actually means.  It means that through the first four months of the year there have been 5 claims.  Given that, you're looking for the probability that all three of the following events occurred:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Exactly 3 claims in the first two months.  &lt;/li&gt;&#xA;&lt;li&gt;Exactly 2 claims in the third month.  &lt;/li&gt;&#xA;&lt;li&gt;Exactly 0 claims in the fourth month.  &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Each claim is equally likely to fall in each month.  So, out of 5 total items, you're trying to fit 3 into a category with a probability of $\frac{1}{2}$, 2 into a category with a probability $\frac{1}{4}$, and 0 into a category with probability $\frac{1}{4}$.  This situation is exactly what a &lt;a href=&quot;http://en.wikipedia.org/wiki/Multinomial_distribution&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;multinomial&lt;/em&gt; distribution&lt;/a&gt; models.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the probability is &#xA;$$\frac{5!}{3! 2! 0!} \left(\frac{1}{2}\right)^3 \left(\frac{1}{4}\right)^2 \left(\frac{1}{2}\right)^0 = \frac{5}{64}.$$&lt;/p&gt;&#xA;" OwnerUserId="2370" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-11-01T04:11:09.677" LastActivityDate="2010-11-01T04:11:09.677" CommentCount="2" />
  <row Id="8483" PostTypeId="2" ParentId="8475" CreationDate="2010-11-01T00:32:27.777" Score="3" ViewCount="" Body="&lt;p&gt;The profinite completion is the inverse limit of the $G/H$ for all normal subgroups $H$ of $G$ of finite index, that is the inverse limit of the $F/I$ for all normal subgroups $I$ of $F$ which are of finite index and contain $R$.&#xA;So you have a continuous morphism $\hat{F} \rightarrow \hat{G}$, it is surjective and its kernel is the intersection of the above $I$, and it is easily seen to be $\bar{R}$.&#xA;Thus you have a continuous bijective morphism $\hat{F}/\bar{R} \rightarrow \hat{G}$, and since $\hat{F}$ is compact (it is profinite) it is a homeomorphism, so it's an isomorphism of topological groups.&lt;/p&gt;&#xA;" OwnerUserId="2660" LastActivityDate="2010-11-01T00:32:27.777" />
  <row Id="8484" PostTypeId="1" AcceptedAnswerId="8486" CreationDate="2010-11-01T00:34:32.947" Score="4" ViewCount="182" Body="&lt;p&gt;Find all positive integers n, such that $$\sum\limits_{k=1}^{n}(k+2)^n=(n+3)^n $$&lt;/p&gt;&#xA;" OwnerUserId="2779" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-01T02:26:46.420" LastActivityDate="2010-11-01T02:26:46.420" Title="Find all positive integers n, such that..." Tags="&lt;number-theory&gt;&lt;diophantine-equations&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="8485" PostTypeId="1" AcceptedAnswerId="8534" CreationDate="2010-11-01T01:32:41.677" Score="2" ViewCount="234" Body="&lt;p&gt;The Hoeffding's inequality is $P(S_n - E[S_n] \geq \epsilon) \leq e^{-2\epsilon^2/k'}$, where $S_n = \sum_{i=1}^{n} X_i$, $X_i$'s are independent bounded random variables, and $k'$ depends on the random variables.&#xA;In the proof of Hoeffding's inequality, an optimization problem of the form is solved:&#xA;$$\min_{s} \ \ e^{-s\epsilon}e^{ks^2}$$ subject to $s &gt; 0$, to obtain a tight upper bound (which in turn yields the Hoeffding's inequality). &#xA;It turns out that $s = \epsilon/2k$ is the value that obtains the Hoeffding's inequality. I don't see how.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Note that $k &gt; 0, \epsilon &gt; 0$.&lt;/p&gt;&#xA;" OwnerUserId="1527" LastEditorUserId="1527" LastEditorDisplayName="" LastEditDate="2010-11-01T03:48:24.260" LastActivityDate="2010-11-01T16:18:05.860" Title="Proof of Hoeffding's inequality" Tags="&lt;probability-theory&gt;&lt;convex-optimization&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="1" />
  <row Id="8486" PostTypeId="2" ParentId="8484" CreationDate="2010-11-01T02:13:15.420" Score="5" ViewCount="" Body="&lt;p&gt;$k^n/(n+3)^n = \left(1-\frac{n+3-k}{n+3}\right)^n$, so using that $\log (1+x) \leq x$ for all $x$,&#xA;\[ \frac{\sum_{k=3}^{n+2} k^n}{(n+3)^n} \leq \sum_{j=1}^n e^{-jn/(n+3)} \leq \frac{1}{e^{n/(n+3)}-1} \]&#xA;This last term is decreasing, and for $n=7$ it is strictly less than $1$, so you are left with only $n \leq 6$ to check.&#xA;It is probably possible to sharpen the inequalities so that there would be less to check.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the end, you get $n=2$ or $3$.&lt;/p&gt;&#xA;" OwnerUserId="2660" LastEditorUserId="2660" LastEditorDisplayName="" LastEditDate="2010-11-01T02:19:10.877" LastActivityDate="2010-11-01T02:19:10.877" />
  <row Id="8487" PostTypeId="1" AcceptedAnswerId="8488" CreationDate="2010-11-01T02:29:09.993" Score="2" ViewCount="214" Body="&lt;p&gt;Please imagine that we have a circular segment with some arc length 's' and chord length 'a' (using notation from &lt;a href=&quot;http://mathworld.wolfram.com/CircularSegment.html&quot; rel=&quot;nofollow&quot;&gt;http://mathworld.wolfram.com/CircularSegment.html&lt;/a&gt;).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Provided only 'a' and 's', and placing the left-hand-side point of the chord at the origin of the Euclidean plane (or a more convenient point), is there sufficient information to write an expression for the height of the circular segment (i.e. the y-axis/&quot;vertical&quot; distance between the chord on the x-axis and the circular arc) as a function of a position on the chord?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's a simple matter to express the chord length in terms of the arc length and 'theta': a = (s) * 2(Sin[theta/2])/theta, or an expression for the arc length in terms of the chord length and theta: s = a/(2(Sin[theta/2])/theta).  And one can write an express for the maximum height as: h = (R - (1/2)*((-a)^2+4*R^2)^(1/2)), where the radius of the circle, 'R' is related to theta as: R = 1/2 Sqrt[a^2/(-1 + Cos[theta/2]^2)]. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there is insufficient information to accomplish the above, I would love to have an intuitive explanation for why this is so. &lt;/p&gt;&#xA;" OwnerUserId="5009" LastEditorUserId="5009" LastEditorDisplayName="" LastEditDate="2010-11-01T02:54:52.730" LastActivityDate="2010-11-01T02:54:52.730" Title="Calculating the height of a circular segment at all points provided only chord and arc lengths" Tags="&lt;geometry&gt;" AnswerCount="1" CommentCount="6" />
  <row Id="8488" PostTypeId="2" ParentId="8487" CreationDate="2010-11-01T02:52:44.400" Score="4" ViewCount="" Body="&lt;p&gt;Using the notation of the figure you have linked to, we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;R \sin \frac{\theta}{2} = \frac{a}{2}&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;we can also write&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;\theta = \frac{s}{R} = \frac{2 s \sin \theta/2}{a}&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From this equation, you can solve for $\theta$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you have solved for $\theta$, you have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;h = R - R \cos(\theta/2)&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $R = a/(2 \sin \theta/2)$, we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;h = \frac{a}{2 \sin \theta/2} \left( 1 - \cos\left(\frac{s \sin\theta/2}{a}\right)\right)&#xA;\end{equation}&lt;/p&gt;&#xA;" OwnerUserId="2641" LastActivityDate="2010-11-01T02:52:44.400" CommentCount="4" />
  <row Id="8489" PostTypeId="1" AcceptedAnswerId="8492" CreationDate="2010-11-01T02:53:14.920" Score="2" ViewCount="67" Body="&lt;p&gt;Let ${a_k}$, ${b_k}$ be two sequences in $\mathbb{R}$. Suppose ${a_k}$ converges to $a$ s.t. $\liminf_{k \to \infty} {a_k} = \lim_{k \to \infty} {a_k} = a$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it then true that $\liminf_{k \to \infty} (a_k - b_k) = \lim_{k \to \infty} a_k + \liminf_{k \to \infty} -b_k$?&lt;/p&gt;&#xA;" OwnerUserId="1736" LastActivityDate="2010-11-01T03:45:16.013" Title="$\liminf$ of difference of two sequences" Tags="&lt;analysis&gt;" AnswerCount="3" />
  <row Id="8490" PostTypeId="2" ParentId="392" CreationDate="2010-11-01T03:08:29.877" Score="1" ViewCount="" Body="&lt;p&gt;As an addendum, you can download the Mathematica notebook from &lt;a href=&quot;http://demonstrations.wolfram.com/GraphingDerivatives/&quot; rel=&quot;nofollow&quot;&gt;Graphing Derivatives&lt;/a&gt;, which allows you to play a little bit with $\text{sin}(x)$, $\text{cos}(x)$ and another function. I think it shows a very obvious but interesting construction of those trigonometric functions.&#xA;In case you don't want to download or install anything, I posted an amateurish &lt;a href=&quot;http://www.youtube.com/watch?v=hBJfoLLZJ8A&quot; rel=&quot;nofollow&quot;&gt;screencast&lt;/a&gt;, so you can see the demonstration. Basically, you draw the $\text{sin}(x)$ function, and in each point $(x,y)$ you calculate/draw the slope. The value of the slope corresponds to the value of the $y$ coordinate of the derivative of the function (in this example, $\text{cos}(x)$), keeping the same $x$ coordinate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is a wonderful exercise to plot some random function, and drawing the derivative of that function based on this procedure, then take a look at the 'true' derivative and see how much your drawing resembles the derivative.&lt;/p&gt;&#xA;" OwnerUserId="2569" LastEditorUserId="2569" LastEditorDisplayName="" LastEditDate="2010-11-01T03:49:15.053" LastActivityDate="2010-11-01T03:49:15.053" />
  <row Id="8491" PostTypeId="2" ParentId="8489" CreationDate="2010-11-01T03:11:51.170" Score="1" ViewCount="" Body="&lt;p&gt;Yes.  Here's an approach to proving this, without full details.  Given $\epsilon\gt0$, take $N$ sufficiently large so that $a_k-b_k$ is within $\epsilon$ of $a-b_k$ for all $k\geq N$.  Then the $\liminf$ of the sequence $a_N-b_N,a_{N+1}-b_{N+1},\ldots$ must lie within $\epsilon$ of that of the sequence $a-b_N,a-b_{N+1},\ldots$.  Removing a finite number of terms from the sequence doesn't change the $\liminf$.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-11-01T03:11:51.170" />
  <row Id="8492" PostTypeId="2" ParentId="8489" CreationDate="2010-11-01T03:19:58.080" Score="3" ViewCount="" Body="&lt;p&gt;Yes. In general you only have subadditivity of the limit inferior (that is, in general you only have $\liminf (a_n+b_n) \geq \liminf(a_n) + \liminf(b_n)$) but if at least one sequence converges, then you have equality.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To see why you have equality, think of $\liminf$ as the infimum of all limits of converging subsequences. If $a_{j_k}+b_{j_k}$ is a converging subsequence of $\{a_n+b_n\}$, then $b_{j_k}$ must also converge (since $a_{j_k}$ converges: every subsequence of a converging sequence converges); and likewise, if $b_{j_k}$ is a converging subsequence of $b_n$, then $a_{j_k}+b_{j_k}$ is a converging subsequence of $a_n+b_n$. Moreover, if $b_{j_k}$ converges to $s$, then $a_{j_k}+b_{j_k}$ converges to $a+s$. Conversely, if $a_{j_k} + b_{j_k}$ converges to $a+t$, then $b_{j_k}$ must converge to $t$. So if $L$ is the set of all points that are limits of subsequences of $b_n$, then the set of all points that are limits of subsequences of $a_n+b_n$ is $a+L = \{a+t \mid t\in L\}$. Therefore&#xA;\[ \liminf (a_n+b_n) = \inf (a+L) = a + \inf L = a + \liminf b_n = \lim_{n\to\infty}a_n + \liminf b_n.\]&#xA;To get the difference, just apply this to $a_n$ and $(-b_n)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Or you can try the same argument with $\limsup$, and use the fact that $\liminf -b_n = -\limsup b_n$; you would think of $\limsup$ as the supremum of all limits of converging subsequences of $b_n$). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now see if you can figure out why you only have an inequality in the general case...&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-01T03:45:16.013" LastActivityDate="2010-11-01T03:45:16.013" CommentCount="3" />
  <row Id="8493" PostTypeId="2" ParentId="8356" CreationDate="2010-11-01T03:31:54.993" Score="0" ViewCount="" Body="&lt;p&gt;Since I can't make comments (haven't got enough points yet), I writing them in the form of answer. The word &quot;pdf&quot; stands for &quot;probability density function&quot;, not for &quot;probability distribution function&quot;. In your particular case, of course, you mean some discrete distributions, which are referred to as &quot;pmf&quot; (probability mass function). So, please &lt;em&gt;do&lt;/em&gt; use the correct terminology. &lt;/p&gt;&#xA;" OwnerUserId="2450" LastActivityDate="2010-11-01T03:31:54.993" />
  <row Id="8494" PostTypeId="2" ParentId="8489" CreationDate="2010-11-01T03:32:35.180" Score="0" ViewCount="" Body="&lt;p&gt;Yes. One possible proof outline:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;If there is a $N$ such that $c_k \geq d_k$ for all $k &gt; N$ then $\liminf c_k \geq \liminf d_k$&lt;/li&gt;&#xA;&lt;li&gt;If $C$ is a constant then $\liminf (C+a_k)=C+\liminf a_k$&lt;/li&gt;&#xA;&lt;li&gt;Let $\lim a_k=a$. For any $\epsilon &gt; 0$ we can find a $N$ such that for all $k &gt; N$&#xA;$$a-\epsilon - b_k \leq a_k - b_k \leq a+\epsilon -b_k$$&#xA;Apply (1) and (2), then make use of the fact that the result is true for all $\epsilon &gt; 0$.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="1195" LastActivityDate="2010-11-01T03:32:35.180" />
  <row Id="8495" PostTypeId="1" AcceptedAnswerId="8499" CreationDate="2010-11-01T03:59:34.213" Score="1" ViewCount="341" Body="&lt;p&gt;The series in question is $\sum\limits_{n=2}^{\infty}\frac{(-1)^n n^2}{( \ln(n))^n}$&#xA;I think this diverges.  So using the divergence test, I am trying to show that limit of the general term of the series is not 0.&#xA;By a few algebraic manipulations we get :&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\lim (-1)^n(n^2)( \ln(n))^{-n}$&lt;/li&gt;&#xA;&lt;li&gt;$\lim (-1)^n(n)/ \ln(n)$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I am not sure how to proceed here.  If you get rid of the alternating, the limit is $\infty$.&#xA;However, I am not sure how to manipulate this to show that (I suspect) the limit does not exist.&lt;/p&gt;&#xA;" OwnerUserId="2533" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-01T04:31:19.870" LastActivityDate="2010-11-04T02:15:55.493" Title="Is $\sum\limits_{n=2}^{\infty} \frac{(-1)^n n^2}{( \ln(n))^n}$ divergent?" Tags="&lt;calculus&gt;&lt;sequences-and-series&gt;" AnswerCount="4" CommentCount="3" FavoriteCount="1" />
  <row Id="8496" PostTypeId="2" ParentId="8495" CreationDate="2010-11-01T04:17:19.370" Score="3" ViewCount="" Body="&lt;p&gt;If $a_n$ is a sequence, then $\lim_{n\to\infty}a_n = 0$ if and only if $\lim_{n\to\infty} |a_n| = 0$. So if the absolute value of the terms does not go to zero, then the terms don't go to zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, if you know that $\lim \frac{n}{\ln (n)}\neq 0$, then it follows that $\lim\frac{(-1)^n n}{\ln n}\neq 0$. (because the former is the limit of the absolute values of the latter). So if your reduction is accurate, and your proof that the limit without the $(-1)^n$ factor is not zero is accurate, then you are done.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That said, I'm not sure how you manage to get to that limit. If you had $(\ln(n^n))^{-1}$, then I would agree; but how do you manage to get rid of the exponent and cancel an $n$ if you have&#xA;\[ \frac{(-1)^n n^2}{(\ln n)^n} \text{?}\]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added:&lt;/strong&gt; Well, looks like the beans have been spilt; your &quot;few algebraic manipulations&quot; were incorrect. Remember that while $\ln(a^k) = k\ln(a)$, it is &lt;strong&gt;not&lt;/strong&gt; true that $(\ln a)^k = k\ln a$. $\ln(a^k)$ is the logarithm of the $k$th power of $a$; $(\ln a)^k$ is the $k$th power of the logarithm of $a$; they are usually different.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-01T04:30:05.960" LastActivityDate="2010-11-01T04:30:05.960" />
  <row Id="8497" PostTypeId="2" ParentId="8495" CreationDate="2010-11-01T04:25:02.233" Score="2" ViewCount="" Body="&lt;p&gt;The series converges by the alternating series test.  Take logs of the absolute values of the terms to get $2\ln(n)-n\ln(\ln(n))$.  Consider the function $f(x)=2\ln(x)-x\ln(\ln(x))$, with derivative $f'(x)=\frac{2}{x}-\ln(\ln(x))-\frac{1}{\ln(x)}$.  Since $f$ goes to $-\infty$ and $f'$ is eventually always negative, continuity and monotonicity of the exponential function tell us that the terms of your sequence have absolute value eventually decreasing to 0, so that the alternating series test applies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second expression you give for the limit does not go to zero, and it is not clear where it came from.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-11-01T04:25:02.233" />
  <row Id="8498" PostTypeId="2" ParentId="8495" CreationDate="2010-11-01T04:26:02.290" Score="1" ViewCount="" Body="&lt;p&gt;You want to prove that $n^2 (\ln (n))^{-n}$ is monotonically decreasing and use Liebniz' test.  As long as $\ln (n)$ is greater than 1, the exponential will dominate the polynomial.  So pick your favorite starting place and fix ln(n) to ln(a) for some a and argue that the actual term is smaller in absolute value than this.  Now show that the version with ln(a) also goes to zero.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-11-01T04:26:02.290" />
  <row Id="8499" PostTypeId="2" ParentId="8495" CreationDate="2010-11-01T04:36:25.760" Score="8" ViewCount="" Body="&lt;p&gt;Applying the root test you can see that the series converges.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$L = \lim_{n\to\infty} |a_{n}|^{1/n}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $L&amp;lt;1$, then the series is absolutely convergent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $L&gt;1$ the series is divergent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $L=1$ then the test is inconclusive. However, if $|a_{n}|^{1/n}\geq 1$ for infinitely many distinct values of n, then the series $\sum_{n=1}^{\infty} a_{n}$ diverges.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your series, you have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$L = \lim_{n\to\infty}\bigg|\displaystyle \frac{(-1)^{n}n^{2}}{\text{ln}^{n}(n)}\bigg|^{\frac{1}{n}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then $L=\lim_{n\to\infty}\bigg|\displaystyle\frac{(-1)n^\frac{2}{n}}{\text{ln}(n)}\bigg|=0$ (because $\lim_{n\to\infty}\ln(n)=\infty$). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, $L&amp;lt;1$ and the series convergences. &lt;/p&gt;&#xA;" OwnerUserId="2569" LastEditorUserId="2569" LastEditorDisplayName="" LastEditDate="2010-11-04T02:15:55.493" LastActivityDate="2010-11-04T02:15:55.493" CommentCount="6" />
  <row Id="8500" PostTypeId="1" AcceptedAnswerId="8533" CreationDate="2010-11-01T04:40:03.430" Score="0" ViewCount="60" Body="&lt;p&gt;Let G be a graph and δ(G) the minimum degree of a peak. Describe an algorithm in pseudocode that, for a given tree T with k&amp;lt;= δ(G) edges, should be build (in polinomial time) a sub-graph H of G so that H is isomorphic with T.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do I even start ? &lt;/p&gt;&#xA;" OwnerUserId="2963" LastActivityDate="2010-11-01T16:12:31.527" Title="The isomorphic subgraph problem." Tags="&lt;homework&gt;&lt;graph-theory&gt;" AnswerCount="1" />
  <row Id="8501" PostTypeId="1" CreationDate="2010-11-01T04:52:13.290" Score="4" ViewCount="679" Body="&lt;p&gt;This may sound silly, but... Suppose an aspiring amateur mathematician wanted to plan to move to another city...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are some cities that are home to some of the largest number of the brightest mathematicians? I'm sure this may depend on university presence, or possibly industry presence, or possibly something surprising. Wondering where the best place to take a non-faculty job at a university and try to make friends with some sharp minds in the computer lab or at the nearby pub might be.&lt;/p&gt;&#xA;" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-01T04:56:54.690" LastActivityDate="2011-02-15T00:58:20.700" Title="Best Cities for Mathematical Study" Tags="&lt;soft-question&gt;&lt;education&gt;" AnswerCount="5" CommentCount="5" FavoriteCount="1" />
  <row Id="8502" PostTypeId="1" CreationDate="2010-11-01T05:47:08.130" Score="0" ViewCount="184" Body="&lt;p&gt;$$\sum_{n=1}^\infty\frac{e^\frac{1}{n}}{n}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I used the limit comparison test with $1/n$, which diverges. So I divide the $n^{th}$ term of my series by $1/n$, which is the same as multiplying by $n$, and all you're left with is $\lim_{n\to\infty}e^\frac{1}{n}$, which is $e^0$, which is $1$, and since $1$ is a constant greater than $0$, and $1/n$ diverges, then the original series diverges as well. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;These were my steps for this problem, but I got 0 points for this (not even partial credit!). What did I do wrong??&lt;/p&gt;&#xA;" OwnerUserId="2985" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2010-11-01T06:01:39.240" LastActivityDate="2010-11-01T06:01:39.240" ClosedDate="2010-11-01T06:32:22.010" Title="Convergence/Divergence using Limit Comparison Test" Tags="&lt;calculus&gt;&lt;sequences-and-series&gt;&lt;convergence&gt;" CommentCount="3" />
  <row Id="8503" PostTypeId="1" AcceptedAnswerId="8507" CreationDate="2010-11-01T06:14:12.360" Score="1" ViewCount="172" Body="&lt;p&gt;Determine all &#xA;    $ N, |N| \le 9 $ such that $ x^2-94y^2=N $ is solvable in positive integers.&lt;/p&gt;&#xA;" OwnerUserId="2505" LastEditorUserId="1778" LastEditorDisplayName="" LastEditDate="2010-11-04T03:54:03.220" LastActivityDate="2010-11-04T03:54:03.220" Title="Pell's equation" Tags="&lt;number-theory&gt;&lt;diophantine-equations&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="8504" PostTypeId="1" AcceptedAnswerId="8505" CreationDate="2010-11-01T06:58:06.847" Score="2" ViewCount="507" Body="&lt;p&gt;Is the subset consisting of all integrable (or square integrable) smooth functions of the set of all integrable (or square integrable) functions dense under the usual Euclidean or integral of absolute difference metric?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By smooth I mean derivatives of all orders exist.&lt;/p&gt;&#xA;" OwnerUserId="2987" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2010-11-02T14:13:12.030" LastActivityDate="2010-11-03T19:54:01.873" Title="Are the smooth functions dense in either L^2 or L^1?" Tags="&lt;real-analysis&gt;&lt;functions&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" />
  <row Id="8505" PostTypeId="2" ParentId="8504" CreationDate="2010-11-01T07:06:57.990" Score="6" ViewCount="" Body="&lt;p&gt;Yes.  In fact, by the Stone-Weierstrass theorem and the existence of smooth bump functions, smooth functions with compact support are uniformly dense in the space of continuous functions with compact support.  Uniform density implies $L^2$ and $L^1$ density for functions with compact (and therefore finite measure) support, and since continuous functions with compact support are dense in $L^2$ and $L^1$, the result follows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you wanted to see this more directly, you can go through the iterations of approximating an arbitrary ($L^1$ or $L^2$) function with a bounded function with bounded support, then with a simple function, then with a step function, and finally approximate the step function with a smooth function using &lt;a href=&quot;http://en.wikipedia.org/wiki/Bump_function&quot; rel=&quot;nofollow&quot;&gt;bump functions&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2010-11-01T15:07:18.127" LastActivityDate="2010-11-01T15:07:18.127" CommentCount="7" />
  <row Id="8506" PostTypeId="2" ParentId="8444" CreationDate="2010-11-01T09:07:10.250" Score="1" ViewCount="" Body="&lt;p&gt;Using the formula for the triangular numbers we note that if&#xA;$m \in I = [2n^2+n+1,2n^2+3n+1]$ for some $n=0,1,2,\ldots$ then $f(m)=m,$ otherwise&#xA;$f(m)=-m.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only possible choice of $n$ is $ \lfloor \sqrt{m/2} \rfloor,$ since if we write&#xA;$l(n) =  2n^2+n+1$ and $u(n) =  2n^2+3n+1$ by writing $\sqrt{m/2} = N + r,$ where $N$ is an integer and $0 \le r &amp;lt; 1$ we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$u \left( \lfloor \sqrt{m/2} \rfloor – 1 \right) = 2N^2 – N &amp;lt; 2N^2+4Nr+r^2 &amp;lt; m,$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and so $m \notin I.$ Similarly &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$l \left( \lfloor \sqrt{m/2} \rfloor + 1 \right) &gt; m,$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so $m \notin I.$ Hence we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f(m) = m \textrm{ when } m \in [2t^2+t+1,2t^2+3t+1] \textrm{ for }&#xA;t =  \lfloor \sqrt{m/2} \rfloor,$$&#xA;otherwise $f(m)=-m.$&lt;/p&gt;&#xA;" OwnerUserId="1301" LastActivityDate="2010-11-01T09:07:10.250" />
  <row Id="8507" PostTypeId="2" ParentId="8503" CreationDate="2010-11-01T09:20:33.417" Score="4" ViewCount="" Body="&lt;p&gt;You are asking for which $N$ there is an element of norm&#xA;$N$ in the quadratic ring $R=\mathbb{Z}[\sqrt{94}]$.&#xA;Now $R$ is the full ring of integers in the field&#xA;$\mathbb{Q}(\sqrt{94})$ which has class number one.&#xA;The problem is almost the same as determining for which&#xA;$N$ there is a norm $N$ ideal in $R$. There is a norm&#xA;$N$ ideal in $R$ iff either $N$ or $-N$ is a norm from $R$.&#xA;But we can't have both $N$ and $-$ being norms, since $-1$&#xA;is not a norm from $K$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the question reduces to&#xA;(i) for which $N$ is there an ideal in $R$ of norm $N$?&#xA;(ii) if there is such an ideal, is the corresponding solution to&#xA;Pell's equation with $+N$ or $-N$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For (i) $N=p_1^{r_1}\cdots p_k^{r_k}$ is a norm from $R$&#xA;iff $r_j$ is even whenever $p_j$ is inert in $R$. Now&#xA;$p$ is inert iff the Legendre symbol $(94/p)=-1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For (ii) we can sometimes use congruence conditions.&#xA;From (i) for instance $x^2-94y^2=\pm3$ is soluble, but&#xA;thinking modulo $8$ we see that $x^2-94y^2=-3$ is insoluble.&#xA;So $x^2-94y^2=3$ is soluble.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-11-01T09:20:33.417" CommentCount="1" />
  <row Id="8508" PostTypeId="1" AcceptedAnswerId="8551" CreationDate="2010-11-01T09:22:34.170" Score="3" ViewCount="185" Body="&lt;p&gt;Hi,&#xA;I define $X_i$ as a random variable that is uniformly distributed between (0,1). What is the expected number of such variables I require to make the sum go just higher than 1. &#xA;Thanks&lt;/p&gt;&#xA;" OwnerUserId="2988" LastActivityDate="2010-11-01T22:42:00.193" Title="Expected number of (0,1) distributed continuous random variables required to sum upto 1." Tags="&lt;probability&gt;" AnswerCount="3" CommentCount="5" FavoriteCount="2" />
  <row Id="8510" PostTypeId="1" AcceptedAnswerId="8514" CreationDate="2010-11-01T09:51:53.300" Score="3" ViewCount="124" Body="&lt;p&gt;Consider the sphere $S^2 = \lbrace (x,y,z) :\ x^2 + y^2 + z^2 = 1 \rbrace$. This is a smooth manifold in $\mathbb{R}^3$, and for a given point $s \in S^2$, one can consider its coordinate neighborhood.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are many ways to put a smooth structure on $S^2$, but all require at least two coordinate neighborhoods. One simple way is to make use of the local coordinates $\theta, \phi$ (azimuthal angle and inclination angle). For $s \in U = S^2 \setminus {(0,0,1), (0,0,-1)}$, define a coordinate map by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ g : (0,2\pi) \times (0,\pi) \rightarrow U $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with formula given by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ g(\theta, \phi) = (\sin(\phi) \cos(\theta), \sin(\phi) \sin(\theta), \cos(\phi)) $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This map is one-to-one, onto, and bicontinuous, so it is a homeomorphism. One can construct a similar map onto $V = S^2 \setminus {(1,0,0),(-1,0,0)}$, so the whole sphere is covered. The fact that this defines a differentiable structure is easy to work out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question I want to ask is, are there any non-constant solutions to the equation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ \frac{\partial^2 f}{\partial \theta^2} + \frac{\partial^2 f}{\partial \phi^2} = 0 $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $f : S^2 \rightarrow R$ is a smooth function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore, what sorts of boundary conditions are necessary in order to ensure uniqueness.&lt;/p&gt;&#xA;" OwnerUserId="1089" LastEditorUserId="1089" LastEditorDisplayName="" LastEditDate="2010-11-01T10:47:38.780" LastActivityDate="2010-11-01T11:49:12.557" Title="Harmonic functions on $S^2$" Tags="&lt;real-analysis&gt;&lt;differential-equations&gt;&lt;differential-geometry&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="1" />
  <row Id="8511" PostTypeId="1" CreationDate="2010-11-01T11:24:55.190" Score="0" ViewCount="103" Body="&lt;p&gt;If you can help me in any of these two questions, I can get the answer to the other:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we have a discrete random variable x and the data pertaining to it in X(n), how in matlab can we determine the prob. mass function pmf(X)? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;OR&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I use CVX to minimize a function log(pmf(X)) ?&lt;/p&gt;&#xA;" OwnerUserId="2858" LastActivityDate="2010-11-01T11:24:55.190" Title="CVX for matlab question" Tags="&lt;matlab&gt;&lt;convex-optimization&gt;" />
  <row Id="8512" PostTypeId="1" CreationDate="2010-11-01T11:34:17.557" Score="-2" ViewCount="201" Body="&lt;p&gt;Let $A$ be and $n\times n$ matrix over $\mathbb{C}$. If $A$ is diagonalizable and $k$ is a positive integer, prove that $A^k$ is diagonalizable. Matrix $A$ $n\times n$ is diagonal, if $a_{ij}=0$, when $i \neq j$. $n\times n$ matrix $A$ is diagonalizable if $A$ is similar to some diagonal matrix. In other words, a matrix is diagonalizable if there is a diagonal matrix $D$ and an invertible matrix $P$ such that $P^{-1}AP$= $D$. How are the eigenvalues and eigenvectors of $A^k$ related to those of $A$?&lt;/p&gt;&#xA;" OwnerUserId="2723" LastEditorUserId="664" LastEditorDisplayName="" LastEditDate="2010-11-01T12:04:34.160" LastActivityDate="2011-02-16T03:59:11.433" Title="prove that $ A^k$ is diagonalizable" Tags="&lt;linear-algebra&gt;" AnswerCount="2" CommentCount="7" />
  <row Id="8513" PostTypeId="2" ParentId="8510" CreationDate="2010-11-01T11:46:57.393" Score="6" ViewCount="" Body="&lt;p&gt;If I understand your question correctly, you are aware that $\frac{\partial^2 f}{\partial \theta^2} + \frac{\partial^2 f}{\partial \phi^2}$ is not actually the Laplacian on a sphere, but you want to know whether there are any nontrivial smooth functions on $S^2$ which are harmonic with respect to this modified Laplacian. The answer is no.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A function $f(\theta,\phi)$ which satisfies $\frac{\partial^2 f}{\partial \theta^2} + \frac{\partial^2 f}{\partial \phi^2} = 0$ is really a harmonic function on the rectangle $[0,2\pi] \times [0,\pi]$, which is mapped onto the sphere via $g^{-1}$. Continuity on the sphere requires that $f(\theta,0) = \text{const}$ (continuity at the north pole), $f(\theta,\pi) = \text{const}$ (continuity at the south pole), and $f(0,\phi) = f(2\pi,\phi) = h(\phi)$ for some $h:[0,\pi]\rightarrow \mathbb{R}$ (continuity across the prime meridian). For any $h$, the boundary conditions are independent of $\theta$, so $f$ is a function of $\phi$ alone. But then $\frac{\partial^2 f}{\partial \phi^2} = 0$ implies that $f$ is linear in $\phi$. If the slope is nonzero, $f$ on the sphere fails to be smooth at the poles. Therefore, for $f$ to be smooth on the sphere and satisfy $\frac{\partial^2 f}{\partial \theta^2} + \frac{\partial^2 f}{\partial \phi^2} = 0$, it has to be constant.&lt;/p&gt;&#xA;" OwnerUserId="856" LastActivityDate="2010-11-01T11:46:57.393" CommentCount="2" />
  <row Id="8514" PostTypeId="2" ParentId="8510" CreationDate="2010-11-01T11:49:12.557" Score="6" ViewCount="" Body="&lt;p&gt;Since $f$ is a smooth function from $S^2\to \mathbb{R}$, it is a smooth function on your coordinate chart. It is easy to see that by periodicity, you also have that $f$ is a smooth function from $S^1\times(0,\pi)\to\mathbb{R}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $f$ is smooth on $S^2$, the limits&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \lim_{\phi \to 0, \pi} \frac{1}{\sin \phi} \frac{\partial}{\partial \theta} f $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;need to exist and be finite: this is the requirement that $f$ is smooth in &quot;the other chart&quot;. Therefore you have that $\partial_\theta f \to 0$ as $\phi$ approach either boundary. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your equation, however, requires that $\partial_\theta f$ be a harmonic function on $S^1\times (0,\pi)$, a compact domain with boundary. Using the maximum principle, you immediately have that $\partial_\theta f = 0$ everywhere. Plugging back into the equation, this implies $\partial^2f/\partial \phi^2 = 0$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And so $f$ must be a linear function in $\phi$. And we now apply the smoothness constraint yet again:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the north pole, smoothness will require that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \lim_{\phi \to 0} \partial_\phi f(\theta,\phi) = \lim_{\phi \to 0} \frac{1}{\sin \phi} \partial_\theta f(\theta + \pi/2, \phi) = 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which implies that the constant slope of the linear function $f(\phi)$ is 0. And hence $f$ must be the constant function. &lt;/p&gt;&#xA;" OwnerUserId="1543" LastActivityDate="2010-11-01T11:49:12.557" />
  <row Id="8515" PostTypeId="2" ParentId="8512" CreationDate="2010-11-01T12:02:49.357" Score="2" ViewCount="" Body="&lt;p&gt;I think robinhoode and Robin are right: is it so difficult to ask things nicely?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, here you have some hints:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Let $v$ be an eigenvector of eigenvalue $\lambda$. That is, $Av = \lambda v$. Compute $A^k v$.&lt;/li&gt;&#xA;&lt;li&gt;Compute $(P^{-1}AP)^k$.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="664" LastActivityDate="2010-11-01T12:02:49.357" CommentCount="1" />
  <row Id="8516" PostTypeId="2" ParentId="8337" CreationDate="2010-11-01T12:32:20.437" Score="5" ViewCount="" Body="&lt;p&gt;This is not really an answer, but rather a long comment prompted by &lt;a href=&quot;http://math.stackexchange.com/questions/8337/different-methods-to-compute-sum-n1-infty-frac1n2/8418#8418&quot;&gt;David Speyer's answer&lt;/a&gt;.&#xA;The proof that David gives seems to be the one in &lt;a href=&quot;http://arxiv.org/abs/math/0701039&quot; rel=&quot;nofollow&quot;&gt;How to compute $\sum 1/n^2$ by solving triangles&lt;/a&gt; by Mikael Passare,&#xA;although that paper uses a slightly different way of seeing that&#xA;the area of the region $U_0$ (in Passare's notation)&#xA;bounded by the positive axes and the curve $e^{-x}+e^{-y}=1$,&#xA;$$\int_0^{\infty} -\ln(1-e^{-x}) dx,$$&#xA;is equal to $\sum_{n\ge 1} \frac{1}{n^2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This brings me to what I really wanted to mention, namely &lt;em&gt;another&lt;/em&gt; curious&#xA;way to see why $U_0$ has that area; I learned this from&#xA;&lt;a href=&quot;http://www.math.chalmers.se/~wastlund/&quot; rel=&quot;nofollow&quot;&gt;Johan Wästlund&lt;/a&gt;.&#xA;Consider the region $D_N$ illustrated below for $N=8$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/6CGRB.png&quot; alt=&quot;A shape with area = sum of reciprocal squares&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Although it's not immediately obvious,&#xA;the area of $D_N$ is $\sum_{n=1}^N \frac{1}{n^2}$.&#xA;&lt;em&gt;Proof&lt;/em&gt;: The area of $D_1$ is 1. To get from $D_N$ to $D_{N+1}$ one removes the boxes along the&#xA;top diagonal, and adds a new leftmost column of rectangles of width $1/(N+1)$&#xA;and heights $1/1,1/2,\ldots,1/N$,&#xA;plus a new bottom row which is the &quot;transpose&quot; of the new column,&#xA;plus a square of side $1/(N+1)$ in the bottom left corner.&#xA;The $k$th rectangle from the top in the new column&#xA;and the $k$th rectangle from the left in the new row (not counting the&#xA;square) have a combined area which exactly matches the $k$th box in the removed diagonal:&#xA;$$ \frac{1}{k} \frac{1}{N+1} + \frac{1}{N+1} \frac{1}{N+1-k} = \frac{1}{k} \frac{1}{N+1-k}. $$&#xA;Thus the area added in the process is just that of the square, $1/(N+1)^2$.&#xA;Q.E.D.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Apparently this shape somehow comes up in connection with the &quot;random&#xA;assignment problem&quot;, where there's an expected value of something which&#xA;turns out to be $\sum_{n=1}^N \frac{1}{n^2}$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now place $D_N$ in the first quadrant, with the lower left corner at the origin.&#xA;Letting $N\to\infty$ gives nothing but the region $U_0$:&#xA;for large $N$ and for $0&amp;lt;\alpha&amp;lt;1$,&#xA;the upper corner of column number $\lceil \alpha N \rceil$ in $D_N$ lies at&#xA;$$ (x,y) =&#xA;   \left(&#xA;    \sum_{n=\lceil (1-\alpha) N \rceil}^N \frac{1}{n},&#xA;    \sum_{n=\lceil \alpha N \rceil}^N \frac{1}{n}&#xA;   \right)&#xA;   \sim&#xA;   \left(\ln\frac{1}{1-\alpha}, \ln\frac{1}{\alpha}\right),$$&#xA;hence (in the limit) on the curve $e^{-x}+e^{-y}=1$.&lt;/p&gt;&#xA;" OwnerUserId="1242" LastEditorUserId="1242" LastEditorDisplayName="" LastEditDate="2010-11-01T13:18:49.870" LastActivityDate="2010-11-01T13:18:49.870" CommentCount="1" />
  <row Id="8517" PostTypeId="2" ParentId="8504" CreationDate="2010-11-01T12:58:28.340" Score="4" ViewCount="" Body="&lt;p&gt;Jonas's argument is good.  Another proof is: given $f \in L^p$ (here $p=1,2$), take the &lt;a href=&quot;http://en.wikipedia.org/wiki/Convolution&quot; rel=&quot;nofollow&quot;&gt;convolution&lt;/a&gt; of $f$ with a sequence of &lt;a href=&quot;http://en.wikipedia.org/wiki/Mollifier&quot; rel=&quot;nofollow&quot;&gt;mollifiers&lt;/a&gt; $\eta_\epsilon$.  Using properties of convolutions, it's easy to check that $f * \eta_\epsilon$ is a smooth function, and that $f * \eta_\epsilon \to f$ in $L^p$ as $\epsilon \to 0$.  This has the advantage of being a little more direct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Edit:&lt;/b&gt; For a reference, see Folland's &lt;em&gt;Real Analysis&lt;/em&gt;, section 8.2.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The smoothness of $f * \eta_\epsilon$ is Proposition 8.10 and comes from differentiating under the integral sign in the convolution (with justification!), and choosing to put the derivative on $\eta_\epsilon$.  Intuitively, it comes from the idea that convolution is an &quot;averaging&quot; operation and tends to smooth, smear, or blur rough areas of $f$ together, and so should be a smoothing operation.  (The wikipedia article has a nice animation illustrating this.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fact that $f * \eta_\epsilon \to f$ in $L^p$ is Folland's Theorem 8.14 (a), and it's pretty elementary.  He also has Proposition 8.17 which proves that $C^\infty_c$ is dense in $L^p$, but it sort of inexplicably starts by using the fact that $C_c$ is dense in $L^p$.  I suppose this is used to get a compactly supported function, so that you can approximate $f \in L^p$ by functions which are not only smooth (which $f * \eta_\epsilon$ is) but also compactly supported (which $f * \eta_\epsilon$ need not be, although $\eta_\epsilon$ is).  But an easier argument would be to first approximate $f$ in $L^p$ norm by a function $g$ which is compactly supported but not necessarily continuous; for example, $g = f 1_{[-N,N]}$ for large $N$ (this works by dominated convergence), and then apply mollifiers to $g$.  Unless, of course, there is some subtlety that I've missed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Edit 2&lt;/b&gt;: Indeed there is.  Folland's 8.14 (a) relies upon the fact that translation is strongly continuous in $L^p$, which uses the density of $C_c$.  So apparently it is not so easy to bypass this step, and that destroys a lot of the &quot;directness&quot; of my argument.&lt;/p&gt;&#xA;" OwnerUserId="822" LastEditorUserId="822" LastEditorDisplayName="" LastEditDate="2010-11-03T19:54:01.873" LastActivityDate="2010-11-03T19:54:01.873" CommentCount="18" />
  <row Id="8518" PostTypeId="1" AcceptedAnswerId="8521" CreationDate="2010-11-01T13:09:50.147" Score="2" ViewCount="183" Body="&lt;p&gt;Given a continuous function $g$ on $\mathbb R$. Is it possible to decompose $\mathbb R$ as the union of a countable collection of intervals $I_n=]a_n, a_{n+1}]$ so that $g$ is monotone on each $I_n$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same question may be asked for a continuous function defined on a bounded interval.&lt;/p&gt;&#xA;" OwnerUserId="2989" LastActivityDate="2010-11-01T18:08:17.563" Title="Decomposition of a continuous function with monotone functions" Tags="&lt;real-analysis&gt;" AnswerCount="3" />
  <row Id="8519" PostTypeId="2" ParentId="8508" CreationDate="2010-11-01T13:26:22.327" Score="1" ViewCount="" Body="&lt;p&gt;I don't know the exact value, but by &lt;a href=&quot;http://en.wikipedia.org/wiki/Wald%27s_equation&quot; rel=&quot;nofollow&quot;&gt;Wald's equation&lt;/a&gt; it is between 2 and 4.&lt;/p&gt;&#xA;" OwnerUserId="822" LastActivityDate="2010-11-01T13:26:22.327" />
  <row Id="8520" PostTypeId="2" ParentId="8518" CreationDate="2010-11-01T13:41:54.037" Score="5" ViewCount="" Body="&lt;p&gt;Whenever someone comes with a guess as to whether continuous functions are &quot;nice&quot; in some way, the answer is almost inevitably &quot;no&quot;. For example this &lt;a href=&quot;http://math.stackexchange.com/questions/7923/are-continuous-functions-always-differentiable/7925&quot;&gt;other question&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And the &lt;a href=&quot;http://en.wikipedia.org/wiki/Weierstrass_function&quot; rel=&quot;nofollow&quot;&gt;Weierstraß function&lt;/a&gt; is likewise a counterexample to your question. Note that it is not of &lt;a href=&quot;http://en.wikipedia.org/wiki/Bounded_variation&quot; rel=&quot;nofollow&quot;&gt;bounded variation&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Made the opening line less... confrontational.&lt;/p&gt;&#xA;" OwnerUserId="1269" LastEditorUserId="1269" LastEditorDisplayName="" LastEditDate="2010-11-01T14:00:47.477" LastActivityDate="2010-11-01T14:00:47.477" />
  <row Id="8521" PostTypeId="2" ParentId="8518" CreationDate="2010-11-01T13:42:10.293" Score="1" ViewCount="" Body="&lt;p&gt;You might think about the &lt;a href=&quot;http://en.wikipedia.org/wiki/Weierstrass_function&quot; rel=&quot;nofollow&quot;&gt;Weierstrass function&lt;/a&gt;.  I think you can prove it is not monotone over any interval by altering the proof that it is not differentiable.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-11-01T13:42:10.293" />
  <row Id="8522" PostTypeId="1" AcceptedAnswerId="8537" CreationDate="2010-11-01T14:01:00.613" Score="0" ViewCount="148" Body="&lt;p&gt;Hello members of MSE,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was wondering what the expansion series of the function &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ f(x) = -\frac{1}{x^3} \cdot \frac{1}{\Gamma(x) \cdot \Gamma(-(\exp(\frac{2}{3}\pi\cdot i))x) \cdot \Gamma(-(\exp(\frac{4}{3}\pi \cdot i))x)} $$ is, at $x = 0$. I'm also interested in the method behind computing the series expansion. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might think: &quot;why don't you just paste this equation in wolframalpha and then see what the expansion series is?&quot; Well, I did exactly that, but wolframalpha couldn't compute it! Does that mean the expansion series doesn't exist or is it more likely that it's just too 'hard' for the computational knowledge engine to find the series?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Max&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;EDIT For more information:  this function is  function (27) at&lt;/em&gt; &lt;a href=&quot;http://mathworld.wolfram.com/InfiniteProduct.html&quot; rel=&quot;nofollow&quot;&gt; this page&lt;/a&gt;, &lt;em&gt;when n=3. I offer my apologies for not stating the question well at first.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="1055" LastEditorUserId="1055" LastEditorDisplayName="" LastEditDate="2010-11-01T17:01:32.820" LastActivityDate="2010-11-01T17:25:33.480" Title="What is the Series Expansion of the following function?" Tags="&lt;analysis&gt;&lt;sequences-and-series&gt;&lt;special-functions&gt;&lt;gamma-function&gt;" AnswerCount="2" CommentCount="6" />
  <row Id="8523" PostTypeId="2" ParentId="8518" CreationDate="2010-11-01T14:21:17.857" Score="2" ViewCount="" Body="&lt;p&gt;The function $f(x) = x \sin(1/x)$ is continuous on $\mathbb{R}$ and is not monotone on any nontrivial interval containing $0$.  For that matter, the same holds for $f_a(x) = x^a \sin(1/x)$ for any positive integer $a$, and this gives a (simple!) example which is highly differentiable: i.e., $k$ times differentiable for each fixed $k$ and all sufficiently large $a$.  &lt;/p&gt;&#xA;" OwnerUserId="299" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2010-11-01T18:08:17.563" LastActivityDate="2010-11-01T18:08:17.563" CommentCount="3" />
  <row Id="8524" PostTypeId="1" CreationDate="2010-11-01T14:27:06.023" Score="10" ViewCount="376" Body="&lt;p&gt;I am a soon-to-be high school graduate with inclination towards mathematics. I enjoy doing math, and am relatively good at it, but I dislike the way I am being taught. I feel like I am being taught methods for solving a lot of problems, but I don't see how any of this fits together.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was considering pursuing a career in mathematics, but I realized I don't really know much about it as a whole as opposed to how to solve specific problems that will appear on some test.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there any good books out there about the &quot;math mindset&quot;, or whatever it is that it's called?&lt;/p&gt;&#xA;" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-01T14:32:00.730" LastActivityDate="2010-11-01T22:57:47.963" CommunityOwnedDate="2010-11-01T15:53:50.130" Title="On the Math Mindset" Tags="&lt;soft-question&gt;" AnswerCount="4" CommentCount="5" FavoriteCount="6" />
  <row Id="8525" PostTypeId="2" ParentId="8522" CreationDate="2010-11-01T14:33:58.057" Score="2" ViewCount="" Body="&lt;p&gt;Up to a constant and a power of $x$, your function is $\frac1{x\Gamma(x)}$. Mathematica tells me that the Taylor series at zero is &#xA;$$&#xA;1+\gamma  x+\left(\frac{\gamma ^2}{2}-\frac{\pi ^2}{12}\right) x^2+\frac{1}{12} x^3 \left(2 \gamma ^3-\gamma  \pi ^2-2 \psi ^{(2)}(1)\right)+\frac{x^4 \left(60 \gamma ^4-60 \gamma ^2 \pi ^2+\pi ^4-240 \gamma  \psi ^{(2)}(1)\right)}{1440}+\dots$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;($\gamma$ is Euler's constant, and the $\psi$'s are poly-gamma functions.)&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2010-11-01T14:33:58.057" CommentCount="9" />
  <row Id="8526" PostTypeId="1" AcceptedAnswerId="8531" CreationDate="2010-11-01T14:36:21.550" Score="1" ViewCount="96" Body="&lt;p&gt;Suppose $f(x,y) = c$ for $0\lt y\lt x\lt 1$ and $0$ outside. What is $P(X+Y \leq 1)$? What is $P(X^2+Y^2 \leq 1)$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So $P(X+Y \leq 1) = \int_{0}^{1} \int_{0}^{1-x} 2 \ dy \ dx$? Likewise, $P(X^2+Y^2 \leq 1) = \int_{0}^{1} \int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}} 2 \ dy \ dx$? This is assuming that $c=2$. &lt;/p&gt;&#xA;" OwnerUserId="2548" LastEditorUserId="2548" LastEditorDisplayName="" LastEditDate="2010-11-01T14:42:42.980" LastActivityDate="2010-11-01T15:32:55.273" Title="Probability and Integrals" Tags="&lt;probability&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="8528" PostTypeId="2" ParentId="8526" CreationDate="2010-11-01T14:56:18.817" Score="1" ViewCount="" Body="&lt;p&gt;No, $P(X+Y \leq 1 ) = \int_0^1 \int_0^{\min (x,1-x)} 2 dy dx = \int_0^{1/2} \int_0^x 2 dy dx + \int_{1/2}^1 \int_0^{1-x} 2 dy dx = 1/4+1/4=1/2$&lt;/p&gt;&#xA;" OwnerUserId="2660" LastActivityDate="2010-11-01T14:56:18.817" />
  <row Id="8529" PostTypeId="1" AcceptedAnswerId="8530" CreationDate="2010-11-01T14:58:25.730" Score="1" ViewCount="140" Body="&lt;p&gt;Is there a formula that can get sunrise at a particular latitude and longitude (and elevation from sea level)? If so, what it is? (Why does it work?)&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Please tag as &quot;earth-science&quot; and any other appropriate tags. &lt;/p&gt;&#xA;" OwnerUserId="2414" LastActivityDate="2010-11-01T15:29:08.210" Title="Formula for getting time of sunrise at a particular location?" Tags="&lt;math-software&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="8530" PostTypeId="2" ParentId="8529" CreationDate="2010-11-01T15:29:08.210" Score="4" ViewCount="" Body="&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Sunrise_equation&quot; rel=&quot;nofollow&quot;&gt;sunrise equation&lt;/a&gt; is what you're looking for. This gives the number of hours of daylight as a function of your latitude and the Sun's declination; the declination is itself a function of the day of the year.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From that one could figure out the time of sunrise (using the longitude and time zone for correcting for the position in the time zone), assuming that noon is exactly halfway between sunrise and sunset.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But ignoring time zones and daylight savings time, the time halfway between sunrise and sunset is not noon! On &lt;i&gt;average&lt;/i&gt; it's noon, but sometimes the sun runs slow and sometimes it runs fast. The correction for this is called the &lt;a href=&quot;http://en.wikipedia.org/wiki/Equation_of_time&quot; rel=&quot;nofollow&quot;&gt;&quot;equation of time&quot;&lt;/a&gt;; the Wikipedia article gives an approximate formula good to within 45 seconds.&lt;/p&gt;&#xA;" OwnerUserId="173" LastActivityDate="2010-11-01T15:29:08.210" />
  <row Id="8531" PostTypeId="2" ParentId="8526" CreationDate="2010-11-01T15:32:55.273" Score="2" ViewCount="" Body="&lt;p&gt;Since the joint density is constant on the support region, you can get your answers by considering areas.  The answer in each case is the area that corresponds to the event in question divided by the total area of the support region.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, to calculate $P(X + Y \leq 1)$, you want the area of the lower triangular region below (i.e., the region for which $x + y \leq 1$), divided by the total area of the triangle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/Cq1RV.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is $$\frac{\frac{1}{4}}{\frac{1}{2}} = \frac{1}{2}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, to calculate $P(X^2 + Y^2 \leq 1)$, you want the area of the circle sector divided by the total area of the triangle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/IdpYU.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is $$\frac{\frac{\pi}{8}}{\frac{1}{2}} = \frac{\pi}{4}.$$&lt;/p&gt;&#xA;" OwnerUserId="2370" LastActivityDate="2010-11-01T15:32:55.273" CommentCount="5" />
  <row Id="8532" PostTypeId="1" AcceptedAnswerId="8819" CreationDate="2010-11-01T15:35:08.983" Score="16" ViewCount="413" Body="&lt;p&gt;Many young, and not so young, mathematicians struggle with how to spend their time. Perhaps this is due to the 90%-10% rule for mathematical insight: 90 pages of work yield only 10 pages of useful ideas. A venerable mathematician once described his career to me as constantly stumbling around in the dark. Of course, this struggle is largely personal...perhaps it is an evolutionary struggle to find one's own way to find something relevant to contribute to mathematics as literature. Such an evolutionary struggle necessitates the artist's turmoil and requires an acknowledgement of &quot;l'importance d'être seul&quot;. Even when I have successfully done some reasonable mathematics, it behooves me to look for a more efficient way to proceed. This sort of searching turns up various nuggets of advice on how to do mathematics, like the following (paraphrased) ones:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Read a paper of a master for a year, and you will get something&lt;/em&gt;- advice given to young mathematicians by Israel Gelfand at Harvard during his 90th birthday conference on the unity of mathematics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Try to imagine a proof, no matter how vague&lt;/em&gt;-Gowers's online essay on the philosophy of mathematics and our relationship to formalism&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I try to colorize it in my mind, to try to see what it's really getting at, rather than simply what it says&lt;/em&gt;-Thurston's online response to Ashley Reiter on reading mathematics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also recall Grothendieck's images of a softening nut and the rising sea...online interviews/documents with suggestions of Atiyah, Singer, Connes, Gromov...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are many other such nuggets, many of which can be found on posts on this site. These all provide very inspiring images, but I would like to be more blunt:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What is your method? How do you go&#xA;  about doing mathematics on a day to&#xA;  day basis?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="941" LastEditorUserId="941" LastEditorDisplayName="" LastEditDate="2010-11-01T17:39:35.800" LastActivityDate="2010-11-12T19:28:24.213" CommunityOwnedDate="2010-11-01T15:53:01.393" Title="What is your method?" Tags="&lt;soft-question&gt;&lt;big-list&gt;&lt;meta-math&gt;&lt;mathematicians&gt;" AnswerCount="2" CommentCount="8" FavoriteCount="10" />
  <row Id="8533" PostTypeId="2" ParentId="8500" CreationDate="2010-11-01T16:12:31.527" Score="1" ViewCount="" Body="&lt;p&gt;Scan the tree $T$ in some arbitrary way, meanwhile tracing out the subgraph $H$ (start at an arbitrary vertex of $G$). Take care not to close cycles. The condition on the minimum degree ensures that you'll never &quot;run out of edges&quot;. &lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-01T16:12:31.527" />
  <row Id="8534" PostTypeId="2" ParentId="8485" CreationDate="2010-11-01T16:18:05.860" Score="1" ViewCount="" Body="&lt;p&gt;Since $\exp$ is monotone increasing, your problem is equivalent to minimizing the quadratic $ks^2 - \epsilon s = k(s - \epsilon/2k)^2 - \epsilon^2/4k$.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-01T16:18:05.860" />
  <row Id="8535" PostTypeId="1" CreationDate="2010-11-01T16:33:42.977" Score="2" ViewCount="73" Body="&lt;p&gt;Suppose $R=S/I$ is the homomorphic image of a local Cohen-Macaulay ring $(S,m)$.&#xA;I know the completion of $R$ as an $S$-module is just&#xA;$R \otimes \hat S$, where $\hat S$ is the $m$-adic completion of $S$.&#xA;Is this also the completion of $R$ as a ring?&#xA;(I think this comes down to showing that the inverse limit of $S/(m^i +I)$&#xA;is the same as the (inverse limit of $S/m^i$) tensored with $S/I$.)&lt;/p&gt;&#xA;" LastEditorUserId="226" LastEditorDisplayName="" LastEditDate="2010-11-01T18:41:29.997" LastActivityDate="2010-11-02T04:14:29.040" Title="Completion of the homomorphic image of a Cohen-Macaulay ring." Tags="&lt;commutative-algebra&gt;" AnswerCount="1" />
  <row Id="8536" PostTypeId="1" CreationDate="2010-11-01T17:04:40.697" Score="2" ViewCount="150" Body="&lt;p&gt;I've wasted a few days trying to prove Thue's lemma with no success! The proof is absolutely easy if you just count. What 'red flags' are useful to give you a warning that counting might be more productive than other approaches?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can start with one point, but it's more of a conceptual one and requires you to see the theorem in a bigger context: If you know that you cannot prove an existential effectively then counting is probably going to be involved. (You might notice that effective solution would imply fast algorithm for factoring or similar).&lt;/p&gt;&#xA;" LastActivityDate="2010-11-12T01:06:52.163" Title="Learning When to count?" Tags="&lt;proof-strategy&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="8537" PostTypeId="2" ParentId="8522" CreationDate="2010-11-01T17:25:33.480" Score="2" ViewCount="" Body="&lt;p&gt;Mathematica can handle (rather easily) that function. For example, the expansion at x=0 to order 5. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f(x) = -1-2 \gamma  x-2 \gamma ^2 x^2+\displaystyle \frac{1}{6} x^3 \left(-8 \gamma ^3-\psi ^{(2)}(1)\right)-$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$-\displaystyle\frac{1}{3} x^4 \left(\gamma  \left(2 \gamma ^3+\psi ^{(2)}(1)\right)\right)+\frac{1}{60} x^5 \left(-16 \gamma ^5-20 \gamma ^2 \psi ^{(2)}(1)+\psi ^{(4)}(1)\right)+O\left(x^6\right)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(same notation than the one used in Mariano's post).&lt;/p&gt;&#xA;" OwnerUserId="2569" LastActivityDate="2010-11-01T17:25:33.480" CommentCount="2" />
  <row Id="8538" PostTypeId="2" ParentId="8461" CreationDate="2010-11-01T17:56:54.300" Score="1" ViewCount="" Body="&lt;p&gt;Hi Mike, thanks for your re-edit...now it makes much more sense and you are right, much cleaner!  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since I was last working on the problem, I have come up with the following alternative solution which yields the same numerical answer as you:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P{X(2) = 3, X(3) = 5 | X(4) = 5}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;= P{X(2) = 3, X(3) = 5, X(4) = 5} / P{X(4) = 5}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;= P{X(2) = 3, X(3) - X(2) = 2, X(4) - X(3) = 0} / P{X(4) = 5} = (using independent increments)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;= P{X(2) =3}.P{X(3) - X(2) = 2}.P{X(4) - X(3) = 0} / P{X(4) = 5&lt;/p&gt;&#xA;&#xA;&lt;p&gt;= {exp((1/3!)-2u(2u)^3).exp((1/2!)-u(u)^2).exp(-u)} / exp((1 / 5!)-4u(4u)^5)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;= 5!(2)^3 / 3!2!1!(4)^5&lt;/p&gt;&#xA;&#xA;&lt;p&gt;= (4 . 5 . 5) / 4^5&lt;/p&gt;&#xA;&#xA;&lt;p&gt;= 5/64&lt;/p&gt;&#xA;&#xA;&lt;p&gt;what do you think of this approach?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;also, sorry if it looks messy everyone, i just dont know how to get fancy formatting using the text entry, can anyone enlighten me on this?&lt;/p&gt;&#xA;" OwnerUserId="2976" LastActivityDate="2010-11-01T17:56:54.300" CommentCount="2" />
  <row Id="8539" PostTypeId="1" AcceptedAnswerId="8542" CreationDate="2010-11-01T18:01:32.527" Score="0" ViewCount="96" Body="&lt;p&gt;$$\sum_{n\geq0} ({\frac{2}{3}})^{2n-1}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I managed to find that this is a geometric series with common ratio of $4/9$, therefore convergent. That means it's possible to calculate the sum of the series. Can I get any tip of how to do the sum of any kind of geometric series that converge? I've been researching, but all articles seem very confusing, specially with English, which I'm not native on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="2950" LastActivityDate="2010-11-01T18:40:29.360" Title="Convergence of a series and respective sum" Tags="&lt;calculus&gt;&lt;sequences-and-series&gt;&lt;homework&gt;&lt;convergence&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="8540" PostTypeId="2" ParentId="8524" CreationDate="2010-11-01T18:10:36.317" Score="8" ViewCount="" Body="&lt;p&gt;Unfortunately, a very popular approach to teaching mathematics at the grade school level is to de-emphasize the concepts and focus more on the rote, procedural aspects of the subject.  I don't think many people really like this type of education but it's one of the simpler approaches to teaching. On the positive side, by focusing on the procedural you have a systematic way with which you can judge a students' progress.   In some sense this is a mode of teaching that is dictated largely by the desire for a convienient means of evaluation, rather than being dictated by holding the students' interest, or teaching them valuable things.  The hope is that by learning enough procedures by rote, the underlying concepts somehow sink-in, as if by magic.  I think this does work for many students but by-and-large it's unsatisfactory to almost all involved.  To teach mathematics in any other way with similar objective measures of success tends to be much more time and energy consuming for all involved, as far as I can tell.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To complement Sam Nead's very nice suggestions, I enjoyed reading Devaney's &quot;Introduction to Chaotic Dynamical Systems&quot;.  I don't remember when exactly I picked it up, might have been grade 12 or 1st year of university.  Parts of it were over my head at the time, but I was able to digest much of the contents, slowly.  In that regard I think it's useful to take a non-linear approach to learning.  It's useful to have as a goal to understand some high-level topic, and to have a mental roadmap for how you're going to get there. You don't want to be too aggressive on this, though.  Around grade 12, Devaney's text was very useful for me.  It's a book that can be used both for graduate courses and for upper-level undergraduate courses.  So you'll probably see it as several steps ahead of what you're used to.  But it's well-written and well-motivated.  Digest what you can, slow down and think about aspects that appear hard, maybe come back to them a few months later.  Attempt some of the more fun-looking problems at the end of sections.  Do some computer experiments.  That kind of thing. &lt;/p&gt;&#xA;" OwnerUserId="642" LastActivityDate="2010-11-01T18:10:36.317" CommunityOwnedDate="2010-11-01T18:10:36.317" CommentCount="1" />
  <row Id="8541" PostTypeId="2" ParentId="8539" CreationDate="2010-11-01T18:11:31.900" Score="2" ViewCount="" Body="&lt;p&gt;$$S = 1 + x + x^2 + x^3 + \cdots, \quad |x|&amp;lt;1. \quad (1)$$&#xA;$$xS = x + x^2 + x^3 + \cdots \quad (2)$$ &#xA;$$(1) - (2) \Rightarrow S-xS = 1.$$&#xA;$$S = \frac{1}{1-x}.$$&lt;/p&gt;&#xA;" OwnerUserId="1301" LastActivityDate="2010-11-01T18:11:31.900" CommentCount="1" />
  <row Id="8542" PostTypeId="2" ParentId="8539" CreationDate="2010-11-01T18:13:54.353" Score="3" ViewCount="" Body="&lt;p&gt;$$ (\frac{2}{3})^{2n-1} = \frac{3}{2}(\frac{4}{9})^n$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we have a geometric progression with common ratio $r$, and sum from 0 to infinity the sum is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$s = \sum_{i=0}^\infty ar^i = \frac{a}{1-r}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;if $|r|&amp;lt;1$ and where $a$ is the &quot;common factor&quot;.&lt;/p&gt;&#xA;" OwnerUserId="2900" LastActivityDate="2010-11-01T18:13:54.353" />
  <row Id="8543" PostTypeId="2" ParentId="8539" CreationDate="2010-11-01T18:30:07.780" Score="1" ViewCount="" Body="&lt;p&gt;$\displaystyle\rm\ \ \sum_{n\ge 0}\ r^{\:a+b\:n}\ =\ \frac{r^a}{1-r^b}\ \ if\ \ |r| &amp;lt; 1\ \ $ e.g. see the Wikipedia page on &lt;a href=&quot;http://en.wikipedia.org/wiki/Geometric_series&quot; rel=&quot;nofollow&quot;&gt;geometric series.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-11-01T18:30:07.780" CommentCount="1" />
  <row Id="8544" PostTypeId="1" CreationDate="2010-11-01T19:17:44.723" Score="3" ViewCount="115" Body="&lt;p&gt;There is a 5 by 5 matrix of points on a plane. How many triangles can be formed using points on this matrix?&lt;/p&gt;&#xA;" OwnerUserId="3004" LastActivityDate="2010-11-01T23:22:10.033" Title="There is a 5 by 5 matrix of points on a plane. How many triangles can be formed using points on this matrix?" Tags="&lt;combinatorics&gt;&lt;puzzle&gt;&lt;permutations&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="8546" PostTypeId="2" ParentId="8524" CreationDate="2010-11-01T19:39:50.333" Score="5" ViewCount="" Body="&lt;p&gt;I have spent a TON of time in the math library at Rutgers just thumbing through all the books, and I never really found a book about the &quot;math mindset&quot; that was half decent.  You could do a lot worse than to read stack exchange every day and become a regular on this forum; everyone here is really helpful and you shouldn't be afraid of posting a &quot;silly&quot; question out of fear of being embarrassed; the only thing that upsets people here is when someone doesn't put a sufficient amount of thought/effort into the problem before asking, or else when the question being asked is clearly off topic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think that talking with mathematicians about specific problems you are working on will give you a far deeper insight into how mathematicians think than reading some low level book for a popular audience will.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most important thing in my opinion is not the book you finally choose, but rather the attitude that you should always try to solve EVERY SINGLE problem in whatever math book you read.  You should spend more time solving the problems than reading the sections.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I recommend you pick up &quot;A survey of modern algebra&quot; by Birkhoff and Mclane.  The difficulty level is pretty low, and the problems are numerous and instructive.  One reason I so highly recommend this particular book for you is that &quot;modern algebra&quot; is not only fundamentally important for contemporary mathematics, it also is not taught in high schools generally, not to mention that many of the results are beautiful.  Another great thing about this specific book is that the author's are famous mathematicians, and you can figure out how they think about math just by seeing the organization of how the material is presented and how the results are derived.  The first chapter in this book deals with number theory, but actually contains a very nice introduction to writing logical and rigorous proofs.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think you should just read through books on specific subjects, and pick the subject you will study based on your current interests and level of knowledge.  Some subjects to start with are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;set theory, metric spaces, complex analysis, group theory, number theory&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One thing you might want to try is just to go ahead and check out some books by the great masters.  For example, there is a great set theory book by Frankel (who has the axioms of set theory named after him), and a great book on number theory by Sierpinski; and I recommend both of these not only because they are really well written and basically at the high school level, but because this will give you an insight into how the great masters thought about math.  Finally, if you are feeling confident go ahead and check out Disquisitions Arithmetic by Gauss, just to try your hand at it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me also mention a compilation of writings for a popular audience by Henri Poincare called &quot;The Value of Science&quot;.  A large portion of this book deals with physics, but there are two particular passages you should pay attention to for understanding the &quot;math mindset&quot;: the first is the discussion of induction, and the second is a fairly famous section where Poincare discusses the internal psychological process he experienced when solving a highly technical problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Best of Luck!&lt;/p&gt;&#xA;" OwnerUserId="1417" LastEditorUserId="1417" LastEditorDisplayName="" LastEditDate="2010-11-01T22:10:37.913" LastActivityDate="2010-11-01T22:10:37.913" CommunityOwnedDate="2010-11-01T19:39:50.333" />
  <row Id="8547" PostTypeId="1" CreationDate="2010-11-01T19:49:40.877" Score="1" ViewCount="91" Body="&lt;p&gt;I've recently come upon the following (seemingly) simple observation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Claim&lt;/strong&gt;: A positive integer $k = d d^{\prime} &gt; 1$ has the opposite parity of $\text{gcd}(d+1,d^{\prime}+1)$ for any pair of distinct divisors $d$ and $d^{\prime}$ of $k$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are substantially simple proofs (see comments), and one proof follows from the formula: &#xA;\begin{eqnarray}&#xA;\text{gcd}(a,b) = a + b - ab + 2 \sum_{k=1}^{a-1} \lfloor \tfrac{kb}{a} \rfloor.&#xA;\end{eqnarray}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Corollary&lt;/strong&gt;: Given a multiplicative arithmetic function $f \colon \mathbb{N} \to \mathbb{N}$ and a pair of coprime integers $a$ and $b$, &#xA;\begin{eqnarray}&#xA;f(ab) + 1 \equiv \text{gcd}(f(a)+1,f(b)+1) \mod 2.&#xA;\end{eqnarray}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question&lt;/strong&gt;: Do either of these simple results have any neat use(s) or interesting application(s)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(NB: The purpose of this post is not so much about finding simple (or simpler) proofs, but rather more about determining possible applications.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks! &lt;/p&gt;&#xA;" OwnerUserId="2720" LastEditorUserId="2720" LastEditorDisplayName="" LastEditDate="2010-11-01T22:50:52.103" LastActivityDate="2010-11-01T22:50:52.103" Title="Parity of $k = d d^{\prime} &gt; 1$ vs $\text{gcd}(d+1,d^{\prime} +1)$ for distinct divisors $d$ and $d^{\prime}$." Tags="&lt;number-theory&gt;" AnswerCount="1" FavoriteCount="2" />
  <row Id="8548" PostTypeId="2" ParentId="8547" CreationDate="2010-11-01T20:02:52.717" Score="2" ViewCount="" Body="&lt;p&gt;I don't have an answer for the applications question (so feel free to vote me down, I guess?), but the proof is much simpler conceptually if you do it in cases:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $k = ab$, and if a is odd and b is even, then the product is even and the divisors swap parity when you add one, so they don't have a factor of two in common. If they're both even, then the product is even but $a+1$ and $b+1$ are odd, so the gcd is odd. If they're both odd, then the product is odd but $a+1$ and $b+1$ are both even so the gcd is even.&lt;/p&gt;&#xA;" OwnerUserId="2954" LastEditorUserId="2954" LastEditorDisplayName="" LastEditDate="2010-11-01T20:20:49.713" LastActivityDate="2010-11-01T20:20:49.713" CommentCount="1" />
  <row Id="8550" PostTypeId="1" CreationDate="2010-11-01T20:08:32.717" Score="1" ViewCount="224" Body="&lt;p&gt;The definition of a Cumulative Distribution Function $(CDF)$ says that&#xA;$$P(X \le x) = F(x)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is all good.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then my text book gives the following theorem without proof:&#xA;$$P(X \lt x) = F(x-)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The book says that the proof is easy, but with my rusted calculus skills, I have trouble to even intuitively understand this theorem. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried to sketch a proof, to improve my understanding:&#xA;$$F(x-) = \lim_{n \to \infty} F(x - \frac{1}{n}) = \lim_{n \to \infty} P(X &amp;lt; x - \frac{1}{n}) = P(X &amp;lt; x).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I assume we can't say $P(X \le x - \frac{1}{n})$, because then $X(\omega) \ge x$, for some $\omega \in \Omega$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this the correct reasoning?&lt;/p&gt;&#xA;" OwnerUserId="2566" LastEditorUserId="2566" LastEditorDisplayName="" LastEditDate="2010-11-01T20:16:21.020" LastActivityDate="2010-11-03T17:06:19.060" Title="Proof of $P(X &lt; x) = F(x−)$" Tags="&lt;calculus&gt;&lt;probability&gt;&lt;probability-theory&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="8551" PostTypeId="2" ParentId="8508" CreationDate="2010-11-01T20:14:03.857" Score="4" ViewCount="" Body="&lt;p&gt;I asked a related question in MathOverflow a while back. Here is the &lt;a href=&quot;http://mathoverflow.net/questions/10567/number-of-uniform-rvs-needed-to-cross-a-threshold&quot; rel=&quot;nofollow&quot;&gt;link&lt;/a&gt;. Let $N$ be the number of $U(0,1)$ random variables needed for the sum to cross $1$. It is in fact possible to derive the pmf of $N$ and compute $E(N)$ as David Bar Moshe has indicated. But there is a nifty derivation if we are only interested in $E(N)$ that involves conditioning on the very first random variable $U_1$ that we pick. I have sketched an outline below. Let me know if any part of the derivation isn't clear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $N(x)$ be expected number of $U(0,1)$ random variables needed for the sum to cross $x$ where $0 \leq x \leq 1$. Then, a recursion can be derived for $f(x) = E(N(x))$ as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f(x) = \int_{0}^{1} E(N(x) \mid U_1 = y) dy$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This integral naturally splits into two - that between $0$ and $x$ and that between $x$ and $1$. If $U_1 &amp;lt; x$, $E(N(x))$ is simply $1 + f(x-U_1)$. If $U_1 &gt; x$, $E(N(x))$ is just $1$. This will give us an integral equation for $f(x)$ that can be solved to get the solution (with suitable initial condition) as $f(x) = e^x$. &lt;/p&gt;&#xA;" OwnerUserId="618" LastActivityDate="2010-11-01T20:14:03.857" />
  <row Id="8552" PostTypeId="1" AcceptedAnswerId="8572" CreationDate="2010-11-01T20:39:30.730" Score="4" ViewCount="254" Body="&lt;p&gt;I am interested in a system of differential equations that is non-linear, but it doesn't seem to be too crazy. I'm not very good at non-linear stuff, so I thought I'd throw it out there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The actual equations I'm looking at have several parameters that'd I'd like to tweak eventually.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;q' = k - m / r&#xA;r' = i - n r - j q&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;i, j, k, m and n are all real-valued constants. I'm guessing that this system would be cyclical in nature, but I'm not sure if it has any explicit solution, so I have produced a version of it with the constants removed to see if that can be solved:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;q' = 1 - 1 / r&#xA;r' = 1 - r - q&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Anyone know if either of these are solvable and what kind of techniques would be needed to solve them if so?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first equation is based on a polar coordinate system where Q (or theta) is the angle and r is radius, and I've made a number of simplifications to make it somewhat tractable.&lt;/p&gt;&#xA;" OwnerUserId="2247" LastActivityDate="2010-11-01T22:37:15.817" Title="Explicit solutions to this nonlinear system of two differential equations" Tags="&lt;differential-equations&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="8553" PostTypeId="2" ParentId="8550" CreationDate="2010-11-01T20:49:05.263" Score="3" ViewCount="" Body="&lt;p&gt;The intuition is that the CDF is just &quot;adding probabilities up&quot;, a little bit at a time.  Remember, the integral is a &lt;b&gt;s&lt;/b&gt;um; that's why $\int$ is an elongated 's'.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So think of your expression $F(x-\frac 1 n)$ as &quot;adding a little bit more&quot;; as $n$ gets larger, you added in the difference&#xA;$$&#xA;\int_{-\infty}^{x-\frac 1 {n+1}} f(x) - \int_{-\infty}^{x-\frac 1 n} f(x) = \int_{x-\frac 1 n}^{x-\frac 1 {n+1}} f(x).&#xA;$$&#xA;Since $f(x) \geq 0$ for all $x$, this difference is also $\geq 0$. The limit is then the sum of the area under the curve $f(x)$ between $x-1/n$ and $x-1/(n+1)$:&#xA;$$&#xA;\sum_{n=1}^\infty \int_{x-\frac 1 n}^{x-\frac 1 {n+1}} f(x).&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if $X$ is a continuous random variable, this sum &lt;i&gt;will&lt;/i&gt; converge to $F(x)$. However, if $X$ has &quot;discrete pieces&quot; things can go wrong and the CDF can jump: see the third picture in &lt;a href=&quot;http://en.wikipedia.org/wiki/File%3aDiscrete_probability_distribution_illustration.png&quot; rel=&quot;nofollow&quot;&gt;this diagram&lt;/a&gt;.  In general, however, the sum will converge to $\lim_{b\to x^-} F(b)$.&lt;/p&gt;&#xA;" OwnerUserId="3009" LastActivityDate="2010-11-01T20:49:05.263" CommentCount="7" />
  <row Id="8554" PostTypeId="2" ParentId="8552" CreationDate="2010-11-01T20:56:37.987" Score="4" ViewCount="" Body="&lt;p&gt;I don't think there's very much hope for an explicit solution. However, you can easily draw a phase portrait and see qualitatively how the solutions behave (in your case, the solutions in the upper half plane $r&gt;0$ spiral towards the stable equilibrium $(q,r)=(0,1)$. &lt;a href=&quot;http://math.rice.edu/~dfield/dfpp.html&quot; rel=&quot;nofollow&quot;&gt;Here's a Java applet by John Polking&lt;/a&gt; for drawing phase portraits (including nullclines); it can also find and analyze equilibria.&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2010-11-01T20:56:37.987" CommentCount="6" />
  <row Id="8555" PostTypeId="2" ParentId="8550" CreationDate="2010-11-01T20:57:59.013" Score="1" ViewCount="" Body="&lt;p&gt;EDIT: I rewrite my answer, since it appears it was not clear enough. I will try to go in more details even though I think this is not really interesting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We define $F(x) := P(X \leq x)$ and we want to prove $F(x-) = P(X &amp;lt; x)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider a sequence $(a_n)$ such that $\lim_{n \rightarrow \infty} a_n = x$, $a_n &amp;lt; x$ for all $n$. For the sake of simplicity, let's admit $a_n$ is increasing ($a_{n+1} \geq a_n$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, let's consider the sequence $(b_n)$ given by $b_n := F(a_n)$. By definition of $F$, we have $b_n = P(X \leq a_n)$. Thus $(b_n)$ is a sequence in [0,1]. Since $b_{n+1} = P(X \leq a_{n+1}) \geq P(X \leq a_n) = b_n$, we have that the sequence $b_n$ is also increasing. Hence, it converges. Now let's calculate the limit (let's call it $b$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know that $P(X \leq a_n) \leq P(X &amp;lt; x)$, so $b_n \leq  P(X &amp;lt; x)$ for all $n$. Thus this should remain true for the limit $b$. You might argue that you are not sure why this should remain true for the limit and why we could not have $b = P(X \leq x)$. Here comes the $\varepsilon$ argument. If  $P(X \leq x) = P(X &amp;lt; x)$ then there is no problem, so suppose $P(X \leq x) \ne P(X &amp;lt; x)$. In that case, let $\varepsilon := \frac{1}{2}\left(P(X \leq x) - P(X &amp;lt; x)\right)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By definition of the limit, you should find infinite members of the sequence in $[b- \varepsilon, b + \varepsilon]$. But this can't be, because all the $b_n$ are smaller than $P(X &amp;lt; x)$ and thus at distance at least 2 $\varepsilon$ from the limit. This proves that if $P(X \leq x) \ne P(X &amp;lt; x)$, then $b \ne P(X \leq x)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ok, so we know $b \leq P(X &amp;lt; x)$. We still need to prove there is an equality here. I will not go into further technical details here, the intuition should tell you that since $a_n \rightarrow x$, you can come as near to $P(X &amp;lt; x)$ as you want with the sequence $b_n$ (consider $\varepsilon = \frac{P(X &amp;lt; x) -b}{2}$ and see what this implies bla-bla-bla).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the beginning I said we could consider $a_n$ to be increasing. Suppose it is not. Then you can construct a new sequence $c_n$, such that $\lim c_n = x$, $c_n \leq a_n$ for all $n$ and $c_n$ is increasing. Then you compare the sequence $d_n := F(c_n)$ with the sequence $b_n$ and bla-bla-bla. This really becomes technical, but it works. So $F(x-) = P(X &amp;lt; x)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;by the way, sorry if I can't write my sequences properly, I can't get { } in math mode, even with 1 backslash, 2 backslaches or 3 backslaches before the {. This really annoys me.&lt;/p&gt;&#xA;" OwnerUserId="2778" LastEditorUserId="2778" LastEditorDisplayName="" LastEditDate="2010-11-03T17:06:19.060" LastActivityDate="2010-11-03T17:06:19.060" CommentCount="2" />
  <row Id="8556" PostTypeId="1" CreationDate="2010-11-01T21:07:39.770" Score="2" ViewCount="160" Body="&lt;p&gt;If I have a stochastic matrix X- the sum of each row is equal to 1 and all elements are non-negative.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given this property, how can I show that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;x'X=x', x&gt;=0 &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Has a non-zero solution?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm assuming this has something to do with proving a feasible dual, but not may be wrong..&lt;/p&gt;&#xA;" OwnerUserId="2014" LastActivityDate="2010-11-03T01:32:29.707" Title="duality with a stochastic matrix" Tags="&lt;linear-algebra&gt;" AnswerCount="3" FavoriteCount="1" />
  <row Id="8557" PostTypeId="1" CreationDate="2010-11-01T21:18:19.470" Score="2" ViewCount="223" Body="&lt;p&gt;$$\int_1^{10}x^xdx$$&#xA;I figured it with a relative error of 1% and I have a response $$≈0.3*10^{10}$$ &#xA;But I don´t know how to accurately calculate it ...&lt;/p&gt;&#xA;" OwnerUserId="2779" LastEditorUserId="2779" LastEditorDisplayName="" LastEditDate="2010-11-09T07:35:01.870" LastActivityDate="2010-11-09T07:35:01.870" Title="Calculate the integral $\int_1^{10}x^xdx$" Tags="&lt;analysis&gt;&lt;homework&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="1" />
  <row Id="8558" PostTypeId="1" AcceptedAnswerId="8560" CreationDate="2010-11-01T21:19:41.560" Score="1" ViewCount="229" Body="&lt;p&gt;I'm trying to use inverse cosine, in the cosine rule to find an angle of a triange when you know 3 sides. I know this formula and have it written down. However I've left my calculator and I'm having to use Google:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.google.co.uk/webhp?hl=en#sclient=psy&amp;amp;hl=en&amp;amp;site=webhp&amp;amp;source=hp&amp;amp;q=arccos((5.9%5E2+%2B+4.1%5E2+-+3.2%5E2)+%2F+(2+&quot; rel=&quot;nofollow&quot;&gt;http://www.google.co.uk/webhp?hl=en#sclient=psy&amp;amp;hl=en&amp;amp;site=webhp&amp;amp;source=hp&amp;amp;q=arccos((5.9%5E2+%2B+4.1%5E2+-+3.2%5E2)+%2F+(2+&lt;/a&gt;&lt;em&gt;+5.9+&lt;/em&gt;+4.1))++in+degrees&amp;amp;aq=f&amp;amp;aqi=&amp;amp;aql=&amp;amp;oq=&amp;amp;gs_rfai=&amp;amp;pbx=1&amp;amp;qscrl=1&amp;amp;fp=3dd9e71350957573&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but it doesn't return the correct answer which makes think theres a difference between arccos and inverse cos.&lt;/p&gt;&#xA;" OwnerUserId="2075" LastActivityDate="2010-11-01T21:37:03.323" Title="Difference between arccos and Cos^-1" Tags="&lt;trigonometry&gt;" AnswerCount="1" />
  <row Id="8559" PostTypeId="1" CreationDate="2010-11-01T21:23:36.407" Score="0" ViewCount="74" Body="&lt;p&gt;If i have $Q = \left\{ y \in \mathbb{R}^n \ \vert \ By=c, y \geq 0 \right\}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and know that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$y_{j} = 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I prove there exists some $p$ where $p'A \geq 0'$, $p'b = 0$ and $(p'A)_{j} &gt; 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At face value this looks like a dual problem, but I'm confused how to proceed without an objective function. Can I arbitrarily assign one, or is there another way to proceed?&lt;/p&gt;&#xA;" OwnerUserId="2014" LastEditorUserId="664" LastEditorDisplayName="" LastEditDate="2010-11-02T04:32:59.017" LastActivityDate="2010-11-02T04:32:59.017" Title="matrix with null variable" Tags="&lt;linear-programming&gt;" CommentCount="5" FavoriteCount="1" />
  <row Id="8560" PostTypeId="2" ParentId="8558" CreationDate="2010-11-01T21:30:15.623" Score="2" ViewCount="" Body="&lt;p&gt;Wolfram Alpha gives about 31.21 degrees.  My calculator gives 31.20576  When I first pasted it into Google, it didn't bring up the calculator, then when I put it into Alpha it lost the s off arccos.  Also between the numbers in the denominator I get spaces instead of asterisks, so maybe it is not multiplying.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some people distinguish between the inverse cosine function (which has the range restricted so there is only one value) and arccos not having the range restricted so it is multivalued, but that doesn't seem to be your problem.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2010-11-01T21:37:03.323" LastActivityDate="2010-11-01T21:37:03.323" CommentCount="4" />
  <row Id="8561" PostTypeId="1" AcceptedAnswerId="8708" CreationDate="2010-11-01T21:42:41.777" Score="6" ViewCount="209" Body="&lt;p&gt;This is the first exercise from Sierpinski's Elementary Theory of Numbers. He gives a proof using induction and I was wondering if this approach was correct as well:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a!b!|(a+b)! \iff \exists c \in \mathbb{N} \text{ such that } (a+b)! = c(a!b!)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming without loss of generality that $a \leq b$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a!b! = \displaystyle\prod\limits_{n=1}^a n^2 \displaystyle\prod\limits_{n=a+1}^b n$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then we define the set S:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$S = \{n \in \mathbb{N} :n &amp;lt; a^2 \wedge \not \exists m \in \mathbb{N}\text{ such that }m^2=n)  \} $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $c = \displaystyle\prod\limits_{n \in D}n \displaystyle\prod\limits_{n=b+1}^{a+b}n$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;then $(a+b)! = c(a!b!)$&lt;/p&gt;&#xA;" OwnerUserId="1523" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-02T02:48:48.593" LastActivityDate="2010-11-03T01:24:07.513" Title="Proving that if a and b are natural numbers, then a!b! | (a+b)!" Tags="&lt;number-theory&gt;" AnswerCount="3" CommentCount="11" FavoriteCount="1" />
  <row Id="8564" PostTypeId="1" AcceptedAnswerId="8571" CreationDate="2010-11-01T21:51:11.843" Score="4" ViewCount="225" Body="&lt;p&gt;Okay, so I'm trying to find $ \int \frac1{\cos x}\mathrm{d}x$ using the substitution $t = \tan\left(\frac{x}{2}\right)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I sub in the trig identity for $\sec$ as $\frac{1+t^2}{1-t^2}$ and then rearrange and substitute $\frac{\mathrm{d}t}{\mathrm{d}x} = \frac12 \left(1+ \tan^2\left(\frac{x}{2}\right)\right)$ so I am left with&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac2{1-t^2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I then used partial fractions to find&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac2{1-t^2} = \frac1{1+t} + \frac1{1-t}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and therefore integrating I get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\int \frac1{\cos x}dx = \ln(t+1) - \ln(t-1)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But subbing in $t = \tan\left(\frac{x}{2}\right)$ doesn't seem to get me anywhere close to the solution that I want to find, which is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ \int \frac1{\cos x}\mathrm{d}x = \ln(\sec x + \tan x) + C$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any help on this would be greatly appreciated. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="3011" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-02T07:09:36.390" LastActivityDate="2010-11-02T13:05:47.003" Title="Integral of $\frac1{\cos x}$ using t substitution" Tags="&lt;calculus&gt;&lt;trigonometry&gt;&lt;integral&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="8565" PostTypeId="2" ParentId="8556" CreationDate="2010-11-01T22:00:38.217" Score="2" ViewCount="" Body="&lt;p&gt;If your matrix $X$ is $n\times n$, then $X$ is the transition matrix of a Markov chain with state space $\lbrace  1 , 2 , \dots , n \rbrace$. The vector $x$ that you are looking for is a stationary measure for the Markov chain. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The existence of such a non-zero $x$ can be proven using probability theory, but a purely linear approach falls under &lt;a href=&quot;http://en.wikipedia.org/wiki/Perron%E2%80%93Frobenius_theorem&quot; rel=&quot;nofollow&quot;&gt;Perron-Frobenius theory&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, the averages ${1\over n}\sum_{k=1}^n X^k$ of the powers of $X$ converge to a matrix $M$ (this statement needs proof!). Then, it is easy to see that any row of $M$ can serve as $x'$ and solves $x'X=x'$. Note that the row sums of $M$ all equal 1, so $x\neq 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you suggest, maybe there is also an approach via linear programming. &lt;/p&gt;&#xA;" OwnerUserId="940" LastActivityDate="2010-11-01T22:00:38.217" />
  <row Id="8566" PostTypeId="2" ParentId="8557" CreationDate="2010-11-01T22:03:12.223" Score="5" ViewCount="" Body="&lt;p&gt;One way to solve this problem is to expand in terms of a power series as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First notice that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x^x = e^{\ln(x) x} = \sum_{n=0}^\infty \frac{(\ln(x) x)^n}{n!} $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Integrating on both sides from $1$ to $10$ gives&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\int_1^{10} x^x dx = \sum_{n=0}^\infty \frac{1}{n!} \int_1^{10} x^n \ln(x)^n dx $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The function $x^n ln(x)^n$ is monotonically increasing, so the integral inside the sum is bounded by:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ \frac{1}{n!} \int_1^{10} x^n \ln(x)^n dx &amp;lt; \frac{9}{n!} \cdot \ln(10)^n \cdot 10^n $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if you want the sum to whatever precision, you can use that expression to determine how small the remainder term is. Next, how on earth do you compute that integral? Well it looks ugly but there is a recursive formula. Let $k, l$ be positive integers. Then using integration by parts&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ \int_1^{10} x^k \ln(x)^l dx = \frac{x^{k+1} ln(x)^l}{k+1} \Bigr \vert_1^{10} - \frac{l}{k+1} \int_1^{10} x^k \ln(x)^{l-1} dx $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since you chose the limits of integration to be from $1$ to $10$ you might have to take quite a few terms in this series in order to compute an accurate value, but that is to be expected as $x^x$ blows up so quickly.&lt;/p&gt;&#xA;" OwnerUserId="1089" LastActivityDate="2010-11-01T22:03:12.223" CommentCount="1" />
  <row Id="8567" PostTypeId="2" ParentId="8564" CreationDate="2010-11-01T22:18:49.200" Score="2" ViewCount="" Body="&lt;p&gt;Try writing $\sec x + \tan x$ in terms of $t$ and see if you can reorganize the two answers to agree.  You may need to tweak the value of $C$ to do this.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-11-01T22:18:49.200" CommentCount="1" />
  <row Id="8569" PostTypeId="2" ParentId="8524" CreationDate="2010-11-01T22:29:30.917" Score="4" ViewCount="" Body="&lt;p&gt;I think certain parts of &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0691118809&quot; rel=&quot;nofollow&quot;&gt;The Princeton Companion to Mathematics&lt;/a&gt; might be just what you need.  While many of the topics will go over your head at the moment, there is a very lucid introduction to the history and basic points of view of modern mathematics in Parts I and II, as well as a lot of interesting historical overviews and essays in Parts VI and VIII.  And of course much of the book is filled with fascinating mathematics.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-11-01T22:29:30.917" CommunityOwnedDate="2010-11-01T22:29:30.917" />
  <row Id="8570" PostTypeId="2" ParentId="8508" CreationDate="2010-11-01T22:32:41.747" Score="3" ViewCount="" Body="&lt;p&gt;This problem is well-known. For an answer, see, for example, the first part of Section 2 in &lt;a href=&quot;http://myweb.facstaff.wwu.edu/curgus/Papers/27Unexpected.pdf&quot; rel=&quot;nofollow&quot;&gt;http://myweb.facstaff.wwu.edu/curgus/Papers/27Unexpected.pdf&lt;/a&gt;, or Equations (7)-(10) in &lt;a href=&quot;http://mathworld.wolfram.com/UniformSumDistribution.html&quot; rel=&quot;nofollow&quot;&gt;http://mathworld.wolfram.com/UniformSumDistribution.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2810" LastEditorUserId="2810" LastEditorDisplayName="" LastEditDate="2010-11-01T22:42:00.193" LastActivityDate="2010-11-01T22:42:00.193" />
  <row Id="8571" PostTypeId="2" ParentId="8564" CreationDate="2010-11-01T22:32:51.420" Score="3" ViewCount="" Body="&lt;p&gt;The problem equivocates to proving that $\frac{t+1}{t-1} = secx + tanx$ since $ln(t+1) - ln(t-1) = ln(\frac{t+1}{t-1})$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the fact that $tan\frac{x}{2} = \frac{1-cosx}{sinx}$, we get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{t+1}{t-1} = \frac{1+sinx-cosx}{1-sinx-cosx} = \frac{secx+tanx-1}{secx-tanx-1}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In addition, $(secx+tanx)(1+secx-tanx) = secx + tanx +(sec^2x-tan^2x) = secx+tanx+1$, so that the result follows.&lt;/p&gt;&#xA;" OwnerUserId="702" LastEditorUserId="702" LastEditorDisplayName="" LastEditDate="2010-11-02T13:05:47.003" LastActivityDate="2010-11-02T13:05:47.003" CommentCount="1" />
  <row Id="8572" PostTypeId="2" ParentId="8552" CreationDate="2010-11-01T22:37:15.817" Score="4" ViewCount="" Body="&lt;p&gt;Taking that second question,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$r' = i - nr - jq$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and differentiating gives&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$r'' = -nr' - jq' = -nr' - j(k-\frac{m}{r})$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or in other words&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$r'' + ar' + \frac{b}{r} = c$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is a much simpler differential equation only one variable. I think that you could probably solve this with power series or clever guessing, but it needs to be worked out.&lt;/p&gt;&#xA;" OwnerUserId="1089" LastActivityDate="2010-11-01T22:37:15.817" CommentCount="6" />
  <row Id="8573" PostTypeId="2" ParentId="8524" CreationDate="2010-11-01T22:57:47.963" Score="3" ViewCount="" Body="&lt;p&gt;Here are some further classics that you may find inspiring and enlightening: Polya: How to solve it; Rademacher and Toeplitz: The enjoyment of mathematics; Kac and Ulam: Mathematics and Logic. &lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-11-01T22:57:47.963" CommunityOwnedDate="2010-11-01T22:57:47.963" />
  <row Id="8574" PostTypeId="2" ParentId="8235" CreationDate="2010-11-01T23:01:49.427" Score="5" ViewCount="" Body="&lt;p&gt;Take $\mathcal{E}=(e_1,\ldots,e_d) \in A^d$ that is a basis of $\mathbb{Q}^d$.&#xA;We choose the norm $|| \cdot ||$ to be $|| \cdot ||_{\infty,\mathcal{E}}$ (sup-norm of the coefficients in the basis $\mathcal{E}$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\mathbb{Z}e_1 + \ldots + \mathbb{Z}e_d$ contains $N \mathbb{Z}^d$ for some positive integer $N$.&#xA;Let $C_0$ be the cone spanned by the $e_i$'s.&#xA;Take $C$ to be the cone spanned by $f_1,\ldots,f_n$ where $f_i = e_i +  \epsilon \sum_{j \neq i} e_j$, choosing $\epsilon&gt;0$ small enough so that $(f_1,\ldots,f_n)$ is free.&#xA;There exists $\eta&gt;0$ (depending on $\epsilon$) such that for every $x \in C$, the coefficients of $x$ in the basis $\mathcal{E}$ are all $\geq \eta ||x||$, so $B(x,\eta ||x||) \subset C_0$ (draw a picture!).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now choose $s_1,\ldots,s_k \in S$ exhausting $\mathbb{Z}^d/N\mathbb{Z}^d$.&#xA;Let $M=\max_i ||s_i||$.&#xA;Take $x \in C \cap \mathbb{Z}^d$, with $||x||&gt;M/\eta$.&#xA;There exists $i$ such that $x-s_i \in N \mathbb{Z}^d$, and $x-s_i \in B(x,M) \subset C_0$, so the coefficients of $x-s_i$ in $\mathcal{E}$ are non-negative integers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a consequence, $x=s_i+(x-s_i)$ is in $S$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just to add rigor to the statement with the &quot;picture&quot;: if $x = \sum_i x_i f_i \in C$, and if $x_{i_0} = \max_i x_i$, then $x = \sum_i y_i e_i$ with $y_{i_0} = \max_i y_i \leq (1+(d-1)\epsilon ) x_{i_0}$ and $y_i \geq \epsilon x_{i_0}$ for each $i$, so that $y_i \geq \epsilon / (1+(d-1)\epsilon) ||x||$.&lt;/p&gt;&#xA;" OwnerUserId="2660" LastEditorUserId="2660" LastEditorDisplayName="" LastEditDate="2010-11-04T11:07:01.237" LastActivityDate="2010-11-04T11:07:01.237" CommentCount="2" />
  <row Id="8575" PostTypeId="1" AcceptedAnswerId="8584" CreationDate="2010-11-01T23:03:41.073" Score="2" ViewCount="79" Body="&lt;p&gt;So suppose I want a path from 0 to $c&gt;0$ on the real line, and I am going to use the function $S(t)$ to get there in (discrete) time $T$.  That is, my position at time 0 is 0, my position at time $T$ is $c$, and my position $P_t$ changes in the following way:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;P_t = (1-\gamma) P_{t-1} + \gamma S(t) &#xA;$$&#xA;where $\gamma&amp;lt;1$ limits fast I can move.  Which can be written&#xA;$$&#xA;\frac{P_t - P_{t-1}}{\gamma} = - P_{t-1} +S(t)&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sadly, I have a drift towards zero of $-P$, and $S$ has to drive me away from zero, towards $c$.  So $S(t)$ is the &quot;size&quot; of my push towards $c$ at date $t$.  The &lt;i&gt;cost&lt;/i&gt; of $S$ is given by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;C(T,S(t)) = \sum_1^{T} (S(t))^2&#xA;$$&#xA;So I am penalized for large movements in my position, and &lt;i&gt;big&lt;/i&gt; pushes towards $c$.  I would like to calculate &#xA;$$&#xA;\min_{T,S(\cdot)} C(T,S(t)) &#xA;$$&#xA;subject to $P_0 =0$, $P_T = c$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is, if I am free to choose the number of steps  it takes to get there, and the path can be whatever I want, what is the cost minimizing route from 0 to $c$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clearly, you could set $S(1) =  c / \gamma$, so I arrive immediately, at cost $( c/\gamma )^2$.  So that is an upper bound on the cost.  I could do it in two steps;  If I step $x$ the first period,  I will need to step &#xA;$$\frac{(c-(1-\gamma) \gamma x)}{\gamma}$$&#xA;the second period, for a total cost of of&#xA;$$&#xA;x^2 + (\frac{(c-(1-\gamma) \gamma x)}{\gamma})^2,&#xA;$$&#xA;Which is minimized for $x=\frac{c (1-\gamma)}{\gamma (2-2 \gamma+\gamma^2 )}$, which therefore costs &#xA;$$&#xA;\frac{c^2 (1-\gamma)^2}{\gamma^2 (2-2 \gamma+\gamma^2)^2}+\frac{(c+\frac{c (1-\gamma) (-1+\gamma)}{2-2 \gamma+\gamma^2})^2}{\gamma^2}&#xA;$$&#xA;which is a lower cost than the one-step option, since $\gamma&amp;lt;1$.  It is not clear to me how to extend this analysis to large $T$ - is there a simple recursive form for the cost minimizing $k$-step path, and the problem reduces to choosing the best $k$?&lt;/p&gt;&#xA;" OwnerUserId="2464" LastActivityDate="2010-11-02T01:38:44.770" Title="Minimizing the cost of a path in a dynamic system" Tags="&lt;calculus&gt;&lt;differential-equations&gt;&lt;optimization&gt;&lt;dynamical-systems&gt;" AnswerCount="1" />
  <row Id="8576" PostTypeId="1" AcceptedAnswerId="8644" CreationDate="2010-11-01T23:19:35.157" Score="2" ViewCount="189" Body="&lt;p&gt;Let &lt;strong&gt;n&lt;/strong&gt; be a natural number &lt;strong&gt;k&lt;/strong&gt; digits. Show that the quantity &lt;strong&gt;Q&lt;/strong&gt; of digits required to write the natural numbers from &lt;strong&gt;1&lt;/strong&gt; to &lt;strong&gt;n&lt;/strong&gt; is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Q = k(n+1) - 111 ... 11&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(111 ... 11 contains k digits equal to 1.)&lt;/p&gt;&#xA;" OwnerUserId="2938" LastEditorUserId="2938" LastEditorDisplayName="" LastEditDate="2010-11-02T10:02:47.987" LastActivityDate="2010-11-02T13:53:06.450" Title="Number of digits from 1 to n" Tags="&lt;number-theory&gt;" AnswerCount="3" CommentCount="3" />
  <row Id="8577" PostTypeId="2" ParentId="8576" CreationDate="2010-11-01T23:21:42.813" Score="0" ViewCount="" Body="&lt;p&gt;Start with the one digit numbers. How many digits are needed for them.  Do you include zero?  How many digits to go from 10 through 99?  And so on...&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-11-01T23:21:42.813" CommentCount="4" />
  <row Id="8578" PostTypeId="2" ParentId="8544" CreationDate="2010-11-01T23:22:10.033" Score="5" ViewCount="" Body="&lt;p&gt;This is the obvious approach (to me, at least), and is mostly a matter of bookkeeping.  To begin with, there are ${{5^2} \choose 3}$ sets of three points.  However, some of these are colinear, which we don't want to count.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are $2 \times 5 \times {5 \choose 3}$ horizontal or vertical sets of three colinear points.  E.g.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; .xxx.   .....&#xA; .....   ...x.&#xA; .....   ...x.&#xA; .....   .....&#xA; .....   ...x.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;There are $3+3+3+3$ sets of three colinear points with rise/run $\in {\pm 2, \pm 1/2}$.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; x....   ..x..&#xA; ..x..   .....&#xA; ....x   ...x.&#xA; .....   .....&#xA; .....   ....x&#xA;&#xA; ...x.   .....&#xA; .....   ....x&#xA; ..x..   ..x..&#xA; .....   x....&#xA; .x...   .....&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;There are $2 \times {5 \choose 3}$ sets of three colinear points with rise/run $\in {\pm 1}$ along the main diagonal or main anti-diagonal.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; x....   .....&#xA; .x...   ...x.&#xA; ..x..   ..x..&#xA; .....   .....&#xA; .....   x....&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;There are $4 \times {4 \choose 3}$ sets of three colinear points along the &quot;shunted&quot; main diagonal and main anti-diagonal (is there a better name for these diagonals?).&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; .x...   ...x.   .....   .....&#xA; .....   ..x..   x....   .....&#xA; ...x.   .....   .x...   ...x.&#xA; ....x   x....   ..x..   ..x..&#xA; .....   .....   .....   .x...&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Finally, there are these $4$ remaining:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; ..x..   ..x..   .....   .....&#xA; .x...   ...x.   .....   .....&#xA; x....   ....x   x....   ....x&#xA; .....   .....   .x...   ...x.&#xA; .....   .....   ..x..   ..x..&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;So there are \[{{5^2} \choose 3}-2 \times 5 \times {5 \choose 3}-(3+3+3+3)-2{5 \choose 3}-4{4 \choose 3}-4=2148\] triangles that can be formed.  I also checked this answer with some code in &lt;a href=&quot;http://www.gap-system.org/&quot; rel=&quot;nofollow&quot;&gt;GAP&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[Here, I have assumed that you want to count congruent triangles separately.]&lt;/p&gt;&#xA;" OwnerUserId="139" LastActivityDate="2010-11-01T23:22:10.033" CommentCount="2" />
  <row Id="8579" PostTypeId="1" AcceptedAnswerId="8580" CreationDate="2010-11-01T23:32:44.143" Score="0" ViewCount="146" Body="&lt;p&gt;I'm having a hard time. I want to generate an equation system with two equations from a second degree equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, get two first degree equations from &lt;/p&gt;&#xA;&#xA;&lt;p&gt;y''=(-k/m)*abs(y')*y'-g&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(k, m and g are constants)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And yes, it is a homework, I just can't figure out how to do it.&lt;/p&gt;&#xA;" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-11-01T23:37:29.923" LastActivityDate="2010-11-01T23:37:29.923" Title="How to transform a second degree differential equation into two first degree linear equations?" Tags="&lt;homework&gt;&lt;differential-equations&gt;" AnswerCount="1" />
  <row Id="8580" PostTypeId="2" ParentId="8579" CreationDate="2010-11-01T23:35:55.417" Score="2" ViewCount="" Body="&lt;p&gt;Set v = y'. Then you get the system:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;v' = (-k/m)*abs(v)*v-g&lt;/p&gt;&#xA;&#xA;&lt;p&gt;v = y'&lt;/p&gt;&#xA;" OwnerUserId="759" LastActivityDate="2010-11-01T23:35:55.417" />
  <row Id="8581" PostTypeId="1" AcceptedAnswerId="8583" CreationDate="2010-11-01T23:48:56.573" Score="0" ViewCount="75" Body="&lt;p&gt;Forgive the total beginner question, it's been a few years since I needed to do this stuff.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the expansion of (1 + x)^n?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: substituted expansion for simplification.&lt;/p&gt;&#xA;" OwnerUserId="3013" LastEditorUserId="3013" LastEditorDisplayName="" LastEditDate="2010-11-01T23:56:23.313" LastActivityDate="2010-11-02T00:04:03.243" Title="Simplify polynomial" Tags="&lt;polynomials&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="8582" PostTypeId="2" ParentId="8576" CreationDate="2010-11-01T23:57:31.990" Score="0" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\ $ Chopping off the unit digit is a 10-to-1 map from k+1 digit onto k-digit naturals. Therefore, since there are 9 naturals with 1-digit, there are 90 with 2-digits, and 900 with 3-digits, etc.&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-11-01T23:57:31.990" CommentCount="4" />
  <row Id="8583" PostTypeId="2" ParentId="8581" CreationDate="2010-11-02T00:04:03.243" Score="2" ViewCount="" Body="&lt;p&gt;$$1 + n$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$1 + 2n + n^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$1 + 3n + 3n^2 + n^3$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$1 + 4n + 6n^2 + 4n^3 + n^4$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$1 + 5n + 10n^2 + 10n^3 + 5n^4 + n^5$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$1 + 6n + 15n^2 + 20n^3 + 15n^4 + 6n^5 + n^6$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See if you can spot a pattern!&lt;/p&gt;&#xA;" LastActivityDate="2010-11-02T00:04:03.243" CommentCount="2" />
  <row Id="8584" PostTypeId="2" ParentId="8575" CreationDate="2010-11-02T01:38:44.770" Score="3" ViewCount="" Body="&lt;p&gt;For fixed $T$, or $k$ which is the same thing, it turns out that you are just minimizing a quadratic function with a linear constraint, and a simple solution exists. However, there is no best $k$ -- you can make the cost arbitrarily small by taking longer and longer to get there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $s$ be the vector with components $s_i = S(i)$ for $i = 1, \ldots, k$. Then if you expand out the recursion in your first equation $P_k = \gamma s_k + (1-\gamma) P_{k-1}$, you will find that&#xA;$$\begin{align}&#xA;P_k &amp;amp;= \sum_{i=1}^k \gamma(1-\gamma)^{k-i} s_i + (1-\gamma)^k P_0 \\&#xA;&amp;amp;= b^Ts + 0&#xA;\end{align}$$ where $b$ is a vector whose components are $b_i = \gamma(1-\gamma)^{k-i}$. So your problem is that of minimizing $C(s) = \sum_{i=1}^k s_i^2 = s^Ts$, subject to the constraint $b^Ts = c$. Using Lagrange multipliers, you can get a straightforward solution. However, you will find that the cost decreases asymptotically with $k$.&lt;/p&gt;&#xA;" OwnerUserId="856" LastActivityDate="2010-11-02T01:38:44.770" CommentCount="1" />
  <row Id="8585" PostTypeId="2" ParentId="8501" CreationDate="2010-11-02T01:51:54.247" Score="1" ViewCount="" Body="&lt;p&gt;I imagine that you could sit in on a lot of classes at Berkeley if you ask the professors. I doubt many would mind if you are very serious about mathematics. &lt;/p&gt;&#xA;" OwnerUserId="2919" LastActivityDate="2010-11-02T01:51:54.247" />
  <row Id="8586" PostTypeId="1" AcceptedAnswerId="8589" CreationDate="2010-11-02T01:52:38.033" Score="3" ViewCount="189" Body="&lt;p&gt;I would really appreciate if someone could just write down for me one example of a second order, or higher, differential equation for which it is known that there are no smooth solutions; and it's fine if it's a partial differential equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At first I thought it would be easy to either come up with an example or else find one by searching google/wiki/arxiv; but now I am not so sure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a thing for non-smooth functions, and it just bothers me that I don't even know a single example of this type of differential equation.  Thanks!&lt;/p&gt;&#xA;" OwnerUserId="1417" LastActivityDate="2010-11-02T15:32:04.277" Title="What is an example of a second order differential equation for which it is known that there are no smooth solutions?" Tags="&lt;differential-equations&gt;" AnswerCount="3" CommentCount="3" />
  <row Id="8587" PostTypeId="2" ParentId="8561" CreationDate="2010-11-02T02:09:41.200" Score="0" ViewCount="" Body="&lt;p&gt;Try $a=b=2$. Then $(a+b)!/a!/b! = 6$ and so $\prod_{n=b+1}^{a+b} n = 12$ cannot divide your $c$.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-02T02:09:41.200" CommentCount="1" />
  <row Id="8588" PostTypeId="1" AcceptedAnswerId="8594" CreationDate="2010-11-02T02:10:48.250" Score="0" ViewCount="71" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have the following formula:&#xA;$\exists y \forall x ((x \geq y) \wedge \neg(y \geq x))$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This essentially boils down to:&#xA;$\exists y \forall x (x &gt; y)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have to check whether this applies to certain universes. But my result depends on whether I can assume that x can also equal y, because in that case the condition will never be true (how could a number be smaller than and at the same time equal to another number?).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you give me a hint?&lt;/p&gt;&#xA;" OwnerUserId="3016" LastEditorUserId="3016" LastEditorDisplayName="" LastEditDate="2010-11-02T02:19:28.807" LastActivityDate="2010-11-02T02:34:04.440" Title="Predicate logic: Two variables - same value allowed?" Tags="&lt;homework&gt;&lt;logic&gt;" AnswerCount="1" CommentCount="7" />
  <row Id="8589" PostTypeId="2" ParentId="8586" CreationDate="2010-11-02T02:20:42.327" Score="8" ViewCount="" Body="&lt;p&gt;There are already &lt;em&gt;first order&lt;/em&gt; linear partial differential equations with smooth coefficients which do not admit any smooth solutions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hans Lewy produced the first &lt;a href=&quot;http://en.wikipedia.org/wiki/Lewy%27s_example#CITEREFMizohata1962&quot; rel=&quot;nofollow&quot;&gt;example&lt;/a&gt; of such a PDE. The equation reads &#xA;$$\left[-i\partial_x+\partial_y-2(x+iy)\partial_z\right]u(x,y,z)=f(x,y,z),\qquad(x,y,z)\in\mathbb R^{3}.$$&#xA;The equation does not have distribution solutions in any neighbourhood of any point in $\mathbb R^3$ provided $f=f(x,y,z)$ is not a real analytic function (it can be smooth though).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The original paper by Lewy is nice, clear and less than 4 pages long (freely available &lt;a href=&quot;http://www.math.ucla.edu/~yhhuang/266.09w/lewy.pdf&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;).   &lt;/p&gt;&#xA;" OwnerUserId="723" LastEditorUserId="723" LastEditorDisplayName="" LastEditDate="2010-11-02T15:32:04.277" LastActivityDate="2010-11-02T15:32:04.277" CommentCount="1" />
  <row Id="8590" PostTypeId="2" ParentId="8586" CreationDate="2010-11-02T02:23:55.457" Score="4" ViewCount="" Body="&lt;p&gt;Consider the partial differential equations associated to the &lt;a href=&quot;http://en.wikipedia.org/wiki/Nash_embedding_theorem&quot; rel=&quot;nofollow&quot;&gt;isometric embedding problem&lt;/a&gt; of the hyperbolic plane into Euclidean 3-space. In $C^1$ there exists a solution by Nash-Kuiper theorem, but it is known classically that there cannot be solutions that are twice or more continuously differentiable. &lt;/p&gt;&#xA;" OwnerUserId="1543" LastActivityDate="2010-11-02T02:23:55.457" CommentCount="2" />
  <row Id="8591" PostTypeId="1" AcceptedAnswerId="8593" CreationDate="2010-11-02T02:24:42.080" Score="1" ViewCount="190" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to understand why polynomial multiplication works the way it is defined.  For example, we know that $(x+1)^2 = x^2+2x+1$, but how can we prove that this actually works?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;More generally, how did we came up with the method used to multiply polynomials?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any insight would be appreciated.&lt;/p&gt;&#xA;" OwnerUserId="2797" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2010-11-02T02:34:34.273" LastActivityDate="2010-11-02T04:50:58.950" Title="Proof that polynomial multiplication works" Tags="&lt;polynomials&gt;" AnswerCount="3" CommentCount="4" />
  <row Id="8592" PostTypeId="2" ParentId="8591" CreationDate="2010-11-02T02:31:48.593" Score="2" ViewCount="" Body="&lt;p&gt;If you have a set $X$ and two real valued functions $f$, $g:X\to\mathbb R$, then there is a standard way to multiply $f$ and $g$: the product is the function $h:X\to\mathbb R$ such that $$h(p)=f(p)g(p)$$ for all elements $p$ of $X$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $X=\mathbb R$ and if $f$ and $g$ are polynomials, then the function $h$ turns out also to be a polynomial, given by the usual formula. This completely determines the multiplication rule.&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2010-11-02T02:31:48.593" CommentCount="7" />
  <row Id="8593" PostTypeId="2" ParentId="8591" CreationDate="2010-11-02T02:32:29.730" Score="8" ViewCount="" Body="&lt;p&gt;We define the property that multiplication has to distribute over addition. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is $A(B+C) = AB + AC$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With polynomials we try to do something similiar, analogous to the real numbers we all know so well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(x+1)^2 = (x+1)(x+1) = (x+1)x + (x+1)1 = (x^2 + x) + (x + 1) = x^2 + 2x + 1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, by the distributive law of multiplication you can see how the defined operation makes sense. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You CAN come up with some binary operation that combines polynomials and doesnt distribute over addition, however this operation would not be called multiplication.&lt;/p&gt;&#xA;" OwnerUserId="1906" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2010-11-02T03:04:59.687" LastActivityDate="2010-11-02T03:04:59.687" CommentCount="0" />
  <row Id="8594" PostTypeId="2" ParentId="8588" CreationDate="2010-11-02T02:34:04.440" Score="3" ViewCount="" Body="&lt;p&gt;I'll put the answer here so the question can be marked as answered.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;For every $x$&quot; means &lt;strong&gt;for every&lt;/strong&gt; $x$, period.  So, yes, there are no restrictions on $x$ not being equal to $y$; any such restrictions would have to be given as predicates (a clause $x\neq y$). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll note (as I did in the comments) that whether the transformation from $(x\geq y) \wedge \neg(y\geq x)$ is equivalent to $x\gt y$ (and whether $\gt$ is areflexive) depends on the interpretation of $\geq$ and $\gt$ on the universe in question. The specification of the universe should not be only the set, but also the meaning of the relational and functional symbols like $\geq$. Only if there is some working convention can you simply assume that the meaning will be &quot;the usual one&quot;.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-11-02T02:34:04.440" />
  <row Id="8595" PostTypeId="2" ParentId="6726" CreationDate="2010-11-02T03:06:26.943" Score="1" ViewCount="" Body="&lt;p&gt;I'd recommend Foundations of Mechanics by Abraham and Marsden. It's a huge book, with a very rigorous treatment of mechanics. Altough Smale is not an author, he's heavily cited in the book (by the way, the book provides an &lt;em&gt;excellent&lt;/em&gt; bibliography).&lt;/p&gt;&#xA;" OwnerUserId="1034" LastActivityDate="2010-11-02T03:06:26.943" />
  <row Id="8596" PostTypeId="1" AcceptedAnswerId="8618" CreationDate="2010-11-02T03:24:21.730" Score="3" ViewCount="253" Body="&lt;p&gt;Is category theory worth learning for the sake of learning it? Can it be used in applied mathematics/probability? I am currently perusing &lt;em&gt;Categories for the Working Mathematician&lt;/em&gt; by Mac Lane. &lt;/p&gt;&#xA;" OwnerUserId="2548" LastEditorUserId="2548" LastEditorDisplayName="" LastEditDate="2010-11-02T03:50:09.653" LastActivityDate="2010-11-02T08:09:14.857" Title="What are the prerequisites for learning category theory?" Tags="&lt;soft-question&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="1" />
  <row Id="8597" PostTypeId="2" ParentId="8591" CreationDate="2010-11-02T03:24:36.707" Score="6" ViewCount="" Body="&lt;p&gt;In addition to crasic's symbolic answer, one can also look at the geometric interpretation of multiplication of two quantities (the area of the rectangle formed with sides of the given length) and find that it is consistent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://1.bp.blogspot.com/_PnLYRqe0k9g/S0nZ-dZsl9I/AAAAAAAAARE/MtAEo96oqp0/s320/multiply-two-binomials.png&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(x+a) * (x+b) = x^2 + ax + bx + ab$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: As pointed out by Douglas, the geometric interpretation is only &lt;em&gt;defined&lt;/em&gt; at all when $x$, $a$, and $b$ are positive, so care must be taken to keep algebraic self-consistency when generalizing to the real numbers.  Geometric &quot;proofs&quot; should only be used as an &quot;intuition check&quot;, in any case.&lt;/p&gt;&#xA;" OwnerUserId="51" LastEditorUserId="51" LastEditorDisplayName="" LastEditDate="2010-11-02T04:50:58.950" LastActivityDate="2010-11-02T04:50:58.950" CommentCount="2" />
  <row Id="8598" PostTypeId="1" AcceptedAnswerId="8601" CreationDate="2010-11-02T03:50:03.023" Score="1" ViewCount="104" Body="&lt;p&gt;A group $G$ is is &lt;em&gt;abelian&lt;/em&gt; if&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ ab = ba $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for all pairs of elements $a,b \in G$, i.e., if multiplication commutes.  Another way of stating this condition is that the &lt;em&gt;identity map&lt;/em&gt; $\mathrm{id}$ gives an isomorphism between left- and right- multiplication.  In other words, we can think of left-and right- multiplication as group actions&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$L: G \rightarrow \mathrm{Sym}(G);\ L(a)\cdot b := ab, $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$R: G \rightarrow \mathrm{Sym}(G);\ R(a)\cdot b := ba. $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In an abelian group we have $L(a)\cdot b = ab = ba = R(a)\cdot b$.  Hence,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$L(\mathrm{id}(a))=R(a).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One might consider a weaker condition: there is &lt;em&gt;some&lt;/em&gt; isomorphism $\varphi:G \rightarrow G$ (not necessarily the identity) such that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$L(\varphi(a))=R(a).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for all $a \in G$.  (In other words, if we &lt;em&gt;really&lt;/em&gt; didn't like multiplying on the right, we could always multiply on the left by $\varphi(a)$ instead without losing anything.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a standard name for this weaker condition?  When can it be satisfied (if ever)?  For some standard examples (e.g., the quaternions) I can work out that no such isomorphism exists, but I don't know what to say about the general case.&lt;/p&gt;&#xA;" OwnerUserId="1612" LastEditorUserId="1612" LastEditorDisplayName="" LastEditDate="2010-11-02T04:05:07.693" LastActivityDate="2010-11-02T04:30:13.420" Title="Can nonabelian groups still have isomorphic left- and right- multiplication?" Tags="&lt;group-theory&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="1" />
  <row Id="8599" PostTypeId="1" CreationDate="2010-11-02T03:54:42.983" Score="3" ViewCount="162" Body="&lt;p&gt;Let '$f$' be a continous but nowhere differentiable function. Is $f$ convolved with mollifier, a smooth function ? &lt;/p&gt;&#xA;" OwnerUserId="2987" LastActivityDate="2010-11-02T08:03:42.223" Title="Let 'f' be a continous but nowhere differentiable function. Is 'f' convolved with mollifier, a smooth function ?" Tags="&lt;analysis&gt;&lt;convolution&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="8600" PostTypeId="2" ParentId="8596" CreationDate="2010-11-02T03:57:12.573" Score="3" ViewCount="" Body="&lt;p&gt;I do not know of any specific applications to probability (&quot;applied mathematics&quot; is a bit too wide a label, though I don't know any specific ones either). I view Category Theory as an interesting unifying language that allows you to speak about ideas that may seem to be isolated from each other in different subjects with the same notions, recognizing the connections between them. The results that I most often actually find useful have to do with adjuctions (specifically, I seem to find very useful the fact that left adjoints commute with colimits and right adjoints commute with limits...).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would say that it is good to know at least the most general ideas, enough so that you can communicate and recognize those connections; of course, I'm a big fan of knowing stuff &quot;just for the sake of knowing it&quot; so I may not be the best advice giver in that respect. &lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-11-02T03:57:12.573" CommunityOwnedDate="2010-11-02T03:57:12.573" />
  <row Id="8601" PostTypeId="2" ParentId="8598" CreationDate="2010-11-02T03:58:32.643" Score="3" ViewCount="" Body="&lt;p&gt;In particular, you would have $L(\varphi(a))\cdot e = R(a)\cdot e$, that is, $\varphi(a)=a$.  So $\varphi=\mathrm{id}$.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-11-02T03:58:32.643" CommentCount="1" />
  <row Id="8602" PostTypeId="2" ParentId="8535" CreationDate="2010-11-02T04:14:29.040" Score="2" ViewCount="" Body="&lt;p&gt;If $S$ is any local ring with maximal ideal $m_S$, and $R = S/I$ is a quotient of $S$, then the maximal ideal $m_R$ of $R$ is $m_S/I$, and its $i$th power (for any&#xA;$i \geq 0$) is equal to $m_S^i + I /I$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The $m_R$-adic completion of $R$ is equal to &lt;/p&gt;&#xA;&#xA;&lt;p&gt;inverse limit of $R/m_R^i$ = inverse limit of $S/(I + m_S^n)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The completion of $R$ as an $S$-module is equal to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;inverse limit of $R\otimes_S (S/m_S^i)$ = inverse limit of $S/(I +m_s^i)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(which by the Artin--Rees theorem is the same as $R\otimes_S \hat{S}$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus the two notions of completion are indeed the same (and there is no need&#xA;to assume that $S$ is Cohen--Macaulay).&lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2010-11-02T04:14:29.040" />
  <row Id="8603" PostTypeId="1" AcceptedAnswerId="8613" CreationDate="2010-11-02T04:18:28.563" Score="2" ViewCount="63" Body="&lt;p&gt;Prove that the Markov-Hurwitz equation $x^2+y^2+z^2=dxyz$ is solvable in positive integers iff d= 1 or 3.  Of course the reverse direction is easy, just set x=y=z=1, d=3.  But I really have no idea how to attack the forward direction.&lt;/p&gt;&#xA;" OwnerUserId="2505" LastActivityDate="2010-11-02T06:36:45.403" Title="Markov-Hurwitz equation" Tags="&lt;number-theory&gt;&lt;diophantine-equations&gt;&lt;analytic-number-theory&gt;" AnswerCount="2" />
  <row Id="8604" PostTypeId="2" ParentId="8596" CreationDate="2010-11-02T04:23:39.853" Score="3" ViewCount="" Body="&lt;p&gt;This issue has been discussed multiple times in MO. For instance &lt;a href=&quot;http://mathoverflow.net/questions/903/resources-for-learning-practical-category-theory&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;, with a view towards programming, or &lt;a href=&quot;http://mathoverflow.net/questions/8543/learning-to-think-categorically/8586#8586&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;, where I proposed a selection of Mac Lane's book.&lt;/p&gt;&#xA;" OwnerUserId="664" LastActivityDate="2010-11-02T04:23:39.853" CommentCount="1" />
  <row Id="8605" PostTypeId="2" ParentId="8598" CreationDate="2010-11-02T04:30:13.420" Score="1" ViewCount="" Body="&lt;p&gt;Note that $L$ is a one-to-one homomorphism from $G$ to $S_G$, while $R$ is a one-to-one homomorphism from $G^{\mathrm{op}}$ to $S_G$ ($R(ab) = R(b)\circ R(a)$, so you have an anti-homomorphism if you work with $G$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could ask more generally if these two subgroups of $S_G$ are conjugate. That is, do there exist permutations $\pi$ and $\psi$ of $G$ such that for every $g\in G$, $\pi^{-1}\circ L(g) \pi = R(\psi(g))$ ? In fact, they always are: take $\pi=\psi$ to be the map that sends every element $x$ to $x^{-1}$ (which is, after all, an isomorphism from $G$ to $G^{op}$!):&#xA;\[ \pi^{-1}(L(g)(\pi(x))) = \pi^{-1}(L(g)(x^{-1})) = \pi^{-1}(gx^{-1}) = xg^{-1}) = R(g^{-1})(x).\]&#xA;So $L(G)$ and $R(G)$ are always conjugate in $S_G$, whether $G$ is abelian or not.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-11-02T04:30:13.420" />
  <row Id="8606" PostTypeId="2" ParentId="8603" CreationDate="2010-11-02T04:36:36.130" Score="0" ViewCount="" Body="&lt;p&gt;For starters, d must be odd.  For if d is even, either all of x,y, and z are even or two are odd and one is even.  If they are all even, there is also a solution $(x/2)^2 + (y/2)^2 + (z/2)^2 = (2d)(x/2)(y/2)(z/2)$ with smaller x,y,z.  If only one of x,y,z is even and d is even, the LHS is 2 mod 4 and the RHS is 0 mod 4.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-11-02T04:36:36.130" />
  <row Id="8607" PostTypeId="2" ParentId="4667" CreationDate="2010-11-02T04:37:37.407" Score="4" ViewCount="" Body="&lt;p&gt;It makes sense to consider some sets of combinatorial objects as spaces (or &lt;a href=&quot;http://en.wikipedia.org/wiki/Polytope&quot; rel=&quot;nofollow&quot;&gt;polytopes&lt;/a&gt;) and therefore discuss dimensionality (e.g. the set of n by n (-1,0,+1)-matrices).  Although, perhaps the word &quot;dimension&quot; could better be described as &quot;&lt;a href=&quot;http://en.wikipedia.org/wiki/Degrees_of_freedom_%28statistics%29&quot; rel=&quot;nofollow&quot;&gt;degrees of freedom&lt;/a&gt;&quot;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Mathematically, degrees of freedom is&#xA;  the dimension of the domain of a&#xA;  random vector, or essentially the&#xA;  number of 'free' components: how many&#xA;  components need to be known before the&#xA;  vector is fully determined.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I suspect that it will be difficult to discuss dimensionality in many combinatorial settings.  For example, imagine constructing a &lt;a href=&quot;http://en.wikipedia.org/wiki/Latin_square&quot; rel=&quot;nofollow&quot;&gt;Latin square&lt;/a&gt;, starting from an empty matrix, placing symbols one-at-a-time in a non-clashing manner.  After placing (say) half of the symbols, we might find: (a) there are still many completions of this partial Latin square, (b) there are no completions of this partial Latin square or (c) there is a unique completion of this partial Latin square.  This seems to go against the notion of dimensionality -- the number of &quot;components&quot; required to determine the Latin square is not fixed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could think of the set of partitions of a set of n elements as having dimension n.  You require n pieces of information to determine the partition (which set each element is in).  But as Qiaochu Yuan points out, who cares?  There's no point in having a notion of &quot;dimension&quot; unless you can use it for something.&lt;/p&gt;&#xA;" OwnerUserId="139" LastActivityDate="2010-11-02T04:37:37.407" />
  <row Id="8608" PostTypeId="2" ParentId="8550" CreationDate="2010-11-02T04:49:00.263" Score="1" ViewCount="" Body="&lt;p&gt;For $x$ fixed, define events $A_n$ by $A_1 = { X \leq x-1 }$ and, for $n \geq 2$, $A_n = { x-\frac{1}{{n - 1}} &amp;lt; X \leq x-\frac{1}{n} }$. Since the disjoint union $A_1 \cup A_2 \cup \ldots$ gives the event ${ X &amp;lt; x }$, we have ${\rm P}(X &amp;lt; x)= {\rm P}(A_1 \cup A_2 \cup \ldots) = \sum\nolimits_{k = 1}^\infty {{\rm P}(A_k )}$. Noting that $\sum\nolimits_{k = 1}^n {{\rm P}(A_k )} = {\rm P}(X \leq x - \frac{1}{n})$, we thus obtain the desired result: ${\rm P}(X &amp;lt; x) = \lim _{n \to \infty } {\rm P}(X \le x - \frac{1}{n}) = \lim _{n \to \infty } F(x - \frac{1}{n}) = F(x - )$.&lt;/p&gt;&#xA;" OwnerUserId="2810" LastEditorUserId="2810" LastEditorDisplayName="" LastEditDate="2010-11-02T17:51:00.887" LastActivityDate="2010-11-02T17:51:00.887" CommentCount="2" />
  <row Id="8609" PostTypeId="1" CreationDate="2010-11-02T05:02:37.767" Score="8" ViewCount="160" Body="&lt;p&gt;A divisor $d$ of $k = p_{1}^{r_{1}} \cdots p_{n}^{r_{n}}$ is &lt;em&gt;unitary&lt;/em&gt; if and only if $d = p_{1}^{\varepsilon_{1}} \cdots p_{n}^{\varepsilon_{n}}$, where each exponent $\varepsilon_{i}$ is either $0$ or $r_{i}$. Let $D_{k} = ${ $d$ } be the subset of unitary divisors of an integer $k &gt; 1$ satisfying $\omega(d) = \omega(k) - 1$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Definition&lt;/strong&gt;. A positive integer $k = p_{1}^{r_{1}} \cdots p_{n}^{r_{n}} &gt; 1$ is &lt;em&gt;hyperbolic&lt;/em&gt; if and only if $\sum_{i = 1}^{n} p_{i}^{-r_{i}} &amp;lt; 1$ or, equivalently, $\sum_{d \in D_{k}} d &amp;lt; k$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, $3$, $10$ and $20$ are hyperbolic, but $30$ and $510510 = 2 \cdot 3 \cdot 5 \cdot 7 \cdot 11 \cdot 13 \cdot 17$ are not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming my calculations are correct, I'll state the following with some confidence:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Indeed, many positive integers are hyperbolic. Of the first $10^{8}$ integers, $70334760$ are hyperbolic. Non-trivial prime powers, squares or higher, are also hyperbolic. If $k$ is a hyperbolic integer, then so are its proper, non-trivial unitary divisors; however, the same cannot be inferred for all divisors (consider the hyperbolic integer $900$ and its non-hyperbolic divisor $30$). In fact, an arbitrary product of any number of non-trivial unitary divisors of a hyperbolic integer is again hyperbolic. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The set of hyperbolic integers is closed under &lt;em&gt;exponentiation&lt;/em&gt;, but not under addition (e.g., $10 + 20 = 30$) or under multiplication (e.g., $3 \times 10 = 30$). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Define the &lt;em&gt;Prime zeta function&lt;/em&gt;, $\zeta_{P}(s) = \sum_{p \text{ prime}} p^{-s}$, which converges absolutely for $\mathsf{Re}(s) &gt; 1$. Recall that the multiplicity of a prime divisor $p$ of $k$ is the largest exponent $r$ such that $p^{r}$ divides $k$ but $p^{r+1}$ does not. If the minimum multiplicity of an integer $k$ is $2$ or greater, then $k$ is hyperbolic as can be seen by the elementary bound&#xA;\begin{eqnarray}&#xA;\sum_{i = 1}^{n} p_{i}^{-r_{i}} &amp;lt; \zeta_{P}(2) \approx 0.452247 ....&#xA;\end{eqnarray}&#xA;Thus, the question of hyperbolicity is non-trivial only for integers with minimum multiplicity $1$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Numerical evidence suggests that the natural density of the hyperbolic integers is greater than $0.988284 \dots$, and I conjecture that &lt;em&gt;almost all&lt;/em&gt; integers are indeed hyperbolic (i.e., the natural density is 1). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; Is anything presently known about such integers? (References welcome!)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; Is there a simple proof showing (or refuting) that almost all integers are hyperbolic? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="2720" LastEditorUserId="2720" LastEditorDisplayName="" LastEditDate="2010-11-02T23:38:18.897" LastActivityDate="2010-11-02T23:38:18.897" Title="Positive integers $k = p_{1}^{r_{1}} \cdots p_{n}^{r_{n}} &gt; 1$ satisfying $\sum_{i = 1}^{n} p_{i}^{-r_{i}} &lt; 1$" Tags="&lt;number-theory&gt;&lt;real-analysis&gt;&lt;analysis&gt;&lt;asymptotics&gt;" AnswerCount="1" FavoriteCount="3" />
  <row Id="8610" PostTypeId="1" AcceptedAnswerId="8805" CreationDate="2010-11-02T05:09:11.257" Score="10" ViewCount="237" Body="&lt;p&gt;Prove that continued fraction of tan(1)=[1;1,1,3,1,5,1,7,1,9,1,11,...].  I tried using the same sort of trick used for finding continued fractions of quadratic irrationals and trying to find a recurrence relation, but that didn't seem to work.&lt;/p&gt;&#xA;" OwnerUserId="2505" LastActivityDate="2010-11-04T02:51:17.377" Title="Continued Fraction expansion of tan(1)" Tags="&lt;number-theory&gt;&lt;analytic-number-theory&gt;&lt;continued-fractions&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="2" />
  <row Id="8611" PostTypeId="1" AcceptedAnswerId="8912" CreationDate="2010-11-02T05:49:23.853" Score="3" ViewCount="341" Body="&lt;p&gt;How can I compute the number of common divisors of two naturals ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example,if we consider &lt;code&gt;(12,24)&lt;/code&gt; the answer is &lt;strong&gt;6&lt;/strong&gt; i.e &lt;code&gt;{1,12,2,6,3,4}&lt;/code&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT : I got an answer &lt;a href=&quot;http://books.google.co.in/books?id=di0WSsXV74wC&amp;amp;pg=PA8&amp;amp;lpg=PA8&amp;amp;dq=%22Number+of+common+divisors%22&amp;amp;source=bl&amp;amp;ots=Wg7wRQKNul&amp;amp;sig=djqKheSaFypwJ_yqmzk_g7yVtO0&amp;amp;hl=en&amp;amp;ei=caTPTNbbH4yivgPYy4DFBg&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=7&amp;amp;ved=0CDIQ6AEwBg#v=onepage&amp;amp;q=%22Number%20of%20common%20divisors%22&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.The solution boils down to finding the number of divisor of the GCD of the two numbers.&lt;/p&gt;&#xA;" OwnerUserId="2109" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-11-02T16:06:23.743" LastActivityDate="2010-11-11T20:11:11.437" Title="Number of common divisors between two given numbers" Tags="&lt;number-theory&gt;&lt;algorithms&gt;" AnswerCount="3" CommentCount="3" />
  <row Id="8612" PostTypeId="2" ParentId="3510" CreationDate="2010-11-02T06:25:38.380" Score="3" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Proof&lt;/strong&gt;:&#xA;Consider the function $f(t) = e^{-it}(\cos t + i \sin t)$ for $t \in \mathbb{R}$. By the quotient rule &#xA;\begin{eqnarray}&#xA;f^{\prime}(t) = e^{-i t}(i \cos t - \sin t) - i e^{-i t}(\cos t + i \sin t) = 0&#xA;\end{eqnarray}&#xA;identically for all $t \in \mathbb{R}$. Hence, $f$ is constant everywhere. Since $f(0) = 1$, it follows that $f(t) = 1$ identically. Therefore, $e^{it} = \cos t + i \sin t$ for all $t \in \mathbb{R}$, as claimed.&lt;/p&gt;&#xA;" OwnerUserId="2720" LastEditorUserId="2720" LastEditorDisplayName="" LastEditDate="2010-11-24T23:30:12.673" LastActivityDate="2010-11-24T23:30:12.673" />
  <row Id="8613" PostTypeId="2" ParentId="8603" CreationDate="2010-11-02T06:36:45.403" Score="5" ViewCount="" Body="&lt;p&gt;See Theorem 6.1 of &lt;a href=&quot;http://www.math.uconn.edu/~kconrad/blurbs/ugradnumthy/descent.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.math.uconn.edu/~kconrad/blurbs/ugradnumthy/descent.pdf&lt;/a&gt;. &#xA;By the way, the point of the case d = 1 and d = 3 is not that there are integral solutions (that is trivial) but that they can be systematically derived from small obvious solutions by a recursive method. See the start of Section 6 at that link for a discussion of that issue and a reference for more information.&lt;/p&gt;&#xA;" OwnerUserId="619" LastActivityDate="2010-11-02T06:36:45.403" CommentCount="1" />
  <row Id="8614" PostTypeId="2" ParentId="8503" CreationDate="2010-11-02T06:48:58.930" Score="6" ViewCount="" Body="&lt;p&gt;Putting the bound on N at 9 seems quite non-random. The continued fraction method for solving x^2 - dy^2 = 1 also works on x^2 - dy^2 = N as long as |N| &amp;lt; sqrt(d), which corresponds exactly to your bound on N when d = 94. Therefore your question almost sounds like it was set by someone who knew continued fractions. How did you come up with your question? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Generally, your questions would be better posed if you explain where they come from instead of writing them in the &quot;Solve this&quot; format, which makes them look suspiciously like homework tasks.&lt;/p&gt;&#xA;" OwnerUserId="619" LastActivityDate="2010-11-02T06:48:58.930" />
  <row Id="8615" PostTypeId="2" ParentId="8586" CreationDate="2010-11-02T07:21:26.830" Score="3" ViewCount="" Body="&lt;p&gt;How about you take the differential equation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ \frac{dy}{dx} = |x|  $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a linear non-homogeneous differential equation, whose solution is $C^1$ but not smooth at $x=0$.&lt;/p&gt;&#xA;" OwnerUserId="1089" LastActivityDate="2010-11-02T07:21:26.830" CommentCount="3" />
  <row Id="8616" PostTypeId="2" ParentId="8611" CreationDate="2010-11-02T07:48:16.513" Score="0" ViewCount="" Body="&lt;p&gt;can u post your algorithm here debanjan , i will be interested to have a looka t it&lt;/p&gt;&#xA;" LastActivityDate="2010-11-02T07:48:16.513" CommentCount="3" />
  <row Id="8617" PostTypeId="2" ParentId="8599" CreationDate="2010-11-02T08:03:42.223" Score="3" ViewCount="" Body="&lt;p&gt;Yes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The key is that when you (as you say in the comments) get the two scenarios:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;f \star (D g) = D(f \star g) = (D f) \star g&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;then &lt;strong&gt;you get to choose which!&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if $D f$ doesn't make sense, then you can ignore it and choose to use the identity $D(f \star g) = f \star (D g)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Taking this to the extreme, you get the - bizarre, in my opinion - result that if $p$ is a polynomial then $f \star p$ is always a polynomial.&lt;/p&gt;&#xA;" OwnerUserId="2907" LastActivityDate="2010-11-02T08:03:42.223" CommentCount="14" />
  <row Id="8618" PostTypeId="2" ParentId="8596" CreationDate="2010-11-02T08:09:14.857" Score="3" ViewCount="" Body="&lt;p&gt;It depends on whether you are talking about &lt;em&gt;Category Theory&lt;/em&gt; as a topic in mathematics (on a par with &lt;em&gt;Geometry&lt;/em&gt; or &lt;em&gt;Probability&lt;/em&gt;) or &lt;em&gt;Category Theory&lt;/em&gt; as a viewpoint on mathematics as a whole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the former, the main prerequisite is that you should have encountered a situation where you wanted to move from one type of &quot;thing&quot; to another type of &quot;thing&quot;: say from a group to its group ring, or from a space to its ring of functions, or from a manifold to its differential graded algebra.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the latter, then there are no prerequisites and it is a Very Good thing to do!  But if the latter, then reading Mac Lane isn't necessarily the best way to go.  However, I'm not sure if there is a textbook (or other) that tries to teach elementary mathematics (of any flavour) from a categorical viewpoint.  I try to teach this way, but I've not written a textbook!  I wrote a bit more on this in response to a question on MO, I copied my answer &lt;a href=&quot;http://www.math.ntnu.no/~stacey/CountingOnMyFingers/TeachingCategories.html&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2907" LastActivityDate="2010-11-02T08:09:14.857" CommentCount="3" />
  <row Id="8619" PostTypeId="2" ParentId="8556" CreationDate="2010-11-02T08:10:27.600" Score="1" ViewCount="" Body="&lt;p&gt;Here is a hint for you:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take $v = (1, 1, 1, \dots, 1)'$, the vector with only ones. By definition of your Matrix, you have $Xv = v$. Thus, $v$ is an eigenvector with eigenvalue 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now change the first basis vector of your vector space to $v$ and keep the other unchanged. In this new basis, what will $X$ looks like?&lt;/p&gt;&#xA;" OwnerUserId="2778" LastActivityDate="2010-11-02T08:10:27.600" CommentCount="2" />
  <row Id="8620" PostTypeId="1" AcceptedAnswerId="8628" CreationDate="2010-11-02T09:06:42.280" Score="3" ViewCount="201" Body="&lt;p&gt;The standard definition of a ring is an abelian group that is a monoid under multiplication (with distributivity). However there are some books that have a weaker definition in that a ring only has to be closed under multiplication (no identity).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a problem in my algebra book asking me to prove that if a ring (defined in the second way) has $p$ elements, where $p$ is prime. If the multiplication is not trivial (i.e. sending everything to 0) then the ring is forced to have a multiplicative identity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Its seems like a trivial proof, but I just can't see what I'm missing. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I have so far:&#xA;Given $R$ is an ring with p elements&#xA;R is an abelian group of prime order, therefore it is cyclically generated, and of characteristic $p$ and isomorphic to $Z/pZ$. Essentially it boils down to showing $Z/pZ$ is forced to have a multiplicative identity, but I just can't see where this comes from (every resource I found seems to take this as fact). Since this is a requirement regardless of multiplicative structure I can't just use the fact that $Z/pZ - 0$  is a group under the typical multiplication.&lt;/p&gt;&#xA;" OwnerUserId="1906" LastEditorUserId="1906" LastEditorDisplayName="" LastEditDate="2010-11-02T09:25:57.020" LastActivityDate="2010-11-02T17:28:42.347" Title="Finite rings of prime order must have a multiplicative identity" Tags="&lt;abstract-algebra&gt;&lt;ring-theory&gt;" AnswerCount="4" />
  <row Id="8621" PostTypeId="1" CreationDate="2010-11-02T09:19:57.960" Score="1" ViewCount="137" Body="&lt;p&gt;Interesting problem... I tried solving it through isosceles triangles, and I've proved it doesn't work on equilateral triangles. Can anyone give me some hints towards a geometric proof?&lt;/p&gt;&#xA;" LastActivityDate="2010-11-02T10:25:42.527" Title="Can two angle bisectors of a triangle be perpendicular?" Tags="&lt;geometry&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="8622" PostTypeId="2" ParentId="8620" CreationDate="2010-11-02T09:22:23.590" Score="0" ViewCount="" Body="&lt;p&gt;Go on a hunt for all the ways in which &lt;em&gt;R&lt;/em&gt; can &lt;em&gt;fail&lt;/em&gt; to be isomorphic to &amp;#8484;/p&amp;#8484;, and demonstrate the failure of any of them to hold.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the subset &lt;em&gt;Z&lt;/em&gt;&amp;nbsp;&amp;sube;&amp;nbsp;&lt;em&gt;R&lt;/em&gt; of elements such that left-multiplication by z&amp;nbsp;&amp;isin;&amp;nbsp;&lt;em&gt;Z&lt;/em&gt; &lt;strong&gt;does&lt;/strong&gt; map everything to zero. We may easily show that &lt;em&gt;Z&lt;/em&gt; is an ideal in &lt;em&gt;R&lt;/em&gt;:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;For y,z&amp;nbsp;&amp;isin;&amp;nbsp;&lt;em&gt;Z&lt;/em&gt;, we have (y+z)r = yr + zr = 0+0 = 0 for any r&amp;nbsp;&amp;isin;&amp;nbsp;&lt;em&gt;R&lt;/em&gt;, so y+r&amp;nbsp;&amp;isin;&amp;nbsp;&lt;em&gt;Z&lt;/em&gt;;&lt;/li&gt;&#xA;&lt;li&gt;For any z&amp;nbsp;&amp;isin;&amp;nbsp;&lt;em&gt;Z&lt;/em&gt; and a,r&amp;nbsp;&amp;isin;&amp;nbsp;&lt;em&gt;R&lt;/em&gt;, we have (az)r = a(zr) = 0 and (za)r = z(ar) = 0; so az,&amp;thinsp;za&amp;nbsp;&amp;isin;&amp;nbsp;&lt;em&gt;Z&lt;/em&gt;.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Well, really, we only needed the first of these two criteria, because it shows that (&lt;em&gt;Z&lt;/em&gt;,+) must be a subgroup of the abelian group (&lt;em&gt;R&lt;/em&gt;,+), which is of prime order. We know that &lt;em&gt;Z&lt;/em&gt;&amp;nbsp;&amp;ne;&amp;nbsp;&lt;em&gt;R&lt;/em&gt;; so it follows that &lt;em&gt;Z&lt;/em&gt;&amp;nbsp;=&amp;nbsp;{0}.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Okay then. If none of the non-zero elements map &lt;em&gt;everything&lt;/em&gt; to zero via multiplication, perhaps some of them map &lt;em&gt;some&lt;/em&gt; things to zero? Select a&amp;nbsp;&amp;isin;&amp;nbsp;&lt;em&gt;R&lt;/em&gt;&amp;nbsp;\&amp;nbsp;{0} arbitrary, and consider the set &lt;em&gt;S&lt;/em&gt;&lt;sub&gt;a&lt;/sub&gt;&amp;nbsp;=&amp;nbsp;{b&amp;nbsp;&amp;isin;&amp;nbsp;&lt;em&gt;R&lt;/em&gt; | ab&amp;nbsp;=&amp;nbsp;0}. Well, again we find that &lt;em&gt;S&lt;/em&gt;&lt;sub&gt;a&lt;/sub&gt; is an abelian subgroup of &lt;em&gt;R&lt;/em&gt;, and again we know that &lt;em&gt;S&lt;/em&gt;&lt;sub&gt;a&lt;/sub&gt; is not allowed to be all of &lt;em&gt;R&lt;/em&gt;; so we know that &lt;em&gt;S&lt;/em&gt;&lt;sub&gt;a&lt;/sub&gt;&amp;nbsp;=&amp;nbsp;{0}. That is, &lt;em&gt;R&lt;/em&gt; contains no zero divisors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, are there any non-zero elements such that multiplication is not invertible? Well, no: any a,b&amp;nbsp;&amp;isin;&amp;nbsp;&lt;em&gt;R&lt;/em&gt; such that ar&amp;nbsp;=&amp;nbsp;br for some non-zero r&amp;nbsp;&amp;isin;&amp;nbsp;&lt;em&gt;R&lt;/em&gt; would have the property (a&amp;nbsp;&amp;minus;&amp;nbsp;b)r&amp;nbsp;=&amp;nbsp;0, so that a&amp;nbsp;=&amp;nbsp;b; so multiplication by elements of &lt;em&gt;R&lt;/em&gt;&amp;nbsp;\&amp;nbsp;{0} yield bijections on  &lt;em&gt;R&lt;/em&gt;&amp;nbsp;\&amp;nbsp;{0}.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This should now enable you to answer your question.&lt;/p&gt;&#xA;" OwnerUserId="439" LastEditorUserId="439" LastEditorDisplayName="" LastEditDate="2010-11-02T10:11:51.240" LastActivityDate="2010-11-02T10:11:51.240" CommentCount="0" />
  <row Id="8623" PostTypeId="1" AcceptedAnswerId="8642" CreationDate="2010-11-02T09:38:35.583" Score="3" ViewCount="262" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Question: Find the distinct number of permutations of four letters of the word MATHEMATICS.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer is 11P4/(2!)^3 = 990&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this correct? This question along with the answer is from a module from my course, so this is all the details that I am able to provide.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My understanding of this question is how many distinct 4 letter words can be formed from the word 'MATHEMATICS'. Therefore, a word like 'MAME' is valid. I know that using 11P4 alone will not suffice because it will say that there are 4 ways to create the 4 letter word MAME from the word 'MATHEMATICS'. So, we I will need to adjust the output of 11P4, and this is where I am stuck. The answer divides 11P4 by (2!)^3. I think this represents the 3 non-distinct letters in the word 'MATHEMATICS' (i.e. M,A,T) which have two copies each. Using Sage, the answer I have is 2,454 which is not the same as the answer provided. Here is the code I wrote:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;#Input according to question&#xA;inputQ='M A T H E M A T I C S'.split(' ')&#xA;&#xA;#Same input, but each letter in input is distinct&#xA;inputTest='M1 A1 T1 H E M2 A2 T2 I C S'.split(' ')&#xA;&#xA;print inputQ&#xA;print inputTest&#xA;&#xA;#Number of letters to form from the given input&#xA;choice=4&#xA;&#xA;ansQ=Permutations(inputQ,choice).list()&#xA;ansTest=Permutations(inputTest,choice).list()&#xA;&#xA;print 'Total distinct permutations of ' + str(choice) + ' letter word(s) from the word ' + ''.join(inputQ) +':' ,len(ansQ)&#xA;print 'If each letter in '+ ''.join(inputQ) + ' were distinct: ', len(ansTest)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;What if I only want 1 letter? Then the answer would be 8, which I got by manually listing down all possibilities. But what it the general formula for this sort of problem?&lt;/p&gt;&#xA;" OwnerUserId="2973" LastEditorUserId="2973" LastEditorDisplayName="" LastEditDate="2010-11-02T13:30:50.023" LastActivityDate="2011-03-03T16:25:04.447" Title="Distinct permutations of x items from n items" Tags="&lt;combinatorics&gt;" AnswerCount="4" CommentCount="6" FavoriteCount="1" />
  <row Id="8624" PostTypeId="1" CreationDate="2010-11-02T09:38:36.730" Score="-1" ViewCount="223" Body="&lt;p&gt;Is pi mean 22/7? If so, how circumference of the circle will be 2*22/7 if radius would be 1.&lt;/p&gt;&#xA;" OwnerUserId="2997" LastActivityDate="2010-11-02T22:34:54.543" Title="How the letter 'pi' came in mathematics?" Tags="&lt;pi&gt;" AnswerCount="4" CommentCount="3" />
  <row Id="8625" PostTypeId="2" ParentId="8624" CreationDate="2010-11-02T09:42:16.053" Score="4" ViewCount="" Body="&lt;p&gt;According to &lt;a href=&quot;http://en.wikipedia.org/wiki/Pi#The_letter_.CF.80&quot; rel=&quot;nofollow&quot;&gt;wikipedia&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;The constant is named &quot;$\pi$&quot; because &quot;$\pi$&quot; is the first letter of the Greek word περίμετρος (perimeter)...&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Questions that can be avoided with a quick look at a wikipedia article are typically frowned upon here.&lt;/p&gt;&#xA;" OwnerUserId="1906" LastActivityDate="2010-11-02T09:42:16.053" CommentCount="4" />
  <row Id="8626" PostTypeId="2" ParentId="8624" CreationDate="2010-11-02T09:43:55.013" Score="2" ViewCount="" Body="&lt;p&gt;$\pi \neq \frac{22}{7}$. First of all $\frac{22}{7}$ is just an approximate value of $\pi$.  Note that $\pi$ is irrational so it cannot be expressed in the form $\frac{p}{q}$ where $p$ and $q \neq 0$ are integers. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;For more information about $\pi$ i would suggest you to read the wiki article: &lt;a href=&quot;http://en.wikipedia.org/wiki/Pi&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Pi&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" LastActivityDate="2010-11-02T09:43:55.013" CommentCount="1" />
  <row Id="8627" PostTypeId="2" ParentId="8624" CreationDate="2010-11-02T09:59:20.437" Score="3" ViewCount="" Body="&lt;p&gt;From &lt;a href=&quot;http://books.google.co.uk/books?id=QFPvAAAAMAAJ&amp;amp;q=Pi%3a+a+biography+of+the+world%27s+most+mysterious+number&amp;amp;dq=Pi%3a+a+biography+of+the+world%27s+most+mysterious+number&amp;amp;hl=en&amp;amp;ei=oN7PTIKkItTR4gaM6-SiBg&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=1&amp;amp;ved=0CC0Q6AEwAABlockquote&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;Pi: A Biography of the World's Most Mysterious Number&lt;/em&gt;&lt;/a&gt; by  Posamentier and  Lehmann:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;According to the well-known mathematics historian Florian Cajori (1859-1930), the symbol $\pi$ was first used in mathematics by William Oughtred (1575-1660) in 1652 when he referred to the ratio of the circumference of a circle to its diameter as $\frac{\pi}{\delta}$, where $\pi$ represented the periphery of a circle and $\delta$ represented the diameter...&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;In 1706 William Jones (1675-1749) published his book &lt;em&gt;Synopsis palmariorum matheseos&lt;/em&gt;, in which he used $\pi$ to represent the ratio of the circumference of a circle to its diameter. This is believed to have been the first time that $\pi$ was used as it is defined today... But not until he [&lt;em&gt;Leonhard Euler - A.R.&lt;/em&gt;] used the symbol $\pi$ in his famous book &lt;em&gt;Introductio in analysin infinitorum&lt;/em&gt; did the use of $\pi$ to represent the ratio of the circumference of a circle to its diameter become widespread.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="723" LastEditorUserId="723" LastEditorDisplayName="" LastEditDate="2010-11-02T10:26:28.073" LastActivityDate="2010-11-02T10:26:28.073" CommentCount="1" />
  <row Id="8628" PostTypeId="2" ParentId="8620" CreationDate="2010-11-02T10:10:14.013" Score="3" ViewCount="" Body="&lt;p&gt;Let $x$ be a nonzero element of the ring. Then $R=\{0,x,2x,3x,\ldots,(p-1)x\}$&#xA;where $2x$ means $x+x$ etc. Then $x^2=jx$ where $1\le j\le p-1$.&#xA;Moreover $(ax)(bx)=abx^2=(abk)x$. All you need to do is to prove that for&#xA;some $a$, $(abk)x=bx$ for all $b$. (It's surely enough to do this for $b=1$).&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-11-02T10:10:14.013" CommentCount="2" />
  <row Id="8629" PostTypeId="2" ParentId="8621" CreationDate="2010-11-02T10:25:42.527" Score="10" ViewCount="" Body="&lt;p&gt;The reason you cannot find a geometric proof is because it is not true.&#xA;Suppose you have a triangle ABC such that two angle bisectors are perpendicular, say the angle bisectors of A and B. Suppose they intersect at some point D, then we have a right-angled triangle ADB. If in ABC the angle at A is a and the angle at B is b, then in ADB the angle at A is a/2 and the angle at B is b/2. ADB right-angled triangle implies a/2+b/2=90, which implies a+b=180, contradiction.&lt;/p&gt;&#xA;" OwnerUserId="2712" LastActivityDate="2010-11-02T10:25:42.527" />
  <row Id="8630" PostTypeId="1" CreationDate="2010-11-02T11:12:01.610" Score="2" ViewCount="149" Body="&lt;p&gt;Given two integers A and B , how to efficiently compute the sum of all the digits of every number int the set  $\{N | A \le N \le B \} $.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example: &lt;strong&gt;A = 100&lt;/strong&gt;  and &lt;strong&gt;B = 777&lt;/strong&gt;, then the required answer is &lt;strong&gt;8655&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am interested in deriving an formula/efficient algorithm for the same.&lt;/p&gt;&#xA;" OwnerUserId="2109" LastEditorUserId="2109" LastEditorDisplayName="" LastEditDate="2010-11-02T11:24:20.617" LastActivityDate="2010-11-02T13:06:36.980" Title="How to (efficiently) compute the sum of the digits of evey number lying between two given numbers ?" Tags="&lt;number-theory&gt;&lt;algorithms&gt;" AnswerCount="1" />
  <row Id="8631" PostTypeId="2" ParentId="8623" CreationDate="2010-11-02T11:49:39.860" Score="3" ViewCount="" Body="&lt;p&gt;After the edit by OP,here is a similar &lt;a href=&quot;http://sg.answers.yahoo.com/question/index?qid=20100302091836AA9Cf9K&quot; rel=&quot;nofollow&quot;&gt;problem and solution&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2109" LastEditorUserId="2109" LastEditorDisplayName="" LastEditDate="2010-11-02T13:58:08.813" LastActivityDate="2010-11-02T13:58:08.813" CommentCount="2" />
  <row Id="8632" PostTypeId="1" CreationDate="2010-11-02T11:51:07.943" Score="-1" ViewCount="80" Body="&lt;p&gt;I have a problem. I wish to understand how following can be proven to hold; $\langle P_S(v), u_m \rangle = \sum_{j=1}^n \frac{ \langle v, u_j \rangle}{||u_j||^2}  \langle u_j,u_m \rangle = \langle v , u_m \rangle $, where $P_S(v) =\sum_{j=1}^n \frac{ \langle v, u_j \rangle}{||u_j||^2} u_j $ and Suppose V is a complex inner product space and $B={u_1, u_2,..., u_n }$ is orthogonal set in $V$ with $u_j&amp;lt;&gt;0$ for all j. also prove that $P_S(v)$ is a linear transformation. &lt;/p&gt;&#xA;" OwnerUserId="2723" LastEditorUserId="2723" LastEditorDisplayName="" LastEditDate="2010-11-02T12:24:21.463" LastActivityDate="2010-11-02T12:48:20.923" Title="explain that $\langle P_S(v), u_m \rangle = \sum_{j=1}^n \frac{ \langle v, u_j \rangle}{||u_j||^2} \langle u_j,u_m \rangle = \langle v, u_m \rangle$" Tags="&lt;discrete-mathematics&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="8633" PostTypeId="1" AcceptedAnswerId="8663" CreationDate="2010-11-02T12:17:48.590" Score="2" ViewCount="295" Body="&lt;p&gt;Fair warning: I am not a math expert (that's why I'm here).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to be able to calculate the probability of rolling a certain side on a die with &lt;em&gt;n&lt;/em&gt; sides where any number of those sides has an unfair advantage over the others.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example: On a 6-sided die, we could assume that the 1 - 6 sides are longer than the other sides due to wear and tear (or someone shaved the 1 or 6) and, therefore, more likely to come face up on a roll. I know that, on a perfect die, the roll is uniformly random and each side has a 1/side chance of coming up.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do I calculate the probability of each side if sides 1 and 6 are longer? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What if sides 1 and 6 are longest, 2 and 5 are second longest, and 3 and 4 are shortest?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What if I'm using a 12-sided die? 10-sided? 20-sided?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm looking for a formula that I can plug numbers into and I'd really, really like an explanation of how it all works, if possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found &lt;a href=&quot;http://books.google.com/books?id=GJP08Lv7vpYC&amp;amp;lpg=PA87&amp;amp;ots=bydhFz2oEh&amp;amp;dq=simulate%2520shaved%2520dice&amp;amp;pg=PA86#v=twopage&amp;amp;q&amp;amp;f=false%20%22this%20link%22&quot; rel=&quot;nofollow&quot;&gt;this link&lt;/a&gt; that talks about using Mathematica to calculate one side shaved on a 6-sided die, but I don't know how this changes when you increase the number of dice, or the syntax being used. I feel like the equation below (from the link) makes sense somewhat (in that &amp;sigma; represents the &lt;em&gt;increase&lt;/em&gt; in probability of a 1 or 6, but I would like to know how to calculate &amp;sigma;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;f = { 1/6 + &amp;sigma;, 1/6 - &amp;sigma;/6, 1/6 - &amp;sigma;/6, 1/6 - &amp;sigma;/6, 1/6 - &amp;sigma;/6, 1/6 + &amp;sigma; };&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could I use the same formula to represent a 20 sided die?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;f = {1/20 + &amp;sigma;, 1/20 - &amp;sigma;/20, ...  ... ... ..., 1/20 - &amp;sigma;/20, 1/20 + &amp;sigma;}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: I took an intro to Statistics in college, but that's all the exposure I've had. I would like to understand more, but this is where I am now. I appreciate any help you can give me and any resources you can point me to.&lt;/p&gt;&#xA;" OwnerUserId="3023" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-11-02T13:03:56.427" LastActivityDate="2010-11-02T16:50:16.397" Title="Probability concerning unfair dice with N sides" Tags="&lt;probability&gt;&lt;statistics&gt;&lt;physics&gt;&lt;applications&gt;" AnswerCount="3" />
  <row Id="8634" PostTypeId="2" ParentId="8632" CreationDate="2010-11-02T12:48:20.923" Score="5" ViewCount="" Body="&lt;p&gt;The first equality follows by definition. Now notice that $$\langle u_j, u_m \rangle = 0$$ if $i\neq j$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus the sum reduces to $$\frac{\langle v, u_m \rangle}{||u_m||^2}\langle u_m, u_m \rangle $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now what is the definition of the norm?&lt;/p&gt;&#xA;" OwnerUserId="2900" LastActivityDate="2010-11-02T12:48:20.923" CommentCount="2" />
  <row Id="8636" PostTypeId="2" ParentId="8633" CreationDate="2010-11-02T12:52:36.513" Score="1" ViewCount="" Body="&lt;p&gt;There are 2 ways to consider your question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The probabilistic way is just to say we have an unfair dice. This means we give probabilites $p_1$, $p_2$, ..., $p_6$. The conditions are that $p_1 + p_2 + \dots + p_6 = 1$ and that $p_i \geq 0$ for $i = 1, 2, ..., 6$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we assign &quot;Probability that the dice will show face number $i$&quot; is $p_i$. For example, a fair dice would have $p_1 = p_2 = \dots = p_6 = \frac{1}{6}$. An unfaire dice like you want (face 1 and 6 bigger, 2 and 5 normal, 3 and 4 smaller) could have $p_1 = \frac{1}{4}, p_2 = \frac{1}{6}, p_3 = \frac{1}{12}, p_4 = \frac{1}{12}, p_5 = \frac{1}{6}, p_6 = \frac{1}{4}$. This is the probabilist approach. You &lt;em&gt;decide&lt;/em&gt; the probabilities you want, and then you do some calculations (for example, you could be interested what is the distribution of the sum of 2 such dices, and so on).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The statistician has a dice and knows/supposes that it is unfair. He &lt;em&gt;cannot&lt;/em&gt; know the exact value $p_i$ for each face. Hovewer, he can try to guess thoses values and create tests to see if these values are probable or not. The usual way to guess the values is to throw the dice many times (say 1000 times) and to count of much 1's, 2's, ... we get. For example, if you count 100 1's, you would say that the probability of having a 1 could be $\frac{100}{1000} = \frac{1}{10}$. After that, you can build tests to verify this guess.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know if I answered your question. But it truly depends on what you want to do. According to your question, you have the unfair dice, and you want to guess the probabilities. This would more be the statistic approach. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means you have a model (like the one you give with $p_i = \frac{1}{n} + \sigma$) and you must create a test to estimate $\sigma$. Basically, you will have to throw the dice several times anyway.&lt;/p&gt;&#xA;" OwnerUserId="2778" LastActivityDate="2010-11-02T12:52:36.513" CommentCount="1" />
  <row Id="8637" PostTypeId="2" ParentId="8633" CreationDate="2010-11-02T12:54:53.423" Score="2" ViewCount="" Body="&lt;p&gt;There are two distinct questions posed here. &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;How to work with probabilities of dice that are biased in some way.  For an $n$ sided die this means that there are some numbers $p_1, p_2, \dots p_n$ with $p_1 + p_2 + \dots p_n = 1$, where $p_i$ is interpreted as the probability of landing on face $i$.  Then the standard methods of probability theory give recipes for calculating the probabilities of outcomes of random processes based on such dice: sums of several dice tosses, results of games, etc.  Calculations are easier for the case where all $p_i$ are equal -- known as a &quot;uniform distribution&quot; or &quot;fair dice&quot; -- but there is nothing fundamentally different about this case compared to that of unequal probabilities.  &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;How to compute the probabilities $p_i$ from the physics of a die and the dice-tossing process.  For coins there is some recent analysis of &quot;dynamical bias in coin tossing&quot; starting from the article with that title by Diaconis, Holmes and Montgomery.  For rectangular dice with uniform mass density (not loaded toward some faces), a simple and not necessarily physically accurate prescription is to assign each face a probability proportional to its surface area, or equivalently, probability equal to its share of the surface area of the entire die.  This can be viewed as the $k=0$ approximation to the probabilities for a thrown die that rolls onto at most $k$ additional faces after landing. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="467" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-11-02T16:02:52.117" LastActivityDate="2010-11-02T16:02:52.117" />
  <row Id="8638" PostTypeId="2" ParentId="8630" CreationDate="2010-11-02T13:06:36.980" Score="3" ViewCount="" Body="&lt;p&gt;You should find the formula $f(n)$ for the sum of digits for numbers from 1 to n, then do $f(B)-f(A)$.  It is not very clean.  The sum of digits of all k digit numbers, that is from $10^{(k-1)} \text { to } 10^k-1$, is $45*(10^k-1)/9$.  Stopping part way is harder, but you can do it by recursion.  If you want the sum of digits of numbers up to 7655, say, you can do sum of digits up to 999, plus 1000*6*5/2 (for the thousands digits up through 6) + 6*sum of digits up to 999 (for the last three digits for 1000 through 6999) + 7*656 (for the thousands in the 7s) + sum of digits up to 655.  The last has one less digit, so you just call your subroutine with that.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-11-02T13:06:36.980" CommentCount="3" />
  <row Id="8639" PostTypeId="1" AcceptedAnswerId="8641" CreationDate="2010-11-02T13:17:54.543" Score="0" ViewCount="82" Body="&lt;p&gt;What is the probability to get $n$ times the same element $k$ choosing randomly from a set $A$ knowing the cardinality $|A| = c$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you in advance,&lt;br&gt;&#xA;rubik&lt;/p&gt;&#xA;" OwnerUserId="2582" LastEditorUserId="822" LastEditorDisplayName="" LastEditDate="2010-11-02T15:06:51.400" LastActivityDate="2010-11-02T15:06:51.400" Title="Probability to get always the same number choosing randomly from a set of $c$ elements" Tags="&lt;probability&gt;&lt;homework&gt;" AnswerCount="2" />
  <row Id="8641" PostTypeId="2" ParentId="8639" CreationDate="2010-11-02T13:23:49.903" Score="2" ViewCount="" Body="&lt;p&gt;It depends: is $k$ fixed in advance? if so then the probability is given by $\frac{1}{c^n}$. If you just ask for the probability to have $n$ times the same element, then it is $\frac{1}{c^{n-1}}$ (since the first time you can have any object). By the way, I supposed you are doing exactly $n$ tries. Of course, the probability changes if you are doing more than $n$ tries.&lt;/p&gt;&#xA;" OwnerUserId="2778" LastEditorUserId="2778" LastEditorDisplayName="" LastEditDate="2010-11-02T13:52:26.183" LastActivityDate="2010-11-02T13:52:26.183" CommentCount="3" />
  <row Id="8642" PostTypeId="2" ParentId="8623" CreationDate="2010-11-02T13:42:50.310" Score="1" ViewCount="" Body="&lt;p&gt;After you edited your question, here is the new answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to solve this question, you should separate into different cases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, let's try to find all the 4-letter words that have different letters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Like Debanjan pointed out, this leads to 8*7*6*5 = 1680 words. The idea for this is just to consider you have 8 possibilities for the first letter, but for the second letter you only have 7 possibilities since it must be different from the first one, and so on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, let's count the word that have 2 &quot;M&quot;, but all other letters are different. You have 7 choices for another letter, and 6 for the last letter. However, the 2 &quot;M&quot; might be anywhere in the word, thus you multiply this by the number of permutation, which is given by $\frac{4!}{2!2!} = 6$ Thus, there are 7*6*6 = 252 such words.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[ EDIT: Here I need to explain something, as Ross commented out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I only count 252 such words and not 504, because in the 42 choices of letters, I can choose &quot;S&quot; and &quot;C&quot;, but I can also choose &quot;C&quot; and then &quot;S&quot;. So either you have to divide the 7*6 by 2, or you only allow permutations that don't change the order of the first other letter chosen and the second other letter chosen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sorry, I didn't want to explain this in details, but it is true it is needed, in order to avoid confusion. ]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are also 252 words with 2 &quot;A&quot; and all other letters different.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are also 252 words with 2 &quot;T&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, there are 6 words with 2 &quot;A&quot; and 2 &quot;M&quot;, 6 words with 2 &quot;A&quot; and 2 &quot;T&quot; and 6 words with 2 &quot;M&quot; and 2 &quot;T&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Total: 1680 + 252 + 252 + 252 + 6 + 6 + 6&lt;/p&gt;&#xA;" OwnerUserId="2778" LastEditorUserId="2778" LastEditorDisplayName="" LastEditDate="2010-11-02T14:44:53.227" LastActivityDate="2010-11-02T14:44:53.227" CommentCount="5" />
  <row Id="8643" PostTypeId="2" ParentId="8323" CreationDate="2010-11-02T13:50:06.387" Score="0" ViewCount="" Body="&lt;p&gt;My numerical work and computing skills are not to be trusted, but the first &quot;near miss&quot; that I recorded was $n=252544447$. &lt;/p&gt;&#xA;" OwnerUserId="940" LastActivityDate="2010-11-02T13:50:06.387" CommunityOwnedDate="2010-11-02T13:50:06.387" CommentCount="3" />
  <row Id="8644" PostTypeId="2" ParentId="8576" CreationDate="2010-11-02T13:53:06.450" Score="1" ViewCount="" Body="&lt;p&gt;As this question is now 14 hours old, I will give an answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First sum the total number of digits of all the natural numbers $ \le k-1.$ Call this sum $S.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are $9 \times 10^{i-1}$ natural numbers of length $i,$ so the total number of digits of  natural numbers of length $i$ is $9i \times 10^{i-1}.$ Hence&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ S=9(1+2.10 + 3.10^2 + \cdots + (k-1)10^{k-2}) = (k-1)10^{k-1} - \frac{10^{k-1}-1}{9}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the first natural number with $k$ digits is $10^{k-1}$ and so there are $n- 10^{k-1}+1$ natural numbers from $10^{k-1}$ to $n$ inclusive. Each of these numbers has $k$ digits and so the number of digits required to write the natural numbers from $1$ to $n$ is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$  k( n- 10^{k-1}+1) + S =  k( n- 10^{k-1}+1) + (k-1)10^{k-1} - \frac{10^{k-1}-1}{9}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ = k(n+1) - \frac{10^k-1}{9}.$$&lt;/p&gt;&#xA;" OwnerUserId="1301" LastActivityDate="2010-11-02T13:53:06.450" CommentCount="1" />
  <row Id="8645" PostTypeId="2" ParentId="8639" CreationDate="2010-11-02T13:59:28.307" Score="1" ViewCount="" Body="&lt;p&gt;Assuming that you draw $n$ times uniformly at random from the set of $c$ elements, it is $\frac{1}{c^{n-1}}$ as Djaian says. You are rolling a $c$-sided die.&lt;/p&gt;&#xA;" OwnerUserId="1527" LastActivityDate="2010-11-02T13:59:28.307" CommentCount="1" />
  <row Id="8646" PostTypeId="2" ParentId="8620" CreationDate="2010-11-02T14:04:54.690" Score="0" ViewCount="" Body="&lt;p&gt;If $R$ is a ring of order $p$, then it injects into the ring $\text{End}(\mathbb{Z}/p\mathbb{Z})$ of endomorphisms of the abelian group $\mathbb{Z}/p\mathbb{Z}$.    How many elements does this ring have?&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-11-02T14:20:56.830" LastActivityDate="2010-11-02T14:20:56.830" CommentCount="2" />
  <row Id="8647" PostTypeId="2" ParentId="8623" CreationDate="2010-11-02T14:13:18.903" Score="3" ViewCount="" Body="&lt;p&gt;Note: This answer presumes that the question is &quot;How many four-letter words can be formed using the letters of the word 'MATHEMATICS'.&quot;  So MAME should be counted as a word.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The provided answer, ${}_{11}P_4 / (2!)^3$, is not correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's temporarily change the question to &quot;How many distinct rearrangements are the of the letters of MATHEMATICS?&quot;, that is, how many 11-letter words can we form.  If the letters were all distinct, the answer would be 11!, that is, ${}_{11}P_11$.  But the letters are not all different, so we must adjust.  We can do so as in the original problem:  First, attach subscripts to identical letters, so that the letters are in fact all distinct, making 11! &quot;words&quot;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we need to erase subscripts.  First take the Ms.  There are two of them, M1 and M2, and swapping these does not change a &quot;word&quot; when we're ignoring subscripts.  So there are 11!/2 words if we ignore the subscripts on M.  Proceeding similarly, we find that there are $11!/(2*2*2)$ distinct words when all the subscripts are erased.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We need factorials when we have more than two of a letter.  Suppose there were three Xs.  Then there are 3! ways of rearranging the Xs, and all these ways lead to the same word when subscripts are erased.  So the adjustment would be to divide by 3!.  (Really this is happening with the Ms too, since 2! = 2.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general:  Suppose we have a total of $n$ letters, including $c_1$ copies of one letter, $c_2$ copies of the second letter, . . . , and $c_i$ copies of the $i^{\rm th}$ letter.  Then the total number of words we can make is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$n! \over { c_1!c_2!\ldots c_i! }$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice that this formula works even for letters that aren't replicated, since 1! = 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So far so good.  The trouble is that the original problem does not ask for words formed from &lt;em&gt;all&lt;/em&gt; the letters, just from four at a time.  And here the trick doesn't work.  We do need to adjust, but we can't simply divide by 2! when there are two copies of a letter, because some of the words don't contain all the copies of that letter!  The adjusting trick, as described, works only when all copies of the letter are necessarily in the word, i.e., when each word contains all letters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To see this even more clearly, suppose I had asked for words of &lt;em&gt;two&lt;/em&gt; letters formed from the letters of MATHEMATICS.  Would the answer be ${}_{11}P_2/(2!)^3$?&#xA;Certainly not, since this expression is not even an integer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a way of phrasing the question so that the provided answer is correct?  I don't see it, and I find it doubtful anyway, for the same reason as in the preceding paragraph.&lt;/p&gt;&#xA;" OwnerUserId="2682" LastEditorUserId="2682" LastEditorDisplayName="" LastEditDate="2010-11-02T14:55:04.923" LastActivityDate="2010-11-02T14:55:04.923" />
  <row Id="8648" PostTypeId="1" CreationDate="2010-11-02T14:14:07.610" Score="-3" ViewCount="101" Body="&lt;p&gt;explain $\langle \sum_{j=1}^n \langle v,u_j \rangle u_j, \sum_{k=1}^n \langle w,u_k \rangle u_k \rangle =\sum_{j=1}^n \langle v,u_j \rangle \sum_{k=1}^n \langle u_j, \langle w, u_k \rangle u_k \rangle$, V is complex inner product space with orthonormal basis $R={u_1, u_2,..., u_n}$&lt;/p&gt;&#xA;" OwnerUserId="2723" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-02T14:40:19.720" LastActivityDate="2010-11-02T20:17:44.127" Title="explain below part of the proof of Parseval's relation" Tags="&lt;analysis&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="8649" PostTypeId="2" ParentId="8648" CreationDate="2010-11-02T14:32:49.783" Score="2" ViewCount="" Body="&lt;p&gt;The inner product is sesquilinear, and each $\langle v,u_j\rangle$ and $\langle w,u_k\rangle$ are scalars. Rewrite $\alpha_j = \langle v,u_j\rangle$ and $\beta_k = \langle w,u_k\rangle$. Then the left hand side is&#xA;$$\left\langle \sum_{j=1}^n\langle v,u_j\rangle u_j, \sum_{k=1}^{n}\langle w,u_k\rangle u_k\right\rangle = \left\langle \sum_{j=1}^n \alpha_ju_j,\sum_{k=1}^n\beta_k u_k\right\rangle.$$&#xA;So by additivity and homogeneity in the first component you have:&#xA;\begin{eqnarray*}&#xA;\left\langle \sum_{j=1}^n\langle v,u_j\rangle u_j, \sum_{k=1}^n\langle w,u_k\rangle u_k\right\rangle &amp;amp; = &amp;amp; \left\langle \sum_{j=1}^n \alpha_ju_j,\sum_{k=1}^n\beta_k u_k\right\rangle\\&#xA;&amp;amp; = &amp;amp; \sum_{j=1}^{n}\left\langle \alpha_j u_j, \sum_{k=1}^n\beta_k u_k\right\rangle\\&#xA;&amp;amp; = &amp;amp; \sum_{j=1}^n\left( \alpha_j\left\langle u_j, \sum_{k=1}^n\beta_k u_k\right\rangle\right).&#xA;\end{eqnarray*}&#xA;Then applying linearity on the second component you get the given expression.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-02T20:17:44.127" LastActivityDate="2010-11-02T20:17:44.127" CommentCount="3" />
  <row Id="8650" PostTypeId="1" AcceptedAnswerId="8666" CreationDate="2010-11-02T15:10:15.733" Score="6" ViewCount="172" Body="&lt;p&gt;what is the number of permutations of a set say {1, 2, 3, 4, 5} that keep at least one integer fixed ?&lt;/p&gt;&#xA;" OwnerUserId="3027" LastEditorUserId="3027" LastEditorDisplayName="" LastEditDate="2010-11-02T15:30:57.347" LastActivityDate="2010-11-02T17:00:59.453" Title="Permutations of a set that keep at least one integer fixed" Tags="&lt;combinatorics&gt;&lt;permutations&gt;" AnswerCount="1" CommentCount="6" />
  <row Id="8651" PostTypeId="1" AcceptedAnswerId="8655" CreationDate="2010-11-02T15:25:13.740" Score="4" ViewCount="297" Body="&lt;p&gt;The 2nd part of the &quot;Fundamental Theorem of Calculus&quot; has never seemed as earth shaking or as &lt;strong&gt;fundamental&lt;/strong&gt; as the first to me. Why is it &quot;fundamental&quot; -- I mean, the &lt;em&gt;mean value theorem&lt;/em&gt;, and the &lt;em&gt;intermediate value theorems&lt;/em&gt; are both pretty exciting by comparison. And after the joyful union of integration and the derivative that we find in the first part, the 2nd part just seems like a yawn. So, what am I missing?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To be clear I'm talking about this:&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Let ƒ be a real-valued function defined on a closed interval [a, b] that admits an antiderivative F on [a,b]. That is, ƒ and F are functions such that for all x in [a, b],&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f(x) = F'(x)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If ƒ is integrable on [a, b] then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\int_a^b f(x)dx = F(b) - F(a).$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I've been through the proof a few times. It makes sense to me.  But, it didn't help me to see the light. To me it just looks like &quot;OK here is how you do the definite integral.&quot; Which dosen't seem like such a big deal, especially when indefinite integrals can be more interesting. &lt;/p&gt;&#xA;" OwnerUserId="2563" LastEditorUserId="2563" LastEditorDisplayName="" LastEditDate="2010-11-02T15:31:52.220" LastActivityDate="2011-04-07T15:25:15.657" Title="The 2nd part of the &quot;Fundamental Theorem of Calculus.&quot;" Tags="&lt;calculus&gt;" AnswerCount="3" CommentCount="4" />
  <row Id="8652" PostTypeId="1" CreationDate="2010-11-02T15:40:33.820" Score="17" ViewCount="1834" Body="&lt;p&gt;I'm a senior undergrad at a top-ish(say, top 15) math school. I'm a solid, not stellar, student. This year I'm taking the qualifying exam grad courses in algebra and analysis and have been taken aback by the &quot;pressure cooker&quot; atmosphere among grad students here. That is, even moreso than in the undergraduate program.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I'm self driven, could going to a &quot;less prestigious&quot; school afford me more space(I mean in a psychological sense) to produce a more solid contribution to math? By &quot;less prestigious&quot;, I mean a school &quot;ranked&quot; significantly lower than the range of schools that I could comfortably get into. For me, &quot;less prestigious&quot; would be ranked around 40-60 on, say, USNews or NRC.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My reasoning is that at such a school, I would be more able to learn the fundamentals at my own pace, as opposed to a pace dictated to me by the program. I know I want to do math, and I think my learning style may be better suited to going at my own pace. Thoughts?&lt;/p&gt;&#xA;" OwnerUserId="3030" LastActivityDate="2010-11-04T04:49:05.350" Title="Does it ever make sense NOT to go to the most prestigious graduate school you can get into?" Tags="&lt;soft-question&gt;" AnswerCount="9" CommentCount="9" FavoriteCount="10" />
  <row Id="8653" PostTypeId="2" ParentId="8651" CreationDate="2010-11-02T15:42:17.037" Score="0" ViewCount="" Body="&lt;p&gt;This is the part of the fundamental theorem that allows you to &lt;em&gt;compute&lt;/em&gt; integrals; then you can compute areas, and with more theory even volumes, surfaces and so on. Exciting enough?&lt;/p&gt;&#xA;" OwnerUserId="2575" LastActivityDate="2010-11-02T15:42:17.037" CommentCount="3" />
  <row Id="8654" PostTypeId="1" AcceptedAnswerId="8657" CreationDate="2010-11-02T15:43:38.890" Score="5" ViewCount="202" Body="&lt;p&gt;Dear MSE-Community,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please let me summarize the method by which L. Euler solved the Basel Problem and how he found the exact value of $\zeta(2n)$ up to $n=13$. Euler used the infinite product&lt;br&gt;&#xA;$$ &#xD;&#xA;\displaystyle   f(x) = \frac{\sin(x)}{x} = \prod_{n=1}^{\infty} \Big(1-\frac{x^2}{n^2\pi^2}\Big) , &#xD;&#xA;$$ &#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Newton%27s_identities&quot; rel=&quot;nofollow&quot;&gt; Newton's identities&lt;/a&gt; and the (Taylor) &lt;a href=&quot;http://www.wolframalpha.com/input/?i=sin(x)/x&quot; rel=&quot;nofollow&quot;&gt; Series Expansion&lt;/a&gt; (at $x=0$) of the sine function divided by $x$ to arrive at &#xA;$$ &#xD;&#xA;1 - \frac{x^2}{\pi^2} \cdot (1 + \frac{1}{4} + \frac{1}{9} + \frac{1}{16} + ... + \frac{1}{n^2}) + x^4(...) = 1 - \frac{x^2}{6} + \frac{x^4}{120} - ... &#xD;&#xA;$$&#xA;Upon subtracting 'one' from both sides, equating the $x^2$ terms to each other and multiplying both sides by $ - \pi^2$, one finds that &#xA;$$&#xD;&#xA;\zeta(2)=\frac{\pi^2}{6}.&#xD;&#xA;$$&#xA;When I first saw this proof and the way it was extended to find the values of the other even zeta-constants, I couldn't help myself thinking: &quot;How could this method be strenghtened to find the values of the &lt;em&gt;odd&lt;/em&gt; zeta-constants?&quot; (And, a little while later, &quot;why hasn't this been done before?&quot;) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I started looking for a similar-looking infinite product, only now I focussed on one of the form &#xA;$$ &#xD;&#xA;\displaystyle   f(x) = \prod_{n=1}^{\infty} \Big(1-\frac{x^a}{k^3 \cdot q}\Big) &#xD;&#xA;$$ &#xA;(for some $ a \in \mathbb{N} , q \in \mathbb{R} $). A little while later I stumbled upon &lt;a href=&quot;http://mathworld.wolfram.com/InfiniteProduct.html&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; website and fixated my eyeballs on equation (27). If we take $n=3$, Prudnikov et al. tell us that &#xA;$$ &#xD;&#xA;\prod_{n=1}^{\infty} \Big(1-\frac{x^3}{k^3}\Big) = - \frac{1}{x^3} \cdot \prod_{k=1}^{2} \frac{1}{\Gamma(-e^{2/3 \pi i k} \cdot x)}. &#xD;&#xA;$$&#xA;Now, I thought that if we could use Newton's Identities again on the left side of the equation and find out what the Taylor Series Expansion of the right-hand side would be, we could find out what the exact value of Apery's Constant and other odd zeta-constants would be. In this &lt;a href=&quot;http://math.stackexchange.com/questions/8522/what-is-the-series-expansion-of-the-following-function&quot;&gt;answer&lt;/a&gt; by Robert Smith, I was told the Series Expansion. So we have &#xA;$$ &#xD;&#xA;1 - x^3(1 + \frac{1}{8} + \frac{1}{27} + ... + \frac{1}{n^3}) = -1 - 2 \cdot \gamma x - 2 \gamma^2 x^2 + \frac{1}{6}x^3(-8\gamma^3 - \psi^{(2)} (1)) - x^4(...)&#xD;&#xA;$$&#xA;Notice that on the left side we only have 'one minus a term with an $x^3$ coefficient', while on the other side we see 'minus one plus $x$, $x^2$, $x^3$ coefficients with their terms'. This is important, because it probably answers the question why the following will not work, but I don't know why and I really would like to know.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I guess you know what I will attempt to do now. We equate the $x^3$ terms with each other, set $x=1$, multiply by minus one and 'find' that &#xA;$$&#xD;&#xA;\zeta(3) = \frac{1}{6}(8\gamma^3 + \psi^{(2)} (1)). &#xD;&#xA;$$&#xA;By combining this with the already known result &#xA;$$ &#xD;&#xA;\zeta(3) = -\frac{1}{2} \psi^{2}(1),&#xD;&#xA;$$&#xA;we 'find' that &#xA;$$&#xD;&#xA;\zeta(3) &amp;#39;=&amp;#39; \gamma^3.&#xD;&#xA;$$&#xA;Obviously, this is wrong. Apery's constant is larger than one, and this value is clearly smaller than one. Could someone please elaborate one &lt;em&gt;where&lt;/em&gt; I went wrong? And does anybody have any sugguestions and/or ideas related to the discussion from above using which we could find &quot;better&quot; values for Apery's Constant and the other odd zeta constants? (For example by pointing out a similar infinite product relation, and by showing that that infinite product has a nicer Series Expansion?) Or could someone point out to me why this approach to finding nicer closed-form representations for these constants clearly won't lead to any results?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Max Muller&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Moderators: If you find any spelling mistakes or errors grammar errors, feel free to correct them. To the rest: $\gamma$ is the Euler-Mascheroni Constant, and it amounts to approximately $0.5772$. The $\psi^{(2)}(x)$ stands for the second logarithmic deriviative of the Gamma-function. As usual, Wikipedia is a pretty good reference for this sort of things.)&lt;/p&gt;&#xA;" OwnerUserId="1055" LastEditorUserId="6179" LastEditorDisplayName="" LastEditDate="2011-05-18T14:28:12.853" LastActivityDate="2011-05-19T12:31:13.897" Title="Why is the following evaluation of Apery's Constant wrong and do you have suggestions on how, if at all, this method could be improved?" Tags="&lt;sequences-and-series&gt;&lt;analysis&gt;&lt;analytic-number-theory&gt;" AnswerCount="1" CommentCount="10" FavoriteCount="1" />
  <row Id="8655" PostTypeId="2" ParentId="8651" CreationDate="2010-11-02T15:49:27.373" Score="9" ViewCount="" Body="&lt;p&gt;It's natural that the Fundamental Theorem of Calculus has two parts, since morally it expresses the fact that differentiation and integration are mutually inverse processes, and this amounts to two statements: (i) integrating and then differentiating and (ii) differentiating and then integrating get us (essentially) back where we started.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, many people have noticed that the two parts are not completely independent: e.g. if $f$ is continuous, then (ii) follows easily from (i).  However, for discontinuous -- but Riemann integrable -- $f$, the theorem still holds, and this is what requires a nontrivial additional argument.  See page 8 of&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://math.uga.edu/~pete/243integrals1.pdf&quot; rel=&quot;nofollow&quot;&gt;http://math.uga.edu/~pete/243integrals1.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for some discussion of this point.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can't tell from your question how squarely this answer addresses it.  If yes, and you have further concerns, please let me know.  &lt;/p&gt;&#xA;" OwnerUserId="299" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2011-04-07T15:25:15.657" LastActivityDate="2011-04-07T15:25:15.657" CommentCount="1" />
  <row Id="8657" PostTypeId="2" ParentId="8654" CreationDate="2010-11-02T16:08:42.200" Score="6" ViewCount="" Body="&lt;p&gt;Your previous question was about the wrong function.  Instead of $\Gamma(x)$ in the denominator you should have $\Gamma(-x)$.  If you fix this you'll probably end up with the same polygamma identity you already knew.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In any case, proving anything about $\zeta(2k+1)$ is known to be quite hard.  If anything simple worked, Euler would have done it, or somebody in the last few centuries anyway.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me also mention that Prudnikov's identity is, if you are willing to accept Euler-style manipulations, trivial.  It is equivalent to a product formula for $\frac{1}{\Gamma(x)}$ which follows (again if you are willing to accept Euler-style manipulations) from an investigation of its roots and does not really tell you anything deep about zeta values.&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-11-02T16:15:45.773" LastActivityDate="2010-11-02T16:15:45.773" CommentCount="6" />
  <row Id="8658" PostTypeId="1" AcceptedAnswerId="8661" CreationDate="2010-11-02T16:25:51.580" Score="3" ViewCount="210" Body="&lt;p&gt;$f(x) = \frac{1}{\cos x}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f'(x) = \frac{\sin(x)}{\cos^2(x)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f''(x) = \frac{2\sin^2(x)+\cos^2(x)}{\cos^3(x)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f^{(3)}(x) = \frac{6\sin^3(x)+5\cos^2(x)\sin(x)}{cos^4(x)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\vdots$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f^{(n)}(x) = \frac{ ?}{cos^{n+1}(x)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some of these are easy: &lt;a href=&quot;http://darkwing.uoregon.edu/~jcomes/251exn.pdf&quot; rel=&quot;nofollow&quot;&gt;http://darkwing.uoregon.edu/~jcomes/251exn.pdf&lt;/a&gt;&#xA;Others are not. Why?&lt;/p&gt;&#xA;" OwnerUserId="2563" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-03T03:08:00.100" LastActivityDate="2010-11-03T03:08:00.100" Title="$f(x) = \frac1{\cos x}$ then $f^{(n)}(x)$ is ...?" Tags="&lt;calculus&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="8659" PostTypeId="2" ParentId="8652" CreationDate="2010-11-02T16:38:55.017" Score="25" ViewCount="" Body="&lt;p&gt;I think that there are several important points.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, you will learn just as much from your fellow&#xA;graduate students, especially in the initial years, than&#xA;you will from your professors, and far more from your&#xA;fellow students if there is a good atmosphere for it. Thus,&#xA;I suggest that you try to land at a school with a&#xA;supportive and studious student environment. You can visit&#xA;the places to find out, and talk to graduate students&#xA;there. From this perspective, it is very helpful to be at a&#xA;place that has many talented graduate students---they will&#xA;help you rise to their level (or you will all rise&#xA;together). Since the top schools tend to have stronger&#xA;students, this can be a good reason to go to a top school&#xA;when it is possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second, most of the pressure on graduate students is&#xA;self-imposed, instantiating their drive to do well&#xA;mathematically. Every school, including the top programs,&#xA;have some students that proceed at a different pace. So you&#xA;can often resist whatever external pressure you imagine is&#xA;there. (One important exception to this is at a school&#xA;where financial support might be withdrawn for slow&#xA;progress---so look into that at the places in which you are&#xA;interested.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Third---and actually I find this to be the most important&#xA;point---you shouldn't look at the school only and make a&#xA;such an important life decision based only on mathematics&#xA;and prestige. Rather, look at all aspects of how the choice&#xA;of a school will affect your life. You must consider the&#xA;city and region as well. For example, do you prefer city&#xA;living or country living? If you like the mountains, hiking&#xA;and snow, then don't go to Florida, and similarly, if big&#xA;city living is your preference, then don't go to small&#xA;town. I find that this kind of consideration is oddly often&#xA;neglected among mathematics students.&lt;/p&gt;&#xA;" OwnerUserId="413" LastActivityDate="2010-11-02T16:38:55.017" CommentCount="2" />
  <row Id="8660" PostTypeId="2" ParentId="8652" CreationDate="2010-11-02T16:41:12.643" Score="15" ViewCount="" Body="&lt;p&gt;There are circumstances that could lead one to be admitted to more prestigious graduate program X and less prestigious graduate program Y and choose to attend Y.  For instance, there was a graduate student at UGA a few years ago who was doing well in the program but at some point revealed to me that he had to care for a sick family member in a different state, so moved to that state and enrolled in a distinctly less prestigious program there.  Of course I don't disagree with his decision -- indeed I admire it -- but there was no question that this was being done for higher reasons than his mathematical well-being and future career.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Treating this question as advice, I would say &lt;strong&gt;No&lt;/strong&gt; -- you should attend the most prestigious graduate program you can get into.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, graduate programs are not linearly ranked: i.e., holding up two graduate programs to each other, one may not be able to identify one as more prestigious than the other.  When this is possible, it is usually the case that one is &lt;em&gt;significantly&lt;/em&gt; more prestigious.  So, to refine the advice of the previous paragraph: you should choose a maximal element of the set of graduate schools you get admitted to, partially ordered by prestige.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Included in this are considerations such as the following: you strongly believe that you want to work with a particular faculty member at a program which is overall less prestigious than some other.  This is a tricky decision that I don't want to enter into here in detail, except to say: if there is any doubt in your mind that you will complete a PhD under Professor X, be very skeptical about trading more than a little in the prestige of the school.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the OP's case, s/he seems to want to choose a non-maximal element in order to take some of the pressure off.  Although I understand the sentiment, I advise strongly against this.  Two points:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Graduate school is a high-pressure, hard-slog situation for everyone.  It has to be.  If you're not fully prepared for that, perhaps you're not fully prepared for a PhD program: think about it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Much of graduate school &lt;em&gt;does&lt;/em&gt; proceed at your own pace.  However, the quality of the students and faculty in your program sets an example and a standard, in my experience the higher the better.  &lt;/p&gt;&#xA;" OwnerUserId="299" LastActivityDate="2010-11-02T16:41:12.643" />
  <row Id="8661" PostTypeId="2" ParentId="8658" CreationDate="2010-11-02T16:45:03.753" Score="10" ViewCount="" Body="&lt;p&gt;This is asking for the $n$-th derivative of the secant function.&#xA;As the derivative of $\sec$ is $\sec\tan$ and that of $\tan$ is&#xA;$\sec^2=1+\tan^2$ then the $n$-th derivative of $\sec$ is&#xA;$\sec f_n(\tan)$ where $f_0(t)=1$ and $f_{n+1}(t)=tf_n(t)+(t^2+1)f_n'(t)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's probably too much to hope to find a nice formula for the&#xA;coefficients of the $f_n$. The constant coefficients of $f_{2m}$&#xA;are essentially the Euler numbers.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-11-02T16:45:03.753" />
  <row Id="8662" PostTypeId="2" ParentId="8652" CreationDate="2010-11-02T16:50:11.197" Score="10" ViewCount="" Body="&lt;p&gt;I really appreciate this question. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many people still hold the out-dated belief that you learn mathematics in the classroom. Sure, you may learn classical formulas, the statement of famous theorems and some folklore surrounding its solver or length of openness (if there is none), but you'll learn &quot;how to be a mathematician&quot; at home or in the library (with [insert your favorite book here]) or &lt;strong&gt;alone&lt;/strong&gt; at the board wrestling with problems. This is the time when you develop your problem solving skills and gauge your limitations. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is nice about the better schools is having a (sound) sounding board for your ideas. If your colleagues are sharp, they will provide quick and clean jugular shots to your questions. However, sometimes it's better to think about questions on your own time and settle them yourself with no interference. This simple and time-honored exercise develops your &quot;internal compass&quot; much more than having access to slick solutions from others. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remember, it is sometimes (perhaps often) the case that we do not have the best vantage point to judge the right course of action in a given (perhaps crucial) point in our lives.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is what I think: &lt;em&gt;Postpone your question for a little while. Go to the best school that you can get into, and find an advisor that suits you and your tempo. Be honest with him/her about your strengths and weaknesses. Develop an on-going dialogue with your colleagues. Work hard but don't go crazy doing so. Ask questions if you get stuck, but not too many. Be respectful. Listen to people that you hold in high regard (and some that you may not). Think for yourself. Find an interesting problem. Write a good thesis. Graduate. Return to the issue of &quot;solid contribution&quot; when you have a better view of the landscape of your chosen field.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="2720" LastEditorUserId="2720" LastEditorDisplayName="" LastEditDate="2010-11-04T04:49:05.350" LastActivityDate="2010-11-04T04:49:05.350" CommentCount="2" />
  <row Id="8663" PostTypeId="2" ParentId="8633" CreationDate="2010-11-02T16:50:16.397" Score="1" ViewCount="" Body="&lt;p&gt;Answering to:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I initially started writing a software program that would simulate uniformly random dice rolls, but then thought about how that wouldn't be realistic because dice get tumbled and smoothed before reaching the consumer (which is why some people have &quot;favorite dice&quot; because they roll 20s often). I realized that I could approximate and fake out the realism, but I got ridiculously curious about how it all actually worked. So, to answer, I'm not measuring real physical objects, but trying to determine how to realistically simulate the rolling of virtual dice.&lt;/em&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you could do is the following. Start with a perfect dice ($p_1 = p_2 = \dots = p_6$). Then, each time you roll it, you can alter the probabilities $p_i$ with an error $\varepsilon_i$. Thus, the probabilities would not be the same in the future. This would simulate the damaging of the dice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Algorithm:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Start with $p_1 = p_2 = \dots = p_6 = \frac{1}{6}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Generate 5 random errors $\varepsilon_1, \varepsilon_2, \dots, \varepsilon_5$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Compute $ \varepsilon_6 = - \varepsilon_1 - \varepsilon_2 - \varepsilon_3 - \varepsilon_4 - \varepsilon_5$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Set new probabilities to $p_1 + \varepsilon_1, p_2 + \varepsilon_2, \dots, p_5 + \varepsilon_5, p_6 + \varepsilon_6$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Roll the dice with this new probabilities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Compute new errors...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I strongly suggest that you choose the way of generating the $\varepsilon_i$ so that they will always be 0 or as near to 0 as possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For fine tuning purpose, it would be good if it was not always $p_6$ that gets the difference of all other errors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This method can easily be adjusted to dices with any number of faces.&lt;/p&gt;&#xA;" OwnerUserId="2778" LastActivityDate="2010-11-02T16:50:16.397" CommentCount="2" />
  <row Id="8664" PostTypeId="2" ParentId="8658" CreationDate="2010-11-02T16:51:11.867" Score="1" ViewCount="" Body="&lt;p&gt;That depends on what you mean by &quot;easy.&quot;  As far as the examples in the .pdf you link to, I claim that the following are true for any reasonable definition of &quot;easy&quot;:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;It is easy to compute the iterated derivatives of powers and logarithms.  &lt;/li&gt;&#xA;&lt;li&gt;It is easy to compute the iterated derivatives of solutions to homogeneous linear ODEs.&lt;/li&gt;&#xA;&lt;li&gt;It is easy to compute the iterated derivatives of $x^n f(x)$ if it is easy to compute the iterated derivatives of $f(x)$.&lt;/li&gt;&#xA;&lt;li&gt;It is easy to compute the iterated derivatives $f(x) + g(x)$ if it is easy to compute the iterated derivatives of $f(x)$ and of $g(x)$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I think that covers all of them.  Otherwise, there is no general reason to expect that it should be easy to compute iterated derivatives.  Even at $x = 0$ they may define complicated and highly nontrivial sequences, for example the &lt;a href=&quot;http://en.wikipedia.org/wiki/Bernoulli_number&quot; rel=&quot;nofollow&quot;&gt;Bernoulli numbers&lt;/a&gt; (which are related to this problem).&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-11-02T16:51:11.867" />
  <row Id="8665" PostTypeId="2" ParentId="8561" CreationDate="2010-11-02T16:58:24.710" Score="0" ViewCount="" Body="&lt;p&gt;It is known that (a+b)!/a!b! represents the number of combinations of (a + b) elements, taken &quot;b to b&quot;. Therefore, it is an integer.&lt;/p&gt;&#xA;" OwnerUserId="2938" LastActivityDate="2010-11-02T16:58:24.710" CommentCount="1" />
  <row Id="8666" PostTypeId="2" ParentId="8650" CreationDate="2010-11-02T17:00:59.453" Score="8" ViewCount="" Body="&lt;p&gt;This is equal to $n!$ minus the number of permutations which keep no elements fixed.  These are known as &lt;a href=&quot;http://en.wikipedia.org/wiki/Derangement&quot; rel=&quot;nofollow&quot;&gt;derangements&lt;/a&gt; and you can find all the relevant formulas at the Wikipedia article.  You can count them, for example, using inclusion-exclusion.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you just want the number of derangements of an $n$-element set, then a nice way to compute it is to round $\frac{n!}{e}$ to the nearest integer.  For example, when $n = 5$ we want to round $\frac{120}{e} \approx 44.1$, so there are $44$ derangements, hence $120 - 44 = 76$ permutations which fix at least one element.  &lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-11-02T17:00:59.453" CommentCount="1" />
  <row Id="8667" PostTypeId="1" AcceptedAnswerId="8728" CreationDate="2010-11-02T17:05:42.457" Score="10" ViewCount="154" Body="&lt;p&gt;Suppose $X$ is a space and $A_1\subseteq A_2\subseteq A_3\subseteq ...\subset X$ is a sequence of subspaces each of which is closed in $X$ and such that $X\cong \varinjlim_{n}A_n$ (i.e. $U$ is open in $X$ if and only if $U\cap A_n$ is open in $A_n$ for each $n$). This topology on $X$ has many names (direct limit, inductive limit, weak topology, maybe more) but I can't seem to find much dealing with separation properties in this general setting. Specifically, I am asking:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $A_n$ is Hausdorff for each $n$, then must $X$ also be Hausdorff?&lt;/p&gt;&#xA;" OwnerUserId="2423" LastActivityDate="2010-11-03T19:16:50.850" Title="Separation in direct limits of closed inclusions" Tags="&lt;general-topology&gt;" AnswerCount="1" FavoriteCount="4" />
  <row Id="8668" PostTypeId="2" ParentId="8652" CreationDate="2010-11-02T17:14:25.197" Score="5" ViewCount="" Body="&lt;p&gt;There are many good comments above, I'd like to add a few other points. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Prestige is a fairly abstract, borderline meaningless word.  It's meant to be an amalgamam of other, &lt;em&gt;actually important&lt;/em&gt; ideas.  And those concepts are the things one really ought to pay attention to.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, when you go to grad school, one thing to keep in mind is &quot;will university X give me the time it takes to do substantial research?&quot;.  It's a good thing to go to a school that can promise you long-term funding with enough teaching work so that you get familiar with the process, but not so much that it distracts substantially from your graduate studies.  Depending on how strictly you use this criterion, you'll find this may narrow down the list of potential graduate schools to something pretty close to the traditional &quot;prestigious&quot; university list.  The above is concerned with the component of prestige associated to &quot;does the university have ample research funding for grad students?&quot; Even among the &quot;top&quot; universities you'll find this varies pretty widely. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another thing to keep in mind is much of the point of going to grad school is to be exposed to as many ideas from diverse areas of mathematics as possible, as most undergraduate programs rarely get beyond the 19th century in mathematical understanding. Smaller universities tend to have less diverse faculty members, less grad students and postdocs, sometimes their teaching load is higher.  The net effect is that if you're at a smaller, more isolated place you're less likely to go to some wonderful talk on a random topic you didn't know anything about.  It's amazing how quickly one can pick up new ideas by going to good random talks, ideas that affect how you look at other subjects, the interconnections between fields, applications to other areas of life, an outsider's perspective on &quot;your&quot; field, and so on.  So going to a &quot;big&quot; place is very useful in that regard. So this is the component of prestige associated to being a large department/university. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are of course other components to prestige, but the above are two fairly substantial ones.&lt;/p&gt;&#xA;" OwnerUserId="642" LastEditorUserId="642" LastEditorDisplayName="" LastEditDate="2010-11-02T19:23:08.137" LastActivityDate="2010-11-02T19:23:08.137" />
  <row Id="8669" PostTypeId="2" ParentId="8652" CreationDate="2010-11-02T17:17:44.173" Score="4" ViewCount="" Body="&lt;p&gt;I think realistically people are going to be trying really hard at even a mid-level PhD program; they just will not have gotten into better schools. They might not be &lt;em&gt;as&lt;/em&gt; intense as those at higher-ranked places, but I seriously doubt things will be at all slow-paced. And I think if you are better prepared than them you may have trouble relating to the other students when going through the various travails of graduate school, which may make it harder to bond with them. And if they're not as strong they won't be useful people to learn from. Not to mention the result of having a thesis advisor at a lesser place, and the greater difficulty being on the job market from a lower-ranked place.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My suggestion would be to go to one of the best places you get into, and then try to work your life around being a grad student there. Find friends who are less miserable, try to not get too stressed out over exams, and so on. Also, once exams are over, no matter where you go things tend not to be that stressed. You just are more acquainted with the ones who are going through that process.&lt;/p&gt;&#xA;" OwnerUserId="3035" LastActivityDate="2010-11-02T17:17:44.173" CommentCount="5" />
  <row Id="8670" PostTypeId="2" ParentId="8620" CreationDate="2010-11-02T17:28:42.347" Score="0" ViewCount="" Body="&lt;p&gt;$\rm R = \{0,a,2a,\cdots,(p-1)a\}\ $ for $\rm\ 0 \ne a\in R\:$. So $\rm\ a^2 = k\:a\ $ and $\rm\ k \ne 0\:$ via multiplication nontrivial. Now $\rm\ j = 1/k\ \:mod\ p\ \Rightarrow\ ia\ ja\ = \ ijka\ =\ ia\ \Rightarrow\ ja = 1\:$.&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-11-02T17:28:42.347" />
  <row Id="8671" PostTypeId="2" ParentId="8652" CreationDate="2010-11-02T17:47:16.517" Score="7" ViewCount="" Body="&lt;p&gt;Maybe you would like an answer from someone who is there. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I went to undergrad at a &quot;Top 100&quot; school. I did extremely well there, but utterly failed to get into the level of grad schools I was aiming at, and am now a grad student at a &quot;Top 50&quot; school. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can only answer based on what I've seen at these schools, so maybe things are different elsewhere. But to me, the situation is kind of sad here. The students know so little and don't seem to really have the right attitude. I think I'm missing a big component of interacting with peers who are learning the same material. I can honestly say I've never had any sort of study session or even just a discussion about mathematics with fellow students that was productive for me. (Although this may more reflect my experience at my undergrad school... I mostly talked to the grad students there, and this was true there, but I'm still new at my graduate school and don't know many people, so maybe this will change.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've never been to one of these prestigious schools, but seeing people online talk about them, it seems like a completely different world. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the one hand, it's nice because I can stand out here. But I can stand out here without even trying very hard. I think this leads to more of a complacent attitude on my part, which is not good. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wonder, what would happen if I were at a top 10 school? I honestly don't know. I might get massacred. I certainly have a much weaker background than entering students at those schools seem to have. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's also the career issue. As far as I can tell, very few students graduating here end up in a tenure track job at the kind of university I would like to be at. It seems like a big challenge to overcome. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's not all negative. I suppose it is nice to not have so much pressure. There are some strong professors who do interesting work I might like to work with. But if you're ambitious (and why get a PhD in math if you're not ambitious?), I wouldn't recommend coming to this type of school if you have other options. &lt;/p&gt;&#xA;" OwnerUserId="3036" LastActivityDate="2010-11-02T17:47:16.517" CommentCount="1" />
  <row Id="8672" PostTypeId="1" AcceptedAnswerId="8700" CreationDate="2010-11-02T17:51:45.140" Score="5" ViewCount="744" Body="&lt;p&gt;Hello.&lt;br&gt;&#xA;Lets say I have a 2 x 2 Matrix (Actually &lt;a href=&quot;http://en.wikipedia.org/wiki/Structure_tensor&quot; rel=&quot;nofollow&quot;&gt;Structure Tensor&lt;/a&gt; of a Discrete Image - I):&#xA;$$ \begin{bmatrix}&#xD;&#xA; \frac{\partial I}{\partial x}\frac{\partial I}{\partial x} &amp;amp; \frac{\partial I}{\partial x}\frac{\partial I}{\partial y} \\\ &#xD;&#xA;    \frac{\partial I}{\partial y}\frac{\partial I}{\partial x} &amp;amp; \frac{\partial I}{\partial y}\frac{\partial I}{\partial y}&#xD;&#xA;\end{bmatrix}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It has 2 properties:  &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Symmetric.&lt;/li&gt;&#xA;&lt;li&gt;Positive Semidefinite.  &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Given those properties, What would be the easiest method to compute (Numerical) its Eigen Vectors (Orthogonal) and Eigen Values?&lt;/p&gt;&#xA;" OwnerUserId="33" LastEditorUserId="33" LastEditorDisplayName="" LastEditDate="2010-12-25T09:57:56.743" LastActivityDate="2010-12-25T09:57:56.743" Title="Eigenvalues and Eigenvectors of 2 x 2 Matrix" Tags="&lt;linear-algebra&gt;&lt;matrices&gt;" AnswerCount="3" CommentCount="5" />
  <row Id="8673" PostTypeId="2" ParentId="8672" CreationDate="2010-11-02T18:03:44.650" Score="5" ViewCount="" Body="&lt;p&gt;Since you work with a $2\times2$ matrix, the corresponding &lt;a href=&quot;http://en.wikipedia.org/wiki/Characteristic_polynomial&quot; rel=&quot;nofollow&quot;&gt;characteristic  polynomial&lt;/a&gt; is quadratic so the eigenvalues can be expressed in closed form in terms of the matrix elements.&#xA;As soon as you get the eigenvalues all you have to do is to solve two systems of linear equations with two unknowns (i.e. the coordinates of the corresponding eigenvector in the standard basis). Again, everything is nice and explicit, no need to apply a fancy numerical method.&lt;/p&gt;&#xA;" OwnerUserId="723" LastEditorUserId="723" LastEditorDisplayName="" LastEditDate="2010-11-02T18:16:36.883" LastActivityDate="2010-11-02T18:16:36.883" CommentCount="3" />
  <row Id="8674" PostTypeId="1" AcceptedAnswerId="8686" CreationDate="2010-11-02T18:25:20.070" Score="5" ViewCount="194" Body="&lt;p&gt;Consider a situation with a bag with infinity number of balls. Each ball is of some color. Number of colors is finite but it is not known. Balls are drawn from the bag one by one and checked for the color.&#xA;We want to stop drawing balls when there is small probability that we will find color that is not already drawn.&#xA;Or more exact, we want to be very sure (with large probability) that there is small portion of balls in bag with colors that are not already drawn.&#xA;Is there a statistical model to describe when to stop?&lt;/p&gt;&#xA;" OwnerUserId="3037" LastActivityDate="2010-11-03T00:54:20.703" Title="Bag with infinite number of colored balls" Tags="&lt;probability-theory&gt;&lt;statistics&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="8676" PostTypeId="2" ParentId="8658" CreationDate="2010-11-02T18:41:24.647" Score="1" ViewCount="" Body="&lt;p&gt;You might want to check out Faa di Bruno's formula: &lt;a href=&quot;http://mathworld.wolfram.com/FaadiBrunosFormula.html&quot; rel=&quot;nofollow&quot;&gt;http://mathworld.wolfram.com/FaadiBrunosFormula.html&lt;/a&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your case you have $f(x) = g(h(x))$ where $g(x) = 1/x$ and $h(x) = \cos(x)$.  The answer is not simple unfortunately, and involves some combinatorics.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively, notice $f(x) \cos(x) = 1$, and thus by taking the $m$th derivative of both sides we get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \sum_{k=0}^m {m \choose k} f^{(k)}(x) \left( \frac{d^{m-k}}{dx^{m-k}} \cos(x) \right) = 0 $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;if $m \geq 1$.  Thus, you can view the vector $(f(x),f'(x),\dots,f^{(n)}(x))$ as the solution to the system of linear equations above with $m=0,1,\dots,n$.  You can invert this matrix numerically, if that if your interest.  A closed form might be possible too. &lt;/p&gt;&#xA;" OwnerUserId="2515" LastActivityDate="2010-11-02T18:41:24.647" />
  <row Id="8677" PostTypeId="2" ParentId="8651" CreationDate="2010-11-02T18:44:29.200" Score="6" ViewCount="" Body="&lt;p&gt;The names &quot;first&quot; and &quot;second&quot; for the two parts of the theorem are meaningless.  More correct names would be &lt;em&gt;existence&lt;/em&gt; and &lt;em&gt;uniqueness&lt;/em&gt;.  It also is not unreasonable to separate the uniqueness statement from the formula relating definite integrals to antiderivatives, which is an algebraic &lt;em&gt;consequence&lt;/em&gt; of the (analytic) uniqueness statement.  The formula could be considered as a third part of the theorem, but numbering pieces of a theorem in a particular order is an uninformative nomenclature -- as is calling theorems &quot;fundamental&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fundamental theorem of calculus asserts existence and uniqueness of antiderivatives (solutions of the differential equation $y' = f(x)$ with given value of $y(x_0)$ at one point).  Apart from purely logical considerations there are several reasons the uniqueness theorem is important.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Indefinite integrals of the form $\int_p^x f(t) dt$, which are what appear in most presentations of the existence part of the theorem, in some cases do not account for all antiderivatives of $f(x)$ as the basepoint $p$ is varied over all real numbers.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;In the more precise presentation $y(x) = y(a) + \int_a^x f(t) dt$ there is still the possibility that other processes, even more magical than integration, might be related to anti-differentiation.  So it is of interest to either find these exotic species, or show that integrals give everything. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;An explicit analysis of uniqueness becomes more pressing when integrating functions with singularities, as in $\int dx/|x|^p$ for $p=1$ and $p=1/2$ (the number of integration constants changes, so this is needed for writing down solution formulas in full generality).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;the algebraic formula implied by uniqueness, $\int_a^b f = F(b)-F(a)$, is important both as a means of computing integrals and as the basis of the notation supporting changes of integration variable (substitutions).  &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="467" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-11-02T19:38:05.627" LastActivityDate="2010-11-02T19:38:05.627" />
  <row Id="8678" PostTypeId="2" ParentId="8652" CreationDate="2010-11-02T18:48:22.187" Score="12" ViewCount="" Body="&lt;p&gt;I happened to have gone to a toppish school (look up my profile if you really care which one). Even though I didn't realize it at the time, there are distinct advantages from the mathematical environment there. But I also think you shouldn't base your decision on graduate schools purely on artifical rankings of prestige. If you are deciding between schools, &lt;em&gt;you have to visit them to get a sense of the environment there&lt;/em&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unlike the undergraduate degree, if you do a graduate degree at a reasonably strong place, you will be spending 4, 5, or maybe 6 years with a close-knit group of people that number from 20-30 to 100+. It is where you will get to know a lot of future colleagues in your field, it is where you pick up new ideas from passing discussions, and it is where you learn how to do research. And the most important factor that goes into your decision of which graduate school to go to is &lt;strong&gt;the people&lt;/strong&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You need to find out if you can work with the faculty there: is there anyone there in your field? Is his or her interest something you want to study? Are they taking students? How do they treat their students? All these are things you should talk to people, especially current students, about when you visit. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You also need to find out if you have fellow students: you don't want to be the only student in the department doing (for example) PDEs when everyone else does algebraic geometry. The students whose research interests are closely related to yours will be the ones you discuss most with (and see most of); they will be the ones you ask for help and they will be your future colleagues. Find out if they have informal student seminars. Arrange your trip so that you can sit in on some of them. Having a couple more senior students who are willing to give you sound advice when you are just starting out will be very useful. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you really must make a decision without first hand experience visiting, then so be it: higher ranked schools tend to have a larger, more diverse pool of faculty. They also tend to attract the better students. So that's where I will put my money were I to bet blindly. But I cannot stress enough the importance of finding out whether the school and environment matches you first hand. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;A little side remark about &quot;more room psychologically&quot;: I would argue that that is actually counter-productive. When I was in graduate school there was significant amount of down time where the research just doesn't go well. Looking back, I am really glad to have had my large and rowdy office: at least one of us will have made some progress at any given time. And with seminars happening now and then, one can at least learn something useful through discussions when the thesis project is stuck. Were not for all these &quot;distractions&quot;, I fear that the ennui caused by the glacial pace of my thesis research may have led to psychological problems. &lt;/p&gt;&#xA;" OwnerUserId="1543" LastActivityDate="2010-11-02T18:48:22.187" CommentCount="2" />
  <row Id="8679" PostTypeId="2" ParentId="8609" CreationDate="2010-11-02T18:53:40.690" Score="4" ViewCount="" Body="&lt;p&gt;Say $k$ has the prime factorization $p_1^{r_1} p_2^{r_2} \cdots p_n^{r_n}$. Then let $H(k)$ be the &quot;hyperbolicity&quot; of this integer, $H(k) = \sum_{i=1}^n p_i^{-r_i}$. Hyperbolic integers are those integers $k$ that have $H(k) &amp;lt; 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, $H(k)$ for some &quot;random&quot; integer $k$ looks like a sum of independent random variables $X_p$, one for each prime $p$, where $P(X_p = p^{-k})$ is the probability that a random integer is divisible by $p^k$ but not $p^{k+1}$. Thus $P(X_p = p^{-k}) = (p-1)/p^{k+1}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular &#xA;$$ E(X_p) = \sum_{k \ge 1} p^{-k} P(X_p = p^{-k}) = \sum_{k \ge 1} {p-1 \over p^{2k+1}} = {1 \over p^2 + p} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Things get a bit tricky in finding the distribution of $\sum_p X_p$; in particular a central limit theorem won't work, because the contribution of each of the small primes is too large.  But there is some limiting distribution, and I suspect it at least bears some resemblance to the limiting distribution of the actual hyperbolicities of integers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Incidentally, my tests seem to point to a natural density of hyperbolic integers somewhere in the neighborhood of 99 percent. I'd conjecture that integers with hyperbolicity less than $k$ have a natural density.  There are similar results on the distribution of $\sigma(n)/n$ where $\sigma(n)$ is the sum of divisors of $n$; see for example Marc Deleglise, &lt;a href=&quot;http://projecteuclid.org/DPubS?service=UI&amp;amp;version=1.0&amp;amp;verb=Display&amp;amp;handle=euclid.em/1048515661&quot; rel=&quot;nofollow&quot;&gt;Bounds for the density of abundant integers&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="173" LastActivityDate="2010-11-02T18:53:40.690" CommentCount="1" />
  <row Id="8681" PostTypeId="2" ParentId="8652" CreationDate="2010-11-02T19:39:50.453" Score="14" ViewCount="" Body="&lt;p&gt;Maybe this is just getting redundant, but I think this is a slightly different point from what people has said above:  while there do exist compelling reasons to go to grad schools that don't maximize prestige, it really doesn't sound to me like you have put forth any.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think on the whole grad students are pretty naive about how big an influence where they go to grad school has on their opportunities after graduating.  Of course, lots of people do manage to go to less fancy schools and go on to have good careers, but if you look at the schools in the 40-60 range you mention, most of their faculty went to more prestigious places.  For example, at Oregon where I'm a professor (US News rank #56), fully half the professors who got Ph.D.s in the US got them at Harvard, MIT, Berkeley or Stanford (all in the top 5 on the USN ranking) and most of the rest at institutions in the low 2 digits (Texas, Maryland, Wisconsin, Penn).  Some of this is for unfair reasons (bias, etc.), some of this is selection reasons, and some is because going to a better school give you more opportunities to be a better mathematician (more exposure to new ideas, more visitors, higher quality peers).  I wouldn't give up on those things because of a nebulous fear of not liking the environment.  Try it out; it's much easier to go to a Berkeley or Michigan for a year and then transfer somewhere else than the other way around.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt; Just as an addendum, I thought I should emphasize here that like Pete said, prestige isn't really well ordered.  There are lots of pairs where it's hard to say really conclusively which is more prestigious, in which case you should go with your gut.  So I'm not talking about things like comparing Chicago and Columbia, and choosing Columbia because you'd rather be in New York, but rather the difference between going to graduate school in the 40-60 range (in USNWR or NRC), and something in, say, the top 20.&lt;/p&gt;&#xA;" OwnerUserId="3041" LastEditorUserId="3041" LastEditorDisplayName="" LastEditDate="2010-11-03T14:16:38.503" LastActivityDate="2010-11-03T14:16:38.503" />
  <row Id="8682" PostTypeId="2" ParentId="7569" CreationDate="2010-11-02T19:42:54.723" Score="0" ViewCount="" Body="&lt;p&gt;My answer on MO:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we are allowed to consider somewhat weaker bounds, both answers depend only on the number of unitary divisors of $k$, which is $2^{\omega(k)}$. By the quoted result of O. Izhboldin and L. Kurliandchik (see Fedor's and Myerson's comments &lt;a href=&quot;http://mathoverflow.net/questions/43326/sharp-upper-bounds-for-sums-of-the-form-sum-p-mid-k-frac1p1&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;), for any set of $n$ positive integers {$a_{1}, \dots, a_{n}$} such that $\sum_{i = 1}^{n} \frac{1}{a_{i}} &amp;lt;1$, we have&#xA;\begin{eqnarray}&#xA;\sum_{i = 1}^{n} \frac{1}{a_{i}} \leq \sum_{i = 1}^{n} \frac{1}{d_{i}} = \frac{d_{n+1} - 2}{d_{n+1} -1} &amp;lt; 1,&#xA;\end{eqnarray}&#xA;where $d_{i}$ is the $i^{\text{th}}$-Euler number, which satisfies the quadratic recurrence $d_{i} = d_{1} \cdots d_{i-1} + 1$ with $d_{1} = 2$. The first few terms of the sequence are 2, 3, 7, 43, 1807....  (A000058). It is relatively straightforward to show that the aforementioned recurrence is equivalent to $d_{i} = d_{i-1}(d_{i-1}-1) + 1$, and it is known from the recurrence that $d_{i} = \lfloor \theta^{2^{i}} + \frac{1}{2} \rfloor$, where $\theta \approx 1.2640$.... Thus,&#xA;\begin{eqnarray}&#xA;\sum_{p \mid k} \frac{1}{p + 1} \leq \sum_{i = 1}^{\omega(k)} \frac{1}{d_{i}} = \frac{d_{\omega(k) + 1} - 2}{d_{\omega(k)+1} - 1} = \frac{\lfloor \theta^{2^{\omega(k) + 1}} + \frac{1}{2} \rfloor - 2}{\lfloor \theta^{2^{\omega(k) + 1}} + \frac{1}{2} \rfloor - 1} &amp;lt; 1.&#xA;\end{eqnarray}&#xA;One can also show that $d_{i} - 1 = \lfloor \vartheta^{2^{i-1}} - \tfrac{1}{2} \rfloor$, where where $\vartheta \approx 1.5979$...., so we have&#xA;\begin{eqnarray}&#xA;1 - \sum_{p \mid k} \frac{1}{p + 1} \geq 1 - \frac{d_{\omega(k)+1} - 2}{d_{\omega(k)+1} - 1} = \frac{1}{d_{\omega(k)+1} - 1} = \lfloor \vartheta^{2^{\omega(k)}} - \tfrac{1}{2} \rfloor^{-1} &gt; 0.&#xA;\end{eqnarray}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Remark&lt;/strong&gt; E. Deutsche points out that the sequence {$d_{i} - 1$} (A007018) counts the number of ordered rooted trees with out-going degree up to 2 with all leaves at top level.  &lt;/p&gt;&#xA;" OwnerUserId="2720" LastEditorUserId="2720" LastEditorDisplayName="" LastEditDate="2010-11-02T23:12:17.947" LastActivityDate="2010-11-02T23:12:17.947" />
  <row Id="8683" PostTypeId="1" AcceptedAnswerId="8696" CreationDate="2010-11-02T19:48:21.317" Score="5" ViewCount="144" Body="&lt;p&gt;A closed curve on a surface can be represented by the boundary it intersects. Suppose the original surface is created by glueing a to A, b to B etc. Pick an orientation, each time it intersects a boundary, add the boundary to the representation. For example in the image, it's &#xA;&lt;img src=&quot;http://i.imgur.com/uNKZX.png&quot; alt=&quot;alt text&quot;&gt;&#xA;The black dot is the starting point to trace the curve.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sequence of letters is called a word.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The length of a word is defined as the amount of letters in the word.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A word W represents a simple closed curve if there exist a simple closed curve that can be represented by W. Let's call those words simple words.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a given surface, are there algorithms to find simple words with length up to L in polynomial time?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The naive algorithm is to generate every word and test which ones are simple, but that is $O(c^L)$ for some constant c. The amount of simple word is $O(L^k)$ for some constant k. As shown in &lt;a href=&quot;http://www.math.princeton.edu/~mmirzakh/data/Growth.pdf&quot; rel=&quot;nofollow&quot;&gt;Growth of the number of simple closed geodesics on hyperbolic surfaces&lt;/a&gt; by Maryam Mirzakhani. The bound given in that paper can be converted into the bound on words.&lt;/p&gt;&#xA;" OwnerUserId="96" LastActivityDate="2010-11-03T09:42:54.867" Title="Algorithms to find simple closed curves on a surface" Tags="&lt;topology&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="8684" PostTypeId="1" AcceptedAnswerId="8704" CreationDate="2010-11-02T19:48:25.807" Score="2" ViewCount="157" Body="&lt;p&gt;I know how to generate solutions to $x^2-Dy^2=\pm 1$ from the fundamental solution to $x^2-Dy^2=\pm 4$.  But how do we know that &lt;strong&gt;all&lt;/strong&gt; the solutions are generated from this fundamental solution?&lt;/p&gt;&#xA;" OwnerUserId="2505" LastEditorUserId="2505" LastEditorDisplayName="" LastEditDate="2010-11-02T21:33:08.523" LastActivityDate="2010-11-03T01:56:28.973" Title="Generalized Pell's Equation: Why is 4 special?" Tags="&lt;number-theory&gt;&lt;diophantine-equations&gt;" AnswerCount="2" CommentCount="0" />
  <row Id="8685" PostTypeId="2" ParentId="8334" CreationDate="2010-11-02T20:10:16.857" Score="1" ViewCount="" Body="&lt;p&gt;I just read &quot;Generalized Feedback Shift Register Pseudorandom Number Algorithm&quot; by T. G. Lewis and W. H. Payne. &#xA;I think that paper settles the question I was raising (going to the source, right?). In essence, the question is &#xA;&quot;What is the correct procedure to use the Generalized Feedback Shift Register Algorithm (GFSR)?&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1.- Start with a sequence and a primitive polynomial $x^{p}+x^{q}+1$. &#xA;For example, $a_{0}=a_{1}=a_{2}=a_{3}=a_{4}=1$ and $x^{5}+x^{2}+1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2.- Elements of the sequence follow $a_{k}=a_{k-p+q}\bigoplus a_{k-p}$ with $k=p, p+1,...$. &#xA;In this example, since we have the first 5 elements of the sequence and according to the polynomial, we are given that $p=5, q=2$.&#xA;Therefore, we can know the next elements of the sequence&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{matrix} a_{6}=a_{3}\bigoplus a_{1}=0 \\ a_{7}=a_{4}\bigoplus a_{2}=0 &#xA;\\ a_{8}=a_{5}\bigoplus a_{3}=1 \\ a_{9}=a_{6}\bigoplus a_{4}=1 \\ ... \\ \end{matrix}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, in this way we construct the rest of the sequence:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\{a_{i}\}_{0}^{30}={1111100011011101010000100101100}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to produce a better random sequence, we apply Kendall's algorithm. Although there are several variations of Kendall's algorithm, the point is to shift the original sequence $1111100011011101010000100|101100$ forwards by 6 bits, that is, $1011001111100011011101010|000100$.&#xA;And again three times more (until we are back with the original sequence). This process gives the following sequence&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{matrix}&#xA;\text{Key} &amp;amp; \text{Sequence} \\&#xA;0 &amp;amp; \|11111\|00011011101010000100|101100\\&#xA;1 &amp;amp; 1011001111100011011101010|000100\\&#xA;2 &amp;amp; 0001001011001111100011011|101010\\&#xA;3 &amp;amp; 101010000100101100111100|011011\\&#xA;4 &amp;amp; 0110111010100001001011001|111100&#xA;\end{matrix}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, we take n-tuples (in this example, 5-tuples are used) which are positioned as the columns of a new array:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{matrix}&#xA;W_{0}: &amp;amp; \|1\|1010 &amp;amp; W_{10}: &amp;amp; 01001&amp;amp; W_{20}: &amp;amp; 00111\\&#xA;W_{1}: &amp;amp; \|1\|0001 &amp;amp; W_{11}: &amp;amp; 10000&amp;amp; W_{21}: &amp;amp; 01111\\&#xA;W_{2}: &amp;amp; \|1\|1011 &amp;amp; W_{12}:&amp;amp; 10110&amp;amp; W_{22}: &amp;amp; 10010\\&#xA;W_{3}: &amp;amp; \|1\|1100 &amp;amp; W_{13}:&amp;amp; 10100&amp;amp; W_{23}: &amp;amp; 01100\\&#xA;W_{4}: &amp;amp; \|1\|0011 &amp;amp; W_{14}:&amp;amp; 01110&amp;amp; W_{24}: &amp;amp; 00101\\&#xA;W_{5}: &amp;amp; 00001 &amp;amp; W_{15}:&amp;amp; 11111&amp;amp; W_{25}: &amp;amp; 10101\\&#xA;W_{6}: &amp;amp; 01101 &amp;amp; W_{16}:&amp;amp; 00100&amp;amp; W_{26}: &amp;amp; 00011\\&#xA;W_{7}: &amp;amp; 01000 &amp;amp; W_{17}:&amp;amp; 11000&amp;amp; W_{27}: &amp;amp; 10111\\&#xA;W_{8}: &amp;amp; 11101 &amp;amp; W_{18}:&amp;amp; 01011&amp;amp; W_{28}: &amp;amp; 11001\\&#xA;W_{9}: &amp;amp; 11110 &amp;amp; W_{19}:&amp;amp; 01010&amp;amp; W_{29}: &amp;amp; 00110&#xA;\end{matrix}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each $W_{i}$ is called a 'word'. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;blockquote&gt;&#xA;    &lt;p&gt;Since each column obeys the recurrence $a_{k}=a_{k-p+q}\bigoplus a_{k-p}$, each word must also obey&#xA;    $W_{k}=W_{k-p+q}\bigoplus W_{k-p}$.&lt;/p&gt;&#xA;  &lt;/blockquote&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As far as I know, that's the correct procedure for using the GFSR algorithm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Corrections or comments will be appreciated.&lt;/p&gt;&#xA;" OwnerUserId="2569" LastEditorUserId="2569" LastEditorDisplayName="" LastEditDate="2010-11-02T20:25:20.667" LastActivityDate="2010-11-02T20:25:20.667" />
  <row Id="8686" PostTypeId="2" ParentId="8674" CreationDate="2010-11-02T20:12:34.703" Score="3" ViewCount="" Body="&lt;p&gt;This problem often comes up in Biology when one is interested in estimating the number of species in an area based on some survey. There are a wide variety of methods developed to suit the specific application in hand. For example, it is very hard to come up with a good estimator for the number of biological species in a rain forest since there are too many &quot;extremely rare&quot; species to take into account. There is no one unique estimator that is globally optimal and you would have to customize your own estimator based on your needs. A good starting point is this &lt;a href=&quot;http://biomet.oxfordjournals.org/content/63/3/435.abstract&quot; rel=&quot;nofollow&quot;&gt;link&lt;/a&gt; and references therein.&lt;/p&gt;&#xA;" OwnerUserId="618" LastActivityDate="2010-11-02T20:12:34.703" CommentCount="1" />
  <row Id="8687" PostTypeId="2" ParentId="8684" CreationDate="2010-11-02T20:33:33.727" Score="3" ViewCount="" Body="&lt;p&gt;I had written out an answer previously, but to make it work I need to assume a little more.  Namely, let me assume that $D$ is &lt;strong&gt;not&lt;/strong&gt; congruent to $1 \pmod 4$.  (I'll think a little more about what's going on in general, and someone else is of course welcome to jump in and help me out.)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of the running assumptions of the Pell equation is that $D$ is squarefree.  In particular, it is not $0 \pmod 4$.  So if you reduce the equation $x^2 - Dy^2 = \pm 4$ &#xA;modulo $4$, you get $x^2 - Dy^2 \equiv 0 \pmod 4$.  Since the squares mod $4$ are $0$ and $1$, running through the four possibilities for $x^2$ and $y^2$ mod $4$ one sees that the only solutions are when $x^2 \equiv y^2 \equiv 0$, i.e., when $x$ and $y$ are even.  [N.B.: here I am using that $D \not \equiv 1 \pmod 4$; otherwise $1^2 - D \cdot 1^2 \equiv 0 \pmod 4$.]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is, every integral solution to $x^2 - Dy^2 = \pm 4$ occurs with $x$ and $y$ even, so we may write $x = 2X$, $y = 2Y$ and substitute, getting &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(2X)^2 - D (2Y)^2 = 4(X^2 - DY^2) = \pm 4$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dividing by $4$, we get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$X^2 - DY^2 = \pm 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The process can also be reversed: if $X^2 - DY^2 = \pm 1$, then $(2X)^2 - D(2Y)^2 = \pm 4$.  Thus we have found a bijection between the sets of solutions to $x^2 - Dy^2 = \pm 4$ and $X^2 - DY^2 = \pm 1$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So this is the most direct answer to your question.  One can still wonder whether this occurs for integers other than $4$, but for other $k &gt; 1$ it can happen that there are solutions to $x^2 - Dy^2 = \pm k$ in which $x$ and $y$ are &lt;em&gt;not&lt;/em&gt; both divisible by $k$.  Try it and see.  So $4$ is special here because the theory of quadratic equations modulo $4$ is especially simple.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Added&lt;/b&gt;: See &lt;a href=&quot;http://sites.google.com/site/tpiezas/008&quot; rel=&quot;nofollow&quot;&gt;http://sites.google.com/site/tpiezas/008&lt;/a&gt; for some further information on going from $x^2 - dy^2 = \pm 4$ to $x^2 - dy^2 = \pm 1$.  &lt;/p&gt;&#xA;" OwnerUserId="299" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2010-11-03T01:21:00.360" LastActivityDate="2010-11-03T01:21:00.360" CommentCount="1" />
  <row Id="8688" PostTypeId="1" AcceptedAnswerId="8729" CreationDate="2010-11-02T20:50:16.143" Score="1" ViewCount="213" Body="&lt;p&gt;I am taking my first real-analysis course this year and have a midterm on friday. I have NO idea what is going on in this class. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if anyone could clarify what the following means.&#xA;The book starts off with talking about Archimedes and his principle which states:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The value of an infinite series, if it&#xA;  exists, is the number T such that&#xA;  given any rational numbers L and M&#xA;  such that L &amp;lt; T &amp;lt; M, all of the finite&#xA;  sums from some point on will be&#xA;  strictly contained in the interval&#xA;  between L and M.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The book then goes on about talking about series that do not converge (ie diverges and how hard it is to get a value for that). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then the book has the following definition:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;An infinite series converges if there is a target value T so that for any L &amp;lt; T and any   M &gt; T, all of the partial sums from some point on are strictly between L and M.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So I have no idea what this means. What does the interval mean? I am more confused about L and M and the interval then I am about the series. Why are they giving me an interval?&lt;/p&gt;&#xA;" OwnerUserId="287" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-02T21:07:11.630" LastActivityDate="2010-11-03T05:01:45.280" Title="Infinite series - Archimedean principle." Tags="&lt;real-analysis&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="2" />
  <row Id="8689" PostTypeId="2" ParentId="8688" CreationDate="2010-11-02T21:06:41.630" Score="2" ViewCount="" Body="&lt;p&gt;&quot;The interval between $L$ and $M$&quot; means all the numbers that are between $L$ and $M$; we call $L$ and $M$ the &quot;endpoints&quot; of the interval. If you imagine the real numbers as the points on the real number line, with $L$ and $M$ located somewhere, the &quot;interval&quot; would be the segment of the line that is between the points $L$ and $M$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Usually, an interval may or may not contain (either or both) of its endpoints. In your case, you are looking at the interval that does &lt;strong&gt;not&lt;/strong&gt; contain either $L$ and $M$ (that is the &quot;strictly between $L$ and $M$&quot; condition in the second quote). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For concreteness, say $L=0$ and $M=1$. Then the interval of numbers strictly between $L$ and $M$ would be all real numbers that are greater than $0$ and smaller than $1$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given a number $x_0$, if $a\lt x_0$ and $x_0\lt b$, then we think of the interval determined by $a$ and $b$ (all real numbers strictly between $a$ and $b$: this includes $x_0$, but will include other things as well) as being a &quot;neighborhood&quot; of $x_0$: they are all points that are within a certain distance from $x_0$, either below $x_0$ or above $x_0$ (no further than $a$ is if they are below $x_0$, and no further than $b$ is if they are above). The definition you quote is basically saying: if you specify how far from $T$ you are willing to tolerate your total (no further than $L$ is from $T$ if the total is less than $T$, and no further than $M$ is from $T$ if the total is more than $T$), then you can always ensure that from some point forward the partial sums will have a total that you is within your margin of tolerance (a total that you can tolerate). This will hold no matter how tolerant or intolerant you are (you are very tolerant if you allow totals that are far away from $T$; that is, both $L$ and $M$ are &quot;far away&quot; from $T$; you are being very intolerant if your $L$ and your $M$ are really, really, really, close to $T$). You can be as intolerant as you wish, but you are required to allow &lt;em&gt;some&lt;/em&gt; leeway.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Intuitively, $T$ is what you think the sum of the series &quot;should&quot; be (if it made sense to add infinitely many numbers). The definition says that the sums you get a long the way will get &quot;as close as you care to specify&quot; to that total $T$, provided you add enough terms of your series, and that once you get there, any further refinments (by adding more terms of your series) will still keep you within your specified 'neighborhood' of $T$. It is very similar to the $\epsilon-\delta$ definition of a limit, if you saw that, or to the &quot;intuitive notion&quot; of limit: as long as you stay close enough to the point (for series, as long as you add up enough terms), the value of the function (for series, the running total) will be as close as you care to &quot;pre-specify&quot; from the limit (for series, as close as you care to &quot;pre-specify&quot; from $T$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Added:&lt;/em&gt; To respond to your query in the comments: &quot;If we have a target value, why add the tolerance?&quot; Keep in mind that we can only add finitely many numbers at a time. We can add two numbers, and then using that we can add three, four, five, etc. But we don't really know what it means to add infinitely many numbers, like the series says we &quot;should&quot; or &quot;want&quot;. So we cannot &lt;em&gt;actually&lt;/em&gt; perform the act of adding up all the (infinitely many) terms. Rather, we can only add up to a certain number of the terms; these are the partial sums. We can (in principle, at least) compute a partial sum to any term we want, but we cannot actually add up all of them (just like we can, in principle, count as high as we want, but we cannot actually count through &quot;all&quot; the natural numbers). So what we have actual access to are the partial sums and only the partial sums. We think that if we &lt;em&gt;could&lt;/em&gt; add up all the terms of the series, then it would make sense to call that sum $T$; so what we do is we compare those partial sums to our target, $T$. But of course, $T$ is supposed to be what we would get after adding up &lt;em&gt;all&lt;/em&gt; the terms of the series (if we could do that), and what we are comparing is &lt;strong&gt;not&lt;/strong&gt; the result of adding all the terms, but only some of them. So the total we have in hand (the partial sum) has no reason to be exactly the same as the target (just like, if you add the first 100 numbers in a list of a thousand numbers, the running total has no reason to be the same as the final total). So we need to add some tolerance to account for the fact that we are not &lt;em&gt;actually&lt;/em&gt; adding up all the terms, but only some of them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(P.S. I've never seen what you quote as the &quot;Archimedean Principle&quot; called that, but that may very well be just me...)&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-02T23:39:36.523" LastActivityDate="2010-11-02T23:39:36.523" CommentCount="3" />
  <row Id="8690" PostTypeId="1" CreationDate="2010-11-02T21:07:00.230" Score="16" ViewCount="635" Body="&lt;p&gt;Hi, I encountered a problem today to prove that (X_n) with X_n = cos(n!) does not have a limit (when n approaches infinity). I have no idea how to do it formally. Could someone help? The simpler the proof (by that i mean less complex theorems are used) the better. Thanks&lt;/p&gt;&#xA;" OwnerUserId="3046" LastEditorUserId="3046" LastEditorDisplayName="" LastEditDate="2010-11-02T21:29:32.567" LastActivityDate="2010-11-03T01:17:56.027" Title="Is there a limit of cos (n!)?" Tags="&lt;analysis&gt;&lt;trigonometry&gt;" AnswerCount="3" CommentCount="7" FavoriteCount="7" />
  <row Id="8691" PostTypeId="2" ParentId="8688" CreationDate="2010-11-02T21:16:50.400" Score="1" ViewCount="" Body="&lt;p&gt;This is really quite a convoluted way to define the limit of a sequence.  I can sympathize with your confusion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first thing to note is that the limit of an infinite series is the limit of its sequence of partial sums.  So, we just need to understand limits of sequences.  The reason for the &lt;em&gt;interval&lt;/em&gt;, is that the series may have positive and negative terms.  So, while a series is busy converging, it may jump to the left or right of its limit.  But the main point is that for any &lt;em&gt;error tolerance&lt;/em&gt; (no matter how small) all the partial sums will &lt;em&gt;eventually&lt;/em&gt; be within this error tolerance of the limit.  Being within a small error tolerance means that you are in a small interval containing the limit (hence the interval business).  Here is how we quantify &lt;em&gt;error tolerance&lt;/em&gt; and &lt;em&gt;eventually&lt;/em&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A sequence $(s_n)$ converges to a limit $s$ if for any $\epsilon&gt;0$ there exists an index $N$ such that $s_n \in (s-\epsilon, s+\epsilon)$ for all $n \geq N$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So think of $\epsilon$ as the &lt;em&gt;error tolerance&lt;/em&gt;, and the $N$ (which depends on $\epsilon$) as a measure of the &lt;em&gt;eventually&lt;/em&gt;.  This definition is mathematically equivalent to the one you gave where $s-\epsilon$ corresponds to $L$ and $s+\epsilon$ corresponds to $M$, but this one is superior for a couple of reasons.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, there is no reason to insist that $L$ and $M$ are rational.  We &lt;em&gt;can&lt;/em&gt; insist that they are rational because rationals approximate real numbers arbitrarily well, but it is confusing to include it in the definition.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Secondly, there is no reason to specify &lt;em&gt;two&lt;/em&gt; error tolerances $\epsilon_1$ and $\epsilon_2$ which is what the confusing definition is doing.  Instead of saying we eventually stray by at most $\epsilon_1$ to the left of the limit and by at most $\epsilon_2$ to the right of the limit, we may as well specify a &lt;em&gt;single&lt;/em&gt; error tolerance $\epsilon:=\min(\epsilon_1, \epsilon_2)$. We can then insist that the sequence eventually strays by at most $\epsilon$ (in either direction) from the limit.  &lt;/p&gt;&#xA;" OwnerUserId="964" LastEditorUserId="964" LastEditorDisplayName="" LastEditDate="2010-11-03T00:06:01.797" LastActivityDate="2010-11-03T00:06:01.797" CommentCount="2" />
  <row Id="8692" PostTypeId="2" ParentId="8690" CreationDate="2010-11-02T21:30:27.503" Score="20" ViewCount="" Body="&lt;p&gt;I suspect that you have miscopied the problem, or your source has an error. I will argue that, while your statement is extremely likely to be true, we probably do not know enough to prove it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is why. Suppose, hypothetically, that &#xA;$$\frac{1}{2 \pi} = \sum_{i \geq 0} \frac{d_i}{i!}$$&#xA;where all the $d_i$ are integers and, for $i$ sufficiently large, each $d_i$ is either $0$ or $1$. This is extraordinarily unlikely. However, if we knew how to prove that this didn't happen, we would probably know enough to show that every digit occurs infinitely often in the decimal expansion of $\pi$, and I believe that is an open problem. So, while this scenario is almost surely false, I don't think we know enough to prove that it is false.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, in this scenario,&#xA;$$\cos n! = \cos (n! 2 \pi \sum_{i \geq 0} \frac{d_i}{i!}) = \cos (2 \pi N + 2 \pi \sum_{i \geq n+1} \frac{d_i }{(n+1)(n+2) \cdots i})$$&#xA;where $N$ is a very large integer. By the periodicity of $\cos$, this is just&#xA;$$\cos (2 \pi\sum_{i \geq n+1} \frac{d_i }{(n+1)(n+2) \cdots i}).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, once $n$ is large enough that $d_i$ is $0$ or $1$ for $i$ larger than $n$, we have&#xA;$$\sum_{i \geq n+1} \frac{d_i n!}{i!} \leq \frac{1}{n+1} + \frac{1}{(n+1)^2} + \frac{1}{(n+1)^3} + \cdots = \frac{1}{n}.$$&#xA;So, in our unlikely scenario, $\lim_{n \to \infty} \sum_{i \geq n+1} \frac{d_i n!}{i!} =0$ and $\lim_{n \to \infty} \cos n! = 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I may try to extract some basic lessons here, questions about trig functions evaluated at integers almost always turn into questions about $\pi$. Often they can be resolved simply by noting that $\pi$ is irrational; occasionally it is important to show that $\pi$ is not well approximated by rationals and sometimes, as in this example, they turn into claims about $\pi$ which are beyond current knowledge. The key point is to work out what they are saying about $\pi$, and then find out whether that fact is known.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Exercise:&lt;/strong&gt; Show that, for any $a$ between $-1$ and $1$, there is a $\theta$ such that $\lim_{n \to \infty} \cos (n! \theta)$ is $a$.&lt;/p&gt;&#xA;" OwnerUserId="448" LastActivityDate="2010-11-02T21:30:27.503" CommentCount="1" />
  <row Id="8693" PostTypeId="2" ParentId="8326" CreationDate="2010-11-02T21:34:09.280" Score="0" ViewCount="" Body="&lt;p&gt;V.O. Manturov,  Knot theory. The first chapter gives an elementaty introduction. To continue, you have to know basical topological concepts, e.g. foundamental group.&lt;/p&gt;&#xA;" OwnerUserId="3047" LastActivityDate="2010-11-02T21:34:09.280" />
  <row Id="8694" PostTypeId="2" ParentId="8326" CreationDate="2010-11-02T21:45:05.527" Score="1" ViewCount="" Body="&lt;p&gt;My intro to knot theory graduate course used &quot;An Introduction to Knot Theory&quot; by Lickorish.  The early chapters on Seifert surfaces and polynomials are quite nice.&lt;/p&gt;&#xA;" OwnerUserId="2838" LastActivityDate="2010-11-02T21:45:05.527" />
  <row Id="8695" PostTypeId="1" AcceptedAnswerId="8702" CreationDate="2010-11-02T21:55:10.973" Score="5" ViewCount="177" Body="&lt;p&gt;Let $A$ be a stochastic matrix.  Thus $A$ has nonnegative entries, and the sum of the elements in each row is 1. This implies that the vector $\begin{bmatrix} 1 &amp;amp; 1 &amp;amp; \cdots &amp;amp; 1 \end{bmatrix}^T$ is an eigenvector corresponding to the eigenvalue 1.  Is it true that there exists a vector $b$ such that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(A - I)x \geq b$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;has no solutions in $x$?  If so, is there a simple proof?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Motivation: I've been trying to construct an answer to &lt;a href=&quot;http://math.stackexchange.com/questions/8556/&quot;&gt;another question&lt;/a&gt; using linear programming duality (as the OP implies he is interested in).  If my reasoning is correct, this is the only step I need to complete the argument.  I feel like this should be an easy question to answer, but I've been working on it for a while with no success.&lt;/p&gt;&#xA;" OwnerUserId="2370" LastActivityDate="2010-11-03T00:39:15.747" Title="No solutions to a matrix inequality?" Tags="&lt;linear-algebra&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="8696" PostTypeId="2" ParentId="8683" CreationDate="2010-11-02T22:32:25.980" Score="3" ViewCount="" Body="&lt;p&gt;Fix a triangulation $T = (t_i)_{i=0}^N$ of the surface $S$.  An arc $\alpha$ embedded in a triangle $t_i$ is &lt;em&gt;normal&lt;/em&gt; if the two points $\partial \alpha$ lie in distinct edges of $t_i$. A single triangle $t_i$ admits three different normal arcs (up to isotopy). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A simple closed curve $\alpha \subset S$ is &lt;em&gt;normal&lt;/em&gt; with respect to $T$ if and only if $\alpha \cap t_i$ is a collection of normal arcs for each $i$.  Exercise: every essential simple curve may be isotoped to be normal.  We may replace any normal curve by a vector of length $3N$ by counting the number of each type of normal arc.  These vectors have non-negative integer entries and satisfy certain &quot;matching equations&quot;.  Conversely, any such vector represents a simple closed multicurve.  (Note that a single isotopy class of curve may have many different representations as a normal curve!)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The matching equations cut a cone out of the positive orthant.  Thus to enumerate all curves up to a fixed length (where here I count the number of normal arcs as the length) it suffices to find a Hilbert basis for this cone and take all combinations up to a fixed size, and so on, and so on.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The result is an algorithm that enumerates all essential simple closed curves up to length $N$ in time polynomial in $N$.  However the precomputation (of the Hilbert basis for the cone, etc) is no joke. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can find a &lt;a href=&quot;http://www.math.wisc.edu/~rkent/normal.ps&quot; rel=&quot;nofollow&quot;&gt;discussion&lt;/a&gt; starting on page 13 of Cameron Gordon's notes for a course on normal surfaces in three-manifolds.  I'll end with a remark: in the case of the (once-holed) torus you can get away with exactly two triangles, the cone is very simple, the basis is very pretty, and everything can be done by hand.  (In fact you can this way rediscover for yourself the classification of simple closed curves in the torus.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: I'll add one more remark.  You are representing curves on the surface via &quot;cutting sequences&quot; which are very closely related to writing out words in terms of the fundamental group.  When working with &lt;em&gt;simple&lt;/em&gt; curves (and the simplicity is crucial here) it can be exponentially more efficient to use the &quot;normal coordinates&quot; I described.  These are very similar to Thurston's &quot;train track&quot; coordinates for simple closed curves.  There is a third way to represent simple curves: Fix a small collection of curves (ie the very short curves) and then act on those by the mapping class group of the surface, as generated by Dehn twists (say).  These &quot;mapping class group coordinates&quot; can again be exponentially more efficient than using cutting sequences.  &lt;/p&gt;&#xA;" OwnerUserId="1307" LastEditorUserId="1307" LastEditorDisplayName="" LastEditDate="2010-11-03T09:42:54.867" LastActivityDate="2010-11-03T09:42:54.867" />
  <row Id="8697" PostTypeId="2" ParentId="8624" CreationDate="2010-11-02T22:34:54.543" Score="0" ViewCount="" Body="&lt;p&gt;Citing &lt;em&gt;Ed Sandifer&lt;/em&gt; preprint on L. Euler and the &lt;a href=&quot;http://www.southernct.edu/~sandifer/Ed/History/Preprints/Talks/NYU%20Basel%20Problem%20Paper.PDF&quot; rel=&quot;nofollow&quot;&gt;Basel Problem&lt;/a&gt; (that was published in &lt;a href=&quot;http://books.google.se/books?id=tK_KRmTf9nUC&amp;amp;pg=PA115&amp;amp;lpg=PA115&amp;amp;dq=Ludolf+Kuelen&amp;amp;source=bl&amp;amp;ots=lum0ovQKcQ&amp;amp;sig=7X4HZ1NrHoeNFM5VWquGShwJQHc&amp;amp;hl=sv&amp;amp;ei=aJHQTODdMsOBOuOupfEE&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=2&amp;amp;ved=0CB0Q6AEwAQ#v=onepage&amp;amp;q=Ludolf%20Kuelen&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;Euler at 300: an appreciation&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;“It is clear that I employ here the letter $\pi$ to indicate the number of Ludolf of Kuelen, $3.14159265$, etc.” By this time, Euler has been using p to denote that constant for several years, but the convention will take many years more before it is universally adopted.&lt;/p&gt;&#xA;" OwnerUserId="1154" LastActivityDate="2010-11-02T22:34:54.543" />
  <row Id="8698" PostTypeId="1" AcceptedAnswerId="8725" CreationDate="2010-11-02T22:57:23.537" Score="4" ViewCount="105" Body="&lt;p&gt;What is the smallest square which contains 11 non-overlapping (except boundary) unit squares?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This question is open but I would like to know a method to verify the best known answer at the moment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm reading paper at &lt;a href=&quot;http://www.emis.de/journals/EJC/Volume_10/PDF/v10i1r8.pdf&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; . In figure 2, there has been said that the best known packing has side length about 3.8772 and tilt angle about 40.182 degrees. How can I verify those values, as another source (Which Way did the Bicycle Go page 105) says that the side length is about 3.877083? I think I have to denote the tilt angle by $\alpha$, a side of the square by $x$ and form two equations but I don't see how to make those equations.&lt;/p&gt;&#xA;" OwnerUserId="2856" LastEditorUserId="2856" LastEditorDisplayName="" LastEditDate="2010-11-03T01:40:30.507" LastActivityDate="2010-11-03T04:21:01.943" Title="Eleven unit squares inside a larger square" Tags="&lt;optimization&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" />
  <row Id="8699" PostTypeId="2" ParentId="8683" CreationDate="2010-11-02T22:58:41.270" Score="0" ViewCount="" Body="&lt;p&gt;An algorithm to determine whether a given element of the fundamental group has a simple representative on surfaces with boundary was described by Birman and Series in their paper&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.ams.org/mathscinet/search/publdoc.html?arg3=&amp;amp;co4=AND&amp;amp;co5=AND&amp;amp;co6=AND&amp;amp;co7=AND&amp;amp;dr=all&amp;amp;pg4=AUCN&amp;amp;pg5=AUCN&amp;amp;pg6=PC&amp;amp;pg7=ALLF&amp;amp;pg8=ET&amp;amp;review_format=html&amp;amp;s4=birman&amp;amp;s5=series&amp;amp;s6=&amp;amp;s7=&amp;amp;s8=All&amp;amp;vfpref=html&amp;amp;yearRangeFirst=&amp;amp;yearRangeSecond=&amp;amp;yrop=eq&amp;amp;r=5&amp;amp;mx-pid=744104&quot; rel=&quot;nofollow&quot;&gt;An algorithm for simple curves on surfaces&lt;/a&gt;. &lt;em&gt;J. London Math. Soc.&lt;/em&gt; (2) 29 (1984), no. 2, 331–342&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the MathSciNet review:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The given algorithm is purely mechanical and easy to check (for finite words). For other algorithms, also for closed surfaces, see the bibliography in the paper and also B. Reinhart [Ann. of Math. (2) 75 (1962), 209--222; MR0150740 (27 #727)].&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;UPDATE&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When writing this answer, I hadn't noticed the 'polynomial time' stipulation.  This won't give you that.&lt;/p&gt;&#xA;" OwnerUserId="2287" LastEditorUserId="2287" LastEditorDisplayName="" LastEditDate="2010-11-03T00:09:59.997" LastActivityDate="2010-11-03T00:09:59.997" CommentCount="5" />
  <row Id="8700" PostTypeId="2" ParentId="8672" CreationDate="2010-11-02T23:20:27.903" Score="7" ViewCount="" Body="&lt;p&gt;As a two-by-two matrix, applying &lt;a href=&quot;http://en.wikipedia.org/wiki/Jacobi_eigenvalue_algorithm&quot; rel=&quot;nofollow&quot;&gt;Jacobi&lt;/a&gt; in fact gives the answer at once!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is known that given a two-by-two matrix $\mathbf{A}$, one can construct an orthogonal matrix $\mathbf{V}=\bigl(\begin{smallmatrix}c&amp;amp;s\\-s&amp;amp;c\end{smallmatrix}\bigr)$ such that $\mathbf{V}^T\mathbf{A}\mathbf{V}$ is diagonal, where the two numbers $c$ and $s$ satisfy $c^2+s^2=1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the off-diagonal elements are not zero (why?), computing $c$ and $s$ is as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\tau=\frac{a_{22}-a_{11}}{2a_{12}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$t=\frac{\mathrm{sgn}(\tau)}{|\tau|+\sqrt{1+\tau^2}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$c=\frac1{\sqrt{1+t^2}}\qquad s=ct$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The eigendecomposition easily follows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(See Golub and Van Loan's excellent &lt;a href=&quot;http://books.google.com/books?id=mlOa7wPX6OYC&amp;amp;pg=PA426&quot; rel=&quot;nofollow&quot;&gt;book&lt;/a&gt; for further details.)&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The point was apparently missed that this method returns an eigensystem with orthogonal eigenvectors, so I suppose I should spell out in code how Jacobi works here, with a &lt;em&gt;Mathematica&lt;/em&gt; example:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;a = N[{{3, -1}, {-1, 2}}, 20];&#xA;&#xA;tau = (a[[2, 2]] - a[[1, 1]])/a[[1, 2]]/2;&#xA;&#xA;t = Sign[tau]/(Abs[tau] + Sqrt[1 + tau^2]);&#xA;&#xA;{c, s} = {1, t}/Sqrt[1 + t^2]&#xA;{0.85065080835203993218, 0.5257311121191336060}&#xA;&#xA;{l1, l2} = {a[[1, 1]] - t a[[1, 2]], a[[2, 2]] + t a[[1, 2]]}&#xA;{3.6180339887498948482, 1.3819660112501051518}&#xA;&#xA;Eigenvalues[a] == {l1, l2}&#xA;True&#xA;&#xA;Eigenvectors[a] == {{c, -s}, {s, c}}&#xA;True&#xA;&#xA;a.{c, -s} == l1 {c, -s}&#xA;True&#xA;&#xA;a.{s, c} == l2 {s, c}&#xA;True&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="498" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-04T00:19:14.947" LastActivityDate="2010-11-04T00:19:14.947" CommentCount="2" />
  <row Id="8701" PostTypeId="1" AcceptedAnswerId="8715" CreationDate="2010-11-02T23:31:18.547" Score="0" ViewCount="139" Body="&lt;p&gt;I need to figure out the Probability that Y &gt; 23 for a chi-square distribution with degrees of freedom = 7.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The book says to do this with some applet. I don't know what it's talking about.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can I do this on my Ti-89 (preferable) or on some website somewhere?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT/UPDATE:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found a program on my ti-89 for chi-square distribution. However it asks me for lower and upper bound and degrees of freedom. I put in 7 for degrees of freedom but i'm not sure how to translate probability Y &gt; 23 into an upper and lower bound?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I put in lower = 0, upper = 23, degrees of freedom = 7 and I said to auto-scale and I got .998295. Is that the right probability?&lt;/p&gt;&#xA;" OwnerUserId="3049" LastEditorUserId="3049" LastEditorDisplayName="" LastEditDate="2010-11-03T00:03:25.807" LastActivityDate="2010-11-03T02:17:13.953" Title="chi-square distribution with v = 7 degrees of freedom Ti-89?" Tags="&lt;probability&gt;&lt;homework&gt;&lt;probability-theory&gt;" AnswerCount="1" CommentCount="9" />
  <row Id="8702" PostTypeId="2" ParentId="8695" CreationDate="2010-11-02T23:49:33.017" Score="5" ViewCount="" Body="&lt;p&gt;Your inequality $(A-I)x \ge b$ has no solutions in $x$ as soon as $b&gt;0$. Indeed, any potential solution would have to satisfy $A x \ge  x + b$ and, since rows of $A$ are nonnegative and sum to one, each element of vector $Ax$ is a convex combination of the components of $x$, which must be less than $x_{max}$, the largest component of $x$. On the other hand, at least one element of $x+b$ is greater than $x_{max}$, which proves the impossibility.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the way, applying &lt;a href=&quot;http://en.wikipedia.org/wiki/Farkas%27_lemma&quot; rel=&quot;nofollow&quot;&gt;Farkas' Lemma&lt;/a&gt; to this impossible system shows that the following always admits solutions in y&#xA;$$y^T (A-I) = 0, y \ge 0 \text{ and } y^T b &gt; 0$$&#xA;which expresses the fact that $A$ necessarily admits a nonnegative left eigenvector with eigenvalue $1$ (the last inequality ensures that $y$ is nonzero). &lt;/p&gt;&#xA;" OwnerUserId="55" LastEditorUserId="55" LastEditorDisplayName="" LastEditDate="2010-11-03T00:39:15.747" LastActivityDate="2010-11-03T00:39:15.747" CommentCount="1" />
  <row Id="8703" PostTypeId="2" ParentId="8690" CreationDate="2010-11-02T23:58:06.797" Score="0" ViewCount="" Body="&lt;p&gt;I tried the following approach where I assumed the limit existed and attempted to get a contradiction. However, the best I could do was to get the same answer as David above that if (in the extremely unlikely case that) the limit exists, it will be 1. I just wanted to make a record of the approach here in case someone sees any potential to this approach.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $x_{n+1} = \cos((n+1)n!)$, it can be written as a $(n+1)$th degree polynomial in $\cos(n!)$. Should the limit exist, it will be a fixed point of this polynomial. Using standard multiple angle formulas (found &lt;a href=&quot;http://mathworld.wolfram.com/Multiple-AngleFormulas.html&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; for ex.), we can show that the limit if it exists should equal $1$.&lt;/p&gt;&#xA;" OwnerUserId="618" LastActivityDate="2010-11-02T23:58:06.797" CommentCount="7" />
  <row Id="8704" PostTypeId="2" ParentId="8684" CreationDate="2010-11-03T00:35:39.200" Score="2" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\ $ The usual descent argument works: $\: $ if $\rm\ x^2-D\:y^2 = 1\ $ has fundamental solution $\rm\:\epsilon^k\:$ and $\rm\:\epsilon^n&gt;0\:$ is another solution then $\rm\:k|n\:$ (else $\rm\:\epsilon^{n\ mod\ k}\:$ is a solution smaller than the fundamental one)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the hint is not detailed enough then you can find full details of this standard argument e.g. in section 1.3 of Jacobson and Williams: Solving the Pell equation. There you'll also find the standard table giving the value of $\rm\ k$ (viz. $\rm\:k|6\:$ and depends on the parity of $\rm x$ and $\rm y$ and the sign of $\pm4)$. Note in particular that the complete proof is not as trivial as the special case in Pete L. Clark's answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason why &quot;4 is special&quot;, i.e. why the equation $\rm\ x^2 - D\:y^2 = \pm 4\ $ proves more fundamental than $\rm\ x^2 - D\:y^2 = \pm 1\ $ is simply because the former equation is the one that arises when checking that an integer has norm $\pm 1$, i.e. $\rm\ N((x+y\sqrt{D})/2) = \pm1$ iff $\rm\ x^2-D\:y^2 = \pm 4\:$; i.e. the 4 arises from clearing &quot;denominators&quot; in the norm equation.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-11-03T01:56:28.973" LastActivityDate="2010-11-03T01:56:28.973" CommentCount="1" />
  <row Id="8705" PostTypeId="2" ParentId="8674" CreationDate="2010-11-03T00:54:20.703" Score="0" ViewCount="" Body="&lt;p&gt;If you assume there are N colored balls, with equal frequency p=1/N, then this should be pretty do-able.  Of course, you're desired confidence interval will affect the answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Without working out all the details, consider that you have drawn the point where you see a few colors 2 or 3 times.  By considering how many balls you have drawn, and how many uniques have been witnessed, you can make an estimate of that N within a confidence bounds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can give the details if you like, but if these assumptions are acceptable you can probably take it from here?&lt;/p&gt;&#xA;" OwnerUserId="2677" LastActivityDate="2010-11-03T00:54:20.703" />
  <row Id="8706" PostTypeId="2" ParentId="8690" CreationDate="2010-11-03T01:17:56.027" Score="0" ViewCount="" Body="&lt;p&gt;Above posters seem to be right, I thought I had this solved for the limit = 1, but realized I was wrong.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If its of any use, here is my faulty reasoning. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;blockquote&gt;&#xA;    &lt;p&gt;As we get more and more factors we will only get closer and closer to a factor that is a multiple of pi.  We know that n! is even for any n &gt;=2, so we get a abritarily closer to 2*pi|n! as n approaches infinity.&#xA;    Sorry of the code is out of form, but:&lt;/p&gt;&#xA;    &#xA;    &lt;blockquote&gt;&#xA;      &lt;p&gt;eps = .00001&lt;br&gt;&#xA;      for i in xrange(2,800000,2):&lt;br&gt;&#xA;      ...     if i*math.pi-math.floor(i*math.pi) &amp;lt; eps: print i&lt;br&gt;&#xA;      ...&lt;br&gt;&#xA;      431230&lt;br&gt;&#xA;      530762&lt;br&gt;&#xA;      630294  &lt;/p&gt;&#xA;      &#xA;      &lt;p&gt;math.cos(math.floor(431230*math.pi)) =&#xA;      0.99999999997167566&lt;br&gt;&#xA;      So any n! for n &gt; 431230*pi will be at least this close, or closer, to 1  &lt;/p&gt;&#xA;    &lt;/blockquote&gt;&#xA;  &lt;/blockquote&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The problem is that product of the factors NOT close to pi is going to grow faster than that epsilon shrinks.  And obviously no integer is going to be an exact multiple of pi (as it is irrational).  Therefore, I am led to agree with other posters that this limit is unlikely.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS - I know this isn't rigorous by any means, but was hoping maybe better minds would have some insightful comments.&lt;/p&gt;&#xA;" OwnerUserId="2677" LastActivityDate="2010-11-03T01:17:56.027" />
  <row Id="8707" PostTypeId="1" AcceptedAnswerId="8709" CreationDate="2010-11-03T01:18:33.007" Score="3" ViewCount="88" Body="&lt;p&gt;Prove or disprove that closed 3-manifolds which are not simply connected cannot be embedded in three-dimensional Euclidean spaces. I am not a mathematics major and I am taking introductory topology this semester. But I need to apply this result for my research. Any help is much appreciated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks,&#xA;Srikanth.&lt;/p&gt;&#xA;" OwnerUserId="3050" LastActivityDate="2010-11-03T01:57:51.563" Title="embedding of closed 3-manifolds." Tags="&lt;topology&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="8708" PostTypeId="2" ParentId="8561" CreationDate="2010-11-03T01:24:07.513" Score="1" ViewCount="" Body="&lt;p&gt;Per fmartin's request, I've collected my above comments into an answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The proposed approach doesn't work. The 2nd last equation is $c = de$, where $e = (b+1)\cdots (a+b)$. Therefore $(a+b)! = d(b+1)\cdots (a+b)a!b! = da!(a+b)!$ implies $1 = d a!$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that Sierpinski's inductive proof is expressed much more clearly by explicitly mentioning the underlying binomial identity that enables the descent, viz. $\binom{a+b}a = \binom{a+b-1}a + \binom{a+b-1}b$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For various integrality proofs of binomial coefficients see also &lt;a href=&quot;http://math.stackexchange.com/questions/2192&quot;&gt;this thread&lt;/a&gt;. There you'll find a very simple proof I discovered that shows how to write a binomial coefficient as a product of fractions whose denominators are all coprime to any given prime p.&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-11-03T01:24:07.513" />
  <row Id="8709" PostTypeId="2" ParentId="8707" CreationDate="2010-11-03T01:25:11.523" Score="4" ViewCount="" Body="&lt;p&gt;If your manifold is closed you must be using some definition that ensures it's compact?  So if $f : M \to \mathbb R^3$ is an embedding, isn't the image simultanously compact and open? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;edit in response to your comment:  if $f : M \to \mathbb R^3$ is an embedding, let $B \subset M$ be an open subset of $M$ which is homeomorphic to an open ball in $\mathbb R^3$.  You need to argue that $f(B)$ is open in $\mathbb R^3$.  If your embedding is smooth there's a big theorem from calculus that gives you the result.  If you're talking about topological embeddings you're going to need a tool. Have you studied the &quot;invariance of dimension&quot; theorem? &lt;/p&gt;&#xA;" OwnerUserId="642" LastEditorUserId="642" LastEditorDisplayName="" LastEditDate="2010-11-03T01:57:51.563" LastActivityDate="2010-11-03T01:57:51.563" CommentCount="2" />
  <row Id="8710" PostTypeId="2" ParentId="8556" CreationDate="2010-11-03T01:32:29.707" Score="1" ViewCount="" Body="&lt;p&gt;Here's the linear programming approach (with considerable help from another user, Fanfan).  Consider the LP&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\min 0^T y$$&#xA;subject to&#xA;$$(X - I)y \geq 1,$$&#xA;where $0$ and $1$ are, respectively, vectors containing all 0's and all 1's.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By Fanfan's &lt;a href=&quot;http://math.stackexchange.com/questions/8695/&quot;&gt;answer to my question&lt;/a&gt; this LP is infeasible.  Thus its dual,&#xA;$$\max 1^T x$$ &#xA;subject to &#xA;$$x^T (X-I) = 0^T,$$&#xA;$$x \geq 0,$$&#xA;is either infeasible or unbounded.  But $x = 0$ is a solution, and so this dual problem is feasible.  Thus it must be unbounded.  But that means there must be a nonzero solution $x_1$ in its feasible region.  Thus we have $x_1^T X = x_1^T$ with $x_1 \geq 0$, $x_1 \neq 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Fanfan's answer to my question also includes another answer to your question - one that uses Farkas's Lemma rather than LP duality.  It ends up being quite similar to my answer here, as, of course, Farkas's Lemma and LP duality are basically equivalent.)&lt;/p&gt;&#xA;" OwnerUserId="2370" LastActivityDate="2010-11-03T01:32:29.707" />
  <row Id="8711" PostTypeId="1" AcceptedAnswerId="8712" CreationDate="2010-11-03T01:34:59.780" Score="5" ViewCount="310" Body="&lt;p&gt;Can a derivative operation commute over an integral operation irrespective of the properties of the function under the integral ?&lt;/p&gt;&#xA;" OwnerUserId="2987" LastActivityDate="2010-11-03T02:02:30.277" Title="Derivative commuting over integral." Tags="&lt;calculus&gt;&lt;integral&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" />
  <row Id="8712" PostTypeId="2" ParentId="8711" CreationDate="2010-11-03T01:57:00.190" Score="7" ViewCount="" Body="&lt;p&gt;Not in general.  I recommend Gelbaum and Olmsted's &lt;em&gt;&lt;a href=&quot;http://books.google.com/books?id=cDAMh5n4lkkC&amp;amp;printsec=frontcover&amp;amp;source=gbs_atb#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;Counterexamples in Analysis&lt;/a&gt;&lt;/em&gt;, which is where I turned to find a counterexample to your question.  Namely, example 15 on page 123 is titled&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A function $f$ for which $d/dx\int_a^b f(x,y)dy\neq\int_a^b[\partial/\partial x f(x,y)]dy$, although each integral is proper.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The example is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f(x,y) = \left\{&#xA;     \begin{array}{lr}&#xA;       \frac{x^3}{y^2}e^{-x^2/y} &amp;amp; : y&gt;0, \\&#xA;       0 &amp;amp; : y=0, &#xA;     \end{array}&#xA;   \right.&#xA;$$&#xA;integrated with respect to $y$ from $0$ to $1$.  Actually, differentiating under the integral sign works here except where $x=0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The function and its partial derivative are not jointly continuous.  When they are jointly continuous, differentiation and integration commute.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2010-11-03T02:02:30.277" LastActivityDate="2010-11-03T02:02:30.277" CommentCount="7" />
  <row Id="8713" PostTypeId="1" AcceptedAnswerId="8714" CreationDate="2010-11-03T02:08:43.687" Score="8" ViewCount="168" Body="&lt;p&gt;This operation is similar to discrete convolution and cross-correlation, but has binomial coefficients:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f(n)\star g(n)=\sum_{k=0}^n \binom{n}{k}f(n-k)g(k) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Particularly,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$a^n\star b^n=(a+b)^n$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;following binomial theorem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I just wonder if there is a name for such operation and where I can read about its properties.&lt;/p&gt;&#xA;" OwnerUserId="2513" LastActivityDate="2010-11-03T03:41:32.010" Title="How this operation is called?" Tags="&lt;combinatorics&gt;&lt;binomial-coefficients&gt;" AnswerCount="3" FavoriteCount="2" />
  <row Id="8714" PostTypeId="2" ParentId="8713" CreationDate="2010-11-03T02:15:10.590" Score="11" ViewCount="" Body="&lt;p&gt;It's called a &lt;em&gt;binomial convolution&lt;/em&gt; in Graham, Knuth, and Patashnik's &lt;em&gt;Concrete Mathematics&lt;/em&gt;.  I don't have that text in front of me (but I bet someone here can give you a page number), but here's a reference on the &lt;a href=&quot;http://fermatslasttheorem.blogspot.com/2006/10/generating-function-for-bernoulli.html&quot; rel=&quot;nofollow&quot;&gt;Fermat's Last Theorem blog&lt;/a&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It would also be worth checking out Section 2.3 of Wilf's &lt;em&gt;Generatingfunctionology&lt;/em&gt;.  This is on exponential generating functions.  The property of interest is that if $F(x)$ and $G(x)$ are the exponential generating functions of $f(n)$ and $g(n)$, respectively, then $F(x)G(x)$ is the exponential generating function of $f(n) \star g(n)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(FYI: You can download the second edition of &lt;em&gt;Generatingfunctionology&lt;/em&gt; from &lt;a href=&quot;http://www.math.upenn.edu/~wilf/DownldGF.html&quot; rel=&quot;nofollow&quot;&gt;Wilf's web site&lt;/a&gt;.)&lt;/p&gt;&#xA;" OwnerUserId="2370" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-11-03T03:41:32.010" LastActivityDate="2010-11-03T03:41:32.010" CommentCount="2" />
  <row Id="8715" PostTypeId="2" ParentId="8701" CreationDate="2010-11-03T02:17:13.953" Score="3" ViewCount="" Body="&lt;p&gt;There's an online calculator &lt;a href=&quot;http://www.fourmilab.ch/rpkp/experiments/analysis/chiCalc.html&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;. Plugging $\chi^2 = 23$ and $df=7$, we get $.0017$. Your calculator calculated the probability that $0 \leq \chi^2 \leq 23$, which is $1-.0017 = .9983$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As I commented above, you could tell that you needed to take $1-p$ instead of $p$ since $\chi^2 &gt; 23$ is an unlikely event, given that the expectation is $7$ and the standard deviation is $\sqrt{14} \approx 4$ (Chebyshev and even Markov would give you that $p &amp;lt; 1/2$).&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-03T02:17:13.953" CommentCount="2" />
  <row Id="8716" PostTypeId="2" ParentId="8713" CreationDate="2010-11-03T02:18:33.313" Score="1" ViewCount="" Body="&lt;p&gt;It corresponds to multiplying exponential generating functions (check &lt;em&gt;generatingfunctionology&lt;/em&gt;).&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-03T02:18:33.313" />
  <row Id="8717" PostTypeId="1" CreationDate="2010-11-03T02:26:13.460" Score="0" ViewCount="73" Body="&lt;p&gt;Most of what I've read defines the support of a distribution topologically if it defines support precisely; e.g the smallest closed subset that has full measure (i.e. the intersection of all closed subsets of full measure). I've seen allusions to purely measure-theoretic definitions, and I think I even came across one once, but I can't find it. Is there a measure-theoretic definition of support that everyone agrees on?&lt;/p&gt;&#xA;" OwnerUserId="2122" LastActivityDate="2010-11-03T12:31:45.490" Title="Non-topological notion of &quot;support&quot; of a probability distribution" Tags="&lt;probability-theory&gt;" AnswerCount="1" />
  <row Id="8718" PostTypeId="1" AcceptedAnswerId="8721" CreationDate="2010-11-03T02:43:42.140" Score="0" ViewCount="135" Body="&lt;p&gt;Let $P \in \mathbb{Z}[x], P(x) = \displaystyle\sum\limits_{j=0}^n a_j x^j, a_n \neq 0$ and $a_0 \neq 0$; if $p/q$ is a root of P (with p and q coprimes) then $p|a_0$ and $q|a_n$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've managed to prove the first part ($p|a_0$) and I suppose I'm not far from proving the second, though I'd really like some feedback since I'm just starting with making proofs of my own.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proof:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$P(x) = a_n(x-p/q)\displaystyle\prod\limits_{j=2}^n (x-r_j)$, with $r_j$ being the other n-1 roots of P(x).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It follows that $a_0 = a_n(-p/q)\displaystyle\prod\limits_{j=2}^n (-r_j)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, $-p/q|a_0$ and obviously $p/q|a_0$. Rephrasing, $a_0 = l\frac{p}{q} = \frac{l}{q} p$ with $l \in \mathbb{Z}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This implies $p|a_0$ if $l/q \in \mathbb{Z}$, but this is trivial since $q|lp$ and q and p are coprimes, so $q|l$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, $p|a_0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the second part, we want to see that $q_i|a_n \forall i \leq n$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We define $d$ as the least common multiple of $\{q_1, q_2,...,q_n\}$. Then, $q_i|a_n \forall i \leq n \iff d|a_n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, it follows that $d|\displaystyle\prod\limits_{j=1}^n q_j$, so we want to see that $a_n = l \displaystyle\prod\limits_{j=1}^n q_j$ with $l \in \mathbb{Z}$. Here's where I have my doubts with the proof as I have no way to show that l is indeed an integer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rearranging the previously given equation for $a_0$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a_n = a_0 \displaystyle\prod\limits_{j=1}^n \frac{-q_j}{p_j}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the previous reasoning, as $p_i|a_0 \forall i \leq n$, then $a_0 = k \displaystyle\prod\limits_{j=1}^n p_j$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Replacing $a_0$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a_n = k \displaystyle\prod\limits_{j=1}^n -q_j$, which is equivalent to $q|a_n$ as shown earlier.&lt;/p&gt;&#xA;" OwnerUserId="1523" LastActivityDate="2010-11-03T04:59:31.007" Title="Proving Gauss' polynomial theorem" Tags="&lt;number-theory&gt;&lt;linear-algebra&gt;&lt;polynomials&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="8719" PostTypeId="2" ParentId="8713" CreationDate="2010-11-03T02:47:13.480" Score="3" ViewCount="" Body="&lt;p&gt;Remark that one can employ the powerful umbral calculus to compute closed forms for many such &lt;strong&gt;binomial convolutions&lt;/strong&gt; of special functions, e.g. from p. 161 of Roman: Umbral Calculus:&lt;br&gt;&#xA;&lt;img src=&quot;http://i.imgur.com/Wt0Ii.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-11-03T02:47:13.480" CommentCount="3" />
  <row Id="8720" PostTypeId="1" AcceptedAnswerId="8722" CreationDate="2010-11-03T02:49:15.527" Score="2" ViewCount="162" Body="&lt;p&gt;I am looking for a simple proof to show that given a countable set of natural numbers $C$ that is closed under addition and whose gcd is 1, there exists two elements $c_1, c_2 \in C$ such that $\gcd(c_1, c_2)=1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I used the Chinese remainder theorem, but my proof is about half a page and I was wondering if there is a shorter one.&lt;/p&gt;&#xA;" OwnerUserId="774" LastEditorUserId="774" LastEditorDisplayName="" LastEditDate="2010-11-03T03:04:17.730" LastActivityDate="2010-11-03T04:25:11.403" Title="Chinese remainder theorem" Tags="&lt;modular-arithmetic&gt;" AnswerCount="2" CommentCount="5" />
  <row Id="8721" PostTypeId="2" ParentId="8718" CreationDate="2010-11-03T02:58:16.417" Score="2" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\displaystyle\rm\ \ 0 = q^n\ P(\frac{p}q)\ =\ a_n\ p^n + q \ (\cdots)\ \Rightarrow\ q\:|\: a_n\:p^n\ \Rightarrow\ q\:|\:a_n\ $ by Euclid's Lemma.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;NOTE&lt;/strong&gt; $\ $  This result is not usually called Gauss' polynomial theorem but, rather, the &lt;a href=&quot;http://en.wikipedia.org/wiki/Rational_root_theorem&quot; rel=&quot;nofollow&quot;&gt;Rational Root Test / Theorem&lt;/a&gt; - which is a special case of &lt;a href=&quot;http://en.wikipedia.org/wiki/Gauss%27s_lemma%28polynomial%29&quot; rel=&quot;nofollow&quot;&gt;Gauss's Lemma&lt;/a&gt; for polynomials. Generally if $\rm\:\alpha\:$ is an algebraic number with primitive minimal polynomial $\rm\: f(x)\in \mathbb Z[x]\:$ and if $\rm\: g(\alpha) = 0\:$ for $\rm\: g(x)\in \mathbb Z[x]\:$ then  $\rm\ f(x)|g(x)\ in\ \mathbb Z[x]$.  The above is the special case $\rm\ \alpha = p/q,\ \ f(x) = q\:x - p,\ \ \gcd(p,q) = 1\:$. Namely, $\rm\ (q\:x-p)\:|\:g\in \mathbb Z[x]\ \Rightarrow\ q|g_n,\ p|g_0\ in\ \mathbb Z\: $.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-11-03T04:59:31.007" LastActivityDate="2010-11-03T04:59:31.007" CommentCount="1" />
  <row Id="8722" PostTypeId="2" ParentId="8720" CreationDate="2010-11-03T03:15:59.653" Score="3" ViewCount="" Body="&lt;p&gt;Because $\gcd(C)=1$ there are $a_1,a_2,\ldots,a_n\in C$ such that $\gcd(a_1,\ldots,a_n)=1$.  Thus there are integers $m_1,\ldots m_n$ such that $m_1a_1+\cdots m_na_n=1$.  Let $\{i_1,\ldots,i_p\}\subset\{1,\ldots,n\}$ be those indices corresponding to positive coefficients $m_{i_k}$ and $\{j_1,\ldots,j_q\}$ those corresponding to negative coefficients $m_{j_k}$.  Because $C$ is closed under addition, $c_1=\sum_{k=1}^pm_{i_k}a_{i_k}$ and $c_2=\sum_{k=1}^q-m_{j_k}a_{j_k}$ are in $C$, and $c_1-c_2=1$.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-11-03T03:15:59.653" CommentCount="2" />
  <row Id="8723" PostTypeId="2" ParentId="8718" CreationDate="2010-11-03T03:23:00.520" Score="1" ViewCount="" Body="&lt;p&gt;If you know how to prove the first part, just apply it to the polynomial $t^n P(t^{-1})$.&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2010-11-03T03:23:00.520" CommentCount="2" />
  <row Id="8724" PostTypeId="1" CreationDate="2010-11-03T03:24:28.007" Score="3" ViewCount="207" Body="&lt;p&gt;The Riemann Zeta Function is defined as $ \displaystyle \zeta(s) = \sum\limits_{n=1}^{\infty} \frac{1}{n^s}$. It is not absolutely convergent or conditionally convergent for $\text{Re}(s) \leq 1$. Using analytic continuation, one can derive the fact that $\displaystyle \zeta(-s) = -\frac{B_{s+1}}{s+1}$ where $B_{s+1}$ are the Bernoulli numbers. Can one obtain this result without resorting to analytic continuation? &lt;/p&gt;&#xA;" OwnerUserId="2548" LastEditorUserId="2548" LastEditorDisplayName="" LastEditDate="2010-11-03T15:11:23.513" LastActivityDate="2010-11-03T15:11:23.513" Title="Riemann Zeta Function and Analytic Continuation" Tags="&lt;complex-analysis&gt;" CommentCount="6" FavoriteCount="2" />
  <row Id="8725" PostTypeId="2" ParentId="8698" CreationDate="2010-11-03T04:15:05.770" Score="0" ViewCount="" Body="&lt;p&gt;It's a pity this example isn't really describe anywhere. Assuming that there's a triple intersection at $(1,2)$, we get the equation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$k - 3 - \sin(\alpha) - \cot(\alpha)&lt;em&gt;(2 + 2&lt;/em&gt;\cos(\alpha) - k) = 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Presumably you can get more equations by assuming that the tilted squares touch each other and corners of the straight squares, but these seem not to be satisfied by the given $k,\alpha$.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;How to find more equations: the tilted squares come in three groups (two, two and one), and all seem to be just touching each other. For each of the groups, one vertex has one known coordinate (the middle group might have a known vertex at $(1,2)$, but I don't know if that's true). You can get equations from the fact that these groups touch certain corners of the straight squares, and from the (apparent) fact that they just touch each other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The simplest way to get these equations is to define vectors $s,t$ for the sides of the tilted square (they can be expressed in terms of $\alpha$), and now all these touchings amount to pairs of points $p,q$ which are parallel to either $s$ or $t$. So $p-q$ is proportional to $s$ (or $t$), and we get an equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, pursuing this route, I get conflicting equations (conflicting in the sense that when I substitute $k$ and $\alpha$, they are contradictory). Maybe you can try and tell us if it works out for you.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2010-11-03T04:21:01.943" LastActivityDate="2010-11-03T04:21:01.943" />
  <row Id="8726" PostTypeId="2" ParentId="8720" CreationDate="2010-11-03T04:25:11.403" Score="0" ViewCount="" Body="&lt;p&gt;(as requested)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose on the contrary that $t_1 \equiv t_2 \equiv 0 \pmod p$ where $p = \gcd(t_1, t_2) &gt; 1$.&#xA;Then, there must be a $t_3 \nequiv 0 \pmod p$ otherwise all elements are divisible by $p &gt; 1$, which is a contradiction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $t_1 = pq$ and $t_2 = pr$.&#xA;By the Chinese remainder theorem, for all $s \ge qr$, there are $c_1, c_2 \in \mathbb{Z}$ such that &#xA;$&#xA;s = c_1q + c_2r&#xA;$.&#xA;Let $s = p^a$ where $a \in \mathbb{N}$ is chosen so that $p^a \ge qr$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then,&#xA;$&#xA;c_1t_1 + c_2t_2 = c_1pq + c_2pr = p(c_1q + c_2r) = ps  = p^{a+1}.&#xA;$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence $\gcd(c_1t_1 + c_2t_2, t_3) = 1$ since the prime factorization of $c_1t_1 + c_2t_2$ consists of $p^{a+1}$ and $t_3$ is relatively prime to $p$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $C$ is closed under addition, $c_1t_1 + c_2t_2\in C$.  Thus, it and $t_3$ are the two elements that have g.c.d. 1.&lt;/p&gt;&#xA;" OwnerUserId="774" LastActivityDate="2010-11-03T04:25:11.403" />
  <row Id="8727" PostTypeId="1" CreationDate="2010-11-03T04:28:29.360" Score="2" ViewCount="230" Body="&lt;p&gt;How would I start off proving that S= (the set of symmetric nxn matrices) is a manifold. &#xA;I tried using the definition directly by saying&#xA;Mn = the space of all nxn matrices&#xA;For every A in Mn there exists open sets U=V=Mn and a bijection F: U-&gt;V by F(A)= A-A^T&#xA;Therefore we have&#xA;F(U ∩ S) = F(S) since S is a subset of Mn&#xA;={0} ∩ Mn&#xA;this is where I get stuck.&#xA;Also, I know that the set of all symmetric nxn matrices is (n^2 + n)/2, therefore that is the dimension of the manifold&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Definition: A set M (subset of R^n) is a k-dimensional manifold if for every x in M there exists open sets U, V and a bijection h:U-&gt;V with &#xA;x in U and &#xA;H(U ∩ M = V ∩ (R^k X {c^(k+1),...,c^n}) for all c's constants&lt;/p&gt;&#xA;" OwnerUserId="2319" LastEditorUserId="2319" LastEditorDisplayName="" LastEditDate="2010-11-03T04:42:11.097" LastActivityDate="2010-11-03T07:16:39.937" Title="The set of symmetric matrices as a manifold" Tags="&lt;differential-geometry&gt;" AnswerCount="1" CommentCount="10" />
  <row Id="8728" PostTypeId="2" ParentId="8667" CreationDate="2010-11-03T04:32:21.173" Score="7" ViewCount="" Body="&lt;p&gt;The answer is no. &lt;a href=&quot;http://cms.math.ca/cmb/v12/cmb1969v12.0337-0338.pdf&quot; rel=&quot;nofollow&quot;&gt;H. Herrlich&lt;/a&gt;  showed, in 1969, that even if you consider each $A_n$ a completely regular space, the direct limit may fail to be Hausdorff. However if all $A_n$ are T$_4$ - spaces then $X$ is a T$_4$ - space (it's not hard to prove this).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A comment about the definition of direct limit. Usually, in category theory, we call &lt;a href=&quot;http://en.wikipedia.org/wiki/Direct_limit&quot; rel=&quot;nofollow&quot;&gt;direct limit&lt;/a&gt; a colimit of a directed family of objects. Using this terminology it's well known that the category of Hausdorff spaces isn't closed under direct limits. You can find some examples in Dugundji's 'Topology' (a shame it's out of print). The definition you are using is very particular, so Herrlich's example is special.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In &lt;a href=&quot;http://dml.cz/bitstream/handle/10338.dmlcz/700787/Toposym_03-1971-1_36.pdf&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; paper D. Hajek and G. Strecker exhibit sufficient conditions for the Hausdorff property to be preserved under direct limits.&lt;/p&gt;&#xA;" OwnerUserId="329" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2010-11-03T19:16:50.850" LastActivityDate="2010-11-03T19:16:50.850" CommentCount="1" />
  <row Id="8729" PostTypeId="2" ParentId="8688" CreationDate="2010-11-03T04:52:21.843" Score="4" ViewCount="" Body="&lt;p&gt;What a spectacularly rubbish way to define convergence. Here's how to think of it in a simple, yet rigorous way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, suppose we have a sequence $S_{n} = \frac{1}{2^n}$. We know that $S_{0} = 1, S_{1} = \frac{1}{2}, S_{2} = \frac{1}{4}$, and so on and so forth, just as you would have defined it in Calculus. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We might now be interested in the sum of $\sum_{n=0}^{\infty} S_{n} = 1 + \frac{1}{2} + \frac{1}{4} + ...$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You may remember that this series will converge to 2. How should we express this analytically?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since we cannot actually sum up an infinite number of terms, we'll never actually get to 2 if we sum this up by hand. However, suppose I only wanted to get a value CLOSE to 2. How close? Let's say we want our sum to be within 0.5 of 2. That is, we want the sum to be between $2-0.5$ and $2+0.5$. This is the interval $(1.5,2.5)$. I don't have to sum up all of the terms to get in this interval. In fact, I need to sum up only 3 terms: $1 + \frac{1}{2} + \frac{1}{4} = 1.75$ to be in this interval.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clearly, $\sum_{n=0}^{3} S_{n}$ is in the interval. We can, if we want to, sum up more than 3 terms, and we will still remain in the interval. The point is that any number of terms more than 3 will still keep us in the interval. We can also say that we need at least the partial sum $S_{2}$ (which is our third term).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, suppose we wanted our interval to be smaller, between $2-0.01$ and $2+0.01$, which is the interval (1.99,2.01). It turns out you need 8 terms for our series to get within this interval. After these eight terms (i.e. after the partial sum $S_7$), your sequence stays in the interval.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In our above examples, we wanted to get within 0.5 or 0.01 of the point of convergence. These are our epsilons. That is $(2-\epsilon, 2 + \epsilon)$ is our interval, and we can set $\epsilon$ to be anything we like. We chose to set epsilon to 0.5 or 0.01, but we could have chosen a smaller number.And therein lies the idea of convergence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You see, no matter how small I chose my epsilon (as long as it is greater than $0$), I can still find a point where my sequence enters the interval $(2-\epsilon, 2 + \epsilon)$, and never leaves it afterwards. When epsilon was 0.5, we needed $S_2$, when epsilon was 0.01, we needed $S_7$. If we can demonstrate that no matter HOW SMALL you make your epsilon, there is a $S_n$ that is in that interval (to stay), you have shown convergence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How would I actually do this? Well, $\sum_{n=0}^{k} \frac{1}{2^n}$ is a geometric series equal to $2^{-k}(2^{k+1}-1)$. There is a formula for this in your textbook, I presume.&#xA;If we want $2^{-k}(2^{k+1}-1) &gt; 2 - \epsilon$ we can solve algebraically to get that $k&gt; -\frac{\log{\epsilon}}{\log{2}}$. If we set $\epsilon=0.01$, we'll get that $k&gt;6.64...$. This means that we need at least $S_7$ as I mentioned above. Either way, the mere existence of this function demonstrates that we can always find a sufficiently large k.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why does this suffice, conceptually? In order for you to argue that the sequence doesn't converge to a particular point, you need to show that the sequence is always some distance away from my point. Yet, no matter how small you make that distance, it is not small enough, for the sequence gets into that interval, and remains there. &lt;/p&gt;&#xA;" OwnerUserId="647" LastEditorUserId="647" LastEditorDisplayName="" LastEditDate="2010-11-03T05:01:45.280" LastActivityDate="2010-11-03T05:01:45.280" CommentCount="5" />
  <row Id="8730" PostTypeId="1" AcceptedAnswerId="8733" CreationDate="2010-11-03T05:04:02.890" Score="2" ViewCount="119" Body="&lt;p&gt;I presume that it is because one needs to take a limit to define the derivative, but is there another reason?&lt;/p&gt;&#xA;" OwnerUserId="3008" LastActivityDate="2010-11-03T06:08:10.527" Title="Why are holomorphic functions defined in reference to an open set?" Tags="&lt;complex-analysis&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="0" />
  <row Id="8731" PostTypeId="1" AcceptedAnswerId="8790" CreationDate="2010-11-03T05:46:54.333" Score="3" ViewCount="186" Body="&lt;p&gt;What are some algorithms used for finding the shortest path between 2 points on a (riemannian) manifold (the manifold may have a smooth boundary)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So far I've had 3 ideas, none of which seem that good:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) (Rubber band principle) Find any path between the two points, then draw the path taut like a stretched rubber band with some damping.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) (Ray-tracing analog) Shoot out geodesic rays from one point in all directions. If a ray hits the target, we're done. If the ray hits the boundary and is not close to tangent, ignore it. If the ray hits the boundary approximately tangent, then follow the geodesic on the boundary in that direction, shooting tangent rays off occasionally.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) (Discretization) Triangulate/tetrahedralize/etc the manifold, and construct a weighted graph where the nodes correspond to the centerpoints of each tetrahedron, the edges correspond to tetrahedra that touch, and the weights correspond to the distance between tetrahedra centers. Then compute the shortest path in the graph.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most of the stuff I've seen in the literature is concerned with the theoretical existence of the path rather than actually computing it. How do you find the shortest path in practice?&lt;/p&gt;&#xA;" OwnerUserId="3060" LastEditorUserId="237" LastEditorDisplayName="" LastEditDate="2010-11-05T00:00:47.273" LastActivityDate="2010-11-05T00:00:47.273" Title="Algorithm for shortest path on manifold" Tags="&lt;geometry&gt;&lt;algorithms&gt;&lt;manifold&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" />
  <row Id="8732" PostTypeId="1" AcceptedAnswerId="8738" CreationDate="2010-11-03T06:05:30.183" Score="2" ViewCount="154" Body="&lt;p&gt;The following two identities comes from my trigonometry module without any sort of proof,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $A + B + C = \pi $ then,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\tan A + \tan B + \tan C = tan A \cdot tan B \cdot tan C$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \tan \frac{A}{2} \cdot \tan \frac{B}{2} + \tan \frac{B}{2} \cdot \tan \frac{C}{2} + \tan \frac{C}{2} \cdot \tan \frac{A}{2} = 1 $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS:I am not much sure about whether the first one is fully correct or not, so if not please suggest the correct one and also I will be grateful if somebody suggest a suitable method (may be using mathematica) to verify an identity like this prior to proving.&lt;/p&gt;&#xA;" OwnerUserId="3744" LastActivityDate="2010-11-03T11:19:19.183" Title="How to prove this trignometrical Identities ?" Tags="&lt;trigonometry&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="8733" PostTypeId="2" ParentId="8730" CreationDate="2010-11-03T06:08:10.527" Score="6" ViewCount="" Body="&lt;p&gt;You may want to take a look at &lt;em&gt;When is a Function that Satisfies the Cauchy-Riemann Equations Analytic?&lt;/em&gt; by J. D. Gray and S. A. Morris, The American Mathematical Monthly, Vol. 85, No. 4, (Apr., 1978), pp. 246-256. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yes, in a sense, we want open domains so limits can be computed, but we have a notion of directional limit, so this is not much of a reason; what I mean is, some justification is needed that we actually lose something otherwise. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The paper above gives examples showing how the Cauchy-Riemann equations alone do not suffice for analyticity, &lt;em&gt;even in the presence of continuity&lt;/em&gt;, and proceeds to look for weak additions to these two requirements. &lt;/p&gt;&#xA;" OwnerUserId="462" LastActivityDate="2010-11-03T06:08:10.527" />
  <row Id="8734" PostTypeId="1" AcceptedAnswerId="8745" CreationDate="2010-11-03T06:27:04.517" Score="5" ViewCount="396" Body="&lt;p&gt;After learning about the binary number system(only 2 symbols i.e. 0 and 1), I just thought why did we adopt the decimal number system (10 symbols) after all?&lt;br&gt;&#xA;I mean if you go to see, it's rather inefficient when compared to the octal(8 symbols) and the hexadecimal(16 symbols)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please tag this appropriately.&lt;/p&gt;&#xA;" OwnerUserId="3062" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-03T07:05:01.787" LastActivityDate="2010-11-03T22:05:08.627" Title="why have we chosen our number system to be decimal (base 10)" Tags="&lt;number-systems&gt;" AnswerCount="5" CommentCount="8" FavoriteCount="1" />
  <row Id="8735" PostTypeId="2" ParentId="8732" CreationDate="2010-11-03T06:32:24.073" Score="1" ViewCount="" Body="&lt;p&gt;Hints:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Prelude.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a. If $\alpha+\beta+\gamma=\pi$, then $\gamma=\pi-\alpha-\beta$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;b. $\tan(\pi-\theta)=-\tan(\theta)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;c. $\tan(\alpha+\beta)=\frac{\tan(\alpha)+\tan(\beta)}{1-\tan(\alpha)\tan(\beta)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;d. $\tan\left(\frac{\pi}{2}-\theta\right)=\frac1{\tan(\theta)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Act I.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\tan(\alpha)+\tan(\beta)-\frac{\tan(\alpha)+\tan(\beta)}{1-\tan(\alpha)\tan(\beta)}=-\tan(\alpha)\tan(\beta)\frac{\tan(\alpha)+\tan(\beta)}{1-\tan(\alpha)\tan(\beta)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can the left-hand side be made to look like the right-hand side?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Act II.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\tan\left(\frac{\alpha}{2}\right)\tan\left(\frac{\beta}{2}\right)+\left(\tan\left(\frac{\alpha}{2}\right)+\tan\left(\frac{\beta}{2}\right)\right)\left(\frac{\tan\left(\frac{\alpha}{2}\right)+\tan\left(\frac{\beta}{2}\right)}{1-\tan\left(\frac{\alpha}{2}\right)\tan\left(\frac{\beta}{2}\right)}\right)^{-1}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Simplify the above expression.&lt;/p&gt;&#xA;" OwnerUserId="498" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-03T11:19:19.183" LastActivityDate="2010-11-03T11:19:19.183" CommentCount="1" />
  <row Id="8737" PostTypeId="2" ParentId="8727" CreationDate="2010-11-03T07:16:39.937" Score="1" ViewCount="" Body="&lt;p&gt;If you don't want to explicitly construct the manifold from charts, you could instead use the regular value theorem. Consider the &quot;how antisymmetric is this matrix&quot; map, f:Rnxn -&gt; R, f(A)=||A-A^t||. If zero is a regular value of f, then the regular value theorem says that the inverse image of 0 (the set of matrices that are symmetric) must be a submanifold of Rnxn. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So one could bound Df(A)h from below with the reverse triangle inequality:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;(||A+sh-(A+sh)^t||-||A-A^t||)/s &#xA;&gt;= (||A-A^t|| - s||h-h^t|| - ||A-A^t||)/s&#xA;= ||h-h^t||&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;(or ... &amp;lt;= -||h-h^t|| depending on sign)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By judiciously choosing the matrix h, you can make this any real number, so Df(A) is onto. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact the result is even stronger since we didn't use the fact that A is symmetric - one could imagine the sets of &quot;equally antisymmetric&quot; matrices (set of all A where ||A-A^t||=c) as submanifolds layered on top of each other that foliate Rnxn.&lt;/p&gt;&#xA;" OwnerUserId="3060" LastActivityDate="2010-11-03T07:16:39.937" />
  <row Id="8738" PostTypeId="2" ParentId="8732" CreationDate="2010-11-03T07:16:43.713" Score="6" ViewCount="" Body="&lt;p&gt;If $A+B+C= \pi \Longrightarrow \tan(A+B) = \tan(\pi -C) =-\tan(C)$. So we have $$\tan(A+B)= \frac{\tan(A) + \tan(B)}{1 - \tan(A)\cdot \tan(B)} = -\tan(C) $$ $$\Longrightarrow \tan(A)+\tan(B) = -\tan(C) \cdot \Bigl[1 - \tan(A)\tan(B)\Bigr]$$ from which the first one follows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And for the second one, we have $\displaystyle\frac{A}{2} + \frac{B}{2} =\frac{\pi}{2}- \frac{C}{2} \Longrightarrow \tan\Bigl(\frac{A+B}{2}\Bigr)= \cot\Bigl(\frac{C}{2}\Bigr)$ Now expanding we have $$\tan\Bigl(\frac{A+B}{2}\Bigr)= \frac{\tan\Bigl(\frac{A}{2}\Bigr) + \tan\Bigl(\frac{B}{2}\Bigr)}{1- \tan\Bigl(\frac{A}{2}\Bigr)\cdot \tan\Bigl(\frac{B}{2}\Bigr)} = \cot\Bigl(\frac{C}{2}\Bigr)$$ Multiplying both sides by $\tan\frac{C}{2}$ we have $$\tan\Bigl(\frac{C}{2}\Bigr) \cdot \Bigl[ \tan\Bigl(\frac{A}{2}\Bigr) + \tan\Bigl(\frac{B}{2}\Bigr) \Bigr] = 1 \cdot \Bigl[ 1 - \tan\Bigl(\frac{A}{2}\Bigr) \cdot \tan\Bigl(\frac{B}{2}\Bigr)\Bigr]$$&lt;/p&gt;&#xA;" LastActivityDate="2010-11-03T10:45:48.867" CommentCount="3" />
  <row Id="8739" PostTypeId="1" AcceptedAnswerId="8741" CreationDate="2010-11-03T08:13:20.627" Score="5" ViewCount="100" Body="&lt;p&gt;While reading about general topology, I started looking at the metric topology. At one point, the interior of a topological space $X$ is defined as such:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$${E}^{\circ}=\{p\in E\ | \ B_p(\epsilon)\subseteq E, \epsilon&gt;0\}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here, $B_p(\epsilon)$ is a ball with center $p$ and radius $\epsilon$. I tried verifying that this met the general properties of an interior, and didn't have much trouble showing ${E}^{\circ}\subseteq E$, $(E\cap F)^{\circ}={E}^{\circ}\cap {F}^{\circ}$, and ${X}^{\circ}=X$. However, I don't see how ${{E}^{\circ}}^{\circ}={E}^{\circ}$. I see that ${{E}^{\circ}}^{\circ}\subseteq {E}^{\circ}$, but tried without success to see the other containment. Is there a reason why this follows simply from the axioms of what a metric is?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Futhermore, apparently another way to approach this topology is to define the family of open sets $\mathcal{T}$ as those which are the union of a family of balls. Again, I tried verifying this for myself. I see that for any $\mathcal{U}\subseteq\mathcal{T}$, $\cup\mathcal{U}$ is a union of sets which each may be represented as the union of a family of balls, so the whole union again may be represented as the union of a family of balls, namely those in each set in $\mathcal{U}$. Also, $X$ may be represented as the union of balls where I take one ball with center $p$ for all $p\in X$. Finally $\emptyset$ is just an empty union of balls. However, I was unable to show that for finite $\mathcal{U}\subseteq\mathcal{T}$, $\cap\mathcal{U}$ is again open. Is there some way to see that the finite intersection could be represented as a union of balls from the properties of the metric?&lt;/p&gt;&#xA;" OwnerUserId="1583" LastActivityDate="2010-11-03T08:30:07.950" Title="Verifying some basics of the metric topology" Tags="&lt;topology&gt;&lt;metric-spaces&gt;" AnswerCount="2" />
  <row Id="8740" PostTypeId="2" ParentId="8739" CreationDate="2010-11-03T08:30:05.797" Score="3" ViewCount="" Body="&lt;p&gt;For the first problem, show that for every point in $E^{\circ}$ there is a ball around it which is also in $E^{\circ}$.  In other words, show that interiors are open.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the second problem, do the same: show that for every point in the intersection there is a ball around it which is also in the intersection.  &lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-11-03T08:30:05.797" CommentCount="1" />
  <row Id="8741" PostTypeId="2" ParentId="8739" CreationDate="2010-11-03T08:30:07.950" Score="2" ViewCount="" Body="&lt;p&gt;A hint for your first question: show that if $p \in E^{\circ}$, then there exists an $\epsilon &gt; 0$ such that $B_p(\epsilon)$ is contained in $E^{\circ}$. This shows that $p$ lies in the interior of $E^{\circ}$, i.e., $p \in (E^{\circ})^{\circ}$.  This comes down to showing that if you have a point $p$ in an open ball $B$, you can find a smaller opne ball $B'$ around $p$ which is entirely contained in $B$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A hint for your second question: it's enough to show that if $B_1$ and $B_2$ are two open balls and $p \in B_1 \cap B_2$ (i.e., $p$ lies in both balls), then there is a small ball $B_3$ around $p$ such that $B_3$ is contained in both $B_1$ and $B_2$.  This is actually quite similar to the exercise of the previous paragraph.  &lt;/p&gt;&#xA;" OwnerUserId="299" LastActivityDate="2010-11-03T08:30:07.950" CommentCount="1" />
  <row Id="8742" PostTypeId="1" AcceptedAnswerId="8757" CreationDate="2010-11-03T10:20:03.050" Score="2" ViewCount="196" Body="&lt;p&gt;It's a really simple question. However I didn't see it in books and I tried to find the answer on the web but failed.&#xA;If I have two independent random variables, X1 and X2, then I define two other random variables Y1 and Y2, where Y1 = f1(X1) and Y2 = f2(X2).&#xA;Intuitively, Y1 and Y2 should be independent, and I can't find a counter example, but I am not sure. Could anyone tell me whether they are independent? Does it depend on some properties of f1 and f2?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you.&lt;/p&gt;&#xA;" OwnerUserId="3066" LastActivityDate="2010-11-03T14:07:55.147" Title="Are functions of independent variables also independent?" Tags="&lt;probability&gt;" AnswerCount="2" />
  <row Id="8743" PostTypeId="2" ParentId="8742" CreationDate="2010-11-03T10:45:41.317" Score="5" ViewCount="" Body="&lt;p&gt;Yes, they are independent. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are studying rigorous probability course with sigma-algebras then you may prove it by noticing that the sigma-algebra generated by $f_{1}(X_{1})$ is smaller than the sigma-algebra generated by $X_{1}$, where $f_{1}$ is borel-measurable function. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are studying an introductory course - then just remark that this theorem is consistent with our intuition: if $X_{1}$ does not contain info about $X_{2}$ then $f_{1}(X_{1})$ does not contain info about $f_{2}(X_{2})$.&lt;/p&gt;&#xA;" OwnerUserId="2481" LastActivityDate="2010-11-03T10:45:41.317" CommentCount="1" />
  <row Id="8744" PostTypeId="2" ParentId="8732" CreationDate="2010-11-03T10:56:05.480" Score="0" ViewCount="" Body="&lt;p&gt;More generally, we have from &lt;a href=&quot;http://en.wikipedia.org/wiki/De_Moivre%27s_formula&quot; rel=&quot;nofollow&quot;&gt;De Moivre's theorem&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\cos(\alpha_1+\alpha_2+\cdots+\alpha_n)=&#xA;\text{Re}\prod_{k=1}^n(\cos \alpha_k + i \sin \alpha_k)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sin(\alpha_1+\alpha_2+\cdots+\alpha_n)=&#xA;\text{Im}\prod_{k=1}^n(\cos \alpha_k + i \sin \alpha_k)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and so&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\tan(\alpha_1+\alpha_2+\cdots+\alpha_n)=&#xA;\frac{ \text{Im}\prod_{k=1}^n ( 1 + i \tan \alpha_k) }{  \text{Re}\prod_{k=1}^n(1 + i \tan\alpha_k) }.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the case $n=3.$&lt;/p&gt;&#xA;" OwnerUserId="1301" LastEditorUserId="1301" LastEditorDisplayName="" LastEditDate="2010-11-03T11:07:10.867" LastActivityDate="2010-11-03T11:07:10.867" />
  <row Id="8745" PostTypeId="2" ParentId="8734" CreationDate="2010-11-03T10:58:22.657" Score="5" ViewCount="" Body="&lt;p&gt;Expanding on the comment by J.M., let me quote from the (highly recommended) book by Georges Ifrah &lt;a href=&quot;http://books.google.com/books?id=FMTI7rwevZcC&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;The Universal History of Numbers&lt;/em&gt;&lt;/a&gt; (Wiley, 2000, pp. 21-22):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Traces of the anthropomorphic origin of counting systems can be found in many languages. In the Ali language (Central Africa), for example, &quot;five&quot; and &quot;ten&quot; are respectively &lt;em&gt;moro&lt;/em&gt; and&#xA;  &lt;em&gt;mbouna&lt;/em&gt;: &lt;em&gt;moro&lt;/em&gt; is actually the word for &quot;hand&quot; and &lt;em&gt;mbouna&lt;/em&gt; is a contraction of &lt;em&gt;moro&lt;/em&gt; (&quot;five&quot;) and &lt;em&gt;bouna&lt;/em&gt;, meaning &quot;two&quot; (thus &quot;ten&quot;=&quot;two hands&quot;).&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;It is therefore very probable that the Indo-European, Semitic and Mongolian words for the first ten numbers derive from expressions related to finger-counting. But this is an unverifiable hypothesis, since the original meanings of the names of the numbers have been lost.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Ifrah then goes on to explain that&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;...the hand makes the two complementary aspects of integers entirely intuitive. It serves as an instrument permitting natural movement between cardinal and ordinal numbering. If you need to show that a set contains three, four, seven or ten elements, you raise or bend &lt;em&gt;simultaneously&lt;/em&gt; three, four, seven or ten fingers, using your hand as cardinal mapping. If you want to count out the same things, then you bend or raise three, four, seven or ten fingers in &lt;em&gt;succession&lt;/em&gt;, using the hand as an ordinal counting tool.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="723" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-03T11:15:30.873" LastActivityDate="2010-11-03T11:15:30.873" CommentCount="4" />
  <row Id="8746" PostTypeId="1" CreationDate="2010-11-03T11:15:51.843" Score="1" ViewCount="56" Body="&lt;p&gt;Consider a cube, with standard topological faces: 6 faces, 12 edges and 8 vertices.&#xA;Define a &lt;em&gt;cube part&lt;/em&gt; as set of faces (possible of different dimension) so that no 2 set elements share same vertex.&#xA;E.g. there is no cube part consisting of 3 2-dimensional faces, because there have to be 2 faces that share a vertex.&#xA;Note, cube part doesn't have to contain all cube vertices.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How many different cube parts there are, counting topologically isomorphic cube parts as same?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think this is application of Redfield–Pólya Theorem, but I'm not familiar with topic and some hints will be usefull.&lt;/p&gt;&#xA;" OwnerUserId="3037" LastActivityDate="2010-11-03T11:15:51.843" Title="Number of sets of cube topological faces" Tags="&lt;combinatorics&gt;" CommentCount="2" />
  <row Id="8747" PostTypeId="2" ParentId="8734" CreationDate="2010-11-03T11:41:34.840" Score="1" ViewCount="" Body="&lt;p&gt;Because it makes the metric system so much simpler :).&lt;/p&gt;&#xA;" OwnerUserId="448" LastActivityDate="2010-11-03T11:41:34.840" CommentCount="2" />
  <row Id="8748" PostTypeId="2" ParentId="8734" CreationDate="2010-11-03T12:01:16.870" Score="6" ViewCount="" Body="&lt;p&gt;I think the answer here might be, that the guys who thought base 10 was a good idea had the largest sticks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If one trusts the wikipedia, the Babylonians had a base 60 system, which can still be felt today with this &quot;60 minutes in an hour&quot; nonesense, and a (related) base 12 system was widely in use too. There are still unique words for &quot;eleven&quot; and &quot;twelve&quot;, as well as expressions as &quot;a dozen&quot;. After all, you can count to twelve using a single hand.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, there was the base 1 latin system, and (wikipedia again) a base 20 system for the mayan, who obvsiously had no shoes, since they could count their toes, too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Something as easy as &quot;base 10 is natural for humans&quot; does not explain it all. =)&lt;/p&gt;&#xA;" OwnerUserId="505" LastActivityDate="2010-11-03T12:01:16.870" CommentCount="3" />
  <row Id="8749" PostTypeId="2" ParentId="8250" CreationDate="2010-11-03T12:11:21.250" Score="9" ViewCount="" Body="&lt;p&gt;If you want an intuition that explains &lt;em&gt;why&lt;/em&gt; Farka's Lemma should be true, you will have to use the geometric interpretation; there's no way around that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want an intuition that shows &lt;em&gt;what&lt;/em&gt; Farka's Lemma achieves (but not why it should be true), there is indeed an algebraic explanation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Namely, Farka's Lemma answers the following question: what is a sufficient and necessary condition for the system of equations&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{array}{rcl}a_{11}x_1 + a_{12}x_2 + \dots + a_{1n}x_n &amp;amp;=&amp;amp; b_1 \\&#xA;a_{21}x_1 + a_{22}x_2 + \dots + a_{2n}x_n &amp;amp;=&amp;amp; b_2 \\&#xA;&amp;amp;\vdots&amp;amp; \\&#xA;a_{m1}x_1 + a_{m2}x_2 + \dots + a_{mn}x_n &amp;amp;=&amp;amp; b_m&#xA;\end{array}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;to have a positive solution $x_i\ge 0$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Oviously, if one of the equations has the form&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ c_1x_1 + c_2x_2 + \dots + c_nx_n = b $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where all the $c_i$ are positive $c_i &gt; 0$, but the right-hand side $b$ is negative $b&amp;lt;0$, then it's clearly impossible to choose all the $x_i$ positive and have them satisfy the equation, because the left-hand side would always be positive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This was only for a single equation, but it is equally obvious that if any &lt;em&gt;linear combination&lt;/em&gt; of the equations of your system&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \left(\sum y_i a_{i1}\right)x_1 + \left(\sum y_i a_{i2}\right)x_2 + \dots + \left(\sum y_i a_{in}\right) x_n = \sum y_i b_i $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;has this form with positive coefficients on the left- and a negative number on the right-hand side, then you can't solve it for $x_i \ge 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, what Farka's Lemma says is that this is in fact the &lt;strong&gt;only bad thing&lt;/strong&gt; that can happen. If the system&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$Ax = b$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;doesn't have a solution $x\ge0$, then this must be because there exists a nasty linear combination of some equations&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(y^TA)x = y^Tb$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;that obstructs solvability because it satisfies $y^TA \ge 0$ but $y^Tb \le 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a sense, Farka's Lemma shows that the non-solvability of the system can be &quot;certified&quot;; the vector $y$ is a &quot;certificate&quot; for the fact that the system cannot be solved. Such certificates, commonly called &lt;strong&gt;obstructions&lt;/strong&gt;, are common in other branches of mathematics, like algebraic geometry (&lt;a href=&quot;http://terrytao.wordpress.com/2007/11/26/hilberts-nullstellensatz/&quot; rel=&quot;nofollow&quot;&gt;Hilbert's Nullstellensatz&lt;/a&gt;) or algebraic topology (homology, cohomology).&lt;/p&gt;&#xA;" OwnerUserId="434" LastActivityDate="2010-11-03T12:11:21.250" />
  <row Id="8750" PostTypeId="1" CreationDate="2010-11-03T12:28:58.783" Score="13" ViewCount="354" Body="&lt;p&gt;The &lt;a href=&quot;http://mathworld.wolfram.com/ReuleauxPolygon.html&quot; rel=&quot;nofollow&quot;&gt;Reuleaux polygons&lt;/a&gt; are analogs of the regular polygons, except that the &quot;sides&quot; are composed of circle arcs instead of lines. It is known that for an odd number of sides, e.g. the Reuleaux triangle, the polygon has constant width.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After reading the paper &lt;a href=&quot;http://www.maa.org/pubs/sampMMA.pdf&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;Roads and Wheels&lt;/em&gt;&lt;/a&gt; by Stan Wagon and Leon Hall, I got curious on how one might construct the appropriate &quot;road&quot; for Reuleaux wheels; i.e., finding the curve such that when a Reuleaux polygon rolls on it, the axle at the centroid of the polygon experiences no vertical displacement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My problem is that it does not seem straightforward, at least to me, how to construct the corresponding differential equation for the road, as presented in the paper. Since circles roll on horizontal lines, and equiangular spirals roll on inclined lines, I would suppose that the road needed for a rolling Reuleaux would not be piecewise linear. &lt;a href=&quot;http://demonstrations.wolfram.com/ARollingReuleauxTriangle/&quot; rel=&quot;nofollow&quot;&gt;This&lt;/a&gt; demonstrates that the &quot;road&quot; cannot be a horizontal line for a Reuleaux triangle, as the axle does not remain level when the curve is rolling.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, how does one construct the road? A solution for just the Reuleaux triangle would be fine, but a general solution is much better.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2011-02-22T10:25:52.817" Title="Reuleaux Rollers" Tags="&lt;geometry&gt;" AnswerCount="2" CommentCount="11" FavoriteCount="4" />
  <row Id="8751" PostTypeId="2" ParentId="8717" CreationDate="2010-11-03T12:31:45.490" Score="1" ViewCount="" Body="&lt;p&gt;Here is a comment made by Dmitri Pavlov for &lt;a href=&quot;http://mathoverflow.net/questions/42310/when-is-l2x-separable/42314#42314&quot; rel=&quot;nofollow&quot;&gt;this MathOverFlow question&lt;/a&gt;.&#xA;Maybe it will be helpful to you.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The support of a measure $m$ on a measurable space is defined as follows. Consider the Boolean algebra of all measurable sets modulo null sets (i.e., two measurable sets are equivalent if their symmetric difference is a measure $0$ set). This Boolean algebra is complete, as described in the link. Now take the supremum of all elements $p$ of this algebra such that $m$ vanishes on $p$. The complement of this supremum is the support of $m$. It is extremely important to factor out the null sets, otherwise the above procedure doesn't make sense. – Dmitri Pavlov Oct 16 at 15:09&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="940" LastActivityDate="2010-11-03T12:31:45.490" CommunityOwnedDate="2010-11-03T12:31:45.490" />
  <row Id="8752" PostTypeId="1" CreationDate="2010-11-03T12:35:37.730" Score="10" ViewCount="589" Body="&lt;p&gt;Four Color Theorem is equivalent to the statement: &quot;Every cubic planar bridgeless graphs is 3-edge colorable&quot;. There is computer assisted proof given by Appel and Haken. Dick Lipton in of his &lt;a href=&quot;http://rjlipton.wordpress.com/2009/04/24/the-four-color-theorem/&quot; rel=&quot;nofollow&quot;&gt;beautiful blogs&lt;/a&gt;&#xA;&lt;a href=&quot;http://rjlipton.wordpress.com/2009/04/24/the-four-color-theorem/&quot; rel=&quot;nofollow&quot;&gt;&lt;/a&gt; posed the following open problem:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Are there non-computer based proofs of the Four Color Theorem?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Surprisingly, While I was reading this paper, &#xA;Anshelevich and Karagiozova, &lt;a href=&quot;http://portal.acm.org/citation.cfm?id=1250790.1250849&quot; rel=&quot;nofollow&quot;&gt;Terminal backup, 3D matching, and covering cubic graphs&lt;/a&gt;&#xA;&lt;a href=&quot;http://portal.acm.org/citation.cfm?id=1250790.1250849&quot; rel=&quot;nofollow&quot;&gt;&lt;/a&gt;, the authors state that Cahit  proved that &quot;every 2-connected cubic planar graph is edge-3-colorable&quot; which is equivalent to the Four Color Theorem (I. Cahit, Spiral Chains: The Proofs of Tait's and Tutte's Three-Edge-Coloring Conjectures. arXiv preprint, math CO/0507127 v1, July 6, 2005).&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Does Cahit's proof resolve the open problem in Lipton's blog by providing non-computer based proof for the Four Color Theorem? Why isn't Cahit's proof widely known and accepted?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Cross posted on cstheory.stackexchange.com as &lt;a href=&quot;http://cstheory.stackexchange.com/questions/2597/human-checkable-proof-of-the-four-color-theorem-closed&quot;&gt;Human checkable proof of the Four Color Theorem?&lt;/a&gt;&#xA;&lt;a href=&quot;http://cstheory.stackexchange.com/questions/2597/human-checkable-proof-of-the-four-color-theorem-closed&quot;&gt;&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="2459" LastActivityDate="2010-12-04T16:44:25.170" Title="Human checkable proof of the Four Color Theorem?" Tags="&lt;graph-theory&gt;" AnswerCount="1" CommentCount="9" />
  <row Id="8753" PostTypeId="1" AcceptedAnswerId="8778" CreationDate="2010-11-03T12:54:54.727" Score="3" ViewCount="147" Body="&lt;p&gt;Let $$\nu(x) = \frac{1}{2} \log{2\pi x} - x + \int\limits_{1}^{x} \frac{\lfloor{t\rfloor}}{t} \ dt$$ then what is $\displaystyle \lim_{x \to \infty} x \cdot \nu(x)$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I really don't know to proceed. Tried something using the expansions of $\log$ and seeing how the integral behaves, but couldn't figure out. &lt;/p&gt;&#xA;" LastActivityDate="2010-11-03T23:32:08.490" Title="Evaluating a given limit" Tags="&lt;number-theory&gt;&lt;analysis&gt;" AnswerCount="1" />
  <row Id="8755" PostTypeId="1" AcceptedAnswerId="8756" CreationDate="2010-11-03T13:22:36.367" Score="3" ViewCount="115" Body="&lt;p&gt;Working on the &lt;a href=&quot;http://en.wikipedia.org/wiki/Birthday_problem&quot; rel=&quot;nofollow&quot;&gt;generalized birthday problem&lt;/a&gt;, where you draw with replacement from $\{1,2,3, \ldots,d\}$ and look for the number of draws $n$ for which you have greater than $1/2$ chance of a match I said&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{d!}{d^d(d-n)!} \lt \frac{1}{2}$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using Stirling's approximation  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{d^d \exp (d-n)} {\exp(d) d^n (d-n)^{(d-n)}} \lt \frac{1}{2}$&lt;br&gt;&#xA;$\frac{1}{\exp(n)}(1+\frac{n}{d-n})^{(d-n)}\lt \frac{1}{2}$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Taking logs and using the Taylor expansion  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(d-n) \ln(1+\frac{n}{d-n})-n \lt -\ln (2)$&lt;br&gt;&#xA;$(d-n) (\frac{n}{d-n}-\frac{n^2}{2(d-n)^2})\lt -\ln(2)$&lt;br&gt;&#xA;$n^2&gt;2 \ln(2) (d-n)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ignoring the n in comparison with d in the last term, because if we wanted it we should keep the next term in the expansion of ln, we get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$n&gt;\sqrt{2 \ln (2) d}$  which agrees with the published result&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;This is fine, but like much math writing it obscures the difficulties of finding the solution.  Before taking logs I wanted to say &#xA;$(1+\frac{n}{d-n})^{(d-n)} \rightarrow \exp(n)$ but that left me with the unhelpful $1 \lt \frac{1}{2}$.  Seeing what happened after taking logs, I need the next term in the expansion as the first one cancels.  I know how to expand a Taylor series as far as I want, but I don't know what the next term in this one is.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2010-11-03T13:36:28.990" LastActivityDate="2010-11-03T14:03:02.883" Title="Next term in $(1+a/n)^n \rightarrow \exp (n)$" Tags="&lt;approximation&gt;&lt;taylor-expansion&gt;" AnswerCount="2" FavoriteCount="2" />
  <row Id="8756" PostTypeId="2" ParentId="8755" CreationDate="2010-11-03T13:31:34.247" Score="4" ViewCount="" Body="&lt;p&gt;$$\left(1+\frac{x}{n}\right)^{n}=e^x\left(1-\frac{x^2}{2n}+\frac{x^3(8+3x)}{24n^2}+\cdots\right).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This &lt;a href=&quot;http://www.maa.org/pubs/Calc_articles/ma052.pdf&quot; rel=&quot;nofollow&quot;&gt;note&lt;/a&gt; might be helpful.&lt;/p&gt;&#xA;" OwnerUserId="723" LastEditorUserId="723" LastEditorDisplayName="" LastEditDate="2010-11-03T13:50:08.433" LastActivityDate="2010-11-03T13:50:08.433" CommentCount="3" />
  <row Id="8757" PostTypeId="2" ParentId="8742" CreationDate="2010-11-03T13:42:20.710" Score="0" ViewCount="" Body="&lt;p&gt;For any two (measurable) sets $A_i$, $i=1,2$, $Y_i \in A_i$ if and only if $X_i \in B_i$, where $B_i$ are the sets { $s : f_i (s) \in A_i$ }. Hence, since the $X_i$ are independent, ${\rm P}(Y_1 \in A_1 , Y_2 \in A_2) = {\rm P}(Y_1 \in A_1) {\rm P}(Y_2 \in A_2)$. Thus, the  $Y_i$ are independent (which is intuitively clear anyway). [We have used here that random variables $Z_i$, $i=1,2$, are independent if and only if ${\rm P}(Z_1 \in C_1 , Z_2 \in C_2) = {\rm P}(Z_1 \in C_1) {\rm P}(Z_2 \in C_2)$ for any two measurable sets $C_i$.]&lt;/p&gt;&#xA;" OwnerUserId="2810" LastEditorUserId="2810" LastEditorDisplayName="" LastEditDate="2010-11-03T14:07:55.147" LastActivityDate="2010-11-03T14:07:55.147" CommentCount="2" />
  <row Id="8758" PostTypeId="2" ParentId="8337" CreationDate="2010-11-03T14:00:17.530" Score="2" ViewCount="" Body="&lt;p&gt;Another variation. We make use of the following identity (proved at the bottom of this note):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{k=1}^n \cot^2 \left( \frac {2k-1}{2n} \frac{\pi}{2} \right) = 2n^2 – n. \quad (1)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now $1/\theta &gt; \cot \theta &gt; 1/\theta - \theta/3 &gt; 0$ for $0&amp;lt; \theta&amp;lt; \pi/2 &amp;lt; \sqrt{3}$ and so&#xA;$$ 1/\theta^2 – 2/3 &amp;lt; \cot^2 \theta &amp;lt; 1/\theta^2. \quad (2)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With $\theta_k = (2k-1)\pi/4n,$ summing the inequalities $(2)$ from $k=1$ to $n$ we obtain   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$2n^2 – n &amp;lt; \sum_{k=1}^n \left( \frac{2n}{2k-1}\frac{2}{\pi} \right)^2 &amp;lt; 2n^2 – n + 2n/3.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\pi^2}{16}\frac{2n^2-n}{n^2} &amp;lt; \sum_{k=1}^n \frac{1}{(2k-1)^2} &amp;lt;&#xA;\frac{\pi^2}{16}\frac{2n^2-n/3}{n^2}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Taking the limit as $n \rightarrow \infty$ we obtain&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$  \sum_{k=1}^\infty \frac{1}{(2k-1)^2} = \frac{\pi^2}{8},$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;from which the result for $\sum_{k=1}^\infty 1/k^2$ follows easily.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To prove $(1)$ we note that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \cos 2n\theta = \text{Re}(\cos\theta + i \sin\theta)^{2n} = &#xA;\sum_{k=0}^n (-1)^k {2n \choose 2k}\cos^{2n-2k}\theta\sin^{2k}\theta.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\cos 2n\theta}{\sin^{2n}\theta} =  \sum_{k=0}^n (-1)^k {2n \choose 2k}\cot^{2n-2k}\theta.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And so setting $x = \cot^2\theta$ we note that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f(x) = \sum_{k=0}^n (-1)^k {2n \choose 2k}x^{n-k}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;has roots $x_j = \cot^2 (2j-1)\pi/4n,$ for $j=1,2,\ldots,n,$ from which $(1)$ follows since&#xA;${2n \choose 2n-2} = 2n^2-n.$&lt;/p&gt;&#xA;" OwnerUserId="1301" LastActivityDate="2010-11-03T14:00:17.530" />
  <row Id="8759" PostTypeId="2" ParentId="8755" CreationDate="2010-11-03T14:03:02.883" Score="2" ViewCount="" Body="&lt;p&gt;$$\rm\quad \bigg(1+\frac{a}n\bigg)^{\:n}\ =\ e^{n\log(1+a/n)}\ =\ e^a\ e^{n\log(1+a/n)-a}\ $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\rm n\:\log \bigg(1+\frac{a}n\bigg)-a\ =\ \frac{a^2}{2\:n} + \frac{a^3}{3\:n^2} +\: \cdots $$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\rm e^{n\log(1+a/n)-a}\ =\ 1 - \frac{a^2}{2\:n} + \frac{3a^4 + 8a^3}{24\: n^2} +\: \cdots $$&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-11-03T14:03:02.883" />
  <row Id="8760" PostTypeId="1" CreationDate="2010-11-03T14:37:32.057" Score="0" ViewCount="235" Body="&lt;p&gt;What are the three cube roots of -1?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Not sure if this is a trick question, But I have been asked this. one of the ansers is -1, what are the other 2?&lt;/p&gt;&#xA;" LastEditorUserId="120" LastEditorDisplayName="" LastEditDate="2010-11-03T14:52:29.903" LastActivityDate="2010-11-03T22:33:41.657" Title="What are the three cube roots of -1?" Tags="&lt;complex-numbers&gt;" AnswerCount="3" CommentCount="3" />
  <row Id="8761" PostTypeId="1" CreationDate="2010-11-03T14:40:48.463" Score="0" ViewCount="69" Body="&lt;p&gt;This is a strange question, it might be easy for all you math wizards or it even may be impossible. If you don't understand what I mean let me know so I can change the way I post the question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Product A can be built using Product B.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To get Product B I have to transform product C or Product D.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a 50% chance to get product B by transforming product C and 25% to get product B by transforming product D.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any way to predict the cost of production for Product A knowing the price for Product C and D&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit 1&lt;/strong&gt;&#xA;There are no transformation costs, only the costs for the materials &lt;/p&gt;&#xA;" OwnerUserId="3068" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-03T16:29:10.010" LastActivityDate="2010-11-03T16:29:10.010" Title="Price Calculation based on probabilities" Tags="&lt;probability&gt;" AnswerCount="1" />
  <row Id="8762" PostTypeId="2" ParentId="8760" CreationDate="2010-11-03T14:44:57.043" Score="2" ViewCount="" Body="&lt;p&gt;Write $-1$ in polar form as $e^{i\pi}$. In general, the cube roots of $r e^{i\theta}$ are given by $r^{1/3}e^{i\theta/3}$, $r^{1/3}e^{i(\theta/3 + 2\pi /3)}$ and $r^{1/3}e^{i(\theta/3 + 4\pi /3)}$. In your case $r = 1$ and $\theta = \pi$, so your cube roots are $e^{i\pi / 3}$, $e^{i\pi}$, and $e^{i 5\pi/ 3}$. Put back into rectangular form, they are ${1 \over 2} + i{\sqrt{3} \over 2}$, $-1$, and ${1 \over 2} - i{\sqrt{3} \over 2}$.&lt;/p&gt;&#xA;" OwnerUserId="3035" LastActivityDate="2010-11-03T14:44:57.043" CommentCount="1" />
  <row Id="8763" PostTypeId="2" ParentId="8760" CreationDate="2010-11-03T14:45:11.680" Score="0" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\ $ Let $\rm\ \ x\ \to\ -x\ \ $ in $\rm\displaystyle\ \ \frac{1-x^3}{1-x}\ =\ 1+x+x^2\:.\ $  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Generally suppose $\rm\:f(x)\:$ is a polynomial over a field with roots $\rm\: a \ne b\:$. Then $\rm\ f(x) = (x-a)\ g(x)\ $  hence $\rm\: f(b) = 0\: \Rightarrow\ (a-b)\:g(b) = 0\ \Rightarrow\ g(b) = 0\ $ i.e. $\rm\:b\:$ is a root of $\rm\ f(x)/(x-a)\:$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From a factorization perspective, the reason that this works is because, over a domain, monic linear polynomials are prime, so the linear factors of a polynomial are unique, i.e. the roots and their multiplicity are unique. e.g. see my &lt;a href=&quot;http://math.stackexchange.com/questions/7622&quot;&gt;post here&lt;/a&gt;. This fails over coefficient rings that are not domains, i.e. have zero-divisors, e.g. $\rm\ x^2-1 = (x-1)(x+1) = (x-4)(x+4)\ $ over $\ \mathbb Z/15\:$. Here, although $4 \ne 1$ is a root of $\rm\ x^2 - 1$ it is not true that 4 is a root of $\rm\ (x^2-1)/(x-1) = x+1\:$. For the example at hand we have $\rm\ x^3 + 1 = (x+1)(x+9)(x-10) = (x+16)(x+22)(x-38)\ $ over $\ \mathbb Z/91\:$.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-11-03T16:00:28.907" LastActivityDate="2010-11-03T16:00:28.907" />
  <row Id="8764" PostTypeId="1" CreationDate="2010-11-03T14:58:36.130" Score="1" ViewCount="77" Body="&lt;p&gt;I'm working to understand some noise on some of our analog-to-digital converter signals. I've analyzed a log of 2^18 samples in MATLAB, and can make the following observations:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;the noise consists of large single-sample spikes on top of my expected signal&lt;/li&gt;&#xA;&lt;li&gt;the FFT spectrum of the noise looks white (hard to say since there is some signal there as well, but all the peaks seem to be at the fundamental + harmonics of my signal)&lt;/li&gt;&#xA;&lt;li&gt;amplitude of the spikes seems equally likely to be positive or negative&lt;/li&gt;&#xA;&lt;li&gt;the distribution of spike amplitudes seems to be approx exp(-K*|x|): that is, if I do either an ascending or descending sort of the data, and plot it on a semilog scale (log-x, linear y) then at the low end of the sort (i.e. the outliers) the distribution looks linear on the semilog plot.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Do any of you know of a random process that might meet/approximate the above criteria? I am looking for some insight into what type of noise this is, so I might better guess where it is coming from.&lt;/p&gt;&#xA;" OwnerUserId="120" LastActivityDate="2010-11-03T22:07:16.063" Title="characterizing noise PDF" Tags="&lt;stochastic-processes&gt;&lt;noise&gt;" AnswerCount="1" />
  <row Id="8765" PostTypeId="1" CreationDate="2010-11-03T15:04:05.663" Score="1" ViewCount="107" Body="&lt;p&gt;((n+1)H&lt;sub&gt;n&lt;/sub&gt; - (n+1)) + H&lt;sub&gt;n+1&lt;/sub&gt; = (n+2)H&lt;sub&gt;n+1&lt;/sub&gt; - (n+2)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need to prove the above statement.&lt;/p&gt;&#xA;" OwnerUserId="3071" LastEditorUserId="3071" LastEditorDisplayName="" LastEditDate="2010-11-03T15:37:08.557" LastActivityDate="2010-11-03T16:38:07.713" Title="I would like a hint on this problem regarding harmonic numbers." Tags="&lt;homework&gt;&lt;discrete-mathematics&gt;&lt;induction&gt;" AnswerCount="2" CommentCount="7" />
  <row Id="8766" PostTypeId="1" AcceptedAnswerId="28067" CreationDate="2010-11-03T15:04:28.713" Score="2" ViewCount="229" Body="&lt;p&gt;I want to find all the roots of a polynomial and decided to compute the eigenvalues of its companion matrix.&#xA;How do I do that?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, if I have this polynomial: $4x^3 - 3x^2 + 9x - 1$, I compute the companion matrix:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{bmatrix} 0&amp;amp;0&amp;amp;\frac{3}{4} \\ 1&amp;amp;0&amp;amp;-\frac{9}{4} \\ 0&amp;amp;1&amp;amp;\frac{1}{4} \end{bmatrix}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now how can I find the eigenvalues of this matrix?&lt;br&gt;&#xA;Thanks in advance,&lt;br&gt;&#xA;rubik&lt;/p&gt;&#xA;" OwnerUserId="2582" LastEditorUserId="2582" LastEditorDisplayName="" LastEditDate="2010-11-03T15:30:00.983" LastActivityDate="2011-05-02T06:12:58.330" Title="Eigenvalues of matrix" Tags="&lt;matrices&gt;&lt;polynomials&gt;&lt;numerical-methods&gt;" AnswerCount="3" CommentCount="8" />
  <row Id="8767" PostTypeId="2" ParentId="8766" CreationDate="2010-11-03T15:16:30.933" Score="3" ViewCount="" Body="&lt;p&gt;Typically, one uses the &lt;a href=&quot;http://dx.doi.org/10.1007/BF02163331&quot; rel=&quot;nofollow&quot;&gt;QR algorithm&lt;/a&gt; on a matrix with a preliminary reduction to upper Hessenberg form (all zeroes below the subdiagonal) for determining the eigenvalues. However, since the Frobenius companion matrix is already Hessenberg to begin with, it can easily be fed into an eigenroutine without preliminary transformation (though a preliminary &lt;a href=&quot;http://dx.doi.org/10.1007/BF02165404&quot; rel=&quot;nofollow&quot;&gt;balancing&lt;/a&gt; might be necessary if the coefficients of the polynomial vary widely in magnitude).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, it is a rather wasteful algorithm: the Frobenius companion matrix is sparse, and subjecting it to the QR algorithm causes the zeroes above the subdiagonal to be filled in quickly. However, there are now &lt;a href=&quot;http://scg.ece.ucsb.edu/publications/oadv2.pdf&quot; rel=&quot;nofollow&quot;&gt;new methods&lt;/a&gt; that exploit the sparsity of the Frobenius companion matrix; if you're sure you do not want to investigate &lt;a href=&quot;http://books.google.com/books?id=4PMqxwG-eqQC&quot; rel=&quot;nofollow&quot;&gt;other methods&lt;/a&gt;, you should look into this new development.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-11-03T15:16:30.933" CommentCount="1" />
  <row Id="8768" PostTypeId="2" ParentId="8766" CreationDate="2010-11-03T15:17:08.653" Score="2" ViewCount="" Body="&lt;p&gt;You may want to take a look at &lt;a href=&quot;http://en.wikipedia.org/wiki/Companion_matrix&quot; rel=&quot;nofollow&quot;&gt;Companion matrix&lt;/a&gt; to be sure that you have the right companion matrix. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to find the eigenvalues of a square matrix, you need to solve&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\text{det}(A-\lambda I)=0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where A is the matrix you are interested in and I is the identity matrix. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can find a couple of examples &lt;a href=&quot;http://en.wikipedia.org/wiki/Eigenvalue,_eigenvector_and_eigenspace#Computation_of_eigenvalues.2C_and_the_characteristic_equation&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2569" LastActivityDate="2010-11-03T15:17:08.653" CommentCount="7" />
  <row Id="8770" PostTypeId="2" ParentId="7454" CreationDate="2010-11-03T15:31:07.193" Score="2" ViewCount="" Body="&lt;p&gt;The only help I can give is to suggest a lower bound.  You have cn students and the study together graph could then have cn(cn-1)/2 edges.  Each session you pick $\frac{cn(n-1)}{2}$ edges to color in and you ask whether after y sessions all the edges are colored.  If you ignore the class grouping, you can do the same problem randomly choosing $\frac{ycn(n-1)}{2}$ edges independently with replacement to color.  I think your case will color the whole graph with higher probability, as no edge can claim more than $y$ of the colorings, but it should be close.  Now this is a nice Poisson distribution.  Each edge is colored with probability $1-\exp(-\lambda)$, where $\lambda$ is the average number of colorings each edge receives, here $\frac{ycn(n-1)}{cn(cn-1)}$, or just about $\frac{y}{c}$.  The chance that all edges are colored is then $(1-\exp(-\lambda))^{\frac{cn(cn-1)}{2}}$, which as you say is very small.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-11-03T15:31:07.193" CommentCount="2" />
  <row Id="8771" PostTypeId="2" ParentId="8765" CreationDate="2010-11-03T15:44:28.530" Score="5" ViewCount="" Body="&lt;p&gt;Hint: Your identity is equivalent to &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(n+1)(H_{n+1}-H_{n})=1,$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is something you can deduce by a simple algebraic manipulation.&lt;/p&gt;&#xA;" OwnerUserId="752" LastActivityDate="2010-11-03T15:44:28.530" CommentCount="1" />
  <row Id="8772" PostTypeId="2" ParentId="8761" CreationDate="2010-11-03T15:51:21.637" Score="2" ViewCount="" Body="&lt;p&gt;You haven't given any information about the cost of turning A into B or turning C or D into B.  Can we assume these are zero?  In any case, it's not too hard to compute the probability distribution of costs of A by just working out similarly to what follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose you go with the strategy &quot;buy C and attempt a transformation to B until you get B&quot;.  Then with probability 1/2, this will cost C plus whatever it costs to turn C into B, which I'll call $B_C$, with probability 1/4, cost $2C+B_C$, with probability $1/2^n$ cost $nC+B_C$.  Thus the expected price will be&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$B_C+C\sum_{n=1}^\infty \frac{n}{2^n}=B_C+2C$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, the strategy &quot;buy D and attempt a transformation to B until you get B&quot; will have expected price&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$B_D+D\sum_{n=1}^\infty \frac{n}{4^n}=B_D+4D/9$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You'll want to compare these two and see which is minimal and run with that strategy.&lt;/p&gt;&#xA;" OwnerUserId="826" LastActivityDate="2010-11-03T15:51:21.637" CommentCount="2" />
  <row Id="8773" PostTypeId="2" ParentId="8337" CreationDate="2010-11-03T15:56:10.233" Score="2" ViewCount="" Body="&lt;p&gt;Here is a complex-analytic proof.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $z\in D=\mathbb{C}\backslash${$0,1$}, let&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$R(z)=\sum\frac{1}{\log^2 z}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where the sum is taken over all branches of the logarithm. Each point in $D$ has a neighbourhood on which the branches of $\log(z)$ are analytic. Since the series converges uniformly away from $z=1$, $R(z)$ is analytic on $D$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now a few observations:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(i) Each term of the series tends to $0$ as $z\to0$. Thanks to the uniform convergence this implies that the singularity at $z=0$ is removable and we can set $R(0)=0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(ii) The only singularity of $R$ is a double pole at $z=1$ due to the contribution of the principal branch of $\log z$. Moreover, $\lim_{z\to1}(z-1)^2R(z)=1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(iii) $R(1/z)=R(z)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By (i) and (iii) $R$ is meromorphic on the extended complex plane, therefore it is rational. By (ii) the denominator of $R(z)$ is $(z-1)^2$. Since $R(0)=R(\infty)=0$, the numerator has the form $az$. Then (ii) implies $a=1$, so that&#xA;$$R(z)=\frac{z}{(z-1)^2}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, setting $z=e^{2\pi i w}$ yields&#xA;$$\sum\limits_{n=-\infty}^{\infty}\frac{1}{(w-n)^2}=\frac{\pi^2}{\sin^2(\pi w)}$$&#xA;which implies that $$\sum\limits_{k=0}^{\infty}\frac{1}{(2k+1)^2}=\frac{\pi^2}{8},$$&#xA;and the identity $\zeta(2)=\pi^2/6$ follows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The proof is due to T. Marshall (&lt;em&gt;American Mathematical Monthly&lt;/em&gt;, Vol. 117(4), 2010, P. 352).&lt;/p&gt;&#xA;" OwnerUserId="723" LastEditorUserId="723" LastEditorDisplayName="" LastEditDate="2010-11-03T18:52:08.830" LastActivityDate="2010-11-03T18:52:08.830" />
  <row Id="8774" PostTypeId="2" ParentId="8734" CreationDate="2010-11-03T15:56:10.447" Score="1" ViewCount="" Body="&lt;p&gt;It is believed that the decimal system evolved mainly due to anthropomorphic reasons (5 digits on each hand) and is thought to be a simplification of the Babylonian sexagesimal (base 60) counting method. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To make this analogy precise, note that the normal hand has 4 fingers (excluding the thumb) with 3 segments, along with 5 digits on the other hand to be used as segment pointers. This gives 3 x 4 x 5 = 60 unique configurations.&lt;/p&gt;&#xA;" OwnerUserId="2720" LastEditorUserId="2720" LastEditorDisplayName="" LastEditDate="2010-11-03T17:09:48.730" LastActivityDate="2010-11-03T17:09:48.730" CommentCount="4" />
  <row Id="8775" PostTypeId="1" AcceptedAnswerId="8786" CreationDate="2010-11-03T16:14:18.130" Score="8" ViewCount="327" Body="&lt;p&gt;Start with the identity&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sum_{i=1}^n i^3 = \left( \sum_{i = 1}^n i \right)^2 = \left(\frac{n(n+1)}{2}\right)^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Differentiate the left-most term with respect to $i$ to get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{d}{di} \sum_{i=1}^n i^3 = 3 \sum_{i = 1}^n i^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Differentiate the right-most term with respect to $n$ to get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{d}{dn} \left(\frac{n(n+1)}{2}\right)^2 = \frac{1}{2}n(n+1)(2n+1)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Equate the derivatives, obtaining&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sum_{i=1}^n i^2 = \frac{1}{6}n(n+1)(2n+1)$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is known to be correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any neat reason why this method happens to get lucky and work for this case?&lt;/p&gt;&#xA;" OwnerUserId="251" LastActivityDate="2010-11-04T00:11:05.833" Title="An incorrect method to sum the first $n$ squares which nevertheless works" Tags="&lt;algebra-precalculus&gt;" AnswerCount="3" CommentCount="6" FavoriteCount="2" />
  <row Id="8777" PostTypeId="2" ParentId="8765" CreationDate="2010-11-03T16:28:58.883" Score="0" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\ $ If $\rm\:f(n)\:$ satsifies a first-order recurrence $\rm\ \ f(n+1) = a\ f(n) + b\ $ then it may be used as a rewrite rule to reduce $\rm\ c_0\ f(n) + c_1\ f(n+1) + \:\cdots\: + c_k\ f(n+k)\ $ to the form $\rm\ d\ f(n) + e\:.\ $ Said in operator form, if $\rm \ S^2 = a\ S + b\ $ then $\rm\ f(S)\ \:mod\:\ (S^2-a\ S-b)\ =\ c\ S + d\ $ has degree 1, since one can eliminate $\rm\:S^n,\ n&gt;2\ $ via the rewrite rule $\rm\: S^2 \to a\ S + b\:$. In essence one is employing the division / Euclidean algorithm to compute a remainder in a polynomial ring of operators. Precisely the same method works for solutions of any (monic) linear recurrence or differential equation.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-11-03T16:38:07.713" LastActivityDate="2010-11-03T16:38:07.713" />
  <row Id="8778" PostTypeId="2" ParentId="8753" CreationDate="2010-11-03T16:44:24.363" Score="3" ViewCount="" Body="&lt;p&gt;[The answer is given at the end.]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First let $n = \left\lfloor x \right\rfloor$. We have $\int_k^{k + 1} {\frac{{\left\lfloor t \right\rfloor }}{t}\,{\rm d}t}  = k[\log (k + 1) - \log (k)]$, $k=1,\ldots,n-1$, and $\int_n^x {\frac{{\left\lfloor t \right\rfloor }}{t}\,{\rm d}t}  = n[\log (x) - \log (n)]$. From this it follows straightforwardly that &#xA;$$&#xA;\nu(x) = \frac{1}{2}\log (2\pi x) - x - \log (n!) + n\log (x).&#xA;$$&#xA;Next, let $\delta_x = x - n$. Thus, $x = n + \delta_x$. Substituting this for $x$ in the above expression, and using &#xA;$$&#xA;\log (M + a) = \log (M) + \frac{a}{M} - \frac{{a^2 }}{{2M^2 }} + O\bigg(\frac{1}{{M^3 }}\bigg), &#xA;$$&#xA;as $M \to \infty$, leads to&#xA;$$&#xA;\nu (x) = \frac{1}{2}\log (2\pi n) - n - \log (n!) + n\log (n) + \frac{{\delta _x }}{{2n}} - \frac{{\delta _x^2 }}{{2n}} + O\bigg(\frac{1}{{n^2 }}\bigg),&#xA;$$&#xA;or&#xA;$$&#xA;\nu (x) = \log \Bigg[\frac{{(\sqrt {2\pi n} )e^{ - n} n^n }}{{n!}}\Bigg] + \frac{{\delta _x }}{{2n}} - \frac{{\delta _x^2 }}{{2n}} + O\bigg(\frac{1}{{n^2 }}\bigg),&#xA;$$&#xA;as $n \to \infty$. Then, from the well-known estimate&#xA;$$&#xA;n! = (\sqrt {2\pi n} )e^{ - n} n^n \bigg(1 + \frac{1}{{12n}} + O\bigg(\frac{1}{{n^2 }}\bigg)\bigg)&#xA;$$&#xA;(and noting that $\log(1/\xi)=-\log(\xi)$), it follows that&#xA;$$&#xA;\nu (x) =  - \log \bigg[1 + \frac{1}{{12n}} + O\bigg(\frac{1}{{n^2 }}\bigg)\bigg] + \frac{{\delta _x }}{{2n}} - \frac{{\delta _x^2 }}{{2n}} + O\bigg(\frac{1}{{n^2 }}\bigg),&#xA;$$&#xA;and in turn,&#xA;$$&#xA;\nu (x) =  - \frac{1}{{12n}} + \frac{{\delta _x }}{{2n}} - \frac{{\delta _x^2 }}{{2n}} + O\bigg(\frac{1}{{n^2 }}\bigg).&#xA;$$&#xA;Finally, since $x = n + \delta_x$, we obtain&#xA;$$&#xA;x \nu(x) = - \frac{1}{{12}} + \frac{{\delta _x }}{{2}} - \frac{{\delta _x^2 }}{{2}} + O\bigg(\frac{1}{{n}}\bigg).$$&#xA;We conclude that $x \nu(x)$ does not have a limit as $x \to \infty$. Rather, since $\delta_x /2 - \delta_x^2 / 2$ is oscillating between $0$ and $1/8$, we have&#xA;$$&#xA;\mathop {\lim \inf }\limits_{x \to \infty } [x\nu (x)] =  - \frac{1}{{12}} \;\;\; {\rm and} \;\;\; \mathop {\lim \sup }\limits_{x \to \infty } [x\nu (x)] =  \frac{1}{{24}}.&#xA;$$&#xA;Nevertheless, we have&#xA;$$&#xA;\mathop {\lim }\limits_{n \to \infty } n\nu (n) =  - \frac{1}{{12}},&#xA;$$&#xA;as $n \to \infty$ (integer).&lt;/p&gt;&#xA;" OwnerUserId="2810" LastEditorUserId="2810" LastEditorDisplayName="" LastEditDate="2010-11-03T23:32:08.490" LastActivityDate="2010-11-03T23:32:08.490" CommentCount="9" />
  <row Id="8779" PostTypeId="1" AcceptedAnswerId="8782" CreationDate="2010-11-03T16:53:02.370" Score="3" ViewCount="199" Body="&lt;p&gt;Prove that $f_n\to f$ in measure on $E$ if and only if given $\varepsilon&gt;0$, there exists $K$ such that |{$x\in E : |f(x)-f_k(x)|&gt;\varepsilon$}|$&amp;lt;\varepsilon$ for $k\ge K$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &quot;only if&quot; direction of this is immediate from the definition of convergence in measure, but the other direction is less obvious to me. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conversely, we suppose that given $\varepsilon &gt;0$, there is a $K$ such that |{$x\in E : |f(x)-f_k(x)|&gt;\varepsilon$}|$&amp;lt;\varepsilon$ for $k\ge K$. My initial thought was to bound the measure of the set in question by, say, $\frac{1}{k}$. But I'm not sure I can do that because $\varepsilon$ not only bounds the measure of the set, but the set also depends on the choice of $\varepsilon$. To show something convergence in measure, I need to show that &lt;em&gt;for every $\varepsilon$&lt;/em&gt; the limit as $k\to\infty$ of the measures of those sets is zero...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[Subquestion: is the use of |$\cdot$| standard for denoting Lebesgue measure? I had never seen it until this course. I had always seen $m(-)$.]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[Sub-subquestion: is there any particular reason that set brackets don't display in math mode? The commands \ { and \ } didn't do anything...]&lt;/p&gt;&#xA;" OwnerUserId="2313" LastEditorUserId="2313" LastEditorDisplayName="" LastEditDate="2010-11-03T17:16:34.853" LastActivityDate="2010-11-03T17:16:34.853" Title="Characterization of convergence in measure" Tags="&lt;real-analysis&gt;&lt;homework&gt;&lt;measure-theory&gt;" AnswerCount="1" CommentCount="6" />
  <row Id="8780" PostTypeId="1" AcceptedAnswerId="8783" CreationDate="2010-11-03T16:57:06.667" Score="6" ViewCount="178" Body="&lt;p&gt;I am wondering if there is a general coordinate-independent way to define a Partial Differential Equation on a Smooth manifold.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is definitely true that in each coordinate neighborhood you could define a function to satisfy a differential equation, but when you change coordinates then the differential equation will most likely have a different form.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, in $\mathbb{R}^3$ one says a function is harmonic if it satisfies&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ \frac{\partial^2}{\partial x^2} f + \frac{\partial}{\partial y^2} f + \frac{\partial}{\partial z^2} f = 0 $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but it is &lt;strong&gt;not&lt;/strong&gt; true that in spherical coordinates a function is harmonic if&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ \frac{\partial^2}{\partial r^2} f + \frac{\partial}{\partial \theta^2} f + \frac{\partial}{\partial \phi^2} f = 0 $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The change of variables to spherical coordinates gives you a much different PDE. So basically, how can you define a differential operator which is coordinate independent on a smooth manifold, so that you can have some notion of a PDE on a manifold.&lt;/p&gt;&#xA;" OwnerUserId="1089" LastActivityDate="2010-11-03T19:01:01.040" Title="PDEs on Manifolds" Tags="&lt;analysis&gt;&lt;differential-equations&gt;&lt;differential-geometry&gt;" AnswerCount="2" FavoriteCount="5" />
  <row Id="8782" PostTypeId="2" ParentId="8779" CreationDate="2010-11-03T17:10:53.703" Score="4" ViewCount="" Body="&lt;p&gt;Convergence in measure means that for all $\varepsilon\gt0$, $|\{x\in E : |f(x)-f_k(x)|&gt;\varepsilon\}|$ goes to $0$, which means that for all $\varepsilon\gt0$, for all $\delta\gt0$, there exists $K$ such that $|\{x\in E : |f(x)-f_k(x)|&gt;\varepsilon\}|&amp;lt;\delta$ for all $k\geq K$.  Taking $\delta=\varepsilon$ gives the &quot;only if&quot;.  To see &quot;if&quot;, given positive $\varepsilon$ and $\delta$, take $K$ such that $k\geq K$ implies that $|\{x\in E : |f(x)-f_k(x)|&gt;\min(\varepsilon,\delta)\}|&amp;lt;\min(\varepsilon,\delta)$.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-11-03T17:10:53.703" CommentCount="4" />
  <row Id="8783" PostTypeId="2" ParentId="8780" CreationDate="2010-11-03T17:15:40.003" Score="5" ViewCount="" Body="&lt;p&gt;You could view the concept of a &quot;Jet bundle&quot; to be precisely what you need to make partial differential equations coordinate-independent on manifolds, as points in jet bundles represent precisely the Taylor polynomials of functions in local coordinates.  Take a look at Chapter 4 in Hirsch's &quot;Differential Topology&quot; textbook for details.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The underlying problem of course is that the tangent bundle of a manifold is not trivial in general so you have to do something new to make sense of partial differential equations.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps the simplest would be to view a vector field as a 1st order differential equation via the directional derivative operator.  From there you can take mixed directional derivatives.  This isn't ideal because you can't express &quot;natural&quot; differential equations like the Laplacian in this manner (for example, think of $S^2$ -- every vector field has a zero...).  edit: as pointed out, on $S^2$ there &lt;em&gt;is&lt;/em&gt; a formulation of the Laplacian in terms of iterated 1st order DE's. I suppose I should have phrased things as &quot;even if you can express a PDE as iterated 1st order DEs on a manifold, it may not be desirable due to the breaking of symmetries brought on by bringing in vector fields.&quot; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most direct approach is to think of partial differential equations as functions on jet bundles.  But you can frequently express interesting differential equations in other manners -- having a connection on a manifold is one of the simplest ways of turning 2nd derivatives into tangential data.  The relevant formulation is called an &lt;a href=&quot;http://en.wikipedia.org/wiki/Ehresmann_connection&quot; rel=&quot;nofollow&quot;&gt;Ehresmann Connection&lt;/a&gt;. But there's all kinds of other structures on manifolds where PDEs can lurk.  Differential forms give you a standard language -- exterior derivative and Hodge star operators on Riemann manifolds give you a language where you can express the Laplacian. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think this is a very good question in that to a large extent the language you need to express a PDE on a manifold very much informs on what you actually know geometrically about the PDE.  So it's a key step in getting insight into the nature of a PDE.  Having a PDE at the jet bundle level basically means you know nothing about it, other than it's a PDE.&lt;/p&gt;&#xA;" OwnerUserId="642" LastEditorUserId="642" LastEditorDisplayName="" LastEditDate="2010-11-03T18:47:53.680" LastActivityDate="2010-11-03T18:47:53.680" CommentCount="13" />
  <row Id="8785" PostTypeId="2" ParentId="8775" CreationDate="2010-11-03T17:34:18.277" Score="9" ViewCount="" Body="&lt;p&gt;Consider &lt;a href=&quot;http://en.wikipedia.org/wiki/Bernoulli_number&quot; rel=&quot;nofollow&quot;&gt;Bernoulli's formula&lt;/a&gt; which gives&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \sum_1^n i^m = \frac{1}{m+1} \sum_0^m {m+1 \choose k} B_k n^{m+1-k} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $B_k$ is the k'th Bernoulli number. Take the derivative with respect to $n$ on the right hand side, you get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \frac{d}{dn} \sum_1^n i^m = \frac{1}{m+1} \sum_0^m \frac{(m+1)\cdot m!}{(m+1 - k)\cdot (m -1 + 1 - k)! k!}\cdot (m+1-k) B_k n^{m-1 + 1-k} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$= \sum_0^m { m \choose k} B_k n^{m-k} = m \sum_1^n i^{m-1} + B_m$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the difference between formally taking the derivative w.r.t. $i$ as you did, and formally taking the derivative w.r.t. $n$, is exactly the Bernoulli constant for the power $m$. Now Bernoulli's constant is equal to 0 for all odd $m$ that is greater than 1. For those cases your observation is also true: the two &quot;derivatives&quot; are equal. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, some random guessing, since I am not an analytic number theorist. The deep connection to Bernoulli numbers probabaly has something to do with analytic number theory and harmonic analysis that I do not know myself. That the formulae work for odd powers is possibly something related to the method of descent, where analytical results in even dimensions can be arrived from results in one higher dimension via some sort of trace. &lt;/p&gt;&#xA;" OwnerUserId="1543" LastActivityDate="2010-11-03T17:34:18.277" />
  <row Id="8786" PostTypeId="2" ParentId="8775" CreationDate="2010-11-03T17:37:20.363" Score="18" ViewCount="" Body="&lt;p&gt;Let $f_k(n)=\sum_{i=1}^n i^k$. We all know that $f_k$ is actually&#xA;a polynomial of degree $k+1$. Also $f_k$ can be characterised by the two&#xA;conditions:&#xA;$$f_k(x)-f_k(x-1)=x^k$$&#xA;and&#xA;$$f_k(0)=0.$$&#xA;Differentiating the first condition gives&#xA;$$f_k'(x)-f_k'(x-1)=k x^{k-1}.$$&#xA;Therefore the polynomial $(1/k)f_k'$ satisfies the &lt;strong&gt;first&lt;/strong&gt; of the two&#xA;conditions that $f_{k-1}$ does. But it may not satisfy the second. But then&#xA;$(1/k)(f_k'(x)-f_k'(0))$ does. So&#xA;$$f_{k-1}(x)=\frac{f_k'(x)-f_k'(0)}k.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The mysterious numbers $f_k'(0)$ are related to the Bernoulli numbers,&#xA;and when $k\ge3$ is odd they obligingly vanish...&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-11-03T17:37:20.363" CommentCount="2" />
  <row Id="8787" PostTypeId="1" AcceptedAnswerId="8789" CreationDate="2010-11-03T17:44:30.137" Score="0" ViewCount="79" Body="&lt;p&gt;I'm interested in finding a recursion or simple representation for a &quot;Hadamard product&quot; of two power series.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The Hadamard Product&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The Hadamard product is defined on generating functions $f(x)$ and $f(y)$ as $f(z)$ below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f(x) = \sum_{i=0}^\infty{c_i x^i}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f(y) = \sum_{i=0}^\infty{d_i y^i}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f(z) = \sum_{i=0}^\infty{c_i \cdot d_i z^i}$&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;My Particular Interest&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now suppose that we work on the series above with only the first N terms as possibly nonzero, and the rest will be considered to be zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take the coefficients of $f(x)$ to be the reciprocals of the first $N$ naturals, e.g.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f(x) = \sum_{i=1}^N{\frac{1}{i} x^{i+1}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now suppose that $f(y)$ is in the form of an expression, i.e.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f(y) = \frac{1-(2y)^{N-1}}{1 - 2y}$ represents the first N terms of the form $(2y)^N$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f(y) = \frac{1-(n+1)x^N+Nx^{N+1}}{(x-1)^2}$ represents the first N naturals in series&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or some similar function...&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;My Question&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Can we find a simple expression, such as a recursion or formula, to represent the &quot;Hadamard product&quot; of these new series?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We could simply write out the $N$ terms in the resulting function $f(z)$.  However, I'm wondering if there is a more compact way of representing this.  I'm particularly interested when $f(x)$ is the series of reciprocals mentioned above.  However, I am interested in finding this result for almost any $f(y)$.  Is there a way to do this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd like to consider both the cases when coefficients of $f(y)$ take on real values and when it is complex-valued.&lt;/p&gt;&#xA;" OwnerUserId="2626" LastActivityDate="2010-11-03T17:52:38.863" Title="Is an easier formulation possible?" Tags="&lt;calculus&gt;&lt;sequences-and-series&gt;&lt;intuition&gt;" AnswerCount="1" />
  <row Id="8789" PostTypeId="2" ParentId="8787" CreationDate="2010-11-03T17:52:38.863" Score="1" ViewCount="" Body="&lt;p&gt;It's $z \int_0^z \frac{f(y)}{y^2} dy$ after you remove the constant and linear terms.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-11-03T17:52:38.863" CommentCount="5" />
  <row Id="8790" PostTypeId="2" ParentId="8731" CreationDate="2010-11-03T18:16:20.580" Score="3" ViewCount="" Body="&lt;p&gt;There are many algorithms for computing shortest paths on polyhedral 2-manifolds.&#xA;With a student I computed the shortest paths shown below with one, the Chen-Han algorithm.&#xA;The algorithms fan out shortest paths to a &lt;em&gt;frontier&lt;/em&gt;, mimicking the structure&#xA;(but not the details) of &lt;a href=&quot;http://en.wikipedia.org/wiki/Dijkstra%27s_algorithm&quot; rel=&quot;nofollow&quot;&gt;Dijkstra's algorithm&lt;/a&gt; for shortest paths in a graph.  Sometimes the method is called the &quot;continuous Dijkstra&quot; method.  These algorithms (all in $\mathbb{R}^3$) are described in many places, including &#xA;the book &lt;em&gt;&lt;a href=&quot;http://www.gfalop.org&quot; rel=&quot;nofollow&quot;&gt;Geometric Folding Algorithms: Linkages, Origami, Polyhedra&lt;/a&gt;&lt;/em&gt;, Chapter 24.&#xA;&lt;br /&gt;&#xA;&lt;img src=&quot;http://i.imgur.com/DCiki.jpg&quot; alt=&quot;Polyhedron&quot;&gt;&#xA;&lt;br /&gt;&#xA;I would assume the same approach will work for triangulated manifolds in arbitrary dimensions, but of course it will be significantly more complicated to implement.&lt;/p&gt;&#xA;" OwnerUserId="237" LastActivityDate="2010-11-03T18:16:20.580" CommentCount="3" />
  <row Id="8791" PostTypeId="1" AcceptedAnswerId="8796" CreationDate="2010-11-03T18:59:27.593" Score="1" ViewCount="155" Body="&lt;p&gt;I got this problem from one of my mates, and i rearranged them, and got it in a summable form. From here could anyone tell me as to how i can sum up this interesting series.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How does one sum the given series: $$S = 1 + \Bigl(1 + \frac{1}{3}\Bigr) \cdot \frac{1}{5} \cdot \frac{1}{3} + \Bigl(1 + \frac{1}{3} + \frac{1}{5}\Bigr)\cdot \frac{1}{5^{2}} \cdot \frac{1}{5} + \Bigl(1 + \frac{1}{3}+ \frac{1}{5} + \frac{1}{7}\Bigr) \cdot \frac{1}{5^{3}} \cdot \frac{1}{7} + \cdots + \text{ad inf}$$&lt;/p&gt;&#xA;" LastActivityDate="2010-11-04T10:10:46.787" Title="Summing a given series" Tags="&lt;analysis&gt;&lt;sequences-and-series&gt;" AnswerCount="3" />
  <row Id="8792" PostTypeId="2" ParentId="8780" CreationDate="2010-11-03T19:01:01.040" Score="9" ViewCount="" Body="&lt;p&gt;Ryan already gave a very good answer. I just want to give a few clarifications on how one should think about PDEs in general. Most importantly: you should &lt;em&gt;not&lt;/em&gt; think a differential operator as defined by &quot;a formula&quot;. What do I mean by that? In your question, you wrote that the Laplacian in Euclidean coordinates is $\frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2} + \frac{\partial^2}{\partial z^2}$, but in Spherical coordinates it is not $\frac{\partial^2}{\partial r^2} + \frac{\partial^2}{\partial \theta^2} + \frac{\partial^2}{\partial \phi^2}$. But why should one have thought that the two expressions are the same? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To illustrate, think of the function on the plane $f(x,y) = x^2 + y^2$. This represents the square of the distance of a point from the origin. In radial coordinates, the same function is written as $f(r,\theta) = r^2$. &lt;em&gt;Not&lt;/em&gt; $r^2 + \theta^2$, which is a very different beast.  Just like how the direct replacement $x \to r$ and $y\to \theta$ changes the &lt;em&gt;function&lt;/em&gt; you are considering, the replacements $\frac{\partial}{\partial x} \to \frac{\partial}{\partial r}$ etc changes the &lt;em&gt;operator&lt;/em&gt; you are considering. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now: to think geometrically (on manifolds), a (real valued) function is an assignment of a number to each point on the manifold. Then in any coordinate system there is a representation of the function by a formula. Since we like to work with formulas, we can interpret the notion of a function as a &quot;map&quot;. This map takes as its input a coordinate system, and outputs a formula that represents your function in that coordinate system. (If you are familiar with category theory, this is similar to how one can think in terms of arrows instead of dots.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, a partial differential operator can be thought of as an object in itself. Our usual way of writing it as partial derivatives in some coordinate system with coefficients is just a convenient representation of the real object upstairs. Therefore, analogous to how a function is merely a &quot;map&quot; from coordinate system to the formulaic representation, you can think of a partial differential operator also as a &quot;map&quot; which takes as its input a coordinate system and outputs an expression which is a sum of partial derivatives with some coefficients. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The jet bundle formulation is just a sophisticated, rigorous way of formulating this idea. The main difficulty is making sure the intuition above is compatible with the change of variables formulas. Suppose we are given a coordinate system $A$, and an expression in terms of partial derivatives which we'll call $L_A$. The question becomes: does there exist some abstract partial differential operator $L$ such that its representation in the coordinate system $A$ is precisely $L_A$? And if there is such an operator, is the fact that, for some function $f$ (written in coordinate system $A$ as $f_A$), $L_A f_A = 0$ invariant under change of coordinate system? (That given any other coordinate system $B$, $L_Bf_B = 0$ also.) The tools of jet bundles allows you to make such a consistent definition/description of partial differential equations. &lt;/p&gt;&#xA;" OwnerUserId="1543" LastActivityDate="2010-11-03T19:01:01.040" CommentCount="1" />
  <row Id="8793" PostTypeId="1" CreationDate="2010-11-03T19:14:30.710" Score="2" ViewCount="180" Body="&lt;p&gt;Let $B \subset [0,2\pi]$ be a Lebesgue measurable set. Prove that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle \lim_{n \to \infty} \int_{B} \cos(nx) dx = 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;OK I did this assuming B is an open interval, this is pretty easy using the fact that the sine function is bounded by 1. Now I'm stuck in the general case, I'm somewhat confused by &quot;Lebesgue measurable&quot; I know this means that the measure is given by the outer measure i.e the infimum of the sum of all measures of open covers of B. But I'm having trouble writing it, I get confused when working with Lebesgue measurable sets. Can you please help me?&lt;/p&gt;&#xA;" OwnerUserId="4219" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-03T19:19:47.510" LastActivityDate="2010-11-03T20:23:01.670" Title="Compute integral of a Lebesgue measurable set" Tags="&lt;measure-theory&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="1" />
  <row Id="8794" PostTypeId="1" AcceptedAnswerId="8803" CreationDate="2010-11-03T19:18:53.260" Score="7" ViewCount="102" Body="&lt;p&gt;&lt;strong&gt;Motivation:&lt;/strong&gt; In odd dimensions, solutions to the wave equation: $u_{tt}(x,t)=\nabla u(x,t)$, $u_t(x,0)=0$, $u(x,0)=f(x)$, ($t\geq 0, x\in \mathbb{R}^n$) have the nice property that the value of $u(x,t)$ only depends on the values $f(y)$ with $|y-x|=t$. For even dimensions, the value $u(x,t)$ depend on all the values $f(y)$ with $|y-x|\leq t$. A consequence of this is, that when you switch a light bulb on and then off (in 3D), there will be a light wave traveling with the speed on light (D'oh) and behind the wave, there will be total darkness. But when you throw a rock into a pond (with a 2D surface), there will be lots of wave traveling outwards from where the rock hit the water and, in theory, the water will never a still again.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; Can anyone give an intuitive explanation of this difference between odd and even dimensions?&lt;/p&gt;&#xA;" OwnerUserId="3078" LastActivityDate="2010-11-03T20:52:15.877" Title="Intuitive explanation of the difference between waves in odd and even dimensions" Tags="&lt;intuition&gt;&lt;differential-equations&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="2" />
  <row Id="8795" PostTypeId="1" AcceptedAnswerId="8801" CreationDate="2010-11-03T19:24:17.623" Score="6" ViewCount="184" Body="&lt;p&gt;The topological $K$-theory of a complex compact manifold $X$ is the commutative monoid $K(X)$ of isomorphism classes of complex vector bundles. Two classes $[E]$ and $[F]$ are equivalent in $K$-theory if they are &lt;i&gt;stably equivalent&lt;/i&gt;: &#xA;there exists a trivial bundle $G$ with &#xA;$$&#xA;(E \oplus G) \cong (F \oplus G).&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In what sense is this the correct notion of equivalence in $K(X)$?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In other words, what does stable equivalence buy you?&lt;/p&gt;&#xA;" OwnerUserId="3008" LastEditorUserId="8246" LastEditorDisplayName="" LastEditDate="2011-04-22T16:16:30.310" LastActivityDate="2011-04-22T16:46:47.410" Title="Why is stable equivalence necessary in topological K-theory?" Tags="&lt;algebraic-topology&gt;&lt;topological-k-theory&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="2" />
  <row Id="8796" PostTypeId="2" ParentId="8791" CreationDate="2010-11-03T19:25:24.503" Score="6" ViewCount="" Body="&lt;p&gt;I presume this is $\sum_{n=0}^\infty a_n 5^{-n}$&#xA;where $(2n+1)a_n=\sum_{k=0}^n 1/(2k+1)$.&#xA;Let&#xA;$$f(x)=\sum_{n=0}^\infty a_n x^{2n+1}$$&#xA;so that $f(0)=0$ and&#xA;$$f'(x)=\sum_{n=0}^\infty x^{2n}\sum_{k=0}^n\frac{1}{2k+1}.$$&#xA;Therefore&#xA;$$f'(x)=\frac{1}{1-x^2}\frac{\tanh^{-1}x}{x}.$$&#xA;Your sum equals&#xA;$$\int_0^{1/5}\frac{\tanh^{-1}x}{x(1-x^2)}dx.$$&#xA;If we let $x=\tanh y$ then $dx=(1-x^2)dy$ and we get&#xA;$$\int_0^t y\coth y\ dy$$&#xA;where $t=\tanh^{-1}(1/5)$.&#xA;I'm getting a nasty feeling that $\int y\coth y\ dy$&#xA;is one of those integrals that can't be done elementarily. :-(&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-11-03T19:25:24.503" CommentCount="2" />
  <row Id="8797" PostTypeId="2" ParentId="3633" CreationDate="2010-11-03T19:26:25.703" Score="2" ViewCount="" Body="&lt;p&gt;Introduce a new coordinate system with a fixed point of $f$ as origin, e.g. the point $\omega:=e^{\pi i/3}$. Writing $x=\omega+\xi$ with a new independent variable $\xi$  one has $\phi(\omega+\xi)=\omega +\psi(\xi)$ for a new unknown function $\psi$ with $\psi(0)=0$. This function $\psi$ satisfies in a neighbourhood of $\xi=0$ the functional equation $\psi(\psi(\xi))=2\omega\xi+\xi^2$. Now you can recursively determine the Taylor coefficients of $\psi$. If you are lucky the resulting formal series actually defines an analytical function in a neighbourhood of $\xi=0$.&lt;/p&gt;&#xA;" OwnerUserId="1303" LastEditorUserId="1303" LastEditorDisplayName="" LastEditDate="2010-11-03T20:25:07.503" LastActivityDate="2010-11-03T20:25:07.503" />
  <row Id="8798" PostTypeId="2" ParentId="8791" CreationDate="2010-11-03T19:27:05.610" Score="0" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\ $ If you massage the odd bisection of the generating function of the harmonic numbers then you should obtain a closed form in terms of (di)logs evaluated in $\:\mathbb Q(\sqrt{5})$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt; $\ $ See Andrey's later answer for some further details of the method I hinted above.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-11-04T01:09:36.053" LastActivityDate="2010-11-04T01:09:36.053" CommentCount="2" />
  <row Id="8799" PostTypeId="2" ParentId="8793" CreationDate="2010-11-03T19:30:50.457" Score="2" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;hint&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Re-write the integral as the following&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \int_0^{2\pi} \chi_B(x) \cos(nx) dx $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\chi_B$ is the characteristic function of the set $B$ (so that it equals 1 on $B$ and 0 else where). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, you've already proven the case where $B$ is an open interval. Now take a sequence of decreasing coverings for $B$ by finitely many disjoint open intervals. For each covering, the result is true by what you've already shown. Take the limit using dominated convergence theorem (if $B \subset C$, then $|\chi_B \cos| \leq |\chi_C \cos|$). &lt;/p&gt;&#xA;" OwnerUserId="1543" LastActivityDate="2010-11-03T19:30:50.457" />
  <row Id="8800" PostTypeId="2" ParentId="8793" CreationDate="2010-11-03T19:36:33.147" Score="1" ViewCount="" Body="&lt;p&gt;Each Lebesgue integrable function can be approximated by a finite step function in norm. That is if $f\in L^1$ and $\varepsilon&gt;0$ then there are intervals $I_1,\ldots,I_N$, and scalars $a_1,\ldots,a_N$ such that &#xA;$$\int |f-\sum_{n=1}^N a_n\chi_n|&amp;lt;\varepsilon.$$&#xA;Where $\chi_n$ is the characteristic function on $I_n$, that is $\chi_n(x)=1$ for $x\in I_n$ and $\chi_n(x)=0$ otherwise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, on each bounded interval $I$ you have already proved that $\int_I \cos(nx)dx\to 0$, hence given $\varepsilon&gt;0$ there is a step function as above and we get&#xA;$$\limsup\left|\int_B\cos(kx)dx\right|= \limsup\left|\int (\chi_B-\sum_{n=1}^N a_n\chi_n +\sum_{n=1}^N a_n\chi_n)\cos(kx)dx\right|$$&#xA;$$\leq \limsup\int |\chi_B-\sum_{n=1}^N a_n\chi_n|dx +\limsup\left|\int\sum_{n=1}^N a_n\chi_n\cos(kx)dx\right|$$&#xA;$$\le\varepsilon + \sum_{n=1}^N |a_n|\cdot \limsup\left|\int\chi_n\cos(kx)dx\right| =\varepsilon +0.$$&#xA;From which you conclude the result since this holds for any $\varepsilon&gt;0$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This works not only for $\chi_B$, but for any $f\in L^1$ - it is called the Riemann-Lebesgue Lemma. &lt;/p&gt;&#xA;" OwnerUserId="1154" LastEditorUserId="1154" LastEditorDisplayName="" LastEditDate="2010-11-03T19:56:59.550" LastActivityDate="2010-11-03T19:56:59.550" />
  <row Id="8801" PostTypeId="2" ParentId="8795" CreationDate="2010-11-03T19:40:07.923" Score="4" ViewCount="" Body="&lt;p&gt;It comes down to whether or not you want $K(X)$ to be a group.  If all you care about is isomorphism classes of vector bundles, that's a perfectly good monoid.  People wouldn't call the monoid $K$-theory because $K$-theory of stuff generally means you've performed some kind of group-completion of a monoid or a category or something. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So there's two ways to respond to your question.  The simplest would be &quot;why don't vector bundles have inverses?&quot; which on the most superficial level has the answer not all vector bundles are $0$-dimensional. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another way to respond would be, what if we performed the formal group-completion of the monoid of isomorphism classes of vector bundles?  Well, you get the same thing.  General nonsense says you have a map from the group completion to $K(X)$ defined your way, since bundles over reasonable spaces have complementary bundles. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this roughly an answer to your question? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;edit in response to your comment: Say $X$ is a finite-dimensional CW-complex, and $\epsilon$ is a finite-dimensional vector bundle over $X$.  Then there is a classifying map for $\epsilon$, this is a map $X \to G_{\infty,k}$ where $k$ is the dimension of $\epsilon$.  $G_{\infty,k}$ is the Grassmannian of $k$-dimensional subspaces of $\mathbb R^\infty$.  A basic theorem about $G_{\infty,k}$ is that any map of a finite-dimensional CW-complex into $G_{\infty,k}$ is homotopic to a map into $G_{m,k}$ for $m$ perhaps very large.  $G_{m,k}$ is canonically isomorphic to $G_{m,m-k}$ (take orthogonal complements).  The bundle classified by the corresponding map $X \to G_{m,m-k}$ is the &lt;em&gt;complementary bundle&lt;/em&gt; to $\epsilon$, denote it $\epsilon&amp;#39;$.  By design, $\epsilon \oplus \epsilon&amp;#39;$ is a trivial bundle as it's the pull-back of the tangent bundle to $\mathbb R^m$.  The fact that maps $X \to G_{\infty,k}$ are homotopic to maps $X \to G_{m,k}$ has many proofs -- one is from the Schubert cell decomposition of $G_{m,k}$ together with cellular approximation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's the idea.  I believe complementary bundles exist for more than bundles over finite-dimensional spaces but that's the most convenient argument that comes to mind (have to run off to teach a class now). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2nd edit:  Let $\mathcal V(X)$ be the group completion of the monoid $\mathcal M(X)$ of isomorphism classes of finite-dimensional vector bundles over $X$.  Let $K(X)$ be the stable isomorphism classes of vector bundles over $X$.  $K(X)$ is a group for a fairly broad class of spaces, by the arguments above.  So by group-completion formal nonsense there is a map $\mathcal V(X) \to K(X)$.   The question is whether this map has an inverse or not. Given an element $[\alpha] \in K(X)$, $\alpha \in \mathcal M(X)$ is a vector bundle over $X$ and is well-defined up to stable equivalence.  To $\alpha$ there is an associated element of $\mathcal V(X)$ and the question is, is the map &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$K(X) \to \mathcal V(X)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;well-defined, where you map $[\alpha]_{K(X)} \longmapsto [\alpha]_{\mathcal V(X)}$?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ah, okay, I see the problem now.  This isn't well-defined.  But things are off only by a tiny thing.  $\mathcal V(X)$ contains a copy of the integers as a direct-summand.  This is because $\mathcal M(X) \to \mathbb Z$ given by taking the dimension of a bundle is a homomorphism of monoids (using addition in $\mathbb Z$).  There is a map back $\{0,1,2,\cdots \} \to \mathcal M(X)$ given by taking trivial bundles over $X$.  So in $\mathcal V(X)$ this gives a splitting $\mathcal V(X) = \mathbb Z \oplus \overline{\mathcal V(X)}$ where $\overline{\mathcal V(X)}$ is the &quot;reduced&quot; group-completion of isomorphism types of vector bundles over $X$.  There &lt;em&gt;is&lt;/em&gt; a well-defined map $K(X) \to \overline{\mathcal V(X)}$ since you can think of $\overline{\mathcal V(X)}$ as being $\mathcal V(X)$ modulo trivial bundles. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does that make sense now?   I haven't thought about this stuff in much detail for years. &lt;/p&gt;&#xA;" OwnerUserId="642" LastEditorUserId="8246" LastEditorDisplayName="" LastEditDate="2011-04-22T16:46:47.410" LastActivityDate="2011-04-22T16:46:47.410" CommentCount="6" />
  <row Id="8802" PostTypeId="2" ParentId="8791" CreationDate="2010-11-03T19:49:38.050" Score="1" ViewCount="" Body="&lt;p&gt;$$S(x) = 1 + \Bigl(1 + \frac{1}{3}\Bigr) \cdot \frac{x^2}{5} \cdot \frac{1}{3} + \Bigl(1 + \frac{1}{3} + \frac{1}{5}\Bigr)\cdot \frac{x^4}{5^{2}} \cdot \frac{1}{5} + \Bigl(1 + \frac{1}{3}+ \frac{1}{5} + \frac{1}{7}\Bigr) \cdot \frac{x^6}{5^{3}} \cdot \frac{1}{7} + \cdots$$&#xA;$$=\sum_{k=0}^{\infty} \Bigl(1 + \frac{1}{3}+\cdots+\frac{1}{2k+1}\Bigr) \cdot \left(\frac{x}{\sqrt 5}\right)^{2k} \cdot \frac{1}{2k+1}$$&#xA;$$=\frac{1}{x}\int_{0}^x\sum_{k=0}^{\infty} \Bigl(1 + \frac{1}{3}+\cdots+\frac{1}{2k+1}\Bigr) \cdot \left(\frac{y}{\sqrt 5}\right)^{2k}dy$$ &#xA;$$=\frac{\sqrt 5}{x}\int_{0}^{x/\sqrt 5}\sum_{k=0}^{\infty} \Bigl(1 + \frac{1}{3}+\cdots+\frac{1}{2k+1}\Bigr) t^{2k+1}\frac{dt}{t}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let&#xA;$$F(t)=\sum_{k=0}^{\infty} \Bigl(1 + \frac{1}{3}+\cdots+\frac{1}{2k+1}\Bigr) t^{2k+1},$$&#xA;then &#xA;$$F(t)+\frac{t}{2}F(t)=H(t):= \sum_{n=0}^{\infty} \Bigl(1 + \frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}\Bigr) t^n.$$&#xA;$H(t)$ is the generating function of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Harmonic_number&quot; rel=&quot;nofollow&quot;&gt;harmonic numbers&lt;/a&gt; and is well known:&#xA;$$H(t)=\frac{-\ln(1-t)}{1-t}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Combining it all together we get that&#xA;$$S(x)=\frac{\sqrt 5}{x}\int_{0}^{x/\sqrt 5} \frac{2\ln(1-t)}{t(t-1)(t+2)}dt$$&#xA;which probably can be calculated in elementary functions.&lt;/p&gt;&#xA;" OwnerUserId="723" LastActivityDate="2010-11-03T19:49:38.050" CommentCount="4" />
  <row Id="8803" PostTypeId="2" ParentId="8794" CreationDate="2010-11-03T20:14:36.753" Score="5" ViewCount="" Body="&lt;p&gt;I had this question sometime ago and was shown the explanation in Balazs 1954 paper &quot;&lt;a href=&quot;http://iopscience.iop.org/0370-1298/68/6/307&quot; rel=&quot;nofollow&quot;&gt;Wave propagation in even and odd dimensional spaces&lt;/a&gt;&quot;.  The singularities of the integrands in the solution vary based on the dimension giving rise to this effect.  &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;When n is even, the integrand has a&#xA;pole in the complex plane, so the&#xA;contour has to be deformed, and the&#xA;result refers to only one time value -- that of the pole.&lt;/li&gt;&#xA;&lt;li&gt;When n is odd, the integrand has a&#xA;branch point, and the contour cannot&#xA;encircle the singularity. &#xA;Consequently the solution contains&#xA;contributions from all times along&#xA;the branch cut.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="1558" LastActivityDate="2010-11-03T20:14:36.753" />
  <row Id="8804" PostTypeId="2" ParentId="8793" CreationDate="2010-11-03T20:23:01.670" Score="0" ViewCount="" Body="&lt;p&gt;Thanks to both. I'm still confused because books define Lebesgue measurable sets in different ways (Caratheodory extension, outer measure, etc). What is the exact definition you are using for Lebesgue measurable set? I'm sure once I understand how it is defined I will be able to understand fully your hint(s). Thanks again.&lt;/p&gt;&#xA;" OwnerUserId="4219" LastActivityDate="2010-11-03T20:23:01.670" CommentCount="3" />
  <row Id="8805" PostTypeId="2" ParentId="8610" CreationDate="2010-11-03T20:50:42.383" Score="8" ViewCount="" Body="&lt;p&gt;We  use the formula given here: &lt;a href=&quot;http://en.wikipedia.org/wiki/Continued_fraction_of_Gauss#Applications&quot; rel=&quot;nofollow&quot;&gt;Gauss' continued fraction&lt;/a&gt; for $\tan z$ and see that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\tan(1) = \cfrac{1}{1 - \cfrac{1}{3 - \cfrac{1}{5 -\dots}}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now use the identity&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\cfrac{1}{a-\cfrac{1}{x}} = \cfrac{1}{a-1 + \cfrac{1}{1 + \cfrac{1}{x-1}}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To transform $$\cfrac{1}{a - \cfrac{1}{b - \cfrac{1}{c - \dots}}}$$ to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\cfrac{1}{a-1 + \cfrac{1}{1 + \cfrac{1}{b-2 + \cfrac{1}{1 + \cfrac{1}{c-2 + \dots}}}}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;to get the expansion for $\displaystyle \tan(1)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The above expansion for $\tan(1)$ becomes&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \cfrac{1}{1-1 + \cfrac{1}{1 + \cfrac{1}{3-2 + \cfrac{1}{1 + \cfrac{1}{5-2 + \dots}}}}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$  = 1 + \cfrac{1}{3-2 + \cfrac{1}{1 + \cfrac{1}{5-2 + \dots}}}$$&#xA;$$= 1 + \cfrac{1}{1 + \cfrac{1}{1 + \cfrac{1}{3 + \cfrac{1}{1 + \cfrac{1}{5 + \dots}}}}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To prove the transformation, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;let $\displaystyle x = b - \cfrac{1}{c - \dots}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \cfrac{1}{a-\cfrac{1}{x}} = \cfrac{1}{a-1 + \cfrac{1}{1 + \cfrac{1}{x-1}}}$$&#xA;$$ = \cfrac{1}{a-1 + \cfrac{1}{1 + \cfrac{1}{b-1 + \cfrac{1}{c - \dots}}}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Applying the identity again to &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\cfrac{1}{b-1 + \cfrac{1}{c - \dots}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;we see that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\cfrac{1}{a-\cfrac{1}{x}} = \cfrac{1}{a-1 + \cfrac{1}{1 + \cfrac{1}{b-2 + \cfrac{1}{1 + \cfrac{1}{c-1 + \cfrac{1}{d - \dots}}}}}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Applying again to $\cfrac{1}{c-1 + \cfrac{1}{d - \dots}}$ etc gives the required CF.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-11-04T02:51:17.377" LastActivityDate="2010-11-04T02:51:17.377" CommentCount="1" />
  <row Id="8806" PostTypeId="2" ParentId="8794" CreationDate="2010-11-03T20:52:15.877" Score="4" ViewCount="" Body="&lt;p&gt;This is a very good question. &lt;a href=&quot;http://www.mathpages.com/home/kmath242/kmath242.htm&quot; rel=&quot;nofollow&quot;&gt;John Baez once gave a very detailed answer to it&lt;/a&gt; which I will not bother to reproduce here. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A not-exactly precise way to think about it is this: a wave propagating in $n$ dimensions can be thought of as a wave propagating in $n+1$ dimensions, but one degree of freedom removed. More precisely: if $f$ solves the wave equation $(\partial_t^2 - \triangle) f(t,x) = 0$ on $\mathbb{R}^d$, then  it also solves the equation $(\partial_t^2 - \partial_y^2 - \triangle) f(t,y,x) = 0$ on $\mathbb{R}^{d+1}$, if you assume that $f(t,y,x) = f(t,0,x) = f(t,x)$ for all $y$. (It is constant in the $y$ direction, so any derivative in that direction is 0.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if a wave were to propagate in, say, 5 dimensions with the property you mentioned. Then in 4 dimensions, the points &quot;inside the light cone&quot; is reachable in 5 dimensions as on the light cone. Or, in other words, assume $f(t,x)$ is a solution to wave equation in 4 dimensions. And let $f(t,y,x)$ be the trivial extension with one dimension added. So $f(t,y,x)$ solves the wave equation in 5 dimensions. Suppose at time $t = 0$, a light bulb turns on at the origin $x = 0$ in 4 dimension space. This corresponds to a bank of lightbulbs turning on in 5 dimensional space along the $y$ axis. The five dimensional principle says that a point at coordinates $(y,x)$ will be illuminated by a lightbulb at $(y_0,0)$ at time $t$ if $t^2 = x^2 + (y-y_0)^2$. But for every $x, y$ such that $|x| &amp;lt; |t|$, you can find two values of $y_0$ such that $t^2 = x^2 + (y-y_0)^2$. And so some light must lag behind. (In the study of partial differential equations, this is also known as the method of descent.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But by this argument, there should still be some light lagging behind for 3 dimensions also! The claim, however, is this: due to the nature of the propagation, there is a destructive interference when you drop two dimensions. That you have destructive interference when you drop two dimensions, but not such problem when you drop only one dimension, is not at all obvious intuitively (at least to me). It however does fall out of the expression for the fundamental solution to the wave equation. &lt;/p&gt;&#xA;" OwnerUserId="1543" LastActivityDate="2010-11-03T20:52:15.877" />
  <row Id="8807" PostTypeId="1" AcceptedAnswerId="8810" CreationDate="2010-11-03T21:02:57.527" Score="3" ViewCount="57" Body="&lt;p&gt;I have a semigroup $G$ (a set with associative binary operation) such that for all $a,b\in G$ there exists $x,y\in G$ such that $ax=ya=b$. Is this property enough to show that $G$ is a group, and if so, how?&lt;/p&gt;&#xA;" OwnerUserId="1818" LastActivityDate="2010-11-03T21:11:54.550" Title="Semigroup with &quot;transitive&quot; operation is a group?" Tags="&lt;group-theory&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="8808" PostTypeId="1" AcceptedAnswerId="8809" CreationDate="2010-11-03T21:04:20.787" Score="10" ViewCount="378" Body="&lt;p&gt;A lot of mathematics as far as I know is interested in the study of Euclidean and locally Euclidean spaces (manifolds).&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;What is the special feature of Euclidean spaces that makes them interesting?&lt;/li&gt;&#xA;&lt;li&gt;Is there a field that studies spaces that are neither globally nor locally Euclidean spaces?&lt;/li&gt;&#xA;&lt;li&gt;If such a field exists, are there any practical uses for it (as in physically existing models)?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="2949" LastActivityDate="2010-12-08T20:00:31.727" Title="Why the interest in locally Euclidean spaces?" Tags="&lt;geometry&gt;&lt;differential-geometry&gt;&lt;euclidean-geometry&gt;" AnswerCount="5" CommentCount="5" FavoriteCount="4" />
  <row Id="8809" PostTypeId="2" ParentId="8808" CreationDate="2010-11-03T21:06:55.873" Score="11" ViewCount="" Body="&lt;p&gt;One of the main reasons for interest is that manifolds are &lt;em&gt;homogeneous&lt;/em&gt;.  So from a geometric point of view, we study manifolds for much the same reason we study groups.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Put in less abstract ways, there is the &quot;classical problem&quot; of whether or not the earth is flat.  The &quot;flat earth problem&quot; deals with the possibility that on small scales something may look linear, but macroscopically it need not be.  Manifolds are the abstract manifestation of the flat earth problem. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;People do study large families of objects that are locally modelled on spaces that aren't Euclidean spaces.  There's infinite-dimensional manifolds, fractal manifolds, orbifolds (locally modelled on euclidean space mod a finite group action), etc... Fibre bundles are maps that are locally projection maps.  This idea resurfaces in mathematics in many different forms and many different ways. &lt;/p&gt;&#xA;" OwnerUserId="642" LastEditorUserId="642" LastEditorDisplayName="" LastEditDate="2010-11-03T21:19:37.703" LastActivityDate="2010-11-03T21:19:37.703" CommentCount="2" />
  <row Id="8810" PostTypeId="2" ParentId="8807" CreationDate="2010-11-03T21:11:54.550" Score="5" ViewCount="" Body="&lt;p&gt;Yes. This is a standard exercise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, let $a\in G$ be arbitrary. Then there exists $e\in G$ such that $ae=a$ (using the first condition with $a=b$). Now let $b\in G$. Then there exists $z\in G$ such that $za=b$. Therefore, $be=(za)e=z(ae) = za = b$. Thus, $e$ is a right identity for $G$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the fact that $ax=e$ is always solvable it follows that every element has a right inverse. Since your semigroup has a right identity and right inverses, then it is a group.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-11-03T21:11:54.550" />
  <row Id="8811" PostTypeId="1" AcceptedAnswerId="8835" CreationDate="2010-11-03T21:30:21.803" Score="5" ViewCount="334" Body="&lt;p&gt;Américo Tavares pointed out in his answer to &lt;a href=&quot;http://math.stackexchange.com/questions/8476/rearrange-the-formula-for-the-sum-of-a-geometric-series-to-find-r&quot;&gt;this question&lt;/a&gt; that finding the ratio of a geometric progression only from knowledge of the sum of its first $n+1$ terms $S = 1+x+x^2+\cdots+x^n$ amounts to solving a polynomial of degree $n$. This suggested to me that there might be up to $n$ real solutions of $x$ for a given sum, but I could not find any. In fact, it turned out that the following fact is true:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $n \ge 1$ and $S \in \mathbb{R}$, the polynomial equation $x^n + x^{n-1} + \cdots + x + 1 = S$ has at most two real solutions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A corollary is that if $n$ is odd, there is exactly one real solution. I was only able to prove this using a rather contrived geometric argument based on the shape of the graph of $y = x^{n+1}$. Is there a simple, direct (and ideally, intuitive) proof of this fact?&lt;/p&gt;&#xA;" OwnerUserId="856" LastActivityDate="2010-11-05T23:00:57.123" Title="Why does the polynomial equation $1 + x + x^2 + \cdots + x^n = S$ have at most two solutions in $x$?" Tags="&lt;sequences-and-series&gt;&lt;polynomials&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="1" />
  <row Id="8812" PostTypeId="2" ParentId="8672" CreationDate="2010-11-03T21:34:16.997" Score="4" ViewCount="" Body="&lt;p&gt;Despite other answers, I thought it might benefit the impatient to see the explicit answer below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $$M = &#xA;\left(&#xA;\begin{array}{cc}&#xA; a &amp;amp; b \\&#xA; b &amp;amp; c&#xA;\end{array}&#xA;\right),$$&#xA;be the input matrix. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Define the discriminant: $\Delta = \sqrt{a^2+4 b^2-2 a c+c^2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, the eigenvalues of $M$ are given by:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\lambda_1 = 0.5(a+c-\Delta)$ and $\lambda_2 = 0.5(a+c+\Delta)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, you can find a matrix $V$ such that &#xA;$$M = V^{-1} \begin{pmatrix}&#xA;\lambda_1 &amp;amp; 0\\&#xA;0 &amp;amp; \lambda_2&#xA;\end{pmatrix}V.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mathematica says that the matrix $V$ is given by&#xA;$$&#xA;V = \begin{pmatrix}&#xA;\frac{a-c-\Delta}{2b} &amp;amp; 1\\&#xA;\frac{a-c+\Delta}{2b} &amp;amp; 1&#xA;\end{pmatrix}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are looking for orthogonal $V$, then the above calculations need some changes.&lt;/p&gt;&#xA;" OwnerUserId="1709" LastActivityDate="2010-11-03T21:34:16.997" CommentCount="3" />
  <row Id="8813" PostTypeId="2" ParentId="8811" CreationDate="2010-11-03T21:36:58.410" Score="11" ViewCount="" Body="&lt;p&gt;The roots are also roots of&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x^{n+1} - Sx + S - 1 = 0$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;which we get by multiplying your equation by $x-1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This polynomial ($x^{n+1} - Sx + S-1$), as we move from $x = -\infty$ to $x = \infty$ is either &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Monotonically increasing, and thus has at most one real root.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Monotonically decreasing, and then monotonically increasing and hence can have at most two real roots.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Monotonically increasing, then decreasing and then again increasing (happens only when $n$ is even). In which case there are at most three real roots, one of which is $1$. So for $S \ne n+1$, the original equation does not have more than two solutions. If $S=n+1$ and $n$ is even, then the turning points are $-1$ and $1$ and the value of the polynomial at $-1$ is positive. So the only roots are $1$ and a root which is $&amp;lt; -1$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This can be seen by looking at its derivative, which is an increasing function for odd $n$, and for even $n$, it is positive, then possibly negative (depending on $S$) and then positive again, as we move from $x = -\infty$ to $x = \infty$.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-11-03T22:28:49.673" LastActivityDate="2010-11-03T22:28:49.673" CommentCount="5" />
  <row Id="8814" PostTypeId="1" CreationDate="2010-11-03T21:39:01.623" Score="17" ViewCount="1448" Body="&lt;p&gt;Here is a funny exercise &#xA;$$\sin(x - y) \sin(x + y) = (\sin x - \sin y)(\sin x + \sin y).$$&#xA;(If you prove it don't publish it here please).&#xA;Do you have similar examples?&lt;/p&gt;&#xA;" OwnerUserId="1154" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2011-01-17T00:57:36.063" LastActivityDate="2011-04-18T02:52:43.850" CommunityOwnedDate="2010-11-04T06:29:02.260" Title="Funny identities" Tags="&lt;trigonometry&gt;&lt;big-list&gt;" AnswerCount="20" CommentCount="10" FavoriteCount="11" />
  <row Id="8815" PostTypeId="2" ParentId="6339" CreationDate="2010-11-03T21:48:52.257" Score="0" ViewCount="" Body="&lt;p&gt;A trivial case when $B$ is semidef is say when the original matrix is elementwise non-negative (and you define sign(0) = 1). In that case, $B=ee^T$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, it might be hard to give a non-trivial set of sufficient conditions on $A$ so that $B$ is semidefinite.&lt;/p&gt;&#xA;" OwnerUserId="1709" LastActivityDate="2010-11-03T21:48:52.257" />
  <row Id="8816" PostTypeId="2" ParentId="8811" CreationDate="2010-11-03T21:57:58.027" Score="8" ViewCount="" Body="&lt;p&gt;A powerful algorithmic way to handle such problems is to employ &lt;a href=&quot;http://en.wikipedia.org/wiki/Sturm%27s_theorem&quot; rel=&quot;nofollow&quot;&gt;Sturm's Theorem&lt;/a&gt;. If you work out the details you will see that it is quite simple for this example. This in turn is a special case of the CAD (cylindrical algebraic decomposition) algorithm  - an effective implementation of Tarski's  quantifier elimination for the first order theory of the reals. The general ideas behind these  methods prove helpful in solving a variety of problems. It's well worth the effort to learn the general methods rather than ad-hoc techniques.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Per Rahul's request, here are further details of applying Sturm's algorithm to the example at hand. We desire to prove that $\rm\ \ g(x) =\ x^n +\:\cdots\: + x^2 + x + 1 - s\ \ $ has at most two distinct real roots. Consider $\rm\ f(x) = (x-1)\ g(x) =\ x^{n+1}-1-s\ (x-1)\:.\ $ Since $\rm\ \ f\:' = (n+1)\ x^n - s\ \ $ we have $\rm\ f\ mod\ f\:'\: =\ f\: - \: x/(n+1)\ f\:'\ =\ a\ x + b\ $ for some $\rm\:a,\:b\in \mathbb R\:$. So the euclidean remainder sequence in the calculation of $\rm\ gcd(f,\:f\:')\ $ has length at most $4$, viz. $\rm\ f,\ f\:',\ a\ x + b,\ c\ $. Thus it has at most $3$ sign changes at any point, so Sturm's theorem implies that $\rm\ f(x)\ $ has at most $3$ distinct real roots. So if $\rm\: x = 1\:$ isn't a multiple root of $\rm \:f(x)\:$ then $\rm\ g(x) = f(x)/(x-1)\ $ has at most $2$ distinct real roots. Else $\rm\ x-1\:|\:gcd(f,\:f\:')\ \Rightarrow\ x-1\:|\:c\ \Rightarrow\ c=0\:$. So in this case  the remainder sequence has length at most $3$, so at most $2$ sign changes, so $\rm\:f\ $ has at most $2$ distinct real roots, therefore ditto for $\rm\:g\:$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Although Sturm's theorem is slightly more work here than Rolle's theorem, it has the added benefit that it allows one to compute the precise number of roots in any interval (versus only bounds using Rolle's theorem).&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-11-05T23:00:57.123" LastActivityDate="2010-11-05T23:00:57.123" CommentCount="3" />
  <row Id="8817" PostTypeId="2" ParentId="8814" CreationDate="2010-11-03T22:01:42.733" Score="9" ViewCount="" Body="&lt;p&gt;The Frobenius automorphism&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(x + y)^p = x^p + y^p$$&lt;/p&gt;&#xA;" LastActivityDate="2010-11-03T22:01:42.733" CommunityOwnedDate="2010-11-04T06:29:02.260" />
  <row Id="8818" PostTypeId="2" ParentId="8734" CreationDate="2010-11-03T22:05:08.627" Score="4" ViewCount="" Body="&lt;p&gt;I don't believe you understand the notion of efficiency in terms of encoding. Informally speaking, you have to keep it mind there are two factors involved: (i) cost of having different symbols (in case of base 10 there as 10 different symbols, in case of base 16 there are 16 different symbols etc) and the length of the resulting string to encode a particular number. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you consider both factors and apply some basic information theory to it, the answer may look a bit surprising: the most efficient encoding has a base $e$ (yes, that very $e = 2.718\dots$). Since we'd rather have some natural number as a base, the best we can get is base 3, and the next is base 2. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, why, then, computers use base 2 (0 and 1) rather than base 3 (say, -1, 0, and 1)? The answer is that it is simple to design the circuits that distinguish between two (rather than three) states. (I do remember reading some of the earlier computers did use base 3, but I can't recall all the details.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, with respect to octals and hexes, those are simply convenient ways to record the binary strings. If you did some machine-level debugging, you probably had a chance to read what's known as &quot;hexadecimal dump&quot; (contents of a memory). Surely it's easy to read than if it were written as binary dump. But what's lurking underneath of that is base 2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(The answer on &quot;why do we use base 10&quot; has been answered elsewhere.)  &lt;/p&gt;&#xA;" OwnerUserId="2450" LastActivityDate="2010-11-03T22:05:08.627" CommentCount="2" />
  <row Id="8819" PostTypeId="2" ParentId="8532" CreationDate="2010-11-03T22:06:21.463" Score="7" ViewCount="" Body="&lt;p&gt;The method is &quot;keep trying&quot;. Try reading the paper again, try reading a different paper, try stopping reading papers and just work it out yourself. Try talking to other people, try struggling on your own, then try talking to people again. Try a different problem, then try going back to the same problem. Try the same method again and again, then try different methods, then go back to the first method and try it one more time. Try just writing out some calculations even if you don't see them going anywhere, try a stupid example, try drawing some pictures. Finally try taking a break, and then also try not taking a break. Try everything once, then try it again!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But enough of that, here's some more practical advice from my own experience (I'm a postdoc by the way).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) For a hard problem, work intensely on it for a few weeks or months, until you feel like you've completely hit a wall. During this period you want to be producing as many calculations, lemmas, pictures, examples, etc as possible. They don't all have to be super-relevant -- your goal is maximum output of material. Then when you finally feel like you're going insane, take a deep breath and move on to something else. After a break of a few weeks or months, which may be spent either on work of some other kind or even on vacation, come back to the problem and see if the fresh perspective helps. It may take multiple cycles like this for a really hard problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Look for a variation of your problem which is easier or which your methods can better handle. Even if the variation is not as strong of a result as what you are ultimately aiming for, it may still &quot;count&quot; and end up being a nicer paper than you realize at first. It's good for your morale to register progress in this way, and if you can get a publication it's good for your CV too. Here is where advisors and mentors can help a lot -- they can help you identify which variation is meaningful, doable, and &quot;interesting&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) Remember that almost nothing ever works. One of the worst feelings to me is when I find a mistake in something I thought I had figured out, and all of a sudden weeks or months of progress begin to unravel. This can be extremely depressing. Try not to take it too hard, and over time you can rebuild and recover. Try also to appreciate the progress you are making, even if it is much less than you wish it was.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Good luck!&lt;/p&gt;&#xA;" LastActivityDate="2010-11-03T22:06:21.463" CommunityOwnedDate="2010-11-03T22:06:21.463" CommentCount="3" />
  <row Id="8820" PostTypeId="2" ParentId="8814" CreationDate="2010-11-03T22:07:04.300" Score="15" ViewCount="" Body="&lt;p&gt;$$\left(\sum\limits_{k=1}^n k\right)^2=\sum\limits_{k=1}^nk^3 .$$&lt;/p&gt;&#xA;" OwnerUserId="723" LastActivityDate="2010-11-03T22:07:04.300" CommunityOwnedDate="2010-11-04T06:29:02.260" />
  <row Id="8821" PostTypeId="2" ParentId="8764" CreationDate="2010-11-03T22:07:16.063" Score="2" ViewCount="" Body="&lt;p&gt;This seems to be Laplace noise. Have a look at: &lt;a href=&quot;http://en.wikipedia.org/wiki/Laplace_distribution&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Laplace_distribution&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is handy for modeling such spikey, sparse noise.&lt;/p&gt;&#xA;" OwnerUserId="1709" LastActivityDate="2010-11-03T22:07:16.063" />
  <row Id="8822" PostTypeId="2" ParentId="8814" CreationDate="2010-11-03T22:12:30.473" Score="5" ViewCount="" Body="&lt;p&gt;$$ \infty! = \sqrt{2 \pi} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It comes from the zeta function.&lt;/p&gt;&#xA;" LastActivityDate="2010-11-03T22:12:30.473" CommunityOwnedDate="2010-11-04T06:29:02.260" CommentCount="3" />
  <row Id="8823" PostTypeId="2" ParentId="8814" CreationDate="2010-11-03T22:18:49.217" Score="4" ViewCount="" Body="&lt;p&gt;\begin{eqnarray}&#xA;\sum_{i_1 = 0}^{n-k} \, \sum_{i_2 = 0}^{n-k-i_1} \cdots \sum_{i_k = 0}^{n-k-i_1 - \cdots - i_{k-1}} 1 = \binom{n}{k}&#xA;\end{eqnarray}&lt;/p&gt;&#xA;" OwnerUserId="2720" LastActivityDate="2010-11-03T22:18:49.217" CommunityOwnedDate="2010-11-04T06:29:02.260" />
  <row Id="8824" PostTypeId="2" ParentId="8814" CreationDate="2010-11-03T22:23:47.103" Score="8" ViewCount="" Body="&lt;p&gt;$$ \sum\limits_{n=1}^{\infty} n =  1 + 2 +3 + \cdots \text{ad inf.} = - \frac{1}{12}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now doesn't this sound funny. You can also see many more on &lt;a href=&quot;http://terrytao.wordpress.com/2010/04/10/the-euler-maclaurin-formula-bernoulli-numbers-the-zeta-function-and-real-variable-analytic-continuation/&quot; rel=&quot;nofollow&quot;&gt;http://terrytao.wordpress.com/2010/04/10/the-euler-maclaurin-formula-bernoulli-numbers-the-zeta-function-and-real-variable-analytic-continuation/&lt;/a&gt;&lt;/p&gt;&#xA;" LastActivityDate="2010-11-03T22:23:47.103" CommunityOwnedDate="2010-11-04T06:29:02.260" CommentCount="5" />
  <row Id="8825" PostTypeId="2" ParentId="8814" CreationDate="2010-11-03T22:26:57.677" Score="2" ViewCount="" Body="&lt;p&gt;\begin{eqnarray}&#xA;\zeta(0) = \sum_{n \geq 1} 1 = -\frac{1}{2}&#xA;\end{eqnarray}&lt;/p&gt;&#xA;" OwnerUserId="2720" LastActivityDate="2010-11-03T22:26:57.677" CommunityOwnedDate="2010-11-04T06:29:02.260" />
  <row Id="8826" PostTypeId="2" ParentId="8814" CreationDate="2010-11-03T22:33:13.333" Score="13" ViewCount="" Body="&lt;p&gt;\begin{eqnarray}&#xA;1^{3} + 2^{3} + 2^{3} + 2^{3} + 4^{3} + 4^{3} + 4^{3} + 8^{3} = (1 + 2 + 2 + 2 + 4 + 4 + 4 + 8)^{2}&#xA;\end{eqnarray}&#xA;More generally, let $D_{k} = ${ $d$ } be the set of unitary divisors of a positive integer $k$, and let $\mathsf{d}^{*} \colon \mathbb{N} \to \mathbb{N}$ denote the number-of-unitary-divisors (arithmetic) function. Then &#xA;\begin{eqnarray}&#xA;\sum_{d \in D} \mathsf{d}^{*}(d)^{3} = \left( \sum_{d \in D} \mathsf{d}^{*}(d) \right)^{2}&#xA;\end{eqnarray}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that $\mathsf{d}^{*}(k) = 2^{\omega(k)}$, where $\omega(k)$ is the number distinct prime divisors of $k$.&lt;/p&gt;&#xA;" OwnerUserId="2720" LastEditorUserId="2720" LastEditorDisplayName="" LastEditDate="2010-11-23T00:42:18.930" LastActivityDate="2010-11-23T00:42:18.930" CommunityOwnedDate="2010-11-04T06:29:02.260" />
  <row Id="8827" PostTypeId="2" ParentId="8760" CreationDate="2010-11-03T22:33:41.657" Score="0" ViewCount="" Body="&lt;p&gt;You may take a look at this previous &lt;a href=&quot;http://math.stackexchange.com/questions/3315/what-is-the-square-root-of-i/3346#3346&quot;&gt;question&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="664" LastActivityDate="2010-11-03T22:33:41.657" />
  <row Id="8828" PostTypeId="2" ParentId="8814" CreationDate="2010-11-03T22:33:44.283" Score="18" ViewCount="" Body="&lt;p&gt;$$\int_0^1\frac{\mathrm{d}x}{x^x}=\sum_{k=1}^\infty \frac1{k^k}$$&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-11-03T22:33:44.283" CommunityOwnedDate="2010-11-03T22:35:34.187" />
  <row Id="8829" PostTypeId="2" ParentId="8814" CreationDate="2010-11-03T22:40:28.237" Score="4" ViewCount="" Body="&lt;p&gt;Two related integrals:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_0^\infty\sin\;x\quad\mathrm{d}x=1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_0^\infty\ln\;x\;\sin\;x\quad \mathrm{d}x=-\gamma$$&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-11-03T22:40:28.237" CommunityOwnedDate="2010-11-03T22:40:28.237" CommentCount="2" />
  <row Id="8830" PostTypeId="1" AcceptedAnswerId="8849" CreationDate="2010-11-03T22:46:20.100" Score="1" ViewCount="99" Body="&lt;p&gt;I am working in the field of machine learning, and I have come across a few papers that show relationships between Gröbner bases and discrete probability. So I come here for help.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you please explain how can Gröbner bases used to describe discrete probability?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have looked at Gröbner bases and I understand the general concepts (and used Maple to calculate a few examples). So it is the link that is missing for me.&lt;/p&gt;&#xA;" OwnerUserId="2949" LastActivityDate="2010-11-04T00:47:21.133" Title="How can Gröbner bases used to describe discrete probability ?" Tags="&lt;probability&gt;&lt;algebra&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="8831" PostTypeId="1" AcceptedAnswerId="8833" CreationDate="2010-11-03T22:56:40.540" Score="2" ViewCount="126" Body="&lt;p&gt;Hello everybody.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basktball player shoots basket for 100 times. On the first throw the ball is always in, on the second - always out. For all consequent shoots the probability of hit is equal to precentage of all scored shoots diveded by total number of shoots done.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the probability the he will score total 50 balls out of 100?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is not a homework, I'm just anxious about solution of this problem. =)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My ruminitions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are random events $E_1, E_2, \dots, E_n$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$E_1$ always happens, $E_2$ - never.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For any $E_n$ with $n &gt; 2$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$p(E_n) = \frac{\sum_{k=1}^{i-1}{x_k}}{i-1}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;${x}$ - random variables, that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x_i = 1$ if event is happened, $0$ otherwise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So,&#xA;$P (\sum_{i=1}^{100}{x_i} = 50) = ?$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Is it true that this problem can be solved with Limit Theorem?&lt;/em&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consequent events are dependant on outcoimes ov previous, can Limit Theorem handle that?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Could you point on how can i solve this problem?&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanx! &#xA;=)&lt;/p&gt;&#xA;" OwnerUserId="3089" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-04T00:36:15.723" LastActivityDate="2010-11-04T00:41:48.083" Title="[Probability Theorem Problem] Limit Theorem applicability" Tags="&lt;probability-theory&gt;" AnswerCount="4" CommentCount="2" FavoriteCount="1" />
  <row Id="8832" PostTypeId="2" ParentId="8652" CreationDate="2010-11-03T22:59:39.887" Score="8" ViewCount="" Body="&lt;p&gt;It depends largely on what your career goals are.  If your goal is to be a professor at a top 100 research school, then you should certainly go to the most prestigious option available as explained by Pete and Ben (and with their caveats that &quot;more&quot; means &quot;significantly more.&quot;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if you have different goals (for example, you want a job at a more teaching oriented institution) then it seems less clear to me that the increase in prestige is always for the best.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While I was a graduate student a professor moved from a top 25 school to Berkeley bringing several of his students along with him.  Those students had the experience of having the same advisor at two different schools, and they found the experience at the two schools to be substantially different.  In particular, at the previous school the professor was not very busy and spent hours talking with his students.  Whereas at the new school there were lots and lots of visitors/postdocs/other profs around all the time and the prof had the opportunity to talk to those people rather than his students.  (He was still a responsible advisor who met regularly with his students.)  So, if having an advisor with lots of extra time who actually seeks you out to talk to you is really important to you then going to a lower ranked school actually can make a significant difference in experience.&lt;/p&gt;&#xA;" OwnerUserId="8" LastActivityDate="2010-11-03T22:59:39.887" CommentCount="2" />
  <row Id="8833" PostTypeId="2" ParentId="8831" CreationDate="2010-11-03T23:24:44.047" Score="3" ViewCount="" Body="&lt;p&gt;The probability is $1/99$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Denote by $P(a,b)$ the probability that after $a+b$ shots, exactly $a$ were in and $b$ were out. Then $P(1,1)=1$ and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$P(a,b) = \frac{a-1}{a+b-1} P(a-1,b) + \frac{b-1}{a+b-1} P(a,b-1)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We prove by induction on $a+b$ that $P(a,b) = 1/(a+b-1)$. When $a+b = 2$, this is true. Otherwise&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$P(a,b) = \frac{a+b-2}{a+b-1} \cdot \frac{1}{a+b-2} = \frac{1}{a+b-1}$&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-03T23:24:44.047" CommentCount="1" />
  <row Id="8834" PostTypeId="2" ParentId="8831" CreationDate="2010-11-03T23:28:04.510" Score="1" ViewCount="" Body="&lt;p&gt;This is problem B1 from the 2002 Putnam exam.  It is useful to generalize and allow for any number of free throws.  We can then prove the stronger result:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Theorem&lt;/strong&gt;. For any number of free throws $n$, and any $i \in [n-1]$ the probability of hitting exactly $i$ free throws after $n$ attempts is $\frac{1}{n-1}$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof.&lt;/strong&gt; We proceed by induction on $n$.  The base case $n=3$ is trivially trivial. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the inductive step, assume the result for $n-1$, and let $i \in [n-1]$.  There are two ways to make exactly $i$ shots after $n$ attempts:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;make $i-1$ shots after $n-1$ attempts and clutch up on your final attempt, or &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;make $i$ shots after $n-1$ attempts and epic fail the final attempt. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;By induction, the required probability is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{1}{n-2} \frac{i-1}{n-1} + \frac{1}{n-2} \frac{n-1-i}{n-1} = \frac{1}{n-1}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the answer to your particular question is $\frac{1}{99}$.  &lt;/p&gt;&#xA;" OwnerUserId="964" LastEditorUserId="964" LastEditorDisplayName="" LastEditDate="2010-11-03T23:35:49.600" LastActivityDate="2010-11-03T23:35:49.600" CommentCount="2" />
  <row Id="8835" PostTypeId="2" ParentId="8811" CreationDate="2010-11-03T23:45:12.717" Score="10" ViewCount="" Body="&lt;p&gt;In between the previous two answers - one very specific and one very general, there is the following one:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Rolle%27s_theorem&quot; rel=&quot;nofollow&quot; title=&quot;Rolle's theoerm&quot;&gt;Rolle's theorem&lt;/a&gt; states that between any two zeros of the function lies a zero of the derivative. So each time you differentiate the number of real roots can not decrease by more than 1. This actually holds with multiplicity as well - if a root is of multiplicity k, then it is of multiplicity (k-1) for the derivative, and there are still roots of the derivative strictly between consecutive distinct roots of the original function. So $x^{n+1}−Sx+S−1$ has at most one root more than $(n+1)x^n-S$ and at most two more than $n(n+1)x^{n-1}$, so not more than three total (counted with multiplicity). One of those is $x=1$, so there are no more than two for the $1+x+\ldots+x^n=S$. QED.&lt;/p&gt;&#xA;" OwnerUserId="2633" LastActivityDate="2010-11-03T23:45:12.717" CommentCount="5" />
  <row Id="8837" PostTypeId="1" AcceptedAnswerId="8840" CreationDate="2010-11-03T23:53:05.037" Score="6" ViewCount="148" Body="&lt;p&gt;I recall seeing a hand-drawn picture of spec of a ring (maybe of Z?) that had been passed around in the early days of the Zariski topology.  Does anyone know where I can find a copy?&lt;/p&gt;&#xA;" OwnerUserId="3009" LastActivityDate="2010-11-04T00:29:59.503" Title="Picture of spec?" Tags="&lt;algebraic-geometry&gt;" AnswerCount="3" FavoriteCount="4" />
  <row Id="8838" PostTypeId="1" AcceptedAnswerId="8848" CreationDate="2010-11-03T23:58:20.123" Score="11" ViewCount="736" Body="&lt;p&gt;Is solving problems quickly an important trait for a mathematician to have? Is solving textbook/olympiad style problems quickly necessary to succeed in math? To make an analogy, is it better to be a sprinter or a marathon runner in mathematics? &lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2011-03-16T23:30:57.807" Title="Is speed an important quality in a mathematician?" Tags="&lt;soft-question&gt;" AnswerCount="5" CommentCount="9" FavoriteCount="9" />
  <row Id="8840" PostTypeId="2" ParentId="8837" CreationDate="2010-11-04T00:09:13.743" Score="6" ViewCount="" Body="&lt;p&gt;I think you mean &lt;a href=&quot;http://www.neverendingbooks.org/wp-content/uploads/2008/12/MumfordsTreasureMap.jpg&quot; rel=&quot;nofollow&quot;&gt;this image&lt;/a&gt; by Mumford. &lt;/p&gt;&#xA;" OwnerUserId="308" LastActivityDate="2010-11-04T00:09:13.743" CommentCount="2" />
  <row Id="8841" PostTypeId="2" ParentId="8775" CreationDate="2010-11-04T00:11:05.833" Score="3" ViewCount="" Body="&lt;p&gt;This is quite obvious if expressed most naturally in terms of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Bernoulli_polynomials&quot; rel=&quot;nofollow&quot;&gt;Bernoulli polynomials&lt;/a&gt;, viz.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\quad\quad\quad\quad\rm\ \ \ \sum n^k\ = \:\frac{1}{k+1} (B_{k+1}(n+1) - B_{k+1}(0))\ $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and $\rm\quad\quad\ B_k^{\:'}(x)\ =\ k\ B_{k-1}(x)\ $ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;thus $\ \rm\ (\sum n^k)'\ =\ k\ \sum n^{k-1} + B_{\:k} $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to &lt;a href=&quot;http://www.ams.org/journals/mcom/1993-61-203/S0025-5718-1993-1197512-7/S0025-5718-1993-1197512-7.pdf&quot; rel=&quot;nofollow&quot;&gt;Knuth's exposition&lt;/a&gt; these identities go all the way back to Jacobi - with special cases known to Faulhaber. Note that the identities in Robin Chapman's post are simply special cases of these well-known identities for Bernoulli polynomials. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Bernoulli polynomials are a special-case of polynomials amenable to study by the powerful techniques of the Umbral Calculus. This is the best way to understand the genesis of their many interesting properties. For a nice introduction see Steven Roman's book.&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-11-04T00:11:05.833" />
  <row Id="8842" PostTypeId="2" ParentId="8837" CreationDate="2010-11-04T00:12:16.347" Score="2" ViewCount="" Body="&lt;p&gt;You are thinking of the picture of Spec($Z[X]$) in Mumford's Red Book of Varieties and Schemes.   A reproduction is downloadable at the link posted in the other answer, and other (less famous) pictures of prime spectra appear in the Eisenbud and Harris introductory book on schemes.&lt;/p&gt;&#xA;" OwnerUserId="467" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-11-04T00:18:14.920" LastActivityDate="2010-11-04T00:18:14.920" />
  <row Id="8843" PostTypeId="2" ParentId="8831" CreationDate="2010-11-04T00:18:34.030" Score="2" ViewCount="" Body="&lt;p&gt;Some additional references may be of interest. This result is a &#xA;well-known feature of Polya's urn model starting with one ball&#xA;each of two different colours.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, this problem is Exercise 5.9 in the &#xA;first edition of &#xA;&lt;em&gt;Introduction to Stochastic Processes&lt;/em&gt; by Lawler.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is also a detailed analysis in section 2.4 of &#xA;&lt;em&gt;Markov Chains and Mixing Times&lt;/em&gt; by Levin, Peres, and Wilmer.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I expect that it can be found in other books on probability and Markov chains. &lt;/p&gt;&#xA;" OwnerUserId="940" LastActivityDate="2010-11-04T00:18:34.030" CommunityOwnedDate="2010-11-04T00:18:34.030" />
  <row Id="8844" PostTypeId="2" ParentId="8837" CreationDate="2010-11-04T00:29:59.503" Score="3" ViewCount="" Body="&lt;p&gt;Be sure not to miss this very interesting paper (in English) on &quot;Mumford's treasure map&quot; and related concepts: Lieven Le Bruyn: &lt;a href=&quot;http://matrix.cmi.ua.ac.be/DATA2/Alev60.pdf&quot; rel=&quot;nofollow&quot;&gt;Un dessins d'enfants&lt;/a&gt;  &lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-11-04T00:29:59.503" />
  <row Id="8846" PostTypeId="2" ParentId="8831" CreationDate="2010-11-04T00:41:48.083" Score="1" ViewCount="" Body="&lt;p&gt;Here is an enumerative proof. We can think of the process as an urn with balls of two colors, say black and white. In the beginning, there is one ball of each color. In each round we pick one of the balls and duplicate it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let us count the number of ways to get $a+1,b+1$ balls after $a+b$ rounds. This involves adding $a$ black balls and $b$ white balls. The first time a black ball is added, there is no choice. The next time, we can choose any of the two existing balls, and so on. In total, there are $a!$ choices for black balls, and $b!$ choices for white balls. We also have to choose in which round we add which color - there are $\binom{a+b}{a} = (a+b)!/a!b!$ such choices. In total, we get $(a+b)!$ choices.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We've shown that the number of ways to get $a+1,b+1$ balls depends only on $a+b$. In fact, we can directly calculate the probability by calculating the total number of possible moves. In the first move we pick one of two balls, then one of three balls, and so on. So in $a+b$ rounds there are $(a+b+1)!$ possible moves. We conclude that each outcome has probability $1/(a+b+1)$.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-04T00:41:48.083" />
  <row Id="8847" PostTypeId="2" ParentId="8838" CreationDate="2010-11-04T00:42:56.070" Score="9" ViewCount="" Body="&lt;p&gt;It is not speed but depth that counts.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-04T00:42:56.070" />
  <row Id="8848" PostTypeId="2" ParentId="8838" CreationDate="2010-11-04T00:44:05.303" Score="27" ViewCount="" Body="&lt;p&gt;To some degree speed matters. If you are going to sit there thinking &quot;hmmmm ... what &lt;em&gt;is&lt;/em&gt; a subspace?&quot; every single time you see the word &quot;subspace&quot; you may never get through even the simplest of linear algebra proofs. So, some knowledge must be there automatically, that's why we all practice so much. That's why we don't just learn definitions but &lt;em&gt;memorize&lt;/em&gt; them. That's the point of knowing a proof by heart (and I do mean &lt;em&gt;knowing&lt;/em&gt; not just memorizing it in that case) so that it falls off of your tongue with ease. That way, when you need to prove something similar you aren't struggling with the mechanics. You have room in your brain to think about the big picture. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In some sense, the process of doing math is all about doing things more quickly. Think of how long it took you to solve a quadratic when you were in grade school. Now you just look at it and you know the solutions are there, and you can probably see the factors, or write the quadratic formula without thinking about it. What once was a big problem is now just a tiny step in much larger problems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That all said, I do think that at times lay people mistake being fast with being good at mathematics. Some of the mathematicians I admire most are careful... even plodding in their thinking, going over each step in a way that seems almost naive. --yet they never seem to &lt;em&gt;miss&lt;/em&gt; anything and in their careful way of reasoning they make many small discoveries that someone who was rushing to the answer might miss. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;These mathematicians whom I admire stop now and then and think carefully about even the solutions to quadratics, they are looking for something deeper in each detail. And they are &lt;em&gt;not self-conscious&lt;/em&gt; about the speed at which they work. &lt;strong&gt;Rushing to impress others is a great source of errors.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm as guilty as anyone else of this. But, I will try to emulate my mentors and not care if it takes me a long time. As long as in the end what I say is logical... and correct.  &lt;/p&gt;&#xA;" OwnerUserId="2563" LastActivityDate="2010-11-04T00:44:05.303" CommentCount="1" />
  <row Id="8849" PostTypeId="2" ParentId="8830" CreationDate="2010-11-04T00:47:21.133" Score="2" ViewCount="" Body="&lt;p&gt;You can find a nice introduction here: &amp;#56256;Pistone, Giovanni and Riccomagno, Eva and Wynn, Henry, Pistone, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pistone, Giovanni; Riccomagno, Eva; Wynn, Henry&lt;br&gt;&#xA;&lt;a href=&quot;http://alexandria.tue.nl/repository/books/542984.pdf&quot; rel=&quot;nofollow&quot;&gt;Computational commutative algebra in discrete statistics&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-11-04T00:47:21.133" CommentCount="1" />
  <row Id="8850" PostTypeId="1" AcceptedAnswerId="8852" CreationDate="2010-11-04T01:28:39.633" Score="0" ViewCount="54" Body="&lt;p&gt;I have a branching process of the form p0=0.1, p1 = 0.6, p2 = 0.3. (any other pn = 0). Z0, the original population is 1. Z1 is the population after 1 timestep, Z2 is the population after 2 timesteps, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I compute the probability mass function of Z3? It would suffice to have an expression I could expand on my TI-89.&lt;/p&gt;&#xA;" OwnerUserId="2617" LastEditorUserId="2617" LastEditorDisplayName="" LastEditDate="2010-11-04T01:35:44.900" LastActivityDate="2010-11-04T01:59:35.763" Title="Given probabilities for a branching process, how do I compute the probability mass function at a time?" Tags="&lt;probability&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="8851" PostTypeId="1" AcceptedAnswerId="8855" CreationDate="2010-11-04T01:56:32.073" Score="0" ViewCount="67" Body="&lt;p&gt;Given that X_i is a partition of the probability space, and Y,Z are events in the probability space such that P[Y intersect X_i] &gt; 0 for all i, how can I prove that P[Z|Y] = sum(from i=1 to infinity) P[Z|Y intersect X_i] * P[X_i | Y] ?&lt;/p&gt;&#xA;" OwnerUserId="2617" LastActivityDate="2010-11-04T02:34:26.073" Title="How can I prove this law of &quot;generalized&quot; total probability?" Tags="&lt;probability&gt;" AnswerCount="1" />
  <row Id="8852" PostTypeId="2" ParentId="8850" CreationDate="2010-11-04T01:59:35.763" Score="3" ViewCount="" Body="&lt;p&gt;The probability of having $k$ individuals after 3 steps is the coefficient of $x^k$ in the expansion of $f(f(f(x)))$, where $f(x) = 0.1 + 0.6x + 0.3x^2$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This generalizes a lot: you can have different types of individuals, each with their own branching probabilities, which can change from step to step. &lt;/p&gt;&#xA;" OwnerUserId="2032" LastActivityDate="2010-11-04T01:59:35.763" CommentCount="1" />
  <row Id="8853" PostTypeId="2" ParentId="8838" CreationDate="2010-11-04T02:08:02.747" Score="17" ViewCount="" Body="&lt;p&gt;Depth of thought is much more important than speed. A good illustration of this is the work of Grigori Perelman, who proved the Poincaré Conjecture. &lt;a href=&quot;http://tinyurl.com/3xga4f5&quot; rel=&quot;nofollow&quot;&gt;In their article about him published in The New Yorker&lt;/a&gt;, Sylvia Nasar and David Gruber wrote:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;'At Leningrad University, which Perelman entered in 1982, at the age of sixteen, he took advanced classes in geometry and solved a problem posed by Yuri Burago, a mathematician at the Steklov Institute, who later became his Ph.D. adviser. “There are a lot of students of high ability who speak before thinking,” Burago said. “Grisha was different. He thought deeply. His answers were always correct. He always checked very, very carefully.” Burago added, “He was not fast. Speed means nothing. Math doesn’t depend on speed. It is about deep.”'&lt;/p&gt;&#xA;" OwnerUserId="1702" LastEditorUserId="1702" LastEditorDisplayName="" LastEditDate="2011-03-16T23:30:57.807" LastActivityDate="2011-03-16T23:30:57.807" CommentCount="3" />
  <row Id="8854" PostTypeId="1" CreationDate="2010-11-04T02:25:52.593" Score="2" ViewCount="388" Body="&lt;p&gt;The classic text of Deligne and Rapoport giving a model for the mod p reduction of the modular curve is found in a &quot;Lecture Notes in Mathematics&quot; volume.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Has it been translated from the original french and/or LaTeX-ed so that it can be searched more easily?&lt;/p&gt;&#xA;" OwnerUserId="2580" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-22T02:10:16.417" LastActivityDate="2010-12-23T00:44:25.973" Title="Has Deligne-Rapoport been translated?" Tags="&lt;reference-request&gt;&lt;modular-forms&gt;" AnswerCount="2" />
  <row Id="8855" PostTypeId="2" ParentId="8851" CreationDate="2010-11-04T02:34:26.073" Score="4" ViewCount="" Body="&lt;p&gt;Because $X_i$ are a partition,&#xA;$$P(Z \cap Y) = \sum_i P(Z \cap Y \cap X_i)$$&#xA;Now apply the definition of conditional probability, once to the left-hand side and twice to the right-hand side, to get your result.&lt;/p&gt;&#xA;" OwnerUserId="1195" LastActivityDate="2010-11-04T02:34:26.073" />
  <row Id="8856" PostTypeId="1" AcceptedAnswerId="8861" CreationDate="2010-11-04T02:56:39.660" Score="10" ViewCount="323" Body="&lt;p&gt;My class had a good time redeveloping the unit circle under the &lt;a href=&quot;http://en.wikipedia.org/wiki/Taxicab_geometry&quot; rel=&quot;nofollow&quot;&gt;taxicab metric&lt;/a&gt;. Now some of them want to do it again with another similar metric. I want to give this to some of my &quot;honors&quot; high-school* calculus I students to work on independently. As such, it can't be too difficult or require too many advanced concepts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What is another metric that I could introduce them to where one can describe a &quot;unit circle&quot; and try to make sense of the sine, cosine and tangent functions?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;*I have a group of high school students enrolled in a college course. The course, as planned, is &lt;em&gt;far&lt;/em&gt; too easy for them. (At the same time, it is quite hard for the college students enrolled. Go figure.) After the high school students do what has been assigned by the online learning system I have them work on other stuff-- which I allow to be pretty self-guided. &lt;/p&gt;&#xA;" OwnerUserId="2563" LastEditorUserId="2563" LastEditorDisplayName="" LastEditDate="2010-11-13T13:15:21.310" LastActivityDate="2010-11-13T18:00:17.880" Title="Developing the unit circle in geometries with different metrics: beyond taxi cabs." Tags="&lt;geometry&gt;&lt;trigonometry&gt;&lt;big-list&gt;&lt;metric-spaces&gt;" AnswerCount="4" CommentCount="3" FavoriteCount="6" />
  <row Id="8857" PostTypeId="2" ParentId="8856" CreationDate="2010-11-04T03:06:50.743" Score="9" ViewCount="" Body="&lt;p&gt;Any norm defines a distance, by $d((a,b),(c,d)) = ||(a-c,b-d)||$. Some common norms:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;There are all the $p$-norms: $||(x,y)||_p = \sqrt[p]{|x|^p + |y|^p}$ (the usual norm occurs with $p=2$); you can do them for any $p$, $0\lt p\lt \infty$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The sup norm: $||(x,y)||_{\infty} = \max\{|x|,|y|\}$. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;You can take a positive linear combination of norms to create a new one. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Given any linear transformation $A\colon\mathbb{R}^2\to\mathbb{R}^2$, and any norm $||\cdot||$, you can define the norm that maps $(x,y)$ to $||A(x,y)||$. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;There is also the discrete metric, though that will create a rather nasty &quot;unit circle&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is actually nice to treat the sup norm, the $p$-norms, and the taxi-cab norm together. If you draw the &quot;unit circles&quot; for all of them, the limit of the $p$ norms as $p\to 0^+$ is the taxicab norm, while the limit as $p\to\infty$ is the sup norm. &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-04T03:47:17.560" LastActivityDate="2010-11-04T03:47:17.560" CommentCount="6" />
  <row Id="8859" PostTypeId="1" AcceptedAnswerId="8873" CreationDate="2010-11-04T03:36:25.853" Score="1" ViewCount="123" Body="&lt;p&gt;This is somewhat of a follow-up question to &lt;a href=&quot;http://math.stackexchange.com/q/8426/2961&quot;&gt;‘Elliptic integrals with parameter outside 0&amp;lt;m&amp;lt;1’&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have an equation that I'm attempting to simplify that has terms that look something like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$&#xA;\Re\bigl( -i (c-1) \mathrm K(c) -2i \mathrm E(c) \bigr)&#xA;$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $c&gt;1$. I.e., they're being evaluated outside their standard parameter range of $0\le m\le 1$. I know the equation has been derived with Mathematica, and I'm suspicious that the $\Re(i\mathrm K(c))$ terms, etc., are more complex than they need to be.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I've taken a look at &lt;a href=&quot;http://dlmf.nist.gov/19.7#i&quot; rel=&quot;nofollow&quot;&gt;Legendre's Relation&lt;/a&gt; which seems to be related to splitting up these integrals into real and imaginary parts, and it doesn't seem to click—I suspect that the expression above is about as simple as I can get it (i.e., even if I manage to replace the integrals by alternate versions that have only real components, the resulting expression will be &lt;em&gt;more&lt;/em&gt; complex than that above). Is there anything I'm missing?&lt;/p&gt;&#xA;" OwnerUserId="2961" LastEditorUserId="2961" LastEditorDisplayName="" LastEditDate="2010-11-04T04:57:13.467" LastActivityDate="2010-11-04T12:57:46.513" Title="Separate elliptic integrals into real and imaginary parts" Tags="&lt;special-functions&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="8860" PostTypeId="1" AcceptedAnswerId="8862" CreationDate="2010-11-04T03:49:40.243" Score="2" ViewCount="127" Body="&lt;p&gt;My number theory library of choice doesn't implement the logarithmic integral for complex values.  I thought that I might take a crack at coding it, but I thought I'd ask here first for algorithmic advice and/or references.  I'm sure there are better methods than naively calculating the integral.&lt;/p&gt;&#xA;" OwnerUserId="1778" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-04T03:52:10.553" LastActivityDate="2010-11-04T04:23:03.603" Title="Efficiently calculating the logarithmic integral with complex argument" Tags="&lt;number-theory&gt;&lt;algorithms&gt;&lt;special-functions&gt;&lt;analytic-number-theory&gt;" AnswerCount="1" />
  <row Id="8861" PostTypeId="2" ParentId="8856" CreationDate="2010-11-04T03:57:06.740" Score="8" ViewCount="" Body="&lt;p&gt;Two funny distances on $\mathbb{R}^2$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The elevator distance.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;d((x_1,y_1),(x_2,y_2)) = &#xA;\begin{cases}&#xA;\vert y_1 - y_2 \vert                  &amp;amp; \text{if}\ x_1 = x_2 \\&#xA;\vert y_1 \vert + \vert x_1 - x_2 \vert + \vert y_2\vert &amp;amp; \text{otherwise} &#xA;\end{cases}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The mail office distance.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;d(p,q) = &#xA;\begin{cases}&#xA;0            &amp;amp; \text{if}\ p = q \\&#xA;\|p\| + \|q\|    &amp;amp; \text{otherwise}&#xA;\end{cases}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A good question to ask: why are they called &quot;elevator&quot; and &quot;mail office&quot; distances, respectively?  :-)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT. As Arturo Magidin points out, with these distances balls centered at the origin are not particularly interesting: you have to try with balls NOT centered at the origin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;MORE EDIT. Don: have you seen Arturo Magidin's comment about making balls centered at the origin &quot;interesting&quot;?&lt;/p&gt;&#xA;" OwnerUserId="664" LastEditorUserId="664" LastEditorDisplayName="" LastEditDate="2010-11-04T06:35:56.103" LastActivityDate="2010-11-04T06:35:56.103" CommentCount="7" />
  <row Id="8862" PostTypeId="2" ParentId="8860" CreationDate="2010-11-04T04:23:03.603" Score="5" ViewCount="" Body="&lt;p&gt;For this answer, I'm assuming the definition&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathrm{li}(z):=\mathrm{PV}\int_0^z\frac{\mathrm{d}u}{\ln\;u}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where we assume the Cauchy principal value (the more common definition in number theory that has a lower limit of 2 differs merely by a constant).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, the first thing you have to note is the identity&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathrm{li}(z)=\mathrm{Ei}(\ln\;z)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\mathrm{Ei}(z)$ is an &lt;a href=&quot;http://dlmf.nist.gov/6.2&quot; rel=&quot;nofollow&quot;&gt;exponential integral&lt;/a&gt;. (Again, I repeat my advice to people who encounter strange functions: you would really do well to check the DLMF first for identities and references.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, $\mathrm{Ei}(z)$ is a slightly more tractable beastie to numerically evaluate, since the singularity at $z=0$ can be confined to a logarithmic part; to wit:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\mathrm{Ei}(z)=\gamma+\ln\;z+\int_0^z \frac{\exp(u)-1}{u}\mathrm{d}u$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where the last portion is an entire function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, depending on which of the left or right half-planes should the exponential integral be evaluated, your strategy will differ (it is a common fact that most special function routines are &lt;em&gt;polyalgorithms&lt;/em&gt;, since their behavior can markedly differ in different regions of the complex plane). I will be vague for the rest of this answer since you did not clarify your region of interest. Suffice it to say that one usually uses a &lt;a href=&quot;http://dlmf.nist.gov/6.9&quot; rel=&quot;nofollow&quot;&gt;continued fraction&lt;/a&gt; for arguments in the left half-plane, a power series for small to medium-sized arguments, and &lt;a href=&quot;http://dlmf.nist.gov/6.12&quot; rel=&quot;nofollow&quot;&gt;asymptotic expansions&lt;/a&gt; for large arguments.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If implementing it yourself is starting to sound daunting (because it is), I will have to point out this &lt;a href=&quot;http://dx.doi.org/10.1145/78928.78934&quot; rel=&quot;nofollow&quot;&gt;paper&lt;/a&gt; and the corresponding &lt;a href=&quot;http://www.netlib.org/toms/683&quot; rel=&quot;nofollow&quot;&gt;FORTRAN subroutine&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope this helps a bit.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-11-04T04:23:03.603" CommentCount="7" />
  <row Id="8863" PostTypeId="2" ParentId="8808" CreationDate="2010-11-04T04:24:07.533" Score="10" ViewCount="" Body="&lt;p&gt;In my opinion (because it really is an opinion), the main reason to care about euclidean spaces is that they come equipped with a lot of particularly nice structures.  In particular, they are finite-dimensional Hilbert spaces.  This means that they are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Finite-dimensional vector spaces (so we can talk about addition and scaling)&lt;/li&gt;&#xA;&lt;li&gt;Complete metric spaces (so we can talk about distances and limits, and limits behave as we would like them to)&lt;/li&gt;&#xA;&lt;li&gt;Inner product spaces (so we can talk about the notion of &quot;angle,&quot; and thereby do all sorts of geometric things.)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;And really, there's just so much geometric intuition that comes along with these ideas -- not to mention an entire calculus apparatus.  After all, there are plenty of topological spaces where the above notions are not defined or otherwise fail to be true.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason (at least to me) to study locally euclidean spaces is that we want to study spaces that are more general than euclidean spaces, yet still retain many of their nice features.  In particular, we want a place where calculus makes sense.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Areas of math which study more abstract spaces include topology and algebraic geometry.  Admittedly, I'm not very well-versed in either just yet, but I'm sure practical uses (and physical models) have been found in both.&lt;/p&gt;&#xA;" OwnerUserId="640" LastActivityDate="2010-11-04T04:24:07.533" CommentCount="3" />
  <row Id="8864" PostTypeId="2" ParentId="8838" CreationDate="2010-11-04T04:46:07.663" Score="2" ViewCount="" Body="&lt;p&gt;Speed is great for brute force calculations and computer simulations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For proofs that rely on abstract ideas and arguments, speed is inconsequential.&lt;/p&gt;&#xA;" OwnerUserId="3017" LastActivityDate="2010-11-04T04:46:07.663" CommentCount="1" />
  <row Id="8865" PostTypeId="1" AcceptedAnswerId="8868" CreationDate="2010-11-04T04:53:44.743" Score="6" ViewCount="351" Body="&lt;p&gt;I'm a bit curious as to how &quot;real&quot; mathematicians would solve this problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Find the least prime number greater than 2000.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, I can always go brute force:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;IF N IS PRIME, OUTPUT N&#xA;ELSE INCREASE N BY 1&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;But that's no fun. (Especially when the number is really high.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there any algorithms/tricks/etc that can help me solve this quickly, and efficiently, especially if I were given large values of N?&lt;/p&gt;&#xA;" OwnerUserId="2736" LastEditorUserId="2736" LastEditorDisplayName="" LastEditDate="2010-11-04T22:50:52.830" LastActivityDate="2010-11-04T22:50:52.830" Title="The least prime greater than 2000" Tags="&lt;number-theory&gt;&lt;homework&gt;&lt;algorithms&gt;&lt;prime-numbers&gt;" AnswerCount="4" CommentCount="7" FavoriteCount="3" />
  <row Id="8867" PostTypeId="2" ParentId="8865" CreationDate="2010-11-04T05:11:31.497" Score="7" ViewCount="" Body="&lt;p&gt;If you want to find several primes at once starting from some point, you can use a &lt;a href=&quot;http://en.wikipedia.org/wiki/Eratosthenes_sieve&quot; rel=&quot;nofollow&quot;&gt;sieve&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-04T05:11:31.497" CommentCount="1" />
  <row Id="8868" PostTypeId="2" ParentId="8865" CreationDate="2010-11-04T05:57:27.523" Score="13" ViewCount="" Body="&lt;p&gt;I expect there are more powerful tools available; but for a simple approach I would apply the Prime Number Theorem (PNT) and a sieve (as Yuval has already suggested).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One useful consequence of the PNT is that around a number &lt;em&gt;N&lt;/em&gt;, approximately one out of every &lt;em&gt;log(N)&lt;/em&gt; numbers is prime.  (By '&lt;em&gt;log&lt;/em&gt;,' number theorists always mean the natural log '&lt;em&gt;ln&lt;/em&gt;'.)  So around 2000, about 1 out of every 7.6 numbers is prime.  Let's just look among the numbers 2001 to 2060 for our next prime-- I'm leaving extra space in case a big prime gap happens to show up around 2000.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we use the sieve-- we can throw out all even numbers in our list, since they are certainly not primes.  Next, we can throw out everything divisible by 3, then everything divisible by 5, by 7, by 11,...  When should we stop sieving?  We can stop when we get to $\sqrt{2060} \approx 45$, since if any number in our list can be factored as &lt;em&gt;M = ab&lt;/em&gt;, then one of the factors must be less than or equal to the square root of the biggest number in our list.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, a computer would be happy to do this process for you.  But, I just wrote down the list and went through my basic sieve (primes up to 43); and I got the prime numbers 2003, 2011, 2017, 2027, 2029, 2039, 2053.&lt;/p&gt;&#xA;" OwnerUserId="1461" LastActivityDate="2010-11-04T05:57:27.523" CommentCount="4" />
  <row Id="8870" PostTypeId="2" ParentId="8865" CreationDate="2010-11-04T06:45:22.180" Score="1" ViewCount="" Body="&lt;p&gt;An integer $p$ is prime if and only if the following equation holds:&#xA;\begin{eqnarray}&#xA;\sum_{n \geq 1} \left \lfloor \frac{p}{n} \right \rfloor - \left \lfloor \frac{p-1}{n} \right \rfloor  = 2&#xA;\end{eqnarray}&#xA;NB: The infinite number of terms of sum is there to accommodate an arbitrarily large integer $p$, but the sum eventually produces a string of $0$s. (See Crandall and Pomerance, &lt;em&gt;Prime Numbers: A Computational Perspective&lt;/em&gt;, New York: Springer, ISBN 0-387-04777-9)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After observing that $2001$ and $2002$ both violate the test early on but $2003$ does not, then you can use any number of primality tests (or choose to prove the sum equals 2 to establish primality directly). Otherwise, you may try something like Wilson's theorem, which states: An integer $p$ is prime if and only if $(p-1)! \equiv -1 \mod p$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mathematica gives the correct result in less than 1 second: $2003$ is the first prime greater than $2000$.&lt;/p&gt;&#xA;" OwnerUserId="2720" LastEditorUserId="2720" LastEditorDisplayName="" LastEditDate="2010-11-04T07:49:17.273" LastActivityDate="2010-11-04T07:49:17.273" CommentCount="5" />
  <row Id="8871" PostTypeId="2" ParentId="8838" CreationDate="2010-11-04T06:50:27.667" Score="1" ViewCount="" Body="&lt;p&gt;What do YOU think?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For mathematics, intelligence is needed.  And intelligence is instantaneous. And something that is instantaneous has no notion of speed; speed requires time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Speed as a measure cannot be applied.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For arithmetic, speed can be applied.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But there is an almost universal confusion between arithmetic and mathematics. Mathematics is the process of digging into the unknown.  Arithmetic is the process of playing with the artefacts that the process of mathematics has uncovered.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is termed mathematics in school is usually nothing more than arithmetic.  It depends not upon the material but upon the teaching.  If the student is actively enquiring,  with the teacher guiding this enquiry only when needed,  then mathematics will ensue.   If the teacher is intent upon stuffing the syllabus into the head of the poor student, the best that can be hoped for is arithmetic.&lt;/p&gt;&#xA;" OwnerUserId="3096" LastActivityDate="2010-11-04T06:50:27.667" />
  <row Id="8873" PostTypeId="2" ParentId="8859" CreationDate="2010-11-04T07:00:36.917" Score="3" ViewCount="" Body="&lt;p&gt;Well, for the simplified case that you now presented in your current version of the question (again, as I said, the complete elliptic integral of the third kind requires a different sort of analysis), the identities in &lt;a href=&quot;http://dlmf.nist.gov/19.7.E3&quot; rel=&quot;nofollow&quot;&gt;formula 19.7.3 in the DLMF&lt;/a&gt; can be rewritten in parameter form as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$K(m)=\frac1{\sqrt{m}}\left(K\left(\frac1{m}\right)-i K\left(1-\frac1{m}\right)\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$E(m)=\sqrt{m}\left(E\left(\frac1{m}\right)-\left(1-\frac1{m}\right)K\left(\frac1{m}\right)+i\left(E\left(1-\frac1{m}\right)-\frac1{m} K\left(1-\frac1{m}\right)\right)\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we apply both to your expression above, it can be simplified to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$2\sqrt{c}E\left(1-\frac1{c}\right)-\frac{1+c}{\sqrt{c}}K\left(1-\frac1{c}\right)$$&lt;/p&gt;&#xA;" OwnerUserId="498" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-04T12:57:46.513" LastActivityDate="2010-11-04T12:57:46.513" CommentCount="3" />
  <row Id="8875" PostTypeId="1" AcceptedAnswerId="12252" CreationDate="2010-11-04T07:14:18.410" Score="1" ViewCount="188" Body="&lt;p&gt;I came across an interesting problem but unable to see how to approach it. How do I use the dominated convergence theorem (LDCT), to show that first derivative of characteristic function of probability distribution at t = 0, $\phi^′(0)=iE[X]$? Any ideas?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;References:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Characteristic_function_(probability_theory&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Characteristic_function_(probability_theory&lt;/a&gt;)&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Dominated_convergence_theorem&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Dominated_convergence_theorem&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="957" LastEditorUserId="957" LastEditorDisplayName="" LastEditDate="2010-11-04T07:42:05.157" LastActivityDate="2010-11-28T22:46:44.740" Title="derivative of characteristic function" Tags="&lt;probability-theory&gt;&lt;measure-theory&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="8876" PostTypeId="1" AcceptedAnswerId="8877" CreationDate="2010-11-04T07:22:58.900" Score="2" ViewCount="170" Body="&lt;p&gt;What does this expression mean - $\lim_{n\rightarrow\infty} E|X_n-X|=0$? $X_n$ is a sequence of random variables and $X$ is a random variable. What does this expression imply? Can I say that the sequence $X_n$ converges to $X$ in probability and almost sure convergence?&lt;/p&gt;&#xA;" OwnerUserId="957" LastActivityDate="2010-11-04T08:33:38.020" Title="convergence of sequence of random variables" Tags="&lt;probability&gt;&lt;probability-theory&gt;&lt;measure-theory&gt;" AnswerCount="2" CommentCount="0" />
  <row Id="8877" PostTypeId="2" ParentId="8876" CreationDate="2010-11-04T07:33:05.227" Score="2" ViewCount="" Body="&lt;p&gt;This is called &lt;em&gt;convergence in the mean&lt;/em&gt; or convergence in the $L^{1}$-norm. In general, if &#xA;\begin{eqnarray}&#xA;\lim_{n \to \infty} \mathbb{E}(| X_{n} - X|^{p}) = 0,&#xA;\end{eqnarray}&#xA;then $X_{n}$ is said to converge to $X$ in the $L^{p}$-norm (provided that $\mathbb{E}(|X_{n}|^{p})$ is finite for all $n \geq 1$). Analytically, there are nice implications of such convergence. For example, convergence in an $L^{p}$-norm implies convergence in an $L^{q}$-norm if $p \geq q$. (See &lt;a href=&quot;http://en.wikipedia.org/wiki/Convergence_of_random_variables&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Convergence_of_random_variables&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Markov's inequality states&#xA;\begin{eqnarray}&#xA;\mathbb{P}(|X_{n} - X| &gt; \epsilon) \leq \epsilon^{-p} \, \mathbb{E}(|X_{n} - X|^{p}).&#xA;\end{eqnarray}&#xA;Thus, $L^{p}$-norm convergence implies convergence in probability.&lt;/p&gt;&#xA;" OwnerUserId="2720" LastEditorUserId="2720" LastEditorDisplayName="" LastEditDate="2010-11-04T07:46:04.093" LastActivityDate="2010-11-04T07:46:04.093" CommentCount="1" />
  <row Id="8878" PostTypeId="2" ParentId="8876" CreationDate="2010-11-04T07:58:49.120" Score="0" ViewCount="" Body="&lt;p&gt;When $\lim _{n \to \infty } E|X_n - X| = 0$, we say that $X_n$ converges in mean to $X$. It is very well known that this implies that $X_n$ converges in probability to $X$, but not that $X_n$ converges almost surely to $X$. Consider a sequence $(X_n)$ of independent random variables such that &#xA;$P(X_n  = 0) = 1 - 1/n$ and $P(X_n  = 1) = 1/n$. Then $E|X_n|=1/n$, and hence $X_n$ converges in mean to $0$. However, since $\sum\nolimits_{n = 1}^\infty  {P(X_n  = 1)}  = \infty $ and the $X_n$ are independent, almost surely the sequence $(X_n)$ contains infinitely many $1$'s; in particular $X_n$ does not converge to $0$ in the almost sure sense.&lt;/p&gt;&#xA;" OwnerUserId="2810" LastEditorUserId="2810" LastEditorDisplayName="" LastEditDate="2010-11-04T08:33:38.020" LastActivityDate="2010-11-04T08:33:38.020" />
  <row Id="8879" PostTypeId="2" ParentId="8875" CreationDate="2010-11-04T08:04:14.123" Score="1" ViewCount="" Body="&lt;p&gt;You need to show that if $(a_n)$ is a sequence of nonzero numbers&#xA;which tends to zero, that $(\phi(a_n)-\phi(0))/a_n\to iE[X]$. Now&#xA;$$\frac{\phi(a)-\phi(0)}{a}=E\left(\frac{e^{iaX}-1}{a}\right).$$&#xA;You need some hypothesis on $X$ for the result to work, for&#xA;instance that $|X|$ has finite expectation. Certainly $(e^{iaX}-1)/a\to iX$&#xA;so to apply dominated convergence we need a function $f(X)$ with finite&#xA;expectation and with $|(e^{iaX}-1)/a|\le f(X)$ at least for $a$ in a&#xA;deleted neighbourhood of $0$. Does $f(X)=|X|$ work?&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-11-04T08:04:14.123" />
  <row Id="8880" PostTypeId="2" ParentId="7142" CreationDate="2010-11-04T08:07:15.277" Score="0" ViewCount="" Body="&lt;p&gt;I think this can easily be formulated as a &lt;a href=&quot;http://en.wikipedia.org/wiki/QCQP&quot; rel=&quot;nofollow&quot;&gt;QCQP&lt;/a&gt;&#xA;that with a Positive Definite matrix. Therefore the problem is convex and can be solved using interior point methods.&lt;/p&gt;&#xA;" LastActivityDate="2010-11-04T08:07:15.277" CommentCount="1" />
  <row Id="8881" PostTypeId="2" ParentId="7391" CreationDate="2010-11-04T08:24:33.093" Score="2" ViewCount="" Body="&lt;p&gt;There is no need for trigonometric identities, complex exponentials or the like. Observe that&#xA;\begin{eqnarray}&#xA;\int_{0}^{\pi} \sin(2 x) \sin (x) dx = \int_{0}^{\pi/2}  \sin(2 x) \sin (x) dx  + \int_{\pi/2}^{\pi}  \sin(2 x) \sin (x) dx &#xA;\end{eqnarray}&#xA;By a change of variables ($x \to \pi - x$) and the oddness of the integrand on the interval $[0,\pi]$, you find that &#xA;\begin{eqnarray}&#xA; \int_{\pi/2}^{\pi}  \sin(2 x) \sin (x) dx = - \int_{0}^{\pi/2}  \sin(2 x) \sin (x) dx ,&#xA;\end{eqnarray}&#xA;which implies that the original integral vanishes identically.&lt;/p&gt;&#xA;" OwnerUserId="2720" LastActivityDate="2010-11-04T08:24:33.093" />
  <row Id="8882" PostTypeId="2" ParentId="8814" CreationDate="2010-11-04T08:36:14.877" Score="1" ViewCount="" Body="&lt;p&gt;$$ \sin \theta \cdot \sin \bigl(60 - \theta \bigr) \cdot \sin \bigl(60 + \theta \bigr) = \frac{1}{4} \sin 3\theta$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \cos \theta \cdot \cos \bigl(60  - \theta \bigr) \cdot \cos \bigl(60 + \theta \bigr) = \frac{1}{4} \cos 3\theta$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \tan \theta \cdot \tan \bigl(60  - \theta \bigr) \cdot \tan \bigl(60 + \theta \bigr) = \tan 3\theta $$&lt;/p&gt;&#xA;" OwnerUserId="2109" LastEditorUserId="1747" LastEditorDisplayName="" LastEditDate="2011-03-09T08:04:23.677" LastActivityDate="2011-03-09T08:04:23.677" CommunityOwnedDate="2010-11-04T08:36:14.877" CommentCount="5" />
  <row Id="8883" PostTypeId="2" ParentId="8814" CreationDate="2010-11-04T08:37:19.310" Score="14" ViewCount="" Body="&lt;p&gt;$$\frac{1}{\sin(2\pi/7)} + \frac{1}{\sin(3\pi/7)} = \frac{1}{\sin(\pi/7)}$$&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2010-11-04T08:37:19.310" CommunityOwnedDate="2010-11-04T08:37:19.310" />
  <row Id="8884" PostTypeId="1" AcceptedAnswerId="8891" CreationDate="2010-11-04T08:53:05.593" Score="12" ViewCount="497" Body="&lt;p&gt;In this &lt;a href=&quot;http://math.stackexchange.com/questions/7583/an-identity-concerning-the-riemann-zeta-function&quot;&gt;post&lt;/a&gt;, David Speyer, actually, gave an expression for $\displaystyle  \frac{t}{e^{t}-1}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is can we sum the given series, using that expression, if not how does one sum this series. $$\sum\limits_{n=1}^{\infty} \frac{n}{e^{2\pi n}-1}=\frac{1}{e^{2\pi}-1} + \frac{2}{e^{4\pi}-1} + \frac{3}{e^{6\pi}-1} + \cdots \text{ad inf}$$&lt;/p&gt;&#xA;" OwnerUserId="9413" LastEditorUserId="9413" LastEditorDisplayName="" LastActivityDate="2010-11-12T04:07:07.623" Title="Summing $\frac{1}{e^{2\pi}-1} + \frac{2}{e^{4\pi}-1} + \frac{3}{e^{6\pi}-1} + \cdots \text{ad inf}$" Tags="&lt;analysis&gt;&lt;sequences-and-series&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="8885" PostTypeId="2" ParentId="8856" CreationDate="2010-11-04T09:04:47.440" Score="4" ViewCount="" Body="&lt;p&gt;One closely related to the two in Agusti's answer is what I call the &lt;strong&gt;bus metric&lt;/strong&gt; (though I change the name to reflect the name of the local bus company - which seems to change each time I teach about metric spaces!).  This is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;d(p,q) = \begin{cases}&#xA;\|p - q\|_2 &amp;amp; \text{if} p, q \text{ are collinear} \\&#xA;\|p\|_2 + \|q\|_2 &amp;amp; \text{otherwise}&#xA;\end{cases}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(In Trondheim (where I am) then the majority of bus routes are radial, so this does link to the students' intuition.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The students could do a nice animation of what happens to a ball of unit length centred at a point $(x,0)$ as $x$ ranges from, say, $-2$ to $2$.&lt;/p&gt;&#xA;" OwnerUserId="2907" LastActivityDate="2010-11-04T09:04:47.440" CommentCount="2" />
  <row Id="8886" PostTypeId="2" ParentId="8814" CreationDate="2010-11-04T09:07:00.950" Score="13" ViewCount="" Body="&lt;p&gt;Machin's Formula:&#xA;\begin{eqnarray}&#xA;\frac{\pi}{4} = 4 \arctan \frac{1}{5} - \arctan \frac{1}{239}.&#xA;\end{eqnarray}&lt;/p&gt;&#xA;" OwnerUserId="2720" LastEditorUserId="2720" LastEditorDisplayName="" LastEditDate="2010-11-05T08:54:51.073" LastActivityDate="2010-11-05T08:54:51.073" CommunityOwnedDate="2010-11-04T09:07:00.950" />
  <row Id="8889" PostTypeId="1" CreationDate="2010-11-04T10:04:05.877" Score="2" ViewCount="105" Body="&lt;p&gt;Let $\mathbf{x} =$ {$x_{i}$} be a set of $n$ positive reals. In every good book on inequalities, one finds the classical result &#xA;\begin{eqnarray}&#xA;AM(\mathbf{x}) \geq GM(\mathbf{x}) \geq HM(\mathbf{x}),&#xA;\end{eqnarray}&#xA;where $AM(\mathbf{x}) = \frac{1}{n} \sum_{i = 1}^{n} x_{i}$ is the arithmetic mean, $GM(\mathbf{x}) = \sqrt[n]{x_{1} \cdots x_{n}}$ is the geometric mean and $HM(\mathbf{x}) = n (\sum_{i = 1}^{n} \frac{1}{x_{i}})^{-1}$ is the harmonic mean of $\mathbf{x}$, respectively. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question&lt;/strong&gt;: I'm curious about sharp bounds of the form: &#xA;\begin{eqnarray}&#xA;HM(\mathbf{x}) \geq f(\mathbf{x}) AM(\mathbf{x}) + g(\mathbf{x}),&#xA;\end{eqnarray}&#xA;where $f$ and $g$ are some functions &lt;em&gt;which do not imply the classical result above or render the inequality trivial&lt;/em&gt;. Do such results exist in the literature (or mathematical folklore)? (References are welcome.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="2720" LastEditorUserId="2720" LastEditorDisplayName="" LastEditDate="2010-11-04T17:26:25.460" LastActivityDate="2010-11-04T17:26:25.460" Title="Sharp (Reverse) Harmonic-Arithmetic Mean Bounds" Tags="&lt;real-analysis&gt;&lt;algebra&gt;&lt;reference-request&gt;&lt;inequality&gt;" CommentCount="6" FavoriteCount="2" />
  <row Id="8890" PostTypeId="1" CreationDate="2010-11-04T10:17:57.210" Score="-5" ViewCount="89" Body="&lt;p&gt;for $z, w \in C$, prove that $2|zw| &amp;lt;= |z|^2+|w|^2$, and deduce that $|z+w|^2 &amp;lt;= 2(|z|^2+|w|^2)$. where $|v|$ is modulus of vector $v \in C$&lt;/p&gt;&#xA;" OwnerUserId="2723" LastEditorUserId="139" LastEditorDisplayName="" LastEditDate="2010-11-22T01:14:43.220" LastActivityDate="2010-11-22T01:14:43.220" Title="for $z, w \in C$, prove that $2|zw| &lt;= |z|^2+|w|^2$" Tags="&lt;complex-numbers&gt;&lt;inequality&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="8891" PostTypeId="2" ParentId="8884" CreationDate="2010-11-04T10:54:51.530" Score="24" ViewCount="" Body="&lt;p&gt;What you require here are the &lt;a href=&quot;http://en.wikipedia.org/wiki/Eisenstein_series&quot; rel=&quot;nofollow&quot;&gt;Eisenstein series&lt;/a&gt;. In particular the evaluation of&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$E_2(\tau) = 1 – 24\sum_{n=1}^\infty \frac{ne^{2\pi i n \tau} }{1 - e^{2\pi i n \tau}},$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;at $\tau = i. $ Rearrange to get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{n=1}^\infty \frac{ne^{2\pi i n \tau} }{1 - e^{2\pi i n \tau} } = \frac{1}{24}(1 – E_2(i) ).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See &lt;a href=&quot;http://en.wikipedia.org/wiki/Lambert_series&quot; rel=&quot;nofollow&quot;&gt;Lambert series&lt;/a&gt; for additional information.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: The function&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$G_ 2(\tau) = \zeta(2) \left(&#xD;&#xA;1 – 24\sum_{n=1}^\infty \frac{ne^{2\pi i n \tau} }{1 - e^{2\pi i n \tau}} \right)&#xD;&#xA;=\zeta(2)E_2(\tau)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;satisfies the quasimodular transformation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$G_ 2\left( \frac{a\tau+b}{c\tau+d} \right) =&#xD;&#xA;(c\tau+d)^2G_ 2(\tau) - \pi i c (c\tau + d).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And so with $a=d=0,$ $c=1$ and $b=-1$ we find $G_ 2(i) = \pi/2.$ Therefore&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$E_2(i) = \frac{ G_ 2( i)}{ \zeta(2)} = \frac{\pi}{2}\frac{6}{\pi^2} = \frac{3}{\pi}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence we obtain&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{n=1}^\infty \frac{n}{e^{2\pi n} – 1} = \frac{1}{24} - \frac{1}{8\pi},$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;as given in the comment to the question by Slowsolver.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a very nice generalisation of the sum in the question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For odd $ m &gt; 1 $ we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{n=1}^\infty \frac{n^{2m-1} }{ e^{2\pi n} -1 } = \frac{B_{2m}}{4m},$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $B_k$ are the Bernoulli numbers defined by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{z}{e^z - 1} = \sum_{k=0}^\infty \frac{B_k}{k!} z^k \quad \textrm{ for }&#xD;&#xA;|z| &amp;lt; 2 \pi.$$&lt;/p&gt;&#xA;" OwnerUserId="1301" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-11-12T04:07:07.623" LastActivityDate="2010-11-12T04:07:07.623" CommentCount="8" />
  <row Id="8892" PostTypeId="1" CreationDate="2010-11-04T11:16:53.877" Score="1" ViewCount="129" Body="&lt;p&gt;I need to find the number of undirected trees on $n$ vertices such that the edges (and not the vertices) are labeled and exactly one label appears twice (i.e. there are $n-2$ possible labels and they all appear).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My reasoning goes like this: From Cayley's formula we know there are $n^{n-2}$ trees with labeled vertices. Erase the label $n$ from its vertex and for every other vertex, move the label from the vertex itself to the edge in the direction of the $n$ vertex. This way we get a tree with $n-1$ labels for the edges. However, this argument is flawed, since for example 0--1--2 and 0--2--1 give the same labeled tree.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now assume that I managed to fix the problem and find the number of edge-labeled trees with $n-1$ edges, and let's denote it by $k$; my next idea is choosing one of the $n-2$ first labels and relabeling $n-1$ to what I choose. This causes double-counting so I think I need to divide by 2 here, but nothing more. Hence the final number is $(n-2)k/2$. Am I correct in this reasoning - and how can I find k? Quite obviously it's not $n^{n-2}$ since otherwise $(n-2)k/2$ might not even be an integer (if $n$ is odd).&lt;/p&gt;&#xA;" OwnerUserId="1818" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-11-04T12:03:08.360" LastActivityDate="2011-06-02T15:09:12.127" Title="Number of undirected trees with labeled edges, one repeating" Tags="&lt;graph-theory&gt;&lt;trees&gt;&lt;combinatorics&gt;" AnswerCount="1" />
  <row Id="8893" PostTypeId="1" AcceptedAnswerId="8896" CreationDate="2010-11-04T11:25:11.120" Score="6" ViewCount="910" Body="&lt;p&gt;I am having trouble expressing the titular question as iterated integrals over a given region. I have tried narrowing down the problem, and have concluded that the simplest way to approach this is to integrate over the XZ plane in the positve octant and multiply by 8, but I am having trouble identifying the bounding functions.&lt;/p&gt;&#xA;" OwnerUserId="3102" LastActivityDate="2010-11-04T14:47:19.917" Title="Volume bounded by cylinders x^2 + y^2 = r^2 and z^2 + y^2 = r^2" Tags="&lt;calculus&gt;&lt;integral&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="8894" PostTypeId="2" ParentId="8892" CreationDate="2010-11-04T11:32:36.813" Score="0" ViewCount="" Body="&lt;p&gt;The correspondence between edge- and vertex-labellings may work better starting from the degree 1 vertices instead of edge $n$ which would be at a random location inside the tree.   For the degree 1 vertices we have a unique choice of label to assign.  Remove these vertices and edges and repeat the process until the tree is a single vertex (0 edges) or a single edge.   These cases correspond to enumeration of trees on labelled vertices, with the &quot;center&quot; of the tree being a vertex or an edge, respectively.&lt;/p&gt;&#xA;" OwnerUserId="467" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-11-04T12:02:02.463" LastActivityDate="2010-11-04T12:02:02.463" />
  <row Id="8895" PostTypeId="1" AcceptedAnswerId="8897" CreationDate="2010-11-04T11:40:52.730" Score="1" ViewCount="67" Body="&lt;p&gt;This one comes from my module without proper explanation.Please suggest how to approach this problem ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS:How do you write degree (in symbol) in math-mode ?&lt;/p&gt;&#xA;" OwnerUserId="3744" LastEditorUserId="3744" LastEditorDisplayName="" LastEditDate="2010-11-04T11:52:56.930" LastActivityDate="2010-11-04T11:54:47.810" Title="In right triangle PQR, ∠Q = $90 ^\circ$ and PR = 2PQ, then ∠R is equal to ?" Tags="&lt;trigonometry&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="8896" PostTypeId="2" ParentId="8893" CreationDate="2010-11-04T11:46:53.200" Score="10" ViewCount="" Body="&lt;p&gt;The solid lies above the region $D$ in the $xy$-plane bounded by the circle $x^{2} + y^{2} = r^{2}$, so the volume is given by the integral $$\int\int\limits_{D} f(x,y) \ dA = \int\limits_{-r}^{r}\int\limits_{-\sqrt{r^{2}-y^{2}}}^{\sqrt{r^{2}-y^{2}}} f(x,y) \ dx dy$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore the required volume of the solid is: $$\int\limits_{-r}^{r}\int\limits_{-\sqrt{r^{2}-y^{2}}}^{\sqrt{r^{2}-y^{2}}} 2\sqrt{r^{2}-y^{2}} \ dx dy = \frac{16}{3}r^{3}$$&lt;/p&gt;&#xA;" LastActivityDate="2010-11-04T11:46:53.200" CommentCount="1" />
  <row Id="8897" PostTypeId="2" ParentId="8895" CreationDate="2010-11-04T11:54:47.810" Score="2" ViewCount="" Body="&lt;p&gt;As Djaian pointed out use the fact that $$\sin\angle{R} = \frac{PQ}{PR}=\frac{PQ}{2PQ}=\frac{1}{2}$$ So $\angle{R} = ?$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also i hope you are aware of the fact that in a right angled triangle $$\sin\theta =\frac{\text{Opposite Side}}{\text{Hypotenuse}}$$&lt;/p&gt;&#xA;" LastActivityDate="2010-11-04T11:54:47.810" />
  <row Id="8898" PostTypeId="1" CreationDate="2010-11-04T13:14:05.177" Score="0" ViewCount="180" Body="&lt;p&gt;Let A and B be nxn matrices over C. How to prove $(AB)^{\ast}=B^{\ast}A^{\ast}$? $A^{\ast}$ is complex conjugate transpose( if matrix $A$ is real then $A^{\ast} = A^{t}$. This is homework and is in the area of linear algebra.&lt;/p&gt;&#xA;" OwnerUserId="2723" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2010-11-04T13:20:48.720" LastActivityDate="2011-01-16T11:09:24.410" Title="How to prove $(AB)^{\ast}=B^{\ast}A^{\ast}$?" Tags="&lt;linear-algebra&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="8899" PostTypeId="2" ParentId="8898" CreationDate="2010-11-04T13:16:54.960" Score="2" ViewCount="" Body="&lt;p&gt;Do you have the theorem that $(AB)^t=B^tA^t$ for real matrices?  Can you follow that proof through for the complex case (using conjugates)?  Without going back to look, I think it works.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-11-04T13:16:54.960" CommentCount="8" />
  <row Id="8900" PostTypeId="1" CreationDate="2010-11-04T13:22:23.917" Score="0" ViewCount="134" Body="&lt;p&gt;Hi all,&#xA;I need some online tool for diagonalizing 2x2 matrices or at least finding the eigenvectors and eigenvalues of it. I don't like to download any stuf because I'm not able to, some online tool will do the job.&#xA;Thanks.&lt;/p&gt;&#xA;" LastActivityDate="2010-11-04T14:12:56.287" Title="Online tool for diagonalizing matrices ?" Tags="&lt;matrices&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="8901" PostTypeId="2" ParentId="8865" CreationDate="2010-11-04T13:42:34.250" Score="8" ViewCount="" Body="&lt;p&gt;Sieving, alone, is a poor approach to the problem unless n is small (which, in your example, it is).  A better approach for large numbers: choose an appropriate interval,* sieve out small primes, test remaining numbers with a probable-prime test, then (if desired) prove primality with ECPP or a similar algorithm (for small primes, APR-CL is better).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many programs can do this search for you, though they generally omit the last step (Pari and Mathematica at least).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;.* If the interval is chosen at random, length 4.5 log x is 95% likely to hold at least one prime, since $4.5e^{-4.5}\approx0.05$.  If your interval is adversarially chosen to have few primes, you could choose an interval as long as about $2e^{-\gamma}\log^2x$.  If performance is critical you can determine the optimal sieving distance based on time and other resources taken for both parts.&lt;/p&gt;&#xA;" OwnerUserId="1778" LastActivityDate="2010-11-04T13:42:34.250" CommentCount="2" />
  <row Id="8902" PostTypeId="1" AcceptedAnswerId="8913" CreationDate="2010-11-04T13:44:08.290" Score="0" ViewCount="53" Body="&lt;p&gt;Let M, K, and L be isomorphic normal subgroups of a group G. Suppose $M \cap K \leq M \cap L$. Can we conclude anything about the index of MK in G and the index of ML in G? I would like it to be the case that MK has smaller index than ML, but I'm having a hard time convincing myself either way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(EDIT: I initially left out the word &quot;normal&quot;.)&lt;/p&gt;&#xA;" OwnerUserId="759" LastEditorUserId="759" LastEditorDisplayName="" LastEditDate="2010-11-04T14:30:46.760" LastActivityDate="2010-11-04T14:36:39.557" Title="Can we say something about the index of these subgroups?" Tags="&lt;group-theory&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="8903" PostTypeId="1" CreationDate="2010-11-04T13:49:11.100" Score="-2" ViewCount="283" Body="&lt;p&gt;How to prove polarization identity?&#xA;For any $u, v \in V$ &#xA;$4 \langle T(u), v \rangle = \langle T(u+v), u+v \rangle - \langle T(u-v), u-v \rangle + i\langle T(u+iv), u+iv \rangle - i\langle T(u-iv), u-iv \rangle$&lt;/p&gt;&#xA;" OwnerUserId="2723" LastEditorUserId="2778" LastEditorDisplayName="" LastEditDate="2010-11-04T15:50:40.333" LastActivityDate="2010-11-21T04:07:39.877" Title="How to prove polarization identity?" Tags="&lt;linear-algebra&gt;&lt;homework&gt;" AnswerCount="1" CommentCount="6" />
  <row Id="8904" PostTypeId="1" AcceptedAnswerId="8915" CreationDate="2010-11-04T13:59:55.313" Score="3" ViewCount="297" Body="&lt;p&gt;From this &lt;a href=&quot;http://math.stackexchange.com/questions/6244/is-there-a-quick-proof-as-to-why-the-vector-space-of-mathbbr-over-mathbbq/6250#6250&quot;&gt;post&lt;/a&gt; we see that $\mathbb{R}$ over $\mathbb{Q}$ is infinite dimensional. Similarly $\mathbb{C}$ over $\mathbb{Q}$ is also infinite dimensional, and i rememeber having solved a problem that $\mathbb{R} \cong \mathbb{C}$ when considered as vector spaces over $\mathbb{Q}$. So i would like to extend this question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose $V$ and $W$ are two vector spaces, over the field $\mathbb{Q}$. Can we say that $V \cong W$ if they have the same dimension over $\mathbb{Q}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I really don't know how to prove or disprove this!&lt;/p&gt;&#xA;" LastActivityDate="2010-11-13T13:21:53.110" Title="Isomorphism of Vector spaces over $\mathbb{Q}$" Tags="&lt;linear-algebra&gt;&lt;algebra&gt;" AnswerCount="4" CommentCount="4" FavoriteCount="1" />
  <row Id="8905" PostTypeId="2" ParentId="8893" CreationDate="2010-11-04T14:04:09.590" Score="7" ViewCount="" Body="&lt;p&gt;This is one of those results in calculus which were anticipated by&#xA;Archimedes. He gave a correct formula for the volume but it is not known exactly &#xA;how Archimedes solved this problem. There is, however, a simple way to&#xA;obtain the answer without much calculus. Let me quote from late Gardner's &lt;a href=&quot;http://books.google.co.uk/books?id=PtBPcAAACAAJ&amp;amp;dq=unexpected+hanging&amp;amp;hl=en&amp;amp;ei=57rSTJywFNSP4ga9j8jcDw&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=2&amp;amp;ved=0CCwQ6AEwAQ&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;The Unexpected Hanging and Other Mathematical Diversions&lt;/em&gt;&lt;/a&gt; (Gardner considers the case $r=1$ but this is not essential, of course):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Imagine a sphere of unit radius inside the volume common&#xA;  to the two cylinders and having as its center the point&#xA;  where the axes of the cylinders intersect. Suppose that the&#xA;  cylinders and sphere are sliced in half by a plane through&#xA;  the sphere's center and both axes of the cylinders. &#xA;  The cross section of the volume common to the cylinders will be a square. &#xA;  The cross section of the sphere will be a circle that fills the square.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/SwXzR.jpg&quot; alt=&quot;alt text&quot;&gt;        &lt;img src=&quot;http://i.imgur.com/vaASe.gif&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Now suppose that the cylinders and sphere are sliced by a&#xA;  plane that is parallel to the previous one but that shaves off&#xA;  only a small portion of each cylinder (&lt;em&gt;have a look at the picture on the left&lt;/em&gt;).&#xA;  This will produce parallel tracks on each cylinder,&#xA;  which intersect as before to form a square cross section of&#xA;  the volume common to both cylinders. Also as before, the&#xA;  cross section of the sphere will be a circle inside the square.&#xA;  It is not hard to see (with a little imagination and pencil&#xA;  doodling) that any plane section through the cylinders, parallel&#xA;  to the cylinders' axes, will always have the same result:&#xA;  a square cross section of the volume common to the cylinders,&#xA;  enclosing a circular cross section of the sphere.&#xA;  Think of all these plane sections as being packed together&#xA;  like the leaves of a book. Clearly, the volume of the sphere&#xA;  will be the sum of all circular cross sections, and the volume&#xA;  of the solid common to both cylinders will be the sum of all&#xA;  the square cross sections. We conclude, therefore, that the&#xA;  ratio of the volume of the sphere to the volume of the solid&#xA;  common to the cylinders is the same as the ratio of the area&#xA;  of a circle to the area of a circumscribed square. A brief calculation&#xA;  shows that the latter ratio is $\pi/4$. This allows the&#xA;  following equation, in which $x$ is the volume we seek:&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;$$\frac{4\pi r^3/3}{x}=\frac{\pi}{4}.$$&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The $\pi$'s drop out, giving $x$ a value of $16r^3/3$. The radius in&#xA;  this case is 1, so the volume common to both cylinders is&#xA;  $16/3$. As Archimedes pointed out, it is exactly $2/3$ the volume&#xA;  of a cube that encloses the sphere; that is, a cube with&#xA;  an edge equal to the diameter of each cylinder.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="723" LastEditorUserId="723" LastEditorDisplayName="" LastEditDate="2010-11-04T14:47:19.917" LastActivityDate="2010-11-04T14:47:19.917" CommentCount="2" />
  <row Id="8906" PostTypeId="2" ParentId="8904" CreationDate="2010-11-04T14:05:08.567" Score="0" ViewCount="" Body="&lt;p&gt;All finite dimensional vector spaces of the same dimension over a field $k$ are isomorphic.  Fix a basis of $V$ and $W$; they each have $n$ elements if $V$ and $W$ are $n$-dimensional.  Then the map that sends basis elements $e_i$ of $V$ to basis elements $f_i$ of $W$ is an isomorphism.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We usually say that if $V$ and $W$ are vector spaces over $k$ of the same dimension, they are isomorphic, but not canonically.  This is because to exhibit an isomorphism requires a &lt;i&gt;choice&lt;/i&gt; of basis.&lt;/p&gt;&#xA;" OwnerUserId="3009" LastActivityDate="2010-11-04T14:05:08.567" CommentCount="5" />
  <row Id="8907" PostTypeId="2" ParentId="8904" CreationDate="2010-11-04T14:12:23.487" Score="1" ViewCount="" Body="&lt;p&gt;Two vector spaces whose dimensions over a field are of the same cardinality are easily seen to be isomorphic by mapping one basis onto the other; i.e. if $V$ has basis ${e_i}_{i\in I}$ and $W$ has basis ${f_j}_{j\in J}$ where $I$ and $J$ are of the same cardinality, pick a bijection $\phi:I\rightarrow J$ and define a linear map by sending $e_i\mapsto f_{\phi(i)}$. Since the $e_i$ map bijectively to the $f_j$, this must be an isomorphism.&lt;/p&gt;&#xA;" OwnerUserId="1306" LastActivityDate="2010-11-04T14:12:23.487" CommentCount="3" />
  <row Id="8908" PostTypeId="2" ParentId="8900" CreationDate="2010-11-04T14:12:56.287" Score="1" ViewCount="" Body="&lt;p&gt;Wolframalpha has an option. Try this: &lt;a href=&quot;http://www.wolframalpha.com/examples/Matrices.html&quot; rel=&quot;nofollow&quot;&gt;http://www.wolframalpha.com/examples/Matrices.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1527" LastActivityDate="2010-11-04T14:12:56.287" />
  <row Id="8909" PostTypeId="1" CreationDate="2010-11-04T14:16:23.030" Score="10" ViewCount="217" Body="&lt;p&gt;I'm interested to know if anyone can point me to a non-calculus way of seeing that the volume of a pyramid = 1/3 x area of base x height. Yes, I've googled.&lt;/p&gt;&#xA;" OwnerUserId="3110" LastActivityDate="2010-11-05T00:00:19.863" Title="How to intuitively see that the volume of a pyramid = 1/3 x area of base x height" Tags="&lt;geometry&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="1" />
  <row Id="8910" PostTypeId="2" ParentId="8909" CreationDate="2010-11-04T14:22:58.393" Score="5" ViewCount="" Body="&lt;p&gt;The easiest intuitive one I know is to take a cube and divide it into three pyramids.  Take one corner and form the pyramid with each of the three faces that don't touch that corner.  They are congruent, so have volume 1/3*base*height as the base is a square the side of the cube and the height is the side of the cube as well.  Then appeal to affine transformations to say this applies to all pyramids.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-11-04T14:22:58.393" CommentCount="2" />
  <row Id="8911" PostTypeId="1" CreationDate="2010-11-04T14:26:32.297" Score="-2" ViewCount="182" Body="&lt;p&gt;Suppose that for all $u \in V, \langle T(u), u \rangle =0$. How to prove that T is the 0 operator, that is $T(v)=0$ for all $v \in V$? Hint: Apply information on my last question about polarization identity to obtain that $\langle T(u), v \rangle = 0$ for all $u, v \in V$ and then let $v=T(u)$. this is a homework in the area of linear algebra.&lt;/p&gt;&#xA;" OwnerUserId="2723" LastEditorUserId="2723" LastEditorDisplayName="" LastEditDate="2010-11-04T15:10:10.670" LastActivityDate="2010-11-05T12:59:11.310" Title="How to prove that T is the 0 operator, that is T(v)=0 for all $u, v \in V$?" Tags="&lt;linear-algebra&gt;&lt;homework&gt;" AnswerCount="1" CommentCount="6" />
  <row Id="8912" PostTypeId="2" ParentId="8611" CreationDate="2010-11-04T14:30:43.547" Score="4" ViewCount="" Body="&lt;p&gt;user2969 asked for the full answer so I'll post it here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first step is to take the gcd of the two numbers.  This can be done with the Euclidean algorithm and is fast: the naive implementation takes quadratic time, which is good enough.  Call this number g.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The next step is to factor g.  This is the hardest step if g is large.  Basic approach: divide g by the primes up to some fixed limit, say 10^5, or until what remains is less than the square of the current prime.  At this point test what remains for primality (first with a Miller-Rabin test then, if desired, with the test of your choice, APR-CL, ECPP, etc.).  If the number is composite, use Pollard's rho and/or Shanks' SQUFOF.  (Whenever you find a factor, check it and its cofactor for primality and return to this step if composite.)  If those fail to find a factor, use ECM to search for a small factor.  At an appropriate crossover point you may wish to use MPQS/SIQS to factor the number instead.  If the number is larger, increase the bounds on ECM.  If no factor is found to ~40 digits, consider using GNFS to factor what remains.  This will take at least a day, depending on the size of the remaining number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you have the factorization it's easy to count the number of divisors: start an accumulator at 1, and for each prime multiply the accumulator by one more than the exponent of that prime.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that for this problem you may be able to short-circuit the factorization for numbers of an appropriate size by trial-dividing up to the cube root of the number and showing that what remains is composite and not a square.  In this case the number of divisors is the number of divisors of the factored part times 4, since the unfactored part is a squarefree semiprime.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Here are some scripts to calculate the value.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://pari.math.u-bordeaux.fr/&quot; rel=&quot;nofollow&quot;&gt;Pari/GP&lt;/a&gt;: &lt;code&gt;common(m,n) = numdiv(gcd(m,n))&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mathematica: &lt;code&gt;Common[m_,n_] := DivisorSigma[0,GCD[m,n]]&lt;/code&gt;&lt;/p&gt;&#xA;" OwnerUserId="1778" LastEditorUserId="1778" LastEditorDisplayName="" LastEditDate="2010-11-04T19:52:26.270" LastActivityDate="2010-11-04T19:52:26.270" CommentCount="1" />
  <row Id="8913" PostTypeId="2" ParentId="8902" CreationDate="2010-11-04T14:36:39.557" Score="6" ViewCount="" Body="&lt;p&gt;It's true for finite groups, since $|MK| = |M| |K|/|M \cap K|$. It will not be true in general for infinite groups. The example that springs to mind is $G={\mathbb Z}^{\infty}$, where it is easy to find subgroups $M,K,L$ all isomorphic to $G$ with $M \cap K = M \cap L = 1$, $ML=G$ and $|G:MK|$ infinite.&lt;/p&gt;&#xA;" OwnerUserId="2604" LastActivityDate="2010-11-04T14:36:39.557" CommentCount="1" />
  <row Id="8914" PostTypeId="2" ParentId="8911" CreationDate="2010-11-04T14:51:24.957" Score="1" ViewCount="" Body="&lt;p&gt;You've left out some important information.  It's not enough to guarantee T(v) is the zero operator just to know T(v) orthogonal to v for all v.  (Think about a right angle rotation in the plane.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course if T(v) is orthogonal to u for all u, then (in particular) T(v) must be zero since &amp;lt; T(v),T(v)&amp;gt; = 0 implies that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So look for the missing extra condition(s)...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edited per Robin Chapman's comments, to incorporate the OP's earlier question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://math.stackexchange.com/questions/8903/how-to-prove-polarization-identity&quot;&gt; How to prove polarization identity?&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The polarization identity expresses &amp;lt;T(v),u&amp;gt; for any v,u in a complex inner product space as a linear combination of terms of the form &amp;lt;T(w),w&amp;gt;.  Thus if all of the latter are zero, so must &amp;lt;T(v),u&amp;gt; be always zero, and the result follows.&lt;/p&gt;&#xA;" OwnerUserId="3111" LastEditorUserId="3111" LastEditorDisplayName="" LastEditDate="2010-11-05T12:59:11.310" LastActivityDate="2010-11-05T12:59:11.310" CommentCount="1" />
  <row Id="8915" PostTypeId="2" ParentId="8904" CreationDate="2010-11-04T14:53:56.173" Score="5" ViewCount="" Body="&lt;p&gt;It is not hard to prove that for &lt;em&gt;any&lt;/em&gt; vector space $\mathbf{V}$, if $\beta$ and $\beta'$ are two bases for $\mathbf{V}$, then the cardinality of $\beta$ equals the cardinality of $\beta'$. For finite dimensional spaces this is standard. For infinite dimensional, for each $x\in\beta$ let $\beta'_x$ be a finite subset of $\beta'$ such that $x\in\mathrm{span}(\beta'_x)$. Then $\cup\beta'_x = \beta'$, so $|\beta'|\leq \aleph_0|\beta|=|\beta|$. Now repeat the process for $y\in\beta'$ to get the other inequality. &lt;strong&gt;Edit:&lt;/strong&gt; Note that this assumes the Axiom of Choice, so that we can talk about the cardinals of $\beta$ and $\beta'$. In the absence of AC, it could be that $\beta$ and $\beta'$ are &quot;incomparable&quot; (no injections going either way), so we cannot conclude that any two bases have the same cardinality without AC.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fix a field $F$, and let $\mathbf{V}$ and $\mathbf{W}$ be vector spaces over $F$. Assuming the &lt;a href=&quot;http://en.wikipedia.org/wiki/Axiom_of_Choice&quot; rel=&quot;nofollow&quot;&gt;Axiom of Choice&lt;/a&gt; we can prove that every vector space has a basis (in fact, &quot;Every vector space over any field has a basis&quot; is equivalent to the Axiom of Choice). So let $\beta=\{v_i\}_{i\in I}$ be a basis for $\mathbf{V}$, and let $\gamma=\{w_j\}_{j\in J}$ be a basis for $\mathbf{W}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $\mathbf{V}$ is isomorphic to $\mathbf{W}$, then let $\varphi\colon\mathbf{V}\to\mathbf{W}$ be an isomorphism. Then $\varphi(\beta)$ is a basis for $\mathbf{W}$, and since $\varphi(\beta)$ and $\gamma$ are both bases of $\mathbf{W}$, and $\varphi$ is a bijection, you have $|\beta|=|\varphi(\beta)|=|\gamma|$. So if $\mathbf{V}$ and $\mathbf{W}$ are isomorphic, then they have bases of the same cardinality.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conversely, suppose $|\beta|=|\gamma|$. Let $f\colon\beta\to\gamma$ be a bijection, and let $\varphi\colon \mathbf{V}\to\mathbf{W}$ be the unique linear transformation that extends $f$ (that is, extend $f$ linearly to all of $\mathbf{V}$). Since $\varphi(\mathbf{V}) = \varphi(\mathrm{span}(\beta)) = \mathrm{span}(\varphi(\beta)) = \mathrm{span}(f(\beta)) = \mathrm{span}(\gamma)=\mathbf{W}$, then $\varphi$ is onto. It is straightforward to verify that $\varphi$ is one-to-one (it takes a basis to a basis). So $\varphi$ is an isomorphism. Thus, if $\mathbf{V}$ and $\mathbf{W}$ have bases of the same cardinality, then $\mathbf{V}$ is isomorphic to $\mathbf{W}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, two vector spaces over the same field are isomorphic if and only if they have bases of the same cardinality, if and only if they have the same dimension (assuming the Axiom of Choice). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, $\mathbb{R}$ and $\mathbb{C}$ are isomorphic as vector spaces over $\mathbb{Q}$, since they both have dimension $2^{\aleph_0}$. But you have to specify over what field you are working: $\mathbb{R}$ and $\mathbb{C}$ are &lt;strong&gt;not&lt;/strong&gt; isomorphic as vector spaces over $\mathbb{R}$ (dimensions 1 and 2, respectively)!. &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-05T03:20:18.040" LastActivityDate="2010-11-05T03:20:18.040" CommentCount="7" />
  <row Id="8916" PostTypeId="2" ParentId="8909" CreationDate="2010-11-04T14:57:49.293" Score="12" ViewCount="" Body="&lt;p&gt;I actually find it easiest to take a cube and divide it into 6 pyramids: just connect all of the corners of a cube to the midpoint, and for each of the 6 faces of the cube, you get a pyramid having that face as its base. If the cube has a side length s, then each pyramid has an s by s base and height s/2, and its volume is 1/6 the volume of the cube, which is 1/3 base times height.&lt;/p&gt;&#xA;" OwnerUserId="759" LastActivityDate="2010-11-04T14:57:49.293" />
  <row Id="8917" PostTypeId="1" CreationDate="2010-11-04T15:43:05.433" Score="1" ViewCount="135" Body="&lt;p&gt;How can we show that every automorphism of S4 is an inner automorphism ?&lt;/p&gt;&#xA;" OwnerUserId="2177" LastActivityDate="2010-11-04T15:50:19.247" Title="Automorphism of S4" Tags="&lt;group-theory&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="8918" PostTypeId="2" ParentId="8917" CreationDate="2010-11-04T15:50:19.247" Score="1" ViewCount="" Body="&lt;p&gt;Note that:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Any automorphism preserves the conjugacy class of transpositions.( By counting size of conjugacy classes)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Any automorphism preserves whether two transpositions share an element (by looking at the order of their product)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Any automorphism permutes the four classes of transpositions {those that move 1}, {those that move 2}, {those that move 3}, {those that move 4}&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;This gives you a permutation on {1,2,3,4} and it is not hard to show that your automorphism is conjugation by this permutation&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" LastActivityDate="2010-11-04T15:50:19.247" />
  <row Id="8919" PostTypeId="2" ParentId="8752" CreationDate="2010-11-04T16:06:58.457" Score="1" ViewCount="" Body="&lt;p&gt;Any proof of the 4CT is human-checkable so long as that human has enough time on their hands.&lt;/p&gt;&#xA;" OwnerUserId="2994" LastActivityDate="2010-11-04T16:06:58.457" CommentCount="2" />
  <row Id="8920" PostTypeId="1" CreationDate="2010-11-04T16:37:00.167" Score="1" ViewCount="301" Body="&lt;p&gt;Inspired by &lt;a href=&quot;http://math.stackexchange.com/questions/8623/distinct-permutations-of-x-items-from-n-items&quot;&gt;this question&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's consider the 11-letter word &quot;MATHEMATICS&quot;. The general question is to find how many words it is possible to have if we can make 4 permutations of letters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sub-question a)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each letter has a position in the word, thus when I say 1 permutation, we exchange the position of 2 letters. For example, with 1 permutation it is possible to have &quot;AMTHEMATICS&quot;, but not &quot;SMATHEMATIC&quot;.&#xA;How much word is it possible to make with 4 permutations, if ...&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;a.1) we want exactly 4 permutations and it is forbidden to have permutation $p$ and its inverse permutation. But it is possible to permute an &quot;M&quot; with another &quot;M&quot;.&lt;/li&gt;&#xA;&lt;li&gt;a.2) same as a.1) but it is forbidden to permute &quot;M&quot; with the other &quot;M&quot;.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;sub-question b) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's say that by permutations, we mean changing the letters in the word in such a way that 4 permutations could lead to &quot;TICSMATHEMA&quot; (when a letter enter a new position, the other letters are pushed).&#xA;How much word is it possible to make with 4 permutations, if we want exactly 4 permutations, that means 4 letters changing positions?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems to me that there is no trivial solution. At least I don't see it. Of course, it would be interesting to have a general solution for the case of a word with $n$ letters (possibly not all different) with $k$ permutations.&lt;/p&gt;&#xA;" OwnerUserId="2778" LastActivityDate="2010-11-04T16:53:52.650" Title="Permutation of 4 letters in a 11-letter word : how many different words can we make?" Tags="&lt;combinatorics&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="8921" PostTypeId="1" CreationDate="2010-11-04T16:43:33.180" Score="4" ViewCount="98" Body="&lt;p&gt;Assume Q is a convex central symmetric curve, whose area is $\displaystyle S$.&#xA;The area of the maximum parallelogram inside Q is $\displaystyle S'$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How to prove the conjecture that $\displaystyle \frac{S'}{S} \ge \frac{2}{\pi}=0.6366\dots$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, If Q is an ellipse, $\displaystyle S'=2ab$, $\displaystyle S=\pi ab$. &#xA;If Q is a regular hexagon, $\displaystyle \frac{S'}{S}= \frac{2}{3}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's trivial that $\displaystyle \frac{S'}{S} \ge \frac{1}{2}$, and I know how to prove $\displaystyle \frac{S'}{S} \ge \frac{4}{4+\pi}=0.56\dots$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From many reason, I believe this conjecture is true.&#xA;  Denote MAP=&quot;Maximum Area Parallelogram&quot;:&#xA;  For any $Q$ and any direction $\theta$, let $P(Q,\theta)$ be the area of MAP which have a corner in this direction.     $S'=\max\{P(Q,\theta)\}$.      In order to make $\frac{S'}{S}$ smallest, We need keep the largest one of $\{P(Q,\theta)\}$ small while S is a constant. Ellipse just keeps everyone in  $\{P(Q,\theta)\}$ average. This is very special, I don't think there will be other curve having this property. On the other hand, distribute equally always lead to the min-max in our knowledge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;About the $\frac{4}{4+\pi}$ lowerbound, the idea is as follows:&#xA;First, use polar function $r(\theta)$ to describe the curve. The condition is that $r(a)r(b)\sin|a-b|\leq C$, and we want to bound is $S=\int_{\theta}{r(\theta)}^2$. &#xA;Second, Without lose of generality, We assume $r(0)=r(90)=1,C=1$, and assume $Q$ is in the boundary of $Z=\{(x,y)|-1\leq x,y\leq 1\}$.&lt;br&gt;&#xA;Third, let $a=r(\theta)$ and $b=r(\theta+90)$ and find a bound for $(a^2+b^2)$ by Cauchy-Inequality. and it will give a bound for the area $S$.&lt;/p&gt;&#xA;" OwnerUserId="3109" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-04T19:19:18.823" LastActivityDate="2010-11-04T22:05:57.670" Title="A conjecture of parallelogram inside convex and central symmetric curve" Tags="&lt;geometry&gt;&lt;convex-optimization&gt;&lt;conjectures&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" />
  <row Id="8922" PostTypeId="2" ParentId="8920" CreationDate="2010-11-04T16:53:52.650" Score="1" ViewCount="" Body="&lt;p&gt;For a.1, there are 11*10/2=55 ways to pick the first permutation.  If you just prohibit inverses(the inverse is the same as the swap), there are then 55*54*53*52/24 ways to pick four permutations.  But this ignores the fact that you may move the same letter twice.  If you want eight letters to move in four swaps, you have 55*36*21*10/24&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-11-04T16:53:52.650" />
  <row Id="8923" PostTypeId="1" AcceptedAnswerId="8997" CreationDate="2010-11-04T17:02:24.447" Score="2" ViewCount="118" Body="&lt;p&gt;Let $X$ denote a metric space. One can assume that $X$ is Polish if that helps, but I was trying to avoid to assume that $X$ is compact. Let $P(X)$ denote the set of Borel probability measures on $X$. The weak*-topology on $P(X)$ is defined as usual: a net $(p_{\alpha})$ in $P(X)$ converges to $p \in P(X)$ iff $|\int fdp_{\alpha}-\int fdp|$ converges to $0$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Question: Is there any metric that metrizes the weak*-topology on $P(X)$ &lt;strong&gt;and&lt;/strong&gt; has convex open balls?  &lt;/p&gt;&#xA;" OwnerUserId="3117" LastActivityDate="2010-11-05T00:16:07.353" Title="Metrization of the weak*-topology on a set of probability measures" Tags="&lt;real-analysis&gt;&lt;probability-theory&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="1" />
  <row Id="8925" PostTypeId="1" CreationDate="2010-11-04T17:08:32.910" Score="1" ViewCount="125" Body="&lt;p&gt;How to show that $e^{x} \geq \left (1 + \frac{x}{n} \right) ^{n}$ holds for each non-negative real $x$ and each integer $n \geq 1$ ? I tried series and induction but got stuck. Can you please help?&lt;/p&gt;&#xA;" OwnerUserId="4219" LastActivityDate="2010-11-05T06:37:15.240" Title="Help inequality involving exponential function" Tags="&lt;real-analysis&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="8927" PostTypeId="1" AcceptedAnswerId="8956" CreationDate="2010-11-04T17:18:57.030" Score="4" ViewCount="201" Body="&lt;p&gt;Wikipedia articles on &quot;unit sphere&quot; and &quot;unit circle&quot; say the radius is 1. Articles on the &quot;unit square&quot; and &quot;unit cube&quot; say the length of the side is 1. Would you expect a unit torus to have major &lt;em&gt;radius&lt;/em&gt; 1 or major &lt;em&gt;diameter&lt;/em&gt; 1?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Admittedly, a torus is a different animal than a sphere, but... It feels most natural to me that the &quot;unit&quot; length should apply to the (major) radius, not the major diameter. Yet I recently came across open source code where someone generated a &quot;unit torus&quot; of major diameter 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is that &quot;wrong enough&quot; that I should change it (in a package of related changes that I'm already preparing to submit)? Can you give me a more solid mathematical basis for advocating that change? Or should I accept the status quo as just a different but legitimate interpretation of &quot;unit torus&quot;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I see from search hits like the following&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.cis.upenn.edu/~ngns/docs/References/CDC09%20RGG%20March06.pdf&quot; rel=&quot;nofollow&quot;&gt;Spectral Analysis of Virus Spreading in Random Geometric Graphs&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.scientificcommons.org/29544141&quot; rel=&quot;nofollow&quot;&gt;Unconditional Proof of the Boltzmann-Sinai Ergodic Hypothesis&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://portal.acm.org/citation.cfm?id=1496770.1496776&quot; rel=&quot;nofollow&quot;&gt;The cover time of random geometric graphs&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://webcache.googleusercontent.com/search?q=cache%3acuTivzCr9Y0J%3aannals.princeton.edu/annals/2006/163-3/annals-v163-n3-p06-p.pdf&quot; rel=&quot;nofollow&quot;&gt;Dimers and amoebae&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;that the term &quot;unit torus&quot; is used in some fields, like dynamical systems and discrete algorithms. But I'm unable to tell from these papers or abstracts what the authors mean exactly by &quot;unit torus&quot;. &lt;a href=&quot;http://webcache.googleusercontent.com/search?q=cache%3acuTivzCr9Y0J%3aannals.princeton.edu/annals/2006/163-3/annals-v163-n3-p06-p.pdf&quot; rel=&quot;nofollow&quot;&gt;Dimers and amoebae&lt;/a&gt; actually gives this definition:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;the unit torus T&lt;sub&gt;2&lt;/sub&gt; = {(z,w) ∈ C&lt;sub&gt;2&lt;/sub&gt; : |z|&#xA;  = |w| = 1}&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This definition appears to give a definite size. But if it's in the two-dimensional vector space over the complex numbers, I don't know how to apply it to $\mathbb{R}^3$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If &quot;unit torus&quot; (in $\mathbb{R}^3$) actually means something that &lt;em&gt;does not have any particular size&lt;/em&gt;, then that would be important to know.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is really not one of programming, but of what this term means in mathematics... including, to what degree is it actually defined (or not) in math?&#xA;I will base my software decisions on that information.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Would tag this &quot;torus&quot; if I could create the tag.)&lt;/p&gt;&#xA;" OwnerUserId="2656" LastEditorUserId="2656" LastEditorDisplayName="" LastEditDate="2010-11-04T20:15:05.630" LastActivityDate="2010-11-04T20:32:52.213" Title="what size is a &quot;unit torus&quot;?" Tags="&lt;geometry&gt;&lt;terminology&gt;" AnswerCount="1" CommentCount="19" FavoriteCount="1" />
  <row Id="8928" PostTypeId="2" ParentId="5052" CreationDate="2010-11-04T17:25:25.570" Score="1" ViewCount="" Body="&lt;p&gt;If you accept enumerative combinatorics as part of number theory, here is a nice application of the Legendre polynomials. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Count the number of ways to insert paired, grammatical and non-repeating parentheses enclosing at least two letters in a word of length $n &amp;gt; 1$. Define $a_1 = 1$ and ignore outmost paratheses. For $n = 2$, there is only one way $(ab) = ab$. For $n = 3$, we have $abc, (ab)c, a(bc)$. For $n = 4$, we have $abcd$, $(ab)cd$, $a(bc)d$, $ab(cd)$, $(ab)(cd)$, $a(bcd)$, $a(b(cd))$, $a((bc)d)$, $(abc)d$, $(a(bc))d$, and $((ab)c)d$. And, so on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sequence continues {$1$, $1$, $3$, $11$, $45$, $197$, $903$, $4279 \dots$ } (A001003) and the terms are called &lt;em&gt;Super Catalan&lt;/em&gt; numbers or &lt;em&gt;Little Schroeder&lt;/em&gt; numbers. &#xA;In general,&#xA;\begin{eqnarray}&#xA;a_n = \tfrac{1}{n} \sum_{k =0}^{n-2} \binom{2n-k-2}{n-1} \binom{n-2}{k} = \tfrac{1}{4n}(3P_{n-1}(3) - P_{n-2}(3)) + \tfrac{1}{2} \epsilon(n),&#xA;\end{eqnarray}&#xA;where $\epsilon$ is the unit arithmetic function, which is $1$ when the argument is $1$ and $0$ otherwise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, there are at least $10$ different combinatorial problems enumerated by this sequence, involving restricted lattice paths, integer compositions and even matrix row-sums. The Online Encyclopedia of Integer Sequences has the present state of knowledge about these fascinating numbers; see there reference section of &lt;a href=&quot;http://www.research.att.com/~njas/sequences/A001003&quot; rel=&quot;nofollow&quot;&gt;http://www.research.att.com/~njas/sequences/A001003&lt;/a&gt; for relevant papers and books.&lt;/p&gt;&#xA;" OwnerUserId="2720" LastEditorUserId="2720" LastEditorDisplayName="" LastEditDate="2010-11-15T21:16:22.587" LastActivityDate="2010-11-15T21:16:22.587" CommentCount="1" />
  <row Id="8929" PostTypeId="2" ParentId="8925" CreationDate="2010-11-04T17:26:16.123" Score="4" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\ $ Consider $\rm\ e^z\ \ge\ 1 + z,\ \ z\ =\ x/n $&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-11-04T17:26:16.123" CommentCount="1" />
  <row Id="8931" PostTypeId="2" ParentId="8925" CreationDate="2010-11-04T18:06:30.607" Score="0" ViewCount="" Body="&lt;p&gt;For each fixed nonnegative $x$, $(1+\frac{x}{n})^n$ increases with $n$.  To see why this is true, note that $(1+\frac{x}{n})^n$ is the amount of money in an account one year after $1$ unit of money is invested at a yearly interest rate of $100x\%$, compounded $n$ times per year.  As $n$ goes to $\infty$, this increasing sequence converges to $e^x$, corresponding to continuously compounded interest. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I do not claim that this is a rigorous proof.  It is merely intended as an explanation.)&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-11-04T18:06:30.607" />
  <row Id="8933" PostTypeId="1" AcceptedAnswerId="8941" CreationDate="2010-11-04T18:25:12.023" Score="4" ViewCount="281" Body="&lt;p&gt;How can I prove that there is no closed form solution to the equation $2^x + 3^x = 10$?&lt;/p&gt;&#xA;" OwnerUserId="2856" LastActivityDate="2010-12-15T03:17:33.547" Title="Proof that there is no closed form solution" Tags="&lt;algebra-precalculus&gt;" AnswerCount="2" CommentCount="7" FavoriteCount="3" />
  <row Id="8934" PostTypeId="2" ParentId="8904" CreationDate="2010-11-04T18:25:17.700" Score="1" ViewCount="" Body="&lt;p&gt;This answer was motivated by some comments of Djaian. As Arturo said, two vector spaces over the same field are isomorphic if and only if they have bases of the same cardinality. The infinite dimensional case is valid in much general context. So the important property of fields is that it is true for finite-dimensional spaces. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To justify this we'll talk about &lt;a href=&quot;http://en.wikipedia.org/wiki/Free_module&quot; rel=&quot;nofollow&quot;&gt;free modules over a ring with identity&lt;/a&gt;. Roughly speaking a module is  a generalization of vector space to arbitrary rings and the word free means that the module has a basis. We have the following result:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If we consider a ring $R$ with identity and a free $R$-module $F$ with an infinite basis $X$, then every basis of $F$ has the same cardinality as $X$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The proof is based on the Cantor–Schröder-Bernstein theorem and is similar to the one Arturo gave for vector spaces. It needs the Axiom of Choice too. Thus if $F_1$ and $F_2$ are two free modules with infinite bases $X_1$ and $X_2$, respectively, then $F_1 \cong F_2$ if and only if $|X_1| = |X_2|$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To see how badly it can be, there is a ring $R$ such that, as a free left $R$-module, $R$ has a basis of $n$ elements, for each positive integer $n$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We say that a ring with identity has &lt;a href=&quot;http://en.wikipedia.org/wiki/Invariant_basis_number&quot; rel=&quot;nofollow&quot;&gt;invariant basis number&lt;/a&gt;  (IBN) if for every free $R$-module $F$, any two bases of $F$ have the same cardinality. The cardinal number of any basis of $F$ is called the rank (or dimension) of $F$ over $R$. It's easy to see that if $R$ is a ring with IBN and $F_1$ and $F_2$ are two free modules over $R$, then $F_1 \cong F_2$ if and only if they have the same rank.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any nonzero commutative ring has IBN. Also any division ring has IBN. Other examples include finite rings, left-Noetherian rings, local rings and group rings.&lt;/p&gt;&#xA;" OwnerUserId="329" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2010-11-13T13:21:53.110" LastActivityDate="2010-11-13T13:21:53.110" CommentCount="6" />
  <row Id="8935" PostTypeId="1" CreationDate="2010-11-04T18:26:52.007" Score="2" ViewCount="87" Body="&lt;p&gt;I am trying to understand the interpretation of &lt;strong&gt;$\sin \frac{A}{2}$&lt;/strong&gt; and &lt;strong&gt;$\cos \frac{A}{2}$&lt;/strong&gt; in terms of  &lt;strong&gt;$\sin A$&lt;/strong&gt; from my book, here is how it is given :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have $ \bigl( \sin \frac{A}{2} + \cos \frac{A}{2} \bigr)^{2} = 1 + \sin A $ and $ \bigl( \sin \frac{A}{2} - \cos \frac{A}{2} \bigr)^{2} = 1 - \sin A $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By adding and subtracting we have $ 2 \cdot \sin \frac{A}{2} = \pm \sqrt{  1 + \sin A } \pm \sqrt{  1 - \sin A } $ ---- (1) and $ 2 \cdot \cos \frac{A}{2} = \mp \sqrt{  1 + \sin A } \mp \sqrt{  1 - \sin A } $ ---(2)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have understood upto this far well, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now they have broke the them into quadrants :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;In 1st quadrant :&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ 2 \cdot \sin \frac{A}{2} =  \sqrt{  1 + \sin A } - \sqrt{  1 - \sin A } $$&#xA; $$ 2 \cdot \cos \frac{A}{2} =  \sqrt{  1 + \sin A } + \sqrt{  1 - \sin A } $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;In 2nd quadrant :&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ 2 \cdot \sin \frac{A}{2} =  \sqrt{  1 + \sin A } + \sqrt{  1 - \sin A } $$&#xA; $$ 2 \cdot \cos \frac{A}{2} =  \sqrt{  1 + \sin A } - \sqrt{  1 - \sin A } $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;In 3rd quadrant :&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ 2 \cdot \sin \frac{A}{2} =  \sqrt{  1 + \sin A } - \sqrt{  1 - \sin A } $$&#xA;  $$ 2 \cdot \cos \frac{A}{2} =  \sqrt{  1 + \sin A } + \sqrt{  1 - \sin A } $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;In 4th quadrant :&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ 2 \cdot \sin \frac{A}{2} =  - \sqrt{  1 + \sin A } - \sqrt{  1 - \sin A } $$&#xA; $$ 2 \cdot \cos \frac{A}{2} =   - \sqrt{  1 + \sin A } + \sqrt{  1 - \sin A } $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, In knew the ALL-SINE-TAN-COSINE rule but still I am not able to figure out how the respective signs are computed in these (above) cases.&lt;/p&gt;&#xA;" OwnerUserId="2109" LastEditorUserId="2109" LastEditorDisplayName="" LastEditDate="2010-11-04T18:51:53.047" LastActivityDate="2010-11-04T18:56:49.053" Title="Expressions of $\sin \frac{A}{2}$ and $\cos \frac{A}{2}$ in terms of $\sin A$" Tags="&lt;trigonometry&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="8936" PostTypeId="1" CreationDate="2010-11-04T18:37:01.020" Score="1" ViewCount="76" Body="&lt;p&gt;Let X be a measure space and let $f_{n}$ be a sequence of functions which converge pointwise to a function f in $L^{p}(X)$ where $p&gt;1$ and suppose $g_{n}$ is a sequence of functions which converge pointwise to a function $g$ in $L^{\frac{p}{p-1}}(X)$. Prove that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle \lim_{n \to \infty} \int_{X} f_{n}(x) g_{n}(x) dx = \int_{X} f(x)g(x) dx$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;No idea how to proceed. Any help? I guess somewhere we need Holder and DCT.&lt;/p&gt;&#xA;" OwnerUserId="4219" LastActivityDate="2010-11-04T19:00:37.907" Title="Convergence of integrals in L^p and L^(p/(p-1))." Tags="&lt;measure-theory&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="8937" PostTypeId="1" CreationDate="2010-11-04T18:44:36.380" Score="4" ViewCount="178" Body="&lt;p&gt;I am currently engaged in independent study of algebraic geometry, using Dan Bump's book. One of the exercises in it outlines a proof of the Krull Intersection Theorem, which [here] is the following:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Let $A$ be a Noetherian local ring with maximal ideal $\mathfrak{m}$, and let $M$ be the intersection of all of the $\mathfrak{m}^n$. Then $M = 0$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The hints direct me to use the Artin-Rees lemma to show that $\mathfrak{m} M = M$, then use Nakayama's lemma to show that $M = 0$ (this second step is easy). I showed this to a professor and he accused the book of using big machinery for no reason, arguing that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathfrak{m} M = \mathfrak{m} \bigcap_{n \ge 0} \mathfrak{m}^n = \bigcap_{n \ge 1} \mathfrak{m}^n = M.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does this argument work? Does Bump apply Artin-Rees because that argument works in some broader context where the above argument fails? &lt;/p&gt;&#xA;" OwnerUserId="3120" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-11-04T18:56:05.200" LastActivityDate="2011-01-30T02:57:38.867" Title="Why is the Artin-Rees lemma used here?" Tags="&lt;algebraic-geometry&gt;&lt;commutative-algebra&gt;" AnswerCount="2" CommentCount="6" />
  <row Id="8938" PostTypeId="2" ParentId="8936" CreationDate="2010-11-04T18:54:22.233" Score="3" ViewCount="" Body="&lt;p&gt;Not true: Let $f_n(x)$ be the a constant times the characteristic function of $[1/n,2/n]$ and $g_n(x)$ be another constant times the characteristic function of $[1/n,2/n]$, the constants defined that the $L^p$ and $L^{p \over p-1}$ norms of&#xA;$f_n$ and $g_n$ respectively are both 1. Then they both converge pointwise to zero, so that the right hand side is zero. But the left integral is always 1. Thus what you're saying isn't even true if $f_n$ and $g_n$ are in $L^p$ and $L^{p \over p-1}$ respectively.&lt;/p&gt;&#xA;" OwnerUserId="3035" LastEditorUserId="3035" LastEditorDisplayName="" LastEditDate="2010-11-04T19:00:37.907" LastActivityDate="2010-11-04T19:00:37.907" CommentCount="1" />
  <row Id="8939" PostTypeId="2" ParentId="8935" CreationDate="2010-11-04T18:56:49.053" Score="4" ViewCount="" Body="&lt;p&gt;The easiest way of computing the signs is to make them match; we know that sin x &amp;gt; 0 if 0 &amp;lt; x &amp;lt; &amp;pi; and that cos x &amp;gt; 0 if -&amp;pi;/2 &amp;lt; x &amp;lt; &amp;pi;/2.  Knowing whether sin A is greater than 0 or less than zero tells you whether $\sqrt{1-\mathrm{sin} A}$ is greater or less than $\sqrt{1+\mathrm{sin} A}$; that in turn lets you figure out what the overall sign on all of the right-hand terms is, and each quadrant corresponds to one of the four positive/negative pairs on the right-hand terms.&lt;/p&gt;&#xA;" OwnerUserId="785" LastActivityDate="2010-11-04T18:56:49.053" />
  <row Id="8940" PostTypeId="1" AcceptedAnswerId="8942" CreationDate="2010-11-04T19:06:20.780" Score="1" ViewCount="407" Body="&lt;p&gt;Dear friends,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like a demonstration of the following theorem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Original question:&lt;/strong&gt;  If a square matrix of order $n\gt 1$ has a row (column) that is linear combination of rows (columns) parallel to it, then the determinant of this matrix is null.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Actual intended question:&lt;/strong&gt; If the determinant of a square matrix of order $n\gt 1$ is zero, then this matrix has some row (column) that is linear combination of rows (columns) parallel to it.&lt;/p&gt;&#xA;" OwnerUserId="2938" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-04T21:13:03.783" LastActivityDate="2010-11-05T03:15:39.630" Title="Determinant equal to zero" Tags="&lt;linear-algebra&gt;" AnswerCount="2" CommentCount="7" FavoriteCount="1" />
  <row Id="8941" PostTypeId="2" ParentId="8933" CreationDate="2010-11-04T19:07:38.330" Score="2" ViewCount="" Body="&lt;p&gt;There is a solution to this equation, but I don't know that you can make a closed-form solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As $\lim_{x\rightarrow -\infty} 2^x + 3^x = 0$ and $\lim_{x \rightarrow \infty} 2^x + 3^x = \infty$ and because $2^x + 3^x$ is continuous there must be a point where&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$2^x + 3^x = 10$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;by the intermediate value theorem.&lt;/p&gt;&#xA;" OwnerUserId="1089" LastActivityDate="2010-11-04T19:07:38.330" CommentCount="1" />
  <row Id="8942" PostTypeId="2" ParentId="8940" CreationDate="2010-11-04T19:13:08.450" Score="2" ViewCount="" Body="&lt;p&gt;If you know the determinant is multilinear (linear in each row/linear in each column), then express the the determinant as a sum of determinants of matrices that have repeated rows/columns. Then show that the determinant of a matrix with two identical rows/columns is always zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you do &lt;em&gt;not&lt;/em&gt; know the determinant is multilinear in the rows/columns, and you know the definition in terms of sums of products over all permutations, then the same argument works to show the determinant is a sum of determinants of matrices with repeated rows/columns.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If your only definition of determinant is inductively (by minors), then prove it by induction, expanding along a row or column that is &lt;em&gt;not&lt;/em&gt; a linear combination of other rows/columns. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; You've edited your question, and now the question is the &lt;em&gt;precise converse&lt;/em&gt; of what you originally posted. This is a bit of bad form, since any answers already posted would automatically look completely wrong-headed. In the future, consider editing to &lt;em&gt;add&lt;/em&gt; the new question to the old, indicating this is an edit or addition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, now you want to show that &lt;em&gt;if&lt;/em&gt; the determinant is zero, then a row/column is a linear combination of other rows/columns. Looking at the tranpose if necessary it suffices to consider the case of rows. Note that applying elementary row operations does not change whether the determinant is zero or nonzero (an elementary row operation will either multiply the determinant by a nonzero constant, or will not change the determinant). So the determinant of the matrix is zero if and only if the determinant of its row-echelon form is zero; the determinant of the row-echelon form is zero if and only if there is a row of zeros. Tracing back the elementary row operations will express the original row as a linar combination of the other rows. &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-04T21:33:03.290" LastActivityDate="2010-11-04T21:33:03.290" CommentCount="1" />
  <row Id="8943" PostTypeId="1" AcceptedAnswerId="8959" CreationDate="2010-11-04T19:23:44.610" Score="1" ViewCount="183" Body="&lt;p&gt;Consider a number say N = 345&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Permutations are :{{3, 4, 5}, {3, 5, 4}, {4, 3, 5}, {4, 5, 3}, {5, 3, 4}, {5, 4, 3}}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The GCD of all of them is &lt;strong&gt;3&lt;/strong&gt; which is the answer.Please suggest a suitable approach for this problem.&lt;/p&gt;&#xA;" OwnerUserId="2109" LastEditorUserId="2109" LastEditorDisplayName="" LastEditDate="2010-11-04T19:36:51.683" LastActivityDate="2011-02-17T05:01:24.393" Title="Computing GCD of all permutations (of the digits) of a given number." Tags="&lt;number-theory&gt;&lt;algorithms&gt;" AnswerCount="4" CommentCount="3" FavoriteCount="1" />
  <row Id="8944" PostTypeId="1" AcceptedAnswerId="8948" CreationDate="2010-11-04T19:25:58.087" Score="9" ViewCount="340" Body="&lt;p&gt;I want to prove that the series&#xA;$$ \sum_{n=1}^{\infty}(-1)^n\left(2^{1/n}-1\right)$$&#xA;converges.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am fairly certain that this converges.  Using the ratio or root test does not seem to work.  Therefore the other tests I am left with are &lt;strong&gt;alternating series test&lt;/strong&gt;, &lt;strong&gt;intergral test&lt;/strong&gt;, &lt;strong&gt;comparison test&lt;/strong&gt; and &lt;strong&gt;limit comparison test&lt;/strong&gt; (that we know about).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Seems like it would be a messy integral and that would not work or I would not be able to solve it.&lt;br&gt;&#xA;I think I could show this using the alternating series test IF the limit of the series is 0.  My problem is formally showing that the limit of this series is 0.  I can't quite get past that.&lt;br&gt;&#xA;My only option left if the alternating series test does not work is the limit comparison test, since the series is not &gt; 0 for large n, and therefore we cannot use the comparison test.  If this is the case I honestly don't have a clue what other series I should compare this with.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you for reading!&lt;/p&gt;&#xA;" OwnerUserId="2533" LastEditorUserId="2533" LastEditorDisplayName="" LastEditDate="2011-03-30T19:14:14.330" LastActivityDate="2011-03-30T19:14:14.330" Title="Prove that $\sum\limits_{n=1}^{\infty}(-1)^n(2^{1/n} - 1)$ is convergent." Tags="&lt;calculus&gt;&lt;homework&gt;&lt;sequences-and-series&gt;" AnswerCount="5" CommentCount="4" />
  <row Id="8945" PostTypeId="1" AcceptedAnswerId="8955" CreationDate="2010-11-04T19:30:05.963" Score="4" ViewCount="205" Body="&lt;p&gt;How to evaluate this integral: $$\int_{0}^{\infty} \biggl\lfloor{\frac{n}{e^{x}}\biggr\rfloor} \ dx, $$where $n \in \mathbb{N}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same integral when asked to evaluate for $n=2$ (say) i can do it by splitting the limits from $x = 0$ to $x = \log{2}$ where $e^{x}$ takes the value 1, and then from $x= \log{2}$ to $x = \infty$. But how to do this for the general case $n$. I thought of two ways:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Using induction on $n$. This will not work!&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Splitting the limits from $x =0$, to $x = \log{n}$ also seems to cause some problems for me. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-04T20:04:53.450" LastActivityDate="2011-04-22T04:06:41.927" Title="Evaluating the integral $\int_{0}^{\infty} \Bigl\lfloor{\frac{n}{e^{x}}\Bigr\rfloor} \ dx $" Tags="&lt;calculus&gt;&lt;analysis&gt;&lt;integral&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" />
  <row Id="8946" PostTypeId="1" AcceptedAnswerId="13407" CreationDate="2010-11-04T19:31:46.867" Score="2" ViewCount="230" Body="&lt;p&gt;I am confused on how to apply Ito's formula on certain problems, especially where expectations are involved. For example, if $W_t$ is a Weiner process and $X_t$ satisfies a below SDE:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$    dX_t = (X_t-\mu)dt + \sigma\sqrt{X_t}dW_t,~~~~~~ X_0 = x_o$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do I find $\partial_t \phi$ or $\partial_\xi \phi$ where $\phi(t,\xi)=E[e^{i\xi X_t}]$ is characteristic function of $X_t$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't quite understand how to approach this problem. Should I first solve the SDE for $X_t$, then compute the expectation $E[e^{i\xi X_t}]$, and then apply Ito's Lemma to find $\partial_t\phi$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Taking it a step further, how would I compute $\partial_t\psi$ where $\psi(t,\xi)=\ln\phi(t,\xi)$ and solve resulting SDE for $\psi(t,\xi)$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reference: &lt;a href=&quot;http://en.wikipedia.org/wiki/It%C5%8D&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/It%C5%8D&lt;/a&gt;'s_lemma&lt;/p&gt;&#xA;" OwnerUserId="957" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-04T19:42:13.760" LastActivityDate="2010-12-07T21:10:49.700" Title="application of Ito's lemma" Tags="&lt;probability-theory&gt;&lt;stochastic-processes&gt;" AnswerCount="1" FavoriteCount="3" />
  <row Id="8947" PostTypeId="2" ParentId="8944" CreationDate="2010-11-04T19:33:16.537" Score="0" ViewCount="" Body="&lt;p&gt;I believe that the alternating series test is the right way to go.  All you have to do is prove that $\lim (2^{1 \over n} - 1)$ is zero.  That is, to employ the test, you &lt;em&gt;don't&lt;/em&gt; need to show that the limit of the &lt;em&gt;series&lt;/em&gt; is zero, but merely that the limit of the &lt;em&gt;summands&lt;/em&gt; is zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[ETA: Arturo reminds us that one must also show that the summands strictly decrease.  But the AST is still the correct approach.]&lt;/p&gt;&#xA;" LastActivityDate="2010-11-04T19:33:16.537" CommentCount="3" />
  <row Id="8948" PostTypeId="2" ParentId="8944" CreationDate="2010-11-04T19:33:42.573" Score="12" ViewCount="" Body="&lt;p&gt;This is an alternating series, so it suggests (nay, &lt;em&gt;screams&lt;/em&gt;) for the use of the Alternating Series Test. You need to show that $\lim\limits_{n\to\infty}(2^{1/n}-1) = 0$, and that for all sufficiently large $n$, $2^{\frac{1}{n+1}}-1 \lt 2^{1/n} - 1$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The latter is equivalent to showing that $2^{\frac{1}{n+1}}\leq 2^{\frac{1}{n}}$; you might want to consider taking logarithms, say base $2$, which respect the order (that is, if $0\lt a\lt b$, then $\log_2(a)\lt\log_2(b)$). As to the former, it amounts to showing that $\lim\limits_{n\to\infty}2^{1/n} = 1$. Since $a^b = e^{b\ln(a)}$, this is equivalent to showing that $\lim\limits_{n\to\infty}e^{\frac{\ln(2)}{n}} = 1$; and since the exponential is continuous, this amounts to checking what happens to the exponent as $n\to\infty$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note also that none of the other tests that you mention are even applicable! The integral test can only be applied to a series $\sum a_n$ in which all terms $a_n$ are nonnegative (which is not the case here), and where you can find a positive, decreasing, and continuous function $f(x)$ such that $f(n)=a_n$ for all $n$ (or at least all sufficiently large $n$). So it is not applicable here. The Comparison Test requires the series to have positive terms, as does the Limit Comparison Test. You &lt;em&gt;could&lt;/em&gt; combine them with the Absolute Convergence Test by checking the convergence of $\sum|a_n|$, and then deducing the convergence of $\sum a_n$ if the series of absolute values converges, and apply one of these tests to the series $\sum|a_n|$. But you cannot apply any of the Integral, Comparison, or Limit Comparison Tests to this series directly.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-13T20:06:52.183" LastActivityDate="2010-11-13T20:06:52.183" CommentCount="6" />
  <row Id="8949" PostTypeId="2" ParentId="8937" CreationDate="2010-11-04T19:34:59.593" Score="5" ViewCount="" Body="&lt;p&gt;There is a proof due to Herstein that is &quot;elementary&quot; in the sense of avoiding the Artin-Rees lemma. Below is Kaplansky's presentation of this proof, from his &quot;Commutative Rings&quot;. A proof using primary decomposition (as in Krull's original proof) can be found in Zariski and Samuel.&#xA;&lt;img src=&quot;http://i.imgur.com/IIOGX.jpg&quot; alt=&quot;alt text&quot;&gt;&#xA;&lt;img src=&quot;http://i.imgur.com/sDQoD.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-11-04T19:34:59.593" />
  <row Id="8950" PostTypeId="2" ParentId="8944" CreationDate="2010-11-04T19:36:26.733" Score="1" ViewCount="" Body="&lt;p&gt;Yes the &lt;strong&gt;Alternating Test&lt;/strong&gt; will work. &lt;/p&gt;&#xA;" LastActivityDate="2010-11-04T19:36:26.733" />
  <row Id="8951" PostTypeId="1" CreationDate="2010-11-04T19:36:49.760" Score="1" ViewCount="81" Body="&lt;p&gt;Suppose $X$ and $Y$ are independent random variables. The mgf for $X$ is $M_{X}(t) = (1-t)^{-1}, \ t&amp;lt;1$. The mgf for $Y$ is $M_{Y}(t) = (1-2t)^{-1}, \ t&amp;lt; 0.5$. Two other random variables $U$ and $V$ are defined by $U = \frac{1}{2}(X+Y)$ and $V = \frac{1}{2}(X-Y)$. Calculate $\text{Cov}(U,V)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So $\text{Cov}(U,V) = E(UV)-E(U)E(V)$. This is the same thing as $\text{Cov}(\frac{1}{2}(X+Y), \frac{1}{2}(X-Y))$. Expanding out, we get &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\text{Cov}(\frac{1}{2}(X+Y), \frac{1}{2}(X-Y)) = \frac{1}{4} \text{Var}(X) -\frac{1}{4} \text{Cov}(X,Y)+\frac{1}{4} \text{Cov}(Y,X)-\frac{1}{4} \text{Var}(Y)$ which equals $\frac{1}{4}(\text{Var}(X)-\text{Var}(Y))$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there an easier way to obtain this without going through all this work? Could one use the fact $U+V = X$ and $U-V = Y$? &lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2010-11-04T19:36:49.760" Title="Covariance and Random Variables" Tags="&lt;probability&gt;" CommentCount="1" />
  <row Id="8952" PostTypeId="2" ParentId="8943" CreationDate="2010-11-04T19:37:34.960" Score="1" ViewCount="" Body="&lt;p&gt;In the case you have given you just need to note that the sum of the digits of 345 is divisible by 3. More generally, my first thoughts are that there is no straightforward answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are of course lots of cases such as numbers like, 198, 5505005, 777777777, 4848448, etc where we can say something more easily.&lt;/p&gt;&#xA;" OwnerUserId="1301" LastActivityDate="2010-11-04T19:37:34.960" />
  <row Id="8953" PostTypeId="2" ParentId="8943" CreationDate="2010-11-04T19:43:52.263" Score="3" ViewCount="" Body="&lt;p&gt;This is Sloane's &lt;a href=&quot;http://oeis.org/classic/A069652&quot; rel=&quot;nofollow&quot;&gt;A069652&lt;/a&gt;. The best way I can think of is to actually permute the digits and calculate -- though checking for 2, 5, and 3/9 first in the obvious way (all digits even, all digits 0 or 5, all digits add to a multiple of 3/9) in hopes that the rest of the number will have gcd 1 and terminate early that way.  Storing numbers in &lt;A href=&quot;http://en.wikipedia.org/wiki/Binary-coded_decimal&quot; rel=&quot;nofollow&quot;&gt;BCD&lt;/a&gt; (or as a vector of digits, or as a byte array) would make this fastest, I imagine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: Building on Gabe's answer, pick two distinct digits &lt;em&gt;m&lt;/em&gt; and &lt;em&gt;n&lt;/em&gt; in the number, if they exist. (If not, return the number.)  The gcd divides 9(m-n). Together with the 2/3/5/9 trick, this should determine most numbers quickly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Otherwise, continue choosing digits until you're gone through all combinations.  The hardest case is numbers where all digits are divisible by 3 and the sum of digits is divisible by 9; you must actually work with the numbers in this case to determine what power of 3 divides the number (at least 2).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In any case here's some naive Math'ca code in case someone wants to check their implementation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;f[n_] := GCD @@ FromDigits /@ Permutations[IntegerDigits[n]]&lt;/code&gt;&lt;/p&gt;&#xA;" OwnerUserId="1778" LastEditorUserId="1778" LastEditorDisplayName="" LastEditDate="2010-11-05T13:10:07.867" LastActivityDate="2010-11-05T13:10:07.867" CommentCount="3" />
  <row Id="8955" PostTypeId="2" ParentId="8945" CreationDate="2010-11-04T19:55:09.173" Score="10" ViewCount="" Body="&lt;p&gt;The integrand will only take on integral values m, and only for a finite measure of x values. The interval of x values in which the integrand equals exactly m is $ (- \ln \frac{m+1}{n} , - \ln \frac{m}{n} ] $. Split the integral over these intervals, evaluate (upper bound of interval minus lower bound times m) and turn them into a summation as $$ \sum_{m=1}^n m \left( \ln \frac{m+1}{m} \right) = \ln \prod_{m=1}^n \left( 1 + \frac{1}{m} \right)^m $$&lt;/p&gt;&#xA;" OwnerUserId="3122" LastActivityDate="2010-11-04T19:55:09.173" CommentCount="3" />
  <row Id="8956" PostTypeId="2" ParentId="8927" CreationDate="2010-11-04T19:55:53.780" Score="2" ViewCount="" Body="&lt;p&gt;My best guess is that the unit $d$-torus is the quotient $\mathbb{R}^d/\mathbb{Z}^d$, endowed with some combination of the following additional structures depending on the field of mathematics in consideration:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The structure of a &lt;a href=&quot;http://en.wikipedia.org/wiki/Topological_group&quot; rel=&quot;nofollow&quot;&gt;topological group&lt;/a&gt; (with the &lt;a href=&quot;http://en.wikipedia.org/wiki/Quotient_space&quot; rel=&quot;nofollow&quot;&gt;quotient topology&lt;/a&gt; and the &lt;a href=&quot;http://en.wikipedia.org/wiki/Quotient_group&quot; rel=&quot;nofollow&quot;&gt;quotient group&lt;/a&gt; structure),&lt;/li&gt;&#xA;&lt;li&gt;The structure of a &lt;a href=&quot;http://en.wikipedia.org/wiki/Riemannian_manifold&quot; rel=&quot;nofollow&quot;&gt;Riemannian manifold&lt;/a&gt; (with the induced &lt;a href=&quot;http://en.wikipedia.org/wiki/Flat_manifold&quot; rel=&quot;nofollow&quot;&gt;flat metric&lt;/a&gt;; this means that it is not curved, by contrast with the induced metric on the usual torus in $\mathbb{R}^3$)&lt;/li&gt;&#xA;&lt;li&gt;The structure of a &lt;a href=&quot;http://en.wikipedia.org/wiki/Probability_space&quot; rel=&quot;nofollow&quot;&gt;probability space&lt;/a&gt;.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;None of these structures depend on an embedding into $\mathbb{R}^n$.  The naming is by analogy with the case $d = 1$, where one gets a description of the usual topological group structure on the circle but, again, &lt;em&gt;without a preferred embedding into $\mathbb{R}^n$.&lt;/em&gt;  None of these structures suggest a good definition of &quot;unit torus&quot; in $\mathbb{R}^3$.  In particular, as Willie notes in the comments, $\mathbb{R}^2/\mathbb{Z}^2$ with the flat metric can't even isometrically embed into $\mathbb{R}^3$&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-11-04T20:32:52.213" LastActivityDate="2010-11-04T20:32:52.213" CommentCount="7" />
  <row Id="8958" PostTypeId="1" AcceptedAnswerId="8962" CreationDate="2010-11-04T20:23:03.687" Score="2" ViewCount="86" Body="&lt;p&gt;Let $f \in End_{F}(V)$ be an Endomorphism of a vector spaces $V$ over the field $F$ with the property $f \circ f=id_{V}$. Show that if $char(F) \neq 2$, the following equotation is true: $$V = Ker(f-id_{V}) \oplus Ker(f+id_{V})$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notes: $\oplus$ is the direct sum, $id_{V}$ is the identity map, Ker is the kernel and $\circ$ is the function composition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My thoughts:&#xA;I think you have to show that every Element in V can be expressed with the direct sum. It is obvious that the direct sum is always an element in V because the Kernel will return an element in V and the sum has to be in V again. But I have no clue where you need that $char(F) \neq 2$&lt;/p&gt;&#xA;" OwnerUserId="3123" LastActivityDate="2010-11-04T20:41:08.360" Title="Equality for subspaces with a linear map" Tags="&lt;linear-algebra&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="8959" PostTypeId="2" ParentId="8943" CreationDate="2010-11-04T20:29:02.900" Score="3" ViewCount="" Body="&lt;p&gt;I've gotta run, so I can't flesh this out quite as much as I want, but here's a start.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that gcd(a, b) = gcd(a, b-a). Thus, if a and b are close to each other, they must have a small gcd.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, consider a number N. Let a consist of all of the digits of N, sorted in ascending order. Now switch the last digit with the largest digit that is different from it, and call this new number b. Then b-a will be &quot;small&quot;. I suspect it's possible to refine this to get a quick answer.&lt;/p&gt;&#xA;" OwnerUserId="759" LastActivityDate="2010-11-04T20:29:02.900" CommentCount="5" />
  <row Id="8960" PostTypeId="1" AcceptedAnswerId="8963" CreationDate="2010-11-04T20:32:06.953" Score="0" ViewCount="75" Body="&lt;p&gt;I have a set of quadratic equations of the form..&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ 2S_0q_0(q_0S_0 + q_1S_1 + ...... + q_iS_n)+k_0 = 0$&lt;br&gt;&#xA;$ 2S_1q_1(q_0S_0 + q_1S_1 + ...... + q_iS_n)+k_1 = 0$&lt;br&gt;&#xA;   .&lt;br&gt;&#xA;   .&lt;br&gt;&#xA;   .&lt;br&gt;&#xA;$ 2S_nq_n(q_0S_0 + q_1S_1 + ...... + q_nS_n)+k_n = 0$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have n equations and n variables $q_0,q_1,q_2.....q_n$. The k and S values are constants.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I analytically reach a solution for this system?&lt;/p&gt;&#xA;" OwnerUserId="2858" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-04T20:38:41.247" LastActivityDate="2010-11-04T20:48:06.637" Title="Solving this set of quadratic equations" Tags="&lt;analysis&gt;&lt;sequences-and-series&gt;&lt;functional-equations&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="1" />
  <row Id="8961" PostTypeId="1" CreationDate="2010-11-04T20:32:23.367" Score="6" ViewCount="261" Body="&lt;p&gt;I have been thinking about the differentials that we use in derivatives and integrals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, I have an equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$${\int{w}{dr}} = \text{other stuff}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The context for this strange equation was:&#xA;$$\begin{align*}&#xD;&#xA;q&amp;#39;(t) &amp;amp;= a - \frac{b}{r(t)},\\&#xD;&#xA;r&amp;#39;(t) &amp;amp;= c - d r(t) - e q(t)\\&#xD;&#xA;r&amp;#39;&amp;#39;(t)&amp;amp;= d r&amp;#39;(t) - e q&amp;#39;(t)\\&#xD;&#xA;r&amp;#39;&amp;#39;(t)&amp;amp;= d r&amp;#39; - e a + \frac{eb}{r}\\&#xD;&#xA;w&amp;amp;=r&amp;#39;\\&#xD;&#xA;w&amp;#39; &amp;amp;= d w - ea + \frac{eb}{r}\\&#xD;&#xA;\frac{dw}{dr} \frac{dr}{dt}&amp;amp;= d w - ea + \frac{eb}{r}\\&#xD;&#xA;w \frac{dw}{dr} &amp;amp;= dw - ea + \frac{eb}{r}&#xD;&#xA;\end{align*}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Integrate both sides with respect to $r$:&#xA;$$\frac{w^2}{2} = \int{d*w dr} - ea r + eb log(r)$$&#xA;Shift the integral to the left hand side:&#xA;$$\int d * w dr = - \frac{w^2}{2} - ea * r - eb * log(r)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Divide by $d$:&#xA;$$\int w dr = \frac{w^2}{2d} - \frac{ear}{d}-\frac{eb}{d} log(r)$$&#xA;And I get to that $w\, dr$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if I can replace $w$ with ${\frac{dr}{dt}}$, and differentials can be multiplied like normal, I would get ${\frac{dr^2}{dt}}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then I have the idea of multiplying by ${\frac{dt}{dt}}$. Can I go ahead and do this? That would give me &#xA;$${\int{\frac{dr^2}{dt^2} dt}}$$&#xA;Which looks like ${w^2}$ to me. But can I do that? &lt;/p&gt;&#xA;" OwnerUserId="2247" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-29T13:52:00.093" LastActivityDate="2011-03-29T13:52:00.093" Title="To what extent can you manipulate differentials like dy and dt like actual values?" Tags="&lt;calculus&gt;&lt;differential-equations&gt;" AnswerCount="3" CommentCount="7" FavoriteCount="1" />
  <row Id="8962" PostTypeId="2" ParentId="8958" CreationDate="2010-11-04T20:32:35.677" Score="3" ViewCount="" Body="&lt;p&gt;You want to show that two things happen:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Every element of $V$ can be written as a sum $u+w$, where $u\in\mathrm{ker}(f-\mathrm{id}_V)$ and $v\in\mathrm{ker}(f+\ker{id}_v)$; and&lt;/li&gt;&#xA;&lt;li&gt;The intersection of $\mathrm{ker}(f-\mathrm{id}_V)$ and $\mathrm{ker}(f+\mathrm{id}_V)$ is trivial (contains only the zero vector).&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;One reason why you need $\mathrm{char}(F)\neq 2$ is that if $\mathrm{char}(F)=2$, then $1=-1$, so $f-\mathrm{id}_V = f+\mathrm{id}_V)$. And the sum of a subspace with itself cannot be a direct sum unless the subspace is trivial (which in this case would require $V$ to be trivial). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second part is fairly easy: suppose $x$ is in the intersection. Then $(f+\mathrm{id}_V)(x) = \mathbf{0}$ and $(f-\mathrm{id}_V)(x) = \mathbf{0}$; the first equation tells you that $f(x)+x=\mathbf{0}$, the second that $f(x)-x=\mathbf{0}$. Now, since $2\neq 0$, you can conclude that $x=\mathbf{0}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So now you &quot;just&quot; need to show that every element of $V$ can be expressed as a sum of vectors in the appropriate place. Now, notice that so far we have not used the fact that $f\circ f=\mathrm{id}_V$, so presumably we are going to need to use it somehow. For one thing, what happens if you take an arbitrary vector $x$, and you look at $x-f(x)$? Then $f(x-f(x)) = f(x)-f(f(x)) = f(x)-x = -(x-f(x))$. So, does $x-f(x)$ lie in one of the two subspaces? What about $x+f(x)$? And, can you express $x$ as a linear combination of $x+f(x)$ and $x-f(x)$ (again, the fact that $2\neq 0$ is going to rear its ugly head...)&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-04T20:41:08.360" LastActivityDate="2010-11-04T20:41:08.360" CommentCount="1" />
  <row Id="8963" PostTypeId="2" ParentId="8960" CreationDate="2010-11-04T20:48:06.637" Score="5" ViewCount="" Body="&lt;p&gt;Hint:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Set $X = q_0S_0 + q_1S_1 + \dots + q_n S_n$ and add up all the equations.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-11-04T20:48:06.637" CommentCount="1" />
  <row Id="8964" PostTypeId="2" ParentId="8944" CreationDate="2010-11-04T20:50:55.290" Score="2" ViewCount="" Body="&lt;p&gt;In many cases expanding the exponential works, in this case:&#xA;$2^{\frac{1}{n}} = \exp ( \frac{\ln 2}{n} ) = 1 + \frac{\ln 2}{n} + c_n $. Where $ \lim_{n \to \infty} n^2 c_n  =  (\ln 2)^2 \times ( \exp ( \frac{\ln 2}{n} ) - 1 - \frac{\ln 2}{n} ) /( \frac{\ln 2}{n})^2 = \frac{(\ln 2)^2}{2}.$ So, $c_n$ converges absolutely.&#xA;Now, $\sum_{ n \geq 1} (-1)^n (2^{\frac{1}{n}} - 1) = \sum_{n \geq 1} (-1)^n \frac{\ln 2}{n} + \sum_{n \geq 1} (-1)^n c_n$, which clearly converges.&lt;/p&gt;&#xA;" OwnerUserId="404" LastActivityDate="2010-11-04T20:50:55.290" />
  <row Id="8965" PostTypeId="1" AcceptedAnswerId="8972" CreationDate="2010-11-04T20:55:20.050" Score="1" ViewCount="66" Body="&lt;p&gt;Let $(X, \prec)$ and $(Y, &amp;lt;)$ be two well-ordered sets. Now, I want to understand the proof that $X$ is isomorphic to an initial segment of $Y$ or $Y$ is isomorphic with an initial segment of $X$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, they define $\mathcal{F}$ to be the set of all mappings with as domain an initial segment of $X$ and $Y$ as codomain and $F: \mathcal{F} \to Y \cup \{Y\}$ as $F(f) = \text{min}(Y \setminus \text{ran} f)$ if $f$ is not surjective and equal to $Y$ if $f$ is surjective. Now, they want to apply the recursion principle to $F$ to get a mapping from $X$ to $Y \cup \{Y\}$ that satisfies $f(x) = F(f|_{\hat{x}})$ for all $x$. Now I wonder how this is possible with the Recursion principle because the codomain is different. (Arturo solved this (very easy...) question in the comments).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Further they begin $x \prec y$ in $X$ and $f(y) \neq Y$ then $f(x) \lt f(y)$, so I assume the above but then I see that (strict) $Y \setminus \text{ran} f|_{\hat{y}} \subset Y \setminus \text{ran} f|_{\hat{x}}$ but why would could the minimum not be equal?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks. &lt;/p&gt;&#xA;" OwnerUserId="1120" LastEditorUserId="1120" LastEditorDisplayName="" LastEditDate="2010-11-04T21:18:23.363" LastActivityDate="2010-11-04T21:26:52.087" Title="Initial segments order-isomorphic." Tags="&lt;set-theory&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="8966" PostTypeId="1" CreationDate="2010-11-04T20:57:51.820" Score="1" ViewCount="338" Body="&lt;p&gt;I wanna know how to solve this equation: $-2x^3 +10x^2 -17x +8=(2x^2)(5x -x^3)^{1/3}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have some trouble to do that and I'd glad with any help I may get.&lt;/p&gt;&#xA;" OwnerUserId="3128" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-11-05T02:17:07.430" LastActivityDate="2010-12-08T08:11:57.290" Title="Solving the equation $-2x^3 +10x^2 -17x +8=(2x^2)(5x -x^3)^{1/3}$" Tags="&lt;algebra&gt;" AnswerCount="5" CommentCount="8" />
  <row Id="8967" PostTypeId="2" ParentId="8966" CreationDate="2010-11-04T21:01:28.490" Score="3" ViewCount="" Body="&lt;p&gt;Cube both sides and collect terms. You should get&#xA;\begin{eqnarray}&#xA;512 - 3264 x + 8856 x^2 - 13457 x^3 + 12702 x^4 - 7794 x^5 + &#xA; 3136 x^6 - 844 x^7 + 120 x^8 = 0&#xA;\end{eqnarray}&#xA;which factorizes into&#xA;\begin{eqnarray}&#xA;(8 - 17 x + 6 x^2) (64 - 272 x + 481 x^2 - 456 x^3 + 258 x^4 - &#xA;   84 x^5 + 20 x^6) = 0.&#xA;\end{eqnarray}&#xA;Since this is an $8^{\text{th}}$-degree polynomial equation, you can use Mathematica or a calculator to determine six of the eight solutions (by solving the $6^{\text{th}}$-degree polynomial equation numerically) and the other two by solving the quadratic equation: $\frac{1}{12}(17 \pm \sqrt{97})$. The 6 solutions are 3 pairs of complex conjugates:&#xA;\begin{eqnarray}&#xA;&amp;amp; &amp;amp; 0.647522 \dots \pm i 2.21209 \dots  \&#xA;&amp;amp; &amp;amp; 0.657656 \dots \pm i 0.218497 \dots  \&#xA;&amp;amp; &amp;amp; 0.794822 \dots \pm i 0.788971 \dots &#xA;\end{eqnarray}&#xA;Of course, you'll have to check that these solutions work. At most one real solution and two complex solutions are spurious.&lt;/p&gt;&#xA;" OwnerUserId="2720" LastEditorUserId="2720" LastEditorDisplayName="" LastEditDate="2010-11-05T06:30:19.867" LastActivityDate="2010-11-05T06:30:19.867" CommentCount="7" />
  <row Id="8968" PostTypeId="2" ParentId="8966" CreationDate="2010-11-04T21:02:19.890" Score="2" ViewCount="" Body="&lt;p&gt;Note that $\frac{1}{12} \left(17+\sqrt{97}\right)$ is not a solution. It just appeared because of cubing.&lt;/p&gt;&#xA;" OwnerUserId="3123" LastEditorUserId="3123" LastEditorDisplayName="" LastEditDate="2010-11-04T21:22:50.000" LastActivityDate="2010-11-04T21:22:50.000" CommentCount="6" />
  <row Id="8969" PostTypeId="1" CreationDate="2010-11-04T21:06:52.010" Score="1" ViewCount="108" Body="&lt;p&gt;If I have two different discrete distributions of random variables X and Y, such that their probability mass functions are related as follows: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$P(X=x_i) = \lambda\frac{P (Y=x_i)}{x_i} $   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;what can I infer from this equation? Any observations or interesting properties that you see based on this relation? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What if,&lt;br&gt;&#xA;  P($X=x_i$) = $\lambda\sqrt{\frac{P (Y=x_i)}{x_i} }$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In both cases, $\lambda$ is a constant.&lt;/p&gt;&#xA;" OwnerUserId="2858" LastEditorUserId="2858" LastEditorDisplayName="" LastEditDate="2010-11-04T23:26:01.930" LastActivityDate="2010-11-05T02:48:44.750" Title="Relationship between these two probability mass functions." Tags="&lt;probability&gt;&lt;probability-theory&gt;" AnswerCount="3" CommentCount="3" />
  <row Id="8970" PostTypeId="2" ParentId="8966" CreationDate="2010-11-04T21:12:40.333" Score="0" ViewCount="" Body="&lt;p&gt;Really? Is it impossible solve it by factorize or trick of algebra? Sorry, I´m just a high school grade studant.&lt;/p&gt;&#xA;" OwnerUserId="3128" LastActivityDate="2010-11-04T21:12:40.333" CommentCount="3" />
  <row Id="8971" PostTypeId="1" CreationDate="2010-11-04T21:21:37.640" Score="1" ViewCount="86" Body="&lt;p&gt;I hope this question is not too general, but I am not completely sure yet how to phrase it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we all know that when we have two sequences of random variables $X_n$ and $Y_n$ for $n \ge 1$, that converge to $X$ and $Y$ (using some measure, almost surely, or with probability, in distribution, etc.) then we have all kind of nice properties such as $X_n + Y_n \rightarrow X + Y$ and $X_n Y_n \rightarrow X Y$ so on (with some regularity conditions).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is: what if we have a general operator on $X$ and $Y$? Are there some mild/not mild conditions on an operator $g(X_n,Y_n)$ that returns a random variable $Z_n$ such that $g(X_n,Y_n) \rightarrow g(X,Y)$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the continuous mapping theorem would state that if $g$ is continuous, then we have that this holds. Are there milder conditions?&lt;/p&gt;&#xA;" OwnerUserId="7076" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-11-05T02:26:02.457" LastActivityDate="2010-11-05T02:26:02.457" Title="Sufficient conditions for convergence of functions of random variables." Tags="&lt;probability&gt;&lt;statistics&gt;" CommentCount="1" />
  <row Id="8972" PostTypeId="2" ParentId="8965" CreationDate="2010-11-04T21:26:52.087" Score="2" ViewCount="" Body="&lt;p&gt;Let $z$ be the least element of $X$ strictly larger than $x$; it exists, since you have at least $y$ among the upper bounds. You cannot have $f(z)=Y$, because you do not have $f(y)=Y$; so $f(z)$ is the least element which is not the image of any element of $\{a\in X\mid a\preceq x\}$. If $z=y$, then this shows that $f(y)$ cannot be equal to $f(x)$ (since $f(y)$ is the least element of the complement of a set that contains $f(x)$); if $z\neq y$, then $f(z)\in \mathrm{ran}f|_{\hat{y}}$; so the minimum of the complement of the range of $f|_{\hat{y}}$ cannot be equal to the minimum of the complement of the range of $f|_{\hat{x}}$: the latter &lt;em&gt;is&lt;/em&gt; $f(z)$, and this is &lt;em&gt;not&lt;/em&gt; in the complement of the former. &lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-11-04T21:26:52.087" />
  <row Id="8973" PostTypeId="2" ParentId="8961" CreationDate="2010-11-04T21:37:32.563" Score="5" ViewCount="" Body="&lt;p&gt;None of your examples make any sense to me. The manipulations with&#xA;differentials which are allowed in calculus are basically only those&#xA;that follow from the chain rule. In particular, an integral can be&#xA;transformed using the change-of-variables formula, but that's it. (OK,&#xA;maybe I'm over-simplifying a little here, but the basic advice is that&#xA;if you're unsure whether something is allowed or not, don't do it.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think much of the confusion surrounding this comes from the fact&#xA;that the notation $dy/dx$ doesn't show at which point the derivative&#xA;is supposed to be evaluated. For example, the derivative of an inverse&#xA;function is given by the rule $dx/dy=1/(dy/dx)$ in this notation,&#xA;which is suggestive and useful, but not very precise. What this really&#xA;says is that if $(a,b)$ is a point on the graph $y=f(x)$ and if the&#xA;slope at that point is $f'(a)=s$, then the point $(b,a)$ lies on the&#xA;graph of the inverse function, $y=f^{-1}(x)$, and the slope of that&#xA;graph &lt;em&gt;at that point&lt;/em&gt; is $1/s$:&#xA;$$ (f^{-1})'(b)=1/s=1/f'(a).$$&#xA;There's no &quot;magic&quot; here, just a simple&#xA;geometric fact which can be illustrated in a picture.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For another example of a similar flavour, see &lt;a href=&quot;http://math.stackexchange.com/q/8040/1242&quot;&gt;this&#xA;question&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: I was a bit too quick. There is something behind your manipulations after all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I understand you correctly, you are looking at a second order ODE of the form $d^2 r/dt^2 = f(r,dr/dt)$; let's call this equation (A). This is equivalent to the first order system&#xA;$dr/dt=w$, $dw/dt=f(r,w)$, which I'll call (B). Any solution $r=R(t)$ of (A)&#xA;corresponds to a solution $(r,w)=(R(t),R'(t))$ of (B).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fix one such solution for the moment.&#xA;(We're trying to solve the equation, so we probably don't know such a solution&#xA;yet, but we can pretend that we have one and see where that leads.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now $(r,w)=(R(t),R'(t))$ is a parametrized curve in the $(r,w)$-plane&#xA;(the &lt;em&gt;phase plane&lt;/em&gt; of the system).&#xA;Provided this curve doesn't have vertical slope, it looks (at least locally)&#xA;like the graph of a function, $w=H(r)$ say.&#xA;This function $H$ satisfies a first-order ODE,&#xA;$$H'(r)=\frac{f(r,H(r))}{H(r)},$$&#xA;which is the useful fact that you learned from Mariano in that other thread.&#xA;[It follows from &quot;$dw/dr=(dw/dt)/(dr/dt)=f(r,w)/w$&quot;&#xA;as explained in the answer that I linked to above.]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The integral you asked about is&#xA;$$ \int_{r_0}^{r_1} H(r) dr.$$&#xA;(I've attached some limits $r_0$ and $r_1$ just because that makes it easier for me to think.)&#xA;Making the change of variables $r=R(t)$, where $R$ is the solution of (A)&#xA;that we used in order to define the function $H$, we get&#xA;$$ \int_{t_0}^{t_1} H(R(t)) R'(t) dt$$&#xA;where $t_0$ and $t_1$ are suitable time values corresponding to $r_0$ and $r_1$&#xA;[that is, $r_0=R(t_0)$ and $r_1=R(t_1)$].&#xA;By the definition of $H$ we have $H(R(t))=R'(t)$,&#xA;since $w=H(r)$ defines a point on the curve $(w,r)=(R(t),R'(t))$.&#xA;Thus the integral becomes&#xA;$$ \int_{t_0}^{t_1} (R'(t))^2 dt,$$&#xA;just like you (and Dan in his answer) said, although expressed in a more&#xA;precise way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, whether this leads to something useful, I don't know.&#xA;But at least it should be a bit clearer what the manipulations &lt;em&gt;mean&lt;/em&gt;, I hope.&lt;/p&gt;&#xA;" OwnerUserId="1242" LastEditorUserId="1242" LastEditorDisplayName="" LastEditDate="2010-11-05T13:17:27.587" LastActivityDate="2010-11-05T13:17:27.587" CommentCount="5" />
  <row Id="8974" PostTypeId="2" ParentId="8969" CreationDate="2010-11-04T21:51:22.067" Score="1" ViewCount="" Body="&lt;p&gt;I don't have enough reputation yet, or this would be a comment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A random variable can have only one probability mass function, so it is not clear what you are asking.  Where do these equations come from?&lt;/p&gt;&#xA;" OwnerUserId="3009" LastActivityDate="2010-11-04T21:51:22.067" CommentCount="1" />
  <row Id="8975" PostTypeId="1" AcceptedAnswerId="9018" CreationDate="2010-11-04T22:05:29.067" Score="0" ViewCount="195" Body="&lt;p&gt;Can someone explain the following definitions of a limit and derivatives. Issac did a good job explaining it in the previous thread: &lt;a href=&quot;http://math.stackexchange.com/questions/8688/infinite-series-archimedean-principle&quot;&gt;http://math.stackexchange.com/questions/8688/infinite-series-archimedean-principle&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The definition of a limit is defined as the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$lim_{x \to a} f(x) = T$, what we actually mean is if we take any number M &gt; T, then we can force $f(x) &amp;lt; M$ by taking $x$ sufficiently close to $a$. Similarly, if we take any $L &amp;lt; T$, then we can force $f(x) &gt; L$ by taking $x$ sufficiently close to $a$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The definition of a derivative is defined as the following:&#xA;The derivative of $f$ at $a$ is the value denoted by $f'(a)$, such that for any $L &amp;lt; f'(a)$ and any $M &gt; f'(a)$, we can force $L &amp;lt; \frac{f(x)-f(a)}{x - a} &amp;lt; M$ simply by taking $x$  sufficiently close  to (but not equal to $a$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is my first real analysis course and I am very confused. If you choose to use episilon delta, please try to make it as simple as possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;thanks&lt;/p&gt;&#xA;" OwnerUserId="287" LastActivityDate="2010-11-05T07:54:54.450" Title="limit and derivatives - precise definitions" Tags="&lt;real-analysis&gt;" AnswerCount="3" CommentCount="5" FavoriteCount="1" />
  <row Id="8976" PostTypeId="2" ParentId="8921" CreationDate="2010-11-04T22:05:57.670" Score="2" ViewCount="" Body="&lt;p&gt;You might find this paper of interest:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fulton C.M., Stein S.K. (1960) Parallelograms inscribed in convex curves. Amer Math Monthly 67: 257–258&lt;/p&gt;&#xA;" OwnerUserId="1369" LastActivityDate="2010-11-04T22:05:57.670" CommentCount="1" />
  <row Id="8977" PostTypeId="1" CreationDate="2010-11-04T22:19:46.550" Score="3" ViewCount="456" Body="&lt;p&gt;How can I calculate the arc between two circles? The arc must be tangent to the two points on the circles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a picture illustrating it. I'm trying to code and calculate the orange arc and the blue arc. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/8GsDV.png&quot; alt=&quot;circle arcs&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The points B, F, D, and E can be changed by the user and dynamically update.&lt;/p&gt;&#xA;" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-04T22:23:44.223" LastActivityDate="2010-11-05T12:52:21.887" Title="Calculating a tangent arc between two points on two circles." Tags="&lt;geometry&gt;&lt;circle&gt;&lt;computational-geometry&gt;" AnswerCount="2" CommentCount="5" FavoriteCount="1" />
  <row Id="8978" PostTypeId="1" CreationDate="2010-11-04T22:43:13.373" Score="10" ViewCount="357" Body="&lt;p&gt;Suppose that you have an exponential generating function.: $E(z)=\sum_{n=0}^{\infty} \frac{a_{n}z^{n}}{n!}$, and that the definition of $a_{n}$ can be reasonably extended to noninteger arguments. (the Catalan numbers $C_{n}$, would be written in terms of the Gamma function thusly: $C_{n} = \frac{\Gamma(2n+1)}{\Gamma(n+2)\Gamma(n+1)}$, for instance), what then is the combinatorial significance of this integral:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$U(z)=\int_{0}^{\infty} \frac{a_{v}z^{v}dv}{\Gamma(v+1)}$$&#xA;?&lt;/p&gt;&#xA;" OwnerUserId="890" LastEditorUserId="6296" LastEditorDisplayName="" LastEditDate="2011-03-11T05:45:32.327" LastActivityDate="2011-04-10T06:39:58.303" Title="What is the combinatoric significance of an integral related to the exponential generating function?" Tags="&lt;combinatorics&gt;&lt;generating-functions&gt;" AnswerCount="1" CommentCount="6" FavoriteCount="5" />
  <row Id="8979" PostTypeId="1" AcceptedAnswerId="8986" CreationDate="2010-11-04T22:44:53.453" Score="0" ViewCount="59" Body="&lt;p&gt;Let f be an increasing function defined on [a,b] and let&lt;br&gt;&#xA;a &amp;lt; x1 &amp;lt; x2 &amp;lt; ... &amp;lt; xn &amp;lt; b be n points in the interior of [a,b]&lt;br&gt;&#xA;Show that the sum of [f(xk+) - f(xk-)] (with k ranging from 1 to n)&lt;br&gt;&#xA;is less than or equal to f(b-) - f(a+)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The proof I have is very wordy and sketchy. I know it would be easier if I could find a telescoping sum but I've got a road block in my mind. :/&lt;/p&gt;&#xA;" OwnerUserId="2319" LastActivityDate="2010-11-04T23:22:04.853" Title="Inequality on an increasing function" Tags="&lt;real-analysis&gt;" AnswerCount="1" />
  <row Id="8980" PostTypeId="1" CreationDate="2010-11-04T22:52:25.637" Score="4" ViewCount="322" Body="&lt;p&gt;Can someone show mathematically how gimbal lock happens when doing matrix rotation with Euler angles for yaw, pitch, roll?  I'm having a hard time understanding what is going on even after reading several articles on Google.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is the only work-around to use quaternions?&lt;/p&gt;&#xA;" OwnerUserId="3134" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-05T11:19:45.247" LastActivityDate="2010-11-05T11:19:45.247" Title="Euler angles and gimbal lock." Tags="&lt;geometry&gt;&lt;trigonometry&gt;&lt;euclidean-geometry&gt;&lt;quaternions&gt;" AnswerCount="2" />
  <row Id="8981" PostTypeId="2" ParentId="8975" CreationDate="2010-11-04T23:00:54.627" Score="0" ViewCount="" Body="&lt;p&gt;I think it's easier if you think about it visually. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;With loss of generality, we can assume as we get closer to 'a' our function is increasing, it could be constant, decreasing, or whatever.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's say you have X1 &amp;lt; a &amp;lt; X2 on the X axis with the corresponding f(X1)=L &amp;lt; T &amp;lt; M=f(X2)&lt;br&gt;&#xA;[Notice that f(a) doesn't necessarily equal T]  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we start squeezing X1 and X2 closer together and hence closer to a, we are forcing f(X1) and f(X2) closer to T. Therefore for x &amp;lt; X2, f(x) &amp;lt; f(X2)=M.&lt;br&gt;&#xA;Similarly, for x &gt; X1, f(x) &gt; f(X1)=L  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hoped this helped. When in doubt, draw a picture.&lt;/p&gt;&#xA;" OwnerUserId="2319" LastActivityDate="2010-11-04T23:00:54.627" />
  <row Id="8982" PostTypeId="2" ParentId="8961" CreationDate="2010-11-04T23:01:22.460" Score="3" ViewCount="" Body="&lt;p&gt;Your calculation looks to me like it produces the right answer provided one interprets it correctly. The only sensible meaning one can attach to $dr^2/dt^2$ should be $(dr/dt)^2$. So it seems your calculation shows that $\int w dr = \int w^2 dt$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My experience from my own undergraduate calculus is that: (i) any calculation I did by multiplying or dividing differentials like this would produce a meaningful and correct answer in the end; (ii) whatever answer was found in the end could also be justified by only using the normal chain rule. In this case, for instance, the change-of-variables formula tells you that $dr = (dr/dt)dt = wdt$, which gives the above formula.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: I realised afterwards that point (i) above might be misleading when one takes a course on calculus in several variables, since $dx \neq \frac{\partial x}{\partial y} dy$ if $x$ depends on more variables than $y$. In principle I would guess that multiplication and division of differentials is still possible as a heuristic but one needs to be very careful to distinguish between d's and $\partial$'s.&lt;/p&gt;&#xA;" OwnerUserId="677" LastEditorUserId="677" LastEditorDisplayName="" LastEditDate="2010-11-05T06:52:06.187" LastActivityDate="2010-11-05T06:52:06.187" />
  <row Id="8983" PostTypeId="1" AcceptedAnswerId="8987" CreationDate="2010-11-04T23:11:10.203" Score="1" ViewCount="227" Body="&lt;p&gt;My question is how do I reduce A'B'C'+AB'C'+ABC' (note that (') stands for a bar over the letter). To get (A+B')C'. I'm so lost just been trying to get it for awhile only using the 10 boolean simplification rules.&lt;/p&gt;&#xA;" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-11-05T02:24:34.327" LastActivityDate="2010-11-05T02:24:34.327" Title="Boolean Simplification of A'B'C'+AB'C'+ABC'" Tags="&lt;boolean-algebra&gt;" AnswerCount="2" />
  <row Id="8984" PostTypeId="2" ParentId="8980" CreationDate="2010-11-04T23:16:23.273" Score="6" ViewCount="" Body="&lt;p&gt;&quot;Euler Angles&quot; you can think of as a function $(S^1)^3 \to SO_3$ or $\mathbb R^3 \to SO_3$.  The derivative of this function does not always have rank 3, so you have degenerate submanifolds where the function is many-to-one.  In this special case that's called &quot;gimbal lock&quot;.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One formalism that avoids this is quaternions.  You can of course use other formalisms, and many other formalisms are naturally related to the quaternion version, so people tend to gravitate to the quaternion version.  One version that's closely related to quaternions would be to use the exponential map for the unit quaternion group.  But this also has &quot;gymbal lock&quot; but of a different kind.  But it does have the rather appealing interpretation as rotations about arbitrary axis -- this is perhaps more useful if you're only interested in rotations that differ from the identity matrix (or some given matrix) by a small amount, they're very natural coordinates on &quot;small scales&quot; in $SO_3$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any special properties you'd like for coordinates on $SO_3$? That might give a sense for where you want to go with this. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit in response to 1st comment: There's various conventions for Euler angles. Let's use what Wikipedia calls &quot;Proper Euler Angles&quot;   &lt;a href=&quot;http://en.wikipedia.org/wiki/Euler_angles&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Euler_angles&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given a unit vector $\vec v \in \mathbb R^3$ and a number $\theta \in \mathbb R$, let $R_{\vec v, \theta}$ be the rotation matrix which fixes $\vec v$ and which rotates counter-clockwise by an angle $\theta$ in the plane orthogonal to $\vec v$.  One version of &quot;Proper Euler angles&quot; would be:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$R_{(1,0,0), \theta_1} R_{(0,1,0), \theta_2} R_{(1,0,0), \theta_3}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &quot;Tait-Bryan&quot; variant would use the three distinct vectors, notice above we only used the standard vectors for the xy-plane.   The Tait-Bryan variant is the one appropriate to pitch/roll/yaw of an aeroplane.  Wikipedia has an excellent picture:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/PcWFU.png&quot; alt=&quot;Pitch/Roll/Yaw&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case you'd have &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$R_{(1,0,0),\theta_1} R_{(0,1,0), \theta_2} R_{(0,0,1),\theta_3} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\theta_3$ would be the roll, $\theta_2$ the pitch and $\theta_1$ the yaw. To relate my coordinates to the picture, $(1,0,0)$ is the direction the plane is pointing. $(0,1,0)$ is the direction of the left wing. $(0,0,1)$ is the direction of the yellow axis sticking out of the top of the plane. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can write out the matrices quite explicitly:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ R_{(1,0,0), \theta} = &#xA; \pmatrix{ 1 &amp;amp; 0 &amp;amp; 0 \cr &#xA;           0 &amp;amp; \cos \theta &amp;amp; -\sin \theta \cr&#xA;           0 &amp;amp; \sin \theta &amp;amp; \cos \theta }$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in this case, one occurance of &quot;gimbal lock&quot; is $\theta_2 = 0$.  In the Tait-Bryan variant it would be when the aeroplane is either pointing straight up or down, which is $\theta_2 = \pm \pi/2$. &lt;/p&gt;&#xA;" OwnerUserId="642" LastEditorUserId="642" LastEditorDisplayName="" LastEditDate="2010-11-05T00:02:03.897" LastActivityDate="2010-11-05T00:02:03.897" CommentCount="2" />
  <row Id="8985" PostTypeId="2" ParentId="8983" CreationDate="2010-11-04T23:18:19.713" Score="0" ViewCount="" Body="&lt;p&gt;First use the distributive law to pull out the C', then work on the As and Bs&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-11-04T23:18:19.713" />
  <row Id="8986" PostTypeId="2" ParentId="8979" CreationDate="2010-11-04T23:22:04.853" Score="6" ViewCount="" Body="&lt;p&gt;First, define $x_0 = a$ and $x_{n+1}=b$.&#xA;Then, for any $k=1,\ldots,n$, we have&#xA;$$&#xA;f(x_k +)-f(x_k -) \leq f\bigg(\frac{{x_k  + x_{k + 1} }}{2}\bigg) - f\bigg(\frac{{x_{k - 1}  + x_k }}{2}\bigg)&#xA;$$&#xA;(since $f$ is increasing). Thus,&#xA;$$&#xA;\sum\limits_{k = 1}^n {[f(x_k  + ) - f(x_k  - )]}  \le f\bigg(\frac{{x_n  + x_{n + 1} }}{2}\bigg) - f\bigg(\frac{{x_0  + x_1 }}{2}\bigg).&#xA;$$&#xA;Now, since $(x_n  + x_{n + 1} )/2 = (x_n  + b)/2 &amp;lt; b$ and $(x_0  + x_1 )/2 = (a  + x_1)/2 &gt; a$, we get&#xA;$$&#xA;\sum\limits_{k = 1}^n {[f(x_k  + ) - f(x_k  - )]}  \le f(b-) - f(a+).&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="2810" LastActivityDate="2010-11-04T23:22:04.853" CommentCount="1" />
  <row Id="8987" PostTypeId="2" ParentId="8983" CreationDate="2010-11-04T23:27:46.040" Score="3" ViewCount="" Body="&lt;pre&gt;&lt;code&gt;A'B'C'+AB'C'+ABC'&#xA;C'(A'B'+AB'+AB)&#xA;C'(A'B'+A(B'+B))&#xA;C'(A'B'+A)&#xA;C'(B'+A)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;It's that last step that used to trip me up.  &lt;code&gt;A'+AB = A'+B&lt;/code&gt;  Forget what that law is called (identity?).&lt;/p&gt;&#xA;" LastActivityDate="2010-11-04T23:41:36.120" CommentCount="4" />
  <row Id="8988" PostTypeId="1" AcceptedAnswerId="8991" CreationDate="2010-11-04T23:30:14.883" Score="1" ViewCount="148" Body="&lt;p&gt;Dear friends,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like a demonstration of the fact below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Being given real numbers &lt;strong&gt;a&lt;/strong&gt; and &lt;strong&gt;b&lt;/strong&gt; (nonzero) and a positive integer &lt;strong&gt;n&lt;/strong&gt;, the order &lt;strong&gt;p&lt;/strong&gt;, that occupies the maximum term (in absolute value)&#xA; of the development of power (a+b)^ n, according to decreasing powers of &lt;strong&gt;a&lt;/strong&gt; is given by:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;p = 1 + integer part of [|b|(n+1)/(|a|+|b|)]&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When &lt;strong&gt;n&lt;/strong&gt; is integer, there are maximum two terms: those of order &lt;strong&gt;p&lt;/strong&gt; and &lt;strong&gt;p-1&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;|a|&lt;/strong&gt; and &lt;strong&gt;|b|&lt;/strong&gt; are the modules of the numbers &lt;strong&gt;a&lt;/strong&gt; and &lt;strong&gt;b&lt;/strong&gt;, respectively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Already, very grateful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Paulo Argolo&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rio de Janeiro, Brazil&lt;/p&gt;&#xA;" OwnerUserId="2938" LastEditorUserId="2938" LastEditorDisplayName="" LastEditDate="2010-11-05T16:23:24.267" LastActivityDate="2010-11-05T16:23:24.267" Title="Maximum term of (a + b) ^ n" Tags="&lt;binomial-coefficients&gt;" AnswerCount="1" />
  <row Id="8989" PostTypeId="2" ParentId="8966" CreationDate="2010-11-04T23:33:14.993" Score="3" ViewCount="" Body="&lt;p&gt;Once someone gives the answer, it becomes easier!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I presume you are only looking for real roots and that $\displaystyle (5x-x^3)^{1/3}$ is the unique &lt;em&gt;real&lt;/em&gt; number whose cube is $\displaystyle 5x - x^3$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First observe that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle (x-2)^3 + 5x - x^3 = -6x^2 + 17x -8$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if $\displaystyle a = (5x-x^3)^{1/3}$ and $\displaystyle b = x-2$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;then the equation becomes&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle -2x^2(x-2) - (-6x^2 - 17x - 8) = 2x^2(5x-x^3)^{1/3}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and thus&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle -2x^2 b - (a^3 + b^3) = 2x^2a$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and so&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle (a+b)(a^2 - ab + b^2 + 2x^2) = 0$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now for any real numbers $\displaystyle a,b,x$ we have that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle a^2 -ab + b^2 \ge 0$ and  $\displaystyle 2x^2 \ge 0$ and thus we must have that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle a + b = 0$ or $\displaystyle 2x^2 = 0$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle a = -b$ can be cubed to give $\displaystyle 6x^2 - 17x + 8 = 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle 2x^2 = 0$ can be easily eliminated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the transformations we did were equivalent, and so &lt;em&gt;both&lt;/em&gt; roots of $\displaystyle 6x^2 - 17x + 8 = 0$ are also roots of the original equation, given the definition of cuberoot at the top of the answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you define $\displaystyle z^{1/3}$ using the principal branch of $\log z$, then the above assumption of $\displaystyle a$ being real is valid only if $\displaystyle 5x - x^3 \ge 0$, which eliminates $\displaystyle \dfrac{17 + \sqrt{97}}{12}$ as a root.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-12-08T08:11:57.290" LastActivityDate="2010-12-08T08:11:57.290" CommentCount="5" />
  <row Id="8990" PostTypeId="1" CreationDate="2010-11-04T23:41:00.410" Score="0" ViewCount="76" Body="&lt;pre&gt;&lt;code&gt;AB+CD+A’BD+A’BC+AB’D+AB’C&#xA;AB+CD+A’B(D+C)+AB’(D+C)&#xA;CD+(AB+A'B(D+C)+AB’(D+C))&#xA;CD+(AB+B(D+C)+A(D+C))&#xA;CD+AB+(B(D+C)+A(D+C))&#xA;CD+AB+(A+B)(D+C)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This is what I got, but I'm not sure if it's correct.  It's as good as I can get it algebraically.  Can it be simplified further?&lt;/p&gt;&#xA;" LastActivityDate="2010-11-05T05:38:12.020" Title="Boolean Algebra: Can this be simplified further?" Tags="&lt;boolean-algebra&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="8991" PostTypeId="2" ParentId="8988" CreationDate="2010-11-04T23:45:55.807" Score="5" ViewCount="" Body="&lt;p&gt;Let $f(k)=C_n^k |a|^{n-k} |b|^k$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then $f(k+1)/f(k)=\frac{|b|}{|a|} \frac{n-k}{k+1} &gt; 1$ iff $k&amp;lt;\frac{|b|(n+1)}{(|b|+|a|)}-1$, so $f(k)$ is increasing until $k=I\left(\frac{|b|(n+1)}{(|b|+|a|)}\right)$ where $I$ denotes the integral part.&lt;/p&gt;&#xA;" OwnerUserId="2660" LastActivityDate="2010-11-04T23:45:55.807" CommentCount="4" />
  <row Id="8992" PostTypeId="1" CreationDate="2010-11-04T23:51:06.830" Score="2" ViewCount="139" Body="&lt;p&gt;Hello. I'm trying to compute the following limit:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle \lim_{n \to \infty} \int_{0}^{\infty} (1+\frac{x}{n})^{(-n)} \sin(x/n) dx$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clearly the integrand converges to 0 pointwise so the trick is to find the dominating function. So for simplicity let ${f_{n}}$ denote the sequence of functions of the integrand: it is not hard to see that $|f_{n}| \leq (1+x/3)^{-3}x$ for all $n\geq 3$ and the latter function is integrable on $(0,\infty)$. But doesn't Lebesgue dominated convergence theorem requires to bound it for all $n\in\mathbb{N}$? What happens if we can't obtain a dominating function for $n=1$ or $n=2$ ? Can you please clarify this? I have read in a book that you can drop some terms, say $n=1,2,3$ and the bound the rest , but why is this possible? doesn't this violates the hypothesis of Lebesgue's theorem?&lt;/p&gt;&#xA;" OwnerUserId="4219" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-05T03:45:26.800" LastActivityDate="2010-11-05T04:30:50.677" Title="Understanding dominating functions" Tags="&lt;measure-theory&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="8993" PostTypeId="1" CreationDate="2010-11-04T23:55:06.997" Score="4" ViewCount="220" Body="&lt;p&gt;The Poincaré conjecture is listed as one of the Millennium Prize Problems and has received significant attention from the media a few years ago when Grigori Perelman presented a proof of this conjecture. But why is this interesting at all? What practical and interesting questions can be answered now that this conjecture is proven?&lt;/p&gt;&#xA;" OwnerUserId="3138" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-05T00:30:05.953" LastActivityDate="2010-11-05T00:31:42.750" Title="What is the importance of the Poincaré conjecture?" Tags="&lt;topology&gt;&lt;manifold&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="8995" PostTypeId="2" ParentId="8909" CreationDate="2010-11-05T00:00:19.863" Score="0" ViewCount="" Body="&lt;p&gt;Put together 6 pyramids to make a cube. You will see that volume of whole cube can be considered as sum of 2 equal smaller volumes (area by half-height of cube). Its also the height of pyramid multiplieds by area. The small volume is also the sum of 3 pyramids, because 3 is half of 6. So single pyramid is 1/3rd of this multiplication.&lt;/p&gt;&#xA;" OwnerUserId="3140" LastActivityDate="2010-11-05T00:00:19.863" CommentCount="2" />
  <row Id="8996" PostTypeId="1" AcceptedAnswerId="9001" CreationDate="2010-11-05T00:15:48.463" Score="3" ViewCount="133" Body="&lt;p&gt;Hi. This is an exercise from Rotman, An Introduction to Homological Algebra, which I've been thinking now and then for a few days and I haven't solved it yet. I've decided to ask here because it is bugging me and I don't have any friends also studying this subject (in case it is of any use, I'm an undergraduate).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $R$ be a commutative noetherian ring. If $A$, $B$ are finitely generated $R$-modules, then $Hom_R(A,B)$ is a finitely generated $R$-module.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's what I've thought as of yet (maybe the problem is that I haven't been trying to apply the useful theorems...):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First of all, since $R$ is noetherian, it suffices to inject $Hom_R(A,B)$ into some finitely generated module (in noetherian rings, submodules of f.g. are f.g.).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, since $R$ is commutative (or since $R$ is noetherian), a finitely generated $R$-module is a quotient of $R^n$ for some $n$. Then let us write $A\simeq \frac{R^n}{I}$ and $B\simeq \frac{R^m}{J}$. Now we've got:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$Hom_R(A,B)\simeq Hom_R\left(\frac{R^n}{I},\frac{R^m}{J}\right)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I had $Hom_R\left(\frac{R^n}{I},\frac{R^m}{J}\right) \hookrightarrow Hom_R(R^n, R^m)$ (this is what I tend to think is the wrong direction, but a interesting question-rising path nevertheless) then it would be over since $Hom_R(R^n, R^m)\simeq R^{nm}$ which is finitely generated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I ask myself: can I see $\frac{R^n}{I}$ as a submodule of $R^n$? Because if I could, then doing the same thing for $R^m$ and passing it to the hom, then it would be over. Now, this would be true if the sequence $$0\to I\hookrightarrow R^n\to \frac{R^n}{I} \to 0$$ split. But this (of course) doesn't always happen. But if I somehow had the third module to be projective, or the first one to be injective, then it would happen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But why would, for example, be $\frac{R^n}{I}$ be projective?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And this is where I got stuck. I'm sorry if this is overly detailed, but I remember reading that it is always good, when asking about a textbook question, to add what you've thought up to that moment. Also, even if my thoughts don't lead to the solution, I would like to know if they are correct!&lt;/p&gt;&#xA;" OwnerUserId="2614" LastEditorUserId="2614" LastEditorDisplayName="" LastEditDate="2010-11-05T02:00:49.527" LastActivityDate="2010-11-05T02:00:49.527" Title="Hom of finitely generated modules over a noetherian ring" Tags="&lt;abstract-algebra&gt;&lt;modules&gt;&lt;homological-algebra&gt;" AnswerCount="1" />
  <row Id="8997" PostTypeId="2" ParentId="8923" CreationDate="2010-11-05T00:16:07.353" Score="3" ViewCount="" Body="&lt;p&gt;On a separable metric space $(X,d)$, weak convergence of probability measures is equivalent to convergence with respect to the Kantorovich-Rubinshtein metric defined by &#xA;$$ \beta(\mu,\nu) = \sup \left( \left|\int_X f\ d(\mu-\nu)\right| : f\in {\rm Lip}(X),\  \sup_{x\in X}|f(x)|\leq 1 \right),$$&#xA;where &#xA;$${\rm Lip}(X):=\left( f:X\to{\mathbb R}, |f(x)-f(y)|\leq d(x,y) {\rm \ for\ all\ } x,y\in X   \right). $$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, see Theorem 11.3.3 of Dudley's &lt;em&gt;Real Analysis and Probability&lt;/em&gt; or Section 8.3 of Bogachev's measure &lt;em&gt;Theory Volume 2&lt;/em&gt;.  &lt;/p&gt;&#xA;" OwnerUserId="940" LastActivityDate="2010-11-05T00:16:07.353" CommentCount="3" />
  <row Id="8998" PostTypeId="2" ParentId="8993" CreationDate="2010-11-05T00:31:42.750" Score="13" ViewCount="" Body="&lt;p&gt;While I'm sure an expert could give a much more informative answer, let me give a naive one.  In mathematics we are always interested in classification results.  If we want to understand an object that occurs in some problem and we have a classification result for that type of object, we can use it to gain traction on the problem.  For example we use the classification of semisimple Lie algebras or the classification of finite simple groups all the time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Poincare conjecture is part of a similar classification effort, but for closed 3-manifolds.  Now, closed 2-manifolds have a well-understood classification in a few senses; there is a &lt;a href=&quot;http://en.wikipedia.org/wiki/Surface#Classification_of_closed_surfaces&quot; rel=&quot;nofollow&quot;&gt;topological classification&lt;/a&gt;, and there is also a &lt;a href=&quot;http://en.wikipedia.org/wiki/Uniformization_theorem#Geometric_classification_of_surfaces&quot; rel=&quot;nofollow&quot;&gt;geometric classification&lt;/a&gt;.  These classification results allow us to tackle many problems involving surfaces, such as the &lt;a href=&quot;http://en.wikipedia.org/wiki/Riemann_surface&quot; rel=&quot;nofollow&quot;&gt;Riemann surfaces&lt;/a&gt; which occur when one analytically continues a holomorphic function.  So it is natural to look for a corresponding classification result in higher dimensions.  (In science journalism this is often justified as saying that topologists are trying to understand the shape of the universe, or something like that; you should feel free to take this justification or leave it.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The topological classification of surfaces shows in particular that a surface is determined up to homeomorphism by its &lt;a href=&quot;http://en.wikipedia.org/wiki/Homology_(mathematics)&quot; rel=&quot;nofollow&quot;&gt;integral homology&lt;/a&gt;, so it's natural to ask whether the same is true for 3-manifolds.  Unfortunately, it's not; there is a &lt;a href=&quot;http://en.wikipedia.org/wiki/Homology_sphere#Poincar.C3.A9_homology_sphere&quot; rel=&quot;nofollow&quot;&gt;famous example&lt;/a&gt; of a 3-manifold with the same homology as a sphere but which is not homeomorphic to it.  One way to prove this is to show that it has a nontrivial &lt;a href=&quot;http://en.wikipedia.org/wiki/Fundamental_group&quot; rel=&quot;nofollow&quot;&gt;fundamental group&lt;/a&gt;, so now the natural question arises as to whether this is enough to uniquely identify the 3-sphere.  (If it is, then the problem of deciding whether a 3-manifold is the 3-sphere is in some sense purely algebraic, and in topology it is always desirable to reduce topological questions to algebraic ones since the latter tend to be easier.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem became famous in part because it was extremely hard and in part because many mathematicians gave incorrect proofs.  This is often how problems become famous; if a problem in a field is hard that indicates that the tools of the field are not adequate to easily address it, so the problem spurs people to improve those tools.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perelman ended up proving the Poincare conjecture by proving a stronger result, the &lt;a href=&quot;http://en.wikipedia.org/wiki/Geometrization_conjecture&quot; rel=&quot;nofollow&quot;&gt;geometrization conjecture&lt;/a&gt;, which is the analogue for 3-manifolds of the uniformization theorem.  So work on the Poincare conjecture has led to a deeper understanding of 3-manifolds in general.  Not only that, but Perelman's work introduced important techniques which can now be used on other problems.  &lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-11-05T00:31:42.750" />
  <row Id="8999" PostTypeId="2" ParentId="8992" CreationDate="2010-11-05T00:43:29.407" Score="5" ViewCount="" Body="&lt;p&gt;The limit of a sequence $\{ a_{n} : n \geq 1\}$ is the same as that of $\{ a_{n+2} : n \geq 1 \}$.&lt;/p&gt;&#xA;" OwnerUserId="404" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-05T04:30:50.677" LastActivityDate="2010-11-05T04:30:50.677" />
  <row Id="9000" PostTypeId="1" CreationDate="2010-11-05T00:52:34.717" Score="0" ViewCount="85" Body="&lt;p&gt;Suppose $G_1$ and $G_2$ are Lie groups with isomorphic Lie algebras.  Then from standard Lie theory we know that there is a simply connected Lie group $G$ such that $G/H_i = G_i$ where $H_i$ is a discrete subgroup of the center of $G$.  I am curious if there is a nice condition on $H_1,H_2$ and how they sit in $G$ that implies $G_1$ and $G_2$ are isomorphic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More generally I guess, when is $G/H = G/K$ for a general group $G$ and normal subgroups $H$ and $K$ (maybe having $H$ and $K$ be in the center makes things easier?).&lt;/p&gt;&#xA;" OwnerUserId="9" LastActivityDate="2010-11-05T00:52:34.717" Title="When is $G/H = G/K$? In particular, when are two Lie groups with isomorphic Lie algebras isomorphic?" Tags="&lt;group-theory&gt;&lt;lie-groups&gt;" CommentCount="2" FavoriteCount="1" />
  <row Id="9001" PostTypeId="2" ParentId="8996" CreationDate="2010-11-05T01:28:10.663" Score="5" ViewCount="" Body="&lt;p&gt;You're nearly there: with your notations, $\mathrm{Hom}_R(A,B)$ injects in $\mathrm{Hom}_R(R^n,B) \simeq B^n$, and $B^n$ is finitely generated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Indeed $\mathrm{Hom}_R(A,B) = \left\{f \in \mathrm{Hom}_R(R^n,B)\ |\ I \subset \ker f \right\}$.&lt;/p&gt;&#xA;" OwnerUserId="2660" LastActivityDate="2010-11-05T01:28:10.663" CommentCount="2" />
  <row Id="9002" PostTypeId="1" AcceptedAnswerId="9003" CreationDate="2010-11-05T01:34:38.660" Score="0" ViewCount="82" Body="&lt;p&gt;I was experimenting with binary operators, and devised a commutative operator with function $f$ such that the operator mapped $f \colon \mathbb{Z} \times \mathbb{Z} \rightarrow \mathbb{R}$ in order to make up some problem (e.g. if $f(3,10)=30$ and $f(4,3)=10$, find $f(4,6)$, etc.).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is, are there any functions $f$ other than $a \otimes b = f(a,b) = qab+r(a+b)$ where $q,r \in \mathbb{R}$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, is it always true that the operator is also distributive if $a \otimes 0 = 0$?&lt;/p&gt;&#xA;" OwnerUserId="1562" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2011-01-17T00:57:58.467" LastActivityDate="2011-01-17T00:57:58.467" Title="Some stuff with binary operators." Tags="&lt;abstract-algebra&gt;" AnswerCount="1" />
  <row Id="9003" PostTypeId="2" ParentId="9002" CreationDate="2010-11-05T01:41:52.723" Score="4" ViewCount="" Body="&lt;p&gt;Every such $f$ has the form $f(a, b) = g(a, b) + g(b, a)$ for some arbitrary function $g : \mathbb{Z} \times \mathbb{Z} \to \mathbb{R}$.  And no.  For example, take $f(a, b) = a^2 b^2$.  &lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-11-05T01:41:52.723" CommentCount="3" />
  <row Id="9004" PostTypeId="2" ParentId="8969" CreationDate="2010-11-05T01:44:33.147" Score="1" ViewCount="" Body="&lt;p&gt;First, you don't have the freedom to choose $\lambda$.  (Maybe you already realize that?)  In order for $P(X = x_i)$ to be a true probability distribution, you must have &#xA;$$\lambda = \left(\sum_{x_i} \frac{P(Y = x_i)}{x_i}\right)^{-1}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given that, if $Y$ is &lt;a href=&quot;http://en.wikipedia.org/wiki/Geometric_distribution&quot; rel=&quot;nofollow&quot;&gt;geometric&lt;/a&gt; with success probability $p$, and $P(X = x_i) = \lambda P(Y = x_i)/x_i$ then $X$ does have the &lt;a href=&quot;http://en.wikipedia.org/wiki/Logarithmic_distribution&quot; rel=&quot;nofollow&quot;&gt;logarithmic distribution&lt;/a&gt; with parameter $q = 1-p$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's why:  $P(Y = k) = (1-p)^{k-1}p$, for $k = 1, 2, \ldots$.  Thus &#xA;$$P(X = k) = \lambda \frac{P(Y = k)}{k} = \lambda \frac{q^{k-1} p}{k} = \frac{\lambda p}{q} \frac{q^k}{k}.$$&#xA;The expression on the right is the logarithmic probability mass function with $\lambda = \frac{-q}{p \ln p}$.&lt;/p&gt;&#xA;" OwnerUserId="2370" LastActivityDate="2010-11-05T01:44:33.147" />
  <row Id="9005" PostTypeId="1" AcceptedAnswerId="9006" CreationDate="2010-11-05T01:56:09.620" Score="0" ViewCount="36" Body="&lt;p&gt;I have a piecewise 2-input function that I would like to repeatedly tile diagonally across a grid. See this image. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/pLaFa.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So here, I know the equations of the pink, blue and green areas in the range 0 - 2. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My problem is that i'm having trouble finding a way to tile the function diagonally across like shown in the image by transforming the inputs. I tried using mod (x1 = x1 mod 2, x2 = x2 mod 2) to transform the input, and it worked somewhat, but it was wrong at grids (2-3, 1-2), (1-2, 2-3), (4-5, 3-4) and (3-4, 4-5). Any help is appreciated!&lt;/p&gt;&#xA;" OwnerUserId="1588" LastActivityDate="2010-11-05T02:17:51.743" Title="Repeating Tiled Piecewise Function Help" Tags="&lt;geometry&gt;&lt;functions&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="9006" PostTypeId="2" ParentId="9005" CreationDate="2010-11-05T02:17:51.743" Score="3" ViewCount="" Body="&lt;p&gt;Let $F$ be the original functions, $f$ the new one. Define&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f(x,y) = F(x-m,y-m)$ where $m = \min(\lfloor x \rfloor, \lfloor y \rfloor)$.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-05T02:17:51.743" CommentCount="1" />
  <row Id="9007" PostTypeId="1" CreationDate="2010-11-05T02:28:29.910" Score="3" ViewCount="168" Body="&lt;p&gt;The &lt;em&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Harmonic_mean&quot; rel=&quot;nofollow&quot;&gt;harmonic mean&lt;/a&gt;&lt;/em&gt; of a finite set of positive real numbers $\{x_1, x_2, \ldots, x_n\}$ is defined to be $$H(\{x_1, x_2, \ldots, x_n\}) = \frac{n}{\frac{1}{x_1} + \frac{1}{x_2} + \cdots + \frac{1}{x_n}}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;em&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Logarithmic_mean&quot; rel=&quot;nofollow&quot;&gt;logarithmic mean&lt;/a&gt;&lt;/em&gt; of two distinct positive real numbers $a$ and $b$ is defined to be $$L(a,b) = \frac{b - a}{\ln b - \ln a}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of the first applications of integration that students often see is the extension of the arithmetic mean of a finite set of real numbers to the arithmetic mean of a function $f(x)$ on a continuous interval $[a,b]$ via $\frac{1}{b-a} \int_a^b f(x) dx.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the same way, you can extend the harmonic mean so that it applies to a positive function $f(x)$ over a continuous interval $[a,b]$.  You get &#xA;$$\frac{b-a}{\int_a^b \frac{dx}{f(x)}} .$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, if you take $f(x) = x$ you obtain the harmonic mean of the continuous interval $[a,b]$.  This is &#xA;$$H([a,b]) = \frac{b-a}{\int_a^b \frac{1}{x}  dx} = \frac{b-a}{\ln b - \ln a} = L(a,b).$$&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;My question is this: Is there an intuitive reason why $H([a,b]) = L(a,b)$?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For comparison purposes, note that if $A$ denotes the arithmetic mean, then $A([a,b]) = A(a,b) = \frac{a+b}{2}$.&lt;/p&gt;&#xA;" OwnerUserId="2370" LastActivityDate="2010-11-05T04:02:30.157" Title="Harmonic mean and logarithmic mean" Tags="&lt;real-analysis&gt;&lt;intuition&gt;&lt;average&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="9008" PostTypeId="1" CreationDate="2010-11-05T02:43:20.417" Score="5" ViewCount="260" Body="&lt;p&gt;Let F be a strictly increasing function on S, a subset of the real line. If you know that F(S) is closed, prove that F is continuous.&lt;/p&gt;&#xA;" OwnerUserId="2319" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2010-11-05T04:28:40.897" LastActivityDate="2010-11-05T22:23:32.930" Title="If F is strictly increasing with closed image, then F is continuous" Tags="&lt;real-analysis&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" />
  <row Id="9009" PostTypeId="2" ParentId="8969" CreationDate="2010-11-05T02:48:44.750" Score="2" ViewCount="" Body="&lt;p&gt;if you rewrite the equation as $P(Y = x) = xP(X = x)/\lambda$, then the distribution of $Y$ is called the length-biased distribution for $X$. it arises, for example, if one has a bunch of sticks in a bag and reaches in and selects one at random - where the probability a particular stick is selected is proportional to its length. if the lengths of the sticks are realizations of the random variable $X$, the distribution of the length of the selected stick is that of $Y$.&lt;/p&gt;&#xA;" OwnerUserId="1547" LastActivityDate="2010-11-05T02:48:44.750" />
  <row Id="9010" PostTypeId="1" AcceptedAnswerId="9020" CreationDate="2010-11-05T02:57:52.910" Score="4" ViewCount="312" Body="&lt;p&gt;I hear people use these words relatively interchangeably.  I'd believe that any differentiable manifold can also be made into a variety (which data, if I understand correctly, implicitly includes an ambient space?), but it's unclear to me whether the only non-varietable manifolds should be those that don't admit smooth structures.  I'd hope there's more to it than that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've heard too that affine schemes are to schemes local coordinates are to manifolds, so maybe my question should be about schemes instead -- I don't even know enough to know...&lt;/p&gt;&#xA;" OwnerUserId="2159" LastActivityDate="2010-11-05T15:02:45.500" Title="What is the difference between a variety and a manifold?" Tags="&lt;geometry&gt;" AnswerCount="4" CommentCount="7" FavoriteCount="1" />
  <row Id="9011" PostTypeId="1" AcceptedAnswerId="9015" CreationDate="2010-11-05T02:58:44.103" Score="0" ViewCount="55" Body="&lt;p&gt;I've been wondering if/when it's possible to &quot;truncate&quot; a series.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Example 1&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For example, the closed form for the series of naturals is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{1}{(x-1)^2}$ = $1 + 2z + 3x^2 + \cdots$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &quot;truncated&quot; form is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{1-(n+1)x^n+n\cdot x^{n+1}}{(x-1)^2}$ = $1 + 2z + 3x^2 + \cdots + n\cdot x^{n-1}$&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Example 2&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The closed form of integers of the form $2^n$ is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{1}{1-2x}$ = $1 + 2x + 4x^2 + \cdots + (2x)^n + \cdots$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &quot;truncated&quot; form is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{1 - (2x)^n}{1 - 2x}$ = $1 + 2x + 4x^2 + \cdots + (2x)^{n-1}$&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The Search&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Could anyone help with finding appropriate resources that address &quot;truncating&quot; a series?  It would be great if there is some work that may include math that relates to finding truncated versions of series.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't have a very good background in math, so anything that doesn't require a lot of background would be very beneficial.  However, I am very interested in learning, and ANY resource that touches upon this topic would be very beneficial and interesting for me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd just like to know where I can learn more about this and related topics.  Again, I'm very interested in being able to produce a closed form or recurrence for a &quot;piece&quot; of a series.&lt;/p&gt;&#xA;" OwnerUserId="2626" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2011-06-03T20:06:15.730" LastActivityDate="2011-06-03T20:06:15.730" Title="Constructing finite versions of arbitrary series" Tags="&lt;sequences-and-series&gt;&lt;intuition&gt;" AnswerCount="1" />
  <row Id="9012" PostTypeId="2" ParentId="9008" CreationDate="2010-11-05T03:08:52.760" Score="2" ViewCount="" Body="&lt;p&gt;Here's an approach by contraposition.  Let $f$ be a strictly increasing function discontinuous at $x\in S$.  Then $f(x)\lt\lim_{y\to x+}f(y)$ or $f(x)\gt\lim_{y\to x-}f(y)$ (or both).  Suppose $f(x)\lt\lim_{y\to x+}f(y)$.  Then you can show that $\lim_{y\to x+}f(y)$ is in $\overline{f(S)}\setminus f(S)$, so $f(S)$ is not closed.  To see that the limit is in the closure of $f(S)$ is a straightforward unwinding of definitions. It's not in $f(S)$ because for every $z\lt x$, $f(z)\lt f(x)\lt\lim_{y\to x+}f(y)$, and for every $z\gt x$, $\lim_{y\to x+}f(y)\lt f(z)$.    (Similarly on the other side.  It may help to keep in mind that $\lim_{y\to x-}f(y)=\sup_{y\lt x}f(y)$ and $\lim_{y\to x+}f(y)=\inf_{y\gt x}f(y)$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a way that doesn't use contraposition (although there is a bit of contradiction).  Let $x$ be an element of $S$, and let $x_1,x_2,\ldots$ be an increasing sequence in $S$ converging to $x$.  Then $f(x_1),f(x_2),\ldots$ is an increasing sequence bounded above by $f(x)$, and hence it converges.  Since $f(S)$ is closed, there is a $z\in S$ such that $f(x_n)\to f(z)$ as $n\to \infty$.  I claim that $z=x$.  If $z$ were bigger than $x$, then we'd have $f(x_n)\leq f(x)\lt f(z)$ for all $n$, making the convergence impossible.  If $z$ were smaller than $x$, we'd have $z$ smaller than $x_n$ for some $n$, so $f(z)\lt f(x_n)\leq f(x_{n+1})\leq\cdots$, again making the convergence impossible.  So $z=x$ as claimed.  This implies that the left-hand limit of $f$ at $x$ exists and equals $f(x)$.  Similarly on the right, so $f$ is continuous.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2010-11-05T03:58:39.427" LastActivityDate="2010-11-05T03:58:39.427" CommentCount="8" />
  <row Id="9013" PostTypeId="2" ParentId="9010" CreationDate="2010-11-05T03:09:17.540" Score="2" ViewCount="" Body="&lt;p&gt;If you are French, a variety &lt;em&gt;is&lt;/em&gt; a manifold.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If not, the connection is a bit more subtle.  The correct analogue of a manifold is a scheme of finite type over a field.  That is, it has a compatible covering by ring spectra (which is similar to the requirement that a manifold have a covering by compatible copies of euclidean space), and it lives over a field (the base space is an honest &quot;point&quot;). Here, I am being intentionally vague about &quot;compatibility&quot;, since this would lead me into a discussion of &lt;em&gt;sheaves&lt;/em&gt;, which I would rather not discuss unless you feel it's necessary. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The classical notion of a variety much more resembles the classical notion of a manifold (see, for instance, Milnor's book &quot;Topology from a differentiable viewpoint&quot;).  The classical notion of a manifold is simply a locally closed subset of $\mathbf{P}^n(\mathbf{R})$ (or just $\mathbf{R}^n$, although looking at locally-closed subsets of projective space makes the analogy that much stronger), and the notion of a variety (quasi-projective, I guess) comes from considering locally closed subsets of a projective space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lastly, there is a theorem of Serre called GAGA (Géométrie algébrique et géométrie analytique), which says that in nice cases, we can &quot;replace&quot; complex varieties with complex-analytic manifolds to &quot;do&quot; cohomology and get a better picture of the real underlying space.  (Here, by &lt;em&gt;complex variety&lt;/em&gt;, we simply mean finite-type schemes over $\mathbf{C}$)&lt;/p&gt;&#xA;" OwnerUserId="126" LastEditorUserId="126" LastEditorDisplayName="" LastEditDate="2010-11-05T15:02:45.500" LastActivityDate="2010-11-05T15:02:45.500" CommentCount="8" />
  <row Id="9014" PostTypeId="2" ParentId="8940" CreationDate="2010-11-05T03:15:39.630" Score="1" ViewCount="" Body="&lt;p&gt;Think of this square n x n matrix as a transformation T on a vector space V. This transformation sends $V^{n} \rightarrow V^{n}$. Now, if the determinant of a matrix is equal to 0, we know it isn't invertible. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since our vector space is finite-dimensional, a transformation that has a non-invertible matrix representation is neither onto nor one-to-one. However, if all our columns were linearly independent vectors in $V^n$, they would span $V^n$ and T would be onto. One of the columns must therefore be a linear combination of the others. &lt;/p&gt;&#xA;" OwnerUserId="647" LastActivityDate="2010-11-05T03:15:39.630" />
  <row Id="9015" PostTypeId="2" ParentId="9011" CreationDate="2010-11-05T03:21:36.543" Score="3" ViewCount="" Body="&lt;p&gt;Truncating to the first $n$ terms is the same as taking the Hadamard product with $\frac{1 - x^n}{1 - x}$, so if you can compute the corresponding integral then you're home free.  In the special case that the original series is rational you can also do it using partial fraction decomposition, and then you are actually done by using appropriate generalizations of the two formulas you've already listed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general I see no reason for truncations to have nice closed forms.  In particular I see no reason for the truncations of a series like $- \ln (1 - x)$ to have nice closed forms, since that would imply that the harmonic series has a nice closed form.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-11-05T03:21:36.543" CommentCount="2" />
  <row Id="9016" PostTypeId="2" ParentId="8966" CreationDate="2010-11-05T03:36:04.160" Score="0" ViewCount="" Body="&lt;p&gt;The algebraic $\frac{1}{12}(17 + \sqrt{97})$ is not a root of the equation&#xA;\begin{eqnarray}&#xA;-2 x^3 + 10 x^2 - 17 x + 8 = (2 x^2) (5 x - x^3)^{1/3}&#xA;\end{eqnarray}&#xA;Plugging it in, you find that the left hand side is real and equal to &#xA;\begin{eqnarray}&#xA;\tfrac{1}{216}(-149 - 37 \sqrt{97}) = -2.37689 \dots&#xA;\end{eqnarray}&#xA;The right side is&#xA;\begin{eqnarray}&#xA;\tfrac{1}{432} (\tfrac{1}{2}( 595 - 61 \sqrt{97})^{1/3} (17 + \sqrt{97})^2 = 1.18844 \dots + i 2.05845 \dots&#xA;\end{eqnarray}&#xA;Note: $595 &amp;lt; 61 \sqrt{97}$. I think the ambiguity lies in the fact that we have not used the third-roots of unity. Numerical computations aside, just plot the two functions. The RHS is a positive function defined only in the I and II quadrants. The LHS is cubic. There is only one real intersection point.&lt;/p&gt;&#xA;" OwnerUserId="2720" LastEditorUserId="2720" LastEditorDisplayName="" LastEditDate="2010-11-05T04:09:41.780" LastActivityDate="2010-11-05T04:09:41.780" CommentCount="4" />
  <row Id="9017" PostTypeId="2" ParentId="9010" CreationDate="2010-11-05T03:37:44.330" Score="2" ViewCount="" Body="&lt;p&gt;A variety is usually defined to be the zero set of some polynomials. I don't know much about this, but I think people usually call manifolds that can be realized as varieties &quot;algebraic manifolds&quot;. Take for example the polynomial $x^2+y^2-1$, then the zero set of this is $x^2+y^2=1$, or the circle $S^1$. So that is an algebraic manifold.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There can be varieties that are not manifolds, for instance, $y^2-x^2(x+1)=0$ is a &quot;nodal cubic&quot; and so it has a singularity at $(0,0)$. It can't be a manifold because it looks like &quot;X&quot; a cross at the origin so is not homeomorphic locally to $\mathbb{R}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a very useful theorem, sometimes called the Regular Level Set Theorem, which says that if you have a smooth map $M\to N$, then the level set of a regular value is a smooth manifold. So in this particular case, our polynomial $f:\mathbb{R}^n \to \mathbb{R}$ is a smooth map, and we care about whether or not zero is a regular value. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This amounts to checking whether or not the matrix $(\frac{\partial f}{\partial x_1}(c), \ldots , \frac{\partial f}{\partial x_n}(c))$ has rank 1 at all $c$ such that $f(c)=0$. i.e. if at least one of the partial derivatives is non-zero. This can easily be extended to work if your variety is defined by a bunch of polynomials $\mathbb{R}^n\to\mathbb{R}^m$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This has a name, called the Jacobian criterion, and for varieties it says that if the Jacobian has full rank, then the variety is non-singular. But these two things are equivalent (Jacobian criterion for non-singular variety and the level set theorem to be a smooth manifold). Thus every non-singular variety is a smooth manifold (Just to be clear, this was considering &quot;variety&quot; to mean a zero set of a collection of polynomials over $\mathbb{R}$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So to reiterate, some varieties are manifolds (if the defining polynomials satisfy a certain condition on partial derivatives) and some are not. Is every smooth manifold a variety? I think not, but it seems harder. A quick Google search turns up a result that seems to say every smooth compact manifold is algebraic.&lt;/p&gt;&#xA;" OwnerUserId="2000" LastActivityDate="2010-11-05T03:37:44.330" CommentCount="1" />
  <row Id="9018" PostTypeId="2" ParentId="8975" CreationDate="2010-11-05T03:48:55.767" Score="3" ViewCount="" Body="&lt;p&gt;Here's how I would explain limits:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Think of the function $f(x) = x^{2}$. We might want to ask ourselves: &quot;What is $\lim_{x \rightarrow 3} f(x)$?&quot; Put simply: As $x$ gets near 3, where happens to our function? Clearly, our function goes to 9, which is $3^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One must understand, however, that a limit is a much more subtle argument. In our case, we could just &quot;plug in&quot; 3. But suppose we wanted to take $\lim_{x \rightarrow 0} \frac{1}{x}$ or $\lim_{x \rightarrow 0} \frac{\sin(x)}{x}$. In these cases, we cannot simply &quot;plug in&quot; our value for x, and we must understand what a limit really is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The arrow $\rightarrow$ in the limit sign means &quot;approaching.&quot; In our first example, we wanted to find out what happens as x approaches 3. Let's look at this ourself:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose $x=1$, then $x^2=1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose $x=2$, then $x^2=4$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose $x=2.5$, then $x^2=6.25$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose $x=2.9$, then $x^2=8.41$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose $x=2.99$, then $x^2=8.9401$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As x gets closer and closer to 3, $x^2$ (our function) gets closer and closer to 9.&#xA;We can also come from the other direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose $x=5$, then $x^2=25$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose $x=4$, then $x^2=16$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose $x=3.5$, then $x^2=12.25$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose $x=3.1$, then $x^2=9.61$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose $x=3.01$, then $x^2=9.0601$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, let me introduce the terminology of epsilons $\epsilon$ and deltas $\delta$. The delta is how far away we are from our chosen limit value for x. In this case, the delta was how far away we were from 3. When we set x=3.5, delta was equal to 0.5. When x=3.1, delta was equal to 0.1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Correspondingly, epsilon was how far away we were from the value of the limit: 9. When we set x=3.1, and our delta was 0.1, we got $f(x) = 9.61$, so the epsilon was 0.61.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And therein lies the fundamental idea of a limit. We can make the epsilon smaller simply by shrinking the delta. The closer your x is to 3, the closer your f(x) is to 9. If you wanted to be within 0.001 of 9 (i.e. $\epsilon = 0.001$), you'll need to be really close to x, and have a really small $\delta$. To be precise, $\delta = 0.00017$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We therefore say that the limit of $f(x) = x^2$ as x approaches 3 is equal to 9, because we can make the output of the function be as close to 9 as we want, simply by constricting the input to be close to 3. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/Bael9.gif&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The above picture provides a great visual description. We have some curvy function $f(x)$, and we want to say that the limit at $x=a$ is equal to $L$. This can be shown because if we narrow the lines $a-\delta$ and $a+\delta$, thereby narrowing our input, we narrow the lines $L+\epsilon$ and $L-\epsilon$, constricting our output. We therefore must conclude that the limit is $L$, because the value of our function approaches $L$ and gets closer and closer to it the closer we get to $x=a$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope this explanation has helped. &lt;/p&gt;&#xA;" OwnerUserId="647" LastEditorUserId="647" LastEditorDisplayName="" LastEditDate="2010-11-05T04:56:10.767" LastActivityDate="2010-11-05T04:56:10.767" CommentCount="4" />
  <row Id="9019" PostTypeId="2" ParentId="8992" CreationDate="2010-11-05T03:49:50.617" Score="3" ViewCount="" Body="&lt;p&gt;There are a couple of ways to think about it; as acarchau says, since you are concerned only about the limit, and the &quot;tail&quot; has the same limit, it is enough to consider the tail. Equivalently: &lt;em&gt;the limit of a sequence depends only on what it is doing &quot;eventually&quot;&lt;/em&gt;. Any amount of &quot;weird&quot; stuff that happens up to a finite bound $N$ is irrelevant (that is, if the sequence is &quot;nice&quot; for all $n\geq N$, then that is enough, because as far as the limit is concerned, we don't care what happens at $n=1$, $n=2,\ldots,n=N-1$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or, consider the sequence of functions $\{g_n\}$, where $g_n(x) = f_{N+n}(x)$. Then the hypothesis of Lebesgue's Theorem hold exactly as written for $g_n(x)$, so we know that we can get the conclusion for the $g_n$. Now rewrite $g_n$ as $f_{N+n}$, and do a change of variable $m=N+n$; then $m\to\infty$ as $n\to\infty$, so the limit of $f_{N+n}$ as $n\to\infty$ is the same as the limit of $f_m$ as $m\to\infty$, giving the result.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But all in all, it's always the same moral: what happens at finitely many terms &quot;at the beginning&quot; does not affect the end behavior of a sequence. Only the &quot;eventual&quot; behavior matters.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-11-05T03:49:50.617" CommentCount="1" />
  <row Id="9020" PostTypeId="2" ParentId="9010" CreationDate="2010-11-05T03:50:24.993" Score="5" ViewCount="" Body="&lt;p&gt;In English (as opposed to French, in which language variety and manifold are synonyms) the word &lt;em&gt;variety&lt;/em&gt; is short for &lt;em&gt;algebraic variety&lt;/em&gt;.  The main differences, then, between (algebraic) varieties and (smooth) manifolds are that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(i) Varieties are cut out in their ambient (affine or projective) space as the zero loci of polynomial functions, rather than simply as the zero loci of smooth functions.   This gives them a more rigid structure.  (Here I am thinking just of quasi-projective varieties; there are objects that people call varieties which can't be immersed into projective space, but there is no need to think about them when you are just learning the subject.  Also, a manifold need not be regarded as lying in an ambient Eulcidean space, but can always be immersed into one, and can then be thought of as being cut out as the zero locus of smooth&#xA;functions.)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rigidity of varieties is reflected in the definition of isomorphism: we define two varieties to be isomorphic if we can find polynomial maps giving rise to mutually inverse bijections from one to the other, while two manifolds are isomorphic (i.e. diffeomorphic) if we can find smooth maps giving rise to mutually inverse bijections between them.   It turns out, for example, that&#xA;the only diffeomorphism invariant of a compact connected surface is its genus $g$, while if we look at smooth connected  projective curves over the complex numbers (which, when we forget the variety structure and just think of them as manifolds, &lt;em&gt;are&lt;/em&gt;&#xA;compact connected surfaces --- note that one complex dimension gives two real&#xA;dimensions) then the genus is not a complete invariant.  For a fixed genus $g \geq 2$,&#xA;there is a $6g-6$-dimensional family of non-isomorphic curves of genus $g$.&#xA;(When $g = 1$ there is a $2$-dimensional family, and when $g = 0$, the&#xA;variety structure is actually uniquely determined.  Also, by &quot;dimension&quot; here&#xA;I mean real dimension; but these families have their own natural algebraic variety structures, of half the dimension --- i.e. there is a $3 g - 3$ dimensional variety parameterizing isomorphism classes of genus $g$ curves&#xA;when $g \geq 2$.  Again the halving of dimension reflects the difference between real and complex dimension.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(ii)  Varieties can admit singularities, whereas we stipulate that manifolds be non-singular (i.e. locally Euclidean).   Here it is useful to think about the fact that the critical locus of a (collection of) smooth function(s) can be pretty nasty, and so if we consider the zero loci of smooth functions and allow singularities, we will allow extremely nasty singularities.  On the other hand,&#xA;the critical locus of a (collection) of polynomial(s) is not so bad (e.g. it is always of codimension at least one in the zero locus), and so allowing singularities in the theory turns out to be okay (and in fact to be more than okay; it turns out to be one of the more powerful features of algebraic geometry).&lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2010-11-05T03:50:24.993" CommentCount="16" />
  <row Id="9021" PostTypeId="2" ParentId="8990" CreationDate="2010-11-05T03:54:38.587" Score="1" ViewCount="" Body="&lt;p&gt;The function is majority (with tie-breaking): the expression is true iff at least two of $A,B,C,D$ are true. This follows from your last term by opening up the parentheses, and can also be checked directly.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-05T03:54:38.587" />
  <row Id="9022" PostTypeId="2" ParentId="9007" CreationDate="2010-11-05T04:02:30.157" Score="3" ViewCount="" Body="&lt;p&gt;The harmonic mean can be generalized to an arbitrary invertible function $w$: define the mean of $x_1,\ldots,x_n$ to be&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$w^{-1}\left(\frac{w(x_1) + \cdots + w(x_n)}{n}\right)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The logarithmic mean can also be generalized re its mean value interpretation for an arbitrary function $f$ such that $f'$ is invertible: define the mean of $x,y$ as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(f')^{-1}\left(\frac{f(x) - f(y)}{x - y}\right)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can generalize your observation by taking $w = f'$. The corresponding mean, before taking $w^{-1}$, is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{1}{y-x}\int_x^y f'(t) \, dt = \frac{f(y)-f(x)}{y-x}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and so the (former) mean of the interval with weight $f'$ is the same as the (latter) mean with respect to $f$.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-05T04:02:30.157" CommentCount="5" />
  <row Id="9023" PostTypeId="1" CreationDate="2010-11-05T04:18:43.593" Score="2" ViewCount="96" Body="&lt;p&gt;This may be a question whose answer is lost in the mists of time, but why is the elliptical integral of the first kind denoted as $F(\pi/2,m)=K(m)$ when that of the second kind has $E(\pi/2,m)=E(m)$? It's not very consistent! Aside from convention, is there anything stopping us from rationalising these names a little?&lt;/p&gt;&#xA;" OwnerUserId="2961" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-05T11:09:47.847" LastActivityDate="2010-11-05T11:09:47.847" Title="Inconsistent naming of elliptic integrals" Tags="&lt;soft-question&gt;&lt;notation&gt;&lt;special-functions&gt;&lt;math-history&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="9024" PostTypeId="2" ParentId="9023" CreationDate="2010-11-05T04:22:13.090" Score="2" ViewCount="" Body="&lt;p&gt;This discrepancy seems to come from the various conventions in defining the nome $q$ of complex lattice $\Lambda_{\tau} = \mathbb{Z} \oplus \tau \mathbb{Z}$ as one of the four $e^{2 \pi i \tau}$, $e^{\pi i \tau}$, $e^{2 i \tau}$ or $e^{i \tau}$, where $\tau \in \mathbb{H}$ is the lattice parameter. Investigate the vast literature on theta functions, and you'll see exactly what I mean by the problem of too many &quot;&lt;em&gt;conventions&lt;/em&gt;&quot;.&lt;/p&gt;&#xA;" OwnerUserId="2720" LastEditorUserId="2720" LastEditorDisplayName="" LastEditDate="2010-11-05T07:26:25.257" LastActivityDate="2010-11-05T07:26:25.257" CommentCount="1" />
  <row Id="9025" PostTypeId="2" ParentId="8977" CreationDate="2010-11-05T04:32:39.590" Score="0" ViewCount="" Body="&lt;p&gt;I'm going to ignore arc AC and address what determines arcs FD and BE based on the points A, B, C, D, E, and F.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If two circles with centers X and Y are tangent at P, then X, Y, and P are collinear.  So, line AF and line CD both pass through the center of the circle containing arc FD and line AB and line CE both pass through the center of the circle containing arc BE.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have coordinates for all of these points, you can write equations of the line and algebraically find their points of intersection.  Given the coordinates of the centers of the circles, you can determine their radii.  Given the coordinates of the centers of the circles and the endpoints of the arcs, you can determine the measure of the central angles that subtend the arcs.  From the radii and central angles, you can determine the arc lengths.  (It's hard to be more specific without introducing a bunch of variables, but I can add more detail if you are not sure how to carry out any of these steps.)&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-11-05T04:32:39.590" />
  <row Id="9026" PostTypeId="1" CreationDate="2010-11-05T04:58:40.620" Score="6" ViewCount="262" Body="&lt;p&gt;I tried to do some math in &lt;a href=&quot;http://www.geekality.net/?p=751&quot; rel=&quot;nofollow&quot;&gt;a blog post&lt;/a&gt; of mine and came to one with a floor function. I wasn't sure how to deal with it so I just ignored it, and then added the ceiling function in my final equation as that seemed to give me the result I wanted. I'm wondering what is the correct way of handling these functions in equations?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I did was this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xA;G(n) &amp;amp;= \left\lfloor n\log{\varphi}-\dfrac{\log{5}}{2}\right\rfloor+1 \\&#xA;n\log{\varphi} &amp;amp;= G(n)+\dfrac{\log{5}}{2}-1 \\&#xA;n &amp;amp;= \left\lceil\dfrac{G(n)+\dfrac{\log{5}}{2}-1}{\log\varphi}\right\rceil&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How should have done this in a correct way? How do I work with the ceiling and floor functions when I shuffle around with equations?&lt;/p&gt;&#xA;" OwnerUserId="3147" LastActivityDate="2010-11-05T17:51:20.293" Title="How do you handle the floor and ceiling function in an equation?" Tags="&lt;algebra&gt;&lt;functions&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="9027" PostTypeId="2" ParentId="9026" CreationDate="2010-11-05T05:26:29.833" Score="5" ViewCount="" Body="&lt;p&gt;You can replace $\lfloor x \rfloor$ with $x - \theta$, where $\theta \in [0,1)$ is some unknown quantity. Similarly, $\lceil x \rceil = x + \theta$ (a different $\theta$ within the same range).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another helpful identity is $\lfloor x \rfloor + n = \lfloor x + n \rfloor$ for any integer $n$.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-05T05:26:29.833" />
  <row Id="9028" PostTypeId="2" ParentId="9026" CreationDate="2010-11-05T05:29:54.760" Score="6" ViewCount="" Body="&lt;p&gt;Your final expression gives you the number you want.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to your blog post, you're looking for the smallest integer $n$ (i.e., the &quot;first Fibonacci number with 1000 digits&quot;) that satisfies &#xA;$$G(n) = \left\lfloor n \log \varphi - \frac{\log 5}{2} \right\rfloor + 1.$$&#xA;There may, of course, be more than one integer $n$ for which this is true.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By definition of the floor function, the values of $n$ that satisfy this are the values that satisfy $$G(n) - 1 \leq n \log \varphi - \frac{\log 5}{2} &amp;lt; G(n),$$&#xA;which, since $\log \phi &gt; 0$, are the values that satisfy&#xA;$$\frac{G(n) + \frac{\log 5}{2}}{\log \varphi} - \frac{1}{\log \varphi} \leq n  &amp;lt; \frac{G(n) + \frac{\log 5}{2}}{\log \varphi}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $\frac{1}{\log \varphi} \approx 4.78$, there are either four or five integers in this interval.  But the smallest one is obtained by taking the ceiling of the lower endpoint of the interval; i.e.,&#xA;$$\left\lceil\frac{G(n) + \frac{\log 5}{2} - 1}{\log \varphi}\right\rceil.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Incidentally, this argument also apparently shows that there are either four or five Fibonacci numbers that have a given number of digits.  (Except in the single-digit case, where there are six (not counting 0).  But your formula for $G(n)$ doesn't hold when $n=1$, so we shouldn't expect this calculation to be true in the single-digit case anyway.)&lt;/p&gt;&#xA;" OwnerUserId="2370" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-11-05T17:51:20.293" LastActivityDate="2010-11-05T17:51:20.293" />
  <row Id="9029" PostTypeId="2" ParentId="8990" CreationDate="2010-11-05T05:38:12.020" Score="2" ViewCount="" Body="&lt;p&gt;To go along with my comment above, the minimal form is what you arrived at. Using a Karnaugh map, you will get $AB+CD+AC+AD+BC+BD$, which factors to your answer.&lt;/p&gt;&#xA;" OwnerUserId="1016" LastActivityDate="2010-11-05T05:38:12.020" />
  <row Id="9030" PostTypeId="2" ParentId="9026" CreationDate="2010-11-05T05:41:31.643" Score="2" ViewCount="" Body="&lt;p&gt;Observe that&#xA;\begin{eqnarray}&#xA;G(n) = \left \lfloor n \log \varphi - \log \sqrt{5} \right \rfloor + 1 = \left \lceil n \log \varphi - \log \sqrt{5} \right \rceil&#xA;\end{eqnarray}&#xA;and write&#xA;\begin{eqnarray}&#xA;\left \lceil \frac{G(n)}{\log \varphi} + \log_{\varphi} \sqrt{5} \right \rceil &amp;amp; = &amp;amp; \left \lceil \tfrac{1}{\log \varphi} \left \lceil n \log \varphi - \log \sqrt{5}  \right \rceil + \log_{\varphi} \sqrt{5} \right \rceil = n.&#xA;\end{eqnarray}&lt;/p&gt;&#xA;" OwnerUserId="2720" LastEditorUserId="2720" LastEditorDisplayName="" LastEditDate="2010-11-05T06:09:30.330" LastActivityDate="2010-11-05T06:09:30.330" />
  <row Id="9031" PostTypeId="2" ParentId="8975" CreationDate="2010-11-05T05:56:06.867" Score="-1" ViewCount="" Body="&lt;p&gt;We say that the limit of $f$ at $a$ is $\ell$ means: For any $x$ in the domain of $f$ other than $a$ itself, $$\forall (\varepsilon &gt;0) \exists (\delta &gt;0) d(x,a)&amp;lt;\delta \Rightarrow d(f(x),\ell)&amp;lt;\varepsilon,$$ where $d$ is the distance function for the metric space $M$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is, for any open neighborhood of $U$ of $\ell$, we can find neighborhood $V$ of $a$ such that  $f(V)\subseteq U$.  Indeed, there is some subtlety here about the domain of $f$ (in particular, $f$ need not be defined &lt;em&gt;at&lt;/em&gt; $a$, only on a subspace whose closure contains $a$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Specifically, let $X$ and $Y$ be topological spaces, and let $\Omega\subseteq X$ be a subspace such that $a\in Cl_X(\Omega-\{a\})$.  Then we say that $\ell$ is &lt;strong&gt;a&lt;/strong&gt; limit point of $f$ at $a$ (in this generality, limits need not be unique) provided that for any open neighborhood $U$ of $\ell$, there exists an open neighborhood $V\subseteq X$ of $a$ such that $f(V\cap \Omega)\subseteq U$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The point here is that no matter how small a neighborhood we give around $\ell$, we can always find a small enough neighborhood of $a$ such that $f$ maps this neighborhood of $a$ into the neighborhood of $\ell$.&lt;/p&gt;&#xA;" OwnerUserId="126" LastEditorUserId="126" LastEditorDisplayName="" LastEditDate="2010-11-05T07:54:54.450" LastActivityDate="2010-11-05T07:54:54.450" CommentCount="4" />
  <row Id="9032" PostTypeId="1" CreationDate="2010-11-05T06:11:44.960" Score="3" ViewCount="168" Body="&lt;p&gt;Suppose I have two tables, each of unknown size, and I'd like to estimate the average of their true sizes. I hire 2 contractors: one guarantees good precision (i.e., her measurement normally-distributed about the true value, with a standard deviation of 5mm), while the other is a dimwit (i.e., his measurement is also unbiased, but with a standard deviation of 100mm). What's the optimal way to combine the two measurements to form a final estimate of their true average size? The formal way to ask this question is, &quot;Given a sample from a normally-distributed random variable with known variance, and another sample from a second normally-distributed random variable with known (but potentially different) variance, what is the best guess as to the expected value of the average of the two random variables?&quot; Is the answer &quot;Just average them.&quot;? Ideally, I want to prove the answer, whatever it is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: I'm not sure if it was clear, but one contractor measures table 1, while the other contractor measures table 2. &lt;/p&gt;&#xA;" OwnerUserId="3298" LastEditorUserId="3298" LastEditorDisplayName="" LastEditDate="2010-11-09T04:54:36.630" LastActivityDate="2010-11-09T05:20:47.153" Title="Optimally combining samples to estimate averages" Tags="&lt;probability&gt;&lt;probability-theory&gt;&lt;statistics&gt;&lt;optimization&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="0" />
  <row Id="9035" PostTypeId="1" AcceptedAnswerId="9046" CreationDate="2010-11-05T06:57:16.197" Score="8" ViewCount="180" Body="&lt;p&gt;Im giving a small talk for a combinatorics class on the &lt;a href=&quot;http://garden.irmacs.sfu.ca/?q=op/erdos_szekeres_conjecture&quot; rel=&quot;nofollow&quot;&gt;Erdos-Szekeres conjecture&lt;/a&gt; regarding the happy ending problem (the paper is focused on recent work regarding the conjecture).  I always find that when the audience is not up for a long technical proof, and they aren't familiar with the field (in this case combinatorial geometry) that it is important to provide strong motivation for why the problem is important, and perhaps mention a couple interesting related results that have applications in other fields, or to other problems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the Happy Ending problem, I am aware of connections to the $n$-hole problem for points in general position, Ramsey theory (via the original paper by erdos and szekeres in 1935), and the generalization of the E&amp;amp;S theorem for convex bodies.  But Im not sure if these are good things to mention since the class does not know any Ramsey theory and convex bodies are probably too abstract to mention offhand.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So my question is: does anyone know of other interesting related results to the &lt;a href=&quot;http://en.wikipedia.org/wiki/Happy_Ending_problem&quot; rel=&quot;nofollow&quot;&gt;Happy Ending Problem, and the corresponding theorem and conjecture.&lt;/a&gt; Or, can you tell me succinctly why or why not this conjecture is important in mathematics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I will accept anything along the lines of similar geometrically or combinatorially, directly related, or mentioned in a paper involving the E&amp;amp;S conjecture.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The original paper from 1935 is a goldmine of things like the ordered pigeonhole principle, but Im interested in things I havent read yet.&lt;/p&gt;&#xA;" OwnerUserId="565" LastActivityDate="2010-11-05T11:23:15.153" Title="Interesting results related to The Happy Ending Problem" Tags="&lt;geometry&gt;&lt;combinatorics&gt;&lt;discrete-geometry&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="9036" PostTypeId="2" ParentId="8501" CreationDate="2010-11-05T07:12:59.970" Score="3" ViewCount="" Body="&lt;p&gt;Without sounding biased in any way, I would say Cambridge/Boston is a good choice for you. In the particular order of funded research/department size there is Harvard University, MIT, Boston University, Boston College, Northeastern University, Brandeis University, Tufts University, Bently University, University of Massachusetts at Boston, Curry College, Eastern Nazarene College, Pine Manor College, Hellenic College, Lesley University, Wheelock College, Lasell College, Simmons University, Cambridge College and Bunker Hill Community College (and many, many more) within the metropolis. See &lt;a href=&quot;http://en.wikipedia.org/wiki/List_of_colleges_and_universities_in_metropolitan_Boston&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/List_of_colleges_and_universities_in_metropolitan_Boston&lt;/a&gt; for a complete list.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A number of these institutions offer extension programs (with open enrollment and classes in the evening or weekends) suitable for life-long learners and aspiring amateur mathematicians. For example, the &lt;em&gt;Masters for Mathematics Teaching Program&lt;/em&gt; at Harvard University offers courses in all major mathematics subject areas, taught by many instructors which hold separate positions in the university (like adjunct/junior faculty, preceptors, senior lecturers, post-doctoral or teaching fellows and even a senior graduate student).&lt;/p&gt;&#xA;" OwnerUserId="2720" LastEditorUserId="2720" LastEditorDisplayName="" LastEditDate="2011-02-02T20:50:13.633" LastActivityDate="2011-02-02T20:50:13.633" CommentCount="2" />
  <row Id="9037" PostTypeId="1" CreationDate="2010-11-05T07:31:37.510" Score="3" ViewCount="208" Body="&lt;p&gt;Is there a surjective map between the (class of) ordinal numbers On and the set No (Conway's surreal numbers) and is it constructable, In Conway's system we have for example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\omega_0 = &amp;lt; 0,1,2,3,... | &gt; $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\epsilon = &amp;lt; 0 | 1, 1/2, 1/4, 1/8, ... &gt; $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(where $\epsilon$ is not the first uncountable ordinal, but the &quot;reciprocal&quot; of $\omega_0$). My question is thus: can you device this for every ordinal, or does Conway's system On eventually run &quot;out of space&quot;).&lt;/p&gt;&#xA;" LastActivityDate="2010-11-05T22:40:45.337" Title="surreal and ordinal numbers" Tags="&lt;set-theory&gt;" AnswerCount="2" />
  <row Id="9038" PostTypeId="2" ParentId="9037" CreationDate="2010-11-05T07:52:09.437" Score="3" ViewCount="" Body="&lt;p&gt;I''m not entirely sure what your question is. In Conway's&#xA;notation &lt;strong&gt;On&lt;/strong&gt; denotes the ordinal numbers (and &lt;strong&gt;No&lt;/strong&gt; denotes the set of&#xA;all surreal Numbers). Basically the elements of &lt;strong&gt;On&lt;/strong&gt; are just von Neumann&#xA;ordinals. In von Neumann's construction an ordinal $\alpha$ is just&#xA;the set $\{\beta:\beta&amp;lt;\alpha\}$ of all preceding ordinals and this&#xA;corresponds to the Conway ordinal $\langle\beta:\beta&amp;lt;\alpha\mid\ \rangle$.&#xA;So the von Neumann and Conway ordinals naturally correspond.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But maybe your question is whether there is a one-to-one correspondence&#xA;between the ordinals and the whole class of surreal numbers. These&#xA;are both proper classes with respect to ZFC. There are subtleties associated&#xA;to defining bijections between proper classes, which I don't really&#xA;understand, but given a strong version of AC I believe that any&#xA;two proper classes can be paired off. (Maybe an experienced set theorist&#xA;could say more?)&lt;/p&gt;&#xA;" OwnerUserId="226" LastEditorUserId="226" LastEditorDisplayName="" LastEditDate="2010-11-05T08:55:34.173" LastActivityDate="2010-11-05T08:55:34.173" CommentCount="2" />
  <row Id="9039" PostTypeId="2" ParentId="9008" CreationDate="2010-11-05T08:05:17.230" Score="2" ViewCount="" Body="&lt;p&gt;Let $f$ be any strictly increasing &lt;s&gt;(not necessarily strictly)&lt;/s&gt; function on $S$.  To show that $f$ is continuous on $S$, it is enough to show that it is continuous at $x$ for every $x \in S$.  If $x$ is an isolated point of $S$, every function is continuous at $x$, so assume otherwise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The key here is that monotone functions can only be discontinuous in a very particular, and simple, way.  Namely, the one-sided limits $f(x-)$ and $f(x+)$ always exist (or rather, the first exists when $x$ is not left-isolated and the second exists when $x$ is not right-isolated): it is easy to see for instance that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f(x-) = \sup_{y &amp;lt; x, \ y \in S} f(y)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore a discontinuity occurs when $f(x-) \neq f(x)$ or $f(x+) \neq f(x)$.  In the first case we have that for all $y &amp;lt; x$, $f(y) &amp;lt; f(x-)$ and for all $y \geq x$, $f(y) &gt; f(x-)$.  Therefore $f(x-)$ is not in $f(S)$.  But by the above expression for $f(x-)$, it is certainly a limit point of $f(S)$.  So $f(S)$ is not closed.  The other case is similar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other nice, related properties of monotone functions include: a monotone function has at most countably many points of discontinuity and a monotone function is a &lt;a href=&quot;http://en.wikipedia.org/wiki/Regulated_function&quot; rel=&quot;nofollow&quot;&gt;regulated function&lt;/a&gt; in the sense of Dieudonné.  In particular the theoretical aspects of integration are especially simple for such functions.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Added&lt;/b&gt;: As Myke notes in the comments below, the conclusion need not be true if $f$ &#xA;is merely increasing (i.e., $x_1 \leq x_2$ implies $f(x_1) \leq f(x_2)$).  A counterexample &#xA;is given by the characteristic function of $[0,\infty)$.&lt;/p&gt;&#xA;" OwnerUserId="299" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2010-11-05T22:23:32.930" LastActivityDate="2010-11-05T22:23:32.930" CommentCount="5" />
  <row Id="9040" PostTypeId="2" ParentId="2949" CreationDate="2010-11-05T08:15:36.600" Score="11" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Exotic Spheres&lt;/strong&gt;. Kervaire and Milnor's proof that there exists 27 distinct differentiable $7$-manifolds that are homeomorphic, but not diffeomorphic, to the standard $7$-sphere (giving $28$ differentiable structures for $S^{7}$).&lt;/p&gt;&#xA;" OwnerUserId="2720" LastEditorUserId="2720" LastEditorDisplayName="" LastEditDate="2011-03-02T14:52:27.747" LastActivityDate="2011-03-02T14:52:27.747" CommunityOwnedDate="2010-11-05T08:15:36.600" CommentCount="0" />
  <row Id="9041" PostTypeId="2" ParentId="8980" CreationDate="2010-11-05T08:15:55.693" Score="5" ViewCount="" Body="&lt;p&gt;This refers to Ryan's answer, but was too long for a comment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was getting confused for a while too, but I think I understand now.&#xA;To avoid misunderstandings, the Tait–Bryan angles should be called&#xA;heading/elevation/bank instead of yaw/pitch/roll. The heading angle&#xA;refers to rotations about the $z$ axis in space, &lt;em&gt;not&lt;/em&gt; about the yaw&#xA;axis which by definition is perpendicular to the &quot;body-wing-plane&quot; of&#xA;the airplane, so that &quot;excellent picture&quot; is in fact a bit misleading&#xA;in this context! The gimbal lock &lt;a href=&quot;http://en.wikipedia.org/wiki/Gimbal_lock#Gimbal_lock_in_three_dimensions&quot; rel=&quot;nofollow&quot;&gt;illustrated on&#xA;Wikipedia&lt;/a&gt;&#xA;occurs when the plane is pointing straight up or down (elevation = 90&#xA;degrees), so that changing the bank is the same thing as changing the&#xA;heading. (That is, there is momentarily no way to change &lt;em&gt;the yaw&lt;/em&gt; by&#xA;manipulating the heading/elevation/bank angles!)&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2010-11-05T08:15:55.693" CommentCount="1" />
  <row Id="9042" PostTypeId="2" ParentId="9032" CreationDate="2010-11-05T08:38:03.997" Score="2" ViewCount="" Body="&lt;p&gt;Just averaging them does not account for the different variances, and therefore does not yield an optimal result.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The best (as in minimum variance, unbiased) estimation of the expected value $\mu$ of those two distributions $X_1, X_2$ with variances $\sigma^2_1,\sigma^2_2$ is $$\hat\mu(x_1, x_2) = \frac{\sigma_1^2x_2+\sigma_2^2x_1}{\sigma_1^2+\sigma_2^2}$$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please correct my notaion. I've learned this stuff with German names. =)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The joint distribution of $X_1, X_2$ forms an exponential family in $T(x_1, x_2) = \sigma_1^2x_2+\sigma_2^2x_1$, since its probability density function has the form &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f(x_1, x_2) = \frac{1}{\sqrt{2\pi\sigma_1^2}\sqrt{2\pi\sigma_2^2}}\exp({-\frac{(x_1-\mu)^2}{2\sigma_1^2}-\frac{(x_2-\mu)^2}{2\sigma_2^2}})$$ $$= h(x)A(\mu)\exp({\frac{\mu(\sigma_1^2x_2+\sigma_2^2x_1)}{2\sigma_1^2\sigma_2^2}}) $$ with $h(x)A(\mu)$ being all the uninteresting stuff. =)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore $T$ is sufficient, $\mu$ depends only on $T$ and is unbiased and the Lehmann–Scheffé theorem together with some mumbling about completeness concludes the proof.&lt;/p&gt;&#xA;" OwnerUserId="505" LastActivityDate="2010-11-05T08:38:03.997" />
  <row Id="9043" PostTypeId="2" ParentId="8814" CreationDate="2010-11-05T08:46:53.767" Score="3" ViewCount="" Body="&lt;p&gt;\begin{eqnarray}&#xA;\sum_{k = 0}^{\lfloor q  - q/p) \rfloor} \left \lfloor \frac{p(q - k)}{q} \right \rfloor = \sum_{k = 1}^{q} \left \lfloor \frac{kp}{q} \right \rfloor&#xA;\end{eqnarray}&lt;/p&gt;&#xA;" OwnerUserId="2720" LastActivityDate="2010-11-05T08:46:53.767" CommunityOwnedDate="2010-11-05T08:46:53.767" />
  <row Id="9044" PostTypeId="2" ParentId="2949" CreationDate="2010-11-05T08:58:16.993" Score="2" ViewCount="" Body="&lt;p&gt;I have had two results in the last year that surprised me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One appeals to my intuition in physics, although it may seem more obvious to those more versed mathematics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The set of pauli matrices (with identity) when multiplied by $i$ form the group of unitary quaternions (under matrix multiplication).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The was surprising to me how you can connect such a physical concept as electron spin to an abstract algebraic concept. Its what led me to dive into group and representation theory as it applies to physics. Now I understand that we can consider spin symmetry as SU(2) , which is injective into SO(3) the group of symmetry of $R^3$ of which the quaternions can be thought of as a representation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second result, which I had to prove in an algebra assignment:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a field $F$ of characteristic $p$ where $p$ is prime $(x+y)^p = x^p + y^p$ for $x,y \in F$. Lovely little result that spits in the face of everything that our grade school teachers taught us in algebra. &lt;/p&gt;&#xA;" OwnerUserId="1906" LastActivityDate="2010-11-05T08:58:16.993" CommunityOwnedDate="2010-11-05T08:58:16.993" />
  <row Id="9045" PostTypeId="1" AcceptedAnswerId="10662" CreationDate="2010-11-05T10:14:07.173" Score="-6" ViewCount="143" Body="&lt;p&gt;How to prove that an nxn matrix A is unitary if and only if $||Az||=||z||$ for all $z \in C^n$? Suggestion: Apply previous question on &quot;How to prove that T is the 0 operator, that is $T(v)=0$ for all $u,v \in V$ ?&quot; to &#xA;$A^{\ast}A-I$, this is a homework in the area of linear algebra.&lt;/p&gt;&#xA;" OwnerUserId="2723" LastActivityDate="2010-11-17T06:17:16.627" Title="How to prove that an nxn matrix A is unitary if and only if $||Az||=||z||$ for all $z \in C^n$?" Tags="&lt;linear-algebra&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="9046" PostTypeId="2" ParentId="9035" CreationDate="2010-11-05T11:23:15.153" Score="4" ViewCount="" Body="&lt;p&gt;You might look at this for some recent related work, and investigate its 14 references:&#xA;&quot;Every Large Point Set contains Many Collinear Points or an Empty Pentagon.&quot; &lt;a href=&quot;http://www.informatik.uni-trier.de/~ley/db/conf/cccg/cccg2009.html#AbelBBCDHKLPW09&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;CCCG&lt;/em&gt; 2009, 99-102&lt;/a&gt;. The main result is exactly what is stated in the title.&lt;/p&gt;&#xA;" OwnerUserId="237" LastActivityDate="2010-11-05T11:23:15.153" CommentCount="3" />
  <row Id="9048" PostTypeId="1" CreationDate="2010-11-05T11:48:43.557" Score="1" ViewCount="227" Body="&lt;p&gt;OK, so I got my ships moving along the bezier in time T, now I need to factor in acceleration, but I don't really know how to do it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm using the DeCasteljau Algorithm, that's what my code looks like:  &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;function linp(d, a, b, t) {&#xA;    d.x = a.x + (b.x - a.x) * t;&#xA;    d.y = a.y + (b.y - a.y) * t;&#xA;}&#xA;&#xA;// a, b, c, d are the 4 controls points, dest will contain the x/y coordinates and t is the time value from 0.0 to 1.0&#xA;Ship.prototype.bezier = function(dest, a, b, c, d, delta) {&#xA;    var ab = {x:0, y: 0}, bc = {x:0, y: 0}, cd = {x:0, y: 0};&#xA;    var abbc = {x:0, y: 0}, bccd = {x:0, y: 0};&#xA;    linp(ab, a, b, t);&#xA;    linp(bc, b, c, t);&#xA;    linp(cd, c, d, t);&#xA;    linp(abbc, ab, bc, t);&#xA;    linp(bccd, bc, cd, t);&#xA;    linp(dest, abbc, bccd, t);&#xA;};&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;So the question is, what is a simple way to factor in the acceleration from the starting speed to the end speed?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update&lt;/strong&gt;:&lt;br&gt;&#xA;Here's a small video how the thing looks when it's not working. I hope this helps.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.youtube.com/watch?v=EIJg1WFbjh4&quot; rel=&quot;nofollow&quot;&gt;http://www.youtube.com/watch?v=EIJg1WFbjh4&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3154" LastEditorUserId="3154" LastEditorDisplayName="" LastEditDate="2010-11-05T13:26:56.943" LastActivityDate="2011-01-10T02:21:16.190" Title="Moving along a bezier curve in time T, factoring in acceleration from X to Y" Tags="&lt;bezier-curve&gt;" AnswerCount="2" CommentCount="9" FavoriteCount="1" />
  <row Id="9049" PostTypeId="1" AcceptedAnswerId="9056" CreationDate="2010-11-05T11:58:44.093" Score="2" ViewCount="113" Body="&lt;p&gt;On what basis the numbering less than 20, the fraction 17/6 (base 10) does not generate a regular tithe?&lt;/p&gt;&#xA;" OwnerUserId="2938" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-05T12:00:34.417" LastActivityDate="2010-11-05T13:11:43.997" Title="Numbering in base other than 10" Tags="&lt;number-systems&gt;" AnswerCount="1" CommentCount="8" FavoriteCount="1" />
  <row Id="9050" PostTypeId="2" ParentId="9010" CreationDate="2010-11-05T12:24:31.177" Score="1" ViewCount="" Body="&lt;p&gt;This is answer to 978...answer (I have not enough reputation to comment).&#xA;He writes &quot;Also, we can be jerks and do something silly: Every compact manifold admits the structure of an affine scheme, by Hochster's Theorem&quot;&#xA;This is not true because Hochster theorem writes that quasicompact opens sets must be base of opens sets and that is completely false for usual manifolds.&lt;/p&gt;&#xA;" OwnerUserId="2872" LastActivityDate="2010-11-05T12:24:31.177" CommentCount="3" />
  <row Id="9051" PostTypeId="2" ParentId="9037" CreationDate="2010-11-05T12:38:49.377" Score="10" ViewCount="" Body="&lt;p&gt;The existence of a bijection between the class of ordinals $On$ and the class of surreal numbers $No$ is independent of the axioms of set theory. There are several interesting possibilities:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;If ZFC is consistent, then there is a model of ZFC in which there is a definable such bijection. This is true in Goedel's constructible universe $L$, for example, for in $L$ there is a definable well-ordering of the universe, and we can use this well-ordering to well-order the surreals, which provides the desired bijection.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;More generally, there is a first-order definable bijection between $On$ and $No$ if and only if the axiom known as $V=HOD$ holds. For the one direction, if $V=HOD$ holds, then there is a definable well-ordering of the universe and hence in particular a definable well-ordering of the surreals. Conversely, under ZFC if there is a definable bijection between $On$ and $No$, then there is a definable well-ordering of $No$. This allows us to construct a definable well-ordering of the class of sets of ordinals, since any set of ordinals determines a transfinite binary sequence of some ordinal length, and we can interpret this sequence as a $\pm 1$ sequence, which determines a unique surreal number by climbing through the tree of left-right cuts. Thus, we can well-order the class of sets of ordinals. But in ZFC every set is coded by a set of ordinals, and so we can construct a well-ordering of the entire universe, by looking for the least ordinal mapping to a surreal whose $\pm 1$ representation codes that set. So in this case, V=HOD holds.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Another way to summarize this argument is to say that if you can well-order $No$---and this is what your bijection to $On$ amounts to---then you can well-order every class.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If you drop the requirement that the bijection be definable, then we should move to the Goedel-Bernays context, in order to treat classes. The assertion that there is a bijection between $On$ and $No$ is equivalent over ZFC+GB to the axiom of Global Choice, which asserts that there is a well-ordering of the universe. This is by the same argument as above. (Note, we need AC for sets in order to make the last step of the argument; the class bijection in effect allows us to sew the set sized well-orderings together into a class well-ordering.) Thus, the theory ZFC+GB+(your bijection) is equivalent to GBC.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Because of this, if ZFC is consistent, then there are models of ZFC that have no bijection between $On$ and $No$, either definable or definable-from-parameters or otherwise. This is because it is known that ZFC does not imply global choice. One can construct such models by performing a class forcing iteration, adding a Cohen subset to every regular cardinal. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Meanwhile, every model of ZFC has a class forcing extension in which there &lt;em&gt;is&lt;/em&gt; a class well-ordering of the universe, simply by forcing to add a global well-ordering, and this forcing extension adds no new sets, only classes. In this sense, it is compatible with every model of ZFC set theory to have the desired bijection as a class, without adding any new sets. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Further, every model of ZFC has a class forcing extension in which there is a &lt;em&gt;definable&lt;/em&gt; bijection between $On$ and $No$, since we can force $V=HOD$. (This forcing, however, does add new sets.) &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Lastly, upon reading your question again, I see that you asked for a surjection from $On$ onto $No$, rather than a bijection. But these are equivalent, since if there is a surjection, then we can remove the redundant ordinals from the domain by only using the least ordinal that maps to a given surreal, and this gives a bijection from a proper class of ordinals to $No$. But every proper class of ordinals is bijection with $On$ simply by collapsing to the order type of the predecessors.&lt;/p&gt;&#xA;" OwnerUserId="413" LastEditorUserId="413" LastEditorDisplayName="" LastEditDate="2010-11-05T22:40:45.337" LastActivityDate="2010-11-05T22:40:45.337" />
  <row Id="9052" PostTypeId="2" ParentId="8814" CreationDate="2010-11-05T12:41:36.190" Score="6" ViewCount="" Body="&lt;p&gt;Well, i don't know whether to classify this as funny or surprising,  but ok it's worth posting.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Let $(X,\tau)$ be a topological space and let $A \subset X$ . By iteratively applying operations of closure and complemention, one cannot produce more than 14 disjoint sets. It's called as the &lt;a href=&quot;http://en.wikipedia.org/wiki/Kuratowski%27s_closure-complement_problem&quot; rel=&quot;nofollow&quot;&gt;Kuratowski's Closure complement problem&lt;/a&gt;. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" LastActivityDate="2010-11-05T12:41:36.190" CommunityOwnedDate="2010-11-05T12:41:36.190" />
  <row Id="9053" PostTypeId="2" ParentId="2949" CreationDate="2010-11-05T12:43:29.277" Score="5" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Kuratowski%27s_closure-complement_problem&quot; rel=&quot;nofollow&quot;&gt;Kuratowski's Complement problem&lt;/a&gt;, is the one which i came across recently, and i was clearly flabbergasted.&lt;/p&gt;&#xA;" LastActivityDate="2010-11-05T12:43:29.277" CommunityOwnedDate="2010-11-05T12:43:29.277" />
  <row Id="9054" PostTypeId="2" ParentId="8977" CreationDate="2010-11-05T12:52:21.887" Score="1" ViewCount="" Body="&lt;p&gt;As others have mentioned in comments, your control points cannot be independent. Nevertheless, if we assume that a given configuration has the properties you want, we can analyze the geometry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll consider the orange arc, $BE$, and I'll assume that both circles $A$ and $C$ overlap the interior of the orange circle, which I'll further assume has not degenerated into a line.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $a = |AB|$, $c = |CE|$, and $x=|AC|$; all of these can be considered known quantities. Let the (unknown) radius of the orange circle be $r = |PB| = |PE|$, where $P$ is the circle's center. Because radii $AB$ and $PB$ are perpendicular to a common tangent line at $B$, these segments lie on the same line; likewise for $CE$ and $PE$; consequently, $P$ lies at the intersection of the two extended radii $AB$ and $CE$, so that the angle $BPE$ is congruent to the angle between the &lt;em&gt;vectors&lt;/em&gt; $AB$ and $CE$. Call the measure of that angle $\theta$; it, too, can be considered a known quantity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, triangle $APC$ has sides of length $x$, $r-a$, and $r-c$ (the last two because of the assumed overlap of circles), with angle $\theta$ between the last two. By the Law of Cosines:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x^2 = (r-a)^2 + (r-c)^2 - 2 (r-a)(r-c) \cos\theta$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solve this quadratic equation for $r$, and you can calculate whatever else you need to know: arc length, location of $P$, equation of the orange circle, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(The equation of the orange circle can be expressed in a form that degenerates into a line equation as $r$ approaches infinity. Note that, in such a degenerate case, $\theta = 0$.)&lt;/p&gt;&#xA;" OwnerUserId="409" LastActivityDate="2010-11-05T12:52:21.887" />
  <row Id="9055" PostTypeId="2" ParentId="2949" CreationDate="2010-11-05T13:09:52.503" Score="9" ViewCount="" Body="&lt;p&gt;1) Exotic structures on $\mathbb{R}^4$ probably puzzles anyone learning about differentiable topology. Even more, the fact that it is &lt;em&gt;only&lt;/em&gt; for $n=4$ is quite remarkable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2)$S^n$ not being a Lie group for all $n$ ...&lt;/p&gt;&#xA;" OwnerUserId="2900" LastActivityDate="2010-11-05T13:09:52.503" CommunityOwnedDate="2010-11-05T13:09:52.503" CommentCount="4" />
  <row Id="9056" PostTypeId="2" ParentId="9049" CreationDate="2010-11-05T13:11:43.997" Score="2" ViewCount="" Body="&lt;p&gt;The terminating decimals (though that term should be reserved for base 10, but I have no better) in any base are precisely the fractions whose denominator has every prime factor represented in the factorization of the base.  So in base 10, it is fractions with denominators of the form 2^x*5^y, where x and y are allowed to be zero.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-11-05T13:11:43.997" CommentCount="2" />
  <row Id="9057" PostTypeId="1" CreationDate="2010-11-05T14:00:38.513" Score="1" ViewCount="194" Body="&lt;p&gt;I am CS major, please be patient if my question is not well-stated. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The dataset is quantitative mass spectrometry (MS) data. By labeling proteins of two different samples A and B, we get the relative abundance of 100 to thousands of proteins in A/B. Alongside with this ratio, we can estimate its variance based on the signal intensities. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wanted: A list of proteins significantly different from the set of all protein ratios.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most proteins remain unchanged between A and B. The population of log-ratios distributes around 1. The histogram shows a bell shape with fat tails. Two-term Gaussian mixture model has been found to provide a good fit to experimental noise. I suppose it would work good for this data - think of experimental and biological noise.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;How to test for significantly different ratios on such a two-term Gaussian mixture model?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Thanks for your responses!&lt;/p&gt;&#xA;" OwnerUserId="3156" LastEditorUserId="3156" LastEditorDisplayName="" LastEditDate="2010-11-08T11:37:50.390" LastActivityDate="2011-05-07T13:50:17.693" Title="How to do hypothesis testing on Gaussian mixture model?" Tags="&lt;statistics&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="9058" PostTypeId="1" AcceptedAnswerId="9067" CreationDate="2010-11-05T14:05:38.057" Score="3" ViewCount="276" Body="&lt;p&gt;How do I solve the following question?&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;You are given the identity    $x^2-ax+144\equiv(x-b)^2$&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Work out the values of $a$ and $b$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Question appears in AQA 43005/1H.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="3157" LastEditorUserId="5363" LastEditorDisplayName="" LastEditDate="2011-04-17T20:12:18.693" LastActivityDate="2011-04-17T20:12:18.693" Title="Work out the values of a and b from the identity $x^2 - ax + 144 \equiv (x-b)^2$" Tags="&lt;algebra&gt;" AnswerCount="5" CommentCount="3" />
  <row Id="9059" PostTypeId="2" ParentId="9058" CreationDate="2010-11-05T14:06:21.393" Score="0" ViewCount="" Body="&lt;p&gt;Posted by &lt;a href=&quot;http://in.answers.yahoo.com/my/profile;_ylt=AuGKDZFCaIla44dugPaZqxXaHwx.;_ylv=3?show=31f743411a42c3a8de21b1c9c1b15871aa&quot; rel=&quot;nofollow&quot;&gt;Pi R Squared&lt;/a&gt; on &lt;a href=&quot;http://in.answers.yahoo.com/question/index?qid=20090510094003AA5Xzgh&quot; rel=&quot;nofollow&quot;&gt;Yahoo Answers&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$a=24$ and $b=12$&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$x^2-24x+144=(x-12)^2$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="3157" LastEditorUserId="3157" LastEditorDisplayName="" LastEditDate="2010-11-05T15:46:28.930" LastActivityDate="2010-11-05T15:46:28.930" CommentCount="6" />
  <row Id="9060" PostTypeId="1" AcceptedAnswerId="9066" CreationDate="2010-11-05T14:12:05.613" Score="3" ViewCount="293" Body="&lt;p&gt;Consider the set &#xA;$$S = [0,1/2] \cup [3/4, 7/8] \cup [15/16, 31/32] \cup \cdots$$&#xA;and the set &#xA;$$T = (0,1/2) \cup  (3/4, 7/8) \cup (15/16, 31/32) \cup \cdots$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is $S$ open, closed or neither? Is $T$ open, closed, or neither?&lt;/p&gt;&#xA;" OwnerUserId="796" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-05T16:53:41.850" LastActivityDate="2010-11-05T19:32:32.223" Title="open and closed sets" Tags="&lt;real-analysis&gt;&lt;general-topology&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="9061" PostTypeId="2" ParentId="9058" CreationDate="2010-11-05T14:14:31.230" Score="5" ViewCount="" Body="&lt;p&gt;You have to note first that $12^2=144$. After this, you have to recall the identity&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(x-b)^2=x^2-2bx+b^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now compare coefficients to get the equations you need to solve for $a$ and $b$.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-11-05T14:14:31.230" CommentCount="1" />
  <row Id="9062" PostTypeId="2" ParentId="9057" CreationDate="2010-11-05T14:33:54.243" Score="0" ViewCount="" Body="&lt;p&gt;When you say the data fits one distribution or another, what are you plotting?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One hypothesis would be that the two samples are drawn from the same distribution.  Then you would expect A/B to be the same over all proteins.  If the distribution is the frequency of A/B for all the different proteins you observe you can test the hypothesis if you have a model of the errors in A and B.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How you model the data should be dependent upon how you will use the model.  It is not clear to me what the model is for.  You can certainly fit any curve through the data.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-11-05T14:33:54.243" CommentCount="1" />
  <row Id="9063" PostTypeId="1" AcceptedAnswerId="9065" CreationDate="2010-11-05T14:45:52.733" Score="6" ViewCount="186" Body="&lt;p&gt;Given a $\sigma$-finite measure $\mu$ on a set $X$ is it possible to formulate a topology on the space of functions $f:X \rightarrow \mathbb{R}$ that gives convergence $\mu$-almost everywhere?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can't seem to find any way to write this and am suspecting that no such topology exists! Is this true?  If so, is there some generalisation of a topological space where one can make sense of convergence without having open sets?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any comments, references or tips would be greatly appreciated.&lt;/p&gt;&#xA;" OwnerUserId="780" LastActivityDate="2010-11-05T15:49:14.060" Title="Characterization of Almost-Everywhere convergence" Tags="&lt;analysis&gt;&lt;measure-theory&gt;&lt;general-topology&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="2" />
  <row Id="9064" PostTypeId="2" ParentId="9060" CreationDate="2010-11-05T15:10:34.630" Score="1" ViewCount="" Body="&lt;p&gt;If we have the standard topology T is the countable union of disjoint open sets and so T is an open set.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The complement of S is $(-\infty, 0) \cup (1/2, 3/4) \cup (7/8, 15/16) \cup \cdots$, etc. if this is open then S is closed. Is it open? Why?&lt;/p&gt;&#xA;" OwnerUserId="2563" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-05T16:01:26.063" LastActivityDate="2010-11-05T16:01:26.063" CommentCount="3" />
  <row Id="9065" PostTypeId="2" ParentId="9063" CreationDate="2010-11-05T15:11:36.710" Score="10" ViewCount="" Body="&lt;p&gt;Given I understand you correctly (topologize almost everywhere convergence), we can show that this is not possible. If we have a topological space, then we have convergence of a sequences if and only if every subsequence has a further convergent subsequence and so on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So pick a sequence that converges in measure but not almost everywhere. There is a theorem that states that every subsequence of this also converges in the same measure. So it has a subsequence that converges almost everywhere (by another theorem). But the original sequence does not converge almost everywhere so we cannot have convergence in a topology.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Should I add more detail?&lt;/p&gt;&#xA;" OwnerUserId="1120" LastEditorUserId="1120" LastEditorDisplayName="" LastEditDate="2010-11-05T15:49:14.060" LastActivityDate="2010-11-05T15:49:14.060" CommentCount="1" />
  <row Id="9066" PostTypeId="2" ParentId="9060" CreationDate="2010-11-05T15:12:09.777" Score="5" ViewCount="" Body="&lt;p&gt;For the sake of the answer I am going to assume this is the standard topology of the real numbers, but I'll try to be as general as possible so it will be easier to generalize.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A union of any number of open sets is still open, so $T$ is open.&#xA;A union of countably many closed sets does not have to be closed, this sort of set is called $F_\sigma$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While some $F_\sigma$ are open sets (in fact in a metric space, such as the real numbers with their standard topology, every open set is $F_\sigma$) this one is not. One of the requirements from an open set (in any topology) is that it will have an open neighborhood surrounding each point inside of it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case above that means there will be an interval around each point, which is a property not being satisfied by $S$, as the point $\frac{1}{2}$ is in $S$ but there is no interval around it contained in $S$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But there is still the case where $S$ could be closed as well, however if we look at the sequence $a_n = \frac{2^{2n-1}-1}{2^{2n-1}}$ (the endpoints of each closed interval within the union) then it is clear that $a_n$ approaches $1$ but since $1 \notin S$ then we have that $S$ is not closed.&lt;/p&gt;&#xA;" OwnerUserId="622" LastEditorUserId="622" LastEditorDisplayName="" LastEditDate="2010-11-05T19:19:13.920" LastActivityDate="2010-11-05T19:19:13.920" CommentCount="2" />
  <row Id="9067" PostTypeId="2" ParentId="9058" CreationDate="2010-11-05T15:42:19.557" Score="1" ViewCount="" Body="&lt;h3&gt;Find the Value of $b$&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;$x^2-ax+144\equiv(x-b)^2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Expand the right-hand side:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x^2-ax+144\equiv{x^2-2bx+b^2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Find the coefficient $(x=0)$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$b^2=144$&lt;br&gt;&#xA;$\Longrightarrow b=\pm{\sqrt{144}}=\pm12$&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Find the Value of $a$&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Substitute $b=\pm12$ back into original equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x^2-ax+144\equiv{x^2\pm{24x}+144}$&lt;br&gt;&#xA;$\Longrightarrow -ax\equiv\pm{24x}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Find the coefficient $(x=1)$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a=\pm{24}$&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Answer&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;$a=24, b=12$ &lt;strong&gt;and&lt;/strong&gt; $a=-24, b=-12$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Thanks to @J. M. and @Djaian.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="3157" LastEditorUserId="3157" LastEditorDisplayName="" LastEditDate="2011-04-17T19:51:34.817" LastActivityDate="2011-04-17T19:51:34.817" CommentCount="2" />
  <row Id="9068" PostTypeId="2" ParentId="9058" CreationDate="2010-11-05T15:43:49.273" Score="7" ViewCount="" Body="&lt;p&gt;The prior answer presumes that you already know the fact that two polynomials functions are identical (if and) only if they have identical coefficients. But the above problem might be posed as motivation before this general fact is proved. In that case you can prove it directly without using this more general fact as follows. Evaluating the identity at $\rm\ x = 0\ $ yields $\rm\ (-b)^2 = 144 = 12^2\ $. So $\rm\ (x-b)^2 = x^2 -2b + 144 \equiv x^2 -ax + 144\:.\ $ RHS $-$ LHS yields $\rm\ (2b-a)\:x \equiv 0\:$. Evaluating this at $\rm\ x = 1\ $ yields $\rm\ a = 2b\:.\ $ And, of course, $\rm\ 0 = 12^2 - (-b)^2 = (12+b)\ (12-b)\ \Rightarrow\ b = \pm12\:$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the above proof yields$\ \ $ (1) the equation is true for all $\rm\: x\ \Rightarrow\ $ (2) the equation is true for $\rm x = 0,\ and\ x = 1\ $ $\rm\ \Rightarrow\ $ (3) $\rm\ a = 2b,\ b^2 = 144\:.\ $ But (3) implies that both sides of the equation have equal coefficients, so it is true for all $\rm\:x\:$, implying (1).$\ $ So all three statements are equivalent. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact generally it is true over $\mathbb C$ that two polynomial functions of degree $\rm\:n\:$ are identical iff they have identical coefficients iff they have equal values at $\rm\:n+1\:$ points (or $\rm\:n\:$ points if they have the same leading coefficient - as above). In fact, this is true not only for polynomials with coefficients in $\mathbb C$ but in any integral domain, i.e any ring without zero-divisors, i.e. $\rm\ ab = 0 \Rightarrow\ a=0\ or\ b=0\:$.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-11-06T16:27:28.487" LastActivityDate="2010-11-06T16:27:28.487" CommentCount="1" />
  <row Id="9069" PostTypeId="1" AcceptedAnswerId="9096" CreationDate="2010-11-05T15:45:22.533" Score="0" ViewCount="162" Body="&lt;p&gt;The probability that a fire will occur is $0.001$. If there is a fire, the amount of damage, $X$, will have a Pareto distribution given by $P(X&gt;x) = \left(\frac{2(10)^6}{2(10)^6+x} \right)^2$. An insurance will pay the excess of the loss over a deductible of $100,000$. For this coverage the one-time insurance premium will be $110 \%$ of the expected payment. Calculate the premium. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the expected payment is $E[W]$ where $W$ denotes the payment. Then $E[W] = E[W| \text{fire}]P(\text{fire})+E[W| \text{no fire}]P(\text{no fire})$. To calculate $E[W| \text{fire}]$, we could use $\int_{0.1}^{\infty} [1-F(x)] \ dx$? This would be: $\int_{0.1}^{\infty} 1-\left[1-\left(\frac{2(10)^6}{2(10)^6+x} \right)^2\right] \ dx$ which equals $\int_{0.1}^{\infty} \left(\frac{2(10)^6}{2(10)^6+x} \right)^2 \ dx$?&lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2010-11-05T20:36:59.363" Title="Probability of Fire" Tags="&lt;probability&gt;" AnswerCount="2" />
  <row Id="9070" PostTypeId="2" ParentId="2949" CreationDate="2010-11-05T16:10:13.553" Score="2" ViewCount="" Body="&lt;p&gt;To me the &lt;a href=&quot;http://en.wikipedia.org/wiki/Cubic_surface#27_lines_on_a_cubic_surface&quot; rel=&quot;nofollow&quot;&gt;Cayley-Salmon theorem&lt;/a&gt; is an example of a result that still strikes me as rather surprising.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Theorem(Cayley-Salmon)&lt;/strong&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;A smooth cubic surface $\mathcal{S} \subset \mathbb{P}^{3}_{k}$ contains exactly 27 lines, where $k$ is an algebraically closed field.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here is a &lt;a href=&quot;http://www-history.mcs.st-and.ac.uk/PrintHT/Cubic_surfaces.html&quot; rel=&quot;nofollow&quot;&gt;link&lt;/a&gt; to some history about it. There's a really nice treatment of this result in chapter 5 of Klaus Hulek's book &lt;a href=&quot;http://books.google.co.cr/books?id=uI_21kSf0iAC&amp;amp;printsec=frontcover&amp;amp;dq=Klaus+hulek&amp;amp;hl=es&amp;amp;ei=fCrUTN-wJoOclge-68XaBQ&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=1&amp;amp;ved=0CCcQ6AEwAA#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;Elementary Algebraic Geometry&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="900" LastActivityDate="2010-11-05T16:10:13.553" CommunityOwnedDate="2010-11-05T16:10:13.553" />
  <row Id="9071" PostTypeId="1" CreationDate="2010-11-05T16:14:49.140" Score="2" ViewCount="91" Body="&lt;p&gt;This comes up in a paper by Beskos and Roberts on exact simulation of diffusions. I'm sure it's easy, but I can't work out what's going on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given a differentiable function $\alpha$ satisfying&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\alpha(u) \geq k_1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\alpha'(u) \leq k_2$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for some constants $k_1$ and $k_2$, show that there exist constants $c_1$ and $c_2$ such that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$c_1 \leq \alpha^2(u) + \alpha'(u) \leq c_2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clearly one condition implies alpha is bounded below and one implies that alpha has at most linear growth, but I don't see how this is enough for the conclusion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit - I'm a fool. They wrote $k_1 \leq \alpha(u), \alpha'(u) \leq k_2$, but there was a line break after the comma. I was reading them as two seperate conditions on $\alpha$ and $\alpha'$. Sorry everyone, thanks for the help. If it wasn't for the counterexamples it would have taken a lot longer to realise that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="3162" LastEditorUserId="3162" LastEditorDisplayName="" LastEditDate="2010-11-05T16:33:41.573" LastActivityDate="2010-11-05T16:33:41.573" Title="simple functional inequality" Tags="&lt;analysis&gt;&lt;inequality&gt;" CommentCount="5" />
  <row Id="9072" PostTypeId="1" CreationDate="2010-11-05T16:39:59.227" Score="2" ViewCount="225" Body="&lt;p&gt;Can someone explain how to picture or construct the Monster Group?&lt;/p&gt;&#xA;" OwnerUserId="3163" LastActivityDate="2010-11-06T11:56:25.063" Title="What is the simplest way to fathom the Monster Group?" Tags="&lt;group-theory&gt;" AnswerCount="1" CommentCount="10" />
  <row Id="9073" PostTypeId="2" ParentId="9069" CreationDate="2010-11-05T16:51:57.140" Score="1" ViewCount="" Body="&lt;p&gt;TO calculate $E(W|fire)$, you first need to calculate the pdf. You can get this by taking derivative of $1-P(X&gt;x)$. Call this function $p(x)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the insurer pays out any amount greater than the deductible of $100000$, the expected payment is given by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;\int_{100000}^\infty (x - 100000) p(x) dx&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The premium should be 110% of the value of the integral.&lt;/p&gt;&#xA;" OwnerUserId="2641" LastActivityDate="2010-11-05T16:51:57.140" CommentCount="2" />
  <row Id="9074" PostTypeId="2" ParentId="9060" CreationDate="2010-11-05T16:52:35.593" Score="4" ViewCount="" Body="&lt;p&gt;To be able to answer these questions you need to know what &quot;open&quot; and &quot;closed&quot; mean. And that depends on context.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the context you are looking at, in which you are dealing with subsets of the real line, without any further information we usually assume that we are dealing with the &lt;em&gt;standard&lt;/em&gt; meanings of &quot;open&quot; and &quot;closed&quot;. The meanings are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;A set is $\mathcal{O}\subseteq\mathbb{R}$ is &lt;strong&gt;open&lt;/strong&gt; if and only if for every $x\in \mathcal{O}$ there is an $\epsilon\gt 0$ (which may depend on $x$) such that $(x-\epsilon,x+\epsilon)\subseteq \mathcal{O}$ (where $(x-\epsilon,x+\epsilon)$ is the set of all real numbers $y$ such that $x-\epsilon \lt y \lt x+\epsilon$). &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;A set $\mathcal{C}\subseteq\mathbb{R}$ is &lt;strong&gt;closed&lt;/strong&gt; if and only if its complement $\mathbb{R}-\mathcal{C}$ is open; if and only if for every point $x\notin \mathcal{C}$, there exists $\epsilon\gt 0$ (which may depend on $x$) such that $(x-\epsilon,x+\epsilon)\cap \mathcal{C}=\emptyset$. Alternatively, a set $\mathcal{C}\subseteq\mathbb{R}$ is &lt;strong&gt;closed&lt;/strong&gt; if and only if for every converging sequence of real numbers $\{x_n\}_{n=1}^{\infty}$, if each $x_n$ is in $\mathcal{C}$, and $\lim\limits_{n\to\infty}x_n = x$, then $x\in\mathcal{C}$. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Some consequences of these definitions are: &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Both the sets $\mathbb{R}$ and $\emptyset$ are open and closed.&lt;/li&gt;&#xA;&lt;li&gt;Every open interval is an open set. Every closed interval is a closed set.&lt;/li&gt;&#xA;&lt;li&gt;If $\mathcal{O}_1$ and $\mathcal{O}_2$ are both open, then their intersection $\mathcal{O}_1\cap\mathcal{O}_2$ is also open (given $x$ in the intersection, find $\epsilon_1\gt 0$ and $\epsilon_2\gt 0$ so that $(x-\epsilon_i,x+\epsilon_i)\subseteq\mathcal{O}_i$, and then let $\epsilon=\min\{\epsilon_1,\epsilon_2\}$ be the witness for the intersection). As a consequence, using De Morgan's laws, if $\mathcal{C}_1$ and $\mathcal{C}_2$ are closed, then $\mathcal{C}_1\cup\mathcal{C}_2$ is closed.&lt;/li&gt;&#xA;&lt;li&gt;Using induction, from 3 it follows that any intersection of &lt;em&gt;finitely many&lt;/em&gt; open sets is open, and any union of &lt;em&gt;finitely many&lt;/em&gt; closed sets is closed.&lt;/li&gt;&#xA;&lt;li&gt;If $\mathcal{O}_i$ is any family (of any size, finite or infinite) of open sets, then $\cup\mathcal{O}_i$ is also open (pick $x$ in the union; then it belongs to some $\mathcal{O}_j$, and since $\mathcal{O}_j$ is open and $x\in\mathcal{O}_j$, there exists $\epsilon\gt 0$ such that $(x-\epsilon,x+\epsilon)\subseteq\mathcal{O}_j\subseteq \cup\mathcal{O}_i$). Consequently, if $\mathcal{C}_i$ is any family (of any size, finite or infinite) of closed sets, then $\cap\mathcal{C}_i$ is closed.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;So, looking back at your examples, it should be clear that $T$ must be open, since it is a union of open intervals; open intervals are open sets, and an arbitrary union of open sets is always open. What is not clear is whether $S$ is closed or not, and that is because:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;In general, an arbitrary intersection of open sets is not necessarily open; it &lt;em&gt;may&lt;/em&gt; be open, or it may be closed (e.g., $\mathop{\cap}\limits_{n=1}^{\infty}(-\frac{1}{n},1+\frac{1}{n}) = [0,1]$), or it may be neither open nor closed. &lt;/li&gt;&#xA;&lt;li&gt;In general, an arbitrary union of closed sets is not necessarily closed; it &lt;em&gt;may&lt;/em&gt; be closed, but it may also be open, or neither.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So... what about $S$? It is somewhat difficult at first thought to try to determine if it is closed using the definition of &quot;complement is open&quot;; but it is not too bad to verify that it does not satisfy the condition I labeled as &quot;equivalent&quot;: if we have a sequence of points in $S$ that converges in $\mathbb{R}$, will the limit be in $S$? Well, take $x_n = 1 - \frac{1}{2^{2n+1}}$; that is $x_1=\frac{7}{8}$, $x_2 = \frac{31}{32}$, $x_3=\frac{127}{128}$, etc. All of these are in $S$, and the sequence converges to $1$. Is $1$ in $S$? $1$ is in $S$ if and only if it is in at leat one of the intervals in the union... is it? Well, no. So $S$ is not closed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is $S$ open, then? Well, if it were open, then since $\frac{1}{2}\in S$, there would have to be some $\epsilon\gt 0$ such that $(\frac{1}{2}-\epsilon,\frac{1}{2}+\epsilon)\subseteq S$. Is there such an $\epsilon$? Well, no; any such interval would contain points smaller than $\frac{1}{2}$, and there are no points smaller than $\frac{1}{2}$ in $S$. So $S$ is not open either.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, if $S$ is not open, but is not closed either... then it's neither.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As Asaf points out, countably infinite unions of closed sets &lt;em&gt;do&lt;/em&gt; show up a lot, so even though they may not be closed, we give them a special name because we often will want to say &quot;this may be neither open nor closed, but &lt;em&gt;is&lt;/em&gt; a countable union of closed sets.&quot; We call them $\mathcal{F}_{\sigma}$ (pronounced &quot;eff-sigma&quot;) sets; $\mathcal{F}$ comes from the French &lt;em&gt;fermé&lt;/em&gt;, for &quot;closed&quot;; $\sigma$ is a letter often used in mathematics to denote that a sum or union is taken a countable, but possibly infinite, number of times (from the French &lt;em&gt;somme&lt;/em&gt; for sum). Dually, we are often interested in sets that are countable intersections of open sets; these sets may be open, closed, or neither, but we call them $\mathcal{G}_{\delta}$ (&quot;gee-delta&quot;); this time, in a fit of internationalism, it comes from the &lt;strong&gt;German&lt;/strong&gt;, specifically &lt;em&gt;Gebiet&lt;/em&gt; (open) and &lt;em&gt;Durchshnitt&lt;/em&gt; (intersection). Your set $S$ is an $\mathcal{F}_\sigma$ set.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More generally, one talks about &lt;strong&gt;Topological spaces&lt;/strong&gt;, generalizing these notions. A topological space is an ordered pair $(X,\mathcal{O})$, where $X$ is any set, and $\mathcal{O}$ is a collection of subsets of $X$; basically, we explicitly specify what sets are going to be the &quot;open&quot; sets. We require that the collection $\mathcal{O}$ satisfy some of the same conditions that the open sets in $\mathbb{R}$ satisfy, namely, we require that:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\emptyset$ and $X$ are both &quot;open&quot; sets (elements of $\mathcal{O}$);&lt;/li&gt;&#xA;&lt;li&gt;If $A$ and $B$ are both &quot;open&quot; sets, then $A\cap B$ is an open set (that is, if $A$ and $B$ are both elements of $\mathcal{O}$, then $A\cap B$ is an element of $\mathcal{O}$); and&lt;/li&gt;&#xA;&lt;li&gt;If $\{A_i\}_{i\in I}$ is a family, finite or infinite, of &quot;open&quot; sets, then $\mathop{\cup}\limits_{i\in I}A_i$ is also an &quot;open&quot; set (if each $A_i$ is an element of $\mathcal{O}$, then so is their union). &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Then we define a subset of $X$ to be closed if and only if its complement is open, and go from there. &lt;em&gt;Caveat&lt;/em&gt;: the description of closed sets via limits of sequences is no longer true; one &lt;em&gt;can&lt;/em&gt; define the notion of convergence, but in arbitrary topological spaces the properties are not as nice as they are in the real numbers (there are spaces where the properties &lt;em&gt;are&lt;/em&gt; nice; they get special names for that). One needs to consider instead a more general notion (called &quot;nets&quot;). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the same set may have many &quot;topologies&quot;, and many notions of what an open or closed set may mean. Which is why people are asking you if it is the &quot;standard&quot; topology or not (assuming you've gotten this far in my response, anyway...) That is, is it the notion I described above, or some other version of a topology? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is all part of the very interesting and deep field of Topology.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-05T19:32:32.223" LastActivityDate="2010-11-05T19:32:32.223" CommentCount="3" />
  <row Id="9075" PostTypeId="1" AcceptedAnswerId="9079" CreationDate="2010-11-05T16:55:29.467" Score="3" ViewCount="395" Body="&lt;p&gt;Q: $\displaystyle \int \frac{1}{\sin x\cos x} dx $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that the correct answer can be obtained by doing:&lt;br&gt;&#xA;$\frac{1}{\sin x\cos x} = \frac{\sin^2(x)}{\sin x\cos x}+\frac{\cos^2(x)}{\sin x\cos x} = \tan(x) + \cot(x)$ and integrating.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, doing the following gets a completely different answer:&#xA;\begin{eqnarray*}&#xA;\int \frac{1}{\sin x\cos x} dx&lt;br&gt;&#xA;&amp;amp;=&amp;amp;\int \frac{\sin x}{\sin^2(x)\cos x} dx\\&#xA;&amp;amp;=&amp;amp;\int \frac{\sin x}{(1-\cos^2(x))\cos x} dx.&#xA;\end{eqnarray*}&#xA;let $u=\cos x, du=-\sin x dx$; then&#xA;\begin{eqnarray*}&#xA;\int \frac{1}{\sin x\cos x} dx&lt;br&gt;&#xA;&amp;amp;=&amp;amp;\int \frac{-1}{(1-u^2)u} du\\&lt;br&gt;&#xA;&amp;amp;=&amp;amp;\int \frac{-1}{(1+u)(1-u)u}du\\ &#xA;&amp;amp;=&amp;amp;\int \left(\frac{-1}{u} - \frac{1}{2(1-u)} + \frac{1}{2(1+u)}\right) du\\&#xA;&amp;amp;=&amp;amp;-\ln|\cos x|+\frac{1}{2}\ln|1-\cos x|+\frac{1}{2}\ln|1+\cos x|+C&#xA;\end{eqnarray*}  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tested both results in Mathematica, and the first method gets the correct answer, but the second method doesn't. Is there any reason why this second method doesn't work?&lt;/p&gt;&#xA;" OwnerUserId="2869" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-05T17:08:30.747" LastActivityDate="2010-11-05T17:48:23.537" Title="Integration of 1/(sinx*cosx)" Tags="&lt;calculus&gt;" AnswerCount="5" CommentCount="4" />
  <row Id="9077" PostTypeId="1" AcceptedAnswerId="9108" CreationDate="2010-11-05T17:09:57.703" Score="1" ViewCount="368" Body="&lt;p&gt;I am working on a &lt;a href=&quot;http://www.chem.uoa.gr/applets/appletsmooth/appl_smooth2.html&quot; rel=&quot;nofollow&quot;&gt;signal-smoothing algorithm&lt;/a&gt; for personal interest. I understand the basic concept of the Savitzky-Golay algorithm but I would like to understand how the &lt;a href=&quot;http://www.statistics4u.info/fundstat_eng/cc_savgol_coeff.html&quot; rel=&quot;nofollow&quot;&gt;coefficients&lt;/a&gt; were discovered. Can someone please explain the the mathematics behind it, so I could calculate the coefficients on my own?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How would the matrix look? For example &lt;a href=&quot;http://www.statistics4u.com/fundstat_eng/cc_filter_savgol_math.html&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; states: &quot;For n=5 and p=3, the filtering coefficients are b= [-0.0857 0.3429 0.4857 0.3429 -0.0857].&quot; However, it does not show the steps to achieve this. If n equal the number of points to sample and p equals the power of the polynomial.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.youtube.com/watch?v=8mAZYv5wIcE&quot; rel=&quot;nofollow&quot;&gt;Least Squares Tutorial&lt;/a&gt; &lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.vias.org/tmdatanaleng/cc_filter_savgol_math.html&quot; rel=&quot;nofollow&quot;&gt;Applying Least Squares&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="3165" LastEditorUserId="3165" LastEditorDisplayName="" LastEditDate="2010-11-06T01:55:30.217" LastActivityDate="2010-11-06T01:55:30.217" Title="Calculating the Savitzky-Golay Coefficients" Tags="&lt;algorithms&gt;&lt;signal-processing&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="9078" PostTypeId="2" ParentId="9075" CreationDate="2010-11-05T17:16:54.823" Score="2" ViewCount="" Body="&lt;p&gt;The second method gives the same answer as the first. By the first method, the answer you get is $-\log(\cos x) + \log(\sin x)$. The first term is the same as what you get by the second method.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you need to show is that $\log(\sin x) = \frac{1}{2}\log(1-\cos x) + \frac{1}{2}\log(1+\cos x)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;\begin{split}&#xA;\frac{1}{2}\log(1-\cos x) + \frac{1}{2}\log(1+\cos x) &amp;amp;= \frac{1}{2}\left( \log(2 \sin^2 \frac{x}{2}) + \log(2 \cos^2 \frac{x}{2})\right) =  \log(2 \sin \frac{x}{2} \cos \frac{x}{2})&#xA;\end{split}&#xA;\end{equation}&lt;/p&gt;&#xA;" OwnerUserId="2641" LastActivityDate="2010-11-05T17:16:54.823" />
  <row Id="9079" PostTypeId="2" ParentId="9075" CreationDate="2010-11-05T17:17:07.770" Score="3" ViewCount="" Body="&lt;p&gt;If I take the derivative of your second answer (call it $g(x)$), I get:&#xA;\begin{eqnarray*}&#xA;\frac{dg}{dx}&#xA;&amp;amp; = &amp;amp; -\frac{-\sin x}{\cos x} + \frac{\sin x}{2(1-\cos x)} + \frac{-\sin x}{2(1+\cos x)}\\&#xA;&amp;amp; = &amp;amp; \frac{\sin x\left(1-\cos^2 x + \frac{1}{2}\cos x(1+\cos x) - \frac{1}{2}\cos x(1-\cos x)\right)}{\cos x(1-\cos x)(1+\cos x)}\\&#xA;&amp;amp; = &amp;amp; \frac{\sin x\left( 1- \cos^2 x + \frac{1}{2}\cos x + \frac{1}{2}\cos^2 x - \frac{1}{2}\cos x + \frac{1}{2}\cos^2 x\right)}{\cos x(1-\cos^2 x)}\\&#xA;&amp;amp; = &amp;amp; \frac{\sin x}{\cos x\&gt;\sin^2 x} = \frac{1}{\cos x\sin x}.&#xA;\end{eqnarray*}&#xA;So I'm not sure why Mathematica says the second method is not &quot;the right answer&quot;. &lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-11-05T17:17:07.770" />
  <row Id="9080" PostTypeId="2" ParentId="9075" CreationDate="2010-11-05T17:17:13.013" Score="3" ViewCount="" Body="&lt;p&gt;This may be an easier method $$\int\frac{1}{\sin{x} \cdot \cos{x}} \ dx = \int\frac{\sec^{2}{x}}{\tan{x}} \ dx$$ by multiplying the numerator and denominator by $\sec^{2}{x}$&lt;/p&gt;&#xA;" LastActivityDate="2010-11-05T17:17:13.013" CommentCount="1" />
  <row Id="9081" PostTypeId="2" ParentId="9032" CreationDate="2010-11-05T17:24:41.663" Score="0" ViewCount="" Body="&lt;p&gt;You can do away with the normality if you restrict yourself to linear estimates. Given $\mu, \sigma_1^2, \sigma_2^2, \sigma_{12}$,, you have random variables $X_1,X_2$, with $E(X_1) = E(X_2) = \mu $, and $\text{Var}(X_1) = \sigma_1^2$ and $\text{Var}(X_2) = \sigma_2^2$, and $\text{Cov}(X_1,X_2) = \sigma_{12}$. You want $l_0,l_1,l_2$ such that $E(l_0+l_1 X_1 + l_2 X_2) = \mu $ (unbiased) and $\text{Var}(l_0 + l_1 x_1 + l_2 X_2)$  is as small as possible (minimum variance unbiased estimator). From the first equation you get $l_0 + l_1 \mu + l_2 \mu = \mu $ for all $\mu$ i.e., $l_0 = \mu ( 1 - l_1 - l_2)$ for all $\mu$. Which is possible iff $l_0 = 0$  and $l_1 + l_2 = 1$. So the second condition reduces to minimizing $\text{Var}(l_1 X_1 + l_2 X_2)^2$ over all $l_1,l_2$ subject to $l_1 + l_2 = 1$. Simple calculus of lagrangian multipliers applied to $\text{Var}(l_1 X_1 + l_2 X_2) - \lambda(l_1 + l_2 - 1)$  leads to&lt;br&gt;&#xA;$&#xA;l_1 \sigma_1^2 + l_2 \sigma_{12} - \lambda = 0&#xA;$&lt;br&gt;&#xA;$&#xA;l_1 \sigma_{12} + l_2 \sigma_{2}^2 - \lambda = 0&#xA;$&lt;br&gt;&#xA;$&#xA;l_1  + l_2  = 1$&lt;br&gt;&#xA;from which we get our estimator as&lt;br&gt;&#xA;$\frac{\alpha_1}{\alpha_1 + \alpha_2} X_1 + \frac{\alpha_2}{\alpha_1 + \alpha_2} X_2$ where $\alpha_1 = \frac{\sigma_2^2 - \sigma_{12}}{\sigma_1^2\sigma_2^2 -\sigma_{12}^2}$ and $\alpha_2 = \frac{\sigma_1^2 - \sigma_{12}}{\sigma_1^2\sigma_2^2 -\sigma_{12}^2}$. Note, if $\sigma_{12} = 0$ then it reduces to $\frac{\sigma_1^2 X_1 + \sigma_2^2 X_2}{\sigma_1^2 + \sigma_2^2}$ as noted above.&lt;/p&gt;&#xA;" OwnerUserId="404" LastActivityDate="2010-11-05T17:24:41.663" CommentCount="1" />
  <row Id="9082" PostTypeId="2" ParentId="9075" CreationDate="2010-11-05T17:27:00.863" Score="2" ViewCount="" Body="&lt;p&gt;Taking log of $\rm\ sin^2(x)\ =\ 1 - cos^2(x)\ = (1-cos(x))\ (1+cos(x))\ $ shows both answers identical&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-11-05T17:27:00.863" />
  <row Id="9083" PostTypeId="2" ParentId="9075" CreationDate="2010-11-05T17:42:13.597" Score="0" ViewCount="" Body="&lt;p&gt;Another way to integrate would be to let $\displaystyle t = \frac{\sin{x}}{\cos{x}}$, then $\displaystyle \frac{dt}{dx} = \frac{1}{\cos^2{x}} \Rightarrow dx = \cos^2{x}\;{dt}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus $ \displaystyle I = \int\frac{\cos^2{x}}{\sin{x}}\;{dt} = \int\frac{\cos{x}}{\sin{x}}\;{dt} = \int\frac{1}{t}\;{dt} = \ln{t}+k = \ln\frac{\sin{x}}{\cos{x}}+k$. &lt;/p&gt;&#xA;" LastActivityDate="2010-11-05T17:48:23.537" CommentCount="1" />
  <row Id="9084" PostTypeId="2" ParentId="9032" CreationDate="2010-11-05T18:18:06.713" Score="0" ViewCount="" Body="&lt;p&gt;More generally, the minimal-variance unbiased linear combination of estimators is their reciprocal-variance-weighted average.&lt;/p&gt;&#xA;" OwnerUserId="875" LastActivityDate="2010-11-05T18:18:06.713" />
  <row Id="9085" PostTypeId="1" AcceptedAnswerId="9087" CreationDate="2010-11-05T18:59:15.770" Score="2" ViewCount="215" Body="&lt;p&gt;I am wondering about particle trajectories for solutions of the Navier-Stokes equation.  Is it possible that there is a Manifold $M$ for which fluid particles move along geodesic's or &quot;straight lines on the manifold&quot;?  If so, would it be fair to say that the Navier-Stokes equations are partial differential equations which &quot;embed&quot; this manifold in $R^3$?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This would be somewhat analogous to the way that the &lt;a href=&quot;http://en.wikipedia.org/wiki/Poincar%C3%A9_disc_model&quot; rel=&quot;nofollow&quot;&gt;Poincare Disk Model&lt;/a&gt; transforms straight lines in the hyperbolic plane into &lt;a href=&quot;http://en.wikipedia.org/wiki/Hyperbolic_geometry&quot; rel=&quot;nofollow&quot;&gt;arcs of circles that are orthogonal to the boundary circle&lt;/a&gt;.  Except in this case, the &quot;straight lines&quot; are transformed into fluid particle trajectories in $R^3$.&lt;/p&gt;&#xA;" OwnerUserId="1417" LastEditorUserId="1543" LastEditorDisplayName="" LastEditDate="2011-04-29T15:26:39.987" LastActivityDate="2011-04-29T15:26:39.987" Title="Manifold interpretation of Navier-Stokes equations" Tags="&lt;differential-geometry&gt;&lt;differential-equations&gt;&lt;manifold&gt;" AnswerCount="2" FavoriteCount="0" />
  <row Id="9086" PostTypeId="1" CreationDate="2010-11-05T19:05:51.967" Score="0" ViewCount="141" Body="&lt;p&gt;The rational root theorem states that, given $P \in \mathbb{Z}[x], P = \displaystyle\sum_{j=0}^n a_j x^j$ with $a_0 \neq 0$ and $a_n \neq 0$, if $P(p/q)=0, p\perp q$ then $p|a_0$ and $q|a_n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a follow-up to &lt;a href=&quot;http://math.stackexchange.com/questions/8718/proving-gauss-polynomial-theorem&quot;&gt;this question&lt;/a&gt;, I'd like to know how can I prove the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a_0 = a_n \displaystyle\prod_{j=1}^n (-r_j)$, with $r_j$ being roots of P(x)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the same as saying:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a_0 = a_n \frac{-p}{q} \displaystyle\prod_{j=2}^n (-r_j)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rearranging this, we get:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a_n = q\frac{a_0}{-p \displaystyle\prod_{j=2}^n (-r_j)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to prove that $\frac{a_0}{-p \displaystyle\prod_{j=2}^n (-r_j)} \in \mathbb{Z}$, without using the rational root theorem (actually, I want to prove the rational root theorem using this identity).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So far I've tried to prove it by contradiction, assuming $a_0 = k(-p)\displaystyle\prod_{j=2}^n (-r_j) + \epsilon$ with $\epsilon \in (0, (-p)\displaystyle\prod_{j=2}^n (-r_j))$ but I don't seem to find anything weird rearranging that equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you manage to find a proof please just post a small hint. Thanks in advance.&lt;/p&gt;&#xA;" OwnerUserId="1523" LastEditorUserId="1523" LastEditorDisplayName="" LastEditDate="2010-11-05T19:28:29.913" LastActivityDate="2010-12-09T05:44:25.243" Title="Proving a statement involving the rational root theorem." Tags="&lt;linear-algebra&gt;&lt;polynomials&gt;" AnswerCount="1" CommentCount="9" />
  <row Id="9087" PostTypeId="2" ParentId="9085" CreationDate="2010-11-05T19:28:03.270" Score="11" ViewCount="" Body="&lt;p&gt;For the Navier Stokes equation, I am rather doubtful, because the presence of the viscosity term makes the equation not time-symmetric. And the geodesic flow is necessarily time-symmetric. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you remove the viscosity term, you are down to the Euler equation, for which there is a well-known &lt;a href=&quot;http://terrytao.wordpress.com/2010/06/07/the-euler-arnold-equation/&quot; rel=&quot;nofollow&quot;&gt;characterisation due to V I Arnold&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1543" LastActivityDate="2010-11-05T19:28:03.270" />
  <row Id="9088" PostTypeId="1" AcceptedAnswerId="9091" CreationDate="2010-11-05T19:42:42.627" Score="1" ViewCount="71" Body="&lt;p&gt;If you've got multiple arrays like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(24,36,28,28,16,27)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(38,38,45,57,35,50)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;every array being 6 integers, each integer in range [0,60]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to find the distance between those 2 arrays.  The problem is related to computer science, however, I would like to interpret these arrays as &quot;planes&quot; so as to find the distance between the two.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What's the best way to find the distance between those 2 arrays?  Can we interpret them as 6 points on a plane?  Or maybe 3 coordinates?&lt;/p&gt;&#xA;" OwnerUserId="1630" LastActivityDate="2010-11-05T19:51:28.813" Title="Problem interpretation - Distance Formula?" Tags="&lt;geometry&gt;&lt;computer-science&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="9089" PostTypeId="2" ParentId="9058" CreationDate="2010-11-05T19:44:51.207" Score="2" ViewCount="" Body="&lt;p&gt;The RHS of &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x^{2}-ax+144\equiv (x-b)^{2}\qquad (1)$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;can be expanded into&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x^{2}-2bx+b^{2}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So identity $(1)$ is equivalent to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x^{2}-ax+144\equiv x^{2}-2bx+b^{2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(2b-a)x+144-b^{2}\equiv 0,$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;which can only be an identity&lt;/em&gt; (&lt;em&gt;i.e. independent of the value of&lt;/em&gt; $x$) &lt;em&gt;if&lt;/em&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$a=2b$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$b^{2}=144.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This system of equations has two solutions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$b=-12,a=-24$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$b=12,a=24.$$&lt;/p&gt;&#xA;" OwnerUserId="752" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2010-11-05T23:17:58.583" LastActivityDate="2010-11-05T23:17:58.583" />
  <row Id="9090" PostTypeId="2" ParentId="9057" CreationDate="2010-11-05T19:46:07.927" Score="0" ViewCount="" Body="&lt;p&gt;If I understand properly, you have a histogram of A/B values for various proteins.  You don't need any model to find the most extreme, just find the 2.5 percentile and 97.5 percentile by counting.  The proteins outside these bounds are the 5% most extreme.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-11-05T19:46:07.927" CommentCount="1" />
  <row Id="9091" PostTypeId="2" ParentId="9088" CreationDate="2010-11-05T19:51:28.813" Score="4" ViewCount="" Body="&lt;p&gt;You can interpret them any way you want, each array as three points in a plane (but using which order), two points in space, one point in six-dimensional space.  There are various distances available.  If you define the points as $a=(a_1,a_2,a_3,a_4,a_5,a_6)$ and similarly for $b$ you can use  $\sum |a_i-b_i|$ or $\sum (a_i-b_i)^2$, for example.  Note I didn't use the dimension at all.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-11-05T19:51:28.813" CommentCount="5" />
  <row Id="9092" PostTypeId="1" CreationDate="2010-11-05T20:01:16.257" Score="5" ViewCount="222" Body="&lt;p&gt;How can I compute the following limit?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle \lim_{a \to 0^+} \left(a \int_{1}^{\infty} e^{-ax}\cos \left(\frac{2\pi}{1+x^{2}} \right)\mathrm dx\right)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any hints you can please give?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cheers&lt;/p&gt;&#xA;" OwnerUserId="4219" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-02T17:51:52.373" LastActivityDate="2010-12-02T18:41:42.373" Title="Compute $\lim_{a \to 0^+} \left(a \int_1^{\infty} e^{-ax}\cos \left(\frac{2\pi}{1+x^{2}} \right)\mathrm dx\right)$" Tags="&lt;calculus&gt;&lt;integral&gt;" AnswerCount="3" CommentCount="3" />
  <row Id="9094" PostTypeId="2" ParentId="9092" CreationDate="2010-11-05T20:13:05.327" Score="2" ViewCount="" Body="&lt;p&gt;Hint: try a change of the independent variable $y:=ax$.&lt;/p&gt;&#xA;" OwnerUserId="723" LastActivityDate="2010-11-05T20:13:05.327" CommentCount="4" />
  <row Id="9095" PostTypeId="1" AcceptedAnswerId="9097" CreationDate="2010-11-05T20:30:02.633" Score="1" ViewCount="223" Body="&lt;p&gt;Hi! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am having some problem wrapping my head around an assignment. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have to find out how many additions, subtractions, multiplications and divisions are used while solving an NxM linear equation, while using Gauss Jordan elimination. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could anyone point me in the right direction?&lt;/p&gt;&#xA;" OwnerUserId="2078" LastActivityDate="2010-11-05T20:41:27.907" Title="Gauss Jordan elimination - count of steps for NxM equation" Tags="&lt;linear-algebra&gt;&lt;algebra&gt;" AnswerCount="1" />
  <row Id="9096" PostTypeId="2" ParentId="9069" CreationDate="2010-11-05T20:36:59.363" Score="0" ViewCount="" Body="&lt;p&gt;So $E[W| \text{fire}] = \int_{0.1}^{\infty} [1-F(x)] \ dx = 4 \int_{0.1}^{\infty} (2+x)^{-2} \ dx = 4/2.1$. Thus $E[W] = (4/2.1)(0.001)+0 = 0.00190476$. So the premium is $(0.0019047)(10^6)(1.1) =2,095$. &lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2010-11-05T20:36:59.363" />
  <row Id="9097" PostTypeId="2" ParentId="9095" CreationDate="2010-11-05T20:41:27.907" Score="3" ViewCount="" Body="&lt;p&gt;It depends on how efficient you want to be, and it also depends on the rank of the matrix. But for a &quot;generic&quot; matrix, you can expect the rank to be $\min(N,M)$, and for &quot;most&quot; entries to be nonzero. (And, do you count multiplying zero by a constant as a &quot;multiplication&quot;? Adding two zeros as an &quot;addition&quot;?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I interpret your question as asking for the number of multiplications and additions of entries, rather than adding rows, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Every time you multiply a row by a constant, you are performing $M$ multiplications (minus the number of entries that are zero if they don't count); every time you add a row to another row you are performing $M$ additions/subtractions (minus the number of entries that are zero in both).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, generically, let us assume that the only zeros you will encounter are the ones you create throught Gauss-Jordan elimination.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, how do you get started? First you pick a pivot; then you use that pivot to eliminate the other entries in the column; there are $N-1$ entries to be eliminated; each elimination requires (i) multiplying your pivot row by a constant; and (ii) adding the pivot row to another row. So you are performing $M$ products and $M$ sums in each step, and you are doing this $N-1$ times. Then, you multiply the first row appropriately to get a $1$ in the pivot, for another $M$ products.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next you pick your next pivot, and you need to eliminate the entries below it on the column; there are $N-2$ entries; each entry requires (i) multiplying your pivot row by a constant, which requires $M-1$ products (first entry is $0$, you can ignore it), and then (ii) adding to the other row, requiring $M-1$ sums (first entry in both is $0$, you can ignore it). You do this $N-2$ times, and then multiply the first row by a constant ($M-1$ products again) to get a $1$ in the pivot.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then you pick your next pivot... etc. Figure out how many products/sums we are doing in each step, how many steps, add them all up. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This will be an &lt;em&gt;upper bound&lt;/em&gt;, since in practice you could end up with lots of zeros, or a matrix that has rank strictly smaller than $\min(M,N)$, so that you get a row of zeroes early on, etc. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reading your title, though, you might be counting &quot;add a row to another row&quot; as a single step, and &quot;multiply a row by a constant&quot; as a single step; in that case, proceed as above but you only count the number of row operations instead of the number of individual numerical operations, and the number of common zeros between the rows is irrelevant.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-11-05T20:41:27.907" CommentCount="1" />
  <row Id="9098" PostTypeId="1" AcceptedAnswerId="9104" CreationDate="2010-11-05T21:21:42.770" Score="0" ViewCount="62" Body="&lt;p&gt;Dear Friends,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like, if possible, to obtain a proof of the theorem below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Being given real numbers &lt;strong&gt;a&lt;/strong&gt; and &lt;strong&gt;b&lt;/strong&gt;, with |a|&gt;|b|, and &lt;strong&gt;m&lt;/strong&gt; is a positive integer, the order &lt;strong&gt;p&lt;/strong&gt;, which occupies the maximum term (in absolute value) the development of power &lt;strong&gt;(a+b)^(-m)&lt;/strong&gt; , according to decreasing powers of &lt;strong&gt;a&lt;/strong&gt;, is given by:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;p = 1 + integer part of [|b|(m-1)/(|a|-|b|)]&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the expression &lt;strong&gt;|b|(m-1)/ (|a|-|b|)&lt;/strong&gt; is an integer, then there are maximum two terms (in absolute value): they are of order &lt;strong&gt;p&lt;/strong&gt; and &lt;strong&gt;p-1&lt;/strong&gt;. &quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Already, most grateful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Paulo Argolo&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rio de Janeiro, Brazil&lt;/p&gt;&#xA;" OwnerUserId="2938" LastActivityDate="2010-11-05T22:40:42.567" Title="Order of the maximum term of (a +b)^(-m)" Tags="&lt;binomial-coefficients&gt;" AnswerCount="1" CommentCount="0" />
  <row Id="9099" PostTypeId="1" CreationDate="2010-11-05T21:22:26.263" Score="5" ViewCount="228" Body="&lt;p&gt;I have an equation system of the form Aix + Biy + Ciz = Di, where (x,y,z) is a unit vector, and (Ai, Bi, Ci, Di) are sets of measurements from a noisy system (with typically 3-5 independant readings). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My first intuition to solve this problem was to pose this as an overdetermined linear equation system AX = B where X = (x,y,z), and to solve for X. However, with that approach, I have no way to enforce that the solution for vector X is a unit vector. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there an elegant (or standard) solution to that problem, or should I simply dive into non-linear equation solving solutions?&lt;/p&gt;&#xA;" OwnerUserId="64" LastActivityDate="2011-01-16T22:49:33.233" Title="Solve a linear equation system, while enforcing a unit vector solution." Tags="&lt;linear-algebra&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" />
  <row Id="9100" PostTypeId="2" ParentId="9099" CreationDate="2010-11-05T22:04:39.523" Score="4" ViewCount="" Body="&lt;p&gt;Unfortunately, the problem is almost certainly innately non-linear, since your solution space is inherently so.  Pragmatically, the best approach is probably to treat your overdetermined linear system as a minimization-problem, least-squares style; this gives you a (positive definite) function to be minimized.  Find the (exact) minimum value in $\mathbb{R}^3$, project that down to $S^2$ (i.e., normalize your vector X) to find the closest point on the sphere, and use that as a starting point for some descent-based method (i.e., find the component of the gradient of your goodness-of-fit function tangent to the sphere, add that to your current point on $S^2$ and renormalize down to the sphere again, and keep doing that until you meet whatever convergence criterion you want to use.  There's no guarantee that this will work if your data is particularly ill-conditioned, but in practice it's likely to be the most straightforward method.&lt;/p&gt;&#xA;" OwnerUserId="785" LastActivityDate="2010-11-05T22:04:39.523" />
  <row Id="9101" PostTypeId="2" ParentId="9099" CreationDate="2010-11-05T22:11:11.257" Score="2" ViewCount="" Body="&lt;p&gt;Your objective is to solve $Ax=b$ subject to the constraint $||x||=1$. One way to do this is to recognize that the solution to $Ax=b$ is an affine space (translated subspace). This means that out of all the solutions to $Ax=b$, there is a unique &quot;smallest&quot; one, i.e. where $||x||$ is as small as possible. To find this solution, you can solve the minimium-norm optimization problem, whose solution is:&#xA;$$&#xA;x_0 = \lim_{\lambda \to 0^+} (A^TA+\lambda I)^{-1}A^Tb&#xA;$$&#xA;In the case where $A$ is skinny and full-rank, this simplifies to $(A^TA)^{-1}A^Tb$. In general, the solution can be found by taking the SVD of $A$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you have verified that $x_0$ is indeed a solution, the next thing to do is check that $||x_0|| \leq 1$, for if that were not the case, then you could confidently assert that there is no solution to $Ax=b$ with norm 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The next step is to find a basis for the nullspace of $A$, i.e. the set of vectors $x$ such that $Ax=0$. This can be done using the SVD again. If $N$ is a matrix whose columns form a basis for the nullspace of $A$, then the set of all solutions to $Ax=b$ is:&#xA;$$&#xA;x = x_0 + Ny&#xA;$$&#xA;where $y$ is an arbitrary vector of appropriate dimension. The key point here is that $x_0$ is orthogonal to all vectors in the nullspace of $A$ (because $x_0$ is the solution point closest to the origin). Thus, you may compute the norm of $x$ using the Pythagorean theorem:&#xA;$$&#xA;||x||^2 = ||x_0||^2 + ||Ny||^2&#xA;$$&#xA;Now we see that the if we want $||x||=1$, we simply need to choose $y$ so that&#xA;$$&#xA;||Ny|| = \sqrt{1 - ||x_0||^2}&#xA;$$&#xA;We may always choose the columns of $N$ to be orthonormal, so let's assume we have done that. Then the set of desired $y$ is precisely the sphere with radius $\sqrt{1 - ||x_0||^2}$. By choosing any $y$ on this sphere, we parametrize all solutions to your problem. If you only care about a particular solution, then you can choose $y$ of the form $[\alpha,0,...,0]$, where $\alpha = \sqrt{1 - ||x_0||^2}$.f&lt;/p&gt;&#xA;" OwnerUserId="1356" LastEditorUserId="1356" LastEditorDisplayName="" LastEditDate="2010-11-06T02:17:56.420" LastActivityDate="2010-11-06T02:17:56.420" CommentCount="7" />
  <row Id="9104" PostTypeId="2" ParentId="9098" CreationDate="2010-11-05T22:40:42.567" Score="0" ViewCount="" Body="&lt;p&gt;The expansion is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sum_{p=0}^{\infty} \binom{-m}{p} a^{-m-p} b^p$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The ratio between the absolute values of the terms corresponding to $p+1$ and to $p$ is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{m+p}{p+1} \cdot \frac{|b|}{|a|}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The $p+1$st term is larger than or equal to the $p$th term if&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$m+p \geq (p+1)\frac{|a|}{|b|}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$p \left(1 - \frac{|a|}{|b|}\right) \geq \frac{|a|}{|b|} - m$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $|a| &amp;lt; |b|$, this is the same as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$p \leq \frac{|a|/|b|-m}{1-|a|/|b|} = \frac{|a|-m|b|}{|b|-|a|}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$p + 1 \leq \frac{(1-m)|b|}{|b|-|a|} = \frac{(m-1)|b|}{|a|-|b|}$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the $p$th term is maximal if&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$p \leq \frac{(m-1)|b|}{|a|-|b|}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$p+1 \geq \frac{(m-1)|b|}{|a|-|b|}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the fraction is an integer $N$, then both $p = N$ and $p = N - 1$ are solutions. Otherwise, $p$ is the floor of the fraction, and I'm off by one wrt your solution.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-05T22:40:42.567" />
  <row Id="9105" PostTypeId="1" AcceptedAnswerId="9107" CreationDate="2010-11-05T22:50:11.627" Score="1" ViewCount="102" Body="&lt;p&gt;I have the following series of equations (n+2 equations n+2 variables):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$k_0q_0+\lambda q_0  + c_0 = 0$&lt;br&gt;&#xA;$k_1q_1+\lambda q_1  + c_1 = 0$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$k_nq_n+\lambda q_n  + c_n = 0$&lt;br&gt;&#xA;$q_1+q_2+....+q_n = 1$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The variables are $q_0,q_1,.....,q_n$ and $\lambda$. k and c are series of constants.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I solve this?&lt;/p&gt;&#xA;" OwnerUserId="2858" LastEditorUserId="2858" LastEditorDisplayName="" LastEditDate="2010-11-06T12:22:54.890" LastActivityDate="2010-11-06T12:22:54.890" Title="Solving Series of equations" Tags="&lt;analysis&gt;&lt;sequences-and-series&gt;&lt;functional-equations&gt;" AnswerCount="3" CommentCount="3" />
  <row Id="9106" PostTypeId="2" ParentId="9105" CreationDate="2010-11-05T22:57:31.847" Score="2" ViewCount="" Body="&lt;p&gt;It's not linear, first of all: you have two variables $\lambda$ and $q_i$ multiplied together. Luckily, your set of equations is algebraic; &lt;a href=&quot;http://mathworld.wolfram.com/GroebnerBasis.html&quot; rel=&quot;nofollow&quot;&gt;Gröbner&lt;/a&gt; may have a shot here.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-11-05T22:57:31.847" CommentCount="2" />
  <row Id="9107" PostTypeId="2" ParentId="9105" CreationDate="2010-11-05T23:10:06.827" Score="4" ViewCount="" Body="&lt;p&gt;$\displaystyle q_i = \frac{-c_i}{k_i + \lambda}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Substituting in the last equation gives us&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle \sum_{i=1}^{n} \frac{-c_i}{k_i + \lambda} = 1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which can be made into a polynomial equation in $\displaystyle \lambda$, which you should be able to solve by standard numerical methods.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you find a root of the above polynomial, substituting $\displaystyle \lambda$ with the value of the root in $\displaystyle q_i = \frac{-c_i}{k_i + \lambda}$ will give the other variables.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-11-05T23:10:06.827" CommentCount="3" />
  <row Id="9108" PostTypeId="2" ParentId="9077" CreationDate="2010-11-05T23:11:47.427" Score="2" ViewCount="" Body="&lt;p&gt;This was supposed to be a comment, but it got too long.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm personally not too enamored of the NR routine for Savitzky-Golay myself; it has to use Gaussian elimination (and thus a two-dimensional array) for what I think is a well-structured problem. Peter Gorry's proposal in &lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/ac00205a007&quot; rel=&quot;nofollow&quot;&gt;this paper&lt;/a&gt; is a better choice, IMHO (a proper implementation merely requires a few one-dimensional arrays). In any event, looking at the &lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/ac60214a047&quot; rel=&quot;nofollow&quot;&gt;original article&lt;/a&gt; by Savitzky and Golay should help you a great deal, as well as F.B. Hildebrand's &lt;a href=&quot;http://books.google.com/books?id=hqucruPBheQC&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;Introduction to Numerical Analysis&lt;/em&gt;&lt;/a&gt;; Hildebrand does not explicitly refer to the Savitzky-Golay coefficients, but he gives the &lt;a href=&quot;http://books.google.com/books?id=hqucruPBheQC&amp;amp;pg=PA357&quot; rel=&quot;nofollow&quot;&gt;general idea&lt;/a&gt; of using least-squares fit polynomials for smoothing data.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-11-05T23:11:47.427" CommentCount="4" />
  <row Id="9109" PostTypeId="1" CreationDate="2010-11-05T23:22:43.680" Score="1" ViewCount="121" Body="&lt;p&gt;How to find a function which satisfies:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$a^x=\lim_{h\to\infty} \left( f_a \left(f_a^{[-1]}(x)+\frac xh\right)\right)^{[h]}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $f^{[n]}(x)$ is the number if iterations of a function (if n=-1 it's inverse function)?&lt;/p&gt;&#xA;" OwnerUserId="2513" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-06T00:54:40.290" LastActivityDate="2010-11-06T00:54:40.290" Title="How to find a function which satisfies such functional equation?" Tags="&lt;calculus&gt;&lt;analysis&gt;&lt;functional-equations&gt;&lt;dynamical-systems&gt;" CommentCount="2" />
  <row Id="9110" PostTypeId="1" CreationDate="2010-11-05T23:35:48.137" Score="0" ViewCount="77" Body="&lt;p&gt;Is there a simple/efficient way of enumerating the elements of a free distributive lattice?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I'm doing some computations with them, and listing their elements in what's probably the least efficient way possible...)&lt;/p&gt;&#xA;" OwnerUserId="274" LastEditorUserId="274" LastEditorDisplayName="" LastEditDate="2010-11-05T23:42:29.520" LastActivityDate="2010-11-06T00:12:14.807" Title="Enumerating the elements of free distributive lattices" Tags="&lt;lattices&gt;" AnswerCount="1" />
  <row Id="9111" PostTypeId="2" ParentId="5008" CreationDate="2010-11-06T00:11:30.410" Score="3" ViewCount="" Body="&lt;p&gt;It is probably worth mentioning that another name for this is the category of abstract (not necessarily separated) algebraic $k$-varieties.  &lt;/p&gt;&#xA;" OwnerUserId="299" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2010-11-06T08:31:51.333" LastActivityDate="2010-11-06T08:31:51.333" CommentCount="2" />
  <row Id="9112" PostTypeId="2" ParentId="9110" CreationDate="2010-11-06T00:12:14.807" Score="1" ViewCount="" Body="&lt;p&gt;It's equivalent to enumerating monotone Boolean functions, e.g. see Fidytek et al. &lt;a href=&quot;http://www.ii.uni.wroc.pl/~lorys/IPL/article79-6-4.pdf&quot; rel=&quot;nofollow&quot;&gt;Algorithms counting monotone Boolean functions&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-11-06T00:12:14.807" CommentCount="1" />
  <row Id="9113" PostTypeId="2" ParentId="9092" CreationDate="2010-11-06T00:25:38.313" Score="1" ViewCount="" Body="&lt;p&gt;Dear Andrey, Thanks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I end up with:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle \lim_{a \to 0+} \int _{a}^{\infty} e^{-y} cos \left( \frac{2\pi}{1+(y/a)^{2}}\right) dy$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then I'm stuck. How do I apply DCT here?&lt;/p&gt;&#xA;" OwnerUserId="4219" LastActivityDate="2010-11-06T00:25:38.313" CommentCount="1" />
  <row Id="9114" PostTypeId="2" ParentId="9105" CreationDate="2010-11-06T01:55:59.887" Score="0" ViewCount="" Body="&lt;p&gt;Define $c_{n+1} = 1$ and $q_{n+1} = 0$. Observe that we can write the above system in the following $(n+2) \times (n + 2)$-matrix form: $K_{\lambda} \mathbf{q} = - \mathbf{c}$, where $K_{\lambda}$ is $\text{diag}(k_{i} + \lambda)$ for the upper $(n+1) \times (n+1)$-submatrix, ($0$ $1$ $1$ $\cdots$ $1$ $0$) for the bottom row and ($0$ $0$ $0$ $\cdots$ $0$ $1$)$^{\top}$ for the right most column. (By enlarging the matrix, we have encoded the constraint $\sum_{i = 1}^{n} q_{i} = 1$.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $\lambda$ is unknown, we can proceed presupposing that $K_{\lambda}$ is not singular (i.e., $\det K_{\lambda} \neq 0$), and solve the constrained system by matrix inversion: $\mathbf{q} =  - K_{\lambda}^{-1} \mathbf{c}$. The result is a vector $\mathbf{q}$ in terms of the unknown $\lambda$ and given constants. To find the allowed values for $\lambda$, numerically (or analytically) solve the polynomial $\mathbf{q} \cdot \mathbf{1} - 1 = 0$.&lt;/p&gt;&#xA;" OwnerUserId="2720" LastEditorUserId="2720" LastEditorDisplayName="" LastEditDate="2010-11-06T02:35:38.043" LastActivityDate="2010-11-06T02:35:38.043" />
  <row Id="9115" PostTypeId="1" AcceptedAnswerId="9117" CreationDate="2010-11-06T02:47:57.693" Score="5" ViewCount="188" Body="&lt;p&gt;So, I'm relearning Group Theory.  And I got the axioms down, I think.  So let's make a concrete example:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The collection of numbers the positive integers less than 7:  1,2,3,4,5,6&lt;/li&gt;&#xA;&lt;li&gt;The • operation will be multiplication mod 7.  &lt;/li&gt;&#xA;&lt;li&gt;Associativity holds.&lt;/li&gt;&#xA;&lt;li&gt;The Identity e is 1.&lt;/li&gt;&#xA;&lt;li&gt;Every element has an inverse: &#xA;&lt;ul&gt;&#xA;&lt;li&gt;1*? mod 7 = 1 --&gt; 1&lt;/li&gt;&#xA;&lt;li&gt;2*? mod 7 = 1 --&gt; 4&lt;/li&gt;&#xA;&lt;li&gt;3*? mod 7 = 1 --&gt; 5&lt;/li&gt;&#xA;&lt;li&gt;4*? mod 7 = 1 --&gt; 2&lt;/li&gt;&#xA;&lt;li&gt;5*? mod 7 = 1 --&gt; 3&lt;/li&gt;&#xA;&lt;li&gt;6*? mod 7 = 1 --&gt; 6&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;But&lt;/strong&gt;!  What is the &lt;em&gt;order&lt;/em&gt; of the group?!  I thought the order would be 7.  But there are 6 elements!  So maybe I was wrong and 0 should be in the group.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But 0 does not have an inverse!  There is no x such that 0*x mod 7 = 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what am I misunderstanding here?  Is it the definition of order?  Is it some other trick about groups?&lt;/p&gt;&#xA;" OwnerUserId="3179" LastActivityDate="2010-11-06T02:57:08.880" Title="What happens to the 0 element in a Finite Group?" Tags="&lt;group-theory&gt;" AnswerCount="3" FavoriteCount="1" />
  <row Id="9116" PostTypeId="1" AcceptedAnswerId="9170" CreationDate="2010-11-06T02:48:00.370" Score="3" ViewCount="317" Body="&lt;p&gt;I am seeking conditions on the distribution of the step sizes that guarantee that a  random walk on the 2D lattice will return to the origin (with probability 1).  Essentially, under what conditions can Pólya's theorem be proved?  Certainly it holds if the steps are of size 1 (and equally probable in all directions), and I believe if the variance of the step sizes is finite, the return probability is still 1.&#xA;But what about distributions with infinite variance, like the &lt;a href=&quot;http://en.wikipedia.org/wiki/L%C3%A9vy_distribution&quot; rel=&quot;nofollow&quot;&gt;Lévy distribution&lt;/a&gt;?&#xA;Does &quot;Lévy flight&quot; return to the origin?  And more generally, are there conditions on&#xA;the distribution that guarantee this return?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is likely all well known, and if so, pointers to the relevant literature would be appreciated.&#xA;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="237" LastEditorUserId="237" LastEditorDisplayName="" LastEditDate="2010-11-06T13:13:47.677" LastActivityDate="2010-11-06T15:27:01.460" Title="Conditions for 2D random walk to return to origin" Tags="&lt;probability-theory&gt;&lt;reference-request&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="2" />
  <row Id="9117" PostTypeId="2" ParentId="9115" CreationDate="2010-11-06T02:51:15.143" Score="9" ViewCount="" Body="&lt;p&gt;The only error is your belief that the order &quot;should&quot; be 7. The order of a finite group is just the number of elements in the group. Your group consists of the positive integers that are smaller than, and relatively prime to, 7. There are six of them, so your group has order 6.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I'm not sure why you thought the order should be 7...)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Indeed, you cannot add $0$ to the mix and still have a group. If you consider the numbers $0,1,\ldots,6$ under multiplication modulo $7$ you do not get a group, you get a &lt;a href=&quot;http://en.wikipedia.org/wiki/Semigroup&quot; rel=&quot;nofollow&quot;&gt;semigroup&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added:&lt;/strong&gt; Ah, Jonas Meyer's reply suggests what is going on; since you say you are &lt;em&gt;relearning&lt;/em&gt; Group Theory, you might have vague memories of the &quot;group of integers modulo $n$&quot; as having order $n$. The group of integers modulo $n$ &lt;em&gt;under addition&lt;/em&gt; has order $n$; but the &lt;strong&gt;multiplicative group&lt;/strong&gt; modulo $n$ consists of the positive integers less than, and relatively prime to, $n$, with the operation being multiplication modulo $n$, and has $\varphi(n)$ elements (Euler's phi function). When $n=7$ (the case you are looking at), the group has $\varphi(7)=6$ elements, as you observed.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-06T02:57:08.880" LastActivityDate="2010-11-06T02:57:08.880" CommentCount="1" />
  <row Id="9118" PostTypeId="2" ParentId="9115" CreationDate="2010-11-06T02:51:25.720" Score="7" ViewCount="" Body="&lt;p&gt;You're right, the group has order 6 because it has six elements.  You can make {0,1,2,3,4,5,6} a group with addition mod 7.  This would be a group of order 7.  &lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-11-06T02:51:25.720" />
  <row Id="9119" PostTypeId="2" ParentId="9115" CreationDate="2010-11-06T02:51:28.473" Score="5" ViewCount="" Body="&lt;p&gt;Good thinking.  0 is not in the group, so the order is 6.  0 is in another group in the multiplicative $\mathbb{Z}/\mathbb{Z}_7$, which is the trivial group (one element).  The groups are disjoint.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-11-06T02:51:28.473" />
  <row Id="9120" PostTypeId="1" AcceptedAnswerId="9121" CreationDate="2010-11-06T03:15:30.023" Score="0" ViewCount="118" Body="&lt;p&gt;If we can say that we're able to determine the weather at 10ms intervals, and if we observe that over a whole second (1000 samples) that it's raining at every sample, what is the accuracy rate within which we've calculated the weather for the second? I'm sure this sounds like homework to some, but it's not. I was reading a statement about sampling rates that claimed a sample every 10ms, and that the result was within 10% accuracy. I am not sure how they came up with this guarantee.&lt;/p&gt;&#xA;" OwnerUserId="3180" LastActivityDate="2010-11-06T03:34:41.717" Title="How can you calculate the accuracy rate of the following set of statements?" Tags="&lt;statistics&gt;" AnswerCount="1" />
  <row Id="9121" PostTypeId="2" ParentId="9120" CreationDate="2010-11-06T03:34:41.717" Score="0" ViewCount="" Body="&lt;p&gt;That depends whether you calculated that it would be raining or not.  The accuracy is 100% if yes, 0% if no.  Flip response aside, it is hard to claim an accuracy rate for a single digital event.  10% accuracy should either refer to an analog prediction, in which case you should be within 10% of the measured value, or a many event digital prediction, in which case the observed rate should be within 10% of your prediction.  But in the popular press, &quot;10% accuracy&quot; has no numeric content, it just means &quot;pretty good, but not perfect&quot;.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-11-06T03:34:41.717" />
  <row Id="9122" PostTypeId="1" AcceptedAnswerId="9124" CreationDate="2010-11-06T05:29:21.267" Score="3" ViewCount="233" Body="&lt;p&gt;I have a generating function that I'm trying to create.  It has a general form to it:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$1 + cx + c(c-1)x^2 + c(c-1)(c-2)x^3 + \dots + c!x^c$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to see a closed form for this function.  I'd like something other than a summation.  I'm particularly interested if someone could walk through how to derive the form for me.  I was hoping that someone could carry on with what I've done and show me how to solve what's left.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;My attempt&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So I try to create a function $A(x)$ (which will be the closed form) using recurrences.  Here I'm using techniques (and especially notation) from &lt;a href=&quot;http://www.math.upenn.edu/~wilf/DownldGF.html&quot; rel=&quot;nofollow&quot;&gt;Wilf's &lt;em&gt;Generatingfunctionology&lt;/em&gt;&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I set $a_0=1$.  I then proceed to multiply by $(c-n)$.  My guess is that the math should be something like the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a_{n+1}=c \cdot a_n - \frac{d A(x)}{dx}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm trying to say that the next term should be $c$ times the previous term minus $n$ times the previous term.  The latter is represented by the derivative, I believe.  So I guess that this could be correct.  Unfortunately, I don't know for sure, and so I'm kind of lost at this point.  I would greatly appreciate if someone could take it from here and explain the rest in detail.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The Differential Equation&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Assuming that the last part is correct, the function becomes a differential equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{A(x)-1}{x}=c \cdot A(x) - \frac{d A(x)}{dx}$&lt;/p&gt;&#xA;" OwnerUserId="2626" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2011-06-03T20:05:56.197" LastActivityDate="2011-06-03T20:05:56.197" Title="Help with a generating function and differential equation..." Tags="&lt;homework&gt;&lt;sequences-and-series&gt;&lt;differential-equations&gt;&lt;generating-functions&gt;" AnswerCount="4" CommentCount="2" />
  <row Id="9123" PostTypeId="1" AcceptedAnswerId="9125" CreationDate="2010-11-06T05:37:46.653" Score="1" ViewCount="145" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;In maths, you can use something as&#xA;  simple as statistical analysis to&#xA;  intuit the theory, and in comp-science&#xA;  you can use simulators.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Is the above statement true for maths ?&lt;/p&gt;&#xA;" OwnerUserId="3027" LastActivityDate="2010-11-06T07:21:40.307" Title="Is the above statement true for maths ?" Tags="&lt;soft-question&gt;&lt;statistics&gt;&lt;math-history&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="9124" PostTypeId="2" ParentId="9122" CreationDate="2010-11-06T06:20:22.180" Score="3" ViewCount="" Body="&lt;p&gt;I believe that your initial attempt is incorrect. If I solve the suggested differential equation, I get&#xA;\begin{eqnarray}&#xA;A(x) = -\frac{1}{cx} \left( 1 - e^{cx} \right) = \sum_{k \geq 1} \frac{(cx)^{k-1}}{k!},&#xA;\end{eqnarray}&#xA;where I have chosen the initial condition to ensure $1$ is the first term of the series expansion. The generating function that you are seeking is the following:&#xA;\begin{eqnarray}&#xA;e^{1/x} \left( x^{c} \Gamma(c+1, \tfrac{1}{x}) - c (-x)^{c} \Gamma(0, \tfrac{1}{x}) (1 -c)^{(c)} \right) = 1 + \sum_{k = 1}^{c} c(c-1) \cdots (c-k+1) \ x^{k},&#xA;\end{eqnarray}&#xA;which may be simplified to&#xA;\begin{eqnarray}&#xA;e^{1/x} \left( \frac{(-x)^{c} \ \Gamma(0, \frac{1}{x})}{\Gamma(-c)} + x^{c} \ \Gamma(c + 1, \tfrac{1}{x}) \right) = 1 - \sum_{k = 1}^{c} c (1 - c)^{(k-1)} \ (-x)^{k},&#xA;\end{eqnarray}&#xA;where $\Gamma(n,x)$ is the incomplete gamma function defined as&#xA;\begin{eqnarray}&#xA;\Gamma(n,x) = \int_{x}^{\infty} t^{n-1} e^{-t} dt = (n-1)! e^{-x} \sum_{k = 0}^{n-1} \frac{x^{k}}{k!}&#xA;\end{eqnarray}&#xA;and $(x)^{(n)}$ is the Pochhammer symbol or rising factorial, $x(x+1)\cdots (x+n-1)$. This form makes no assumption on the integrality of $c$. However, if $c$ is a positive integer, then the formula simplifies greatly. For example, if $c = 3$, the left side specializes to &#xA;\begin{eqnarray}&#xA;e^{1/x} x^3 \Gamma(4, \tfrac{1}{x}) = 1 + 3 x + 6 x^{2} + 6 x^{3}. &#xA;\end{eqnarray}&#xA;NB: There is no extra term. The last exponent of $x$ is indeed $c$ with coefficient $c!$. In fact, the last coefficient is repeated because $c(c-1) \cdots 2 = c(c-1) \cdots 2 \cdot 1 = c!$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, the generating function for integer $c$ is &#xA;\begin{eqnarray}&#xA;e^{1/x} x^{c} \ \Gamma(c+1, \tfrac{1}{x}).&#xA;\end{eqnarray}&lt;/p&gt;&#xA;" OwnerUserId="2720" LastEditorUserId="2720" LastEditorDisplayName="" LastEditDate="2010-11-08T05:30:16.450" LastActivityDate="2010-11-08T05:30:16.450" CommentCount="5" />
  <row Id="9125" PostTypeId="2" ParentId="9123" CreationDate="2010-11-06T07:21:40.307" Score="2" ViewCount="" Body="&lt;p&gt;If I'm understanding the question correctly, I suppose the OP is aiming to posit that in mathematics we &lt;em&gt;verify&lt;/em&gt; a mathematical truth in nature by experimentation and statistical data analysis versus how we verify a computational truth by seeing it in practice through simulations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Not to be overly pedantic, but to me a mathematical or computational result is true when it is the conclusion of a rigorous application of sound thought through a sequence of logical steps from an initial set of assumptions within a particular (logical) framework. I see no distinction between mathematical or computational truths.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In practice, however, when we try to apply mathematical or computational rules to nature, the results aren't always so easy to interpret, because &lt;em&gt;truth&lt;/em&gt; isn't well-defined. This is why statistics is so applicable -- it gives us a way to quantify the grey area. Through simulations we can test ideas in an ideal setting, so the grey area is as big as we allow it to be. &lt;/p&gt;&#xA;" OwnerUserId="2720" LastActivityDate="2010-11-06T07:21:40.307" />
  <row Id="9126" PostTypeId="2" ParentId="9122" CreationDate="2010-11-06T08:05:26.663" Score="2" ViewCount="" Body="&lt;p&gt;The general form of your sum can be expressed succinctly as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum (-c)_k (-x)^k$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $(a)_k$ is a Pochhammer symbol.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We recognize at once that this is a hypergeometric series; specifically, it is a ${}_2 F_0$ :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$${}_2 F_0 \left(-c,1;;-x\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which can be rewritten as a &lt;a href=&quot;http://dlmf.nist.gov/13.6.vi&quot; rel=&quot;nofollow&quot;&gt;Tricomi confluent hypergeometric function&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac1{x}U\left(1,c+2,\frac1{x}\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which can be re-expressed as an &lt;a href=&quot;http://dlmf.nist.gov/8.5.E3&quot; rel=&quot;nofollow&quot;&gt;incomplete gamma function&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x^c\exp\left(\frac1{x}\right)\Gamma\left(c+1,\frac1{x}\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the differential equation, it can be obtained from &lt;a href=&quot;http://functions.wolfram.com/07.33.13.0011.01&quot; rel=&quot;nofollow&quot;&gt;this formula&lt;/a&gt; to give:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x^2\frac{\mathrm{d}^2 y}{\mathrm{d}x^2}-\left(c-2-\frac1{x}\right)x\frac{\mathrm{d}y}{\mathrm{d}x}-cy=0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where the solution of interest satisfies the initial conditions $y(0)=1$ and $y^{\prime}(0)=c$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll leave others to flesh out the details.&lt;/p&gt;&#xA;" OwnerUserId="498" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-06T09:58:24.620" LastActivityDate="2010-11-06T09:58:24.620" CommentCount="5" />
  <row Id="9127" PostTypeId="1" CreationDate="2010-11-06T08:06:29.767" Score="1" ViewCount="262" Body="&lt;p&gt;Say I have a function    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$V(x)=A(x-x_1)(x-x_2)(x-x_3)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $x_1$, $x_2$, $x_3$ are the three roots in increasing order and $A$ is positive. Clearly $V(x)$ is positive at large $x &amp;gt; x_3$, negative between $x_2$ and $x_3$, and so on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, say I wish to evaluate&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_{x_2}^{x_3}\sqrt{-V(x)}\mathrm dx$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do I do it?&lt;/p&gt;&#xA;" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2011-05-05T07:38:20.533" LastActivityDate="2011-05-05T07:38:20.533" Title="Integrals of the square root of a cubic polynomial" Tags="&lt;calculus&gt;&lt;integral&gt;" AnswerCount="2" />
  <row Id="9128" PostTypeId="2" ParentId="9127" CreationDate="2010-11-06T08:16:01.663" Score="3" ViewCount="" Body="&lt;p&gt;This is an &lt;a href=&quot;http://en.wikipedia.org/wiki/Elliptic_integral&quot; rel=&quot;nofollow&quot;&gt;elliptic integral&lt;/a&gt; and can be expressed&#xA;in terms of the complete elliptic integral of the second kind.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-11-06T08:16:01.663" />
  <row Id="9129" PostTypeId="2" ParentId="9127" CreationDate="2010-11-06T08:18:13.450" Score="7" ViewCount="" Body="&lt;p&gt;Well, you can't integrate it in terms of elementary functions like the arctangent or the logarithm in general (unless there is some special configuration of your roots). Having an integrand that contains the square root of a cubic will result in the use of so-called &quot;&lt;a href=&quot;http://dlmf.nist.gov/19&quot; rel=&quot;nofollow&quot;&gt;elliptic integrals&lt;/a&gt;&quot;.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I'll evaluate the integral explicitly here for reference; most of the current computing environments seem to be unable to return reasonable expressions for elliptic integrals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For this solution, I will assume $A=1$; you can multiply the result of the following by $\sqrt{A}$ afterwards.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, we consider first the &lt;em&gt;indefinite&lt;/em&gt; integral (let's worry about the limits later)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int\sqrt{(x_3-x)(x-x_2)(x-x_1)}\;\mathrm dx,\qquad x_3 &amp;gt; x_2 &amp;gt; x_1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first step here is to perform an appropriate Möbius (rational) transformation, which is equivalent to projecting your cubic to a quartic or quadratic, which hopefully has a nice factorization. I'll skip the details on how to obtain it, and point out that the required substitution is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x=\frac{x_2-x_1\frac{x_3-x_2}{x_3-x_1}u}{1-\frac{x_3-x_2}{x_3-x_1}u}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Performing this substitution yields&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{(x_3-x_2)^2(x_2-x_1)^2}{(x_3-x_1)\sqrt{x_3-x_1}}\int\frac1{\left(1-\frac{x_3-x_2}{x_3-x_1}u\right)^3}\sqrt{\frac{u(1-u)}{1-\frac{x_3-x_2}{x_3-x_1}u}}\mathrm du$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is at this point that we require the services of the &lt;a href=&quot;http://dlmf.nist.gov/22&quot; rel=&quot;nofollow&quot;&gt;Jacobian elliptic functions&lt;/a&gt; $\mathrm{sn}(v|m)$, $\mathrm{cn}(v|m)$, and $\mathrm{dn}(v|m)$. The elliptic function identities pertinent to this problem are the two Pythagorean relations&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathrm{cn}^2(v|m)+\mathrm{sn}^2(v|m)=1,\qquad \mathrm{dn}^2(v|m)+m\,\mathrm{sn}^2(v|m)=1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and the differential relation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\mathrm d}{\mathrm dv}\mathrm{sn}(v|m)=\mathrm{cn}(v|m)\mathrm{dn}(v|m)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;thus, letting $u=\mathrm{sn}^2(v|m)$ (and letting $\Delta$ be the constant in front to avoid clutter), we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$2\Delta\int\frac{\mathrm{sn}(v|m)\mathrm{cn}(v|m)\mathrm{dn}(v|m)}{\left(1-\frac{x_3-x_2}{x_3-x_1}\mathrm{sn}^2(v|m)\right)^3}\sqrt{\frac{\mathrm{sn}^2(v|m)(1-\mathrm{sn}^2(v|m))}{1-\frac{x_3-x_2}{x_3-x_1}\mathrm{sn}^2(v|m)}}\mathrm dv$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and if we let $m=\frac{x_3-x_2}{x_3-x_1}$, both Pythagorean relations can be applied to yield&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$2\Delta\int\frac{\mathrm{sn}^2(v|m)\mathrm{cn}^2(v|m)}{\mathrm{dn}^6(v|m)}\mathrm dv$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At this point, we now insert the proper limits for the definite integral by performing the inverses of the last two transformations on the limits $x=x_2$ and $x=x_3$. The new limits are seen to be $v=0$ and $v=K(m)=K\left(\frac{x_3-x_2}{x_3-x_1}\right)$, where $K(m)$ is the complete elliptic integral of the first kind. The definite integral is now&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$2\Delta\int_0^{K(m)}\frac{\mathrm{sn}^2(v|m)\mathrm{cn}^2(v|m)}{\mathrm{dn}^6(v|m)}\mathrm dv$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0387053182&quot; rel=&quot;nofollow&quot;&gt;formula 361.18 in Byrd/Friedman&lt;/a&gt; (and after much tears and algebra), we finally arrive at the result&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align*}&#xD;&#xA;\frac2{15}\sqrt{x_3-x_1}&amp;amp;\left(2(x_1^2+x_2^2+x_3^2-x_1 x_2-x_2 x_3-x_1 x_3)E\left(\frac{x_3-x_2}{x_3-x_1}\right)-\right.\\ &#xD;&#xA; &amp;amp;\left.\quad(x_2-x_1)(x_3-x_1+x_2-x_1)K\left(\frac{x_3-x_2}{x_3-x_1}\right)\right)&#xD;&#xA;\end{align*}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $E(m)$ is the complete elliptic integral of the second kind.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0387053182&quot; rel=&quot;nofollow&quot;&gt;Byrd/Friedman&lt;/a&gt; would be one of the best references on this subject; they have a comprehensive listing of formulae for reducing elliptic integrals to the Legendre-Jacobi forms.&lt;/p&gt;&#xA;" OwnerUserId="498" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2011-05-05T07:35:05.810" LastActivityDate="2011-05-05T07:35:05.810" CommentCount="1" />
  <row Id="9130" PostTypeId="1" AcceptedAnswerId="9166" CreationDate="2010-11-06T08:29:24.350" Score="3" ViewCount="171" Body="&lt;p&gt;Let $a_n$ be a sequence of real numbers such that $\exp(it\cdot a_n)$ converges to 1 for all real t.&#xA;Show that $a_n$ converges to 0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One can show this by letting $X_n$ be degenerate random variables which take value $a_n$ with prob 1 and then using results about characteristic functions and convergence.&#xA;But this seems like overkill and it doesn't provide any intuition for what's going on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems like there should be a simpler proof based on basic properties of continuous functions and convergent sequences.&#xA;Any ideas ?&lt;/p&gt;&#xA;" OwnerUserId="3182" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-06T14:51:31.320" LastActivityDate="2010-11-06T17:41:56.290" Title="Elementary solution of convergence problem ?" Tags="&lt;real-analysis&gt;&lt;sequences-and-series&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="0" />
  <row Id="9131" PostTypeId="1" AcceptedAnswerId="9136" CreationDate="2010-11-06T09:29:29.877" Score="2" ViewCount="100" Body="&lt;p&gt;Hi! What do I miss in the following?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $R$ be a commutative Noetherian ring with unit. A map $f:M\to N$ of $R$-modules is injective/surjective iff the associated map $f_p:M_p\to N_p$ on the localization is injective/surjective for every prime ideal $p$ of the ring $R$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is an isomorphism $M_p=M\otimes_R R_p$. A sequence&#xA;$$0\to K\xrightarrow{k} M\xrightarrow{f} N\to 0$$&#xA;is exact iff $k$ is injective and $s$ is surjective. Hence the above sequence is exact iff for every prime ideal $p$ of $R$ the sequence&#xA;$$0\to K\otimes_R R_p\xrightarrow{k'} M\otimes_R R_p\xrightarrow{f'} N\otimes_R R_p\to 0$$&#xA;is exact. But $R_p$ is not flat over $R$ in general, is it? What am I missing? &#xA;Is the sequence still exact if I tensor with the residue field $R_p/m_p$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you!&lt;/p&gt;&#xA;" OwnerUserId="3183" LastActivityDate="2010-11-06T16:33:34.480" Title="Flatness of local rings" Tags="&lt;algebra&gt;&lt;algebraic-geometry&gt;&lt;commutative-algebra&gt;" AnswerCount="2" CommentCount="5" FavoriteCount="1" />
  <row Id="9132" PostTypeId="1" AcceptedAnswerId="9155" CreationDate="2010-11-06T09:34:04.653" Score="1" ViewCount="113" Body="&lt;p&gt;Dear friends,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any known bound on sum of independent but not identically distributed geometric random variables?&#xA;I have to show that the tail of the sum drops exponentially (like in the Chernoff bounds for the sum of iid geom. variables).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Formally, if $X_i \sim Geom(p_i)$, $X = \sum_{i=1}^n X_i$, and $E[X]=\Theta(n)$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it possible to show that $\Pr(X &amp;lt; 2E[X]) &gt; 1 - \delta ^n$, where $\delta &amp;lt; 1$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you in advance,&#xA;Michael.&lt;/p&gt;&#xA;" OwnerUserId="3184" LastEditorUserId="3184" LastEditorDisplayName="" LastEditDate="2010-11-08T10:35:08.367" LastActivityDate="2010-11-08T10:35:08.367" Title="Probability Theory, Chernoff Bounds, Sum of Independent (but not identically distributed) r.v." Tags="&lt;probability&gt;" AnswerCount="1" />
  <row Id="9133" PostTypeId="1" CreationDate="2010-11-06T09:41:50.567" Score="1" ViewCount="182" Body="&lt;p&gt;I was asked to prove the next formula - &#xA;(A+BC'+C)C' = ABC'+AB'C'+A'BC'&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i need to show all the stages of the simplification, i have all of the rules/identities.&lt;br&gt;&#xA;i have tried many times and i did not make it, i always get something else &#xA;thank you&lt;/p&gt;&#xA;" LastActivityDate="2010-11-06T11:50:41.527" Title="Boolean algebra, ABC'+AB'C'+A'BC' = (A+BC'+C)C', can anyone show it please" Tags="&lt;boolean-algebra&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="9134" PostTypeId="1" AcceptedAnswerId="9135" CreationDate="2010-11-06T09:54:49.203" Score="2" ViewCount="238" Body="&lt;p&gt;Please suggest suitable approach for this problem&lt;/p&gt;&#xA;" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-06T11:00:10.720" LastActivityDate="2010-11-06T11:00:10.720" Title="Find the maximum and minimum value of $ \sin ^4 \theta + \cos ^4 \theta $ ?" Tags="&lt;trigonometry&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="9135" PostTypeId="2" ParentId="9134" CreationDate="2010-11-06T10:03:11.197" Score="6" ViewCount="" Body="&lt;p&gt;Note that&#xA;$$1 = (\sin^2\theta + \cos^2\theta)^2$$&#xA;and use $\sin2\theta = 2\sin\theta\cos\theta.$&lt;/p&gt;&#xA;" OwnerUserId="1301" LastActivityDate="2010-11-06T10:03:11.197" CommentCount="1" />
  <row Id="9136" PostTypeId="2" ParentId="9131" CreationDate="2010-11-06T10:08:29.520" Score="1" ViewCount="" Body="&lt;p&gt;Your characterisation of an exact sequence is missing exactness at $M$.&lt;/p&gt;&#xA;" OwnerUserId="638" LastActivityDate="2010-11-06T10:08:29.520" CommentCount="1" />
  <row Id="9137" PostTypeId="1" AcceptedAnswerId="9165" CreationDate="2010-11-06T10:23:43.480" Score="0" ViewCount="46" Body="&lt;p&gt;Hi!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have this table:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&#xA;x    |     0     1     2     3     4     5&#xA;------------------------------------------&#xA;f(x) |    .7    .2   .05   .03   .01   .01&#xA;F(x) |    .7    .9   .95   .98   .99     1&#xA;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Where f(x) is the density function for x and F(x) is the cumulative distribution function for x.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;hr/&gt;&#xA;And I am being asked to use F to verify that P[X=3]=0.03&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My guess is that I should start with P[X=3]=F(2) ... something ... something, since F(x)=P(X&amp;lt;=x) but I am not sure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any suggestions?&lt;/p&gt;&#xA;" OwnerUserId="713" LastActivityDate="2010-11-06T14:40:23.590" Title="How to use the cumulative dist. fun to verify an exact density (Probability)" Tags="&lt;probability&gt;&lt;statistics&gt;" AnswerCount="1" />
  <row Id="9138" PostTypeId="1" AcceptedAnswerId="9185" CreationDate="2010-11-06T10:38:57.543" Score="6" ViewCount="537" Body="&lt;p&gt;Hello alltogether&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dealing with a physics Problem I get the following equation to solve for $\alpha$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$1 = m \; \text{cos}(\alpha) + \text{sin}(\alpha)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Putting this in Mathematica gives the result:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a==2 \text{ArcTan}\left[\frac{1-m}{1+m}\right]$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However I am unable to get this result myself. No matter what I try normal equation transformations or rewriting the equation with the complex e-Function ..., everything fails. Even going the other Direction from Mathematica's Solution to my original equation resulted in nothing sensible. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any help of how to do do this transformation is very much appreciated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance&lt;/p&gt;&#xA;" OwnerUserId="3188" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-11-11T17:06:00.067" LastActivityDate="2010-11-11T17:06:00.067" Title="Solve trigonometric equation: $1 = m \; \text{cos}(\alpha) + \text{sin}(\alpha)$" Tags="&lt;trigonometry&gt;" AnswerCount="7" CommentCount="0" />
  <row Id="9139" PostTypeId="2" ParentId="9138" CreationDate="2010-11-06T10:48:30.203" Score="8" ViewCount="" Body="&lt;p&gt;So you have $1 = m\cos{\alpha} + \sin{\alpha}$. Now you have $1+m = m(\cos{\alpha}+1) + \sin\alpha$ and $1-m = m(\cos\alpha -1) + \sin{\alpha}$. So then you have $$\frac{1-m}{1+m} = \frac{m(\cos\alpha-1) + \sin\alpha}{m(\cos\alpha +1) + \sin\alpha} = \frac{-m \tan\frac{\alpha}{2} + 1}{m\cot\frac{\alpha}{2} + 1}$$ using the half angle formulae's. Then cross multiplying and solving you should be able to get the value of $\alpha$&lt;/p&gt;&#xA;" OwnerUserId="9413" LastEditorUserId="9413" LastEditorDisplayName="anonymous" LastEditDate="2010-11-07T09:39:50.823" LastActivityDate="2010-11-07T09:39:50.823" CommentCount="7" />
  <row Id="9140" PostTypeId="2" ParentId="9138" CreationDate="2010-11-06T10:49:16.317" Score="10" ViewCount="" Body="&lt;p&gt;Consider the equation&#xA;$$a\cos t+b\sin t=c$$&#xA;as an equation for $t$. Here $a$ and $b$ are real numbers, not both zero.&#xA;The trick to solve this is to divide by $\sqrt{a^2+b^2}$ to get&#xA;$$\frac{a}{\sqrt{a^2+b^2}}\cos t+\frac{b}{\sqrt{a^2+b^2}}\sin t&#xA;=\frac{c}{\sqrt{a^2+b^2}},$$&#xA;equivalently&#xA;$$a'\cos t+b'\sin t=c'$$&#xA;where now $a'^2+b'^2=1$. This means that $(a',b')$ lies on the unit circle.&#xA;There is then an angle $u$ so that $(a',b')=(\cos u,\sin u)$.&#xA;The equation becomes&#xA;$$\cos u\cos t+\sin u\sin t=c',$$&#xA;that is&#xA;$$\cos(t-u)=c'.$$&#xA;We see there is a solution for $t$ iff $|c'|\le1$. In general when $|c'|&amp;lt;1$&#xA;there are exactly two solutions for $t$ modulo $2\pi$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(All of this is standard A-level maths.)&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-11-06T10:49:16.317" CommentCount="6" />
  <row Id="9141" PostTypeId="1" AcceptedAnswerId="9144" CreationDate="2010-11-06T11:44:47.157" Score="0" ViewCount="199" Body="&lt;p&gt;Dear friends,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could someone provide, please, a proof of the theorem below?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Being &lt;strong&gt;x&lt;/strong&gt; and &lt;strong&gt;b&lt;/strong&gt; integers greater than 1, which can not be represented as powers of the same basis(positive integer) and &lt;strong&gt;integer exponent&lt;/strong&gt;, then the logarithm of &lt;strong&gt;x&lt;/strong&gt;, in base &lt;strong&gt;b&lt;/strong&gt;, is an irrational number.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well... Assuming that the logarithm is a rational number p/q (p and q are relatively prime integers), I know I can write &lt;strong&gt;x^q&lt;/strong&gt; = &lt;strong&gt;b^p&lt;/strong&gt;, but I can not conclude from this fact that &lt;strong&gt;b&lt;/strong&gt; e &lt;strong&gt;x&lt;/strong&gt; are powers of a same integer (and with integer exponent).&lt;/p&gt;&#xA;" OwnerUserId="2938" LastEditorUserId="2938" LastEditorDisplayName="" LastEditDate="2010-11-07T11:33:33.857" LastActivityDate="2010-11-07T11:33:33.857" Title="About irrational logarithms" Tags="&lt;logarithms&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="9142" PostTypeId="2" ParentId="9133" CreationDate="2010-11-06T11:50:41.527" Score="1" ViewCount="" Body="&lt;p&gt;opening the brackets is done as with &quot;regular&quot; algebra, which yields:  &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;AC' + BC'C' + CC'&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;from your identites:  &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;C'C' = C'  &#xA;CC' = 0  &#xA;X + 0 = X  &lt;/pre&gt;&#xA;&#xA;&lt;p&gt;which imply that the above expression can be simplified to:  &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;AC' + BC'&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;from here you should look to expand the expression by the following identities:  &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;X*1 = X  &#xA;X + X' = 1  &lt;/pre&gt;&#xA;&#xA;&lt;p&gt;which yields:  &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;AC'(B + B') + BC'(A + A')&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;expand the brackets as usual, change the order of multiplication with the identity XY = YX, cancel common terms based on X + X = X, and youre done...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;someone should add a h/w tag to this question. i dont have any points...&lt;/p&gt;&#xA;" OwnerUserId="2881" LastActivityDate="2010-11-06T11:50:41.527" />
  <row Id="9143" PostTypeId="2" ParentId="9072" CreationDate="2010-11-06T11:56:25.063" Score="6" ViewCount="" Body="&lt;p&gt;I think the answer really depends on what you mean by &quot;fathom&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want a short construction of the Monster, there is a sketch by Conway in one of the later chapters of &lt;em&gt;Sphere Packings, Lattices, and Groups&lt;/em&gt; (occasionally available on Google Books).  The construction there goes through a nice progression of increasingly complicated exceptional objects, like the Golay code and the Leech lattice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to understand a &quot;natural&quot; object on which the Monster acts by symmetries, you should read up about vertex algebras.  At our current state of knowledge (and depending on who you ask), the most natural object on which the monster acts is the monster vertex algebra $V^\natural$ (also known as the moonshine module, or the monster VOA), which is a graded vector space, together with some extra structure like a rather complicated multiplication operation $V^\natural \otimes V^\natural \to V^\natural((z))$.  The construction of $V^\natural$ is given in the book &lt;em&gt;Vertex operator algebras and the Monster&lt;/em&gt; by I. Frenkel, Lepowsky, and Meurman.  A string theorist might say that it is given by orbifolding the Leech lattice CFT.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to consider some basic facts about the Monster, you can have a look at the ATLAS of finite groups, or play around with the software GAP.  Both have the character table, and orders of centralizers of elements, etc.  Wilson showed that the monster is a Hurwitz group, so Ryan Budney's comment about acting on a Riemann surface holds for the minimum possible genus (about $10^{52}$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to understand finer points about the structure of the Monster group, you're pretty much out of luck.  It's big enough that there are plenty of explicit questions whose answers we don't know.  For example, the conjugacy classes of homomorphisms from $\mathbb{Z} \times \mathbb{Z}$ (i.e., pairs of commuting elements) are not classified, and $H^4$ with coefficients in $\mathbb{Z}$ is still unknown (very annoying).&lt;/p&gt;&#xA;" OwnerUserId="3189" LastActivityDate="2010-11-06T11:56:25.063" />
  <row Id="9144" PostTypeId="2" ParentId="9141" CreationDate="2010-11-06T11:57:00.697" Score="4" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Hint:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $\log_b\;x$ were a rational number $\frac{p}{q}$, then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x^q=b^p$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, go back to your initial assumption...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edited to add: &lt;a href=&quot;http://www.southalabama.edu/mathstat/personal_pages/feldvoss/logintirrat.pdf&quot; rel=&quot;nofollow&quot;&gt;This&lt;/a&gt; might be of interest.&lt;/p&gt;&#xA;" OwnerUserId="498" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-06T12:08:48.683" LastActivityDate="2010-11-06T12:08:48.683" CommentCount="2" />
  <row Id="9145" PostTypeId="1" AcceptedAnswerId="9856" CreationDate="2010-11-06T12:16:30.500" Score="0" ViewCount="86" Body="&lt;p&gt;Hey guys, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider n loaded dice with differing probability distributions. I have constructed an equation that tells me the probability mass, the number of times a certain number is expected to be thrown. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now i want to extend the numerical range from a line onto a plane. How do i get the expected number of times a number/number combination is thrown now?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The two dicerolls are independent. Can i compute the probability mass for the combination using my equation for computing the probability mass for the line? Since the probability mass is not a probability i cannot just multiply those. What will i have to change?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for the advice&lt;/p&gt;&#xA;" OwnerUserId="2362" LastActivityDate="2010-11-11T14:05:45.213" Title="Gaussian Distributions and probability mass" Tags="&lt;probability&gt;&lt;probability-theory&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="9146" PostTypeId="2" ParentId="9122" CreationDate="2010-11-06T12:28:34.407" Score="0" ViewCount="" Body="&lt;p&gt;I found a differential equation which gives the correct result under &lt;em&gt;Mathematica 7.0.1&lt;/em&gt;.  The differential equation is the same one I derived in the question (see my comment on adding x):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{A(x)-1}{x}=c \cdot A(x) - \frac{d A(x)}{dx}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The result it gives as the solution (besides the series) is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$e^{-1/x}x^{-c}(C_1 - x^{c-1}E_c(-\frac{1}{x}))$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $E_c(z)$ is the Exponential Integral Function, for which the following equation holds: $-x^{c-1}E_c(b)$ = $b^c x^c \Gamma[1-c,b]$ and $b=-\frac{1}{x}$. $C_1$ is the constant of integration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please note my derivation assumed $c \in \mathbb{N}$.&lt;/p&gt;&#xA;" OwnerUserId="2626" LastActivityDate="2010-11-06T12:28:34.407" />
  <row Id="9147" PostTypeId="1" AcceptedAnswerId="9153" CreationDate="2010-11-06T12:42:24.743" Score="2" ViewCount="103" Body="&lt;p&gt;This is an exercise from Apostol's number theory book. How does, one prove that $$ \frac{\sigma(n)}{n} &amp;lt; \frac{n}{\varphi(n)} &amp;lt; \frac{\pi^{2}}{6} \frac{\sigma(n)}{n} \quad \text{if} \ n \geq 2$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I thought of using the formula $$\frac{\varphi(n)}{n} = \prod\limits_{p \mid n} \Bigl(1 - \frac{1}{p}\Bigr)$$ but couldn't get anything further. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Notations:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;$\sigma(n)$ stands for the sum of divisors&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$\varphi(n)$ stands for the Euler's Totient Function.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" LastActivityDate="2010-11-06T13:52:57.037" Title="Proving $ \frac{\sigma(n)}{n} &lt; \frac{n}{\varphi(n)} &lt; \frac{\pi^{2}}{6} \frac{\sigma(n)}{n}$" Tags="&lt;number-theory&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="9148" PostTypeId="1" AcceptedAnswerId="9151" CreationDate="2010-11-06T13:04:10.253" Score="0" ViewCount="63" Body="&lt;p&gt;&lt;a href=&quot;http://stackoverflow.com/questions/4113133/changing-a-positive-value-to-a-negative-one&quot;&gt;http://stackoverflow.com/questions/4113133/changing-a-positive-value-to-a-negative-one&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;why? is this just a computer science problem , or is this a mathematical property.&lt;/p&gt;&#xA;" OwnerUserId="3191" LastEditorUserId="171" LastEditorDisplayName="" LastEditDate="2010-11-06T18:50:32.200" LastActivityDate="2010-11-06T18:50:32.200" Title="Why does 0-N invert's the sign." Tags="&lt;number-theory&gt;" AnswerCount="1" />
  <row Id="9149" PostTypeId="1" CreationDate="2010-11-06T13:04:14.807" Score="1" ViewCount="97" Body="&lt;p&gt;For a mapping algorithm I'm working on, I'm trying out the effect of sigmoid weightings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Right now I'm using $$y = \frac{1}{1+x^n}$$ where n is the steepness of the sigmoid function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This graph shows values of x between 0 and 256 and the following values of n: 2, 4 and 32. X values have been scaled to a range of -10 to 10. Also plotted is equivalent $y=x$ line.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/Fu0Ev.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The effect I would like to achieve is a function that lies somewhere between the $y=x$ and the $n=2$ functions. When I set n to be between 1 and 2 however, the location of the asymptotes changes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I use values of n such that $1 &amp;lt; n &amp;lt; 2$ and keep the asymptotes at 0 and 1?&lt;/p&gt;&#xA;" OwnerUserId="3190" LastActivityDate="2010-11-06T15:11:51.273" Title="Is there a general (logistic?) function for sigmoids over a given range?" Tags="&lt;functions&gt;" AnswerCount="2" />
  <row Id="9150" PostTypeId="1" AcceptedAnswerId="9181" CreationDate="2010-11-06T13:14:10.350" Score="0" ViewCount="138" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Banach_fixed_point_theorem&quot; rel=&quot;nofollow&quot;&gt;Banach's fixed point theorem&lt;/a&gt; gives us a sufficient condition for a function in a complete metric space to have  a fixed point, namely it needs be a contraction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm interested in how to calculate the limit of the sequence $x_0 = f(x), x_1 = f(x_0), \ldots, x_n = f(x_{n-1})$ for a fixed $x$. I couldn't figure out a way to do this limit with ordinary limits calculations. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only thing I have at my disposal is the proof of the theorem, from which we see that the sequence $x_n$ is a Cauchy sequence; from this, I'm able to say, for example, that $\left|f(f(f(x))) - f(f(f(f(x))))\right| \leq \left|f(x_0)-f(x_1)\right| ( \frac{k^3}{1-k})$, where $k$ is the contraction constant, but I can't get any further in the calculations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;My question is&lt;/em&gt;: how should I procede to calculate this limit exactly? If there are non-numerical (read: analytical) way to do this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Remark&lt;/strong&gt;: I'm interested in functions $\mathbb{R} \rightarrow \mathbb{R}$ (as it can be seen from my use of the euclidean metric in $\mathbb{R}$)&lt;/p&gt;&#xA;" OwnerUserId="1415" LastActivityDate="2010-11-06T17:33:23.030" Title="Finding the fixed points of a contraction" Tags="&lt;sequences-and-series&gt;&lt;limit&gt;&lt;metric-spaces&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="9151" PostTypeId="2" ParentId="9148" CreationDate="2010-11-06T13:18:23.310" Score="4" ViewCount="" Body="&lt;p&gt;I'm not sure what level of answer you're looking for, so (at the risk of appearing stupid) here's my explanation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;0 is nothing. A negative number is the absence of something, a deficit. Subtraction is the removal of one quantity from another quantity. Therefore subtracting something from nothing results in a deficit, and absence, a negative number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Put visually:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;-5  -4  -3  -2  -1   0   1   2   3   4   5&#xA;         ^-----------^+++++++++++^&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;If I want to make 3 a negative number, I can subtract 3 from 0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a bonus, you can also make a number negative by multiplying it by -1.&lt;/p&gt;&#xA;" OwnerUserId="3190" LastActivityDate="2010-11-06T13:18:23.310" />
  <row Id="9152" PostTypeId="1" CreationDate="2010-11-06T13:19:39.173" Score="2" ViewCount="243" Body="&lt;p&gt;Possibly not a Math problem in the true sense of the word, but here goes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Three Nines - and various math operators = can be turned into various simple numbers. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(9-9)*9 = 0&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(9/9)^9 = 1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(9+9)/9 = 2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;sqrt (9+9-9) = 3&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(sqrt(9) +9/9) = 4&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(sqrt(9!) - 9/9 = 5&lt;/p&gt;&#xA;&#xA;&lt;p&gt;sqrt (9*9) - sqrt (9) = 6     (or 9 - (9/Sqrt (9) ) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;7 I am stumped on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;9 - (9/9) = 8&lt;/p&gt;&#xA;&#xA;&lt;p&gt;9 * (9/9) = 9&lt;/p&gt;&#xA;&#xA;&lt;p&gt;9 +(9/9) = 10&lt;/p&gt;&#xA;&#xA;&lt;p&gt;99/9 = 11&lt;/p&gt;&#xA;&#xA;&lt;p&gt;9 + (9/sqrt(9)) = 12&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ANy answers for 7, and anything for 13 and beyond?&lt;/p&gt;&#xA;" OwnerUserId="3193" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2010-11-06T14:53:35.557" LastActivityDate="2010-11-07T07:08:25.120" Title="The Three Nines" Tags="&lt;puzzle&gt;" AnswerCount="4" CommentCount="5" FavoriteCount="1" />
  <row Id="9153" PostTypeId="2" ParentId="9147" CreationDate="2010-11-06T13:28:32.520" Score="9" ViewCount="" Body="&lt;p&gt;Writing $n=p_1^{a_1}p_2^{a_2}\ldots p_k^{a_k}$ for primes $p_i$ we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sigma (n) = \prod_{i=1}^k \frac{p_i^{a_i+1} – 1}{p_i – 1}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and so&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sigma(n)\phi(n) = n^2 \prod_{i=1}^k \left( 1 - \frac{1}{p_i^{a_i+1}} \right) \qquad (1)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;from which both inequalities follow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the RH inequality we note that the expansion of the reciprocal of&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \prod_{i=1}^k \left( 1 - \frac{1}{p_i^{a_i+1}} \right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is $&amp;lt; \sum_{r=1}^\infty 1/r^ 2 = \pi^2/6.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sigma(n) = \prod_{i=1}^k (1+p_i +p_i^2 + \cdots + p_ i^{a_i}),$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is where the formula for $\sigma(n)$ comes from and to obtain $(1)$ we've just multiplied this by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\phi(n) = n \prod_{i=1}^k \left( 1 - \frac{1}{p_i} \right),$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and factored out all the $p_i^{a_i}.$&lt;/p&gt;&#xA;" OwnerUserId="1301" LastEditorUserId="1301" LastEditorDisplayName="" LastEditDate="2010-11-06T13:52:57.037" LastActivityDate="2010-11-06T13:52:57.037" />
  <row Id="9154" PostTypeId="2" ParentId="9152" CreationDate="2010-11-06T13:44:22.137" Score="2" ViewCount="" Body="&lt;p&gt;Here's a quick cheat for 7.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$7 = 9 - 9^0 - 9^0.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Depending on what functions one's allowed they are easy to come by. Here are some more examples for $7.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ d(9)! + \log_9 9 = \lfloor \sqrt {\sigma(9)} \rfloor ! + \frac{9}{9}$$&#xA;$$= \lfloor \log \sqrt{9!} \rfloor +  \frac{9}{9} =&#xA;\lfloor \sqrt{ \log 9^9 } \rfloor + \sqrt{9} = 7.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As usual, $d(n)$ is the number of divisors of $n$&#xA;and $\sigma(n)$ is the sum of the divisors of $n.$&lt;/p&gt;&#xA;" OwnerUserId="1301" LastEditorUserId="1301" LastEditorDisplayName="" LastEditDate="2010-11-06T19:55:14.677" LastActivityDate="2010-11-06T19:55:14.677" CommentCount="4" />
  <row Id="9155" PostTypeId="2" ParentId="9132" CreationDate="2010-11-06T13:44:45.177" Score="0" ViewCount="" Body="&lt;p&gt;Hoeffding's inequality can be used but the random variables need to be bounded to get meaningful bounds. &lt;a href=&quot;http://en.wikipedia.org/wiki/Hoeffding%27s_inequality&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Hoeffding%27s_inequality&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1527" LastActivityDate="2010-11-06T13:44:45.177" CommentCount="1" />
  <row Id="9156" PostTypeId="1" CreationDate="2010-11-06T13:45:44.290" Score="0" ViewCount="134" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have curve, that is created from circles in axis X (only x coordinate changes, y is still 0). My curve is given by this equation:&#xA;(x + R * sin(a))^2 + y^2 = (r * sin(a))^2 .... a = parameter&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Basicly, it creates Spiric sections)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now i have got parametric (on any other kind of equtaion) of line&#xA;X = X0 + D * t   ..... t = parameter&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need to calculate intersection of &quot;curve&quot; x line. How can i do that ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" LastActivityDate="2010-11-06T13:45:44.290" Title="Parametric curve vs line intersection" Tags="&lt;geometry&gt;" CommentCount="2" />
  <row Id="9157" PostTypeId="1" AcceptedAnswerId="9202" CreationDate="2010-11-06T13:59:24.387" Score="1" ViewCount="96" Body="&lt;p&gt;hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This problem has me scared to my wits, mostly because there are more like it!&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;The mean of a random sample of n observations drawn from a N $ ( \mu ,\sigma^{2} ) $ distribution is denoted by $  \overline{\chi} $. Given that P $ (| \overline{\chi} - \mu| &gt; 0.5\sigma ) \lt 0.05 $ &lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Find the smallest value of n.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I tried:&lt;br&gt;&#xA;Since $\overline{\chi}$ ~ N $ \left ( \mu ,\frac{\sigma^2}{n} \right )  $&lt;br&gt;&#xA;P $ (| \overline{\chi} - \mu| &gt; 0.5\sigma ) \lt 0.05 $  Should be:&lt;br&gt;&#xA;$$ \frac{0.5\sigma - \mu}{\sqrt{\sigma^{2}/n}} \lt 0.05$$&#xA;$$1-\phi\left(\frac{0.5\sigma - \mu}{\sqrt{\sigma^{2}/n}}\right) \lt 0.05 $$&lt;br&gt;&#xA;($\phi$ is the normal distribution function, I lookup a value from the normal dist. table.)&#xA;$$ \frac{0.5\sigma - \mu}{\sqrt{\sigma^{2}/n}} \lt 1.645$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now where do I go from here!?&#xA;This is totally confusing me, especially that darn modulus. How do I use my result so far with this?&lt;br&gt;&#xA;P $ (| \overline{\chi} - \mu| &gt; 0.5\sigma )$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&#xA;Gideon&lt;/p&gt;&#xA;" OwnerUserId="2576" LastEditorUserId="2576" LastEditorDisplayName="" LastEditDate="2010-11-07T03:27:48.550" LastActivityDate="2010-11-07T03:27:48.550" Title="finding n with N $\left ( \mu ,\frac{\sigma^2}{n} \right ) $ with given probability" Tags="&lt;homework&gt;&lt;probability-theory&gt;&lt;statistics&gt;&lt;central-limit-theorem&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="9158" PostTypeId="2" ParentId="9150" CreationDate="2010-11-06T14:00:47.250" Score="5" ViewCount="" Body="&lt;p&gt;The limit is the fixpoint, so you &quot;just&quot; need to solve the equation $x=f(x)$. Whether this can be done in closed form or not depends on what $f$ looks like.&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2010-11-06T14:00:47.250" CommentCount="8" />
  <row Id="9159" PostTypeId="1" AcceptedAnswerId="9162" CreationDate="2010-11-06T14:08:23.033" Score="6" ViewCount="112" Body="&lt;p&gt;I'm trying to parametrize the space curve determined by the boundary of a standard orange peel: for example, the one on &lt;a href=&quot;http://www.dreamstime.com/stock-photo-orange-peel-image14340&quot; rel=&quot;nofollow&quot;&gt;this photo&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, the ideal curve would be inside the unit cube; have only one point of intersection with every horizontal plane $z=k$, when $k\in [-1,1]$; would start in (0,0,-1) and end in (0,0,1), wrapping itself around them; and touch the boundary of the cube when z=0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's sort of a standard helix, compressed. I hope I was clear.&lt;/p&gt;&#xA;" OwnerUserId="2614" LastActivityDate="2010-11-06T14:25:13.817" Title="How to parameterize an orange peel" Tags="&lt;geometry&gt;" AnswerCount="1" />
  <row Id="9160" PostTypeId="2" ParentId="9150" CreationDate="2010-11-06T14:11:40.027" Score="1" ViewCount="" Body="&lt;p&gt;In addition to what Hans Lundmark has said about solving $x = f(x)$, you could also try writing a simple computer programme to read a number $n$ and a starting value $x_0$, and compute the result of applying f to $x_0$ $n$ times, thus delivering an approximation to the root that you are seeking. The value of $n$ may have to be fairly large in some cases. &lt;/p&gt;&#xA;" OwnerUserId="3195" LastEditorUserId="3195" LastEditorDisplayName="" LastEditDate="2010-11-06T14:19:07.163" LastActivityDate="2010-11-06T14:19:07.163" CommentCount="2" />
  <row Id="9161" PostTypeId="1" AcceptedAnswerId="9173" CreationDate="2010-11-06T14:16:30.070" Score="5" ViewCount="388" Body="&lt;p&gt;Reading the Wikipedia page for &lt;a href=&quot;http://en.wikipedia.org/wiki/Random_walk&quot; rel=&quot;nofollow&quot;&gt;Random Walk&lt;/a&gt;, I was wondering what is the definition for a general random walk as a random process, so that all the concepts such as &lt;a href=&quot;http://en.wikipedia.org/wiki/Random_walk#Lattice_random_walk&quot; rel=&quot;nofollow&quot;&gt;random walk on $\mathbb{Z}^d$&lt;/a&gt;, &lt;a href=&quot;http://en.wikipedia.org/wiki/Random_walk#Gaussian_random_walk&quot; rel=&quot;nofollow&quot;&gt;Gaussian random walk&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/Random_walk#Random_walk_on_graphs&quot; rel=&quot;nofollow&quot;&gt;random walk on a graph&lt;/a&gt; can fall into this general definition?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) In Ross's book for stochastic process, he defined a random walk to be the sequence of partial sums of a sequence of i.i.d. real-valued random variables. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So is a Gaussian random walk the sequence of partial sums of a sequence of i.i.d. real-valued random variables with a common Gaussian distribution?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How does a random walk on a graph fit into Ross's definition?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) I was thinking maybe the concept of random walk is equivalent to that of discrete-time Markov process, where the state space can be either continuous or discrete or even non-numeric (such as for a random walk on a graph)? Is it correct?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So can I say the graph in &quot;a random walk on a graph&quot; is just the &lt;a href=&quot;http://en.wikipedia.org/wiki/Graphical_model&quot; rel=&quot;nofollow&quot;&gt;graphical model&lt;/a&gt; for the distribution of the random walk as a stochastic process?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also the graph in &quot;a random walk on a graph&quot; has nothing to do with &lt;a href=&quot;http://en.wikipedia.org/wiki/Random_graph&quot; rel=&quot;nofollow&quot;&gt;random graph&lt;/a&gt;, right?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(3) if you can come up with a definition for random walk, please also explain how various special kinds of random walks fit into the definition you provided.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks and regards!&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;EDIT:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks to Dr Schmuland for the reply! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Although I have chosen the best answer, I am still not clear about the following questions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) Is a random walk (including random walk on a group such as $\mathbb{Z}^d$ and random walk on a graph) equivalent to a discrete-time Markov process? If not, what makes random walks different from general discrete time Markov processes?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) For a random walk defined on a group such as $\mathbb{Z}^d$, does it really require the increments between every two consecutive indices be i.i.d. (seems to be the definition in Ross's Stochastic processes) or just independence will be enough and identical distribution is not necessary? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Reply to Update of Dr Schmuland&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks, Dr Schmuland, for providing a big picture that I have not clearly realized before! It took me some while to understand. Learning more makes me feel better.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Generally when the state space is additive group, are these two properties - markov property and increment-independence - equivalent? Must time-homogeniety be defined only for Markov process? If it can be define for a general process, is it equivalent to the property that the increments over same length period have the same distribution?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also wonder if, for a random walk on a graph, at each vertex, the transition probabily must be uniform? If yes, is the transition probability uniform at a vertex over a set of the neighbours and itself or just over a set of its neighbours (itself excluded, which means it is impossible to stay at the vertex)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As your last exercise, I think time-homogeniety and identical-increment-distribution are equivalent, and Markov property and increment-independence are equivalent, so every time-homogeneous Markov chain will be a random walk on the group {1,2,3}. But my answer will be different for it to be a random walk on the graph derived for {1,2,3} as you mentioned, the transition probability from a vertex to other vertices has to be uniform.&lt;/p&gt;&#xA;" OwnerUserId="1281" LastEditorUserId="1281" LastEditorDisplayName="" LastEditDate="2010-11-09T16:06:43.700" LastActivityDate="2010-11-09T16:06:43.700" Title="Concept of Random Walk" Tags="&lt;stochastic-processes&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="2" />
  <row Id="9162" PostTypeId="2" ParentId="9159" CreationDate="2010-11-06T14:25:13.817" Score="7" ViewCount="" Body="&lt;p&gt;Well, you seem to have a lot of options; there are a number of spherical spirals that would do. The &lt;a href=&quot;http://mathworld.wolfram.com/SphericalSpiral.html&quot; rel=&quot;nofollow&quot;&gt;loxodrome&lt;/a&gt; is one (the spherical analogue of the equiangular spiral), and &lt;a href=&quot;http://mathworld.wolfram.com/SeiffertsSphericalSpiral.html&quot; rel=&quot;nofollow&quot;&gt;Seiffert's spiral&lt;/a&gt; is another.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-11-06T14:25:13.817" CommentCount="1" />
  <row Id="9163" PostTypeId="1" AcceptedAnswerId="9174" CreationDate="2010-11-06T14:32:56.923" Score="1" ViewCount="265" Body="&lt;p&gt;Quoted from &lt;a href=&quot;http://en.wikipedia.org/wiki/Wiener_process&quot; rel=&quot;nofollow&quot;&gt;Wikepedia&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The Wiener process can be constructed as the scaling limit of a random walk, or other discrete-time stochastic processes with stationary independent increments. This is known as Donsker's theorem.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I was wondering if the scaling limit of any &lt;a href=&quot;http://en.wikipedia.org/wiki/Random_walk&quot; rel=&quot;nofollow&quot;&gt;random walk&lt;/a&gt; is always a Wiener process, or just the scaling limits of some special kinds of random walks are, such as &lt;a href=&quot;http://en.wikipedia.org/wiki/Random_walk#Gaussian_random_walk&quot; rel=&quot;nofollow&quot;&gt;Gaussian random walk&lt;/a&gt;?&lt;/p&gt;&#xA;" OwnerUserId="1281" LastActivityDate="2010-11-06T16:25:35.493" Title="Wiener Process and Random Walk" Tags="&lt;stochastic-processes&gt;" AnswerCount="1" />
  <row Id="9164" PostTypeId="2" ParentId="9152" CreationDate="2010-11-06T14:37:52.433" Score="8" ViewCount="" Body="&lt;p&gt;$7=(\sqrt{9})!+\frac{9}{9}$.  Also for 5 you need to move the ! outside the parens.&lt;br&gt;&#xA;$18=\sqrt{(\sqrt{9})!!+9}-9=\sqrt{9}^{\sqrt{9}}-9$&lt;br&gt;&#xA;and you can sqrt the last 9 and add and subtract to get a few more&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$13=\frac{9}{.9}+\sqrt{9}$&lt;br&gt;&#xA;If you allow the overline for a repeating decimal or the floor function  (floor(sqrt(9)) you can convert a 9 into a 1.  This gives new flexibility&lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2010-11-06T15:30:26.957" LastActivityDate="2010-11-06T15:30:26.957" />
  <row Id="9165" PostTypeId="2" ParentId="9137" CreationDate="2010-11-06T14:40:23.590" Score="2" ViewCount="" Body="&lt;p&gt;The CDF $F(x)$ is the sum of the probabilities $f(y)$ for $y \leq x$.  Since $F(x+1) - F(x) = f(x+1)$, you can verify that $0.98 - 0.95 = 0.3$. &lt;/p&gt;&#xA;" OwnerUserId="3009" LastActivityDate="2010-11-06T14:40:23.590" CommentCount="1" />
  <row Id="9166" PostTypeId="2" ParentId="9130" CreationDate="2010-11-06T14:42:22.830" Score="5" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Step 1&lt;/strong&gt; The sequence $(a_n)$ cannot be bounded, if it doesn't converge to 0. Suppose the contrary, then up to a subsequence $(a_{\sigma(n)})$ converges to some value $A$. Let $t = \pi/A$. Then $\exp i t a_\sigma(n) \to -1$ contradicting convergence assumption. (Another way to see this step is to note that if $(a_n)$ were bounded, then the sequence $(\exp i t a_n)$ is equicontinuous, and on any compact interval the convergence must be uniform.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Step 2&lt;/strong&gt; The sequence cannot be unbounded. Assume the contrary, then up to a subsequence we can choose $a_{\sigma(n+1)} &gt; 3 a_{\sigma(n)}$. Without loss of generality we can assume the subsequence is positive. Let $t_n = 2\pi a_{\sigma(n)}^{-1}$. Define a sequence of closed intervals recursively. Let $I_1 = [t_1/4, 3t_1/4]$. By construction $\exists m_2 \in \mathbb{Z}$ such that $m_2t_2, (m_2+1)t_2 \in I_1$. Let $I_2 = [ (m_2 + 1/4) t_2, (m_2 + 3/4) t_2]$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;At each step $I_n$ has length $t_n / 2$. Hence there exists $m_{n+1}$ such that $m_{n+1}t_{n+1}, (m_{n+1}+1)t_{n+1} \in I_n$. And so we construct $I_{n+1}$ analogously. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, we have $I_n \supset I_{n+1} \supset I_{n+2} \ldots$. Furthermore, by construction, $\Re e^{it a_{\sigma(n)}} \leq 0$ for $t \in I_n$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then for $t\in \cap_1^\infty I_n$, we have that $\exp i t a_j$ is bounded away from 1 on a subsequence, and hence contradicts the assumption of convergence. &lt;/p&gt;&#xA;" OwnerUserId="1543" LastActivityDate="2010-11-06T14:42:22.830" CommentCount="3" />
  <row Id="9167" PostTypeId="2" ParentId="9149" CreationDate="2010-11-06T14:43:21.347" Score="2" ViewCount="" Body="&lt;p&gt;I think it would help if you quantified what you mean by the &lt;em&gt;location&lt;/em&gt; of the asymptote.  In your graph, it &lt;em&gt;looks&lt;/em&gt; like $y$ takes on the values 0 or 1 at different places for different values of $n$, but $y \ne 0$ for any $x &gt; 0$ and $y \ne 1$ for any finite $x$ no matter the value of $n&gt;0$. You could define the location of the asymptotes, for example, as the values of $x$ where $y$ equals 0.001 and 0.999.  Or you could define the locations of the asymptotes as places where the derivative $y'$ equals 0.001, for example.&lt;/p&gt;&#xA;" OwnerUserId="247" LastActivityDate="2010-11-06T14:43:21.347" />
  <row Id="9168" PostTypeId="2" ParentId="9149" CreationDate="2010-11-06T15:11:51.273" Score="1" ViewCount="" Body="&lt;p&gt;You can also use $y = \frac{1}{1+ax^n}$ to give yourself another parameter to play with.  It will still go from 0 to 1 as x goes from -infinity to infinity, but you can change how quickly it gets close.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-11-06T15:11:51.273" CommentCount="1" />
  <row Id="9169" PostTypeId="1" CreationDate="2010-11-06T15:21:26.193" Score="0" ViewCount="140" Body="&lt;p&gt;I am trying to evaluate this integral or at least get bounds for its aboslute value.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have where $\tau \to \infty$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_{1}^{\infty} f(t) \frac{\tau \sin(\tau\log t)}{t^{\sigma+1}} dt$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f(t) = (t + 1/2) - \lfloor t + 1/2 \rfloor$ (where $\lfloor.\rfloor$ is the floor function)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Integrating by parts gives me,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_{1}^{\infty} \frac{f(t)}{t^\sigma} d(-\cos(\tau \log t)) = \left[ -\frac{f(t)}{t^\sigma}\cos(\tau \log t) \right]_{1}^{\infty} + B$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$B = \int_{1}^{\infty} \cos(\tau\log t) g'(t) dt$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $g(t) = f(t)/t^\sigma$ and is piecewise continuous&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using Riemann-Lebesgue lemma we have $B = 0$ as $\tau \to \infty$, giving us,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_{1}^{\infty} \frac{f(t)}{t^\sigma} d(-\cos(\tau \log t)) = \left[ -\frac{f(t)}{t^\sigma}\cos(\tau \log t) \right]_{1}^{\infty}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, since, $0 \leq f(t) \leq 1$, I think it makes the absolute value of the integral $\leq 1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, are these steps correct?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any further insights will be highly appreciated...&lt;/p&gt;&#xA;" OwnerUserId="2320" LastEditorUserId="2320" LastEditorDisplayName="" LastEditDate="2010-11-06T15:41:00.173" LastActivityDate="2010-11-06T15:41:00.173" Title="Is this the right way to evaluate this integral?" Tags="&lt;calculus&gt;&lt;analysis&gt;&lt;integral&gt;" CommentCount="3" />
  <row Id="9170" PostTypeId="2" ParentId="9116" CreationDate="2010-11-06T15:27:01.460" Score="2" ViewCount="" Body="&lt;p&gt;You are correct. The &quot;crude&quot; criterion for recurrence of a 2d random walk is $\mu=0$ and $\sigma^2&amp;lt;\infty$ for the jump distribution. The jump sizes are otherwise unrestricted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &quot;detailed&quot; criterion involves the characteristic function $\phi$ of the jump distribution, i.e., its Fourier transform. It says that 2d random walk is transient or recurrent as the real part of $(1-\phi(\theta))^{-1}$ is Lebesgue integrable on a neighborhood of the origin. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;These results are from Section 8 of Spitzer's &lt;em&gt;Principles of Random Walk (2e)&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Spitzer gives a detailed example of &lt;strong&gt;symmetric&lt;/strong&gt; one-dimensional random walks, and shows that their recurrence or transience depends on the size of the tail of the jump distribution. That is, he supposes that &#xA;$$0&amp;lt;\lim_{|x|\to\infty} |x|^{1+\alpha}P(0,x)=c&amp;lt;\infty,$$ and concludes that this walk is recurrent when $\alpha\geq 1$ and transient when $\alpha&amp;lt;1$.    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, somehat unexpectedly, there exist symmetric transient random walks in one dimension. Their jump distribution has such large tails that the walk leaps back and forth with large jumps and satisfies $\liminf_n X_n=-\infty$ and $\limsup_n X_n=+\infty$ without a guarantee of returning to the origin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It should be possible to modify his arguments to the two dimensional case.&lt;/p&gt;&#xA;" OwnerUserId="940" LastActivityDate="2010-11-06T15:27:01.460" CommentCount="3" />
  <row Id="9171" PostTypeId="1" CreationDate="2010-11-06T15:51:20.450" Score="0" ViewCount="139" Body="&lt;p&gt;What is the minimum for this function of $x_1,x_2,...,x_n$:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sum_{i=1}^{n}{c_i \log x_i} + \lambda (\sum_{i=1}^{n}{d_i x_i})$&#xA;   , where $\lambda$, c and d series are positive constants, $x_i \in (0.02,1]$ and $\sum x_i = 1$.&lt;/p&gt;&#xA;" OwnerUserId="2858" LastEditorUserId="2858" LastEditorDisplayName="" LastEditDate="2010-11-06T16:47:18.867" LastActivityDate="2011-01-05T17:44:36.090" Title="Minimum for this function" Tags="&lt;analysis&gt;&lt;sequences-and-series&gt;&lt;algebra&gt;" AnswerCount="1" CommentCount="6" />
  <row Id="9172" PostTypeId="2" ParentId="9122" CreationDate="2010-11-06T15:51:29.630" Score="1" ViewCount="" Body="&lt;p&gt;If you know about the &lt;a href=&quot;http://en.wikipedia.org/wiki/Incomplete_gamma_function&quot; rel=&quot;nofollow&quot;&gt;incomplete Gamma function&lt;/a&gt; then the solution is immediate&lt;/p&gt;&#xA;&#xA;&lt;p&gt;since $\rm\quad\quad\quad\quad\quad\ \  \Gamma(s,x)\ =\ (s-1)\ \Gamma(s-1,x) + x^{s-1}\ e^{-x} $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;hence $\rm\displaystyle\quad e^x\ \Gamma(s+1,x)\ =\ \sum_{k=0}^s\ \frac{s!}{k!}\: x^k\ \ $ for $\rm\:s\in \mathbb N$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your method of summing the lower-factorical coefficient recurrence $\rm\ c_{(k+1)}\ = \ (c-k)\ c_{(k)}\ $ will yield a hypergeometric differential equation with the above solution - presuming that you know how to solve such hypergeometric differential equations. But that's overkill compared to the above.&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-11-06T15:51:29.630" />
  <row Id="9173" PostTypeId="2" ParentId="9161" CreationDate="2010-11-06T15:53:27.803" Score="4" ViewCount="" Body="&lt;p&gt;It's a good question. The naive answer is that the two concepts are different. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Random walks on ${\mathbb Z}^d$ or ${\mathbb R}$, etc. are examples of a random walk on an additive group $G$. There is a fixed distribution $\mu$ on $G$, and at each time you randomly select an element of the group (using distribution $\mu$) and add it to the current state.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In contrast, for a random walk on a (locally finite) graph, you randomly choose an edge  uniformly and follow that edge to the next state (node). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the concepts may be related at a deeper level. The first chapter of Woess's &lt;em&gt;Random Walks on Infinite Graphs and Groups&lt;/em&gt; explains how a random walk on a graph is related to a random walk on a quotient of its automorphism group, and also how a random walk on a group is related to a walk on its Cayley graph. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I haven't read the book in detail, but my impression is that, nevertheless, the two concepts are not exactly equivalent.  &lt;hr&gt;&#xA;&lt;strong&gt;Update (Nov. 7 2010)&lt;/strong&gt; &lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any stochastic process $(X_n)$ taking values in a countable state space&#xA;can be built up out of the following pieces:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) an initial distribution $P(X_0=x)$ that tells us&#xA;  how to choose the starting point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) transition kernels that tell us how to choose the next&#xA;state given the history of the process up to the present, that is,&#xA; $P(X_{n+1}=x_{n+1} | X_n=x_n, \dots, X_0=x_0)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, a process is called Markov if the kernels only depend on $n$, $x_{n+1}$ and $x_n$,&#xA;that is,&#xA;$P(X_{n+1}=x_{n+1} | X_n=x_n, \dots, X_0=x_0) = P(X_{n+1}=x_{n+1} | X_n=x_n)$.&#xA;This is a big assumption and a huge simplification. But enough models of&#xA;real world random behaviour have this &quot;memoryless&quot; property to make Markov&#xA;processes wide spread and extremely useful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A further simplification is to assume that the Markov process is time homogeneous,&#xA;so that the transition rules only depend on the states and not on time. That is,&#xA;we assume that $P(X_{n+1}=y | X_n=x)=P(X_1=y | X_0=x)$ for all $x,y$.&#xA;The crucial point here is that $p(x,y):=P(X_{n+1}=y | X_n=x)$ is the same for all $n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's now assume that the state space is an additive group.&#xA;The transition kernels for a general process can be rewritten in terms of increments:&#xA;$$P(X_{n+1}=x_{n+1} | X_n=x_n, \dots, X_0=x_0)$$&#xA;can be written &#xA;$$P(X_{n+1}-X_n=x_{n+1}-x_n | X_n-X_{n-1}=x_n-x_{n-1}, \dots, X_0=x_0).$$&#xA;If the increments are independent, then this becomes&#xA;$$P(X_{n+1}-X_n=x_{n+1} -x_n)=:\mu_n(x_{n+1}-x_n).$$&#xA;In this case, the process is automatically Markov and&#xA;the transition kernels are an arbitrary  sequence $(\mu_n)$ of probabilities on the state space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, for a random walk we assume &lt;strong&gt;both&lt;/strong&gt; time homogeneity and independent increments.&#xA;Thus, the transition kernel $p(x,y)$ is represented by a single measure: $p(x,y)=\mu(y-x)$.&#xA;This imposes  a kind of homogeneity in space and in time.&#xA;We are down to an extremely special, yet still&#xA;interesting process, that only depends on the initial distribution and&#xA;the single measure $\mu$. This is why Ross insists that for random walks the increments are&#xA;identically distributed; just being independent is not enough. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because random walks are so special, they can be analyzed by &#xA; mathematical tools that don't work for most Markov processes. In particular,&#xA; tools from harmonic analysis are very fruitful here.&#xA;A standard reference, though a little old, is  &lt;em&gt;Principles of Random Walk (2nd ed)&lt;/em&gt;&#xA;by Frank Spitzer. A more recent treatment is given in &lt;em&gt;Random Walk: A Modern Introduction&lt;/em&gt;&#xA; by Gregory Lawler and Vlada Limic (2010).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's an exercise to convince you that random walks are only a small part of&#xA;all Markov chains. Take the state space to be the corners of a triangle, labelled $\{ 0, 1, 2\}$, &#xA;considered as the group ${\mathbb Z}$ mod 3. Any 3 by 3 stochastic matrix&#xA; (non-negative entries and row sums all equal to 1) defines a time homogeneous Markov&#xA;chain on this state space. Which of these are random walks?&lt;/p&gt;&#xA;" OwnerUserId="940" LastEditorUserId="940" LastEditorDisplayName="" LastEditDate="2010-11-07T20:32:08.440" LastActivityDate="2010-11-07T20:32:08.440" CommentCount="4" />
  <row Id="9174" PostTypeId="2" ParentId="9163" CreationDate="2010-11-06T16:25:35.493" Score="3" ViewCount="" Body="&lt;p&gt;In light of your &lt;a href=&quot;http://math.stackexchange.com/questions/9161/concept-of-random-walk&quot;&gt;other question&lt;/a&gt;, one should say there are a lot of things referred to by the name &quot;random walk&quot;.  In its simplest form, Donsker's theorem is about a process $X_n$ in $\mathbb{R}^d$ whose increments $X_n - X_{n-1}$ are iid with any distribution that has zero mean and finite variance.  In particular, the increments do not have to be Gaussian.  It then asserts that the scaling limit of the process $X_n$ is $d$-dimensional Brownian motion (aka Wiener process).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are lots of other stochastic processes that have Brownian motion as their scaling limits, and still other processes that have a scaling limit that is not Brownian motion.  Is there something specific you're interested in?&lt;/p&gt;&#xA;" OwnerUserId="822" LastActivityDate="2010-11-06T16:25:35.493" CommentCount="2" />
  <row Id="9175" PostTypeId="2" ParentId="9171" CreationDate="2010-11-06T16:32:29.493" Score="1" ViewCount="" Body="&lt;p&gt;Let $ x_i=1/m $ certainly $ x_i \in (0,1] $ for all $i$ and all natural numbers $m$.&#xA;Then your function becomes&#xA;$-\log m \sum_{i=1}^{n}{c_i} + \frac{\lambda}{m} (\sum_{i=1}^{n}{d_i})$ which can be written as $-m A + B/m $ where $A,B$ are positive constants.&#xA;This can be made arbitrarily large and negative, so there is no minimum.&#xA;If this is part of some work on Lagrangian multipliers you need to think again about your restrictions on $\lambda$.&lt;/p&gt;&#xA;" OwnerUserId="2712" LastActivityDate="2010-11-06T16:32:29.493" CommentCount="4" />
  <row Id="9176" PostTypeId="2" ParentId="9131" CreationDate="2010-11-06T16:33:34.480" Score="1" ViewCount="" Body="&lt;p&gt;Exact sequence of $R=\mathbb Z$ modules $0\to \mathbb Z \to \mathbb Q$ becomes not exact after choice $p=2\mathbb Z$  and tensor with $R_p/m_p =\mathbb Z /(2) $ because it becomes    $0 \to  \mathbb Z /(2)\to 0$&lt;/p&gt;&#xA;" OwnerUserId="2872" LastActivityDate="2010-11-06T16:33:34.480" />
  <row Id="9178" PostTypeId="1" AcceptedAnswerId="9179" CreationDate="2010-11-06T16:40:23.447" Score="1" ViewCount="209" Body="&lt;p&gt;I'm looking to strengthen my understanding of the math that is directly useful to practical computer science, as opposed to unsolved computer science problems. In other words, the kind of math that would be useful in day to day algorithmic analysis, and come up during senior level interviews at software companies. My math knowledge is very rudimentary. I want to say it's almost at a 9th or 10th grade level. Will I need to go back and refresh on algebra, or can I jump right into combinatorics? What are the prerequisites?&lt;/p&gt;&#xA;" OwnerUserId="3180" LastActivityDate="2010-11-06T22:07:45.527" Title="What are the prerequisites for combinatorics?" Tags="&lt;number-theory&gt;&lt;combinatorics&gt;&lt;algorithms&gt;&lt;computer-science&gt;&lt;discrete-mathematics&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="9179" PostTypeId="2" ParentId="9178" CreationDate="2010-11-06T16:53:35.007" Score="2" ViewCount="" Body="&lt;p&gt;In my opinion the &quot;prerequisite&quot; knowledge for Combinatorics/Discrete mathematics is primarily basic logic. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many people would say that this falls under discrete math, but I think the main things that you will need to understand are Truth Tables, De Morgan, Induction, set theory and such. Many discrete books cover these topics, so don't be too afraid. If you really want to get another book to cover this material, the first chapter of &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0471368792&quot; rel=&quot;nofollow&quot;&gt;Herstein&lt;/a&gt; does this very well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as algorithm analysis, you will probably want to understand some basic asymptotics, big O, and recursion, all of these things can be found in most introductions to discrete math, such as the often recommended &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0201558025&quot; rel=&quot;nofollow&quot;&gt;Concrete Mathematics.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are particularly worrying about Geometry, Trig and Calc, I wouldn't. A few things from PreCalc/Calc will be useful, but I don't thing overwhelmingly so. I don't expect you to see much Trig, and Geometry will mostly used to provide examples.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other question I linked in the comment will also be useful probably.&lt;/p&gt;&#xA;" OwnerUserId="16" LastActivityDate="2010-11-06T16:53:35.007" />
  <row Id="9180" PostTypeId="2" ParentId="9178" CreationDate="2010-11-06T16:53:41.613" Score="0" ViewCount="" Body="&lt;p&gt;To &quot;strengthen your understanding of the math that is directly useful to practical computer science&quot; I think your best bet is simply to master the material in  Graham; Knuth; Patashnik. Concrete Mathematics and Rosen: Discrete mathematics and its applications.&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-11-06T16:53:41.613" CommentCount="1" />
  <row Id="9181" PostTypeId="2" ParentId="9150" CreationDate="2010-11-06T17:33:23.030" Score="3" ViewCount="" Body="&lt;p&gt;@Andy (in reply to your comment/question &quot;Could you provide some example that has a closed form and explain if (and how) it is possible to find the fixed point without solving x = f(x) but trying to calculate the limit of x_n?&quot;: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe that you would be hard-pressed to achieve this, since your function $f$ is a continuous function (being a contraction map in the first place); and if you then take limits of both sides of $x_n = f(x_{n-1})$, you will get:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\lim_{n \rightarrow \infty} x_n = \lim_{n \rightarrow \infty} f(x_{n-1})$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which (by continuity) leads to:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\lim_{n \rightarrow \infty} x_n = f (\lim_{n \rightarrow \infty} x_{n-1})$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$l = f(l)$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;with $l = \lim_{n \rightarrow \infty} x_n$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means that you will have to solve $l = f(l)$, which was what you wanted to avoid in the first place!&lt;/p&gt;&#xA;" OwnerUserId="3195" LastActivityDate="2010-11-06T17:33:23.030" CommentCount="2" />
  <row Id="9182" PostTypeId="1" AcceptedAnswerId="9184" CreationDate="2010-11-06T17:37:29.217" Score="1" ViewCount="75" Body="&lt;p&gt;In Hatcher's &lt;em&gt;Algebraic Topology,&lt;/em&gt; section 3.2, during the computation of the cohomology ring of a $n$-torus, the following assertion is made. Let $Y$ be a space and $R$ a commutative ring. Then the natural map in cohomology&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ H^n(I \times Y, R) \to H^n( \partial I \times Y, R)$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;is a split injection. This is likely straightforward, and I believe it follows directly from the Kunneth theorem. However, I don't see a direct proof, which is probably what Hatcher is indicated. Can someone explain to me what I'm missing?&lt;/p&gt;&#xA;" OwnerUserId="536" LastActivityDate="2010-11-06T17:43:27.117" Title="Why is $H^n(I \times Y, R) \to H^n( \partial I \times Y, R)$ a split injection?" Tags="&lt;algebraic-topology&gt;" AnswerCount="1" />
  <row Id="9184" PostTypeId="2" ParentId="9182" CreationDate="2010-11-06T17:43:27.117" Score="5" ViewCount="" Body="&lt;p&gt;Isn't $I$ just the closed unit interval? Then $I\times Y$ is homotopy&#xA;equivalent to&#xA;$Y$ and $\partial I\times Y$ is just two copies of $Y$. It's&#xA;clear that your map then is just the diagonal map from&#xA;$H^n(Y)$ to $H^n(Y)\oplus H^n(Y)$.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-11-06T17:43:27.117" CommentCount="1" />
  <row Id="9185" PostTypeId="2" ParentId="9138" CreationDate="2010-11-06T17:44:15.123" Score="12" ViewCount="" Body="&lt;p&gt;Added: As explained in the comments, certain trigonometric equations such as the linear equations in $\sin x$ and $\cos x$ can be solved by a resolvent quadratic equation. One method is to write the $\sin x$ and $\cos x$ functions in terms of the same trigonometric function. Since all [direct] trigonometric functions of the simple angle can be expressed rationally as a function of the $\tan$ of the half-angle, such a conversion is adequate for these equations.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Since $$\cos \alpha =\frac{1-\tan ^{2}\frac{\alpha }{2}}{1+\tan ^{2}\frac{%&#xA;\alpha }{2}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sin \alpha =\frac{2\tan \frac{\alpha }{2}}{1+\tan ^{2}%&#xA;\frac{\alpha }{2}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;your equation &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$m\cos \alpha +\sin \alpha =1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is equivalent to &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$m-m\tan ^{2}\frac{\alpha }{2}+2\tan \frac{\alpha }{2}=1+\tan ^{2}\frac{\alpha }{2}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One may set $x=\tan \frac{\alpha }{2}$ ($\alpha =2\arctan x$), and thus get&#xA;the quadratic equation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left( 1+m\right) x^{2}-2x+1-m=0.$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Its solutions are: $x=\frac{1}{m+1}\left( -m+1\right) $ (if $m\neq -1$) or $%&#xA;x=1$ (if $m=-1$), which gives &lt;/p&gt;&#xA;&#xA;&lt;p&gt;i) If $m\neq -1$, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\alpha =2\arctan x=2\arctan \frac{1-m}{m+1},$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ii) If $m=-1,$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\alpha =2\arctan 1=\frac{\pi }{2}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A different technique to solve a linear equation in $\sin \alpha $ and $\cos&#xA;\alpha $ is to use an auxiliary angle $\varphi $. If you set $m=\tan \varphi &#xA;$, your equation takes the form&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sin \alpha +\tan \varphi \cdot \cos \alpha =1$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;or&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sin (\alpha +\varphi )=\cos \varphi =\frac{1}{\pm \sqrt{1+\tan ^{2}\varphi &#xA;}}=\pm \sqrt{\frac{1}{1+m^{2}}},$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and obtain &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\alpha =\pm \arcsin \sqrt{\frac{1}{1+m^{2}}}-\arctan m.$$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Detailed derivation: from $m\cos \alpha +\sin \alpha =1$ and $m=\tan \varphi &#xA;$, we get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sin \alpha +\tan \varphi \cdot \cos \alpha =1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sin \alpha +\dfrac{\sin \varphi }{\cos \varphi }\cdot \cos \alpha =1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\dfrac{\sin \alpha \cdot \cos \varphi +\sin \varphi \cdot \cos \alpha }{\cos&#xA;\varphi }=1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\dfrac{\sin \left( \alpha +\varphi \right) }{\cos \varphi }=1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sin \left( \alpha +\varphi \right) =\cos \varphi $.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The identity&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\cos \varphi =\pm \sqrt{\dfrac{1}{1+\tan ^{2}\varphi }}$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;can be obtained as follows&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sin ^{2}\varphi +\cos ^{2}\varphi =1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\dfrac{\sin ^{2}\varphi }{\cos ^{2}\varphi }+1=\dfrac{1}{\cos ^{2}\varphi }$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\tan ^{2}\varphi +1=\dfrac{1}{\cos ^{2}\varphi }$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\cos ^{2}\varphi =\dfrac{1}{1+\tan ^{2}\varphi }$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sin \left( \alpha +\varphi \right) =\pm \sqrt{\dfrac{1}{1+\tan ^{2}\varphi }%&#xA;}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\alpha +\varphi =\arcsin \left( \pm \sqrt{\dfrac{1}{1+\tan ^{2}\varphi }}%&#xA;\right) $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;($m=\tan \varphi $, $\varphi =\arctan m$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\alpha +\arctan m=\arcsin \left( \pm \sqrt{\dfrac{1}{1+m^{2}}}\right) $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and finally&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\alpha =\arcsin \left( \pm \sqrt{\dfrac{1}{1+m^{2}}}\right) -\arctan m.$&lt;/p&gt;&#xA;" OwnerUserId="752" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2010-11-07T12:50:06.913" LastActivityDate="2010-11-07T12:50:06.913" CommentCount="10" />
  <row Id="9186" PostTypeId="1" CreationDate="2010-11-06T17:51:29.743" Score="3" ViewCount="153" Body="&lt;p&gt;This is a follow-up (refinement?) of &lt;a href=&quot;http://math.stackexchange.com/questions/9010/what-is-the-difference-between-a-variety-and-a-manifold&quot;&gt;this question&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In learning some algebraic topology, I've learned to think of an affine scheme as spec $R$.  (I've been told that this is a legitimate use of terminology because it provides an embedding of $Rings^{op}$ into a larger category.)  One construction we can do, for example, is create the &quot;tangent scheme&quot;, which is obtained by localizing and completing.  The examples I've been looking at are $\hat{\mathbb{G}}_a$, $\hat{\mathbb{G}}_m$, $(\mathbb{Z}[[t]],F)$ (FGLs (1-dimensional, commutative) more generally), $T_1C^k(\hat{\mathbb{G}}_a,\hat{\mathbb{G}}_m)\cong C^k(\hat{\mathbb{G}}_a,\hat{\mathbb{G}}_a)$ (&quot;commutative $k$-variate FGLs satisfying the 2-cocycle condition&quot;), etc.  We then have exponential maps, which e.g. in the last case is $exp:C^k(\hat{\mathbb{G}}_a,\hat{\mathbb{G}}_a) \rightarrow C^k(\hat{\mathbb{G}}_a,\hat{\mathbb{G}}_m)$ given by $g\mapsto 1+g$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just to test the waters, here is my question (although please feel free to push it further or in a different direction).  Presumably these exponential maps are not always injective.  Once we apply this picture to a particular ring (or perhaps even before?), can we translate differential-geometric ideas like conjugate points, geodesics, &quot;$exp$ is a local isomorphism&quot;, etc. into algebro-geometric language?  When we can, which theorems for manifolds carry over to schemes and which must we discard?&lt;/p&gt;&#xA;" OwnerUserId="2159" LastEditorUserId="2159" LastEditorDisplayName="" LastEditDate="2010-11-06T21:01:40.413" LastActivityDate="2010-12-26T09:31:35.473" Title="To what extent do the stories on manifolds carry over to schemes?" Tags="&lt;algebraic-geometry&gt;&lt;differential-geometry&gt;" AnswerCount="1" CommentCount="7" />
  <row Id="9187" PostTypeId="2" ParentId="9138" CreationDate="2010-11-06T17:55:15.110" Score="6" ViewCount="" Body="&lt;p&gt;Because you're familiar with complex exponentials, all you needed to succeed is to observe that by setting $w = m - i$ your equation becomes&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$w \exp(i \alpha) + \bar{w} \exp(-i \alpha) = 2.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(&lt;strong&gt;Edit&lt;/strong&gt; What might motivate this choice of $w$?  Well, if you're thinking of complex numbers you're thinking that $\cos(\alpha) + i \sin(\alpha) = \exp(i \alpha)$.  To obtain $m \cos(\alpha) + \sin(\alpha)$ from this you need to multiply $\cos(\alpha)$ by $m$ and $i \sin(\alpha)$ by $-i$.  Thus it's automatic to set out by computing&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(m - i)(\cos(\alpha) + i \sin(\alpha)) = \left(m \cos(\alpha) + \sin(\alpha)\right) + \left(\text{real stuff}\right) i.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The real part of this expression is exactly what you need, whence the original equation reads&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$1 = \mathbb{Re}\left(w \exp(i \alpha))\right) = \frac{w \exp(i \alpha) + \overline{w \exp(i \alpha)}}{2} = \frac{w \exp(i \alpha) + \bar{w} \exp(-i \alpha)}{2}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Multiply by 2 to clear the fraction.  &lt;strong&gt;End of edit&lt;/strong&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Setting $z = \exp(i \alpha)$ to simplify the notation yields&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$w z^2 - 2 z + \bar{w} = 0,$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with two generic solutions by the quadratic formula:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$z = \frac{1 + \sqrt{1 - |w|^2}}{w}, \quad z = \frac{1 - \sqrt{1 - |w|^2}}{w}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because $\sqrt{1 - |w|^2} = \pm m i$, the solutions simplify to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$z = \frac{i + m}{1 + m i}, \quad z = i.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Once again we see a connection to fractional linear transformations, as per &lt;a href=&quot;http://math.stackexchange.com/questions/6695/ways-to-evaluate-int-sec-theta-d-theta/6702#6702&quot;&gt;@Qiaochu Yuan's comment&lt;/a&gt;.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By definition, $\alpha$ is the argument (polar coordinate) of a solution $z$.  Writing $\tan(\theta) = m$ (which defines $\theta$ up to an integral multiple of $\pi$), the argument of the denominator is $\theta$ while the argument of the numerator is equal to $\pi/2 - \theta$.  Whence&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\alpha = \pi/2 - 2\theta \quad \text{or} \quad \alpha = \pi/2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(both modulo an integral multiple of $2 \pi$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you prefer, in the first case you may write&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\alpha = \pi/2 - 2\arctan(m).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second case obviously gives the trivial solutions $1 = m \cos(\pi/2 + 2 n \pi) + \sin(\pi/2 + 2 n \pi)$ $= m 0 + 1.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first solution appears to be a new one, but it's equivalent to the others that have been obtained.  As a check, we may assume $\theta$ is not an odd multiple of $\pi/2$, whence $\alpha$ is not an odd multiple of $\pi/2$, and compute&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$m \cos(\alpha) + \sin(\alpha)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ = \tan(\theta) \cos(\pi/2 - 2 \theta) + \sin(\pi/2 - 2\theta) = \tan(\theta) \sin(2 \theta) + \cos(2 \theta)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$= \frac{\sin(\theta)}{\cos(\theta)}\left(2 \sin(\theta) \cos(\theta)\right) + \left(\cos(\theta)^2 - \sin(\theta)^2\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$= \sin(\theta)^2 + \cos(\theta)^2 = 1.$$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Although for a problem like this I would ordinarily take the route described by @Robin Chapman, I like this complex number approach because it produces a simpler solution (which actually is defined for $m = -1$, btw) by means of straightforward techniques and it explicitly produces &lt;em&gt;all&lt;/em&gt; solutions.  When $m \lt -1$ and we use the standard branch of the arctangent it differs by $2 \pi$ from the one &lt;em&gt;Mathematica&lt;/em&gt; obtained, but that's of no consequence.  (It's curious that &lt;em&gt;Mathematica&lt;/em&gt; cannot tell these two solutions are the same, even when you restrict the domain to real nonnegative $m$.  A plot of their difference will remove any lingering doubts.)&lt;/p&gt;&#xA;" OwnerUserId="1489" LastEditorUserId="1489" LastEditorDisplayName="" LastEditDate="2010-11-06T20:45:46.643" LastActivityDate="2010-11-06T20:45:46.643" CommentCount="3" />
  <row Id="9188" PostTypeId="1" AcceptedAnswerId="9196" CreationDate="2010-11-06T18:23:03.233" Score="1" ViewCount="589" Body="&lt;p&gt;In this &lt;a href=&quot;http://math.stackexchange.com/questions/8904/isomorphism-of-vector-spaces-over-mathbbq&quot;&gt;post&lt;/a&gt; we saw isomorphism of vector spaces over $\mathbb{Q}$. Just came across this question:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Is $\mathbb{Q}(\sqrt{2}) \cong \mathbb{Q}(\sqrt{3})$?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In know these as $\mathbb{Q}$-Vector spaces, are isomorphic from the  linked post. But as fields are they isomorphic? I neither know how to prove it nor how to disprove it.&lt;/p&gt;&#xA;" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-06T19:47:59.120" LastActivityDate="2010-11-07T01:01:30.787" Title="Is $\mathbb{Q}(\sqrt{2}) \cong \mathbb{Q}(\sqrt{3})$?" Tags="&lt;abstract-algebra&gt;&lt;field-theory&gt;" AnswerCount="3" CommentCount="8" FavoriteCount="1" />
  <row Id="9189" PostTypeId="1" AcceptedAnswerId="9211" CreationDate="2010-11-06T18:26:40.497" Score="0" ViewCount="196" Body="&lt;p&gt;Hi, I was reading John Lee's Introduction to Smooth manifolds, and I came across this question:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Let $M$ be a smooth manifold, and let $\delta : M \rightarrow \mathbb{R}$ be a positive continuous function. Using a partition of unity, show that there is a smooth function $\tilde{\delta} : M \rightarrow \mathbb{R}$ such that $0 &amp;lt; \tilde{\delta}(x) &amp;lt; \delta(x)$ for all $x \in M$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I thought about it for a while, and I'm pretty stuck on it. Does anyone have any ideas?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Edit: It has been pointed out that one can basically assume that $M = \mathbb{R}$, because the proof should be the same in both cases. If you do not know anything about smooth manifolds, feel free to do this!)&lt;/p&gt;&#xA;" OwnerUserId="1089" LastEditorUserId="1089" LastEditorDisplayName="" LastEditDate="2010-11-06T20:40:46.790" LastActivityDate="2010-11-06T22:22:01.867" Title="Question on Partitions of Unity" Tags="&lt;real-analysis&gt;&lt;analysis&gt;&lt;differential-geometry&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="9190" PostTypeId="2" ParentId="9188" CreationDate="2010-11-06T18:37:54.027" Score="5" ViewCount="" Body="&lt;p&gt;To prove this: Suspect the fields are not isomorphic, then we can attempt to find a property which holds inside one and does not the other - but whose truth is preserved by isomorphism.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the field $\mathbb{Q}(\sqrt{2})$ there is an element which satisfies the field property $x^2=2$. There is no element in $\mathbb{Q}(\sqrt{3})$ which satisfies this, but suppose for a contradiction that there was an isomorphism $\psi : \mathbb{Q}(\sqrt{2}) \rightarrow \mathbb{Q}(\sqrt{3})$ we would have $\psi(x^2) = \psi(2)$ which is equivalent to $\psi(x)^2 = \psi(1)+\psi(1)$ and since $\psi(1) = 1$ we have an element of $\mathbb{Q}(\sqrt{3})$ which, squared, is 2.&lt;/p&gt;&#xA;" LastActivityDate="2010-11-06T18:37:54.027" CommentCount="8" />
  <row Id="9191" PostTypeId="1" CreationDate="2010-11-06T18:50:57.950" Score="0" ViewCount="149" Body="&lt;p&gt;Assume function $f$ has a discontinuity (jump discontinuity) at $x_0$. This discontinuity should be buried to level $k$, meaning (let $f$ is modified to $g$ after an operation) $g$, $g'$, $g'',\ldots g^{(k-1)}$ ($g^{(i)}$ means $i$th derivative of $g$) should all be continuous at $x_0$. What is the operation on $f$ by which this an be achieved (for example convolving with a suitable function, etc.,) ? For a slightly different question, assume $f$ is continous at $x_0$ but not differentiable.   &lt;strong&gt;EDIT&lt;/strong&gt;: $f$ has a compact support or atleast belong to $L^1$ or $L^2$. same should be the case with $g$.&lt;/p&gt;&#xA;" OwnerUserId="2987" LastEditorUserId="2987" LastEditorDisplayName="" LastEditDate="2010-11-06T20:56:57.453" LastActivityDate="2010-11-07T02:41:46.037" Title="How to bury a discontinuity to a higher level ?" Tags="&lt;calculus&gt;&lt;real-analysis&gt;" AnswerCount="1" CommentCount="27" />
  <row Id="9192" PostTypeId="1" AcceptedAnswerId="9194" CreationDate="2010-11-06T18:55:29.543" Score="1" ViewCount="103" Body="&lt;p&gt;I have a velocity vector on X, Y and Z axes. The values are represented as M/S (meters per second).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What algorithm should I use to convert this vector into a general MPH (miles per hour) value that takes into account all axes of the vector?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This may (or may not) be a very simple question. But I am far from being a genius at math! So apologies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks very much.&lt;/p&gt;&#xA;" OwnerUserId="3198" LastActivityDate="2010-11-06T19:04:04.773" Title="Converting a velocity vector into MPH?" Tags="&lt;vector-space&gt;" AnswerCount="1" />
  <row Id="9193" PostTypeId="2" ParentId="9188" CreationDate="2010-11-06T18:59:11.353" Score="2" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\ $ Compare discriminants: $\rm\displaystyle\ \big\{\big(\alpha-\alpha'\big)^2:\ \alpha \in \mathbb Q(\sqrt 2)\big\}\: =\ 2\ \mathbb Q^{\:2}\ \ $ vs. $\rm\ \ 3\ \mathbb Q^{\:2}\ $ for $\rm\ \mathbb Q(\sqrt 3)\ $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that if $\rm\ \alpha,\: \alpha'\ \not\in\mathbb Q\ $ are conjugate then they remain so under any field isomorphism since their minimal polynomial $\rm\ (x-\alpha)\ (x-\alpha')\ $ is in $\rm\:\mathbb Q[x]\:$ so it is fixed by any isomorphism.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact quadratic fields are characterized uniquely be their &lt;strong&gt;discriminant&lt;/strong&gt;.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-11-06T23:09:37.403" LastActivityDate="2010-11-06T23:09:37.403" CommentCount="4" />
  <row Id="9194" PostTypeId="2" ParentId="9192" CreationDate="2010-11-06T19:04:04.773" Score="2" ViewCount="" Body="&lt;p&gt;Multiply each coordinate by 2.23693629.  In general, to convert units of a vector, you just convert units of each coordinate.  This works because scalar multiplication by positive constants preserves direction, and because lengths of vectors and scalar multiplication are related by $\|c v\|=|c|\|v\|$.  Scalar multiplication in this case just means multiplying each coordinate by the conversion factor.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-11-06T19:04:04.773" CommentCount="7" />
  <row Id="9195" PostTypeId="2" ParentId="9138" CreationDate="2010-11-06T19:11:24.167" Score="8" ViewCount="" Body="&lt;p&gt;Since we are adding different proofs, here is a geometric one, for the special case when $m &gt; 0$ and $\alpha$ is acute.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We use the &lt;a href=&quot;http://en.wikipedia.org/wiki/Law_of_tangents&quot; rel=&quot;nofollow&quot;&gt;Law of Tangents&lt;/a&gt; which states, that in a triangle&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/gkGVS.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the identity&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \frac{a-b}{a+b} = \frac{\tan\frac{\alpha - \beta}{2}}{\tan\frac{\alpha+\beta}{2}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;holds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now consider the below triangle (pardon the crude figure)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/1ivPA.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have $\displaystyle \angle{BCA} = \alpha$, $\displaystyle  \angle{CBA} = \pi/2$ and $\displaystyle \angle{CAB} = \pi/2 - \alpha$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ABC is a right angled triangle with $\displaystyle AC = 1 = m\cos\alpha + \sin \alpha$. Point $\displaystyle D$ is chosen on $\displaystyle AC$ such that $\displaystyle AD = \sin \alpha$ and so $\displaystyle DC = m\cos \alpha$. We also have $\displaystyle BC = \cos \alpha$ and $AB = \sin \alpha$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now $\displaystyle AD = \sin \alpha = AB$ and so $\triangle ABD$ is isoceles and hence $\displaystyle \angle{ABD}  = \angle{ADB} = \pi/4 + \alpha/2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus we have that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle x = \angle{CDB} = 3\pi/4 - \alpha/2$ and $\displaystyle  y = \angle{CBD} = \pi/4-\alpha/2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now apply the law of tangents to $\displaystyle \triangle BCD$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\cos \alpha - m\cos\alpha}{\cos\alpha + m\cos\alpha} = \frac{\tan (x/2 - y/2)}{\tan(x/2 + y/2)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ = \frac{\tan(\pi/4)}{\tan(\pi/2 - \alpha/2)} = \tan(\alpha/2)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{1-m}{1+m} = \tan(\alpha/2)$$&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-11-06T19:16:55.993" LastActivityDate="2010-11-06T19:16:55.993" CommentCount="3" />
  <row Id="9196" PostTypeId="2" ParentId="9188" CreationDate="2010-11-06T19:33:29.927" Score="11" ViewCount="" Body="&lt;p&gt;More generally: suppose $d$ and $d'$ are both squarefree integers, both different from $1$, and consider $F_1 = \mathbb{Q}(\sqrt{d})$ and $F_2 = \mathbb{Q}(\sqrt{d'})$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;They are both isomorphic as $\mathbb{Q}$-vector spaces, since they are both of dimension $2$; or more explicitly, every element of $F_1$ can be written uniquely as $a+b\sqrt{d}$ with $a,b\in\mathbb{Q}$ (unique because  $\sqrt{d}\notin\mathbb{Q}$), and every element of $F_2$ can be written uniquely as $x+y\sqrt{d'}$ with $x,y\in\mathbb{Q}$. The map $f\colon F_1\to F_2$ given by $f(a+b\sqrt{d}) = a + b\sqrt{d'}$ is additive and $\mathbb{Q}$-homogeneous, clearly bijective, so $F_1$ and $F_2$ are isomorphic as vector spaces over $\mathbb{Q}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, they are never isomorphic as fields; clearly, $d$ is a square in $F_1$. I claim $d$ can only be a square in $F_2$ if $d=d'$. Indeed, if$(x+y\sqrt{d'})^2 = d$. That means that $x^2 + d'y^2 + 2xy\sqrt{d'} = d$, hence $2xy = 0$ and $x^2+d'y^2=d$. If $x=0$, then $d=d'y^2$, so clearing denominators you get $da^2 = d'b^2$ for some $a,b\in\mathbb{Z}$, $\gcd(a,b)=1$; since both $d$ and $d'$ are squarefree, it follows that $|a|=|b|=1$, so $d=d'$. If $y=0$, then $d=x^2$, so $d$ is the square of a rational, contradicting the fact that it is a squarefree integer different from $1$. Thus, of $d$ is a square in $F_2$, then $d=d'$. Hence, if $F_1\cong F_2$, then $d=d'$ (converse is immediate).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, since every quadratic extension of $\mathbb{Q}$ is equal to $\mathbb{Q}(\sqrt{d})$ for some squarefree integer $d$ different from $1$, you conclude that any two quadratic extensions are either identical or not isomorphic as fields. &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-06T20:08:49.083" LastActivityDate="2010-11-06T20:08:49.083" CommentCount="6" />
  <row Id="9197" PostTypeId="1" CreationDate="2010-11-06T19:55:03.010" Score="2" ViewCount="126" Body="&lt;p&gt;Let $f$ be in $L^{1}(\mathbb{R})$, $\mathbb{R}$ the real numbers. Show that for every $\varepsilon &gt; 0$ there exists $A \subseteq R$ , measurable, such that $m(A) &amp;lt; \infty$ , $f$ is bounded on $A$ and $ \int_{\mathbb{R}} |f|  &amp;lt; \int_{A} |f| + \varepsilon$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we take $A$ as the support of the simple function which approximates $f$ in the $L^{1}$ norm then this has finite measure and it satisfies the other conditions. But I don't see why $f$ must be bounded on it. Any ideas?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you.&lt;/p&gt;&#xA;" OwnerUserId="4219" LastActivityDate="2010-11-06T21:46:08.253" Title="Measurable set on which a function is bounded" Tags="&lt;measure-theory&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="2" />
  <row Id="9198" PostTypeId="2" ParentId="9197" CreationDate="2010-11-06T20:05:09.190" Score="3" ViewCount="" Body="&lt;p&gt;$|f|I_{|f|\leq n} $ increases to $|f|I_{|f| &amp;lt; \infty} = |f|$ (a.e.) as $f$ is in $L^{1}(\mathcal{R})$, the result follows from monotone convergence theorem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Update : This answer is incomplete, please see Jonas's answer.&lt;/p&gt;&#xA;" OwnerUserId="404" LastEditorUserId="404" LastEditorDisplayName="" LastEditDate="2010-11-06T21:46:08.253" LastActivityDate="2010-11-06T21:46:08.253" CommentCount="1" />
  <row Id="9199" PostTypeId="1" AcceptedAnswerId="9203" CreationDate="2010-11-06T20:09:25.667" Score="7" ViewCount="404" Body="&lt;p&gt;This is motivated by &lt;a href=&quot;http://math.stackexchange.com/questions/8933/proof-that-there-is-no-closed-form-solution&quot;&gt;this question&lt;/a&gt; and the fact that I have no access to the Timothy Chow's paper &lt;a href=&quot;http://www.jstor.org/pss/2589148&quot; rel=&quot;nofollow&quot;&gt;What Is a Closed-Form Number?&lt;/a&gt; indicated there by&#xA;Qiaochu Yuan.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If an equation $f(x)=0$ has no closed form solution, what does it normally&#xA;mean? Added: $f$ may depend (and normally does) on parameters. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To me this is equivalent to say that one cannot solve it for $x$ in&#xA;the sense that there is no elementary expression $g(c_{1},c_{2},\ldots&#xA;,c_{p})$ consisting only of a finite number of polynomials, rational&#xA;functions, roots, exponentials, logarithmic and trigonometric functions,&#xA;absolute values, integer and fractional parts, such that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f(g(c_{1},c_{2},\ldots ,c_{p}))=0$.&lt;/p&gt;&#xA;" OwnerUserId="752" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2010-11-07T14:46:49.140" LastActivityDate="2010-11-07T14:46:49.140" Title="What does closed form solution usually mean?" Tags="&lt;soft-question&gt;&lt;closed-form&gt;" AnswerCount="2" CommentCount="9" FavoriteCount="1" />
  <row Id="9200" PostTypeId="2" ParentId="9197" CreationDate="2010-11-06T20:11:57.560" Score="4" ViewCount="" Body="&lt;p&gt;You can do this as a &quot;$2\epsilon$-proof&quot; (or $\epsilon/2$ if you prefer).  First, since $(\int_{-n}^n|f|)_{n=1}^\infty$ converges to $\int_{\mathbb{R}}|f|$, there is an $n$ such that $\int_{\mathbb{R}}|f|\lt\int_{-n}^n|f|+\epsilon/2$.  Then, since $(\int_{-n}^n|f|\cdot\chi_{\{|f|\leq m\}})_{m=1}^\infty$ converges to $\int_{-n}^n|f|$, there is an $m$ such that $\int_{-n}^n|f|\lt\int_{-n}^n|f|\cdot\chi_{\{|f|\leq m\}}+\epsilon/2$.  Take $A=[-n,n]\cap\{x:|f(x)|\leq m\}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rather than using simple functions to show this, I would use this as a first step to showing that $f$ can be approximated by simple functions, because now $f$ can be uniformly approximated by simple functions on $A$.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2010-11-06T20:34:56.377" LastActivityDate="2010-11-06T20:34:56.377" CommentCount="4" />
  <row Id="9201" PostTypeId="2" ParentId="9138" CreationDate="2010-11-06T20:13:10.737" Score="8" ViewCount="" Body="&lt;p&gt;Yet another approach, which is very specific to this problem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{eqnarray}&#xA;1 - \sin\alpha &amp;amp;=&amp;amp; m \cos\alpha \\&#xA;\left( 1 - \sin\alpha \right)^2 &amp;amp;=&amp;amp; m^2 \cos^2\alpha \\&#xA;\left( 1 - \sin\alpha \right)^2 &amp;amp;=&amp;amp; m^2 ( 1 - \sin\alpha )( 1 + \sin\alpha ) \\&#xA;\left( 1 - \sin\alpha \right)\left((1 - \sin\alpha ) - m^2(1+\sin\alpha)\right) &amp;amp;=&amp;amp; 0 \\&#xA;\left( 1 - \sin\alpha \right)\left( (1-m^2) - ( 1 + m^2 )\sin\alpha \right) &amp;amp;=&amp;amp; 0&#xA;\end{eqnarray}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, $\sin\alpha=1$ or $\sin\alpha=\frac{1-m^2}{1+m^2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that, if $\sin\alpha=1$, then your original equation reduces to the $m$-less tautology, &quot;$0=0$&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the second case, a little manipulation shows relations between $m$ and other trig functions of $\alpha$. For instance, you could substitute back into your original equation to get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$1 = m\cos\alpha + \frac{1-m^2}{1+m^2} \;\;\; \Longrightarrow \;\;\; \cos\alpha=\frac{2m}{1+m^2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;from which you can get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\tan\alpha = \frac{\sin\alpha}{\cos\alpha} = \frac{1-m^2}{2m}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\tan\frac{\alpha}{2}=\frac{1-\cos\alpha}{\sin\alpha} = \frac{(1-m)^2}{1-m^2}=\frac{1-m}{1+m}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;whichever you prefer.&lt;/p&gt;&#xA;" OwnerUserId="409" LastActivityDate="2010-11-06T20:13:10.737" CommentCount="3" />
  <row Id="9202" PostTypeId="2" ParentId="9157" CreationDate="2010-11-06T20:33:23.493" Score="2" ViewCount="" Body="&lt;p&gt;This is not a good question because it actually has nothing to do with sampling theory (nor with the Chebyshev inequality, for that matter): it's quite artificial.  Nevertheless, you have grasped its intent well.  A good way to work through it is to reason in rather basic terms, rather than relying solely on mathematical manipulation; to wit:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;The standard deviation of $\bar{\chi}$ is $1/\sqrt{n}$ times the SD of the parent distribution, $\sigma$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Therefore we should express the cutoff $0.5 \sigma$ in terms of the SD of $\bar{\chi}$ itself:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$0.5 \sigma = 0.5 \sqrt{n} \left(\sigma / \sqrt{n}\right) = 0.5 \sqrt{n} SD_{\bar{\chi}}.$$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;We know (or can look up in a table) that $5$% of the time any normally distributed variate is further than $1.96$ times its standard deviation from its mean.  (The value $1.645$ corresponds to a probability of $10$%: the absolute value requires you to look at both tails, not just the right hand tail.)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Thus the question is merely asking you to solve the equation&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;$$0.5 \sqrt{n} \ge 1.96.$$&lt;/p&gt;&#xA;" OwnerUserId="1489" LastActivityDate="2010-11-06T20:33:23.493" CommentCount="1" />
  <row Id="9203" PostTypeId="2" ParentId="9199" CreationDate="2010-11-06T20:39:27.777" Score="9" ViewCount="" Body="&lt;p&gt;I would say it very much depends on the context, and what tools are at your disposal. For instance, telling a student who's just mastered the usual tricks of integrating elementary functions that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int\frac{\exp{u}-1}{u}\mathrm{d}u$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int\sqrt{(u+1)(u^2+1)}\mathrm{d}u$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;have no closed form solutions is just the fancy way of saying &quot;no, you can't do these integrals yet; you don't have the tools&quot;. To a working scientist who uses exponential and elliptic integrals, however, they do have closed forms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a similar vein, when we say that nonlinear equations, whether algebraic ones like $x^5-x+1=0$ or transcendental ones like $\frac{\pi}{4}=v-\frac{\sin\;v}{2}$ have no closed form solutions, what we're really saying is that we can't represent solutions to these in terms of functions that we know (and love?). (For the first one, though, if you know hypergeometric or theta functions, then yes, it has a closed form.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe it is fair to say that for as long as we haven't seen the solution to an integral, sum, product, continued fraction, differential equation, or nonlinear equation frequently enough in applications to give it a &lt;em&gt;standard&lt;/em&gt; name and notation, we just cop out and say &quot;nope, it doesn't have a closed form&quot;.&lt;/p&gt;&#xA;" OwnerUserId="498" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-07T02:09:12.013" LastActivityDate="2010-11-07T02:09:12.013" CommentCount="6" />
  <row Id="9204" PostTypeId="1" AcceptedAnswerId="9219" CreationDate="2010-11-06T21:07:12.483" Score="1" ViewCount="155" Body="&lt;p&gt;You are allowed to use +, -, / and * (plus, minus, division and multiplication) signs and bracketing. These signs you can put between the numbers&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1963&lt;/p&gt;&#xA;&#xA;&lt;p&gt;to form mathematical expressions. You must put at least one sign between two numbers and – cannot be used as “negative”, thus -1+9+6+3 is not allowed, but 1-9+6+3 is allowed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is: what is the smallest natural number that cannot be expressed in this way? How to show (elegantly, without computer) that it is impossible to get 14?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found myself the following representations:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;0=1*9-6-3&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1=(1/9)*(6+3)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2=1+9/(6+3)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3=1*9/(6-3)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4=1+9/(6-3)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;5=(1+9)/(6/3)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;6=1*9-6+3&lt;/p&gt;&#xA;&#xA;&lt;p&gt;7=1+9-6+3&lt;/p&gt;&#xA;&#xA;&lt;p&gt;8=1+9-6/3&lt;/p&gt;&#xA;&#xA;&lt;p&gt;9=1*(9-6)*3&lt;/p&gt;&#xA;&#xA;&lt;p&gt;10=1+(9-6)*3&lt;/p&gt;&#xA;&#xA;&lt;p&gt;11=1*9+6/3&lt;/p&gt;&#xA;&#xA;&lt;p&gt;12=1+9+6/3&lt;/p&gt;&#xA;&#xA;&lt;p&gt;13=1+9+6-3&lt;/p&gt;&#xA;" OwnerUserId="2856" LastEditorUserId="2856" LastEditorDisplayName="" LastEditDate="2010-11-07T11:30:22.800" LastActivityDate="2010-11-07T11:30:22.800" Title="A problem related to the number 1963" Tags="&lt;recreational-mathematics&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="9205" PostTypeId="2" ParentId="9138" CreationDate="2010-11-06T21:30:58.060" Score="5" ViewCount="" Body="&lt;p&gt;Another:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{eqnarray}&#xA;\frac{1-\sin\alpha}{\cos\alpha} &amp;amp;=&amp;amp; m \\&#xA;\frac{1-\cos\left(\frac{\pi}{2}-\alpha\right)}{\sin\left(\frac{\pi}{2}-\alpha\right)} &amp;amp;=&amp;amp;m \\&#xA;\tan\frac{\frac{\pi}{2}-\alpha}{2} = \tan\left(\frac{\pi}{4}-\frac{\alpha}{2}\right) &amp;amp;=&amp;amp; m \\&#xA;\frac{\tan\frac{\pi}{4}-\tan\frac{\alpha}{2}}{1+\tan\frac{\pi}{4} \tan\frac{\alpha}{2}} = \frac{1-\tan\frac{\alpha}{2}}{1+\tan\frac{\alpha}{2}} &amp;amp;=&amp;amp; m \\&#xA;\Longrightarrow \tan\frac{\alpha}{2} &amp;amp;=&amp;amp; \frac{1-m}{1+m}&#xA;\end{eqnarray}&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="409" LastActivityDate="2010-11-06T21:30:58.060" CommentCount="1" />
  <row Id="9206" PostTypeId="1" AcceptedAnswerId="9215" CreationDate="2010-11-06T21:41:30.660" Score="2" ViewCount="74" Body="&lt;p&gt;The following inequality is from the proof that the $L^p$ norm is Gâteaux differentiable for $ 1 &amp;lt; p&amp;lt;\infty$ (from &quot;Analysis&quot; by Lieb and Loss). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $a$, $b\in\mathbb{C}$ and $-1\leq t\leq 1$, $t\not=0.$ Then &#xA;$$|a|^p-|a-b|^p\leq\frac{1}{t}(|a+tb|^p-|a|^p) \leq |a+b|^p-|a|^p.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I managed to prove the second inequality for positive $t$ by writing $a+tb=(1-t)a+t(a+b)$and using the convexity of $\cdot^p$. From this the first inequality follows for negative $t$ by substituting $-b$ for $b$. The same trick would finish the proof, if I could prove either the second inequality for negative $t$ or the first inequality for positive $t$.   &lt;/p&gt;&#xA;" OwnerUserId="1878" LastActivityDate="2010-11-06T23:35:14.570" Title="The complex numbers inequality $(|a+tb|^p-|a|^p)/t \leq |a+b|^p-|a|^p$" Tags="&lt;analysis&gt;&lt;inequality&gt;" AnswerCount="1" />
  <row Id="9207" PostTypeId="1" CreationDate="2010-11-06T21:44:42.513" Score="0" ViewCount="259" Body="&lt;p&gt;Hi, I need to prove or disprove whether non-singular matrices commute. Can you help me and tell me how? What should i use cos i really don't know, I am beginner in this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance.&lt;/p&gt;&#xA;" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-07T00:24:48.680" LastActivityDate="2010-11-07T00:24:48.680" Title="Nonsingular matrices commute - prove or disprove" Tags="&lt;linear-algebra&gt;&lt;homework&gt;&lt;matrices&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="9208" PostTypeId="2" ParentId="9178" CreationDate="2010-11-06T22:07:45.527" Score="0" ViewCount="" Body="&lt;p&gt;Graham et al. and Rosen are both excellent books but they are also very comprehensive. If you are looking for something that is shorter (definitely not comprehensive) and more like an &quot;outline&quot; you might find this small paperback useful:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Essential Discrete Mathematics for Computer Science by Todd Feil and Joan Krone, Prentice-Hall, 2003.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The chapters are somewhat independent and include pseudo-code for some algorithms. &lt;/p&gt;&#xA;" OwnerUserId="1369" LastActivityDate="2010-11-06T22:07:45.527" />
  <row Id="9209" PostTypeId="1" AcceptedAnswerId="9214" CreationDate="2010-11-06T22:09:26.510" Score="1" ViewCount="155" Body="&lt;p&gt;According to the definition of stochastic process in &lt;a href=&quot;http://en.wikipedia.org/wiki/Stochastic_process#Definition&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Given a probability space $(\Omega, \mathcal{F}, P)$, a stochastic process (or random process) with state space $X$ is a collection of $X$-valued random variables indexed by a set $T$ (&quot;time&quot;). That is, a stochastic process $F$ is a collection&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$\{ F_t : t \in T \}$&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;where each $F_t$ is an $X$-valued random variable.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Does that mean for all values of the index $t$, the state space $X$ and the $\sigma$-algebra on it must be same? For a stochastic process, Can it be allowed that the state space and the $\sigma$-algebra on it is different for different value of the index?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it correct that, as a stochastic process, &quot;a random walk on a graph&quot; seems to have different state spaces at different stages/times? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks and regards!&lt;/p&gt;&#xA;" OwnerUserId="1281" LastEditorUserId="1281" LastEditorDisplayName="" LastEditDate="2010-11-06T22:48:13.013" LastActivityDate="2011-01-15T22:00:27.077" Title="state space of stochastic process" Tags="&lt;probability-theory&gt;&lt;stochastic-processes&gt;" AnswerCount="2" />
  <row Id="9210" PostTypeId="1" CreationDate="2010-11-06T22:16:02.890" Score="7" ViewCount="197" Body="&lt;p&gt;Is there a general formula to determine the probability of unbounded, cumulative dice rolls hitting a specified number? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For Example, with a D6 and 14:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;5 + 2 + 3 + 4 = 14 : success&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1 + 1 + 1 + 6 + 5 + 4 = 17 : failure &lt;/p&gt;&#xA;" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-11-06T23:13:35.827" LastActivityDate="2010-11-07T01:09:01.290" Title="Probability of cumulative dice rolls hitting a number" Tags="&lt;number-theory&gt;&lt;probability&gt;" AnswerCount="2" CommentCount="9" FavoriteCount="1" />
  <row Id="9211" PostTypeId="2" ParentId="9189" CreationDate="2010-11-06T22:22:01.867" Score="1" ViewCount="" Body="&lt;p&gt;Have to do it with 2 coverings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Covering  1: For each $m\in M$ chose open $V_m \subset M$ and $\epsilon_m &gt;0$ with $\epsilon_m &amp;lt; inf \delta(V_m)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Covering 2: Take locally finite covering $(U_i)_{i\in I}$ finer than preceding covering. For each  $i\in I$ chose $m\in M$ with  $U_i\subset V_m$ and put $\delta_i =\epsilon_m$  . Remark that on $U_i$ we have $\delta_i &amp;lt; inf  \delta (U_i).$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then associate partition of unity $(\phi_i)_{i\in I}$ with covering $(U_i)_{i\in I}$ and wanted function is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\tilde \delta=\sum \delta_i \phi_i $$&lt;/p&gt;&#xA;" OwnerUserId="2872" LastActivityDate="2010-11-06T22:22:01.867" CommentCount="3" />
  <row Id="9212" PostTypeId="2" ParentId="9199" CreationDate="2010-11-06T22:36:32.030" Score="3" ViewCount="" Body="&lt;p&gt;To better understand closed forms, you may want to familiarize yourself with what's called Differential Algebra.  Just as number theory relies on abstract structures such as rings, fields, ideals, etc to express roots algebraic equations using elementary numbers, similarly there is a parallel apparatus for expressing functions (i.e. solutions of differential equations) using differential rings, fields, ideals  called &lt;a href=&quot;http://en.wikipedia.org/wiki/Differential_algebra&quot; rel=&quot;nofollow&quot;&gt;Differential Algebra&lt;/a&gt;.  It is this underlying mechanism that defines which functions can be expressed as &quot;closed forms&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Parallels:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Similar to splitting fields for&#xA;algebraic equations, there is a&#xA;parallel Galois theory with&#xA;Picard-Vessiot extensions and what&#xA;not.&lt;/li&gt;&#xA;&lt;li&gt;Similar to correspondence between&#xA;subfields of number fields and&#xA;Galois subgroups, on the&#xA;differential side, there is a correspondence between differential subfields and subgroups of algebraic groups.&lt;/li&gt;&#xA;&lt;li&gt;Just as algebraic equations can be&#xA;determined to be solvable by&#xA;radicals, similarly linear&#xA;differential equations can be&#xA;determined to be solvable by&#xA;exponentials, Liouvillian functions,&#xA;etc.  There is an ascending tower of differential fields which can be built &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;There is more...I am no expert in this differential algebra field but if you want some freely available references, see &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Seiler &lt;a href=&quot;http://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.47.6596&quot; rel=&quot;nofollow&quot;&gt;Computer Algebra and&#xA;differential equations&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;Van der Put &lt;a href=&quot;http://iwi.eldoc.ub.rug.nl/FILES/root/1999/JSymbCompvdPut/1999JSymbCompvdPut.pdf&quot; rel=&quot;nofollow&quot;&gt;Galois theory of&#xA;differential equations, algebraic&#xA;groups and Lie algebras&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;Papers by &lt;a href=&quot;http://www4.ncsu.edu/~singer/ms_papers.html&quot; rel=&quot;nofollow&quot;&gt;Michael F Singer&lt;/a&gt; are good. See for example Galois theory of linear differential equations.&lt;/li&gt;&#xA;&lt;li&gt;Check the &lt;a href=&quot;http://www.sci.ccny.cuny.edu/~ksda/posted.html&quot; rel=&quot;nofollow&quot;&gt;Kolchin seminar in&#xA;Differential Algebra&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="1558" LastActivityDate="2010-11-06T22:36:32.030" />
  <row Id="9213" PostTypeId="2" ParentId="9210" CreationDate="2010-11-06T22:42:39.250" Score="8" ViewCount="" Body="&lt;p&gt;Assuming the order matters (i,e 1+2 is a different outcome from 2+1)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The probability of getting the sum $n$ with dice numbered $1,2,\dots,6$ is the coefficient of $x^n$ in&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{j=0}^{\infty}(\frac{x+x^2+x^3+x^4+x^5+x^6}{6})^j = \frac{6}{6-x-x^2-x^3-x^4-x^5-x^6}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Writing it as partial fractions (using roots of $6-x-x^2-x^3-x^4-x^5-x^6=0$)  or using Cauchy's integral formula to find the coefficient of $x^n$, Taylor series, etc should work.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-11-06T22:50:50.050" LastActivityDate="2010-11-06T22:50:50.050" CommentCount="5" />
  <row Id="9214" PostTypeId="2" ParentId="9209" CreationDate="2010-11-06T22:44:40.297" Score="2" ViewCount="" Body="&lt;p&gt;Yes, the measurable space $(X, \mathcal{M})$ ($M \subset 2^X$ being a $\sigma$-algebra) should be the same for all $t$ in the index set.  If you want the process to occupy different sets $X_t$ at different times, then the right things to do is to take $X$ to be the disjoint union of all the $X_t$.  There is no requirement that $F_t$ be surjective, so there is no harm if some of the state space is not used at some times.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a random walk on a graph, the state space $X$ is just the vertex set of the graph, typically with the discrete $\sigma$-algebra.  Then for each $t \in T$, $F_t$ is some random element of $X$, i.e. a random vertex of the graph.  Formally, if your underlying probability space is $(\Omega, \mathcal{F}, P)$, then $F$ is a map from $\Omega \times T$ into $X$, with the property that for each $t \in T$, the map $\Omega \ni \omega \mapsto F(\omega, t) \in X$, denoted $F_t$, is measurable.&lt;/p&gt;&#xA;" OwnerUserId="822" LastActivityDate="2010-11-06T22:44:40.297" />
  <row Id="9215" PostTypeId="2" ParentId="9206" CreationDate="2010-11-06T23:35:14.570" Score="1" ViewCount="" Body="&lt;p&gt;The same basic idea works for negative values of $t$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you want to prove is equivalent to $|a + tb|^p - |a|^p \geq t|a + b|^p - t|a|^p$. This in turn is the same as $(1 - t)|a|^p \leq |a + tb|^p + (-t)|a + b|^p$, which in turn is equivalent to $|a|^p \leq {1 \over 1 - t} |a + tb|^p + {-t \over 1 - t} |a + b|^p$. Note that $a = {1 \over 1 - t} (a + tb) + {-t \over 1 - t} (a + b)$, so you can use convexity as you used before.&lt;/p&gt;&#xA;" OwnerUserId="3035" LastActivityDate="2010-11-06T23:35:14.570" />
  <row Id="9216" PostTypeId="2" ParentId="9207" CreationDate="2010-11-06T23:40:57.100" Score="-1" ViewCount="" Body="&lt;p&gt;This is one of those problems where one example that contradicts the claim is enough to disprove it (also known as giving a counterexample).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The phrase non-singular matrices indicates we are talking about square matrices, and presumably you want them to be of equal size (since otherwise their product is not defined in either order).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So as the commentators have suggested, we should try a couple of &quot;random&quot; two-by-two matrices, as these will &quot;usually&quot; not commute.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are a pair of non-commuting non-singular matrices:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;1  0&#xA;0 -1&#xA;&#xA;1  1&#xA;0  1&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Multiplying by the first of these matrices on the left scales the second row by a factor of -1 (and otherwise leaves the entries of the first row unchanged), while multiplying by it on the right scales the second column by a factor of -1 (and otherwise leaves the entries of the first column unchanged).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Try and you'll quickly see that the pair of matrices above do not commute (but both are non-singular, e.g. because their determinants are nonzero).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;regards, hm&lt;/p&gt;&#xA;" OwnerUserId="3111" LastActivityDate="2010-11-06T23:40:57.100" CommentCount="2" />
  <row Id="9217" PostTypeId="2" ParentId="8186" CreationDate="2010-11-07T00:04:55.157" Score="0" ViewCount="" Body="&lt;p&gt;This follows easily by Bézout's lemma (as Qiaochu notes, x and y only need to be coprime to generate the unit ideal (indeed, this is the proper generalization of the term &quot;coprime&quot;)).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can ever so-slightly strengthen this to show that $\mathbf{Z}$ is a PID (this is the true content of Bézout's lemma).&lt;/p&gt;&#xA;" OwnerUserId="126" LastEditorUserId="126" LastEditorDisplayName="" LastEditDate="2010-11-07T00:54:29.040" LastActivityDate="2010-11-07T00:54:29.040" />
  <row Id="9219" PostTypeId="2" ParentId="9204" CreationDate="2010-11-07T00:17:17.140" Score="0" ViewCount="" Body="&lt;pre&gt;&#xA;from fractions import Fraction&#xA;&#xA;def Jaska(L):&#xA;    if len(L) == 1:&#xA;        yield L[0], Fraction(L[0])&#xA;    else:&#xA;        for l in range(len(L) - 1):&#xA;            for e1, v1 in Jaska(L[:l+1]):&#xA;                for e2, v2 in Jaska(L[l+1:]):&#xA;                    yield '(%s + %s)' % (e1, e2), v1 + v2&#xA;                    yield '(%s - %s)' % (e1, e2), v1 - v2&#xA;                    yield '(%s * %s)' % (e1, e2), v1 * v2&#xA;                    if v2 != 0:&#xA;                        yield '(%s / %s)' % (e1, e2), v1 / v2&#xA;&#xA;def JaskaAll(L):&#xA;    solutions = dict([])&#xA;    for e, v in Jaska(L):&#xA;        if v.denominator == 1:&#xA;            v = v.numerator&#xA;        if v not in solutions:&#xA;            solutions[v] = []&#xA;        solutions[v].append(e[1:-1])&#xA;    return solutions&#xA;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;You can check &lt;pre&gt;14 in JaskaAll([1,9,6,3])&lt;/pre&gt; to make sure.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-07T00:17:17.140" CommentCount="1" />
  <row Id="9221" PostTypeId="2" ParentId="9210" CreationDate="2010-11-07T01:09:01.290" Score="3" ViewCount="" Body="&lt;p&gt;The short answer is not really, but it doesn't matter.  If you're interested in computing the exact probability there is a way to do it reasonably quickly by writing down a certain matrix and using binary exponentiation, but I'll let you in on a secret: for large $n$, the answer is exponentially close to $\frac{1}{6}$ (that is, it's exactly what you would naively expect).  A very similar result is proven in the MO thread Michael linked to.  So if that's all you wanted to know, then you know it.  If you really wanted to estimate how close it was to $\frac{1}{6}$ then you'd need to estimate the next largest root of the characteristic polynomial $x^6 = \frac{x^5 + ... + 1}{6}$.  &lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-11-07T01:09:01.290" />
  <row Id="9222" PostTypeId="1" AcceptedAnswerId="9234" CreationDate="2010-11-07T01:14:00.053" Score="3" ViewCount="243" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm having trouble with the following exercise in do Carmo's Riemannian geometry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $X$ and $Y$ be differentiable vector fields on a Riemannian manifold $M$. Let $p \in M$ and let $c: I \to M$ be an integral curve of $X$ through $p$, i.e. $c(t_0) = p$ and $\frac{dc}{dt} = X(c(t))$. &#xA;Prove that the Riemannian connection of $M$ is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(\nabla_XY \ )(p) = \frac{d}{dt} (P^{-1}_{c,t_0,t}(Y(c(t)))\ |_{t=t_0}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $P^{-1}_{c,t_0,t}: T_{c(t_0)}M \to T_{c(t)}M$ is the parallel transport along $c$, from $t_0$ to $t$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I guess, I don't have enough understanding of how to handle the parallel transport (since it is only given as the unique solution to a differential equation).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any hints would be greatly appreciated!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you,&#xA;S. L.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: Do Carmo first introduced the notion of an affine connection &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\nabla: \text{Vect}(M) \times \text{Vect}(M) \to \text{Vect}(M)$,   $(X,Y) \mapsto \nabla_XY$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With the following properties:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\nabla_{fX + gY}Z = f\nabla_XZ + g \nabla_YZ$&lt;/li&gt;&#xA;&lt;li&gt;$\nabla_X(Y+Z) = \nabla_XY + \nabla_XZ$&lt;/li&gt;&#xA;&lt;li&gt;$\nabla_X(fY) = f\nabla_XY + X(f)Y$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;for $X,Y,Z \in \text{Vect}(M)$ and $f,g \in C^\infty(M)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And then showed that there is a unique correspondence which associates to a vector field $V$ along the differentiable curve $c: I \to M$ another vector field $\frac{DV}{dt}$ along c, called covariant derivative of $V$ along $c$, with three more properties:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\frac{D}{dt}(V+W) =  \frac{D}{dt}V +  \frac{D}{dt}W$&lt;/li&gt;&#xA;&lt;li&gt;$ \frac{D}{dt}(fV) =  \frac{df}{dt}V +  f\frac{D}{dt}V$&lt;/li&gt;&#xA;&lt;li&gt;If $V$ is induced by a vector field $Y \in \text{Vect}(M), then  \frac{D}{dt}V = \nabla_{dc/dt}Y$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Then he showed existence and uniqueness of the parallel transport along a curve, and went on to prove existence and uniqueness of Levi-Civita connection.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope this makes things clearer? Thanks for the quick answer!&lt;/p&gt;&#xA;" OwnerUserId="3208" LastEditorUserId="3208" LastEditorDisplayName="" LastEditDate="2010-11-07T01:45:05.440" LastActivityDate="2010-11-07T19:14:11.043" Title="Expression for Levi-Civita Connection" Tags="&lt;differential-geometry&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="2" />
  <row Id="9223" PostTypeId="1" AcceptedAnswerId="9224" CreationDate="2010-11-07T01:48:02.990" Score="0" ViewCount="127" Body="&lt;p&gt;The table of integrals says that&#xA;$\int \frac{dx}{a^{2}+x^{2}}=\frac{1}{a}\arctan\frac{x}{a}+C$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What's wrong with my proof?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{align*}&#xA;y &amp;amp;= \arctan\frac{x}{a} \\&#xA;a\tan y&amp;amp;=x \\&#xA;\tan y &amp;amp;= \frac{x}{a} \\&#xA;\frac{d}{dx} \tan y &amp;amp;= \frac{d}{dx} \frac{x}{a} \\&#xA;\sec^{2}y \frac{dy}{dx} &amp;amp;= \frac{1}{a} \\&#xA;\frac{dy}{dx} &amp;amp;= \frac{1}{a} \frac{1}{\sec^{2}y} \\&#xA;&amp;amp;= \frac{1}{a} \frac{1}{1 + \tan^{2} y} \\&#xA;\frac{dy}{dx}&amp;amp;= \frac{1}{a} \frac{1}{1 + (\frac{x}{a})^{2}} \\&#xA;\int \frac{dy}{dx}\,dx &amp;amp;= \int \frac{1}{a} \frac{1}{1 + (\frac{x}{a})^{2}}\,dx\\&#xA;\arctan\frac{x}{a} &amp;amp;= \frac{1}{a} \int \frac{1}{1 + (\frac{x}{a})^{2}}\,dx \\&#xA;\end{align*}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it right so far?&#xA;Any help appreciated!&lt;/p&gt;&#xA;" OwnerUserId="1958" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-07T01:54:21.160" LastActivityDate="2010-11-07T03:42:45.530" Title="Proving an Integral" Tags="&lt;trigonometry&gt;&lt;integral&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="9224" PostTypeId="2" ParentId="9223" CreationDate="2010-11-07T01:58:41.343" Score="2" ViewCount="" Body="&lt;p&gt;Other than having unnecessary steps (why did you move $a$ to the left hand side in the second step, only to bring it back to the right hand side in the next step?), and missing the constant of integration on the left hand side in the last equation (as well as the missing &quot;$dx$&quot; which I fixed), nothing wrong so far. Now perform the operation inside the integral to get $\frac{a^2}{a^2+x^2}$, pull out the constant, simplify, and move it the left hand side to get the formula.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, if what you want is to show that the integral of $\frac{1}{a^2+x^2}$ is really equal to $\frac{1}{a}\arctan(\frac{x}{a}) + C$, then why not simply differentiate the latter and see if you get the integrand?&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-11-07T01:58:41.343" CommentCount="3" />
  <row Id="9225" PostTypeId="1" CreationDate="2010-11-07T02:40:09.467" Score="1" ViewCount="197" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;On a small island there are 25&#xA;  inhabitants. One of these inhabitants,&#xA;  named Jack, starts a rumor which&#xA;  spreads around the isle. Any person&#xA;  who hears the rumor continues&#xA;  spreading it until he or she meets&#xA;  someone who has heard the story&#xA;  before. At that point, the person&#xA;  stops spreading it, since nobody likes&#xA;  to spread stale news.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;a) Do you think all 25 inhabitants will eventually hear the rumor or will the rumor die out before that happens? Estimate the proportion of inhabitants who will hear the rumor.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;How would one go about calculating the probability? I would try to block it out, but there are so many potential branches. This problem is in the simulations portion of our statistics book, but I'm curious if there is any other way that doesn't involve simulating the situation.&lt;/p&gt;&#xA;" OwnerUserId="104" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-07T02:50:09.783" LastActivityDate="2010-11-07T04:28:57.070" Title="Probability of the propagation of a rumor problem." Tags="&lt;probability&gt;&lt;homework&gt;&lt;statistics&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="1" />
  <row Id="9226" PostTypeId="2" ParentId="9191" CreationDate="2010-11-07T02:41:46.037" Score="3" ViewCount="" Body="&lt;p&gt;If $f(x)$ has a jump discontinuity at $x_0$, then $g_k(x) = f(x)(x-x_0)^{k+1}$ has a continuous $k$th derivative at $x_0$ but is not $(k+1)$ times differentiable at $x_0$.  &lt;/p&gt;&#xA;" OwnerUserId="299" LastActivityDate="2010-11-07T02:41:46.037" CommentCount="3" />
  <row Id="9227" PostTypeId="2" ParentId="9222" CreationDate="2010-11-07T02:44:07.607" Score="3" ViewCount="" Body="&lt;p&gt;This is a cultural comment, rather than an answer; but it is a bit long for the comment box, which is why I am writing it here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My comment is the following: giving a connection, and giving the associated parallel transport, are essentially the same thing, and so it should be possible to establish the formula you are trying to prove by reasonably conceptual, high-level, thinking, rather than mucking about too much with complicated differential equations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me elaborate a little: how would you define the derivative of a vector field along a curve?  Well, the idea is that you want to form the usual Newton quotient&#xA;$$ \dfrac{V\bigl(c(t+\epsilon)\bigr) - V\bigl(c(t)\bigr)}{\epsilon}$$&#xA;and then let $\epsilon$ go to zero, to compute that derivative of $V$ at the&#xA;point $c(t)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only problem is that the subtraction in this formula doesn't make sense,&#xA;because the tangent vectors being subtracted are based at different points:&#xA;one is based at $c(t+\epsilon)$ and the other at $c(t)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Enter parallel transport: suppose we have an agreed upon way to shift, or &lt;em&gt;transport&lt;/em&gt;, tangent&#xA;vectors along a curve.  Then we can apply this so as to transport &#xA;$V\bigl(c(t+\epsilon)\bigr)$ from $c(t+\epsilon)$ to $c(t)$, and then form the&#xA;above Newton quotient, and proceed to compute a derivative.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An agreed upon method for shifting vectors along curves is formally referred to as a choice of &lt;em&gt;parallel transport&lt;/em&gt;, because if you have such a method, and if you define a vector field along a curve $c(t)$ by choosing some $V_0$ at&#xA;$c(t_0)$ and then using the agreed upon parallel transport to define&#xA;$V\bigl(c(t)))$ by tranporting $V_0$ along $c$ to $c(t)$, you obtain a vector&#xA;field along $c(t)$ whose derivative at each point will be zero (by construction!).  Thus this vector field does not change along $c(t)$,&#xA;and so consists of parallel vectors, hence the name &lt;em&gt;parallel tranport&lt;/em&gt;.&#xA;(Of course, the notions of &lt;em&gt;change&lt;/em&gt; and &lt;em&gt;parallel&lt;/em&gt; are not intrinsict to&#xA;the vector field; they depend upon our particular choice of parallel transport.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, our choice of parallel transport should satisfy some axioms (it should be linear; it should be smooth (so that the vector fields along curves&#xA;that it gives rise to are smooth); and so on).  If you look in the right text,&#xA;you will find these axioms written down.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As we've seen, a choice of parallel transport on our manifold gives a way&#xA;of differentiating vector fields along curves.  But since a tangent vector&#xA;is just an infinitesimal curve, one sees that we in fact have a way of &#xA;differentiating a vector field in the direction of a given vector field: extend that vector field to a curve, and then differentiate along that curve.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus a choice of parallel transport determines an affine connection, i.e. a way&#xA;of differentiating vector fields along tangent vectors.  Conversely, this is enough information to determine the parallel transport: we parallel transport a tangent vector along a curve by making sure that its derivative (using the given affine connection) at every point of the curve, in the tangent direction of the curve, vanishes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we have gone full circle, from parallel transport, to affine connection, back to parallel transport.  The text you are reading is spelling out the second half of this circle, but is perhaps a little scanty on details regarding the first half; in fact, the exercise you are trying to solve is exactly about the first half of the circle, and its goal is to verify that you truly are going around in a circle: i.e. that you are ending up where you started from.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There really is something to verify here (i.e. the exercise is not trivial if you have not done it before, and are learning these ideas for the first time), but I hope that the above discussion may help shed some light on its meaning,&#xA;and also make it seem less daunting (and perhaps more conceptual and less computational) than it might otherwise appear.&lt;/p&gt;&#xA;" OwnerUserId="221" LastEditorUserId="221" LastEditorDisplayName="" LastEditDate="2010-11-07T19:00:21.193" LastActivityDate="2010-11-07T19:00:21.193" CommentCount="2" />
  <row Id="9228" PostTypeId="2" ParentId="9225" CreationDate="2010-11-07T03:00:20.303" Score="0" ViewCount="" Body="&lt;p&gt;Take a look at the &lt;a href=&quot;http://en.wikipedia.org/wiki/Logistic_function&quot; rel=&quot;nofollow&quot;&gt;logistic function&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="664" LastActivityDate="2010-11-07T03:00:20.303" />
  <row Id="9229" PostTypeId="2" ParentId="704" CreationDate="2010-11-07T03:13:54.083" Score="1" ViewCount="" Body="&lt;p&gt;I encountered this situation when dealing with permutations; we phrased it as follows.  If $\sigma$ is a permutation of $\mathbb{Z}_n$ and $d$ is a divisor of $n$, then it is impossible for precisely two of the following statements to hold for some $i,j \in \mathbb{Z}_n$:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$i \equiv j \pmod d$,&lt;/li&gt;&#xA;&lt;li&gt;$\sigma(i) \equiv \sigma(j) \pmod d$ and&lt;/li&gt;&#xA;&lt;li&gt;$\sigma(i)-i \equiv \sigma(j)-j \pmod d$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I suspect a modification of this would suffice in most cases.&lt;/p&gt;&#xA;" OwnerUserId="139" LastActivityDate="2010-11-07T03:13:54.083" />
  <row Id="9230" PostTypeId="1" CreationDate="2010-11-07T03:25:39.717" Score="2" ViewCount="231" Body="&lt;p&gt;I see people like Terry Tao and others take extensive notes. But is this really necessary? When I do this I feel like I am rewriting a textbook.&lt;/p&gt;&#xA;" LastActivityDate="2010-11-07T04:04:52.273" Title="What are examples of mathematicians who dont take many notes?" Tags="&lt;soft-question&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" />
  <row Id="9231" PostTypeId="1" AcceptedAnswerId="9237" CreationDate="2010-11-07T03:31:38.067" Score="0" ViewCount="90" Body="&lt;p&gt;I've read a few articles explaining the way to use the Minkowski reduced basis in a lattice in order to measure the uniformity of the output of a random number generator. However, I can't prove a conclusion drawn from the following statement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given a set of linearly independent vectors $\{v_{1}, v_{2}, ... ,v_{d}\}$ in $\mathbb{R^{d}}$, a lattice is the set of vectors $w$ of the form $\sum_{i=1}^{d}z_{i}v_{i}$, where $z_{i}$ are integers. The set of vectors $\{v_{i}\}$ is a basis for the lattice. The basis is a Minkowski reduced basis if&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$||v_{k}||\leq||\sum_{i=1}^{d}z_{i}v_{i}||, \qquad \text{for} \quad 1\leq k \leq d,$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for all sets of $d$ integers $z_{i}$ such that the greatest common divisor of the set $\{z_{k},z_{k+1},...,z_{d}\}$ is $1$. Accordingly, this implies $|v_{1}|\leq |v_{2}| \leq ... \leq |v_{d}|$. In the condition for a Minkowski reduced basis, $||v||$ denotes the Euclidean norm of the vector $v$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why can we infer that $|v_{1}|\leq |v_{2}| \leq ... \leq |v_{d}|$)?. That's what I'd like to know. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as I know, $||v_{1}||$ (for example), differs only from $||v_{2}||$ in having the restriction $\text{gcd}(\{z_{1},z_{2}\})=1$, while the second vector has $\text{gcd}(\{z_{2}\})=1$ in $\mathbb{R^{2}}$, however, I don't know how can I reach the conclusion from that fact. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ideas?&lt;/p&gt;&#xA;" OwnerUserId="2569" LastEditorUserId="2569" LastEditorDisplayName="" LastEditDate="2010-11-07T04:48:59.150" LastActivityDate="2010-12-07T05:44:25.140" Title="Proof for Minkowski reduced basis" Tags="&lt;random&gt;&lt;lattices&gt;" AnswerCount="1" />
  <row Id="9232" PostTypeId="2" ParentId="9230" CreationDate="2010-11-07T03:36:11.457" Score="5" ViewCount="" Body="&lt;p&gt;Everyone learns a bit differently.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some people learn better by listening; others by writing. I find that the best way for me to understand someone else's proof is to write it out in careful detail, rather than simply reading it.  For me, taking notes is a way to help me keep up with the material while digesting and internalising it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other people don't learn that way. George Bergman (my advisor), for example, throughly dislikes lecturing (in the traditional way) because he says that it is a method for transferring information from the lecturer's notes to the student's notebook without passing through the brain of either (an opinion he has held since he took his first graduate course; so clearly he derived little benefit from taking notes the way you describe). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If all you are doing is writing and this process does not help you think about what you are writing, then it is probably not in your best interest to write notes. Better to keep the pad for any sudden ideas or insights you might want to come back later, and trust the textbook to provide the information you are interested in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So... whether it is &quot;necessary&quot; is not an absolute, but rather a function of your own personal preference. In short: &quot;whatever floats &lt;em&gt;your&lt;/em&gt; boat&quot;.  &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-07T04:04:07.910" LastActivityDate="2010-11-07T04:04:07.910" CommentCount="4" />
  <row Id="9233" PostTypeId="2" ParentId="9223" CreationDate="2010-11-07T03:42:45.530" Score="1" ViewCount="" Body="&lt;p&gt;A better way to prove it would be to let $\displaystyle x = a\tan{\theta}$, then $\displaystyle \frac{dx}{d\theta} = a\sec^2{\theta} \Rightarrow dx = a\sec^2{\theta}\;{d\theta}$. So:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle \int\frac{1}{x^2+a^2}\;{dx} = \int\frac{a\sec^2{\theta}}{a^2\tan^2{\theta}+a^2}\;{d\theta} = \frac{1}{a}\int\frac{a\sec^2{\theta}}{a(\tan^2{\theta}+1)}\;{d\theta}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle  = \frac{1}{a}\int\frac{a\sec^2{\theta}}{a\sec^2{\theta}}\;{d\theta} = \frac{1}{a}\int\;{d\theta} = \frac{1}{a}\theta+k = \frac{1}{a}\arctan\frac{x}{a}+k$.&lt;/p&gt;&#xA;" OwnerUserId="3207" LastActivityDate="2010-11-07T03:42:45.530" CommentCount="4" />
  <row Id="9234" PostTypeId="2" ParentId="9222" CreationDate="2010-11-07T03:46:44.623" Score="3" ViewCount="" Body="&lt;p&gt;Let $\{e_i\}\subseteq T_{c(t_0)}M$ be a basis.  Define $E_i(t)$ as the parallel translation of $e_i$ along $c(t)$.  Prove to yourself that $\{E_i(t)\}$ forms a basis of $T_{c(t)}$ for all $t$ (hint, use the uniqueness part of solving linear ODEs).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, we can write $Y(c(t)) = \sum_i a_i(t) E_i(t)$.  Then, since the $E_i(t)$ are parallel, and since $P_{c,t_0,t}$ is linear, we have $P^{-1}_{c,t_0, t}(Y(c(t)) = \sum_i a_i(t) e_i$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From here, by actually computing the limit, it's not too hard to see that $\frac{d}{dt} P^{-1}_{c,t_o,t}Y(c(t))|_{t=t_0} = \sum_i a_i'(t_0)e_i$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, the goal is to show that $\nabla_X Y(p)$ can also be written like this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But $\nabla_X Y(p) = \nabla_X \sum_i a_i(t)E_i(t)|_{t=t_0} = \sum_i a_i(t_0) \nabla_X E_i(t)|_{t=t_0} + \sum_i a_i'(t_0) E_i(t_0)$.  (The second equality is just the Leibniz rule every connection must satisfy).  However, $\nabla_X E_i(t)$ is $\frac{D}{dt} E_i(t)$ and this is 0 since the $E_i$ are parallel.  Hence, the first term of the sum vanishes, so we get the desired result.&lt;/p&gt;&#xA;" OwnerUserId="331" LastEditorUserId="331" LastEditorDisplayName="" LastEditDate="2010-11-07T19:14:11.043" LastActivityDate="2010-11-07T19:14:11.043" CommentCount="1" />
  <row Id="9235" PostTypeId="2" ParentId="9230" CreationDate="2010-11-07T03:56:08.727" Score="2" ViewCount="" Body="&lt;p&gt;When I go to a lecture I try to discern the general idea and condense the talk into a small summary -- sometimes half a page, sometimes just a couple of lines.  Sometimes I get technical details out of a talk, sometimes it's more of a big picture.  Sometimes I miss the point. Sometimes I latch onto an idea I don't understand and spend the entire talk trying to work that out, or I try to prove the speaker wrong.  But as Arturo mentions, different people have different styles. &lt;/p&gt;&#xA;" OwnerUserId="642" LastEditorUserId="642" LastEditorDisplayName="" LastEditDate="2010-11-07T04:04:52.273" LastActivityDate="2010-11-07T04:04:52.273" />
  <row Id="9236" PostTypeId="2" ParentId="9225" CreationDate="2010-11-07T04:00:14.737" Score="3" ViewCount="" Body="&lt;p&gt;An &lt;a href=&quot;http://en.wikipedia.org/wiki/Epidemic_model#The_SIR_Model&quot; rel=&quot;nofollow&quot;&gt;SIR model&lt;/a&gt; is also appropriate because of this aspect of the problem: &quot;Any person who hears the rumor continues spreading it until he or she meets someone who has heard the story before.  At that point, the person stops spreading it, since nobody likes to spread stale news.&quot;  Here, susceptible = hasn't heard rumor, infected = has heard rumor but is still willing to spread it, and recovered = has heard rumor but is no longer willing to spread it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I see you're in 9th grade, and so the differential equations in the SIR model may not be something you've seen yet.  Still, you may get something out of reading the explanation of the model on the Wikipedia page.  Also, the standard SIR model doesn't quite fit this scenario; the equations for the infected and recovered populations for your scenario should be &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{dI}{dt} = \beta SI - \gamma RI,$$&#xA;$$\frac{dR}{dt} = \gamma RI.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know you said you only wanted hints, but I'm giving you a fuller explanation of the SIR model because, according to &lt;a href=&quot;http://www.math.vt.edu/people/gboquet/papers/discrete.pdf&quot; rel=&quot;nofollow&quot;&gt;this paper&lt;/a&gt;, &quot;While the formal model is simple, in general it cannot be solved analytically.&quot;  So this may just be one of those problems for which simulation is the best solution we have now.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;O.K., I have to add this: For fun, you should also check out the &lt;a href=&quot;http://cse.ucdavis.edu/~chaos/courses/nlp/Projects2009/DanielSpielman/Report.pdf&quot; rel=&quot;nofollow&quot;&gt;SIRZ&lt;/a&gt; (susceptible-infected-recovered-zombie) model.  We can all be thankful that the authors say, &quot;Most of the analysis concluded that, under reasonable circumstances, a zombie apocalypse is unlikely.&quot;&lt;/p&gt;&#xA;" OwnerUserId="2370" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-11-07T04:28:57.070" LastActivityDate="2010-11-07T04:28:57.070" />
  <row Id="9237" PostTypeId="2" ParentId="9231" CreationDate="2010-11-07T04:02:56.460" Score="1" ViewCount="" Body="&lt;p&gt;Just consider the sets $\{z_1,\ldots,z_d\} = \{0,\ldots,0,1,0,\ldots 0 \}$ in the definition of Minkowski reduced bases, where for $v_k$ you put a 1 at the $k+1$-st place.&lt;/p&gt;&#xA;" OwnerUserId="3212" LastActivityDate="2010-11-07T04:02:56.460" CommentCount="8" />
  <row Id="9238" PostTypeId="2" ParentId="9152" CreationDate="2010-11-07T04:05:30.523" Score="1" ViewCount="" Body="&lt;p&gt;I recommend &lt;a href=&quot;http://mrob.com/pub/ries/index.html&quot; rel=&quot;nofollow&quot;&gt;ries&lt;/a&gt; which finds solutions to this problem, related problems with different digits and operators, and in general finds operations that end up with a specified number.&lt;/p&gt;&#xA;" OwnerUserId="1778" LastActivityDate="2010-11-07T04:05:30.523" />
  <row Id="9239" PostTypeId="1" AcceptedAnswerId="9246" CreationDate="2010-11-07T04:22:57.923" Score="4" ViewCount="111" Body="&lt;p&gt;A powder can be compressed by packing it down. Each time it is packed down it loses $\frac{1}{2}$ then $\frac{1}{4}$ then $\frac{1}{8}$ ... etc. of it's total volume.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This powder is placed in to a container of unit volume and packed down. Then the remaining space is filled with fresh powder and packed down again. &lt;strong&gt;The packing action acts on the powder multiple times.&lt;/strong&gt; So, for example, here is the volume of powder after the first few packings:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(1)(\frac12) = \frac12$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(1)(\frac12)(\frac34) + (\frac12)(\frac12) =  \frac58$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(1)(\frac12)(\frac34)(\frac78) + (\frac12)(\frac12)(\frac34) + (\frac38)(\frac12) = \frac{45}{64}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(1)(\frac12)(\frac34)(\frac78)(\frac{15}{16}) + (\frac12)(\frac12)(\frac34)(\frac78) + (\frac38)(\frac12)(\frac34) + (\frac{19}{64})(\frac12)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\vdots$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;I would like to know how much powder is in the container after it has been packed n times, in terms of n.&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="2563" LastEditorUserId="2563" LastEditorDisplayName="" LastEditDate="2010-11-07T04:44:56.917" LastActivityDate="2010-11-07T05:49:36.717" Title="Packing powder series, suspect it is easy, yet..." Tags="&lt;calculus&gt;&lt;sequences-and-series&gt;" AnswerCount="3" CommentCount="3" />
  <row Id="9240" PostTypeId="2" ParentId="9225" CreationDate="2010-11-07T04:23:44.853" Score="4" ViewCount="" Body="&lt;p&gt;Mike Spivey's remark that there are only three classes of people is quite relevant if you want to simulate this.  At the start, there are 24 susceptible people and one infected.  Call the number of each category S, I, and R.  If two I meet, they both become R, so I goes down by 2 and R goes up by 2.  If an S meets an I, the S becomes I, so S goes down by 1 and I goes up by 1.  If R meets I, I goes down by 1 and R goes up by 1.  Other meetings have no result.  My guess is that somebody doesn't hear, because as S falls more and more meetings deplete the I's.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-11-07T04:23:44.853" CommentCount="1" />
  <row Id="9241" PostTypeId="2" ParentId="9239" CreationDate="2010-11-07T04:40:22.983" Score="2" ViewCount="" Body="&lt;p&gt;Not much help, but the limiting packing density is a factor 3.462746619.  The &lt;a href=&quot;http://oldweb.cecm.sfu.ca/projects/ISC/ISCmain.html&quot; rel=&quot;nofollow&quot;&gt;inverse symbolic calculator&lt;/a&gt; finds the compression (limiting volume of the initial unit volume) as 0.2887880950866024 with the correct formula.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-11-07T04:40:22.983" CommentCount="1" />
  <row Id="9242" PostTypeId="1" AcceptedAnswerId="9268" CreationDate="2010-11-07T04:52:37.307" Score="5" ViewCount="205" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After thinking about it for a while and consulting other students, no one seems to be able to find an example of the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given the PDE &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\dfrac{\partial f}{\partial x} = 0 \quad $      on $U = { (x,y) \in \mathbb R^2 ; y&gt;0, 1 &amp;lt; x^2 + y^2 &amp;lt; 4}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am looking for a solution $f$ that does not only depend on $y$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can this be?!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The exercise is taken form Lee's &quot;Introduction to smooth manifolds&quot;, p. 517 at the end of the chapter on the Frobenius theorem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Note: According to the errata, the condition on $U$ is $y &gt; 0$, not $x &gt; 0$ as your copy of the book might state).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;S. L.&lt;/p&gt;&#xA;" OwnerUserId="3208" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-11-07T13:58:39.843" LastActivityDate="2010-11-07T15:25:39.687" Title="Counterintuitive PDE" Tags="&lt;differential-equations&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="1" />
  <row Id="9243" PostTypeId="1" AcceptedAnswerId="9247" CreationDate="2010-11-07T04:56:42.657" Score="1" ViewCount="74" Body="&lt;p&gt;$f$ is smooth (derivatives of all orders exist) except at $x_1$,$x_2$,$x_3$ where derivatives exist only upto orders of $k_1$,$k_2$,$k_3$ respectively, where $k_i$ is a natural number. Same is the case with another function $g$ which is smooth except at $t_1$,$t_2$,$t_3$ where derivatives exist only upto orders of $l_1$,$l_2$,$l_3$ respectively. Both $f$,$g$ belong to either $L^1$ or $L^2$ or have compact support. What is the behaviour of the function $h=f*g$&lt;/p&gt;&#xA;" OwnerUserId="2987" LastEditorUserId="2987" LastEditorDisplayName="" LastEditDate="2010-11-07T05:39:21.700" LastActivityDate="2010-11-07T06:00:39.117" Title="behaviour of discontinuities after convolution" Tags="&lt;calculus&gt;&lt;convolution&gt;" AnswerCount="1" />
  <row Id="9244" PostTypeId="2" ParentId="9239" CreationDate="2010-11-07T05:03:04.837" Score="2" ViewCount="" Body="&lt;p&gt;The best I can come up with is horribly recursive (not so bad if you store intermediary values, as mine does below), but it would work to use Mathematica or some other software to calculate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Define $f(0) = 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f(n)=\displaystyle\sum_{i=0}^{n-1}\bigg(\displaystyle\prod_{j=1}^{n-i}(1-\frac{1}{2^j})\bigg)(1-f(i))$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mathematica code:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;f[0] = 0;&#xA;f[n_] := f[n] = Sum[Product[(1 - 1/2^j), {j, 1, n - i}] (1 - f[i]), {i, 0, n - 1}]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="1016" LastEditorUserId="1016" LastEditorDisplayName="" LastEditDate="2010-11-07T05:16:11.620" LastActivityDate="2010-11-07T05:16:11.620" />
  <row Id="9245" PostTypeId="1" AcceptedAnswerId="9257" CreationDate="2010-11-07T05:03:15.347" Score="6" ViewCount="183" Body="&lt;p&gt;Imagine that we have something like the rubik's cube on $U=I^{3}=[0,1]^{3}$, and that we say that all the points on the xy, yz, and zx faces are black, and that all the points on the other three faces are white.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Define a legal move to be a rotation by an integral multiple of $\frac{\pi}{4}$ around one of the lines going through the center of each pair of opposite faces, for any square subset of $U$ orthogonal to the respective line. Note that there are an uncountable number of them on each axes. Points retain their color after legal moves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a sequence of legal moves that will produce a circle -- some points one color and some points another color? -- on one of the faces of $U'$? (where $U'$ is meant to denote $U$ after said sequence of moves has been performed?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;addition&lt;/strong&gt;: I strongly suspect that no single countably infinite set of legal moves works, just on the basis of cardinality arguments -- a circle would have a continuum of points, and any countably infinite sequence of legal moves would only be able to move a countably infinite set of points.&lt;/p&gt;&#xA;" OwnerUserId="890" LastEditorUserId="890" LastEditorDisplayName="" LastEditDate="2010-11-07T08:30:57.717" LastActivityDate="2010-11-07T12:40:01.403" Title="continuous rubik's cube circle drawing" Tags="&lt;group-theory&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="2" />
  <row Id="9246" PostTypeId="2" ParentId="9239" CreationDate="2010-11-07T05:49:36.717" Score="2" ViewCount="" Body="&lt;p&gt;The amount of powder remaining at each point is $(1-1/2),(1-1/4),\ldots$, and so the constant you're looking for (in the limit) is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\prod_{n=1}^{\infty} (1 - 2^{-n}) \approx 0.288788095086602$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is also the limiting probability that an $n \times n$ matrix over $GF(2)$ be regular.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Denote the exact result for $n$ by $R_n$. We want to estimate the error $R_n - R_\infty$. Note that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$R_n = \frac{R_\infty}{\prod_{m=n+1}^{\infty} (1 - 2^{-m})}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The logarithm of the denominator is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$-\Theta(\sum_{m=n+1}^\infty 2^{-m}) = -\Theta(2^{-n})$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and so the denominator itself is $1 - \Theta(2^{-n})$. Thus&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$R_n = R_\infty (1 + \Theta(2^{-n})) = R_\infty + \Theta(2^{-n})$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The product can be converted to a sum (exercise!)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$R_\infty = \sum_{n=2}^\infty \frac{(-1)^n}{\prod_{k=2}^n (2^k-1)} = \frac{1}{3} - \frac{1}{3\cdot 7} + \frac{1}{3 \cdot 7 \cdot 15} - \cdots$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using this sum, it's easy to prove that this constant is irrational (hint: same as the usual proof that $e$ is irrational using its series representation $\sum_{n=0}^\infty 1/n!$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can also obtain excellent rational approximations this way:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$1/3, 2/7, 13/45, 188/651, 5731/19845, \ldots$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the series is Leibniz, you know that the error after summing $k$ terms is at most the next term. For example, $5731/19845$ is at most $1/78129765$ too large than the actual constant.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-07T05:49:36.717" CommentCount="2" />
  <row Id="9247" PostTypeId="2" ParentId="9243" CreationDate="2010-11-07T06:00:39.117" Score="1" ViewCount="" Body="&lt;p&gt;Assume $f$ and $g$ has compact support, and put $h=f*g$, then there is no problem for the derivatives to be integrable. If $D^{n_1}f,D^{n_2}g\in L^1$ then we have&#xA;$$D^n h = D^{n_1}f*D^{n_2}g, \qquad \text{where } n_1+n_2=n.$$&#xA;Hence we certainly have that $h$ is differentiable of order $n=n_1+n_2$ where $n_1=\min(k_1,k_2,k_3)$ and $n_2= \min(l_1,l_2,l_3)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Depending on the discontinuity  at $x_i$ the integrability of $D^if$ might be lost ($i=1,2,3$) and the same holds for $g$, so we can not do better than this.&lt;/p&gt;&#xA;" OwnerUserId="1154" LastActivityDate="2010-11-07T06:00:39.117" CommentCount="1" />
  <row Id="9248" PostTypeId="2" ParentId="9242" CreationDate="2010-11-07T07:00:02.550" Score="0" ViewCount="" Body="&lt;p&gt;$f$ can be locally but not globally independent of $x$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $|y| &amp;lt; 1$ the function can have different values on the left and right sides of the annulus.  The restriction of the function to the left or the right side depends only on $y$.  For example, $f = {\rm sgn(x)}(1-y^2)$ for $|y| \leq 1$ and $0$ in the rest of the region.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(restored following correction of problem.)&lt;/p&gt;&#xA;" OwnerUserId="467" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-11-07T15:25:39.687" LastActivityDate="2010-11-07T15:25:39.687" />
  <row Id="9249" PostTypeId="2" ParentId="9152" CreationDate="2010-11-07T07:08:25.120" Score="5" ViewCount="" Body="&lt;p&gt;As Derek points out, we can blow the question wide open by allowing operations not included on basic calculators. In fact, even a single nine could be quite productive if we allow (say) floor, ceiling, factorial, and the natural logarithm:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\lfloor\ln\ln9\rfloor=0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\lceil\ln\ln9\rceil=1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\lfloor\ln 9\rfloor=2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\lceil\ln 9\rceil=3$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\lceil\ln\ln\lceil\ln(9!)\rceil!\rceil=4$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\lfloor\ln\ln\lceil\ln\lceil\ln\lceil\ln(9!)\rceil!\rceil!\rceil!\rfloor&#xA;= 5$$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\lceil\ln 9\rceil!=6$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\lceil\ln\lceil\ln9\rceil!!\rceil= 7$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\vdots$$&lt;/p&gt;&#xA;" OwnerUserId="2474" LastActivityDate="2010-11-07T07:08:25.120" CommentCount="1" />
  <row Id="9250" PostTypeId="2" ParentId="9242" CreationDate="2010-11-07T08:15:10.680" Score="7" ViewCount="" Body="&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Mean_value_theorem&quot; rel=&quot;nofollow&quot;&gt;Mean Value Theorem&lt;/a&gt; implies that any function f differentiable on U must be constant w.r.t. x for any fixed y.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is an error in the text -- it should say y&gt;0 instead of x&gt;0.  (see &lt;a href=&quot;http://www.math.washington.edu/~lee/Books/Smooth/errata.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.math.washington.edu/~lee/Books/Smooth/errata.pdf&lt;/a&gt;)&lt;/p&gt;&#xA;" OwnerUserId="139" LastActivityDate="2010-11-07T08:15:10.680" CommentCount="2" />
  <row Id="9251" PostTypeId="1" CreationDate="2010-11-07T09:24:17.217" Score="0" ViewCount="89" Body="&lt;p&gt;I have Ubuntu 10.10 installed and need to plot 3D graphic. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have installed several free applications available in the repository, like QtiPlot and GNU Octave. I have found out and created the following graphic.&#xA;&lt;img src=&quot;http://i.imgur.com/88XMt.png&quot; alt=&quot;Gaussian function&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I have to show in the same graphic the position of my experiment results, which consist of elements with three parameters: X, Y and Z coordinates which had been calculated with the same function as above graphic. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any idea to do that?  Would be better if you propose solution in free apps,because there are several proprietary apps like Maple or MATLAB.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you very much.&lt;/p&gt;&#xA;" OwnerUserId="2698" LastActivityDate="2010-11-07T09:24:17.217" Title="Gaussian function" Tags="&lt;functions&gt;&lt;applications&gt;" CommentCount="2" />
  <row Id="9252" PostTypeId="1" AcceptedAnswerId="9360" CreationDate="2010-11-07T10:47:44.250" Score="6" ViewCount="180" Body="&lt;p&gt;Happened to get this from a friend.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;How does one show that for all pairs of real numbers $x$ and $y$, there is $n \in \mathbb{N}$ such that $$ x &amp;lt; \frac{\varphi(n+2)}{\varphi(n)} &amp;lt; y$$ where $\varphi$ is the Euler's totient function.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" LastActivityDate="2010-11-08T02:57:11.390" Title="Showing $x &lt; \frac{\varphi(n+2)}{\varphi(n)} &lt; y$ for $x,y \in \mathbb{R}$." Tags="&lt;number-theory&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="3" />
  <row Id="9253" PostTypeId="2" ParentId="154" CreationDate="2010-11-07T11:04:57.770" Score="7" ViewCount="" Body="&lt;p&gt;Even though quite a few people have already contributed towards answering your question, I would like to offer some thoughts of mine that may help you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not going to give you ways of showing that complex numbers are necessary and meaningful, rather an idea about why I think people find them &lt;em&gt;meaningless&lt;/em&gt; and how this can be resolved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am prompted by the part of your question that says: &quot;...most non-mathematicians find it hard to accept that such a number is meaningful. In contrast, they feel that real numbers have an obvious and intuitive meaning.&quot; I have thought about this a lot in the past and have come to the conclusion that the problem of finding complex numbers meaningful lies in the meaning we have attached to the number systems &quot;preceding&quot; them in the hierachical chain: $\mathbb{N}, \mathbb{Z}, \mathbb{Q}, \mathbb{R}$. The meaning we have attached is that of quantity! Numbers all the way up to and including the reals are scalar quantities: We use them for distance, area, volume, weight, speed, intensity and so on. However, when you get to the complex numbers all that has to go away. There is no such thing as $2 + 3i$ kilogrammes or $-4i$ dollars or $1-5i$ centimetres... Yet we ask the learner to call them numbers!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the learner is faced with an apparent contradiction in terms: They are asked to think of these strange entities as &lt;em&gt;numbers&lt;/em&gt; while at the same time these new additions to the realm don't behave as the numbers of old. And you should take into account that the older numbers have been around for a long time. That's where I think the problem lies. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To circumvent this, I think that what one needs to do is explain to the learner that - from now on - numbers are going to take on a much wider role than before: They are going to be used not only to denote quantities, but also to denote &lt;em&gt;directions&lt;/em&gt;, which is precisely what complex numbers do on the two-dimensional plane. Every complex number can be represented by a vector and this automatically suggests a direction. The &quot;older&quot; meaning is not lost, since a complex number has a modulus in addition to its real and imaginary parts and these three are &lt;em&gt;quantities&lt;/em&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, one might say &quot;But real numbers denote directions as well: +1 denotes a unit displacement to the right of 0 along the x-axis while -1 denotes a displacement to the left&quot;. This is indeed true, however, I sincerely doubt that the untrained mind of a learner, who encounters complex numbers for the first time, will delve in that direction. And, if it does, then that's fine, because it shows that the direction concept was already lurking in the number hierarchy ever since negative integers were introduced!!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope all this is of help to you.&lt;/p&gt;&#xA;" OwnerUserId="3195" LastEditorUserId="3195" LastEditorDisplayName="" LastEditDate="2011-03-25T16:41:10.590" LastActivityDate="2011-03-25T16:41:10.590" CommunityOwnedDate="2011-04-15T20:45:29.080" CommentCount="1" />
  <row Id="9254" PostTypeId="2" ParentId="3405" CreationDate="2010-11-07T11:37:32.217" Score="0" ViewCount="" Body="&lt;p&gt;I think these observations are probably already in other answers, or even too simple to have been listed, but I wanted to add my working anyway:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;p(2n+2)=(n+1)/n(p(2n)-2p(n)/(2n+1))&lt;/p&gt;&#xA;&#xA;&lt;p&gt;thus if k(n)= np(n)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;k(2n+2)=(n+1)^2/n^2/(2n+1)((2n+1)k(2n)-4k(n))&lt;/p&gt;&#xA;" OwnerUserId="820" LastActivityDate="2010-11-07T11:37:32.217" />
  <row Id="9255" PostTypeId="1" CreationDate="2010-11-07T12:01:43.480" Score="0" ViewCount="108" Body="&lt;p&gt;How can I render the &lt;em&gt;connected&lt;/em&gt; sum of $n$ topological spaces (or other quite strange topological operations) in a &quot;product-like&quot; manner? What I'm looking for is writing, for example&#xA;$$&#xA;{\Large{*}}_{i=1}^n G_i&#xA;$$&#xA;for the free product of n groups. How can I render this with the # operation in topological spaces? (Yes, I'm studying Hatcher's AT...)&lt;/p&gt;&#xA;" OwnerUserId="685" LastEditorUserId="685" LastEditorDisplayName="" LastEditDate="2010-11-07T12:56:02.210" LastActivityDate="2010-11-08T10:08:38.463" Title="Define strange operators" Tags="&lt;soft-question&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="9256" PostTypeId="1" CreationDate="2010-11-07T12:13:36.553" Score="1" ViewCount="162" Body="&lt;p&gt;Prove that, where a,b,...,e are real numbers and a≠0,&#xA; if ax + by = c&#xA;has the same solution set as&#xA; ax+dy=e&#xA;then they are the same equation. What if a=0?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: If a≠0 then the solution set of the first equation is {(x,y)|x=c-by/a}&lt;/p&gt;&#xA;" OwnerUserId="3215" LastActivityDate="2010-11-08T20:02:35.127" Title="Prove that ax + by = c has same solution set as ax+dy=e" Tags="&lt;homework&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="9257" PostTypeId="2" ParentId="9245" CreationDate="2010-11-07T12:21:16.377" Score="5" ViewCount="" Body="&lt;p&gt;What a great question!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is my solution. I am aiming for a black circle inscribed on a&#xA;white face (where the boundary edge points are invisible,&#xA;as mentioned in the comments).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pick one of the white faces to be the desired final face F.&#xA;Now, consider any given horizontal row on that face, which&#xA;is all white. We want to add two black dots to such a row (except for top/bottom/center, which are handled already by the edge point color condition). These rows come in upper/lower symmetric dual&#xA;pairs. Call such a pair the&#xA;current working pair of rows. We make a quarter turn to&#xA;each of these working rows to the side. Now, operating on&#xA;that side face S, we make two moves parallel to F, which&#xA;will not upset F, in order to make the working rows on S&#xA;have exactly two black dots each in the correct positions.&#xA;This is possible since the squares containing the working&#xA;rows have not been moved previously, and the corresponding&#xA;columns containing the desired black dots have also not&#xA;been moved previously, since those dots will only be moved&#xA;for this pair of working rows. That is, it is precisely the&#xA;black dot opposite the dual white dot that we desire, and&#xA;by working in dual pairs, we put all four black dots into&#xA;position with two moves. (Is it clear?) Thus, on the side&#xA;face, we have the two working rows looking exactly like we&#xA;want. So we turn them each a quarter turn back to the main&#xA;face F, and proceed with another pair of working rows. The&#xA;center row and top/bottom rows do not need adjusting,&#xA;because of the invisible edge points. So this seems to do&#xA;it!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That said, there are problematic issues in setting up your&#xA;problem. This kind of problem, where one wants to describe&#xA;a task involving infinitely many steps, is known as a&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Supertask&quot; rel=&quot;nofollow&quot;&gt;supertask&lt;/a&gt;, and is&#xA;open to a number of interesting paradoxes and problematic&#xA;issue, some of which I describe in &lt;a href=&quot;http://mathoverflow.net/questions/22033/infinite-cpu-clock-rate-and-hotel-hilbert/22038#22038&quot; rel=&quot;nofollow&quot;&gt;this MathOverflow&#xA;answer&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this problem, we can imagine carrying out my solution in&#xA;a transfinite sequence of moves. The fact that each row on the desired face is&#xA;handled with finitely many moves makes my particular&#xA;solution less problematic than other situations one can&#xA;imagine. In general, for example, if a square has been&#xA;moved infinitely many times, what position is it in at the&#xA;limit? You haven't said, and it isn't clear what it should&#xA;be. But solutions in which every given square is moved only&#xA;finitely often seem to avoid this problem, particularly when the order that the finite sequences of moves are made in doesn't matter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Edit.&lt;/em&gt; I notice now that you didn't say that the diameter of the circle should be the same as the diameter of the square. You can easily modify my solution for any size circle on any face. In this case, you do have to handle the top/bottom/middle rows of the circle, but this is no problem and is handled the same as the other rows. The top/bottom rows of the circle form a dual pair of working rows just like the others, and the center row has no dual, and is handled by itself.&lt;/p&gt;&#xA;" OwnerUserId="413" LastEditorUserId="413" LastEditorDisplayName="" LastEditDate="2010-11-07T12:40:01.403" LastActivityDate="2010-11-07T12:40:01.403" />
  <row Id="9258" PostTypeId="2" ParentId="9256" CreationDate="2010-11-07T12:29:58.453" Score="0" ViewCount="" Body="&lt;p&gt;Hint:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $ax + by = c$ has the same solution set as $ax + dy = e$ then any pair $(x,y)$ solving the first equation solves the second too. Let $(x,y)$ and $(x',y')$ with $y \neq y'$ be solutions. Subtracting the two we have $(b - d)y = c - e$ and $(b - d)y' = c - e$ but the only way for $(b - d)y = (b - d)y'$ to hold is when $b = d$, because $y$ is not equal to $y'$. Clearly then, $0 = c - e$ and this proves that $c = e$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason this proof does not go through when $a = 0$ is because there are no solutions with $y$ distinct.&lt;/p&gt;&#xA;" LastActivityDate="2010-11-08T10:16:34.537" CommentCount="2" />
  <row Id="9259" PostTypeId="1" AcceptedAnswerId="9305" CreationDate="2010-11-07T12:32:21.163" Score="-1" ViewCount="464" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How to solve this problem, I can not figure it out:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we list all the natural numbers below 10 that are multiples of 3 or 5, we get 3, 5, 6 and 9. The sum of these multiples is 23.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Find the sum of all the multiples of 3 or 5 below 1000.&lt;/p&gt;&#xA;" OwnerUserId="652" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2010-11-07T18:02:06.653" LastActivityDate="2010-11-07T19:03:23.657" Title="Find the sum of all the multiples of 3 or 5 below 1000." Tags="&lt;number-theory&gt;&lt;homework&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="9260" PostTypeId="2" ParentId="9259" CreationDate="2010-11-07T12:36:00.893" Score="3" ViewCount="" Body="&lt;p&gt;The multiples of 3 are 3,6,9,12,15,18,21,24,27,30,....&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The multiples of 5 are 5,10,15,20,25,30,35,40,45,....&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The intersection of these two sequences is 15,30,45,...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sum of the first numbers 1+2+3+4+...+n is n(n+1)/2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sum of the first few multiples of k, say k+2k+3k+4k+...+nk must be kn(n+1)/2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now you can just put these ingredients together to solve the problem.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;To find n use 1000/3 = 333 + remainder, 1000/5 = 200 + remainder, 1000/15 = 66 + remainder and then sum multiples of 3: $3\cdot 333(333+1)/2 = 166833$. multiples of 5: $5\cdot 200(200+1)/2 = 100500$ and subtract multiples of 15 $15\cdot 66(66+1)/2 = 33165$ to get 234168.&lt;/p&gt;&#xA;" LastActivityDate="2010-11-07T12:36:00.893" />
  <row Id="9261" PostTypeId="2" ParentId="275" CreationDate="2010-11-07T12:37:26.820" Score="1" ViewCount="" Body="&lt;p&gt;The Shape of Space by Jeffrey Weeks is really great.&lt;/p&gt;&#xA;" OwnerUserId="3060" LastActivityDate="2010-11-07T12:37:26.820" CommunityOwnedDate="2010-11-07T12:37:26.820" />
  <row Id="9262" PostTypeId="1" AcceptedAnswerId="9263" CreationDate="2010-11-07T12:38:26.633" Score="0" ViewCount="196" Body="&lt;p&gt;Hi&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a simple formula for the n-th derivative of some expression. I'm suppose to prove that it's true for all N, by using induction. I set n=1, and show that the base case works. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The next step, were I assume that it works a K is fine. But how is that shown on paper? And the that I don't understand at all, is how to show algebraically that this is true. Do you just &quot;plug-in&quot;, replace all the k's with k+1 and then what? Or? Well as you see, I don't get it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you so much for any input :)  &lt;/p&gt;&#xA;" OwnerUserId="2230" LastActivityDate="2010-11-08T19:19:36.513" Title="Induction, not what, but how it's done?" Tags="&lt;homework&gt;&lt;induction&gt;" AnswerCount="4" CommentCount="1" />
  <row Id="9263" PostTypeId="2" ParentId="9262" CreationDate="2010-11-07T12:53:28.080" Score="6" ViewCount="" Body="&lt;p&gt;The principle of induction says that if you have some $\mathbb N$ indexed family of propositions $P(n)$, that you can prove the theorem $\forall n \in \mathbb N, P(n)$ just by proving:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$P(1)$&lt;/li&gt;&#xA;&lt;li&gt;$\forall n \in \mathbb N. P(n) \Rightarrow P(n+1)$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Intuitively this is quite simple, for any given number say 3051 the statement P(3051) must be true, since P(0) is true, P(1) is true, P(2) is true ... all the way up to P(3051). Mathematically this is just reflecting the nature of $\mathbb N$ so it's a tautology.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;It is probably not clear what a proposition is though, because nobody ever talks about them. Well if you are familiar with functions, like $f(x) = x^2-1$, given any number say 3 you find that f(3) is a number too: 8. A proposition is like a function except that it computes a mathematical statement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I could define the following proposition: $S(x) :\iff \exists n, n^2 = x$ and that would hold for all square numbers for example $S(9) \iff \exists n, n^2 = 9$ and so we can prove $S(9)$ by proving $\exists n, n^2 = 9$ by exhibiting the witness 3 then showing that $3^2 = 9$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A much more interesting example might be the proposition $\text{Strong}_P(n)$ which is defined like so:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;$\text{Strong}_P(1) :\iff P(1)$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$\text{Strong}_P(2) :\iff P(1) \text{ and } P(2)$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$\text{Strong}_P(3) :\iff P(1) \text{ and } P(2) \text{ and } P(3)$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;...&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$\text{Strong}_P(n+1) :\iff \text{Strong}_P(n+1) \text{ and } P(n+1)$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;It is relatively easy to verify that $\text{Strong}_P(n) \iff (\forall k \leq n, P(k))$ so this really is the principle of &quot;strong&quot; induction. Furthermore you can prove the definition holds for all $n$ using normal induction.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;A very beautiful theorem is that the sum of cubes is the square of the sum, e.g.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$1^3 + 2^3 + 3^3 + 4^3 + 5^3 = (1 + 2 + 3 + 4 + 5)^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So let $P(n)$ be the statement &quot;sum of cubes up to n = square of sum up to n&quot;. Obviously it holds for P(1) since 1 = 1, now we would like to prove $\forall n \in \mathbb N. P(n) \Rightarrow P(n+1)$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for all n, given that the sum of cubes up to n equals the square of the sum of to n, then &quot;the same statement for n+1&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we have the assumption $1^3 + 2^3 + \cdots + n^3 = (1 + 2 + \cdots + n)^2$ and need to prove $1^3 + 2^3 + \cdots + n^3 + (n+1)^3 = (1 + 2 + \cdots + n + n+1)^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we use algebra, the right hand side is $(1 + 2 + \cdots + n)^2 + 2(n+1)(1 + 2 + \cdots + n) + (n+1)^2$. So all we have to show now is that $(n+1)^3 = 2(n+1)(1 + 2 + \cdots + n) + (n+1)^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Multiplying out some things $n^3 + 3 n^2 + 3 n + 1 = 2(n+1)(1 + 2 + \cdots + n) + n^2 + 2 n + 1$ cancelling some things gives: $n (n + 1)^2 = 2(n+1)(1 + 2 + \cdots + n)$ and we quickly notice it being equivalent to the formula $(1 + 2 + \cdots + n) = n (n + 1)/2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Shall we try to prove this by induction also! Let $T(n)$ be the statement that $(1 + 2 + \cdots + n) = n (n + 1)/2$. Clearly T(1) holds, since 1 = 1. Now we should like to prove that for any given $n$ the hypothesis $T(n)$, that is $(1 + 2 + \cdots + n) = n (n + 1)/2$ implies that $(1 + 2 + \cdots + n + n+1) = (n + 1) (n + 2)/2$. Algebraically we can just substitute the hypothesis in to get $n (n + 1)/2 + n+1 = (n + 1) (n + 2)/2$ doubling this $n (n + 1) + 2 n + 2 = (n + 1) (n + 2)$ and this is obviously true by multiplying out both sides.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(this was a stupid example though beacuse you can just check the first 4 cases and not do any induction).&lt;/p&gt;&#xA;" LastActivityDate="2010-11-08T19:19:36.513" CommentCount="9" />
  <row Id="9264" PostTypeId="1" AcceptedAnswerId="9272" CreationDate="2010-11-07T12:53:47.580" Score="-2" ViewCount="182" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's say I have this series of numbers &lt;code&gt;2, 5, 10, 17&lt;/code&gt;. Now somebody told me that next number is 26. He used this function for that:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;f(n) = n*n + 1&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Can anyone explain how does that function solve the problem to come up with 26?&lt;/p&gt;&#xA;" OwnerUserId="652" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-07T13:45:13.070" LastActivityDate="2010-11-07T13:59:23.520" Title="How to solve number series with f(n) = n*n + 1" Tags="&lt;recreational-mathematics&gt;" AnswerCount="1" CommentCount="15" />
  <row Id="9265" PostTypeId="1" CreationDate="2010-11-07T13:01:56.213" Score="1" ViewCount="119" Body="&lt;p&gt;I understand that we use the chain rule to differentiate the $y$ part of it. Like if it stood $xy$, I would use the product rule, that say that the derivative of $y$ is $dy/dx$. But what if it's just $y$ alone or with a constant in front off it? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you! &lt;/p&gt;&#xA;" OwnerUserId="2230" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-14T07:12:42.610" LastActivityDate="2010-11-14T07:57:46.033" Title="Question about implicit differentiation" Tags="&lt;calculus&gt;&lt;homework&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="9268" PostTypeId="2" ParentId="9242" CreationDate="2010-11-07T13:38:37.183" Score="4" ViewCount="" Body="&lt;p&gt;I will use the corrected version mentioned by Douglas, i.e.  $U$ will be the domain defined by $y&gt;0$ and $1&amp;lt; x^2 +y^2&amp;lt; 4$. Consider a $C^\infty$ function $\phi (y)$ which is equal to $1$ for negative $y$, $0$ for $y\geq 1/2$ and strictly decreasing for $0 &amp;lt; y &amp;lt; 1/2$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The required function $f$ is then defined by:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f(x,y) =-\phi (y)$ if $(x,y)\in U$ and $x\leq 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f(x,y) =+\phi (y)$ if $(x,y)\in U$ and $x\geq 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It does not only depend on $y$ since $f(-3/2,1/4)&amp;lt;0$ and $f(+3/2,1/4)&gt;0$. Nevertheless we do have $\dfrac {\partial f}{\partial x}=0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;PS&lt;/strong&gt; The Mean Value Theorem doesn't apply because the segment joining the two points $(-3/2,1/4)$ and $(+3/2,1/4)$ (for example) is &lt;em&gt;not&lt;/em&gt; entirely included in $U$.&lt;/p&gt;&#xA;" OwnerUserId="3217" LastEditorUserId="3217" LastEditorDisplayName="" LastEditDate="2010-11-07T14:32:03.167" LastActivityDate="2010-11-07T14:32:03.167" CommentCount="5" />
  <row Id="9269" PostTypeId="1" AcceptedAnswerId="9288" CreationDate="2010-11-07T13:47:11.603" Score="0" ViewCount="124" Body="&lt;p&gt;If I have a quadratic form like $y^2 - x^2 - x = k$ none of the techniques I know work because of the nasty $x$. Note that homogenizing doesn't work because a solution of $Y^2 - X^2 - X Z = k Z^{(2)}$ does not lead to a solution of the original equation in integers, at least as far as I have been able to determine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How does one understand the solution set of these equation? How can I solve them?&lt;/p&gt;&#xA;" LastActivityDate="2010-11-07T18:42:35.650" Title="How to solve inhomogeneous quadratic forms in integers?" Tags="&lt;number-theory&gt;&lt;diophantine-equations&gt;&lt;quadratic-forms&gt;" AnswerCount="2" />
  <row Id="9270" PostTypeId="2" ParentId="275" CreationDate="2010-11-07T13:49:39.430" Score="1" ViewCount="" Body="&lt;p&gt;Have a look at &lt;a href=&quot;http://mathoverflow.net/questions/8609/favorite-popular-math-book&quot; rel=&quot;nofollow&quot;&gt;http://mathoverflow.net/questions/8609/favorite-popular-math-book&lt;/a&gt; for other opinions&lt;/p&gt;&#xA;" OwnerUserId="2710" LastActivityDate="2010-11-07T13:49:39.430" CommunityOwnedDate="2010-11-07T13:49:39.430" />
  <row Id="9271" PostTypeId="2" ParentId="9269" CreationDate="2010-11-07T13:52:38.537" Score="4" ViewCount="" Body="&lt;p&gt;These all reduce to equations&#xA;$$f(x,y)=k$$&#xA;where $x$ and $y$ are integers, possibly satisfying some&#xA;congruence conditions, and $f$ is a homogeneous&#xA;quadratic form. In two variables this basically&#xA;comes down to solving norm equations in the rings&#xA;of integers in quadratic fields.&#xA;For your example you get&#xA;$$4y^2-4x^2-4x-1=4k-1$$&#xA;that is&#xA;$$y'^2-x'^2=4k-1$$&#xA;where $x'=2x-1$ is odd and $y'=2y$ is even. This is atypically&#xA;easy, since it factorizes as $(y'-x')(y'+x')=4k-1$. More typically&#xA;you'd get a Pell-type equation.&lt;/p&gt;&#xA;" OwnerUserId="226" LastEditorUserId="226" LastEditorDisplayName="" LastEditDate="2010-11-07T18:42:35.650" LastActivityDate="2010-11-07T18:42:35.650" CommentCount="3" />
  <row Id="9272" PostTypeId="2" ParentId="9264" CreationDate="2010-11-07T13:59:23.520" Score="5" ViewCount="" Body="&lt;p&gt;In general given any length $n$ sequence of numbers there exists a polynomial $p(x)$ of degree $n$ which gives this sequence. It's easy to find such a polynomial so I will not discuss how to do this. Note that as a consequence of this, if we are given a length $n$ sequence we can find a polynomial for the length $n+1$ sequence which gives any value we want after giving the values in the initial sequence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Futhermore, there are many different types of function other than polynomials - many of which can fit any given sequence. For this reason it is impossible to really be &quot;right&quot; when answering a question of the form &quot;What's the next number ...?&quot;. Any prediction is just as (in)valid without putting some constraints on it.&lt;/p&gt;&#xA;" LastActivityDate="2010-11-07T13:59:23.520" CommentCount="2" />
  <row Id="9273" PostTypeId="2" ParentId="275" CreationDate="2010-11-07T14:35:21.353" Score="2" ViewCount="" Body="&lt;p&gt;If you are willing to invest a bit of time, a lot of Conway's books are very good. I especially recommend&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Winning_Ways_for_your_Mathematical_Plays&quot; rel=&quot;nofollow&quot;&gt;Winning ways for your mathematical plays&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://plus.maths.org/content/symmetries-things&quot; rel=&quot;nofollow&quot;&gt;Symmetries of things&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;A bit less &quot;story&quot; but still quite a lot of fun is Robert Lang's &lt;a href=&quot;http://www.langorigami.com/pubs/books/books.php4&quot; rel=&quot;nofollow&quot;&gt;Origami design secrets&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also recommend James R. Newman's &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0486432688&quot; rel=&quot;nofollow&quot;&gt;The world of mathematics&lt;/a&gt;. But be warned, while the content is not very technical (indeed, many articles in the collection are from public lectures of famous scientists), it can get a little bit dry at times. If you are patient enough for it, it is a very good companion to Courant and Robbins' &lt;em&gt;What is mathematics&lt;/em&gt;. (It is sort of like the Princeton Companion, but older and slightly more down-to-earth). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lastly, you can also try the various books and articles by &lt;a href=&quot;http://bit-player.org/&quot; rel=&quot;nofollow&quot;&gt;Brian Hayes&lt;/a&gt;. &lt;/p&gt;&#xA;" OwnerUserId="1543" LastActivityDate="2010-11-07T14:35:21.353" CommunityOwnedDate="2010-11-07T14:35:21.353" />
  <row Id="9274" PostTypeId="1" AcceptedAnswerId="9348" CreationDate="2010-11-07T15:19:57.743" Score="0" ViewCount="536" Body="&lt;p&gt;Hi, I need to find an isomorphism between [0,1] to (0,1) can you help me with this please?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;thanks. benny&lt;/p&gt;&#xA;" OwnerUserId="3755" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2010-11-26T21:22:48.343" LastActivityDate="2010-11-28T09:19:44.817" Title="Isomorphism between [0,1] to (0,1)" Tags="&lt;homework&gt;&lt;elementary-set-theory&gt;" AnswerCount="6" CommentCount="2" />
  <row Id="9275" PostTypeId="2" ParentId="9274" CreationDate="2010-11-07T15:27:16.357" Score="1" ViewCount="" Body="&lt;p&gt;Hello Benny,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What do you mean by &quot;isomorphism&quot;? Are you considering these sets with a given group or ring structure (if then, which one?), or as topological spaces (in which case you'd be looking for a &quot;homeomorphism&quot;), or just as sets?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know of a canonical group structure on either set, but if you want your map between the two to be continuous you might have some luck by looking at the notion of compactness and how it behaves under continuous functions.&lt;/p&gt;&#xA;" OwnerUserId="3225" LastActivityDate="2010-11-07T15:27:16.357" CommentCount="3" />
  <row Id="9276" PostTypeId="2" ParentId="9274" CreationDate="2010-11-07T15:36:04.413" Score="0" ViewCount="" Body="&lt;p&gt;The bijection 1/x turns (0,1) into (1,$\infty$) which you can chop up into countably many &lt;code&gt;(]&lt;/code&gt;'s. Also you can split &lt;code&gt;[]&lt;/code&gt; into &lt;code&gt;[)(]&lt;/code&gt; and these can be divide countably many times to bisect with the &lt;code&gt;(]&lt;/code&gt;'s but it flips the ordering around.&lt;/p&gt;&#xA;" LastActivityDate="2010-11-07T15:36:04.413" />
  <row Id="9277" PostTypeId="1" AcceptedAnswerId="9290" CreationDate="2010-11-07T15:38:31.720" Score="0" ViewCount="75" Body="&lt;p&gt;I thought of writing this question &lt;a href=&quot;http://math.stackexchange.com/questions/9171/minimum-for-this-function&quot;&gt;Minimum for this function&lt;/a&gt; in a different way, if it helps.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to minimize,&#xA; $\sum_{i=1}^{n}{a_ix_i} +\nu \sum_{i=1}^{n}{b_i 2^{x_i}}$&lt;br&gt;&#xA;where,&lt;br&gt;&#xA;$a_i \epsilon [0,1],\quad b_i \epsilon (0,\infty),\quad x_i \epsilon [xmin,0)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;br&gt;&#xA;$\sum_{i=1}^{n}(2^{x_i}) = 1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;xmin, $\nu$, $a_i$  and  $b_i$ are constants.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I guess the tricky part is to minimize the function while ensuring all $2^{x_i}$ sum upto 1 for these n variables.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you for your patience and help.&lt;/p&gt;&#xA;" OwnerUserId="2858" LastActivityDate="2010-11-07T16:44:53.333" Title="Minimum for this function" Tags="&lt;algebra&gt;" AnswerCount="2" />
  <row Id="9278" PostTypeId="2" ParentId="9262" CreationDate="2010-11-07T15:42:18.020" Score="0" ViewCount="" Body="&lt;p&gt;muad's answer is clear, but one could emphasize a couple of points:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mathematical induction is essentially a &lt;strong&gt;logical chain reaction&lt;/strong&gt;. &#xA;It helps to visualize the goal ``$\forall n\ P(n)$'' as an infinite number of separate assertions $P(1), P(2), \ldots$  The aim of the induction step is to establish &lt;em&gt;logical links&lt;/em&gt; between successive assertions:&#xA;$$&#xA;P(1)\Rightarrow P(2)\Rightarrow \ldots \Rightarrow P(17)\Rightarrow P(18)\Rightarrow\ldots&#xA;$$ &#xA;That is, $\forall n\in\mathbb N\ P(n)\Rightarrow P(n+1)$. These links are &lt;em&gt;purely hypothetical inferences&lt;/em&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The handout &lt;a href=&quot;http://www.math.cmu.edu/~bobpego/21131/lcr.pdf&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; provides an expanded explanation with examples.&lt;/p&gt;&#xA;" OwnerUserId="2947" LastActivityDate="2010-11-07T15:42:18.020" CommentCount="2" />
  <row Id="9279" PostTypeId="1" CreationDate="2010-11-07T15:43:46.393" Score="2" ViewCount="75" Body="&lt;p&gt;I have $X_1,...$ Bernoulli variables with probability $p$ of success.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to get an $n$ such that with probability $\delta$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$P(\sum_{i=1}^n X_i \ge k) \ge \delta$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This $n$ would of course depend on $k$, $p$ and $\delta$. Can I get an upper bound for that $n(k,\delta,p)$? (meaning, I want some $n(k,\delta,p)$ for which the above holds, the smaller $n$ is, the better.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="7076" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-07T16:04:08.640" LastActivityDate="2010-11-07T17:06:55.270" Title="can I get an upper bound on the tail of a binomial variable?" Tags="&lt;probability&gt;&lt;statistics&gt;" AnswerCount="2" />
  <row Id="9280" PostTypeId="2" ParentId="9274" CreationDate="2010-11-07T15:55:21.937" Score="5" ViewCount="" Body="&lt;p&gt;You can identify a countably infinite collection of points starting with 0 and converging to some limit less than 1/2.  Just shift each point one down the line.  Now repeat from the top end.  All points not involved go to themselves.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-11-07T15:55:21.937" CommentCount="2" />
  <row Id="9281" PostTypeId="1" CreationDate="2010-11-07T15:59:15.477" Score="2" ViewCount="175" Body="&lt;p&gt;Let $k$ be an algebraic closed field. Let $x$ be a point in $X=P_k^1$. What is $O_{X,x}$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, if I have $x=(t-a)\in \text{Spec }k[t]$. Looking $x$ inside $P_k^1$, does $O_{X,x}=k[t]_{(t-a)}$? I'm confused when I have to deal with the sheaf of rings.&lt;/p&gt;&#xA;" OwnerUserId="1971" LastActivityDate="2011-01-06T22:44:39.320" Title="Stalks on Projective Scheme" Tags="&lt;algebraic-geometry&gt;&lt;schemes&gt;&lt;projective-schemes&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="9282" PostTypeId="2" ParentId="9279" CreationDate="2010-11-07T16:08:31.830" Score="3" ViewCount="" Body="&lt;p&gt;Consult a discussion of Chernoff bounds like &lt;a href=&quot;http://en.wikipedia.org/wiki/Chernoff_bound&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Chernoff_bound&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1527" LastActivityDate="2010-11-07T16:08:31.830" />
  <row Id="9283" PostTypeId="2" ParentId="9277" CreationDate="2010-11-07T16:18:55.593" Score="0" ViewCount="" Body="&lt;p&gt;I don't think rewriting this way helps. In the first place, how would you show that if $y$ is the optimal solution of $\min_y f(y)$, then $x = 2^y$ is the optimal solution to $\min_x f(\log x)$?&lt;/p&gt;&#xA;" OwnerUserId="1527" LastActivityDate="2010-11-07T16:18:55.593" />
  <row Id="9284" PostTypeId="2" ParentId="9281" CreationDate="2010-11-07T16:20:13.097" Score="2" ViewCount="" Body="&lt;p&gt;The stalk of the generic point is, of course, the rational function field $k(t)$. Now $\mathbb{P}^1$ is covered by two copies of $\mathbb{A}^1$. If a closed point $x$ corresponds to the maximal ideal generated by $(t-a)$ for some $a \in k$, then the stalk is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$O_{\mathbb{A}^1,x} = k[t]_{(t-a)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and this is isomorphic to $k[t]_{(t)}$.&lt;/p&gt;&#xA;" OwnerUserId="1650" LastActivityDate="2010-11-07T16:20:13.097" CommentCount="1" />
  <row Id="9285" PostTypeId="2" ParentId="9274" CreationDate="2010-11-07T16:27:06.080" Score="7" ViewCount="" Body="&lt;p&gt;The maps $(0,1) \to [0,1], x \mapsto x$ and $[0,1] \to (0,1), x \mapsto 1/4 + x/2$ are injective. The theorem of Cantor-Schröder-Bernstein implies that there is a bijection. It is actually possible to work out the proof of the theorem in these simple examples and get an explicit bijection, namely&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$g : [0,1] \to (0,1), x \mapsto 1/2 \pm 1/2^{n+2}$ if $x = 1/2 \pm 1/2^{n+1}$ for some $n$, and $x \mapsto x$ else.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See also &lt;a href=&quot;http://matheplanet.com/matheplanet/nuke/html/article.php?sid=873&quot; rel=&quot;nofollow&quot;&gt;this article&lt;/a&gt; if you know german. ;)&lt;/p&gt;&#xA;" OwnerUserId="1650" LastActivityDate="2010-11-07T16:27:06.080" />
  <row Id="9286" PostTypeId="1" AcceptedAnswerId="9292" CreationDate="2010-11-07T16:33:19.933" Score="5" ViewCount="749" Body="&lt;p&gt;How to prove&#xA; $$\int_{0}^{+\infty} e^{-x^2} dx = \frac{\sqrt \pi}{2}$$&lt;/p&gt;&#xA;" OwnerUserId="1382" LastEditorUserId="1382" LastEditorDisplayName="" LastEditDate="2010-11-07T17:23:58.123" LastActivityDate="2010-11-22T01:21:02.573" Title="Proving $\int_{0}^{+\infty} e^{-x^2} dx = \frac{\sqrt \pi}{2}$" Tags="&lt;calculus&gt;&lt;integral&gt;" AnswerCount="5" CommentCount="3" FavoriteCount="5" />
  <row Id="9287" PostTypeId="2" ParentId="9274" CreationDate="2010-11-07T16:37:05.817" Score="1" ViewCount="" Body="&lt;p&gt;If you just want a bijection there are many ways to do it. Here's a rather contrived one:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $\alpha$ be a &lt;a href=&quot;http://en.wikipedia.org/wiki/Limit_ordinal&quot; rel=&quot;nofollow&quot;&gt;limit ordinal&lt;/a&gt; of the same cardinality as $(0,1)$. By the &lt;a href=&quot;http://en.wikipedia.org/wiki/Cantor%E2%80%93Bernstein%E2%80%93Schroeder_theorem&quot; rel=&quot;nofollow&quot;&gt;Cantor-Bernstein-Schröder theorem&lt;/a&gt; there exists a bijective map $f: (0,1) \to \alpha$. Let $S$ denote the &lt;a href=&quot;http://en.wikipedia.org/wiki/Successor_ordinal&quot; rel=&quot;nofollow&quot;&gt;successor &quot;function&quot;&lt;/a&gt; (I know it's not a function - hence the quotes). Note that $\alpha$ does not have a maximal element, so it makes sense to define $\phi: [0,1] \to (0,1)$ by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\phi(x) = \begin{cases}&#xA;(f^{-1} \circ S \circ S \circ f)(x) &amp;amp; \text{ for $x \in (0,1)$}, \\&#xA;f^{-1}(1) &amp;amp; \text{ for $x = 1$}, \\&#xA;f^{-1}(0) &amp;amp; \text{ for $x = 0$}.&#xA;\end{cases}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Showing that $\phi$ is a bijection is an easy exercise. It's also completely useless and nonconstructive. &lt;/p&gt;&#xA;" OwnerUserId="1269" LastActivityDate="2010-11-07T16:37:05.817" CommentCount="1" />
  <row Id="9288" PostTypeId="2" ParentId="9269" CreationDate="2010-11-07T16:43:57.677" Score="4" ViewCount="" Body="&lt;p&gt;Over 200 years ago Lagrange solved the general binary quadratic Diophatine equation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\rm a\ x^2 + b\ xy + c\ y^2 + d\ x + e\ y + f = 0 $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It reduces to a Pell equation: put $\rm\ D = b^2-4ac,\ E = bd-2ae,\ F = d^2-4af\:.\ $ Then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\rm  D\ Y^2\ =\ (D\ y + E)^2 + D\ F - E^2,\quad\quad Y\ =\ 2ax + by + d $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore if we put $\rm\quad\ \ X = D\ y + E,\quad\ \ N = E^2 - D\ F\quad\ \ $ we have the Pell equation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\rm X^2 - D\ Y^2\ =\ N $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dario Alpern has a web page &lt;a href=&quot;http://www.alpertron.com.ar/QUAD.HTM&quot; rel=&quot;nofollow&quot;&gt;Quadratic two integer variable equation solver&lt;/a&gt; that will solve any such equation - with complete descriptions of the methods involved. For some  recent optimizations of Lagrange's algorithm see this paper H. C. Williams et al. &lt;a href=&quot;http://www.springerlink.com/content/58614q7346xg2650/fulltext.pdf&quot; rel=&quot;nofollow&quot;&gt;A new look at an old equation&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-11-07T16:43:57.677" />
  <row Id="9289" PostTypeId="2" ParentId="9262" CreationDate="2010-11-07T16:44:09.573" Score="0" ViewCount="" Body="&lt;p&gt;So you get it conceptually (assume $P$ true for an arbitrary $n$, show that that implies  $P$ true for $n+1$) (and yes that seems weird but if you parenthesize the formula with quantifiers, it works out that you're not really assuming the consequent).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But in practice...that's what you want to know how to do it. Examples are best. Let's try a simple one that you've probably seen before but specify exactly what is happening. For&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sum_{k=0}^{n-1} 2k+1 = n^2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;in the inductive case, assume the above is true, then show that the LHS with $n+1$ substituted for $n$ equals the RHS with the same substitution. That is, show &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sum_{k=0}^{n} 2k+1 = (n+1)^2.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The strategy is to do algebraic manipulation so that you see where you can apply the induction hypothesis. You can do the manipulation on either side, but you have to follow the rules (here specifically when proving an equational identity, i.e. 'LHS = RHS', you have to make sure you keep them separate, that is show a chain of manipulation leading from one to the other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can start from either side: the RHS obviously expands to $n^2 + 2n + 1$. Now use your pattern matching skills to find something that looks like something in the induction hypothesis. Here it would be the $n^2$ part. This looks like the RHS of the induction hypothesis; you can replace it with the LHS (of the induction hypothesis), to get &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(\sum_{k=0}^{n-1} 2k+1)  + 2n + 1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(extra parens put in to show what got replaced) And the extra part outside of the summation, the $2n+1$, is the same as extending the top of the range of the summation to $n$ (here you have to be familiar with the summation notation). So this last part can be converted immediately into &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sum_{k=0}^{n} 2k+1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But  that is the end of the equational chain, going from  $(n+1)^2$ to $\sum_{k=0}^{n} 2k+1$, maintaining equality the  the whole way, most of the time just doing the algebraic manipulation, but necessarily using the induction hypothesis somewhere along the line, that is, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(n+1)^2 = ... = ... = \sum_{k=0}^{n} 2k+1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd suspect a lot of the difficulty here is just keeping separate what you know already and what you have to prove (what you don't actually know yet -in- the proof). And ...yeah...just keeping track of all the symbols.&lt;/p&gt;&#xA;" OwnerUserId="1919" LastActivityDate="2010-11-07T16:44:09.573" />
  <row Id="9290" PostTypeId="2" ParentId="9277" CreationDate="2010-11-07T16:44:53.333" Score="2" ViewCount="" Body="&lt;p&gt;First let $x = (x_1, \dots, x_n)$ and write $f(x) = \sum_i^n a_i x_i + \nu\sum_i^n b_i 2^{x_i}$ and $g(x) = \sum_i^n 2^{x_i} - 1$. Now the method of &lt;a href=&quot;http://en.wikipedia.org/wiki/Lagrange_multipliers&quot; rel=&quot;nofollow&quot;&gt;Lagrange's multipliers&lt;/a&gt; tells you that you have a solution $x'$ iff it is an extremal point of the function $f(x) + \lambda g(x)$ with $\lambda$ a parameter to be determined. So you get equations $a_i + (\nu b_i + \lambda) 2^{x'_i} \log{2} = 0$ for all $1 \leq i \leq n$. It's now easy to express $x'_i$ as a function of lambda and plugging this into the remaining constraint $g(x') = 0$ to find the lambda and complete the solution. Hope this helps.&lt;/p&gt;&#xA;" OwnerUserId="3223" LastActivityDate="2010-11-07T16:44:53.333" CommentCount="4" />
  <row Id="9292" PostTypeId="2" ParentId="9286" CreationDate="2010-11-07T16:47:10.920" Score="19" ViewCount="" Body="&lt;p&gt;This is an old favorite of mine.&lt;br&gt;&#xA;Define $$I=\int_{-\infty}^{+\infty} e^{-x^2} dx$$ &#xA;Then $$I^2=\bigg(\int_{-\infty}^{+\infty} e^{-x^2} dx\bigg)\bigg(\int_{-\infty}^{+\infty} e^{-y^2} dy\bigg)$$  &#xA;$$I^2=\int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty}e^{-(x^2+y^2)} dxdy$$  &#xA;Now change to polar coordinates&lt;br&gt;&#xA;$$I^2=\int_{0}^{+2 \pi}\int_{0}^{+\infty}e^{-r^2} rdrd\theta$$  &#xA;The $\theta$ integral just gives $2\pi$, while the $r$ integral succumbs to the substitution $u=r^2$  &#xA;$$I^2=2\pi\int_{0}^{+\infty}e^{-u}du/2=\pi$$ &#xA;So $$I=\sqrt{\pi}$$ and your integral is half this by symmetry&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have always wondered if somebody found it this way, or did it first using complex variables and noticed this would work.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2010-11-20T23:43:16.607" LastActivityDate="2010-11-20T23:43:16.607" CommentCount="13" />
  <row Id="9293" PostTypeId="1" AcceptedAnswerId="9294" CreationDate="2010-11-07T16:54:51.377" Score="0" ViewCount="79" Body="&lt;p&gt;Hello,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $R\subset A\times A$ a binary relation, and note $a\sim b$ for $(a,b)\in R$. We define&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(a,b)\in R\circ R\Leftrightarrow  \exists b\in A : a\sim b,  b\sim c$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and $R^n=R\circ\dots\circ R$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My books says that $(a,b)\in R^n$ iff there is a path of length $n$ in the graph of $R$. Isn't it false ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, if $(a,b)\in R^3$, there exists $c\in A$ such that $(a,c)\in R^2$, $(c,b)\in R^2$ $\Leftrightarrow$ there exists $d, f\in A$ such that &#xA;$$a\sim d\sim c\sim f\sim b$$&#xA;which is a path of length 4, not 3.&lt;/p&gt;&#xA;" OwnerUserId="524" LastActivityDate="2010-11-07T16:56:45.730" Title="Binary relation powers and path length" Tags="&lt;graph-theory&gt;&lt;discrete-mathematics&gt;&lt;relations&gt;" AnswerCount="1" />
  <row Id="9294" PostTypeId="2" ParentId="9293" CreationDate="2010-11-07T16:56:45.730" Score="2" ViewCount="" Body="&lt;p&gt;Your book is right. Note that $R^3 = R \circ R \circ R = R^2 \circ R$, not $R^2 \circ R^2$.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-07T16:56:45.730" CommentCount="1" />
  <row Id="9295" PostTypeId="2" ParentId="9279" CreationDate="2010-11-07T17:06:55.270" Score="1" ViewCount="" Body="&lt;p&gt;The following observation, which is straightforward to derive, may be very useful here (also with regard to Naga's answer):&#xA;$$&#xA;{\rm P}\bigg(\sum\limits_{i = 1}^n {(1 - X_i ) \ge n - k} \bigg) \le 1 - \delta &#xA;$$&#xA;implies &#xA;$$&#xA;{\rm P}\bigg(\sum\limits_{i = 1}^n {X_i  \ge k} \bigg) \ge \delta. &#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="2810" LastActivityDate="2010-11-07T17:06:55.270" />
  <row Id="9297" PostTypeId="1" CreationDate="2010-11-07T17:45:30.773" Score="0" ViewCount="73" Body="&lt;p&gt;I need to turn the equation L=12/x + x into a quadratic forumla to solve. i have been given the answers of 0.619m 0r 19.38m if L = 20m&lt;/p&gt;&#xA;&#xA;&lt;p&gt;what iv go so far&lt;/p&gt;&#xA;&#xA;&lt;p&gt;0= 12/x + x -20&#xA;but i dont know how to turn 12/x to a x^2 value&lt;/p&gt;&#xA;" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2010-11-07T18:00:12.563" LastActivityDate="2010-11-07T18:00:12.563" Title="Quadratic equation rearranging" Tags="&lt;algebra-precalculus&gt;&lt;homework&gt;" AnswerCount="1" />
  <row Id="9298" PostTypeId="2" ParentId="9256" CreationDate="2010-11-07T17:46:50.500" Score="2" ViewCount="" Body="&lt;p&gt;Follow the supplied hint: evaluate $\rm\ x = (c-b\: y)/a = (e-d\:y)/a\ $ at $\rm\ y = 0\ $ and $\rm\ y = -1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More generally, see also this closely related &lt;a href=&quot;http://math.stackexchange.com/questions/9058/work-out-the-values-of-a-and-b-from-the-identity-x-ax-144-x-b/9068#9068&quot;&gt;recent question.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-11-08T20:02:35.127" LastActivityDate="2010-11-08T20:02:35.127" />
  <row Id="9299" PostTypeId="2" ParentId="9297" CreationDate="2010-11-07T17:49:51.547" Score="2" ViewCount="" Body="&lt;p&gt;Hi Robin. Multiply the equation by $x$ and you'll get the polynomial you want.&lt;/p&gt;&#xA;" OwnerUserId="592" LastActivityDate="2010-11-07T17:49:51.547" />
  <row Id="9300" PostTypeId="2" ParentId="9286" CreationDate="2010-11-07T17:52:16.620" Score="7" ViewCount="" Body="&lt;p&gt;A variation on Ross Millikan's answer. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can start again with the observation&#xA;$$\left(\int_{-\infty}^{\infty}e^{-x^2}dx\right)\left(\int_{-\infty}^{\infty}e^{-y^2}dy\right)=\left(\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{-(x^2+y^2)}dxdy\right)=V.$$&#xA;Now $V$ is simply the volume of the body &#xA;$$-\infty &amp;lt; x,y &amp;lt; \infty,\qquad 0 &amp;lt; z &amp;lt; e^{-(x^2+y^2)},$$&#xA;or, equivalently,&#xA;$$0 &amp;lt; x^2+y^2 &amp;lt; -\ln z,\qquad  0 &amp;lt; z &amp;lt; 1.$$&#xA;This implies that the body is a &lt;a href=&quot;http://en.wikipedia.org/wiki/Solid_of_revolution&quot; rel=&quot;nofollow&quot;&gt;solid of revolution&lt;/a&gt;. Using the&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Disk_integration&quot; rel=&quot;nofollow&quot;&gt;disk integration&lt;/a&gt; formula, we have&#xA;$$V=\int_{0}^{1}\pi(-\ln z)dz=[\pi(z-z\ln z)]_{0}^1=\pi.$$&lt;/p&gt;&#xA;" OwnerUserId="723" LastEditorUserId="723" LastEditorDisplayName="" LastEditDate="2010-11-07T20:10:12.667" LastActivityDate="2010-11-07T20:10:12.667" CommentCount="4" />
  <row Id="9301" PostTypeId="2" ParentId="9262" CreationDate="2010-11-07T18:07:45.717" Score="0" ViewCount="" Body="&lt;p&gt;Here is a very simple illustrative example.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;LEMMA&lt;/strong&gt; $\rm\quad D^n\ e^{c\:x}\ =\ c^n\ e^{c\:x }\quad$ for all $\rm\:n\in \mathbb N$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Base step:&lt;/strong&gt; $\ $ It is true for $\rm\: n = 0\ $ since $\rm\ D^0 = I, \ \ c^0 = 1\:$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Induction step:&lt;/strong&gt; $\ $ If it is true for $\rm\: n\:$ then differentiating both sides yields&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\quad\quad\quad \quad D^{n+1}\ e^{c\:x}\ =\ D^n\: (c^n\ e^{c\:x })\ =\ c^{n+1}\ e^{c\:x}  $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;hence it is true for $\rm\: n+1\:$. Therefore it is true for all $\rm\:n\in \mathbb N\:$ by induction.&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-11-07T18:07:45.717" />
  <row Id="9302" PostTypeId="1" AcceptedAnswerId="9303" CreationDate="2010-11-07T18:28:45.753" Score="2" ViewCount="490" Body="&lt;p&gt;Say I have a symmetric matrix. I have the concept of 2-norm as defined on wikipedia. Now I want to prove (disprove?) that the norm of a symmetric matrix is maximum absolute value of its eigenvalue. I would really appreciate if this can be done only using simple concepts of linear algebra.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am quite new to mathematics. &lt;/p&gt;&#xA;" OwnerUserId="1674" LastEditorUserId="1674" LastEditorDisplayName="" LastEditDate="2011-01-20T15:23:19.963" LastActivityDate="2011-01-20T15:23:19.963" Title="Norm of a symmetric matrix?" Tags="&lt;linear-algebra&gt;&lt;norm&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="9303" PostTypeId="2" ParentId="9302" CreationDate="2010-11-07T18:34:42.670" Score="5" ViewCount="" Body="&lt;p&gt;Given a symmetric matrix, you have a complete set of eigenvalues and eigenvectors.  So any vector can be represented as a linear combination of the eigenvectors.  Multiply your matrix by an arbitrary unit vector decomposed into the eigenvectors.  Then note that the maximum length of the resultant vector is achieved when the input vector is along the eigenvector associated with the largest eigenvalue.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-11-07T18:34:42.670" CommentCount="8" />
  <row Id="9304" PostTypeId="2" ParentId="9286" CreationDate="2010-11-07T18:55:11.317" Score="6" ViewCount="" Body="&lt;p&gt;An alternative derivation is to show that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_{0}^{\infty}xe^{-x^{2}y^{2}}\; dy=I,$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $I$ is your integral:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$I=\int_{0}^{\infty}e^{-x^2}\; dx,$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and then evaluate $I^2$ by reversing the order of integration. If $x&gt;0$, then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_{0}^{\infty}xe^{-x^{2}y^{2}}\; dy=x\int_{0}^{\infty}e^{-{(xy)}^2}\; dy=x\int_{0}^{\infty}e^{-u^2}\dfrac{du}{x}=\int_{0}^{\infty}e^{-u^2}\; du=I.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$I^2=\int_{0}^{\infty}e^{-x^2}\; dx\int_{0}^{\infty}xe^{-x^{2}y^{2}}\; dy=\displaystyle\int_{0}^{\infty}dy\int_{0}^{\infty}xe^{-x^2}e^{-x^{2}y^{2}}\; dx$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$=\int_{0}^{\infty}dy\int_{0}^{\infty}xe^{-x^{2}(1+y^2)}\; dx=\int_{0}^{\infty}dy\dfrac{1}{2\left( 1+y^{2}\right) }\left[ -e^{-x^{2}\left( 1+y^{2}\right) }\right] _{x=0}^{\infty }$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$=\int_{0}^{\infty }\dfrac{1}{2\left( 1+y^{2}\right) }\; dy=\dfrac{1}{2}\left[ \arctan y\right] _{y=0}^{\infty }=\dfrac{\pi}{4}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$I=\dfrac{\sqrt{\pi}}{2}.$$&lt;/p&gt;&#xA;" OwnerUserId="752" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2010-11-07T21:54:53.343" LastActivityDate="2010-11-07T21:54:53.343" CommunityOwnedDate="2010-11-07T18:55:11.317" CommentCount="0" />
  <row Id="9305" PostTypeId="2" ParentId="9259" CreationDate="2010-11-07T18:58:11.493" Score="4" ViewCount="" Body="&lt;p&gt;The previously posted answer isn't correct. The statement of the problem is to sum the multiples of 3 and 5 below 1000, not up to and equal 1000. The correct answer is&#xA;\begin{eqnarray}&#xA;\sum_{k_{1} = 1}^{333} 3k_{1} + \sum_{k_{2} = 1}^{199} 5 k_{2} - \sum_{k_{3} =1}^{66} 15 k_{3} = 166833 + 99500 - 33165 = 233168,&#xA;\end{eqnarray}&#xA;where we have the used the identity&#xA;\begin{eqnarray}&#xA;\sum_{k = 1}^{n} k = \tfrac{1}{2} n(n+1).&#xA;\end{eqnarray}&lt;/p&gt;&#xA;" OwnerUserId="2720" LastEditorUserId="2720" LastEditorDisplayName="" LastEditDate="2010-11-07T19:03:23.657" LastActivityDate="2010-11-07T19:03:23.657" />
  <row Id="9306" PostTypeId="1" AcceptedAnswerId="9307" CreationDate="2010-11-07T19:09:15.907" Score="2" ViewCount="180" Body="&lt;p&gt;According to the solution manual:&#xA;$\int \frac{x}{\sqrt{1-x^{4}}}dx = \frac{1}{2}\arcsin x^{2}+C$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My solution doesn't seem to be working. I know another way of solving it (setting $u=x^{2}$) but the fact that this way of solving it doesn't work bothers me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\text{set }u=1-x^{4}\text{ so } dx=\frac{du}{-4x^{3}}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{align*}&#xA;\int \frac{x}{\sqrt{1-x^{4}}}dx &amp;amp;= \int \frac{x}{\sqrt{u}}dx  \\&#xA;&amp;amp;= \int \frac{xdu}{-4x^{3}\sqrt{u}} \\&#xA;&amp;amp;= -\frac{1}{4} \int \frac{du}{x^{2}\sqrt{u}} \\&#xA;\end{align*}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\text{set } v=\sqrt{u} \text{ so }du=2\sqrt{u}\,dv&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align*}&#xA;-\frac{1}{4} \int \frac{du}{x^{2}\sqrt{u}} &amp;amp;= -\frac{1}{2} \int \frac{dv}{x^{2}}  \\&#xA;&amp;amp;= -\frac{1}{2} \int \frac{dv}{\sqrt{1-v^{2}}} \\&#xA;&amp;amp;= -\frac{1}{2} \arcsin (v) + C \\&#xA;&amp;amp;= -\frac{1}{2} \arcsin (\sqrt {1-x^{4}}) + C \\&#xA;\end{align*}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll be happy to clarify any steps I took. Thanks!&lt;/p&gt;&#xA;" OwnerUserId="1958" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-07T19:13:50.270" LastActivityDate="2010-11-07T19:28:15.543" Title="Can't Solve an Integral" Tags="&lt;calculus&gt;&lt;integral&gt;" AnswerCount="1" />
  <row Id="9307" PostTypeId="2" ParentId="9306" CreationDate="2010-11-07T19:22:38.563" Score="8" ViewCount="" Body="&lt;p&gt;Your solution is &lt;em&gt;an&lt;/em&gt; antiderivative of the original function. You can always check whether your solution is correct by taking its derivative. This also implies that the book solution and your solution differ by a constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For this specific problem, imagine the right triangle with sides $x^2$ and $\sqrt{1-x^4}$ and hypotenuse $1$. Then $\arcsin\sqrt{1-x^4} = \frac{\pi}{2} - \arcsin x^2$, and it should be easy to see from there how both solutions are related.&lt;/p&gt;&#xA;" OwnerUserId="1016" LastEditorUserId="1016" LastEditorDisplayName="" LastEditDate="2010-11-07T19:28:15.543" LastActivityDate="2010-11-07T19:28:15.543" />
  <row Id="9311" PostTypeId="1" AcceptedAnswerId="9315" CreationDate="2010-11-07T19:42:17.930" Score="13" ViewCount="512" Body="&lt;p&gt;This should be an easy exercise: Given a finite odd abelian group $G$, prove that $\prod_{g\in G}g=e$. Indeed, using Lagrange's theorem this is trivial: There is no element of order 2 (since the order must divide the order of $G$, but it is odd), and so every element except $e$ has a unique inverse which is different from it. Hence both the element and its inverse participate in the product and cancel each other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My problem is simple - I need to solve this &lt;strong&gt;without&lt;/strong&gt; Lagrange's theorem. So either there's a smart way to prove the nonexistance of an element of order 2 in an odd abelian group, or I'm missing something even more basic...&lt;/p&gt;&#xA;" OwnerUserId="1818" LastEditorUserId="8575" LastEditorDisplayName="" LastEditDate="2011-05-22T06:12:32.120" LastActivityDate="2011-05-22T06:12:32.120" Title="Product of all elements in an odd finite abelian group is 1" Tags="&lt;group-theory&gt;&lt;finite-groups&gt;" AnswerCount="6" CommentCount="2" FavoriteCount="2" />
  <row Id="9312" PostTypeId="2" ParentId="9311" CreationDate="2010-11-07T19:55:03.817" Score="1" ViewCount="" Body="&lt;p&gt;Let $G$ be a finite abelian group of odd order. Then we have &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\prod_{g \in G}g \prod_{g \in G} g = &#xA;\prod_{g \in G} g \prod_{g \in G} g^{-1} = \prod_{g \in G} gg^{-1} = e.&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence $x:= \prod_{g \in G}\;g$ is its own inverse. Now you just need to show that the only element that's its own inverse in $G$ is $e$.&lt;/p&gt;&#xA;" OwnerUserId="1269" LastActivityDate="2010-11-07T19:55:03.817" CommentCount="3" />
  <row Id="9313" PostTypeId="1" AcceptedAnswerId="9317" CreationDate="2010-11-07T20:09:40.097" Score="2" ViewCount="232" Body="&lt;p&gt;Say I have a binomial coefficient $y=\binom{5n+3}{n+2}$ or $y=\binom{n^2+4}{3n}$ something of the sorts in terms of the variable $n$. How can I determine $f$ so that $y = O(f)$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a general method for this sort of thing.  I have only ever seen $O$-notation in a computer science setting and its treatment was not concerned with deriving anything except the bounds for &quot;simple&quot; functions like logarithms and polynomials.  As a side question, what would be a good resource to learn further methods to find $O$ of more complex functions?  I understand the formal definition, but Im unsure how to use it practically.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My apologies if this is a stupid question. Its not homework.&lt;/p&gt;&#xA;" OwnerUserId="565" LastEditorUserId="565" LastEditorDisplayName="" LastEditDate="2010-11-07T23:10:32.693" LastActivityDate="2010-11-07T23:10:32.693" Title="How can I determine asymptotic growth of binomial coefficients?" Tags="&lt;binomial-coefficients&gt;&lt;asymptotics&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="9314" PostTypeId="2" ParentId="9311" CreationDate="2010-11-07T20:12:39.857" Score="3" ViewCount="" Body="&lt;p&gt;Here is one argument (although surely there are simpler ones): since $G$ is abelian, the elements of order dividing 2 form a subgroup $H$ of $G$.  On the other hand, an abelian group every element of which is of order dividing $2$ can be thought of as a vector space over the field of 2 elements, and so (since $H$ is finite, and hence has finite dimension) we see that either $H$ is trivial (if its dimension over the field of 2 elements is zero), or else that the number of elements in  $H$ is even (if its dimension is positive).  One element of $H$ is the identity $e$, and so either $H$ is trivial, i.e. $G$ contains no elements of exact order $2$, or else the number of elements in $H \setminus \{e\}$, i.e. the number of elements of exact order 2 in $G$, is odd.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, it is easy to see that when $G$ is odd, the number of elements of exact order 2 is even.  (Count the elements of $G$ by thinking about the orbits of the map $g \mapsto g^{-1}$.)  Thus when $G$ is odd, the group $H$ must indeed be trivial, and so $G$ contains no elements of order 2.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In short, we have avoided an appeal to Lagrange's theorem by instead appealing to a somewhat coarser counting argument, together with linear algebra over the field of 2 elements.  Whether or not this is absurd, the readers can decide!&lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2010-11-07T20:12:39.857" />
  <row Id="9315" PostTypeId="2" ParentId="9311" CreationDate="2010-11-07T20:14:15.273" Score="10" ViewCount="" Body="&lt;p&gt;If don't exist and element with order 2 than you are done. Supose $g\in G$ such that $g^2=e$. Since ${g_1,\ldots, g_n}={gg_1,\ldots,gg_n}$ (the brackets should be here), than $\prod g_i = g^n \prod g_i$ and $g^n=e$. Putting $n=2k+1$, $e=g^{2k+1}=g^{2k}g=g$.&lt;/p&gt;&#xA;" OwnerUserId="1971" LastEditorUserId="1971" LastEditorDisplayName="" LastEditDate="2010-11-07T20:19:31.137" LastActivityDate="2010-11-07T20:19:31.137" CommentCount="5" />
  <row Id="9316" PostTypeId="2" ParentId="9311" CreationDate="2010-11-07T20:17:24.693" Score="3" ViewCount="" Body="&lt;p&gt;It's kind of easy to prove Lagrange's theorem in this case. For any $h \in G$,  $h^{|G|}\prod_{g \in G} g$ $= \prod_{g \in G} hg$. Since $g \rightarrow hg$ permutes the elements of the group, $= \prod_{g \in G} hg = \prod_{g \in G} g$ and&#xA;thus $h^{|G|} = e$. It's a short step from here to say that the order of $h$ divides $|G|$. In any event if $|h|$ were 2, one could write $1 = |G| - 2k$ for some $k$ and get that $h^1 = h^{|G|} (h^2)^{-k} = e$. Surely a Lagrange-free proof wouldn't be much simpler...&lt;/p&gt;&#xA;" OwnerUserId="3035" LastActivityDate="2010-11-07T20:17:24.693" />
  <row Id="9317" PostTypeId="2" ParentId="9313" CreationDate="2010-11-07T20:18:06.520" Score="5" ViewCount="" Body="&lt;p&gt;Use &lt;a href=&quot;http://en.wikipedia.org/wiki/Stirling%27s_approximation&quot; rel=&quot;nofollow&quot;&gt;Stirling's approximation&lt;/a&gt; in $\binom{n}{k} = \frac{n!}{k!(n-k)!}$.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-07T20:18:06.520" />
  <row Id="9318" PostTypeId="2" ParentId="9311" CreationDate="2010-11-07T20:19:41.143" Score="1" ViewCount="" Body="&lt;p&gt;By the classification of finite abelian groups, any abelian group $G$ can be written as $Z_{n_1} \oplus \cdots \oplus Z_{n_k}$ where the $n_i$ are prime powers.  For odd abelian groups, none of the $n_i$ are powers of two. Write each element of the group $G$ as a $k$-tuple $(m_1,\ldots, m_k)$ where $0 \le m_i &amp;lt; n_i$.  Let $N = n_1 n_2 \cdots n_k$ be the order of the group.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then the $j$th component of $\sum_{g \in G} g$ is &#xA;$$ {N \over n_j} \left( \sum_{m_j=0}^{n_j-1} m_j \right) $$&#xA;and the sum is of course $n_j(n_j-1)/2$. This simplifies to $N(n_j-1)/2$. Since $n_j$ is odd this is a multiple of $N$ (as an integer); therefore it's a multiple of $n_j$ (as an integer) and thus $0$ (as an element of $Z_{n_j}$). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I don't have an algebra text at hand; does the proof of the classification use Lagrange's theorem?)&lt;/p&gt;&#xA;" OwnerUserId="173" LastActivityDate="2010-11-07T20:19:41.143" CommentCount="3" />
  <row Id="9319" PostTypeId="1" AcceptedAnswerId="9333" CreationDate="2010-11-07T20:31:36.627" Score="4" ViewCount="317" Body="&lt;p&gt;We were given a challenge by our calculus professor and I've been stuck on it for a while now. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Show that the set of subsequence limits of $A_n=\sin(n)$ is $[-1, 1]$ (another way to phrase this would be: $\forall r\in [-1,1]$ there exists a subsequence of $A_n=\sin(n)$ that converges to $r$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What would be a good way to start?&lt;/p&gt;&#xA;" OwnerUserId="3237" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-07T21:33:21.763" LastActivityDate="2010-11-08T20:28:43.747" Title="sin(n) subsequence limits set" Tags="&lt;calculus&gt;" AnswerCount="2" CommentCount="19" FavoriteCount="1" />
  <row Id="9320" PostTypeId="2" ParentId="9313" CreationDate="2010-11-07T20:33:47.087" Score="2" ViewCount="" Body="&lt;p&gt;The answer to your second question is to read Graham, Knuth, and Patashnik's &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0201558025&quot; rel=&quot;nofollow&quot;&gt;Concrete Mathematics&lt;/a&gt;, at least for simpler practical examples.  For complicated sequences it can be extremely difficult and in many cases open to find asymptotics, e.g. for Ramsey-theoretic sequences, so you'll have to narrow the scope of your question a little to get a reasonable answer.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-11-07T20:33:47.087" CommentCount="1" />
  <row Id="9321" PostTypeId="1" CreationDate="2010-11-07T20:50:55.397" Score="1" ViewCount="235" Body="&lt;p&gt;Hi. I'm reading page number 4 here. In particular the section where it deals with the case $p=\infty$, that is , showing that $L^{\infty}$ is complete.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.core.org.cn/NR/rdonlyres/Mathematics/18-125Fall2003/5E3917E2-C212-463B-9EDB-671486133388/0/18125_lec15.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.core.org.cn/NR/rdonlyres/Mathematics/18-125Fall2003/5E3917E2-C212-463B-9EDB-671486133388/0/18125_lec15.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Two questions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Why is the convergence uniform? where it says &quot;for $x \in N^{c}$ , $f_{n}$ is a Cauchy sequence of complex numbers. Thus $f_{n} \rightarrow f$ uniformly. Clearly we have pointwise convergence but why is it uniform?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) I don't see why $||f_{n} - f||_{\infty} \rightarrow 0$. Can you please explain this step?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="4219" LastActivityDate="2010-11-07T21:22:03.697" Title="Understanding proof of completeness of $L^{\infty}$" Tags="&lt;measure-theory&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="9322" PostTypeId="2" ParentId="9321" CreationDate="2010-11-07T21:22:03.697" Score="3" ViewCount="" Body="&lt;p&gt;The key is that we are working in $N^c$, where we have, morally speaking, defined $N$ to be the set of points where things go wrong. In detail, $N$ is the set of points $x$ where either 1) $f(x)$ is bigger than the limsup $\|f\|$, or 2) the series $(f_n)$ is not Cauchy at $x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Outside of $N$, the terms $f_n$ are bounded and the series $(f_n)$ is Cauchy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For question 1), use the fact that $(f_n)$ is (uniformely!) Cauchy outside of $N$ and that $\mathbb C$ is complete to define a limit function $f$ on the set $N^c$. It should follow, pretty much by definition of $f$, that $f_n \to f$ uniformely on $N^c$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For question 2), then we know from question 1 that $\|f_n - f\|\infty \to 0$ on $N^c$. Next, extend the limit function $f$ to the whole set by setting it equal to 0 on $N$. What is the measure of $N$? How does that play into the definition of the $\| \cdot \|_\infty$ norm, and thus the question about limits?&lt;/p&gt;&#xA;" OwnerUserId="3225" LastActivityDate="2010-11-07T21:22:03.697" />
  <row Id="9323" PostTypeId="1" AcceptedAnswerId="9326" CreationDate="2010-11-07T21:27:43.913" Score="2" ViewCount="177" Body="&lt;p&gt;Suppose that $S_1$ and $S_2$ are the vanishing sets of a system of polynomial equations in n variables over a field $\mathbb{k}$ (ideal in $\mathbb{k}[X_1,\dots,X_n]$) and a system of polynomial equations in m variables over a field $\mathbb{k}$ (ideal in $\mathbb{k}[Y_1,\dots,Y_m]$). We can give $S_1$ and $S_2$ the Zariski topology by letting all algebraic subsets of $S_1$ and $S_2$ to be closed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Surely, if $Y_j\circ f\colon S_1\to\mathbb{k}$ is a polynomial in $\mathbb{k}[X_1,\dots,X_n]/I(S_1)$, for every $j$, then $f$ is continuous in the Zariski topology.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The condition that $Y_j\circ f$ be polynomial, however, seems too strict: as long as the vanishing set of $Y_j\circ f$ were equal to the vanishing set of a polynomial function, $f$ would be continuous, which means that the usual morphisms from $S_1$ to $S_2$ are not the same as the continuous maps between them as (Zariski-)topological spaces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question then is of two parts.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;Is it true that there exist Zarsiki-continuous maps that are not polynomial? &lt;/li&gt;&#xA;  &lt;li&gt;If yes, then what is the geometric way of thinking about the extra structure that polynomial maps preserve, i.e. continuous maps squash subsets of $S_1$ in such a way that non-algebraic subsets never get squashed into algebraic ones, but some algebraic ones may get squashed into non-algebraic ones, but polynomials also give what features to the geometric squashing?&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="400" LastActivityDate="2010-11-07T21:40:46.947" Title="Are there Zariski-continuous maps between algebraic sets that are not polynomial maps?" Tags="&lt;algebraic-geometry&gt;" AnswerCount="3" FavoriteCount="1" />
  <row Id="9324" PostTypeId="1" AcceptedAnswerId="9330" CreationDate="2010-11-07T21:28:10.080" Score="1" ViewCount="121" Body="&lt;p&gt;I've been trying to google the answer to this question, but have had no luck, so I'm asking this question here. Let's say the origin is at (0, 0), body 1 with mass m1 is at (-r1, 0), and body 2 with mass m2 is at (r2, 0). How can I calculate the initial velocities v1 and v2 for the bodies so that they trace circular orbits around each other?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason I'm asking is because I've only been able to get the answer for the specific problem where m1 = m2, but I can't explain it very well, and because I can't explain it very well I can't generalize it to bodies of different masses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for any assistance!&lt;/p&gt;&#xA;" LastEditorUserId="723" LastEditorDisplayName="" LastEditDate="2010-11-07T21:30:27.973" LastActivityDate="2010-11-07T21:59:40.740" Title="two-body problem circular orbits" Tags="&lt;physics&gt;&lt;classical-mechanics&gt;" AnswerCount="2" />
  <row Id="9325" PostTypeId="1" AcceptedAnswerId="32987" CreationDate="2010-11-07T21:30:57.333" Score="4" ViewCount="299" Body="&lt;p&gt;I was wondering if equilibrium distribution, steady-state distribution, stationary distribution and limiting distribution mean the same, or there are differences between?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I learn them in the context of Discrete-time Markov Chain as far as I know. Or they also appear in other situations of stochastic processes and probability?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the Wikipedia page for &lt;a href=&quot;http://en.wikipedia.org/wiki/Markov_chain&quot; rel=&quot;nofollow&quot;&gt;Markov Chain&lt;/a&gt;, there seems not very clear to me in defining and using these concepts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="1281" LastActivityDate="2011-04-14T15:03:53.320" Title="equilibrium distribution, steady-state distribution, stationary distribution and limiting distribution" Tags="&lt;stochastic-processes&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="0" />
  <row Id="9326" PostTypeId="2" ParentId="9323" CreationDate="2010-11-07T21:36:41.907" Score="5" ViewCount="" Body="&lt;p&gt;If I understand your question correctly, then (1) yes, there are going to be lots of Zariski continuous maps that are not polynomial (although, if one asks that all higher products of the map preserve the Zariski topology on the corresponding sources and targets, this is a much stronger restriction, I think --- does anyone know the details?); and (2) the usual extra structure that one adds is the sheaf of regular functions on source and target; polynomial maps pullback regular functions on the target to regular functions on the source.  Thus varieties are naturally thought of as being objects&#xA;in the category of locally ringed spaces (i.e. spaces equipped with a certain structure sheaf of rings) rather than just in the category of topological spaces.&lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2010-11-07T21:36:41.907" />
  <row Id="9327" PostTypeId="2" ParentId="9323" CreationDate="2010-11-07T21:37:33.480" Score="1" ViewCount="" Body="&lt;p&gt;Consider the affine line and Look for an example there yourself.&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2010-11-07T21:37:33.480" />
  <row Id="9328" PostTypeId="2" ParentId="9323" CreationDate="2010-11-07T21:40:46.947" Score="1" ViewCount="" Body="&lt;p&gt;Yes, there exists Zariski continous maps between algebraic varieties that are not polynomials. For example all  bijection between $\mathbb A ^1$ and itself is Zariski continous but not polynomial in general: think of bijection that permutes $0$ and $1$ and leaves other points fixed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Geometric way of thinking is algebraic subsets must be preserved . Only use polynomials and rationnal functions.&lt;/p&gt;&#xA;" OwnerUserId="2872" LastActivityDate="2010-11-07T21:40:46.947" CommentCount="5" />
  <row Id="9329" PostTypeId="2" ParentId="9324" CreationDate="2010-11-07T21:53:04.220" Score="0" ViewCount="" Body="&lt;p&gt;Hello Bob,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Silly question, but have you read the Wikipedia page on the two-body problem (http://en.wikipedia.org/wiki/Two-body_problem)? If I understand your problem correctly, then it seems that it is fairly well treated on that page.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You say you want to determine the set of initial vectors $v_1$ and $v_2$ such that the two bodies trace circular orbits around each other. If $x_1(t)$ and $x_2(t)$ denote the positions of the bodies at time $t$, then do you mean that you want the distance $|x_1(t) - x_2(t)|$ between the two to be equal to some fixed radius $r$? If so, then this seems to be treated in the &quot;Displacement vector motion&quot; section of the Wikipedia page (if the radius $r$ is constant, then its derivative w/r/t time is ...?).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The WP page seems to solve two one-body problems to arrive at a complete description of the system of solutions. In contrast, you only ask for a partial description of the system (i.e. what inital vectors give circular orbits). An alternative solution in your case could be to redefine the origin of your coordinate system to be given by $x_1(t)$ at time $t$; then your job is just to determine which initial vectors give circular orbits of one mass in orbit around another fixed one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second method actually sounds like it's more instructive and fun to do than working through the WP page. I'd think of Isaac Newton and try the second method to start with, then turn to the WP page (or here) if I got stuck. Good luck.&lt;/p&gt;&#xA;" OwnerUserId="3225" LastActivityDate="2010-11-07T21:53:04.220" CommentCount="1" />
  <row Id="9330" PostTypeId="2" ParentId="9324" CreationDate="2010-11-07T21:59:40.740" Score="2" ViewCount="" Body="&lt;p&gt;I will assume the phrase &quot;circular orbits around each other&quot; means that you are looking for a solution where the two bodies remain constant distance apart. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The easiest thing to do is to work in a coordinate system adapted to the centre of mass. The total separation between the two masses is $R = r_1 + r_2$. The centre of mass sits at $(R_0,0)$ where $R_0 = (-m_1 r_1 + m2 r_2)/(m_1 + m_2)$. Without loss of generality, we'll assume that the centre of mass remains immobile (else we can subtract the velocity of the centre of mass from both bodies and use the fact that Galilean motions leave the mechanics invariant). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the distance from $m_1$ to the centre of mass is $|-r_1 - R_0| = R m_2 / (m_1 + m_2)$, similarly the distance from $m_2$ to the centre of mass will be $R m_1/(m_1 + m_2)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, let us write down the equations of motion. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;The assumption that the centre of mass remains immobile implies that the total momentum is 0. This means that $m_1 v_1 + m_2 v_2 = 0$. Furthermore, that $R$ is constant requires that the initial velocities be such that the two masses start out moving perpendicular to the $x$ axis. So hereon we can assume that $v_1, v_2$ are scalars, which represent the initial speed in the $y$ direction for the masses respectively. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Relative to the (immobile) centre of mass, the first mass is travelling in a circular orbit of radius $R m_2 / (m_1 + m_2)$. This requires force $ m_1 v_1^2 / (R m_2 / (m_1 + m_2) ) = v_1^2 \cdot \frac{m_1 (m_1 + m_2)}{R m_2}$. We have also a similar expression for the force acting on $m_2$. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The gravitational attraction between the two masses is $G m_1 m_2 / R^2$. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So now we solve the system&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ m_1 v_1 + m_2 v_2 = 0 $$&#xA;$$ \frac{G m_1 m_2}{R^2} = v_1^2 \frac{m_1 (m_1 + m_2)}{m_2 R} = v_2^2 \frac{m_2(m_1 + m_2)}{m_1R} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which gives&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ v_1 = \sqrt{ \frac{G m_2^2}{R(m_1 + m_2)} }, v_2 = \sqrt{ \frac{G m_1^2}{R(m_1 + m_2)} }$$&lt;/p&gt;&#xA;" OwnerUserId="1543" LastActivityDate="2010-11-07T21:59:40.740" CommentCount="1" />
  <row Id="9332" PostTypeId="2" ParentId="9281" CreationDate="2010-11-07T22:02:36.203" Score="2" ViewCount="" Body="&lt;p&gt;For topological space $X$, open subset $U \subset X$ and point $x\in U$ we have for all sheaf $F$ on $X$ : $F_x = (F|U)_x$. Apply to $X=\mathbb P ^1, U=\mathbb A^1, x=(t-a)$&lt;/p&gt;&#xA;" OwnerUserId="2872" LastActivityDate="2010-11-07T22:02:36.203" CommentCount="2" />
  <row Id="9333" PostTypeId="2" ParentId="9319" CreationDate="2010-11-07T22:04:56.157" Score="0" ViewCount="" Body="&lt;p&gt;Following Qiaochu hint, I'll try to elaborate a bit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Note:&lt;/strong&gt; This is a complete rewrite of the proof to fix a flaw pointed out by Qiaochu and make it overall clearer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First some notation. Let $S^1 = [0, 2\pi)$ with 0 and $2\pi$ identified. For two points $a, b \in S^1$ we'll denote by $a \oplus b = a + b \mod 2\pi$ (and likewise for $\ominus$) and let $A_n = \{n \mod 2\pi\}_{n=1}^{\infty}$. Also denote by $U_{\varepsilon}(p)$ a punctured $\varepsilon$-neighborhood of $p \in S^1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Our strategy will be to show that $\{A_n\}$ has at least one limit point in $S^1$. From this we will conclude that 0 is also a limit point and finally we will use this fact to show that every point in $S_1$ is a limit point of $A_n$ (this means that $\{A_n\}$ is dense in S^1). Having established this, we will use continuity of $\sin$ to finally resolve the problem.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a preparation we will note the relation $A_{n+m} = A_n \oplus A_m \quad (1)$. As a corollary of this we have that the sequence $\{A_n\}$ is injective (for if not, there would exist $n, m \in \mathbb{N}, k \in \mathbb{Z} \quad n&gt;m$  such that $A_n = A_m$ and so $n - m = 2 k \pi$, a contradiction with irrationality of $\pi$). This implies simple but crucial fact that the image of the sequence contains infinitely many points.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, to establish the density of ${A_n}$ in $S^1$ we will first show that there exists at least one limit point $p \in S^1$. This is established by a standard argument: consider intervals $I = [0, \pi]$ and $J = [\pi, 2\pi]$ and take the one which contains infinitely many points of $\{A_n\}$ (if both do, take the &quot;bottom&quot; one, i.e. $I$). Call this interval $I_0$. Now again divide it in two intervals of same length and let $I_1$ be the one with infinitely many points. Continue in this way to obtain an infinite sequance of intervals $I_0 \supset I_1 \supset \cdots$. Then the set  $K = \cap_{n=0}^{\infty} I_n$ is non-empty (this should be covered in standard calculus course, I hope) and any point $p \in K$ is surely a limit point (by construction of $\{I_n\}$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks to the above we now know that for each $\varepsilon &gt; 0$ there exist infinitely many points in $U_\varepsilon(p)$, i.e. infinitely many $n, m \quad n &gt; m$ such that $|A_n - A_m - 2k\pi| &amp;lt; 2\varepsilon$ for some $k \in \mathbb{Z}$. But from (1) we get that $|A_{n-m} - 2k\pi| &amp;lt; 2\varepsilon$. Because we identify 0 and $2\pi$ in $S^1$ we can see that these differences are concentrated around 0. Therefore we have shown that 0 is also a limit point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we will show that every $p \in S^1$ is a limit point. Suppose we are given $\varepsilon &gt; 0$. We just take any $A_k$ from $U_{\varepsilon}(0)$ and note that for $l = \lfloor {p \over A_k} \rfloor$ we have $A_{lk} \in U_{\varepsilon}(p)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To conclude the answer we will show that any $r \in [-1, 1]$ is indeed a limit point of $\{\sin(n)\}$. Take $s = \arcsin (r) \mod 2\pi$. Then by the above, we know that $s$ is a limit point of $\{A_n\}$ so there is a subsequence $\{A_{n_k}\} \to s$ and observe by continuity of $\sin$ that $\{\sin(A_{n_k})\} = \{\sin(n_k)\} \to sin(s) = r$.&lt;/p&gt;&#xA;" OwnerUserId="3223" LastEditorUserId="3223" LastEditorDisplayName="" LastEditDate="2010-11-08T18:55:36.713" LastActivityDate="2010-11-08T18:55:36.713" CommentCount="15" />
  <row Id="9335" PostTypeId="1" CreationDate="2010-11-07T22:28:11.550" Score="1" ViewCount="213" Body="&lt;p&gt;Hey,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How to prove $\limsup(\{A_n \cup B_n\}) = \limsup(\{A_n\}) \cup \limsup(\{B_n\})$? Thanks!&lt;/p&gt;&#xA;" LastEditorUserId="630" LastEditorDisplayName="" LastEditDate="2010-11-08T20:16:58.143" LastActivityDate="2010-11-08T20:16:58.143" Title="Proof: Limit superior intersection" Tags="&lt;homework&gt;&lt;limit&gt;&lt;elementary-set-theory&gt;" AnswerCount="2" CommentCount="7" />
  <row Id="9336" PostTypeId="2" ParentId="9335" CreationDate="2010-11-07T22:33:20.690" Score="3" ViewCount="" Body="&lt;p&gt;Use the definition, and double inclusion; that is, show that every element of $\limsup(A\cup B)$ must be either an element of $\limsup(A)$ or of $\limsup(B)$; then show that every element of $\limsup(A)$ must be in $\limsup(A\cup B)$ and that every element of $\limsup(B)$ must be in $\limsup(A\cup B)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, one must assume that you mean your &quot;$A$&quot; to be a sequence of sets and your &quot;$B$&quot; to likewise be a sequence of sets... Otherwise, what you write does not really make much sense  (limit superior and limit inferior of a single set is not usually defined).&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-07T22:42:52.880" LastActivityDate="2010-11-07T22:42:52.880" CommentCount="4" />
  <row Id="9337" PostTypeId="2" ParentId="9335" CreationDate="2010-11-07T22:46:54.280" Score="0" ViewCount="" Body="&lt;p&gt;Another nice way is to use characteristic functions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The map $\chi : \mathcal{P}(\Omega) \to \{0,1\}^\Omega$ assigns to every subset of $\Omega$ its characteristic function.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\chi$ is bijective.&lt;/li&gt;&#xA;&lt;li&gt;$\chi$ is continuous, i.e. $\chi_{\lim\sup_{n\to\infty} A_n} = \lim\sup_{n\to\infty}\, \chi_{A_n}$ (pointwise limit)&lt;/li&gt;&#xA;&lt;li&gt;$\chi$ is a homomorphism, i.e. $\chi_{A \cup B} = \chi_A + \chi_B - \chi_A \chi_B$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Now your question reduces to the computation of an ordinary limit.&lt;/p&gt;&#xA;" OwnerUserId="474" LastActivityDate="2010-11-07T22:46:54.280" />
  <row Id="9338" PostTypeId="1" CreationDate="2010-11-07T22:49:02.610" Score="1" ViewCount="197" Body="&lt;p&gt;Say I have the following numbers:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;10&#xA;10&#xA;9&#xA;5&#xA;4&#xA;4&#xA;4&#xA;1&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;How do i go about calculating the percentile for each score?  Is there a standard formula for figuring this out?&lt;/p&gt;&#xA;" OwnerUserId="2490" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-08T05:56:11.100" LastActivityDate="2010-12-08T05:56:11.100" Title="How do you calculate percentile?" Tags="&lt;statistics&gt;&lt;algorithms&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="9339" PostTypeId="2" ParentId="9338" CreationDate="2010-11-07T22:54:12.370" Score="2" ViewCount="" Body="&lt;p&gt;The $x$th percentile is the point where $x$ percent of the data lies below. Here you have 8 data points; $1$ is in the bottom $12.5\%$, so it is the 12.5-percentile. $4$ is the (top of the) bottom $50\%$ so it corresponds to the 50 percentile. $5$ is at the top of the bottom $62.5\%$, so it corresponds to the 62.5 percentile. Etc.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-11-07T22:54:12.370" />
  <row Id="9340" PostTypeId="1" CreationDate="2010-11-07T23:09:10.670" Score="0" ViewCount="160" Body="&lt;p&gt;Hello, I am trying to reproduce the following bound:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sum_{1\leq m\leq x, m\neq n}\frac{1}{|\log(m/n)|}=O(x\log(x))$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for $x\geq 2$ and some $n$, $1\leq n\leq x$ (the implicit constant shouldn't depend on $n$). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have tried to bound each term by $\frac{1}{|\log(1\pm 1/x)|}$, but this doesn't seem to be good enough. Splitting the sum at $\log(x)$ also didn't help. Any ideas? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you very much for your comments or hints.&lt;/p&gt;&#xA;" OwnerUserId="3244" LastEditorUserId="3244" LastEditorDisplayName="" LastEditDate="2010-11-08T00:39:43.287" LastActivityDate="2011-01-14T05:05:47.730" Title="Bounding the series $\sum_{m\leq x,m\neq n}\frac{1}{|\log(m/n)|}$" Tags="&lt;sequences-and-series&gt;&lt;analytic-number-theory&gt;&lt;logarithms&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="9343" PostTypeId="2" ParentId="9340" CreationDate="2010-11-07T23:49:16.007" Score="1" ViewCount="" Body="&lt;p&gt;This answer corresponds to the original question, where $n$ was assumed fixed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Considering the corresponding integral $\int_a^x {\frac{{du}}{{\log (u/n)}}}  = n\int_{a/n}^{x/n} {\frac{{du}}{{\log (u)}}} $ (where $a$ is some constant) and the fact that the logaritmic integral function li(x) is $O(x/\log(x))$ as $x \to \infty$, your bound should be $O(x/\log(x))$ (and not $O(x\log(x))$.&lt;/p&gt;&#xA;" OwnerUserId="2810" LastEditorUserId="2810" LastEditorDisplayName="" LastEditDate="2010-11-08T00:54:02.823" LastActivityDate="2010-11-08T00:54:02.823" CommentCount="1" />
  <row Id="9344" PostTypeId="2" ParentId="9302" CreationDate="2010-11-07T23:50:35.250" Score="4" ViewCount="" Body="&lt;p&gt;Using the &lt;a href=&quot;http://en.wikipedia.org/wiki/Spectral_theorem#Hermitian_matrices&quot; rel=&quot;nofollow&quot;&gt;spectral theorem&lt;/a&gt; to obtain an orthogonal basis of eigenvectors for the matrix is probably the best approach, similar to what Ross said, given your criterion of &quot;simple concepts of linear algebra&quot;.  Here is another approach I thought worth mentioning.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The largest of the absolute values of the eigenvalues of a matrix $A$ is called the spectral radius of $A$, which I'll denote $r(A)$.  In general one allows complex eigenvalues (in part because there are real matrices with no real eigenvalues, like &#xA;$\begin{bmatrix} 0 &amp;amp; 1 \\ -1 &amp;amp; 0 \end{bmatrix}$), but fortunately it turns out that real symmetric matrices can have only real eigenvalues, so it makes no difference here.  The inequality $r(A)\leq \|A\|$ is easy, because if $\lambda$ is an eigenvalue for $A$ and $v$ is a corresponding (nonzero) eigenvector, then $\|Av\|=\|\lambda v\| = |\lambda|\|v\|$, which shows that $\|A\|\geq |\lambda|$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remarkably, there is a &lt;a href=&quot;http://en.wikipedia.org/wiki/Spectral_radius#Theorem_.28Gelfand.27s_formula.2C_1941.29&quot; rel=&quot;nofollow&quot;&gt;formula&lt;/a&gt; for the spectral radius in terms of the norm, $r(A)=\lim_{k\to\infty}\|A^k\|^{1/k}$.  I'll note before going on that everything I've said so far is independent of the norm on the space on which matrices act (that is, independent of which &lt;a href=&quot;http://en.wikipedia.org/wiki/Operator_norm&quot; rel=&quot;nofollow&quot;&gt;operator norm&lt;/a&gt; you choose for the matrices).  But now, assuming the Euclidean norm (which is implicit in your question), the matrix norm has the property that $\|A^2\|=\|A\|^2$ for a real symmetric matrix $A$.  Inductively you get $\|A^{2^k}\|=\|A\|^{2^k}$, and applying the spectral radius formula yields $r(A)=\lim_{k\to\infty}\|A^{2^k}\|^{1/2^k}=\lim_{k\to\infty}\|A\|=\|A\|$.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2010-11-07T23:59:01.480" LastActivityDate="2010-11-07T23:59:01.480" />
  <row Id="9345" PostTypeId="2" ParentId="9311" CreationDate="2010-11-07T23:50:41.807" Score="6" ViewCount="" Body="&lt;p&gt;If finite abelian group $\rm G$ has an elt $\rm\: j\: $ of order $\:2\:$ then $\rm\:\ \:g\ \to\ j\: g\ \ $ pairs its elts so $\rm G$ has even order. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a special case of the often useful fact that the cardinalities of a finite set and its fixed-point set under an involution have equal parity, since the non-fixed points are paired by the involution. Hence, as above, when there are no fixed points ($\rm\: j\ne 1\: \Rightarrow\: j\:g\ne g\: $) the set has even cardinality.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Such simple symmetries often lie at the heart of elegant proofs, e.g. the famous &lt;a href=&quot;http://mathoverflow.net/questions/31113/zagiers-one-sentence-proof-of-fermats-theorem&quot; rel=&quot;nofollow&quot;&gt;Heath-Brown-Zagier proof&lt;/a&gt; that every prime $\rm\:\equiv 1\ (mod\ 4)\ $ is a sum of two squares.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-11-08T01:38:39.527" LastActivityDate="2010-11-08T01:38:39.527" />
  <row Id="9348" PostTypeId="2" ParentId="9274" CreationDate="2010-11-08T00:36:13.643" Score="0" ViewCount="" Body="&lt;p&gt;The easiest way (if you can apply Cantor-Bernstein-Schroder theorem) is to find a bijection from $[0,1]$ onto a subset, say $[1/4,3/4]$; that is easy to do. Since the identity map from $(0,1)$ to $[0,1]$ is a bijection, it follows from the theorem that the two sets are isomorphic.&lt;/p&gt;&#xA;" OwnerUserId="3249" LastActivityDate="2010-11-08T00:36:13.643" CommentCount="1" />
  <row Id="9349" PostTypeId="1" CreationDate="2010-11-08T00:40:58.947" Score="1" ViewCount="93" Body="&lt;p&gt;Hi. Can someone simply explain to me how to calculate linear integral linke below?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_{L} 5y \mathrm{d}L$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where L is line segment from (0;0) to (0,2;0,2).&lt;/p&gt;&#xA;" OwnerUserId="3250" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2010-11-08T05:38:28.693" LastActivityDate="2010-11-08T05:38:28.693" Title="Liner integral ?" Tags="&lt;homework&gt;&lt;integral&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" />
  <row Id="9350" PostTypeId="1" CreationDate="2010-11-08T00:45:52.497" Score="1" ViewCount="104" Body="&lt;p&gt;Is there some increasing function $f(n)$ that grows slower than $n^{c}$ for some $c &gt; 1$ such that $\sum_{n=1}^{\infty} \frac{1}{f(n)}$ converges? &lt;/p&gt;&#xA;" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-08T00:50:21.790" LastActivityDate="2010-11-08T00:52:31.807" Title="Smallest function whose inverse converges" Tags="&lt;real-analysis&gt;&lt;sequences-and-series&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="9351" PostTypeId="1" CreationDate="2010-11-08T00:47:00.327" Score="2" ViewCount="215" Body="&lt;p&gt;This is a bit of a silly question, but I've been puzzled on how this can be done, so I ask here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are given an implicit Cartesian equation like&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$4 x^4-4 x^3+8 y^2 x^2-27 x^2+12 y^2 x+4 y^4-27 y^2+27=0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x^6+3 y^2 x^4-3 x^4+3 y^4 x^2+21 y^2 x^2+3 x^2+y^6-3 y^4+3 y^2-1=0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and are told that you are not allowed to graph the equation (which makes the determination of geometric properties a bit difficult), but are allowed to do manipulations to the equation. &lt;strong&gt;Is it possible to determine the symmetry of the graph of a curve only from its equation?&lt;/strong&gt; Or to be more precise, can one determine $n$ in the sentence &quot;the curve with implicit Cartesian equation so-and-so has n-fold symmetry&quot; just from the implicit equation?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For some curves, one quick way of determining symmetry is to transform to polar coordinates: for instance,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(x^2+y^2)^2=x^2-y^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;transforms into the polar equation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$r^2=\cos\;2\theta$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and due to the periodicity of the cosine, it is seen that it has two-fold symmetry (i.e., invariant under rotations of $\frac{2\pi}{2}=\pi$ radians.) However, we all know that not all algebraic curves can undergo such a neat conversion, and that this method presumes that the origin is &quot;in the vicinity&quot; of the curve. (On the other hand, if you determine through the usual methods that the curve has the origin as a singular point, conversion to polar coordinates is especially attractive!)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, I would like to see if a more general method is available.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Tangentially, we know that only the circle has infinitely many axes of symmetry, so if the method can also demonstrate that only the circle has this property, that would be a plus.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Last but not the least (and since I feel it does not deserve to be asked as a separate question): is there a bounded algebraic curve that has no axes of symmetry whatsoever? Most of the examples I see have at least two-fold symmetry.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-11-08T01:18:13.343" Title="Determining the symmetry of a plane algebraic curve from its Cartesian equation." Tags="&lt;algebraic-geometry&gt;&lt;plane-curves&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="9352" PostTypeId="2" ParentId="9350" CreationDate="2010-11-08T00:49:38.407" Score="4" ViewCount="" Body="&lt;p&gt;Yes, $f(n)=n(\log(n))^2$.  (Of course you can ignore the first term where the reciprocal is undefined, and the finitely many terms where it may be bigger than $n^c$ for $c$ close to $1$.)&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-11-08T00:49:38.407" />
  <row Id="9353" PostTypeId="2" ParentId="9350" CreationDate="2010-11-08T00:52:31.807" Score="4" ViewCount="" Body="&lt;p&gt;The answer to your title question, however, is no.  That is, given an increasing function $f(n) : \mathbb{N} \to \mathbb{R}$ such that $\sum \frac{1}{f(n)}$ converges, there exists an increasing function $g(n)$ such that &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\sum \frac{1}{g(n)}$ converges, and&lt;/li&gt;&#xA;&lt;li&gt;$\lim_{n \to \infty} \frac{g(n)}{f(n)} = 0$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;(Unfortunately the proof escapes me at the moment.)&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-11-08T00:52:31.807" CommentCount="4" />
  <row Id="9354" PostTypeId="1" AcceptedAnswerId="9371" CreationDate="2010-11-08T01:04:26.440" Score="4" ViewCount="85" Body="&lt;p&gt;In several &quot;proof-theory-for-dummies&quot;-type texts I have read, $I\Delta_0+Exp$ or a theory equivalent to it shows up as a &quot;base&quot; theory, though in what sense it is minimal is not clearly addressed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I realise this is a vague question but if I had the knowledge to make it more specific, I wouldn't need to ask it in the first place.&lt;/p&gt;&#xA;" LastActivityDate="2010-11-08T03:39:55.447" Title="What is special about elementary recursive arithmetic?" Tags="&lt;logic&gt;" AnswerCount="1" />
  <row Id="9355" PostTypeId="1" AcceptedAnswerId="10079" CreationDate="2010-11-08T01:06:52.317" Score="8" ViewCount="319" Body="&lt;p&gt;Consider a random walk on an undirected, non-bipartite graph. Let $\pi$ be the stationary distribution of this process, and let the hitting time $H(s,t)$ be the expected time until a walk beginning at node $s$ reaches node $t$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I learned from &lt;a href=&quot;http://www.cs.yale.edu/publications/techreports/tr1029.pdf&quot; rel=&quot;nofollow&quot;&gt;Random walks on graphs: a survey,&lt;/a&gt; by L. Lovasz, that the quantity &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \sum_{t} \pi(t) H(s,t)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is independent of $s$. In the Lovasz survey, this falls out as a byproduct of a lengthy calculation. I have two questions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(i) Is there a simple (calculation-free, combinatorial) proof of this statement?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(ii) In what generality does this statement hold? Random walks on undirected graphs are a very particular kind of Markov process. What conditions on the probability transition matrix of the process do you need for the above to hold?&lt;/p&gt;&#xA;" OwnerUserId="2120" LastActivityDate="2010-11-13T21:59:40.457" Title="About a weighted sum of hitting times for random walks on graphs" Tags="&lt;probability&gt;&lt;stochastic-processes&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="2" />
  <row Id="9356" PostTypeId="2" ParentId="9351" CreationDate="2010-11-08T01:12:46.820" Score="2" ViewCount="" Body="&lt;p&gt;I'll assume we're talking about Euclidean symmetries.  Then we're looking at transformations of the form $x \mapsto ax + by + c, y = -bx + ay + d$ where $a^2 + b^2 = 1$ (or the analogous set of equations for orientation-reversing symmetries).  The condition $k f(x, y) = f(ax + by + c, -bx + ay + d)$ for some nonzero $k$ (which can be determined in terms of $a, b, c, d$ by evaluating both sides at a point where $f(x, y) \neq 0$), when equating coefficients, into a system of polynomial equations in $a, b, c, d, k$.  Put this together with the equation $a^2 + b^2 = 1$ and you have a system you can plug into a CAS which will, using Grobner basis methods, tell you what the solutions look like.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I am assuming that $f$ has a certain nondegeneracy property here, namely that $f$ generates the ideal of polynomial functions vanishing on $f(x, y) = 0$ in $\mathbb{R}[x, y]$.)&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-11-08T01:18:13.343" LastActivityDate="2010-11-08T01:18:13.343" CommentCount="1" />
  <row Id="9357" PostTypeId="1" AcceptedAnswerId="9358" CreationDate="2010-11-08T01:54:17.537" Score="4" ViewCount="214" Body="&lt;p&gt;Further to my understanding &lt;a href=&quot;http://math.stackexchange.com/questions/7743/getting-better-at-proofs&quot;&gt;What makes a proof&lt;/a&gt;, I'm doing exercises.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The purpose of this question is to compare the proof a certain textbook provides with the one that is &lt;em&gt;logical to me&lt;/em&gt;, that I made up.  If my proof is weak, &lt;em&gt;why is it weak&lt;/em&gt;?  Why is the proof from the book acceptable (other than it finds a way to involve $\lim_{h \to 0}$ :)...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Without further ado the question:  Prove &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$D_x(\int_a^x f(t)dt) = f(x)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My proof is this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $\int f(t)dt = F(t)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then $D_x( \int_a^x{f(t)dt} ) = D_x( \left[ F(t) \right]_a^x )$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$= D_x( F(x) - F(a) )$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But since F(a) is constant with respect to x, it goes to 0 when derived:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$= F'(x) $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$= f(x) $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is that good?  If not, specifically what is wrong with it?  I know proofs have to do with &quot;your audience&quot;, but a lot of proofs are written in assembly language -- they're so low level!&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The proof in the book I'm using says&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Let $h(x) = \int_a^x{f(t)dt}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$h(x + \Delta x) - h(x) = \int_a^{x+\Delta x}{f(t)dt} - \int_a^x{f(t)dt}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$= \int_a^x{f(t)dt} + \int_x^{x+\Delta x}{f(t)dt} - \int_a^x{f(t)dt}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$= \int_x^{x+\Delta x}{f(t)dt}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$= \Delta x (f(x*))$ for some x* between x and x + $\Delta$ x by the mean value theorem for integrals&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, dividing both sides by $\Delta x$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{h(x + \Delta x) - h(x)}{\Delta x} = f(x*)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore $D_x(\int_a^x f(t)dt) = D_x(h(x))$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$= \lim_{\Delta x \to 0} \frac{h(x + \Delta x) - h(x)}{\Delta x} $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$= \lim_{\Delta x \to 0} f(x*) $ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But as $\Delta x \to 0$, $x + \Delta x \to x$ so $x* \to x$ (since $x*$ is between $x$ and $x + \Delta x$).  Since f is continuous, $\lim_{x \to 0} f(x*) = f(x)$&lt;/p&gt;&#xA;" OwnerUserId="1069" LastActivityDate="2010-11-08T13:05:59.287" Title="Prove $D_x(\int_a^x f(t)dt) = f(x)$" Tags="&lt;calculus&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="9358" PostTypeId="2" ParentId="9357" CreationDate="2010-11-08T02:02:06.640" Score="8" ViewCount="" Body="&lt;p&gt;Your first line has a serious problem: $\int f(t)dt$ is not a function of $t$; the indefinite integral is a &lt;em&gt;family&lt;/em&gt; of functions, not a single function. So saying &quot;Let $\int f(t)dt= F(t)$&quot; does not make sense. So as soon as you try to get started, you are already in trouble. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, you can say, &quot;well, &lt;em&gt;pick&lt;/em&gt; some antiderivative instead of taking the whole family.&quot; But if you try to define $F$ as &quot;pick some antiderivative of $f$&quot;, then your problem is that you have no way to guarantee that there &lt;em&gt;is&lt;/em&gt; such a thing in the first place! The whole point of this result is showing that there &lt;em&gt;is&lt;/em&gt; an antiderivative, so you cannot assume there is one to start with.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, if you assume the First Fundamental Theorem of Calculus &lt;em&gt;and&lt;/em&gt; that $f(t)$ has an antiderivative, then this result is very easy, exactly as you do it: $\int_a^x f(t)dt = F(x)-F(a)$, so the derivative with respect to $x$ is $F'(x)=f(x)$. But this assumes that there is an antiderivative for $f$ and that the FTC (Part 1) holds; but since this theorem is often used to prove Part 1, that could also make your argument circular.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, really, the major flaw is that you are assuming that there is an antiderivative for $f$ in the first place (you can prove FTC (Part 1) without this result, so the circularity problem is not fatal). &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-08T05:13:29.953" LastActivityDate="2010-11-08T05:13:29.953" />
  <row Id="9359" PostTypeId="2" ParentId="9349" CreationDate="2010-11-08T02:14:00.127" Score="3" ViewCount="" Body="&lt;p&gt;Assuming you mean to calculate the line integral $\int_L {5y \, {\rm d}s}$, where $L$ is the segment from $(0,0)$ to $(0.2,0.2)$, then define a bijective parametrization $r:[0,0.2] \to L$ by $r(t)=(t,t)$. Also let $f(x,y)=5y$. Then, &#xA;$$&#xA;\int_L {f\,{\rm d}s}  = \int_0^{0.2} {f(r(t))|r'(t)|\,{\rm d}t}. &#xA;$$&#xA;Here, $f(r(t)) = f(t,t) = 5t$ and $|r'(t)| = |(1,1)| = \sqrt{2}$.&#xA;Thus, $\int_L {5y \, {\rm d}s} = \sqrt 2 \int_0^{0.2} {5t \, {\rm d}t} $.&lt;/p&gt;&#xA;" OwnerUserId="2810" LastEditorUserId="2810" LastEditorDisplayName="" LastEditDate="2010-11-08T02:42:18.763" LastActivityDate="2010-11-08T02:42:18.763" CommentCount="1" />
  <row Id="9360" PostTypeId="2" ParentId="9252" CreationDate="2010-11-08T02:17:24.157" Score="8" ViewCount="" Body="&lt;p&gt;Okay, here is an answer assuming plausible conjectures about linear equations in primes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Lemma:&lt;/strong&gt; For any real numbers $(a,b)$ with $0 &amp;lt; a &amp;lt; b &amp;lt; 1$, there is a positive integer $n$ such that $\phi(n)/n \in (a,b)$. Moreover, for any integer $k$, we can arrange that $n$ is relatively prime to $k$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof:&lt;/strong&gt; Let $N$ be larger than $k$, and large enough that $b(1-1/N)&gt;a$. Consider the product $\prod_{p &gt;N} (1-1/p)$, where the product runs over primes greater than $N$. This infinite product is $0$, because $\sum 1/p=\infty$. So there is some $M$ such that $\prod_{N &amp;lt; p &amp;lt; M} p$ is less than $b$. Choose the first such $M$ for which this happens. Then $\prod_{N &amp;lt; p &amp;lt; M} p &gt; b (1-1/M) &gt; a$ by the choice of $M$. Take $n=\prod_{N&amp;lt; p &amp;lt; M} p$. QED&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, to answer your question. Choose intervals $(a_1, b_1)$ and $(a_2, b_2)$ with $0 &amp;lt; a_i &amp;lt; b_i &amp;lt; 1$ such that $x &amp;lt; a_2/b_1 &amp;lt; b_2/a_1 &amp;lt; y$. Using the lemma, find $n_1$ and $n_2$ such that $\phi(n_i)/n_i \in (a_i, b_i)$, the $n_i$ are both odd, and $GCD(n_1, n_2) = 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if you believe the &lt;a href=&quot;http://arxiv.org/abs/math/0606088&quot; rel=&quot;nofollow&quot;&gt;generalized Hardy-Littlewood conjecture&lt;/a&gt; then there are infinitely many pairs  $(q_1, q_2)$ of prime numbers such that $n_1 q_1 +2 = n_2 q_2$. For $q_1$ large enough, $n:=n_1 q_1$ has the desired property.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, this equation is one of the ones to which Green and Tao's results do not apply. (In their terminology, $\psi_1$ and $\psi_2$ are linearly dependent.) You might be able to dodge this issue by asking for $q_1$ and $q_2$ to be almost prime rather than prime. (Almost prime meaning a product of two large primes.) I'm not enough of an expert to know whether that trick would work in this setting; if you seriously care about the question, you might try e-mailing Terry Tao.&lt;/p&gt;&#xA;" OwnerUserId="448" LastEditorUserId="448" LastEditorDisplayName="" LastEditDate="2010-11-08T02:57:11.390" LastActivityDate="2010-11-08T02:57:11.390" />
  <row Id="9361" PostTypeId="1" AcceptedAnswerId="9362" CreationDate="2010-11-08T02:21:09.180" Score="2" ViewCount="107" Body="&lt;p&gt;I had hoped that this would be easier.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have power series that I'm using.  I want to know what simple structure describes the operations that I'm performing on them.  It would be better if I could find something that non-mathematicians (actually computer scientists) would understand, but I'd like to know the technical terminology too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, let me define three power series, $f(x)$, $g(x)$, and $h(x)$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f(x) = \sum_{i=m_1}^{n_1}{c_i x^i}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$g(x) = \sum_{i=m_1}^{n_1}{d_i x^i}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$h(y) = \sum_{i=m_2}^{n_2}{e_i y^i}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then the two operations that I perform are as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) Subtraction $r(x) = f(x) - g(x)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$r(x) = \sum_{i=m_1}^{n_1}{c_i - d_i}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) Multiplication $r(x,y) = f(x)h(y)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$r(x,y) = \sum_{i=m_1}^{n_1}{\sum_{j=m_2}^{n_2}{c_i \cdot e_j}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All of the coefficients and variables are integers. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm writing a paper, and I'm searching for a more &quot;definite&quot; description than what I have.  It is somewhat ackward to describe, and I want to convey this information exactly and concisely.  Plus, I don't want the reader to have any doubts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've been considering that I should just write out the math, since it seems to leave less doubt.&lt;/p&gt;&#xA;" OwnerUserId="2626" LastEditorUserId="2626" LastEditorDisplayName="" LastEditDate="2010-11-08T03:23:03.410" LastActivityDate="2010-11-08T03:33:45.303" Title="What structure is this?" Tags="&lt;calculus&gt;&lt;soft-question&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="1" />
  <row Id="9362" PostTypeId="2" ParentId="9361" CreationDate="2010-11-08T02:24:32.317" Score="0" ViewCount="" Body="&lt;p&gt;The first is $\rm\ (f-g)(1)\ =\ f(1)-g(1)\ $ and the second is $\rm\ (f\:h)(1)\ =\ f(1)\ h(1)$&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-11-08T03:33:45.303" LastActivityDate="2010-11-08T03:33:45.303" CommentCount="5" />
  <row Id="9363" PostTypeId="2" ParentId="154" CreationDate="2010-11-08T03:05:26.850" Score="0" ViewCount="" Body="&lt;p&gt;Here is a possible line of reasoning.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Say, we start with natural numbers, $\mathbb{N}$, then add $0$, making them the whole numbers $\mathbb{N}_0$, then add negative numbers, making them integers $\mathbb{Z}$, then expand them to rationals $\mathbb{Q}$, and finally to reals $\mathbb{R}$. In some respect, we keep filling up the line, till there are no gaps left. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the question: why do we call them &quot;numbers&quot;? What are the properties we expect from some object so we can call it a number. In fact, just a few: addition and multiplication must be defined for all of them, and we want these operations to be commutative and associative.  The distributive law of multiplication over addition must hold. It's nice to have a notion of ordering, so we can say $a \lt b$, and multiplication by some positive number must be compatible with that notion: if $a \lt b$ and $x$ is positive, then $ax \lt bx$. Those are the minimal requirements, met by $\mathbb{N}$. As we expand our numbering system, we get more and more useful properties, and the structure (both algebraic and topological) gets more complicated: from semigroup in case of $\mathbb{N}$ to complete ordered field in case of $\mathbb{R}$, but those basic ones always hold. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the next step in our &quot;quest for expanding our system&quot; would be to look at $\mathbb{R}^2$, $\mathbb{R}^3$ etc. In general, the objects (&quot;points&quot;) populating the space $\mathbb{R}^n$ are called &quot;vectors&quot; and they differ from numbers in one important respect: we can't define vector multiplication so it meets our requirements for numbers (commutativity, associativity and having multiplication distributed over addition). So we don't treat vectors in $\mathbb{R}^n$ as the numbers, in particular, we don't think of them as atomic objects, but rather in terms of the components or coordinates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's one important exception, however: i is $\mathbb{R}^2$. In fact we &lt;em&gt;can&lt;/em&gt; define multiplication of two objects (points, vectors) in $\mathbb{R}^2$ so it meets all our requirements and we can call these objects &quot;numbers&quot;. To be precise, all except one: the ordering. Multiplication we define is no longer compatible with the ordering. So we agree that probably not having that feature isn't such a big deal. Once we introduced multiplication, we end up with complete field (but not ordered field) of entities which we have a right to call the numbers. So we call them the complex numbers and designate them as $\mathbb{C}$. There are some similarities between $\mathbb{R}^2$ and $\mathbb{C}$, but there are also some difference, the most important one being the fact that multiplication is defined so $\mathbb{C}$ is a field rather than vector space. In fact we can treat the elements of $\mathbb{C}$ as the scalars and use them to build a vector space, a complex matrices, etc. Well, this is roughly, why we can call complex numbers &quot;numbers&quot;. They exhibit the same behavior (except for ordering) as the numbers we're familiar with do. One of my favorite examples: when we define the derivative in vector analysis, the formulas look quite different from those we got used in Calculus. But when we define the derivative for complex-valued functions, the formula is exactly the same, because multiplication and hence division) is defined. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, why complex numbers are important is a different question. What I wanted to focus is the fact that we have a right to call them &quot;numbers&quot;. But, just like we can think of rational numbers as a special case of reals, and when move from reals to rationals, we loose the important property of completeness, we &lt;em&gt;can&lt;/em&gt; think of real numbers as a special case of complex numbers, and when we move from complex to reals we loose some nice properties as well. In many respects, studying complex numbers helps us to better understand real numbers. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a sidenote, again, when we move from $\mathbb{R}$ to $\mathbb{C}$, we loose the ordering. Complex numbers form a complete field, but this is not an ordered field. (There is only one complete ordered field, up to isomorphism.) Now suppose we want to explore $\mathbb{R}^n$ further. That's the next best? $\mathbb{R}^4$. We can define multiplication of objects (vectors) in $\mathbb{R}^4$ which is associative, but - alas - not commutative. It's kind of hard to think of numbers whose multiplication is not commutative. So we no longer call them (quaternions, $\mathbb{H}$) numbers. Finally, the next interesting space is $\mathbb{R}^8$, where we still can define multiplication, but it is no longer associative (in addition to being non-commutative). It's &lt;em&gt;really&lt;/em&gt; hard to deal with something which is not associative, and certainly it's hard to think of those entities as numbers. &lt;/p&gt;&#xA;" OwnerUserId="2450" LastActivityDate="2010-11-08T03:05:26.850" CommunityOwnedDate="2011-04-15T20:45:29.080" />
  <row Id="9365" PostTypeId="1" CreationDate="2010-11-08T03:19:06.777" Score="1" ViewCount="169" Body="&lt;p&gt;In a Cartesian system, I've got the slope, start point and distance of a line segment.  What's the formula to find the endpoint?&lt;/p&gt;&#xA;" OwnerUserId="3259" LastActivityDate="2010-11-08T04:07:50.827" Title="Endpoint of a line knowing slope, start and distance" Tags="&lt;geometry&gt;" AnswerCount="2" />
  <row Id="9366" PostTypeId="1" AcceptedAnswerId="9368" CreationDate="2010-11-08T03:20:52.150" Score="1" ViewCount="72" Body="&lt;p&gt;Suppose $A$ is a set of $2n$ positive integers { $a_{i}$ } with $a_{i} \geq 2$ which can be partitioned into at most $r \leq n$ coprime pairs. The following fact is true: For all $A$ and every $n \geq 1$,&#xA;$$&#xA;\prod_{i = 1}^{2n} (a_{i} - 1) \equiv 0 \pmod{2^{r}}. &#xA;$$&#xA;For example, the set {$2,3,5,6$} can be partitioned into two coprime pairs $(2,3)$ and $(5,6)$, so $2^{2} = 4$ divides $1 \cdot 2 \cdot 4 \cdot 5 = 40$. The set {$2,4,8,16$} cannot be partitioned into coprime pairs (i.e., $r = 0$), so the result above is not particularly interesting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; Can this result be sharpened? References are welcome.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(&lt;strong&gt;Update&lt;/strong&gt;) I think the above result can be generalized to the following:&#xA;Suppose $A$ is a set of $2n$ positive integers { $a_{i}$ } with $a_{i} \geq 2$ which can be partitioned into at most $r \leq n$ pairs with &lt;em&gt;odd greatest common divisor&lt;/em&gt;. Then $2^{r}$ divides $\prod_{i = 1}^{2n} (a_{i} - 1)$.&lt;/p&gt;&#xA;" OwnerUserId="2720" LastEditorUserId="2720" LastEditorDisplayName="" LastEditDate="2010-11-08T04:32:07.207" LastActivityDate="2010-11-08T04:32:07.207" Title="Power of $2$ Divisibility Problem" Tags="&lt;number-theory&gt;&lt;number&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" />
  <row Id="9367" PostTypeId="2" ParentId="9365" CreationDate="2010-11-08T03:22:44.073" Score="2" ViewCount="" Body="&lt;p&gt;If the point is $(a,b)$, then the distance from $(a,b)$ to $(x,y)$ is&#xA;$$\sqrt{(x-a)^2 + (y-b)^2}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the point is $(a,b)$, then the points that lie on the line through $(a,b)$ with slope $m$ are the points of the form&#xA;$$(a,b) + k(1,m)$$&#xA;where $k$ is a constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Putting the two together, if you know the start point $(a,b)$, and the slope $m$, and the distance $d$, then find the (two) values of $k$ that will give you a distance of $d$ by plugging in and solving for $k$.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-08T03:29:17.250" LastActivityDate="2010-11-08T03:29:17.250" CommentCount="1" />
  <row Id="9368" PostTypeId="2" ParentId="9366" CreationDate="2010-11-08T03:25:40.500" Score="3" ViewCount="" Body="&lt;p&gt;Each pair contains at most one even number, since the two numbers have odd gcd. That means that at least one $a_i-1$ for each pair is even, so the product is divisible at least $2^s$, where $s$ is the number of pairs with odd gcd.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added:&lt;/strong&gt; As far as sharpening the result, take a set of $2n$ &lt;a href=&quot;http://en.wikipedia.org/wiki/Fermat_numbers&quot; rel=&quot;nofollow&quot;&gt;Fermat numbers&lt;/a&gt;, numbers of the form $F_k=2^{2^k}+1$. If $k\neq\ell$, then $F_k$ and $F_{\ell}$ share no common factor, so if you take $2n$ of them, say $F_{k_1},\ldots,F_{k_{2n}}$ with $k_1\lt k_2\lt\cdots\lt k_{2n}$, then you can divide them into $n$ coprime pairs; your product will equal $2^{2^{k_1}+\cdots+2^{k_{2n}}}$. So you can make this product divisible by arbitrarily large powers of $2$ for a fixed value of $n$ and $r$.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-08T04:08:42.257" LastActivityDate="2010-11-08T04:08:42.257" CommentCount="4" />
  <row Id="9370" PostTypeId="1" CreationDate="2010-11-08T03:39:10.250" Score="1" ViewCount="219" Body="&lt;p&gt;In one of the programming work I am doing, I encountered a set of linear eqns. with modulo. I am putting it in simple format with only 2 variables,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(a11x + a12y) mod 8 = b1&lt;br&gt;&#xA;(a21x + a22y) mod 8 = b2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In case of simple eqns. with out modulo I can use: AX=B, X=inv(A)*B. But how to handle this modulo? Any clue or pointers will be very helpful.&lt;/p&gt;&#xA;" OwnerUserId="3260" LastEditorUserId="1562" LastEditorDisplayName="" LastEditDate="2010-11-08T05:06:39.147" LastActivityDate="2010-11-08T05:06:39.147" Title="how to solve linear equations involving modulo?" Tags="&lt;algebra&gt;&lt;modular-arithmetic&gt;" AnswerCount="2" />
  <row Id="9371" PostTypeId="2" ParentId="9354" CreationDate="2010-11-08T03:39:55.447" Score="4" ViewCount="" Body="&lt;p&gt;The main use of this theory is that it is strong enough to prove many truly basic facts about natural numbers, while still being weak enough that it carries little &quot;baggage&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$I\Delta_0 + \text{Exp}$ is strong enough to define functions that encode and decode finite sequence of numbers (Goedel numbering) and, similarly, it can encode and decode finite sets of numbers as single numbers. This makes it possible to formalize many basic number-theoretic and proof-theoretic results in this theory. Those results, in turn, make it possible to use $I\Delta_0 + \text{Exp}$ as a base over which we can prove stronger statements are equivalent. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Harvey Friedman has made a somewhat polemical &quot;&lt;a href=&quot;http://cs.nyu.edu/pipermail/fom/1999-April/003014.html&quot; rel=&quot;nofollow&quot;&gt;grand conjecture&lt;/a&gt;&quot; that &lt;em&gt;every&lt;/em&gt; fact of elementary number theory proven in &lt;em&gt;Annals of Mathematics&lt;/em&gt; can be proven in $I\Delta_0 + \text{Exp}$. This includes, for example, Fermat's Last Theorem. This is not to say that the original proofs will work - the proofs in $I\Delta_0 + \text{Exp}$ may be much more lengthy and complex. But the fact that this conjecture is even plausible speaks to the strength of $I\Delta_0 + \text{Exp}$. (The status of FLT here is not known.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$I\Delta_0 + \text{Exp}$ is very weak in other senses. Its proof theoretic ordinal is $\omega^3$, which is extremely smaller than the ordinal of Peano arithmetic and even much smaller than the ordinal of the &quot;next&quot; canonical theory, $I\Sigma_1$. This is philosophically relevant in certain instances where we wish to keep our assumptions relatively weak (in terms of their proof-theoretic ordinal). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$I\Delta_0 + \text{Exp}$ is not &quot;minimal&quot; in a formal sense. We could work with &lt;em&gt;ad hoc&lt;/em&gt; theories that are weaker, but still include the facts of number theory we need (perhaps as axioms). However, $I\Delta_0 + \text{Exp}$ has a relatively concrete and natural definition, which makes it more attractive to work with than some concocted system.  (Of course, it may be that $I\Delta_0 + \text{Exp}$ is equivalent to some axiom $T$ over some even weaker base theory, in which case $I\Delta_0 + \text{Exp}$ would be minimal as an extension of the base theory that proves $T$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another commonly used weak theory is $I\Sigma_1$. This theory extends $I\Delta_0 + \text{Exp}$ (modulo some definitional extensions) but $I\Sigma_1$ is still relatively weak (its ordinal is $\omega^\omega$). Many basic facts of number theory are easier to prove in $I\Sigma_1$, making it more convenient than $I\Delta_0 + \text{Exp}$ for various purposes, at the cost of a larger proof theoretic ordinal. &lt;/p&gt;&#xA;" OwnerUserId="630" LastActivityDate="2010-11-08T03:39:55.447" />
  <row Id="9372" PostTypeId="1" AcceptedAnswerId="9379" CreationDate="2010-11-08T03:41:17.643" Score="5" ViewCount="494" Body="&lt;p&gt;I am teaching a Calc II course and came across the following series when finding the interval of convergence for the Taylor series of $f(x)=\sqrt{x}$ centered at $x=1$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\sum_{n=2}^\infty \frac{1\cdot 3\cdot 5\cdot 7\cdots (2n-3)}{2^nn!}.&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that it doesn't converge, but how do you show it?  I would like to be able to use only the technology of a Calc II course, but any answer would be enlightening.&lt;/p&gt;&#xA;" OwnerUserId="2838" LastActivityDate="2010-11-08T16:08:10.400" Title="Why Doesn't This Series Converge?" Tags="&lt;calculus&gt;&lt;teaching&gt;" AnswerCount="6" CommentCount="1" FavoriteCount="1" />
  <row Id="9373" PostTypeId="2" ParentId="9370" CreationDate="2010-11-08T03:50:31.663" Score="2" ViewCount="" Body="&lt;p&gt;You can do Gauss-Jordan elimination performing operations modulo $8$. Just be mindful that &quot;division&quot; is only defined for odd integers (you are really multiplying by the inverse modulo $8$). &lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-11-08T03:50:31.663" />
  <row Id="9375" PostTypeId="2" ParentId="9365" CreationDate="2010-11-08T04:07:50.827" Score="2" ViewCount="" Body="&lt;p&gt;An equivalent way to Arturo's answer is as follows: from the slope $m$, you can determine the cosine and the sine of the angle from the horizontal axis of a line with that slope:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$c=\frac{1}{\sqrt{1+m^2}} \qquad s=\frac{m}{\sqrt{1+m^2}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(exercise: verify that they are the cosine and sine of a certain angle)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From this construction, you can easily determine the two points at a distance r from your starting point $(h,k)$ as $(h,k)\pm r(c,s)$.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-11-08T04:07:50.827" />
  <row Id="9376" PostTypeId="2" ParentId="9370" CreationDate="2010-11-08T04:09:22.520" Score="1" ViewCount="" Body="&lt;p&gt;You can apply Cramer's rule if the determinant is odd (so invertible mod 8). For an introduction to linear algebra over commutative rings see Wm. C. Brown: Matrices over commutative rings.&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-11-08T04:09:22.520" />
  <row Id="9377" PostTypeId="1" AcceptedAnswerId="9380" CreationDate="2010-11-08T04:20:55.520" Score="1" ViewCount="219" Body="&lt;p&gt;Suppose a set S has 11 elements. How many subsets of S have an even&#xA;number of elements? Express your answer as a summation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not really sure how to approach this. I think the general formula for this type of question is n!/k!(n-k)! where n is 11 and k is the number we're choosing. So would we just take the summation starting at 0 to 5 of 11!/2i!(n-2i)! Or what?&lt;/p&gt;&#xA;" OwnerUserId="2985" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-28T04:07:37.737" LastActivityDate="2010-11-28T04:07:37.737" Title="Combinations and Permutations Question" Tags="&lt;combinatorics&gt;&lt;homework&gt;&lt;discrete-mathematics&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="9378" PostTypeId="1" AcceptedAnswerId="9396" CreationDate="2010-11-08T04:30:10.903" Score="8" ViewCount="344" Body="&lt;p&gt;Where, or do there exist, good visualizations of sheaves, stalks, stacks, and/or schemes? I'm a better visual thinker than I am a symbolic thinker, and it would be easier for me to follow some of the algebraic geometry papers that have no pictures whatsoever if I had some kind of visual intuition of what these objects look like. &lt;/p&gt;&#xA;" OwnerUserId="890" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-08T05:17:23.877" LastActivityDate="2011-02-07T06:30:25.880" Title="Visualizations of some of the abstractions of algebraic geometry." Tags="&lt;algebraic-geometry&gt;&lt;visualization&gt;" AnswerCount="6" CommentCount="3" FavoriteCount="2" />
  <row Id="9379" PostTypeId="2" ParentId="9372" CreationDate="2010-11-08T04:32:11.280" Score="16" ViewCount="" Body="&lt;p&gt;I don't think the series diverges.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$  \frac{1 \cdot 3 \cdot 7 \cdots (2n-3)}{2 \cdot 4 \cdot 6 \cdots 2n}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ = \frac{{2n \choose n}}{(2n-1)4^n} =  \frac{1}{(2n-1)\sqrt{\pi n}}(1 + O(\frac{1}{n}))$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;using the approximation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{ {2n \choose n}}{4^n} = \frac{1}{\sqrt{\pi n}}(1 + O(\frac{1}{n}))$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and so the series must converge! Perhaps you have a mistake in your computation?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a more elementary proof of convergence, see the end of the answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe you should be able to compute it using the series expansion for&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{1}{\sqrt{1-x^2}} = \sum_{k=0}^{\infty} \frac{{2k \choose k} x^{2k}}{4^k}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(you will need to subtract some terms, divide by $x^2$ and integrate).&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;An elementary proof of convergence.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We will show that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;if $\displaystyle S_n = \frac{1 \cdot 3 \cdots (2n-1)}{2 \cdot 4 \cdots 2n}$ then&#xA;$\displaystyle S_n \le \frac{1}{\sqrt{n+1}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This proves the partial term of your series, which is $\displaystyle \sum_{k=2}^{n} \frac{S_k}{2k-1} &amp;lt; \sum_{k=2}^{\infty} \frac{1}{(2k-1)\sqrt{k+1}} &amp;lt; C$ (for some constant $C$), and thus is bounded above. Since the series is monotonically increasing and bounded above, it is convergent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We will prove that $\displaystyle S_n \le \frac{1}{\sqrt{n+1}}$ by induction on $n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $n=1$ it is clearly true.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now $S_{n+1} = S_n \frac{2n+1}{2n+2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider $\displaystyle  1 - \frac{2n+1}{2n+2} = \frac{1}{2n+2}  \ge \frac{1}{\sqrt{n+2}(\sqrt{n+2} + \sqrt{n+1})} = \frac{\sqrt{n+2} - \sqrt{n+1}}{\sqrt{n+2}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus $\displaystyle \frac{2n+1}{2n+2} \le \frac{\sqrt{n+1}}{\sqrt{n+2}}$ and so&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle S_{n} \le \frac{1}{\sqrt{n+1}} \Rightarrow S_{n+1} \le \frac{1}{\sqrt{n+2}}$&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-11-08T15:23:06.247" LastActivityDate="2010-11-08T15:23:06.247" CommentCount="10" />
  <row Id="9380" PostTypeId="2" ParentId="9377" CreationDate="2010-11-08T04:34:53.200" Score="6" ViewCount="" Body="&lt;p&gt;Hi fprime. If you can use the identity: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sum_{k = 0}^{n} (-1)^{k} {n \choose k} = 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;then you can show that the cardinality of the set of subsets with even cardinality is equal to the cardinality of the set of subsets with odd cardinality.&lt;/p&gt;&#xA;" OwnerUserId="592" LastActivityDate="2010-11-08T04:34:53.200" CommentCount="8" />
  <row Id="9381" PostTypeId="2" ParentId="9372" CreationDate="2010-11-08T04:49:46.853" Score="2" ViewCount="" Body="&lt;p&gt;Using Stirling formula $n!\sim \sqrt {2\pi n}(n/e)^n$, I also get that&#xA;the $n$th term of the series is $O(1/n^{3/2})$. So the series is convergent. In the derivation I used that $$\lim_{n\to\infty}\left(1-\frac{1}{n}\right)^n=e^{-1}.$$&lt;/p&gt;&#xA;" OwnerUserId="3249" LastActivityDate="2010-11-08T04:49:46.853" />
  <row Id="9382" PostTypeId="2" ParentId="9372" CreationDate="2010-11-08T04:50:18.210" Score="2" ViewCount="" Body="&lt;p&gt;In fact the sum converges. You can find a closed form by massaging the formula that I posted on &lt;a href=&quot;http://math.stackexchange.com/questions/4744/formal-power-series-coefficient-multiplication/4750#4750&quot;&gt;Hadamard products&lt;/a&gt; viz. $\rm\ (1-4x)^{-1/2} \;=\; \sum\ \binom{2n}{n}\ x^n\:.\ $ See also &lt;a href=&quot;http://math.stackexchange.com/questions/7778/proving-2n-choose-n-2n-frac-1-cdot-3-cdot-5-cdots-2n-1n/7802#7802&quot;&gt;this post&lt;/a&gt;.  Alternatively:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/GktjF.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-11-08T15:52:02.743" LastActivityDate="2010-11-08T15:52:02.743" CommentCount="1" />
  <row Id="9384" PostTypeId="2" ParentId="9372" CreationDate="2010-11-08T05:06:18.447" Score="7" ViewCount="" Body="&lt;p&gt;You can use Taylor approximations here. Note that the ratio between consecutive terms is ${2n - 3 \over 2n} = \exp(\ln(1 - 3/2n)) = \exp(-{3 \over 2n} + O(1/n^2))$. So the product is comparable to $\exp(-{3 \over 2} \sum_{i = 2}^n {1 \over n} + O(1/n))$, which in turn is comparable to $\exp(-{3 \over 2} \ln(n))$ or $n^{-{3 \over 2}}$. Thus the series converges.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Comment: Seeing others computing the actual value... once you know it converges (absolutely) you can use the original power series to get the actual value. For $|x| &amp;lt; 1$, $(1 - x)^{1 \over 2} = 1 - {x \over 2} - {1 \over 2} \sum_{n=2}^{\infty} A_nx^n$, where $A_n$ is the $n$ term of the sum. Since all terms of the sum are positive and it converges for $x = 1$, if you take limits as $x$ goes to 1 you get the sum $S$ we're looking at. In other words, 0 = 1 - 1/2 - ${S \over 2}$ or $S = 1$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And actually if you think about it.. if the series did diverge, taking limits of the series as $x$ goes to 1 from below would give infinity since all terms are positive. Thus just taking limits of both sides as $x$ goes to 1 gives both convergence and the correct value. &lt;/p&gt;&#xA;" OwnerUserId="3035" LastEditorUserId="3035" LastEditorDisplayName="" LastEditDate="2010-11-08T16:04:04.973" LastActivityDate="2010-11-08T16:04:04.973" CommentCount="1" />
  <row Id="9385" PostTypeId="2" ParentId="9338" CreationDate="2010-11-08T05:18:42.350" Score="3" ViewCount="" Body="&lt;p&gt;well, if they come sorted like that, you can:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;calculate the length of the list (in this case, 8)&lt;/li&gt;&#xA;&lt;li&gt;start going up in ascending order (from 1, for example)&lt;/li&gt;&#xA;&lt;li&gt;once you hit a number bigger than the number you want, count how many numbers are before it (if you're looking for 5, that would be 5 numbers because you stopped at 9)&lt;/li&gt;&#xA;&lt;li&gt;divide what you found in #3 by the length of the list in #1, multiply it by 100, and there's your percentile.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;otherwise, just count how many numbers are &amp;lt;= the number you're looking for, then divide it by the total. the above is just faster for when they come sorted.&lt;/p&gt;&#xA;" OwnerUserId="1562" LastActivityDate="2010-11-08T05:18:42.350" />
  <row Id="9386" PostTypeId="1" AcceptedAnswerId="9388" CreationDate="2010-11-08T05:20:49.353" Score="2" ViewCount="125" Body="&lt;p&gt;If x,y,z are natural numbers, how many solutions are there to x+y+z=25?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How would I figure this out? I cant even begin dissecting this problem. Where do I begin?&lt;/p&gt;&#xA;" OwnerUserId="2985" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2010-11-08T05:33:52.950" LastActivityDate="2010-11-08T06:03:43.980" Title="Number of possible solutions for an equation" Tags="&lt;number-theory&gt;&lt;homework&gt;&lt;discrete-mathematics&gt;" AnswerCount="3" CommentCount="0" />
  <row Id="9387" PostTypeId="2" ParentId="9386" CreationDate="2010-11-08T05:26:01.077" Score="1" ViewCount="" Body="&lt;p&gt;This is called the &quot;partition function&quot; of 25. See here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Partition_function_%28number_theory%29&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Partition_function_%28number_theory%29&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's not an easy-to-compute-directly function - the best bet is simply writing a small script that counts solutions using a double loop.&lt;/p&gt;&#xA;" OwnerUserId="1818" LastActivityDate="2010-11-08T05:26:01.077" CommentCount="5" />
  <row Id="9388" PostTypeId="2" ParentId="9386" CreationDate="2010-11-08T05:33:01.157" Score="6" ViewCount="" Body="&lt;p&gt;Suppose that &quot;natural number&quot; doesn't include 0. Take 25 balls, and put a wall between any two. A partition $a+b+c=25$ is the same as a choice of two walls, and there are $\binom{24}{2} = 276$ of these.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If zeroes are allowed, then $(x+1)+(y+1)+(z+1) = 28$, and so the answer is $\binom{27}{2} = 351$.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-08T05:33:01.157" CommentCount="6" />
  <row Id="9389" PostTypeId="1" AcceptedAnswerId="9390" CreationDate="2010-11-08T05:40:26.280" Score="2" ViewCount="63" Body="&lt;p&gt;let $f$:&lt;strong&gt;R&lt;/strong&gt;-&gt;&lt;strong&gt;N&lt;/strong&gt; such that given any $x_1$,$x_2$ belong to &lt;strong&gt;R&lt;/strong&gt; such that $x_1 &amp;lt; x_2$ then there exists a $x_3$ belongs to $(x_1,x_2)$ such that $f(x_3)&gt; max(f(x_1),f(x_2))$. How many such mappings are possible ?&lt;/p&gt;&#xA;" OwnerUserId="2987" LastActivityDate="2010-11-08T05:48:05.797" Title="a Mapping problem" Tags="&lt;real-analysis&gt;&lt;functions&gt;" AnswerCount="1" />
  <row Id="9390" PostTypeId="2" ParentId="9389" CreationDate="2010-11-08T05:48:05.797" Score="2" ViewCount="" Body="&lt;p&gt;There are $\aleph_0^\aleph$ such mappings. The upper bound is trivial. For the lower bound, pick any function $f\colon \mathbb{R} \longrightarrow \mathbb{N}$ such that $f(\pm \frac{p}{q}) = q$.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-08T05:48:05.797" CommentCount="7" />
  <row Id="9391" PostTypeId="1" AcceptedAnswerId="9393" CreationDate="2010-11-08T05:48:34.370" Score="3" ViewCount="248" Body="&lt;p&gt;According to &lt;a href=&quot;http://en.wikipedia.org/wiki/Trinomial_expansion&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;, the number of terms in (x+y+z)^30 is 496. I'm assuming this is before like terms are added up. How many terms would there be if like terms were combined? How would I go about figuring that out?&lt;/p&gt;&#xA;" OwnerUserId="2985" LastEditorUserId="2985" LastEditorDisplayName="" LastEditDate="2010-11-08T05:58:32.397" LastActivityDate="2010-11-08T23:16:09.737" Title="Number of terms in a trinomial expansion" Tags="&lt;number-theory&gt;&lt;homework&gt;&lt;discrete-mathematics&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="9392" PostTypeId="2" ParentId="9391" CreationDate="2010-11-08T05:57:47.280" Score="4" ViewCount="" Body="&lt;p&gt;First note that every $x^ay^bz^c$ with $a + b + c = 30$ appears. One way of seeing this is observing ${\partial^{30}(x+y+z)^{30} \over \partial_x^a\partial_y^b\partial_z^c} $ at $(0,0,0)$ is $30!$, which is not zero. So for a given value of $a$, every possible $(b,c)$ with $b + c = 30 - a$ can happen. Since $b$ goes from $0$ to $30 - a$, there are $31 - a$ possibilities for $b$, each of which forces $c$ to have the single value $30 - a - b$. Thus for a given $a$ there are &#xA;$31 - a$ different possible $(b,c)$. Adding this over all $a$ from 0 to 31 this gives the sum of $31 + 30 + ... + 0 = 496$.&lt;/p&gt;&#xA;" OwnerUserId="3035" LastEditorUserId="3035" LastEditorDisplayName="" LastEditDate="2010-11-08T23:16:09.737" LastActivityDate="2010-11-08T23:16:09.737" CommentCount="2" />
  <row Id="9393" PostTypeId="2" ParentId="9391" CreationDate="2010-11-08T05:59:12.863" Score="8" ViewCount="" Body="&lt;p&gt;No, the 496 is the number of terms after like terms are combined.  Before like terms are combined there are $3^{30}$ terms.  This is because you have 30 different factors, and so the number of terms you get before combining is the number of ways to choose 30 elements when there are three choices for each.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Zaricuse's answer is hinting at how to derive the formula on the Wikipedia page.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's another way to look at the formula on the Wikipedia page: The number of terms in the expansion of $(x+y+z)^n$ after combining is the number of ways to choose $n$ elements &lt;em&gt;with replacement&lt;/em&gt; (since you can choose $x,y,z$ more than once) in which order does not matter from a set of 3 elements.  This formula is known to be &#xA;$$\binom{3+n-1}{n} = \binom{n+2}{n} = \frac{(n+1)(n+2)}{2}.$$ &#xA;See, for example, MathWorld's entry on &lt;a href=&quot;http://mathworld.wolfram.com/BallPicking.html&quot; rel=&quot;nofollow&quot;&gt;Ball Picking&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2370" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-11-08T22:06:29.063" LastActivityDate="2010-11-08T22:06:29.063" CommentCount="4" />
  <row Id="9394" PostTypeId="2" ParentId="9386" CreationDate="2010-11-08T06:03:43.980" Score="0" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\rm\ \ (x,y,z)\ \to\ \{x,\ x+y\}\ $ bijects solutions with two elt subsets of $\{1,2,\cdots,24\} $&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-11-08T06:03:43.980" />
  <row Id="9395" PostTypeId="1" AcceptedAnswerId="9401" CreationDate="2010-11-08T06:34:06.817" Score="0" ViewCount="181" Body="&lt;p&gt;What is the cardinality of set of all smooth functions belonging to $L^1$ or $L^2$ ? What is that of set of all integrable or square integrable functions ? &lt;/p&gt;&#xA;" OwnerUserId="2987" LastActivityDate="2010-11-08T12:32:07.567" Title="what is the cardinality of set of all smooth functions in $L^1$ ?" Tags="&lt;real-analysis&gt;" AnswerCount="2" CommentCount="17" FavoriteCount="1" />
  <row Id="9396" PostTypeId="2" ParentId="9378" CreationDate="2010-11-08T06:44:43.767" Score="3" ViewCount="" Body="&lt;p&gt;Mumford's The Red Book of Varieties and Schemes is quite famous for its pictures and the emphasis on geometric intuition.&lt;/p&gt;&#xA;" OwnerUserId="3212" LastActivityDate="2010-11-08T06:44:43.767" />
  <row Id="9397" PostTypeId="2" ParentId="9377" CreationDate="2010-11-08T06:49:58.933" Score="4" ViewCount="" Body="&lt;p&gt;Here is a very easy way of seeing that the number of sets of even cardinality is equal to the number of sets of odd cardinality: if the total number of elements is odd (like in your case) then the map that sends a set to its complement establishes a bijection between sets of even and of odd cardinality. If the total number of elements is even, then there is the same bijection between the sets of even cardinality not containing the element 1 and the sets of odd cardinality not containing 1. Also, there is the obvious bijection between sets of even cardinality not containing 1 and those of odd cardinality containing 1, so again you have a bijection between odd and even sets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, you just have to compute the total number of subsets, which I am sure you can do.&lt;/p&gt;&#xA;" OwnerUserId="3212" LastEditorUserId="3212" LastEditorDisplayName="" LastEditDate="2010-11-08T07:04:24.240" LastActivityDate="2010-11-08T07:04:24.240" />
  <row Id="9398" PostTypeId="1" AcceptedAnswerId="9399" CreationDate="2010-11-08T06:53:09.550" Score="1" ViewCount="267" Body="&lt;p&gt;I have a function given implicitly, you know. X og Y on both sides. Than it says, assume y=y(x). That's fine. And find y'(0), that I should be able to do. But what about y''(0)? How do you treat the dy/dx parts when taking the second derivative? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: I would also like to follow the tip in the book, that says when I'm after actual values. We can just insert the value instead solving for dy/dx. &lt;/p&gt;&#xA;" OwnerUserId="2230" LastEditorUserId="1543" LastEditorDisplayName="" LastEditDate="2011-04-18T12:07:40.277" LastActivityDate="2011-04-18T12:07:40.277" Title="Find the second derivative of some implicit function?" Tags="&lt;calculus&gt;&lt;homework&gt;" AnswerCount="2" />
  <row Id="9399" PostTypeId="2" ParentId="9398" CreationDate="2010-11-08T06:56:03.710" Score="6" ViewCount="" Body="&lt;p&gt;You simply replace all the $\frac{\mathrm{d}y}{\mathrm{d}x}$ terms in your second derivative with the expression you got for $\frac{\mathrm{d}y}{\mathrm{d}x}$ through implicit differentiation.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;As an explicit example, suppose we wanted to find $\frac{\mathrm{d}^2 y}{\mathrm{d}x^2}$ for the expression $x^2+y^2-r=0$ ($r$ here is a constant). We find&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\mathrm{d}y}{\mathrm{d}x}=-\frac{x}{y}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and differentiating again gives&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\mathrm{d}^2 y}{\mathrm{d}x^2}=-\frac{1+\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2}{y}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we substitute the expression we got for the first derivative into the second derivative, we get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\mathrm{d}^2 y}{\mathrm{d}x^2}=-\frac{1+\left(-\frac{x}{y}\right)^2}{y}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which gives&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\mathrm{d}^2 y}{\mathrm{d}x^2}=-\frac{x^2+y^2}{y^3}$$&lt;/p&gt;&#xA;" OwnerUserId="498" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-08T07:04:16.367" LastActivityDate="2010-11-08T07:04:16.367" />
  <row Id="9400" PostTypeId="2" ParentId="9372" CreationDate="2010-11-08T07:00:21.383" Score="4" ViewCount="" Body="&lt;p&gt;Let's start with the formula&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle \sum_{n=1}^\infty \frac{\binom{2n}{n}}{(2n-1)4^n}.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle \frac{\binom{2n}{n}}{2n-1} = \frac{(2n)!}{(2n-1)n!n!} = 2 \frac{(2n-2)!}{n!(n-1)!} = 2C_{n-1},$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $C_n$ is the $n$th Catalan number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now $C_n$ counts the number of paths of length $2(n+1)$ that reach the origin for the first time at their end, always being to the &lt;i&gt;right&lt;/i&gt; of the origin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since a one-dimensional random walk will get back to the origin w.p. $1$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle \sum_{n=1}^\infty \frac{C_{n-1}}{4^n} = \frac{1}{2}.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore the original sum equals&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle \sum_{n=1}^\infty \frac{2C_{n-1}}{4^n} = 1.$&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-08T07:00:21.383" />
  <row Id="9401" PostTypeId="2" ParentId="9395" CreationDate="2010-11-08T07:29:32.983" Score="3" ViewCount="" Body="&lt;p&gt;A continuous function is determined by its value on the rational points, so there are at most $\aleph^{\aleph_0} = \aleph$ of them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conversely, it's not difficult to find $\aleph$ smooth (integrable) functions in $L_1 \cap L_2$, just take any such non-zero function $f$ and consider $\{rf : r \in \mathbb{R}\}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: For definiteness, $x \mapsto e^{-x}$ is $C^\infty$ and in $\bigcap_{p&gt;0} L_p$.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2010-11-08T07:51:12.397" LastActivityDate="2010-11-08T07:51:12.397" CommentCount="7" />
  <row Id="9402" PostTypeId="1" AcceptedAnswerId="9409" CreationDate="2010-11-08T07:40:58.583" Score="8" ViewCount="423" Body="&lt;p&gt;Suppose that we do not know anything about the complex analysis (numbers). In this case, how to calculate the following integral in closed form?&#xA;$$\int_0^\infty\frac{\cos\;x}{1+x^2}\mathrm{d}x$$&lt;/p&gt;&#xA;" OwnerUserId="2323" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-11-08T17:33:10.077" LastActivityDate="2010-11-08T17:33:10.077" Title="Calculating the integral $\int_{0}^{\infty} \frac{\cos x}{1+x^2}$ without using complex analysis" Tags="&lt;real-analysis&gt;&lt;integral&gt;" AnswerCount="4" CommentCount="2" FavoriteCount="4" />
  <row Id="9403" PostTypeId="2" ParentId="9402" CreationDate="2010-11-08T07:53:15.237" Score="-2" ViewCount="" Body="&lt;p&gt;You can try using the Taylor expansion of $\cos$.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-08T07:53:15.237" />
  <row Id="9404" PostTypeId="1" CreationDate="2010-11-08T08:00:37.030" Score="2" ViewCount="159" Body="&lt;p&gt;Let $f \colon (\mathbb{C}^{n},\mathbf{0}) \to (\mathbb{C},0)$ be a complex analytic function with isolated critical point at the origin. Define the singular hypersurface $V_{f, \kappa} = f^{-1}(\kappa)$ for small $\kappa &gt; 0$. Let $\pi \colon \tilde{V} \to V$ be a resolution of $V = V_{f,0}$, and let $p_{g} = \dim H^{n-2}(\tilde{V}, \mathcal{O})$ denote the corresponding &lt;em&gt;geometric genus&lt;/em&gt; of $V$, which is independent of the chosen resolution. (This is the same geometric genus in the Yau-Durfee Theorem: $\mu \geq n! \ p_{g}$ for $n &gt; 2$, See &lt;a href=&quot;http://www.intlpress.com/JDG/archive/1993/37-2-375.pdf&quot; rel=&quot;nofollow&quot;&gt;Y93&lt;/a&gt; or &lt;a href=&quot;http://www.mathjournals.org/mrl/2006-013-006/2006-013-006-006.pdf&quot; rel=&quot;nofollow&quot;&gt;Y06&lt;/a&gt;.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Define the map $\varphi_{f} = f/\| f \| \colon S_{\epsilon}^{2n-1} \setminus V_{f, \kappa} \to S^{1}$, where $\epsilon &gt; 0$ is sufficiently small. The intersection of $V_{\kappa}$ with a small sphere $S_{\epsilon}^{2n-1}$ is an algebraic link. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the Brieskorn-Pham polynomial $f = \sum_{i = 1}^{n} z_{i}^{a_{i}}$ with $a_{i} \geq 2$. In particular, for $f = z_{1}^{p} + z_{2}^{q}$, the intersection above is a torus knot $T_{p,q}$ if $p$ and $q$ are coprime. In 1968, Milnor conjectured that the unknotting number $u(T_{p,q})$ is related to the dimension of the local algebra $A_{f} = \mathbb{C}[x,y] / \langle x^{p-1}, y^{q-1} \rangle$, which is also the rank of the middle homology group $H_{1}(F_{f,0}; \mathbb{Z})$ of the corresponding fiber $F_{f,0} = \varphi_{f}^{-1}(e^{i \theta})$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In 1928, Brauner proved $\mu(f) = \text{rank} \ H_{1}(F_{f,0}, \mathbb{Z}) = (p-1)(q-1)$ for $f = x^{p} + y^{q}$. In 1994, Kronheimer and Mrowka proved the Thom conjecture, and as a consequence, the Milnor conjecture followed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Edited) &lt;strong&gt;Question(s)&lt;/strong&gt;: Is the genus $g$ of the fiber $F_{f,0}$ related to the geometric genus $p_{g}$ of $V$ for $n &gt; 2$? Is there a geometric genus-like object for $n = 2$? If so, how might it relate to invariants of the torus knots?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Also posted on MO.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="2720" LastEditorUserId="2720" LastEditorDisplayName="" LastEditDate="2010-11-08T20:49:09.397" LastActivityDate="2010-11-08T20:49:09.397" CommunityOwnedDate="2010-11-08T20:49:09.397" Title="The Milnor Conjecture on the Unknotting Number of a Torus Knot" Tags="&lt;complex-analysis&gt;&lt;algebraic-geometry&gt;&lt;algebraic-curves&gt;" CommentCount="7" FavoriteCount="2" />
  <row Id="9405" PostTypeId="1" CreationDate="2010-11-08T08:02:36.410" Score="0" ViewCount="140" Body="&lt;p&gt;Let G be the group U(24) and H the cyclic subgroup generated by the element 7.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Exhibit all of the distinct cosets of H in G.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am coming up with this.&#xA;(I created the Cayley table for U(24) - using the elements 1,5,7,11,13,17,19,23.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&amp;lt;1&gt; = {1}&#xA;&amp;lt;5&gt; = (5*5 = 1)  {5}&#xA;&amp;lt;7&gt; = (7*7 = 1)  {7}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and so forth since each element times itself is equal to the identity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this correct? there are 8 elements and therefore 8 distinct cosets for U(24).&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Next piece of the problem I'm working thru.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is H a normal subgroup of G?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not quite sure what H is... exactly to compare &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm guessing &amp;lt;7&gt; = H&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so by looking at the left and right cosets&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&amp;lt;1&gt;H  &amp;lt;-&gt;  H&amp;lt;1&gt;&#xA;&amp;lt;5&gt;H (5*7 = 11) &amp;lt;-&gt; H&amp;lt;5&gt; (7*5 = 11)&#xA;&amp;lt;7&gt;H (7*7 = 1) &amp;lt;-&gt; H&amp;lt;7&gt; (7*7 = 1)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and then I would stop there and say it was normal subgroup of G.&#xA;with order 1 and the index of H in G is 2?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But Lagrange Theorem states the order of G = (order of H)* index of H in G... this doesn't end up working in this case (Makes me wonder if I am doing this part wrong)&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Exhibit distinct elements of the quotient group of G/H... (So I can construct a Cayley Table for G/H.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am just starting to work on this... some help to get me started would be nice :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" LastActivityDate="2010-11-08T08:16:02.950" Title="Exhibiting the distinct cosets of a cyclic subgroup H in group G?" Tags="&lt;abstract-algebra&gt;&lt;homework&gt;" AnswerCount="1" />
  <row Id="9406" PostTypeId="2" ParentId="9405" CreationDate="2010-11-08T08:16:02.950" Score="1" ViewCount="" Body="&lt;p&gt;The group $H$ has two elements $\{7, 1\}$ and therefore its index in $G$ is 4. Also note that right cosets coincide with left cosects so that $gH = Hg$ (this comes from the fact that G is abelian and also answers your second question). So e.g. $5 \{7, 1\} = \{11, 5\} = 11 \{7, 1\}$. You'll get 4 cosets like this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Third question is really the same thing as the first question (thanks to $H$ being normal).&lt;/p&gt;&#xA;" OwnerUserId="3223" LastActivityDate="2010-11-08T08:16:02.950" CommentCount="6" />
  <row Id="9407" PostTypeId="1" AcceptedAnswerId="9410" CreationDate="2010-11-08T08:16:41.640" Score="3" ViewCount="116" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;what are some &lt;em&gt;natural&lt;/em&gt; examples of integral domains $A$ together with sequences of elements satisfying $... | f_2 | f_1 | f_0$, but so that $f_i$ divides no power of $f_{i+1}$? So in the spectrum there will be an ascending chain of basic-open subsets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course we could just take the union of the $k[y^{-n} x,y]$ in $k(x,y)$ or the union of the $\mathbb{Z}[2^{-n} x]$ in $\mathbb{Q}(x)$, or a valuation domain whose value group has arbitrary small positive elements, but I don't call these examples natural. What about rings of holomorphic functions on a connected open subset of $\mathbb{C}$?&lt;/p&gt;&#xA;" OwnerUserId="1650" LastActivityDate="2010-11-08T21:27:49.347" Title="Examples of infinite divisor sequences" Tags="&lt;commutative-algebra&gt;&lt;ring-theory&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="9408" PostTypeId="2" ParentId="9402" CreationDate="2010-11-08T08:18:20.660" Score="2" ViewCount="" Body="&lt;p&gt;The previous answer is not correct. If you use the Taylor expansion of cosine and integrate termwise you consider integrals of the following form:&#xA;\begin{eqnarray}&#xA;\int_{0}^{\infty} \frac{x^{a} \ dx}{1 + x^{2}} = \tfrac{\pi}{2} \sec (\tfrac{\pi a}{2})&#xA;\end{eqnarray}&#xA;which is only well-defined if $-1 &amp;lt; a &amp;lt; 1$.&lt;/p&gt;&#xA;" OwnerUserId="2720" LastActivityDate="2010-11-08T08:18:20.660" />
  <row Id="9409" PostTypeId="2" ParentId="9402" CreationDate="2010-11-08T08:18:54.573" Score="14" ViewCount="" Body="&lt;p&gt;This can be done by the useful technique of differentiating under the integral sign.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, this is exercise 10.23 in the second edition of &quot;Mathematical Analysis&quot; by Tom Apostol.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is the brief sketch (as laid out in the exercise itself).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $\displaystyle F(y) = \int_{0}^{\infty} \frac{\sin xy}{x(1+x^2)} \ dx$ for $y &gt; 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Show that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle F''(y) - F(y) + \pi/2 = 0$ and hence deduce  that $\displaystyle F(y) = \frac{\pi(1-e^{-y})}{2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Use this to deduce that for $y &gt; 0$ and $a &gt; 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle  \int_{0}^{\infty} \frac{\sin xy}{x(x^2 + a^2)} \ dx = \frac{\pi(1-e^{-ay})}{2a^2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle  \int_{0}^{\infty} \frac{\cos xy}{x^2 + a^2} \ dx = \frac{\pi e^{-ay}}{2a}$&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-11-08T08:18:54.573" CommentCount="5" />
  <row Id="9410" PostTypeId="2" ParentId="9407" CreationDate="2010-11-08T08:30:08.633" Score="4" ViewCount="" Body="&lt;p&gt;In rings of analytic functions (real or complex), you could take any function with an infinite product expansion, and remove one factor at a time.  E.g., you could take $f_0(z)=\sin(\pi z)$, and $f_k(z)=\frac{\sin(\pi z)}{(z-1)(z-2)\cdots(z-k)}$.  As indicated in this example, you don't really need to think about infinite products.  You can take any function with infinitely many zeros, and divide out by factors corresponding to those zeros (including the multiplicity of each zero, so that taking powers won't add them back in).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(It might be nicer to take $f_0(z)=\sin(\pi z)$ and $f_k(z)=\frac{\sin(\pi z)}{\left(1-\frac{z^2}{1^2}\right)\left(1-\frac{z^2}{2^2}\right)\cdots\left(1-\frac{z^2}{k^2}\right)}$, so that you're actually removing factors from the product expansion.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can do this on any connected open subset of $\mathbb{C}$ or $\mathbb{R}$ (connected because you want an integral domain), but another &quot;natural&quot; class of examples that comes to mind is the set of &lt;a href=&quot;http://eom.springer.de/b/b016630.htm&quot; rel=&quot;nofollow&quot;&gt;Blaschke products&lt;/a&gt; with infinitely many zeros on the open unit disk.  In this case, you could take&#xA;$$f_k(z)=\frac{f_0(z)}{(\phi_{a_1}(z))^{m_1}(\phi_{a_2}(z))^{m_2}\cdots(\phi_{a_k}(z))^{m_k}},$$&#xA;where $a_1,a_2,\ldots$ are the distinct zeros of the infinite Blaschke product $f_0$, $m_k$ is the multiplicity of $a_k$, and $\phi_{a_k}$ is a holomorphic automorphism of the disk that sends $a_k$ to $0$ (unique up to a constant multiple of modulus $1$).  Thus each $f_k$ is a Blaschke product and has one less zero than $f_{k-1}$.  &lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2010-11-08T21:27:49.347" LastActivityDate="2010-11-08T21:27:49.347" CommentCount="1" />
  <row Id="9411" PostTypeId="1" CreationDate="2010-11-08T09:09:42.717" Score="0" ViewCount="135" Body="&lt;p&gt;If I am given a Cayley Table of 8 elements (an octic table). How would I go about finding a subgroup that satisfies the following conditions?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Find H is a normal subgroup of G&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) G is a normal subgroup of octic group&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) G is NOT normal subgroup of octic group&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;1) If aH = Ha for every a in G, then H is said to be a normal subgroup. So I must prove this is a fact, by trial and error picking out a subgroup of order 2 that will satisfy this statement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2 &amp;amp; 3) Not sure what it means to show that G is a subgroup of the octic group. Isn't G the actual octic group?&lt;/p&gt;&#xA;" LastActivityDate="2010-11-08T09:09:42.717" Title="If I am given a Cayley Table of 8 elements how can I derive some facts about it?" Tags="&lt;abstract-algebra&gt;&lt;homework&gt;" CommentCount="5" />
  <row Id="9412" PostTypeId="1" AcceptedAnswerId="9418" CreationDate="2010-11-08T09:54:28.433" Score="1" ViewCount="184" Body="&lt;p&gt;Hi there, In a calculus course I just completed I struggled to understand the Taylor and Maclaurin series. I got the basics but not some of the harder concepts. Could anyone point me in the right direction? Books, tutorials, etc?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many thanks.&lt;/p&gt;&#xA;" OwnerUserId="2897" LastActivityDate="2010-11-08T12:12:06.810" Title="Learning Taylor/Maclaurin Series" Tags="&lt;sequences-and-series&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="2" />
  <row Id="9413" PostTypeId="2" ParentId="9255" CreationDate="2010-11-08T10:08:38.463" Score="1" ViewCount="" Body="&lt;p&gt;This question was asked over at tex.se, &lt;a href=&quot;http://tex.stackexchange.com/questions/5042/define-strange-operators&quot;&gt;Define strange operators&lt;/a&gt;, with an accepted answer by frabjous:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The &lt;code&gt;\mathop{..}&lt;/code&gt; command from amsmath makes its argument behave like a &quot;large operator&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;with some example code.&lt;/p&gt;&#xA;" OwnerUserId="100" LastActivityDate="2010-11-08T10:08:38.463" CommunityOwnedDate="2010-11-08T10:08:38.463" />
  <row Id="9414" PostTypeId="1" AcceptedAnswerId="9415" CreationDate="2010-11-08T10:37:53.200" Score="0" ViewCount="107" Body="&lt;p&gt;I have posted about induction on it's own so I think I understand that part. But I'm slightly confused when it comes to using induction to show that n+1 is also the derivative. The base case if fine, I just set n=1. And show that is equal to the derivative not using the formula. But how about k+1? Can I just choose k=the-basecase. And than prove k+1 by just finding f''(x)? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;f&lt;sup&gt;(n)&lt;/sup&gt;(x) = a&lt;sup&gt;n-1&lt;/sup&gt;e&lt;sup&gt;ax&lt;/sup&gt;(ax+n) &lt;/p&gt;&#xA;" OwnerUserId="2230" LastActivityDate="2010-11-08T17:30:42.227" Title="Induction to prove a simple formula for calculating the n-th derivative?" Tags="&lt;calculus&gt;&lt;induction&gt;" AnswerCount="2" />
  <row Id="9415" PostTypeId="2" ParentId="9414" CreationDate="2010-11-08T11:06:48.040" Score="3" ViewCount="" Body="&lt;p&gt;We have $f^n(x)= a^{n-1}e^{ax}(ax+n)$ and so&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{d}{dx} f^n(x) = a^{n-1} \frac{d}{dx} \lbrace e^{ax}(ax+n) \rbrace&#xA;= a^{n-1} \left( ae^{ax}(ax+n) + ae^{ax} \right) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$= a^n e^{ax}(ax + n+1) = f^{n+1}(n).$$&lt;/p&gt;&#xA;" OwnerUserId="1301" LastActivityDate="2010-11-08T11:06:48.040" />
  <row Id="9416" PostTypeId="1" CreationDate="2010-11-08T11:27:55.503" Score="3" ViewCount="286" Body="&lt;p&gt;Say I pass 512 samples into my FFT&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My microphone spits out data at 10KHz,  so this represents 1/20s.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(So the lowest frequency FFT would pick up would be 40Hz).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The FFT will return an array of 512 frequency bins&#xA; - bin 0: [0 - 40Hz)&#xA; - bin 1: [40 - 80Hz)&#xA;etc&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if my original sound contained energy at say 115Hz,  how can I accurately retrieve this frequency?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is going to lie in bin #2,  but very close to bin #3.   so I would expect both bins to contain something nonzero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Question: how about the bins either side of this? Would they be guaranteed to be zero if there are no other frequencies close in the original signal?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Main question:  is there some algorithm for deciphering the original frequency given the relative bin strengths?&lt;/p&gt;&#xA;" OwnerUserId="3096" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-08T11:36:27.200" LastActivityDate="2011-03-19T03:07:44.947" Title="Extracting exact frequencies from FFT output" Tags="&lt;fourier-analysis&gt;&lt;signal-processing&gt;" AnswerCount="4" CommentCount="3" FavoriteCount="3" />
  <row Id="9417" PostTypeId="1" CreationDate="2010-11-08T11:41:35.493" Score="3" ViewCount="183" Body="&lt;p&gt;Please find a closed form for partial sum of a function&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f(x)=a^{1/x}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want it to be expressed in terms of bounded number of elementary functions and/or well known special functions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;No computer algebra systems I have tried so far could find a satisfactory solution. I believe that the expression can exist in the terms of incomplete Gamma function or its generalizations because indefinite integral  of this function can be expressed in terms of incomplete Gamma function:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int f(x) dx= x\sqrt[x]{a}-\operatorname{Ei}\left(\frac{\ln a}{x}\right)\ln a$$&lt;/p&gt;&#xA;" OwnerUserId="2513" LastActivityDate="2010-11-09T11:19:41.607" Title="Find a closed form for $\sum_{k=1}^{x-1} a^{1/k}$" Tags="&lt;analysis&gt;&lt;sequences-and-series&gt;" AnswerCount="1" CommentCount="6" FavoriteCount="1" />
  <row Id="9418" PostTypeId="2" ParentId="9412" CreationDate="2010-11-08T12:12:06.810" Score="7" ViewCount="" Body="&lt;p&gt;I think it's great that you're seeking a deeper understanding of an important topic-- even after the class is over!  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you haven't done so, go back to your calculus text and carefully read the sections on Taylor and MacLaurin series.  (I know math textbooks often seem obtuse and unhelpful, but it's a valuable skill to be able to read slowly and learn with just the text as your guide.)  Pay close attention to the examples, because they are acutely chosen to help you see the ideas and connections you need.  If there's a gap you can't fill in, ask!  --the more detailed your question, the more direct our help will be.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My best guess is that solving some problems by hand would make you more comfortable working with power series.  Here are some problems to try.  (It may take a lot of writing, but as long as you have one idea of what to do next, try it!  There's a lot to keep track of with power series, and practice is important.)&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Prove, by expanding as MacLaurin series, that $e^{ix} = \cos x + i \sin x$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Start with the MacLaurin series for $(\cos x)$ and $(\sin x)$, square them, then add to show that $\cos^2 x + \sin^2 x = 1$.  (To start, just work up to $x^4$, but go further if you can.)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Start with the MacLaurin series for $e^x$ and $\ln (1-x)$ and show that $e^{\ln (1-x)} = 1-x$.  (Again, just work up to $x^4$ or so.  You could also try the same thing with $e^{-\ln(1-x)}$.)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Prove, using MacLaurin series that $\frac{d}{dx} e^x = e^x$, $\frac{d}{dx} \sin x = \cos x$, and $\frac{d}{dx} \cos x = - \sin x$.  (For these, don't stop at $x^4$.  You can use summation notation to work with all terms, from 0 to infinity, at the same time.)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Write the MacLaurin series for $\frac{1}{1-x}$, then take the integral of the series.  (What do you get?)  Take the derivative of the series.  (What do you get?)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Find the MacLaurin series for $(\arctan x)$ by taking its derivative, finding the MacLaurin series expansion, then integrating that expansion.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I don't have other references for you, but power series expansions can be used in solving limits (replacing L'Hospital's rule), in differential equations, in combinatorics (generating functions), and in complex analysis.  If any of those topics piques your interest, I'm sure people here or in your department can give some references to check out.&lt;/p&gt;&#xA;" OwnerUserId="1461" LastActivityDate="2010-11-08T12:12:06.810" CommentCount="1" />
  <row Id="9419" PostTypeId="2" ParentId="9395" CreationDate="2010-11-08T12:32:07.567" Score="2" ViewCount="" Body="&lt;p&gt;The second part of the question asks for the cardinality of the set of integrable or square-integrable functions.  I will assume you actually mean &quot;functions&quot; rather than &quot;equivalence classes of functions under the relation of equality almost everywhere&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $\beta$ be the cardinality of the set of all functions from $\mathbb{R}$ to $\mathbb{R}$; by standard set theory this is the same as the cardinality of the powerset of the real numbers: $\beta = 2^{|\mathbb{R}|} = 2^{2^{\aleph_0}}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Certainly the set of integrable functions, and the set of square integrable functions, can have cardinality no more than $\beta$. It turns out this is exactly the cardinality.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $E$ be a Cantor set; the key properties are that $|E| = |\mathbb{R}|$ and the measure of $E$ is $0$. Consider the set of all functions that are $0$ for every $x$ that is not in $E$. All of these functions are both integrable and square integrable, because the measure of $E$ is zero. The cardinality of this set is the cardinality of the set of functions from $E$ to $\mathbb{R}$, which is exactly $\beta$ because $|E| = |\mathbb{R}|$. &lt;/p&gt;&#xA;" OwnerUserId="630" LastActivityDate="2010-11-08T12:32:07.567" />
  <row Id="9420" PostTypeId="1" AcceptedAnswerId="9474" CreationDate="2010-11-08T12:40:32.783" Score="2" ViewCount="203" Body="&lt;p&gt;Suppose $h$ is a hash function, $h$ : { 0, 1 } &lt;sup&gt;*&lt;/sup&gt; $\rightarrow$ { 0, 1 } &lt;sup&gt;n&lt;/sup&gt; and for all $w_1$, $w_2$ it holds: $h(w_1 \oplus w_2) =&#xA;h(w_1) \oplus h(w_2)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;$\oplus$ is the XOR operation.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why isn't $h$ a cryptographically good hash function?&lt;/p&gt;&#xA;" OwnerUserId="2411" LastActivityDate="2010-11-09T23:47:11.487" Title="Why is hash function $h$ ($h(w_1 \oplus w_2) = h(w_1) \oplus h(w_2)$) not good?" Tags="&lt;homework&gt;&lt;cryptography&gt;" AnswerCount="2" CommentCount="8" />
  <row Id="9421" PostTypeId="2" ParentId="7481" CreationDate="2010-11-08T12:49:21.850" Score="2" ViewCount="" Body="&lt;p&gt;In what follows I assume that a map $f:\ ]0,1[\to \mathbb R^2$ is wanted. $$ $$ There are continuous surjective maps $p:[0,1]\to[0,1]^2$; they are called Peano curves. One may choose $p$ in such a way that the &quot;curve&quot; begins at $(0,0)$ and ends at $(1,1)$. Therefore it is possible to construct a continuous map $f:\ ]0,1]\to\mathbb R^2$ such that the restrictions of $f$ to the intervals $[{1\over n+1},{1\over n}]$ are Peano curves covering an increasing sequence of squares $Q_n$ whose union is $\mathbb R^2$. At the end, the point $1$ may safely be omitted from the domain of $f$.&lt;/p&gt;&#xA;" OwnerUserId="1303" LastActivityDate="2010-11-08T12:49:21.850" />
  <row Id="9422" PostTypeId="1" AcceptedAnswerId="9444" CreationDate="2010-11-08T12:53:07.197" Score="6" ViewCount="327" Body="&lt;p&gt;Could you provide a geometric explanation of taylor expansion? &lt;/p&gt;&#xA;" OwnerUserId="1382" LastActivityDate="2010-11-08T15:32:30.663" Title="Intuition explanation of taylor expansion?" Tags="&lt;intuition&gt;&lt;taylor-expansion&gt;" AnswerCount="5" CommentCount="2" FavoriteCount="2" />
  <row Id="9423" PostTypeId="1" CreationDate="2010-11-08T12:58:43.300" Score="4" ViewCount="91" Body="&lt;p&gt;I read a little remark in a topology text that the &lt;em&gt;direct&lt;/em&gt; image of an open set under a continuous mapping is not necessarily open. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the definition of &lt;em&gt;direct&lt;/em&gt; image in this case? I tried googling to no avail, and only found references to sheaf theory on wikipedia, which I don't believe to be what I'm looking for. Thanks.&lt;/p&gt;&#xA;" LastActivityDate="2010-11-08T13:00:32.597" Title="Definition of direct image?" Tags="&lt;terminology&gt;&lt;definition&gt;" AnswerCount="1" />
  <row Id="9424" PostTypeId="2" ParentId="9423" CreationDate="2010-11-08T13:00:32.597" Score="4" ViewCount="" Body="&lt;p&gt;They just mean &quot;image&quot;. An example here is the image of the set $(0, 2\pi)$ under the map $\sin(x)$, which is $(0, 1]$. &lt;/p&gt;&#xA;" OwnerUserId="630" LastActivityDate="2010-11-08T13:00:32.597" CommentCount="2" />
  <row Id="9425" PostTypeId="1" AcceptedAnswerId="9428" CreationDate="2010-11-08T13:05:32.693" Score="0" ViewCount="168" Body="&lt;p&gt;&lt;strong&gt;Question&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Determine all positive integers &lt;strong&gt;x&lt;/strong&gt; and &lt;strong&gt;y&lt;/strong&gt; that satisfy the equation &#xA;&lt;strong&gt;x^2 + 2 = 5y&lt;/strong&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2938" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-08T13:42:12.597" LastActivityDate="2010-11-08T14:45:14.250" Title="x^2 + 2 = 5y (being x and y positive integers)" Tags="&lt;number-theory&gt;&lt;homework&gt;&lt;diophantine-equations&gt;" AnswerCount="2" />
  <row Id="9426" PostTypeId="2" ParentId="9357" CreationDate="2010-11-08T13:05:59.287" Score="1" ViewCount="" Body="&lt;p&gt;The integral is a priori defined as the area beneath a curve (assuming Riemann-Stieltjes integral).&#xA;Hence, you should not assume that it is the anti-derivative of a function,&#xA;that is $\int f dx $ has nothing AT ALL to do with $\int_a^b f dx$&#xA;the first expression means anti-derivative, the second means area under a curve.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your proof is supposed to give a connection between the two notions, antiderivative, and area under graph.&lt;/p&gt;&#xA;" OwnerUserId="934" LastActivityDate="2010-11-08T13:05:59.287" />
  <row Id="9427" PostTypeId="1" CreationDate="2010-11-08T13:08:51.010" Score="1" ViewCount="70" Body="&lt;p&gt;Given an $R$-algebra $S$, is $Hom_{R-Alg}(\prod_{i=1}^n R, S) = Hom_{S-Alg}(\prod_{i=1}^n S, S)$?&lt;/p&gt;&#xA;" OwnerUserId="3267" LastEditorUserId="664" LastEditorDisplayName="" LastEditDate="2010-11-09T03:26:19.663" LastActivityDate="2010-11-09T03:26:19.663" Title="$Hom_{R-Alg}(\prod R, S)$" Tags="&lt;commutative-algebra&gt;" AnswerCount="2" CommentCount="0" />
  <row Id="9428" PostTypeId="2" ParentId="9425" CreationDate="2010-11-08T13:13:31.150" Score="-2" ViewCount="" Body="&lt;p&gt;Hint: When is $5y - 2$ a square? Modulo 5 this is $5y - 2 = 3$, but squares mod 5 are always 0, 1 or 4 so the equation has no solutions.&lt;/p&gt;&#xA;" LastActivityDate="2010-11-08T14:10:17.227" CommentCount="1" />
  <row Id="9430" PostTypeId="2" ParentId="9378" CreationDate="2010-11-08T13:36:11.437" Score="6" ViewCount="" Body="&lt;p&gt;David Einsenbud and Joe Harris' book &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0387986375&quot; rel=&quot;nofollow&quot;&gt;The Geometry of Schemes&lt;/a&gt; has a large number of really nice pictures which could help you gain a little bit of geometric intuition. That being said, it is also an excellent book which you should check out if you are interested in schemes.&lt;/p&gt;&#xA;" OwnerUserId="900" LastActivityDate="2010-11-08T13:36:11.437" />
  <row Id="9431" PostTypeId="1" AcceptedAnswerId="9439" CreationDate="2010-11-08T13:36:35.807" Score="1" ViewCount="186" Body="&lt;p&gt;&lt;strong&gt;Question&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Show that $n^2 + 3n + 5$ is not divisible by $121$, where $n$ is an integer.&lt;/p&gt;&#xA;" OwnerUserId="2938" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-09T04:47:42.287" LastActivityDate="2010-11-09T04:47:42.287" Title="$n^2 + 3n +5$ is not divisible by $121$" Tags="&lt;number-theory&gt;&lt;homework&gt;&lt;modular-arithmetic&gt;&lt;divisibility&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="9432" PostTypeId="1" CreationDate="2010-11-08T13:36:38.130" Score="1" ViewCount="96" Body="&lt;p&gt;The Clifford product of a pair of vectors $a,b$ is an associative operation defined by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ ab = a \cdot b + a \wedge b.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In sufficiently low dimensions I am used to being able to define the Clifford product on arbitrary $k$-vectors by repeatedly applying the vector definition.  For instance, suppose that I build a Clifford algebra over $\mathbb{R}^3$ with the usual (positive) Euclidean inner product.  Then I can easily write out the Clifford product of &lt;em&gt;any&lt;/em&gt; pair of basis bivectors.  For instance,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ e_{12}e_{13} = e_1 e_2 e_1 e_3 = -e_2(e_1 e_1)e_3 = -e_2 (1) e_3 = -e_{23}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In four dimensions I get stuck, because it's possible that two of the indices don't &quot;cancel,&quot; and then I don't know how to apply the product:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$e_{12}e_{34} = e_1 e_2 e_3 e_4.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where do I go from here?  I strongly suspect that this equals just $e_{1234}$, but I don't know how to show (in an explicit, pedantic, algebraic way) that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$e_1 e_2 e_3 e_4 = e_1 \wedge e_2 \wedge e_3 \wedge e_4.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="3268" LastActivityDate="2010-11-08T18:43:09.847" Title="How do I evaluate the Clifford product in dimensions greater than 3?" Tags="&lt;exterior-algebra&gt;" AnswerCount="2" CommentCount="5" />
  <row Id="9434" PostTypeId="1" CreationDate="2010-11-08T13:56:16.213" Score="2" ViewCount="211" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was working on a Stock Market Analysis website as a college project.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was keen on knowing some algorithms which I could potentially use for analysing trends in the stock market.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would be glad if some one could help.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank You.&lt;/p&gt;&#xA;" OwnerUserId="3269" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-08T14:03:17.140" LastActivityDate="2010-11-08T23:21:36.050" Title="Examples for Algorithms that can be used for Stock Market Analysis" Tags="&lt;algorithms&gt;&lt;finance&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" />
  <row Id="9435" PostTypeId="1" AcceptedAnswerId="9440" CreationDate="2010-11-08T14:00:00.710" Score="-3" ViewCount="129" Body="&lt;p&gt;I admit, the question is a little provocative, but is asked in all earnestness.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is inspired by this one: &lt;a href=&quot;http://math.stackexchange.com/questions/11/does-99999-1&quot;&gt;Does .99999... = 1?&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many of the arguments here point to the fact that 0.999... and 1 are arbitrarily close together. Put differently this means that &quot;there is no gap in-between the two&quot; - which concludes that these numbers are just the same one number: 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In another field, concerning the &lt;a href=&quot;http://en.wikipedia.org/wiki/Transcendental_number&quot; rel=&quot;nofollow&quot;&gt;transcendental numbers&lt;/a&gt;, it is proved that they are complete. So again put differently there are &quot;no gaps in the number-line&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Interestingly enough this is not seen as a proof that all transcendental numbers are only just one number. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the moment I can't find a good explanation why &quot;no gaps&quot; here means &quot;is equal&quot; and there it means &quot;they are all different&quot; (=complete). Could anybody please enlighten me?&lt;/p&gt;&#xA;" OwnerUserId="346" LastActivityDate="2010-11-08T14:38:27.567" Title="Does the &quot;no gaps&quot;-property with transcendental numbers mean that there is only &quot;one number&quot;?" Tags="&lt;transcendental-numbers&gt;" AnswerCount="2" CommentCount="6" FavoriteCount="1" />
  <row Id="9436" PostTypeId="2" ParentId="9427" CreationDate="2010-11-08T14:03:28.343" Score="3" ViewCount="" Body="&lt;p&gt;If $R$ is a commutative ring and $T$ an $R$-algebra, a morphism $\phi:R\times \cdots\times R\to T$ of $R$-algebras is in particular left $R$-linear, so there exist $a_1$, $\dots$, $a_n\in T$ such that $\phi(x_1,\dots,x_n)=x_1a_1+\cdots x_ny_n$. Since $\phi$ is unitary, $\phi(1,\dots,1)=1$, so $a_1+\cdots+a_n=1$. Let $e_i=\phi(0,\dots,0,1,0,\dots,1)$, with the one in the $i$th position. Then  multiplicativity implies that $e_i^2=e_i$ for all $i$ and $e_ie_j=0$ if $i\neq j$. This gives a concrete description of the set $\hom_{R\mathrm{-alg}}(R\times \cdots\times R, T)$, and you should be able to use it to answer your question.&lt;/p&gt;&#xA;" OwnerUserId="274" LastEditorUserId="274" LastEditorDisplayName="" LastEditDate="2010-11-08T14:09:26.607" LastActivityDate="2010-11-08T14:09:26.607" />
  <row Id="9437" PostTypeId="2" ParentId="9422" CreationDate="2010-11-08T14:08:32.167" Score="5" ViewCount="" Body="&lt;p&gt;I give it a try: If you want to know where you will be in x time driving a car you can find out by separating the different components: position at the moment, speed, acceleration, jolt and so on and add them all together.&lt;/p&gt;&#xA;" OwnerUserId="346" LastActivityDate="2010-11-08T14:08:32.167" CommentCount="1" />
  <row Id="9438" PostTypeId="2" ParentId="9422" CreationDate="2010-11-08T14:12:41.093" Score="6" ViewCount="" Body="&lt;p&gt;Think of a Taylor series not as one entity but as a sequence of approximations. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first term gives a constant approximation: f(x + h) is approximately f(x). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first two terms give a linear approximation: f(x + h) is approximately f(x) plus a trend term, h f'(x).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first three terms include a constant approximation, a linear trend, and a curvature term to account for the change in the linear trend: f(x + h) is approximately f(x) + h f'(x) + h^2 f''(x)/2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next you add a term to account for the change in the curvature, etc.&lt;/p&gt;&#xA;" OwnerUserId="247" LastActivityDate="2010-11-08T14:12:41.093" />
  <row Id="9439" PostTypeId="2" ParentId="9431" CreationDate="2010-11-08T14:15:57.893" Score="3" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\rm\quad\ m\ =\ n^2 + 3\:n+5\ \equiv\ (n-4)^2\ \:(mod\ 11)\ \Rightarrow\ n\ =\ 4+11\:k \ \Rightarrow\ m = \ldots\ (mod\ 11^2)$&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-11-08T14:15:57.893" />
  <row Id="9440" PostTypeId="2" ParentId="9435" CreationDate="2010-11-08T14:22:31.000" Score="4" ViewCount="" Body="&lt;p&gt;The property of the real numbers which formalizes these informal statements about gaps is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Least-upper-bound_property&quot; rel=&quot;nofollow&quot;&gt;least upper bound property&lt;/a&gt;.  It states that any set of real numbers that has an upper bound has a least upper bound, i.e. a supremum.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The set $\{ .9, .99, .999, ... \}$ has an upper bound, so it has a least upper bound.  The least upper bound cannot be less than $1$, so it must be exactly $1$.  On the other hand, for any reasonable definition of $.999...$, this number is also an upper bound of the same set, and it is less than or equal to $1$.  So the two are equal (because least upper bounds are unique).  The &quot;gap&quot; which does not exist here is the nonexistent gap between two least upper bounds, and its nonexistence is a purely formal property given the existence of least upper bounds, so the nonexistence of this gap has nothing to do with completeness or the least upper bound property; it's the existence of the least upper bound that is important.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The least upper bound property implies that Cauchy sequences of real numbers converge, which is usually what people mean when they say that there are no &quot;gaps&quot; in the number line.  Another way to formalize this statement is that the real line is &lt;a href=&quot;http://en.wikipedia.org/wiki/Connected_space&quot; rel=&quot;nofollow&quot;&gt;connected&lt;/a&gt;, which implies, among other things, the &lt;a href=&quot;http://en.wikipedia.org/wiki/Intermediate_value_theorem&quot; rel=&quot;nofollow&quot;&gt;intermediate value theorem&lt;/a&gt;.  (Compare with the case of $\mathbb{Q}$, which is totally disconnected.)&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-11-08T14:22:31.000" />
  <row Id="9441" PostTypeId="2" ParentId="8357" CreationDate="2010-11-08T14:36:03.183" Score="1" ViewCount="" Body="&lt;p&gt;If $a &gt; b &gt; 0$ then $a^2 &gt; b^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a &gt; b$ means there is a positive $k$ such that $a = b + k$. Squaring this equation we have $a^2 = b^2 + (2bk + k^2)$ but $2bk + k^2$ is just another positive so $a^2 &gt; b^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason we know $2bk + k^2$ is positive is because of the ordered field axioms, one says if $x$ and $y$ are positive so is $xy$ and another says that $x+y$ is positive. That is why we need $b$ to be positive.&lt;/p&gt;&#xA;" LastActivityDate="2010-11-08T14:36:03.183" />
  <row Id="9442" PostTypeId="2" ParentId="9435" CreationDate="2010-11-08T14:38:27.567" Score="2" ViewCount="" Body="&lt;p&gt;1/ To say that a topological set is &quot;complete&quot; has a mathematical meaning. For example, one definition of completeness is that all Cauchy-sequences converge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A number is not complete (or at least I have never heard of such a definition for numbers).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We say that the real numbers are complete, because they satisfy the definition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2/ When we have a sequence, we can (sometimes) see if this sequence converge or not. Again, the notion of convergence has a mathematical meaning, and suppose that the sequence obey a certain definition. When a sequence converges, we name an element &quot;limit&quot;, this is an element that obey a certain definition too. It is straightforward from the definitions of convergence and of limit, that if a limit exists, it is unique.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example we can study the sequence given by 0, 0.9, 0.99, 0.999, ... This sequence converges and has limit 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or we can study another sequence $\frac{p_n}{q_n}$ given by the conditions $q_n = 10^n$ and $\left(\frac{p_n}{q_n}\right)^2 \leq 2 \leq \left(\frac{p_n+1}{q_n}\right)^2$. This sequence will also converge, and the limit will be $\sqrt{2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if observe that this last example is a sequence composed only by elements in $\mathbb{Q}$, we might be surprised to see that the limit is actually not in $\mathbb{Q}$. Mathematicians don't like the fact of not having the limit within the working topological set. Therefore they build $\mathbb{R}$ which is the set all (classes of equivalence of) Cauchy-sequences of elements in $\mathbb{Q}$. Hence $\mathbb{R}$ satisfy the definition of completeness.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again, saying that $\mathbb{R}$ is complete is just a matter of definition. It has nothing to do with the word &quot;gap&quot;. As far as I know, &quot;gap&quot; has no mathematical meaning.&lt;/p&gt;&#xA;" OwnerUserId="2778" LastActivityDate="2010-11-08T14:38:27.567" CommentCount="1" />
  <row Id="9443" PostTypeId="2" ParentId="9425" CreationDate="2010-11-08T14:45:14.250" Score="1" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\rm\ \  x^2 \equiv\ \ldots\ \:(mod\ 5)\ $ versus $\rm\ \mathbb Z^2\ \equiv\ \{0,\:\pm1,\:\pm2\}^2\ \equiv\ \ldots \ \:(mod\ 5)$&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-11-08T14:45:14.250" />
  <row Id="9444" PostTypeId="2" ParentId="9422" CreationDate="2010-11-08T14:48:13.990" Score="4" ViewCount="" Body="&lt;p&gt;We know that the higher the degree of an equation, the more &quot;turning points&quot; it &lt;em&gt;may&lt;/em&gt; have. For example, a parabola has one &quot;turning point.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/D6fEV.jpg&quot; alt=&quot;alt text&quot;&gt;&#xA;(A parabola has an equation of the form $y=ax^2 + bx +c$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A cubic of the form $y=ax^3 + bx^2 +cx +d$ can have up to two &quot;turning points,&quot; though it may have fewer. In general, an equation of degree $n$ may have up to $n-1$ turning points.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/35376.gif&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Here is the polynomial $f(x) = 2x^4 - x^3 -3x^2 + 7x - 13$. It is degree 4 and it has the maximum number of turning points, 4-1=3. But, keep in mind, some degree 4 polynomials have only one or two turning points. The degree gives us the MAXIMUM number as $n=1$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is important because, if you want to use a polynomial to approximate a function, you will want to use a polynomial of high enough degree to match the &quot;features&quot; of the function. The Taylor series will let you do this with functions that are &quot;infinitely differentiable&quot; since it uses the derivatives of the function to approximate the functions behavior.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are Taylor polynomials of increasing degree and the sine curve. Notice how they are &quot;wrapping around&quot; the sine curve, giving an approximation that fits better and better over more of the curve as the degree of the Taylor polynomial increases. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/j9caf.gif&quot; alt=&quot;alt text&quot;&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Source for this image: &lt;a href=&quot;http://202.38.126.65/navigate/math/history/Mathematicians/Taylor.html&quot; rel=&quot;nofollow&quot;&gt;http://202.38.126.65/navigate/math/history/Mathematicians/Taylor.html&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the sine curve has &lt;em&gt;so many&lt;/em&gt; turning points it is easy to see that to match all of the features of the sine curve we will need to take the limit of the $n^{th}$ degree Taylor polynomial as $n \rightarrow \infty$.*&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's the intuition behind the Taylor series. The higher the degree, the better the &quot;fit.&quot;  Why? Because higher degree curve have more &quot;turning points&quot; so they can better match the shape of things like the sine function. (As long as the function we are approximating is differentiable.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;*Side note: A function may have only a few turning points and still need infinitely many terms of the Taylor polynomial. Take the &lt;a href=&quot;http://en.wikipedia.org/wiki/Catenary&quot; rel=&quot;nofollow&quot;&gt;catenary&lt;/a&gt;, for example, which only has one turning point since it looks like a parabola. The Taylor series for the catenary will not have any terms where the coefficients are zero, since the derivatives of the catenary are hyperbolic sinusoidal functions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, even with the catenary, higher degree polynomials give a better approximation.   &lt;/p&gt;&#xA;" OwnerUserId="2563" LastEditorUserId="2563" LastEditorDisplayName="" LastEditDate="2010-11-08T15:01:34.370" LastActivityDate="2010-11-08T15:01:34.370" CommentCount="1" />
  <row Id="9446" PostTypeId="1" CreationDate="2010-11-08T14:57:14.103" Score="7" ViewCount="215" Body="&lt;p&gt;Why is every group the quotient of a free group by a normal subgroup?&lt;/p&gt;&#xA;" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-08T15:43:10.620" LastActivityDate="2010-11-08T19:09:51.407" Title="Every group is the quotient of a free group by a normal subgroup" Tags="&lt;group-theory&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="1" />
  <row Id="9447" PostTypeId="2" ParentId="9422" CreationDate="2010-11-08T14:59:33.993" Score="0" ViewCount="" Body="&lt;p&gt;We are approximating a function by polynomials at a point.As a first approximation, we give a give polynomial whose value at that point is same as the functions. In the second step, we make the first derivative equal too. In the third step,the second derivative is made equal and so on... &lt;/p&gt;&#xA;" OwnerUserId="3271" LastActivityDate="2010-11-08T14:59:33.993" />
  <row Id="9448" PostTypeId="2" ParentId="9432" CreationDate="2010-11-08T15:13:46.473" Score="2" ViewCount="" Body="&lt;p&gt;It's not easy to give a proof without knowing exactly what your&#xA;definitions of the Clifford and exterior algebras are, so the&#xA;following argument is still a little handwaving. But let's say that we&#xA;somehow have defined what the exterior algebra of a vector space $V$&#xA;is; we know that its elements are multivectors, and the rule which&#xA;generates everything is that $x \wedge y + y \wedge x=0$ if $x$ and&#xA;$y$ are vectors in $V$. The Clifford algebra has the same&#xA;elements as the exterior algebra, and the same linear space structure,&#xA;but the multiplication is different: it is generated by the rule&#xA;$xy+yx=Q(x,y)$ where $Q$ is the inner product on $V$. Since&#xA;&quot;orthogonal&quot; means that $Q(e_i,e_j)=0$, it shouldn't be too hard to&#xA;believe that the Clifford and exterior multiplications agree for&#xA;orthogonal vectors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe &lt;a href=&quot;http://en.wikipedia.org/wiki/Clifford_algebra#Relation_to_the_exterior_algebra&quot; rel=&quot;nofollow&quot;&gt;this section on Wikipedia&lt;/a&gt; can be of some help too?&lt;/p&gt;&#xA;" OwnerUserId="1242" LastEditorUserId="1242" LastEditorDisplayName="" LastEditDate="2010-11-08T16:04:53.183" LastActivityDate="2010-11-08T16:04:53.183" />
  <row Id="9449" PostTypeId="2" ParentId="9422" CreationDate="2010-11-08T15:32:30.663" Score="1" ViewCount="" Body="&lt;p&gt;Predict global while computing local!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A Taylor expansion of a function $f$ around some value $x_0$ is similar to a prediction of the function at a neighboring value $x$ knowing progressively more about the variation of $f$ at the point $x_0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First step: easiest prediction: nothing changed, that is, $f(x) = f(x_0)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second step: we know the first derivative, so we predict the function was linear between $x_0$ and $x$ : $f(x) = f(x_0) + (x-x_0)f'(x_0)$. See, everything still local as the derivate is given at $x_0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The next step give a generalization of this predictions for higher derivatives. The different forms give bounds to the error or more knowledge of the residual.&lt;/p&gt;&#xA;" OwnerUserId="1090" LastActivityDate="2010-11-08T15:32:30.663" />
  <row Id="9450" PostTypeId="2" ParentId="9446" CreationDate="2010-11-08T15:42:52.493" Score="5" ViewCount="" Body="&lt;p&gt;Given any group G, let F(G) be the free group whose generators are the elements of G. Define a group homomorphism f from F(G) to G in the obvious way: by sending the generator g of F(G) to the element g of G. This is a surjective homomorphism, so G is isomorphic to F(G) / ker f.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, let M be the set of words of the form $g_1 g_2 h^{-1}$ where $g_1 g_2 = h$ in G -- in other words, M encodes the multiplication table for G. Let N be the normal closure of M. Then G is isomorphic to F(G) / N.&lt;/p&gt;&#xA;" OwnerUserId="759" LastActivityDate="2010-11-08T15:42:52.493" />
  <row Id="9451" PostTypeId="2" ParentId="7892" CreationDate="2010-11-08T15:43:35.800" Score="4" ViewCount="" Body="&lt;p&gt;From &lt;em&gt;&lt;a href=&quot;http://books.google.co.uk/books?id=Kx2cjyzTIYkC&amp;amp;printsec=frontcover&amp;amp;dq=proofs+without+words&amp;amp;hl=en&amp;amp;ei=7RnYTLrRHZnc4wbR-on9Bw&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=1&amp;amp;ved=0CDEQ6AEwAA#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;Proofs without Words&lt;/a&gt;&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/0SBs6.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="723" LastEditorUserId="723" LastEditorDisplayName="" LastEditDate="2010-11-08T16:11:57.947" LastActivityDate="2010-11-08T16:11:57.947" CommentCount="1" />
  <row Id="9452" PostTypeId="2" ParentId="9446" CreationDate="2010-11-08T15:47:13.820" Score="8" ViewCount="" Body="&lt;p&gt;It is a consequence of the universal property of free groups. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $X$ is a set, then the &lt;strong&gt;free group on&lt;/strong&gt; $X$, $F(X)$, is a group $F(X)$ that contains $X$ as a subset, and such that for every group $G$ and every set-theoretic map $f\colon X\to G$ there exists a unique group homomorphism $\mathcal{F}\colon F(X)\to G$ such that $\mathcal{F}(x) = f(x)$ for all $x\in X$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $G$ is a group, let $X$ be the underlying set of $G$ (the set of all elements of $G$, considered merely as a set), and let $F(X)$ be the free group on $X$. The set-theoretic map $i\colon X\to G$ that maps each element of $G$ to itself induces a homomorphism $\mathcal{F}\colon F(X)\to G$ such that $\mathcal{F}(x)=i(x)$ for all $x\in X$. Since the image of $\mathcal{F}$ includes at least the image of $i$, the map $\mathcal{F}$ is onto. By the First Fundamental Theorem of Homomorphisms, the image of $\mathcal{F}$ (which is $G$) is isomorphic to $F(X)/\mathrm{ker}(\mathcal{F})$. So $G$ is isomorphic to the quotient of a free group by a normal subgroup.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can replace the underlying set of $G$ with any generating set of $G$ and the proof goes through.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Added:&lt;/em&gt; Of course, this answer requires that you know that a free group exists on any set, and that it has the appropriate universal property. A good exposition, with motivation and three different constructions of the free group, is in &lt;a href=&quot;http://math.berkeley.edu/~gbergman/245/&quot; rel=&quot;nofollow&quot;&gt;Chapter 2 of George Bergman's book&lt;/a&gt;, &lt;em&gt;An invitation to General Algebra and Universal Constructions&lt;/em&gt;. Here's a link to a &lt;a href=&quot;http://math.berkeley.edu/~gbergman/245/Ch.2.ps&quot; rel=&quot;nofollow&quot;&gt;Postscript version&lt;/a&gt;. &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-08T17:14:41.037" LastActivityDate="2010-11-08T17:14:41.037" />
  <row Id="9453" PostTypeId="2" ParentId="8357" CreationDate="2010-11-08T16:06:32.617" Score="1" ViewCount="" Body="&lt;p&gt;Yes, when &lt;strong&gt;a&lt;/strong&gt; and &lt;strong&gt;b&lt;/strong&gt; are positive real numbers. In this case, we can write:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a&gt;b =&gt; a-b&gt;0 =&gt; (a+b).(a-b)&gt;0 =&gt; (a^2)-(b^2)&gt;0&lt;/p&gt;&#xA;" OwnerUserId="2938" LastActivityDate="2010-11-08T16:06:32.617" />
  <row Id="9454" PostTypeId="2" ParentId="9372" CreationDate="2010-11-08T16:08:10.400" Score="5" ViewCount="" Body="&lt;p&gt;Further to the other answers posted, not only does the sum converge, but you can conclude that it converges precisely because of the way it was derived. As you state, the terms $C_n=\frac{1\cdot3\cdot5\cdot7\cdots(2n-3)}{2^nn!}$ arise as the Taylor expansion of $f(x)=\sqrt{x}$ about $x=1$,&#xA;$$&#xA;\sqrt{1-x}=1-\frac{x}{2}-\sum_{n=2}^\infty C_nx^n.&#xA;$$&#xA;As $\sqrt{1-x}$ is a well-defined analytic function for $\Vert x\Vert &amp;lt; 1$, this has radius of convergence 1. Also, as the square root is well defined at 0, &lt;a href=&quot;http://en.wikipedia.org/w/index.php?title=Monotone_convergence_theorem&amp;amp;oldid=388896065#Convergence_of_a_monotone_series&quot; rel=&quot;nofollow&quot;&gt;monotone convergence&lt;/a&gt; gives&#xA;$$&#xA;\sum_{n=2}^\infty C_n=\lim_{x\to1}\sum_{n=2}^\infty C_nx^n=\lim_{x\to1}\left(1-\frac{x}{2}-\sqrt{1-x}\right)=\frac12.&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="1321" LastActivityDate="2010-11-08T16:08:10.400" />
  <row Id="9455" PostTypeId="1" AcceptedAnswerId="9457" CreationDate="2010-11-08T16:27:02.170" Score="0" ViewCount="104" Body="&lt;p&gt;If G is a finite nilpotent group, then every minimal normal subgroup of $G$ is contained in the center of $G$ and has prime order.&lt;/p&gt;&#xA;" OwnerUserId="2177" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-08T16:59:36.290" LastActivityDate="2010-11-08T17:02:28.897" Title="A question about nilpotent group" Tags="&lt;group-theory&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="9456" PostTypeId="2" ParentId="9402" CreationDate="2010-11-08T17:01:36.917" Score="7" ViewCount="" Body="&lt;p&gt;Since &#xA;$$\frac{x}{1+x^2}=\int_{0}^{\infty}e^{-y}\sin (xy)dy,$$&#xA;we have that&#xA;$$I=\int_{0}^{\infty}\frac{\cos bx}{1+x^2}dx=\int_{0}^{\infty}\frac{\cos bx}{x}dx\int_{0}^{\infty}e^{-y}\sin (xy)dy.$$&#xA;Changing the order of integration (which can be justified by the standard method) yields&#xA;$$I=\int_{0}^{\infty}e^{-y}dy\int_{0}^{\infty}\frac{\sin xy}{x} \cos bx dx.$$&#xA;The calculation of the integral (a.k.a. the discontinuous Dirichlet factor)&#xA;$$\int_{0}^{\infty}\frac{\sin xy}{x} \cos bx dx=\begin{cases}0, &amp;amp; 0 &amp;lt; y &amp;lt; b&#xA;\\ \ \\ \pi/2, &amp;amp; 0 &amp;lt; b &amp;lt; y, \end{cases}$$&#xA;can be easily reduced to the calculation of the standard &lt;a href=&quot;http://en.wikipedia.org/wiki/Dirichlet_integral&quot; rel=&quot;nofollow&quot;&gt;Dirichlet integral&lt;/a&gt;.&#xA;Therefore,&#xA;$$I=\frac{\pi}{2}\int_{b}^{\infty}e^{-y}dy=\frac{\pi}{2}e^{-b}.$$&lt;/p&gt;&#xA;" OwnerUserId="723" LastActivityDate="2010-11-08T17:01:36.917" CommentCount="3" />
  <row Id="9457" PostTypeId="2" ParentId="9455" CreationDate="2010-11-08T17:02:28.897" Score="3" ViewCount="" Body="&lt;p&gt;A finite nilpotent group is a product of $p$-groups. So you can do a very quick computation to show that you can reduce to the case where $G$ is a $p$-group.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then look at &lt;a href=&quot;http://math.stackexchange.com/questions/8389/a-question-about-p-group/8400#8400&quot;&gt;this question&lt;/a&gt; to answer Steve D's comment query. &lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-11-08T17:02:28.897" CommentCount="2" />
  <row Id="9458" PostTypeId="2" ParentId="9414" CreationDate="2010-11-08T17:30:42.227" Score="0" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;LEMMA&lt;/strong&gt; $\rm\quad\quad\ \  (f_0\:e^{\:g})^{(n)}\ \ \ \ =\ \ f_n\ e^g\ \: $ when $\rm\: \ g'\ f_n + f_n'\: =\ f_{n+1} $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Base case:&lt;/strong&gt; $\rm\quad\ (f_0\:e^{\:g})^{(0)}\ \ \ \ =\ \ f_0\ e^g\ \ $ is true.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Induction:&lt;/strong&gt; $\rm\quad (f_0\:e^{\:g})^{(n+1)} = \ (f_n\ e^g)'\ \:=\  (g'\ f_n + f_n')\ e^g \ =\ f_{n+1}\ e^g\quad\ $  &lt;strong&gt;QED&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now $\rm\ g = ax,\ f_{n+1} = a^n\:(ax+n+1)\ =\ a\ (a^{n-1} (ax + n))+ a^n\ =\ g'\ f_n + f_n'$&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-11-08T17:30:42.227" />
  <row Id="9459" PostTypeId="2" ParentId="7892" CreationDate="2010-11-08T17:31:15.737" Score="3" ViewCount="" Body="&lt;p&gt;Another proof uses the fact that $\displaystyle \pi &gt; e$ and that $e^x &gt; 1 + x$ for $x &gt; 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have $$e^{\pi/e -1} &gt; \pi/e$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$e^{\pi/e} &gt; \pi$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and thus&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$e^{\pi} &gt; \pi^e$$&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-11-08T17:31:15.737" />
  <row Id="9460" PostTypeId="1" AcceptedAnswerId="9490" CreationDate="2010-11-08T17:37:54.460" Score="2" ViewCount="194" Body="&lt;p&gt;Hello,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $g: \mathbb{C}\times[a,b]\to\mathbb{R}$ a continuous function and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$g(t)=g(h_0,t)=\lim_{h\to h_0} g(h,t) \ \forall t\in \mathbb{R}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is the following result/reasoning correct ? If we have a sequence of functions, we need uniform convergence to pass the limit inside the integral, but it seems it works everytime here. Or there is a flaw ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\lim_{h\to h_0} \int_a^b g(h,t)dt= \int_a^b g(t)dt$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;because&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Idea 1:&lt;/strong&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$$\forall t\in[a,b], \ \forall&#xA;  \epsilon &gt;0, \exists \delta(t)&gt;0\text{&#xA;  such that }|g(h,t)-g(t)|&amp;lt;\epsilon \&#xA;  \forall |h-h_0|&amp;lt;\delta(t)$$&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Then if&#xA;  $|h-h_0|&amp;lt;\delta=\inf_{t\in[a,b]}&#xA;  \delta(t)$, &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$$|g(h,t)-g(t)|&amp;lt;\epsilon \ \forall&#xA;  |h-h_0|&amp;lt;\delta(t), \ \forall&#xA;  t\in[a,b]$$&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Thus&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$$\left|\int_a^b&#xA;  (g(h,t)-g(t))dt\right|\le \int_a^b&#xA;  |g(h,t)-g(t)|dt\le (b-a)\epsilon$$&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;Doesn't work if $\inf_{t\in[a,b]}&#xA;  \delta(t)=0$, see George's comment.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;~&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Idea 2:&lt;/strong&gt; (added after George's comment)&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Let $\overline{B(h_0, \gamma)}\subset&#xA;  C$ a closed ball. Resctrict $g$ to $&#xA;  \overline{B(h_0, \gamma)}\times[a,b]$.&#xA;  Because it is continuous on a compact,&#xA;  this function will be uniformly&#xA;  continuous. Thus&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$$\forall \epsilon &gt;0 \ \exists&#xA;  \delta&gt;0 \ : \&#xA;  |g(h,t)-g(h_0,t)|&amp;lt;\epsilon \ \forall&#xA;  |h-h_0|&amp;lt;\min(\delta, \gamma)$$&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;and&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$$\left|\int_a^b&#xA;  (g(h,t)-g(t))dt\right|\le \int_a^b&#xA;  |g(h,t)-g(t)|dt\le (b-a)\epsilon$$&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;if $|h-h_0|&amp;lt;\min(\delta, \gamma)$.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Does this one work ;) ?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Thank you for any help :)&lt;/p&gt;&#xA;" OwnerUserId="524" LastEditorUserId="524" LastEditorDisplayName="" LastEditDate="2010-11-08T19:50:02.707" LastActivityDate="2010-11-08T21:59:50.420" Title="Limit of integral" Tags="&lt;real-analysis&gt;&lt;integral&gt;&lt;limit&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="1" />
  <row Id="9461" PostTypeId="1" AcceptedAnswerId="9465" CreationDate="2010-11-08T18:08:21.630" Score="1" ViewCount="191" Body="&lt;p&gt;A set M $\in$ $\mathbb{Z}^2$ is defined by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Rule 1) $(0, 1)$ $\in$ M&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Rule 2) If $(x, y)$ $\in$ M, then $(x + 1, y + 2x + 3)$ $\in$ M&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(a) (2 points) Starting with $(0, 1)$, write out the six pairs with the smallest first coordinates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(b) (2 points) State the (simple) relationship that holds between the first and second coordinates of all pairs in M.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ok for a I got:&#xA;$(0,1)$,$(1,4)$,$(2,9)$,$(3,16)$,$(4,25)$,$(5,36)$,$(6,49)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However for b, I cant see the relationship. Can anyone give me any hints?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ok I found the relation: $y=(x+1)^2$. However, I must now prove this claim using structural induction. I can get up to the inductive hypothesis, but after that I'm not really sure what to do.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Base case: $(0,1), 1=(0+1)^2 = 1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Inductive Hypothesis: For any element $x,y$ if $x\in M$ then $y=(x+1)^2$. We must show that $y+1=(x+1)^2 + 1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not really sure that my IH is correct either.&lt;/p&gt;&#xA;" OwnerUserId="2985" LastEditorUserId="2985" LastEditorDisplayName="" LastEditDate="2010-11-08T18:45:48.473" LastActivityDate="2010-11-08T19:33:58.363" Title="Find a pattern in a recursive definition" Tags="&lt;number-theory&gt;&lt;homework&gt;&lt;discrete-mathematics&gt;&lt;number&gt;" AnswerCount="3" CommentCount="3" />
  <row Id="9462" PostTypeId="2" ParentId="9446" CreationDate="2010-11-08T18:23:09.907" Score="4" ViewCount="" Body="&lt;p&gt;This is one of the most intuitive observations in all of group theory, and it illustrates the quotient operation in the most fundamental way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll provide two separate answers. The first is fully intuitive; the second is a formalized version of the first.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;First answer:&lt;/strong&gt; Take a group $G$. A relation on $G$ is an equation satisfied by some of the elements. For instance, $eg = g$ where $e$ is the identity is a relation satisfied by all group elements $g \in G$. Because we can always multiply by inverses in a group, we can rewrite this relation as $egg^{-1} = gg^{-1} = e$, i.e., $e = e$. This can be applied to any relation. If $G$ is abelian, then $ab = ba$ for all $a,b \in G$, and we can rewrite this as $aba^{-1}b^{-1} = e$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, a relation asserts that some product of group elements coincides with the identity, so the only information we need to understand the relation is the product which occurs on the left side of the equals sign.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now every group has a few relations which are implied directly by the group axioms. $aa^{-1} = e$ is one of them. We can ask whether the group has any &lt;em&gt;extra&lt;/em&gt; relations which are not implied by the group axioms. If no such relations exist, i.e., if the only relations which hold are those which must hold by virtue of the group axioms, then the group is said to be &lt;strong&gt;free&lt;/strong&gt;; the group is &quot;free of additional relations.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have a group $G$, one natural thing to do is to introduce new relations into it and to thereby create new groups. But you can't just introduce completely random relations because (a) the relations can't contradict each other or pre-exising relations and (b) the resulting structure must again be a group. Now we saw earlier that a relation can be specified as a product of group elements. In order that the relations satisfy (a) and (b), it turns out it is necessary and sufficient that the corresponding products form a &lt;em&gt;normal subgroup&lt;/em&gt; $N$. The result of introducing the collection of relations $N$ into the group $G$ is the quotient $G/N$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any group $G$ can be obtained in this manner. You start with the free group $F$ whose generators are elements of $G$ &lt;em&gt;considered as a set&lt;/em&gt;. And then you look at all the additional relations satisfied by elements of $G$ and assemble them into a normal subgroup $N$. Then $G = F/N$ by the above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Second answer:&lt;/strong&gt; Given any set $S$, the free group on $S$ is that group $F(S)$ for which every function $f : S \rightarrow G$ from $S$ to an arbitrary group $G$ extends to a unique &lt;em&gt;homomorphism&lt;/em&gt; $\tilde{f} : F(S) \rightarrow G$. There are various ways of constructing $F(S)$ explicitly. For instance, you may take $F(S)$ to consist of words over the alphabet whose letters are elements of $S$ and $S'$, where $S'$ has the letter $s^{-1}$ (a symbol at the moment) for each symbol $s \in S$. It's important to notice that $F(S)$ actually contains &lt;em&gt;equivalence classes&lt;/em&gt; of words, because we introduce the obvious cancellation rules; e.g., $abb^{-1}c$ can be reduced via cancellation to $ac$. It must be proved that all possible algorithms for reduction yield the same reduced word; I'll omit that step.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You also have to prove that this group $F(S)$ satisfies the stated universal property. I won't prove this in detail, but it is more or less intuitive. Since $\tilde{f}$ has to be a homomorphism, we find, for instance, that $\tilde{f}(ab) = \tilde{f}(a) \tilde{f}(b) = f(a)f(b)$. In general, since $f$ is defined for all elements of $S$, $\tilde{f}$ is thereby defined uniquely for all elements of $F(S)$. [It is via similar reasoning that you may determine that it is sufficient to know the values of a linear operator on the elements of a basis of a vector space.]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we start with our group $G$ which we would like to write as a quotient of a free group. Which free group? That free group whose generators are the symbols from $G$. So we pick $F(G)$. Now we need to introduce the needed relations in order to collapse $F(G)$ into $G$. How do we carry it out? By the first answer, we could easily accomplish this if only we knew the normal subgroup $N$ of relations, but it seems that in this general case we don't really know $N$ concretely.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, we can figure out $N$ as follows. We can take the identity map $f : G \rightarrow G$ and extend it to a homomorphism $\tilde{f} : F(G) \rightarrow G$. The extension $\tilde{f}$ is in general not injective, and its kernel is precisely the group of relations $N$! (Formally this is an application of one of the standard theorems on homomorphisms.) Then $G = F(G)/N$ as before.&lt;/p&gt;&#xA;" OwnerUserId="251" LastEditorUserId="251" LastEditorDisplayName="" LastEditDate="2010-11-08T19:09:51.407" LastActivityDate="2010-11-08T19:09:51.407" CommentCount="2" />
  <row Id="9463" PostTypeId="2" ParentId="9432" CreationDate="2010-11-08T18:43:09.847" Score="2" ViewCount="" Body="&lt;p&gt;It appears that you are thinking in terms of &quot;geometric algebra&quot;.&#xA;For those unfamliar with this doctrine, it's a way of regarding&#xA;the Clifford algebra on an inner product space $(V,Q)$ and the exterior algebra&#xA;on the $V$ as the same set, but with different product operations.&#xA;Let's stick to a ground field of characteristic zero. Then the Clifford&#xA;algebra $C$ of $(V,Q)$ is generated by the elements $v\in V$ with relations&#xA;$vv=Q(v)$. Now for $v_1,\ldots,v_k\in V$ one can define wedge product&#xA;$$v_1\wedge v_2\wedge\cdots\wedge v_k&#xA;=\frac1{k!}\sum_{\pi\in S_k}(-1)^{\text{sgn}(\pi)}v_{\pi(1)}v_{\pi(2)}\cdots&#xA;v_{\pi(k)}\in C.$$&#xA;This wedge product identifies $C$ with the exterior algebra $\bigwedge(V)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With this definition, Hans's comments are absolutely right. If&#xA;$v_1,\ldots,v_k$ are pairwise orthogonal then the Clifford product&#xA;$v_1v_2\cdots v_k$ equals the exterior product&#xA;$v_1\wedge v_2\wedge\cdots\wedge v_k$, and this is certainly the case&#xA;for orthogonal basis vectors: $e_1 e_2 e_3 e_4&#xA;=e_1\wedge e_2\wedge e_3\wedge e_4$.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-11-08T18:43:09.847" />
  <row Id="9464" PostTypeId="2" ParentId="9427" CreationDate="2010-11-08T18:44:35.647" Score="1" ViewCount="" Body="&lt;p&gt;Tensor products commute with finite products (of modules, but the same isom. gives it for rings), in particular $(\prod_{i=1}^{n} R) \otimes_R S = \prod_{i=1}^{n} S$.&lt;/p&gt;&#xA;" OwnerUserId="1650" LastActivityDate="2010-11-08T18:44:35.647" CommentCount="4" />
  <row Id="9465" PostTypeId="2" ParentId="9461" CreationDate="2010-11-08T18:47:43.383" Score="1" ViewCount="" Body="&lt;p&gt;Your inductive step is to show that if $(x,(x+1)^2) \in M$, then $(x+1,(x+2)^2) \in M$.  So apply Rule 2 to $(x,(x+1)^2)$&lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2010-11-08T19:04:11.107" LastActivityDate="2010-11-08T19:04:11.107" CommentCount="3" />
  <row Id="9466" PostTypeId="2" ParentId="9461" CreationDate="2010-11-08T18:52:51.660" Score="1" ViewCount="" Body="&lt;p&gt;Your inductive hypothesis is wrong. You should be showing that if $(x,(x+1)^2)\in M$, then $(x+1, ((x+1)+1)^2) \in M$. From there you should be able to see how to get the result (just use Rule 2).&lt;/p&gt;&#xA;" OwnerUserId="1016" LastEditorUserId="1016" LastEditorDisplayName="" LastEditDate="2010-11-08T18:58:25.373" LastActivityDate="2010-11-08T18:58:25.373" />
  <row Id="9467" PostTypeId="2" ParentId="9461" CreationDate="2010-11-08T19:04:33.643" Score="1" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\rm\ \ f\ (x,\:y)\ = \ (x+1,\ y+2x+3)\ \Rightarrow\ f\ (x,\ (x+1)^2)\ =\ (x+1,\ (x+2)^2) $&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-11-08T19:33:58.363" LastActivityDate="2010-11-08T19:33:58.363" />
  <row Id="9468" PostTypeId="1" AcceptedAnswerId="9506" CreationDate="2010-11-08T19:12:31.090" Score="5" ViewCount="161" Body="&lt;p&gt;&lt;a href=&quot;http://mathworld.wolfram.com/CatalansConstant.html&quot; rel=&quot;nofollow&quot;&gt;Quite&lt;/a&gt; a  &lt;a href=&quot;http://mathworld.wolfram.com/MertensConstant.html&quot; rel=&quot;nofollow&quot;&gt;few&lt;/a&gt; mathematical constants are known to arise in several branches of mathematics (more &lt;a href=&quot;http://en.wikipedia.org/wiki/Euler_product&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;). I have no doubt that they are useful to the understanding of some mathematical structures and add to the whole body of knowledge of mathematics as a whole. Sometimes, however, I find myself looking at these constants and I feel unsatisfied. I suspect that more constants can be expressed in more 'fundamental' constants like $\pi, e$ and $\gamma$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Although Euler probably came close to be a living one, mathematicians of the past did not have computers. We can now use computers to determine te value of the afore mentioned constants with great precision. We can also try to conjecture the exact value of these constants by finding the value of some arithmetic combination of $\pi, e$ and $\gamma$ and the real numbers that corresponds to one of the 'unevaluated' constants with high precision. My question is: have experimental mathematicians 'found' the value(s) of some of the constants I described (by the the method I described or a similar one)?   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Max&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S. If someone knows of a reference to a paper/book that summarizes some results in this subield of a field, I would be grateful to him/her.&lt;/p&gt;&#xA;" OwnerUserId="1055" LastEditorUserId="1055" LastEditorDisplayName="" LastEditDate="2010-11-19T15:46:30.380" LastActivityDate="2010-11-19T16:12:56.710" Title="Suggesting closed-form representations of mathematical constants by means of experimental mathematics?" Tags="&lt;number-theory&gt;" AnswerCount="5" CommentCount="7" />
  <row Id="9470" PostTypeId="2" ParentId="9468" CreationDate="2010-11-08T19:18:58.530" Score="2" ViewCount="" Body="&lt;p&gt;Have you tried the &lt;a href=&quot;http://oldweb.cecm.sfu.ca/projects/ISC/ISCmain.html&quot; rel=&quot;nofollow&quot;&gt;Inverse Symbolic Calculator&lt;/a&gt;?&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-08T19:18:58.530" CommentCount="7" />
  <row Id="9471" PostTypeId="2" ParentId="457" CreationDate="2010-11-08T19:28:31.727" Score="1" ViewCount="" Body="&lt;p&gt;Well, one Stokes theorem surely deserves to be viewed in physical terms. It's hard not to think about these matters in terms of physical flows (be it fluid, or electrical fields), sinks, sources, etc. and this can give intuition when dealing with arbitrary differential forms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another physical concept often used in abstract mathematics is conservation of energy, or more generally integrals of motion. Used all the time in analysis of PDE, as far as I know.&lt;/p&gt;&#xA;" OwnerUserId="3223" LastActivityDate="2010-11-08T19:28:31.727" CommunityOwnedDate="2010-11-08T19:28:31.727" />
  <row Id="9472" PostTypeId="2" ParentId="9468" CreationDate="2010-11-08T19:33:26.010" Score="3" ViewCount="" Body="&lt;p&gt;The rational constant for an Apery like series for $$\zeta(4)$$ was found experimentally, using continued fractions.&lt;/p&gt;&#xA;" LastActivityDate="2010-11-08T19:33:26.010" CommentCount="2" />
  <row Id="9473" PostTypeId="2" ParentId="9468" CreationDate="2010-11-08T19:38:19.547" Score="0" ViewCount="" Body="&lt;p&gt;Gauss discovered the AGM/elliptic integral connection through the value $$\frac{\omega}{\pi}$$ by experimental mathematics.&lt;/p&gt;&#xA;" LastActivityDate="2010-11-08T19:38:19.547" />
  <row Id="9474" PostTypeId="2" ParentId="9420" CreationDate="2010-11-08T19:52:26.433" Score="2" ViewCount="" Body="&lt;p&gt;The function $h$ violates the one-wayness property of cryptographically good hash functions, because it's not computationally infeasible to recover the message $w$ from the hash $h(w)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we can generate (obtain) a set of message-hash pairs, then we can use the XOR operations on some of the hashes to get the $h(w)$ hash. If we use the same XORing procedure on the corresponding messages, we can recover the message $w$ as well. The whole process is computationally feasible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Kudos to &lt;a href=&quot;http://math.stackexchange.com/users/232/qiaochu-yuan&quot;&gt;Qiaochu Yuan&lt;/a&gt; for the help.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;(be free to comment or edit)&lt;/sup&gt;&lt;/p&gt;&#xA;" OwnerUserId="2411" LastActivityDate="2010-11-08T19:52:26.433" />
  <row Id="9475" PostTypeId="1" AcceptedAnswerId="9477" CreationDate="2010-11-08T20:03:18.823" Score="4" ViewCount="230" Body="&lt;p&gt;The usual symbol for the cardinality of the continuum (i.e. the real numbers) is Fraktur $\mathfrak{c}$. However, I recall some sources also using $\aleph$ (with no subscript). This usage is not mentioned in Wikipedia or Mathworld, but I found some support for it over Google.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is the $\aleph$ notation standard?&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-08T20:31:24.067" Title="Symbol for the cardinality of the continuum" Tags="&lt;set-theory&gt;&lt;notation&gt;" AnswerCount="2" CommentCount="6" />
  <row Id="9477" PostTypeId="2" ParentId="9475" CreationDate="2010-11-08T20:12:22.290" Score="2" ViewCount="" Body="&lt;p&gt;I have seen the use of $\aleph$ and $\mathfrak{c}$ as well the explicit $2^{\aleph_0}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're uncertain, it's best just to add &quot;We denote the cardinality of the continuum by ...&quot;.&lt;/p&gt;&#xA;" OwnerUserId="622" LastActivityDate="2010-11-08T20:12:22.290" />
  <row Id="9479" PostTypeId="2" ParentId="9319" CreationDate="2010-11-08T20:28:43.747" Score="1" ViewCount="" Body="&lt;p&gt;Suppose you want to find positive integers $n, k$ such that $|n - 2 \pi k - \arcsin r| &amp;lt; \frac{1}{m}$ for some large positive integer $m$ and $r \in [-1, 1]$.  Then I claim that some value of $n$ less than $2\pi m$ works.  This is the pattern I was trying to get you to see, although I wasn't doing a very good job of it (I should've asked you to plot $k$ instead of $n$).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately I think this is a little hard to prove except when $r = 0$; fortunately, using the second half of Marek's argument you can prove the same result with a weaker bound on $n$ for general $r$ using the case of $r = 0$, so I would encourage you to try that case first.  To do this, try plotting the fractional parts of the numbers $\frac{n}{2\pi}$ in $[0, 1)$.  &lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-11-08T20:28:43.747" CommentCount="1" />
  <row Id="9480" PostTypeId="2" ParentId="9475" CreationDate="2010-11-08T20:31:24.067" Score="3" ViewCount="" Body="&lt;p&gt;I have never seen $\aleph$ used without a subscript in any treatise of Set Theory.&lt;/p&gt;&#xA;" OwnerUserId="3195" LastActivityDate="2010-11-08T20:31:24.067" />
  <row Id="9481" PostTypeId="2" ParentId="6981" CreationDate="2010-11-08T20:33:25.327" Score="1" ViewCount="" Body="&lt;p&gt;The complex exponential solutions make sense for linear ODEs because you can get a characteristic polynomial since the exponential is the eigenfunction of the derivative operator. The reason why a complex solution times any constant is still a complex solution is due to the linearity in the linear ODE. If $x(t)$ is a solution and $i*y(t)$ is a solution then any linear combination is also a solution. Additivity and homogeneity are not properties of the solutions of a non-linear ODE and so it makes no sense to assume that the solution is of the form $e^{st}$ .&lt;/p&gt;&#xA;" OwnerUserId="2256" LastActivityDate="2010-11-08T20:33:25.327" />
  <row Id="9482" PostTypeId="1" AcceptedAnswerId="9485" CreationDate="2010-11-08T20:44:09.120" Score="1" ViewCount="102" Body="&lt;p&gt;The implicit expression $(b-a)=(a+b)^3$ looks like it could be written explicitly for $a$ as a function of $b$. The only region of interest is for $a,b&gt;0$ Here is what the plot looks like:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/jPQ4Z.gif&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="2256" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-08T21:17:07.270" LastActivityDate="2010-11-08T21:20:11.760" Title="Express implicit equation explicitly" Tags="&lt;algebra&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="9483" PostTypeId="1" AcceptedAnswerId="9805" CreationDate="2010-11-08T20:44:39.367" Score="2" ViewCount="134" Body="&lt;p&gt;Let $(X, \prec)$ and $(Y, &amp;lt;)$ be two well-ordered sets. I want to show that if $X$ is isomorphic to a subset of Y then $X$ is isomorphic with an initial segment of $Y$. (The other direction is of course true).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would appreciate some hints. Does this require the recursion theorem to construct the isomorphism?&lt;/p&gt;&#xA;" OwnerUserId="1120" LastActivityDate="2010-11-11T02:40:05.730" Title="Order-isomorphic with a subset iff order-isomorphic with an initial segment." Tags="&lt;set-theory&gt;" AnswerCount="1" CommentCount="6" />
  <row Id="9485" PostTypeId="2" ParentId="9482" CreationDate="2010-11-08T21:05:42.730" Score="5" ViewCount="" Body="&lt;p&gt;Writing $p=a+b$ you have the cubic equation,&#xA;$$&#xA;p^3+p-2b=0.&#xA;$$&#xA;This is already in &quot;depressed cubic&quot; form (no $p^2$ term), so it can be solved directly by &lt;a href=&quot;http://en.wikipedia.org/w/index.php?title=Cubic_function&amp;amp;oldid=395526018#Roots_of_a_cubic_function&quot; rel=&quot;nofollow&quot;&gt;standard methods&lt;/a&gt;. The coefficient of p is positive, so it is strictly increasing and there will be a single real root.&#xA;$$&#xA;p = \sqrt[3]{\sqrt{b^2+1/27}+b}-\sqrt[3]{\sqrt{b^2+1/27}-b}&#xA;$$&#xA;or,&#xA;$$&#xA;a = \sqrt[3]{\sqrt{b^2+1/27}+b}-\sqrt[3]{\sqrt{b^2+1/27}-b}-b.&#xA;$$&#xA;Alternatively, using the &lt;a href=&quot;http://en.wikipedia.org/w/index.php?title=Cubic_function&amp;amp;oldid=395526018#Trigonometric_.28and_hyperbolic.29_method&quot; rel=&quot;nofollow&quot;&gt;hyperbolic method&lt;/a&gt;,&#xA;$$&#xA;a=\frac{2}{\sqrt{3}}\sinh\left(\frac13\sinh^{-1}\left(3\sqrt{3}b\right)\right)-b.&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="1321" LastEditorUserId="1321" LastEditorDisplayName="" LastEditDate="2010-11-08T21:13:18.303" LastActivityDate="2010-11-08T21:13:18.303" CommentCount="13" />
  <row Id="9488" PostTypeId="1" AcceptedAnswerId="9497" CreationDate="2010-11-08T21:44:51.287" Score="2" ViewCount="337" Body="&lt;p&gt;Inspired by the recent post &quot;Limit of integral&quot;, I propose the following problem (hoping it will not turn out to be too easy). Suppose that $g:[0,1] \times [0,1] \to {\bf R}$ is continuous in both variables separately. Is it true that, for all $x_0 \in [0,1]$,&#xA;$$&#xA;\mathop {\lim }\limits_{x \to x_0 } \int_0^1 {g(x,y)\,{\rm d}y}  = \int_0^1 {g(x_0 ,y)\,{\rm d}y} .&#xA;$$ &lt;/p&gt;&#xA;" OwnerUserId="2810" LastActivityDate="2010-11-12T06:30:18.900" Title="Limit of integral - part 2" Tags="&lt;real-analysis&gt;&lt;integral&gt;&lt;limit&gt;" AnswerCount="3" FavoriteCount="1" />
  <row Id="9490" PostTypeId="2" ParentId="9460" CreationDate="2010-11-08T21:59:50.420" Score="1" ViewCount="" Body="&lt;p&gt;So, the second idea was the right one: considering $g$ on a compact to have it uniformly continuous. See my message above.&lt;/p&gt;&#xA;" OwnerUserId="524" LastActivityDate="2010-11-08T21:59:50.420" CommentCount="0" />
  <row Id="9491" PostTypeId="1" AcceptedAnswerId="9492" CreationDate="2010-11-08T22:02:14.007" Score="4" ViewCount="113" Body="&lt;p&gt;If $a$ and $b$ are positive real numbers, and if $f(x)$ has the following asymptotic property&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f(x) = \frac{a + O(\frac{1}{\sqrt{x}})}{b + O(\frac{1}{\sqrt{x}})}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;then is the following true?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f(x) = \frac{a}{b} + O(\frac{1}{\sqrt{x}})$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This might look like homework but it isn't.&lt;/p&gt;&#xA;" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-11-08T22:49:05.267" LastActivityDate="2010-11-08T22:49:05.267" Title="function asymptotic where $f(x) = \frac{a + O(\frac{1}{\sqrt{x}})}{b + O(\frac{1}{\sqrt{x}})}$" Tags="&lt;asymptotics&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="9492" PostTypeId="2" ParentId="9491" CreationDate="2010-11-08T22:18:41.580" Score="5" ViewCount="" Body="&lt;p&gt;Yes.  One way to see this is to actually do the long division (like the kind you learned in elementary school)!  Unfortunately, typesetting that in full on this forum will overtax my LaTeX powers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, dividing $b + O\left(\frac{1}{\sqrt{x}}\right)$ into $a + O\left(\frac{1}{\sqrt{x}}\right)$ yields $\frac{a}{b}$ with a remainder of $O\left(\frac{1}{\sqrt{x}}\right)$.  So we have &#xA;$$\frac{a + O\left(\frac{1}{\sqrt{x}}\right)}{b + O\left(\frac{1}{\sqrt{x}}\right)} = \frac{a}{b} + \frac{O\left(\frac{1}{\sqrt{x}}\right)}{b + O\left(\frac{1}{\sqrt{x}}\right)} = \frac{a}{b} + O\left(\frac{1}{\sqrt{x}}\right),$$&#xA;since $b + O\left(\frac{1}{\sqrt{x}}\right) = O(1).$&lt;/p&gt;&#xA;" OwnerUserId="2370" LastActivityDate="2010-11-08T22:18:41.580" />
  <row Id="9493" PostTypeId="1" CreationDate="2010-11-08T22:25:47.307" Score="0" ViewCount="143" Body="&lt;p&gt;I have a table in my database that records, 48 times per day, the total users, total active users within 2 weeks, total users logged in within the past 24 hours , and total users that logged in within the last 24 hours  that had visited the site prior to the last 24 hours.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm trying to use this data to calculate the churn rate of my users, but in some cases I am getting a negative number so I assume I am doing it incorrectly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My attempt:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;(Yesterday's active in past 24 hours&#xA;  [returning] - Todays's active in past&#xA;  24 hours [returning] + Change in&#xA;  Active Users) / Todays's active in&#xA;  past 24 hours [returning]&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I assume I have the correct data somewhere with all of the data I am saving, I'm just not using it the right way.&lt;/p&gt;&#xA;" OwnerUserId="3284" LastActivityDate="2010-11-08T22:25:47.307" Title="How to calculate daily user churn rate?" Tags="&lt;analysis&gt;&lt;computer-science&gt;" CommentCount="3" />
  <row Id="9494" PostTypeId="2" ParentId="9491" CreationDate="2010-11-08T22:28:16.750" Score="6" ViewCount="" Body="&lt;p&gt;It is true.  In the spirit of epsilon/delta, you are challenged to prove that $|f(x)-\frac{a}{b}| \lt \frac{M}{\sqrt{x}}$ for $x\gt x_0$ where your challenger gives M and you have to find an $x_0$ that works.  But you get to challenge back saying the numerator should be within $\frac{N}{\sqrt{x}}$ of $a$ and similarly the denominator should be within $\frac{P}{\sqrt{x}}$ of $b$.  So take $N=\frac{M}{2b}$ and $P=\frac{aM}{2b^2}$ and take the larger of the $x_0$'s that come back.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-11-08T22:28:16.750" CommentCount="4" />
  <row Id="9495" PostTypeId="2" ParentId="9488" CreationDate="2010-11-08T22:34:43.707" Score="1" ViewCount="" Body="&lt;p&gt;This is true if there is a (measurable) function $f(y)$ such that $$|g(x,y)|\le f(y), \mbox{ for }0\le x\le 1 $$  and $$\int_0^1 f(y)\,dy&amp;lt;\infty.$$&#xA;This follows from the Lebesgue dominated convergence theorem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: Every continuous function is measurable.&lt;/p&gt;&#xA;" OwnerUserId="3249" LastEditorUserId="3249" LastEditorDisplayName="" LastEditDate="2010-11-08T23:45:20.760" LastActivityDate="2010-11-08T23:45:20.760" CommentCount="4" />
  <row Id="9497" PostTypeId="2" ParentId="9488" CreationDate="2010-11-08T22:44:33.180" Score="5" ViewCount="" Body="&lt;p&gt;The answer is no. In fact, it is related to &lt;a href=&quot;http://en.wikipedia.org/wiki/Leibniz_integral_rule&quot; rel=&quot;nofollow&quot;&gt;Leibniz's rule&lt;/a&gt;. The counterexample below is from 'Counterexamples in Analysis' and was recently posted &lt;a href=&quot;http://math.stackexchange.com/questions/8711/derivative-commuting-over-integral/8712#8712&quot;&gt;here&lt;/a&gt; by Jonas Meyer. Consider the function &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ g(x,y) = \left\{&#xA;     \begin{array}{lr}&#xA;       \left (\frac{3x^2}{y^2} - \frac{2x^4}{y^3} \right)e^{-x^2/y} &amp;amp;  y&gt;0, \\&#xA;       0 &amp;amp; y=0. &#xA;     \end{array}&#xA;   \right.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's easy to see that $g$ is continuous in both variables separately. We have that $\int_{0}^1 g(0,y) dy = 0 $ and $$ \int_{0}^1 g(x,y) dy = e^{-x^2}(1-2x^2)$$ if $ x \ne 0$. So $$\lim_{x \rightarrow 0} \int_{0}^1 g(x,y) dy = 1 \ne \int_{0}^1 g(0,y) dy.$$&lt;/p&gt;&#xA;" OwnerUserId="329" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2010-11-08T23:37:40.667" LastActivityDate="2010-11-08T23:37:40.667" CommentCount="6" />
  <row Id="9498" PostTypeId="2" ParentId="7499" CreationDate="2010-11-08T23:04:06.657" Score="0" ViewCount="" Body="&lt;p&gt;Check out Andrew Ng's website for a tutorial on SVMs.&#xA;www.stanford.edu/class/cs229/notes/cs229-notes3.pdf&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yeah, in this case I believe that the feature space and what you are representing are two different dimensions.&lt;/p&gt;&#xA;" LastActivityDate="2010-11-08T23:04:06.657" />
  <row Id="9499" PostTypeId="2" ParentId="9398" CreationDate="2010-11-08T23:13:35.200" Score="1" ViewCount="" Body="&lt;p&gt;If you can use partial derivatives, then you can do the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First you find $dy/dx$, say $$\frac{dy}{dx}=g(x,y).$$ Then by chain rule &#xA;$$\frac{d^2y}{dx^2}=\frac{\partial g}{\partial x}+\frac{\partial g}{\partial y}\frac{dy}{dx}=\frac{\partial g}{\partial x}+\frac{\partial g}{\partial y}g(x,y).$$&lt;/p&gt;&#xA;" OwnerUserId="3249" LastActivityDate="2010-11-08T23:13:35.200" />
  <row Id="9500" PostTypeId="2" ParentId="9488" CreationDate="2010-11-08T23:15:05.283" Score="4" ViewCount="" Body="&lt;p&gt;I think I have another counterexample.  Define $f(x)=\int_{0}^x e^{-1/t}dt$ for $x\gt0$.  This is chosen because it goes to zero as $x$ goes to zero from the right and because experimentation led me to the differential equation $\frac{f''(x)}{f'(x)}=\frac{1}{x^2}$ as a sufficient condition for the following to work.  Define &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$g(x,y) = \left\{&#xD;&#xA;     \begin{array}{lr}&#xD;&#xA;       \frac{xy}{f(x)+y^2} &amp; \text{if } x\gt0, \\&#xD;&#xA;       0 &amp; \text{if } x=0, &#xD;&#xA;     \end{array}&#xD;&#xA;   \right.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for $(x,y)$ in $[0,1]\times[0,1]$, and let $x_0=0$.  The right hand side of your tentative equation is obviously $0$.  The left hand side is &#xA;$$\lim_{x\to0+}\frac{x}{2}\log\left(1+\frac{1}{f(x)}\right),$$&#xA;which comes out to $\frac{1}{2}$ after $2$ applications of l'Hôpital's rule, if I did it correctly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I started by playing with the standard example of a discontinuous but separately continuous function on $\mathbb{R}^2$, $f(x,y)=\frac{xy}{x^2+y^2}$ when $x$ or $y$ is nonzero, $f(x,y)=0$ when $x=y=0$. Then I tried to see how the $x^2$ in the denominator could be modified to give a counterexample here, by replacing it with an unknown $f(x)$ that goes to $0$ at $0$ and seeing what further properties of $f(x)$ would make it work.  As I mentioned, this led in particular to the sufficient condition $\frac{f''(x)}{f'(x)}=\frac{1}{x^2}$, and hence to this example.  Unfortunately, I can't offer any real intuition.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I decided to look at this a little more, and came up with an example simpler than the other one I gave.  With $$g(x,y)=\frac{-y}{\log(x/2)(x+y^2)}$$ &#xA;for $x\gt0$ and $0$ otherwise, the same result as above holds.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2010-11-12T06:30:18.900" LastActivityDate="2010-11-12T06:30:18.900" CommentCount="6" />
  <row Id="9501" PostTypeId="2" ParentId="9434" CreationDate="2010-11-08T23:21:36.050" Score="2" ViewCount="" Body="&lt;p&gt;Just a comment up front: Obviously, you're not going to find the perfect algorithm to predict stock prices.  Such a thing would the Holy Grail of Wall Street. :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That said, during my brief stint as a part-time business school prof here are the basic methods we discussed in the forecasting unit.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Moving average&lt;/li&gt;&#xA;&lt;li&gt;Weighted moving average&lt;/li&gt;&#xA;&lt;li&gt;Exponential smoothing&lt;/li&gt;&#xA;&lt;li&gt;Exponential smoothing with trend&lt;/li&gt;&#xA;&lt;li&gt;Least-squares linear regression (with a nod to BBischof!)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Each has pros and cons.  The first four are discussed in the Wikipedia page on &lt;a href=&quot;http://en.wikipedia.org/wiki/Exponential_smoothing&quot; rel=&quot;nofollow&quot;&gt;exponential smoothing&lt;/a&gt;, although exponential smoothing with trend is called &quot;double exponential smoothing&quot; there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me emphasize again that these are &lt;em&gt;basic&lt;/em&gt; methods.  If you want something fancier, you could try &lt;a href=&quot;http://en.wikipedia.org/wiki/Geometric_Brownian_motion&quot; rel=&quot;nofollow&quot;&gt;geometric Brownian motion&lt;/a&gt;.  Be aware, though, that there are LOTS of other algorithms out there.  You can probably imagine how many finance companies are trying to come up with a better way of predicting stock prices.&lt;/p&gt;&#xA;" OwnerUserId="2370" LastActivityDate="2010-11-08T23:21:36.050" />
  <row Id="9502" PostTypeId="1" CreationDate="2010-11-08T23:33:34.510" Score="1" ViewCount="111" Body="&lt;p&gt;Hello, I am interested in a method to find the roots of the following equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;f(t) = \sum_{i=1}^n \alpha_i e^{\beta_i t} + \gamma t + \delta = 0.&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For my application, coefficients $\alpha_i$, $\beta_i$, $\gamma$, and $\delta$ are real.  $n$ is typically a small integer, say 10.  In particular I am interested in the smallest positive real root of $f$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For those interested, this equation arises when attempting to compute the point of intersection between the solution to the linear ODE&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;\dot x(t) &amp;amp;= Ax(t) + b &amp;amp; (A = A^T) \\ &#xA;x(0) &amp;amp;= x_0&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and the boundary of a set of linear constraints&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;Cx(t) \ge d.&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The initial point is always feasible $(Cx_0\ge d)$.  For my purposes, all matrices and vectors are real.&lt;/p&gt;&#xA;" OwnerUserId="3285" LastActivityDate="2010-11-08T23:52:09.063" Title="solving a univariate equation with a sum of exponentials" Tags="&lt;linear-algebra&gt;&lt;algebra&gt;&lt;differential-equations&gt;&lt;roots&gt;" AnswerCount="1" />
  <row Id="9503" PostTypeId="2" ParentId="9378" CreationDate="2010-11-08T23:41:25.957" Score="1" ViewCount="" Body="&lt;p&gt;I've just started with this stuff, so take this for what it's worth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What's been working for me is to imagine a sheaf as the sheaf of continuous functions from the reals to the reals. The restriction maps are just the normal restrictions of functions to open sets. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For presheaves I think of the presheaf of bounded continuous functions from reals to reals. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Stalks are just the germs thereof. I draw an element of a stalk by drawing coordinate axes, choosing a point x on the horizontal axis, and then draw the graph of a continuous function on a tiny neighborhood of x. The neighborhood is supposed to be &quot;infinitesimal.&quot; The stalk is the set of all such graphs on that neighborhood. &lt;/p&gt;&#xA;" OwnerUserId="566" LastActivityDate="2010-11-08T23:41:25.957" CommentCount="4" />
  <row Id="9504" PostTypeId="2" ParentId="9502" CreationDate="2010-11-08T23:52:09.063" Score="1" ViewCount="" Body="&lt;p&gt;Assuming $\alpha_i, \beta_i, \gamma$, and $\delta $ are givens, any numeric root finder should make quick work of this.  There won't be an algebraic solution.  There is good info at &lt;a href=&quot;http://nr.com/&quot; rel=&quot;nofollow&quot;&gt;Numerical Recipes&lt;/a&gt; chapter 9.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-11-08T23:52:09.063" CommentCount="3" />
  <row Id="9505" PostTypeId="1" AcceptedAnswerId="9515" CreationDate="2010-11-09T00:23:57.653" Score="10" ViewCount="883" Body="&lt;p&gt;We know that $2^4 = 4^2$ and $(-2)^{-4}$ = $(-4)^{-2}$. Is there another pair of integers $(x, y)$, with different $x$ and $y$, which satisfies the equality  $x^y = y^x$?&lt;/p&gt;&#xA;" OwnerUserId="2938" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-11-09T01:54:53.973" LastActivityDate="2010-11-19T19:49:10.333" Title="$x^y = y^x$ (x and y are integers)" Tags="&lt;number-theory&gt;&lt;diophantine-equations&gt;" AnswerCount="4" FavoriteCount="1" />
  <row Id="9506" PostTypeId="2" ParentId="9468" CreationDate="2010-11-09T00:24:39.610" Score="4" ViewCount="" Body="&lt;p&gt;I can't resist giving a couple of examples of (contrived) numerical coincidences (both are from &lt;a href=&quot;http://books.google.co.uk/books?id=csfnE14Kj-cC&amp;amp;printsec=frontcover&amp;amp;dq=experimentation+in+mathematics+computational+paths+to+discovery&amp;amp;hl=en&amp;amp;ei=A5TYTJu1E8mG4Aap0OXqCA&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=1&amp;amp;ved=0CDMQ6AEwAA#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;Experimentation in Mathematics&lt;/em&gt;&lt;/a&gt; by J.M. Borwein, D.H. Bailey and R. Girgensohn).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_{0}^{\infty}\cos(2x)\prod_{k=1}^{\infty}\cos(x/k)\ dx=\frac{\pi}{8}-\epsilon,$$&#xA;where &#xA;$0&amp;lt;\epsilon&amp;lt;10^{-41}.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example 2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum\limits_{k=1}^{\infty}e^{-(k/10)^2}\approx5\sqrt\pi-\frac{1}{2}=8.362269254527580...$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, they agree through 427 (four hundred twenty seven) digits yet they are not equal.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;A moral. Make sure you understand the context and use your inverse symbolic calculator with caution.&lt;/p&gt;&#xA;" OwnerUserId="723" LastEditorUserId="723" LastEditorDisplayName="" LastEditDate="2010-11-09T05:41:37.797" LastActivityDate="2010-11-09T05:41:37.797" CommentCount="4" />
  <row Id="9507" PostTypeId="2" ParentId="9505" CreationDate="2010-11-09T00:34:10.720" Score="19" ViewCount="" Body="&lt;p&gt;For every integer $n$, $x = y = n$ is a solution. So assume $x \neq y$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose $n^m = m^n$. Then $n^{1/n} = m^{1/m}$. Now the function $x \mapsto x^{1/x}$ reaches its maximum at $e$, and is otherwise monotone. Thus (assuming $n &amp;lt; m$) we must have $n &amp;lt; e$, i.e. $n = 1$ or $n = 2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $n = 1$ then $n^m = 1$ and so $m = 1$, so it's a trivial solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $n = 2$ then $n^m$ is a power of $2$, and so (since $m &gt; 0$) $m$ must also be a power of $2$, say $m = 2^k$. Then $n^m = 2^{2^k}$ and $m^n = 2^{2k}$, so that $2^k = 2k$ or $2^{k-1} = k$. Now $2^{3-1} &amp;gt; 3$, and so an easy induction shows that $k \leq 2$. If $k = 1$ then $n = m$, and $k = 2$ corresponds to $2^4 = 4^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Up till now we considered $n,m&amp;gt;0$. We now go over all other cases. The solution $n = m = 0$ is trivial (whatever value we give to $0^0$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $n=0$ and $m \neq 0$ then $n^m = 0$ whereas $m^n = 1$, so this is not a solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $n &amp;gt; 0$ and $m &amp;lt; 0$ then $0 &amp;lt; n^m \leq 1$ whereas $|m^n| \geq 1$. Hence necessarily $n^m = 1$ so that $n = 1$. It follows that $m^1 = 1^m = 1$. In particular, there's no solution with opposite signs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $n,m &amp;lt; 0$ then $(-1)^m (-n)^m = n^m = m^n = (-1)^n (-m)^n$, so that $n,m$ must have the same parity. Taking inverses, we get $(-n)^{-m} = (-m)^{-n}$, so that $-n,-m$ is a solution for positive integers. The only non-trivial positive solution $2,4$ yields the only non-trivial negative solution $-2,-4$.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2010-11-16T18:28:27.290" LastActivityDate="2010-11-16T18:28:27.290" CommentCount="4" />
  <row Id="9508" PostTypeId="1" AcceptedAnswerId="9512" CreationDate="2010-11-09T00:35:25.027" Score="5" ViewCount="211" Body="&lt;p&gt;I need to write a coupon code system but I do not want to save each coupon code in the database. (For performance and design reasons.) Rather I would like to generate codes subsequent that are watermarked with another code.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They should like kind of fancy and random. Currently they look like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1: AKFCU2, 2: AKFDU2, 3: AKFDW2, 4: AKHCU2, 5: AKHCW2, ..., 200: CLFSU2, 201: CLFSW2, ...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's obvious that subsequent codes look very similar as I just converted my code ingredients (the watermark and the integer in the front) to binary base and permutated the order by a fixed scheme. But to prevent people to easily guess another valid code or even accidently enter another valid code (thus making another code invalid by using it) I would prefer something more chaotic, like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1: FIOJ32, 2: X9NIU2, 3: SIUUN7, 4: XTVV4S, ...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the end the problem is to find a bijective, discrete function on the domain {0,1}^27 (or alternatively {0,1,2,3,4, ..., [10^(8.5)]}) that is far away from being continuous. Also it should be as simple as possible to implement. (EDIT: I also need to implement the reverse function.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any suggestions for such a function?&lt;/p&gt;&#xA;" OwnerUserId="3290" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-09T08:36:55.387" LastActivityDate="2010-11-10T02:53:58.737" Title="Looking for a bijective, discrete function that behaves as chaotically as possible" Tags="&lt;discrete-mathematics&gt;&lt;computer-science&gt;&lt;cryptography&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="9509" PostTypeId="2" ParentId="7499" CreationDate="2010-11-09T00:36:34.167" Score="2" ViewCount="" Body="&lt;p&gt;As mentioned, the kernel tricks embeds your original points to a higher dimensional space (in fact, in some cases infinite dimensional - but of course the linear subspace generated by your actual points is finite dimensional).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an example, using the embedding $(x,y) \mapsto (x,y,x^2,y^2)$ (that actually corresponds to a quadratic kernel, I think) then the equation of an arbitrary ellipse becomes linear.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-09T00:36:34.167" />
  <row Id="9510" PostTypeId="1" CreationDate="2010-11-09T00:38:51.410" Score="26" ViewCount="489" Body="&lt;p&gt;Winding yarn into a ball suggests some mathematical questions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/neiZH.jpg&quot; alt=&quot;Yarn&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Under some natural model, what paths should the yarn follow to achieve the most dense ball?&#xA;One model is that used by Henryk Gerlach and Heiko von der Mosel in their paper&#xA;&quot;On sphere-filling ropes&quot; &lt;a href=&quot;http://arxiv.org/abs/1005.4609&quot; rel=&quot;nofollow&quot;&gt;arXiv:1005.4609v1 (math.GT)&lt;/a&gt;.&#xA;(This is the same model I suggested in a &lt;a href=&quot;http://mathoverflow.net/questions/26525/coiling-rope-in-a-box&quot; rel=&quot;nofollow&quot;&gt;MO question&lt;/a&gt;.)&#xA;I expect it would be difficult to extend the optimal solutions of the above paper to the multiple layers for a ball of yarn. (Below is shown part of Fig.6 from their paper.)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/0kn2O.jpg&quot; alt=&quot;Rope&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;What paths should the yarn follow to achieve the least dense ball?&#xA;Here I imagine layers forming a grid that suspends the yarn above as much empty space as possible.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Random winding.&#xA;Typical instructions for how to do this by hand say, &quot;Change directions every once in a while while you are winding,&quot; or &quot;As you wrap, slowly rotate the ball counterclockwise to keep the distribution even.&quot; What density is achieved by random winding?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I ask these questions primarily out of curiosity. Perhaps there is an analogous process (winding the interior of a golf ball or baseball?) that has been studied mathematically.&lt;/p&gt;&#xA;" OwnerUserId="237" LastEditorUserId="237" LastEditorDisplayName="" LastEditDate="2010-11-09T00:55:46.600" LastActivityDate="2011-04-21T22:58:52.007" Title="Optimal yarn balls" Tags="&lt;geometry&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="3" />
  <row Id="9511" PostTypeId="1" AcceptedAnswerId="9514" CreationDate="2010-11-09T00:43:08.677" Score="2" ViewCount="132" Body="&lt;p&gt;The unit doublet is a symbolic object whose convolution with a differentiable function is supposed to give the derivative:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(x * u_1)(t) = \frac{dx(t)}{dt}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See also: &lt;a href=&quot;http://en.wikipedia.org/wiki/Unit_doublet&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Unit_doublet&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can this equation be made rigorous, in a way that it would be valid for all $x\in C^1(\mathbb R)$?  I.e., is there a definition of the unit doublet as a distribution or a measure, and a version of the convolution operation that would make sense for all $C^1(\mathbb R)$ functions?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is related to my question on MO: &lt;a href=&quot;http://mathoverflow.net/questions/2969/&quot; rel=&quot;nofollow&quot;&gt;http://mathoverflow.net/questions/2969/&lt;/a&gt; , but the theorems referred in the answers to that question include only a small fraction of $C^1(\mathbb R)$.&lt;/p&gt;&#xA;" OwnerUserId="145" LastEditorUserId="145" LastEditorDisplayName="" LastEditDate="2010-11-09T01:02:10.617" LastActivityDate="2010-11-09T01:38:31.383" Title="Rigorous definition of convolution with the unit doublet" Tags="&lt;fourier-analysis&gt;&lt;signal-processing&gt;" AnswerCount="1" />
  <row Id="9512" PostTypeId="2" ParentId="9508" CreationDate="2010-11-09T00:47:24.937" Score="2" ViewCount="" Body="&lt;p&gt;Take any odd $a$ and calculate $x \mapsto ax + b \pmod{2^{27}}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Here's a more sophisticated suggestion. The following functions are all invertible and easy to implement:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt; Multiplication by an odd number modulo $2^{27}$.&#xA;&lt;li&gt; Addition of an arbitrary number modulo $2^{27}$.&#xA;&lt;li&gt; XOR of an arbitrary 27-bit number.&#xA;&lt;li&gt; Rotation of the bits (can be implemented using two shifts).&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;If you compose them repeatedly they become harder to break (but probably still easy for someone who already knows the general form of your cipher).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By composing, I mean you apply several of them in succession (you can apply a given mapping more than once with same or different parameters).&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Another suggestion is to use a random permutation. It's not difficult to generate a random permutation, and given a permutation to calculate its inverse. You can store both permutations in a file and load them to memory (if you have enough - it's 1GB for both) each time your main program starts.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2010-11-09T07:36:53.853" LastActivityDate="2010-11-09T07:36:53.853" CommentCount="7" />
  <row Id="9513" PostTypeId="1" AcceptedAnswerId="9518" CreationDate="2010-11-09T00:57:58.813" Score="3" ViewCount="96" Body="&lt;p&gt;I am trying to prove the following fact for a homework assignment in algebraic geometry:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let R be a local ring, and X a prescheme. Show that there is a one-one correspondence between R-valued points on X and pairs (x, g) of points x on X and local homomorphisms $g: O_{X, x} \to R$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, given f: Spec R -&gt; X, and writing x for f(M) where M is the maximal ideal of R, the induced map on stalks gives us a map $g: O_{X,x} \to R_M = R$. Then it remains to show that given a point x of X and a map $g: O_{X,x} \to R$, that there is a map f: Spec R -&gt; X such that f(M) = x and such that the induced map is g.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I haven't had any luck yet with associating a map f with a given g. Here are my thoughts and some of the things I've tried:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;When R is a field, then Spec R has only a single point, and thus the map f is completely determined by where it sends the zero ideal. This induces a local homomorphism from $O_{X,x}$ to k, which is to say that it induces a map from $O_{X,x} / M_x \to k$ since the maximal ideal $M_x$ of $O_{X,x}$ is contained in the kernel of the local homomorphism. Anyway, as long as there is some local homomorphism from $O_{X,x}$ to k, then the map f sending (0) to x is a well-defined morphism. Now, when R is again just a local ring, it seems unlikely that f is completely determined by where it sends M, and I can't see how to work with this.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I've tried extending the map g to a map of sheaves, thinking that maybe from there I'd be able to find my map of preschemes. But here I get a little lost, and I'm not sure how to make it work.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;We know a lot about the topology of Spec R, and this forces some properties of X. For example, the only open neighborhood of M in Spec R is the whole space. Therefore, f(Spec R) covers any open neighborhood of f(M). I haven't pursued this way very far yet, as it seems a little further afield.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I just need a little hint to get me going again -- I've been thinking about this one for a while and I can't seem to get unstuck.&lt;/p&gt;&#xA;" OwnerUserId="759" LastActivityDate="2010-11-13T11:40:11.503" Title="How to find a bijection between R-valued points of X and local ring homomorphisms?" Tags="&lt;homework&gt;&lt;algebraic-geometry&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="1" />
  <row Id="9514" PostTypeId="2" ParentId="9511" CreationDate="2010-11-09T01:28:24.360" Score="3" ViewCount="" Body="&lt;p&gt;For $C^1(\mathbb{R})$ it is a bit delicate. But for $C^1(\mathbb{R})$ with compact support (or rapid decay a la Schwartz functions), you can define $u_1$ as (up to a negative sign) the distributional derivative of the Dirac distribution. It is a distribution of compact support, and so also a tempered distribution. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the fact that it has compact support, you should be able to convolve it against any distribution (see Hormander, Analysis of Linear Partial Differential Operator, Volume 1, Chapter 4). A function in $C^1$ is locally integrable, and hence a distribution. So the object $x*u_1$ is well defined as a distribution. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So at least distributionally the equation you wrote make sense: that acting on any smooth function with compact support, the two sides act the same way as linear functionals. (Since for any test function $\psi \in C^\infty_c(\mathbb{R})$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \langle(x*u_1),\psi\rangle = -\langle x,(\psi*u_1)\rangle = - \langle x,\psi'\rangle = \langle x',\psi\rangle $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;you see that the two sides are equally defined as distributions, using that a convolution of a test function against a distribution is always well defined and also a smooth function.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the equation you wrote down is meaningful in the sense of distributions. And hence you can take $dx/dt$ as a continuous representative (in the equivalent class of distributions) of $x*u_1$. Which I think is about as rigorous as you need it. &lt;/p&gt;&#xA;" OwnerUserId="1543" LastEditorUserId="1543" LastEditorDisplayName="" LastEditDate="2010-11-09T01:38:31.383" LastActivityDate="2010-11-09T01:38:31.383" CommentCount="1" />
  <row Id="9515" PostTypeId="2" ParentId="9505" CreationDate="2010-11-09T01:40:19.037" Score="13" ViewCount="" Body="&lt;p&gt;This is a classic (and well known problem).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The general solution of $x^y = y^x$ is given by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x = (1+1/u)^u, y = (1+1/u)^{u+1}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It can be shown that if $x$ and $y$ are rational, then $u$ must be an integer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For more details, see here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.mathematik.uni-bielefeld.de/~sillke/PUZZLES/x%5Ey-x%5Ey&quot; rel=&quot;nofollow&quot;&gt;http://www.mathematik.uni-bielefeld.de/~sillke/PUZZLES/x%5Ey-x%5Ey&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.qbyte.org/puzzles/p048s.html&quot; rel=&quot;nofollow&quot;&gt;http://www.qbyte.org/puzzles/p048s.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-11-09T05:03:27.990" LastActivityDate="2010-11-09T05:03:27.990" CommentCount="1" />
  <row Id="9516" PostTypeId="1" CreationDate="2010-11-09T02:06:18.103" Score="1" ViewCount="77" Body="&lt;p&gt;Is there a way of enumerating necklaces, in the spirit in which the factorial number system ( &lt;a href=&quot;http://en.wikipedia.org/wiki/Factorial_number_system&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Factorial_number_system&lt;/a&gt; ) can be used to enumerate permutations?&lt;/p&gt;&#xA;" OwnerUserId="2809" LastActivityDate="2010-11-09T02:34:07.540" Title="Enumerating necklaces" Tags="&lt;combinatorics&gt;&lt;reference-request&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" />
  <row Id="9517" PostTypeId="2" ParentId="9516" CreationDate="2010-11-09T02:34:07.540" Score="2" ViewCount="" Body="&lt;p&gt;There is code here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.csse.monash.edu.au/~lloyd/tildeAlgDS/Recn/Necklaces/&quot; rel=&quot;nofollow&quot;&gt;http://www.csse.monash.edu.au/~lloyd/tildeAlgDS/Recn/Necklaces/&lt;/a&gt;&lt;/p&gt;&#xA;" LastActivityDate="2010-11-09T02:34:07.540" />
  <row Id="9518" PostTypeId="2" ParentId="9513" CreationDate="2010-11-09T03:12:57.397" Score="3" ViewCount="" Body="&lt;p&gt;For any point $x \in X$, there is a canonical morphism $\mathrm{Spec} \mathcal{O}_{X, x} \to X$. To prove this, first take an affine neighborhood of $x$. You may also want to look at the definition of a morphism of locally ringed spaces / schemes, specifically, the requirement on stalks.&lt;/p&gt;&#xA;" OwnerUserId="455" LastActivityDate="2010-11-09T03:12:57.397" CommentCount="1" />
  <row Id="9519" PostTypeId="2" ParentId="6437" CreationDate="2010-11-09T03:33:41.100" Score="1" ViewCount="" Body="&lt;p&gt;To connect all computers to each other you need n-1 connections, i.e., 4. To connect 3 printers you need exactly 3 connections. So the number connections required is 7. &lt;/p&gt;&#xA;" OwnerUserId="2513" LastActivityDate="2010-11-09T03:33:41.100" />
  <row Id="9520" PostTypeId="1" CreationDate="2010-11-09T03:46:56.433" Score="5" ViewCount="142" Body="&lt;p&gt;Is this a known special function:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_0^1 a^p(1-a)^{1-p}\,b^{1-p}\,(1-b)^p dp\qquad ?$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am really only interested in maximizing this over $(a,b)$ in $[0,1] \times [0,1]$, so a pointer to a nice numerical evaluation is appreciated as much or more so than an unstable exact formula.&lt;br&gt;&#xA;Thanks for any help&lt;/p&gt;&#xA;" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-09T03:57:26.503" LastActivityDate="2010-11-09T04:46:34.663" Title="Is this a known special function?" Tags="&lt;integral&gt;&lt;optimization&gt;&lt;special-functions&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="9521" PostTypeId="2" ParentId="9355" CreationDate="2010-11-09T03:54:15.427" Score="0" ViewCount="" Body="&lt;p&gt;To answer (ii), I think it needs to be ergodic, i.e., aperiodic and recurrent.&lt;/p&gt;&#xA;" OwnerUserId="774" LastActivityDate="2010-11-09T03:54:15.427" />
  <row Id="9522" PostTypeId="1" AcceptedAnswerId="9535" CreationDate="2010-11-09T03:55:38.047" Score="6" ViewCount="194" Body="&lt;p&gt;Given a row of 16 houses where 10 are red and 6 are blue, what is the expected number of neigbors of a different color?&lt;/p&gt;&#xA;" OwnerUserId="2809" LastActivityDate="2010-11-10T17:52:51.113" Title="Expected number of neighbors" Tags="&lt;combinatorics&gt;&lt;probability&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="9523" PostTypeId="1" AcceptedAnswerId="14092" CreationDate="2010-11-09T03:58:12.667" Score="5" ViewCount="244" Body="&lt;p&gt;My text gives a much more complicated proof of this result, which makes me wonder if the argument I have in my head for this has something wrong with it.  Does this work, or have I made a bad assumption somewhere along the line?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $U \subseteq \mathbb{C}$ be open, $\overline{D}(z_0, r) \subset U$, and $f : U \rightarrow \mathbb{C}$ be holomorphic with a zero of order $n$ at $z_0$ and no other zeroes in $U$. Taking a power series expansion at $z_0$,&#xA;$$f(z) = a_n(z-z_0)^n + o((z-z_0)^{n+1}).$$&#xA;Differentiating,&#xA;$$f'(z) = n a_n(z-z_0)^{n-1} + o((z-z_0)^n),$$&#xA;so we have&#xA;$$\lim_{z \rightarrow z_0} \frac{(z-z_0) f'(z)}{f(z)} = \lim_{z \rightarrow z_0} \frac{n a_n(z-z_0)^n + o((z-z_0)^{n+1})}{a_n(z-z_0)^n + o((z-z_0)^{n+1})} = n.$$&#xA;Define the function $g : U \rightarrow \mathbb{C}$ by&#xA;$$g(z) = \begin{cases}&#xA;(z-z_0) f'(z)/f(z) &amp;amp; \text{if } z \neq z_0 \\&#xA;n &amp;amp; \text{otherwise}&#xA;\end{cases}$$&#xA;By the Cauchy integral formula,&#xA;$$n = g(z_0) = \frac{1}{2 \pi i} \oint_{\partial\overline{D}(z_0, r)} \frac{g(z)}{z - z_0}  dz =  \frac{1}{2 \pi i} \oint_{\partial\overline{D}(z_0, r)} \frac{(z-z_0)f'(z)}{(z-z_0)f(z)} dz=  \frac{1}{2 \pi i} \oint_{\partial\overline{D}(z_0, r)} \frac{f'(z)}{f(z)} dz$$&#xA;as desired.&lt;/p&gt;&#xA;" OwnerUserId="3296" LastEditorUserId="3296" LastEditorDisplayName="" LastEditDate="2010-11-09T05:44:52.713" LastActivityDate="2010-12-12T22:14:13.440" Title="Proof that $\frac{1}{2\pi i}\oint f'(z)/f(z) \, dz = n$" Tags="&lt;complex-analysis&gt;" AnswerCount="2" CommentCount="6" />
  <row Id="9524" PostTypeId="1" AcceptedAnswerId="14902" CreationDate="2010-11-09T04:01:42.107" Score="10" ViewCount="572" Body="&lt;p&gt;What are some good ways to motivate the material on infinite series that appears at the end of a typical American Calculus II course?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My students in this course are generally from biochemistry, computer science, economics, business, and physics (with a few humanities folks taking the course for fun) - not just math majors.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have struggled some in the past to motivate the infinite series material to these students.  For one, it doesn't fit with the rest of Calc II, which is on the integral.  Over the years I have &quot;converged&quot; on telling them that the main point of the unit is Taylor series and that the rest of the material is there primarily so that we have the tools we need in order to understand Taylor series.  Then I illustrate some of the many uses of Taylor series (mainly function approximation, at this level).  This approach works better than anything I've come up with thus far with respect to getting my students to care about infinite series, but I feel a little like I'm selling the rest of the material short by subordinating it to Taylor series.  Does anyone have other ways of motivating infinite series that they would like to share?  (Again, only a small percentage of the students in my class are math majors.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Background: The material in this unit typically consists of sequences, basic series (like geometric and telescoping ones), a slew of tests for convergence (e.g., integral test, ratio test, root test), an introduction to power series, Taylor and Maclaurin series, and maybe binomial series.  &lt;/p&gt;&#xA;" OwnerUserId="2370" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-12-19T23:33:36.320" LastActivityDate="2011-01-06T23:22:44.117" Title="Motivating infinite series" Tags="&lt;calculus&gt;&lt;sequences-and-series&gt;&lt;teaching&gt;" AnswerCount="9" CommentCount="15" FavoriteCount="3" />
  <row Id="9525" PostTypeId="1" AcceptedAnswerId="9568" CreationDate="2010-11-09T04:05:48.133" Score="0" ViewCount="152" Body="&lt;p&gt;This is in relation to a problem dealing with the three-dimensional analogue of Pell's Equation.  I would like to factor&#xA;$ x^3+Dy^3+D^2z^3-3Dxyz $ into $\frac{1}{2}(x+Dy+D^2y)$ and another factor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like just a hint on how to go about doing this as I'm pretty much stumped.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;: I think the factor I was looking for is $\frac{1}{2}(x+Ny+N^2z)$ where $N=D^3$.&lt;/p&gt;&#xA;" OwnerUserId="2505" LastEditorUserId="2505" LastEditorDisplayName="" LastEditDate="2010-11-09T23:22:51.400" LastActivityDate="2010-11-09T23:32:53.793" Title="Factoring Multiple Variable Polynomials" Tags="&lt;number-theory&gt;&lt;algebra&gt;&lt;diophantine-equations&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="9527" PostTypeId="2" ParentId="9523" CreationDate="2010-11-09T04:09:47.367" Score="6" ViewCount="" Body="&lt;p&gt;If $f$ has a zero of order $n$ at $z_0$, then there is an function $h$ holomorphic near $z_0$ and such that $f(z)=(z-z_0)^nh(z)$ near $z_0$ and $h(z_0)\neq0$. Computing, $$\frac{f'}{f}=\frac{n}{z-z_0}+\frac{h'(z)}{h(z)}.$$ If you use the Cauchy formula now over a sufficiently small circle around $z_0$ you get (a corrected version of) what you want.&lt;/p&gt;&#xA;" OwnerUserId="274" LastEditorUserId="274" LastEditorDisplayName="" LastEditDate="2010-11-09T04:18:26.970" LastActivityDate="2010-11-09T04:18:26.970" CommentCount="3" />
  <row Id="9530" PostTypeId="2" ParentId="9520" CreationDate="2010-11-09T04:17:31.080" Score="8" ViewCount="" Body="&lt;p&gt;You can get a closed-form answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_0^1 a^p (1-a)^{1-p} b^{1-p} (1-b)^p dp = b(1-a) \int_0^1 \left(\frac{a(1-b)}{b(1-a)}\right)^p dp = \left. \frac{b(1-a)}{\ln \frac{a(1-b)}{b(1-a)}} \left(\frac{a(1-b)}{b(1-a)}\right)^p \right|_0^1 $$&#xA;$$= \frac{a(1-b) - b(1-a)}{\ln a + \ln (1-b) - \ln b - \ln (1-a)} = \frac{a-b}{\ln a + \ln (1-b) - \ln b - \ln (1-a)}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This holds if $a \neq b$ and if neither of $a$ or $b$ is 0 or 1.  If $a = b$, then instead we have $$a(1-a) \int_0^1 dp = a - a^{2}.$$&#xA;And, of course, if $a$ or $b$ is 0 or 1 then the value of the integral is 0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, as far as maximizing, you can use the usual approach of finding where both partial derivatives are 0.  I haven't worked through the calculations, but I strongly suspect that because of the symmetry in $a$ and $b$ that the maximum value will occur at $a = b$.&lt;/p&gt;&#xA;" OwnerUserId="2370" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-11-09T04:46:34.663" LastActivityDate="2010-11-09T04:46:34.663" CommentCount="5" />
  <row Id="9531" PostTypeId="2" ParentId="9524" CreationDate="2010-11-09T04:19:16.330" Score="3" ViewCount="" Body="&lt;p&gt;While I somewhat struggle to relate to the problem (series was possibly the most interesting section of Calc II to me), I can share some insight on how my teacher was able to make them seem interesting. She used Mathematica demonstrations throughout every section of our calculus classes (I also had her for Calc III), and I found that the visualization of terms and partial sums helped to visually see the asymptotic behavior of each towards zero and the sum, respectively. Taylor and Maclaurin series were also very &quot;unimportant&quot; to me (in the sense that I did not at the time grasp their usefulness), until I was shown how the polynomials converged to the function. Some of her files are available on demonstrations.wolfram.com:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://demonstrations.wolfram.com/SeriesAFewExamples/&quot; rel=&quot;nofollow&quot;&gt;http://demonstrations.wolfram.com/SeriesAFewExamples/&lt;/a&gt;&#xA;&lt;a href=&quot;http://demonstrations.wolfram.com/SeriesStepsOnANumberLine/&quot; rel=&quot;nofollow&quot;&gt;http://demonstrations.wolfram.com/SeriesStepsOnANumberLine/&lt;/a&gt;&#xA;&lt;a href=&quot;http://demonstrations.wolfram.com/GraphsOfTaylorPolynomials/&quot; rel=&quot;nofollow&quot;&gt;http://demonstrations.wolfram.com/GraphsOfTaylorPolynomials/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, a key idea that has always stuck with me, is that we only know how to perform the basic arithmetic operations, and more complex functions like $e^x$, $\sin x$, etc. can all be computed using just addition, subtraction, multiplication, and division.&lt;/p&gt;&#xA;" OwnerUserId="1016" LastActivityDate="2010-11-09T04:19:16.330" CommentCount="5" />
  <row Id="9532" PostTypeId="1" AcceptedAnswerId="9533" CreationDate="2010-11-09T04:27:54.823" Score="3" ViewCount="416" Body="&lt;p&gt;arising out of comment made by Yuval Filmus in &lt;a href=&quot;http://math.stackexchange.com/questions/9395/what-is-the-cardinality-of-set-of-all-smooth-functions-in-l1&quot;&gt;what is the cardinality of set of all smooth functions in $L^1$ ?&lt;/a&gt; I got this idea (forgive me for my ignorance for if it is nothing but an elementary definition/result in real analysis). The idea is like this                        &lt;em&gt;Let $f:X\to Y$ is a mapping, where $X$ is a complete metric space (not sure if its strictly needed or whether a looser condition would do). If $f$ is a continuous mapping then $f$ is uniquely specified by a mapping $g:E\to Y$ where $E$ is a dense subset of $X$.&lt;/em&gt; What are the condition under which it is valid ? also  the validity of the converse statement.&lt;/p&gt;&#xA;" OwnerUserId="2987" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-10T20:24:20.237" LastActivityDate="2010-11-13T20:37:23.970" Title="Link between a Dense subset and a Continuous mapping" Tags="&lt;real-analysis&gt;&lt;general-topology&gt;" AnswerCount="1" />
  <row Id="9533" PostTypeId="2" ParentId="9532" CreationDate="2010-11-09T04:41:16.033" Score="11" ViewCount="" Body="&lt;p&gt;If $Y$ is a Hausdorff topological space, then the value of a continuous function $f\colon X\to Y$ is completely determined by the value of $f$ on a dense subset $E$ of $X$; to see this, suppose $f$ and $g$ are two functions that agree on a dense subset $E$ of $X$, and let $u\in X\setminus E$. If $f(u)\neq g(u)$, then there are open neighborhoods $U$ and $V$ of $f(u)$ and $g(u)$, respectively, such that $U\cap V=\emptyset$. Then $f^{-1}(U)$ is an open neighborhood of $u$, as is $g^{-1}(V)$. Their intersection is an open neighborhood of $u$, and therefore must contain elements of $E$; but then any $e\in E$ in the intersection has $f(e)=g(e)$, with $f(e)\in U$ and $g(e)\in V$, contradicting that $U\cap V=\emptyset$. Therefore, $f(u)=g(u)$, hence $f=g$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this situation, it does not matter if $X$ is a complete metric space (or even a metric space); the key is $Y$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To see that the key is $Y$, consider the extreme case in which $Y$ has the indiscrete topology (the only open sets are the empty set and the set $Y$). Then &lt;em&gt;any&lt;/em&gt; function into $Y$ is continuous, so you can make your $X$ anything you want, and have two functions that agree on any subspace you care to specify and yet differ somewhere else.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added:&lt;/strong&gt; Note that any metric space is necessarily Hausdorff, so if your maps are between metric spaces, then the property holds (as in the case of Yuval's answer). This because you have the property that $d(x,y)\geq 0$ and $d(x,y)=0$ if and only if $x=y$. Thus, given $x,y\in Y$, $x\neq y$, let $\epsilon=d(x,y)\gt 0$. Then $B(x,\frac{\epsilon}{2})$ and $B(y,\frac{\epsilon}{2})$ are open neighborhoods of $x$ and $y$ (respectively) that are disjoint: if $z$ were in the intersection, then $d(x,z)\lt \frac{\epsilon}{2}$, $d(z,y)\lt\frac{\epsilon}{2}$, and by the triangle inequality we would conclude that $d(x,y)\lt\epsilon$, a contradiction. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Final Addition:&lt;/strong&gt; See &lt;a href=&quot;http://math.stackexchange.com/questions/9825/does-this-property-characterize-a-space-as-hausdorff&quot;&gt;here&lt;/a&gt; for a discussion of the converse. &lt;/p&gt;&#xA;&#xA;&lt;hr/&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added 2:&lt;/strong&gt; A result towards a possible converse: certainly we need some separation on $Y$. Suppose that there exist points $x,y\in Y$, $x\neq y$, such that every open subset that contains $x$ also contains $y$ (so $Y$ could be $T_0$, but cannot be $T_1$). Let $X$ be the Sierpinski space, $X=\{a,b\}$ with topology $\tau = \{\emptyset, \{b\}, X\}$. Let $f,g\colon X\to Y$ be the following maps: $f$ is the constant function that maps everything to $x$; $g$ is the function that maps $a\mapsto x$, $b\mapsto y$. The constantfunction is certainly continuous. For $g$, if $U$ is an open subset that contains $y$ but not $x$, then $g^{-1}(U)=\{b\}$, which is open; so $g$ is continuous. And $g$ and $f$ agree on the dense set $\{a\}$, but are not equal. &#xA;I'm still trying to figure out the $T_1$ case (for all $x,y\in Y$, $x\neq y$, there exist open sets $U,V$ such that $x\in U-V$ and $y\in V-U$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added 3:&lt;/strong&gt; Another step: $T_1$ is not sufficient; a colleague came up with this one (I kept trying the cofinite topology on $\mathbb{N}$ and not getting anywhere): take two copies of the real line and identify every point except the origin; this is $Y$. The result is a $T_1$ space, but not Hausdorff since no neighborhoods of the two copies of the origin are disjoint.  Now let $X$ be the real line, and let $E = (-\infty,0)\cup(0,\infty)$ be the dense subset. The two obvious injections, one mapping $0$ to the first copy in $Y$ and the other mapping it to the other copy, are both continuous and agree on $E$ but not on all of $X$, so $T_1$ does not suffice for the property. At least, then, in the hierarchy of $T$-spaces, the first level at which we are guaranteed the property is Hausdorff. This does not, however, establish whether the converse property characterises Hausdorff-ness. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added 4.&lt;/strong&gt; I almost have the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Conjecture.&lt;/strong&gt; Assuming the Axiom of Choice, the following are equivalent for a topological space $Y$:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$Y$ is Hausdorff.&lt;/li&gt;&#xA;&lt;li&gt;For every topological space $X$, every dense subset $E$ of $X$, and every pair of continuous maps $f,g\colon X\to Y$, if $f$ and $g$ agree on $E$, then $f=g$.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Argument.&lt;/em&gt; Suppose that $Y$ is not Hausdorff. If there exist points $x,y\in Y$, $x\neq y$, such that every open neighborhood of $x$ contains $y$, then the map from the Sierpinski space indicated above shows that $Y$ does not have property 2. So we may assume that $Y$ is at least a $T_1$ space. But since $Y$ is not Hausdorff, there exist points $s,t\in Y$, $s\neq t$, such that for every neighborhoods $U$ of $s$ and $V$ of $t$ such that $s\in U-V$ and $t\in V-U$, we have $U\cap V\neq \emptyset$. Let $\mathfrak{U}\_s$ be the family of open neighborhoods of $s$ that do not contain $t$, and let $\mathfrak{V}\_t$ be the family of open neighborhoods of $t$ that do not contain $s$. Let $P=\mathfrak{U}\_s\times \mathfrak{V}\_t$, and partially order $P$ by letting $(U,V)\leq (U&amp;#39;,V&amp;#39;)$ if and only if $U&amp;#39;\subseteq U$ and $V&amp;#39;\subseteq V$. This makes $P$ into a directed partially ordered set (given any $(U,V),(R,S)\in P$, there exists $(U&amp;#39;,V&amp;#39;)\in P$ such that $(U,V)\leq (U&amp;#39;,V&amp;#39;)$ and $(R,S)\leq (U&amp;#39;,V&amp;#39;)$. Now, for each $(U,V)\in P$, we are assuming that $U\cap V\neq\emptyset$, so using the Axiom of Choice let $y_{(U,V)}\in Y$ be an element of $U\cap V$. Note that $y_{(U,V)}\neq s$ and $y_{(U,V)}\neq t$ for all $(U,V)\in P$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let $X = \{y_{(U,V)}\mid (U,V)\in P\}\cup\{s,t\}$, and give $X$ the induced topology from $Y$, so that the inclusion maps $X\hookrightarrow Y$ is continuous. Note that the set $E=\{y_{(U,V)}\mid (U,V)\in P\}$ is dense in $X$: for every open neighborhood $B$ of $s$ in $X$, there exists an open set $\mathcal{O}\_B\in Y$ such that $\mathcal{O}\_B\cap X = B$; in particular, $\mathcal{O}\_B$ is a neighborhood of $s$; let $\mathcal{V}$ be any open neighborhood of $t$ that does not contain $S$, and let $B&amp;#39;=\mathcal{V}\cap X$; let $\mathcal{U}$ be any open neighborhood of $s$ that does not contain $t$. Then $\mathcal{U}\cap\mathcal{O}\_B$ is open, hence $B&amp;#39;=\mathcal{U}\cap\mathcal{O}\_B\cap X$ is an open subset of $X$ that is contained in $B$. Consider now $y_{(\mathcal{U}\cap\mathcal{O}\_B,\mathcal{V})}$. This is in $\mathcal{U}\cap\mathcal{O}_B\cap\mathcal{V}\cap X \subseteq B&amp;#39;$, and is plainly in $E$. In particular, in $X$ we have that $B&amp;#39;\cap E\neq\emptyset$, and hence $B\cap E\neq \emptyset$. Thus, every open neighborhood of $s$ in $X$ contains points of $E$, hence $x$ lies in the closure of $E$. A symmetric argument holds for $t$. Thus, $E$ is dense in $X$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now consider the maps $f,g\colon X\to Y$ defined as follows: $f$ and $g$, restricted to $E$, are the identity; $f(s)=f(t)=s$; and $g(s)=g(t)=t$. I claim that $f$ and $g$ are both continuous. Indeed, let $\mathcal{O}$ be an open set in $Y$. If $\mathcal{O}\cap\{s,t\}=\emptyset$ or $\{s,t\}\subseteq\mathcal{O}$, there is nothing to do: the inverse image under both $f$ and $g$ is just the intersection with $X$, hence open in $X$. So assume without loss of generality that $s\in\mathcal{O}$ but $t\notin \mathcal{O}$. Note that $g^{-1}(\mathcal{O}) = (\mathcal{O}-\{s\})\cap X$, and since $Y$ is $T_1$ removing a single point from an open set results in an open set, so $g^{-1}(\mathcal{O})$ is open. So we just need to show that $f^{-1}(\mathcal{O}) = (\mathcal{O}\cap X)\cup\{t\}$ is open in $X$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And that is where I am a bit stuck at present. Can anyone either verify or falsify this?&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-13T20:37:23.970" LastActivityDate="2010-11-13T20:37:23.970" CommentCount="8" />
  <row Id="9534" PostTypeId="2" ParentId="9524" CreationDate="2010-11-09T05:03:40.453" Score="2" ViewCount="" Body="&lt;p&gt;I believe that power series were the main motivation for the formal study of infinite series; for instance, lots and lots of Euler's work consists of just multiplying out formal power series without any explicit concern for convergence.  Perhaps demonstrating e.g. Euler products in number theory would show why we'd be interested in having a firm basis on which to rest such an argument.&lt;/p&gt;&#xA;" OwnerUserId="3296" LastActivityDate="2010-11-09T05:03:40.453" CommentCount="4" />
  <row Id="9535" PostTypeId="2" ParentId="9522" CreationDate="2010-11-09T05:12:59.257" Score="8" ViewCount="" Body="&lt;p&gt;The answer is 7.5.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $X$ be the number of neighbors of a different color.  Let $X_i$ be 1 if houses $i$ and $i+1$ are different colors and 0 otherwise.  Then $X = \sum_{i=1}^{15} X_i.$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the expected number of neighbors of a different color is &#xA;$$E\left[\sum_{i=1}^{15} X_i\right] = \sum_{i=1}^{15} E[X_i] = \sum_{i=1}^{15} P(X_i = 1).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, $$P(X_i = 1) = P(i \text{ is red})P(i+1 \text{ is blue}|i \text{ is red}) + P(i \text{ is blue})P(i+1 \text{ is red}|i \text{ is blue}) $$&#xA;$$ = \frac{10}{16} \frac{6}{15} + \frac{6}{16} \frac{10}{15} = \frac{120}{16(15)} = \frac{1}{2}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus $$E[X] = \sum_{i=1}^{15} \frac{1}{2} = 7.5.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More generally, using indicator variables is often an effective way to calculate expected values.  The linearity of the expectation operator allows you to sidestep all the nasty dependencies you would otherwise be forced to deal with.  For lots of interesting examples using this approach, see Section 7.2 of Sheldon Ross's &lt;em&gt;A First Course in Probability&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given that the answer comes out so nicely I would not be surprised if there is a cleaner argument than mine.  If someone comes up with one I would love to see it. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Moron's generalization has inspired me to generalize mine. :)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose you have $m$ red houses and $n$ blue houses.  Then the expected number of neighbors of a different color is $$\frac{2mn}{m+n}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similar to my argument above,&#xA;$$P(X_i = 1) = P(i \text{ is red})P(i+1 \text{ is blue}|i \text{ is red}) + P(i \text{ is blue})P(i+1 \text{ is red}|i \text{ is blue}) $$&#xA;$$ = \frac{m}{m+n} \frac{n}{m+n-1} + \frac{n}{m+n} \frac{m}{m+n-1} = \frac{2mn}{(m+n)(m+n-1)}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus $$E[X] = \sum_{i=1}^{m+n-1} \frac{2mn}{(m+n)(m+n-1)} = \frac{2mn}{m+n}.$$&lt;/p&gt;&#xA;" OwnerUserId="2370" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-11-10T17:52:51.113" LastActivityDate="2010-11-10T17:52:51.113" CommentCount="2" />
  <row Id="9537" PostTypeId="2" ParentId="9524" CreationDate="2010-11-09T05:27:19.150" Score="3" ViewCount="" Body="&lt;p&gt;I think series do fit in with the integral quite nicely - the integral can be seen a &quot;series on a continuous set of summands&quot; while series are the discrete analog. They are both generalizations of finite everyday sums. The techniques used in the study of infinite series are similar to the ones used for improper integrals, and the integral convergence test is another reason why knowing integrals before studying series can be helpful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for why studying series is important, I think you should use motivating examples from the fields your students know, if possible. There is one basic &quot;global&quot; example I can think of - the paradoxes of Zeno.&lt;/p&gt;&#xA;" OwnerUserId="1818" LastActivityDate="2010-11-09T05:27:19.150" CommentCount="2" />
  <row Id="9538" PostTypeId="2" ParentId="9086" CreationDate="2010-11-09T05:35:03.803" Score="2" ViewCount="" Body="&lt;p&gt;Given $f(x) = a_n x^n + a_{n-1} x^{n-1} + \ldots + a_0$ has roots $r_1, r_2, \ldots r_n$, we have&#xA;$$a_n x^n + a_{n-1} x^{n-1} + \ldots + a_0 = a_n (x - r_1)(x - r_2) ... (x-r_n);$$&#xA;dividing through by $a_n$ gives&#xA;$$x^n + \frac{a_{n-1}}{a_n} x^{n-1} + \ldots + \frac{a_0}{a_n} = (x - r_1)(x - r_2) ... (x-r_n).$$&#xA;Multiplying out the RHS, the constant term is $\prod_{i=1}^n (-r_i)$.  This must match up with the constant term on the left, $a_0 / a_n$, which establishes the identity you wanted to come up with.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for actually proving the rational root theorem, it's not hard, but this isn't the way to go about it.  Just assume that $f(x) = a_n x^n + a_{n-1} x^{n-1} + \ldots + a_0$ as before and that $f(p/q) = 0$, where $p/q$ is in lowest terms (i.e., where $p$ and $q$ are relatively prime.)  Then we have&#xA;$$0 = f(p/q) = a_n \frac{p^n}{q^n} + a_{n-1} \frac{p^{n-1}}{q^{n-1}} + \ldots + a_0$$&#xA;Multiplying each side by $q^n$, we have&#xA;$$0 = a_n p^n + a_{n-1} p^{n-1} q + \ldots + a_0 q^n;$$&#xA;rearranging,&#xA;$$- a_n p^n = q(a_{n-1} p^{n-1} +  a_{n-2} p^{n-2} q +\ldots + a_0 q^{n-1}).$$&#xA;Since the RHS is divisible by $q$, so is the LHS.  But $p$ is relatively prime to $q$, which means that $a_n$ must be divisible by $q$.  Going back and rearranging things in a different way,&#xA;$$p(a_n p^{n-1} + a_{n-1} p^{n-2} q + \ldots a_1) = - a_0 q^n;$$&#xA;since the LHS is divisible by $p$, the RHS must be as well.  Again, $p$ is relatively prime to $q$, so $a_0$ is divisible by $p$.&lt;/p&gt;&#xA;" OwnerUserId="3296" LastActivityDate="2010-11-09T05:35:03.803" CommentCount="2" />
  <row Id="9539" PostTypeId="2" ParentId="7966" CreationDate="2010-11-09T06:16:17.710" Score="6" ViewCount="" Body="&lt;p&gt;I really like the way Bott &amp;amp; Tu develop spectral sequences in &quot;Differential Forms in Algebraic Topology&quot;.  It would be an especially good choice if you happen to have some familiarity with de Rham cohomology; they begin by just sort of easing you into the whole idea by proving the isomorphism of de Rham cohomology and Cech cohomology by constructing a double complex whose cohomology is isomorphic to both of those.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, they only give uses for spectral sequences in topology.  I think they mainly introduce them to be able to give the spectral sequences for the homology and cohomology of the total space of a fibration.  In this application there is particularly nice geometric intuition for the algebraic machinery, which I found very useful for trying to get to the bottom of what these things really do.&lt;/p&gt;&#xA;" OwnerUserId="2159" LastActivityDate="2010-11-09T06:16:17.710" CommentCount="1" />
  <row Id="9540" PostTypeId="1" CreationDate="2010-11-09T06:23:34.190" Score="4" ViewCount="45" Body="&lt;p&gt;This question was firstly posted on &lt;a href=&quot;http://mathoverflow.net/questions/45252/does-order-isomorphism-of-linear-extensions-of-two-partially-ordered-sets-imply-o&quot; rel=&quot;nofollow&quot;&gt;Mathoverflow&lt;/a&gt;. Two answers are pretty interesting. The potential counter-example given in the second answer is really interesting, but it is not surely a counter-example.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider two partially ordered sets $A = {a&amp;lt; b,a&amp;lt; c}$, $B={x&amp;lt; z,y&amp;lt; z}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Their linear extensions (here we allow equality in linear extensions) for $A, B$ are &#xA;$A_L={A_1={a&amp;lt; b&amp;lt; c}, A_2={a&amp;lt; b= c}, A_3={a&amp;lt; c&amp;lt; b}}$&#xA;and&#xA;$B_L = { B_1 = {x&amp;lt; y&amp;lt; z}, B_2={y&amp;lt; x&amp;lt; z}, B_3 ={x= y&amp;lt; z}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We may define $f_1: A_1\to B_1$ by $f_1(a)=x, f_1(b)=y, f_1(c)=z$ and $f_2:A_3\to B_2$ by $f_2(a)=y, f_2(b)=x, f_2(c)=z$, but there are no bijective mappings from $A_2$ to $B_3$ s.t. $t&amp;lt; s\Longleftrightarrow f_3(t)&amp;lt; f_3(s)$ and $t=s \Longleftrightarrow f_3(t)=f_3(s)$.  (And it's easy to see that no other pairing of the $A_i$ with the $B_j$ will allow such maps to be chosen.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are no order isomorphisms from $A$ to $B$, either.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is the following conjecture true?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For any partially ordered sets $A,B$, if there exist isomorphisms from $A$'s linear extensions to $B$'s, then there exists an isomorphism from $A$ to $B$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If not, could you give me a counter example? &lt;/p&gt;&#xA;" OwnerUserId="3303" LastEditorUserId="3303" LastEditorDisplayName="" LastEditDate="2010-11-09T06:31:18.890" LastActivityDate="2010-11-09T06:31:18.890" Title="Does order isomorphism of linear extensions of two partially ordered sets imply order isomorphism of themselves?" Tags="&lt;combinatorics&gt;&lt;homological-algebra&gt;&lt;order-theory&gt;" FavoriteCount="1" />
  <row Id="9541" PostTypeId="1" CreationDate="2010-11-09T07:51:04.227" Score="2" ViewCount="94" Body="&lt;p&gt;Given $xy=C$ where $x, y$ are integer variables  and $C$ is integer constant. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What is the most efficient algorithm that finds $x,y$ such that $x+y$ is minimum?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Providing references is highly appreciated. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: Input integers are reasonably encoded.&lt;/p&gt;&#xA;" OwnerUserId="2459" LastEditorUserId="2459" LastEditorDisplayName="" LastEditDate="2010-11-09T08:00:20.923" LastActivityDate="2010-11-09T08:00:20.923" Title="Hardness of perimeter minimization?" Tags="&lt;computer-science&gt;" CommentCount="6" FavoriteCount="1" />
  <row Id="9542" PostTypeId="1" AcceptedAnswerId="9555" CreationDate="2010-11-09T08:03:10.800" Score="3" ViewCount="188" Body="&lt;p&gt;I began reading Paul Halmos' &quot;Naive Set Theory&quot;, and encountered the &quot;Axiom of Specification&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;Blockquote&gt; To every set $A$ and to every condition $S(x)$ there corresponds a set $B$ whose elements are exactly those elements $x$ of $A$ for which $S(x)$ holds.&#xA;&lt;/Blockquote&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Earlier in the same section, I learned that statements in set theory should be &quot;sentences&quot;.  A sentence was defined by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;Blockquote&gt; There are two basic types of sentences, namely, assertions of belonging, &#xA;$x \in A$,&#xA;and assertions of equality,&#xA;$A = B$;&#xA;all other sentences are obtained from such &lt;i&gt;atomic&lt;/i&gt; sentences by repeated applications of the usual logical operators...&lt;/Blockquote&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A more complete definition of a sentence follows, which can be read on Google Books here: &lt;a href=&quot;http://goo.gl/XvK2B&quot; rel=&quot;nofollow&quot;&gt;http://goo.gl/XvK2B&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried to translate the axioms and theorems in the book into sentences, but it seems like the Axiom of Specification is not a sentence.  It refers to &quot;every condition&quot;, but I have no way to build a sentence that refers to &quot;every condition&quot; because the atomic sentences only refer to sets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is the Axiom of Specification a sentence?  If not, does that mean that statements about set theory do not need to be sentences?  What other sorts of statements are allowed?  (I'm using &quot;statement&quot; colloquially since I don't know the technical term.)&lt;/p&gt;&#xA;" OwnerUserId="3305" LastEditorUserId="630" LastEditorDisplayName="" LastEditDate="2010-11-09T14:51:15.590" LastActivityDate="2010-11-09T14:51:15.590" Title="How can I write the Axiom of Specification as a sentence?" Tags="&lt;set-theory&gt;&lt;logic&gt;&lt;axioms&gt;" AnswerCount="2" />
  <row Id="9543" PostTypeId="2" ParentId="9542" CreationDate="2010-11-09T08:17:09.713" Score="-2" ViewCount="" Body="&lt;p&gt;$$\forall A, \forall S, \exists B ; x \in A, x \in S \Rightarrow x \in B \text{ and } x \in B \Rightarrow x \in A, x \in S$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Read it as: &quot;For all $A$, for all $S$, there exists $B$ such that, if $x$ is in $A$ and $x$ is in $S$, then $x$ is also in $B$ and, if $x$ is in $B$, then $x$ is in $A$ and $x$ is in $S$.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Saying that a property $S(x)$ is true is usually translated as $x \in S$). &lt;/p&gt;&#xA;" OwnerUserId="2778" LastActivityDate="2010-11-09T08:17:09.713" CommentCount="5" />
  <row Id="9544" PostTypeId="2" ParentId="9508" CreationDate="2010-11-09T08:25:07.537" Score="0" ViewCount="" Body="&lt;p&gt;you could make sth inspired by RSA, where you do not need two keys. like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;calculate everything modulo $p$, where $p$ is prim. that will then be the GF[$p$]  &lt;a href=&quot;http://en.wikipedia.org/wiki/Finite_field&quot; rel=&quot;nofollow&quot;&gt;(GaloisField)&lt;/a&gt;.&#xA;now instead of finding a generator $g$ for that GF[$p$] and calculating anything in powers of $g$, you could use any large numbers for multiplication and addition. this allone will then be very unsafe, but it will be very chaotic in combination with some bit-operations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the encoding and decoding functions can be combined out of simple unsafe functions. the combination makes them more unpredictable, especially with the bit-operations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(calculate once the reciprocal ($m^{-1}\equiv m^{p-2}$) of every $m$, that you use to multiply; you'll need it for decryption.)&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;let's define functions enc1 and dec1 based on multiplication:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$enc1_m(x)=x·m$ (mod p)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$dec1_m(y)=y·m^{-1}$ (mod p)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;..................................................&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and another pair based on addition:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$enc2_a(x)=x+a$ (mod p)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$dec2_a(y)=y-a$ (mod p)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;..................................................&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and another pair based on bit-shuffling, but that need to be invertible and never generate a value $\geq p$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;let $(\odot,\oplus,\otimes,\neg)$ be the bitwise (and,or,xor,not) operations&lt;/p&gt;&#xA;&#xA;&lt;p&gt;let $permutateAndXorLowestNBits_{s}(x,n)$ change the lowest n bits of x so that it is invertible. in other words, this condition holds:&#xA;$$\forall x,s,n: permutateAndXorLowestNBits_{s}^{-1}(permutateAndXorLowestNBits_{s}(x,n),n)=x$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;let N(x) be the number of lowest bits of x that can be &quot;safely&quot; changed so that after changing those bits, N will result in the same number and the value never exceeds p. in other words, these two conditions hold: $$x \oplus (2^{N(x)}-1) &amp;lt; p$$   $$\forall y. x \odot \neg(2^{N(x)}-1) \leq y \leq x \oplus (2^{N(x)}-1) \Rightarrow N(y)=N(x)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$enc3_s(x) = permutateAndXorLowestNBits_{s}(x,N(x))$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$dec3_s(y) = permutateAndXorLowestNBits_{s}^{-1}(y,N(y))$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;..................................................&lt;/p&gt;&#xA;&#xA;&lt;p&gt;now, the resulting key will be (m,a,s) and the functions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$enc_{m,a,s}(x)=enc1_m(enc2_a(enc3_s(x)))$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$dec_{m,a,s}(y)=dec3_s(dec2_a(dec1_m(y)))$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but any other more complec combination would be valid, too. remember, that any combination of several enc1s and enc2s can be expressed as one single combination of enc1 and enc2; so, alternate it with enc3.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;i do not know, how hard it will be to break this if only p is known. without the bit-operations, a linear equation would break this, but with those bit-operations it will be very chaotic. does anyone know that? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;otherwise, you could still make sth like a &lt;a href=&quot;http://en.wikipedia.org/wiki/Feistel_cipher&quot; rel=&quot;nofollow&quot;&gt;Feistel cipher&lt;/a&gt;, but i do not know wether it would be more secure or how to measure that.&lt;/p&gt;&#xA;" OwnerUserId="3226" LastEditorUserId="3226" LastEditorDisplayName="" LastEditDate="2010-11-10T02:53:58.737" LastActivityDate="2010-11-10T02:53:58.737" CommentCount="2" />
  <row Id="9545" PostTypeId="1" CreationDate="2010-11-09T10:01:37.287" Score="1" ViewCount="134" Body="&lt;p&gt;Let $\Sigma$ be an alphabet of size $s$, with which we build strings of length $n$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;strong&gt;Hamming ball&lt;/strong&gt; of radius $d$ centered at $x\in\Sigma^n$ is the set $B(x, d)$ of words in $\Sigma^n$ that differ from $x$ in &lt;em&gt;at most&lt;/em&gt; $d$ positions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, the &lt;strong&gt;Hamming circle&lt;/strong&gt; of radius $d$ centered at $x\in\Sigma^n$ is the set $C(x, d)$ of words in $\Sigma^n$ that differ from $x$ in &lt;em&gt;exactly&lt;/em&gt; $d$ positions (note: &quot;Hamming ball&quot; is a standard term, but I do not know whether the expression &quot;Hamming circle&quot; exists at all).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there closed expressions for $|B(x, d_1)\cap B(y, d_2)|$ and $|C(x, d_1)\cap C(y, d_2)|$, given any two distinct strings $x,y\in\Sigma^n$ and any two $d_1,d_2\in\mathbb{N}$?&lt;/p&gt;&#xA;" OwnerUserId="689" LastActivityDate="2010-11-09T14:21:55.200" Title="Intersections of Hamming balls and &quot;circles&quot;" Tags="&lt;combinatorics&gt;&lt;counting&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="0" />
  <row Id="9546" PostTypeId="1" AcceptedAnswerId="9557" CreationDate="2010-11-09T10:05:55.067" Score="0" ViewCount="65" Body="&lt;p&gt;Let $A \in \mathbb{R}^{n \times n}$ be a positive definite matrix and let $a &gt; 0$ be some constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am interested in the following quantity&#xA;$$&#xA;\hat x = {\rm arg} \max_{x \in R^n}\ x' \left( \frac{1}{a + \|x\|^2_2} A \right) x &#xA;$$&#xA;Is it possible to relate $\|\hat x\|_2^2$ to the eigenvalue and $\hat x$ to the eigenvector of $A$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore, suppose that $E \in \mathbb{R}^{p \times p}$ is such that $A + E$ is still positive definite and $\|E\|$ is small in some norm (say, spectral). Consider &#xA;$$&#xA;\hat y = {\rm arg} \max_{y \in R^n}\ y' \left( \frac{1}{a + \|y\|^2_2} (A+E) \right) y&#xA;$$&#xA;Are there known results that can be used to quantify $| \|\hat x\|_2^2 - \|\hat y\|_2^2 |$? I suppose the norms should be close if the perturbation $E$ is small, but don't know how to quantify this notion. &lt;/p&gt;&#xA;" OwnerUserId="2384" LastEditorUserId="2384" LastEditorDisplayName="" LastEditDate="2010-11-09T11:58:30.370" LastActivityDate="2010-11-09T12:20:23.417" Title="How to relate the following problem to known results on the eigenvectors?" Tags="&lt;linear-algebra&gt;&lt;matrices&gt;&lt;perturbation-theory&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="9547" PostTypeId="1" AcceptedAnswerId="9551" CreationDate="2010-11-09T10:47:17.603" Score="0" ViewCount="58" Body="&lt;p&gt;This is from Calculus I test, that I failed to solve correctly. This really bugs me, because for &lt;code&gt;Solve[y == Log[E, 3 - 2x] / Sqrt[x + 1], x]&lt;/code&gt;, the Mathematica is telling me that: &quot;The equations appear to involve the variables to be solved for in an essentially non-algebraic way.&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/Eyjzo.gif&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I assume that the answer is $x \in \mathbb{R}$&lt;/p&gt;&#xA;" OwnerUserId="2317" LastEditorUserId="2317" LastEditorDisplayName="" LastEditDate="2010-11-09T11:11:58.967" LastActivityDate="2010-11-09T11:24:32.947" Title="Finding Range of: $y=\log_e(3-2x)/\sqrt{x+1}$?" Tags="&lt;calculus&gt;" AnswerCount="1" />
  <row Id="9548" PostTypeId="1" AcceptedAnswerId="9565" CreationDate="2010-11-09T10:53:11.240" Score="5" ViewCount="258" Body="&lt;p&gt;$f(a,b) = f(a-1, b) + f(a-1, b-1) + f(a, b-1), ab \neq 0$ &lt;br&gt;&#xA;$f(a,b) = 1, ab = 0$ &lt;br&gt;&#xA;So what is $f(a, b)$?&lt;/p&gt;&#xA;" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2011-06-03T23:46:35.713" LastActivityDate="2011-06-04T11:25:25.263" Title="General expression of $f(a, b)$ if $f(a, b)=f(a-1,b) + f(a, b-1) + f(a-1, b-1)$?" Tags="&lt;recurrence-relations&gt;" AnswerCount="4" />
  <row Id="9549" PostTypeId="2" ParentId="9417" CreationDate="2010-11-09T11:12:52.663" Score="2" ViewCount="" Body="&lt;p&gt;Just for fun here are some bounds for $\sum_{k=1}^n a^{1/k}$ for $ a \ge 1.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$a^{1/k} = 1 + \frac{\log a}{1! k} + \frac{(\log a)^2}{2! k^2}&#xA;+ \frac{(\log a)^3}{3! k^3} + \cdots$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and so&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{k=1}^n a^{1/k} = \zeta_n(0) + (\log a)\zeta_n(1) + \frac{(\log a)^2}{2!}\zeta_n(2) +&#xA;\frac{(\log a)^3}{3!}\zeta_n(3) + \cdots$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\zeta_n(r) = \sum_{k=1}^n 1/k^r.$&#xA;Thus $\zeta_n(0)=n$ and $\zeta_n(1)=H_n = 1 + 1/2 + 1/3 + \cdots + 1/n.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now $H_n= \log(n + 1/2) + \gamma + \epsilon(n),$ where $0&amp;lt; \epsilon(n)&amp;lt; 1/24n^2$ and $\gamma$ is the Euler-Mascheroni constant, &#xA;and (by comparing the sum with the integral of $1/x^r$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \frac{1}{(r-1)(n+1)^{r-1}} &amp;lt; \sum_{k=n+1}^\infty \frac{1}{k^r} &amp;lt;&#xA; \frac{1}{(r-1)n^{r-1}}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \frac{1}{(r-1)(n+1)^{r-1}} &amp;lt; \zeta(r) - \zeta_n(r) &amp;lt;&#xA; \frac{1}{(r-1)n^{r-1}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and so we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ l(a) \le \sum_{k=1}^n a^{1/k} \le u(a) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$l(a) = n + (\log a) \left( \log (n+ 1/2) + \gamma \right)&#xA;+\frac{(\log a)^2}{2!} \left( \zeta(2) - \frac{1}{n} \right)$$&#xA;$$+\frac{(\log a)^3}{3!} \left( \zeta(3) - \frac{1}{2n^2} \right)&#xA;+\frac{(\log a)^4}{4!} \left( \zeta(4) - \frac{1}{3n^3} \right) + \cdots$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$u(a) = n + (\log a) \left( \log (n+ 1/2) + \gamma + \frac{1}{24n^2} \right)&#xA;+\frac{(\log a)^2}{2!} \left( \zeta(2) - \frac{1}{n+1} \right)$$&#xA;$$+\frac{(\log a)^3}{3!} \left( \zeta(3) - \frac{1}{2(n+1)^2} \right)&#xA;+\frac{(\log a)^4}{4!} \left( \zeta(4) - \frac{1}{3(n+1)^3} \right) + \cdots.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Providing $a$ is not enormous and $n$ is sufficiently large, we only need a few terms for a good approximation to our sum.&lt;/p&gt;&#xA;" OwnerUserId="1301" LastEditorUserId="1301" LastEditorDisplayName="" LastEditDate="2010-11-09T11:19:41.607" LastActivityDate="2010-11-09T11:19:41.607" CommentCount="4" />
  <row Id="9550" PostTypeId="2" ParentId="9548" CreationDate="2010-11-09T11:17:45.587" Score="7" ViewCount="" Body="&lt;p&gt;I don't know about a formula for $f(a,b)$, but the generating function is&#xA;$$\sum_{a,b=0}^\infty f(a,b)x^ay^b=\frac1{1-x-y-xy}.$$&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-11-09T11:17:45.587" />
  <row Id="9551" PostTypeId="2" ParentId="9547" CreationDate="2010-11-09T11:24:32.947" Score="2" ViewCount="" Body="&lt;p&gt;Just observe that at $\lim_{x\to -1^+} y(x) = \infty$ and $\lim_{x \to 3/2^-} y(x) = -\infty$ and $y(x)$ is continuous on $x \in (-1, 3/2)$. Therefore the range is indeed $(-\infty,\infty)$.&lt;/p&gt;&#xA;" OwnerUserId="3223" LastActivityDate="2010-11-09T11:24:32.947" />
  <row Id="9552" PostTypeId="1" AcceptedAnswerId="9558" CreationDate="2010-11-09T11:29:17.660" Score="0" ViewCount="175" Body="&lt;p&gt;So, &lt;a href=&quot;http://en.wikipedia.org/wiki/Affine_space&quot; rel=&quot;nofollow&quot;&gt;wikipedia&lt;/a&gt; has a page about &quot;affine space&quot;, where such a space is an object (set, topological space, blah) with a free transitive (blah, continuous, blah) action by the additive group of a free module.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At least when the base ring is an algebraically closed field, we can identify the top. sp. of $\mathbf{A}^n_k$, with such an affine space (I don't care enough to check whether or not the topology is compatible).  (The point sets agree by the Nullstellensatz)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case that k is a field but not alg. closed, is there any way to make sense of the discrepancy between the definitions, or is calling the affine variety $\mathbf{A}^n_k$ &quot;affine n-space&quot; just a relic of the ultraclassical algebraically closed case where the identification actually makese sense?&lt;/p&gt;&#xA;" OwnerUserId="126" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2010-11-19T01:09:12.577" LastActivityDate="2010-11-19T04:11:35.230" Title="Axiomatic affine n-space and the (classical) affine variety $A^n_k$ only agree for K an alg. closed field?" Tags="&lt;geometry&gt;" AnswerCount="1" CommentCount="7" />
  <row Id="9553" PostTypeId="2" ParentId="9548" CreationDate="2010-11-09T11:32:08.457" Score="4" ViewCount="" Body="&lt;p&gt;I doubt there's a nice closed form in both entries.  For example, the diagonal entries $f(n, n)$ have generating function&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{n=0}^{\infty} f(n, n) x^n = \frac{1}{\sqrt{1 - 6x - x^2}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which can be proven using the methods described &lt;a href=&quot;http://qchu.wordpress.com/2009/10/07/extracting-the-diagonal/&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;, but suggests that $f(n, n)$ doesn't have a product formula (for example like the Catalan numbers do).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For fixed $n$, it turns out that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{m=0}^{\infty} f(m, n) x^m = \frac{(1 + x)^n}{(1 - x)^{n+1}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which translates into the binomial identity&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f(m, n) = \sum_{k=0}^{n} {n \choose k} {m + n-k \choose n}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but this identity isn't terribly useful for large $n$.  &lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="226" LastEditorDisplayName="" LastEditDate="2010-11-09T14:02:43.353" LastActivityDate="2010-11-09T14:02:43.353" CommentCount="2" />
  <row Id="9554" PostTypeId="1" AcceptedAnswerId="9556" CreationDate="2010-11-09T12:02:18.400" Score="0" ViewCount="955" Body="&lt;p&gt;I'd heard of propositional logic for years, but until I came across &lt;a href=&quot;http://math.stackexchange.com/questions/4043/what-are-good-resources-for-learning-predicate-logic-predicate-calculus&quot;&gt;this question&lt;/a&gt;, I'd never heard of predicate logic.  Moreover, the fact that &lt;em&gt;Introduction to Logic: Predicate Logic&lt;/em&gt; and &lt;em&gt;Introduction to Logic: Propositional Logic&lt;/em&gt; (both by Howard Pospesel) are distinct books leads me to believe there are significant differences between the two fields.  What distinguishes predicate logic from propositional logic?&lt;/p&gt;&#xA;" OwnerUserId="506" LastActivityDate="2010-11-09T15:11:06.870" Title="What's the difference between predicate and propositional logic?" Tags="&lt;logic&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="9555" PostTypeId="2" ParentId="9542" CreationDate="2010-11-09T12:02:50.743" Score="6" ViewCount="" Body="&lt;p&gt;The axiom of specification is not a sentence. It's an &quot;axiom scheme&quot;, which is to say that it is a family of sentences. This is one of the things that becomes more clear when you move to axiomatic set theory, instead of naive set theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For each sentence $S(x)$ that does not mention $B$, the axiom of specification includes the axiom&#xA;$$&#xA;\forall A \exists B \forall x ( x \in B \Leftrightarrow x \in A \land S(x)). &#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What that axiom says, informally, is that given a set $A$ and a definition $S$ of a subset of $A$, that subset actually exists.   The scheme is slightly more general than my previous formula, because the scheme allows sentences with &quot;parameters&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The restriction that $S$ does not mention $B$ is to avoid paradoxes. Otherwise we would have as an axiom (letting $A = \{0\}$ and letting $S$ be &quot;$x \not \in B$&quot;)&#xA;$$&#xA;\exists B \forall x ( x \in B \Leftrightarrow x \in \{0\} \land x \not \in B).&#xA;$$&#xA;That set is paradoxical - it contains 0 if and only if it doesn't contain $0$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason that we cannot quantify over sentences is that set theory is formalized using the logical system of &quot;first order logic&quot;. That system is not able to quantify over sentences. This isn't an arbitrary choice; the inability to quantify over sentences is a necessary result of certain logical properties of first-order logic that are desirable. There are other logics in which one can quantify over sentences, but these logics do not have nice properties (and some have argued these logics themselves include set theory). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;All of this is explained, in great detail, in books on axiomatic set theory. One reasonable book is Levy's &lt;em&gt;Basic set theory&lt;/em&gt;. The standard graduate textbook is Kunen's &lt;em&gt;Set theory: an introduction to independence proofs&lt;/em&gt;, and it can be used to learn axiomatic set theory, but it is somewhat terse at the beginning and is better as a second book on axiomatic set theory in my opinion. &lt;/p&gt;&#xA;" OwnerUserId="630" LastEditorUserId="630" LastEditorDisplayName="" LastEditDate="2010-11-09T12:17:02.713" LastActivityDate="2010-11-09T12:17:02.713" CommentCount="1" />
  <row Id="9556" PostTypeId="2" ParentId="9554" CreationDate="2010-11-09T12:10:50.753" Score="2" ViewCount="" Body="&lt;p&gt;Propositional logic (also called sentential logic) is the logic the includes sentence letters (A,B,C) and logical connectives, but not quantifiers. The semantics of propositional logic uses truth assignments to the letters to determine whether a compound propositional sentence is true. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Predicate logic is usually used as a synonym for first-order logic, but sometimes it is used to refers to other logics that have similar syntax. Syntactically, first-order logic has the same connectives as propositional logic, but it also has variables for individual objects, quantifiers, symbols for functions, and symbols for relations. The semantics include a domain of discourse for the variables and quantifiers to range over, along with interpretations of the relation and function symbols.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many undergrad logic books will present both propositional and predicate logic, so if you find one it will have much more info. A couple well-regarded options that focus directly on this sort of thing are Mendelson's book or Enderton's book. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This set of &lt;a href=&quot;http://www.math.psu.edu/simpson/courses/math557/logic.pdf&quot; rel=&quot;nofollow&quot;&gt;lecture notes&lt;/a&gt; by Stephen Simpson is free online and has a nice introduction to the area. &lt;/p&gt;&#xA;" OwnerUserId="630" LastActivityDate="2010-11-09T12:10:50.753" />
  <row Id="9557" PostTypeId="2" ParentId="9546" CreationDate="2010-11-09T12:20:23.417" Score="2" ViewCount="" Body="&lt;p&gt;Let $\lambda$ be the biggest eigenvalue of A and $x_{\lambda}$ a corresponding eigenvector. Since $x' A x \leq \lambda \left\| x \right\|^2$, we obtain&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;x' \left( \frac{1}{a + \left\| x \right\|_2^2} A \right) x&#xA;\leq \frac{\lambda \left\| x \right\|^2}{a + \left\| x \right\|^2} &amp;lt; \lambda&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But for $x = t x_{\lambda}$ and $t \to \infty$, the expression goes to $\lambda$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means the maximum is never attained.&lt;/p&gt;&#xA;" OwnerUserId="960" LastActivityDate="2010-11-09T12:20:23.417" />
  <row Id="9558" PostTypeId="2" ParentId="9552" CreationDate="2010-11-09T12:44:34.593" Score="3" ViewCount="" Body="&lt;p&gt;Affine space means a vector space in which the origin is not preferred.  Its automorphisms are the group of affine linear transformations over $k$ (the usual linear transformations with respect to some (arbitrary) choice of origin, together with translations).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In algebraic geometry, one often encodes geometric objects in slightly complicated ways, and the preferred method of encoding is time-dependent, e.g. in that what Weil meant by a variety, and what it typically meant now, are not literally the same mathematical object, although the are both supposed to refer to the same Platonic reality.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, one shouldn't think about $\mathbb A^n_k$ too literally in terms of its points when comparing with the Wikipedia discussion of affine space (since what its points are will depend on what foundations you use!).  It does encode the geometric concept of affine space (as in the linked Wikipedia article) in the context of algebraic geometry.  For example, its automorphisms are precisely the group of affine linear transformations over $k$. [Correction: As Mariano points out in a comment below, there are many non-linear automorphisms of $\mathbb A^n$; what I should have written is that the automorphisms of $\mathbb A^n$ which extend to $\mathbb P^n$ --- so these are the automorphisms of $\mathbb A^n$ that extend to the hyperplane at infinity in some reasonable sense --- are precisely the affine linear automorphisms.]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you do insist on making the connection precise on the level of points, then&#xA;one way to say it is that if $\Omega$ is &lt;em&gt;any&lt;/em&gt; extension field of $k$, then&#xA;the $\Omega$-valued points of $\mathbb A^n_k$ are naturally an $n$-dimensional affine space over $\Omega$.&lt;/p&gt;&#xA;" OwnerUserId="221" LastEditorUserId="221" LastEditorDisplayName="" LastEditDate="2010-11-19T04:11:35.230" LastActivityDate="2010-11-19T04:11:35.230" CommentCount="4" />
  <row Id="9561" PostTypeId="2" ParentId="9554" CreationDate="2010-11-09T14:08:37.047" Score="1" ViewCount="" Body="&lt;p&gt;Propositional logic is an axiomatization of Boolean logic.  As such predicate logic includes propositional logic.  Both systems are known to be consistent, e.g. by exhibiting models in which the axioms are satisfied.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Propositional logic is decidable, for example by the method of truth tables:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Truth_table&quot; rel=&quot;nofollow&quot;&gt; [Truth table -- Wikipedia]&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and &quot;complete&quot; in that every tautology in the sentential calculus (basically a Boolean expression on variables that represent &quot;sentences&quot;, i.e. that are either True or False) can be proven in propositional logic (and conversely).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Predicate logic (also called predicate calculus and first-order logic) is an extension of propositional logic to formulas involving terms and predicates.  The full predicate logic is undecidable:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/First-order_logic&quot; rel=&quot;nofollow&quot;&gt; [First-order logic -- Wikipedia]&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is &quot;complete&quot; in the sense that all statements of the predicate calculus which are satisfied in every model can be proven in the &quot;predicate logic&quot; and conversely.  This is a famous theorem by Gödel (dissertation,1929):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/G%C3%B6del_completeness_theorem&quot; rel=&quot;nofollow&quot;&gt; [Gödel's completeness theorem -- Wikipedia]&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;regards, hardmath&lt;/p&gt;&#xA;" OwnerUserId="3111" LastEditorUserId="3111" LastEditorDisplayName="" LastEditDate="2010-11-09T15:11:06.870" LastActivityDate="2010-11-09T15:11:06.870" />
  <row Id="9562" PostTypeId="2" ParentId="7966" CreationDate="2010-11-09T14:09:41.900" Score="7" ViewCount="" Body="&lt;p&gt;I like the exposition in the appendix to Eisenbud's &lt;em&gt;Commutative Algebra,&lt;/em&gt; which does everything in a kind of friendly way via exact couples. It &lt;em&gt;doesn't&lt;/em&gt; do the Grothendieck spectral sequence, though, but that's in many other places (e.g. Lang's &lt;em&gt;Algebra&lt;/em&gt;), and essentially follows (after some homological stuff) from the spectral sequence of a double complex. (I should confess that I've not myself needed to know too much about spectral sequences at the moment (beyond the Leray spectral sequence and the Cech-to-derived functor one in sheaf cohomology), and my knowledge of them is correspondingly limited.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EGA III has several applications of spectral sequences in algebraic geometry. The very first result, that the higher quasi-coherent cohomology vanishes on an affine scheme, implicitly uses the Cech-to-derived functor spectral sequence since the computation given is that of the Cech cohomology (but he doesn't actually go through the details--I think they're in Godement's &lt;em&gt;Theorie de faisceaux&lt;/em&gt;, but the deduction of it as a special case of the Grothendieck's spectral sequence is in Milne's online notes on etale cohomology). Another application in EGA III is the proof of the proper mapping theorem: that a proper map of noetherian schemes preserves coherence. The argument is very pretty, and starts in the standard case by doing this for a &lt;em&gt;projective&lt;/em&gt; morphism (which is essentially the calculation, as in Hartshorne, of the cohomology of the standard line bundles on $\mathbb{P}^n$). But to deduce it for a proper morphism, the real fun starts when you &quot;approximate&quot; a proper $f: X \to Y$  morphism by a projective one (using Chow's lemma) and then show that there is a &quot;significant&quot; sheaf on $X$ which pushes forward to $Y$ coherently using the projective morphism and the Grothendieck spectral sequence between various push-forwards. (There's also a devissage argument that I'm skipping here.)&lt;/p&gt;&#xA;" OwnerUserId="536" LastActivityDate="2010-11-09T14:09:41.900" />
  <row Id="9563" PostTypeId="2" ParentId="9545" CreationDate="2010-11-09T14:21:55.200" Score="2" ViewCount="" Body="&lt;p&gt;To get started, let us work on the circle intersection where the distances are the same:&#xA;$|C(x,d)\cap C(y,d)|$.  Let the distance between $x$ and $y$ be $m$.  Then we pick $i$ places of the disagreement where we agree with $x$, have to have $i$ of the disagreement that we agree with $y$, and $d-i$ of the others where we have to disagree with both to get the correct distance.  Each of those $d-i$ can have $s-2$ choices, as they have to disagree with $x$ and $y$, as can the $m-2i$ that we didn't choose to agree with x or y.  So $$|C(x,d)\cap C(y,d)|=\sum_{i=0}^{\frac{m}{2}} \binom{m}{i} \binom{m-i}{i} \binom{n-m}{d-i} (d+m-3i)^{(s-2)}$$  If you can sum this, you are better than I.  For the ball, we can just sum over $m$.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-11-09T14:21:55.200" />
  <row Id="9564" PostTypeId="1" CreationDate="2010-11-09T14:23:51.660" Score="2" ViewCount="34" Body="&lt;p&gt;How can I prove that the Cartier dual of $\alpha_p$ is again $\alpha_p$ (using the Yoneda lemma)? It should be something like $\alpha_p(R) \to (\alpha_p(R) \to \mu_p(R)), x \mapsto (y \mapsto exp_{p-1}(x + y)$, where $exp_{p-1}$ is the truncated exponential sequence. My problem is that this isn't a homomorphism.&lt;/p&gt;&#xA;" OwnerUserId="3267" LastActivityDate="2010-11-09T14:23:51.660" Title="Cartier dual of alpha_p" Tags="&lt;algebraic-geometry&gt;" FavoriteCount="1" />
  <row Id="9565" PostTypeId="2" ParentId="9548" CreationDate="2010-11-09T14:24:02.127" Score="9" ViewCount="" Body="&lt;p&gt;These numbers actually have a name.  They're called &lt;a href=&quot;http://www.research.att.com/~njas/sequences/A008288&quot; rel=&quot;nofollow&quot;&gt;Delannoy numbers&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The link here is to the On-Line Encyclopedia of Integer Sequences.  Generally, when you have a sequence of integers that you want to learn more about the OEIS is a great place to start looking.  For instance, both Robin Chapman's and Qiaochu Yuan's formulas are given at the link provided, as well as some other formulas and a large number of references and related problems.&lt;/p&gt;&#xA;" OwnerUserId="2370" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-11-09T14:35:42.677" LastActivityDate="2010-11-09T14:35:42.677" />
  <row Id="9567" PostTypeId="1" CreationDate="2010-11-09T14:55:04.790" Score="0" ViewCount="161" Body="&lt;p&gt;how to find the center of an arc given start and end points&lt;/p&gt;&#xA;" LastActivityDate="2010-11-10T22:35:41.070" Title="how to calculate an arc center" Tags="&lt;analytic-geometry&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="9568" PostTypeId="2" ParentId="9525" CreationDate="2010-11-09T15:25:21.630" Score="7" ViewCount="" Body="&lt;p&gt;You've got the factor wrong.  There is a well-known identity&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$a^3 + b^3 + c^3 - 3abc = \frac{1}{2} \left( a + b + c \right) \left( (a - b)^2 + (b - c)^2 + (c - a)^2 \right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which you were probably thinking of, but to get the correct factorization you should take $a = x, b = \sqrt[3]{D} y, c = \sqrt[3]{D^2} z$. &lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-11-09T15:25:21.630" CommentCount="1" />
  <row Id="9569" PostTypeId="1" AcceptedAnswerId="39948" CreationDate="2010-11-09T15:55:56.270" Score="3" ViewCount="167" Body="&lt;p&gt;Ok, that's a theorem. As it says in &lt;a href=&quot;http://math.stackexchange.com/questions/6148/if-a-graph-can-be-colored-with-max-4-colors-is-it-planar&quot;&gt;this question&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I just cannot understand how mathematics can prove such a thing; what does exactly this theorem say? How to prove it mathematically?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="3314" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-05-18T20:31:59.610" LastActivityDate="2011-05-18T20:39:27.713" Title="Every planar graph can be colored with 4 colors max." Tags="&lt;graph-theory&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="9570" PostTypeId="2" ParentId="9522" CreationDate="2010-11-09T16:00:52.453" Score="3" ViewCount="" Body="&lt;p&gt;The chances of a particular neighbour pair being the same colour is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \frac{{14 \choose 8} + {14 \choose 10}}{{16 \choose 6}} = \frac{4004}{8008} = \frac{1}{2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence the answer is $\displaystyle 7.5$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is happening because $\displaystyle {14 \choose 8}, {14 \choose 9}, {14 \choose 10}$ are in arithmetic progession: The number of ways of being same colour is $\displaystyle {14 \choose 8} + {14 \choose 10}$ and the number of ways of being different is $\displaystyle 2{14 \choose 9}$. The probability is $\displaystyle \frac{1}{2}$ if these two are equal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Interestingly, $\displaystyle {n \choose r}, {n \choose r+1}, {n \choose r+2}$ are in arithmetic progression if and only if $\displaystyle n+2$ is a perfect square and $\displaystyle r$ is given by $\displaystyle r = \frac{n-2 \pm \sqrt{n+2}}{2}$ (see the end of the answer for a proof).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So for instance, the whole bunch of problems:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;15 red, 10 blue&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;21 red, 15 blue&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;etc give rise to this neat probability of being $\displaystyle \frac{1}{2}$.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof that n+2 is a perfect square&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle {n \choose r}, {n \choose r+1}, {n \choose r+2}$ are in arithmetic progression iff&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle 2{n \choose r+1} = {n \choose r} + {n \choose r+2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i.e&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle 2 = \frac{r+1}{n-r} + \frac{n-r-1}{r+2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Doing some manipulations gives us&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle (n-2r-2)^2 = n+2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle r = \frac{n-2 \pm \sqrt{n+2}}{2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which has an integer solution iff $\displaystyle n+2$ is a perfect square.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-11-09T18:24:03.647" LastActivityDate="2010-11-09T18:24:03.647" CommentCount="3" />
  <row Id="9571" PostTypeId="1" CreationDate="2010-11-09T16:06:24.540" Score="1" ViewCount="91" Body="&lt;p&gt;From Wikipedia's triangle center article: &quot;Thus every point is potentially a triangle center. However the vast majority of triangle centers are of little interest, just as most continuous functions are of little interest. The Encyclopedia of Triangle Centers is an ever-expanding list of interesting ones.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hyperbolic Barycentric Coordinates are used in:&#xA;&lt;a href=&quot;http://ajmaa.org/searchroot/files/pdf/v6n1/v6i1p18.pdf&quot; rel=&quot;nofollow&quot;&gt;http://ajmaa.org/searchroot/files/pdf/v6n1/v6i1p18.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What makes a triangle center interesting, and why would hyperbolic triangle centers be interesting ? Wouldn't they just be copies of the euclidean triangle centers ?&lt;/p&gt;&#xA;" OwnerUserId="3316" LastEditorUserId="3316" LastEditorDisplayName="" LastEditDate="2010-11-10T10:05:33.940" LastActivityDate="2010-11-10T10:05:33.940" Title="Triangle centers" Tags="&lt;geometry&gt;" />
  <row Id="9572" PostTypeId="1" CreationDate="2010-11-09T16:09:16.430" Score="1" ViewCount="84" Body="&lt;p&gt;During my studies of real analysis I've come across the definition of a field (or algebra) of sets. My question is: What is the connection between this structure and the structure of a field or algebra known from abstract algebra?&lt;/p&gt;&#xA;" OwnerUserId="766" LastActivityDate="2010-11-09T20:51:37.433" Title="Field of sets versus a field as an algebraic structure" Tags="&lt;real-analysis&gt;&lt;abstract-algebra&gt;&lt;definition&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" />
  <row Id="9573" PostTypeId="1" AcceptedAnswerId="9581" CreationDate="2010-11-09T16:13:07.893" Score="2" ViewCount="114" Body="&lt;p&gt;We have a perfect binary tree on 2^k-1 nodes. Every node in the tree is marked with probability 1/2, and a node is either marked or unmarked. We want to find a marked node and return it. Our algorithm is as follows&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;procedure find(v);&#xA; if (v is marked)&#xA;   print(v)&#xA; else if (v is not a leaf)&#xA;   v1 is the left child of v&#xA;   v2 is the right child of v&#xA;   find(v1) ( recursive call )&#xA;   find(v2) ( recursive call )&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;So the algorithm basically starts from the top. Searches down the tree in a depth first fashion, and returns at least one marked node if there are any. Note that the algorithm can easily return multiple marked nodes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now what I want to do is to say something about the expected number of nodes the algorithm will consider on a random tree (1/2 probability of marking nodes), when the algorithm is run on the root of the tree.&lt;/p&gt;&#xA;" OwnerUserId="3317" LastEditorUserId="3317" LastEditorDisplayName="" LastEditDate="2010-11-09T17:32:29.010" LastActivityDate="2010-11-09T23:54:41.857" Title="Expected time of tree search algorithm on random input" Tags="&lt;probability-theory&gt;&lt;algorithms&gt;&lt;random&gt;&lt;probability&gt;" AnswerCount="3" CommentCount="4" />
  <row Id="9574" PostTypeId="2" ParentId="9572" CreationDate="2010-11-09T16:32:30.607" Score="5" ViewCount="" Body="&lt;p&gt;The term &quot;field&quot; in &quot;field / algebra of sets over A&quot; has no relation to the term &quot;field&quot; as used in ring theory. Rather it denotes a Boolean subalgebra of the power set of A, i.e. a collection of subsets of A closed under union, intersection and complement. The term is well-known due to its frequent use in the famous Stone representation theorem: $\ $every Boolean algebra is isomorphic to a field of sets. For further information see &lt;a href=&quot;http://en.wikipedia.org/wiki/Field_of_sets&quot; rel=&quot;nofollow&quot;&gt;this Wikipedia page&lt;/a&gt; and &lt;a href=&quot;http://groups.google.com/group/sci.logic/browse_frm/thread/20373f3ea47d79c0/cceccb1e3469a031&quot; rel=&quot;nofollow&quot;&gt;this sci.math thread&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: Boolean algebras can be considered as &lt;a href=&quot;http://en.wikipedia.org/wiki/Boolean_ring&quot; rel=&quot;nofollow&quot;&gt;Boolean rings&lt;/a&gt;, i.e. rings where every element is idempotent, i.e. $\: x^2 = x\:$. However, except for the field of two elements, such rings are never fields: $\ 0 = x^2 - x = x\ (x-1)\ \Rightarrow\ x = 0\ \ {\rm or}\ \ x = 1\ $ in a field, i.e. fields have only trivial idempotents.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-11-09T20:51:37.433" LastActivityDate="2010-11-09T20:51:37.433" />
  <row Id="9575" PostTypeId="1" AcceptedAnswerId="9576" CreationDate="2010-11-09T16:42:44.653" Score="3" ViewCount="274" Body="&lt;p&gt;If there is none, why?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And for the other side, what about open set (0, 1) to closed set [0, 1] with a continuous function?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="3318" LastActivityDate="2010-11-09T16:53:02.680" Title="Is there a continuous function from [0, 1] onto (0, 1)?" Tags="&lt;analysis&gt;&lt;topology&gt;" AnswerCount="2" />
  <row Id="9576" PostTypeId="2" ParentId="9575" CreationDate="2010-11-09T16:44:49.640" Score="5" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT:&lt;/strong&gt; For the first one use the fact that, Continuous image of a compact set is compact.&lt;/p&gt;&#xA;" LastActivityDate="2010-11-09T16:44:49.640" CommentCount="4" />
  <row Id="9578" PostTypeId="2" ParentId="9575" CreationDate="2010-11-09T16:53:02.680" Score="3" ViewCount="" Body="&lt;p&gt;For the other side consider $f: (0,1) \to [0,1]$ defined as $f(x)= |\cos(2\pi x)|^{2}$&lt;/p&gt;&#xA;" OwnerUserId="832" LastActivityDate="2010-11-09T16:53:02.680" CommentCount="1" />
  <row Id="9579" PostTypeId="2" ParentId="9573" CreationDate="2010-11-09T17:09:21.853" Score="0" ViewCount="" Body="&lt;p&gt;Idea: calculate recursively the expected number of nodes the algorithm visits on a complete binary tree of height $n$ &lt;i&gt;assuming at least one node is marked&lt;/i&gt;. Note that this assumptions alters slightly the probabilities of relevant events.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: This idea only makes sense if you stop after the first marked node is found, in which case JeffE's solution is much much better.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2010-11-09T19:06:52.483" LastActivityDate="2010-11-09T19:06:52.483" />
  <row Id="9580" PostTypeId="1" CreationDate="2010-11-09T17:18:59.470" Score="2" ViewCount="131" Body="&lt;p&gt;if $X_n$ converges to $X$ and $Y_n$ converges to $Y$ in distribution, what about $X_n + Y_n $ would that converge to $X+Y$ in distribution ?&#xA;any ideas how i could prove or disprove this &lt;/p&gt;&#xA;" OwnerUserId="3321" LastActivityDate="2010-11-09T18:40:17.147" Title="Linearity of convergence in distribution of random variables" Tags="&lt;probability-theory&gt;" AnswerCount="2" />
  <row Id="9581" PostTypeId="2" ParentId="9573" CreationDate="2010-11-09T17:20:36.680" Score="1" ViewCount="" Body="&lt;p&gt;Let's assume as Yuval suggests that 2^k - 1 nodes are meant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can get a recursive expression for the expected number of nodes N(k), where the basis case is N(1) = 1, i.e. that if we have one node, we have to visit it (whether or not it turns out to be marked).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If k &gt; 1, then with probability 1/2 the top node or root of the binary tree will be marked, and thus only the top node will be visited.  In the alternative we have to visit both left and right subtrees.  Thus:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;N(k+1) = 1/2 + (1/2)*[N(k) + N(k)] = 1/2 + N(k)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The solution of this recurrence relation + initial condition is simply:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;N(k) = (k+1)/2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;regardless of whether there are or are not marked nodes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;regards, hm&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Correction:&lt;/b&gt;  As Ross points out (see comment), the recurrence is N(k) = 1 + N(k), with solution N(k) = k, due to the fact that we will always visit the top node (before finding out whether it's marked).&lt;/p&gt;&#xA;" OwnerUserId="3111" LastEditorUserId="3111" LastEditorDisplayName="" LastEditDate="2010-11-09T23:54:41.857" LastActivityDate="2010-11-09T23:54:41.857" CommentCount="1" />
  <row Id="9583" PostTypeId="1" CreationDate="2010-11-09T17:29:01.487" Score="2" ViewCount="100" Body="&lt;p&gt;I am looking for an algorithm that &quot;describes&quot; the following situation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;15 - 22 people are combined into groups of  4 or 5 based upon stated preferences of who they wish to be grouped with.  Each person will be grouped with at least one of their preferences.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This isn't &quot;homework&quot;.&lt;/p&gt;&#xA;" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-11-09T21:03:42.897" LastActivityDate="2010-11-11T20:14:07.673" Title="Looking for an algorithm for grouping based on preferences" Tags="&lt;combinatorics&gt;&lt;algorithms&gt;&lt;graph-theory&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" />
  <row Id="9584" PostTypeId="2" ParentId="9573" CreationDate="2010-11-09T17:32:34.017" Score="2" ViewCount="" Body="&lt;p&gt;Assuming nodes are marked by independent coin flips, then the order that you visit nodes has no effect whatsoever.  So the expected number of nodes visited is exactly the expected number of fair coin flips to get heads: 2.&lt;/p&gt;&#xA;" OwnerUserId="475" LastActivityDate="2010-11-09T17:32:34.017" CommentCount="2" />
  <row Id="9587" PostTypeId="2" ParentId="9580" CreationDate="2010-11-09T17:55:04.670" Score="1" ViewCount="" Body="&lt;p&gt;Convergence in distribution is a pretty weak concept.. Suppose you consider probability distributions on $[0,1]$. Let $X = Y = X_n$ for all $n$ have a density supported on $[0,1/2]$ alone, and let $Y_n$ be the same distribution except shifted to the right by $1/2$. Then all these random variables have the same distribution, so convergence in distribution is automatic. But also each $X_n + Y_n$ is the same, but different from $X + Y$, so you won't get convergence in distribution. &lt;/p&gt;&#xA;" OwnerUserId="3035" LastActivityDate="2010-11-09T17:55:04.670" />
  <row Id="9589" PostTypeId="2" ParentId="6428" CreationDate="2010-11-09T17:59:53.627" Score="5" ViewCount="" Body="&lt;p&gt;There's a canonical example of fibrations used in algebraic topology which are not fibre bundles.  That is, given a continuous function $f : X \to Y$ there is a homotopy-equivalence $\phi : X' \to X$ and a fibration $f' : X' \to Y$ such that $f \circ \phi$ is homotopic to $f'$.  I believe this idea goes back to Serre (or perhaps earlier).  The fibre of $f'$ is called the homotopy-fibre of $f$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A common usage of this construction is with the Postnikov Tower of a space.  In his dissertation Jean-Pierre Serre used this (and some &quot;closure&quot; observations of the Serre spectral sequence of a fibration, then it was called Serre $\mathcal C$-theory, nowadays this technology is subsumed in localization) to show that most of the homotopy-groups of the spheres are finite.  &lt;/p&gt;&#xA;" OwnerUserId="642" LastActivityDate="2010-11-09T17:59:53.627" />
  <row Id="9591" PostTypeId="2" ParentId="9580" CreationDate="2010-11-09T18:31:25.663" Score="1" ViewCount="" Body="&lt;p&gt;Suppose that $X_n$ converges in distribution to $X$ where $X$ is a symmetric random variable, say $X \sim N(0,1)$. Then, trivially, $X_n$ also converges in distribution to $-X$ (since $X$ and $-X$ are identically distributed). However, $X_n + X_n$ does not converge in distribution to $X+(-X)=0$.&lt;/p&gt;&#xA;" OwnerUserId="2810" LastEditorUserId="2810" LastEditorDisplayName="" LastEditDate="2010-11-09T18:40:17.147" LastActivityDate="2010-11-09T18:40:17.147" />
  <row Id="9592" PostTypeId="1" AcceptedAnswerId="9599" CreationDate="2010-11-09T18:37:19.987" Score="5" ViewCount="135" Body="&lt;p&gt;I have an interesting math question I've developed while playing cards with my family and wonder if anyone has a way to approach the problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Say I have n people who each have a deck of cards. Each deck has a different number of cards in it. We are shuffling the cards for a game and in the end we all want to have the same number of cards to shuffle. Each individual can do one of two actions per round of shuffling. They can shuffle their deck, or they can split their deck in half and swap one half of their deck with half of another person's deck. My question is, is there an equation based on n to describe a method of swapping that will eventually give all n players the same amount of cards? Furthermore, how many rounds does it take? (In my mind, I define a round as a period of time where everyone performs one action. A player can only perform one action per round and the round ends when all players have performed one action)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To make the problem simpler, you can assume that each individual can only do one action: split their deck and swap with any other player.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is trivial when n is a power of 2. For example if we have 2 people, they split their decks, swap, and they have the same number of cards. It takes one round.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For 4 people, you split them into 2 groups of 2 people and perform the swap internally in the group. Then you make 2 groups with one person from each group in round one and have them swap. This takes 2 rounds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to see if there is an simple elegant solution when n is not a power of 2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(You can also assume that each deck splits evenly every time, which is a big assumption but should give a good basis for how to solve the problem. A similar problem would be to replace the cards with cups of liquid which gets rid of passing half of cards around.)&lt;/p&gt;&#xA;" OwnerUserId="3324" LastEditorUserId="3324" LastEditorDisplayName="" LastEditDate="2010-11-09T20:43:46.027" LastActivityDate="2010-11-09T22:34:06.730" Title="Card shuffling to Even Decks" Tags="&lt;number-theory&gt;&lt;combinatorics&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" />
  <row Id="9593" PostTypeId="2" ParentId="9592" CreationDate="2010-11-09T18:44:12.747" Score="2" ViewCount="" Body="&lt;p&gt;It depends on the number of cards each person holds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance with 3 people, one holding 8 cards, one 16 and the other 32 you can never get each to have the same number, as the total is not divisible by 3.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps I misunderstood your intent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the cards were &quot;liquid&quot; (or basically arbitrary real numbers), even then, there is no procedure (independent of the cards each person holds), which will end in finite number of steps.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, if you start with 1,1,2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After a finite number of steps each will have a finite representation when written in base 2, but $4/3$ does not have a finite representation in base 2.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-11-09T19:09:46.077" LastActivityDate="2010-11-09T19:09:46.077" CommentCount="6" />
  <row Id="9594" PostTypeId="1" CreationDate="2010-11-09T18:54:51.987" Score="3" ViewCount="260" Body="&lt;p&gt;I am writing a program to solve a Rubik's cube, and would like to know the answer to this question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are 12 ways to make one move on a Rubik's cube. How many ways are there to make a sequence of six moves?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From my project's specification: up to six moves may be used to scramble the cube. My job is to write a program that can return the cube to the solved state. I am allowed to use up to 90 moves to solve it. Currently, I can solve the cube, but it takes me over 100 moves (which fails the objective)... so I ask this question to figure out if a brute force method is applicable to this situation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the number of ways to make six moves is not overly excessive, I can just make six random moves, then check to see if the cube is solved. Repeat if necessary.&lt;/p&gt;&#xA;" OwnerUserId="3325" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2011-05-01T13:33:28.920" LastActivityDate="2011-05-01T13:33:28.920" Title="How many ways can I make six moves on a Rubik's cube?" Tags="&lt;combinatorics&gt;&lt;recreational-mathematics&gt;" AnswerCount="3" CommentCount="3" />
  <row Id="9595" PostTypeId="1" AcceptedAnswerId="9601" CreationDate="2010-11-09T18:56:47.543" Score="1" ViewCount="231" Body="&lt;p&gt;I'm considering a parametrized &lt;a href=&quot;http://en.wikipedia.org/wiki/Sigmoid_function&quot; rel=&quot;nofollow&quot;&gt;sigmoid&lt;/a&gt; function such as the following logistic:&lt;br&gt;&#xA;$$f(x)=\frac{e^{a+bx}}{1+e^{a+bx}}$$&lt;br&gt;&#xA;And I'm interested only in the interval $\displaystyle x &gt;= 0$ and $\displaystyle x &amp;lt; x_{max}$ (with a given $x_{max}$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Two tiny questions regarding its properties:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Is the &lt;strong&gt;sum of sigmoid functions always a sigmoid function itself&lt;/strong&gt;?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Is the log function (i.e. $\displaystyle f(x) = log_{a}(bx^{c})$ or similar) &lt;strong&gt;a particular case of a sigmoid function&lt;/strong&gt;? In other words, can I pick parameters for the logistic function above so that it behaves like a log function in the given interval?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="2916" LastEditorUserId="2916" LastEditorDisplayName="" LastEditDate="2010-11-09T21:40:16.653" LastActivityDate="2010-11-10T00:58:17.697" Title="Properties of sigmoid functions" Tags="&lt;statistics&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="9596" PostTypeId="1" CreationDate="2010-11-09T19:01:47.023" Score="3" ViewCount="355" Body="&lt;p&gt;I have always taken for granted that expected value is a linear operator. For &lt;em&gt;any&lt;/em&gt; random variables $X$ and $Y$: $E(aX + bY) = aE(X) + bE(Y)$. Can anyone point me to a rigorous proof of this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, I know that generally median $Med()$ is not a linear operator, meaning $Med(aX + bY)$ might not be equal to $a Med(X) + b Med(Y)$. Are there absolute criteria / rules when $Med$ is a linear operator, and when it is not?&lt;/p&gt;&#xA;" OwnerUserId="3326" LastEditorUserId="3326" LastEditorDisplayName="" LastEditDate="2010-11-09T22:37:03.357" LastActivityDate="2010-11-11T21:58:06.740" Title="Expected value is a linear operator? Under what conditions is median also a linear operator?" Tags="&lt;probability-theory&gt;" AnswerCount="5" CommentCount="4" />
  <row Id="9597" PostTypeId="2" ParentId="9596" CreationDate="2010-11-09T19:03:00.513" Score="1" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT:&lt;/strong&gt; Use the definition of $$ E(X) = \sum\limits_{x} x P(X)$$&lt;/p&gt;&#xA;" LastActivityDate="2010-11-09T19:03:00.513" CommentCount="2" />
  <row Id="9598" PostTypeId="1" AcceptedAnswerId="9600" CreationDate="2010-11-09T19:47:53.720" Score="2" ViewCount="207" Body="&lt;p&gt;How does one prove this identity: $$\lim_{n \to \infty} 2^{2n-1} \sqrt{n} \frac{ \Gamma(n)^{2}}{\Gamma(2n)} = \sqrt{\pi}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Taken from &lt;em&gt;Gamelin : Complex Analysis&lt;/em&gt;&lt;/p&gt;&#xA;" LastActivityDate="2010-11-10T02:17:06.250" Title="Proving $\lim_{n \to \infty} 2^{2n-1} \sqrt{n} \frac{ \Gamma(n)^{2}}{\Gamma(2n)} = \sqrt{\pi}$" Tags="&lt;complex-analysis&gt;&lt;special-functions&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="9599" PostTypeId="2" ParentId="9592" CreationDate="2010-11-09T19:48:09.663" Score="6" ViewCount="" Body="&lt;p&gt;There is a standard problem (I often assign it in linear algebra, and I was assigned the problem in representation theory); seems like it might be related.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose you have $n$ people sitting in a circular table, each of them with two neighbors. They each have a bowl of porridge in front of them, and a spoon in each hand. Unhappy with their individual portions, they all simultaneously grab half the porridge from their left-hand neighbor, half the porridge from the right-hand neighbor, and they put it in their bowl (which has been emptied by his neighbors doing the same). What happens to the distribution of porridge over time?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Edit: More explicit, applied to the cards.&lt;/em&gt; So, let's consider the case of three eaters/shufflers. I'm assuming for convenience now that the cards are, as Moron suggests, &quot;liquid&quot;; we'll get back to the discrete case presently. Let $x_i(n)$ be the number of cards that the $i$th person has after $n$ steps, with $\mathbf{x}(0) = (a_0,b_0,c_0)$ being the initial distribution of cards (so $a_0,b_0,c_0\geq 0$). Then the procedure can be represented as&#xA;$$\begin{array}{rcrcrcl}&#xA;x_1(n+1) &amp;amp;=&amp;amp;  &amp;amp; &amp;amp;\frac{1}{2}x_2(n) &amp;amp; + &amp;amp; \frac{1}{2}x_3(n)\\&#xA;x_2(n+1) &amp;amp; = &amp;amp; \frac{1}{2}x_1(n) &amp;amp; &amp;amp; &amp;amp;+&amp;amp; \frac{1}{2}x_3(n)\\&#xA;x_3(n+1) &amp;amp; = &amp;amp; \frac{1}{2}x_1(n) &amp;amp; + &amp;amp; \frac{1}{2}x_2(n)&#xA;\end{array}$$&#xA;So we can view this as the linear system $\mathbf{x}(n+1) = A\mathbf{x}(n)$, where&#xA;$$A = \left(\begin{array}{lll}&#xA;0 &amp;amp; 0.5 &amp;amp; 0.5\\&#xA;0.5 &amp;amp; 0 &amp;amp; 0.5\\&#xA;0.5 &amp;amp; 0.5 &amp;amp; 0&#xA;\end{array}\right).$$&#xA;The characteristic polynomial of $A$ is $-t(t-0.5)^2$. Since the rows all add up to $1$, then $(1,1,1)^T$ is an eigenvector of $\lambda=1$ (that is, if everyone starts with $1$, then after applying the procedure they all end up again with $1$: $(1,1,1)^T$ is mapped to one times itself). A basis for the eigenspace corresponding to $\lambda=-\frac{1}{2}$ is $(1,-1,0)^T$ and $(1,0,-1)^T$. If, for instance, player one began with one card, and player two with one &quot;anti-card&quot; (a card made of antimatter, say), then after the first step, player three got half a card from 1 and half an anti-card from 2, which vanished in a puff of smoke; player 2 now has half a card he got from 1, and player 1 has half an anti-card he got from 2. So $(1,-1,0)^T$ is mapped to $(-\frac{1}{2},\frac{1}{2},0)$. Similarly with $(1,0,-1)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, because $(1,1,1), (1,-1,0), (1,0,-1)$ is a basis for $\mathbb{R}^3$, our original distribution $\mathbf{x}(0)$ can be written uniquely as&#xA;\[ \mathbf{x}(0) = a(1,1,1) + b(1,-1,0) + c(1,0,-1),\]&#xA;for some $a$, $b$, and $c$. Note that the number of cards among all three players is $3a$. If we apply the procedure $n$ times to get $\mathbf{x}(n)$, we will get:&#xA;\begin{align*}&#xA;\mathbf{x}(n) &amp;amp;= A^n\mathbf{x}(0)\\&#xA;&amp;amp;= A^n(a(1,1,1)^T + b(1,-1,0)^T + c(1,0,-1)^T)\\&#xA;&amp;amp;= (a,a,a)^T + (-0.5)^n(b,-b,0)^T + (-0.5)^n(c,0,-c)^T\\&#xA;&amp;amp;= (a + (-0.5)^n(b+c), a-(0.5)^nb, a-(0.5)^nc)^T.&#xA;\end{align*}&#xA;Now, in reality we are interested in the floor of these quantities (since we are really dealing with actual cards, no negative and fractional cards). As soon as $(0.5)^nb$ and $(0.5)^nc$ are both smaller than $\frac{1}{4}$, the given distribution will be essentially the even distribution $(a,a,a)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For an explicit example: suppose that we start with Player 1 having $30$ cards, Player $2$ with $16$ cards, and player $3$ with $22$ cards (I picked the numbers at random). First, we write $(30,16,22)$ as a linear combination of $(1,1,1)$, $(1,-1,0)$, and $(1,0,-1)$. Solving the corresponding system we get:&#xA;$$(30,16,22) = \frac{68}{3}(1,1,1) + \frac{20}{3}(1,-1,0) + \frac{2}{3}(1,0,-1).$$&#xA;So after $n$ applications of the procedure, we will have&#xA;$$&#xA;\mathbf{x}(n) = \frac{68}{3}(1,1,1) + \frac{(-1)^n20}{3(2^n)}(1,-1,0) + \frac{(-1)^n2}{3(2^n)}(1,0,-1).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So each player will have exactly one third of the cards, modified by a bit. &#xA;If $n=2$, then the coefficient of $(1,-1,0)$ is $1.67$ and the coefficient of $(1,0,-1)$ is $0.167$; so you would expect the first player to have the nearest integer to  $\frac{68}{3}+1.67$ cards, namely about $24$; the second player should have the nearest integer to $\frac{68}{3} -1.67$ cards, so about $21$ cards; and the third player the nearest integer to $\frac{68}{3}-0.167$ cards; this is almost $22.5$, so he'll have $23$ cards (to complete the tally). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $n=3$, you get about $22$ cards for the first player, $23$ for the second, and $23$ for the third). If $n=4$, the first player should have approximately $23$ cards; the second player should have approximately $22$ cards; and the third player should have approximately $23$ cards. After this point, you'll just be shuffling around who is one card short.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A similar phenomenon occurs with any odd number of people. This may not be the fastest procedure for dividing them among an odd number of players, though. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you try to do this with $n=4$ people, then the matrix you get is:&#xA;$$ A = \left(\begin{array}{llll}&#xA;0 &amp;amp; 0.5 &amp;amp; 0 &amp;amp; 0.5\\&#xA;0.5 &amp;amp; 0 &amp;amp; 0.5 &amp;amp; 0\\&#xA;0 &amp;amp; 0.5 &amp;amp; 0 &amp;amp; 0.5\\&#xA;0.5 &amp;amp; 0 &amp;amp; 0.5 &amp;amp; 0&#xA;\end{array}\right).$$&#xA;This time, the characteristic polynomial is $t^2(t-1)(t+1)$, so you have one eigenvalue $\lambda=1$ (with corresponding eigenvector $(1,1,1,1)$, which is associated to &quot;the cards are evenly distributed&quot;); you get one eigenvalue $\lambda=-1$, with corresponding eigenvector $(1,-1,1,-1)$; and you get two eigenvectors mapping to $0$; for instance $(0,1,0,-1)$ and $(1,0,-1,0)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you express your original distribution as a linear combination of these,&#xA;\[ (x_1(0),x_2(0),x_3(0),x_4(0)) = \alpha(1,1,1,1) + \beta(1,-1,1,-1) + \gamma(0,1,0,-1) + \delta(1,0,-1,0),\]&#xA;the last two components don't matter after the first go around, so you end up with &#xA;\[ \mathbf{x}(n) = \bigl( \alpha + (-1)^n\beta, \alpha - (-1)^n\beta, \alpha+(-1)^n\beta, \alpha-(-1)^n\beta\bigr).\]&#xA;So depending you are going to have a certain portion of the cards evenly distributed (corresponding to $\alpha$), and a certain portion of the cards (corresponding to $\beta$) swapping between even- and odd-numbered players at each step.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Edit: Another way of looking at the $n=4$ case:&lt;/em&gt; A better way of seeing what is happening in this case is to replace the vector $(1,-1,1,-1)$ with the vector $(2,0,2,0)$ (the sum of the eigenvector and $(1,1,1,1)$); we still have a basis, so we can write any initial distribution as&#xA;$$(x_1(0),x_2(0),x_3(0),x_4(0)) = \alpha(1,1,1,1) + \beta(2,0,2,0) + \gamma(0,1,0,-1)$ + \delta(1,0,-1,0).$$&#xA;But now, when you apply $A$, the vector $(2,0,2,0)$ transforms into the vector $(0,2,0,2)$, and this vector is in turn transformed back into $(2,0,2,0)$. The total number of cards is $4(\alpha+\beta)$. Of these, $4\alpha$ of the cards will eventually end up evenly distributed among the players, but the remaining $4\beta$ cards will end up evenly distributed between two non-adjacent players, and then swapped over to the other two players, then swapped back. If you imagine you are in a bridge table, first the North-South partnership has the extra $4\beta$ cards, $2\beta$ cards each, then they go to East-West, then back to North-South, then back ot East-West, etc. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A similar phenomenon occurs with any even number of players.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added:&lt;/strong&gt; &lt;em&gt;Back to the discrete case.&lt;/em&gt; Okay, above we were dealing with the &quot;liquid cards&quot;/&quot;porridge&quot; case, where the cards are divisible any way we want. What happens if we are playing actual cards that we don't want to split up? Notice that each player is only required to divide his cards in half; so the only problem arises if a player has an odd number of cards. In that case, the player can just give the extra card to whichever of his neighbors he likes the most, or randomly; the final distribution in this case is very close to the ideal &quot;continuous&quot; distribution: each of his neighbors was supposed to get $x$-and-a-half cards from him, but got either $x$ or $x+1$; so the most that any one person can be off from the &quot;ideal&quot; result of applying the procedure is $1$ card (half a card from one neighbor, half a card from the other). This vector will be fairly close to the &quot;ideal&quot; continuous vector, and so the result of applying the procedure to this distribution will also result in a distribution which is fairly close to what the ideal distribution is. (Linear maps are continuous, and in this case the eigenvalues are all of absolute value less than or equal to $1$, so the approximation will not get worse as time goes on; it will either remain the same, or get better). You can see this in the example I worked out for $n=3$. If you start with $(30,16,22)$, then after the first step you get $(19,26,23)$; at the next step, the &quot;predicted&quot; distribution is $(24,21,23)$. The actual distributions you get are either $(24,22,22)$, $(25,21,22)$, $(24,21,23)$, or $(25,20,23)$, depending on how the extra cards get distributed. The sup-distance to the predicted distribution is 1 in all cases. Do it a third time, and depending on how things break you will end up in either $(22,24,22)$, $(21,24,23)$, $(22,23,23)$, $(21,23,24)$, $(23,23,22)$, or $(21,25,22)$. The &quot;worst&quot; distribution (the last one) is still better than the worst distribution in the previous step. And so on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;In summary:&lt;/strong&gt; the &quot;give half your cards to each of your neighbors and get half of each of their cards&quot; will yield an even distribution after a fairly small-ish number of steps if there is an odd number of players, but will not necessarily yield and even distribution if there is an even number of players. In the latter case, part of the deck will get evenly distributed, while the other part ends up swapping between adjacent &quot;teams&quot;.  One advantage of this, at least for an odd number of players, is that it is a single procedure that is repeated a number of times (much like riffle-shuffling a deck) in order to achieve uniformity, rather than a more complex algorithm that involves different actions at different steps.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-09T22:34:06.730" LastActivityDate="2010-11-09T22:34:06.730" CommentCount="11" />
  <row Id="9600" PostTypeId="2" ParentId="9598" CreationDate="2010-11-09T19:50:50.147" Score="6" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Stirling%27s_approximation&quot; rel=&quot;nofollow&quot;&gt;Stirling's approximation&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="723" LastActivityDate="2010-11-09T19:50:50.147" />
  <row Id="9601" PostTypeId="2" ParentId="9595" CreationDate="2010-11-09T19:58:39.573" Score="4" ViewCount="" Body="&lt;p&gt;I am assuming that a sigmoid function is bounded, and its graph has exactly one inflection point (among other properties).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For (1), then, the answer is no.  If $f(x)=\frac{e^{3x}}{1+e^{3x}}$ and $g(x)=\frac{e^{2x-5}}{1+e^{2x-5}}$, then $h(x)=f(x)+g(x)$ can be seen to have two points of inflection, so h(x) is not sigmoid. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For (2), your function is unbounded on $0\le x \le x_{\mbox{max}}$, but sigmoid functions are bounded.  So these functions have decidedly different behavior.  Though, it does depend somewhat on what you mean by &quot;like&quot;.&lt;/p&gt;&#xA;" OwnerUserId="2937" LastActivityDate="2010-11-09T19:58:39.573" CommentCount="5" />
  <row Id="9602" PostTypeId="1" AcceptedAnswerId="9603" CreationDate="2010-11-09T20:10:48.530" Score="2" ViewCount="99" Body="&lt;p&gt;What are the open sets of a Stone space of a Boolean algebra B?&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Stone_space&quot; rel=&quot;nofollow&quot;&gt;According to Wikipedia&lt;/a&gt;, they are the ultrafilters on B that contain an element of B.  However, this cannot be the case since an ultrafilter is a set of subsets of B and contains no elements of B.  Using the alternate definition of the Stone space as the space of homomorphisms from B to {0,1} doesn't offer much help, either.&lt;/p&gt;&#xA;" LastActivityDate="2010-11-09T20:28:45.933" Title="Topology on Stone Spaces of Boolean Algebra" Tags="&lt;topology&gt;&lt;logic&gt;&lt;boolean-algebra&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="9603" PostTypeId="2" ParentId="9602" CreationDate="2010-11-09T20:18:14.460" Score="5" ViewCount="" Body="&lt;p&gt;I think you are confused by the two meanings of ultrafilter. In the context of a Boolean algebra $B$, an ultrafilter is just a maximal proper filter, that is, a subset of $B$ that is closed under $\wedge$, upward closed, and does not contain $0$. The Stone space $M(B)$ consists of all such maximal filters, and its usual topology is generated by basic open sets of the form&#xA;$$&#xA;N_p = \{ f \in M(B) : p \in f \}&#xA;$$&#xA;with one such basic open set $N_p$ for each $p \in B$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a second, incompatible meaning of ultrafilter: an ultrafilter on an arbitrary set $S$ is a maximal filter in the above sense on the lattice of subsets of $S$. This is what people mean when they say &quot;an ultrafilter on $\omega$&quot;, for example.  When $S$ is already being treated as a lattice, the former sense is usually what is meant, but you can only tell by context. &lt;/p&gt;&#xA;" OwnerUserId="630" LastEditorUserId="630" LastEditorDisplayName="" LastEditDate="2010-11-09T20:28:45.933" LastActivityDate="2010-11-09T20:28:45.933" />
  <row Id="9605" PostTypeId="1" AcceptedAnswerId="9609" CreationDate="2010-11-09T20:48:42.753" Score="1" ViewCount="108" Body="&lt;p&gt;As an exercise in Python programming I've written a script that finds amicable pairs, and I've had fun doing so. But I'm wondering if amicable pairs have a use, in mathematics or elsewhere? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a reminder:&lt;br&gt;&#xA;Amicable numbers are two different numbers so related that the sum of the proper divisors of each is equal to the other number. (A proper divisor of a number is a positive integer divisor other than the number itself. For example, the proper divisors of 6 are 1, 2, and 3.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, the smallest pair of amicable numbers is (220, 284); for the proper divisors of 220 are 1, 2, 4, 5, 10, 11, 20, 22, 44, 55 and 110, of which the sum is 284; and the proper divisors of 284 are 1, 2, 4, 71, and 142, of which the sum is 220.&#xA;(from the Wikipedia &lt;a href=&quot;http://en.wikipedia.org/wiki/Amicable_number&quot; rel=&quot;nofollow&quot;&gt;article&lt;/a&gt;)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first few amicable pairs, according to &lt;em&gt;Math World&lt;/em&gt;, are:&#xA;(220, 284), (1184, 1210), (2620, 2924) (5020, 5564), (6232, 6368), (10744, 10856), (12285, 14595), (17296, 18416), (63020, 76084)&lt;/p&gt;&#xA;" OwnerUserId="1964" LastActivityDate="2010-11-09T20:58:56.720" Title="Amicable pairs: any use for them yet?" Tags="&lt;number-theory&gt;" AnswerCount="1" />
  <row Id="9606" PostTypeId="1" CreationDate="2010-11-09T20:54:06.547" Score="1" ViewCount="125" Body="&lt;p&gt;I have a set of functions that map square matrices with $n$ rows and columns to square matrices with $k &amp;lt; n$ rows and columns.  Is there a name for this property?  I know that 'projection' would be the wrong word.  Also I don't mean that the function &lt;em&gt;only&lt;/em&gt; changes the size, it might also do other things.&lt;/p&gt;&#xA;" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-09T21:59:35.210" LastActivityDate="2010-11-11T20:22:07.407" Title="What is the name for a function of a matrix that changes the matrix size?" Tags="&lt;terminology&gt;&lt;matrices&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="9608" PostTypeId="1" AcceptedAnswerId="9643" CreationDate="2010-11-09T20:55:28.983" Score="1" ViewCount="265" Body="&lt;p&gt;I have a stochastic variable x with this property: if it's measured at &#xA;t1 and again at t2, then x(t2)-x(t1) has a normal distribution with &#xA;mean 0 and standard deviation Sqrt[t2-t1]. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to find the distribution of the maximum value this variable &#xA;reaches between t1 and t2, or confirm my intuition that this is not &#xA;well-defined. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My approach: break [t1,t2] into multiple intervals and integrate the &#xA;following (PDF[NormalDistribution[x,y]] = normal probability &#xA;distribution with mu of x and standard deviation of y): &lt;/p&gt;&#xA;&#xA;&lt;p&gt;PDF[NormalDistribution[0,Sqrt[1/4]]][x0]* &#xA;PDF[NormalDistribution[x0,Sqrt[1/4]]][x1]* &#xA;PDF[NormalDistribution[x1,Sqrt[1/4]]][x2]* &#xA;PDF[NormalDistribution[x2,Sqrt[1/4]]][x3] &lt;/p&gt;&#xA;&#xA;&lt;p&gt;where each xi is integrated from -Infinity to m. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This specific example computes the probability that the maximum on &#xA;[0,1] is less than m by breaking [0,1] into 4 parts. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Breaking [0,1] into more parts should yield more accurate results, &#xA;although I slightly suspect that the limit diverges. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mathematica slows to a crawl even breaking [0,1] into 5 or more parts. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've tried replacing the normal distribution with others (uniform, &#xA;DeltaDirac, C/(1+x^2), etc), with no better luck. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Googling yields many results (this appears to be a &quot;Wiener Process&quot;), &#xA;but I can't find the actual distribution of the maximum anywhere (nor &#xA;does it say anywhere that such a maximum doesn't exist). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ultimate goal is to price box options:&#xA;&lt;a href=&quot;http://money.stackexchange.com/questions/4312/calculating-fair-value-of-an-oanda-com-box-option&quot;&gt;http://money.stackexchange.com/questions/4312/calculating-fair-value-of-an-oanda-com-box-option&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="2469" LastActivityDate="2010-11-10T18:00:28.353" Title="Determining distribution of maximum of dependent normal variables" Tags="&lt;integral&gt;&lt;normal-distribution&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="9609" PostTypeId="2" ParentId="9605" CreationDate="2010-11-09T20:58:56.720" Score="6" ViewCount="" Body="&lt;p&gt;Apparently, John Conway says no.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;For example, John Conway says they are not very useful mathematically:&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The only application or use for these numbers is the original one -&#xA;     you insert a pair of amicable numbers into a pair of amulets, of &#xA;     which you wear one yourself and give the other to your beloved!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;From &lt;a href=&quot;http://mathforum.org/library/drmath/view/52560.html&quot; rel=&quot;nofollow&quot;&gt;Dr. Math&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2937" LastActivityDate="2010-11-09T20:58:56.720" />
  <row Id="9610" PostTypeId="2" ParentId="9598" CreationDate="2010-11-09T20:59:23.833" Score="4" ViewCount="" Body="&lt;p&gt;You can also use &lt;a href=&quot;http://en.wikipedia.org/wiki/Wallis_product&quot; rel=&quot;nofollow&quot;&gt;Wallis product&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/Qxr9R.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which can be shown by considering $\displaystyle \int_{0}^{\pi/2} \sin^n x$ (as Wallis originally did) or using the product expansion for $\sin x$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle \sin x = x\prod_{n=1}^{\infty} (1- \frac{x^2}{\pi^2 n^2})$ and setting $\displaystyle x=\pi/2$&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-11-09T20:59:23.833" />
  <row Id="9611" PostTypeId="2" ParentId="9596" CreationDate="2010-11-09T21:13:02.193" Score="0" ViewCount="" Body="&lt;p&gt;For the second question, we want to show that $\text{Med}(aX+bY)$ might not equal $a \text{Med}(X)+ b \text{Med}(Y)$. Suppose $f(x)$ is the density function of $X$ and $g(y)$ is the density function of $Y$. Let $m_1$ be the median of $X$ and $m_2$ be the median of $Y$. Then $m_1$ is the value for which $\int_{0}^{m_1} f(x) \ dx = 0.5$. Likewise, $m_2$ is the value for which $\int_{0}^{m_2} g(y) \ dy = 0.5$. I guess you could use this fact to show that in general, $\text{Med}(aX+bY) \neq a \text{Med}(X)+ b \text{Med}(Y)$. &lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2010-11-09T21:13:02.193" />
  <row Id="9612" PostTypeId="1" CreationDate="2010-11-09T21:18:38.870" Score="2" ViewCount="157" Body="&lt;p&gt;How to solve the recursion: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$p[n,m] = p[n-1,m-1] + p[n+1,m-1] + p[n,m-1]$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ideally in general, but if you need base cases: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$p[n,0] = 0 \text{ (for } n \neq 0),$$ &#xA;$$p[0,0] = 1$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've asked a similar question previously and believe $n^{m-2}$ is &#xA;somehow involved. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would turning this into a DFQ help? &lt;/p&gt;&#xA;" OwnerUserId="2469" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2011-06-03T20:05:22.280" LastActivityDate="2011-06-03T20:05:22.280" Title="Solve recursion $p[n,m] = p[n-1,m-1] + p[n+1,m-1] + p[n,m-1]$" Tags="&lt;recurrence-relations&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="0" />
  <row Id="9613" PostTypeId="2" ParentId="9606" CreationDate="2010-11-09T21:20:07.137" Score="2" ViewCount="" Body="&lt;p&gt;Well, &lt;em&gt;projection&lt;/em&gt; could be the right word, depending on what you want to do with the matrices. Note that matrices with $n$ rows and $m$ columns form a vector space of dimension $n \times m$. If you now restrict to matrices of $k &amp;lt; n$ rows, you can treat them as a $k \times m$-dimensional subspace of the original vector space, having entries in rows $i &gt; k$ set to zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also note that the dual operation of reducing the number columns might be called a &lt;em&gt;restriction&lt;/em&gt;. This comes from the fact that you can treat $n \times m$ dimensional matrix as a representation of an operator mapping $m$-dimensional vector space to a $n$-dimensional vector space and reducing number of columns restricts this operator to a subspace of the original vector space.&lt;/p&gt;&#xA;" OwnerUserId="3223" LastActivityDate="2010-11-09T21:20:07.137" />
  <row Id="9614" PostTypeId="2" ParentId="9612" CreationDate="2010-11-09T21:31:00.293" Score="2" ViewCount="" Body="&lt;p&gt;See OEIS &lt;a href=&quot;http://www.research.att.com/~njas/sequences/A027907&quot; rel=&quot;nofollow&quot;&gt;A027907&lt;/a&gt; and references on trinomial coefficients.  If you just start making the triangle in a spreadsheet and type the numbers into OEIS you will find it.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-11-09T21:31:00.293" />
  <row Id="9615" PostTypeId="2" ParentId="9548" CreationDate="2010-11-09T21:33:36.843" Score="2" ViewCount="" Body="&lt;p&gt;Following Robin Chapman's answer, $$\sum_{a,b \geq 0} f(a,b)x^a y^b = \frac{1}{1-x-y-xy} = \sum_{n \geq 0} (x+y+xy)^n = \sum_{n \geq 0} \sum_{i+j+k=n} \frac{n!}{i!j!k!} x^i y^j (xy)^k$$&#xA;so $$f(a,b)=\sum_{k=0}^{\min (a,b)} \frac{(a+b-k)!}{(a-k)!(b-k)!k!}$$&lt;/p&gt;&#xA;" OwnerUserId="2660" LastEditorUserId="6179" LastEditorDisplayName="" LastEditDate="2011-06-04T11:25:25.263" LastActivityDate="2011-06-04T11:25:25.263" />
  <row Id="9616" PostTypeId="1" AcceptedAnswerId="9618" CreationDate="2010-11-09T21:42:44.793" Score="0" ViewCount="173" Body="&lt;p&gt;Take a Multiplicative group G, defined by a prime p.  Now take an element t from it.  Its inverse is t&lt;sup&gt;-1&lt;/sup&gt;.  To calculate the inverse you can use the extended Euclidian algorithm or this trick, &lt;a href=&quot;http://en.wikipedia.org/wiki/Modular_multiplicative_inverse#Direct_Modular_Exponentiation&quot; rel=&quot;nofollow&quot;&gt;which I found on wikipedia&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;t&lt;sup&gt;-1&lt;/sup&gt; = t&lt;sup&gt;p-2&lt;/sup&gt; mod p&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now this makes some sense to me, but the following, which I found in some code, does not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take t, and raise it to u (also in G):  t&lt;sup&gt;u&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(t&lt;sup&gt;u&lt;/sup&gt;)&lt;sup&gt;-1&lt;/sup&gt; = t&lt;sup&gt;p-1-u&lt;/sup&gt; mod p&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is this trick called, and why does it work?&lt;/p&gt;&#xA;" OwnerUserId="3179" LastActivityDate="2010-11-10T00:15:09.273" Title="Why does this trick to find the inverse of an element in a Multiplicative Cyclic Group work?" Tags="&lt;group-theory&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="9617" PostTypeId="2" ParentId="9596" CreationDate="2010-11-09T21:43:40.207" Score="4" ViewCount="" Body="&lt;p&gt;First question: By definition, on a probability space ($\Omega,\mathcal{F},P$), the expected value of a random variable $X:\Omega\to \mathbb{R}$ is defined as&#xA;$$E(X)=\int_\Omega X(\omega) dP(\omega).$$ Note that it is only well-defined if the integral converges absolutely, i.e. $$\int_\Omega |X(\omega)| dP(\omega)&amp;lt;\infty$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(The integrals above are Lebesgue integrals. If $X$ is discrete, $P$ is the point measure and this integral turns into a sum.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, if the expected values of both $X$ and $Y$ exist, then $E(X+Y)$ exists (triangle inequality) and, for $a,b\in \mathbb{R}$, we can use the linearity of the Lebesgue integral to conclude&#xA;$$E(aX+bY)=\int_\Omega aX +bYdP=a\int_\Omega X dP+b\int_\Omega YdP=aE(X)+bE(Y).$$&lt;/p&gt;&#xA;" OwnerUserId="3244" LastActivityDate="2010-11-09T21:43:40.207" CommentCount="2" />
  <row Id="9618" PostTypeId="2" ParentId="9616" CreationDate="2010-11-09T21:46:10.070" Score="3" ViewCount="" Body="&lt;p&gt;The &quot;trick&quot; is that $\displaystyle t^{p-1} = 1 \mod p$ always.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So $\displaystyle t^{p-1-u} \cdot t^u = 1 \mod p$ and &lt;/p&gt;&#xA;&#xA;&lt;p&gt;so&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle t^u$ and $\displaystyle t^{p-1-u}$ are inverses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also for a general group $G$ and $a, b \in G$, $a^b$ does not make much sense as a power of $a$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As Arturo points out, it in fact is used to denote the element $b^{-1}ab$.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-11-09T21:56:22.383" LastActivityDate="2010-11-09T21:56:22.383" CommentCount="5" />
  <row Id="9619" PostTypeId="2" ParentId="9616" CreationDate="2010-11-09T21:52:43.223" Score="1" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\rm\displaystyle\quad\quad  A^N = 1\ \ \Rightarrow\ \ A^{-K}\ =\ \frac{1}{A^K}\ =\ \frac{A^N}{A^K}\ =\ A^{N-K} $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's not really a &quot;trick&quot; nor does it have any standard name. It's simply a consequence of the cyclic structure, viz $\rm\ -K \ \equiv\ N-K\ \:(mod\ N)\:$. This is already quite familiar to everyone in the form of clock arithmetic, e.g. 1:50  = 10 before 2,$\ $ i.e. $\rm\: -10\ \equiv\ 50\ \ (mod\ 60)\:.\:$  Note that, as a consequence, there can be no concept of sign (positive/negative) in cyclic groups, i.e. they cannot be ordered. &lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-11-10T00:15:09.273" LastActivityDate="2010-11-10T00:15:09.273" />
  <row Id="9620" PostTypeId="1" CreationDate="2010-11-09T22:07:10.980" Score="2" ViewCount="186" Body="&lt;p&gt;I get confused in the following problem. Actually I don't know how to derive the derivative of the Jacobian. Could anybody help me?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given a smooth vector field $\vec{b}$ on $R^{n}$, let $\vec{X}(s)=\vec{X}(s,x,t)$solves the ODE:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\dot{\vec{X}}=\vec{b}(\vec{X})(s \in R)$, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\vec{X}(t)=\vec{X}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Define the Jacobian $J(s,x,t)=det D_{x}\vec{X}(s,x,t)$,  derive Euler's formula:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$J_{s}=(div\vec{b}(\vec{X}))J$&lt;/p&gt;&#xA;" OwnerUserId="2855" LastActivityDate="2011-01-08T22:44:40.600" Title="A problem in Evan's PDE 2nd edition. (Problem 6 on Pg 162)" Tags="&lt;differential-equations&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="9621" PostTypeId="2" ParentId="9596" CreationDate="2010-11-09T22:14:31.650" Score="4" ViewCount="" Body="&lt;p&gt;EXTENDED/REVISED ANSWER&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some general points concerning the second question. By definition, $m$ is a median of $X$ if ${\rm P}(X \ge m) \geq 1/2$ and ${\rm P}(X \le m) \geq 1/2$. While a median is uniquely determined for any common example of a continuous random variable, it is not uniquely determined in general. For example, any number $m \in [-1,1]$ is a median for random variable $X$ with ${\rm P}(X=1) = {\rm P}(X=-1) = 1/2$. Hence my previous answer to this question (see below), where I assumed that $m(X)=0$ since $X$ is symmetric, should be revised. This is done simply as follows. We define $X$ and $Y$ exactly as before, and introduce another random variable $\tilde X$ defined to be equal to $X$ with probability $1-1/n$ and to $0$ with probability $1/n$. It is immediately checked that the symmetric random variables $\tilde X$ and $Y$ have a unique median, equal to $0$. Thus $m(\tilde X) + m(Y) = 0$, as required. On the other hand, one easily verifies that ${\rm P}(\tilde X + Y = 1) \to 3/4$ as $n \to \infty$ (cf. my previous answer), which implies that $\tilde X + Y$ has a unique median, equal to $1$. So, $m(\tilde X + Y) \neq m(\tilde X) + m(Y)$, as required. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In view of this example, we now give a counterexample for the case where $X$ and $Y$ are independent. Let $X$ and $Y$ be i.i.d. random variables with common probability mass function given by $p(2)= p(-1) = \frac{1}{2}(1 - \frac{1}{n})$, $p(0) = \frac{1}{n}$. Then, $X$ and $Y$ have a unique median, equal to $0$. On the other hand, one verifies that both ${\rm P}(X+Y \geq 1)$ and  ${\rm P}(X+Y \leq 1)$ tend to $3/4$ as $n \to \infty$; hence, $X + Y$ has a unique median, equal to $1$. So, $m(X + Y) \neq m(X) + m(Y)$, as required. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In view of the preceding examples, we finally consider the case where $X$ and $Y$ are both symmetric and independent. Assuming both $X$ and $Y$ have a unique median, it must obviously be equal to $0$. For any fixed numbers $a$ and $b$, $aX + bY$ is also symmetric. Moreover, $aX + bY$ has a unique median, equal to $0$. This can be carried out straightforwardly, upon observing that ${\rm P}(X \in (-\varepsilon,\varepsilon), Y \in (-\varepsilon,\varepsilon)) &gt; 0$ for any $\varepsilon &gt; 0$. Hence, $m(aX+bY)=am(X)+bm(Y)=0$. From this, it easy to establish the following generalization. Suppose that $X$ and $Y$ are independent and symmetric around arbitrary points, say $m_1$ and $m_2$, respectively. Assume that both $X$ and $Y$ have a unique median (these medians are necessarily given by $m(X)=m_1$ and $m(Y)=m_2$). Then, for any fixed numbers $a$ and $b$, $m(aX+bY)$ has a unique median, equal to $am(X)+bm(Y)=am_1+bm_2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PREVIOUS ANSWER&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the second question, let us show that even if $X$ and $Y$ are symmetric random variables, then $m(X+Y)$ might be different from $m(X)+m(Y)$ (where $m$ denotes median). Suppose that ${\rm P}(X=1) = {\rm P}(X=-1) = 1/2$; hence $X$ is symmetric. Define $Y$ as follows: if $X=1$ then $Y=0$, whereas if $X=-1$ then $Y=2$ or $-2$ with probability $1/2$ each. Then, ${\rm P}(Y=0)=1/2$, ${\rm P}(Y=2)=1/4$, and ${\rm P}(Y=-2)=1/4$. Hence $Y$ is symmetric and, in turn, $m(X)+m(Y)=0+0=0$. However, ${\rm P}(X+Y=1) = 3/4$ (and ${\rm P}(X+Y=-3) = 1/4)$. In particular, $m(X+Y)=1$ (since ${\rm P}(X+Y=1) \geq 1/2$).&lt;/p&gt;&#xA;" OwnerUserId="2810" LastEditorUserId="2810" LastEditorDisplayName="" LastEditDate="2010-11-11T21:58:06.740" LastActivityDate="2010-11-11T21:58:06.740" CommentCount="3" />
  <row Id="9622" PostTypeId="1" AcceptedAnswerId="9729" CreationDate="2010-11-09T22:18:04.183" Score="1" ViewCount="116" Body="&lt;p&gt;Reading a book on fractional calculus reminded me that I'd like to know more on the following method/idea.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given an integral: $\displaystyle \int_{-\pi}^\pi{\frac{\displaystyle\sum_{j=a}^b{e^{i\cdot c_j \cdot t}}}{\displaystyle\sum_{k=a}^b{e^{i \cdot d_k \cdot t}}} dt}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can use recurrences and generating functions to find all of the derivatives.  Using all of the derivatives (evaluated at a point), I can essentially create another function (for example, the function at point $n$ is the $n$th derivative).  The function brings up a curiousity:  how accurate could it be at predicting the integral itself?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;EDIT&lt;/em&gt;  An additional thought:  If this method fails, perhaps the derivatives could still be used to aid in a very good spline extrapolation using only a few points.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I'm wondering what problems arise when trying to extrapolate/interpolate an integral of the form above using derivatives.  I'd like to know any studies of this problem if they exist, and what I can expect.  I realize I'm being a bit vague, but I'm not exactly sure what the good questions are regarding this issue.  I would like to explore this possibility as much as possible, and hope that answers arrive with hope and how to evaluate the integral this way.&lt;/p&gt;&#xA;" OwnerUserId="2626" LastEditorUserId="2626" LastEditorDisplayName="" LastEditDate="2010-11-09T23:40:34.020" LastActivityDate="2010-11-10T16:21:59.640" Title="Using derivatives to evaluate an integral of exponentials" Tags="&lt;calculus&gt;&lt;intuition&gt;&lt;numerical-methods&gt;&lt;approximation&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="9623" PostTypeId="1" AcceptedAnswerId="9628" CreationDate="2010-11-09T22:20:03.730" Score="6" ViewCount="372" Body="&lt;p&gt;I know the general Black-Scholes formula for Option pricing theory (for calls and puts), however I want to know the other solutions to the Black-Scholes PDE and its various boundary conditions. Can someone start from the B-S PDE and derive its various solutions based on different boundary conditions? Even if you could provide some links/sources where it is done, I'll appreciate that. The point is that I want to know various other solutions and their boundary conditions which are derived from Black-Scholes PDE. Thank you.&lt;/p&gt;&#xA;" OwnerUserId="754" LastEditorUserId="8246" LastEditorDisplayName="" LastEditDate="2011-05-10T22:42:04.393" LastActivityDate="2011-05-11T00:17:25.553" Title="Black Scholes PDE and its many solutions" Tags="&lt;reference-request&gt;&lt;differential-equations&gt;&lt;finance&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" />
  <row Id="9624" PostTypeId="1" CreationDate="2010-11-09T22:28:28.493" Score="0" ViewCount="61" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Possible Duplicate:&lt;/strong&gt;&lt;br&gt;&#xA;  &lt;a href=&quot;http://math.stackexchange.com/questions/9606/what-is-the-name-for-a-function-of-a-matrix-that-changes-the-matrix-size&quot;&gt;What is the name for a function of a matrix that changes the matrix size?&lt;/a&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;p&gt;What are some examples of named functions that change the size of a matrix?  For example, I guess a 'projection' does this under certain circumstances?  What are other examples?&lt;/p&gt;&#xA;" LastEditorUserId="1016" LastEditorDisplayName="" LastEditDate="2010-11-09T22:51:00.357" LastActivityDate="2010-11-09T22:51:00.357" ClosedDate="2010-11-11T11:08:05.473" Title="names of functions which change the size of a matrix" Tags="&lt;terminology&gt;&lt;big-list&gt;&lt;matrices&gt;" CommentCount="1" />
  <row Id="9625" PostTypeId="2" ParentId="9583" CreationDate="2010-11-09T22:34:47.303" Score="2" ViewCount="" Body="&lt;p&gt;This appears to be a variation of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Stable_marriage_problem&quot; rel=&quot;nofollow&quot;&gt;marriage problem&lt;/a&gt;. You might want to consider work related to this one.&lt;/p&gt;&#xA;" OwnerUserId="3330" LastActivityDate="2010-11-09T22:34:47.303" />
  <row Id="9626" PostTypeId="2" ParentId="9620" CreationDate="2010-11-09T22:41:57.293" Score="1" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The derivative of the determinant can be evaluated using the &lt;a href=&quot;http://en.wikipedia.org/wiki/Determinant#Derivative&quot; rel=&quot;nofollow&quot;&gt;adjugate&lt;/a&gt;. Then it remains to compute &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \frac{d}{ds} D_x X(s,x,t) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for which you need to use the ODE.&lt;/p&gt;&#xA;" OwnerUserId="1543" LastActivityDate="2010-11-09T22:41:57.293" CommentCount="1" />
  <row Id="9627" PostTypeId="1" CreationDate="2010-11-09T22:58:07.070" Score="4" ViewCount="136" Body="&lt;p&gt;Say I have two locally free sheaves $F,G$ on projective variety $X$. I know the cohomology groups $H^i(X,F)$ and $H^i(X,G)$. Is this enough to give me information about $H^i(X,F\otimes G)$? In particular, if $H^i(X,F)=0$, what conditions on $G$ guarantee that also $H^i(X,F\otimes G)=0$?&lt;/p&gt;&#xA;" OwnerUserId="3332" LastActivityDate="2010-11-09T22:58:07.070" Title="Cohomology of a tensor product of sheaves" Tags="&lt;algebraic-geometry&gt;" CommentCount="2" FavoriteCount="2" />
  <row Id="9628" PostTypeId="2" ParentId="9623" CreationDate="2010-11-09T23:05:46.703" Score="9" ViewCount="" Body="&lt;p&gt;The Black-Scholes &lt;em&gt;formula&lt;/em&gt; involving the standard normal distribution is specific to call or put options.  The Black-Scholes &lt;em&gt;formalism&lt;/em&gt;, relating the prices to random walks and PDE, works for pricing a European option with arbitrary payoff.  For any boundary condition (except some artificial ones with  incredibly rapid growth that makes the random walk expectations diverge) the price is the expected value of the option value at the time of maturity, the expectation taken over price paths of the underlying security whose &quot;risk-adjusted&quot; drift and volatility are calculated from the parameters assumed to govern the &quot;physical&quot; price movement.  So one does not need the whole theory of random walks, just the probability density of where the risk-adjusted path will land at the time of maturity, and integrating the European option payoff against this density will give the answer.  The Black-Scholes theory shows that the result satisfies the PDE.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively, Black Scholes is, after a change of variables, equivalent to a &quot;backwards&quot; heat diffusion equation -- one whose time parameter is $(-t)$, or better, $(T_{m} - t)$ where $T_m$ is the maturity time of the option.  So your question is the same as asking for all solutions of the heat equation. I assume that means heat diffusion with arbitrary boundary conditions.  As long as the boundary conditions do not grow extremely fast the expectation of a random walk gives the solution. &lt;/p&gt;&#xA;" OwnerUserId="467" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-11-10T16:45:47.060" LastActivityDate="2010-11-10T16:45:47.060" />
  <row Id="9629" PostTypeId="1" AcceptedAnswerId="9630" CreationDate="2010-11-09T23:11:26.617" Score="1" ViewCount="137" Body="&lt;p&gt;Let M  be an (n-1)-manifold in R^n . Let M(e)  be the set of end-points of normal vectors (in both directions) of length e  and suppose e  is small enough so that M(e) is also an (n-1)-manifold. Show that M(e)  is orientable (even if M  is not)&lt;/p&gt;&#xA;" OwnerUserId="3333" LastActivityDate="2010-11-09T23:19:02.167" Title="Orientation on a Manifold" Tags="&lt;differential-geometry&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="9630" PostTypeId="2" ParentId="9629" CreationDate="2010-11-09T23:19:02.167" Score="2" ViewCount="" Body="&lt;p&gt;Each point $p$ of $M(\epsilon)$ comes from a point $q$ in $M$. Consider the vector $X$ field on $M(\epsilon)$ which on $p$ takes the value $\vec{qp}$. Then $N$ is a non-zero normal field on $M(\epsilon)$ (maybe one needs to consider the projection of $N$ onto the normal line to $M(\epsilon)$ at each point, but that projection is surely non-zero and, moreover, to normalize this projection)&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2010-11-09T23:19:02.167" />
  <row Id="9631" PostTypeId="2" ParentId="9525" CreationDate="2010-11-09T23:32:53.793" Score="0" ViewCount="" Body="&lt;p&gt;Thanks to Qiaochu Yuan's identity I was able to factor into&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ \frac{1}{2}(x+Ny+N^2z)[(x-Ny)^2+(Ny-N^2z)^2+(N^2z-x)^2] $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which shows that if &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ x^3+Dy^3+D^2z^3-3Dxyz=1 $ and $ x,y,z $ are very large positive, then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ x-Dy $ and $ y-Dz $ must be small.&lt;/p&gt;&#xA;" OwnerUserId="2505" LastActivityDate="2010-11-09T23:32:53.793" />
  <row Id="9632" PostTypeId="2" ParentId="9612" CreationDate="2010-11-09T23:41:55.893" Score="1" ViewCount="" Body="&lt;p&gt;Given your boundary conditions, if we open up the recurrence relation then we discover an $m$-step process, where at each step we change $n$ by $0$ or $\pm 1$. For every walk that ends up at $0$, we &quot;score&quot; one point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How many such walks exist? There is an equal number $k$ of $+1$s and $-1$s, and so we get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle \sum_{k=0}^{\lfloor m/2 \rfloor} \binom{m}{k,k} = \sum_{k=0}^{\lfloor m/2 \rfloor} \frac{m!}{k!k!(m-2k)!}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to estimate it, let's calculate the largest coefficient, assuming $m$ is divisible by $3$. It is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle \frac{m!}{(m/3)!} \approx \frac{\sqrt{2\pi m}(m/e)^m}{\sqrt{2\pi (m/3)}^3 (m/3e)^m} = \frac{3^m}{(2\pi/3^{1.5})m}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Roughly, we expect about $O(\sqrt{m})$ of the summands to be of comparable size, so a conjectured upper bound would be $O(3^m/\sqrt{m})$.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-09T23:41:55.893" />
  <row Id="9633" PostTypeId="2" ParentId="9420" CreationDate="2010-11-09T23:47:11.487" Score="0" ViewCount="" Body="&lt;p&gt;In WEP, encryption is done by XORing some stream cipher (RC4), and CRC is used as a signature (can't remember if CRC is applied before or after encryption). Since both operations commute, you can modify a packet by XORing and then fix the signature, without ever knowing the key (or the parts of the packet you didn't modify).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;CRC is often used as a hash function, although nowadays MD5 is more common. CRC is linear, and WEP is a case where that's a problem (even checksum would've been better).&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-09T23:47:11.487" />
  <row Id="9634" PostTypeId="2" ParentId="9596" CreationDate="2010-11-09T23:56:40.117" Score="3" ViewCount="" Body="&lt;p&gt;Intuitive (and somewhat informal) justification for the linearity of expectation. Suppose that random variables $X$ and $Y$ have expectation ${\rm E}(X)$ and ${\rm E}(Y)$, respectively. Suppose also that $(X_n,Y_n)$ is a sequence of i.i.d. random vectors having the (joint) distribution of $(X,Y)$. Then, in particular, $X_n$, $Y_n$, and $aX_n + bY_n$ are sequences of i.i.d. random variables having the distribution of $X$, $Y$, and $aX + bY$, respectively. Thus, from&#xA;$$&#xA;a{\rm E}(X) + b{\rm E}(Y) \stackrel{{\rm a.s.}}{=} a\mathop {\lim }\limits_{n \to \infty } \frac{{\sum\nolimits_{i = 1}^n {X_i } }}{n} + b\mathop {\lim }\limits_{n \to \infty } \frac{{\sum\nolimits_{i = 1}^n {Y_i } }}{n},&#xA;$$&#xA;we conclude that&#xA;$$&#xA;{\rm E}(aX + bY) \stackrel{{\rm a.s.}}{=} \mathop {\lim }\limits_{n \to \infty } \frac{{\sum\nolimits_{i = 1}^n {(aX_i  + bY_i )} }}{n}  \stackrel{{\rm a.s.}}{=} a{\rm E}(X) + b{\rm E}(Y),&#xA;$$&lt;br&gt;&#xA;where $\stackrel{{\rm a.s.}}{=}$ stands for `almost surely equal', and where we have used the strong law of large numbers, namely, if $Z_n$ is a sequence of i.i.d. random variables having the distribution of a random variable $Z$ with finite expectation, then ${\rm E}(Z) \stackrel{{\rm a.s.}}{=} \mathop {\lim }\limits_{n \to \infty } \frac{{\sum\nolimits_{i = 1}^n {Z_i } }}{n}$. &lt;/p&gt;&#xA;" OwnerUserId="2810" LastActivityDate="2010-11-09T23:56:40.117" />
  <row Id="9635" PostTypeId="2" ParentId="9595" CreationDate="2010-11-10T00:58:17.697" Score="0" ViewCount="" Body="&lt;p&gt;It is incorrect to speak of &quot;the sigmoid function&quot;.  The Wikipedia link shows that there are many to choose from.  For (2), note that $\displaystyle f(x) = \log_{a}(bx^{c})=c\log_{a}{x}+\log_{a}{b}=d\ln{x}+e$ by absorbing the constants into the change of base of log.  This is definitely not sigmoidal.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-11-10T00:58:17.697" />
  <row Id="9636" PostTypeId="1" CreationDate="2010-11-10T02:00:53.490" Score="0" ViewCount="151" Body="&lt;p&gt;Let $L^{\infty}$ denote the set of all essentially bounded functions and suppose that&#xA;$f \in L^{\infty}$ and $g \in L^{\infty}$. Then the product $fg$ is in $L^{\infty}$. Now the question is: prove or disprove: $\|fg\|_\infty \leq \|f\|_\infty \|g\|_\infty$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any hints?&lt;/p&gt;&#xA;" OwnerUserId="4219" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-10T16:15:01.740" LastActivityDate="2010-11-10T16:15:01.740" Title="Prove or disprove $L^{\infty}$ inequality" Tags="&lt;measure-theory&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="9637" PostTypeId="2" ParentId="9598" CreationDate="2010-11-10T02:17:06.250" Score="0" ViewCount="" Body="&lt;p&gt;Wolfram|Alpha shows an &lt;a href=&quot;http://www.wolframalpha.com/input/?i=limit+as+n+goes+to+infinity+of+%28+%28power%282%2C2n-1%29%29+%28sqrt%28n%29%29+%28power%28gamma%28n%29%2C2%29+%2F+gamma%282n%29%29+%29&quot; rel=&quot;nofollow&quot;&gt;expansion&lt;/a&gt;.&lt;/p&gt;&#xA;" LastActivityDate="2010-11-10T02:17:06.250" />
  <row Id="9638" PostTypeId="1" CreationDate="2010-11-10T02:24:24.893" Score="3" ViewCount="145" Body="&lt;p&gt;Our math prof gave us this and we are not sure what to make of the notation, can someone give us a hand? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/r5Fvf.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2010-11-10T05:36:11.640" LastActivityDate="2010-11-10T05:36:11.640" Title="Notation help. What do $\mathcal{P}(A)$ and $\chi_{-}$ mean here?" Tags="&lt;homework&gt;&lt;notation&gt;&lt;elementary-set-theory&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="9639" PostTypeId="2" ParentId="9636" CreationDate="2010-11-10T02:26:47.593" Score="1" ViewCount="" Body="&lt;p&gt;What Qiaochu said, and use the definition of the $L^\infty$-norm along with a similar result regarding absolute values.&lt;/p&gt;&#xA;" OwnerUserId="2527" LastActivityDate="2010-11-10T02:26:47.593" CommentCount="3" />
  <row Id="9640" PostTypeId="2" ParentId="9638" CreationDate="2010-11-10T02:32:30.580" Score="4" ViewCount="" Body="&lt;p&gt;$\mathcal{P}(A)$ is the power set of $A$, i.e., the set of subsets of $A$, and $2^A$ is the set of functions from $A$ to $\{0,1\}$.  The characteristic function $\chi_S$ of a subset $S$ of $A$ is the function $\chi_S:A\to\{0,1\}$ defined by $\chi_S(x)=1$ if $x$ is in $S$ and $\chi_S(x)=0$ if $x$ is not in $S$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general it is a good idea to ask your professor to clarify notation on the homework if it is not easy to find in the notes or text.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-11-10T02:32:30.580" CommentCount="4" />
  <row Id="9641" PostTypeId="2" ParentId="9638" CreationDate="2010-11-10T02:38:02.560" Score="6" ViewCount="" Body="&lt;p&gt;$\mathcal{P}(A)$ is the power set of $A$; it is a set whose elements are precisely the subsets of $A$. For example, if $A=\{1,2,3\}$, then $\mathcal{P}(A) = \{\emptyset, \{1\}, \{2\}, \{3\}, \{1,2\}, \{1,3\}, \{2,3\}, \{1,2,3\}\}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given any two sets $A$ and $B$, $A^B$ represents the set of all functions from $B$ to $A$. So $2^A$, with &quot;$2$&quot; meaning the set $\{0,1\}$ means the set of all functions $f\colon A\to \{0,1\}$; all functions whose domain is $A$, and which take only the values $0$ and $1$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $A$ is a set and $S$ is a subset, then $\chi_S$ is the characteristic function of $S$. It is a function $\chi_S\colon A\to \{0,1\}$ defined by:&#xA;$$\chi_S(a) = \left\{\begin{array}{ll}&#xA;0 &amp;amp;\mbox{if $a\not\in S$,}\\&#xA;1 &amp;amp; \mbox{if $a\in S$.}&#xA;\end{array}\right.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By &quot;$\chi_{-}$&quot; he means the function that takes each subset $S$ to the characteristic function $\chi_S$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is no new notation in part (b).&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-11-10T02:38:02.560" />
  <row Id="9642" PostTypeId="1" CreationDate="2010-11-10T02:38:17.190" Score="2" ViewCount="113" Body="&lt;p&gt;Assume a capacitated network flow problem with graph $G = (N,A)$ and capacities $u_{ij}$ (finite), costs for each path of $c_{ij}$ and I/O of $b_{i}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can it be shown there is an equivalent transportation problem with a source node for every arc $(i,j)$ in $N$ having supply $u_{ij}$, and a sink node with demand $\sum _{k | (i,k) \in A}  u_{ik}-b_{i}$ for every node $i$ in $N$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At every supply node $(i,j)$ there are two outgoing infinite capacity arcs: one goes to demand node $i$, and its cost coefficient is 0; the other goes to demand node $j$ and its cost coefficient is $c_{ij}$.&lt;/p&gt;&#xA;" OwnerUserId="2014" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-11-14T17:22:11.077" LastActivityDate="2010-11-14T17:33:41.537" Title="Convert capacitated network flow problem" Tags="&lt;network-flow&gt;" AnswerCount="1" FavoriteCount="0" />
  <row Id="9643" PostTypeId="2" ParentId="9608" CreationDate="2010-11-10T02:43:41.063" Score="5" ViewCount="" Body="&lt;p&gt;You have a stochastic process $\lbrace{X(t):t \geq 0\rbrace}$ with the property that $X(t_2)-X(t_1) \sim N(0,t_2 - t_1)$, which is quite obviously supposed to be Brownian motion (BM). Suppose first that you want to find the distribution function of the running maximum $M(t)=\mathop {\max }\limits_{0 \le s \le t} X(s)$ (the maximum exists, since BM has continuous sample paths). There is a very simple formula for that, namely:&#xA;$$&#xA;{\rm P}(M(t) \le x) = \sqrt {\frac{2}{{\pi t}}} \int_0^x {e^{ - u^2 /(2t)} {\rm d}u}, \;\; x \geq 0. &#xA;$$&#xA;The situation is a little more complicated if you want to find the distribution function of $&#xA;M(t_1 ,t_2 ) = \mathop {\max }\limits_{t_1  \le s \le t_2 } X(s)$ (i.e., the maximum of $X$ over the time interval $[t_1,t_2]$). For this purpose, we need to condition on the initial value $X(t_1)$. Since $X(t_1) \sim N(0,t_1)$, it has density function $f(u;t_1) = \frac{1}{{\sqrt {2\pi t_1 } }}e^{ - u^2 /(2t_1 )}$, and by the law of total probability we have&#xA;$$&#xA;{\rm P}(M(t_1 ,t_2 ) \le x) = \int_{ - \infty }^\infty  {{\rm P}(M(t_1 ,t_2 ) \le x|X_{t_1 }  = u)f(u;t_1) {\rm d}u}. &#xA;$$&#xA;Now, as follows from basic properties of BM, conditioned on $X_{t_1}=u$, $M(t_1 ,t_2 )$ can be replaced by $u + M(0,t_2 - t_1)$, i.e. by $u + M(t_2 - t_1)$ (more precisely, by $u$ plus an independent copy of $M(t_2 - t_1)$, which is independent of $X_{t_1}$). This leads to&#xA;$$&#xA;{\rm P}(M(t_1 ,t_2 ) \le x) = \int_{ - \infty }^\infty  {{\rm P}(M(t_2  - t_1 ) \le x - u)f(u;t_1 ){\rm d}u}. &#xA;$$&#xA;Finally, since $M(t_2  - t_1 )$ cannot be negative, we have to integrate only from $-\infty$ to $x$. That is,&#xA;$$&#xA;{\rm P}(M(t_1 ,t_2 ) \le x) = \int_{ - \infty }^x {{\rm P}(M(t_2  - t_1 ) \le x - u)f(u;t_1 ) {\rm d}u}, \;\; x \in {\bf R}.&#xA;$$&#xA;So, we have a double integral with an elementary integrand. Maybe one can simplify it. Also, maybe one can find the result in the literature. &lt;/p&gt;&#xA;" OwnerUserId="2810" LastEditorUserId="2810" LastEditorDisplayName="" LastEditDate="2010-11-10T17:54:31.510" LastActivityDate="2010-11-10T17:54:31.510" CommentCount="6" />
  <row Id="9644" PostTypeId="2" ParentId="9636" CreationDate="2010-11-10T03:14:57.393" Score="0" ViewCount="" Body="&lt;p&gt;Isn't it true that $$  ||fg||_{\mathrm{sup}} \leq ||f||_{\mathrm{sup}}||g||_{\mathrm{sup}} $$ for any two bounded function f and g? I do not seem to understand the subtlety here. Doesn't this prove the question (since f and g can be unbounded only over a measure zero set)&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-10T04:51:27.783" LastActivityDate="2010-11-10T04:51:27.783" CommentCount="1" />
  <row Id="9645" PostTypeId="1" AcceptedAnswerId="9646" CreationDate="2010-11-10T03:34:48.483" Score="2" ViewCount="208" Body="&lt;p&gt;When I first learned linear algebra, both the professor and the book used brackets like &lt;code&gt;[&lt;/code&gt; and &lt;code&gt;]&lt;/code&gt; to enclose matrices. However, in my current differential equations textbook, matrices are enclosed by parenthesis, and I suddenly realize everybody else are using them too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So are brackets/parenthesis for enclosing matrices always totally interchangeable?&lt;/p&gt;&#xA;" OwnerUserId="3143" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2010-11-10T03:54:57.467" LastActivityDate="2010-11-10T04:05:32.813" Title="Parenthesis vs brackets for matrices" Tags="&lt;matrices&gt;&lt;notation&gt;" AnswerCount="2" />
  <row Id="9646" PostTypeId="2" ParentId="9645" CreationDate="2010-11-10T03:44:22.780" Score="6" ViewCount="" Body="&lt;p&gt;It is just a question of notation and what you are comfortable with. The only thing you might want to be aware is |A| stands for determinant of the matrix A and hence cannot be used in place of () or [] to represent the matrix&lt;/p&gt;&#xA;" OwnerUserId="3302" LastActivityDate="2010-11-10T03:44:22.780" />
  <row Id="9647" PostTypeId="2" ParentId="9645" CreationDate="2010-11-10T04:05:32.813" Score="5" ViewCount="" Body="&lt;p&gt;They're interchangeable. I think a lot of people tend to use parentheses just because they're easier to write by hand.&lt;/p&gt;&#xA;" OwnerUserId="173" LastActivityDate="2010-11-10T04:05:32.813" CommentCount="3" />
  <row Id="9648" PostTypeId="1" AcceptedAnswerId="9674" CreationDate="2010-11-10T04:26:19.850" Score="10" ViewCount="206" Body="&lt;p&gt;To motivate my question, recall the following well-known fact: Suppose that $p\equiv 1\pmod 4$ is a prime number. Then the equation $x^2\equiv -1\pmod p$ has a solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One can show this as follows: Consider the following polynomial in ${\mathbb Z}_p[x]$: $x^{4k}-1$, where $p=4k+1$. The roots of this polynomial are precisely the elements of ${\mathbb Z}_p^*$, each one with multiplicity 1. The polynomial factors as $(x^{2k}-1)((x^k)^2+1)$, and it follows that if $a$ is any element of ${\mathbb Z}_p^*$ with $a^{2k}\ne 1$ (and there are precisely $(p-1)/2$ possible such $a$), then $b=a^k$ satisfies $b^2\equiv -1\pmod p$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, there are other arguments, but I am interested in pursuing this line of reasoning. My question is the following:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;From the quadratic reciprocity law, we&#xA;  have that $x^2\equiv -2\pmod p$ has a&#xA;  solution iff $p\equiv 1$ or $3\pmod&#xA;  8$. Is there a proof of the&#xA;  right-to-left implication using some&#xA;  polynomial and appropriate counting of&#xA;  roots, as in the case shown above?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="3340" LastActivityDate="2010-11-10T16:38:25.087" Title="A particular case of the quadratic reciprocity law" Tags="&lt;number-theory&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="4" />
  <row Id="9649" PostTypeId="1" AcceptedAnswerId="9727" CreationDate="2010-11-10T04:46:11.093" Score="2" ViewCount="229" Body="&lt;p&gt;This is a real-analysis homework question so I of course have to be very precise and justify anything or any theorem I use.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $b_n$ is a bounded sequence and $\lim(a_n) = 0$, show that $\lim(a_nb_n) = 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Intuitively, since $b_n$ is bounded, then sup($b_n$) is some finite number and therefore we can take an $N$ natural number as large as we need such that for all $n\gt N$ $b_na_n$ approaches $0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At first I thought to use the limit theorems, but since $a_n$ is not bounded, the general limit theorems do not reply. (I am referring to $\lim(X + Y) = \lim X + \lim Y$ for $X,Y$ sequences etc).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was thinking then to use the definition of the limit somehow to show that since $b_n$ is bounded we can take as intuitively stated above $N$ large enough to show the statement is true.  I'm not sure how to proceed with this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you for your replies in advance!&lt;/p&gt;&#xA;" OwnerUserId="2533" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-10T04:52:40.623" LastActivityDate="2010-11-10T16:20:13.650" Title="If $b_n$ is a bounded sequence and $\lim a_n = 0$, show that $\lim(a_nb_n) = 0$" Tags="&lt;real-analysis&gt;&lt;homework&gt;" AnswerCount="5" CommentCount="1" />
  <row Id="9650" PostTypeId="2" ParentId="9649" CreationDate="2010-11-10T04:51:57.030" Score="1" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT:&lt;/strong&gt; Let $N(\epsilon)$ be such that $\frac{1}{N(\epsilon)}&amp;lt;\frac{\epsilon}{M}$, where $|b_{n}| \leq M$ ( since $b_{n}$ is bounded.) Then think of $|a_{n}b_{n} - 0|$ for $n \geq N(\epsilon)$.&lt;/p&gt;&#xA;" LastActivityDate="2010-11-10T04:51:57.030" />
  <row Id="9651" PostTypeId="2" ParentId="9649" CreationDate="2010-11-10T04:52:34.737" Score="3" ViewCount="" Body="&lt;p&gt;Hint: If $|a_n|\leq\varepsilon$, then $|a_nb_n|\leq\varepsilon |b_n|\leq \varepsilon \sup_k|b_k|$.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-11-10T04:52:34.737" />
  <row Id="9652" PostTypeId="2" ParentId="9649" CreationDate="2010-11-10T04:53:39.920" Score="2" ViewCount="" Body="&lt;p&gt;Since the $b_n$ are bounded, there is a constant $C$ such that $-C\lt b_n\lt C$ for all $n$. Fix $n$. Suppose first $a_n\gt 0$. Then $-C a_n&amp;lt; a_nb_n &amp;lt; Ca_n$. Similarly, if $a_n\lt 0$ then $Ca_n &amp;lt; a_n b_n &amp;lt; -C a_n$. In short: $|a_n b_n|\lt C|a_n|$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, for any sequence $c_n$, $c_n\to 0$ iff $|c_n|\to 0$. Right? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But $C|a_n|\to 0$ since $a_n\to 0$. Since $|a_nb_n|\le C|a_n|$, we must also have $|a_n b_n|\to 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(This is just a sketch. Let me know if you need me to clarify something.)&lt;/p&gt;&#xA;" OwnerUserId="3340" LastActivityDate="2010-11-10T04:53:39.920" />
  <row Id="9653" PostTypeId="2" ParentId="9649" CreationDate="2010-11-10T04:55:52.213" Score="3" ViewCount="" Body="&lt;p&gt;You write *... but since $a_n$ is not bounded...*&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Presumably you mean that you have not been explicitly told that the sequence $a_n$ is bounded. That is true; you have not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, as it happens, the following holds:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Theorem.&lt;/strong&gt; Let $c_n$ be any sequence of real numbers. If $c_n$ converges, then it is bounded.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Intuitively, why is this true? Well, for large enough $n$, the values of $c_n$ must be close to the limit $L$; so only the &quot;early&quot; values of $c_n$ can get away from the limit, and since there are only finitely many of them, they can only go so far. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, try to prove this carefully, so you can conclude that $a_n$ &lt;strong&gt;is&lt;/strong&gt; bounded, so that the theorems you want to apply &lt;em&gt;can&lt;/em&gt; be applied.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-11-10T04:55:52.213" CommentCount="2" />
  <row Id="9654" PostTypeId="1" CreationDate="2010-11-10T04:59:16.333" Score="0" ViewCount="145" Body="&lt;p&gt;Suppose $\mu(X) &amp;lt; \infty$ and $0 &amp;lt; p&amp;lt; q \leq \infty$. I'm reading the following proof:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $q=\infty$ then $L^{q}(\mu) \subseteq L^{p}(\mu)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proof:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $q=\infty$ then $\int |f|^{p} \leq (||f||_{\infty})^{p}  \int 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why we have the above inequality?? &lt;/p&gt;&#xA;" OwnerUserId="4219" LastActivityDate="2010-11-10T06:45:42.513" Title="Please explain inequality" Tags="&lt;measure-theory&gt;" AnswerCount="3" CommentCount="7" />
  <row Id="9655" PostTypeId="1" AcceptedAnswerId="9658" CreationDate="2010-11-10T05:00:02.030" Score="4" ViewCount="133" Body="&lt;p&gt;Hi&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $M$ be a compact, connected smooth manifold. If $p, q$ are points in $M$, is there always a geodesic that goes from $p$ to $q$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that this is certainly not true if $M$ is not compact, but I couldn't find a counterexample for the compact case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anybody help me out?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;S.&lt;/p&gt;&#xA;" OwnerUserId="3208" LastActivityDate="2010-12-07T20:37:25.857" Title="Geodesics on Compact Manifolds" Tags="&lt;differential-geometry&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="9656" PostTypeId="2" ParentId="9654" CreationDate="2010-11-10T05:03:53.357" Score="1" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT:&lt;/strong&gt; Note that $-||f|\|_{\infty} \leq f(x)\leq ||f||_{\infty}$ almost everywhere. Hence...&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-11-10T05:03:53.357" />
  <row Id="9657" PostTypeId="2" ParentId="9654" CreationDate="2010-11-10T05:05:54.870" Score="1" ViewCount="" Body="&lt;p&gt;Let's just write $\|f\|$ for the infinity norm. We have that $\|f\|$ is defined as the essential sup of the values that $f$ takes. But then $|f(x)|^p\le\|f\| ^p$ for any $x\in X$ except in a set of measure 0. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, what you have is $\int|f|^p\le \int\|f\| ^p$, because the set where the inequality between $|f|$ and $\|f\|$ is reversed is negligible. And, since $\|f\| ^p$ is a constant, we can move it out of the integral. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Note $\|f\|$ could very well be infinite, but in that case the inequality you need is obvious, because anything is $\le\infty$. So, only the case where $\|f\| &amp;lt;\infty$ needs arguing.)&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Ok, I was using the wrong norm. You have $\|f\|$ defined as the infimum of $\|h\|_ {\rm sup}$ where $h$ ranges over all (measurable) functions that agree with $f$ a.e.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fix any such $h$. Then $|f|=|h|$ a.e., so $|f|^p=|h|^p\le \|h\|^ p_ {\rm sup}$ a.e. So $\int |f|^p=\int |h|^p\le \|h\|^p_ {\rm sup}\int 1$, as above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now take the infimum of the right hand side.&lt;/p&gt;&#xA;" OwnerUserId="3340" LastEditorUserId="3340" LastEditorDisplayName="" LastEditDate="2010-11-10T05:28:32.113" LastActivityDate="2010-11-10T05:28:32.113" CommentCount="4" />
  <row Id="9658" PostTypeId="2" ParentId="9655" CreationDate="2010-11-10T05:08:10.353" Score="7" ViewCount="" Body="&lt;p&gt;Yes. This is part of the classical Hopf-Rinow theorem, &lt;em&gt;q.v.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2010-11-10T05:08:10.353" CommentCount="1" />
  <row Id="9659" PostTypeId="1" AcceptedAnswerId="9664" CreationDate="2010-11-10T05:37:42.280" Score="8" ViewCount="730" Body="&lt;p&gt;I am a bit confused about the concept of proper subsets,precisely whether to include one or both of the &lt;strong&gt;void set&lt;/strong&gt; and &lt;strong&gt;the set itself&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An extract from my module goes like this :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&lt;strong&gt;Obviously,every set is the subset of itself and the void set $\phi$ is the subset of every set. These two subsets are called improper subsets.&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It also includes a theorem which states that &quot;&lt;strong&gt;&lt;em&gt;Let A be a finite set having n elements. Then the total number of subsets of A is ($2^n$) and the number of proper subsets of A is ($2^{n}-1).$&lt;/em&gt;&lt;/strong&gt;&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then again in a sample solution of this problem &quot;&lt;strong&gt;&lt;em&gt;If A = {a,b,c},then the number of proper subsets of A is ?&lt;/em&gt;&lt;/strong&gt;&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&lt;strong&gt;Total no of subsets of {a,b,c} = $2^3$ = 8.&#xA;But each set have two improper subset, so number of improper subsets are 6.&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this solution correct ? If so please explain the concept.&lt;/p&gt;&#xA;" OwnerUserId="2109" LastEditorUserId="2109" LastEditorDisplayName="" LastEditDate="2010-11-10T05:59:24.867" LastActivityDate="2011-05-06T04:26:41.910" Title="Is the void set (∅) a proper subset of every set ?" Tags="&lt;terminology&gt;&lt;elementary-set-theory&gt;" AnswerCount="6" FavoriteCount="1" />
  <row Id="9660" PostTypeId="2" ParentId="9659" CreationDate="2010-11-10T05:43:44.180" Score="2" ViewCount="" Body="&lt;p&gt;A set A is a subset of a set B if every element of A is also an element of B.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A set A is a proper subset of a set B if A is a subset of B and there is at least one element of B that's not an element of A.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, the void set is a subset of all sets, and it's a proper subset of every set except itself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, notice that we notate the void set using $\emptyset$, not $\phi$.&lt;/p&gt;&#xA;" OwnerUserId="1523" LastActivityDate="2010-11-10T05:43:44.180" />
  <row Id="9662" PostTypeId="2" ParentId="9659" CreationDate="2010-11-10T05:43:59.967" Score="0" ViewCount="" Body="&lt;p&gt;The null set is a proper subset of any set other than itself. A set $A$ is a proper set of $B$ if $\forall x \in A, x \in B$ and $\exists x \in B$ such that $x \notin A$. Clearly $\phi$ satisfies this provided the superset is not $\phi$ itself&lt;/p&gt;&#xA;" OwnerUserId="3302" LastActivityDate="2010-11-10T05:43:59.967" />
  <row Id="9664" PostTypeId="2" ParentId="9659" CreationDate="2010-11-10T05:45:01.770" Score="11" ViewCount="" Body="&lt;p&gt;Calling $\emptyset$ and $A$ &quot;improper&quot; subsets of a set $A$ is not universal, and it is confusing in this case, because the meaning of &quot;proper&quot; is not the same.  It is standard to say that $S$ is a proper subset of $A$ if (and only if) every element of $S$ is an element of $A$, but $S$ is not equal to $A$, i.e., at least one element of $A$ is not in $S$.  Under this definition, $\emptyset$ is a proper subset of every &lt;em&gt;nonempty&lt;/em&gt; set, even though it is &quot;improper&quot; according to the convention you were also given.  Just remember that mathematical terminology varies and isn't always logical.  Here &quot;improper&quot; does not mean &quot;not proper&quot;.  (For this reason I would personally not use the convention of calling the empty set &quot;improper&quot;.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When finding all proper subsets, you should count the empty set.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-11-10T05:45:01.770" CommentCount="12" />
  <row Id="9665" PostTypeId="2" ParentId="9659" CreationDate="2010-11-10T05:46:23.977" Score="3" ViewCount="" Body="&lt;p&gt;Given a set $A$, a proper subset is any set $B$ such that $B\subseteq A$ and $B\neq A$; that is, $B$ is contained in $A$ but is not equal to $A$. This is denoted by $B\subset A$ in some texts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So while $A$ is a subset of itself, it is &lt;em&gt;not&lt;/em&gt; a proper subset of itself. And this is true for &lt;em&gt;any&lt;/em&gt; set, even the empty set (or void set, as you call it).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Speaking of which, the empty set $\emptyset$ is not only a subset of any set, but also a proper subset of any non-empty set.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edited to include a solution to OP's new question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have the answer in front of you. If $A$ has cardinality $n$, then the number of subsets is $2^n$ and the number of proper subsets is $2^n-1$, because the only set we have to &quot;throw out&quot; is $A$ itself in order to get all the proper subsets. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if $A=\{a,b,c\}$, then there are $2^3-1=7$ proper subsets. We can just list them:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\{a,b\}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\{a,c\}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\{b,c\}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\{a\}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\{b\}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\{c\}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\emptyset$&lt;/p&gt;&#xA;" OwnerUserId="2313" LastEditorUserId="2313" LastEditorDisplayName="" LastEditDate="2010-11-10T05:52:15.913" LastActivityDate="2010-11-10T05:52:15.913" CommentCount="1" />
  <row Id="9666" PostTypeId="2" ParentId="9654" CreationDate="2010-11-10T05:48:15.610" Score="0" ViewCount="" Body="&lt;p&gt;Note that $|f| \leq ||f||_{\infty}$ a.e. This follows from the definition of $||f||_{\infty}$. Hence, $\int |f|^{p} d \mu \leq \int (||f||_{\infty})^{p} d \mu = \int ||f||_{\infty}^{p} \int d \mu$. Thus it follows.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-10T06:45:42.513" LastActivityDate="2010-11-10T06:45:42.513" />
  <row Id="9667" PostTypeId="1" CreationDate="2010-11-10T05:54:34.423" Score="1" ViewCount="110" Body="&lt;p&gt;Working on probability theory in an economics class. This was on of the practice problems that I'm struggling to back up with math. Any help would be appreciated! :)&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;An accident insurance company has reserves to cover claims on no more than 3 accidents. If all accidents have a probability of 0.01 of occurring, if all accidents are independent, and if the insurance company writes 100 policies, what&#xA;is the probability that they will not have enough reserves?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2010-11-10T06:03:44.977" LastActivityDate="2010-11-10T12:10:15.203" Title="Economics Practice Homework Problem" Tags="&lt;probability&gt;&lt;homework&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="9668" PostTypeId="1" CreationDate="2010-11-10T06:18:01.713" Score="1" ViewCount="80" Body="&lt;p&gt;As usual denote $L^p$ the quotient space where two integrable functions are identified if they are equal almost everywhere. So I'm using the definition written here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Lp_space&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Lp_space&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then we have the following result: for each $p \geq 1$ we have $L^{\infty}(X) \subseteq L^{p}(X)$ where X is a finite measure space and $L^{\infty}$ denotes the set of all essentially bounded functions endowed with the $||f||_{\infty}$ pseudonorm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I took $f \in L^{\infty}(X)$ then by definition there is some bounded function $g$ such that $g=f$ a.e. But then $f=g$. So:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\int |f|^{p} = \int |g|^{p} \leq \int (||g||_{\sup})^{p} &amp;lt; \infty$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Questions: Is the above correct? Why do we need $p \geq 1$. Why wouldn't $p&gt;0$ work? Is it because we need $p \geq 1$ in the case q is not $\infty$ or where exactly?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you.&lt;/p&gt;&#xA;" OwnerUserId="4219" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2010-11-10T06:36:07.727" LastActivityDate="2010-11-10T06:36:07.727" Title="Quotients , L^p" Tags="&lt;measure-theory&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="9669" PostTypeId="1" AcceptedAnswerId="9671" CreationDate="2010-11-10T06:22:58.947" Score="1" ViewCount="91" Body="&lt;p&gt;Suppose $X$ is a Banach space and $T: X \to X$ is a bounded linear operator. If $T$ is onto, is it necessarily 1-1? &lt;/p&gt;&#xA;" OwnerUserId="1736" LastActivityDate="2010-11-10T06:29:20.170" Title="Surjectivity implying injectivity for bounded linear operators" Tags="&lt;functional-analysis&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="9670" PostTypeId="2" ParentId="9668" CreationDate="2010-11-10T06:26:21.723" Score="1" ViewCount="" Body="&lt;p&gt;If p&amp;lt;1, then the integral $(\int |f|^p d \mu)^{\frac{1}{p}}$ no longer defines a norm. That is why we need $p \geq 1$ typically. However, for the conclusion you want to draw, I guess it will still work if $p \geq 0$.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastActivityDate="2010-11-10T06:26:21.723" CommentCount="15" />
  <row Id="9671" PostTypeId="2" ParentId="9669" CreationDate="2010-11-10T06:29:20.170" Score="3" ViewCount="" Body="&lt;p&gt;No.  This is true in the finite dimensional case, where a linear transformation is one-to-one iff it is onto iff it is invertible.  Some counterexamples for your question are the &quot;backward shifts&quot; on sequence spaces like $c_0$ or $\ell^p$ with $1\leq p\leq \infty$.  Let $X$ be one of these spaces, and let $T(a_0,a_1,a_2,\ldots)=(a_1,a_2,a_3,\ldots)$.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-11-10T06:29:20.170" />
  <row Id="9672" PostTypeId="2" ParentId="9594" CreationDate="2010-11-10T06:37:16.893" Score="0" ViewCount="" Body="&lt;p&gt;I would leave this as a comment, but I do not have the reputation ...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What do you consider a move? The normal way of understanding the Rubik's cube, is that you can do 18 or 27 moves at each step. That is; you can move each side 90  degrees in two different directions, or make a 180 degree move. In total 3 moves per side, and you have 6 or 9 sides depending on whether you count the &quot;middle slices&quot; of the cube as valid sides or not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you arrive at 12 legal moves, do you only consider the 90 degree moves on the &quot;outer&quot; sides as legal?&lt;/p&gt;&#xA;" OwnerUserId="1823" LastActivityDate="2010-11-10T06:37:16.893" CommentCount="2" />
  <row Id="9673" PostTypeId="1" AcceptedAnswerId="9675" CreationDate="2010-11-10T06:56:56.627" Score="3" ViewCount="154" Body="&lt;p&gt;How do I show&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\mathrm{d}^2 \log(\Gamma(z))}{\mathrm{d}z^2} = \displaystyle \sum_{n = 0}^{\infty} \frac{1}{(z+n)^2}$$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Gamma(z)$ is the gamma function.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-10T16:08:10.767" LastActivityDate="2010-11-10T19:46:51.240" Title="$\frac{\mathrm{d}^2 \log(\Gamma (z))}{\mathrm{d}z^2} = \displaystyle \sum_{n = 0}^{\infty} \frac{1}{(z+n)^2}$" Tags="&lt;complex-analysis&gt;" AnswerCount="2" CommentCount="9" FavoriteCount="1" />
  <row Id="9674" PostTypeId="2" ParentId="9648" CreationDate="2010-11-10T07:25:40.830" Score="12" ViewCount="" Body="&lt;p&gt;If $p=8k+1$ then consider $a$ with $a^{4k}\equiv-1$ (mod $p$).&#xA;Let $b=a^k-a^{7k}$. Then&#xA;$$b^2=a^{2k}-2a^{8k}+a^{14k}\equiv a^{2k}(1+a^{4k})-2&#xA;\equiv-2\pmod{p}.$$&#xA;For $p\equiv3$ (mod $8$) this argument will also work but you need to use the&#xA;Galois field $GF(p^2)$ rather than the integers modulo $p$.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-11-10T07:25:40.830" CommentCount="1" />
  <row Id="9675" PostTypeId="2" ParentId="9673" CreationDate="2010-11-10T07:59:14.597" Score="8" ViewCount="" Body="&lt;p&gt;Use the hadamard product formula&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Gamma(z) = \frac{e^{-\gamma z}}{z} \prod_{k=1}^\infty \left( 1 + \frac{z}{k} \right)^{-1} e^{z/k} $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, note that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{d \log(\Gamma(z))} {dz} = \frac{\Gamma'(z)}{\Gamma(z)} $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For an infinite product, there is an easy way to compute this expression. If&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ f(z) = \prod f_n(z)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;then it is not hard to prove that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ \frac{f'(z)}{f(z)} = \sum \frac{f_n'(z)}{f_n(z)} $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;applying this to the Gamma function gives&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ \frac{\Gamma'(z)}{\Gamma(z)} = -\gamma - \frac{1}{z} + \sum_{k=1}^\infty \frac{-1}{k(1 + z/k)} + \frac{1}{k} $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then we have to take one more derivative to get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ \frac{d^2 \log(\Gamma(z))} {dz^2} = \frac{1}{z^2} + \sum_{k=1}^\infty \frac{1}{(k + z)^2} = \sum_{n=0}^\infty \frac{1}{(z+n)^2} $&lt;/p&gt;&#xA;" OwnerUserId="1089" LastActivityDate="2010-11-10T07:59:14.597" />
  <row Id="9677" PostTypeId="2" ParentId="9673" CreationDate="2010-11-10T08:06:05.890" Score="2" ViewCount="" Body="&lt;p&gt;Let's assume the Gauss formula &#xA;$$\frac{\Gamma'(a)}{\Gamma(a)}+\gamma=\int_{0}^{1}\frac{1-t^{a-1}}{1-t}dt$$&#xA;holds (where $\gamma$ is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Euler%E2%80%93Mascheroni_constant&quot; rel=&quot;nofollow&quot;&gt;Euler–Mascheroni constant&lt;/a&gt;). Integrating the identity &#xA;$$\frac{1-t^{a-1}}{1-t}=\sum_{k=0}^{\infty}(t^{k}-t^{a+k-1})$$&#xA;yields the series&#xA;$$\frac{d\ln \Gamma(a)}{da}+\gamma=\sum_{k=0}^{\infty}\left(\frac{1}{k+1}-\frac{1}{a+k}\right)$$&#xA;which converges uniformly on finite intervals $a\in[0,A]$. Now we can differentiate the latter series in $a$ to obtain that&#xA;$$\frac{d^2\ln\Gamma(a)}{da^2}=\sum_{k=0}^{\infty}\frac{1}{(a+k)^2}.$$&#xA;The differentiation is valid since the resulting series converges uniformly for $a\geq 0.$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Derivation of the Gauss formula.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the basic properties of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Beta_function&quot; rel=&quot;nofollow&quot;&gt;beta function&lt;/a&gt;, we get&#xA;$$\Gamma(b)-B(a,b)=\Gamma(b)-\frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}=\frac{b\Gamma(b)(\Gamma(a+b)-\Gamma(a))}{b\Gamma(a+b)}$$&#xA;$$=\frac{\Gamma(b+1)}{\Gamma(a+b)}\cdot\frac{\Gamma(a+b)-\Gamma(a)}{b}.$$&#xA;Passing to the limit $b\to0$ yields&#xA;$$\frac{d\ln \Gamma(a)}{da}=\frac{\Gamma'(a)}{\Gamma(a)}=\lim\limits_{b\to 0}(\Gamma(b)-B(a,b)),$$&#xA;or &#xA;$$\frac{\Gamma'(a)}{\Gamma(a)}=\lim\limits_{b\to 0}\int_{0}^{\infty}x^{b-1}\left(e^{-x}-\frac{1}{(1+x)^{a+b}}\right)dx=\int_{0}^{\infty}\left(e^{-x}-\frac{1}{(1+x)^{a}}\right)\frac{dx}{x}.\qquad(1)$$&#xA;Identity (1) can be used to define the Euler constant $\gamma$&#xA;$$\qquad\qquad\qquad\qquad\qquad\qquad-\gamma:=\frac{\Gamma'(1)}{\Gamma(1)}=\int_{0}^{\infty}\left(e^{-x}-\frac{1}{1+x}\right)\frac{dx}{x}.\qquad\qquad\qquad\qquad\qquad\qquad(2)$$&#xA;Subtracting (2) from (1) and using the substitution $t=\frac{1}{1+x}$ we obtain that&#xA;$$\frac{\Gamma'(a)}{\Gamma(a)}-\frac{\Gamma'(1)}{\Gamma(1)}=\int_{0}^{1}\frac{1-t^{a-1}}{1-t}dt.$$&lt;/p&gt;&#xA;" OwnerUserId="723" LastEditorUserId="723" LastEditorDisplayName="" LastEditDate="2010-11-10T19:46:51.240" LastActivityDate="2010-11-10T19:46:51.240" />
  <row Id="9679" PostTypeId="1" AcceptedAnswerId="9682" CreationDate="2010-11-10T09:14:24.257" Score="0" ViewCount="67" Body="&lt;p&gt;I am doing some pre study/revision in maths before i take my next education steps and am working through a &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0201624427&quot; rel=&quot;nofollow&quot;&gt;hefty engineering maths book&lt;/a&gt; from the beginning. The first chapter is a review of algebraic techniques using indices.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is to simplify this&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;\frac{2xy^2}{(2xy)^2}&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and the solution given is this&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;\frac{1}{2x}&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;what are the intermediate steps to arrive at this solution?&lt;/p&gt;&#xA;" OwnerUserId="676" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2010-11-10T14:20:11.963" LastActivityDate="2010-11-10T14:20:11.963" Title="What are the steps to perform this algebraic simplification" Tags="&lt;algebra-precalculus&gt;" AnswerCount="1" />
  <row Id="9682" PostTypeId="2" ParentId="9679" CreationDate="2010-11-10T09:33:57.943" Score="2" ViewCount="" Body="&lt;p&gt;$$\frac{2xy^{2}}{\left( 2xy\right) ^{2}}=\frac{2xy^{2}}{2^{2}x^{2}y^{2}}=%&#xA;\frac{2xy^{2}/\left( 2xy^{2}\right) }{2^{2}x^{2}y^{2}/\left( 2xy^{2}\right) }%&#xA;=\frac{1}{2x}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;provided that $2xy^{2}\neq 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Justification&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\left( 2xy\right) ^{2}=2^{2}x^{2}y^{2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\dfrac{2^{2}x^{2}y^{2}}{2xy^{2}}=2x$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\dfrac{2xy^{2}}{2xy^{2}}=1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{2xy^{2}}{\left( 2xy\right) ^{2}}=2xy^{2}\left( 2xy\right)&#xA;^{-2}=2xy^{2}2^{-2}x^{-2}y^{-2}=2^{1-2}x^{1-2}y^{2-2}=2^{-1}x^{-1}y^{0}=\frac{1}{2x}$$&lt;/p&gt;&#xA;" OwnerUserId="752" LastActivityDate="2010-11-10T09:33:57.943" CommentCount="1" />
  <row Id="9683" PostTypeId="1" CreationDate="2010-11-10T09:45:13.173" Score="1" ViewCount="121" Body="&lt;p&gt;Hello..is it possible to calculate shadow areas of buildings or simulate shadows of buildings in a city, using the heights of these buildings and the sun angle and azimuth? the basic light tracing concept using trigonometry is known. But the Problem is how to find away to come up with simulated shadow of group of buildings in one place.  &lt;/p&gt;&#xA;" LastActivityDate="2010-11-10T11:51:44.910" Title="shadow simulation from buildings" Tags="&lt;geometry&gt;&lt;algorithms&gt;&lt;computer-science&gt;&lt;matlab&gt;&lt;geometric-construction&gt;" AnswerCount="2" />
  <row Id="9684" PostTypeId="1" AcceptedAnswerId="9704" CreationDate="2010-11-10T09:52:38.190" Score="0" ViewCount="87" Body="&lt;p&gt;Suppose we are given two real numbers: $a, b \in R$, $b &gt; a$. Find $m \in [1, 2, 2.5, 5]$ and $k \in Z$, satisfying the following condition: $\frac{b-a}{m10^k} \in [5; 10]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How to solve this problem without using brute force?&lt;/p&gt;&#xA;" OwnerUserId="1041" LastActivityDate="2010-11-10T14:24:49.677" Title="How to solve this problem programmatically?" Tags="&lt;nonlinear-optimization&gt;&lt;convex-optimization&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="9685" PostTypeId="1" AcceptedAnswerId="9689" CreationDate="2010-11-10T10:05:29.043" Score="2" ViewCount="83" Body="&lt;p&gt;When f is a Lebesgue measurable function, is $\Vert f \Vert_p$ continuous in p, when p is in a set such that f belongs to $L_p$? If yes, how to show that? Thanks!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yes. It is an exercise taken from Rudin's book. I have spent quite a while thinking but got no idea.&lt;/p&gt;&#xA;" OwnerUserId="3348" LastEditorUserId="3348" LastEditorDisplayName="" LastEditDate="2010-11-10T10:22:29.097" LastActivityDate="2010-11-10T10:38:51.277" Title="$\Vert f \Vert_p$ is continuous in p ?" Tags="&lt;real-analysis&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" />
  <row Id="9687" PostTypeId="2" ParentId="9685" CreationDate="2010-11-10T10:23:36.693" Score="1" ViewCount="" Body="&lt;p&gt;So you're asking about $p \mapsto \left( \int_X |f|^p d\mu \right)^{1/p}$. Isn't this clearly a composition of continuous functions?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First $p \mapsto |f|^p$ (no hints, but this should be pretty easy), then the Lebesgue integral (hint: it's a linear map between the Banach space $L^p(X,\mu)$ and $\mathbb{R}$. Which linear maps are continuous?), and finally $x \mapsto x^{1/p}$. &lt;/p&gt;&#xA;" OwnerUserId="1269" LastEditorUserId="1269" LastEditorDisplayName="" LastEditDate="2010-11-10T10:38:25.030" LastActivityDate="2010-11-10T10:38:25.030" CommentCount="1" />
  <row Id="9688" PostTypeId="1" AcceptedAnswerId="9739" CreationDate="2010-11-10T10:29:21.157" Score="0" ViewCount="92" Body="&lt;p&gt;For a Lebesgue measurable function f, $\Vert f \Vert_p$ converges to $\Vert f \Vert_\inf$ as $p$ goes to infinity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was wondering if $\Vert f \Vert_p$ converges to $\exp(\int \log |f|)$ as $p$ goes to 0 for probability measure? How to prove this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="3348" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-10T15:57:00.157" LastActivityDate="2010-11-10T17:19:50.370" Title="convergence of $\Vert f \Vert_p$ with respect to p" Tags="&lt;real-analysis&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="9689" PostTypeId="2" ParentId="9685" CreationDate="2010-11-10T10:38:51.277" Score="1" ViewCount="" Body="&lt;p&gt;This may not be the answer you are looking for, but:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using Holder's inequality you can show that the map $p^{-1}\mapsto \|f\|_p$ is &lt;strong&gt;log-convex&lt;/strong&gt;. Now just use the fact that any convex function on an interval is continuous. &lt;/p&gt;&#xA;" OwnerUserId="1543" LastActivityDate="2010-11-10T10:38:51.277" CommentCount="1" />
  <row Id="9690" PostTypeId="1" AcceptedAnswerId="9782" CreationDate="2010-11-10T10:43:41.600" Score="3" ViewCount="150" Body="&lt;p&gt;In a parallelogram ABCD. M is the midpoint of CD. Line BM intersects AC at L and it also intersects AD extended at E. Prove that EL=2BL&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS: This is not a homework problem. I was solving geometry for fun. I'm unable to solve this. :(&lt;/p&gt;&#xA;" OwnerUserId="2264" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-11-11T15:12:59.007" LastActivityDate="2010-11-11T15:12:59.007" Title="A parallelogram and a line joining a vertex to the midpoint of opposite side." Tags="&lt;geometry&gt;&lt;high-school&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="1" />
  <row Id="9691" PostTypeId="2" ParentId="9683" CreationDate="2010-11-10T10:44:45.607" Score="3" ViewCount="" Body="&lt;p&gt;I'll assume that the buildings are made up of polygons.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Project each polygon onto the ground to find its shadow.&lt;/li&gt;&#xA;&lt;li&gt;Merge these shadow polygons (e.g. using &lt;a href=&quot;http://en.wikipedia.org/wiki/Weiler%E2%80%93Atherton_clipping_algorithm&quot; rel=&quot;nofollow&quot;&gt;Weiler-Atherton&lt;/a&gt;).&lt;/li&gt;&#xA;&lt;li&gt;Find the area of the merged polygon.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="3350" LastActivityDate="2010-11-10T10:44:45.607" CommentCount="1" />
  <row Id="9692" PostTypeId="1" AcceptedAnswerId="9741" CreationDate="2010-11-10T10:52:20.273" Score="2" ViewCount="195" Body="&lt;p&gt;As we know that $L_p \subseteq L_q$ when $0 &amp;lt; p &amp;lt; q$ for probability measure, I was wondering when $L_p = L_q$ is true and why. Is it to impose some restriction on the domain space? Thanks!&lt;/p&gt;&#xA;" OwnerUserId="3348" LastActivityDate="2010-11-10T23:28:06.377" Title="When $L_p = L_q$ ?" Tags="&lt;real-analysis&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="2" />
  <row Id="9694" PostTypeId="2" ParentId="9692" CreationDate="2010-11-10T11:49:30.433" Score="1" ViewCount="" Body="&lt;p&gt;For probability measure $ L_p \subset L_q $ .&#xA;Equality is unreachable, see &lt;a href=&quot;http://en.wikipedia.org/wiki/Lp_space&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Lp_space&lt;/a&gt;. &lt;/p&gt;&#xA;" OwnerUserId="1035" LastEditorUserId="1035" LastEditorDisplayName="" LastEditDate="2010-11-10T11:55:49.233" LastActivityDate="2010-11-10T11:55:49.233" CommentCount="1" />
  <row Id="9695" PostTypeId="2" ParentId="9683" CreationDate="2010-11-10T11:51:44.910" Score="-1" ViewCount="" Body="&lt;p&gt;if you have the data on the buildings, model the whole scene as tall boxes in OpenGL, render it with a suitable light source, see the shadow.&lt;/p&gt;&#xA;" OwnerUserId="589" LastActivityDate="2010-11-10T11:51:44.910" CommentCount="2" />
  <row Id="9696" PostTypeId="2" ParentId="9688" CreationDate="2010-11-10T12:03:51.017" Score="2" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Approximate by non-negative step functions (if $f$ vanishes on a set of positive measure, using a Holder inequality you can argue that $\|f\|_p \leq C |\mathrm{supp}(f)|^{\frac1p - 1} \searrow 0$). For step functions, that $\|f\|_p\geq \exp \int \log |f|$ is a consequence of the arimetic-geometric mean inequality. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The AM-GM inequality becomes an equality when all its terms are equal. For a non-negative step function $f$, as $p\to 0$, you have that $f \to 1$. By continuity you can quantify the difference between AM-GM. So you have that the convergence is true for step functions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To finish off you use the usual diagonal argument to extract a subsequence from the approximating sequence of step functions, together with a suitable sequence of $p$'s. &lt;/p&gt;&#xA;" OwnerUserId="1543" LastActivityDate="2010-11-10T12:03:51.017" CommentCount="1" />
  <row Id="9697" PostTypeId="2" ParentId="9667" CreationDate="2010-11-10T12:10:15.203" Score="2" ViewCount="" Body="&lt;p&gt;Let's $X_i$ be a Bernoulli variable that is 1 if an accident occured with the policy number $i$, 0 otherwise. Thus we have:&#xA;$$P(X_i = 1) = 0.01, P(X_i = 0) = 0.99$$&#xA;This is true for all $i$ since we assume independance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $Y = X_1 + X_2 + \dots + X_{100}$ the sum of all $X_i$. $Y$ counts the number of accidents that occured.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That means you are looking for $P(Y &gt; 3)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if you are having probability class, you should have seen what a Bernoulli variable is, and what the distribution of a variable that is the sum of independant Bernoulli variables is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you don't know, look &lt;a href=&quot;http://en.wikipedia.org/wiki/Binomial_distribution&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please note that according to your question I had to made the assumption that the probability you give (0.01), is the probability that for 1 policy there will be 1 accident and that there cannot be more than 1 accident for the same policy. Usually, we give a probability that in a given amount of time there will be an accident. &lt;/p&gt;&#xA;" OwnerUserId="2778" LastActivityDate="2010-11-10T12:10:15.203" />
  <row Id="9698" PostTypeId="2" ParentId="9692" CreationDate="2010-11-10T12:17:07.890" Score="4" ViewCount="" Body="&lt;p&gt;Equality only when $p = q$. Suppose $p &amp;lt; q$, and let $r = \frac{p+q}{2}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $B_n$ be a sequence of disjoint measurable subsets of your probability space with the property that their measures $|B_n| = m_n \leq 2^{-n}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $\chi_n$ denote the characteristic function of $B_n$. Consider the sequence of functions &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f_n = \sum_1^n \frac{1}{m_k^{1/r}} \chi_k$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is easy to check that this sequence is bounded in $L^p$ and Cauchy, so converges to a limit function $f$. But the sequence is unbounded in $L^q$. Hence the limit function $f$ is an element of $L^p$ that is not in $L^q$. &lt;/p&gt;&#xA;" OwnerUserId="1543" LastEditorUserId="1543" LastEditorDisplayName="" LastEditDate="2010-11-10T23:28:06.377" LastActivityDate="2010-11-10T23:28:06.377" CommentCount="4" />
  <row Id="9699" PostTypeId="2" ParentId="9690" CreationDate="2010-11-10T12:17:31.063" Score="4" ViewCount="" Body="&lt;p&gt;Let BD and AC intersects at point O.&#xA;We have triangle BCD. BO=OD because ABCD is parallelogram.&#xA;In this triangle CO and BM are medians.&#xA;Medians intersects in one point and divide each other in relation 2:1.&#xA;So  BL = 2LM.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also BM = ME thus triangle MBC = triangle MDE because they are congruent and DM = CM.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So BL = 2/3 BM =&gt; BL = 2EL.    &lt;/p&gt;&#xA;" OwnerUserId="1035" LastActivityDate="2010-11-10T12:17:31.063" />
  <row Id="9701" PostTypeId="1" AcceptedAnswerId="9714" CreationDate="2010-11-10T12:45:34.250" Score="0" ViewCount="191" Body="&lt;p&gt;A Steinmetz solid is the intersection of two cylinders. My question is how to find the surface area of one with both cylinders of radius 1, without parameterizing the solid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I parameterized it; that is, used $r(u,v)=\langle \sqrt{1-v^2}, u \sqrt{1-v^2}, v \rangle$, $u,v \in [-1,1]$, and integrated the magnitude of the  cross product of the partials: $$\int_{-1}^1 \int_{-1}^1 |r_u \times r_v| dR$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I make the solid into a function $z=f(x,y)$ rather than parameterizing it?&lt;/p&gt;&#xA;" OwnerUserId="1562" LastActivityDate="2010-11-10T14:55:25.087" Title="Surface Area of a Steinmetz Solid" Tags="&lt;calculus&gt;&lt;homework&gt;" AnswerCount="1" CommentCount="0" />
  <row Id="9703" PostTypeId="1" AcceptedAnswerId="9788" CreationDate="2010-11-10T14:22:38.950" Score="10" ViewCount="1117" Body="&lt;p&gt;My daughter is stuck on the concept that 2&lt;sup&gt;0&lt;/sup&gt; = 1, having the intuitive expectation that it be equal to zero. I have tried explaining it, but I guess not well enough. How would you explain the concept to a child, other than the teachers &quot;that is just the rule&quot; approach?&lt;/p&gt;&#xA;" OwnerUserId="3356" LastActivityDate="2011-01-12T22:56:28.417" Title="How do I explain 2 to the power of zero equals 1 to a child" Tags="&lt;arithmetic&gt;&lt;teaching&gt;" AnswerCount="12" CommentCount="11" FavoriteCount="6" />
  <row Id="9704" PostTypeId="2" ParentId="9684" CreationDate="2010-11-10T14:24:49.677" Score="2" ViewCount="" Body="&lt;p&gt;This is a &quot;normalization&quot; type operation.  For simplicity we can replace the two parameters a,b with the single one c = b - a &gt; 0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming the interpretation proposed by Eric above, you want c*(10^-k) to fall into an interval m*[5,10) where m is in {1,2,2.5,5}.  In other words we want to normalize c by a power of 10 so that it belongs to (at least) one of these intervals:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[5,10), [10,20), [12.5,25), [25,50)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While the middle pair of these intervals has nonempty overlap, the union of all four intervals is [5,50) and covers exactly one order of magnitude (power of 10).  It follows that the exponent k is unique, although the factor m may or may not be unique.  Note that this is equivalent to pigeonholing c/5 into one of four subintervals covering [1,10).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How to find these &quot;without using brute force&quot; ?  The best programmatic approach depends on what language and hardware is available.  The IEEE standard for floating-point arithmetic (IEEE-754) allows some fairly portable assumptions to be made, but the challenge is to get a &quot;leading digit&quot; and characteristic exponent in base 10 when the usual representation of floating-point numbers is base 2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we take the &quot;common logarithm&quot; log_10(c/5) and break it into an integer part and a fractional part:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;log_10(c/5) = iPart + fPart  where fPart in [0,1)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now iPart here is the same value we want for k in your original notation for the problem, and we can determine m from fPart by doing a bit of testing:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;if ( fPart &amp;lt; log_10(2) ) then m = 1&lt;br&gt;&#xA;else if ( fPart &amp;lt; log_10(2.5) ) then m = 2&lt;br&gt;&#xA;    else if ( fPart &amp;lt; log_10(5) ) then m = 2.5&lt;br&gt;&#xA;        else m = 5  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;regards, hm&lt;/p&gt;&#xA;" OwnerUserId="3111" LastActivityDate="2010-11-10T14:24:49.677" />
  <row Id="9705" PostTypeId="1" CreationDate="2010-11-10T14:28:52.747" Score="2" ViewCount="265" Body="&lt;p&gt;How many subgroups with index 2 are there of a free group on two generators? What are their generators?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All I know is that the subgroups should have 2*2 + 1 - 2 = 3 generators. &lt;/p&gt;&#xA;" LastActivityDate="2010-11-10T15:40:17.790" Title="Finding subgroups of a free group with a specific index" Tags="&lt;group-theory&gt;" AnswerCount="3" FavoriteCount="2" />
  <row Id="9706" PostTypeId="2" ParentId="9705" CreationDate="2010-11-10T14:30:58.613" Score="2" ViewCount="" Body="&lt;p&gt;Let $F$ be free on $a$ and $b$, and let $N$ be a subgroup of index $2$. Then $N$ is normal, and there is a map $\phi:F\to\mathbb Z/2\mathbb Z$ with kernel precisely $N$. What are the possible images of $a$ and $b$? For each possible choice you can describe the corresponding generators of $N$ as a normal subgroup, and with a little extra work, generators as subgroup.&lt;/p&gt;&#xA;" OwnerUserId="274" LastEditorUserId="274" LastEditorDisplayName="" LastEditDate="2010-11-10T14:46:59.597" LastActivityDate="2010-11-10T14:46:59.597" />
  <row Id="9707" PostTypeId="2" ParentId="9703" CreationDate="2010-11-10T14:32:27.887" Score="2" ViewCount="" Body="&lt;p&gt;I would appeal to the laws of exponents.  Show for $a,b&gt;0$ that $2^a* 2^b=2^{a+b}$, then extend it to negative $b$, then set $a=b$.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-11-10T14:32:27.887" CommentCount="1" />
  <row Id="9709" PostTypeId="2" ParentId="9703" CreationDate="2010-11-10T14:34:20.040" Score="5" ViewCount="" Body="&lt;p&gt;This is a question answered all over the Internet, here's an OK answer:&#xA;&lt;a href=&quot;http://mathforum.org/dr.math/faq/faq.number.to.0power.html&quot; rel=&quot;nofollow&quot;&gt;http://mathforum.org/dr.math/faq/faq.number.to.0power.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2900" LastActivityDate="2010-11-10T14:34:20.040" CommentCount="1" />
  <row Id="9710" PostTypeId="1" AcceptedAnswerId="9715" CreationDate="2010-11-10T14:47:06.003" Score="1" ViewCount="125" Body="&lt;p&gt;I received some homework (calculus), which I can't prove:&#xA;&lt;code&gt;f(n) * g(n) &amp;gt;= f(n) + g(n)&lt;/code&gt; for some n&gt;=1 is always true &lt;code&gt;(f(n),  g(n) &amp;gt; = 1)&lt;/code&gt;, I think that this is true, so I need to prove that there is some &lt;code&gt;n&amp;gt;=1&lt;/code&gt; (n is a natural number) for which this inequality will be true always, thanks in advance for any idea&lt;/p&gt;&#xA;" LastActivityDate="2010-11-11T19:31:40.807" Title="prove or refute" Tags="&lt;calculus&gt;&lt;homework&gt;" AnswerCount="3" CommentCount="4" />
  <row Id="9711" PostTypeId="2" ParentId="9705" CreationDate="2010-11-10T14:51:05.960" Score="1" ViewCount="" Body="&lt;p&gt;I like to approach this sort of problem using graphs. The free group on two generators is the fundamental group of a wedge of two circles $R_2$, which I picture as a red oriented circle and a black oriented circle. A subgroup of index $ k$ corresponds to a covering map $G\to R_2$ of index $k$. $G$ can be pictured as a (Edit: basepointed) $k$-vertex connected graph with red and black oriented edges such that at every vertex there is one incoming and one outgoing edge of each color. In the case $k=2$, it's not hard to write down all such graphs. I count three myself.&lt;/p&gt;&#xA;" OwnerUserId="2941" LastEditorUserId="2941" LastEditorDisplayName="" LastEditDate="2010-11-10T15:17:04.470" LastActivityDate="2010-11-10T15:17:04.470" CommentCount="9" />
  <row Id="9712" PostTypeId="1" AcceptedAnswerId="10401" CreationDate="2010-11-10T14:52:40.500" Score="1" ViewCount="34" Body="&lt;p&gt;Let $C$ be a convex subset of a Banach space $X$ and $T:C\to C$ a (norm) continuous affine map, i.e. $$T(tx+(1-t)y)=tT(x)+(1-t)T(y)$$ for $0\le t\le 1$. Is $T$ weakly continuous, i.e continuous as a map from $(C,\tau)$ to $(C,\tau)$ where $\tau$ is the topology induced by the weak topology of $X$. &lt;/p&gt;&#xA;" OwnerUserId="3249" LastActivityDate="2010-11-15T14:29:30.687" Title="Weak Continuity of Affine maps" Tags="&lt;analysis&gt;&lt;functional-analysis&gt;&lt;convex-analysis&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="9713" PostTypeId="1" CreationDate="2010-11-10T14:54:08.113" Score="5" ViewCount="197" Body="&lt;p&gt;I'm teaching first semester calculus and trying to find a way to explain why each hypothesis in l'Hostpial's rule is needed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $f$ and $g$ are real differentiable functions on an interval containing a point $c$ then there are three hypotheses one needs to check in order to apply l'Hospitals rule to compute $\lim_{x \to c} f(x)/g(x)$ :&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$g'(x) \neq 0$ on a neighborhood of $c$, (the wikipedia page misses this)&lt;/li&gt;&#xA;&lt;li&gt;$\lim_{x \to c} f'(x)/g'(x)$ exists (in the extended sense including $\pm \infty$),&lt;/li&gt;&#xA;&lt;li&gt;Either $\lim_{x \to c} g(x) = \pm \infty$ or $\lim_{x \to c} f(x) = \lim_{x \to c} g(x) = 0$.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;What I am looking for is a pair of functions $f$ and $g$ so that 2 and 3 both hold and you can reasonably compute $\lim_{x \to c} f(x)/g(x)$ but get a different answer than computing the limit of the quotient of the derivatives. Bonus points if you can make $\lim_{x \to c} f(x) = \pm \infty$ as well in the case that $g(x) \to \pm \infty$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was shown an example like this a long time ago in my analysis sequence but it would be lost on the students I think:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $f(x) = 2x + \sin(2x)$ and let $g(x) = (2x+\sin(2x))e^{-\sin(x)}$. Then $f/g = e^{\sin x}$ on $(0,\infty)$ so $\lim_{x \to \infty} f(x)/g(x)$ does not exist. Both functions will go to $\infty$ as $x \to \infty$ and the limit of the quotients of the derivatives actually goes to zero. The issue is that $g'(x)$ has infinitely many zeros as $x \to \infty$ so you cannot use l'Hospital.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any clue on how to make this more inviting to a freshman calculus student?&lt;/p&gt;&#xA;" OwnerUserId="3341" LastActivityDate="2010-11-13T16:00:24.710" Title="Non-vanishing of derivative hypothesis in l'Hospital's rule" Tags="&lt;calculus&gt;&lt;real-analysis&gt;" AnswerCount="1" CommentCount="9" />
  <row Id="9714" PostTypeId="2" ParentId="9701" CreationDate="2010-11-10T14:55:25.087" Score="2" ViewCount="" Body="&lt;p&gt;Taking the axes of the cylinders along $x$ and $y$, you have $z=\sqrt{1-(\max{(|x|,|y|)})^2}$.  So you have to integrate $$\int_{-1}^1 \int_{-x}^x f(x) dx dy + \int_{-1}^1 \int_{-y}^y f(y) dx dy$$  where $f$ is the area element.  The first integral gets x&gt;y and the second gets y&gt;x.  By symmetry they are the same.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-11-10T14:55:25.087" />
  <row Id="9715" PostTypeId="2" ParentId="9710" CreationDate="2010-11-10T14:57:13.747" Score="4" ViewCount="" Body="&lt;p&gt;Hint: $2.25 &amp;lt; 3$.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-11-10T14:57:13.747" CommunityOwnedDate="2010-11-10T14:57:13.747" CommentCount="4" />
  <row Id="9716" PostTypeId="2" ParentId="9705" CreationDate="2010-11-10T14:58:56.360" Score="3" ViewCount="" Body="&lt;p&gt;Mariano's answer is quite right, but relies on the 'accident' that all subgroups of index two are normal.  If you wanted to find all the subgroups of index three, say, you would need to try another approach. Here's one idea.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The free group of rank two, $F$, is the fundamental group of the figure-eight graph $\Gamma$, which has one vertex and two edges.  Fix the vertex as the base point.  The subgroups of index $k$ correspond, via covering-space theory, to connected covering spaces $\widehat{\Gamma}\to\Gamma$ of degree $k$, with a choice of base vertex $\hat{v}\in\widehat{\Gamma}$.  (If, on the other hand, you only wanted to count &lt;em&gt;conjugacy classes&lt;/em&gt; of subgroups, you could forget about $\hat{v}$.)  Because the covering map $p:\widehat{\Gamma}\to\Gamma$ has degree $k$, the graph $\widehat{\Gamma}$ has exactly $k$ vertices.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's decorate the graph $\Gamma$ so that we can see $F=\langle a,b\rangle$ more clearly.  Label each edge with the corresponding generator $a$ or $b$.  Furthermore, orient each edge to indicate in which direction the generator goes around it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can use the covering map $p$ to pull the labels and orientations back from $\Gamma$ to $\widehat{\Gamma}$.  That is, if $\hat{e}$ is an edge of $\widehat{\Gamma}$ and $p(\hat{e})=e$ is labelled $a$, then you should label $\hat{e}$ with $a$ also, and orient $\hat{e}$ so that $p$ sends the orientation on $\hat{e}$ to the orientation on $e$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With a little thought, it's not too hard to see that this decoration on $\widehat{\Gamma}$---the labels and orientation---are enough to reconstruct the map $p$: they tell you where to send each edge, and there's only one choice of where to send each vertex. The statement that $p$ is  a covering map translates into a nice condition on the decoration of $\widehat{\Gamma}$: for each vertex $\hat{u}$, you see exactly one edge with label going into $\hat{u}$, and exactly one edge with each label going out of $\hat{u}$.  I'll call this condition 'the covering condition'.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This discussion turns counting the subgroups of index $k$ into the combinatorial problem of counting the connected, decorated, based graphs $\widehat{\Gamma}$ with $k$ vertices that satisfy the covering condition.  With a little thought, one can write down a formula for this.  &lt;a href=&quot;http://www.ams.org/mathscinet/search/publdoc.html?arg3=&amp;amp;co4=AND&amp;amp;co5=AND&amp;amp;co6=AND&amp;amp;co7=AND&amp;amp;dr=all&amp;amp;pg4=AUCN&amp;amp;pg5=AUCN&amp;amp;pg6=PC&amp;amp;pg7=ALLF&amp;amp;pg8=ET&amp;amp;review_format=html&amp;amp;s4=hall%2C%20m%2a&amp;amp;s5=&amp;amp;s6=&amp;amp;s7=&amp;amp;s8=All&amp;amp;vfpref=html&amp;amp;yearRangeFirst=&amp;amp;yearRangeSecond=&amp;amp;yrop=eq&amp;amp;r=181&amp;amp;mx-pid=28836&quot; rel=&quot;nofollow&quot;&gt;Marshall Hall Jr&lt;/a&gt; did just this (though I don't think he thought in terms of the covering spaces of graphs) and came up with the following formula. Let $N(k,r)$ be the number of subgroups of index $k$ in the free group of rank $r$.  Then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$N(k,r)=k(k!)^{r-1}-\sum_{i=1}^{k-1}((k-1)!)^{r-1}N(i,r)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For some related things, I wrote a blog post about proving theorems about free groups by thinking about covering spaces of graphs &lt;a href=&quot;http://ldtopology.wordpress.com/2008/12/01/in-memoriam-j-r-stallings-topology-of-finite-graphs/&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively, if you don't like covering spaces, an equivalent point of view is to count transitive actions by permutation groups with $r$ generators on based sets of size $k$.  This turns out to be the same combinatorial problem.&lt;/p&gt;&#xA;" OwnerUserId="2287" LastEditorUserId="2287" LastEditorDisplayName="" LastEditDate="2010-11-10T15:40:17.790" LastActivityDate="2010-11-10T15:40:17.790" CommentCount="3" />
  <row Id="9717" PostTypeId="2" ParentId="9648" CreationDate="2010-11-10T15:02:14.187" Score="10" ViewCount="" Body="&lt;p&gt;Similar arguments work for other cases as well. For example, if $p = 1 \mod 3k$, find $a$ such that $a^{3k}=1$ and $a^k \neq 1$. Set $z=a^k$, then $z^2+z+1=0$ and we deduce that $(z - z^2)^2 = -3$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a larger example, if $p=5k+1$, then there is an $a$ in $\mathbb{F}_p$ with $a^{5k}=1$ and $a^k \neq 1$. Let $z=a^k$, so $z^4+z^3+z^2+z+1=0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $b = z - z^2 - z^3 + z^4$. Then &#xA;$$b^2 = z^8 -  2 z^7 - z^6 + 4 z^5 - z^4 - 2 z^3 + z^2=$$&#xA;$$-z^4 - z^3 - z^2 - z + 4 = 5.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So $p \equiv 1 \mod 5$ implies that $\left( \frac{5}{p} \right)=1$. Just as in Robin's answer, you can also prove that $p \equiv 4 \mod 5$ implies $\left( \frac{5}{p} \right)=1$ in this manner, but it is more difficult: the element $z$ will live in $\mathbb{F}_{p^2}$, and you will need to check that $b$ lives in $\mathbb{F}_{p}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, for every squarefree $M$, let $\Delta$ be $|M|$ if $M \equiv 1 \mod 4$ and $\Delta=4 |M|$ otherwise. Then you can prove in this manner the statement &quot;if $p = 1 + \Delta k$ then $\left( \frac{\Delta}{p} \right) = 1$.&quot; The first steps are the same in each case: find $a$ in $\mathbb{F}_p$ such that $a^{\Delta k}=1$ but $a^{d k} \neq 1$ for any proper divisor $d$ of $\Delta$. Let $z=a^k$, so $z$ is a primitive $\Delta$ root of unity in $\mathbb{F}_p$. At this point, some magic polynomial $g_M$ is introduced, such that $g_M(z)^2 = M$. If you want to prove the full quadratic reciprocity law, you can do so using this approach as well, but you will need to work in finite fields of non-prime order.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I often thought it would be fun to assign a class the job of finding the magic polynomials $g_M$, for various values of $M$, and see if they could guess the general pattern. Hint: $g$ stands for &quot;Gauss sum&quot;.&lt;/p&gt;&#xA;" OwnerUserId="448" LastEditorUserId="448" LastEditorDisplayName="" LastEditDate="2010-11-10T16:16:29.227" LastActivityDate="2010-11-10T16:16:29.227" CommentCount="6" />
  <row Id="9718" PostTypeId="2" ParentId="9703" CreationDate="2010-11-10T15:02:16.320" Score="40" ViewCount="" Body="&lt;p&gt;I will give a different answer than the answer I gave in the other thread which tries to appeal to intuition.  I am sure your daughter has no problem accepting that $2*0 = 0$.  Intuitively this is because if you add $2$ to itself zero times, you get zero.  Or, to be concrete, if someone gives you two apples zero times, you have zero apples.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For repeatedly adding $2$, talking about collections of apples is a good model.  But for repeatedly multiplying by $2$, it isn't necessarily, since you can't multiply apples and apples (at least, not in a way that makes sense to a child).  But you can multiply apples by numbers; that is, you can start with $1$ apple, then double the number of apples you have to get $2$ apples, then double the number of apples you have to get $4$ apples, and so forth.  In general if you double your apples $n$ times, you have $2^n$ apples.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What happens if you double your apples zero times?  Well, that means you haven't started doubling them yet, so you still have $1$ apple.  If you want your notation to be consistent, then you should say $2^0 = 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a subtly different argument from the argument I gave before.  It's intuitive what it means to add different amounts of apples, and it's intuitive what it means to have zero apples.  But the twos I am now working with aren't numbers of apples, they're just abstract numbers; in other words, they're unitless, so it's harder to get a grip on them.  What $2^n$ really represents above is an endomorphism of the free commutative monoid on an apple, which is much less concrete than an apple.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a way to gain intuition here which sort of involves units, but I don't know if you can convince your daughter that it makes sense.  One way to interpret $2^n$ is that it is the &quot;size&quot; of an $n$-cube of side length $2$ in dimension $n$.  For example, the length of a segment of side length $2$ is $2$, the area of a square of side length $2$ is $4$, and so forth.  One way to say this is that $2^n$ is the number of $n$-cubes of side length $1$ that fit into an $n$-cube of side length $2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To get a meaningful interpretation of the above when $n = 0$ we need to decide what $0$-dimensional objects are.  Well, if $2$-dimensional space is a plane and $1$-dimensional space is a line, then $0$-dimensional space must be... a point.  In particular, a $0$-cube, of any side length, is a point, and so exactly one $0$-cube of side length $1$ fits into a $0$-cube of side length $2$.  Hence $2^0 = 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I'm really curious what her response to this argument will be, actually.  Could you report back on this?)&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-11-10T15:02:16.320" CommentCount="7" />
  <row Id="9719" PostTypeId="1" AcceptedAnswerId="9721" CreationDate="2010-11-10T15:12:18.483" Score="1" ViewCount="122" Body="&lt;p&gt;I have proved this using Venn diagram but when I am trying to prove this using the rule that &lt;strong&gt;&lt;em&gt;&quot;If $ A \subset B \text{ and } B \subset A $ then $ A = B $&quot;&lt;/em&gt;&lt;/strong&gt;, I am having some problems with my understanding of the same,here is how I  did so far :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $x \in A \cap (B-C) \Rightarrow x \in A  \text{ and }  x \in (B-C) \Rightarrow x \in A \text{ and } (x \in B \text{ and } x \notin C)  $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How to proceed next ? Since if I am do something like this : &lt;strong&gt;$ x \in A \text{ and } x \in B \text{ and } x \in A \text{ and } x \notin C $&lt;/strong&gt;,it's not giving the correct results, what exactly I am missing here ?&lt;/p&gt;&#xA;" OwnerUserId="2109" LastEditorUserId="6460" LastEditorDisplayName="" LastEditDate="2011-05-03T14:41:13.450" LastActivityDate="2011-05-03T14:41:13.450" Title="Proving : $A \cap (B-C) = (A \cap B) - (A \cap C) $" Tags="&lt;elementary-set-theory&gt;" AnswerCount="1" />
  <row Id="9721" PostTypeId="2" ParentId="9719" CreationDate="2010-11-10T15:17:13.220" Score="6" ViewCount="" Body="&lt;p&gt;Hint: $x \notin C \Rightarrow x \notin D \cap C$ for any set $D$.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-11-10T15:17:13.220" CommentCount="1" />
  <row Id="9722" PostTypeId="2" ParentId="9710" CreationDate="2010-11-10T15:34:39.840" Score="0" ViewCount="" Body="&lt;p&gt;If f and g are both constant functions, then it's easy to see that the claim isn't true in general. Let's assume that g is not constant, and since this problem came from a calculus class, let's assume that f and g are differentiable. Compare the derivative of f(x) * g(x) with the derivative of f(x) + g(x). What can you say about the relative size of the derivatives? What does that tell you about the functions?&lt;/p&gt;&#xA;" OwnerUserId="759" LastActivityDate="2010-11-10T15:34:39.840" CommentCount="3" />
  <row Id="9723" PostTypeId="1" AcceptedAnswerId="9725" CreationDate="2010-11-10T15:38:30.200" Score="1" ViewCount="68" Body="&lt;p&gt;Find a necessary and sufficient condition for the linear equation Ax = b to have&#xA;no solution. (hint: Use duality to find a strong alternative to Ax = b).&lt;/p&gt;&#xA;" OwnerUserId="2014" LastActivityDate="2010-11-10T15:53:26.217" Title="Use duality to find a strong alternative" Tags="&lt;linear-programming&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="0" />
  <row Id="9724" PostTypeId="1" AcceptedAnswerId="9744" CreationDate="2010-11-10T15:42:04.527" Score="0" ViewCount="145" Body="&lt;p&gt;Which of the following is the unbounded set ?&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;$ X = \{ x\mid x = \frac{1}{n},n \in \mathbb{N} \} $ &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$ Y = \{ x\mid x = \frac{1}{2^n},n \in  \mathbb{N} \} $ &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$ Z = \{ x\mid x = 2^n,n \in \mathbb{N} \} $ &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$ W = \{ x\mid x \in \mathbb{N}, x \lt 4532 \} $ &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;After reading the definition from &lt;a href=&quot;http://en.wikipedia.org/wiki/Bounded_set&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt; , I can only convince myself that it can't be W but am not getting which is correct from the other three ?&lt;/p&gt;&#xA;" OwnerUserId="2109" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-17T05:50:57.830" LastActivityDate="2010-11-17T05:50:57.830" Title="A question on Unbounded set" Tags="&lt;elementary-set-theory&gt;" AnswerCount="2" CommentCount="6" />
  <row Id="9725" PostTypeId="2" ParentId="9723" CreationDate="2010-11-10T15:48:25.317" Score="1" ViewCount="" Body="&lt;p&gt;Solution exists iff $b$ is orthogonal to the null space of $A^{*}$. This is typically referred to as the Fundamental Theorem of Linear Algebra.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2010-11-10T15:53:26.217" LastActivityDate="2010-11-10T15:53:26.217" CommentCount="4" />
  <row Id="9726" PostTypeId="2" ParentId="9724" CreationDate="2010-11-10T16:13:36.853" Score="5" ViewCount="" Body="&lt;p&gt;A set in $\mathbb{R}$ is bounded if you can find two numbers, $a$ and $b$ such that all elements of the set are between $a$ and $b$.  If you want to claim a set is unbounded, whatever $a$ and $b$ I give you, you should be able to find at least one element outside the interval $(a,b)$.  Which of your sets satisfies that?&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-11-10T16:13:36.853" CommentCount="1" />
  <row Id="9727" PostTypeId="2" ParentId="9649" CreationDate="2010-11-10T16:20:13.650" Score="4" ViewCount="" Body="&lt;p&gt;You want to show that the absolute value of (b_n)(a_n) can be made arbitrarily small. But, for all n, the absolute value of b_n does not exceed some positive constant M since the sequence b is bounded. Thus |b_na_n| does not exceed M|a_n|. Since the sequence a converges to 0, this is all you should need. The argument is basically the Squeeze Theorem from elementary calculus.&lt;/p&gt;&#xA;" OwnerUserId="2933" LastActivityDate="2010-11-10T16:20:13.650" CommentCount="1" />
  <row Id="9728" PostTypeId="1" CreationDate="2010-11-10T16:21:35.273" Score="4" ViewCount="390" Body="&lt;p&gt;Let $g:[0,1] \rightarrow \mathbb{R}$ be twice differentiable with $g&amp;#39;&amp;#39;(x)\gt 0$ for all $x\in[0,1]$. If $g(0)&amp;gt;0$ and $g(1)=1$, show that $g(d)=d$ for some point $d\in(0,1)$ if and only if $g&amp;#39;(1)\gt 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proof. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose there exits $d\in(0,1)$ such that $g(d)=d$. Then by the MVT applied on $[d,1]$, $f'(c)(d-1)=g(d)-g(1) $for some $c\in(d,1)$. But $g(d)=d$ and $g(1)=1$ so $f&amp;#39;(c)=1$. Now, since $g&amp;#39;&amp;#39;(x)\gt 0$ for $x\in[0,1]$, $g&amp;#39;$ is increasing on this interval, therefore $1\gt d$ implies $g&amp;#39;(1)-g&amp;#39;(d)\gt 0$. &#xA;Assume $g&amp;#39;(1)\gt 1$ and let $f(x)=g(x)-x$. Then $f(0)=g(0)-0\gt 0$ by hypothesis. I want to show $f(x)\lt 0$ for some $x$ in a neighborhood of $1$. $g&amp;#39;(1)=  \lim\limits_{x\to1}\frac{g(x)-g(1)}{x-1}$ so there exists $\delta\gt 0$ such that given $\epsilon\gt 0$ with $x\in(1-\delta,1)$ implies $|g&amp;#39;(1)-\frac{g(x)-1}{x-1}|\lt \epsilon$. If $\epsilon=g&amp;#39;(1)-1\gt 0$, then $g&amp;#39;(1)-\epsilon\lt \frac{g(x)-1}{x-1}$ which implies $g(x)\lt x$ for $x\in(1-\delta,1)$. The result follows by the IVT.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any comments, corrections or different solutions are welcome.&lt;/p&gt;&#xA;" OwnerUserId="2467" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-17T17:45:54.180" LastActivityDate="2011-01-29T10:08:48.443" Title="Twice differentiable function, show there is a fixed point." Tags="&lt;real-analysis&gt;&lt;homework&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" />
  <row Id="9729" PostTypeId="2" ParentId="9622" CreationDate="2010-11-10T16:21:59.640" Score="1" ViewCount="" Body="&lt;p&gt;If you can calculate all the derivatives at a point, say t=0, you can create a Laurent series around zero.  As long as all the poles of the function are outside a circle of radius $\pi$, this will converge to the function.  You can then integrate term by term to get the integral.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-11-10T16:21:59.640" />
  <row Id="9730" PostTypeId="1" AcceptedAnswerId="9734" CreationDate="2010-11-10T16:22:49.767" Score="0" ViewCount="94" Body="&lt;p&gt;Imagine I had two sets of samples. The only information I have about them is their&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Size&lt;/li&gt;&#xA;&lt;li&gt;Average&lt;/li&gt;&#xA;&lt;li&gt;Standard deviation&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Is it possible to calculate the standard deviation of a third set that is composed by the union of all samples in both sets?&#xA;And what if the original sets had the same size?&lt;/p&gt;&#xA;" OwnerUserId="3364" LastEditorUserId="3364" LastEditorDisplayName="" LastEditDate="2010-11-10T16:58:56.460" LastActivityDate="2010-11-10T16:58:56.460" Title="Is it possible to calculate the standard deviation without the samples?" Tags="&lt;statistics&gt;&lt;standard-deviation&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="9731" PostTypeId="1" AcceptedAnswerId="9743" CreationDate="2010-11-10T16:30:51.340" Score="3" ViewCount="564" Body="&lt;p&gt;I'm stuck with this induction proof:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, far, given:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;T(1)  = 2&lt;br&gt;&#xA;T(n)  = 2T(n/2)+2&lt;br&gt;&#xA;      = 2(2T(n/[2^2])+2) + 2&lt;br&gt;&#xA;      = [2^2]T(n/[2^2]) + [2^2] + 2&lt;br&gt;&#xA;      = 2^2 + [2^2] + 2&lt;br&gt;&#xA;      = [2^3]T(n/[2^3]) + [2^3] + [2^2] + 2&lt;br&gt;&#xA;      = [2^3]T(n/[2^3]) + 2{[2^2] + [2^1] + 1} &#xA;.&lt;br&gt;&#xA;.&lt;br&gt;&#xA;.&lt;br&gt;&#xA;      = [2^k]T(n/[2^k]) + 2{2^(k) - 1}&#xA;How then do I show this to be correct (the proof). So far I have:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let (n/[2^k]) = 1&lt;br&gt;&#xA;=&gt; n = 2^k&lt;br&gt;&#xA;So, T(n) = nT(1) + 2(n  - 1)&#xA;T(n) = 4n - 2&#xA;//This is where I'm stuck&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proof (by induction):&#xA;When n = 1, T(1) = 2.&#xA;Assume T(k) is true [T(n) = 4n - 2]&#xA;//THIS IS WHERE I'M STUCK&lt;/p&gt;&#xA;" OwnerUserId="3366" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2010-11-10T23:27:34.727" LastActivityDate="2010-11-10T23:27:34.727" Title="Prove by induction (T(n) = 2T(n/2) + 2)" Tags="&lt;induction&gt;" AnswerCount="3" CommentCount="3" />
  <row Id="9732" PostTypeId="2" ParentId="9703" CreationDate="2010-11-10T16:32:05.263" Score="2" ViewCount="" Body="&lt;p&gt;Why not just take out a calculator, let her punch in any number she likes, and then let her hit square root over and over.  The limit is 1, and to show $2^0=1$ will take some explaining, but i think it's the best way to think about this in the long term.  Actually this sequence will allow you to later on introduce $log$ and $e$ in a simple way.&lt;/p&gt;&#xA;" OwnerUserId="1417" LastActivityDate="2010-11-10T16:32:05.263" />
  <row Id="9733" PostTypeId="2" ParentId="9703" CreationDate="2010-11-10T16:32:35.240" Score="18" ViewCount="" Body="&lt;p&gt;How about this: There's always an implicit 1 in the expansion:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2^3 = 2 * 2 * 2 * 1 = 8&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2^2 = 2 * 2 * 1 = 4&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2^1 = 2 * 1 = 2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2^0 = 1 = 1&lt;/p&gt;&#xA;" OwnerUserId="3367" LastActivityDate="2010-11-10T16:32:35.240" CommentCount="1" />
  <row Id="9734" PostTypeId="2" ParentId="9730" CreationDate="2010-11-10T16:37:25.803" Score="1" ViewCount="" Body="&lt;p&gt;Yes, it is possible.  The equation should be in your book, or you can look at &lt;a href=&quot;http://en.wikipedia.org/wiki/Standard_deviation&quot; rel=&quot;nofollow&quot;&gt;the Wikipedia page&lt;/a&gt; under the heading Population-based statistics&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-11-10T16:37:25.803" />
  <row Id="9735" PostTypeId="2" ParentId="9648" CreationDate="2010-11-10T16:38:25.087" Score="4" ViewCount="" Body="&lt;p&gt;If I remember correctly, this approach with a Gauss sum (associated to an 8th root of 1 in a finite field) is used in the first page or two of Serre's &lt;em&gt;Cours d'Arithmetique&lt;/em&gt; to determine the &quot;supplementary law&quot; for $\left( \frac{2}{p} \right)$.  The magical algebraic identity in this case is that if $a^4 = -1$, then $(a \pm 1/a)^2 = \pm 2$.&lt;/p&gt;&#xA;" OwnerUserId="467" LastActivityDate="2010-11-10T16:38:25.087" />
  <row Id="9736" PostTypeId="2" ParentId="9730" CreationDate="2010-11-10T16:41:27.543" Score="0" ViewCount="" Body="&lt;p&gt;As Ross suggested, from Wikipedia:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Standard deviations of non-overlapping (X ∩ Y = ∅) sub-populations can be aggregated as follows if the size (actual or relative to one another) and means of each are known:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://upload.wikimedia.org/math/9/5/1/951ab01cabd025deadd5bf497bae971f.png&quot; rel=&quot;nofollow&quot;&gt;http://upload.wikimedia.org/math/9/5/1/951ab01cabd025deadd5bf497bae971f.png&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3364" LastActivityDate="2010-11-10T16:41:27.543" />
  <row Id="9737" PostTypeId="2" ParentId="9731" CreationDate="2010-11-10T16:50:41.003" Score="1" ViewCount="" Body="&lt;p&gt;If you calculate $T(n)$ you can observe that $T(n)=4n-2$.  To prove that by induction, observe that it is true for $n=1$ as $T(1)=2=4*1-2$.  Then assume it is true for $T(n)$ and prove that it is true for $T(2n)$.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2010-11-10T17:32:24.167" LastActivityDate="2010-11-10T17:32:24.167" CommentCount="2" />
  <row Id="9738" PostTypeId="2" ParentId="9703" CreationDate="2010-11-10T17:08:18.947" Score="8" ViewCount="" Body="&lt;p&gt;I would try to explain it in terms of exponents&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$2=2^{1}=2^{1+0}=2^{1}\times 2^{0}=2\times 2^{0}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and by a division&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\dfrac{2}{2}=\dfrac{2}{2}\times 2^{0}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$1=1\times 2^{0}=2^{0}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: Applying the same argument to any number different from $0$ gives the same result.&lt;/p&gt;&#xA;" OwnerUserId="752" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2010-11-10T18:59:22.567" LastActivityDate="2010-11-10T18:59:22.567" />
  <row Id="9739" PostTypeId="2" ParentId="9688" CreationDate="2010-11-10T17:19:50.370" Score="2" ViewCount="" Body="&lt;p&gt;As my approach is different to that of Willie Wong, I give it here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume $f$ is in &lt;em&gt;some&lt;/em&gt; $L^{p_0}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $(p_n)$ be such that $0 &amp;lt; p_n &amp;lt; p_0$ and $p_n \downarrow 0$. Define:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$g_n(x) = \frac{1}{p_0} (|f(x)|^{p_0} - 1) - \frac{1}{p_n} (|f(x)|^{p_n} - 1).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Recall that $\frac{1}{p} (x^p - 1) \downarrow \log x$ as $p \downarrow 0$ for all $x &gt; 0$. So let the $g_n \to g$. Then by the Monotone Convergence Theorem we have that&#xA;$$\int g_n \uparrow \int g.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This implies that &#xA;$$\int \frac{1}{p_n} (|f|^{p_n} -1) \downarrow \int \log |f|.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, by Jensen's inequality we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\text{exp} \left ( \int \log |f| \right ) \leq \left ( \int |f|^{p_n} \right )^\frac{1}{p_n}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and by the inequality $|t| \leq \text{exp}(|t| - 1)$ we get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left ( \int |f|^{p_n} \right )^\frac{1}{p_n} \leq \text{exp} \left ( \int \frac{1}{p_n} (|f|^{p_n} - 1) \right ).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, this together with our previous claim proves the theorem.&lt;/p&gt;&#xA;" OwnerUserId="1120" LastActivityDate="2010-11-10T17:19:50.370" />
  <row Id="9740" PostTypeId="2" ParentId="9608" CreationDate="2010-11-10T17:30:15.913" Score="0" ViewCount="" Body="&lt;p&gt;OK, I think I've found a closed form (someone double check my math here).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the CDF of x (for x&gt;0) is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\text{erf}\left(\frac{x}{\sqrt{2} \sqrt{t}}\right)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the PDF (for x&gt;0) is clearly:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{\sqrt{\frac{2}{\pi }} e^{-\frac{x^2}{2 t}}}{\sqrt{t}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To compute the probability that the maximum between t1 and t2 is exactly x, we compute the probability that x[t1] == y (normal distribution w SD of Sqrt[t1]):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{e^{-\frac{y^2}{2 \text{t1}}}}{\sqrt{2 \pi } \sqrt{\text{t1}}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and then multiply the probability that the max will be x-y between t1 and t2 (only valid for y&amp;lt;=x):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{\sqrt{\frac{2}{\pi }} e^{-\frac{(x-y)^2}{2&#xA;(\text{t2}-\text{t1})}}}{\sqrt{\text{t2}-\text{t1}}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we integrate this wrt y from -Infinity to x:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{e^{-\frac{(x-y)^2}{2 (\text{t2}-\text{t1})}-\frac{y^2}{2\text{t1}}}}{\pi  \sqrt{\text{t1}} \sqrt{\text{t2}-\text{t1}}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and finally get:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{e^{-\frac{x^2}{2 \text{t2}}} \left(\text{erf}\left(\frac{x\sqrt{\frac{1}{\text{t1}}-\frac{1}{\text{t2}}}}{\sqrt{2}}\right)+1\right)}{\sqrt{2 \pi } \sqrt{\text{t2}}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I realize all of this was in the previous post, but I wanted to play around with LaTex.&lt;/p&gt;&#xA;" OwnerUserId="2469" LastEditorUserId="2469" LastEditorDisplayName="" LastEditDate="2010-11-10T18:00:28.353" LastActivityDate="2010-11-10T18:00:28.353" CommentCount="1" />
  <row Id="9741" PostTypeId="2" ParentId="9692" CreationDate="2010-11-10T17:31:24.697" Score="7" ViewCount="" Body="&lt;p&gt;This is essentially Exercise 6.5 in Folland's &lt;em&gt;Real Analysis&lt;/em&gt;.  Let $(X,\mathcal{F},\mu)$ be a measure space, and let $$m = \inf\{\mu(A) : A \in \mathcal{F}, \mu(A) &gt; 0\}$$&#xA;$$M = \sup\{\mu(A) : A \in \mathcal{F}, \mu(A) &amp;lt; \infty\}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $0 &amp;lt; p &amp;lt; q &amp;lt; \infty$, it is a fact that $L^p(\mu) \subset L^q(\mu)$ iff $m &gt; 0$, and $L^q(\mu) \subset L^p(\mu)$ iff $M &amp;lt; \infty$ (which in particular holds when $\mu$ is finite).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So for a finite measure $\mu$, it is necessary and sufficient that for some $m &gt;  0$ every set either has measure 0 or has measure at least $m$.  This is going to force your space to be &quot;effectively&quot; finite in some sense.&lt;/p&gt;&#xA;" OwnerUserId="822" LastActivityDate="2010-11-10T17:31:24.697" CommentCount="1" />
  <row Id="9742" PostTypeId="2" ParentId="9731" CreationDate="2010-11-10T17:31:45.553" Score="0" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\: $ From the first few values we guess $\rm\ T(2^n)\ =\ 2^{n+2}-2\ $ and induction confirms it:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\rm T(2^{n+1})\ =\ 2\ T(2^n) + 2 \ =\ 2\ (2^{n+2}-2) +\ 2\ =\ 2^{n+3} - 2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One can extend $\rm\:T\:$ to $\:\mathbb N\:$ by defining $\rm\ T(2k+1) = 2\ T(k+1)-2 $ and now one easily proves by induction that $\rm\ T(k) = 4\:k-2\ $ since&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\quad\quad\quad\quad\quad\quad T(2k+1)\ =\ 2\ T(k+1)-2\ =\ 2\ (4k+2)-2\ =\ 4(2k+1)-2 $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\quad\quad\quad\quad\quad\quad\quad\quad\ T(2k)\ =\ 2\ \ \ \ T(k)\ \ +\ \ \: 2\ =\ 2\ (4k-2) + 2\ =\ 4 (2k) - 2 $&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-11-10T19:16:05.943" LastActivityDate="2010-11-10T19:16:05.943" />
  <row Id="9743" PostTypeId="2" ParentId="9731" CreationDate="2010-11-10T17:36:19.217" Score="1" ViewCount="" Body="&lt;p&gt;Here is how I attempt this from the point you left off:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know that &lt;code&gt;T(1) = 2&lt;/code&gt;. We are trying to prove that &lt;code&gt;T(n) = 4n-2&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is trivially true for n = 1:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;T(1) = 4(1) - 2&#xA;     = 4 - 2&#xA;     = 2&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Assume&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;T(k) = 4k - 2&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;From the original definition:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;T(k+1) = 2T( [k+1] / 2 ) + 2&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;//since we assumed up to T(k) is correct and (k+1)/2 is less than T(k), we substitute:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, We have &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;2( 4( (k+1) /2) - 2  ) + 2&#xA;= 4(k+1)- 4 + 2&#xA;= 4(k+1) - 2&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Proven.&lt;/p&gt;&#xA;" OwnerUserId="3371" LastEditorUserId="3371" LastEditorDisplayName="" LastEditDate="2010-11-10T20:01:19.530" LastActivityDate="2010-11-10T20:01:19.530" CommentCount="3" />
  <row Id="9744" PostTypeId="2" ParentId="9724" CreationDate="2010-11-10T18:01:51.110" Score="2" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT:&lt;/strong&gt; When do you say that a set is bounded. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;em&gt;A set $X \subset \mathbb{R}$ is said to be bounded if you can find a $M \in \mathbb{R}$ such that $|x| \leq M$ for all $x \in X$.&lt;/em&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Now consider your cases:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;$X = \{ x | x = \frac{1}{n} ; \ n \in \mathbb{N}\}$ *&lt;em&gt;NOTE:&lt;/em&gt;* $\frac{1}{n} \leq 1$ for all $n \in \mathbb{N}$&#xD;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Similarly the Second one.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Now , the third one is unbounded. (Why?). Assume $2^{n} \leq M$ for all $n \in \mathbb{N}$ and obtain a contradiction.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Fourth one follows from the definition.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" LastActivityDate="2010-11-17T02:56:52.863" CommentCount="1" />
  <row Id="9745" PostTypeId="1" CreationDate="2010-11-10T18:32:59.247" Score="1" ViewCount="108" Body="&lt;p&gt;Let $f \in L^{1}(\mathbb{R})$ such that $f(x) &gt;0$ for almost all $x \in \mathbb{R}$. Let A be a Lebesgue measurable set such that $\int_{A} f =0$. Prove $m(A) = 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;wlog we can assume $f(x)&gt;0$ everywhere define $A_{n}={x \in A: f(x) &gt; \frac{1}{n}}$&#xA;so since $f(x) &gt; 0$ everywhere, $\int_{A_n} f dm \leq \int_{A} fdm = 0$. Then it is easy to see that A is a countable union of sets of measure 0, so A has measure 0 in this case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is: why do we need A to be Lebesgue measurable? doesn't the above proof works for any measurable set? I still don't see the difference.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="4219" LastActivityDate="2010-11-10T18:43:29.333" Title="Integral, Lebesgue measure" Tags="&lt;measure-theory&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" />
  <row Id="9746" PostTypeId="1" AcceptedAnswerId="9806" CreationDate="2010-11-10T18:43:28.880" Score="1" ViewCount="140" Body="&lt;p&gt;It is claimed in some logic books that the compactness theorem of first-order logic can be proven using Tychnoff's theorem from topology.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, to me this feels very strange because I consider logic as something &quot;lower&quot; in the sense that the theorems of topology follow from the axioms and theorems in logic. How can we avoid circularities then?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another small question I have is that if we have $A \iff B$ and we prove both directions with contraposition, does there exist a direct proof? You could say we could just follow the steps backwards, but is this always possible?&lt;/p&gt;&#xA;" OwnerUserId="1120" LastActivityDate="2010-11-11T02:47:55.600" Title="Tychonoff and compactness (logic) and another small logic question." Tags="&lt;topology&gt;&lt;logic&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="9747" PostTypeId="2" ParentId="9745" CreationDate="2010-11-10T18:43:29.333" Score="1" ViewCount="" Body="&lt;p&gt;When they write $L^1(\mathbb{R})$, the measure on $\mathbb{R}$ hasn't been specified, so it's assumed to be Lebesgue measure.  If you had $L^1(\mu)$ for an arbitrary measure $\mu$ your argument would work just as well.&lt;/p&gt;&#xA;" OwnerUserId="3296" LastActivityDate="2010-11-10T18:43:29.333" CommentCount="1" />
  <row Id="9748" PostTypeId="1" CreationDate="2010-11-10T19:03:33.667" Score="0" ViewCount="80" Body="&lt;p&gt;Let X be a sigma finite measure space with $p &gt; 0$. The book states the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There exists a map $f \in L^{p}(X)$ such that $f&gt;0$ and f is bounded by $1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proof: Let $C_{n}$ be a sequence of disjoint sets of finite measure such that X is the union&#xA;of these sets. Now just put $f(x) = (\frac{2^{-n}}{1+\mu(C_{n})})^{\frac{1}{p}}$ whenever $x \in C_{n}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is pretty clear the boundedness of f. How do we show that $f \in L^{p}(X)$ ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it because $\int_{X} f \leq \mu(C_{n}) &amp;lt; \infty$ ?&lt;/p&gt;&#xA;" OwnerUserId="4219" LastEditorUserId="4219" LastEditorDisplayName="" LastEditDate="2010-11-10T19:47:18.573" LastActivityDate="2010-11-10T19:47:22.303" Title="Sigma finite and construction of map" Tags="&lt;measure-theory&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="9749" PostTypeId="1" AcceptedAnswerId="9751" CreationDate="2010-11-10T19:08:25.143" Score="4" ViewCount="353" Body="&lt;p&gt;What are some interesting applications of the Mean Value Theorem for derivatives (both the 'extended' or 'non-extended' versions as seen &lt;a href=&quot;http://en.wikipedia.org/wiki/Mean_value_theorem&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; are of interest)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So far I've seen some trivial applications like finding the number of roots of a polynomial equation. What are some more interesting applications of it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm asking this as I'm not exactly sure why MVT is so important - so examples which focus on explaining that would be appreciated.&lt;/p&gt;&#xA;" OwnerUserId="181" LastActivityDate="2010-11-13T12:34:04.600" Title="Applications of the Mean Value Theorem" Tags="&lt;calculus&gt;&lt;applications&gt;" AnswerCount="7" CommentCount="2" FavoriteCount="1" />
  <row Id="9750" PostTypeId="2" ParentId="9749" CreationDate="2010-11-10T19:20:24.837" Score="2" ViewCount="" Body="&lt;p&gt;There are applications.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For an important one, Taylor series proof relies on it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An other application I like is to quickly come up with and prove inequalities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example 1)&#xA;$\displaystyle |\cos x - \cos y| \le |x - y|$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example 2)&#xA;$\displaystyle \frac{1}{2\sqrt{n+1}} &amp;lt; \sqrt{n+1} - \sqrt{n} &amp;lt; \frac{1}{2\sqrt{n}}$&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-11-10T19:20:24.837" />
  <row Id="9751" PostTypeId="2" ParentId="9749" CreationDate="2010-11-10T19:23:49.383" Score="4" ViewCount="" Body="&lt;p&gt;There are several applications of the Mean Value Theorem. It is one of the most important theorems in analysis and is used all the time. I've listed $5$ important results below. I'll provide some motivation to their importance if you request.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$1)$ If $f: (a,b) \rightarrow \mathbb{R}$ is differentiable and $f'(x) = 0$ for all $x \in (a,b)$, then $f$ is constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$2)$ &lt;strong&gt;Leibniz's rule:&lt;/strong&gt; Suppose $ f : [a,b] \times [c,d] \rightarrow \mathbb{R}$ is a continuous function with $\partial f/ \partial x$ continuous. Then the function $F(x) = \int_{c}^d f(x,y)dy$ is derivable with derivative $$ F'(x) = \int_{c}^d \frac{\partial f}{\partial x} (x,y)dy.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$3)$ &lt;strong&gt;L'Hospital's rule&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$4)$ If $A$ is an open set in $\mathbb{R}^n$ and $f:A \rightarrow \mathbb{R}^m$ is a function with continuous partial derivatives, then $f$ is differentiable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$5)$ &lt;strong&gt;Symmetry of second derivatives:&lt;/strong&gt; If $A$ is an open set in $\mathbb{R}^n$ and $f:A \rightarrow \mathbb{R}$ is a function of class $C^2$, then for each $a \in A$, $$\frac{\partial^2 f}{\partial x_i \partial x_j} (a) = \frac{\partial^2 f}{\partial x_j \partial x_i} (a)$$&lt;/p&gt;&#xA;" OwnerUserId="329" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2010-11-10T20:06:45.493" LastActivityDate="2010-11-10T20:06:45.493" />
  <row Id="9752" PostTypeId="2" ParentId="9748" CreationDate="2010-11-10T19:27:51.897" Score="1" ViewCount="" Body="&lt;p&gt;Note that $X = \displaystyle \cup_{n=1}^{\infty} C_n$ such that $\mu(C_n) &amp;lt; \infty$. (This is because $X$ is a sigma finite measure space under $\mu$. Also, note that we can choose $C_n$ such that $C_n$'s are mutually disjoint. So, we can assume that $C_i \cap C_j = \phi$. (We need this to define $f(x)$ uniquely. You could of course circumvent this by defining $f$ uniquely on the intersection incase they are not disjoint)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $f(x) = (\frac{2^{-n}}{1+\mu(C_n)})^{1/p}$, whenever $x \in C_n$. (If $C_n$'s are not disjoint, define $f$ to be the infimum of $(\frac{2^{-n}}{1+\mu(C_n)})^{1/p}$ on the intersection of the sets). Clearly, $f(x) &gt; 0$ (Since each of the $\mu(C_n) &amp;lt; \infty$). Also, $f(x)$ is bounded above by $1$. (By construction of $f(x)$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence $\int_X |f|^p d\mu = \int_{ \displaystyle \cup_{n=1}^{\infty} C_n} |f|^p d\mu \leq \displaystyle \sum_{n=1}^{\infty} \int_{C_n} |f|^p d\mu = \displaystyle \sum_{n=1}^{\infty} \int_{C_n} (\frac{2^{-n}}{1+\mu(C_n)}) d\mu$&#xA;$ = \displaystyle \sum_{n=1}^{\infty} (\frac{2^{-n}}{1+\mu(C_n)}) \mu(C_n) &amp;lt; \displaystyle \sum_{n=1}^{\infty} \frac{1}{2^n} = 1$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus $f \in \mathcal{L}^p(X)$.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2010-11-10T19:47:22.303" LastActivityDate="2010-11-10T19:47:22.303" CommentCount="4" />
  <row Id="9753" PostTypeId="2" ParentId="9749" CreationDate="2010-11-10T19:28:12.967" Score="3" ViewCount="" Body="&lt;p&gt;Some more applications:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;If the derivative of a function $f$ is everywhere strictly positive, then f is a strictly increasing function&lt;/em&gt;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;Suppose $f$ is differentiable on whole of $\mathbb{R}$, and $f'(x)$ is a constant. Then $f$ is linear.&lt;/em&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;Mean Value theorem plays an important role in the proof of &lt;a href=&quot;http://en.wikipedia.org/wiki/Fundamental_theorem_of_calculus&quot; rel=&quot;nofollow&quot;&gt;Fundamental Theorem of Calculus&lt;/a&gt;.&lt;/em&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;Suppose $f$ is continuous on $[a,b]$ and $f'$ exists and is bounded on the interior, then $f$ is of &lt;a href=&quot;http://en.wikipedia.org/wiki/Bounded_variation&quot; rel=&quot;nofollow&quot;&gt;Bounded Variation&lt;/a&gt; on $[a,b]$&lt;/em&gt;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" LastActivityDate="2010-11-10T19:43:08.847" />
  <row Id="9754" PostTypeId="2" ParentId="9749" CreationDate="2010-11-10T19:47:12.880" Score="1" ViewCount="" Body="&lt;p&gt;The generalization of the Mean Value Theorem known as &lt;a href=&quot;http://en.wikipedia.org/wiki/Cauchy%27s_mean_value_theorem#Cauchy.27s_mean_value_theorem&quot; rel=&quot;nofollow&quot;&gt;Cauchy's Mean Value Theorem&lt;/a&gt; can be used to prove L'Hopital's Rule. See for instance the following &lt;a href=&quot;http://en.wikipedia.org/wiki/L%27H%C3%B4pital%27s_rule#Proof_of_l.27H.C3.B4pital.27s_rule&quot; rel=&quot;nofollow&quot;&gt;proof&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="900" LastActivityDate="2010-11-10T19:47:12.880" />
  <row Id="9755" PostTypeId="2" ParentId="9703" CreationDate="2010-11-10T19:52:45.870" Score="3" ViewCount="" Body="&lt;p&gt;I think it's quiet simple to understand:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The base point is 1 (I'm not talking about the base... read on...).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;from this initial point of 1, you are applying the base to the result n times:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2^3 = 1 * (2 *2 *2) = 8&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2^-3 = ((( 1 / 2) /2) / 2)  = 1/8&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The base point is always 1 so&#xA;when you are in the 0 point (n^0) you don't have to multiply the very basic 1 so you have &quot;1&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the starting point is the &quot;1&quot; not the base.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;edit: I've just noticed that Zarkonnen already gave this answer, yet I think that the way I presented the answer is easier for a child to understand- So Zarkonnen, I hope you are ok with this post ;)&lt;/p&gt;&#xA;" OwnerUserId="3375" LastEditorUserId="3375" LastEditorDisplayName="" LastEditDate="2010-11-10T20:03:24.203" LastActivityDate="2010-11-10T20:03:24.203" />
  <row Id="9756" PostTypeId="2" ParentId="9703" CreationDate="2010-11-10T20:12:16.600" Score="7" ViewCount="" Body="&lt;p&gt;I'd demonstrate this using a pattern.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$2^3 = 8$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$2^2 = 4$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$2^1 = 2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$2^0 = 1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$2^{-1} = 1/2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$2^{-2} = 1/4$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you decrease the exponent, you divide by 2. So, when you go from 2&lt;sup&gt;1&lt;/sup&gt; to 2&lt;sup&gt;0&lt;/sup&gt;, of course you divide by 2, which gives you 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From there, you can segue into negative exponents, if you'd like. Just keep dividing by 2.&lt;/p&gt;&#xA;" OwnerUserId="2994" LastActivityDate="2010-11-10T20:12:16.600" CommentCount="1" />
  <row Id="9757" PostTypeId="2" ParentId="9746" CreationDate="2010-11-10T20:14:45.803" Score="1" ViewCount="" Body="&lt;p&gt;For the second question, suppose that in some proof system you can prove $p(x) \Leftrightarrow q(x)$ for some expressions $p,q$, and only $p(x)\Rightarrow q(x)$ requires proof by contradiction. Then to prove $p(x) \land q(y) \Leftrightarrow q(x) \land p(y)$ you probably need proof by contradiction on both sides.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you only know that $p(x) \Rightarrow q(x)$ requires proof by contradiction ($p(x) \Leftarrow q(x)$ not necessarily being true), then $p(x) \Rightarrow p(x) \land q(x)$ also probably requires proof by contradiction, and this time both sides are equivalent.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-10T20:14:45.803" CommentCount="2" />
  <row Id="9758" PostTypeId="1" AcceptedAnswerId="9761" CreationDate="2010-11-10T20:17:32.560" Score="4" ViewCount="332" Body="&lt;p&gt;What is an intuitive explanation of a positive-semidefinite matrix? Or a simple example which gives more intuition for it rather than the bare definition. Say $x$ is some vector in space and $M$ is some operation on vectors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The definition is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A $n$ × $n$ Hermitian matrix M is called &lt;em&gt;positive-semidefinite&lt;/em&gt; if&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x^{*} M x \geq 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for all $x \in \mathbb{C}^n$ (or, all $x \in \mathbb{R}^n$ for the real matrix), where $x^*$ is the conjugate transpose of $x$.&lt;/p&gt;&#xA;" OwnerUserId="915" LastEditorUserId="915" LastEditorDisplayName="" LastEditDate="2010-11-10T20:29:49.630" LastActivityDate="2011-03-17T02:34:32.563" Title="Intuitive explanation of a positive-semidefinite matrix" Tags="&lt;linear-algebra&gt;&lt;matrices&gt;" AnswerCount="4" CommentCount="3" FavoriteCount="2" />
  <row Id="9759" PostTypeId="2" ParentId="9749" CreationDate="2010-11-10T20:21:34.317" Score="1" ViewCount="" Body="&lt;p&gt;If $f$ and $g$ are two differentiable functions on an interval $I$ and for&#xA;all $x\in I$, $f^{\prime }(x)=g^{\prime }(x)$, then the difference $f(x)-g(x)&#xA;$ is constant on $I$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Physical interpretation (assuming the movement is uninterruptible): If the average speed of a car between two locations was $V$ km/h, then there was at least one instant where the speed indicator displayed $V$ km/h. &lt;/p&gt;&#xA;" OwnerUserId="752" LastActivityDate="2010-11-10T20:21:34.317" />
  <row Id="9760" PostTypeId="1" CreationDate="2010-11-10T20:22:58.653" Score="12" ViewCount="289" Body="&lt;p&gt;We have three counters, $i, j, k$, all initialized to zero. Each step consists of adding or subtracting one from one of the counters, so $(\Delta i, \Delta j, \Delta k)$ is selected among $(\pm1, 0, 0), (0, \pm1, 0), (0, 0, \pm1)$, each with probability 1/6. What is the expected number of steps before all the counters are N modulo 2N at the same time?&lt;/p&gt;&#xA;" OwnerUserId="2649" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-11-12T04:33:41.310" LastActivityDate="2010-11-13T20:45:14.273" Title="Expected number of steps before three counters reach N modulo 2N at the same time" Tags="&lt;probability&gt;" AnswerCount="4" CommentCount="8" FavoriteCount="9" />
  <row Id="9761" PostTypeId="2" ParentId="9758" CreationDate="2010-11-10T20:24:34.173" Score="6" ViewCount="" Body="&lt;p&gt;One intuitive definition is as follows. Multiply any vector with a positive definite matrix. The angle between the original vector and the resultant vector will always be less than $\frac{\pi}{2}$. In some sense the positive definite matrix tries to keep the vector within a certain half space of the vector. This is analogous to what a positive number does to a real variable. Multiply it and it only stretches or contracts the number but never reflects it about the origin.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastActivityDate="2010-11-10T20:24:34.173" CommentCount="1" />
  <row Id="9762" PostTypeId="1" AcceptedAnswerId="9764" CreationDate="2010-11-10T20:25:23.327" Score="1" ViewCount="95" Body="&lt;p&gt;Injective functions are not the same thing as injective &lt;em&gt;correspondences&lt;/em&gt;, correct? Injective functions are a subset of the injective &lt;em&gt;correspondences&lt;/em&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example in $y^2 = x$, y is not a function of x, but this is still a correspondence between $\mathbb{R} \rightarrow \mathbb{R}$. Can we say that the correspondence, $C:X \rightarrow Y$ is injective? (Since, if $C(x_1) = C(x_2)$ then $x_1 = x_2$.) Can we say it is also a surjective &lt;em&gt;correspondence&lt;/em&gt; since every y-value has at least one (mostly two) pre-images?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or am I way off base applying these definitions to things that are not functions?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Last question. &quot;Mappings&quot; are, as the Wikipedia suggests, the same thing as functions. (For some reason I thought mappings were correspondences. In fact, it is in my old notes. My old notes are wrong, right? )&lt;/p&gt;&#xA;" OwnerUserId="2563" LastEditorUserId="2563" LastEditorDisplayName="" LastEditDate="2010-11-10T20:37:03.127" LastActivityDate="2010-11-10T20:44:05.773" Title="Injective functions vs. correspondences." Tags="&lt;abstract-algebra&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="9763" PostTypeId="2" ParentId="9758" CreationDate="2010-11-10T20:40:17.200" Score="2" ViewCount="" Body="&lt;p&gt;First I'll tell you how I think about Hermitian positive-definite matrices.  A Hermitian positive-definite matrix $M$ defines a sesquilinear inner product $\langle Mv, w \rangle = \langle v, Mw \rangle$, and in fact every inner product on a finite-dimensional inner product space $V$ has this form.  In other words it is a way of computing angles between vectors, or a way of projecting vectors onto other vectors; over the real numbers it is the key ingredient to doing Euclidean geometry.  An inner product can be recovered from the norm $\langle Mv, v \rangle = \langle v, Mv \rangle$ it induces, and a norm in turn can be recovered from its unit sphere $\{ v : \langle Mv, v \rangle = 1 \}$.  This unit sphere is a distorted version of the usual unit sphere; the distortions will occur along axes corresponding to the eigenvectors of $M$, and the amount of distortion corresponds to the (inverses of the) corresponding eigenvalues.  For example when $\dim V = 2$ it is an ellipse and when $\dim V = 3$ it is an ellipsoid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A Hermitian positive-semidefinite matrix $M$ no longer describes an inner product because it is not necessarily positive-definite, but it still defines a sesquilinear form.  It also defines a function $\langle Mv, v \rangle$ which is no longer a norm because it is not necessarily positive-definite; some people call these &quot;pseudonorms,&quot; I think.  The corresponding unit sphere $\{ v : \langle Mv, v \rangle = 1 \}$ might now be lower-dimensional than the usual unit sphere, depending on how many eigenvalues are equal to zero; for example if $\dim V = 3$ it might be an ellipsoid, or an ellipse, or two points.  &lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-11-10T20:40:17.200" CommentCount="3" />
  <row Id="9764" PostTypeId="2" ParentId="9762" CreationDate="2010-11-10T20:44:05.773" Score="2" ViewCount="" Body="&lt;p&gt;What you're describing is universally known as a &quot;relation.&quot;  It is true that you can have an injective relation which is not a function.  Unfortunately, there are different uses for the term &quot;correspondence&quot; -- used alone, it can denote a general relation, but &quot;one-to-one correspondence&quot; can also denote a bijection.  The safest thing to do is probably to avoid using the word altogether.&lt;/p&gt;&#xA;" OwnerUserId="3296" LastActivityDate="2010-11-10T20:44:05.773" CommentCount="2" />
  <row Id="9765" PostTypeId="2" ParentId="9758" CreationDate="2010-11-10T20:44:56.280" Score="3" ViewCount="" Body="&lt;p&gt;Let's consider the set $E$ of all vectors $y=Mx$, where $x\in\mathbb R^n$ belongs to the unit sphere (i.e. $\|x\|=1$). In other words, $E$ is the image of the unit sphere under the linear transformation. If the matrix is non-degenerate, $E$ is an $n$-dimensional ellipsoid. But if we assume additionally that $M$ is symmetric then we can say much more about the structure of $E$. Namely, the directions of the axes of the ellipsoid are pairwise orthogonal and represented by the eigenvectors of the matrix. Moreover, the lengths of the semiaxes are given by the corresponding eigenvalues.&lt;/p&gt;&#xA;" OwnerUserId="723" LastEditorUserId="723" LastEditorDisplayName="" LastEditDate="2010-11-10T23:50:20.640" LastActivityDate="2010-11-10T23:50:20.640" />
  <row Id="9766" PostTypeId="1" CreationDate="2010-11-10T20:56:28.677" Score="2" ViewCount="129" Body="&lt;p&gt;Hello&#xA;i have to solve the following problem&#xA;find an $\displaystyle f(k)$ where $\displaystyle S_k=\theta(f(k))$&#xA;where $\displaystyle S_k =\sum_{n=1}^{k^2-1} \sqrt{n}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried first of all to calculate or &quot;limit&quot; my sum using integral rule so i came up with&#xA;$\displaystyle \frac{2(k^2-1)^{3/2}}{3} \leq S_k \leq \frac{2(k^3-1)}{3}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but after that i am in a dead end as i do not know $\displaystyle S_k$ so i can not simplify anything&#xA;can anyone help how to proceed this problem ?&#xA;thanks &lt;/p&gt;&#xA;" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-11-10T21:02:28.970" LastActivityDate="2010-11-10T21:28:37.123" Title="Problem calculating theta notation" Tags="&lt;asymptotics&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="9767" PostTypeId="2" ParentId="9766" CreationDate="2010-11-10T21:04:28.237" Score="5" ViewCount="" Body="&lt;p&gt;As far as I see you already have the solution there. In Theta-notation only the fastest growing terms are important. And that is the same for the upper and lower bound.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, you have: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Theta(k^3) \leq S_k \leq \Theta(k^3).$$&lt;/p&gt;&#xA;" OwnerUserId="915" LastEditorUserId="915" LastEditorDisplayName="" LastEditDate="2010-11-10T21:28:37.123" LastActivityDate="2010-11-10T21:28:37.123" CommentCount="5" />
  <row Id="9768" PostTypeId="1" CreationDate="2010-11-10T21:10:06.663" Score="4" ViewCount="139" Body="&lt;p&gt;This may be embarassingly simple, but I can't see it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $M$ be a Riemannian manifold of dimension $n$; fix $x \in M$, and let $B(x,r)$ denote the geodesic ball in $M$ of radius $r$ centered at $x$.  Let $V(r) = \operatorname{Vol}(B(x,r))$ be the Riemannian volume of $B(x,r)$.  It seems to be the case that for small $r$, $V(r) \sim r^n$, i.e. $V(r)/r^n \to c$ with $0 &amp;lt; c &amp;lt; \infty$.  How is this proved, and where can I find it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given a neighborhood $U \ni x$ and a chart $\phi : U \to \mathbb{R}^n$, certainly $\phi$ has nonvanishing Jacobian, hence (making $U$ smaller if necessary) bounded away from 0.  So $\operatorname{Vol}(\phi^{-1}(B_{\mathbb{R}^n}(\phi(x), r))) \sim r^n$.  But I do not see how to relate the pullback $\phi^{-1}(B_{\mathbb{R}^n}(\phi(x), r))$ of a Euclidean ball to a geodesic ball in $M$.&lt;/p&gt;&#xA;" OwnerUserId="822" LastActivityDate="2010-11-10T22:43:59.873" Title="Volume of a geodesic ball" Tags="&lt;differential-geometry&gt;" AnswerCount="2" />
  <row Id="9769" PostTypeId="2" ParentId="9703" CreationDate="2010-11-10T21:21:09.797" Score="2" ViewCount="" Body="&lt;p&gt;When I was a kid, I had to teach this concept (and fractions) to some kids in a lower grade. The best method is to work backward with division.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2^3 = 8&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2^2 = (2^3)/2 = 8/2 = 4&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2^1 = (2^2)/2 = 4/2 = 2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2^0 = (2^1)/2 = 2/2 = 1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And so on. This is also the only way to wrap your head around negative powers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Incidentally, the best way to teach fractions is by having pizza for dinner.&lt;/p&gt;&#xA;" OwnerUserId="3385" LastEditorUserId="367" LastEditorDisplayName="" LastEditDate="2011-01-12T22:56:28.417" LastActivityDate="2011-01-12T22:56:28.417" CommentCount="1" />
  <row Id="9770" PostTypeId="1" AcceptedAnswerId="9783" CreationDate="2010-11-10T21:27:10.203" Score="5" ViewCount="336" Body="&lt;p&gt;Greetings!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am trying to understand what it means to have an imaginary number in an exponent. What does $x^{i}$ where $x$ is real mean?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've read a &lt;a href=&quot;http://mathforum.org/library/drmath/view/52251.html&quot; rel=&quot;nofollow&quot;&gt;few&lt;/a&gt;  &lt;a href=&quot;http://www.newton.dep.anl.gov/newton/askasci/1995/math/MATH057.HTM&quot; rel=&quot;nofollow&quot;&gt;pages&lt;/a&gt; on this issue, and they all seem to boil down to the same thing:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Any real number $x$ can be written as $e^{\ln{x}}$ (seems obvious enough.)&lt;/li&gt;&#xA;&lt;li&gt;Mumble mumble mumble&lt;/li&gt;&#xA;&lt;li&gt;This is equivalent to $e^{\cos{x} + i\sin{x}}$ &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Clearly I'm missing something in step 2. I understand (at least I think I do) how the complex number $\cos{x} + i\sin{x}$ maps to a point on the unit circle in a complex plane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I am missing, I suppose, is how this point is related to the natural log of $x$. Moreover, I don't understand what complex exponentiation &lt;em&gt;is&lt;/em&gt;. I can understand integer exponentiation as simple repeated multiplication, and I can understand other things (like fractional or negative exponents) by analogy with the operations that undo them. But what does it mean to repeat something $i$ times?&lt;/p&gt;&#xA;" OwnerUserId="3384" LastActivityDate="2011-05-18T18:25:32.603" Title="Understanding imaginary exponents" Tags="&lt;trigonometry&gt;&lt;complex-numbers&gt;&lt;logarithms&gt;" AnswerCount="5" CommentCount="5" FavoriteCount="1" />
  <row Id="9771" PostTypeId="2" ParentId="9768" CreationDate="2010-11-10T21:49:54.190" Score="1" ViewCount="" Body="&lt;p&gt;A Google search for the title of the question finds &lt;a href=&quot;http://projecteuclid.org/DPubS?service=UI&amp;amp;version=1.0&amp;amp;verb=Display&amp;amp;handle=euclid.mmj/1029001150&quot; rel=&quot;nofollow&quot;&gt;this article&lt;/a&gt; where the first five coefficients in the series expansion of the volume (in powers of $r$) are computed. The first term is the same as the Euclidean volume (proportional to $r^n$, in other words); then come higher-order corrections depending on the curvature.&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2010-11-10T21:49:54.190" CommentCount="2" />
  <row Id="9772" PostTypeId="2" ParentId="9770" CreationDate="2010-11-10T21:54:27.420" Score="2" ViewCount="" Body="&lt;p&gt;It is a way to extend your real variable function into the complex domain in a &quot;nice&quot; (read nice as holomorphic) way. You might ask the question why do we need to even extend in the first place? The need for extension comes from the fact that a $n^{th}$ order polynomial with real co-efficients need not have all its roots as real numbers. So you end up with these complex numbers. Once we find that we have these new numbers, we begin to wonder why should we not try to feed these numbers into functions which are defined on the real variables. But then we find that we have a zillion ways to define these functions for these new complex numbers. A &quot;nice&quot; way to extend these functions is to do in a way that the extended function is holomorphic. That is why you extend $e^{i \theta}$ to the complex plane as $\cos(\theta) + i \sin(\theta)$. Once you do this extension then all the other things related to logarithms and exponentiation follows in a logically coherent way. For instance, we can now define $x^i$ (say $x&amp;gt;0$ or else rewrite $x$ as $-y$ where $y&amp;gt;0$ and write $-1$ as $e^{i \pi}$) as $(e^{\log (x)})^{i} = e^{i \log(x)} = \cos(\log(x)) + i \sin(\log(x))$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is how I understand it.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2011-05-18T18:25:32.603" LastActivityDate="2011-05-18T18:25:32.603" CommentCount="2" />
  <row Id="9773" PostTypeId="2" ParentId="9770" CreationDate="2010-11-10T21:56:18.660" Score="2" ViewCount="" Body="&lt;p&gt;Lets say you want to figure out what $x^{a + ib}$ is, then like you mentioned, you start by writing&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x^{a + ib} = e^{a \ln(x) + i b \ln(x)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This can be split up though as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ e^{a \ln(x) + i b \ln(x)} = e^{a \ln(x)} e^{i b \ln(x)} = x^a e^{i b \ln(x)} $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and then you can use euler's formula to take care of the imaginary exponent, so that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x^{a + ib} = x^a \cos(b \ln(x)) + i x^a \sin(b \ln(x)) $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This formula does not provide much intuition as to what is really happening though. You mentioned that you can understand integer exponentiation as simple repeated multiplication, but I don't think that that is the right way to look at it in complex analysis. I think it is much better to look at exponentiation geometrically.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When exponentiating a real number by a complex value, we found that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x^{a + ib} = x^a e^{i b \ln(x)} $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so exponentiating $x$ by $a + ib$ gives you the point in the plane with magnitude $x^a$ and angle $b \ln(x)$.&lt;/p&gt;&#xA;" OwnerUserId="1089" LastActivityDate="2010-11-10T21:56:18.660" CommentCount="2" />
  <row Id="9774" PostTypeId="2" ParentId="9770" CreationDate="2010-11-10T22:07:37.547" Score="0" ViewCount="" Body="&lt;p&gt;The quantity x^i is well-defined if x is a &lt;b&gt;positive&lt;/b&gt; real number.  For negative (or complex) values of x, the evaluation is open to multiple interpretations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have the following.  The natural logarithm of positive real numbers is well-defined, so:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;x^i = exp(i*ln(x))&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bear in mind that for real x &gt; 0, all one knows about ln(x) is that it is real.  It might be positive, negative, or zero depending on the comparison of x to 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the function exp can be defined by a power series that converges everywhere in the complex plane.  As it turns out, that definition of the exponential function implies the following when the argument is purely imaginary (as here):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;x^i = exp(i*ln(x)) = cos(ln(x)) + i*sin(ln(x))&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For more see:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Euler%27s_formula&quot; rel=&quot;nofollow&quot;&gt; [Euler's formula -- Wikipedia]&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;regards, hm&lt;/p&gt;&#xA;" OwnerUserId="3111" LastActivityDate="2010-11-10T22:07:37.547" CommentCount="3" />
  <row Id="9775" PostTypeId="2" ParentId="9594" CreationDate="2010-11-10T22:10:19.700" Score="1" ViewCount="" Body="&lt;p&gt;12^6 is just under 3 million.  So it would probably not work to randomly try six unscrambles.  But it wouldn't be too hard to make a data file of all the positions and their unscramble twists if you can find a reasonable way to search it, like some hash function on a description of the position.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-11-10T22:10:19.700" CommentCount="2" />
  <row Id="9776" PostTypeId="1" AcceptedAnswerId="9778" CreationDate="2010-11-10T22:10:39.067" Score="2" ViewCount="233" Body="&lt;p&gt;How do I calculate the outcome of taking one complex number to the power of another, ie  (a + bi)^(c + di)?&lt;/p&gt;&#xA;" OwnerUserId="3367" LastActivityDate="2010-11-11T11:02:25.487" Title="A complex number to the power of another complex number" Tags="&lt;complex-numbers&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="1" />
  <row Id="9777" PostTypeId="2" ParentId="9770" CreationDate="2010-11-10T22:15:11.397" Score="4" ViewCount="" Body="&lt;p&gt;The complex exponential $e^z$ for complex $z=x+iy$ preserves the law of exponents of the real exponential and satisfies $e^0=1$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;By definition &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$e^z=e^{x+iy}=e^xe^{iy}=e^x(\cos y+\sin y)$$ &#xD;&#xA;&#xD;&#xA;which agrees with the real exponential function when $y=0$. &#xD;&#xA;&#xD;&#xA;The [principal logarithm][1] of $z$ is the complex number &#xD;&#xA;&#xD;&#xA;$$w=\text{Log }z=\log |z|+i\arg z$$&#xD;&#xA;&#xD;&#xA;so that $e^w=z$, where $\arg z$ (the principal argument of $z$) is the real number in $-\pi\lt \arg z\le \pi$, with $x=|z|\cos (\arg z)$ and $y=|z|\sin (\arg z)$. &#xD;&#xA;&#xD;&#xA; &#xD;&#xA;&#xD;&#xA;The complex power is &#xD;&#xA;&#xD;&#xA;$$z^w=e^{w\text{ Log} z}.$$ &#xD;&#xA;&#xD;&#xA;In your example $z=x,w=i$ is therefore $x^i=e^{i \text{ Log}x}$. &#xD;&#xA;&#xD;&#xA;If $x&gt;0$, $\text{Log }x=\log x$. If $x&amp;lt;0$, $\text{Log }x=\log |x|+i\pi$. &#xD;&#xA;&#xD;&#xA;Examples:&#xD;&#xA;&#xD;&#xA;$(-1)^i=e^{i\text{Log }(-1)}=e^{i(i\pi)}=e^{-\pi}$.&#xD;&#xA;&#xD;&#xA;$2^i=e^{i\text{Log }(2)}=e^{i\log 2}=\cos (\log 2)+i\sin (\log 2)$.&#xD;&#xA;&#xD;&#xA;$(-2)^i=e^{i\text{Log }(-2)}=e^{i(\log 2+i\pi)}=e^{i\log 2}e^{-\pi}=(\cos (\log 2)+i\sin (\log 2))e^{-\pi}.$&lt;/p&gt;&#xA;" OwnerUserId="752" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2010-11-11T11:39:49.023" LastActivityDate="2010-11-11T11:39:49.023" CommentCount="2" />
  <row Id="9778" PostTypeId="2" ParentId="9776" CreationDate="2010-11-10T22:16:55.910" Score="7" ViewCount="" Body="&lt;p&gt;First you need to realize that this is a multi-valued function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let us choose the principal logarithm for convenience. So our argument $\theta$ will lie between $(-\pi,\pi]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We then write $a+ib = r e^{i \theta} = e^{(ln(r) + i \theta)}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, we get $(a+ib)^{c+id} = e^{(ln(r) + i \theta)(c+id)}$. Do the necessary algebraic manipulations in the exponent to get $e^{(cln(r) - d \theta) + i(d ln(r) + c \theta)}$. You might also want to take a look at the previous question asked on a similar &lt;a href=&quot;http://math.stackexchange.com/questions/9770/understanding-imaginary-exponents&quot;&gt;topic&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2010-11-10T23:54:07.967" LastActivityDate="2010-11-10T23:54:07.967" />
  <row Id="9779" PostTypeId="2" ParentId="9776" CreationDate="2010-11-10T22:21:01.890" Score="5" ViewCount="" Body="&lt;p&gt;Well, assuming principal values of the complex logarithm (otherwise much craziness ensues):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(a+bi)^{c+di}=\exp((c+di)\ln(a+bi))$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$=\exp((c+di)(\ln|a+bi|+i\arg(a+bi)))$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$=\exp((c\ln|a+bi|-d\arg(a+bi))+i(c\arg(a+bi)+d\ln|a+bi|))$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$=\exp((c\ln|a+bi|-d\arg(a+bi)))\exp(i(c\arg(a+bi)+d\ln|a+bi|))$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and I'll let you finish it at this point, using the fact that $\exp(ix)=\cos\;x+i\sin\;x$&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-11-10T22:21:01.890" />
  <row Id="9780" PostTypeId="2" ParentId="9567" CreationDate="2010-11-10T22:35:41.070" Score="4" ViewCount="" Body="&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/LENum.png&quot; alt=&quot;circles through 2 points&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As Américo Tavares and J. M. said in their comments, two points do not uniquely determine an arc.  Shown above are several circles that pass through two given points.  However, the center of any circle that passes through those two points is on the perpendicular bisector of the line segment with those points as endpoints.&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-11-10T22:35:41.070" CommentCount="1" />
  <row Id="9781" PostTypeId="2" ParentId="9768" CreationDate="2010-11-10T22:43:59.873" Score="1" ViewCount="" Body="&lt;p&gt;It's simple, all right.  As I realized not long after posting (and as Hans also suggested), the key is the exponential map.  The tangent space $T_x M$ gets an inner product space structure from the Riemannian metric; we can isometrically identify it with $\mathbb{R}^n$.  Now $\exp_x : \mathbb{R}^n \to M$ is a diffeomorphism on some small ball $B_{\mathbb{R}^n}(0,\epsilon)$; on this ball, straight lines map to length-minimizing geodesics (see Do Carmo, &lt;em&gt;Riemannian Geometry&lt;/em&gt;, Proposition 3.6), and thus Euclidean balls map to geodesic balls of the same radius.  Taking $\epsilon$ smaller if necessary, we can assume the Jacobian of $\exp_x$ is bounded away from $0$ and $\infty$ on $B_{\mathbb{R}^n}(0, \epsilon)$; thus for $r &amp;lt; \epsilon$ we have that $\operatorname{Vol}(B(x,r))$ is comparable to $\operatorname{Vol}(B_{\mathbb{R}^n}(0,r)) \sim r^n$.&lt;/p&gt;&#xA;" OwnerUserId="822" LastActivityDate="2010-11-10T22:43:59.873" />
  <row Id="9782" PostTypeId="2" ParentId="9690" CreationDate="2010-11-10T22:44:19.577" Score="1" ViewCount="" Body="&lt;p&gt;Here is a different way to look at it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Make two copies of the parallelogram, sharing the side CD.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/czbHh.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second copy is A'B'DC.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We will show that B'=E.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider $\triangle{B'AB}$ and $\triangle{B'DM}$. These are similar and since B'D = AD, we have that DM = AB/2. M is the midpoint of CD and thus B' = E.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus B'L' = BL.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now consider $\triangle{B'DL'}$ and $\triangle{B'AL}$. These are similar and so B'L' = LL', as B'D = AD.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus EL = EL' + LL' = 2B'L' = 2BL.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-11-10T23:10:27.057" LastActivityDate="2010-11-10T23:10:27.057" />
  <row Id="9783" PostTypeId="2" ParentId="9770" CreationDate="2010-11-10T22:58:32.723" Score="5" ViewCount="" Body="&lt;p&gt;Consider a real number $A$, and take it to the power $i$.  If our system of complex numbers is to be consistent, then $A^i$ must be a complex number; in other words, there must be two real numbers $x$ and $y$, which depend on $A$, such that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$A^i=x+iy$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore, we can write $A^{-i}=x-iy$ for the same $x$ and $y$.  Hence:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x^2+y^2=(x+iy)(x-iy)=A^iA^{-i}=A^{i-i}=A^0=1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have shown that for any real number $A$, $|A^i|=1$, and therefore $A^i$ corresponds to a complex number which lies some angle $\theta$ along the unit circle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now consider the sine and cosine functions for extremely small angles $\epsilon$.  A tiny angle $\epsilon$ cuts out a slice of the unit circle, and the curvature of the circumference over this small angle is negligible.  We can therefore think of this slice as a right triangle with angle $\epsilon$, and the hypotenuse and adjacent sides are both length one since they correspond to the radius of the unit circle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the formula for the arc length of a circle, it's easy to determine that in the right triangle formed by the small angle approximation, the length of the side opposite to the angle $\epsilon$ is equal to $\epsilon$.  We can read off the $(x,y)$ coordinates from this diagram (which are $(cos(\epsilon),sin(\epsilon))$), and therefore we conclude that for very small angles $\epsilon$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$sin(\epsilon) \approx \epsilon \hspace{10mm} cos(\epsilon) \approx 1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;therefore $cos(\epsilon) + isin(\epsilon) \approx 1+i\epsilon$, and hence for real numbers $A$ which are extremely close to one (so that $lnA$ is small), the complex number $A^i$ lies approximately at an angle $lnA$ along the unit circle, since $A^i=e^{ilnA}\approx 1+i(lnA)$.&lt;/p&gt;&#xA;" OwnerUserId="1417" LastEditorUserId="1417" LastEditorDisplayName="" LastEditDate="2010-11-10T23:34:26.873" LastActivityDate="2010-11-10T23:34:26.873" CommentCount="4" />
  <row Id="9784" PostTypeId="1" CreationDate="2010-11-10T23:05:35.173" Score="7" ViewCount="171" Body="&lt;p&gt;If A and B are two points on the earth, how could I find any arbitrary point between them along the shortest distance side of their great circle path?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Points are in radians&lt;br&gt;&#xA;longitude = 0 to 2pi&lt;br&gt;&#xA;latitude = 0 to pi, 0 being at north pole&lt;br&gt;&#xA;Points are not antipodal&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I desire something where I specify a range 0.0 to 1.0, with 0.0 being point A and 1.0 being point B and 0.5 being the midpoint between them, with all other values being their corresponding points.  Thanks!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note:  This is not homework.  I'm 41yrs old and this is for a personal project I'm working on.&lt;/p&gt;&#xA;" LastActivityDate="2010-11-11T00:19:17.247" Title="Spherical geometry: Arbitrary point between two points" Tags="&lt;spherical-geometry&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" />
  <row Id="9785" PostTypeId="1" CreationDate="2010-11-10T23:12:41.093" Score="2" ViewCount="164" Body="&lt;p&gt;Let $\Omega = $ {$\omega_{i}$} be an ordered set of $n$ positive reals in the unit interval, $\omega_{1} \leq \cdots \leq \omega_{n} \leq 1$. Define the $n$-simplex $\Delta(\Omega; (\mathbb{R}^{+})^{n})$ by the non-negative points $(x_{1}, \dots, x_{n}) \subset (\mathbb{R}^{+})^{n}$ which satisfy the inequality &#xA;\begin{eqnarray}&#xA;\omega_{1} x_{1} + \cdots + \omega_{n} x_{n} \leq 1.&#xA;\end{eqnarray}&#xA;Let $X$ be a non-trivial subset of the integers $\mathbb{Z}^{n}$. Define $\Delta(\Omega, X) = X \cap \Delta(\Omega, (\mathbb{R}^{+})^{n})$. It is well-known that&#xA;\begin{eqnarray}&#xA;|\Delta(\Omega, \mathbb{N}^{n})| \leq \frac{1}{n!} \prod_{i = 1}^{n} \frac{1}{\omega_{i}} \quad \text{and} \quad |\Delta(\Omega, (\mathbb{Z}^{+})^{n})| \leq \frac{1}{n!} \left(1 + \sum_{i = 1}^{n} \omega_{i} \right)^{n} \prod_{i = 1}^{n} \frac{1}{\omega_{i}},&#xA;\end{eqnarray}&#xA;where $\mathbb{Z}^{+}$ denotes the set of non-negative integers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question(s)&lt;/strong&gt;: For the given bounds above, are any sharper bounds known? Given the similarity in form, are there formulas for other $X$ sets, say for integers greater than some arbitrary integer $c$ or integers satisfying some congruence condition (e.g., $a \equiv b$ mod $d$)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(&lt;em&gt;Update&lt;/em&gt;) The theory of Ehrhart polynomials is relevant to the question above. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; Suppose I'd like to use the Ehrhart machinery to count the number of non-negative integer solutions of $a_{1} x_{1} + \cdots + a_{n} x_{n} \leq r$ for a non-negative integer $r$ and positive integers {$a_{i}$}. How does one proceed?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="2720" LastEditorUserId="2720" LastEditorDisplayName="" LastEditDate="2010-11-16T21:48:26.607" LastActivityDate="2011-01-15T22:44:52.193" Title="Upper Bounds on the Number of Lattice Points in an $n$-Simplex" Tags="&lt;number-theory&gt;&lt;combinatorics&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="2" />
  <row Id="9786" PostTypeId="1" AcceptedAnswerId="9787" CreationDate="2010-11-10T23:24:03.673" Score="1" ViewCount="70" Body="&lt;p&gt;I think the following picture helps explain it easier than words:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/lggXy.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know the coordinates of point a, which is at a corner of the triangle, as well as the coordinates of point b, which is somewhere along the edge of the opposite side.  I also know the y-coordinate of point c, and I'm trying to find its x-coordinate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I've got so far is that as the y-coordinate of b moves up to c, the x coordinate of c approaches the x coordinate of b.  And as the y-coordinate of b moves down to a, the x coordinate of c moves to half-way between the x coordinates of a and b.  But I don't know the direct relationship.&lt;/p&gt;&#xA;" OwnerUserId="3392" LastActivityDate="2010-11-10T23:35:17.233" Title="How to solve for the x-coordinate of the top of an isosceles triangle" Tags="&lt;geometry&gt;&lt;trigonometry&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="9787" PostTypeId="2" ParentId="9786" CreationDate="2010-11-10T23:35:17.233" Score="3" ViewCount="" Body="&lt;p&gt;Assuming the base is horizontal as you have drawn it, we have enough information.  Let us call the coordinates $(xa,ya), (xb,yb),$ and $(xc,yc)$.  Then $$\frac{bx-cx}{by-cy}=-\frac{ax-cx}{ay-cy}$$ because the slopes have to be negatives of each other.  You have five of the six variables.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-11-10T23:35:17.233" CommentCount="2" />
  <row Id="9788" PostTypeId="2" ParentId="9703" CreationDate="2010-11-10T23:43:58.760" Score="8" ViewCount="" Body="&lt;p&gt;I want to extend the answer by @Qiaochu Yuan. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I assume the kid accepts $2*0 = 0$. In other terms:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Some number &lt;em&gt;times&lt;/em&gt; 0 yields the no-changer of &lt;em&gt;plus&lt;/em&gt;&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Analoguously:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Some number &lt;em&gt;to the power&lt;/em&gt; 0 yields the no-changer of &lt;em&gt;times&lt;/em&gt;&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By no-changer I refer, of course, to the unit element. That this can be added/multiplied to anything without resulting in a change should be accepted. I am unsure wether this approach helps understanding the hierarchy of arithmetic operators or wether you need the hierarchy for understanding the approach.&lt;/p&gt;&#xA;" OwnerUserId="3330" LastActivityDate="2010-11-10T23:43:58.760" CommentCount="2" />
  <row Id="9789" PostTypeId="2" ParentId="9785" CreationDate="2010-11-10T23:55:12.863" Score="1" ViewCount="" Body="&lt;p&gt;If $\omega_1 = \cdots = \omega_n = 1/k$ then the equation translates to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle x_1 + \cdots + x_n \leq k$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and so the number of non-negative integer solutions is $\binom{n+k-1}{k}$, which is roughly $n^k/k!$, compared to your estimate $k^n/n!$ (that's assuming $k$ is much smaller; if we assume $n$ is much smaller, we get your estimate).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the way, how do you derive your own upper bound? I assume you start with the volume of the simplex $1/n!$ and then blow up dimension $i$ by $1/\omega_i$, getting your expression. How do you then relate it to the number of lattice points? You can easily construct a convex polytope of arbitrarily small area with arbitrarily many lattice points.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-10T23:55:12.863" />
  <row Id="9790" PostTypeId="2" ParentId="9784" CreationDate="2010-11-11T00:06:44.403" Score="3" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://williams.best.vwh.net/avform.htm#Dist&quot; rel=&quot;nofollow&quot;&gt;This&lt;/a&gt; aviation website has the information that you were looking for. The formula presented there returns the latitude and longitude of a point that is a fraction $f$ between points A and B except when they are antipodal just as you mentioned in the question.&lt;/p&gt;&#xA;" OwnerUserId="618" LastEditorUserId="618" LastEditorDisplayName="" LastEditDate="2010-11-11T00:19:17.247" LastActivityDate="2010-11-11T00:19:17.247" CommentCount="1" />
  <row Id="9791" PostTypeId="1" AcceptedAnswerId="10301" CreationDate="2010-11-11T00:18:06.203" Score="8" ViewCount="339" Body="&lt;p&gt;The definition of small is that they have O(lg n) bits. One way is just to test the integers 2,3,... for primality and keep the first n primes, but this takes at least O(n log n) time (times the cost of primality testing) using a naive algorithm. Is it possible to do it in linear time?&lt;/p&gt;&#xA;" OwnerUserId="2634" LastActivityDate="2011-04-25T17:00:51.787" Title="How can one efficiently generate n small relatively prime integers?" Tags="&lt;number-theory&gt;&lt;algorithms&gt;&lt;prime-numbers&gt;" AnswerCount="3" CommentCount="8" FavoriteCount="0" />
  <row Id="9792" PostTypeId="1" CreationDate="2010-11-11T00:18:22.383" Score="1" ViewCount="51" Body="&lt;p&gt;I am looking for an algorithm to compute the N shortest lattice vectors in R^2. The problem statement is as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given a lattice $L: \{ m \vec{u}+n\vec{v} \} \in \mathcal{R}^2$, a norm $\left\lVert\cdot\right\rVert$, and an integer $N&gt;0$, compute a set of $N$ shortest lattice vectors $S$ such that if $\vec{x}\in S$, then for all $\vec{y} \notin S$, $\left\lVert\vec{y}\right\rVert \ge \left\lVert\vec{x}\right\rVert$. One may assume that $\vec{u}$ and $\vec{v}$ are reduced basis vectors returned from Gauss' algorithm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ideally, the algorithm can take an existing $S$ and expand it by adding in the next &quot;ring&quot; of vectors with the next greatest norm. I have a rough idea of how to go about doing this for the Euclidean norm; you can move along the perimeter of the already-found vectors and add in new vectors of minimal norm, but I am wondering if it is possible to generalize this to any norm. Even generalizing to $L_1$ or $L_\infty$ norms would be good.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;: Description of my algorithm for the $L_2$ norm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The algorithm seeks to find the next ring of points with the next largest norm (radius from the origin) given an existing ring. We accomplish this by scanning along the &quot;perimeter&quot; of the existing ring, in only one of the dimensions since the other length in the other dimension is roughly known based on the equation of a circle.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Assume you are given the reduced lattice basis vectors $\vec{u}$ and $\vec{v}$, integer coefficients $(a,b)$ such that $a \vec{u} + b\vec{v}$ is one of the points on the existing ring of radius $r$.&lt;/li&gt;&#xA;&lt;li&gt;Determine the maximum extent we need to scan in the $\vec{u}$ direction:&#xA;Compute $m&gt;0$ such that $\left\lVert m\vec{u}+n\vec{v} \right\rVert = r$ has no solution for any $n$. This is a simple quadratic equation. Round $m$ away from zero by one to be safe. The amount of time it will take to complete the scan is proportional to $m$.&lt;/li&gt;&#xA;&lt;li&gt;Initialize $r_b = \infty$, the radius of the best next candidate ring. Initialize the set $S$ of next ring points to empty.&lt;/li&gt;&#xA;&lt;li&gt;For each $i$ in $-m \le i \le m$, determine $j$ so that $r_c = \left\lVert i\vec{u}+j\vec{v} \right\rVert &gt; r$. If $r_c \lt r_b$, then we found an even smaller ring: set $r_b = r_c$ and set $S = \{i,j\}$. Otherwise if $r_c = r_b$ we found another point on the ring and add $\{i,j\}$ to $S$. Otherwise, if $r_c \gt r_b$, do nothing.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;A little thought should convince you that this works. This only works for the $L_2$ norm since it explicitly makes use of the shape of the equi-norm contours. For a general norm, the shape is not known ahead of time.&lt;/p&gt;&#xA;" OwnerUserId="398" LastEditorUserId="398" LastEditorDisplayName="" LastEditDate="2010-11-11T06:35:59.310" LastActivityDate="2010-11-11T06:35:59.310" Title="The N-shortest lattice vectors problem in R^2" Tags="&lt;geometry&gt;&lt;lattices&gt;" CommentCount="4" />
  <row Id="9793" PostTypeId="2" ParentId="9776" CreationDate="2010-11-11T00:19:27.077" Score="3" ViewCount="" Body="&lt;p&gt;I transcribe part of &lt;a href=&quot;http://math.stackexchange.com/questions/9770/understanding-imaginary-exponents/9777#9777&quot;&gt;my answer&lt;/a&gt; to &lt;a href=&quot;http://math.stackexchange.com/questions/9770/understanding-imaginary-exponents&quot;&gt;this question&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The complex exponential $e^z$ for complex $z=x+iy$ preserves the law of exponents of the real exponential and satisfies $e^0=1$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;By definition &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$e^z=e^{x+iy}=e^xe^{iy}=e^x(\cos y+\sin y)$$ &#xD;&#xA;&#xD;&#xA;which agrees with the real exponential function when $y=0$. The [principal logarithm][3] of $z$ is the complex number &#xD;&#xA;&#xD;&#xA;$$w=\text{Log }z=\log |z|+i\arg z$$&#xD;&#xA;&#xD;&#xA;so that $e^w=z$, where $\arg z$ (the principal argument of $z$) is the real number in $-\pi\lt \arg z\le \pi$, with $x=|z|\cos (\arg z)$ and $y=|z|\sin (\arg z)$. &#xD;&#xA;&#xD;&#xA;&#xD;&#xA;The complex power is &#xD;&#xA;&#xD;&#xA;$$z^w=e^{w\text{ Log} z}.$$ &#xD;&#xA;&#xD;&#xA;In your case you have: $z=a+bi,w=c+di$&#xD;&#xA;&#xD;&#xA;$$(a+bi)^{c+di}=z^{w}=e^{w\text{Log }z}=e^{(c+di)\text{Log }(a+bi)}$$&#xD;&#xA;&#xD;&#xA;$$=e^{(c+di)(\log |a+bi|+i\arctan \frac{d}{c}))}$$&#xD;&#xA;&#xD;&#xA;Since&#xD;&#xA;&#xD;&#xA;$$(c+di)\left(\log |a+bi|+i\arctan \frac{d}{c}\right)=(c+di)\log |a+bi|+(c+di)i\arctan \frac{d}{c}$$&#xD;&#xA;&#xD;&#xA;$$=c\log |a+bi|-d\arctan \frac{d}{c}+i\left( d\log |a+bi|+c\arctan \frac{d}{c}%&#xD;&#xA;\right) $$&#xD;&#xA;&#xD;&#xA;we obtain&#xD;&#xA;&#xD;&#xA;$$e^{(c+di)\left(\log |a+bi|+i\arctan \frac{d}{c})\right)}=e^{c\log |a+bi|-d\arctan &#xD;&#xA;\frac{d}{c}}e^{i\left( d\log |a+bi|+c\arctan \frac{d}{c}\right) }$$&#xD;&#xA;&#xD;&#xA;and&#xD;&#xA;&#xD;&#xA;$$e^{i\left( d\log |a+bi|+c\arctan \frac{d}{c}\right) }=$$&#xD;&#xA;&#xD;&#xA;$$=\cos \left( d\log |a+bi|+c\arctan \frac{d}{c}\right)+i\sin \left( d\log |a+bi|+c\arctan \frac{d}{c}\right).$$&lt;/p&gt;&#xA;" OwnerUserId="752" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2010-11-11T11:02:25.487" LastActivityDate="2010-11-11T11:02:25.487" CommentCount="1" />
  <row Id="9794" PostTypeId="2" ParentId="9048" CreationDate="2010-11-11T00:30:41.167" Score="1" ViewCount="" Body="&lt;p&gt;OK, got it figured out my self, before including the acceleration I first had to parameterize the curve itself, you can read all the stuff over at my gamedev question:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://gamedev.stackexchange.com/questions/5373/moving-ships-between-two-planets-along-a-bezier-missing-some-equations-for-accel&quot;&gt;http://gamedev.stackexchange.com/questions/5373/moving-ships-between-two-planets-along-a-bezier-missing-some-equations-for-accel&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3154" LastActivityDate="2010-11-11T00:30:41.167" />
  <row Id="9795" PostTypeId="1" CreationDate="2010-11-11T00:38:32.687" Score="0" ViewCount="98" Body="&lt;p&gt;$G$ and $G \times G$ where $G = Z\_2 \times Z\_2 \times Z\_2 \times Z\_2 \times  Z\_2 \times \cdots$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer says yes but I cannot figure out what homomorphism function I could use.&lt;/p&gt;&#xA;" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-11T00:59:01.083" LastActivityDate="2010-11-11T01:11:43.607" Title="Are they isomorphic?" Tags="&lt;abstract-algebra&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="9796" PostTypeId="2" ParentId="9795" CreationDate="2010-11-11T00:45:45.567" Score="3" ViewCount="" Body="&lt;p&gt;Think of $$G = Z\_{2_1}\times Z\_{2_2} \times Z\_{2_3} \times Z\_{2_4} \times Z\_{2_5} \times \ldots$$ and $$G \times G= (Z\_{2_1}\times Z\_{2_3} \times Z\_{2_5} \times \ldots) \times (Z\_{2_2}\times Z\_{2_4} \times Z\_{2_6} \times \ldots)$$&lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2010-11-11T01:11:43.607" LastActivityDate="2010-11-11T01:11:43.607" CommentCount="4" />
  <row Id="9797" PostTypeId="1" AcceptedAnswerId="9798" CreationDate="2010-11-11T01:14:13.043" Score="0" ViewCount="72" Body="&lt;p&gt;Ok this is a really silly question and I should know this, but I cant seem to figure something out:&lt;img src=&quot;http://i.imgur.com/F0U3t.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for the last step, how do they know that 0&amp;lt;=x&amp;lt;=4? If we use the minimum value of theta, which is $-pi/2$, and plug that into $x=cos(theta)$, then we get 0, and same for pi/2.&lt;/p&gt;&#xA;" OwnerUserId="2985" LastActivityDate="2010-11-11T03:46:28.687" Title="Parametric Equation Question" Tags="&lt;calculus&gt;" AnswerCount="2" CommentCount="0" />
  <row Id="9798" PostTypeId="2" ParentId="9797" CreationDate="2010-11-11T01:18:28.440" Score="3" ViewCount="" Body="&lt;p&gt;Because over $-\pi /2\leq \theta \leq \pi /2, 0\leq\cos{\theta}\leq1$.  At $\theta =0, \cos{\theta}=1$&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-11-11T01:18:28.440" />
  <row Id="9799" PostTypeId="1" AcceptedAnswerId="9801" CreationDate="2010-11-11T02:16:42.727" Score="5" ViewCount="238" Body="&lt;p&gt;I know that Banach Tarski Paradox gives that &quot;A three-dimensional Euclidean ball is equidecomposable with two copies of itself&quot; and I have read that doubling the ball can be accomplished with five pieces. My question is :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Is it possible to construct (mathematically) these five sets or is the proof more of an existence result and not a construction?&quot; If it is possible to construct, I would really appreciate if someone can show the construction here.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastActivityDate="2010-11-11T03:23:57.847" Title="Banach Tarski Paradox" Tags="&lt;set-theory&gt;&lt;measure-theory&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" />
  <row Id="9800" PostTypeId="2" ParentId="9760" CreationDate="2010-11-11T02:21:17.550" Score="3" ViewCount="" Body="&lt;p&gt;This question in its full generality is probably best tackled by some generating function approach as Qiaochu points out. I toyed around with the $N=1$ case and got some interesting results that I am sharing below. For the $N=1$ case, we don't have to distinguish between the $+1$ and the $-1$s. In fact, a nice visualization is to think of a row of bulbs one of which is randomly toggled at each stage. Question: Given that they are all off initially, how long does it take on an average for all of them to light up?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $f_k$ be the expected number of steps for all bulbs to light up starting from $k$ lighted bulbs out of a total of $n$ bulbs. Then, we have the recursions&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f_0 = 1 + f_1$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f_k = 1 + \frac{k}{n} f_{k-1} + \frac{n-k}{n} f_{k+1}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f_n = 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We need to solve for $f_0$. I cheated on solving the recursion by numerically solving it for various $n$ and appealing to OEIS. The answer turns out to be &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f_0 = n \sum_{j=0}^{n-1} \frac{2^j}{j+1}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of $3$ bulbs, the answer is $10$. I don't know how to extend this approach to the case of arbitrary $N$ without making it messy. I am not very comfortable with generating function based solution approaches. If someone were to post one for this problem, I would love to learn the techniques of &quot;generating functionology&quot; through that.&lt;/p&gt;&#xA;" OwnerUserId="618" LastEditorUserId="618" LastEditorDisplayName="" LastEditDate="2010-11-11T02:26:34.103" LastActivityDate="2010-11-11T02:26:34.103" CommentCount="2" />
  <row Id="9801" PostTypeId="2" ParentId="9799" CreationDate="2010-11-11T02:23:05.170" Score="9" ViewCount="" Body="&lt;p&gt;No, it is not possible to explicitly construct the pieces. The pieces are, by necessity, non-measurable. This means that they cannot be Borel sets, which covers essentially every &quot;explicit construction&quot; technique you might think of. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moreover, there are models of ZF set theory (without the axiom of choice) in which the Banach-Tarski paradox fails. So the construction must, necessarily, make use of some form of the axiom of choice.  This means that an even wider range of construction techniques - those that can be carried out in ZF - are insufficient to form the decomposition. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After clarification, it seems that one part of the question is to find an explicit proof that (using the axiom of choice) it is possible to get a decomposition using exactly 5 pieces. A proof of this is given by Francis Su's thesis on the paradox &lt;a href=&quot;http://www.math.hmc.edu/%7Esu/papers.dir/banachtarski.pdf&quot; rel=&quot;nofollow&quot;&gt;(PDF)&lt;/a&gt;. Theorem 20 gives the proof of the five-piece decomposition, and by tracing back the previous results you can work out exactly what the pieces are. That thesis is, by the way, a wonderful reference for many other aspects of the paradox as well. &lt;/p&gt;&#xA;" OwnerUserId="630" LastEditorUserId="630" LastEditorDisplayName="" LastEditDate="2010-11-11T03:23:57.847" LastActivityDate="2010-11-11T03:23:57.847" CommentCount="10" />
  <row Id="9802" PostTypeId="1" CreationDate="2010-11-11T02:35:00.560" Score="1" ViewCount="95" Body="&lt;p&gt;If | | denotes the absolute value then is it always true $|x^{s}|=|x|^{s}$ whenever s is any &lt;em&gt;positive&lt;/em&gt; real number? I know this holds for any positive integer. What about positive reals?&lt;/p&gt;&#xA;" OwnerUserId="4219" LastActivityDate="2010-11-11T03:12:00.123" Title="Elementary question power absolute values" Tags="&lt;algebra-precalculus&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="9803" PostTypeId="1" AcceptedAnswerId="9857" CreationDate="2010-11-11T02:35:11.860" Score="0" ViewCount="210" Body="&lt;p&gt;Let V be a vector space with a non-trivial subspaces U, W. $ U \neq W$. Consider $u_0 \in V$, with $u_0 \neq 0$ and $w_0 \in V$ with $w_0 \neq 0$ then we have $u_0 + U \in V/U$ and $w_0 + W \in V/W$. Is there any sensible way to define:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(u_0 + U) + (w_0 + W)$? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm thinking &quot;no&quot; $(2,4) + (2,6,7)$ makes no sense. So, why should this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I mean, we would need to find some new vector space that had all of the elements in $V/U$ and in $V/W$ and their sums and scalar products. Maybe $V/U + V/W$ -- but then we need both $V/U$ and $V/W$ to be subspaces of some larger vector space. Is there a way to make sense of this at all?&lt;/p&gt;&#xA;" OwnerUserId="2563" LastEditorUserId="2563" LastEditorDisplayName="" LastEditDate="2010-11-11T15:24:40.260" LastActivityDate="2010-11-11T15:24:40.260" Title="Bad ideas: Adding vectors from different vector spaces?" Tags="&lt;linear-algebra&gt;&lt;vector-space&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="9804" PostTypeId="2" ParentId="9703" CreationDate="2010-11-11T02:36:19.667" Score="5" ViewCount="" Body="&lt;p&gt;I think that it is worth pointing out that, strictly speaking, explaining “2&lt;sup&gt;0&lt;/sup&gt; equals 1” is not really what people do in this situation.  The best we can do is to convince a child of the following facts:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;If 2&lt;sup&gt;0&lt;/sup&gt; is any number, it makes more sense to consider that 2&lt;sup&gt;0&lt;/sup&gt;=1 than considering 2&lt;sup&gt;0&lt;/sup&gt; as any other numbers (such as 0).&lt;/li&gt;&#xA;&lt;li&gt;It is more interesting to consider 2&lt;sup&gt;0&lt;/sup&gt; to be 1 than giving up.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Some of the other answers provide good ways to convince a child of these facts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the reason that 2&lt;sup&gt;0&lt;/sup&gt; equals 1 rather than 2&lt;sup&gt;0&lt;/sup&gt; is undefined is really conventions and experiences: it is much more convenient to define 2&lt;sup&gt;0&lt;/sup&gt;=1 than leaving 2&lt;sup&gt;0&lt;/sup&gt; undefined.  I do not think that it is possible to convince a child of this fact.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Compare this to the following.  Let &lt;i&gt;f&lt;/i&gt;(&lt;i&gt;x&lt;/i&gt;) = sin &lt;i&gt;x&lt;/i&gt; / &lt;i&gt;x&lt;/i&gt;.  What is &lt;i&gt;f&lt;/i&gt;(0)?&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;If &lt;i&gt;f&lt;/i&gt;(0) were any number, it would make more sense to consider that &lt;i&gt;f&lt;/i&gt;(0)=1 than considering &lt;i&gt;f&lt;/i&gt;(0) as any other values.&lt;/li&gt;&#xA;&lt;li&gt;It would be more interesting to consider &lt;i&gt;f&lt;/i&gt;(0) to be 1 than giving up.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;However, these “evidences” do not make &lt;i&gt;f&lt;/i&gt;(0)=1 under the usual way of mathematical writing.  I think that the only things that differentiate these two cases are conventions and experiences.&lt;/p&gt;&#xA;" OwnerUserId="926" LastActivityDate="2010-11-11T02:36:19.667" />
  <row Id="9805" PostTypeId="2" ParentId="9483" CreationDate="2010-11-11T02:40:05.730" Score="3" ViewCount="" Body="&lt;p&gt;Jonas: What you need to show is that if $Y$ is well-ordered, then any subset $Z$ of $Y$ is isomorphic to an initial segment of $Y$. How would you define such an isomorphism $f$? You do not have many possibilities: The least element of $Z$ must be mapped by $f$ to the smallest element of $Y$. The next element of $Z$ must be mapped by $f$ to the next element of $Y$, and so on. You just need to check that this works.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In more detail, say that $g$ is an &lt;em&gt;approximation&lt;/em&gt; iff its domain is an initial segment of $Z$, its range is an initial segment of $Y$, and $g$ is an order isomorphism from its domain onto its range.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One checks (by considering least disagreements) that any two approximations coincide on the common part of their domains, so we can let $f$ be the common function resulting from pasting all the approximations together.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We just need to see that $f$ exhausts $Z$. For this, one can check that $f(x)\le x$ for all $x$, so one &quot;does not run out of room.&quot; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a slightly more elegant (and formal) way of presenting this argument, by appealing to the recursion theorem. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $x$ in $Z$, define $f(x)=\sup_Y({\rm succ}_Y f(y)\mid y\in Z, y\lt x)$, where ${\rm succ}_Y a$ means the successor in $Y$ of $a$. Then $f$ is precisely the map we wanted. (One still needs a little argument to see that it works).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A trickier question for another day is whether the recursion theorem is &lt;em&gt;needed&lt;/em&gt; for the argument. This may be formalized by asking whether Zermelo set theory proves the result (so we do not have access to replacement), but there may be finer formulations. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(The above was written quickly. Let me know of infelicities or plain falsehoods and I'll try to edit.)&lt;/p&gt;&#xA;" OwnerUserId="462" LastActivityDate="2010-11-11T02:40:05.730" CommentCount="2" />
  <row Id="9806" PostTypeId="2" ParentId="9746" CreationDate="2010-11-11T02:47:55.600" Score="2" ViewCount="" Body="&lt;p&gt;When we study mathematical logic, we use the same mathematical methods that would be used to study any other area (algebra, analysis, topology, etc.).  This includes the use of set-theoretical techniques.  Otherwise, we would be needlessly hamstringing ourselves.  In principle, the results that we obtain could be built up entirely from primitive axioms, just like the rest of mathematics; but just like the rest of mathematics complete formalization is not usually the goal. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, even ignoring those practical considerations, it's not hard to see that none of the axioms of point set topology or set theory used in topology presuppose that the compactness theorem of logic holds. So there is no circularity if we use topological methods to prove the compactness theorem. &lt;/p&gt;&#xA;" OwnerUserId="630" LastActivityDate="2010-11-11T02:47:55.600" />
  <row Id="9807" PostTypeId="1" CreationDate="2010-11-11T02:48:13.867" Score="2" ViewCount="137" Body="&lt;p&gt;The question is: Given a licence plate that can have either 2 or 3 letters followed by either 2 or 3 numbers, how many different license plates can be printed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My math is as such $(26^2 + 26^3)*(9^2+9^3)$ which comes to 14,784,120 but the book says the answer is 20,077,200. What am I doing wrong here?&lt;/p&gt;&#xA;" OwnerUserId="3071" LastActivityDate="2010-11-22T08:11:11.890" Title="Issue with permutation problem" Tags="&lt;homework&gt;&lt;discrete-mathematics&gt;&lt;permutations&gt;" AnswerCount="1" CommentCount="7" />
  <row Id="9808" PostTypeId="2" ParentId="9791" CreationDate="2010-11-11T02:49:36.860" Score="-1" ViewCount="" Body="&lt;p&gt;Check out Chapter 1 of Paul Pollack's book &quot;&lt;a href=&quot;http://books.google.com/books?id=MI7zfW7CkzIC&amp;amp;lpg=PP1&amp;amp;ots=4uNSVP2yUZ&amp;amp;dq=pollack%20%22not%20always%20buried%20deep%22&amp;amp;pg=PP1#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;Not Always Buried Deep&lt;/a&gt;&quot;.  He lists several elementary proofs for the infinitude of primes, some of which can be adapted to form simple iterative algorithms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Sorry, I can't get Latex to work here.   Isn't it like MathOverflow?  Anyway..)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let S = {p1...pk} be the set of previously generated coprime integers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And take M = product of integers in S&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now do any of the following to get new integer coprime to the set S.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;em&gt;(Euclid)&lt;/em&gt;: M + 1 &lt;/li&gt;&#xA;&lt;li&gt;&lt;em&gt;(Stieltjes)&lt;/em&gt;: Factor M as A.B in some way, and take A + B&lt;/li&gt;&#xA;&lt;li&gt;&lt;em&gt;(Euler)&lt;/em&gt; Totient(M)&lt;/li&gt;&#xA;&lt;li&gt;&lt;em&gt;(Braun and Metrod)&lt;/em&gt;: N = M/p1 + ... + M/p_k&lt;/li&gt;&#xA;&lt;li&gt;&lt;em&gt;(Goldbach)&lt;/em&gt;: N = 2 + M &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;There is also Saidak's proof from the same book:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Take N1=n; N2 = N1(N1+1); N3 =&#xA;N2(N2+1)... Nk=Nk(Nk+1) and so on. &#xA;This a relative prime sequence.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I am probably missing a couple of other proofs.  Anyway, check the book!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT1 : **I am going to retain the answer despite the negative votes, in the hope that it might spur someone else to invent a better algorithm ! *&lt;/p&gt;&#xA;" OwnerUserId="1558" LastEditorUserId="1558" LastEditorDisplayName="" LastEditDate="2010-11-11T14:13:13.260" LastActivityDate="2010-11-11T14:13:13.260" CommentCount="7" />
  <row Id="9809" PostTypeId="1" AcceptedAnswerId="9813" CreationDate="2010-11-11T02:50:53.067" Score="3" ViewCount="105" Body="&lt;p&gt;I've been reading Vakil's notes on algebraic geometry (on my own -- this is not part of a class), and I'm stuck on one problem (number 6.1.H).  It goes as follows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $X$ be a scheme.  Prove that $X$ is quasicompact and quasiseparated if and only if $X$ can be covered by a finite number of affine open subsets, any two of which have intersection also covered by a finite number of affine open subsets.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's not hard to show one direction, namely that if $X$ is quasicompact and quasiseparated then it has a cover of the indicated form.  It's also not hard to prove that if $X$ has a cover of the indicated form, then $X$ is quasicompact.  I'm having difficulty with the &quot;quasiseparated&quot; part.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you very much for any help.&lt;/p&gt;&#xA;" OwnerUserId="3400" LastActivityDate="2010-11-12T18:04:59.373" Title="Quasiseparated if finitely covered by affines in appropriate way" Tags="&lt;algebraic-geometry&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="9810" PostTypeId="2" ParentId="9802" CreationDate="2010-11-11T02:54:15.763" Score="1" ViewCount="" Body="&lt;p&gt;Yes, even when $x \in \mathbb{C}$.  To see this, write $x = |x| e^{i \theta}$, in which case $$|x^s| = ||x|^s e^{i s \theta}| = |x|^s |e^{i s \theta}| = |x|^s.$$&lt;/p&gt;&#xA;" OwnerUserId="3296" LastActivityDate="2010-11-11T02:54:15.763" CommentCount="1" />
  <row Id="9811" PostTypeId="2" ParentId="9802" CreationDate="2010-11-11T03:02:56.210" Score="0" ViewCount="" Body="&lt;p&gt;This is a fairly standard argument proceeding along the lines of first extending the relation from $\mathbb{Z}$ to $\mathbb{Q}$ and finally to $\mathbb{R}$ (of course only for the values where the relation makes sense).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the first extension use the fact that every rational $r \in \mathbb{Q}$ can be written as $r = p/q$ with $p, q \in \mathbb{Z}$. For the second, use the continuity of the norm and density of $\mathbb{Q}$ in $\mathbb{R}$.&lt;/p&gt;&#xA;" OwnerUserId="3223" LastActivityDate="2010-11-11T03:02:56.210" />
  <row Id="9812" PostTypeId="2" ParentId="9802" CreationDate="2010-11-11T03:12:00.123" Score="2" ViewCount="" Body="&lt;p&gt;Assuming you're talking about real $x$, the only time you would be concerned with whether the equality holds is if $x$ is negative.  (Otherwise, $|x|=x$ and $|x^s|=x^s$, so there's nothing to show.)  Now, how do you define $x^s$ if $x$ is negative and $s$ is irrational? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One answer is given in more generality by user3296.  You can use complex numbers, in which case it works out well.  However, if you only want to work with real values, then trying to take irrational powers of negative numbers is best avoided.  To see this, think first about how $x^s$ can be defined when $x$ is positive.  One way is by taking rational numbers $p/q$ approaching $s$, and taking a limit of $x^{p/q}=\sqrt[q]{x}^p$.  But this definition won't work if $x$ is negative, because for some choices of $p/q$ the result will be imaginary, for others negative, and for others positive.  Another way is using the definition $x^s=e^{s\ln{x}}$, where $\ln$ is the natural logarithm defined on the positive reals.  To extend this definition to negative $x$, you need a way of defining a logarithm of negative numbers.  This actually works (although there is not a unique choice), but the result is that you end up with complex numbers rather than real numbers.  For example, you can define $\log$ on $(-\infty,0)$ by $\log(x)=\ln(-x)+\pi i$, so that $x^s=e^{s(\ln(-x)+\pi i)}=e^{s\pi i}(-x)^s$.  But this is not real when $s$ is not an integer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As long as you have no problem using complex exponentials (and logarithms, at least implicitly), then the answer is yes, as user3296 says.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-11-11T03:12:00.123" />
  <row Id="9813" PostTypeId="2" ParentId="9809" CreationDate="2010-11-11T03:14:13.943" Score="1" ViewCount="" Body="&lt;p&gt;Hint: Starting from your open affine cover of $X$, construct an open affine cover&#xA;of $X \times X$, and then show that the preimage of each member of that open affine cover is a union of finitely many affines.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another hint, more directly relevant to Vakil's definition of quasi-separated:&#xA;I am implicitly cheating somewhat, because I am still secretely thinking about quasi-separated as meaning that the diagonal map is quasi-compact.  If I do that,&#xA;then the exercise at hand translates into the following: show that if a morphism&#xA;$f$ has the property that the target has an open affine cover, the preimage of each member of which is quasi-compact, then $f$ is quasi-compact (i.e. each quasi-compact in the target has quasi-compact preimage).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you should certainly remind yourself how to prove this fact, because the tools used there will be the same tools you need in your present exercise.  (Even though you may not be working explicitly with the formulation in terms of the diagonal morphism, the underlying argument will have to be essentially the same.)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;E.g. first convince yourself that you only have to check that the intersection of any two open affines is quasi-compact.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, an important point in all these kinds of arguments: if $f: X \to Y$&#xD;&#xA;with $X$ and $Y$ affine, say $X =$ Spec $B$ and $Y =$ Spec $A$, and $a \in A$, so that Spec $A_a$ is a distiguished open, then the inverse image of Spec $A_a$&#xD;&#xA;is a distinguished open affine in $X$ (it is Spec $B_a$).  This a fundamental tool: it is the one way that you have to construct arbitrarily small open affine neighbourhoods in $Y$ whose preimages in $X$ are again affine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you will want to use the idea of the preceding paragraph, but now &quot;inverse image&quot; will be replaced by &quot;intersection&quot;.  (When you take inverse images of product neighbourhoods along the diagonal map, you are computing intersections,&#xA;so this replacement makes sense; but again, you won't have to think about the diagonal map explicitly if you dont' want to.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's probably enough of a rambling hint for now.  If you have those ideas in mind and put everything you know together, you hopefully can find your way to a complete proof.&lt;/p&gt;&#xA;" OwnerUserId="221" LastEditorUserId="221" LastEditorDisplayName="" LastEditDate="2010-11-12T18:04:59.373" LastActivityDate="2010-11-12T18:04:59.373" CommentCount="3" />
  <row Id="9814" PostTypeId="2" ParentId="9760" CreationDate="2010-11-11T03:26:05.783" Score="9" ViewCount="" Body="&lt;p&gt;Still working on a general solution, but here are a few exact solutions for small $N$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $N = 1..8 $ the exact values are 10, $\frac{416}{5}$, $\frac{10294}{35}$, $\frac{18481664}{25823}$, $\frac{235115486250}{165493097}$, $\frac{1690479776164}{681135455}$, $\frac{25081308000562315334439801586}{6314587098176165651428105}$, $\frac{282872889746351048352862015165111214080}{47431617359482015647230177875265569}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Numerical values: &lt;code&gt;10, 83.2, 294.114, 715.706, 1420.7, 2481.86, 3971.96, 5963.8.&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These were computed &quot;the hard way&quot; by making a large state diagram, exploiting a lot of symmetry, and then solving a system of linear equations. There are 165 states, for example, in the case of $N=8$, which means solving for 165 equations in 165 unknowns. There is undoubtedly a more clever way to compute this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I offer the bound $c \cdot N^3$ without proof; I believe the expected number of steps is asymptotic to this value with $c \approx 12$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added 13-Nov-2010&lt;/strong&gt;:&#xA;The following table provides strong evidence for the $12N^3$ asymptotic bound given earlier.  &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; N        E(N)    12(N^3)&#xA;&#xA; 1        10.0        12&#xA; 2        83.2        96&#xA; 3       294.1       324&#xA; 4       715.7       768&#xA; 5      1420.7      1500&#xA; 6      2481.9      2592&#xA; 7      3972.0      4116&#xA; 8      5963.8      6144&#xA; 9      8530.2      8748&#xA;10     11743.8     12000&#xA;11     15677.6     15972 &#xA;12     20404.2     20736&#xA;13     25996.5     26364&#xA;14     32527.3     32928&#xA;15     40069.3     40500&#xA;16     48695.3     49152&#xA;17     58478.1     58956&#xA;18     69490.5     69984&#xA;19     81805.3     82308&#xA;20     95495.3     96000&#xA;21    110633.2    111132&#xA;22    127291.9    127776&#xA;23    145544.0    146004&#xA;24    165462.6    165888&#xA;25    187120.2    187500&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Edit 11-Nov-2010:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is the system of linear equations for $N=3$. Ostensibly there are $6^3 = 216$ states but by symmetry this can be reduced to $20$. The three subscripts on each variable represent the distance from the goal state in each dimension. In accordance with the problem statement we start at the farthest node from the goal. Thus we are solving for $x_{3,3,3}$ here. (The answer is $\frac{10294}{35}$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I show this tedium in hopes someone can figure out how to generalize.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x_{0,0,0}=0 $ &lt;br /&gt;&lt;br /&gt;&#xA;$x_{0,0,1}=\frac{1}{6} x_{0,0,0}+\frac{1}{6} x_{0,0,2}+\frac{2}{3} x_{0,1,1}+1 $ &lt;br /&gt;&lt;br /&gt;&#xA;$x_{0,0,2}=\frac{1}{6} x_{0,0,1}+\frac{1}{6} x_{0,0,3}+\frac{2}{3} x_{0,1,2}+1 $ &lt;br /&gt;&lt;br /&gt;&#xA;$x_{0,0,3}=\frac{1}{3} x_{0,0,2}+\frac{2}{3} x_{0,1,3}+1 $ &lt;br /&gt;&lt;br /&gt;&#xA;$x_{0,1,1}=\frac{1}{3} x_{0,0,1}+\frac{1}{3} x_{0,1,2}+\frac{1}{3} x_{1,1,1}+1 $ &lt;br /&gt;&lt;br /&gt;&#xA;$x_{0,1,2}=\frac{1}{6} x_{0,0,2}+\frac{1}{6} x_{0,1,1}+\frac{1}{6} x_{0,1,3}+\frac{1}{6} x_{0,2,2}+\frac{1}{3} x_{1,1,2}+1 $ &lt;br /&gt;&lt;br /&gt;&#xA;$x_{0,1,3}=\frac{1}{6} x_{0,0,3}+\frac{1}{3} x_{0,1,2}+\frac{1}{6} x_{0,2,3}+\frac{1}{3} x_{1,1,3}+1 $ &lt;br /&gt;&lt;br /&gt;&#xA;$x_{0,2,2}=\frac{1}{3} x_{0,1,2}+\frac{1}{3} x_{0,2,3}+\frac{1}{3} x_{1,2,2}+1 $ &lt;br /&gt;&lt;br /&gt;&#xA;$x_{0,2,3}=\frac{1}{6} x_{0,1,3}+\frac{1}{3} x_{0,2,2}+\frac{1}{6} x_{0,3,3}+\frac{1}{3} x_{1,2,3}+1 $ &lt;br /&gt;&lt;br /&gt;&#xA;$x_{0,3,3}=\frac{2}{3} x_{0,2,3}+\frac{1}{3} x_{1,3,3}+1 $ &lt;br /&gt;&lt;br /&gt;&#xA;$x_{1,1,1}=\frac{1}{2} x_{0,1,1}+\frac{1}{2} x_{1,1,2}+1 $ &lt;br /&gt;&lt;br /&gt;&#xA;$x_{1,1,2}=\frac{1}{3} x_{0,1,2}+\frac{1}{6} x_{1,1,1}+\frac{1}{6} x_{1,1,3}+\frac{1}{3} x_{1,2,2}+1 $ &lt;br /&gt;&lt;br /&gt;&#xA;$x_{1,1,3}=\frac{1}{3} x_{0,1,3}+\frac{1}{3} x_{1,1,2}+\frac{1}{3} x_{1,2,3}+1 $ &lt;br /&gt;&lt;br /&gt;&#xA;$x_{1,2,2}=\frac{1}{6} x_{0,2,2}+\frac{1}{3} x_{1,1,2}+\frac{1}{3} x_{1,2,3}+\frac{1}{6} x_{2,2,2}+1 $ &lt;br /&gt;&lt;br /&gt;&#xA;$x_{1,2,3}=\frac{1}{6} x_{0,2,3}+\frac{1}{6} x_{1,1,3}+\frac{1}{3} x_{1,2,2}+\frac{1}{6} x_{1,3,3}+\frac{1}{6} x_{2,2,3}+1 $ &lt;br /&gt;&lt;br /&gt;&#xA;$x_{1,3,3}=\frac{1}{6} x_{0,3,3}+\frac{2}{3} x_{1,2,3}+\frac{1}{6} x_{2,3,3}+1 $ &lt;br /&gt;&lt;br /&gt;&#xA;$x_{2,2,2}=\frac{1}{2} x_{1,2,2}+\frac{1}{2} x_{2,2,3}+1 $ &lt;br /&gt;&lt;br /&gt;&#xA;$x_{2,2,3}=\frac{1}{3} x_{1,2,3}+\frac{1}{3} x_{2,2,2}+\frac{1}{3} x_{2,3,3}+1 $ &lt;br /&gt;&lt;br /&gt;&#xA;$x_{2,3,3}=\frac{1}{6} x_{1,3,3}+\frac{2}{3} x_{2,2,3}+\frac{1}{6} x_{3,3,3}+1 $ &lt;br /&gt;&lt;br /&gt;&#xA;$x_{3,3,3}=x_{2,3,3}+1 $&#xD;&lt;/p&gt;&#xA;" OwnerUserId="130" LastEditorUserId="130" LastEditorDisplayName="" LastEditDate="2010-11-13T20:45:14.273" LastActivityDate="2010-11-13T20:45:14.273" />
  <row Id="9815" PostTypeId="1" AcceptedAnswerId="9826" CreationDate="2010-11-11T03:33:28.383" Score="2" ViewCount="369" Body="&lt;p&gt;This is a homework question so specifically I am looking for a direction (help). Not an absolute answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand how to solve 2nd order nonhomogeneous (and I think 3rd order is similiar) recurrence relations. Whats kinking me up with this one is a term with n.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;h[n] = 4h[n- 1] - 4h[n-2] + 3n + 1&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Here I am using square brackets to indicate subscripts.&#xA;Now I can reorganize this equation into:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;0 = h[n] - 4h[n-1] + 4h[n-2] -3n-1&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;From which I can derive the characteristic equation:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;0 = x^3 - 4x^2 + 4x - 3n -1&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Normally at this point I would solve the roots and be on my way, but here I have that extra n. How do I deal with this?&lt;/p&gt;&#xA;" OwnerUserId="1670" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2010-11-11T04:51:01.967" LastActivityDate="2010-11-11T14:56:08.317" Title="How to solve 3rd order nonhomogeneous recurrence relation with n in equation?" Tags="&lt;combinatorics&gt;&lt;homework&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="9816" PostTypeId="2" ParentId="9797" CreationDate="2010-11-11T03:46:28.687" Score="0" ViewCount="" Body="&lt;p&gt;Well, y^2 has to be positive, so the minimum that y^2/25 could be is 0.  So, x^2/16 is 1 at the maximum =&gt; x≤4.  The same argument applies for x = 0&lt;/p&gt;&#xA;" LastActivityDate="2010-11-11T03:46:28.687" CommentCount="1" />
  <row Id="9817" PostTypeId="2" ParentId="9815" CreationDate="2010-11-11T04:11:52.497" Score="2" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\ $ Find a particular polynomial solution $\rm\:h[n]\:$ by undetermined coefficients (notice that the degree can be at most 1).$\ $ The general solution is then the sum of any particular solution and the general solution of the associated homogeneous equation $\rm\ h[n] -4\ h[n-1] + 4\ h[n-2]\ =\ 0\:$.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-11-11T14:56:08.317" LastActivityDate="2010-11-11T14:56:08.317" CommentCount="2" />
  <row Id="9819" PostTypeId="1" CreationDate="2010-11-11T05:09:15.967" Score="1" ViewCount="118" Body="&lt;p&gt;A triangle on a sphere is comprised of points A, B and C. How to determine its area?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know the formula:&lt;br&gt;&#xA;A = E * R^2, where R is radius of sphere, and E is the excess angle of (a + b + c - pi), but how to determine the angles between ABC, ACB and BAC?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks! :-)&lt;/p&gt;&#xA;" LastActivityDate="2010-11-11T05:58:22.240" Title="Area of a spherical triangle" Tags="&lt;spherical-geometry&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="9820" PostTypeId="1" CreationDate="2010-11-11T05:32:31.367" Score="1" ViewCount="121" Body="&lt;p&gt;According to the book the following is true:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $f$ be a measurable function and suppose $|f| \leq h$ where $h$ is a real-valued function&#xA;and $h$ is integrable. Then f is integrable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is: suppose instead we have the inequality $ |f| \leq h$ but almost everywhere and of course $h$ assumed to be integrable. Is it still true that f is integrable?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would this follow because the sets of measure zero &quot;don't count&quot; when integrating?&lt;/p&gt;&#xA;" OwnerUserId="4219" LastActivityDate="2010-11-11T19:53:19.313" Title="Doubt inequality almost everywhere" Tags="&lt;measure-theory&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="9821" PostTypeId="2" ParentId="9820" CreationDate="2010-11-11T05:43:23.833" Score="3" ViewCount="" Body="&lt;p&gt;Yes on both counts. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $E$ be set of measure zero such that $|f|\leq h$ on $X-E$. Then let $g=f\chi_{X-E}$ (that is, redefine $f$ to be zero in $E$). Then $|g|\leq h$ everywhere, so $g$ is integrable. Then you have&#xA;\begin{align*}&#xA;\int_X g\,d\mu &amp;amp;= \int_{(X-E)\cup E}g\,d\mu = \int_(X-E)g\,d\mu + \int_E g\,d\mu\\&#xA;&amp;amp;= \int_{X-E}f\,d\mu + 0 = \int_{X-E}f\,d\mu = \int_{X-E}f\,d\mu + \int_E f\,d\mu\\&#xA;&amp;amp;= \int_{X}f\,d\,\mu,&#xA;\end{align*}&#xA;because $\int_E fd\,\mu$ exists and is equal to $0$, since $\mu(E)=0$ (so any sequence of simple functions that converges to $f$ will always have integral equal to $0$ over $E$). &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-11T19:53:19.313" LastActivityDate="2010-11-11T19:53:19.313" CommentCount="10" />
  <row Id="9822" PostTypeId="2" ParentId="9819" CreationDate="2010-11-11T05:44:47.017" Score="2" ViewCount="" Body="&lt;p&gt;The angle between two great circles is equal to the angle between the planes containing them. Such a plane always passes through the centre of the sphere, say $O$. So, for example, angle $a$ is the angle between two planes, one of which passes through $A$, $B$, and $O$, and the other passes through $A$, $C$, and $O$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The angle between two planes, in turn, equals the angle between their normals, and you can determine the normal from the three points by using the cross product.&lt;/p&gt;&#xA;" OwnerUserId="856" LastEditorUserId="856" LastEditorDisplayName="" LastEditDate="2010-11-11T05:49:53.063" LastActivityDate="2010-11-11T05:49:53.063" />
  <row Id="9823" PostTypeId="2" ParentId="9819" CreationDate="2010-11-11T05:58:22.240" Score="3" ViewCount="" Body="&lt;p&gt;The area of a spherical triangle can also be calculated using the lengths of its sides, as in&#xA;this &lt;a href=&quot;http://mathforum.org/library/drmath/view/65316.html&quot; rel=&quot;nofollow&quot;&gt;Dr.Math link.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1301" LastActivityDate="2010-11-11T05:58:22.240" />
  <row Id="9824" PostTypeId="1" AcceptedAnswerId="9832" CreationDate="2010-11-11T06:08:30.027" Score="2" ViewCount="121" Body="&lt;p&gt;You have three bags, A, B, and C. Bags A and B hold x red and y blue marbles each; bag C holds 2x red marbles and 2y blue marbles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let us say that we pull out z marbles from bag A, z marbles from bag B, and 2z marbles from bag C. Will the expected number of red marbles pulled from bag A and bag B together be the same as the expected number of marbles pulled from bag C?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe the answer is yes, but I am having trouble figuring out why.&lt;/p&gt;&#xA;" OwnerUserId="796" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-11-15T22:26:41.697" LastActivityDate="2010-11-15T22:26:41.697" Title="Probability question" Tags="&lt;probability&gt;&lt;combinatorics&gt;" AnswerCount="2" />
  <row Id="9825" PostTypeId="1" AcceptedAnswerId="10083" CreationDate="2010-11-11T06:15:04.560" Score="34" ViewCount="806" Body="&lt;p&gt;As a result of &lt;a href=&quot;http://math.stackexchange.com/questions/9532/link-between-a-dense-subset-and-a-continuous-mapping&quot;&gt;this question&lt;/a&gt;, I've been thinking about the following condition on a topological space $Y$:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;For every topological space $X$, $E\subseteq X$, and continuous maps $f,g\colon X\to Y$, if $E$ is dense in $X$, and $f$ and $g$ agree on $E$ (that is, $f(e)=g(e)$ for all $e\in E$), then $f=g$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If $Y$ is Hausdorff, then $Y$ satisfies this condition. The question is whether the converse holds: if $Y$ satisfies the above condition, will it necessarily be Hausdorff? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $Y$ is not at least $T_1$, then $Y$ does not have the property: if $u,v\in Y$ are such that $u\neq v$ and every open neighborhood of $u$ contains $v$, then let $X$ be the Sierpinski space, $X=\{a,b\}$, $a\neq b$, with topology $\tau=\{\emptyset,\{b\},X\}$, $E=\{b\}$, let $f,g\colon X\to Y$ be given by $f(a)=f(b)=v$, and $g(a)=u$, $g(b)=v$. Then both $f$ and $g$ are continuous, agree on the dense subset $E$, but are distinct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My attempt at a proof of the converse assumes the Axiom of Choice and proceeded as follows: assume $Y$ is $T_1$ but not $T_2$; let $u$ and $v$ be witnesses to the fact that $Y$ is not $T_2$, let $\mathcal{U}\_s$ and $\mathcal{V}\_t$ be the collection of all open nbds of $s$ that do not contain $t$, and all open nbds of $t$ that do not contain $s$, respectively. Construct a net with index set $\mathcal{U}\_s\times\mathcal{V}\_t$ (ordered by $(U,V)\leq (U&amp;#39;,V&amp;#39;)$ if and only if $U&amp;#39;\subseteq U$ and $V'\subseteq V$) by letting $y_{(U,V)}$ be a point in $U\cap V$ (this is where AC comes in). Let $E=\{y_{(U,V)}\mid (U,V)\in\mathcal{U}\_s\times\mathcal{V}\_t\}$, and let $X=E\cup\{s\}$. Give $X$ the induced topology; let $f\colon X\to Y$ be the inclusion map, and let $g\colon X\to Y$ be the map that maps $E$ to itself identically, but maps $s$ to $t$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only problem is I cannot quite prove that $g$ is continuous; the difficulty arises if I take an open set $\mathcal{O}\in \mathcal{V}_t$; the inverse image under $g$ is equal to $((\mathcal{O}\cap X)-\{t\})\cup\{s\}$, and I have not been able to show that this is open in $X$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Does the condition above characterize Hausdorff spaces? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If not, I would appreciate a counterexample. If it &lt;em&gt;does&lt;/em&gt; characterize Hausdorff, then ideally I would like a way to finish off my proof, but if the proof is unsalvageable (or nobody else can figure out how to finish it off either) then any proof will do. &lt;/p&gt;&#xA;&#xA;&lt;hr/&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added:&lt;/strong&gt; A little digging turned up  this question raised in the Problem Section of the &lt;strong&gt;American Mathematical Monthly&lt;/strong&gt; back in 1964 by Alan Weinstein. The &lt;a href=&quot;http://www.jstor.org/stable/2315068&quot; rel=&quot;nofollow&quot;&gt;solution by Sim Lasher&lt;/a&gt; gives a one paragraph proof that does not require one to consider $T_1$ and non-$T_1$ spaces separately. &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-14T21:08:25.210" LastActivityDate="2010-11-14T21:08:25.210" Title="Does this property characterize a space as Hausdorff?" Tags="&lt;general-topology&gt;" AnswerCount="2" CommentCount="13" FavoriteCount="4" />
  <row Id="9826" PostTypeId="2" ParentId="9815" CreationDate="2010-11-11T06:27:30.523" Score="2" ViewCount="" Body="&lt;p&gt;Your 'characteristic equation' doesn't work, as you noted, because of the presence of the variable $n$; it leaves the equation making no sense.  While the particular-solution method works well, there's another general technique that can be used (in principle) to solve recurrence relations that are linear in previous terms and polynomial in $n$: generating functions.  Consider the generating function $H(x) = \sum_{n=0}^{\infty} h[n]x^n$ and use your equation for $h[n]$ to come up with a corresponding equation for $H(x)$. Note that the generating function for $h[n-1]$ is simply $xH(x)$; the 'generating function' for a constant $c$ is $c\over 1-x$, and the generating function for $n$ is $x\over(1-x)^2$ (as can be found by taking the derivative of the generating function for 1 and multiplying by $x$).  Once you come up with a formula you can use it to solve for $H(x)$ and then use the usual partial-fraction methods to turn this back into an explicit formula for $h[n]$.&lt;/p&gt;&#xA;" OwnerUserId="785" LastActivityDate="2010-11-11T06:27:30.523" />
  <row Id="9827" PostTypeId="2" ParentId="9824" CreationDate="2010-11-11T06:33:57.250" Score="2" ViewCount="" Body="&lt;p&gt;Yes.  The expected number of red marbles from A is $\frac{xz}{x+y}$ and the same for B.  The expected number from C is $\frac{4xz}{2x+2y}$  The 4 in the numerator is 2*2, one from the number of pulls, one from the number of red marbles.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-11-11T06:33:57.250" />
  <row Id="9828" PostTypeId="2" ParentId="457" CreationDate="2010-11-11T06:37:42.627" Score="1" ViewCount="" Body="&lt;p&gt;I recall what I now remember as &quot;the Metereological theorem&quot; from college:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;At any point in time there exist two&#xA;  antipodal points on the equator that&#xA;  have the same temperature.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Trivial to prove. Only requires the mean value theorem and continuity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could demonstrate it by stating that you can not draw a simple closed curve around a circle without having two points 180 degrees apart being exactly the same distance from the circle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same is true of any continuous physical property like pressure, humidity, etc. Also, it can be extended to any three metereological properties on a sphere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's rather amazing what you can do with just continuity and the mean value theorem. &lt;/p&gt;&#xA;" OwnerUserId="3403" LastEditorUserId="3403" LastEditorDisplayName="" LastEditDate="2010-11-11T06:48:32.227" LastActivityDate="2010-11-11T06:48:32.227" CommunityOwnedDate="2010-11-11T06:37:42.627" />
  <row Id="9829" PostTypeId="2" ParentId="9803" CreationDate="2010-11-11T06:43:45.077" Score="0" ViewCount="" Body="&lt;p&gt;Certainly you can add them as elements of $V$.  They should all have the same number of components as the dimension of $V$.  But if $u_0 \in U$ why isn't $u_0+U$ just $U$?  $U$ is a vector space, so is closed under addition.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-11-11T06:43:45.077" CommentCount="2" />
  <row Id="9830" PostTypeId="1" AcceptedAnswerId="9855" CreationDate="2010-11-11T06:46:09.450" Score="2" ViewCount="150" Body="&lt;p&gt;$K$ numbers are chosen with replacement form the numbers $1,2,3, \cdots N$.In how many ways we can choose the numbers so that the maximum number chosen is exactly $R \le N $ ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The explanation given in my module is like this :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&lt;strong&gt;Since the maximum number is $R$, each of $K$ numbers can be chosen in $R$ ways.so, the number be can be chosen in $R^K$ ways.These includes the $(r-1)^k$ choices in which none of the $K$chosen numbers is $R$. Hence, the required number is $R^K - (r-1)^k $.&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not getting this explanation,and after reading &lt;a href=&quot;http://math.stackexchange.com/users/742/arturo-magidin&quot;&gt;Arturo Magidin&lt;/a&gt;'s answer I am somewhat more confused now.&lt;/p&gt;&#xA;" OwnerUserId="2109" LastEditorUserId="2109" LastEditorDisplayName="" LastEditDate="2010-11-11T15:11:28.467" LastActivityDate="2010-11-11T15:11:28.467" Title="Number of ways of choosing K numbers from $\mathbb{N}$ which doesn't exceed a maximum" Tags="&lt;combinatorics&gt;&lt;algebra-precalculus&gt;" AnswerCount="2" CommentCount="7" />
  <row Id="9831" PostTypeId="2" ParentId="9830" CreationDate="2010-11-11T06:50:16.607" Score="4" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt; My original answer was incorrect even with the interpretation I had in mind. Here is the correction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I originally gave a formula for the number of ways to pick $K$ numbers so that the maximum is &lt;em&gt;at most&lt;/em&gt; $R$, rather than exactly $R$. Moreover, I interpreted the question as asking for the number of different selections, so that if you first picked $1$ and then $2$, this was the same &quot;pick&quot; as first picking $2$ and then $1$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we interpret it that way, so that we are asking for different multisets of choices, then we are asking about combinations with repetitions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The number of ways of choosing $r$ objects from among $n$ objects with replacement is $\binom{n+r-1}{r}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the number of ways to pick $K$ from among the $R$ that are less than or equal to $R$ is $\binom{R+K-1}{K}$. However, some of them will not include the number $R$; how many? Well, $\binom{(R-1)+K-1}{K}=\binom{R-K-2}{K}$, since these are the number of choices from among the first $R-1$ instead. Thus, the total with this interpretation is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation*}&#xA;\binom{R+K-1}{K} - \binom{R+K-2}{K} = \binom{R+K-2}{K-1}.&#xA;\end{equation*}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively, reserve one spot for for $R$, and then pick the remaining $K-1$ any way you want from among the first $R$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the explanation posted tells us that in fact they are thinking of the &lt;em&gt;order&lt;/em&gt; of the choices as important: so picking $1$ first and $2$ second is different from picking $2$ first and $1$ second.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In that case, we can pick any of the first $R$ numbers at each step, $K$ steps, so that gives $R^K$ choices. In some of them we never picked $R$, however; how many? Well, if we can only pick from among $1,\ldots,R-1$, then by the same argument we have $(R-1)^K$ possibilities. So the total is&#xA;$R^K - (R-1)^K$ as explained there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Added:&lt;/em&gt; Since, as you say, you've never seen the formula for combinations with repetitions, and since it is dear to my heart (I figured it out from scratch during my first semester, when considering the question of how many different rolls with $n$ $r$-sided dice there are by constructing a big table, eventually recognizing it as Pascal's Triangle on its side, then using the combinatorial formula for the entry to get the formula, and finally coming up with a proof of it) (and that was probably more than you needed to know) here goes:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose you have $n$ elements; without loss of generality, assume they are the numbers $1$ through $n$. Now, we want to pick $r$ elements form them with repetitions, so we will have $b_1,\ldots,b_r$. Order them so that $b_1\leq b_2\leq\cdots\leq b_r$. Now consider the vector $(b_1,b_2+1,b_3+2,\ldots,b_{r}+r-1)$. Then the entries of these vector are all distinct, and each entry is a number between $1$ and $n+r-1$. Different choices lead to different vectors, so the number of possible choices with repetitions is at most the number of ways to pick $r$ objects from $n+r-1$ possibilities &lt;em&gt;without&lt;/em&gt; repetition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conversely, suppose you pick $r$ numbers from among $1,2,\ldots,n+r-1$ without repetition, $a_1,a_2,\ldots,a_r$. Order them so that $a_1\lt a_2\lt\cdots\lt a_r$. Then consider $(a_1, a_2-1, a_3-2,\ldots,a_r-r+1)$. Since $a_i\geq i$, then $a_i-(i-1)\geq 1$, and $a_i\lt a_{i+1}$ so $a_i-i+1 \leq a_{i+1}-i$. Thus, this new vector corresponds, as before, to a choice of $r$ numbers from among $1,2,\ldots,n$ &lt;em&gt;with&lt;/em&gt; repetitions allowed.  Different choices lead to different vectors, so the number of possible ways to pick $r$ objects from $n+r-1$ possibilities without repetition is at most the number of ways to pick $r$ objects from among $n$ possibilities &lt;em&gt;with&lt;/em&gt; repetions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, the two quantities are equal. Since the number of ways to pick $r$ objects from among $n+r-1$ possibilities without repetition is $\binom{n+r-1}{r}$, the formula follows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(When I proved it to myself, I took my original set and added symbols &quot;R1&quot;, &quot;R2&quot;,$\ldots$,R(r-1), which stood for &quot;repeat position 1&quot;, &quot;repeat position 2&quot;, etc. Then I picked from among $1,2,\ldots,n$,R1,R2,$\ldots$,R(r-1) without repetition, and wrote them in order, with $1\lt 2\lt\cdots\lt n\lt$ R1$\lt\cdots\lt$ R(r-1); this then gave me a &quot;pick with repetitions&quot;.)&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-11T07:26:28.857" LastActivityDate="2010-11-11T07:26:28.857" CommentCount="5" />
  <row Id="9832" PostTypeId="2" ParentId="9824" CreationDate="2010-11-11T06:59:11.680" Score="2" ViewCount="" Body="&lt;p&gt;The expected number of red balls pulled from A is $z\frac{x}{x+y}$. This is because the probability that any of the $z$ individual balls is red is $\frac{x}{x+y}$. To see that, consider the experiment of taking $z$ balls. Number each ball you've taken - the first ball number $1$, the second ball $2$, and so on. Each of the numbers on its own is uniformly distributed, so the probability that the $k$th ball is red is exactly $\frac{x}{x+y}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rest is an easy calculation, as in Ross's answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the way, the number of red balls is distributed according to a Hypergeometric distribution (q.v.).&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-11T06:59:11.680" CommentCount="4" />
  <row Id="9833" PostTypeId="1" AcceptedAnswerId="9834" CreationDate="2010-11-11T07:00:41.563" Score="2" ViewCount="168" Body="&lt;p&gt;If $X$ is a Banach space, and $T:X \to X$ is a bounded linear operator with norm &amp;lt; $1$, then $I-T$ has a bounded inverse defined by $(I-T)^{-1} = \sum_{n=0}^\infty T^n$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thinking in terms of a converse, if $T$ is any bounded linear operator defined on $X$, then does the existence of a bounded inverse $S=(I-T)^{-1}$ imply that $S$ can be represented as $S=\sum_{n=0}^\infty T^n$?   &lt;/p&gt;&#xA;" OwnerUserId="1736" LastEditorUserId="1736" LastEditorDisplayName="" LastEditDate="2010-11-11T07:08:04.890" LastActivityDate="2010-11-11T20:55:07.647" Title="If $(I-T)^{-1}$ exists, can it always be written in a series representation?" Tags="&lt;functional-analysis&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="9834" PostTypeId="2" ParentId="9833" CreationDate="2010-11-11T07:23:20.593" Score="11" ViewCount="" Body="&lt;p&gt;No, not even in the finite-dimensional case. If $T$ is a linear map&#xA;from $\mathbb{R}^n$ to itself with all eigenvalues $&gt;1$ in absolute&#xA;value, then $I-T$ is invertible, but $\sum T^n$ certainly does not&#xA;converge.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-11-11T07:23:20.593" CommentCount="1" />
  <row Id="9835" PostTypeId="1" AcceptedAnswerId="9870" CreationDate="2010-11-11T07:31:19.450" Score="1" ViewCount="120" Body="&lt;p&gt;i was reading wikipedia article on &lt;a href=&quot;http://en.wikipedia.org/wiki/Naive_Bayes_classifier&quot; rel=&quot;nofollow&quot;&gt;Naive Bayes classifier&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but i didn't understand some of the calculations.&#xA;can anybody explain. how they have calculated this&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P(height|male) = 1.5789 (A probability distribution over 1 is OK. It is the area under the bell curve that is equal to 1.)&#xA;P(weight|male) = 5.9881e-06&#xA;P(foot size|male) = 1.3112e-3&lt;/p&gt;&#xA;" OwnerUserId="1093" LastActivityDate="2010-11-11T17:22:39.470" Title="Naive Bayes classifier" Tags="&lt;probability-theory&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="9836" PostTypeId="1" AcceptedAnswerId="9839" CreationDate="2010-11-11T07:48:58.937" Score="1" ViewCount="92" Body="&lt;p&gt;How does someone approximate something like &#xA;$$\frac{1}{n}\sum_{k=1}^n e^{-k I(k)^2}$$&#xD;&#xA;where I(k) is equidistributed on $[0,1]$. Ideally I would like to show its about $K/\sqrt{n}$ for some constant $K$. This is not homework, but a question I stumbled upon &quot;in nature&quot; and I don't really have any ideas on how to tackle it.&lt;/p&gt;&#xA;" OwnerUserId="899" LastActivityDate="2010-11-11T08:24:28.610" Title="Summing something with random and non-random parts." Tags="&lt;probability&gt;&lt;analysis&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="9837" PostTypeId="2" ParentId="9833" CreationDate="2010-11-11T07:56:32.453" Score="3" ViewCount="" Body="&lt;p&gt;Well there are slightly weaker condition that improves the result you cite. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose $T$ is a bounded operator with spectral radius $r(T)&amp;lt; 1$&#xA;Then by the spectral radius formula we have &#xA;$$\lim_{n\to\infty}\|T^n\|^{1/n} =\inf_n\|T^n\|^{1/n}=r(T)&amp;lt; 1$$ &#xD;&#xA;which ensure the convergence of &#xD;&#xA;$$\sum_{n=0}^\infty\|T^n\|$$&#xD;&#xA;which in turn, by the triangle inequality, bounds &#xD;&#xA;$$\left\|\sum_{n=0}^\infty T^n\right\|$$ &#xD;&#xA;Now, it is a standard exercise to show that  &#xD;&#xA;$$\sum_{n=0}^\infty T^n=(I-T)^{-1}$$&#xD;&#xA;If there is any doubt at all do not hesitate to ask..&#xD;&#xA;&#xD;&#xA;**Edit:** By the Banach algebra inequality we have&#xD;&#xA;$\|T^n\|\le\|T\|^n$, which means that $$r(T)=\inf_k\|T^k\|^{1/k}\le\|T^n\|^{1/n}\le\|T\|$$&#xD;&#xA; Hence $\|T\|&amp;lt;1$ implies not only $r(T)&amp;lt;1$, but also $r(T)\le\|T\|&amp;lt;1$. &#xD;&#xA;&#xD;&#xA;Also, this is not the case in the example of Robin above, because we also have $r(T)=\sup{|\lambda|:\lambda\in\sigma(T)}$ where $\sigma(T)$ is the spectrum of $T$ (the set of all $\lambda\in\mathbb{C}$ such that $\lambda I-T$ is not invertible) and the eigenvaules of $T$ is certainly in the spectrum. (Note that $r(T)$ is the radius of the smallest closed disc that contain $\sigma(T)$ - hence the name &lt;em&gt;spectral radius&lt;/em&gt;). &lt;/p&gt;&#xA;" OwnerUserId="1154" LastEditorUserId="1154" LastEditorDisplayName="" LastEditDate="2010-11-11T20:55:07.647" LastActivityDate="2010-11-11T20:55:07.647" CommentCount="10" />
  <row Id="9839" PostTypeId="2" ParentId="9836" CreationDate="2010-11-11T08:24:28.610" Score="5" ViewCount="" Body="&lt;p&gt;To get a rough idea for your sum, I tried the following, don't know if it is correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First compute the expected value of $e^{-k x^2}$ as&#xA;$$\int_0^1 e^{-k x^2} dx = \frac{\sqrt{\pi}\mbox{erf}(\sqrt{k})}{2\sqrt{k}}$$&#xD;&#xA;&#xD;&#xA;Now, I used the cheap bound that $\mbox{erf}(\sqrt{k}) &amp;lt; 1$, to see that in expected value each term of your sum is approximately $c/(2\sqrt{k})$, for some constant $c$. Thus, the your sum is approximately:&#xD;&#xA;$$\sum_{k=1}^n \frac{c}{2\sqrt{k}} \approx K\sqrt{n},$$&#xD;&#xA;&#xD;&#xA;so that after using the normalizing $1/n$ outside the sum, the approximate value of your sum is indeed $K/\sqrt{n}$.&lt;/p&gt;&#xA;" OwnerUserId="1709" LastActivityDate="2010-11-11T08:24:28.610" CommentCount="1" />
  <row Id="9840" PostTypeId="1" AcceptedAnswerId="9852" CreationDate="2010-11-11T08:54:22.300" Score="0" ViewCount="315" Body="&lt;p&gt;The formula for finding the roots of a polynomial is as follows&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x = \frac {-b \pm \sqrt{ b^2 - 4ac }}{2a} $$&#xD;&#xA;&#xD;&#xA;what happens if you want to find the roots of a polynomial like this simplified one&#xD;&#xA;&#xD;&#xA;$$ 3x^2 + x + 24 = 0 $$&#xD;&#xA;&#xD;&#xA;then the square root value becomes&#xD;&#xA;&#xD;&#xA;$$ \sqrt{ 1^2 - 4\cdot3\cdot24 } $$&#xD;&#xA;&#xD;&#xA;&#xD;&#xA;$$ = \sqrt{ -287 } $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is the square root of a negative number, which isn't allowed. What do you do in this case? I know there are other methods, i.e. factorisation and completing the square, but does this mean that this formula can only be used in specialised cases or have i gone wrong somewhere along the path?&lt;/p&gt;&#xA;" OwnerUserId="676" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-11T10:34:28.400" LastActivityDate="2010-11-12T03:20:11.593" Title="Finding roots of polynomials, negative square root" Tags="&lt;algebra-precalculus&gt;&lt;polynomials&gt;" AnswerCount="3" CommentCount="4" />
  <row Id="9841" PostTypeId="1" AcceptedAnswerId="9847" CreationDate="2010-11-11T08:57:03.860" Score="0" ViewCount="118" Body="&lt;p&gt;Is the average of i.i.d. random variables with zero mean and finite variance convergent to 0 in probability or $L^2$? How do I show this? I am just beginning to learn convergence of random variables, so an example would be helpful.&lt;/p&gt;&#xA;" OwnerUserId="957" LastActivityDate="2010-11-11T11:30:33.433" Title="convergence of average of iid random variables" Tags="&lt;probability-theory&gt;&lt;measure-theory&gt;&lt;probability&gt;&lt;convergence&gt;" AnswerCount="2" />
  <row Id="9842" PostTypeId="2" ParentId="9840" CreationDate="2010-11-11T08:59:48.753" Score="3" ViewCount="" Body="&lt;p&gt;Yeah so have you read about complex numbers. The root will be &#xA;$$x = \frac{-1 \pm{i} \sqrt{287}}{2 \times 3}$$&#xD;&#xA;&#xD;&#xA;where $i=\sqrt{-1}$. Read more about complex numbers and when a polynomial can have complex roots, that happens when the *Discriminant* factor $b^{2}-4ac &amp;lt;0$. &lt;/p&gt;&#xA;" LastActivityDate="2010-11-11T08:59:48.753" CommentCount="4" />
  <row Id="9843" PostTypeId="2" ParentId="9840" CreationDate="2010-11-11T09:04:37.770" Score="4" ViewCount="" Body="&lt;p&gt;Hi Aran,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;100% correct, and good observation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To solve this, we define $\sqrt{-1}=i$ where $i$ is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Imaginary_unit&quot; rel=&quot;nofollow&quot;&gt;imaginary unit&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then $\sqrt{-287}=\sqrt{287}i$, and we can solve as per the general quadratic formula. Numbers of the form $a+bi$ are known as complex numbers and are extremely useful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general the term $b^2-4ac$ is known as the discriminant of the quadratic equation. It should be clear that if $b^2-4ac&gt;0$ there exists two real solutions, if $b^2-4ac=0$ there is one solution (the repeated root) and if &#xA;$$b^2-4ac \lt 0$$ there are two complex solutions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The quadratic formula is the most general way to solve the quadratic equation - so you are doing the right thing.&lt;/p&gt;&#xA;" OwnerUserId="2219" LastEditorUserId="2219" LastEditorDisplayName="" LastEditDate="2010-11-12T03:20:11.593" LastActivityDate="2010-11-12T03:20:11.593" CommentCount="4" />
  <row Id="9844" PostTypeId="1" CreationDate="2010-11-11T10:44:42.673" Score="0" ViewCount="75" Body="&lt;p&gt;Suppose $h$ is a function on $Z$ that is periodic with period N, that is $h(n+N)=h(n)$ for all $n$. How to prove: for any $m \in Z$, &#xA;$\sum_{n=m}^{m+N-1} h(n) =\sum_{n=0}^{N-1} h(n)$. In other words, any sum over an interval of length $N$ yields the same result.&#xA;This is problem from the area of linear algebra and is a homework&lt;/p&gt;&#xA;" OwnerUserId="2723" LastEditorUserId="2723" LastEditorDisplayName="" LastEditDate="2010-11-11T11:26:27.087" LastActivityDate="2010-11-11T13:58:18.350" Title="How to prove: for any $m \in Z$, $\sum_{n=m}^{m+N-1} h(n) =\sum_{n=0}^{N-1} h(n)$" Tags="&lt;linear-algebra&gt;&lt;homework&gt;" AnswerCount="3" CommentCount="4" />
  <row Id="9845" PostTypeId="2" ParentId="9841" CreationDate="2010-11-11T10:52:55.617" Score="0" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Law_of_large_numbers&quot; rel=&quot;nofollow&quot;&gt;Law of large numbers&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="3223" LastActivityDate="2010-11-11T10:52:55.617" />
  <row Id="9846" PostTypeId="1" AcceptedAnswerId="23299" CreationDate="2010-11-11T10:56:53.273" Score="4" ViewCount="321" Body="&lt;p&gt;One way to tessellate a 3D sphere is by iterated subdivision of an icosahedron. I am wondering whether this method gives a homogeneous surface density of vertices. To the eye, it seems to do so, and logic indicates that too (each face has the same are in the icosahedron, and faces in each subdivision are created of equal area), but is there some bias that I'm not thinking of?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Otherwise, what tessellation method can yield a constant density of vertices?&lt;/p&gt;&#xA;" OwnerUserId="3406" LastActivityDate="2011-02-22T23:55:57.167" Title="Which tessellation of the sphere yields a constant density of vertices?" Tags="&lt;geometry&gt;" AnswerCount="3" CommentCount="4" FavoriteCount="1" />
  <row Id="9847" PostTypeId="2" ParentId="9841" CreationDate="2010-11-11T10:58:23.457" Score="2" ViewCount="" Body="&lt;p&gt;Suppose that $X_1,X_2,\ldots$ is a sequence of i.i.d. random variables with ${\rm E}(X_i)=0$ and ${\rm Var }(X_i) = \sigma^2$. Then, by the strong law of large numbers, the average $\bar X_n  = \frac{{\sum\nolimits_{i = 1}^n {X_i } }}{n}$ converges to $0$ almost surely, hence, in particular, in probability. However, this can also be shown directly as follows. By definition of converges in probability, we need to show that, for any $\varepsilon &gt; 0$, ${\rm P}(|\bar X_n - 0| \geq \varepsilon) \to 0$ as $n \to \infty$. We have ${\rm E}(\bar X_n) = 0$ and ${\rm Var}(\bar X_n) = \sigma^2/n$. Hence, by Chebyshev's inequality, &#xA;$$&#xD;&#xA;{\rm P}(|\bar X_n - 0|  \ge \varepsilon ) \le \frac{{{\rm Var}(\bar X_n )}}{{\varepsilon ^2 }} \to 0.&#xD;&#xA;$$&#xD;&#xA;The convergence in probability is thus proved. As for the convergence to $0$ in $L^2$, this follows straight from the definition. Indeed,&#xD;&#xA;$$&#xD;&#xA;{\rm E}(|\bar X_n  - 0|^2 ) = {\rm Var}(\bar X_n ) \to 0.&#xD;&#xA;$$  &lt;/p&gt;&#xA;" OwnerUserId="2810" LastEditorUserId="2810" LastEditorDisplayName="" LastEditDate="2010-11-11T11:30:33.433" LastActivityDate="2010-11-11T11:30:33.433" CommentCount="1" />
  <row Id="9848" PostTypeId="2" ParentId="9844" CreationDate="2010-11-11T11:07:11.267" Score="1" ViewCount="" Body="&lt;p&gt;Let $m \in \{0 \dots, N-1\}$. Then we have&#xA;$$\sum_{n=m}^{m+N-1} h(n) = \sum_{n=m}^{N-1} h(n) + \sum_{n=0}^{m-1} h(n + N) = \sum_{n=m}^{N-1} h(n) + \sum_{n=0}^{m-1} h(n) = \sum_{n=0}^{N-1} h(n)$$&#xD;&#xA;&#xD;&#xA;Now for $m \in \mathbb{Z}$ write $m = kN + r$ with $k \in \mathbb{Z}$ and $r \in \{0, \dots, N-1\}$. We clearly have $$\sum_{n=kN+r}^{k(N+1) + r -1} h(n) = \sum_{n=r}^{N+r-1} h(n+kN) = \sum_{n=r}^{N+r-1} h(n)$$&#xD;&#xA;Now the last expression equals $\sum_{n=0}^{N-1} h(n)$ by the first part of the argument.&lt;/p&gt;&#xA;" OwnerUserId="3223" LastEditorUserId="3223" LastEditorDisplayName="" LastEditDate="2010-11-11T11:38:22.503" LastActivityDate="2010-11-11T11:38:22.503" CommentCount="3" />
  <row Id="9849" PostTypeId="1" CreationDate="2010-11-11T11:12:37.300" Score="0" ViewCount="73" Body="&lt;p&gt;How to prove that $N^{-1/2} W_N$ is an unitary matrix( Suggestion: Use lemma 2.2 and lemma 1.105 ii.) \&#xA;Lemma 1.105 &#xA;Let $A$ be an $nxn$ matrix over $C$. Then the following statements are equivalent: &#xA;i. A is unitary&#xA;ii. The columns of A form an orthonormal basis for $C^n$&#xA;iii. The rows of A form an orthonormal basis for $C^n$&#xA;iv. A preserves inner products, that is, $\langle Az, Aw \rangle =\langle z, w \rangle $ for all $z,w \in C^n$&#xA;v. $||Az|| = ||z||$, for all $z \in C^n$ \&#xA;lemma 2.2&#xA;The set ${E_0,...,E_{N-1} }$ is an orthonormal basis for $l^2(Z_N)$.&#xA;Also $W_N = \left(\begin{array}{ccccccc}&#xD;&#xA;1 &amp; 1 &amp; 1 &amp; 1 &amp; . &amp; . &amp; 1  \\&#xD;&#xA;1 &amp; \omega_N &amp; \omega_N^2 &amp; \omega_N^3 &amp; . &amp; . &amp; \omega_N^{N-1} \\&#xD;&#xA;1 &amp; \omega_N^2 &amp; \omega_N^4 &amp; \omega_N^6 &amp; . &amp; . &amp; \omega_N^{2(N-1)} \\&#xD;&#xA;1 &amp; \omega_N^3 &amp; \omega_N^6 &amp; \omega_N^9 &amp; . &amp; . &amp; \omega_N^{3(N-1)} \\&#xD;&#xA;. &amp; . &amp; . &amp; . &amp; . &amp; . &amp; . \\&#xD;&#xA;. &amp; . &amp; . &amp; . &amp; . &amp; . &amp; . \\&#xD;&#xA;1 &amp; \omega_N^{N-1} &amp; \omega_N^{2(N-1)} &amp; \omega_N^{3(N-1)} &amp; . &amp; . &amp; \omega_N^{(N-1)(N-1)} \\&#xD;&#xA;\end{array}\right)$, &#xA;Where $\omega_N = e^{-2 \pi i/N}$ then &#xA;$\omega_N^{mn} = e^{-2 \pi i mn/N}$ &#xA;This a homework in the area of linear algebra. And I have no idea how to procede in the proof, because it might become quite technical proof if you would just make straight computation, but I hope it is done by lemma 1.105 and lemma 2.2.&lt;/p&gt;&#xA;" OwnerUserId="2723" LastEditorUserId="1154" LastEditorDisplayName="" LastEditDate="2010-11-11T11:33:08.173" LastActivityDate="2010-12-11T14:44:24.980" Title="How to prove that $N^{-1/2} W_N$ is an unitary matrix?" Tags="&lt;linear-algebra&gt;&lt;homework&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="9850" PostTypeId="2" ParentId="9844" CreationDate="2010-11-11T11:28:13.733" Score="1" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT:&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Expand&#xA;$$\sum_{n=m}^{m+N-1}h(n)$$&#xD;&#xA;What do you know about $h$?&lt;/p&gt;&#xA;" OwnerUserId="1154" LastActivityDate="2010-11-11T11:28:13.733" />
  <row Id="9851" PostTypeId="1" AcceptedAnswerId="10791" CreationDate="2010-11-11T11:39:20.567" Score="10" ViewCount="265" Body="&lt;p&gt;If for a function $f(x)$ only its absolute value $|f(x)|$ and the absolute value $|\tilde f(k)|$ of its Fourier transform $\tilde f(k)=N\int f(x)e^{-ikx} dx$ is known, can $f(x) = |f(x)|e^{i\phi(x)}$ and thus the phase function $\phi(x)$ be extracted? (with e.g. $N=1/(2\pi)$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As Marek already &lt;a href=&quot;http://math.stackexchange.com/questions/9851/can-the-phase-of-a-function-be-extracted-from-only-its-absolute-value-and-its-fou/9854#9854&quot;&gt;stated&lt;/a&gt;, this is even not uniquely possible for $f(x)=c\in\mathbb C$, since the global phase cannot be re-determined. So please let me extend the question to&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Under what circumstances is the phase-retrieval (up to a global phase) uniquely possible, and what ambiguities could arise otherwise?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="163" LastEditorUserId="163" LastEditorDisplayName="" LastEditDate="2010-11-12T09:36:31.587" LastActivityDate="2010-12-22T09:17:55.817" Title="Can the phase of a function be extracted from only its absolute value and its Fourier transform's absolute value?" Tags="&lt;analysis&gt;&lt;functional-analysis&gt;&lt;fourier-analysis&gt;" AnswerCount="3" CommentCount="8" FavoriteCount="2" />
  <row Id="9852" PostTypeId="2" ParentId="9840" CreationDate="2010-11-11T11:56:29.627" Score="9" ViewCount="" Body="&lt;p&gt;The other answers are nice, so I won't reiterate them.  It's worth pointing out, though, that when you say &quot;I know there are other methods, i.e. factorisation and completing the square, but does this mean that this formula can only be used in specialised cases&quot;, you seem to be implying that these other methods might yield different results.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This isn't the case.  If the quadratic formula fails to give you real-valued answers, then the other two methods (and every applicable method) will also fail, because the quadratic polynomial itself simply doesn't &lt;em&gt;have&lt;/em&gt; any real-valued roots.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can visualize this easily: as you may know, the graph of a quadratic polynomial is a parabola, and the real roots of a polynomial correspond to its graphs' zeros, i.e. points where the graph crosses the $x$-axis.  So if you imagine a parabola with its vertex above the $x$-axis and opening upwards, you can see that it will &lt;em&gt;never&lt;/em&gt; cross the $x$-axis -- it has no real zeros at all!&#xA;&lt;img src=&quot;http://i.imgur.com/wNzW3.png&quot; alt=&quot;alt graph of quadratic polynomial&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is exactly the situation with $y = 3x^2 + x + 24$ and every other quadratic polynomial for which $b^2 - 4ac$ (the discriminant) is negative.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lastly (since it seems as though this may be one of your first introductions to complex numbers) let me clarify something that confuses my students greatly: don't let the words &quot;real&quot; and &quot;imaginary&quot; mislead you into thinking that real numbers are somehow genuine or actual and imaginary numbers are somehow fake.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just like a &quot;negative&quot; number isn't sad, cynical, or irritated, and just as an &quot;irrational&quot; number isn't my crazy ex-girlfriend, &quot;real&quot; and &quot;imaginary&quot; are technical terms, and not meant to carry with them their English connotations.  Although they earned those names due to prior misunderstandings about numbers, today we know that imaginary numbers are just as actual and valid as real numbers, and in fact complex (not &quot;complicated&quot;) numbers have huge real-world applications and consequences.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Along these same lines: you say that taking the square root of a negative number &quot;isn't allowed&quot;.  More accurate would be to say that the square root of a negative number isn't a real number (&quot;real&quot;, again, in the technical sense).  If you require the result of taking the square root to be a real number, then you're right, taking the square root of a negative isn't allowed.  But otherwise it's perfectly fine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By analogy: you're allowed to divide the whole number 6 by the whole number 3, and the result is the whole number 2.  But if you divide the whole number 1 by the whole number 2, then the result, $\frac{1}{2}$, isn't a whole number.  If you require your operations on a certain class of numbers to produce a result from that same class of numbers, then lots of operations (like division, and subtraction too, for that matter) will have situations in which they aren't &quot;allowed&quot; to occur.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This requirement is sometimes very desirable, and there are certainly contexts in which you will want to impose it.  But you should understand that imposing this requirement is a choice that you make because it fits the context of the problems you're working on.  It's not divine law.  If you find yourself in a situation in which you would like to take the square root of a negative number and produce the corresponding imaginary number, you are definitely &quot;allowed&quot; to do so.&lt;/p&gt;&#xA;" OwnerUserId="506" LastEditorUserId="506" LastEditorDisplayName="" LastEditDate="2010-11-12T02:58:22.973" LastActivityDate="2010-11-12T02:58:22.973" CommentCount="3" />
  <row Id="9853" PostTypeId="2" ParentId="9844" CreationDate="2010-11-11T13:28:47.857" Score="1" ViewCount="" Body="&lt;p&gt;The important point here is that any set of $N$ consecutive integers is a complete residue system modulo $N$. See &lt;a href=&quot;http://mathrefresher.blogspot.com/2008/10/complete-residue-system.html&quot; rel=&quot;nofollow&quot;&gt;complete residue system.&lt;/a&gt; We consider the index modulo $N$ since $h(n)$ has period $N.$ i.e. the numbers&#xA;$ \lbrace m,m+1,m+2,\ldots,m+N-1 \rbrace $ modulo $N$ are $ \lbrace 0,1,2,\ldots,N-1 \rbrace $ in some order.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your identity follows immediately from this fact.&lt;/p&gt;&#xA;" OwnerUserId="1301" LastEditorUserId="1301" LastEditorDisplayName="" LastEditDate="2010-11-11T13:58:18.350" LastActivityDate="2010-11-11T13:58:18.350" />
  <row Id="9854" PostTypeId="2" ParentId="9851" CreationDate="2010-11-11T13:45:48.963" Score="5" ViewCount="" Body="&lt;p&gt;The short answer is no. Take constant function $f(x) \equiv C \in \mathbb{C}$. Disregarding normalization, we have $\hat{f} = C \delta$ (in the sense of distributions). Clearly, there is no way to recover the phase of $C$ once we take the absolute value on both sides.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To make this a little more explicit, consider a lot easier version of the problem on the group $G = \mathbb{Z} / N\mathbb{Z}$. Its dual is $\hat{G} = G$. If you'll write out the Fourier transform equations (i.e. $\hat{f}(k) = \sum_{n=0}^{N-1} f(n) \exp(-{i k n \over 2 \pi})$), you'll obtain $2N$ real equations for $2N$ coefficients ($N$ Fourier phases and $N$ original phases). The properties of this system of equations are not clear to me, but the case $N=1$ (this is the same as in the first paragraph, but here we don't need to talk about distributions) already shows that the solutions need not be unique.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope someone else can provide more information, I'd be also interested to see what conditions on $f$ one needs to assume to get a unique solution. Even for the case $G = \mathbb{Z} / N\mathbb{Z}$ this looks interesting enough.&lt;/p&gt;&#xA;" OwnerUserId="3223" LastEditorUserId="3223" LastEditorDisplayName="" LastEditDate="2010-11-11T14:00:17.033" LastActivityDate="2010-11-11T14:00:17.033" CommentCount="5" />
  <row Id="9855" PostTypeId="2" ParentId="9830" CreationDate="2010-11-11T14:04:26.663" Score="2" ViewCount="" Body="&lt;p&gt;The explanation in your module is a common technique in combinatorics.  If you want to count how many there are of something, count some set that includes all of them and some more, then subtract out the members of the set that don't belong.  We can work a small example (unfortunately the numbers get large quickly). Let K=4, R=2.  N doesn't matter in this problem as we just throw away all the numbers $\{R+1, R+2, R+3,\ldots,N\}$.  There are $2^4=16$ strings of four numbers with each number being 1 or 2.  You can list them-they look like the binary numbers from 0 to 15 with the zeros changed to twos.  But one =1^4 of them is all ones, so we shouldn't count it.  The answer is $15=2^4-1^4=R^K-(R-1)^K$&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-11-11T14:04:26.663" CommentCount="1" />
  <row Id="9856" PostTypeId="2" ParentId="9145" CreationDate="2010-11-11T14:05:45.213" Score="3" ViewCount="" Body="&lt;p&gt;What you want is called the &quot;Joint Distribution Function&quot; of the dice.  That is, for two dice with random variables $X$ and $Y$ measuring the number showing on the faces, it is $P(X=x,Y=y)$.  Since the dice are independent, the JDF &lt;i&gt;is&lt;/i&gt; the product of the individual distribution functions of $X$ and $Y$ (what you call the mass function):&#xA;$$P(X=x,Y=y) = P(X=x)P(Y=y).$$&lt;/p&gt;&#xA;" OwnerUserId="3009" LastActivityDate="2010-11-11T14:05:45.213" />
  <row Id="9857" PostTypeId="2" ParentId="9803" CreationDate="2010-11-11T14:25:07.493" Score="1" ViewCount="" Body="&lt;p&gt;Well, let's go back to the definitions. I am sorry if I go too far into the details, but it is difficult to judge what you exactly understand based only on your question. So maybe I explain too much, but don't take this as an offense.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What exactly is $V/U$ ? I will explain this a little, since it might be the source of the problem. $V/U$ is the set of the class of equivalences of vectors from $V$. The equivalence relation is given by $x \sim y :\iff x-y \in U$. This set can naturally be considered as a vector space if you define:&#xA;$$[x] + [y] := [x+y] \text{ and } \lambda[x] := [\lambda x]$$&#xD;&#xA;&#xD;&#xA;Let's take an example: $V = \mathbb{R}^4$, $U = \mathbb{R}^2$, $W = \mathbb{R}$. First, what does it mean to consider $\mathbb{R}^2$ and $\mathbb{R}$ as subspaces of $\mathbb{R}^4$ ? Well, we need to actually consider $\mathbb{R}^2 \times \{(0,0)\}$ and $\mathbb{R} \times \{(0,0,0)\}$. Thus, actually $U = \{ x = (x_1, x_2, x_3, x_4) \in \mathbb{R}^4 | x_3 = x_4 = 0\}$ and $W = \{ x = (x_1, x_2, x_3, x_4) \in \mathbb{R}^4 | x_2 = x_3 = x_4 = 0\}$.&#xD;&#xA;&#xD;&#xA;So a first equivalence relation $\sim_A$ can be defined for $V/U$:&#xD;&#xA;$$x \sim_A y :\iff x_3 = y_3 \text{ and } x_4 = y_4$$&#xD;&#xA;&#xD;&#xA;Same for $W/U$:&#xD;&#xA;$$x \sim_B y :\iff x_2 = y_2, x_3 = y_3, x_4 = y_4$$&#xD;&#xA;&#xD;&#xA;Thus let's write, $V/U = \{[x]_A \}$ and $V/W = \{[x]_B \}$, where obviously $[x]_A = \{y \in \mathbb{R}^4 | y_3 = x_3, y_4 = x_4\}$ and $[x]_B = \{y \in \mathbb{R}^4 | x_2 = y_2, x_3 = y_3, y_4 = x_4\}$.&#xD;&#xA;&#xD;&#xA;Note that $V/U$ has dimension 2 (a base is given by $\{[(0,0,1,0)]_A , [(0,0,0,1)]_A\}$) and that $V/W$ has dimension 3 (a base is given by $\{[(0,1,0,0)]_B , [(0,0,1,0)]_B, [(0,0,0,1)]_B\}$ ).&#xD;&#xA;&#xD;&#xA;Now, how can we define an addition $[u]_A + [v]_B$ ? This would be trying to add sets. You can try to define $[u]_A + [v]_B$ in a natural way but you won't manage it: simply saying $[u]_A + [v]_B := (u_1+v_1, u_2 + v_2 , u_3 + v_3, u_4 + v_4)$ won't work, the addition will not be well-defined (because if you change the representant of $[u]_A$, $u_1$ and $u_2$ can be anything, and if you change representant $[v]_B$, $v_1$ can be anything).&#xD;&#xA;&#xD;&#xA;&#xD;&#xA;&#xD;&#xA;Furthermore, your intuition should tell you that since both dimensions of $V/U$ and $V/W$ are 2 and 3, such an addition can only be defined properly in a vector space of at least dimension 5 (otherwise you lose information). Too bad $\mathbb{R}^4$ has only dimension 4. So you should be convinced that it is impossible to define a natural addition properly in $V$ for this example.&#xD;&#xA;&#xD;&#xA;&#xD;&#xA;(Caution: I don't say it is impossible to define an addition inside $V$. Actually, it is possible to define several additions, but it won't be &quot;natural&quot;)&#xD;&#xA;&#xD;&#xA;&#xD;&#xA; &#xD;&#xA;Hence, the only way to do it properly is to consider $V/U \oplus V/W$. Of course, in special cases, where $\dim(V/U) + \dim(V/W) \leq \dim(V)$, then it is possible to rearrange things so that you consider $V/U \oplus V/W$ as a subspace of $V$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sorry for the long answer, that depending on your level of understanding might not be very clear.&lt;/p&gt;&#xA;" OwnerUserId="2778" LastEditorUserId="2778" LastEditorDisplayName="" LastEditDate="2010-11-11T14:33:11.090" LastActivityDate="2010-11-11T14:33:11.090" CommentCount="2" />
  <row Id="9858" PostTypeId="2" ParentId="9849" CreationDate="2010-11-11T14:34:25.097" Score="1" ViewCount="" Body="&lt;p&gt;Following Slowsolver,note that the entries along a row or column are in geometric progression.  So just write down $W_N^*W_N$ from the definition of a matrix multiplication.  It will be a sum of N terms in geometric progression, which you can sum.  The powers in each sum off the main diagonal will form a permutation of $\{0,1,2,\ldots,N-1\}$.  As they are all the $N^{th}$ roots of unity, they sum to zero.  Along the main diagonal you will get all 1's, which sum to N.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-11-11T14:34:25.097" />
  <row Id="9859" PostTypeId="2" ParentId="9846" CreationDate="2010-11-11T14:58:04.783" Score="2" ViewCount="" Body="&lt;p&gt;See the links below. They're not about tesselation, but about evenly distributing points on a sphere (from which you can extract a tesselation):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://local.wasp.uwa.edu.au/~pbourke/geometry/spherepoints/&quot; rel=&quot;nofollow&quot;&gt;http://local.wasp.uwa.edu.au/~pbourke/geometry/spherepoints/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.math.vanderbilt.edu/~esaff/texts/161.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.math.vanderbilt.edu/~esaff/texts/161.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www2.research.att.com/~njas/icosahedral.codes/&quot; rel=&quot;nofollow&quot;&gt;http://www2.research.att.com/~njas/icosahedral.codes/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://sitemason.vanderbilt.edu/page/hmbADS#code&quot; rel=&quot;nofollow&quot;&gt;http://sitemason.vanderbilt.edu/page/hmbADS#code&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://eqsp.sourceforge.net/&quot; rel=&quot;nofollow&quot;&gt;http://eqsp.sourceforge.net/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.math.niu.edu/~rusin/known-math/index/spheres.html&quot; rel=&quot;nofollow&quot;&gt;http://www.math.niu.edu/~rusin/known-math/index/spheres.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="589" LastActivityDate="2010-11-11T14:58:04.783" />
  <row Id="9860" PostTypeId="2" ParentId="9846" CreationDate="2010-11-11T15:05:30.220" Score="1" ViewCount="" Body="&lt;p&gt;I've got my answer from one of the links (http://www.math.vanderbilt.edu/~esaff/texts/161.pdf) given by lhf, in the last paragraph of page 9 and first of page 10:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What's more, they are not&#xA;  asymptotically uniformly distributed.&#xA;  To see this, it is enough to observe&#xA;  that after the first step, the points&#xA;  are centered at 80 spherical triangles&#xA;  which do not all have the same area&#xA;  (the projection process in- creases&#xA;  the areas of the &quot;middle triangles&quot;&#xA;  more than the rest). As the further&#xA;  steps in the process yield the same&#xA;  number of points in each of the 80&#xA;  triangles, asymptotic uniform&#xA;  distribution cannot hold&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Thanks for the link!&lt;/p&gt;&#xA;" OwnerUserId="3406" LastActivityDate="2010-11-11T15:05:30.220" CommentCount="1" />
  <row Id="9861" PostTypeId="1" CreationDate="2010-11-11T15:11:07.510" Score="2" ViewCount="48" Body="&lt;p&gt;I have two sets (a,b) of data and I am restricted to using the &#xA;values in the sets once and only once when they are combined.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, the 2nd highest value possible from &#xA; a data set of a{3,2} and b{2,0} is a1+b2=3. &#xA;There are two possible ways of combining these two data sets: &#xA; (1) a1+b1 and a2+b2 which results in values of 5 and 2, and &#xA; (2) a1+b2 and a2+b1 which results in values of 3 and 4. &#xA;The (2) combination results in the highest (or maximum) 2nd-high value &#xA;possible of 3. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ergo, the 3rd highest value possible from &#xA; a data set of a{3,2,1} and b{4,3,0} is a1+b3=3. &#xA;There are six possible ways of combining these two data sets: &#xA;(1) a1+b1 followed by a2+b2  and a3+b3, which results in values of 7, 5 and 1, and &#xA;(2) a1+b1 followed by a2+b3  and a3+b2, which results in values of 7, 2 and 4, and &#xA;(3) a1+b2 followed by a2+b1  and a3+b3, which results in values of 6, 6 and 1, and &#xA;(4) a1+b2 followed by a2+b3  and a3+b1, which results in values of 5, 2 and 5, and &#xA;(5) a1+b3 followed by a2+b1  and a3+b2, which results in values of 3, 6 and 4, and &#xA;(6) a1+b3 followed by a2+b2  and a3+b1, which results in values of 3, 5 and 5. &#xA;The (5) and (6) combinations both result in a maximum 3rd-highest value of 3, &#xA;  from a1+b3. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;By brute force and analogy, I have determined the maximum 1st through nth highest combination of ai+bi in two data sets can be determined as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maximum             Formula &#xA;1st-high            a1+b1 &#xA;2nd-high            min(a1+b2, a2+b1) &#xA;3rd-high            min(a1+b3, a2+b2, a3+b1) &#xA;4th-high            min(a1+b4, a2+b3, a3+b2, a4+b1) &#xA;... &#xA;8th-high            min(a1+b8, a2+b7, a3+b6, a4+b5, a5+b4, a6+b3, a7+b2, a8+b1) &#xA;... &#xA;nth-high            min(a1+bn, a2+b(n-1), ... , a(n-1)+b2, an+b1) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a handle on the formula, what I want to know is &#xA;(a) where the answer to this question has been published, and&#xA;(b) what it is formally described by in mathematics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am sure this is not the first time this has come up.&#xA;Is it addressed in a text book or in the literature?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" LastActivityDate="2010-11-11T15:27:42.270" Title="Maximum combined 1st through nth highest combinations of ai + bi" Tags="&lt;algebra-precalculus&gt;&lt;statistics&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="9862" PostTypeId="2" ParentId="9861" CreationDate="2010-11-11T15:27:42.270" Score="1" ViewCount="" Body="&lt;p&gt;Let $a = 10,9,0$ and $b = 10,8,0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First highest is $a_1+b_1 = 20$, second highest is $a_2+b_1 = 19$, third highest is $a_1+b_2 = 18$. Third highest invalidates your formula.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-11T15:27:42.270" CommentCount="1" />
  <row Id="9863" PostTypeId="1" AcceptedAnswerId="9897" CreationDate="2010-11-11T15:35:29.257" Score="1" ViewCount="119" Body="&lt;p&gt;I am trying to understand (simplify) the following space&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$Z = S^2 \times S^2 / E$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $(a_1 , a_2 , a_3) \in S^2$ and $(n_1 , n_2 , n_3) \in S^2$. Here ${a_1}^2 + {a_2}^2 + {a_3}^2 = 1$ and ${n_1}^2 + {n_2}^2 + {n_3}^2 = 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The equivalence relations $E$ are the following&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) $[(a_1, a_2, a_3);(n_1, n_2, n_3)] \sim [(-a_1, -a_2, -a_3);(n_1, n_2, n_3)]$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) $[(a_1, a_2, a_3);(n_1, n_2, n_3)] \sim [(a_1, a_2, a_3); (\pi; a_1,a_2,a_3)*(-n_1, -n_2, -n_3)]$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $(\pi; a_1,a_2,a_3)$ is a rotation of angle $\pi$ along the axis $(a_1, a_2, a_3)$. So the point $(\pi; a_1,a_2,a_3)*(-n_1, -n_2, -n_3)$ is the mirror image of $(n_1, n_2, n_3)$ in the plane whose normal is $(a_1, a_2, a_3)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The objective is to simplify the equivalence relation (2) by a homeomorphism of $S^2 \times S^2$. For example, the approach I am trying is to rotate the second $S^2$ such that the mirror plane becomes horizontal (XY plane). Since the normal of the mirror plane is $(a_1, a_2, a_3)$, the amount by which the second $S^2$ has to be rotated depends on the $(a_1, a_2, a_3)$ it corresponds to. And for this transformation to be a homeomorphism the rotation operations should vary continuously. The transformation should not complicate the first equivalence relation; one way this condition could be imposed is by ensuring that the rotation operation corresponding to $(a_1, a_2, a_3)$ and $(-a_1, -a_2, -a_3)$ are be the same. I am not able to come up with such a transformation. All my attempts to simplify (2) end up complicating (1). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To be more specific, I want to get a homeomorphism between the space $Z$ and the following space:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$Z^' = S^2 \times S^2 / E^' $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The equivalence relations $E^'$ are the following&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) $[(a_1, a_2, a_3);(n_1, n_2, n_3)] \sim [(-a_1, -a_2, -a_3);(n_1, n_2, n_3)]$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) $[(a_1, a_2, a_3);(n_1, n_2, n_3)] \sim [(a_1, a_2, a_3); (n_1, n_2, -n_3)]$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any ideas are appreciated. Please let me know if you have any questions.&lt;/p&gt;&#xA;" OwnerUserId="3050" LastEditorUserId="3050" LastEditorDisplayName="" LastEditDate="2010-11-11T18:02:00.360" LastActivityDate="2010-11-11T19:56:08.360" Title="Equivalence relations on $S^2 \times S^2$" Tags="&lt;topology&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="2" />
  <row Id="9864" PostTypeId="2" ParentId="8814" CreationDate="2010-11-11T15:53:30.517" Score="9" ViewCount="" Body="&lt;p&gt;$\displaystyle\big(a^2+b^2\big)\cdot\big(c^2+d^2\big)=\big(ac \mp bd\big)^2+\big(ad \pm bc\big)^2$&lt;/p&gt;&#xA;" OwnerUserId="1747" LastActivityDate="2010-11-11T15:53:30.517" CommunityOwnedDate="2010-11-11T15:53:30.517" CommentCount="3" />
  <row Id="9865" PostTypeId="2" ParentId="457" CreationDate="2010-11-11T16:50:36.673" Score="1" ViewCount="" Body="&lt;p&gt;I think one that comes up often in discussions like this is &lt;a href=&quot;http://en.wikipedia.org/wiki/Brouwer_fixed_point_theorem&quot; rel=&quot;nofollow&quot;&gt;Brouwers Fixed point theorem&lt;/a&gt; which I think can be demonstrated by putting a map of your country on a table and asserting that one of the points on the map is directly above the corresponding point in your country (ie where you are right now).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another way i think is simpler is to take two sheets of paper, crumple one up and place it on top of the other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I remember someone telling me that when you stir a cup of coffee, one point in the fluid will always end up at the same place as it started by this theorem, but im not convinced this is really a continuous map since the particles in the fluid are discrete objects.&lt;/p&gt;&#xA;" OwnerUserId="565" LastActivityDate="2010-11-11T16:50:36.673" CommunityOwnedDate="2010-11-11T16:50:36.673" />
  <row Id="9866" PostTypeId="2" ParentId="457" CreationDate="2010-11-11T16:59:54.103" Score="1" ViewCount="" Body="&lt;p&gt;Physical interpretation of the Mean Value Theorem: If the average speed of a car between two locations was V km/h, then there was at least one instant where the speed indicator &#xA;displayed V km/h.&lt;/p&gt;&#xA;" OwnerUserId="752" LastActivityDate="2010-11-11T16:59:54.103" CommunityOwnedDate="2010-11-11T16:59:54.103" />
  <row Id="9867" PostTypeId="1" CreationDate="2010-11-11T17:12:35.117" Score="0" ViewCount="291" Body="&lt;p&gt;it is a question Convergence/Divergence of calculus II! Please give me a hand!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Determine convergence or divergence using any method covered so far.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{n = 1}^{\infty} \sin (1/n)$$&lt;/p&gt;&#xA;" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-11-11T17:44:32.653" LastActivityDate="2010-11-11T17:44:32.653" Title="Convergence/Divergence of $\sum_{n=1}^{\infty} \sin(1/n)$" Tags="&lt;calculus&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="9868" PostTypeId="1" CreationDate="2010-11-11T17:13:44.007" Score="0" ViewCount="119" Body="&lt;p&gt;Hi. Suppose given two manifolds $X$ and $Y$, both orientable of dimension $n$, and a map $f:X\to Y$. Is there a relationship between the degree of $f$ calculated with respect to homology (by which I mean the induced map on the top homology groups) and the degree of $f$ calculated with respect to cohomology (by which I mean the induced map on the top cohomology groups)? Thanks!&lt;/p&gt;&#xA;" OwnerUserId="2272" LastActivityDate="2010-11-11T17:13:44.007" Title="Computing degree of map" Tags="&lt;algebraic-topology&gt;" CommentCount="3" />
  <row Id="9869" PostTypeId="2" ParentId="9867" CreationDate="2010-11-11T17:17:48.817" Score="3" ViewCount="" Body="&lt;p&gt;Hint: $\sin(x) / x \to 1$ as $x \downarrow 0$.&lt;/p&gt;&#xA;" OwnerUserId="2810" LastActivityDate="2010-11-11T17:17:48.817" />
  <row Id="9870" PostTypeId="2" ParentId="9835" CreationDate="2010-11-11T17:22:39.470" Score="1" ViewCount="" Body="&lt;p&gt;These are the probability densities of the assumed distributions, evaluated at the sample data. In this case, the distributions of height, weight and foot size are assumed to follow independent Gaussian distributions whose parameters are given in the second table. If you don't know the probability density function of a Gaussian distribution, look &lt;a href=&quot;http://en.wikipedia.org/wiki/Normal_distribution&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, for the height, the assumed distribution is a Gaussian with mean $\mu = 5.855$ and variance $\sigma^2 = 3.5033\times 10^{-2}$. As the sample's value is $x = 6$, you can plug that in to the probability density function of the distribution to get &lt;a href=&quot;http://www.wolframalpha.com/input/?i=f+%3D+e%5E%28-%28x-mu%29%5E2%2F%282+sigma%5E2%29%29%2F%28sqrt%282+pi+sigma%5E2%29%29%2C+x+%3D+6%2C+mu+%3D+5.855%2C+sigma%5E2+%3D+3.5033e-2&quot; rel=&quot;nofollow&quot;&gt;$f(x) = 1.5789$&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="856" LastActivityDate="2010-11-11T17:22:39.470" />
  <row Id="9871" PostTypeId="2" ParentId="9867" CreationDate="2010-11-11T17:23:08.227" Score="2" ViewCount="" Body="&lt;p&gt;You need to determine convergence for $\sum_{n=1}^\infty\sin(1/n)$. The series diverges. The two hints below may guide you when trying to justify this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint 1: $ \lim_{\theta\to0}\sin(\theta)/\theta=1$ and $1/n\to 0$ as $n\to\infty$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint 2: $\sin(1/n)$ is positive. So you may attempt a limit comparison test.&lt;/p&gt;&#xA;" OwnerUserId="462" LastEditorUserId="462" LastEditorDisplayName="" LastEditDate="2010-11-11T17:40:23.970" LastActivityDate="2010-11-11T17:40:23.970" CommentCount="4" />
  <row Id="9872" PostTypeId="1" CreationDate="2010-11-11T17:40:52.367" Score="1" ViewCount="145" Body="&lt;p&gt;I think it is desirable to have that $M_{m\times n}\left(\mathbb{K}\right)\not=M_{m'\times n'}\left(\mathbb{K}\right)$ if $m\not=m'$ or $n\not=n'$. In other words, the set of all $m\times n$ matrices on $\mathbb{K}$ should be different from the set of all $m'\times n'$ matrices on the same field when $m\not=m'$ or $n\not=n'$. But if I define $M_{m\times n}\left(\mathbb{K}\right)$ as $\left(\mathbb{K}^n\right)^m$ then $M_{0\times n}\left(\mathbb{K}\right)=M_{0\times n'}\left(\mathbb{K}\right)$ for every $n$ and $n'$. That is, two matrices with zero rows are equal, no matter how many columns they have, because $X^0$ is the set containing the empty tuple, for every set $X$. I could have defined $M_{m\times n}\left(\mathbb{K}\right)$ as $\left(\mathbb{K}^m\right)^n$ instead, but then the problem is with $M_{m\times 0}\left(\mathbb{K}\right)$ and $M_{m'\times 0}\left(\mathbb{K}\right)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Two questions:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Am I defining matrices correctly?&lt;/li&gt;&#xA;&lt;li&gt;Are $m\times 0$ or $0\times n$ matrices that important?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="2128" LastEditorUserId="2128" LastEditorDisplayName="" LastEditDate="2010-11-11T17:49:44.727" LastActivityDate="2010-11-11T19:58:36.507" Title="$0\times n$ matrices" Tags="&lt;matrices&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="9873" PostTypeId="1" AcceptedAnswerId="9874" CreationDate="2010-11-11T17:52:28.437" Score="1" ViewCount="165" Body="&lt;p&gt;How to convert the following partial differential equation (pde) &#xA;$$\frac{\partial V}{\partial t}=aV-as\frac{\partial V}{\partial s}-b^2s^2\frac{\partial^2 V}{\partial s^2}$$&#xD;&#xA;into a pde of the form&#xD;&#xA;$$\frac{\partial u}{\partial t}=x^2\frac{\partial^2 u}{\partial x^2}+cx\frac{\partial u}{\partial x}$$&#xD;&#xA;by some change of variables?&#xD;&#xA;Here $a,b,c$ are constants. (A hint should be enough.)&lt;/p&gt;&#xA;" OwnerUserId="3249" LastActivityDate="2010-11-11T17:57:22.127" Title="Change of variables in a partial differential equation" Tags="&lt;analysis&gt;&lt;differential-equations&gt;" AnswerCount="1" />
  <row Id="9874" PostTypeId="2" ParentId="9873" CreationDate="2010-11-11T17:57:22.127" Score="3" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;hint&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you send $t\mapsto -t$, it will fix the sign. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;An exponential integrating factor factor of $e^{At}$ will get rid of the $aV$ term. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then you just need to send $s\mapsto\lambda x$ and compute the correct scaling factors. &lt;/p&gt;&#xA;" OwnerUserId="1543" LastActivityDate="2010-11-11T17:57:22.127" CommentCount="1" />
  <row Id="9875" PostTypeId="1" AcceptedAnswerId="9894" CreationDate="2010-11-11T18:31:17.747" Score="3" ViewCount="114" Body="&lt;p&gt;Let $$\cdots \rightarrow A_{n+1}\rightarrow^{f_{n+1}} A_n \rightarrow^{f_{n}} A_{n-1}\rightarrow \cdots $$ be an inverse system of finite dimensional vector spaces with the property that the $A_i$ are 'eventually constant', i.e., there is an $m$ such that the maps $f_i$ are isomorphisms for every $i\ge m$. Does it follow that $$\lim_{\leftarrow} A_n \simeq A_m?$$&lt;/p&gt;&#xA;" OwnerUserId="3332" LastActivityDate="2010-11-11T20:12:24.623" Title="inverse limit of isomorphic vector spaces" Tags="&lt;commutative-algebra&gt;&lt;homological-algebra&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="9876" PostTypeId="1" CreationDate="2010-11-11T18:32:01.383" Score="2" ViewCount="45" Body="&lt;p&gt;Let $A,B \in \mathbb{R}^{n \times n}$ be two positive semi-definite matrices and let $a &gt; 0$ be a constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the following maximization problem&#xA;$$&#xD;&#xA;\max_{x \in \mathbb{R}^n, \gamma}\ x^T \left(\frac{\gamma}{a + \gamma} A - \gamma B \right)x&#xD;&#xA;\qquad \text{subject to } \gamma &gt; 0, \|x\| = 1&#xD;&#xA;$$&#xD;&#xA;and suppose that we know that $\hat x$ and $\hat \gamma$ maximize the expression. &#xD;&#xA;In this case, we have that $\hat x$ is the largest eigenvector of the matrix $\frac{\hat \gamma}{a + \hat \gamma} A - \hat \gamma B$. &#xD;&#xA;&#xD;&#xA;Suppose that we perturb the matrix A by another matrix $E$, which is supposed to be small in some norm (say operator). Consider again the following optimization problem&#xD;&#xA;$$&#xD;&#xA;\max_{x \in \mathbb{R}^n, \gamma}\ x^T \left(\frac{\gamma}{a + \gamma} (A+E) - \gamma B \right)x&#xD;&#xA;\qquad \text{subject to } \gamma &gt; 0, \|x\| = 1&#xD;&#xA;$$&#xD;&#xA;and denote $\tilde x$ and $\tilde \gamma$ the maximizers. &#xD;&#xA;&#xD;&#xA;The question is how to bound the difference $|\hat \gamma - \tilde \gamma|$ as some function of $E$? Intuitively, if $E$ is small then $\hat \gamma$ and $\tilde \gamma$ should also be close. Any pointers to the relevant literature would be also appreciated. &lt;/p&gt;&#xA;" OwnerUserId="2384" LastActivityDate="2010-11-11T18:32:01.383" Title="How to bound the change between optimal points when perturbing an objective function?" Tags="&lt;linear-algebra&gt;&lt;matrices&gt;&lt;optimization&gt;&lt;perturbation-theory&gt;" FavoriteCount="1" />
  <row Id="9877" PostTypeId="1" AcceptedAnswerId="9878" CreationDate="2010-11-11T18:37:09.800" Score="2" ViewCount="153" Body="&lt;p&gt;I understand that for a number $a$ to have a multiplicative inverse in mod $n$, it must be coprime to $n$; therefore, all numbers that do not have a multiplicative inverse mod $n$ must share a factor with $n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, my prediction was that if $n$ is even, then the sum of all numbers that do not have a multiplicative inverse mod $n$ is $\frac{n}{2}\bmod{n}$, and if $n$ is odd, then the sum is $0\bmod{n}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Originally my proof was constructed something like this: if $n = p_1^{\alpha_1} p_2^{\alpha_2} \cdots p_k^{\alpha_k}$, then the sum of all the elements of this array:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xD;&#xA;\begin{matrix}&#xD;&#xA;p_1 &amp;amp; p_2 &amp;amp; \cdots &amp;amp; p_k \\ 2p_1 &amp;amp; 2p_2 &amp;amp; \cdots &amp;amp; 2p_k \\ \vdots &amp;amp; \vdots &amp;amp; \cdots &amp;amp; \vdots \\ n-2p_1 &amp;amp; n-2p_2 &amp;amp; \cdots &amp;amp; n-2p_k \\ n-p_1 &amp;amp; n-p_2 &amp;amp; \cdots &amp;amp; n-p_k \end{matrix}&#xD;&#xA;\end{align}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;would cancel out in the correct way. However, while trying to prove this, I hit a roadblock  because I wasn't sure how to account for the multiplicity of the factors of $n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I understand &lt;a href=&quot;http://en.wikipedia.org/wiki/Coprime#Properties&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt; correctly, this is the same as finding the sum of all numbers that are not units in the ring of integers modulo $n$, $\mathbb{Z}/n\mathbb{Z}$. Is there any way to do this? If the result is so nice, I feel like there is also a nice way to prove this.&lt;/p&gt;&#xA;" OwnerUserId="1562" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-11-12T15:38:13.110" LastActivityDate="2010-11-12T15:55:10.930" Title="Prove the sum of all numbers that do not have a multiplicative inverse mod $n$" Tags="&lt;algebra&gt;&lt;ring-theory&gt;&lt;modular-arithmetic&gt;" AnswerCount="4" />
  <row Id="9878" PostTypeId="2" ParentId="9877" CreationDate="2010-11-11T18:41:20.217" Score="4" ViewCount="" Body="&lt;p&gt;If $k$ is not coprime with $n$, then $n-k$ is not coprime with $n$ either.&#xA;So you can group those $k$ by pairs $(k,n-k)$, and if $n$ is even and $\neq 2$ (and only in this case), $n/2$ is not coprime with $n$ and &quot;alone&quot;.&lt;/p&gt;&#xA;" OwnerUserId="2660" LastActivityDate="2010-11-11T18:41:20.217" CommentCount="1" />
  <row Id="9879" PostTypeId="1" AcceptedAnswerId="9887" CreationDate="2010-11-11T18:42:05.940" Score="1" ViewCount="165" Body="&lt;p&gt;I wanted to use this fact in a proof, it seems obvious, but I should probably prove this as well.  Any proof will do.  Also for the future is there a way to search for these proofs google does not work very well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Ah, it may also be good to show that the n-th root of any real number a s.t. $0\lt a\lt1$ is also $\lt1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="2533" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-11T19:33:45.953" LastActivityDate="2010-11-11T20:46:02.327" Title="Prove that if $a$ is a real number, and $0\lt a \lt 1$, then $\sqrt{a} \gt a$" Tags="&lt;real-analysis&gt;&lt;algebra&gt;" AnswerCount="4" CommentCount="2" />
  <row Id="9880" PostTypeId="2" ParentId="9877" CreationDate="2010-11-11T18:46:50.270" Score="2" ViewCount="" Body="&lt;p&gt;Note that if $(a,n) = 1$, then $(n-a,n)=1$.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastActivityDate="2010-11-11T18:46:50.270" />
  <row Id="9882" PostTypeId="2" ParentId="9872" CreationDate="2010-11-11T18:48:52.370" Score="1" ViewCount="" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;No,I guess the only sensible way of defining $\mathbb{K}^0$ when $\mathbb{K}$ is a field is by $\mathbb{K}^0=0$, since it should be a 0-dimensional vector space.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;No, I have never seen $m\times 0$-matrices used anywhere, other than here.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="3332" LastActivityDate="2010-11-11T18:48:52.370" />
  <row Id="9883" PostTypeId="2" ParentId="9879" CreationDate="2010-11-11T18:49:14.720" Score="2" ViewCount="" Body="&lt;p&gt;Depending on the level of the audience, it might be seen as obvious.  I would just argue that $\sqrt{x}$ is monotonic and $\gt 0$ on $(0,1)$, so $\sqrt{a}\lt1$ then multiply both sides by $\sqrt{a}$&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-11-11T18:49:14.720" />
  <row Id="9886" PostTypeId="2" ParentId="9879" CreationDate="2010-11-11T19:05:17.877" Score="3" ViewCount="" Body="&lt;p&gt;We argue by contradiction. Suppose $a \in (0,1)$ and $\sqrt{a} \leq a$. Then &#xA;\begin{eqnarray}&#xA;\frac{1}{\sqrt{a}} = \frac{\sqrt{a}}{a} \leq 1, &#xA;\end{eqnarray}&#xA;which implies that $\sqrt{a} \geq 1$ and therefore $a \geq 1$, which is absurd by construction.&lt;/p&gt;&#xA;" OwnerUserId="2720" LastActivityDate="2010-11-11T19:05:17.877" />
  <row Id="9887" PostTypeId="2" ParentId="9879" CreationDate="2010-11-11T19:13:33.983" Score="4" ViewCount="" Body="&lt;p&gt;If you know that whenever $a\lt b$ and $c\gt 0$ then $ac\lt bc$, then for any $x$ with $0\lt x\lt 1$, multiplying through by $x$ gives &#xA;\begin{equation*}&#xA;0 \lt x^2 \lt x \lt 1.&#xA;\end{equation*}&#xA;(The last inequality is &quot;inherited&quot; from the fact that you already know that $x\lt 1$). Repeat to get&#xA;\begin{equation*}&#xA;0\lt x^3 \lt x^2 \lt x \lt 1.&#xA;\end{equation*}&#xA;And so on; you get $0\lt x^n \lt x^{n-1}\lt\cdots \lt x \lt 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Working the other way, if $1\lt y$, then multiplying through by $y$ you get &#xA;\begin{equation*}&#xA;1\lt y\lt y^2&#xA;\end{equation*}&#xA;(with the first inequality &quot;inherited from your original assumption) and repeating this process leads to&#xA;\begin{equation*}&#xA;1\lt y\lt y^2 \lt\cdots \lt y^n&#xA;\end{equation*}&#xA;So: if $0\lt z\lt 1$, then $0\lt z^n\lt 1$. If $1\lt z$, then $1\lt z^n$. And of course, if $z=1$ then $z^n = 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, start with $0\lt a\lt 1$. Then $\sqrt[n]{a}$ cannot be greater than $1$ (then $a = \left(\sqrt[n]{a}\right)^n \gt 1$ by the above), so $\sqrt[n]{a}\lt 1$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now replace $x$ with $\sqrt[n]{a}$ to get the desired inequality.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-11T19:31:57.623" LastActivityDate="2010-11-11T19:31:57.623" CommentCount="3" />
  <row Id="9888" PostTypeId="2" ParentId="9872" CreationDate="2010-11-11T19:27:30.790" Score="6" ViewCount="" Body="&lt;p&gt;As Laurent S points out, the sensible way to define $\mathbb{K}^0$ is as the one element set (either $\{0\}$ or $\{\emptyset\}$). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could likewise argue that since you want $M_{m\times n}(\mathbb{K})$ to correspond to linear transformations from $\mathbb{K}^n$ to $\mathbb{K}^m$, and since $\mathbb{K}^0$ is a perfectly good vector space, then we &lt;em&gt;should&lt;/em&gt; define the matrices $M_{0\times n}(\mathbb{K})$ and $M_{m\times 0}(\mathbb{K})$. Then you are certainly forced to define both as the trivial set, since only the trivial linear transformation exists in either case (the zero map for $m=0$, and the trivial inclusion for $n=0$).   In that respect, they would fit with the general theory and not require exceptions when describing vector spaces (where you would need to explicitly exclude the zero-dimensional vector space, for instance).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why, then, are $M_{0\times n}(\mathbb{K})$ and $M_{0\times n'}(\mathbb{K})$ both equal even when $n\neq n'$, if the first &quot;is&quot; a linear transformation from $\mathbb{K}^n$ to $\{0\}$ and the second &quot;is&quot; a linear transformation from $\mathbb{K}^{n'}$ to $\{0\}$, and hence different? The key is the &lt;code&gt;&quot;&lt;/code&gt;. They are both &lt;em&gt;identified&lt;/em&gt; with those maps, but as a vector space itself they are in fact isomorphic, so there is no horrible universe-shattering error that crept in by having the &quot;matrix&quot; rings be equal. Of course, now your theorem that $M_{m\times n}(\mathbb{K})$ is uniquely determined (in some way) by the pair $(m,n)$ is no longer true: you need to say that they are equal if and only if either $(m,n)=(m',n')$ or $mn=m'n'=0$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Such is life: there is often a special case that needs to be dealt with separately, and your choice is whether you exclude it by design or you exclude it by mention. Here, your choice is between specifying &quot;nonzero&quot; in a lot of theorems about vector spaces, or specifying &quot;or $mn=m'n'=0$&quot; in the theorem about matrices. The latter is usually less disruptive. (In set theory, one often has to deal with the emptyset in a special manner; that corresponds to $0$ here).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As to their interest: they have very little structure to them, so they are not generally interesting in and of themselves. They can show up as part of a general scheme (as above, fitting into the role of representing linear transformations between the standard finite dimensional vector spaces) so that exceptions don't need to be made, but as such they play more the role of &quot;exception-avoiders&quot; than &quot;interesting in and of themselves.&quot;&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-11T19:58:36.507" LastActivityDate="2010-11-11T19:58:36.507" />
  <row Id="9889" PostTypeId="2" ParentId="9710" CreationDate="2010-11-11T19:31:40.807" Score="0" ViewCount="" Body="&lt;p&gt;I know its been a while since I got to exercise the old brain matter, but to disprove this equation, all you need to do is find one example which makes it false.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you can take any n in which f(n) or g(n) returns 1.  And take any n in which f(n) or g(n) does not return 1, say it returns x.  Which would make the equation false, 1*x &amp;lt; 1+x.&lt;/p&gt;&#xA;" LastActivityDate="2010-11-11T19:31:40.807" />
  <row Id="9890" PostTypeId="1" AcceptedAnswerId="9892" CreationDate="2010-11-11T19:33:25.737" Score="3" ViewCount="120" Body="&lt;p&gt;I am learning the basics of category theory, so this question is probably obvious to anyone who knows the subject.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The resources I've seen all take the following approach:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;0) A category is a collection of objects and morphisms between those objects that satisfy some rules.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) A functor is a morphism in the category of categories.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) A natural transformation is a morphism in the category of functors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But they all stop right there. What about:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) the morphisms in the category of natural transformations? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;4) Or the &quot;morphisms in the category of the morphisms in the category of natural transformations&quot; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;5) ... &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are these uninteresting? Why does the &quot;meta-ness&quot; stop at 2 levels deep?&lt;/p&gt;&#xA;" OwnerUserId="3060" LastActivityDate="2011-03-30T02:21:19.247" Title="Morphisms in the category of natural transformations ?" Tags="&lt;category-theory&gt;&lt;morphism&gt;" AnswerCount="2" />
  <row Id="9891" PostTypeId="1" AcceptedAnswerId="9988" CreationDate="2010-11-11T19:34:17.147" Score="1" ViewCount="144" Body="&lt;p&gt;Could someone provide a good reference for learning matrix calculus? I've recently moved to a more engineering-oriented field where it's commonly used and don't have much experience with it.&lt;/p&gt;&#xA;" OwnerUserId="3162" LastActivityDate="2010-11-12T09:53:04.770" Title="Reference for matrix calculus" Tags="&lt;calculus&gt;&lt;matrices&gt;&lt;reference-request&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="9892" PostTypeId="2" ParentId="9890" CreationDate="2010-11-11T19:40:01.667" Score="6" ViewCount="" Body="&lt;p&gt;They are not uninteresting. Just look up the term &lt;em&gt;n-category&lt;/em&gt; (e.g. &lt;a href=&quot;http://math.ucr.edu/home/baez/n_categories/index.html&quot; rel=&quot;nofollow&quot;&gt;Baez's introduction&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nevertheless, it's true that functors and natural transformations already suffice for most common ideas and constructions used across mathematics, so this is why one usually stops there.&lt;/p&gt;&#xA;" OwnerUserId="3223" LastActivityDate="2010-11-11T19:40:01.667" />
  <row Id="9893" PostTypeId="2" ParentId="9891" CreationDate="2010-11-11T19:41:32.247" Score="1" ViewCount="" Body="&lt;p&gt;Gene Howard Golub, Charles F. Van Loan book on &quot;Matrix Computations&quot; is regarded as the &quot;Bhagavad Gita&quot; for Matrix Algorithms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://books.google.com/books?id=mlOa7wPX6OYC&amp;amp;printsec=frontcover&quot; rel=&quot;nofollow&quot;&gt;http://books.google.com/books?id=mlOa7wPX6OYC&amp;amp;printsec=frontcover&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is also another book by &quot;Gene Howard Golub, Gerard Meurant&quot; on &quot;Matrices, moments, and quadrature with applications&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://books.google.com/books?id=IZvkFET3LlwC&amp;amp;printsec=frontcover&quot; rel=&quot;nofollow&quot;&gt;http://books.google.com/books?id=IZvkFET3LlwC&amp;amp;printsec=frontcover&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, &quot;Numerical Linear Algebra&quot; by Trefethen and Bau is well-written and easy to read.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://books.google.com/books?id=bj-Lu6zjWbEC&amp;amp;printsec=frontcover&quot; rel=&quot;nofollow&quot;&gt;http://books.google.com/books?id=bj-Lu6zjWbEC&amp;amp;printsec=frontcover&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would highly recommend Trefethen and Bau since I have read it completely. I feel it is ideal for self-study or for a one quarter course. Once you are done with this you can take a look at Golubs' book. Golubs' book is really good for reference.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-11-11T19:44:06.050" LastActivityDate="2010-11-11T19:44:06.050" CommentCount="1" />
  <row Id="9894" PostTypeId="2" ParentId="9875" CreationDate="2010-11-11T19:51:03.713" Score="2" ViewCount="" Body="&lt;p&gt;Remember that if your directed set has a maximum, then the inverse limit is just the maximum; here this is &quot;essentially&quot; what you have, since after a certain point you are not really getting anything new: you can identify all $A_i$ with $i\geq m$, collapsing the &quot;left tail&quot; of your inverse system into a terminating one, so the inverse system is just that maximum. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, this works whether you are working with finite dimensional vector spaces or any kind of structure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But of course, &quot;essentially&quot; is not the same as &quot;exactly.&quot; So you use the universal property. I claim that $A_m$ has the desired property. Let  $f_{ij}$ with $i\geq j$ be defined as the identity if $i=j$, and as the composition $f_i\circ f_{i-1}\circ\cdots\circ f_j$ if $i\gt j$. Then the projection maps $\pi_j\colon A_m\to A_j$ are defined by $\pi_j=f_{mj}$ if $j\leq m$, and as $\pi_j = f_{jm}^{-1}$ if $j\gt m$; note that this makes sense since $f_{jm}$ is a composition of isomorphisms when $j\gt m$, so it is itself an isomorphism and has an inverse. Note that for any $i\gt j$ we have $f_{ij}\circ \pi_j = \pi_i$, since you have $f_{ij}\circ f_{jk} = f_{ik}$ whenever $i\geq j \geq k$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let $P$ be any object together with maps $p_j\colon P\to A_j$ such that for all $i\gt j$ you have $f_{ij}\circ p_i = p_j$. Then $p=p_m\colon P\to A_m$ has the appropriate property, that is, $p_j = \pi_j\circ p$ for all $j$: if $j\leq m$, then this because $\pi_j = f_{mj}$, so $\pi_j\circ p = f_{mj}\circ p_m = p_j$, and if $j\gt m$ then $f_{jm}\circ p_j = p_m$, so pre-composing with $f_{jm}^{-1}$ we get $p_j = (f_{jm})^{-1}\circ p_m = \pi_j\circ p$. The map is unique: if $f\colon P\to A_m$ has the same property, then $f=f_m = p$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, $(A_m, \{\pi_j\})$ has the desired universal property, and so &quot;is&quot; the inverse limit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You'll note that the fact that we are dealing with finite dimensional vector spaces is completely immaterial; what matters is the properties of the functions in play. This is typical of universal constructions.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-11T20:12:24.623" LastActivityDate="2010-11-11T20:12:24.623" CommentCount="2" />
  <row Id="9895" PostTypeId="2" ParentId="8611" CreationDate="2010-11-11T19:54:01.793" Score="1" ViewCount="" Body="&lt;p&gt;In response to user2969 :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At first you need to computed the &lt;a href=&quot;http://en.wikipedia.org/wiki/Greatest_common_divisor&quot; rel=&quot;nofollow&quot;&gt;gcd&lt;/a&gt;, it is always best to use &lt;a href=&quot;http://en.wikipedia.org/wiki/Euclidean_algorithm&quot; rel=&quot;nofollow&quot;&gt;Euclidean algorithm&lt;/a&gt; for this,my C++ implementation for the same :&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;int E_GCD(int a,int b){&#xA;   return b ? gcd(b,a%b):a;&#xA;}&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Now we need to compute the number of divisors of &lt;code&gt;N = E_GCD(A,B)&lt;/code&gt;, this can be done efficiently by first computing the factorization using any fast algorithm like &lt;a href=&quot;http://en.wikipedia.org/wiki/Pollard%27s_rho_algorithm&quot; rel=&quot;nofollow&quot;&gt;Pollard rho integer factorization&lt;/a&gt; or &lt;a href=&quot;http://en.wikipedia.org/wiki/Lenstra_elliptic_curve_factorization&quot; rel=&quot;nofollow&quot;&gt;ECM&lt;/a&gt; then using the standard number theory $\text{trick}^*$ to for finding the number of factors,however in this problem if the values of &lt;code&gt;A&lt;/code&gt; and &lt;code&gt;B&lt;/code&gt;  are small (say:$0 \lt A,B \le 10^6$),then we don't need really need to use any of such sophisticated algorithms,instead you could build up your own algorithm simply based on the fact the the factors of a number occur in pair that is if say &lt;code&gt;i&lt;/code&gt; is a factor of &lt;code&gt;N&lt;/code&gt; then so is $N/i$ using this key observation I build up the algorithm,here is my C++ implementation of this idea :&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;  N = E_GCD(A,B);&#xA;  int ans = 0;&#xA;  int sqt = (int)sqrt(N);&#xA;  for(int i = 1 ; i &amp;lt;= sqt; i++){&#xA;      if(N % i == 0){&#xA;          ans += 2 ;&#xA;          if(i == N/i) ans--;&#xA;      }&#xA;  }&#xA;  printf(&quot;%d\n&quot;,ans);&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Notice the fact that a number cannot a have a divisor which is greater than the square root of the number,rest of the code I believe is self explanatory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;*$\text{If }N = a^p \cdot b^q  \cdot c^r \cdots \text{,a,b,c ... are primes,number of factors of N is :}(p+1) \cdot (q+1)\cdot(r+1)\cdots$ &lt;/p&gt;&#xA;" OwnerUserId="2109" LastEditorUserId="2109" LastEditorDisplayName="" LastEditDate="2010-11-11T20:11:11.437" LastActivityDate="2010-11-11T20:11:11.437" />
  <row Id="9896" PostTypeId="1" AcceptedAnswerId="9908" CreationDate="2010-11-11T19:55:55.553" Score="7" ViewCount="520" Body="&lt;p&gt;In naieve set theory, you quickly run into existence trouble if you try to do meta-things things like take &quot;the set of all sets&quot;. For example, does the set of all sets that don't contain themselves contain itself? (see, Russell)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, people have no trouble casually talking about the &quot;category of all categories&quot;, etc. How do we know there isn't a contradiction lurking somewhere here - especially since many books define categories in terms of sets?&lt;/p&gt;&#xA;" OwnerUserId="3060" LastActivityDate="2011-01-14T16:07:08.213" Title="Category of all categories vs. Set of all sets" Tags="&lt;set-theory&gt;&lt;category-theory&gt;" AnswerCount="4" CommentCount="1" FavoriteCount="3" />
  <row Id="9897" PostTypeId="2" ParentId="9863" CreationDate="2010-11-11T19:56:08.360" Score="1" ViewCount="" Body="&lt;p&gt;This isn't quite what you asked for, but notice that (1) simply gives us $\mathbb{RP}^2 \times S^2$.  I'm not sure I understand your notation for (2); it seems like it doesn't depend at all on $\pi$.  If it's true that you're just reflecting $\vec{n}$ through the plane perpendicular to $\vec{a}$, then you're just getting a copy of $D^2$ over every point of $\mathbb{RP}^2$.  This is a fiber bundle, and if you think through it I think you'll find that it's the disk bundle of the tangent bundle $T\mathbb{RP}^2$.&lt;/p&gt;&#xA;" OwnerUserId="2159" LastActivityDate="2010-11-11T19:56:08.360" CommentCount="8" />
  <row Id="9898" PostTypeId="1" CreationDate="2010-11-11T20:01:02.567" Score="1" ViewCount="92" Body="&lt;p&gt;Given a diffusion $dX_t = \mu(X_t)dt + \sigma(X_t)dW_t$, if one applies the transform&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f(x) = \int_a^x \frac{1}{\sigma(u)}du$, one can use Ito's lemma to show that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$df(X_t) = \left( \frac{\mu(X_t)}{\sigma(X_t)} - \frac{\sigma'(X_t)}{2} \right)dt + dW_t$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a way to explicitly invert this transform in the general case, assuming $\sigma$ is sufficiently nice?&lt;/p&gt;&#xA;" OwnerUserId="3162" LastActivityDate="2010-11-11T20:53:52.640" Title="How do I invert this transform?" Tags="&lt;probability-theory&gt;&lt;stochastic-processes&gt;&lt;transformation&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="9899" PostTypeId="2" ParentId="9877" CreationDate="2010-11-11T20:01:29.290" Score="0" ViewCount="" Body="&lt;p&gt;Since you were explicitly looking for a different proof:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $\displaystyle S_n$ be the sum of all numbers $k$ such that $\displaystyle 1 \leq k$ &amp;lt; $\displaystyle n$ and $\displaystyle (n,k) = 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have that $\displaystyle S_2 = 1, S_3 = 3 = 0 \mod 3$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The claim is that $\displaystyle S_n = 0 \mod n$ if $\displaystyle n &amp;gt; 2$ (which can easily be seen by pairing $\displaystyle k$ with $\displaystyle n-k$ as noted by other answers, but we will give a different proof).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To see this differently, notice that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle S_n + \sum_{d \mid n, d &amp;gt; 1} (d \times S_{n/d}) = n(n-1)/2$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is got by partitioning the numbers $\displaystyle \{1,2,\cdots, n-1 \}$ by their gcd with $\displaystyle n$ and adding up the individual partitions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The partition $\displaystyle \{k : (k,n) = 1\}$ sums to $\displaystyle S_n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The partition $\displaystyle \{k : (k,n) = d\}$ is same as the set $\displaystyle \{k : (k/d, n/d) = 1\}$ and so sums to $\displaystyle  d \times S_{n/d}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also notice that $\displaystyle n(n-1)/2 = 0 \mod n$ if $\displaystyle n$ is odd, and $\displaystyle n(n-1)/2 = n/2 \mod n$ if $\displaystyle n$ is even.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus by total induction on $\displaystyle n$, $S_n = 0 \mod n$ if $\displaystyle n &amp;gt; 2$ as $\displaystyle S_{n/d} = 0 \mod n/d$ implies $\displaystyle d \times S_{n/d} = 0 \mod n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, what you are looking for is $\displaystyle \sum_{d \mid n, d &amp;gt; 1} (d \times S_{n/d})$ which is $\displaystyle 0 \mod n$ if $\displaystyle n$ is odd, and $\displaystyle n/2 \mod n$ if $\displaystyle n$ is even: the only non-zero term $\displaystyle \mod n$ corresponds to $\displaystyle n/d = 2$.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-11-12T15:12:45.317" LastActivityDate="2010-11-12T15:12:45.317" />
  <row Id="9901" PostTypeId="2" ParentId="9896" CreationDate="2010-11-11T20:05:07.710" Score="1" ViewCount="" Body="&lt;p&gt;You can talk about &quot;the category of all categories&quot; with the proviso &quot;in a given universe&quot;. &quot;Our universe&quot; is big enough to include the set of natural numbers and all the usual operations you do with sets (union, intersection...) See Mac Lane's &quot;Categories for the working mathematician&quot;, first chapter, for the details. &lt;/p&gt;&#xA;" OwnerUserId="664" LastActivityDate="2010-11-11T20:05:07.710" />
  <row Id="9902" PostTypeId="2" ParentId="6387" CreationDate="2010-11-11T20:08:36.847" Score="4" ViewCount="" Body="&lt;p&gt;Make the observation that $|x| = \theta(x) x - \theta(-x) x$ for all real $x$, where $\theta$ is the Heaviside function (evaluating to $1$ if the argument is positive and $0$ otherwise). It is known that&#xA;\begin{eqnarray}&#xA;\int \theta(x) \ x \ dx = \theta(x) \frac{x^{2}}{2} + C \quad \text{and} \quad \int \theta(-x) \ x \ dx = \theta(-x) \frac{x^{2}}{2} + C^{\prime},&#xA;\end{eqnarray}&#xA;where $C$ and $C^{\prime}$ are constants of integration. The identity for $n = 1$ follows by subtraction and the representation of $|x|$ above. With $n$ integrations, we have&#xA;\begin{eqnarray}&#xA;\int \cdots \int |x| \ dx = \theta(x) \int \cdots \int x \ dx  - \theta(-x) \int \cdots \int x \ dx = \frac{|x| x^{n}}{(n+1)!} + P_{n},&#xA;\end{eqnarray}&#xA;where $P_{n}$ is a polynomial in $x$, as claimed.&lt;/p&gt;&#xA;" OwnerUserId="2720" LastActivityDate="2010-11-11T20:08:36.847" />
  <row Id="9903" PostTypeId="2" ParentId="9877" CreationDate="2010-11-11T20:09:48.447" Score="1" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\ $ It's a special case of Wilson's theorem for groups - see &lt;a href=&quot;http://math.stackexchange.com/questions/9311/product-of-all-elements-in-an-odd-finite-abelian-group-is-1/9345#9345&quot;&gt;my answer here&lt;/a&gt; - which highlights the key role played by symmetry (here a negation reflection / involution). Namely, since your set is closed under negation, its non-fixed points $\rm -k\not\equiv k\:$ pair up and contribute zero to the sum, leaving only the sum of the fixed points $\rm - k\equiv k\ \iff\ 2\:k\equiv 0\:,\ $ so $\rm\ k\equiv 0\ $ if $\rm\: n\:$ is odd, else $\rm\ k \equiv 0,\ n/2\:$.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-11-12T15:55:10.930" LastActivityDate="2010-11-12T15:55:10.930" />
  <row Id="9904" PostTypeId="2" ParentId="9606" CreationDate="2010-11-11T20:22:07.407" Score="0" ViewCount="" Body="&lt;p&gt;If the space of nxn and kxk matrices are viewed as vector spaces, then functions mapping between them would be called &quot;operators&quot;. That's probably not as specific as you want since it doesn't take into account the matrix structure, only the vector space structure.&lt;/p&gt;&#xA;" OwnerUserId="3060" LastActivityDate="2010-11-11T20:22:07.407" />
  <row Id="9905" PostTypeId="2" ParentId="9896" CreationDate="2010-11-11T20:24:19.073" Score="12" ViewCount="" Body="&lt;p&gt;In general there are two ways to make foundations of category theory precise. The first one is to make use of &lt;a href=&quot;http://en.wikipedia.org/wiki/Grothendieck_universe&quot; rel=&quot;nofollow&quot;&gt;Grothendieck universes&lt;/a&gt;, which is just a way to limit the number of sets we are considering. So when we say the category of sets (groups, modules, ...) is implicit that we are restricting ourselves to some universe which is a set in the usual sense. This approach isn't all nice, since we cannot prove in ZFC that every set $S$ belongs to some universe $\mathcal{U}$, so we need to take it as an axiom. However people seems to prefer this one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second one is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Von_Neumann-Bernays-Godel_set_theory&quot; rel=&quot;nofollow&quot;&gt;Neumann–Bernays–Gödel set theory&lt;/a&gt; (NBG). It is an axiomatic set theory where class is the primitive concept. Then we say that a class $S$ is a set if there is a class $C$ such that $A \in C$. Thus a set is a particular kind of class and a class that is not a set is called proper class. Using this axiomatic, the collection of all sets is a proper class. So when we say the category of sets (rings, topological spaces, ...) we have a proper class of objects. However we cannot speak of the class of all classes and we introduce the name small category to refer to categories with a &lt;strong&gt;&lt;em&gt;set&lt;/em&gt;&lt;/strong&gt; of objects. Then we can speak of the category of all small categories and functors between them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;NBG does not mess the other things we already have in mathematics since it is a &lt;a href=&quot;http://en.wikipedia.org/wiki/Conservative_extension&quot; rel=&quot;nofollow&quot;&gt; conservative extension &lt;/a&gt; of ZFC. Meaning that all theorems we already know in ZFC are still valid in NBG and if we prove a theorem in NBG using only the language of ZFC then it is a theorem in ZFC.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But you don't need to worry too much with these things. In my opinion they are very technical for a first course and in most cases they are not really important. Just treat your categories in the same naive way almost everyone treats sets and remember that things can be done precisely.&lt;/p&gt;&#xA;" OwnerUserId="329" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2010-11-16T00:49:04.417" LastActivityDate="2010-11-16T00:49:04.417" />
  <row Id="9906" PostTypeId="2" ParentId="9825" CreationDate="2010-11-11T20:27:27.203" Score="4" ViewCount="" Body="&lt;p&gt;Here's a sketch of my idea for proving that $g$ is continuous:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Show that $s$ and $t$ are witnesses that $Y$ is not $T_2$ iff any eventually nonconstant net converging to $s$ or $t$ converges to &lt;em&gt;both&lt;/em&gt; $s$ and $t$.&lt;/li&gt;&#xA;&lt;li&gt;Observe that $g$ is continuous on $E$ (why?).&lt;/li&gt;&#xA;&lt;li&gt;Show that $g$ is continuous at $s$ by using the convergent net definition as follows: Let $(x_\alpha)$ be a net that converges to $s$. We split into cases:&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$(x_\alpha)$ is eventually constant. Here we need to use that $Y$ is $T_1$. Because of this, $x_\alpha$ is eventually equal to $s$, so $g(x_\alpha)$ is eventually equal to $t$ and hence $g(x_\alpha) \to t = g(s)$. &lt;/li&gt;&#xA;&lt;li&gt;$(x_\alpha)$ is not eventually constant and converges to $s$. By the first subresult, we have $g(x_\alpha) \to t = g(s)$.&lt;/li&gt;&#xA;&lt;/ol&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;(NB! My topology class used Munkres' &quot;Topology&quot; which outside of a few exercises never mentions nets. So it might be wrong and full of holes to fill out)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I spent a lot of time trying to figure out some cleverer $X$ than the one you came up with, but it was really hard. It seems obvious though that it has to be constructed from $Y$ somehow. Some of my ideas involved starting with $2^Y$, i.e. the set of indicator functions on $Y$, giving it the product topology and identifying with $\mathcal{P}(Y)$ and from there you can pass to a topology on $E = \lbrace U \cap V \mid U \in \mathcal{U}_s, V \in \mathcal{V}_t \rbrace$, but that effectively boils down to what you did. It doesn't even avoid choice as you still have to use that to define $f$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another idea I tried was to consider only the case where $Y$ is infinite and then constructing something using the cofinite topology, but that too didn't pan out.&lt;/p&gt;&#xA;" OwnerUserId="1269" LastActivityDate="2010-11-11T20:27:27.203" CommentCount="2" />
  <row Id="9907" PostTypeId="2" ParentId="9879" CreationDate="2010-11-11T20:46:02.327" Score="3" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\rm \ \sqrt a - a \ =\ \sqrt a\ \:(1 - \sqrt a)\ &gt; 0\ $ since both factors are $&gt; 0$  on the interval $(0,1)$.&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-11-11T20:46:02.327" />
  <row Id="9908" PostTypeId="2" ParentId="9896" CreationDate="2010-11-11T20:50:07.333" Score="8" ViewCount="" Body="&lt;p&gt;If you base your mathematics on Set Theory, then you run into the same foundational issues in Category Theory as you do in Set Theory (since Categories will have to be modeled &quot;with&quot; Set Theory in some sense).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can go the other way, though: you can base your mathematics with Category Theory, and develop the Category of Sets through it, as Lawvere does &lt;a href=&quot;http://www.tac.mta.ca/tac/reprints/articles/11/tr11abs.html&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;. Then your primitive notions, the bedrock of your theory, is not Set Theory but Categories. There are precise ways in which you can talk about &quot;Category of all categories&quot;, analogous to the way in which you can talk about &quot;the class of all sets&quot; if you start with Set Theory instead.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do we know there are no problems lurking in this viewpoint? Well, we don't, just like we don't know that there are no serious problems lurking in Set Theory (e.g., that it is not consistent). &lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-11-11T20:50:07.333" CommentCount="1" />
  <row Id="9909" PostTypeId="2" ParentId="9898" CreationDate="2010-11-11T20:53:52.640" Score="2" ViewCount="" Body="&lt;p&gt;if you take $\mu(u)=\sigma(u)=\alpha+u^2$ then the image diffusion is the same whatever $\alpha$.&lt;/p&gt;&#xA;" OwnerUserId="2905" LastActivityDate="2010-11-11T20:53:52.640" />
  <row Id="9910" PostTypeId="2" ParentId="154" CreationDate="2010-11-11T21:03:43.360" Score="2" ViewCount="" Body="&lt;p&gt;I usually say: &quot;Believe it or not, electricity and radio waves actually &lt;strong&gt;do&lt;/strong&gt; behave like complex numbers. You don't see that in high school, but electric and electronics engineers do.&quot;&lt;/p&gt;&#xA;" OwnerUserId="2612" LastActivityDate="2010-11-11T21:03:43.360" CommunityOwnedDate="2011-04-15T20:45:29.080" />
  <row Id="9911" PostTypeId="1" AcceptedAnswerId="9913" CreationDate="2010-11-11T21:20:10.393" Score="3" ViewCount="172" Body="&lt;p&gt;We all know that $\displaystyle \sum_{n=1}^{\infty} \frac{1}{n^s}$ converges for $s&gt;1$ and diverges for $s \leq 1$ (Assume $s \in \mathbb{R}$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was curious to see till what extent I can push the denominator so that it will still diverges.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I took $\displaystyle \sum_{n=2}^{\infty} \frac{1}{n\log n}$ and found that it still diverges. (This can be checked by using the well known test that if we have a monotone decreasing sequence, then $\displaystyle \sum_{n=2}^{\infty} a_n$ converges iff $\displaystyle \sum_{n=2}^{\infty} 2^na_{2^n}$ converges).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;No surprises here. I expected it to diverge since $\log n$ grows slowly than any power of $n$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, when I take $\displaystyle \sum_{n=2}^{\infty} \frac{1}{n(\log n)^s}$, I find that it converges $\forall s&gt;1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(By the same argument as previous). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This doesn't make sense to me though. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If this were to converge, then I should be able to find a $s_1 &gt; 1$ such that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle \sum_{n=2}^{\infty} \frac{1}{n^{s_1}}$ is greater than $\displaystyle \sum_{n=2}^{\infty} \frac{1}{n (\log n)^s}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Doesn't this mean that in some sense $\log n$ grows faster than a power of $n$? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(or)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How should I make sense of (or) interpret this result?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I am assuming that my convergence and divergence conclusions are right).&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2010-11-11T21:54:40.240" LastActivityDate="2010-11-11T23:37:56.930" Title="$\displaystyle \sum_{n=1}^{\infty} \frac{1}{n\log^sn}$" Tags="&lt;number-theory&gt;&lt;real-analysis&gt;" AnswerCount="3" CommentCount="6" />
  <row Id="9913" PostTypeId="2" ParentId="9911" CreationDate="2010-11-11T21:36:09.030" Score="5" ViewCount="" Body="&lt;p&gt;Yes $\displaystyle \sum_{n=2}^{\infty} \dfrac{1}{n (\log n)^s}$ is convergent if $\displaystyle s &gt; 1$, we can see that by comparing with the corresponding integral.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As to your other question, if&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle \sum_{n=2}^{\infty} \frac{1}{n^{s_1}}$ is greater than $\displaystyle \sum_{n=2}^{\infty} \frac{1}{n (\log n)^s}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;does not imply $\log n$ grows faster than a power of $\displaystyle n$. You cannot compare them term by term.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What happens is that the first &quot;few&quot; terms of the series dominate (the remainder goes to 0). For a small enough $\displaystyle \epsilon$,  we have that $\log n &gt; n^{\epsilon}$ for a sufficient number of (initial) terms, enough for the series without $\log n$ to dominate the other.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-11-11T22:08:08.337" LastActivityDate="2010-11-11T22:08:08.337" CommentCount="1" />
  <row Id="9915" PostTypeId="2" ParentId="9911" CreationDate="2010-11-11T22:01:23.423" Score="3" ViewCount="" Body="&lt;p&gt;This is a special case of classic general logarithmic convergence tests. Some early work in asymptotics was motivated by attempts to determine the &quot;boundary of convergence&quot; in terms of various functions, e.g. log-exp functions. Below is a a very interesting excerpt from Hardy's classic &quot;A course in pure mathematics&quot;. For further work see his &quot;Orders of infinity&quot; and see especially this very interesting &lt;a href=&quot;http://www.gfisher.org/fisher%20-%20the%20infinite%20and%20infinitesimal%20quantities%20of%20du%20bois-reymond%20and%20their%20reception.pdf&quot; rel=&quot;nofollow&quot;&gt;paper by G. Fisher&lt;/a&gt; on Paul du Bois-Reymond's work on the boundary between convergence and divergence (where he discovered &lt;strong&gt;diagonalization&lt;/strong&gt; before Cantor).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note in particular the following very general result that is presented in the excerpt: the following series and integral are convergent if $\rm\ s &gt; 1\ $ and divergent if $\rm\: s\le 1\ $ where $\rm\: n_0,\ a\ $ are any numbers large enough to ensure positivity of $\rm\ log_k x := \log \log \cdots \log x\:,\ $ iterated $\rm\:k\:$ times.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\rm \sum_{n_0}^\infty \frac{1}{n \log n\ \log_2 n\ \cdots\ \log_{k-1} n\ (\log_k n)^s } $$&#xD;&#xA;&#xD;&#xA;$$\rm \int_{a}^\infty \frac{dx}{x \log x\ \log_2 x\ \cdots\ \log_{k-1} x\ (\log_k x)^s } $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/cChMu.jpg&quot; alt=&quot;alt text&quot;&gt;&#xA;&lt;img src=&quot;http://i.imgur.com/5Ge4P.jpg&quot; alt=&quot;alt text&quot;&gt;&#xA;&lt;img src=&quot;http://i.imgur.com/VUS3e.jpg&quot; alt=&quot;alt text&quot;&gt;&#xA;&lt;img src=&quot;http://i.imgur.com/S5njR.jpg&quot; alt=&quot;alt text&quot;&gt;&#xA;&lt;img src=&quot;http://i.imgur.com/FEDXp.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-11-11T23:37:56.930" LastActivityDate="2010-11-11T23:37:56.930" />
  <row Id="9916" PostTypeId="1" AcceptedAnswerId="9920" CreationDate="2010-11-11T22:25:03.430" Score="1" ViewCount="338" Body="&lt;p&gt;I know this means that there are two cosets. I also know that one must be H itself.&#xA;This means that the remaining coset(s) must be equal for right and left. Also since there is only one possible coset, this means that for all elements, a,b in G, then aH = bH and Ha = Hb, since each element in G acting on the subgroup must produce the same set. This means aH = Ha I believe. From there H = aHa-1, which given the information I am not sure if this good or how to use it. Any help would good.&lt;/p&gt;&#xA;" OwnerUserId="3417" LastActivityDate="2011-01-14T05:13:37.427" Title="H is a subgroup of index 2 in finite group G, then every left coset of H is a right coset as well" Tags="&lt;group-theory&gt;" AnswerCount="3" CommentCount="4" FavoriteCount="1" />
  <row Id="9918" PostTypeId="2" ParentId="9916" CreationDate="2010-11-11T22:32:38.427" Score="2" ViewCount="" Body="&lt;p&gt;You say &quot;Also since there is only one possible coset, this means that for all elements, a,b in G, then aH = bH and Ha = Hb, since each element in G acting on the subgroup must produce the same set.&quot; but this is not true.  Let a be the identity and b not be in H.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But earlier you say &quot;I know this means that there are two cosets. I also know that one must be H itself.&quot;  And if all cosets are the same size, what can you say about the one that isn't H?&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-11-11T22:32:38.427" />
  <row Id="9919" PostTypeId="2" ParentId="9911" CreationDate="2010-11-11T22:35:09.970" Score="3" ViewCount="" Body="&lt;p&gt;Du Bois-Reymond was interested in this question, positing some sort of series &quot;between&quot; all convergent series and all divergent series. Hausdorff didn't like that idea, and showed that given any countable sequences of convergence series converging &quot;ever more slowly&quot; and divergent series diverging &quot;ever more slowly&quot;, you can find a convergent and a divergent series &quot;in between&quot;. Here the order relation is the one given by the (conclusive) ratio test.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bill's remark is related to the recursive Elias codes. The first few codes correspond to the convergent series&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle \frac{1}{n(\log n)^2}, \frac{1}{n\log n (\log \log n)^2}, \ldots$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;He then diagonalizes to obtain his $\omega$ code (Elias omega code, if you want to look it up), which corresponds to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle \frac{1}{n\log n\log \log n \cdots (\log^* n)^2},$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\log^* n$ is the number of times you need to apply $\log$ until you get below some constant (the other logs continue up to this constant); confusingly, coding theorists use $\log^* n$ to mean the $\log n \log\log n \cdots$ part. We can continue the iteration like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle \frac{1}{n\log n\log\log n \cdots \log^*n (\log \log^* n)^2}, \frac{1}{n\log n\log\log n \cdots \log^*n \log \log^* n (\log\log \log^* n)^2}, \cdots$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and so on. We can diagonalize again to obtain an $\omega + \omega$ code, and so on at least for $n \omega + m$. We an probably continue even to $\omega^2$ and beyond.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The corresponding divergent series are&#xA;$\displaystyle \frac{1}{n\log n}, \frac{1}{n\log n\log\log n}, \ldots, \frac{1}{n\log n\log\log n \cdots \log^* n}, \frac{1}{n\log n\log\log n \cdots \log^* n \log\log^* n}, \ldots$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One can ask whether there is a \emph{scale} for convergent series, that is a chain of (mutually comparable) series converging more and more slowly, such that for each convergent series there's one in the chain converging even more slowly; you can ask the same question about divergent series. Surprisingly, the existence of these is independent of ZFC (they exist given CH, and models in which they don't exist can be constructed using forcing).&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-11T22:35:09.970" CommentCount="2" />
  <row Id="9920" PostTypeId="2" ParentId="9916" CreationDate="2010-11-11T22:41:17.443" Score="3" ViewCount="" Body="&lt;p&gt;The (left) cosets of $H$ partition the group $G$; since there are two cosets and one of them is $H$, that means that, whatever the other one is (call it $\mathcal{C}$), &lt;em&gt;as a set&lt;/em&gt;, you have that $H\cap\mathcal{C}=\emptyset$ and $H\cup&#xD;&#xA;\mathcal{C}=G$. These two together tell you that the other one &lt;strong&gt;must&lt;/strong&gt; be equal, as a &lt;em&gt;set&lt;/em&gt;, to the complement of $H$ inside of $G$; that is, $\mathcal{C}=G-H$. So the two cosets are $H$ and $G-H$. Of course, $G-H$ can be written as $xH$ for &lt;em&gt;some&lt;/em&gt; $x$; in fact, for any $x\notin H$ you have $xH = G-H$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the exact same argument holds for &lt;em&gt;right&lt;/em&gt; cosets: one of them is $H$, and the other one must be, as a set, the complement of $H$. So the two right cosets are $H$ and $G-H$, and $G-H$ can be written as $Hy$ for some $y$; in fact, for any $y\notin H$ you have $Hy=G-H$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now note that in both cases, the cosets are:&#xA;    1. $H$, the coset of the elements of $H$; and&#xA;    2. $G-H$, the coset of the elements not in $H$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So each of the two left cosets is also a right coset and vice-versa.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2010-11-11T22:49:28.630" LastActivityDate="2010-11-11T22:49:28.630" CommentCount="1" />
  <row Id="9922" PostTypeId="1" CreationDate="2010-11-11T23:22:42.497" Score="2" ViewCount="191" Body="&lt;p&gt;Find all continuous functions on R satisfying f(x), that f(x)=f(sinx) for all x.&lt;/p&gt;&#xA;" OwnerUserId="2779" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-11-11T23:32:13.287" LastActivityDate="2010-11-12T07:37:50.637" Title="Find all continuous functions on R satisfying f(x), that f(x)=f(sinx) for all x." Tags="&lt;functional-equations&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="1" />
  <row Id="9923" PostTypeId="2" ParentId="9922" CreationDate="2010-11-11T23:28:19.657" Score="2" ViewCount="" Body="&lt;p&gt;For each $x$, $\sin(\sin(\sin...\sin(x)...))$ (n-times) goes to zero as $n$ goes to infinity, so by continuity &lt;strike&gt; f(x)=0 &lt;/strike&gt; $f(x)=f(0)$.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-11-11T23:28:19.657" CommentCount="2" />
  <row Id="9924" PostTypeId="2" ParentId="9922" CreationDate="2010-11-11T23:28:38.517" Score="8" ViewCount="" Body="&lt;p&gt;$f(x)$ must be constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We must have that if $x_1 = x$ and $x_{n+1} = \sin x_n$, then $f(x_{n+1}) = f(x_n) = f(x_{n-1}) \dots = f(x)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now $x_n \to 0$ as $n \to \infty$. Thus $f(x) = f(0)$&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-11-11T23:28:38.517" />
  <row Id="9925" PostTypeId="2" ParentId="154" CreationDate="2010-11-11T23:36:09.973" Score="3" ViewCount="" Body="&lt;p&gt;I have just discovered this web page and your question.  This is probably a late response.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A lot of people think that at some point someone came along and simply asserted &quot;yes there is a square root of -1 after all&quot; and called it 'i'. That's a common misconception, and, unfortunately, seems to be what is often &quot;taught&quot; in &quot;high schools&quot; in the US.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The set of Real Numbers that you are used to is an example of what mathematicians call a &quot;field&quot;. That means you can add, subtract, multiply, and divide according to familiar properties. The field C contains the Reals: it is just the two-dimensional plane endowed with a very natural addition (vectors) and a really cool multiplication, which takes a little effort to understand.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A complex number is just a point on the plane, an ordered pair. The absolute value of a complex number is its distance to the origin, and let's call its &quot;angle&quot; the angle it forms with the positive x-axis. The Real numbers are just the x-axis, and &quot;i&quot; is just (0,1). So the real number 1 is (1,0) and -1 is (-1,0). Then multiplying complex numbers multiplies their absolute values and adds the angles. That's why (0,1) times itself is -1 = (-1,0). 90 degrees plus 90 degrees = 180 degrees.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is essentially no other way to imbed R (the reals) in a field in which all numbers have square roots. If you want to be technical, any other such field is isomorphic to C. Furthermore, there is no way to make any higher R^n (R^3 = space for example) into a field in a natural way. The quaternions are a way to make R^4 into a division ring, close but not a field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, in short, the field of complex numbers is extremely real and concrete.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Last March Steven Strogatz wrote a column about C = the field of complex numbers in the New York Times. You might enjoy this exchange of emails I had with someone a few months later, after Strogatz's article appeared in the New York Times.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://home.bway.net/lewis/Complex.htm&quot; rel=&quot;nofollow&quot;&gt;http://home.bway.net/lewis/Complex.htm&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1837" LastActivityDate="2010-11-11T23:36:09.973" CommunityOwnedDate="2011-04-15T20:45:29.080" CommentCount="1" />
  <row Id="9927" PostTypeId="1" CreationDate="2010-11-12T00:23:58.440" Score="0" ViewCount="99" Body="&lt;p&gt;Hi all,&#xA;I wanted help in problem 7.4 of  Gilbarg and Trudinger . &#xA;The question says( in my language) that prove the following product formula for differentiation&lt;br&gt;&#xA;D(uv) = uDv + vDu for all u , v  in W^1{\omega} such that uv, uDv + vDu are locally integrable.&#xA;Some help will be appreciated.&lt;/p&gt;&#xA;" LastActivityDate="2010-11-12T00:23:58.440" Title="product rule ( sobolev functions)" Tags="&lt;real-analysis&gt;" CommentCount="1" />
  <row Id="9928" PostTypeId="1" CreationDate="2010-11-12T00:47:19.327" Score="3" ViewCount="117" Body="&lt;p&gt;In the derived category $D(C)$ of an abelian category $C$, one formally inverts quasi-isomorphisms.  In the context of model categories, one inverts weak equivalences.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What does one gain by doing so?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Is there a big-picture way to think about what is being done?&lt;/p&gt;&#xA;" OwnerUserId="3008" LastActivityDate="2010-11-12T00:47:19.327" Title="Yoga of localization in categories?" Tags="&lt;category-theory&gt;" CommentCount="3" FavoriteCount="2" />
  <row Id="9929" PostTypeId="2" ParentId="8536" CreationDate="2010-11-12T01:06:52.163" Score="1" ViewCount="" Body="&lt;p&gt;The short answer, in my opinion, is that counting is one of the first things you should try when you approach a problem and it looks even remotely possible to use.  If counting doesn't work, &lt;em&gt;then&lt;/em&gt; you should try all your other tools.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-11-12T01:06:52.163" />
  <row Id="9930" PostTypeId="2" ParentId="8415" CreationDate="2010-11-12T01:32:51.243" Score="1" ViewCount="" Body="&lt;p&gt;Here's an idea that might lead somewhere.  Let $F(x) = \sum_{n \ge 0} \frac{f(n)}{n!} x^n$, let $g(n) = \frac{1}{2^n} \sum_{k=0}^{n} {n \choose n-k} f(k)$, and let $G(x) = \sum_{n \ge 0} \frac{g(n)}{n!} x^n$.  Then it's not hard to see that $G(x) = e^{\frac{x}{2}} F \left( \frac{x}{2} \right)$, so we are looking for functions $F$ such that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$F(x) - e^{\frac{x}{2}} F \left( \frac{x}{2} \right) = A(x)$$&#xD;&#xA;&#xD;&#xA;for some $A(x)$ with Taylor coefficients tending to some nonzero limit.  Let $H(x) = e^x F(x)$.  Now we want &#xD;&#xA;&#xD;&#xA;$$H(x) - H \left( \frac{x}{2} \right) = e^{-x} A(x).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This condition is easier to work with on the left now but harder to work with on the right.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-11-12T01:32:51.243" CommentCount="1" />
  <row Id="9931" PostTypeId="1" CreationDate="2010-11-12T01:52:46.237" Score="2" ViewCount="63" Body="&lt;p&gt;Let $G = \hbox{proj.lim.}_{\alpha} \{ G_{\alpha} , \varphi^{\alpha}_{\beta} \}$ be a projective limit of simple groups (i.e., each $\varphi^{\alpha}_{\beta}\colon G_{\alpha}\to G_{\beta}$ is a surjective group homomorphism between simple groups).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is clear that if $\varphi^{\alpha}_{\beta}$ is not the trivial map ($x\mapsto 1$), then $\varphi^{\alpha}_{\beta}$ is an isomorphism. I believe $G$ is always a simple group. But I don't know how to prove this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any suggestions?&lt;/p&gt;&#xA;" OwnerUserId="3426" LastEditorUserId="8575" LastEditorDisplayName="" LastEditDate="2011-05-19T08:58:30.367" LastActivityDate="2011-05-19T08:58:30.367" Title="projective limit of simple groups" Tags="&lt;group-theory&gt;&lt;algebraic-topology&gt;&lt;simple-groups&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" />
  <row Id="9933" PostTypeId="1" AcceptedAnswerId="9942" CreationDate="2010-11-12T02:11:02.340" Score="9" ViewCount="622" Body="&lt;p&gt;Someone recently asked me why a negative * a negative is positive, and why a negative * a positive is negative, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I went ahead and gave them a proof by contradiction like so:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume (-x) * (-y) = -xy&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then divide both sides by (-x) and you get (-y) = y&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since we have a contradiction, then our first assumption must be incorrect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm guessing I did something wrong here...since the conclusion of (-x) * (-y) = (xy) is hard to derive from what I wrote.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a better way to explain this?  Is my proof incorrect?  Also, what would an intuitive way to explain the negation concept be, if there is one?&lt;/p&gt;&#xA;" OwnerUserId="1630" LastActivityDate="2011-03-09T19:50:04.113" Title="Why negative times negative = positive?" Tags="&lt;teaching&gt;" AnswerCount="11" CommentCount="4" />
  <row Id="9934" PostTypeId="1" CreationDate="2010-11-12T02:14:17.193" Score="11" ViewCount="251" Body="&lt;p&gt;Cyclic groups have at most one subgroup of any given finite index. Can we describe the class of all groups having such property?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you!&lt;/p&gt;&#xA;" OwnerUserId="3427" LastActivityDate="2010-11-28T15:01:03.837" Title="Groups having at most one subgroup of any given finite index" Tags="&lt;group-theory&gt;" AnswerCount="3" CommentCount="4" FavoriteCount="1" />
  <row Id="9935" PostTypeId="1" AcceptedAnswerId="9943" CreationDate="2010-11-12T02:21:20.957" Score="6" ViewCount="172" Body="&lt;p&gt;I know a proof that $\frac{(30n)!n!}{(15n)!(10n)!(6n)!}$ is an integer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The proof goes as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If a prime $p$ divides $(15n)!(10n)!(6n)!$, then the power of the prime dividing $(15n)!(10n)!(6n)!$ is lesser than the $(30n)!n!$. It is relatively easy to prove this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\textbf{My question is there a counting argument to prove that this is an integer?}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By this, I mean does $\frac{(30n)!n!}{(15n)!(10n)!(6n)!}$ count something?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I have a gut feeling that this should count something but I have not thought in depth about this).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance $\frac{n!}{r!(n-r)!}$ counts the number of ways of choosing $r$ objects out of $n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, are there other examples similar to this? (I tried searching for other such examples in vain)&lt;/p&gt;&#xA;" OwnerUserId="3302" LastActivityDate="2010-11-12T02:46:47.287" Title="Does $\frac{(30n)!n!}{(15n)!(10n)!(6n)!}$ count something?" Tags="&lt;number-theory&gt;&lt;combinatorics&gt;" AnswerCount="1" CommentCount="6" FavoriteCount="1" />
  <row Id="9936" PostTypeId="2" ParentId="9933" CreationDate="2010-11-12T02:34:40.947" Score="0" ViewCount="" Body="&lt;p&gt;This come from ring axioms. (As Qiaochu pointed out, we do not need the Field Axioms for this.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First prove $0 \times y = 0$, $\forall y$ in the ring and where $0$ is the additive identity in the ring.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then prove $(-x) \times y = - (x \times y)$, $\forall x,y$ in the ring.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Use these two and then manipulate to get the result $(-x) \times (-y) = x \times y$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me know if you run into any difficulty.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2010-11-12T02:51:27.670" LastActivityDate="2010-11-12T02:51:27.670" CommentCount="3" />
  <row Id="9937" PostTypeId="2" ParentId="9933" CreationDate="2010-11-12T02:36:43.920" Score="4" ViewCount="" Body="&lt;p&gt;Here's a proof.  First, for all $x$, $x\cdot 0=x\cdot(0+0)=x\cdot 0 +x\cdot 0$.  Subtracting $x\cdot0$ from each side, $x\cdot0=0$.  Now, for all $x$ and $y$, $0=x\cdot0=x\cdot(-y+y)=x\cdot(-y)+x\cdot y$.  Subtracting $x\cdot y$ from both sides, $x\cdot(-y)=-(x\cdot y)$.  Applying this twice along with the identity $-(-a)=a$, $(-x)\cdot(-y)=-(-x)\cdot y=-(-(x\cdot y))=x\cdot y$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your proof implicitly uses the fact that $-xy=(-x)y$, and assumes that there are only two possibilities, $xy$ or $-xy$, then shows that the latter is impossible.  These seem like plausible assumptions, but I tried to be very careful in my proof above (thus using $-(x\cdot y)$ rather than simply $-xy$ to not be confused with $(-x)\cdot y$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I only have a vague intuitive notion that I probably can't explain well, but I sometimes think of a negative number like $-5$ as being &quot;$5$ in the other direction&quot;, and so multiplying by $-5$ means &quot;multiply by $5$ and switch direction&quot;, i.e., sign.  This means if you multiply $-5$ by a negative number, you should switch its direction back to positive.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-11-12T02:36:43.920" CommentCount="1" />
  <row Id="9938" PostTypeId="2" ParentId="9933" CreationDate="2010-11-12T02:37:50.750" Score="2" ViewCount="" Body="&lt;p&gt;I think the &lt;em&gt;x&lt;/em&gt; and &lt;em&gt;y&lt;/em&gt; get in the way a bit; you can see the crucial steps using just 1 and -1.  What you've really shown is that (-1)(-1)=-1 leads to a contradiction.  If we divide by -1, we get -1=1, which is not true!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Getting the right answer, (-1)(-1)=1, uses a couple more steps:  First, you must agree that (1)+(-1)=0, (1)(-1)=-1, and (0)(-1)=0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, we multiply the first equation by (-1) and use the distributive property to get (-1)(-1)+(-1)(1)=(-1)(0).  Now, we simplify the parts we know to get (-1)(-1)+(-1)=0.  Solve for (-1)(-1), and you get (-1)(-1)=1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, we must have (-1)(-1)=1 if we accept basic rules of arithmetic: 0 is the additive identity, 1 is the multiplicative identity, -1 is the additive inverse of 1, and multiplication distributes over addition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One physical explanation people often like for negative*negative=positive is multiplying rates.  You can film someone walking forwards (positive rate) or walking backwards (negative rate).  Now, play the film back, but in reverse (another negative rate).  What do you see if you play a film backwards of someone walking backwards?  You see the person walking forwards, because negative*negative=positive!&lt;/p&gt;&#xA;" OwnerUserId="1461" LastActivityDate="2010-11-12T02:37:50.750" />
  <row Id="9939" PostTypeId="1" AcceptedAnswerId="9953" CreationDate="2010-11-12T02:38:24.027" Score="8" ViewCount="289" Body="&lt;p&gt;$H$ is a subgroup of $G$ and $a,b$ are in $G$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was rather lost for a while but I think I may have&#xA;actually proved it, though I am unsure. . Does this make sense? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$aH = bH$ &#xA;$\Rightarrow$ $b$ is an element of $aH$ &#xA;$\Rightarrow$ there exists an $h$ in $H$ s.t. $ah = b$&#xA;$\Rightarrow$ $a^{-1}ah = a^{-1}b$ &#xA;$\Rightarrow$ $h = a^{-1}b$&#xA;$\Rightarrow$ $hb^{-1} = a^{-1}bb^{-1}$&#xA;$\Rightarrow$ $hb^{-1} = a^{-1}$&#xA;This means $a^{-1}$ is an element of $Hb^{-1}$, so they are equal. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wasn't quite sure I could do those operations.&#xA;Can anyone tell me where to learn to do the special symbols, like quantifiers and relations b/c it would make it simpler.&lt;/p&gt;&#xA;" OwnerUserId="3417" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-12T03:37:36.290" LastActivityDate="2010-11-12T04:57:21.617" Title="If $aH = bH$ then $Ha^{-1} = Hb^{-1}$, prove or find a counter example." Tags="&lt;group-theory&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="9940" PostTypeId="2" ParentId="9933" CreationDate="2010-11-12T02:40:03.480" Score="3" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\rm\ \ x\ y\ =\ (-x)\ (-y)\ $ by rewriting $\rm\ x\ y + x\ (-y) + (-x)\ (-y)\ $ in two different ways.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or, $\rm\ x\ y,\ (-x)\ (-y)\ $ are both inverses of $\rm\ x\ (-y)\ $ so they're equal by &lt;strong&gt;uniqueness of inverses&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As I've &lt;a href=&quot;http://math.stackexchange.com/questions/3510/how-to-prove-eulers-formula-expi-t-costi-sint/3514#3514&quot;&gt;emphasized before&lt;/a&gt;, &lt;strong&gt;uniqueness theorems&lt;/strong&gt; are powerful tools for proving equalities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More generally the &lt;strong&gt;Law of Signs&lt;/strong&gt; holds for any &lt;strong&gt;odd&lt;/strong&gt; functions &#xA;under composition, e.g. polynomials with all terms having odd power. Indeed we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\ \ (-f)\ (-g)\  =\ -(f(-g))\ =\ f(g)\ \iff\ f(-g)\ = -(f(g))\ =\ (-f)(g) $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm \phantom{\ \ (-f)\ (-g)\  =\ -(f(-g))\ =\ f(g)\ }\iff \ f(-x)\ = -f(x)\:,\ $ ie. $\rm\:f\:$ odd, $\ $ via $\rm\ g = x = $ identity &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Generally such functions enjoy only a weaker &lt;a href=&quot;http://en.wikipedia.org/wiki/Nearring&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;near&lt;/em&gt;-ring&lt;/a&gt; structure. &#xA;In above ring case, distributivity implies that multiplication &#xA;is linear hence odd (here one views the ring in Cayley-style &#xA;as the ring of endormorphisms of its abelian additive group, &#xA;i.e. represent each ring elt $\rm\ r\ $ by the linear map  $\rm\ x \to r\ x\:,\ $ &#xA;i.e. as a 1-dim matrix).&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-11-12T03:00:44.017" LastActivityDate="2010-11-12T03:00:44.017" CommentCount="1" />
  <row Id="9941" PostTypeId="1" AcceptedAnswerId="10163" CreationDate="2010-11-12T02:40:48.850" Score="2" ViewCount="138" Body="&lt;p&gt;I'm trying to find a general solution for $\displaystyle\iint_{S}[[x+y]] dA$ where $S=[a,b] \times [c,d]$ and $[[x]]$ is the greatest integer function. I know that if $a,b,c,d \in \mathbb{Z}$, then the answer is just $\frac{(d-c)(b-a)(a+b+c+d-1)}{2}$ (proven below), but how do I account for the extra volume when $a,b,c,d\in \mathbb{R}$?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;By looking at the function itself, it's clear that the volume of the region above $S=[a,a+1]\times[b,b+1]$ is $a+b+\frac{1}{2}$ (the volume of the rectangular prism is $a+b$ while the volume of the triangular prism on top of it is $\frac{1}{2}$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, the total volume is:&#xA;$$\displaystyle\sum_{y=c}^{d-1}\sum_{x=a}^{b-1} x+y+\frac{1}{2}=\sum_{y=c}^{d-1}\frac{b^2-a^2}{2}+(b-a)y=\frac{(d-c)(b-a)(a+b+c+d-1)}{2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One other question I'd like to confirm is that $\displaystyle\underset{S}{\int \cdots \int} [[x_1+x_2+\cdots+x_n]] dx_1 \cdots dx_n$ over $S=[i_1,i_2] \times [i_3,i_4] \times \cdots \times [i_{2n-1},i_{2n}]$ is:&#xA;$$\displaystyle\frac{1}{2}\left(\sum_{j=1}^{2n}i_j\right)\left(\prod_{k=1}^{n}(i_{2k}-i_{2k-1})\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this true if $i_k\in\mathbb{Z}$ for all $k\in[0,n]$? Can this, too, be extended to all reals?&lt;/p&gt;&#xA;" OwnerUserId="1562" LastEditorUserId="1562" LastEditorDisplayName="" LastEditDate="2010-11-12T03:45:49.650" LastActivityDate="2010-11-13T20:32:13.310" Title="How to find a closed form solution to $\displaystyle\iint_{S}[[x+y]] dA$" Tags="&lt;calculus&gt;" AnswerCount="2" />
  <row Id="9942" PostTypeId="2" ParentId="9933" CreationDate="2010-11-12T02:42:58.317" Score="4" ViewCount="" Body="&lt;p&gt;Well if I were to explain this in an intuitive way to someone (or at least try), I would like to think of an analogy with walking over the real line, by agreeing that walking left will be walking in the negative direction and walking right in the positive direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then I will try to convey the idea that if you are multiplying to numbers (let's suppose they are integers to make things easier to picture) then a product as $2*3$ would just mean that you have to walk right (in the positive direction) a distance of $2$ (say miles for instance) three times, that is, first you walk $2$ miles, then another $2$ miles and finally another $2$ miles to the right.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now you picture where you're at? Well, you're at the right of the origin so you are in the positive section. But in the same way you can play this idea with a negative times a positive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With the same example in mind, what would $-2*3$ mean? First, suppose that the $-2$ just specifies that you will have to walk left a distance of $2$ miles. Then how many times you will walk that distance? Just as before $3$ times and in the end you'll be $6$ miles to the left of the origin so you'll be in the negative section.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally you'll have to try to picture what could (-2)*(-3) mean. Maybe you could think of the negative sign in the second factor just making you change direction, that is, it makes you turn around and start walking the specified distance. So in this case the $-2$ tells you to walk left a distance of $2$ miles but the $-3$ tells you to fist turn around and then walk $3$ times the $2$ miles in the other direction, so you'll end up walking right and end in the point that is $6$ miles to the right of the origin, so you'll be in the positive section, and $(-2)*(-3) = 6$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know if this will help but it's the only way I can think of this in some intuitive sense.&lt;/p&gt;&#xA;" OwnerUserId="900" LastActivityDate="2010-11-12T02:42:58.317" />
  <row Id="9943" PostTypeId="2" ParentId="9935" CreationDate="2010-11-12T02:46:47.287" Score="4" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://mathoverflow.net/questions/26336/integer-valued-factorial-ratios&quot; rel=&quot;nofollow&quot;&gt;This MO thread&lt;/a&gt; has a lot of good information.  I think we should be pessimistic.  Pietro Majer's answer and the comments therein suggest that this would be hard, at least in the sense that one should not expect a proof anything like the proof for binomial coefficients.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-11-12T02:46:47.287" CommentCount="1" />
  <row Id="9944" PostTypeId="2" ParentId="9939" CreationDate="2010-11-12T03:06:39.307" Score="4" ViewCount="" Body="&lt;p&gt;$aH=bH$ does not imply that $a$ or $b \in H$.  It just implies they are in the same coset.  It does imply that $b^{-1}a \in H$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To get the nice formatting, enclose your formulas in dollar signs.  If you right click on a formula, there is a menu item Show Source that will show you the text (without the dollar signs) that created it.  Double dollar signs get display mode.  When you are typing, an interpreted version is visible below.  Handy references for LaTeX are &lt;a href=&quot;http://web.ift.uib.no/Teori/KURS/WRK/TeX/symALL.html&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; and &lt;a href=&quot;http://www.artofproblemsolving.com/Wiki/index.php/LaTeX%3aCommands&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-11-12T03:06:39.307" />
  <row Id="9945" PostTypeId="1" CreationDate="2010-11-12T03:09:03.317" Score="1" ViewCount="89" Body="&lt;p&gt;Why is the following inequality true: if $x \geq 0$ then $(1-\frac{x}{n})^{n} \leq e^{-x}$ ? here $n$ is a positive integer. Is there a quick way to see this?&lt;/p&gt;&#xA;" OwnerUserId="4219" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-12T04:05:03.053" LastActivityDate="2010-11-12T04:05:03.053" Title="$(1-\frac{x}{n})^n\lt \exp(-x)$" Tags="&lt;calculus&gt;" AnswerCount="4" CommentCount="2" />
  <row Id="9946" PostTypeId="2" ParentId="9945" CreationDate="2010-11-12T03:15:12.207" Score="0" ViewCount="" Body="&lt;p&gt;You can look at it this way:&#xA;$(1 - \frac{x}{n})^n = \sum_{k=0}^{n} C(n,k) (-1)^k \frac{x^k}{n^k}$&#xA;and&#xA;$e^{-x} = \sum_{k=0}^{+\infty} (-1)^k \frac{x^k}{k!}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it clearer now ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(hint: one of the sums is finite, the other not, and one of the terms grows quicker than the other)&lt;/p&gt;&#xA;" OwnerUserId="1865" LastActivityDate="2010-11-12T03:15:12.207" CommentCount="2" />
  <row Id="9947" PostTypeId="2" ParentId="9945" CreationDate="2010-11-12T03:20:50.500" Score="1" ViewCount="" Body="&lt;p&gt;You might want to prove this for $0 \leq x \leq n$. Else it might not be always true. (For instance, consider the case when $x$ is really large and $n$ is even)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the function $f(y) = e^{y} \times (1-y)$. Note that $f(0) = 1$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We will now prove that $f(y)$ is a decreasing function $\forall y \geq 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{df(y)}{dy} = e^{y} - ye^{y} - e^{y} = -ye^{y} \leq 0$, $\forall y \geq 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence $f(y)$ is a decreasing function $\forall y \geq 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, we have $f(y) \leq f(0) = 1$, $\forall y \geq 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, we get $e^{y} \times (1-y) \leq 1$, which implies $(1-y) \leq e^{-y}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Replace $y = \frac{x}{n}$ and raise both sides to the $n^{th}$ power where $n \in \mathbb{R}^{+}$. The inequality remains the same since $0 \leq x \leq n$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, we get $(1-\frac{x}{n})^n \leq e^{-x}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is also useful to see that $\displaystyle \lim_{n \rightarrow \infty} (1-\frac{x}{n})^n = e^{-x}$ from below.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2010-11-12T03:40:13.557" LastActivityDate="2010-11-12T03:40:13.557" CommentCount="1" />
  <row Id="9948" PostTypeId="2" ParentId="9945" CreationDate="2010-11-12T03:24:33.913" Score="1" ViewCount="" Body="&lt;p&gt;consider the case where $x = 4$ and $n=2$. Here, LHS &gt; RHS. So the inequality does not always hold.&lt;/p&gt;&#xA;" OwnerUserId="2641" LastActivityDate="2010-11-12T03:24:33.913" />
  <row Id="9949" PostTypeId="2" ParentId="9933" CreationDate="2010-11-12T03:25:31.387" Score="5" ViewCount="" Body="&lt;p&gt;The elementary intuition behind the product of two negatives can be thought of as follows. You have a bank account. You pay 3 bills for 40 dollars each, $3 X (-40) = -120$ is added to your account. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The opposite of being billed would be billing someone else. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, if you bill 3 people for 40 dollars each, $(-3) X (-40)$ is added to your account. This value should be positive since it results in you receiving money. &lt;/p&gt;&#xA;" OwnerUserId="2563" LastActivityDate="2010-11-12T03:25:31.387" />
  <row Id="9950" PostTypeId="2" ParentId="9939" CreationDate="2010-11-12T03:25:45.770" Score="1" ViewCount="" Body="&lt;p&gt;You're on the right track. $aH = bH$ is equivalent to $a^{-1}b \in H$, and $Ha^{-1} = Hb^{-1}$ is equivalent to $a^{-1} (b^{-1})^{-1} = a^{-1}b \in H$, which is the same thing. One approach that should work would be to turn your arguments into statements of this nature. &lt;/p&gt;&#xA;" OwnerUserId="3035" LastActivityDate="2010-11-12T03:25:45.770" />
  <row Id="9951" PostTypeId="1" CreationDate="2010-11-12T03:29:40.600" Score="-1" ViewCount="206" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Possible Duplicate:&lt;/strong&gt;&lt;br&gt;&#xA;  &lt;a href=&quot;http://math.stackexchange.com/questions/9141/about-irrational-logarithms&quot;&gt;About irrational logarithms&lt;/a&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;p&gt;Please help proving that $\log_{10}(2)$ is irrational.&lt;/p&gt;&#xA;" OwnerUserId="3430" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-12T03:39:38.677" LastActivityDate="2010-11-14T20:15:10.407" ClosedDate="2010-11-15T07:48:25.783" Title="prove $\log_{10} (2)$ is irrational .." Tags="&lt;logarithms&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="9952" PostTypeId="2" ParentId="9945" CreationDate="2010-11-12T03:32:02.597" Score="0" ViewCount="" Body="&lt;p&gt;First thing: you need $n$ to be even for this if  1 - x/n is negative.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your statement is the same as $1 - {x \over n} \leq e^{-{x \over n}}$. Letting $t = {x \over n}$ this follows from $1 - t \leq e^{-t}$. This can be proven using calculus.&lt;/p&gt;&#xA;" OwnerUserId="3035" LastActivityDate="2010-11-12T03:32:02.597" CommentCount="2" />
  <row Id="9953" PostTypeId="2" ParentId="9939" CreationDate="2010-11-12T03:37:03.403" Score="2" ViewCount="" Body="&lt;p&gt;For formatting, see &lt;a href=&quot;http://meta.math.stackexchange.com/questions/107/what-should-go-in-the-math-stackexchange-faq/255#255&quot;&gt;here&lt;/a&gt; and some of the links there. You need to know a bit of LaTeX.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As to your argument, it is (mostly) correct. My only quibbles would be of presentation. From $aH=bH$ you conclude that there must exist $h\in H$ such that $b=ah$. That's fine. Your next step should be that &quot;there exists $h\in H$ such that $a^{-1}b=a^{-1}ah = h$&quot;. Then &quot;There exists $h\in H$ such that $a^{-1}=a^{-1}bb^{-1}=hb^{-1}$&quot;, hence $a^{-1}$ is in the coset $Hb^{-1}$. Your final statement &quot;hence they are equal&quot; is technically wrong, since $a^{-1}$ is not equal to the coset $Hb^{-1}$; what you &lt;em&gt;meant&lt;/em&gt; to say is that the since $a^{-1}$ is in $Hb^{-1}$, then the coset $Ha^{-1}$ is equal to the coset $Hb^{-1}$. Other that these quibbles, the argument seems correct to me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, to see if you understand it well, try to figure out which, if any, of your implications are reversible, to see if the &lt;em&gt;converse&lt;/em&gt; also holds.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-12T04:57:21.617" LastActivityDate="2010-11-12T04:57:21.617" />
  <row Id="9954" PostTypeId="2" ParentId="9951" CreationDate="2010-11-12T03:40:35.663" Score="7" ViewCount="" Body="&lt;p&gt;Hint: if $10^{a/b} = 2$ then $10^a = 2^b$. &lt;/p&gt;&#xA;" OwnerUserId="630" LastActivityDate="2010-11-12T03:40:35.663" />
  <row Id="9956" PostTypeId="2" ParentId="9951" CreationDate="2010-11-12T03:41:37.940" Score="2" ViewCount="" Body="&lt;p&gt;Prove by contradiction. So start by assuming&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\log_{10}{2}=\frac{m}{n}$$, where $m$ and $n$ are integers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$10^\frac{m}{n}=2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$10^m=2^n$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and derive a contradiction.&lt;/p&gt;&#xA;" OwnerUserId="2219" LastActivityDate="2010-11-12T03:41:37.940" CommentCount="2" />
  <row Id="9957" PostTypeId="2" ParentId="9933" CreationDate="2010-11-12T03:47:56.043" Score="3" ViewCount="" Body="&lt;p&gt;Quite a good explanation is that one wants the distributive law to work in general with &lt;em&gt;positive&lt;/em&gt; quantities when you add (smaller) negative ones:&#xA;If $x&gt;a\ge0$ and $y&gt;b\ge0$ then&#xA;$$&#xD;&#xA;(x-a)(y-b)=(x+(-a))(y+(-b))=xy+(-a)y+x(-b)+(-a)(-b)&#xD;&#xA;$$&#xA;For this to always work, one needs $(-a)y=-(ay)$ in case $b=0$, $x(-b)=-xb$ in case $a=0$, and $(-a)(-b)=ab$.&lt;/p&gt;&#xA;" OwnerUserId="2947" LastActivityDate="2010-11-12T03:47:56.043" />
  <row Id="9958" PostTypeId="1" CreationDate="2010-11-12T03:49:28.350" Score="8" ViewCount="654" Body="&lt;p&gt;I would like this to be a community wiki. But, I don't know how to do that here. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to know sources, and examples of good &quot;challenge&quot; problems for students who have studied pre-calculus and some calculus. (differentiation and the very basics of integration.) Topics could be related to things such as:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Taylor Series.&lt;/li&gt;&#xA;&lt;li&gt;Product Rule, Quotient Rule, Chain Rule.&lt;/li&gt;&#xA;&lt;li&gt;Simple limits.&lt;/li&gt;&#xA;&lt;li&gt;Delta Epsilon Proofs.&lt;/li&gt;&#xA;&lt;li&gt;Induction proofs for the sum of the first n, integer, squares, etc.&lt;/li&gt;&#xA;&lt;li&gt;Integration by substitution.&lt;/li&gt;&#xA;&lt;li&gt;Other topics...&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;What I have found so far are too many problems that are just a bit too difficult. The problem can have a &quot;trick&quot; but it needs to be something a freshman could do. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is one problem that I thought was just at the right level:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $f(x) = \frac{x}{x+\frac{x}{x+ \frac{x}{x+ \vdots}}}$, find $f'(x)$*&lt;/p&gt;&#xA;&#xA;&lt;p&gt;*To be honest this problem makes me a little nervous. Still, I like it.&lt;/p&gt;&#xA;" OwnerUserId="2563" LastActivityDate="2010-11-12T19:26:59.030" CommunityOwnedDate="2010-11-12T08:33:04.083" Title="Interesting calculus problems of medium difficulty?" Tags="&lt;calculus&gt;&lt;algebra-precalculus&gt;&lt;big-list&gt;" AnswerCount="16" CommentCount="2" FavoriteCount="9" />
  <row Id="9959" PostTypeId="2" ParentId="9934" CreationDate="2010-11-12T03:54:05.687" Score="13" ViewCount="" Body="&lt;p&gt;Since $G$ has exactly one subgroup of each finite index, and the index of a conjugate of $H$ equals the index of $H$, then every subgroup of finite index is normal. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $G$ is finite, then &lt;em&gt;every&lt;/em&gt; subgroup is normal, so the group must be a &lt;a href=&quot;http://en.wikipedia.org/wiki/Hamiltonian_group&quot; rel=&quot;nofollow&quot;&gt;Dedekind group&lt;/a&gt; (also known as Hamiltonian groups). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;All such groups that are nonabelian are of the form $G = Q_8 \times B \times D$, where $Q_8$ is the quaternion group of $8$ elements, $B$ is a direct sum of copies of the cyclic group of order $2$, and $D$ is an abelian group of odd order. Any of the factors may be missing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $Q_8$ contains several subgroups of index $4$, if a factor of $Q_8$ appears then $G$ would have several subgroups of the same index, hence $G$ must in fact be an abelian group. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $G$ is finite and abelian, it is isomorphic to a direct sum of cyclic groups, $G = C_{a_1}\oplus\cdots\oplus C_{a_k}$, where $1\lt a_1|a_2|\cdots|a_k$. If $k\gt 1$, then $G$ contains at least two subgroups of order $a_{k-1}$; thus $k=1$ so $G$ is in fact cyclic. So the only finite groups with the desired property are the cyclic groups.&#xA;If $G$ is infinite, you can have other possibilities. One example is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Pr%C3%BCfer_group&quot; rel=&quot;nofollow&quot;&gt;Prufer group&lt;/a&gt;, &lt;strong&gt;Added:&lt;/strong&gt; but only by vacuity: it has no proper subgroups of finite index.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, if $H$ if a subgroup of finite index in $G$ then $H$ is normal, as above, and $G/H$ also has the desired property and is finite; thus, $G/H$ is cyclic for every subgroup of finite index by the argument above. I'm sure there's more to be said, but I'll think about it a bit first...&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-13T02:32:26.113" LastActivityDate="2010-11-13T02:32:26.113" CommentCount="11" />
  <row Id="9960" PostTypeId="1" AcceptedAnswerId="9973" CreationDate="2010-11-12T03:54:37.590" Score="1" ViewCount="91" Body="&lt;p&gt;How does one show for complex numbers u and v, and for p&gt;1 that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$||u|^{p-1}u - |v|^{p-1}v|\leq c_p |u-v|(|u|^{p-1}+|v|^{p-1})$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $c_p$ is some constant dependent on p.  My intuition is to use some version of the mean value theorem with $F(u) = |u|^{p-1}u$, but I'm not sure how to make this work for complex-valued functions.  Plus there seems to be an issue with the fact that $F$ may not smooth near the origin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For context, this shows up in Terry Tao's book Nonlinear Dispersive Equations: Local and Global Analysis on pg. 136, where it is stated without proof as an &quot;elementary estimate&quot;.&lt;/p&gt;&#xA;" OwnerUserId="1692" LastActivityDate="2010-11-12T05:34:05.237" Title="Complex inequality $||u|^{p-1}u - |v|^{p-1}v|\leq c_p |u-v|(|u|^{p-1}+|v|^{p-1})$" Tags="&lt;complex-numbers&gt;&lt;inequality&gt;" AnswerCount="2" />
  <row Id="9961" PostTypeId="2" ParentId="9951" CreationDate="2010-11-12T03:57:01.947" Score="1" ViewCount="" Body="&lt;p&gt;We argue by contradiction. Suppose $\log_{10} 2 = \frac{p}{q}$ is rational with $q &amp;gt; p &amp;gt; 0$. Then $ 2^{q} = 10^{p} = 2^{p} \ 5^{p}$, so $2^{q - p} = 5^{p}$. Since $q &amp;gt; p$ and $p &amp;gt; 0$, it follows that $2^{q - p} \equiv 0 \mod 5$, which is impossible since no power of $2$ ends in $0$ or $5$. Hence $\log_{10} 2$ is irrational. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This method generalizes easily to prove the irrationality of many reals of the form $\log_{b} \ a$ for $a, b \in \mathbb{N}$.&lt;/p&gt;&#xA;" OwnerUserId="2720" LastEditorUserId="2720" LastEditorDisplayName="" LastEditDate="2010-11-14T20:15:10.407" LastActivityDate="2010-11-14T20:15:10.407" CommentCount="2" />
  <row Id="9963" PostTypeId="2" ParentId="9931" CreationDate="2010-11-12T04:14:49.053" Score="2" ViewCount="" Body="&lt;p&gt;Suppose that for every $\beta$, there is $\alpha \geq \beta$ such that $\phi_{\beta}^{\alpha}$ is trivial.  Then the projective limit is trivial.&#xA;(Because if $(x_{\beta})$ is an element of the projective limit, then&#xA;for each $\beta$, we have $x_{\beta} = \phi_{\beta}^{\alpha}(x_{\alpha}) = 1,$&#xA;if we choose $\alpha$ for the given $\beta$ so that $\phi_{\beta}^{\alpha}$ is&#xA;trivial --- as we may, by assumption.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, suppose that for some $\beta$, the maps $\phi_{\beta}^{\alpha}$&#xA;are isomorphisms for all $\alpha \geq \beta$.  If $A$ is the directed set indexing the projective limit, then replacing $A$ by its subset $A_{\geq \beta}&#xD;&#xA;:= \{\alpha \in A \, | \ \alpha \geq \beta\}$ doesn't change the projective&#xA;limit (since $A_{\geq \beta}$ is cofinal in $A$), and now all the transition&#xA;maps $\phi_{\gamma}^{\alpha}$ for $\alpha, \gamma \in A_{\geq \beta}$ are isomorphisms. (Since each of $\phi_{\beta}^{\gamma}$ and  $\phi_{\beta}^{\alpha} = \phi_{\beta}^{\gamma}\circ \phi_{\gamma}^{\alpha}$ is an isomorphism, by assumption.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the projective limit of the $G_{\alpha}$ (for $\alpha \in A_{\geq &#xD;&#xA;\beta}$) has all its transition maps being isomorphisms, and so is isomorphic to $G_{\beta}$,&#xA;and hence is simple.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus the projective limit is either trivial or simple, depending on whether one is in the first or second case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Added in response to Arturo Magidin's comment below: I hadn't paid attention to the fact that the transition maps are assumed surjective.  This implies that in my first case, all the $G_{\beta}$ are trivial (!), since they are (by assumption) the image of some $G_{\alpha}$ under the trivial map.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So unless all the $G_{\beta}$ are trivial, then one must be in the second case, when the $\phi^{\alpha}_{\beta}$ are all eventually isomorphisms.&lt;/p&gt;&#xA;" OwnerUserId="221" LastEditorUserId="221" LastEditorDisplayName="" LastEditDate="2010-11-12T04:23:51.993" LastActivityDate="2010-11-12T04:23:51.993" CommentCount="6" />
  <row Id="9964" PostTypeId="1" AcceptedAnswerId="9969" CreationDate="2010-11-12T04:23:26.490" Score="2" ViewCount="65" Body="&lt;p&gt;Given a sequence $(x_n)$ in a metric space $M$, call it &lt;em&gt;uniformly separated&lt;/em&gt; if all pairwise distances $d(x_n,x_m)$ between distinct terms are uniformly bounded away from zero.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose that a given sequence $(x_n)$  has the property that it has no Cauchy subsequence. Must it have a uniformly separated subsequence?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A couple of my students asserted this in homework without watertight proof, and proof or disproof is eluding me. Clearly any counterexample cannot be in ${\mathbb R}^n$. Enjoy!&lt;/p&gt;&#xA;" OwnerUserId="2947" LastActivityDate="2010-11-12T04:33:05.353" Title="If no Cauchy subsequence exists, must a uniformly separated subsequence exist?" Tags="&lt;convergence&gt;&lt;metric-spaces&gt;" AnswerCount="1" />
  <row Id="9965" PostTypeId="2" ParentId="9958" CreationDate="2010-11-12T04:24:02.363" Score="1" ViewCount="" Body="&lt;p&gt;I recommend the following sources (big advantage: free access):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Mathematical Reflections: &lt;a href=&quot;http://awesomemath.org/mathematical-reflections/&quot; rel=&quot;nofollow&quot;&gt;http://awesomemath.org/mathematical-reflections/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Through the problem column, we challenge students to develop their creative problem solving and reasoning skills by devising solutions to the proposed questions. Exceptional solutions will be published, with the intent of encouraging students to formally write out and submit their work to be showcased in print&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) School Science and Mathematics: &lt;a href=&quot;http://ssmj.tamu.edu/problems.php&quot; rel=&quot;nofollow&quot;&gt;http://ssmj.tamu.edu/problems.php&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;This section offers readers an opportunity to exchange interesting mathematical problems and solutions. Proposals are always welcomed...&#xA;The editors encourage undergraduate and pre-college students to submit solutions. Teachers can help by assisting their students in submitting solutions. Student solutions should include the class and school name&quot;.&lt;/p&gt;&#xA;" OwnerUserId="2810" LastActivityDate="2010-11-12T04:24:02.363" CommunityOwnedDate="2010-11-12T08:33:04.083" />
  <row Id="9966" PostTypeId="2" ParentId="9958" CreationDate="2010-11-12T04:27:51.170" Score="2" ViewCount="" Body="&lt;p&gt;The &lt;a href=&quot;http://people.missouristate.edu/lesreid/POTW.html&quot; rel=&quot;nofollow&quot;&gt;Missouri State problems&lt;/a&gt; are often very good&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-11-12T04:27:51.170" CommunityOwnedDate="2010-11-12T08:33:04.083" />
  <row Id="9967" PostTypeId="2" ParentId="9760" CreationDate="2010-11-12T04:29:17.103" Score="3" ViewCount="" Body="&lt;p&gt;Here's another direction of attack.  The expected number of steps before a single, given counter reaches $N$ is exactly $3N^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As in I. J. Kennedy's approach, let $x_k$ denote the expected time before a given counter reaches $N$ given that the counter starts $k$ steps away from $N$.  Then the $x_k$'s satisfy, for $1 \leq k \leq N-1,$&#xA;$$x_k = \frac{1}{6} (1 + x_{k+1}) + \frac{4}{6} (1 + x_k) + \frac{1}{6} (1 + x_{k-1}),$$&#xA;with boundary conditions&#xA;$x_0 = 0$ and $x_N = \frac{2}{3} (1 + x_N) + \frac{1}{3}(1 + x_{N-1}),$ the second of which simplifies to $x_N = x_{N-1} + 3$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The difference equation reduces to $$x_{k+1} - 2x_k + x_{k-1} = -6,$$ and thus is linear, nonhomogeneous, and second-order.  The solution technique is very similar to that for the corresponding differential equation.  The characteristic equation for the homogeneous difference equation is $r^2 -2r + 1 = 0$, which means we have the double root $r = 1$.  Thus the general solution to the homogeneous equation is $y_k = A(1)^k + B k (1)^k = A + Bk$.  We then use the method of undetermined coefficients with $z_k = Ck^2$ as our candidate to find a particular solution to the nonhomogeneous equation.  This yields $C = -3$.  Thus the general solution to the difference equation is &#xA;$$x_k = A + Bk - 3k^{2}.$$&#xA;Now, we use the boundary conditions to find $A$ and $B$.  These yield $A = 0$ and &#xA;$$A + BN - 3N^2 = A + B(N-1) - 3(N-1)^2 + 3.$$&#xA;Solving this, we have $B = 6N$.  Thus our solution is &#xA;$$x_k = 6Nk - 3k^2,$$&#xA;which means &#xA;$$x_N = 3N^{2}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't have time right now to pursue this further, but maybe someone else can use this to find a reasonable upper bound for the time until all three counters hit $N$ simultaneously like the OP wants.  In particular, though, if it takes on the order of $N^2$ steps on average for a given counter to reach $N$ for the first time it seems that it should take more than the order of $N^3$ steps on average for all three of them to hit $N$ together.&lt;/p&gt;&#xA;" OwnerUserId="2370" LastActivityDate="2010-11-12T04:29:17.103" />
  <row Id="9968" PostTypeId="1" CreationDate="2010-11-12T04:29:19.270" Score="4" ViewCount="101" Body="&lt;p&gt;Let we have a competitive survival game in which a player has choice between different resources to earn. The question here is which resource should he prefer to maximize the chance of survival. I tried to find out if there already established measures of such sort, but so far without success.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some thoughts led me to the following measure of the resource' value:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$V=\int_{t_0}^{t_1} e^{\int_0^t \log (p_1(u))du}\log \frac{p_1(t)}{p_0(t)} dt$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $p_0(t)$ is the probability density of survival without the resource, $p_1(t)$ is the probability density of survival with the resource and $(t_0,t_1)$ is the period of time through which the resource affects the probability of survival (the time for which the value is evaluated considered t=0)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to know if there any similar measures already proposed and about the possible drawbacks of this proposed value. I also wonder to which extent this can be applied to real economics.&lt;/p&gt;&#xA;" OwnerUserId="2513" LastEditorUserId="2513" LastEditorDisplayName="" LastEditDate="2010-11-12T10:32:37.817" LastActivityDate="2010-11-12T10:32:37.817" Title="Measure of value of resources in a competitive game" Tags="&lt;probability&gt;&lt;game-theory&gt;&lt;economics&gt;" CommentCount="2" />
  <row Id="9969" PostTypeId="2" ParentId="9964" CreationDate="2010-11-12T04:33:05.353" Score="3" ViewCount="" Body="&lt;p&gt;A proof can be inferred from the answers to an earlier question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://math.stackexchange.com/questions/7210/which-metric-spaces-are-totally-bounded/&quot;&gt;Which metric spaces are totally bounded?&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="467" LastActivityDate="2010-11-12T04:33:05.353" CommentCount="1" />
  <row Id="9970" PostTypeId="2" ParentId="9958" CreationDate="2010-11-12T04:37:19.490" Score="1" ViewCount="" Body="&lt;p&gt;Variational calculus is always fun (for example, derive from Newton's law the shape of a chain left hanging from its two ends, you end up with cosh) and is a nice introduction to analytical mechanics. Or pursuit problems (http://mathworld.wolfram.com/PursuitCurve.html) were a real challenge, but were quite enjoyable. However, those a more dealing with differential equations. &lt;/p&gt;&#xA;" OwnerUserId="1034" LastActivityDate="2010-11-12T04:37:19.490" CommunityOwnedDate="2010-11-12T08:33:04.083" />
  <row Id="9971" PostTypeId="2" ParentId="9760" CreationDate="2010-11-12T05:04:31.830" Score="5" ViewCount="" Body="&lt;p&gt;Here's a sketch of another derivation of the hitting time for one counter (re Mike's answer).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the usual $\pm 1$ random walk on the integer line. It is well known that the expected time it takes to reach distance $N$ is $N^2$. In our case reaching distance $N$ and reaching the value $N$ modulo $2N$ is the same, since we can't get to $\pm 3N$ without moving through $\pm N$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We move the specified counter in expectation once every three steps, and therefore it's $3N^2$ and not $N^2$ (because the direction of movement and this &quot;waiting for our turn&quot; are independent).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This argument shows that the case of one counter is rather special.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-12T05:04:31.830" CommentCount="2" />
  <row Id="9972" PostTypeId="2" ParentId="9960" CreationDate="2010-11-12T05:18:07.613" Score="1" ViewCount="" Body="&lt;p&gt;Without loss, you can assume $|u|\le|v|$ and replace $u$ by $uv$ to reduce the problem to the situation $v=1$ and $|u|\le1$. Unless you need $c_p$ explicitly, then it's clear, yes?&lt;/p&gt;&#xA;" OwnerUserId="2947" LastActivityDate="2010-11-12T05:18:07.613" />
  <row Id="9973" PostTypeId="2" ParentId="9960" CreationDate="2010-11-12T05:34:05.237" Score="2" ViewCount="" Body="&lt;p&gt;Suppose without loss of generality that $|u| \geq |v| &gt; 0$. Then you can divide the equation through by $|v|^p$ and your task it to prove $||w|^{p-1}w - 1| \leq c_p|w - 1|(|w|^{p-1} + 1)$, where $w = u/v$. Note that &#xA;$$||w|^{p-1}w - 1| =  ||w|^{p-1}w - |w|^{p-1} + |w|^{p-1} - 1| $$&#xA;$$\leq ||w|^{p-1}w - |w|^{p-1}| + ||w|^{p-1} - 1|$$&#xA;Note the first term is $|w|^{p-1}|w - 1|$ is automatically bounded by your right hand side. So you're left trying to show that $||w|^{p-1} - 1|$ is bounded by your right hand side. For this it suffices to show that&#xA;$$||w|^{p-1} - 1| \leq c_p||w| - 1|| (|w|^{p-1} + 1)$$&#xA;Since $|w| \geq 1$ by the assumption that $|u| \geq |v|$, it suffices to show&#xA;that for all real $r \geq 1$ one has&#xA;$$r^{p-1} - 1 \leq c_p(r - 1)(r^{p-1} + 1)$$&#xA;Now use the mean value theorem as you originally wanted to.&lt;/p&gt;&#xA;" OwnerUserId="3035" LastActivityDate="2010-11-12T05:34:05.237" CommentCount="1" />
  <row Id="9974" PostTypeId="2" ParentId="9958" CreationDate="2010-11-12T05:41:12.580" Score="3" ViewCount="" Body="&lt;p&gt;I like this problem, and my students seem to like it also. Technically, they should use density of the rationals, but you can skirt around this.(At least the solution I am thinking of).&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Find all continuous functions $f:\mathbb{R}\rightarrow\mathbb{R}$ such that $f(x+y)=f(x)+f(y)$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and related:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Show a function is linear iff $f(x)=f(x-a)+f(x-b)-f(x+a+b)$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16" LastActivityDate="2010-11-12T05:41:12.580" CommunityOwnedDate="2010-11-12T08:33:04.083" CommentCount="2" />
  <row Id="9975" PostTypeId="2" ParentId="9958" CreationDate="2010-11-12T05:44:45.697" Score="2" ViewCount="" Body="&lt;p&gt;Another one that is kinda silly, but a little fun is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $f:\mathbb{R}\rightarrow\mathbb{R}$ be a function such that;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\int_0^{\pi/4}f\left(x\right)dx=\int_0^{\pi/8}f\left(x\right)dx=\int_0^{\pi/16}f\left(x\right)dx=\int_0^{\pi/32}f\left(x\right)dx=\dots$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Show that $f\left(\sqrt[3]{t}e^t\right)=-\left(3t\right)f\left(\sqrt[3]{t}e^t\right)$ for infinately many $t$.&lt;/p&gt;&#xA;" OwnerUserId="16" LastActivityDate="2010-11-12T05:44:45.697" CommunityOwnedDate="2010-11-12T08:33:04.083" />
  <row Id="9976" PostTypeId="2" ParentId="9958" CreationDate="2010-11-12T05:48:36.097" Score="1" ViewCount="" Body="&lt;p&gt;You got me started...&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Assume $f\left(x\right)$ is a function ith a continuous derivative and $f\left(1\right)=0$. Show $\exists c\in\mathbb{R}$ such that $c^2f'\left(c\right)+2cf\left(c\right)=0.$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16" LastActivityDate="2010-11-12T05:48:36.097" CommunityOwnedDate="2010-11-12T08:33:04.083" />
  <row Id="9977" PostTypeId="2" ParentId="9958" CreationDate="2010-11-12T05:51:59.607" Score="1" ViewCount="" Body="&lt;p&gt;Ok last one for a while, I promise. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You probably all know this one as it is quite famous:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $f:\left[0,1\right]\rightarrow\left[0,1\right]$, then define &lt;em&gt;a point of period n,&lt;/em&gt; for $n\in\mathbb{N}$ to be a point $x_1$ such that &#xA;            \begin{equation}&#xA;                f^n\left(x_1\right)=\underbrace{f\circ f\circ\dots\circ f}_{n-times}\left(x_1\right)=x_1 ,&#xA;            \end{equation}&#xA;        and that for no $m$ less than $n$, &#xA;$f^m\left(x_1\right)=x_1$. That is, $n$ is the smallest natural number such that $f^n\left(x_1\right)=x_1$. So a point, $x_1$, has period $2$ if  $f\left(f\left(x_1\right)\right)=f\left(x_2\right)=x_1$ and $x_1\neq x_2$. Now define a &lt;em&gt;fixed point of f&lt;/em&gt; to be a point $x_0$ such that $f\left(x_0\right)=x_0$; i.e. it has period one.   Show that if $f$ has a point of period $2$, then it has a fixed point.&lt;/p&gt;&#xA;" OwnerUserId="16" LastEditorUserId="3223" LastEditorDisplayName="" LastEditDate="2010-11-12T11:27:20.750" LastActivityDate="2010-11-12T11:27:20.750" CommunityOwnedDate="2010-11-12T08:33:04.083" />
  <row Id="9978" PostTypeId="1" AcceptedAnswerId="9979" CreationDate="2010-11-12T06:02:41.353" Score="0" ViewCount="231" Body="&lt;p&gt;I have been sick for my real analysis class and I am a having a tough time getting the book.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could someone explain the what open, closed and compact sets are and how they are related with uniform continuity? I think the hinging point was a poor understanding of accumulation points. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="3417" LastActivityDate="2010-11-12T21:00:39.260" Title="Real Analysis Continuity" Tags="&lt;real-analysis&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="9979" PostTypeId="2" ParentId="9978" CreationDate="2010-11-12T06:25:23.400" Score="5" ViewCount="" Body="&lt;p&gt;I assume you're talking about open, closed, and compact subsets of a metric space.  If you're just talking about the real line $\mathbb{R}$, then ignore the commentary about abstract metric spaces and just replace the term &quot;neighborhood&quot; with &quot;open interval.&quot;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;In metric spaces, a set $O$ is &lt;strong&gt;open&lt;/strong&gt; if every point in $O$ is an interior point of $O$.  In other words, every point of $O$ can be surrounded by an open ball (neighborhood) which itself lies in $O$.  So, for instance, the open intervals $(a,b)$ are open subsets of $\mathbb{R}$, and discs $\{x^2 + y^2 \lt R^2 \}$ are open subsets of $\mathbb{R}^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, one thing to keep in mind is that in more abstract metric spaces, the open sets can be a little strange-looking.  For instance, there are metric spaces where every subset is an open set!&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Now, a set $F$ is &lt;strong&gt;closed&lt;/strong&gt; if every accumulation point of $F$ is in $F$.  Accumulation points are points $x$ such that every open neighborhood around $x$ must intersect the set in question.  So, for instance, the accumulation points of the open disc $\{x^2 + y^2 \lt R^2\}$ are precisely the set of points in the closed disc $\{x^2 + y^2 \leq R^2\}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One way to think about closed sets is that they are closed under the operation of taking limits.  That is, if one were to take a sequence of points in a closed set, then their limit would still have to be in the set.  (This can be made rigorous and should be proven at some point.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As with open sets, the closed sets of a metric space can still be rather strange-looking, and again, there are examples of spaces in which every subset is closed.  Nevertheless, it is always true that finite subsets of metric spaces are closed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You should note also that if $(X,d)$ is a metric space, then the sets $\emptyset$ and $X$ are always going to be both open and closed.  In general, the subsets of a metric space may be open, closed, both or neither.  (So just because a set is not open does not make it closed!)&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Compact sets are a little less intuitive.  The rigorous definition is that a set $K$ is &lt;strong&gt;compact&lt;/strong&gt; if every open cover of $K$ has a finite subcover.  Since I don't want to babble for too long, I won't go into what this means in detail.  However, I've always intuited compact sets as being &quot;super closed&quot; sets -- they're like closed sets, but even stronger.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is a fact (whose proof you should learn or create) that every compact subset of a metric space is closed and bounded, but the converse is not necessarily true.  In the euclidean spaces $\mathbb{R}^n$, the converse is, in fact, true, but this is not obvious, and is known as the Heine-Borel Theorem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unlike open and closed sets, in a metric space $(X,d)$, the set $X$ may not be compact.  Therefore, it makes sense to talk about &quot;compact metric spaces,&quot; whereas it doesn't make sense to talk about &quot;open&quot; or &quot;closed&quot; metric spaces (since they're all open and closed subsets of themselves).&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Uniform continuity&lt;/strong&gt; is a stronger condition than continuity.  That is, every uniformly continuous function is continuous, but not necessarily conversely.  It is a theorem (Heine-Cantor) that on compact sets, the converse will hold.  So on compact sets, continuity is equivalent to uniform continuity.  Again, this is an important theorem whose proof you should learn.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Essentially, uniform continuity of a function $f$ on a set $E$ can be thought of as saying that the &quot;degree of continuity&quot; is the same for every point $x$ in the set $E$.  This can be made slightly more precise by saying that the $\delta &gt; 0$ that one gets in the $\epsilon-\delta$ definition of continuity can be chosen to be the same for all $x$ in $E$ -- that is, it will depend only on $\epsilon$, and not on the point $x$.  For plain old continuous functions, the $\delta$ will depend on both $\epsilon$ and $x$.&lt;/p&gt;&#xA;" OwnerUserId="640" LastEditorUserId="640" LastEditorDisplayName="" LastEditDate="2010-11-12T06:40:04.380" LastActivityDate="2010-11-12T06:40:04.380" CommentCount="3" />
  <row Id="9980" PostTypeId="2" ParentId="9978" CreationDate="2010-11-12T06:28:42.787" Score="1" ViewCount="" Body="&lt;p&gt;Difficult one to answer, given the numerous definitions of open sets. Also - more information on what you know, etc, would be useful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll try not to be overly formal - someone will probably be more so than me. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take a point $x$ in a set. If for every point $x$ we can find another point arbitrarily close then this is an open set. From the standard wikipedia definition:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A subset $U$ of the Euclidean n-space $\mathbb{R}^n$ is called open if, given any point $x$ in $U$, there exists a real number $\epsilon &gt; 0$ such that, given any point $y$ in $\mathbb{R}^n$ whose Euclidean distance from $x$ is smaller than $\epsilon$, $y$ also belongs to $U$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now (again see Wikipedia) you can replace the Euclidean space with any metric space, and come up with an equivalent definition. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Closed sets are sets where the complement is open. One of the most confusing aspects is there exists sets that are both open and closed. These are known as Clopen sets. The definition of closed sets you are after may be the one involving accumulation points. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Definition: $x$ is an accumulation points of $U$ if every $\epsilon$ neighborhood of $x$ contains a points ($\ne x$) in $U$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A subset of $U$ is closed if if it contains all its accumulation points.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I see Jesse has posted a nice answer - I will leave this as an alternative reference. &lt;/p&gt;&#xA;" OwnerUserId="2219" LastActivityDate="2010-11-12T06:28:42.787" />
  <row Id="9981" PostTypeId="2" ParentId="9978" CreationDate="2010-11-12T06:31:19.930" Score="2" ViewCount="" Body="&lt;p&gt;So I'm going to try to explain what you asked for, though you should probably get your book ASAP or reference a &lt;a href=&quot;http://kr.cs.ait.ac.th/~radok/math/mat6/calc12.htm#1.6%20Further%20Discussion%20of%20the%20Concept%20of%20Limit&quot; rel=&quot;nofollow&quot;&gt;free analysis book&lt;/a&gt; and Wikipedia, and probably most importantly, get someone to take notes for you in your class. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A point p is an accumulation/limit point of a set E if &lt;strong&gt;every&lt;/strong&gt; neighborhood around that point has an element of the set E not equal to p. Think of a neighborhood as an ball around a point. A neighborhood in $\mathbb{R}$ (the real line) is just an interval. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A set is closed if every limit point of a set E is inside E. An example of where you could have a limit point that is not an element of a set is given by the set $\mathbb{Q}$ (the rationals). You can build a series of rational numbers that converge to say for example $\pi$, and hence $\pi$ is a limit point of $\mathbb{Q}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is to say if you have an any interval around $\pi$, you'll have an element of $\mathbb{Q}$. However, pi is not an element of $\mathbb{Q}$, hence the rationals aren't a closed set (Actually every irrational number is a limit point of $\mathbb{Q}$... you may want to think about why).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A set is open if every point of that set is an interior point. A point is an interior point if there exists at least one open ball around that point such that that open ball is contained entirely in the set.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For your usage, assuming that you'll be studying $\mathbb{R}$, a compact set is one which is closed and bounded. A set is bounded if you can but an open ball at its center that covers the entire set. Think [-1,1], with the ball centered at 0, and radius 1.&#xA;Note that (-1,1) is not closed, since both endpoints are limit points of the set, but not elements of the set (ie. you can have an infinite series that converges to 1, and has elements that are in (-1,1), meaning 1 is a limit point, yet 1 is not an element of the set).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The relationship between compactness and uniform continuity is given by &lt;a href=&quot;http://en.wikipedia.org/wiki/Heine%E2%80%93Cantor_theorem&quot; rel=&quot;nofollow&quot;&gt;Heine-Cantor&lt;/a&gt;.&#xA;Hope that helps. &lt;/p&gt;&#xA;" OwnerUserId="2919" LastEditorUserId="2919" LastEditorDisplayName="" LastEditDate="2010-11-12T21:00:39.260" LastActivityDate="2010-11-12T21:00:39.260" CommentCount="1" />
  <row Id="9982" PostTypeId="2" ParentId="9958" CreationDate="2010-11-12T06:51:09.063" Score="8" ViewCount="" Body="&lt;p&gt;If $C_0 + C_1/2 + \ldots + C_n/(n+1) = 0$, where each $C_i \in \mathbb{R}$, then prove that the equation $C_0 + C_1x + \ldots + C_nx^n = 0$ has at least one real root between 0 and 1.  (Rudin, Ch 5 Exercise 4)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $|f(x)| \leq |x|^2$ for all $x \in \mathbb{R}$, then prove that $f$ is differentiable at $x = 0$.  (Spivak?) (OK, so this one is easier than &quot;medium&quot; perhaps, but I like it anyway.  It illustrates nicely that growth conditions can have an impact on smoothness.)&lt;/p&gt;&#xA;" OwnerUserId="640" LastActivityDate="2010-11-12T06:51:09.063" CommunityOwnedDate="2010-11-12T08:33:04.083" CommentCount="1" />
  <row Id="9983" PostTypeId="2" ParentId="9958" CreationDate="2010-11-12T07:02:00.047" Score="0" ViewCount="" Body="&lt;p&gt;Suppose $\displaystyle f:\mathbb{R} \to \mathbb{R}$ and $\displaystyle g:\mathbb{R} \to \mathbb{R}$ are differentiable functions such that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f'(x) = 3f(x) + 7g(x)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$g'(x) = 2f(x) + 3g(x)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Find all such $f$ and $g$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint: Consider $\displaystyle h(x) = 2g(x) - f(x)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given the level they are at, I suppose this is the right level of difficulty you are looking at.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A little more advanced, they can differentiate again and construct a second order differential equation in $\displaystyle f$ (or $\displaystyle g$), which will work with arbitrary coefficients.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More advanced than that, they can consider a matrix differential equation.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-11-12T07:02:00.047" CommunityOwnedDate="2010-11-12T07:04:58.953" />
  <row Id="9984" PostTypeId="2" ParentId="9958" CreationDate="2010-11-12T07:04:11.247" Score="2" ViewCount="" Body="&lt;p&gt;Since they know &quot;basic&quot; integration:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Find the integer part of &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{n=1}^{10000} \dfrac{1}{\sqrt[4]{n}}$$&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-11-12T07:04:11.247" CommunityOwnedDate="2010-11-12T07:05:14.337" CommentCount="1" />
  <row Id="9985" PostTypeId="2" ParentId="9958" CreationDate="2010-11-12T07:15:53.637" Score="0" ViewCount="" Body="&lt;p&gt;Suppose $\displaystyle f:[0,1] \to \mathbb{R}$ is a function such that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) $\displaystyle g(t) = \lim_{x \to t} f(x) $ exists $\displaystyle \forall t \in [0,1]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) $\displaystyle f(x) + g(x) = 2x \ \ \forall x \in [0,1]$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Find all such $\displaystyle f$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The trick is to realize that $\displaystyle g$ must be continuous (perhaps this is near the medium/difficult border).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For extra credit, they can try showing that 1) is enough to show that there is at least one point where $\displaystyle f$ is continuous. (See this: &lt;a href=&quot;http://math.stackexchange.com/questions/3777/&quot;&gt;Is there a function with a removable discontinuity at every point?&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact it can be shown that if 1) is true, the set of discontinuities of $\displaystyle f$ is countable (See the same question above).&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-11-12T07:21:08.537" LastActivityDate="2010-11-12T07:21:08.537" CommunityOwnedDate="2010-11-12T07:15:53.637" />
  <row Id="9986" PostTypeId="2" ParentId="9922" CreationDate="2010-11-12T07:37:50.637" Score="3" ViewCount="" Body="&lt;p&gt;For those of you who wonder why the sequence $x_{n+1} = \sin x_n$ tends to zero, note that after the first application of sine, we get a number in $\pm 1$, and from there on, the numbers keep their sign, and monotonically approach zero (since $|\sin x| \leq |x|$). So the sequence must approach a limit, which is also a fixed point of the sine function. The only fixed point of sine is zero.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-12T07:37:50.637" />
  <row Id="9987" PostTypeId="1" CreationDate="2010-11-12T09:09:17.797" Score="1" ViewCount="123" Body="&lt;p&gt;I apologize for posting such an untechnical question, but with responses it could surely be posed in a better form.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm a math noob, but I've seen (as we all have) a few examples of &quot;connections&quot; between different areas of math by way of an equality or some other relating device.  An example is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;e^iπ + 1 = 0&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These five constants serve different functions in a variety mathematical tasks.  It seems to me that any constant is a &quot;measurement&quot; of some kind, which then begs the question &quot;What is it a measurement of?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you think of some places to start?  If you have a &quot;bridge&quot; you like, post it.&lt;/p&gt;&#xA;" OwnerUserId="1825" LastActivityDate="2010-11-12T16:42:11.993" Title="Making meaning of mathematical &quot;bridges&quot;" Tags="&lt;philosophy&gt;&lt;meta-math&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="9988" PostTypeId="2" ParentId="9891" CreationDate="2010-11-12T09:53:04.770" Score="1" ViewCount="" Body="&lt;p&gt;Actually the books cited above by Sivaram are excellent for numerical stuff. If you want &quot;matrix calculus&quot; then the following books might be helpful:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Matrix Differential Calculus with Applications in Statistics and Econometrics by Magnus and Neudecker&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Functions of Matrices by N. Higham&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Calculus on Manifolds by Spivak&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Some classic, but very useful material can also be found in&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Introduction to Matrix Analysis by Bellman.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;As a simple example, the books will teach (unless you already know it) how to compute, say, the derivative of $f(X) = \log\det(X)$ for an invertible matrix $X$.&lt;/p&gt;&#xA;" OwnerUserId="1709" LastActivityDate="2010-11-12T09:53:04.770" />
  <row Id="9989" PostTypeId="2" ParentId="9958" CreationDate="2010-11-12T10:14:41.033" Score="1" ViewCount="" Body="&lt;p&gt;This is a classic one due to Arnold (A Mathematical Trivium):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$${1 \over 2\pi} \int_{0}^{2\pi} \sin^{100}(x) dx$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They should also try to find a numeric value (to within one-percent error, say).&lt;/p&gt;&#xA;" OwnerUserId="3223" LastActivityDate="2010-11-12T10:14:41.033" CommunityOwnedDate="2010-11-12T10:14:41.033" />
  <row Id="9990" PostTypeId="2" ParentId="9958" CreationDate="2010-11-12T11:13:51.697" Score="1" ViewCount="" Body="&lt;p&gt;Hello,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The best site I have come across when it comes to these kind of problems, is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.artofproblemsolving.com/Forum/index.php&quot; rel=&quot;nofollow&quot;&gt;http://www.artofproblemsolving.com/Forum/index.php&lt;/a&gt;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regards&lt;/p&gt;&#xA;" OwnerUserId="3438" LastActivityDate="2010-11-12T11:13:51.697" CommunityOwnedDate="2010-11-12T11:13:51.697" CommentCount="1" />
  <row Id="9991" PostTypeId="1" CreationDate="2010-11-12T11:54:06.670" Score="4" ViewCount="143" Body="&lt;p&gt;A graph G is planar if and only if xxx.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What can xxx be substituted for? Note that this is from a topological POV so a graph is a 1-dim cw complex and I guess the fundamental group should be used somehow. &lt;/p&gt;&#xA;" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-12T12:02:16.113" LastActivityDate="2010-11-12T14:33:01.243" Title="When is a graph planar?" Tags="&lt;topology&gt;&lt;graph-theory&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="9992" PostTypeId="2" ParentId="9991" CreationDate="2010-11-12T11:59:44.587" Score="3" ViewCount="" Body="&lt;p&gt;Please refer Kuratowski's theorem on Planar graphs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Kuratowski:&lt;/strong&gt; A graph $G$ is &lt;em&gt;planar&lt;/em&gt; iff $G$ does not contain a sub division of $K_{5}$ or $K_{3,3}$.&lt;/p&gt;&#xA;" LastActivityDate="2010-11-12T12:07:04.483" CommentCount="2" />
  <row Id="9993" PostTypeId="1" AcceptedAnswerId="10041" CreationDate="2010-11-12T12:22:46.150" Score="4" ViewCount="127" Body="&lt;p&gt;Solving a heat equation with central symmetry i got the following result:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is to find a sphere center temperature vs. time given that the surface of the sphere is kept constant at $$T_1$$ and at initial moment t=0 the temperature inside the sphere is distributed uniformly with value $$T_0$$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A solution is following:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$T(t)=T_1+2 (T_0-T_1)\sum_{k=1}^\infty (-1)^{k+1} e^{-t(\frac{k\pi a}{r})^2}$$  &#xA;Here r is the radius of the sphere and a=const is a coefficient of thermal conductivity.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the solution must satisfy the initial condition $$T(0)=T_0$$:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$T(0)=T_1+2 (T_0-T_1)\sum_{k=1}^\infty (-1)^{k+1} $$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is satisfied only if the following holds:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1-1+1-1+1-1+ ... =1/2  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, we found this divergent sum  through physical arguments? Is this appropriate for the pure mathematician?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lets go a step further and differentiate the solution by t:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac {d T(t)}{dt}=-2 (T_0-T_1)(\frac{\pi a}{r})^2\sum_{k=1}^\infty (-1)^{k+1} k^2 e^{-t(\frac{k\pi a}{r})^2}$$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, at initial moment t=0 we have:     &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac {d T(0)}{dt}=-2 (T_0-T_1)(\frac{\pi a}{r})^2\sum_{k=1}^\infty (-1)^{k+1} k^2 $$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously  $$\frac {d T(0)}{dt}=0$$  because  the maximum speed in the universe is limited.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus the following must be valid:   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1^2 - 2^2 + 3^2 - 4^2 + ... =0  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here we can say what has been said previously, i think. But at the moment i'm not sure about the following divergent sum:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1 - 2 + 3 - 4 + ... &lt;/p&gt;&#xA;" OwnerUserId="2323" LastActivityDate="2010-11-12T20:27:01.177" Title="Summing divergent series based on physics" Tags="&lt;sequences-and-series&gt;&lt;physics&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="9994" PostTypeId="1" CreationDate="2010-11-12T12:30:13.440" Score="3" ViewCount="168" Body="&lt;p&gt;I am not good with vector spaces so I would be grateful for any help.&lt;br&gt;&#xA;As I've been told I need to take $v \in \mathrm{Im}(T)$, $v\neq 0$, and show that if $T(v) = \mathbf{0}$ then $T^2 = 0$. But if $T(v) \neq \mathbf{0}$, then need to show that if $\{e_1,\ldots,e_{n-1}\}$ is a basis of $\mathrm{Ker}(T)$, then $\{e_1,\dots,e_{n-1},v\}$ is a basis of $V$. Then deduce that $T$ is diagonalisable.  I don't really know how to do that though...&lt;/p&gt;&#xA;" OwnerUserId="2401" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-12T15:32:28.460" LastActivityDate="2010-11-12T15:32:28.460" Title="Let $V$ be a $k$-vector space of dimension $n$ and $T: V \to V$ a linear map of rank 1. Show that either $T^2 = 0$ or that $T$ is diagonalisable?" Tags="&lt;linear-algebra&gt;&lt;vector-space&gt;" AnswerCount="4" />
  <row Id="9995" PostTypeId="2" ParentId="9993" CreationDate="2010-11-12T12:35:23.260" Score="0" ViewCount="" Body="&lt;p&gt;It was good enough for Euler and Ramanujan apparently.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/1_%E2%88%92_2_%2B_3_%E2%88%92_4_%2B_%C2%B7_%C2%B7_%C2%B7&quot; rel=&quot;nofollow&quot;&gt;1-2+3-4+...&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/1+2+3+4+...&quot; rel=&quot;nofollow&quot;&gt;1+2+3+4+...&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1269" LastActivityDate="2010-11-12T12:35:23.260" />
  <row Id="9996" PostTypeId="1" AcceptedAnswerId="9998" CreationDate="2010-11-12T12:46:38.293" Score="1" ViewCount="149" Body="&lt;p&gt;I'm trying to prove that $\frac{\prod \mathbb{Z_p}}{\bigoplus \mathbb{Z_p}}$ is a divisible $\mathbb{Z}$-module (p is prime, and the direct sum and direct product are taken over the set of all primes). It is an exercise from Rotman, An Introduction to Homological Algebra. Here's what I've done so far:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\bigoplus \mathbb{Z_p}$ is the torsion submodule of $\prod \mathbb{Z_p}$, so the quotient is torsion-free. Since $\mathbb{Z}$ is a PID, then the quotient is flat. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can flatness help me prove divisibility? Well, since $\frac{\prod \mathbb{Z_p}}{\bigoplus \mathbb{Z_p}}$ is flat then $Hom_\mathbb{Z} \left( \frac{\prod \mathbb{Z_p}}{\bigoplus \mathbb{Z_p}}, \frac{\mathbb{Q}}{\mathbb{Z}}\right)$, the character module, is injective.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And I don't know how to continue. I don't think this is the way to go, but it's what I've tried. Another thing I've observed is that again, since $\mathbb{Z}$ is a PID, then $\frac{\prod \mathbb{Z_p}}{\bigoplus \mathbb{Z_p}}$ is divisible iff it is injective, but once again, I don't know what to do with this.&lt;/p&gt;&#xA;" OwnerUserId="2614" LastActivityDate="2010-11-12T13:17:13.413" Title="$\frac{\prod \mathbb{Z_p}}{\bigoplus \mathbb{Z_p}}$ is a divisible abelian group" Tags="&lt;modules&gt;&lt;homological-algebra&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="9997" PostTypeId="2" ParentId="9958" CreationDate="2010-11-12T13:10:41.460" Score="1" ViewCount="" Body="&lt;p&gt;A great source of problems I have used in the past is, &quot;Problems in calculus of one variable&quot;, by I A Maron. &lt;/p&gt;&#xA;" OwnerUserId="3431" LastActivityDate="2010-11-12T13:10:41.460" CommunityOwnedDate="2010-11-12T13:10:41.460" />
  <row Id="9998" PostTypeId="2" ParentId="9996" CreationDate="2010-11-12T13:15:47.520" Score="8" ViewCount="" Body="&lt;p&gt;what about the direct approach?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose you have $(a_i)\in \prod \mathbb{Z}_p$ and an integer n, then modulo $\bigoplus \mathbb{Z_p}$ you may assume that $a_p=0$ for all the primes dividing n. n is invertible in all the rest of $\mathbb{Z}_p$ so you can find $\frac{a_p}{n}$ in them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So define $b_p = 0 $ for $p\mid n$ and $b_p = \frac{a_p}{n}$ for $p\not\mid n$ and then&#xA;$n (b_p)+\bigoplus \mathbb{Z_p} = (a_n) + \bigoplus \mathbb{Z_p}$&lt;/p&gt;&#xA;" OwnerUserId="2125" LastActivityDate="2010-11-12T13:17:13.413" CommentCount="1" />
  <row Id="9999" PostTypeId="1" AcceptedAnswerId="10001" CreationDate="2010-11-12T13:17:52.120" Score="13" ViewCount="944" Body="&lt;p&gt;The standard way (other than generating up to $N$) is to check if $(5*N*N + 4)$ or $(5*N*N - 4)$ is a perfect square.  What is the mathematical logic behind this? Also, is there any other way for checking the same?  &lt;/p&gt;&#xA;" OwnerUserId="2109" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-12-25T21:41:17.843" LastActivityDate="2011-06-03T01:49:55.897" Title="Checking if a number is a Fibonacci or not?" Tags="&lt;number-theory&gt;&lt;fibonacci-number&gt;" AnswerCount="7" CommentCount="3" FavoriteCount="4" />
  <row Id="10000" PostTypeId="2" ParentId="9994" CreationDate="2010-11-12T13:33:49.017" Score="0" ViewCount="" Body="&lt;p&gt;EDITED after Prometheus's comment to use the characterization using sum of dimension of eigenspaces instead of counting eigenvalues. This argument should be correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously $\dim V = n \geq 1$. Consider $T$'s restriction to its image. This is a linear map on a 1-dimensional space, so we can write $T\vert_{T(V)}(v) = cv$ for some $c \in k$. There are now two possibilities. Either $c = 0$ or $c \neq 0$.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$c=0$: Then $T^2 = 0$ (why?) and all of $T$'s eigenvalues are $0$ since we must have $0 = T(T(v))  =T(\lambda v) = \lambda T(v) = \lambda^2 v$ for any eigenvalue $\lambda$ and any eigenvector $v$. We conclude that $T$ can't be diagonalizable since all of its eigenvalues are $0$, but the eigenspace corresponding to $0$ is $\ker T$ which has dimension $n-1$.&lt;/li&gt;&#xA;&lt;li&gt;$c \neq 0$: First note that clearly $T^2 \neq 0$. We show that the sum of the dimensions of $T$'s eigenspaces is $n$, so $T$ must be diagonalizable [&lt;a href=&quot;http://en.wikipedia.org/wiki/Diagonalizable_matrix&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;]. And this is very easy: $T(V)$ is a 1-dimensional eigenspace with eigenvalue $c$, and $\ker T$ is an n-1-dimensional eigenspace with eigenvalue $0$. Thus the sum of the dimensions of the eigenspaces is at least $n$, but it can't be greater than $n$, so it must be exactly $n$ (not 100% sure on this... LinAlg is a long time ago...)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1269" LastEditorUserId="1269" LastEditorDisplayName="" LastEditDate="2010-11-12T15:14:05.630" LastActivityDate="2010-11-12T15:14:05.630" CommentCount="2" />
  <row Id="10001" PostTypeId="2" ParentId="9999" CreationDate="2010-11-12T13:42:31.977" Score="12" ViewCount="" Body="&lt;p&gt;Binet's formula tells you that $F_n$ is asymptotic to $\frac{1}{\sqrt{5}} \phi^n$, where $\phi = \frac{1 + \sqrt{5}}{2}$ is the golden ratio.  So to check if a number $x$ is a Fibonacci number you should compute $n = \frac{\log (\sqrt{5} x)}{\log \phi}$.  If this number is very close to an integer then $x$ should be very close to the Fibonacci number $F_{[n]}$ where $[n]$ denotes the closest integer to $n$.  In fact I think $x$ is a Fibonacci number if and only if the error is less than about $\frac{\log \phi}{5x^2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if you don't trust that computation, you can compute $F_{[n]}$ in $O(\log n)$ time, for example by using &lt;a href=&quot;http://en.wikipedia.org/wiki/Exponentiation_by_squaring&quot; rel=&quot;nofollow&quot;&gt;binary exponentiation&lt;/a&gt; to compute powers of the matrix $\left[ \begin{array}{cc} 1 &amp;amp; 1 \\\ 1 &amp;amp; 0 \end{array} \right]$, and then you can check whether this number equals $x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your first question can be answered using the theory of (generalized) &lt;a href=&quot;http://en.wikipedia.org/wiki/Pells_equation&quot; rel=&quot;nofollow&quot;&gt;Pell equations&lt;/a&gt;.  &lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-11-12T14:33:36.313" LastActivityDate="2010-11-12T14:33:36.313" CommentCount="12" />
  <row Id="10002" PostTypeId="2" ParentId="9999" CreationDate="2010-11-12T13:58:53.067" Score="13" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\rm\ n\:$ is a Fibonacci number iff the interval $\rm\ [\phi\ n - 1/n,\ \phi\ n + 1/n]\ $ contains a positive integer (the next Fibonacci number for $\rm\ n &gt; 1\:$). This follows from basic properties of continued fractions. For one proof see e.g. $\ $ T. Komatsu: &lt;a href=&quot;http://www.fq.math.ca/Scanned/41-1/komatsu.pdf&quot; rel=&quot;nofollow&quot;&gt;The interval associated with a fibonacci number&lt;/a&gt;. $\ $ This is a reasonably efficient test since it requires only a few multiplications and additions. $\ $ For example, $\rm\ 2 \to [2.7,\ 3.7],\ \ 3\to [4.5,\ 5.2],\ \ 5 \to [7.9,\ 8.3]\ \ldots$&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2011-03-04T03:06:02.327" LastActivityDate="2011-03-04T03:06:02.327" CommentCount="7" />
  <row Id="10003" PostTypeId="2" ParentId="9994" CreationDate="2010-11-12T14:03:13.503" Score="3" ViewCount="" Body="&lt;p&gt;Suppose that $0\neq v\in Im(T)$, then $T(v)\in Im(T)$. $dim(Im(T))=1$ so $Im(T)=span(v)=F\cdot v$ and you get that there is a $\lambda$ such that $T(v)=\lambda v$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $T(v)=0$ then $T^2(V) \subseteq T(Im(T)) = T(F\cdot v)=0$, so $T^2 = 0$. Assume now that $T(v)\neq 0$, then $Im(T) \cap \ker(T) = {0}$ so a basis for the kernel (which is of dimension n-1) is a set of eigenvectors of eigenvalue 0 and $v$ is the n-th eigenvector of eigenvalue $\lambda\neq 0$&lt;/p&gt;&#xA;" OwnerUserId="2125" LastActivityDate="2010-11-12T14:03:13.503" CommentCount="7" />
  <row Id="10004" PostTypeId="2" ParentId="9991" CreationDate="2010-11-12T14:05:58.497" Score="1" ViewCount="" Body="&lt;p&gt;There are lots of characterizations of planar graphs, e.g. Kuratowski's&#xA;theorem as already mentioned. Another is Whitney's theorem that a finite&#xA;graph $G$ is planar if and only if the dual matroid to the matroid of $G$&#xD;&#xA;is graphic (also comes from a graph).&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-11-12T14:05:58.497" />
  <row Id="10005" PostTypeId="2" ParentId="9994" CreationDate="2010-11-12T14:24:40.770" Score="1" ViewCount="" Body="&lt;p&gt;First, if the rank of T is 1, then the dimension of its kernel is n-1. That means 0 is an eigenvalue of order n-1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, as Prometheus said, you just have to find out whether you will find that last eigenvalue or not. To find it out, just take a vector $v \notin Ker(T)$ (thus, $v \in Im(T)$), and look at $T(v)$. Either you'll again fall on $0$. What does it mean ?&#xA;You have taken a vector from the only dimension not belonging to the kernel of $T$. It means it belongs to $Im(T)$. Any other vector there can be obtained by multiplying $v$ by some coefficient $t \in \mathbb{R}$ (or $\mathbb{C}$, ...). Since T is linear, the same will go for the image of $v$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if $T(v) = 0$, all the vectors from that dimension will fall on $0$ when going through $T$. So for any vector $u \in Im(T)$, $T(u) = 0$. That means that for any vector $v \in V$, $T(T(v)) = 0$, that is, $T^2(v) = 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now if $T(v)$ isn't zero, there will be some vector $u \in V$ (but different from $0$) such that $T(v) = u$. Thus, for any $t \in \mathbb{R}$, $T(t \cdot v) = t \cdot T(v) = t \cdot u$. And $u$ is in $Im(T)$ thus colinear to $v$. Now you have v being an eigenvector of $T$. Taking a basis of $Ker(T)$ and adding $v$ to it provides you a basis of $V$ on which $T$ is diagonalisable.&lt;/p&gt;&#xA;" OwnerUserId="1865" LastActivityDate="2010-11-12T14:24:40.770" CommentCount="4" />
  <row Id="10006" PostTypeId="2" ParentId="9991" CreationDate="2010-11-12T14:33:01.243" Score="1" ViewCount="" Body="&lt;p&gt;A very good treatment of when a graph can be embedded in the plane and more generally into other surfaces is given by the excellent book:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Graphs on Surfaces by Bojan Mohar and Carsten Thomassen (John Hopkins U. Press, 2001)&lt;/p&gt;&#xA;" OwnerUserId="1369" LastActivityDate="2010-11-12T14:33:01.243" />
  <row Id="10007" PostTypeId="1" AcceptedAnswerId="10012" CreationDate="2010-11-12T15:02:12.513" Score="3" ViewCount="115" Body="&lt;p&gt;This question is a follow-up to &lt;a href=&quot;http://math.stackexchange.com/questions/9996/frac-prod-mathbbz-p-bigoplus-mathbbz-p-is-a-divisible-abelian-group&quot;&gt;this one&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I'm trying to prove that the torsion submodule of $\prod \mathbb{Z}_p$ is not a direct summand of $\prod \mathbb{Z}_p$. I found this interesting because it's an example that over a PID, a non-finitely generated module may fail to be decomposed as torsion $\oplus$ something free, showing even that it can't be decomposed as torsion $\oplus$ anything.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, it is easily proven that the torsion submodule of $\prod \mathbb{Z}_p$ is $\bigoplus \mathbb{Z}_p$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if we could decompose $\prod \mathbb{Z}_p$ as torsion $\oplus$ something it would have to be $\prod \mathbb{Z}_p = \bigoplus \mathbb{Z}_p \oplus \frac{\prod \mathbb{Z}_p}{\bigoplus \mathbb{Z}_p}$ (this is correct, right?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, using the question linked above, we know the right summand is divisible ($\iff$ injective), besides it is torsion-free ($\iff$ flat).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I think) I've learned my lesson from my previous question, so I also tried proving that there is a member in $\prod \mathbb{Z}_p$ which can't be written as a sum of something in $\bigoplus \mathbb{Z}_p$ plus something in $\frac{\prod \mathbb{Z}_p}{\bigoplus \mathbb{Z}_p}$, but I couldn't advance any further in this direction either.&lt;/p&gt;&#xA;" OwnerUserId="2614" LastActivityDate="2010-11-12T15:25:59.360" Title="The torsion submodule of $\prod \mathbb{Z}_p$ is not a direct summand of $\prod \mathbb{Z}_p$" Tags="&lt;abstract-algebra&gt;&lt;modules&gt;&lt;homological-algebra&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="10008" PostTypeId="1" CreationDate="2010-11-12T15:02:18.363" Score="0" ViewCount="161" Body="&lt;p&gt;I have a problem that I cannot solve regarding interest payments, and payment holidays.&#xA;The scenario is this&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I purchase a product which is paid for over a 12 month period, I need to calculate the monthly interest payments. However, the complexity here, is that I have the option to take one month payment holiday (is optional, but can be any month over the 12 month period).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So far, I have&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;annual interest (a) = (e.g for 5%, a=5) &#xA;&#xA;monthly interest (m) = ((1 + a/100)^(1/12) - 1) x 100&#xA;&#xA;balance = balance*(1+m) - instalment&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;How do I calculate the instalment, especially when including payment holiday?&lt;/p&gt;&#xA;" OwnerUserId="3442" LastActivityDate="2011-04-11T16:40:07.623" Title="Instalment APR Calculation Formula" Tags="&lt;finance&gt;" AnswerCount="1" />
  <row Id="10009" PostTypeId="2" ParentId="9994" CreationDate="2010-11-12T15:16:34.123" Score="1" ViewCount="" Body="&lt;p&gt;Prometheus' answer is essentialy correct. I'd just like to point something out to you so that you can gain a little intuition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know that the rank of $T$ is one-dimensional, which means that there exists ($n-1$)-dimensional subspace $W \subset V$ such that $T(W) = 0$ (imagine this in three dimensions as a plane that is being sent to zero). Now pick any $u \in V$ not lying in $W$. We have $T(u) = v \neq 0$ (because otherwise the rank of $T$ would be zero). So what can we say about $v$? Well, it either lies in $W$ and therefore for any $u&amp;#39; \in V$, with decomposition $u&amp;#39; = u + w&amp;#39;$ for some $w&amp;#39; \in W$, we have $T^2(u&amp;#39;) = T^2(u + w&amp;#39;) = T^2(u) + T^2(w&amp;#39;) = T(v) + 0 = 0$ (because both $v$ and $w&amp;#39;$ are in $W$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So assume that $v \notin W$. But then we know $T(v) = \lambda v$ because the span of $v$ is one-dimensional. Therefore we can write $T = \lambda E_v \oplus 0_W$ which is the desired diagonalization.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Try thinking about it in two dimensions and matrices and convince yourself that this problem is all about this:&#xA;$$\pmatrix{0&amp;amp;1\cr0&amp;amp;0} \quad \pmatrix{\lambda &amp;amp; 0\cr 0 &amp;amp;0}$$&lt;/p&gt;&#xA;" OwnerUserId="3223" LastActivityDate="2010-11-12T15:16:34.123" CommentCount="1" />
  <row Id="10010" PostTypeId="1" AcceptedAnswerId="10015" CreationDate="2010-11-12T15:22:06.720" Score="6" ViewCount="109" Body="&lt;p&gt;By the mean value theorem it's easy to show that $|a_{n+1}-a_{n}| \leq \frac{5}{6}|a_{n}-a_{n-1}|$ for every n. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next, I thought of saying $|a_{n+1}-a_{n}| \leq ... \leq (\frac{5}{6})^{n}|a_{1}| \to 0$ and somehow show that ** if $M_{n}$ is the closed interval whose end points are $a_{n}$ and $a_{n-1}$ then $a_{n+1} \in M_{n}$ which implies $M_{n+1} \subseteq M_{n}$ and then to finish with &lt;a href=&quot;http://en.wikipedia.org/wiki/Cantor%27s_intersection_theorem&quot; rel=&quot;nofollow&quot;&gt;Cantor's intersection theorem&lt;/a&gt; that gives us convergence of $a_{n}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I'm not even sure if ** is correct and I haven't even used the fact that $a_{0} = 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Following the tip and some more thought I've come up with the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For every $m\gt n$: $|a_{m}-a_{n}|\leq|a_{m}-a_{m-1}+a_{m-1}-...+a_{n-1}-a_{n}|\leq$ &lt;br&gt;&#xA;$\leq\sum_{k=n}^{m-1}|a_{k+1}-a_{k}|\leq|a_{1}|\sum_{k=n}^{m-1}(\frac{5}{6})^{k}\le$ &lt;br&gt;&#xA;$\le|a_{1}|\sum_{k=n}^{\infty}(\frac{5}{6})^{k}=|a_{1}|\frac{(\frac{5}{6})^{n}}{\frac{1}{6}}=6|a_{1}|(\frac{5}{6})^{n} \to 0$ and from here it's easy to show that the sequence is Cauchy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please correct me if I made an error.&lt;/p&gt;&#xA;" OwnerUserId="3237" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-11-12T18:25:51.330" LastActivityDate="2010-11-12T18:25:51.330" Title="Convergence of $a_{0} = 0, a_{n}=f(a_{n-1})$ when $|f'(x)|\leq \frac{5}{6}$" Tags="&lt;calculus&gt;&lt;sequences-and-series&gt;&lt;homework&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="10011" PostTypeId="2" ParentId="10007" CreationDate="2010-11-12T15:24:14.150" Score="2" ViewCount="" Body="&lt;p&gt;Ask yourself this question: if $x \in \prod_p \mathbb{Z}_p$ is such that $x \in n \prod_p \mathbb{Z}_p$ for each $n \in \mathbb{Z} \setminus \{0\}$, then what can we say about $x$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You should be able to conclude that there is no embedding from $\frac{\prod_p\mathbb{Z}_p}{\otimes_p \mathbb{Z}_p}$ to $\prod_p \mathbb{Z}_p$&lt;/p&gt;&#xA;" OwnerUserId="2660" LastActivityDate="2010-11-12T15:24:14.150" />
  <row Id="10012" PostTypeId="2" ParentId="10007" CreationDate="2010-11-12T15:25:59.360" Score="4" ViewCount="" Body="&lt;p&gt;Let $G=\prod_p Z_p$ and $H=\bigoplus_p Z_p$ and regard $H$&#xD;&#xA;as a subgroup of $G$. You want to know if $H$ is a direct factor&#xA;of $G$. If it was there would be a subgroup $K$ of $G$&#xD;&#xA;isomorphic to $G/H$. There isn't. You know that $G/H$&#xD;&#xA;is divisible, but there are no nonzero elements $x\in G$ having the property that&#xA;$x=p y$ for some $y\in G$ for all primes $p$.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-11-12T15:25:59.360" CommentCount="0" />
  <row Id="10013" PostTypeId="2" ParentId="10008" CreationDate="2010-11-12T15:31:53.483" Score="1" ViewCount="" Body="&lt;p&gt;There will not be a closed form, as it will vary with the month of the holiday.  If you want to come out exactly even you have to choose in advance the month of the holiday.  One approach is to assume the holiday is the first month (which will lead to the highest payment-do you see why?).  Then you can apply one month's interest to the original balance and start an 11 month repayment with no holiday allowed.  You should be able to use your usual formulas for this.  Then as long as no holiday is taken, apply the interest to the current balance and take off the payment as with repaying a loan in variable installments.  If no holiday is taken the eleventh month payment will be slightly smaller and no payment will be due in month twelve. Alternately you could bill the first month assuming the holiday will be the second month, then if no holiday is taken in the first bill the second as if the holiday will be the third, etc.  This will be quite messy.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-11-12T15:31:53.483" />
  <row Id="10014" PostTypeId="1" AcceptedAnswerId="10016" CreationDate="2010-11-12T15:40:08.477" Score="1" ViewCount="95" Body="&lt;p&gt;Ok, so this might not be a permutation or combination but I'm curious as to what the answer is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Question: There are 3 groups and 5 Objects in each group. How many combinations are there if your choosing one from each group?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yes this is in reference to Call of Duty: Black Ops perks lol.&lt;/p&gt;&#xA;" OwnerUserId="3444" LastActivityDate="2010-11-12T15:48:50.387" Title="Simple Combination/Permutation Question" Tags="&lt;combinatorics&gt;&lt;permutations&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="10015" PostTypeId="2" ParentId="10010" CreationDate="2010-11-12T15:42:42.120" Score="4" ViewCount="" Body="&lt;p&gt;You presumably want $|f&amp;#39;(x)|\le 5/6$. It's not the case that $a_{n+1}$&#xD;&#xA;need lie in the interval between $a_{n-1}$ and $a_n$. What can you say&#xA;about $|a_n-a_m|$? (The value of $a_0$ isn't relevant).&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-11-12T15:42:42.120" CommentCount="1" />
  <row Id="10016" PostTypeId="2" ParentId="10014" CreationDate="2010-11-12T15:43:47.817" Score="3" ViewCount="" Body="&lt;p&gt;You have five choices from the first group, times five from the second, times five from the third, for a total of 125 choices total.  To see why this is the case, imagine writing out the possibilities:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1,1,1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1,1,2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1,1,3&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1,1,4&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1,1,5&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1,2,1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;...&lt;/p&gt;&#xA;" OwnerUserId="3296" LastActivityDate="2010-11-12T15:43:47.817" CommentCount="1" />
  <row Id="10017" PostTypeId="2" ParentId="10014" CreationDate="2010-11-12T15:45:38.527" Score="1" ViewCount="" Body="&lt;p&gt;How many combinations are there if there is only one group?  For each of those, how many ways can you choose from the second group?  And the third?  Multiplication is your friend.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-11-12T15:45:38.527" />
  <row Id="10019" PostTypeId="2" ParentId="9958" CreationDate="2010-11-12T16:08:38.130" Score="0" ViewCount="" Body="&lt;p&gt;Here are some interesting problems:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;$\displaystyle \int\limits_{0}^{\frac{\pi}{2}} \sin{x}^{\cos{x}} \Bigl[ \cos{x}\cot{x} - \log(\sin{x})^{\sin{x}}\Bigr] \ dx$&#xD;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$\displaystyle \int \tan{x} \cdot \tan{2x} \cdot \tan{3x} \ dx$ : Why i say this is interesting because, many people would attempt to solve it directly but one can use the fact that $\tan(3x)=\tan(x + 2x)$&#xD;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;There are many resources available in the web. The &lt;em&gt;Harvard - MIT Mathematics&lt;/em&gt; tournament problems should be challenging.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" LastActivityDate="2010-11-12T16:08:38.130" CommunityOwnedDate="2010-11-12T16:08:38.130" />
  <row Id="10020" PostTypeId="1" AcceptedAnswerId="10027" CreationDate="2010-11-12T16:15:11.483" Score="0" ViewCount="58" Body="&lt;p&gt;I'm trying to show that the expression&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\omega_i^2 \mid \omega \mid^{-\beta}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(where $\mid \omega \mid = ( \omega_1^2 + \omega_2^2)^{\frac{1}{2}}$ and $i = [1, 2]$) is defined at the origin under certain conditions. Here's my attempt. I'd like to know if its correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because $\mid \omega \mid^2 = \omega_1^2 + \omega_2^2 \Rightarrow \omega_i^2 \le \mid \omega \mid^2$ for any $i$. Therefore, $\omega_i^2 \mid \omega \mid^{-\beta} \le \mid \omega \mid^2 \mid \omega \mid^{-\beta} = \mid \omega \mid^{2 - \beta}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore the expression is defined at the origin if $2 - \beta &amp;gt; 0$ or $\beta &amp;lt; 2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: in addition can it be shown that because $\omega_i \le \mid \omega \mid$ (I assume this is obvious), $\omega_i^3 \le \mid \omega \mid^3$ and by combining both inequalities it is possible to show in general that $\omega_i^n \le \mid \omega \mid^n$ for odd and even values of $n$.&lt;/p&gt;&#xA;" OwnerUserId="1752" LastEditorUserId="1752" LastEditorDisplayName="" LastEditDate="2010-11-12T17:14:20.347" LastActivityDate="2010-11-12T17:53:30.603" Title="For what values of $\beta$ does $\omega_i^2 \mid \omega \mid^{-\beta}$ not blow up at the origin?" Tags="&lt;real-analysis&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="10021" PostTypeId="1" CreationDate="2010-11-12T16:23:53.707" Score="-3" ViewCount="175" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Possible Duplicate:&lt;/strong&gt;&lt;br&gt;&#xA;  &lt;a href=&quot;http://math.stackexchange.com/questions/11/does-99999-1&quot;&gt;Does .99999&amp;hellip; = 1?&lt;/a&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;p&gt;Help solve an argument: is 1 greater than 0.9 (recurring)?&lt;/p&gt;&#xA;" OwnerUserId="3446" LastActivityDate="2010-11-12T16:46:33.907" ClosedDate="2010-11-12T18:05:34.510" Title="Is 1 greater than 0.9 (recurring)?" Tags="&lt;number-theory&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="10022" PostTypeId="1" AcceptedAnswerId="10032" CreationDate="2010-11-12T16:28:49.770" Score="2" ViewCount="127" Body="&lt;p&gt;We are playing a game with my friend and trying to determine who is the better player.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The game has an luck element to it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The current score is 9-7.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The following questions have risen:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) How many games we should play to get statistically significant result that one is better than the other? We can assume p=9/16 or p=0,5 and something standard for z.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) What can we infer from the current game score? We tried &lt;a href=&quot;http://en.wikipedia.org/wiki/Binomial_proportion_confidence_interval#Normal_approximation_interval&quot; rel=&quot;nofollow&quot;&gt;binomial proportion confidence interval&lt;/a&gt; (we used Wilson score, because N&amp;lt;30). &lt;a href=&quot;http://bit.ly/aM259u&quot; rel=&quot;nofollow&quot;&gt;Here are the results&lt;/a&gt;. Not sure how to interpret them.. does it mean that with 95%, the probability, at which the game score is 9-7, is within 34%-76%?&lt;/p&gt;&#xA;" OwnerUserId="1759" LastActivityDate="2010-11-12T21:06:32.370" Title="How many games should we play to determine the better player statistically significantly?" Tags="&lt;probability&gt;&lt;card-games&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="10024" PostTypeId="2" ParentId="9987" CreationDate="2010-11-12T16:42:11.993" Score="1" ViewCount="" Body="&lt;p&gt;This is discussed at length in the book &lt;a href=&quot;http://en.wikipedia.org/wiki/Where_Mathematics_Comes_From&quot; rel=&quot;nofollow&quot;&gt;Where Mathematics Comes From&lt;/a&gt; by Lakoff and Núñez.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-12T16:42:11.993" />
  <row Id="10025" PostTypeId="2" ParentId="10021" CreationDate="2010-11-12T16:46:33.907" Score="-3" ViewCount="" Body="&lt;p&gt;$1 - 0.9 = 0.1 &amp;gt; 0$.&#xA;You're welcome.&lt;/p&gt;&#xA;" OwnerUserId="1865" LastActivityDate="2010-11-12T16:46:33.907" />
  <row Id="10026" PostTypeId="1" AcceptedAnswerId="10028" CreationDate="2010-11-12T17:45:54.217" Score="2" ViewCount="58" Body="&lt;p&gt;I want to prove that:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If $(x_{k})\to L$ and $\forall x_{i}\in (x_{k})$, $x_{i}$ is a subsequential limit of $a_{n}$ then $L$ is also a subsequential limit of $a_{n}$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I came up with the following: Let $\epsilon\gt0$; if $(x_{k})\to L$ then we simply pick $x_{i}\in(L-\epsilon, L+\epsilon)$ and because $x_{i}$ is a subsequential limit of $a_{n}$ we know that in every neighborhood of $L$ there are infinite elements of $a_{n}$ and we conclude that $L$ is also a subsequential limit of $a_{n}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This seems a bit clumsy, is there a better way to show this? Perhaps with Bolzano-Weistrass?&lt;/p&gt;&#xA;" OwnerUserId="3237" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-12T17:53:56.597" LastActivityDate="2010-11-12T19:23:19.670" Title="If $(x_{k})\to L$ and $\forall x_{i}\in (x_{k})$, $x_{i}$ is a subsequential limit of $a_{n}$ then..." Tags="&lt;calculus&gt;&lt;sequences-and-series&gt;&lt;homework&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="10027" PostTypeId="2" ParentId="10020" CreationDate="2010-11-12T17:53:30.603" Score="1" ViewCount="" Body="&lt;p&gt;Yes, with your edit, your proof is fine. I think what you are trying to show is that the function $f : \omega \mapsto \omega_i^2\lvert\omega\rvert^{-\beta}$ has a well-defined and finite limit at the origin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To make your proof fully rigorous, all you need to say is that $f$ is bounded between two continuous functions $g : \omega \mapsto 0$ and $h : \omega \mapsto \lvert\omega\rvert^{2-\beta}$ whose gap tends to zero at the origin, therefore $f$ must also be continuous at the origin. And that's a perfectly good way to show that a function is continuous.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can also show that the condition is necessary, i.e. that if $\beta \ge 2$ then $f$ is not continuous at the origin, simply by taking limits approaching the origin along two different directions, for example one along $\omega_1 = 0$ and the other along $\omega_2 = 0$.&lt;/p&gt;&#xA;" OwnerUserId="856" LastActivityDate="2010-11-12T17:53:30.603" CommentCount="2" />
  <row Id="10028" PostTypeId="2" ParentId="10026" CreationDate="2010-11-12T18:21:28.057" Score="1" ViewCount="" Body="&lt;p&gt;I'm as mystified as Jonas Meyer on why you think this is &quot;clumsy&quot;. It follows exactly along the intuition: I can get arbitrarily close to $L$ using the $x_i$, and I can find subsequence of $(a_i)$ that gets arbitrarily close to the $x_i$, so I can find subsequences that get arbitrarity close to things that get arbitrarily close.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But perhaps what you want is some idea of which subsequence that might be? Well, we can get it done as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is an $N_1$ such that if $k\geq N_1$, then $|x_k-L|\lt 1$. And since $x_k$ is the limit of a subsequence of $(a_n)$, there is an $n_1$ such that $|a_{n_1}-x_{N_1}|\lt 1$. In particular, $|a_{n_1}-L|\lt 2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, there is an $N_2\gt N_1$ such that for all $k\geq N_2$, $|x_k-L|\lt\frac{1}{2}$. Since $x_{N_2}$ is the limit of a subsequence of $(a_n)$, there is an $n_2$, $n_2\gt n_1$, such that $|a_{n_2}-x_{N_2}|\lt \frac{1}{2}$; in particular, $|a_{n_2}-L|\lt 1$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Continue this way; assume that we have found $N_k$, $N_k\gt\cdots\gt N_1$ such that $|x_{N_i}-L|\lt \frac{1}{2^{i-1}}$, and $n_1\lt n_2\lt\cdots\lt n_k$ with $|x_{n_i}-x_{N_i}|\lt \frac{1}{2^{i-1}}$, so $|x_{n_i}-L|\lt \frac{1}{2^{i-2}}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then there is an $N_{k+1}\gt N_k$ such that for all $j\geq N_{k+1}$, $|x_{j}-L|\lt \frac{1}{2^k}$. Since $x_{N_{k+1}}$ is the limit of a subsequence of $(a_n)$, there is an $n_{k+1}\gt n_k$ such that $|a_{n_{k+1}}-x_{N_{k+1}}|\lt \frac{1}{2^k}$, and in particular $|a_{n_{k+1}}-L|\lt \frac{1}{2^{k-1}}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Inductively, we get a subsequence $(a_{n_k})$ of $(a_n)$. I claim this subsequence converges to $L$. Let $\epsilon\gt 0$; find $k$ such that $0\lt \left(\frac{1}{2}\right)^{k-2}\lt \epsilon$. Then for all $\ell\geq k$ we have&#xA;\begin{equation*}&#xA;|a_{n_{\ell}} - L|\lt \frac{1}{2^{\ell-2}} \lt \frac{1}{2^{k-2}}\lt \epsilon.&#xA;\end{equation*}&#xA;Thus, the sequence converges to $L$, as claimed. QED&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Personally, I don't think this is particularly &quot;elegant&quot;, but I don't think it is clumsy either. It is exactly the intuition: get very close to $L$ using the $x_i$, then get very close to $x_i$ using some $a_j$, and this gives you an $a_j$ that is very close to $L$. Just keep doing it and you get a subsequence converging to $L$. &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-12T19:23:19.670" LastActivityDate="2010-11-12T19:23:19.670" />
  <row Id="10029" PostTypeId="1" AcceptedAnswerId="10035" CreationDate="2010-11-12T18:28:11.403" Score="8" ViewCount="597" Body="&lt;p&gt;I along with one of my friends were just discussing some basic things in group theory, when this question came up:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;What are the &lt;em&gt;fundamental results&lt;/em&gt; in group theory?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;We happened to list out some:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Fundamental Theorem of Group Homomorphism:&lt;/strong&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Cayley's Theorem&lt;/strong&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Sylow's Theorem&lt;/strong&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;There may be many more, but as far as my little knowledge is concerned, i think these are very important. Then we started explaining why each one of the above results were more powerful. I could explain as to how the &lt;em&gt;Fundamental Theorem of Group Homomorphism&lt;/em&gt;  can be used to derive some good results, in group theory, and also i could show my friends the power of the &lt;em&gt;Sylow Theorems&lt;/em&gt; just by considering groups of order $pq$. But i could never illustrate him as to how powerful Cayley's theorem is. Can anyone explain the significance of &lt;em&gt;Cayley's theorem&lt;/em&gt; and why it plays a central role in group theory. I am also curious to know whether any important results proved in Group theory using &lt;strong&gt;Cayley's theorem.&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="9413" LastEditorUserId="9413" LastEditorDisplayName="" LastActivityDate="2010-11-12T19:22:51.807" Title="Importance of Cayley's theorem" Tags="&lt;abstract-algebra&gt;&lt;group-theory&gt;&lt;algebra&gt;" AnswerCount="4" CommentCount="5" FavoriteCount="1" />
  <row Id="10030" PostTypeId="2" ParentId="10029" CreationDate="2010-11-12T18:42:13.657" Score="0" ViewCount="" Body="&lt;p&gt;It is central to the theory of group representations. It allows to prove that any group $G$ of cardinal $n$ has a representation (a vector space $E$ such that $G$ is isomorphic to the linear group of $E$) of dimension $n$.&lt;/p&gt;&#xA;" OwnerUserId="441" LastActivityDate="2010-11-12T18:42:13.657" CommentCount="3" />
  <row Id="10031" PostTypeId="2" ParentId="10029" CreationDate="2010-11-12T18:43:08.323" Score="5" ViewCount="" Body="&lt;p&gt;It shows that the axiomatization of transformation groups is correct.  Any transformation group is an axiomatic group, and any axiomatic group -- a structure, or more precisely a &quot;model&quot;, satisfying the group axioms -- is a group of transformations.  Historically, concrete groups of transformation appeared before abstract groups defined by algebraic axioms, so there is a question whether the algebra captures all (or possibly, more than) the intended examples.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, Cayley's theorem with its initially strange but retroactively instinctive idea of considering a structure's internal &quot;action on itself&quot;, promoting symmetries of an object to the status of an object in their own (collective) right, is an early example of the formal style typical of later algebra and it helps prepare the ground psychologically for working in an abstract or axiomatized mode that was new in the 19th century.&lt;/p&gt;&#xA;" OwnerUserId="467" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-11-12T18:49:52.543" LastActivityDate="2010-11-12T18:49:52.543" CommentCount="5" />
  <row Id="10032" PostTypeId="2" ParentId="10022" CreationDate="2010-11-12T18:52:44.270" Score="1" ViewCount="" Body="&lt;p&gt;I'm not sure I understand your notation, but if we assume that one player wins 9/16 of the time, and we want a confidence of 99.5% (p=0.5%), then we can write the probability that the ratio of 9/16 would arise by chance as a function of the number of games, then solve for the number of games that gives a probability of 0.5%.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: I wrote a function to calculate this for 9/16 and p=5%.  The result is that it takes 265 trials (149 wins by one player) to exceed confidence of 95%.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For part (2), you can infer that you can't yet say who is better with 95% confidence.  All you can say with 95% confidence is that the &quot;true&quot; outcome is somewhere between 34% and 76%.  As long as your confidence interval includes 50%, you can't say who's better.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's my sample code in Python (you need the scipy.stats package):&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;s = 9&#xA;n = 16&#xA;&#xA;while binom_test(s, n) &amp;gt; 0.05:&#xA;    n += 1&#xA;    s += 9.0/16&#xA;&#xA;print n&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="3378" LastEditorUserId="3378" LastEditorDisplayName="" LastEditDate="2010-11-12T21:06:32.370" LastActivityDate="2010-11-12T21:06:32.370" CommentCount="1" />
  <row Id="10033" PostTypeId="1" AcceptedAnswerId="10403" CreationDate="2010-11-12T18:59:57.760" Score="2" ViewCount="137" Body="&lt;p&gt;In Stein-Shakarchi's book Fourier Analysis (p. 170), the solution of the Black-Scholes equation&#xA;$$\frac{\partial V}{\partial t}+rs\frac{\partial V}{\partial s}+\frac{\sigma^2s^2}{2}\frac{\partial^2 V}{\partial s^2}-rV=0$$ for &#xA;$0\lt t\lt T$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with boundary condition $V(s,T)=F(s)$, is given as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$V(s,t)=\frac{e^{-r(T-t)}}{\sqrt{2\pi\sigma^2(T-t)}}\int_0^\infty e^{-\frac{(\log(s/s^*)+(r-\sigma^2/2)(T-t))^2}{2\sigma^2(T-t)}} F(s^*)\,ds^*.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My own derivation shows that $ds^*$ in the formula should be replaced by $$\frac{ds^*}{s^*}.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this a typo in the book?&lt;/p&gt;&#xA;" OwnerUserId="3249" LastEditorUserId="1778" LastEditorDisplayName="" LastEditDate="2010-11-15T17:22:18.757" LastActivityDate="2010-11-15T17:22:18.757" Title="Black-Scholes equation" Tags="&lt;analysis&gt;&lt;differential-equations&gt;&lt;finance&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="10034" PostTypeId="2" ParentId="10029" CreationDate="2010-11-12T19:05:08.840" Score="8" ViewCount="" Body="&lt;p&gt;Cayley's Theorem was &lt;em&gt;very&lt;/em&gt; important historically. Groups originally arose from considering groups of permutations (specifically, the action of some functions on the roots of some polynomials). Every group was really a collection of permutations of some set of specific objects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then Cayley introduced the notion of an &quot;abstract&quot; group; the idea that you simply had some &lt;em&gt;things&lt;/em&gt; that you could &quot;compose&quot; in some way giving you an associative operation, with an &quot;identity&quot; and inverses. He points out that all the things that people had been considering up to that point were &quot;groups&quot; in this sense. But he also wanted to make the point that he was not introducing a new class of objects, but that every object that satisfied his definition would also be a &quot;group&quot; in the old sense. That is, all he was doing was to recast the old notion, rather than expanding the class (at least, as applied to &lt;em&gt;finite&lt;/em&gt; groups; for infinite groups there was no consensus on what &quot;permutation of an infinite set&quot; meant [whether it meant what we now think of as &quot;bijection&quot;, or whether it meant what we now call &quot;a bijection with finite support&quot;, that is, that only moves finitely many objects]). What Cayley's Theorem says is &quot;Every &lt;em&gt;thing&lt;/em&gt; that satisfies this definition can be thought of as a &quot;group&quot; in the old sense of a collection of permutations that is closed under composition, inverses, and includes the identity.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The idea of the proof turns out to be useful in other contexts as well, as indicated by T. So I would classify Cayley's theorem as more historically important than important today.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-11-12T19:05:08.840" />
  <row Id="10035" PostTypeId="2" ParentId="10029" CreationDate="2010-11-12T19:17:50.293" Score="6" ViewCount="" Body="&lt;p&gt;Well, it gives (part of) a proof of the first of Sylow's theorems: it is quite easy to prove that if $G$ is a finite group that admits a $p$-Sylow subgroup $S$, and if $H$ is a subgroup of $G$, then $H$ also admits a $p$-Sylow subgroup ($H$ acts on $G/S$, whose cardinal is prime to $p$, so there must be an orbit of cardinal prime to $p$, and $\mathrm{Stab}(gS)=H \cap gSg^{-1}$, so you find that the $\mathrm{Stab}$ of any element of this orbit is a $p$-Sylow of $H$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, using Cayley's theorem, you can embed any finite group $H$ in $G=\mathrm{GL}_n(\mathbb{Z}/p \mathbb{Z})$ (think permutation matrices), where $n= \mathrm{card} H$, and $G$ admits a $p$-Sylow: the subgroup of upper-triangular unipotent matrices (recall $\mathrm{card} G = (p^n-1)(p^n-p) \ldots (p^n-p^{n-1})=p^{n(n-1)/2} m$ with $m$ prime to $p$).&lt;/p&gt;&#xA;" OwnerUserId="2660" LastEditorUserId="2660" LastEditorDisplayName="" LastEditDate="2010-11-12T19:22:51.807" LastActivityDate="2010-11-12T19:22:51.807" CommentCount="1" />
  <row Id="10036" PostTypeId="2" ParentId="9958" CreationDate="2010-11-12T19:26:59.030" Score="1" ViewCount="" Body="&lt;p&gt;Let $m$ be an integer $\geq 2$. Define $u_0=e^{1/m}$, and $u_{n+1}=\frac{u_n}{u_n-I(u_n)}$ where $I$ denotes integral part.&#xA;Prove that $I(u_n)=mn$ for $n \geq 1$.&lt;/p&gt;&#xA;" OwnerUserId="2660" LastActivityDate="2010-11-12T19:26:59.030" CommunityOwnedDate="2010-11-12T19:26:59.030" CommentCount="3" />
  <row Id="10037" PostTypeId="2" ParentId="8532" CreationDate="2010-11-12T19:28:24.213" Score="0" ViewCount="" Body="&lt;p&gt;I think in complement to the above answer, it is useful to consider the comments of Littlewood in &lt;a href=&quot;http://www.springerlink.com/content/558154440wkt1833/fulltext.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.springerlink.com/content/558154440wkt1833/fulltext.pdf&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wonder what the current opinion of this is?&lt;/p&gt;&#xA;" OwnerUserId="941" LastActivityDate="2010-11-12T19:28:24.213" CommunityOwnedDate="2010-11-12T19:28:24.213" />
  <row Id="10038" PostTypeId="1" AcceptedAnswerId="10056" CreationDate="2010-11-12T20:07:31.840" Score="2" ViewCount="148" Body="&lt;p&gt;One proof of the halting problem goes by contradiction like this :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume there is a Turing machine $H$ that can decide the halting problem, then construct a Turing machine $Q$ that takes as input a Turing machine $M$ and an input $I$ to $M$, such that if $H$ decides that $M(I)$ halts, then $Q$ doesn't halt, and vice versa. Then $Q(Q)$ should not halt if $Q(Q)$ halts. A contradiction, then $H$ couldn't have existed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However there is a small technical details. The input to $Q$ required to generate this contradiction is not well-founded. That is, $Q$ takes a machine $Q$ and it's input $I$ (i.e. a pair $(Q,I)$). Evidently, $I$ is a also pair $(Q,I&amp;#39;)$, where $I&amp;#39;=(Q,I&amp;#39;&amp;#39;)$, and so on. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is, let $I=(Q,I&amp;#39;)$ be equivalently written as a set $I={(1,Q),I&amp;#39;}={(1,Q),{(1,Q),I&amp;#39;&amp;#39;}}=...$; showing that $I$ is not well-founded. So does the contradiction in the proof arises because $H$ couldn't exist or because $I$ is not well-founded ?&lt;/p&gt;&#xA;" OwnerUserId="558" LastActivityDate="2010-11-13T07:58:32.290" Title="Does this proof for the undecidability of the halting problem violate the axiom of regularity ?" Tags="&lt;computer-science&gt;&lt;computability&gt;" AnswerCount="2" CommentCount="9" />
  <row Id="10039" PostTypeId="1" CreationDate="2010-11-12T20:16:54.383" Score="1" ViewCount="167" Body="&lt;p&gt;Find all functions $f:\mathbb{R} \to \mathbb{R}$, that are continuous at the point $x=0$ and satisfy:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f (x+y)=f (x)+f (y)+xy (x+y) \ \ \forall x,y \in \mathbb{R} $$ &lt;/p&gt;&#xA;" OwnerUserId="2779" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-11-12T20:42:15.420" LastActivityDate="2010-11-12T20:44:04.750" Title="Solve the functional equation $f (x+y)=f (x)+f (y)+xy (x+y)$, $f$ continuous at $0$" Tags="&lt;functions&gt;&lt;functional-equations&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" />
  <row Id="10040" PostTypeId="2" ParentId="10039" CreationDate="2010-11-12T20:26:52.597" Score="7" ViewCount="" Body="&lt;p&gt;Let $\displaystyle g(x) = f(x) - x^3/3$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then we have $\displaystyle g(x+y) = g(x) + g(y)$ which by continuity at $\displaystyle 0$ implies $\displaystyle g(x) = kx$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This gives $\displaystyle f(x) = x^3/3 + kx$ which satisfies the original equation for any $k$.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;For a proof that $\displaystyle g(x+y) = g(x) + g(y)$ and $\displaystyle g$ continuous at $\displaystyle 0$ implies $\displaystyle g(x) = kx$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First notice that $\displaystyle g(0) = 0$ and that continuity at $\displaystyle 0$ implies continuity everywhere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then by induction, we can prove that for any integer $\displaystyle n$, $\displaystyle g(n) = ng(1)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which can then be extended to the rationals: $\displaystyle g(r) = rg(1) \ \ \forall r \in \mathbb{Q}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given any real $\displaystyle x$, pick a sequence of rationals $\displaystyle \{r_n\}$ converging to $\displaystyle x$ and use the continuity of $\displaystyle g$ at $\displaystyle x$ and that $\displaystyle g(r_n) = r_n g(1)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a classic and you should find plenty of literature on it.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-11-12T20:44:04.750" LastActivityDate="2010-11-12T20:44:04.750" CommentCount="3" />
  <row Id="10041" PostTypeId="2" ParentId="9993" CreationDate="2010-11-12T20:27:01.177" Score="4" ViewCount="" Body="&lt;p&gt;(Background for other readers: the derivation of the solution can be found&#xA;&lt;a href=&quot;http://books.google.com/books?id=eHANhZwVouYC&amp;amp;pg=PA89&quot; rel=&quot;nofollow&quot;&gt;on Google books&lt;/a&gt;,&#xA;see equations (6.18) and (6.19).)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a good and very natural question, but your reasoning is flawed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The solution formula that you quote for $T(t)$ is only valid for $t&amp;gt;0$,&#xA;and it does satisfy $\lim_{t\to 0} T(t) = T_0$;&#xA;it's just that taking the limit $t\to 0$ termwise isn't justified.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(That the limit is $T_0$ can be seen by tracing all the steps in&#xA;the derivation carefully and using knows facts about convergence of&#xA;Fourier series. Unfortunately, explaining this would require drawing&#xA;more figures than I can do here with a reasonable amount of effort...&#xA;Someone who knows more than I about theta functions can probably&#xA;say something about how to compute the limit directly.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your argument &quot;because the maximum speed in the universe is limited&quot; is useless,&#xA;since the heat equation is not the real world; in fact the heat equation&#xA;admits &lt;em&gt;infinite&lt;/em&gt; speed of propagation!&#xA;(A solution with compact support at $t=0$ will not have compact support for any $t&amp;gt;0$.)&#xA;Moreover, taking the limit $t\to 0$ termwise is even less justified in the&#xA;differentiated sum than it was in the original sum.&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2010-11-12T20:27:01.177" />
  <row Id="10042" PostTypeId="1" AcceptedAnswerId="10296" CreationDate="2010-11-12T20:39:07.693" Score="5" ViewCount="126" Body="&lt;p&gt;Take projective real space $\mathbb P_n (\mathbb R)$ of ODD dimension. It is easy to proof that all his Stiefel-Whitney numbers are zero . So according Thom theorem&#xA;there must exists manifold $M$ with boundary such that boundary is &#xA;$\partial M= \mathbb P_n (\mathbb R)$. I should like  to see directly such $M$, without using Thom Theorem . For example if $n=1$  evident choice is $M=$ closed disk.&lt;br&gt;&#xA;I have no idea in general case. Can some one help please ?&lt;/p&gt;&#xA;" OwnerUserId="2872" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2010-11-13T14:29:15.803" LastActivityDate="2010-11-14T20:47:31.820" Title="For what manifold is boundary given odd-dimensional projective space?" Tags="&lt;algebraic-topology&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="10043" PostTypeId="2" ParentId="10038" CreationDate="2010-11-12T20:46:30.423" Score="3" ViewCount="" Body="&lt;p&gt;You got the proof wrong in the details.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let us denote with $\uparrow$ resp $\downarrow$ nontermination resp. termination.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume $H$ is the Turing machine (i.e. its code) that can solve the halting problem, that is $H(M,x) = 1 \Leftrightarrow M(x) \downarrow$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Define $Q(M) = \begin{cases}1, &amp;amp;H(M,M) = 0 \\\\ \uparrow, &amp;amp;H(M,M)=1\end{cases}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, $Q(Q) \downarrow\ \Leftrightarrow\ Q(Q) \uparrow$ which yields the contradiction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you can see, everything is well defined, no problem at all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; Please see Arturo's answer for a more verbose explanation how you should understand arguments in $\mathbb{N}^k, k&amp;gt;1$, for Turing machines.&lt;/p&gt;&#xA;" OwnerUserId="3330" LastEditorUserId="3330" LastEditorDisplayName="" LastEditDate="2010-11-13T07:55:41.480" LastActivityDate="2010-11-13T07:55:41.480" CommentCount="1" />
  <row Id="10044" PostTypeId="2" ParentId="9933" CreationDate="2010-11-12T21:02:12.210" Score="0" ViewCount="" Body="&lt;p&gt;Perhaps some intuition can be gained by plotting each number's position on the number line.  Taking the inverse of any number is visualized by taking the mirror-image of the original plot.  So the inverse of a positive number (a point to the right of zero) is a negative number (a point to the left of zero, at the same distance from zero).  Likewise, the inverse of a negative number is a positive number.  If we agree that multiplying a number by -1 is the same as finding the inverse, then we can see that the product of two negatives must be a positive, because the mirror-image of a mirror-image is the original image.&lt;/p&gt;&#xA;" OwnerUserId="3378" LastActivityDate="2010-11-12T21:02:12.210" />
  <row Id="10045" PostTypeId="1" AcceptedAnswerId="10050" CreationDate="2010-11-12T21:11:19.807" Score="2" ViewCount="140" Body="&lt;p&gt;For what value of $\alpha$ is $W_t^{\alpha}$ a martingale where $\alpha \in \mathbb{Z}$? ($W_t$ standard Brownian motion)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can prove for $\alpha = 0$ and $\alpha = 1$, it is a martingale. (rather it is obvious). But my question is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How to prove that $W_t^{\alpha}$ is not a martingale for $\alpha \in \mathbb{Z}\cap\{0,1\}^{c}$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(This question was asked in an phone interview. They were only interested in the answer which I told correctly.)&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2010-11-12T21:33:37.730" LastActivityDate="2010-11-12T21:43:57.250" Title="For what value of $\alpha$ is $W_t^{\alpha}$ a martingale?" Tags="&lt;probability-theory&gt;&lt;stochastic-processes&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="2" />
  <row Id="10046" PostTypeId="1" AcceptedAnswerId="10064" CreationDate="2010-11-12T21:12:26.630" Score="8" ViewCount="382" Body="&lt;p&gt;In exercise 36 &lt;em&gt;Miscellaneous Taylor Coefficients using Bernoulli&#xA;numbers&lt;/em&gt; on pages 88-89 of Louis Comtet's &lt;em&gt;&lt;a href=&quot;http://books.google.pt/books?id=C0HPgWhEssYC&amp;amp;printsec=frontcover&amp;amp;dq=Louis+Comtet+Advanced+Combinatorics&amp;amp;source=bl&amp;amp;ots=rEhBo3Xuv8&amp;amp;sig=rTARizLlaQKIHQNZcHAlHKRUKUc&amp;amp;hl=pt-PT&amp;amp;ei=jaHdTJnVI5OKhQeWqpiJDQ&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=1&amp;amp;ved=0CBkQ6AEwAA#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;Advanced Combinatorics&lt;/a&gt;&lt;/em&gt;, 1974,&#xA;one is asked to obtain the following explicit formula for the Bernoulli numbers:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$B_{2n}=(-1)^{n-1}\dfrac{1+\left[ \varphi _{n}\right] }{2(2^{2n}-1)}$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\varphi _{n}=\dfrac{2(2^{2n}-1)(2n)!}{2^{2n-1}\pi ^{2n}}\displaystyle\sum_{k=1}^{3n}\dfrac{1}{k^{2n}}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(with $\displaystyle\sum_{n\geq 0}B_{n}\dfrac{t^{n}}{n!}=\dfrac{t}{e^{t}-1}$),&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and to prove, among other sums, that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle\sum_{n=1}^{\infty }\dfrac{1}{\dbinom{2n}{n}}=\dfrac{1}{3}+\dfrac{2\pi\sqrt{3}}{27}\qquad (\ast ).$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alfred van der Poorten wrote &lt;a href=&quot;http://www.springerlink.com/content/3866u5593p65l016/&quot; rel=&quot;nofollow&quot;&gt;here (section 10)&lt;/a&gt;: &lt;em&gt;seeing that&lt;/em&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle\sum_{n=1}^{\infty}\dfrac{x^{2n}}{n^{2}\dbinom{2n}{n}}=2\arcsin^{2}\left( \dfrac{x}{2}\right) \qquad (\ast \ast )$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(...) &lt;em&gt;formula&lt;/em&gt; [ $(\ast )$ ] &lt;em&gt;become&lt;/em&gt;[s] &lt;em&gt;quite accessible to proof&lt;/em&gt;.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not able to show formula $(\ast \ast )$ neither how it can be used to&#xA;prove $(\ast )$.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question&lt;/strong&gt;: Could you provide (a) more detailed hint(s) on how and/or&#xA;different ways in which formula $(\ast )$ can be derived?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Added: For information the other sums are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle\sum_{n=1}^{\infty }\dfrac{1}{n\dbinom{2n}{n}}=\dfrac{\pi \sqrt{3}}{9},\quad$ $\displaystyle\sum_{n=2}^{\infty }\dfrac{1}{n^{2}\dbinom{2n}{n}}=\dfrac{\pi ^{2}}{18},\quad$ $\displaystyle\sum_{n=2}^{\infty }\dfrac{1}{n^{4}\dbinom{2n}{n}}=\dfrac{17\pi ^{4}}{3240}.$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="752" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2010-11-13T14:55:51.003" LastActivityDate="2010-11-14T23:14:47.127" Title="Exercise from Comtet's Advanced Combinatorics: prove $27\sum_{n=1}^{\infty }1/\binom{2n}{n}=9+2\pi \sqrt{3}$" Tags="&lt;combinatorics&gt;&lt;sequences-and-series&gt;&lt;binomial-coefficients&gt;" AnswerCount="5" CommentCount="1" FavoriteCount="3" />
  <row Id="10047" PostTypeId="1" AcceptedAnswerId="10052" CreationDate="2010-11-12T21:16:36.953" Score="4" ViewCount="134" Body="&lt;p&gt;Let $S$ be a sheaf over $X$ and $r$ an element in $S_x$ for some $x$ in $X$. Must there exist a section $s$ in $S(X)$ such that such that $s$ equals $r$ when mapped to $S_x$ by the canonical map?  &lt;/p&gt;&#xA;" LastActivityDate="2010-11-13T22:04:06.697" Title="Can every element in the stalk be represented by a section in the top space?" Tags="&lt;algebraic-geometry&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="1" />
  <row Id="10048" PostTypeId="2" ParentId="10046" CreationDate="2010-11-12T21:21:50.720" Score="6" ViewCount="" Body="&lt;p&gt;$1/\binom{2n}{n} = (2n+1) \int_0^1 x^n(1-x)^n dx$.  Sum these over $n$ to get the integral of a rational function.&lt;/p&gt;&#xA;" OwnerUserId="467" LastActivityDate="2010-11-12T21:21:50.720" CommentCount="4" />
  <row Id="10049" PostTypeId="2" ParentId="9933" CreationDate="2010-11-12T21:22:10.070" Score="0" ViewCount="" Body="&lt;p&gt;There's also the somewhat dissatisfactory answer &quot;because the definition of an ordered field says so&quot;. Recall that an &lt;em&gt;ordered field&lt;/em&gt; is a field $K$ such that there exists $P \subset K$ such that&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;If $x,y \in P$ then $x+y \in P$ and $xy \in P$.&lt;/li&gt;&#xA;&lt;li&gt;For all $x \in K$ *exactly one* of the following holds. $x \in P$, $-x \in P$ or $x = 0$.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;And the elements of $P$ are called positive. &lt;a href=&quot;http://en.wikipedia.org/wiki/Ordered_field&quot; rel=&quot;nofollow&quot;&gt;Obligatory Wikipedia link&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So how do you show that a negative real number times a negative real number is positive? Well either you give the student that $\mathbb{R}$ is an ordered field (typical for a first course in real analysis) and that essentially boils down to &quot;negative times negative is positive &lt;em&gt;because I said so, da#&amp;amp;it&lt;/em&gt;&quot;, or you go the long and arduous route and construct the reals - typically using Dedekind cuts or equivalence classes of rational Cauchy sequences - and show that they are in fact an ordered field.&lt;/p&gt;&#xA;" OwnerUserId="1269" LastEditorUserId="1269" LastEditorDisplayName="" LastEditDate="2010-11-12T21:27:19.080" LastActivityDate="2010-11-12T21:27:19.080" CommentCount="2" />
  <row Id="10050" PostTypeId="2" ParentId="10045" CreationDate="2010-11-12T21:31:26.570" Score="2" ViewCount="" Body="&lt;p&gt;First, the question doesn't even make sense if $\alpha$ is not an integer, because $W^\alpha$ is undefined as a real valued process when W goes negative. So, in that case, I'll assume that W is started from a positive value and stopped when it hits zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's two methods: Use Ito's formula for $X=W^\alpha$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;dX = \alpha W^{\alpha-1}\,dW + \frac12\alpha(\alpha-1) W^{\alpha-2}\,dt.&#xD;&#xA;$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first term on the right is a local martingale. So the only way that X can be a martingale is for $\alpha(\alpha-1)\int W_t^{\alpha-2}\,dt$ to be a local martingale. It is a standard result that continuous finite variation processes cannot be a local martingale unless they are constant, giving $\alpha(\alpha-1)=0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or, use Jensen's inequality. As $f(x)=x^\alpha$ is strictly convex for $\alpha\not\in[0,1]$ and strictly concave for $\alpha\in(0,1)$ (restricting to the positive reals), the function&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$t\mapsto\mathbb{E}[W_t^\alpha]$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is strictly increasing for $\alpha\not\in[0,1]$ and strictly decreasing for $\alpha\in(0,1)$. So, $W^\alpha$ is not a martingale in these cases. In fact, it is a submartingale for $\alpha \not\in[0,1]$ and a supermartingale for $\alpha\in(0,1)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case, Jensen's inequality is the simplest method. However, Ito's formula applies much more generally, to any twice differentiable function of a semimartingale.&lt;/p&gt;&#xA;" OwnerUserId="1321" LastEditorUserId="1321" LastEditorDisplayName="" LastEditDate="2010-11-12T21:43:57.250" LastActivityDate="2010-11-12T21:43:57.250" CommentCount="1" />
  <row Id="10051" PostTypeId="2" ParentId="10045" CreationDate="2010-11-12T21:39:35.487" Score="2" ViewCount="" Body="&lt;p&gt;Assume that $\alpha&amp;gt;0$. &#xA;Apply the optional stopping theorem to the process $W^\alpha_t$ using the hitting time $T$ of $\{-1 , 2\}$. If $W^\alpha_t$ were a martingale, we'd have &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$0=E(W^\alpha_0)=E(W^\alpha_T)=(2/3)(-1)^\alpha+(1/3)2^\alpha.$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It follows that  $2^{\alpha-1}=1$, so $\alpha=1$.   &lt;/p&gt;&#xA;" OwnerUserId="940" LastActivityDate="2010-11-12T21:39:35.487" CommentCount="6" />
  <row Id="10052" PostTypeId="2" ParentId="10047" CreationDate="2010-11-12T21:39:56.247" Score="10" ViewCount="" Body="&lt;p&gt;No: take sheaf of holomorphic functions $S=\mathcal O$ on $X=\mathbb C$.Then stalk at $x=1$ of $1/z$ does not come from $\mathcal O(\mathbb C)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;@Arturo: no you are not misremembering. Here is an example. Take $X=\mathbb P_n (\mathbb C)$ and $S=\mathcal O (-1)$= tautologic bundle.Then every stalk $S_x$ is isomorphic with $\mathcal O_x$ so is not zero but $S(\mathbb P_n (\mathbb C))=0$. (By GAGA principle of Serre you can chose to think in holomorphic way or algebraic way in this example)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;@Robin Chapman: Yes, there are sheaves $S$ with all morphisms $S(X) \to S_x$ surjective but that are not flabby. For example, take $X=\mathbb R$ and $S=C$ = sheaf of continous functions. Then  for $x\in \mathbb R$ and $f_x \in C_x$ take bump function $\phi$ equal $1$ in vicinity of $x$ and support smaller than domain of definition of $f$. Then $\phi f$ (extended by zero) is global function in $C(\mathbb R)$ and his germ at $x$ is equal $f_x$. But of course $C$ is not flabby.&lt;/p&gt;&#xA;" OwnerUserId="2872" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-13T22:04:06.697" LastActivityDate="2010-11-13T22:04:06.697" CommentCount="2" />
  <row Id="10053" PostTypeId="2" ParentId="9933" CreationDate="2010-11-12T21:40:33.130" Score="0" ViewCount="" Body="&lt;p&gt;It might be easiest to explain using whole numbers. Suppose $P$ is some positive number. Then $-P$ is negative. Now $-2P$ is $P$ subtracted from $-P$, so is still negative. Subtract another $P$ and you get $-3P$, which is still negative. Similarly for $-4P, -5P$, and so on. So negative times positive is positive. Same idea for positive times negative. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When it comes to negative times negative, it's a little harder... But how about... $-P$ is negative, so $-(-P)$ is now positive, flipping around $0$. So $-2(-P)$, which is $-(-P)$ added to itself, is still positive. In general adding $-(-P)$ to itself $Q$ times gives $(-Q)(-P)$, which is therefore positive as well.&lt;/p&gt;&#xA;" OwnerUserId="3035" LastActivityDate="2010-11-12T21:40:33.130" />
  <row Id="10056" PostTypeId="2" ParentId="10038" CreationDate="2010-11-12T22:04:48.327" Score="3" ViewCount="" Body="&lt;p&gt;This is essentially Raphael's answer expanded, and taking into account what I was (probably unsuccessfully) trying to communicate in the comments. As I recall, the reason the Universal Machine is important is that it is through the UTM that we can actually explain how the &quot;construction&quot; of the modified machines come about; but I could have messed that up and if so I apologize. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Added:&lt;/em&gt; I guess the point is that you are technically correct that the proof you present, which glosses over some important details, is incorrect. The machine $Q$ you give cannot be asked about the behavior of $Q$ with $Q$ as an input, given the way you describe $Q$. Technically the reason is that if you try to formalize it along the lines I give below, you end up unable to formalize the input you should put into $Q$. The argument is &quot;morally&quot; correct in that a few technical (yet standard) tricks will solve that issue by using a slight modification of $Q$ instead of $Q$ directly. Then again, it is clear that the proof is not trying to be technically accurate, but only &quot;morally&quot; accurate, so one might forgive slight blurrings. It's similar to what happens in the usual description of Cantor's 2nd proof of uncountability of the reals, where the issue of distinct representations of reals is glossed over and can lead to a techncially incorrect argument.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Turing machines each have a code $C$, and take numbers as input. (We can use, for example, Goedel numbering to translate specific sentences or problems into numbers). Let us write $C(m)$ to mean the giving the machine with cod $C$ the input $m$.We can take a pair $(C,m)$, where both $C$ and $m$ are numbers, and obtain a number $N$ that &quot;corresponds&quot; (in a precise and well-defined way) with a pair $(C,m)$. Let us write that as $N\sim (C,m)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose you have a machine $H$ that solves the Halting Problem; we can tweak it so that it will accept any in put and, when given an input $N$, will do the following:&#xA;\begin{equation*}&#xA;H(N) = \left\{\begin{array}{ll}&#xA;1 &amp;amp; \mbox{if $N\sim (C,m)$, and $C$ halts when $m$ is its input;}\\&#xA;0 &amp;amp; \mbox{if $N\sim (C,m)$ and $C$ does not halt when $m$ is its input;}\\&#xA;0 &amp;amp; \mbox{if $N\not\sim(C,m)$ for any $C$ and $m$.}&#xA;\end{array}\right.&#xA;\end{equation*}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $H$ exists, then we can &quot;construct&quot; (that is, there also exists) a Turing Machine $Q$ such that:&#xA;\begin{equation*}&#xA;Q(N) = \left\{\begin{array}{ll}&#xA;1 &amp;amp; \mbox{if $H(M)=0$, where $M\sim(N,N)$;}\\&#xA;\mbox{does not halt} &amp;amp; \mbox{if $H(M)=1$, where $M\sim(N,N)$.}&#xA;\end{array}\right.&#xA;\end{equation*}&#xA;Note that $M$ is well-defined, since it is just the code for the pair $(N,N)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The key is that $Q$ does not take as an input a pair of the form $(C,k)$; it takes a single number $k$, and then figures out what $H(k,k)$ would have been. It's much more restrictive machine than one that &quot;does the opposite&quot; of what $H$ does. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, $Q$ being a Turing machine has a specific code $C$. So the question is, what is $Q(C)$? By definition, $Q(C) = 1$ if and only if $H(M)=0$, where $M\sim(C,C)$; but $H(M)=0$ means either that $M$ is not the code of a valid pair, or else that $C$ does not halt when given the number $C$ as an input. However, $M$ *is* the code of a pair, because that is how we defined it. So that means that $Q(C)=1$ if and only if the machine with code $C$ does not halt when given $C$ as an input. But the machine with code $C$ *is* $Q$, so $Q(C)=1$ if and only if the result of doing $Q(C)$ is non-terminating. Therefore, we cannot have $Q(C)=1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When is $Q(C)\neq 1$? $Q(C)\neq 1$ if and only if $Q(C)$ does not halt; and $Q(C)$ does not halt if and only if $H(M)=1$, where $M\sim(C,C)$. But $H(M)=1$ in this case if and only if the machine with code $C$ halts when it is given the input $C$. But the machine with code $C$ *is* $Q$, so $Q(C)$ fails to halt if and only if $Q(C)$ halts, a contradiction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the key here is two-fold: first, that the inputs to Turing Machines are just numbers, rather than actual pairs or sets. We are identifying what in the meta-language corresponds to a pair with a single number in order to do the input, so we cannot run into problems of foundation (inputs are not pairs of pairs of pairs... etc, they are just &lt;em&gt;numbers&lt;/em&gt;). The issue &lt;em&gt;could&lt;/em&gt; arise if we could not define the number because it requires a recursive process that &quot;goes on indefinitely&quot; and leads to an &quot;infinite number&quot;. And &lt;strong&gt;second&lt;/strong&gt;: you do not construct $Q$ so that it will halt if $H$ says 'doesn't halt' with that same input, or will halt if $H$ says 'it &lt;em&gt;does&lt;/em&gt; halt' with the same input. Instead, $Q$ transforms its input in a precise and well defined way (in this case, taking $N$ to the code for $(N,N)$) and only &lt;em&gt;then&lt;/em&gt; asking whether $H$ will halt with &lt;em&gt;that&lt;/em&gt; new input or not. That is, $Q$ need not be quite as general as you make it out to be in your post, which is why we can actually use $Q$ to ask about itself. &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-13T07:58:32.290" LastActivityDate="2010-11-13T07:58:32.290" CommentCount="4" />
  <row Id="10057" PostTypeId="2" ParentId="10046" CreationDate="2010-11-12T22:05:32.033" Score="4" ViewCount="" Body="&lt;p&gt;You want to get rid of the factor of $n^2$ in the denominator.  So differentiate with respect to $x$, that gets rid of one factor.  If you modify what you get you can differentiate again to get rid of the other.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-11-12T22:05:32.033" CommentCount="1" />
  <row Id="10059" PostTypeId="1" AcceptedAnswerId="10063" CreationDate="2010-11-12T22:13:35.230" Score="1" ViewCount="141" Body="&lt;p&gt;I'm sorry for this kind of specific question, I'd love if you could link to resources (prime lists, etc) that can answer similar questions more generically.&lt;/p&gt;&#xA;" LastActivityDate="2010-11-12T22:32:25.297" Title="What is the largest prime less than 2^31?" Tags="&lt;number-theory&gt;&lt;prime-numbers&gt;&lt;online-resources&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="10061" PostTypeId="2" ParentId="10059" CreationDate="2010-11-12T22:19:10.117" Score="7" ViewCount="" Body="&lt;p&gt;It is $2^{31}-1$. You might want to check Mersenne prime for similar details.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Mersenne_prime&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Mersenne_prime&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3302" LastActivityDate="2010-11-12T22:19:10.117" CommentCount="5" />
  <row Id="10062" PostTypeId="2" ParentId="9524" CreationDate="2010-11-12T22:31:40.350" Score="3" ViewCount="" Body="&lt;p&gt;I want to give a more elaborate example for computer scientists.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given a context free grammar, we can derive (in well-behaved cases) its structure function with a (often) simple calculation. This function's coefficients $a_n$ are the number of derivation trees for words of length $n$ (or, for unambiguous grammars, the number of words). With additional tricks, we can get a number of properties of the generated language.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Simple example: Let a grammar with the set of rules $S \rightarrow aS | bS | \varepsilon$. This directly translates to $S(z) = 2zS(z) + 1$. Solving this equation is simple; we obtain $S(z) = (1-2z)^{-1}$. The coefficients $[z^n]S(z) = 2^n$ are, of course, the number of different words of length $n$ this (unambiguous) grammar can generate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This directly yields a convenient tool to check wether a grammar is unambiguous if we know the number of words (and we can solve the equation system and obtain the series expansion).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are many further extensions and applications.&lt;/p&gt;&#xA;" OwnerUserId="3330" LastActivityDate="2010-11-12T22:31:40.350" />
  <row Id="10063" PostTypeId="2" ParentId="10059" CreationDate="2010-11-12T22:32:25.297" Score="1" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://www.prime-numbers.org/prime-number-2147480000-2147485000.htm&quot; rel=&quot;nofollow&quot;&gt;http://www.prime-numbers.org/prime-number-2147480000-2147485000.htm&lt;/a&gt; tells you that it's 2147483647 (about 2/3rds of the way down, third column).  This website seems like a good resource if you're looking for lots of primes.&lt;/p&gt;&#xA;" OwnerUserId="1887" LastActivityDate="2010-11-12T22:32:25.297" CommentCount="1" />
  <row Id="10064" PostTypeId="2" ParentId="10046" CreationDate="2010-11-12T22:50:35.830" Score="6" ViewCount="" Body="&lt;p&gt;Another one, which has been mentioned multiple times on this site (search for Wallis' product) is the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle \int_{0}^{\pi/2} \sin^{2n-1} x \ dx = \dfrac{2 \cdot 4 \cdot 6 \cdots (2n-2)}{1 \cdot 3 \cdot 5 \cdots (2n-1)} \ \ n \gt 1$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now $$\displaystyle \dfrac{2 \cdot 4 \cdot 6 \cdots (2n-2)}{1 \cdot 3 \cdot 5 \cdots (2n-1)} = \dfrac{2^{2n}}{2n{2n \choose n}} = \dfrac{2^{2n-1}}{n{2n \choose n}}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle \int_{0}^{\pi/2} n\left(\dfrac{\sin x}{2}\right)^{2n-1} \ dx = \dfrac{1}{{2n \choose n}}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the sum can be found easily.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The formula by Alfred Van der Pooten can be proved using this approach too. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A proof of Wallis Product is given here: &lt;a href=&quot;http://crypto.stanford.edu/pbc/notes/pi/wallis.xhtml&quot; rel=&quot;nofollow&quot;&gt;http://crypto.stanford.edu/pbc/notes/pi/wallis.xhtml&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a (closely related) problem on this site using this, see here: &lt;a href=&quot;http://math.stackexchange.com/questions/3502/summing-the-series-1k-frac2k2k1-a2k1&quot;&gt;Summing the series $(-1)^k \frac{(2k)!!}{(2k+1)!!} a^{2k+1}$&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-11-12T23:04:00.223" LastActivityDate="2010-11-12T23:04:00.223" CommentCount="1" />
  <row Id="10065" PostTypeId="1" AcceptedAnswerId="10067" CreationDate="2010-11-12T23:08:57.350" Score="6" ViewCount="388" Body="&lt;p&gt;I started by showing that $1\leq a_{n} \leq n$ (by induction) and then $\frac{1}{n}\leq \frac{a_{n}}{n} \leq 1$ which doesn't really get me anywhere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On a different path I showed that $a_{n} \to \infty$ but can't see how that helps me.&lt;/p&gt;&#xA;" OwnerUserId="3237" LastEditorUserId="3237" LastEditorDisplayName="" LastEditDate="2010-11-12T23:18:15.990" LastActivityDate="2010-11-15T18:19:39.677" Title="Given $a_{1}=1, \ a_{n+1}=a_{n}+\frac{1}{a_{n}}$, find $\lim_{n\to\infty}\frac{a_{n}}{n}$" Tags="&lt;calculus&gt;&lt;sequences-and-series&gt;&lt;homework&gt;" AnswerCount="7" CommentCount="3" FavoriteCount="1" />
  <row Id="10066" PostTypeId="2" ParentId="10065" CreationDate="2010-11-12T23:16:03.930" Score="7" ViewCount="" Body="&lt;p&gt;Now that the homework is solved, here is some more investigation into this interesting sequence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think we can show that $$\displaystyle a_{n}^2 \sim 2n + \dfrac{\log n}{2} - C$$ for some constant $\displaystyle C \gt 0$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By $\displaystyle x_n \sim y_n$ I mean $\displaystyle \lim_{n \to \infty} (x_n - y_n) = 0$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider $b_n = a_{n}^2 - 2n$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then we have that $\displaystyle b_{n+1} = b_n + \dfrac{1}{b_n + 2n}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice that $b_3 \gt 0$ and thus for sufficiently large $\displaystyle n$, $\displaystyle b_n \gt 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is also easy to show that $\displaystyle b_n \lt 2n$. In fact, we can easily show that $b_n \lt \log n$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we have that, for sufficiently large $\displaystyle m,n$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle  b_{m+1} - b_n = \sum_{k=n}^{m} \dfrac{1}{b_k + 2k}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $\displaystyle 0 \lt b_k \lt \log k$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;we have that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle  \sum_{k=n}^{m} \dfrac{1}{2k} \gt b_{m+1} - b_n \gt \sum_{k=n}^{m} \dfrac{1}{2k}(1- \dfrac{b_k}{2k})$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Here we used $\displaystyle \dfrac{1}{1+x} \gt \ \ 1-x, 1 \gt x \gt 0$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now Since $b_k \lt \log k$, we have that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle  \sum_{k=n}^{m} \dfrac{1}{2k} \gt b_{m+1} - b_n \gt \sum_{k=n}^{m} \dfrac{1}{2k} - \sum_{k=n}^{m} \dfrac{\log k}{4k^2}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the fact that $\displaystyle H_m - H_n = \log(\dfrac{m+1}{n}) + O(\dfrac{1}{n} - \dfrac{1}{m})$, where $\displaystyle H_n = \sum_{k=1}^{n} \dfrac{1}{k}$ is the $\displaystyle n^{th}$ harmonic number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We see that, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;if $c_n = b_n - \dfrac{\log n}{2}$, then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle O(\dfrac{1}{n} -\dfrac{1}{m}) \gt c_{m+1} - c_n \gt O(\dfrac{1}{n} -\dfrac{1}{m}) -\sum_{k=n}^{m} \dfrac{\log k}{4k^2}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now $\displaystyle \sum_{k=1}^{\infty} \dfrac{\log k}{k^2}$ is convergent and so by the Cauchy convergence criteria, we have that $\displaystyle  c_n$ is convergent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus the sequence $\displaystyle a_{n}^2 - 2n - \dfrac{\log n}{2}$ converges and hence, for some $\displaystyle C$ we have that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle a_{n}^2 \sim 2n + \dfrac{\log n}{2} - C$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or in other words&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle a_{n} \sim \sqrt{2n + \dfrac{\log n}{2} - C}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A quick (possibly incorrect) computer simulation seems to show a very slow convergence to $\displaystyle C = 1.47812676429749\dots$ &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The previous answer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint: Consider $(a_n)^2$ and try to apply similar reasoning as you did for $a_n$.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-11-15T18:19:39.677" LastActivityDate="2010-11-15T18:19:39.677" CommentCount="18" />
  <row Id="10067" PostTypeId="2" ParentId="10065" CreationDate="2010-11-12T23:16:36.757" Score="12" ViewCount="" Body="&lt;p&gt;Use $a_{n+1} = \frac{1}{a_n} + \frac{1}{a_{n-1}} + \cdots + 1$ and $a_n \longrightarrow \infty$.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-12T23:16:36.757" CommentCount="4" />
  <row Id="10068" PostTypeId="1" AcceptedAnswerId="10137" CreationDate="2010-11-12T23:18:01.070" Score="0" ViewCount="156" Body="&lt;p&gt;Let $f(x)$ continuous function on $R$ wich can be in different signs. Prove, that there is exists an arithmetic progression $a, b, c   (a&amp;lt;b&amp;lt;c)$,  such that $f(a)+f(b)+f(c)=0$.&lt;/p&gt;&#xA;" OwnerUserId="2779" LastEditorUserId="1269" LastEditorDisplayName="" LastEditDate="2010-11-13T09:19:56.870" LastActivityDate="2010-11-13T15:58:16.427" Title="Continuous function of one variable" Tags="&lt;real-analysis&gt;&lt;homework&gt;" AnswerCount="3" CommentCount="3" />
  <row Id="10069" PostTypeId="1" CreationDate="2010-11-12T23:22:06.260" Score="3" ViewCount="130" Body="&lt;p&gt;Say we have the function $$G(z) = \sum_{n \geq 0} g_n z^n.$$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a name for the transform T defined so that $$(T(G))(z) = \sum_{n \geq 0} g_n z^{n^2}?$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there anything known about this kind of transformation?&lt;/p&gt;&#xA;" OwnerUserId="3456" LastActivityDate="2010-11-12T23:22:06.260" Title="What is known about the transformation of a power series in which $z^n$ is replaced with $z^{n^2}$?" Tags="&lt;transformation&gt;&lt;power-series&gt;" CommentCount="10" FavoriteCount="1" />
  <row Id="10070" PostTypeId="2" ParentId="10065" CreationDate="2010-11-12T23:26:53.727" Score="2" ViewCount="" Body="&lt;p&gt;Hint: prove by induction that $a_n \leq 2\sqrt{n}$.&lt;/p&gt;&#xA;" OwnerUserId="223" LastActivityDate="2010-11-12T23:26:53.727" />
  <row Id="10071" PostTypeId="2" ParentId="10065" CreationDate="2010-11-12T23:28:11.217" Score="5" ViewCount="" Body="&lt;p&gt;As $a_{n+1}^2 = a_n^2 + 2 + \frac{1}{a_n^2}$, we know that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ a_{n+1}^2 \leq a_n^2 + 3 $&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $a_n^2 \leq 3n$, then $a_{n+1}^2 \leq 3(n+1) $, and since $a_1^2 = 1 \leq 3$, by the induction hypothesis $a_n^2 \leq 3n$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, $\frac{a_n^2}{n^2} \leq 3/n$ for all $n$, and hence&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\lim_{n \rightarrow \infty} \frac{a_n^2}{n^2} = 0$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;from this it follows though that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\lim_{n \rightarrow \infty} \frac{a_n}{n} = 0$.&lt;/p&gt;&#xA;" OwnerUserId="1089" LastActivityDate="2010-11-12T23:28:11.217" CommentCount="0" />
  <row Id="10072" PostTypeId="1" AcceptedAnswerId="10075" CreationDate="2010-11-12T23:30:25.213" Score="2" ViewCount="112" Body="&lt;p&gt;I have a book that uses the notation $l^\infty(\mathcal{F})$, where $\mathcal{F}$ is a set of measurable functions, $f \in \mathcal{F}$ then $f \colon \mathcal{X} \rightarrow \mathbb{R}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not sure what $l^\infty(\mathcal{F})$ is. I know what $l^\infty$ is by itself, the set of bounded sequences (Wikipedia), but not sure about this one.&lt;/p&gt;&#xA;" OwnerUserId="7076" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-12T23:39:00.287" LastActivityDate="2010-11-13T00:18:58.537" Title="what could this notation mean?" Tags="&lt;real-analysis&gt;&lt;probability&gt;&lt;statistics&gt;&lt;notation&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="10073" PostTypeId="1" AcceptedAnswerId="10076" CreationDate="2010-11-12T23:40:09.507" Score="2" ViewCount="260" Body="&lt;p&gt;In complex analysis, we care about the functions of a complex variable. Are such functions just special case of the complex valued function defined in Rudin's Principles of Mathematical Analysis, since they have many similar properties such as uniformly convergence in series. Moreover, I wonder the relationship between the functions of two real variables and the functions of a complex variable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="2177" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-25T09:16:38.353" LastActivityDate="2010-11-25T09:16:38.353" Title="A question about complex analysis" Tags="&lt;soft-question&gt;&lt;complex-analysis&gt;" AnswerCount="4" CommentCount="4" FavoriteCount="1" />
  <row Id="10074" PostTypeId="2" ParentId="10068" CreationDate="2010-11-12T23:43:33.747" Score="3" ViewCount="" Body="&lt;p&gt;Try defining the following function on $R \times R^+$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ g(x,h) = f(x-h) + f(x) + f(x+h) $&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Argue that if $f$ can take positive and negative values, then so can $g$.  Then, use the intermediate value theorem.  &lt;/p&gt;&#xA;" OwnerUserId="2515" LastEditorUserId="2515" LastEditorDisplayName="" LastEditDate="2010-11-12T23:51:18.037" LastActivityDate="2010-11-12T23:51:18.037" />
  <row Id="10075" PostTypeId="2" ParentId="10072" CreationDate="2010-11-13T00:18:58.537" Score="5" ViewCount="" Body="&lt;p&gt;Did you check page 34 (which I got by checking the List of Symbols on page 506)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://books.google.com/books?id=seH8dMrEgggC&amp;amp;pg=PA34&quot; rel=&quot;nofollow&quot;&gt;http://books.google.com/books?id=seH8dMrEgggC&amp;amp;pg=PA34&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/QlpEf.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-11-13T00:18:58.537" />
  <row Id="10076" PostTypeId="2" ParentId="10073" CreationDate="2010-11-13T00:21:39.810" Score="7" ViewCount="" Body="&lt;p&gt;$\newcommand{\Hol}{\mathcal{H}}$&#xD;&#xA;Complex analysis (usually) refers specifically to questions involving differentiable functions starting with the deceptively simply looking case of functions of one varible on open subsets of the complex plane. Already here there are many differences from the theory of differentiable functions $f: (a,b) \to \mathbb{R}$. Some examples:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If $G \subset \mathbb{C}$ is open and $f: G \to \mathbb{C}$ is once differentiable or has an antiderivative, it is infinitely often differentiable. Such functions are called &quot;holomorphic&quot; or &quot;holomorphic on $G$&quot; if one wants to be precise (sometimes written $f \in \Hol(G)$). This is in stark contrast to the real case where a differentiable function may fail to have even a continuous derivative.&lt;/li&gt;&#xA;&lt;li&gt;If $f \in \Hol(G)$, then for any $z \in G$ there exists an open ball around $z$ contained in $G$ on which $f$ is equal to its own Taylor series at $z$. Even smooth real functions are not this nice. Wikipedia has an article on &lt;a href=&quot;http://en.wikipedia.org/wiki/Non-analytic_smooth_function&quot; rel=&quot;nofollow&quot;&gt;the standard example&lt;/a&gt; - though I also like the &lt;a href=&quot;http://en.wikipedia.org/wiki/Fabius_function&quot; rel=&quot;nofollow&quot;&gt;Fabius function&lt;/a&gt;.&lt;/li&gt;&#xA;&lt;li&gt;Cauchy's integral formula which roughly states that if you know the values of a holomorphic function on the boundary of a disk, you know its values everywhere inside the disk.&lt;/li&gt;&#xA;&lt;li&gt;Liouville's theorem which states that if $f \in \Hol(\mathbb{C})$ is bounded, then $f$ is constant. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Essentially: When you're wondering whether the theory of holomorphic functions is &quot;nice&quot; in some way, the answer is usually yes. Whereas the theory of even smooth functions is a lot more messy. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have not provided links to Wikipedia's articles on the relevant theorems. Instead I recommend &lt;a href=&quot;http://www.math.ku.dk/noter/filer/koman-10.pdf&quot; rel=&quot;nofollow&quot;&gt;these free notes&lt;/a&gt; by Christian Berg from the University of Copenhagen. They are in English and have been in use in the university's first course in complex analysis for close to a decade (to the best of my knowledge) and have had multiple updates and corrections over the years.&lt;/p&gt;&#xA;" OwnerUserId="1269" LastActivityDate="2010-11-13T00:21:39.810" CommentCount="1" />
  <row Id="10077" PostTypeId="2" ParentId="10073" CreationDate="2010-11-13T00:51:55.360" Score="1" ViewCount="" Body="&lt;p&gt;For the second part of your question, one can think of complex functions as continuous functions of two real variables that satisfy some differential equations. The differential equations are the Cauchy-Riemann equations mentioned in the answer Jonas Meyer linked to: &lt;a href=&quot;http://math.stackexchange.com/questions/5108/how-is-mathbbc-different-than-mathbbr2/5111#5111&quot;&gt;How is $\mathbb{C}$ different than $\mathbb{R}^2$?&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This gives straight away a reason why all functions of two real variables are not functions of a single complex variable, since not all functions of two real variables satisfy those differential equations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A somewhat deeper understanding of what is going on can be obtained by looking at the complex coordinates themselves. In two real variables we often denote our coordinates by $(x,y)$. Let's agree to identify the complex plane $\mathbb C$ with $\mathbb R^2$ in the usual way (i.e. via $z = x + iy$). Then we can perform a coordinate change; instead of looking at $(x,y)$, we will consider the coordinates $(z,\bar z)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a very subtle change in point of view, and merits your attention. Please take the time to convince yourself that this amounts to a change of basis of the $\mathbb R$-vector space $\mathbb R^2$: if $e_1 = (1,0)$ and $e_2 = (0,1)$ is the usual basis, then the new basis is given by $e_1 + e_2 = (1,1)$ and $e_1 - e_2 = (1,-1)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What this means is that any function $f$ of the real variables $(x,y)$ can be expressed as a function of the new variables $(z,\bar z)$. Now one can say that just as in the case of real analysis we start by considering the case of one real variable, then we want to consider one complex variable to begin with in the case of complex analysis. This leads us to consider the functions $f$ which only depend on the variable $z$ (or $\bar z$, the choice amounting to a choice of a square root of $-1$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, we consider those functions $f$ of the variables $(z,\bar z)$ which do not depend on $\bar z$. Another way to express this property is to say that we require $f$ to satisfy the differential equation $\partial f/\partial \bar z = 0$, but once you write out what this means in the original coordinates $(x,y)$ this equation turns out to be equivalent to the Cauchy-Riemann equations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, when we investigate what it means for a function to depend on a single complex variable, we stumble upon the Cauchy-Riemann equations, and thus upon the definition of a holomorphic function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wholeheartedly recommend the first chapter of Greene and Krantz's &quot;Function theory of one complex variable&quot; for a more in depth discussion of this point of view. They take some lovely examples with polynomials in $(x,y)$ and then in $z$ which show quite explicitly what is going on.&lt;/p&gt;&#xA;" OwnerUserId="3225" LastActivityDate="2010-11-13T00:51:55.360" CommentCount="1" />
  <row Id="10078" PostTypeId="1" CreationDate="2010-11-13T00:56:09.783" Score="0" ViewCount="97" Body="&lt;p&gt;If I have a distribution mass function p(Y) , and values of $X_0,X_1,...,X_i$, I want to assign values to $X_{i+1},...,X_n$ such that the series of X follows p(Y) distribution where Y is the set of all distinct values of X. Is there any technique to do this?&lt;/p&gt;&#xA;" OwnerUserId="2858" LastEditorUserId="2858" LastEditorDisplayName="" LastEditDate="2010-11-13T03:18:33.493" LastActivityDate="2010-11-18T01:31:52.727" Title="Creating data set to match a particular distribution." Tags="&lt;probability-theory&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="10079" PostTypeId="2" ParentId="9355" CreationDate="2010-11-13T01:29:49.997" Score="6" ViewCount="" Body="&lt;p&gt;This holds for &lt;em&gt;any&lt;/em&gt; irreducible finite state Markov chain. I'm not sure about a pure combinatorial proof, but it can be shown quite quickly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $M(t)$ be the expected time the chain takes to return to its initial state, if started at $t$. As it spends, on average, a time $M(t)-1$ outside the state $t$ after each visit to $t$, the fraction of its time spent at $t$ is $1/M(t)$. So, the stationary distribution is $\pi(t)=1/M(t)$ (this is standard, see &lt;a href=&quot;http://en.wikipedia.org/w/index.php?title=Markov_chain&amp;amp;oldid=394586440#Steady-state_analysis_and_limiting_distributions&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now $X_n$ be a Markov chain with the given transition matrix and, for a fixed state $t$, consider the process $H(X_n,t)$. This is just the expected time remaining before it next hits $t$. On average, at each step, this will decrease by 1 when $X_n\not=t$ and increase by $M(t)-1=1/\pi(t)-1$ when $X_n=t$. So, $Y_n\equiv\sum_tH(X_n,t)\pi(t)$ stays the same on average across each step, regardless of the value of $X_n$. That, is it is a &lt;a href=&quot;http://en.wikipedia.org/wiki/Martingale_%28probability_theory%29&quot; rel=&quot;nofollow&quot;&gt;martingale&lt;/a&gt;. Consider what happens when $Y_n$ reaches its maximum value. As it can't increase further and its expected value remains constant, $Y_n$ must remain constant. So, $\sum_tH(s,t)\pi(t)$ is constant.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The bit above the line answers your question. However, the calculations in the linked paper follow through for all irreducible Markov chains. In particular the formula (3.3) holds. This is the product of the lengthy calculation that you mention. I would like to show this now, which will involve going through some calculations. Let $M_{ij}=\mathbb{P}(X_{n+1}=j\vert X_n=i)$ be the transition matrix, $H_{ij}=H(i,j)$ be the hitting time matrix, $I$ be the identity matrix, $J_{ij}=1$ be the matrix of ones, and $F$ be the diagonal matrix with entries being the mean return times $F_{ii}=1/\pi(i)$. The description above of the average change in $H(X_n,t)$ across each time step can be written in matrix form &#xA;$$&#xD;&#xA;(M-I)H=F-J.\qquad\qquad{\rm(1)}&#xD;&#xA;$$&#xD;&#xA;Note that $F\pi=J\pi={\bf 1}$. So, (1) gives $(M-I)H\pi=0$, which is just the martingale property described above. The only eigenvalues of M with eigenvector 1 are proportional to ${\bf 1}$. So, $H\pi$ is proportional to ${\bf 1}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can solve (1). Let $M^*={\bf 1}\pi^{\rm T}$. Then $MM^*=M^*M=M^*$ and $M^*F=M^*J=J$. Trying $A=(I-M+M^*)^{-1}(J-F)$ in place of H in (1),&#xA;$$&#xD;&#xA;\begin{align}&#xD;&#xA;(M-I)A &amp;amp;=(I-M+M^*)^{-1}(M-I)(J-F)\\\\&#xD;&#xA;&amp;amp;=(I-M+M^*)^{-1}(M-M^*-I)(J-F)\\\\&#xD;&#xA;&amp;amp;=F-J.&#xD;&#xA;\end{align}&#xD;&#xA;$$&#xD;&#xA;Also, the only square matrices satisfying $(M-I)A=0$ are the ones with constant columns, so of the form ${\bf 1}v^{\rm T}$ for a vector v. So, (1) has the general solution&#xA;$$&#xD;&#xA;H = (I-M+M^*)^{-1}(J-F)+{\bf 1}v^{\rm T}&#xD;&#xA;$$&#xD;&#xA;and, using the condition $H_{ii}=0$ gives&#xA;$$&#xD;&#xA;v_i=\left((1-M+M^*)^{-1}(F-J)\right)_{ii}.&#xD;&#xA;$$&#xD;&#xA;Next, $J\pi=F\pi={\bf 1}$ so,&#xA;$$&#xD;&#xA;H\pi = (v^{\rm T}\pi){\bf 1}&#xD;&#xA;$$&#xD;&#xA;has constant entries of value&#xA;$$&#xD;&#xA;\begin{align}&#xD;&#xA;v^{\rm T}\pi &amp;amp;= \sum_i\left((I-M+M^*)^{-1}(F-J)\right)_{ii}\pi_i\\\\&#xD;&#xA;&amp;amp;= {\rm Tr}\left[(I-M+M^*)^{-1}(F-J)F^{-1}\right]\\\\&#xD;&#xA;&amp;amp;= {\rm Tr}\left[(I-M+M^*)^{-1}(I-M^*)\right].&#xD;&#xA;\end{align}&#xD;&#xA;$$&#xD;&#xA;If the eigenvalues of M are $\lambda_1,\ldots,\lambda_n$ with $\lambda_1=1$, then it can be seen that the corresponding eigenvalues of $(I-M+M^*)^{-1}(I-M^*)$ are $0,(1-\lambda_2)^{-1},(1-\lambda_3)^{-1},\ldots,(1-\lambda_n)^{-1}$ giving&#xA;$$&#xD;&#xA;(H\pi)_i=\sum_{j=2}^n\frac{1}{1-\lambda_j}.&#xD;&#xA;$$&#xD;&#xA;This is independent of i, and is the same as equation (3.3) in the linked paper.&lt;/p&gt;&#xA;" OwnerUserId="1321" LastEditorUserId="1321" LastEditorDisplayName="" LastEditDate="2010-11-13T21:59:40.457" LastActivityDate="2010-11-13T21:59:40.457" CommentCount="1" />
  <row Id="10080" PostTypeId="2" ParentId="10065" CreationDate="2010-11-13T01:47:51.817" Score="0" ViewCount="" Body="&lt;p&gt;You might also want to take another sequence $b_n = \frac{a_n}{n}$ and use the recurrence to try to prove that $\displaystyle \lim_{n \rightarrow \infty} b_n = 0$&lt;/p&gt;&#xA;" OwnerUserId="3302" LastActivityDate="2010-11-13T01:47:51.817" />
  <row Id="10082" PostTypeId="1" CreationDate="2010-11-13T02:29:30.847" Score="0" ViewCount="102" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Possible Duplicate:&lt;/strong&gt;&lt;br&gt;&#xA;  &lt;a href=&quot;http://math.stackexchange.com/questions/51/the-cow-in-the-field-problem-intersecting-circular-areas&quot;&gt;The cow in the field problem (intersecting circular areas)&lt;/a&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;p&gt;A farmer ties a cow with a rope to a pole at the edge of a round field of radius &lt;strong&gt;r&lt;/strong&gt;. How long must the rope be compared to &lt;strong&gt;r&lt;/strong&gt; if the farmer wants to ensure that the cow can roam exactly half of the field?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The point is to find a &lt;a href=&quot;http://en.wikipedia.org/wiki/Closed-form_expression&quot; rel=&quot;nofollow&quot;&gt;closed-form expression&lt;/a&gt;, not solve this with numerical analysis etc. Is that possible?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;There's a nice depiction of the problem here:&#xA;_http://www11.speedyshare.com/files/25162122/download/Cow.png&lt;/p&gt;&#xA;" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-13T04:58:10.480" LastActivityDate="2010-11-13T04:58:10.480" ClosedDate="2010-11-13T19:08:43.160" Title="Is it possible to express the area of the intersection of 2 circles as a closed-form expression?" Tags="&lt;geometry&gt;" CommentCount="2" />
  <row Id="10083" PostTypeId="2" ParentId="9825" CreationDate="2010-11-13T03:12:34.047" Score="8" ViewCount="" Body="&lt;p&gt;[This answer moved from &lt;a href=&quot;http://math.stackexchange.com/questions/9532/link-between-a-dense-subset-and-a-continuous-mapping&quot;&gt;this question&lt;/a&gt; on Arturo's advice]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the following goes a long way towards proving a converse:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $(Y, T)$ be any T1 topological space with at least two points and let $a$ and $b$ be distinct points in $Y$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $X = Y\setminus\{b\}$.&#xA;Let $f: X \to Y$ be the inclusion of $X$ in $Y$.&#xA;Let $g: X \to Y$ agree with $f$ on $X\setminus\{a\}$ and $g(a) = b$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, define the topology on $X$ to be the coarsest topology that makes both $f$ and $g$ continuous.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With these assumptions it turns out that $X\setminus\{a\}$ is dense in $X$ if and only if $a$ and $b$ do not have disjoint neighbourhoods in $Y$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To see that this is true, let us construct a base of the topology on $X$.&#xA;To make $f$ continuous we only need to take the subspace topology. Since&#xA;X is open in Y this is $S_1 = \{ G \in T \mid b \notin G \}$. To also make $g$ continuous we need to add the open neighbourhoods of $b$, with $b$ replaced by $a$. This gives $S_2 = \{ (H\setminus\{b\} \cup \{a\} \mid H \in T, b \in H \}$.&#xA;Now $S_1 \cup S_2$ is a subbase of the topology on $X$.&#xA;Since $S_1$ and $S_2$ are already closed under finite intersection, and each covers $X$, we can say that&#xA;$B = \{ G \cap H \mid G \in S_1, H \in S_2 \}$ is a base of the topology.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then (remembering that finite sets are closed in Y) we find that the following are all equivalent:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$X\setminus\{a\}$ is not dense in $X$&#xD;&lt;/li&gt;&#xA;&lt;li&gt;$\{a\}$ is open in $X$&#xD;&lt;/li&gt;&#xA;&lt;li&gt;$\{a\} \in B$&#xD;&lt;/li&gt;&#xA;&lt;li&gt;there are $G \in S_1, H \in S_2$ such that $G \cap H = \{a\}$&#xD;&lt;/li&gt;&#xA;&lt;li&gt;there are $G \in S_1, H \in S_2$ such that $a \in G$ and $G \cap (H\setminus\{a\}\cup\{b\}) = \oslash$&#xD;&lt;/li&gt;&#xA;&lt;li&gt;$a$ and $b$ have disjoint neighbourhoods in $(Y, T)$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="3457" LastEditorUserId="3457" LastEditorDisplayName="" LastEditDate="2010-11-14T09:26:26.920" LastActivityDate="2010-11-14T09:26:26.920" CommentCount="1" />
  <row Id="10085" PostTypeId="1" AcceptedAnswerId="10087" CreationDate="2010-11-13T03:55:34.867" Score="17" ViewCount="755" Body="&lt;p&gt;I am not a math major but more of an amateur mathematician working in computational math. So the question I ask may not make complete sense.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am comfortable with the different sizes of infinities and Cantor's &quot;diagonal argument&quot; to prove that the set of all subsets of an infinite set has cardinality strictly greater than the set itself. So (assuming continuum hypothesis) if we have a set $\Omega$ and $|\Omega| = \aleph_i$, then the cardinality of $2^{\Omega}$ is $|2^{\Omega}| = \aleph_{i+1} &amp;gt; \aleph_i$ and we have $\aleph_{i+1} = 2^{\aleph_i}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am fine with these argument. (I am assuming what I have written down makes sense!).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I don't understand is why should there be a smallest $\infty$? Is there a proof that there exists a smallest $\infty$ and that this is what we address as &quot;countably infinite&quot;? or to rephrase the question why can't I find a set whose power set gives us $\mathbb{N}$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason why I am asking this question is in &quot;some sense&quot; $\aleph_i = \log_2(\aleph_{i+1})$. I do not completely understand why this process should stop when $\aleph_i = \aleph_0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Though coming to think about it I can feel that if I take an infinite set with $\log_2 \aleph_0$ elements I can still put it in one-to-one correspondence with the Natural number set. So Is $\log_2 \aleph_0 = \aleph_0$ (or) am I just confused? If $n \rightarrow \infty$, then $2^n \rightarrow \infty$ faster while $\log (n) \rightarrow \infty$ slower and $\log (\log (n)) \rightarrow \infty$ even slower and $\log(\log(\log (n))) \rightarrow \infty$ even &quot;more&quot; slower and so on).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a clean (and relatively elementary) way to explain this to me?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I am totally fine if you direct me to some paper (or) webpage. I tried googling but could not find an answer to my question)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again, I apologize if this question is too trivial or doesn't make sense.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2011-03-14T04:39:07.547" LastActivityDate="2011-03-14T13:08:28.293" Title="Why is $\omega$ the smallest $\infty$?" Tags="&lt;logic&gt;&lt;set-theory&gt;&lt;infinity&gt;&lt;philosophy&gt;" AnswerCount="4" CommentCount="4" FavoriteCount="2" />
  <row Id="10086" PostTypeId="1" AcceptedAnswerId="22026" CreationDate="2010-11-13T03:57:54.960" Score="3" ViewCount="154" Body="&lt;p&gt;If my understanding is correct, naive set theory needs to be restricted in order to avoid paradoxes including the Russell paradox. Typically, the restriction is expressed in terms of size. For example, the set of all sets must be excluded.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I recall that I came across a paper in Arxiv some time ago which explained that a useful restriction may be expressed in terms of symmetry conditions rather than size. Can anyone explain to me the concept and/or provide a link to the paper in question?&lt;/p&gt;&#xA;" OwnerUserId="2949" LastEditorUserId="622" LastEditorDisplayName="" LastEditDate="2010-11-13T07:18:37.563" LastActivityDate="2011-02-14T17:40:01.423" Title="Limiting set theory using symmetry" Tags="&lt;set-theory&gt;&lt;reference-request&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="10087" PostTypeId="2" ParentId="10085" CreationDate="2010-11-13T04:26:16.483" Score="21" ViewCount="" Body="&lt;p&gt;First, let me clear up a misunderstanding.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Question:  Is $2^\omega = \aleph_1$?  More generally is $2^{\aleph_\alpha} = \aleph_{\alpha+1}$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer of &quot;yes&quot; to the first question is known as the continuum hypothesis (CH), while an answer of &quot;yes&quot; to the second is known as the generalized continuum hypthothesis (GCH).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Answer:  Both are undecidable using ZFC.  That is, Godel has proven that if you assume the answer to CH is &quot;yes&quot;, then you don't add any new contradictions to set theory.  In particular, this means it's impossible to prove the answer is &quot;no&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Later, Cohen showed that if you assume the answer is &quot;no&quot;, then you don't add any new contradictions to set theory.  In particular, this means it's impossible to prove the answer is &quot;yes&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer for GCH is the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All of this is just to say that while you are allowed to assume an answer of &quot;yes&quot; to GCH (which is what you did in your post), there is no way you can prove that you are correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With that out of the way, let me address your actual question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yes, there is a proof that $\omega$ is the smallest infinite cardinality.  It all goes back to some very precise definitions.  In short, when one does set theory, all one really has to work with is the &quot;is a member of&quot; relation $\epsilon$.  One defines an &quot;ordinal number&quot; to be any transitive set $X$ such that $(X,\epsilon)$ is a well ordered set.  (Here, &quot;transitive&quot; means &quot;every element is a subset&quot;.  It's a weird condition which basically means that $\epsilon$ is a transitive relation).  For example, if $X = \emptyset$ or $X=\{\emptyset\}$ or $X = \{\{\emptyset\},\emptyset\}$, then $X$ is an ordinal.  However, if $X=\{\{\emptyset\}\}$, then $X$ is not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are 2 important facts about ordinals.  First, EVERY well ordered set is (order) isomorphic to a unique ordinal.  Second, for any two ordinals $\alpha$ and $\beta$, precisely one of the following holds:  $\alpha\epsilon\beta$ or $\beta\epsilon\alpha$ or $\beta = \alpha$.  In fact, it turns out the the collection of ordinals is well ordered by $\epsilon$, modulo the detail that there is no set of ordinals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, cardinal numbers are simply special kinds of ordinal numbers.  They are ordinal numbers which can't be bijected (in, perhaps NOT an order preserving way) with any smaller ordinal.  It follows from this that the collection of all cardinal numbers is also well ordered.  Hence, as long as there is one cardinal with a given property, there will be a least one.  One example of such a property is &quot;is infinite&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, let me just point out that for FINITE numbers (i.e., Finite natural numbers), one usually cannot find a solution $m$ to $m = log_2(n)$.  Thus, as least from an inductive reasoning point of view, it's not surprising that there are infinite cardinals which can't be written as $2^k$ for some cardinal $k$.&lt;/p&gt;&#xA;" OwnerUserId="331" LastActivityDate="2010-11-13T04:26:16.483" />
  <row Id="10088" PostTypeId="2" ParentId="10085" CreationDate="2010-11-13T04:30:28.747" Score="13" ViewCount="" Body="&lt;p&gt;Suppose $A$ is an infinite set. In particular, it is not empty, so there exists a $x_1\in A$. Now $A$ is infinite, so it is not $\{x_1\}$, so there exists an $x_2\in A$ such that $x_2\neq x1$. Now $A$ is infinite, so it is not $\{x_1,x_2\}$, so there exists an $x_3\in A$ such that $x_3\neq x_1$ and $x_3\neq x_2$. Now $A$ is infinite, so it is not $\{x_1,x_2,x_3\}$, so there exists an $x_4\in A$ such that $x_4\neq x_1$, $x_4\neq x_2$  and $x_4\neq x_3$... And so on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This was you can construct a sequence $(x_n)_{n\geq1}$ of elements of $A$ such that $x_i\neq x_j$ if $i\neq j$. If you define a function $f:\mathbb N\to A$ so that $f(i)=x_i$ for all $i\in\mathbb N$, then $f$ is injective.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By definition, then, the cardinal of $A$ is greater or equal to that of $\mathbb N$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $\mathbb N$ is itself infinite, this shows that the smallest infinite cardinal is $\aleph_0=|\mathbb N|$.&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2010-11-13T04:30:28.747" CommentCount="2" />
  <row Id="10089" PostTypeId="2" ParentId="10085" CreationDate="2010-11-13T04:36:18.180" Score="4" ViewCount="" Body="&lt;p&gt;If you assume both the continuum hypothesis and the (weaker) axiom of countable choice, then no infinite set can have cardinality strictly smaller than that of $\mathbb{N}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, assume the continuum hypothesis, that there is no set of cardinality strictly between $\aleph_{0} = | \mathbb{N} |$ and $2^{\aleph_{0}} = \aleph_{1}$. Suppose $X$ is an infinite set of cardinality strictly smaller than $\aleph_{0} = |\mathbb{N}|$. Then the corresponding power sets must satisfy $|\mathcal{P}(X)| = 2^{|X|} \leq 2^{|\mathbb{N}|} = \aleph_{1}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $|\mathcal{P}(X)| = \aleph_{1}$, then it follows that $2^{|X|} = 2^{|\mathbb{N}|}$, hence $|X| = |\mathbb{N}|$, which is impossible by construction. Thus, $|\mathcal{P}(X)|$ &amp;lt; $\aleph_{1}$, and by CH, we have $|\mathcal{P}(X)|$ &amp;lt; $\aleph_{0}$. Certainly, $|X|$ &amp;lt; $|\mathcal{P}(X)|$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To summarize, the set $X$ has cardinality strictly less than a set (its power set) that has cardinality strictly less than $\aleph_{0}$. Continuing in this fashion, we find an infinite sequence of decreasing transfinite cardinals, and $X$ is an infinite set with arbitrarily small transfinite cardinality. (If we also assume the generalized continuum hypothesis, then there are no infinite sets with cardinality between these sets either.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The axiom of countable choice says that there is a least transfinite cardinal, and no such pathological set $X$ can therefore exist.&lt;/p&gt;&#xA;" OwnerUserId="2720" LastActivityDate="2010-11-13T04:36:18.180" CommentCount="15" />
  <row Id="10090" PostTypeId="2" ParentId="10068" CreationDate="2010-11-13T05:29:14.563" Score="0" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Think about the intermediate value theorem. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Think about some $x$ and some $y$ with $f(x)\gt0$ and $f(y)\lt0$. &lt;/p&gt;&#xA;" OwnerUserId="1154" LastEditorUserId="1154" LastEditorDisplayName="" LastEditDate="2010-11-13T14:30:54.847" LastActivityDate="2010-11-13T14:30:54.847" CommentCount="2" />
  <row Id="10092" PostTypeId="1" AcceptedAnswerId="10093" CreationDate="2010-11-13T07:38:36.263" Score="3" ViewCount="179" Body="&lt;p&gt;Are all dense open sets in R cocountable?&#xA;That is, are all dense open sets in R such that their complements are at most countable?&#xA;It would seem like they must be since the closed sets that are uncountable are all intervals, so their complements are not dense.&lt;/p&gt;&#xA;" LastActivityDate="2010-11-28T20:09:30.240" Title="Dense Open Sets and Cocountability" Tags="&lt;topology&gt;&lt;general-topology&gt;" AnswerCount="3" CommentCount="0" />
  <row Id="10093" PostTypeId="2" ParentId="10092" CreationDate="2010-11-13T07:48:14.050" Score="5" ViewCount="" Body="&lt;p&gt;No, the cardinality of their complements can be as big as that of all of the real numbers.  For an explicit example, the complement of the Cantor set is dense and open.  It consists of all of the &quot;middle third&quot; open intervals removed from $[0,1]$ together with the unbounded intervals $(-\infty,0)$ and $(1,\infty)$.  The Cantor set is a closed uncountable set whose connected components are points.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The complement of a dense open set can be &quot;big&quot;in respects other than cardinality.  Let the rational numbers be listed as $r_1,r_2,r_3,\ldots$, and for each $k$ let $I_k$ be an open interval of length $1/2^k$ containing $r_k$.  Then $U=I_1\cup I_2\cup I_3\cup\cdots$ is a dense open subset of $\mathbb{R}$, but $U$ is a union of intervals the sum of whose lengths is $1/2 + 1/2^2 + 1/2^3+\cdots=1$.  So in a sense the complement of this dense open set must take up &quot;most of the space&quot; in the real numbers.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2010-11-13T08:03:23.457" LastActivityDate="2010-11-13T08:03:23.457" CommentCount="0" />
  <row Id="10094" PostTypeId="2" ParentId="10092" CreationDate="2010-11-13T07:49:27.310" Score="2" ViewCount="" Body="&lt;p&gt;No; the complement of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Cantor_Set&quot; rel=&quot;nofollow&quot;&gt;Cantor set&lt;/a&gt; is a dense open set in $\mathbb{R}$ that is not cocountable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Cantor set is closed and uncountable, so its complement is open. By the by, the Cantor set is a closed set that is uncountable but not an interval; in fact, the Cantor set has empty interior, so it cannot contain any intervals. Every open neighborhood of every point in the Cantor set contains both points of the Cantor set and points not in the Cantor set, so the complement is dense.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-11-13T07:49:27.310" />
  <row Id="10095" PostTypeId="1" AcceptedAnswerId="10101" CreationDate="2010-11-13T08:58:48.923" Score="1" ViewCount="259" Body="&lt;p&gt;Hi, I'm very n00b in maths, and I don't understand what I can do with a generating function.&lt;br&gt;&#xA;For example, if I want to compute the first spread polynomial, how can I do that from the generating function:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/TBMoJ.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or the exponential one:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/bCswn.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance,&#xA;rubik&lt;/p&gt;&#xA;" OwnerUserId="2582" LastEditorUserId="2582" LastEditorDisplayName="" LastEditDate="2010-11-14T08:47:48.170" LastActivityDate="2010-11-14T08:47:48.170" Title="How to find an explicit formula from a generating function" Tags="&lt;polynomials&gt;&lt;generating-functions&gt;" AnswerCount="3" CommentCount="5" FavoriteCount="1" />
  <row Id="10096" PostTypeId="1" CreationDate="2010-11-13T09:03:46.283" Score="5" ViewCount="232" Body="&lt;p&gt;One of the answers to &lt;a href=&quot;http://math.stackexchange.com/questions/2502/what-are-large-cardinals-for&quot;&gt;this question&lt;/a&gt; indicates that large cardinals are useful for destructive testing of set theory. That aside, and not assuming GCH, are there any sets known that have a cardinality of $\aleph_{n}$, $n&amp;gt;0$, or that of any of the large cardinals? There are a few examples of sets that have these cardinalities on &lt;a href=&quot;http://en.wikipedia.org/wiki/Aleph_number&quot; rel=&quot;nofollow&quot;&gt;the wikipedia page&lt;/a&gt;, but they are meager and few compared to the examples on the page for &lt;a href=&quot;http://en.wikipedia.org/wiki/Beth_number&quot; rel=&quot;nofollow&quot;&gt;Beth cardinalities&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="890" LastEditorUserId="622" LastEditorDisplayName="" LastEditDate="2011-05-18T18:14:41.653" LastActivityDate="2011-05-18T18:14:41.653" Title="Examples of sets whose cardinalities are $\aleph_{n}$, or any large cardinal. (not assuming GCH)" Tags="&lt;elementary-set-theory&gt;&lt;cardinals&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="10097" PostTypeId="1" CreationDate="2010-11-13T09:30:09.040" Score="1" ViewCount="266" Body="&lt;p&gt;Let $f:\mathbb{R} \to \mathbb{R}$ is a function with these special properties. $f$ is continuous everywhere. $f$ is not smooth (not infinitely differentiable). $f$ is differentiable only finitely many times everywhere. $f$ belongs to $L^p$. For any $k$ belongs to $\mathbb{N}$ let $E$ be the set of all points where $f$ is differentiable exactly $k$ times then $E$ is not dense in any open subset of $\mathbb{R}$.Is such a function feasible? I need suggestions on ways to construct such a special function.  &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;This is the final version. not going to change it further. I apologize for the inconvinience&lt;/p&gt;&#xA;" OwnerUserId="2987" LastEditorUserId="2987" LastEditorDisplayName="" LastEditDate="2010-11-13T11:03:52.330" LastActivityDate="2010-11-17T18:14:55.360" Title="Construction of a special function" Tags="&lt;real-analysis&gt;&lt;functions&gt;" AnswerCount="2" CommentCount="9" />
  <row Id="10098" PostTypeId="2" ParentId="154" CreationDate="2010-11-13T10:05:31.230" Score="5" ViewCount="" Body="&lt;p&gt;I think some confusion comes from calling the numbers that only have imaginary part &quot;imaginary numbers&quot;. I believe if mathematicians had other name to give to them (but I am really not thinking on a better name), there would be less confusion.&lt;/p&gt;&#xA;" OwnerUserId="3467" LastActivityDate="2010-11-13T10:05:31.230" CommunityOwnedDate="2011-04-15T20:45:29.080" CommentCount="1" />
  <row Id="10100" PostTypeId="2" ParentId="10097" CreationDate="2010-11-13T10:26:30.700" Score="2" ViewCount="" Body="&lt;p&gt;I deleted this temporarily while I was working on an update. Also... whoever upvoted this, please remove your upvote as this answer is incorrect. The older answers are below this update. As for why it took me so long to edit this to reflect that it's wrong, that's because there has been water in the telephone cable or something. This area has been without phones and internet for about 48 hours.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instead of trying to answer this question, I'll try and point out some examples of odd behaviour. The essential problem is that an exhibit of a function with similar properties as those you stated is probably going to be as a limit of differentiable functions. But uniform limits and differentiability don't play that nicely together - q.v.  [&lt;a href=&quot;http://en.wikipedia.org/wiki/Uniform_convergence#to_Differentiability&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;].&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's start simple. Consider $f_1 \in C^1([-1,1])$ and define $f_n$ as follows:&#xA;$$ f_n(x) = \begin{cases}&#xD;&#xA;g_n(x) &amp;amp; \text{ if } -\tfrac{1}{n} \lt x \lt \tfrac{1}{n} \\\\&#xD;&#xA;f_{n-1}(x) &amp;amp; \text{ otherwise,}&#xD;&#xA;\end{cases} $$&#xD;&#xA;where $g_n \in C^n([-\tfrac{1}{n},\tfrac{1}{n}])$ such that $g_n^{(k)}(\tfrac{\pm 1}{n}) = f_{n-1}^{(k)}(\tfrac{\pm 1}{n})$, $k = 0,\ldots n-1$ and $\sup_{x \in [-\tfrac{1}{n},\tfrac{1}{n}]} \lvert g_n (x)- f_{n-1}(x) \rvert &amp;lt; 2^{-n}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By construction this is a Cauchy sequence in $C[-1,1]$, so the limit, $f$ is continuous. Moreover for any $X \subset [-1,1]$ that doesn't contain a neighbourhood of $0$, the sequence $(f_n|_X)$ is eventually constant. So we know everything about $f$'s behaviour with respect to differentiability everywhere except at $0$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clearly it's possible for $f$ to be smooth at $0$ - for example if $(f_n)$ is eventually constant on some neighbourhood of $0$. But $g_n$ can have pretty much any behaviour on $[-\tfrac{1}{2n},\tfrac{1}{2n}]$, so $(g_n^{(k)}(0))$ can be &lt;em&gt;any&lt;/em&gt; sequence of real numbers for each $k$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This demonstrates that limits of real functions can be truly bizarre. The $f_n$ get nicer and nicer around $0$, but the limit can have any behaviour at all at $0$: Not differentiable, exactly $k$ times differentiable or smooth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure this next idea says anything at all about the problem (because the limit might not be differentiable at all), but I find it a bit of a waste to not write it down. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We shall say that a function $f$ is &quot;$F^k$ on $U$&quot; if $f$ is continuous and $k$ times differentiable on $U$ and nowhere $k+1$ times differentiable in $U$. Obviously the requirement that $f$ be continuous isn't needed for $k \neq 0$ ($f$ differentiable $\Rightarrow$ $f$ continuous), but stating it twice isn't a problem. We are mostly going to be concerned about intervals, so for endpoints of intervals we shall in addition require that the one-sided derivatives exist.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let $f_0$ be an $F^0$ function on $[0,1]$, asumme that we have defined the sequence up to $f_{n-1}$. We define $f_{n}$ in the following way: For each $k = 0, \ldots n-1$ and for each maximal open subinterval where $f_{n-1}$ is $F^k$, set $f_n$ to be an $F^{k+1}$ function on the two outer thirds of the interval such that the $k$ derivatives of $f_{n-1}$ and $f_n$ agree on the endpoints and such that $|f_n(x) - f_{n-1}(x)| &amp;lt; 2^{-n}$ for each $x$ in the interval (so the sequence is Cauchy). On the inner one third of the intervals we set $f_n = f_{n-1}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This simple ASCII diagram should help a bit with grokking the idea:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&#xA;     0                             F^0                             1&#xA;f_0: |-------------------------------------------------------------|&#xA;              F^1                  F^0                  F^1&#xA;f_1: |-------------------)(-------------------)(-------------------|&#xA;       F^2    F^1    F^2    F^1    F^0    F^1    F^2    F^1    F^2 &#xA;f_2: |-----)(-----)(-----)(-----)(-----)(-----)(-----)(-----)(-----]&#xA;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;If this makes you think of the Cantor set and the Towers of Hanoi, it's not a coincidence - that's exactly how I got the idea. Anyway as I said, the limit of the $f_n$ might not be differentiable anywhere, but it seems to me that it's at least a starting point for constructing a sequence which has a limit that's $F^k$ on a nowhere dense set in $[0,1]$ for all $k &amp;gt; 0$.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;This answer is based on a misreading of the question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proposition.&lt;/strong&gt; There exists no $f \in C(\mathbb{R})$ such that:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$f \notin C^\infty(\mathbb{R})$,&lt;/li&gt;&#xA;&lt;li&gt;If $k \in \mathbb{N}_{0}$, then $E_k := \lbrace x \in \mathbb{R} \mid f \text{ is exactly } k \text{ times differentiable at } x\rbrace$ is nowhere dense.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof.&lt;/strong&gt; Consider $\cup_{k=0}^\infty E_k$. This has to be all of $\mathbb{R}$ since $f$ is continuous. But then we'd have written $\mathbb{R}$ as a countable union of nowhere dense sets contradicting that $\mathbb{R}$ is a Baire space.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;This is my second answer which is also wrong&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proposition.&lt;/strong&gt; There exists no function, $f$ such that:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$f \in C(\mathbb{R})$,&lt;/li&gt;&#xA;&lt;li&gt;for all $k \in \mathbb{N}$ the set of points where $f$ is exactly $k$ times differentiable is nowhere dense,&lt;/li&gt;&#xA;&lt;li&gt;$f$ is nowhere smooth.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof.&lt;/strong&gt; Recall that if $f$ is differentiable at a point $x_1$, there exists an entire open interval (UPDATE: no there doesn't. The &lt;a href=&quot;http://en.wikipedia.org/wiki/Cantor_function&quot; rel=&quot;nofollow&quot;&gt;Cantor function&lt;/a&gt; is a counterexample), $I_1$ around $x_1$ where $f$ is differentiable. Find a closed interval, $J_1$ containing $x_1$ inside $I_1$. Since the set of points where $f$ is exactly $k$ times differentiable is nowhere dense, there must be a point $x_2 \in J_1$ where $f$ is twice differentiable. Like before we get a closed interval $J_2$ contained in $J_1$ where $f$ is twice differentiable. Proceed by induction and we get a nested sequence of closed intervals, $J_1 \supset J_2 \supset \dotsb$. By the nested interval theorem, $\cap_{\mathbb{N}} J_n$ is nonempty and it's easy to see that $f$ has to be smooth on all the points in the intersection, contradicting our assumption that $f$ was nowhere smooth.&lt;/p&gt;&#xA;" OwnerUserId="1269" LastEditorUserId="1269" LastEditorDisplayName="" LastEditDate="2010-11-17T18:14:55.360" LastActivityDate="2010-11-17T18:14:55.360" CommentCount="12" />
  <row Id="10101" PostTypeId="2" ParentId="10095" CreationDate="2010-11-13T10:32:56.997" Score="4" ViewCount="" Body="&lt;p&gt;If you know what Taylor expansion is, then you should know that $S_n(s)$ in e.g.f. is just its $n$th derivative at zero. For the ordinary g.f., it's almost the same, except that you have to divide the derivative by $n!$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, if you want to do more with g.f. than this, I strongly advise you to at least look at this &lt;a href=&quot;http://en.wikipedia.org/wiki/Generating_function&quot; rel=&quot;nofollow&quot;&gt;wikipedia&lt;/a&gt; page and perhaps even into the &lt;a href=&quot;http://www.math.upenn.edu/~wilf/DownldGF.html&quot; rel=&quot;nofollow&quot;&gt;generatingfunctionology&lt;/a&gt; book, which in my opinion does a pretty good job of introducing g.f. to a beginner.&lt;/p&gt;&#xA;" OwnerUserId="3223" LastEditorUserId="3223" LastEditorDisplayName="" LastEditDate="2010-11-13T12:40:59.810" LastActivityDate="2010-11-13T12:40:59.810" CommentCount="1" />
  <row Id="10102" PostTypeId="1" AcceptedAnswerId="10132" CreationDate="2010-11-13T10:44:00.693" Score="11" ViewCount="311" Body="&lt;p&gt;A friend of mine is a differential geometer, and keeps insisting that he doesn't need the axiom of choice for the things he does. I'm fairly certain that's not true, though I haven't dug into the details of Riemannian geometry (or the real analysis it's based on, or the topology, or the theory of vector spaces, etc...) to try and find a theorem or construction that uses the axiom of choice, or one of its logical equivalences.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So do you know of any result in Riemannian geometry that needs the axiom of choice? They should be there somewhere, I particularily suspect that one or more is hidden in the basic topology results one uses.&lt;/p&gt;&#xA;" OwnerUserId="3225" LastActivityDate="2010-11-13T16:18:41.227" Title="Where do we need the axiom of choice in Riemannian geometry?" Tags="&lt;topology&gt;&lt;logic&gt;&lt;differential-geometry&gt;&lt;axiom-of-choice&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="3" />
  <row Id="10103" PostTypeId="2" ParentId="10097" CreationDate="2010-11-13T10:44:53.003" Score="0" ViewCount="" Body="&lt;p&gt;Here is a construction:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the operator $T:X\to Y$ where $X=L^1\cap L^\infty$ defined by &#xA;$$T(f)(x)=f*g(x)$$ where $g(x)=e^{-x^2}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some properties of $Y$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Young's inequality on convolutions we see that $T(f)\in L^r$ where $1/r=1/p+1/q-1$ provided $f\in L^p$ and $g\in L^q$ -  in our case this is true for all $r$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, $T(f)$ is smooth (we talked about that in a previous thread of yours).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we by some reason wish to destroy the smoothness, we can if wish use $g(x)=e^{-x^2}h(x)$ where $h\in C^n\cap L^\infty$ instead of the previous $g$.&lt;/p&gt;&#xA;" OwnerUserId="1154" LastActivityDate="2010-11-13T10:44:53.003" CommentCount="6" />
  <row Id="10104" PostTypeId="1" AcceptedAnswerId="10117" CreationDate="2010-11-13T11:24:23.543" Score="1" ViewCount="42" Body="&lt;p&gt;Let any Amicable Pair be represented by (a, b), with a &amp;lt; b. In a search for Amicable Pairs, let (n, m) with n &amp;lt; m, be a candidate to be tested for being an amicable pair. Given n, is there way to calculate both the least upper bound and greatest lower bound for the m that would make (n, m) an Amicable Pair?&lt;/p&gt;&#xA;" OwnerUserId="1964" LastActivityDate="2010-11-13T13:03:59.263" Title="Amicable Pair (a, b): given a, what are limits on size of b?" Tags="&lt;number-theory&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="10105" PostTypeId="2" ParentId="10047" CreationDate="2010-11-13T11:29:34.290" Score="1" ViewCount="" Body="&lt;p&gt;This is only true in pathological cases. Neither in complex analysis (see above) nor in algebraic geometry: If we take the structure sheaf $\mathcal{O}_{Spec(A)}$ of an affine scheme, then the question is whether the localzation maps $A \to A_{\mathfrak{p}}$ are surjective, which is almost never true.&lt;/p&gt;&#xA;" OwnerUserId="1650" LastActivityDate="2010-11-13T11:29:34.290" CommentCount="2" />
  <row Id="10106" PostTypeId="2" ParentId="10047" CreationDate="2010-11-13T11:38:08.403" Score="1" ViewCount="" Body="&lt;p&gt;This raises another question. If the restriction maps from $S(X)$ to $S(U)$&#xD;&#xA;are surjective for all open sets $U$ (a sheaf with this&#xA;property is called &lt;em&gt;flasque&lt;/em&gt; or &lt;em&gt;flabby&lt;/em&gt;) then certainly the maps&#xA;from $S(X)$ to each stalk $S_x$ are surjective. So are there sheaves&#xA;$S$ with the property that each map $S(X)\to S_x$ is surjective but which are&#xA;not flabby? I expect there are, but cannot see an example immediately.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-11-13T11:38:08.403" CommentCount="1" />
  <row Id="10107" PostTypeId="2" ParentId="9513" CreationDate="2010-11-13T11:40:11.503" Score="0" ViewCount="" Body="&lt;p&gt;Let $s$ be the special point of $Spec(R)$, i.e. the one which corresponds to the unique maximal ideal of $R$. Then every point $p$ of $Spec(R)$ generalizes to $s$, i.e. $s \in \overline{\{p\}}$. If $f : Spec(R) \to X$ is a continuous map, we thus have $f(s) \in \overline{\{f(p)\}}$, i.e. every open neighborbood of $f(s)$ contains $f(p)$. In other words, if $U$ is an open neighborhood of $f(s)$, the map $f$ actually factors as $Spec(R) \to U \to X$. If $f$ is a morphism of schemes, this is also a factorization in the category of schemes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus a morphism $f : Spec(R) \to X$ mapping $s \mapsto x$ corresponds to a equivalence class of morphisms $Spec(R) \to U$ mapping $s \mapsto x$, where $U$ is an affine open neighboord of $x$. Such a morphism corresponds to a homomorphism $\mathcal{O}_X(U) \to R$, and since we consider equivalence classes, we get a homomorphism $\mathcal{O}_{X,x} \to R$, whose locality corresponds to the condition $s \mapsto x$ above.&lt;/p&gt;&#xA;" OwnerUserId="1650" LastActivityDate="2010-11-13T11:40:11.503" />
  <row Id="10108" PostTypeId="2" ParentId="9749" CreationDate="2010-11-13T11:50:39.437" Score="0" ViewCount="" Body="&lt;p&gt;An application which is used a lot in any calculus course:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The derivative of a differentiable function vanishes at an extremum(ie maximum or minimum) point.&lt;/p&gt;&#xA;" OwnerUserId="1119" LastActivityDate="2010-11-13T11:50:39.437" />
  <row Id="10109" PostTypeId="1" AcceptedAnswerId="10111" CreationDate="2010-11-13T12:07:51.213" Score="1" ViewCount="69" Body="&lt;p&gt;When operating with sigma algebras, what does it mean when we talk about a countably infinite set? And, what are then closed countably infinite interesections?&lt;/p&gt;&#xA;" OwnerUserId="3468" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2010-11-13T14:00:17.587" LastActivityDate="2010-11-13T14:00:17.587" Title="What's a closed countably infinite intersection?" Tags="&lt;measure-theory&gt;&lt;elementary-set-theory&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="10110" PostTypeId="2" ParentId="10073" CreationDate="2010-11-13T12:21:29.893" Score="0" ViewCount="" Body="&lt;p&gt;Holomorphicity is a stronger condition compared to differentiability in real variables. Here are some theorems of complex analysis, which are not true in real analysis.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;If two holomorphic functions agree on an open set, they have to agree everywhere. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Any bounded function holomorphic on all of $\mathbb C$ is necessarily constant.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;A nonconstant holomorphic function can miss at most one value.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;You can see from these theorems that one has much more control over a holomorphic function, as opposed to a differentiable function of two real variables.&lt;/p&gt;&#xA;" OwnerUserId="1119" LastActivityDate="2010-11-13T12:21:29.893" />
  <row Id="10111" PostTypeId="2" ParentId="10109" CreationDate="2010-11-13T12:21:34.263" Score="4" ViewCount="" Body="&lt;p&gt;I expect that you have read that a sigma algebra $\mathcal{A}$ has&#xA;to be &quot;closed under countably infinite intersections&quot;. What this means&#xA;is that if you have a family $(A_n)$ of elements of $\mathcal{A}$&#xD;&#xA;indexed by the natural numbers $\mathbb{N}$, then the intersection&#xA;$\bigcap_{n\in\mathbb{N}}A_n$ of all the $A_n$ is also an element of $\mathcal{A}$.&lt;/p&gt;&#xA;" OwnerUserId="226" LastEditorUserId="226" LastEditorDisplayName="" LastEditDate="2010-11-13T12:27:45.993" LastActivityDate="2010-11-13T12:27:45.993" CommentCount="0" />
  <row Id="10112" PostTypeId="1" AcceptedAnswerId="10135" CreationDate="2010-11-13T12:31:42.080" Score="3" ViewCount="203" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The thing is, since I do a job, and work on mathematics in my spare time, I do not have any connection with the academia. Secondly, the internet is the only way I can interact and reach out to other mathematicians or math enthusiasts like me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I need to discuss my paper with someone, where do I do so over the internet? (This is not about peer review, or about publishing my paper.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thats my question, more clearly:&#xA;&lt;strong&gt;Where do I find people on the internet who can advise me on the papers that I write?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;OR&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Where I can discuss my papers on the internet?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More specifically to give some background: I recently released a paper about a new zero-free region for Riemann Zeta Function &lt;a href=&quot;http://arxiv.org/pdf/0911.5572v14&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/pdf/0911.5572v14&lt;/a&gt;. I wish to get feedback about that. Do you know of any forums/groups, etc. where I can get a good and guiding response? Any links, forums, groups relevant to this topic would be highly appreciated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Roupam&lt;/p&gt;&#xA;" OwnerUserId="2320" LastActivityDate="2010-11-14T05:48:01.437" Title="About finding advisors on the internet" Tags="&lt;soft-question&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="10113" PostTypeId="2" ParentId="9749" CreationDate="2010-11-13T12:34:04.600" Score="0" ViewCount="" Body="&lt;p&gt;MVT is very important. In calculus and analysis, of course. But it's important in other areas too, like applied mathematics an even number theory. For example, for showing Liouville numbers are transcendental.&lt;/p&gt;&#xA;" OwnerUserId="1971" LastActivityDate="2010-11-13T12:34:04.600" />
  <row Id="10114" PostTypeId="2" ParentId="9378" CreationDate="2010-11-13T12:40:53.943" Score="0" ViewCount="" Body="&lt;p&gt;The books &quot;Geometry of Schemes&quot; and &quot;The red book of Varieties and Schemes&quot; are great to get intuition about this. &lt;/p&gt;&#xA;" OwnerUserId="1971" LastEditorUserId="1971" LastEditorDisplayName="" LastEditDate="2011-02-07T06:30:25.880" LastActivityDate="2011-02-07T06:30:25.880" />
  <row Id="10115" PostTypeId="2" ParentId="10065" CreationDate="2010-11-13T12:42:44.087" Score="3" ViewCount="" Body="&lt;p&gt;Lets rewrite:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{a_{n+1}-a_n}{n+1-n}=\frac{1}{a_n}$$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, as n goes to infinity lets denote $$a_n=f$$ .Then approximately:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{d f}{dn}=\frac{1}{f}$$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;After integrating:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{f^2}{2}=n+c=&amp;gt;\frac{f^2}{n^2}=\frac{2}{n}+\frac{c}{n^2}$$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="2323" LastActivityDate="2010-11-13T12:42:44.087" CommentCount="1" />
  <row Id="10116" PostTypeId="2" ParentId="9378" CreationDate="2010-11-13T12:52:00.460" Score="1" ViewCount="" Body="&lt;p&gt;Algebraic varieties are a lot like complex analytic spaces, which are special kinds of manifolds. So if you develop intuition about geometrical objects equipped with complex analytic functions, then it would help you a great deal in intuition on varieties.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another way is to get intuition on curves, which are simpler than the rest. And smooth complex projective curves are like compact Riemann surfaces, which can be visualized.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could again look at pictures such as that of $Spec\ \mathbb Z$ and $Spec\ \mathbb Z[x]$ given in Mumford's Red book.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The functor of points approach is also quite a nice way to think. Roughly, a scheme can be specified by its $R$-valued points for every ring $R$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A concrete example of a stalk is the notion of germs of holomorphic functions, which might be familiar from complex analysis.&lt;/p&gt;&#xA;" OwnerUserId="1119" LastActivityDate="2010-11-13T12:52:00.460" />
  <row Id="10117" PostTypeId="2" ParentId="10104" CreationDate="2010-11-13T13:03:59.263" Score="1" ViewCount="" Body="&lt;p&gt;An Amicable pair of numbers $(m,n)$ is such that each is equal to the sum of the proper divisors of the other, so if you know $n$ then $m=\sigma(n)-n,$ where $\sigma(n)$ is the usual sum of divisors function. Hence you are essentially asking for bounds on $\sigma(n).$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now $\sigma(n)-n=1$ when $n$ is prime and so for a lower bound $m \ge 1,$ since $\sigma(n) \ge n+1$ for all $n \ge 2.$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The following upper bound for $\sigma(n)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sigma(n) &amp;lt; &#xD;&#xA;e^\gamma  n \log \log n + \frac{ 0.6482 n}{\log \log n}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;was given on &lt;a href=&quot;http://mathoverflow.net/questions/33755/erdos-kac-for-sum-of-divisors&quot; rel=&quot;nofollow&quot;&gt;mathoverflow.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1301" LastActivityDate="2010-11-13T13:03:59.263" CommentCount="4" />
  <row Id="10118" PostTypeId="2" ParentId="10102" CreationDate="2010-11-13T13:56:33.363" Score="6" ViewCount="" Body="&lt;p&gt;It is very possible that your colleague generally avoids using the axiom of choice, because manifolds are relatively concrete objects.  Assuming that your friend is studying only separable manifolds, or better yet only compact ones, that makes it possible to do many constructions very explicitly, without using the axiom of choice. This is similar to the way that various principles of analysis that require the axiom of choice in general do not require the axiom of choice when they are applied to Euclidean spaces. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you're right that it may be easier to find AC in the background results.  The trouble is that many of these background results are studied in far more generality than they are used. For example, suppose an analyst uses the fact that $[0,1]\times[0,1]$ is compact. This fact follows from Tychonoff's theorem, which for general topological spaces does require the axiom of choice. But we could prove the compactness of the unit square more directly, avoiding the axiom of choice altogether (this relies on the separability of the square, in particular). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So sometimes the axiom of choice is used for convenience, by the invocation of a very general result, but it could be avoided if necessary. If we were to study only smooth manifolds that had already been embedded into Euclidean space, I suspect that we would be able to do pretty much everything in Zermelo-Fraenkel set theory without the axiom of choice. But it would take careful attention in the proofs to make sure that we replace choice-based techniques with alternative methods. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know this isn't a direct answer to the question, but I think it's relevant since it explains a caveat with possible answers: just because a general result requires AC doesn't mean that AC is required for all consequences of that result. &lt;/p&gt;&#xA;" OwnerUserId="630" LastActivityDate="2010-11-13T13:56:33.363" CommentCount="3" />
  <row Id="10119" PostTypeId="1" CreationDate="2010-11-13T14:14:52.020" Score="2" ViewCount="115" Body="&lt;p&gt;During lesson we were given a starter activity which was to try and factorise polynomials and see what happened.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The polynomials were&#xA;$x^3-8$,&#xA;$x^3-3x^2+ x -3$&#xD;&#xA;and&#xA;$x^4 - 16$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I could not work out what happened to them, and it's bugging me. If anyone can explain then it'd be helpful! Thank you.&lt;/p&gt;&#xA;" OwnerUserId="1968" LastEditorUserId="1242" LastEditorDisplayName="" LastEditDate="2010-11-13T15:09:54.000" LastActivityDate="2010-11-14T03:41:15.440" Title="Factorising certain polynomials." Tags="&lt;polynomials&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="10120" PostTypeId="1" AcceptedAnswerId="10130" CreationDate="2010-11-13T14:16:16.450" Score="0" ViewCount="132" Body="&lt;p&gt;Hi, I've got a solution to a economy class calculation, but I can't figure out how they came to the answer. Could you help me?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-5p + 80 = 8p +-50&lt;/p&gt;&#xA;&#xA;&lt;p&gt;.......? = ......? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;120 = 12p&lt;/p&gt;&#xA;&#xA;&lt;p&gt;p = 10&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Extra: do you know which rule they are using to get from the first line to the 3d?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many thx&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Math beginner,&#xA;Frank&lt;/p&gt;&#xA;" OwnerUserId="2605" LastActivityDate="2010-11-14T05:25:22.077" Title="quick basic math equation question" Tags="&lt;arithmetic&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="10121" PostTypeId="1" AcceptedAnswerId="10127" CreationDate="2010-11-13T14:20:18.443" Score="1" ViewCount="50" Body="&lt;p&gt;This is actually a generalized version I wrote of a homework question that intrigued me:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Let $f$ be continuous in $[0,1]$ and&#xA;  $\forall x\in [0,1], \ f(x)=f(1-x)$.&#xA;  If $r\in [0,1]$ and $1-r$ are the only&#xA;  sub limits of $a_{n}$ then $f(a_{n})$&#xD;&#xA;  converges.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(the original question states $r=\frac{1}{3}$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I solved the original question using a method shown in class and tried to implement it to prove the generalized version (see my own answer). I was wondering if there's a better way.&lt;/p&gt;&#xA;" OwnerUserId="3237" LastActivityDate="2010-11-13T15:21:45.737" Title="if $r\in (0,1), 1-r$ are the only sub limits of $a_{n}$ then $f(a_{n})$ converges when..." Tags="&lt;calculus&gt;&lt;sequences-and-series&gt;" AnswerCount="3" CommentCount="3" />
  <row Id="10122" PostTypeId="2" ParentId="10121" CreationDate="2010-11-13T14:21:17.053" Score="0" ViewCount="" Body="&lt;p&gt;Here's my own attempt at solving this (comments are appreciated!):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We denote $b_{n}=f(a_{n})$ and by Bolzano-Weistrauss, $b_{n_{k}}\to b$.&#xA;The subsequence $a_{n_{k}}$ might not converge, but again by Bolzano-Weistrauss there exists $a_{n_{k_{m}}}$ that converges to either $r$ or $1-r$ (because those are the only sublimits of $a_{n}$).&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;if $a_{n_{k_{m}}}\to r$ then $b_{n_{k_{m}}}=f(a_{n_{k_{m}}})\to f(r)$ (because $f$ is continuous in $[0,1]$).&lt;/li&gt;&#xA;&lt;li&gt;Similarly, if $a_{n_{k_{m}}}\to 1-r$ then $b_{n_{k_{m}}}=f(a_{n_{k_{m}}})\to f(1-r)$.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;But since $\forall x\in [0,1], \ f(x)=f(1-x)$, we get $f(r)=f(1-r)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To finish the argument we note that $b_{n_{k_{m}}}$ is a subsequence of $b_{n_{k}}\to b$, thus $b_{n_{k_{m}}}\to b$ and we conclude that $b=f(r)=f(1-r)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $b$ was chosen arbitrarily we get that $b_{n}$ has a single sublimit which implies $b_{n}\to f(r)$ &lt;/p&gt;&#xA;" OwnerUserId="3237" LastActivityDate="2010-11-13T14:21:17.053" />
  <row Id="10123" PostTypeId="2" ParentId="10096" CreationDate="2010-11-13T14:23:34.303" Score="4" ViewCount="" Body="&lt;p&gt;There are very few examples where you directly prove in ZFC that a certain set must have size $\aleph_n$. This is because most of the sets we construct are defined in terms of power sets. This means that we can compute the size of these sets in terms of ℶ numbers pretty easily, but we can't compute them in terms of ℵ numbers. The difficulty is related to the unprovability of the continuum hypothesis. It turns out that ZFC can say very, very little about the cardinalities of ℶ numbers. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One way to get sets of a fixed cardinality is to talk directly about well orderings. For example, $\aleph_1$ is exactly the set of order types of well orderings of $\omega$ (regardless of what $\beth_1$ is). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For large cardinals, there is no way to explicitly compute their cardinality. For example, any inaccessible cardinal number $\kappa$ will have the property that $|\kappa| = \aleph_\kappa$, so you will not be able to make progress by trying to compute its ℵ number. &lt;/p&gt;&#xA;" OwnerUserId="630" LastEditorUserId="630" LastEditorDisplayName="" LastEditDate="2010-11-13T14:30:01.057" LastActivityDate="2010-11-13T14:30:01.057" CommentCount="3" />
  <row Id="10124" PostTypeId="1" AcceptedAnswerId="10213" CreationDate="2010-11-13T14:23:42.650" Score="6" ViewCount="245" Body="&lt;p&gt;Suppose I have a set $\mathcal{S}$ of $N$ distinct items. Now consider the set $\mathcal{P}$ of all possible pairs that I can draw from $S$. Naturally, $|\mathcal{P}| = \binom{N}{2}$. Now when I draw $k$ items (pairs) from $\mathcal{P}$ with a uniform distribution, what is the expected number of distinct items from $S$ in those $k$ pairs?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S.: I also asked this question over at &lt;a href=&quot;http://stats.stackexchange.com/questions/4453/&quot;&gt;stats&lt;/a&gt;, but got no answers so far, so I am trying here. Thanks for your time!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt; I pick the pairs without replacement.&lt;/p&gt;&#xA;" OwnerUserId="3470" LastEditorUserId="3470" LastEditorDisplayName="" LastEditDate="2010-11-13T22:47:51.273" LastActivityDate="2010-11-14T16:14:21.260" Title="How to calculate the expected number of distinct items when drawing pairs?" Tags="&lt;combinatorics&gt;&lt;probability&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="0" />
  <row Id="10125" PostTypeId="2" ParentId="10120" CreationDate="2010-11-13T14:25:10.233" Score="2" ViewCount="" Body="&lt;p&gt;Add $5p+50$ to both sides of the = sign, and then simplify each side.&#xA;(If you wish you can do this in two steps: Add $5p$ and then 50 to each side of the equality sign.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt; &#xA;The above leads to the equation $130 = 13p$ which can be solved directly by dividing both sides of the equality sign by 13: $130/13 = 13p/13$, which after simplification gives us&#xA;$10=p$ -- which is the same as saying $p=10$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: To reach the equation $120=12p$, that is contained in the original question, is not necessary in order to solve the equation.&lt;/p&gt;&#xA;" OwnerUserId="1154" LastEditorUserId="1154" LastEditorDisplayName="" LastEditDate="2010-11-14T05:25:22.077" LastActivityDate="2010-11-14T05:25:22.077" CommentCount="4" />
  <row Id="10126" PostTypeId="1" CreationDate="2010-11-13T14:49:19.913" Score="3" ViewCount="122" Body="&lt;p&gt;This is a follow-up to the question &lt;a href=&quot;http://mathoverflow.net/questions/45378/undecidability-in-conways-game-of-life&quot; rel=&quot;nofollow&quot;&gt;Undecidability in Conway’s Game of Life&lt;/a&gt; I posted at mathoverflow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For some cellular automata it can be proven that they can simulate a Turing machine, normally by explicit construction. For some of them it can be proven that they can &lt;em&gt;not&lt;/em&gt; simulate a Turing machine.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Are there computable sufficient conditions on a cellular automaton to be Turing-complete (besides of being one of the known cases)? (My guess is: No.)&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;If not: Why not?&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;If not: Can &lt;em&gt;non&lt;/em&gt;-computable sufficient conditions be named?&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;What about necessary conditions?&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;What's the &quot;standard&quot; way of showing that a cellular automaton is &lt;em&gt;not&lt;/em&gt; Turing-complete?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1792" LastActivityDate="2011-04-03T23:41:39.627" Title="(Semi-)Decidability of Turing-completeness of cellular automata" Tags="&lt;computability&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="10127" PostTypeId="2" ParentId="10121" CreationDate="2010-11-13T14:54:06.850" Score="0" ViewCount="" Body="&lt;p&gt;We claim that $f(a_n) \to \xi = f(r) = f(1-r)$. If $r = \tfrac{1}{2}$ then $(a_n)$ must be convergent (why?). Since $f$ is continuous $f(a_n)$ converges too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So assume that $r \neq \tfrac{1}{2}$. WLOG we can assume that $r \lt \tfrac{1}{2}$. Consider the subsequence $(a_{n_k})$ consisting of the elements of $(a_n)$ that are less than $\tfrac{1}{2}$ and the subsequence $(a_{n_l})$ consisting of the elements of $(a_n)$ that are greater than or equal to $\tfrac{1}{2}$ and convince yourself that they are in fact subsequences and that they exhaust the $a_n$. In particular $a_{n_k} \to r$ and $a_{n_l} \to 1-r$. Now all that's left to do is to observe that since $f$ is continuous, $f(a_{n_k})$ and $f(a_{n_l})$ are eventually in any open interval around $\xi$. Since they exhaust $(a_n)$ we must have $f(a_n) \to \xi$.&lt;/p&gt;&#xA;" OwnerUserId="1269" LastActivityDate="2010-11-13T14:54:06.850" />
  <row Id="10128" PostTypeId="2" ParentId="10119" CreationDate="2010-11-13T15:07:20.353" Score="1" ViewCount="" Body="&lt;p&gt;It's not clear exactly what's your problem, but the last polynomial factorizes like this:&#xA;$$x^4-16=(x^2)^2-4^2=(x^2+4)(x^2-4)=(x^2+4)(x+2)(x-2).$$&#xD;&#xA;For the first one, it's easy to find the root $x=2$, which means that you can factor out $x-2$.&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2010-11-13T15:07:20.353" CommentCount="1" />
  <row Id="10129" PostTypeId="2" ParentId="10124" CreationDate="2010-11-13T15:10:43.257" Score="4" ViewCount="" Body="&lt;p&gt;You are picking edges from a complete graph and looking for the expected number of vertices that get picked.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the expected number of vertices that &lt;em&gt;don't&lt;/em&gt; get picked.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The probability that a vertex gets picked in 1 try is $\dfrac{2}{n}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus it does not get picked in all tries is $(1-\dfrac{2}{n})^k$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus the expected number of vertices that don't get picked is $n(1-\dfrac{2}{n})^k$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and thus the expected number of vertices that get picked is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$n(1 - (1-\dfrac{2}{n})^k)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As Ross points out, this assumes you pick the pairs with replacement.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-11-13T15:41:23.857" LastActivityDate="2010-11-13T15:41:23.857" CommentCount="4" />
  <row Id="10130" PostTypeId="2" ParentId="10120" CreationDate="2010-11-13T15:11:27.653" Score="2" ViewCount="" Body="&lt;p&gt;-5p+80=8p-50&lt;/p&gt;&#xA;&#xA;&lt;p&gt;80=5p+8p-50&lt;/p&gt;&#xA;&#xA;&lt;p&gt;80=13p-50&lt;/p&gt;&#xA;&#xA;&lt;p&gt;80+50=13p&lt;/p&gt;&#xA;&#xA;&lt;p&gt;130=13p (to get 120=12p at this step, you can multiply both sides by 12/13, but it's not necessary to arrive at the final answer)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;p=10&lt;/p&gt;&#xA;" OwnerUserId="3431" LastActivityDate="2010-11-13T15:11:27.653" CommentCount="1" />
  <row Id="10131" PostTypeId="2" ParentId="10102" CreationDate="2010-11-13T15:14:22.003" Score="6" ViewCount="" Body="&lt;p&gt;Your friend will probably weasel out of this example by claiming that it isn't strictly &quot;Riemannian&quot; geometry, but it's definitely differential geometry.  The example is the Hodge theorem, which asserts that every cohomology class of a Riemannian manifold is represented by a unique harmonic form (where &quot;harmonic&quot; is with respect to the Laplace-Beltrami operator).  I highly doubt that a proof can be crafted without some reasonably serious functional analysis; the standard approach uses elliptic theory and Sobolev theory which requires the Banach-Alaoglu theorem which in turn requires the Tychonoff theorem.  In general I would wager that any result which involves geometric PDE theory (including some results in, say, minimal surface theory which genuinely are Riemmannian) is going to demand the axiom of choice at some level.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other than that, you might check out the work of Alexander Nabutovsky.  He has obtained some really serious results about the structure of geodesics and on the moduli space of Riemannian metrics using techniques from logic and computability theory - I wouldn't be surprised if AC is hiding somewhere.&lt;/p&gt;&#xA;" OwnerUserId="1509" LastActivityDate="2010-11-13T15:14:22.003" />
  <row Id="10132" PostTypeId="2" ParentId="10102" CreationDate="2010-11-13T15:21:19.093" Score="5" ViewCount="" Body="&lt;p&gt;It looks to me like the &lt;a href=&quot;http://en.wikipedia.org/wiki/Arzel%C3%A0%E2%80%93Ascoli_theorem&quot; rel=&quot;nofollow&quot;&gt;Arzel&amp;agrave;--Ascoli theorem&lt;/a&gt; needs at least some weak form of choice.  (I have started &lt;a href=&quot;http://mathoverflow.net/questions/45928/does-arzela-ascoli-require-choice&quot; rel=&quot;nofollow&quot;&gt;an MO question&lt;/a&gt; to clarify this.)  One often uses this in geometry; for example, to guarantee the existence of minimizing geodesics connecting pairs of points.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: See Andres Caicedo's answer on MO (at above link).  The answer is affirmative.  Also, the database list of equivalents he mentions contains some very innocuous-looking statements that I bet your friend has never thought twice about using.&lt;/p&gt;&#xA;" OwnerUserId="822" LastEditorUserId="822" LastEditorDisplayName="" LastEditDate="2010-11-13T16:18:41.227" LastActivityDate="2010-11-13T16:18:41.227" CommentCount="1" />
  <row Id="10133" PostTypeId="2" ParentId="10121" CreationDate="2010-11-13T15:21:45.737" Score="0" ViewCount="" Body="&lt;p&gt;It is quite obvious that if $r$ and $1-r$ are the only sublimits of $\{a_n\}$ in the interval $[0,l]$, then for any $\delta \gt 0$, $(r-\delta, r+\delta)\cup (1-r-\delta,1-r+\delta)$ contains all but finitely many terms of the sequence. &#xA;Let $\epsilon&amp;gt;0$. Choose $\delta&amp;gt;0$ such that $|f(x)-f(r)|\lt \epsilon$ for all $x\in (r-\delta, r+\delta)$. Then by symmetry of $f$,&#xA;also $|f(x)-f(r)|\lt \epsilon$ for all $x\in (1-r-\delta, 1-r+\delta)$. It follows that&#xA;$$|f(a_n)-f(r)|\lt \epsilon$$ for all but finitely many $n$'s. This proves that $\lim_{n\to\infty} f(a_n)=f(r)$.&lt;/p&gt;&#xA;" OwnerUserId="3249" LastActivityDate="2010-11-13T15:21:45.737" />
  <row Id="10134" PostTypeId="1" CreationDate="2010-11-13T15:23:21.523" Score="0" ViewCount="145" Body="&lt;p&gt;Let $f:[0,1] \rightarrow [0,\infty)$ be a measurable function such that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\mu (\{x \in [0,1]: f(x) &amp;gt; t \}) \leq \frac{1}{t(ln(t))^{2}}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;holds for each $t&amp;gt;3$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Show $f$ is an integrable map.&lt;/p&gt;&#xA;" OwnerUserId="4219" LastActivityDate="2010-11-19T11:00:14.900" Title="Prove f is an integrable map" Tags="&lt;measure-theory&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" />
  <row Id="10135" PostTypeId="2" ParentId="10112" CreationDate="2010-11-13T15:24:16.420" Score="3" ViewCount="" Body="&lt;p&gt;I think randomly emailing professors is not going to earn any replies. Professors usually have a lot on their plate and they get hundreds of emails each day. Unsolicited emails are usually ignored. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One possibility you can explore is trying to establish email contact with a few graduate students in related areas. Once you have done this, you may send copies of your work to some of them and try to get feedback. You are more likely to find graduate students who are responsive and some of the advanced ones are actually used to reviewing papers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: I found another useful piece on this subject by Henry Cohn&#xA;&lt;a href=&quot;http://research.microsoft.com/en-us/um/people/cohn/Thoughts/advice.html&quot; rel=&quot;nofollow&quot;&gt;http://research.microsoft.com/en-us/um/people/cohn/Thoughts/advice.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3431" LastEditorUserId="3431" LastEditorDisplayName="" LastEditDate="2010-11-14T05:48:01.437" LastActivityDate="2010-11-14T05:48:01.437" CommentCount="1" />
  <row Id="10136" PostTypeId="1" CreationDate="2010-11-13T15:50:34.480" Score="1" ViewCount="115" Body="&lt;p&gt;How do I show that a monomorphism $F \rightarrow G$ of sheaves induces an injection on stalks? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When showing that monomorphism is an injection on sets one uses the maps $x \mapsto a$ and $x \mapsto b$ where $a,b$ are some fix elements such that they have the same image under the monomorphism. Then by using the monomorphism property we must have $a = b$. But since we are dealing with sheaves we are supposed to construct a skyscraper sheaf and two maps that map &quot;almost all&quot; section of the skyscraper sheaf to $a$ and $b$ (respectively), or at least elements that are &quot;connected&quot; to $a$ and $b$ (since these are representations of elements in the stalk there are other options depending on which open set we are taking). When I say &quot;almost all&quot; and &quot;connected&quot; it is just my visualization of how it should be.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not able to neither define the correct skyscraper sheaf nor the maps to $F$ (especially such that the maps commute with the restriction maps). I was hoping for &lt;a href=&quot;http://math.stackexchange.com/questions/10047/can-every-element-in-the-stalk-be-represented-by-a-section-in-the-top-space&quot;&gt;this&lt;/a&gt; being true which might have helped me but now I am stuck and need some help with this problem. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; I am still interested in how the problem will be solved by using a skyscraper sheaf. &lt;/p&gt;&#xA;" OwnerUserId="3473" LastEditorUserId="3473" LastEditorDisplayName="" LastEditDate="2010-11-17T17:20:18.467" LastActivityDate="2010-11-17T17:20:18.467" Title="Monomorphisms of sheaves gives an injection of stalks" Tags="&lt;algebraic-geometry&gt;" AnswerCount="2" />
  <row Id="10137" PostTypeId="2" ParentId="10068" CreationDate="2010-11-13T15:58:16.427" Score="0" ViewCount="" Body="&lt;p&gt;Let's ponder like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At some point $x$ $f(x)&amp;gt;0$, therefore, in the vicinity of this point there is an increasing arithmetic progression&#xA;$a_{0}, \ b_{0}, \ c_{0}$ that, $f(a_{0})+f(b_{0})+f(c_{0})&amp;gt;0$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Like this one will be found increasing arithmetic progression of &#xA;$a_{1}, \ b_{1}, \ c_{1}$ that, $f(a_{1})+f(b_{1})+f(c_{1})&amp;lt;0$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For all values of parametr $t[0,1]$ сonsider the arithmetic progression &#xA;$a(t), \ b(t), \ c(t)$, where &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a(t)=a_{0}(1-t)+a_{1}t$, &lt;br&gt; &#xA;$b(t)=b_{0}(1-t)+b_{1}t$, &lt;br&gt;&#xA;$c(t)=c_{0}(1-t)+c_{1}t$. &lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Function $F(t)=f(a(t))+f(b(t))+f(c(t))$ continuously depends on $t$, at $t=0 \ F(t)&amp;gt;0$, and at $t=1 \ F(t)&amp;lt;0$. It means that in some $t \ F(t)=0$ and the corresponding progress  $a(t), \ b(t), \ c(t)$  is required. &lt;/p&gt;&#xA;" OwnerUserId="2779" LastActivityDate="2010-11-13T15:58:16.427" />
  <row Id="10138" PostTypeId="2" ParentId="9713" CreationDate="2010-11-13T16:00:24.710" Score="1" ViewCount="" Body="&lt;p&gt;Strictly speaking we cannot satisfy (2) exists lim&lt;sub&gt;x→c&lt;/sub&gt;f′(x)/g′(x), even in the extended sense, unless (1) g'(x) is nonzero in a (deleted) neighborhood of x = c.  That is required for f'(x)/g'(x) to be defined there (in the deleted neighborhood).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Such gaps in the definition of f'(x)/g'(x) only matter if they accumulate at x = c, since we can otherwise work around them by shrinking the neighborhood to avoid them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can entertain the possibility that those gaps all turn out to be removable discontinuities in f'(x)/g'(x), so that a continuous extension of f'(x)/g'(x) as x approaches c is uniquely determined.  For that to happen we would need f'(x) to have roots of equal or greater multiplicity as g'(x) at those gaps.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But this amounts to &quot;multiplying top and bottom by the same thing&quot;, in the sense that f'(x) and g'(x) would have a common factor containing all the roots of g'(x) sufficiently near x = c.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The proposed example f(x) = 2x + sin(2x) and g(x) = f(x)e&lt;sup&gt;-sin(x)&lt;/sup&gt; works out in this way.  The quotient f'(x)/g'(x) is undefined when g'(x) = 0, but:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;g'(x) = [f'(x) - cos(x) f(x)] e&lt;sup&gt;-sin(x)&lt;/sup&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;f'(x) = 2(1 + cos(2x)) = 4 cos&lt;sup&gt;2&lt;/sup&gt;(x)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A common factor containing all the roots of g'(x) for sufficiently large x is then cos(x).  If we &quot;cancel&quot; that factor, removing roots in the denominator, the resulting &quot;simplified&quot; f'(x)/g'(x) has a bounded numerator and a denominator that tends to -&amp;infin; as x grows (since f(x) tends to &amp;infin;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Evidently we cannot blithely patch over removable singularities in f'(x)/g'(x) for the sake of l'Hopital's rule.&lt;/p&gt;&#xA;" OwnerUserId="3111" LastActivityDate="2010-11-13T16:00:24.710" />
  <row Id="10139" PostTypeId="2" ParentId="10119" CreationDate="2010-11-13T16:03:37.417" Score="1" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\rm\ \ x-a \ $ divides $\rm\ x^n - a^n\ $ handles the 1st and 3rd. The 2nd is $\rm\ x^2\ (x-3) + (x-3)\ =\ \ldots$&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-11-13T16:03:37.417" />
  <row Id="10140" PostTypeId="2" ParentId="10136" CreationDate="2010-11-13T16:23:50.923" Score="2" ViewCount="" Body="&lt;p&gt;There is a nice proof of this on page 7 of Kiran Kedlaya's notes (and also that surjectivity carries over to stalks)&#xA;&lt;a href=&quot;http://ocw.mit.edu/courses/mathematics/18-726-algebraic-geometry-spring-2009/lecture-notes/MIT18_726s09_lec03_sheaves.pdf&quot; rel=&quot;nofollow&quot;&gt;http://ocw.mit.edu/courses/mathematics/18-726-algebraic-geometry-spring-2009/lecture-notes/MIT18_726s09_lec03_sheaves.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3431" LastActivityDate="2010-11-13T16:23:50.923" CommentCount="1" />
  <row Id="10141" PostTypeId="2" ParentId="10136" CreationDate="2010-11-13T16:27:13.813" Score="4" ViewCount="" Body="&lt;p&gt;In fact, more is true. If $U$ is an open set, then $F(U) \to G(U)$ is a monomorphism for $F \to G$ a monomorphism of sheaves. To see this, use the constant sheaf $\mathbb{Z}_U$ over $U$. To hom from $\mathbb{Z}_U$ into a sheaf $H$ is the same thing as giving a section of $H$ over $U$. This implies injectivity on the stalks because filtered colimits are exact in the category of abelian groups (or modules over a ring).&lt;/p&gt;&#xA;" OwnerUserId="536" LastActivityDate="2010-11-13T16:27:13.813" CommentCount="1" />
  <row Id="10142" PostTypeId="2" ParentId="10095" CreationDate="2010-11-13T16:40:21.133" Score="2" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\quad\quad $ With $\rm\ \ \alpha\ =\ 1 - 2\ s + 2\ \sqrt{s^2 - s}\ \ $ we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \rm \frac{s\ x\ (1+x)}{(1-x)^3 + 4\ s\ x\ (1-x)}\ =\ \frac{\alpha}{4\ (x-\alpha)} +  \frac{\alpha&amp;#39;}{4\ (x-\alpha&amp;#39;)} - \frac{1}{2\ (x-1)} $$&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-11-13T16:40:21.133" CommentCount="0" />
  <row Id="10144" PostTypeId="2" ParentId="10095" CreationDate="2010-11-13T16:48:06.413" Score="1" ViewCount="" Body="&lt;p&gt;Here it's easier to work with the exponential generating function.  You just need to find the Taylor series of each term on the RHS.  You already know the Taylor series of one term.  To find the Taylor series of the other term, use the fact that $\cos t = \frac{e^{it} + e^{-it}}{2}$.  &lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-11-13T16:48:06.413" />
  <row Id="10145" PostTypeId="2" ParentId="9941" CreationDate="2010-11-13T17:22:29.380" Score="1" ViewCount="" Body="&lt;p&gt;Although this is not an answer, I thought that the following identity (pulled out of the book &quot;Concrete Mathematics&quot;) might be useful for your analysis (or rather its proof actually that should be useful):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_0^1\cdots\int_0^1 f(\lfloor x_1+\cdots+x_n\rfloor)dx_1\cdots dx_n = \sum_k \genfrac{\langle}{\rangle}{0pt}{}{n}{k} \frac{f(k)}{n!},$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\genfrac{\langle}{\rangle}{0pt}{}{n}{k}$ is the Euler-number (first-kind).&lt;/p&gt;&#xA;" OwnerUserId="1709" LastActivityDate="2010-11-13T17:22:29.380" />
  <row Id="10146" PostTypeId="1" AcceptedAnswerId="10149" CreationDate="2010-11-13T17:30:20.840" Score="1" ViewCount="343" Body="&lt;p&gt;Often i have wondered about this question, but today i had a chance to recollect it and hence i am posting it here. During &lt;em&gt;high-school&lt;/em&gt; days one generally learns &lt;em&gt;Integration&lt;/em&gt; and i still loving doing problems on &lt;em&gt;Integration.&lt;/em&gt; But i have never understood this idea of &lt;em&gt;integration by parts&lt;/em&gt;. Why do we adopt the &lt;strong&gt;ILATE&lt;/strong&gt; or &lt;strong&gt;LIATE&lt;/strong&gt; rule to do problems and what was the reason behind thinking of such a rule. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Reference:&lt;/strong&gt; &lt;a href=&quot;http://en.wikipedia.org/wiki/Integration_by_parts&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Integration_by_parts&lt;/a&gt;&lt;/p&gt;&#xA;" LastActivityDate="2010-11-13T19:32:36.610" Title="Intuition behind the ILATE rule" Tags="&lt;calculus&gt;&lt;real-analysis&gt;&lt;analysis&gt;" AnswerCount="2" CommentCount="6" FavoriteCount="0" />
  <row Id="10147" PostTypeId="1" AcceptedAnswerId="10152" CreationDate="2010-11-13T17:39:31.003" Score="2" ViewCount="290" Body="&lt;p&gt;How to prove that finite additivity follows from countable additivity!!?&lt;/p&gt;&#xA;" OwnerUserId="953" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2010-11-20T01:45:56.867" LastActivityDate="2010-11-20T01:45:56.867" Title="Finite additivity follows from countable additivity!" Tags="&lt;measure-theory&gt;&lt;elementary-set-theory&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" />
  <row Id="10149" PostTypeId="2" ParentId="10146" CreationDate="2010-11-13T17:49:15.727" Score="7" ViewCount="" Body="&lt;p&gt;The way I see it, when you differentiate an inverse trigonometric function, you don't get another inverse trigonometric function. Instead you get &quot;simpler&quot; funtions like $1/(1 + x^2)$ or $1/\sqrt{1-x^2}$. This does not typically happen with the antiderivative of such functions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, when you differentiate a logarithmic funtion, the logarithm disappears.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, when using integration by parts $\int u dv = uv - \int v du$, it makes sense to select the inverse trigonometric or logarithmic function to be the one that is the u term.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of algebraic, trigonometric and exponential functions both integration and differentiation don't change the nature of the function, so they come later in the ILATE order.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, this is just intuition and there are examples where you can violate this so called rule and still integrate by parts without any problems&lt;/p&gt;&#xA;" OwnerUserId="2641" LastEditorUserId="2641" LastEditorDisplayName="" LastEditDate="2010-11-13T19:32:36.610" LastActivityDate="2010-11-13T19:32:36.610" CommentCount="1" />
  <row Id="10150" PostTypeId="2" ParentId="10146" CreationDate="2010-11-13T17:50:41.463" Score="3" ViewCount="" Body="&lt;p&gt;As a technique for explicitly integrating functions given by formulas as usually seen in calculus classes, integration by parts works because $u&amp;#39;v$ can be easier to integrate than $uv&amp;#39;$.  There are multiple ways an integrand can be considered as a product of the form $u&amp;#39;v$, but some choices lead to $uv&amp;#39;$ being of no use, perhaps even harder to integrate.  The ILATE mnemonic you mention (or whatever it is) gives some rules that approximate the best guesses as to what will work well.  Better than this mnemonic (which I've never thought about) is to just have enough experience doing integrals to be able to see what will work and why (at least for the type of integrals one typically sees in a calculus class).&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-11-13T17:50:41.463" />
  <row Id="10151" PostTypeId="2" ParentId="8856" CreationDate="2010-11-13T18:00:17.880" Score="6" ViewCount="" Body="&lt;p&gt;It is a lovely result of Hermann Minkowski that any plane centrally symmetric convex set can serve as the &quot;unit ball&quot; of a distance function.&lt;/p&gt;&#xA;" OwnerUserId="1369" LastActivityDate="2010-11-13T18:00:17.880" />
  <row Id="10152" PostTypeId="2" ParentId="10147" CreationDate="2010-11-13T18:02:16.723" Score="10" ViewCount="" Body="&lt;p&gt;I guess this question arises because in the definition of countable additivity it is assumed that there are infinitely many events.  But if there are finitely many events, you can just throw in countably many empty events.  That is, given $A_1,\ldots,A_n$, define $A_{n+1}=\emptyset$, $A_{n+2}=\emptyset$, and so on, then apply countable additivity.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-11-13T18:02:16.723" />
  <row Id="10153" PostTypeId="2" ParentId="10147" CreationDate="2010-11-13T18:20:49.417" Score="2" ViewCount="" Body="&lt;p&gt;I assume that you want to show that finite additivity follows from countably infinite additivity. In general, though countable means both finite and countably infinite.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We are given that $\mu(\displaystyle \cup_{i=1}^\infty A_i) = \displaystyle \sum_{i=1}^\infty \mu(A_i)$ whenever $A_i \cap A_j = \emptyset$, $\forall i \neq j$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we now want to show $\mu(\displaystyle \cup_{i=1}^n A_i) = \displaystyle \sum_{i=1}^n \mu(A_i)$, $\forall n \in \mathbb{N}$, whenever $A_i \cap A_j = \emptyset$, $\forall i \neq j$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now consider $A_{i} = \emptyset$, $\forall i &amp;gt; n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the way we have defined, note that $A_i \cap A_j = \emptyset$, $\forall i \neq j$. (Note: $A \cap \emptyset = \emptyset$, $\forall A$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By countably infinite additivity, $\mu(\displaystyle \cup_{i=1}^\infty A_i) = \displaystyle \sum_{i=1}^\infty \mu(A_i)$ and note that $\mu(A_i) = 0$, $\forall i &amp;gt; n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also note that $\displaystyle \cup_{i=1}^\infty A_i = \displaystyle \cup_{i=1}^n A_i$, since $A_i = \emptyset$, $\forall i &amp;gt; n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, $\mu(\displaystyle \cup_{i=1}^n A_i) = \displaystyle \sum_{i=1}^n \mu(A_i)$ whenever $A_i \cap A_j = \emptyset$, $\forall i \neq j$.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2010-11-13T19:13:14.060" LastActivityDate="2010-11-13T19:13:14.060" CommentCount="6" />
  <row Id="10154" PostTypeId="1" AcceptedAnswerId="10158" CreationDate="2010-11-13T18:35:57.950" Score="0" ViewCount="124" Body="&lt;p&gt;I've written Haskell function, that helps me to compute $x^{-1} \mod m$.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;invM x m = invMhelp x 0 m; --modular multiplicative inverse x^-1 (mod m) = inv x m  &#xA;invMhelp x i m = if (x*i `mod` m == 1) then i else (invMhelp x (i+1) m);&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;For example:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Main&amp;gt; 13 `invM` 2109&#xA;649&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;That means $13^{-1} = 649 \mod 2109$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;How can I compute $x^{-2} \mod m$?&lt;/strong&gt; ($x^{-2} \mod m = (x^{-1})^2 \mod m$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Can I simply use the power of two of the multiplicative inverse and then apply the modulo m again?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example: $13^{-1} = 649 \mod 2109$, $649^2 \mod 2109 = 1510$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is $13^{-2} \mod 2109$ equal to $1510$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sorry for the noob question, but I'm not sure about this. Thanks for answers.&lt;/p&gt;&#xA;" OwnerUserId="2411" LastEditorUserId="2411" LastEditorDisplayName="" LastEditDate="2010-11-13T18:41:47.897" LastActivityDate="2010-11-13T21:18:23.977" Title="How do I compute x^(-2) mod m?" Tags="&lt;number-theory&gt;&lt;modular-arithmetic&gt;" AnswerCount="2" />
  <row Id="10155" PostTypeId="1" AcceptedAnswerId="10157" CreationDate="2010-11-13T18:39:21.067" Score="0" ViewCount="199" Body="&lt;p&gt;Please observe the triangle shown in the figure below(In red).I consider the angle between 5 and 4.The base should be 5 ,hypotenuse 4 and perpendicular 3,but according to the solution manual the base is 4 and the hypotenuse 5.Please tell me how.Thanks!&lt;img src=&quot;http://i.imgur.com/pdKkI.gif&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="2857" LastEditorUserId="2857" LastEditorDisplayName="" LastEditDate="2010-11-13T18:53:15.313" LastActivityDate="2010-11-16T14:07:18.193" Title="How to solve this triangle?" Tags="&lt;geometry&gt;" AnswerCount="3" CommentCount="5" />
  <row Id="10156" PostTypeId="2" ParentId="10154" CreationDate="2010-11-13T18:42:40.800" Score="3" ViewCount="" Body="&lt;p&gt;Yes. That will work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a \equiv b \pmod{m} \Rightarrow  a^k \equiv b^k \pmod{m}$, whenever $k \in \mathbb{N}$&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2010-11-13T18:49:50.810" LastActivityDate="2010-11-13T18:49:50.810" CommentCount="2" />
  <row Id="10157" PostTypeId="2" ParentId="10155" CreationDate="2010-11-13T18:51:48.500" Score="6" ViewCount="" Body="&lt;p&gt;This is not a complete answer to your question, but here's a hint. (I cannot post comments yet).&#xA;(3,5,4) is not a Pythagorean triplet &lt;em&gt;i.e.&lt;/em&gt; there are no right angled triangle with base 5, perpendicular 3 and hypoteneuse 4.&lt;/p&gt;&#xA;" OwnerUserId="3431" LastActivityDate="2010-11-13T18:51:48.500" />
  <row Id="10158" PostTypeId="2" ParentId="10154" CreationDate="2010-11-13T18:55:58.620" Score="5" ViewCount="" Body="&lt;p&gt;$\rm\: a\equiv b^{-1} \Rightarrow\ a^2 \equiv b^{-2}\: $ via &lt;strong&gt;congruences are preserved by multiplication&lt;/strong&gt; (so too by squaring). If you don't yet know that property then you can instead prove the sought result as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\rm a\equiv b^{-1}\ \:(mod\ m)\ \Rightarrow\ m\:|\: ab-1\ \Rightarrow\ m|(ab-1)(ab+1)\ =\ a^2 b^2 - 1\ \Rightarrow\ a^2 \equiv b^{-2}\ \:(mod\ m)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For completeness, here is the proof that congruences are preserved by mutliplication&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;LEMMA&lt;/strong&gt; $\rm\ \ A\equiv a,\ B\equiv b\ \Rightarrow\ AB\equiv ab\ \:(mod\ m)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof&lt;/strong&gt; $\rm\ \ m\: |\: A-a,\:\:\ B-b\ \Rightarrow\ m\ |\ (A-a)\ B + a\ (B-b)\ =\ AB - ab $&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This &lt;strong&gt;congruence product rule&lt;/strong&gt; is at the heart of many other similar product rules, for example Leibnitz's product rule for derivatives in calculus, e.g. see my &lt;a href=&quot;http://math.stackexchange.com/questions/3101/derivative-of-a-product-and-derivative-of-quotient-of-functions-theorem-i-dont/3178#3178&quot;&gt;post here.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-11-13T21:18:23.977" LastActivityDate="2010-11-13T21:18:23.977" CommentCount="1" />
  <row Id="10159" PostTypeId="1" CreationDate="2010-11-13T19:33:52.253" Score="1" ViewCount="186" Body="&lt;p&gt;Can somebody please explain continuity and discontinuity in fourier series?&lt;/p&gt;&#xA;" LastActivityDate="2010-11-14T08:17:27.197" Title="Continuity and discontinuity in fourier series?" Tags="&lt;fourier-series&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="10160" PostTypeId="1" CreationDate="2010-11-13T19:42:26.177" Score="5" ViewCount="200" Body="&lt;p&gt;I watched the excellent interview with Richard Feynman: &lt;a href=&quot;http://www.youtube.com/watch?v=PsgBtOVzHKI&quot; rel=&quot;nofollow&quot;&gt;http://www.youtube.com/watch?v=PsgBtOVzHKI&lt;/a&gt; In the interview Feynman mention that he at young age re-invented operator calculus.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have searched for &quot;operator calculus&quot; and have not found any accessible references that introduce the topic. Maybe operator calculus go under another name today, than at the time of the interview?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you give me a reasonably simple explanation of operator calculus, and also give some references to books on the subject?&lt;/p&gt;&#xA;" OwnerUserId="3479" LastEditorUserId="1120" LastEditorDisplayName="" LastEditDate="2010-11-13T19:55:36.393" LastActivityDate="2010-11-13T20:56:21.267" Title="What is operator calculus?" Tags="&lt;reference-request&gt;&lt;operator-theory&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="2" />
  <row Id="10161" PostTypeId="2" ParentId="10160" CreationDate="2010-11-13T20:11:16.703" Score="10" ViewCount="" Body="&lt;p&gt;At the heart the key idea is quite simple. Namely, one views functions, endomorphisms, etc. as &quot;numbers&quot; and manipulates them as such (where valid). E.g. for a derivative $\rm D = d/dx\ $ we have that $\rm D^2 - c^2\ =\ (D - c)(D + c) $ just as for numbers, as long $\rm\:D\:$ commutes with $\rm\:c\:$, i.e. $\rm\:c\:$ is constant. So we can solve constant coefficient linear differential / difference equations by simply factoring their operator (&quot;characteristic&quot;) polynomials into linear factors over $\mathbb C$. One can also similarly perform transcendental operator manipulations such as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\rm f(D)\ e^{t\:x}\ =\ f(t)\ e^{t\ x}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\rm e^{\:t\ D}\ f(x)\ =\ f(x+t) $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\rm t^{x\ D}\ f(x)\ =\ f(t\: x) $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and the Generalized Leibniz Rule&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\rm g(D)\ f(x)\ = \ \sum_{n=0}^\infty\ \frac{f^{(n)}(x)\ g^{(n)}(D)}{n!}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For further details see Roman: Umbral Calculus and Rainville: Special Functions and especially Rota: Finite operator calculus.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-11-13T20:56:21.267" LastActivityDate="2010-11-13T20:56:21.267" CommentCount="1" />
  <row Id="10162" PostTypeId="2" ParentId="10159" CreationDate="2010-11-13T20:19:51.840" Score="7" ViewCount="" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;If you have a removable discontinuity at a point, the Fourier series will converge to the limit of the function at the point.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If you have a jump discontinuity at some point, then the Fourier series will converge to the average of the values of left and right limits. But the higher harmonics are significant, resulting in the &quot;Gibbs phenomenon&quot;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;For more complicated discontinuities, we really cannot say anything. You can have a look at the &lt;a href=&quot;http://en.wikipedia.org/wiki/Convergence_of_Fourier_series&quot; rel=&quot;nofollow&quot;&gt;wikipedia page&lt;/a&gt; for more details.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The best one can say about the convergence of Fourier series is that it always converges almost everywhere, according to &lt;a href=&quot;http://en.wikipedia.org/wiki/Carleson%27s_theorem&quot; rel=&quot;nofollow&quot;&gt;Carleson's theorem&lt;/a&gt;. But this kind of theorems involve more sophisticated levels of mathematical analysis.&lt;/p&gt;&#xA;" OwnerUserId="1119" LastEditorUserId="1119" LastEditorDisplayName="" LastEditDate="2010-11-14T08:17:27.197" LastActivityDate="2010-11-14T08:17:27.197" CommentCount="3" />
  <row Id="10163" PostTypeId="2" ParentId="9941" CreationDate="2010-11-13T20:32:13.310" Score="1" ViewCount="" Body="&lt;p&gt;Moving the rectangle $j$ units to the right changes the integral by $j|S|$. So we may assume $a+c\geq0$. Parametrize $S$ by $x=(u-v)/2$, $y=(u+v)/2$. The corresponding domain $S&amp;#39;$ in the $(u,v)$-plane is a rectangle with $45^\circ$ slopes of its sides. Cut up $S&amp;#39;$ by two vertical lines into two triangles and a parallelogram. The integral in question is a linear combination of integrals of the following kind:&#xA;$\int_0^q \lfloor u\rfloor du$, $ \int_0^q u \lfloor u\rfloor du$.&#xA; These integrals have to be precomputed, and in the end some simplifications should be possible.&lt;/p&gt;&#xA;" OwnerUserId="1303" LastActivityDate="2010-11-13T20:32:13.310" />
  <row Id="10164" PostTypeId="1" AcceptedAnswerId="10192" CreationDate="2010-11-13T20:44:10.220" Score="0" ViewCount="125" Body="&lt;p&gt;Let $g(t)$ be a real function of a real variable such that &#xA;$$\sup |p(t)g(t)|\lt \infty$$ for any polynomial $p$. Is it true that for any nonngetive&#xA;integer $k$, one has $$\sup_{(x,y)\in\mathbb{R}^2}(\frac{|x|}{1+|y|})^k |g(x-y)|\le A_k, $$&#xD;&#xA;for some positive constant $A_k\lt \infty$?&lt;/p&gt;&#xA;" OwnerUserId="3249" LastEditorUserId="3249" LastEditorDisplayName="" LastEditDate="2010-11-13T21:55:29.343" LastActivityDate="2010-11-14T01:24:10.357" Title="A real function question" Tags="&lt;real-analysis&gt;&lt;functions&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="10165" PostTypeId="2" ParentId="10164" CreationDate="2010-11-13T21:07:35.003" Score="1" ViewCount="" Body="&lt;p&gt;Hint 1: What kind of function must $g(t)$, satisfying the first inequality, be? Can you think of some simple such function yourself?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint 2: What can you say about the function $y \mapsto \left({1 \over 1 + |y|}\right)^k$? Can it spoil the second inequality?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint 3: Let $x = y + t$. Triangle inequality is your friend.&lt;/p&gt;&#xA;" OwnerUserId="3223" LastActivityDate="2010-11-13T21:07:35.003" CommentCount="4" />
  <row Id="10166" PostTypeId="1" AcceptedAnswerId="10169" CreationDate="2010-11-13T21:17:55.507" Score="0" ViewCount="211" Body="&lt;p&gt;I was wondering if this question should be posted in this forum or in Stack Exchange, but reading both &quot;What kind of questions can I ask here?&quot;, in Math I found&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Software that mathematicians use&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So I guess it should fit here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a TI-89 (a --powerful, I guess-- Texas Instruments calculator) which I've had since high-school days; I very rarely use it anymore. Of course, for more advanced mathematical computations there is great software for the PC, but maybe some of it (the most simple of it) has been done for the TI-89, and I thought it might be cool finding some use for it again.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most notably while studying Galois theory lately, I've been using Wolfram Alpha to quickly determine the discriminant of a polynomial of degree &gt;= 3. Of course one could just use the formula (it's not that cumbersome for degrees 3 and 4), but I thought there might be a program that would do it for you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the question is: what good, useful &quot;advanced&quot; (advanced undergraduate and forth) math software do you know for the TI-89?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, do you know of one to compute the discriminant of a polynomial?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope the question is well-posed and suited for this forum. If not, please redirect me.&lt;/p&gt;&#xA;" OwnerUserId="2614" LastActivityDate="2010-11-13T21:23:56.123" Title="Advanced math software for the TI-89 (discriminant of a general polynomial?)" Tags="&lt;math-software&gt;&lt;calculator&gt;" AnswerCount="1" />
  <row Id="10167" PostTypeId="1" AcceptedAnswerId="10175" CreationDate="2010-11-13T21:19:23.213" Score="1" ViewCount="232" Body="&lt;p&gt;Let $A$ be a finitely generated Abelian group. Let $tA$ denote the torsion subgroup. Prove that $A$ has a subgroup isomorphic to $A/tA$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that $A/tA$ is torsion free, so my thinking so far has been to take the non-torsion elements of $A$ and consider the subgroup generated by them. Then show that subgroup and $A/tA$ are isomorphic. (I'm doing all this in additive notation, just so you know.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $A&amp;#39;=\{a\in A\,|\, na\neq 0 \text{ for all } n\in \mathbb{Z}^+\}$ and consider the subgroup $B=\langle A&amp;#39;\rangle$. Since $A$ is finitely generated, so is $B$. Let the generators be $\{a_1, \ldots , a_n\}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Restrict the natural projection $\pi: A\to A/tA$ to a map $\varphi: B\to A/tA$. Since $\pi$ is a well-defined homomorphism, so is $\phi$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I'd like to show that $\varphi$ is bijective, but I'm a little stuck. First of all, I haven't convinced myself it's even true. Secondly, for injectivity, does it suffice to only consider two generators $a_i$ and $a_j$ and suppose their images are distinct and go from there? Or do I have to take two arbitrary elements of $B$? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any help would be great, thanks!&lt;/p&gt;&#xA;" OwnerUserId="2313" LastActivityDate="2010-11-15T17:36:14.650" Title="Quotient of an Abelian group with its torsion subgroup" Tags="&lt;group-theory&gt;" AnswerCount="3" CommentCount="4" />
  <row Id="10168" PostTypeId="2" ParentId="8944" CreationDate="2010-11-13T21:21:56.407" Score="2" ViewCount="" Body="&lt;p&gt;A general suggestion: There are a few problems asking for convergence of divergence of series of the form $\sum_n (-1)^ng(n)$ or $\sum_n g(n)$ for some function $g$ for which one can easily rewrite $g(n)$ as&#xA;$$f(1/n)-f(0)$$ for some $f$ differentiable at 0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In all these cases one can try to replace $g(n)=f(1/n)-f(0)$ with $f&amp;#39;(0)/n$ and use elementary estimates and inequalities for the remainder error, if needed (sometimes a limit comparison test will already suffice).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, in the problem here, $g(n)=2^{1/n}-1$ so blatantly we can try $f(x)=2^x$. We have $f&amp;#39;(0)=\ln 2$, which is the estimate that appears in acarchau's answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For another example, $\sum_n\sin(1/n)$ diverges: Take $f(x)=\sin(x)$. &lt;/p&gt;&#xA;" OwnerUserId="462" LastActivityDate="2010-11-13T21:21:56.407" />
  <row Id="10169" PostTypeId="2" ParentId="10166" CreationDate="2010-11-13T21:23:56.123" Score="1" ViewCount="" Body="&lt;p&gt;The discriminant can be computed as a (Sylvester) determinant (or resultant), e.g. see &lt;a href=&quot;http://en.wikipedia.org/wiki/Discriminant&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt; Thus you need merely find a program that computes determinants.&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-11-13T21:23:56.123" CommentCount="4" />
  <row Id="10170" PostTypeId="2" ParentId="10167" CreationDate="2010-11-13T21:25:22.903" Score="1" ViewCount="" Body="&lt;p&gt;Unfortunately the whole group will be generated by the nontorsion elements if the group is infinite. One thing you can do is use the classification theorem for finitely generated abelian groups which says that a finitely generated abelian group is of the form $Z_{m_1} \times ... \times Z_{m_n} \times Z \times ... \times Z$, with finitely many $Z$'s appearing on the right. In this case $tA$ is $Z_{m_1} \times ... \times Z_{m_n}$ and $A$ / $tA$ will be the $Z \times ... \times Z$ which is clearly a subgroup of $A$.&lt;/p&gt;&#xA;" OwnerUserId="3035" LastActivityDate="2010-11-13T21:25:22.903" CommentCount="2" />
  <row Id="10171" PostTypeId="1" CreationDate="2010-11-13T21:26:04.963" Score="0" ViewCount="76" Body="&lt;p&gt;If I understand correctly, the Euclidean Dirac operator is given by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$D=\sum_{i=1}^n e_i \frac{\partial}{\partial x_i},$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $e_i$ are bases for $Cl_{0,n}(\mathbb{R})$, i.e., the $n$-dimensional Clifford algebra with negative-definite signature over the reals (so $e_i^2=-1$), and $x_i$ are the corresponding coordinates.  Several sources state that $D^2 = -\Delta_n$ where $\Delta_n$ is the standard Euclidean Laplace operator&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Delta_n = \sum_{i=1}^n \frac{\partial^2}{\partial x_i^2}.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I write out $D^2 f$ explicitly for some function $f:\mathbb{R}^n \rightarrow \mathbb{R}$, scalar terms from the Laplacian certainly appear, e.g.,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$e_1 \frac{\partial}{\partial x_1}\left( e_1 \frac{\partial}{\partial x_1} f \right) = e_1 \left( e_1 \frac{\partial^2}{\partial x_1^2}f + (\frac{\partial}{\partial x_1}e_1)\frac{\partial}{\partial x_1}f \right)=e_1^2 \frac{\partial^2}{\partial x_1^2}f = -\frac{\partial^2}{\partial x_1^2}f.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I also end up with bivector cross terms that shouldn't be there:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$e_1 \frac{\partial}{\partial x_1}\left( e_2 \frac{\partial}{\partial x_2} f \right) = e_1 \left( e_2 \frac{\partial^2}{\partial x_1 \partial x_2}f + (\frac{\partial}{\partial x_1}e_2)\frac{\partial}{\partial x_2}f \right)=e_1 e_2 \frac{\partial^2}{\partial x_1 \partial x_2}f = e_{12}\frac{\partial^2}{\partial x_1 \partial x_2}f.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Should I only be considering the scalar part of $D^2$, or am I simply doing something wrong here?&lt;/p&gt;&#xA;" OwnerUserId="2330" LastActivityDate="2010-11-13T21:26:04.963" Title="How do you show that the Laplacian is the square of the (Euclidean) Dirac operator?" Tags="&lt;differential-geometry&gt;&lt;vector-calculus&gt;" CommentCount="2" />
  <row Id="10172" PostTypeId="1" AcceptedAnswerId="10226" CreationDate="2010-11-13T21:36:54.673" Score="1" ViewCount="62" Body="&lt;p&gt;What are some families of graphs with known &lt;a href=&quot;http://en.wikipedia.org/wiki/Tree_decomposition&quot; rel=&quot;nofollow&quot;&gt;tree width&lt;/a&gt; k?&#xA;For instance, for tree k=1, for cycle k=2, for series-parallel graph which is not a tree k=2, what are some other examples?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm looking for some easy to construct families (or famous graphs which I can find in Mathematica's GraphData or on the internet) for testing tree decomposition code.&lt;/p&gt;&#xA;" OwnerUserId="998" LastActivityDate="2010-11-14T07:55:38.470" Title="Families of graphs with treewidth k" Tags="&lt;graph-theory&gt;&lt;mathematica&gt;" AnswerCount="1" />
  <row Id="10173" PostTypeId="1" AcceptedAnswerId="10176" CreationDate="2010-11-13T21:41:16.417" Score="0" ViewCount="137" Body="&lt;p&gt;All,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given any number, how can we retrieve each digit of that number?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i.e. For number = 4321, how can I get individual digits 4,3,2 and 1?&lt;/p&gt;&#xA;" OwnerUserId="2659" LastEditorUserId="630" LastEditorDisplayName="" LastEditDate="2010-11-13T22:09:22.273" LastActivityDate="2010-11-14T01:34:17.143" Title="Extract each digit of a number" Tags="&lt;number-theory&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="10174" PostTypeId="2" ParentId="10173" CreationDate="2010-11-13T21:46:43.350" Score="2" ViewCount="" Body="&lt;p&gt;1) Take N modulo 10.&#xA;2) Set N = [N/10] ([] is the greatest integer funtion)&#xA;3) GO back to step 1) until N = 0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope I haven't misunderstood your question here.&lt;/p&gt;&#xA;" OwnerUserId="2641" LastActivityDate="2010-11-13T21:46:43.350" />
  <row Id="10175" PostTypeId="2" ParentId="10167" CreationDate="2010-11-13T21:47:40.087" Score="4" ViewCount="" Body="&lt;p&gt;In fact more is true. Every finitely generated abelian group is a direct sum of its torsion part and the quotient by its torsion part. The structure theorem of finitely generated abelian groups is even stronger than this. &#xA;You can find a proof of the weaker result in Peter May's notes (Theorem 1.11)&#xA;&lt;a href=&quot;http://www.math.uchicago.edu/~may/TQFT/Lecture2.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.math.uchicago.edu/~may/TQFT/Lecture2.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3431" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-13T22:12:46.967" LastActivityDate="2010-11-13T22:12:46.967" CommentCount="4" />
  <row Id="10176" PostTypeId="2" ParentId="10173" CreationDate="2010-11-13T21:48:22.630" Score="4" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\rm\displaystyle\quad 3\ =\ \bigg\lfloor \frac{4321}{100}\bigg\rfloor \ mod\ 10 \ =\ 43\ mod\ 10\ \ $ i.e. rightshift then chop off least significant digit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or, if you need all digits &quot;unwind&quot; the Horner representation $\rm\ N\ =\ d_0 + 10\ (d_1 + 10\ (d_2 + \:\cdots\: ))\:$ by using division by 10 with remainder, viz $\rm\ d_0 = N\ mod\ 10,\ \ d_1 = (N-d_0)/10\ mod\ 10,\ \ldots$&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-11-13T22:11:55.923" LastActivityDate="2010-11-13T22:11:55.923" />
  <row Id="10177" PostTypeId="1" AcceptedAnswerId="10248" CreationDate="2010-11-13T21:52:21.503" Score="4" ViewCount="189" Body="&lt;p&gt;I'm trying to follow a proof about immersing/embedding $\mathbb{RP}^n$ into $\mathbb{R}^{n+1}$, which goes roughly as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Write $\tau=T\mathbb{RP}^n$.  The normal bundle $\nu$ has rank 1, so its Steifel-Whitney class is $w(\nu)=1$ or $w(\nu)=1+x$.  In every case, we need $w(\nu)\cdot w(\tau) = w(\nu \oplus \tau) = w(\epsilon^{n+1})=1$.  If $w(\nu)=1$, then $w(\tau)=(1+x)^{n+1}=1$, so $n+1=2^r$.  If $w(\nu)=1+x$, then similarly $(1+x)^{n+2}=1$ so $n+2=2^r$.  &lt;strong&gt;If the immersion is an embedding, the former case must hold.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is this true?  I feel like there should be an easy reason, but none of the people I talked with were able to nail down anything precise.  This could be wrong, but it seems like this is tacitly saying that a codimension-1 embedding of a closed manifold must be in fact of an orientable manifold, which is the same as saying that the the normal line bundle has trivial $w_1$ (since line bundles are totally classified by their orientability, i.e. by $w_1$).  Is this true?&lt;/p&gt;&#xA;" OwnerUserId="2159" LastEditorUserId="2159" LastEditorDisplayName="" LastEditDate="2010-11-14T05:46:38.250" LastActivityDate="2010-11-14T20:23:22.630" Title="Does having a codimension-1 embedding of a closed manifold $M^n \subset \mathbb{R}^{n+1}$ require $M$ to be orientable?" Tags="&lt;differential-topology&gt;" AnswerCount="3" CommentCount="5" FavoriteCount="3" />
  <row Id="10178" PostTypeId="2" ParentId="10173" CreationDate="2010-11-13T21:52:34.700" Score="2" ViewCount="" Body="&lt;p&gt;For the k'th digit (counting left from the unit's place), divide the number by $10^k$ and call the remainder r. Then, $[r/(10^{k-1})]$ gives the k'th digit.&lt;/p&gt;&#xA;" OwnerUserId="3431" LastEditorUserId="3431" LastEditorDisplayName="" LastEditDate="2010-11-14T01:34:17.143" LastActivityDate="2010-11-14T01:34:17.143" />
  <row Id="10179" PostTypeId="2" ParentId="10155" CreationDate="2010-11-13T21:59:40.213" Score="3" ViewCount="" Body="&lt;p&gt;By definition, the hypothenuse is the longest side of a right triangle; so in &lt;em&gt;any&lt;/em&gt; case you could not possibly have a triangle in which the &quot;base&quot; is larger than the hypothenuse as you claim it &quot;should&quot; be here. In any case, &quot;base&quot; depends on orientation (so it is possible for the hypothenuse to &lt;em&gt;be&lt;/em&gt; the base). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your picture is misleading you somehow into thinking that the side opposite the right angle is the side of length $4$; it cannot be. &lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-11-13T21:59:40.213" CommentCount="2" />
  <row Id="10180" PostTypeId="1" AcceptedAnswerId="10195" CreationDate="2010-11-13T22:22:14.820" Score="1" ViewCount="103" Body="&lt;p&gt;I am reading some handwritten notes, and scribbled next to a restriction of scalars functor, are the words &quot;a pullback&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't understand why this might be the case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, consider a field $k$ with a map $\varphi:k\rightarrow A$ for $A$ an associative unital $k$-algebra. Then we consider the induced functor $\varphi_*:A-mod\rightarrow k-mod$ restriction of scalars.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Is $\varphi_*$ the pullback of some diagram in the category of functors from $A$-mod to $k$-mod?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I know this a fairly strange question, but I just cant see what the scribbler had in mind.&lt;/p&gt;&#xA;" OwnerUserId="16" LastEditorUserId="16" LastEditorDisplayName="" LastEditDate="2010-11-13T23:38:40.793" LastActivityDate="2010-11-14T00:12:49.563" Title="Is restriction of scalars a pullback?" Tags="&lt;category-theory&gt;&lt;modules&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="1" />
  <row Id="10181" PostTypeId="1" AcceptedAnswerId="10691" CreationDate="2010-11-13T22:28:25.217" Score="3" ViewCount="240" Body="&lt;p&gt;I'd like to learn formal math. Preferably, though not necessarily, starting with predicate logic/first order logic rather than higher order logic. I am trying to find resources (papers, books etc.) for doing this, but I haven't found anything I really like. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are lots of resources for predicate and first order logic, but most do not approach the topics in a very formal way. For example, many text don't seem to try to define what they mean by &quot;variables&quot; or mention substitution as an important concept. Tries to explicitly describe as many of the rules of the game as it can. Many texts bring up &quot;truth tables&quot; without having formal rules for what you're allowed to do with those tables. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anyone have resources that fit these criteria?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;: many of the answers are good and helpful, but I feel like I should add some clarifying remarks:&#xA;Many texts mention that you can view math as merely manipulation of symbols. I don't doubt that this can be done, but I would like to see it done. A resource that explains the process of producing proofs explicitly in terms of manipulating symbols rather than in terms of functions, statements etc (at least without first defining these terms) would be helpful. I'd like to be able to pretend I was a person who didn't know any math and was just acting as a human computer, producing proofs. I'd like a resource that explains producing proofs like I was such a computer (not necessarily ONLY like that).&lt;/p&gt;&#xA;" OwnerUserId="3480" LastEditorUserId="3480" LastEditorDisplayName="" LastEditDate="2010-11-18T04:40:19.263" LastActivityDate="2010-11-18T04:40:19.263" Title="Resources for learning formal math?" Tags="&lt;logic&gt;&lt;reference-request&gt;&lt;formal-systems&gt;" AnswerCount="4" CommentCount="1" FavoriteCount="4" />
  <row Id="10182" PostTypeId="2" ParentId="10181" CreationDate="2010-11-13T22:44:56.813" Score="1" ViewCount="" Body="&lt;p&gt;Aside from general wikipedia and searches from google I would advise going to either the public or collegiate library. When I wanted to learn more about that I got &quot;Introduction to Mathematical Structures and Proofs&quot; by Larry J. Gerstein. It was in a shelf of free books outside my college's library. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In regard to the truth tables they should have. You will see they are generated from the ideas of Boolean algebra. In fact, &lt;a href=&quot;http://en.wikipedia.org/wiki/Truth_tables&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Truth_tables&lt;/a&gt; explains it quite well. I believe the necessary part is to do problems, and for that all you need is to go through definitions and see their consequences. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for texts for important concepts you may wish to pick up a book on the philosophy of mathematics, though you can see simply about variables from &lt;a href=&quot;http://en.wikipedia.org/wiki/Variable_%28mathematics%29&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Variable_%28mathematics%29&lt;/a&gt;. Some can be rather dense, I would suggest an introduction. I am reading &quot;The Philosophy of Mathematics&quot; by W.D. Hart. It is a collection of short essays pretty much on important philosophical issues and notions in mathematics. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Have fun. &lt;/p&gt;&#xA;" OwnerUserId="3417" LastActivityDate="2010-11-13T22:44:56.813" />
  <row Id="10183" PostTypeId="1" CreationDate="2010-11-13T22:54:35.183" Score="0" ViewCount="129" Body="&lt;p&gt;Hello,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have problem calculating the following summation :&#xA;$$S = \sum_{j=1}^{k^2-1} \lfloor \sqrt{j}\rfloor.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as I understand the mean of that summation it will be something like $$1+1+1+2+2+2+2+2+3+3+3+3+3+3+3+\cdots$$&#xD;&#xA;and I suspect that the last summation number will be $(k-1)^2$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I really can find the pattern of the equal simpler summation.   &lt;/p&gt;&#xA;" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-13T23:04:36.577" LastActivityDate="2010-11-14T20:21:25.753" Title="Summation problem" Tags="&lt;sequences-and-series&gt;" AnswerCount="4" CommentCount="0" />
  <row Id="10184" PostTypeId="2" ParentId="10183" CreationDate="2010-11-13T23:00:48.240" Score="1" ViewCount="" Body="&lt;p&gt;Try replacing it by a sum of the form $\sum_1^k k \cdot c_k$, where $c_k$ is the number of times that $\lfloor \sqrt{j} \rfloor = k$.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-13T23:00:48.240" />
  <row Id="10185" PostTypeId="2" ParentId="10183" CreationDate="2010-11-13T23:04:20.587" Score="6" ViewCount="" Body="&lt;p&gt;Since the last value for $j$ is $k^2-1$, none of the terms of the sum are $k$; they are all between $1$ and $k-1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How many $1$'s will be in the sum? Well, we'll get $1$ when $j$ is any number between $1^2$ and $2^2-1$; then we'll get $2$ for each number between $2^2$ and $3^2-1$. Then we'll get $3$ for each number between $3^2$ and $4^2-1$. Etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, if $n\leq k-1$, how many times does it show up in the sum? It shows up exactly the number of times that there are integers between $n^2$ and $(n+1)^2-1$, inclusively. This is&#xA;$$(n+1)^2 - n^2 = n^2 + 2n + 1 - n^2 = 2n+1.$$&#xD;&#xA;So your sum has $2(1)+1 = 3$ ones; $2(2)+1 = 5$ twos; $2(3)+1=7$ threes; etc. Up to $k-1$, which appears exactly $2(k-1)+1 = 2k-1$ times.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we get that&#xA;$$S = \sum_{r=1}^{k-1} r(2r+1) = 2\left(\sum_{r=1}^{k-1}r^2\right) + \sum_{r=1}^{k-1}r.$$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-11-13T23:04:20.587" CommentCount="1" />
  <row Id="10186" PostTypeId="1" AcceptedAnswerId="10193" CreationDate="2010-11-13T23:05:46.813" Score="3" ViewCount="222" Body="&lt;p&gt;So I found this&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\prod_{k=3}^{\infty} 1 - \tan( \pi/2^k)^4$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://people.missouristate.edu/lesreid/Advanced.html&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have only ever done tests for convergence of infinite sums.&#xA;At &lt;a href=&quot;http://en.wikipedia.org/wiki/Infinite_product&quot; rel=&quot;nofollow&quot;&gt;this link&lt;/a&gt; it shows a way to convert but&#xA;in this case an is less than one for all k. I can see that $ f(k) = 1 - \tan( \pi/2^k)^4 $ is a strictly increasing function and in fact it is increasing very quickly however it is bound [ $ 1 - \tan( \pi/2^3)^4$, 1) where the lower bound is quite nearly one. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This seems to mean it will converge to something slightly less than the lower bound of the function. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I see that is converges. I want to know if there analytic method to provide a closed from of the solution such as with an alternating series. I figure I can simply do it computationally with MatLab and it would be very simple but I thought there must another way.&lt;/p&gt;&#xA;" OwnerUserId="3417" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-12-17T01:45:01.407" LastActivityDate="2010-12-17T01:45:01.407" Title="Finding a closed form for an infinite product" Tags="&lt;calculus&gt;&lt;trigonometry&gt;" AnswerCount="3" CommentCount="3" />
  <row Id="10187" PostTypeId="2" ParentId="10177" CreationDate="2010-11-13T23:07:26.960" Score="6" ViewCount="" Body="&lt;p&gt;Any compact, connected manifold $M$ of dimension $n$ embedded in $\mathbb{R}^{n+1}$ is orientable.&#xA;This follows from the fact that there exists a smooth function $f$ on $\mathbb{R}^{n+1}$ s.t. $M$ is the zero locus of $f$ and the derivative of $f$ does not vanish on $M$ (so the gradient of $f$ gives a smooth normal vector field).&#xA;This also proves Jordan-Brouwer's theorem ($M$ cuts $\mathbb{R}^{n+1}$ in two parts: $f&amp;gt;0$ and $f&amp;lt;0$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The existence of $f$ is a &quot;technical lemma&quot; ;)&#xA;First, using compactness, you can show that there is an $\epsilon &amp;gt; 0$ and a covering $\left( V_x \right)_{x \in M}$ of $M$ such that $B=\left\{ y \in \mathbb{R}^{n+1}\ |\ d(y,M) \lt \epsilon \right\}$ is the disjoint union of the $\left\{ x + t n_x \ |\ -\epsilon \lt t \lt \epsilon \right\}$ ($n_x$ being a vector of norm one orthogonal to $T_x M$), and so $B$ is locally $V_x \times ]-\epsilon,\epsilon[$.&#xA;From there you can deduce that $\mathbb{R^{n+1}}$ has an open covering $\left(A_i\right)_i$ with smooth functions $f_i$ on $A_i$ s.t. whenever $A_i \cap A_j \neq \emptyset$, $f_i = \pm f_j$ locally on $A_i \cap A_j$, and $\cup_i f_i^{-1}(0) = M$ has empty interior (for this you need to take a smooth  non-decreasing function $\lambda : \mathbb{R} \rightarrow \mathbb{R}$ s.t. $f(x)=-1$ when $x \lt -\epsilon/2$, $f(x)=1$ when $x \gt \epsilon/2$, and $f$ is increasing inbetween. then for $(y,t) \in V_x \times ]-\epsilon,\epsilon[$, define $f_i(y,t)=\lambda(t)$, and take the $f_i$s to be $1$ outside the $\epsilon/2$-neighbourhood of $M$).&#xA;This, together with the simple connectedness of $\mathbb{R}^{n+1}$, gives a smooth function $f$ on $\mathbb{R}^{n+1}$ locally equal to the $\pm f_i$, and you are done.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now you have a long exercise to solve your technical problem ;)&#xA;I'm sorry I couldn't just give you a reference, but the only one I have (and from which I took the above sketch of proof) is a french book: Thèmes d'analyse pour l'agrégation, Calcul différentiel by Stéphane Gonnord and Nicolas Tosel (p. 100).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: actually they give a reference: Elon L. Lima, The Jordan-Brouwer separation theorem for smooth hypersurfaces, American Mathematical Monthly, Volume 95 Issue 1, Jan. 1988&lt;/p&gt;&#xA;" OwnerUserId="2660" LastEditorUserId="2660" LastEditorDisplayName="" LastEditDate="2010-11-13T23:22:58.433" LastActivityDate="2010-11-13T23:22:58.433" CommentCount="16" />
  <row Id="10188" PostTypeId="2" ParentId="10124" CreationDate="2010-11-13T23:10:45.117" Score="0" ViewCount="" Body="&lt;p&gt;You can calculate the exact probability of drawing $t$ unique vertices using inclusion-exclusion. We can estimate this probability as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\binom{n}{t} \binom{\binom{t}{2}}{k}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but you're overcounting instances when less than $ t $ vertices actually appear; you can fix that using inclusion exclusion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given that, you can try to calculate the expectation directly.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;You can also try to work your way from Moron's formula. His formula gives the expectation when &lt;i&gt;k&lt;/i&gt; edges are drawn with replacement. We know that distribution of the number of distinct edges drawn, so we can express the expectation when &lt;i&gt;k&lt;/i&gt; edges are drawn with replacement using the expectations when $k_0$ unique edges are drawn for $k_0 \leq k$. We get linear equations and maybe we can solve them to find the probability when &lt;i&gt;k&lt;/i&gt; unique edges are drawn.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-13T23:10:45.117" />
  <row Id="10189" PostTypeId="2" ParentId="10186" CreationDate="2010-11-13T23:13:39.607" Score="3" ViewCount="" Body="&lt;p&gt;Convert the product to a sum by taking logarithms. For $x$ small, $\tan x \approx x$ and $\log (1 - x) \approx -x$ and so after taking logs, your product is roughly like the convergent geometric series $\sum \pi 2^{-k}$. So it converges.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-13T23:13:39.607" />
  <row Id="10190" PostTypeId="2" ParentId="10183" CreationDate="2010-11-13T23:14:46.547" Score="2" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\ $ Your displayed sum has $\rm\ (2^2 - 1^2)\ \ 1&amp;#39;s,\ \ (3^2-2^2)\ \ 2&amp;#39;s,\ \ (4^2-3^2)\ \ 3&amp;#39;s,\ \ldots $&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-11-13T23:14:46.547" />
  <row Id="10191" PostTypeId="2" ParentId="10186" CreationDate="2010-11-13T23:15:06.460" Score="6" ViewCount="" Body="&lt;p&gt;First of all I do not understand why you say the product converges to 1. In fact the product must me strictly less than the first term namely, $(1-\tan^4(\frac{\pi}{8}))$. You can however say this function converges since $\displaystyle \sum_{k=1} \tan^4(\frac{\pi}{2^k}) &amp;lt; \infty$. Hence the product converges absolutely.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Note: $\displaystyle \Pi_{k=1}^n (1+a_k) \leq \exp(\displaystyle \sum_{k=1}^n a_k)$. So convergence of $\displaystyle \sum_{k=1}^{\infty} a_k$ implies the convergence of the infinite product $\displaystyle \Pi_{k=1}^{\infty} (1+a_k)$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried to find an approximate answer using Matlab and then make an educated guess about the limit and then try to prove it. However, the limit of the first $10^7$ terms give me $0.968946146259369$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am unable to make an educated guess (something based on $\pi$,$e$ or the golden ratio). It will be interesting to see what this product converges to!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\textbf{EDIT:}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is just for the sake of completion and the solution is due to $\textbf{Moron}$. Following the idea suggested by Moron, and since we know the product has to converge let&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$p_n = \displaystyle \Pi_{k=1}^n (1-\tan^4(\frac{\pi}{2^{k+2}})) = \displaystyle \Pi_{k=1}^n \frac{\cos(\frac{\pi}{2^{k+1}})}{\cos^4(\frac{\pi}{2^{k+2}})} = \frac{\cos(\frac{\pi}{4})}{\cos(\frac{\pi}{2^{n+2}})} \frac{1}{\displaystyle \Pi_{k=1}^n \cos^3(\frac{\pi}{2^{k+2}})}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$p_n = \frac{\cos(\frac{\pi}{4})}{\cos(\frac{\pi}{2^{n+2}})} \frac{8^n \times \sin^3(\frac{\pi}{2^{n+2}})}{\sin^3(\frac{\pi}{4})}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take the limit as $n \rightarrow \infty$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, $p = \frac{\cos(\frac{\pi}{4})}{1} \times \frac{(\frac{\pi}{4})^3}{\sin^3(\frac{\pi}{4})}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, $p = 2 \times (\frac{\pi}{4})^3 = \frac{\pi^3}{32} \approx 0.968946146259369$&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2010-11-14T01:35:46.243" LastActivityDate="2010-11-14T01:35:46.243" CommentCount="3" />
  <row Id="10192" PostTypeId="2" ParentId="10164" CreationDate="2010-11-13T23:37:11.247" Score="3" ViewCount="" Body="&lt;p&gt;By change of variable $x=y+t$, we need to bound the expression $$(\frac{|y+t|}{1+|y|})^k |g(t)|.$$ Since $$\frac{|y+t|}{1+|y|}\le \frac{|y|+|t|}{1+|y|}\le 1+|t|,$$ and &#xA;$$\frac{(1+|t|)^k}{1+t^{2k}}$$ and $$(1+t^{2k})|g(t)|$$ are bounded, say by $B_k$ and $C_k$, we find that&#xA;$$(\frac{|y+t|}{1+|y|})^k |g(t)|\le (1+|t|)^k|g(t)|\le B_kC_k.$$&lt;/p&gt;&#xA;" OwnerUserId="3249" LastActivityDate="2010-11-13T23:37:11.247" CommentCount="3" />
  <row Id="10193" PostTypeId="2" ParentId="10186" CreationDate="2010-11-13T23:38:08.017" Score="8" ViewCount="" Body="&lt;p&gt;Write $\displaystyle 1- \tan^4 \theta$ as $\dfrac{\cos^2 \theta - \sin^2 \theta}{\cos^4 \theta} = \dfrac{\cos 2\theta}{\cos^4 \theta}$ and make repeated use of the following trick:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To evaluate $\displaystyle \prod_{k=1}^{n}\cos \dfrac{\theta}{2^k}$, multiply and divide by $\sin \dfrac{\theta}{2^n}$ and use $\displaystyle 2\sin \theta \cos \theta = \sin 2\theta$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This should give you a closed form formula for the product of first $\displaystyle n$ terms, which I believe can be evaluated easily as $\displaystyle n \to \infty$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For more information on this you can refer: &lt;a href=&quot;http://en.wikipedia.org/wiki/Vi%C3%A8te%27s_formula&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Vi%C3%A8te's_formula&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-11-13T23:50:11.557" LastActivityDate="2010-11-13T23:50:11.557" CommentCount="3" />
  <row Id="10194" PostTypeId="1" AcceptedAnswerId="10196" CreationDate="2010-11-14T00:04:29.133" Score="1" ViewCount="96" Body="&lt;p&gt;Suppose $64$ students are taking an exam. The exam has $16$ questions in multiple choice format. Students must fill an oval for every one of the $16$ questions. They can't revise their answers. The time taken for a students to answer a question is exponentially distributed with mean $8$ minutes. So $f(t) = \frac{1}{8}e^{-t/8}$. The times for answering questions are independent. Using the normal approximation calculate the probability that the average time taken by the students is more than $132$ minutes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we want to find $P[T_1+T_2 + \cdots + T_{64} &amp;gt; 132(64)]$. What would be the distribution of a sum of exponential random variables? Would it be a gamma distribution? If we let $S = T_{1} + T_{2} + \cdots + T_{64}$, then $P[S &amp;gt; 132(64)] = 1-P[S \leq 132(64)]$. &lt;/p&gt;&#xA;" OwnerUserId="2548" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-11-14T00:25:47.663" LastActivityDate="2010-11-14T00:40:18.057" Title="Questions on a Test" Tags="&lt;probability&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="10195" PostTypeId="2" ParentId="10180" CreationDate="2010-11-14T00:12:49.563" Score="2" ViewCount="" Body="&lt;p&gt;The sense in which &quot;pullback&quot; is being used here is the same sense it's being used in &lt;a href=&quot;http://en.wikipedia.org/wiki/Pullback&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; Wikipedia article; that is, it's being used to refer to the process of &quot;pulling back&quot; a morphism $A \to \text{End}(R)$ (an $A$-module) along the morphism $k \to A$ to get a morphism $k \to \text{End}(R)$ (a $k$-module) by precomposition.  This is different from the categorical pullback.  I don't know why they're named the same.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-11-14T00:12:49.563" />
  <row Id="10196" PostTypeId="2" ParentId="10194" CreationDate="2010-11-14T00:24:11.923" Score="3" ViewCount="" Body="&lt;p&gt;Yes. It will be a Gamma distribution. But here as the problem says you can approximate it by a normal distribution. If you look at the variable $T = \displaystyle \sum_{k=1}^{64} T_i$, it has mean $64 \times 16 \times 8$ and the variance is again $64 \times 16 \times 8^2$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Note that $T_i$ has mean $16 \times 8$ and has variance $16 \times 8^2$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence all you need now is to look at the random variable $\frac{T-64 \times 16 \times 8}{\sqrt{64 \times 16 \times 8^2}}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, $P(T \geq 132 \times 64) = P(\frac{T-64 \times 16 \times 8}{32 \times 8} \geq \frac{132 \times 64 - 128 \times 64}{256}) \approx  P(Z \geq 1) = 0.158655$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $Z$ is the standard normal random variable with zero mean and unit variance. Look up the table for this value.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Also, Kindly check my arithmetic calculations)&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2010-11-14T00:40:18.057" LastActivityDate="2010-11-14T00:40:18.057" CommentCount="0" />
  <row Id="10197" PostTypeId="2" ParentId="10194" CreationDate="2010-11-14T00:26:41.887" Score="1" ViewCount="" Body="&lt;p&gt;The sum of i.i.d exponential random variables follows the Erlang distribution (which is a special case of the Gamma distribution). For details, check out &lt;a href=&quot;http://en.wikipedia.org/wiki/Erlang_distribution&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Erlang_distribution&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="2641" LastActivityDate="2010-11-14T00:26:41.887" />
  <row Id="10198" PostTypeId="1" AcceptedAnswerId="10199" CreationDate="2010-11-14T00:53:39.297" Score="1" ViewCount="40" Body="&lt;p&gt;Suppose $f(x,y)$ is bivariate normal. You are given $E(X)$, $E(Y)$, $\text{Var}(X)$ and $\text{Var}(Y)$. You are also given $\rho$ the correlation coefficient. How would one find $\text{Var}(Y|X=k)$? Would you use &lt;a href=&quot;http://math.stackexchange.com/questions/7040/conditional-and-total-variance&quot;&gt;the total law of variance&lt;/a&gt;? We know that $\rho = \frac{\text{Cov}(X,Y)}{\sigma_x \sigma_y}$. &lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2010-11-14T00:59:34.160" Title="Bivariate Normal Distribution" Tags="&lt;probability&gt;" AnswerCount="1" />
  <row Id="10199" PostTypeId="2" ParentId="10198" CreationDate="2010-11-14T00:59:34.160" Score="1" ViewCount="" Body="&lt;p&gt;Check the &lt;a href=&quot;http://en.wikipedia.org/wiki/Multivariate_normal_distribution#Conditional_distributions&quot; rel=&quot;nofollow&quot;&gt;Wikipedia article&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-14T00:59:34.160" />
  <row Id="10200" PostTypeId="2" ParentId="10164" CreationDate="2010-11-14T01:24:10.357" Score="1" ViewCount="" Body="&lt;p&gt;Trying to be systematic here... your condition is equivalent to saying $|g(x)| \leq C_n{1 \over 1 + |x|^n}$ for all $n$. So if your inequality is going to be true, we'd expect for some $l$ depending on $k$ we would have the following for all $(x,y)$:&#xA;$${|x|^k \over 1 + |y|^k}{1 \over 1 + |x - y|^l} \leq C_k$$&#xD;&#xA;And if we can show this then your inequality follows. A natural place to start would be choosing $l = k$, in which case what is needed is&#xA;$$|x|^k \leq C_k (1 + |y|^k)(1 + |x - y|^k)$$&#xD;&#xA;Notice though that since $x = y + (x-y)$, $|x| \leq |y| + |x-y|$ and thus $|x|^k \leq A_k(|y|^k + |x - y|^k)$ by an easy argument. These two terms are part of the right hand side of the above, so that's all you need.&lt;/p&gt;&#xA;" OwnerUserId="3035" LastActivityDate="2010-11-14T01:24:10.357" CommentCount="1" />
  <row Id="10201" PostTypeId="1" AcceptedAnswerId="10206" CreationDate="2010-11-14T02:01:23.647" Score="0" ViewCount="89" Body="&lt;p&gt;How to prove that the power &lt;strong&gt;b ^ n&lt;/strong&gt; is an irrational number, where &lt;strong&gt;b&lt;/strong&gt; is an integer greater than 1 and &lt;strong&gt;n&lt;/strong&gt; is an irrational number?&lt;/p&gt;&#xA;" OwnerUserId="2938" LastActivityDate="2010-11-14T03:11:57.327" Title="b^n is irrational, where b&gt;1 is a integer and n is an irrational" Tags="&lt;number-theory&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="10202" PostTypeId="1" AcceptedAnswerId="10205" CreationDate="2010-11-14T02:31:37.860" Score="6" ViewCount="113" Body="&lt;p&gt;Suppose I have a 2-variable function $g(k,n)$ and I know that $g(k,n)=O(n^{f(k)})$, for fixed $k$ as $n \rightarrow \infty$, for some function $f=f(k)$.  Suppose I also know that $f(k)=O(\log k)$.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Is it legitimate to write $g(k,n)=O(n^{O(\log k)})$?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="139" LastEditorUserId="139" LastEditorDisplayName="" LastEditDate="2010-11-14T02:41:37.707" LastActivityDate="2010-11-24T22:21:22.187" Title="Is it legitimate to write nested big-Os in an asymptotic formula for a multivariable function?" Tags="&lt;asymptotics&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="1" />
  <row Id="10203" PostTypeId="1" AcceptedAnswerId="10214" CreationDate="2010-11-14T02:37:28.513" Score="2" ViewCount="57" Body="&lt;p&gt;I'm dealing with tables that show connectivity within a network. Can anyone point me toward a technique for ordering the rows and columns so that nodes that are connected tend to be close to each other in the table?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, if I had a network like this (connections are directional), the structure isn't immediately obvious; it just looks random.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;  1 2 3 4 5&#xA;1     x   x&#xA;2   x   &#xA;3     x  x&#xA;4   x  x &#xA;5        x &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;But if you re-order the columns and rows this way, the structure is clear; you can see some   organisation in the network.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;  2 4 5 3 1&#xA;2 x    &#xA;4 x x   &#xA;5   x   &#xA;3     x x &#xA;1     x x &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Of course, I'm dealing with larger tables, with ~1024 nodes. I could just code up something that tries to find the arrangement with the minimal average distance between connected nodes by some general-purpose method like gradient descent. But perhaps there's a technique that's particularly suited to this type of problem?&lt;/p&gt;&#xA;" OwnerUserId="3483" LastActivityDate="2010-11-14T05:29:03.673" Title="Technique for &quot;untangling&quot; directed networks?" Tags="&lt;analysis&gt;&lt;matrices&gt;" AnswerCount="1" CommentCount="7" />
  <row Id="10204" PostTypeId="2" ParentId="10202" CreationDate="2010-11-14T02:38:34.020" Score="2" ViewCount="" Body="&lt;p&gt;It is legitimate provided you interpret it in the right sense. However, I do not understand what purpose it will serve though, since (obviously) you cannot write that it is equal to $\mathcal{O}(n^{log(k)})$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\textbf{EDIT:}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was reading a blog by TerryTao and he has used this notation. So I assume, you can use it provided you give the right interpretation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://terrytao.wordpress.com/2010/11/20/the-guth-katz-bound-on-the-erdos-distance-problem/&quot; rel=&quot;nofollow&quot;&gt;http://terrytao.wordpress.com/2010/11/20/the-guth-katz-bound-on-the-erdos-distance-problem/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2010-11-24T22:21:22.187" LastActivityDate="2010-11-24T22:21:22.187" CommentCount="1" />
  <row Id="10205" PostTypeId="2" ParentId="10202" CreationDate="2010-11-14T02:50:20.440" Score="3" ViewCount="" Body="&lt;p&gt;In this particular case, you can simplify and conclude that $g(k,n) = n^{O(\log k)}$, since you can put fold the first constant into the second (as long as $n,k \geq 1 + \epsilon$). This fails if you think of $n$ as tending to infinity, but $k$ could be arbitrarily close to $1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: My answer assumes that the constant in $g(k,n) = O(n^{f(k)})$ doesn't depend on &lt;i&gt;k&lt;/i&gt;, which is the usual convention in computer science. If it does, we'd write $g(k,n) = O_k(n^{f(k)})$ to emphasize this (or sometimes just emphasize it in words).&lt;/p&gt;&#xA;" OwnerUserId="1277" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2010-11-14T03:17:33.953" LastActivityDate="2010-11-14T03:17:33.953" CommentCount="3" />
  <row Id="10206" PostTypeId="2" ParentId="10201" CreationDate="2010-11-14T03:11:57.327" Score="3" ViewCount="" Body="&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Gelfond%E2%80%93Schneider_theorem&quot; rel=&quot;nofollow&quot;&gt;Gelfond-Schneider theorem&lt;/a&gt; shows that if you assume that $n$ is algebraic, then $b^n$ is indeed irrational, in fact transcendental! You can even take $b$ to be algebraic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we can ask whether there's an easy proof that $ b^n $ is irrational for &lt;i&gt;b&lt;/i&gt; integral and &lt;i&gt;n&lt;/i&gt; an &lt;i&gt;algebraic&lt;/i&gt; irrational.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-14T03:11:57.327" />
  <row Id="10207" PostTypeId="2" ParentId="10202" CreationDate="2010-11-14T03:12:55.517" Score="2" ViewCount="" Body="&lt;p&gt;&lt;strike&gt;No (if you mean what I think you mean by that statement).&lt;/strike&gt;  If you have more than one variable in play it's always a good idea to keep track of which variables the implied constant for each big-O depends on.  &lt;strike&gt;In your example, the implied constant for the outer big-O depends on $k$, but the implied constant for the inner big-O doesn't, so it seems misleading to use notation which pretends they are the same.  And if the implied constant grows fast enough, the final statement you want is false (if an unadormed big-O means it doesn't depend on any of the variables which appear on the LHS).  Consider, as I said in my comment to Yuval Filmus, $g(k, n) = 2^k n^{\log k}, f(k) = \log k$.  In addition to $f(k) = O(\log k)$ you need to know that the implied constant doesn't depend on $n$.&lt;/strike&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt;  The above was nonsense.  Actually you are fine.  You have a constant $C_1$ such that $f(k) \le C_1 \log k$ and another constant $C_2$ such that $g(k) \le C_2 n^{f(k)}$, hence $g(k) \le C_2 n^{C_1 \log k}$, or as Yuval Filmus points out, $g(k) = n^{O(\log k)}$.  &lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-11-14T12:23:28.557" LastActivityDate="2010-11-14T12:23:28.557" CommentCount="1" />
  <row Id="10208" PostTypeId="2" ParentId="10119" CreationDate="2010-11-14T03:41:15.440" Score="1" ViewCount="" Body="&lt;p&gt;a^3 - b^3 = (a - b)(a^2 + ab + b^2). So x^3 - 8 = (x - 2)(x^2 + 2x + 4).&lt;br&gt;&#xA;For x^3 - 3x^2 + x - 3, factor by grouping the first two terms together and taking out the GCF and grouping the second two and doing the same. This should give you&lt;br&gt;&#xA;x^2(x - 3) + 1(x - 3). This gives you a greatest common factor of (x - 3) that you can now factor out.&lt;br&gt;&#xA;(x - 3)(x^2 + 1). And you are done.&lt;/p&gt;&#xA;" OwnerUserId="134" LastActivityDate="2010-11-14T03:41:15.440" />
  <row Id="10209" PostTypeId="1" CreationDate="2010-11-14T03:57:25.963" Score="6" ViewCount="421" Body="&lt;p&gt;I am going to a Computer Science Course in University next year. I heard that Discrete Mathematics is whats required for Comp Sci so, I am looking for resources/books that I can read to get started with. I picked up &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0071615865&quot; rel=&quot;nofollow&quot;&gt;Schaum's Outline of Discrete Mathematics&lt;/a&gt;. I find that some things are alittle hard to understand still. I have also looked at some videos from &lt;a href=&quot;http://www.youtube.com/user/khanacademy?blend=1&amp;amp;ob=4&quot; rel=&quot;nofollow&quot;&gt;Khan Academy on YouTube&lt;/a&gt;, I find the videos easy to understand. Except that I don't really know where am I supposed to start 1st. I am on algebra now, but can someone tell me whats the sequence I should go from here, eg. topics/playlists?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An you know, for maths, there must be alot of practice. So where can I find lots of practice questions? Schaum's Outline of Discrete Mathematics has quite little practice questions. &lt;/p&gt;&#xA;" OwnerUserId="2006" LastEditorUserId="7810" LastEditorDisplayName="" LastEditDate="2011-04-06T02:47:28.590" LastActivityDate="2011-04-06T02:47:28.590" Title="Resources/Books for Discrete Mathematics" Tags="&lt;discrete-mathematics&gt;&lt;big-list&gt;&lt;books&gt;&lt;online-resources&gt;" AnswerCount="6" CommentCount="3" FavoriteCount="7" />
  <row Id="10210" PostTypeId="2" ParentId="10209" CreationDate="2010-11-14T03:59:40.220" Score="0" ViewCount="" Body="&lt;p&gt;&quot;Algorithm Design&quot; by Authors Kleinberg and Tardos is a good comprehensive one! There is a good amount of theory and also a good number of problems at the end of each chapter.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastActivityDate="2010-11-14T03:59:40.220" />
  <row Id="10211" PostTypeId="2" ParentId="10209" CreationDate="2010-11-14T04:07:30.940" Score="5" ViewCount="" Body="&lt;p&gt;I think Kenneth Rosen's book on Discrete Mathematics does a good job of being exhaustive. Although I have heard mixed reviews of the book from others, I myself, began studying discrete mathematics from this book and was very happy with it. &lt;/p&gt;&#xA;" OwnerUserId="3431" LastActivityDate="2010-11-14T04:07:30.940" />
  <row Id="10212" PostTypeId="1" AcceptedAnswerId="10215" CreationDate="2010-11-14T04:29:36.513" Score="8" ViewCount="165" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Abel%27s_identity&quot; rel=&quot;nofollow&quot;&gt;Abel's identity&lt;/a&gt; states that if $X(t)$ and $A(t)$ are $n\times n$ matrix-valued functions such that $X&amp;#39;(t)=A(t)X(t)$, then $\frac{d}{dt}(\det X(t)) = \mathrm{tr}\,A(t) \cdot \det X(t)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is whether there's a nice high-brow basis-independent way to see this. Given that you can state the problem without ever referring to matrix entries, I want prove it without ever referring to matrix entries. I'd expect the identity $\det e^A = e^{\mathrm{tr}(A)}$ to play a central role in the proof, but I have yet to come up with such an argument.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;You can prove this without too much trouble in a bare-hands way, but I don't see how to turn this into a basis-free proof. Suppose $X_i(t)$ is the matrix you get from $X(t)$ by taking the derivative of every entry in the $i$-th row. Then $\frac{d}{dt}(\det X(t)) = \sum_{i=1}^n \det(X_i(t))$. Using the relation $X&amp;#39;(t)=A(t)X(t)$ you can express the derivative of the $i$-th row of $X&amp;#39;(t)$ in terms of the entries of $A$ and $X$. When you do this, you find that $\det X_i(t)$ is simply $\det X(t)$ multiplied by the $(i,i)$-th entry of $A(t)$.&lt;/p&gt;&#xA;" OwnerUserId="60" LastActivityDate="2010-11-14T06:31:52.377" Title="Is there a basis-independent proof of Abel's identity?" Tags="&lt;differential-equations&gt;&lt;matrices&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="10213" PostTypeId="2" ParentId="10124" CreationDate="2010-11-14T05:11:31.547" Score="6" ViewCount="" Body="&lt;p&gt;For choosing without replacement, here is an exact answer.  Assuming $n \geq 2$, so that there is at least one pair, and $1 \leq k \leq \binom{n}{2}$, so that you're choosing at least one pair but not more than the total number of pairs, the expected value is  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$n - \left(\frac{n^2 - 3n - 2k + 4}{n-1}\right) \frac{\binom{\binom{n}{2} - n + 1}{k-1}}{\binom{\binom{n}{2} - 1}{k-1}}.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can assume that we are choosing pairs in order.  Let $X_k$ be the number of distinct items from $S$ through $k$ pairs.  Let $Y_i$ be the number of items in the $i$th pair that did not appear in any of the previous pairs.  So $X_k = \sum_{i=1}^k Y_i$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, $Y_i$ is either 0, 1, or 2.  Since there are $\binom{n}{2} - n + 1$ pairs that do not contain a given item and $\binom{n}{2} - 2n + 3$ pairs that do not contain either of two given items, we have&#xA;$$P(Y_i = 1) = \frac{\binom{\binom{n}{2} - n + 1}{i-1} + \binom{\binom{n}{2} - n + 1}{i-1} - 2 \binom{\binom{n}{2} - 2n + 3}{i-1}}{\binom{\binom{n}{2} - 1}{i-1}}$$&#xD;&#xA;and&#xA;$$P(Y_i = 2) = \frac{\binom{\binom{n}{2} - 2n + 3}{i-1}}{\binom{\binom{n}{2} - 1}{i-1}}.$$&#xD;&#xA;Thus &#xA;$$E[Y_i] = 2\frac{\binom{\binom{n}{2} - n + 1}{i-1}}{\binom{\binom{n}{2} - 1}{i-1}}.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It can be proved by induction that &#xA;$$\sum_{i=0}^k \frac{\binom{M}{i}}{\binom{N}{i}} = \frac{(N+1)\binom{N}{k} - (M-k)\binom{M}{k}}{(N+1-M)\binom{N}{k}}.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus &#xA;$$E[X_k] = \sum_{i=1}^k E[Y_i] = 2\sum_{i=1}^k \frac{\binom{\binom{n}{2} - n + 1}{i-1}}{\binom{\binom{n}{2} - 1}{i-1}} $$ &#xA;$$= 2\frac{(\frac{n(n-1)}{2}-1+1)\binom{\binom{n}{2}}{k-1} - (\frac{n(n-1)}{2} - n + 1 - k + 1)\binom{\binom{n}{2} - n + 1}{k-1}}{(\binom{n}{2} - 1+1-\binom{n}{2} + n - 1)\binom{\binom{n}{2} - 1}{k-1}}$$&#xD;&#xA;$$= \frac{n(n-1)\binom{\binom{n}{2}}{k-1} - (n^2 - 3n - 2k + 4)\binom{\binom{n}{2} - n + 1}{k-1}}{(n - 1)\binom{\binom{n}{2} - 1}{k-1}}$$&#xD;&#xA;$$= n -\frac{(n^2 - 3n - 2k + 4)\binom{\binom{n}{2} - n + 1}{k-1}}{(n - 1)\binom{\binom{n}{2} - 1}{k-1}}.$$&lt;/p&gt;&#xA;" OwnerUserId="2370" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-11-14T16:14:21.260" LastActivityDate="2010-11-14T16:14:21.260" CommentCount="4" />
  <row Id="10214" PostTypeId="2" ParentId="10203" CreationDate="2010-11-14T05:29:03.673" Score="1" ViewCount="" Body="&lt;p&gt;Rahul's link to the Wikipedia article on topological sorting led me to &lt;strong&gt;tsort&lt;/strong&gt;, which is a Unix command to do topological sorting. The networks I'm dealing with might have some cycles, but it seems that &lt;strong&gt;tsort&lt;/strong&gt; can deal with cycles gracefully, and report the cycles to stderr. It looks like that will do the job I need. Thank you, Rahul!&lt;/p&gt;&#xA;" OwnerUserId="3483" LastActivityDate="2010-11-14T05:29:03.673" />
  <row Id="10215" PostTypeId="2" ParentId="10212" CreationDate="2010-11-14T05:31:04.903" Score="5" ViewCount="" Body="&lt;p&gt;Please forgive this argument that plays rather fast and loose with infinitesimals. If you ignore second-order terms, an infinitesimal change $\delta t$ in $t$ takes $X(t)$ to $\big(I + \delta t A(t)\big) X(t)$. Since the determinant distributes over multiplication, all you need to show is that $\det\big(I + \delta t A(t)\big)$ is $1 + \delta t \operatorname{tr}A$ for infinitesimal $\delta t $. More formally, you need $\frac{d}{d\tau} \det(I + \tau A) = \operatorname{tr}A$. Since $I + \tau A$ agrees with $e^{\tau A}$ to first order (according to the Taylor series), you can differentiate the identity $\det e^{\tau A} = e^{\operatorname{tr} \tau A}$ with respect to $\tau$ and get the desired result.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update:&lt;/strong&gt; For a high-brow view, I think it makes much more sense to cast this in a geometric light, as it has much to do with &lt;a href=&quot;http://mathoverflow.net/questions/13526/geometric-interpretation-of-trace/13530#13530&quot; rel=&quot;nofollow&quot;&gt;the geometric interpretation of the trace&lt;/a&gt;. Consider how the vector ODE $x&amp;#39;(t) = A(t) x(t)$ acts on Euclidean $n$-space. The identity $\det e^A = e^{\operatorname{tr}A}$ is a statement of the fact that for constant $A$, the volume of any region grows (or shrinks) exponentially at the rate $\operatorname{tr}A$ under this ODE. (This is fine for varying $A(t)$ too as long as we don't care about second-order effects.) Now think of $X(t) $ as defining a parallelepiped in $n$-space as it flows around under the action of the ODE. The volume of the parallelepiped is $\det X(t)$, and its rate of change is therefore $\operatorname{tr}A $ times that. I think this is the answer I should have given in the first place.&lt;/p&gt;&#xA;" OwnerUserId="856" LastEditorUserId="856" LastEditorDisplayName="" LastEditDate="2010-11-14T06:31:52.377" LastActivityDate="2010-11-14T06:31:52.377" />
  <row Id="10216" PostTypeId="1" CreationDate="2010-11-14T05:39:10.997" Score="2" ViewCount="313" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Companion_matrix&quot; rel=&quot;nofollow&quot;&gt;The companion matrix&lt;/a&gt; of a monic polynomial in 1 variable over a field $F$ plays an important role in understanding the structure of finite dimensional $F[x]$-modules.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is an important fact that the characteristic polynomial and the minimal polynomial of $C(f)$ are both equal to $f$. This can be seen quite easily by induction on the degree of $f$ .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anyone know a different proof of this fact? I would love to see a graph theoretic proof or a non inductive algebraic proof, but I would be happy with anything that makes it seem like more than a coincidence!&lt;/p&gt;&#xA;" OwnerUserId="1725" LastActivityDate="2010-11-14T22:45:51.437" Title="The characteristic and minimal polynomial of a companion matrix" Tags="&lt;combinatorics&gt;&lt;algebra&gt;&lt;reference-request&gt;&lt;commutative-algebra&gt;" AnswerCount="3" />
  <row Id="10217" PostTypeId="2" ParentId="10216" CreationDate="2010-11-14T05:59:13.283" Score="3" ViewCount="" Body="&lt;p&gt;Suppose your matrix is over a field $\mathbb{F}$. Look at $G = F[x]/f$, where $f$ is your polynomial of degree $n$. Then $G$ is a vector space over $\mathbb{F}$, and $C(f)$ is the matrix (with respect to the basis $1,x,x^2,\ldots,x^{n-1}$) corresponding to the linear operator $g \mapsto x \cdot g$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $f = 0$ in $G$, also $fx^i = 0$ in $G$, and so $f$ is a polynomial of degree $n$ such that $f(C(f)) = 0$; thus it must be the characteristic polynomial of $C(f)$. Moreover, any polynomial $g$ of smaller degree does not reduce to $0$ in $G$, so in particular $g(C(f))$ applied to the vector $1$ does not equal the zero vector. So $f$ is the minimal polynomial of $C(f)$.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-14T05:59:13.283" CommentCount="2" />
  <row Id="10218" PostTypeId="1" AcceptedAnswerId="10219" CreationDate="2010-11-14T06:05:19.193" Score="1" ViewCount="63" Body="&lt;p&gt;Say I have an equation where the value is $m-4$ when $m \ge 5, n=3,4$; $n-4$ when $n \ge 5, m=3,4$; and $m+n-9$ when $m,n \ge 5$. How can I combine these all into one statement? How can I combine them with further ones, like $m-6$ where $m \ge 7, n \in [3,6]$, etc?&lt;/p&gt;&#xA;" OwnerUserId="1562" LastEditorUserId="1562" LastEditorDisplayName="" LastEditDate="2010-11-14T06:11:49.940" LastActivityDate="2010-11-14T07:47:47.833" Title="How to combine cases" Tags="&lt;algebra&gt;" AnswerCount="3" CommentCount="7" />
  <row Id="10219" PostTypeId="2" ParentId="10218" CreationDate="2010-11-14T06:25:44.917" Score="1" ViewCount="" Body="&lt;p&gt;I think for the first statement the following formula should work with the given domain:&#xA;$f(m,n)=(m-4)H[m-5]+(n-4)H[n-5]+H[m-5]H[n-5]$&#xD;&#xA;where H is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Heaviside_step_function&quot; rel=&quot;nofollow&quot;&gt;heaviside&lt;/a&gt; step function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the phrasing for the second part can be improved, judging from your comments above. From what I gather you can add terms such as $(m-6)H[m-7]$ for the later conditions.&lt;/p&gt;&#xA;" OwnerUserId="3431" LastEditorUserId="3431" LastEditorDisplayName="" LastEditDate="2010-11-14T06:36:36.417" LastActivityDate="2010-11-14T06:36:36.417" />
  <row Id="10220" PostTypeId="2" ParentId="9265" CreationDate="2010-11-14T06:44:54.127" Score="2" ViewCount="" Body="&lt;p&gt;Note that $\frac{d(f(x) \times y)}{dx} = \frac{d(f(x))}{dx} \times y + f(x) \times  \frac{dy}{dx}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So for instance, if $f(x) = x$, we get $\frac{d(x \times y)}{dx} = \frac{d(x)}{dx} \times y + x \times  \frac{dy}{dx} = 1 \times y + x \times \frac{dy}{dx} = y + x \times \frac{dy}{dx}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, if $f(x) = c$, where $c$ is a constant, then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{d(c \times y)}{dx} = \frac{d(c)}{dx} \times y + c \times \frac{dy}{dx} = 0 \times y + c \times \frac{dy}{dx} = c \times \frac{dy}{dx}$&lt;/p&gt;&#xA;" OwnerUserId="3302" LastActivityDate="2010-11-14T06:44:54.127" />
  <row Id="10221" PostTypeId="2" ParentId="10218" CreationDate="2010-11-14T06:48:27.003" Score="2" ViewCount="" Body="&lt;p&gt;For your first question, it turns out that on the values you've specified, $f(m,n) = \max(m-5,0)+\max(n-5,0) + 1$. In general, if you're trying to get a handle on a function of many variables, it's a good idea to see if it can be separated into a combination of functions of fewer variables.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For your second question, if what you mean is that you want to add $m - 6$ to the output if certain conditions are satisfied, and leave it alone otherwise, you could define $g(m,n) = f(m,n) + h(m,n)$, where $h(m,n)$ is either $m-6$ or zero depending on etc etc. Or you could just write out all the cases of $g(m,n)$, which is the systematic but mind-numbing way.&lt;/p&gt;&#xA;" OwnerUserId="856" LastActivityDate="2010-11-14T06:48:27.003" />
  <row Id="10223" PostTypeId="1" CreationDate="2010-11-14T07:38:57.790" Score="0" ViewCount="86" Body="&lt;p&gt;$\max(f(n), g(n)) = O(f(n) + g(n))$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do I prove this? Also I'd appreciate the markup being corrected, thanks.&lt;/p&gt;&#xA;" OwnerUserId="538" LastEditorUserId="538" LastEditorDisplayName="" LastEditDate="2010-11-14T07:46:41.360" LastActivityDate="2010-11-14T07:46:41.360" Title="Max of functions are in big O of the sum of their individual big Os" Tags="&lt;asymptotics&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="10224" PostTypeId="2" ParentId="10223" CreationDate="2010-11-14T07:43:54.527" Score="2" ViewCount="" Body="&lt;p&gt;You can use $\max(f(n),g(n)) \leq f(n) + g(n)$, given that both functions are non-negative.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-14T07:43:54.527" CommentCount="2" />
  <row Id="10225" PostTypeId="2" ParentId="10218" CreationDate="2010-11-14T07:47:47.833" Score="0" ViewCount="" Body="&lt;p&gt;$$ f(n,m) = \begin{cases} m-4 &amp;amp; m \geq 5,\, n=3,4, \\ n-4 &amp;amp; n \geq 5,\, m=3,4, \\ m+ n - 9 &amp;amp; m,n \geq 5. \end{cases}$$&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-14T07:47:47.833" />
  <row Id="10226" PostTypeId="2" ParentId="10172" CreationDate="2010-11-14T07:55:38.470" Score="1" ViewCount="" Body="&lt;p&gt;Have you tried &lt;a href=&quot;http://www.treewidth.com/&quot; rel=&quot;nofollow&quot;&gt;this website&lt;/a&gt;? They have a small collection of graphs and a library for computing their exact tree-width, which you can compare against your code.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can also try constructing &quot;random&quot; graphs by following the definition, though in that case you only know an upper bound on the tree-width; probably every reasonable way of doing this will actually result in a graph of &quot;most probably&quot; the tree-width you're aiming at.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-14T07:55:38.470" CommentCount="1" />
  <row Id="10227" PostTypeId="2" ParentId="9265" CreationDate="2010-11-14T07:57:46.033" Score="0" ViewCount="" Body="&lt;p&gt;If it's just $y$ alone then you don't need to use implicit differentiation, just differentiate the other side of the equation. If it's $cy$ then you can just divide both sides by $c$.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-14T07:57:46.033" />
  <row Id="10228" PostTypeId="1" CreationDate="2010-11-14T08:33:33.490" Score="0" ViewCount="144" Body="&lt;p&gt;after traversing a wireless error filled channel packets arrive at a network node according to a Poisson process with rate λ. these packets could have corrupted bits due to the wireless channel which has a bit error probability &lt;em&gt;p&lt;/em&gt;. A packet is said to arrive at the node successfully if no bits are in error. the number of bits in a given packet is Poisson distributed with mean μ.Assume packets are not transmitted . find the rate at which successful packets arrive at the network node&lt;/p&gt;&#xA;" OwnerUserId="3490" LastEditorUserId="3490" LastEditorDisplayName="" LastEditDate="2010-11-14T08:52:30.610" LastActivityDate="2011-01-13T10:44:50.373" Title="What is the rate at which successful packets arrive?" Tags="&lt;probability&gt;&lt;homework&gt;&lt;stochastic-processes&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="10229" PostTypeId="1" CreationDate="2010-11-14T08:59:50.390" Score="1" ViewCount="89" Body="&lt;p&gt;Which are the parallel lines?&#xA;&lt;img src=&quot;http://i.imgur.com/IalyM.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I prove that a and b are parallel but can't prove that c and d are.&#xA;The angle ,which is 135 degrees and the 45 one are alternate angles.They aren't equal ,so c &lt;em&gt;isn't&lt;/em&gt; parallel to d but the answer is that c is parallel to d.Am I right that they aren't?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;:&#xA;They are a little inclined like these:&#xA;&lt;img src=&quot;http://i.imgur.com/q7buJ.jpg&quot; alt=&quot;http://www.math10.com/f/download/file.php?id=1169&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1795" LastEditorUserId="1795" LastEditorDisplayName="" LastEditDate="2010-11-14T10:10:42.993" LastActivityDate="2010-11-14T15:12:18.103" Title="Parallel straight lines" Tags="&lt;geometry&gt;" AnswerCount="1" CommentCount="9" />
  <row Id="10230" PostTypeId="2" ParentId="10183" CreationDate="2010-11-14T09:25:15.643" Score="1" ViewCount="" Body="&lt;p&gt;Let $\lfloor \sqrt{n} \rfloor = a$, then the following sum holds:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{0\le k &amp;lt; n} \lfloor \sqrt{k} \rfloor = na - \frac{a^3}{3} -&#xD;&#xA;    \frac{a^2}{2} - \frac{a}{6}.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might also enjoy the slightly more tricky sum:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{0\le k &amp;lt; n} \lfloor k^{1/3} \rfloor = nb - \frac{b^2}{4} - \frac{b^3}{2} - \frac{b^4}{4},$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where this time $b =\lfloor n^{1/3} \rfloor$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One way to obtain these answers is to merely apply the summation identity that I &lt;a href=&quot;http://math.stackexchange.com/questions/8336/proving-sum-limits-k1n-bigl-lfloor-fracka-bigr-rfloor-bigl-lfloo/8368#8368&quot;&gt;mentioned here&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1709" LastEditorUserId="1709" LastEditorDisplayName="" LastEditDate="2010-11-14T20:21:25.753" LastActivityDate="2010-11-14T20:21:25.753" />
  <row Id="10231" PostTypeId="1" AcceptedAnswerId="10235" CreationDate="2010-11-14T09:35:07.517" Score="2" ViewCount="210" Body="&lt;p&gt;How to solve this manually ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt; In my module the answer is given as &lt;strong&gt;$0$&lt;/strong&gt; but when I used mathematica &lt;code&gt;N[Sin[10 Degree] + Sin[20 Degree] + Sin[30 Degree] - Sin[360 Degree],50]&lt;/code&gt; gives &lt;strong&gt;$1.0156683209925990818958162414 \cdots$&lt;/strong&gt; (truncated)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I guess there is some mistake in the problem statement.&lt;/p&gt;&#xA;" LastActivityDate="2010-11-16T19:58:25.793" Title="Find the value of $\sin 10^\circ + \sin 20^\circ + \sin 30^\circ - \sin 360^\circ $" Tags="&lt;trigonometry&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="10232" PostTypeId="1" CreationDate="2010-11-14T09:40:25.313" Score="2" ViewCount="43" Body="&lt;p&gt;It's best explained with an image of course:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/j2MpR.gif&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;L1, L2, h1 &amp;amp; h2 are all given, and I would like to calculate the length of L3. Is it possible, and how?&lt;/p&gt;&#xA;" OwnerUserId="3491" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2010-11-14T09:52:18.397" LastActivityDate="2010-11-14T09:58:22.093" Title="How to calculate the length of line in a trapezium, given some other lines." Tags="&lt;geometry&gt;" AnswerCount="1" />
  <row Id="10233" PostTypeId="1" CreationDate="2010-11-14T09:49:19.677" Score="19" ViewCount="489" Body="&lt;p&gt;I want to motivate the quadratic reciprocity theorem, which at first glance does not look too important to justify it being one of Gauss' favorites. So far I can think of two uses that are basic enough to be shown immediately when presenting the theorem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) With the QRT, it is immediate to give a simple, efficient algorithm (that can be done even by hand) for computing Legendre symbols.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) In Euler's proof of Fermat's claim on the conditions in which a prime $p$ is of the form $x^2+ny^2$ (for certain small values of $n$) the proof is reduced to finding the conditions under which $p$ divides $x^2+ny^2$ for some $x,y$, hence to the question under which conditions is $-n$ a quadratic residue modulo $p$, which leads immediately to the QTR (for example, for $n=3$, where we get that $p\equiv_3 1$). I really like this example since it begins with an &quot;historic&quot; problem and proceeds to &quot;discover&quot; the QTR through special cases (which is what Euler did in practice - see Cox's book on &quot;Primes of the form $x^2+ny^2$&quot;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I am sure there are many more examples (and I'm especially curious as to how Gauss reached the theorem himself). I'd love to hear about them and receive references for further reading.&lt;/p&gt;&#xA;" OwnerUserId="1818" LastActivityDate="2010-11-18T18:27:13.693" Title="Uses of quadratic reciprocity theorem" Tags="&lt;number-theory&gt;&lt;quadratic-reciprocity&gt;&lt;famous-theorems&gt;&lt;motivation&gt;" AnswerCount="7" CommentCount="2" FavoriteCount="5" />
  <row Id="10234" PostTypeId="1" AcceptedAnswerId="10239" CreationDate="2010-11-14T09:50:48.013" Score="3" ViewCount="124" Body="&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Group_%28mathematics%29&quot; rel=&quot;nofollow&quot;&gt;group&lt;/a&gt; axioms demand that one has an inverse element. When it comes to multiplication zero has no inverse element because one cannot reconstruct the original number after it was multiplied by zero: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is an irreversible loss of information!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore all numbers cannot form a group under multiplication as long as zero is included.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When it comes to matrices some matrices have an inverse, some don't have one. &lt;a href=&quot;http://en.wikipedia.org/wiki/Singular_matrix&quot; rel=&quot;nofollow&quot;&gt;Singular matrices&lt;/a&gt; have &lt;a href=&quot;http://en.wikipedia.org/wiki/Determinant&quot; rel=&quot;nofollow&quot;&gt;determinant&lt;/a&gt; zero and don't have one. When the column vectors are linear dependent the determinant is zero and the matrix is singular.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My question&lt;/strong&gt;&lt;br&gt;&#xA;Where does the loss of information happen here? How is 'more' information lost when a vector is multiplied by a singular matrix (compared to multiplying it by an invertible matrix)?&lt;/p&gt;&#xA;" OwnerUserId="346" LastActivityDate="2010-11-14T10:06:51.683" Title="Loss of information when multiplying by singular matrix" Tags="&lt;linear-algebra&gt;&lt;group-theory&gt;" AnswerCount="3" FavoriteCount="1" />
  <row Id="10235" PostTypeId="2" ParentId="10231" CreationDate="2010-11-14T09:52:28.927" Score="7" ViewCount="" Body="&lt;p&gt;The last two terms have &lt;a href=&quot;http://en.wikipedia.org/wiki/Exact_trigonometric_constants&quot; rel=&quot;nofollow&quot;&gt;simple well-known values&lt;/a&gt;.&#xA;The first two are less appetizing; see &lt;a href=&quot;http://mathworld.wolfram.com/TrigonometryAnglesPi18.html&quot; rel=&quot;nofollow&quot;&gt;here for 10 degrees&lt;/a&gt; and &lt;a href=&quot;http://mathworld.wolfram.com/TrigonometryAnglesPi9.html&quot; rel=&quot;nofollow&quot;&gt;here for 20 degrees&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2010-11-14T09:52:28.927" />
  <row Id="10236" PostTypeId="2" ParentId="10233" CreationDate="2010-11-14T09:57:19.160" Score="0" ViewCount="" Body="&lt;p&gt;There is a very nice historical introduction at &lt;a href=&quot;http://en.wikipedia.org/wiki/Quadratic_reciprocity&quot; rel=&quot;nofollow&quot;&gt;wikipedia&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This shows how mathematicians started to think in the direction of quadratic reciprocity after seeing the work of Fermat, and later how Legendre &quot;almost&quot; proved it, etc..&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A lot of credit has to be attributed to Legendre. It is not Gauss alone who made contributions to arithmetic during that period. In fact, in the preface to his &quot;Disquisitiones&quot;, Gauss acknowledges the work of Legendre.&lt;/p&gt;&#xA;" OwnerUserId="1119" LastActivityDate="2010-11-14T09:57:19.160" CommentCount="1" />
  <row Id="10237" PostTypeId="2" ParentId="10232" CreationDate="2010-11-14T09:58:22.093" Score="1" ViewCount="" Body="&lt;p&gt;Denote the angle of the triangle by $\alpha$. Express all segments of the hypotenuse using $\alpha$ and $h_2,L_1,L_2$, then write an equation for $\alpha$ using $h_1$. Solve for $\alpha$, and find $L_3$. &lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-14T09:58:22.093" />
  <row Id="10238" PostTypeId="2" ParentId="10234" CreationDate="2010-11-14T10:01:07.210" Score="6" ViewCount="" Body="&lt;p&gt;The loss of information is measured by the nullity (co-rank) of the matrix. A matrix with nullity $k$ is similar to a block matrix with regular first block and $0_{k\times k}$ second block. So there's a basis where $k$ of the basis vectors are &quot;thrown away&quot;.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-14T10:01:07.210" />
  <row Id="10239" PostTypeId="2" ParentId="10234" CreationDate="2010-11-14T10:04:03.387" Score="7" ViewCount="" Body="&lt;p&gt;Let $A$ and $x$ be given, and set $b=Ax$. If $A$ is invertible, $x$ can be uniquely recovered from $b$ as $x=A^{-1}b$ (no information is lost, provided that we know $A$). But if $A$ is singular, someone who knows $A$ and $b$ cannot say for sure what $x$ was, since there are infinitely many solutions of the linear system $Ax=b$.&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2010-11-14T10:04:03.387" CommentCount="2" />
  <row Id="10240" PostTypeId="2" ParentId="10228" CreationDate="2010-11-14T10:05:47.817" Score="1" ViewCount="" Body="&lt;p&gt;Calculate the probability that a (single) random packet arrives successfully. Conclude the sought rate.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-14T10:05:47.817" CommentCount="3" />
  <row Id="10241" PostTypeId="2" ParentId="10234" CreationDate="2010-11-14T10:06:51.683" Score="3" ViewCount="" Body="&lt;p&gt;When multiplying by an invertible matrix $A$, you might say that no information is lost because you can recover $B$ from $AB$ by multiplying by $A^{-1}$.  Another way to put it is that the map $B\mapsto AB$ is injective.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $A$ is singular, then information is lost because the map $B\mapsto AB$ is no longer injective.  One explanation is that $A$ has nontrivial nullspace, and $AB=0$ for each matrix $B$ whose column space lies in the null space of $A$.  It follows that $AB=AC$ when the column space of $B-C$ lies in the null space of $A$.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-11-14T10:06:51.683" CommentCount="3" />
  <row Id="10242" PostTypeId="2" ParentId="10233" CreationDate="2010-11-14T10:27:15.407" Score="9" ViewCount="" Body="&lt;p&gt;This is a good question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My own take on motivating quadratic reciprocity is recorded &lt;a href=&quot;http://math.uga.edu/~pete/4400qrlaw.pdf&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; (these are lecture notes from an undergraduate course on introductory number theory).  If you look there, you will find that most of what I have said is an elaboration of the two points you bring up.  I think a crisp way of explaining what QR does for you is in the idea of the &quot;direct&quot; and &quot;inverse&quot; problems attached to the Legendre symbol $(\frac{n}{p})$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Namely, for the &lt;strong&gt;direct problem&lt;/strong&gt; we fix $p$ and ask which integers $n$ are squares modulo $p$.  This is clearly a finite problem.  On the other hand there is the &lt;strong&gt;inverse problem&lt;/strong&gt;: we fix an integer $n$ and ask for which primes $p$ we have that &#xA;$n$ is a square modulo $p$.  This is, &lt;em&gt;a priori&lt;/em&gt;, an infinite problem.  However, it is one of great relevance to classical number theory: e.g. all of the many proofs I have seen of Fermat's Two Squares theorem pass through the fact that $-1$ is a square modulo an odd prime $p$ iff $p \equiv 1 \pmod 4$.  More generally, if you look at the Diophantine equation $x^2 - n y^2 = p$, for $n$ a nonzero integer and $p$ a prime with $\operatorname{gcd}(p,n) = 1$, then reducing modulo $p$ gives the necessary condition &#xA;$(\frac{n}{p}) = 1$.  Quadratic reciprocity to the rescue!  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Secondly, as you also say, quadratic reciprocity gives an efficient algorithm for answering whether a particular $n$ is a square modulo $p$, much faster than computing all $\frac{p-1}{2}$ squares modulo $p$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In my experience, this is more than enough for students to appreciate the usefulness of Gauss' &lt;em&gt;aureum theorema&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="299" LastActivityDate="2010-11-14T10:27:15.407" />
  <row Id="10243" PostTypeId="2" ParentId="10046" CreationDate="2010-11-14T10:33:39.517" Score="2" ViewCount="" Body="&lt;p&gt;Just as additional information, the inverse of these binomial coefficients satisfy several wonderful identities. For example, on &lt;a href=&quot;http://mathworld.wolfram.com/CentralBinomialCoefficient.html&quot; rel=&quot;nofollow&quot;&gt;Mathworld&lt;/a&gt; I saw the following amazing identity (valid for $k\ge 1$):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{n \ge 1}\frac{1}{n^k \binom{2n}{n}} = \frac{1}{2}{}_{k+1}F_k(\underbrace{1,\ldots,1}_{k+1}; \frac{3}{2}, \underbrace{2,\ldots,2}_{k-1}; \frac{1}{4}),$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where ${}_{k+1}F_k$ is a generalized hypergeometric function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How to derive the above result? I don't recall the answer immediately, but perhaps browsing &lt;a href=&quot;http://www.math.rutgers.edu/~zeilberg/&quot; rel=&quot;nofollow&quot;&gt;Zeilberger's website&lt;/a&gt; might be helpful.&lt;/p&gt;&#xA;" OwnerUserId="1709" LastEditorUserId="1709" LastEditorDisplayName="" LastEditDate="2010-11-14T22:29:50.287" LastActivityDate="2010-11-14T22:29:50.287" CommentCount="4" />
  <row Id="10244" PostTypeId="2" ParentId="10216" CreationDate="2010-11-14T10:49:25.830" Score="3" ViewCount="" Body="&lt;p&gt;This is essentially Yuval's answer expressed in a slightly different way.&#xA;Let your companion matrix be&#xA;$$C=\pmatrix{0&amp;amp;1&amp;amp;0&amp;amp;\cdots&amp;amp;0\\\\&#xD;&#xA;0&amp;amp;0&amp;amp;1&amp;amp;\cdots&amp;amp;0\\\\&#xD;&#xA;\vdots&amp;amp;\vdots&amp;amp;\vdots&amp;amp;\ddots&amp;amp;\vdots\\\\&#xD;&#xA;0&amp;amp;0&amp;amp;0&amp;amp;\cdots&amp;amp;1\\\\&#xD;&#xA;-a_0&amp;amp;-a_1&amp;amp;-a_2&amp;amp;\cdots&amp;amp;-a_{n-1}}.$$&#xD;&#xA;Then for the vector $v=(1\,\,0\,\,0\cdots 0)$,&#xA;$$v\sum_{j=0}^{n-1} b_j C^j=&#xD;&#xA;\pmatrix{b_0&amp;amp;b_1&amp;amp;b_2&amp;amp;\cdots&amp;amp;b_{n-1}}$$&#xD;&#xA;so that $g(C)\ne0$ for all nonzero polynomials $g$ of degree less than $n$.&#xA;So the minimal polynomial has degree $n$, and equals the&#xA;characteristic polynomial (via Cayley-Hamilton).&#xA;But $vC^n=(-a_0\,\, {-a_1}\,\, {-a_2}\cdots{-a_{n-1}})$&#xD;&#xA;and for $v(C^n+\sum_{j=0}^{n-1}b_j C^j)=0$ we need $a_j=b_j$. So the minimal&#xA;and characteristic polynomials both equal $f$.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-11-14T10:49:25.830" CommentCount="2" />
  <row Id="10245" PostTypeId="2" ParentId="10155" CreationDate="2010-11-14T10:56:02.447" Score="2" ViewCount="" Body="&lt;p&gt;This is a transcrition of my comment:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My interpretation of the picture is: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) the origine of the coordinate system $x,y,z$ is the centre of the piece &quot;eye&quot; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) force $F_1$ (with a magnitude of 80 N) lies in the $x,z$ plane &lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) its direction is $(x,y,z)=(4,0,3)$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;4) the angle between the force components $F_{1_{x}},F_{1_{z}}$ is $π/2$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;5) the force component $F_{1_{z}}=0$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;6) the side (of the right triangle) with length 5 is opposite the right angle formed by the sides with lengths 3 and 4.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In addition:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The triangle lies in the  $x,z$ plane. Its vertices are $P(0,0,0)$, $Q(4,0,0)$ and $R(4,0,3)$. The  hypotenuse is $PR$ and the base $PQ$.  &lt;/p&gt;&#xA;" OwnerUserId="752" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2010-11-16T14:07:18.193" LastActivityDate="2010-11-16T14:07:18.193" />
  <row Id="10246" PostTypeId="1" AcceptedAnswerId="10247" CreationDate="2010-11-14T11:11:26.550" Score="1" ViewCount="94" Body="&lt;p&gt;A power type &lt;strong&gt;b^[(a^/(1/n)]&lt;/strong&gt; where &lt;strong&gt;b&lt;/strong&gt;, &lt;strong&gt;a&lt;/strong&gt; and &lt;strong&gt;n&lt;/strong&gt; are positive integers greater than 1 and a^(1/n) is irrational, always represents an irrational number?&lt;/p&gt;&#xA;" OwnerUserId="2938" LastActivityDate="2010-11-14T11:14:00.310" Title="b^[a^(1/n)]: is it an irrational number?" Tags="&lt;number-theory&gt;" AnswerCount="1" />
  <row Id="10247" PostTypeId="2" ParentId="10246" CreationDate="2010-11-14T11:14:00.310" Score="6" ViewCount="" Body="&lt;p&gt;Yes, it is in fact transcendental by the &lt;a href=&quot;http://en.wikipedia.org/wiki/Gelfond%E2%80%93Schneider_theorem&quot; rel=&quot;nofollow&quot;&gt;Gelfond-Schneider theorem&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1119" LastActivityDate="2010-11-14T11:14:00.310" />
  <row Id="10248" PostTypeId="2" ParentId="10177" CreationDate="2010-11-14T11:45:15.483" Score="4" ViewCount="" Body="&lt;p&gt;The normal bundle of a codimension 1 embedding of a compact closed manifold $M$ in $\mathbb R^{n+1}$ is indeed trivial. Otherwise, you could find a simple closed curve in $\mathbb R^{n+1}$ that intersects $M$ in a single point. This implies that both the curve and $M$ represent nontrivial mod $2$ homology classes in $\mathbb R^{n+1}$; this is because the intersection product is dual to the cup product and cannot be nonzero on trivial homology classes. However there are no nontrivial homology classes in $\mathbb R^{n+1}$ since it is contractible. So we get a contradiction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once the normal bundle is seen to be trivial, one can use the ambient orientation of $\mathbb R^{n+1}$ to locally orient the manifold, since there is a well defined positive normal direction.&lt;/p&gt;&#xA;" OwnerUserId="2941" LastActivityDate="2010-11-14T11:45:15.483" CommentCount="6" />
  <row Id="10249" PostTypeId="2" ParentId="10209" CreationDate="2010-11-14T12:27:57.717" Score="7" ViewCount="" Body="&lt;p&gt;If you are going to study computer science, you want to read &lt;a href=&quot;http://en.wikipedia.org/wiki/Concrete_Mathematics&quot; rel=&quot;nofollow&quot;&gt;Graham, Knuth, Patashnik: Concrete Mathematics&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Look into solving recursion, combinatorics and maybe a little stochastics. Getting the hand of infinite sequences and series is also advisable. Above stated book covers that all because it is made for computer scientists. You should have math courses covering all that, though.&lt;/p&gt;&#xA;" OwnerUserId="3330" LastActivityDate="2010-11-14T12:27:57.717" />
  <row Id="10250" PostTypeId="2" ParentId="10233" CreationDate="2010-11-14T12:31:42.837" Score="12" ViewCount="" Body="&lt;p&gt;Quadratic reciprocity allows you to make precise certain intuitions about the primes.  More precisely, it tells you that for every finite set $p_1, p_2, ... p_n$ of primes and every function $f : \{ 1, 2, ... n \} \to \{ -1, 1 \}$ there exists an arithmetic progression such that any prime $q$ in that progression satisfies $\left( \frac{p_i}{q} \right) = f(i)$.  In other words, you can get primes to behave locally independently (with respect to being or not being a quadratic residue).  You can use this idea to give one proof that, for example, $\mathbb{Q}(\sqrt{2}, \sqrt{3}, ... \sqrt{p_n})$ has the degree over $\mathbb{Q}$ that you think it does; this is described &lt;a href=&quot;http://qchu.wordpress.com/2009/07/02/square-roots-have-no-unexpected-linear-relationships/&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-11-14T12:31:42.837" />
  <row Id="10251" PostTypeId="1" AcceptedAnswerId="10312" CreationDate="2010-11-14T15:10:15.830" Score="1" ViewCount="134" Body="&lt;p&gt;How the following conversion is justified ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\cos \biggl( \theta + \frac{2\pi}{3} \biggr) + \cos \biggl( \theta + \frac{4\pi}{3} \biggr) = \cos \biggl(  \frac{\pi}{2} - \theta \biggr) - \cos \biggl(  \frac{\pi}{3} +\theta \biggr) $$&lt;/p&gt;&#xA;" OwnerUserId="2109" LastActivityDate="2010-11-14T23:30:25.673" Title="Explain this step" Tags="&lt;trigonometry&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="10252" PostTypeId="2" ParentId="10229" CreationDate="2010-11-14T15:12:18.103" Score="1" ViewCount="" Body="&lt;p&gt;This is a problem with extra information. Part of solving problems is knowing what to ignore! You are thinking about 4 lines, but to show two lines are parallel, you only really need 3.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, you don't need to use &lt;strong&gt;all&lt;/strong&gt; of the information given to solve this problem. In fact, you can &lt;em&gt;ignore line d&lt;/em&gt; and just focus on how line c interacts with the lines you want to prove are parallel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next look at the theorem about parallel lines and their &lt;a href=&quot;http://www.mathnstuff.com/math/spoken/here/2class/260/trans.htm&quot; rel=&quot;nofollow&quot;&gt;transverse&lt;/a&gt; again.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, you should have your proof.&lt;/p&gt;&#xA;" OwnerUserId="2563" LastActivityDate="2010-11-14T15:12:18.103" />
  <row Id="10253" PostTypeId="2" ParentId="10065" CreationDate="2010-11-14T15:17:41.997" Score="7" ViewCount="" Body="&lt;p&gt;I completely forgot about the &lt;a href=&quot;http://en.wikipedia.org/wiki/Stolz%E2%80%93Ces%C3%A0ro_theorem&quot; rel=&quot;nofollow&quot;&gt;Stolz-Cesàro theorem&lt;/a&gt;, from which we get:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\lim_{n\to \infty}=\frac{a_{n+1}-a_{n}}{n+1-n}=\lim_{n\to \infty}=\frac{\frac{1}{a_{n}}}{1}=\lim_{n\to \infty}=\frac{1}{a_{n}}=0$ (the same technique works for $\frac{(a_{n})^2}{n}$)&lt;/p&gt;&#xA;" OwnerUserId="3237" LastActivityDate="2010-11-14T15:17:41.997" CommentCount="1" />
  <row Id="10255" PostTypeId="1" AcceptedAnswerId="10259" CreationDate="2010-11-14T15:36:44.623" Score="5" ViewCount="233" Body="&lt;p&gt;Is there a common name for curves, obtained from dragging a point along another curve, similar to how tractrix is obtained by dragging a point along a line?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is a parametric equation of such curve given the parametric equation of a curve along which the master goes?&lt;/p&gt;&#xA;" OwnerUserId="2513" LastActivityDate="2011-04-10T15:41:19.533" Title="Tractrix-like curves" Tags="&lt;geometry&gt;&lt;plane-curves&gt;&lt;analytic-geometry&gt;" AnswerCount="2" />
  <row Id="10256" PostTypeId="1" AcceptedAnswerId="10303" CreationDate="2010-11-14T15:47:09.413" Score="3" ViewCount="170" Body="&lt;p&gt;Let $A$ be a finite dimensional vector space, $\cdots \rightarrow A_{n+1}\rightarrow^{f_{n+1}} A_n \rightarrow^{f_{n}} A_{n-1}\rightarrow \cdots $ be an inverse system of finite dimensional vector spaces, and let $\phi_i:A\to A_i$ be linear maps inducing a map $\phi:A\to\lim\limits_{\leftarrow} A_i$. If $\phi$ is an injection (resp. surjection), does it follow that $\phi_i$ is an injection (resp. surjection) for some $i\gg 0$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My guess is that it does, since the dimensions $\dim ker \phi_i$ and $\dim coker \phi_i$ should be non-increasing sequences. &lt;/p&gt;&#xA;" OwnerUserId="3332" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-14T21:16:21.533" LastActivityDate="2010-11-15T00:38:44.463" Title="limits of finite dimensional vector spaces" Tags="&lt;linear-algebra&gt;&lt;homological-algebra&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="10257" PostTypeId="1" AcceptedAnswerId="10320" CreationDate="2010-11-14T15:53:40.110" Score="6" ViewCount="325" Body="&lt;p&gt;I have changed the title (replaced &quot;well-behaved&quot; by &quot;càdlàg&quot;), since it seems that &quot;a well-behaved function&quot; might be interpreted as &quot;a function of bounded variation&quot; (rather than &quot;a càdlàg function&quot;, which I actually meant).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $f:[0,1] \to {\bf R}$ have a following property: $f$ is continuous except at the points $x_{k,n} = \frac{{2k - 1}}{{2^n }}$, $k=1,\ldots,2^{n-1}$, $n=1,2,3,\ldots$, where $\lim _{x \downarrow x_{k,n} } f(x) = f(x_{k,n} )$ but $f(x_{k,n} ) -&#xD;&#xA;\lim _{x \uparrow x_{k,n} } f(x) = a_n  &amp;gt; 0$. Can such a function exist if $a_n &amp;gt; 1/ \log(n)$ for all sufficiently large $n$?&lt;/p&gt;&#xA;" OwnerUserId="2810" LastEditorUserId="2810" LastEditorDisplayName="" LastEditDate="2010-11-14T23:37:49.887" LastActivityDate="2010-11-15T01:25:57.267" Title="How much a càdlàg (i.e., right-continuous with left limits) function can jump?" Tags="&lt;real-analysis&gt;" AnswerCount="1" CommentCount="11" />
  <row Id="10258" PostTypeId="2" ParentId="10255" CreationDate="2010-11-14T16:05:50.593" Score="0" ViewCount="" Body="&lt;p&gt;Some of them are called pursuit curves: &lt;a href=&quot;http://mathworld.wolfram.com/PursuitCurve.html&quot; rel=&quot;nofollow&quot;&gt;http://mathworld.wolfram.com/PursuitCurve.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="890" LastActivityDate="2010-11-14T16:05:50.593" CommentCount="1" />
  <row Id="10259" PostTypeId="2" ParentId="10255" CreationDate="2010-11-14T16:16:01.183" Score="6" ViewCount="" Body="&lt;p&gt;Generalized tractrices (tractories) exist, see &lt;a href=&quot;http://www.2dcurves.com/derived/generalized%20tractrix.html&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; or &lt;a href=&quot;http://www-cabri.imag.fr/abracadabri/Courbes/Tract/Tract3.html&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; or &lt;a href=&quot;http://www.mathcurve.com/courbes2d/tractrice/tractoire.shtml&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt;. (the last two are in French, but with slightly more detail than the first one.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This &lt;a href=&quot;http://books.google.com/books?id=93rvAAAAMAAJ&amp;amp;pg=PA127&quot; rel=&quot;nofollow&quot;&gt;old book&lt;/a&gt; ought to be of interest as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This &lt;a href=&quot;http://books.google.com/books?id=f9lAHgWV0vIC&amp;amp;pg=PA176&quot; rel=&quot;nofollow&quot;&gt;book&lt;/a&gt; describes Euler's treatment of the problem of the tractory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a note, the problem of finding the generating curve, given the tractory, is a much easier problem (hint: use the tangent vector of a curve) than finding the tractory corresponding to a generating curve.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;For those who have difficulty reading French, the third link I mentioned gives the prescription for generating the corresponding tractory from a generating curve with parametric equations $(f(t)\quad g(t))$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The parametric equations for the tractory of $(f(t)\quad g(t))$ (in vector form) is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{pmatrix}f(t)\\g(t)\end{pmatrix}-\frac{a}{\sqrt{f^{\prime}(t)^2+g^{\prime}(t)^2}}\begin{pmatrix}\cos\;\alpha(t)&amp;amp;\sin\;\alpha(t)\\-\sin\;\alpha(t)&amp;amp;\cos\;\alpha(t)\end{pmatrix}\cdot\begin{pmatrix}f^{\prime}(t)\\g^{\prime}(t)\end{pmatrix}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or explicitly&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align*}x&amp;amp;=f(t)-\frac{a}{\sqrt{f^{\prime}(t)^2+g^{\prime}(t)^2}}(f^{\prime}(t)\cos\;\alpha(t)+g^{\prime}(t)\sin\;\alpha(t))\\y&amp;amp;=g(t)-\frac{a}{\sqrt{f^{\prime}(t)^2+g^{\prime}(t)^2}}(g^{\prime}(t)\cos\;\alpha(t)-f^{\prime}(t)\sin\;\alpha(t))\end{align*}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where the function $\alpha(t)$ satisfies the differential equation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\mathrm d\alpha}{\mathrm dt}=\frac{f^{\prime}(t)g^{\prime\prime}(t)-g^{\prime}(t)f^{\prime\prime}(t)}{f^{\prime}(t)^2+g^{\prime}(t)^2}-\frac{\sin\alpha}{a}\sqrt{f^{\prime}(t)^2+g^{\prime}(t)^2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and $a$ is the length of the segment running through the generating curve.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an example, here is an animation showing the curve $(3\cos\;t-2\cos^3 t+\cos 2t\quad 2\sin^3 t+\sin 2t)$ and its tractory with segment length 1:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/AeY1U.gif&quot; alt=&quot;nephroid tractory&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(This is a less fancy version of the last bicycle animation in the third French link.)&lt;/p&gt;&#xA;" OwnerUserId="498" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2011-04-10T15:41:19.533" LastActivityDate="2011-04-10T15:41:19.533" CommentCount="15" />
  <row Id="10260" PostTypeId="2" ParentId="10256" CreationDate="2010-11-14T16:45:38.703" Score="0" ViewCount="" Body="&lt;p&gt;Like you said, we have $\ker\phi_{i+1}\subseteq \ker\phi_{i}$. Hence the decreasing sequence of kernels either stabilizes at some nontrivial subspace of $A$, in which case $\phi$ is not injective, or it decreases to $0$. But it has to decrease to $0$ after finitely many $i$, so after that, $\phi_i$ is injective.&lt;/p&gt;&#xA;" OwnerUserId="2941" LastActivityDate="2010-11-14T16:45:38.703" />
  <row Id="10261" PostTypeId="1" AcceptedAnswerId="10262" CreationDate="2010-11-14T17:02:56.153" Score="4" ViewCount="375" Body="&lt;p&gt;I feel like finding the inverse of $y=xe^x$ should have an easy answer but can't find it.&lt;/p&gt;&#xA;" OwnerUserId="1958" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-01T17:44:29.703" LastActivityDate="2010-12-01T17:44:29.703" Title="Inverse of $y=xe^x$" Tags="&lt;special-functions&gt;" AnswerCount="3" />
  <row Id="10262" PostTypeId="2" ParentId="10261" CreationDate="2010-11-14T17:07:45.197" Score="12" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Lambert_W_function&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Lambert_W_function&lt;/a&gt;, see the section Applications.&lt;/p&gt;&#xA;" OwnerUserId="2810" LastActivityDate="2010-11-14T17:07:45.197" CommentCount="1" />
  <row Id="10263" PostTypeId="2" ParentId="9642" CreationDate="2010-11-14T17:33:41.537" Score="2" ViewCount="" Body="&lt;p&gt;What you're looking for is described on pages 247-249 of Robert Vanderbei's &lt;a href=&quot;http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.111.1824&amp;amp;rep=rep1&amp;amp;type=pdf&quot; rel=&quot;nofollow&quot;&gt;Linear Programming: Foundations and Extensions&lt;/a&gt; (2nd edition), although Vanderbei has some of his sign conventions opposite of yours.  Take a close look in particular at Figure 14.4 on page 249 (&quot;Adding a new node to accommodate an arc $(i,j)$ having an upper bound $u_{ij}$ on its flow capacity&quot;).  &lt;/p&gt;&#xA;" OwnerUserId="2370" LastActivityDate="2010-11-14T17:33:41.537" />
  <row Id="10264" PostTypeId="1" AcceptedAnswerId="10270" CreationDate="2010-11-14T17:35:43.267" Score="11" ViewCount="401" Body="&lt;p&gt;I'm just reviewing for my exam tomorow looking at old exams, unfortunately I don't have solutions.  Here is a question I found : determine if the series converges or diverges.  If it converges find it's limit.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle \sum\limits_{n=1}^{\infty}\dfrac{\sin(n-\sqrt{n^2+n})}{n}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've ruled down possible tests to the limit comparison test, but I feel like I've made a mistake somewhere.&lt;br&gt;&#xA;&lt;strong&gt;divergence test&lt;/strong&gt; - limit is 0 by the squeeze theorem&lt;br&gt;&#xA;&lt;strong&gt;integral test&lt;/strong&gt; - who knows how to solve this&lt;br&gt;&#xA;&lt;strong&gt;comparison test&lt;/strong&gt; - series is not positive&lt;br&gt;&#xA;&lt;strong&gt;ratio root tests&lt;/strong&gt; - on the absolute value of the series, this wouldn't work out&lt;br&gt;&#xA;&lt;strong&gt;alternating series test&lt;/strong&gt; - would not work, the series is not decreasing or alternating  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any ideas what to compare this series here with or where my mistake is on my reasoning above?&lt;/p&gt;&#xA;" OwnerUserId="2533" LastEditorUserId="2533" LastEditorDisplayName="" LastEditDate="2011-03-30T19:13:45.873" LastActivityDate="2011-03-30T19:13:45.873" Title="Does the series $\sum\limits_{n=1}^{\infty}\frac{\sin(n-\sqrt{n^2+n})}{n}$ converge?" Tags="&lt;calculus&gt;&lt;homework&gt;&lt;sequences-and-series&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" />
  <row Id="10270" PostTypeId="2" ParentId="10264" CreationDate="2010-11-14T17:59:45.080" Score="18" ViewCount="" Body="&lt;p&gt;The key here is that $n - \sqrt{n^2 + n}$ converges to $-{1 \over 2}$ as $n$ goes to infinity:&#xA;$$n - \sqrt{n^2 + n}=  (n - \sqrt{n^2 + n}) \times  {n + \sqrt{n^2 + n} \over n + \sqrt{n^2 + n}}$$&#xD;&#xA;$$= {n^2 - (n^2 + n) \over n + \sqrt{n^2 + n}} = -{n \over  n + \sqrt{n^2 + n}}$$&#xD;&#xA;$$= -{1 \over 1 + \sqrt{1 + {1 \over n}}}$$&#xD;&#xA;Take limits as $n$ goes to infinity to get $-{1 \over 2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence $\sin(n - \sqrt{n^2 + n})$ converges to $\sin(-{1 \over 2})$, and the series diverges similarly to ${1 \over n}$, using the limit comparison test for example.&lt;/p&gt;&#xA;" OwnerUserId="3035" LastEditorUserId="3035" LastEditorDisplayName="" LastEditDate="2010-11-14T18:05:30.553" LastActivityDate="2010-11-14T18:05:30.553" CommentCount="2" />
  <row Id="10271" PostTypeId="1" AcceptedAnswerId="10272" CreationDate="2010-11-14T18:03:30.463" Score="0" ViewCount="101" Body="&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/1jymi.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sorry if this is a little basic, but my math is pretty rusty.  I know the point coordinates of the two blue dots.  I also know the distances from the middle blue dot to the edges of the square.  How can I calculate the coordinates of the red dot?&lt;/p&gt;&#xA;" OwnerUserId="3496" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-11-14T18:08:11.283" LastActivityDate="2010-11-14T18:10:11.780" Title="Calculate point on hypotenuse of right-angled triangle." Tags="&lt;trigonometry&gt;" AnswerCount="1" />
  <row Id="10272" PostTypeId="2" ParentId="10271" CreationDate="2010-11-14T18:10:11.780" Score="1" ViewCount="" Body="&lt;p&gt;The two right angled triangles in the picture are &lt;a href=&quot;http://www.gcseguide.co.uk/similar_triangles.htm&quot; rel=&quot;nofollow&quot;&gt;similar&lt;/a&gt;. Hence the ratio of the corresponding sides are the same. You can use this to get the remaining coordinates.&lt;/p&gt;&#xA;" OwnerUserId="3431" LastActivityDate="2010-11-14T18:10:11.780" />
  <row Id="10273" PostTypeId="2" ParentId="10261" CreationDate="2010-11-14T18:20:34.727" Score="8" ViewCount="" Body="&lt;p&gt;Inversion of $xe^{-x}$ is a problem with combinatorial structure, from which one can also invert $xe^x$ and derive most of the properties at the Wikipedia page.  At least for theoretical understanding it is easier to treat the inverse of $x\exp(-x)$ as the basic object and then translate it into &quot;Lambert W&quot; terms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Inverse of $f(x) = x \exp(-x)$ is the generating function for rooted trees $\Sigma n^{n-1} X^n/n!$.  That the series represents the inverse is equivalent to a simple combinatorial fact (&quot;removing the root and its edges from a tree, gives a collection of rooted trees&quot;).  The inverse of $xe^x$ is then $W(x)= -f^{-1}(-x)$, which is the same series with minus signs.  The radius of convergence $1/e$ follows from Stirling's approximation to $n!$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The coefficients of $W(x)^r$ given in the Wikipedia article are, up to some signs, Abel polynomials.  This also follows from the combinatorial interpretation: $(f^{-1}(x))^r$ is the generating function whose $x^n$ coefficient is the number maps from a set of size $n$ to a set of $r$ elements with a rooted tree on each fiber of the map.&lt;/p&gt;&#xA;" OwnerUserId="467" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-11-14T18:57:19.797" LastActivityDate="2010-11-14T18:57:19.797" />
  <row Id="10274" PostTypeId="1" CreationDate="2010-11-14T18:30:13.967" Score="1" ViewCount="290" Body="&lt;p&gt;A ring $R$ is a Boolean ring provided that $a^2=a$ for every $a \in R$. How can we show that every Boolean ring is commutative?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance.&lt;/p&gt;&#xA;" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-05-21T02:20:29.137" LastActivityDate="2011-05-21T02:20:29.137" Title="How to show that every Boolean ring is commutative?" Tags="&lt;ring-theory&gt;" AnswerCount="5" />
  <row Id="10275" PostTypeId="2" ParentId="10274" CreationDate="2010-11-14T18:35:11.537" Score="2" ViewCount="" Body="&lt;p&gt;Plug $a = x + y$.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-14T18:35:11.537" />
  <row Id="10276" PostTypeId="2" ParentId="10261" CreationDate="2010-11-14T18:37:20.413" Score="-1" ViewCount="" Body="&lt;p&gt;Asymptotically, $x \approx \log y - \log\log y$.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-14T18:37:20.413" />
  <row Id="10277" PostTypeId="2" ParentId="10274" CreationDate="2010-11-14T18:42:39.553" Score="8" ViewCount="" Body="&lt;p&gt;Every Boolean ring is of characteristic 2, since&#xA;$a+a=(a+a)^2=a^2+a^2+a^2+a^2=a+a+a+a$&#xD;&#xA;Now, for any $x,y$ in the ring&#xA;$x+y=(x+y)^2=x^2+xy+yx+y^2=x+y+xy+yx$&#xD;&#xA;So that $xy+yx=0$ and hence $xy=-yx$. But since the ring has characteristic 2, $yx=-yx=xy$&lt;/p&gt;&#xA;" OwnerUserId="3431" LastActivityDate="2010-11-14T18:42:39.553" />
  <row Id="10278" PostTypeId="2" ParentId="10274" CreationDate="2010-11-14T18:43:34.597" Score="0" ViewCount="" Body="&lt;p&gt;As Yuval points out $(x+y)^{2} = x+y$ which implies $x^{2} + y^{2} + x \cdot y + y \cdot x = x+y$. Now from this you have $x \cdot y + y \cdot x =0$.&lt;/p&gt;&#xA;" LastActivityDate="2010-11-14T18:43:34.597" CommentCount="1" />
  <row Id="10279" PostTypeId="2" ParentId="10274" CreationDate="2010-11-14T18:53:49.673" Score="0" ViewCount="" Body="&lt;p&gt;We want to show that $xy = yx$ for all $x,y \in R$. We know that $(x+y)^2 = x+y$. So $(x+y)^2 = (x+y)(x+y) = xx+xy+yx+yy = x+xy+yx+y = x+x^2y^2+y^2x^2+y$. This equals $x+(xy)+(yx) + y = x+y$ so that $xy = yx$.&lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2010-11-14T18:53:49.673" />
  <row Id="10280" PostTypeId="1" AcceptedAnswerId="10309" CreationDate="2010-11-14T18:54:15.883" Score="0" ViewCount="191" Body="&lt;p&gt;Let $x \in \mathbb{R} \backslash \mathbb{Q}, x&amp;gt;0$ and $q \in \mathbb{N}, q&amp;gt;0$, prove that there is an $r \in \mathbb{N}, r&amp;gt;0$ with:&#xA;     $r \cdot x - \left\lfloor r \cdot x \right\rfloor &amp;lt; \frac{1}{q}$ or $1-(r \cdot x - \left\lfloor r \cdot x \right\rfloor) &amp;lt; \frac{1}{q}.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was given the hint to divide the sequence $a_r := r \cdot x - \left\lfloor r \cdot x \right\rfloor \in [0,1)$ into q intervals $[0,\frac{1}{q}),[\frac{1}{q},\frac{2}{q}),\ldots,[\frac{q-1}{q},1)$ and use the pigeonhole principle, but I cannot see how this would help to the problem.&lt;/p&gt;&#xA;" OwnerUserId="3123" LastEditorUserId="3249" LastEditorDisplayName="" LastEditDate="2011-01-03T04:04:41.297" LastActivityDate="2011-01-03T04:04:41.297" Title="Proving that an equation is solvable, Floor function." Tags="&lt;real-analysis&gt;" AnswerCount="2" CommentCount="6" />
  <row Id="10281" PostTypeId="2" ParentId="10280" CreationDate="2010-11-14T19:18:29.073" Score="0" ViewCount="" Body="&lt;p&gt;Let's use the notation $\{x\} = x - \lfloor x \rfloor$. We can think of $\{\cdot\}$ as a function $\mathbb{R} \longrightarrow \mathbb{R}/\mathbb{Z}$, indeed a homomorphism of groups (with addition as the group operator). That means that $\{x + y\} = \{x\} + \{y\}$, or what matters more in your case, $\{x - y\} = \{x\} - \{y\}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now suppose $\{r_1x\},\{r_2x\}$ are close, what does this say about $\{(r_1-r_2)x\}$?&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-14T19:18:29.073" CommentCount="4" />
  <row Id="10282" PostTypeId="1" AcceptedAnswerId="10374" CreationDate="2010-11-14T19:22:31.127" Score="13" ViewCount="814" Body="&lt;p&gt;As a physics student, I've come across mathematical objects called &lt;strong&gt;tensors&lt;/strong&gt; in several different contexts. Perhaps confusingly, I've also been given both the mathematician's and physicist's definition, which I believe are slightly different.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I currently think of them in the following ways, but have a tough time reconciling the different views:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;An extension/abstraction of scalars, vectors, and matrices in mathematics.&lt;/li&gt;&#xA;&lt;li&gt;A multi-dimensional array of elements.&lt;/li&gt;&#xA;&lt;li&gt;A mapping between vector spaces that represents a co-ordinate independent transformation.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In fact, I'm not even sure how correct these three definitions are. Is there a particularly relevant (rigorous, even) definition of tensors and their uses, that might be suitable for a mathematical physicist?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Direct answers/explanations, as well as links to good introductory articles, would be much appreciated.&lt;/p&gt;&#xA;" OwnerUserId="56" LastActivityDate="2010-11-29T15:55:04.040" Title="An Introduction to Tensors" Tags="&lt;tensors&gt;" AnswerCount="5" CommentCount="11" FavoriteCount="10" />
  <row Id="10283" PostTypeId="1" AcceptedAnswerId="10285" CreationDate="2010-11-14T19:35:01.253" Score="0" ViewCount="57" Body="&lt;p&gt;$10 \%$ of people in a population suffer from high blood pressure. Calculate the probability that the twelfth person tested will be the second one with high blood pressure. So this is a negative binomial distribution. Now $P(X=k) = \binom{k+r-1}{k} (1-p)^{r}p^{k}$. So this is the probability of $k$ successes after $r$ failures. So $P(X=2) = \binom{11}{2}(0.9)^{10}(0.1)^2$? &lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2010-11-14T20:15:08.267" Title="Success and Failure" Tags="&lt;probability&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="10284" PostTypeId="1" CreationDate="2010-11-14T19:41:03.790" Score="0" ViewCount="124" Body="&lt;p&gt;Say we have a set of 12 numbers - (0,1,2,0,3,0,4,0,5,6,0).&#xA;We go thru this set one number at a time, and choose a number with a probability 1/3; we sample. What is the probability we end up with 3 0's out of the 4 numbers we will likely choose(on average)? &#xA;Is there a closed-form solution for this?&lt;/p&gt;&#xA;" LastActivityDate="2010-11-14T21:18:32.603" Title="Probability Question" Tags="&lt;probability&gt;" AnswerCount="4" CommentCount="10" />
  <row Id="10285" PostTypeId="2" ParentId="10283" CreationDate="2010-11-14T19:45:26.750" Score="1" ViewCount="" Body="&lt;p&gt;The 12th person tested must be one with high blood pressure. Thus only one of the first eleven can have high blood pressure. So the total probability is the probability that exactly one of the first eleven has high blood pressure multiplied by the probability that the 12th person has high blood pressure. Hence $$P = \bigg({11 \choose 1}(0.9)^{10}(0.1) \bigg) \cdot (0.1) = 11(0.9)^{10} (0.1)^2$$&lt;/p&gt;&#xA;" OwnerUserId="1016" LastActivityDate="2010-11-14T19:45:26.750" />
  <row Id="10286" PostTypeId="2" ParentId="10282" CreationDate="2010-11-14T19:55:14.113" Score="1" ViewCount="" Body="&lt;p&gt;The most general notion I know is that tensor product of modules. You can read about this here&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Tensor_product_of_modules&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Tensor_product_of_modules&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since vector spaces are modules, this definition specializes to vector spaces. The tensor product of elements in these vector spaces that one usually sees in engineering and physics texts (frequently matrices) is basically an element in the tensor product of the corresponding vector spaces.&lt;/p&gt;&#xA;" OwnerUserId="3431" LastEditorUserId="3431" LastEditorDisplayName="" LastEditDate="2010-11-14T20:37:40.520" LastActivityDate="2010-11-14T20:37:40.520" CommentCount="4" />
  <row Id="10287" PostTypeId="2" ParentId="10284" CreationDate="2010-11-14T19:56:00.197" Score="3" ViewCount="" Body="&lt;p&gt;If I understand the problem, you have a list of 12 objects, 5 of which are special (the 0s).  (I use the word &quot;list&quot; instead of &quot;set&quot; since the latter implies there is no repetition among the objects).  Each of the 12 objects will be independently considered, and &quot;chosen&quot; with probability 1/3.  You want to know the probability of 3 of the special objects being chosen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In that case, the 7 non-special objects are irrelevant.  You just want to know the probability that exactly 3 out of 5 independent events occur, when each one occurs with probability 1/3.  This is the probability of a &lt;a href=&quot;http://en.wikipedia.org/wiki/Binomial_distribution&quot; rel=&quot;nofollow&quot;&gt;binomial random variable&lt;/a&gt; with $n=5$ and $p=1/3$ taking on the value $3$.  The linked Wikipedia page has a formula for computing this.&lt;/p&gt;&#xA;" OwnerUserId="822" LastActivityDate="2010-11-14T19:56:00.197" CommentCount="1" />
  <row Id="10288" PostTypeId="2" ParentId="10284" CreationDate="2010-11-14T19:56:30.130" Score="-1" ViewCount="" Body="&lt;p&gt;The probability that you want is $(1/3)^4(2/3)^8 N$, where $N$ is the number of subsets of length $4$ containing exactly $3$ zeroes.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-14T19:56:30.130" CommentCount="2" />
  <row Id="10289" PostTypeId="2" ParentId="2981" CreationDate="2010-11-14T20:12:40.690" Score="0" ViewCount="" Body="&lt;p&gt;There is no such thing as a &quot;countable closure of this process&quot;, as there is no a priori definition of the set of functions envisaged here, let alone a topology on this set. All we can speak of is the ${\it union}$ of the larger and larger sets of functions constructed in the process.&lt;/p&gt;&#xA;" OwnerUserId="1303" LastActivityDate="2010-11-14T20:12:40.690" CommentCount="1" />
  <row Id="10290" PostTypeId="2" ParentId="10283" CreationDate="2010-11-14T20:15:08.267" Score="1" ViewCount="" Body="&lt;p&gt;Yes, you can view this as a negative binomial distribution, but I think you are misunderstanding what the negative binomial represents. A negative binomial  with parameters $r$ and $p$  is the trial number which produces the $r$th success, where each success has probability $p$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your case, you have a negative binomial random variable $X$ with parameters $r=2$ and $p=0.1$, and you should be looking for the probability that the 2nd success occurs on 12th trial, i.e $P\{X = 12\}$. I also think you have the formula for the probability mass function slightly wrong.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you think about it the probability for the 12th trial to be a success is $0.1$. For the other 11 trials, we need to have exactly 1 out of 11 to be a success, so $(0.1)(0.9)^{10}$ for each case with $11 \choose 1$ cases. The formula should be&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$P(X=k)={k-1 \choose r-1}(1−p)^{k-r}p^r$ Use that and you should end up with what Brandon and kahen came up with.&lt;/p&gt;&#xA;" OwnerUserId="2596" LastActivityDate="2010-11-14T20:15:08.267" />
  <row Id="10291" PostTypeId="2" ParentId="10282" CreationDate="2010-11-14T20:16:50.827" Score="8" ViewCount="" Body="&lt;p&gt;In mathematics, tensors are one of the first objects encountered which cannot be fully understood without their accompanying universal mapping property.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Before talking about tensors, one needs to talk about the tensor product of vector spaces. You are probably already familiar with the direct sum of vector spaces. This is an addition operation on spaces. The tensor product provides a multiplication operation on vector spaces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The key feature of the tensor product is that it replaces &lt;em&gt;bilinear maps&lt;/em&gt; on a cartesian product of vector spaces with &lt;em&gt;linear maps&lt;/em&gt; on the tensor product of the two spaces. In essence, if $V,W$ are vector spaces, there is a bijective correspondence between the set of bilinear maps on $V\times W$ (to &lt;em&gt;any&lt;/em&gt; target space) and the set of &lt;em&gt;linear&lt;/em&gt; maps on $V\otimes W$ (the tensor product of $V$ and $W$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This can be phrased in terms of a universal mapping property. Given vector spaces $V,W$, a tensor product $V\otimes W$ of $V$ and $W$ is a space together with a map $\otimes : V\times W \rightarrow V\otimes W$ such that for any vector space $X$ and any &lt;em&gt;bilinear&lt;/em&gt; map $f : V\times W \rightarrow X$ there exists a unique &lt;em&gt;linear&lt;/em&gt; map $\tilde{f} : V\otimes W \rightarrow X$ such that $f = \tilde{f}\circ \otimes$. In other words, every bilinear map on the cartesian product factors uniquely through the tensor product.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It can be shown using a basic argument that the tensor product is unique up to isomorphism, so you can speak of &quot;the&quot; tensor product of two spaces rather than &quot;a&quot; tensor product, as I did in the previous paragraph.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A tensor is just an element of a tensor product.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One must show that such a tensor product &lt;em&gt;exists&lt;/em&gt;. The standard construction is to take the free vector space over $V\times W$ and introduce various bilinearity relations. See my link at the bottom for an article that does this explicitly. In my experience, however, the key is to be able to use the above mapping property; the particular construction doesn't matter much in the long run. The map $\otimes : V\times W \rightarrow V\otimes W$ sends the pair $(v,w) \in V\times W$ to $v\otimes w \in V\otimes W$. The image of $\otimes$ is the space of so-called elementary tensors, but a general element of $V\otimes W$ is &lt;em&gt;not&lt;/em&gt; an elementary tensor but rather a linear combination of elementary tensors. (In fact, due to bilinearity, it is enough to say that a general tensor is a sum of elementary tensors with the coefficients all being 1.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most generic reason why tensors are useful is that the tensor product is a machine for replacing bilinear maps with linear ones. In much of mathematics and physics, one seeks to find linear approximations to things; tensors can be seen as one tool for this, although exactly how they accomplish it is less clear than many other tools in the same vein. Here are some more specific reasons why they are useful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For finite-dimensional spaces $V,W$, the tensor product $V^*\otimes W$ is isomorphic to the space of homomorphisms $\text{Hom}(V,W)$. So in other words every linear map $V \rightarrow W$ has a tensor expansion, i.e., a representation as a tensor in $V^* \otimes W$. For instance, if $\{v_i\}$ is a basis of $V$ and $\{x_i\}$ is the dual basis of $V^*$, then $\sum x_i \otimes v_i \in V^* \otimes V$ is a tensor representation of the identity map on $V$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Tensor products tend to appear in a lot of unexpected places. For instance, in analyzing the linear representations of a finite group, once the irreducible representations are known it can be of benefit to construct also a &quot;tensor product table&quot; which decomposes the tensor products of all pairs of irreducible representations as direct sums of irreducible representations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In physics, one often talks about a rank $n$ tensor being an assembly of numbers which transform in a certain way under change of coordinates. What one is really describing here is all the different coordinate representations of an abstract tensor in a tensor power $V^{\otimes n}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If one takes the direct sum of all tensor powers of a vector space $V$, one obtains the tensor algebra over $V$. In other words, the tensor algebra is the construction $k\oplus V\oplus (V\otimes V) \oplus (V\otimes V\otimes V) \oplus \dots$, where $k$ is the base field.  The tensor algebra is naturally graded, and it admits several extremely useful quotient algebras, including the well-known exterior algebra of $V$. The exterior algebra provides the natural machinery for differential forms in differential geometry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's an example of the exterior algebra in practice. Suppose one wishes to classify all nonabelian two-dimensional Lie algebras $\mathfrak{g}$. The Lie bracket $[\cdot,\cdot]$ is antisymmetric and bilinear, so the machinery of tensor products turns it into a &lt;em&gt;linear&lt;/em&gt; map $\bigwedge^2 V \rightarrow V$, where $V$ is the underlying vector space of the algebra. Now $\bigwedge^2 V$ is one-dimensional and since the algebra is nonabelian the Lie bracket is not everywhere zero; hence as a linear map the Lie bracket has a one-dimensional image. Then one can choose a basis $\{X,Y\}$ of $V$ such that $[X,Y] = X$, and we conclude that there is essentially only one nonabelian Lie algebra structure on a two-dimensional vector space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A fantastic reference on tensor products of modules was written by Keith Conrad: &lt;a href=&quot;http://www.math.uconn.edu/~kconrad/blurbs/linmultialg/tensorprod.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.math.uconn.edu/~kconrad/blurbs/linmultialg/tensorprod.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="251" LastEditorUserId="251" LastEditorDisplayName="" LastEditDate="2010-11-29T15:55:04.040" LastActivityDate="2010-11-29T15:55:04.040" CommentCount="2" />
  <row Id="10292" PostTypeId="1" AcceptedAnswerId="10294" CreationDate="2010-11-14T20:22:16.703" Score="1" ViewCount="157" Body="&lt;p&gt;Hi, &#xA;so studying for my midterm on Tuesday (intro to abstract algebra). The topics on the exam are Division Algorithm, Divisibility, Prime Numbers, FTA, Congruency, Congruent Classes and very brief introduction to rings. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was reading a few theorems about Congruency and have a couple of questions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to know what a &quot;congruent class&quot; is. My notes say &quot;the congruence class of a modulo n&quot; is a set: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ \left\{ \text{all } b \in \mathbb{Z} | b \equiv a \pmod{n} \right\}  $ which is also saying&#xA;$ \left\{ \text{all } a + kn \in \mathbb{Z}  | k \in \mathbb{Z} \right\}  $&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;okay so got that. I just wrote it for some people who might need a refreshed (it is a 3rd undergrad course after all). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in my notes our professor has a following example:&#xA;$\left[ 60 \right]_{17} = \left[ 43\right]_{17}$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) &lt;strong&gt;so the way I figured this out is that to check if they are equivalent, we subtract 60-43 and see if that is a multiple of n = 17. Is this how you can check if they are equal classes?&lt;/strong&gt;  If not, is there a better way to do so? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) A certain theorem states: Let $n \in \mathbb{Z}_+; a, b \in \mathbb{Z}$ and $gcd(a,n) = d$ then $[a]x=[b]$ has exactly $d$ solutions. &lt;strong&gt;My question here is that is x a congruence class or a random integer? What is x and how do I solve for it?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) Is it true that if we are in $\mathbb{Z}_{12}$ then $[7]x=[11]$ can be rewritten as $ 7x \equiv 11 \pmod{12}$? If so, &lt;strong&gt;would finding the solution be similar to &lt;a href=&quot;http://math.stackexchange.com/questions/2991/not-understanding-simple-modulus-congruency&quot;&gt;solution in this question&lt;/a&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thankyou. I am just very confused about congurency and stuff. I understand the theorems but I am hoping someone would give me an &quot;easy&quot; explanation of what is going on. I still don't know the difference between circle plus and regular plus except that circle plus has to satisfy certain axioms. Am I right?&lt;/p&gt;&#xA;" OwnerUserId="287" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-14T21:58:07.597" LastActivityDate="2010-11-14T23:17:41.867" Title="Congruency and Congruent Classes" Tags="&lt;abstract-algebra&gt;&lt;elementary-number-theory&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="10293" PostTypeId="2" ParentId="10177" CreationDate="2010-11-14T20:23:22.630" Score="3" ViewCount="" Body="&lt;p&gt;Here's a nice solution I just thought of, which may in fact be logically equivalent to Jim's (if it's even correct!).  I welcome comments addressing that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Compactify $\mathbb{R}^{n+1}$ to $S^{n+1}$, so we consider $M \subset S^{n+1}$.  By Alexander duality, $$\tilde{H}_0(S^{n+1} \backslash M ; \mathbb{Z}/2) \cong \tilde{H}^{(n+1)-0-1}(M;\mathbb{Z}/2) = \tilde{H}^n(M;\mathbb{Z}/2)=\mathbb{Z}/2,$$ so $H_0(S^{n+1}\backslash M;\mathbb{Z}/2)=\mathbb{Z}/2 \oplus \mathbb{Z}/2$.  Hence $M$ separates $S^{n+1}$.&lt;/p&gt;&#xA;" OwnerUserId="2159" LastActivityDate="2010-11-14T20:23:22.630" CommentCount="4" />
  <row Id="10294" PostTypeId="2" ParentId="10292" CreationDate="2010-11-14T20:32:48.750" Score="1" ViewCount="" Body="&lt;p&gt;There is a much more general definition of congruence classes, but I shall restrict to the one that is sufficient for your course. Given any integer $n$ the only possible remainders that we can get when we divide an integer $a$ by $n$ are $0,1,...,n-1$. &#xA;Each set of integers which leave the same remainder on division by $n$ form what is called as a congruence class modulo $n$. All such congruence classes are mutually disjoint since a number can leave only one remainder on division by $n$. Their union is the set of all integers. Each integer in any given congruence class is said to be a representative of the class.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To check whether two integers are in the same class, we check their difference and see if it is divisible by $n$ (since the remainders of these integers cancel out when we divide by $n$). So your approach to your first question is right.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the second question, $x$ is indeed a congruence class, since otherwise the equation does not make sense. We can define operations on the congruence classes by the correponding operations on their representatives. Its easy to check these are well defined.&lt;/p&gt;&#xA;" OwnerUserId="3431" LastActivityDate="2010-11-14T20:32:48.750" CommentCount="3" />
  <row Id="10295" PostTypeId="2" ParentId="10274" CreationDate="2010-11-14T20:38:56.230" Score="0" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\rm\quad\ \ A = X+Y\ \ \Rightarrow\ \ X\ Y = - Y\ X\:.\ $ But $\rm -1 = 1\ $ via $\rm\ A = -1$&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-11-14T20:38:56.230" />
  <row Id="10296" PostTypeId="2" ParentId="10042" CreationDate="2010-11-14T20:40:19.490" Score="4" ViewCount="" Body="&lt;p&gt;See the equivalent question on mathoverflow: &lt;a href=&quot;http://mathoverflow.net/questions/8829/what-manifolds-are-bounded-by-rpodd&quot; rel=&quot;nofollow&quot;&gt;What manifolds are bounded by RP^odd?&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Since it seems there is a good reason to have the answer recorded as such (and, borrowing from the suggestion &lt;a href=&quot;http://meta.math.stackexchange.com/questions/1148/what-should-one-do-when-ones-question-has-been-answered-in-the-comments&quot;&gt;here&lt;/a&gt;), I'm moving my comment here.  Since all I'm doing it linking to another place, I don't want to gain reputation for this, so I'm making it community wiki.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, in an effort to personally gain &lt;i&gt;something &lt;/i&gt; from this, I'll provide a link to a similar &lt;a href=&quot;http://mathoverflow.net/questions/14698/what-manifold-has-mathbbhpodd-as-a-boundary&quot; rel=&quot;nofollow&quot;&gt;question&lt;/a&gt; I asked on MO which still hasn't been answered.  The question is: What manifold has $\mathbb{H}P^{odd}$ as a boundary?  Incidentally, the case of $\mathbb{C}P^{odd}$ is covered in my question.&lt;/p&gt;&#xA;" OwnerUserId="331" LastEditorUserId="331" LastEditorDisplayName="" LastEditDate="2010-11-14T20:47:31.820" LastActivityDate="2010-11-14T20:47:31.820" CommunityOwnedDate="2010-11-14T20:40:19.490" CommentCount="2" />
  <row Id="10297" PostTypeId="1" AcceptedAnswerId="10305" CreationDate="2010-11-14T20:55:00.653" Score="1" ViewCount="116" Body="&lt;p&gt;Suppose Jill and Jane are individually painting separate rooms. Each of the rooms has $4$ walls. The time taken to paint each wall is exponentially distributed with mean $1$ hour. What is the probability that Jill will finish painting $3$ walls before Jane finishes $2$ walls? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This seems to look somewhat similar to a negative binomial distribution. Is this the correct distribution to use?  &lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2010-11-15T20:31:17.400" Title="Painting Walls and Probability" Tags="&lt;probability&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" />
  <row Id="10298" PostTypeId="2" ParentId="10284" CreationDate="2010-11-14T21:00:23.987" Score="2" ViewCount="" Body="&lt;p&gt;If I understand the problem, you have a sequence of independent and identically distributed random variables $X_1,X_2,\ldots,X_n$ (in your example $n=11$, as noted by Yuval), such that ${\rm P}(X_i = j) = p_j$ (in your example $j \in \lbrace 0,1,\ldots,6 \rbrace $); you sample each $X_i$ with probability $p$ (so in average you pick $pn$ numbers; in your example $p=1/3$), and wants to find the probability $P(n,j,m)$ of obtaining $0 \leq m \leq n$ occurrences of the number $j$. Then, this probability can be computed, using the law of total probability, as follows:&#xA;$$&#xD;&#xA;P(n,j,m) = \sum\limits_{k = m}^n {{n \choose k}p^k (1 - p)^{n - k} {k \choose m} p_j^m (1 - p_j )^{k - m} }. &#xD;&#xA;$$&#xD;&#xA;The term ${n \choose k}p^k (1 - p)^{n - k}$ corresponds to sampling $k$ $X_i$'s out of $n$, and the term&#xA;${k \choose m} p_j^m (1 - p_j )^{k - m} $ corresponds to obtaining $m$ occurrences of the number $j$ given that $k (\geq m)$ $X_i$'s were sampled. The rest is left for you.&lt;/p&gt;&#xA;" OwnerUserId="2810" LastActivityDate="2010-11-14T21:00:23.987" />
  <row Id="10299" PostTypeId="2" ParentId="10284" CreationDate="2010-11-14T21:18:32.603" Score="3" ViewCount="" Body="&lt;p&gt;In your original question it sounded like you wanted the conditional probability $P\{\text{we obtain 3 0&amp;#39;s}|\text{4 events are sampled}\}$ which is the number of size 4 subsets that contain 3 0's divided by the total number of size 4 subsets which would be, if $k$ is the number of 0's total, $$\frac{{k \choose 3}{12-k \choose 1}}{{12 \choose 4}}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general it sounds like what you really want is the probability that any single sample is either a high frequency event or a low frequency event. Say you have an event that occurs with a high frequency, but that each trial is independent, we can model this by giving it a probability, say $p_h=\frac{n}{100}$ where n is the average # that occur in any 100 trials.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now say you sample each trial independently with probability $p_s$. Then the probability that any single sample is one of the high frequency events is the conditional probability $$P\{\text{event is the high frequency event}| \text{event is sampled}\} = p_h $$&#xD;&#xA;This is because, presumably, your sampling and your event occurrences are independent&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It also sounds like you would like the expected value of the number of high frequency events sampled given a certain number of total events, $n$. Consider any single event. In order for it to be a &quot;success&quot; it has to be both a high frequency event and be sampled. $$P\{\text{an event is high frequency and is sampled}\} = p_h p_s$$&#xD;&#xA;The probability to have $k$ high frequency events sampled when you have $n$ total events is a binomial distribution with parameter $p_hp_s$.&#xA;$$P\{k \text{ of our sampled events are high frequency events}\} = {n \choose k} (p_sp_h)^k(1-p_sp_h)^{n-k}$$&#xD;&#xA;The expected number of sampled events that are high frequency is then just $np_sp_h$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Everything works the same with a low frequency event, just use an appropriate $p_l$ instead of $p_h$.&lt;/p&gt;&#xA;" OwnerUserId="2596" LastActivityDate="2010-11-14T21:18:32.603" />
  <row Id="10300" PostTypeId="2" ParentId="10282" CreationDate="2010-11-14T21:19:25.087" Score="11" ViewCount="" Body="&lt;p&gt;Once you understand what a &lt;a href=&quot;http://en.wikipedia.org/wiki/Tensor_product&quot; rel=&quot;nofollow&quot;&gt;tensor product&lt;/a&gt; is and what a &lt;a href=&quot;http://en.wikipedia.org/wiki/Dual_space&quot; rel=&quot;nofollow&quot;&gt;dual space&lt;/a&gt; is, then a tensor of type $(n, m)$ is an element of $V^{\ast \otimes m} \otimes V^{\otimes n}$ where $V$ is some vector space.  This is the same thing as a multilinear map $V^m \to V^{\otimes n}$ or, if you don't like the asymmetry, a multilinear map $V^{\ast n} \times V^{m} \to F$ (where $F$ is the underlying field).  Examples:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;A tensor of type $(0, 0)$ is a scalar.&lt;/li&gt;&#xA;&lt;li&gt;A tensor of type $(1, 0)$ is a vector.&lt;/li&gt;&#xA;&lt;li&gt;A tensor of type $(0, 1)$ is a covector.&lt;/li&gt;&#xA;&lt;li&gt;A tensor of type $(1, 1)$ is a linear transformation.&lt;/li&gt;&#xA;&lt;li&gt;A tensor of type $(0, 2)$ is a bilinear form, for example an inner product.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;When you pick a basis of $V$, you can write tensors in terms of the natural basis on $V^{\ast \otimes n} \otimes V^{\otimes m}$ coming from taking products of the basis on $V$ with the corresponding dual basis on $V^{\ast}$.  This is where the &quot;multidimensional array&quot; definition of a tensor comes from, since this is the natural generalization of writing a matrix as a square array (which is equivalent to writing an element of $V^{\ast} \otimes V$ in terms of the basis $e_i^{\ast} \otimes e_j$ where $\{ e_i \}$ is a basis).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When a physicist says &quot;tensor,&quot; sometimes they mean a &lt;a href=&quot;http://en.wikipedia.org/wiki/Tensor_field&quot; rel=&quot;nofollow&quot;&gt;tensor field&lt;/a&gt;.  This is a &quot;globalization&quot; of the above definition: it is a compatible set of choices, for each tangent space $V = T_p(M)$ of a smooth manifold $M$, of a tensor of type $(n, m)$ as defined above.  Note that $V^{\ast}$ is the cotangent space.  Examples:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;A tensor field of type $(0, 0)$ is a smooth function.&lt;/li&gt;&#xA;&lt;li&gt;A tensor field of type $(1, 0)$ is a vector field.&lt;/li&gt;&#xA;&lt;li&gt;A tensor field of type $(0, 1)$ is a differential $1$-form.&lt;/li&gt;&#xA;&lt;li&gt;A tensor field of type $(1, 1)$ is a morphism of vector fields.&lt;/li&gt;&#xA;&lt;li&gt;A tensor field of type $(0, 2)$ which is symmetric and nondegenerate is a &lt;a href=&quot;http://en.wikipedia.org/wiki/Metric_tensor&quot; rel=&quot;nofollow&quot;&gt;metric tensor&lt;/a&gt;.  If it is also positive-definite, it is a &lt;a href=&quot;http://en.wikipedia.org/wiki/Riemannian_manifold&quot; rel=&quot;nofollow&quot;&gt;Riemannian metric&lt;/a&gt;.  If it has signature $(1, n-1)$, it is a &lt;a href=&quot;http://en.wikipedia.org/wiki/Pseudo-Riemannian_manifold#Lorentzian_manifold&quot; rel=&quot;nofollow&quot;&gt;Lorentzian metric&lt;/a&gt;.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-11-15T11:24:37.547" LastActivityDate="2010-11-15T11:24:37.547" CommentCount="8" />
  <row Id="10301" PostTypeId="2" ParentId="9791" CreationDate="2010-11-14T21:24:13.267" Score="4" ViewCount="" Body="&lt;p&gt;You can find the first &lt;em&gt;n&lt;/em&gt; primes in time $n\log n/\log\log n$ with the Atkin-Bernstein sieve.  This is better than your naive $n\log^5n$ or so algorithm (or drop the exponent by 1 using pretesting with M-R), but still superlinear.&lt;/p&gt;&#xA;" OwnerUserId="1778" LastEditorUserId="1778" LastEditorDisplayName="" LastEditDate="2010-11-15T18:07:32.193" LastActivityDate="2010-11-15T18:07:32.193" CommentCount="1" />
  <row Id="10302" PostTypeId="2" ParentId="10297" CreationDate="2010-11-14T21:32:00.060" Score="1" ViewCount="" Body="&lt;p&gt;Let $T$ denote the period of time until Jill finishes painting $3$ walls. Then, $T$ has gamma density $f_T (t) = t^2 {\rm e}^{-t}/2$, $t &amp;gt; 0$. Viewing the number of walls painted by Jane up to time $t$ as a Poisson process (with unit rate), the question amounts to calculate&#xA;$$&#xD;&#xA;\int_0^\infty  {{\rm P}(N_t  &amp;lt; 2|T = t)f_T (t)\,{\rm d}t}&#xD;&#xA;$$&#xD;&#xA;(by virtue of the law of total probability, conditioning on $T$), where $\lbrace N_t : t \geq 0 \rbrace$ is a unit rate Poisson process. The required probability then follows to be $0.3125 = 5/16$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A complete generalization:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose Jill and Jane are individually painting separate rooms. Each of the rooms has $n \ge \max \lbrace m_1 , m_2 \rbrace$ walls. The time it takes Jill and Jane to paint each wall is exponentially distributed with mean $1/\lambda_1$ and $1/\lambda_2$ hour, respectively. What is the probability that Jill will finish painting $m_1$ walls before Jane finishes $m_2$ walls? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer is&#xA;$$&#xD;&#xA;\bigg(\frac{{\lambda _1 }}{{\lambda _1  + \lambda _2 }}\bigg)^{m_1 } \sum\limits_{k = 0}^{m_2  - 1} {{k + m_1  - 1 \choose k} \bigg(\frac{{\lambda _2 }}{{\lambda _1  + \lambda _2 }}\bigg)^k }.&#xD;&#xA;$$&#xD;&#xA;This can be proved easily either by using Mike's method, noting that ${\rm P}(X_1  &amp;lt; X_2 ) = \frac{{\lambda _1 }}{{\lambda _1  + \lambda _2 }}$ for independent exponential rv's $X_1$ and $X_2$ with respective means $1/\lambda_1$ and $1/\lambda_2$, or by using my method, noting that the time $T$ it takes Jill to paint $m_1$ walls has gamma density $f_T (t) = \frac{{\lambda _1^{m_1 } {\rm e}^{ - \lambda _1 t} t^{m_1  - 1} }}{{(m_1  - 1)!}}$, $t &amp;gt; 0$. (Letting $\lambda_1 = \lambda_2 = 1$, $m_1=3$, and $m_2=2$, this formula gives $5/16$, the same as for the original problem.)&lt;/p&gt;&#xA;" OwnerUserId="2810" LastEditorUserId="2810" LastEditorDisplayName="" LastEditDate="2010-11-15T20:31:17.400" LastActivityDate="2010-11-15T20:31:17.400" />
  <row Id="10303" PostTypeId="2" ParentId="10256" CreationDate="2010-11-14T21:53:27.463" Score="3" ViewCount="" Body="&lt;p&gt;If $\mathbf{v}\in A$ is in the kernel of $\phi_i$, then for all $j\leq i$ you have $\phi_j(\mathbf{v}) = f_{ij}\phi_j(\mathbf{v}) = f_{ij}(\mathbf{0}) = \mathbf{0}$ (where $f_{ij}\colon A_i\to A_j$ is the map obtained by composing the $f_k$). Therefore, $\mathrm{ker}(\phi_i)\subseteq \mathrm{ker}(\phi_j)$ for all $i\geq j$. Hence you have a descending sequence&#xA;\begin{equation*}&#xA;\cdots \subseteq \mathrm{ker}(\phi_{i+1})\subseteq \mathrm{ker}(\phi_i)\subseteq \cdots \subseteq \mathrm{ker}(\phi_0).&#xA;\end{equation*}&#xA;Since $\dim(A)\lt\infty$, the sequence must stabilize, so there exists $n_0$ such that for all $n\geq n_0$, $\mathrm{ker}(\phi_n)=\mathrm{ker}(\phi_{n_0})$. Let $\mathbf{N}$ be that subspace of $A$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $\mathbf{v}\in\mathbf{N}$, then $\phi_i(\mathbf{v})=\mathbf{0}$ for all $i$, hence $\phi(\mathbf{v})=\mathbf{0}$, so $\mathbf{N}\subseteq \mathrm{ker}(\phi)$. Conversely, anything in $\mathrm{ker}(\phi)$ lies in the intersection of all kernels of the $\phi_i$. Thus, $\mathbf{N}=\mathrm{ker}(\phi)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In conclusion, $\phi$ is one to one if and only if the $\phi_i$ are eventually one-to-one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For surjection, you are going to be out of luck. Consider the system with $A_i = \mathbf{F}^2$ for all $i$, $f_i\colon\mathbf{F}^2\to\mathbf{F}^2$ given by $f_i(a,b)=(a,0)$. I claim that the inverse limit of this system is $\mathbf{V}=\mathbf{F}$ with structure maps $\mathfrak{f}_i\colon\mathbf{V}\to\mathbf{F}^2$ given by $\mathfrak{f}(a) = (a,0)$. First, note that this is a consistent system of maps from $\mathbf{V}$ to the $A_i$. Next, suppose that $\mathbf{W}$ is any vector space and that you have maps $g_i\colon\mathbf{W}\to\mathbf{F}^2$ for all $i$, with the property that for all $i\geq j$, $g_j(\mathbf{w}) = f_{ij}(g_i(\mathbf{w}))$. If $g_i(\mathbf{w}) = (\mathbf{w}_{1i},\mathbf{w}_{2i})$, then this means that $g_j(\mathbf{w}) = (\mathbf{w}_{1i}, 0)$. In particular, for all $j$ we have that $\mathrm{Im}(g_j)$ is contained in the subspace $\{(x,0)\mid x\in\mathbf{F}\}$ of $A_j$, and since the $f_i$ are isomorphisms when restricted to those subspaces, then it follows that for each $\mathbf{w}\in\mathbf{W}$ there is an $a_{\mathbf{w}}\in\mathbf{F}$ such that $g_j(\mathbf{w}) = (a_{\mathbf{w}},0)$ for all $j$. Hence, we have a map $\mathfrak{g}\colon\mathbf{W}\to \mathbf{V}$ with $\mathfrak{f}_j\circ\mathfrak{g}=g_j$ for all $j$, namely $\mathfrak{g}(\mathbf{w}) = a_{\mathbf{w}}$, and the map is clearly unique. Thus, $(\mathbf{V},\mathfrak{f}_i)$ is the inverse limit of the system. Then $\mathbf{V}$ itself with the structure maps gives a counterexample: the induced map to the inverse limit is onto (in fact, it is the identity), but the structure maps are never onto, so it is false that $\phi_i$ is onto for sufficiently large $i$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added:&lt;/strong&gt; However, if your structure maps $f_i$ are &lt;em&gt;eventually surjective&lt;/em&gt;, then your conclusion will follow. To see this, assume that $f_i$ is surjective for all $i\geq N$; I claim that the inverse limit maps $\mathfrak{f}_j\colon\lim\limits_{\leftarrow}A_i\to A_j$ are surjective for all $j\geq N$. Fix $j\geq N$, and let $\mathbf{v}\in A_j$. Let $\mathbf{v}_{j+1}\in A_{j+1}$ be any preimage of $\mathbf{v}$, and inductively define $\mathbf{v}_{j+k+1}$ to be any preimage of $\mathbf{v}_{j+k}$. Then define maps $g_i\colon\mathbf{F}\to A_i$ by letting $g_i(1) = \mathbf{v}_{j+m}$ if $i=j+m\geq N$, and $g_i(1)=f_{ji}(\mathbf{v})$ if $i\lt N$. This induces a map $\mathfrak{g}\colon\mathbf{F}\to\lim\limits_{\leftarrow}A_i$, and $g_i = \mathfrak{f}_i\circ\mathfrak{g}$. In particular, $\mathbf{v}\in\mathrm{Im}(\mathfrak{f}_j)$. Thus, $\mathfrak{f}_j$ is onto for all $j\geq N$. Then, if you have $\phi\colon A\to A_i$ as in your statement surjective, then $\phi_j = \mathfrak{f}_j\circ\phi\colon A\to A_j$ is a composition of surjective functions for alL $j\geq N$, so the $\phi_j$ are eventually surjective, as desired. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added 2:&lt;/strong&gt; To see how I came up with the example, notice that if you let $B_{i-1} = \mathrm{Im}(f_i)$, then you get an inverse system with the $B_i$ instead of the $A_i$; the inverse limit of this system is isomorphic to the inverse limit of the original system (try to prove this). This is because a consistent system of homomorphism $\phi_i\colon A\to A_i$ has to have $\mathrm{Im}(\phi_i)\subseteq B_i$. So what really matters is what is going on in the images of the structure maps $f_i$, hence surjectivity into the limit depends only on how your maps $f_i$ relate to the $B_i$, not how they relate to the $A_i$. That is why if the maps $f_i$ are eventually surjective you do get the implication.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-15T00:38:44.463" LastActivityDate="2010-11-15T00:38:44.463" />
  <row Id="10304" PostTypeId="1" AcceptedAnswerId="10306" CreationDate="2010-11-14T21:53:34.850" Score="2" ViewCount="91" Body="&lt;p&gt;Suppose we have an infinite sequence of intervals $[a_n,b_n]$, where for all n, $a_n&amp;lt;b_n$. Then is there some $x\in \mathbb{R}$ such that for each n there is some $k\in\mathbb{Z}$ such that $kx\in [a_n,b_n]$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems unlikely, but I'm not sure where to begin with the proof. Thank you.&lt;/p&gt;&#xA;" OwnerUserId="2934" LastActivityDate="2010-11-15T13:44:39.623" Title="For a sequence of intervals can we find some x such that each interval contains a multiple of x?" Tags="&lt;real-analysis&gt;" AnswerCount="1" />
  <row Id="10305" PostTypeId="2" ParentId="10297" CreationDate="2010-11-14T21:56:15.767" Score="3" ViewCount="" Body="&lt;p&gt;Here's another way to get the answer of $\frac{5}{16}$.  The number of possible ways for Jill to finish three walls before Jane finishes two is small enough to just list.  They are (with $X$ denoting Jill finishing a wall and $Y$ denoting Jane finishing a wall)&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;XXX     &#xA;YXXX    &#xA;XYXX    &#xA;XXYX&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Because of the memorylessness of the exponential distribution, starting from any point in time the probability that the next wall will be finished by Jill is $\frac{1}{2}$.  Thus the four listed events have probabilities $\frac{1}{8}$, $\frac{1}{16}$, $\frac{1}{16}$, and $\frac{1}{16}$, respectively.  So the probability that Jill finishes three before Jane finishes two is $\frac{5}{16}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Update: This answer assumes that you're looking for the probability that Jill finishes &lt;em&gt;at least&lt;/em&gt; 3 walls before Jane finishes 2.  If you want the probability that Jill finishes &lt;em&gt;exactly&lt;/em&gt; 3 walls before Jane finishes 2, then Yuval Filmus is correct that you want $P(X = 3)$, where $X$ is a negative binomial random variable with success probability $p = \frac{1}{2}$ and number of failures $r = 2$.  Here, $P(X = 3) = \frac{1}{8}$.)&lt;/p&gt;&#xA;" OwnerUserId="2370" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-11-15T05:12:23.920" LastActivityDate="2010-11-15T05:12:23.920" CommentCount="4" />
  <row Id="10306" PostTypeId="2" ParentId="10304" CreationDate="2010-11-14T22:05:15.383" Score="9" ViewCount="" Body="&lt;p&gt;Sketch of a proof: Take $a_n = 2^{-n}$ and $b_n = 2^{-n} + 2^{-n-2}$. Suppose such $x$ exists, it must be non-zero since none of the intervals intersect the origin. WLOG assume it is positive. Then take $N$ sufficiently large such that $2^{-N + 1} &amp;lt; x$. Then there cannot be any integer $k$ such that $a_N &amp;lt; k x$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Edit&lt;/em&gt; Since the OP asked about sequences of intervals bounded away from 0. Clearly there exists sequences such that the desired property is true: just take $[a_n,b_n]$ to be intervals containing $n$. But there can also be counter examples. Let $2&amp;gt; C_n &amp;gt; 1$ be a decreasing sequence of numbers, and $\delta_n &amp;lt; C_n - C_{n+1}$ a decreasing sequence of numbers tending to 0. $C_n$ will be prescribed at the end of the proof. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the intervals $a_n = C_n - \delta_n$, $b_n = C_n$. By construction the intervals do not intersect. Let $m$ be an integer. If there were to exist some $x$ such that there exists $k,k&amp;#39;\in \mathbb{Z}$ such that $kx \in [a_m,b_m]$ and $k&amp;#39;x \in [a_{m+1},b_{m+1}]$, this will require &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ x \in [a_m / k, b_m/k] \cap [a_{m+1}/k&amp;#39;, b_{m+1}/k&amp;#39;] \subset [a_{m+1}/k, b_m/k] \cap [a_{m+1}/k&amp;#39;, b_m/k&amp;#39;] $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we are led to consider intersections of annuli with rescalings of itself. Now, if $\frac{b_m}{a_{m+1}} - 1 &amp;lt; 1 / N$, then for any $k,k&amp;#39; &amp;lt; N$, the intersection $[a_{m+1}/k, b_m/k] \cap [a_{m+1}/k&amp;#39;, b_m/k&amp;#39;] = \emptyset$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, let $C_n = \frac{n+1}{n}$. Then &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$b_m - a_{m+1} &amp;lt; b_m - b_{m+2} = \frac1m \frac2{m+2} &amp;lt; \frac1m b_{m+1} &amp;lt; \frac1m a_{m+1}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So this means that for $x$ to have multiples belonging to both $[a_m,b_m]$ and $[a_{m+1},b_{m+1}]$, said multiplication factor needs to be at least $m$ (actually at least $m^2$, but $m$ suffices for the proof). But since $1 &amp;lt; a_m,b_m &amp;lt; 2$, this implies that $x &amp;lt; 2/m$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now take $m\to \infty$, we see that $x$ must equal to 0, which is absurd. &lt;/p&gt;&#xA;" OwnerUserId="1543" LastEditorUserId="1543" LastEditorDisplayName="" LastEditDate="2010-11-15T13:44:39.623" LastActivityDate="2010-11-15T13:44:39.623" CommentCount="2" />
  <row Id="10307" PostTypeId="2" ParentId="10292" CreationDate="2010-11-14T22:36:50.577" Score="1" ViewCount="" Body="&lt;p&gt;Okay, let's start from the beginning. You define the congruence class of $a$ as:&#xA;\begin{equation*}&#xA;[a]_n = \{ b\in\mathbb{Z}\mid a\equiv a\pmod{n}\}.&#xA;\end{equation*}&#xA;Now, what does $a\equiv b\pmod{n}$ mean? It means that $n$ divides $a-b$, or equivalently, that $a$ and $b$ leave the same remainder when you divide them by $n$. Considering the last of the conditions, notice that:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Since each number leaves the same remainder as itself when divided by $n$, $a\equiv a \pmod{n}$ for all $a$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If $a$ leaves the same remainder as $b$ when divided by $n$, then $b$ leaves the same remainder as $a$ when divided by $n$. So if $a\equiv b\pmod{n}$, then $b\equiv a \pmod{n}$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If $a$ leaves the same remainder as $b$, and $b$ leaves the same remainder as $c$, then $a$ and $c$ leave the same remainder as well. So if $a\equiv b\pmod{n}$ and $b\equiv c\pmod{n}$, then $a\equiv c\pmod{n}$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;That means that:&#xA;\begin{equation*}&#xA;[a]_{n} = [b]_n\Longleftrightarrow a\equiv b\pmod{n}.&#xA;\end{equation*}&#xA;Why? Well, suppose the congruence classes are the same. Since $a\in[a]_n=[b]_n$, that means that $b\equiv a \pmod{n}$ (since $a\in[b]_n$), so $a\equiv b\pmod {n}$. That proves that if the classes are equal, then $a\equiv b\pmod{n}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conversely, suppose that $a\equiv b\pmod{n}$. How do we prove that $[a]_{n}=[b]_{n}$? Since they are sets, the usual way is to show that each is contained in the other. If $c\in[a]_n$, then $a\equiv c\pmod{n}$ by definition; since we also have $b\equiv a\pmod{n}$ (since $a\equiv b\pmod{n}$ is our assumption), then $b\equiv c\pmod{n}$, so $c\in[b]_n$. Therefore, every that is in $[a]_{n}$ is also in $[b]_n$, so $[a]_n\subseteq [b]_n$. For the converse inclusion, if $c\in[b]_n$, then $b\equiv c\pmod{n}$, and since we also have $a\equiv b\pmod{n}$, then $a\equiv c \pmod{n}$ so $c\in[a]_n$. Therefore, $[b]_n\subseteq [a]_n$. Since we have the two inclusions, we conclude that $[a]_n=[b]_n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We also have the following, which is perhaps more surprising:&#xA;\begin{equation*}&#xA;[a]_n\cap[b]_n\neq\emptyset\Longleftrightarrow [a]_n=[b]_n.&#xA;\end{equation*}&#xA;That is: the only way the class of $a$ and the class of $b$ have &lt;em&gt;anything&lt;/em&gt; in common is if they are identical. Why? Well, if they are identical they certainly have nonempty intersection, since the class of $a$ always includes at least $a$. And conversely, if $c\in[a]_n\cap[b]_n$, then $a\equiv c\pmod{n}$ and $b\equiv c\pmod{n}$, from which we conclude that $a\equiv b\pmod{n}$, so $[a]_n=[b]_n$ by what we just finished proving.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On to your questions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;(1)&lt;/strong&gt; How do we check if $[a]_{17} = [b]_{17}$? Precisely the way you did it: the classes are equal if and only if $a\equiv b \pmod{17}$. How do you check if $a\equiv b\pmod{17}$? By checking to see if $17$ divides $a-b$. So how do you check in general whether $[a]_n=[b]_n$? You check to see if $a-b$ is a multiple of $n$. If it is, then the classes are equal. If $a-b$ is not a multiple of $n$, then they are not equal, and in fact they are disjoint.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are other ways: for example, if you could show that $[60]_{17}$ and $[43]_{17}$ have &lt;em&gt;any&lt;/em&gt; element in common, then you would be able to conclude that they are equal. &lt;em&gt;Sometimes&lt;/em&gt; it may be simpler to see that $[a]_n$ and $[b]_n$ have some element in common than to check if $n$ divides $a-b$; but the &lt;em&gt;standard&lt;/em&gt; way of checking is to see whether $n$ divides $a-b$, just as you did.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;(2)&lt;/strong&gt; What you have is an equation in which $x$ is an unknown. You are really looking for all solutions to the congruence&#xA;\begin{equation*}&#xA;ax \equiv b \pmod {n}&#xA;\end{equation*}&#xA;That is, all integers $x$ that make the congruence true. For example, $x=3$ is a solution to&#xA;\begin{equation*}&#xA;2x \equiv 1 \pmod{5}&#xA;\end{equation*}&#xA;because $(2)(3)= 6\equiv 1 \pmod{5}$. In fact, because $3$ is a solution, so is $3+5k$ for any integer $k$; that is, &lt;em&gt;any&lt;/em&gt; element in $[3]_5$ is a solution, since $3$ is a solution. So the solutions will &lt;em&gt;actually&lt;/em&gt; be a collection of congruence classes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moreover, if we replace $2$ in the equation with $7$, so that we are trying to solve $7x\equiv 1 \pmod{5}$, then anything which was a solution to $2x\equiv 1\pmod{5}$ is still a solution, and any solution to the first one is still a solution to the new congruence; why? because $2x\equiv 7x\pmod{5}$ for &lt;em&gt;all&lt;/em&gt; $x$, as $2x-7x = -5(x)$ is always a multiple of $5$. So we can replace $2$ with &lt;em&gt;any&lt;/em&gt; element of $[2]_5$ and not change the solutions. Likewise, we can replace $1$ with any element of $[1]_5$ and not change the solutions. So it's almost as if instead of trying to solve the single congruence&#xA;\begin{equation*}&#xA;2x \equiv 1 \pmod{5}&#xA;\end{equation*}&#xA;we are trying to solve the equation&#xA;\begin{equation*}&#xA;[2]_5 x = [1]_5&#xA;\end{equation*}&#xA;So that's why you have written $[a]x=[b]$. That really means the congruence $ax\equiv b \pmod{n}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you wrote, however, is incorrect. You are missing a clause: the congruence will have $d$ solutions if $d$ divides $b$. Otherwise, it's not going to have any. For an easy example, consider the congruence $2x\equiv 1 \pmod{4}$. Then $\gcd(2,4)=2$, but there are no solutions to the congruence, because $2x$ is always even, so $2x-1$ is always odd, so $4$ never divides $2x-1$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, how do you solve a congruence $ax\equiv b\pmod{n}$ when $\gcd(a,n)$ divides $b$? let $d=\gcd(a,n)$. Then we can write $a=da&amp;#39;$, $b=db&amp;#39;$, and $n=dn&amp;#39;$. Then $ax\equiv b\pmod{n}$ if and only if $n$ divides $ax-b$. But $ax-b = d(a&amp;#39;x - b&amp;#39;)$, and $dn&amp;#39;$ divides $d(a&amp;#39;x-b&amp;#39;)$ if and only if $n&amp;#39;$ divides $a&amp;#39;x-b&amp;#39;$, if and only if $a&amp;#39;x\equiv b&amp;#39;\pmod{n&amp;#39;}$. So if we can solve a congruence when $\gcd(a,n)=1$, then we can solve any congruence where $\gcd(a,n)$ divides $b$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, how do you solve a system $ax \equiv b\pmod{n}$ when $\gcd(a,n)=1$ (which will &lt;em&gt;of course&lt;/em&gt; divide $n$)? Since $\gcd(a,n)=1$, then we can write $n$ as a linear combination of $a$ and $n$, $1=ar+ns$ for some integers $r$ and $s$ (for example, using the Euclidean algorithm). Multiplying through by $b$ you get $b=a(rb)+ n(sb)$. That means that $n(sb) = a(rb)-b$, so $a(rb)\equiv b \pmod{n}$, which means that $x=rb$ is a solution. If $y$ is any other solution, then $ay\equiv b\pmod{n}$, and $ax\equiv b\pmod{n}$, so $ax\equiv ay\pmod{n}$, hence $n$ divides $ax-ay=a(x-y)$, and since $\gcd(a,n)=1$, then $n$ divides $x-y$; so if $y$ is any other solution, then $x\equiv y\pmod{n}$. So the only &lt;em&gt;congruence class&lt;/em&gt; that is a solution is $[x]_n$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, what about general congruences? Suppose you have $[a]_nx=[b]_n$ and $\gcd(a,n)=d$ divides $b$. Write $a=da&amp;#39;$, $b=db&amp;#39;$, and $n=dn&amp;#39;$. Notice that $\gcd(a&amp;#39;,n&amp;#39;)=1$. Find a solution $x_0$ to $[a&amp;#39;]_{n&amp;#39;}x=[b&amp;#39;]_{n&amp;#39;}$. Then $n&amp;#39;$ divides $a&amp;#39;x_0 -b&amp;#39;$, so $dn&amp;#39;=n$ divides $d(a&amp;#39;x_0-b&amp;#39;) = da&amp;#39;x_0-db&amp;#39; = ax_0-b$. So $x_0$ is also a solution to the original problem. If $y$ is any other solution, then as before we get that $n$ divides $a(x_0-y)$; so $dn&amp;#39;$ divides $da&amp;#39;(x_0-y)$, hence $n&amp;#39;$ divides $a&amp;#39;(x_0-y)$, and since $\gcd(a&amp;#39;,n&amp;#39;)=1$, then $n&amp;#39;$ divides $x_0 - y$; that is, $y\equiv x_0\pmod{n&amp;#39;}$, so $y=x_0+kn&amp;#39;$ for some $k$. Conversely, if $y = x_0+\ell n&amp;#39;$ for some $\ell$, then &#xA;\begin{equation*}&#xA;ay = a(x_0+\ell n') = ax_0 + \ell an' = ax_0 + \ell (da')n' = ax_0 + \ell a'(dn') = ax_0 + (\ell a')n,&#xA;\end{equation*} &#xA;which says that $ay \equiv ax_0 \pmod{n}$, so $y$ is also a solution. So what are the different solutions? Well, we have $x_0$, $x_0+n&amp;#39;$, $x_0+2n&amp;#39;$, $x_0+3n&amp;#39;,\ldots, x_0+(d-1)n&amp;#39;$. All of these are &lt;em&gt;not&lt;/em&gt; congruent to one another modulo $n$; the &quot;next&quot; one, however, will be $x_0+dn&amp;#39; = x_0+n$, which &lt;strong&gt;is&lt;/strong&gt; congruent to $x_0$ modulo $n$. So the only distinct &lt;em&gt;congruence classes&lt;/em&gt; that are solutions are $[x_0]_n$, $[x_0+n&amp;#39;]_n,\ldots,[x_0+(d-1)n&amp;#39;]_n$, giving you exactly $d$ distinct classes that are solutions to $[a]_nx=[b]_n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To see this in practice, take $n= 77$, $a=21$, and $b=14$. We want to solve the system $[21]_{77} x = [9]_{77}$ Since $\gcd(77,21)=7$ divides $b$, the system has solutions; in fact, it has $7$ different congruence classes modulo $77$ as solutions.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, we write $21 = 7\cdot 7$, $14 = 7\cdot 2$, $77 = 7\cdot 11$. And we first solve the system $[3]_{11}x=[2]_{11}$. To do this, we write $1$ as a linear combination of $3$ and $11$, like so: $1 = 3(4) - 11(1)$. Then multiply by $2$ to get $2 = 3(8)-11(2)$. So $x_0=8$ is a solution (indeed, $3(8)=24\equiv 2\pmod{11}$). Now going back to the original congruence, we take $x_0$ and we add multiples of $11$ (why $11$? Because $n=77$ is $7\cdot 11$, and $7$ is the $d$ from before) until we have our $7$ different solutions. So the solutions are: $[8]_{77}$, $[19]_{77}$, $[30]_{77}$, $[41]_{77}$, $[52]_{77}$, $[63]_{77}$, and $[74]_{77}$. You can verify that they all are solutions to $[21]_{77}x = [14]_{77}$. For example, $21\cdot 41 = 861$, and $[861]_{77}=[14]_{77}$ because $861-14 = 847 = 77\cdot 11$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;(3)&lt;/strong&gt; Yes: that is exactly the meaning of $[a]_nx=[b]_n$; it means $ax\equiv b\pmod{n}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope this helps, despite the length. &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-14T23:17:41.867" LastActivityDate="2010-11-14T23:17:41.867" />
  <row Id="10308" PostTypeId="2" ParentId="10216" CreationDate="2010-11-14T22:45:51.437" Score="0" ViewCount="" Body="&lt;p&gt;I have been thinking about the the problem a bit today. What Robin and Yuval have both shown is that if the Cayley hamilton theorem is true, then the characteristic and minimal polynomial of $C(f)$ are both equal to $f$ .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conversely, assume that for all $f \in F[x]$, the the characteristic and minimal polynomial of $C(f)$ are both equal to $f$ .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $V$ be a finite dimensional $F$-vector space and $T : V \to V$ a linear transformation. we know from the classification theorem of modules over PIDs that there exists a basis $B$ of $V$ and $f_1, \dots, f_s \in F[x]$ such that $f_1 \mid \dots \mid f_s$ and &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ [T]_B = \begin{pmatrix} C(f_1) &amp;amp; &amp;amp; \\ &amp;amp; \ddots &amp;amp; \\ &amp;amp; &amp;amp; C(f_s)  \\ \end{pmatrix} := M$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is clear that the characteristic polynomial of $M$ is the product of the characteristic polynomials of all of the $C(f_i)$s and the minimal polynomial of $M$ is the least common multiple of the minimal polynomials of all the $C(f_i)$s. We see form the assumption that the characteristic polynomial of $T$ is $f_1 f_2 \dots f_s$ and the minimal polynomial of $T$ is $f_s$. This proves the Cayley hamilton theorem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This shows that the Cayley Hamilton theorem is equivalent to the fact &quot;for all $f \in F[x]$, the the characteristic and minimal polynomial of $C(f)$ are both equal to $f$&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proving the Cayley hamilton theorem without assuming knowledge of modules over PIDs or companion matrices is quite delicate (from what I remember from first year university).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This seems to support the idea that in order prove to the Cayley hamilton theorem (or the fact about companion matrices), you need to at some point get your hands dirty (whether it be directly computing the minimal and characteristic polynomials of a companion matrix or going through a delicate proof of the Cayley hamilton theorem).&lt;/p&gt;&#xA;" OwnerUserId="1725" LastActivityDate="2010-11-14T22:45:51.437" CommentCount="9" />
  <row Id="10309" PostTypeId="2" ParentId="10280" CreationDate="2010-11-14T23:09:06.213" Score="2" ViewCount="" Body="&lt;p&gt;Write $\{ x\}$ for $x-\lfloor x\rfloor$.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Prove that $\{ rx\}$ for $r=0,1,\cdots,q$ are all distinct.&#xA;(Here is where you use the condition that $x$ is irrational.)&lt;/li&gt;&#xA;&lt;li&gt;Using pigeonhole principle,  find  integers $r,s, 0\le r&amp;lt;s,$ such that&lt;br&gt;&#xA;$\{ sx \},\{ rx\}$ belong to the same subinterval.&lt;/li&gt;&#xA;&lt;li&gt;Then prove that $s-r$ is your solution.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="3249" LastEditorUserId="3249" LastEditorDisplayName="" LastEditDate="2010-11-15T14:18:16.597" LastActivityDate="2010-11-15T14:18:16.597" />
  <row Id="10310" PostTypeId="2" ParentId="10046" CreationDate="2010-11-14T23:14:47.127" Score="1" ViewCount="" Body="&lt;p&gt;Here is some more useful info related to your first sum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Recall that when summing $\sum_{n\ge 0} t_n$ we can use the idea of hypergeometrics, by computing the ratio $t_{n+1}/t_n$, and then reading-off from it the Hypergeometric function that solves the summation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your case, the ratio is&#xA;$$\frac{t_{n+1}}{t_n} = \frac{(n+1)(n+1)}{(n+1/2)4},$$ so the sum over all $t_n$ ranging from $n\ge 0$ instead of $n\ge 1$ as you ask, is nothing but the hypergeometric function&#xA;$${}_2F_1(1, 1; \frac{1}{2}; \frac{1}{4})$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which may be verified using some of the ideas above to be equal to &#xA;$$\frac{2}{27} \left(18+\sqrt{3} \pi \right)$$&lt;/p&gt;&#xA;" OwnerUserId="1709" LastActivityDate="2010-11-14T23:14:47.127" />
  <row Id="10312" PostTypeId="2" ParentId="10251" CreationDate="2010-11-14T23:30:25.673" Score="2" ViewCount="" Body="&lt;p&gt;First, notice that &#xA;$$\cos(x+\pi) = \cos(x)\cos(\pi) - \sin(x)\sin(\pi) = \cos(x)(-1)-\sin(x)(0) = -\cos(x).$$&#xD;&#xA;So, rewriting, using the above, and using that $\cos(-x)=\cos x$ you have:&#xA;\begin{align*}&#xA;\cos\left(\theta+\frac{2\pi}{3}\right) + \cos\left(\theta+\frac{4\pi}{3}\right) &amp;amp;= \cos\left(\left(\theta-\frac{\pi}{3}\right)+\pi\right) + \cos\left(\left(\theta+\frac{\pi}{3}\right)+\pi\right)\\&#xA;&amp;amp;= -\cos\left(\theta-\frac{\pi}{3}\right) - \cos\left(\theta+\frac{\pi}{3}\right)\\&#xA;&amp;amp;= -\cos\left(\frac{\pi}{3}-\theta\right) - \cos\left(\frac{\pi}{3}+\theta\right).&#xA;\end{align*}&#xA;So: the equality as written cannot hold: you would need $\cos(\frac{\pi}{2}+\theta) = - \cos(\frac{\pi}{3}-\theta)$ for all $\theta$, and plugging in $\theta=0$ shows this cannot hold. I suspect your $\frac{\pi}{2}$ should have been a $\frac{\pi}{3}$, and that you are missing a minus sign before the first term. If this is not the case, then please edit the question to give the equality you are actually trying to prove. &lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-11-14T23:30:25.673" CommentCount="1" />
  <row Id="10314" PostTypeId="1" AcceptedAnswerId="10316" CreationDate="2010-11-14T23:31:54.750" Score="0" ViewCount="272" Body="&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/ULjJd.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answers are already given after the = sign of each question. But I dont know how they arrived at these answers. What does it mean to say f(A)=1 and so on? I cant find the connection..&lt;/p&gt;&#xA;" OwnerUserId="2985" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-15T00:39:20.513" LastActivityDate="2010-11-15T00:47:35.543" Title="Determining if two graphs are isomorphic" Tags="&lt;graph-theory&gt;&lt;trees&gt;" AnswerCount="1" CommentCount="7" />
  <row Id="10316" PostTypeId="2" ParentId="10314" CreationDate="2010-11-14T23:41:38.670" Score="3" ViewCount="" Body="&lt;p&gt;A graph can be thought of as a pair $(V,E)$, where $V$ are the vertices and $E$ are the edges. An &lt;strong&gt;isomorphism&lt;/strong&gt; between two graphs, $G_1 = (V_1,E_1)$ and $G_2=(V_2,E_2)$ means a function $f\colon V_1\to V_2$ that is bijective between the vertices (one-to-one and onto, so every vertex in $G_1$ is mapped to a vertex of $G_2$, and each vertex in $G_2$ is the image of one and only one vertex in $G_1$) &lt;strong&gt;and&lt;/strong&gt; for vertices $v_1,v_2$ of $G_1$, $v_1$ is adjacent to $v_2$ in $G_1$ if and only if $f(v_1)$ is isomorphic to $f(v_2)$ in $G_2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the answer they give is a description of an isomorphism between the second graph and the first: it is a function from the set of vertices of the second graph, $\{A,B,C,D,G,H,I,J\}$, and the vertices of the first graph, $\{1,2,3,4,5,6,7,8\}$, with the property that two vertices in the second graph $x$ and $y$, are adjacent if and only if $f(x)$ is adjacent to $f(y)$ in the second graph. For instance, $A$ and $I$ are adjacent, so $f(A)=1$ needs to be adjacent to $f(I)=4$ (which it is), and $A$ and $J$ are not adjacent, so $f(A)=1$ should &lt;strong&gt;not&lt;/strong&gt; be adjacent to $f(J)=7$ (which it is not).  You can check that the function described actually &lt;em&gt;is&lt;/em&gt; an isomorphism by checking that every pair of adjacent vertices are sent to adjacent vertices, and every pair of non-adjacent vertices are sent to non-adjacent vertices.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now... how did they come up with $f$ in the first place? Well, let's think about the graphs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the second graph, each vertex is adjacent to exactly three vertices, so if there is going to be an isomorphism between the two graphs, then every vertex in the first graph needs to be adjacent to exactly three vertices. This is indeed true, which is a good start for a possible isomorphism (if it wasn't true, we would have &lt;em&gt;no&lt;/em&gt; hope of having an isomorphism, and if fact we would know that no isomorphism exists).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, the second graph is &quot;bipartite&quot;: you can divide the vertices into two groups in such a way that all edges are between the two groups, with no edges between vertices in the same group. If we are going to find an isomorphism between this graph and the first, we need to be able to &lt;em&gt;also&lt;/em&gt; separate the vertices of the first graph into two groups of four, so that all edges are between the groups and no edge goes from one vertex in a group to another vertex in the &lt;em&gt;same&lt;/em&gt; group.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can we divide the vertices in the first graph into two sets of four so that all edges are between the sets? Well, let's see: if $1$ is in one set, then $2$, $4$, and $5$ should be on the other set because there can be no edges between two vertices in the same set (remember: we are trying to check if there is an isomorphism). Notice that if we do this, then there better not be any edges between $2$, $4$, and $5$, because we are going to put them in the same set (the set which is not the set with $1$ in it). Good, no edges between any of $2$, $4$, and $5$. And $6$ and $3$ need to be with $1$, because they are both adjacent to $2$, and so they cannot be in the same set with $2$; the only possibility left is the set with $1$ in it. That means that $1$, $3$, and $6$ need to be in the same set of vertices, which means there should be no edges between them (no edges between any two vertices in the same set, remember?). Good, no edges between any of $1$, $3$, and $6$. And $8$ needs to be with $1$, because it is adjacent to $4$. And so $7$ needs to be in with $2$, $4$, and $5$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, if we can find an isomorphism between the two graphs, then we can divide the vertices of the first graph into two sets, just like the ones in the second graph, and one set needs to have $1$, $3$, $6$, and $8$, and the other set needs to have $2$, $4$, $5$, and $7$. And all edges should be between the sets, not no edges between vertices in the same set. This all holds, so we are on the right track!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, can we define an isomorphism based on this? There is a lot of symmetry, so it probably does not matter where we send $A$, just so long as we do things right. So let's send $A$ to $1$: $f(A)=1$. Because $G$, $H$, and $I$ are adjacent to $S$, we need $f(G)$, $f(H)$, and $f(I)$ to be adjacent to $f(A)=1$; so $f(G)$, $f(H)$, and $f(I)$ need to be $2$, $4$, and $5$ in some order. Say $f(G)=2$. Then $f(H)$ needs to be either $4$ or $5$. If we make it $5$, then since $2=f(G)$ and $5=f(H)$ have only two mutually adjacent vertex (namely $1$ and $6$), and $G$ and $H$ have only two mutually adjacent vertices (namely $A$ and $B$), and $A$ already maps to $1$, then we need $B$ to map to $6$. So so far we have $f(A)=1$, $f(B)=6$, $f(G)=2$, and $f(H)=5$. $B$ is adjacent to $J$, which is not adjacent to $A$, so $f(J)$ needs to be adjacent to $f(B)=6$, but not to $f(A)=1$; and it needs to be in the same group as $2$ (so it has to be one of $2$, $4$, $5$, or $7$); $5$ and $2$ are already taken, $4$ is adjacent to $1$, so we need to make $f(J)=7$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So so far we have $f(A)=1$, $f(B)=6$, $f(G)=2$, $f(H)=5$, and $f(J)=7$. What about $C$? $f(C)$ needs to be adjacent to $f(G)=2$ and to $f(J)=7$, but not to $f(H)=5$; the only possibility not yet taken is $f(C)=3$. Continuing this way you get the isomorphism they give.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: this is not the only possible isomorphism; other choices lead to other isomorphisms, of which there are many because the graphs are so symmetric. &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-15T00:47:35.543" LastActivityDate="2010-11-15T00:47:35.543" CommentCount="3" />
  <row Id="10317" PostTypeId="1" AcceptedAnswerId="10324" CreationDate="2010-11-14T23:47:20.670" Score="1" ViewCount="72" Body="&lt;p&gt;Suppose $P(A) = 0.6$ and $P(B) = 0.7$. Find the smallest possible value of $P(A \cap B)$. Instead of deducing this from $P(A \cup B) = P(A)+P(B)-P(A \cap B)$, could we just use the fact that $P(A|B)P(B) = P(A \cap B)$ and $P(B|A)P(A) = P(A \cap B)$? I notice that $(0.6)(0.7)^2 = 0.294$ which is close to $0.3$, the smallest value of $P(A \cap B)$. &lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2010-11-15T00:35:49.743" Title="Conditional Probability" Tags="&lt;probability&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="10318" PostTypeId="1" AcceptedAnswerId="10321" CreationDate="2010-11-15T00:14:18.120" Score="1" ViewCount="517" Body="&lt;p&gt;Let X and Y be independent random variables,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$  P(X = k) = P(Y = k) = p(1 - p)^{k-1} $&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do you show that the pmf of $ Z = X + Y $, is negative binomial, and how do you find&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ P(X = Y) $?&lt;/p&gt;&#xA;" OwnerUserId="2976" LastActivityDate="2010-11-15T05:51:13.513" Title="Sum of two independent geometric random variables" Tags="&lt;probability&gt;&lt;statistics&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="10319" PostTypeId="1" AcceptedAnswerId="10323" CreationDate="2010-11-15T00:15:24.270" Score="2" ViewCount="106" Body="&lt;p&gt;Suppose $f$ is an unknown polynomial of degree $n$ (in one indeterminate) but the sequence $\{ f(k) \}_{k \in \mathbb{N}}$ is given. It is a nice exercise to show that one needs only the first $n+1$ terms of the sequence to determine the coefficients of $f$. That is, simply solve the matrix equation $A\mathbf{x} = b$, where $\mathbf{b} = (f(0), \dots, f(n))^{\top}$, $A$ is the Vandermonde matrix of $(i^{j})_{i,j = 0, \dots, n}$ and $\mathbf{x} = (c_{0}, \dots, c_{n})^{\top}$ (the unknown coefficients of $f$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question&lt;/strong&gt;: Is there a closed form expression for the coefficients of a finite degree polynomial $f$ in terms of the sequence $\{ f(k) \}_{k \in \mathbb{N}}$ that doesn't involve matrix inversion or differentiation or explicitly calculating the polynomial in question?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(&lt;em&gt;Motivation&lt;/em&gt;) The Ehrhart polynomial counts the number of integer lattice points intersecting a dilate of a polytope and can be calculated by the residue of an associated complex rational function (see M. Beck's articles on the subject). Some of the coefficients of the Ehrhart polynomial can be related to an $n$-volume, a relative area and the euler characteristic of said polytope. However, computing coefficients of the Ehrhart polynomial is not a particularly easy task. Having simple formulas for them, say in terms of the residues above would be nice to have at one's disposal. A reasonable starting point is answering the question above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="2720" LastEditorUserId="2720" LastEditorDisplayName="" LastEditDate="2010-11-15T02:57:13.520" LastActivityDate="2010-11-15T02:57:13.520" Title="Determining Coefficients of a Finite Degree Polynomial $f$ from the Sequence $\{f(k)\}_{k \in \mathbb{N}}$" Tags="&lt;linear-algebra&gt;&lt;algebra&gt;" AnswerCount="1" FavoriteCount="2" />
  <row Id="10320" PostTypeId="2" ParentId="10257" CreationDate="2010-11-15T00:17:34.640" Score="5" ViewCount="" Body="&lt;p&gt;If $f\colon[0,1]\to\mathbb{R}$ is cadlag (&lt;a href=&quot;http://en.wikipedia.org/wiki/C%C3%A0dl%C3%A0g&quot; rel=&quot;nofollow&quot;&gt;continu &amp;agrave; droite, limites &amp;agrave; gauche&lt;/a&gt;) then we can ask what the possible jumps are. That is, for what functions $g\colon(0,1]\to\mathbb{R}$ is there a cadlag function $f$ with $g(x)=\Delta f(x)\equiv f(x)-f(x-)$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer is that $g$ occurs as the jumps of a cadlag function if and only if the set $\{x\in(0,1]\colon\vert g(x)\vert &amp;gt; \epsilon\}$ is finite for each $\epsilon &amp;gt; 0$. In particular, there is a cadlag function with jumps as you describe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, the necessity: If $S=\{x\in(0,1]\colon\vert g(x)\vert &amp;gt; \epsilon\}$ was not finite, then it would contain a strictly increasing or strictly decreasing sequence $x_n$. Then, $\vert f(x_n)-f(y_n)\vert &amp;gt; \epsilon$ for some $y_n$ chosen arbitrarily close to $x_n$. Replacing $x_n$ by $y_n$ where necessary gives a strictly increasing or strictly decreasing sequence $x_n\in[0,1]$ such that $\vert f(x_{n+1})-f(x_n)\vert &amp;gt; \epsilon/2$. However, as f is cadlag and has left and right limits everywhere, this contradicts the requirement that $f(x_n)$ tends to a limit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, we can show sufficiency: Set $\epsilon_n=2^{-n}$ for $n\ge1$ and $\epsilon_0=\infty$. For each $n\ge1$, consider the finite set $S_n=\{x\in(0,1]\colon \epsilon_{n-1}\ge\vert g(x)\vert &amp;gt; \epsilon_n\}$. We can construct a function $f_n\colon[0,1]\to\mathbb{R}$ such that $\Delta f_n(x)=1_{\{x\in S_n\}}g(x)$ and $\vert f_n(x)\vert\le\epsilon_{n-1}$. The idea is to take $f_n(x)=g(x)$ and $f_n(x-)=0$ for points in $S_n$, and linearly interpolate between these.&#xA;More precisely, if $S_n$ is empty we set $f_n=0$. Otherwise,&#xA;$$&#xD;&#xA;f_n(x)= g(a)(b-x)/(b-a)&#xD;&#xA;$$&#xD;&#xA;for $a\le x &amp;lt; b$. Here, $a &amp;lt; b$ are consecutive points of $S_n$. Also set $f_n(x)=0$ for $x$ less than the minimum of $S_n$ and $f_n(x)=g(c)$ for $x\ge c=\max S_n$. This has the required jumps $\Delta f_n(x)=1_{\{x\in S_n\}}g(x)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally set $f(x)=\sum_{n=1}^\infty f_n(x)$. As $\vert f_n\vert \le 2^{1-n}$ for $n\ge 2$ this converges uniformly and,&#xA;$$&#xD;&#xA;\Delta f(x)=\sum_n\Delta f_n(x)=\sum_n1_{\{x\in S_n\}}g(x)=g(x).&#xD;&#xA;$$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="1321" LastEditorUserId="1321" LastEditorDisplayName="" LastEditDate="2010-11-15T01:25:57.267" LastActivityDate="2010-11-15T01:25:57.267" CommentCount="2" />
  <row Id="10321" PostTypeId="2" ParentId="10318" CreationDate="2010-11-15T00:20:23.887" Score="2" ViewCount="" Body="&lt;p&gt;$P(Z=r)=\Sigma_{k=1}^r P(X=k)P(Y=r-k)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$P(X=Y)=\Sigma_{k=1}^{\infty} P(X=k)P(Y=k)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am assuming $k$ is a positive integer. Otherwise, the summations need to be modified accordingly.&lt;/p&gt;&#xA;" OwnerUserId="3431" LastActivityDate="2010-11-15T00:20:23.887" CommentCount="2" />
  <row Id="10322" PostTypeId="1" AcceptedAnswerId="10338" CreationDate="2010-11-15T00:24:15.697" Score="2" ViewCount="125" Body="&lt;p&gt;in trying to find this limit, all the (x,y) paths i check approaching (0,0) produce the same result, namely, the function approaches 0. consequently, i came up with the simplest proof i could that the limit indeed exists and is zero. however wolframalpha tells me that the limit doesnt exist. assuming im wrong, can anyone shed light on where my proof is flawed, and possibly provide a path on which the function doesnt approach 0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;\lim_{(x,y) \rightarrow (0,0)}&#xA;\frac { x \cdot \sin(\frac{y}{\sqrt{x}}) }&#xA;{\sqrt{ x^2 + y^2 }}&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;provided $x&amp;gt;0$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;my proof:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;lets take $\varepsilon &amp;gt; 0$ and prove the existence of some $\delta &amp;gt; 0$ such that&lt;br&gt;&#xA;$\forall (x,y) \in B(0, \delta ) : |f(x,y)| &amp;lt; \varepsilon$  &#xA;(where f is our function above, and as stated $x&gt;0$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;taking the x to the denominator, we get:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;\frac { \sin(\frac{y}{\sqrt{x}}) }&#xA;{\sqrt{ 1 + (\frac{y}{x})^2 }}&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;sin is bounded between -1 and 1, so from here i intend from here to show that i can choose a relevant $\delta$ where either the denominator is large enough to make the expression smaller than our $\varepsilon$, and if not, the numerator is smaller than our $\varepsilon$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;we find the relevant denominator bound by:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;\frac {1}&#xA;{\sqrt{ 1 + (\frac{y}{x})^2 }} &amp;lt; \varepsilon&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is simply:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;\frac {y}{x} &gt; \sqrt{\frac{1}{\varepsilon^2} - 1}&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;clearly, if this inequality holds, then the absolute value of the function is indeed bound by our $\varepsilon$ as required.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so lets deal with the seconds case:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;\frac {y}{x} \leq \sqrt{\frac{1}{\varepsilon^2} - 1}&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;multiplying through by $\sqrt {x}$ we get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;\frac {y}{\sqrt{x}}&#xA;\leq \sqrt{x} \cdot \sqrt{\frac{1}{\varepsilon^2} - 1}&#xA;\leq \sqrt{\delta} \cdot \sqrt{\frac{1}{\varepsilon^2} - 1}&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the last inequality is because $x$ is obivously bound by the $\delta$ we choose&lt;br&gt;&#xA;it should be clear that the right hand side of the inequality is arbitrarily small, although we will formalise somewhat:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;take $\delta &amp;lt; \varepsilon^4$:&#xA;\begin{equation}&#xA;\frac {y}{\sqrt{x}}&#xA;\leq \sqrt{\delta} \cdot \sqrt{\frac{1}{\varepsilon^2} - 1}&#xA;&amp;lt; \sqrt{\varepsilon^4} \cdot \sqrt{\frac{1}{\varepsilon^2}}&#xA;= \varepsilon^2 \cdot \frac{1}{\varepsilon}&#xA;= \varepsilon&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;without too much formality, we say that for sufficiently small $\delta$, $\sin x \approx x$ and therefore&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;{\sin{ \frac {y}{\sqrt{x}}}&#xA;\approx&#xA;\frac {y}{\sqrt{x}}&#xA;\leq \varepsilon}&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;of course, i was lazy and didnt add in the necessary stages to prove the approximation will hold in the inequality, although it does, because i can take the $\delta$ to be such that the difference between the sin and the value itself is less than $\frac{\varepsilon}{3}$ and also such that $\frac{y}{\sqrt {x}} &amp;lt; \frac{\varepsilon}{3}$ which means the sin is indeed less than $\varepsilon$ as required. this takes care of the second case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;QED&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(i didnt take much care with whether y is positive or negative, although the function is odd, so we can work with positives and then just prove the result for the negatives)&lt;/p&gt;&#xA;" OwnerUserId="2881" LastActivityDate="2010-11-15T01:18:56.493" Title="multivariable limit" Tags="&lt;limit&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="10323" PostTypeId="2" ParentId="10319" CreationDate="2010-11-15T00:32:13.097" Score="4" ViewCount="" Body="&lt;p&gt;Yes; you can take &lt;a href=&quot;http://en.wikipedia.org/wiki/Finite_difference&quot; rel=&quot;nofollow&quot;&gt;finite differences&lt;/a&gt;.  Let $\Delta f(x) = f(x+1) - f(x)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Theorem:&lt;/strong&gt;  $$f(x) = \sum_{k=0}^{n} \Delta^k f(0) {x \choose k}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where ${x \choose k}$ is the polynomial $\frac{x(x-1)...(x-(k-1))}{k!}$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Proof.&lt;/em&gt;  Observe that $\Delta {x \choose k} = {x \choose k-1}$.  Take $k$ finite differences of both sides and set $x = 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In practice, this means you can work out what $f$ is by writing down a finite difference table.  This is really easy to do by hand, and then the top row of the table tells you what the coefficients $\Delta^k f(0)$ are above.  In linear algebra terms, what we are doing is using a basis with respect to which the matrix $A$ is upper-triangular, and this makes life much easier.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is also a known explicit formula for the inverse of $A$ which is equivalent to the &lt;a href=&quot;http://en.wikipedia.org/wiki/Lagrange_polynomial&quot; rel=&quot;nofollow&quot;&gt;Lagrange interpolation formula&lt;/a&gt;.  This is sometimes useful for theoretical reasons (e.g. as a way to control the behavior of the interpolation polynomial or to deduce certain identities).&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;You can extract a &quot;closed form&quot; for the coefficients of $f$ using either of the two methods above, although I think using finite differences is slightly nicer.  We have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$${x \choose k} = \frac{1}{k!} \sum_{i=0}^{k} s(k, i) x^i$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $s(k, i)$ are the &lt;a href=&quot;http://en.wikipedia.org/wiki/Stirling_numbers_of_the_first_kind&quot; rel=&quot;nofollow&quot;&gt;Stirling numbers of the first kind&lt;/a&gt;.  This gives&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$[x^i] f(x) = \sum_{k=i}^{n} \Delta^k f(0) \frac{s(k, i)}{k!}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Delta^k f(0) = \sum_{j=0}^{k} (-1)^{k-j} {k \choose j} f(j).$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So $[x^i] f(x) = \sum_{j=0}^{n} a_{i,j} f(j)$ where&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$a_{i,j} = \sum_{k=j}^{n} (-1)^{k-j} {k \choose j} \frac{s(k, i)}{k!}.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know whether this sum can be simplified further.&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-11-15T01:57:35.113" LastActivityDate="2010-11-15T01:57:35.113" CommentCount="0" />
  <row Id="10324" PostTypeId="2" ParentId="10317" CreationDate="2010-11-15T00:35:49.743" Score="1" ViewCount="" Body="&lt;p&gt;From the equation in the second line&#xA;$$P(A \cup B) = P(A) + P(B) - P(A \cap B)$$&#xD;&#xA;we have&#xA;$$P(A \cap B) = P(A) + P(B) - P(A \cup B)$$&#xD;&#xA;In order to minimize $P(A \cap B)$, we should maximize $P(A \cup B)$ since both $P(A)$ and $P(B)$ are fixed. The maximum value for $P(A \cup B)$ is $1$, so we have the minimum value of $P(A \cap B)$ as&#xA;$$P(A \cap B) = 0.6 + 0.7 - 1 = 0.3$$&lt;/p&gt;&#xA;" OwnerUserId="1016" LastActivityDate="2010-11-15T00:35:49.743" />
  <row Id="10326" PostTypeId="1" AcceptedAnswerId="10331" CreationDate="2010-11-15T00:42:31.290" Score="0" ViewCount="183" Body="&lt;p&gt;Let $X_{i} = 1$ if &lt;em&gt;ith&lt;/em&gt; person gets his or her own hat back and $0$ otherwise. Let $S_{n} = \sum_{i=1}^{n}X_{i}$, then $S_{n}$ is the total number of people who get their own hats back. Show that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a) $E(X_{i}^{2}) = \frac{1}{n}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;b) $E(X_{i}.X_{j}) = \frac{1}{n(n - 1)}$ for $i\neq j$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;c) $E(S_{n}^{2}) = 2$ (using (a) and (b)).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;d) $var(S_{n}) = 1$&lt;/p&gt;&#xA;" OwnerUserId="2976" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2010-11-15T20:07:35.533" LastActivityDate="2010-11-15T20:07:35.533" Title="Suppose that n people have their hats returned at random, find the following" Tags="&lt;probability&gt;&lt;statistics&gt;" AnswerCount="2" CommentCount="5" />
  <row Id="10327" PostTypeId="1" AcceptedAnswerId="10344" CreationDate="2010-11-15T00:44:02.297" Score="6" ViewCount="284" Body="&lt;p&gt;Is it algebraic the number 0.2468101214 ...? (After point, the natural numbers are juxtaposed pairs).&lt;/p&gt;&#xA;" OwnerUserId="2938" LastEditorUserId="2938" LastEditorDisplayName="" LastEditDate="2010-11-15T00:49:44.323" LastActivityDate="2010-11-15T02:10:41.477" Title="0.246810121416...: Is it a algebraic number?" Tags="&lt;number-theory&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="10328" PostTypeId="2" ParentId="10326" CreationDate="2010-11-15T00:55:58.623" Score="1" ViewCount="" Body="&lt;p&gt;I'll give some hints:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Notice that $X_i^2 = X_i$, since $X_i$ is $0$ or $1$.  Then, &#xA;$E X_i = 0 \cdot \mathrm{P}(X_i=0) + 1 \cdot \mathrm{P}(X_i = 1) = \mathrm{P}(X_i = 1)$.  What is this probability?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2)  Again, $X_i X_j$ is $0$ or $1$.  Notice that $X_i X_j = 1$ if and only if person $i$ gets their hat AND person $j$ gets their hat.  What is the probability of this?  Let $A$ be the event person $i$ gets his hat and $B$ be the event person $j$ does.  Then &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \mathrm{P}(X_i X_j = 1) = \mathrm{P}(A \cap B) = \mathrm{P}(A) \mathrm{P}(B|A) $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are $\mathrm{P}(A)$ and $\mathrm{P}(B|A)$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3)  After some algebra,  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$E S_n^2 = E(\sum_{i=1}^n X_i )^2 = E(\sum_{i=1}^n X_i^2 +  \sum_{i \neq j} X_i X_j)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Distribute the $E$ through. Use 1) and 2). How many ways can $i \neq j$? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;4)  $\mathrm{Var} S_n = E S_n^2 - (E S_n)^2$  &lt;/p&gt;&#xA;" OwnerUserId="2515" LastActivityDate="2010-11-15T00:55:58.623" />
  <row Id="10329" PostTypeId="1" AcceptedAnswerId="10345" CreationDate="2010-11-15T00:56:46.890" Score="2" ViewCount="218" Body="&lt;p&gt;Let X be an infinite-dimensional Banach space, and T a compact operator from X to X. Why must 0 then be a spectral value for T? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe this is equivalent to saying that T is not bijective, but I am not sure how to show that injectivity implies the absence of surjectivity and the other way around (or if this is even the right way to approach the problem).&lt;/p&gt;&#xA;" OwnerUserId="1736" LastActivityDate="2010-11-15T02:16:51.390" Title="Invertibility of compact operators in infinite dimensions" Tags="&lt;functional-analysis&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="10331" PostTypeId="2" ParentId="10326" CreationDate="2010-11-15T01:01:26.243" Score="2" ViewCount="" Body="&lt;p&gt;Please note:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;a) $X_{i}=1$ probability $\frac{1}{n}$ and 0, otherwise. Thus $X_{i}^{2}=1$ with probability $\frac{1}{n}$ and 0, otherwise, which &lt;em&gt;implies&lt;/em&gt; $E(X_{i}^{2})=\frac{1}{n}$&#xD;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;b) The probability that both $X_{i}$ and $X_{j}$ are 1 is $\frac{1}{n(n-1)}$ (Why?)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;c) $E(S_{n}^{2}) = \sum\limits_{i} E[X_{i}^{2}] + \sum\limits_{i}\sum\limits_{j \neq i} E[X_{i}X_{j}= n \cdot \frac{1}{n} + n(n-1) \cdot \frac{1}{n(n-1)}$&#xD;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;d) $Var(S_{n}) = E[S_{n}^{2}] - (E[S_{n}])^{2}$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" LastActivityDate="2010-11-15T01:01:26.243" />
  <row Id="10332" PostTypeId="2" ParentId="10322" CreationDate="2010-11-15T01:02:00.650" Score="1" ViewCount="" Body="&lt;p&gt;Wolframalpha is not infallible. Your proof looks OK to me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One trick that sometimes helps for problems like this is to write $y = ax$ and then examine your expression. In this case you get ${1 \over \sqrt{1 + a^2}} \sin (a \sqrt{x})$. Then you can use that $|\sin(\theta)| \leq |\theta|$ for all $\theta$, to get that it is bounded in absolute value by ${a \over \sqrt{1 + a^2}} \sqrt{x}$. Since $\big|{a \over \sqrt{1 + a^2}}\big|$ is at most 1, your expression's absolute value is at most $\sqrt{x}$, which goes to zero as $(x,y)$ approaches the origin.&lt;/p&gt;&#xA;" OwnerUserId="3035" LastEditorUserId="3035" LastEditorDisplayName="" LastEditDate="2010-11-15T01:15:15.853" LastActivityDate="2010-11-15T01:15:15.853" CommentCount="2" />
  <row Id="10333" PostTypeId="1" AcceptedAnswerId="10334" CreationDate="2010-11-15T01:02:34.673" Score="0" ViewCount="99" Body="&lt;p&gt;Does it make sense to say that there are more irrational numbers than rational?&lt;/p&gt;&#xA;" OwnerUserId="2938" LastActivityDate="2010-11-15T01:09:04.720" Title="Are there many more irrational numbers than rational?" Tags="&lt;number-theory&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="10334" PostTypeId="2" ParentId="10333" CreationDate="2010-11-15T01:03:19.767" Score="4" ViewCount="" Body="&lt;p&gt;Yes, in the sense of cardinality. In the same sense that one can say that there are more real numbers than there are rationals (or integers; or natural numbers).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rational numbers are countable, but the irrationals are uncountable. &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-15T01:09:04.720" LastActivityDate="2010-11-15T01:09:04.720" CommentCount="2" />
  <row Id="10335" PostTypeId="1" AcceptedAnswerId="10340" CreationDate="2010-11-15T01:12:24.920" Score="1" ViewCount="73" Body="&lt;p&gt;Hey guys, second post for today. I have a midterm on tuesday so I am just practicing some proofs from the textbook. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question states: &#xA;If $a \in \mathbb{Z}$, prove that $a^2$ is not congruent to 2 modulo 4 or 3 modulo 4. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Our work:&#xA;It's an if then proof, so I am thinking we can do a contradiction proof. &#xA;Suppose $a^2 \equiv 2 \pmod{4}$ and $a^2 \equiv 3 \pmod{4}$. &#xA;Then you can write by definition &#xA;\begin{align*}&#xA;a^2 &amp;amp;= 4k + 2 \\&#xA; a^2 &amp;amp;= 4k + 3&#xA;\end{align*}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but hold on, another way to write 2 mod 4 is $[2]_{4}$ which means the infinite set of all the integers in the form of $2 + 4k$ and $[3]_{4}$ which means the infinite set of all the integers in the form of $3 + 4k$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;so we one set is a set of even integers and one set is of odd integers. but our a^2 can't be in both right? &lt;/p&gt;&#xA;" OwnerUserId="287" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-15T01:18:32.807" LastActivityDate="2010-11-15T02:03:43.810" Title="Congruency proof" Tags="&lt;abstract-algebra&gt;&lt;elementary-number-theory&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="1" />
  <row Id="10336" PostTypeId="1" AcceptedAnswerId="10364" CreationDate="2010-11-15T01:14:54.623" Score="1" ViewCount="88" Body="&lt;p&gt;I figured out this order on $\omega^2$: elements of different columns are ordered by their column number; within column $m$, the order is $0 \gt 1 \gt 2 \gt \cdots \gt n$, where $n$ is the first such that $P(m,n)$ is true; then $0 \lt n+1 \lt n+2 \lt \cdots$. The column is an infinite descending sequence if there is no such $n$. This order is well-founded iff for all $m$ there is an $n$ such that $P(m,n)$ is true. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I can reduce any $\Pi_{2}^{0}$ sentence to a computable ordering which is well-founded iff the sentence is true. I don't see how to do this for $\Sigma_{2}^{0}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now Wikipedia tells me Kleene's $O$ is $\Pi_{1}^{1}$ complete, so I'm still a long way away from what I should be able to do. Rather than being told what the full reduction is, if instead I could just get a hint for the $\Sigma_{2}^{0}$ case, then maybe I can pretend to myself that I figured &quot;most&quot; of it out on my own?&lt;/p&gt;&#xA;" LastEditorUserId="630" LastEditorDisplayName="" LastEditDate="2010-11-15T04:54:39.820" LastActivityDate="2010-11-15T04:54:39.820" Title="Hint on Kleene's O" Tags="&lt;algorithms&gt;&lt;logic&gt;&lt;computability&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="10337" PostTypeId="1" AcceptedAnswerId="10513" CreationDate="2010-11-15T01:17:35.980" Score="2" ViewCount="333" Body="&lt;p&gt;The title is somewhat deceptive: I know that $|\mathbb{Q}|=|\mathbb{Z}|.$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But suppose I wanted to compare the sets, knowing that they are of the same cardinality but still wondering if there was some sense in which Q could be larger.  There are two usual ways of thinking about this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\mathbb{Z}\subset\mathbb{Q}$ but $\mathbb{Q}\not\subset\mathbb{Z}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(sorry, there's no \subsetneq symbol on math.stackexachange) and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;'For any distinct integers m,n there are infinitely many rationals between m and n.'&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These methods both rely on the natural mapping between integers and rationals, to wit: $n\mapsto\frac n1$.  What if we didn't have this correspondence, or were working with an application where it wasn't sensible?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is, we still have the usual order &amp;lt; on Q and Z, and the usual operations + and *, but we no longer have a reason to privilege the usual mapping over, say, $n\mapsto\frac2n$ or a diagonal mapping.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm much less concerned about the 'answer' part of the answers than I am of the 'question' part: the main point is 'how could this question be made precise?' rather than answering such question.&lt;/p&gt;&#xA;" OwnerUserId="1778" LastActivityDate="2010-11-16T04:08:33.300" Title="Are there many more rational numbers than integers?" Tags="&lt;set-theory&gt;" AnswerCount="3" CommentCount="3" />
  <row Id="10338" PostTypeId="2" ParentId="10322" CreationDate="2010-11-15T01:18:56.493" Score="5" ViewCount="" Body="&lt;p&gt;Your proof is correct, although somewhat long-winded. It's a good practice, once you have a proof, to go back and see if you can simplify it as much as possible, aiming to find the most direct path to the desired result. In the process, one often gains a better understanding of the problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since you ended up using the fact that $\sin \theta \le \theta$ for all $\theta &amp;gt; 0$, let's see if we can use this from the outset. Since the function is odd with respect to $y$, it suffices to only consider $y \ge 0$ as you noticed. Then we have $$\frac{x\sin\frac{y}{\sqrt{x}}}{\sqrt{x^2+y^2}} \le \frac{y\sqrt{x}}{\sqrt{x^2+y^2}} \le \sqrt{x}$$&#xD;&#xA;because $y \le \sqrt{x^2+y^2}$. In other words, the function is positive and bounded above by $\sqrt{x}$. From here, you can quite easily get to where you want.&lt;/p&gt;&#xA;" OwnerUserId="856" LastActivityDate="2010-11-15T01:18:56.493" />
  <row Id="10339" PostTypeId="2" ParentId="10335" CreationDate="2010-11-15T01:22:10.603" Score="0" ViewCount="" Body="&lt;p&gt;You cannot assume &lt;em&gt;both&lt;/em&gt; that $a^2\equiv 2\pmod{4}$ *and* that $a^2\equiv 3\pmod{4}$. Your &quot;assumption&quot; should be &quot;either $a^2\equiv 2\pmod{4}$ *OR* $a^2\equiv 3\pmod{4}$.&quot; That is, you need to do it by cases. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This because your assumption for a proof by contradiction should be &lt;em&gt;exactly&lt;/em&gt; the negation of your conclusion, not something else.  Your conclusion here is &#xA;$$\neg\left( a^2\equiv 2\pmod{4} \text{ or } a^2\equiv 3\pmod{4}\right)$$&#xD;&#xA;so the negation is just the statement that &lt;strong&gt;either&lt;/strong&gt; $a^2$ is $2$ *&lt;em&gt;or&lt;/em&gt;* $3$ modulo $4$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, no; you're not doing it right here, because your &quot;assumption for contradiction&quot; is not the correct one.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-15T01:27:33.450" LastActivityDate="2010-11-15T01:27:33.450" CommentCount="4" />
  <row Id="10340" PostTypeId="2" ParentId="10335" CreationDate="2010-11-15T01:22:17.457" Score="3" ViewCount="" Body="&lt;p&gt;Hint: There are only 4 possible choices for $a$ modulo $4$. You can explicitly look at each one and calculate $a^2$ modulo $4$.&lt;/p&gt;&#xA;" OwnerUserId="1016" LastActivityDate="2010-11-15T01:22:17.457" CommentCount="4" />
  <row Id="10341" PostTypeId="2" ParentId="10337" CreationDate="2010-11-15T01:40:36.537" Score="6" ViewCount="" Body="&lt;p&gt;The two statements you give relate to two different issues; the first relates to the fact that there is a natural embedding of $\mathbb{Z}$ in $\mathbb{Q}$ arising from their constructions, and it makes sense to ask whether $\mathbb{Z}$ is a proper subset of $\mathbb{Q}$. This is certainly a sensible question whenever we have two sets $A$ and $B$ and $A$ is &lt;em&gt;given&lt;/em&gt; as a subset of $B$. For arbitrary sets it does not make sense, as your comments about &quot;other mappings&quot; show.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second statement has to do with &lt;em&gt;order&lt;/em&gt;, but again with the &lt;em&gt;natural&lt;/em&gt; or &lt;em&gt;canonical&lt;/em&gt; ordering that both $\mathbb{Z}$ and $\mathbb{Q}$ inherit from the order in $\mathbb{N}$ (which is defined in terms of the successor function). The order of $\mathbb{Q}$ is dense, while the order of $\mathbb{Z}$ is not: $\mathbb{Z}$ is order-isomorphic to subsets of $\mathbb{Q}$, but not the other way around. If you &quot;order&quot; the linearly ordered sets by saying that $(P,\leq)$ is smaller than or equal to $(Q,\preceq)$ if and only if there is an order-preserving injection $P\to Q$, then you get equivalence classes of linearly ordered sets (much in the same way as with cardinals), and in &lt;em&gt;that&lt;/em&gt; way of ordering linearly ordered sets, $\mathbb{Z}$ with its natural ordering &lt;em&gt;is&lt;/em&gt; strictly smaller than $\mathbb{Q}$, also with its natural ordering. But if you think there is nothing special about the usual embedding $\mathbb{Z}\hookrightarrow \mathbb{Q}$, then presumably you also think there is nothing special about the natural ordering of $\mathbb{Z}$ or $\mathbb{Q}$, and then you can give them orderings that will make it come out &quot;the other way&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you can certainly make the question &quot;precise&quot; if you specify more information (&quot;as ordered sets with their usual ordering&quot;; &quot;as subsets of $\mathbb{R}$&quot;; etc). But that extra context needs to be explicit to make it make sense.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-15T01:46:07.263" LastActivityDate="2010-11-15T01:46:07.263" CommentCount="9" />
  <row Id="10342" PostTypeId="1" AcceptedAnswerId="10391" CreationDate="2010-11-15T02:02:04.803" Score="2" ViewCount="113" Body="&lt;p&gt;Consider a $G$-equivariant map $\pi:X\to Y$ for $G$ an affine algebraic group, such that $\pi$ is a good categorical quotient.  Is there any relationship between $H^*_G(X)$ and $H^*(Y)$?  Is there if $\pi$ is a good geometric quotient, or if the quotient space is smooth?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: A categorical quotient is an equivariant map $\pi:X\to Y$ that is constant on $G$-orbits.  It's good if the topology on $Y$ is induced by $X$ ($\pi$ is a surjective open submersion) and the map from the functions on any affine $V \subset Y$ to $G$-invariant functions on $\pi^{-1}(V)$ is an isomorphism.  It's a good geometric quotient if the $G$-orbits closed in $Y$.&lt;/p&gt;&#xA;" OwnerUserId="3008" LastEditorUserId="3008" LastEditorDisplayName="" LastEditDate="2010-11-15T14:45:42.750" LastActivityDate="2010-11-15T14:45:42.750" Title="Quotient spaces and equivariant cohomology" Tags="&lt;algebraic-topology&gt;&lt;cohomology&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="10343" PostTypeId="2" ParentId="10335" CreationDate="2010-11-15T02:03:43.810" Score="1" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\rm\quad (2\:n)^2\ =\ 4\:n^2,\quad (2\:n+1)^2\ =\ 4\:(n^2+n)+1\ \ $ so squares are $\rm\equiv 0,\: 1\ (mod\ 4)$&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-11-15T02:03:43.810" />
  <row Id="10344" PostTypeId="2" ParentId="10327" CreationDate="2010-11-15T02:10:41.477" Score="13" ViewCount="" Body="&lt;p&gt;No, this number is transcendental. The proof by Mahler mentioned in a comment shows this. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A good reference to learn about basic transcendental number theory is the book &quot;Making transcendence transparent: an intuitive approach to classical transcendental number theory&quot;, by Edward Burger and Robert Tubbs, Springer-Verlag (2004).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In chapter 1 of the book the proof of the transcendence of Mahler's constant $0.1234\dots$ is discussed. The idea is to show that the &quot;obvious&quot; rational approximations actually are very very close, to the point that they would contradict easy estimates (due to Liouville) for how quickly rational numbers can approximate irrational algebraic numbers. The Wikipedia entry on &lt;a href=&quot;http://en.wikipedia.org/wiki/Liouville_number&quot; rel=&quot;nofollow&quot;&gt;Liouville numbers&lt;/a&gt; discusses Liouville's approximation theorem and related results:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $\alpha$ is algebraic of degree $d\ge 2$ then there is a constant $C$ such that for any rational $p/q$ with $q&amp;gt;0$, we have $$ \left|\alpha-\frac pq\right|&amp;gt;\frac{C}{q^d}. $$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Actually, there is a bit of work needed here. The estimates the book discusses together with a &lt;em&gt;strengthening&lt;/em&gt; of Liouville's theorem give the proof for Mahler's constant, and the same argument works for the number you are asking. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Thue%E2%80%93Siegel%E2%80%93Roth_theorem&quot; rel=&quot;nofollow&quot;&gt;strengthening&lt;/a&gt; we need is due to Klaus Roth in 1955, and he was awarded the Fields medal in 1958 for this result.&lt;/p&gt;&#xA;" OwnerUserId="462" LastActivityDate="2010-11-15T02:10:41.477" />
  <row Id="10345" PostTypeId="2" ParentId="10329" CreationDate="2010-11-15T02:16:51.390" Score="1" ViewCount="" Body="&lt;p&gt;You are correct that it is the same as saying that $T$ is not bijective, because it follows from the open mapping theorem that a bounded operator on a Banach space has a bounded inverse if it is bijective.  However, more straightforward answers for your question can be given without explicitly thinking in these terms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can show that if $T$ is compact and $S$ is bounded, then $ST$ is compact.  If $T$ were invertible, this would imply that $I=T^{-1}T$ is compact.  This in turn translates to saying that the closed unit ball of $X$ is compact.  One way to see that this is impossible in the infinite dimensional case is implicit in &lt;a href=&quot;http://math.stackexchange.com/questions/4815/fitting-an-infinite-collection-of-balls-in-an-infinite-dimensional-unit-ball&quot;&gt;this question&lt;/a&gt;.  There is an infinite sequence of points in the unit ball whose pairwise distances are bounded below, no subsequence of which is Cauchy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Compact operators on infinite dimensional spaces can be injective, but they can never be surjective.  The closed subspaces of the range of a compact operator are finite dimensional.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, what Qiaochu said in his comment above.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-11-15T02:16:51.390" />
  <row Id="10346" PostTypeId="1" AcceptedAnswerId="10605" CreationDate="2010-11-15T02:31:51.857" Score="6" ViewCount="247" Body="&lt;p&gt;Combining the answers given by me and Ralth to the probability question at &lt;a href=&quot;http://math.stackexchange.com/questions/10284/probability-question&quot;&gt;Probability Question&lt;/a&gt;, we get the following identity:&#xA;$$&#xD;&#xA;\sum\limits_{k = m}^n {{n \choose k}p^k (1 - p)^{n - k} {k \choose m} p_j^m (1 - p_j )^{k - m} } = {n \choose m} (p p_j)^m (1 - p p_j)^{n-m},&#xD;&#xA;$$&#xD;&#xA;where $m \geq 0$ and $n \geq m$ are arbitrary integers, and $p$ and $p_j$ are arbitrary probabilities (in Ralth's notation $m$ is $k$, $p$ is $p_s$, and $p_j$ is $p_h$). Can you prove this identity directly (i.e., in a non-probabilistic setting)? Whether you'll find this identity interesting or not, at least Ralth's answer may now gain its due recognition.&lt;/p&gt;&#xA;" OwnerUserId="2810" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-11-15T07:34:18.140" LastActivityDate="2010-11-16T21:17:10.470" Title="Non-probabilistic proofs of a binomial coefficient identity from a probability question" Tags="&lt;probability&gt;&lt;combinatorics&gt;&lt;algebra&gt;&lt;binomial-coefficients&gt;" AnswerCount="3" FavoriteCount="1" />
  <row Id="10347" PostTypeId="1" AcceptedAnswerId="10357" CreationDate="2010-11-15T02:40:01.817" Score="0" ViewCount="138" Body="&lt;p&gt;How would I go about creating a polynomial $f$ such that&lt;br&gt;&#xA;\begin{align*}&#xA;f(x) &amp;amp;= A_1\\&lt;br&gt;&#xA;f(y) &amp;amp;= A_2\\&lt;br&gt;&#xA;f'(x) &amp;amp;= B_1\\&lt;br&gt;&#xA;f'(y) &amp;amp;= B_2&#xA;\end{align*}&lt;br&gt;&#xA;With $x$ not equal to $y$ and $A_1$, $A_2$, $B_1$, $B_2$ real numbers?&lt;/p&gt;&#xA;" OwnerUserId="3333" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-15T04:08:40.783" LastActivityDate="2010-11-15T06:15:30.360" Title="Create a function" Tags="&lt;real-analysis&gt;" AnswerCount="2" CommentCount="21" />
  <row Id="10348" PostTypeId="1" AcceptedAnswerId="10354" CreationDate="2010-11-15T03:05:08.443" Score="2" ViewCount="107" Body="&lt;p&gt;I have a finite alphabet A and a set S of $n$ ordered pairs $(a, b)$ with $a,b\in A.$  I want to find out how many sequences $(a_1, b_1),\ldots,(a_n,b_n)$ there are where adjacent elements have either $a_{k-1}=a_k$ or $b_{k-1}=b_k.$ (Since S is a set, they can't have both.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any ideas on how to go about finding this?  Formulas, generating functions, algorithms, or whatever you have.  In my particular application I have $|A|=10$ and $|S|=21$, but the question seems general.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, in my application $(a_1,b_1)$ and $(a_k,b_k)$ are &lt;em&gt;not&lt;/em&gt; adjacent. (They can have, but need not have, one member different.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; For clarification, I'm &lt;em&gt;not&lt;/em&gt; allowing all elements from $A\times A$, just those in $S$.  No valid answer can ignore S.  For example, I might have S = {(1, 1), (3, 3)} in which case there are no valid orderings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For those interested in using a matrix approach (not sure if this can work; the path needs to go through each pair exactly once), here's the adjacency matrix:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;[0 1 1 1 0 0 1 0 1 0 0 0 0 1 0 1 0 0 0 0 0]&#xA;[1 0 1 1 1 0 0 0 0 1 0 1 0 0 0 0 1 0 1 0 0]&#xA;[1 1 0 1 0 0 0 1 0 0 1 0 0 0 1 0 0 0 0 0 1]&#xA;[1 1 1 0 0 1 0 0 0 0 0 0 1 0 0 0 0 1 0 1 0]&#xA;[0 1 0 0 0 1 0 0 0 1 0 1 0 0 0 0 1 0 1 0 0]&#xA;[0 0 0 1 1 0 0 0 0 0 0 0 1 0 0 0 0 1 0 1 0]&#xA;[1 0 0 0 0 0 0 1 1 0 0 0 0 1 0 1 0 0 0 0 0]&#xA;[0 0 1 0 0 0 1 0 0 0 1 0 0 0 1 0 0 0 0 0 1]&#xA;[1 0 0 0 0 0 1 0 0 1 1 0 0 1 0 1 0 0 0 0 0]&#xA;[0 1 0 0 1 0 0 0 1 0 1 1 0 0 0 0 1 0 1 0 0]&#xA;[0 0 1 0 0 0 0 1 1 1 0 0 0 0 1 0 0 0 0 0 1]&#xA;[0 1 0 0 1 0 0 0 0 1 0 0 1 0 0 0 1 0 1 0 0]&#xA;[0 0 0 1 0 1 0 0 0 0 0 1 0 0 0 0 0 1 0 1 0]&#xA;[1 0 0 0 0 0 1 0 1 0 0 0 0 0 1 1 0 0 0 0 0]&#xA;[0 0 1 0 0 0 0 1 0 0 1 0 0 1 0 0 0 0 0 0 1]&#xA;[1 0 0 0 0 0 1 0 1 0 0 0 0 1 0 0 1 1 0 0 0]&#xA;[0 1 0 0 1 0 0 0 0 1 0 1 0 0 0 1 0 1 1 0 0]&#xA;[0 0 0 1 0 1 0 0 0 0 0 0 1 0 0 1 1 0 0 1 0]&#xA;[0 1 0 0 1 0 0 0 0 1 0 1 0 0 0 0 1 0 0 1 0]&#xA;[0 0 0 1 0 1 0 0 0 0 0 0 1 0 0 0 0 1 1 0 0]&#xA;[0 0 1 0 0 0 0 1 0 0 1 0 0 0 1 0 0 0 0 0 0]&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Kudos to anyone who recognizes it (and thus my goal).&lt;/p&gt;&#xA;" OwnerUserId="1778" LastEditorUserId="1778" LastEditorDisplayName="" LastEditDate="2010-11-15T17:39:51.543" LastActivityDate="2010-11-15T20:02:06.317" Title="Sequences of pairs where adjacent elements differ in exactly one component" Tags="&lt;linear-algebra&gt;&lt;combinatorics&gt;&lt;graph-theory&gt;" AnswerCount="4" CommentCount="6" />
  <row Id="10349" PostTypeId="1" AcceptedAnswerId="10356" CreationDate="2010-11-15T03:13:38.293" Score="1" ViewCount="92" Body="&lt;p&gt;I have trouble understanding an argument in the proof of the Kummer-Dedekind theorem.&#xA;I am referring to a proof given in Peter Stevenhagen's notes.&#xA;&lt;a href=&quot;http://websites.math.leidenuniv.nl/algebra/ant.pdf&quot; rel=&quot;nofollow&quot;&gt;http://websites.math.leidenuniv.nl/algebra/ant.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is theorem 3.1 on page 27. For the proof of the second part, the first statement says that since $p_i$ contains $pR$, it is invertible. I am not sure what the author is invoking here. For every ideal contains a principal ideal, but that is not sufficient for invertibility. Also, the fact that $p_i$ contains $pR$ should follow from the first part of the theorem since $p_i=pR+g_i(\alpha)R$. So I don't see what fact is used in proving invertibility of $p_i$ based on its containment of $pR$. Any help would be very appreciated.&lt;/p&gt;&#xA;" OwnerUserId="3431" LastEditorUserId="1778" LastEditorDisplayName="" LastEditDate="2010-11-15T17:20:33.813" LastActivityDate="2010-11-15T17:20:33.813" Title="Invertibility of prime ideals in a number ring lying over prime numbers" Tags="&lt;number-theory&gt;&lt;prime-numbers&gt;&lt;commutative-algebra&gt;&lt;algebraic-number-theory&gt;" AnswerCount="1" />
  <row Id="10350" PostTypeId="2" ParentId="10346" CreationDate="2010-11-15T03:20:00.787" Score="3" ViewCount="" Body="&lt;p&gt;Consider the following two-step process: Start with $n$ elements. Pick each one with probability $p$. Among the $k$ elements chosen in the first phase, pick each one with probability $p_j$. The left-hand side describes the probability that the final set is of size $m$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can also calculate this probability directly: every element is chosen to the final set with probability $pp_j$, whence the right-hand side.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-15T03:20:00.787" CommentCount="2" />
  <row Id="10351" PostTypeId="2" ParentId="10348" CreationDate="2010-11-15T03:24:48.350" Score="-1" ViewCount="" Body="&lt;p&gt;Perhaps I'm misunderstanding your problem. But if you want to count how many sequences of pairs there are which where each two conscutive pairs have at least one element in common, the answer seems easy. Just count how many sequences there are that have not that property, and substract from the total.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I get, then, $a^{2n} - a^2 (a-1)^{2(n-1)} = a^2 [ a^{2(n-1)} - (a-1)^{2(n-1)}]$ (where $a = |A|$)&lt;/p&gt;&#xA;" OwnerUserId="312" LastEditorUserId="312" LastEditorDisplayName="" LastEditDate="2010-11-15T03:30:34.790" LastActivityDate="2010-11-15T03:30:34.790" CommentCount="4" />
  <row Id="10352" PostTypeId="2" ParentId="10348" CreationDate="2010-11-15T03:28:33.933" Score="0" ViewCount="" Body="&lt;p&gt;EDIT: This answer assumes that $S = A \times A$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given a pair $(a,b)$, there are $2|A|-1$ pairs $(a&amp;#39;,b&amp;#39;)$ such that $a=a&amp;#39;$ or $b=b&amp;#39;$ (or both). Therefore if we don't care about the relation between first and last element, we get $|A|^2(2|A|-1)^{n-1}$ sequences.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2010-11-15T20:02:06.317" LastActivityDate="2010-11-15T20:02:06.317" CommentCount="1" />
  <row Id="10353" PostTypeId="1" AcceptedAnswerId="10386" CreationDate="2010-11-15T03:31:58.640" Score="1" ViewCount="131" Body="&lt;p&gt;Two vectors $\mathbf{x}=(x_1,x_2)$ and $\mathbf{y}=(y_1,y_2)$ are orthogonal if their scalar product is zero, i.e. $x_1 y_1 + x_2 y_2=0$. What surface does this equation describe in the four-dimensional $(x_1,x_2,y_1,y_2)$ space? How about the orthogonality condition for $3$-dimensional (or higher dimensional) vectors? What is the corresponding surface and how would one visualize it?&#xA;Thanks for the answers.&lt;/p&gt;&#xA;" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-15T04:05:09.603" LastActivityDate="2010-11-15T22:35:45.837" Title="Surface described by orthogonality condition for vectors" Tags="&lt;vector-calculus&gt;" AnswerCount="2" />
  <row Id="10354" PostTypeId="2" ParentId="10348" CreationDate="2010-11-15T03:35:36.640" Score="4" ViewCount="" Body="&lt;p&gt;Let $m = \lvert A\rvert$, and denote by $L_{m,m}$ what is variously called the $m\times m$ &lt;a href=&quot;http://en.wikipedia.org/wiki/Rook%27s_graph&quot; rel=&quot;nofollow&quot;&gt;rook's graph&lt;/a&gt; or &lt;a href=&quot;http://mathworld.wolfram.com/LatticeGraph.html&quot; rel=&quot;nofollow&quot;&gt;lattice graph&lt;/a&gt;: it has $m\times m$ vertices arranged in a square, with an edge connecting two vertices if they are in the same row or column. There is a natural mapping between the vertices of $L_{m,m}$ and ordered pairs of elements from $A$, of which $S$ is a subset. Then a sequence of length $n$ in your problem is equivalent to a &lt;a href=&quot;http://en.wikipedia.org/wiki/Hamiltonian_path&quot; rel=&quot;nofollow&quot;&gt;Hamiltonian path&lt;/a&gt; in the subgraph of $L_{m,m}$ induced by $S$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The number of such paths will certainly depend on the elements of $S$, and not just their cardinality. For example, let $m = 2$ and $n = \lvert S\rvert = 2$; if the two elements of $S$ are adjacent, then there is one Hamiltonian path, otherwise none.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The simplest algorithm I can think of is to simply generate all permutations of the $n$ vertices and check whether each one is a valid path. &lt;a href=&quot;http://en.wikipedia.org/wiki/Hamiltonian_path_problem&quot; rel=&quot;nofollow&quot;&gt;Finding a &lt;em&gt;single&lt;/em&gt; Hamiltonian path is NP-complete&lt;/a&gt;, so I'm not sure you can do much better. You could do a search for &quot;count number of hamiltonian paths&quot; and see what you get.&lt;/p&gt;&#xA;" OwnerUserId="856" LastEditorUserId="856" LastEditorDisplayName="" LastEditDate="2010-11-15T18:25:46.680" LastActivityDate="2010-11-15T18:25:46.680" CommentCount="8" />
  <row Id="10355" PostTypeId="2" ParentId="10347" CreationDate="2010-11-15T03:38:07.267" Score="3" ViewCount="" Body="&lt;p&gt;This would be a typical question given in the beginning of a linear algebra class. You have the information that $f(x)$ passes through the points $(X,A_1)$ and $(Y,A_2)$ and $f&amp;#39;(x)$ passes through the points $(X, B_1)$ and $(Y,B_2)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you have a set of 4 linear equations for 4 unknowns. Try setting up a linear system and solving it.&#xA;Hint: The first condition could be written as&#xA;$$[X^3 \;\; X^2 \;\; X \;\;\; 1] \left[\begin{array}{c}a\\ b\\ c\\ d\\\end{array} \right] =A_1 $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you put together all 4 of your equations you end up with an augmented matrix which you can put in RREF form to find your answer&lt;/p&gt;&#xA;" OwnerUserId="2596" LastEditorUserId="2596" LastEditorDisplayName="" LastEditDate="2010-11-15T03:44:14.393" LastActivityDate="2010-11-15T03:44:14.393" CommentCount="1" />
  <row Id="10356" PostTypeId="2" ParentId="10349" CreationDate="2010-11-15T03:42:12.127" Score="1" ViewCount="" Body="&lt;p&gt;Principal ideals are invertible, and so too are divisors of invertible ideals, so  $\mathfrak{p}\:|\:pR\ \Rightarrow\ \mathfrak{p}\ $ invertible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the ring in the theorem is not necessarily a a Dedekind domain so it need not be true that $\ I \supset J\ \Rightarrow\ I|J$&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-11-15T03:50:20.213" LastActivityDate="2010-11-15T03:50:20.213" CommentCount="6" />
  <row Id="10357" PostTypeId="2" ParentId="10347" CreationDate="2010-11-15T04:16:13.943" Score="5" ViewCount="" Body="&lt;p&gt;In general, there are many possible answers. But, suppose your polynomial looks like this:&#xA;$$f(X) = a_n X^n + \cdots + a_1 X + a_0$$&#xD;&#xA;with $a_0,\ldots,a_n$ real numbers. Then &#xA;$$f&amp;#39;(X) = na_n X^n + \cdots + a_1.$$&#xD;&#xA;Evaluating at $x$ and at $y$ will give you a bunch of equations that they need to satisfy:&#xA;\begin{align*}&#xA;a_n x^n + \cdots + a_1x + a_0 &amp;amp; = A_1\\&#xA;na_n x^{n-1} + \cdots + a_1 &amp;amp;= B_1\\&#xA;a_n y^n + \cdots  a_1y + a_0 &amp;amp;= A_2\\&#xA;na_n y^{n-1} + \cdots a_1 &amp;amp;= B_2.&#xA;\end{align*}&#xA;You cannot really to get an answer if the degree of the polynomial is $0$ or $1$: too many equations with too few unknowns. Degree $2$ would give you four equations with three unknowns, which may be too restrictive as well. So your best hope of having it be solvable and not too hard is for the degree (for $n$) to be $3$. Then you'll have a bunch of equations like so:&#xA;$$\begin{array}{rcrcrcrcl}&#xD;&#xA;a_3 x^3 &amp;amp;+&amp;amp; a_2 x^2 &amp;amp;+&amp;amp; a_1 x &amp;amp;+&amp;amp; a_0 &amp;amp;=&amp;amp; A_1\\&#xD;&#xA;a_3 y^3 &amp;amp;+&amp;amp; a_2 y^2 &amp;amp;+&amp;amp; a_1 y &amp;amp;+&amp;amp; a_0 &amp;amp;=&amp;amp; A_2\\&#xD;&#xA;3a_3 x^2 &amp;amp;+&amp;amp; 2a_2 x &amp;amp;+&amp;amp; a_1   &amp;amp; &amp;amp;     &amp;amp;=&amp;amp; B_1\\&#xD;&#xA;3a_3 y^2 &amp;amp;+&amp;amp; 2a_2 y &amp;amp;+&amp;amp; a_1   &amp;amp; &amp;amp;     &amp;amp;=&amp;amp; B_2.&#xD;&#xA;\end{array}$$&#xD;&#xA;So you go ahead and solve this system in one of the standard ways; for example, back substitution if you don't know the methods from linear algebra.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added:&lt;/strong&gt; How to solve these if you don't know linear algebra? The less sophisticated method is probably to &quot;solve and backsubstitute&quot;. Using the last equation, you can write $a_1$ using $y$, $a_2$, $a_3$, and $B_2$. Substitute that value into each of the previous three equations, and then each of those equations only has three unknowns. Then use one of them to write $a_2$ in terms of the other two unknowns, and substitute that into the remaining two equations. Then use one of them to &quot;solve&quot; for one unknown in terms of the other, and substitute into the remaining one. Solve for the unknown, and then work backwards.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, suppose that $x=2$ and $y=3$, $A_1 = -2$, $A_2=1$, $B_1 = 0$, and $B_2 = 1$ (I just picked those at random). So the equations become:&#xA;$$\begin{array}{rcrcrcrcl}&#xD;&#xA;8a_3 &amp;amp;+&amp;amp; 4a_2  &amp;amp;+&amp;amp; 2a_1  &amp;amp;+&amp;amp; a_0 &amp;amp;=&amp;amp; -2\\&#xD;&#xA;27a_3 &amp;amp;+&amp;amp; 9a_2  &amp;amp;+&amp;amp; 3a_1  &amp;amp;+&amp;amp; a_0 &amp;amp;=&amp;amp; 1\\&#xD;&#xA;12a_3  &amp;amp;+&amp;amp; 4a_2  &amp;amp;+&amp;amp; a_1   &amp;amp; &amp;amp;     &amp;amp;=&amp;amp; 0\\&#xD;&#xA;27a_3  &amp;amp;+&amp;amp; 6a_2  &amp;amp;+&amp;amp; a_1   &amp;amp; &amp;amp;     &amp;amp;=&amp;amp; 1.&#xD;&#xA;\end{array}$$&#xD;&#xA;From the last equation, you have that $a_1 = 1 - 27a_3 - 6a_2$. Plugging that into each of the other three equations you get:&#xA;$$\begin{array}{rcrcrcl}&#xD;&#xA;-36a_3 &amp;amp; - &amp;amp; 8a_2 &amp;amp; + &amp;amp; a_0 &amp;amp;=&amp;amp; -4\\&#xD;&#xA;-54a_3 &amp;amp; - &amp;amp; 9a_2 &amp;amp; + &amp;amp; a_0 &amp;amp; = &amp;amp; -2\\&#xD;&#xA;-15a_3 &amp;amp; - &amp;amp; 2a_2 &amp;amp;   &amp;amp;     &amp;amp; = &amp;amp; -1.&#xD;&#xA;\end{array}$$&#xD;&#xA;Then we can use the first equation to write $a_0 = 36a_3 + 8a_2 - 4$, and substituting into the second equation we get&#xA;$$\begin{array}{rcrcl}&#xD;&#xA;-18a_3 &amp;amp; - &amp;amp; a_2 &amp;amp; = &amp;amp; 2\\&#xD;&#xA;-15a_3 &amp;amp; - &amp;amp; 2a_2 &amp;amp; = &amp;amp; -1.&#xD;&#xA;\end{array}$$&#xD;&#xA;From the first equation you get $a_2 = -18a_3-2$, which plugged into the second equation gives $21a_3=-3$, or $a_3 = -\frac{1}{7}$. This gives you the value for $a_2$, since you know $a_2 = -18a_3-2$; once you have $a_2$ and $a_3$, you use $a_0 = 36a_3+8a_2-4$ to find $a_0$; once you have $a_0$, $a_2$, and $a_3$, you use $a1 = 1 - 27a_3 - 6a_2$ to find $a_1$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can do the back-substitution in any order; this is &lt;em&gt;not&lt;/em&gt; the best way of doing it, but it's certainly &lt;em&gt;a&lt;/em&gt; way that may have occurred to you if you have ever solved a system of two equations with two unknowns. It's called &quot;back-substitution&quot; because you use one equation to isolate a variable, then you &quot;substitute it&quot; into the equations that are &quot;at the bacK&quot;, the other equations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Trying to do this for more equations with more unknowns quickly becomes far too annoying, which is why there are better methods, such as Gaussian Elimination. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added 2:&lt;/strong&gt; If, as your comments in the question suggest, you are supposed to come up with &lt;em&gt;generic&lt;/em&gt; formulas for arbitrary numbers $A_1, A_2,B_1,B_2$, $x$, and $y$, then this method is difficult, though you &lt;em&gt;can&lt;/em&gt; push it through if you are careful and are really comfortable with algebra. You have to remember that $x$, $y$, $A_1$, $A_2$, $B_1$, and $B_2$ are supposed to be &lt;em&gt;numbers&lt;/em&gt;, not variables, so if you can write $a_i$ in terms only of these, then you've &quot;solved&quot; $a_i$. Your final answer would express $a_0$, $a_1$, $a_2$, and $a_3$ in terms of $A_1$, $A_2$, $B_1$, $B_2$, $x$, and $y$. &lt;/p&gt;&#xA;&#xA;&lt;hr/&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added 3:&lt;/strong&gt; Your comments &lt;em&gt;clearly&lt;/em&gt; say now that you are trying to come up with formulas that you can just plug $x$, $y$, $A_1$, $A_2$, $B_1$, and $B_2$ into to get your polynomial. &lt;em&gt;You are not going to be able to come up with them using the methods discussed so far.&lt;/em&gt; Not because it's impossible, but  because you are not exhibiting the kind of comfort with algebra that you would need for that. It would involve carrying around fairly ugly expressions around in a comfortable way, and it seems clear you cannot do this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So let's try a slightly different tack. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Start slow. A simple way to get a polynomial that will give you $A_1$ when evaluated at $x$ and $A_2$ when evaluated at $y$ is to &quot;cheat&quot;: make it a sum of two polynomials, one zero at $x$ and the other zero at $y$, and then just adjust things so they come out right. So you can take, for example, something like&#xA;$$f(X) = K(X-x) + M(X-y).$$&#xD;&#xA;Then $f(x) = M(x-y)$ and $f(y)=K(y-x)$. So you get easy formulas for $M$ and $K$: since $M(x-y)=A_1$, then $M = A_1/(x-y)$ (which you can do because $x\neq y$); and since $K(y-x) = A_2$, then $K=A_2/(y-x)$. So we get&#xA;$$f(X) = \frac{A_2}{y-x}(X-x) + \frac{A_1}{x-y}(X-y).$$&#xD;&#xA;Now, this is not going to do it for the derivatives, though. So we are going to have to add things to this polynomial to get the derivatives to come out right. How can we add things so as not to mess up what we've already achieved (that the first two equations are satisfied)? Well, we can make sure we don't mess up what we already have if we make sure that whatever we add has value $0$ at both $x$ and $y$. So we'll try something like:&#xA;$$f(X) = \frac{A_2}{y-x}(X-x) + \frac{A_1}{x-y}(X-y) + (X-x)(X-y)g(X)$$&#xD;&#xA;for some $g(X)$. What is the derivative of such an $f(X)$? Remembering that $x$, $y$, $A_1$, and $A_2$ are &lt;em&gt;numbers&lt;/em&gt;, then applying the product rule we have:&#xA;$$f&amp;#39;(X) = \frac{A_2}{y-x} + \frac{A_1}{x-y} + (X-y)g(X) + (X-x)g(X) + (X-x)(X-y)g&amp;#39;(X).$$&#xD;&#xA;If you evaluate at $x$, we get&#xA;$$B_1 = f&amp;#39;(x) = \frac{A_2}{y-x}+\frac{A_1}{x-y}+(x-y)g(x).$$&#xD;&#xA;Everything is numbers that we'll know, except for $g(x)$. So that means that we need&#xA;$$g(x) = \frac{B_1}{x-y} - \frac{A_2}{(y-x)(x-y)} - \frac{A_1}{(x-y)^2}.$$&#xD;&#xA;And if we evaluate $f&amp;#39;(X)$ at $y$ we get&#xA;$$B_2 = \frac{A_2}{y-x} + \frac{A_1}{x-y} + (y-x)g(y).$$&#xD;&#xA;Again, everything except $g(y)$ are numbers we already are supposed to know. So we get:&#xA;$$g(y) = \frac{B_2}{y-x} - \frac{A_2}{(y-x)^2} - \frac{A_1}{(x-y)(y-x)}.$$&#xD;&#xA;Notice that the derivative of $g(X)$ turns out to not play any role at all after all! So we can make $g(X)$ as simple as we possibly can; say, a linear polynomial.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can we achieve this? Well, we discussed above how to make a polynomial of degree $1$ that has specific values at two different points, right? So you can just use the same trick we used for $f(x)$ and $f(y)$, but with $g(x)$ and $g(y)$, using the (complicated) expressions we just found for $g(x)$ and for $g(y)$ instead of the $A_1$ and $A_2$ that we used for $f(X)$. So go to it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Note:&lt;/em&gt; You can keep playing with this if you have still more points-and-values; it gets progresssively more difficult. You would need to start with polynomials of degree $n-1$ if you know the value at $n$ points; this leads you to Lagrange Interpolation polynomials. The advantage of the linear algebra method is that the complexity grows in a bit more manageable fashion than this method.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-15T06:15:30.360" LastActivityDate="2010-11-15T06:15:30.360" CommentCount="8" />
  <row Id="10358" PostTypeId="1" AcceptedAnswerId="10361" CreationDate="2010-11-15T04:17:30.847" Score="2" ViewCount="105" Body="&lt;p&gt;There are two equivalent definitions of a locally convex topological vector space. Note that since the vector space is topological, addition by a fixed vector x and multiplication by a fixed scalar r are both homeomophisms, which means it is enough to give a neighborhood basis at the vector 0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first definition is that $0$ has a local base consisting of open sets that are &lt;em&gt;convex&lt;/em&gt; (if $x$ and $y$ are in $S$, then $\lambda x+(1-\lambda)y$ is also in $S$ for every $\lambda\in[0,1]$), &lt;em&gt;balanced&lt;/em&gt;, (if $x\in S$, then $r x\in S$ for every $r\in\mathbb{F}$ with $|r|=1$, where $\mathbb F$ is either the real or complex numbers), and &lt;em&gt;absorbing&lt;/em&gt; (for any vector $y$ there exists a $\lambda\in (0,\infty)$ such that $\lambda S$ contains $y$, i.e. $S$ absorbs $y$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second definition is that $0$ has a local base given by the ''balls'' of radius $r$ for each $r$ of each semi-norm in some fixed collection of semi-norms (a semi-norm is just like a norm except that non-zero vectors can have a non-zero semi-norm). The two definitions are equivalent in one direction because &quot;balls&quot; of semi-norms are convex, balanced and absorbing, and in the other because we can define for any convex, balanced, absorbing set $S$ the semi-norm $f_S$ by assigning to $x$ the &quot;least&quot; (infimum) $\lambda$ for which $x$ is in $\lambda S$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is: what are some examples of topological vector spaces for which $0$ does have a local base of convex open sets, but that no local base consists of convex sets that are also balanced and absorbing?&lt;/p&gt;&#xA;" OwnerUserId="400" LastEditorUserId="400" LastEditorDisplayName="" LastEditDate="2010-11-15T04:47:05.207" LastActivityDate="2010-11-15T04:47:05.207" Title="What is an example of a topological vector space which contains a non-absorbent or non-balanced open convex set around the origin?" Tags="&lt;functional-analysis&gt;" AnswerCount="2" />
  <row Id="10359" PostTypeId="2" ParentId="10358" CreationDate="2010-11-15T04:21:07.977" Score="0" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;NB:&lt;/strong&gt; &lt;em&gt;As Matt notes in a comment below, the following does not really work!&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider a vector space $V$ of dimension at least $2$, pick a non-zero proper subspace $W\subseteq V$, and consider the unique linear topology on $V$ for which $\{W\}$ is a basis of neighborhoods of $0\in V$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want an Hausdorff example, let now $V$ be an infinite dimensional vector space, and consider a decreasing sequence $(W_n)_{n\geq1}$ of non-zero subspaces of $V$ such that $\bigcap_{n\geq1}W_n=0$. Now put on $V$ the unique linear topology which has $\{V_n\}_{n\geq1}$ as a basis of neighborhoods of $0\in V$.&lt;/p&gt;&#xA;" OwnerUserId="274" LastEditorUserId="274" LastEditorDisplayName="" LastEditDate="2010-11-15T04:39:28.797" LastActivityDate="2010-11-15T04:39:28.797" CommentCount="2" />
  <row Id="10361" PostTypeId="2" ParentId="10358" CreationDate="2010-11-15T04:39:53.527" Score="3" ViewCount="" Body="&lt;p&gt;Unless I am misunderstanding the situation, there is no such example.  That is, if $V$ is a topological vector space (over $\mathbb R$ or $\mathbb C$) which has a basis of neighbourhoods of the origin consisting of convex sets,  then it also has a basis of neighbourhoods of the origin consisting of balanced convex sets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, any neighbourhood of $0$ is automatically absorbent (because of continuity of scalar multiplication).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My reference is Robertson and Roberston, &lt;em&gt;Topological Vector Spaces&lt;/em&gt; (Cambridge Tracts in Mathematics and Mathematical Physics, 53), which is a kind of summary of Bourbaki's Topological Vector Spaces volume.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The proof is not hard: one first uses continuity of scalar multiplication to show&#xA;that any neighbourhood of $0$ contains a balanced neighbourhood.  Now if $U$ is a convex n.h of $0$, and $W$ is a balanced n.h. of $0$ contained in $U$, then $V:= \cap_{|x| \geq 1} x U$ contains $W$, so is a n.h., is convex (being the intersection of convex sets), is balanced (by construction), and lies in $U$ (again by construction --- consider $x  = 1$).  Thus $V$ is a convex balanced n.h. of $0$ contained in $U$.&lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2010-11-15T04:39:53.527" />
  <row Id="10362" PostTypeId="1" AcceptedAnswerId="10363" CreationDate="2010-11-15T04:42:46.957" Score="3" ViewCount="152" Body="&lt;p&gt;I'm going through an examples section (on improper integrals) but I got lost at this bit:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\lim_{t\to-\infty} \frac{t}{e^{-t}} = \lim_{t\to-\infty}\frac{1}{-e^{-t}}.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think it's a simple algebra trick but I don't see how the right hand side came to be. How did it become $1/-e^{-t}$?&lt;/p&gt;&#xA;" OwnerUserId="3514" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-15T05:42:31.053" LastActivityDate="2010-11-15T17:19:52.387" Title="Help seeing algebra in calculus" Tags="&lt;calculus&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="10363" PostTypeId="2" ParentId="10362" CreationDate="2010-11-15T04:45:05.340" Score="4" ViewCount="" Body="&lt;p&gt;It is an application of &lt;a href=&quot;http://en.wikipedia.org/wiki/L%27H%C3%B4pital%27s_rule&quot; rel=&quot;nofollow&quot;&gt;l'Hôpital's rule&lt;/a&gt;. &lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-11-15T04:45:05.340" CommentCount="1" />
  <row Id="10364" PostTypeId="2" ParentId="10336" CreationDate="2010-11-15T04:53:17.637" Score="1" ViewCount="" Body="&lt;p&gt;The &quot;usual proof&quot; I know for $\Sigma^0_2$ is the same as for general $\Pi^1_1$ formulas. For $\Sigma^0_2$ formulas in particular, this is one possible method: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Take the negation of your original formula to get a $\Pi^0_2$ formula. Construct a partial order on a subset of $\omega$ such that the partial order has an infinite descending sequence if and only if the original formula is false.&lt;/li&gt;&#xA;&lt;li&gt;Find a way to turn this partial order into a linear order while keeping the relevant properties.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This is intentionally just a hint. The usual method by which these things are proved is not particularly obvious because it relies on a clever relationship between computable linear orders and computable subtrees of $\omega^{&amp;lt;\omega}$ (this relationship goes by the name &quot;Kleene-Brouwer order&quot;). &lt;/p&gt;&#xA;" OwnerUserId="630" LastActivityDate="2010-11-15T04:53:17.637" CommentCount="1" />
  <row Id="10365" PostTypeId="2" ParentId="10362" CreationDate="2010-11-15T04:57:12.420" Score="1" ViewCount="" Body="&lt;p&gt;The ratio of functions which you provide are an example of an indeterminate form at the limit point. L'Hospital's rule allows you to differentiate the numerator and denominator independently and then take the limit, giving the intended limit. For your example,&#xA;\begin{eqnarray}&#xA;\lim_{t \to -\infty} \frac{t}{e^{-t}} \stackrel{L.'H.}{=} \lim_{t \to -\infty} \frac{1}{-e^{-t}} = 0.&#xA;\end{eqnarray}&lt;/p&gt;&#xA;" OwnerUserId="2720" LastEditorUserId="2720" LastEditorDisplayName="" LastEditDate="2010-11-15T17:19:52.387" LastActivityDate="2010-11-15T17:19:52.387" CommentCount="2" />
  <row Id="10367" PostTypeId="2" ParentId="10346" CreationDate="2010-11-15T05:22:55.737" Score="2" ViewCount="" Body="&lt;p&gt;Writing the left-hand side as&#xA;$$&#xD;&#xA;\sum\limits_{k = 0}^{n - m} {{n \choose k + m} p^{k + m} (1 - p)^{n - (k + m)} {k + m \choose m} p_j^m (1 - p_j )^{k + m - m} } &#xD;&#xA;$$&#xD;&#xA;we find, after some algebra, that the original equation is equivalent to&#xA;$$&#xD;&#xA;\sum\limits_{k = 0}^l {{l \choose k}p^k (1 - p)^{l - k} (1 - p_j )^k }  = (1 - pp_j )^l ,&#xD;&#xA;$$&#xD;&#xA;where $l (=n-m) \geq 0$. Since the case $p=0$ is satisfied ($0^0=1$), we assume that $p = x/y$, with $0 &amp;lt; x \leq y$. Substituting this leads straightforwardly to the equivalent equation&#xA;$$&#xD;&#xA;\sum\limits_{k = 0}^l {{l \choose k}(x - x p_j)^k (y-x)^{l-k}  = (y - xp_j)^l}.&#xD;&#xA;$$&#xD;&#xA;By the binomial theorem, the left-hand side is equal to $[(x - xp_j ) + (y - x)]^l$, completing the proof.&lt;/p&gt;&#xA;" OwnerUserId="2810" LastActivityDate="2010-11-15T05:22:55.737" />
  <row Id="10371" PostTypeId="2" ParentId="10318" CreationDate="2010-11-15T05:51:13.513" Score="1" ViewCount="" Body="&lt;p&gt;While Timothy Wagner's answer is correct, I thought you might like to see another way to answer your first question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Often the simplest way to prove that the sum of independent random variables has a particular distribution is to use &lt;a href=&quot;http://en.wikipedia.org/wiki/Moment-generating_function&quot; rel=&quot;nofollow&quot;&gt;moment-generating functions&lt;/a&gt;.  This is because 1) if $X$ and $Y$ are independent with mgf's $M_X(t)$ and $M_Y(t)$, then $M_{X+Y}(t) = M_X(t) M_Y(t)$ and 2) moment-generating functions (when they exist) characterize distributions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Applying this to your problem, a geometric $(p)$ random variable has mgf $$\frac{pe^t}{1 - (1-p)e^t}.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus $$M_{X+Y}(t) = \left(\frac{pe^t}{1 - (1-p)e^t}\right)^{2}.$$&#xD;&#xA;Since this is the mgf of a negative binomial random variable, $X+Y$ must have a negative binomial distribution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(There are different conventions for defining negative binomial and geometric random variables, so depending on the convention used in a particular reference the mgf's there may be slightly different from the ones I give here.)&lt;/p&gt;&#xA;" OwnerUserId="2370" LastActivityDate="2010-11-15T05:51:13.513" />
  <row Id="10372" PostTypeId="2" ParentId="9933" CreationDate="2010-11-15T06:41:49.017" Score="2" ViewCount="" Body="&lt;p&gt;&#xA;&#xA;Informal justification of positive x negative = negative:&lt;br&gt;&#xA;&lt;br&gt;&#xA;Continue the pattern:&lt;br&gt;&#xA;2x3=6&lt;br&gt;&#xA;2x2=4&lt;br&gt;&#xA;2x1=2&lt;br&gt;&#xA;2x0=0&lt;br&gt;&#xA;2x-1= ? (Ans=-2)&lt;br&gt;&#xA;2x-2= ? (Ans=-4)&lt;br&gt;&#xA;2x-3= ? (Ans=-6)&lt;br&gt;&#xA;&lt;br&gt;&#xA;The number on the right-hand side keeps decreasing by 2.&lt;br&gt;&#xA;&lt;br&gt;&#xA;Informal justification of negative x negative = positive:&lt;br&gt;&#xA;&lt;br&gt;&#xA;Continue the pattern:&lt;br&gt;&#xA;2x-3=-6&lt;br&gt;&#xA;1x-3=-3&lt;br&gt;&#xA;0x-3=0&lt;br&gt;&#xA;-1x-3= ? (Ans=3)&lt;br&gt;&#xA;-2x-3= ? (Ans=6)&lt;br&gt;&#xA;-3x-3= ? (Ans=9)&lt;br&gt;&#xA;&lt;br&gt;&#xA;The number on the right-hand side keeps increasing by 3.&lt;br&gt;&#xA;&#xA;&lt;/p&gt;&#xA;" OwnerUserId="3515" LastActivityDate="2010-11-15T06:41:49.017" />
  <row Id="10373" PostTypeId="1" CreationDate="2010-11-15T07:54:51.813" Score="2" ViewCount="301" Body="&lt;p&gt;How to describe all ring homomorphism from $Z \times Z \to Z$ ?&lt;/p&gt;&#xA;" LastActivityDate="2011-05-03T09:53:24.517" Title="All homomorphisms from the ring $Z \times Z \to Z$" Tags="&lt;ring-theory&gt;" AnswerCount="3" CommentCount="4" />
  <row Id="10374" PostTypeId="2" ParentId="10282" CreationDate="2010-11-15T08:19:43.697" Score="11" ViewCount="" Body="&lt;p&gt;At least to me, it is helpful to think in terms of bases.&#xA;(I'll only be talking about tensor products of finite-dimensional vector&#xA;spaces here.)&#xA;This makes the universal mapping property that Zach Conn talks about&#xA;a bit less abstract (in fact, almost trivial).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First recall that if $L:V \to W$ is a linear map, then $L$ is completely determined&#xA;by what it does to a basis $\{ e_i \}$ for $V$:&#xA;$$L(x)=L(\sum x_i e_i) = \sum x_i L(e_i).$$&#xD;&#xA;(The coefficients of $L(e_i)$ in a basis for $W$ gives the $i$th column&#xA;in the matrix for $L$ with respect to the given bases.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Tensors come into the picture when one studies &lt;em&gt;multilinear&lt;/em&gt; maps.&#xA;If $B:V \times U \to W$ is a bilinear map, then $B$&#xD;&#xA;is completely determined by the values $B(e_i,f_j)$ where&#xA;$\{ e_i \}$ is a basis for $V$&#xD;&#xA;and $\{ f_j \}$ is a basis for $U$:&#xA;$$B(x,y) = B(\sum x_i e_i,\sum y_j f_j) = \sum\sum x_i y_j B(e_i,f_j).$$&#xD;&#xA;For simplicity, consider the particular case when $W=\mathbf{R}$;&#xA;then the values $B(e_i,f_j)$&#xD;&#xA;make up a set of $N=nm$ real numbers (where $n$ and $m$ are the&#xA;dimensions of $V$ and $U$), and these numbers are all that we need to keep&#xA;track of in order to know everything about the bilinear map $B:V \times U \to \mathbf{R}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice that in order to compute $B(x,y)$ we don't really need to know the&#xA;individual vectors $x$ and $y$, but rather the $N=mn$ numbers $\{ x_i y_j \}$.&#xA;Another pair of vectors $v$ and $u$ with $v_i u_j = x_i y_j$ for all $i$ and $j$&#xD;&#xA;will satisfy $B(v,u)=B(x,y)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This leads to the idea of splitting the computation of $B(x,y)$ into two stages.&#xA;Take an $N$-dimensional vector space $T$ (they're all isomorphic so it doesn't matter&#xA;which one we take) with a basis $(g_1,\dots,g_N)$.&#xA;Given $x=\sum x_i e_i$ and $y=\sum y_j f_j$,&#xA;first form the vector in $T$&#xD;&#xA;whose coordinates with respect to the basis $\{ g_k \}$ are given by the column vector&#xA;$$(x_1 y_1,\dots,x_1 y_m,x_2 y_m,\dots,x_2 y_m,\dots,x_n y_1,\dots,x_n y_m)^T.$$&#xD;&#xA;Then run this vector through the &lt;em&gt;linear&lt;/em&gt; map $\tilde{B}:T\to\mathbf{R}$ whose matrix&#xA;is the row vector&#xA;$$(B_{11},\dots,B_{1m},B_{21},\dots,B_{2m},\dots,B_{n1},\dots,B_{nm}).$$&#xD;&#xA;This gives, by construction, $\sum\sum B_{ij} x_i y_j=B(x,y)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We'll call the space $T$ the tensor product of the vector spaces $V$ and $U$&#xD;&#xA;and denote it by $T=V \otimes U$;&#xA;it is &quot;uniquely defined up to isomorphism&quot;.&#xA;The vector in $T$ that we formed from $x\in V$ and $y\in U$ in the first stage above&#xA;will be denoted $x \otimes y$;&#xA;it's a &quot;bilinear mixture&quot; of $x$ and $y$ which doesn't allow us to&#xA;reconstruct $x$ and $y$ individually,&#xA;but still contains exactly all the information needed&#xA;in order to compute $B(x,y)$ for any bilinear map $B$;&#xA;we have $B(x,y)=\tilde{B}(x \otimes y)$.&#xA;This is the &quot;universal property&quot;; any bilinear map $B$ from $V \times U$&#xD;&#xA;can be computed by taking a &quot;detour&quot; through $T$, and this detour&#xA;is unique, since the map $\tilde{B}$ is constructed uniquely from&#xA;the values $B(e_i,f_j)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To tidy this up, one would like to make sure that the definition is&#xA;basis-independent. One way is to check that everything transforms&#xA;properly under changes of bases. Another way is to do the construction&#xA;by forming a much bigger space and taking a quotient with respect to&#xA;suitable relations (without ever mentioning bases).&#xA;Then, by untangling definitions, one can for&#xA;example show that a bilinear map $B:V \times U \to \mathbf{R}$ can be&#xA;canonically identified with an element of the space $V^* \otimes U^*$,&#xA;and dually an element of $V \otimes U$ can be identified with a&#xA;bilinear map $V^* \otimes U^* \to \mathbf{R}$.&#xA;Yet other authors find this a convenient &lt;em&gt;starting&lt;/em&gt; point, so that they&#xA;instead &lt;em&gt;define&lt;/em&gt; $V \otimes U$ to be the space of bilinear maps $V^*&#xD;&#xA;\otimes U^* \to \mathbf{R}$.&#xA;So it's no wonder that one can become a little confused when trying&#xA;to compare different definitions...&lt;/p&gt;&#xA;" OwnerUserId="1242" LastEditorUserId="1242" LastEditorDisplayName="" LastEditDate="2010-11-15T09:22:31.033" LastActivityDate="2010-11-15T09:22:31.033" />
  <row Id="10375" PostTypeId="1" CreationDate="2010-11-15T09:09:06.967" Score="4" ViewCount="149" Body="&lt;p&gt;Where is the well-pointedeness assumption of the Elementary theory of the category of sets (Lawvere's category-theoretic axiomatization of set theory) used in everyday math?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Specifically, if you have a topos with natural numbers object (assume choice if you want to), what familiar theorems don't hold? I've heard that showing the Dedekind reals are the same as the Cauchy reals is one. Where in the arguments is well-pointedness used? It seems hard to find examples of this.&lt;/p&gt;&#xA;" OwnerUserId="3518" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-14T03:20:35.457" LastActivityDate="2011-02-14T03:20:35.457" Title="Where is the well-pointedness assumption of ETCS used in everyday math?" Tags="&lt;logic&gt;&lt;category-theory&gt;&lt;topos-theory&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="10377" PostTypeId="1" AcceptedAnswerId="10381" CreationDate="2010-11-15T10:14:42.020" Score="2" ViewCount="209" Body="&lt;p&gt;For a sequence defined by a formula normally the usual limit rules allows&#xA;one to find its limit. But for a sequence defined by a recurrence, up to now&#xA;I have only seen some refined ad hoc methods, mostly in Problems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Trick explains that &quot;&lt;a href=&quot;http://www.tricki.org/article/I_have_a_problem_about_the_convergence_of_a_sequence&quot; rel=&quot;nofollow&quot;&gt;There is one trick that is (...) first to prove&#xA;that a limit exists, and then to use the recurrence to determine what the&#xA;limit must be&lt;/a&gt;&quot; illustrated by the example of the recurrence $&#xD;&#xA;a_{n+1}=a_{n}/2+1/a_{n}$, $a_{0}=2$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question&lt;/strong&gt;: Are there relatively general methods to find the limit of a&#xA;sequence defined by a recurrence?  &lt;/p&gt;&#xA;" OwnerUserId="752" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2011-06-03T20:05:05.027" LastActivityDate="2011-06-03T20:05:05.027" Title="Methods to find the limit of a sequence defined by a recurrence" Tags="&lt;sequences-and-series&gt;&lt;limit&gt;&lt;recurrence-relations&gt;" AnswerCount="1" CommentCount="6" />
  <row Id="10378" PostTypeId="1" CreationDate="2010-11-15T10:35:12.080" Score="0" ViewCount="50" Body="&lt;p&gt;I'm in a little bit of a pickle here, can someone tell me if it is actually possible to solve these? From the general rule of thumb, 4 variables, 4 equations should be able to right?&lt;/p&gt;&#xA;" LastActivityDate="2010-11-15T11:10:37.977" Title="Is it possible to solve: xa = 4200, xb = 4410, ya=4000, yb = 4200 for x, y, a and b?" Tags="&lt;linear-algebra&gt;&lt;homework&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="10379" PostTypeId="1" CreationDate="2010-11-15T10:37:38.807" Score="1" ViewCount="152" Body="&lt;p&gt;It is known that, in the universe of complex numbers, the only root of the equation x² - 2x + 1 = 0 is 1.  Could we say that the equation has two equal real roots? Or should we say that the equation has one real root with multiplicity 2?&lt;/p&gt;&#xA;" OwnerUserId="2938" LastEditorUserId="2938" LastEditorDisplayName="" LastEditDate="2010-11-15T10:43:34.090" LastActivityDate="2010-11-15T22:49:37.140" Title="A root? Or two roots?" Tags="&lt;algebra&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="10381" PostTypeId="2" ParentId="10377" CreationDate="2010-11-15T10:47:06.160" Score="6" ViewCount="" Body="&lt;p&gt;Not really. Even once you've shown that a limit exists, a limit of, say, $x_{n+1} = f(x_n)$ is the same as a fixed point of $f$, or a solution to $f(x) - x = 0$. Needless to say it does not make much sense to ask whether there are general methods to find solutions to equations. For example, one can write down solutions to differential equations as fixed points, and there are no general methods to solve differential equations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In that sense all methods to find limits of recurrences are &quot;ad hoc&quot; in the same way that all methods to solve differential equations or Diophantine equations are &quot;ad hoc.&quot;  We do what we can.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-11-15T10:47:06.160" />
  <row Id="10382" PostTypeId="2" ParentId="10378" CreationDate="2010-11-15T11:10:37.977" Score="1" ViewCount="" Body="&lt;p&gt;The best you can do is $20x = 21y$ and $21a = 20b.$&#xD;&#xA;If you write your second pair of equations as&#xA;$$a \left( \frac{20}{21} x \right) = 4000,$$&#xD;&#xA;$$b \left( \frac{20}{21} x \right) = 4200$$&#xD;&#xA;we note that they are equivalent to your first pair&#xA;$$ax = 4200$$&#xD;&#xA;$$bx = 4410,$$&#xD;&#xA;simply by multipling up by the $20/21.$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for your general rule of thumb, note that a system of equations&#xA;can be &lt;a href=&quot;http://www.mathwords.com/i/inconsistent_system_of_equations.htm&quot; rel=&quot;nofollow&quot;&gt;inconsistent.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1301" LastActivityDate="2010-11-15T11:10:37.977" CommentCount="3" />
  <row Id="10383" PostTypeId="1" AcceptedAnswerId="10388" CreationDate="2010-11-15T11:13:54.137" Score="0" ViewCount="170" Body="&lt;p&gt;Hey guys,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm stuck with this problem I'm trying to solve from about an hour. Here's the question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the remainder when (3^202)+137 is divided by 101?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are 4 options -&gt; 36, 45, 56, 11&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to know the answer of the question with proper and possibly easiest method to solve the problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance, waiting for reply. :)&lt;/p&gt;&#xA;" OwnerUserId="712" LastActivityDate="2010-11-18T00:31:54.393" Title="What will be the remainder?" Tags="&lt;arithmetic&gt;" AnswerCount="3" />
  <row Id="10384" PostTypeId="2" ParentId="10383" CreationDate="2010-11-15T11:20:44.413" Score="9" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT:&lt;/strong&gt; Fermat's Little Theorem, which says that $a^{p-1} \equiv 1 \ (\text{mod} \ p)$ , where $p$ is a prime and $101$ is a prime. Remainder should be $45$&lt;/p&gt;&#xA;" LastActivityDate="2010-11-15T11:31:04.090" CommentCount="1" />
  <row Id="10385" PostTypeId="2" ParentId="10373" CreationDate="2010-11-15T11:52:58.200" Score="7" ViewCount="" Body="&lt;p&gt;To sum up all the comments\answers so far:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As Chandru wrote, if $f:\mathbb{Z}\times \mathbb{Z} \rightarrow \mathbb{Z}$ is a ring homomorphism, then it is also an abelian group homomorphism (for the addition operation), so $f([m,n]) = m \cdot f([1,0]) + n \cdot (f([0,1])$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since you also want the function to preserve multiplication, then you have other conditions.&#xA;For example, as Robin commented f preserves idempotent -$f([0,1])=f([0,1][0,1])=f([0,1])f([0,1])$ so $x=f([0,1])$ is an element of $\mathbb{Z}$ that satisfies $x^2=x$ - therefore it must be 1 or 0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;f also preserves zero divisors so $0=f([0,0])=f([0,1][1,0])=f([0,1])f([1,0])$ which means that at least one of them is zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;you are now left with three possibilities -&#xA;$f_1([m,n])=m,\; \; \;f_2([m,n])= n,\; \; \;f_3([m,n])=0$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="2125" LastActivityDate="2010-11-15T11:52:58.200" />
  <row Id="10386" PostTypeId="2" ParentId="10353" CreationDate="2010-11-15T11:53:58.760" Score="2" ViewCount="" Body="&lt;p&gt;This is only a first approximation to your answer, but I think it's a good start.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $x$ and $y$ be vectors in $\mathbb{R}^n$ and assume $x\cdot y = 0$.  Now, I'm going to add an additional condition that neither $x$ nor $y$ is $0$.  To give away the punchline, it will turn out there's a nice description of &lt;i&gt;these&lt;/i&gt; points.  The remaining points (where either $x=0$ or $y=0$ or both) will be degenerate in a way, because then the dot product doesn't tell you anything.  It turns out these remaining points destroy the &quot;niceness&quot; of the others (at least, how I usually define nice, i.e., getting a smooth manifold).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, let $X =\{(x,y)\in \mathbb{R}^{2n}|$ $x\neq 0$, $y\neq 0,$ and $x\cdot y = 0\}$.  The goal is to understand the shape of $X$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first thing to notice is that if $(x,y)\in X$, then so is $(\lambda x,y)$ and $(x,\mu y)$ for any $\lambda, \mu &amp;gt; 0$.  Further, if we set $Y =\{(x,y)\in X|\text{  } |x|=|y|=1 \}$, then it's clear that every point in $X$ is of the form $(\lambda x, \mu y)$ for a unique choice of $\lambda, \mu &amp;gt; 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It follows from this that $X = Y\times (0,\infty)\times (0,\infty)$.  That is, $X$ has the shape of $Y$ extended in 2 directions.  But what is the shape of $Y$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the moment, let's take an apparent detour.  How do you describe a sphere in $\mathbb{R}^n$?  It's $S^n = \{ x\in\mathbb{R}^{n+1}| \text{  }|x|=1\}$.  (Note that the $n$ in $S^n$ is one less than the $n$ in $\mathbb{R}^{n+1}$)  What is a tangent vector on $S^n$ at a point $x\in S^n$?  More generally, what is the tangent plane through a point $x\in S^n$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It shouldn't be too hard to convince yourself by looking at this on the circle and usual sphere that the tangent plane/line through a point $x$ in $S^n$ is the plane with normal vector $x$ going through the point $x$ (so $x$ does double duty as both a point and a vector).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Saying the normal vector is $x$ means that for any vector $y$ in the tangent plane, we must have $x\cdot y = 0$.  What if we take a unit vector $y$ in the tangent plane?  Well, we just find that $|x| = |y| = 1$ and $x\cdot y = 0$.  But these were exactly our conditions on belonging to $Y$!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means that $Y$ is the so called Unit Tangent Bundle of the sphere - it's the collection of all $(p,v)$ where $p$ is a point on the sphere (i.e., $|p| = 1$, $v$ is a tangent vector to there sphere at $p$ (i.e., $p\cdot v = 0$) where $v$ has unit length (i.e., $|v| = 1$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means that, in particular, you can think of these spaces as &quot;the collection of points on a sphere paired with a choice of unit length tangent vector&quot;.  It turns out this shape is fairly well understood.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, when $n=1$, one gets the unit tangent bundle of a circle.  It is know that the tangent bundle of a circle is trivial - that is, is diffeomorphic to $S^1\times \mathbb{R}$.  From here it follows that the unit tangent bundle is diffeomorphic to $S^1\times S^1$.  This shape is called a Torus and is best physically described as a hollow donut.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This, in particular, almost answers your first question about what we get in $\mathbb{R}^4$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $n =2$, one gets the unit tangent bundle of $S^2$.  It's known that this is diffeomorphic to $\mathbb{R}P^3$, which can almost be visualized.  For $n=3$, one gets the unit tangent bundle of $S^3$, which is known to be diffeomorphic to $S^3\times S^2$.  I don't know of a suitably nice description of any other unit tangent bundles except for $n=7$, one gets $S^7\times S^6$.&lt;/p&gt;&#xA;" OwnerUserId="331" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2010-11-15T22:35:45.837" LastActivityDate="2010-11-15T22:35:45.837" CommentCount="1" />
  <row Id="10387" PostTypeId="1" AcceptedAnswerId="10394" CreationDate="2010-11-15T12:23:34.437" Score="1" ViewCount="57" Body="&lt;p&gt;I'm stuck with the dual of an exercise in Atiyah-MacDonald.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's easily seen that tensoring over $R$ with $-\otimes M$ the sequence&#xA;$$&#xD;&#xA;0\to \mathfrak a\to R\to R/\mathfrak a\to 0&#xD;&#xA;$$&#xD;&#xA;one gets the isomorphism $R/\mathfrak a\otimes M\cong M/\mathfrak aM$.&#xA;What can I deduce from &quot;&lt;em&gt;Hom&lt;/em&gt;(-,M)ing&quot; the same sequence? I have, let's say&#xA;$$&#xD;&#xA;0\to \mathfrak a\xrightarrow{\alpha} R\xrightarrow{\beta} R/\mathfrak a\to 0&#xD;&#xA;$$&#xD;&#xA;and &#xA;$$&#xD;&#xA;0\to \hom(R/\mathfrak a,M) \xrightarrow{\beta^*} \hom(R,M)\cong M\xrightarrow{\alpha^*} \hom(\mathfrak a,M)&#xD;&#xA;$$&#xD;&#xA;but now what can I do? What is $\hom(R/\mathfrak a,M)$ like?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know, it's easy, but...&lt;/p&gt;&#xA;" OwnerUserId="685" LastActivityDate="2010-11-15T13:43:37.640" Title="How to &quot;Hom&quot; the canonical sequence in a ring" Tags="&lt;commutative-algebra&gt;" AnswerCount="2" CommentCount="0" />
  <row Id="10388" PostTypeId="2" ParentId="10383" CreationDate="2010-11-15T12:24:02.227" Score="1" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;I think Chandru1 is correct. The answer is 45.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How did I get it?&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;1) The remainder of 3^30 / 101 = 6&#xA;2) 6^6 = 46656.&#xA;3) 46656 * (The remainder of 3^22 / 101) = 2286144.&#xA;4) 2286144 + 137 = 2286281&#xA;5) The remainder of 2286281 / 101 = 45&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Note in step 2 and 3 that 30 * 6 = 180. And 180 + 22 = 202&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A simpler example: The reaminder when (2^10)+7 is divided by 6 (the answer is 5).&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;1) The reaminder of 2^4 / 6 = 4&#xA;2) 4^2 = 16&#xA;3) 16 * (The remainder of 2^2 / 6) = 64&#xA;4) 64 + 7 = 71&#xA;5) The remainder of 71 / 6 = 5&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="3314" LastActivityDate="2010-11-15T12:24:02.227" CommentCount="1" />
  <row Id="10389" PostTypeId="1" CreationDate="2010-11-15T12:32:03.033" Score="2" ViewCount="382" Body="&lt;p&gt;I have to solve the following recurrence :$\displaystyle T(n)=2T(n/2)+T(n/3)+\theta(n^2)$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have done the whole tree analyses and  now I have to prove that $\displaystyle T(n) \leq dn^{2}\log_{2}(n)$ but I can not find mathematical way to prove that, is there any trick with logarithms that I can use?&lt;/p&gt;&#xA;" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2011-06-03T20:04:07.023" LastActivityDate="2011-06-03T20:04:07.023" Title="Recurrence trouble: $T(n)=2T(n/2)+T(n/3)+\theta(n^2)$" Tags="&lt;asymptotics&gt;&lt;recurrence-relations&gt;" AnswerCount="3" CommentCount="8" FavoriteCount="1" />
  <row Id="10390" PostTypeId="2" ParentId="10282" CreationDate="2010-11-15T13:01:54.610" Score="6" ViewCount="" Body="&lt;p&gt;Mathematicians and physicists use very different languages when they talk about tensors. Fortunately, they are talking about the same thing, but unfortunately, this is not obvious at all. Let me explain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For simplicity, I'm going to focus on &lt;strong&gt;covariant 2-tensors&lt;/strong&gt;, since this case already contains the main intuition. Also, I'm not going to talk about the distinction between covariant and contravariant, but I'll get all the indices right for future study.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Physicist's definition&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Definition: A covariant 2-tensor is a set of numbers $t_{ij}$ with two indices that transforms in a particular way under a change of coordinates… Wait, wait, coordinates in &lt;em&gt;what space&lt;/em&gt;? Physicists usually don't mention it, but they mean coordinates in a given &lt;em&gt;vector space&lt;/em&gt; $V$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More precisely, let $\{\vec e_i\}$ be a basis of the vector space $V$. Then, every &lt;strong&gt;vector&lt;/strong&gt; $\vec v$ can be expressed in terms of its &lt;em&gt;coordinates&lt;/em&gt; $v^i$ as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\vec v = \sum_i v^i \vec e_i .$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, there are two objects: the vector $\vec v$ which I think of as &quot;solid&quot; or &quot;fundamental&quot;, and its coordinates $v^i$, which are &quot;ephemeral&quot;, since I have to choose a basis $\vec e_i$ before I can talk about them at all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore, in a &lt;strong&gt;different basis&lt;/strong&gt; $\{\vec e&amp;#39;_i\}$ of our vector space, the coordinates of one and the same vector $\vec v$ are very different numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \vec v = \sum_i v^i \vec e_i = \sum_i v&amp;#39;^i \vec e&amp;#39;_i .$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but $v^i ≠ v&amp;#39;^i$. So, the vector is the fundamental thing. Its coordinates are useful for calculations, but they are ephemeral and heavily depend on the choice of basis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, when defining a &lt;strong&gt;covariant 2-tensor&lt;/strong&gt;, physicists do something very mysterious: they define a fundamental object (= the 2-tensor) not by describing it directly, but only by specifying how its ephemeral coordinates look like and change when switching to a different basis. Namely, a change of basis&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \vec e&amp;#39;_i = \sum_j R_i^a \vec e_a $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;will change the coordinates $t_{ij}$ of the tensor via&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ t&amp;#39;_{ij} = \sum_{ab} R^a_i R^b_j t_{ab} .$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If that is not completely unintuitive, I don't know what is.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Mathematician's definition&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Mathematicians define tensors differently. Namely, the give a direct, fundamental description of what a 2-tensor is and only then ponder how it looks like in different coordinate systems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is the definition: a &lt;strong&gt;covariant 2-tensor $t$ is a bilinear map&lt;/strong&gt; $t : V\times V \to \mathbb{R}$. That's it. (Bilinear = linear in both arguments).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, a covariant 2-tensor $t$ is a thing that eats two vectors $\vec v$, $\vec w$ and returns a number $t(\vec v, \vec w) \in\mathbb{R}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, what does this thing look like in coordinates? Choosing a basis $\lbrace \vec e_i \rbrace$, bilinearity allows us to write&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ t(\vec v, \vec w) = t(\sum_i v^i \vec e_i, \sum_j w^j \vec e_j) = \sum_{ij} v^iw^j t(\vec e_i,\vec e_j) .$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, we simply call the numbers $t_{ij} = t(\vec e_i, \vec e_j)$ the &lt;strong&gt;coordinates of the tensor $t$&lt;/strong&gt; in the basis $\vec e_i$. You can calculate that these numbers will behave just like the physicists tell us when you change the basis to $\vec e_i&amp;#39;$. So, the physicist's tensor and the mathematician's tensor are one and the same thing.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Tensor product&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Actually, mathematicians do something more advanced, they define a so called &lt;strong&gt;tensor product&lt;/strong&gt; of vector spaces. The previous definition as a bilinear map is still correct, but mathematicians like to write this as &quot;$t\in V^*\otimes V^*$&quot; instead of &quot;$t: V\times V \to \mathbb{R}$ and $t$ bilinear&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, for a first understanding of the physicist's vs the mathematician's definition, it is not necessary to understand the mathematical tensor product.&lt;/p&gt;&#xA;" OwnerUserId="434" LastActivityDate="2010-11-15T13:01:54.610" CommentCount="1" />
  <row Id="10391" PostTypeId="2" ParentId="10342" CreationDate="2010-11-15T13:03:10.670" Score="2" ViewCount="" Body="&lt;p&gt;I don't know what a good categorical quotient is, but if $X$ is a CW complex, and the $G$ action is free and cellular, then $H^*_G(X)\cong H^*(X/G)$. See Brown's &quot;Cohomology of Groups,&quot; p.173.&lt;/p&gt;&#xA;" OwnerUserId="2941" LastActivityDate="2010-11-15T13:03:10.670" CommentCount="0" />
  <row Id="10392" PostTypeId="1" AcceptedAnswerId="10420" CreationDate="2010-11-15T13:25:41.353" Score="5" ViewCount="337" Body="&lt;p&gt;Let's say I have a truth table like this:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;X  Y   A&#xA;t  t  f&#xA;t  f  t&#xA;f  t  t&#xA;f  f  f&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Now, I have to find the formula for A. This case is rather easy because I can immediately see that this looks like the inverted values of X~B, thus ¬(X~B). Now, for a more complicated problem I do not know the method.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;X  Y  Z  A&#xA;t  t  t  f&#xA;t  t  f  f&#xA;t  f  t  t&#xA;t  f  f  f&#xA;f  t  t  f&#xA;f  t  f  f&#xA;f  f  t  t&#xA;f  f  f  f&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Is there an approach i'm missing here?&lt;/p&gt;&#xA;" OwnerUserId="3527" LastEditorUserId="3527" LastEditorDisplayName="" LastEditDate="2010-11-15T13:31:40.023" LastActivityDate="2010-11-23T18:21:44.007" Title="How to find the logical formula for a given truth table?" Tags="&lt;logic&gt;" AnswerCount="5" />
  <row Id="10393" PostTypeId="2" ParentId="10387" CreationDate="2010-11-15T13:27:28.597" Score="0" ViewCount="" Body="&lt;p&gt;Asking «what is $M/\operatorname{im}\alpha^*$?» is &lt;em&gt;not&lt;/em&gt; the dual of the A-M exercise. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you really want a dual statement, you should be looking for a description of $\hom(R/\mathfrak a,M)$.&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2010-11-15T13:27:28.597" CommentCount="1" />
  <row Id="10394" PostTypeId="2" ParentId="10387" CreationDate="2010-11-15T13:32:21.570" Score="3" ViewCount="" Body="&lt;p&gt;Hint: Any $R$-module homomorphism $f$ from $R/\mathfrak{a}$ to $M$ is determined once we know $f(1)$. But every such homomorphism is killed by $\mathfrak{a}$ since $\mathfrak{a}f(1)=f(\mathfrak{a})=f(0)=0$. So one can show that, $Hom_R(R/\mathfrak{a},M)\cong Ann_{\mathfrak{a}}(M)$ &lt;/p&gt;&#xA;" OwnerUserId="3431" LastEditorUserId="3431" LastEditorDisplayName="" LastEditDate="2010-11-15T13:43:37.640" LastActivityDate="2010-11-15T13:43:37.640" CommentCount="3" />
  <row Id="10395" PostTypeId="2" ParentId="10392" CreationDate="2010-11-15T13:32:58.527" Score="1" ViewCount="" Body="&lt;p&gt;Solve first the problem of finding the formula for truth tables which have exactly one T on the rightmost table. Then go to the general case.&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2010-11-15T13:32:58.527" />
  <row Id="10396" PostTypeId="2" ParentId="10392" CreationDate="2010-11-15T13:35:21.223" Score="10" ViewCount="" Body="&lt;p&gt;I believe &lt;a href=&quot;http://en.wikipedia.org/wiki/Karnaugh_map&quot; rel=&quot;nofollow&quot;&gt;Karnaugh Maps&lt;/a&gt; are useful for this purpose.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-11-15T13:35:21.223" CommentCount="1" />
  <row Id="10397" PostTypeId="2" ParentId="10389" CreationDate="2010-11-15T13:37:11.343" Score="3" ViewCount="" Body="&lt;p&gt;Try applying the &lt;a href=&quot;http://en.wikipedia.org/wiki/Akra-Bazzi_theorem&quot; rel=&quot;nofollow&quot;&gt;Akra-Bazzi&lt;/a&gt; theorem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By Akra-Bazzi, if $p$ satisfies&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle 2(1/2)^p + (1/3)^p = 1$, (notice that $\displaystyle 1 \lt p \lt 2$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle T(x) = \theta(x^p(1 + \int_{1}^{x} x^{1-p})) = \theta(x^p(1+x^{2-p}))$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;since $\displaystyle 1 \lt p \lt 2$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;we must have that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle T(x)  = \theta(x^2) = O(x^2 \log_{2} x)$.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-11-15T14:26:07.273" LastActivityDate="2010-11-15T14:26:07.273" CommentCount="8" />
  <row Id="10398" PostTypeId="2" ParentId="10353" CreationDate="2010-11-15T13:47:27.977" Score="0" ViewCount="" Body="&lt;p&gt;Cut your hypersurface $\{(x_1,x_2,x_3,x_4)\in\mathbb R^4:x_1x_2+x_3x_4=0\}$ with the hyperplanes $H_t=\{(x_1,x_2,x_3,x_4)\in\mathbb R^4:x_1=t\}$, for $t\in\mathbb R$, and visualize those intersections. &lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2010-11-15T13:47:27.977" />
  <row Id="10399" PostTypeId="1" AcceptedAnswerId="10407" CreationDate="2010-11-15T14:14:54.323" Score="2" ViewCount="51" Body="&lt;p&gt;I believe I've found all Amicable Pairs (n, m) such that n is in the closed interval [1, 34000000]. Here's my &lt;a href=&quot;http://tutoree7.pastebin.com/fg3mVNmF&quot; rel=&quot;nofollow&quot;&gt;list&lt;/a&gt;.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;By inspection I believe that for all n in the list, n is odd iff its last digit is 5. Can it be proved that this is true for all Amicable Pairs? &lt;/p&gt;&#xA;" OwnerUserId="1964" LastActivityDate="2010-11-15T14:54:10.467" Title="True?: Let (n, m) be an arbitrary Amicable Pair. Then n is odd iff its last digit is 5" Tags="&lt;number-theory&gt;" AnswerCount="2" />
  <row Id="10400" PostTypeId="1" AcceptedAnswerId="10406" CreationDate="2010-11-15T14:21:37.407" Score="0" ViewCount="221" Body="&lt;p&gt;Let $R$ be a commutative ring and $I_1, \cdots, I_n$ pairwise comaximal ideals in $R$, e.g. $I_i + I_j = R$ for $i \neq j$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why are the ideals $I_1^{n_1}, ... , I_r^{n_r}$ (for any $n_1,...,n_r \in\mathbb N$) also comaximal?&lt;/p&gt;&#xA;" LastActivityDate="2011-04-11T22:41:19.193" Title="Comaximal ideals in a commutative ring" Tags="&lt;algebra&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="10401" PostTypeId="2" ParentId="9712" CreationDate="2010-11-15T14:29:30.687" Score="1" ViewCount="" Body="&lt;p&gt;I believe the answer given in my comments is correct. &lt;/p&gt;&#xA;" OwnerUserId="3249" LastActivityDate="2010-11-15T14:29:30.687" />
  <row Id="10403" PostTypeId="2" ParentId="10033" CreationDate="2010-11-15T14:35:42.240" Score="1" ViewCount="" Body="&lt;p&gt;I have checked my derivation thoroughly, and I believe now that it is a typo.&lt;/p&gt;&#xA;" OwnerUserId="3249" LastActivityDate="2010-11-15T14:35:42.240" />
  <row Id="10404" PostTypeId="2" ParentId="10383" CreationDate="2010-11-15T14:37:57.587" Score="2" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\  101\ $ is prime so a little Fermat $\rm\ \Rightarrow\ \ 3^{101}\ \equiv\ 3\ \ \Rightarrow\ \ 3^{202}\ \equiv\ \ldots\ (mod\ 101)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since your comment reveals you are not familiar with modular arithmetic, here is an alternative.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;By Fermat's little theorem $101$ divides $\: 3^{101}-3\: $ so it also divides $\rm\ (3^{101}-3)\ (3^{101}+3)\ =\ 3^{202}-9$&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-11-18T00:31:54.393" LastActivityDate="2010-11-18T00:31:54.393" CommentCount="2" />
  <row Id="10405" PostTypeId="2" ParentId="10399" CreationDate="2010-11-15T14:38:59.363" Score="1" ViewCount="" Body="&lt;p&gt;How about 87633 and 69615?  Found 87633 in &lt;a href=&quot;http://oeis.org/A063990&quot; rel=&quot;nofollow&quot;&gt;OEIS A063990&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-11-15T14:38:59.363" CommentCount="1" />
  <row Id="10406" PostTypeId="2" ParentId="10400" CreationDate="2010-11-15T14:40:00.533" Score="3" ViewCount="" Body="&lt;p&gt;It is sufficient to prove this for the case two comaximal ideals, say $I,J$. Need to show, $I^m+J^n=R$ for any positive integers $m,n$. Now, $R=R^{m+n-1}=(I+J)^{m+n-1}\subseteq I^m+J^n$ since in the binomial expansion of $(I+J)^{m+n-1}$, in every term, either the power of $I$ is at least $m$ or the power of $J$ is at least $n$ by pigeonhole principle.&lt;/p&gt;&#xA;" OwnerUserId="3431" LastEditorUserId="3431" LastEditorDisplayName="" LastEditDate="2010-11-15T14:57:05.397" LastActivityDate="2010-11-15T14:57:05.397" />
  <row Id="10407" PostTypeId="2" ParentId="10399" CreationDate="2010-11-15T14:54:10.467" Score="2" ViewCount="" Body="&lt;p&gt;I think the first counterexample (n,m) with n &amp;lt; m where n is odd and not congruent to 5 mod 10 is (34765731, 36939357), which by misfortune is just outside your search region.&lt;/p&gt;&#xA;" OwnerUserId="3531" LastActivityDate="2010-11-15T14:54:10.467" CommentCount="3" />
  <row Id="10408" PostTypeId="1" AcceptedAnswerId="10412" CreationDate="2010-11-15T14:57:34.677" Score="2" ViewCount="185" Body="&lt;p&gt;I want to describe the polynomials with integer coefficients and the property that f'(x)|f(x) (the derivate divides the polynomial).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I know that f(x) divides g(x) if all of f(x)'s roots are roots of g(x). I also know that f'(x) has degree one less than f(x). Finally, f'(x) detects multiple roots. If a is a root of f(x) and f'(x), it is at least a double root. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using these facts I've conjectured that f(x), if it has degree n, has 1 root with multiplicity n or 2 roots, one with multiplicity n-1. But I can't prove it and I fear that I am making a mistake thinking just about linear terms.&lt;/p&gt;&#xA;" OwnerUserId="2298" LastActivityDate="2010-11-15T16:02:39.687" Title="Polynomials, derivatives and repeated roots" Tags="&lt;polynomials&gt;" AnswerCount="3" CommentCount="6" />
  <row Id="10409" PostTypeId="2" ParentId="10373" CreationDate="2010-11-15T15:00:49.867" Score="1" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\ $ hom's preserve orthogonal idempotents, so $\ \{\:(0,1), (1,0)\:\}\to \{0,1\}\ $ if nontrivial&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-11-15T15:00:49.867" />
  <row Id="10410" PostTypeId="2" ParentId="10167" CreationDate="2010-11-15T15:02:57.573" Score="2" ViewCount="" Body="&lt;p&gt;Here is the way I hinted at the solution: proving first that a finitely generated torsion-free abelian group is free, then use the fact that free groups are projective.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The argument for the first part is that used by Lang in his &lt;em&gt;Algebra&lt;/em&gt;: suppose $A$ is a finitely generated torsion-free abelian group; we may assume it is nonzero. Pick $x_1,\ldots,x_n\in A$ that are maximal with respect to the property that if $\alpha_1x_1+\cdots +\alpha_nx_n=0$, then $\alpha_i=0$ for all $i$. Let $B$ be the subgroup generated by the $x_i$; note that $B$ is free abelian.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $y_1,\ldots,y_k$ be a generating set for $A$. For each $i$, &#xA;there exists an integer $d_i\neq 0$ and  integers $\beta_1,\ldots,\beta_n$, not all zero, such that $d_iy +\beta_1x_1 + \cdots +\beta_n x_n=0$; that is, $d_iy_i\in B$. Letting $d=\mathrm{lcm}(d_1,\ldots,d_k)$, we have that $dA$ is contained in $B$. Since $A$ is torsion free, the map $A\to B$ given by $x\mapsto dx$ is an injection, so $A$ is isomorphic to a subgroup of a free abelian group, hence $A$ is free abelian.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, if $A$ is a finitely generated group, and $tA$ is its torsion group, we know that $A/tA$ is finitely generated and torsion-free, hence free. Let $x_1,\ldots,x_k$ be a basis for $A/tA$. For each $i$, let $a_i\in A$ be any element that maps to $x_i$ under the canonical projection $\pi\colon A\to A/tA$. Since $A/tA$ is free, we have a (unique) group homomorphism $\rho\colon A/tA\to A$ that maps $x_i$ to $a_i$. Since $\pi\circ\rho = \mathrm{id}_{A/tA}$, then $\rho$ must be one-to-one, hence $A/tA$ is isomorphic to its image under $\rho$ inside of $A$. Thus, $A$ has a subgroup that is isomorphic to $A/tA$. &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-15T17:36:14.650" LastActivityDate="2010-11-15T17:36:14.650" />
  <row Id="10411" PostTypeId="2" ParentId="10408" CreationDate="2010-11-15T15:04:39.160" Score="1" ViewCount="" Body="&lt;p&gt;This is assuming you are talking about polynomials over $\mathbb{C}$ or $\mathbb{R}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume $f$ has leading coefficient $1$ and is degree $n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$nf(x) = f&amp;#39;(x)(x-a)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f&amp;#39;(x)/nf(x) = 1/(x-a)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Integrate&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\log(f(x)) = n\log(x-a) + C$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus $f(x) = K(x-a)^n$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-11-15T15:39:09.180" LastActivityDate="2010-11-15T15:39:09.180" CommentCount="4" />
  <row Id="10412" PostTypeId="2" ParentId="10408" CreationDate="2010-11-15T15:11:51.243" Score="7" ViewCount="" Body="&lt;p&gt;I'll assume you want the divisibility to be in $\mathbb{Q}[x]$.  Write $\displaystyle f(x) = \prod_{i=1}^{k} (x - a_i)^{m_i}$ where the $a_i$ are distinct (in $\mathbb{C}$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Lemma:&lt;/strong&gt;  $\displaystyle \frac{f&amp;#39;(x)}{f(x)} = \sum_{i=1}^{k} \frac{m_i}{x - a_i}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a direct consequence of the product rule and you do not need to use any calculus to prove it.  (That is, you don't need to take logarithms.  The derivative can be defined as a totally formal operation on polynomials and this identity holds in $\mathbb{Q}(x)$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $f&amp;#39;(x)$ divides $f(x)$, then $\frac{f&amp;#39;(x)}{f(x)} = \frac{n}{x - a}$ for some $a$.  Now, the functions $\frac{1}{x - a}$ are linearly independent (you also do not need to use any calculus to prove this) in $\mathbb{C}(x)$.  It follows that this is possible if and only if $f$ has $a$ as a repeated root of multiplicity $n$, so $a$ is the only root.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It follows that $f$ has the form $c(px - q)^n$ for some integers $c, p, q$ with $a, p \neq 0$.  (I'm ignoring the degenerate cases.)&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-11-15T15:18:10.690" LastActivityDate="2010-11-15T15:18:10.690" CommentCount="3" />
  <row Id="10413" PostTypeId="1" AcceptedAnswerId="10434" CreationDate="2010-11-15T15:13:17.183" Score="1" ViewCount="78" Body="&lt;p&gt;Given a $1$ dimensional domain $R$ with fraction field $K$. Suppose $\mathfrak{m}$ is a maximal ideal of $R$. Then, I need to show that $R_m$ is a maximal subring of $K$ in the sense that if $x\in K-R_\mathfrak{m}$, then, $R_m[x]=K$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1)Is this true? How would you go about showing this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2)If this is true, can any of the hypothesis be relaxed (domain, $1$ dimensional,&quot;at a maximal ideal&quot;)?&lt;/p&gt;&#xA;" OwnerUserId="3431" LastEditorUserId="3431" LastEditorDisplayName="" LastEditDate="2010-11-15T15:20:33.493" LastActivityDate="2010-11-15T19:31:05.610" Title="Localization of a 1 dimensional domain at a maximal ideal is a maximal subring of the fraction field" Tags="&lt;commutative-algebra&gt;" AnswerCount="1" CommentCount="9" />
  <row Id="10414" PostTypeId="1" CreationDate="2010-11-15T15:26:21.017" Score="18" ViewCount="447" Body="&lt;p&gt;Imagine we have a wizard that knows a few spells. Each spell has 3 attributes:&#xA;Damage, cooldown time, and a cast time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cooldown time: the amount of time (t) it takes before being able to cast that spell again. A spell goes on &quot;cooldown&quot; the moment it begins casting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cast time: the amount of time (t) it takes to use a spell. While the wizard is casting something another spell cannot be cast and it cannot be canceled.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is:&#xA;How would you maximize damage given different sets of spells?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is easy to calculate the highest damage per cast time. But what about in situations where it is better to wait then to get &quot;stuck&quot; casting a low damage spell when a much higher one is available: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, &lt;/p&gt;&#xA;&#xA;&lt;h1&gt;1) 100 damage, 1 second cast time, 10 second cool down.&lt;/h1&gt;&#xA;&#xA;&lt;h1&gt;2) 10 damage, 4 second cast time, 0 second cool down.&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;So, in this case you would cast #1, #2, wait. Cast #1&lt;/p&gt;&#xA;" OwnerUserId="3534" LastEditorUserId="3534" LastEditorDisplayName="" LastEditDate="2010-11-15T16:52:12.767" LastActivityDate="2010-11-19T08:14:00.693" Title="Math Wizardry - Formula for selecting the best spell." Tags="&lt;optimization&gt;" AnswerCount="5" CommentCount="9" FavoriteCount="3" />
  <row Id="10415" PostTypeId="2" ParentId="10379" CreationDate="2010-11-15T15:52:20.070" Score="2" ViewCount="" Body="&lt;p&gt;&quot;Two equal roots&quot; and &quot;a root of multiplicity (at least) two&quot; mean the same thing. &lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-11-15T15:52:20.070" />
  <row Id="10416" PostTypeId="2" ParentId="10400" CreationDate="2010-11-15T15:55:43.607" Score="3" ViewCount="" Body="&lt;p&gt;If you are familar  with ideal radicals then, as I mentioned on &lt;a href=&quot;http://groups.google.com/group/sci.math/msg/b24d9f96f53a0b92&quot; rel=&quot;nofollow&quot;&gt;sci.math&lt;/a&gt;, the proof is a one-liner:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\rm rad\ (I^m + \cdots\ + J^n) \ \supset\ I +\cdots+ J = 1\ \ \Rightarrow\ \ I^m +\cdots + J^n\ =\ 1 $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively, and much more generally, it maybe viewed as an immediate consequence of the &lt;strong&gt;Freshman's Dream&lt;/strong&gt; binomial theorem $\rm\ (A + B)^n = A^n + B^n\ $. This theorem is true for both arithmetic of GCDs and (invertible) ideals simply because, &#xA;in both cases, multiplication is cancellative and addition is idempotent, &#xA;i.e. $\rm\ A + A = A\ $ for ideals and $\rm\ (A,A) = A\ $ for GCDs. Combining this with the &#xA;associative, commutative, distributive laws of addition and multiplication &#xA;we obtain an extremely elementary high-school-level proof of the Freshman's Dream: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\quad (A + B)^4 \ =\  A^4  +  A^3 B  +  A^2 B^2 + AB^3 + B^4 $&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\quad\phantom{(A + B)^4 }\ =\  A^2\ (A^2 + AB + B^2) + (A^2 + AB + B^2)\ B^2 $&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\quad\phantom{(A + B)^4 }\ =\ (A^2 + B^2)\ \:(A + B)^2 $&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So $\rm\ {(A + B)^2 }\ =\ \  A^2 + B^2\ $ if $\rm\ A+B\ $ is cancellative, e.g. if $\rm A+B = 1$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same proof works generally since, as above&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\quad (A + B)^{2n}\ =\  A^n\ (A^n + \cdots+ B^n) + (A^n +\cdots+ B^n)\ B^n $&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\quad\phantom{(A + B)^{2n}}\ =\ (A^n + B^n)\ (A + B)^n  $&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the GCD case  $\rm\ A+B\ := (A,B) = \gcd(A,B)\ $ for $\rm\:A,B\:$ in a GCD-domain, &#xA;i.e. a domain where $\rm\: \gcd(A,B)\:$ exists for all $\rm\:A,B \ne 0,0\:$.  Hence &#xA;Dream is true since $\rm\:(A,B)\:$ is cancellable, being nonzero in a domain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a domain, nonzero principal ideals are cancellable, so Dream is true &#xA;for ideals in a PID (e.g. $\mathbb Z\:$), or f.g. (finitely generated) ideals in a Bezout domain. &#xA;More generally, Dream also holds true in any Dedekind domain (e.g. &#xA;any number ring) since nonzero ideals are invertible hence cancellable. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact this &quot;Freshman's Dream&quot; is true for all f.g. ideals in domain $\rm\:D\:$ &#xA;iff every nonzero f.g. ideal is invertible. Such domains are known as &#xA;Prufer domains. They're non-Noetherian generalizations of Dedekind &#xA;domains. Moreover they form an important class of domains because &#xA;they may also be equivalently characterized by a large number of &#xA;other important properties, e.g. they are precisely the domains &#xA;satisfying CRT (Chinese Remainder Theorem); $\ $ Gauss's Lemma: the &#xA;content ideal $\rm\ \ c(fg) = c(f)\ c(g)\:$;$\ $  nonzero f.g. ideals are cancellable; &#xA;$\ $ f.g. ideals satisfy contains $\Rightarrow$ divides; $\: $ etc. It's been estimated &#xA;that there are close to a hundred such characterizations known. &#xA;See my&lt;a href=&quot;http://groups.google.com/group/sci.math/msg/ba5ac69dddfc4a67&quot; rel=&quot;nofollow&quot;&gt; sci.math post&lt;/a&gt; for about thirty such characterizations.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-11-15T18:13:58.270" LastActivityDate="2010-11-15T18:13:58.270" />
  <row Id="10417" PostTypeId="2" ParentId="10408" CreationDate="2010-11-15T16:02:39.687" Score="2" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Rolle%27s_theorem&quot; rel=&quot;nofollow&quot; title=&quot;Rolle's theorem&quot;&gt;Rolle's theorem.&lt;/a&gt; Between two zeroes of function lies a zero of its derivative. If f has two distinct zeroes, there is a zero of f' which is not a zero of f. &lt;/p&gt;&#xA;" OwnerUserId="2633" LastActivityDate="2010-11-15T16:02:39.687" CommentCount="1" />
  <row Id="10418" PostTypeId="2" ParentId="10414" CreationDate="2010-11-15T16:47:53.417" Score="5" ViewCount="" Body="&lt;p&gt;I don't have an answer, but I just want to point out that the greedy algorithm fails.  That is, if we choose to cast, at any point, the available spell which maximizes $\frac{ \text{damage} }{ \text{cast time} }$, we don't actually maximize our damage output because of cooldown.  Consider the pair of spells $(300, 5, 12)$ and $(50, 3, 0)$ (where the three numbers are damage, cast time in second times, and cooldown time in seconds): the greedy algorithm suggests the casting schedule&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Cast spell 1 ($300$ damage, $5$ seconds),&lt;/li&gt;&#xA;&lt;li&gt;Cast spell 2 ($50$ damage, $3$ seconds),&lt;/li&gt;&#xA;&lt;li&gt;Cast spell 2 ($50$ damage, $3$ seconds),&lt;/li&gt;&#xA;&lt;li&gt;Cast spell 2 ($50$ damage, $3$ seconds),&lt;/li&gt;&#xA;&lt;li&gt;Repeat&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;which gives about $32.1$ damage per second.  However, the casting schedule&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Cast spell 1 ($300$ damage, $5$ seconds),&lt;/li&gt;&#xA;&lt;li&gt;Cast spell 2 ($50$ damage, $3$ seconds),&lt;/li&gt;&#xA;&lt;li&gt;Cast spell 2 ($50$ damage, $3$ seconds),&lt;/li&gt;&#xA;&lt;li&gt;Wait $1$ second,&lt;/li&gt;&#xA;&lt;li&gt;Repeat&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;gives about $33.3$ damage per second.  So a correct algorithm must take into account which spells are about to finish cooling down.  &lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-11-15T16:47:53.417" />
  <row Id="10419" PostTypeId="1" AcceptedAnswerId="10425" CreationDate="2010-11-15T16:55:24.327" Score="0" ViewCount="68" Body="&lt;p&gt;Hi&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have a function, implicitly defined, how would understand the following; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume $y=y(x)$, find $y&amp;#39;(0)$ and $y&amp;#39;&amp;#39;(0)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;After I implicitly derivate it the first time, I have an expression with $dy/dx$. Do I just fill in zero for $x$, remove the parts that gets canceled out? I'm not sure if the answer is a number or a function?&lt;/p&gt;&#xA;" OwnerUserId="2230" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-15T17:37:40.703" LastActivityDate="2010-11-15T19:42:19.387" Title="Easy; understanding question about implicit diff?" Tags="&lt;calculus&gt;&lt;homework&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="10420" PostTypeId="2" ParentId="10392" CreationDate="2010-11-15T17:08:35.637" Score="4" ViewCount="" Body="&lt;p&gt;Expanding on Mariano's answer: to get a formula for a truth table that has exactly one &lt;code&gt;t&lt;/code&gt; and the rest of the lines are &lt;code&gt;f&lt;/code&gt;, look at the line and write down the values of the variables in that line. For instance, if you wanted a formula for a truth table with three variables as in your second example which has a &lt;code&gt;t&lt;/code&gt; in the third line (corresponding to $X$ and $Z$ true, and $Y$ false) and &lt;code&gt;f&lt;/code&gt; in all the other ones, then since that line is &quot;$X$ is true, $Y$ is false, and $Z$ is true&quot;, then you use the formula $X \wedge (\neg Y)\wedge Z$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, suppose you have a formula with &lt;code&gt;t&lt;/code&gt;'s in &lt;em&gt;two&lt;/em&gt; rows and &lt;code&gt;f&lt;/code&gt;s everywhere else. Say, three variables, a &lt;code&gt;t&lt;/code&gt; in row three, a &lt;code&gt;t&lt;/code&gt; in row five, and &lt;code&gt;f&lt;/code&gt;'s everywhere else. A formula that has a &lt;code&gt;t&lt;/code&gt; in just row three is, as before, $X\wedge (\neg Y)\wedge Z$. A formula that has a &lt;code&gt;t&lt;/code&gt; in just row five is $(\neg X)\wedge (\neg Y)\wedge Z$. So a formula that as a &lt;code&gt;t&lt;/code&gt; in just row three &lt;em&gt;or&lt;/em&gt; row five is:&#xA;$$\Bigl( X\wedge (\neg Y)\wedge Z\Bigr) \vee \Bigl((\neg X)\wedge (\neg Y)\wedge Z\Bigr).$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For an arbitrary number of &lt;code&gt;t&lt;/code&gt;'s, just take the disjunction of enough formulas, each of which corresponds to a table with just one &lt;code&gt;t&lt;/code&gt;. &lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-11-15T17:08:35.637" CommentCount="2" />
  <row Id="10422" PostTypeId="1" AcceptedAnswerId="10458" CreationDate="2010-11-15T17:12:18.763" Score="3" ViewCount="229" Body="&lt;p&gt;Hi.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given a finite abelian group $G$. $H$ and $K$ subgroups of $G$ such that $G/H$ and $G/K$ are primary cyclic. $H$ and $K$ intersect trivially ($H\cap K=1$),  and $H\neq 1$,  $K\neq 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it then true that $HK=G$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My attempt so far: It's easy to show that $[G:HK] | ([G:H], [G:K])$. So if $G/H$ and $G/K$ are primary cyclic for different primes, then we're done. But I don't know what to do if they're not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you!&lt;/p&gt;&#xA;" OwnerUserId="3533" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-15T17:15:33.960" LastActivityDate="2010-11-16T03:45:32.357" Title="$G$ finite abelian. $G/H$, $G/K$ primary cyclic. $H\cap K=1$. $H\neq 1\neq K$. Is $HK=G$?" Tags="&lt;group-theory&gt;" AnswerCount="2" CommentCount="6" />
  <row Id="10423" PostTypeId="2" ParentId="10419" CreationDate="2010-11-15T17:15:02.810" Score="0" ViewCount="" Body="&lt;p&gt;There are lot of resources on the web and i think you can figure out your problem by reading some sources. Here i give some files in which the concept of &lt;em&gt;implicit differentiation&lt;/em&gt; has been written nicely.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;http://www.mash.dept.shef.ac.uk/Resources/web-implicit.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.mash.dept.shef.ac.uk/Resources/web-implicit.pdf&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;http://www.youtube.com/watch?v=ximF06lmqPM&quot; rel=&quot;nofollow&quot;&gt;http://www.youtube.com/watch?v=ximF06lmqPM&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Regarding your problem suppose assume you want to calculate the &lt;em&gt;derivative&lt;/em&gt; of the function $z=f(x,y)$, (say $z=x^{2}+y^{2}$) &lt;em&gt;where  $y$ is a function of $x$&lt;/em&gt;, then by chain rule you have $$\frac{dz}{dx} = 2x + 2y \cdot \frac{dy}{dx}$$&lt;/p&gt;&#xA;" LastActivityDate="2010-11-15T17:15:02.810" CommentCount="2" />
  <row Id="10424" PostTypeId="2" ParentId="6314" CreationDate="2010-11-15T17:24:41.013" Score="1" ViewCount="" Body="&lt;p&gt;The answer to the question as stated is no, as others have explained. However, if we relax the hypothesis from disjoint to non-overlapping, then the answer is yes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Two intervals $I_1$ and $I_2$ are non-overlapping if $I_1^{\circ}\cap I_2^{\circ}=\emptyset$; that is, if their &lt;em&gt;interiors&lt;/em&gt; are disjoint. If the intervals are closed and non-overlapping, then they intersect at most in their boundaries. For example, in $\mathbb{R}$, the intervals $\left[0,\frac{1}{2}\right]$ and $\left[\frac{1}{2},1\right]$ are non-overlapping, but clearly not disjoint as they share the point $\frac{1}{2}$.&lt;/p&gt;&#xA;" OwnerUserId="2313" LastActivityDate="2010-11-15T17:24:41.013" />
  <row Id="10425" PostTypeId="2" ParentId="10419" CreationDate="2010-11-15T17:47:54.730" Score="2" ViewCount="" Body="&lt;p&gt;If you have a formula for $\frac{dy}{dx}$ and you want to find the value of $y&amp;#39;(0)$, then &lt;em&gt;first&lt;/em&gt; you plug in $x=0$ to the implicit formula defining $y$, and evaluate to get the value of $y$. Then you plug in the value of $y$ you found and $x=0$ into the formula for $\frac{dy}{dx}$ to get the value of $y&amp;#39;(0)$.  Similarly for $y&amp;#39;&amp;#39;(0)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, suppose you are looking at the circle of radius $1$, $x^2+y^2=1$. Differentiating implicitly, we have:&#xA;\begin{align*}&#xA;\frac{d}{dx}(x^2+y^2) &amp;amp;= \frac{d}{dx}1\\&#xA;2x + 2y\frac{dy}{dx} &amp;amp;= 0\\&#xA;2y\frac{dy}{dx} &amp;amp;= -2x\\&#xA;\frac{dy}{dx} &amp;amp;= -\frac{x}{y}.&#xA;\end{align*}&#xA;To find $y&amp;#39;(0)$, we first plug in $x=0$ into $x^2+y^2=1$ to find the value of $y$ that corresponds to $x=0$; we get $y^2=1$, so $y=\pm 1$. Thus, we have two points, $(0,1)$ and $(0,-1)$. Then we plug in these values into the formula for $\frac{dy}{dx}$, and we get that in both cases we have $y&amp;#39;(0) = -\frac{0}{\pm 1}=0$. So $y&amp;#39;(0) = 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the second derivative, we again take derivatives:&#xA;\begin{align*}&#xA;\frac{dy}{dx} &amp;amp;= -\frac{x}{y}\\&#xA;\frac{d}{dx}\left(\frac{dy}{dx}\right) &amp;amp;= -\frac{d}{dx}\left(\frac{x}{y}\right)\\&#xA;\frac{d^2y}{dx^2} &amp;amp;= -\left(\frac{y(x)' - x(y')}{y^2}\right)\\&#xA;\frac{d^2y}{dx^2} &amp;amp;= \frac{x\frac{dy}{dx} - y}{y^2}\\&#xA;\frac{d^2y}{dx^2} &amp;amp;= \frac{x\left(-\frac{x}{y}\right) - y}{y^2}.&#xA;\end{align*}&#xA;The last equality, because $\frac{dy}{dx}=-\frac{x}{y}$. So now we plug in $x=0$, $y=\pm 1$ into &lt;em&gt;this&lt;/em&gt; equation, and we get that $y&amp;#39;&amp;#39;(0) = -\frac{y}{y^2} = -\frac{1}{y} = -\frac{1}{\pm 1}=\mp 1$. That is, $y&amp;#39;&amp;#39;(0)=1$ on $(0,-1)$, and $y&amp;#39;&amp;#39;(0)=-1$ on $(0,1)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, if your implicit curve has a single point on the vertical line $x=0$, then you only have one value of $y$ to contend with, so the process will be simpler (as far as evaluating). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;But&lt;/strong&gt;: Notice that you don't take the derivative of the function you get &lt;em&gt;after&lt;/em&gt; evaluating at $x=0$ to get the second derivative! You have to take the entire function and only &lt;em&gt;after&lt;/em&gt; differentiating &lt;em&gt;again&lt;/em&gt; do you once again plug in $x=0$ and the value of $y$ in order to find $y&amp;#39;&amp;#39;(0)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added:&lt;/strong&gt; If all you want is to find the values of $y&amp;#39;&amp;#39;(0)$ and $y&amp;#39;(0)$, you can avoid some complications in the computation of $\frac{d^2y}{dx^2}$ by starting from before you solved for $y&amp;#39;$. In the example above, after we find that when $x=0$ we have $y=\pm 1$ and $y&amp;#39;(0)=0$, we can start from the second line in the computation of $y&amp;#39;$:&#xA;$$2x + 2y\frac{dy}{dx} = 0,$$&#xD;&#xA;and after simplifying by cancelling the $2$, we can take derivatives again: just be careful with the product rule:&#xA;\begin{align*}&#xA;x + 2yy' &amp;amp;= 0\\&#xA;\frac{d}{dx}(x+2yy') &amp;amp;= 0\\&#xA;1 + 2(y'y' + yy'') &amp;amp;= 0\\&#xA;2(y')^2 + yy'' &amp;amp;= -1.&#xA;\end{align*}&#xA;Now, plugging in $x=0$, $y&amp;#39;(0)=0$, and $y(0)=\pm 1$, we get $\pm 1y&amp;#39;&amp;#39;(0)=-1$, hence $y&amp;#39;&amp;#39;(0) = \mp 1$; i.e., $y&amp;#39;&amp;#39;(0)=1$ when $y=-1$ and $y&amp;#39;&amp;#39;(0)=-1$ when $y=1$. Then you don't have to worry about fractions until you start solving. &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-15T19:42:19.387" LastActivityDate="2010-11-15T19:42:19.387" CommentCount="2" />
  <row Id="10427" PostTypeId="1" CreationDate="2010-11-15T18:26:12.963" Score="3" ViewCount="242" Body="&lt;p&gt;What is $x$ in closed form if $2x-\sin2x=\pi/2$, $x$ in the first quadrant?&lt;/p&gt;&#xA;" OwnerUserId="3538" LastEditorUserId="3538" LastEditorDisplayName="" LastEditDate="2010-11-15T19:21:46.793" LastActivityDate="2010-11-15T21:24:00.893" Title="Solving $2x - \sin 2x = \pi/2$ for $0 &lt; x &lt; \pi/2$" Tags="&lt;trigonometry&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="10429" PostTypeId="1" AcceptedAnswerId="10430" CreationDate="2010-11-15T18:41:23.470" Score="4" ViewCount="268" Body="&lt;p&gt;All power sets contain an empty set, so does this make the power set itself an empty set? Or am I mistaken?&lt;/p&gt;&#xA;" OwnerUserId="3540" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2010-11-20T01:54:07.327" LastActivityDate="2010-11-20T01:54:07.327" Title="Is a power set null because it contains an empty set?" Tags="&lt;elementary-set-theory&gt;" AnswerCount="4" CommentCount="3" />
  <row Id="10430" PostTypeId="2" ParentId="10429" CreationDate="2010-11-15T18:48:14.757" Score="8" ViewCount="" Body="&lt;p&gt;The short answer to your question is &quot;A power set is &lt;em&gt;never&lt;/em&gt; null (or empty), even if it is the power set of the empty set.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A power set $\mathcal{P}(X)$ is the &lt;em&gt;set of subsets&lt;/em&gt; of a given set $X$. The elements of $\mathcal{P}(X)$ are themselves sets, trivial or not. Since the empty set $\emptyset$ -- the set containing no elements -- is a subset of every set $X$, then $\emptyset \in \mathcal{P}(X)$ as a &lt;em&gt;set&lt;/em&gt;. Thus, every power set is non-trivial and has at least one element, namely the empty set. For example, if $X =$ {$a, b,c$}, then $\mathcal{P}(X)$ has 8 elements: the empty set $\emptyset$, the singletons {$a$}, {$b$} and {$c$}, all pairs {$a,b$}, {$a,c$} and {$b,c$} and the entire set $X =$ {$a, b,c$}. In fact, if $X$ has $|X|$ elements, one can prove that $|\mathcal{P}(X)| = 2^{|X|}$, even in the case that $X$ is infinite. When $X$ is finite, the proof follows from the identity &#xA;\begin{eqnarray}&#xA;\sum_{k = 0}^{n} \binom{n}{k} = 2^{n},&#xA;\end{eqnarray}&#xA;which is the sum of all ways to choose $k$ objects from $n$ objects as $k$ increases from $0$ to $n$. I'll leave it as an exercise to show how this counts $|\mathcal{P}(X)|$ with $|X| = n$.&lt;/p&gt;&#xA;" OwnerUserId="2720" LastEditorUserId="2720" LastEditorDisplayName="" LastEditDate="2010-11-15T20:57:33.710" LastActivityDate="2010-11-15T20:57:33.710" />
  <row Id="10431" PostTypeId="2" ParentId="10427" CreationDate="2010-11-15T18:49:01.140" Score="4" ViewCount="" Body="&lt;p&gt;The solution is given by $$\displaystyle  x = \pi/4 + D/2$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\displaystyle D$ is the root of $\cos y = y$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The root of $\displaystyle \cos y = y$ is nowadays known as the &lt;a href=&quot;http://mathworld.wolfram.com/DottieNumber.html&quot; rel=&quot;nofollow&quot;&gt;Dottie Number&lt;/a&gt; and apparently has no known &quot;closed form&quot; solution. If you consider this number to be part of your constants, then the above can be considered a closed form solution.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;For a proof:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $\displaystyle y = \sin(2x)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;then we have that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle 2x  = \pi/2 + y$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle y = \sin 2x = \sin (\pi/2 + y) = \cos y$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The root of $$\displaystyle y = \cos y$$ is $\displaystyle y = 0.739085\dots$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice that $\displaystyle \pi/2 &amp;gt; x \gt \pi/4$ (as $\displaystyle f(x) = 2x - \sin 2x$ is increasing in $\displaystyle [0,\pi/2]$), so if $\displaystyle x = \pi/4 + z$ then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle \sin(2x) = \sin(\pi/2 + 2z) = \cos 2z = 0.739085\dots$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And thus $\displaystyle z = \dfrac{0.739085\dots}{2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus $$\displaystyle x \sim \pi/4 + \dfrac{0.739085}{2} \sim 1.154940730005\dots$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See Also: &lt;a href=&quot;http://oeis.org/A003957&quot; rel=&quot;nofollow&quot;&gt;A003957&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-11-15T19:38:44.353" LastActivityDate="2010-11-15T19:38:44.353" CommentCount="1" />
  <row Id="10432" PostTypeId="2" ParentId="10429" CreationDate="2010-11-15T18:49:08.457" Score="14" ViewCount="" Body="&lt;p&gt;You seem to be saying that if a set contains the empty set, it is itself empty. This not true. In fact the opposite is true:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A set is empty if and only it contains no elements. If a set contains the empty set, it has at least one element (the empty set) and is thus, by definition, not empty.&lt;/p&gt;&#xA;" OwnerUserId="11" LastActivityDate="2010-11-15T18:49:08.457" />
  <row Id="10433" PostTypeId="2" ParentId="10348" CreationDate="2010-11-15T18:52:27.657" Score="2" ViewCount="" Body="&lt;p&gt;Rubin [1] gives an algorithm for counting Hamiltonian paths.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mathematica has tools for working with Hamiltonian paths, but the naive&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Length[HamiltonianCycles[G, All]]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;gives an out-of-memory exception for this particular graph.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[1] Frank Rubin, &lt;a href=&quot;http://portal.acm.org/citation.cfm?id=321854&quot; rel=&quot;nofollow&quot;&gt;A search procedure for Hamilton paths and circuits&lt;/a&gt; (1974)&lt;/p&gt;&#xA;" OwnerUserId="1778" LastEditorUserId="1778" LastEditorDisplayName="" LastEditDate="2010-11-15T19:23:00.677" LastActivityDate="2010-11-15T19:23:00.677" />
  <row Id="10434" PostTypeId="2" ParentId="10413" CreationDate="2010-11-15T18:52:36.267" Score="1" ViewCount="" Body="&lt;p&gt;No, I am afraid this maximality is not true.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let us always suppose that $R$ is a domain: else there is no fraction field (although there are substitutes). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, the result is certainly not true for higher dimensional rings. For example take&#xA;$R=\mathbb C[x,y] , m=(x,y) , K=\mathbb C(x,y)$ . &#xA;Then you have strict inclusions&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathbb C[x,y]_m \subset \mathbb C[x,y]_m [1/x] \subset \mathbb C (x,y)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence we see that the result in your case is not purely formal and we must use dimension one.  But unfortunately the result may not be true even then, which answers  question 1) negatively. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Indeed,  just observe   in the case $R=\mathbb C[[t^2,t^3]], \; m=(t^2,t^3), \; R=R_m$ the strict inclusions &#xA;$$R=\mathbb C[[t^2,t^3]] \subset \mathbb C [[t]]\subset \mathbb C ((t))$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To end on a cheerful note, your result is indeed true if $R_m$ is a discrete valuation ring.&#xA;Here is the proof. Take $a\in K\setminus R_m$ and consider $R_m\subset R_m [a]\subset K$ ( the first inclusion is strict). Since $a$ is not in $R_m$, we know that $1/a$ *is* in $R_m$: this is characteristic of valuation rings. Hence $1/a$ is &lt;em&gt;a fortiori&lt;/em&gt; in $R_m[a]$ so that $1\in R[a]$ and we have proved $R[a]=K$.  &lt;/p&gt;&#xA;" OwnerUserId="3217" LastEditorUserId="3217" LastEditorDisplayName="" LastEditDate="2010-11-15T19:31:05.610" LastActivityDate="2010-11-15T19:31:05.610" CommentCount="1" />
  <row Id="10435" PostTypeId="1" AcceptedAnswerId="10442" CreationDate="2010-11-15T18:57:47.547" Score="0" ViewCount="254" Body="&lt;p&gt;Let a statement P is &quot;X is true if and only if Y is true&quot;. What is the negation of P? I am little confused. It seems that digital equivalent of this statement is P = X and Y. Hence negation of P is (not X) or (not Y) i.e. Either X or Y is false. Am I right guys?   &lt;/p&gt;&#xA;" OwnerUserId="1674" LastActivityDate="2010-11-15T20:47:28.680" Title="Negation of if and only if?" Tags="&lt;logic&gt;" AnswerCount="5" CommentCount="1" FavoriteCount="1" />
  <row Id="10436" PostTypeId="2" ParentId="10429" CreationDate="2010-11-15T18:59:41.827" Score="1" ViewCount="" Body="&lt;p&gt;To add on the previous answers, I just want to clarify that as it was remarked before the power set of $A$, usually denoted by $\mathcal{P}(A)$ is the collection of all the subsets of $A$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Namely, if $A = \emptyset$ then it has only one subset - itself, as the empty set is a subset of every given set. Therefore, $\mathcal{P}(\emptyset) = \{\emptyset \}$, and again as remarked before - it means that it has one element, and more specifically that it is non empty.&lt;/p&gt;&#xA;" OwnerUserId="622" LastActivityDate="2010-11-15T18:59:41.827" />
  <row Id="10437" PostTypeId="2" ParentId="10435" CreationDate="2010-11-15T19:04:30.670" Score="7" ViewCount="" Body="&lt;p&gt;The digital equivalent is P = X XNOR Y, and thus the negation is (not P) = X XOR Y. In other words, P is false when X is true but Y is false, or when X is false but Y is true.&lt;/p&gt;&#xA;" OwnerUserId="759" LastActivityDate="2010-11-15T19:04:30.670" CommentCount="1" />
  <row Id="10439" PostTypeId="2" ParentId="10435" CreationDate="2010-11-15T19:04:54.610" Score="1" ViewCount="" Body="&lt;p&gt;Yes, but you have to be very precise here, because the negation of implication is &lt;em&gt;exclusive&lt;/em&gt; (not inclusive) OR. So the answer is &quot;Either $X$ or $Y$ is false, but not both&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, if you are confused, start with a truth table for implication and then negate it. Resulting table matches XOR (exclusive OR). &lt;/p&gt;&#xA;" OwnerUserId="2450" LastActivityDate="2010-11-15T19:04:54.610" />
  <row Id="10440" PostTypeId="2" ParentId="10435" CreationDate="2010-11-15T19:07:36.467" Score="1" ViewCount="" Body="&lt;p&gt;You are not right.  Let $X = (\ell$ is even) and $Y = (\ell$ is not odd).  Then clearly $X \Leftrightarrow Y$, but &quot;($\ell$ is not even) or ($\ell$ is odd)&quot; is strictly weaker; you want &quot;($\ell$ is not even) $\Leftrightarrow$ ($\ell$ is odd)&quot; to be true.&lt;/p&gt;&#xA;" OwnerUserId="3008" LastActivityDate="2010-11-15T19:07:36.467" />
  <row Id="10441" PostTypeId="1" AcceptedAnswerId="10445" CreationDate="2010-11-15T19:09:12.910" Score="0" ViewCount="79" Body="&lt;p&gt;I'm working through an exercise in Hartschorne concerning direct limits of sheaves on noetherian topological spaces. It seems natural to use the the fact that every open subset of a noetherian topological space is the quasi-compact in its induced topology. I'm some trouble proving what I need when I only use this property. I'm guessing that I'll have to use noetherian-ness more directly, but I thought I would ask if anyone knew an example offhand.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Is there a non-notherian topological space which has this property?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="756" LastActivityDate="2010-11-16T23:00:22.443" Title="Are there non-noetherian topological spaces in which every open subset is quasi-compact?" Tags="&lt;topology&gt;&lt;algebraic-geometry&gt;" AnswerCount="2" />
  <row Id="10442" PostTypeId="2" ParentId="10435" CreationDate="2010-11-15T19:15:00.453" Score="10" ViewCount="" Body="&lt;p&gt;$X\leftrightarrow Y$ is the conjunction of $X\leftarrow Y$ and $X\rightarrow Y$. The negation of a conjunction is the disjunction of the negations; the negation of $P\rightarrow Q$ is $P\wedge \neg Q$. So we have:&#xA;\begin{align*}&#xA;\neg(X\leftrightarrow Y) &amp;amp;\Longleftrightarrow \neg\Bigl( (X\rightarrow Y)\wedge (Y\rightarrow X)\Bigr)\\&#xA;&amp;amp;\Longleftrightarrow \neg(X\rightarrow Y)\vee \neg(Y\rightarrow X)\\&#xA;&amp;amp;\Longleftrightarrow (X\wedge \neg Y) \vee (Y\wedge \neg X).&#xA;\end{align*}&#xA;So the negation of &quot;$X$ is true if and only if $Y$ is true&quot; is &quot;Either $X$ is true and $Y$ is false, or $X$ is false and $Y$ is true.&quot; &lt;em&gt;Added&lt;/em&gt;: as it happens, as noted by Rahul Narain in his comment, this is in turn equivalent to &quot;$X$ is true if and only if $Y$ is false&quot; (just compare the cases when they are each true). So you also get that&#xA;$$\neg(X\leftrightarrow Y) \Longleftrightarrow X\leftrightarrow \neg Y \Longleftrightarrow \neg X\leftrightarrow Y.$$&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-15T20:01:28.413" LastActivityDate="2010-11-15T20:01:28.413" CommentCount="6" />
  <row Id="10443" PostTypeId="1" AcceptedAnswerId="10448" CreationDate="2010-11-15T19:17:27.760" Score="7" ViewCount="647" Body="&lt;p&gt;In school and at university we never had eigenvalues nor differential equations, so these concepts were really giving me a hard time. Now I developed some intuition for both concepts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I learned that both are connected in some way, i.e. there is an eigenvector/eigenvalue approach  to solve differential equations. Unfortunately most of the text I find are way beyond my understanding and seem to require very extensive study.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a possibility to provide me with an intuition of the connection anyway and perhaps give an easy example of how to use this approach to solve a differential equation?&lt;/p&gt;&#xA;" OwnerUserId="346" LastActivityDate="2010-11-15T21:17:38.093" Title="How are eigenvectors/eigenvalues and differential equations connected?" Tags="&lt;linear-algebra&gt;&lt;intuition&gt;&lt;differential-equations&gt;&lt;examples&gt;&lt;eigenvectors&gt;" AnswerCount="5" CommentCount="2" FavoriteCount="3" />
  <row Id="10444" PostTypeId="2" ParentId="10441" CreationDate="2010-11-15T19:18:07.953" Score="1" ViewCount="" Body="&lt;p&gt;No.  Noetherianness is equivalent to the condition that every open subset is compact, and is in fact equivalent to the condition that every subset is compact.  This is a nice exercise.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-11-15T19:18:07.953" />
  <row Id="10445" PostTypeId="2" ParentId="10441" CreationDate="2010-11-15T19:20:55.267" Score="2" ViewCount="" Body="&lt;p&gt;If every open subspace of a space is quasi compact, then the space is Noetherian. &#xA;This is exercise 6.5 in Atiyah Macdonald. It's an easy proof. Let me know if you need more help with it.&#xA;Hint: If you consider an increasing sequence of open sets, its union is open and hence quasi compact. &lt;/p&gt;&#xA;" OwnerUserId="3431" LastEditorUserId="3431" LastEditorDisplayName="" LastEditDate="2010-11-16T23:00:22.443" LastActivityDate="2010-11-16T23:00:22.443" CommentCount="1" />
  <row Id="10446" PostTypeId="2" ParentId="10443" CreationDate="2010-11-15T19:24:29.910" Score="2" ViewCount="" Body="&lt;p&gt;A good reference is here: &lt;a href=&quot;http://www.sosmath.com/diffeq/system/linear/eigenvalue/eigenvalue.html&quot; rel=&quot;nofollow&quot;&gt;http://www.sosmath.com/diffeq/system/linear/eigenvalue/eigenvalue.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3431" LastActivityDate="2010-11-15T19:24:29.910" />
  <row Id="10447" PostTypeId="2" ParentId="10427" CreationDate="2010-11-15T19:25:07.847" Score="1" ViewCount="" Body="&lt;p&gt;To get a graphical impression of the solution have a look here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.wolframalpha.com/input/?i=solve+2+x+-+sin+(2+x)+%3D+Pi/2+for+x&quot; rel=&quot;nofollow&quot;&gt;http://www.wolframalpha.com/input/?i=solve+2+x+-+sin+(2+x)+%3D+Pi/2+for+x&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: That WA doesn't give a closed form is at least a hint that there might be none (very probable) &lt;/p&gt;&#xA;" OwnerUserId="346" LastActivityDate="2010-11-15T19:25:07.847" CommentCount="4" />
  <row Id="10448" PostTypeId="2" ParentId="10443" CreationDate="2010-11-15T19:26:53.210" Score="5" ViewCount="" Body="&lt;p&gt;You could think of eigenvectors as giving you the coordinate system in which your problem has the simplest form. It's analogous to working physical problems by turning your coordinate system the right way to make some term go away.&lt;/p&gt;&#xA;" OwnerUserId="247" LastActivityDate="2010-11-15T19:26:53.210" />
  <row Id="10449" PostTypeId="2" ParentId="10379" CreationDate="2010-11-15T19:34:09.913" Score="0" ViewCount="" Body="&lt;p&gt;If you want to be totally precise, you have to say that it has one real root with multiplicity 2. But &quot;two equal real roots&quot; is almost universally understood as a shortcut for saying that.&lt;/p&gt;&#xA;" OwnerUserId="173" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-15T22:49:37.140" LastActivityDate="2010-11-15T22:49:37.140" />
  <row Id="10450" PostTypeId="1" AcceptedAnswerId="10451" CreationDate="2010-11-15T19:38:56.710" Score="3" ViewCount="79" Body="&lt;p&gt;I was reviewing quantum factorization and am slightly unclear on a classical detail of order-finding. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given a (suitably nice) periodic function $f$ with unknown period $r$ and a power of two $N &amp;gt; r^2$, the quantum subroutine yields (with bounded probability) a result of the form $\lfloor jr/N \rceil$ for $1 \le j \le r-1$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moreover, $\left \lvert \lfloor jr/N \rceil/N - j/r \right \rvert \le 1/2N$, so since $N &amp;gt; r^2$ a theorem assures us that $j/r$ is a convergent in the continued fraction expansion of $\lfloor jr/N \rceil/N$. Now I see lots of hand-waving at this point in most discussions of the algorithm. If the length of the CFE is $poly(\log(N))$, then this hand-waving can be trivially dispensed with (perhaps at the cost of a slight inefficiency). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, does the CFE of $M/N$ have length $poly(\log(N))$?&lt;/p&gt;&#xA;" OwnerUserId="241" LastActivityDate="2010-11-15T19:49:54.683" Title="What is the length of a continued fraction expansion of a rational number?" Tags="&lt;continued-fractions&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" />
  <row Id="10451" PostTypeId="2" ParentId="10450" CreationDate="2010-11-15T19:49:54.683" Score="4" ViewCount="" Body="&lt;p&gt;The length of the CF of a rational number $\displaystyle \dfrac{a}{b}$ is closely related to the number of steps it takes in the Euclidean algorithm for finding gcd of $\displaystyle a,b$, which can, in the worst case, be shown to be proportional to the number of digits in $\displaystyle b$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the claim is true.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See: &lt;a href=&quot;http://www.albany.edu/~hammond/gellmu/examples/confrac.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.albany.edu/~hammond/gellmu/examples/confrac.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also: &lt;a href=&quot;http://en.wikipedia.org/wiki/Euclidean_algorithm#Number_of_steps&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Euclidean_algorithm#Number_of_steps&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-11-15T19:49:54.683" />
  <row Id="10452" PostTypeId="1" CreationDate="2010-11-15T19:54:01.687" Score="0" ViewCount="176" Body="&lt;p&gt;If $a_1,\ldots,a_n$ are in arithmetic progression and $a_i\gt 0$ for all $i$, then I would like to prove the following two identities:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$$ \frac{1}{\sqrt{a_1} + \sqrt{a_2}} + \frac{1}{\sqrt{a_2} + \sqrt{a_3}} + \cdots + \frac{1}{\sqrt{a_{n-1}} + \sqrt{a_n}} = \frac{n-1}{\sqrt{a_1} + \sqrt{a_n}}$$&#xD;&lt;/li&gt;&#xA;&lt;li&gt;$$ \frac{1}{a_1 \cdot a_n} + \frac{1}{a_2 \cdot a_{n-1}} + \frac{1}{a_3 \cdot a_{n-2}}+ \cdots + \frac{1}{a_n \cdot a_1} = \frac{2}{a_1 + a_n} \biggl( \frac{1}{a_1} + \frac{1}{a_2} + \cdots +  \frac{1}{a_n} \biggr)$$&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="2109" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-15T20:18:20.577" LastActivityDate="2010-11-16T18:29:21.037" Title="Prove an identity involving terms in arithmetic progression." Tags="&lt;algebra-precalculus&gt;&lt;sequences-and-series&gt;" AnswerCount="3" CommentCount="17" />
  <row Id="10453" PostTypeId="1" AcceptedAnswerId="10455" CreationDate="2010-11-15T19:59:55.163" Score="2" ViewCount="74" Body="&lt;p&gt;Hello,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a newbie question, but I would be grateful for any reference that you could give.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;let $f(x) \in \mathbb{A}$, where $x \in \mathbb{A}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a symbol to indicate the repeated application of $f$ unto its results, in a manner similar to $\sum$ and $\prod$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you!&lt;/p&gt;&#xA;" OwnerUserId="272" LastActivityDate="2010-11-15T20:04:11.983" Title="Newbie question about the re-application of a function on its result" Tags="&lt;analysis&gt;&lt;notation&gt;" AnswerCount="2" CommentCount="9" />
  <row Id="10454" PostTypeId="2" ParentId="10453" CreationDate="2010-11-15T20:03:26.117" Score="5" ViewCount="" Body="&lt;p&gt;It is usually indicated by $f\circ f$ or if there are more than two repeated applications, simply $f^n(x)$&lt;/p&gt;&#xA;" OwnerUserId="3431" LastActivityDate="2010-11-15T20:03:26.117" CommentCount="3" />
  <row Id="10455" PostTypeId="2" ParentId="10453" CreationDate="2010-11-15T20:04:11.983" Score="5" ViewCount="" Body="&lt;p&gt;The symbol $\circ$ is used for composition of functions: $g\circ h(x)$ means $g(h(x))$. For composing a function with itself, many people use $f^{\circ n}(x)$, defined inductively by&#xA;\begin{align*}&#xA;f^{\circ 1}(x) &amp;amp;= f(x),\\&#xA;f^{\circ(k+1)}(x) &amp;amp;= f\Bigl( f^{\circ k}(x)\Bigr).&#xA;\end{align*}&#xA;So $f^{\circ 2}(x) = f(f(x))$; $f^{\circ 3}(x) = f(f^{\circ 2}(x)) = f(f(f(x)))$, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Timothy Wagner also notes that $f^n(x)$ is often used; that is true, but it has the disadvantage of often being ambiguous. For example, $\sin^2(x)$ is (almost) always understood to mean $(\sin(x))^2$, and not $\sin(\sin(x))$. &lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-11-15T20:04:11.983" />
  <row Id="10457" PostTypeId="2" ParentId="10452" CreationDate="2010-11-15T20:15:26.617" Score="4" ViewCount="" Body="&lt;p&gt;This question is now old enough for some more complete answers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For number 1: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{k=1}^{n-1} \frac{1}{ \sqrt{a_k}+ \sqrt{a_{k+1}}} = &#xD;&#xA;\sum_{k=1}^{n-1} \frac{\sqrt{a_{k+1}} - \sqrt{a_k}}{d}$$&#xD;&#xA;where $d$ is the common difference,&#xA;$$ = \frac{1}{d} \left( \sqrt{a_n} - \sqrt{a_1} \right)&#xD;&#xA;= \frac{a_n - a_1}{d(\sqrt{a_n} + \sqrt{a_1})}$$&#xD;&#xA;$$= \frac{n-1}{\sqrt{a_1} + \sqrt{a_n}}.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And for number 2:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ S = 2 \sum_{k=1}^n \frac{1}{a_k} = &#xD;&#xA;\left( \frac{1}{a_1} + \frac{1}{a_n} \right) + &#xD;&#xA;\left( \frac{1}{a_2} + \frac{1}{a_{n-1}} \right) + \cdots +&#xD;&#xA;\left( \frac{1}{a_n} + \frac{1}{a_1} \right)$$&#xD;&#xA;$$ = \sum_{k=1}^n \frac{a_{n-k+1}+ a_k}{a_k a_{n-k+1}}.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now $ a_{n-k+1}+ a_k = 2a_1 + (n-1)d = a_1 + a_n$ and so&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ S = (a_1+a_n) \sum_{k=1}^n \frac{1}{a_k a_{n-k+1}}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;from which the result follows.&lt;/p&gt;&#xA;" OwnerUserId="1301" LastEditorUserId="1301" LastEditorDisplayName="" LastEditDate="2010-11-16T10:55:17.520" LastActivityDate="2010-11-16T10:55:17.520" />
  <row Id="10458" PostTypeId="2" ParentId="10422" CreationDate="2010-11-15T20:16:58.440" Score="4" ViewCount="" Body="&lt;p&gt;If $a\in G$ is of order $p^n$ for some prime $p$ and $a \notin H$ then $a+H$ is of order $p^d$ for some $d\leq n$ so $G/H$ must be a p-group. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;With this argument we also know that $H$ must contain all elements that are not of order $p^m$ for some m. So if |G| has at least two different prime divisors, then $G/H$ and $G/K$ must be p-group and q-group for different primes (because $H\cap K= 1$). Therefore all the p-sylow groups are contained in at least one of H or K so HK=G.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we are now remained with the case where $|G|=p^n$. I tried to prove this case (for too much time), but it appears that this is not true. For example, take $G=\mathbb{Z}_2 \times \mathbb{Z}_4$ and let H be the subgroup $\{ (1,0), (0,0) \}$ and $K=\{ (1,2), (0,0) \}$, then $G/H\cong G/K \cong \mathbb{Z}_4 $ and $|HK|=4 &amp;lt; |G|=8$&lt;/p&gt;&#xA;" OwnerUserId="2125" LastActivityDate="2010-11-15T20:16:58.440" CommentCount="2" />
  <row Id="10459" PostTypeId="2" ParentId="10389" CreationDate="2010-11-15T20:18:30.210" Score="0" ViewCount="" Body="&lt;p&gt;For suitable constant $c$, $f(n) = (T(n)/n^2) + c$ satisfies a recurrence $f(n) = Af(n/2)+Bf(n/3)$ which has solutions (for continuous-valued $n$) of the form $n^\alpha$.  This should give a guess about the asymptotics for integer $n$ and one could probably incorporate lower order terms (fluctuations from the complex solutions with real part less than that of $\alpha$) by analyzing the zeros of $G(x)=1 - Ax^{\log 2} - Bx^{\log 3}$ or the generating function with real exponents (generalized power series expansion) for $1/G(x)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think this type of exercise for the large $n$ asymptotics and translation to the continuous problem is in books such as (e.g.) Flajolet and Sedgewick's text on asymptotic combinatorics.  However, for this specific recurrence it might be possible to go beyond the standard asymptotics and handle integer-valued $n$ directly, because the operations of replacing $n$ by $P(n)=[n/2]$ and $Q(n)=[n/3]$ commute.  One could look at base 6 expansions of $n$, or classify integers according to which operations $P^a Q^b$ take them to $0$ and $1$, and sums over the $(a,b)$ would give an integer-valued expansion formula for $f(n)$, similar to the binomial theorem or the expansion of a rational generating function $1/(1-H(x))$ as a geometric series.&lt;/p&gt;&#xA;" OwnerUserId="467" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-11-15T23:44:30.360" LastActivityDate="2010-11-15T23:44:30.360" />
  <row Id="10461" PostTypeId="2" ParentId="10452" CreationDate="2010-11-15T20:24:20.257" Score="2" ViewCount="" Body="&lt;p&gt;For the first one, use induction (or) note that $\frac{1}{\sqrt{a_k}+\sqrt{a_{k+1}}} = \frac{\sqrt{a_{k+1}}-\sqrt{a_k}}{d}$, where $d$ is the common difference between the successive terms. Now use the telescopic summation to cancel out the terms in the numerator and massage it to get the final expression on the right hand side.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the second one, try to write each term on the Left Hand Side as a difference of two terms and proceed.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2010-11-16T02:52:57.120" LastActivityDate="2010-11-16T02:52:57.120" CommentCount="1" />
  <row Id="10462" PostTypeId="2" ParentId="10443" CreationDate="2010-11-15T20:36:53.050" Score="3" ViewCount="" Body="&lt;p&gt;The collection of all functions that are differentiable at least $n$ times (let's call it $\mathcal{F}_n$ just for the purposes of this post) form a vector space; and the operator $D$ defined by $D(f) = f&amp;#39;$ is linear ($D(f+g)=D(f)+D(g)$, and $D(\alpha f) = \alpha D(f)$ for all $f,g$ and scalars $\alpha$). So $D$ is a linear transformation from the vector space of functions that can be differentiated at least $n$ times, and functions that can be differentiated at least $n-1$ times.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, suppose you have a &lt;em&gt;homogeneous&lt;/em&gt; linear differential equation: &#xA;$$ f_n(x) y^{(n)}(x) + \cdots + f_1(x)y&amp;#39;(x) + f_0(x)y(x) = 0.$$&#xD;&#xA;Then the collection of all function $y(x)$ that are solutions to this system form a subspace of $\mathcal{F}_n$: because if $y_1$ and $y_2$ are solutions, then so is $y_1+\alpha y_2$ for any scalar $\alpha$, and the zero function is certainly a solution. It is, in fact, the nullspace of a certain linear transformation, namely, the linear transformation $L$ given by&#xA;$$L(y) = f_n(x)y^{(n)}(x)+\cdots + f_1(x)y&amp;#39;(x) + f_0(x)y(x).$$&#xD;&#xA;So one can bring some linear algebra to bear on this problem; e.g., determining the dimension of the solution space, etc. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, in analogy to the case of systems of linear equations, consider a *&lt;em&gt;non&lt;/em&gt;*homogeneous linear differential equation:&#xA;$$f_n(x) y^{(n)}(x) + \cdots + f_1(x)y&amp;#39;(x) + f_0(x)y(x) = g(x).$$&#xD;&#xA;Suppose you could find a particular solution $y_p$ to this equation. Then, if $y_h$ is any solution to the corresponding homogeneous equation, then $y_p+y_h$ is a solution to the nonhomogeneous equation as well; and if $z_p$ is another solution to the nonhomogeneous equation, then $y_p-z_p$ is a solution to the homogeneous equation. So &lt;em&gt;every&lt;/em&gt; solution to the nonhomogeneous equation is of the form $y_p + y_h$, where $y_p$ is the particular solution to found, and $y_h$ is a solution to the associated homogeneous equation. This is exactly the same thing that happens with systems of linear equations (where the unknowns are numbers). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In many cases, you can show that any solutions to your equations will actually be infinitely differentiable (e.g., if your $f$s are infinitely differentiable). Then you can think of the linear transformation $L$ as a linear operator on the space of infinitely differentiable functions. Then eigenvectors and eigenvalues can come into play: eigenvectors and eigenvalues give you simpler ways of thinking about a linear transformation, so they give you simpler ways of thinking about &lt;em&gt;this&lt;/em&gt; particular linear transformation (which happens to correspond to solutions of a differential equation). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, systems of linear differential equations very naturally lead to linear transformations where the eigenvectors and eigenvalues play a key role in helping you solve the system, because they &quot;de-couple&quot; the system, by allowing you to think of a complex system in which each of the variables affects the derivative of the others as a system in which you have some new variables that are completely independent of one another (or in the case of generalized eigenvectors, easily dependent on only some of the others). This makes the system easier to solve.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-11-15T20:36:53.050" />
  <row Id="10463" PostTypeId="2" ParentId="10427" CreationDate="2010-11-15T20:38:45.800" Score="2" ViewCount="" Body="&lt;p&gt;Repeating the same thing as the other answers, but with a moderately more elegant one:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $y = 2x - \frac{\pi}{2}$, then substituting:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$y = sin(y+\frac{\pi}{2})$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$y = cos(y)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$y = D$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x = \frac{y+\frac{\pi}{2}}{2} = \frac{D}{2} + \frac{\pi}{4}$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="272" LastEditorUserId="272" LastEditorDisplayName="" LastEditDate="2010-11-15T21:24:00.893" LastActivityDate="2010-11-15T21:24:00.893" />
  <row Id="10464" PostTypeId="2" ParentId="10375" CreationDate="2010-11-15T20:41:25.423" Score="0" ViewCount="" Body="&lt;p&gt;Ah, I forgot about the Yoneda lemma.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, I think it's impossible to show that a strictly increasing function into a partial order is monic, because the normal proof uses the fact that monic on global elements implies monic. In fact, this lemma is probably used a lot.&lt;/p&gt;&#xA;" OwnerUserId="3518" LastActivityDate="2010-11-15T20:41:25.423" />
  <row Id="10465" PostTypeId="2" ParentId="10435" CreationDate="2010-11-15T20:47:28.680" Score="0" ViewCount="" Body="&lt;p&gt;The assertion $X \leftrightarrow Y$ can also be written as $X = Y$. So its negation is $X \neq Y$, which is the same as $X = \overline{Y}$ (since $X,Y \in \{0,1\}$), which is the same as $\overline{X} \leftrightarrow Y$.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-15T20:47:28.680" />
  <row Id="10466" PostTypeId="2" ParentId="10443" CreationDate="2010-11-15T20:55:31.357" Score="4" ViewCount="" Body="&lt;p&gt;Suppose $\sum a_n y^{(n)} = 0$ is a homogeneous linear differential equation with constant coefficients.  Let $D$ be the linear operator defined on, say, the space $C^{\infty}(\mathbb{R})$ of smooth functions $\mathbb{R} \to \mathbb{R}$ which sends a function $y$ to its derivative $y&amp;#39;$.  Then the solutions to the above differential equation are the nullspace of the operator $\sum a_n D^n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It turns out that this nullspace is finite-dimensional, so we can use finite-dimensional linear algebra on it.  It is not just a vector space, though: it comes equipped with an action of $D$, which commutes with $\sum a_n D^n$, and which therefore splits the nullspace into eigenspaces of $D$.  In the simplest case, these eigenspaces will all be one-dimensional.  Now, the eigenvectors of $D$ are precisely the functions satisfying $Dy = \lambda y$ for some eigenvector $\lambda$, which are precisely the exponential functions $e^{\lambda y}$.  And it's not hard to see that such a function is a solution of the differential equation if and only if $\lambda$ is a root of the &lt;strong&gt;characteristic polynomial&lt;/strong&gt; $\sum a_n \lambda^n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Example.&lt;/em&gt;  &lt;a href=&quot;http://en.wikipedia.org/wiki/Simple_harmonic_motion&quot; rel=&quot;nofollow&quot;&gt;Simple harmonic motion&lt;/a&gt; is governed by the differential equation $m D^2 + k D^0 = 0$, which has characteristic polynomial $m \lambda^2 + k = 0$.  The roots of this polynomial are complex, but if we allow ourselves to work with complex numbers (formally, in the above situation we tensor with $\mathbb{C}$) we find that the eigenvalues are $\lambda = \pm i \sqrt{ \frac{k}{m} }$, so the set of solutions is all functions of the form&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x = A e^{ i \sqrt{ \frac{k}{m} } t } + B e^{-i \sqrt{ \frac{k}{m} } t}.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By Euler's formula, if we restrict our solutions to be real we get the familiar periodic sine and cosine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general the eigenspaces will not be one-dimensional and then the theory of Jordan normal form applies.  This occurs, for example, when finding the general form of &lt;a href=&quot;http://en.wikipedia.org/wiki/Damping&quot; rel=&quot;nofollow&quot;&gt;damped harmonic motion&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-11-15T20:55:31.357" />
  <row Id="10467" PostTypeId="1" AcceptedAnswerId="10471" CreationDate="2010-11-15T20:59:12.770" Score="1" ViewCount="122" Body="&lt;p&gt;In &lt;em&gt;Classical Fourier Analysis&lt;/em&gt; by Loukas Grafakos we have in Proposition 2.3.25 the following definition for $\mathcal{S}_\infty(\mathbf{R}^n)$, namely that these are all the Schwartz functions $\phi$ such that for all multi-indices $\alpha$ we have that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_{\mathbf{R}^n} x^\alpha \phi(x) \, dx = 0.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I'm trying to find non-trivial functions in this space. I know that the Fourier transform maps the Schwartz-functions to itself, so I note that the requirement is actually that the Fourier transform evaluated in $0$ of $x^\alpha \phi(x)$ is $0$. So, $x^\alpha \mapsto i^\alpha d/dx^\alpha$ in the Fourier domain, so we actually want a function $\phi$ such that (let's take $n = 1$):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left . \frac{d}{dx^\alpha} \widehat{\phi}(x) \right |_{x = 0} = 0.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For all $n \geq 0$ An obvious candidate is $f(x) = \text{exp}(-1/x^2)$ for $x &amp;gt; 0$ and $0$ otherwise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if I now compute the Fourier transform (or the inverse) of this function (with Maple) I get another function say $g$, but if I plot the real part of $g$ it is not smooth (it has a cusp in 0), how is this possible? Further the integral which I want to be zero is only zero for odd $n$, for even $n$ it is complex! What goes wrong? &lt;/p&gt;&#xA;" OwnerUserId="1120" LastEditorUserId="1120" LastEditorDisplayName="" LastEditDate="2010-11-15T21:05:21.297" LastActivityDate="2010-11-15T21:08:42.213" Title="Fourier transform of a special Schwartz function." Tags="&lt;fourier-analysis&gt;&lt;harmonic-analysis&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="10468" PostTypeId="1" CreationDate="2010-11-15T20:59:47.717" Score="0" ViewCount="172" Body="&lt;p&gt;Hi there,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i'm a CompSci university student and i want some help for my math questions.&#xA;Therefore i am looking for a good discussion board for &lt;strong&gt;math homework questions&lt;/strong&gt;.&#xA;I don't know any (good), because i come from an &lt;em&gt;non-native-english-speaking&lt;/em&gt; country.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;any ideas ?&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;note&lt;/strong&gt; : I know that i could post my questions here, but:&lt;br&gt;&#xA;  1. this is not a discussion board, where i can discuss &lt;strong&gt;about&lt;/strong&gt; my&#xA;  question&lt;br&gt;&#xA;  2. i won't annoy the experts with &quot;simple&quot; algebra question :)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="3549" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-15T22:51:08.417" LastActivityDate="2010-11-23T16:46:43.467" Title="Which discussion board is good for homework questions ?" Tags="&lt;homework&gt;" AnswerCount="4" CommentCount="1" />
  <row Id="10469" PostTypeId="1" AcceptedAnswerId="10682" CreationDate="2010-11-15T21:00:39.633" Score="5" ViewCount="112" Body="&lt;p&gt;Hi, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm reading Hatcher's notes on spectral sequences and he mentions that steenrod squares commute with the coboundary operator for pairs (X,A) which would then explain why these operations commute with the transgression. It says it's because &#xA;that coboundary operator can be defined in terms of suspension and we know steenrod operations commute with suspension. Does anyone know the details of this reasoning?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So...&#xA;Assuming the standard axioms of steenrod operations, how do we prove that it commutes with the coboundary operator for pairs?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you,&lt;/p&gt;&#xA;" OwnerUserId="3550" LastActivityDate="2010-11-17T09:07:44.160" Title="Why steenrod commute with transgression" Tags="&lt;algebraic-topology&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" />
  <row Id="10470" PostTypeId="2" ParentId="10468" CreationDate="2010-11-15T21:03:41.503" Score="2" ViewCount="" Body="&lt;p&gt;Have you looked at this:&#xA;&lt;a href=&quot;http://www.artofproblemsolving.com/&quot; rel=&quot;nofollow&quot;&gt;http://www.artofproblemsolving.com/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.artofproblemsolving.com/Forum/index.php&quot; rel=&quot;nofollow&quot;&gt;http://www.artofproblemsolving.com/Forum/index.php&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3431" LastActivityDate="2010-11-15T21:03:41.503" />
  <row Id="10471" PostTypeId="2" ParentId="10467" CreationDate="2010-11-15T21:08:42.213" Score="1" ViewCount="" Body="&lt;p&gt;As $x\to\infty$, $\exp(-1/x^2)\to1$ so this isn't in the Schwartz&#xA;space. Why not multiply this by $\exp(-x)$ to get it decaying to&#xA;$0$ at infinity rapidly enough?&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-11-15T21:08:42.213" CommentCount="1" />
  <row Id="10472" PostTypeId="2" ParentId="10443" CreationDate="2010-11-15T21:17:38.093" Score="5" ViewCount="" Body="&lt;p&gt;If you have a linear system on matrix form, $dX/dt=AX$,&#xA;where $X(t)$ is a vector in $\mathbf{R}^n$ and $A$ is an&#xA;$n\times n$ constant real matrix,&#xA;then $X(t) = \exp(\lambda t) V$ is a solution to the system&#xA;if $V$ is an eigenvector of $A$ with eigenvalue $\lambda$.&#xA;(This works since $\exp(\lambda t)$ is an eigenfunction of&#xA;the differential operator $d/dt$ with eigenvalue $\lambda$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $A$ is diagonalizable, then the general&#xA;solution is a linear combination of terms like above:&#xA;$X(t)=\sum_{k=1}^n c_k \exp(\lambda_k t) V_k$.&#xA;(If the eigenvalues are complex, this is the general complex-valued solution;&#xA;one can perform some standard tricks to extract the general real-valued solution&#xA;from that.)&#xA;Nondiagonalizable matrices $A$ cause a bit more trouble, but&#xA;that can be handled too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Of course, this is related to what has been said in other answers already,&#xA;but with a system of this type one sees clearly that the eigenvectors play a role&#xA;too, not just the eigenvalues.)&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2010-11-15T21:17:38.093" />
  <row Id="10473" PostTypeId="2" ParentId="10468" CreationDate="2010-11-15T21:19:20.993" Score="0" ViewCount="" Body="&lt;p&gt;Have a look at &lt;a href=&quot;http://en.allexperts.com/c/science/8.htm&quot; rel=&quot;nofollow&quot;&gt;http://en.allexperts.com/c/science/8.htm&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2810" LastActivityDate="2010-11-15T21:19:20.993" />
  <row Id="10474" PostTypeId="2" ParentId="10468" CreationDate="2010-11-15T21:25:27.040" Score="1" ViewCount="" Body="&lt;p&gt;Here are some:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;http://www.mathhelpforum.com&quot; rel=&quot;nofollow&quot;&gt;http://www.mathhelpforum.com&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;http://mathforum.org/dr/math/&quot; rel=&quot;nofollow&quot;&gt;http://mathforum.org/dr/math/&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" LastActivityDate="2010-11-15T21:25:27.040" />
  <row Id="10475" PostTypeId="1" AcceptedAnswerId="10481" CreationDate="2010-11-15T21:30:39.987" Score="3" ViewCount="116" Body="&lt;p&gt;While reading about Moufang loops in the book &quot;An introduction to Quasigroups and their Representations&quot; by Smith, I've encountered the following statement:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The set $ S^7 $ of nonzero octonions of norm 1 forms a Moufang loop under multiplication. Geometrically, this set is a 7-sphere.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;While I understand why this set indeed forms a Moufang loop, I'm not sure how it is viewed as a sphere, or what is the general connection between Moufang loops and this geometrical point of view. Could anyone elaborate?&lt;/p&gt;&#xA;" OwnerUserId="766" LastEditorUserId="766" LastEditorDisplayName="" LastEditDate="2010-11-19T10:11:36.787" LastActivityDate="2010-11-19T10:11:36.787" Title="Nonzero Octonions as a 7-sphere" Tags="&lt;abstract-algebra&gt;&lt;octonions&gt;&lt;loops&gt;&lt;quasigroups&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="10476" PostTypeId="2" ParentId="6716" CreationDate="2010-11-15T21:31:26.457" Score="1" ViewCount="" Body="&lt;p&gt;Perhaps consider a full subcategory of Set with just the functions you are allowed to create given ZF. Then find some function that in normal set theory requires AC. This should do the trick. Or use the topos of constructible sets and functions, same thing.&lt;/p&gt;&#xA;" OwnerUserId="3518" LastActivityDate="2010-11-15T21:31:26.457" CommentCount="1" />
  <row Id="10477" PostTypeId="1" CreationDate="2010-11-15T21:38:11.007" Score="1" ViewCount="245" Body="&lt;p&gt;Calculate N = 9 + 99 +999 + ... + 999 ... 99 (The last part consists of 999 digits equal to 9.)&lt;/p&gt;&#xA;" OwnerUserId="2938" LastActivityDate="2010-11-15T23:33:54.590" Title="N= 9 + 99 + 999 + ... 999...99" Tags="&lt;number-theory&gt;" AnswerCount="4" CommentCount="6" />
  <row Id="10478" PostTypeId="2" ParentId="10477" CreationDate="2010-11-15T21:45:00.623" Score="5" ViewCount="" Body="&lt;p&gt;$\displaystyle N = \sum_{k=1}^{999} (10^k - 1) = \dfrac{10(10^{999}-1)}{9} - 999 $&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-11-15T21:45:00.623" />
  <row Id="10479" PostTypeId="2" ParentId="10477" CreationDate="2010-11-15T21:46:30.487" Score="2" ViewCount="" Body="&lt;p&gt;$N = 9 + 99 + 999 + ... = $&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(10 - 1) + (100 - 1) + (1000 - 1)+ ... =$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(10 + 100 + 1000 + ...) - 999 = $&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$111...1110 - 999 = $&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$111...10111$ &lt;/p&gt;&#xA;" OwnerUserId="272" LastActivityDate="2010-11-15T21:46:30.487" />
  <row Id="10480" PostTypeId="2" ParentId="10477" CreationDate="2010-11-15T21:49:41.060" Score="2" ViewCount="" Body="&lt;p&gt;The answer is&#xA;\begin{eqnarray}&#xA;N = \sum_{k = 1}^{999} 10^{k} - 1 = \frac{10(10^{999}-1)}{10-1}  -999,&#xA;\end{eqnarray}&#xA;which consists of nine-hundred and ninety-nine $1$s and one $0$ in the thousands place.&lt;/p&gt;&#xA;" OwnerUserId="2720" LastEditorUserId="2720" LastEditorDisplayName="" LastEditDate="2010-11-15T21:56:58.743" LastActivityDate="2010-11-15T21:56:58.743" />
  <row Id="10481" PostTypeId="2" ParentId="10475" CreationDate="2010-11-15T21:50:24.360" Score="6" ViewCount="" Body="&lt;p&gt;The real  octonions are a &lt;a href=&quot;http://en.wikipedia.org/wiki/Normed_division_algebra&quot; rel=&quot;nofollow&quot;&gt;normed division algebra&lt;/a&gt;, and in such a thing you always have a &lt;em&gt;unit sphere&lt;/em&gt;. The unit sphere, because of multiplicativity of the norm, is always closed under the product.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The title of your question, though, refers to the &lt;em&gt;nonzero&lt;/em&gt; octonions, and that is surly not a sphere (it has the homotopy type of a sphere, and in fact the unit sphere is a strong deformation retract---but I guess that if you know what this term means then you also are aware of this fact!)&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2010-11-15T21:50:24.360" />
  <row Id="10482" PostTypeId="2" ParentId="10379" CreationDate="2010-11-15T21:52:12.010" Score="1" ViewCount="" Body="&lt;p&gt;Both are correct, as well as saying that the root is &lt;em&gt;degenerate&lt;/em&gt; (with multiplicity of two).&lt;/p&gt;&#xA;" OwnerUserId="272" LastActivityDate="2010-11-15T21:52:12.010" />
  <row Id="10483" PostTypeId="2" ParentId="10422" CreationDate="2010-11-15T21:53:32.470" Score="2" ViewCount="" Body="&lt;p&gt;I think this is a counterexample (assuming I haven't messed up something; I spent a fair amount of time trying to prove the result is true, reducing and reducing and reducing, and then the distillation suggested this; I &lt;em&gt;really&lt;/em&gt; hope I didn't mess it up):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take $G=C_4\times C_{16}$, with $C_4$ generated by $x$ and $C_{16}$ generated by $y$. Take $H=\langle (x,y^2)\rangle$ and $K=\langle(x,y^4)\rangle$. Then $G/H$ is generated by the images of $(x,1)$ and $(1,y)$. But $(x,1)H = (1,y^{-2})H=(1,y)^{-2}H$, so $G/H$ is generated by $(1,y)H$ and hence is cyclic (and since $G$ is a $p$-group, necessarily primary cyclic). Similarly, $(x,1)K = (1,y^{-4})K$, so $G/K$ is generated by $(1,y)K$ and hence is cyclic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose that $(x,y^2)^r = (x,y^4)^s$ for some integers $r$ and $s$; then $r\equiv s\pmod{4}$ and $2r\equiv 4s\pmod{16}$. But if $2r\equiv 4s\pmod{16}$, then $r\equiv 2s\pmod{8}$, hence $s\equiv r\equiv 2s\pmod {4}$.  The only possibility is $s\equiv 0 \pmod{4}$, hence $x^s=1$. But we also now have $4s\equiv 0 \pmod{16}$, so $y^{4s}=1$. Therefore, $(x,y^4)^s = (x^s,y^{4s}) = (1,1)$. Therefore, $H\cap K=\{(1,1)\}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, $H$ and $K$ are both contained in $\langle(x,1),(1,y^2)\rangle$, so $HK$ is a proper subgroup of $G$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added:&lt;/strong&gt; Some comments: your hypothesis imply in any case that $G$ is a product of two primary cyclic groups. To see this, note that if $G/H$ is $p$-primary, and $G/K$ is $q$-primary, then $H$ must contain all $p&amp;#39;$-parts of $G$, and $K$ must contain all $q&amp;#39;$-primary parts of $G$, so $H\cap K=1$ implies that there is no prime which is different to both $p$ and $q$. If $p\neq q$, then the $p$-part of $G$ is a direct sum of $p$-primary cyclic groups, the $q$-part a sum of $q$-primary cyclic groups. Say there are two $p$-primary cyclic groups; then $H$ must contain the entire $q$-part as a subgroup, and have a nontrivial intersection with the $p$-part; since $K$ contains the $p$-part, $H\cap K$ would be nontrivial. Symmetrically with $q$. Thus, if $p\neq q$, then $G=C_{p^a}\times C_{q^b}$, and $H=C_{q^b}$, $K=C_{p^a}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $p=q$, then $G$ is a $p$-group. Then $K \cong K/(H\cap K) = HK/H$, which is a subgroup of $G/H$, so $K$ is cyclic. Symmetrically, $H$ is cyclic. Thus, $G$ can be generated by $2$ elements, so the decomposition of $G$ into a direct factors of cyclic groups has at most two factors. But $G$ cannot be cyclic and and $p$-group, because then any two nontrivial subgroups have nontrivial intersection, contradicting the hypothesis. Thus, $G=C_{p^a}\times C_{p^b}$ in this case as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, the conditions require that $G=C_{p^a}\times C_{q^b}$ with $p$ and $q$ primes, possibly equal, and $a$ and $b$ positive.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-16T03:45:32.357" LastActivityDate="2010-11-16T03:45:32.357" />
  <row Id="10484" PostTypeId="1" CreationDate="2010-11-15T22:03:10.377" Score="0" ViewCount="47" Body="&lt;p&gt;Let $R$ be a commutative semiring and let $I$ be an $R$-ideal.  Suppose $X \subseteq R$ satisfies the following conditions:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$I \subseteq X$&#xD;&lt;/li&gt;&#xA;&lt;li&gt;$a \in X\ \&amp;amp;\ b \in X \Leftrightarrow ab \in X$&#xD;&lt;/li&gt;&#xA;&lt;li&gt;$a+b \in X\ \&amp;amp;\ b \in I \Rightarrow a \in X$.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Does such a subset have a standard name?  &lt;/p&gt;&#xA;" LastActivityDate="2010-11-15T22:24:55.203" Title="Does this semiring-theoretic concept have a name?" Tags="&lt;abstract-algebra&gt;" CommentCount="4" />
  <row Id="10486" PostTypeId="2" ParentId="10475" CreationDate="2010-11-15T22:22:57.097" Score="2" ViewCount="" Body="&lt;p&gt;Mariano's answer is of course right on the money.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since I myself happen to currently have my library's copy of Conway and Smith's book &lt;em&gt;Quaternions and Octonions&lt;/em&gt;, let me recommend this book to you as well.  In general, throughout the book they continually emphasize a geometric approach to the algebraic structures mentioned in the title, so it would be a good place to read up on the topic of your question.  &lt;/p&gt;&#xA;" OwnerUserId="299" LastActivityDate="2010-11-15T22:22:57.097" />
  <row Id="10487" PostTypeId="2" ParentId="10414" CreationDate="2010-11-15T22:24:24.813" Score="1" ViewCount="" Body="&lt;p&gt;How about this algorithm?  At each moment your decision is which spell to cast next.  Having decided, if the cool down time for that one has expired, then go ahead and cast it.  If not, wait until it has expired, then cast it.  To make the decision, rank the spells in damage/time where time is from now to completion (including any cool down time left).  Then see if you can fit in one of the other spell without delaying the best one.  Taking Qiochu's example, at the start spell 1 is 300/5=60 dam/sec and spell 2 is 50/3=16.7dam/sec.  Having cast spell 1, it is now 300/17=17.6 dam/sec, still better than 2.  But we can get 2 done twice before 1 is available, so we should do that.  The question would be whether you can change the parameters so that you should use the weaker spell even though it will delay the strong one by just a little bit.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-11-15T22:24:24.813" CommentCount="1" />
  <row Id="10488" PostTypeId="2" ParentId="8814" CreationDate="2010-11-15T22:34:47.463" Score="4" ViewCount="" Body="&lt;p&gt;Ah, this is one identity which comes into use for proving the &lt;em&gt;Euler's Partition Theorem&lt;/em&gt;. The identity is as follows: $$ (1+x)(1+x^{2})(1+x^{3}) \cdots  = \frac{1}{(1-x)(1-x^{3})(1-x^{5}) \cdots}$$&lt;/p&gt;&#xA;" LastActivityDate="2010-11-15T22:34:47.463" CommunityOwnedDate="2010-11-15T22:34:47.463" />
  <row Id="10489" PostTypeId="2" ParentId="10429" CreationDate="2010-11-15T22:40:04.807" Score="1" ViewCount="" Body="&lt;p&gt;An empty set contains nothing. A power set of some set $A$, $\mathscr{P}(A)$  always contains something, at least $A$ itself, thus $\mathscr{P}(A)$ is not empty. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason I decided to add my two cents is to point out to the frequent source of confusion which is associating empty set with zero, and then associating zero with &quot;nothing&quot;. Then, as the (faulty) reasoning goes, the set that contains only the empty set contains nothing, and therefore it is the empty set. The major cause of such confusion is mostly due to notation: designating the empty set as $\emptyset$, which looks almost like $0$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My rule of thumb: if in doubt, replace $\emptyset$ with $\{ \, \}$ and then think of $\{ \, \}$ as a container that contains nothing. Empty container. Clearly, the empty set. But, say, we have some other set $A$ that contains only the empty set. Again, $A$ is a container, but now it contains another container that contains nothing: $\{ \ \{ \, \} \ \}$. Clearly, a set $A$ contains &lt;em&gt;something&lt;/em&gt;, so it's not empty. However writing the same thing as $\{ \ \emptyset \ \}$ may cause some confusion to those exploring the set theory: is that zero the set  $A$ contains? Looks like zero, anyway! But zero is nothing! So the set contains nothing. So it is the empty set. &lt;/p&gt;&#xA;" OwnerUserId="2450" LastActivityDate="2010-11-15T22:40:04.807" />
  <row Id="10490" PostTypeId="1" AcceptedAnswerId="10493" CreationDate="2010-11-15T22:40:49.673" Score="12" ViewCount="539" Body="&lt;p&gt;&lt;strong&gt;From WIKI:&lt;/strong&gt; In calculus and other branches of mathematical analysis, an indeterminate form is an algebraic expression obtained in the context of limits. Limits involving algebraic operations are often performed by replacing subexpressions by their limits; if the expression obtained after this substitution does not give enough information to determine the original limit, it is known as an indeterminate form. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The indeterminate forms include $0^{0},\frac{0}{0},(\infty - \infty),1^{\infty}, \ \text{etc}\cdots$&#xD;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;My question is can anyone give me a &lt;em&gt;nice explanation&lt;/em&gt; of why $1^{\infty}$ is considered to be an indeterminate form? Because, i don't see any justification of this fact. I am still perplexed.&lt;/p&gt;&#xA;" OwnerUserId="9413" LastEditorUserId="9413" LastEditorDisplayName="" LastActivityDate="2010-11-16T16:28:16.430" Title="Why is $1^{\infty}$ considered to be an indeterminate form" Tags="&lt;calculus&gt;&lt;real-analysis&gt;&lt;analysis&gt;" AnswerCount="3" CommentCount="6" FavoriteCount="3" />
  <row Id="10491" PostTypeId="2" ParentId="10414" CreationDate="2010-11-15T22:41:18.587" Score="0" ViewCount="" Body="&lt;p&gt;Let's rephrase to: what's the maximal damage that I can cause in, say, 100 seconds? The constraints then become:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Can't spend more than 100 seconds casting&lt;/li&gt;&#xA;&lt;li&gt;Can't cast a single spell more than $\frac{100}{cast\ time + cool\ time}$ times&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Let $n_i$ be the number of times I cast spell i. I want to maximize $\sum_i n_i damage_i$ subject to $\sum_i n_i cast_i &amp;lt; 100$ and each $n_i(cast_i + cool_i) &amp;lt; 100$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This can be rewritten to a linear programming problem as follows:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The first constraint is $[cast_0 \dots cast_m] [n_0 \dots n_m]^T&amp;lt;100$&#xD;&lt;/li&gt;&#xA;&lt;li&gt;The other constraints are $[0 \dots (cast_i + cool_i) \dots 0] [n_0 \dots n_m]^T &amp;lt; 100$&#xD;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;EDIT&lt;/em&gt;: My assumptions are&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;You cannot cast two spells at the same time. Constraint #1 ensures that this is true.&lt;/li&gt;&#xA;&lt;li&gt;You can only cast a spell once every $cast + cool$ period. Constraint #2 ensures this is true. (Note that this does &lt;em&gt;not&lt;/em&gt; say you cannot cast a &lt;em&gt;different&lt;/em&gt; spell in this period; only that you cannot cast the &lt;em&gt;same&lt;/em&gt; spell in this period.)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="2549" LastEditorUserId="2549" LastEditorDisplayName="" LastEditDate="2010-11-16T01:36:56.903" LastActivityDate="2010-11-16T01:36:56.903" CommentCount="12" />
  <row Id="10492" PostTypeId="2" ParentId="10389" CreationDate="2010-11-15T22:44:33.120" Score="2" ViewCount="" Body="&lt;p&gt;Do it by induction. First, let's set the constant in $\theta$ to 1, so we get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$T(n)=2T(n/2)+T(n/3)+n^2$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, assume $T(m) \leq k m^2$ for some constant $k$ for all $m &amp;lt; n$. Then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$T(n) = 2 T(n/2) + T(n/3) + n^2 \leq k n^2/2 + k n^2/9 + n^2 \leq (11 k /18 +1) n^2 $&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and if $k = 3$, say, we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$T(n) \leq (11/6+1) n^2 \leq 3 n^2.$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so if our bound holds for all $m &amp;lt; n$, then our bound holds for $n$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To make this a real proof, you would need to take the case when $n$ is not a multiple of 6 into account, but the details will work out when you do them carefully.&lt;/p&gt;&#xA;" OwnerUserId="2501" LastActivityDate="2010-11-15T22:44:33.120" />
  <row Id="10493" PostTypeId="2" ParentId="10490" CreationDate="2010-11-15T22:44:54.653" Score="30" ViewCount="" Body="&lt;p&gt;Forms are indeterminate because, depending on the specific expressions involved, they can evaluate to different quantities.  For example, all of the following limits are of the form $1^{\infty}$, yet they all evaluate to different numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\lim_{n \to \infty} \left(1 + \frac{1}{n^2}\right)^n = 1$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\lim_{n \to \infty} \left(1 + \frac{1}{n}\right)^n = e$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\lim_{n \to \infty} \left(1 + \frac{1}{\ln n}\right)^n = \infty$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To expand on this some (and this thought process can be applied to other indeterminate forms, too), one way to think about it is that there's a race going on between the expression that's trying to go to 1 and the expression that's trying to go to $\infty$.  If the expression that's going to 1 is in some sense faster, then the limit will evaluate to 1.  If the expression that's going to $\infty$ is in some sense faster, then the limit will evaluate to $\infty$.  If the two expressions are headed toward their respective values at essentially the same rate, then the two effects sort of cancel each other out and you get something strictly between 1 and $\infty$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are some other cases, too, like &#xA;$$\lim_{n \to \infty} \left(1 - \frac{1}{\ln n}\right)^n = 0,$$&#xD;&#xA;but this still has the expression going to $\infty$ &quot;winning.&quot;  Since $1 - \frac{1}{\ln n}$ is less than 1 (once $n &gt; 1$), the exponentiation forces the limit to 0 rather than $\infty$.&lt;/p&gt;&#xA;" OwnerUserId="2370" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-11-16T16:28:16.430" LastActivityDate="2010-11-16T16:28:16.430" CommentCount="3" />
  <row Id="10494" PostTypeId="1" AcceptedAnswerId="10523" CreationDate="2010-11-15T22:55:34.077" Score="0" ViewCount="104" Body="&lt;p&gt;so I dont know if this is really what the website is meant for so let me know.  I have a midterm on abstract algebra (intro) and I've been doing practice problems and I just need to know whether my answers are correct and if my way of doing them is right. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Does the following have a solution: $x^2 \equiv 1 \pmod{3}$&#xD;&#xA;Well we know that a solution exists if $\mathrm{gcd}(a, n) | b$ where $a = x^2$, $b = 1$ and $n = 3$.&#xA;we can rewrite the equation as $1 \equiv x^2 \pmod{3}$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) If $a \equiv 3 \pmod{4}$ prove that there are no integers c, d such that $a = c^2 + d^2$&#xD;&#xA;By definition we can write this as $a = 4n + 3, n \in \mathbb{Z}$. so does this prove it? since $4n + 3$ can never be in the form of $c^2 + d^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) Show that no perfect square has 2, 3, 7, 8 as its last digit. The way we did this was we listed out all the classes from in $\mathbb{Z}_{10}$ which are $[0]_{10}..[9]_{10}$ and showed that $a^2 (0, 1, 4, 16, 25 ...)$ are never an element of the $[2], [3], [7], [8]$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4) If $a \in \mathbb{Z}$, prove that last digit of $a^4$ is $0, 1, 5, \text{or } 6$&lt;/p&gt;&#xA;" OwnerUserId="287" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-16T03:55:56.670" LastActivityDate="2010-11-16T04:25:43.700" Title="Congruency proofs" Tags="&lt;abstract-algebra&gt;&lt;elementary-number-theory&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="10495" PostTypeId="1" AcceptedAnswerId="10503" CreationDate="2010-11-15T22:57:31.910" Score="9" ViewCount="166" Body="&lt;h3&gt;Background&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Recall that an integral scheme $X$ is a scheme which is both irreducible and reduced; equivalently, its ring of functions is an integral domain on every open subset.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given any point $p$, there is a local ring $R_p$ at $p$, which is given by localizing the ring of functions $R_U$ on any affine neighborhood $U$ of $p$ at the prime corresponding to $p$.  This local ring then has a residue field $K_p/p$.  The &lt;strong&gt;characteristic&lt;/strong&gt; of $p$ is then the characteristic of the residue field at $p$.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Question&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Its not too hard to come up with integral schemes where the characteristic of points jumps around.  The standard example of $Spec(\mathbb{Z})$ has a closed point of every positive characteristic, and the unique non-closed point (the generic point) has characteristic 0 (since the residue field is $\mathbb{Q}$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we don't require connectedness, then the disjoint union of $Spec(\mathbb{Q})$ and $Spec(\mathbb{Z}/2)$ has closed points of order 0 and 2, respectively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I have been playing with examples, and I can't seem to come up with an example of an integral scheme that has &lt;em&gt;closed&lt;/em&gt; points of both kinds.  &lt;strong&gt;Can an integral scheme have closed points of both positive and zero characteristic?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would be curious to see such an example, since there is a wide gap in my intuition between schemes whose closed points have positive characteristic (which are inherently arithmetic) and schemes whose closed points have characteristic zero (which are either geometric, or certain arithmetic localizations).&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Algebraic Version of Question (No Schemes Needed)&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;I should mention, though my motivation and curiosity are geometric in origin, the problem has an algebraic version.  &lt;strong&gt;Is there an integral domain $R$, and two maximal ideals $m$ and $n$, such that $R/m$ has positive characteristic, and $R/n$ has zero characteristic?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="1062" LastActivityDate="2010-11-16T01:07:26.177" Title="Can an integral scheme have closed points of both positive and zero characteristic?" Tags="&lt;algebraic-geometry&gt;&lt;schemes&gt;" AnswerCount="1" FavoriteCount="2" />
  <row Id="10497" PostTypeId="2" ParentId="10477" CreationDate="2010-11-15T23:33:54.590" Score="1" ViewCount="" Body="&lt;p&gt;$\rm\phantom{\ 10\ S -} S\  =\ \ 9\cdots 9\ +\ \cdots\ +\ 999\ +\ 99\ +\ 9$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\ 10\ S - S\ =\ \ 9\cdots 90\ +\ \cdots +\ (990-999)\ +\ (90-99)\ -\ 9  $&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad \Rightarrow\ \  9\ S\ =\ \ \: 9\cdots90\  +\ 999\ (-9)$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad \Rightarrow\quad\ \ S\ =\ \ 1\cdots10\ - \ 999$&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-11-15T23:33:54.590" />
  <row Id="10498" PostTypeId="1" AcceptedAnswerId="10522" CreationDate="2010-11-15T23:34:49.423" Score="0" ViewCount="75" Body="&lt;p&gt;So some of you may have noticed my algebra exam is tomorrow!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is: If $[a]_{n} = [1]_{n}$ prove that $\mathrm{gcd}(a, n) = 1$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well I tried it one way. We know the above means the following: $a \equiv 1 \pmod{n}$. This is also written as $1 \equiv a \pmod{n}$&#xD;&#xA;Then the gcd of these are  as follows:&#xA;$$\mathrm{gcd}(a, n) := d  $$ and &#xA;$$\mathrm{gcd}(1, n) := d&amp;#39; $$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;At this point I thought $ d = d&amp;#39;$ which is NOT the case as pointed out by roommate. but lets assume it does so I finished the proof as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathrm{gcd}(a,n) = \mathrm{gcd}(1, n)$$&#xD;&#xA;$$\mathrm{gcd}(a, n) = 1$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but then obviously that's not the case (or is it). I then tried the following:&#xA;$$d = as + nt,  \quad s, t \in \mathbb{Z}$$ &#xA;$$d&amp;#39; = as&amp;#39; + nt&amp;#39;,  \quad s&amp;#39;, t&amp;#39; \in \mathbb{Z}$$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And now I'm stuck again :(&lt;/p&gt;&#xA;" OwnerUserId="287" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-16T03:48:11.560" LastActivityDate="2010-11-16T03:53:03.677" Title="relating GCD, congruency and congruent classes." Tags="&lt;abstract-algebra&gt;&lt;elementary-number-theory&gt;" AnswerCount="2" />
  <row Id="10499" PostTypeId="2" ParentId="10494" CreationDate="2010-11-15T23:42:13.077" Score="0" ViewCount="" Body="&lt;p&gt;For 1) you have just rewritten the equation backwards, so have not proved anything.  You can use the same approach as 3)-square each number mod 3 and see if any square to 1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For 2) why do you know 4n+3 cannot be the sum of two squares?  You need to list the numbers mod 4 that are squares, then find that no pair of them sums to 3&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For 4) do it just like 3, but using the fourth power.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-11-15T23:42:13.077" CommentCount="2" />
  <row Id="10500" PostTypeId="2" ParentId="10498" CreationDate="2010-11-15T23:43:52.737" Score="2" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\rm\ \ d\:|\:n,\:1+kn\ \Rightarrow\ d\:|\: 1$&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-11-15T23:43:52.737" CommentCount="2" />
  <row Id="10501" PostTypeId="2" ParentId="10490" CreationDate="2010-11-15T23:48:06.637" Score="1" ViewCount="" Body="&lt;p&gt;Here is some intuitive explanation, suitable also for non-mathematicians. Suppose that an imaginary basketball player has a probability $p = 0.999$ of making a free throw. The probability that he makes $10000$ free throws in a row is very small, the probability that he makes $100$ is high, and the probability that he makes $1000$ is approximately $e^{-1}$.&lt;/p&gt;&#xA;" OwnerUserId="2810" LastActivityDate="2010-11-15T23:48:06.637" CommentCount="2" />
  <row Id="10502" PostTypeId="2" ParentId="10494" CreationDate="2010-11-15T23:55:47.340" Score="2" ViewCount="" Body="&lt;p&gt;1) prime $\rm\: p\:|\:(x-1)(x+1)\ \Rightarrow\ p|(x-1)\ \ or\ \ p|(x+1)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) $\rm\ even^2\equiv 0,\ odd^2\equiv 1\ (mod\ 4)\ $ so two squares can't have sum $\equiv 3$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4) By 3) $\rm\ a^2\not\equiv \{\pm2,\:\pm3\}\ \Rightarrow\ a^2 \equiv \{0,\:\pm1,\:\pm4,\:5\}\ \Rightarrow\ a^4\equiv \{0,1,6,5\}\ \:(mod\ 10)$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-11-15T23:55:47.340" />
  <row Id="10503" PostTypeId="2" ParentId="10495" CreationDate="2010-11-16T00:24:17.553" Score="9" ViewCount="" Body="&lt;p&gt;Yes, but you're not going to like it.  Let $R = \mathbb{Z}[x_1, x_2, ... ]$.  Let $q_1, q_2, ... $ be an enumeration of the rationals; then the map $R \to \mathbb{Q}$ which sends $x_i$ to $q_i$ is surjective, so its kernel is a maximal ideal $n$ such that $R/n$ has characteristic zero.  On the other hand, let $m = (p, x_1, x_2, ...)$.  Then $R/m \simeq \mathbb{F}_p$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt;  Okay, so here is a Noetherian example.  Let $R&amp;#39;$ denote the localization of $\mathbb{Z}$ at $p$ (since $\mathbb{Z}_p$ generally means something else) and let $R = R&amp;#39;[x]$.  Then the map $R \to \mathbb{Q}$ which sends $x$ to $\frac{1}{p}$ is surjective, and so is the map $R \to \mathbb{F}_p$ which sends $x$ to, say, $1$.  It seems interesting to try to picture $\text{Spec } R$; it might help to stare at &lt;a href=&quot;http://matrix.cmi.ua.ac.be/DATA2/mumforddrawing.jpg&quot; rel=&quot;nofollow&quot;&gt;Mumford's picture of $\text{Spec } \mathbb{Z}[x]$&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-11-16T01:07:26.177" LastActivityDate="2010-11-16T01:07:26.177" CommentCount="7" />
  <row Id="10504" PostTypeId="1" AcceptedAnswerId="10508" CreationDate="2010-11-16T00:37:41.837" Score="0" ViewCount="105" Body="&lt;p&gt;If $X,Y$ and $Z$ have identical distributions and are independent, and can assume $0$ or $1$, what is $\text{Var}(XYZ)$? So this is either $0$ or $1$. It's $1$ with probability $1/8$ and $0$ with probability $7/8$. This means that $\text{Var}(XYZ) = 7/16$? Or is this not right? How does this use the definition of variance? We could note that $E(X) = E(X^2)$, etc....?&lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2010-11-16T00:47:12.873" Title="Variance and identical random variables" Tags="&lt;probability&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="10506" PostTypeId="2" ParentId="10504" CreationDate="2010-11-16T00:45:18.973" Score="2" ViewCount="" Body="&lt;p&gt;You imply that $X, Y$, and $Z$ have equal chance of being 0 or 1.  In that case $E((XYZ)^2)=\frac{1}{8}$ and $(E(XYZ))^2=\frac{1}{64}$.  So the variance would be $\frac{7}{64}$&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-11-16T00:45:18.973" />
  <row Id="10507" PostTypeId="1" CreationDate="2010-11-16T00:46:01.780" Score="2" ViewCount="131" Body="&lt;p&gt;Given the natural numbers bbb...b (n digits equal to b) and bbbb...b(m digits equal to b), show that the greatest common divisor of these numbers is bbb ... b (M digits equal to b, where M is the greatest common divisor of n and m).&lt;/p&gt;&#xA;" OwnerUserId="2938" LastActivityDate="2010-11-16T02:46:02.697" Title="gcd(bbb...b, bbb...b)" Tags="&lt;number-theory&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="10508" PostTypeId="2" ParentId="10504" CreationDate="2010-11-16T00:47:12.873" Score="3" ViewCount="" Body="&lt;p&gt;$Var(X) = E[X^2] - E[X]^2$. Assuming each of the variables take 0 or 1 equally likely, as you noted, $E[XYZ] = E[X^2Y^2Z^2] = 1/8$. So $Var(X) = 1/8 - 1/64 = 7/64$.&lt;/p&gt;&#xA;" OwnerUserId="1527" LastActivityDate="2010-11-16T00:47:12.873" />
  <row Id="10509" PostTypeId="2" ParentId="10507" CreationDate="2010-11-16T00:49:13.140" Score="5" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\ $ The Euclidean algorithm computing their GCD mimicks that for $\rm\: gcd(m,n)\:$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Put $\rm\ 1^{[n]}\: :=\ 1\cdots 1\ \ (n\ 1&amp;#39;s)\ =\ (10^n-1)/9\:.\ $ Then your gcd is $\rm\ b\ gcd(1^{[m]},\:1^{[n]}) $ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But $\rm\ 1^{[m]}-1^{[n]}\ =\  1^{[m-n]}\cdot 10^n\ \ \ $ [e.g. $\rm\ 11111 - 111\ =\ 11000\ =\ 1^{[2]}\cdot 10^3\ $]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and $\rm\ gcd(10,1^{[n]}) = 1\ $ since $\rm\ 9\cdot 1^{[n]}\ =\ 10^n-1\ $ and $\rm\ \rm\ gcd(10,\:10^n-1) = 1\:$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, choosing notation so that $\rm\:m\ge n\:,\ $ and applying the above observations we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ \rm\quad\quad\:\ gcd(1^{[m]},1^{[n]})\ =\ gcd(1^{[m]}-1^{[n]},\:1^{[n]})\ =\ gcd(1^{[m-n]}\cdot 10^n,\:1^{[n]})\ =\  gcd(1^{[m-n]}\:,\:1^{[n]})$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So $\rm\ \ gcd(1^{[m]},1^{[n]})\ =\ gcd(1^{[m-n]}\:,\:1^{[n]})\ $ just like $\rm\ gcd(m,n)\ =\ gcd(m-n,n)\: $. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So $\rm\ \ gcd(1^{[m]},1^{[n]})\ =\ 1^{[gcd(m,n)]}\ $ follows from the above by a simple induction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a full proof and some further insight see my posts in this &lt;a href=&quot;http://math.stackexchange.com/questions/7473/number-theory-proving-question&quot;&gt;prior question.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-11-16T02:46:02.697" LastActivityDate="2010-11-16T02:46:02.697" />
  <row Id="10510" PostTypeId="2" ParentId="10414" CreationDate="2010-11-16T01:07:03.400" Score="15" ViewCount="" Body="&lt;p&gt;It is worth noting that, in extreme special cases, this problem degenerates to the &lt;a href=&quot;http://en.wikipedia.org/wiki/Knapsack_problem&quot; rel=&quot;nofollow&quot;&gt;Knapsack Problem&lt;/a&gt;, which is NP-complete to solve exactly.  To see this, imagine that there is one spell, henceforth called the Megaspell, which does a very, very large amount of damage, has zero casting time, and has some positive cooldown $n$.  If all the other spells do much less damage than the Megaspell, it will clearly be optimal to cast the Megaspell every $n$ seconds and then optimize the cooldown time with the weaker spells.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, assume all the other spells also have cooldown $n$.  If one optimizes a given $n$-second downtime with these spells, then the same spell-sequence will also be possible in the next $n$-second downtime, and so we can assume the solution is $n$-periodic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem then reduces to optimizing the $n$ available seconds with the lesser spells, each of which may only be cast once.  If one replaces casting time with 'weight' and damage with 'value', one recovers the Knapsack Problem for maximum weight n.&lt;/p&gt;&#xA;" OwnerUserId="1062" LastActivityDate="2010-11-16T01:07:03.400" CommentCount="8" />
  <row Id="10512" PostTypeId="1" AcceptedAnswerId="10517" CreationDate="2010-11-16T01:22:39.143" Score="6" ViewCount="205" Body="&lt;p&gt;Let $G$ be a (finite?) group. By definition, the Eilenberg-MacLane space $K(G,1)$ is a CW complex such that $\pi_1(K(G,1)) = G$ while the higher homotopy groups are zero. One can consider the singular cohomology of $K(G,1)$, and it is a theorem that this is isomorphic to the group cohomologies $H^*(G, \mathbb{Z})$. According to one of my teachers, this can be proved by an explicit construction of $K(G, 1)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, it seems like there ought to be a categorical argument. $K(G, 1)$ is the object that represents the functor $X \to H^1(X, \mathbb{Z})$ in the category of pointed CW complexes, say, while the group cohomology consists of the universal $\delta$-functor that begins with $M \to M^G$ for $M$ a $G$-module. In particular, I would be interested in a deeper explanation of this &quot;coincidence&quot; that singular cohomology on this universal object happens to equal group cohomology.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there one?&lt;/p&gt;&#xA;" OwnerUserId="536" LastActivityDate="2010-12-02T03:49:56.953" Title="Why is the cohomology of a $K(G,1)$ group cohomology?" Tags="&lt;algebraic-topology&gt;&lt;group-cohomology&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="3" />
  <row Id="10513" PostTypeId="2" ParentId="10337" CreationDate="2010-11-16T01:37:28.773" Score="7" ViewCount="" Body="&lt;p&gt;As I mentioned in the comments section of Arturo's post, $\mathbb{Q}$ is as large as possible among countable linearly ordered sets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Said another way, if $X$ is countable an linearly ordered, then there is an order preserving injection $X\rightarrow\mathbb{Q}$.  So, if one (partially) orders the set of all countable linearly ordered sets by $X\leq Y$ iff there is a 1-1 order preserving map from $X$ to $Y$, then $\mathbb{Q}$ is the largest element of this set.  (To be frank, I'm not positive this satisfies the antisymmetry condition of being a partial order, but whatever).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The proof is quite constructive.  The key properties of $\mathbb{Q}$ are that it has no first or last element, and inbetween any two elements of $\mathbb{Q}$ there is another one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using these properties, the proof goes as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;X is countable so list the elements $X=\{x_1,x_2,...\}$.  We will inductively define a function $f:X\rightarrow \mathbb{Q}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To start the induction, let $f(x_1)$ be any element of $\mathbb{Q}$.  Next, assume $f$ has been defined on $\{x_1,...,x_n\}$ with $f$ injective and order preserving.  We now define $f(x_{n+1})$.  If $x_{n+1}$ is smaller than all of the previous $x_i$, define $f(x_{n+1}) = q$ where $q$ is any rational number smaller than all of $f(x_1),...,f(x_n)$.  Such a $q$ certainly exists since the set $\{f(x_1),...,f(x_n)\}$ is finite.  Likewise, if $x_{n+1}$ is bigger than all the previous $x_i$, choose $f(x_{n+1})$ bigger than all of the $f(x_i)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, we assume $x_i &amp;lt; x_{n+1} &amp;lt; x_j$ for some previously some $x_i$ and $x_j$ in $\{x_1,...,x_n\}$.  Further, we choose $x_i$ from $\{x_1,...,x_n\}$ as large as possible but still less than that $x_{n+1}$, and likewise, choose $x_j$ as small as possible.  Then we define $f(x_{n+1})$ to be any rational number in between $f(x_i)$ and $f(x_j)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By induction, $f$ is defined for all $x_i$, and this $f$ is the desired injection.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With more care, one can show that if $X$ is dense in itself with no first or last element, then we can arrange for $f$ to be surjective, showing that these properties uniquely characterize $\mathbb{Q}$ up to isomorphism!   (The &quot;more care&quot; is to well order $\mathbb{Q}$ and at any stage where a choice is made, to choose the minimal rational number satisfies whatever it has to).&lt;/p&gt;&#xA;" OwnerUserId="331" LastActivityDate="2010-11-16T01:37:28.773" CommentCount="2" />
  <row Id="10514" PostTypeId="1" AcceptedAnswerId="10630" CreationDate="2010-11-16T01:38:36.183" Score="2" ViewCount="93" Body="&lt;p&gt;Hello,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $T$ be a distribution, $f$ be smooth. It is known that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$T(f(x - \cdot) ) \in C ^ \infty$ .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, for $\phi$ a test function&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$( T \ast f ) (\phi) = T( f( - \cdot ) \ast \phi ) = T( \int f( x - \cdot ) \phi( \cdot ) )$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;because the convolution is still a distribution. However, with regard to the first result we expect&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ T( \int f( x - \cdot ) \phi( \cdot ) ) = \int T ( f( x - \cdot ) \phi( \cdot ) )$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This intuitively makes sense, as $T$ is a distribution in the $x$-variable, whereas we integrate over the unnamed second variable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nevertheless, I don't know a proof of the last equality, and I am not familiar enough with the topologies of the smooth functions, test functions and distribution, to use approximative arguments. Can you help?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="3557" LastEditorUserId="3557" LastEditorDisplayName="" LastEditDate="2010-11-16T19:25:59.247" LastActivityDate="2010-11-16T22:57:00.057" Title="When to exchange integration and application of a distribution" Tags="&lt;functional-analysis&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="10515" PostTypeId="2" ParentId="10490" CreationDate="2010-11-16T01:40:01.297" Score="6" ViewCount="" Body="&lt;p&gt;Look at the logarithm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More specifically, consider $f(x)^{g(x)}$ as $x \to \infty$, where $\lim_{x \to \infty} g(x) = \infty$ and $\lim_{x \to \infty} f(x) = 1$. (This is something of form $1^\infty$.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now say $f(x) = e^{h(x)}$, so $h(x) = \log f(x)$. Then $\lim_{x \to \infty} h(x) = \lim_{x \to \infty} \log f(x) = \log \lim_{x \to \infty} f(x) = \log 1 = 0$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then &#xA;$$\lim_{x \to \infty} f(x)^{g(x)} = \lim_{x \to \infty} \exp (g(x) \log f(x)) = \exp \lim_{x \to \infty} (g(x) \log f(x)) $$&#xD;&#xA;and since the limit of a product is the product of the limits, that's&#xA;$$ \exp [ (\lim_{x \to \infty} g(x)) \cdot (\lim_{x \to \infty} \log f(x))] $$&#xD;&#xA;or&#xA;$$ \exp [ (\lim_{x \to \infty} g(x)) \cdot (\lim_{x \to \infty} h(x)) ]. $$&#xD;&#xA;But the first limit is infinity, and the second is zero. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the indeterminacy of $1^\infty$ follows directly from the indeterminancy of $\infty \cdot 0$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(The indeterminacy of $\infty^0$ actually follows in the same way, by taking the factors in the other order.)&lt;/p&gt;&#xA;" OwnerUserId="173" LastEditorUserId="173" LastEditorDisplayName="" LastEditDate="2010-11-16T01:51:21.923" LastActivityDate="2010-11-16T01:51:21.923" CommentCount="3" />
  <row Id="10516" PostTypeId="1" AcceptedAnswerId="10540" CreationDate="2010-11-16T02:02:15.710" Score="5" ViewCount="186" Body="&lt;p&gt;Let $\bar{\mathcal{P}}$ denote the closed, convex polytope with vertices at the origin and the positive rational points $(b_{1}, \dots, 0), \dots, (0, \dots, b_{n})$. Define the Ehrhart quasi-polynomial $L_{\mathcal{P}}(t) = |t \bar{\mathcal{P}} \cap \mathbb{Z}^{n}|$, which has the form: $\sum c_{k}(t) \ t^{k}$ with periodic functions $c_{k}(t)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question 1:&lt;/strong&gt; When is the Ehrhart quasi-polynomial a polynomial, i.e., the functions $c_{k}(t)$ are constants? Does this phenomenon occur only when vertices are at integral lattice points (i.e., $b_{i}$ are positive integers)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question 2:&lt;/strong&gt; Suppose I have an Ehrhart polynomial in an indeterminate $t$. What is the significance of the value of the function at rational (non-integral) $t$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question 3:&lt;/strong&gt; Suppose I'd like to count &lt;em&gt;positive solutions&lt;/em&gt; (instead of &lt;em&gt;non-negative solutions&lt;/em&gt;) of $\sum \frac{x_i}{b_i} \leq t$ with $t$ positive and fixed. Assuming that $b_{i}$ are positive integers, what is the corresponding &quot;Ehrhart-like&quot; polynomial in $t$ which enumerates the (positive) integral points in the $t$-dilate $t\bar{\mathcal{P}}$? Does it follow from a simple variable change in $t$ or $b_{i}$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(&lt;em&gt;Update&lt;/em&gt;) Here is an example of what I'm trying to do. Suppose I'd like to calculate the non-negative integer solutions of &#xA;\begin{eqnarray}&#xA;21 x_{1} + 14 x_{2} + 6 x_{3} \leq 1&#xA;\end{eqnarray}&#xA;(corresponding to the number of positive integer solutions of $21 x_{1} + 14 x_{2} + 6 x_{3} \leq 42$). Equivalently, by dividing through by the product $6 \cdot 14 \cdot 21 = 1764$, we can consider &#xA;\begin{eqnarray}&#xA;\frac{x_{1}}{84} + \frac{x_2}{126} + \frac{x_3}{294} \leq \frac{1}{1764}.&#xA;\end{eqnarray}&#xA;Here, $\mathbf{b} = (84, 126,294)$, so the corresponding polytope is &lt;em&gt;integral&lt;/em&gt;. The Ehrhart polynomial for $t$-dilates is &#xA;\begin{eqnarray}&#xA;L_{\bar{\mathcal{P}}}(t) = 1 + 231 t + 18522 t^{2} + 518616 t^{3},&#xA;\end{eqnarray}&#xA;but setting $t = \frac{1}{1764}$ gives a meaningless answer. My initial impression is that along with violating the requirement that $t$ must be an integer, what I am actually calculating is the number of lattice points in the $t$-dilate of the polytope defined by $\frac{1}{1764}$ replaced with $1$. Is there an interpolation scheme to correctly calculate the number of non-negative solutions of the first equation by finding the values of $L_{\bar{\mathcal{P}}}(0)$ and $L_{\bar{\mathcal{P}}}(1)$? Thoughts?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="2720" LastEditorUserId="2720" LastEditorDisplayName="" LastEditDate="2010-11-16T21:13:05.680" LastActivityDate="2010-11-16T21:45:51.680" Title="Counting Lattice Points with Ehrhart Polynomials" Tags="&lt;number-theory&gt;&lt;combinatorics&gt;" AnswerCount="2" FavoriteCount="2" />
  <row Id="10517" PostTypeId="2" ParentId="10512" CreationDate="2010-11-16T02:43:54.480" Score="7" ViewCount="" Body="&lt;p&gt;Akhil, you're thinking of this the opposite of how I think group cohomology was discovered.  The concept of group cohomology originally centered around the questions about the (co)homology of $K(\pi,1)$-spaces, by people like Hopf (he called them aspherical rather than $K(\pi,1)$ spaces, and Hopf preferred homology to cohomology at that point).  I think the story went that Hopf observed his formula for $H_2$ of a $K(\pi,1)$, which was a description of $H_2$ entirely in terms of the fundamental group of the space. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This motivated people to ask to what extent (co)homology is an invariant of the fundamental group of a $K(\pi,1)$-space. This was resolved by Eilenberg and Maclane.  Eilenberg and Maclane went the extra step to show that one can define cohomology of a group directly in terms of a group via what nowadays would be called a &quot;bar construction&quot; (ie skipping the construction of the associated $K(\pi,1)$-space).  Bar constructions exist topologically and algebraically and they all have a similar feel to them.  On the level of spaces, bar constructions are ways of constructing classifying spaces.  For groups they construct the cohomology groups of a group. The latter follows from the former -- if you're comfortable with the concept of the &quot;nerve of a category&quot;, this is how you construct an associated simplicial complex to a group (a group being a category with one object).  The simplicial (co)homology of this object is your group (co)homology.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dieudonne's &quot;History of Algebraic and Differential Topology&quot; covers this in sections V.1.D and V.3.B.   I don't think that answers all your questions but it answers some. &lt;/p&gt;&#xA;" OwnerUserId="642" LastActivityDate="2010-11-16T02:43:54.480" CommentCount="1" />
  <row Id="10518" PostTypeId="1" AcceptedAnswerId="10542" CreationDate="2010-11-16T03:09:07.210" Score="1" ViewCount="183" Body="&lt;p&gt;I'm approximating an integral with only exponentials.  i.e., it is equal to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle \int_{-\pi}^\pi{\frac{\displaystyle\sum_{j=a}^b{c_j e^{i\cdot d_j \cdot t}}}{\displaystyle\sum_{k=a}^b{r_k e^{i \cdot s_k \cdot t}}} dt}$&#xD;&#xA;where $c_j, d_j, r_k, s_k \in \mathbb{Z}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, it's a fraction of exponentials over exponentials.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I realized that we can integrate the individual terms of the power series expansion to integrate the expression.  I also found that for the particular integrals that I'm working with, only the first $N$ terms (for some $N$) significantly contribute to the value of the integral.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;My idea&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We can use feedback and recurrences to approximate this integral.  For example, the first two terms of the power series expansion of the function will be something like $c_0 + c_1x$.  If we attempt to integrate these terms, we should get $c_0x + \frac{c_1}{2}x$ evaluated at the two endpoints of the interval of integration.  But we can approximate this second function.  Suppose we take the original power series expansion ($c_0 + c_1x$), rewrite $x$ as $x/2$, and then multiply by $x$.  Bingo!  We have the integrated function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we can make &quot;guesses&quot; at the integral using the rewriting technique above.  Without going into detail, I then have a way to determine the error of the guess.  So we can introduce a feedback system to get closer and closer to the actual integral.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;My Question&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Does a method like this already exist?  I've seen techniques for approximating integrals using intervals, but this technique attempts to do it using feedback, instead.  Has a similar technique been used?  I would like to see the method, if it exists.&lt;/p&gt;&#xA;" OwnerUserId="2626" LastActivityDate="2010-11-16T08:58:37.310" Title="Is this integration approximation method known/used?" Tags="&lt;calculus&gt;&lt;intuition&gt;&lt;integral&gt;&lt;numerical-methods&gt;&lt;approximation&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" />
  <row Id="10519" PostTypeId="1" AcceptedAnswerId="10532" CreationDate="2010-11-16T03:40:39.923" Score="0" ViewCount="197" Body="&lt;p&gt;We have points A, B, C. On 2d plane. How having points coordinates (x, y) calculate area of triangle formed by them?&lt;/p&gt;&#xA;" OwnerUserId="2065" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-16T10:16:14.013" LastActivityDate="2010-11-16T10:16:14.013" Title="How to calculate area of triangle, having its points 2d coordinates?" Tags="&lt;geometry&gt;&lt;linear-algebra&gt;&lt;euclidean-geometry&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="10520" PostTypeId="2" ParentId="10518" CreationDate="2010-11-16T03:43:29.003" Score="2" ViewCount="" Body="&lt;p&gt;You can read about Robmerg Integration in section 4.3 of &lt;a href=&quot;http://www.nrbook.com/c/&quot; rel=&quot;nofollow&quot;&gt;Numerical Recipes&lt;/a&gt; or in &lt;a href=&quot;http://en.wikipedia.org/wiki/Romberg_integration&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt; and the references.  It is a more formal approach to what you are describing.  You talk about using feedback, but it is not clear where you get the feedback from.  If your rewriting is really term-by-term integration (see &lt;a href=&quot;http://math.stackexchange.com/questions/9622/using-derivatives-to-evaluate-an-integral-of-exponentials&quot;&gt;your previous question and my answer&lt;/a&gt;) you do have feedback from that.  The Romberg routine follows the changes in the integral as the order rises and makes a guess based on that.  As explained earlier in the book regarding fitting functions, there are risks in this-maybe the function goes wild in a small area.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-11-16T03:43:29.003" CommentCount="1" />
  <row Id="10521" PostTypeId="2" ParentId="10519" CreationDate="2010-11-16T03:45:36.587" Score="1" ViewCount="" Body="&lt;p&gt;If by square you mean the area, &lt;a href=&quot;http://en.wikipedia.org/wiki/Heron%27s_formula&quot; rel=&quot;nofollow&quot;&gt;Heron's formula&lt;/a&gt; is your friend.  Just calculate the side lengths and the semiperimeter.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-11-16T03:45:36.587" CommentCount="2" />
  <row Id="10522" PostTypeId="2" ParentId="10498" CreationDate="2010-11-16T03:53:03.677" Score="1" ViewCount="" Body="&lt;p&gt;Yes, the GCDs are the same, but your argument shows a lot of confusion, as Bill points out. First, when you claim that $d=d&amp;#39;$, the problem is that you are simply &lt;em&gt;asserting&lt;/em&gt; a result that is &lt;em&gt;stronger&lt;/em&gt; than what you are trying to prove. You are trying to prove the simpler case that if $a\equiv 1 \pmod{n}$ then $\gcd(a,n)=1$, and you are invoking the result that if $a\equiv b\pmod{n}$ then $\gcd(a,n)=\gcd(b,n)$. That's &lt;strong&gt;more&lt;/strong&gt; than what you want to prove, so you cannot simply &lt;em&gt;assume&lt;/em&gt; it's true.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But really, think about $[a]_n = [1]_n$. That means that $1 = a+kn$ for some integer $k$. That means that you can write $1$ as a linear combination of $a$ and $n$. That means that $\gcd(a,n)$ divides $1$. That means...&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-11-16T03:53:03.677" CommentCount="4" />
  <row Id="10523" PostTypeId="2" ParentId="10494" CreationDate="2010-11-16T04:03:10.653" Score="1" ViewCount="" Body="&lt;p&gt;For (1), you confused the condition for a congruence of the form $ax\equiv b\pmod{n}$ to have a solution with the congruence you have here. This congruence is not a linear congruence, so the condition does not apply.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;No, that does not even begin to work; it's completely different kinds of congruences (linear vs. quadratic). And you did not determine if it had solutions or not, you just said that solving $x^2\equiv 1 \pmod{3}$ is the same as solving $1\equiv x^2\pmod{3}$. If I asked you to figure out if there is an integer such that $x^2=4$, would saying &quot;that's the same as finding an integer such that $4=x^2$&quot; constitute a valid answer that solves the problem of &lt;em&gt;determining&lt;/em&gt; if there is or is not such an integer? No.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The simplest way to solve &lt;em&gt;this&lt;/em&gt; one is to check: there are only three possible values for $x$ modulo $3$, namely $0$, $1$, or $2$. Do any of them give you a solution to the congruence?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More generally, you can use the ideas hinted at in Bill Dubuque's answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For (2), you aren't proving anything; you are just repeating what you have been asked to prove. Proving it requires more than just repeating the statement with conviction. &lt;em&gt;Why&lt;/em&gt; is it that $4k+3$ can never be written in the form $a^2+b^2$? You never say, you just assert it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instead, think about what $a^2+b^2$ would be modulo $4$. What are the possible values of $a^2$ modulo $4$? Of $b^2$? So what are the possible values of $a^2+b^2$ modulo $4$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(3) is essentially correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For (4), try the same approach as (3). &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-16T04:25:43.700" LastActivityDate="2010-11-16T04:25:43.700" />
  <row Id="10524" PostTypeId="2" ParentId="10337" CreationDate="2010-11-16T04:08:33.300" Score="1" ViewCount="" Body="&lt;p&gt;I think the intuitive point here is captured in the idea that between -n and n there are only finitely many integers and infinitely many rationals.  So over any finite range (and intuitively therefore over the entire range because it is uniform) the density of integers in the rationals is zero.  The effort is to carry over the idea of &lt;a href=&quot;http://en.wikipedia.org/wiki/Natural_density&quot; rel=&quot;nofollow&quot;&gt;natural density&lt;/a&gt; in the sense that there are fewer squares than naturals because the natural density of squares is zero. Natural density is a well defined operation, but may still not capture all the comparisons one would like to make.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In essence, this is captured in a remark on a previous post that cardinality is a coarse way to compare sets.  Natural density is finer but demands an ordering.  It allows more distinctions to be made, most (all?) of which are in line with our intuition.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-11-16T04:08:33.300" CommentCount="2" />
  <row Id="10525" PostTypeId="2" ParentId="2384" CreationDate="2010-11-16T04:37:06.367" Score="0" ViewCount="" Body="&lt;p&gt;I have developed some educational freeware to teach students how to write simple, though thoroughly rigorous mathematical proofs. The software verifies every line of proof as it is entered. It also provides some important insights into the nature of modern mathematics. One of the worked examples in the tutorial is to prove the set of all things cannot exist. (It would be too big, and would include at least one very troublesome object that would make the whole thing blow up!) I think your budding mathematicians would enjoy it. Download the software and tutorial at my website: &lt;a href=&quot;http://www.dcproof.com&quot; rel=&quot;nofollow&quot;&gt;http://www.dcproof.com&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="3515" LastActivityDate="2010-11-16T04:37:06.367" CommunityOwnedDate="2010-11-16T04:37:06.367" />
  <row Id="10526" PostTypeId="2" ParentId="10512" CreationDate="2010-11-16T04:50:45.147" Score="2" ViewCount="" Body="&lt;p&gt;Well, you'll probably want a more conceptual proof, but one thing you can do is check they are computed by the same chain complex: for $K(G,1)$ take the simplicial construction of the classifying space $BG$ and compute its cohomology in the usual way for simplicial sets (using the dual to the complex of formal linear combinations of simplices); for the group cohomology take the free resolution of the trivial $\mathbb{Z}[G]$-module $\mathbb{Z}$ where the nth term is $\mathbb{Z}[G^n]$, which amounts to the same complex as before.&lt;/p&gt;&#xA;" OwnerUserId="1070" LastActivityDate="2010-11-16T04:50:45.147" />
  <row Id="10527" PostTypeId="1" AcceptedAnswerId="10528" CreationDate="2010-11-16T05:15:28.640" Score="3" ViewCount="111" Body="&lt;p&gt;Suppose $r_i$, $1 \le i \le n$, and $c_j$, $1 \le j \le m$, are nonnegative integers.  When does there exist an $n \times m$ matrix in $\text{Mat}_{n \times m} (\mathbb{Z}^+)$, i.e. nonnegative entries, such that $r_i$ is the sum of the entries in its $i$th row and $c_j$ is the sum of the entries in its $j$th columns?&lt;/p&gt;&#xA;" OwnerUserId="314" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-11-16T05:49:43.337" LastActivityDate="2010-11-16T05:49:43.337" Title="Constraints on sum of rows and columns of matrix" Tags="&lt;combinatorics&gt;&lt;algebra-precalculus&gt;&lt;matrices&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="10528" PostTypeId="2" ParentId="10527" CreationDate="2010-11-16T05:20:03.367" Score="5" ViewCount="" Body="&lt;p&gt;There is a survey paper by Alexander Barvinok on this:&#xA;&lt;a href=&quot;http://www.math.lsa.umich.edu/~barvinok/linalg.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.math.lsa.umich.edu/~barvinok/linalg.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3431" LastActivityDate="2010-11-16T05:20:03.367" CommentCount="2" />
  <row Id="10529" PostTypeId="1" CreationDate="2010-11-16T05:50:59.983" Score="0" ViewCount="137" Body="&lt;p&gt;what is the sigma field created by X(t,w) where t belongs to [0,1] and X(t,w) =1 if t=w and zero otherwise.&lt;/p&gt;&#xA;" LastActivityDate="2010-11-16T11:31:34.993" Title="sigma algebra stochastic process" Tags="&lt;sigma-algebra&gt;" AnswerCount="1" />
  <row Id="10531" PostTypeId="2" ParentId="1730" CreationDate="2010-11-16T06:03:17.430" Score="0" ViewCount="" Body="&lt;p&gt;Why not use a real &quot;function machine,&quot; which each of your students should have -- a scientific calculator? After all, most, if not all the functions in your course will be numerical examples. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a function of one variable, use any of the trigonometric, squaring, cubing, square roots, or log functions. Use square root, inverse trig and log functions for examples of restricted domains of definition. You don't have to explain what these functions mean at first, just emphasize that you put in one number, and get another number back. Have students make a table of values using the calculator to reinforce this fact. Be sure to include examples that leave values unchanged, e.g. sqrt(0)=0, sqrt(1)=1, sin(0)=0, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For functions of two variables, use the ordinary arithmetic operations: add, subtract, multiply, divide or exponent functions.&lt;/p&gt;&#xA;" OwnerUserId="3515" LastActivityDate="2010-11-16T06:03:17.430" />
  <row Id="10532" PostTypeId="2" ParentId="10519" CreationDate="2010-11-16T06:08:48.063" Score="3" ViewCount="" Body="&lt;p&gt;To make Rahul's comment more explicit, the determinant formula&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac12 \begin{vmatrix}x_1&amp;amp;y_1&amp;amp;1\\x_2&amp;amp;y_2&amp;amp;1\\x_3&amp;amp;y_3&amp;amp;1\end{vmatrix}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where the $(x_i,y_i)$ are the coordinates of the corners, gives the (signed) area of the triangle. For a guaranteed positive result, the points are to be taken anticlockwise.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-11-16T06:08:48.063" CommentCount="2" />
  <row Id="10533" PostTypeId="1" CreationDate="2010-11-16T06:24:27.363" Score="2" ViewCount="255" Body="&lt;p&gt;Hello...in a game of contract bridge, partner and me together have 11 cards in a suite. The remaining 2 cards in the same suite can be distributed amongst our opponents as either 1-1 or 2-0. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the probability that it will be distributed as 1-1 and what is the probability it will be distributed as 2-0?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once the above is solved, how can you extend it if more cards are missing. That is, let us assume that partner and me have 8 cards between us. How do you calculate the probability for the distributions (5-0, 4-1, 3-2)?&lt;/p&gt;&#xA;" OwnerUserId="1791" LastActivityDate="2010-11-16T15:55:02.370" Title="Probability in a game of Contract Bridge" Tags="&lt;probability&gt;&lt;card-games&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="10534" PostTypeId="2" ParentId="10533" CreationDate="2010-11-16T06:30:55.410" Score="1" ViewCount="" Body="&lt;p&gt;The number of cards in suit a given opponent gets is distributed &lt;a href=&quot;http://en.wikipedia.org/wiki/Hypergeometric_distribution&quot; rel=&quot;nofollow&quot;&gt;hypergeomtrically&lt;/a&gt;. In your particular case, $1-1$ is slightly more likely, with probability $13/25 = 1/2 + 1/50$. This is not surprising, since once you got one of the cards, the probability you'll get another drops, since then there's only one of them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: The hypergeometric distribution with parameters $(N,m,n)$ concerns a situation in which the universe consists of $N$ cards, $m$ of which are special, and you are drawing $n$ of them. The probability that the number of special cards you draw is $k$ is exactly $$\frac{\binom{m}{k}\binom{N-m}{n-k}}{\binom{N}{n}},$$ since there are $\binom{m}{k}\binom{N-m}{n-k}$ choices for a hand with exactly $k$ special cards, and $\binom{N}{n}$ total hands.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to find the distribution of $1-1$, you plug in $k=1$. In order to find the distribution of $2-0$, you plug in $k=0$ and $k=2$, and add, since you don't care which player gets none and which gets both.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2010-11-16T15:55:02.370" LastActivityDate="2010-11-16T15:55:02.370" CommentCount="1" />
  <row Id="10535" PostTypeId="1" AcceptedAnswerId="10538" CreationDate="2010-11-16T06:39:47.603" Score="0" ViewCount="126" Body="&lt;p&gt;The area of a pyramid with a square base with side length $L$ and height $h$ is calculated as follows: &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;In $\mathbb{R}^3$ place the pyramid upright with one side flush with the z-axis so that the corners of the base are at $(0,0,\frac{1}{L})$, $(0,0,-\frac{1}{L})$, $(L,0,\frac{1}{L})$ and $(L,0,-\frac{1}{L})$.&#xA;&lt;img src=&quot;http://i.imgur.com/gNKcT.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;We will slice parallel to the yz-plane from $x=0$ up to $x=\frac{L}{2}$. (We can then double it to get the whole pyramid.) &lt;strong&gt;The slices are rectangles&lt;/strong&gt; that all have a base equal to L and height given by $f(x)=\frac{2hx}{L}$. The cross-sectional area is then, $A(x) = L\frac{2hx}{L}=2hx$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;So, we have $2 \int_{0}^{\frac L2} (2hx) dx \neq \frac13 L^2 h$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;What has gone wrong?    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;ANS: The slices are NOT rectangles. DUH. Thanks, Rahul Narain. &lt;/p&gt;&#xA;" OwnerUserId="2563" LastEditorUserId="2563" LastEditorDisplayName="" LastEditDate="2010-11-16T07:00:59.963" LastActivityDate="2010-11-16T07:20:46.677" Title="How to explain why this method of slicing a solid to find volume fails?" Tags="&lt;calculus&gt;&lt;geometry&gt;&lt;integral&gt;" AnswerCount="1" CommentCount="6" />
  <row Id="10536" PostTypeId="1" AcceptedAnswerId="10537" CreationDate="2010-11-16T06:52:22.817" Score="1" ViewCount="75" Body="&lt;p&gt;Suppose that $\mathcal{B}$ is a base for a topology on a space $X$. Is there a nice way of thinking about how we can modify $\mathcal{B}$ (for instance, to simplify computations) without changing the topology it generates? It seems non-trivial to compute the topology generated by a base, but maybe some &quot;small enough&quot; changes to the base should be safe. The situation is delicate:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If I have a fixed open set $U$, consider the family of sets $\mathcal{B&amp;#39;} = \{ B \in \mathcal{B} | B \subset U \text{ or } B \subset (X \setminus U) \}$&#xD;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This does not generate the same topology as $\mathcal{B}$, since $\mathcal{B&amp;#39;}$ won't, in general, be a base. For instance, if $X = \mathbb{R}$, and $U = (0,1)$, then $1$ is not contained in any element of $\mathcal{B&amp;#39;}$.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Cover $X$ by open sets $U_i$, and set $\mathcal{B&amp;#39;} = \{ B \in \mathcal{B} | B \subset U_i \text{ for some } i \}$&#xD;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This, on the other hand, does sometimes work. For instance, consider $X = \text{Spec} A$ with the Zariski topology, and suppose that $X$ is covered by $U_i = \text{Spec} A_i$. A base for the Zariski topology on $X$ is given by sets of the form $D(f) = \{ \mathfrak{p} \subseteq A | f \not \in \mathfrak{p} \}$. We can restrict this base to only include those $D(f)$ that lie in some $U_i$, and we obtain the same topology. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suppose my question is the following:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If $\mathcal{B}$ is a base for a topology $\mathcal{T}$, are there some nice types of subbases of $\mathcal{B}$ (along the lines of the second example above) of $\mathcal{B}$ that will always generate $\mathcal{T}$?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I think this sort of thing comes up when checking that various properties of scheme morphisms are affine local, so I've also tagged this with [algebraic-geometry].&lt;/p&gt;&#xA;" OwnerUserId="576" LastActivityDate="2010-11-16T06:57:56.520" Title="When does a subbase of a base generate the same topology?" Tags="&lt;topology&gt;&lt;algebraic-geometry&gt;" AnswerCount="1" />
  <row Id="10537" PostTypeId="2" ParentId="10536" CreationDate="2010-11-16T06:57:56.520" Score="2" ViewCount="" Body="&lt;p&gt;Since a topology generated by a base consists of open sets that are union of basic open sets, you may drop, from a given base, any open set that is a union of open sets in the same base and get a smaller base.&lt;/p&gt;&#xA;" OwnerUserId="3431" LastActivityDate="2010-11-16T06:57:56.520" />
  <row Id="10538" PostTypeId="2" ParentId="10535" CreationDate="2010-11-16T07:20:46.677" Score="2" ViewCount="" Body="&lt;p&gt;This is an elaboration of my comment, so the question can be marked as answered.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't see how you are placing your pyramid. &quot;One side flush with the z-axis&quot; is not enough information, as one may rotate the shape about the $z$-axis while keeping it flush. If the base lies on the $xz$-plane, its corners have $z$-coordinate $0$ or $L$, not $1/L$, and the slices parallel to the $yz$-plane will not be rectangles but trapeziums (a.k.a. trapeziods), as one can see by considering the slice at $L/2$ which is clearly a triangle (a degenerate trapezium) and not a rectangle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A slice at a given $x$-coordinate $x &amp;lt; L/2$ is a trapezium with height $2hx/L$, as you noticed, but with one parallel size of length $L$ and the other of length $L-2x$. The area of the trapezium is therefore $A = 2hx/L\cdot(L - x) = 2hx\cdot(1-x/L)$, and the volume of the pyramid is $V = 2\int_{x=0}^{L/2} A dx = hL^2/3$ as expected.&lt;/p&gt;&#xA;" OwnerUserId="856" LastActivityDate="2010-11-16T07:20:46.677" />
  <row Id="10539" PostTypeId="1" CreationDate="2010-11-16T07:25:59.637" Score="3" ViewCount="137" Body="&lt;p&gt;Hi all,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Currently I am reading&#xA;Logic an Structure by Dirk van Dalen (2008).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As I am missing some basics I try to find related lectures on youtube.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I frequently watch MIT, Stanford, and University of Nottinham lectures on youtube.&#xA;www.youtube.com/user/MIT&#xA;www.youtube.com/user/StanfordUniversity&#xA;www.youtube.com/user/NottmUniversity&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Howevery, I have a hard time to find lectures on topics like predicate logic, high order logic, and  intuitionistic logic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anyone happen to have pointers?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="3377" LastActivityDate="2010-11-16T07:25:59.637" Title="logic lectures on youtube" Tags="&lt;soft-question&gt;&lt;logic&gt;&lt;higher-order-logic&gt;" CommentCount="2" />
  <row Id="10540" PostTypeId="2" ParentId="10516" CreationDate="2010-11-16T07:32:11.903" Score="5" ViewCount="" Body="&lt;p&gt;For question 3, by the reciprocity law $(-1)^{n}P(-t)$&#xD;&#xA;counts positive integer solutions of $\sum x_i/b_i&amp;lt;t$&#xD;&#xA;where $P$ is the Ehrhart plynomial of the $n$-simplex&#xA;with vertices $(0,\ldots,0)$ and the $(0,\ldots,b_i,\ldots,0)$.&#xA;This is short of your target by the number of positive integer&#xA;solutions of $\sum x_i/b_i=t$. But that is $(-1)^{n-1}Q(-t)$&#xD;&#xA;where $Q$ is the Ehrhart polynomial of the $(n-1)$-simplex&#xA;with vertices the $(0,\ldots,b_i,\ldots,0)$.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-11-16T07:32:11.903" CommentCount="0" />
  <row Id="10541" PostTypeId="1" CreationDate="2010-11-16T08:30:44.107" Score="1" ViewCount="36" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;who of you in this forum uses R (http://www.r-project.org/) with the multicore, SNOW or CUDA packages, so for advanced calculations that need more power than a workstation CPU? On which hardware do you compute these scripts? At home/ at work or do you have data center access somewhere?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The background of these questions is the following: I am currently writing my M.Sc. thesis about R and High-Performance-Computing and need a strong knowledge about who actually uses R. I read that R had 1 million users in 2008, bu thats more or less the only user statistics I could find on this topic - so I hope for your answers!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sincerely Heinrich&lt;/p&gt;&#xA;" OwnerUserId="3561" LastActivityDate="2010-11-16T08:30:44.107" Title="Who uses R with multicore, SNOW or CUDA package for resource intense computing?" Tags="&lt;statistics&gt;&lt;computer-science&gt;&lt;applications&gt;" CommentCount="2" />
  <row Id="10542" PostTypeId="2" ParentId="10518" CreationDate="2010-11-16T08:58:37.310" Score="2" ViewCount="" Body="&lt;p&gt;Are you aware that integral of this type can be computed exactly by means of the residue theorem? The price to pay is that you have to find the roots of the equation $\sum_a^b r_k z^{s_k}=0$ in the unit disk. (You may assume all $s_k\ge0$.)&lt;/p&gt;&#xA;" OwnerUserId="1303" LastActivityDate="2010-11-16T08:58:37.310" CommentCount="1" />
  <row Id="10543" PostTypeId="1" CreationDate="2010-11-16T09:11:51.597" Score="4" ViewCount="253" Body="&lt;p&gt;Hello,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I heard about &lt;a href=&quot;http://en.wikipedia.org/wiki/Yakov_Perelman#Books&quot; rel=&quot;nofollow&quot;&gt;Yakov Perelman&lt;/a&gt; and his books. I just finished reading his two volumes of &lt;strong&gt;Physics for Entertainment&lt;/strong&gt;. What a delightful read! What a splendid author. This is the exact book I've been searching for. I can use it to develop interest for science (math &amp;amp; physics) in my students.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;His math books:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Mathematics Can Be Fun&lt;/li&gt;&#xA;&lt;li&gt;Figures for Fun&lt;/li&gt;&#xA;&lt;li&gt;Arithmetic for entertainment&lt;/li&gt;&#xA;&lt;li&gt;Geometry for Entertainment&lt;/li&gt;&#xA;&lt;li&gt;Lively Mathematics&lt;/li&gt;&#xA;&lt;li&gt;Fun with Maths &amp;amp; Physics&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;His physics books:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Physics for Entertainment (1913)&lt;/li&gt;&#xA;&lt;li&gt;Physics Everywhere&lt;/li&gt;&#xA;&lt;li&gt;Mechanics for entertainment&lt;/li&gt;&#xA;&lt;li&gt;Astronomy for entertainment&lt;/li&gt;&#xA;&lt;li&gt;Tricks and Amusements&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I want to get all the above books. Because books from author like this cannot be disappointing. But unfortunately not all of them are available. :(&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also read another amazing book &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/069111966X&quot; rel=&quot;nofollow&quot;&gt;How to Solve It: A New Aspect of Mathematical Method by G.Poyla&lt;/a&gt;. This books actually teaches you &lt;em&gt;how to think&lt;/em&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the similar lines if you have any book suggestions (with very practical problems &amp;amp; case studies) for physics &amp;amp; Math (Please don't differentiate between math &amp;amp; physics here. If I can develop interest in one of the subject he'll gain interest in other.) please contribute.&lt;/p&gt;&#xA;" OwnerUserId="2264" LastActivityDate="2010-11-16T15:58:38.537" Title="Books that develop interest &amp; critical thinking among high school students." Tags="&lt;books&gt;&lt;teaching&gt;&lt;high-school&gt;" AnswerCount="3" CommentCount="4" FavoriteCount="4" />
  <row Id="10544" PostTypeId="1" AcceptedAnswerId="10570" CreationDate="2010-11-16T09:22:58.160" Score="2" ViewCount="185" Body="&lt;p&gt;I have a set and its cardinality. How can I calculate the cardinality of its complementary set?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;: Gotta apologize. I was talking about well-defined sets with a fairly low positive integer as cardinality. Qwirk comes pretty close when talking about subsets and a given superset.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So see it this way. I have a certain set of numbers. Take some extra condition to get a subset, know its cardinality and now try to calculate the cardinality of a subset with the negated condition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sorry for being unclear.&lt;/p&gt;&#xA;" OwnerUserId="3016" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-16T14:34:08.780" LastActivityDate="2010-11-16T19:11:09.470" Title="Cardinality of complementary set" Tags="&lt;homework&gt;&lt;elementary-set-theory&gt;" AnswerCount="3" CommentCount="5" />
  <row Id="10545" PostTypeId="1" AcceptedAnswerId="10550" CreationDate="2010-11-16T09:52:09.963" Score="2" ViewCount="222" Body="&lt;p&gt;If in a $\displaystyle\bigtriangleup$ ABC, $\displaystyle\cot \biggl( \frac{A}{2} \biggr) = \frac{b+c}{a} $, then $\displaystyle\bigtriangleup$ ABC is of which &lt;a href=&quot;http://en.wikipedia.org/wiki/Triangle#By_relative_lengths_of_sides&quot; rel=&quot;nofollow&quot;&gt;type&lt;/a&gt; ? &lt;/p&gt;&#xA;" OwnerUserId="2109" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-11-16T20:55:06.180" LastActivityDate="2010-11-17T07:42:15.903" Title="What type of triangle satisfies: $\cot \biggl( \frac{A}{2} \biggr) = \frac{b+c}{a} $" Tags="&lt;geometry&gt;&lt;trigonometry&gt;&lt;euclidean-geometry&gt;" AnswerCount="5" CommentCount="4" />
  <row Id="10546" PostTypeId="2" ParentId="10544" CreationDate="2010-11-16T10:36:00.583" Score="2" ViewCount="" Body="&lt;p&gt;Possible slight rephrasing of the question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given a set $X$ and a subset $S$ then calculate the cardinality of the complement of $S$ with respect to $X$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, this can be uncountably infinite, countably infinite or even have finite cardinality!&lt;/p&gt;&#xA;" OwnerUserId="2219" LastActivityDate="2010-11-16T10:36:00.583" CommentCount="3" />
  <row Id="10547" PostTypeId="1" AcceptedAnswerId="10567" CreationDate="2010-11-16T10:46:27.520" Score="3" ViewCount="210" Body="&lt;p&gt;This is extracted from my module :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a $\displaystyle\bigtriangleup ABC$,&#xA;$$\displaystyle \frac{\sin (A) + \sin(B)}{\sin(A) -\sin(B)} = \frac{k \cdot(\sin (A) + \sin(B))}{k \cdot(\sin(A) -\sin(B))} = \frac{a+b}{a-b}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $A,B,C$ are the angles and  $a,b,c$ are the opposite sides,I can't figure out how this step holds valid?&lt;/p&gt;&#xA;" OwnerUserId="2109" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-11-16T18:38:12.960" LastActivityDate="2011-06-01T11:07:05.467" Title="Why is $\frac{k \cdot(\sin (A) + \sin(B))}{k \cdot(\sin(A) -\sin(B))} = \frac{a+b}{a-b}$ valid in a triangle?" Tags="&lt;trigonometry&gt;" AnswerCount="4" CommentCount="4" FavoriteCount="1" />
  <row Id="10548" PostTypeId="2" ParentId="10529" CreationDate="2010-11-16T11:31:34.993" Score="2" ViewCount="" Body="&lt;p&gt;$X_{t}$ is a stochastic process. Maybe you mean a natural filtration?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sigma algebra generated by $X_{t}$ is $\sigma(X_{t})=\{\emptyset,[0;1],t,[0;t)\cup (t;1]\}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The natural filtration $\mathcal{F}_{t}$ is the minimal sigma algebra which contains all $\sigma(X_{u})$ for $u\in [0;t]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So $\mathcal{F}_{t}$ consists of all countable subsets of $[0;t]$ and their complements.&lt;/p&gt;&#xA;" OwnerUserId="2481" LastActivityDate="2010-11-16T11:31:34.993" CommentCount="3" />
  <row Id="10549" PostTypeId="2" ParentId="10545" CreationDate="2010-11-16T11:42:34.860" Score="3" ViewCount="" Body="&lt;p&gt;The cosine formula states&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$a^2 = b^2+ c^2 – 2bc \cos A$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and since&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\cos A = \frac{ \cot^2 (A/2) – 1}{ \cot^2(A/2)+1}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$a^2 = b^2 + c^2 -2bc \frac{ (b+c)^2 – a^2}{(b+c)^2 + a^2},$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which reduces to $a^4 = (b^2 – c^2)^2.$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And so taking the square root $a^2 =b^2 – c ^2,$ where $b$ is the larger side.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence $b^2 = c^2+a^2$ and so we have a right-angled triangle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And so it's the angle opposite the larger of $b$ and $c$ which is the right-angle.&lt;/p&gt;&#xA;" OwnerUserId="1301" LastActivityDate="2010-11-16T11:42:34.860" />
  <row Id="10550" PostTypeId="2" ParentId="10545" CreationDate="2010-11-16T11:43:08.213" Score="7" ViewCount="" Body="&lt;p&gt;So by &lt;em&gt;Law of Sines&lt;/em&gt; we have $$ \frac{a}{\sin{A}} = \frac{b}{\sin{B}} = \frac{c}{\sin{C}} =k (\text{say})$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From this your equation becomes, $$\frac{\cos\frac{A}{2}}{\sin\frac{A}{2}} = \frac{k(\sin{B} + \sin{C})}{k \sin{A}} = \frac{\sin{B}+\sin{C}}{\sin{A}} = \frac{\cos\frac{A}{2} \cos\frac{B-C}{2}}{\sin\frac{A}{2} \cos\frac{A}{2}}$$ From this you get $$\frac{\pi}{2} - \Bigl(\frac{B+C}{2}\Bigr)= \pm \frac{B-C}{2}$$ and you can do it then.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This gives $B=\frac{\pi}{2}$ or $C =\frac{\pi}{2}$, which means the triangle is &lt;em&gt;right-angled&lt;/em&gt;. Hurray!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;NOTE:&lt;/strong&gt; The keen idea when you see problems of this type is to use the &lt;em&gt;Sine rule&lt;/em&gt;. Do you think that this problem would have been difficult, if you had applied the sine rule as i did in the first step. Very often in &lt;em&gt;Mathematics&lt;/em&gt; starting the solution is the difficult part. Once you figure that out, then the solutions simply flows. &lt;/p&gt;&#xA;" OwnerUserId="9413" LastEditorUserId="9413" LastEditorDisplayName="anonymous" LastEditDate="2010-11-17T07:42:15.903" LastActivityDate="2010-11-17T07:42:15.903" CommentCount="7" />
  <row Id="10551" PostTypeId="2" ParentId="10545" CreationDate="2010-11-16T11:43:13.433" Score="3" ViewCount="" Body="&lt;p&gt;OK, first rewrite it as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \cos(A/2) = \frac{b+c}{a} \sin(A/2) $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Multiply both sides by $2 \cos(A/2)$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ 2 \cos^2(A/2) = \frac{b+c}{a} \sin(A) $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now use the rule $a/\sin(A)=b/\sin(B)=c/\sin(C)$ to transform this to:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ 2 \cos^2(A/2) = \sin(B)+\sin(C) $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using Simpson's rule on the second hand of the equation to sum the sines gives:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ 2 \cos^2(A/2) = 2\sin((B+C)/2)\cos((B-C)/2) $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since in a triangle $A+B+C=\pi$ we have $ 2\sin((B+C)/2) = 2\sin((\pi-A)/2) = 2\cos(A/2) $ and thus&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \cos(A/2) = \cos((B-C)/2) $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or $A+C=B$ (If $B&gt;C$). Combining this with $A+B+C=\pi$ this gives $B=\pi/2$.&lt;/p&gt;&#xA;" OwnerUserId="3567" LastEditorUserId="3567" LastEditorDisplayName="" LastEditDate="2010-11-16T12:24:12.010" LastActivityDate="2010-11-16T12:24:12.010" />
  <row Id="10552" PostTypeId="2" ParentId="10547" CreationDate="2010-11-16T11:46:41.487" Score="1" ViewCount="" Body="&lt;p&gt;Geometrically, this is not too hard to see. Presumably, you understand why&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\sin(A) + \sin(B)}{\sin(A) - \sin(B)} = \frac{k(\sin(A) + \sin(B))}{k(\sin(A) - \sin(B))}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let k be the length of the normal from C onto AB. What is then k sin(A) and k sin(B)?&lt;/p&gt;&#xA;" OwnerUserId="3536" LastActivityDate="2010-11-16T11:46:41.487" />
  <row Id="10553" PostTypeId="2" ParentId="10545" CreationDate="2010-11-16T11:53:28.710" Score="3" ViewCount="" Body="&lt;p&gt;Begin with the Law of Cosines to compute $\cot^2\frac{A}{2}$ in terms of the lengths of the sides of the triangle:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\cos{A} = \frac{b^2+c^2-a^2}{2bc}$$&#xD;&#xA;$$\Rightarrow \begin{cases}&#xD;&#xA;\cos^2{\frac{A}{2}}=\frac{1+\cos{A}}{2}=\frac{2bc+b^2+c^2-a^2}{4bc} = \frac{(b+c)^2-a^2}{4bc} \\&#xD;&#xA;\sin^2{\frac{A}{2}}=\frac{1-\cos{A}}{2}=\frac{2bc-(b^2+c^2-a^2)}{4bc} = \frac{a^2-(b-c)^2}{4bc}&#xD;&#xA;\end{cases}$$&#xD;&#xA;$$\Rightarrow \cot^2\frac{A}{2}=\frac{\cos^2{\frac{A}{2}}}{\sin^2{\frac{A}{2}}}=\frac{(b+c)^2-a^2}{a^2-(b-c)^2}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, work in the condition from the problem, and let Algebra bring you home ...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{eqnarray}&#xD;&#xA;\frac{(b+c)^2}{a^2} &amp;amp;=&amp;amp; \frac{(b+c)^2-a^2}{a^2-(b-c)^2} \\&#xD;&#xA;(b+c)^2\left(a^2-(b-c)^2\right)&amp;amp;=&amp;amp; a^2\left((b+c)^2-a^2\right) \\&#xD;&#xA;a^2(b+c)^2-(b+c)^2(b-c)^2 &amp;amp;=&amp;amp; a^2 (b+c)^2-a^4 \\&#xD;&#xA;a^4-(b+c)^2(b-c)^2 &amp;amp;=&amp;amp; 0 \\&#xD;&#xA;a^4-(b^2-c^2)^2 &amp;amp;=&amp;amp; 0 \\&#xD;&#xA;\left(a^2+(b^2-c^2)\right)\left(a^2-(b^2-c^2)\right) &amp;amp;=&amp;amp; 0 \\&#xD;&#xA;\end{eqnarray}$$&#xD;&#xA;$$\Rightarrow \hspace{0.25in} a^2+b^2=c^2 \hspace{0.25in}\text{or}\hspace{0.25in} a^2+c^2=b^2$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nice little problem. I suspect there's a more-direct path to the solution, though. (&lt;strong&gt;Edit&lt;/strong&gt; I see a couple such paths had been posted when I was working on my answer! :)&lt;/p&gt;&#xA;" OwnerUserId="409" LastEditorUserId="409" LastEditorDisplayName="" LastEditDate="2010-11-16T12:22:50.560" LastActivityDate="2010-11-16T12:22:50.560" />
  <row Id="10554" PostTypeId="1" AcceptedAnswerId="10561" CreationDate="2010-11-16T12:19:46.390" Score="3" ViewCount="277" Body="&lt;p&gt;&lt;strike&gt;I claim that it is commonly believed that&lt;/strike&gt; Mathematical objects can be seen as genuinely static, with no &quot;Platonic&quot; time in which they do genuinely evolve.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nevertheless time has its place in mathematics:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;An endomorphism of a set (seen as a set of states of a system) into itself can be seen as evolution of the system in discrete time steps.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;For a function of a totally ordered set into a set (seen as above) the ordered set can be seen as &quot;time&quot;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;as the time-like component in Minkowski space &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Questions (slightly modified after Qiaochu's comment and Vhailor's answer):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Which other constructs do give you a &quot;time feeling&quot; or give rise to &quot;dynamic intuition&quot; &lt;strike&gt;admit a comparable straight-forward&#xA;  interpretation as &quot;time&quot;&lt;/strike&gt;?&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The examples above are set-theoretical (&quot;concrete&quot;). Is there a more abstract modelling of &quot;time&quot;, maybe in category theory?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1792" LastEditorUserId="1792" LastEditorDisplayName="" LastEditDate="2010-11-16T14:06:29.940" LastActivityDate="2011-03-10T19:05:42.250" Title="Time in Mathematics" Tags="&lt;soft-question&gt;&lt;philosophy&gt;" AnswerCount="3" CommentCount="3" />
  <row Id="10555" PostTypeId="1" AcceptedAnswerId="10602" CreationDate="2010-11-16T13:01:53.193" Score="2" ViewCount="168" Body="&lt;p&gt;(1). I was wondering about what are the relation and differences between formal and informal logic?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What topics does each of them have?&#xA;For example, topics such as Meaning and Definition, Syllogistic Logic, Inductive Logic, Probabilistic/Statistical Reasoning, Deductive Logic&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is symbolic logic same as formal logic?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) what is the relation between informal/formal logic and deductive/non-deductive reasoning/inference?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;must formal logic be only for deduction, not for non-deductive reasoning/inference?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(3) what are other usual/better ways to categorize various logic topics?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks and regards!&lt;/p&gt;&#xA;" OwnerUserId="1281" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-11-16T19:42:18.627" LastActivityDate="2010-11-16T19:42:18.627" Title="categorization of logic" Tags="&lt;logic&gt;&lt;computer-science&gt;&lt;philosophy&gt;" AnswerCount="3" />
  <row Id="10556" PostTypeId="1" CreationDate="2010-11-16T13:19:19.807" Score="1" ViewCount="107" Body="&lt;p&gt;According to the wikipedia article: &lt;a href=&quot;http://en.wikipedia.org/wiki/Levenberg_Marquardt&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Levenberg_Marquardt&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;--&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$S(\boldsymbol\beta+\boldsymbol\delta) \approx \|\mathbf{y} - \mathbf{f}(\boldsymbol\beta) - \mathbf{J}\boldsymbol\delta\|^2$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Taking the derivative with respect to δ and setting the result to zero gives:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(J^{T}J)\boldsymbol \delta  = J^{T} [y - f(\boldsymbol \beta)])$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;--&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My attempt to derive the equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\|\mathbf{y} - \mathbf{f}(\boldsymbol\beta) - \mathbf{J}\boldsymbol\delta\|^2&#xD;&#xA;= (\mathbf{y} - \mathbf{f}(\boldsymbol\beta) - \mathbf{J}\boldsymbol\delta)^T(\mathbf{y} - \mathbf{f}(\boldsymbol\beta) - \mathbf{J}\boldsymbol\delta)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;using product rule:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{\partial \|\mathbf{y} - \mathbf{f}(\boldsymbol\beta) - \mathbf{J}\boldsymbol\delta\|^2}{\partial \boldsymbol\delta} = (-J^{T})(\mathbf{y} - \mathbf{f}(\boldsymbol\beta) - \mathbf{J}\boldsymbol\delta) + (\mathbf{y} - \mathbf{f}(\boldsymbol\beta) - \mathbf{J}\boldsymbol\delta)^T(-J)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The dimensions of the left and right side don't match.  I believe there might be something wrong with my differentiation.  There seems to be a transpose missing, but I'm not sure what would cause a transpose in the differentiation operation.&lt;/p&gt;&#xA;" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-16T13:23:56.733" LastActivityDate="2010-11-16T15:21:56.960" Title="How to derive the Levenberg–Marquardt algorithm with matrix calculus" Tags="&lt;calculus&gt;&lt;linear-algebra&gt;&lt;optimization&gt;&lt;numerical-methods&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="10557" PostTypeId="2" ParentId="10544" CreationDate="2010-11-16T13:21:43.477" Score="0" ViewCount="" Body="&lt;p&gt;To elaborate on Qwirk's answer, look at $\mathbb{R}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\mathbb{R} \setminus \mathbb{R}^+$ is uncountably infinite, even though $\mathbb{R}^+$ is uncountably infinite. On the other hand, let $T$ be the set of all real transcendental numbers. The set $\mathbb{R} \setminus T$ is exactly the set of all algebraic numbers, which we know to be countable. Finally, let $S = \mathbb{R} \setminus \{0, 1, \dots, n\}$ for some $n \in \mathbb{N}$. Taking the complement of $S$ in $\mathbb{R}$, you see that you can also get a finite set.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, in general, you can not say anything about the cardinality of the complement of a set. However, if the cardinality of a subset $S \subset X$ is strictly less than that of $X$, then the cardinality of $X - S$ must be the same as $X$. You should try to see this for yourself.&lt;/p&gt;&#xA;" OwnerUserId="1016" LastActivityDate="2010-11-16T13:21:43.477" CommentCount="1" />
  <row Id="10558" PostTypeId="2" ParentId="10533" CreationDate="2010-11-16T13:24:43.070" Score="2" ViewCount="" Body="&lt;p&gt;You can use the technique of &lt;em&gt;Vacant Spaces&lt;/em&gt; to calculate probabilities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;During play, if you know that there is space for $\displaystyle x$ unknown cards with West and $\displaystyle y$ unknown cards with East, the probability that a specific card is with west is $\displaystyle x/(x+y)$. This assumes you ignore the carding of the opponents so far. This gives accurate results when computing the &lt;em&gt;apriori&lt;/em&gt; probabilities and gives reasonable values during the play of the hand.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, say you want to calculate the apriori probability of a 3-2 split.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First consider a specific holding say (Q32 with West and 54 with East) and try to calculate the chances.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Chances that West gets the Q = 13/26. &lt;br/&gt;&#xA;Chances that West gets the 3 = 12/25.&lt;br/&gt;&#xA;Chances that West gets the 2 = 11/24. &lt;br/&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Chances that East gets the 5 = 13/23. &lt;br/&gt;&#xA;Chances that East gets the 4 = 12/22. &lt;br/&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus the chances of the specific split Q32-54 is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$13/26 \times 12/25 \times 11/24 \times 13/23 \times 12/22$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are 5 choose 3  + 5 choose 2 = 20 such splits.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus the probability of 3-2 split is 20 times&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$13/26 \times 12/25 \times 11/24 \times 13/23 \times 12/22$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$= 0.678\dots$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus there is a 67.8% chance of a 3-2 break.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A good rule of thumb to remember is that if there are an odd number of cards outstanding, then apriori chances are that they will split as evenly as possible. So with 5 cards outstanding, 3-2 is split is more likely than 4-1 or 5-0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there are an even number of cards outstanding, then they tend to break unevenly. i.e with 2 cards outstanding, 2-0 split is more likely than 1-1. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;See:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An article on Vacant Spaces: &lt;a href=&quot;http://www.clairebridge.com/textes/vacantplaces.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.clairebridge.com/textes/vacantplaces.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pavlicek's Suit Break Calculator: &lt;a href=&quot;http://www.rpbridge.net/xsb2.htm&quot; rel=&quot;nofollow&quot;&gt;http://www.rpbridge.net/xsb2.htm&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-11-16T13:34:31.530" LastActivityDate="2010-11-16T13:34:31.530" CommentCount="6" />
  <row Id="10559" PostTypeId="2" ParentId="10543" CreationDate="2010-11-16T13:26:02.737" Score="1" ViewCount="" Body="&lt;p&gt;At the high school level, I like &lt;em&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0935218084&quot; rel=&quot;nofollow&quot;&gt;Thinking Physics&lt;/a&gt;&lt;/em&gt; by Lewis Carroll Epstein very much. It's entertaining and illustrated by cartoons, and it delves deeply into the right and wrong answers to basic physics problems. An example: Can you float a battleship in a bathtub? &lt;/p&gt;&#xA;" OwnerUserId="940" LastActivityDate="2010-11-16T13:26:02.737" CommentCount="1" />
  <row Id="10560" PostTypeId="2" ParentId="460" CreationDate="2010-11-16T13:45:33.067" Score="0" ViewCount="" Body="&lt;p&gt;Excerpts from Grillet’s Semigroups Chapter 1, Page 1,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;“There are 1160 distinct … semigroups of order 5; &#xA;15793 semigroups of order 6; 836,021 semigroups&#xA;of order 7…”&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Granted, many of those semigroups are not that interesting.&#xA;However, transformation semigroups (not monoids unless &#xA;you count the identity map in) are interesting because most&#xA;transformations(functions) are not bijective. Thus, transformation&#xA;semigroups are more natural than permutation groups.&lt;/p&gt;&#xA;" OwnerUserId="3568" LastActivityDate="2010-11-16T13:45:33.067" CommunityOwnedDate="2010-11-16T13:45:33.067" />
  <row Id="10561" PostTypeId="2" ParentId="10554" CreationDate="2010-11-16T13:55:52.753" Score="2" ViewCount="" Body="&lt;p&gt;I know of two other concepts that have a &quot;time&quot; feeling to them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The definition for homotopy involves a parameter which can very intuitively be interpreted as time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A lie group also has to me some feeling of time, since it adds on top of classical geometry the idea that isometries must be a sort of continuous motion through time, not just a &quot;teleportation&quot; between two states.&lt;/p&gt;&#xA;" OwnerUserId="887" LastActivityDate="2010-11-16T13:55:52.753" CommentCount="2" />
  <row Id="10562" PostTypeId="2" ParentId="10543" CreationDate="2010-11-16T13:56:02.583" Score="2" ViewCount="" Body="&lt;p&gt;I would like to add that books that cultivate interest and critical thinking do not have to be primarily about math (or physics).  It is very important to offer young students a broad view of the culture of mathematics and science in general.  To this end I recommend Feynman's &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0393316041&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;Surely You're Joking, Mr. Feynman&lt;/em&gt;&lt;/a&gt;, James Watson's &lt;a href=&quot;http://www.amazon.com/Double-Helix-James-D-Watson/dp/B000O8VDNG/ref=sr_1_2?s=books&amp;amp;ie=UTF8&amp;amp;qid=1289915557&amp;amp;sr=1-2&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;The Double Helix&lt;/em&gt;&lt;/a&gt;, and P. Yourgrau's &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0465092942&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;A World Without Time, The Forgotten Legacy of Godel and Einstein&lt;/em&gt;&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These books (and many others) offer a glimpse into the lives of some very prolific modern scientists.  I am certain that you will find others that can be described this way, and your students will find them on their own, once they know what to look for.&lt;/p&gt;&#xA;" OwnerUserId="121" LastActivityDate="2010-11-16T13:56:02.583" />
  <row Id="10563" PostTypeId="1" AcceptedAnswerId="10569" CreationDate="2010-11-16T13:59:00.960" Score="3" ViewCount="121" Body="&lt;p&gt;Let $f$ be a distribution on $\mathbf{R}^n$ (in the Schwartz sense) such that&#xA;$$\frac{\partial f}{\partial x_i} = 0 \text{ for $i = 1, \ldots, n$.}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then how to prove that $f$ is a constant? I had this exercise in a class last year but I couldn't find how to do the induction step (for $n = 1$ it is clear of course).&lt;/p&gt;&#xA;" OwnerUserId="1120" LastEditorUserId="1120" LastEditorDisplayName="" LastEditDate="2010-11-16T15:24:52.557" LastActivityDate="2010-11-16T16:24:02.230" Title="Derivatives distribution." Tags="&lt;fourier-analysis&gt;&lt;harmonic-analysis&gt;&lt;distribution-theory&gt;" AnswerCount="3" FavoriteCount="1" />
  <row Id="10564" PostTypeId="1" AcceptedAnswerId="12760" CreationDate="2010-11-16T14:00:52.317" Score="1" ViewCount="90" Body="&lt;p&gt;Hey there,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In 1D projective geometry,&#xA;I want to compute the 2x2 Homography matrix $H$ (in homogeneous coordinates), given 3 pairs of corresponding points.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i.e. I want to find H such that:&#xA;$$\left(\begin{array}{cc}&#xD;&#xA;h_{11} &amp;amp;  h_{12}\\&#xD;&#xA;h_{21} &amp;amp; h_{22}\end{array}\right)\left(\begin{array}{ccc}&#xD;&#xA;0 &amp;amp; a &amp;amp; a+b\\&#xD;&#xA;1 &amp;amp; 1 &amp;amp; 1&#xD;&#xA;\end{array}\right)   = &#xD;&#xA;\left(\begin{array}{ccc}&#xD;&#xA;0 &amp;amp;a&amp;#39; &amp;amp;a&amp;#39;+b&amp;#39;\\&#xD;&#xA;1 &amp;amp; 1 &amp;amp; 1&#xD;&#xA;\end{array}\right).$$   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I've got 6 equations here and only 3 unknowns. &#xA;(dof(H) = 4 elements less one for scaling = 3).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I thought about 3 scaling factors that would add up to 6 unknowns, s.t. we would have a unique solution. But how exactly do I insert the scaling factors into the matrices and how can I compute H then?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do you have a clue?&lt;/p&gt;&#xA;" OwnerUserId="3573" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-16T15:09:28.383" LastActivityDate="2010-12-02T16:25:39.533" Title="How to compute a 2x2 Homography out of 3 corresponding points?" Tags="&lt;linear-algebra&gt;&lt;projective-geometry&gt;" AnswerCount="2" />
  <row Id="10565" PostTypeId="2" ParentId="10547" CreationDate="2010-11-16T14:08:08.977" Score="4" ViewCount="" Body="&lt;p&gt;I'm not sure what $k$ is supposed to be, but another way to see that the first and third expressions are equal is to rewrite the first one as&#xA;$$\frac{\frac{\sin(A)}{\sin(B)}+1}{\frac{\sin(A)}{\sin(B)}-1}$$&#xD;&#xA;and then note that the law of sines tells you&#xA;$$\frac{\sin(A)}{\sin(B)} = \frac{a}{b}$$.&lt;/p&gt;&#xA;" OwnerUserId="822" LastActivityDate="2010-11-16T14:08:08.977" />
  <row Id="10567" PostTypeId="2" ParentId="10547" CreationDate="2010-11-16T14:38:14.857" Score="12" ViewCount="" Body="&lt;p&gt;From the $\sin $ rule, you know that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{a}{\sin A}=\frac{b}{\sin B}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;In a proportion such as&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{p}{q}=\frac{r}{s},$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;we have&lt;/em&gt; (&lt;em&gt;by a property of proportions, which you can check algebraically&lt;/em&gt;): &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{p+r}{p-r}=\frac{q+s}{q-s}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Setting $p=a$, $q=\sin A$, $r=b$, $s=\sin B$ gives you&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{a+b}{a-b}=\frac{\sin A+\sin B}{\sin A-\sin B}.$$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Added: Algebraic deduction:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\dfrac{p+r}{p-r}=\dfrac{q+s}{q-s}\Leftrightarrow \left( p+r\right) \left( q-s\right) -\left( p-r\right)\left( q+s\right) =0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Leftrightarrow 2qr-2ps=0\Leftrightarrow qr=ps\Leftrightarrow \dfrac{p}{q}=\dfrac{r}{s}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And clearly, for $k\neq 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\dfrac{q+s}{q-s}=\dfrac{k\left( q+s\right) }{k\left( q-s\right) }.$$&lt;/p&gt;&#xA;" OwnerUserId="752" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2011-06-01T11:07:05.467" LastActivityDate="2011-06-01T11:07:05.467" />
  <row Id="10568" PostTypeId="2" ParentId="10563" CreationDate="2010-11-16T14:40:22.763" Score="1" ViewCount="" Body="&lt;p&gt;Suppose that $(x_1,\ldots,x_n) \neq (y_1,\ldots,y_n)$. If we fix $x_2,\ldots,x_n$, then $f(x_1,x_2,\ldots,x_n)$ is constant with respect to $x_1$. Hence, $f(x_1,x_2,\ldots,x_n) = f(y_1,x_2,\ldots,x_n)$. Next, fix $y_1,x_3,\ldots,x_n$. Then, $f(y_1,x_2,\ldots,x_n)$ is constant with respect to $x_2$. Hence, $f(y_1,x_2,\ldots,x_n) = f(y_1,y_2,\ldots,x_n)$. Continue this way to conclude that $f(x_1,\ldots,x_n) = f(y_1,\ldots,y_n)$, that is, $f$ is constant.&lt;/p&gt;&#xA;" OwnerUserId="2810" LastActivityDate="2010-11-16T14:40:22.763" CommentCount="4" />
  <row Id="10569" PostTypeId="2" ParentId="10563" CreationDate="2010-11-16T14:50:21.667" Score="3" ViewCount="" Body="&lt;p&gt;For $\partial f/\partial x_i$ to vanish, then $f$&#xD;&#xA;has to vanish on the space of test functions of the form&#xA;$\partial \phi_i/\partial x_i$. You effectively need to&#xA;show the subspace spanned by all of these (as $i$ varies)&#xA;is of codimension one in $C_c(\mathbb{R}^n)$. This subspace&#xA;has to be the space of all test functions integrating to zero.&#xA;To go from&#xA;$n-1$ to $n$, take a test function $\phi$ with integral $0$ and let&#xA;$\psi(x_1,\ldots,x_{n-1})=\int\phi(x_1,\ldots,x_n)dx_n$.&#xA;Then $\psi$ is a test function on $\mathbb{R}^{n-1}$&#xD;&#xA;with integral $0$ so is a sum of functions&#xA;$\partial \psi_i/\partial x_i$ where the $\psi_i$ are test&#xA;functions on $\mathbb{R}^{n-1}$. Extend to test functions&#xA;$\phi_i$ on $\mathbb{R}^n$ by letting $\phi_i(x_1,\ldots,x_n)&#xD;&#xA;=\psi_i(x_1,\ldots,x_{n-1})h(x_n)$ where $h$ is a fixed bump function&#xA;with integral $1$. Then $\psi-\sum\partial\phi_i/\partial x_i$&#xD;&#xA;has zero integral along all lines in the $x_n$-direction,&#xA;so is the $x_n$-partial derivative of a test function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added&lt;/strong&gt;&#xA;I should add that the above is effectively the proof that&#xA;$H_c^n(\mathbb{R}^n)=\mathbb{R}$ where $H_c^*$ denotes de Rham&#xA;cohomology with compact supports.&lt;/p&gt;&#xA;" OwnerUserId="226" LastEditorUserId="226" LastEditorDisplayName="" LastEditDate="2010-11-16T16:24:02.230" LastActivityDate="2010-11-16T16:24:02.230" CommentCount="1" />
  <row Id="10570" PostTypeId="2" ParentId="10544" CreationDate="2010-11-16T15:01:26.880" Score="5" ViewCount="" Body="&lt;p&gt;So, you have a given set $X$, and a specific subset $A$. You know the cardinality/size of $A$, and want to know the cardinality of $X\setminus A$, the (relative) complement of $A$ in $X$. Use $|X|$ to denote the size/cardinality of the set. I will assume the Axiom of Choice so that all sets have comparable cardinalities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In two cases, this is completely determined by the cardinalities of $A$ and of $X$. They are:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;If $X$ is finite, of size $n$; then $A$ is finite, and the size of $X\setminus A$ is $|X|-|A|$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If $X$ is infinite, and the cardinality of $A$ is &lt;em&gt;strictly smaller&lt;/em&gt; than the cardinality of $X$, then the cardinality of $X\setminus A$ equals the cardinality of $X$.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;em&gt;Reason:&lt;/em&gt; If $\kappa$ and $\lambda$ are cardinals, and at least one of them is infinite, then $\kappa+\lambda = \kappa\lambda = \max\{\kappa,\lambda\}$. Here, $\kappa+\lambda$ is the cardinality of the disjoint union of a set of cardinality $\kappa$ and a set of cardinality $\lambda$; $\kappa\lambda$ is the cardinality of the set $X\times Y$, where $|X|=\kappa$ and $|Y|=\lambda$.&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;For example, $|\mathbb{R}\setminus\mathbb{Q}|=|\mathbb{R}|$, because $|\mathbb{Q}|=\aleph_0 \lt 2^{\aleph_0}=|\mathbb{R}|$. If $A$ is a finite subset of $|\mathbb{N}|$, then $|\mathbb{N}\setminus A|=\aleph_0 = |\mathbb{N}|$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, this is all you can say. If $X$ is infinite and $|A|=|X|$, then the cardinality of the complement $X\setminus A$ could be anything from $0$ and up to $|X|$. To get a set of size $n$ with $n$ finite, pick any subset $B$ of $|X|$ of size $n$, and take $A=X\setminus B$. To get a set with complement of cardinality $\kappa$ for any $\kappa\lt|X|$, biject $X$ with the &lt;em&gt;cardinal&lt;/em&gt; $|X|$, which has a subset of cardinality $\kappa$; take the complement. To get a set of cardinality $|X|$, use the fact that $|X|=|X\times X|$. Then, if $x\in X$ is a particular element, then the subset that corresponds under a given bijection to $X\times\{x\}$ has complement of size $|X|$. For specific sets $X$ it is possible to find specific examples. For $\mathbb{N}$, the even numbers have complement of size $|\mathbb{N}|$. In the real numbers, the unit interval has complement of size $|\mathbb{R}|$. And so on.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-16T19:11:09.470" LastActivityDate="2010-11-16T19:11:09.470" CommentCount="4" />
  <row Id="10571" PostTypeId="2" ParentId="10556" CreationDate="2010-11-16T15:21:56.960" Score="1" ViewCount="" Body="&lt;p&gt;Yes, I had a chance to see that.  But my issue is more related to matrix calculus than levenberg-marquardt.  To put it in simpler form, what's the difference between:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{\partial f(x)}{\partial x}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{\partial f(x)^T}{\partial x}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where f(x) is a R^n -&gt; R^n vector function.&lt;/p&gt;&#xA;" LastActivityDate="2010-11-16T15:21:56.960" CommentCount="3" />
  <row Id="10572" PostTypeId="1" AcceptedAnswerId="10591" CreationDate="2010-11-16T15:27:41.497" Score="5" ViewCount="102" Body="&lt;p&gt;Are there necessary and sufficient conditions for the existence of a circuit, or a disjoint set of circuits, that passes through each vertex once in a directed graph?&lt;/p&gt;&#xA;" OwnerUserId="314" LastActivityDate="2010-11-16T17:17:53.670" Title="Existence of circuit passing through each vertex in a directed graph" Tags="&lt;graph-theory&gt;" AnswerCount="3" />
  <row Id="10574" PostTypeId="2" ParentId="10564" CreationDate="2010-11-16T15:39:10.187" Score="1" ViewCount="" Body="&lt;p&gt;I believe the result should be in the form&#xA;$$\left(\begin{array}{ccc}&#xD;&#xA;0 &amp;amp; k_2a&amp;#39; &amp;amp; k_3(a&amp;#39;+b&amp;#39;)\\&#xD;&#xA;k_1 &amp;amp; k_2 &amp;amp; k_3&#xD;&#xA;\end{array}\right)$$&#xD;&#xA;with $k_1,k_2,k_3$ nonzero, to account for the homogeneous coordinates. Remember that the projective point $[x:y]$ is the same as the point $[kx:ky]$. So the middle answer doesn't have to look like $[a&amp;#39;:1]$, it can be any of the other representations of the point, hence $[k_2a&amp;#39;:k_2]$. That gives you three &quot;scaling&quot; factors. Is that what you were hoping for?&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-11-16T15:39:10.187" />
  <row Id="10575" PostTypeId="2" ParentId="2694" CreationDate="2010-11-16T15:41:08.240" Score="-2" ViewCount="" Body="&lt;p&gt;KCd,&#xA;I wouldn't agree with you about your assessment of a problem as important or unimportant--though in context and not in content. Whether or not something is important has been a question of the interests and attention of the particular researcher(s). If a large body of mathematical researchers find a mainstream topic interesting, then the questions arising from it become important to them. As a matter of fact, the importance of a mathematical problem is not REALISED until enormous effort from experts (and sometimes amateurs) have been expended to no avail. Fermat's last theorem was in the same state until the leading mathematiarans of the 19th century failed to settle it and some became &quot;ashamed&quot; that such a &quot;simple&quot; problem had not succumbed to the then prowess of the human mind. Until Kummer did a groundbreaking attack on FLT, which introduced &quot;new ideas&quot;, mathematicians wouldn't have elevated the status of FLT.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a RULE OF THUMB, you must know that EVERY DIFFICULT QUESTION is IMPORTANT until PROVEN OTHERWISE.  For as long as its proof is undiscovered, its importance is shroude in a mystery that makes many to naively refer to it as UNIMPORTANT. What would you say, if someone comes with an ingeniously simple and &quot;isolated&quot; proof of FLT? Would that diminish the importance of Wile's proof? Perhaps yes, afterall then FLT could be settled without the need of advanced machinery from cohomology, commutative algebra, L-functions, p-adic, etc &lt;/p&gt;&#xA;" OwnerUserId="3579" LastActivityDate="2010-11-16T15:41:08.240" CommunityOwnedDate="2010-11-16T15:41:08.240" />
  <row Id="10576" PostTypeId="2" ParentId="10572" CreationDate="2010-11-16T15:48:09.817" Score="1" ViewCount="" Body="&lt;p&gt;I don't think any are known. A quick search on the web gave &lt;a href=&quot;http://ajc.maths.uq.edu.au/pdf/23/ocr-ajc-v23-p115.pdf&quot; rel=&quot;nofollow&quot;&gt;a paper by Gutin and Yeo&lt;/a&gt;, which also directs to Chapter 1 of &lt;strong&gt;Digraphs - Theory, Algorithms, and Applications&lt;/strong&gt; by Bang-Jensen and Gutin (Springer Monographs, ISBN: 978-1-84800-997-4).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In any case, if there were necessary and sufficient conditions, then taking any graph and replacing each edge by a pair of directed edges going back-and-forth would give you a characterization of Hamiltonian graphs, if I'm not very confused, which is not known.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-11-16T15:48:09.817" />
  <row Id="10577" PostTypeId="1" AcceptedAnswerId="10607" CreationDate="2010-11-16T15:50:50.080" Score="3" ViewCount="116" Body="&lt;p&gt;I've come up with the following variation of a logarithm, intended to work on values that can be 0, or can grow exponentially from zero in either positive or negative direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$myLog(x) =&#xD;&#xA;\begin{cases}&#xD;&#xA; \log(x+1), &amp;amp; \mbox{if }x \geq 0 \\&#xD;&#xA; -\log(-x+1), &amp;amp; \mbox{if }x \lt 0&#xD;&#xA;\end{cases}&#xD;&#xA;$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a name for this formula or something similar to it? Has it been studied before? I'd like to read up on it if there is anything out there, but without knowing a name for it I haven't had much success with Google.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a graph of the function. It resembles arctangent, but it doesn't have asymptotes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/jWyYl.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="477" LastEditorUserId="477" LastEditorDisplayName="" LastEditDate="2010-11-16T16:02:55.680" LastActivityDate="2010-11-16T19:45:21.680" Title="Name for logarithm variation that works on non-positive values?" Tags="&lt;functions&gt;&lt;logarithms&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="10578" PostTypeId="2" ParentId="10543" CreationDate="2010-11-16T15:58:38.537" Score="1" ViewCount="" Body="&lt;p&gt;As a kid I really enjoyed &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0760723613&quot; rel=&quot;nofollow&quot;&gt;The Mathematical Tourist&lt;/a&gt; by Ivars Peterson (at the time, the 1988 edition).  The best part is that it gave just enough detail (for example, on how to compute a particular type of fractal) to inspire me to implement a computer program of my own to carry out the calculation or simulation.  In this way it really was a good guidebook for my own mathematical/computational explorations.&lt;/p&gt;&#xA;" OwnerUserId="2191" LastActivityDate="2010-11-16T15:58:38.537" />
  <row Id="10579" PostTypeId="1" CreationDate="2010-11-16T16:02:22.027" Score="3" ViewCount="136" Body="&lt;p&gt;What is the difference between:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) $\frac{\partial (\textbf{x}^{T}A)}{\partial \textbf{x}}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) $\frac{\partial (A\textbf{x})}{\partial \textbf{x}}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where A is a nxn matrix and x is a n sized column vector.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the definition of a Jacobian on wikipedia (http://en.wikipedia.org/wiki/Jacobian_matrix_and_determinant)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My answers are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) $A^{T}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) $A$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However this is not correct because:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\textbf{x}^{T}A = (A^{T}\textbf{x})^{T}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;performing $\frac{\partial}{\partial \textbf{x}}$ on each side results&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$A^T=A$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is not true.&#xA;Does $\frac{\partial}{\partial \textbf{x}}$  become $\frac{\partial}{\partial \textbf{x}^{T}}$ when moved inside the transpose on the right side?&lt;/p&gt;&#xA;" LastActivityDate="2010-11-17T04:01:02.673" Title="First order differential rule for matrices" Tags="&lt;calculus&gt;&lt;linear-algebra&gt;" AnswerCount="2" FavoriteCount="0" />
  <row Id="10580" PostTypeId="2" ParentId="10563" CreationDate="2010-11-16T16:02:36.047" Score="2" ViewCount="" Body="&lt;p&gt;If $f(x)$ is a smooth function, one gets that $f(x)$ is constant relatively easily as in Shai Covo's comment. For a general distribution, one can convolve with approximations to the identity: Let $\phi(x)$ be a fixed smooth function with compact support, and let $\phi_k(x) =  k^n\phi(kx)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Recall $f \ast \phi_k$ is a smooth function (the integral of $&amp;lt;f, \phi_y&amp;gt;$ with respect to $y$, where $\phi_y$ is the translate of $\phi$ by $y$) and is defined by $&amp;lt;f \ast \phi_k, s&amp;gt; = &amp;lt;f(x) , s \ast \phi_k(x)&amp;gt;$ for $C_c(R^n)$ functions $s(x)$. Since $s \ast \phi_k(x)$ converges to $s(x)$ uniformly, with the same property holding for derivatives of $s(x)$, $f \ast \phi_k$ converges to $f$ as distributions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But each $f \ast \phi_k(x)$ is a smooth function, and $&amp;lt;f \ast \phi_k, \partial_i s(x)&amp;gt;$ $=$  $&amp;lt;f(x), \partial_i s \ast \phi_k(x)&amp;gt;$ $=$  $&amp;lt;f(x), \partial_i (s \ast \phi_k(x))&amp;gt;$ $= 0$ by assumption. Hence by the smooth function case $f \ast \phi_k(x)$ is a constant. Thus $f(x)$ is the distributional limit of constants and must be constant itself.&lt;/p&gt;&#xA;" OwnerUserId="3035" LastActivityDate="2010-11-16T16:02:36.047" CommentCount="2" />
  <row Id="10582" PostTypeId="2" ParentId="10555" CreationDate="2010-11-16T16:33:20.230" Score="1" ViewCount="" Body="&lt;p&gt;Unfortunately I'm not versed in philosophical logic, so let's start with some &quot;different categories of logic&quot;.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt; Scope &lt;/h3&gt;&#xA;&#xA;&lt;p&gt;The first relevant category of logic is &lt;i&gt;scope&lt;/i&gt;, what is the logic talking about. &lt;b&gt;Propositional logic&lt;/b&gt; talks about nonparametric propositions. A typical rule (modus ponens) allows you to derive a proposition B given that you know A and that A implies B.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Predicate logic&lt;/b&gt; talks about parametric propositions, for example &quot;X is a poster&quot;. We can derive &quot;There's X such that X is a poster&quot; from &quot;Tim is a poster&quot;, whereas in propositional logic &quot;Tim is a poster&quot; is monolithic and cannot be broken up to &quot;Tim&quot; and &quot;X is a poster&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Predicate logic is also known as first-order logic. If we quantify over predicates then we get &lt;b&gt;Second order logic&lt;/b&gt;. This allows us to use concepts such as &quot;There's a predicate P such that P(X) iff X=Tim&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The scope is enlarged in a different direction by &lt;b&gt;Modal logic&lt;/b&gt;, where we can talk about events (propositions) eventually happening, happening until some other event happens, and so on.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt; Rules &lt;/h3&gt;&#xA;&#xA;&lt;p&gt;The rules of logic people are usually taught constitute &lt;b&gt;Classical logic&lt;/b&gt;. In classical logic you're allowed to use proof by contradiction. However, some mathematicians (well, logicians) dislike such rules since they're &lt;i&gt;non-constructive&lt;/i&gt; (this is like set-theorists who consider what happens when you're not allowed to use AC, the axiom of choice).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By modifying (restricting) the rules, we arrive at &lt;b&gt;Intuitionistic logic&lt;/b&gt; (or &lt;b&gt;Constructive mathematics&lt;/b&gt;), in which every proposition that we prove has a constructive proof from the givens. In particular, you can't use a proof of the sort &quot;if X, then Y; if not X, then also Y&quot; unless you can decide whether X is in fact true. The resulting world is nice - for example all functions are continuous. On the other hand, not every theorem you can prove in classical logic is true in the constructive sense, although surprisingly many are (for example, the fundamental theorem of algebra is still provable).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a different direction, in &lt;b&gt;Infinitary logic&lt;/b&gt; the proof is infinite and there are logical rules with infinitely many premises. For example, you could have a rule deducing &quot;P(n) for every natural n&quot; from the (countably) many propositions P(0), P(1), ... In general these logics are not well-behaved, but if you choose the parameters correctly, they are (look for $L_{\omega_1,\omega}$).&lt;/p&gt;&#xA;&#xA;&lt;h3&gt; Goals &lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Different areas of logic have different goals. For example, in &lt;b&gt;Structural Proof Theory&lt;/b&gt; one goal is to understand how simple a proof system can be made, and what are the consequences. Using this syntactic approach, you can show that certain statements are not provable in certain systems by converting them into a very simple form, bounding the resulting size of the simple proof, and show that it is too short to prove the given statement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A complementary direction is &lt;b&gt;Model Theory&lt;/b&gt;, which is all about what a given set of axioms describes. You can prove for example that a set of first-order axioms cannot characterize the size of the universe (if it's infinite). The model-theoretic approach to showing that some theorem is unprovable is to exhibit a model of the axioms in which the theorem isn't true.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In computer science, people are interested in &lt;b&gt;Proof Complexity&lt;/b&gt;, which studies how long it takes to prove statements if you're only allowed to use basic means. The holy grail is proving a statement which is even stronger than $P \neq NP$. A different sub-field constructs systems in which every predicate proven to exist is efficiently computable.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt; Categorical logic &lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Unfortunately I don't know anything about &lt;i&gt;categorical logic&lt;/i&gt;. Apparently there are some kinds of categories (in the sense of category theory) which describe &quot;universes of set theory&quot; along with their associated logic, which is often constructive. These things are called &lt;b&gt;Topoi&lt;/b&gt; (singular &lt;b&gt;Topos&lt;/b&gt;).&lt;/p&gt;&#xA;&#xA;&lt;h3&gt; Philosophers' logic &lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Philosophers are interested in different aspects of logic. Unfortunately, I don't know much about these, so I'll only give some sporadic examples.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is some interest in nice, succinct systems of logic whose axioms are &quot;as natural as possible&quot;. In mathematical logic, by and large people aren't interested in the exact form of the axioms but in what they describe and what can be proven about the system using them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other interest is in what the allowable rules of logic should be. I described classical logic and intuitionistic logic. There are many systems in between (and even systems weaker than the latter), which might be natural given some philosophical stance. Most mathematicians don't even bother with intuitionistic logic, although some logicians (and even people from other branches) are intrigued by it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's also some interest in making sense of paradoxes - given a statement which is paradoxical (for example, the Liar paradox), what should its truth value be? The mathematical solution is to avoid self-reference in some systematic way, since mathematicians are more interested (in this case) in the well-being of the system rather than these &quot;monstrous examples&quot;.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-16T16:33:20.230" />
  <row Id="10583" PostTypeId="2" ParentId="10572" CreationDate="2010-11-16T16:46:58.850" Score="2" ViewCount="" Body="&lt;p&gt;A collection of disjoint cycles that cover all vertices of an (undirected) graph is known as a 2-factor. &lt;a href=&quot;http://www.is.kochi-u.ac.jp/~fujisawa/papers/08AFS.pdf&quot; rel=&quot;nofollow&quot;&gt;This paper&lt;/a&gt; references some criterion of Tutte for the existence of these, which, paraphrasing the authors, in some sense completely solves the problem. It's Theorem 1 on page 2, which points to &quot;The factors of graphs&quot; by Tutte (1952).&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-16T16:46:58.850" CommentCount="4" />
  <row Id="10584" PostTypeId="2" ParentId="10579" CreationDate="2010-11-16T16:49:03.767" Score="2" ViewCount="" Body="&lt;p&gt;There is actually a hugh difference. For one, when you say $\frac{\partial (x^TA)}{\partial x}$, the numerator is a $1 \times n$ vector. Whereas in case of $\frac{\partial (Ax)}{\partial x}$, the numerator is a $n \times 1$ vector.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Look up this [wiki][1] page for details on how matrix derivatives should be done and the rules.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You may want to look up &lt;a href=&quot;http://matrixcookbook.com/&quot; rel=&quot;nofollow&quot;&gt;http://matrixcookbook.com/&lt;/a&gt; which gives you a lot of matrix identities like these.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[1]: &lt;a href=&quot;http://en.wikipedia.org/wiki/Matrix_calculus&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Matrix_calculus&lt;/a&gt; page&lt;/p&gt;&#xA;" OwnerUserId="3302" LastActivityDate="2010-11-16T16:49:03.767" />
  <row Id="10585" PostTypeId="2" ParentId="10547" CreationDate="2010-11-16T16:56:37.890" Score="4" ViewCount="" Body="&lt;p&gt;The sine rule for a triangle goes as &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{a}{\sin(A)} = \frac{b}{\sin(B)} = \frac{c}{\sin(C)} = 2R$, where $R$ is the radius of the circumcircle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From this, we get $\sin(A) = \frac{a}{2R}$ and $\sin(B) = \frac{b}{2R}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem has $\frac{\sin(A) + \sin(B)}{\sin(A) - \sin(B)}$. Plug in the expressions for $\sin(A)$ and $\sin(B)$ from the above and we get,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{\frac{a}{2R}+\frac{b}{2R}}{\frac{a}{2R}-\frac{b}{2R}} = \frac{\frac{a+b}{2R}}{\frac{a-b}{2R}} = \frac{a+b}{a-b}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, $\frac{\sin(A) + \sin(B)}{\sin(A) - \sin(B)} = \frac{a+b}{a-b}$&lt;/p&gt;&#xA;" OwnerUserId="3302" LastActivityDate="2010-11-16T16:56:37.890" />
  <row Id="10586" PostTypeId="2" ParentId="7195" CreationDate="2010-11-16T17:08:14.367" Score="0" ViewCount="" Body="&lt;p&gt;Observe this formulas:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \sin^2 u = \frac{1 - \cos 2u}{2} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \cos^2 u = \frac{1 + \cos 2u}{2} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then:&lt;br&gt;&#xA;$$ \int \frac{1}{1 - \cos 2u} \, du = - \frac{\cot u}{2} $$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using of this formulas and partial integration formula you can easily get last part &lt;a href=&quot;http://integrals.wolfram.com/index.jsp?expr=%281-m%2ASin%5Bx%5D%5E2%29%5E%281%2F2%29%2FSin%5Bx%5D%5E2&amp;amp;random=false&quot; rel=&quot;nofollow&quot; title=&quot;mathematica online integrator&quot;&gt;of the answer&lt;/a&gt;.&lt;br&gt;&#xA;Maybe this will help you.&lt;/p&gt;&#xA;" OwnerUserId="1035" LastEditorUserId="45" LastEditorDisplayName="" LastEditDate="2011-03-27T05:34:02.060" LastActivityDate="2011-03-27T05:34:02.060" />
  <row Id="10587" PostTypeId="2" ParentId="1730" CreationDate="2010-11-16T17:08:44.147" Score="0" ViewCount="" Body="&lt;p&gt;One way I heard a lecturer describe functions recently was that of the CD player analogy.  &lt;/p&gt;&#xA;" OwnerUserId="9516" LastActivityDate="2010-11-16T17:08:44.147" />
  <row Id="10588" PostTypeId="1" AcceptedAnswerId="10590" CreationDate="2010-11-16T17:13:37.003" Score="3" ViewCount="225" Body="&lt;p&gt;I need to prove that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$P(A \backslash B) = P(A) - P(A\cap B)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$P(A\cup B) = P(A) + P(B) - P(A\cap B)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;using the axioms of probability but can't see where to start. Note: this is homework, so a hint only would be nice :)&lt;/p&gt;&#xA;" OwnerUserId="1973" LastEditorUserId="1973" LastEditorDisplayName="" LastEditDate="2010-11-16T17:41:01.567" LastActivityDate="2010-11-16T17:41:01.567" Title="Proofs using the Axioms of Probability" Tags="&lt;probability&gt;&lt;homework&gt;&lt;axioms&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="10589" PostTypeId="2" ParentId="10588" CreationDate="2010-11-16T17:17:04.340" Score="2" ViewCount="" Body="&lt;p&gt;Well,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $B \subset A$, then $A = B \cup (A \setminus B)$ which is a disjoint union so...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now,&#xA;$$A \cup B = (A \setminus (A \cap B)) \cup (A \cap B) \cup (B \setminus (A \cap B))$$&#xD;&#xA;which is again a disjoint union. So...&lt;/p&gt;&#xA;" OwnerUserId="1120" LastActivityDate="2010-11-16T17:17:04.340" />
  <row Id="10590" PostTypeId="2" ParentId="10588" CreationDate="2010-11-16T17:17:09.817" Score="3" ViewCount="" Body="&lt;p&gt;Hint: Think &quot;countable additivity&quot;. What can you say about $A\setminus B$ and $A\cap B$ if you are hoping to use this axiom.&lt;/p&gt;&#xA;" OwnerUserId="3431" LastActivityDate="2010-11-16T17:17:09.817" />
  <row Id="10591" PostTypeId="2" ParentId="10572" CreationDate="2010-11-16T17:17:53.670" Score="3" ViewCount="" Body="&lt;p&gt;According to the classic text, &quot;Computers and Intractability, A Guide to the Theory of NP-Completeness&quot; by Garey and Johnson, the following problem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Partition into Hamiltonian Subgraphs&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Given a directed graph $G=(V,A)$, can&#xA;  the vertices be partitioned into&#xA;  disjoint sets $V_1, V_2, \dots, V_k$&#xD;&#xA;  for some $k$, such that each $V_i$&#xD;&#xA;  contains at least three vertices and&#xA;  induces a subgraph of $G$ that&#xA;  contains a Hamiltonian Circuit.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;is NP-Complete, by a reduction from 3SAT.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The book also mentions that if we allow each $V_i$ to contain at least &lt;em&gt;two&lt;/em&gt; vertices, then this is solvable in polynomial time using Matching techniques.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-11-16T17:17:53.670" />
  <row Id="10592" PostTypeId="2" ParentId="10588" CreationDate="2010-11-16T17:24:15.400" Score="1" ViewCount="" Body="&lt;p&gt;The second one as written is false.  I'm sure you meant&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$P(A \cup B) = P(A) + P(B) - P(A \cap B)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Jonas' answer is a good hint on how to apply the first part to prove the second part.&lt;/p&gt;&#xA;" OwnerUserId="3537" LastActivityDate="2010-11-16T17:24:15.400" />
  <row Id="10593" PostTypeId="1" CreationDate="2010-11-16T17:35:58.997" Score="1" ViewCount="268" Body="&lt;p&gt;A manifold M is simply connected if for every pair of 1-cubes c1,c2: [0,1] -&gt; M with&lt;br&gt;&#xA;c1(0) = c1(1) = c2(0) = c2(1) = t&lt;br&gt;&#xA;there is a 2-cube b such that&lt;br&gt;&#xA;1) b1,0 = c1 and b1,1 = c2&lt;br&gt;&#xA;2) for all p in [0,1] b2,0(p) = b2,1(p) = t  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is, if I'm given a manifold M,&lt;br&gt;&#xA;for example choose M = S^2 = {(x,y,z) in R^3 such that x^2 + y^2 + z^2 = 1}&lt;br&gt;&#xA;how do I prove that it is simply connected?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, how would you prove that a manifold is not simply connected?&lt;/p&gt;&#xA;" OwnerUserId="3333" LastEditorUserId="3333" LastEditorDisplayName="" LastEditDate="2010-11-17T02:12:01.550" LastActivityDate="2010-12-08T07:07:36.730" Title="How to prove a manifold is simply connected" Tags="&lt;differential-geometry&gt;" AnswerCount="2" CommentCount="14" />
  <row Id="10595" PostTypeId="1" AcceptedAnswerId="10600" CreationDate="2010-11-16T18:01:38.653" Score="3" ViewCount="193" Body="&lt;p&gt;Given a countable family of semi-norms $p_i$, we can define a metric&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$d(f,g) = \sum \limits_{i=0}^{\infty} 2^{-i} \frac{ p_i(f-g) }{ 1 + p_i(f-g) }$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have the locally convex topology induced by the semi-norms as above, as well as the topology induced by the metric.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How does the proof work to show their equality?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am known to the proof of Rudin (Functional analysis), but utilizes a different metric:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$d(f,g) = \max \limits_{i \in \mathbb N}  2^{-i} \frac{ p_i(f-g) }{ 1 + p_i(f-g) }$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whereas the proof for this metric is fairly easy - you can handle value of the sequence on its own - i do not see how a similar proof might work for the first metric.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One guess would be to show equivalence of both metrics, but I don't see even that, as on $l^1$, the sum-norm-topology is strictly finer than the max-norm-topology.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you help me?&lt;/p&gt;&#xA;" OwnerUserId="3557" LastEditorUserId="3557" LastEditorDisplayName="" LastEditDate="2010-11-16T19:08:59.770" LastActivityDate="2010-11-17T09:25:20.047" Title="Proof that Frechet-metric generates same topology as the semi-norms" Tags="&lt;functional-analysis&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="10596" PostTypeId="2" ParentId="10555" CreationDate="2010-11-16T18:15:43.487" Score="2" ViewCount="" Body="&lt;p&gt;Any formal system consists of:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;a list of symbols (a vocabulary or alphabet),&lt;br&gt;&lt;/li&gt;&#xA;&lt;li&gt;a list of of typographical rules to connect these symbols to write statements (a grammar or syntax), and&lt;br&gt;&lt;/li&gt;&#xA;&lt;li&gt;a list typographical rules for deriving new statements from other statements (axioms or rules of inference). &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;By typographical rules, I mean rules that tell you only how to manipulate the symbols without reference to any meaning or interpretation that these symbols might have. Consider for example a rule of inference in propositional logic that I call  the Join Rule:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If X and Y are true statements, then so is &quot;(X) &amp;amp; (Y)&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice that I make no reference in the statement of this rule to any possible meaning of the &quot;&amp;amp;&quot; symbol. (You can probably guess that it is the AND-operator.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can play around with my own implementation of formal propositional logic using a freeware program I have developed, available at my website &lt;a href=&quot;http://www.dcproof.com&quot; rel=&quot;nofollow&quot;&gt;http://www.dcproof.com&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3515" LastActivityDate="2010-11-16T18:15:43.487" />
  <row Id="10597" PostTypeId="2" ParentId="10452" CreationDate="2010-11-16T18:29:21.037" Score="1" ViewCount="" Body="&lt;p&gt;Both identities can be proved quite easily with inductions. Let $d$ be the common difference, i.e. $d=a_{n+1}-a_n$.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Use induction. So we need to prove&#xA;$$ \frac{n-2}{\sqrt{a_1}+\sqrt{a_ {n-1}}}+\frac{1}{\sqrt{a_{n-1}}+\sqrt{a_ n}}=\frac{n-1}{\sqrt{a_1}+\sqrt{a_ n}}.$$&#xD;&#xA;Rationalize the denominators and substitute $$a_{n-1}-a_1=(n-2)d, a_n-a_{n-1}=d, a_n-a_1=(n-1)d.$$ Upon cancellations, we find&#xA;$$\sqrt{a_{n-1} }-\sqrt{a_ 1}+\sqrt{a_n }-\sqrt{a_{n-1} }=\sqrt{a_ n}-\sqrt{a_ 1}$$ on the left and the same on the right.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Use induction, assuming true for $m&amp;lt;n$. Canceling the sum $$ \frac{1}{a_1a_n}+\frac{1}{a_na_1}$$ from the left with the term $$\frac{2}{a_1+a_n}\left(\frac{1}{a_1}+\frac{1}{a_n}\right)$$ on the right and using the induction hypothesis on the remaining terms $$\frac{1}{a_2a_{n-1}}+\cdots +\frac{1}{a_{n-1}a_2},$$ we find that  we need to prove  $$\frac{2}{a_2+a_{n-1}}$$ on the left is equal to $$\frac{2}{a_1+a_{n}}$$ on the right. And this is true because $a_2+a_{n-1}=a_1+a_n=2a_1+(n-1)d$. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="3249" LastActivityDate="2010-11-16T18:29:21.037" />
  <row Id="10598" PostTypeId="2" ParentId="10595" CreationDate="2010-11-16T18:37:48.123" Score="1" ViewCount="" Body="&lt;p&gt;Assuming that you forgot a $2^{-i}$ in the definition of $d$ we argue as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know that a Hausdorff TVS is metrizable iff zero has a countable neighborhood basis. If this is true then the topology is even generated by a translation invariant metric.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another theorem states that a topology on a TVS is locally convex iff it is generated by a family of seminorms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, if we combine these and use that your metric is translation invariant, we obtain the result. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, this is indirect, probably there is a direct proof.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;: Oh, I don't think the direct proof is so hard, I will see if I have time to add it. You should just observe that the neighborhood $\{x : p_n(x) \leq 2^{-k}\}$ contains the set $\{x : |x| \leq 2^{-m-k-1} \}$ and that the set $\{x : |x| \leq 2^{-k} \}$ contains the neighborhood $\{x : p_{k + 1}(x) \leq 2^{-k - 2} \}$. Where $|x| := d(x,0)$.&lt;/p&gt;&#xA;" OwnerUserId="1120" LastEditorUserId="1120" LastEditorDisplayName="" LastEditDate="2010-11-16T23:39:36.793" LastActivityDate="2010-11-16T23:39:36.793" CommentCount="3" />
  <row Id="10599" PostTypeId="1" CreationDate="2010-11-16T18:38:10.997" Score="2" ViewCount="70" Body="&lt;p&gt;Let $p:E\rightarrow B$ be a Serre fibration of path connected spaces with fiber $F$. Are the connecting homomorphisms $\partial:\pi_{n+1}(B)\rightarrow \pi_{n}(F)$ in the long exact sequence of $p$ induced by a &lt;strong&gt;continuous&lt;/strong&gt; map $\Omega B\rightarrow F$?&lt;/p&gt;&#xA;" OwnerUserId="2423" LastActivityDate="2010-11-16T19:34:10.190" Title="Connecting Homomorphism in LES of fibration" Tags="&lt;algebraic-topology&gt;" AnswerCount="1" />
  <row Id="10600" PostTypeId="2" ParentId="10595" CreationDate="2010-11-16T18:49:57.670" Score="2" ViewCount="" Body="&lt;p&gt;To prove equivalence of the topologies you can use the fact that both topologies are characterized by their convergent sequences.  This is true in any first countable space.  Metric spaces are first countable, and the topology induced by a countable family of seminorms is first countable.  Here, since you have translation invariance, it is enough to check sequences converging to 0 (although this simplification is not essential).  That is, you can show that if $x_1,x_2,\ldots$ is a sequence in your vector space, then $d(x_n,0)\to 0$ as $n\to\infty$ if and only if for all $i$, $p_i(x_n)\to0$ as $n\to\infty$.  For the left to right direction, you can use the fact that $\frac{p_i}{1+p_i}\leq 2^id$.  For the other direction you can first bound the tail, then work with the remaining finitely many terms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is no actual need to consider sequences.  You basically want to show that the identity map is a homeomorphism, and this reduces to showing it is continuous at 0 in both directions, which in any case involves the same estimates you would make when working with sequences. (In other words, considering first countability a priori isn't necessary.  If you really want to, you could instead work with nets.)  Note that Jonas T's answer suggests a complementary approach (that was given around the same time), working directly with neighborhoods of $0$ instead of with sequences.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2010-11-17T09:25:20.047" LastActivityDate="2010-11-17T09:25:20.047" CommentCount="3" />
  <row Id="10601" PostTypeId="2" ParentId="1099" CreationDate="2010-11-16T19:08:05.717" Score="0" ViewCount="" Body="&lt;p&gt;Barnsley's book FRACTALS EVERYWHERE has many examples of this.  I believe some edition of that book (long ago...) came with software that lets you make your own examples, too.&lt;/p&gt;&#xA;" OwnerUserId="442" LastActivityDate="2010-11-16T19:08:05.717" />
  <row Id="10602" PostTypeId="2" ParentId="10555" CreationDate="2010-11-16T19:08:50.810" Score="2" ViewCount="" Body="&lt;p&gt;Informal logic is, well, informal.  As such it is not unambiguously defined what the subject matter is, how different statements are related to each other, or (more ambitiously) what statements &quot;mean&quot;.   This type of activity is more popular in philosophy, linguistics and other areas that are not specifically mathematical.  Modern interest in informal reasoning is often about how to formalize it, as in AI, natural language processing or data mining.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Formalized logic is as described in the other answers.  Essentially, there are many types of formal logic, and each one is a game with its own well-defined rules (in principle, performable by a computer) for taking a set of statements -- informally thought of as &quot;premises&quot; -- and adding to it some additional statements, informally thought of as &quot;conclusions&quot;.  In spirit it is very close to computer programming and its study as a field resembles computer science. Some branches such as model theory or set theory are studied purely as mathematics in the sense that the explicitly combinatorial/linguistic aspect is in the background and the sentences are studied in a &quot;semantic&quot; approach that speaks in terms of the structures, objects and their properties (the material supposedly described by the logic) rather than the permissible syntactic operations in a formal logical system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mathematics is intermediate between these two extremes.  It uses logic(s) that could be called &lt;em&gt;potentially&lt;/em&gt; (or presumably, or in-principle) &lt;em&gt;formalizable&lt;/em&gt;.  Steps of the game are not codified precisely enough to be programmed on a computer, but they are standardized and sufficiently well-defined to correspond to known genuine moves in some actual formal systems, so there is a presumption that any correct non-formalized proof is robust enough to have many slightly different expansions into a very detailed formal proof in several of the formal systems suggested as a &quot;foundation&quot; for mathematics.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You mentioned probabilistic or statistical reasoning.   This is a somewhat separate question because any deterministic setting can be enlarged a probabilistic one (e.g., instead of measuring attributes or their absence by $1$ or $0$, allow any numerical value between $1$ and $0$) and the logical rules modified accordingly in a known way.  Similarly, any setting that uses ordinary or probabilistic logic on objects thought of as &quot;data&quot; can be placed in the context of statistics, where one considers how the data might have been generated in addition to the data itself, and the rules for reasoning and speaking about this richer picture can again be updated in a routine way.   But these probabilistic/fuzzy/statistical enrichments are a kind of fixed recipe where the &quot;logic&quot; and &quot;probabilistic&quot; ingredients are cooked separately and mixed in an understood way.  You could discuss the result as a new form of logic but really it is a repackaging of what you started with, not a fundamentally different mode of reasoning.&lt;/p&gt;&#xA;" OwnerUserId="467" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-11-16T19:34:01.657" LastActivityDate="2010-11-16T19:34:01.657" />
  <row Id="10603" PostTypeId="1" AcceptedAnswerId="10674" CreationDate="2010-11-16T19:24:48.690" Score="10" ViewCount="261" Body="&lt;p&gt;What do the finite-dimensional continuous complex representations of the additive group $\mathbb{Q}$ with the usual topology look like?  With the discrete topology?  Which representations are indecomposable?  Irreducible?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only ones I can think of are of the form $t \mapsto e^{tA}$ for some $A \in \mathcal{M}_n(\mathbb{C})$.  I would be willing to believe that these are the only ones in the first case, but I'm less sure in the second case.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-11-17T09:25:13.387" Title="Representation theory of the additive group of the rationals?" Tags="&lt;representation-theory&gt;" AnswerCount="3" FavoriteCount="1" />
  <row Id="10604" PostTypeId="2" ParentId="10599" CreationDate="2010-11-16T19:28:30.387" Score="3" ViewCount="" Body="&lt;p&gt;Consider the path fibration $PB \to B$, assuming you've chosen a base-point in $B$.  This map is null homotopic.  A constant map $PB \to B$ lifts to unique base-point preserving map $PB \to E$ (constant the base-point in $F$).  So provided your fibration satisfies the homotopy lifting property for the special case of the path fibration over $B$, you get a map $PB \to E$ covering the path fibration $PB \to E$.  Restricting this to the loop space $\Omega B \subset PB$ gives a map $\Omega B \to F$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I presume it's known and pretty easy to check whether or not Serre fibrations have this property, but certainly general fibrations have them (say, taking the fibration definition in Hatcher's book). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Were you interested in this technical detail on the distinction between fibrations and Serre fibrations, or were you more interested in the general question of when the connecting homomorphism is induced by an actual map of spaces? &lt;/p&gt;&#xA;" OwnerUserId="642" LastEditorUserId="642" LastEditorDisplayName="" LastEditDate="2010-11-16T19:34:10.190" LastActivityDate="2010-11-16T19:34:10.190" CommentCount="2" />
  <row Id="10605" PostTypeId="2" ParentId="10346" CreationDate="2010-11-16T19:34:54.637" Score="5" ViewCount="" Body="&lt;p&gt;We can prove this using the &lt;a href=&quot;http://en.wikipedia.org/wiki/Multinomial_theorem&quot; rel=&quot;nofollow&quot;&gt;Multinomial Theorem&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have, using the Multinomial Theorem, that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle (a+b+c)^n = \sum_{m=0}^{n} \sum_{k=m}^{n} {n \choose k}{k \choose m} a^m b^{k-m} c^{n-k}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Set $\displaystyle a = pp_j x$, $b = p(1-p_j)$, $c = 1-p$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle (pp_jx + p(1-p_j) + 1-p )^n = \sum_{m=0}^{n} \sum_{k=m}^{n} {n \choose k}{k \choose m} (pp_jx)^m (p(1-p_j))^{k-m} (1-p)^{n-k}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i.e.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle (pp_jx + 1 - pp_j)^n = \sum_{m=0}^{n} \sum_{k=m}^{n} {n \choose k}{k \choose m} p_j^m(1-p_j)^{k-m} p^{k}(1-p)^{n-k} x^m$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Expanding the left hand side using binomial theorem and comparing the coefficients of $\displaystyle x^m$ gives the result.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-11-16T21:17:10.470" LastActivityDate="2010-11-16T21:17:10.470" />
  <row Id="10606" PostTypeId="1" AcceptedAnswerId="10621" CreationDate="2010-11-16T19:44:47.747" Score="1" ViewCount="125" Body="&lt;p&gt;I'm familiar with the notation $\mathrm{ord}_a(x)$, when $x$ is an integer.  However, I'm reading a book where this notation is used with $x$ rational.  I'm not sure how to interpret this?&lt;/p&gt;&#xA;" OwnerUserId="2212" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-16T23:43:43.170" LastActivityDate="2010-11-16T23:43:43.170" Title="Notation: $\mathrm{ord}_a(x)$ for $x\in \mathbb{Q}$" Tags="&lt;number-theory&gt;&lt;notation&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="1" />
  <row Id="10607" PostTypeId="2" ParentId="10577" CreationDate="2010-11-16T19:45:21.680" Score="8" ViewCount="" Body="&lt;p&gt;Given your requirements, here's how you could have invented the inverse hyperbolic sine, even if you had never heard of it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You want a monotonic function that grows logarithmically at both infinities. You try adding, subtracting, and multiplying logarithmic functions, but that doesn't work. However, since the function you want is monotonic, you can think about the &lt;em&gt;inverse&lt;/em&gt; of your function, which must grow &lt;em&gt;exponentially&lt;/em&gt; at both infinities. And adding together exponentials does do the job for that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the inverse of the function you invented, which I'll call $f$. Its inverse is $$f^{-1}(y) = \begin{cases}e^y - 1 &amp;amp; \text{if } y \ge 0, \\ 1 - e^{-y} &amp;amp; \text{if } y \lt 0.\end{cases}$$ Since $e^{-y} \le 1$ when $e \ge 0$, and $e^y &amp;lt; 1$ when $y &amp;lt; 0$, the above inverse function is quite close to the much more nicely-behaved $$\begin{align}g(y) &amp;amp;= e^y - e^{-y} \\ &amp;amp;\equiv 2 \sinh y.\end{align}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in some sense, $f(x)$ behaves like $\sinh^{-1} (x/2)$. As it turns out, they agree at $x = 0$ (although their derivatives differ by a factor of two), and as $x \to \infty$, they both approach $\ln x$ and the difference between them decreases as $1/x$.&lt;/p&gt;&#xA;" OwnerUserId="856" LastActivityDate="2010-11-16T19:45:21.680" CommentCount="1" />
  <row Id="10608" PostTypeId="2" ParentId="2694" CreationDate="2010-11-16T19:49:13.757" Score="35" ViewCount="" Body="&lt;p&gt;Most of the answers so far have been along the general lines of 'Why hard problems are important', rather than 'Why the Collatz conjecture is important'; I will try to address the latter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the basic question being touched on is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In what ways does the prime factorization of $a$ affect the prime factorization of $a+1$?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Of course, one can always multiply out the prime factorization, add one, and then factor again, but this throws away the information of the prime factorization of $a$.  Note that this question is also meaningful in other UFDs, like $\mathbb{C}[x]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems very hard to come up with answers to this question that don't fall under the heading of 'immediate', such as distinct primes in each factorization.  This seems to be in part because a small change in the prime factorization for $a$ (multiplication by a prime, say) can have a huge change in the prime factorization for $a+1$ (totally distinct prime support perhaps).  Therefore, it is tempting to regard the act of adding 1 as an essentially-random shuffling of the prime factorization.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most striking thing about the Collatz conjecture is that it seems to be making a deep statement about a subtle relation between the prime factorizations of $a$ and $a+1$.  Note that the Collatz iteration consists of three steps; two of which are 'small' in terms of the prime factorization, and the other of which is adding one:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;multiplying by 3 has a small effect on the factorization.&lt;/li&gt;&#xA;&lt;li&gt;adding 1 has a (possibly) huge effect on the factorization.&lt;/li&gt;&#xA;&lt;li&gt;factoring out a power of 2 has a small effect on the factorization (in that it doesn't change the other prime powers in the factorization).&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So, the Collatz conjecture seems to say that there is some sort of abstract quantity like 'energy' which is cannot be arbitrarily increased by adding 1.  That is, no matter where you start, and no matter where this weird prime-shuffling action of adding 1 takes you, eventually the act of pulling out 2s takes enough energy out of the system that you reach 1.  I think it is for reasons like this that mathematicians suspect that a solution of the Collatz conjecture will open new horizons and develop new and important techniques in number theory.&lt;/p&gt;&#xA;" OwnerUserId="1062" LastActivityDate="2010-11-16T19:49:13.757" CommunityOwnedDate="2010-11-16T19:49:13.757" CommentCount="4" />
  <row Id="10609" PostTypeId="2" ParentId="10603" CreationDate="2010-11-16T19:52:12.413" Score="4" ViewCount="" Body="&lt;p&gt;In the usual topology, each continuous&#xA;representation of $(\mathbb{Q},{+})$ will extend to a continuous&#xA;representation of $(\mathbb{R},{+})$ so you'll only get the obvious ones.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the discrete topology there are non-obvious representations even&#xA;in dimension one. Consider the map&#xA;$\alpha:\mathbb{Q}\to \mathbb{Q}_p\to\mathbb{Q}_p/\mathbb{Z}_p&#xD;&#xA;\cong \mathbb{Z}[1/p]/\mathbb{Z}$. Then $t\mapsto\exp(2\pi i\alpha(t))$&#xD;&#xA;is a character of $\mathbb{Q}$, but not of the form $t\mapsto e^{at}$.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-11-16T19:52:12.413" />
  <row Id="10610" PostTypeId="2" ParentId="10606" CreationDate="2010-11-16T19:53:06.580" Score="5" ViewCount="" Body="&lt;p&gt;For any rational number $x$ and any prime integer $p$, we can write $x=p^r\frac{m}{n}$ uniquely (where $r$ is an integer thanks to Yuval for clarifying that), where $m,n$ are integers relatively prime to $p$. Then, we write $ord_p(x)=r$&lt;/p&gt;&#xA;" OwnerUserId="3431" LastEditorUserId="3431" LastEditorDisplayName="" LastEditDate="2010-11-16T22:59:18.887" LastActivityDate="2010-11-16T22:59:18.887" CommentCount="2" />
  <row Id="10611" PostTypeId="2" ParentId="10231" CreationDate="2010-11-16T19:58:25.793" Score="1" ViewCount="" Body="&lt;p&gt;No, the answer is definitely not zero. But, of course, you have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \sum_{k=0}^{36} \sin(n\times 10&amp;#176;) = 0 $$&lt;/p&gt;&#xA;" OwnerUserId="3406" LastActivityDate="2010-11-16T19:58:25.793" />
  <row Id="10612" PostTypeId="2" ParentId="2694" CreationDate="2010-11-16T20:01:18.167" Score="8" ViewCount="" Body="&lt;p&gt;I believe the $3x+1$ problem is considered a test case for ergodic theory, i.e., proving that certain probabilistic expectations true are true for orbits of a specific system.  There are papers by Sinai, Lagarias and others giving probabilistic models, similar to the asymptotic predictions made in questions about distribution of prime numbers, where the predictions are reliable but proving them in any particular case (twin primes, $n^2+1$, Goldbach, ...) is a centuries-old open problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is analogous to transcendence or irrationality proofs of specific numbers: proving that what is &quot;true with probability 1&quot; really holds in a particular instance is extremely hard and pushes the theory forward.  Other than that there are no problems outside the $3x+1$ conjecture that use the same iteration, so it is a sink and not a source in the applications-of-theory graph.&lt;/p&gt;&#xA;" OwnerUserId="467" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-11-16T20:06:52.173" LastActivityDate="2010-11-16T20:06:52.173" CommunityOwnedDate="2010-11-16T20:01:18.167" CommentCount="1" />
  <row Id="10613" PostTypeId="2" ParentId="10545" CreationDate="2010-11-16T20:15:29.453" Score="1" ViewCount="" Body="&lt;p&gt;I tried a geometry proof for this and thought it was worth sharing it here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am unable to draw the figure. I would appreciate if some one could take the effort to draw the picture or let me know what is the easiest way to draw such pictures.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the triangle $ABC$. Let the angle bisector of $A$ meet $BC$ at $D$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By angle bisector theorem, we get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{AB}{AC} = \frac{DC}{DB}$. Hence, we get $\frac{b}{c} = \frac{DC}{DB}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Add one to both sides, we get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{b+c}{c} = \frac{DC+DB}{DB} = \frac{a}{DB}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rearranging we get $\frac{c}{DB} = \frac{b+c}{a} = \cot(\frac{A}{2})$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A similar argument (or using the fact that $\frac{c}{DB} = \frac{b}{DC}$) yields, $\frac{b}{DC} = \frac{b+c}{a} = \cot(\frac{A}{2})$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, we are almost done. (Note that if $\angle{B} = \frac{\pi}{2}$ or $\angle{C} = \frac{\pi}{2}$, the above is true.) However, we still need to prove that for no other possibility this is satisfied.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To prove the above. Draw a perpendicular to $AB$ at $B$ and similarly draw a perpendicular to $AC$ at $C$. Let these two will intersect $AD$ at $D&amp;#39;$ and $D&amp;#39;&amp;#39;$. Now note that $BD&amp;#39; = BD$ since $\frac{BD}{AB} = \cot(\frac{A}{2}) = \frac{BD&amp;#39;}{AB}$ (Since $D&amp;#39;BA$ is a right angled triangle). Similarly $CD&amp;#39;&amp;#39; = CD$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But this is not possible unless $D&amp;#39; = D$ or $D&amp;#39;&amp;#39; = D$, which implies $\angle{B} = \frac{\pi}{2}$ or $\angle{C} = \frac{\pi}{2}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again, I would appreciate if some one could take the effort to draw the picture or let me know what is the easiest way to draw such pictures. The argument I hope however gives you the idea behind the method.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastActivityDate="2010-11-16T20:15:29.453" />
  <row Id="10614" PostTypeId="1" CreationDate="2010-11-16T20:55:06.290" Score="2" ViewCount="149" Body="&lt;p&gt;I figured that I can create a deterministic &quot;random&quot; numbers generator by utilizing a bit of &quot;magic&quot; that I picked up from some cryptography. However I seem to have missed a detail.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically the function $\displaystyle f(n) = p^n \mod q$ should for $\displaystyle n$ in the range $\displaystyle [1,q-1]$ produce all the numbers in the same range, but in a scrambled order, however that is only true for some pairs of $\displaystyle p$ and $\displaystyle q$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How exactly do I find and verify such pairs?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Update:&lt;br&gt;&#xA;I now know why I missed that detail, I'm think my computer will be done finding a few suitable number pairs by brute force long before I will be done understanding a more efficient approach.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think I'll run two of these generators in tandem, using their sum mod limit to generate a number in the field $[0;limit-1]$. For each pair I'll choose $q$ close to 2^32, $p$ close to 2^16, and assert that $p$ is prime, $q$ is prime, $(q-1)/2$ is prime and that $p$ is a primitive root of $q$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for your input, but I'll leave the question open to anyone who can explain a computationally efficient method of asserting primitive rootness status of two numbers in a way that I understand.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Update:&lt;br&gt;&#xA;I have found the pairs 65539 (mod 4294967087) and 65537 (mod 4294965887), this gives the following JavaScript function:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;function seed(state1,state2){&#xA;    var mod1=4294967087&#xA;    var mul1=65539&#xA;    var mod2=4294965887&#xA;    var mul2=65537&#xA;    state1=state1%mod1&#xA;    state2=state2%mod2&#xA;    function random(limit){&#xA;        state1=(state1*mul1)%mod1&#xA;        state2=(state2*mul2)%mod2&#xA;        return (state1+state2)%limit&#xA;    }&#xA;    return random&#xA;}&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="3584" LastEditorUserId="3584" LastEditorDisplayName="" LastEditDate="2010-11-17T16:11:06.517" LastActivityDate="2010-11-17T16:11:06.517" Title="Deterministic random numbers generator using $p^n \mod q$" Tags="&lt;prime-numbers&gt;&lt;random&gt;" AnswerCount="2" />
  <row Id="10615" PostTypeId="2" ParentId="10614" CreationDate="2010-11-16T21:02:13.357" Score="4" ViewCount="" Body="&lt;p&gt;Assuming $\displaystyle q$ is a prime, you need $\displaystyle p$ to be a &lt;a href=&quot;http://mathworld.wolfram.com/PrimitiveRoot.html&quot; rel=&quot;nofollow&quot;&gt;Primitive Root&lt;/a&gt; of $\displaystyle q$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is known that every prime $\displaystyle q$ has a primitive root. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact $\displaystyle q$ can also be a power of an odd prime or twice the power of an odd prime, see here: &lt;a href=&quot;http://www.math.uiuc.edu/~hildebr/453/notes/nt-notes5.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.math.uiuc.edu/~hildebr/453/notes/nt-notes5.pdf&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It (according to the wiki link below) has been shown that the least primitive root of $\displaystyle q$, $\displaystyle g_q$ satisfies $\displaystyle g_q &amp;lt; C\sqrt[4]{q}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and by assuming the generalized Riemann Hypothesis, in fact that $\displaystyle g_p = O(\log^6 q)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See here: &lt;a href=&quot;http://en.wikipedia.org/wiki/Primitive_root_modulo_n#Order_of_magnitude_of_primitive_roots&quot; rel=&quot;nofollow&quot;&gt;Magnitude of Primitive Roots&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So given a $\displaystyle q$, you should be able to simply do a scan to find the least primitive root relatively quickly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are interested in $\displaystyle p$ also being prime, see also: &lt;a href=&quot;http://en.wikipedia.org/wiki/Artin%27s_conjecture_on_primitive_roots&quot; rel=&quot;nofollow&quot;&gt;Artin's Conjecture on Primitive Roots&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-11-16T21:14:57.413" LastActivityDate="2010-11-16T21:14:57.413" />
  <row Id="10616" PostTypeId="1" AcceptedAnswerId="10628" CreationDate="2010-11-16T21:03:07.290" Score="4" ViewCount="396" Body="&lt;p&gt;How can I prove that if G is an abelian group with elements a and b with orders m and n, respectively, then G contains an element whose order is the least common multiple of m and n?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's an exercise from Hungerford's book, but it's not homework. I could not solve it when I take the course on groups and I think it should be easy. There is a hint that says to first prove when m and n are coprimes. I did this part. But I have no idea how to solve the general case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="3588" LastActivityDate="2010-11-17T02:51:10.800" Title="Order of elements in abelian groups" Tags="&lt;group-theory&gt;" AnswerCount="4" FavoriteCount="1" />
  <row Id="10617" PostTypeId="2" ParentId="10616" CreationDate="2010-11-16T21:11:34.373" Score="2" ViewCount="" Body="&lt;p&gt;Hi--&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is also one of the starred problems in &lt;em&gt;Herstein's: Topics in Algebra&lt;/em&gt; book. Take a look into page 3 of this file:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.csie.ntu.edu.tw/~b89053/herstein.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.csie.ntu.edu.tw/~b89053/herstein.pdf&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In case you aren't able to view the solution i type it down as i am not able to download this file, but able to view it in Google Docs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Solution:&lt;/strong&gt; Let $a$ and $b$ be elements of order $m$ and $n$ respectively.  Write $\text{lcm} (m,n)= \prod\limits_{i=1}^{t}p_{i}^{\alpha_{i}}$, where the $p_{i}$'s are distinct primes, and $\alpha_{i} &amp;gt; 0$ for each $i$. For any $i \in [1,2, \cdots,t ]$, $p_{i}^{\alpha_{i}}$ divides, either $m$ or $n$. If say, $p_{i}^{\alpha_{i}} \mid  m$, then, $a^{m/p_{i}^{\alpha_{i}}}$ is of order $p_{i}^{\alpha_{i}}$. In this way we obtain elements $x_{1},\cdots,x_{t}$ of orders $p_{1}^{\alpha_{1}}, \cdots, p_{t}^{\alpha_{t}}$ respectively. Now consider elements in $G$ of the form $$x_{1}^{\beta_{1}}, \cdots, x_{t}^{\beta_{t}} \quad \cdots (3)$$ where $\beta_{i} \in \mathbb{N}$, and $i \in [1,2,\cdots,t]$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Our next goal, is to show that, the single elements, $(x_{1},\cdots,x_{t})$ generates all elements of the form $(3)$. For this purpose, we prove that for all $\beta_{i} \in \mathbb{N}$,  $i \in [1,2,\cdots t]$, the system $$ k = \beta_{1} \ (\text{mod} \ p_{1}^{\alpha_{1}})$$ $$ \cdot $$ $$\cdot $$ $$\cdot $$ $$k = \beta_{t} \ (\text{mod} \ p_{t}^{\alpha_{t}}$$ has a solution in $k$. This follows by setting $$ k =\sum\limits_{i=1}^{t} \beta_{i}c_{i} p_{1}^{\alpha_{1}} \cdots p_{i-1}^{\alpha_{i-1}}p_{i+1}^{\alpha_{i+1}} \cdots p_{t}^{\alpha_{t}}$$ where $c_{i}$ satisfies $$ c_{i} p_{1}^{\alpha_{1}} \cdots p_{i-1}^{\alpha_{i-1}}p_{i+1}^{\alpha_{i+1}} \cdots p_{t}^{\alpha_{t}} \equiv 1 \ (\text{mod} \ p_{i}^{\alpha_{i}})$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's not hard to see that the elements of the form in Eq.(3) form a subgroup $H$ of $G$. Hence for each $i$ the order $p_{i}^{\alpha_{i}}$ of the subgroup generated by $x_{i}$ divides $|H|$ by &lt;em&gt;Lagrange's theorem.&lt;/em&gt;  This gives $|H| \geq \prod\limits_{i=1}^{t} p_{i}^{\alpha_{i}}$. Clearly $|H| \leq \prod\limits_{i=1}^{t} p_{i}^{\alpha_{i}}$ also holds. Since the element, $(x_{1}, \cdots, x_{t})$ generates exactly $H$ it has order $|H|$&lt;/p&gt;&#xA;" LastActivityDate="2010-11-16T21:34:36.870" CommentCount="1" />
  <row Id="10618" PostTypeId="2" ParentId="10616" CreationDate="2010-11-16T21:18:59.357" Score="3" ViewCount="" Body="&lt;p&gt;Let $n = \prod_i p_i^{n_i}$ and $m = \prod_i p_i^{m_i}$, where the $p_i$s are the same. Thus $\mathrm{lcm}(n,m) = \prod_i p_i^{\max(n_i,m_i)}$. Let $N = \{ i : n_i \geq m_i \}$ and $M = \{ i : n_i &amp;lt; m_i \}$. Define $n&amp;#39; = \prod_{i \in N} p_i^{n_i}$ and $m&amp;#39; = \prod_{i \in M} p_i^{m_i}$. Now $a&amp;#39; = a^{n/n&amp;#39;}$ is of order $n&amp;#39;$, and $b&amp;#39; = b^{m/m&amp;#39;}$ is of order $m&amp;#39;$. Since $n&amp;#39;,m&amp;#39;$ are coprime, we know that there is an element of order $n&amp;#39;m&amp;#39; = \mathrm{lcm}(n,m)$.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;You can also prove it using the structure theorem for abelian groups, but that's an overkill.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-16T21:18:59.357" CommentCount="2" />
  <row Id="10619" PostTypeId="2" ParentId="10614" CreationDate="2010-11-16T21:32:54.020" Score="3" ViewCount="" Body="&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Primitive_root_modulo_n#Finding_primitive_roots&quot; rel=&quot;nofollow&quot;&gt;wiki article&lt;/a&gt; has an algorithm for testing primitive roots; it requires you to factor $q-1$.  But once you've done that, together with the GRH bound mentioned in Moron's answer you should at least be able to locate a primitive root in polynomial time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But this is a really inadvisable way of generating &quot;random&quot; numbers.  Given even two consecutive values of the sequence you can predict all of the others.  There are much better methods described, for example, &lt;a href=&quot;http://en.wikipedia.org/wiki/Pseudorandom_number_generator&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-11-16T21:32:54.020" CommentCount="1" />
  <row Id="10620" PostTypeId="2" ParentId="10516" CreationDate="2010-11-16T21:45:51.680" Score="2" ViewCount="" Body="&lt;p&gt;(&lt;em&gt;Partial Answer to Question 1&lt;/em&gt;) No, this phenomenon isn't restricted to integral convex polytopes. There are examples of non-integral rational convex polytopes with Ehrhart quasi-polynomials that are polynomials. See this &lt;a href=&quot;http://adsabs.harvard.edu/abs/2003math......9329D&quot; rel=&quot;nofollow&quot;&gt;reference&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2720" LastActivityDate="2010-11-16T21:45:51.680" />
  <row Id="10621" PostTypeId="2" ParentId="10606" CreationDate="2010-11-16T21:48:23.970" Score="2" ViewCount="" Body="&lt;p&gt;The notation is defined and explained in some detail &lt;a href=&quot;http://math.uga.edu/~pete/4400intro.pdf&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;: see pages 11-13.&lt;/p&gt;&#xA;" OwnerUserId="299" LastActivityDate="2010-11-16T21:48:23.970" CommentCount="4" />
  <row Id="10622" PostTypeId="1" AcceptedAnswerId="10633" CreationDate="2010-11-16T21:48:39.773" Score="3" ViewCount="124" Body="&lt;p&gt;Given an open set $U \subset \mathbb R ^n $, there exists an exhaustion by compact sets, i.e. a sequence of compact sets $K_i$, s.t.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\cup _{i=0}^{\infty} K_i = U$ and $\forall i \in \mathbb N : K_i \subset K_{i+1} ^{\circ}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can imagine that different exhaustions by compact sets 'propagate' at different speeds for the various parts $U$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let us call an exhaustion $(K_i)_i$ 'stronger' than another exhaustions $(L_i)_i$, whenever we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\forall i \in \mathbb N \exists j \in \mathbb N : L_i \subset K_j$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We call two exhaustions equivalent, if each one is stronger than the other - i.e. for each compact set in the first, we have a compact superset in the second, and vice versa.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question&lt;/strong&gt;: Are all exhaustions of $U$ equivalent?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Purpose: You encounter various settings, where you define a certain structure by such an exhaustion. If the above question has a positive answer, this would spare from wondering whether your construction is independent of such an exhaustion.&lt;/p&gt;&#xA;" OwnerUserId="3557" LastActivityDate="2010-11-17T10:24:04.733" Title="'Equivalent' Exhaustion by compact sets" Tags="&lt;topology&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="0" />
  <row Id="10623" PostTypeId="1" AcceptedAnswerId="10632" CreationDate="2010-11-16T21:59:31.080" Score="1" ViewCount="71" Body="&lt;p&gt;I'm working on a program that displays a bunch of thumbnails on the bottom of the screen and I'm using the mouse's position to slide the images from side to side. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The center of the screen is 0 on a number line. Every frame, the mouse's X position is either added or subtracted from the x position of each thumbnail. (So if the mouse is at -15, each thumbnail will move -15 pixels.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The images should stop moving if they are too far from the center. &quot;Too far&quot; means that they are off the screen to a particular side, or if there are more images than fits on the screen, the image should stop sliding when there is enough room for the last image to appear on the opposite side.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I guess you could say that the x position of each thumbnail is a function of the mouse position, what number thumbnail it is and the the number of thumbnails. The screen width is 800 here (or -400 to 400)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What would be the mathematical formula for the above scenario?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT, a (perhaps) clearer explanation of what I want:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want the images to scroll horizontally. If the mouse is on the right, they should move towards the left, and if the mouse is on the left, the images should scroll towards the right. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The images should stop scrolling so that they never completely scroll offscreen. (If there are more images than the screen can show, allow images to scroll off just enough to show the other images.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The speed of the scrolling depends on the distence of the mouse cursor from the center.&lt;/p&gt;&#xA;" OwnerUserId="2414" LastEditorUserId="2414" LastEditorDisplayName="" LastEditDate="2010-11-16T22:20:37.813" LastActivityDate="2010-11-16T22:52:50.920" Title="Formula for positioning each image in a row of images" Tags="&lt;functions&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" />
  <row Id="10626" PostTypeId="2" ParentId="10603" CreationDate="2010-11-16T22:15:36.697" Score="1" ViewCount="" Body="&lt;p&gt;Taking Pontrjagin duals of the short exact sequence of discrete abelian groups &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$0 \rightarrow \mathbb{Z} \rightarrow \mathbb{Q} \rightarrow \mathbb{Q}/\mathbb{Z} \rightarrow 0$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;gives&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$0 \rightarrow \hat{\mathbb{Z}} \rightarrow \mathbb{Q}^{\vee} \rightarrow S^1 \rightarrow 0$, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;-- here $\hat{\mathbb{Z}}$ is the profinite completion of $\mathbb{Z}$ -- so this classifies the one-dimensional &quot;unitary&quot; continuous representations of $\mathbb{Q}$.  I believe that the exact sequence above is split; of all places, this came up at a dinner party I attended last week.  (&lt;b&gt;Added&lt;/b&gt;: no, it is not split as a sequence of topological groups or even as a sequence of groups: see Matt E's comment below.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;N.B.: If you ask a number theorist (broadly construed) what the &quot;usual topology&quot; on $\mathbb{Q}$ is, she will say that it is the discrete topology.  This is the topology it inherits from the topology on the adele ring $A_{\mathbb{Q}}$.  &lt;/p&gt;&#xA;" OwnerUserId="299" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2010-11-17T09:25:13.387" LastActivityDate="2010-11-17T09:25:13.387" CommentCount="4" />
  <row Id="10627" PostTypeId="1" CreationDate="2010-11-16T22:28:08.083" Score="8" ViewCount="141" Body="&lt;p&gt;Functions from $\mathbb{C} \rightarrow \mathbb{C}$ are hard to visualize because of their 4-dimensional nature. One nice way of looking at them is by what's called &lt;a href=&quot;http://en.wikipedia.org/wiki/Domain_coloring&quot; rel=&quot;nofollow&quot;&gt;domain coloring&lt;/a&gt;. An example from the wiki article is shown below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When we look at the graph of a real function ($\mathbb{R} \rightarrow \mathbb{R}$), we can get a feel for some of the function's properties: where its zeros are, if it's continuous, if it's differentiable, etc. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is what kinds of properties of complex functions can we &quot;see&quot; by looking at a domain coloring? In the example below, we can obviously see where the zeros are and where it blows up. I'm wondering, can we tell if a function is analytic? A rational function? Can we estimate an integral like we could by looking at a real graph?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://upload.wikimedia.org/wikipedia/commons/thumb/d/d3/Color_complex_plot.jpg/200px-Color_complex_plot.jpg&quot; /&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f(z) = \frac{(z^2 − 1)(z − 2 − i)^2}{(z^2 + 2 + 2i)}$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="130" LastActivityDate="2010-11-16T23:26:31.397" Title="What can be gleaned from looking at a domain-colored graph of a complex function?" Tags="&lt;complex-analysis&gt;&lt;intuition&gt;&lt;visualization&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="10628" PostTypeId="2" ParentId="10616" CreationDate="2010-11-16T22:29:12.933" Score="5" ViewCount="" Body="&lt;p&gt;Below is an inductive proof excerpted from my post in this prior &lt;a href=&quot;http://math.stackexchange.com/questions/6852/maximum-order-of-integers-coprime-to-a-prime-p/6856#6856&quot;&gt;thread here.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;LEMMA&lt;/strong&gt; $\ $ A finite abelian group $\rm\:G\:$ has an  lcm-closed  order set, i.e. with $\rm\: o(X) = $ order of $\rm\: X$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\quad\quad\quad\quad\quad\ X,Y \in G\ \ \Rightarrow\ \ \exists\ Z \in G:\  o(Z) = lcm(o(X),o(Y))$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof&lt;/strong&gt; $\ \ $  By induction on  $\rm o(X)\ o(Y)\:.\ $ If it's $\:1\:$ then trivially $\rm\:Z = 1\:$. $\ $ Otherwise&lt;/p&gt;&#xA;&#xA;&lt;p&gt;write  $\rm\ o(X)\ =\ AP,\: \ \ o(Y) = BP&amp;#39;,\ \ \ P&amp;#39;|P = p^m &amp;gt; 1\:,\ \ $  prime $\rm\: p\:$ coprime to $\rm\: A,B$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then  $\rm\: o(X^P) = A,\ \ o(Y^{P&amp;#39;}) = B\:.\ $  By induction there's a $\rm\: Z\:$ with $\rm \: o(Z) = lcm(A,B)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so  $\rm\ \ o(X^A\: Z)\: =\: P\ lcm(A,B)\: =\: lcm(AP,BP&amp;#39;)\: =\: lcm(o(X),o(Y))\:$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;NOTE&lt;/strong&gt; $\ \ $ This is the element-wise form of what's known as &quot;Herstein's hardest problem&quot;, viz. 2.5.11, p. 41 in the first edition of Herstein's popular textbook &quot;Topics in Algebra&quot;. In the 2nd edition Herstein added  the following note (problem 2.5.26, p.48) &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Don't be discouraged if you don't get this problem with what you know of group theory up to this stage. I don't know anybody, including myself, who has done it subject to the restriction of using material developed so far in the text.  But it is fun to try. I've had more correspondence about this problem than about any other point in the whole book.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-11-16T23:01:32.987" LastActivityDate="2010-11-16T23:01:32.987" CommentCount="6" />
  <row Id="10630" PostTypeId="2" ParentId="10514" CreationDate="2010-11-16T22:31:40.850" Score="2" ViewCount="" Body="&lt;p&gt;I'll do it for when $f(x)$ has compact support; the general Schwartz function case is a modification of this. I use the fact that the Schwatrz functions are sequentially dense in the distributions, meaning that there is a sequence of Schwartz functions $g_n$ such that $\int g_n(x) \alpha(x)$ converges to $T(\alpha)$ as $n$ goes to infinity for any Schwartz function $\alpha(x)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you replace $T$ by $g_n(x)$, then your identity holds by Fubini's theorem. So one takes limits now as $n$ goes to infinity. The left hand side converges to $T \int f(x - \circ)\phi(\circ) d\circ$ immediately.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the right-hand side, for each $x$ one has that $g_n(f(x - \circ) \phi(\circ))$ converges to $T(f(x - \circ) \phi(\circ))$. Since the $g_n(f(x - \circ) \phi(\circ))$ will be uniformly bounded (the construction of the $g_n$ is quite explicit and readily implies this) and since $f(x - \circ) \phi(\circ)$ is the zero distribution for $x$ outside a compact set, the dominated convergence theorem gives that the right hand side converges to $\int T (f(x - \circ) \phi(\circ)) dx$ as needed.&lt;/p&gt;&#xA;" OwnerUserId="3035" LastEditorUserId="3035" LastEditorDisplayName="" LastEditDate="2010-11-16T22:57:00.057" LastActivityDate="2010-11-16T22:57:00.057" />
  <row Id="10631" PostTypeId="2" ParentId="10616" CreationDate="2010-11-16T22:35:58.183" Score="4" ViewCount="" Body="&lt;p&gt;Both responses you have received provide correct answers. In fact, you don't need to go so far as to obtain elements of relatively prime orders; all you need to worry about are primes with the property that they divide $nm$, and the largest power of $p$ that divides $n$ is equal to the largest power of $p$ that divides $m$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Lemma.&lt;/strong&gt; Let $x$ and $y$ be commuting elements of a group $G$, and assume $x$ and $y$ are of finite orders $n$ and $m$, respectively. Suppose that for every prime $p$ that divides $nm$, the largest power of $p$ that divides $n$ is different from the largest power of $p$ that divides $m$. Then the order of $xy$ in $G$ is $\mathrm{lcm}(n,m)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Proof.&lt;/em&gt; Let $\ell=\mathrm{lcm}(n,m)$. It is easy to verify that $(xy)^{\ell}=1$, so we just need to show that $\ell$ is the smallest positive integer $k$ such that $(xy)^k=1$. Suppose that $(xy)^k = 1$, with $0\lt k\leq \ell$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $(xy)^k = x^ky^k = 1$, then $x^k = y^{-k}$, and in particular $x^k$ has the same order as $y^k$. The order of $x^k$ is $n/\gcd(n,k)$, and the order of $y^k$ is $m/\gcd(n,k)$. Let $p$ be a prime that divides $\ell$, and let $a=\mathrm{ord}_p(n)$, $b=\mathrm{ord}_p(m)$, and $c=\mathrm{ord}_p(k)$. Assume $b\lt a$. If $c\lt a$, then $\mathrm{ord}_p(n/\gcd(n,k)) = a-c$, and $\mathrm{ord}_p(m/\gcd(m,k))=\max(0,b-c)\lt a-c$, which is impossible. Thus, $c=a$. A symmetric argument shows that if $a\lt b$, then $c=b$. That is, for all primes $p$ that divide $\ell$, we have $\mathrm{ord}_p(k) = \max(\mathrm{ord}_p(n),\mathrm{ord}_p(n)) = \mathrm{ord}_p(\ell)$. Hence $k=\ell$. &lt;strong&gt;QED&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So now assume that $a$ is an element of order $n$ in an abelian group, and let $b$ be an element of order $m$. Let $p_1,\ldots,p_k$ be the primes that divide $nm$ and for which the $p$-order of $p_i$ in $m$ and in $n$ are equal. Then&#xA;$$\mathrm{lcm}(n,m) = \mathrm{lcm}\left(n, \frac{m}{p_1\cdots p_k}\right).$$&#xD;&#xA;Since $m/(p_1\cdots p_k)$ is the order of $b^{p_1\cdots p_k}$, then it follows from the lemma that $ab^{p_1\cdots p_k}$ has order $\mathrm{lcm}(n,m)$, as desired. &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-17T02:51:10.800" LastActivityDate="2010-11-17T02:51:10.800" CommentCount="4" />
  <row Id="10632" PostTypeId="2" ParentId="10623" CreationDate="2010-11-16T22:52:50.920" Score="1" ViewCount="" Body="&lt;p&gt;You're probably looking for something like a logit or an expit or even a sine function with the right parameters put in.&#xA;On a more practical level though, it might be easier to program in a piecewise-linear relationship b/w your mouse position and the strip movement based on where the strip is.&lt;/p&gt;&#xA;" LastActivityDate="2010-11-16T22:52:50.920" />
  <row Id="10633" PostTypeId="2" ParentId="10622" CreationDate="2010-11-16T22:59:57.530" Score="5" ViewCount="" Body="&lt;p&gt;Yes.  Let $\{L_i\}_{i=1}^\infty$, $\{K_i\}_{i=1}^\infty$ be two exhaustions.  It follows that $\bigcup_{i=1}^\infty K_i^\circ = U$, so for each $n$, $\{K_i^\circ\}_{i=1}^\infty$ is an open cover of $L_n$.  By compactness, it has a finite subcover $\{K_{i_1}^\circ, \dots, K_{i_l}^\circ\}$.  But if $m = \operatorname{max}(i_1, \dots, i_l)$, we have $K_{i_j}^\circ \subset K_{i_j + 1} \subset K_{m+1}$ for all $j=1,\dots, l$.  Thus $L_n \subset K_{m+1}$, and so $K_i$ is stronger than $L_i$.  By symmetry they are equivalent.&lt;/p&gt;&#xA;" OwnerUserId="822" LastActivityDate="2010-11-16T22:59:57.530" CommentCount="5" />
  <row Id="10634" PostTypeId="1" AcceptedAnswerId="10641" CreationDate="2010-11-16T23:14:18.827" Score="1" ViewCount="135" Body="&lt;p&gt;If we have a module M over a ring Z, and we consider the the canonical module homomorphism M→Q⊗M over Z, is it true that the kernel of this map is the torsion submodule of M? Why?&lt;/p&gt;&#xA;" OwnerUserId="3595" LastActivityDate="2010-11-17T01:23:59.550" Title="Torsion submodule equal to kernel of canonical map" Tags="&lt;modules&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="10635" PostTypeId="2" ParentId="10627" CreationDate="2010-11-16T23:26:31.397" Score="1" ViewCount="" Body="&lt;p&gt;Hans Lundmark's pages on &lt;a href=&quot;http://www.mai.liu.se/~halun/complex/domain_coloring-unicode.html&quot; rel=&quot;nofollow&quot;&gt;domain coloring&lt;/a&gt;, and &lt;a href=&quot;http://www.mai.liu.se/~halun/complex/elliptic/&quot; rel=&quot;nofollow&quot;&gt;his page on Elliptic functions&lt;/a&gt; is good places to start. (Picard's great theorems, Lucas's theorem, Weierstrass $\wp$ function)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't think we can see that a function is rational  -- just moving the roots in the function that you have displayed above by $\pi/100$ in some random direction would be sufficient to make it not a rational function, but the picture would more or less look the same.&lt;/p&gt;&#xA;" OwnerUserId="890" LastActivityDate="2010-11-16T23:26:31.397" CommentCount="1" />
  <row Id="10636" PostTypeId="1" CreationDate="2010-11-16T23:48:38.097" Score="2" ViewCount="155" Body="&lt;p&gt;In the proof of Stone-Weierstrass theorem provided in Rudin's Principles of Mathematical Analysis, why do we only need to show that there exists a sequence of polynomials $P_n$ that converges uniformly to the continuous complex function on $[0,1]$?&lt;/p&gt;&#xA;" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-16T23:50:43.850" LastActivityDate="2010-11-17T00:00:42.733" Title="Stone-Weierstrass Theorem" Tags="&lt;analysis&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="10637" PostTypeId="2" ParentId="10636" CreationDate="2010-11-17T00:00:42.733" Score="3" ViewCount="" Body="&lt;p&gt;I think you are just asking why, &quot;We may assume, without loss of generality, that $[a, b] = [0,1]$.&quot; The reason is that if $f$ is continuous on $[a,b]$, then we can uniformly approximate $f(a+(b-a)x)$ on $[0,1]$ with a sequence of polynomials $(p_n)_{n=1}^\infty$, and then $(p_n((x-a)/(b-a)))_{n=1}^\infty$ will converge uniformly to $f$.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-11-17T00:00:42.733" />
  <row Id="10639" PostTypeId="2" ParentId="10622" CreationDate="2010-11-17T00:17:23.330" Score="0" ViewCount="" Body="&lt;p&gt;Ok, one version on my own, more technical.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $V$ be any compact set included in $U$. It remains to show $\exists i \in \mathbb N : V \subset K_i$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that $\forall i : V \cap K_i$ bounded again.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose, there is no index $i$ s.t. $V \subset K_i$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then there is a nested sequence $V_i$ of non-empty bounded sets, $V = V_0, \forall i : V_i \cap K_i = \{\}$, .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $x_i \in V_i$ be a sequence of points, which exists by assumption. Then this sequence is bounded and w.l.o.g it is convergent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence let $x_i \rightarrow x$. By assumption, $\exists k \in \mathbb N : \exists \epsilon &amp;gt; 0 : B_{\epsilon}(x) \subset {K_k}^{\circ}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As $x_i$ convergence, we skip a leading number of elements and w.l.o.g. $\forall i : x_i \in B_{\epsilon/2}(x) \subset B_{\epsilon}(x) \subset K_k$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After reindexing the $V_i$ according to our choices of subsequences, we obtain a contradiction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, such there is an $i$ s.t. $V \subset K_i$. As $V$ has been any compact set, we are done.&lt;/p&gt;&#xA;" OwnerUserId="3557" LastEditorUserId="3557" LastEditorDisplayName="" LastEditDate="2010-11-17T10:24:04.733" LastActivityDate="2010-11-17T10:24:04.733" />
  <row Id="10640" PostTypeId="1" CreationDate="2010-11-17T00:30:24.343" Score="0" ViewCount="127" Body="&lt;p&gt;So in algebraic geometry, the prime spectrum of the ring of polynomials over a field k is isomorphic to the set of k-homomorphisms from the ring of polynomials to k, and we do this by using the fact that the residue field (localise w.r.t. some prime ideal, then mod out the ideal generated by this in the localisation) is isomorphic to k, and hence any polynomial goes to an element of k, via the equivalence relations. This all makes sense as the ideal generated by a prime ideal of the polynomial ring in its localisation is clearly maximal, but I'm stuck on a very basic example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let C[x,y] be the polynomial ring over the complex numbers and (x - 1) be the prime ideal. What number does y go to? A C-algebra homomorphism is defined completely by its action on the generators of the polynomial ring (so just the n variables, and taking n = 2 we get x |--&gt; 1, but y |--&gt; ?), but I can't figure out how this works in this simple case. It should maybe be zero (otherwise how do you invert 1 + y?), but I don't know. Can anyone explain this?&lt;/p&gt;&#xA;" OwnerUserId="3597" LastActivityDate="2010-11-17T00:35:02.530" Title="The prime spectrum in multiple variables - a basic example/question" Tags="&lt;algebraic-geometry&gt;&lt;commutative-algebra&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="10641" PostTypeId="2" ParentId="10634" CreationDate="2010-11-17T00:33:06.900" Score="2" ViewCount="" Body="&lt;p&gt;Hint: If $m\in M$ is a torsion element, it is killed by some integer, say $r$. Then, $1\otimes m=\frac{r}{r}\otimes m=\frac{1}{r}\otimes rm=\frac{1}{r}\otimes 0=0$ where the second equality follows from the bilinearity of the tensor product.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was working under the assumption that $Z$ is the ring of integers and $Q$ the ring of rationals. As Yuval points out, everything works the same way if you think of $Z$ as an arbitrary domain and $Q$ as its fraction field.&lt;/p&gt;&#xA;" OwnerUserId="3431" LastEditorUserId="3431" LastEditorDisplayName="" LastEditDate="2010-11-17T01:23:59.550" LastActivityDate="2010-11-17T01:23:59.550" CommentCount="5" />
  <row Id="10642" PostTypeId="2" ParentId="10640" CreationDate="2010-11-17T00:35:02.530" Score="3" ViewCount="" Body="&lt;p&gt;The &lt;em&gt;maximal&lt;/em&gt; ideals in a polynomial ring of an algebraically closed field $k$ is isomorphic to the set of $k$-homomorphisms from the polynomial ring into $k$. (This is the Nullstellensatz.) This is not true for arbitrary prime ideals, nor is it true for nonalgebraically closed fields. (Consider for instance the ideal $(X^2 +1) \subset \mathbb{R}[X]$; this morally wants to correspond to the points $\pm i$, but it doesn't, since those don't live in the reals. Nonetheless, it happens that the &lt;em&gt;scheme&lt;/em&gt; $\mathrm{Spec} \mathbb{R}[X]$ includes all prime ideals, so it includes a copy of $\pm i$ glued together, if you want to call it that.) In the terminology of algebraic geometry, one would say that the point you described is not &quot;rational over $\mathbb{C}$.&quot; Such points do exist, and make schemes of even finite type over an algebraically closed field different from varieties.&#xA;(Fortunately, it turns out that closed points alone determine many topological properties---the precise statement is that the closed points (or maximal ideals) are dense in any constructible sets.)&lt;/p&gt;&#xA;" OwnerUserId="536" LastActivityDate="2010-11-17T00:35:02.530" />
  <row Id="10643" PostTypeId="1" AcceptedAnswerId="10644" CreationDate="2010-11-17T00:37:51.050" Score="4" ViewCount="150" Body="&lt;p&gt;I am looking for the proof of the following theorem:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Let $(a_n)$ be a sequence of real&#xA;  numbers.  Then there exists a function&#xA;  $f$ which is infinitely differentiable&#xA;  at 0, and  $$ \frac{d^nf}{dx^n}(0) =&#xD;&#xA; a_n, \ \ \text{for all } n.$$&#xD;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I would appreciate either a sketch of the proof or an online reference to it.  A general case is listed as &lt;a href=&quot;http://en.wikipedia.org/wiki/Borel%27s_lemma&quot; rel=&quot;nofollow&quot;&gt;Borel's lemma&lt;/a&gt; in Wikipedia, without proof.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The hard part is when the power series $\sum_n \frac{a_n}{n!}x^n$ has a zero radius of convergence. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: Thanks for the answers! &lt;/p&gt;&#xA;" OwnerUserId="145" LastEditorUserId="145" LastEditorDisplayName="" LastEditDate="2010-11-17T01:17:33.417" LastActivityDate="2010-11-17T01:17:33.417" Title="Every real sequence is the derivative sequence of some function" Tags="&lt;calculus&gt;" AnswerCount="2" FavoriteCount="2" />
  <row Id="10644" PostTypeId="2" ParentId="10643" CreationDate="2010-11-17T00:44:30.800" Score="5" ViewCount="" Body="&lt;p&gt;As you mentioned, this is a result of Borel. Proofs can be found in several sources, see for example the book on &quot;Complex variables&quot; by Berenstein and Gay (it may just be an exercise there, but at least there is a &quot;hint&quot;). The idea is to try a power series. Of course, this may not converge, so you use the kind of functions that come in constructions of smooth partitions of unity to help the convergence; the point is that you can ensure these functions decay to zero sufficiently fast. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; There is a proof in Wikipedia, actually. See &lt;a href=&quot;http://en.wikipedia.org/wiki/Non-analytic_smooth_function&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="462" LastEditorUserId="462" LastEditorDisplayName="" LastEditDate="2010-11-17T00:49:52.697" LastActivityDate="2010-11-17T00:49:52.697" />
  <row Id="10645" PostTypeId="2" ParentId="10643" CreationDate="2010-11-17T00:46:26.807" Score="4" ViewCount="" Body="&lt;p&gt;This is a famous theorem of Borel. If you have Hormander's &quot;The analysis of partial differential operators I&quot; it's Theorem 1.2.6 there. He proves the result in any dimension. The basic idea is to write a sum of functions $\sum_n {a_n \over n!}\phi(m_n x)x^n$ where $\phi(x)$ is $1$ near $x = 0$ and is equal to zero outside of a small set containing $0$. If the $m_n$ are chosen carefully then the sum will have the desired properties. &lt;/p&gt;&#xA;" OwnerUserId="3035" LastActivityDate="2010-11-17T00:46:26.807" CommentCount="0" />
  <row Id="10646" PostTypeId="1" AcceptedAnswerId="10659" CreationDate="2010-11-17T00:59:01.017" Score="5" ViewCount="302" Body="&lt;p&gt;Hello, I've been trying to prove the following statement, but the converse has been giving me trouble. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A topological space $X$ is regular if and only if every closed subset $Z\subseteq X$ is the intersection of all open sets $U\subseteq X$ which contain it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here, the defintion of regular is that for any point $p\in X$ and any closed subset $Z\subset X$ not containing $p$, there exist open sets $E,F\subset X$ such that $p\in E$, $Z\subseteq F$, and $E\cap F=\emptyset$. The space does not necessarily have to be a $T_1$ space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe I showed the forward statement correctly. I suppose $X$ is regular, and clearly $Z\subseteq\cap\mathcal{U}$, where $\mathcal{U}$ is the family of all open sets containing $Z$. Then if I take $p\in\cap\mathcal{U}$, then $p$ is in every open subset containing $Z$. Towards a contradiction, I assume that $p$ is not in $Z$. Then since $X$ is regular, there exist open sets $E,F\subset X$ such that $p\in E$, $Z\subseteq F$, and $E\cap F=\emptyset$. Thus $p\not\in F$, but $F$ is an open set containing $Z$, and thus must contain $p$, a contradiction. Thus $p\in Z$, and so $Z=\cap\mathcal{U}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The backwards step has stumped me for a while. I take any point $p$ and a closed set $Z$ which does not contain $p$. So $Z=\cap\mathcal{U}$, and since $p\not\in Z$, there exists an open set $F$ containing $Z$ which does not contain $p$. The only observation I've been able to make is that $F^c$ is a closed set containing $p$, and then $F^c=\cap\mathcal{W}$ where $\mathcal{W}$ is every open set containing $F^c$. Hence $p$ is in every open set containing $F^c$. Taking any open sets $E$ and $F$ such that $p\in E$ and $Z\subseteq F$, and I haven't been able to see a way that there exists two such open sets that are disjoint, to show the space is regular. Showing $F^c$ is an open set would work, but I'm not sure if that's even true. I was hoping someone could point out what I'm missing, thank you.&lt;/p&gt;&#xA;" OwnerUserId="1583" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2010-11-20T01:51:23.927" LastActivityDate="2010-11-20T01:51:23.927" Title="A space is regular if each closed set $Z$ is the intersection of all open sets containing $Z$?" Tags="&lt;topology&gt;&lt;general-topology&gt;" AnswerCount="3" CommentCount="4" />
  <row Id="10647" PostTypeId="1" AcceptedAnswerId="10665" CreationDate="2010-11-17T01:29:43.157" Score="0" ViewCount="244" Body="&lt;p&gt;I was reading Rudin's Principles of Mathematical Analysis, and I came across the definition 7.19, where it says that a sequence of functions $f_n(x)$ is pointwise bounded on E if there exists a finite-valued function $\phi$ defined on E such that &#xA;$$ |f_n(x)| &amp;lt; \phi(x) $$&#xD;&#xA;for x element of E, n = 1, 2 ,3 ... &#xA;While $f_n$ is uniformly bounded on E if there exists a number M s.t. &#xA;$|f_n(x)| &amp;lt; M$&#xD;&#xA;for x element of E, n = 1, 2 , 3 ... &#xA;But if we define the set U as the values of $\phi(x)$ from our first definition and define the sup of the set as R, then don't we get the second definition. Wouldn't that mean that pointwise bounded implies uniform bounded? &lt;/p&gt;&#xA;" OwnerUserId="2919" LastActivityDate="2010-11-17T06:21:45.373" Title="Why doesn't pointwise bounded imply uniform bounded?" Tags="&lt;real-analysis&gt;&lt;analysis&gt;" AnswerCount="1" CommentCount="6" />
  <row Id="10648" PostTypeId="1" AcceptedAnswerId="10650" CreationDate="2010-11-17T02:08:44.310" Score="30" ViewCount="1240" Body="&lt;p&gt;When you draw a circle in a plane you can perfectly surround it with 6 other circles of the same radius. This works for any radius. What's the significance of 6? Why not some other number?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm looking for an answer deeper than &quot;there are 6*60=360 degrees in a circle, so you can picture it&quot;.&lt;/p&gt;&#xA;" OwnerUserId="2809" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-11-17T20:10:49.727" LastActivityDate="2010-11-18T05:59:44.103" Title="Why is a circle in a plane surrounded by 6 other circles." Tags="&lt;geometry&gt;&lt;differential-geometry&gt;&lt;discrete-geometry&gt;&lt;circle&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="8" />
  <row Id="10649" PostTypeId="1" AcceptedAnswerId="10652" CreationDate="2010-11-17T02:16:27.703" Score="1" ViewCount="169" Body="&lt;p&gt;I am given two random variables Y1 and Y2 and:&#xA;E(Y1) = 4&#xA;E(Y2) = -1&#xA;V(Y1) = 2&#xA;V(Y2) = 8&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am asked to find Cov(Y1,Y1)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know Cov(Y1,Y2) = E(Y1Y2)-E(Y1)E(Y2)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure if it is a typo, but it says Cov(Y1,Y1) NOT Cov(Y1,Y2)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer is 2, but i'm not sure how to get E(Y1Y2) from the information given, or why it's asking for the Cov of the same variable?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for any help!&lt;/p&gt;&#xA;" OwnerUserId="3049" LastActivityDate="2010-11-17T02:42:43.850" Title="find cov(Y1,Y1) given expected value and variance" Tags="&lt;homework&gt;&lt;statistics&gt;&lt;probability&gt;" AnswerCount="2" />
  <row Id="10650" PostTypeId="2" ParentId="10648" CreationDate="2010-11-17T02:23:25.467" Score="46" ViewCount="" Body="&lt;p&gt;The short answer is &quot;because they don't work,&quot; but that's kind of a copout.  This is actually quite a deep question.  What you're referring to is &lt;a href=&quot;http://en.wikipedia.org/wiki/Sphere_packing&quot; rel=&quot;nofollow&quot;&gt;sphere packing&lt;/a&gt; in two dimensions, specifically the &lt;a href=&quot;http://en.wikipedia.org/wiki/Kissing_number&quot; rel=&quot;nofollow&quot;&gt;kissing number&lt;/a&gt;, and sphere packing is actually quite a sophisticated and active field of mathematical research (in arbitrary dimensions).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's one answer, which isn't complete but which tells you why $6$ is a meaningful number in two dimensions.  The packing you refer to is a special type of packing called a &lt;em&gt;lattice&lt;/em&gt; packing, which means it comes from an arrangement of regularly spaced points; in this case, the &lt;a href=&quot;http://en.wikipedia.org/wiki/Hexagonal_lattice&quot; rel=&quot;nofollow&quot;&gt;hexagonal lattice&lt;/a&gt;.  The number $6$ appears here because the hexagonal lattice has $6$-fold symmetry.  So a natural question might be whether one can find lattices in two dimensions with, say $7$-fold or $8$-fold symmetry, since these might correspond to circle packings with more circles around a given circle.  (Intuitively, we expect more symmetric lattices to give rise to denser packings and to packings where each circle has more neighbors.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer is no: $6$-fold symmetry is the best you can do!  This is a consequence of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Crystallographic_restriction_theorem&quot; rel=&quot;nofollow&quot;&gt;crystallographic restriction theorem&lt;/a&gt;.  The generalization of the theorem to $n$ dimensions says this: it is possible for a lattice to have $d$-fold symmetry only if $\phi(d) \le n$, where $\phi$ is &lt;a href=&quot;http://en.wikipedia.org/wiki/Euler%27s_totient_function&quot; rel=&quot;nofollow&quot;&gt;Euler's totient function&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The generalization implies that you still cannot do better than $6$-fold symmetry in $3$ dimensions.  There are two natural lattice packings in $3$ dimensions, which &lt;a href=&quot;http://en.wikipedia.org/wiki/Hexagonal_close_packing#Fcc_and_hcp_lattices&quot; rel=&quot;nofollow&quot;&gt;both occur&lt;/a&gt; in molecules and crystals in nature and which both have $6$-fold symmetry, and &lt;a href=&quot;http://en.wikipedia.org/wiki/Kepler_conjecture&quot; rel=&quot;nofollow&quot;&gt;it turns out&lt;/a&gt; that these are the densest sphere packings in $3$ dimensions.  It also turns out that they give the correct kissing number in $3$ dimensions, which is $12$ (see the wiki article).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In $4$ dimensions, the kissing number is $24$, and I believe the corresponding packing is a lattice packing coming from a lattice with $8$-fold symmetry, which is possible in $4$ dimensions.  In higher dimensions, only two other kissing numbers are known: $8$ dimensions, where the &lt;a href=&quot;http://en.wikipedia.org/wiki/E8_lattice&quot; rel=&quot;nofollow&quot;&gt;$E_8$ lattice&lt;/a&gt; gives kissing number $240$, and $24$ dimensions, where the &lt;a href=&quot;http://en.wikipedia.org/wiki/Leech_lattice&quot; rel=&quot;nofollow&quot;&gt;Leech lattice&lt;/a&gt; gives kissing number $196560$!  These lattices are really mysterious objects and are related to a host of other mysterious objects in mathematics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A great reference for this stuff, although it is a little dense, is Conway and Sloane's &lt;a href=&quot;http://www2.research.att.com/~njas/doc/splag.html&quot; rel=&quot;nofollow&quot;&gt;Sphere Packing, Lattices, and Groups&lt;/a&gt;.  &lt;strong&gt;Edit:&lt;/strong&gt;  And for a very accessible and engaging introduction to symmetry in the plane and in general, I highly recommend Conway, Burgiel, and Goodman-Strauss's &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/1568812205&quot; rel=&quot;nofollow&quot;&gt;The Symmetries of Things&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-11-18T05:59:44.103" LastActivityDate="2010-11-18T05:59:44.103" CommentCount="6" />
  <row Id="10651" PostTypeId="2" ParentId="10646" CreationDate="2010-11-17T02:34:04.830" Score="5" ViewCount="" Body="&lt;p&gt;Nate's example is a good one and actually solves the question as it is. But suppose you think that the Cromarty property, as Nate called it, is not good enough since 'almost all' spaces satisfy it. Then you try to substitute it for a stronger condition. You can require that every closed subset $Z \subseteq X$ is a countable intersection of open sets. That's exactly what a &lt;a href=&quot;http://en.wikipedia.org/wiki/G%CE%B4_space&quot; rel=&quot;nofollow&quot;&gt;$G_{\delta}$ space&lt;/a&gt; is. In this situation the example Nate gave still works, you just need to require that the space $X$ is countable. In fact, if $F$ is a closed subset of $X$ with the cofinite topology then $X-F$ is the countable union of the closed sets $\{x\}$, with $x \in X - F$. So &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$F = \bigcap_{x \in X-F}  (X - \{x\}).$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, a countable set $X$ with the cofinite topology is a $G_{\delta}$ space. But then you conclude that your condition isn't strong enough. Now you try to impose some kind of separation. The question becomes: is there a Hausdorff $G_{\delta}$ space that isn't regular?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer is yes. Let $K = \{ 1/n : n \in \mathbb{N} \}$ and consider the collection $\mathcal{B}$ of all open intervals $(a,b)$, along with all sets of the form $(a,b)-K$. It's easy to show that $\mathcal{B}$ is a basis for a topology on $\mathbb{R}$, which is called the &lt;a href=&quot;http://en.wikipedia.org/wiki/K-topology&quot; rel=&quot;nofollow&quot;&gt;K-topology&lt;/a&gt;. When $\mathbb{R}$ is endowed with this topology it's denoted by $\mathbb{R}_K$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The K-topology is finer than the standard topology on the real line. So it's a Hausdorff space (actually a &lt;a href=&quot;http://en.wikipedia.org/wiki/Urysohn_space&quot; rel=&quot;nofollow&quot;&gt;completely Hausdorff space&lt;/a&gt;). Also, $\mathbb{R}_K$ is second countable, since the colletion $\mathcal{B&amp;#39;}$ of all open intervals $(a,b)$, along with all sets of the form $(a,b)-K$ and with $a,b \in \mathbb{Q}$ is a basis for the K-topology. $\mathbb{R}_K$ isn't regular, since the closed set and the point $0$ cannot be separated by disjoint open sets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let's see why $\mathbb{R}_K$ is a $G_{\delta}$ space. It's easier to prove that it's an $F_{\sigma}$ space, that is, a space such that every open set can be written as a countable union of closed sets, which is equivalent to prove that it's a $G_{\delta}$ space. Since $\mathbb{R}_K$ is second countable, every open set can be written as a countable union of basis elements. If we prove that every basis element is an $F_{\sigma}$ set, then will follow that $\mathbb{R}_K$ is an $F_{\sigma}$ space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sets $(a,b)$ are $F_{\sigma}$, since $(a,b) = \bigcup_{n=1}^{\infty} [a + \frac{1}{n}, b - \frac{1}{n}]$. Now consider $(a,b) - K$; we do not need to worry with the end point $b$, so suppose $b &amp;gt; 1$. If $a&amp;gt;0$, let $n_0$ be the largest natural such that $a &amp;lt; 1/n_0$. Then &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ (a,b)-K = (a, \frac{1}{n_0}) \cup \bigcup_{n=1}^{n_0-1} (\frac{1}{n+1}, \frac{1}{n}) \cup (1,b).$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each of the sets in the decomposition above are $F_{\sigma}$, thus $ (a,b)-K$ is $F_{\sigma}$. If $a=0$, then &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ (0,b)-K = \bigcup_{n=1}^{\infty} (\frac{1}{n+1}, \frac{1}{n}) \cup (1,b),$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so $(0,b)-K$ is $F_{\sigma}$. If $a&amp;lt;0$, then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ (a,b)-K = (a,0] \cup ((0,b)-K) = \bigcup_{n=1}^{\infty} [a + \frac{1}{n}, 0] \cup ((0,b)-K),$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so $(a,b)-K$ is an $F_{\sigma}$ and we conclude that $\mathbb{R}_K$ is an $F_{\sigma}$ space.&lt;/p&gt;&#xA;" OwnerUserId="329" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2010-11-17T23:53:44.933" LastActivityDate="2010-11-17T23:53:44.933" CommentCount="4" />
  <row Id="10652" PostTypeId="2" ParentId="10649" CreationDate="2010-11-17T02:35:22.097" Score="2" ViewCount="" Body="&lt;p&gt;$cov(Y_1,Y_1)=var(Y_1)=2$ &#xA;This follows from the definition of covariance:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$cov(Y_1,Y_1)=E(Y_1\cdot Y_1)-E(Y_1)E(Y_1)=E(Y_1^2)-E(Y_1)^2$&lt;/p&gt;&#xA;" OwnerUserId="3431" LastEditorUserId="3431" LastEditorDisplayName="" LastEditDate="2010-11-17T02:42:43.850" LastActivityDate="2010-11-17T02:42:43.850" CommentCount="2" />
  <row Id="10653" PostTypeId="2" ParentId="10649" CreationDate="2010-11-17T02:41:27.097" Score="2" ViewCount="" Body="&lt;p&gt;Cov$(Y_1,Y_2) = \text{Cov}((Y_1-E[Y_1])(Y_2-E[Y_2])) = E[Y_1 Y_2]-E[Y_1]E[Y_2]$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let $Y_2 = Y_1$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then Cov$(Y_1,Y_2) = E[{Y_1}^2] -(E[Y_1])^2$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This equation should look familiar...the variance of $Y_1$&lt;/p&gt;&#xA;" OwnerUserId="2596" LastActivityDate="2010-11-17T02:41:27.097" CommentCount="1" />
  <row Id="10654" PostTypeId="2" ParentId="7064" CreationDate="2010-11-17T02:47:50.947" Score="0" ViewCount="" Body="&lt;p&gt;US notation  - which is taught in Australia&lt;/p&gt;&#xA;" OwnerUserId="2219" LastActivityDate="2010-11-17T02:47:50.947" CommunityOwnedDate="2010-11-18T06:46:45.763" />
  <row Id="10655" PostTypeId="2" ParentId="10646" CreationDate="2010-11-17T03:40:42.593" Score="0" ViewCount="" Body="&lt;p&gt;Isn't the trivial (indiscrete) topology a counterexample?  Perhaps there are more assumptions somewhere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: I had a different definition of &quot;regular&quot; in mind from the asker.  The question has been clarified, and the trivial topology is not a counterexample.&lt;/p&gt;&#xA;" OwnerUserId="822" LastEditorUserId="822" LastEditorDisplayName="" LastEditDate="2010-11-17T05:23:30.763" LastActivityDate="2010-11-17T05:23:30.763" CommentCount="5" />
  <row Id="10656" PostTypeId="2" ParentId="10579" CreationDate="2010-11-17T04:01:02.673" Score="0" ViewCount="" Body="&lt;p&gt;Thank for the matrixcookbook, its a useful source.  I found the identity which confuses me:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{\partial \textbf{x}^{T} B \textbf{x}}{\partial \textbf{x}} = (B + B^{T})\textbf{x}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;via product rule:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{\partial \textbf{x}^{T} B \textbf{x}}{\partial \textbf{x}} = \frac{\partial \textbf{x}^{T}}{\partial \textbf{x}}B\textbf{x} + \textbf{x}^{T}\frac{\partial B \textbf{x}}{\partial \textbf{x}}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At this point, I'm stuck because I'm not sure if $\frac{\partial \textbf{x}^{T}}{\partial \textbf{x}}$ is identity $I$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I assume this definition of the Jacobian:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\begin{bmatrix} \dfrac{\partial y_1}{\partial x_1} &amp;amp; \cdots &amp;amp; \dfrac{\partial y_1}{\partial x_n} \\ \vdots &amp;amp; \ddots &amp;amp; \vdots \\ \dfrac{\partial y_m}{\partial x_1} &amp;amp; \cdots &amp;amp; \dfrac{\partial y_m}{\partial x_n}  \end{bmatrix}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And if I write $\frac{\partial B \textbf{x}}{\partial \textbf{x}}$ component by component with the definition above, it becomes simply $B$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{\partial \textbf{x}^{T} B \textbf{x}}{\partial \textbf{x}} = B\textbf{x} + \textbf{x}^{T}B$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is not correct, as you stated the left side is a nx1 vector and the right side is a nx1 vector.  I'm not sure where my faulty assumption is.&lt;/p&gt;&#xA;" LastActivityDate="2010-11-17T04:01:02.673" CommentCount="3" />
  <row Id="10657" PostTypeId="2" ParentId="10181" CreationDate="2010-11-17T04:15:22.113" Score="1" ViewCount="" Body="&lt;p&gt;There is a good list on &lt;a href=&quot;http://hbpms.blogspot.com/2008/05/stage-4-foundations-and-discrete.html&quot; rel=&quot;nofollow&quot;&gt;this website.&lt;/a&gt; I can personally recommend the Walicki notes (free). It includes a good introduction to modal logic, so that's always nice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of the books, Geoffrey Hunter's Metalogic is very good and covers all bases.&lt;/p&gt;&#xA;" OwnerUserId="3564" LastActivityDate="2010-11-17T04:15:22.113" CommentCount="2" />
  <row Id="10658" PostTypeId="1" AcceptedAnswerId="10668" CreationDate="2010-11-17T04:26:05.377" Score="4" ViewCount="327" Body="&lt;p&gt;Suppose that there is a certain collected works of plays that is N symbols long in the following sense: a &quot;symbol&quot; is one of the 26 letters of the alphabet, a line break, period, space, or a colon; in other words there are 30 possible symbols.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If &quot;a monkey&quot; randomly &quot;types&quot; 1 of these 30 symbols at a rate of one per second, how long will it take M monkeys working at this rate, on average, for one of them to randomly write this specific N symbol long collected works?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For clarity let me state that I am assuming each monkey ceaselessly types random symbols at this rate, and unless a monkey immediately types the right things, the collected works will be preceded by gibberish.&lt;/p&gt;&#xA;" OwnerUserId="1417" LastEditorUserId="1417" LastEditorDisplayName="" LastEditDate="2010-11-17T05:57:12.297" LastActivityDate="2010-11-17T06:25:56.077" Title="Monkeys and Typewriters" Tags="&lt;probability-theory&gt;" AnswerCount="2" FavoriteCount="0" />
  <row Id="10659" PostTypeId="2" ParentId="10646" CreationDate="2010-11-17T05:46:23.483" Score="6" ViewCount="" Body="&lt;p&gt;Ok, let's try another counterexample.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For simplicity, say a space with your property (any closed set is the intersection of its neighborhoods) is &quot;&lt;em&gt;Cromarty&lt;/em&gt;&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $X$ be an infinite set with the cofinite topology (the closed sets are all the finite sets and $X$ itself).  If $Z \subset X$ is closed (hence finite), let $\mathcal{U}$ be the collection of all open sets containing $Z$.  For any $x \notin Z$, $\{x\}^c$ is open (because it is cofinite) and contains $Z$, so $\{x\}^c \in \mathcal{U}$, and thus $x \notin \bigcap \mathcal{U}$; hence $Z^C \subset \bigcap \mathcal{U}$, so $\bigcap \mathcal{U}=Z$.  So $X$ is Cromarty.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, $X$ is not regular, since every pair of nonempty open sets has nonempty (in fact, infinite) intersection.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: To amplify, it looks like the Cromarty property is equivalent to being &lt;a href=&quot;http://en.wikipedia.org/wiki/R0_space&quot; rel=&quot;nofollow&quot;&gt;$R_0$&lt;/a&gt;, which means: if $x$ has a neighborhood not containing $y$, then $y$ has a neighborhood not containing $x$.  This is of course strictly weaker than being regular (as the cofinite topology shows).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose $X$ is $R_0$, and $Z$ is closed.  Suppose $x \notin Z$, $y \in Z$.  Then $Z^c$ is a neighborhood of $x$ not containing $y$, so $y$ has a neighborhood $U_y$ that does not contain $x$.  Now $U = \bigcup_{y \in Z} U_y$ is a neighborhood of $Z$ that does not contain $x$.  As above, it follows that $Z$ is the intersection of its neighborhoods, so $X$ is Cromarty.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conversely, suppose $X$ is Cromarty.  Suppose a point $x$ has a neighborhood $U$ not containing $y$.  $U^c$ is closed, hence the intersection of its neighborhoods, so $U^c$ has a neighborhood $V$ not containing $x$.  But $V$ is also a neighborhood of $y$, so $X$ is $R_0$.&lt;/p&gt;&#xA;" OwnerUserId="822" LastEditorUserId="822" LastEditorDisplayName="" LastEditDate="2010-11-17T13:58:16.870" LastActivityDate="2010-11-17T13:58:16.870" CommentCount="5" />
  <row Id="10660" PostTypeId="2" ParentId="10658" CreationDate="2010-11-17T05:48:08.830" Score="7" ViewCount="" Body="&lt;p&gt;Unfortunately, an exact answer will depend on the specific sequence of $N$ symbols.  To see why, take a simple example in which you only have two possible symbols, A and B, with $N = 2$, and only one monkey.  Compare the expected times until the monkey types the sequence AA vs. the sequence BA.  Unless the monkey types AA at the very beginning (with probability $\frac{1}{4}$), the first time AA appears the BA sequence &lt;em&gt;must&lt;/em&gt; have already occurred.  So the latter sequence will have the shorter expected time.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are lots of these sorts of counterintuitive results about the likelihood of seeing a certain sequence before another certain sequence and about the expected time until a sequence is first seen.  For more information, see MathWorld's entry on &lt;a href=&quot;http://mathworld.wolfram.com/CoinTossing.html&quot; rel=&quot;nofollow&quot;&gt;Coin Tossing&lt;/a&gt;, or, for even more information, the article &quot;&lt;a href=&quot;http://bact.mathcircles.org/files/Summer2010/PenneyAnte.pdf&quot; rel=&quot;nofollow&quot;&gt;Penney Ante&lt;/a&gt;&quot; that appeared in the &lt;em&gt;UMAP Journal&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if $N$ is large and the symbols are assumed to be evenly distributed in the target sequence, maybe you can avoid the problems with the two-symbol and unevenly distributed example I gave here.  Or, if you force each monkey to stop after typing exactly $N$ symbols, check to see if they have typed the target sequence, and then have them start over again from scratch if they have not, then you can definitely avoid these problems.  (For more on the latter, see this article &quot;&lt;a href=&quot;http://plus.maths.org/content/infinite-monkey-business&quot; rel=&quot;nofollow&quot;&gt;Infinite Monkey Business&lt;/a&gt;&quot;.)&lt;/p&gt;&#xA;" OwnerUserId="2370" LastActivityDate="2010-11-17T05:48:08.830" />
  <row Id="10661" PostTypeId="1" AcceptedAnswerId="10667" CreationDate="2010-11-17T05:56:57.727" Score="2" ViewCount="115" Body="&lt;p&gt;How to find the value of&#xA;$$\displaystyle\sqrt{3} \cdot \cot (20^{\circ}) - 4 \cdot \cos (20^{\circ})$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;manually ?&lt;/p&gt;&#xA;" OwnerUserId="2109" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-17T05:59:24.433" LastActivityDate="2010-11-17T06:24:06.243" Title="Find the value of $\displaystyle\sqrt{3} \cdot \cot (20^{\circ}) - 4 \cdot \cos (20^{\circ}) $" Tags="&lt;trigonometry&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="10662" PostTypeId="2" ParentId="9045" CreationDate="2010-11-17T06:10:04.373" Score="0" ViewCount="" Body="&lt;p&gt;I suppose that your definition of unitary is that $A^*A=AA^*=I$. &lt;br/&gt;&#xA;Let $A$ be unitary and $x\in V$. Then $$\|Ax\|^2=\langle Ax,Ax\rangle=\langle x, A^*Ax\rangle =\langle x,Ix\rangle=\langle x, x\rangle=\|x\|^2.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conversely, suppose $\|Ax\|=\|x\|$ for all $x\in V$. Then for all $x,y\in V$, one has&#xA;$$\|A(x-y)\|^2=\|Ax-Ay\|^2=\|Ax\|^2-2\langle Ax,Ay\rangle+\|Ay\|^2=\|x\|^2-2\langle Ax,Ay\rangle+\|y\|^2$$ and $$ \|x-y\|^2=\|x\|^2-2\langle x,y\rangle+\|y\|^2.$$ &#xA;Equating $\|A(x-y)\|^2$ and $\|x-y\|^2$ yields $\langle Ax,Ay\rangle=\langle x,y\rangle$. Thus, for all $x,y$,&#xA;$$\langle x,(A^*A-I)y\rangle=\langle x,A^*Ay\rangle-\langle x,y\rangle=\langle Ax,Ay\rangle-\langle x,y\rangle=0$$&#xD;&#xA;Then it follows from your cited previous question that $A^*A-I=0$. &lt;/p&gt;&#xA;" OwnerUserId="3249" LastEditorUserId="3249" LastEditorDisplayName="" LastEditDate="2010-11-17T06:17:16.627" LastActivityDate="2010-11-17T06:17:16.627" />
  <row Id="10663" PostTypeId="1" AcceptedAnswerId="10671" CreationDate="2010-11-17T06:14:31.963" Score="5" ViewCount="150" Body="&lt;p&gt;In a $\displaystyle\bigtriangleup$ ABC,R is circumradius and $\displaystyle 8R^2 = a^2 + b^2 + c^2 $ , then $\displaystyle\bigtriangleup$ ABC is of which &lt;a href=&quot;http://en.wikipedia.org/wiki/Triangle#By_internal_angles&quot; rel=&quot;nofollow&quot;&gt;type&lt;/a&gt; ?&lt;/p&gt;&#xA;" OwnerUserId="2109" LastActivityDate="2010-11-17T07:47:44.397" Title="What type of triangle satisfies: $8R^2 = a^2 + b^2 + c^2 $ ?" Tags="&lt;geometry&gt;&lt;trigonometry&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="10664" PostTypeId="1" AcceptedAnswerId="10669" CreationDate="2010-11-17T06:18:46.987" Score="1" ViewCount="201" Body="&lt;p&gt;I tried solving this, and I am pretty sure I am integrating this correctly, however, my solution manual shows -1 in the equation when doing this and I do not know why.  The answer in the solution manual is correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Problem: Find the corresponding distribution function and use it to determine the probability that a random variable having the distribution function will take on a value between 0.4 and 1.6.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;f(x) = x for 0 &amp;lt; x &amp;lt; 1&#xA;       2-x for 1 &amp;lt;= x &amp;lt; 2&#xA;       0 elsewhere&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;so for F(0.4 &amp;lt; x &amp;lt; 1.6) I did after integrating:&lt;br&gt;&#xA;2(1.6) - [(1.6)^2 / 2] - [(0.4)^2 / 2] = 1.84&lt;br&gt;&#xA;however the correct answer is 0.84.  The solution manual has a -1 in their equation, but I do not know how they got it.&lt;/p&gt;&#xA;" OwnerUserId="3601" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-11-17T06:42:12.680" LastActivityDate="2010-11-17T06:42:12.680" Title="Calculating probability with a piecewise density function" Tags="&lt;homework&gt;&lt;statistics&gt;&lt;integral&gt;&lt;probability&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="10665" PostTypeId="2" ParentId="10647" CreationDate="2010-11-17T06:21:45.373" Score="1" ViewCount="" Body="&lt;p&gt;The short answer is: there need not be a real number that is the supremum of the values of $\phi(x)$. You may have $\sup\{\phi(x)\mid x \in E\} = \infty$. If that is the case, you're out of luck.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-11-17T06:21:45.373" />
  <row Id="10667" PostTypeId="2" ParentId="10661" CreationDate="2010-11-17T06:24:06.243" Score="8" ViewCount="" Body="&lt;p&gt;$$\displaystyle \sin(60^{\circ}-20^{\circ}) = \sin 40^{\circ} = 2 \sin 20^{\circ} \cos 20^{\circ}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle \frac{\sqrt{3}}{2} \cos 20^{\circ} - \frac{1}{2} \sin 20^{\circ} = 2 \sin 20^{\circ} \cos 20^{\circ}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle \frac{\sqrt{3}}{2} \cos 20^{\circ}  - 2 \sin 20^{\circ} \cos 20^{\circ}  = \frac{1}{2} \sin 20^{\circ}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Multiply by $\displaystyle \frac{2}{\sin 20^{\circ}}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle \sqrt{3} \cot 20^{\circ} - 4 \cos 20^{\circ} = 1$$&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-11-17T06:24:06.243" CommentCount="5" />
  <row Id="10668" PostTypeId="2" ParentId="10658" CreationDate="2010-11-17T06:25:56.077" Score="4" ViewCount="" Body="&lt;p&gt;We will estimate the probability for a &quot;generic&quot; string. The number of occurrences of the string in any given monkey's output is roughly distributed Poisson with $\lambda = 30^{-N}$. The time until the first event happens is thus roughly distributed exponentially with rate $\lambda = 30^{-N}$. The minimum of $M$ such processes is also distributed exponentially with rate $\lambda = M/30^N$. Thus the expected time is roughly $30^N/M$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same estimate can be obtained if we calculate the expected number of appearances. The expected number of appearances in any given monkey's stream for the first $t+M-1$ characters is $t/30^N$. For $M$ monkeys, it is $tM/30^N$. This is $1$ for $t = 30^N/M$, and this gives a rough estimate for the actual expectation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, assuming that the string &quot;doesn't overlap with itself&quot;, we can get an exact expression for the expectation (depending only on $N$ and $M$) using Theorem 1.1 in &quot;String overlaps, pattern matching, and nontransitive games&quot; (Guibas and Odlyzko '81), which gives a generating function for the probability that a given monkey is not done after $t$ steps.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The paper also gives an expression for &quot;non-generic&quot; strings and for multiple strings, but the collected works are not going to overlap themselves; even if they do, it will probably have only a slight effect on the probabilities.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-17T06:25:56.077" />
  <row Id="10669" PostTypeId="2" ParentId="10664" CreationDate="2010-11-17T06:35:00.050" Score="2" ViewCount="" Body="&lt;p&gt;Since $f(t)$ is defined piecewise, you have to be careful with the integral.  This is the source of your mistake.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $0 &amp;lt; x &amp;lt; 1$, &#xA;$$F(x) = \int_0^x f(t) dt = \int_0^x t dt = \left.\frac{1}{2}t^2\right|_0^x = \frac{1}{2}x^2,$$&#xD;&#xA;which you have.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if $1 \leq x &amp;lt; 2$, then you have to break the integral that yields $F(x)$ up into pieces.  This is &#xA;$$F(x) = \int_0^x f(t) dt = \int_0^1 t dt + \int_1^x (2-t) dt = \left.\frac{1}{2}t^2\right|_0^1 + \left[2t - \frac{1}{2}t^2\right]_1^x $$&#xD;&#xA;$$= \frac{1}{2} + 2x - \frac{1}{2}x^2 - 2 + \frac{1}{2} = 2x - \frac{1}{2}x^2 - 1.$$&#xD;&#xA;Here's where the $-1$ comes in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a common mistake.  If it makes you feel any better, my probability students trip up over this all the time. :)&lt;/p&gt;&#xA;" OwnerUserId="2370" LastActivityDate="2010-11-17T06:35:00.050" CommentCount="1" />
  <row Id="10670" PostTypeId="2" ParentId="10664" CreationDate="2010-11-17T06:35:26.027" Score="1" ViewCount="" Body="&lt;p&gt;First, probabilities are from $0$ to $1$, so you're certainly wrong.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second, you calculated the cdf incorrectly for the interval $1 \leq x &amp;lt; 2$. It should be $$F(x) = \int_0^1 t \, \mathrm{d}t + \int_1^x 2-t \, \mathrm{d}t.$$ You forgot the first part, and integrated the second part from $0$ instead of from $1$. Since $$\int_0^1 t \, \mathrm{d}t = 1/2$$ whereas $$\int_0^1 2-t \, \mathrm{d}t = 3/2,$$ you counter $1/2$ too little and $3/2$ too much, for a total gain of $+1$.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-17T06:35:26.027" />
  <row Id="10671" PostTypeId="2" ParentId="10663" CreationDate="2010-11-17T06:36:57.193" Score="7" ViewCount="" Body="&lt;p&gt;This essentially means $\sin^2(A)+\sin^2(B)+\sin^2(C)=2$. (This follows from $\sin$ rule.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Replace $C = \pi - (A+B)$ to get $\sin^2(A+B) = \cos^2(A) + \cos^2(B)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Expand $\sin(A+B)$ and do the manipulations to get &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$2\cos^2(A)\cos^2(B) = 2\sin(A)\sin(B)\cos(A)\cos(B)$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;which means $\cos(A) = 0$ or $\cos(B) = 0$ or $\cos(A)\cos(B) = \sin(A)\sin(B) \Rightarrow \cos(A+B) = 0 \Rightarrow \cos(C) = 0$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence either $A = \pi/2$ or $B = \pi/2$ or $C = \pi/2$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the triangle is a right-angled triangle.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastActivityDate="2010-11-17T06:36:57.193" />
  <row Id="10672" PostTypeId="1" CreationDate="2010-11-17T06:54:15.730" Score="2" ViewCount="281" Body="&lt;p&gt;This is from the book Principles of Mathematical Analysis by Rudin,  number 4 of chapter 7. It says consider &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ f(x) = \sum\limits_{n=1}^{\infty}{ 1/(1+ n^2 x)  } $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question asks: &#xA;(1) For what values of x does the series converge absolutely. We got that the series converges when x $\not=$ 0 &amp;amp; x $\not= -1/k^2 $ when k is an integer since, there is a discontinuity when n reaches the value $k^2$ .&lt;br&gt;&#xA;However we don't understand how to do any of the following questions asked. Any hints would be greatly appreciated. We were told that this problem was suppose to be fairly hard for its position in the problem set (ie. 4th question in the Rudin book).&lt;br&gt;&#xA;(2) What interval does it converge uniformly?&lt;br&gt;&#xA;(3) On what intervals does it fail to converge uniformly ?&lt;br&gt;&#xA;(4) Is f continuous wherever the series converges?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strike&gt;(5) Is f bounded?&lt;/strike&gt;&lt;/p&gt;  &#xA;" OwnerUserId="2919" LastEditorUserId="2919" LastEditorDisplayName="" LastEditDate="2010-11-17T22:08:00.610" LastActivityDate="2010-11-17T22:08:00.610" Title="Convergence of a particular Series Uniformly (Homework)" Tags="&lt;real-analysis&gt;&lt;analysis&gt;&lt;homework&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="10673" PostTypeId="2" ParentId="7064" CreationDate="2010-11-17T06:56:49.590" Score="0" ViewCount="" Body="&lt;p&gt;In Israel I was taught (in the 90's) the US notation, with the divisor on the right (might have something to do with the fact that we write Hebrew right to left).&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-17T06:56:49.590" CommunityOwnedDate="2010-11-18T06:46:45.763" />
  <row Id="10674" PostTypeId="2" ParentId="10603" CreationDate="2010-11-17T07:05:10.070" Score="12" ViewCount="" Body="&lt;p&gt;One way to think of $\mathbb Q$ is the direct limit over positive integers $n$ of $\frac{1}{n} \mathbb Z$.  Thus giving a character of $\mathbb Q$ is the same&#xA;as giving an element in the projective limit of the character groups of&#xA;$\frac{1}{n}\mathbb Z$.  In particular, if we restrict to unitary characters,&#xA;we obtain the $\mathbb Q^{\vee}$ is the projective limit of circle groups $S^1$ under the $n$th power maps.  This object is (I think) called a solenoid; to number theorists it is better known as the adele class group $\mathbb A/\mathbb Q$.  (Here and throughout I am using the discrete topology; if one instead considers the&#xA;induced topology from $\mathbb R$, then, as Robin explains, one just gets &#xA;characters of $\mathbb R$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The exact sequence $0 \to \hat{\mathbb Z} \to \mathbb Q^{\vee} \to S^1 \to 0$&#xD;&#xA;in Pete's answer arises from the map taking the solenoid to the base $S^1$; the fibres of this map are copies of $\hat{\mathbb Z}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we wanted not necessarily unitary characters, we would instead get&#xA;the projective limit of copies of $\mathbb C^{\times}$ under the $n$th power maps.  Since $\mathbb C^{\times} = \mathbb R_{&amp;gt; 0} \times S^1$, and since&#xA;$\mathbb R_{&amp;gt; 0}$ is uniquely divisible, this projective limit is simply &#xA;$\mathbb R_{&amp;gt; 0}$ times the solenoid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On a slightly tangential note, let me remark that&#xA;the relationship with the adeles is important (e.g. it is the first step in Tate's thesis):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the adeles are the (restricted) product of $\mathbb R$ and each $\mathbb Q_p$, and since these are all self-dual, it is easy to see that $\mathbb A$ is self-dual.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One then has the exact sequence&#xA;$$0 \to \mathbb Q \to \mathbb A \to \mathbb A/\mathbb Q \to 0$$&#xD;&#xA;which is again self-dual (the duality swaps $\mathbb Q$ and the solenoid&#xA;$\mathbb A/\mathbb Q$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One should compare this with the exact sequence&#xA;$$0 \to \mathbb Z \to \mathbb R \to \mathbb R/\mathbb Z = S^1 \to 0.$$&#xD;&#xA;This is again self-dual ($\mathbb R$ is self-dual,&#xA;and duality swaps the integers and the circle).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This brings out the important intuition that the adeles are to $\mathbb Q$ as&#xA;$\mathbb R$ is to $\mathbb Z$.&lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2010-11-17T07:05:10.070" CommentCount="1" />
  <row Id="10675" PostTypeId="2" ParentId="10663" CreationDate="2010-11-17T07:14:26.483" Score="3" ViewCount="" Body="&lt;p&gt;In general, in $\triangle{ABC}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $\sin^2 A + \sin^2 B + \sin^2 C \gt 2$ then $\triangle{ABC}$ is acute angled.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $\sin^2 A + \sin^2 B + \sin^2 C  = 2$ then $\triangle{ABC}$ is right angled.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $\sin^2 A + \sin^2 B + \sin^2 C \lt 2$ then $\triangle{ABC}$ is obtuse angled.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume $A \le B \lt \pi/2$ and $ A \le B \le C$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically, if $k = \sin^2 A + \sin^2 B + \sin^2 C$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;then we have that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$3-2k = \cos 2A + \cos 2B + \cos (2A+2B)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i.e&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$3-2k = 2\cos(A+B)\cos(A-B) + 2\cos^2(A+B) -1 $&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i.e&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$4-2k = 4\cos(A+B)\cos A\cos B$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if $k &amp;gt; 2$, then $\cos(A+B) \lt 0$ hence acute.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$k = 2$, then $\cos(A+B) = 0$ hence right triangled.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$k &amp;lt; 2$, then $\cos(A+B) \gt 0$, hence obtuse.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, we can go further and show that the maximum possible value of $k$ is $k = \frac{9}{4}$ which corresponds to $\triangle{ABC}$ being equilateral, as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$4\sin^2 A + 4\sin^2 B + 4\sin^2 C = 9  + \delta$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i.e.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(2 - 2\cos2A) + (2-2\cos 2B) + 4(1- \cos^2 (A+B)) = 9 + \delta$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i.e.&#xA;$1 + 2\cos2A + 2\cos 2B + 4\cos^2(A+B) = -\delta$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i.e.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$1 + 4\cos(A+B)\cos(A-B) + 4\cos^2(A+B) = -\delta$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i.e.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sin^2(A-B) + \cos^2(A-B) + 4\cos(A+B)\cos(A-B) + 4\cos^2(A+B) = -\delta$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i.e.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sin^2(A-B) + (\cos (A-B) + 2\cos(A+B))^2 = -\delta$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence $\delta \le 0$ and so $\sin^2 A + \sin^2 B + \sin^2 C \le \frac{9}{4}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The case $\delta = 0$ gives us $\sin(A-B) = 0$ and $\cos(A+B) = \frac{-1}{2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence $A=B=C$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus the max value of $\sin^2 A + \sin^2 B + \sin^2 C$ is $\frac{9}{4}$ and is achieved when $A=B=C$.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-11-17T07:47:44.397" LastActivityDate="2010-11-17T07:47:44.397" />
  <row Id="10676" PostTypeId="2" ParentId="10648" CreationDate="2010-11-17T07:25:20.713" Score="17" ViewCount="" Body="&lt;p&gt;The question is equivalent to asking why 6 equilateral triangles fit together exactly around a point, with no additional room left over.  The answer is &quot;because the Euclidean plane is flat&quot;, a condition equivalent to triangles having angle sum of 180 degrees (half the angle around a point), so that each vertex of a symmetrical triangle has 1/3 of half of a full rotation = 1/6.  That the six-circle arrangement exists for any radius is also a special feature of Euclidean geometry: scale invariance.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For flat surfaces such as a cylinder (rolled up plane) or flat torus (as in the Asteroids video game) the perfect 6-circle configurations exist only for small enough radius of the circles.  These geometries are, in small regions, the same as the Euclidean plane but differ &quot;globally&quot;, e.g. there is a maximum distance between points of the torus.  So the magic number of 6 is really about local flatness (absence of curvature) and having this configuration for circles of all size is a global question about the space in which the circles live.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In hyperbolic geometry there are tesselations of the plane by equilateral triangles with angle $180/n$ at each vertex, for any $n \geq 7$.  In the picture of the $n=7$ triangular tesselation at &lt;a href=&quot;http://www.plunk.org/~hatch/HyperbolicTesselations/3_7_trunc0_512x512.gif&quot; rel=&quot;nofollow&quot;&gt;http://www.plunk.org/~hatch/HyperbolicTesselations/3_7_trunc0_512x512.gif&lt;/a&gt; (triangles in white with the dual tesselation by heptagons shown in blue) if at each triangle vertex you draw a circle inscribed in its heptagon, there will be 7 circles exactly surrounding each circle, with all circles of the same radius.  The same type of configuration exists for any $n$ and suitable radius of the circles.  So the flatness condition is necessary; the theorem is false in negatively curved two-dimensional geometry.  Six is also not the correct number for spherical geomety, and both spherical and hyperbolic geometry lack a radius-independent &quot;number of circles that fit around one circle&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the geometry of surfaces, having 5-or-fewer as the local number of circles that can be fit around a single circle characterizes positive curvature, as in spherical geometry.  Having more-than-6 fit, or extra room when surrounding one circle by six, is a characterization of negative curvature, as in hyperbolic geometry. This is a statement about the local geometry of general 2-dimensional surfaces, and does not assume the surface has the same amount of curvature at all points, as would be the case for the spherical and hyperbolic analogues of Euclidean plane geometry, where there is a homogeneity assumption that &quot;geometry is the same at all points&quot;.  Having exactly 6 circles fit perfectly is a characterization of locally Euclidean (that is, zero curvature) geometry.  If you don't know what curvature is in a technical sense, for purposes of this discussion it is (for surfaces) a number that can be associated to any point on the surface, and curvature being zero in a region of surface is equivalent to the ability to make a distance-preserving planar map of that region.  Impossibility of doing this for surfaces of nonzero curvature, such as the sphere which is positively curved, was Gauss' &lt;em&gt;Theorema Egregium&lt;/em&gt; which ruled out perfect flat cartography of the Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Flatness plays the same role in the higher dimensional, sphere-packing interpretation of the question that Qiaochu suggested.  The necessary packings exist only in a limited set of dimensions.  The reason is that exactly surrounding a sphere of radius $r$ by $k$ spheres of that radius means more than a kissing number of $k$, the optimal kissing configuration should also be &lt;em&gt;rigid&lt;/em&gt; (tight enough that the spheres cannot be moved).  Rigidity is false in three dimensions and is believed to be false in most other dimensions but in those cases where it is known or suspected to be true (i.e., dimensions $d = 2, 8, 24$ and possibly a few others) the existence of a configuration with the same set of equalities between various interpoint distances as in the optimal kissing arrangement (in flat Euclidean space $\mathbb{R}^d$)should force a homogeneous $d$-dimensional geometry to be flat.  This is because deforming the curvature of the space in the positive direction would reduce, and negative curvature would increase, the freedom to position such spheres around a central sphere.&lt;/p&gt;&#xA;" OwnerUserId="467" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-11-17T21:12:13.910" LastActivityDate="2010-11-17T21:12:13.910" />
  <row Id="10677" PostTypeId="1" AcceptedAnswerId="10684" CreationDate="2010-11-17T08:18:15.290" Score="1" ViewCount="142" Body="&lt;p&gt;Hello everyone.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a question about a transformation of a matrix. Lets say we have the following matrix&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$&#xD;&#xA;M =&#xD;&#xA;\left[ {\begin{array}{cc}&#xD;&#xA; 4 &amp;amp; 3  \\&#xD;&#xA; 4 &amp;amp; 3  \\&#xD;&#xA; \end{array} } \right]&#xD;&#xA;$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then I want to transform all numbers in the matrix with the following function:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f(x) = (x - 4)^2 + 2x -4$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What would be the correct notation for this?&lt;/p&gt;&#xA;" OwnerUserId="3605" LastActivityDate="2010-11-17T10:29:37.393" Title="Matrix transformation notation" Tags="&lt;matrices&gt;&lt;notation&gt;&lt;transformation&gt;" AnswerCount="2" />
  <row Id="10678" PostTypeId="2" ParentId="10672" CreationDate="2010-11-17T08:41:02.677" Score="2" ViewCount="" Body="&lt;p&gt;For strictly positive $x$ you can already see that each term of the series, except the first one, is bounded by:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{1}{1+n^2 x} &amp;lt; \frac{1}{(n-1)^2 x}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The series formed by the bounds converges, so by the Weierstrass M-test, you have uniform convergence for $x&amp;gt;0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $x&amp;lt;-1$ you can make the same story:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left|\frac{1}{1+n^2 x}\right| &amp;lt; \frac{1}{(n+1)^2 |x|}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The series formed by the bounds converges, so by the Weierstrass M-test, you have uniform convergence for $x&amp;lt;-1$ as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The hard part is of course what happens on the open interval $]-1,0[$, you have already excluded the endpoints as well as all the negative inverses of squares of integers within that interval. &lt;/p&gt;&#xA;" OwnerUserId="3567" LastActivityDate="2010-11-17T08:41:02.677" CommentCount="2" />
  <row Id="10680" PostTypeId="2" ParentId="10672" CreationDate="2010-11-17T09:02:06.350" Score="2" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have seen a number of discontinuities at certain points.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Can you use the M-test between points of discontinuities? &#xA;1.a If you can then it is good.&#xA;1.b If you can not do that - why? Can you do estimates if you remove a part around these points?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Can you show that the convergence is NOT uniform near these points? &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="1154" LastActivityDate="2010-11-17T09:02:06.350" />
  <row Id="10681" PostTypeId="2" ParentId="10677" CreationDate="2010-11-17T09:06:08.823" Score="1" ViewCount="" Body="&lt;p&gt;Could use the &quot;Hadamard Product&quot;... I don't know whether there is a consistently accepted notation for such a product, however a few examples &lt;a href=&quot;http://planetmath.org/encyclopedia/HadamardProduct.html&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; and on Wikipedia seem to use an open circle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In that sense, your expression would take the form&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(M-4O)\circ(M-4O)+2M-4O$ where $O= \left(  \begin{array} {cc} 1 &amp;amp;1 \\ 1&amp;amp;1 \end{array}  \right) $&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In MATLAB, use .* for element-wise multiplication.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are probably other possibilities.&lt;/p&gt;&#xA;" OwnerUserId="3564" LastActivityDate="2010-11-17T09:06:08.823" />
  <row Id="10682" PostTypeId="2" ParentId="10469" CreationDate="2010-11-17T09:07:44.160" Score="1" ViewCount="" Body="&lt;p&gt;I realized that your question wasn't exactly about the Steenrod axioms themselves, but about the definition of the coboundary operator involving suspension.  In reduced homology, the boundary operator $\partial$ for the pair $(X,A)$ (where the inclusion $i:A\rightarrow X$ is a cofibration) can be defined to come from the &quot;topological boundary map&quot; $\partial^!$ followed by the inverse of the suspension isomorphism.  The former is itself a composition&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \partial^! = \pi \circ \psi^{-1}: X/A \rightarrow Ci \rightarrow \Sigma A, $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $Ci$ is the mapping cone of $i$, $\psi^{-1}$ is a homotopy inverse of the quotient $\psi: Ci \rightarrow Ci/CA=X/A$, and $\pi: Ci \rightarrow Ci/X=\Sigma A$.  So&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \partial = (\Sigma_*)^{-1} \circ \partial^!_* : \tilde{H}_q(X/A) \rightarrow \tilde{H}_q(\Sigma A) \rightarrow \tilde{H}_{q-1}(A) .$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, this is true for any reduced homology theory.  See May's &quot;Concise Course&quot; for details, pp. 106-7.  I'm pretty sure that the situation for cohomology is very similar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bottom line: In this formulation, the coboundary operator is the composition of a map induced from an actual map on spaces and the (inverse of the (?)) suspension isomorphism.  Steenrod squares commute with both of these, so they commute with the coboundary operator.&lt;/p&gt;&#xA;" OwnerUserId="2159" LastActivityDate="2010-11-17T09:07:44.160" CommentCount="3" />
  <row Id="10683" PostTypeId="2" ParentId="10233" CreationDate="2010-11-17T09:28:27.033" Score="1" ViewCount="" Body="&lt;p&gt;The most interesting applications requiring the services of Quadratic Reciprocity  are, in my opinion, related to the field of Mathematical Cryptography!!  As mentioned above, Quadratic Reciprocity is a particularly useful tool when you want to see if a number is a square mod p (p prime.)  One cryptosystem in particular that requires the help of Quadratic Reciprocity is the Goldwasser-Micali public key cryptosystem; this is the case because it poses the following question based on the following information:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let p, q be (secret) primes and let N=pq be given.  For a given integer a, determine whether a is a square mod N, ie, determine if there exists an integer u satisfying u^2 = a mod N.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, it is especially easy for Bob, the receiver of the message who knows how to factor N, to solve this problem because a is a square mod pq iff (a/p)=1 and (a/q)=1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is all from the Springer book entitled &lt;em&gt;An Introduction to Mathematical Cryptography&lt;/em&gt; by Hoffstein, Pipher, and Silverman.  This is such a great book (it was the course text for a course on Cryptography I took last spring) and Id recommend it to anyone who is interested in learning more about the practical and totally was AWESOME applications of Algebraic Number Theory.  There is so much more to be said about Quadratic Reciprocity (mentioned in the end of the 3rd chapter in this book) so be sure to check it out!&lt;/p&gt;&#xA;" OwnerUserId="3608" LastActivityDate="2010-11-17T09:28:27.033" />
  <row Id="10684" PostTypeId="2" ParentId="10677" CreationDate="2010-11-17T10:29:37.393" Score="3" ViewCount="" Body="&lt;p&gt;Maybe I don't understand your question, but: what's wrong with this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;\begin{pmatrix}&#xD;&#xA;f(4) &amp;amp; f(3) \\\&#xD;&#xA;f(4) &amp;amp; f(3)&#xD;&#xA;\end{pmatrix}&#xD;&#xA;$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More generally, if you have a matrix $(a^i_j)$, you could write the resulting matrix as $(f(a^i_j))$.&lt;/p&gt;&#xA;" OwnerUserId="664" LastActivityDate="2010-11-17T10:29:37.393" CommentCount="1" />
  <row Id="10685" PostTypeId="1" AcceptedAnswerId="10813" CreationDate="2010-11-17T10:33:29.033" Score="3" ViewCount="116" Body="&lt;p&gt;At Wikipedia I found that &quot;&lt;a href=&quot;http://en.wikipedia.org/wiki/Asymmetric_graph&quot; rel=&quot;nofollow&quot;&gt;according to a strengthened version of Frucht's theorem, there are infinitely many asymmetric cubic graphs&lt;/a&gt;&quot;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Are there infinitely many asymmetric&#xA;  &lt;em&gt;planar&lt;/em&gt; cubic graphs, too?&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;If so, does it follow that there is an &lt;em&gt;infinite&lt;/em&gt; asymmetric planar cubic graph?&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;If so, how could this graph be characterized?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Background&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am looking for an homogeneous and isotropic (in the large) regular graph that could &quot;mimick&quot; a discretized plane (without distinguished directions as in a grid). So an infinite asymmetric 4-regular graph would be even better (reflecting dimension 2).&lt;/p&gt;&#xA;" OwnerUserId="1792" LastEditorUserId="1792" LastEditorDisplayName="" LastEditDate="2010-11-17T11:15:29.260" LastActivityDate="2010-11-18T05:12:56.353" Title="Asymmetric planar cubic graphs" Tags="&lt;graph-theory&gt;" AnswerCount="2" />
  <row Id="10686" PostTypeId="1" AcceptedAnswerId="10692" CreationDate="2010-11-17T11:12:08.157" Score="4" ViewCount="270" Body="&lt;p&gt;Reposting from stackoverflow :) - Told to go here :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I've got a issue with a math assignment, and I were hoping someone could at least point me to what I should be doing, because currently I'm just wasting my time, thats for sure. - Please don't solve it all, thats up for me to do so :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The assignment sounds like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A swimming pool is circular with a 10-metre diameter. The depth is constant along east-west lines and increases linearly from 1meter at the south end to 2meters at the north end. - Find the volume of water in the pool.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I figured out that its about using double integrals on polar coordinates, however from there I kinda stalled, first of all; (1) if the pool is circular and has a 10-metre diameter, am I certain that the diameter is 10metres everywhere on the pool? - or just on a single spot? - That is could the pool be oval? (2) How far goes a 'side' alike west, east, north and south, on such a circular pool? - if its circular when will I know when I'm in the east line, and what about the north-east? (3) Would it seem likely that I'm going to describe the pool as a polar function first, before even considering the use of Integrals?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know this might not be the correct fora, for math questions, however stackoverflow has always seemed like a nice helpfull community :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;//Skeen&lt;/p&gt;&#xA;" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-11-17T16:38:34.637" LastActivityDate="2010-11-17T16:38:34.637" Title="Finding the volume of water in a swimming pool" Tags="&lt;homework&gt;&lt;integral&gt;&lt;polar-coordinates&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="10687" PostTypeId="1" AcceptedAnswerId="10688" CreationDate="2010-11-17T11:16:06.377" Score="1" ViewCount="99" Body="&lt;p&gt;$G$ is a finite abelian group. $A \times B$ can be embedded in $G$.&#xA;Does this mean there exist $C$, $D$ such that $G=C \times D$ and $A$ can be embedded in $C$ and $B$ can be embedded in $D$?&lt;/p&gt;&#xA;" OwnerUserId="3533" LastActivityDate="2010-11-17T12:57:18.393" Title="$G$ finite abelian. $A \times B$ embedded in $G$. Is $G=C \times D$ such that $A$ embedded in $C$, $B$ embedded in $D$?" Tags="&lt;group-theory&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="10688" PostTypeId="2" ParentId="10687" CreationDate="2010-11-17T11:28:58.053" Score="3" ViewCount="" Body="&lt;p&gt;Yes, this reduces to the case of finite $p$-groups. Use the lemma&#xA;that if $G=Z_{p^{r_1}}\times Z_{p^{r_2}}\times\cdots\times Z_{p^{r_k}}$&#xD;&#xA;where $r_1\ge r_2\ge\cdots \ge r_k$&#xD;&#xA;then each subgroup of $G$ is isomorphic to&#xA;$Z_{p^{s_1}}\times Z_{p^{s_2}}\times\cdots\times Z_{p^{s_k}}$&#xD;&#xA;where each $s_i\le r_i$.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-11-17T11:28:58.053" CommentCount="4" />
  <row Id="10689" PostTypeId="2" ParentId="10686" CreationDate="2010-11-17T11:51:43.363" Score="2" ViewCount="" Body="&lt;p&gt;To address the question that stalled you: a circular pool isn't oval.  If the problem meant for you to consider an oval pool, it would have said &quot;an elliptical pool with (say) horizontal axis $a$ and vertical axis $b$,&quot; or something like that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You don't need polar coordinates for this, nor do you need a double integral.  Single-variable calculus will do the trick just fine.  For starters, orient the pool so that the circle which defines it is centered at the origin, and look down at the pool from overhead.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Think of the water in it as if it was a big block of jello.  Now imagine slicing that block of jello horizontally infinitely many times, so that you have a stack of jello slices, each of which is infinitely thin, and each of which has its own width and depth, and each of which is located at its own $y$ coordinate (that is to say, for each slice, the points that make up the slice all have the same $y$-coordinate).  Each slice is essentially a rectangular box with volume width times depth times &quot;thinness&quot;.  In the figure below, the blue lines represent the individual jello slices, with the thick one being the slice located at $y = -3$.&#xA;&lt;img src=&quot;http://i.imgur.com/1v00U.png&quot; alt=&quot;figure of pool&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the equation of the circle (which, since it's centered at the origin, is $x^2 + y^2 = r^2$, where $r$ is the radius of the circle), you can figure out the width of each jello slice.  Knowing that the depth increases linearly, you can also figure out the depth of each slice as a function of its $y$ coordinate. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;[Actually, it's not important that the depth increase linearly, per se, just that it changes north-south (i.e. as a function of $y$) and not east-west (i.e. as a function of $x$).  If it changed east-west instead, you'd slice the jello vertically instead of horizontally.  If it changed in all directions, then you'd need a double integral.].&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, once you know the width and the depth, you can compute the volume of each slice: $$V_{\mbox{slice}} = \mbox{width}_{\mbox{slice}} \cdot \mbox{depth}_{\mbox{slice}} \cdot dy$$ where $dy$ represents the infinite &quot;thinness&quot; of the slice.  Now it's just a question of adding them up, which is exactly the point of a definite integral.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the total volume is:&#xA;$$V_{\mbox{total}} = \int_{-r}^{-r} V_{\mbox{slice}} = \int_{-r}^{r} \mbox{width}_{\mbox{slice}} \cdot \mbox{depth}_{\mbox{slice}} \cdot dy$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll leave you to figure out the appropriate expressions (in terms of $y$) for the width and the depth of each slice.&lt;/p&gt;&#xA;" OwnerUserId="506" LastEditorUserId="506" LastEditorDisplayName="" LastEditDate="2010-11-17T12:13:03.533" LastActivityDate="2010-11-17T12:13:03.533" />
  <row Id="10690" PostTypeId="2" ParentId="10687" CreationDate="2010-11-17T12:57:18.393" Score="2" ViewCount="" Body="&lt;p&gt;My first reaction was that the answer is no, but I had misinterpreted the question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A different question is: If $A$ and $B$ are subgroups of the abelian group $G$ such that $A \times B$ embeds in $G$, is it true that there are subgroups $C$ and $D$ of $G$ with $A \subseteq C$, $B \subseteq D$ and $G \cong C \times D$.&lt;/p&gt;&#xA;" OwnerUserId="2604" LastActivityDate="2010-11-17T12:57:18.393" />
  <row Id="10691" PostTypeId="2" ParentId="10181" CreationDate="2010-11-17T12:57:26.453" Score="2" ViewCount="" Body="&lt;p&gt;For an undergraduate-level book, none of these three can do you wrong:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Mendelson, &lt;em&gt;Introduction to mathematical logic&lt;/em&gt;&lt;/li&gt;&#xA;&lt;li&gt;Enderton, &lt;em&gt;A mathematical introduction to logic&lt;/em&gt;&lt;/li&gt;&#xA;&lt;li&gt;Boolos, Burgess, and Jeffrey, &lt;em&gt;Computability and logic&lt;/em&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;They certainly go into the details, and they will leave you in a position where you could go further. If you can only look at one, try Mendelson.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;An older text, which is now reprinted by Dover, is Kleene's &lt;em&gt;Mathematical logic&lt;/em&gt;. It is also very thorough, and has the advantage of being inexpensive. (Be aware that Kleene's &lt;em&gt;Introduction to metamathematics&lt;/em&gt; is a completely different book, which is much more advanced.)&lt;/p&gt;&#xA;" OwnerUserId="630" LastActivityDate="2010-11-17T12:57:26.453" />
  <row Id="10692" PostTypeId="2" ParentId="10686" CreationDate="2010-11-17T12:58:50.337" Score="7" ViewCount="" Body="&lt;p&gt;Imaging one more inverted pool of same size below our original pool. They two pools will exactly fit in and will together form a cylinder of diameter 10 metres and height 3 metres.&lt;/p&gt;&#xA;" OwnerUserId="3615" LastActivityDate="2010-11-17T12:58:50.337" CommentCount="1" />
  <row Id="10693" PostTypeId="1" AcceptedAnswerId="11345" CreationDate="2010-11-17T13:15:41.040" Score="2" ViewCount="337" Body="&lt;p&gt;Define a subprobability measure to be a measure on Borel sigma algebra of the real line $\mathbb{R}$  with the measure for the whole real line less or equal than 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was wondering about the definition of vague convergence of a sequence of subprobability measures $\{ \mu_n, n\geq 1 \}$ to another subprobablity measure $\mu$. The convergence can be defined in slightly two different ways as in &lt;a href=&quot;http://books.google.com/books?id=zu80w7wd40UC&amp;amp;printsec=frontcover&amp;amp;dq=Chung%27s+probability+theory&amp;amp;source=bl&amp;amp;ots=AnNrU86FDf&amp;amp;sig=EAzRmK6PIIdg0qLjvUiUHlFblCM&amp;amp;hl=en&amp;amp;ei=UNjjTOC0IoX6lwe_leWbDQ&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=3&amp;amp;ved=0CCsQ6AEwAg&quot; rel=&quot;nofollow&quot;&gt;Chung's probability theory book&lt;/a&gt; p85 and p90:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) if there exists a dense subset D of the real line $\mathbb{R}$  so that $ \forall a \text{ and } b \in D \text{ with } a &amp;lt;b, \mu_n((a,b]) \rightarrow \mu((a,b])$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) if there exists a dense subset D of the real line $\mathbb{R}$  so that $ \forall a \text{ and } b \in D \text{ with } a &amp;lt;b, \mu_n((a,b)) \rightarrow \mu((a,b))$ (the original text just says converges, not mention that converges to $ \mu((a,b))$ which I guess can be added?).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How to show these two definitions are equivalent?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A side question: &#xA;is there a definition of vague congergence for general measures on more general sigma algebra with more general underlying space?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you so much!&lt;/p&gt;&#xA;" OwnerUserId="3616" LastEditorUserId="3616" LastEditorDisplayName="" LastEditDate="2010-11-17T13:38:33.710" LastActivityDate="2010-11-22T13:08:27.933" Title="Vague convergence of measures" Tags="&lt;probability-theory&gt;&lt;measure-theory&gt;" AnswerCount="4" CommentCount="3" />
  <row Id="10694" PostTypeId="1" CreationDate="2010-11-17T13:30:21.210" Score="0" ViewCount="36" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I receive 800 data per day. I have to spot failures in thoses data.&#xA;I use average and std dev to detect site where there is a problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there some model for that kind of purpose ? Do you have books reference abou that ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance&lt;/p&gt;&#xA;" OwnerUserId="3617" LastActivityDate="2010-11-17T13:30:21.210" Title="Statistic failure detection" Tags="&lt;statistics&gt;" CommentCount="1" />
  <row Id="10695" PostTypeId="2" ParentId="10693" CreationDate="2010-11-17T13:41:57.257" Score="0" ViewCount="" Body="&lt;p&gt;Try $(a,b] = \bigcap_{c \in D, c &amp;gt; b} (a,c)$ and $(a,b) = \bigcup_{c \in D, c &amp;lt; b} (a,c]$.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-17T13:41:57.257" CommentCount="1" />
  <row Id="10696" PostTypeId="1" AcceptedAnswerId="10712" CreationDate="2010-11-17T14:03:12.540" Score="0" ViewCount="133" Body="&lt;p&gt;Is it possible to define distribution function for a measure on a general measurable space, which generalizes the definition of distribution function for a probability measure on the extended real line? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If yes, how? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If no, how general/special can the measure and the measurable space be in order to define its distribution function?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="3616" LastActivityDate="2010-11-17T16:35:24.057" Title="Define distribution function on a general measure space." Tags="&lt;probability-theory&gt;&lt;measure-theory&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="10697" PostTypeId="1" AcceptedAnswerId="10724" CreationDate="2010-11-17T15:03:22.310" Score="6" ViewCount="471" Body="&lt;p&gt;YARFMO (Yet another reposting from Mathoverflow) ;-)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The more you know about math the more you find conceptions previously thought correct to be false:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1.) math is not as exact as many believe - in many cases, as Gowers points out in his amazing book &quot;Mathematics. A very short intro&quot;, it is mainly about approximations, boundaries or even &quot;only&quot; existence. Example: The Riemann hypothesis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2.) math is not &quot;true&quot; (whatever that means) in its own right - it is only true within an axiomatic system (deductive approach). Example: The parallel axiom.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3.) closed form solutions are not as closed as many think: &quot;closed-form&quot; is just shorthand for &quot;popular enough to be given a name and notation.&quot; Examples: $\pi, e, \mathrm{Si}(x), \mathrm{li}(x)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My question&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do you have further common misconceptions about math (ideally also with examples)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think that the ones given above only scratch the surface because in the end the picture of math also changes historically the further we proceed on this never-ending endeavor of human mind.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(As should be clear already, I am talking about the &quot;meta'-level, so please don't give examples like $(x+y)^2=x^2+y^2$ - this is already covered &lt;a href=&quot;http://mathoverflow.net/questions/23478&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; and &lt;a href=&quot;http://math.stackexchange.com/questions/7864&quot;&gt;here&lt;/a&gt;.)&lt;/p&gt;&#xA;" OwnerUserId="346" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-17T15:20:32.130" LastActivityDate="2011-02-16T02:17:07.077" CommunityOwnedDate="2010-11-18T17:24:01.440" Title="Common misconceptions about math" Tags="&lt;soft-question&gt;&lt;big-list&gt;&lt;education&gt;&lt;philosophy&gt;&lt;meta-math&gt;" AnswerCount="5" CommentCount="4" FavoriteCount="1" />
  <row Id="10698" PostTypeId="2" ParentId="10593" CreationDate="2010-11-17T15:34:15.453" Score="0" ViewCount="" Body="&lt;p&gt;Proving that a manifold $M$ is not simply connected (if true) is more difficult than proving that $M$ is simply connected (if true), because the first requires an &quot;impossibility proof&quot; (like proving that you cannot trisect an angle with ruler and compass). For an impossibility proof you need a &quot;higher theory&quot;. E.g., in order to prove that an annulus $R$  in $\mathbb R^2$ is not simply connected you show that $R$ carries a closed vector field (namely $\nabla\arg $) which is not the gradient of a globally defined function. Such a thing would not exist on a simply connected domain.&lt;/p&gt;&#xA;" OwnerUserId="1303" LastActivityDate="2010-11-17T15:34:15.453" CommentCount="1" />
  <row Id="10699" PostTypeId="1" CreationDate="2010-11-17T15:36:41.443" Score="1" ViewCount="126" Body="&lt;p&gt;I was wondering what is wrong in the following proof:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proposition. Let $\{f_{n}\}$ be a sequence of integrable functions such that $f_{n}$ converges pointwise to a function $f$. Show that if:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\lim \int |f_{n} - f| d\mu =0$ then $\int |f| d\mu = \lim \int |f_{n}| d\mu$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well I used the fact that $||f_{n}| - |f| | \leq |f_{n} - f|$ and integrating both sides and using the assumption that $\lim \int |f_{n} - f| d\mu =0$ I get the result. But I never used the fact that $f_{n}$ converges pointwise to $f$. Why do we need this assumption or what is wrong?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="4219" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-17T17:36:13.100" LastActivityDate="2010-11-17T18:45:30.613" Title="Doubt convergence of integrals" Tags="&lt;measure-theory&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="10700" PostTypeId="1" CreationDate="2010-11-17T15:46:17.867" Score="0" ViewCount="79" Body="&lt;p&gt;We are suppose to compute the quaternion which performs 1/5 of the rotation of this quaternion:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;[ 0.965  (0.149  -0.149   0.149)]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The answer provided is shown as below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://i.imgur.com/FIRJY.jpg&quot; rel=&quot;nofollow&quot;&gt;Picture of the answer&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm stuck at the second last row of the ans. How do you change from the '0.998 q0 + 0.202 q' to [0.998 0.03 -0.03 0.03]&lt;/p&gt;&#xA;" LastActivityDate="2011-03-17T17:33:35.373" Title="Question regarding practical SLERP" Tags="&lt;quaternions&gt;" AnswerCount="1" />
  <row Id="10701" PostTypeId="1" AcceptedAnswerId="10705" CreationDate="2010-11-17T15:49:07.583" Score="2" ViewCount="93" Body="&lt;p&gt;Let $f: A \rightarrow B$ be a ring homomorphism between commutative rings with identity. Then there exists an induced map $f&amp;#39; : Spec(B) \rightarrow Spec(A)$. If $f&amp;#39;$ is surjective, then clearly every prime ideal of $A$ is a contracted ideal. Now my question is, is the converse true?&lt;/p&gt;&#xA;" OwnerUserId="3620" LastActivityDate="2011-03-28T06:38:26.687" Title="If every prime ideal is a contracted ideal, does it imply that the induced map between spectrums is surjective?" Tags="&lt;commutative-algebra&gt;" AnswerCount="1" />
  <row Id="10702" PostTypeId="1" CreationDate="2010-11-17T15:49:27.090" Score="2" ViewCount="121" Body="&lt;p&gt;If $\sin(A+B) + \sin(B+C) + \cos(C-A) = \frac{3}{2}$ show that, $$1.\sin A + \cos B + \sin C = 0$$ $$2. \cos A + \sin B +\cos C = 0$$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="2109" LastActivityDate="2010-11-17T18:53:17.450" Title="Proving trignometrical identities:" Tags="&lt;trigonometry&gt;" AnswerCount="1" CommentCount="10" />
  <row Id="10703" PostTypeId="2" ParentId="10699" CreationDate="2010-11-17T15:51:12.987" Score="3" ViewCount="" Body="&lt;p&gt;Nothing's wrong, your proof is fine.&lt;/p&gt;&#xA;" OwnerUserId="3035" LastActivityDate="2010-11-17T15:51:12.987" />
  <row Id="10704" PostTypeId="2" ParentId="10685" CreationDate="2010-11-17T15:54:31.930" Score="1" ViewCount="" Body="&lt;p&gt;I am not sure if this gets at what you mean for the 4-valent case:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take the infinite graph 4-valent which arises from tiling the plane with 4-gons (squares). Pick one vertex and remove that vertex and the 4 edges attached to it. (We now have four 3-valent vertices.) Join pairs of these 3-valent vertices with two edges each having slope 1. We now have a 4-valent infinite graph with two new 3-gons and a single 6-gon.  We can get an asymmetric infinite graph be carrying out this process in various unsymmetrical locations of the original tiling. By starting with irregularly spaced lines (rectangles rather than squares) the two new added edges will have very different directions.&lt;/p&gt;&#xA;" OwnerUserId="1369" LastActivityDate="2010-11-17T15:54:31.930" />
  <row Id="10705" PostTypeId="2" ParentId="10701" CreationDate="2010-11-17T15:59:19.440" Score="2" ViewCount="" Body="&lt;p&gt;Let $P$ be a prime ideal of $A$, and suppose that&#xA;$P=f^{-1}(I)$ where $I$ is some ideal of $B$.&#xA;Let $S=A-P$, a multiplicatively closed subset&#xA;of $A$, and $T=f(S)$. Then $T$ is a multiplicatively&#xA;closed subset of $B$ and is disjoint from $I$.&#xA;By Zorn's lemma, there is an ideal $Q$ of $B$&#xD;&#xA;maximal with respect to the conditions that $Q$&#xD;&#xA;contains $I$ and $Q$ is disjoint from $T$. By&#xA;a standard argument, $Q$ is prime. Moreover $f^{-1}(Q)=P$.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-11-17T15:59:19.440" CommentCount="1" />
  <row Id="10706" PostTypeId="2" ParentId="10181" CreationDate="2010-11-17T16:03:58.397" Score="0" ViewCount="" Body="&lt;p&gt;May I humbly suggest some educational freeware I have developed to teach formal logic and basic proof-writing skills. It is a downloadable, PC-based proof assistant (editor and logic-checker). Being a computer program, it is of necessity a purely formal system. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Included with the download is a tutorial that can be used as an introduction to formal mathematics. It includes ten worked examples of formal proofs, plus exercises with hints and full solutions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For testimonials, and a free full-function download, visit my website: &lt;a href=&quot;http://www.dcproof.com&quot; rel=&quot;nofollow&quot;&gt;http://www.dcproof.com&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3515" LastActivityDate="2010-11-17T16:03:58.397" CommentCount="2" />
  <row Id="10707" PostTypeId="2" ParentId="10700" CreationDate="2010-11-17T16:11:25.970" Score="1" ViewCount="" Body="&lt;p&gt;Your answer omits what $q$ and $q_0$ are. I presume&#xA;this is just a real linear combination of quaternions;&#xA;in your peculiar notation:&#xA;$$t[a\ \ (b\ \ c\ \ d)]+t&amp;#39;[a&amp;#39;\ \ (b&amp;#39;\ \ c&amp;#39;\ \ d&amp;#39;)]&#xD;&#xA;=[ta+t&amp;#39;a&amp;#39;\ \ (tb+t&amp;#39;b&amp;#39;\ \ tc+t&amp;#39;c&amp;#39;\ \ td+t&amp;#39;d&amp;#39;)].$$&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-11-17T16:11:25.970" CommentCount="1" />
  <row Id="10708" PostTypeId="2" ParentId="10233" CreationDate="2010-11-17T16:20:47.243" Score="2" ViewCount="" Body="&lt;p&gt;Davenport, in &quot;The Higher Arithmetic&quot; Section III.5 asserts that the law of quadratic reciprocity, in its original form as conjectured by Euler, was the following assertion:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $a$ be any natural number and $p,q$ any primes such that $p\equiv q$ mod $4a$ or $p\equiv -q$ mod $4a$. Then $a$ is a square mod $p$ if and only of $a$ is a square mod $q$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Davenport proves that  the usual quadratic reciprocity formula for odd primes is equivalent to this assertion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This suggests what to me is the most impressive of all applications of quadratic reciprocity: That the  prime divisors of the values of quadratic polynomials fall into residue classes. One is immediately led to speculate that, more generally, the prime divisors of the values of polynomials have an intelligible structure which is somehow related to residue classes. This is really most surprising, given that polynomials combine addition and multiplication arbitrarily.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As we know, to follow up this idea it was necessary to generalize the idea of ``residue class'' to rings of algebraic integers, giving birth to a big chunk of modern number theory along the way. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe to understand the importance of quadratic reciprocity, the main thing is not to accumulate a catalogue of specific applications, but to see quadratic reciprocity as a historical gateway to modern Number Theory. &lt;/p&gt;&#xA;" OwnerUserId="746" LastActivityDate="2010-11-17T16:20:47.243" />
  <row Id="10709" PostTypeId="2" ParentId="10693" CreationDate="2010-11-17T16:22:38.790" Score="1" ViewCount="" Body="&lt;p&gt;Here is the general notion of vague convergence, taken from Olav Kallenberg's &lt;em&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0387953132&quot; rel=&quot;nofollow&quot;&gt;Foundations of Modern Probability (2nd edition)&lt;/a&gt;&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $S$ be a locally compact, second countable Hausdorff space equipped with its Borel $\sigma$-field ${\cal S}$. Let $\hat{\cal S}$ be the ring of relatively compact Borel sets, and ${\cal M}(S)$ the space of locally finite non-negative measures.&#xA;Locally finite means that $\mu(B)&amp;lt;\infty$ when $B\in \hat{\cal S}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The vague topology on ${\cal M}(S)$ is the topology generated by the mappings &#xA;$\mu\mapsto \int f\ d\mu$ for every $f$ a non-negative continuous function with compact support.  &lt;/p&gt;&#xA;" OwnerUserId="940" LastActivityDate="2010-11-17T16:22:38.790" CommentCount="1" />
  <row Id="10710" PostTypeId="1" AcceptedAnswerId="10720" CreationDate="2010-11-17T16:22:50.537" Score="2" ViewCount="86" Body="&lt;p&gt;Given $1&amp;lt;q\leq2$ and $0\leq p\leq1$, let us consider the following function:&#xA;$$\phi\left(\alpha\right)=p\times\left|1-\alpha\right|^{q}+\left(1-p\right)\times\left|1+\alpha\right|^{q}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The minimum over $\mathbb{R}$ I have found is:&#xA;$$\alpha\left(p\right)=\frac{p^{\frac{1}{q-1}}-\left(1-p\right)^{\frac{1}{q-1}}}{p^{\frac{1}{q-1}}+\left(1-p\right)^{\frac{1}{q-1}}}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to show that: $\exists c&amp;gt;0,\exists\gamma\in\left]0;1\right],\forall p\in\left[0;1\right],\left|1-2p\right|\leq c\left(1-\phi\left(\alpha\left(p\right)\right)\right)^{\gamma}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I asked my question, I was mistaken: there was $\alpha\left(p\right)$ instead of $\phi\left(\alpha\left(p\right)\right)$ in the right hand side above. Actually, you just have to show the following to conclude:&#xA;$$\forall p\in\left[0;1\right],\left(1-2p\right)^{2} = 1-4p\left(1-p\right)\leq 1-\phi\left(\alpha\left(p\right)\right)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And it is simple indeed! :)&lt;/p&gt;&#xA;" OwnerUserId="2380" LastEditorUserId="2380" LastEditorDisplayName="" LastEditDate="2010-11-18T19:01:52.787" LastActivityDate="2010-11-18T19:01:52.787" Title="Inequality based on the minimum of a function" Tags="&lt;homework&gt;&lt;functions&gt;&lt;optimization&gt;&lt;inequality&gt;" AnswerCount="1" FavoriteCount="0" />
  <row Id="10711" PostTypeId="1" CreationDate="2010-11-17T16:23:54.433" Score="0" ViewCount="169" Body="&lt;p&gt;Prove that X is a Random Variable IFF sigma field generated by X is countably generated.&lt;/p&gt;&#xA;" LastActivityDate="2010-11-17T16:45:21.867" Title="countably generated sigma algebra" Tags="&lt;probability&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="10712" PostTypeId="2" ParentId="10696" CreationDate="2010-11-17T16:35:24.057" Score="2" ViewCount="" Body="&lt;p&gt;There is a well-known definition of cumulative distribution function $F$ for measures on finite dimensional Euclidean space ${\mathbb R}^d$. A function $F$ is the distribution function of a measure on ${\mathbb R}^d$ if and only if $F$ is right continuous and has non-negative increments, where these notions are defined appropriately for multidimensional space. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can find details about this in Chapter 3 of Olav Kallenberg's &lt;em&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0387953132&quot; rel=&quot;nofollow&quot;&gt;Foundations of Modern Probability (2nd edition)&lt;/a&gt;&lt;/em&gt;, for example.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not aware of useful extensions of the idea of distribution functions to other partially ordered spaces, but they may exist. &lt;/p&gt;&#xA;" OwnerUserId="940" LastActivityDate="2010-11-17T16:35:24.057" CommentCount="1" />
  <row Id="10713" PostTypeId="1" AcceptedAnswerId="10722" CreationDate="2010-11-17T16:45:18.310" Score="1" ViewCount="75" Body="&lt;p&gt;Find modules $M,N$ and a ring $A$ s.t. $M\otimes_A N\not \cong M\otimes_{\mathbb{Z}}N$.&lt;/p&gt;&#xA;" OwnerUserId="2856" LastActivityDate="2010-11-17T18:28:10.057" Title="Non-isomorphic tensorproducts" Tags="&lt;modules&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="10714" PostTypeId="2" ParentId="10711" CreationDate="2010-11-17T16:45:21.867" Score="1" ViewCount="" Body="&lt;p&gt;&quot;If and only if&quot; doesn't make sense here, but you should be able to prove that for any map $X:\Omega\to {\mathbb R}$ the $\sigma$-field generated by $X$, that is, $X^{-1}({\cal B}(\mathbb R))$,  is countably generated. Hint: The $\sigma$-field of Borel sets of $\mathbb R$ is countably generated.    &lt;/p&gt;&#xA;" OwnerUserId="940" LastActivityDate="2010-11-17T16:45:21.867" CommentCount="1" />
  <row Id="10715" PostTypeId="2" ParentId="10713" CreationDate="2010-11-17T16:52:08.380" Score="1" ViewCount="" Body="&lt;p&gt;Let $A=\mathbb Z[X]$ the polynomial ring, let $M=\mathbb Z[X,X^{-1}]$ be the ring of Laurent polynomials views as an $A$ module, and let $N=A/(X)$. As $\mathbb Z$-modules, both $M$ and $N$ are free $\mathbb Z$-modules. You can easily check that $M\otimes_AN=0$ yet $M\otimes_\mathbb ZN\neq0$.&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2010-11-17T16:52:08.380" CommentCount="1" />
  <row Id="10716" PostTypeId="2" ParentId="10233" CreationDate="2010-11-17T17:08:07.363" Score="3" ViewCount="" Body="&lt;p&gt;I guess this might as well be a separate answer.  You can use quadratic reciprocity to give elementary proofs of certain special cases of &lt;a href=&quot;http://en.wikipedia.org/wiki/Dirichlets_theorem_on_arithmetic_progressions&quot; rel=&quot;nofollow&quot;&gt;Dirichlet's theorem&lt;/a&gt;.  First, you should be aware of the following nice result and its &quot;Euclidean&quot; proof.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Lemma:&lt;/strong&gt;  Let $f(x) \in \mathbb{Z}[x]$ and let $P_f$ be the set of primes $p$ such that $p | f(n)$ for some $n$.  Then $P_f$ is infinite.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Proof.&lt;/em&gt;  If $f(0) = 0$ then this is obvious, so suppose otherwise.  Let $p_1, p_2, ... p_n$ be a finite set of primes in $P_f$.  Then for any $k$, $\frac{1}{f(0)} f(k f(0) p_1 ... p_n)$ must be divisible by a prime which is not one of the $p_i$, and choosing $k$ sufficiently large we can find a new prime $p_{n+1}$ in $P_f$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An alternate proof is given &lt;a href=&quot;http://qchu.wordpress.com/2009/09/02/some-remarks-on-the-infinitude-of-primes/&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.  Using this lemma and properties of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Cyclotomic_polynomial#Applications&quot; rel=&quot;nofollow&quot;&gt;cyclotomic polynomials&lt;/a&gt;, you can prove that there exist infinitely many primes congruent to $1 \bmod n$ for any $n$ without any heavy machinery, so I will skip these cases.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using quadratic reciprocity, you can prove that the following arithmetic progressions also contain infinitely many primes:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;$11 \bmod 12$:  Let $f(x) = 3x^2 - 1$.  Then $p | f(n)$ implies $\left( \frac{3}{p} \right) = 1$.  However, we can modify the proof of the lemma to prove that infinitely many of the primes dividing $f$ must be congruent to $3 \bmod 4$.  To see this, let $p_1, .. p_n$ be finitely many primes with this property and consider $f(2 p_1 ... p_n) \equiv 3 \bmod 4$.  It follows that infinitely many primes $p$ satisfy $\left( \frac{3}{p} \right) = 1$ and $p \equiv 3 \bmod 4$, so by quadratic reciprocity $\left( \frac{p}{3} \right) = -1$, so $p \equiv 2 \bmod 3$.  Hence $p \equiv 11 \bmod 12$.  In particular, there are infinitely many primes congruent to $2 \bmod 3$ and infinitely many primes congruent to $3 \bmod 4$.  &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$4 \bmod 5$:  Let $f(x) = x^2 - 5$.  Then $p | f(n)$ implies $\left( \frac{5}{p} \right) = 1$.  Again, we can modify the proof of the lemma to prove that infinitely many of these primes are not congruent to $1 \bmod 5$.  To see this, let $p_1, ... p_n$ be finitely many primes with this property, none of which are equal to $5$, and consider either $f(p_1 ... p_n)$ or $f(2 p_1 ... p_n)$, one of which is not congruent to $1 \bmod 5$ and which therefore has a prime factor which is not congruent to $1 \bmod 5$.  So infinitely many primes $p$ satisfy $\left( \frac{5}{p} \right) = 1$ and $p \not \equiv 1 \bmod 5$.  By quadratic reciprocity this gives $\left( \frac{p}{5} \right) = 1$, hence $p \equiv 4 \bmod 5$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$3 \bmod 8$:  Let $f(x) = x^2 + 2$.  Then $p | f(n)$ implies $\left( \frac{-2}{p} \right) = 1$.  Again, we can modify the proof of the lemma to prove that infinitely many of these primes are not congruent to $1 \bmod 8$.  To see this, let $p_1, ... p_n$ be finitely many primes with this property, all of which are odd, and consider $f(2p_1 ... p_n) \equiv 6 \bmod 8$.  So infinitely many primes $p$ satisfy $\left( \frac{-2}{p} \right) = 1$ and $p \not \equiv 1 \bmod 8$.  By quadratic reciprocity this gives $p \equiv 3 \bmod 8$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;And so on.  For what progressions is it possible to give these kinds of proofs?  It turns out this is possible for the progression $a \bmod n$ if and only if $a^2 \equiv 1 \bmod n$.  For details, see Keith Conrad's &lt;a href=&quot;http://www.math.uconn.edu/~kconrad/blurbs/gradnumthy/dirichleteuclid.pdf&quot; rel=&quot;nofollow&quot;&gt;Euclidean proofs of Dirichlet's theorem&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More generally, quadratic reciprocity is the key to writing down the &lt;a href=&quot;http://en.wikipedia.org/wiki/Dedekind_zeta_function&quot; rel=&quot;nofollow&quot;&gt;Dedekind zeta functions&lt;/a&gt; of quadratic number fields explicitly, and trying to generalize this leads you into class field theory and so forth.&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-11-17T17:13:40.803" LastActivityDate="2010-11-17T17:13:40.803" />
  <row Id="10717" PostTypeId="1" AcceptedAnswerId="10718" CreationDate="2010-11-17T17:16:02.907" Score="2" ViewCount="175" Body="&lt;p&gt;This is a problem in baby Rudin. Chapter 4, problem 15. I'm just looking at examples at the moment. It says that any continuous open map is monotone. Recall, an open map f: X -&gt; Y is such that any open set U in X will haves its image, f(U) open in Y.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So,&#xA;Let $$ f: \mathbb{R} \to [-1, 1] $$ take $$ x \mapsto sin(x) $$ and consider the open interval $$ (\frac{\pi}{3}, \frac{7\pi}{12}) $$ The image of this interval in [-1, 1] is $$ (\frac{\sqrt{3}}{2}, 1] $$ which is open in [-1, 1] (its complement is [-1, sqrt(3)/2], certainly closed). The problem is, the function is not monotonic on this interval. It is continuous, so it must not be an open map. I'm pretty sure I have the image right, so, where am I making a mistake?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I really hope I'm not making a dumb error!&lt;/p&gt;&#xA;" OwnerUserId="2465" LastEditorUserId="2465" LastEditorDisplayName="" LastEditDate="2010-11-17T17:36:49.807" LastActivityDate="2010-11-17T18:13:07.397" Title="continuous open mappings are monotonic" Tags="&lt;real-analysis&gt;&lt;homework&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="10718" PostTypeId="2" ParentId="10717" CreationDate="2010-11-17T17:27:47.293" Score="1" ViewCount="" Body="&lt;p&gt;The mistake is misinterpreting for which codomain the statement holds.  Continuous open maps from $\mathbb{R}$ to $\mathbb{R}$ are monotone.  You have changed the codomain to the image with subspace topology, and this changes the meaning of open map.  Since the interval $\left(\frac{\sqrt{3}}{2},1\right]$ is not open in $\mathbb{R}$, your example demonstrates the fact that this non-monotone continuous map cannot be an open map.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Athough in this case you have a map that is not open, note that if you are trying to show that a map &lt;em&gt;is&lt;/em&gt; open, it is not enough to give an example of an open set that is mapped to an open set.  You would have to prove that &lt;em&gt;every&lt;/em&gt; open set is mapped to an open set.  On the other hand, to show that a map is not open, you only need to demonstrate that there exists an open set that is not mapped to an open set, and this fact will come in handy if you approach this problem using contraposition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A side remark: You will at some point have to invoke completeness of $\mathbb{R}$ (or a theorem that uses completeness).  The function $f(x)=x^3-6x$ as a map from $\mathbb{Q}$ to $\mathbb{Q}$ is continuous and open but not monotone.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2010-11-17T18:13:07.397" LastActivityDate="2010-11-17T18:13:07.397" CommentCount="2" />
  <row Id="10719" PostTypeId="2" ParentId="10233" CreationDate="2010-11-17T18:06:37.897" Score="4" ViewCount="" Body="&lt;p&gt;The quadratic reciprocity law in any of its forms shows that there is an un-obvious correlation between different primes.  The $(p,q)$ symbol constrains the $(q,p)$ symbol.   This is astonishing compared to other more &quot;linear&quot; theorems about congruences or unique factorization.  In its 20th-century reformulations quadratic reciprocity is seen as an avatar of other reciprocity laws in geometry (reciprocity for tame symbols) and even geometric topology (linking numbers where knots play the role of primes) and although these other theorems are in some ways easier to prove, the analogies between all of them are mysterious.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically, if you are not shocked by this theorem, you don't completely understand it.   Historically it was a hard-won, prize result and not an inevitable universal discovery like the Pythagorean formula or other theorems that were difficult in their time but found independently in many times and places.  Many cultures had knowledge of basic number theoretic facts but quadratic reciprocity is one of the first signs of number theory as a science.&lt;/p&gt;&#xA;" OwnerUserId="467" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-11-18T18:27:13.693" LastActivityDate="2010-11-18T18:27:13.693" CommentCount="2" />
  <row Id="10720" PostTypeId="2" ParentId="10710" CreationDate="2010-11-17T18:07:15.823" Score="2" ViewCount="" Body="&lt;p&gt;If $\alpha$ is large (of either sign), $\phi$ increases without bound essentially as $\alpha ^q$.  Is $\alpha \in (-1,1)$ in which case we can delete the absolute value bars?&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-11-17T18:07:15.823" CommentCount="3" />
  <row Id="10721" PostTypeId="2" ParentId="10697" CreationDate="2010-11-17T18:25:47.110" Score="8" ViewCount="" Body="&lt;p&gt;Perhaps more meta than you wanted since it's not about how you perceive mathematics coming from mathematics, but rather how the rest of the world views it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's that whole misconception that if you're good at math you're good at arithmetic. You know... that thing where someone says something like &quot;Hey, you're good at math! Can you do this calculation (typically menial multiplication) for me in your head?&quot; Sure I &lt;em&gt;could&lt;/em&gt;, but that's because I can keep a lot of stuff in my head at once and make sure I do it correctly. It's not as if that says anything about mathematical ability.&lt;/p&gt;&#xA;" OwnerUserId="1269" LastActivityDate="2010-11-17T18:25:47.110" CommunityOwnedDate="2010-11-18T17:24:01.440" CommentCount="2" />
  <row Id="10722" PostTypeId="2" ParentId="10713" CreationDate="2010-11-17T18:28:10.057" Score="3" ViewCount="" Body="&lt;p&gt;Well, $A\otimes_A A$ is certainly isomorphic to $A$. So all we need&#xA;is a ring for which $A\otimes_{\mathbb{Z}} A$ isn't isomorphic to $A$.&#xA;An obvious choice is a ring like the Gaussian integers $A=\mathbb{Z}[i]$&#xD;&#xA;which is free of rank $2$ as a $\mathbb{Z}$-module, for then&#xA;$A\otimes_{\mathbb{Z}} A$ is free of rank $4$ as a $\mathbb{Z}$-module.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-11-17T18:28:10.057" />
  <row Id="10723" PostTypeId="2" ParentId="10699" CreationDate="2010-11-17T18:45:30.613" Score="1" ViewCount="" Body="&lt;p&gt;Zaricuse has already given the correct answer, but I want to add a few remarks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even if you don't assume pointwise convergence, the hypothesis $\lim \int |f_{n} - f| d\mu =0$ implies that $(f_n)$ converges to $f$ in measure, which implies that there is a subsequence that converges to $f$ pointwise almost everywhere.  But as you have shown, this is not needed in the proof.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you consider the normed space $L^1(\mu)$ of absolutely $\mu$ integrable functions with norm $\|f\|=\int|f|d\mu$, what you have shown is that the norm is continuous as a function from $L^1(\mu)$ with the norm topology to $[0,\infty)$.  The same is true in any normed space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As Qiaochu mentioned in a comment, pointwise convergence (almost everywhere) allows you to prove the converse, which follows from a generalized form of the dominated convergence theorem.  &lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-11-17T18:45:30.613" CommentCount="4" />
  <row Id="10724" PostTypeId="2" ParentId="10697" CreationDate="2010-11-17T18:47:56.960" Score="15" ViewCount="" Body="&lt;p&gt;Here are a couple of misconceptions that people who don't know about mathematics commonly hold. Firstly, that it's immutable, fixed, unchanging; surely it's all been discovered? It's a belief that even a highly educated non-mathematician might hold.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This point was illustrated to me not long ago when a friend, who speaks at least 10 languages fluently, asked why I found mathematics interesting. She believed that it was cold and lifeless and thus inherently less appealing than language, which she viewed as living and dynamic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried to persuade her that mathematics grows and changes too. Areas such as Group Theory  were born from ideas used in Number Theory and Geometry. And old ways of doing things can be surpassed. The history of cryptography is a good example.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The conversation also touched on a second misconception: that mathematicians simply follow set rules. I explained that imagination was very important in doing mathematics. And when I said that  there was great beauty in mathematics, I was faced with almost total incomprehension. Surely, mathematics was the very antithesis of beauty?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're reading this I don't need to convince you that these are indeed misconceptions, but unfortunately I don't think I convinced my friend!&lt;/p&gt;&#xA;" OwnerUserId="1301" LastActivityDate="2010-11-17T18:47:56.960" CommunityOwnedDate="2010-11-17T18:47:56.960" CommentCount="2" />
  <row Id="10725" PostTypeId="2" ParentId="10702" CreationDate="2010-11-17T18:53:17.450" Score="6" ViewCount="" Body="&lt;p&gt;The problem seems to be missing some assumptions, as noted by Americo.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, If $B = 0$ and $A=C$ are acute angles, such that $\sin A  = 1/4$ the we have that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sin(A+B) + \sin(B+C) + \cos(A-C) = 3/2$, but none of&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sin A + \cos B + \sin C$ or $\cos A + \sin B + \cos C$ are $0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In any case, this looks like a perfect problem for using complex numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $B&amp;#39; = \pi/2 - B$ and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$z_1 = \cos A + i \sin A$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$z_2 = \cos B&amp;#39; + i \sin B&amp;#39;$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$z_3 = \cos C + i \sin C$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The given identity is $\cos (A-B&amp;#39;) + \cos (C - B&amp;#39;) + \cos (A-C) = 3/2$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i.e.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{z_1}{z_2} +\frac{z_2}{z_1} +\frac{z_3}{z_2} +\frac{z_2}{z_3} +\frac{z_1}{z_3} +\frac{z_3}{z_1}  = 3$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The two identities&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;$\sin A + \cos B + \sin C = 0$&#xD;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$\cos A + \sin B +\cos C = 0$&#xD;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;are equivalent to showing that $z_1 + z_2 + z_3 = 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Eq 1, says that the imaginary part of $z_1 + z_2 + z_3$ is $0$ and Eq 2 says that the real part is $0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope that helps.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-11-17T18:53:17.450" CommentCount="5" />
  <row Id="10726" PostTypeId="1" AcceptedAnswerId="10769" CreationDate="2010-11-17T18:53:36.220" Score="2" ViewCount="108" Body="&lt;p&gt;Consider the following arithmetic function, $f(a,b) = (a+1, \lceil a- \frac{a}{b} \rceil , \lceil a- \frac{2a}{b} \rceil, \dots, \lceil \frac{a}{b} \rceil, 1)$, where all terms except the first and last are computed by the application of a ceiling function. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For any pair of positive integers $(a,b)$, the vector $f(a,b)$ consists of the terms of an integer partition of length $a+1$. Moreover, $f(a,b)$ and $f(b,a)$ represent dual partitions and have reflected Ferrers Diagrams. This can be proved by recalling the identity&#xA;\begin{eqnarray}&#xA;\sum_{j = 0}^{b-1} \left \lceil \tfrac{a(b-j)}{b} \right \rceil = \sum_{j = 0}^{a-1} \left \lceil \tfrac{b(a-j)}{a} \right \rceil.&#xA;\end{eqnarray}&#xA;(Going in the other direction, one can use the methods outlined here to give a neat combinatorial proof of this ceiling summation identity.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, these integer partitions contain no $0$s, contain at least one $1$ and at most $a$ $1$s.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question&lt;/strong&gt;: Are these partitions a special (read &lt;em&gt;named&lt;/em&gt;) subset of the standard partitions?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Problem 1&lt;/strong&gt;: Determine all pairs $(a,b)$ such that $f(a,b)$ are the terms of an integer partition of $k$ for a fixed $k &amp;gt; 1$. Determine a simple way to find $k$ from $a$ and $b$ (and not by simply adding the terms of the defining relation).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Problem 2&lt;/strong&gt;: Find other symmetries like $f(a,b) \leftrightarrow f(b,a)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(&lt;em&gt;Motivation&lt;/em&gt;) For example, $f(2,3) = (4,2,1)$, $f(3,2) = (3,2,1,1)$, $f(1,5) = (6,1)$ and $f(5,1) = (2,1,1,1,1,1)$, which are &lt;em&gt;dual&lt;/em&gt; integer partitions of $7$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Problem 3&lt;/strong&gt;: Give an interpretation to the sequence {$f(a,b)$} in terms of $a$ and $b$ for fixed $a$ or $b$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(&lt;em&gt;Motivation&lt;/em&gt;) The sequence {$f(1,b)$} consists of terms of the form $(2,1, \dots, 1)$ with $b + 1$ $1$s, which are integer partitions of $b+2$; {$f(2,b)$} consists of terms of the form $(3, 1), (3,1,1), (3,2,1)$ or $(3, 2, \dots, 2, 1, \dots, 1)$ which are each integer partitions of those integers greater than 3 not divisible by $3$; {$f(3,b)$} consists of special integer partitions of integers congruent to {$1,2,5$} mod $6$. Similarly, {$f(4,b)$} consists of integer partitions of integers congruent to {$1, 2, 3$} mod $5$ or {$1,2,6,8$} mod $10$; {$f(5,b)$} consists of integer partitions of integers congruent to {$1, 2, 4$} mod $6$, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is some Mathematica code that might be useful:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;f[a_, b_] := &#xA;  Flatten[{a + 1, Table[Ceiling[a - j a/b], {j, 1, b - 1}], 1}];&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a = 2;&#xA;Table[Total@f[a, b], {b, 1, 100}]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="2720" LastEditorUserId="2720" LastEditorDisplayName="" LastEditDate="2010-11-17T22:30:56.940" LastActivityDate="2010-11-17T23:03:33.783" Title="A Vector Arithmetic Function, Special Integer Partitions and Residue Sequences" Tags="&lt;number-theory&gt;&lt;combinatorics&gt;" AnswerCount="1" FavoriteCount="2" />
  <row Id="10727" PostTypeId="1" AcceptedAnswerId="10730" CreationDate="2010-11-17T19:01:17.643" Score="0" ViewCount="84" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This question is based on the diagram&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/Ko6HT.gif&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;taken from this &lt;a href=&quot;http://www.ies.co.jp/math/products/trig/applets/yogen_auto/yogen_auto.html&quot; rel=&quot;nofollow&quot;&gt;link&lt;/a&gt;. I don't understand why the areas of A6 and A5 adds up to $2bc\cos(A)$.&lt;/p&gt;&#xA;" OwnerUserId="2973" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-17T19:23:32.030" LastActivityDate="2010-11-17T19:24:50.437" Title="Law of Cosines Proof" Tags="&lt;trigonometry&gt;" AnswerCount="3" CommentCount="3" />
  <row Id="10728" PostTypeId="2" ParentId="10727" CreationDate="2010-11-17T19:18:19.383" Score="1" ViewCount="" Body="&lt;p&gt;The other side of A6 has length $c \cos A$ and that of A5 length $b \cos A.$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So their total area is $b \times c \cos A + c \times b \cos A = 2bc \cos A.$&lt;/p&gt;&#xA;" OwnerUserId="1301" LastActivityDate="2010-11-17T19:18:19.383" />
  <row Id="10729" PostTypeId="2" ParentId="10727" CreationDate="2010-11-17T19:24:44.680" Score="0" ViewCount="" Body="&lt;p&gt;First note that A5 = A6. (The visual explains it using the fact that parallelogram area between the parallel lines are the same). And note that A5 is nothing but the side of the square times the length along the side $AB$. The side of the square is $c$. The length along the side $AB$ is $b \cos(A)$. Hence $A5 = c b \cos(A)$. Hence $A5 + A6 = 2bc \cos(A)$.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastActivityDate="2010-11-17T19:24:44.680" />
  <row Id="10730" PostTypeId="2" ParentId="10727" CreationDate="2010-11-17T19:24:50.437" Score="0" ViewCount="" Body="&lt;p&gt;If you look at A5. It is a rectangle with side $b$ and, say, $x$.&#xA;Now $\cos A= x/c$, so $x=c\cos A$. Hence the area of A5 is $bx=bc\cos A$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A similar argument tells you that the area of A6 is $cb\cos A$.&lt;/p&gt;&#xA;" OwnerUserId="1154" LastActivityDate="2010-11-17T19:24:50.437" />
  <row Id="10731" PostTypeId="1" AcceptedAnswerId="10805" CreationDate="2010-11-17T19:34:56.667" Score="2" ViewCount="90" Body="&lt;p&gt;Let $K \subset R^n$ be compact, and let $C_0^{\infty}(K)$ be the space of smooth functions on with support in $K$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $p \in [1,\infty), \alpha$ multiindex, let $|f|_{\alpha,p} = (\int_K |D^{\alpha}f|^p)^{\frac{1}{p}}$, and $|f|_{\alpha,\infty} = sup_K |D^{\alpha}f|$ - as usual.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $p \in [1,\infty]$, let $\tau_p$ be the locally convex topology generated on $C_0^{\infty}(K)$ by the seminorms $(|f|_{\alpha,p})_{\alpha}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Claim:&lt;/strong&gt; All the $\tau_p$ are the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One proof of this might utilize the Sobolev theory (Kondrashow embedding theorem) which is too heavy as tool for my taste.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, I don't know a more elementary proof. Furthermore I guess a beautiful proof of this might work similarly for other spaces like the Schwartz-space or the space of smooth functions on the whole of $R^n$. Do you have any good idea?&lt;/p&gt;&#xA;" OwnerUserId="3557" LastActivityDate="2010-11-18T19:54:53.957" Title="p-seminorms on smooth functions are equivalent" Tags="&lt;analysis&gt;&lt;functional-analysis&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="10733" PostTypeId="2" ParentId="10648" CreationDate="2010-11-17T19:48:49.283" Score="0" ViewCount="" Body="&lt;p&gt;The centres of three circles of radius r just touching one another will form an equilateral triangle of side length 2r. Exactly 6 such non-overlapping equilateral triangles will be formed in this way as you keep adding more circles around the edge of the central circle. Why 6? These equilateral triangles will have a common vertex at the centre, each forming a 60 degree angle there. And as, you pointed out, &quot;there are 6*60=360 degrees in a circle.&quot;&lt;/p&gt;&#xA;" OwnerUserId="3515" LastEditorUserId="3515" LastEditorDisplayName="" LastEditDate="2010-11-17T19:56:23.347" LastActivityDate="2010-11-17T19:56:23.347" CommentCount="7" />
  <row Id="10734" PostTypeId="2" ParentId="10697" CreationDate="2010-11-17T19:59:16.743" Score="1" ViewCount="" Body="&lt;p&gt;I've noticed it's quite common among algebraists dealing with finite structures (like finite groups) to think of mathematics (limited to their area, of course) as the absolute God given truth, harmonious, non-contradictory, sort of Platonic ideal world.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This point of view can be possible accepted, as long we limit ourselves to finite mathematics, but what happens when we introduce the notion of infinity? The whole harmonious God given world goes out the window and the mathematics becomes what it really is: a kind of human activity which reflects both the power of our intelligence as well as its limitations. &lt;/p&gt;&#xA;" OwnerUserId="2450" LastActivityDate="2010-11-17T19:59:16.743" CommunityOwnedDate="2010-11-18T17:24:01.440" CommentCount="1" />
  <row Id="10735" PostTypeId="1" AcceptedAnswerId="10742" CreationDate="2010-11-17T20:11:16.077" Score="0" ViewCount="150" Body="&lt;p&gt;I have to do this for an upcoming exam, but cannot find anywhere (in the textbook or online) how to do this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I only really need to know a couple points to plot it... when x = 0, and then the earliest crossing of the x-axis.&lt;/p&gt;&#xA;" OwnerUserId="3325" LastActivityDate="2010-11-17T20:40:07.083" Title="How to sketch a sinc function by hand?" Tags="&lt;calculus&gt;&lt;fourier-analysis&gt;&lt;signal-processing&gt;" AnswerCount="3" />
  <row Id="10736" PostTypeId="2" ParentId="10735" CreationDate="2010-11-17T20:14:08.663" Score="0" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://www.wolframalpha.com/input/?i=plot+sinc%28x%29&quot; rel=&quot;nofollow&quot;&gt;http://www.wolframalpha.com/input/?i=plot+sinc%28x%29&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-17T20:14:08.663" CommentCount="1" />
  <row Id="10738" PostTypeId="2" ParentId="10697" CreationDate="2010-11-17T20:24:03.320" Score="4" ViewCount="" Body="&lt;p&gt;Following up on Kahen's comment, most people who only take math through High School think math is about calculation. For higher math, description is more correct than calculation, but that still doesn't capture the flavor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most people think math is about numbers. Some is, some isn't.&lt;/p&gt;&#xA;" OwnerUserId="1312" LastActivityDate="2010-11-17T20:24:03.320" CommunityOwnedDate="2010-11-18T17:24:01.440" />
  <row Id="10740" PostTypeId="2" ParentId="10735" CreationDate="2010-11-17T20:32:40.027" Score="1" ViewCount="" Body="&lt;p&gt;I think its helpful also to graph it with $\frac{1}{x}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.wolframalpha.com/input/?i=plot+sinc%28x%29%2C+1%2Fx&quot; rel=&quot;nofollow&quot;&gt;http://www.wolframalpha.com/input/?i=plot+sinc%28x%29%2C+1%2Fx&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you write the function as $f(x) = \frac{\sin(x)}{x}$ for $x \not=0$ and $1$ if $x=0$  you can see that it will be zero at exactly when $\sin(x)$ is zero (except at $0$ where it is $1$). &lt;/p&gt;&#xA;" OwnerUserId="2596" LastActivityDate="2010-11-17T20:32:40.027" />
  <row Id="10741" PostTypeId="1" CreationDate="2010-11-17T20:32:54.007" Score="14" ViewCount="1113" Body="&lt;p&gt;In answering a question I mentioned the Asteroids video game as an example -- at one time, the canonical example -- of a locally flat geometry that is globally different from the Euclidean plane.  It might be out of date in 2010.  This raises its own question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;are there other real-life examples of geometries formed by identifications?  We know the cylinder and Moebius strip and there are probably some interesting equivalents of those.  Origami are coverings of a punctured torus and I heard that there are crocheted examples of complicated 2-d and 3-d objects.  &lt;em&gt;Are there simple, Asteroids-like conversational examples for flat surfaces formed as quotients?&lt;/em&gt;  Punctures and orbifold points and higher dimensional examples all would be interesting, but I am looking less for mathematically sophisticated than conversationally relevant examples, such as a famous game or gadget that makes a cellphone function as a torus instead of a rectangle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(edit:  Pac-Man, board games such as Chutes and Ladders, or any game with magic portals that transport you between different locations, all illustrate identification of points or pieces of the space, but they lead to non-homogeneous geometries.  The nice thing about Asteroids was that it was clearly the whole uniform geometry of the torus.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;edit-2: the Flying Toasters screen-saver would have been an example of what I mean, except that video of it exists online and shows it to be a square window onto motion in the ordinary Euclidean plane.)&lt;/p&gt;&#xA;" OwnerUserId="467" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-11-18T03:39:51.813" LastActivityDate="2010-11-18T05:10:31.957" Title="Quotient geometries known in popular culture, such as &quot;flat torus = Asteroids video game&quot;." Tags="&lt;geometry&gt;&lt;topology&gt;&lt;teaching&gt;" AnswerCount="5" CommentCount="8" FavoriteCount="6" />
  <row Id="10742" PostTypeId="2" ParentId="10735" CreationDate="2010-11-17T20:40:07.083" Score="1" ViewCount="" Body="&lt;p&gt;You know the value at zero, it goes to zero at $x = n \pi$ and in-between reaches the maximum and minimum at close to $x = (n+ \frac{1}{2}) \pi$ (when $n&amp;gt;0$ or $n&amp;lt;-1$) with the value at these points being $\frac{(-1)^n}{(n+ \frac{1}{2}) \pi}$. So knowing these you should be able to roughly sketch the function.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastActivityDate="2010-11-17T20:40:07.083" />
  <row Id="10743" PostTypeId="1" AcceptedAnswerId="10751" CreationDate="2010-11-17T20:47:26.387" Score="1" ViewCount="192" Body="&lt;p&gt;If I have a value of 25%, and I want the value as it would be on the opposite side of the halfway point (75% in this case), what is a formula that can calculate this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know the appropriate tags for this so feel free to adjust them.  However, it is a programming-related question.&lt;/p&gt;&#xA;" OwnerUserId="3636" LastEditorUserId="1543" LastEditorDisplayName="" LastEditDate="2011-05-03T10:12:17.327" LastActivityDate="2011-05-03T10:12:17.327" Title="How to reverse percentage?" Tags="&lt;arithmetic&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="1" />
  <row Id="10744" PostTypeId="1" AcceptedAnswerId="10756" CreationDate="2010-11-17T20:47:52.190" Score="1" ViewCount="110" Body="&lt;p&gt;After asking a few questions earlier, I think I've been able to describe my main issue.  Could someone show me why this identity is true:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\partial \textbf{x}^{T} B \textbf{x}}{\partial \textbf{x}} = (B + B^{T})\textbf{x}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've gone through things like Matrixcookbook which seem to just show this identity without any derivation of it.  I have tried to unsuccessfully derive the formula below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Via product rule:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{\partial \textbf{x}^{T} B \textbf{x}}{\partial \textbf{x}} = \frac{\partial \textbf{x}^{T}}{\partial \textbf{x}}B\textbf{x} + \textbf{x}^{T}\frac{\partial B \textbf{x}}{\partial \textbf{x}}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I realize product rule for scalars is not the same for matrices, but it still holds as long as order is preserved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At this point, I'm stuck because I'm not sure if $\frac{\partial \textbf{x}^{T}}{\partial \textbf{x}}$ is identity $I$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I assume this definition of the Jacobian:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\begin{bmatrix} \dfrac{\partial y_1}{\partial x_1} &amp;amp;; \cdots &amp;amp;; \dfrac{\partial y_1}{\partial x_n} \\ \vdots &amp;amp;; \ddots &amp;amp;; \vdots \\ \dfrac{\partial y_m}{\partial x_1} &amp;amp;; \cdots &amp;amp;; \dfrac{\partial y_m}{\partial x_n}  \end{bmatrix}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And if I write $\frac{\partial B \textbf{x}}{\partial \textbf{x}}$ component by component with the definition above, it becomes simply $B$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{\partial \textbf{x}^{T} B \textbf{x}}{\partial \textbf{x}} = B\textbf{x} + \textbf{x}^{T}B$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is not correct, the left side is a nx1 vector and the right side is a nx1 vector.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure where my faulty assumption is.  I've gone through various sources which don't explain the situation I showed above.  Am I missing some very simple point that those books assume?&lt;/p&gt;&#xA;" LastActivityDate="2010-11-17T21:25:17.760" Title="Correct method to differentiate a first order vector/matrix" Tags="&lt;linear-algebra&gt;&lt;vector-calculus&gt;" AnswerCount="3" />
  <row Id="10745" PostTypeId="1" CreationDate="2010-11-17T20:48:21.847" Score="1" ViewCount="128" Body="&lt;p&gt;Hey,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Problem: Calculate limit of $\frac{1}{\sqrt{n+1}} + \frac{1}{\sqrt{n+2}} + \frac{1}{\sqrt{n+3}} + \cdots + \frac{1}{\sqrt{n+n^2}}$ as $n$ approaches infinity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solution: Denote the above some as $X$, then we can bound it:&#xA;$$ \infty\longleftarrow\frac{1}{\sqrt{n+n^2}} \lt X \lt \frac{n^2}{\sqrt{n+1}} \lt \frac{n^2}{\sqrt{n}} = \sqrt{\frac{n^4}{n}}\longrightarrow \infty.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, from the Squeeze Principle, $\lim X = \infty$. Am I doing the right thing?&lt;/p&gt;&#xA;" OwnerUserId="3637" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-17T21:41:43.823" LastActivityDate="2010-11-17T21:41:43.823" Title="Question about Squeeze Principle" Tags="&lt;calculus&gt;" AnswerCount="5" CommentCount="1" FavoriteCount="1" />
  <row Id="10746" PostTypeId="2" ParentId="10741" CreationDate="2010-11-17T20:49:56.133" Score="4" ViewCount="" Body="&lt;p&gt;In (some of) the &lt;a href=&quot;http://en.wikipedia.org/wiki/Civilization_%28series%29&quot; rel=&quot;nofollow&quot;&gt;Civilization games&lt;/a&gt; you can decide whether you want to play on a (bounded) plane, or a cylinder, or even a torus, I think.&lt;/p&gt;&#xA;" OwnerUserId="367" LastActivityDate="2010-11-17T20:49:56.133" />
  <row Id="10747" PostTypeId="2" ParentId="10745" CreationDate="2010-11-17T20:54:05.227" Score="0" ViewCount="" Body="&lt;p&gt;The term $\frac{1}{\sqrt{n+n^2}}$ goes to zero as $n \rightarrow \infty$, so that side doesn't work.  You should think about how many terms you are adding.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2010-11-17T21:02:55.007" LastActivityDate="2010-11-17T21:02:55.007" CommentCount="2" />
  <row Id="10748" PostTypeId="2" ParentId="10743" CreationDate="2010-11-17T20:54:21.870" Score="2" ViewCount="" Body="&lt;p&gt;Unless I completely missed the point of the question, the answer is rather trivially $100\% - p$.&lt;/p&gt;&#xA;" OwnerUserId="11" LastActivityDate="2010-11-17T20:54:21.870" />
  <row Id="10749" PostTypeId="2" ParentId="5420" CreationDate="2010-11-17T20:57:10.527" Score="1" ViewCount="" Body="&lt;p&gt;There is an &lt;a href=&quot;http://en.wikipedia.org/wiki/Binary_logarithm#Real_number&quot; rel=&quot;nofollow&quot;&gt;algorithm&lt;/a&gt; for computing $\log_2(x)$ that might suit you.&#xA;Combine that with the spigot algorithm for $e$, and you can get $\ln(x)$.&#xA;From there, you can use Newton-Raphson to get $\exp(x)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know if this roundabout way ends up doing any better than just recomputing.&lt;/p&gt;&#xA;" OwnerUserId="3638" LastActivityDate="2010-11-17T20:57:10.527" />
  <row Id="10750" PostTypeId="2" ParentId="10743" CreationDate="2010-11-17T20:59:14.890" Score="1" ViewCount="" Body="&lt;p&gt;Never mind, it's:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(100 - 25)&#xA;or&#xA;(100 - 34.04)&lt;/p&gt;&#xA;" OwnerUserId="3636" LastActivityDate="2010-11-17T20:59:14.890" CommentCount="2" />
  <row Id="10751" PostTypeId="2" ParentId="10743" CreationDate="2010-11-17T21:00:50.880" Score="3" ViewCount="" Body="&lt;p&gt;It's unclear to me what it is you want. One possible interpretation is that you know how much 25\% is, and you want to know how much 75\% is. For this situation it is very easy: if $c$ is 25% of some unknown quantity $y$, and you want to know how much 75% of $y$ is, it is $3c$. For instance, if you know that $27$ is 25% of something, then $3(27)=81$ is 75% of that same something.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Cross-multiplication#Rule_of_Three&quot; rel=&quot;nofollow&quot;&gt;Rule of Three&lt;/a&gt; will let you figure out how much $q$% is if you know what $p$% is. If you know that $c$ is $p$% of an unknown quantity, and you want to know how much $q$% of that same quantity will be, then you can set it up as a cross-multiplication problem:&#xA;$$\begin{array}{ccc}&#xD;&#xA;c &amp;amp; \text{---} &amp;amp; p\\&#xD;&#xA;x &amp;amp; \text{---} &amp;amp; q&#xD;&#xA;\end{array}$$&#xD;&#xA;so that $x=qc/p$. For example, if you know that $23$ is $17$% of some unknown quantity $x$, and you would like to know who much 77% of that same quantity would be, it is $23(77/17)  \approx 104.176$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But after writing all of the above it strikes me that the question may be even more basic: you have a certain quantity $p$ between $0$ and $50$, which you interpret as a percentage, and you want to know what quantity $q$ is just as far from $50$ but &quot;on the other side&quot;. Since you want $q-50 = 50-p$, then you want $q=100-p$. &lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-11-17T21:00:50.880" CommentCount="3" />
  <row Id="10752" PostTypeId="2" ParentId="10744" CreationDate="2010-11-17T21:03:39.837" Score="1" ViewCount="" Body="&lt;p&gt;$\frac{\partial}{\partial x}$ is just the gradient.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$xBx$ is scalar, namely $\sum_i x_i \sum_j B_{ij} x_j = \sum_i \sum_j B_{ij} x_i x_j$. This can be simplified to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$sum_i B_{ii} x_i^2 + \sum_{i&amp;lt;j} x_i x_j ( B_{ij} + B_{ji})$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For any index $i$, the derivative with respect to $i$ becomes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$2 B_{ii} x_i + \sum_{j \neq i} = x_j ( B_{ij} + B_{ji})$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But for any $i$, the above term is just the i-th component of the vector&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(B + B^T)x$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="3557" LastActivityDate="2010-11-17T21:03:39.837" />
  <row Id="10753" PostTypeId="2" ParentId="10745" CreationDate="2010-11-17T21:07:14.717" Score="1" ViewCount="" Body="&lt;p&gt;Yes, you are doing the right thing, assuming you meant $n^2 / \sqrt{n+n^2} &amp;lt; X$ rather than $1 / \sqrt{n+n^2} &amp;lt; X$.&lt;/p&gt;&#xA;" OwnerUserId="2810" LastActivityDate="2010-11-17T21:07:14.717" CommentCount="1" />
  <row Id="10754" PostTypeId="2" ParentId="10745" CreationDate="2010-11-17T21:13:43.063" Score="0" ViewCount="" Body="&lt;p&gt;Each term is bounded from below by the last term of your sum. Therefore a lower bound for the sum is $n^2$ times the last term. Show that this lower bound diverges as $n$ goes to infinity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case, you only really need a lower bound. Don't bother with an upper bound.&lt;/p&gt;&#xA;" OwnerUserId="3567" LastActivityDate="2010-11-17T21:13:43.063" />
  <row Id="10755" PostTypeId="2" ParentId="10745" CreationDate="2010-11-17T21:18:01.780" Score="2" ViewCount="" Body="&lt;p&gt;In proving something diverges to $\infty$, you don't need to squeeze it both from below &lt;em&gt;and&lt;/em&gt; from above: it is enough to &quot;half-squeeze&quot; it from below by something that goes to $\infty$ (because the limit you want is getting &quot;pushed up&quot; to $\infty$). So here, you can just note that each summand is greater than or equal to $\frac{1}{\sqrt{n+n^2}}$, so the sum of $n^2$ of them is at least as large as $\frac{n^2}{\sqrt{n+n^2}}$, and since $\lim\limits_{n\to\infty}\frac{n^2}{\sqrt{n+n^2}}=\infty$ and $\frac{n^2}{n+n^2}\leq X(n)$ for each $n$ (notice that what you called $X$ actually &lt;em&gt;varies&lt;/em&gt; with each $n$, so you should not give it a name that makes it look like it is fixed), then $\lim\limits_{n\to\infty}X(n)=\infty$ as well.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-11-17T21:18:01.780" CommentCount="3" />
  <row Id="10756" PostTypeId="2" ParentId="10744" CreationDate="2010-11-17T21:25:00.523" Score="1" ViewCount="" Body="&lt;p&gt;Here is a somewhat simplified view of things to get you started (one can avoid&#xA;all of the un-necessary elementwise notation down there, but it seems that to&#xA;help you start, it might still prove to be useful):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose $x \in \mathbb{R}^n$. Further assume the convention that $x$ is a&#xA;&lt;strong&gt;column vector&lt;/strong&gt; so that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x = \begin{bmatrix} x_1\\\\ x_2\\\\ \vdots\\\\ x_n&#xD;&#xA;\end{bmatrix}&#xD;&#xA;$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now suppose that we have some function $f: \mathbb{R}^n \to \mathbb{R}$ (i.e., a&#xA;&lt;strong&gt;scalar&lt;/strong&gt; valued function of the vector $x$). We &lt;em&gt;define&lt;/em&gt; the partial&#xA;derivatives of $f(x)$ as the vector&#xA;$$\frac{\partial f(x)}{\partial x} = &#xD;&#xA;\begin{bmatrix}&#xD;&#xA;  \frac{\partial f(x)}{\partial x_1}\\\\&#xD;&#xA;  \frac{\partial f(x)}{\partial x_2}\\\\&#xD;&#xA;  \vdots\\\\&#xD;&#xA;  \frac{\partial f(x)}{\partial x_n}&#xD;&#xA;\end{bmatrix}&#xD;&#xA;$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let us look at your function $f(x) = x^TBx$. To simplify things for our&#xA;brute-force attack, we rewrite this is:&#xA;$$f(x) = \sum_{ij} x_iB_{ij}x_j$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let us try to compute the partial derivative wrt the $p$-th coordinate of&#xA;$x$, i.e., $\partial f(x) / \partial x_p$. This is given by&#xA;$$\frac{\partial f(x)}{\partial x_p} = \frac{\partial}{\partial x_p}\sum_{ij}&#xD;&#xA;x_i B_{ij} x_j$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$= \sum_{j} B_{pj}x_j + \sum_i x_iB_{ip}$$&#xD;&#xA;using the product rule for differentiation. But recognize that &#xA;$$\sum_i B_{ip}x_i = \sum_i B^T_{pi}x_i = (B^Tx)_p$$&#xD;&#xA;while&#xA;$$\sum_j B_{pj}x_j = (Bx)_p$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, we conclude that $$\frac{\partial f(x)}{\partial x_p} = (Bx)_p +&#xD;&#xA;(B^Tx)_p$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Collecting derivatives for $p=1,\ldots,n$ yields the desired identity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now you can follow this approach to obtain derivatives for other functions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another point that might help you is this: If $f$ is a function as defined&#xA;above, i.e., $f$ is a function of a column-vector $x$, then $\partial f&#xD;&#xA;/ \partial x$ is a &lt;strong&gt;column-vector&lt;/strong&gt;, while $\partial f / \partial x^T$ is a&#xA;&lt;strong&gt;row-vector&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please have a look at the matrix-calculus book by Magnus that I have cited in&#xA;this answer:&lt;a href=&quot;http://math.stackexchange.com/questions/9891/reference-for-matrix-calculus/9988#9988&quot;&gt;here&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1709" LastActivityDate="2010-11-17T21:25:00.523" CommentCount="1" />
  <row Id="10757" PostTypeId="2" ParentId="10744" CreationDate="2010-11-17T21:25:17.760" Score="0" ViewCount="" Body="&lt;p&gt;You're right about suspecting that $\frac{ \partial \mathbf x^T }{\partial \mathbf x} \neq I$ because the dimensions no longer match up. I think you might be wading into tensor territory with such operations.&lt;/p&gt;&#xA;" OwnerUserId="3638" LastActivityDate="2010-11-17T21:25:17.760" />
  <row Id="10758" PostTypeId="2" ParentId="10745" CreationDate="2010-11-17T21:25:23.190" Score="1" ViewCount="" Body="&lt;p&gt;First  $\frac{1}{\sqrt{n+n^{2}}} \rightarrow 0 $ if $n\rightarrow \infty $.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Futhermore, in the infinite case the squeeze principle used so:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;if $a_{n}\leq b_{n}$ and $a_{n}\rightarrow \infty$ then $b_{n}\rightarrow \infty$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also you sequence i do not understand, &lt;/p&gt;&#xA;" OwnerUserId="3421" LastActivityDate="2010-11-17T21:25:23.190" />
  <row Id="10759" PostTypeId="2" ParentId="10741" CreationDate="2010-11-17T21:58:25.237" Score="16" ViewCount="" Body="&lt;p&gt;You could view a kaleidoscope as a picture of a pattern in the quotient of the plane by a suitable group action. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/Z15gn.jpg&quot; alt=&quot;pic&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, some Escher work is a picture of life inside hyperbolic surfaces -- quotients of the hyperbolic plane. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/rFcU4.jpg&quot; alt=&quot;Escher&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the more extreme end, $\mathbb R^2$ is a quotient of $\mathbb R$, so presumably you have a stricter question in mind than the one you've actually written, as you can get all kinds of things as quotients. $SO_3$ (rotations in 3-space) are a quotient of the unit quaternions.  This is used in computer graphics, among other things. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the kaleidoscope, you're really viewing the quotient as an orbifold with its natural geometric structure, rather than just as a topological space. &lt;/p&gt;&#xA;" OwnerUserId="642" LastActivityDate="2010-11-17T21:58:25.237" CommentCount="3" />
  <row Id="10760" PostTypeId="1" AcceptedAnswerId="10803" CreationDate="2010-11-17T21:59:39.963" Score="1" ViewCount="173" Body="&lt;p&gt;I was looking for alternate proofs of the theorem that &quot;a vector space $V$ of dimension greater than $1$ over an infinite field is not a union of fewer than $|\mathbf{F}|$ proper subspaces&quot; and possible generalizations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A simple measure-theoretic proof over $\mathbb{R}$ is as follows: By countable additivity the sum of the measures of any collection of subspaces is zero since the measure of each subspace is zero, which is a contradiction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to look at proofs over arbitrary infinite fields and would like to know if similar statements hold for say modules (finitely generated or otherwise) over infinite rings.&lt;/p&gt;&#xA;" OwnerUserId="3431" LastEditorUserId="3431" LastEditorDisplayName="" LastEditDate="2010-11-17T22:35:45.317" LastActivityDate="2010-11-18T05:43:06.063" Title="A vector space over $R$ is not a countable union of proper subspaces" Tags="&lt;real-analysis&gt;&lt;linear-algebra&gt;&lt;algebra&gt;" AnswerCount="3" CommentCount="6" />
  <row Id="10761" PostTypeId="1" AcceptedAnswerId="22728" CreationDate="2010-11-17T22:07:15.927" Score="3" ViewCount="121" Body="&lt;p&gt;Let $\mu (r)&amp;gt;2$ be the &lt;a href=&quot;http://mathworld.wolfram.com/IrrationalityMeasure.html&quot; rel=&quot;nofollow&quot;&gt;irrationality measure&lt;/a&gt; of a transcendental number $r$, and consider the following set of points $P \in\mathbb{R}$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$P=\{r\in \mathbb{R}: \mu(r)=Constant\}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this set a fractal, and If so, then what is it's dimension?&lt;/p&gt;&#xA;" OwnerUserId="1417" LastEditorUserId="1417" LastEditorDisplayName="" LastEditDate="2010-11-17T22:12:18.343" LastActivityDate="2011-02-19T00:18:48.333" Title="Sets of Constant Irrationality Measure" Tags="&lt;fractals&gt;&lt;transcendental-numbers&gt;" AnswerCount="1" CommentCount="10" FavoriteCount="2" />
  <row Id="10762" PostTypeId="2" ParentId="10760" CreationDate="2010-11-17T22:08:19.860" Score="1" ViewCount="" Body="&lt;p&gt;Have a look here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://mathoverflow.net/questions/26/can-a-vector-space-over-an-infinite-field-be-a-finite-union-of-proper-subspaces&quot; rel=&quot;nofollow&quot;&gt;http://mathoverflow.net/questions/26/can-a-vector-space-over-an-infinite-field-be-a-finite-union-of-proper-subspaces&lt;/a&gt;&lt;/p&gt;&#xA;" LastActivityDate="2010-11-17T22:08:19.860" CommunityOwnedDate="2010-11-17T22:08:19.860" CommentCount="1" />
  <row Id="10763" PostTypeId="1" AcceptedAnswerId="10786" CreationDate="2010-11-17T22:16:59.413" Score="1" ViewCount="91" Body="&lt;p&gt;A commutative ring $R$ with unity is Hermite if for all $x,y\in R$ there exists $t,u,v\in R$ such that $x=tu$, $y=tv$ and $(u,v)=(1)$. Is there a finite commutative ring with unity that is not Hermite?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This characterisation is taken from theorem 3 of:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some Remarks About Elementary Divisor Rings, Leonard Gillman and Melvin Henriksen, Transactions of the American Mathematical Society, Vol. 82, No. 2 (Jul., 1956), pp. 362-365 &lt;/p&gt;&#xA;" OwnerUserId="3643" LastEditorUserId="3643" LastEditorDisplayName="" LastEditDate="2010-12-20T14:06:24.047" LastActivityDate="2010-12-20T14:06:24.047" Title="Is there a finite non-Hermite commutative ring with unity?" Tags="&lt;ring-theory&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="10764" PostTypeId="1" AcceptedAnswerId="10765" CreationDate="2010-11-17T22:26:49.663" Score="2" ViewCount="181" Body="&lt;p&gt;Let $f(x)$  be a Lipschitz  real-valued function defined on a closed interval $I$. The derivative $f &amp;#39;(x)$ exists a.e. since $f$ is absolutely continuous. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is: Is $f &amp;#39;(x)$ necessarily square integrable, i.e. in $L^2(I)$ ?&lt;/p&gt;&#xA;" OwnerUserId="3249" LastActivityDate="2010-11-17T22:55:28.290" Title="Is the derivative of a Lipschitz function square integrable?" Tags="&lt;real-analysis&gt;&lt;measure-theory&gt;&lt;functional-analysis&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="10765" PostTypeId="2" ParentId="10764" CreationDate="2010-11-17T22:29:03.633" Score="4" ViewCount="" Body="&lt;p&gt;Yes, because the derivative is bounded.  If $M$ is the Lipschitz constant, then $|f&amp;#39;(x)|\leq M$ everywhere it exists.  This follows from the fact that each of the difference quotients has this bound.  Lipschitz continuous functions on $I$ are precisely the indefinite integrals of bounded measurable functions on $I$, whereas AC functions are the indefinite integrals of integrable functions on $I$.  (Thus you find AC counterexamples to this by integrating integrable but not square integrable functions.)&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2010-11-17T22:55:28.290" LastActivityDate="2010-11-17T22:55:28.290" />
  <row Id="10766" PostTypeId="1" AcceptedAnswerId="10775" CreationDate="2010-11-17T22:33:16.037" Score="0" ViewCount="151" Body="&lt;p&gt;How does one show that any open interval in n-dimensional Euclidean space is connected?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is homework, and I have been stuck on it for a few hours now. Seems like it should be easy, but I can't get a proof that I am comfortable with.&lt;/p&gt;&#xA;" OwnerUserId="3644" LastActivityDate="2010-11-17T23:28:55.937" Title="Any open interval in an n-dimensional Euclidean space is connected" Tags="&lt;real-analysis&gt;&lt;homework&gt;" AnswerCount="2" CommentCount="7" />
  <row Id="10767" PostTypeId="2" ParentId="10741" CreationDate="2010-11-17T22:49:42.913" Score="9" ViewCount="" Body="&lt;p&gt;Many role-playing games have toroidal worlds. Especially Dragon Quest and Final Fantasy games. It's a pity that few go through the trouble of implementing a true spheric topology. I don't mind other topologies if they fit the game world, but in RPGs it's mostly understood that the world should be Earth-like.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And then, there is also this marvelous site:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.geometrygames.org/HyperbolicGames/&quot; rel=&quot;nofollow&quot;&gt;http://www.geometrygames.org/HyperbolicGames/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3567" LastActivityDate="2010-11-17T22:49:42.913" />
  <row Id="10768" PostTypeId="1" CreationDate="2010-11-17T22:59:56.757" Score="0" ViewCount="91" Body="&lt;p&gt;Suppose $f$ is a real-valued function and $f(x) \geq 1$ for every real $x \in [0,1]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why we can always find a unique positive integer $n$ such that $2^{n} \leq f(x) &amp;lt; 2^{n+1}$ ?&lt;/p&gt;&#xA;" OwnerUserId="4219" LastActivityDate="2010-11-17T23:20:38.910" Title="Existence of integer" Tags="&lt;real-analysis&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="10769" PostTypeId="2" ParentId="10726" CreationDate="2010-11-17T23:03:33.783" Score="2" ViewCount="" Body="&lt;p&gt;Here's an answer to the second part of Problem 1: $$k = \frac{(a+1)(b+1) - \gcd(a,b) + 3}{2}.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to your definition,&#xA;$$k = 1 + (a+1) + \sum_{j=1}^{b-1} \left\lceil \frac{ja}{b} \right\rceil.$$ &#xA;Now, $\left\lceil \frac{ja}{b} \right\rceil = \left\lfloor \frac{ja + b - 1}{b} \right\rfloor.$  (See, for example, Exercise 3.12, p.96, in &lt;em&gt;Concrete Mathematics&lt;/em&gt;, 2nd edition.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is known that $$\sum_{j=0}^{b-1} \left\lfloor \frac{aj + x}{b} \right\rfloor = d \left\lfloor \frac{x}{d} \right\rfloor + \frac{(a-1)(b-1)}{2} + \frac{d-1}{2},$$&#xD;&#xA;for any real number $x$, where $d = \gcd(a,b)$.  (See, for example, Eq. 3.32, p. 94, of &lt;em&gt;Concrete Mathematics&lt;/em&gt;, 2nd edition.)  We have $x = b-1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then $$d \left\lfloor \frac{b-1}{d} \right\rfloor = d\frac{b-1}{d} - d \left\{\frac{b-1}{d}\right\} = b-1 - d \left\{\frac{b}{d} - \frac{1}{d}\right\} = b - 1  - d \frac{d-1}{d} = b-d,$$&#xD;&#xA;where $\{x\}$ denotes the fractional part of $x$.  (The second-to-last step holds because $d | b$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Putting everything together yields $$k = a + 2 + b - d + \frac{(a-1)(b-1)}{2} + \frac{d-1}{2},$$&#xD;&#xA;and then combining terms gets the rest.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;HR&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The elements in $f(a,b)$ are closely related to Spec$(\frac{a}{b}) = \{\lfloor \frac{a}{b}\rfloor, \lfloor\frac{2a}{b}\rfloor,  \lfloor\frac{3a}{b}\rfloor, \ldots \}$, although you want ceilings, and you want to stop when the multiplier is $b-1$.  You may find the discussion of Spec$(\alpha)$ in &lt;em&gt;Concrete Mathematics&lt;/em&gt; helpful.  (In fact, when I looked at your problem I thought, &quot;This reminds me of the discussion of spectra in &lt;em&gt;Concrete Math&lt;/em&gt;,&quot; and then using results in there it wasn't too difficult to find the expression for $k$.)&lt;/p&gt;&#xA;" OwnerUserId="2370" LastActivityDate="2010-11-17T23:03:33.783" />
  <row Id="10770" PostTypeId="2" ParentId="10768" CreationDate="2010-11-17T23:09:00.267" Score="0" ViewCount="" Body="&lt;p&gt;You can't do it for $f(x)=3x+1$.&lt;/p&gt;&#xA;" OwnerUserId="589" LastActivityDate="2010-11-17T23:09:00.267" CommentCount="2" />
  <row Id="10771" PostTypeId="2" ParentId="10768" CreationDate="2010-11-17T23:20:38.910" Score="1" ViewCount="" Body="&lt;p&gt;The intervals $[2^n,2^{n+1})$ cover the entire real half (existence) line $y\geq 1$ and no two of them intersect (uniqueness). So any point on the graph of the function must intersect one of these since $f(x)\geq 1$. You don't need any hypothesis on the domain of the function. &lt;/p&gt;&#xA;" OwnerUserId="3431" LastActivityDate="2010-11-17T23:20:38.910" />
  <row Id="10772" PostTypeId="1" AcceptedAnswerId="10777" CreationDate="2010-11-17T23:27:14.380" Score="1" ViewCount="78" Body="&lt;p&gt;I'm reading Number Theory I, by Kato, et al.  The authors claim that for $x$, $y\in \mathbb{Q}$ and $m$ a square free positive integer, if $2x$, $x^2-my^2 \in \mathbb{Z}$, then for any odd prime $p$ $\mathrm{ord}_p(m)+2\mathrm{ord}_p(y) \geq 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I get that $\mathrm{ord}_p(my^2)=\mathrm{ord}_p(m)+2\mathrm{ord}_p(y)$ and $\mathrm{ord}_p(x)\geq 0$.  I'm just not sure that I can say that $\mathrm{ord}_p(x^2-my^2)=\min\{\mathrm{ord}_p(x^2), \mathrm{ord}_p(my^2)\}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm having a bit of trouble seeing this.  Does anyone know what to do here?&lt;/p&gt;&#xA;" OwnerUserId="2212" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-18T00:03:52.490" LastActivityDate="2010-11-18T00:10:42.903" Title="Proof for a fact involving $\mathrm{ord}_p(x)$" Tags="&lt;number-theory&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="10773" PostTypeId="2" ParentId="10766" CreationDate="2010-11-17T23:27:43.843" Score="0" ViewCount="" Body="&lt;p&gt;Can you show that an interval is connected in $\mathbb{R}$?  Then can you say (a,b) is connected to (c,b) is connected to (c,d)?&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-11-17T23:27:43.843" />
  <row Id="10774" PostTypeId="1" CreationDate="2010-11-17T23:27:53.927" Score="4" ViewCount="197" Body="&lt;p&gt;In Andy's answer to the question &quot;&lt;a href=&quot;http://mathoverflow.net/questions/12045/what-are-fixed-points-of-the-fourier-transform&quot; rel=&quot;nofollow&quot;&gt;What are fixed points of the Fourier Transform&lt;/a&gt;&quot; on Math Overflow, he shows that the Fourier Transform has eigenvalues {+1, +i, -1, -i} and that the projections of any function onto the corresponding four eigenspaces may be found through some simple linear algebra.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to get a better feeling for these four eigenspaces of the fourier transform.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;How can I find some interesting members of each of these eigenspaces?&lt;/li&gt;&#xA;&lt;li&gt;How can I show that Hermite-Gaussians are in one (or more?) of the eigenspaces?&lt;/li&gt;&#xA;&lt;li&gt;How can one define usable projection operators onto these eigenspaces?&lt;/li&gt;&#xA;&lt;li&gt;The wikipedia &lt;a href=&quot;http://en.wikipedia.org/wiki/Fourier_transform#Eigenfunctions&quot; rel=&quot;nofollow&quot;&gt;article&lt;/a&gt; on the Fourier Transform mentions that Wiener defined the Fourier Transform via these projections.  What exactly was Wiener's approach?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="2191" LastActivityDate="2010-11-18T12:33:56.453" Title="How do I compute the eigenfunctions of the Fourier Transform?" Tags="&lt;linear-algebra&gt;&lt;fourier-analysis&gt;" AnswerCount="1" />
  <row Id="10775" PostTypeId="2" ParentId="10766" CreationDate="2010-11-17T23:28:55.937" Score="3" ViewCount="" Body="&lt;p&gt;OK. This is a basic result which states that &lt;em&gt;Cartesian&lt;/em&gt; product of connected sets are connected. You may look here for the proof:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://planetmath.org/encyclopedia/ProofOfProductsOfConnectedSpaces.html&quot; rel=&quot;nofollow&quot;&gt;http://planetmath.org/encyclopedia/ProofOfProductsOfConnectedSpaces.html&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" LastActivityDate="2010-11-17T23:28:55.937" />
  <row Id="10777" PostTypeId="2" ParentId="10772" CreationDate="2010-11-17T23:47:44.087" Score="1" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\rm\ \ 2x\in \mathbb Z\ \Rightarrow\ ord_p(x)\ge 0\ \ $  so $\rm\ ord_p (x^2-my^2) \ge 0\ \Rightarrow\ ord_p(my^2) \ge 0$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Recall the &lt;strong&gt;principle of domination:&lt;/strong&gt;  $\rm\ ord_p\ Y &amp;lt; ord_p\ X\ \Rightarrow\ ord_p\:(X+Y) = ord_p\ Y\:$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-11-18T00:10:42.903" LastActivityDate="2010-11-18T00:10:42.903" CommentCount="4" />
  <row Id="10778" PostTypeId="1" CreationDate="2010-11-17T23:47:52.373" Score="1" ViewCount="157" Body="&lt;p&gt;Now, suppose the transformation(in 2d) I am working with has two&lt;br&gt;&#xA;separate functions&#xA;for x and y.&lt;br&gt;&#xA;That is, the transformation for x is of the form&lt;br&gt;&#xA;x'=x/(x+y)&lt;br&gt;&#xA;and the transformation of y is&lt;br&gt;&#xA;y'=y/(x+y)&lt;br&gt;&#xA;Each is an LFT(The schwarzian derivatives are 0) but is&lt;br&gt;&#xA;the transform as a whole still considered an LFT?&lt;/p&gt;&#xA;" LastEditorUserId="1543" LastEditorDisplayName="" LastEditDate="2011-04-17T20:32:00.883" LastActivityDate="2011-05-17T20:58:25.167" Title="is this a linear fractional transformation(LFT)?" Tags="&lt;complex-analysis&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="10779" PostTypeId="1" AcceptedAnswerId="10789" CreationDate="2010-11-18T00:04:35.267" Score="2" ViewCount="159" Body="&lt;p&gt;I want to prove that for non-negative random variables with distribution F:&#xA;$$E(X^{n}) = \int_0^\infty n x^{n-1} P(\{X≥x\}) dx$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is the following proof correct?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$R.H.S = \int_0^\infty n x^{n-1} P(\{X≥x\}) dx = \int_0^\infty n x^{n-1} (1-F(x)) dx$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;using integration by parts:&#xA;$$R.H.S = [x^{n}(1-F(x))]_0^\infty + \int_0^\infty x^{n} f(x) dx = 0 + \int_0^\infty x^{n} f(x) dx = E(X^{n})$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If not correct, then how to prove it?&lt;/p&gt;&#xA;" OwnerUserId="953" LastEditorUserId="953" LastEditorDisplayName="" LastEditDate="2010-11-18T00:20:13.273" LastActivityDate="2010-11-18T01:15:17.577" Title="Moments and non-negative random variables?" Tags="&lt;probability&gt;" AnswerCount="5" FavoriteCount="1" />
  <row Id="10780" PostTypeId="1" AcceptedAnswerId="10790" CreationDate="2010-11-18T00:11:40.313" Score="4" ViewCount="258" Body="&lt;p&gt;Let $f$ be the $2\pi$ periodic function which is the even extension of $$x^{1/n}, 0 \le x \le \pi,$$ where $n \ge 2$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am looking for a general theorem that implies that the Fourier series of $f$ converges to $f$, pointwise, uniformly or absolutely. &lt;/p&gt;&#xA;" OwnerUserId="3249" LastEditorUserId="3249" LastEditorDisplayName="" LastEditDate="2010-11-18T13:35:32.977" LastActivityDate="2010-11-18T13:35:32.977" Title="Convergence of a Fourier series" Tags="&lt;analysis&gt;&lt;fourier-analysis&gt;&lt;fourier-series&gt;" AnswerCount="4" FavoriteCount="1" />
  <row Id="10781" PostTypeId="2" ParentId="10779" CreationDate="2010-11-18T00:13:16.117" Score="0" ViewCount="" Body="&lt;p&gt;Only if $\lim_{x\rightarrow \infty} x^n(1-F(x)) = 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It doesn't look like that's true in general.&lt;/p&gt;&#xA;" OwnerUserId="3638" LastActivityDate="2010-11-18T00:13:16.117" CommentCount="0" />
  <row Id="10782" PostTypeId="1" AcceptedAnswerId="10785" CreationDate="2010-11-18T00:14:01.280" Score="4" ViewCount="185" Body="&lt;p&gt;Here is how I have worked it out so far:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\log _2(x-4)+\log(x+2)=4$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\log _2((x-4)(x+2)) = 4$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(x-4)(x+2)=2^4$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(x-4)(x+2)=16$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do I proceed from here?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x^2+2x-8 = 16$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x^2+2x = 24$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x(x+2) = 24$ Which I know is not the right answer&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x^2+2x-24 = 0$ Can't factor this&lt;/p&gt;&#xA;" OwnerUserId="3647" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-18T00:47:26.460" LastActivityDate="2010-11-18T00:47:26.460" Title="Solving $\log _2(x-4) + \log _2(x+2) = 4$" Tags="&lt;algebra-precalculus&gt;&lt;logarithms&gt;" AnswerCount="3" CommentCount="3" />
  <row Id="10783" PostTypeId="2" ParentId="10779" CreationDate="2010-11-18T00:16:53.040" Score="0" ViewCount="" Body="&lt;p&gt;As a matter of notation, you should write the expectation as $E(X^n)$ instead of $E(x^n)$. The expectation is not a function of $x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your proof would work if $\lim_{x \to \infty} x^n (1-F(x)) = 0$. I'm not sure this is true for any distribution.&lt;/p&gt;&#xA;" OwnerUserId="2641" LastActivityDate="2010-11-18T00:16:53.040" CommentCount="1" />
  <row Id="10784" PostTypeId="2" ParentId="10782" CreationDate="2010-11-18T00:19:28.977" Score="1" ViewCount="" Body="&lt;p&gt;After $(x-4)(x+2)=16$, you get $x^2-2x-24=0$ (the coefficient of $x$ is $-2$ not $2$). So $x=\frac{2\pm \sqrt{100}}{2}$ by the quadratic formula. So $x=6$ or $x=-4$  &lt;/p&gt;&#xA;" OwnerUserId="3431" LastEditorUserId="3431" LastEditorDisplayName="" LastEditDate="2010-11-18T00:28:27.017" LastActivityDate="2010-11-18T00:28:27.017" CommentCount="3" />
  <row Id="10785" PostTypeId="2" ParentId="10782" CreationDate="2010-11-18T00:19:38.557" Score="5" ViewCount="" Body="&lt;p&gt;It is $x^2-2x-8 = 16$ my friend. So you get $x^2 - 2x -24 = 0$, which factors as $(x-6)(x+4) = 0$. Hence, $x=6$ or $x = -4$.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastActivityDate="2010-11-18T00:19:38.557" CommentCount="2" />
  <row Id="10786" PostTypeId="2" ParentId="10763" CreationDate="2010-11-18T00:19:54.493" Score="1" ViewCount="" Body="&lt;p&gt;Yes, $\mathbb{F}_2[x,y]/(x,y)^2$.&lt;/p&gt;&#xA;" OwnerUserId="2660" LastActivityDate="2010-11-18T00:19:54.493" CommentCount="2" />
  <row Id="10787" PostTypeId="2" ParentId="10782" CreationDate="2010-11-18T00:31:35.470" Score="1" ViewCount="" Body="&lt;p&gt;The quadratic formula:&#xA;$$ x = \frac{ -b \pm \sqrt{b^2 - 4 a c} }{2a} = \frac{-2 \pm \sqrt{2^2 - 4 \times 1 \times (-24) }}{2 \times 1} $$&#xD;&#xA;4 + 96 in the square root.&lt;/p&gt;&#xA;" OwnerUserId="3638" LastActivityDate="2010-11-18T00:31:35.470" />
  <row Id="10788" PostTypeId="2" ParentId="10779" CreationDate="2010-11-18T00:45:01.177" Score="3" ViewCount="" Body="&lt;p&gt;A better way to prove it would be to use Fubini's theorem to change the order of integration. This also gives you a condition when the result you have is true.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider $I = \displaystyle \int_0^{\infty}n x^{n-1} (1-F(x))dx$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the fact that $\displaystyle \int_x^{\infty} f(y)dy = 1 - F(x)$, we get $I = \displaystyle \int_0^{\infty}n x^{n-1} \int_x^{\infty} f(y) dy dx$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we first integrate with respect to $y$ (the inner integral) and $y$ goes from $x$ to $\infty$ and then integrate with respect to $x$ (the outer integral), $x$ goes from $0$ to $\infty$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Change the order of integration. i.e. integrate with respect to $x$ first and then with respect to $y$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that this can be done provided the integral $I &amp;lt; \infty$ (See Fubini's theorem). This is the condition svenkatr and trutheality get as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Changing the order of integration, we get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$I = \displaystyle \int_0^{\infty} \displaystyle \int_{0}^{y} nx^{n-1}f(y)dxdy$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that now $x$ in the inner integral goes from $0$ to $y$ and $y$ goes from $0$ to $\infty$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the inner integral with respect to $x$ can performed easily and now we get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$I = \displaystyle \int_0^{\infty} y^{n}f(y)dy = E[X^n]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, we have $\displaystyle \int_0^{\infty}n x^{n-1} (1-F(x))dx = E[X^n]$.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2010-11-18T00:50:25.087" LastActivityDate="2010-11-18T00:50:25.087" />
  <row Id="10789" PostTypeId="2" ParentId="10779" CreationDate="2010-11-18T00:53:05.457" Score="3" ViewCount="" Body="&lt;p&gt;Here's another way.  (As the others point out, the statement is true if $E[X^n]$ actually exists.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $Y = X^n$.  $Y$ is non-negative if $X$ is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know &#xA;$$E[Y] = \int_0^{\infty} P(Y \geq t) dt,$$&#xD;&#xA;so&#xA;$$E[X^n] = \int_0^{\infty} P(X^n \geq t) dt.$$&#xD;&#xA;Then, perform the change of variables $t = x^n$.  This immediately yields &#xA;$$E[X^n] = \int_0^{\infty} n x^{n-1} P(X^n \geq x^n) dx = \int_0^{\infty} n x^{n-1} P(X \geq x) dx.$$&lt;/p&gt;&#xA;" OwnerUserId="2370" LastActivityDate="2010-11-18T00:53:05.457" />
  <row Id="10790" PostTypeId="2" ParentId="10780" CreationDate="2010-11-18T01:06:43.587" Score="2" ViewCount="" Body="&lt;p&gt;Perhaps you can apply the one found here: &lt;a href=&quot;http://books.google.com/books?id=XqqNDQeLfAkC&amp;amp;pg=PA84&quot; rel=&quot;nofollow&quot;&gt;http://books.google.com/books?id=XqqNDQeLfAkC&amp;amp;pg=PA84&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Snapshot:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/ySQ0q.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-11-18T01:06:43.587" CommentCount="5" />
  <row Id="10791" PostTypeId="2" ParentId="9851" CreationDate="2010-11-18T01:09:28.220" Score="3" ViewCount="" Body="&lt;p&gt;Do the &lt;a href=&quot;http://digitus.itk.ppke.hu/~matyi/optika/Phase_Diversity/AO82_PRComparison1.pdf&quot; rel=&quot;nofollow&quot;&gt;usual numerical phase retrieval algorithms&lt;/a&gt; not work?&lt;/p&gt;&#xA;" OwnerUserId="2191" LastActivityDate="2010-11-18T01:09:28.220" CommentCount="1" />
  <row Id="10792" PostTypeId="2" ParentId="10780" CreationDate="2010-11-18T01:13:18.283" Score="2" ViewCount="" Body="&lt;p&gt;I looked it up on Wikipedia. Assuming that the articles are correct, it seems that your function satisfies some &lt;a href=&quot;http://en.wikipedia.org/wiki/H%C3%B6lder_condition&quot; rel=&quot;nofollow&quot;&gt;Hölder condition&lt;/a&gt; and thus by the &lt;a href=&quot;http://en.wikipedia.org/wiki/Convergence_of_Fourier_series#Pointwise_convergence&quot; rel=&quot;nofollow&quot;&gt;Dirichlet-Dini chriterion&lt;/a&gt; its Fourier series converges pointwise to &lt;em&gt;f&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could also show that your function is Lipschitz and then you'd have &lt;a href=&quot;http://en.wikipedia.org/wiki/Convergence_of_Fourier_series#Absolute_convergence&quot; rel=&quot;nofollow&quot;&gt;absolute convergence&lt;/a&gt; of its Fourier series.&lt;/p&gt;&#xA;" OwnerUserId="1269" LastActivityDate="2010-11-18T01:13:18.283" CommentCount="1" />
  <row Id="10793" PostTypeId="2" ParentId="10779" CreationDate="2010-11-18T01:15:17.577" Score="2" ViewCount="" Body="&lt;p&gt;If ${\rm E}(X^n) &amp;lt; \infty$, then $\int_0^\infty  {x^n f(x){\rm d}x}  &amp;lt; \infty $, and in turn, $\int_M^\infty  {x^n f(x){\rm d}x}  \to 0$ as $M \to \infty$. Since $\int_M^\infty  {x^n f(x){\rm d}x} \geq \int_M^\infty  {M^n f(x){\rm d}x}  = M^n [1 - F(M)]$, we have that $M^n [1 - F(M)] \to 0$ as $M \to \infty$. Hence, your solution is correct (assuming that ${\rm E}(X^n) &amp;lt; \infty$).&lt;/p&gt;&#xA;" OwnerUserId="2810" LastActivityDate="2010-11-18T01:15:17.577" CommentCount="1" />
  <row Id="10794" PostTypeId="2" ParentId="10209" CreationDate="2010-11-18T01:25:01.640" Score="3" ViewCount="" Body="&lt;p&gt;Mathematics - A Discrete Introduction  by Edward R. Scheinerman is the best book for learning discrete math that I have come across. &#xA;Kenneth Rosen's book blows; sorry, but it's true - I had to suffer through it back in college. It wasn't fun. &#xA;Scheinerman's book made discrete math interesting again. I think it was the uncluttered presentation and clear explanations.&lt;/p&gt;&#xA;" LastActivityDate="2010-11-18T01:25:01.640" />
  <row Id="10795" PostTypeId="2" ParentId="10078" CreationDate="2010-11-18T01:31:52.727" Score="1" ViewCount="" Body="&lt;p&gt;As long as the sequence doesn't fall outside of the support of the function, it is a sequence that could be generated by the function (the all-heads sequence for a fair coin toss is a good example).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you can try to do is minimize the distance b/w the empirical CDF of the data and the CDF of your distribution.&#xA;You have to decide what distance metric to use (maximum absolute difference is one obvious choice) and once you have that it boils down to solving an optimization problem with each new data point added.&lt;/p&gt;&#xA;" OwnerUserId="3638" LastActivityDate="2010-11-18T01:31:52.727" />
  <row Id="10796" PostTypeId="1" AcceptedAnswerId="10798" CreationDate="2010-11-18T01:32:52.687" Score="0" ViewCount="62" Body="&lt;p&gt;The number of dropped connections per call follows a Poisson distribution. From four calls, the number of dropped connections is 2, 0 , 3, 1.&lt;br&gt;&#xA;Obtain maximum likelihood estimate that the next two calls will be completed without any accidental drops.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know the maximum likelihood estimate of lamda is 1.5.&#xA;I think I am suppose to use this formula:&#xA;$$f(x|\lambda) = \frac{\lambda^x e^{-\lambda}}{x!}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I do not know what x should be.  The answer should be 0.0498&lt;/p&gt;&#xA;" OwnerUserId="3601" LastEditorUserId="3601" LastEditorDisplayName="" LastEditDate="2010-11-18T01:39:23.010" LastActivityDate="2010-11-18T01:39:23.010" Title="Obtain maximum likelihood estimate" Tags="&lt;homework&gt;&lt;statistics&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="10797" PostTypeId="2" ParentId="10780" CreationDate="2010-11-18T01:36:38.410" Score="2" ViewCount="" Body="&lt;p&gt;You can actually do your example pretty explicitly. For $f(x) = |x|^{\alpha}$ you actually have that $\hat{f}(n) = c_{\alpha} n^{-1 - \alpha}  + O(n^{-2})$, so the cosine series is absolutely convergent. If a Fourier series of a continuous function converges, it has to converge to the original function; I refer you to a Fourier analysis text for this fact though.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To get the above expression for $\hat{f}(n)$, recall that $\hat{f}(n)$ is defined as&#xA;$\hat{f}(n) = 2\int_0^1x^{\alpha} \cos(n\pi x)dx$. Integrating by parts, this is the same as ${2 \alpha \over \pi}n^{-1}\int_0^1x^{\alpha - 1} \sin(n\pi x)dx$. This in turn can be written as ${2 \alpha \over \pi}n^{-1} \int_0^\infty x^{\alpha - 1} \sin(n\pi x)dx - &#xD;&#xA;{2 \alpha \over \pi}n^{-1} \int_1^\infty x^{\alpha - 1} \sin(n\pi x)dx$. (These integrals are convergent improper integrals). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;By changing variables $x$ to $nx$ the first integral becomes the main term $c_{\alpha}n^{-1 - \alpha}$. By doing one more integration by parts and then taking absolute values, the second term is bounded by $C n^{-2}$ as needed.&lt;/p&gt;&#xA;" OwnerUserId="3035" LastActivityDate="2010-11-18T01:36:38.410" />
  <row Id="10798" PostTypeId="2" ParentId="10796" CreationDate="2010-11-18T01:36:54.530" Score="1" ViewCount="" Body="&lt;p&gt;x is the number of dropped connections, right? So $x=0$ for no dropped connections.&lt;/p&gt;&#xA;" OwnerUserId="3638" LastActivityDate="2010-11-18T01:36:54.530" CommentCount="3" />
  <row Id="10800" PostTypeId="2" ParentId="10741" CreationDate="2010-11-18T02:22:38.327" Score="4" ViewCount="" Body="&lt;p&gt;A running track is a good one-dimensional example.  We start with the real line and identify all points {x + n*400meter}, or equivalently we take a 400meter long line segment and glue the ends together.   Now, we can sit in the stands and watch an entire 800m, 1600m, or 3200m race, because we have a nice &quot;compact&quot; track!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you explain how tiling the real line with 400m segments leads to a running track, it may be easier to explain how two-dimensional tilings of the plane (or sphere or hyperbolic plane) can lead to quotient spaces.  Of course, we need to know how to glue the edges, and that point may be a little vague in a casual conversation-- but if Asteroids is not available, I think that tilings are one of the best ways to visualize quotient spaces.&lt;/p&gt;&#xA;" OwnerUserId="1461" LastActivityDate="2010-11-18T02:22:38.327" />
  <row Id="10801" PostTypeId="1" AcceptedAnswerId="10802" CreationDate="2010-11-18T02:29:52.660" Score="2" ViewCount="128" Body="&lt;p&gt;You are dealt five cards from 52 card deck that is thoroughly shuffled. What is the probability that two out of five cards are seven?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it:&#xA;$${4C2 . 12C3 . (4C1)^3} / 52C5$$&lt;/p&gt;&#xA;" OwnerUserId="953" LastEditorUserId="622" LastEditorDisplayName="" LastEditDate="2010-11-18T05:43:05.843" LastActivityDate="2010-11-18T05:43:05.843" Title="Probability that two out of five cards are seven?" Tags="&lt;probability&gt;&lt;combinatorics&gt;&lt;counting&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="10802" PostTypeId="2" ParentId="10801" CreationDate="2010-11-18T02:33:00.333" Score="5" ViewCount="" Body="&lt;p&gt;Kindly consult the Wikipedia page on Hypergeometric distribution. The correct formula is $$\frac{\binom{4}{2}\binom{48}{3}}{\binom{52}{5}}.$$&#xD;&#xA;So $\binom{12}{3} \binom{4}{1}^3$ is incorrect.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-18T02:33:00.333" />
  <row Id="10803" PostTypeId="2" ParentId="10760" CreationDate="2010-11-18T02:59:55.177" Score="2" ViewCount="" Body="&lt;p&gt;Here's a proof of the finite dimensional case over any field (finite or infinite):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Theorem.&lt;/strong&gt; Let $\mathbf{V}$ be a nonzero finite dimensional vector space over $\mathbf{F}$. If $\mathbf{V}$ is a union of $\kappa$ proper subspaces, then $\kappa\geq|\mathbf{F}|$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Proof.&lt;/em&gt; Write $\mathbf{V}=\bigcup\limits_{k\in\kappa} W_{k}$, with $W_k$ a proper subspace of $\mathbf{V}$. By enlarging the $W_k$ as necessary, we may assume that $\dim(\mathbf{V}/W_i) = 1$ for all $i$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The proof is by induction on $\dim(\mathbf{V})$. The result is trivially true if $\dim(\mathbf{V})=1$, because $\mathbf{V}$ is never the union of proper subspaces in this case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the case of $\dim(\mathbf{V})=2$, let $\{w_1\}$ be a basis for $W_1$, and let $v\notin W_1$. For each $\alpha\in \mathbf{F}$ there exists $j_{\alpha}\in\kappa$ such that $w_1+\alpha v\in W_{j_{\alpha}}$. Moreover, if $\alpha\neq\beta$, then $w_1+\alpha v$ and $w_1+\beta v$ are linearly independent, since $\{w_1,v\}$ are a basis for $\mathbf{V}$. Thus, no $W_k$ contains more than one $w_1+\alpha v$. This gives an injection from $\mathbf{F}$ to $\kappa$, proving that $\kappa\geq|\mathbf{F}|$, as required.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume the result holds for $n$-dimensional vector spaces, and let $\mathbf{V}$ be $(n+1)$-dimensional. Let $\{w_1,\ldots,w_n\}$ be a basis for $W_1$, and $v\notin W_1$. For each $\alpha\in\mathbf{F}$, consider the subspace $W_{\alpha}=\mathrm{span}(w_1+\alpha v,w_2,\ldots,w_n)$. If $W_{\alpha}$ is contained in some $W_k$, then $W_{\alpha}=W_k$ by dimension considerations; and if $W_{\alpha}=W_{\beta}$, then $\alpha=\beta$, for otherwise we would be able to find a nontrivial linear combination involving $w_1,\ldots,w_n,v$. So there is again an injection from the set&#xA;$$S=\{\alpha\in\mathbf{F} \mid W_{\alpha}=W_k\text{ for some }k\in \kappa\}$$&#xD;&#xA;to $\kappa$. If the set has cardinality $|\mathbf{F}|$ we are done. Otherwise, let $\alpha_0\in\mathbf{F}\setminus S$, and look at $W_{\alpha_0}$. For each $k\in \kappa$, $W_{\alpha_0}\cap W_{\kappa}\neq W_{\alpha_0}$; since&#xA;$$W_{\alpha_0} = W_{\alpha_0}\cap\mathbf{V} = W_{\alpha_0}\cap\left(\bigcup_{k\in\kappa}W_k\right) = \bigcup_{k\in\kappa}(W_{\alpha_0}\cap W_k),$$&#xD;&#xA;then by the induction hypothesis we have that $\kappa\geq|\mathbf{F}|$. &lt;strong&gt;QED&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strike&gt;I suspect a similar argument can be made in the infinite dimensional case; certainly, we can construct the analogous set to $S$, and if $|S|=|\mathbf{F}|$ then we are done. But if not, then $\dim(W_{\alpha})=\dim(\mathbf{V})$, so we cannot really use a reduction argument. But I think there may be a way to tweak this.&lt;/strike&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As Pete L. Clark has noted, the result does not hold in the infinite dimensional case.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-18T05:43:06.063" LastActivityDate="2010-11-18T05:43:06.063" CommentCount="1" />
  <row Id="10804" PostTypeId="2" ParentId="10209" CreationDate="2010-11-18T03:07:33.170" Score="4" ViewCount="" Body="&lt;p&gt;It may be worth having a look at the following &lt;a href=&quot;http://hbpms.blogspot.com/2008/05/stage-1-introductory-discrete.html&quot; rel=&quot;nofollow&quot;&gt;blog&lt;/a&gt; for a selection of resources for introductory discrete mathematics resources, a lot of them free.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are looking to buy a book, &lt;em&gt;Discrete Mathematics with Applications&lt;/em&gt; by Susanna S. Epp is particularly good. Plenty of exercises and examples.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I really dislike Rosen's book. I don't find his examples instructive at all.&lt;/p&gt;&#xA;" OwnerUserId="3564" LastActivityDate="2010-11-18T03:07:33.170" CommentCount="1" />
  <row Id="10805" PostTypeId="2" ParentId="10731" CreationDate="2010-11-18T03:38:24.490" Score="1" ViewCount="" Body="&lt;p&gt;Here's a remark added to original remarks below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As mentioned below, Hölder's inequality reduces this to showing that $\tau_1$ is finer than $\tau_\infty$.  This means showing that the $\tau_1$ to $\tau_\infty$ identity map  is continuous, while we already know that its inverse is continuous.  Hence if you prove that $\tau_1$ and $\tau_\infty$ are complete, this follows from the &lt;a href=&quot;http://en.wikipedia.org/wiki/Open_mapping_theorem_%28functional_analysis%29#Generalizations&quot; rel=&quot;nofollow&quot;&gt;open mapping theorem&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I'm not sure this is worth posting, because I can only finish the easy case $n=1$, but here's a proof for that case from someone ignorant of Sobolev embedding.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $1\leq p\leq q\leq\infty$, then because $K$ has finite measure, Hölder's inequality yields $|f|_{\alpha,p}\leq C|f|_{\alpha,q}$ for a constant $C$ depending only on $p$, $q$, and the measure of $K$.  Thus $\tau_q$ is finer than $\tau_p$, and the work is in showing that $\tau_1$ is finer than $\tau_\infty$.  Here, as Willie Wong pointed out, it is no longer possible to do so by directly comparing seminorms with the same multi-index.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose that $(f_k)$ is a sequence of functions such that $|f_k|_{\alpha,1}\to 0$ for all $\alpha$ as $k\to\infty$.  This implies that the same holds for the sequence $(D^{\alpha_0}f_k)$ for each fixed multi-index $\alpha_0$, so it is enough to check that $|f_k|_{0,\infty}\to0$, that is, $(f_k)$ converges uniformly to $0$.  Here's a way to see this when $n=1$.  Let $y$ be a point where all of the functions vanish, say $y=\inf K$ for definiteness.  For each $x\in K$, $|f_k(x)|=|f_k(x)-f_k(y)|=|\int_y^x f_k&amp;#39;(t)dt|\leq \int_K|f_k&amp;#39;(t)|dt=|f_k|_{1,1}$, which goes to $0$ independent of $x$.  Hence $(f_k)$ converges uniformly to $0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Actually, with slightly more work the proof that $\tau_1$ is finer than $\tau_\infty$ for $n=1$ can be extended to the case of smooth functions with arbitrary support.  With the same assumptions on $(f_k)$, for each $x\in\mathbb{R}$, $|f_k(x)-f_k(0)|=|\int_0^x f_k&amp;#39;(t)dt|\leq \int|f_k&amp;#39;(t)|dt=|f_k|_{1,1}$, which goes to $0$ independent of $x$.  This means that there is a sequence $(\varepsilon_k)$ of positive numbers converging to $0$ such that each $f_k$ is within $\varepsilon_k$ of the constant function with value $f_k(0)$.  Since $|f_k|_{0,1}$ goes to $0$, the corresponding constants must converge to $0$, and this implies that $(f_k)$ converges uniformly to $0$.  As Willie Wong also pointed out, you don't get $\tau_q$ finer than $\tau_p$ for $q\gt p$ in this case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To extend this argument to $n\gt 1$, I would need bounds on $|f(x)-f(y)|$ in terms of volume integrals of sums of (a finite number of) partial derivatives of $f$, and this is where my ignorance keeps me from going further.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2010-11-18T19:54:53.957" LastActivityDate="2010-11-18T19:54:53.957" CommunityOwnedDate="2010-11-18T19:54:53.957" CommentCount="2" />
  <row Id="10806" PostTypeId="1" AcceptedAnswerId="10807" CreationDate="2010-11-18T04:14:49.887" Score="1" ViewCount="160" Body="&lt;p&gt;Hi,&#xA;I'm out of practice with algebra, and I'm having a mind blank on how to solve for $t$ in the following equation. It's for some collision detection if you're wondering.&#xA;$$\Bigl( \bigl(a-y-(qt)\bigr) \bigl(b-x-(rt)\bigr) \Bigr) - \Bigl( \bigl(c-x-(rt)\bigr) \bigl(d-y-(qt)\bigr) \Bigr) = z$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm really not seeing it at the moment. Thanks for any help.&lt;/p&gt;&#xA;" OwnerUserId="3652" LastEditorUserId="1583" LastEditorDisplayName="" LastEditDate="2010-11-18T04:42:17.027" LastActivityDate="2010-11-18T04:42:17.027" Title="Solving a tricky equation" Tags="&lt;algebra-precalculus&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="10807" PostTypeId="2" ParentId="10806" CreationDate="2010-11-18T04:24:20.717" Score="5" ViewCount="" Body="&lt;p&gt;You could denote the differences $(a-y)$, $(b-x)$, $(c-x)$, $(d-y)$ as a single symbol to simplify computation. That is, $(a-y)=A$, $(b-x)=B$, $(c-x)=C$, $(d-y)=D$. Then you have &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(A-qt)(B-rt)-(C-rt)(D-qt)=z$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(AB-qBt-rAt+qrt^2)-(CD-rDt-qCt+qrt^2)=z$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice that the $qrt^2$ terms cancel upon distributing the negative, and then you have a linear equation for $t$, so $t$ will be easy to isolate, and then you can resubstitute the values for $A$, $B$, $C$, and $D$. I hope my algebra has been correct.&lt;/p&gt;&#xA;" OwnerUserId="1583" LastActivityDate="2010-11-18T04:24:20.717" CommentCount="2" />
  <row Id="10809" PostTypeId="2" ParentId="10741" CreationDate="2010-11-18T04:51:58.160" Score="5" ViewCount="" Body="&lt;p&gt;The original Sonic the Hedgehog games (on the Sega Genesis and similar consoles) provide good examples. Many of the levels wrap around vertically, so that there is no bottom or top. They are equivalent to cylinders, I suppose.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are some examples from Sonic 3:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://info.sonicretro.org/File:Ic1map.PNG&quot; rel=&quot;nofollow&quot;&gt;http://info.sonicretro.org/File:Ic1map.PNG&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://info.sonicretro.org/File:Mg1map.PNG&quot; rel=&quot;nofollow&quot;&gt;http://info.sonicretro.org/File:Mg1map.PNG&lt;/a&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The designers used this to make the giant hill at the beginning of the ice cap level that is many times taller than the level itself. Also, the Sonic 3 Special stages are toroidal, just like Asteroids, though in the game they are presented with an apparent spherical curvature, which had me terribly lost and confused until I googled it. In hindsight, I should have realized that you can't cover a sphere with squares, but Sonic doesn't exactly give you a lot of time for thinking.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://info.sonicretro.org/File:S%26KSS1.png&quot; rel=&quot;nofollow&quot;&gt;http://info.sonicretro.org/File:S%26KSS1.png&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://www.youtube.com/watch?v=LpDbSlbhP5M&quot; rel=&quot;nofollow&quot;&gt;http://www.youtube.com/watch?v=LpDbSlbhP5M&lt;/a&gt;  &lt;/p&gt;&#xA;" OwnerUserId="3653" LastEditorUserId="3653" LastEditorDisplayName="" LastEditDate="2010-11-18T05:10:31.957" LastActivityDate="2010-11-18T05:10:31.957" />
  <row Id="10810" PostTypeId="1" AcceptedAnswerId="10811" CreationDate="2010-11-18T04:57:07.050" Score="4" ViewCount="147" Body="&lt;p&gt;I'm trying to derive the integral form of the Bessel function by finding&#xA;the $k$th coefficient of the Laurent series expansion of the function&#xA;$f(z) =\exp [\lambda(z-\frac{1}{z})]$.&#xA;I managed to get it down to the form&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$&#xD;&#xA;J_k(\lambda) =\frac{1}{2\pi}\int_{0}^{2 \pi} e^{i(\lambda \sin\theta - k \theta)} d\theta&#xD;&#xA;=\frac{1}{2\pi}\int_{0}^{2 \pi} [\cos(\lambda \sin\theta - k \theta)&#xD;&#xA;+i\sin(\lambda \sin\theta - k \theta)]d\theta&#xD;&#xA;$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, I need to show that this is equivalent to&#xA;$&#xD;&#xA;\frac{1}{2\pi}\int_{0}^{2 \pi} \cos(\lambda \sin\theta - k \theta)d\theta&#xD;&#xA;$.&#xA;In other words, I need to show that&#xA;$&#xD;&#xA;\int_{0}^{2 \pi}\sin(\lambda \sin\theta - k \theta)d\theta =0&#xD;&#xA;$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I can't figure out how to do this. I tried expanding using trig identities and then&#xA;writing sin and cos as Taylor Series and integrating term by term, but no luck. What am I missing?&lt;/p&gt;&#xA;" OwnerUserId="2029" LastActivityDate="2010-11-18T17:54:25.187" Title="Deriving Bessel's Integrals" Tags="&lt;homework&gt;&lt;complex-analysis&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="10811" PostTypeId="2" ParentId="10810" CreationDate="2010-11-18T04:59:45.637" Score="1" ViewCount="" Body="&lt;p&gt;Split the integral involving $\sin$ as $\int_0^{\pi}$ + $\int_{\pi}^{2\pi}$. Change variable in the second integral as $\phi = \theta - \pi$. Do some trigonometric manipulations like $\sin(\theta + n \pi) = (-1)^n \sin(\theta)$ and you will find that the second integral turns out to be negative of the first.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2010-11-18T17:54:25.187" LastActivityDate="2010-11-18T17:54:25.187" CommentCount="3" />
  <row Id="10812" PostTypeId="2" ParentId="10810" CreationDate="2010-11-18T05:01:17.787" Score="3" ViewCount="" Body="&lt;p&gt;Use $\displaystyle \int_{a}^{b} f(x)\ dx = \int_{a}^{b} f(a+b-x) \ dx$ &lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-11-18T05:01:17.787" />
  <row Id="10813" PostTypeId="2" ParentId="10685" CreationDate="2010-11-18T05:12:56.353" Score="3" ViewCount="" Body="&lt;p&gt;Unless I'm mistaken, you can add a sufficiently large &quot;cyclic ladder&quot; to the Frucht graph as in: &lt;img src=&quot;http://i.imgur.com/ZreVp.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would argue that, in an automorphism, the right-hand side would need to be mapped to itself.  Therefore, an automorphism of the above graph would imply an automorphism of the Frucht graph (and therefore must be trivial).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An infinite version could be constructed by appending two infinite ladders instead (actually, I'm pretty sure you could attach one infinite ladder).&lt;/p&gt;&#xA;" OwnerUserId="139" LastActivityDate="2010-11-18T05:12:56.353" CommentCount="3" />
  <row Id="10814" PostTypeId="2" ParentId="10760" CreationDate="2010-11-18T05:22:39.007" Score="1" ViewCount="" Body="&lt;p&gt;I wrote a &lt;a href=&quot;http://math.uga.edu/~pete/coveringnumbersv2.pdf&quot; rel=&quot;nofollow&quot;&gt;short note&lt;/a&gt; on exactly this problem.  It will appear in the Monthly one of these years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note in particular that your quoted statement is not necessarily true for infinite-dimensional vector spaces: an infinite-dimensional vector space over any field can be covered by $\aleph_0$-many proper subspaces.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You ask also about modules over rings.  Yes, there has been work on that.  I believe that Apoorva Khare, a mathematician who independently proved most of the results in my note, also has some work on the case of modules over various rings.  If you google for &quot;covering numbers&quot;, you'll find that many, many papers on this topic have been written.&lt;/p&gt;&#xA;" OwnerUserId="299" LastActivityDate="2010-11-18T05:22:39.007" CommentCount="2" />
  <row Id="10816" PostTypeId="2" ParentId="10780" CreationDate="2010-11-18T05:32:17.310" Score="2" ViewCount="" Body="&lt;p&gt;I found the following theorems  from the book &quot;Introduction to classical real analysis&quot; by Karl R. Stromberg, 1981.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;(Zygmund) If $f$ satisfies a Holder (also called Lipschitz) condition of order $\alpha\gt 0$ and $f$ is of bounded variation on $[0,2\pi]$, then the Fourier series of $f$ converges absolutely (and hence uniformly).  p. 521.&lt;br/&gt;&#xA;This applies to the example in my  question.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If $f$ is absolutely continuous, then the Fourier series of $f$ converges uniformly  but not necessarily absolutely.  p. 519 Exercise 6(d) and p.520 Exercise 7c.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;(Bernstein) If $f$ satisfies a Holder condition of order $\alpha\gt 1/2$ , then the  Fourier series of $f$ converges absolutely (and hence uniformly).  p.520 Exercise  8 (f)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;(Hille) For each $0&amp;lt;\alpha\le 1/2$, there exists a function that satisfies a Holder condition of order $\alpha$ whose Fourier series converges uniformly, but not absolutely.  p.520 Exercise  8 (f)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="3249" LastActivityDate="2010-11-18T05:32:17.310" />
  <row Id="10817" PostTypeId="1" AcceptedAnswerId="10887" CreationDate="2010-11-18T06:39:57.267" Score="1" ViewCount="236" Body="&lt;p&gt;On p.28 (&lt;a href=&quot;http://books.google.com/books?id=KvKdSkPwwmMC&amp;amp;lpg=PP1&amp;amp;dq=reid%20undergraduate%20algebraic%20geometry&amp;amp;pg=PA28#v=onepage&amp;amp;q=no%20rational%20parametrisation&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;) Reid proves that the curve $y^2=x(x-1)(x-\lambda)$ for $\lambda\neq 0,1$ has no rational paramaterisation. At one point (on p.29), he has the equation $$r^2=ap(p-q)(p-\lambda q)$$ where $p$ and $q$ are polynomials over a field $k$, and $a$, $\lambda$ are constants in the field. He then says that &quot;by considering factorisation into primes, there exist nonzero constants $b,c,d\in k$ such that $$bp,c(p-q),d(p-\lambda q)$$ are all squares in $k[t]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; $p$ and $q$ are known to be coprime.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I feel like this should be obvious, but I don't follow his argument. Why do we know that those factors are squares (up to units)?&lt;/p&gt;&#xA;" OwnerUserId="3656" LastEditorUserId="3656" LastEditorDisplayName="" LastEditDate="2010-11-18T06:49:38.307" LastActivityDate="2010-11-18T20:08:28.730" Title="Proof from Reid's Undergraduate Algebraic Geometry: help with minor details" Tags="&lt;homework&gt;&lt;algebraic-geometry&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" />
  <row Id="10818" PostTypeId="1" AcceptedAnswerId="10864" CreationDate="2010-11-18T07:38:45.867" Score="2" ViewCount="150" Body="&lt;p&gt;If $A$ is a positive definite matrix, then the solution to the minimization problem $(1/2)x^TAx - b^Tx$ is given by $A^{-1}b$.  I'm interested in the generalization of this to a Hilbert space.  What conditions on $A$ should be required for this to be true in a Hilbert space?  Also is there a reference that includes this? I searched Google books, but couldn't find it. &lt;/p&gt;&#xA;" OwnerUserId="145" LastEditorUserId="145" LastEditorDisplayName="" LastEditDate="2010-11-18T21:05:36.203" LastActivityDate="2010-11-18T21:05:36.203" Title="Quadratic minimization in a Hilbert space" Tags="&lt;functional-analysis&gt;&lt;optimization&gt;&lt;hilbert-spaces&gt;&lt;calculus-of-variations&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="10819" PostTypeId="1" AcceptedAnswerId="10844" CreationDate="2010-11-18T09:21:26.283" Score="1" ViewCount="267" Body="&lt;p&gt;A rectangle can be cut into two connected pieces of equal area so that they can be rearranged to get a new rectangle with different side lengths than the original one. Show that this can be done in countable many ways. Can this be done in uncountable many ways?&lt;/p&gt;&#xA;" OwnerUserId="2856" LastEditorUserId="2159" LastEditorDisplayName="" LastEditDate="2010-11-18T17:06:47.193" LastActivityDate="2010-11-18T17:06:47.193" Title="Splitting rectangle into two parts" Tags="&lt;geometry&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="10820" PostTypeId="1" AcceptedAnswerId="10822" CreationDate="2010-11-18T09:36:20.920" Score="1" ViewCount="66" Body="&lt;p&gt;I have a unit sphere, on the surface of which are defined spherical caps. I typically characterize the caps by the unit vector $n$ from the center of the sphere to the top of the cap, and the angle $\theta$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is: given a pair of spherical caps, how can I determine whether they intersect?&lt;/p&gt;&#xA;" OwnerUserId="3406" LastActivityDate="2010-11-18T10:12:08.593" Title="How to test whether spherical caps intersect?" Tags="&lt;spherical-geometry&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="10821" PostTypeId="1" AcceptedAnswerId="10823" CreationDate="2010-11-18T09:50:18.043" Score="0" ViewCount="94" Body="&lt;p&gt;Today, I read these two new Logarithmic identity $\displaystyle$ $$a^{\log_a m} = m $$ $$\log_{a^q}{m^p} = \frac{p}{q} \log_a m$$ Both of them seems new to me,so even after solving some problems (directly) based on thesm I haven't fully understood how they holds good,Could anybody show me how to prove them ?&lt;/p&gt;&#xA;" OwnerUserId="2109" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-18T10:05:36.137" LastActivityDate="2010-11-18T10:28:26.197" Title="Proving Logarithmic identity" Tags="&lt;algebra-precalculus&gt;&lt;logarithms&gt;" AnswerCount="1" CommentCount="10" />
  <row Id="10822" PostTypeId="2" ParentId="10820" CreationDate="2010-11-18T10:12:08.593" Score="5" ViewCount="" Body="&lt;p&gt;The angle between $n_1$ and $n_2$ is the analogue of the &quot;distance&quot; between the centres of the two caps. The caps intersect if this angle is less than the sum of $\theta_1$ and $\theta_2$, assuming the $\theta_i$ play the role of &quot;radius&quot; and not &quot;diameter&quot; in your notation.&lt;/p&gt;&#xA;" OwnerUserId="856" LastActivityDate="2010-11-18T10:12:08.593" />
  <row Id="10823" PostTypeId="2" ParentId="10821" CreationDate="2010-11-18T10:21:16.420" Score="3" ViewCount="" Body="&lt;p&gt;Just so this doesn't remain unanswered:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;This is a statement of the fact that the functions $a^x$ and $\log_a\;x$ are inverses of each other; thus, $\log_a\;a^x=a^{\log_a\;x}=x$&#xD;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Letting c be a positive real number not equal to one:&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;$\log_{a^q}\;m^p=\frac{\log_c\;m^p}{\log_c\;a^q}=\frac{p\;\log_c\;m}{q\;\log_c\;a}=\frac{p}{q}\log_a\;m$&lt;/p&gt;&#xA;" OwnerUserId="498" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-18T10:28:26.197" LastActivityDate="2010-11-18T10:28:26.197" />
  <row Id="10824" PostTypeId="1" CreationDate="2010-11-18T10:48:14.147" Score="0" ViewCount="175" Body="&lt;p&gt;The &lt;a href=&quot;http://www.math-help-ace.com/Logarithm-Tables.html&quot; rel=&quot;nofollow&quot;&gt;rule&lt;/a&gt;: &lt;strong&gt;&quot;Characteristic of a number($\lt$ 1) with n zeros after decimal and before a significant digit  = $\overline{( n + 1 )}$&quot;&lt;/strong&gt; This rule is mentioned in my module,Goggle lead me to this &lt;a href=&quot;http://www.math-help-ace.com/Logarithm-Tables.html&quot; rel=&quot;nofollow&quot;&gt;link&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I am trying to verify with some random numbers like &lt;code&gt;Log[10, 0.0004356]&lt;/code&gt; its giving me $-3.36091$ and &lt;code&gt;Log[10, 0.725]&lt;/code&gt; gives me $-0.139662$. Giving $-3$ and $-0$ as characteristic respectivley, however shouldn't the characteristic be $-4$ and $-1$ respectively,consistent to the rule ? what exactly I am missing here ?&lt;/p&gt;&#xA;" OwnerUserId="2109" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-18T15:51:20.280" LastActivityDate="2011-01-24T19:30:01.713" Title="Characteristic of Logarithms of positive numbers $\lt 1$" Tags="&lt;logarithms&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="10825" PostTypeId="1" CreationDate="2010-11-18T10:48:36.927" Score="1" ViewCount="97" Body="&lt;p&gt;Suppose that K is a finite extension over $\mathbb{Q}$. Let $p$ be a prime in $\mathbb{Q}$, let $p\mathcal{O}_K=\mathfrak{P}_1\ldots\mathfrak{P}_n$ be its prime decomposition in $\mathcal{O}_K$, then is it true that any prime ideal in $\mathcal{O}_K$ above $p$ is a factor in the decomposition of $p$?&lt;/p&gt;&#xA;" LastActivityDate="2010-11-18T14:06:36.230" Title="prime decomposition in galois extensions" Tags="&lt;algebraic-number-theory&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="2" />
  <row Id="10826" PostTypeId="1" CreationDate="2010-11-18T10:49:13.427" Score="2" ViewCount="103" Body="&lt;p&gt;I have $n$ random elements from 0 to $k$. What is the probability that there is some element which appears more than $n/2$ times (or $n/2$ times).&lt;/p&gt;&#xA;" OwnerUserId="3317" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2010-11-19T05:58:08.070" LastActivityDate="2010-11-19T05:58:08.070" Title="Probability of having a majority element" Tags="&lt;probability&gt;&lt;probability-theory&gt;" AnswerCount="2" FavoriteCount="0" />
  <row Id="10827" PostTypeId="1" CreationDate="2010-11-18T11:10:13.843" Score="0" ViewCount="68" Body="&lt;p&gt;Suppose that q is a rational prime, and K is a p-extension over Q where p is prime, then q ramifies in K if and only if q is p or q is congruent to 1 mod p. I don't know why this is true, can someone convince me of this?&lt;/p&gt;&#xA;" LastActivityDate="2010-11-19T12:49:47.727" Title="ramification in p-extensions" Tags="&lt;algebraic-number-theory&gt;" AnswerCount="3" FavoriteCount="1" />
  <row Id="10828" PostTypeId="2" ParentId="10827" CreationDate="2010-11-18T11:21:59.320" Score="2" ViewCount="" Body="&lt;p&gt;This can't possibly be true: by a &lt;a href=&quot;http://en.wikipedia.org/wiki/Dirichlet%27s_theorem_on_arithmetic_progressions&quot; rel=&quot;nofollow&quot;&gt;famous theorem of Dirichlet&lt;/a&gt;, there are infinitely many prime numbers $q$ such that $q \equiv 1 \pmod p$.  However, only finitely many &#xA;primes ramify in any finite extension $K/\mathbb{Q}$: those which divide the discriminant.  &lt;/p&gt;&#xA;" OwnerUserId="299" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2010-11-18T11:49:00.020" LastActivityDate="2010-11-18T11:49:00.020" />
  <row Id="10829" PostTypeId="2" ParentId="10825" CreationDate="2010-11-18T11:31:05.190" Score="2" ViewCount="" Body="&lt;p&gt;Yes.  I suppose that the definition of a prime ideal $\mathcal{P}$ being &quot;above $p$&quot; you are using is $\mathcal{P} \cap \mathbb{Z} = (p)$.  From this it is clear that if &#xA;$\mathcal{P}$ lies above $p$, then it contains $p\mathbb{Z}$, and -- since it is an ideal -- hence also $p \mathcal{O}_K$.  But recall that in a Dedekind domain &quot;to contain is to divide&quot;, hence $\mathcal{P}$ divides $p\mathcal{O}_K$ and appears in the (unique!) prime factorization of $\mathcal{O}_K$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that this does not require that $K/\mathbb{Q}$ be Galois (as appears in the title but not the body of the question), and it has nothing to do with algebraic number fields &lt;em&gt;per se&lt;/em&gt;: in general, for a finite extension of Dedekind domains $S/R$, the prime ideals of $S$ lying over a given prime $\mathfrak{p}$ of $R$ are exactly the ones appearing in &#xA;the factorization of $\mathfrak{p}S$.  &lt;/p&gt;&#xA;" OwnerUserId="299" LastActivityDate="2010-11-18T11:31:05.190" CommentCount="2" />
  <row Id="10830" PostTypeId="2" ParentId="10774" CreationDate="2010-11-18T11:40:31.950" Score="2" ViewCount="" Body="&lt;p&gt;One method is to apply the &lt;a href=&quot;http://planetmath.org/encyclopedia/BargmannTransform.html&quot; rel=&quot;nofollow&quot;&gt;Bargmann transform:&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$B_f(z) = \int_{-\infty}^{\infty}f(x)exp(2 \pi x z - \pi x^2 - \frac{\pi}{2}z^2) dy$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;to the eigenvalue equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f(x) = \lambda \int_{-\infty}^{\infty}f(y)exp(-2 \pi ix y) dy$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;to obtain the following relation for the Bargmann transformed eigenfunctions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$B_f(z) = \lambda B_f(-iz)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;whose solutions are the monomials:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$B_f(z) = z^n$, corresponding to the eigenvalues $i^{n_{mod4}}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the inverse Bargmann transform of the monomials are just the Hermite functions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remark: The application of the Bargmann transform to the eigenvalue equation entails &#xA;a change of the integration order which is possible because the Hermite functions are bounded.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The projection operators onto the n-th subspace in the Bargamann &#xA;representation are given by the kernels:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$P(v,\bar{z})=\frac{(v\bar{z})^n}{n!}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which act on the Bargmann transformed functions according to:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$Pf(v) = \int_{\mathbb{C}}P(v,\bar{z}) f(z) exp(-z\bar{z})dzd\bar{z}$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the time domain the projection kernels are just $P(t, \tau) = H_n(\tau) H_n(t)$ due to &#xA;the orthonormality of the Hermite funtions.&lt;/p&gt;&#xA;" OwnerUserId="1430" LastEditorUserId="1430" LastEditorDisplayName="" LastEditDate="2010-11-18T12:33:56.453" LastActivityDate="2010-11-18T12:33:56.453" CommentCount="2" />
  <row Id="10831" PostTypeId="2" ParentId="10827" CreationDate="2010-11-18T11:41:07.283" Score="2" ViewCount="" Body="&lt;p&gt;If $K/\mathbb{Q}$ is a degree $p$ extension of $\mathbb{Q}$&#xD;&#xA;then any particular prime can ramify: consider $K=\mathbb{Q}(q^{1/p})$.&#xA;Suppose that $K$ is a Galois and so cyclic degree $p$ extension.&#xA;Then only finitely many primes ramify in $\mathbb{Q}$ and they&#xA;can only have the form $q=p$ or $q\equiv1$ (mod $p$).&#xA;(I'm sure this is the result you are thinking of).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same is true for $p$-power degree Galois extensions&#xA;(consider the ramification groups at their localizations).&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-11-18T11:41:07.283" />
  <row Id="10832" PostTypeId="1" AcceptedAnswerId="10841" CreationDate="2010-11-18T11:42:19.020" Score="0" ViewCount="131" Body="&lt;p&gt;Hi&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was given a simple function, very simple. f(x) = bla bla. With a constant and an X. The point was to find for which values of a(the constant) the function is not one-to-one. The domain of the function is limited from zero and upwards. Here is what I put down on the paper: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As long as a is negative, it will not be one-to-one. If a = -2. Then f(0) = 4 and f(4)=4. Thereby concluding that it's not one-to-one in it's domain as long as a &amp;lt; 0. And stating the point of one-to-one is that we shouldn't be able to reach the same output with different input. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I didn't get full score on this, what am I missing? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;edit: f(x) = (x+a)^2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;edit: I've added it! I not able to comment on my own post :( &lt;/p&gt;&#xA;" OwnerUserId="2230" LastEditorUserId="2230" LastEditorDisplayName="" LastEditDate="2010-11-18T11:49:20.590" LastActivityDate="2010-11-18T15:47:20.520" Title="Calculus, given function. Say when it's not one-to-one? (What did i miss)" Tags="&lt;calculus&gt;&lt;homework&gt;" AnswerCount="3" CommentCount="5" />
  <row Id="10833" PostTypeId="2" ParentId="10832" CreationDate="2010-11-18T11:58:27.293" Score="0" ViewCount="" Body="&lt;p&gt;Try to draw its graphic for $a=0$: maybe you'll recognise something familiar, won't you? Then, try to figure out what happens when you put some $a\neq 0$ (and you'll see that this doesn't change anything at all about injectivity).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then you could try to think about this equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;(x_1 +a )^2 = (x_2 + a)^2 \ .&#xD;&#xA;$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again: start with the case $a=0$, then see what happens for $a \neq 0$ (nothing new).&lt;/p&gt;&#xA;" OwnerUserId="664" LastActivityDate="2010-11-18T11:58:27.293" CommentCount="2" />
  <row Id="10834" PostTypeId="2" ParentId="10693" CreationDate="2010-11-18T12:29:41.023" Score="1" ViewCount="" Body="&lt;p&gt;We will only prove here that definition (1) implies definition (2) (this is the easy part).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose that the condition in definition (1) holds. That is, $\mu_n ((a,b]) \to \mu ((a,b])$ for any $a,b \in D$, $a &amp;lt; b$, where $D$ is a dense subset of $R$. Define $\tilde D = \lbrace x \in R:\mu (\lbrace x \rbrace ) = 0 \rbrace$. Then $\tilde D$ is dense in $R$, since its complement is at most countable. For any $a,b \in \tilde D$, $a &amp;lt; b$, and $a&amp;#39;,a&amp;#39;&amp;#39;,b&amp;#39;,b&amp;#39;&amp;#39; \in D$, $a&amp;#39; &amp;lt; b&amp;#39;$, $a&amp;#39;&amp;#39; &amp;lt; b&amp;#39;&amp;#39;$, such that $a&amp;#39;&amp;#39; &amp;lt; a &amp;lt; a&amp;#39;$ and $b&amp;#39; &amp;lt; b &amp;lt; b&amp;#39;&amp;#39;$, we have&#xA;$$&#xD;&#xA;\mu _n ((a&amp;#39;,b&amp;#39;]) \le \mu _n ((a,b)) \le \mu _n ((a&amp;#39;&amp;#39;,b&amp;#39;&amp;#39;]).&#xD;&#xA;$$&#xD;&#xA;By assumption, $\mu _n ((a&amp;#39;,b&amp;#39;]) \to \mu ((a&amp;#39;,b&amp;#39;])$ and $\mu _n ((a&amp;#39;&amp;#39;,b&amp;#39;&amp;#39;] \to \mu ((a&amp;#39;&amp;#39;,b&amp;#39;&amp;#39;])$ as $n \to \infty$. Then, since $a&amp;#39;,a&amp;#39;&amp;#39;$ and $b&amp;#39;,b&amp;#39;&amp;#39;$ can be chosen arbitrarily close to $a$ and $b$, respectively, we can conclude that &#xA;$$&#xD;&#xA;\mu ((a,b)) \le \mathop {\lim \inf }\limits_{n \to \infty } \mu _n ((a,b)) \le \mathop {\lim \sup }\limits_{n \to \infty } \mu _n ((a,b)) \le \mu ([a,b]).&#xD;&#xA;$$&#xD;&#xA;However, by assumption $\mu (\lbrace a \rbrace) = \mu (\lbrace b \rbrace) = 0$, and hence $\lim _{n \to \infty } \mu _n ((a,b)) = \mu ((a,b))$. Thus, the condition in definition (2) is satisfied (with $\tilde D$ playing the role of $D$ in that definition).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is instructive to consider here the following simple example. As is customary, denote by $\delta_x$ the probability measure concentrated at $x$. Define $\mu_n = \delta_{-1/n}$, $n \geq 1$, and $\mu = \delta_0$. Then, for any $a,b \in R$, $a &amp;lt; b$, we have $\mu _n ((a,b]) \to \mu ((a,b])$, and hence, by definition (1), $\mu_n$ converges vaguely to $\mu$. On the other hand, $\mu _n ((-2,0)) = 1$ for any $n \geq 1$, whereas $\mu ((-2,0)) = 0$. However, if we define $\tilde D$ as above, then $\tilde D = R - \lbrace 0 \rbrace$, and for any $a,b \in \tilde D$, we do have $\mu _n ((a,b)) \to \mu ((a,b))$; hence, $\mu_n$ converges vaguely to $\mu$ also according to definition (2).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, as for the OP's remark (concerning definition (2)) that &quot;the original text just says converges, not mention that converges to $ \mu((a,b))$ which I guess can be added?&quot;, it is clear that we are deliberately only given that $\mu_n ((a,b))$ converges -- this is what makes the converse implication (from definition (2) to (1)) a substantially more difficult problem.&lt;/p&gt;&#xA;" OwnerUserId="2810" LastEditorUserId="2810" LastEditorDisplayName="" LastEditDate="2010-11-18T12:42:17.250" LastActivityDate="2010-11-18T12:42:17.250" CommentCount="3" />
  <row Id="10835" PostTypeId="1" CreationDate="2010-11-18T12:39:54.123" Score="3" ViewCount="94" Body="&lt;p&gt;Two numbers are equivalent if they have the same digits ex 2271 and 1227 are equivalent. How many equivalent classes are there between 1000 &amp;amp; 9999? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure I fully understand how this works, is the answer to this question (9*10*10*10)/4!? First calculating the total numbers formed and divindg by 4! to get rid of the repetitions.&lt;/p&gt;&#xA;" LastActivityDate="2010-11-18T17:36:00.353" Title="Combinations question" Tags="&lt;combinatorics&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="0" />
  <row Id="10836" PostTypeId="2" ParentId="10826" CreationDate="2010-11-18T12:51:33.573" Score="2" ViewCount="" Body="&lt;p&gt;This is related to the problem of finding the integral of a normal distribution, in short this means that there is no simple formula that can solve your problem, you will have to do it as a sum of probabilities for a number of simpler cases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The probability of a specific element appearing exactly $m$ times in a group of $n$ elements chosen randomly from a range of $k$ elements is: $$f(m, n, k) = \frac{(1/k)^m\ ((k-1)/k)^{n-m}\ n!}{m!\ (n-m)!}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the probability of a specific element appearing $m$ or more times is: $$\sum_{p=roundup(n/2)}^n f(p, n, k)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the case of any element appearing $m$ or more times things get tricky, I can't think of any simple way of taking the correlation between the numbers of different elements into account, but for your specific question less will do. You can simply multiply the result of the previous formula by $k$, but for even values of $n$ you will have to account for the special case where half of the elements is one value and half is another, since that will have been counted twice, so you will have to subtract that probability: $$\frac{k\ (k-1)\ (1/k)^n\ n!}{2\ (n/2)!^2}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that my $k$ is different from your $k$, since there are $k+1$ numbers from $0$ to $k$ you will have to add 1 to your $k$ before using it in the formulas.&lt;/p&gt;&#xA;" OwnerUserId="3584" LastActivityDate="2010-11-18T12:51:33.573" />
  <row Id="10838" PostTypeId="2" ParentId="10832" CreationDate="2010-11-18T12:54:03.700" Score="1" ViewCount="" Body="&lt;p&gt;The key here is that your domain is $[0,\infty)$. In your solution you gave a very loose reason why it is not one-to-one for $a &amp;lt; 0$, and did not say anything about the case where $a \geq 0$.&#xA;For $a &amp;lt; 0$, consider $f(0)$ and $f(-2a)$. Both are well-defined, as they are contained in the domain. Then&#xA;$$f(0) = (0 + a)^2 = a^2 = (-2a + a)^2 = f(-2a)$$&#xD;&#xA;Since $0 \neq -2a$ by assumption, $f$ is not one-to-one for $a &amp;lt; 0$.&#xA;To finish, you need to show that $f$ is one-to-one for $a \geq 0$. Hint: Derivatives should be useful here...&lt;/p&gt;&#xA;" OwnerUserId="1016" LastActivityDate="2010-11-18T12:54:03.700" />
  <row Id="10839" PostTypeId="1" AcceptedAnswerId="10849" CreationDate="2010-11-18T13:36:42.663" Score="3" ViewCount="206" Body="&lt;p&gt;This question is about &lt;a href=&quot;http://projecteuler.net/index.php?section=problems&amp;amp;id=113&quot; rel=&quot;nofollow&quot;&gt;Project Euler 113&lt;/a&gt;: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Working from left-to-right if no digit&#xA;  is exceeded by the digit to its left&#xA;  it is called an increasing number; for&#xA;  example, 134468.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Similarly if no digit is exceeded by&#xA;  the digit to its right it is called a&#xA;  decreasing number; for example, 66420.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;We shall call a positive integer that&#xA;  is neither increasing nor decreasing a&#xA;  &quot;bouncy&quot; number; for example, 155349.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;As n increases, the proportion of&#xA;  bouncy numbers below n increases such&#xA;  that there are only 12951 numbers&#xA;  below one-million that are not bouncy&#xA;  and only 277032 non-bouncy numbers&#xA;  below 10^10.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;How many numbers below a googol&#xA;  (10 pow 100) are not bouncy?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is obviously not something you can brute force. I found a &lt;a href=&quot;http://freelancersunite.net/project_euler/project-euler-problem-113/&quot; rel=&quot;nofollow&quot;&gt;mathematical solution&lt;/a&gt; that does the trick in no time, the problem is, I don't understand it! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can someone point me in the right direction? Not having a math background, I'm not even sure where to begin! &lt;/p&gt;&#xA;" OwnerUserId="3668" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-11-18T19:56:51.413" LastActivityDate="2010-11-18T19:56:51.413" Title="Combinatorial counting" Tags="&lt;combinatorics&gt;&lt;project-euler&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="10840" PostTypeId="2" ParentId="10825" CreationDate="2010-11-18T14:06:36.230" Score="0" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\rm\ \ P \supset \mathfrak{P}_1\ldots\mathfrak{P}_n\ \Rightarrow\ P \supset \mathfrak P_i\ $ for some $\rm\:i\:.\ $ This is true for any prime ideal $\rm\ P\ $ in any ring. Further $\rm\ P = \mathfrak P_i\ $ when the $\ \mathfrak P_i\ $ are maximal, e.g. in one dimensional domains, i.e. domains where nonzero prime ideals are maximal.&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-11-18T14:06:36.230" />
  <row Id="10841" PostTypeId="2" ParentId="10832" CreationDate="2010-11-18T14:12:08.827" Score="4" ViewCount="" Body="&lt;p&gt;There are two things you are missing: you are missing the case in which $a\geq 0$; and you did not really justify your assertion when $a\lt 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Technically, the only case you actually established was the case $a=-2$. That leaves quite a few values of $a$ untouched!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When one asserts that something happens &quot;for all numbers less than $0$&quot;, it is almost never good enough to check what happens in &lt;em&gt;one&lt;/em&gt; number less than $0$. That's called &quot;argument by example&quot;, and examples almost never establish general cases. It would be like trying to argue that &lt;em&gt;everyone&lt;/em&gt; in your class is going to fail by pointing to one student in the class who happens to be failing. It's certainly possible that all will fail, but just showing one person who will is not going to establish this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So: first, you need to finish off the case $a\lt 0$: if $a$ is &lt;strong&gt;any&lt;/strong&gt; number less than $0$, why is $f(x)=(x+a)^2$ not one-to-one on $[0,\infty)$? Because, &lt;em&gt;no matter what $a$ is&lt;/em&gt;, you can always find at least two distinct points in $[0,\infty)$ on which the function takes the same value: $x=0$, and $x=-2a$ (which is positive because $a\lt 0$, and so both in the domain and different from $0$). Indeed, $f(0)=(0+a)^2 = a^2$, and $f(-2a) = (-2a+a)^2 = (-a)^2 = a^2=f(0)$. You'll note that this takes care of &lt;em&gt;all&lt;/em&gt; the values of $a$ that are negative. (&lt;em&gt;Note:&lt;/em&gt; These are not the only points you can use; for instance, $\frac{-a}{2} and $\frac{-3a}{2}$ will also have equal images. But to show that something is &lt;em&gt;not&lt;/em&gt; one-to-one it &lt;em&gt;does&lt;/em&gt; suffice to give a single example of two distinct points that map to the same thing, because the assertion &quot;is not one-to-one&quot; means &quot;there is &lt;strong&gt;at least one&lt;/strong&gt; pair $a,b$, with $a\neq b$ and $f(a)=f(b)$). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, we also need to deal with the case $a\geq 0$. What happens then? If $a\geq 0$, then $x+a\geq 0$ (since $x\geq 0$). If $f(x)=f(y)$, then we have:&#xA;\begin{align*}&#xA;f(x) &amp;amp;= f(y)\\&#xA;(x+a)^2 &amp;amp;= (y+a)^2\\&#xA; \sqrt{(x+a)^2} &amp;amp;= \sqrt{(y+a)^2}\\&#xA;|x+a|&amp;amp;=|y+a|&#xA;\end{align*}&#xA;(remember that $\sqrt{r^2}=|r|$). Now, because $x$, $y$, and $a$ are all nonnegative, so are $x+a$ and $y+a$, so they equal their absolute value: $|x+a|=x+a$ and $|y+a|=y+a$. So we get $x+a=y+a$, and so we deduce $x=y$. That is, if $a\geq 0$, then for $f(x)$ to equal $f(y)$ we &lt;em&gt;must&lt;/em&gt; have $x=y$. This means that in this case, $f$ is one-to-one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In summary: you forgot to check $a\geq 0$. And you did not fully establish $a\lt 0$. It's not enough to just &quot;it is not one-to-one when $a\lt 0$&quot;, because this is only says what happens if $a\lt 0$; your answer is not presumed to be exhaustive (that is, we don't assume that you are telling us &lt;em&gt;all&lt;/em&gt; the values in which it is not one-to-one) because you are not saying it is exhaustive (to do so, you would have to say &quot;it is not one-to-one &lt;em&gt;if and only if&lt;/em&gt; $a\lt 0$&quot;, and even then you would have to discuss what happens when $a\geq 0$).&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-18T15:47:20.520" LastActivityDate="2010-11-18T15:47:20.520" />
  <row Id="10842" PostTypeId="2" ParentId="10839" CreationDate="2010-11-18T14:23:23.930" Score="0" ViewCount="" Body="&lt;p&gt;First, you are trying to count increasing and decreasing numbers.  They are related in that you only reverse the digits of one to get the other.  The only difference is that increasing numbers cannot start with 0, but decreasing ones can end with 0.  So let's just work on counting increasing ones.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How many one digit increasing numbers are there? Nine.  How many one digit increasing numbers are there that end in 1? One.  So make a function f(x,n) where f(x,n) is the number of n digit increasing numbers that end in x, you should be able to make a recurrence for each one.  That is, (6,n)=some function of the f(x,n-1).  The increasing numbers under a googol are just the sum of f(x,100) over x.  Then you do the same for decreasing numbers but have to add f(0,n) to the mix.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I didn't look at the program you cite to see if it works anything like this.  But this approach worked for me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are many counting problems that yield to techniques like this.  Instead of keeping a list of all the increasing numbers (which would grow too big), just keep track of how many there are of a given size and find a recurrence to get to the next size up.  The next three problems in the list can be attacked this way as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is another approach.  Think of a series of 101 zeros.  Then all the increasing numbers can be constructed by putting up to nine &quot;step up&quot; marks between the digits, with one of them required to be between the first and second zeros (to avoid leading zeros).  You can even make it exactly nine steps if you allow some to be at the end.  I'll let somebody else make this more explicit&lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2010-11-18T15:17:31.473" LastActivityDate="2010-11-18T15:17:31.473" />
  <row Id="10843" PostTypeId="2" ParentId="10835" CreationDate="2010-11-18T14:33:05.557" Score="0" ViewCount="" Body="&lt;p&gt;One approach is to work by cases of the number of matching digits.  If no digits match and none is zero, there are in fact 4!=24 permutations.  So this accounts for 9*8*7*6/24 equivalence classes.  If no digits match and one is zero, the zero cannot be the first digit, so there are only 3*3*2*1=18 permutations and 1*9*8*7/18 equivalence classes.  If you have one pair of matching digits and no zero, there are how many permutations and how many equivalence classes?&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-11-18T14:33:05.557" />
  <row Id="10844" PostTypeId="2" ParentId="10819" CreationDate="2010-11-18T15:13:40.633" Score="2" ViewCount="" Body="&lt;p&gt;Let n &amp;ge; 1 be an integer.  Divide the width of an H-by-W rectangle into n+1 equal parts, and the height into n equal parts.  Separate the rectangle into two connected and congruent pieces along a path stepped as necessary from a side divided at the 1/(n+1) part of its length, down a distance of 1/n of its height, and so on alternating until the opposite side of the rectangle is reached.  The two parts may then be &quot;slid&quot; across one another to form a rectangle of height (n+1)/n H and width n/(n+1) W.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That gives a countable number of ways to rearrange the rectangle, at most one of which could be congruent to the original (i.e. when W = (n+1)/n H for some n, or conversely).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe these are the only such dissections of a rectangle into two pieces, and if so there are not uncountably many ways to do it (though of course the roles of height and width can be exchanged).  The locations of the four original corners (right angles) are constrained so that they can only appear on the boundary of the rearranged rectangle, and the rearrangement works both ways.  The dissection must thus operate so as to separate the boundary of both rectangles into two parts (because an intact boundary of a rectangle cannot be encompassed by a different rectangle of equal area).  A bit of case analysis should then show that the two rectangles share a pair of opposite corners and that the path dividing the pieces must enter and exit at the &quot;new&quot; pair of corners.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;regards, hm&lt;/p&gt;&#xA;" OwnerUserId="3111" LastActivityDate="2010-11-18T15:13:40.633" CommentCount="1" />
  <row Id="10845" PostTypeId="1" AcceptedAnswerId="10847" CreationDate="2010-11-18T15:22:52.667" Score="1" ViewCount="82" Body="&lt;p&gt;Original equation: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$g(x) = \frac{(x+1)(x^2+2)(x^3+3)}{\sqrt{x^4+3}}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I take ln on both sides, and than differentiate I get this:   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{1}{g(x)} = \frac{1}{x+1}+\frac{1}{x^2+2}+\frac{1}{x^3+3}-\frac{1}{2}* \frac{1}{x^4+3} $&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this good so far? I don't know what to do about the left side?&lt;/p&gt;&#xA;" OwnerUserId="2230" LastActivityDate="2010-11-18T15:27:05.847" Title="Calculus, differentiation, but first taking ln on both sides?" Tags="&lt;calculus&gt;&lt;homework&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="10846" PostTypeId="1" AcceptedAnswerId="10871" CreationDate="2010-11-18T15:26:29.500" Score="5" ViewCount="242" Body="&lt;p&gt;What is an example of a  &lt;strong&gt;continuous&lt;/strong&gt;, or even better, differentiable, $2\pi$ (or 1) periodic function whose Fourier series converges pointwise but not uniformly? (Such function cannot be of Holder class, or absolutely continuous.) &lt;/p&gt;&#xA;" OwnerUserId="3249" LastEditorUserId="3249" LastEditorDisplayName="" LastEditDate="2010-11-18T19:56:47.777" LastActivityDate="2010-11-18T19:56:47.777" Title="Pointwise but not uniform convergence of a Fourier series" Tags="&lt;analysis&gt;&lt;fourier-analysis&gt;&lt;fourier-series&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="10847" PostTypeId="2" ParentId="10845" CreationDate="2010-11-18T15:27:05.847" Score="4" ViewCount="" Body="&lt;p&gt;Close, but you forgot to apply the chain rule!&lt;/p&gt;&#xA;" OwnerUserId="1306" LastActivityDate="2010-11-18T15:27:05.847" CommentCount="4" />
  <row Id="10848" PostTypeId="1" AcceptedAnswerId="10854" CreationDate="2010-11-18T15:39:02.120" Score="5" ViewCount="372" Body="&lt;p&gt;Under the isomorphism of Hilbert spaces $L^2(S^1)\to\ell^2(\mathbb Z),\quad e^{2\pi i n t}\mapsto e_n$, smooth functions on the circle are mapped to rapidly decaying sequences (see &lt;a href=&quot;http://en.wikipedia.org/wiki/Fourier_series#Properties&quot; rel=&quot;nofollow&quot;&gt;wikipedia&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is the converse also true? That is, does every rapidly decaying sequence in $\ell^2(\mathbb Z)$ represent a smooth function?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Motivation: For certain arguments, it would be really nice to read off smoothness from the Fourier coefficients.&lt;/p&gt;&#xA;" OwnerUserId="367" LastEditorUserId="367" LastEditorDisplayName="" LastEditDate="2010-11-18T15:47:47.700" LastActivityDate="2011-03-22T06:33:44.597" Title="Does rapid decay of Fourier coefficients imply smoothness?" Tags="&lt;fourier-analysis&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="4" />
  <row Id="10849" PostTypeId="2" ParentId="10839" CreationDate="2010-11-18T15:49:53.917" Score="7" ViewCount="" Body="&lt;p&gt;Suppose you wanted to find the number of decreasing numbers of exactly $\displaystyle k$ digits.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now if you knew that the digits were distinct, you chould choose k distinct digits and sort them, giving you 10 choose k  = $\displaystyle {10 \choose k}$ numbers. If the number was in base $n$, the corresponding number would be $\displaystyle {n \choose k}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now given a non-decreasing number like&lt;/p&gt;&#xA;&#xA;&lt;p&gt;5543220&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can construct a &quot;sorted&quot; number with distinct digits in a &lt;em&gt;different&lt;/em&gt; base, by adding 0 to the rightmost digit, adding 1 to the digit to the left of it, 2 to the digit to the left of that, and so on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;5543220 becomes BA86430 in base 10 + 6 = 16.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A decreasing number with $\displaystyle k$ digits correponds in a 1-1 fashion to a sorted number of $k$ digits, in base $\displaystyle 10+k-1 = 9+k$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As we saw earlier the number of &quot;sorted&quot; numbers of $\displaystyle k$ digits in base $\displaystyle n$ is $\displaystyle {n \choose k}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus the number of decreasing numbers is $\displaystyle {9+k \choose k}  = {9+k \choose 9}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the total number of decreasing numbers $\displaystyle \lt 10^{100}$ is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle \sum_{k=1}^{99} {9+k \choose 9}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using $\displaystyle {n \choose r} - {n-1 \choose r} = {n-1 \choose r-1}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the above becomes&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle \sum_{k=1}^{99} {10+k \choose 10} - {9+k \choose 10}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is a telescoping sum and becomes&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle  = {10+99 \choose 10} - {9+1 \choose 10} = {109 \choose 10} - 1$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do something similar for the increasing numbers.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-11-18T15:49:53.917" CommentCount="5" />
  <row Id="10850" PostTypeId="1" AcceptedAnswerId="10852" CreationDate="2010-11-18T15:49:56.607" Score="0" ViewCount="112" Body="&lt;p&gt;I am facing some trouble in solving this equation : $$\displaystyle3 \cdot x^{\log_5 2} + 2^{log_5 x} = 64$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Give me some hints to proceed on this.&lt;/p&gt;&#xA;" OwnerUserId="2109" LastActivityDate="2010-11-18T16:38:16.850" Title="Solving an equation" Tags="&lt;algebra-precalculus&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="10851" PostTypeId="2" ParentId="10824" CreationDate="2010-11-18T15:54:12.273" Score="3" ViewCount="" Body="&lt;p&gt;Read carefully: you take the characteristic as $-(n+1)$, but then you &lt;strong&gt;add a positive number&lt;/strong&gt; (the mantissa); so the actual number you will get will be $-(n+1)+c$, with $c\gt 0$, and so you will get a number $\ell$ that satisfies&#xA;$$-(n+1) \leq \ell \lt -n.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in your example $0.0004356$, you start with a characteristic of $-4$, but then you add a positive number to get a number strictly larger than $-4$, but less than $-3$. With $0.725$, you start with $-1$, then add a positive number to get a number strictly larger than $-1$ but less than $0$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Added:&lt;/em&gt; Basically, what you are doing is taking your number and multiplying by a sufficiently large power of $10$ so that you &lt;em&gt;just&lt;/em&gt; get it to $1$ or over. If the first significant digit occurs in position $n+1$ after $n$ zeroes, then $10^{n+1}x$ will &quot;shift&quot; the decimal point $n+1$ spaces to the right, so you will get a number of the form $d_{n+1}.d_{n+2}d_{n+3}d_{n+4}\ldots$ (where your original number was $x = 0.0\ldots0d_{n+1}d_{n+2}d_{n+3}\ldots$, with $n$ zeroes after the decimal). You already know how to figure out these logarithms: they have characteristic $0$, and a certain mantissa $m$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what you get then is that $\log_{10}(10^{n+1}x) = 0.m$. But by the properties of the logarithm, you know that &#xA;$$0.m = \log_{10}(10^{n+1}x) = \log_{10}(10^{n+1}) + \log_{10}(x) = (n+1) + \log_{10}(x).$$ &#xA;Since $\log_{10}(x)$ is what you &lt;em&gt;really&lt;/em&gt; want to know, that tells you that &#xA;$$\log_{10}(x) = -(n+1)+0.m$$&#xD;&#xA;So, we take $-(n+1)$ as the &quot;characteristic&quot;, and then add the &lt;em&gt;positive&lt;/em&gt; number $m$. So we get $-(n+1)\leq \log_{10}(x) \lt -n$.You get equality if and only if $m=0$, which happens if and only if $d_{n+2}=d_{n+3}=\cdots = 0$, which is exactly when $x=10^{-(n+1)}$. Otherwise, you get a number strictly between $-(n+1)$ and $-n$. But these numbers, when written in decimal, have $-n$ before the decimal, not $-(n+1)$. That's why you are getting $-3$ and $0$, and not $-4$ and $-1$, in your examples.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-18T21:22:51.663" LastActivityDate="2010-11-18T21:22:51.663" />
  <row Id="10852" PostTypeId="2" ParentId="10850" CreationDate="2010-11-18T16:03:38.923" Score="6" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\rm\ \ x^{\: log_5 2}\ =\ 5^{\: log_5 x\ log_5 2}\ =\ 2^{\: log_5 x}\ \ $  (or take $\rm\:log_5\:$ of both sides if that's clearer to you)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence the equation reduces to $\rm\ 2^{\: \log_5 x}\ =\ 2^4\ \Rightarrow\ log_5\: x\ =\ 4\ \Rightarrow\ x\ =\ \ldots$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;TIP&lt;/strong&gt; $\ $ Typically such exponential equations will be solvable in closed form only if the exponentials are all linearly dependent, so you should always check for that first.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-11-18T16:38:16.850" LastActivityDate="2010-11-18T16:38:16.850" CommentCount="5" />
  <row Id="10853" PostTypeId="2" ParentId="10826" CreationDate="2010-11-18T16:05:45.890" Score="1" ViewCount="" Body="&lt;p&gt;If we interpret the $n$ random elements from $0$ to $k$ to mean that each different selection is equally likely, then this question does have a very nice closed form answer: see (1) below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To be clear, say we are selecting two objects from the set $ \lbrace 0,1 \rbrace $ then by each selection being equally likely we mean $ \lbrace 00 \rbrace, $ $ \lbrace 01 \rbrace $ and&#xA;$ \lbrace 11 \rbrace $ each have probability $1/3.$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in the case where $k=1$ and $n=2,$ the probability that some object appears&#xA;$ &amp;gt; \lceil n/2 \rceil $ i.e. $2$ times, is $2/3.$ This is given by $(1)$ below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, when $k=3$ and $n=3$ we're looking at the probability that some selection contains $3$ objects all the same. This can be seen to be $1/5$ by noting that if we select from&#xA;$ \lbrace 0,1,2,3 \rbrace $ then we have $4$ possibilities of the form&#xA;$ \lbrace 000 \rbrace $ (all the same), $12$ of type $ \lbrace 001 \rbrace $ (two the same and one of another) and $4$ of type $ \lbrace 012 \rbrace $ (all different).  As a check, $(1)$ gives the probability as $1/5.$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So here's the derivation of $(1).$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The total number of ways we can place $n$ identical objects into $k+1$ boxes is&#xA;${n+k \choose k}.$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So now we need to find the number of ways of placing the $n$ identical objects into the $k+1$ boxes where one box contains $ &amp;gt; \lceil n/2 \rceil $ objects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose there are $ \lceil n/2 \rceil + i $ objects in one box,&#xA;$ 1 \le i \le \lfloor n/2 \rfloor,$ then the other $ \lfloor n/2 \rfloor - i $ objects can be distributed amongst the other $k$ boxes in&#xA;$$ { \lfloor n/2 \rfloor - i + k -1 \choose k – 1 }$$&#xD;&#xA;ways.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence the total number of ways that one box can contain $ &amp;gt; \lceil n/2 \rceil $ objects is&#xA;$$ (k+1) \sum_{i=1}^{ \lfloor n/2 \rfloor}{ \lfloor n/2 \rfloor - i + k -1 \choose k – 1 }$$&#xD;&#xA;$$ = \frac {k+1}{k} (\lfloor n/2 \rfloor + k -1){ \lfloor n/2 \rfloor + k - 2 \choose k – 1 }.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence the probability that one box contains $ &amp;gt; \lceil n/2 \rceil $ objects is given by&#xA;( $ k \ge 1, n \ge 1 $ )&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \frac {k+1}{k} \frac{ (\lfloor n/2 \rfloor + k -1){ \lfloor n/2 \rfloor + k - 2 \choose k – 1 } }&#xD;&#xA;{ {n+k \choose k} }. \qquad (1)$$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="1301" LastActivityDate="2010-11-18T16:05:45.890" CommentCount="3" />
  <row Id="10854" PostTypeId="2" ParentId="10848" CreationDate="2010-11-18T16:06:52.880" Score="6" ViewCount="" Body="&lt;p&gt;Yes. Smoothness is equivalent to the Fourier coefficients forming a sequence that decays rapidly (faster than any polynomial). To see the direction you asked about, note that if $\{c_n\}$ is a rapidly decaying sequence, then the sum $\sum c_n e^{in x}$ will converge uniformly as will every derivative. So the sum will represent a  smooth function. (The continuous analog of this is that the Fourier transform induces an isomorphism of the &lt;a href=&quot;https://secure.wikimedia.org/wikipedia/en/wiki/Schwartz_space&quot; rel=&quot;nofollow&quot;&gt;Schwartz space&lt;/a&gt; onto itself.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, one can use the Fourier coefficients to give an $L^2$ definition of differentiability. Morally, a function ought to be $C^k$ if its Fourier coefficients decay like $|n|^{-k}$. This is not quite accurate, but one can use this to define the &lt;a href=&quot;https://secure.wikimedia.org/wikipedia/en/wiki/Sobolev_space&quot; rel=&quot;nofollow&quot;&gt;Sobolev spaces&lt;/a&gt;&#xA;on the circle (more generally on the torus, or using the Fourier transform on euclidean space, and by an extension process on compact manifolds) that describe how many &quot;weak&quot; derivatives a function has. This sort of argument shows that $L^2$ differentiability is roughly comparable with normal differentiability, and using such Sobolev spaces turns out to be integral when one tries to prove explicit bounds about things such as &lt;a href=&quot;https://secure.wikimedia.org/wikipedia/en/wiki/Elliptic_regularity&quot; rel=&quot;nofollow&quot;&gt;elliptic regularity&lt;/a&gt;. There, working with Fourier coefficients (or transforms) and a multiplication operator is significantly easier than working in the original function space with the correspondingly more complicated differential operator. (I am using the fact that constant-coefficient differential operators correspond under the Fourier transform to multiplication by a polynomial.)&lt;/p&gt;&#xA;" OwnerUserId="536" LastEditorUserId="536" LastEditorDisplayName="" LastEditDate="2010-11-18T16:12:16.900" LastActivityDate="2010-11-18T16:12:16.900" CommentCount="1" />
  <row Id="10855" PostTypeId="2" ParentId="10850" CreationDate="2010-11-18T16:09:41.050" Score="0" ViewCount="" Body="&lt;p&gt;Hint: 3y + y = 64&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Further (basic) hint: $a^{\log _b c}  = (e^{\log a} )^{\log c/\log b} $&lt;/p&gt;&#xA;" OwnerUserId="2810" LastEditorUserId="2810" LastEditorDisplayName="" LastEditDate="2010-11-18T16:21:04.053" LastActivityDate="2010-11-18T16:21:04.053" CommentCount="1" />
  <row Id="10856" PostTypeId="1" AcceptedAnswerId="10870" CreationDate="2010-11-18T16:48:37.803" Score="2" ViewCount="117" Body="&lt;p&gt;I am given $n$ points in 2D.(Each of say approximately equal weight). I want to partition it into $m$ clusters ($m$ can be anything and it is input by the user) in such a way that the center of mass of each cluster is &quot;far&quot; from center of mass of all other clusters. What is a good heuristic approach (it should also be quick and easy to implement) for this? My current approach is to set up a binary tree at each step. What I am doing now is that the line I choose to separate cluster at each step which maximizes the moment of inertia of the set of points in the cluster I am splitting. Any suggestion welcome!&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-11-18T17:38:38.303" LastActivityDate="2010-11-18T18:43:58.177" Title="Partitioning a graph (clustering of point sets in 2 dimensions)" Tags="&lt;algorithms&gt;&lt;graph-theory&gt;&lt;computer-science&gt;" AnswerCount="3" />
  <row Id="10857" PostTypeId="2" ParentId="10856" CreationDate="2010-11-18T17:17:29.790" Score="3" ViewCount="" Body="&lt;p&gt;You might find &lt;a href=&quot;http://en.wikipedia.org/wiki/K-means_clustering&quot; rel=&quot;nofollow&quot;&gt;K-Means Clustering&lt;/a&gt; (and related) helpful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is one of the classic machine learning problems, you should find plenty of literature on this.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-11-18T17:17:29.790" />
  <row Id="10858" PostTypeId="2" ParentId="10856" CreationDate="2010-11-18T17:23:42.193" Score="4" ViewCount="" Body="&lt;p&gt;The keyword is &quot;clustering&quot; as mentioned in Moron's answer.  Any problem of this type will be NP-complete.  In practice, K-means is not bad in its runtime or (depending very much on the application) its results.  Like the simplex algorithm for linear programming, it can take exponential time in the worst case, but its practical complexity is much lower.  The worst-case bound was proven only very recently.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, partitioning a &lt;em&gt;graph&lt;/em&gt; is a different problem.  Here you are partitioning a set of points and distances are used but not any graph structure. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(added:)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is the smoothed analysis.  K-means has polynomial runtime when averaged over (Gaussian) random perturbations of the input, which is not surprising considering the practical efficiency:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://arxiv.org/abs/0904.1113&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/0904.1113&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="467" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-11-18T17:35:47.603" LastActivityDate="2010-11-18T17:35:47.603" CommentCount="1" />
  <row Id="10859" PostTypeId="1" CreationDate="2010-11-18T17:23:54.017" Score="3" ViewCount="115" Body="&lt;p&gt;In &quot;A Short Course on Spectral Theory&quot;, page 10, William Arveson asserts that the &quot;ax+b group&quot;, ie. the group generated by all dilations and translations of the real line, is isomorpic to the group of all (real) 2x2 matrices of the form &lt;/p&gt;&#xA;&#xA;&lt;p&gt;a     b&lt;/p&gt;&#xA;&#xA;&lt;p&gt;0     1/a&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a&gt;0, b real.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is very easy to check that the ax+b group is isomorphic to the group of all matrices of the form &lt;/p&gt;&#xA;&#xA;&lt;p&gt;a     b&lt;/p&gt;&#xA;&#xA;&lt;p&gt;0     1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a&gt;0, b real.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So these two matrix groups should be isomorphic. Is this correct? Can someone give me the isomorphism? I've tried for a while and can't seem to get it.&lt;/p&gt;&#xA;" LastActivityDate="2010-11-18T19:24:25.173" Title="matrix representation of ax+b group" Tags="&lt;group-theory&gt;&lt;functional-analysis&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="10860" PostTypeId="2" ParentId="10835" CreationDate="2010-11-18T17:30:25.303" Score="2" ViewCount="" Body="&lt;p&gt;Represent each equivalence class by the number which has it's digits sorted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now this is uniquely determined by the number of 0s, number of 1s, ..., number of 9s.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus you need to find the total number of tuples $(n_0, n_1, \dots, n_9)$ such that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle \sum_{i=0}^{9} n_i = 4$ such that $n_i \ge 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is given by $\displaystyle {13 \choose 4}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since this also counts $0000$, the answer for your problem is $\displaystyle {13 \choose 4}-1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, the number of disitnct $n$-tuples of non-negative integers which sum to $k$ is given by $\displaystyle {n+k-1 \choose k}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See here: &lt;a href=&quot;http://en.wikipedia.org/wiki/Stars_and_bars_%28probability%29&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Stars_and_bars_%28probability%29&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-11-18T17:36:00.353" LastActivityDate="2010-11-18T17:36:00.353" />
  <row Id="10861" PostTypeId="1" AcceptedAnswerId="10881" CreationDate="2010-11-18T17:41:17.070" Score="0" ViewCount="92" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;If set is closed under some operation,&#xA;  what does that tell us the about structure&#xA;  of the groupoid?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It's a question from one of the previous math exams which was shown to me by one of my friends.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, I can't find a good answer to it. As far as I know, the answer is nothing, because of the definition of groupoid we use. Here's what (among other things) my math textbook has to say about groupoids:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Ordered pair (S,*), where arbitrary&#xA;  binary operation closed in  set S&#xA;  is marked with *, is called groupoid.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="348" LastEditorUserId="348" LastEditorDisplayName="" LastEditDate="2010-11-18T17:55:09.073" LastActivityDate="2011-05-19T08:45:34.333" Title="If set is closed under some operation, what does that tell us about the structure of the groupoid?" Tags="&lt;group-theory&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="1" />
  <row Id="10862" PostTypeId="2" ParentId="10859" CreationDate="2010-11-18T17:49:04.240" Score="2" ViewCount="" Body="&lt;p&gt;The isomorphism is given by mapping $\begin{pmatrix} a &amp;amp; b \\ 0 &amp;amp; a^{-1} \end{pmatrix}$ to $\begin{pmatrix} a^2 &amp;amp; a b \\ 0 &amp;amp; 1 \end{pmatrix}$.&#xA;The inverse is given by $\begin{pmatrix} a &amp;amp; b \\ 0 &amp;amp; 1 \end{pmatrix}&#xD;&#xA;\mapsto \begin{pmatrix} a^{1/2} &amp;amp; a^{-1/2} b \\ 0 &amp;amp; a^{-1/2}\end{pmatrix}.$&lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2010-11-18T17:49:04.240" />
  <row Id="10863" PostTypeId="1" AcceptedAnswerId="10865" CreationDate="2010-11-18T17:50:41.790" Score="1" ViewCount="90" Body="&lt;p&gt;Find a general expression for the Laurent Series expansion of&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f(z) = (z-\alpha)^{-n}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;when $|z|&amp;gt;|\alpha|$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've tried expanding the function using the binomial theorem, and I've tried&#xA;integrating with the general formula for Laurent Series coefficients &#xA;using Cauchy's Integral formula, but I can't get the solution,&#xA;which I know is supposed to be&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Sigma_{j=n}^{\infty} \alpha^{j-n} \frac{(j-1)!}{(n-1)!(j-n)!}z^{-j}$&lt;/p&gt;&#xA;" OwnerUserId="2029" LastActivityDate="2010-11-18T18:06:39.233" Title="Expanding Binomial using Laurent Series" Tags="&lt;homework&gt;&lt;complex-analysis&gt;" AnswerCount="1" />
  <row Id="10864" PostTypeId="2" ParentId="10818" CreationDate="2010-11-18T17:53:29.807" Score="2" ViewCount="" Body="&lt;p&gt;What you need is to look at some texts on functional analysis, especially about &quot;Direct Methods&quot; in &quot;Calculus of Variations&quot;. I'll give a quick primer here to flesh out what Rahul Narain said in the comments, but it will be by no means complete.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First let us review how the minimization problem is solved in finite dimensional spaces. We will consider a finite dimensional vector space $V$ with an inner product $\langle,\rangle$. Let $A$ be a self-adjoint linear operator $V\to V$, and $b$ some element of $V$. We consider the minimization problem with the non-linear functional $J(x) = \frac12 \langle x,Ax\rangle - \langle b,x\rangle$. There are two steps used in finding the minimizer. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Step 1) &lt;strong&gt;Existence of a minimizer&lt;/strong&gt;.  First we need to show that $J(x)$ is bounded from below. This is where we first use that $A$ has only non-negative eigenvalues. If $A$ were to have negative eigenvalues, let $v$ be its eigenvector, and look at $J(kv)$ for $k$ a scalar. Letting $k\nearrow\infty$ we see that $J(kv)\searrow -\infty$ and there cannot be an minimum. Next, if $A$ were to have a non-empty kernel, such that there exists some vector $v$ with the property $Av = 0$ and $\langle b,v\rangle \neq 0$, then again taking $J(kv)$ and sending $k$ to either $\pm\infty$, we will have that the functional $J$ is not bounded below. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So $A$ must only have positive eigenvalues. Now, let $\lambda$ be the smallest of $A$'s eigenvalues. Then we have $\langle x,Ax\rangle &amp;gt; \lambda \|x\|^2$. Using Cauchy-Schwartz inequality, we then have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ J(x) &amp;gt; \frac12 \lambda \|x\|^2 - \|b\|\|x\| $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we see that for $\|x\| &amp;gt; 2\|b\| / \lambda$, $J(x) &amp;gt; 0$. And by continuity, $J(x)$ is bounded below. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, let $j = \inf J &amp;lt; 0$. (The infimum is negative since by choosing $w$ a sufficiently small multiple of $b$, we see that $J(w) &amp;lt; 0$.) By the definition of infimum, there exists a sequence of vectors $(v_k)$ such that $J(v_k) \searrow j$. Since the infimum is less than 0, by our previous computations we know that the sequence $(v_k)$ must all satisfy $\|v_k\| \leq 2\|b\|/\lambda$. Now, this set is a bounded closed ball in a finite dimensional vector space. In particular, it is compact. Hence up to a subsequence $v_k$ must converge. Now using continuity of $J$, you have that the limit element $v_\infty$ of the sequence &lt;em&gt;attains&lt;/em&gt; the minimum. That is $J(v_\infty) = j$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Step 2) &lt;em&gt;Computing the minimizer&lt;/em&gt;. Here we just use the Euler-Lagrange equations. At the minimizer, the functional is stationary. That is $(\frac{\delta}{\delta x}J)(v_\infty) = 0$. Taking the variation of the functional $J$, we get, using again that $A$ is self-adjoint, that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \frac{\delta}{\delta x}J = 0 \iff Ax - b = 0 $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $A$ only has positive eigenvalues, it is invertible, so we can solve $v_\infty = A^{-1}b$. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Formally, we can try to do the same thing in infinite dimension. But there are a few caveats. First is that it is not enough that $A$ only has positive eigenvalues. You need that $A$ is &lt;em&gt;coercive&lt;/em&gt; in the following sense:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \langle x, Ax\rangle &amp;gt; \lambda \|x\|^2$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for some $\lambda$. In finite dimensions that $A$ is positive implies the existence of this $\lambda$. In infinite dimensions this is not true. Imagine the $\ell^2$ Hilbert space with standard inner product and the standard basis $(e_k)$. Let $Ae_k = \frac1k e_k$. Then $A$ has only positive eigenvalues. But by taking larger and larger $k$'s, we can make the eigenvalue approach 0 as closely as we want, and violate coercivity. Coercivity allows us to guarantee that the functional $J(x)$ has a bottom. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second problem is that a closed bounded set in a Hilbert space is not necessarily compact. So we cannot conclude immediately that a minimizing sequence converges in the norm topology. However, using Banach-Alaoglu theorem, the minimizing sequence must have a convergent subsequence in the weak topology of your Hilbert space. So it is sufficient to show that your functional $J(x)$ is what is called &quot;sequentially weakly lower-semicontinuous&quot; to get the desired conclusion. (That condition just guarantees that for any weakly converging sequence $w_k\to_w w_\infty$, $\lim\inf J(w_k) \geq J(w_\infty)$.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In our particular case, things can be made slightly easier. If we in addition assume that $A$ is a bounded linear operator. Then we can use $(x,y) = \langle x,Ay\rangle$ as an equivalent inner product on our &lt;em&gt;Hilbert space&lt;/em&gt;. Then $A^{-1}b$ exists by the Riesz representation theorem, and our functional can be written as $J(x) = \frac12 (x,x) - (A^{-1}b, x)$. Now we run through our previous arguments again. Now $w_k\to_w w_\infty$, and therefore $(A^{-1}b,w_k) \to (A^{-1}b,w_\infty)$ strongly by the definition of weak limits. And we also use the property of weak limits that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \lim\inf (w_k,w_k) \geq (w_\infty,w_\infty)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which gurantees lower semi-continuity as desired. So now we've shown that the minimizer exists, and computing the variation we see that the minimizer must satisfy $w_\infty = A^{-1}b$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So to repeat, a sufficient condition is that $A$ is self-adjoint, bounded, &lt;em&gt;and&lt;/em&gt; coercive. &lt;/p&gt;&#xA;" OwnerUserId="1543" LastActivityDate="2010-11-18T17:53:29.807" CommentCount="1" />
  <row Id="10865" PostTypeId="2" ParentId="10863" CreationDate="2010-11-18T18:06:39.233" Score="2" ViewCount="" Body="&lt;p&gt;$$(z-\alpha)^{-n} = \frac1{z^n} \left(1-\frac\alpha z\right)^{-n}.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now expand the (...) term using the &lt;a href=&quot;http://en.wikipedia.org/wiki/Binomial_theorem#Newton.27s_generalized_binomial_theorem&quot; rel=&quot;nofollow&quot;&gt;generalized binomial theorem&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="171" LastActivityDate="2010-11-18T18:06:39.233" />
  <row Id="10866" PostTypeId="1" AcceptedAnswerId="10869" CreationDate="2010-11-18T18:36:36.790" Score="1" ViewCount="189" Body="&lt;p&gt;This actually triggered me in my mind from &lt;a href=&quot;http://math.stackexchange.com/questions/10850/solving-an-equation&quot;&gt;here&lt;/a&gt;. After some playing around I notice that the relation $a^{\log_b x} = x^{\log_b a}$ is true for any valid value of $a,b$ and $x$. I am very inquisitive to see how this holds ?&lt;/p&gt;&#xA;" OwnerUserId="2109" LastActivityDate="2010-11-19T02:02:27.203" Title="How $a^{\log_b x} = x^{\log_b a}$ ?" Tags="&lt;algebra-precalculus&gt;&lt;logarithms&gt;" AnswerCount="4" CommentCount="5" />
  <row Id="10867" PostTypeId="2" ParentId="803" CreationDate="2010-11-18T18:37:38.170" Score="3" ViewCount="" Body="&lt;p&gt;I really like this one (image taken from &lt;a href=&quot;http://www.cut-the-knot.org/pythagoras/&quot; rel=&quot;nofollow&quot;&gt;cut-the-knot&lt;/a&gt; #4)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/NRfbG.gif&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;along with &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(a+b)^2 = 4\cdot\frac{1}{2}a b + c^2$$&#xD;&#xA;$$\Leftrightarrow c^2 = a^2 + b^2$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's so clear and easy ...&lt;/p&gt;&#xA;" OwnerUserId="917" LastActivityDate="2010-11-18T18:37:38.170" CommunityOwnedDate="2010-11-18T18:37:38.170" />
  <row Id="10868" PostTypeId="1" AcceptedAnswerId="10878" CreationDate="2010-11-18T18:41:21.983" Score="2" ViewCount="302" Body="&lt;p&gt;I am quite a baffled now, I am not getting by &lt;em&gt;how&lt;/em&gt; it can be written that :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align*}&#xA;&amp;amp;\log_{10} \tan 40^\circ \cdot \log_{10} \tan 41^\circ \cdot \log_{10} \tan 42^\circ \cdot\log_{10} \tan 43^\circ \cdots \log_{10} \tan 50^\circ  \\&#xA;&amp;amp;= \log_{10} \tan 40^\circ + \log_{10} \tan 41^\circ + \log_{10} \tan 42^\circ + \log_{10} \tan 43^\circ + \cdots + \log_{10} \tan 50^\circ &#xA;\end{align*}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it even valid ? If yes,how ?&lt;/p&gt;&#xA;" OwnerUserId="2109" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-12-17T01:48:09.500" LastActivityDate="2010-12-17T01:48:09.500" Title="Baffling identity: $\prod \log_{10} \tan = \sum \log_{10} \tan$" Tags="&lt;trigonometry&gt;&lt;logarithms&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="10869" PostTypeId="2" ParentId="10866" CreationDate="2010-11-18T18:41:21.910" Score="6" ViewCount="" Body="&lt;p&gt;For any positive $r$ and any $s$, you have &#xA;$$r^s = b^{\log_b(r^s)} = b^{s\log_b(r)}.$$&#xD;&#xA;So, taking $r=a$ and $s=\log_b(x)$, we have:&#xA;\begin{align*}&#xA;a^{\log_b(x)} &amp;amp;= b^{\log_b(x)\log_b(a)}\\&#xA;&amp;amp;= b^{\log_b(a)\log_b(x)}\\&#xA;&amp;amp; = b^{\log_b(x^{\log_b(a)})}\\&#xA;&amp;amp;= x^{\log_b(a)}.&#xA;\end{align*}&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-11-18T18:41:21.910" />
  <row Id="10870" PostTypeId="2" ParentId="10856" CreationDate="2010-11-18T18:43:58.177" Score="3" ViewCount="" Body="&lt;p&gt;You might want to look at this survey paper on clustering methods:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://cchen1.csie.ntust.edu.tw:8080/students/2011/Survey%20of%20clustering%20algorithms.pdf&quot; rel=&quot;nofollow&quot;&gt;http://cchen1.csie.ntust.edu.tw:8080/students/2011/Survey%20of%20clustering%20algorithms.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1369" LastActivityDate="2010-11-18T18:43:58.177" />
  <row Id="10871" PostTypeId="2" ParentId="10846" CreationDate="2010-11-18T18:44:03.263" Score="3" ViewCount="" Body="&lt;p&gt;Consider:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f_{n,N}(z) = \sin(Nx) \sum_{k = 1}^n \frac{\sin(kx)}{k}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now consider&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_k \frac{1}{k^2} f_{2^{k^3}, 2^{k^3 - 1}}(z)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now for $x = \pi / (4n)$ and $N = 2n$ we get that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sin(\pi/4) \sum_1^n \frac{1}{k} &amp;gt; \frac{1}{\sqrt{2}} \log n$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we have for some $x$ that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$|s_{n_k + 1} - s_{n_k}| \geq \frac{1}{\sqrt{2}} \frac{1}{k^2} \log n_k$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we cannot have uniform convergence. I believe this is due to Hugo Steinhaus.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope I didn't make a mistake, but it is along these lines, I can correct it if I made an error.&lt;/p&gt;&#xA;" OwnerUserId="1120" LastActivityDate="2010-11-18T18:44:03.263" CommentCount="6" />
  <row Id="10872" PostTypeId="2" ParentId="10866" CreationDate="2010-11-18T18:45:31.997" Score="3" ViewCount="" Body="&lt;p&gt;Using log properties, we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a^{\log_b(x)} = b^{\log_b\left(a^{\log_b(x)}\right)} = b^{\log_b(x)\log_b(a)} = \left(b^{\log_b(x)}\right)^{\log_b(a)} = x^{\log_b(a)}$&lt;/p&gt;&#xA;" OwnerUserId="3537" LastActivityDate="2010-11-18T18:45:31.997" CommentCount="1" />
  <row Id="10873" PostTypeId="2" ParentId="803" CreationDate="2010-11-18T18:46:16.503" Score="0" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://www.math.ntnu.no/~hanche/pythagoras/&quot; rel=&quot;nofollow&quot;&gt;http://www.math.ntnu.no/~hanche/pythagoras/&lt;/a&gt;&lt;/p&gt;&#xA;" LastActivityDate="2010-11-18T18:46:16.503" CommunityOwnedDate="2010-11-18T18:46:16.503" />
  <row Id="10874" PostTypeId="1" AcceptedAnswerId="10886" CreationDate="2010-11-18T18:49:47.623" Score="2" ViewCount="242" Body="&lt;p&gt;In algebra, we learn that if a function $ f(x) $ has a one-to-one mapping, then we can find the inverse function $ f^{-1}(x) $. The method that I have seen taught is the &quot;horizontal line test&quot;: if any horizontal line touches the graph of the function more than once, then it must not be one-to-one. This method is not exactly rigorous, since any function with a non-finite range can not be completely viewed on a graph. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there an analytic method to determine if a function is one-to-one? Is this possible in elementary algebra or calculus?&lt;/p&gt;&#xA;" OwnerUserId="3384" LastActivityDate="2010-11-18T20:18:22.383" Title="Analytic method for determining if a function is one-to-one" Tags="&lt;calculus&gt;&lt;algebra-precalculus&gt;&lt;algebra&gt;&lt;functions&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="10876" PostTypeId="2" ParentId="10874" CreationDate="2010-11-18T18:53:40.707" Score="0" ViewCount="" Body="&lt;p&gt;A function is invertible if and only if it is bijective, that is surjective (onto) and injective (one-to-one), so your statement is not correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to determine that if a function is injective, you assume $f(x) = f(y)$ and derive $x = y$, alternatively you can assume $x \neq y$ and show that $f(x) \neq f(y)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if you have a linear function $f$, then we see that we need to show that if $f(x) = f(y)$ we have that $f(x - y) = 0$, so in other words we need to show that $f(x) = 0$ implies $x = 0$. This is usually easier.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case you want to show a function is bijective (and hence invertible) you first show that it is injective and then that if $f: X \to Y$ that for every $y \in Y$ you can find an $x \in X$ such that $f(x) = y$.&lt;/p&gt;&#xA;" OwnerUserId="1120" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-18T19:28:21.570" LastActivityDate="2010-11-18T19:28:21.570" CommentCount="3" />
  <row Id="10877" PostTypeId="2" ParentId="10859" CreationDate="2010-11-18T18:55:47.087" Score="1" ViewCount="" Body="&lt;p&gt;The isomorphism arises simply from factoring $\rm\ \ aa\ X + b\ \  =\ \ a\ (ax+b/a)\:,\ $ namely&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\ \ \ (aa\ X + b)\ \: \circ\: \ (AA\ X+B)\quad =\quad a\ a\ (A\ A\ \ X\ \ +\ \ B)\ \ +\ \ b\quad\ \ \  =\quad\ \ \  aa\ \ AA\ X\ +\ aaB + b$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\displaystyle\ a\bigg(aX+\frac{b}a\bigg)\circ \:A\:\bigg(AX+\frac{B}A\bigg)\ =\ aA\:\bigg(aX+\frac{b}{aA}\bigg)\circ\bigg(AX+\frac{B}A\bigg)\ =\ aA\bigg(aA\ X\ + \frac{aaB + b}{aA}\bigg)\ $&#xD;&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-11-18T19:24:25.173" LastActivityDate="2010-11-18T19:24:25.173" />
  <row Id="10878" PostTypeId="2" ParentId="10868" CreationDate="2010-11-18T18:56:01.867" Score="9" ViewCount="" Body="&lt;p&gt;Yes it is valid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\log_{10} \tan 40^{\circ} + \log_{10} \tan 50^{\circ} = \log_{10} \tan 40^{\circ} + \log_{10} \cot 40^{\circ} = \log_{10} 1 = 0$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly combine 41 and 49, 42 and 48 etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The product on the left side is $0$ as $\log_{10} \tan 45^{\circ} = 0$.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-11-18T18:56:01.867" CommentCount="7" />
  <row Id="10879" PostTypeId="2" ParentId="10868" CreationDate="2010-11-18T18:57:40.183" Score="4" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT:&lt;/strong&gt; Observe that $$\log_{10} \tan{40} + \log_{10} \tan{50} = \log_{10} \sin{40} - \log_{10} \cos{40} + \log_{10} \sin{50} - \log_{10} \cos{50}$$ Now use the fact that $\sin(90 - x) =\cos{x}$ and see that they cancel out.&lt;/p&gt;&#xA;" LastActivityDate="2010-11-18T18:57:40.183" />
  <row Id="10881" PostTypeId="2" ParentId="10861" CreationDate="2010-11-18T19:24:24.813" Score="3" ViewCount="" Body="&lt;p&gt;Assuming &quot;the grupoid&quot; means the grupoid (using the definition you quote) made up of that specific set and that specific operation (as opposed to some other operation), then the answer is &quot;Nothing beyond the fact that it is a grupoid.&quot; Because any property you care to name that is a property of &lt;em&gt;some&lt;/em&gt; but &lt;strong&gt;not all&lt;/strong&gt; grupoids will arise from a particular example of a &quot;set that is closed under some operation,&quot; but will also fail to arise from another particular example. That is, absent more information, you can't say anything else. &lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-11-18T19:24:24.813" CommentCount="1" />
  <row Id="10882" PostTypeId="1" AcceptedAnswerId="10889" CreationDate="2010-11-18T19:29:31.913" Score="1" ViewCount="250" Body="&lt;p&gt;In MATLAB, when you plot something, there's a tool available which is called &quot;curve fitting&quot;. And if you have a set of data points and a linear correlation, this tool will easily come up with an equation on the form y = ax + c. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to do exactly that, but I need to do it programmatically. I know that I can use &lt;code&gt;polyfit&lt;/code&gt; to find the coefficient, and &lt;code&gt;polyval&lt;/code&gt; to evaluate as many data points as I want, but what I can't figure out is how to find the constant c. This is very important, since I need to be able to plot the line as well next to my data points.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you.&lt;/p&gt;&#xA;" OwnerUserId="3680" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-18T23:16:44.157" LastActivityDate="2010-11-18T23:16:44.157" Title="Find equation of curve fit programmatically in Matlab?" Tags="&lt;statistics&gt;&lt;math-software&gt;&lt;matlab&gt;&lt;regression&gt;" AnswerCount="2" />
  <row Id="10883" PostTypeId="2" ParentId="10882" CreationDate="2010-11-18T19:32:53.673" Score="0" ViewCount="" Body="&lt;p&gt;Use the equations &lt;a href=&quot;http://en.wikipedia.org/wiki/Simple_linear_regression&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-18T19:32:53.673" CommentCount="1" />
  <row Id="10884" PostTypeId="2" ParentId="10866" CreationDate="2010-11-18T19:45:03.643" Score="5" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\rm\ \ \ log(A^{\log X})\ =\ log\ X\ \ log\ A\ =\ log(X^{\log A})\:,\ $ where $\rm\ log := log_b$&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-11-18T19:45:03.643" />
  <row Id="10885" PostTypeId="1" AcceptedAnswerId="10901" CreationDate="2010-11-18T19:53:49.433" Score="7" ViewCount="348" Body="&lt;p&gt;Hi, do you have an example of a non separable locally compact connected metric space?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you&lt;/p&gt;&#xA;" OwnerUserId="3681" LastActivityDate="2010-11-19T13:44:22.690" Title="Non separable locally compact connected metric space" Tags="&lt;topology&gt;&lt;metric-spaces&gt;" AnswerCount="3" FavoriteCount="1" />
  <row Id="10886" PostTypeId="2" ParentId="10874" CreationDate="2010-11-18T19:59:49.630" Score="8" ViewCount="" Body="&lt;p&gt;First, let me answer your question; but please keep reading, because there's lots more to say.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yes: there is an analytic way to see if a function is one-to-one. For this, you need the &quot;analytic definition&quot; of being one-to-one. The definition is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f\text{ is one-to-one if and only if for all }a,b\text{ if }a\neq b\text{ then } f(a)\neq f(b).$$&#xD;&#xA;Logically, this is equivalent to:&#xA;$$f\text{ is one-to-one if and only if for all }a,b\text{ if }f(a)=f(b)\text{ then }a=b.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So this provides a way to check if the function is one-to-one: if you can find $a\neq b$ such that $f(a)=f(b)$, then $f$ is not one-to-one; and this is essentially the best way of doing it: &lt;strong&gt;exhibit&lt;/strong&gt; a pair of distinct numbers that map to the same thing. To &lt;em&gt;prove&lt;/em&gt; that $\sin(x)$ is not one-to-one, all I need to do is say: &quot;Look, $0$ and $\pi$ are different, but $\sin(0)=\sin(\pi)$.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To prove that a function &lt;em&gt;is&lt;/em&gt; one-to-one, you can do it in any of two (equivalent) ways: show that if $a$ and $b$ are &lt;em&gt;any&lt;/em&gt; numbers with the property that $f(a)=f(b)$, then it must be the case that $a=b$; or show that if $a\neq b$, then $f(a)$ must be different from $f(b)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, to show that $f(x)=x^3$ is one to one, we can note that if $f(a)=f(b)$, then $a^3=b^3$, and taking cubic roots we conclude that $a=\sqrt[3]{a^3} = \sqrt[3]{b^3} = b$. So if $f(a)=f(b)$, then $a=b$. QED. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or we can argue that if $a\neq b$, then either $a\lt b$ or $a\gt b$. If $a\lt b$, then $a^3\lt b^3$ (you can prove this easily using the properties of real numbers and inequalities), so $f(a)\neq f(b)$. If $a\gt b$, then $f(a)=a^3\gt b^3=f(b)$, hence $f(a)\neq f(b)$. Either way, $a\neq b$ implies $f(a)\neq f(b)$, so $f$ is one-to-one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the standard way of showing, analytically, that a function is one-to-one. How one establishes the implication will depend on the function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That said, there are some things you may want to remember:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First: to specify a function, we usually need to specify at least two things: the domain of the function, and the value of the function at any point in the domain. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;More often, we are interested in functions between two specific sets. In that case, we actually need to specify &lt;em&gt;three&lt;/em&gt; things: the domain, the set in which the images will lie, and the value of the functions at any point of the domain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, if we say that we have a function $f\colon X\to Y$ between two sets, then we mean that:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;For &lt;em&gt;every&lt;/em&gt; $x\in X$, there is an element of $y\in Y$ such that $f(x)=y$; and&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;For each $x\in X$, there is only one $y\in Y$ with $f(x)=y$ (unique value).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;So, we need &lt;strong&gt;every&lt;/strong&gt; element of $X$ to have a unique image. Different elements of $X$ may have the same image, but a single element should not have multiple images. We call $X$ the &lt;strong&gt;domain&lt;/strong&gt; of $f$, and we call $Y$ the &lt;strong&gt;codomain&lt;/strong&gt; of $f$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In calculus, however, we are usually a bit sloppy. We almost never mention either the domain or the codomain! Instead, we agree that if we will either specify the domain explicitly, or else we will mean &quot;the natural domain&quot;. And we almost never mention the codomain at all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if you have a function $f\colon X\to Y$, then we say that a function $g\colon Y\to X$ &quot;going the other way&quot; is &quot;the inverse of $f$&quot; if and only if two things happen:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;For every $x\in X$, $g(f(x)) = x$  ($g$ &quot;un-does&quot; what $f$ does); and&lt;/li&gt;&#xA;&lt;li&gt;For every $y\in Y$, $f(g(y)) = y$  ($f$ &quot;un-does&quot; what $g$ does).&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Now, for this to actually work, we need two things:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Given an element $y\in Y$, there can be at most one $x\in X$ such that $y=f(x)$. This is the &quot;one-to-one&quot; condition. If you think in terms of calculus and the graph, it is precisely the &quot;horizontal line test&quot;: for each value of $y$ (each horizontal line), there is at most one point in the domain where $f$ takes value $y$ (it cuts the graph in at most one point).&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;em&gt;Reason:&lt;/em&gt; If you had $x\neq x&amp;#39;$ but with $y=f(x)=f(x&amp;#39;)$, then you would need $g(y)=x$ so that $x=g(y)=g(f(x))$, but you would also need $g(y)=x&amp;#39;$ because $x&amp;#39;=g(y) = g(f(x&amp;#39;))$. But a function cannot have two different values at the same point, so this is an impossible situation for $g$. The only way to solve this quandry is for either $g$ not to exist, or for $x$ and $x&amp;#39;$ not to exist. &lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Given any element $y\in Y$, there is &lt;em&gt;at least&lt;/em&gt; one $x\in X$ such that $y=f(x)$. This is the &quot;onto&quot; or &quot;surjective&quot; part people have mentioned.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;em&gt;Reason:&lt;/em&gt; Since we also want $f(g(y))=y$ for every $y\in Y$, we need there to be an element of $X$, namely $g(y)$, that maps to $y$.&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So: if $f\colon X\to Y$ has an inverse, then it must be &lt;em&gt;both&lt;/em&gt; one-to-one, &lt;strong&gt;and&lt;/strong&gt; onto the set $Y$. This is &lt;em&gt;necessary.&lt;/em&gt; In fact, it is also sufficient, and one can show that if there is an inverse, then there is one and only one inverse, so we call it $f^{-1}$ instead of $g$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, here's the thing: if your function satisfies the &lt;em&gt;first&lt;/em&gt; condition (one-to-one), but &lt;em&gt;not&lt;/em&gt; the second, then you can &quot;cheat&quot;: instead of thinking of $f$ as a function from $X$ to $Y$, we let $Y&amp;#39;=\mathrm{Image}(f)$, and then look at the function $\mathfrak{f}\colon X\to Y&amp;#39;$, with $\mathfrak{f}(x)=f(x)$ for every $x$; the only thing we changed is what we want the set $Y$ to be. This is not really the same as $f$: $\mathfrak{f}$ one &lt;em&gt;is&lt;/em&gt; both one-to-one and onto, so $\mathfrak{f}$ *does* have an inverse, even though $f$ does not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, if you think of the function $f\colon\mathbb{R}\to \mathbb{R}$ given by $f(x)=e^x$, then $f$ is one-to-one, but is not onto ($f(x)$ does not take negative values). So this function is not invertible. However, if we tweak the function and think of it instead as $\mathfrak{f}\colon\mathbb{R}\to (0,\infty)$, given by $\mathfrak{f}(x)=e^x$, then $\mathfrak{f}$ *is* onto, is also one-to-one, so $\mathfrak{f}$ *is* invertible. The inverse is a function $f^{-1}\colon (0,\infty)\to\mathbb{R}$, so the only valid inputs are positive numbers. (You may know who $f^{-1}$ is: it's the natural logarithm). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, notice that to check if a function has an inverse, you need to know &lt;strong&gt;both&lt;/strong&gt; what $X$ and what $Y$ is. But in Calculus we almost never mention $Y$. So what can we do?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, we agree that we will take $Y$ to be &quot;the image of $f$&quot;; that is, the collection&#xA;$$\{ f(x) \mid x\text{ is in the domain of }f\}.$$&#xD;&#xA;That means that we always &quot;automatically&quot; assume our functions are &quot;onto&quot;. We say this by saying that the function is &quot;onto its image&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given that agreement, in order to figure out if $f$ has an inverse, we just need to know if $f$ is one-to-one, which is why your calculus book says things like &quot;a function has an inverse if and only if it is one to one, if and only if it passes the horizontal line test.&quot; They are refering exclusively to functions whose codomain is &lt;em&gt;always&lt;/em&gt; taken to be the image. &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-18T20:18:22.383" LastActivityDate="2010-11-18T20:18:22.383" CommentCount="3" />
  <row Id="10887" PostTypeId="2" ParentId="10817" CreationDate="2010-11-18T20:08:28.730" Score="4" ViewCount="" Body="&lt;p&gt;Because $p$ and $q$ are coprime, then $p$ and $p-q$ are coprime: any irreducible factor of both would divide $p-(p-q)=q$, so it would divide both $p$ and $q$. Likewise, since $\lambda\neq 0$, $p$ and $p-\lambda q$ are coprime (same reason). And since $\lambda\neq 1$, then $p-\lambda q$ and $p-q$ are coprime; any shared factor between them is a factor of $\lambda(p-q)-(p-\lambda q) = (\lambda-1)p$ (hence of $p$) and of $(p-q)-(p-\lambda q) = (\lambda-1)q$ (hence of $q$), but $p$ and $q$ have no common factors other than units.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So $p$, $(p-q)$, and $(p-\lambda q)$ are pairwise coprime as well. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;At this point, the argument is the same as in any UFD (e.g., the integers): you have $r^2 = xyz$, with $x$, $y$ and $z$ relatively prime; looking at each irreducible factor of $r$, and at each irreducible factor of $x$, of $y$, and of $z$, you get that each irreducible factor of $x$, of $y$, and of $z$ must appear to an even exponent, so each of $x$, $y$, and $z$ are squares.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-11-18T20:08:28.730" CommentCount="1" />
  <row Id="10888" PostTypeId="1" AcceptedAnswerId="10890" CreationDate="2010-11-18T20:30:36.250" Score="2" ViewCount="305" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All, I am having some trouble in checking whether a sequence of functions &lt;em&gt;converges pointwise or uniformly&lt;/em&gt;. The problem is, sometimes, my intuition is right and sometimes its wrong. Finding the &lt;em&gt;limit function&lt;/em&gt; is the most difficult task I would say.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am able to understand some simple examples, such as a &lt;em&gt;pointwise convergence&lt;/em&gt; of $f_{n}(x)= \frac{x}{n}$ over $\mathbb{R}$. But when given complex problems, I am having trouble in guessing the limit function. For example, I had trouble in solving this.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Test the convergence of $f_{n}(x) = n \Bigl[ (x)^{1/n} - 1 \Bigr]$ over $X=[1,t]$ for $t&amp;gt;1$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Given such functions, how can I improve my ability of guessing the limit function, and then proceeding to check whether the convergence is uniform or not. One method which I found extremely useful is the following theorem:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$f_{n} \to f$ uniformly on $X$ iff $y_{n} =\sup \{|f_{n}(x)-f(x)| : x \in X \} \to 0$. But as I see with some examples, this result is not always successful. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-18T20:37:29.037" LastActivityDate="2010-11-18T20:56:56.860" Title="Common Problems while showing Uniform Convergence of functions" Tags="&lt;real-analysis&gt;&lt;analysis&gt;&lt;sequences-and-series&gt;&lt;convergence&gt;" AnswerCount="1" CommentCount="6" FavoriteCount="1" />
  <row Id="10889" PostTypeId="2" ParentId="10882" CreationDate="2010-11-18T20:39:18.987" Score="3" ViewCount="" Body="&lt;p&gt;I'm going to give two answers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;HR&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Answer 1: If you already know $a$, then $c = \bar{y} - a\bar{x}$, where $\bar{x}$ and $\bar{y}$ are the means of the $x$ and $y$ values, respectively.  (This is actually in that pile of formulas in the link provided by Yuval Filmus.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;HR&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Answer 2: At least in my version of Matlab, polyfit gives both $a$ and $c$.  For example, if you generate a set of data via (I'm modifying the example given in the Matlab help)&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;x = (0: 0.1: 2.5)';&#xA;y = erf(x); &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;and then call polyfit with $n=1$ (for a linear curve fit, or linear regression)&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;p = polyfit(x,y,1)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The output is &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;0.3554    0.3191&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;which means your linear equation is $y = 0.3554x + 0.3191$.&lt;/p&gt;&#xA;" OwnerUserId="2370" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-11-18T22:21:16.490" LastActivityDate="2010-11-18T22:21:16.490" CommentCount="1" />
  <row Id="10890" PostTypeId="2" ParentId="10888" CreationDate="2010-11-18T20:46:46.733" Score="8" ViewCount="" Body="&lt;p&gt;Intuition is developed by doing examples and becoming familiar with these things, so in that respect you are heading in the right direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as &quot;guessing the limit function&quot;, that's not what we want to do! We don't want to guess, we want to &lt;em&gt;figure it out&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Certainly if there is going to be uniform convergence, then there &lt;em&gt;has&lt;/em&gt; to be pointwise convergence. So you can begin by considering the limit of $f_n(x)$ for fixed $x$; if it does not exist, you're done: no convergence, pointwise &lt;em&gt;or&lt;/em&gt; uniform. If it always exists, then the limit function &lt;em&gt;must&lt;/em&gt; be the function&#xA;$$f(x_0) = \lim_{n\to\infty}f_n(x_0)$$&#xD;&#xA;for each $x_0$. I hope that by &quot;guessing the function&quot; you actually mean &quot;recognize it as a function I may already know&quot;. This is not always possible in any case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the functions $f_n(x)=n\left((x)^{1/n}-1\right)$, you will first want to consider the convergence at a given point $a$ that lies in $[1,t]$. Does the limit exist for a fixed $a$? That is, does&#xA;$$\lim_{n\to\infty} n\left(a^{1/n}-1\right)$$&#xD;&#xA;exist? If so, you want to try to express it as a function of $a$. This is a problem in sequences (or limits). This is an $\infty\times 0$ indeterminate, so you can try using L'Hopital's Rule to find it:&#xA;\begin{align*}&#xA;\lim_{n\to\infty}n\left(a^{1/n}-1\right) &amp;amp;= \lim_{x\to\infty}\frac{a^{1/x}-1}{\frac{1}{x}} = \lim_{x\to\infty}\frac{e^{\ln(a)/x}-1}{\frac{1}{x}}\\&#xA;&amp;amp;= \lim_{x\to\infty}\frac{-\frac{1}{x^2}\ln(a)e^{\ln(a)/x}}{-\frac{1}{x^2}}\\&#xA;&amp;amp;= \lim_{x\to\infty}\ln(a)e^{\ln(a)/x} = \ln(a).&#xA;\end{align*}&#xA;So there &lt;em&gt;is&lt;/em&gt; a pointwise limit, and $\lim\limits_{n\to\infty}f_n(a) = \ln(a)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This tells you that the function you are converging &lt;em&gt;pointwise&lt;/em&gt; to is $f(x)=\ln(x)$. Now that you have a target function, you can check to see if the convergence is uniform in any of the many ways we have of checking of the convergence is uniform.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Added:&lt;/em&gt; Mind you, just like we have ways of determining that a sequence converges without actually determining what it converges &lt;em&gt;to&lt;/em&gt; (for example, showing that the sequence is Cauchy), we also have ways of determining if a sequence of functions converges pointwise without actually determining &lt;em&gt;to what&lt;/em&gt;. &lt;strong&gt;And&lt;/strong&gt; we have ways of determining if a sequence of functions converges &lt;em&gt;uniformly&lt;/em&gt; to &lt;em&gt;something&lt;/em&gt; without having to first determine if it converges pointwise, or &lt;em&gt;what&lt;/em&gt; it converges to. You could show that the sequence is &quot;uniformly Cauchy&quot; or &quot;Cauchy in the sup norm&quot;: you define &#xA;$$||f_n-f_m||_{\infty} = \sup\{ |f_n(x)-f_m(x)|\}.$$&#xD;&#xA;We say the sequence $\{f_n\}$ is Cauchy in the sup norm if and only if for every $\epsilon\gt 0$ there exists $N\gt 0$ such that if $n,m\geq N$, then $||f_n-f_m||_{\infty}\lt \epsilon$. It is straightforward to show that if $\{f_n\}$ is Cauchy in this sense, then for each $a$ you have $\{f_n(a)\}$ is Cauchy (in the usual sense), so there must be pointwise convergence (to &lt;em&gt;something&lt;/em&gt;); and you can also prove that if the sequence $\{f_n\}$ is Cauchy in the sup norm, then the convergence &lt;em&gt;will&lt;/em&gt; be uniform (to whatever it is that it converges). This is to figuring it if $\{f_n\}$ converges uniformly to a given $f$ like showing a sequence $\{a_n\}$ is Cauchy is to figuring out if it converges to a given $L$.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-18T20:56:56.860" LastActivityDate="2010-11-18T20:56:56.860" CommentCount="5" />
  <row Id="10891" PostTypeId="2" ParentId="10885" CreationDate="2010-11-18T21:05:53.103" Score="0" ViewCount="" Body="&lt;p&gt;This isn't so much an answer (so please don't vote this up) so much as a pointer on some things that don't work. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whatever one cooks up it can't be an infinite-dimensional Banach space since those aren't locally compact. Uncountably many copies of the unit interval each horizontal line segment some fixed distance away from all others doesn't work either since it's not connected.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This slightly more clever construction is also no good: Consider the set $[0,1] \times [0,1]$ with the pseudometric $d&amp;#39;$ given by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ d&amp;#39;((x_1,y_1),(x_2,y_2)) = \begin{cases}&#xD;&#xA;\vert x_1 - x_2 \vert &amp;amp; \text{ if } y_1 = y_2 \lor x_1 = 0 \lor x_2 = 0  \\\\&#xD;&#xA;2                     &amp;amp; \text{ if } y_1 \neq y_2 \land x_1,x_2 \neq 0,&#xD;&#xA;\end{cases} $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and let $X$ be its Kolmogorov quotient and $d$ its associated metric. It satisfies all the desired properties except local compactness - $\{0\}\times [0,1]$ doesn't have a compact neighbourhood.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The difficulty in conjuring up an example seems to come from the fact that you want a nonseparable connected space where every point has a separable neighbourhood (compact metric implies 2nd-countable).&lt;/p&gt;&#xA;" OwnerUserId="1269" LastActivityDate="2010-11-18T21:05:53.103" CommentCount="1" />
  <row Id="10892" PostTypeId="1" CreationDate="2010-11-18T21:20:13.603" Score="1" ViewCount="173" Body="&lt;p&gt;How to prove the following theorem?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Theorem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the binomial numbers $C(n,p)$ and $C(n,q)$ are equal, with $p$ different from $q$, then $p+q = n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;($n$, $p$ and $q$ are natural numbers)&lt;/p&gt;&#xA;" OwnerUserId="2938" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-11-18T22:19:50.610" LastActivityDate="2010-11-19T16:56:38.343" Title="$C(n,p) = C(n,q) =&gt; p + q= n$" Tags="&lt;binomial-coefficients&gt;" AnswerCount="5" />
  <row Id="10893" PostTypeId="2" ParentId="10892" CreationDate="2010-11-18T21:21:33.990" Score="6" ViewCount="" Body="&lt;p&gt;The binomial coefficients are &lt;a href=&quot;http://faculty.salisbury.edu/~despickler/pascgalois/classroom/NumberTheory/Fisher/Binomial.pdf&quot; rel=&quot;nofollow&quot;&gt;Unimodal&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-11-18T21:21:33.990" CommentCount="1" />
  <row Id="10894" PostTypeId="2" ParentId="7064" CreationDate="2010-11-18T21:22:15.223" Score="0" ViewCount="" Body="&lt;p&gt;This is so interesting, especially the idea of language dictating direction!  I grew up in Argentina and attended a bilingual school...so I learned both European and US notation but am stronger in the US because we used it more often.  Now I'm teaching language and will do a lesson on the different &quot;formats&quot; for simple math.  Gracias a todos!&lt;/p&gt;&#xA;" LastActivityDate="2010-11-18T21:22:15.223" CommunityOwnedDate="2010-11-18T21:22:15.223" CommentCount="1" />
  <row Id="10895" PostTypeId="2" ParentId="2949" CreationDate="2010-11-18T21:27:47.037" Score="3" ViewCount="" Body="&lt;p&gt;Apart from the above sets of interesting and surprising results, this one should also be mentioned.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gaussian quadrature rule:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An $n$ point Gaussian quadrature rule, yields an exact result for polynomials of degree $2n − 1$ or less.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastActivityDate="2010-11-18T21:27:47.037" CommunityOwnedDate="2010-11-18T21:27:47.037" CommentCount="1" />
  <row Id="10896" PostTypeId="1" AcceptedAnswerId="10915" CreationDate="2010-11-18T21:28:32.963" Score="2" ViewCount="160" Body="&lt;p&gt;Here's the probability (I think) that a particle in Brownian motion &#xA;(w/ standard deviation $\sqrt{t}$) will exceed $m$ between times $t_1$ and $t_2$: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac1{2\sqrt{2\pi}}\int_{-\infty }^m \frac1{\sqrt{t_1}}e^{-\frac{x^2}{2t_1}}\left(1+\mathrm{erf}\left(\frac{m-x}{2\sqrt{t_2-t_1}}\right)\right)\mathrm{d}x$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or, in Mathematica (slightly different form): &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;p[m_,t1_,t2_] := Integrate[ &#xA; PDF[NormalDistribution[0,Sqrt[t1]]][x]* &#xA; CDF[NormalDistribution[x, Sqrt[2]*Sqrt[t2-t1]]][m], &#xA;{x,-Infinity,m}, Assumptions -&amp;gt; { &#xA; t2 &amp;gt;= t1 &amp;gt;= 0, Element[m,Reals], Element[t2, Reals], Element[t1, Reals]} &#xA;] &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Mathematica&lt;/em&gt; can numerically integrate this for specific values of &lt;code&gt;m&lt;/code&gt;, &#xA;&lt;code&gt;t1&lt;/code&gt;, and &lt;code&gt;t2&lt;/code&gt;, but it's not superfast. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I now want to find the partial derivatives of &lt;code&gt;p[]&lt;/code&gt; with respect to each &#xA;of its variables. Ideally in closed-form (&lt;em&gt;Mathematica&lt;/em&gt; can't find one), &#xA;but a good approximation if not. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've tried power series and a few other techniques, but I've found &#xA;nothing good enough for a wide range of values for &lt;code&gt;m&lt;/code&gt;, &lt;code&gt;t1&lt;/code&gt;, and &lt;code&gt;t2&lt;/code&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solving this problem will calculate the &quot;Greeks&quot; for box options:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://money.stackexchange.com/questions/4312/calculating-fair-value-of-an-oanda-com-box-option&quot;&gt;http://money.stackexchange.com/questions/4312/calculating-fair-value-of-an-oanda-com-box-option&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2469" LastEditorUserId="346" LastEditorDisplayName="" LastEditDate="2010-11-19T16:19:59.153" LastActivityDate="2010-11-19T16:19:59.153" Title="Derivatives of Brownian motion or Box Options Greeks" Tags="&lt;calculus&gt;&lt;numerical-methods&gt;&lt;finance&gt;&lt;computer-algebra-systems&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" />
  <row Id="10897" PostTypeId="1" AcceptedAnswerId="10899" CreationDate="2010-11-18T21:30:08.097" Score="1" ViewCount="125" Body="&lt;p&gt;I need to re-arrange the following equation to make $d$ the subject [i.e., solve for $d$]:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$a = \sqrt{\frac{d}{be}}.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The square root part is really confusing me, any help would be much appreciated.&lt;/p&gt;&#xA;" OwnerUserId="3686" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-19T00:15:15.650" LastActivityDate="2010-11-19T00:15:15.650" Title="Re-arranging an equation with a square root" Tags="&lt;algebra-precalculus&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="10899" PostTypeId="2" ParentId="10897" CreationDate="2010-11-18T21:36:12.230" Score="2" ViewCount="" Body="&lt;p&gt;Ok. Squaring both sides you obtain $$a^{2} = \frac{d}{be} \Longrightarrow d = (be) \times a^{2}$$&lt;/p&gt;&#xA;" LastActivityDate="2010-11-18T21:56:47.377" CommentCount="9" />
  <row Id="10900" PostTypeId="2" ParentId="10892" CreationDate="2010-11-18T21:42:31.100" Score="4" ViewCount="" Body="&lt;p&gt;It is suffices to show that $q! (n-q)! = p! (n-p)!$ implies $p+q=n$ ($p \ne q$). Assume without loss of generality that $p &amp;gt; q$. Then, we need to show that $p!/q! = (n-q)!/(n-p)!$ implies $p+q=n$, or $(q+1) (q+2) \cdots p = (n-p+1) (n-p+2) \cdots (n-q)$ implies $p+q=n$. Since both sides of the last equality have identical number of terms ($p-q$), we must have $q+1 = n-p+1$, that is $p+q=n$.&lt;/p&gt;&#xA;" OwnerUserId="2810" LastEditorUserId="2810" LastEditorDisplayName="" LastEditDate="2010-11-18T21:57:59.487" LastActivityDate="2010-11-18T21:57:59.487" />
  <row Id="10901" PostTypeId="2" ParentId="10885" CreationDate="2010-11-18T21:48:21.550" Score="11" ViewCount="" Body="&lt;p&gt;Well, no one will be able to provide an example, since it doesn't exist. In the appendix A to chapter $1$ of his book 'A Comprehensive Introduction to Differential Geometry, volume $1$' Spivak proves that a connected, locally compact, paracompact space is $\sigma$-compact. This last term means that you can write your space as a countable union of compact subspaces. It's easy to see that every $\sigma$-compact space is Lindelöf.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here you are considering a connected, locally compact metric space, so it is $\sigma$-compact and then Lindelöf. But, in metric spaces, it is equivalent to being separable.&lt;/p&gt;&#xA;" OwnerUserId="329" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2010-11-19T13:44:22.690" LastActivityDate="2010-11-19T13:44:22.690" CommentCount="1" />
  <row Id="10902" PostTypeId="2" ParentId="2949" CreationDate="2010-11-18T22:46:36.460" Score="7" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Probabilistic_method&quot; rel=&quot;nofollow&quot;&gt;Erdös' Probabilistic Method&lt;/a&gt; because it is so elegant.&lt;/p&gt;&#xA;" OwnerUserId="3330" LastActivityDate="2010-11-18T22:46:36.460" CommunityOwnedDate="2010-11-18T22:46:36.460" CommentCount="1" />
  <row Id="10903" PostTypeId="1" CreationDate="2010-11-18T22:52:28.040" Score="3" ViewCount="102" Body="&lt;p&gt;For a university project, I am implementing an automated regression analysis tool.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I have very little background in statistics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what books / articles / material would you suggest I could use to brush up on this topic, based on your experiences?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-18T23:17:26.573" LastActivityDate="2010-11-19T03:40:53.907" Title="Recommended Reading on Regression Analysis?" Tags="&lt;statistics&gt;&lt;reference-request&gt;&lt;regression&gt;" AnswerCount="2" />
  <row Id="10904" PostTypeId="1" AcceptedAnswerId="10926" CreationDate="2010-11-18T22:56:04.143" Score="1" ViewCount="123" Body="&lt;p&gt;Question:&lt;br&gt;&#xA;Let $a$ be a quadratic residue to a prime modulus $p$. Prove $a^{(p-1)/2} \equiv 1 \pmod{p}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My attempt at a solution:&lt;br&gt;&#xA;\begin{align*}&#xA;&amp;amp;a\text{ is a quadratic residue}\\&lt;br&gt;&#xA;&amp;amp;\Longrightarrow a\text{ is a residue class of $p$ which has even index $c$ relative to a primitive root $g$}\\ &#xA;&amp;amp;\Longrightarrow a \equiv g^c \pmod{p}\\&lt;br&gt;&#xA;&amp;amp;\Longrightarrow a \equiv g^{2k} \pmod{p}\text{ where $2k=c$}\\&lt;br&gt;&#xA;&amp;amp;\Longrightarrow g^{2kv} \equiv g^{c} \pmod{p}\text{ for some natural number $v$}\\&lt;br&gt;&#xA;&amp;amp;\Longrightarrow 2kv \equiv c \pmod{p-1}\text{ (by a proof in class)}\\&lt;br&gt;&#xA;&amp;amp;\Longrightarrow 2kv \equiv 2k \pmod{p-1}\\&#xA;&amp;amp;\Longrightarrow kv \equiv k \pmod{(p-1)/2}\\&lt;br&gt;&#xA;&amp;amp;\Longrightarrow v \equiv k(k^{-1}) \pmod{(p-1)/2}\text{ since $\gcd(2k, p-1)$ is not equal to 1}\\&#xA;&amp;amp;\Longrightarrow k^{-1} \text{ (k inverse exists)}\\&lt;br&gt;&#xA;&amp;amp;\Longrightarrow v \equiv 1 \pmod{(p-1)/2}.&#xA;\end{align*}  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe this implies that $g^{(p-1)/2} \equiv 1 \pmod{p}$, is this correct?&lt;br&gt;&#xA;Although what I was required to show was $a^{(p-1)/2} \equiv 1 \pmod{p}$, am I on the right track, how do I show this, I've spent quite some time on this and looked over all the proofs in my notes, I can't seem to find out how.&lt;/p&gt;&#xA;" OwnerUserId="2588" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-19T04:38:39.400" LastActivityDate="2010-11-19T04:45:08.737" Title="Let a be a quadratic residue to a prime modulus p. Prove a^((p-1)/2) = 1 mod p" Tags="&lt;number-theory&gt;&lt;modular-arithmetic&gt;" AnswerCount="2" CommentCount="6" />
  <row Id="10905" PostTypeId="1" AcceptedAnswerId="10908" CreationDate="2010-11-18T23:11:41.337" Score="1" ViewCount="81" Body="&lt;p&gt;My algebra is sooo rusty so this should be an easy one!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Area of a rectangle is&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;A = W x H&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;If you're given A (say 150) and a ratio of W:H (say 3:2) how can you calculate W and H?&lt;/p&gt;&#xA;" OwnerUserId="3690" LastActivityDate="2010-11-18T23:54:01.680" Title="Calculate width and Height of rectangle containing given area and conforming to given ratio." Tags="&lt;algebra&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="10906" PostTypeId="2" ParentId="10903" CreationDate="2010-11-18T23:13:32.067" Score="4" ViewCount="" Body="&lt;p&gt;Off the top of my head:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0072472278&quot; rel=&quot;nofollow&quot;&gt;Bevington &lt;em&gt;Data Reduction and Error Analysis for the Physical Sciences&lt;/em&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0471170828&quot; rel=&quot;nofollow&quot;&gt;Draper, Smith &lt;em&gt;Applied Regression Analysis&lt;/em&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0471746967&quot; rel=&quot;nofollow&quot;&gt;Chatterjee, Hadi &lt;em&gt;Regression Analysis by Example&lt;/em&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0471691178&quot; rel=&quot;nofollow&quot;&gt;Belsley, Kuh, Welsch &lt;em&gt;Regression Diagnostics: Identifying Influential Data and Sources of Collinearity&lt;/em&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0521791685&quot; rel=&quot;nofollow&quot;&gt;Monahan &lt;em&gt;Numerical Methods of Statistics&lt;/em&gt;&lt;/a&gt; (not entirely about regression, but the chapter on regression has pointers on implementation)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm sure there are other nice refs, but I'll let the others point them out.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-11-18T23:13:32.067" CommentCount="2" />
  <row Id="10907" PostTypeId="2" ParentId="10892" CreationDate="2010-11-18T23:15:33.880" Score="1" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\ $ In order to prove that $\rm\ \binom{n}{p}\ =\ \binom{n}{q}\ \:\Rightarrow\ \  \ p+q = n\ \ \ if\ \ \ p \ne q\ \ $ for $\rm\ \ p,q\in [0, n]$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;consider how to prove that $\rm\ \ \ sin(p)\ =\ sin(q)\ \Rightarrow\ p+q = \pi\ \ \ if\ \ \ p \ne q\ \ $ for $\rm\ \ p,q\in [0,\pi]$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;MORAL&lt;/strong&gt; $\ $ A proper choice of $\rm \:sin\:$ can get you &quot;over the hump&quot;.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-11-18T23:40:38.043" LastActivityDate="2010-11-18T23:40:38.043" />
  <row Id="10908" PostTypeId="2" ParentId="10905" CreationDate="2010-11-18T23:30:02.880" Score="3" ViewCount="" Body="&lt;p&gt;Right so  you have $W:H$ is $3:2$, which means $$150 = 3x \times 2x$$ So this is just a simple equation which you can solve for. That is $6x^{2}=150$ which says $x^{2} =25$ and hence $x =\sqrt{25}$ that is $x =5 $. So $W = 3 \times 5 =15 $ and $H= 2 \times 5 =10$ and you can see that they have ratio $3 : 2$.&lt;/p&gt;&#xA;" LastActivityDate="2010-11-18T23:49:34.907" CommentCount="3" />
  <row Id="10909" PostTypeId="2" ParentId="10904" CreationDate="2010-11-18T23:32:14.123" Score="6" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\ $ Put $\rm\ a = b^2\ $ and use Fermat's little theorem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;NOTE&lt;/strong&gt; $\ $ The converse is also true - it is known as &lt;a href=&quot;http://en.wikipedia.org/wiki/Euler%27s_criterion&quot; rel=&quot;nofollow&quot;&gt;&lt;strong&gt;Euler's criterion&lt;/strong&gt;.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2010-11-19T03:50:13.880" LastActivityDate="2010-11-19T03:50:13.880" />
  <row Id="10910" PostTypeId="1" CreationDate="2010-11-18T23:51:54.423" Score="0" ViewCount="131" Body="&lt;p&gt;I'm writing a program to solve this proof, but I don't know how to go about solving it.&#xA;If anyone has some insight it would be great help.  Thanks&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;For every odd integer $n$, $3 \leq n \leq 199$, there exists an integer $m \geq 0$, and a prime number $p$, such that $n = 2^{m} + p$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="3691" LastActivityDate="2010-11-18T23:54:46.883" Title="Method to solving this proof with a java app" Tags="&lt;prime-numbers&gt;" CommentCount="6" />
  <row Id="10911" PostTypeId="2" ParentId="10905" CreationDate="2010-11-18T23:54:01.680" Score="2" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\rm\displaystyle\ \ \frac{W}H = \frac{3}2\ \Rightarrow\ 2\:W = 3\:H\:,\ $ so $\rm\ 300 = 2\:W\:H = 3\:H\:H\ \Rightarrow\ H = \ldots $&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-11-18T23:54:01.680" CommentCount="2" />
  <row Id="10912" PostTypeId="2" ParentId="10885" CreationDate="2010-11-19T00:34:27.023" Score="2" ViewCount="" Body="&lt;p&gt;Meanwhile, the &lt;a href=&quot;http://en.wikipedia.org/wiki/Long_line_%28topology%29&quot; rel=&quot;nofollow&quot;&gt;long line&lt;/a&gt; is an example of a space that is non-separable, locally compact, connected  and &lt;em&gt;locally&lt;/em&gt; metrizable, but not metrizable. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The long line is the space $[0,1)\times\omega_1$, consisting of $\omega_1$ many copies of the half-open unit interval, using the (inverse lexical) order topology, where $\omega_1$ is the first uncountable ordinal. These intervals fit together neatly, so that every initial segment of the order looks exactly like an interval in the real line. Thus, the long line is locally compact, locally metrizable and connected. But the long line is non-separable, since every countable subset of $\omega_1$ and hence of the long line is bounded. Lastly, it is not metrizable, since otherwise this would contradict Nuno's answer.&lt;/p&gt;&#xA;" OwnerUserId="413" LastEditorUserId="413" LastEditorDisplayName="" LastEditDate="2010-11-19T00:43:03.973" LastActivityDate="2010-11-19T00:43:03.973" CommentCount="2" />
  <row Id="10913" PostTypeId="2" ParentId="2949" CreationDate="2010-11-19T00:38:33.103" Score="4" ViewCount="" Body="&lt;p&gt;I was very surprised to learn about the Cantor set, and all of its amazing properties. The first one I learnt is that it is uncountable (I would never have told), and that it has measure zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was shown this example as a freshman undergraduate, for an example of a function that is Riemann-integrable but whose set of points of discontinuity is uncountable. (equivalently, that this set has measure zero). This came more as a shock to me, since I had already studied some basic integrals in high school, and we had defined the integral only for continuous functions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Later, after learning topology and when learning measure theory, I was extremely shocked to see that this set can be modified to a residual set of measure zero! I think the existence of such sets and the disconnectednes of topology and measure still gives me the creeps...&lt;/p&gt;&#xA;" OwnerUserId="2614" LastActivityDate="2010-11-19T00:38:33.103" CommunityOwnedDate="2010-11-19T00:38:33.103" />
  <row Id="10915" PostTypeId="2" ParentId="10896" CreationDate="2010-11-19T01:26:12.523" Score="4" ViewCount="" Body="&lt;p&gt;In view of your previous &lt;a href=&quot;http://math.stackexchange.com/questions/9608/determining-distribution-of-maximum-of-dependent-normal-variables&quot;&gt;post&lt;/a&gt;,&#xA;the probability you are considering is actually&#xA;$$&#xD;&#xA;\bar P(m,t_1,t_2) = 1 - \frac{1}{{\sqrt {2\pi t_1 } }}\int_{ - \infty }^m {{\rm erf}\bigg(\frac{{m - x}}{{2\sqrt {t_2  - t_1 } }}\bigg)e^{ - x^2 /(2t_1 )} {\rm d}x},\;\; m \in {\bf R}.&#xD;&#xA;$$&#xD;&#xA;This is the complement of the probability considered in your previous post.&#xA;So, we may consider instead the partial derivatives of &#xA;$$&#xD;&#xA;P(m,t_1,t_2) = \frac{1}{{\sqrt {2\pi t_1 } }}\int_{ - \infty }^m {{\rm erf}\bigg(\frac{{m - x}}{{2\sqrt {t_2  - t_1 } }}\bigg)e^{ - x^2 /(2t_1 )} {\rm d}x}.&#xD;&#xA;$$&#xD;&#xA;Now, erf(x) has derivative ${\rm erf}&amp;#39;(x)= \frac{2}{{\sqrt \pi  }}e^{ - x^2 }$, $x &amp;gt; 0$. &#xA;Then, the partial derivative with respect to $t_2$ is obtained as follows:&#xA;$$&#xD;&#xA;\frac{{\partial}}{{\partial t_2 }}P(m,t_1 ,t_2 ) = \frac{1}{{\sqrt {2\pi t_1 } }}\int_{ - \infty }^m {\frac{\partial }{{\partial t_2 }}{\rm erf}\bigg(\frac{{m - x}}{{2\sqrt {t_2  - t_1 } }}\bigg)e^{ - x^2 /(2t_1 )} {\rm d}x},&#xD;&#xA;$$&#xD;&#xA;and explicitly,&#xA;$$&#xD;&#xA;\frac{{\partial}}{{\partial t_2 }}P(m,t_1 ,t_2 ) = &#xD;&#xA;\frac{{ - 1}}{{\sqrt {8 \pi ^2 t_1 (t_2  - t_1 )^3 } }}\int_{ - \infty }^m {\exp \bigg[ - \frac{{(m - x)^2 }}{{4(t_2  - t_1 )}} - \frac{{x^2 }}{{2t_1 }}\bigg](m - x){\rm d}x}.&#xD;&#xA;$$&#xD;&#xA;(The probability that the maximum is smaller than $m$ decreases as $t_2$ increases, hence the minus sign in this equation.) To find the partial derivative with respect to $m$ corresponds to taking the derivative of a convolution, according to the rule $(f * g)&amp;#39; = f&amp;#39; * g$. Specifically, this says that&#xA;$$&#xD;&#xA;\frac{{\partial}}{{\partial m }}P(m,t_1 ,t_2 ) = \frac{1}{{\sqrt {2\pi t_1 } }}\int_{ - \infty }^m {\frac{\partial }{{\partial m }}{\rm erf}\bigg(\frac{{m - x}}{{2\sqrt {t_2  - t_1 } }}\bigg)e^{ - x^2 /(2t_1 )} {\rm d}x},&#xD;&#xA;$$&#xD;&#xA;and explicitly,&#xA;$$&#xD;&#xA;\frac{{\partial}}{{\partial m }}P(m,t_1 ,t_2 ) =&#xD;&#xA;\frac{{1}}{{\sqrt {2  \pi ^2 t_1 (t_2  - t_1 ) } }}\int_{ - \infty }^m {\exp \bigg[ - \frac{{(m - x)^2 }}{{4(t_2  - t_1 )}} - \frac{{x^2 }}{{2t_1 }}\bigg]{\rm d}x}.&#xD;&#xA;$$&#xD;&#xA;Maybe you can simplify further. The partial derivative with respect to $t_1$ is apparently a quite complicated expression.&lt;/p&gt;&#xA;" OwnerUserId="2810" LastActivityDate="2010-11-19T01:26:12.523" CommentCount="8" />
  <row Id="10916" PostTypeId="1" CreationDate="2010-11-19T01:42:05.760" Score="3" ViewCount="238" Body="&lt;p&gt;I learned of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Plate_trick&quot; rel=&quot;nofollow&quot;&gt;plate trick&lt;/a&gt; via Wikipedia, which states that this is a demonstration of the fact that SU(2) double-covers SO(3). It also offers a &lt;a href=&quot;http://gregegan.customer.netspace.net.au/APPLETS/21/21.html&quot; rel=&quot;nofollow&quot;&gt;link&lt;/a&gt; to an animation of the &quot;belt trick&quot; which is apparently equivalent to the plate trick. Since I've thought most about the belt version, I'll phrase my question in terms of the belt trick.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not clear on how the plate/belt trick relates to the double covering. Specifically, I am looking for a sort of translation of each step of the belt trick into the Lie group setting. For example, am I correct in interpreting the initial twisting of the belt as corresponding to the action of a point in SU(2)? Which point? Do I have the group right? &lt;/p&gt;&#xA;" OwnerUserId="3694" LastEditorUserId="3694" LastEditorDisplayName="" LastEditDate="2010-11-19T16:54:32.150" LastActivityDate="2010-11-19T16:54:32.150" Title="Can somebody explain the plate trick to me?" Tags="&lt;recreational-mathematics&gt;&lt;lie-groups&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="10917" PostTypeId="2" ParentId="10866" CreationDate="2010-11-19T02:02:27.203" Score="2" ViewCount="" Body="&lt;p&gt;Otherwise :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Changing the base of logarithm, we have :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\log_{b}a= \displaystyle\frac{\log_{x}a}{\log_{x}b}$$ and&#xA;$$\log_{x}b= \displaystyle\frac{\log_{b}b}{\log_{b}x} = \frac{1}{\log_{b}x} $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By combine these two ecuations, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\log_{b}a= (\log_{b}x)(\log_{x}a) \Leftrightarrow (\log_{b}a)(\log_{x}x)= (\log_{b}x)(\log_{x}a) \Leftrightarrow \log_{x}x^{\log_{b}a} = \log_{x}a^{\log_{b}x}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By last, canceling $\log_{x}$ on both sides, we have :&#xA;$$x^{\log_{b}a} = a^{\log_{b}x}$$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="3421" LastActivityDate="2010-11-19T02:02:27.203" />
  <row Id="10918" PostTypeId="2" ParentId="10916" CreationDate="2010-11-19T02:52:49.787" Score="6" ViewCount="" Body="&lt;p&gt;A 360 degree rotation of the arm results in a very sore arm unless you unwind it or wind it a second time. Think about the rotation of the hand as being a path in the space of all rotations of 3-space. Or if you really want to rotate something, tape a belt to a soccer ball and tape the other end of the belt to the wall. Just rotate the ball a full rotation, the belt gets a twist. Now rotate again; the belt has two full twists. The twists in the belt can be undone by looping the ball back through the belt as in the animation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The diagram in the lower right shows the paths in the space $SO(3)$ which is the $3$-ball with its antipodal points on its boundary identified. That is why the doubled path appears to be broken; it is passing through antipodal points. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might play with a Mobius band and the pencil loop which maps twice around to gain more intuition. The point of thinking of the Mobius band is that it is embedded in the space of rotations. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is really worth thinking this through while going to sleep or while riding the bus!&lt;/p&gt;&#xA;" OwnerUserId="722" LastActivityDate="2010-11-19T02:52:49.787" CommentCount="2" />
  <row Id="10919" PostTypeId="2" ParentId="10903" CreationDate="2010-11-19T03:40:53.907" Score="3" ViewCount="" Body="&lt;ul&gt;&#xA;&lt;li&gt;Thisted, &lt;em&gt;Elements of Statistical&#xA;Computing&lt;/em&gt; &lt;/li&gt;&#xA;&lt;li&gt;Rao, &lt;em&gt;Linear Statistical&#xA;Inference and Its Applications&lt;/em&gt;&lt;/li&gt;&#xA;&lt;li&gt;Chapter 3 and 4 of Hastie, Tibshirani and Friedman, &lt;em&gt;Elements of Statistical Learning&lt;/em&gt; (&lt;a href=&quot;http://www-stat.stanford.edu/~tibs/ElemStatLearn/&quot; rel=&quot;nofollow&quot;&gt;downloadable&lt;/a&gt;)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1195" LastActivityDate="2010-11-19T03:40:53.907" CommentCount="1" />
  <row Id="10920" PostTypeId="1" AcceptedAnswerId="10925" CreationDate="2010-11-19T04:01:37.127" Score="0" ViewCount="94" Body="&lt;p&gt;Let a be a natural number.&lt;br&gt;&#xA;Suppose  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a^4 \not\equiv 1 \pmod {29}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a^7 \not\equiv 1 \pmod {29}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a^{14} \not\equiv 1 \pmod {29}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Calculate the order of a modulo 29.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using Euler's Criterion, I have $a^{14} = a^{(29-1)/2} \not\equiv 1 \pmod {29}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How would I go about solving this problem, I have not seen this kind of problem before.&lt;/p&gt;&#xA;" OwnerUserId="2588" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-19T04:12:07.643" LastActivityDate="2010-11-19T22:04:15.720" Title="Let a be a natural number, Suppose $a^4, a^7, a^{14} \not\equiv 1 \pmod{29}$, calculate the order of a modulo 29." Tags="&lt;number-theory&gt;&lt;modular-arithmetic&gt;" AnswerCount="4" />
  <row Id="10921" PostTypeId="2" ParentId="2949" CreationDate="2010-11-19T04:03:54.773" Score="1" ViewCount="" Body="&lt;p&gt;I remember a homework question in elementary school. Something like this: Billy and Jane's house is x blocks east and y blocks north of school. Billy walks home by walking east for x blocks and then north for y blocks. Jane decides to take a short cut: she walks alternately a block north and a block east. (There is a picture: Jane's route is a step-like hypotenuse.) Is Jane's route really a short cut?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course it is exactly the same distance, but I found this really hard to digest. I knew that in the triangle the sum of the two sides would exceed the hypotenuse. &lt;/p&gt;&#xA;" OwnerUserId="1728" LastActivityDate="2010-11-19T04:03:54.773" CommunityOwnedDate="2010-11-19T04:03:54.773" CommentCount="4" />
  <row Id="10922" PostTypeId="2" ParentId="10920" CreationDate="2010-11-19T04:07:55.770" Score="3" ViewCount="" Body="&lt;p&gt;First we assume $(a,29) = 1$. Else order doesn't exist.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Use Fermat's little theorem (or) Euler's theorem $a^{\phi(n)} \equiv 1 \pmod n$, whenever $(a,n) = 1$. $\phi(29) = 28$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $a^l \equiv 1 \pmod n$, then order of $a$ divides $l$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So order has to divide $28$ but not $4$,$7$,$14$. So the order has to be $28$.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2010-11-19T04:09:48.843" LastActivityDate="2010-11-19T04:09:48.843" CommentCount="1" />
  <row Id="10923" PostTypeId="2" ParentId="10920" CreationDate="2010-11-19T04:08:22.453" Score="2" ViewCount="" Body="&lt;p&gt;What do you know about the multiplicative group modulo 29?  What are the possible orders of nonzero elements mod 29?  (It's not hard to answer if you know the structure of the group; it's also possible to compute directly, since there are only 28 elements to check.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you know the structure, you should be able to work out the answer.&lt;/p&gt;&#xA;" OwnerUserId="1461" LastActivityDate="2010-11-19T04:08:22.453" />
  <row Id="10924" PostTypeId="1" AcceptedAnswerId="10935" CreationDate="2010-11-19T04:12:27.157" Score="2" ViewCount="92" Body="&lt;p&gt;I'm not sure if this is the proper place to ask this, but I am looking for an English translation of Voronoy's paper on the generalized Pell's equation.  It might be his doctoral dissertation; I'm not sure.  I believe the title is &quot;On a generalization of the Algorithm of Continued Fractions&quot;.&lt;/p&gt;&#xA;" OwnerUserId="2505" LastEditorUserId="2505" LastEditorDisplayName="" LastEditDate="2010-11-19T04:18:09.650" LastActivityDate="2010-11-19T06:56:01.820" Title="English translation of Voronoy paper on generalized Pell's equation." Tags="&lt;number-theory&gt;&lt;diophantine-equations&gt;" AnswerCount="1" />
  <row Id="10925" PostTypeId="2" ParentId="10920" CreationDate="2010-11-19T04:18:15.853" Score="3" ViewCount="" Body="&lt;p&gt;Presumably, $\gcd(a,29)=1$; else, $a\equiv 0\pmod{29}$, and all your incongruences hold, and there is no order as Sivaram points out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From Fermat's Little Theorem, you know that $a^{28}\equiv 1\pmod{29}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, what is the &lt;em&gt;smallest&lt;/em&gt; positive integer $j$ such that $a^j\equiv 1\pmod{29}$? It is at most $28$. If we divide $28$ by $j$ with remainder, $28 = qj + r$, $0\leq r\lt j$, we have that&#xA;$$a^r = a^{28-qj} \equiv a^{28}(a^{j})^{-q} \equiv 1 \pmod{29}.$$&#xD;&#xA;Since $j$ is the &lt;em&gt;smallest&lt;/em&gt; positive integer such that $a^j\equiv 1 \pmod{29}$, but $r$ is nonnegative and smaller, you must have $r=0$. (This is a standard argument, shows up all the time; it is really the fact that the set of exponents $r$ for which $a^r\equiv 1 \pmod{29}$ form an &lt;a href=&quot;http://en.wikipedia.org/wiki/Ideal_%28ring_theory%29&quot; rel=&quot;nofollow&quot;&gt;ideal of the integers&lt;/a&gt;). So $j$ divides $28$. That is, the order must divide $28$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what can $j$ be? It must be positive, it must divide $28=2^2\times 7$. So it must be either $1$, $2$, $4$, $7$, $14$, or $28$. Given the information you have, what must $j$ be?&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-19T22:04:15.720" LastActivityDate="2010-11-19T22:04:15.720" CommentCount="1" />
  <row Id="10926" PostTypeId="2" ParentId="10904" CreationDate="2010-11-19T04:45:08.737" Score="2" ViewCount="" Body="&lt;p&gt;Bill has succintly told you how to prove the result. But you were also asking for comments on your proposed argument. I will address that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In line 5, where did the $v$ come from, and what is its role? Notice that you can take $v=1$ and what you write is true. So how is this giving you any information?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In line 9, you are already using an inverse of $k$, even though you only assert its existence in the next line. You cannot do that: in order to use it, you must first show it exists, and you haven't done it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But assuming it does exist, and that your entire chain of argument holds, you'll notice that all you concluded was that $v\equiv 1\pmod{(p-1)/2}$. This is of course natural: you have $a=g^c=g^{2k}=g^{2kv}$; you can always take $v=1$ and that will work regardless of $k$, $g$, $a$... And you probably know now that it does not lead to a proof.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you had not actually proven anything. You've only written $a$ as an even power of a primitive root, and &lt;em&gt;that's it&lt;/em&gt;. Lines 1 through 4 are correct; but from line 5 through the end, you are just spinning your wheels and not getting any closer to the result you want.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-11-19T04:45:08.737" CommentCount="3" />
  <row Id="10927" PostTypeId="1" AcceptedAnswerId="10930" CreationDate="2010-11-19T04:54:42.533" Score="10" ViewCount="340" Body="&lt;p&gt;This is motivated by looking at the calculus exams of some of my undergraduate students. A recurring mistake is assuming that the derivative of the product of functions is a product of derivatives and the derivative of the quotient of two functions is the quotient of their derivatives. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This might be an ill formed question, but do their exist a pair of functions such that these rules hold. Explicitly, do there exist $f(x),g(x)$ satisfying all of:&#xA;\begin{align*}&#xA;f&amp;amp;\neq 0;\\&#xA;\frac{d(fg)}{dx}&amp;amp;=\frac{df}{dx}\cdot \frac{dg}{dx}\\&#xA;\text{and }\frac{d(f/g)}{dx}&amp;amp;=\frac{df}{dx}/ \frac{dg}{dx}\text{ ?}&#xA;\end{align*}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Just to clarify my question, looking at the responses so far. I am looking for functions satisfying all three conditions above simultaneously.&lt;/p&gt;&#xA;" OwnerUserId="3431" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-11-19T16:29:34.687" LastActivityDate="2010-11-19T16:29:34.687" Title="When do the Freshman's dream product and quotient rules for differentiation hold?" Tags="&lt;calculus&gt;&lt;differential-equations&gt;" AnswerCount="3" CommentCount="7" FavoriteCount="3" />
  <row Id="10928" PostTypeId="2" ParentId="10920" CreationDate="2010-11-19T04:59:50.950" Score="0" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\rm\quad\quad\quad D\:|\:28,\ \ \ D\:\nmid\:4,\:14\ \ \Rightarrow\ \ D = 28\:.\ $ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Generally $\rm\ \ \ D\ |\ N = P_1^{e_1}\:\cdots\ P_k^{e_k}\:,\ \ \ \forall\ i:\ D\:\nmid N/P_i\ \ \Rightarrow\ \ D = N\:,\ \ \ assuming\ all\  P_j\ are\ prime\:$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This idea is at the heart of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Pocklington_primality_test&quot; rel=&quot;nofollow&quot;&gt;&lt;strong&gt;Pocklington-Lehmer primality test.&lt;/strong&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-11-19T06:49:28.677" LastActivityDate="2010-11-19T06:49:28.677" CommentCount="2" />
  <row Id="10930" PostTypeId="2" ParentId="10927" CreationDate="2010-11-19T05:22:19.897" Score="10" ViewCount="" Body="&lt;p&gt;I don't think there is a solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f&amp;#39;g&amp;#39;=f&amp;#39;g+g&amp;#39;f$ implies $\frac{f&amp;#39;}{f}=\frac{g&amp;#39;}{g&amp;#39;-g}$ where the denominators are nonzero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{f&amp;#39;}{g&amp;#39;}=\frac{f&amp;#39;g-g&amp;#39;f}{g^2}$ implies $\frac{f&amp;#39;}{f}=\frac{(g&amp;#39;)^2}{g(g&amp;#39;-g)}$ where the denominators are nonzero.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $f$ and $g$ are nonzero and $g&amp;#39;-g$ is nonzero, this implies that $\frac{g&amp;#39;}{g&amp;#39;-g}=\frac{(g&amp;#39;)^2}{g(g&amp;#39;-g)}$, which implies $g&amp;#39;=g$.  Thus $g=g&amp;#39;$ if $f$ and $g$ are nonzero, so $g(x)=ce^x$.  But now from $f&amp;#39;g&amp;#39;=f&amp;#39;g+g&amp;#39;f$ we have $ce^xf(x)=0$, which implies $f=0$.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;For solutions to just the product rule, the equations above suggest taking $g$ such that $\frac{g&amp;#39;}{g&amp;#39;-g}$ is integrable, and &#xA;$$f(x)=\exp\left(\int_a^x\frac{g&amp;#39;(t)}{g&amp;#39;(t)-g(t)}dt\right),$$ which in particular works for $f(x)=e^{bx}$ and $g(x)=e^{cx}$ with $bc=b+c$ (as seen in a now deleted answer).  Similarly, for just the quotient rule, one could try finding $g$ such that $\frac{(g&amp;#39;)^2}{g(g&amp;#39;-g)}$ is integrable, and then take &#xA;$$f(x)=\exp\left(\int_a^x\frac{g&amp;#39;(t)^2}{g(t)(g&amp;#39;(t)-g(t))}dt\right).$$  &#xA;For example, $f(x)=e^{bx}$ and $g(x)=e^{cx}$ with $bc=b+c^2$.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2010-11-19T05:57:09.663" LastActivityDate="2010-11-19T05:57:09.663" CommentCount="4" />
  <row Id="10931" PostTypeId="2" ParentId="10927" CreationDate="2010-11-19T05:29:40.147" Score="6" ViewCount="" Body="&lt;p&gt;The product rule equality gives $f&amp;#39;g+g&amp;#39;f = f&amp;#39;g&amp;#39;$, or $-fg&amp;#39; = f&amp;#39;g-f&amp;#39;g&amp;#39;$. From this, using the quotient rule equality we get&#xA;$$&#xD;&#xA;\frac{f&amp;#39;}{g&amp;#39;} = \frac{gf&amp;#39;-fg&amp;#39;}{g^2} = \frac{gf&amp;#39; +f&amp;#39;g - f&amp;#39;g&amp;#39;}{g^2}&#xD;&#xA;= f&amp;#39;\left(\frac{2g-g&amp;#39;}{g^2}\right).&#xD;&#xA;$$&#xD;&#xA;If $f$ is not constant, then $f&amp;#39;\neq 0$. So we get $\frac{1}{g&amp;#39;} = \frac{2g-g&amp;#39;}{g^2}$, or $g^2 - 2gg&amp;#39;+(g&amp;#39;)^2 = 0$, or $(g-g&amp;#39;)^2 = 0$; hence $g=g&amp;#39;$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, from the product rule again we have $f&amp;#39;g&amp;#39;=f&amp;#39;g+fg&amp;#39;$, or $f&amp;#39;g = f&amp;#39;g+fg$, so $fg=0$. This requires $g=0$, which makes the quotient rule impossible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $f$ is constant, then the product rule equality requires $g$ constant, and then the quotient rule equality cannot hold. &lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-11-19T05:29:40.147" CommentCount="5" />
  <row Id="10932" PostTypeId="1" CreationDate="2010-11-19T05:38:14.587" Score="0" ViewCount="49" Body="&lt;p&gt;Is there any partial differential equation which doesn't have any classical as well as generalized fuction solution?&lt;/p&gt;&#xA;" LastActivityDate="2010-11-19T05:38:14.587" Title="classical function as well as generalized function" Tags="&lt;functions&gt;" CommentCount="1" />
  <row Id="10933" PostTypeId="2" ParentId="10927" CreationDate="2010-11-19T05:45:16.930" Score="7" ViewCount="" Body="&lt;p&gt;Another answer to your question from a different angle is this: They hold in a different kind of calculus.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you swap subtraction for division and division for taking roots in the usual definition of the derivative, you get &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f^*(x) = \lim_{\Delta x \to 0} \left(\frac{f(x+\Delta x)}{f(x)}\right)^{\frac{1}{\Delta x}}.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's not too hard to prove that $(fg)^* = f^* g^*$ and $(f/g)^* = f^*/g^*$ under this definition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The calculus that results from this definition of the derivative (and the corresponding definition of the integral) goes by various names, such as &quot;&lt;a href=&quot;http://en.wikipedia.org/wiki/Multiplicative_calculus&quot; rel=&quot;nofollow&quot;&gt;multiplicative calculus&lt;/a&gt;,&quot; &quot;non-Newtonian calculus,&quot; and &quot;product calculus.&quot;  Many of the standard results in the usual calculus (e.g., Fundamental Theorem, Mean Value Theorem, l'Hopital-type rules, Taylor's Theorem) can be redone for this multiplicative calculus.  The idea goes back at least to Vito Volterra in the late 1800s.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to the Wikipedia page, &quot;Opinions differ as to the usefulness of the multiplicative calculi,&quot; but there are some applications.  For instance, the product derivative measures the multiplicative rate of change of some function and so is useful when you're interested in looking at, say, growth rates of stock prices.  The product integral has some corresponding applications; my favorite is that it can be used to calculate geometric means (in the same way that the usual integral can find arithmetic means).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Multiplicative calculus also has a close relationship to the usual calculus.  For instance, the derivatives are related via &#xA;$$f^*(x) = \exp\left(\frac{d}{dx}\ln |f(x)|\right) = e^{f&amp;#39;(x)/f(x)}.$$&#xD;&#xA;The product integral and the usual integral have a similar relationship.  In fact, I think part of the reason some folks don't find this multiplicative calculus all that interesting is that product derivatives and integrals can so easily be expressed in terms of the usual ones.  The question, I suppose, is whether there is really anything new and different going on here.  (Note also that this relationship I just gave means that the product derivative is just $\exp$ of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Logarithmic_derivative&quot; rel=&quot;nofollow&quot;&gt;logarithmic derivative&lt;/a&gt; of $f$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, a few years ago I wrote a &lt;a href=&quot;http://math.pugetsound.edu/~mspivey/ProdCalc.pdf&quot; rel=&quot;nofollow&quot;&gt;survey paper&lt;/a&gt; and had a student do a &lt;a href=&quot;http://math.pugetsound.edu/~mspivey/ProdInt.pdf&quot; rel=&quot;nofollow&quot;&gt;summer research project&lt;/a&gt; on this before I realized how much was already out there and before the Wikipedia page appeared.  There are lots of other references and information on the Wikipedia page, so if you find this interesting you should check it out.  The page on the &lt;a href=&quot;http://en.wikipedia.org/wiki/Product_integral&quot; rel=&quot;nofollow&quot;&gt;product integral&lt;/a&gt; is also a good source.&lt;/p&gt;&#xA;" OwnerUserId="2370" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-11-19T16:28:56.747" LastActivityDate="2010-11-19T16:28:56.747" CommentCount="3" />
  <row Id="10934" PostTypeId="2" ParentId="653" CreationDate="2010-11-19T06:01:04.227" Score="2" ViewCount="" Body="&lt;p&gt;Homotopy theory / algebraic topology was born out of applications rather than abstract nonsense considerations. So there's plenty of applications, as that's how the subject began. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps the first topological proof would be the bridges of Konigsberg problem: &lt;a href=&quot;http://en.wikipedia.org/wiki/Seven_Bridges_of_K%C3%B6nigsberg&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Seven_Bridges_of_K%C3%B6nigsberg&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where algebraic topology started getting off the ground was in the work of Poincare.  The Poincare-Hopf Index theorem: &lt;a href=&quot;http://en.wikipedia.org/wiki/Poincar%C3%A9%E2%80%93Hopf_theorem&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Poincar%C3%A9%E2%80%93Hopf_theorem&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;was a landmark.  In its natural setting it was a relationship between Euler characteristic, tangent bundles and intersection theory.  But from the perspective of a differential equator it's a fundamental tool that allows you to determine whether or not differential equations have fixed points.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Applications have piled-up over the years.  Some of the more modern ones are listed in other people's responses.  The birth of topological dynamics in the mid 20-th century was of course a big one. &lt;/p&gt;&#xA;" OwnerUserId="642" LastActivityDate="2010-11-19T06:01:04.227" />
  <row Id="10935" PostTypeId="2" ParentId="10924" CreationDate="2010-11-19T06:01:53.227" Score="3" ViewCount="" Body="&lt;p&gt;If you seek a detailed exposition of Voronoi's algorithm then see below. Based on what they wrote, the authors may have an English translation of some of his work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.jstor.org/stable/2006105&quot; rel=&quot;nofollow&quot;&gt;&lt;strong&gt;Calculation of the Regulator of a Pure Cubic Field&lt;/strong&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By H. C. Williams, G. Cormack and E. Seah&lt;br&gt;&#xA;Mathematics of Computation, Vol. 34, No. 150 (Apr., 1980), pp. 567-6&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Abstract&lt;/strong&gt;. A description is given of a modified version of Voronoi's algorithm for obtaining the regulator of a pure cubic field $\mathbb Q(\sqrt[3]{D})$. This new algorithm has the advantage of executing relatively rapidly for large values of $D$. It also eliminates a computational problem which occurs in almost all algorithms for finding units in algebraic number fields. This is the problem of performing calculations involving algebraic irrationals by using only approximations of these numbers. &lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-11-19T06:56:01.820" LastActivityDate="2010-11-19T06:56:01.820" CommentCount="3" />
  <row Id="10936" PostTypeId="1" AcceptedAnswerId="10939" CreationDate="2010-11-19T06:11:29.823" Score="5" ViewCount="118" Body="&lt;p&gt;So, we are considering the subset&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ S = \{(x, y) \in \mathbb{R^2} | (x \text{ and } y \in \mathbb{Q}) \text{ or }  (x \text{ and } y \notin \mathbb{Q})\} $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And consider its complement $$ T = \mathbb{R^2} \backslash S $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The set T is disconnected, actually I am fairly certain it is totally disconnected. I am just having problems showing that rigorously. I was trying to show it using straight lines but I don't think I was getting anywhere. I know that a totally disconnected set's only connected sets are the one point sets. I've been trying to show that given two arbitrary points, that a separation exists between them. It is more difficult since this is in the plane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any hints at all would be a great help. Maybe I'm making mountains of molehills.&lt;/p&gt;&#xA;" OwnerUserId="2465" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-19T06:36:43.183" LastActivityDate="2010-11-19T06:41:37.377" Title="showing a set is totally disconnected" Tags="&lt;real-analysis&gt;&lt;homework&gt;&lt;topology&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="10937" PostTypeId="1" AcceptedAnswerId="12083" CreationDate="2010-11-19T06:20:02.563" Score="0" ViewCount="70" Body="&lt;p&gt;Suppose you want to find $E(X)$ for a non-negative random variable $X$. So you can use $\int_{0}^{\infty} [1-F(x)] \ dx$. To get $1-F(x)$, all you have to do is integrate as follows: $\int_{x}^{\infty} f(t) \ dt$? In other words, instead of integrating from $t=0$ to $t= x$, we can just integrate from $t = x$ to $t = \infty$? &lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2010-11-27T17:01:22.133" Title="Expected Value and Distribution Functions" Tags="&lt;probability&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="10939" PostTypeId="2" ParentId="10936" CreationDate="2010-11-19T06:32:59.553" Score="2" ViewCount="" Body="&lt;p&gt;I had written out a full solution, but since this is homework, I've removed it and replaced it with this suggestion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One way to prove that $T$ is totally disconnected is to show that whenever $p$ and $q$ are distinct points of $T$, then there are open sets $A$ and $B$ of $\mathbb{R}^2$ such that $T\subseteq A\cup B$, $(A\cap B)\cap T=\emptyset$, $p\in A$, and $q\in B$. This will show that there is a disconnection of $T$ in which $p$ and $q$ are in distinct components. In particular, $p$ and $q$ cannot be in the same connected component of $T$. If this holds for &lt;em&gt;all&lt;/em&gt; pairs of points $p$ and $q$, then the connected components of $T$ must be single points.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, pick two distinct points $p$ and $q$ in $T$. Try to find a line that is completely contained in $S$ and which separates $p$ and $q$. One way to achieve a line completely contained in $S$ is to have it go through a rational point with rational slope. Then throw away the line to get your two sets $A$ and $B$.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-19T06:41:37.377" LastActivityDate="2010-11-19T06:41:37.377" />
  <row Id="10940" PostTypeId="1" CreationDate="2010-11-19T06:33:59.070" Score="1" ViewCount="149" Body="&lt;p&gt;Hi,&#xA;I have some questions about Sasaki -and Sasaki-Einstein geometry.&#xA;I hope, you can answer me some of these questions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1.) Let $(M, g)$ a Riemannian manifold and $(C(M)=\mathbb{R}_{&amp;gt;0} \times M, \overline{g}=dr^{2}+r^{2}g)$ denote its metric &lt;/p&gt;&#xA;&#xA;&lt;p&gt;cone. In the paper &quot;Sasaki-Einstein Manifolds&quot; from Lames Sparks, I find that the Sasaki manifold $(M, g)$ is naturally isometrically embedded into the cone via the inclusion&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;M=\lbrace r=1\rbrace=\lbrace1 \times M\rbrace \subset C(M).&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I see this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2.) Let $\overline{\nabla}$ the Levi-Civita connection of $\overline{g}$. Is then $\overline{\nabla}_{\partial_{r}}\partial_{r}=0$&#xD;&#xA;or &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\overline{\nabla}_{\partial_{r}}\partial_{r}=\partial_{r}$ with &#xA;$\partial_{r}:=\frac{\partial}{\partial r}$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3.) How can I show, that the Lie-derivative of $J$ in direction from $r \partial_{r}$ is equal to zero?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4.) If I define $\xi=r \partial_{r}$, how can I show that $\mathcal{L}_{\xi} \overline{g}=0$, i.e. $\xi$ is a Killing vectorfield. With $\mathcal{L}$ I mean the Lie-derivative.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;5.) Let $\eta=d^{c} \log r=i(\partial-\overline{\partial}) \log r$, where $\partial$ and $\overline{\partial}$ the usual Dolbeault operators. How can I show that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;\eta(X)=\frac{1}{r^{2}} \overline{g}(J(r\partial_{r}), X)&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and the Kähler 2-form on $C(M)$ is $\omega=\frac{1}{2}d(r^{2}\eta)=\frac{1}{2} i \partial \overline{\partial} r^{2}$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;6.) We define $\Phi(X)$ as $\Phi(X)=\nabla_{X} \xi$. How can I show that $(\nabla_{X}\Phi) Y=g(\xi, Y)X-g(X, Y)\xi$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;7.) &lt;a href=&quot;http://arxiv.org/PS_cache/hep-th/pdf/9810/9810250v1.pdf&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/PS_cache/hep-th/pdf/9810/9810250v1.pdf&lt;/a&gt;, Proposition 1.1.2, page number 4. Can someone proof the equivalence (i) $\Leftrightarrow$ (ii)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;8.) Let $M=S^{2m+1} \cong \mathbb{C}^{m+1}$. This is a Sasaki manifold. But what is $\xi$ oder $\Phi$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;9.) Analogous: Let $M=\mathbb{R}^{2m+1}$, it is also Sasaki, but what is $\xi$ or $\Phi$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;10.) I have read the following about 3-Sasaki-manifolds. The definition of 3-Sasaki-manifolds (cone is hyperkähler) implies &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\text{Hol}(\overline{g}) \subset Sp(k) \subset SU(2k)$ and then that 3-Sasaki manifolds are also Sasaki-Einstein. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I prove this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And the last question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;11.) Let $(M, g)$ a Sasaki-manifold of dimension $2m+1$. Then the following are equivalent:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i) $(M, g)$ is Sasaki with $\text{Ric}_{g}=2m \cdot g$.&#xA;ii) $(C(M), \overline{g})$ is Ricci flat.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the sketch of proof?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks a lot and best regards&#xA;Ronald&lt;/p&gt;&#xA;" LastActivityDate="2010-11-19T06:41:37.577" Title="11 questions about Sasaki -and Sasaki-Einstein geometry" Tags="&lt;differential-geometry&gt;" CommentCount="4" />
  <row Id="10941" PostTypeId="2" ParentId="1181" CreationDate="2010-11-19T07:03:26.147" Score="1" ViewCount="" Body="&lt;p&gt;You can quickly reduce to the case of a transitive action, in which case we just want to explain why the total number of times that something gets fixed is exactly the size of the group.  But in this case everything is symmetric at all points in the (unique) orbit.  So to count all the times something gets fixed, we can just count how many times a particular x gets fixed, and multiply by the size of the orbit.  Now we've reduced to the fact that the size of an orbit is the index of the stabilizer.&lt;/p&gt;&#xA;" OwnerUserId="1392" LastEditorUserId="1392" LastEditorDisplayName="" LastEditDate="2010-11-19T07:22:51.490" LastActivityDate="2010-11-19T07:22:51.490" />
  <row Id="10942" PostTypeId="1" AcceptedAnswerId="10954" CreationDate="2010-11-19T07:48:07.137" Score="6" ViewCount="126" Body="&lt;p&gt;I heard about it sometime somewhere and want to read about it now, but I can't recall what the name is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Start with $a_1 = \ldots =a_n=1$. Choose a number between 1 and $n$ with probability $a_i/(a_1+ \ldots + a_n)$ to choose $i$. If $i_0$ is the number chosen, increase $a_i$ by 1 and now choose another number and so on indefinitely.&lt;/p&gt;&#xA;" OwnerUserId="3533" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-11-19T16:14:56.903" LastActivityDate="2010-11-19T21:15:13.410" Title="What's the name of this stochastic process?" Tags="&lt;probability&gt;" AnswerCount="2" FavoriteCount="0" />
  <row Id="10943" PostTypeId="2" ParentId="10414" CreationDate="2010-11-19T08:14:00.693" Score="0" ViewCount="" Body="&lt;p&gt;I was able to solve the problem with a computer algorithm but am still not certain how it can be done mathematically. It was pointed out by @Greg Muller that the knapsack problem is applicable but I just don't have the mathematical prowess to apply it. If someone could show how that can be done please do.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Will share my logic here, hopefully it is useful to someone out there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first step is to determine the spell with the most damage per cast time. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This spell becomes the &quot;baseline&quot; spell since it will guarantee the highest damage per second. Meaning, you should always cast this spell if the following 2 conditions are met:&#xA;1) The baseline spell is available (not on cooldown).&#xA;2) You are not currently casting a spell.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So it then becomes a matter of filling in other spells while the baseline spell is on cooldown. Between (cast time) and (cooldown - cast time). However,  some overlapping can occur (rule 2 above is false).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It then becomes a matter of recursing through all non-baseline spells to find all sequences of spells which do not violate the 2 rules.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For spells which DO overlap you must penalize them for potential damage the baseline spell could have done (up to its maximum damage).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take for example, 2 spells &lt;/p&gt;&#xA;&#xA;&lt;h1&gt;1: 300 damage, 3s cast time, 10s cooldown&lt;/h1&gt;&#xA;&#xA;&lt;h1&gt;2: 290 damage, 3s cast time,  3s cooldown&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;The most damage comes from the sequence 1 - 2 - 2 - 2. Which causes an overlap of 2 seconds into a potential #1 cast. However, this is still beneficial since if you dont cast the third spell (ie. 1 - 2 - 2) you will do 880 damage with 1 second to spare. If you cast the extra #2 spell you will do 1170 - 2 second of #1 which is 200. So 970 damage is your relative damage.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This algorithm is significantly faster than other algorithms which look for sequences that match a target goal: ie. time limit or damage.&lt;/p&gt;&#xA;" OwnerUserId="3534" LastActivityDate="2010-11-19T08:14:00.693" CommentCount="2" />
  <row Id="10944" PostTypeId="1" CreationDate="2010-11-19T09:22:02.837" Score="0" ViewCount="89" Body="&lt;p&gt;Is there a simple algorithm for efficient decomposition on this special kind of graph?&lt;/p&gt;&#xA;" OwnerUserId="998" LastActivityDate="2010-11-19T09:22:02.837" Title="Tree decomposition for planar nearest neighbor graphs" Tags="&lt;graph-theory&gt;" CommentCount="0" />
  <row Id="10946" PostTypeId="1" CreationDate="2010-11-19T09:53:08.977" Score="1" ViewCount="58" Body="&lt;p&gt;I'm looking for an algorithm to solve this problem:&#xA;Given set of n points on 2D euclidean space, create a net of rectilinear edges, so that:&#xA;1. Every two points are connected with shortest edge.&#xA;2. Sum of all edges is minimal (minimized).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's clearly a NP-hard problem and it's similar to Steiner tree problem but I can't use any of the common aproaches because of constraint 1. Moreover, in my case, every algorithm to solve that problem is feasible as long as it has polynomial complexity - the function of the objective is here the key.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any good ideas how to solve this? My current aproach is rather naive: connect every two points with an edge and then merge the edges that are overlapping.&lt;/p&gt;&#xA;" OwnerUserId="3704" LastActivityDate="2010-11-19T09:53:08.977" Title="Minimum rectilinear net" Tags="&lt;topology&gt;&lt;algorithms&gt;&lt;optimization&gt;" CommentCount="6" />
  <row Id="10947" PostTypeId="2" ParentId="7201" CreationDate="2010-11-19T09:59:39.423" Score="0" ViewCount="" Body="&lt;p&gt;Perhaps what you're after is on page 31 in Wilf's &lt;a href=&quot;http://www.math.upenn.edu/~wilf/DownldGF.html&quot; rel=&quot;nofollow&quot;&gt;generatingfunctionology&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Proposition&lt;/strong&gt;. A formal power series $f=\sum_{n \geq 0} a_n x^n$ has a reciprocal if and only if $a_0 \neq 0$.  In that case the reciprocal is unique.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="139" LastActivityDate="2010-11-19T09:59:39.423" CommentCount="1" />
  <row Id="10948" PostTypeId="1" CreationDate="2010-11-19T10:00:34.853" Score="1" ViewCount="56" Body="&lt;p&gt;Suppose T is an elliptic Möbius transformation that fixes p and q.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a. Prove that the pole and inverse pole lie on the perpendicular bisector of he line segment pq.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;b. Show that T is the composition of two inversions about cline hat contain p and q.&lt;/p&gt;&#xA;" OwnerUserId="2888" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-19T10:27:42.693" LastActivityDate="2010-11-19T10:27:42.693" Title="Elliptic Möbius transformations" Tags="&lt;geometry&gt;&lt;homework&gt;&lt;complex-analysis&gt;" />
  <row Id="10949" PostTypeId="1" AcceptedAnswerId="11014" CreationDate="2010-11-19T10:06:51.457" Score="8" ViewCount="258" Body="&lt;p&gt;Is it known whether every finite abelian group is isomorphic to the ideal class group of the ring of integers in some number field? If so, is it still true if we consider only imaginary quadratic fields?&lt;/p&gt;&#xA;" OwnerUserId="546" LastEditorUserId="546" LastEditorDisplayName="" LastEditDate="2010-11-19T10:11:56.160" LastActivityDate="2010-11-20T12:30:20.480" Title="Finite abelian groups as class groups" Tags="&lt;number-theory&gt;&lt;algebraic-number-theory&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="4" />
  <row Id="10950" PostTypeId="2" ParentId="8001" CreationDate="2010-11-19T10:34:56.583" Score="2" ViewCount="" Body="&lt;p&gt;I'm sorry I didn't see your question earlier.  The answer is in fact quite easy: Just permute the symbols with 3-&gt;6, 4-&gt;3 and 6-&gt;4 (or its inverse if you want to go in the other direction).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, you can permute the symbols arbitrarily in a Latin square and it will preserve row-completeness and column-completeness (or lack thereof).&lt;/p&gt;&#xA;" OwnerUserId="139" LastEditorUserId="139" LastEditorDisplayName="" LastEditDate="2010-11-19T10:50:15.237" LastActivityDate="2010-11-19T10:50:15.237" CommentCount="1" />
  <row Id="10951" PostTypeId="2" ParentId="2949" CreationDate="2010-11-19T10:36:33.417" Score="2" ViewCount="" Body="&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/L8M7q.jpg&quot; alt=&quot;Burning ship fractal&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fractals, especially the ones related to simple dynamical processes like the Mandelbrot set or this eerie Burning ship fractal really still inspire me with awe.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;It's not really a mathematical result, but after seeing all the nice entries here, I thought this lighter one would fit in well:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Young man, in mathematics you don't understand things. You just get used to them.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I see all the examples here, this dictum by von Neumann comes to mind. I'm always remembered of how true it is.&lt;/p&gt;&#xA;" OwnerUserId="3567" LastEditorUserId="3567" LastEditorDisplayName="" LastEditDate="2011-04-05T14:58:18.313" LastActivityDate="2011-04-05T14:58:18.313" CommunityOwnedDate="2010-11-19T10:36:33.417" />
  <row Id="10952" PostTypeId="1" AcceptedAnswerId="10959" CreationDate="2010-11-19T10:43:52.690" Score="2" ViewCount="205" Body="&lt;p&gt;In Chapter 1 of &lt;em&gt;&lt;a href=&quot;http://www.springer.com/mathematics/algebra/book/978-3-642-03979-9&quot; rel=&quot;nofollow&quot;&gt;Polynomials&lt;/a&gt;&lt;/em&gt; by Victor Prasolov, Springer, 2001, the following theorem is proved. (p.3)&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Theorem 1.1.4 (Ostrovsky). Let&#xA;  $f(x)=x^{n}-b_{1}x^{n-1}-\cdots -b_{n}$,&#xA;  where all the numbers $b_{i}$ are non-negative and at least one of them&#xA;  is nonzero. If the greatest common&#xA;  divisor of the indices of the positive&#xA;  coefficients $b_{i}$ is equal to&#xA;  $1$, then $f$ has a unique positive&#xA;  root $p$ and the absolute value of&#xA;  the other roots are $&amp;lt;$ p.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The following is one of the Problems to Chapter 1 (p.41).&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Problem 1.5 - Find the number of real&#xA;  roots of the following polynomials&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;a) ...&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;b) $nx^{n}-x^{n-1}-\cdots -1$&#xD;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question: How to solve this Problem?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Added: $nx^{n}-x^{n-1}-\cdots -1=0$ $\Leftrightarrow x^{n}-\dfrac{1}{n}x^{n-1}-\cdots -\dfrac{1}{n}=0$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Added 2: &lt;a href=&quot;http://planetmath.org/encyclopedia/SturmsTheorem.html&quot; rel=&quot;nofollow&quot;&gt;Sturm's Theorem.&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="752" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2010-11-20T00:00:03.113" LastActivityDate="2010-11-20T00:00:03.113" Title="Problem from Victor Prasolov's Polynomials -- Finding the number of real roots of $nx^{n}-x^{n-1}-\cdots -1$" Tags="&lt;polynomials&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="10953" PostTypeId="2" ParentId="10134" CreationDate="2010-11-19T11:00:14.900" Score="1" ViewCount="" Body="&lt;p&gt;For the sake of having less open questions and the fact that the final is probably already done I will show that if $f \in L^p(X,\mu)$, $0 &amp;lt; p &amp;lt; \infty$ then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\|f\|_p^p = p \int_0^\infty x^{p - 1} \mu\{z : |f(z)| &amp;gt; x\} \, dx$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align*}&#xA;p \int_0^\infty &amp;amp;x^{p - 1} \mu\{z : |f(z)| &gt; x\} \, dx\\&#xA;&amp;amp;= p \int_0^\infty \int_X 1_{{z : |f(z)| &gt; x}} \, d\mu(z) \, dx \text{ writing the measure as an integral.}\\&#xA;&amp;amp;= \int_X \int_0^{|f(z)|} p x^{p - 1} \, dx \, d\mu(z) \text{ Fubini,}\\&#xA;&amp;amp;= \int_X |f(x)|^p \, d\mu(x)&#xA;\end{align*}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, you have $p = 1$, so&#xA;$$\int_{2}^\infty \mu \{x \in [0,1] : |f(x)| &amp;gt; t \} \, dt \leq \int_{2}^\infty \frac{1}{t \log^2 t} \, dt = \frac{1}{\log(2)}.$$&lt;/p&gt;&#xA;" OwnerUserId="1120" LastActivityDate="2010-11-19T11:00:14.900" CommentCount="2" />
  <row Id="10954" PostTypeId="2" ParentId="10942" CreationDate="2010-11-19T11:54:28.483" Score="6" ViewCount="" Body="&lt;p&gt;For this process and a substantial generalization, see &lt;a href=&quot;http://www.combinatorics.net/aoc/toc/v7n2/Chung/7_2_141.pdf&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt;.&#xA;The case $p=0$ and $\gamma = 1$ corresponds to the process you described.&lt;/p&gt;&#xA;" OwnerUserId="2810" LastActivityDate="2010-11-19T11:54:28.483" CommentCount="1" />
  <row Id="10955" PostTypeId="1" AcceptedAnswerId="10976" CreationDate="2010-11-19T11:55:38.263" Score="1" ViewCount="228" Body="&lt;p&gt;I want to motivate the theory of $C_0$-semigroups to someone, and the following question was asked:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What is an example of a non-separable linear PDE?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Preferably a simple homogeneous one.&lt;/p&gt;&#xA;" OwnerUserId="1120" LastEditorUserId="1120" LastEditorDisplayName="" LastEditDate="2010-11-19T12:40:54.963" LastActivityDate="2010-11-19T16:22:59.473" Title="Non-separable linear PDE." Tags="&lt;differential-equations&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="10956" PostTypeId="2" ParentId="5595" CreationDate="2010-11-19T12:17:15.240" Score="0" ViewCount="" Body="&lt;p&gt;@crasic, Can you outline the formal proof out here?&lt;/p&gt;&#xA;" OwnerUserId="2565" LastActivityDate="2010-11-19T12:17:15.240" CommentCount="2" />
  <row Id="10957" PostTypeId="2" ParentId="2949" CreationDate="2010-11-19T12:38:49.397" Score="1" ViewCount="" Body="&lt;p&gt;I was very surprised when I discovered that $$0.\overline{99} = 1$$&lt;/p&gt;&#xA;" OwnerUserId="2411" LastEditorUserId="2411" LastEditorDisplayName="" LastEditDate="2010-11-20T09:03:55.323" LastActivityDate="2010-11-20T09:03:55.323" CommunityOwnedDate="2010-11-19T12:38:49.397" />
  <row Id="10958" PostTypeId="2" ParentId="10827" CreationDate="2010-11-19T12:49:47.727" Score="0" ViewCount="" Body="&lt;p&gt;For the case $K=\mathbb{Q}(q^{1/p})$ a $p$-extension over $\mathbb{Q}$ the discriminant of $K/\mathbb{Q}$ is $p^{3}q^{p-1}$ so both $p$ and $q$ ramify. I don't understand why $q$ should be congruent to 1 mod $p$.&lt;/p&gt;&#xA;" LastActivityDate="2010-11-19T12:49:47.727" />
  <row Id="10959" PostTypeId="2" ParentId="10952" CreationDate="2010-11-19T13:00:23.703" Score="8" ViewCount="" Body="&lt;p&gt;I think there is some value here in knowing how to do such problems &quot;by hand.&quot;  The proof in this case is quite simple.  If $|x| &amp;gt; 1$, then $|x^n| &amp;gt; |x^k|$ for $k &amp;lt; n$, hence&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$n |x|^n &amp;gt; |x|^{n-1} + |x|^{n-2} + ... + |1| \ge |x^{n-1} + x^{n-2} + ... + 1|$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;by the triangle inequality, so this polynomial $f(x)$ has no roots of absolute value greater than $1$.  It follows that any real roots lie in $[-1, 1]$.  By inspection $x = 1$ is a root and $x = 0, -1$ are not, so any remaining roots lie in $(0, 1)$ or $(-1, 0)$.  If $x \in (0, 1)$, then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x^{n-1} + x^{n-2} + ... + 1 &amp;gt; nx^n$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so there are no roots in $(0, 1)$.  To find any remaining roots in $(-1, 0)$, let&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$g(x) = f(x) (x - 1) = nx^n(x - 1) - x^n + 1 = nx^{n+1} - (n+1) x^n + 1.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then $g&amp;#39;(x) = n(n+1) x^n - n(n+1) x^{n-1} = n(n+1) x^{n-1}(x - 1)$ has roots $x = 0, 1$, hence $g$ is monotonic on $(-1, 0)$, so to determine if there are roots on this interval it suffices to compute $g(-1)$ and $g(0)$.  We have $g(0) = 1$ and $g(-1) = -2n$ if $n$ is even and $g(-1) = 2n+2$ if $n$ is odd.  In the first case there is one real root in $(-1, 0)$ by the IVT and in the second case there are none.&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-11-19T14:13:28.693" LastActivityDate="2010-11-19T14:13:28.693" CommentCount="2" />
  <row Id="10960" PostTypeId="1" AcceptedAnswerId="10961" CreationDate="2010-11-19T13:30:13.820" Score="2" ViewCount="67" Body="&lt;p&gt;Suppose $0.6$ of students submit assigned homework on time. In a typical class of $40$ students a professor has assigned homework. If a student submits homework on time, he gets $5$ points. Otherwise he gets $0$ points. Find the variance of the total number of points by all students. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So this is a binomial distribution. The variance is $40(0.6)(0.4)$? &lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2010-11-19T13:36:46.593" Title="Binomial Distribution" Tags="&lt;probability&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="10961" PostTypeId="2" ParentId="10960" CreationDate="2010-11-19T13:36:46.593" Score="2" ViewCount="" Body="&lt;p&gt;Close, but the variance has to depend upon the number of points given.  You have the variance in the number of students that turn in the work.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-11-19T13:36:46.593" CommentCount="3" />
  <row Id="10963" PostTypeId="1" CreationDate="2010-11-19T14:16:32.193" Score="1" ViewCount="80" Body="&lt;p&gt;I have an interval $[2,t]$ containing some number of primes. I now want to divide this interval into two intervals $a=[2,m]$ and $b=]m,t]$ such that the number of primes in $a$ and $b$ is almost the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am aware of the approximation $\pi(t)=\frac{t}{2\ln(t)}$, but can not seem to get any good and usable result out of this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is for a computer implementation, so avoiding Lambert functions and likewise functions would be much appreciated.&lt;/p&gt;&#xA;" OwnerUserId="3317" LastActivityDate="2011-03-28T09:38:27.670" Title="Dividing an interval, such that the primes get divided even" Tags="&lt;prime-numbers&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="10964" PostTypeId="2" ParentId="10963" CreationDate="2010-11-19T14:41:12.310" Score="2" ViewCount="" Body="&lt;p&gt;The approximation $\pi(x) \sim \frac{x}{\ln x}$ given by the prime number theorem tells you there are about $\frac{t}{\ln t}$ primes in $[2, t]$.  Now, the PNT also tells you that the $n^{th}$ prime $p_n$ behaves asymptotically like $p_n \sim n \ln n$.  We want $n \sim \frac{t}{2 \ln t}$, hence&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$m \sim \frac{t}{2 \ln t} \left( \ln t - \ln 2 - \ln \ln t \right).$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What this tells you is that for reasonably large values of $t$ (say a hundred digits) the approximation $m \sim \frac{t}{2 \ln t}$ should work quite well.  For small values of $t$ you are probably better off using the sieve of Erathosthenes to explicitly count primes.  You shouldn't really trust the other two terms anyway because the two approximations I gave above are only compatible up to ignoring them.&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-11-19T14:47:44.050" LastActivityDate="2010-11-19T14:47:44.050" />
  <row Id="10965" PostTypeId="1" CreationDate="2010-11-19T14:45:53.723" Score="1" ViewCount="116" Body="&lt;p&gt;Hi,&#xA;i have a question to special cases of Sasaki-manifolds. Let $(M, g, \xi, \eta, \Phi)$ a Sasaki-manifold.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In special case maybe $M=S^{2m+1} \cong \mathbb{C}^{m+1}$. This is a Sasaki manifold. But what is $\Phi$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Analogous: Let $M=\mathbb{R}^{2m+1}$, it is also Sasaki, but what is $\xi$ or $\Phi$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks and best regards&#xA;Ronald&lt;/p&gt;&#xA;" LastActivityDate="2011-03-21T12:33:46.050" Title="Special types of Sasaki manifolds" Tags="&lt;differential-geometry&gt;" AnswerCount="0" CommentCount="2" />
  <row Id="10967" PostTypeId="1" AcceptedAnswerId="10974" CreationDate="2010-11-19T15:29:02.330" Score="1" ViewCount="160" Body="&lt;p&gt;I recently read an article in which the author describes how to find some functions that obey to certain recursion relationships. If we want to find a function that satisfies, for example, &#xA;$f(x^a) = x \cdot f(x)$, then the author explains we can proceed as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider $ x &amp;gt; a &amp;gt; 1$ . Then $f(x) = f((x^{(1/a)})^a) = x^{1/a}f(x^{(1/a)})$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$= x^{1/a}f((x^{(1/a^2)})^a) = x^{1/a + 1/a^2}f(x^{(1/a^2)}) = x^{1/a + 1/a^2 + ... + 1/a^n}f(x^{1/a^n})$ .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know that the limit of $1/a^n$ is zero, when $n$ tends to infinity. The equality $ r + r^2 + r^3 + ... = r/(1-r)$ is also useful. When we finally set $f(1) = 1$, we may write:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f(x) = x^{\frac{1/a}{1-1/a}} f(x^0) = x^{\frac1{a-1}}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the author and I wonder if a function can be found that satisifies the recurrence relationship $f(\log(x)) = x \cdot f(x)$ . For me, the main motivator for asking this question is plain curiosity. As always, pointers to relevant literature are very much appreciated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Max&lt;/p&gt;&#xA;&#xA;&lt;p&gt;NB: log(x) is the natural logarithm of x. &lt;/p&gt;&#xA;" OwnerUserId="1055" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-19T15:43:53.813" LastActivityDate="2010-11-19T17:23:15.133" Title="Find a function such that $f(\log(x)) = x \cdot f(x) $" Tags="&lt;functional-equations&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="10968" PostTypeId="1" AcceptedAnswerId="10971" CreationDate="2010-11-19T15:35:35.317" Score="1" ViewCount="84" Body="&lt;p&gt;$M$, $N$ are sets, $f: M \rightarrow N$ is a function, $R \subseteq M\times M$ is an equivalence relation on $M$, $\pi: M \rightarrow M/R$ is the canonical projection (i.e. $\pi(x)=[x]_R$), and $\ker f \subseteq M\times M$ is defined as $\{(x,y) \in M \times M: f(x)=f(y)\}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Show that $R \subseteq \ker f \Leftrightarrow (\exists F: M/R \rightarrow N)(f=F\circ \pi)$ and that there is exactly one $F$, if both statements in the logical equivalence are true.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, let's see what I've come up with so far:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;$\Rightarrow$&quot;: assume $R \subseteq \ker f$ and construct a function $F: M/R \rightarrow N$ which works like this: given a $c \in M/R$, choose an $x \in M$ with $c=[x]_R$ and let $F(c)=f(x)$. This works exactly because we know that $(\forall a,a&amp;#39; \in [x]_R)(f(a) = f(a&amp;#39;))$. Now let $x \in M$ and $y=\pi(x)=[x]_R$, then $F(y)=f(x)$ and $F(\pi(x))=(F\circ \pi)(x)=f(x)$ for all $x \in M$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;$\Leftarrow$&quot;: assume $\exists F: M/R \rightarrow N$ with $f=F\circ \pi$: then $(\forall c \in \{[x]_R: x\in M\})(F(c)=f(x))$. Let $a \in M/R$ and $x,x&amp;#39; \in a$, then $\pi(x)=\pi(x&amp;#39;)$. We can then apply $F$ to both sides and see that $f(x)=F(\pi(x))=F(\pi(x&amp;#39;))=f(x&amp;#39;)$, i.e. $R\subseteq M\times M$ can be defined as $\{(x,y)\in M\times M: f(x)=f(y)\}$ and therefore $R\subseteq \ker f$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Have I missed something important or is this correct? If so, how would I go about proving that there is exactly one F satisfying the given attributes?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a side-note: I'm very new to mathematics (and I'm no native speaker either), so any hints regarding style/language are very welcome.&lt;/p&gt;&#xA;" OwnerUserId="3546" LastEditorUserId="630" LastEditorDisplayName="" LastEditDate="2010-11-19T17:14:44.307" LastActivityDate="2010-11-19T17:14:44.307" Title="Equivalence relation/kernel/projection: exactly one $F$ such that $f=F\circ \pi$" Tags="&lt;homework&gt;&lt;elementary-set-theory&gt;&lt;relations&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="10969" PostTypeId="1" AcceptedAnswerId="10973" CreationDate="2010-11-19T15:36:01.133" Score="1" ViewCount="136" Body="&lt;p&gt;I was wondering if there are any general results to this problem. Explicitly, given an $n$-cube and $n$ distinct colors, in how many ways can we paint the faces of the cube. Here we consider two colorings the same if they can be obtained from one another by rotations. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $n=3$ this number is $57$ by Burnside's lemma. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there any results for higher dimensions?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: Thanks for all the comments and sorry for the errors in my question. There is no intersection condition. &lt;/p&gt;&#xA;" OwnerUserId="3431" LastEditorUserId="3431" LastEditorDisplayName="" LastEditDate="2010-11-19T17:13:47.887" LastActivityDate="2010-11-19T17:13:47.887" Title="n-coloring an n-cube" Tags="&lt;combinatorics&gt;" AnswerCount="1" CommentCount="7" />
  <row Id="10970" PostTypeId="2" ParentId="10892" CreationDate="2010-11-19T15:46:18.353" Score="0" ViewCount="" Body="&lt;p&gt;Suppose that for a fixed $n$ we have the equality $\binom{n}{p} = \binom{n}{q}$ with $p \neq q$. Observe that&#xA;\begin{eqnarray}&#xA;0 = \frac{\pi}{2^{n-1}} \left[ \binom{n}{p} - \binom{n}{q} \right] = \int_{-\pi}^{\pi} \left( \cos (2p - n)x - \cos (2q - n)x \right) \ \cos^{n} x \ dx.&#xA;\end{eqnarray}&#xA;The cosine difference identity&#xA;\begin{eqnarray}&#xA;\cos A - \cos B = -2 \sin (\tfrac{A + B}{2}) \ \sin (\tfrac{A - B}{2})&#xA;\end{eqnarray}&#xA;implies &#xA;\begin{eqnarray}&#xA;\int_{-\pi}^{\pi} \left( \sin (p + q - n)x \  \sin (p- q)x \right) \ \cos^{n} x \ dx = 0 .&#xA;\end{eqnarray}&#xA;I'll leave it as an exercise to finish the proof and conclude the result.&lt;/p&gt;&#xA;" OwnerUserId="2720" LastActivityDate="2010-11-19T15:46:18.353" />
  <row Id="10971" PostTypeId="2" ParentId="10968" CreationDate="2010-11-19T15:46:25.957" Score="2" ViewCount="" Body="&lt;p&gt;Your &quot;$\Leftarrow$&quot; proof is a muddled. First, $R$ is &lt;em&gt;not&lt;/em&gt; necessarily equal to $\mathrm{ker} f$, but that is what you are asserting is happening when you say &quot;$R$ can be defined as...&quot;.  Second, I don't see you showing that every element of $R$ is in $\mathrm{ker} f$, which is what you need to do.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In that direction, you want to show that if $f=F\circ \pi$ for some function $F\colon M/R \to N$, then $R\subseteq \mathrm{ker} f$. Since you are trying to prove an inclusion, what you want to do is start by taking an element $(a,b)\in R$, and show that $(a,b)\in \mathrm{ker}f$. So you want to show that $f(a)=f(b)$. We know that $f(a) = F\circ\pi(a) = F(\pi(a)) = F([a]_R)$. And then...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To show that the function $F$ is unique, assume you have a function $G\colon M/R\to N$ with $f=G\circ\pi$. You want to show that $F=G$ (this will prove uniqueness). They have the same domain and they have the same codomain, so you only need to show that for every $[x]_R\in M/R$, you have $G([x]_{R})=F([x]_{R})$. Now, $G([x]_R) = G(\pi(x)) = f(x)$. What about $F([x]_{R})$?&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-11-19T15:46:25.957" CommentCount="1" />
  <row Id="10972" PostTypeId="1" CreationDate="2010-11-19T15:56:54.843" Score="5" ViewCount="413" Body="&lt;p&gt;Hello people, two integrals that got my attention because I really don't know how to solve them. They are a solution to the CDW equation below critical temperature of a 1D strongly correlated electron-phonon system. The second one  is used in the theory of superconductivity, while the first is a more complex variation in lower dimensions. I know the result for the second one, but without the whole calculus, it is meaningless. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \int_0^b \frac{\tanh(c(x^2-b^2))}{x-b}\mathrm{d}x $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \int_0^b \frac{\tanh(x)}{x}\mathrm{d}x \approx \ln\frac{4e^\gamma b}{\pi} \text{as} \ b \to \infty$$&#xD;&#xA;where $\gamma = 0.57721...$ is Euler's constant&lt;/p&gt;&#xA;" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-11-22T18:24:28.817" LastActivityDate="2010-11-23T00:41:44.187" Title="Integral with Tanh: $\int_{0}^{b} \tanh(x)/x \mathrm{d} x$" Tags="&lt;calculus&gt;" AnswerCount="3" CommentCount="10" />
  <row Id="10973" PostTypeId="2" ParentId="10969" CreationDate="2010-11-19T16:06:15.277" Score="3" ViewCount="" Body="&lt;p&gt;The answer is $1$ in all dimensions, if you mean what I think you mean by &quot;faces.&quot;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The underlying graph you're trying to color is the graph whose vertices are the points $(\pm 1, 0, 0, ...), (0, \pm 1, 0, ...), ...$ where a vertex is adjacent to every other vertex except the opposite one (and note that by &quot;vertex&quot; here I mean &quot;face of an $n$-cube&quot;); in other words it is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Tur%C3%A1n_graph&quot; rel=&quot;nofollow&quot;&gt;Turan graph&lt;/a&gt; $T_2(2n)$.  In particular the subgraph formed by $(1, 0, 0, ...), (0, 1, 0, ...), ...$ is a $K_n$, so we must use each of the $n$ colors exactly once to color each of these vertices.  But now it's not hard to see that, since a vertex and its opposite share the same neighbors, they must have the same color.  So the only valid colorings are those which use each of the $n$ colors exactly twice each, once for each pair of opposite vertices.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now every permutation of the set of pairs of opposite vertices is realizable by a rotation, and it follows that the coloring is unique up to rotation.   &lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-11-19T16:25:09.700" LastActivityDate="2010-11-19T16:25:09.700" CommentCount="3" />
  <row Id="10974" PostTypeId="2" ParentId="10967" CreationDate="2010-11-19T16:08:27.507" Score="3" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Claim&lt;/strong&gt; given any function $g:(0,1]\to \mathbb{R}$, there exists a unique extension $f:\mathbb{R}\to\mathbb{R}$ satisfying your relation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Proof&lt;/em&gt;. For $x &amp;gt; 1$, there exists a unique $n\in\mathbb{N}$ such that taken the $n$-fold logarithm of $x$ gives you a number in $(0,1]$. And for $x \leq 0$, $e^x \in (0,1]$. So by iteration the function is well-defined. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Edit&lt;/em&gt; Okay, let me do this explicitly. Fix your favourite function $g(x)$ on $(0,1]$. Mine happens to be the &lt;a href=&quot;http://en.wikipedia.org/wiki/Cantor_function&quot; rel=&quot;nofollow&quot;&gt;Cantor function&lt;/a&gt;. It doesn't matter at all for the construction what this function is. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let your $f(x)$ be defined piecewise. For $x\in (0,1]$ define $f(x) = g(x)$. &#xA;For $x\leq 0$, define $f(x) = e^{x} g(e^{x})$. Notice that $e^x$ for $x\leq 0$ is a number in $(0,1]$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $1 &amp;lt; x \leq e$, define $f(x) = \frac1x g(\log x)$. For $e &amp;lt; x \leq e^e$, let $f(x) = \frac1x f(\log x) = \frac1{x\log x} g(\log \log x)$. For $e^e &amp;lt; x \leq e^{e^e}$, let $f(x) = \frac1x f(\log x) = \frac1{x \cdot \log x \cdot \log\log x} g(\log \log \log x)$. And so on. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If your favourite function is $g(x) = 0$, then when you run this procedure you get Chandru's example where $f(x) = 0$ everywhere. If your favourite function is $g(x) = 1$, you have that $f(x) = e^{x}$ for $x \leq 0$, $f(x) = 1$ for $0 &amp;lt; x \leq 1$, $f(x) = \frac1x$ for $1 &amp;lt; x \leq e$, $f(x) = \frac{1}{x\log x}$ for $e &amp;lt; x \leq e^e$ and so on and so forth. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For whatever function you choose to start with as $g(x)$ defined on $(0,1]$, you get one corresponding function $f(x)$ that solves your recurrence relation. Since there are uncountably many functions on $(0,1]$, there are also uncountably many possible solutions to your recurrence relation. &lt;/p&gt;&#xA;" OwnerUserId="1543" LastEditorUserId="1543" LastEditorDisplayName="" LastEditDate="2010-11-19T17:23:15.133" LastActivityDate="2010-11-19T17:23:15.133" CommentCount="7" />
  <row Id="10975" PostTypeId="2" ParentId="9468" CreationDate="2010-11-19T16:12:56.710" Score="0" ViewCount="" Body="&lt;p&gt;Not many people know that WA is also able of giving you all kinds of possible closed forms concerning fundamental constants, e.g.:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.wolframalpha.com/input/?i=2.9299372410244&quot; rel=&quot;nofollow&quot;&gt;http://www.wolframalpha.com/input/?i=2.9299372410244&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can also click &quot;more&quot;...&lt;/p&gt;&#xA;" OwnerUserId="346" LastActivityDate="2010-11-19T16:12:56.710" />
  <row Id="10976" PostTypeId="2" ParentId="10955" CreationDate="2010-11-19T16:22:59.473" Score="1" ViewCount="" Body="&lt;p&gt;See my comments above on some general ideas. For specific examples, perhaps a good starting point would be L. P. Eisenhart, &quot;Enumeration of potentials for which one-particle Schrodinger equations are separable,&quot; Phys. Rev. 74, 87-89 (1948). So just take an arbitrary potential function $V(x)$ that is &lt;em&gt;not&lt;/em&gt; in Eisenhart's list, and consider the Schroedinger equation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ [-i \partial_t + \triangle + V(x)]\psi(t,x) = 0$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is Linear, homogeneous, and aside from the trivial separability of the $t$ variable, satisfies your requirements. &lt;/p&gt;&#xA;" OwnerUserId="1543" LastActivityDate="2010-11-19T16:22:59.473" CommentCount="1" />
  <row Id="10978" PostTypeId="2" ParentId="10892" CreationDate="2010-11-19T16:56:38.343" Score="2" ViewCount="" Body="&lt;p&gt;Mike Spivey correctly points out that unimodality does not suffice. Here's a quick way to fill the gap:  Consider the set $S = \{p,q,n-p,n-q\}$. Now $n\choose x$ assumes only one value for $x\in S$, by hypothesis and the fact that ${n\choose p}={n\choose n-p}$. But $p\neq q$ and $p+q\neq n$ together imply that $S$ has at least three distinct values.  This plus unimodality gives a contradiction.&lt;/p&gt;&#xA;" OwnerUserId="2682" LastActivityDate="2010-11-19T16:56:38.343" CommentCount="2" />
  <row Id="10979" PostTypeId="2" ParentId="9505" CreationDate="2010-11-19T17:02:03.867" Score="3" ViewCount="" Body="&lt;p&gt;Although this thing has already been answered, here a shorter proof&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because $x^y = y^x $ is symmetric we first demand that $x&amp;gt;y$&#xD;&#xA;Then we proceed simply this way:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$ x^y = y^x $       &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$ x = y^{\frac x y } $         &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$ \frac x y = y^{\frac x y -1} $       &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$ \frac x y -1 = y^{\frac x y -1} - 1 $       &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now we expand the rhs into its well-known exponential-series      &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$ \frac x y -1 =  \ln(y)*(\frac x y -1) + \frac {((\ln(y)*(\frac x y -1))^2}{2!} + ... $       &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here by the definition x&gt;y the lhs is positive, so if $ \ln(y) $ &gt;=1 we had lhs $\lt$ rhs&#xA;Thus $ \ln(y) $ must be smaller than 1, and the only integer y&gt;1 whose log is smaller than 1 is y=2, so there is the only possibility $y = 2$ and we are done.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;[update] Well, after having determined $y=2$ the same procedure can be used to show, that after manyally checking $x=3$ (impossible) $x=4$ (possible) no $x&amp;gt;4$ can be chosen.    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We ask for $x=4^{1+\delta}   ,\delta &amp;gt; 0 $ inserting the value 2 for y:    &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$ 4^{(1+\delta)*2}=2^{4^{(1+\delta)}} $      &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Take log to base 2:    &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$ (1+\delta)*4=4^{(1+\delta)} $      &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$ \delta =4^{\delta} - 1 $      &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$ \delta = \ln(4)*\delta  + \frac { (\ln(4)*\delta)^2 }{2!} + \ldots $      &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$ 0 = (\ln(4)-1)*\delta  + \frac { (\ln(4)*\delta)^2 }{2!} + \ldots $      &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because $ \ln(4)-1 &amp;gt;0 $ this can only be satisfied if $ \delta =0 $    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So indeed the only solutions, assuming x&gt;y, is $ (x,y) = (4,2)$ .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[end update]&lt;/p&gt;&#xA;" OwnerUserId="1714" LastEditorUserId="1714" LastEditorDisplayName="" LastEditDate="2010-11-19T18:11:54.043" LastActivityDate="2010-11-19T18:11:54.043" />
  <row Id="10980" PostTypeId="1" AcceptedAnswerId="10986" CreationDate="2010-11-19T17:04:02.763" Score="3" ViewCount="211" Body="&lt;p&gt;Let $A$ be a commutative ring with identity. If $A$ has finite number of prime ideals $p_1,...p_n$ and moreover $\prod_{i=1}^n p_i^{k_i} = 0$ for some $k_i$. Are the prime ideals necessarily maximal? &lt;/p&gt;&#xA;" OwnerUserId="3620" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-19T17:49:26.193" LastActivityDate="2010-11-19T18:14:19.817" Title="Ring with finitely many prime ideals with an extra condition. Are they maximal?" Tags="&lt;commutative-algebra&gt;&lt;ring-theory&gt;" AnswerCount="1" />
  <row Id="10982" PostTypeId="1" AcceptedAnswerId="10983" CreationDate="2010-11-19T17:17:51.827" Score="2" ViewCount="117" Body="&lt;p&gt;The rank of a finitely-generated abelian group is the size of the maximal $\mathbb{Z}$-linearly-independent subset; loosely this means the number of distinct copies of $\mathbb{Z}$ contained as direct summands. I have come across a situation where a notion somewhat like rank may be useful for finite abelian groups (which obviously all have rank zero). I'm looking at graphs embedded on surfaces and I have a procedure for associated certain finite abelian groups to a graph embedding. I've found that the groups I'm getting this way are all quotients of $\mathbb{Z}^{2g}$, where $g$ is the genus of the surface. So for example, I can realize $(\mathbb{Z}/2\mathbb{Z})^2$ as a group coming from a graph embedded on a torus, but I can't get $(\mathbb{Z}/2\mathbb{Z})^3$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's my question: for a finite abelian group $G$, define the &quot;finite rank&quot; of $G$ to be the minimal rank of a free abelian group that surjects onto $G$. Is this a common notion? Although there is the possibility this could be an argumentative question, is it a &lt;em&gt;useful&lt;/em&gt; notion (or conversely, can you make the case for it being &lt;em&gt;useless&lt;/em&gt;)? &lt;/p&gt;&#xA;" OwnerUserId="3694" LastEditorUserId="3694" LastEditorDisplayName="" LastEditDate="2010-11-19T17:32:55.283" LastActivityDate="2010-11-20T06:57:30.257" Title="Analogue of rank for finite abelian groups?" Tags="&lt;group-theory&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="10983" PostTypeId="2" ParentId="10982" CreationDate="2010-11-19T17:26:24.630" Score="5" ViewCount="" Body="&lt;p&gt;Unless I am mistaken, the &quot;finite rank&quot; is just the minimal size of a generating set for $G$. Indeed, if $X$ is a generating set for $G$, then the free abelian group on $X$ surjects onto $G$ by the map induced by the embedding of $X$ into $G$, so the &quot;finite rank&quot; is at most the smallest size of a generating set. And if a free abelian group $F$ of rank $k$ surjects onto $G$, then the image of a free generating set for $F$ maps to a generating set for $G$ (possibly not injectively), so $G$ has a generating set with at most $k$ elements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This number, in an arbitrary group, is usually written $d(G)$; I don't know of any special name for it, and none of my books seems to have any name attached to it. In $p$-groups, it is the index of the Frattini subgroup, $[G:\Phi(G)]$. &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-19T22:02:26.860" LastActivityDate="2010-11-19T22:02:26.860" CommentCount="2" />
  <row Id="10984" PostTypeId="2" ParentId="10972" CreationDate="2010-11-19T17:41:51.003" Score="3" ViewCount="" Body="&lt;p&gt;I get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle \int_{0}^{b}\dfrac{\tanh(x)}{x} \approx C + \log b$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle C = \int_{0}^{1}(\dfrac{\tanh(x)}{x} -\dfrac{2}{x(e^{2/x}+1)} \ )\text{dx} = 0.81878\dots$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have $\displaystyle \log \frac{4e^{\gamma}}{\pi} = 0.81878\dots$, &lt;strike&gt;so your formula could be right! &lt;/strike&gt; It is in fact correct. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;See Derek's excellent &lt;a href=&quot;http://math.stackexchange.com/questions/10972/integral-with-tanh-int-0b-tanhx-x/11358#11358&quot;&gt;answer&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Derivation&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have $\displaystyle \dfrac{\tanh(x)}{x} = \dfrac{1}{x} - \dfrac{2}{x(e^{2x}+1)}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming $b &amp;gt; 1$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle f(b) = \int_{0}^{b}\dfrac{\tanh(x)}{x} \ \text{dx} = \int_{0}^{1}\dfrac{\tanh(x)}{x} \ \text{dx}  + \int_{1}^{b} (\dfrac{1}{x} - \dfrac{2}{x(e^{2x}+1)})\ \text{dx}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle f(b) =  \int_{0}^{1}\dfrac{\tanh(x)}{x} \ \text{dx}  + \log b - \int_{1}^{b}\dfrac{2}{x(e^{2x}+1)}\ \text{dx}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now&#xA;$$\displaystyle \int_{1}^{b}\dfrac{2}{x(e^{2x}+1)}\ \text{dx} = \int_{1}^{\infty}\dfrac{2}{x(e^{2x}+1)}\ \text{dx} - \int_{b}^{\infty}\dfrac{2}{x(e^{2x}+1)}\ \text{dx}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Subsitute $t = \dfrac{1}{x}$ in the first integral, we get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle \int_{1}^{\infty}\dfrac{2}{x(e^{2x}+1)}\ \text{dx} = \int_{0}^{1}\dfrac{2}{t(e^{2/t}+1)}\ \text{dt}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle f(b) =  \int_{0}^{1}(\dfrac{\tanh(x)}{x} -\dfrac{2}{x(e^{2/x}+1)} \ )\text{dx}  + \log b + \int_{b}^{\infty}\dfrac{2}{x(e^{2x}+1)}\ \text{dx}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, as $\displaystyle b \to \infty$, $\displaystyle \int_{b}^{\infty}\dfrac{2}{x(e^{2x}+1)}\ \text{dx} \to 0$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus we have that,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle \int_{0}^{b}\dfrac{\tanh(x)}{x} \approx \log b + \int_{0}^{1}(\dfrac{\tanh(x)}{x} -\dfrac{2}{x(e^{2/x}+1)} \ )\text{dx} $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle \int_{0}^{b}\dfrac{\tanh(x)}{x} \approx C + \log b$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle C = \int_{0}^{1}\left(\dfrac{\tanh(x)}{x} -\dfrac{2}{x(e^{2/x}+1)} \ \right)\text{dx} = 0.81878\dots$$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-22T20:23:11.670" LastActivityDate="2010-11-22T20:23:11.670" CommentCount="6" />
  <row Id="10985" PostTypeId="1" CreationDate="2010-11-19T18:09:54.047" Score="5" ViewCount="86" Body="&lt;p&gt;Can I draw a net for a Steinmetz solid from two cylinders with a compass?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is, can we flatten the net? I often make a model using paper and a compass-- it looks about right... is it really a valid method of construction?&lt;/p&gt;&#xA;" OwnerUserId="2563" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-20T12:31:00.400" LastActivityDate="2010-12-20T12:31:00.400" Title="Can I draw a net for a Steinmetz solid with a compass?" Tags="&lt;geometry&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="10986" PostTypeId="2" ParentId="10980" CreationDate="2010-11-19T18:14:19.817" Score="3" ViewCount="" Body="&lt;p&gt;No, but the counterexample is trivial. Take any integral domain with finitely many prime ideals which is not a field. For example, the localization $\mathbb{Z}_{(p)}$ of the integers at a prime p. The zero ideal is non-maximal and prime so, trivially, $\prod_{i=1}^np_i=0$. Maybe this isn't exactly what you were meaning to ask?&lt;/p&gt;&#xA;" OwnerUserId="1321" LastActivityDate="2010-11-19T18:14:19.817" CommentCount="6" />
  <row Id="10987" PostTypeId="1" AcceptedAnswerId="10988" CreationDate="2010-11-19T18:32:14.573" Score="0" ViewCount="116" Body="&lt;p&gt;Let $$ \theta(x) = \sum\limits_{p \leq x} \log{p} \quad \ ; \ \psi(x)=\sum\limits_{n=1}^{\infty} \theta(x^{1/n})$$ then how does one prove $$e^{\psi(x)}= \text{lcm}[ 1,2,\cdots , \lfloor{x\rfloor}]$$&lt;/p&gt;&#xA;" LastActivityDate="2010-11-19T18:49:45.387" Title="Showing $e^{\psi(x)}= \text{lcm}[ 1,2,\cdots , \lfloor{x\rfloor}]$" Tags="&lt;number-theory&gt;&lt;analytic-number-theory&gt;" AnswerCount="1" />
  <row Id="10988" PostTypeId="2" ParentId="10987" CreationDate="2010-11-19T18:43:54.660" Score="3" ViewCount="" Body="&lt;p&gt;$e^{\psi(x)} = \exp({\displaystyle \sum_{n=1}^{\infty} \theta(x^{1/n})}) = \displaystyle \Pi_{n=1}^{\infty} \exp(\theta (x^{1/n})) = \displaystyle \Pi_{n=1}^{\infty} \exp(\displaystyle \sum_{p \leq x^{1/n}} \log(p))$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ = \Pi_{n=1}^{\infty} (\Pi_{p \leq x^{1/n}} p) = (\Pi_{p \leq x} p) \times (\Pi_{p \leq x^{1/2}} p) \times (\Pi_{p \leq x^{1/3}} p) ... = lcm[1,2,3,...,[x]]$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\textbf{EDIT:}$ To make the last equality a bit more clear, the highest power of $p$ dividing $lcm[1,2,3,...,[x]]$ is given by $[\log_p([x])]$&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2010-11-19T18:49:45.387" LastActivityDate="2010-11-19T18:49:45.387" />
  <row Id="10989" PostTypeId="2" ParentId="10985" CreationDate="2010-11-19T18:55:42.467" Score="1" ViewCount="" Body="&lt;p&gt;Imagine keeping one of the cylinders intact, but merely marking on it the curves that would have been the edges of the Steinmetz solid. If you unfold the cylinder, the curves must be periodic in the azimuthal direction, so they cannot be circles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you solve $(x,y,z) = (\cos \theta, \sin \theta, z) = (x, \cos \phi, \sin \phi)$, you'll find that the curves are actually sinusoids, $z = \pm \cos \theta$.&lt;/p&gt;&#xA;" OwnerUserId="856" LastActivityDate="2010-11-19T18:55:42.467" />
  <row Id="10990" PostTypeId="1" CreationDate="2010-11-19T18:56:04.087" Score="1" ViewCount="203" Body="&lt;p&gt;Dear MSE-Community,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I was trying to find closed-form representations for odd zeta-values, I used &#xA;$$ \Gamma(z) = \frac{e^{-\gamma \cdot z}}{z} \prod_{n=1}^{\infty} \Big( 1 + \frac{z}{n} \Big)^{-1} e^{\frac{z}{n}} $$ &#xA;and rearranged it to &#xA;$$ \frac{\Gamma(z)}{e^{-\gamma \cdot z}} = \prod_{n=1}^{\infty} \Big( 1 + \frac{z}{n} \Big)^{-1} e^{z/n}. $$&#xD;&#xA;As we know that $$\prod_{n=1}^{\infty} e^{z/n} = e^{z + z/2 + z/3 + \cdots + z/n} = e^{\zeta(1) z},$$ &#xA;we can state that &#xA;$$\prod_{n=1}^{\infty} \Big( 1 + \frac{z}{n} \Big) = \frac{e^{z(\zeta(1) - \gamma)}}{z\Gamma(z)}\qquad\text{(1)}$$&#xD;&#xA;I then stumbled upon the Wikipedia page of &lt;a href=&quot;http://en.wikipedia.org/wiki/Ramanujan_summation&quot; rel=&quot;nofollow&quot;&gt;Ramanujan Summation&lt;/a&gt; (see the bottom of the page), which I used to set $\zeta(1) = \gamma$ (which was, admittedly, a rather dangerous move. Remarkably, things went well eventually. Please don't stop reading). The $z^3$ -coefficient of both sides can now be obtained. Consider &#xA;\begin{align*}&#xA;(1-ax)(1-bx) &amp;amp;= 1 - (a+b)x + abx^2\\&#xA;&amp;amp;= 1-(a+b)x + (1/2)((a+b)^2-(a^2+b^2))&#xA;\end{align*}&#xA;and&#xA;\begin{align*}(1-ax)(1-bx)(1-cx) &amp;amp;= 1 - (a + b + c)x\\&#xA;&amp;amp;\qquad + (1/2)\Bigl((a + b + c)^2 - (a^2 + b^2 + c^2)\Bigr)x^2\\&#xA;&amp;amp;\qquad -(abc)x^3.&#xA;\end{align*}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can also set &#xA;\begin{align*}&#xA;(abc)x^3 &amp;amp;= (1/3)\Bigl((a^3 + b^3 + c^3) - (a + b + c)\Bigr)\\&#xA;&amp;amp;\qquad + (1/2)(a + b + c)^3 -(a + b + c)(a^2 + b^2 + c^2).&#xA;\end{align*}&#xA;It can be proved by induction that the x^3 term of $(1-ax)(1-bx)\cdots(1-nx)$ is equal to &#xA;\begin{align*}&#xA;(1/3)&amp;amp;\Bigl((a^3 + b^3 + c^3 +\cdots + n^3) - (a + b + c + \cdots + n^3))\\&#xA;&amp;amp;\qquad+ (1/2)(a + b + c + \cdots + n)^3 -(a + b + c + \cdots + n)(a^2 + b^2 + c^2 + \cdots + n^2).\qquad\text{(2)}&#xA;\end{align*}&#xA;On the right side of equation (1), the $z^3$-term can be found by looking at the z^3 term of the Taylor expansion series of $1/(z \Gamma(z))$, which is $(1/3)\zeta(3) + (1/2)\zeta(2) + (1/6)\gamma^3$. We then use (2) to obtain the equality&#xA;$$ (1/6)\gamma^3 - (1/2)\gamma \pi^2 - (1/6) \psi^{(2)}(1) = 1/3)\zeta(3) + (1/2)\zeta(2) + (1/6)\gamma^3$$ &#xA;to find that &#xA;$$\zeta(3) = - (1/2) \psi^{(2)} (1),$$ (3)&#xA;which is a true result that has been &#xA;known (&lt;em&gt;known&lt;/em&gt; should be a hyperlink but it isn't for some reason) for quite a long time. The important thing here is that I used $\zeta(1) = \gamma$, which isn't really true. Ramanujan assigned a summation value to the harmonic series (again, see Ramanujan Summation), and apparently it can be used to verify results and perhaps to prove other conjectures/solve problems. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My first question is: Is this a legitimate way to prove (3) ? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Generalizing this question: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When and how are divergent series and their summation values used in mathematics? What are the 'rules' when dealing with summed divergent series and using them to (try to) find new results?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Max&lt;/p&gt;&#xA;" OwnerUserId="1055" LastEditorUserId="1055" LastEditorDisplayName="" LastEditDate="2010-11-22T22:20:58.833" LastActivityDate="2010-11-22T22:20:58.833" Title="Uses of Divergent Series and their summation-values in mathematics ?" Tags="&lt;analysis&gt;&lt;sequences-and-series&gt;&lt;big-list&gt;" AnswerCount="1" CommentCount="12" FavoriteCount="2" />
  <row Id="10991" PostTypeId="2" ParentId="2949" CreationDate="2010-11-19T19:06:04.113" Score="2" ViewCount="" Body="&lt;p&gt;Fintushel and Stern's construction of exotic K3's by surgery on torus fibered knots in S^3.  If the Alexander polynomial of the knot is not monic then the smooth structure doesn't admit a symplectic structure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://arxiv.org/pdf/dg-ga/9612014.pdf&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/pdf/dg-ga/9612014.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's also very beautifully explained in the last chapter of Scorpan's &quot;The Wild World of 4-manifolds.&quot; The entire construction is available in the Google books preview.&lt;/p&gt;&#xA;" OwnerUserId="3722" LastActivityDate="2010-11-19T19:06:04.113" CommunityOwnedDate="2010-11-19T19:06:04.113" />
  <row Id="10992" PostTypeId="1" CreationDate="2010-11-19T19:25:40.927" Score="1" ViewCount="1020" Body="&lt;p&gt;Alright, I'm trying to figure out how to calculate a  critical value using t-distribution in Microsoft Excel... ex. a one-tailed area of 0.05 with 39 degrees of freedom: t=1.685&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know the answer, but how do I get this? I've tried TDIST() TINV() and TTEST() but they all give me different answers. This web calculator: &lt;a href=&quot;http://www.danielsoper.com/statcalc/calc10.aspx&quot; rel=&quot;nofollow&quot;&gt;http://www.danielsoper.com/statcalc/calc10.aspx&lt;/a&gt; always gives me what I'm looking for but I cannot manage to get Excel to do the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any help would be greatly appreciated!&lt;/p&gt;&#xA;" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-11-20T00:04:26.697" LastActivityDate="2010-11-20T00:09:10.717" Title="Finding critical value using t-distribution in Excel" Tags="&lt;statistics&gt;&lt;math-software&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="10993" PostTypeId="1" CreationDate="2010-11-19T19:27:51.577" Score="3" ViewCount="309" Body="&lt;p&gt;Suppose $x \geq 0$, $y \geq 0$ and $0&amp;lt;p&amp;lt;1$. Why is the following inequality true?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$|x^{p}-y^{p}| \leq |x-y|^p$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="4219" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2010-11-20T00:21:43.673" LastActivityDate="2010-11-20T05:32:44.807" Title="Please explain inequality $|x^{p}-y^{p}| \leq |x-y|^p$" Tags="&lt;real-analysis&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" />
  <row Id="10994" PostTypeId="2" ParentId="10992" CreationDate="2010-11-19T19:37:02.567" Score="1" ViewCount="" Body="&lt;p&gt;Non-answer: Export the data to CSV and use software that is suited for statistical work, e.g. R.&lt;/p&gt;&#xA;" OwnerUserId="3330" LastActivityDate="2010-11-19T19:37:02.567" />
  <row Id="10995" PostTypeId="2" ParentId="5595" CreationDate="2010-11-19T19:40:17.960" Score="4" ViewCount="" Body="&lt;p&gt;I don't really have the intuition for this, but I know the formal proof.  This is equivalent to showing that the probability that in a permutation of $[n]$ chosen uniformly at random, two elements chosen uniformly at random are in the same cycle is $1/2$.  By symmetry, it's enough to show that the probability that $1$ and $2$ are in the same cycle is $1/2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are many ways to show this fact. For example: the probability that $1$ is in a cycle of length $k$ is $1/n$, for $1 \le k \le n$. This is true because the number of possible $k$-cycles containing $1$ is ${n-1 \choose k-1} (k-1)! = (n-1)!/(n-k)!$, and the number of ways to complete a permutation once a $k$-cycle is chosen is $(n-k)!$. So there are $(n-1)!$ permutations of $[n]$ in which $1$ is in a $k$-cycle.  Now the probability that $2$ is in the same cycle as $1$, given that $1$ is in a $k$-cycle, is $(k-1)/(n-1)$. So the probability that $2$ is in the same cycle as $1$ is&#xA;$$ \sum_{k=1}^n {k-1 \over n-1} {1 \over n} = {1 \over n(n-1)} \sum_{k=1}^n (k-1) = {1 \over n(n-1)} {n(n-1)\over 2} = 1/2. $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively, the &lt;a href=&quot;http://en.wikipedia.org/wiki/Chinese_restaurant_process&quot; rel=&quot;nofollow&quot;&gt;Chinese restaurant process&lt;/a&gt; with $\alpha = 0, \theta = 1$ generates a uniform random permutation of $[n]$ at the $n$th step; $2$ is paired with $1$ at the second step with probability $1/2$. This is a bit more elegant but requires some understanding of the CRP.&lt;/p&gt;&#xA;" OwnerUserId="173" LastActivityDate="2010-11-19T19:40:17.960" />
  <row Id="10996" PostTypeId="2" ParentId="9505" CreationDate="2010-11-19T19:49:10.333" Score="2" ViewCount="" Body="&lt;p&gt;Say $x^y = y^x$, and $x &amp;gt; y &amp;gt; 0$. Taking logs, $y \log x = x \log y$; rearranging, $(\log x)/x = (\log y)/y$. Let $f(x) = (\log x)/x$; then this is $f(x) = f(y)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, $f^\prime(x) = (1-\log x)/x^2$, so $f$ is increasing for $x&amp;lt;e$ and decreasing for $x&amp;gt;e$. So if $x^y = y^x$ has a solution, then $x &amp;gt; e &amp;gt; y$.  So $y$ must be $1$ or $2$. But $y = 1$ doesn't work. $y=2$ gives $x=4$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I've always thought of this as the ``standard'' solution to this problem and I'm a bit surprised nobody has posted it yet.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $x &amp;gt; 0 &amp;gt; y$, then $0 &amp;lt; x^y &amp;lt; 1$ and $y^x$ is an integer, so there are no such solutions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $0 &amp;gt; x &amp;gt; y$, then $x^y = y^x$ implies $x$ and $y$ must have the same parity. Also, taking reciprocals, $x^{-y} = y^{-x}$. Then $(-x)^{-y} = (-y)^{-x}$ since $x$ and $y$ have the same parity. (The number of factors of $-1$ we introduced to each side differs by $x-y$, which is even.) So solving the problem where $x$ and $y$ are negative reduces to solving it when $x$ and $y$ are positive.&lt;/p&gt;&#xA;" OwnerUserId="173" LastActivityDate="2010-11-19T19:49:10.333" />
  <row Id="10997" PostTypeId="2" ParentId="10993" CreationDate="2010-11-19T19:53:05.853" Score="10" ViewCount="" Body="&lt;p&gt;Assume w.lo.g $x&amp;gt;y$. Then the statement becomes $x^p - y^p \leq (x-y)^p$. Since $y &amp;gt; 0$, divide through out by $y^p$. So we need to show $(\frac{x}{y})^p - 1 \leq (\frac{x}{y}-1)^p$ whenever $x &amp;gt; y &amp;gt;0$ and $0&amp;lt;p&amp;lt;1$. Let $t = \frac{x}{y}$. So we need to show $t^p - 1 \leq (t-1)^p$ whenever $t &amp;gt; 1$ and $0&amp;lt;p&amp;lt;1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a calculus problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $f(t) = (t-1)^p - (t^p -1) $ where $0&amp;lt;p&amp;lt;1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Show that the function $f(t)$ is increasing for $t \geq 1$ and when $0 &amp;lt; p &amp;lt; 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So $f(t) \geq f(1)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and $f(1) = 0$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, we get the desired result.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\textbf{EDIT:}$ To show $f(t)$ is increasing for $t \geq 1$ and when $0 &amp;lt; p &amp;lt; 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We need to show $0 &amp;lt; f&amp;#39;(t) = p (t-1)^{p-1} - pt^{p-1}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and since $p&amp;gt;0$, all we need to show is that $(t-1)^{p-1}&amp;gt;t^{p-1}$, $\forall t &amp;gt; 1$ and $0&amp;lt;p&amp;lt;1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $0&amp;lt;p&amp;lt;1$, we need to show $t^{1-p} &amp;gt; (t-1)^{1-p}$ which is true since $p&amp;lt;1$ and $t&amp;gt;t-1&amp;gt;0$.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2010-11-19T20:19:13.020" LastActivityDate="2010-11-19T20:19:13.020" CommentCount="2" />
  <row Id="10998" PostTypeId="1" AcceptedAnswerId="10999" CreationDate="2010-11-19T20:01:45.803" Score="8" ViewCount="180" Body="&lt;p&gt;A vector $v \in \mathbb{Z}^n$ is primitive if there does not exist some vector $v&amp;#39; \in \mathbb{Z}^n$ and some $k \in \mathbb{Z}$ such that $v = k v&amp;#39;$ and $k \geq 2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a paper I'm writing right now, I'd like to know that a &quot;random&quot; vector in $\mathbb{Z}^n$ is primitive.  Let me make this precise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $\|\cdot\|_{1}$ be the $L^{1}$ norm on $\mathbb{Z}^n$, so $\|v\|_1 = \sum_{i=1}^n |v_i|$, where the $v_i$ are the components of $v$.  Define $\mathcal{V}_k$ to be the number of vectors $v$ in $\mathbb{Z}^n$ such that $\|v\|_1 \leq k$.  Define $\mathcal{P}_k$ to be the number of primitive vectors $v$ in $\mathbb{Z}^n$ such that $\|v\|_1 \leq k$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I then want&#xA;$$\lim_{k \rightarrow \infty} \frac{\mathcal{P}_k}{\mathcal{V}_k} = 1.$$&#xD;&#xA;Assuming this is true, is there any nice estimate as to how fast it approaches $1$?&lt;/p&gt;&#xA;" OwnerUserId="3400" LastActivityDate="2010-11-20T13:25:22.573" Title="Probability that a vector in $\mathbb{Z}^n$ is primitive" Tags="&lt;linear-algebra&gt;&lt;probability&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="10999" PostTypeId="2" ParentId="10998" CreationDate="2010-11-19T20:06:17.490" Score="12" ViewCount="" Body="&lt;p&gt;In the case where $n = 2$, you're asking for the &quot;probability&quot; that two integers are relatively prime; this is well-known to be $6/\pi^2$, not 1. In the general-$n$ case, the probability that $n$ integers are relatively prime is $1/\zeta(n)$.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Reference:&lt;/strong&gt; &lt;a href=&quot;http://en.wikipedia.org/wiki/Coprime&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Coprime&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="173" LastActivityDate="2010-11-19T20:22:53.200" CommentCount="4" />
  <row Id="11000" PostTypeId="1" AcceptedAnswerId="11006" CreationDate="2010-11-19T20:35:01.747" Score="4" ViewCount="163" Body="&lt;p&gt;In 'An Introduction to Probability Theory and Applications' by W. Feller I encountered this apparently innocuous problem.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A throw of twelve dice can result in&#xA;  $6^{12}$ different outcomes, to all of&#xA;  which we attribute equal&#xA;  probabilities. The event that each&#xA;  face appears twice can occur in as&#xA;  many ways as twelve dice can be&#xA;  arranged in six groups of two each.&#xA;  Hence the probability of the event is&#xA;  $\displaystyle \frac{12!}{2^{6}6^{12}}=0.003438$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The reasoning is that by doing that you're grouping two 1's, two 2's, ..., two 6's (each group using one partition in the multinomial) and the result is the number of different partitions that can be found with that particular characteristic. However, I had doubts about that answer. To understand better that problem, I did a simpler example with $4$ dice instead of $12$ (in this case, the event is the number of ways in which two faces appear twice).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the same result I get as the probability $\displaystyle \frac{4!}{2^{2}6^{4}}=$ &lt;s&gt;0.46&lt;/s&gt; $0.0046$. Then, to see if that's true I ran a little simulation of this case in Mathematica: &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;dice = Table[Table[Random[Integer, {1, 6}], {i, 1, 4}], &#xA;{j, 1, 1000000}]; i=0;&#xA;f[{a_, b_, c_, d_}] := Which[a === b &amp;amp;&amp;amp; c === d &amp;amp;&amp;amp; a != c, &#xA;i++, a === c &amp;amp;&amp;amp; b === d &amp;amp;&amp;amp; a != b, &#xA;i++, a === d &amp;amp;&amp;amp; b === c &amp;amp;&amp;amp; a != b, i++;];&#xA;Map[f, dice, {1}];&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;After $1000000$ steps I got $69687$ cases in which two faces appear twice. This is equivalent to a probability of $0.069687$. Far smaller than what I expected based on the calculation above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since this latter example is much more manageable than the one with twelve dice, I did the following&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With four dice we have the following:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Partition $r_{1}$ contains the first and second dice and partition $r_{2}$ contains the third and fourth dice.&lt;/li&gt;&#xA;&lt;li&gt;Partition $r_{1}$ contains the second and fourth dice and partition $r_{2}$ contains the first and fourth third.&lt;/li&gt;&#xA;&lt;li&gt;Partition $r_{1}$ contains the first and fourth dice and partition $r_{2}$ contains the second and third dice.&lt;/li&gt;&#xA;&lt;li&gt;Partition $r_{2}$ contains the first and second dice and partition $r_{1}$ contains the third and fourth dice.&lt;/li&gt;&#xA;&lt;li&gt;Partition $r_{2}$ contains the second and fourth dice and partition $r_{1}$ contains the first and fourth third.&lt;/li&gt;&#xA;&lt;li&gt;Partition $r_{2}$ contains the first and fourth dice and partition $r_{1}$ contains the second and third dice.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;For each case, we can have $30$ outcomes in which two faces appear two times. For example, for the first case we have $1122, 1133, 1144, 1 155, 1166, 2211, 2233, 2244,2255, 2266,... 6611, 6622, 6633, 6644, 6655$. However, such outcomes are repeated twice (particularly, the outcomes of case 1 repeat the outcomes of case 4, etc). Therefore, the number of different outcomes which produce two faces appearing two times are $\frac{6}{2}6*5=90$. Since there are $6^{4}$ outcomes, we have a probability of $\frac{90}{6^{4}}=0.0694444$, which is the result that produces the simulation in Mathematica.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it wrong the first reasoning? If so, is there a general approach to use the multinomial coefficient to solve this kind of problems. For instance, it appears that this only happens for $r_{1}=r_{2}=r_{k}$. Otherwise, there are not repeated outcomes.&lt;/p&gt;&#xA;" OwnerUserId="2569" LastEditorUserId="2569" LastEditorDisplayName="" LastEditDate="2010-11-19T21:44:52.753" LastActivityDate="2010-11-19T22:12:39.000" Title="Probability for twelve dice" Tags="&lt;probability&gt;" AnswerCount="2" />
  <row Id="11001" PostTypeId="2" ParentId="10998" CreationDate="2010-11-19T20:43:34.930" Score="9" ViewCount="" Body="&lt;p&gt;Further to Michael's answer, not only does $\mathcal{P}_k/\mathcal{V}_k\to1/\zeta(n)$, but we can calculate a bound for the rate of convergence. I'll also give an argument which is a little different from the one given in his Wikipedia reference.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Noting that the set $\lbrace v\in\mathbb{R}^n\colon\Vert v\Vert_1\le k\rbrace$ has volume $ck^n$ (for a constant c depending only on the dimension n) and surface area proportional to $k^{n-1}$ gives&#xA;$$&#xD;&#xA;\mathcal{V}_{k}-1=ck^n + O(k^{n-1}).\qquad\qquad{\rm(1)}&#xD;&#xA;$$&#xD;&#xA;The '-1' on the left hand side is not relevant for large k as it can be absorbed into the &lt;em&gt;O&lt;/em&gt;(&lt;i&gt;k&lt;/i&gt;&lt;sup&gt;&lt;em&gt;n&lt;/em&gt;&amp;minus;1&lt;/sup&gt;) error term, and is just there so that (1) is also valid for small &lt;em&gt;k&lt;/em&gt;&amp;nbsp;&amp;lt;&amp;nbsp;1.&#xA;Noting that every nonzero $v\in\mathbb{Z}^n$ decomposes uniquely as $v=mv^\prime$ for integer $m\ge1$ and primitive $v^\prime\in\mathbb{Z}^n$ leads to the following relation between $\mathcal{P}_k$ and $\mathcal{V}_k$,&#xA;$$&#xD;&#xA;\mathcal{V}_k-1=\sum_{m=1}^\infty\mathcal{P}_{\frac{k}{m}}.&#xD;&#xA;$$&#xD;&#xA;This can be inverted via the &lt;a href=&quot;http://en.wikipedia.org/wiki/M%C3%B6bius_function&quot; rel=&quot;nofollow&quot;&gt;M&amp;ouml;bius function&lt;/a&gt; &amp;mu;,&#xA;$$&#xD;&#xA;\mathcal{P}_k=\sum_{m=1}^\infty\mu(m)(\mathcal{V}_{\frac{k}{m}}-1).&#xD;&#xA;$$&#xD;&#xA;In dimension $n &amp;gt; 2$, substituting (1) into this expression gives&#xA;$$&#xD;&#xA;\mathcal{P}_k=\sum_{m=1}^\infty \mu(m)c k^n m^{-n} + O(k^{n-1}).\qquad\qquad{(2)}&#xD;&#xA;$$&#xD;&#xA;The $O(k^{n-1})$ comes from the sum $\sum_m (k/m)^{n-1}$ from the remainder term of (1) which, for $n &amp;gt; 2$, gives $k^{n-1}$ multiplied by a convergent sum.&#xA;Dividing through by $\mathcal{V}_k$,&#xA;$$&#xD;&#xA;\mathcal{P}_k/\mathcal{V}_k=\sum_{m=1}^\infty\mu(m)m^{-n}+O(1/k)=1/\zeta(n)+O(1/k).&#xD;&#xA;$$&#xD;&#xA;Edit: The case for $n=2$ is actually a little bit different, and we do not obtain such a good convergence rate. As the sum $\sum_m(k/m)^{n-1}$ does not converge, the error term in (2) does not apply. Instead, we can use $O(1_{\lbrace k\ge1\rbrace}k+1_{\lbrace k &amp;lt; 1\rbrace}k^2)$ for the error term in (1). This leads to an error of order $k\sum_{m\le k}m^{-1}+k^2\sum_{m &amp;gt; k}m^{-2}\sim k\log k$ in (2), giving&#xA;$$&#xD;&#xA;\mathcal{P}_k/\mathcal{V}_k=1/\zeta(2)+O(\log k/k).&#xD;&#xA;$$&#xD;&#xA;You can also look at the paper &lt;a href=&quot;http://dx.doi.org/10.1016/0022-314X%2872%2990038-8&quot; rel=&quot;nofollow&quot;&gt;On the probability that &lt;em&gt;k&lt;/em&gt; positive integers are relatively prime&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1321" LastEditorUserId="1321" LastEditorDisplayName="" LastEditDate="2010-11-20T13:25:22.573" LastActivityDate="2010-11-20T13:25:22.573" />
  <row Id="11002" PostTypeId="1" CreationDate="2010-11-19T21:14:35.510" Score="4" ViewCount="243" Body="&lt;p&gt;Can we determine if a binomial coefficient $C(n,p)$ is even or odd, without calculating its value? &#xA;($p\lt n$, $p$ and $n$ are positive integers)&lt;/p&gt;&#xA;" OwnerUserId="2938" LastEditorUserId="6296" LastEditorDisplayName="" LastEditDate="2011-03-14T03:17:02.153" LastActivityDate="2011-03-14T03:17:02.153" Title="$C(n,p)$: even or odd?" Tags="&lt;binomial-coefficients&gt;&lt;parity&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="11003" PostTypeId="2" ParentId="10942" CreationDate="2010-11-19T21:15:13.410" Score="2" ViewCount="" Body="&lt;p&gt;Thanks for the pointer, Shai Covo. The name of this specific process is &lt;a href=&quot;http://en.wikipedia.org/wiki/Preferential_attachment&quot; rel=&quot;nofollow&quot;&gt;preferential attachment&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2380" LastActivityDate="2010-11-19T21:15:13.410" />
  <row Id="11005" PostTypeId="2" ParentId="11002" CreationDate="2010-11-19T21:18:56.540" Score="7" ViewCount="" Body="&lt;p&gt;Yes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We only need to know how many powers of two appear in the prime factorization of C(n,p). The number of powers of two that appear in the prime factorization of $n!$ is $\lfloor n/2 \rfloor + \lfloor n/4 \rfloor + \lfloor n/8 \rfloor + \cdots$ -- count all the multiples of two, of four, of eight, and so on. This turns out to be $n$ minus the number of ones in the binary expansion of $n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It turns out that this reduces to the following: C(n,p) is odd if and only if every bit of the binary expansion of $p$ is less than the corresponding bit of the binary expansion of $n$. This is due to Lucas.&lt;/p&gt;&#xA;" OwnerUserId="173" LastActivityDate="2010-11-19T21:18:56.540" />
  <row Id="11006" PostTypeId="2" ParentId="11000" CreationDate="2010-11-19T21:21:26.833" Score="7" ViewCount="" Body="&lt;p&gt;There are two things going on here.  First, $\displaystyle \frac{4!}{2^{2}6^{4}}$ is about $0.0046$, not $0.46$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second, the problem described by Feller and your simplification to four dice are not the same problem.  Feller's problem requires that &lt;em&gt;each&lt;/em&gt; face appear twice.  Your four-dice problem requires that $some$ two faces appear twice.  There's only one way that Feller's event can be satisfied; namely, that each of 1, 2, 3, 4, 5, and 6 appear exactly twice.  There are many more ways that some two faces could appear twice.  For instance, you could have two 1's and two 2's, two 1's and two 3's, etc.  In fact, there are $\binom{6}{2} = 15$ different ways to choose the two faces that will appear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And that solves your problem.  Taking the correct result from Feller, $\displaystyle \frac{4!}{2^{2}6^{4}} \approx 0.0046$, and multiplying it by the $15$ different ways to choose the faces gives $\displaystyle \frac{90}{6^4} \approx 0.0694444$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;HR&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With respect to your question about the multinomial coefficient, $\displaystyle \binom{n}{k_1, k_2, \ldots, k_m}$ calculates the number of ways to partition a set of $n$ items into $m$ groups with $k_1$ items in the first group, $k_2$ items in the second group, and so forth.  So in Feller's problem $\displaystyle \binom{12}{2, 2, 2, 2, 2, 2}$ is calculating the number of ways to put two dice in the 1's group, two dice in the 2's group, and so forth.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the four-dice problem you describe, you haven't specified a particular partition of dice into groups; there are several partitions that will satisfy &quot;two faces appearing twice.&quot;  So, in the four-dice problem, $\displaystyle \binom{4}{2, 2}$ counts the number of ways to put two dice in a 1's group and two dice in a 2's group.  It also counts the number of ways to put two dice in a 1's group and two dice in a 3's group, and it counts the number of ways to put two dice in a 1's group and two dice in a 4's group, and so forth.  So to calculate the total number of ways of obtaining &quot;two faces appearing twice&quot; you need to multiply $\displaystyle \binom{4}{2, 2}$ by the number of ways to pick two of the faces out of six possible faces, which is $\binom{6}{2}$.  Then you mutiply by $\frac{1}{6^4}$ to get the probability you want.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want a general answer, the probability of throwing $n$ $d$-sided dice and having exactly $m$ faces appear $\frac{n}{m}$ times would be &#xA;$$\binom{n}{m} \frac{n!}{((\frac{n}{m})!)^m d^n} .$$&lt;/p&gt;&#xA;" OwnerUserId="2370" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-11-19T22:12:39.000" LastActivityDate="2010-11-19T22:12:39.000" CommentCount="12" />
  <row Id="11008" PostTypeId="2" ParentId="11000" CreationDate="2010-11-19T21:27:27.200" Score="2" ViewCount="" Body="&lt;p&gt;The case of four dice is not so hard. You require XXYY in some order, where X != Y. So let the first throw be X. Then there are three possibilities: XXYY, XYXY, and XYYX. Each has probability 1/6 x 5/6 x 1/6 = 5/216. So we get 5/72, which is about 0.0694. This agrees very well with your simulation!&lt;br&gt;&#xA;The question is, where did you go wrong in your calculation? You calculated the probability of a &lt;em&gt;specific&lt;/em&gt; XXYY -- say, two ones and two fives. Now you have to multiply by the number of such (X,Y) pairs, which is 15. And 15 x 4! / (2^2 x 6^4) is 5/72.&lt;/p&gt;&#xA;" OwnerUserId="1508" LastActivityDate="2010-11-19T21:27:27.200" CommentCount="1" />
  <row Id="11009" PostTypeId="2" ParentId="11002" CreationDate="2010-11-19T21:30:04.267" Score="8" ViewCount="" Body="&lt;p&gt;There is a beautiful result of Kummer that says that if $q$ is a prime, then the largest power of $q$ that divides $\binom{m+n}{n}$, $m$ and $n$ nonnegative, is the number of carries when $m$ and $n$ are added in base $q$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a corollary, we get the Theorem of Lucas mentioned by Michael Lugo: we get a carry in the addition if and only if there is a $k$ such that the $k$th digit of $m+n$ in base $q$ is smaller than the $k$th digit in the base $q$ expansion of $n$, so $\binom{m+n}{n}$ is divisible by $q$ if and only if there is a $k$ such that the $k$th digit of the base $q$ expansion of $m+n$ is smaller than the $k$th digit of the base $q$ expansion of $n$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, to determine if $\binom{n}{p}$ is odd, you just need to know whether there is at least one carry when adding $p$ and $n-p$ in base $2$; you get a carry exactly when the $k$th binary digit of both $n-p$ and $p$ is $1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So just write $p$ and $n-p$ in binary, one digit at a time (from least significant to most significant digit), until you get that both are $1$; or write $p$ and $n$ one binary digit at a time, until you get a $1$ in $p$ and a $0$ in $n$. If this happens, $\binom{n}{p}$ is even; otherwise, it is odd. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Andrew Granville's paper &lt;a href=&quot;http://www.cecm.sfu.ca/organics/papers/granville/Binomial/toppage.html&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;Arithmetic properties of binomial coefficients&lt;/em&gt;&lt;/a&gt; is chock full of interesting stuff about binomial coefficients and their divisibility. The proof of Kummer's Theorem can be found &lt;a href=&quot;http://www.cecm.sfu.ca/organics/papers/granville/paper/binomial/html/node3.html#SECTION00030000000000000000&quot; rel=&quot;nofollow&quot;&gt;there&lt;/a&gt;. There are also &lt;a href=&quot;http://www.cecm.sfu.ca/organics/papers/granville/&quot; rel=&quot;nofollow&quot;&gt;other formats available&lt;/a&gt;.  &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-20T05:25:14.973" LastActivityDate="2010-11-20T05:25:14.973" CommentCount="3" />
  <row Id="11010" PostTypeId="1" AcceptedAnswerId="11011" CreationDate="2010-11-19T21:42:36.717" Score="0" ViewCount="217" Body="&lt;p&gt;Which of the following mappings are linear?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(a)$\varphi : \mathbb{R} \rightarrow \mathbb{R}^{2}, \varphi(x) = (1 - x, (1 - x)^{2}),$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(b)$\varphi : \mathcal{F}(\mathbb{R},\mathbb{R}) \rightarrow \mathbb{R}, \varphi(f) = f(1),$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(c)$\varphi : \mathbb{R}[t] \rightarrow \mathbb{R}[t], \varphi(f) = t \cdot f', $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(d)Let $V$ be the $\mathbb{R}$-Vectorspace of converging sequences $(a_{n})$ of real numbers. Let $\varphi \rightarrow \mathbb{R}, \varphi((a_{n})) = a^{3}_{1} +$ lim $a_{n}.$&lt;/p&gt;&#xA;" OwnerUserId="3711" LastEditorUserId="3711" LastEditorDisplayName="" LastEditDate="2010-11-24T00:58:57.707" LastActivityDate="2010-11-24T17:38:46.187" Title="which of the following are linear mappings?" Tags="&lt;linear-algebra&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="11011" PostTypeId="2" ParentId="11010" CreationDate="2010-11-19T21:45:50.647" Score="9" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT:&lt;/strong&gt; A mapping is said to be linear if $f(x+y)=f(x)+f(y)$ and $f(ax)=af(x)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Verify whether the maps which you have given satisfy the above 2 conditions. If they satisfy then conclude that the map is linear. A simple example would be to consider $\varphi : \mathbb{R} \to \mathbb{R}$ defined by $\varphi(x)=cx$, note that $\varphi(x+y)=c(x+y)=cx+cy=\varphi(x)+\varphi(y)$ and $\varphi(ax)= a(cx)=a \varphi(x)$. Hence $\varphi$ is a linear map. With this example in mind, proceed for your question.&lt;/p&gt;&#xA;" LastActivityDate="2010-11-19T21:51:08.537" CommentCount="9" />
  <row Id="11012" PostTypeId="1" AcceptedAnswerId="21247" CreationDate="2010-11-19T21:48:43.227" Score="4" ViewCount="143" Body="&lt;p&gt;Assume $M$ is a (finitely generated) $A$-module such that $\wedge^n M$ is free of rank $1$ for some $n \geq 1$. Does it follow that $M$ is free of rank $n$? Or at least locally free of rank $n$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, is there a way of characterizing local free modules via exterior powers and tensor products? &lt;/p&gt;&#xA;" OwnerUserId="1650" LastActivityDate="2011-02-09T20:45:52.683" Title="characterizing free modules by exterior power" Tags="&lt;commutative-algebra&gt;" AnswerCount="3" />
  <row Id="11013" PostTypeId="2" ParentId="10949" CreationDate="2010-11-19T21:51:41.947" Score="3" ViewCount="" Body="&lt;p&gt;At least as of 1999, this was still an open question, according to the MathReview of a paper by Marc Perret (&lt;em&gt;On the ideal class group problem for global fields&lt;/em&gt;, J. Number Theory &lt;strong&gt;77&lt;/strong&gt; (1999), no. 1, pages 27-35; MR1695698 (2000d:11135), review by Bruno Anglès). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Luther Claborn proved in 1966 that every abelian group is the ideal class group of &lt;em&gt;some&lt;/em&gt; Dedekind domain (&lt;em&gt;Every abelian group is a class group&lt;/em&gt;, Pacific J. Math. &lt;strong&gt;18&lt;/strong&gt; (1966), pages 219-222, MR0195889 (33 #4085)). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gary Cornell (&lt;em&gt;Abhyankar's lemma and the class group&lt;/em&gt;. Number theory, Carbondale 1979 (Proc. Southern Illinois Conf., Southern Illinois Univ., Carbondale, Ill., 1979), pp. 82–88, Lecture Notes in Math., 751, Springer, Berlin, 1979, MR0564924 (82c:12007)) proved that every finite abelian group is a &lt;em&gt;subgroup&lt;/em&gt; of the ideal class group of a cyclotomic extension of $\mathbb{Q}$.  There have been some refinements, but I did not find anything in MathSciNet that would suggest the problem has been settled in the interim.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-19T22:01:13.447" LastActivityDate="2010-11-19T22:01:13.447" CommentCount="1" />
  <row Id="11014" PostTypeId="2" ParentId="10949" CreationDate="2010-11-19T22:33:38.823" Score="7" ViewCount="" Body="&lt;p&gt;Virtually nothing is known about the question of which abelian groups can be the ideal class group of (the full ring of integers of) some number field.  So far as I know, it is a plausible conjecture that all finite abelian groups (up to isomorphism, of course) occur in this way.  Conjectures and heuristics in this vein have been made, but unfortunately for me I'm not so familiar with them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The situation for imaginary quadratic fields is different.  Here there is an absolute bound on the size of an integer $k$ such that the class group of an imaginary quadratic field can be isomorphic to $(\mathbb{Z}/2\mathbb{Z})^k$.  Conditionally on the Generalized Riemann Hypothesis, the largest such $k$ is $4$.  This has do to with &lt;strong&gt;idoneal numbers&lt;/strong&gt;, of which the following paper provides a very fine survey:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.mast.queensu.ca/~kani/papers/idoneal-f.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.mast.queensu.ca/~kani/papers/idoneal-f.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Actually the truth is slightly stronger: let $H_D$ be the class group of the imaginary quadratic field $\mathbb{Q}(\sqrt{-D})$.  Then, as $D$ tends to negative infinity through squarefree numbers, the size of $2H_D$ (the image of multiplication by $2$) tends to infinity.  See for instance &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://arxiv.org/PS_cache/arxiv/pdf/0811/0811.0358v2.pdf&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/PS_cache/arxiv/pdf/0811/0811.0358v2.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for some recent explicit bounds on this.  &lt;/p&gt;&#xA;" OwnerUserId="299" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2010-11-20T12:30:20.480" LastActivityDate="2010-11-20T12:30:20.480" CommentCount="2" />
  <row Id="11015" PostTypeId="2" ParentId="10993" CreationDate="2010-11-19T22:50:52.363" Score="5" ViewCount="" Body="&lt;p&gt;This follows from the more general inequality&#xA;\begin{equation}(1)\qquad\qquad|x+y|^p\le |x|^p+|y|^p\qquad\qquad\text{(for $x,y\in\mathbb{C}$ and $0\lt p\le1$)}\end{equation}&#xA;Indeed, if we replace $x$ by $x-y$ in (1) we get &#xA;$$|x|^p\le |x-y|^p+|y|^p$$&#xD;&#xA;which imply&#xA;$$|x|^p-|y|^p\le |x-y|^p$$&#xD;&#xA;To prove (1), first note&#xA;$$|x+y|^p\le(|x|+|y|)^p$$&#xD;&#xA;Hence it is sufficient to prove (1) for $x,y\ge0$ in which case we may apply Sivaram's argument in the previous answer.&lt;/p&gt;&#xA;" OwnerUserId="1154" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-20T05:32:44.807" LastActivityDate="2010-11-20T05:32:44.807" CommentCount="5" />
  <row Id="11016" PostTypeId="1" AcceptedAnswerId="11017" CreationDate="2010-11-19T23:01:19.047" Score="3" ViewCount="131" Body="&lt;p&gt;I know it is unknown if there are infinitely many primes of the form $n^2+1$. Is it known if there is a positive integer $k$ such that $|\{n\in\mathbb{Z}:n^2+1 \text{ has at most k prime factors}\}|=\infty$?&lt;/p&gt;&#xA;" OwnerUserId="2856" LastActivityDate="2010-11-19T23:06:21.037" Title="Prime factors of $n^2+1$" Tags="&lt;number-theory&gt;" AnswerCount="2" />
  <row Id="11017" PostTypeId="2" ParentId="11016" CreationDate="2010-11-19T23:05:42.090" Score="8" ViewCount="" Body="&lt;p&gt;Yes, Iwaniec, &quot;Almost-primes represented by quadratic polynomials&quot;, Inventiones Math., 47:171–188, 1978, proves that there exist inﬁnitely many &#xA;integers $n$ such that $n^2 + 1$ is either prime or the product of two &#xA;primes. &lt;/p&gt;&#xA;" OwnerUserId="462" LastActivityDate="2010-11-19T23:05:42.090" CommentCount="3" />
  <row Id="11018" PostTypeId="2" ParentId="11016" CreationDate="2010-11-19T23:06:21.037" Score="1" ViewCount="" Body="&lt;p&gt;Please see  Iwaniec, &quot;Almost-primes represented by quadratic polynomials&quot;, Inventiones Math&lt;/p&gt;&#xA;" LastActivityDate="2010-11-19T23:06:21.037" />
  <row Id="11019" PostTypeId="1" CreationDate="2010-11-19T23:08:19.103" Score="1" ViewCount="73" Body="&lt;p&gt;Let ${f_{n}}$ be a sequence in $L^{2}(X,\mu)$ such that $||f_{n}||_{2} \rightarrow 0$ as $n \rightarrow \infty$. How to show that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle \lim_{n \to \infty} \int_{X} |f_n(x)| \log(1+|f_{n}(x)|) d\mu = 0$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="4219" LastActivityDate="2011-03-28T05:38:27.020" Title="Compute limit of integral" Tags="&lt;measure-theory&gt;" AnswerCount="1" />
  <row Id="11020" PostTypeId="2" ParentId="11019" CreationDate="2010-11-19T23:12:10.163" Score="2" ViewCount="" Body="&lt;p&gt;Use the fact that $\log (1+y) \leq y$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we get $\displaystyle \int_{X} |f_n(x)| \log(1+|f_n(x)|) d \mu \leq \displaystyle \int_{X} |f_n(x)| |f_n(x)| d \mu = \displaystyle \int_{X} |f_n(x)|^2 d \mu = ||f_n||_2^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, note that $\displaystyle \int_{X} |f_n(x)| \log(1+|f_n(x)|) d \mu \geq 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, $0 \leq \displaystyle \lim_{n \rightarrow \infty} \displaystyle \int_{X} |f_n(x)| \log(1+|f_n(x)|) d \mu \leq \lim_{n \rightarrow \infty} ||f_n||_2^2 = 0$&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2010-11-19T23:36:36.903" LastActivityDate="2010-11-19T23:36:36.903" CommentCount="3" />
  <row Id="11023" PostTypeId="1" CreationDate="2010-11-19T23:44:53.220" Score="0" ViewCount="89" Body="&lt;p&gt;If $f \in L(0,1)$ show that $x^{n} f(x) \in L(0,1)$ for $n \in \mathbb{N}$ and that $\displaystyle \int_{0}^{1} x^{n} f(x) dx \rightarrow 0$ as $n \to \infty$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is the following attempt correct?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $0&amp;lt;x&amp;lt;1$ then $0&amp;lt;x^{n}&amp;lt;1$ also $x^{n}$ is measurable (being cts) and $f$ is measurable because it is in $L(0,1)$ so $x^{n} f(x)$ is measurable. Now certainly $|x^{n}f(x)| \leq |f(x)|$ so this implies that $x^{n} f(x)$ is integrable so in $L(0,1)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now to finish the second part can we simply say that: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x^{n} f(x) \rightarrow 0$ as $n \rightarrow \infty$ pointwise so we may apply convergence theorem.&lt;/p&gt;&#xA;" OwnerUserId="4219" LastActivityDate="2010-11-19T23:49:49.557" Title="Elementary problem in integration theory" Tags="&lt;measure-theory&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="11024" PostTypeId="2" ParentId="11023" CreationDate="2010-11-19T23:49:49.557" Score="1" ViewCount="" Body="&lt;p&gt;I am assuming that by $L(0,1)$, you mean $\mathcal{L}^1 (0,1)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have $x^nf(x)$ dominated by $f(x)$ in the interval $(0,1)$ and $f \in \mathcal{L}^1 (0,1)$. Hence, you can use dominated convergence theorem to swap the limit and the integral.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And hence you will get the desired result since $\displaystyle \lim_{n \rightarrow \infty} x^n f(x) = 0$ almost surely.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastActivityDate="2010-11-19T23:49:49.557" CommentCount="0" />
  <row Id="11025" PostTypeId="2" ParentId="10949" CreationDate="2010-11-19T23:59:46.147" Score="6" ViewCount="" Body="&lt;p&gt;The smallest abelian group which is not the class group of an imaginary quadratic field is $(\mathbf{Z}/3 \mathbf{Z})^3$. There are six other groups of order&#xA;$&amp;lt; 100$ which do not occur in this way, of orders&#xA;$32$, $27$, $64$, $64$, $81$, and $81$ respectively.&#xA;The groups $(\mathbf{Z}/3 \mathbf{Z})^2$ and $(\mathbf{Z}/2 \mathbf{Z})^4$ occur as class groups of imaginary quadratic fields exactly once, for $D = -4027$ and $-5460$ respectively. &#xA;These results follow from the &quot;class number $100$&quot; problem, solved by Mark Watkins.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you restrict to the $p$-part of the class group, then the answer (for general number fields) is positive. That is, for any abelian $p$-group $A$, there exists a number field $K$ with class group $C$ such that $C \otimes \mathbf{Z}_p = A$.&#xA;There is even a non-abelian analog of this. Namely, for any finite $p$-group $G$, there exists a number field $K$ such that the maximal Galois $p$-extension $L/K$ unramified everywhere has Galois group $G$. This is a very recent result of Ozaki.&lt;/p&gt;&#xA;" OwnerUserId="3728" LastActivityDate="2010-11-19T23:59:46.147" CommentCount="5" />
  <row Id="11026" PostTypeId="1" AcceptedAnswerId="11030" CreationDate="2010-11-20T00:00:57.793" Score="0" ViewCount="68" Body="&lt;p&gt;It is a well-known fact that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f\colon a \mapsto \left(\begin{array}{cc}&#xD;&#xA;1 &amp;amp; 1\\&#xD;&#xA;0 &amp;amp; 1\\&#xD;&#xA;\end{array}\right), ~ b \mapsto \left(\begin{array}{cc}&#xD;&#xA;1 &amp;amp; 0\\&#xD;&#xA;1 &amp;amp; 1\\&#xD;&#xA;\end{array}\right)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is a monomorphism from the free monoid generated by $a$ and $b$ to the matrix monoid $\mathbb{Z}^{2\times 2}$.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Is there an efficient algorithm which computes the length of $f^{-1}(X)$ from $X \in \mathbb{Z}^{2\times 2}$ (that is, the number of symbols in corresponding word $x$, if $f^{-1}(X) = \{x\}$)?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Is there an efficient algorithm which computes $f^{-1}(X)$ from $X \in \mathbb{Z}^{2\times 2}$?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Is there a standard reference for this problem?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" LastActivityDate="2010-11-20T13:41:53.387" Title="Inverse of a monoid homomorphism" Tags="&lt;algebra&gt;&lt;reference-request&gt;" AnswerCount="1" />
  <row Id="11027" PostTypeId="2" ParentId="10992" CreationDate="2010-11-20T00:03:53.703" Score="1" ViewCount="" Body="&lt;p&gt;The right function is &lt;code&gt;TINV()&lt;/code&gt;, but the problem is that &lt;code&gt;TINV()&lt;/code&gt; assumes that you're inputting the two-tailed probability, not the one-tailed probability.  In other words, &lt;code&gt;TINV(p,d)&lt;/code&gt; outputs the value of $x$ such that $P(|X| \geq x) = p$, where $X$ has a $t$ distribution with $d$ degrees of freedom.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus if you want  &lt;code&gt;TINV()&lt;/code&gt; to give you the critical value associated with a one-tailed probability, you have to double the probability first to account for the other tail.  Thus, for example, &lt;code&gt;TINV(0.1,39)&lt;/code&gt; yields an output of &lt;code&gt;1.684875122&lt;/code&gt;, which is the critical value you want.&lt;/p&gt;&#xA;" OwnerUserId="2370" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-11-20T00:09:10.717" LastActivityDate="2010-11-20T00:09:10.717" />
  <row Id="11028" PostTypeId="1" CreationDate="2010-11-20T00:07:04.263" Score="2" ViewCount="197" Body="&lt;p&gt;Stuck with this problem from Zgymund's book. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose that $f_{n} \rightarrow f$ almost everywhere and that $f_{n}, f \in L^{p}$ where $1&amp;lt;p&amp;lt;\infty$. Assume that $||f_{n}||_{p} \leq M &amp;lt; \infty$. Prove that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\int f_{n}g \rightarrow \int fg$ as $n \rightarrow \infty$ for all $g \in L^{q}$ such that $\frac{1}{p} + \frac{1}{q} = 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Right, so I estimate the difference of the integrals and using Hölder end up with:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$|\int f_{n} g - \int fg| \leq ||g||_{q} ||f_{n} - f||_{p}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From here I'm stuck because we are not assuming convergence in the seminorm but just pointwise convergence almost everywhere. How to proceed?&lt;/p&gt;&#xA;" OwnerUserId="4219" LastEditorUserId="4219" LastEditorDisplayName="" LastEditDate="2010-11-20T00:35:56.297" LastActivityDate="2010-11-21T02:47:11.463" Title="Convergence of integrals in L^p" Tags="&lt;measure-theory&gt;" AnswerCount="3" CommentCount="12" FavoriteCount="1" />
  <row Id="11029" PostTypeId="1" AcceptedAnswerId="11032" CreationDate="2010-11-20T00:09:03.883" Score="2" ViewCount="51" Body="&lt;p&gt;For a 2D Normal distribution $N(0, \left[ \begin{array}{cc} 1 &amp;amp; -1/4 \\ -1/4 &amp;amp; 1 \end{array} \right])$, I am now trying to maximize its density function over $\{ x\geq 10, y \geq 10 \}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My intuition is that the optimal solution is $(10,10)$. My problem is that I am not sure which way to find the optimal solution. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="1281" LastActivityDate="2010-11-20T00:26:57.090" Title="Maximize normal density function over a subset" Tags="&lt;probability&gt;&lt;optimization&gt;" AnswerCount="1" />
  <row Id="11030" PostTypeId="2" ParentId="11026" CreationDate="2010-11-20T00:09:46.863" Score="0" ViewCount="" Body="&lt;p&gt;Yes; see the Wikipedia article on &lt;a href=&quot;http://en.wikipedia.org/wiki/Smith_normal_form&quot; rel=&quot;nofollow&quot;&gt;Smith normal form&lt;/a&gt;.  In two dimensions you are essentially using the Euclidean algorithm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt;  The basic observation here is that if $M = \left[ \begin{array}{cc} x &amp;amp; y \\\ z &amp;amp; w \end{array} \right]$ is an integer matrix, then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$M f(a^{-1}) = \left[ \begin{array}{cc} x &amp;amp; y - z \\\ z &amp;amp; w - z \end{array} \right]$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and similarly&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$M f(b^{-1}) = \left[ \begin{array}{cc} x - y &amp;amp; y \\\ z - w &amp;amp; w \end{array} \right].$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If in addition $\det M = 1$ and $x, y, z, w$ are non-negative integers, then exactly one of these operations will give a matrix with the same properties (this is equivalent to $f$ being a monomorphism).  So, as I said, essentially one is performing the Euclidean algorithm on the columns (or, if you multiply from the left instead, on the rows).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an example, if $M = \left[ \begin{array}{cc} 8 &amp;amp; 5 \\\ 3 &amp;amp; 2 \end{array} \right]$, then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$M f(b^{-1}) = \left[ \begin{array}{cc} 3 &amp;amp; 5 \\\ 1 &amp;amp; 2 \end{array} \right]$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$M f(b^{-1} a^{-1}) = \left[ \begin{array}{cc} 3 &amp;amp; 2 \\\ 1 &amp;amp; 1 \end{array} \right]$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$M f(b^{-1} a^{-1} b^{-1}) = \left[ \begin{array}{cc} 1 &amp;amp; 2 \\\ 0 &amp;amp; 1 \end{array} \right] = f(aa)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;hence $M = f(aabab)$ as desired.&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-11-20T13:41:53.387" LastActivityDate="2010-11-20T13:41:53.387" CommentCount="5" />
  <row Id="11031" PostTypeId="1" CreationDate="2010-11-20T00:22:32.650" Score="1" ViewCount="168" Body="&lt;p&gt;I am trying to do the following stochastic integral&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;\int_0^T \mu(B_s) dB_s - \frac{\int_0^T (\mu(B_s))^2 ds}{2}&#xD;&#xA;$$&#xD;&#xA;where $ \{ B_t \}$ is a standard Brownian motion, and $\mu(x) = \frac{\beta}{2x} (x^2 - \frac{4\gamma}{\sigma^2} - \frac{1}{ \beta})$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I currently only know how to integrate when the integrands are constant, i.e. $\mu$.is constant function. I was wondering how to integrate under more complicated case as above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks and regards!&lt;/p&gt;&#xA;" OwnerUserId="1281" LastEditorUserId="1281" LastEditorDisplayName="" LastEditDate="2010-11-23T02:27:51.583" LastActivityDate="2011-03-17T11:06:50.930" Title="Compute a stochastic integral" Tags="&lt;stochastic-processes&gt;&lt;stochastic-integrals&gt;" AnswerCount="2" CommentCount="5" FavoriteCount="1" />
  <row Id="11032" PostTypeId="2" ParentId="11029" CreationDate="2010-11-20T00:26:57.090" Score="4" ViewCount="" Body="&lt;p&gt;The density of your distribution is&#xA;$$ \exp \left( {-1 \over 2(1-1/16)} \left[ x^2 + y^2 + {xy \over 2} \right] \right)$$&#xD;&#xA;multiplied by a normalizing constant. (See &lt;a href=&quot;http://en.wikipedia.org/wiki/Multivariate_normal_distribution&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;; you have $\mu_x = \mu_y = 0, \sigma_x = \sigma_y = 1, \rho = -1/4$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So your problem reduces to minimizing the quadratic $x^2 + y^2 + xy/2$ over $\{ x \ge 10, y \ge 10 \}$. This function is increasing in both $x$ and $y$ and so is minimized at $x = y = 10$. &lt;/p&gt;&#xA;" OwnerUserId="173" LastActivityDate="2010-11-20T00:26:57.090" />
  <row Id="11033" PostTypeId="2" ParentId="11031" CreationDate="2010-11-20T00:31:19.877" Score="4" ViewCount="" Body="&lt;p&gt;You can do this by different methods. One way I like is to use Ito's formula.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, consider $B_t^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$d(B_t^2) = 2B_t dB_t + dt$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, $\int_{0}^{T} B_t dB_t = \frac{1}{2} \int_{0}^{T} (d(B_t^2) - dt) = \frac{1}{2} (B_T^2 - T)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, consider $B_t^3$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Use Ito's formula,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$d(B_t^3) = 3B_t^2 dB_t + 3 B_t dt$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, $\int_{0}^{T} B_t^2 dB_t = \frac{1}{3} \int_{0}^{T} (d(B_t^3) - 3 B_t dt) = \frac{1}{3} (B_T^3 - 3 \int_{0}^{T} B_t dt)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Use these to get your desired answer. (Plug in for $\mu(B_s)$ and then simplify to get integrals of the form $\int_{0}^{T} B_t^2 dB_t$ and $\int_{0}^{T} B_t dB_t$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope this helps.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastActivityDate="2010-11-20T00:31:19.877" />
  <row Id="11034" PostTypeId="2" ParentId="11002" CreationDate="2010-11-20T00:33:04.650" Score="3" ViewCount="" Body="&lt;p&gt;I already commented with one Wikipedia link above, but I just want to add that if you run the Pascal's triangle recurrence $\bmod 2$ what you get is a sequence of approximations to the &lt;a href=&quot;http://en.wikipedia.org/wiki/Sierpinski_triangle&quot; rel=&quot;nofollow&quot;&gt;Sierpinski triangle&lt;/a&gt;.  So the result that comes out of Lucas' theorem is an explicit nonrecursive algorithm for plotting the Sierpinski triangle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also want to take this opportunity to mention a paper which discusses generalizations of this problem, and which is both quite entertaining to read and perhaps has one of the strangest titles of any math paper: Granville's &lt;a href=&quot;http://www.gianpierobiancoli.it/wp-content/uploads/2009/10/beeb.pdf&quot; rel=&quot;nofollow&quot;&gt;Zaphod Beeblebrox's Brain and the Fifth-Ninth Row of Pascal's Triangle&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-11-20T00:33:04.650" />
  <row Id="11035" PostTypeId="2" ParentId="11028" CreationDate="2010-11-20T00:56:09.103" Score="4" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By Egorov's theorem, convergence a.e. implies for every $\epsilon$ there exists $B$ with $|B| &amp;lt; \epsilon$ such that $f_n\to f$ uniformly on $X\setminus B$ (where $X$ is &quot;almost&quot; the whole space).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Split $\int (f_n - f)g$ in two pieces, one over $B$ and one over $X\setminus B$. On $X\setminus B$ uniform convergence implies the integral can be made as small as you want. Holder's inequality implies the integral on $B$ is controlled by $(M + \|f\|_p)\|g\|_{L^q(B)}$. Taking $B\searrow$ a measure zero set, then the integral on $B$ of $g$ goes to zero. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finish by taking a diagonalizing sequence as usual. &lt;/p&gt;&#xA;" OwnerUserId="1543" LastActivityDate="2010-11-20T00:56:09.103" CommentCount="8" />
  <row Id="11036" PostTypeId="1" AcceptedAnswerId="11046" CreationDate="2010-11-20T00:59:16.760" Score="3" ViewCount="275" Body="&lt;p&gt;For every odd integer $n$, $3\leq n\leq 199$, there exists an integer $m\geq 0$, and a prime number $p$, such that $n=2^m+p$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I created a software program in Java to solve this problem.  The statement is actually false.  The first odd number it fails at is 127.  I really appreciate everyone's input on this.&lt;/p&gt;&#xA;" OwnerUserId="3691" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-21T03:27:11.740" LastActivityDate="2010-11-21T03:27:11.740" Title="Can I solve this with induction?" Tags="&lt;discrete-mathematics&gt;&lt;proof-strategy&gt;&lt;induction&gt;" AnswerCount="3" CommentCount="4" />
  <row Id="11038" PostTypeId="2" ParentId="11036" CreationDate="2010-11-20T01:32:04.000" Score="4" ViewCount="" Body="&lt;p&gt;Induction can't work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Edit&lt;/em&gt;.  As Arturo points out, I made too strong of a statement here.  Perhaps a more correct way to state it is &quot;Induction in the usual sense can't work&quot; or &quot;There is likely no natural inductive proof&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it DID work, then there wouldn't be an upper bound of of n = 199.  However, the case n = 251 provides a counterexample since&lt;/p&gt;&#xA;&#xA;&lt;p&gt;251 = 1+ 2*125&lt;/p&gt;&#xA;&#xA;&lt;p&gt;251 = 2 + 3*83&lt;/p&gt;&#xA;&#xA;&lt;p&gt;251 = 4 + 13*19&lt;/p&gt;&#xA;&#xA;&lt;p&gt;251 = 8 + 3*81&lt;/p&gt;&#xA;&#xA;&lt;p&gt;251 = 16 + 5*49&lt;/p&gt;&#xA;&#xA;&lt;p&gt;251 = 32 + 3*73&lt;/p&gt;&#xA;&#xA;&lt;p&gt;251 = 64 + 11*17&lt;/p&gt;&#xA;&#xA;&lt;p&gt;251 = 128 + 3*41&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And all other powers of 2 are greater than 251.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I didn't check if this is the smallest counterexample or not)&lt;/p&gt;&#xA;" OwnerUserId="331" LastEditorUserId="331" LastEditorDisplayName="" LastEditDate="2010-11-20T05:04:12.220" LastActivityDate="2010-11-20T05:04:12.220" CommentCount="1" />
  <row Id="11039" PostTypeId="2" ParentId="11036" CreationDate="2010-11-20T01:40:20.127" Score="1" ViewCount="" Body="&lt;p&gt;Induction is not valid, because it is used to show propositions for all integers positive or integers infinite. The condition $n\Rightarrow n+1$ involves integers infinites. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe that the solution is prove each for find $m$ and $p$ in each case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$173= 2^{6}+109$$&#xD;&#xA;$$153=2 ^{6}+89$$&#xD;&#xA;$$117=2^{4}+101$$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="3421" LastEditorUserId="3421" LastEditorDisplayName="" LastEditDate="2010-11-20T01:50:35.410" LastActivityDate="2010-11-20T01:50:35.410" />
  <row Id="11040" PostTypeId="1" AcceptedAnswerId="11042" CreationDate="2010-11-20T02:23:30.837" Score="3" ViewCount="265" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have taken a look at the following topics:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://math.stackexchange.com/questions/3069/number-of-infinite-sets-with-different-cardinalities&quot;&gt;number of infinite sets with different cardinalities&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://math.stackexchange.com/questions/1467/cardinality-of-all-cardinalities&quot;&gt;Cardinality of all cardinalities&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://math.stackexchange.com/questions/8013/are-there-uncountably-infinite-orders-of-infinity&quot;&gt;Are there uncountably infinite orders of infinity?&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://math.stackexchange.com/questions/5378/types-of-infinity&quot;&gt;Types of infinity&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But still can't quite find/understand the answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) What is the easiest way to prove (if possible, without using ordinals etc. as my current math understanding of set theory counts only cardinals, and countable &amp;amp; uncountable sets) that the number of cardinalities that exists is not countable (that is, can't be put into bijection with N)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) What exactly does it mean that the set of all cardinals is so big that it's not even a set, but a class? Where does contradiction that does not allow it to be a set arise? I have read Pete's notes at &lt;a href=&quot;http://math.uga.edu/~pete/settheorypart1.pdf&quot; rel=&quot;nofollow&quot;&gt;http://math.uga.edu/~pete/settheorypart1.pdf&lt;/a&gt;, but am not quite sure how #20 leads up to that conclusion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks a lot! &lt;/p&gt;&#xA;" OwnerUserId="3731" LastEditorUserId="3731" LastEditorDisplayName="" LastEditDate="2010-12-27T06:59:29.477" LastActivityDate="2010-12-27T06:59:29.477" Title="Cardinality of a set that consists of all existing cardinalities" Tags="&lt;set-theory&gt;&lt;infinity&gt;&lt;cardinals&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="2" />
  <row Id="11042" PostTypeId="2" ParentId="11040" CreationDate="2010-11-20T02:34:16.637" Score="6" ViewCount="" Body="&lt;p&gt;There is no &quot;number of cardinalities&quot;. As you say, there are so many that they cannot form a set. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose that ${\mathcal A}$ is a set whose elements are sets, with the property that if $A,B\in{\mathcal A}$ and $A\ne B$, then $|A|\ne|B|$, i.e., $A,B$ have different cardinalities. Let $C=\bigcup{\mathcal A}$, i.e., $C=\bigcup_{A\in{\mathcal A}}A$. Clearly, $|C|\ge|A|$ for each $A\in{\mathcal A}$. Let $D={\mathcal P}(C)$ be the power set of $C$. Then $|D|&amp;gt;|C|$ so $|D|&amp;gt;|A|$ for any $A\in{\mathcal A}$. This proves that there cannot be a set of all cardinalities, because given any such set, we just found a new cardinality different from all the ones in the set.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, if ${\mathcal A}$ is countable, this shows that the ''number'' of cardinalities is not countable. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;There is a small remark that may be worth making. The argument works just as well if we do not require that all sets in ${\mathcal A}$ have different cardinalities, but simply that for any prescribed cardinality we want to consider, there is at least one set in ${\mathcal A}$ of that size (but there may more than one). This is slightly more general, but there is also a technical advantage, namely, in this form, the argument does not depend in any version of the axiom of choice. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Finally: I just checked Pete's nice note that is linked to in the body of the question. His fact 20 there is essentially the argument I've shown here.)&lt;/p&gt;&#xA;" OwnerUserId="462" LastEditorUserId="462" LastEditorDisplayName="" LastEditDate="2010-11-20T04:32:35.753" LastActivityDate="2010-11-20T04:32:35.753" CommentCount="2" />
  <row Id="11043" PostTypeId="2" ParentId="2130" CreationDate="2010-11-20T02:51:11.153" Score="3" ViewCount="" Body="&lt;p&gt;(Arriving late on the scene.)  One important thing about small categories which hasn't been mentioned yet is that they’re what you use to define small limits and colimits.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You want to say, for instance, that the category $\mathbf{Top}$ of topological spaces is complete, in some sense.  But if you tried to say “&lt;em&gt;all&lt;/em&gt; (co)limits exist”, then you could take something like “the coproduct of $\mathrm{ob}(\mathbf{Top})$-many copies of the point”, which would get you into the territory of Russell’s paradox.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you &lt;em&gt;can&lt;/em&gt; show is statements like “the category of &lt;em&gt;small&lt;/em&gt; spaces has all &lt;em&gt;small&lt;/em&gt; (co)limits”.  (Or replace small with “$n$-small” for the $n$th Grothendieck universe, or with “$\lambda$-bounded” if you prefer set-theoretic language, and so on.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So this exemplifies an important thing about small categories: they’re the categories which play nicely in algebraic constructions with small objects of other kinds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yet another reason for wanting to talk about small categories is because we obviously want to discuss some kind of category of categories, $\mathbf{Cat}$.  But versions of Russell’s paradox tell us that this category has to be bigger than all the categories we put into it; so to form it as a category itself, we have to accept a size limitation of some kind.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or, from another point of view (a different logical formalism, to be precise), if we want to really talk about the category of &lt;em&gt;all&lt;/em&gt; categories, we have to let ourselves talk about this as a proper class — in other words, step back to a perspective where we can see larger things than we could before; at which point, for want of a better term, we start calling the things we could see before “small”.&lt;/p&gt;&#xA;" OwnerUserId="2439" LastActivityDate="2010-11-20T02:51:11.153" />
  <row Id="11044" PostTypeId="1" AcceptedAnswerId="11053" CreationDate="2010-11-20T03:14:23.077" Score="3" ViewCount="116" Body="&lt;p&gt;My PhD supervisor called the following result something along the lines: &quot;The theorem in the theory of Latin squares with the most incorrect proofs&quot; (then, to my embarrassment, proceeded to point out an error in my &quot;proof&quot; at the time).  I'll pose proving it as a problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Definition&lt;/strong&gt;:  A $k \times n$ *Latin rectangle* is a $k \times n$ matrix with symbols from $\{1,2,\ldots,n\}$ for which each symbol occurs exactly once in each row and at most once in each column.  For example:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;1 2 4 3&#xA;2 3 1 4&#xA;4 1 3 2&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;is a $3 \times 4$ Latin rectangle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Definition&lt;/strong&gt;:  A Latin rectangle is called &lt;em&gt;reduced&lt;/em&gt; if the first row is $(1,2,\ldots,n)$ and the first column is $(1,2,\ldots,k)^T$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The above Latin rectangle is not reduced, but this one is:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;1 2 3 4&#xA;2 3 4 1&#xA;3 4 1 2&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Let $L_{k,n}$ be the number of $k \times n$ Latin rectangles.  Let $R_{k,n}$ be the number of reduced $k \times n$ Latin rectangles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Theorem&lt;/strong&gt;:  For all $1 \leq k \leq n$, \[L_{k,n}=\frac{n!(n-1)!}{(n-k)!}R_{k,n}.\]&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Problem&lt;/strong&gt;:  Prove it (and double-check your proof).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="139" LastActivityDate="2010-11-25T04:06:19.677" Title="What is the number of Latin rectangles given the number of reduced Latin rectangles?" Tags="&lt;combinatorics&gt;&lt;latin-square&gt;" AnswerCount="2" FavoriteCount="3" />
  <row Id="11045" PostTypeId="2" ParentId="11040" CreationDate="2010-11-20T03:26:52.013" Score="1" ViewCount="" Body="&lt;p&gt;If the OP or anyone else would like to suggest an addition or modification to what is written in my notes on this point, I would be glad to modify them accordingly.  (These notes exist for people to read and understand them and not for any other reason!)  To me, what I wrote looks essentially the same as Andres' answer above, but as I said, it doesn't matter much what I think -- there's obviously some room for improvement here.&lt;/p&gt;&#xA;" OwnerUserId="299" LastActivityDate="2010-11-20T03:26:52.013" CommunityOwnedDate="2010-11-20T03:26:52.013" />
  <row Id="11046" PostTypeId="2" ParentId="11036" CreationDate="2010-11-20T03:27:58.347" Score="8" ViewCount="" Body="&lt;p&gt;I sort of disagree with the statements saying that induction &quot;cannot work&quot;; but it is a subtle disagreement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One can use &quot;induction&quot; to prove a statement for a restricted range of numbers. The most common type of such restricted ranges are statements that apply &quot;eventually&quot;: for every $n\geq N$ for some $N$. The two standard ways of doing so is to either do the &quot;base case&quot; with $n=N$ instead of $n=0$ (or $1$, depending on whether you are a set theorist or a number theorist&lt;sup&gt;1&lt;/sup&gt;); or to establish the equivalent proposition for $N+n$ (or $N-1+n$ if you are a number theorist) and do &quot;induction on $n$&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But one can take a statement such as the one you give and establish, &lt;em&gt;by mathematical induction&lt;/em&gt;, a proposition which is essentially what you want. It's just that it is almost always a wasteful exercise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To see what I mean, let us take the statement you want to prove: for every odd integer $n$, $3\leq n\leq 199$, there is an integer $m$ and a prime number $p$ such that $n=2^m+p$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now consider the following statement P:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;for all natural numbers $n$, either $n$ is even, or $n\lt 3$, or $n\gt 199$, or $n=2^m+p$ for some integer $m$ and some prime $p$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This is a statement about &lt;em&gt;all&lt;/em&gt; natural numbers, and as such could be proven by induction. What is more, the statement is &lt;strong&gt;equivalent&lt;/strong&gt; to the one you want to prove: if the statement above is true, then given an odd integer $n$, $3\leq n\leq 199$, the proposition will hold for $n$; but $n$ is not even, is not less than $3$, and is not greater than $199$, so therefore the clause &quot;$n=2^m+p$ for some integer $m$ and some prime $p$&quot; must hold for $n$, exactly what you want. And if the proposition you want to prove is true, then the statement above is true, since the other clauses of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Disjunction&quot; rel=&quot;nofollow&quot;&gt;disjunction&lt;/a&gt; will be trivially true for all &quot;other&quot; natural numbers. So in a sense, proving your statement is &quot;the same&quot; as proving the statement P I give above. So, in that sense, you &lt;em&gt;can&lt;/em&gt; prove your statement by induction, because you can prove the &lt;em&gt;equivalent&lt;/em&gt; statement P, which is something about all natural numbers, by induction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's just that the induction argument would be very convoluted, and almost certainly a waste of time, because the inductive step would have to break up into cases depending on $k$: if $k\geq 199$, then the proof proceeds one way; if $k$ is odd, another; and so on, until you get down to showing what happens if $2\leq k\leq 198$, with $k$ even. At that point, your argument would almost certainly devolve into simply a direct proof of the fact, so that &quot;Induction&quot; is really a smokescreen that only confuses the issue.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, in a sense, &quot;yes, you could prove it by induction, but it is almost certainly a very, very bad idea to try to do so.&quot; The same is true for other statements in which $n$ is bounded above. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Induction is particularly powerful when you are trying to prove something holds for &lt;em&gt;infinitely many&lt;/em&gt; natural numbers, because that is not something that we can verify by checking the natural numbers one at a time. When we are trying to show something holds for finitely many natural number, then induction is seldom the way to go.&lt;/p&gt;&#xA;&#xA;&lt;hr/&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;1&lt;/sup&gt; The standard way of defining the natural numbers within set theory is to construct them as the smallest &lt;a href=&quot;http://en.wikipedia.org/wiki/Inductive_set_%28axiom_of_infinity%29&quot; rel=&quot;nofollow&quot;&gt;inductive set&lt;/a&gt;; then the set contains $\emptyset$, which is identified with $0$; or through &lt;a href=&quot;http://en.wikipedia.org/wiki/Peano_axioms&quot; rel=&quot;nofollow&quot;&gt;the Peano axioms&lt;/a&gt;, which usually include $0$ (though they need not do so). So for set theorists (and logicians), $\mathbb{N}$ contains $0$. But historically, $\mathbb{N}$ represented the counting numbers, and so &quot;began&quot; with $1$. So number theorists usually consider $\mathbb{N}$ to consist only of the positive integers.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-20T03:55:56.457" LastActivityDate="2010-11-20T03:55:56.457" CommentCount="7" />
  <row Id="11047" PostTypeId="1" CreationDate="2010-11-20T03:44:24.323" Score="0" ViewCount="102" Body="&lt;p&gt;How many committees of 5 can be chosen from 10 people if each committee includes Alice and&#xA;excludes Bob?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How many committees of 5 can be chosen from 12 people if each committee&#xA;(i) includes Alice;&#xA;(ii) excludes Bob;&#xA;(iii) includes Alice and excludes Bob;&#xA;(iv) includes at least one of either Alice or Bob&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please help answers this problem and include the explaination of how to get the answers. Thanks you. &lt;/p&gt;&#xA;" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2010-11-20T04:02:44.077" LastActivityDate="2010-11-20T04:12:10.990" Title="Principle of counting problem" Tags="&lt;homework&gt;&lt;counting&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="11048" PostTypeId="2" ParentId="11047" CreationDate="2010-11-20T03:51:05.027" Score="2" ViewCount="" Body="&lt;p&gt;1) To form a committee of $5$ people which includes Alice, we need to choose just $4$ more members. Since it excludes Bob, we only have $8$ out of the $10$ people to choose from (Alice and Bob have been accounted for). So this can be done in ${8}\choose{4}$ ways.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2)Inclusion into a committee is to be regarded as one less person to choose and exclusion from a committee should be thought as one less person to choose from. The various cases are just a reapplication of these ideas once or more.&lt;/p&gt;&#xA;" OwnerUserId="3431" LastEditorUserId="3431" LastEditorDisplayName="" LastEditDate="2010-11-20T04:12:10.990" LastActivityDate="2010-11-20T04:12:10.990" CommentCount="1" />
  <row Id="11049" PostTypeId="2" ParentId="11047" CreationDate="2010-11-20T04:00:24.797" Score="0" ViewCount="" Body="&lt;p&gt;@Timothy Wagner: here is my suggestion.  Thanks for not doing the hardest one. There is a wide range in how people respond to homework questions, so don't take mine as gospel-make up your own mind.  But the differences between us are small.  It is in how big a hint to give.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I bet this is homework.  Please tag as such.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1)You have 10 people to choose from.  Alice is required and Bob is excluded.  So there are really only 8 to think about.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2)iii is the same (with 10-&gt;12) as the first.  i and ii and not so different&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-11-20T04:00:24.797" CommentCount="8" />
  <row Id="11050" PostTypeId="1" AcceptedAnswerId="11051" CreationDate="2010-11-20T04:46:13.610" Score="0" ViewCount="203" Body="&lt;p&gt;Let $\{f_n\}$ be a sequence of smooth functions which converges to a function $f$.  If the convergence is not uniform at a point $a$ the $f$ is discontinuous at $a$.&#xA;Is there any different type of convergence where if it happens at $a$ then $f$ is continuous at $a$ but is not differentiable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $\{f_n\}$ be a sequence of smooth functions which converges to a function $f$.If there  is a discontinuity in $f$ at some point $a$ then the convergence is nonuniform.Is there any different type of convergence needed for $f$ to be continuous at some point $a$ but not differentiable ?&lt;/p&gt;&#xA;" OwnerUserId="2987" LastEditorUserId="2987" LastEditorDisplayName="" LastEditDate="2010-11-20T18:37:22.323" LastActivityDate="2010-11-20T18:37:22.323" Title="Convergence and differentiability" Tags="&lt;real-analysis&gt;" AnswerCount="1" CommentCount="23" FavoriteCount="1" />
  <row Id="11051" PostTypeId="2" ParentId="11050" CreationDate="2010-11-20T05:10:39.233" Score="1" ViewCount="" Body="&lt;p&gt;P[a,b], the set of polynomials on [a,b] (obviously a subset of the smooth functions on [a,b]) is dense in C[a,b]. In addition the set of functions that are differentiable at (at least) a single point in [a,b] are of the first category in C[a,b]. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In some sense &quot;most&quot; convergent sequences of smooth functions converge to nowhere differentiable functions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Semi-related, but you might in interested in the fact that $C^k(\Omega)$, $\Omega \subset \mathbb{R}^n$ open, can be made into a Fréchet space for any $k = 1,2,\ldots, \infty$. See Wikipedia's article on &lt;a href=&quot;http://en.wikipedia.org/wiki/Fr%C3%A9chet_space#Examples&quot; rel=&quot;nofollow&quot;&gt;Fréchet spaces&lt;/a&gt;, the topology on $C^k(\Omega)$ being uniform convergence of $f_n$ and all of its derivatives up to order $k$ (q.v. &lt;a href=&quot;http://en.wikipedia.org/wiki/Multiindex&quot; rel=&quot;nofollow&quot;&gt;multiindex&lt;/a&gt;) &lt;/p&gt;&#xA;" OwnerUserId="1269" LastEditorUserId="1269" LastEditorDisplayName="" LastEditDate="2010-11-20T05:21:23.963" LastActivityDate="2010-11-20T05:21:23.963" CommentCount="1" />
  <row Id="11053" PostTypeId="2" ParentId="11044" CreationDate="2010-11-20T06:36:47.237" Score="1" ViewCount="" Body="&lt;p&gt;Consider the following reduction algorithm:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Move the unique column in which $1$ is the first element to the left (extract it and put it as the first column).&lt;/li&gt;&#xA;&lt;li&gt;Apply the unique injection that maps the $j$th element in the first column to $j$.&lt;/li&gt;&#xA;&lt;li&gt;Apply the unique permutation on the $n-1$ columns to the right of the first column which makes the first row $1\ldots n$.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Taking our steps in reverse to unscramble a reduced form, we have $(n-1)!$ choices in the third step, $(n-1)^{\underline{k-1}} = (n-1)!/(n-k)!$ choices in the second step, and $n$ choices in the first step, to a total of $$(n-1)!\frac{(n-1)!}{(n-k)!}n = \frac{n!(n-1)!}{(n-k)!}.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More formally, we could define&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$A_{k,n}$ as the number of Latin rectangles whose upper left value is $1$.&lt;/li&gt;&#xA;&lt;li&gt;$B_{k,n}$ as the number of Latin rectangles whose left-most column is reduced.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Then we show the following:&#xA;$$\begin{align*} L_{k,n} &amp;amp;= n A_{k,n}, \\ A_{k,n} &amp;amp;= \frac{(n-1)!}{(n-k)!} B_{k,n}, \\ B_{k,n} &amp;amp;= (n-1)R_{k,n}. \end{align*}$$&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-20T06:36:47.237" CommentCount="2" />
  <row Id="11054" PostTypeId="2" ParentId="8180" CreationDate="2010-11-20T06:56:50.723" Score="1" ViewCount="" Body="&lt;p&gt;In my experience I haven't seen the gain go above one, but theoretically, it shouldn't go above one either, because then the product of the gain and innovation will lie outside the range of the predicted value and the measurement received. This will be incorrect, because the estimated value should be in between the prediction and the measurement received.&lt;/p&gt;&#xA;" OwnerUserId="3737" LastActivityDate="2010-11-20T06:56:50.723" />
  <row Id="11056" PostTypeId="1" CreationDate="2010-11-20T07:02:49.970" Score="4" ViewCount="133" Body="&lt;p&gt;I was reading Martin Gardner's &lt;em&gt;Mathematical Games&lt;/em&gt; column on the Ulam spiral which appeared in the March 1964 issue of &lt;em&gt;Scientific American&lt;/em&gt;.  (The spiral actually featured on the cover of that issue.)  Gardner makes the following statement: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The grid on the cover suggests that throughout the entire number series expressions of this form are likely to vary markedly from those &quot;poor&quot; in primes to those that are &quot;rich,&quot; and that on the rich lines an unusual amount of clumping occurs.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;By &quot;this form&quot; Gardner means the form $4x^2+bx+c$.  I'm curious - and a little bit skeptical - about his last statement concerning clumping.  I know that the existence of prime-rich and prime-poor polynomials is a longstanding conjecture, going back to Euler's discovery that polynomials such as $x^2-x+41$ generate unusually many primes, and that Hardy and Littlewood and also Bateman and Horn made concrete proposals as to what the density of primes in such polynomials ought to be.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is whether there is any evidence, either numerical or heuristic, that there should be a large amount of clumping in the primes of the form $x^2-x+41$.  Famously, the first 40 values of $x$ all give primes, but if one goes to higher values of $x$ are there more long clusters of primes than one would expect if the primes were randomly distributed?&lt;/p&gt;&#xA;" OwnerUserId="3736" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-20T07:14:24.540" LastActivityDate="2010-11-20T17:00:55.627" Title="Ulam spiral: Is there an &quot;unusual amount of clumping&quot; in prime-rich quadratic polynomials?" Tags="&lt;number-theory&gt;&lt;probability&gt;&lt;prime-numbers&gt;" AnswerCount="1" />
  <row Id="11057" PostTypeId="2" ParentId="10972" CreationDate="2010-11-20T07:03:19.470" Score="1" ViewCount="" Body="&lt;p&gt;For $x$ large, $\tanh x$ is very close to $1$. Therefore for large $b$, $$\int_0^b \frac{\tanh x}{x} \, \mathrm{d}x \approx C + \int^b \frac{\mathrm{d}x}{x} = C&amp;#39; + \log b.$$ You can prove it rigorously and obtain a nice error bound if you wish. Your post indicates a specific value of $C&amp;#39;$, but for large $b$, any two &quot;close&quot; constants $C_1,C_2$ will satisfy $$\log b + C_1 \approx \log b + C_2,$$ so probably $\gamma + \log (4/\pi)$ has no significance other than being a number close to $C&amp;#39;$ and having a nice form.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we do the estimation rigorously, we will probably find out that $C&amp;#39;$ is well defined (i.e. the error in the first $\approx$ is $o(b)$), and then one can ask for its value. It probably has no nice closed form.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: In fact $\gamma + \log (4/\pi)$ is the correct constant, as shown in Derek's answer.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2010-11-23T00:41:44.187" LastActivityDate="2010-11-23T00:41:44.187" CommentCount="0" />
  <row Id="11058" PostTypeId="2" ParentId="6863" CreationDate="2010-11-20T07:03:25.303" Score="0" ViewCount="" Body="&lt;p&gt;I'm not entirely sure of what parameters you're taking as an input, but as far as my experience goes with the H matrix, it is used to focus on a particular input, irrespective of the weightage you intend to give it (I've seen the H matrix in other forms too, but do not have the knowledge to comment on it). &#xA;What you seem to be looking for, is an algorithm to fuse the inputs of two sensors. The best algorithm available for it is IMM (interacting multiple model). It internally uses Kalman filters, and has steps for estimate fusion and mixing probabilities. A transition probability matrix will help you decide on emphasizing the positive properties and minimising the negative. The models correct each other, which plays a large role in getting a more accurate estimate.&lt;/p&gt;&#xA;" OwnerUserId="3737" LastActivityDate="2010-11-20T07:03:25.303" />
  <row Id="11059" PostTypeId="1" CreationDate="2010-11-20T07:17:32.633" Score="0" ViewCount="56" Body="&lt;p&gt;In a Kalman filter taking into account NCT, the state matrix is [x; vx; y; vy; omega]. Where omega is the angular velocity.&#xA;What I don't understand is that when the F matrix for NCT is&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;F=[ blah blah blah blah 0; &#xA;      blah blah blah blah 0; &#xA;      blah blah blah blah 0;&#xA;      blah blah blah blah 0;&#xA;      0    0    0    0    1; ];&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;How can the F matrix or any other matrix directly contribute to the omega value? How can the Kalman filter take into account the angular velocity of a particle if there is no contribution to omega, from any of the matrices in the filter? (the P matrix contributes a little, but that causes omega to become 1e-31).&#xA;Am I doing something wrong? Is there a correct way to use the NCT model?&lt;/p&gt;&#xA;" OwnerUserId="3737" LastActivityDate="2011-05-19T09:59:05.157" Title="How does the NCT model take angular velocity into account? (Nearly coordinated turn)" Tags="&lt;matrices&gt;&lt;discrete-mathematics&gt;&lt;simulation&gt;" AnswerCount="1" />
  <row Id="11060" PostTypeId="2" ParentId="11059" CreationDate="2010-11-20T08:01:16.727" Score="0" ViewCount="" Body="&lt;p&gt;The whole idea of a &quot;coordinated turn&quot; is that the turn is executed at constant angular velocity. The &quot;nearly&quot; is there to indicate that the only contribution to a change in omega is from noise.&lt;/p&gt;&#xA;" OwnerUserId="3638" LastActivityDate="2010-11-20T08:01:16.727" CommentCount="1" />
  <row Id="11061" PostTypeId="2" ParentId="10990" CreationDate="2010-11-20T08:03:27.953" Score="1" ViewCount="" Body="&lt;p&gt;Since your last question looks much general: you might be interested in the nice introductory paper &lt;a href=&quot;http://www.math.nmsu.edu/~davidp/euler2k2.pdf&quot; rel=&quot;nofollow&quot;&gt;&quot;Dances between discrete and continuous&quot;&lt;/a&gt;. Here D. Pengelley discusses some early motivations (and techniques how) to deal with divergent series (focused on L.Euler). I liked that paper much when I started reading about divergent summation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: Thinking a second time about your qualified question I can't imagine you don't know this all (basic reference G.H.Hardy, &quot;divergent series&quot; et al.) and you are looking for a more modern historical overview over that subject - so my answer might be somehow misfocused...  &lt;/p&gt;&#xA;" OwnerUserId="1714" LastEditorUserId="1714" LastEditorDisplayName="" LastEditDate="2010-11-20T08:10:18.260" LastActivityDate="2010-11-20T08:10:18.260" CommentCount="2" />
  <row Id="11062" PostTypeId="1" AcceptedAnswerId="11067" CreationDate="2010-11-20T08:06:31.297" Score="2" ViewCount="162" Body="&lt;p&gt;How to find the domain of the function $\sqrt{ \log_{\frac{1}{2}} x}$ ?&lt;/p&gt;&#xA;" OwnerUserId="3744" LastActivityDate="2010-11-20T11:24:49.823" Title="Domain Problem: $\sqrt{ \log_{\frac{1}{2}} x}$" Tags="&lt;algebra-precalculus&gt;&lt;logarithms&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="3" />
  <row Id="11064" PostTypeId="2" ParentId="11012" CreationDate="2010-11-20T08:43:30.793" Score="3" ViewCount="" Body="&lt;p&gt;This isn't a complete answer, but just an idea about your&#xA;first question. Let $P$ be a rank one projective module&#xA;over the commutative ring $A$, and $P^*=\mathrm{Hom}_A(P,A)$&#xD;&#xA;be its dual. Then for $M=P\oplus P^*$, $\bigwedge^2 M\cong&#xD;&#xA;P\otimes_A P^*\cong A$ is free. There must be $P$ for which $M$&#xD;&#xA;isn't free, but I can't think of any off the top of my head.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $A$ is a Dedekind domain then $M$ is free. Taking $A=C^\infty(N)$&#xD;&#xA;where $N$ is a smooth manifold, then $P$ would correspond to&#xA;a line bundle on $N$. If $M$ is free then the direct sum of this&#xA;line bundle and its dual would be trivial. Surely there are manifolds&#xA;and line bundles for which this isn't true?&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-11-20T08:43:30.793" CommentCount="2" />
  <row Id="11065" PostTypeId="1" CreationDate="2010-11-20T09:01:54.330" Score="0" ViewCount="44" Body="&lt;p&gt;I've got a Python script that grinds out Amicable Pairs by trying out each integer: i, i+1, i+2, ... . I've now found all APs with 1 &amp;lt;= n &amp;lt;= 65 million. At this point I'm finding that it takes 24 minutes per each additional million. To get into much larger n I'll need to find a way to filter out substantial numbers of n in advance. Have any been discovered?   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My list of the first 202 APs: &lt;a href=&quot;http://tutoree7.pastebin.com/dpEc0RbZ&quot; rel=&quot;nofollow&quot;&gt;http://tutoree7.pastebin.com/dpEc0RbZ&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1964" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-20T11:13:29.880" LastActivityDate="2010-11-20T11:13:29.880" Title="Let (n, m), n &lt; m be an Amicable Pair. Looking for large sets of integers that cannot be n." Tags="&lt;number-theory&gt;" CommentCount="2" />
  <row Id="11066" PostTypeId="1" CreationDate="2010-11-20T09:11:33.357" Score="1" ViewCount="211" Body="&lt;p&gt;Just today I had a bet with my friend over the following problem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How many winning configurations can you have in a nxn Tic-Tac-Toe game where players win if a they get n/2 in either a row or column, consecutively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;n is even.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, in a 4x4 game, players win if a they get 2 of their symbols in either a row or column, consecutively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I bet the figure to be &quot;2 * ( 2 * n ) * ( 3 ** ( n / 2 ) )&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do I win?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How to proceed if we were to count only draws? ( how many board configurations can there be so that they are always draws - i.e. no one wins )&lt;/p&gt;&#xA;" OwnerUserId="3740" LastEditorUserId="3740" LastEditorDisplayName="" LastEditDate="2010-11-21T02:59:34.243" LastActivityDate="2010-11-21T05:55:12.600" Title="How many winning configurations can you have in a nxn Tic-Tac-Toe game where players win if a they get n/2 in either a row or column, consecutively." Tags="&lt;combinatorics&gt;" AnswerCount="2" CommentCount="7" />
  <row Id="11067" PostTypeId="2" ParentId="11062" CreationDate="2010-11-20T10:19:38.080" Score="6" ViewCount="" Body="&lt;p&gt;Assuming the result is real, we must have $\log _{\frac{1}{2}}x\geq 0$.&#xA;Since $\log _{\frac{1}{2}}x\geq 0\Leftrightarrow 0&amp;lt;x\leq 1$, the domain is $%&#xD;&#xA;0&amp;lt;x\leq 1$, with $x\in \mathbb{R}$.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Added 2: $\log _{\frac{1}{2}}x=\frac{\log x}{\log \frac{1}{2}}=\frac{\log x}{\log&#xD;&#xA;1-\log 2}=\frac{\log x}{0-\log 2}=-\frac{\log x}{\log 2}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\log _{\frac{1}{2}}x\geq 0\Leftrightarrow-\frac{\log x}{\log 2}\geq 0\Leftrightarrow\log x\le 0\Leftrightarrow 0&amp;lt;x\leq 1.$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Added: plot of $\log_{\frac{1}{2}}x$ (green) and $\sqrt{\log_{\frac{1}{2}}x}$ (blue).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/8PsLd.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="752" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2010-11-20T11:24:49.823" LastActivityDate="2010-11-20T11:24:49.823" CommentCount="6" />
  <row Id="11068" PostTypeId="1" CreationDate="2010-11-20T10:24:42.697" Score="0" ViewCount="204" Body="&lt;p&gt;How to determine the width of a Gaussian membership function?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/mp1D5.jpg&quot; alt=&quot;definition&quot;&gt;&lt;/p&gt;&#xA;" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-20T11:04:26.453" LastActivityDate="2011-05-19T18:59:18.080" Title="How to determine the width of a Gaussian membership function?" Tags="&lt;functions&gt;" AnswerCount="1" CommentCount="7" />
  <row Id="11069" PostTypeId="2" ParentId="2949" CreationDate="2010-11-20T11:05:41.553" Score="6" ViewCount="" Body="&lt;p&gt;The solution to &lt;a href=&quot;http://en.wikipedia.org/wiki/Hilbert%27s_tenth_problem&quot; rel=&quot;nofollow&quot;&gt;&lt;strong&gt;Hilbert's 10th problem&lt;/strong&gt;&lt;/a&gt;, i.e. the &lt;a href=&quot;http://en.wikipedia.org/wiki/MRDP_theorem&quot; rel=&quot;nofollow&quot;&gt;&lt;strong&gt;MRDP theorem&lt;/strong&gt;&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Number theorists were trying to find a general method to solve Diophantine equations. Special cases of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Diophantine_equation&quot; rel=&quot;nofollow&quot;&gt;Diophantine equations&lt;/a&gt; were/are studied extensively and the theorems are quite nice. Learning the fascinating fact that there is no general method (algorithm) to solve arbitrary Diophantine equations was surprising for me.&lt;/p&gt;&#xA;" OwnerUserId="468" LastActivityDate="2010-11-20T11:05:41.553" CommunityOwnedDate="2010-11-20T11:05:41.553" />
  <row Id="11070" PostTypeId="1" AcceptedAnswerId="11071" CreationDate="2010-11-20T11:11:05.943" Score="1" ViewCount="92" Body="&lt;p&gt;I've been having some rather morbid thoughts lately, so, naturally, I decided to share:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose I have a deadly disease, which has a chance of killing me every day that it is left uncured. Also, suppose that there's this operation I can go through, but it is not without its risks: If it is successful, I won't have to worry about the disease killing me ever again. If it's a failure, I die there and then. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;More formally, let $q : N \rightarrow [0,1]$ be the probability that the disease kills me at any given day $t \in N$. Also, let $p : N \rightarrow [0,1]$ be the probability the that the operation fails if I have it on day $t \in N$. Assume that both $p$ and $q$ are monotonically increasing. The question is: At what day $t_0$ do I have to go through the operation in order to maximise my expected lifespan?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, to solve this: It is obvious that at any day that $p(t) \leq q(t)$ you're better off having the operation because if you don't, you'd have the same chance of dying and no &#xA;chance of being cured. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, that can't possibly be the whole story. What if $p(t) &amp;gt; q(t)$ for every $t \in N$? You would still have to do the operation at some point, otherwise you'll eventually die because of the accumulated probability of all those days you're left uncured. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In that light, let $f(t) = \prod_{k=0}^{t}(1-q(k))$ be the probability that you survive $t$ days without going through the operation. Then, all you need to do is to solve the inequality $f(t_0) &amp;lt; f(t_0-1) \cdot (1-p(t_0))$. Is that correct? Or am I still missing something?&lt;/p&gt;&#xA;" OwnerUserId="3742" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-11-20T20:27:49.363" LastActivityDate="2010-11-20T20:27:49.363" Title="To operate or not to operate?" Tags="&lt;probability&gt;&lt;probability-theory&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="0" />
  <row Id="11071" PostTypeId="2" ParentId="11070" CreationDate="2010-11-20T11:34:51.220" Score="1" ViewCount="" Body="&lt;p&gt;Elaborating on my comments above, the formula for the expected lifespan in the case of not taking the operation is&#xA;$$\sum_{t&amp;lt;T}t\bigl(f(t)-f(t-1)\bigr),$$&#xD;&#xA;while the expected lifespan in case of taking the operation is&#xA;$$\sum_{t&amp;lt;t_0}t\bigl(f(t)-f(t-1)\bigr)+T\cdot f(t_0).$$&#xD;&#xA;Here $T$ should be the age that you will die if you take the operation, survive it, and die a natural death.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now you just have to plug in and see which number is larger.&lt;/p&gt;&#xA;" OwnerUserId="367" LastEditorUserId="367" LastEditorDisplayName="" LastEditDate="2010-11-20T12:38:02.247" LastActivityDate="2010-11-20T12:38:02.247" />
  <row Id="11072" PostTypeId="1" CreationDate="2010-11-20T11:40:24.690" Score="0" ViewCount="57" Body="&lt;p&gt;Goodmorning at all .&#xA;I'm a software development and i have a small question in mathematic. &#xA;I have a two variables function in a interval [(0,0),(6,6)]. I wont divide this interval in two equal subinterval . So i think [(0,0),(3,3)] and [(3,3),(6,6)]&#xA;It is correct ? &#xA;Which algorithm i should apply? &#xA;Thanks &lt;/p&gt;&#xA;" LastActivityDate="2010-11-20T11:40:24.690" Title="divide an interval of function in two variables" Tags="&lt;calculus&gt;&lt;analysis&gt;&lt;soft-question&gt;&lt;functions&gt;" CommentCount="1" />
  <row Id="11073" PostTypeId="1" AcceptedAnswerId="11115" CreationDate="2010-11-20T11:46:36.697" Score="1" ViewCount="173" Body="&lt;p&gt;Hello,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $S^1$ the unit circle in $\mathbb{R}^2$ and &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$d: S^1\times S^1\to\mathbb{R}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$d(\theta_1,\theta_2) = \left\{&#xD;&#xA;    \begin{array}{ll}&#xD;&#xA;        |\theta_1-\theta_2| &amp;amp; \mbox{if } |\theta_1-\theta_2|\le \pi \\&#xD;&#xA;        2\pi-|\theta_1-\theta_2| &amp;amp; \mbox{else}&#xD;&#xA;    \end{array}&#xD;&#xA;\right.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm trying to prove that this function satisfies the triangle inequality&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$d(\theta_1,\theta_3)\le d(\theta_1,\theta_2)+d(\theta_2,\theta_3)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are three possible cases:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$|\theta_1-\theta_2|, |\theta_2-\theta_3|\le\pi$&#xD;&lt;/li&gt;&#xA;&lt;li&gt;$|\theta_1-\theta_2|\le\pi, |\theta_2-\theta_3|&amp;gt;\pi$ or $|\theta_2-\theta_3|\le\pi, |\theta_1-\theta_2|&amp;gt;\pi$&#xD;&lt;/li&gt;&#xA;&lt;li&gt;$|\theta_1-\theta_2|, |\theta_2-\theta_3|&amp;gt;\pi$&#xD;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I proved the first two ones using the triangle inequality for the absolute value, but I'm stuck for the third.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is equivalent to prove that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$|\theta_1-\theta_3|+|\theta_1-\theta_2|+|\theta_2-\theta_3|\le 4\pi\text{ if }|\theta_1-\theta_3|\le\pi$$&#xD;&#xA;and&#xA;$$|\theta_1-\theta_2|+|\theta_2-\theta_3|-|\theta_1-\theta_3|\le 2\pi\text{ if }|\theta_1-\theta_3|&amp;gt;\pi$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could you give me a hint ? Or did I misdefine $d$ or any other error ?&lt;/p&gt;&#xA;" OwnerUserId="524" LastActivityDate="2010-11-21T13:40:09.720" Title="Unit circle metric" Tags="&lt;geometry&gt;&lt;metric-spaces&gt;&lt;circle&gt;" AnswerCount="4" FavoriteCount="1" />
  <row Id="11074" PostTypeId="2" ParentId="11073" CreationDate="2010-11-20T12:26:11.873" Score="2" ViewCount="" Body="&lt;p&gt;Apologies: this should be a comment, but I am not &quot;reputable&quot; enough to leave comments.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you proved 1, then you should have no problems with 3. The key to the exercise is to understand this metric geometrically. Note that if you draw two radii on the unit circle, then there are two obvious angles between them, that you could measure. The two-case-rule just says &quot;always measure the smaller one of the two angles&quot;. In other words, the distance between two angles is always at most $\pi$, which should allow you to easily complete the proof.&lt;/p&gt;&#xA;" OwnerUserId="3212" LastActivityDate="2010-11-20T12:26:11.873" CommentCount="2" />
  <row Id="11076" PostTypeId="1" AcceptedAnswerId="11079" CreationDate="2010-11-20T13:06:42.590" Score="6" ViewCount="518" Body="&lt;p&gt;It is nature to understand the need for scalar(numbers), but why do we invent vectors? Who invent it and for what?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: As George Lowther pointed out the problem is too broad, I added the following questions as concrete supplements.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;It's easy for our human to understand the law of addition of scalar number, but why the addition of vector follow the parallelogram rule, why doesn't it follow some other laws?&#xA;link to physics: How do we know that the addition of forces follow the parallelogram law?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The length of 2-dimensional vector is the hypotenuse of the triangle whose is constructed from the two component of the vector and the length of 3-dimensional vector is also follows this way. But what about 4-dimensional vector?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Why do we define dot production law of vector like it is now? Is it because of its physical essence or its equivalence to the law of cosine?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="1382" LastEditorUserId="1382" LastEditorDisplayName="" LastEditDate="2010-11-20T13:54:46.573" LastActivityDate="2010-12-20T10:52:40.393" Title="Why do we need vectors and who invented it?" Tags="&lt;vector-calculus&gt;" AnswerCount="6" CommentCount="2" FavoriteCount="3" />
  <row Id="11077" PostTypeId="2" ParentId="11073" CreationDate="2010-11-20T13:20:40.893" Score="1" ViewCount="" Body="&lt;p&gt;This is not a formal proof, but gives, I think, the geometric intuition:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/ObvOd.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="752" LastActivityDate="2010-11-20T13:20:40.893" CommentCount="2" />
  <row Id="11079" PostTypeId="2" ParentId="11076" CreationDate="2010-11-20T13:37:46.463" Score="14" ViewCount="" Body="&lt;p&gt;Vectors should be thought of, at a first approximation, as &quot;numbers with direction&quot;. For physical phenomena which carries a direction, such as velocity and displacement, vectors are immensely useful. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The concept of a number with direction most likely dates to antiquity, as the making of maps and sign-posts already implicitly incorporates the notion. The modern representation of a vector/point in space with an ordered triplet of numbers is often attributed to the advent of &lt;a href=&quot;http://en.wikipedia.org/wiki/Analytic_geometry&quot; rel=&quot;nofollow&quot;&gt;analytical geometry&lt;/a&gt; due to the philosopher &lt;a href=&quot;http://ualr.edu/lasmoller/descartes.html&quot; rel=&quot;nofollow&quot;&gt;Rene Descartes&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A different notion of vectors also arose with the &quot;discovery&quot; of the complex numbers by Jerome Cardan: the imaginary numbers can be thought of as living on a different direction as the real scalars (so the complex numbers form a real vector space). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Over the past 400 years or so the notion of vector gradually evolved to become what we know today, with contributions from branches of mathematics that developed into modern analysis and algebra. A nice summary of that period of development is available &lt;a href=&quot;http://web.archive.org/web/20040126161844/http://www.nku.edu/~curtin/crowe_oresme.pdf&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;. See Michael Crowe's &lt;a href=&quot;http://en.wikipedia.org/wiki/A_History_of_Vector_Analysis&quot; rel=&quot;nofollow&quot;&gt;book&lt;/a&gt; for a fuller description of also the Greek contributions and the influences from the 16th century in this matter. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In short, vectors shouldn't be thought of as being &quot;invented&quot;, nor should it be attributed to one person alone.&lt;/p&gt;&#xA;" OwnerUserId="1543" LastActivityDate="2010-11-20T13:37:46.463" CommentCount="1" />
  <row Id="11080" PostTypeId="2" ParentId="11076" CreationDate="2010-11-20T13:48:20.963" Score="4" ViewCount="" Body="&lt;p&gt;A case study of a particular application:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Vectors are absolutely necessary for game development, for instance. How is a 3D model, which may be positioned at any point in space and rotated in any direction and is viewed by a camera pointing in an arbitrary direction, actually rendered?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, the model is stored in some local coordinate system. This is the data that actually fills the file from which the model is loaded. Although the exact contents will differ across different rendering systems, at the very least you will have the coordinates of the vertices relative to some arbitrarily chosen center point. These are viewed as vectors, and this viewpoint is crucial for what comes next.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now all the models are loaded by the rendering system, but they need to be placed into a global coordinate system. This includes putting them in their actual position in the world, rotating them, etc. This stage is accomplished by transforming each of the vectors in local coordinates according to a world transformation matrix--this matrix translates and rotates the model. [You may be wondering how translations are accomplished in this way. That's because most rendering systems store vertices in four-dimensional homogeneous coordinates, and the transformation matrices are $4\times 4$--it's a worthwhile exercise to see that this allows you to perform affine transformations on $\mathbb{R}^3$.]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next another matrix transformation is applied to all the vertex data to take into account the position and orientation of the camera.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally yet another transformation is applied which projects all the data onto the viewing screen--this is not an orthogonal projection but a perspective projection, so that objects in the distance seem smaller than objects close to the camera.&lt;/p&gt;&#xA;" OwnerUserId="251" LastActivityDate="2010-11-20T13:48:20.963" />
  <row Id="11081" PostTypeId="1" AcceptedAnswerId="11092" CreationDate="2010-11-20T13:55:12.470" Score="1" ViewCount="145" Body="&lt;p&gt;$\lim_{x\to\infty} \frac{e^x}{x^n}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;n is any natural number. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using L'hopital doesn't make much sense to me. I did find this in the book: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;In a struggle between a power and an exp, the exp wins.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can I refer that line as an answer? If the fraction would have been flipped, then the limit would be zero. But in this case I the limit is actually $\infty$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="2230" LastActivityDate="2010-11-20T16:17:50.117" Title="Calculus, find the limit, Exp vs Power?" Tags="&lt;calculus&gt;&lt;homework&gt;&lt;limit&gt;" AnswerCount="5" CommentCount="1" />
  <row Id="11082" PostTypeId="2" ParentId="11076" CreationDate="2010-11-20T14:00:34.260" Score="8" ViewCount="" Body="&lt;p&gt;Willie covered a lot of what I wanted to say; however I'd like to make a little historical digression: before we ever had the concept of a vector, there was the quaternion, William Rowan Hamilton's generalization of the usual complex numbers. They proved very convenient for physical applications, and thus the use of quaternions took off. In fact the electromagnetic equations of Maxwell were first couched in quaternion notation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It was Josiah Gibbs and independently Oliver Heaviside who looked at decomposing the quaternion into a scalar (real) part, and a vector (imaginary) part, and found that the manipulations in this new formulation were &quot;cleaner&quot;. Vector analysis took off, and quaternions became less prominent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd say more, but a &lt;a href=&quot;http://books.google.com/books?id=y5-S5dmVqGIC&quot; rel=&quot;nofollow&quot;&gt;book&lt;/a&gt; has already been written about this matter, so I refer you to it.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-11-20T14:00:34.260" CommentCount="3" />
  <row Id="11083" PostTypeId="2" ParentId="11081" CreationDate="2010-11-20T14:00:40.063" Score="1" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT:&lt;/strong&gt; One way of looking at this would be: $$\frac{1}{x^{n}} \biggl[ \biggl(1 + \frac{x}{1!} + \frac{x^{2}}{2!} + \cdots + \frac{x^{n}}{n!}\biggr) + \frac{x^{n+1}}{(n+1)!} + \cdots \biggr]$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope you understand why i put the brackets inside those terms.&lt;/p&gt;&#xA;" LastActivityDate="2010-11-20T14:00:40.063" CommentCount="1" />
  <row Id="11084" PostTypeId="2" ParentId="11081" CreationDate="2010-11-20T14:10:38.933" Score="1" ViewCount="" Body="&lt;p&gt;Why isn't L'Hôpital a good solution? Just use induction.&lt;/p&gt;&#xA;" OwnerUserId="3567" LastActivityDate="2010-11-20T14:10:38.933" CommentCount="2" />
  <row Id="11085" PostTypeId="1" AcceptedAnswerId="11086" CreationDate="2010-11-20T14:22:42.477" Score="3" ViewCount="442" Body="&lt;p&gt;How do I calculate:&#xA;The height of a regular tetrahedron, side length 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just to be completely clear, by height I mean if you placed the shape on a table, how high up would the highest point be from the table.&lt;/p&gt;&#xA;" OwnerUserId="3746" LastActivityDate="2010-11-28T22:38:33.123" Title="Height of a tetrahedron" Tags="&lt;geometry&gt;" AnswerCount="4" CommentCount="2" FavoriteCount="1" />
  <row Id="11086" PostTypeId="2" ParentId="11085" CreationDate="2010-11-20T14:27:33.737" Score="7" ViewCount="" Body="&lt;p&gt;The first thing you need to do is to note that the apex of a regular tetrahedron lies directly above the center of the bottom triangular face. Thus, find the length of the segment connecting the center of an equilateral triangle with unit length to a corner, and use the Pythagorean theorem with the length of an edge as the hypotenuse, and the length you previously derived as one leg. The height you need is the other leg of the implied right triangle.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Here's a view of the geometry:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/Cj5bL.png&quot; alt=&quot;tetrahedron&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and here's a view of the bottom face:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/0QvQD.png&quot; alt=&quot;triangle&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the second diagram, the face is indicated by dashed lines, and the (isosceles) triangle formed by the center of the triangle and two of the corners is indicated by solid lines.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Knowing that the short sides of the isosceles triangle bisect the 60° angles of the equilateral triangle, we find that the angles of the isosceles triangle are 30°, 30° and 120°.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the law of cosines and the knowledge that the longest side of the isosceles triangle has unit length, we have the equation for the length $\ell$ of the short side (the length from the center of the bottom face to the nearest vertex):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$1=2\ell^2-2\ell^2\cos 120^{\circ}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solving for $\ell$, we find that the length from the center of the bottom face to the nearest vertex is $\frac{1}{\sqrt{3}}$, as indicated &lt;a href=&quot;http://mathworld.wolfram.com/EquilateralTriangle.html&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From this, the Pythagorean theorem says that the height $h$ (the length from the center of the bottom face) satisfies&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$h^2+\left(\frac{1}{\sqrt{3}}\right)^2=1$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solving for $h$ in the above equation, we now find the height to be $\sqrt{\frac23}=\frac{\sqrt{6}}{3}$, as mentioned &lt;a href=&quot;http://mathworld.wolfram.com/Tetrahedron.html&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="498" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-20T22:00:33.340" LastActivityDate="2010-11-20T22:00:33.340" CommentCount="4" />
  <row Id="11087" PostTypeId="2" ParentId="11081" CreationDate="2010-11-20T14:51:09.867" Score="1" ViewCount="" Body="&lt;p&gt;I believe the problem is tailor-made for repeated application of L'Hopital's Rule, but here are some thoughts ...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could note that $e^{x} = (e^{x/n})^n$, and consider $\left( \lim \frac{e^{x/n}}{x}\right)^n$, so that you are comparing an exponential to a single power of $x$, which might be a bit less daunting for you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A bit more cleanly, and to make the numerator and denominator match better, define $y := \frac{x}{n}$. Then $$\frac{e^{x}}{x^n}=\frac{e^{ny}}{(ny)^n}=\frac{\left(e^{y}\right)^n}{n^n y^n}=\frac{1}{n^n}\frac{\left(e^{y}\right)^n}{y^{n}}=\frac{1}{n^n}\left(\frac{e^y}{y}\right)^n$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $n$ is a constant, you can direct your limiting attention to $\frac{e^y}{y}$ (as $y \to \infty$, of course).&lt;/p&gt;&#xA;" OwnerUserId="409" LastActivityDate="2010-11-20T14:51:09.867" />
  <row Id="11088" PostTypeId="1" AcceptedAnswerId="11093" CreationDate="2010-11-20T15:07:58.707" Score="2" ViewCount="120" Body="&lt;p&gt;Let $f:\mathbb{R}\to \mathbb{C}$ be a Lebesgue measurable function. For each $z\in \mathbb{C}$ let $Arg (z)$ be the principal argument of $z$ (define it to be $0$ if $z=0$). Define $g(x)= \sqrt{|f(x)|}exp(\frac{1}{2} i Arg (f(x))$. Then $g(x)$ is measurable and $(g(x))^2=f(x)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am sure the above statement is correct, but just wanted to confirm with you all. The only part of concern is, of course, whether the function $exp(\frac{1}{2} i Arg (f(x))$ is measurable. &lt;/p&gt;&#xA;" OwnerUserId="3249" LastEditorUserId="3249" LastEditorDisplayName="" LastEditDate="2010-11-20T15:42:43.187" LastActivityDate="2010-11-20T17:40:20.023" Title="Square root of a complex measurable function" Tags="&lt;real-analysis&gt;&lt;measure-theory&gt;" AnswerCount="1" />
  <row Id="11089" PostTypeId="2" ParentId="11081" CreationDate="2010-11-20T15:12:12.557" Score="2" ViewCount="" Body="&lt;p&gt;Use the fact that $e^x \ge \left( 1 + \frac{x}{n} \right)^n$ for any $n &amp;gt; 0$.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-11-20T15:12:12.557" />
  <row Id="11090" PostTypeId="2" ParentId="11085" CreationDate="2010-11-20T15:15:46.940" Score="3" ViewCount="" Body="&lt;p&gt;Consider the tetrahedron inscribed in the unit cube, with vertices at (0,0,0), (1,1,0), (0,1,1), (1,0,1). Its height is the distance from (0,0,0) to the centre of the opposite face, which is given by the equation $x+y+z = 2$. Thus its height is $\frac{2}{\sqrt 3}$, and since the edges of this tetrahedron have length $\sqrt 2$, the height of a regular tetrahedron with side $x$ is $x \sqrt{\frac{2}{3}}$.&lt;/p&gt;&#xA;" OwnerUserId="3747" LastEditorUserId="3747" LastEditorDisplayName="" LastEditDate="2010-11-28T22:38:33.123" LastActivityDate="2010-11-28T22:38:33.123" CommentCount="1" />
  <row Id="11091" PostTypeId="1" CreationDate="2010-11-20T15:16:06.970" Score="5" ViewCount="870" Body="&lt;p&gt;I greatly enjoyed the Lights Out game described &lt;a href=&quot;http://en.wikipedia.org/w/index.php?title=Lights_Out_%28game%29&amp;amp;oldid=395986075&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; (I am sorry I had to link to an older page because some wikidiot keeps deleting most of the page). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Its mathematical analysis is &lt;a href=&quot;http://www.jaapsch.net/puzzles/lomath.htm&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; (it's just linear algebra)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I just discovered a hexagonal version:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://cwynn.com/games/lightsoutmobile.htm&quot; rel=&quot;nofollow&quot;&gt;http://cwynn.com/games/lightsoutmobile.htm&lt;/a&gt;&lt;br&gt;&#xA;  (and hopefully soon) &lt;a href=&quot;https://sites.google.com/site/beyondlightsout/&quot; rel=&quot;nofollow&quot;&gt;https://sites.google.com/site/beyondlightsout/&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Are there any mathematical results for this version of the game, before I dig in and start the analysis myself. I have the major references including  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;em&gt;Turning Lights Out with Linear Algebra&lt;/em&gt;, by M. Anderson and T. Feil (1998). &lt;em&gt;Math Magazine&lt;/em&gt;, vol. 71, no. 4, October 1998, pp. 300-303.  It's &lt;a href=&quot;http://www.math.ksu.edu/~dmaldona/math551/lights_out.pdf&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; (thanks to J.M.).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Update: I've been playing with the iPhone T-Lights hexagonal versions. I can get most of them, except for the ones where the &quot;template&quot; is a Y shape. Any ideas?&lt;/p&gt;&#xA;" OwnerUserId="2809" LastEditorUserId="2809" LastEditorDisplayName="" LastEditDate="2010-11-30T02:41:33.813" LastActivityDate="2010-11-30T02:41:33.813" Title="Lights out game on hexagonal grid" Tags="&lt;linear-algebra&gt;&lt;combinatorics&gt;&lt;discrete-mathematics&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="1" />
  <row Id="11092" PostTypeId="2" ParentId="11081" CreationDate="2010-11-20T15:21:37.597" Score="3" ViewCount="" Body="&lt;p&gt;Repeated use of &lt;a href=&quot;http://en.wikipedia.org/wiki/L%27H%C3%B4pital%27s_rule&quot; rel=&quot;nofollow&quot;&gt;L'Hôpital's rule&lt;/a&gt; ($n$ times):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\underset{x\rightarrow \infty }{\lim }\dfrac{e^{x}}{x^{n}}=\underset{%&#xD;&#xA;x\rightarrow \infty }{\lim }\dfrac{e^{x}}{nx^{n-1}}=\underset{x\rightarrow&#xD;&#xA;\infty }{\lim }\dfrac{e^{x}}{n(n-1)x^{n-2}}=\cdots =$ $=\cdots=\underset{x\rightarrow \infty }{\lim }\dfrac{e^{x}}{n(n-1)\cdots&#xD;&#xA;3\cdot x^{2}}=\underset{x\rightarrow \infty }{\lim }\dfrac{e^{x}}{%&#xD;&#xA;n(n-1)\cdots 3\cdot 2x}=$ $=\underset{x\rightarrow \infty }{\lim }\dfrac{e^{x}}{%&#xD;&#xA;n(n-1)\cdots 3\cdot 2\cdot 1}=\underset{x\rightarrow \infty }{\lim }\dfrac{%&#xD;&#xA;e^{x}}{n!}=\infty$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Added: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To convince yourself: If you had $\underset{x\rightarrow \infty }{\lim }\dfrac{e^{x}}{x^{10}}$ you would have to apply L'Hôpital's rule ten times.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Added 2: Plot of $\dfrac{e^{x}}{x^{3}}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/nGHEu.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="752" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2010-11-20T16:17:50.117" LastActivityDate="2010-11-20T16:17:50.117" />
  <row Id="11093" PostTypeId="2" ParentId="11088" CreationDate="2010-11-20T15:24:02.757" Score="2" ViewCount="" Body="&lt;p&gt;Rewriting my answer to improve clarity:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\exp({1 \over 2}i(Arg(f(x)))$ can be written as $h(g(f(x)))$, where $h(z) = e^{{1 \over 2} iz}$ and $g(z) = Arg(z)$. By Rudin thm 1.12d) for example, a measurable function followed by a Borel measurable function is measurable. So it suffices to show $g$ and $h$ are Borel measurable. For $h(z)$ this follows from the fact that $h$ is continuous. For $g(z)$, regardless of how you define $Arg(z)$,&#xA;the set $ \{z: a &amp;lt; Arg(z) &amp;lt; b\}$ is the union of at most two wedges which are Borel sets. So $\{z: g(z) \in (a,b) \}$ is a Borel set. This means $g(z)$ is Borel measurable as well.&lt;/p&gt;&#xA;" OwnerUserId="3035" LastEditorUserId="3035" LastEditorDisplayName="" LastEditDate="2010-11-20T17:40:20.023" LastActivityDate="2010-11-20T17:40:20.023" CommentCount="10" />
  <row Id="11094" PostTypeId="1" AcceptedAnswerId="11100" CreationDate="2010-11-20T15:51:13.020" Score="1" ViewCount="53" Body="&lt;p&gt;I have two random variables $X$ and $Y$, both receiving values between 0 and 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that $E[X - Y] \ge 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can I get any inequality of the form:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$P(X - Y \ge \delta) \le F(\delta,X,Y)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $F(\delta,X,Y)$ is a (reasonable) function of $\delta$ and $E[X-Y]$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Markov inequality would be good here, for example, by setting $F(\delta,X,Y) = E[X-Y]/\delta$. However, Markov inequality would require $X-Y \ge 0$, and I do not necessarily have that.&lt;/p&gt;&#xA;" OwnerUserId="7076" LastActivityDate="2010-11-20T19:22:33.233" Title="can I get a bound on the probability of deviation, similar to Markov inequality?" Tags="&lt;probability&gt;&lt;statistics&gt;&lt;inequality&gt;" AnswerCount="3" />
  <row Id="11095" PostTypeId="1" AcceptedAnswerId="11101" CreationDate="2010-11-20T16:02:50.690" Score="1" ViewCount="267" Body="&lt;p&gt;I've read about many attempts to find patterns in prime numbers.  First, is there a mathematical way to prove there is not a pattern to prime numbers?  Since there are ways to check if a number is prime or not, can these methods be combined into some large function that produces prime numbers?  And in case it isn't obvious, no, I never got passed pre-Cal.&lt;/p&gt;&#xA;" OwnerUserId="3748" LastActivityDate="2010-12-29T00:46:47.917" Title="Patterns in Prime numbers, and the null hypothesis" Tags="&lt;prime-numbers&gt;" AnswerCount="4" CommentCount="7" />
  <row Id="11096" PostTypeId="2" ParentId="11076" CreationDate="2010-11-20T16:12:05.323" Score="8" ViewCount="" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;As with most anything in physics, the addition of forces follows the parallelogram law exactly because it agrees with every experiment mankind has ever made---in other words, ultimately it's an assumption we make based on what we see. But, it's perfectly intuitive that this is how forces should behave: say you're at the origin on the xy-plane and have a force pushing you in the positive x-direction and one pushing you in the positive y-direction, both of equal magnitude. Then it should physically and intuitively clear that the net result will be a force pushing you in the direction of (1,1). If the force in the x-direction was stronger, then the net result would still push you in some &quot;diagonal&quot; direction, but aligned more with the x-axis. This is exactly the parallelogram law.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Same exact thing. The problem is that we can't really picture what a right triangle in 4 dimensions looks like, but it's exactly the same idea.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Of course, there are many possible &quot;dot products&quot; that can be defined on vectors, and in many applications it is indeed useful to use one different from the standard one. But I would say the importance of the standard dot product indeed comes from its relation with the cosine of the angle between them: again if you try to think about this physically, this is a precise measure of how much the two vectors (or forces if you like) are working &quot;with&quot; or &quot;against&quot; each other. Note in particular that the sign of the dot product solely depends on whether or not the angle between your two vectors is less than, equal to, or greater than 90 degrees.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="1914" LastActivityDate="2010-11-20T16:12:05.323" />
  <row Id="11097" PostTypeId="2" ParentId="11095" CreationDate="2010-11-20T16:13:21.963" Score="3" ViewCount="" Body="&lt;p&gt;It really depends on what you mean by patterns. Legendre showed that there is no rational algebraic function that outputs only primes. You can read more about prime generating functions here&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://mathworld.wolfram.com/Prime-GeneratingPolynomial.html&quot; rel=&quot;nofollow&quot;&gt;http://mathworld.wolfram.com/Prime-GeneratingPolynomial.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The famous Green-Tao theorem states that there are arbitrarily large arithmetic progressions in primes&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Green%E2%80%93Tao_theorem&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Green%E2%80%93Tao_theorem&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3431" LastActivityDate="2010-11-20T16:13:21.963" />
  <row Id="11098" PostTypeId="2" ParentId="11068" CreationDate="2010-11-20T16:23:05.353" Score="0" ViewCount="" Body="&lt;p&gt;You don't say what data you have.  If you have (x,y) pairs, you can just use a multidimensional fitting routine like &lt;a href=&quot;http://nr.com/&quot; rel=&quot;nofollow&quot;&gt;Numerical Recipes&lt;/a&gt; or Excel's goal seek function to get all the parameters.  If you have random samples it may be hard to distinguish the effects of m and s unless you have a lot of them.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-11-20T16:23:05.353" />
  <row Id="11099" PostTypeId="2" ParentId="11012" CreationDate="2010-11-20T16:29:31.427" Score="2" ViewCount="" Body="&lt;p&gt;Following variation of Robin Chapman idea works in algebraic geometry.&#xA;Take $X$ elliptic curve minus point : it is affine variety. For general point $P \in X$ take $L=\mathcal O(P)$. Then $M=L\oplus L^*=\mathcal O(P) \oplus \mathcal O(-P)$ has properties:&#xA;A) $\Lambda ^2 (M)=\mathcal O$ is free of rank one.&#xA;B) $M$ is not free because it has no sections except zero (since  $\mathcal O(P)$&#xD;&#xA; and $\mathcal O(-P)$ have no sections except zero)&lt;/p&gt;&#xA;" OwnerUserId="2872" LastActivityDate="2010-11-20T16:29:31.427" />
  <row Id="11100" PostTypeId="2" ParentId="11094" CreationDate="2010-11-20T16:31:43.097" Score="3" ViewCount="" Body="&lt;p&gt;Since $0 \le X,Y \le 1$, you have $X-Y \ge -1$, so applying the Markov inequality to $X-Y+1$ gives&#xA;$$P(X-Y \ge \delta) = P(X-Y+1 \ge \delta+1) \le (E[X-Y]+1)/(\delta+1)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Without any restrictions on the range of $X,Y$, no such bound is possible.  For any numbers $\mu, \delta \in \mathbb{R}$, $p \in (0,1)$, there exists a random variable $Z$ (here $Z=X-Y$) with $P(Z \ge \delta) = p$ and $EZ = \mu$.  Thus $p$ cannot be bounded by any function of $\mu$ and $\delta$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, let $Z=a$ with probability $p$ and $Z=b$ with probability $1-p$.  Then we have $\mu = EZ = pa + (1-p)b$, so $b = (\mu-pa)/(1-p)$.  Since $b \to -\infty$ as $a \to \infty$, you can choose $a$ so large that $a &amp;gt; \delta$ and $b &amp;lt; \delta$, and then $P(Z \ge \delta) = p$.&lt;/p&gt;&#xA;" OwnerUserId="822" LastEditorUserId="822" LastEditorDisplayName="" LastEditDate="2010-11-20T16:39:54.057" LastActivityDate="2010-11-20T16:39:54.057" CommentCount="2" />
  <row Id="11101" PostTypeId="2" ParentId="11095" CreationDate="2010-11-20T16:49:59.673" Score="3" ViewCount="" Body="&lt;p&gt;I like Dirichlet's Theorem, which states that for relatively prime $a,d \in \mathbb{Z}^+$, there are infinitely many primes in the progression $\{a + nd \mid n \in \mathbb{Z^+}\}$.  Further, the proportion of primes in any relatively prime residue class of $d$ is about $1/\phi(d)$, where $\phi$ is the Euler phi function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In essence, the primes can be seen as somewhat evenly distributed.  You can see more below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Dirichlet%27s_theorem_on_arithmetic_progressions&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Dirichlet%27s_theorem_on_arithmetic_progressions&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3537" LastActivityDate="2010-11-20T16:49:59.673" />
  <row Id="11102" PostTypeId="2" ParentId="11085" CreationDate="2010-11-20T16:51:17.207" Score="1" ViewCount="" Body="&lt;p&gt;You can also use trig based on the dihedral angle between two faces of the tetrahedron.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Writing $ABC$ for the base triangle, $O$ for the apex, $K$ for the center of $ABC$ (the foot of the perpendicular dropped from $O$), and $M$ for the midpoint of (for instance) side $BC$, we have a right triangle $OKM$ with right angle at $K$. So,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\text{height of tetrahedron} = |OK| = |OM|\sin{M}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$OM$ is the height of the (equilateral) face $OBC$, measuring $\frac{\sqrt{3}}{2}s$, where $s$ is the length of a side.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the measure of angle $M$ ... Note that this is the dihedral angle between faces $OBC$ and $ABC$; it is also the angle between (congruent) segments $OM$ and $AM$ in triangle $OMA$. We can use the Law of Cosines as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{eqnarray}&#xD;&#xA;|OA|^2 &amp;amp;=&amp;amp; |OM|^2 + |AM|^2 - 2 |OM||AM|\cos{M} \\&#xD;&#xA;s^2 &amp;amp;=&amp;amp; \left(\frac{\sqrt{3}}{2}s\right)^2 + \left(\frac{\sqrt{3}}{2}s\right)^2 - 2 \left(\frac{\sqrt{3}}{2}s\right)\left(\frac{\sqrt{3}}{2}s\right) \cos{M} \\&#xD;&#xA;s^2 &amp;amp;=&amp;amp; \frac{3}{4} s^2 + \frac{3}{4}s^2 - 2 \frac{3}{4} s^2 \cos{M} \\&#xD;&#xA;1 &amp;amp;=&amp;amp; \frac{3}{2} - \frac{3}{2} \cos{M} \\&#xD;&#xA;\frac{-1}{2} &amp;amp;=&amp;amp; - \frac{3}{2} \cos{M} \\&#xD;&#xA;\frac{1}{3} &amp;amp;=&amp;amp; \cos{M} \;\;\; (**)\\&#xD;&#xA;\Rightarrow \sqrt{1-\left(\frac{1}{3}\right)^2} = \frac{\sqrt{8}}{3} =\frac{2\sqrt{2}}{3}&amp;amp;=&amp;amp; \sin{M}&#xD;&#xA;\end{eqnarray}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\text{height of tetrahedron} = |OK| = |OM|\sin{M} = \frac{\sqrt{3}}{2} s \cdot \frac{2\sqrt{2}}{3} = \frac{\sqrt{6}}{3}s$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(**) This cosine is the reason I posted this approach. It's sometimes handy to know (as in this problem); even better, it's easy to remember, because it turns out that it fits a simple pattern (which might be more-likely to impress interviewers):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{eqnarray}&#xD;&#xA;\cos\left({\text{angle between two sides of a regular triangle}}\right) &amp;amp;=&amp;amp; \frac{1}{2}\\&#xD;&#xA;\cos\left({\text{angle between two faces of a regular tetrahedron}}\right) &amp;amp;=&amp;amp; \frac{1}{3}\\&#xD;&#xA;\cos\left({\text{angle between two facets of a regular n-simplex}}\right) &amp;amp;=&amp;amp; \frac{1}{n}&#xD;&#xA;\end{eqnarray}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Who would've suspected, upon first encountering it, that the &quot;$2$&quot; in &quot;$\cos{60^{\circ}}=\frac{1}{2}$&quot; was actually a reference to the dimension of the triangle?)&lt;/p&gt;&#xA;" OwnerUserId="409" LastActivityDate="2010-11-20T16:51:17.207" CommentCount="3" />
  <row Id="11103" PostTypeId="2" ParentId="11056" CreationDate="2010-11-20T17:00:55.627" Score="3" ViewCount="" Body="&lt;p&gt;You will find of interest Fund and Williams, &quot;Quadratic polynomials which have a high density of prime values&quot;, Mathematics of Computation, 55:191, July 1990, 345-353.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.jstor.org/pss/2008810&quot; rel=&quot;nofollow&quot;&gt;Jstor.org link&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2937" LastActivityDate="2010-11-20T17:00:55.627" CommentCount="1" />
  <row Id="11104" PostTypeId="1" AcceptedAnswerId="11105" CreationDate="2010-11-20T17:04:40.953" Score="5" ViewCount="198" Body="&lt;p&gt;Hi,&#xA;Is $L^2(\mathbb{R})$ a Banach algebra, with convolution?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am pretty sure the answer is no, because I think that&#xA;$f,g \in L^2(\mathbb{R})$ does not imply that $f*g \in L^2(\mathbb{R})$. However, I can't seem to find a counterexample, or a proof that this is not the case, using Fourier Transform for example.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can someone give a hint?&#xA;Thank you&lt;/p&gt;&#xA;" OwnerUserId="3752" LastEditorUserId="367" LastEditorDisplayName="" LastEditDate="2010-11-20T17:28:43.983" LastActivityDate="2010-11-21T19:46:11.597" Title="Is $L^2(\mathbb{R})$ with convolution a Banach Algebra?" Tags="&lt;functional-analysis&gt;&lt;fourier-analysis&gt;&lt;convolution&gt;" AnswerCount="3" FavoriteCount="1" />
  <row Id="11105" PostTypeId="2" ParentId="11104" CreationDate="2010-11-20T17:19:33.153" Score="5" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Hint&lt;/strong&gt;.  The Fourier transform preserves norms on $L^2$, i.e. $\|f\|_2 = \|\hat f\|_2$.  Also $\hat{(f * g)} = \hat f  \hat g$.  So it is enough that you find a function which is in $L^2$ but whose square is not in $L^2$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit after the comment.&lt;/strong&gt; The proof of $\hat{(f * g)} = \hat f \hat g$ with $f,\ g \in L^2$ requires extending the Fourier transform to the space of tempered distributions.  But you can actually prove what you want with its $L^1$ version and a contradiction.  Assume that convolution is a Banach algebra, and take $f,\ g  \in L^1 \cap L^2$.  Then by assumption $\|f*g\|_2\leq \|f\|_2\|g\|_2$.  From this it follows by isometry that $\|\hat f \hat g\|_2 \leq \|\hat f\|_2\|\hat g\|_2$.  &lt;strike&gt;But since $L^1\cap L^2$ is dense in $L^2$, this must be true for all $f,\ g \in L^2$, for which you can find counter-examples.&lt;/strike&gt; In particular, taking $g = f$, one gets $\|\hat f ^2\|_2 \leq \|\hat f\|_2^2$ i.e.,  $\|\hat f \|_4 \leq \|\hat f\|_2$.  But a counter example can be produced for this by taking any function $f \in L^1\cap L^2$ and forming a sequence $f_n$ by scaling and normalizing it accordingly so that $ \|\hat f_n\|_2$ stays constant but  $\|\hat f_n\|_4$ blows up. &lt;/p&gt;&#xA;" OwnerUserId="145" LastEditorUserId="145" LastEditorDisplayName="" LastEditDate="2010-11-21T19:46:11.597" LastActivityDate="2010-11-21T19:46:11.597" CommentCount="4" />
  <row Id="11106" PostTypeId="2" ParentId="11028" CreationDate="2010-11-20T17:20:07.770" Score="0" ViewCount="" Body="&lt;p&gt;I'm not really convinced by Willie Wong's argument, although I don't doubt that it is correct, but I cannot completely fill in the details (especially the Egoroff part) so I made my own (inspired by his).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First note that the hypothesis that $f \in L^p$ is redundant by Fatou's lemma. Now note that we could as well consider $f_n \to 0$ weakly (the conclusion actually means weak-convergence) by linearity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now note that:&#xA;\begin{align*}&#xA;F_n &amp;amp;: L^q \to \mathbf{K}\\&#xA;g &amp;amp; \mapsto \int f_n g.&#xA;\end{align*}&#xA;defines a bounded operator linear operator on $L^q$ by Hölder with operatornorm smaller or equal to $M$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remember that the stepfunctions are dense in $L^q$. So by linearity and continuity it is sufficient to consider indicatorfunctions $1_A$ with $A$ having finite measure. Now we have a finite measure space so we can use Egoroff.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We want to show that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_A f_n \to 0.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, now find a $B$ with $|B| \leq \epsilon$ such that $f_n$ converges uniformly to $0$ on $A \setminus B$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, for sufficiently large $n$ we have that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_{A \setminus B} |f_n| \leq \epsilon |A \setminus B| \leq \epsilon |A|.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_{B} |f_n| \leq |B|^{1/q} \|f_n\|_p \leq \epsilon^{1/q} \|f_n\|_p.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;KTHXBYE.&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="1120" LastActivityDate="2010-11-20T17:20:07.770" />
  <row Id="11107" PostTypeId="2" ParentId="11104" CreationDate="2010-11-20T17:22:47.470" Score="4" ViewCount="" Body="&lt;p&gt;You can take any $f,g\in L^2(\mathbb R)$ such that their pointwise product $fg$ does not belong to $L^2(\mathbb R)$ and use the fact that the Fourier transform of a product is a convolution (and that the Fourier transform is an isometry of $L^2(\mathbb R)$ onto itself). &lt;/p&gt;&#xA;" OwnerUserId="723" LastActivityDate="2010-11-20T17:22:47.470" />
  <row Id="11108" PostTypeId="1" CreationDate="2010-11-20T17:25:56.213" Score="1" ViewCount="139" Body="&lt;p&gt;$g(x) =\frac{3-5x}{5-x}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that I am retarded for not being able to do this on my own. Algebraically you just have to solve for x. Even so, I can't do it. &lt;/p&gt;&#xA;" OwnerUserId="2230" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-20T21:45:19.563" LastActivityDate="2010-11-20T21:45:19.563" Title="Find the inverse of the following function" Tags="&lt;homework&gt;&lt;algebra-precalculus&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="11109" PostTypeId="2" ParentId="11108" CreationDate="2010-11-20T17:28:03.640" Score="6" ViewCount="" Body="&lt;p&gt;Hint: $y=\frac{3-5x}{5-x}$ so $y(5-x)=3-5x$. Now collect terms involving $x$ on one side and those not involving $x$ on the other and solve for $x$.&lt;/p&gt;&#xA;" OwnerUserId="3431" LastActivityDate="2010-11-20T17:28:03.640" CommentCount="3" />
  <row Id="11110" PostTypeId="2" ParentId="2949" CreationDate="2010-11-20T17:35:35.880" Score="11" ViewCount="" Body="&lt;p&gt;In class of number theory, identities of Ramanujan(continued fractions).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $\alpha, \beta &amp;gt;0$ with $\alpha\beta=\frac{1}{5}$, then:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\left\{ \left(\displaystyle\frac{\sqrt{5}+1}{2} \right)^{5}+ \displaystyle\frac{e^{-\frac{2\pi\alpha}{5}}}{1+\displaystyle\frac{e^{-2\pi\alpha}}{1+\displaystyle\frac{e^{-4\pi\alpha}}{1+...}}}\right\}\cdot\left\{ \left(\displaystyle\frac{\sqrt{5}+1}{2} \right)^{5}+ \displaystyle\frac{e^{-\frac{2\pi\beta}{5}}}{1+\displaystyle\frac{e^{-2\pi\beta}}{1+\displaystyle\frac{e^{-4\pi\beta}}{1+...}}}\right\} = 5\sqrt{5}\left(\displaystyle\frac{\sqrt{5}+1}{2} \right)^{5}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Beautiful result!!!&lt;/p&gt;&#xA;" OwnerUserId="3421" LastActivityDate="2010-11-20T17:35:35.880" CommunityOwnedDate="2010-11-20T17:35:35.880" CommentCount="2" />
  <row Id="11111" PostTypeId="2" ParentId="11108" CreationDate="2010-11-20T17:44:46.147" Score="3" ViewCount="" Body="&lt;p&gt;Maybe you could start with a less complicated example and build up from there. Why don't you try finding the inverse function of $f(x) = x + 3$ using the idea that Timothy gave you of putting $y = x + 3$ and then solving for $x$. Then you can try $f(x) = 7x - 2$ for instance. Now you could also try $f(x) = \frac{3}{x}$ for example. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Something that is sometimes very useful is to consider &quot;easier&quot; problems related to your original problem and trying to solve those.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe the point is to gain some experience on what the particular difficulty may be, and that the &quot;harder&quot; problem may combine some of the difficulties that you will have treated in the particular, less complicated examples that you considered. This advice is treated by Polya in his excellent book about problem solving.&lt;/p&gt;&#xA;" OwnerUserId="900" LastActivityDate="2010-11-20T17:44:46.147" CommentCount="1" />
  <row Id="11112" PostTypeId="2" ParentId="11066" CreationDate="2010-11-20T17:48:37.673" Score="2" ViewCount="" Body="&lt;p&gt;For the 4x4 case there are only two drawn configurations-a checkerboard.  As there are 12870 evenly filled boards, there are 12868 winning ones.  This is not the 144 you calculated.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-11-20T17:48:37.673" />
  <row Id="11113" PostTypeId="1" AcceptedAnswerId="11160" CreationDate="2010-11-20T17:55:08.553" Score="5" ViewCount="201" Body="&lt;p&gt;Suppose $R=k[[x_1,...,x_n]]$ is a formal power series ring over the field $k$, what can we say about the structure of $R/p$ if $p$ is a prime ideal of $R$ such that dim$(R/p)=1$. In particular, are these subrings of power series rings in one variable?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Motivation: Given a subring of a power series ring in one variable which can be written as, $k[[x^{a_1},...,x^{a_n}]]$ where $a_1,...,a_n$ are distinct positive integers, we have a homomorphism $k[[x_1,...,x_n]]\to k[[x^{a_1},...,x^{a_n}]]$ which sends $x_i\to x^{a_i}$. The kernel of this homomorphism must be a prime ideal since $k[[x^{a_1},...,x^{a_n}]]$ is a domain. Moreover, the quotient modulo the kernel must be one dimensional since the image ring is one dimensional. I was wondering if we have a (partial) converse to this result. (I do realize the second paragraph would analogously work with polynomial rings, but it seems power series rings and much nicer (they are regular and local), so I was hoping for a nicer description of the quotient in the power series case).&lt;/p&gt;&#xA;" OwnerUserId="3431" LastEditorUserId="3431" LastEditorDisplayName="" LastEditDate="2010-11-20T22:59:08.533" LastActivityDate="2010-11-22T03:31:02.417" Title="One dimensional quotients of formal power series rings" Tags="&lt;abstract-algebra&gt;&lt;commutative-algebra&gt;&lt;ring-theory&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" />
  <row Id="11114" PostTypeId="2" ParentId="1002" CreationDate="2010-11-20T18:11:51.367" Score="5" ViewCount="" Body="&lt;p&gt;It took me quite a while to understand what exactly is meant by &lt;em&gt;Fourier transform&lt;/em&gt; since it can refer to various algorithms, operations and results. Though I'm quite new in this topic, I'll try to give a short but hopefully intuitive overview on what I came up with (feel free to correct me):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's say you have a function $f(t)$ that maps some time value $t$ to some value $f(t)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we'll try to approximate $f$ as the sum of simple harmonic oscillations, i.e. sine waves of certain frequencies $\omega$. Of course, there are some frequencies that fit well to $f$ and some that approximate it less well. Thus we need some value $\hat{f}(\omega)$ that tells us &lt;em&gt;how much&lt;/em&gt; of a given oscillation with frequency $\omega$ is present in the approximation of $f$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take for example the red function from &lt;a href=&quot;http://de.wikipedia.org/w/index.php?title=Datei%3aSin%28x%29%2B0,13sin%283x%29.svg&amp;amp;filetimestamp=20100814211526&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/d9hST.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is defined as &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f(t) = \sin(t)+0.13\sin(3t)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The green oscillation with $\omega=1$ has the biggest impact on the result, so let's say $$\hat{f}(1)=1$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The blue sine wave ($\omega=3$) has at least some impact, but it's amplitude is much smaller. Thus we say $$\hat{f}(3)=0.13$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other frequencies may not be present in the approximation at all, thus we would write $$\hat{f}(\omega)  = 0$$ for these.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now if we knew $\hat{f}(\omega)$ not only for some but all possible frequencies $\omega$, we could perfectly approximate our function $f$. And that's what the &lt;em&gt;continuous Fourier transform&lt;/em&gt; does.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It takes some function $f(t)$ of time and returns some other function $\hat{f}(\omega) = \mathcal{F}(f)$, it's &lt;em&gt;Fourier transform&lt;/em&gt;, that describes how much of any given &lt;em&gt;frequency&lt;/em&gt; is present in $f$. It's just another representation of $f$, of equal information but with a completely different domain. Often though, problems can be solved much easier in this other representation (which is like finding the appropriate coordinate system).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But given a Fourier transform, we can integrate over all frequencies, put together the weighted sine waves and get our $f$ again, which we call inverse Fourier transform $\mathcal{F}^{-1}$.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Now why should one want to do that?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most importantly, the Fourier transform has many nice mathematical properties (i.e. convolution is just multiplication). It's often much easier to work with the Fourier transforms than with the function itself. So we transform, have an easy job with filtering, transforming and manipulating sine waves and transform back after all. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's say we want to do some noise reduction on a digital image. Rather than manipulating a function $\text{image} : \text{Pixel} \to \text{Brightness}$, we transform the whole thing and work with $\mathcal{F}(\text{image}) : \text{Frequency} \to \text{Amplitude}$. &#xA;Those party of high frequency that cause the noise can simply be cut off - $\mathcal{F}(\text{image})(\omega) = 0, \omega &amp;gt; ...Hz$. We transform back et voilà.&lt;/p&gt;&#xA;" OwnerUserId="917" LastActivityDate="2010-11-20T18:11:51.367" CommunityOwnedDate="2011-04-14T11:45:07.350" />
  <row Id="11115" PostTypeId="2" ParentId="11073" CreationDate="2010-11-20T18:14:46.610" Score="1" ViewCount="" Body="&lt;p&gt;I would say that w.l.o.g. $\theta_1 \leq \theta_3$ and break it up into the cases of $\theta_1 \leq \theta_2 \leq \theta_3$, $\theta_2 \leq \theta_1 \leq \theta_3$, and $\theta_1 \leq \theta_3 \leq \theta_2$. That allows you to use more than just the triangle inequality on the real line.&lt;/p&gt;&#xA;" OwnerUserId="3638" LastActivityDate="2010-11-20T18:14:46.610" CommentCount="1" />
  <row Id="11116" PostTypeId="1" AcceptedAnswerId="11127" CreationDate="2010-11-20T18:17:56.747" Score="0" ViewCount="89" Body="&lt;p&gt;Hello,&lt;br&gt;&#xA;I have a probability space&#xA;$\omega = 2^{\{1,\ldots,n\}}$&#xD;&#xA;$\sigma$-algebra $2^\omega$&#xD;&#xA;and $P(\{s\})=(p^{|s|})*(1-p)^{(n-|s|)}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I assume that $n=2k$,$k$ natural number&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need to find a random variable that will distribute like $\mathrm{Bin}(k,p^2)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you help me with this please?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;br&gt;&#xA;benny&lt;/p&gt;&#xA;" OwnerUserId="3755" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-21T05:10:38.163" LastActivityDate="2010-12-21T05:10:38.163" Title="random variable with binomial distribution" Tags="&lt;homework&gt;&lt;probability-theory&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="11117" PostTypeId="2" ParentId="8113" CreationDate="2010-11-20T18:20:02.643" Score="0" ViewCount="" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;crasic has taken the words out of my mouth.  It's the same as the number of proper subsets of {2, 3, 4, 5, 6}, i.e. 2^5 - 1 = 31.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Only 63% like cheese, so there cannot be more than 63% who like both cheese and apples.  So B is correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;37% don't like cheese, and 27% don't like apples (which I personally find hard to imagine).  The lower bound for x is found in the case where these two sets are disjoint - 64% either don't like cheese or don't like apples, leaving 36%.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="3754" LastActivityDate="2010-11-20T18:20:02.643" />
  <row Id="11118" PostTypeId="1" AcceptedAnswerId="11143" CreationDate="2010-11-20T18:23:20.110" Score="1" ViewCount="125" Body="&lt;p&gt;(a)Is there a linear mapping $f : \mathbb{R}^{3} \rightarrow \mathbb{R}^3$ with&#xA;$ f(1,1,1)=(3,2,7), f(0,2,1)=(2,1,-1), f(2,0,1)=(1,0,0)$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(b)Is there a linear mapping $f :\mathbb{R}^{3} \rightarrow \mathbb{R}^3$ with &#xA;$ f(0,2,-1)=(3,-1,3), f(1,0,2)=(0,-2,1), f(-2,2,-5)=(3,3,1)$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Substantiate your statements and if it exists, provide such a linear mapping.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My attempt so far:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(a) is it first important to determine whether $(1,1,1), (0,2,-1), (2,0,1)$ (called vectors, right??) are a basis for $\mathbb{R}^{3}$? in that case my understanding is a basis is linear independent vectors which span a vectorspace. Since $2(1,1,1)-(0,2,1)=(2,0,1)$ i would like to conclude that these vectors are linearly dependent $\Rightarrow$ do not form a basis for $\mathbb{R}^{3} \Rightarrow$ there is no linear map for these vectors. In case my reasoning is correct, I just wanted to double check is that i could have reached the linear dependence conclusion by turning the vectors into a matrix and seeing if that yielded a row of zeros or a free variable, right? (i happened to just look at it and recognize that this time).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(b)actually after writing my (a) attempt i realize that (b) would also be linearly dependent, correct? i still haven't learned to create a matrix with latex yet, but in case someone is skeptical that it is linearly dependent i could show you my steps. however, first i'd prefer to know if i am even on the right track ;)&#xA;as a note i'm always happy to hear if i am using terminology improperly. thanks in advance for all your help!&lt;/p&gt;&#xA;" OwnerUserId="3711" LastActivityDate="2010-11-21T19:41:15.753" Title="Are there linear mappings for the following vectors, if so, what are they?" Tags="&lt;linear-algebra&gt;" AnswerCount="6" CommentCount="1" FavoriteCount="1" />
  <row Id="11119" PostTypeId="2" ParentId="11095" CreationDate="2010-11-20T18:29:33.043" Score="4" ViewCount="" Body="&lt;p&gt;I especially like this characterization of primes made by Don Zagier&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;There are two facts about the&#xA;  distribution of prime numbers which I&#xA;  hope to convince you so overwhelmingly&#xA;  that they will be permanently engraved&#xA;  in your hearts.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The first is that despite their simple&#xA;  definition and role as the building&#xA;  blocks of the natural numbers, the&#xA;  prime numbers... grow like weeds among&#xA;  the natural numbers, seeming to obey&#xA;  no other law than that of chance, and&#xA;  nobody can predict where the next one&#xA;  will sprout.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The second fact is even more&#xA;  astonishing, for it states just the&#xA;  opposite: that the prime numbers&#xA;  exhibit stunning regularity, that&#xA;  there are laws governing their&#xA;  behaviour, and that they obey these&#xA;  laws with almost military precision.&quot;&#xA;  Don Zagier, Bonn University inaugural lecture&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is taken from &lt;a href=&quot;http://empslocal.ex.ac.uk/people/staff/mrwatkin/isoc/quotes.htm&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; site where you can find many more quotations from prominent mathematicians.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Formula_for_primes&quot; rel=&quot;nofollow&quot;&gt;Here&lt;/a&gt; you can find some formulas for primes.&lt;/p&gt;&#xA;" OwnerUserId="1747" LastEditorUserId="1747" LastEditorDisplayName="" LastEditDate="2010-11-20T19:45:42.617" LastActivityDate="2010-11-20T19:45:42.617" />
  <row Id="11120" PostTypeId="2" ParentId="11118" CreationDate="2010-11-20T18:33:43.137" Score="2" ViewCount="" Body="&lt;p&gt;Hint: Suppose we have a linear map $f:\mathbb{R^3}\to \mathbb{R^3}$. Say, $f_1(x,y,z)=a_1x+b_1y+c_1z$, $f_2(x,y,z)=a_2x+b_2y+c_2z$, $f_3(x,y,z)=a_3x+b_3y+c_3z$ where $f_1,f_2,f_3$ are the $x,y,z$ components of $f$. Substituting the values $(1,1,1),(0,2,1),(2,0,1)$ in this map, we get the corresponding values of the output if such a function exists. Now this gives us $9$ equations in $9$ unknowns, $a_i,b_i,c_i$ with $i=1,2,3$. If these equations form a determined system, we have a unique solution and hence a map. If the system is overdetermined, i.e. one of the equations is a linear combination of the others, then we have several such maps. If the system is inconsistent, i.e. the equations cannot be satisfied simulatenously, then we have no solution and no such map. So it just boils down to solving these equations simultaneously.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: Adrian makes a useful remark, see his comment below.&lt;/p&gt;&#xA;" OwnerUserId="3431" LastEditorUserId="3431" LastEditorDisplayName="" LastEditDate="2010-11-20T19:37:14.197" LastActivityDate="2010-11-20T19:37:14.197" CommentCount="5" />
  <row Id="11121" PostTypeId="2" ParentId="11118" CreationDate="2010-11-20T18:33:55.457" Score="0" ViewCount="" Body="&lt;p&gt;A brute-force approach: try to write using different notation: e.g., say you are searching for a matrix $A$ that satisfies $AX=Y$, where the columns of $X$ are given by the different vectors that are arguments to your $f$, and the columns of $Y$ are the different vectors that you obtain on applying $f$. Then, see if you can &quot;solve&quot; to get $A=XY^{-1}$.&lt;/p&gt;&#xA;" OwnerUserId="1709" LastActivityDate="2010-11-20T18:33:55.457" />
  <row Id="11122" PostTypeId="1" AcceptedAnswerId="11124" CreationDate="2010-11-20T18:35:17.733" Score="1" ViewCount="156" Body="&lt;p&gt;Let $a$ and $b$ be real numbers and $p&amp;gt;0$. What is the best way to prove that $|a+b|^p&amp;lt;2^p(|a|^p+|b|^p)$? &lt;/p&gt;&#xA;" OwnerUserId="1971" LastActivityDate="2011-05-11T20:23:30.737" Title="Yet another inequality" Tags="&lt;real-analysis&gt;&lt;inequality&gt;" AnswerCount="4" FavoriteCount="2" />
  <row Id="11123" PostTypeId="2" ParentId="11122" CreationDate="2010-11-20T18:40:58.060" Score="4" ViewCount="" Body="&lt;p&gt;If $0&amp;lt;p\leq1$ you don't need any powers of $2$.  This came up in another &lt;a href=&quot;http://math.stackexchange.com/questions/10993/please-explain-inequality-xp-yp-leq-x-yp&quot;&gt;recent question&lt;/a&gt;.  If $1\leq p$ you can strengthen the inequality to $|a+b|^p\leq 2^{p-1}(|a|^p+|b|^p)$ by applying convexity of the function $t\mapsto t^p$.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2011-05-11T20:23:30.737" LastActivityDate="2011-05-11T20:23:30.737" />
  <row Id="11124" PostTypeId="2" ParentId="11122" CreationDate="2010-11-20T18:41:24.887" Score="2" ViewCount="" Body="&lt;p&gt;If you divide $2^p$ on both sides, then you get $(|a+b|/2)^p &amp;lt; |a|^p + |b|^p$. Can you see why that would be true? &lt;/p&gt;&#xA;" OwnerUserId="1736" LastActivityDate="2010-11-20T18:41:24.887" CommentCount="1" />
  <row Id="11125" PostTypeId="2" ParentId="11118" CreationDate="2010-11-20T18:42:02.130" Score="1" ViewCount="" Body="&lt;p&gt;If you define $f(x,y,z)=A\cdot \left(&#xD;&#xA;                        \begin{array}{}&#xD;&#xA;                         x \\&#xD;&#xA;                         y \\&#xD;&#xA;                         z \\&#xD;&#xA;                        \end{array}&#xD;&#xA;                      \right)$, where the matrix $A$ is given for  the 9 ecuations as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ A\cdot \left(&#xD;&#xA;                        \begin{array}{}&#xD;&#xA;                         1 \\&#xD;&#xA;                         1 \\&#xD;&#xA;                         1 \\&#xD;&#xA;                        \end{array}&#xD;&#xA;                      \right) = \left(&#xD;&#xA;                        \begin{array}{}&#xD;&#xA;                         3 \\&#xD;&#xA;                         2 \\&#xD;&#xA;                         7 \\&#xD;&#xA;                        \end{array}&#xD;&#xA;                      \right)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can find all coefficients of the matrix $A$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So each relation as the previous we find 3 ecuation, in total 9 ecuations with 9 unknowns, then you problem is solves .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But remember this depends of that ecuations system can be solved !!!!&lt;/p&gt;&#xA;" OwnerUserId="3421" LastEditorUserId="3421" LastEditorDisplayName="" LastEditDate="2010-11-20T18:57:50.550" LastActivityDate="2010-11-20T18:57:50.550" CommentCount="1" />
  <row Id="11126" PostTypeId="2" ParentId="11118" CreationDate="2010-11-20T18:43:32.097" Score="1" ViewCount="" Body="&lt;p&gt;For (a), your reasoning is incomplete. For example consider the problem of finding $f$ when&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f(1,0,0) = (0,0,4)$, $f(2,0,0) = (0,0,8)$ and $f(3,0,0) = (0,0,12)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is clearly a linear function $f$ that satisfies this (infinitely many, in fact), even though the vectors $(1,0,0), (2,0,0), (3,0,0)$ are linearly dependent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you show that the &quot;output&quot; vectors are linearly independent, however, your justification would be good.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Timothy Wagner's answer is slightly incomplete: you have to solve the 3 equations in 3 unknowns problem 3 times: once for each output dimension.&lt;/p&gt;&#xA;" OwnerUserId="3638" LastEditorUserId="3638" LastEditorDisplayName="" LastEditDate="2010-11-20T18:51:58.403" LastActivityDate="2010-11-20T18:51:58.403" />
  <row Id="11127" PostTypeId="2" ParentId="11116" CreationDate="2010-11-20T18:45:42.003" Score="1" ViewCount="" Body="&lt;p&gt;Hint: $2 = n/k$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another hint: $p^2$ has the same exponent $2$ as $n/k$.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2010-11-20T19:29:47.240" LastActivityDate="2010-11-20T19:29:47.240" CommentCount="2" />
  <row Id="11128" PostTypeId="2" ParentId="11118" CreationDate="2010-11-20T18:51:38.593" Score="1" ViewCount="" Body="&lt;p&gt;Given the linear dependence you observed in the input vectors in (a), you could check if the same dependence applies to the output vectors.  As a linear map requires that f(ax+b)=af(x)+f(b), if the output vectors do not satisfy the same dependence there is no map.  This would save you trying to solve the equations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, if you prove the input vectors independent, there is guaranteed to be a linear map.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2010-11-20T20:00:58.130" LastActivityDate="2010-11-20T20:00:58.130" CommentCount="0" />
  <row Id="11129" PostTypeId="1" CreationDate="2010-11-20T18:57:33.530" Score="1" ViewCount="134" Body="&lt;p&gt;The task: &#xA;$\lim_{x\to\infty} \sqrt{x^2+1} -x $&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've multiplied with the conjugate expression ($\sqrt{x^2+1} +x$), then I get this&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\lim_{x\to\infty} \frac{1}{\sqrt{x^2+1} +x} $&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this correct so far? And what would be the next step? &lt;/p&gt;&#xA;" OwnerUserId="2230" LastEditorUserId="2230" LastEditorDisplayName="" LastEditDate="2010-11-20T19:11:25.593" LastActivityDate="2010-11-20T23:38:57.483" Title="Another limit task, I've multiplied by the conjugate, now what?" Tags="&lt;calculus&gt;&lt;homework&gt;" AnswerCount="6" CommentCount="2" FavoriteCount="1" />
  <row Id="11130" PostTypeId="2" ParentId="11129" CreationDate="2010-11-20T18:59:31.047" Score="2" ViewCount="" Body="&lt;p&gt;The multiplication of conjugates is not correct: $(\sqrt{x^2+1}+x)(\sqrt{x^2+1}-x)=(x^2+1)-x^2=1$&lt;/p&gt;&#xA;" OwnerUserId="3431" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2010-11-20T19:02:16.910" LastActivityDate="2010-11-20T19:02:16.910" CommentCount="2" />
  <row Id="11131" PostTypeId="2" ParentId="2949" CreationDate="2010-11-20T19:03:25.863" Score="3" ViewCount="" Body="&lt;p&gt;How about the Anti-Calculus Proposition (Erdös): Suppose f is analytic throughout a closed disc and assumes its maximum modulus at the boundary point a. Then f´(a) is not equal to 0 unless f is constant. (Source Bak/Newman: Complex Analysis 2nd ed.)&lt;/p&gt;&#xA;" OwnerUserId="3757" LastActivityDate="2010-11-20T19:03:25.863" CommunityOwnedDate="2010-11-20T19:03:25.863" />
  <row Id="11132" PostTypeId="2" ParentId="11129" CreationDate="2010-11-20T19:03:26.753" Score="1" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT:&lt;/strong&gt; What you have to do is this : $$ \sqrt{x^{2}+1} -x \times \frac{\sqrt{x^{2}+1} + x}{\sqrt{x^{2}+1} + x} = \frac{1}{\sqrt{x^{2}+1} +x}$$ then proceed to evaluate your limit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now observe that $\lim_{x \to \infty} \frac{1}{x} = 0$. &lt;/p&gt;&#xA;" LastActivityDate="2010-11-20T19:14:22.453" />
  <row Id="11133" PostTypeId="2" ParentId="11129" CreationDate="2010-11-20T19:07:23.080" Score="1" ViewCount="" Body="&lt;p&gt;$$\underset{x\rightarrow \infty }{\lim }\sqrt{x^{2}+1}-x=\underset{%&#xD;&#xA;x\rightarrow \infty }{\lim }\frac{\left( \sqrt{x^{2}+1}-x\right) \left( &#xD;&#xA;\sqrt{x^{2}+1}+x\right) }{\sqrt{x^{2}+1}+x}=\underset{x\rightarrow \infty }{%&#xD;&#xA;\lim }\frac{1}{\sqrt{x^{2}+1}+x}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Added&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\underset{x\rightarrow \infty }{\lim }{\sqrt{x^{2}+1}+x}=\infty$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and (see &lt;a href=&quot;http://en.wikipedia.org/wiki/Limit_of_a_function&quot; rel=&quot;nofollow&quot;&gt;Wikipedia, limits involving infinity and properties&lt;/a&gt;) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\underset{x\rightarrow \infty }{\lim }\frac{1}{\sqrt{x^{2}+1}+x}=\dfrac{\underset{x\rightarrow \infty }{\lim }1}{\underset{x\rightarrow \infty }{\lim }{\sqrt{x^{2}+1}+x}}=\dfrac{1}{\infty}=0$$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="752" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2010-11-20T21:01:06.323" LastActivityDate="2010-11-20T21:01:06.323" CommentCount="5" />
  <row Id="11135" PostTypeId="2" ParentId="11091" CreationDate="2010-11-20T19:15:09.670" Score="4" ViewCount="" Body="&lt;p&gt;Yes there are linear algebra and graph theoretic results for this game, for any arrangement of switches and starting from all off position.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice that you can construct a simple graph out of the arrangement of lights, by taking the lights to be vertices, and making them adjacent in the graph if they are neighbours.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Graph theoretic:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(This can found here: Problem 17 on &lt;a href=&quot;http://books.google.com/books?id=e99fXXYx9zcC&amp;amp;pg=PA42&quot; rel=&quot;nofollow&quot;&gt;http://books.google.com/books?id=e99fXXYx9zcC&amp;amp;pg=PA42&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proposition A:&lt;/strong&gt; Given a simple undirected graph $G(V,E)$, with vertex set $V$ and edge set $E$, we can partition $V$ into two sets $V_1$ and $V_2$ such that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\forall y \in V_1, \ \  |\{ x: \{x,y\} \in E \wedge x \in V_1\}| = 0 \mod 2$ &lt;br/&gt;&#xA;$\forall y \in V_2, \ \  |\{ x: \{x,y\} \in E \wedge x \in V_1\}| = 1 \mod 2$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i.e. any vertex in $V_1$ is incident to an &lt;em&gt;even&lt;/em&gt; number of vertices in $V_1$ and any vertex not in $V_1$ (i.e. in $V_2$) is incident to an &lt;em&gt;odd&lt;/em&gt; number of vertices in $V_1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\circ$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So selecting the vertices in $V_1$ will turn on all the lights, if we start from the all off position.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We tackle a closely related problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We first show that: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proposition B&lt;/strong&gt;: Given a simple undirected graph $G(V,E)$, there is a partition of the vertex set of $G$, $V = V_1 \cup V_2$ such that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\forall y \in V_1, \ \  |\{ x: \{x,y\} \in E \wedge x \in V_1\}| = 0 \mod 2$ &lt;br/&gt;&#xA;$\forall y \in V_2, \ \  |\{ x: \{x,y\} \in E \wedge x \in V_2\}| = 0 \mod 2$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i.e. Any vertex in $V_1$ is adjacent to an &lt;em&gt;even&lt;/em&gt; number of vertices in $V_1$ and any vertex in $V_2$ is adjacent to an &lt;em&gt;even&lt;/em&gt; number of vertices in $V_2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\circ$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The proof of Proposition B is by induction on the number of vertices in $G$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Base cases are easy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose $G$ has $n+1$ vertices. If all the vertices of $G$ have even degree, we are done, by taking $V_2 = \emptyset$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose $o$ is a vertex with odd degree and let $o_1, o_2, \cdots, o_k$ be the neighbours of $o$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Form a graph $G&amp;#39;$ as follows: If $o_i$ and $o_j$ are adjacent in $G$, they are not adjacent in $G&amp;#39;$ and vice versa. Also delete $o$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$G&amp;#39;$ is a graph with $n$ vertices.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Apply the induction hypothesis to $G&amp;#39;$. Say we get a partition $V&amp;#39; = X \cup Y$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now $o$ must be incident to an even number of vertices in one of $X$ or $Y$. Say $X$. Add $o$ to $X$ to get a partition for $G$. We can easily show that this partition of $G$ is what we need.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\square$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now to apply this to our problem (Proposition A):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Add a new vertex $N$ to G and make it adjacent to every vertex of even degree in $G$ to get a new graph $G&amp;#39;$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Apply Proposition B to $G&amp;#39;$. Suppose the partition satisfying proposition B is $V = X \cup Y$. We can easily show that this is also the partition in $G$ (ignoring $N$) for proposition A.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Note:&lt;/strong&gt; This not only proves the existence of a solution, but the induction proof actually gives you an O(n^2) time &lt;strong&gt;algorithm&lt;/strong&gt; to &lt;em&gt;find&lt;/em&gt; the solution.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Linear Algebra Proof due to Noga Alon:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Over $\mathbb{F}_2$, let $A$ by any $n\times n$ symmetric matrix. We can show that if $d$ is the diagonal vector of $A$, then $d$ is in the space spanned by the rows of $A$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This follows from the identity (valid over $\mathbb{F}_2$) that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$v^{T}Av = d^{T}v$, where $u^{T}$ is the transpose of $u$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus if $v$ is in the null space of $A$, then $d$ is orthogonal to $v$ and as a consequence, $d$ is in the row space of $A$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We consider the matrix A to be $B+I$ where $B$ is the adjacency matrix of the corresponding graph we get, from the arrangement of the lights.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-11-29T03:41:55.760" LastActivityDate="2010-11-29T03:41:55.760" CommentCount="14" />
  <row Id="11136" PostTypeId="2" ParentId="11129" CreationDate="2010-11-20T19:15:11.783" Score="3" ViewCount="" Body="&lt;p&gt;Remember that $a^{2}-b^{2}=(a-b)(a+b)$, put $a=\sqrt{x^{2}+1}$ and $b=x$ get:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(\sqrt{x^{2}+1}+x)(\sqrt{x^{2}+1}-x)=(\sqrt{x^{2}+1})^{2}-(x)^{2} = x^{2}+1-x^{2}= 1.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The end...&lt;/p&gt;&#xA;" OwnerUserId="3421" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-20T22:01:56.897" LastActivityDate="2010-11-20T22:01:56.897" />
  <row Id="11137" PostTypeId="2" ParentId="11094" CreationDate="2010-11-20T19:19:14.873" Score="2" ViewCount="" Body="&lt;p&gt;Nate's bound is optimal. Consider $X$ to be a fair coin toss and $Y = 1-X$. Then $EX=EY$ and $P(X-Y \geq 1) = 1/2$.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-20T19:19:14.873" />
  <row Id="11138" PostTypeId="2" ParentId="11094" CreationDate="2010-11-20T19:22:33.233" Score="1" ViewCount="" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;I would like the bound to go to 0 as E[X-Y] goes to 0 as well, and it won't happen with this bound. I should have specified it explicitly when I wrote &quot;reasonable&quot;, perhaps it was not clear. Any ideas about that?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I don't think such a bound is possible. Consider $X,Y$ i.i.d uniform. $E[X-Y] = 0$ yet $P(X-Y \geq \delta) &amp;gt; 0$ for all $\delta &amp;lt; 1$.&lt;/p&gt;&#xA;" OwnerUserId="3638" LastActivityDate="2010-11-20T19:22:33.233" CommentCount="0" />
  <row Id="11139" PostTypeId="2" ParentId="11091" CreationDate="2010-11-20T19:28:44.260" Score="2" ViewCount="" Body="&lt;p&gt;Here is another version of this game. You have a symmetric matrix $A$ over $GF(2)$ (i.e. $0,1$ with all computations done modulo $2$). Then you can show that the diagonal of $A$ is always in the range of $A$, and this is best possible in the sense that the range of $A$ could consist of only the zero vector and the diagonal of $A$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was actually given this &quot;game&quot; as a riddle, so I wouldn't ruin it for you with an answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: &lt;a href=&quot;http://www.cs.toronto.edu/~yuvalf/Range.pdf&quot; rel=&quot;nofollow&quot;&gt;Here&lt;/a&gt;'s my answer, which is probably very similar to Moron's proof. Of course Noga's proof is much neater, but it's not algorithmic (in any other respect it's much better).&lt;/p&gt;&#xA;" OwnerUserId="1277" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2010-11-20T22:43:17.477" LastActivityDate="2010-11-20T22:43:17.477" CommentCount="0" />
  <row Id="11140" PostTypeId="2" ParentId="11129" CreationDate="2010-11-20T19:51:39.353" Score="2" ViewCount="" Body="&lt;p&gt;We have $$\sqrt{1 + \epsilon} = 1 + \frac{\epsilon}{2} + O(\epsilon^2)$$ using the Taylor expansion. So $$\sqrt{x^2+1} = x\sqrt{1+x^{-2}} = x\left(1 + \frac{1}{2x^2} + O(x^{-4})\right) = x + \frac{1}{2x} + O(x^{-3}).$$ Hence $$\sqrt{x^2+1} - x = \frac{1}{2x} + O(x^{-3}) \longrightarrow 0.$$&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-20T19:51:39.353" />
  <row Id="11141" PostTypeId="1" AcceptedAnswerId="11165" CreationDate="2010-11-20T20:14:14.713" Score="2" ViewCount="141" Body="&lt;p&gt;Hello,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a question about solving a system of geometric equations. I really hope someone here can help me, it's been several months since I try to solve the problem but without success. As I am not an expert in maths, I count on your help. Thank you in advance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Problem:&lt;/strong&gt;&lt;br&gt;&#xA;I have 3 cercles C1, C2, C3. For example (&lt;a href=&quot;http://www.scribd.com/doc/43416793?secret_password=tz701s0bwws9k9jqzl4&quot; rel=&quot;nofollow&quot;&gt;refer to this figure&lt;/a&gt;)&lt;br&gt;&#xA;C1: $(x+1)^2+(y−4)^2=9$ &lt;br&gt;&#xA;C2: $(x+4)^2+y^2=25$ &lt;br&gt;&#xA;C3: $(x−2)^2+y^2=16$    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;N2 and N3 are two points:&lt;br&gt;&#xA;N2 is located at the intersection of C1 and C2, and is outside C3.&lt;br&gt;&#xA;N3 is located at the intersection of C1 and C3, and is outside C2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;T1, T2 and T3 are three points (defining the blue triangle in the figure):&lt;br&gt;&#xA;T1 is located at C1 (and inside C2 and C3).&lt;br&gt;&#xA;T2 is located at the intersection of C2 and the line that passes through T1 and N2.&lt;br&gt;&#xA;T3 is located at the intersection of C3 and the line that passes through T1 and N3.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt;&lt;br&gt;&#xA;Given &lt;strong&gt;any&lt;/strong&gt; three circles C1, C2, and C3, is there a point T1 (defined as above) such that distance(T1,T2)=α∗distance(T1,T3) (α is some constant, for example equal to 1) ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried Maple, it gives me the solution, but not the way how to calculate it. Any idea ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit: What I need is a proof of calculability.&#xA;For example, in a system of linear equation with $x$ variables, we know that $x-1$ equations are needed, otherwise there is an infinity of possible solutions. For my system, I know how to prove that if a solution exists, it is necessarily unique. But I don't know an algorithm that can calculate it because the equations are not linear.&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you.&lt;/p&gt;&#xA;" OwnerUserId="3749" LastEditorUserId="3749" LastEditorDisplayName="" LastEditDate="2010-11-20T22:00:10.050" LastActivityDate="2010-11-21T01:43:17.597" Title="A system of geometric equations" Tags="&lt;geometry&gt;&lt;analysis&gt;" AnswerCount="1" CommentCount="12" FavoriteCount="1" />
  <row Id="11142" PostTypeId="2" ParentId="11122" CreationDate="2010-11-20T21:29:45.400" Score="4" ViewCount="" Body="&lt;p&gt;Well,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align*}&#xA;|a + b|^p \leq (|a| + |b|)^p &amp;amp;\leq 2^p \text{max}\{|a|^p, |b|^p\}\\&#xA;&amp;amp;\leq 2^{p - 1} (|a|^p + |b|^p - |a^p - b^p|)\\&#xA;&amp;amp;\leq 2^{p - 1} (|a|^p + |b|^p)&#xA;\end{align*}&lt;/p&gt;&#xA;" OwnerUserId="1120" LastActivityDate="2010-11-20T21:29:45.400" CommentCount="1" />
  <row Id="11143" PostTypeId="2" ParentId="11118" CreationDate="2010-11-20T21:46:58.903" Score="3" ViewCount="" Body="&lt;p&gt;(i) You are sort of on the right track, but you are trying to (1) prove too much; and (2) drawing the wrong conclusion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have a bunch of linearly independent vectors, then you can &lt;em&gt;always&lt;/em&gt; find (at least one) linear transformation that maps them to whatever you want. This because we can always do it with a basis (decide where you want the basis vectors to go, and that gives you a linear transformation), and we can always take any collection of linearly independent vectors and complete it to a basis. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;If&lt;/em&gt; your three vectors were linearly independent (hence a basis for $\mathbb{R}^3$, then you would know that there &lt;em&gt;is&lt;/em&gt; a linear transformation with the desired values, because you can always find a linear transformation that does whatever you want to a basis. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if the three vectors are &lt;em&gt;not&lt;/em&gt; linearly independent, this does not preclude the possibility that there is a linear transformation with the desired properties! For instance, $(1,0,0)$, $(0,1,0)$, and $(1,1,0)$ do not form a basis for $\mathbb{R}^3$, but I can nonetheless find a linear transformation such that $f(1,0,0)=(2,1,-1)$, $f(0,1,0) = (0,1,7)$, and $f(1,1,0)=(2,2,6)$ (in fact, many); for instance, $f(x,y,z) = (2x,x+y,7y-x)$ will do it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem with your argument is that you misusing statement. You know that &quot;If they are a basis, then a linear transformation exists&quot;. You are trying to use this by saying &quot;If they are not a basis, then there is no linear transformation&quot;, but that is false. What you &lt;em&gt;can&lt;/em&gt; conclude is that &quot;if there is no linear transformation, then they are not a basis&quot; (this is called the &quot;contrapositive&quot;). In general, if you know that &quot;If P then Q&quot;, then from &quot;P is false&quot; you cannot conclude &lt;em&gt;anything&lt;/em&gt;. Q may or may not hold. For instance, &quot;If I fall into the pool, I will get wet.&quot; Does that mean that if I don't fall into the pool, I will be dry? No; I could get caught in a rainshower. If I know I didn't fall into the pool, I just don't know whether I will be wet or dry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For this problem, thoguh, you have actually put your finger on exactly what you need to put your finger on: the fact that one of your vectors is a linear combination of the other two. Since $(2,0,1) = 2(1,1,1)-(0,2,1)$, that means that for &lt;strong&gt;any&lt;/strong&gt; linear transformation $T$, you &lt;em&gt;must&lt;/em&gt; have&#xA;$$T(2,0,1) = T\bigl(2(1,1,1)-(0,2,1)\bigr) = 2T(1,1,1) - T(0,2,1).$$&#xD;&#xA;So you would need $f(2,0,1)=(1,0,0)$ to be equal to $$2f(1,1,1)-f(0,2,1)=2(3,2,7)-(2,1,-1) = (6-2, 4-1, 14+1) = (4,3,15).$$&#xD;&#xA;Since $(1,0,0)\neq (4,3,15)$, there can be no linear transformation with the desired properties.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Added:&lt;/em&gt; The key here is that if you have vectors that satisfy some linear dependence, then any linear transformation must &lt;em&gt;also&lt;/em&gt; respect that linear dependence. Since $(2,0,1) = 2(1,1,1)-(0,2,1)$, then &lt;strong&gt;any&lt;/strong&gt; linear transformation $f$ will necessarily satisfy that $f(2,0,1) = 2f(1,1,1)-f(0,2,1)$. Since the values you have in (a) &lt;em&gt;don't&lt;/em&gt; satisfy this, then you know there is no such linear transformation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What if they &lt;em&gt;do&lt;/em&gt; satisfy it, as in (b)? Here you have that $(0,2,-1)=2(1,0,2)+(-2,2,-5)$, so if $f$ is a linear transformation, then you must have&#xA;$$f(0,2,-1) = 2f(1,0,2) + f(-2,2,-5).$$&#xD;&#xA;Here, this &lt;em&gt;does&lt;/em&gt; happen, since &#xA;$$2f(1,0,2) + f(-2,2,-5) = 2(0,-2,1) + (3,3,1) = (3, -1, 3) = f(0,2,-1).$$&#xD;&#xA;So, since they satisfy the equations, does there exist a linear transformation that does this? Yes! In fact, there are &lt;em&gt;infinitely many&lt;/em&gt; different linear transformations that do this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can you find one? Well, this is where you can try doing systems of linear equations. You want to find a linear transformation&#xA;$$f(x,y,z) = (ax+by+cz, \ell x+my+nz, rx+sy+tz)$$&#xD;&#xA;such that if you plug in $x=0$, $y=2$, and $z=-1$, you get $(3,-1,3)$; and if you plug in $x=1$, $y=0$, $z=2$, you get $(0,-2,1)$. This gives you six equations in nine unknowns (the unknowns being $a$, $b$, $c$, $\ell$, $m$, $n$, $r$, $s$, and $t$), and it will have a solution. (The other equation you get when you plug in $x=-2$, $y=2$, and $z=-5$ is actually a &lt;em&gt;consequence&lt;/em&gt; of the six we already had; taht is what we checked to begin with). So you can try solving that system and getting one possible map.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(When the three vectors you have are a basis, you will get nine equations in nine unknowns with a single possible solution, so that will be the linear transformation you want.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But there is a way here to save yourself a lot of work: take your favorite two vectors from this set of three, and then find a fourth vector that, together with the first two, makes up a basis. Then define the linear transformation by picking an arbitrary value for the image of that one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, start with $(0,2,-1)$ and $(1,0,2)$. Notice that $(1,0,0)$ is not a linear combination of these two, so $(0,2,-1)$, $(1,0,2)$, and $(1,0,0)$ makes a basis. Since we can always find a linear transformation that maps $(0,2,-1)$ to $(3,-1,3)$, $(1,0,2)$ to $(0,-2,1)$, and $(1,0,0)$ to anything (say $(0,0,0)$ because it makes things &lt;em&gt;easier&lt;/em&gt;), that one will do. Notice that, if $f(1,0,0) = (0,0,0)$, that tells you that $a=\ell=r=0$, so now we only have three more unknowns. In fact, if we can figure out what $f(0,1,0)$ must be, that will gives us automatically $b$, $m$, and $s$; and $f(0,0,1)$ will give us $c$, $n$, and $t$. Can we do that? Yes: because $(0,2,-1)$, $(1,0,2)$, and $(1,0,0)$ are a basis: we have&#xA;$$(0,1,0) = \frac{1}{2}(0,2,-1) + \frac{1}{4}(1,0,2) -\frac{1}{4}(1,0,0).$$&#xD;&#xA;So&#xA;$$f(0,1,0) = \frac{1}{2}(3,-1,3) + \frac{1}{4}(0,-2,1) - \frac{1}{4}(0,0,0).$$&#xD;&#xA;Then to find the values of $c$, $n$, and $t$, we use the fact that&#xA;$$(0,0,1) = \frac{1}{2}(1,0,2) - \frac{1}{2}(1,0,0).$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This also works when you are trying to find your function $f$ explicitly when the three vectors you have &lt;em&gt;are&lt;/em&gt; a basis: try to write $(1,0,0)$, $(0,1,0)$, and $(0,0,1)$ as linear combinations of the three vectors (this amounts to solving three systems of three linear equations with three unknowns first; or to finding the inverse of a matrix; so it looks like the same amount of work, but because there are a lot of zeros it is often easier to solve these than the general one with $a$, $b,\ldots,t$). And then use the value that $f$ must have at $(1,0,0)$, $(0,1,0)$, and $(0,0,1)$ to figure out the &quot;formula&quot; for $f$. &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-21T19:41:15.753" LastActivityDate="2010-11-21T19:41:15.753" CommentCount="7" />
  <row Id="11144" PostTypeId="2" ParentId="11104" CreationDate="2010-11-20T21:48:52.570" Score="2" ViewCount="" Body="&lt;p&gt;As seen in the other answers $L^2(\mathbb{R})$ is not a Banach algebra. Instead we may consider a &lt;em&gt;weighted&lt;/em&gt; $L^2$-space, let $w$ be a positive continuous function defined on $\mathbb{R}$ and $L^2_w= L^2_w(\mathbb{R})$ be the space of all measurable functions $f$ such that&#xA;$fw\in L^2(\mathbb{R})$ - the norm is defined by&#xA;$$&#xD;&#xA;\|f\|_{L^2_w}=\|fw\|_{L^2}.&#xD;&#xA;$$&#xD;&#xA;It is an exercise to show that $L^2_w$ is a Banach space. If we defined convolution as in the non-weighted space, that is&#xA;$$f*g(x)=\int_\mathbb{R} f(y) g(x-y)dy\qquad\text{(note no weight here)} $$&#xD;&#xA;we will get a Banach algebra if $w^{-2}*w^{-2}\le w^{-2}$. Unfortunately, I do not know if this condition can be improved. See this thread for more details, and a proof of the last statement:  &lt;a href=&quot;http://math.stackexchange.com/questions/2952/when-is-the-weighted-space-ellp-mathbbz-omega-a-banach-algebra-p1&quot;&gt;When is the weighted space $\ell^p(\mathbb{Z},\omega)$ a Banach algebra ($p&gt;1$)?&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1154" LastActivityDate="2010-11-20T21:48:52.570" />
  <row Id="11145" PostTypeId="1" AcceptedAnswerId="11290" CreationDate="2010-11-20T22:03:37.527" Score="13" ViewCount="367" Body="&lt;p&gt;Let $\ell^n_p$ denote $\mathbb{R}^n$ with the $p$-norm; assume $n&amp;gt;1$.  Now it's well-known that for $p\ne q$, $\ell^n_p$ and $\ell^n_q$ are never isometric unless $n=2$ and $\{p,q\}=\{1,\infty\}$, but I realized recently I didn't actually know how to prove this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can prove most of it - the only case I still need to rule out is the case where $n=2$ and $q=p&amp;#39;$ (i.e. $\frac{1}{1-1/p}$).  Now of course, it is indeed possible that $q=p&amp;#39;$, because $p$ could be $2$ (so $p=q$) or $1$ or $\infty$ (the sole exceptional case).  But the problem then is showing that $p$ must actually be one of those three.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore I can show that if there is such an isometry, then, after applying a translation and some signed permutation matrix, it must be $\lambda R$, where $\lambda=2^{1/p-1/2}=2^{1/2-1/p&amp;#39;}$ and $R$ is rotation by $\pi/4$.  So actually all we need to do is, given $p\notin \{1,2,\infty\}$, to find a point $v$ such that $||\lambda R v||_{p&amp;#39;}\ne ||v||_p$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, graphing things, we can see that in fact &lt;em&gt;any&lt;/em&gt; point that is not on the axes or the lines $y=\pm x$ will work; but I do not know how to prove it for even a single specified point, even, say, $(1,2)$, because the resulting equation for $p$ (assuming $p\ne 1,\infty$, of course) is just absolutely terrible, and attempting to, say, determine when each side is larger by differentiating seems to only make things worse.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Addendum Nov 21&lt;/strong&gt;: I should perhaps point out - obviously the point can vary with $p$, but I have no idea how given $p$ one might pick a specific point that makes it easier to prove - especially since it is after all true for nearly any point!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So how is this case handled? And is there by any chance a nicer approach that I'm missing?&lt;/p&gt;&#xA;" OwnerUserId="2884" LastEditorUserId="2884" LastEditorDisplayName="" LastEditDate="2010-11-21T22:30:00.317" LastActivityDate="2010-11-23T10:38:59.083" Title="How to show $\ell^2_p$ not isometric to $\ell^2_{p'}$ unless $p\in\{1,2,\infty\}$?" Tags="&lt;linear-algebra&gt;&lt;analysis&gt;" AnswerCount="3" CommentCount="5" FavoriteCount="9" />
  <row Id="11147" PostTypeId="2" ParentId="11122" CreationDate="2010-11-20T22:19:11.407" Score="1" ViewCount="" Body="&lt;p&gt;Here is an alternative for the case $p\ge1$ we may also use Hölder's inequality&#xA;$$|a+b|=|1\cdot a +1\cdot b|\leq (1^q+ 1^q)^{1/q}(|a|^p+|b|^p)^{1/p}=2^{1/q}(|a|^p+|b|^p)^{1/p}$$&#xD;&#xA;where $1/q+1/p=1$, which reduces to&#xA;$$|a+b|^p\le 2^{p/q}(|a|^p+|b|^p)=2^{p(1-1/p)}(|a|^p+|b|^p)=2^{p-1}(|a|^p+|b|^p).$$&#xD;&#xA;For $0&amp;lt;p&amp;lt;1$, please see my answer in the thread &lt;a href=&quot;http://math.stackexchange.com/questions/10993/please-explain-inequality-xp-yp-leq-x-yp&quot;&gt;&quot;Please explain inequality $|x^p−y^p|\le|x−y|^p$&quot;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1154" LastActivityDate="2010-11-20T22:19:11.407" />
  <row Id="11148" PostTypeId="1" CreationDate="2010-11-20T22:34:39.103" Score="5" ViewCount="146" Body="&lt;p&gt;A &lt;em&gt;diagonal&lt;/em&gt; of a &lt;a href=&quot;http://en.wikipedia.org/wiki/Latin_square&quot; rel=&quot;nofollow&quot;&gt;Latin square&lt;/a&gt; is a selection of n entries in which no two entries occur in the same row or column.  For example: the entries marked with an asterisk below form a diagonal.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;1  2* 3  4&#xA;2  3  4  1*&#xA;3  4  1* 2&#xA;4* 1  2  3&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Theorem&lt;/strong&gt;:  Every Latin square contains a diagonal in which no symbol appears thrice (or more).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The asterisked diagonal in the above example is a diagonal in which no symbol appears twice.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Problem&lt;/strong&gt;: Prove the above theorem.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is quite a fun problem to solve, but there is a trap.&lt;/p&gt;&#xA;" OwnerUserId="139" LastActivityDate="2011-04-15T19:17:24.863" Title="Does every Latin square contain a diagonal in which no symbol appears thrice?" Tags="&lt;combinatorics&gt;&lt;latin-square&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="2" />
  <row Id="11149" PostTypeId="2" ParentId="9286" CreationDate="2010-11-20T22:55:16.273" Score="4" ViewCount="" Body="&lt;p&gt;I know this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Define $f$ and $g$ as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f(x):=\left(\int_{0}^{x}e^{-t^{2}}dt\right)^{2} \ \  \ and  \  \ \  g(x):=\left(\int_{0}^{1}\frac{e^{-x^{2}(t^{2}+1)}}{t^{2}+1}dt\right)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, $$f&amp;#39;(x)=2e^{-x^{2}}\int_{0}^{x}e^{-t^{2}}dt$$ and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$g&amp;#39;(x)=\int_{0}^{1}\frac{\partial}{dx}\left[\frac{e^{-x^{2}(t^{2}+1)}}{t^{2}+1}\right]dt = -2xe^{-x^{2}}\int_{0}^{1}e^{-x^{2}t^{2}}dt$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So putting $t=tx$, get  $\displaystyle\int_{0}^{1}e^{-x^{2}t^{2}}dt= \frac{1}{x}\displaystyle\int_{0}^{x}e^{-t^{2}}dt$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then we get; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$g&amp;#39;(x)=-2e^{-x^{2}}\int_{0}^{x}e^{-t^{2}}dt$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus $f&amp;#39;(x)+g&amp;#39;(x)=0$ for all $x$, then $f(x)+g(x)$ is an constant function. Also $$f(0)+g(0)=\displaystyle\int_{0}^{1}\frac{1}{t^{2}+1}dt = \displaystyle\frac{\pi}{4}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then $f(x)+g(x)=\displaystyle\frac{\pi}{4}$ for all $x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now $\lim_{x \to{+}\infty}{g(x)}=0$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So $$\displaystyle\frac{\pi}{4} = \lim_{x \to{+}\infty}{f(x)+g(x)}=\lim_{x \to{+}\infty}{f(x)}= \left(\int_{0}^{\infty}e^{-t^{2}}dt\right)^{2}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_{0}^{\infty}e^{-t^{2}}dt=\sqrt{\frac{\pi}{4}}= \frac{\sqrt{\pi}}{2}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The end.&lt;/p&gt;&#xA;" OwnerUserId="3421" LastEditorUserId="3421" LastEditorDisplayName="" LastEditDate="2010-11-22T01:21:02.573" LastActivityDate="2010-11-22T01:21:02.573" />
  <row Id="11150" PostTypeId="1" AcceptedAnswerId="11155" CreationDate="2010-11-20T23:02:43.643" Score="20" ViewCount="779" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could someone provide me with good explanation of why $0^0 = 1$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My train of thought:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x &amp;gt; 0$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$0^x = 0^{x-0} = 0^x/0^0$, so&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$0^0 = 0^x/0^x = ?$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Possible answers:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$0^0 * 0^x = 1 * 0^x$, so  $0^0 = 1$&#xD;&lt;/li&gt;&#xA;&lt;li&gt;$0^0 = 0^x/0^x = 0/0 = \text{undefined}$&#xD;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Thank you&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS. I've read the &lt;a href=&quot;http://mathforum.org/dr.math/faq/faq.0.to.0.power.html&quot; rel=&quot;nofollow&quot;&gt;explanation&lt;/a&gt; on mathforum.org, but it isn't clear to me.&lt;/p&gt;&#xA;" OwnerUserId="3761" LastEditorUserId="3761" LastEditorDisplayName="" LastEditDate="2010-11-21T09:37:28.377" LastActivityDate="2010-11-22T11:49:17.377" Title="Zero to zero power" Tags="&lt;discrete-mathematics&gt;" AnswerCount="6" CommentCount="5" FavoriteCount="3" />
  <row Id="11151" PostTypeId="2" ParentId="11150" CreationDate="2010-11-20T23:06:05.717" Score="10" ViewCount="" Body="&lt;p&gt;$0^0$ is undefined. It is an Indeterminate form.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might want to look at this post.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://math.stackexchange.com/questions/10490/why-is-1-infty-considered-to-be-an-indeterminate-form&quot;&gt;Why is $1^{\infty}$ considered to be an indeterminate form&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you said, $0^0$ has many possible interpretations and hence it is an indeterminate form.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle \lim_{x \rightarrow 0^{+}} x^{x} = 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle \lim_{x \rightarrow 0^{+}} 0^{x} = 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle \lim_{x \rightarrow 0^{-}} 0^{x} = $ not defined.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle \lim_{x \rightarrow 0} x^{0} = 1$.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2010-11-22T11:49:17.377" LastActivityDate="2010-11-22T11:49:17.377" CommentCount="2" />
  <row Id="11152" PostTypeId="2" ParentId="11150" CreationDate="2010-11-20T23:08:18.603" Score="4" ViewCount="" Body="&lt;p&gt;Some indeterminates forms $0^{0}, \displaystyle\frac{0}{0}, 1^{\infty}, \infty − \infty, \displaystyle\frac{\infty}{\infty}, 0 &amp;#215; \infty, $ and $\infty^{0}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Futhermore,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\lim_{ x \rightarrow 0+ }x^{0}=1$$ &#xA;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\lim_{ x \rightarrow 0+ }0^{x}=0$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See &lt;a href=&quot;http://en.wikipedia.org/wiki/Indeterminate_form&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Indeterminate_form&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3421" LastEditorUserId="3421" LastEditorDisplayName="" LastEditDate="2010-11-20T23:43:10.867" LastActivityDate="2010-11-20T23:43:10.867" />
  <row Id="11153" PostTypeId="2" ParentId="11150" CreationDate="2010-11-20T23:10:07.073" Score="9" ViewCount="" Body="&lt;p&gt;This is merely a definition, and can't be proved via standard algebra. However, two examples of places where it is convenient to assume this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) The binomial formula: $(x+y)^n=\sum_{k=0}^n {n\choose k}x^ky^{n-k}$. When you set $y=0$ (or $x=0$) you'll get a term of $0^0$ in the sum, which should be equal to 1 for the formula to work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) If $A,B$ are finite sets, then the set of all functions from $B$ to $A$, denoted $A^B$, is of cardinality $|A|^{|B|}$. When both $A$ and $B$ are the empty sets, there is still one function from $B$ to $A$, namely the empty function (a function is a collection of pairs satisfying some conditions; an empty collection is a legal function if the domain $B$ is empty).&lt;/p&gt;&#xA;" OwnerUserId="1818" LastActivityDate="2010-11-20T23:10:07.073" CommentCount="1" />
  <row Id="11155" PostTypeId="2" ParentId="11150" CreationDate="2010-11-20T23:14:47.343" Score="33" ViewCount="" Body="&lt;p&gt;In Calculus and Analysis, there is no good answer as to what $0^0$ &quot;should&quot; be, so it is usually left undefined.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically, if you consider $x^y$ as a function of two variables, then there is no limit as $(x,y)\to(0,0)$ (with $x\geq 0$): if you approach along the line $y=0$, then you get $\lim\limits_{x\to 0^+} x^0 = \lim\limits_{x\to 0^+} 1 = 1$; so perhaps we should define $0^0=1$? Well, the problem is that if you approach along the line $x=0$, then you get $\lim\limits_{y\to 0^+}0^y = \lim\limits_{y\to 0^+} 0 = 0$. So should we define it $0^0=0$? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, if you approach along other curves, you'll get other answers. Since $x^y = e^{y\ln(x)}$, if you approach along the curve $y=\frac{1}{\ln(x)}$, then you'll get a limit of $e$; if you approach along the curve $y=\frac{\ln(7)}{\ln(x)}$, then you get a limit of $7$. And so on. There is just no good answer from the analytic point of view. So, for calculus, algebra, we just don't want to give it any value, we just declare it undefined.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;However&lt;/strong&gt;, from a set-theory point of view, there actually is one and only one sensible answer to what $0^0$ should be! In set theory, $A^B$ is the set of all functions from $B$ to $A$; and when $A$ and $B$ denote &quot;size&quot; (cardinalities), then the &quot;$A^B$&quot; is defined to be the size of the set of all functions from $A$ to $B$. In this context, $0$ is the empty set, so $0^0$ is the collection of all functions from the empty set to the empty set. And, as it turns out, there &lt;em&gt;is&lt;/em&gt; one (and only one) function from the empty set to the empty set: the empty function. So the set $0^0$ has one and only one element, and therefore we &lt;em&gt;must&lt;/em&gt; define $0^0$ as $1$. So if we are talking about &lt;em&gt;cardinal exponentiation&lt;/em&gt;, then the only possible definition is $0^0=1$, and we define it that way, period. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added 2:&lt;/strong&gt; the same holds in Discrete Mathematics, when we are mostly interested in &quot;counting&quot; things. In Dicrete Mathematics, $n^m$ represents the number of ways in which you can make $m$ selections out of $n$ possibilities, when repetitions are allowed and the order matters. (This is &lt;em&gt;really&lt;/em&gt; the same thing as &quot;maps from $\{1,2,\ldots,m\}$ to $\{1,2,\ldots,n\}$&quot; when interpreted appropriately, so it is again the same thing as in set theory). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what should $0^0$ be? It should be the number of ways in which you can make no selections when you have no things to choose from. Well, there is exactly &lt;strong&gt;one&lt;/strong&gt; way of doing that: just sit and do nothing! So we  make $0^0$ equal to $1$, because that is the correct number of ways in which we can do the thing that $0^0$ represents. (This, as opposed to $0^1$, say, where you are required to make $1$ choice with nothing to choose from; in that case, you &lt;em&gt;cannot&lt;/em&gt; do it, so the answer is that $0^1=0$). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your &quot;train of thoughts&quot; don't really work: If $x\neq 0$, then $0^x$ means &quot;the number of ways to make $x$ choices from $0$ possibilities&quot;. This number is $0$. So for &lt;em&gt;any&lt;/em&gt; number $k$, you have $k\cdot 0^x = 0 = 0^x$, hence you cannot say that the equation $0^0\cdot 0^x = 0^x$ suggests that $0^0$ &quot;should&quot; be $1$. The second argument also doesn't work because you cannot divide by $0$, which is what you get with $0^x$ when $x\neq 0$. So it really comes down to what you want $a^b$ to mean, and in discrete mathematics, when $a$ and $b$ are nonnegative integers, it's a count: it's the number of distinct ways in which you can do a certain thing (described above), and that leads necessarily to the definition that makes $0^0$ equal to $1$: because $1$ &#xA;&lt;em&gt;is&lt;/em&gt; the number of ways of making no selections from no choices.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Coda.&lt;/strong&gt; In the end, it is a matter of definition and utility. In Calculus, in analysis, in algebra, there is no reasonable definition (the closest you can come up with is trying to justify it via the binomial theorem, which I personally think is a bit weak), and it is far more useful to leave it undefined or indeterminate, since otherwise it would lead to all sorts of exceptions when dealing with the limit laws. In set theory, in discrete mathematics, etc., the definition $0^0=1$ is both useful and natural, so we define it that way in that context. For other contexts (such as the one mentioned in mathforum, when you are dealing exclusively with analytic functions where the problems with limits do not arise) there may be both natural and useful definitions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We basically define it (or fail to define it) in whichever way it is most useful and natural to do so for the context in question. For Discrete Mathematics, there is no question what that &quot;useful and natural&quot; way &lt;em&gt;should&lt;/em&gt; be, so we define it that way.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-21T20:53:48.097" LastActivityDate="2010-11-21T20:53:48.097" CommentCount="15" />
  <row Id="11156" PostTypeId="1" AcceptedAnswerId="11157" CreationDate="2010-11-20T23:15:25.243" Score="7" ViewCount="153" Body="&lt;p&gt;I'm familiar with the &quot;standard&quot; proof using Galois theory that there is no general formula for solving an equation of fifth (or higher) degree using radicals (i.e. arithmetic and root-taking). However, now I'm wondering if other proofs of different nature were found (in particular ones relying on analysis rather than algebra). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What sparked my interest was seeing a description of the solution of the 2nd, 3rd and 4th degree equations via something that looked like discrete Fourier transform.&lt;/p&gt;&#xA;" OwnerUserId="1818" LastActivityDate="2010-11-20T23:21:33.467" Title="A different proof of the insolubility of the quintic?" Tags="&lt;abstract-algebra&gt;&lt;galois-theory&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" />
  <row Id="11157" PostTypeId="2" ParentId="11156" CreationDate="2010-11-20T23:21:33.467" Score="4" ViewCount="" Body="&lt;p&gt;This is the content of the Abel-Ruffini theorem (whose proof predates Galois')&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Abel%E2%80%93Ruffini_theorem&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Abel%E2%80%93Ruffini_theorem&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A topological proof may be found in this paper.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.tmna.ncu.pl/files/v16n2-02.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.tmna.ncu.pl/files/v16n2-02.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3431" LastActivityDate="2010-11-20T23:21:33.467" />
  <row Id="11159" PostTypeId="2" ParentId="11129" CreationDate="2010-11-20T23:38:57.483" Score="0" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\rm\ \ \ \ $ Put $\rm\ f = x,\ g = 1\ $ in the following simple&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;LEMMA&lt;/strong&gt; $\rm\displaystyle\ \ \ \sqrt{f^2+g}\ -\ f\ \to\ 0 \ \ \ if\ \ \ \frac{f}g \to\ \infty\ \ \ and\ \ \frac{1}g\ \to\ \infty\ \ or\ \ r\in \mathbb R$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof&lt;/strong&gt; $\rm\displaystyle\quad\quad\ \ \sqrt{f^2+g}\ -\ f\ \ =\ \frac{g}{\sqrt{f^2+g}\ +\ f}\ =\ \frac{1}{\sqrt{(\frac{f}g)^2+\frac{1}g}\ +\ \frac{f}g}$&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-11-20T23:38:57.483" />
  <row Id="11160" PostTypeId="2" ParentId="11113" CreationDate="2010-11-20T23:46:12.923" Score="2" ViewCount="" Body="&lt;p&gt;The integral closure of a complete Noetherian local domain is also complete, local and Noetherian. So the integral closure of $R/p$ is a complete regular local ring containing a field (normal rings in dimension $1$ are regular by Serre criterion), and you win! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: add more details, by Tymothy's comments (now deleted): So let $S$ be the integral closure of $R/p$ in its quotient field. Then by definition, $S$ is integrally closed (normal), thus regular because $\dim S=1$. Clearly $R/p$ is a subring of $S$. By Cohen Structure theorem, $S$ is a power series ring with one variable. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fact about integral closure in my first sentence is well-known. Check out  Theorem 4.3.4 of the book available online by Huneke-Swanson, or Eisenbud Chapter 13, or Matsumura somewhere (probably the finiteness of integral closure is true for excellent rings as well). &lt;/p&gt;&#xA;" OwnerUserId="223" LastEditorUserId="223" LastEditorDisplayName="" LastEditDate="2010-11-22T03:31:02.417" LastActivityDate="2010-11-22T03:31:02.417" CommentCount="2" />
  <row Id="11161" PostTypeId="1" AcceptedAnswerId="11164" CreationDate="2010-11-21T01:17:53.913" Score="2" ViewCount="244" Body="&lt;p&gt;Why does X divided by zero not equal X? After all, X is not being divided by anything.&lt;/p&gt;&#xA;" OwnerUserId="2414" LastActivityDate="2010-11-21T08:12:54.487" Title="Why does X divided by zero not equal X?" Tags="&lt;arithmetic&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="11163" PostTypeId="2" ParentId="11161" CreationDate="2010-11-21T01:23:14.290" Score="3" ViewCount="" Body="&lt;p&gt;I would say that if you're dividing by 1, then you're &quot;not really dividing&quot;, which is why $X/1=X$.  Dividing by $0$, however, is undefined.  Here's a way to see this based on basic properties of arithmetic.  Suppose we could define, say $3/0$ as a real number $r$.  Then multiplying both sides of the equation $3/0=r$ by $0$ yields $3=0\cdot r$.  But $0\cdot r=0$ for all real numbers $r$, so this is impossible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Informally, dividing $X$ by $n$ tells you how big to make the pieces of pie if you have to serve them to $n$ people.  This means that $n$ pieces of size $X/n$ give a total amount of $X$.  But if $n=0$, then no matter the size of each piece, $0$ of them can never give a total amount of $X$.  (If $X=0$, then the reasoning becomes more intricate.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(This reminds me of another confusion involving $0$ and language.  I have seen a temptation to replace &quot;the solution is 0&quot; with &quot;there is no solution,&quot; and vice versa among precalculus or beginning calculus students.)&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps this problem of contrasting dividing by $0$ and &quot;not dividing by anything&quot; can be further clarified by comparison with what happens when adding and subtracting.  When you add $0$, you're &quot;not adding anything&quot;, and sure enough $X+0=X$ holds for all $X$.  Similarly for subtraction: $X-0=X$, and here the interpretation &quot;not subtracting anything&quot; works out.  This works because $0$ is a &lt;em&gt;neutral element&lt;/em&gt; with respect to addition (by definition).  Changing operations to multiplication, $0$ is no longer the neutral element; $1$ is.  Multiplying by $1$ is what translates to &quot;don't change anything&quot;: $X\cdot 1=X$.  Thus &quot;undoing&quot; multiplication by $1$, naming dividing by $1$, also doesn't change anything.  But since $0$ no longer has this innocuous behavior with respect to multiplication, things similarly change for division.  It changes drastically, because multiplication by $0$ becomes something that &lt;em&gt;can't be undone&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2010-11-21T08:12:54.487" LastActivityDate="2010-11-21T08:12:54.487" CommentCount="2" />
  <row Id="11164" PostTypeId="2" ParentId="11161" CreationDate="2010-11-21T01:23:39.897" Score="6" ViewCount="" Body="&lt;p&gt;Think of division as an &quot;inverse&quot; operation to multiplication.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, let us look at $\frac{a}{b}$ means.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $x = \frac{a}{b}$. What it means is when $x$ is multiplied by $b$ it gives $a$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, if you want to make sense of $\frac{a}{0}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Say $\frac{a}{0}$ is $y$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, we want $y$ such that $y \times 0$ gives us $a$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But we know that $y \times 0 = 0$, $\forall y$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So there is a logical inconsistency here.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastActivityDate="2010-11-21T01:23:39.897" />
  <row Id="11165" PostTypeId="2" ParentId="11141" CreationDate="2010-11-21T01:43:17.597" Score="4" ViewCount="" Body="&lt;p&gt;Here's an approach that leverages the geometry of the situation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $K_i$ be the center of circle $C_i$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Points $N_2$ and $N_3$ on $C_1$ can be considered &quot;known&quot;. Hence, the measure of the central angle $N_2 K_1 N_3$ (corresponding to the &quot;top&quot; arc $N_2 N_3$ in your diagram) is also known, and easily computable with dot products; call that $2\theta_1$. (Why the &quot;2&quot;? Read on...) By the Inscribed Angle Theorem, the angle $N_2 T_1 N_3$ --&lt;em&gt;no matter where $T_1$ is on &quot;bottom&quot; arc $N_2 N_3$&lt;/em&gt;-- is half the size of the central angle; that is $N_2 T_1 N_3$ has measure $\theta_1$. This must also be the measure of angle $T_2 T_1 T_3$, at the top of your blue triangle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let the blue triangle's angle at $T_2$ have (unknown) measure $\theta_2$. Let $P_2$ be the (other) point at which line $T_2 T_3$ meets circle $C_2$. The angle $N_2 T_2 P_2$ has measure $\theta_2$ (because it's the &quot;same angle&quot; as $T_1 T_2 T_3$). By the Inscribed Angle Theorem again, the &lt;em&gt;central&lt;/em&gt; angle $N_2 K_2 P_2$ (measured along the &quot;outer&quot; arc $N_2 P_2$) must be twice as big: $2\theta_2$. Therefore, if we know the size of $\theta_2$, then we know how far around circle $C_2$ we have to travel from $N_2$ to get to $P_2$.  Likewise, with (unknown) $\theta_3$ the blue triangle's angle at $T_3$ and $P_3$ the (other) point where line $T_2 T_3$ meets $C_3$, we have that angle $N_3 K_3 P_3$ has measure $2\theta_3$, which tells us how to get to $P_3$ from $N_3$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, if we can find formulas for $\theta_2$ and $\theta_3$ in terms of $\theta_1$ and $a$, then we'll effectively know the locations of $P_2$ and $P_3$ relative to $N_2$ and $N_3$. The line joining the $P$s determines the points $T_2$ and $T_3$, which in turn give us the location of $T_1$ (as the intersection of lines $T_2 N_2$ and $T_3 N_3$ with circle $C_1$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's get to it ...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Law of Sines, and your assumption about the ratio of side lengths, tells us that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\sin{\theta_3}}{\sin{\theta_2}}=\frac{|T_1 T_2|}{|T_1 T_3|}=a$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$a \sin{\theta_2} = \sin{\theta_3}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But $\theta_3=\pi-\left(\theta_1+\theta_2\right)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$a\sin{\theta_2} = \sin{\left(\pi-\left(\theta_1+\theta_2\right)\right)}=\sin{\left(\theta_1+\theta_2\right)}=\sin\theta_1 \cos\theta_2+\cos\theta_1 \sin\theta_2$$&#xD;&#xA;$$\sin{\theta_2}\left(a-\cos\theta_1\right)=\sin\theta_1 \cos{\theta_2}$$&#xD;&#xA;$$\sin^2{\theta_2}\left(a-\cos\theta_1\right)^2=\sin^2\theta_1 \cos^2{\theta_2}$$&#xD;&#xA;$$\left( 1 - \cos^2{\theta_2} \right)\left(a- \cos\theta_1\right)^2 = \sin^2\theta_1 \cos^2\theta_2$$&#xD;&#xA;$$\left(a-\cos\theta_1\right)^2 = \cos^2{\theta_2}\left(a^2 - 2 a \cos\theta_1+ \cos^2\theta_1+\sin^2\theta_1\right)= \cos^2{\theta_2}\left(1 +a^2 - 2a \cos\theta_1\right)$$&#xD;&#xA;$$\cos^2{\theta_2}=\frac{\left(a-\cos\theta_1\right)^2}{1 + a^2 - 2 a \cos\theta_1}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{1+\cos\left(2\theta_2\right)}{2}=\frac{\left(a-\cos\theta_1\right)^2}{1 + a^2 - 2 a \cos\theta_1}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\cos\left(2\theta_2\right)=\frac{a^2-1-2a\cos\theta_1+2\cos^2\theta_1}{1 + a^2 - 2 a \cos\theta_1}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, swapping $\theta_2 \leftrightarrow \theta_3$ and $a \leftrightarrow \frac{1}{a}$ ...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\cos\left(2\theta_3\right)=\frac{1-a^2-2a\cos\theta_1+2a^2\cos^2\theta_1}{1 + a^2 - 2 a \cos\theta_1}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, use $N_2$ and $N_3$ and $2\theta_2$ and $2\theta_3$ to locate $P_2$ and $P_3$, then $T_2$ and $T_3$, and finally $T_1$. I doubt the formulas are particularly elegant, but each step in finding them should be relatively straightforward.&lt;/p&gt;&#xA;" OwnerUserId="409" LastActivityDate="2010-11-21T01:43:17.597" CommentCount="1" />
  <row Id="11166" PostTypeId="2" ParentId="9286" CreationDate="2010-11-21T01:59:55.537" Score="6" ViewCount="" Body="&lt;p&gt;It might be worth mentioning that one also can use spherical coordinates in 3-dimensions analogously to the polar coordinates Ross Millikan used above: If $I$ denotes $\int_{-\infty}^{\infty} e^{-x^2}dx$, then we have&#xA;$$I^3 =  \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \int_{-\infty}^{\infty}e^{-x^2 - y^2 - z^2}\,dx\,dy\,dz$$&#xD;&#xA;Switching to spherical coordinates this becomes&#xA;$$\int_0^{2\pi}\int_0^{\pi}\int_0^{\infty}\sin(\phi) \rho^2 e^{-\rho^2}\,d\rho\,d\phi\,d\theta$$&#xD;&#xA;Doing the theta and $\phi$ integrations this becomes &#xA;$$I^3 = 4\pi\int_0^{\infty}\rho^2 e^{-\rho^2}\,d\rho$$&#xD;&#xA;One can then integrate parts in this, differentiating $\rho$ and integrating $\rho e^{-\rho^2}$. This leads us to&#xA;$$I^3 = 2\pi \int_0^{\infty} e^{-\rho^2}\,d\rho$$&#xD;&#xA;Note the right-hand side is exactly $2\pi * {I \over 2} = \pi I$. Thus $I^3 = \pi I$ and thus $I = \sqrt{\pi}$ as needed. Obviously polar coordinates are faster. Just sayin'...&lt;/p&gt;&#xA;" OwnerUserId="3035" LastActivityDate="2010-11-21T01:59:55.537" />
  <row Id="11167" PostTypeId="2" ParentId="11161" CreationDate="2010-11-21T02:08:33.080" Score="8" ViewCount="" Body="&lt;p&gt;I just want to mention that in certain situations it is perfectly valid to think of something nonzero divided by something zero as &quot;infinity.&quot;  For example, the slope of the line between the points $(a, b)$ and $(c, d)$ is $\frac{d - b}{c - a}$, unless $c - a = 0$ and $d - b \neq 0$, where this expression stops making sense.  However, the corresponding line is vertical, and it is perfectly reasonable to think of its slope as being &quot;infinity.&quot;  More precisely, the space of slopes is a gadget called the &lt;a href=&quot;http://en.wikipedia.org/wiki/Projective_line&quot; rel=&quot;nofollow&quot;&gt;projective line&lt;/a&gt; $\mathbb{P}^1(\mathbb{R})$ and if all one cares about is the geometric structure of this space (that is, one is not too concerned with adding and multiplying) it is perfectly valid to think about &quot;infinity&quot; in the same way as any other element, since they're all taken to each other by projective transformations.  These ideas are very important in certain branches of mathematics, such as algebraic geometry.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A more concrete application is that the tangent function can be thought of as taking values in $\mathbb{P}^1(\mathbb{R})$ instead of in $\mathbb{R}$, and then it no longer has &quot;singularities&quot;; in fact, it defines a homeomorphism from the circle to $\mathbb{P}^1(\mathbb{R})$.  Doing the same thing over $\mathbb{C}$ allows you to think of meromorphic functions as functions to the &lt;a href=&quot;http://en.wikipedia.org/wiki/Riemann_sphere&quot; rel=&quot;nofollow&quot;&gt;Riemann sphere&lt;/a&gt; $\mathbb{P}^1(\mathbb{C})$, and this idea also has many fruitful applications.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-11-21T02:08:33.080" CommentCount="1" />
  <row Id="11168" PostTypeId="1" AcceptedAnswerId="11170" CreationDate="2010-11-21T02:24:36.227" Score="0" ViewCount="233" Body="&lt;p&gt;A homework problem states:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An aquarium 5 ft long, 2ft wide, and 3ft deep is full of water. Part C: find the hydrostatic force on one end of the aquarium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Questions (I've completed part a and b. B asked for the force on the bottom which was 1875 lbs.:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;What exactly does one end of the aquarium mean?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Is there a standard to applying long and wide in mathematics problems? It may not matter in calculating area, but in other instances it might. For example, calculating the area of a vertical standing slice would either have to be 2dx or 5dx.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;My solutions manual shows that part C requires setting up an integral with a range from 0 to 3. This means that the &quot;one end&quot; of the aquarium is the top side? How else could the range be 0 to 3?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Here's my drawing of the aquarium:&#xA;&lt;img src=&quot;http://i.imgur.com/IOgTo.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="3514" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-21T03:28:32.623" LastActivityDate="2010-11-21T03:28:32.623" Title="Hydrostatic Force and Integration" Tags="&lt;calculus&gt;&lt;homework&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="11169" PostTypeId="2" ParentId="11028" CreationDate="2010-11-21T02:47:11.463" Score="1" ViewCount="" Body="&lt;p&gt;Here is a proof which is not based on Egoroff's theorem. As Jonas T points out in an other answer, Fatou's Lemma imply&#xA;$$\int|f|^p=\int\lim |f_n|^p =\int\liminf |f_n|^p  \le \liminf\int |f_n|^p \le M^p$$&#xD;&#xA;and hence $f\in L^p$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;At this point we may assume $f=0$ (consider the problem with $f_n$ replaced by $f_n-f$), and we must show $\int f_ng\to 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fix a number $A\gt0$, and let $E_n=\{x: |f_n(x) g(x)|\le A|g(x)|^q\}$ where $q=p/(p-1)$ (the conjugate to $p$). By dominated convergence we have&#xA;$$\int_{E_n}f_ng\,dx\to0$$&#xD;&#xA;On the complement, $E_n^c =X-E_n$, we have $|g|^{q}\le A^{-1}|f_n g|$. Hence, by Hölder's inequality&#xA;$$\int_{E_n^c}|f_ng|\,dx\le\left(\int_{E_n^c} |f_n|^p\right)^{1/p}\left(\int_{E_n^c} |g|^q\right)^{1/q}\le M A^{-1/q}\left(\int_{E_n^c}|f_ng|\,dx\right)^{1/q}$$&#xD;&#xA;In other words&#xA;$$\left(\int_{E_n^c}|f_ng|\,dx\right)^{1-1/q}\le MA^{-1/q}$$&#xD;&#xA;that is &#xA;$$\limsup_n\int_{E_n^c}|f_ng|\,dx\le M^pA^{-p/q}.$$&#xD;&#xA;In total we have shown &#xA;$$\limsup_n\int|f_ng|\,dx\le M^pA^{-p/q}$$&#xD;&#xA;for all $A\gt0$, from which be conclude $$\lim_n\int|f_ng|=0.$$&lt;/p&gt;&#xA;" OwnerUserId="1154" LastActivityDate="2010-11-21T02:47:11.463" CommentCount="1" />
  <row Id="11170" PostTypeId="2" ParentId="11168" CreationDate="2010-11-21T02:50:38.443" Score="1" ViewCount="" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Objects are assumed to be longer than they are wide (unless a front and back are explicitly specified).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;You integrate over depth to get the force onto one side because at each horizontal cros-section the pressure is different (since it's a function of depth).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="3638" LastActivityDate="2010-11-21T02:50:38.443" />
  <row Id="11171" PostTypeId="1" CreationDate="2010-11-21T03:51:54.140" Score="1" ViewCount="87" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am trying to do filter design using MATLAB, in particular there is one step where I form an expression &lt;code&gt;H&lt;/code&gt; and then use &lt;code&gt;[N D] = numden(H)&lt;/code&gt; to get the numerator and denominator for input into a &lt;code&gt;freqz(N, D)&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But it seems like the &lt;code&gt;N&lt;/code&gt; and &lt;code&gt;D&lt;/code&gt; do not reduce to lowest form (divide by common factors). How can I do that in MATLAB?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance!&lt;/p&gt;&#xA;" OwnerUserId="3766" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-21T05:05:31.563" LastActivityDate="2010-11-21T05:05:31.563" Title="MATLAB numden() does not simplify by lowest common factor." Tags="&lt;math-software&gt;&lt;matlab&gt;" CommentCount="2" />
  <row Id="11172" PostTypeId="2" ParentId="8903" CreationDate="2010-11-21T04:07:39.877" Score="1" ViewCount="" Body="&lt;p&gt;Expand RHS, using distributive property and the property that $\langle u,cv\rangle=\bar{c}\langle u,v\rangle$. Simplify and cancel many terms to get the LHS.&lt;/p&gt;&#xA;" OwnerUserId="3249" LastActivityDate="2010-11-21T04:07:39.877" />
  <row Id="11173" PostTypeId="1" AcceptedAnswerId="11174" CreationDate="2010-11-21T04:19:34.987" Score="1" ViewCount="102" Body="&lt;p&gt;Show that the set of the R-homomorphism  $A\rightarrow B$  with factorization through an injective module is an subgroup of $Hom_{R}(A,B)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's a question from  my last test in  abstract algebra, that I can't solve.&lt;/p&gt;&#xA;" OwnerUserId="3421" LastEditorUserId="3421" LastEditorDisplayName="" LastEditDate="2010-11-21T15:51:24.193" LastActivityDate="2010-11-21T15:51:24.193" Title="Help with $Hom_{R}(A,B)$" Tags="&lt;abstract-algebra&gt;&lt;modules&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="11174" PostTypeId="2" ParentId="11173" CreationDate="2010-11-21T04:29:38.400" Score="2" ViewCount="" Body="&lt;p&gt;Hint: This just follows from the fact that a direct sum of two injective modules is injective. &lt;/p&gt;&#xA;" OwnerUserId="3431" LastEditorUserId="3431" LastEditorDisplayName="" LastEditDate="2010-11-21T04:49:30.230" LastActivityDate="2010-11-21T04:49:30.230" CommentCount="1" />
  <row Id="11175" PostTypeId="1" CreationDate="2010-11-21T04:43:59.833" Score="2" ViewCount="156" Body="&lt;p&gt;Let $F \to E \to B$ be a fibration with $B$ simply connected (more generally, such that $\pi_1(B)$ acts trivially on the homology of $F$). Then there is a Serre spectral sequence $H_p(B, H_q(F)) \to H_{p+q}(E)$.  One can do the same for singular cohomology. However, for reasonable spaces (specifically, locally contractible spaces, e.g. CW complexes), singular cohomology is the same as sheaf cohomology of the constant sheaf $\mathbb{Z}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But there is another spectral sequence for sheaf cohomology: the Leray spectral sequence. Given spaces $X, Y$ and $f: X \to Y$, and a sheaf $\mathcal{F}$ on $X$, there is a spectral sequence $H^p(Y, R^q_f(\mathcal{F})) \to H^{p+q}(X, \mathcal{F})$.&#xA;The Wikipedia &lt;a href=&quot;http://en.wikipedia.org/wiki/Leray_spectral_sequence&quot; rel=&quot;nofollow&quot;&gt;article&lt;/a&gt; hints that the topological implications of this include in particular the Serre spectral sequence. I would be interested in this, because I like the machinery of the Grothendieck spectral sequence (from which the Leray spectral sequence easily follows), and would be curious if the Serre spectral sequence could be obtained as a corollary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this possible?&lt;/p&gt;&#xA;" OwnerUserId="536" LastActivityDate="2010-11-21T05:07:16.803" Title="Is the Serre spectral sequence a special case of the Leray spectral sequence?" Tags="&lt;algebraic-topology&gt;" AnswerCount="1" CommentCount="10" FavoriteCount="1" />
  <row Id="11176" PostTypeId="2" ParentId="11175" CreationDate="2010-11-21T05:07:16.803" Score="1" ViewCount="" Body="&lt;p&gt;Yes.  In fact, the result is basically obvious if you use Czech cohomology on the base.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Serre really had two key insights.  First, sheaf cohomology is a pain to compute, but if there is no fundamental group then for fiber bundles the Leray spectral sequence is really just using normal old-fashioned untwisted cohomology.  Second, you don't really need to work with fiber bundles -- all you need are Serre fibrations, and those are easy to construct.  In particular, you have the standard Serre fibration $\Omega X \rightarrow PX \rightarrow X$, where $\Omega X$ is the loop space of $X$ and $PX$ is the space of paths starting at the basepoint of $X$ and the map $PX \rightarrow X$ is &quot;evaluation at the endpoint&quot;.  Clearly $PX$ is contractible!  An amazing amount of milage can be had from this silly observation!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Serre also really developed many of the key algebraic tricks one needs to work with spectral sequences.  For instance, he had the amazing idea that one can work modulo &quot;Serre classes&quot;, and thus ignore things like torsion.  It's like pretending to localize spaces long before Sullivan and Quillen realized you could do so for real!&lt;/p&gt;&#xA;" LastActivityDate="2010-11-21T05:07:16.803" CommentCount="14" />
  <row Id="11177" PostTypeId="1" AcceptedAnswerId="11180" CreationDate="2010-11-21T05:13:35.040" Score="7" ViewCount="297" Body="&lt;p&gt;Hi all,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am looking to expand my knowledge on set theory (which is pretty poor right now -- basic understanding of sets, power sets, and different (infinite) cardinalities). Are there any books that come to your mind that would be useful for an undergrad math student who hasn't taken a set theory course yet? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks a lot for your suggestions!&lt;/p&gt;&#xA;" OwnerUserId="3731" LastEditorUserId="3731" LastEditorDisplayName="" LastEditDate="2010-12-27T06:59:08.917" LastActivityDate="2010-12-27T12:43:26.587" Title="What are good books/other readings for elementary set theory?" Tags="&lt;reference-request&gt;&lt;elementary-set-theory&gt;" AnswerCount="6" CommentCount="2" FavoriteCount="7" />
  <row Id="11178" PostTypeId="2" ParentId="11177" CreationDate="2010-11-21T05:25:09.780" Score="3" ViewCount="" Body="&lt;p&gt;Enderton's book should be a gentle, easy read for an undergraduate&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0122384407&quot; rel=&quot;nofollow&quot;&gt;http://www.amazon.com/Elements-Set-Theory-Herbert-Enderton/dp/0122384407/ref=pd_sim_b_26&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3431" LastActivityDate="2010-11-21T05:25:09.780" />
  <row Id="11179" PostTypeId="2" ParentId="7503" CreationDate="2010-11-21T05:26:28.373" Score="1" ViewCount="" Body="&lt;p&gt;Let $\chi$ be the indicate function of $[-1,1]$. Then the Fourier transform of $\chi$ is $$g(\xi)=\frac{\sin(2\pi\xi)}{\pi\xi}.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Compute the triple convolution $h(x)$ of $\chi$ and evaluate it at 0. That will be 3. From Fourier inverse formula one gets $$h(x)=\int_{-\infty}^\infty (g(\xi))^3e^{2\pi ix\xi}d\xi.$$ Evaluating at $x=0$ yields&#xA;$$ 3=\int_{-\infty}^\infty \frac{\sin^3(2\pi\xi)}{\pi^3\xi^3}d\xi,$$&#xD;&#xA;from which the result follows by substitution.&lt;/p&gt;&#xA;" OwnerUserId="3249" LastActivityDate="2010-11-21T05:26:28.373" />
  <row Id="11180" PostTypeId="2" ParentId="11177" CreationDate="2010-11-21T05:27:37.170" Score="9" ViewCount="" Body="&lt;p&gt;I recommend Naive Set Theory by Halmos.  It's a friendly, thin and fun to read introduction to set theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://books.google.com/books?id=x6cZBQ9qtgoC&amp;amp;printsec=frontcover&amp;amp;source=gbs_ge_summary_r&amp;amp;cad=0#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;http://books.google.com/books?id=x6cZBQ9qtgoC&amp;amp;printsec=frontcover&amp;amp;source=gbs_ge_summary_r&amp;amp;cad=0#v=onepage&amp;amp;q&amp;amp;f=false&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="145" LastActivityDate="2010-11-21T05:27:37.170" />
  <row Id="11181" PostTypeId="2" ParentId="11177" CreationDate="2010-11-21T05:33:15.187" Score="2" ViewCount="" Body="&lt;p&gt;Halmos's book mentioned above is very gentle and easy, and you should look there first.  Afterwards, when I was an undergraduate I remember learning a lot from Set Theory for the Working Mathematician by Krzysztof Ciesielski.  In particular, it has a long chapter showing how transfinite induction can be used to construct all sorts of odd subsets of $\mathbb{R}^n$ and what not.&lt;/p&gt;&#xA;" LastActivityDate="2010-11-21T05:33:15.187" />
  <row Id="11182" PostTypeId="2" ParentId="11066" CreationDate="2010-11-21T05:33:54.157" Score="8" ViewCount="" Body="&lt;p&gt;I believe the answer is 643888.  Here's how I arrived at this number:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Let $C_k$ be the set of $4 \times 4$ $(0,1)$-matrices with $k$ 1's for which there do not exist two neighbouring 1's.&lt;/li&gt;&#xA;&lt;li&gt;Let $B_k$ be the set of $4 \times 4$ $(0,1)$-matrices with $k$ 1's for which there exists two neighbouring 1's, and for which deletion of some 1 implies membership in $C_{k-1}$.&lt;/li&gt;&#xA;&lt;li&gt;For any two (0,1)-matrices, $L$ and $M$, let $\chi(L,M)=1$ if $L$ and $M$ do not have a 1 in the same position and $0$ otherwise.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;We need the sets $B_k$ and $C_k$ to avoid multiple win situations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The number you seek is given by \[\sum_{k=0}^{n^2/2} \Big(\sum_{L \in B_k} \sum_{M \in C_{k-1}} \chi(L,M) + \sum_{L \in C_k} \sum_{M \in B_k} \chi(L,M)\Big).\]  The first term counts when player 1 wins, and the second term counts when player 2 wins.  I implemented this formula in GAP with the code below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This function determines if a (0,1)-matrix has n/2 lines in a horizontal or vertical line.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;user3740Win:=function(L,n)&#xA;  local m,i,j;&#xA;  if(n mod 2=1) then return fail; fi;&#xA;  m:=n/2;&#xA;  for i in [1..n] do for j in [1..m+1] do&#xA;    if(ForAll([0..m-1],k-&amp;gt;L[i][j+k]=1)) then return true; fi;&#xA;    if(ForAll([0..m-1],k-&amp;gt;L[j+k][i]=1)) then return true; fi;&#xA;  od; od;&#xA;  return false;&#xA;end;;&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This function determines if a (0,1)-matrix has n/2 lines in a horizontal or vertical line, and that deletion of one of the 1's results in falsifying user3740Win(L,n).&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;user3740StrongWin:=function(L,n)&#xA;  local M,i,j;&#xA;  if not(user3740Win(L,n)) then return false; fi;&#xA;  M:=List([1..n],i-&amp;gt;List([1..n],j-&amp;gt;L[i][j]));&#xA;  for i in [1..n] do for j in [1..n] do&#xA;    M[i][j]:=0;&#xA;    if(user3740Win(M,n)=false) then return true; fi;&#xA;    M[i][j]:=L[i][j];&#xA;  od; od;&#xA;  return false;&#xA;end;;&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This function generates the (0,1)-matrices with at most n^2/2 1's.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;user3740MatricesStep1:=function(n)&#xA;  local A,B,L,M,i,j;&#xA;  A:=Filtered(Tuples([0,1],n^2),i-&amp;gt;Sum(i)&amp;lt;=n^2/2);&#xA;  B:=[];&#xA;  for L in A do&#xA;    M:=List([1..n],i-&amp;gt;List([1..n],j-&amp;gt;0));&#xA;    for i in [1..n] do for j in [1..n] do&#xA;      M[i][j]:=L[n*(i-1)+j];&#xA;    od; od;&#xA;    Append(B,[M]);&#xA;  od;&#xA;  return B;&#xA;end;;&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Starting from A (the set of (0,1)-matrices with at most n^2/2 1's), I run through k=0..n^2/2.  B is $B_k$ above.  C1 is $C_{k-1}$ above.  C2 is $C_k$ above.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;user3740Count:=function(n)&#xA;  local A,B,C1,C2,S,i,j,k,L,M;&#xA;  A:=user3740MatricesStep1(n);&#xA;  S:=[];&#xA;  for k in [0..n^2/2] do&#xA;    B:=Filtered(A,L-&amp;gt;user3740StrongWin(L,n) and Sum(Sum(L))=k);&#xA;    C1:=Filtered(A,L-&amp;gt;(not user3740Win(L,n)) and Sum(Sum(L))=k-1);&#xA;    C2:=Filtered(A,L-&amp;gt;(not user3740Win(L,n)) and Sum(Sum(L))=k);&#xA;    Print(&quot;k=&quot;,k,&quot; &quot;,Size(B),&quot; &quot;,Size(C1),&quot; &quot;,Size(C2),&quot;\n&quot;);&#xA;    for L in B do for M in C1 do&#xA;      if(not 2 in Union(L+M)) then Append(S,[L+2*M]); fi;&#xA;    od; od;&#xA;    for L in C2 do for M in B do&#xA;      if(not 2 in Union(L+M)) then Append(S,[L+2*M]); fi;&#xA;    od; od;&#xA;    Print(&quot;count=&quot;,Size(S),&quot;\n&quot;);&#xA;  od;&#xA;  return S;&#xA;end;;&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Just so the reader can independently confirm these calculations, here is the output:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;gap&amp;gt; S:=user3740Count(2);;&#xA;k=0 0 0 1&#xA;count=0&#xA;k=1 4 1 0&#xA;count=4&#xA;k=2 0 0 0&#xA;count=4&#xA;gap&amp;gt; S:=user3740Count(4);;&#xA;k=0 0 0 1&#xA;count=0&#xA;k=1 0 1 16&#xA;count=0&#xA;k=2 24 16 96&#xA;count=2072&#xA;k=3 284 96 276&#xA;count=58808&#xA;k=4 1200 276 405&#xA;count=315836&#xA;k=5 2276 405 304&#xA;count=571892&#xA;k=6 2032 304 114&#xA;count=638352&#xA;k=7 848 114 20&#xA;count=643752&#xA;k=8 156 20 2&#xA;count=643888&#xA;gap&amp;gt; Random(S);&#xA;[ [ 0, 0, 0, 0 ], [ 2, 0, 1, 2 ], [ 0, 1, 1, 0 ], [ 1, 2, 0, 0 ] ]&#xA;gap&amp;gt; Random(S);&#xA;[ [ 2, 0, 2, 1 ], [ 1, 1, 0, 2 ], [ 0, 0, 1, 0 ], [ 1, 2, 0, 0 ] ]&#xA;gap&amp;gt; Random(S);&#xA;[ [ 0, 1, 1, 0 ], [ 1, 2, 0, 2 ], [ 0, 0, 0, 1 ], [ 2, 1, 0, 2 ] ]&#xA;gap&amp;gt; Random(S);&#xA;[ [ 0, 2, 1, 0 ], [ 0, 0, 0, 0 ], [ 1, 1, 1, 2 ], [ 2, 0, 0, 0 ] ]&#xA;gap&amp;gt; Random(S);&#xA;[ [ 0, 2, 0, 2 ], [ 1, 1, 2, 0 ], [ 1, 0, 0, 1 ], [ 0, 1, 2, 0 ] ]&#xA;gap&amp;gt; Size(Filtered(S,L-&amp;gt;Sum(Sum(L))=4));&#xA;336&#xA;gap&amp;gt;&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The numbers are k (as in the above formula), |B|, |C1|, |C2|, then &quot;count&quot; is the running sum.&lt;/p&gt;&#xA;" OwnerUserId="139" LastEditorUserId="139" LastEditorDisplayName="" LastEditDate="2010-11-21T05:55:12.600" LastActivityDate="2010-11-21T05:55:12.600" CommentCount="1" />
  <row Id="11183" PostTypeId="2" ParentId="8890" CreationDate="2010-11-21T05:37:04.460" Score="1" ViewCount="" Body="&lt;p&gt;The first inequality follows from expanding $$(|z|-|w|)^2.$$ The second inequality follows from expanding&#xA;$$(z+w)(\bar{z}+\bar{w}).$$&#xD;&#xA;Note that $\bar{z}w+z\bar{w}=2\Re(\bar{z}w)$ and $\Re(c)\le |c|$. Then apply the first part.&lt;/p&gt;&#xA;" OwnerUserId="3249" LastActivityDate="2010-11-21T05:37:04.460" />
  <row Id="11184" PostTypeId="2" ParentId="11177" CreationDate="2010-11-21T05:41:37.247" Score="2" ViewCount="" Body="&lt;p&gt;I recently bought the book &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0821827316&quot; rel=&quot;nofollow&quot;&gt;Basic Set Theory&lt;/a&gt; by A. Shen and N.K. Vereshchagin and it has been a really nice read. It is very accessible and has a lot of exercises. It covers the basics and is very short, about a 100 pages or so. I would recommend it sincerely, although I'm not sure if it will be too basic for what you already know.&lt;/p&gt;&#xA;" OwnerUserId="900" LastActivityDate="2010-11-21T05:41:37.247" CommentCount="2" />
  <row Id="11185" PostTypeId="1" AcceptedAnswerId="11186" CreationDate="2010-11-21T06:11:57.147" Score="0" ViewCount="103" Body="&lt;p&gt;Proof in progress.&lt;/p&gt;&#xA;" OwnerUserId="3740" LastEditorUserId="3740" LastEditorDisplayName="" LastEditDate="2010-11-23T19:46:56.743" LastActivityDate="2010-11-23T19:47:28.787" Title="In a Lattice, x ∧ y ≤ x and x ∧ y ≤ y" Tags="&lt;order-theory&gt;" AnswerCount="2" CommentCount="6" />
  <row Id="11186" PostTypeId="2" ParentId="11185" CreationDate="2010-11-21T06:30:45.377" Score="2" ViewCount="" Body="&lt;p&gt;I do not see why you assert that $x\wedge y$ must be either $x$ or $y$. This is not true in an arbitrary lattice: for instance, in the lattice of subsets of a given set $X$ (with $\leq$ corresponding to inclusin, $\wedge$ corresponding to intersection, and $\vee$ corresponding to union) it is false that $x\cap y$ must be either $x$ or $y$. So you should really be feeling not comfortable about your assertions well before you actually do. But even if you got to that step, you are definitely in trouble: it does &lt;em&gt;not&lt;/em&gt; follow that $y\leq x$ &quot;as $L$ is a partially ordered set&quot;. That assertion has no justification whatsoever.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As to your &quot;proof by contradiction&quot;, a proof by contradiction does not begin by negating your &quot;definition&quot; (as you claim earlier). A proof by contradiction begins by negating the &lt;strong&gt;proposed conclusion&lt;/strong&gt;. If you wanted to do this by contradiction, you would have to being by assuming that your conclusion, &quot;$x\wedge y \leq x$ and $x\wedge y \leq y$&quot; is false. So you would need to assume that &quot;$x\wedge y\not\leq x$ OR $x\wedge y \not\leq y$&quot; is true.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And even if you get to your third line, you have it exactly backwards: it does &lt;em&gt;not&lt;/em&gt; follow that $x\leq x\wedge y$ and $y\leq x\wedge y$. What makes you think that?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So... how do you prove this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You say you are defining the lattince operations in terms of the partial order. That is, you &quot;know&quot; what $\leq$ means, and then you define $\wedge$ and $\vee$ using $\leq$. The definition would be that $x\wedge y$ is the greatest lower bound of $\{x,y\}$ (which must exist if you call the poset a lattice) and $x\vee y$ is the least upper bound of $\{x,y\}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, using that definition, $x\wedge y$ is the greatest lower bound of the set $\{x,y\}$. In particular, it is a lower bound of $\{x,y\}$. What is the definition of &quot;lower bound&quot;? What does that tell you?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is also possible that it is the other way around: you &quot;know&quot; what $\wedge$ and $\vee$ mean, and you define a partial order in terms of them. If you define $\leq$ in terms of the lattice operations, then, explicitly, you &lt;em&gt;define&lt;/em&gt; the partial order associated to the lattice by saying that for $a,b\in L$, &#xA;$$a\leq b \Longleftrightarrow  a\wedge b=a,$$&#xD;&#xA;or else you define it by&#xA;$$a \leq b \Longleftrightarrow a\vee b = b.$$ &#xA;In fact, both definitions are equivalent, as can be shown using the properties of the two operations in a lattice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in this case, what you want to show is that $(x\wedge y)\vee x = x$ and that $(x\wedge y)\vee y = y$. (Or if you defined it using $\wedge$, that $(x\wedge y)\wedge x = x\wedge y$ and that $(x\wedge y)\wedge y = x\wedge y$). These will follow directly from the properties that operations that make up a lattice must satisfy (commutativity, associativity, idempotency, and the two absorption laws). &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-21T07:10:44.983" LastActivityDate="2010-11-21T07:10:44.983" CommentCount="7" />
  <row Id="11187" PostTypeId="2" ParentId="11185" CreationDate="2010-11-21T07:42:57.463" Score="0" ViewCount="" Body="&lt;p&gt;Proof in progress.&lt;/p&gt;&#xA;" OwnerUserId="3740" LastEditorUserId="3740" LastEditorDisplayName="" LastEditDate="2010-11-23T19:47:28.787" LastActivityDate="2010-11-23T19:47:28.787" CommentCount="6" />
  <row Id="11188" PostTypeId="1" CreationDate="2010-11-21T08:08:52.600" Score="0" ViewCount="89" Body="&lt;p&gt;&lt;strong&gt;Added later:&lt;/strong&gt; Proof now in progress.&lt;/p&gt;&#xA;&#xA;&lt;hr/&gt;&#xA;&#xA;&lt;p&gt;Let ( L, ≤ ) be a lattice, x, y, z ∈ L. Prove that $x ≤ y$ and $x ≤ z &amp;lt;=&amp;gt; x ≤ y ∧ z$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I proceed to prove as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The statement can be split into two implications :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(i) $x ≤ y$ and $x ≤ z =&amp;gt; x ≤ y ∧ z$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(ii) $x ≤ y ∧ z =&amp;gt; x ≤ y$ and $x ≤ z$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(i) &lt;em&gt;Approach 1 : Proceeding by use of definitions:&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x ≤ y$ and $x ≤ z =&amp;gt; x$ is a LB in L&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $w = y ∧ z$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since w is the GLB, by defn. x ≤ w, i.e. x ≤ y ∧ z&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Approach 2 : Proof by contradiction:&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $x ≤ y$ and $x ≤ z =&amp;gt; x$ not $≤ y ∧ z$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;x ≤ y and x ≤ z =&gt; x is a LB in L&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $w = y ∧ z$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since w is the GLB, by defn. x ≤ w, i.e. x ≤ y ∧ z&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a contradiction has been reached, the original assertion is true.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(ii) &lt;em&gt;Approach 1 : Proceeding by use of definitions:&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x ≤ y ∧ z =&amp;gt; x$ is a LB in L&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By definition of LB, x ≤ for all w ∈ L, i.e. x ≤ y and x ≤ z&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Approach 2 : Proof by contradiction:&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $x ≤ y ∧ z =&amp;gt;$ not x ≤ y or not x ≤ z&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x ≤ y ∧ z =&amp;gt; x$ is a LB in L&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By definition of LB, x ≤ for all w ∈ L, i.e. x ≤ y and x ≤ z&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a contradiction has been reached, the original assertion is true.&lt;/p&gt;&#xA;" OwnerUserId="3740" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-23T20:31:50.677" LastActivityDate="2010-11-23T21:11:35.103" Title="In a Lattice, x ≤ y and x ≤ z &lt;=&gt; x ≤ y ∧ z" Tags="&lt;order-theory&gt;" AnswerCount="1" CommentCount="7" />
  <row Id="11189" PostTypeId="1" AcceptedAnswerId="11207" CreationDate="2010-11-21T08:42:39.337" Score="1" ViewCount="128" Body="&lt;p&gt;I read that one can prove by AM-GM-inequality that for all $a,b,c\in\mathbb{R}_+$ we have that&#xA;$$11(a^6 + b^6 + c^6) + 40abc(ab^2 + bc^2 + ca^2) \ge 51abc(a^2b + b^2c + c^2a)$$&#xD;&#xA;How this can be done? Is it possible to prove stronger inequalities like&#xA;$$10(a^6 + b^6 + c^6) + 41abc(ab^2 + bc^2 + ca^2) \ge 51abc(a^2b + b^2c + c^2a),$$&#xD;&#xA;or even &#xA;$$a^6 + b^6 + c^6 + 50abc(ab^2 + bc^2 + ca^2) \ge 51abc(a^2b + b^2c + c^2a)$$&#xD;&#xA;without computer?&lt;/p&gt;&#xA;" OwnerUserId="2856" LastActivityDate="2010-11-21T13:27:22.837" Title="Inequality of three variable polynomial" Tags="&lt;polynomials&gt;&lt;inequality&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" />
  <row Id="11190" PostTypeId="1" AcceptedAnswerId="11192" CreationDate="2010-11-21T09:30:23.737" Score="1" ViewCount="184" Body="&lt;p&gt;Let $r \in \mathbb{N}$. Observe the equivalence relation $\sim \subseteq \mathbb{Z} \times \mathbb{Z}$ defined as $x \sim y :\Leftrightarrow (\exists k \in \mathbb{Z})(y=x+kr)$. Show that an operation $&amp;#39;+&amp;#39;$ on $\mathbb{Z}/\sim$  given by $[x]_\sim + [y]_\sim := [x+y]_\sim$ is well-defined.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My attempt at a proof:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $x\sim x&amp;#39;, y\sim y&amp;#39;$, then we know that $(\exists k_1 \in \mathbb{Z})(x=x&amp;#39;+k_1r)$ and $(\exists k_2 \in \mathbb{Z})(y=y&amp;#39;+k_2r)$. Adding $x$ and $y$, we see that $x+y=x&amp;#39;+y&amp;#39;+(k_1+k_2)r$. Because $\mathbb{Z}$ is closed under addition, $k_1+k_2 \in \mathbb{Z}$, and eventually $x+y\sim x&amp;#39;+y&amp;#39;$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, I'm not even sure whether I'm on the right track here. As far as I can see, I've shown that adding two elements of (different) equivalence classes yields an element, whose equivalence class is in $\mathbb{Z}/\sim$. Is that sufficient to show that $&amp;#39;+&amp;#39;$ is well-defined? &lt;/p&gt;&#xA;" OwnerUserId="3546" LastActivityDate="2010-11-21T10:20:22.543" Title="Equivalence relation: showing that an operation is well-defined" Tags="&lt;homework&gt;&lt;relations&gt;" AnswerCount="1" />
  <row Id="11192" PostTypeId="2" ParentId="11190" CreationDate="2010-11-21T10:07:19.903" Score="2" ViewCount="" Body="&lt;p&gt;Yes, you're on the right track.  &quot;Well-defined&quot; just means that it should be an actual operation that takes 2 equivalence classes and sends them to a (uniquely determined) equivalence class.  The way the operation is written makes use of specific representatives of the equivalence classes.  $[x+y]$ makes sense as an element of $\mathbb{Z}/\sim$, but you need to know that it is determined by the classes of $x$ and $y$, rather than depending on the choices of representatives.  To do so, you suppose that another arbitrary choice of representatives is made (your $x&amp;#39;$ and $y'$), and show that the sum $[x&amp;#39;+y&amp;#39;]$ is the same.  That is what your work shows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The language in your second to last sentence isn't entirely accurate.  What you've shown is that adding 2 equivalence classes determines an equivalence class.  The key word is &quot;an&quot; (singular); you've shown that you couldn't get different equivalence classes by &quot;adding&quot; in different ways.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To contrast, suppose $r=3$ and we want to &quot;define&quot; an operation by $[x]^{[y]}=[x^y]$.  If we try to combine $[2]$ and $[1]$ using this &quot;operation&quot;, we get $[2]^{[1]}=[2^1]=[2]$.  But $[1]=[4]$, so we also get $[2]^{[1]}=[2]^{[4]}=[2^4]=[16]=[1]$.  Thus this &quot;operation&quot; is not well-defined.  (Of course, there are other problems, like defining powers of $0$ or negative powers, but the point here is the non-uniqueness, even if it did make sense.)  Better examples of non-well-defined &quot;operations&quot; arise when you study groups.  There you study equivalence classes called cosets of subgroups and try to combine them analogously to how equivalence classes mod $r$ are added in your problem.  You find that unless the subgroup satisfies a certain property called being &quot;normal&quot;, the &quot;operation&quot; is not well-defined.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2010-11-21T10:20:22.543" LastActivityDate="2010-11-21T10:20:22.543" CommentCount="1" />
  <row Id="11193" PostTypeId="1" AcceptedAnswerId="11291" CreationDate="2010-11-21T10:30:35.343" Score="3" ViewCount="175" Body="&lt;p&gt;What is the :&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;most useful&lt;/li&gt;&#xA;&lt;li&gt;prettiest&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;way to visualise extra dimensions in shapes and charts?&lt;/p&gt;&#xA;" OwnerUserId="3746" LastEditorUserId="3746" LastEditorDisplayName="" LastEditDate="2010-11-21T16:47:20.277" LastActivityDate="2010-11-22T02:27:06.600" CommunityOwnedDate="2010-11-21T14:26:01.350" Title="Visualising extra dimensions" Tags="&lt;geometry&gt;" AnswerCount="4" CommentCount="2" FavoriteCount="2" />
  <row Id="11194" PostTypeId="1" CreationDate="2010-11-21T10:35:12.407" Score="1" ViewCount="210" Body="&lt;p&gt;$$T(n)=1+2\sum_{i=1}^{n-1}T(i) , \quad n &amp;gt; 1$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$T(1)=1$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;any hint or how to solve?&lt;/p&gt;&#xA;" OwnerUserId="3741" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-21T11:46:22.160" LastActivityDate="2010-11-21T20:16:20.090" Title="Solving the recurrence relation that contains summation of nth term." Tags="&lt;discrete-mathematics&gt;" AnswerCount="5" CommentCount="3" />
  <row Id="11196" PostTypeId="2" ParentId="11194" CreationDate="2010-11-21T10:45:51.463" Score="3" ViewCount="" Body="&lt;p&gt;Hint: Consider $T(n) - T(n-1)$ in case the $1$ is outside of the sum. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider $\frac{T(n)}{n-1} - \frac{T(n-1)}{n-2}$ if you sum also over the $1$.&lt;/p&gt;&#xA;" OwnerUserId="3330" LastEditorUserId="3330" LastEditorDisplayName="" LastEditDate="2010-11-21T10:58:21.110" LastActivityDate="2010-11-21T10:58:21.110" CommentCount="2" />
  <row Id="11198" PostTypeId="2" ParentId="11073" CreationDate="2010-11-21T11:21:10.870" Score="1" ViewCount="" Body="&lt;p&gt;If the three points $z_1$, $z_2$, $z_3\in S^1$ can be covered by a semicircle $H$ then the restriction $d\restriction H$ acts on the $z_i$ like the usual metric on $\mathbb R$, if not, then the sum of any two distances between the $z_i$  is $&amp;gt;\pi$ and the third distance is $&amp;lt;\pi$.&lt;/p&gt;&#xA;" OwnerUserId="1303" LastEditorUserId="1303" LastEditorDisplayName="" LastEditDate="2010-11-21T13:40:09.720" LastActivityDate="2010-11-21T13:40:09.720" />
  <row Id="11199" PostTypeId="1" CreationDate="2010-11-21T11:21:40.247" Score="2" ViewCount="271" Body="&lt;p&gt;While experimenting with certain sums, I came to the following sum:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$S_n = \sum_{k=0}^n \frac{n-k+1}{(n+1) (n+2) (n-k+2)}.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After rewriting the summand as&#xA;$$\frac{n-k+1}{(n+1) (n+2) (n-k+2)} = \frac{n^{\underline{k}}}{(n+2)^{\underline{k}}(n-k+2)^2}$$&#xD;&#xA;and then feeding the sum to Mathematica, i obtained&#xA;$$S_n = \frac{7}{6(n+1)(n+2)}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But directly summing the first expression above does not yield this closed-form result, because this &quot;result&quot; is wrong! (see edits below) Where is the bug or what am i missing?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt; There seems to be some bug / feature in Mathematica which is leading to the erroneous computation above; the correct answer is of course, as also observed in two answers below. If someone can explain the bug, then he / she should feel free to report it to Wolfram.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(To be specific, the answer above is using $7/6 = (1+2)-H_{1+2}$, instead of $n+2 - H_{n+2}$ as it should)&lt;/p&gt;&#xA;" OwnerUserId="1709" LastEditorUserId="1709" LastEditorDisplayName="" LastEditDate="2010-11-23T18:26:33.307" LastActivityDate="2010-11-23T19:47:54.120" Title="Bug in mathematica: computing the sum of the ratios $(n-k+1)/(n-k+2)$" Tags="&lt;sequences-and-series&gt;" AnswerCount="3" CommentCount="13" FavoriteCount="1" />
  <row Id="11200" PostTypeId="2" ParentId="11199" CreationDate="2010-11-21T11:38:23.267" Score="2" ViewCount="" Body="&lt;p&gt;You can write the sum as&#xA;$$\frac1{(n+1)(n+2)} \sum_{k=0}^{n} \frac{n+1-k}{n+2-k},$$&#xD;&#xA;so it's enough to compute the last part of it. This is just&#xA;$$\frac{n+1}{n+2} + \frac{n}{n+1} + \dots + \frac{n+1-n}{n+2-n} = 1 - \frac1{n+2} + 1 - \frac1{n+1} + \dots + 1 - \frac1{n+2-n}$$&#xD;&#xA;$$= n + 2 - H_{n+2},$$&#xD;&#xA;where $H_n = \sum_{k=1}^n \frac1{k}$ is the $n$:th Harmonic number.&#xA;Your sum is then&#xA;$$\frac{n + 2 - H_{n+2}}{(n+1)(n+2)}.$$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="3776" LastEditorUserId="3776" LastEditorDisplayName="" LastEditDate="2010-11-21T13:17:11.520" LastActivityDate="2010-11-21T13:17:11.520" CommentCount="1" />
  <row Id="11201" PostTypeId="1" AcceptedAnswerId="11202" CreationDate="2010-11-21T12:12:51.010" Score="0" ViewCount="75" Body="&lt;p&gt;I should prove or give a counterexample for:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\partial (\bigcup_{i \in \mathbb{N}} A_i) \subset \bigcup_{i \in \mathbb{N}} (\partial A_i)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $\partial$ is the boundary and (X,d) is a metric space with $A_{i} \subset X $ for each $i \in \mathbb{N}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I cannot make up a counterexample for finite sets $A_{i}$ but I think there might be one for infinite ones, or am I wrong?&lt;/p&gt;&#xA;" OwnerUserId="3123" LastEditorUserId="3123" LastEditorDisplayName="" LastEditDate="2010-11-21T12:25:57.077" LastActivityDate="2010-11-21T12:28:33.090" Title="Boundary in metric spaces" Tags="&lt;topology&gt;&lt;metric-spaces&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="11202" PostTypeId="2" ParentId="11201" CreationDate="2010-11-21T12:27:39.490" Score="1" ViewCount="" Body="&lt;p&gt;you can take the metric space $\mathbb{R}$ and the sets $A_i = \{\frac{1}{i}\}$.&#xA;the boundary of $\bigcup A_i = \{\frac{1}{i} \mid i\in \mathbb{N}\}=A$ is $A\cup \{0\}$ while $\partial A_i = A_i$ so $\bigcup \partial A_i = A $ &lt;/p&gt;&#xA;" OwnerUserId="2125" LastActivityDate="2010-11-21T12:27:39.490" />
  <row Id="11203" PostTypeId="2" ParentId="11201" CreationDate="2010-11-21T12:28:33.090" Score="0" ViewCount="" Body="&lt;p&gt;Let $X = \{ 1/i , i \in \mathbb N \}$ considered as a metric sub-space of $\mathbb R$. Let $A_i = \{ 1/i\}$. The RHS is empty whereas LHS is the set $\{ 0\}$.&lt;/p&gt;&#xA;" OwnerUserId="1119" LastActivityDate="2010-11-21T12:28:33.090" />
  <row Id="11204" PostTypeId="1" CreationDate="2010-11-21T12:33:45.863" Score="0" ViewCount="88" Body="&lt;p&gt;Let $\Omega \subset R^n$ be open, and let us equip $\mathcal E = C^\infty$ with the locally convex topology induced by the following family of semi-norms:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $K \subset \Omega$ and $\alpha, \beta \in \mathbb N _ 0 ^n$, let $|f|_{K,\alpha,\beta} = sup_{x \in K} | \partial_x^{\alpha} f(x) |$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is easy to see that compact sets are closed and bounded within that topology. But how does the opposite direction work?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you!&lt;/p&gt;&#xA;" OwnerUserId="3557" LastActivityDate="2010-11-21T15:19:42.647" Title="Why is $\mathcal E$ a Montel-space?" Tags="&lt;functional-analysis&gt;&lt;general-topology&gt;" AnswerCount="1" CommentCount="0" />
  <row Id="11205" PostTypeId="2" ParentId="11199" CreationDate="2010-11-21T12:36:49.420" Score="2" ViewCount="" Body="&lt;p&gt;Here is the same answer with a variation on the calculations &#xA;$\begin{eqnarray*}&#xD;&#xA;S_{n} &amp;amp;=&amp;amp;\sum_{k=0}^{n}\dfrac{n-k+1}{(n+1)(n+2)(n-k+2)} \\&#xD;&#xA;&amp;amp;=&amp;amp;\dfrac{1}{(n+1)(n+2)}\sum_{k=0}^{n}\dfrac{n-k+1}{n-k+2}\qquad (n+1)(n+2)%&#xD;&#xA;\text{ is independent of }k \\&#xD;&#xA;&amp;amp;=&amp;amp;\dfrac{1}{(n+1)(n+2)}\sum_{m=2}^{n+2}\frac{m-1}{m}\qquad \text{substitution }m=n-k+2 \\&#xD;&#xA;&amp;amp;=&amp;amp;\dfrac{1}{(n+1)(n+2)}\sum_{m=2}^{n+2}\left( 1-\dfrac{1}{m}\right)  \\&#xD;&#xA;&amp;amp;=&amp;amp;\dfrac{1}{(n+1)(n+2)}\left(\sum_{m=2}^{n+2}1\right)-\dfrac{1}{(n+1)(n+2)}\sum_{m=2}^{n+2}%&#xD;&#xA;\dfrac{1}{m} \\&#xD;&#xA;&amp;amp;=&amp;amp;\dfrac{n+2-2+1}{(n+1)(n+2)}-\dfrac{1}{(n+1)(n+2)}\left( \sum_{i=1}^{n+2}%&#xD;&#xA;\dfrac{1}{i}\right) +\dfrac{1}{(n+1)(n+2)} \\&#xD;&#xA;&amp;amp;=&amp;amp;\dfrac{n+1}{(n+1)(n+2)}-\dfrac{H_{n+2}}{(n+1)(n+2)}+\dfrac{1}{(n+1)(n+2)} \\&#xD;&#xA;&amp;amp;=&amp;amp;\dfrac{n+2-H_{n+2}}{(n+1)(n+2)},&#xD;&#xA;\end{eqnarray*}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$H_{n+2}=\displaystyle\sum_{i=1}^{n+2}\dfrac{1}{i}.$&lt;/p&gt;&#xA;" OwnerUserId="752" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2010-11-21T21:58:56.970" LastActivityDate="2010-11-21T21:58:56.970" CommentCount="1" />
  <row Id="11206" PostTypeId="1" AcceptedAnswerId="11254" CreationDate="2010-11-21T12:48:20.977" Score="0" ViewCount="200" Body="&lt;p&gt;How to find center of ellipse from two points(these are just points on the ellipse, not related to foci), and two radiuses (rx and ry, from standard definition of the ellipse x^2/rx^2 + y^2/ry^2=1) of that ellipse? (there will be two centers, I assume)&lt;/p&gt;&#xA;" OwnerUserId="3777" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-22T00:16:41.210" LastActivityDate="2010-11-22T00:16:41.210" Title="Calculating center of the ellipse" Tags="&lt;geometry&gt;&lt;conic-sections&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="11207" PostTypeId="2" ParentId="11189" CreationDate="2010-11-21T13:09:54.293" Score="0" ViewCount="" Body="&lt;p&gt;You want to apply AM-GM to some of the terms on the LHS to get some of the terms on the RHS.  A naive way to do this such as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$b^6 + c^6 + 2a^2 b^3 c \ge 4 a b^3 c^2$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;uses too many of the $11$ terms relative to the $40$ terms and only proves the weaker inequality&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$a^6 + b^6 + c^6 + abc(ab^2 + bc^2 + ca^2) \ge 2abc(a^2 b + b^2 c + c^2 a)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;after cyclic summation.  A slightly better combination of terms is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$5a^6 + c^6 + 12a^2 b^3 c \ge 18 a^3 b^2 c$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but this still uses too many of the $11$ terms relative to the $40$ terms and proves the inequality&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$a^6 + b^6 + c^6 + 2abc(ab^2 + bc^2 + ca^2) \ge 3abc(a^2 b + b^2 c + c^2 a)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;after cyclic summation.  You get a strong enough result by mixing some of the $40$ terms with each other to get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$4a^6 + 12a^2 b^3 c + 4a^3 b c^2 \ge 20 a^3 b^2 c$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which proves&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$a^6 + b^6 + c^6 + 4abc(ab^2 + bc^2 + ca^2) \ge 5abc(a^2 b + b^2 c + c^2 a).$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now multiply by $10$ and apply AM-GM to the remaining terms to conclude.  The question of whether one can do better than this using AM-GM is a geometric question about using convex linear combinations of the vectors $(6, 0, 0), (0, 6, 0), (0, 0, 6), (2, 3, 1), (3, 1, 2), (1, 2, 3)$ to obtain the vectors $(3, 2, 1), (2, 1, 3), (1, 3, 2)$ and one might be able to prove that the above result is optimal.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-11-21T13:09:54.293" />
  <row Id="11208" PostTypeId="1" CreationDate="2010-11-21T13:13:22.837" Score="0" ViewCount="240" Body="&lt;p&gt;By using the exponential formula of $y=ab^t$, we can find doubling time by $2^{t/T}= b^t$; another formula for the exponential is $y=ae^{kt}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems to be correct and also I am quite confused with how can we estimate the doubling time or halving time using graphs. I am quite confused; please help. Show me in both graphical form and using a formula. Giving any example would be fine as I am not quite a mathematician and can hardly make up a proper example, but please help give me the clear detail and idea. &lt;/p&gt;&#xA;" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2010-11-22T23:15:31.377" LastActivityDate="2010-11-22T23:15:31.377" Title="Finding doubling time and halving time using graphs and exponential formula" Tags="&lt;algebra-precalculus&gt;" CommentCount="1" />
  <row Id="11209" PostTypeId="2" ParentId="11189" CreationDate="2010-11-21T13:27:22.837" Score="1" ViewCount="" Body="&lt;p&gt;I don't have enough karma to comment, so I'll just drop this here:&#xA;&lt;a href=&quot;http://www.artofproblemsolving.com/Forum/viewtopic.php?t=154225&quot; rel=&quot;nofollow&quot;&gt;http://www.artofproblemsolving.com/Forum/viewtopic.php?t=154225&lt;/a&gt;&#xA;discusses the problem.&lt;/p&gt;&#xA;" OwnerUserId="3776" LastActivityDate="2010-11-21T13:27:22.837" CommentCount="0" />
  <row Id="11210" PostTypeId="1" AcceptedAnswerId="11228" CreationDate="2010-11-21T13:34:30.273" Score="3" ViewCount="193" Body="&lt;p&gt;Investigating this sequence led me to the following conclusions:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;if $a_{1}\geq 2$ then $a_{n}\geq 2$ and $a_{n}$ is decreasing, thus it converges and is bounded.&lt;/li&gt;&#xA;&lt;li&gt;if $-1\lt a_{1} \lt 2$ then $a_{n}$ is increasing, thus it converges and is bounded.&lt;/li&gt;&#xA;&lt;li&gt;if $a_{1}=-1$ or $a_{1}=2$ the sequence is constant and is bounded.&lt;/li&gt;&#xA;&lt;li&gt;if $a_{1}\lt -1$ then for some $k,~a_{k}\gt 0$ and then we're in one of the previous options.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;But this seems overly complicated and very long to prove.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I thought of trying the following argument: define $f(x)=\frac{4x+2}{x+3}$. We see that $\displaystyle \lim_{x\to \infty}f(x)=4$ and $f(x)$ is continuous in the interval $[0, \infty)$ therefore  it's bounded there. Then we show that if $a_{1}&amp;lt;0$ then $\exists N,~\forall n\gt N,~a_{n}&amp;gt;0$ which gives us $a_{n+1}=f(a_{n})$ is bounded.&lt;/p&gt;&#xA;" OwnerUserId="3237" LastActivityDate="2010-11-22T10:31:59.510" Title="How can I show the sequence $a_{n+1}=\frac{4a_{n}+2}{a_{n}+3}$ is bounded for every $a_{1}\in \mathbb{R}$?" Tags="&lt;calculus&gt;&lt;homework&gt;&lt;sequences-and-series&gt;" AnswerCount="4" CommentCount="4" />
  <row Id="11211" PostTypeId="2" ParentId="11150" CreationDate="2010-11-21T13:54:35.360" Score="5" ViewCount="" Body="&lt;p&gt;$0^{0}$ is just one instance of an &lt;a href=&quot;http://en.wikipedia.org/wiki/Empty_product&quot; rel=&quot;nofollow&quot;&gt;empty product&lt;/a&gt;, which means it is the multiplicative identity 1.&lt;/p&gt;&#xA;" OwnerUserId="2380" LastActivityDate="2010-11-21T13:54:35.360" />
  <row Id="11212" PostTypeId="1" CreationDate="2010-11-21T13:55:55.343" Score="2" ViewCount="234" Body="&lt;p&gt;I am a student taking engineering course and wish to learn more about Fourier Transforms. It seems very useful. Would highly appreciate it if anyone could advise me where to start.&lt;/p&gt;&#xA;" LastActivityDate="2010-11-22T13:28:11.667" Title="Recommended books/links for Fourier Transform beginners?" Tags="&lt;fourier-analysis&gt;" AnswerCount="4" CommentCount="3" FavoriteCount="1" />
  <row Id="11213" PostTypeId="2" ParentId="11212" CreationDate="2010-11-21T14:05:07.953" Score="1" ViewCount="" Body="&lt;p&gt;I first learned Fourier techniques from &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0073039381&quot; rel=&quot;nofollow&quot;&gt;Bracewell's &lt;em&gt;The Fourier Transform and Its Applications&lt;/em&gt;&lt;/a&gt;, but I'm digging the non-conventional treatment in &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/3527404414&quot; rel=&quot;nofollow&quot;&gt;Meikle's &lt;em&gt;A New Twist to Fourier Transforms&lt;/em&gt;&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-11-21T14:05:07.953" CommentCount="5" />
  <row Id="11214" PostTypeId="1" CreationDate="2010-11-21T14:06:30.120" Score="5" ViewCount="203" Body="&lt;p&gt;My consideration might be total nonsense (as a high school student, I lack the mathematical knowledge to really check my idea), but I was just wondering whether one could find a continuous generalization of the Taylor series comparable to the continous Fourier transform.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just like the Fourier transform decomposes a function $f(t)$ into a sum of trigonometric functions with a result $\hat{f}(\omega)=\mathcal{F}(f)(\omega)$ with the domain of &lt;em&gt;frequencies&lt;/em&gt;, could one define a Taylor transform $\mathcal{T}(f)$ that would operate on $n$th derivatives?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instead of the trigonometric functions in Fourier transform, $f$ would be decomposed into a sum of polynomials of form $\frac{1}{n!}x^n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So is it a valid operation to generalize the Taylor series for continuous values of $n$ (i.e. using integrals instead of sums)? My naive approach would be&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f(t) = \intop_0^{\infty} \mathcal{T}(f)(n)\cdot \frac{t^n}{\Gamma(n+1)} \mathrm{d}n$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in fact, the transform would define (or at least require) &lt;strong&gt;non-integer derivatives&lt;/strong&gt; of a function. Do these exist (like a non-integer &lt;a href=&quot;http://en.wikipedia.org/wiki/Function_iteration&quot; rel=&quot;nofollow&quot;&gt;iterates&lt;/a&gt; of functions can be defined too)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example:&#xA;Assuming that $\frac{\mathrm{d}^ne^x}{\mathrm{d}x^n}(0) = 1$ for any $x\in \mathbb{R}$, I'd come up with&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$e^t = \intop_0^{\infty} \frac{t^n}{\Gamma(n+1)} \mathrm{d}n$$ as a continuous version of&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$e^t = \sum_{n=0}^{\infty} \frac{t^n}{n!}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus $$\mathcal{T}(e^t)(n) = 1,\,n\in\mathbb{R}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is a transform like I explained possible or - in case it is - even somehow useful?&lt;/p&gt;&#xA;" OwnerUserId="917" LastActivityDate="2011-03-18T20:44:29.710" Title="Is there a continuous Taylor/MacLaurin transform (like the Fourier transform)?" Tags="&lt;calculus&gt;&lt;analysis&gt;" AnswerCount="3" CommentCount="6" FavoriteCount="1" />
  <row Id="11215" PostTypeId="2" ParentId="11210" CreationDate="2010-11-21T14:07:09.790" Score="0" ViewCount="" Body="&lt;p&gt;What about $a_1 = -3$? The function don't even exists there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, your argument is valid: if $a_n = f(n)$ and $\lim_{x\to \infty} f(x) = L$ then $\lim_{n\to \infty} a_n = L$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You never care about the first terms of the sequence, because when you've convergence to the infinity, you can just pick a maximum value and set it as limit. For example, if your sequence converges to 4, you can set a bound as $\max (4, a_1, a_2, ..., a_k)$ for any k you like.&lt;/p&gt;&#xA;" OwnerUserId="3635" LastActivityDate="2010-11-21T14:07:09.790" CommentCount="1" />
  <row Id="11216" PostTypeId="2" ParentId="11212" CreationDate="2010-11-21T14:11:30.970" Score="1" ViewCount="" Body="&lt;p&gt;For a general engineering perspective, Erwin Kreyszig's  book &quot;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0470458364&quot; rel=&quot;nofollow&quot;&gt;Advanced Engineering Mathematics&lt;/a&gt;&quot; would have some chapters on Fourier and other integral transforms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a more mathematical approach, but still with applications in mind, Sneddon's book &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0486685225&quot; rel=&quot;nofollow&quot;&gt;Fourier Transforms&lt;/a&gt; is recommended. It has a lot of physics applications.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The book of Taub and Schilling on &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0070629552&quot; rel=&quot;nofollow&quot;&gt;Principles of Communication Systems&lt;/a&gt; is very good from an electrical engineering point of view. I particularly liked this last book.&lt;/p&gt;&#xA;" OwnerUserId="1119" LastEditorUserId="1119" LastEditorDisplayName="" LastEditDate="2010-11-21T14:19:20.913" LastActivityDate="2010-11-21T14:19:20.913" />
  <row Id="11217" PostTypeId="1" AcceptedAnswerId="11221" CreationDate="2010-11-21T14:13:32.950" Score="1" ViewCount="150" Body="&lt;p&gt;I am trying to simplify this expression by as usual the expansion way,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\biggl(\frac{ 1+x^2}{1-x^2}\biggr)^2 = \frac{1}{1-y^2}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After some steps I am getting:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$4x^2 - y^2 - 2x^2y^2 - x^4y^2 = 0$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer suggested in my module is $x^2y = 2x - y$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the answer to be correct I think what I should get is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$4x^2 - y^2 - 4xy - x^4y^2 = 0$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What exactly I am doing wrong ? I tried to find an error in my solution, but unable to spot any(yet).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt; For reference I am adding the other options mentioned the question (and now the question too):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;if $4\biggl[\frac{x^2}{1} + \frac{x^{6}}{3}+ \frac{x^{10}}{5} + \cdots \biggr] = y^2 + \frac{y^4}{2} + \frac{y^6}{3} + \cdots $, then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x^2y = 2x+y \text{ or } x = 2y^2 - 1 \text{ or } x^2y = 2x + y^2$$&lt;/p&gt;&#xA;" OwnerUserId="2109" LastEditorUserId="2109" LastEditorDisplayName="" LastEditDate="2010-11-21T14:42:05.480" LastActivityDate="2010-11-21T17:07:56.117" Title="Simplification : $\biggl(\frac{ 1+x^2}{1-x^2}\biggr)^2 = \frac{1}{1-y^2}$" Tags="&lt;algebra-precalculus&gt;&lt;algebra&gt;&lt;logarithms&gt;" AnswerCount="2" CommentCount="5" />
  <row Id="11218" PostTypeId="1" AcceptedAnswerId="11219" CreationDate="2010-11-21T14:19:41.807" Score="0" ViewCount="87" Body="&lt;p&gt;How to prove :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \ln \biggl(\frac{x+1}{x-1}\biggr) = 2\biggl[\frac{1}{x} +  \frac{1}{3x^3} +  \frac{1}{5x^5} + \cdots \biggr]$$ where $|x| \gt 1$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not able to get how to proceed on this one ?&lt;/p&gt;&#xA;" OwnerUserId="2109" LastEditorUserId="2109" LastEditorDisplayName="" LastEditDate="2010-11-21T14:24:48.543" LastActivityDate="2010-11-22T01:19:06.663" Title="A proof based on Logarithmic series:" Tags="&lt;algebra-precalculus&gt;&lt;algebra&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="11219" PostTypeId="2" ParentId="11218" CreationDate="2010-11-21T14:30:39.450" Score="3" ViewCount="" Body="&lt;p&gt;Note that $\frac{x+1}{x-1} = \frac{1 + 1/x }{1 - 1/x}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the series for $\log (1+x)$ converges absolutely when $|x| &amp;lt;1$. Therefore it is permissible to rearrange the series in any way you like.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now note that $\log \left(\frac{1+1/x}{1-1/x}\right) = \log (1+1/x) - \log (1-1/x)$. Expand both logarithmic series and group the terms.&lt;/p&gt;&#xA;" OwnerUserId="1119" LastActivityDate="2010-11-21T14:30:39.450" CommentCount="2" />
  <row Id="11220" PostTypeId="1" AcceptedAnswerId="11225" CreationDate="2010-11-21T14:37:18.007" Score="2" ViewCount="131" Body="&lt;p&gt;Let $f_n,f \in L^p$ and $f_n \to f$ uniformly. Further let $g \in L^q$. Now, clearly $\int (f_n - f) g$ is defined. How do I show that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int |f_n - f||g|$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;can be made arbitrary small if I choose $n$ large enough? If we just naively apply Hölder we will not be able to make it small but just finite. Further, we cannot just replace $|f_n - f|$ by the bound &quot;$\epsilon$&quot; because that is larger than this. Does someone have a hint?&lt;/p&gt;&#xA;" OwnerUserId="1120" LastActivityDate="2010-11-21T15:05:14.313" Title="Uniform convergence and $L^p$-convergence." Tags="&lt;measure-theory&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" />
  <row Id="11221" PostTypeId="2" ParentId="11217" CreationDate="2010-11-21T14:49:28.820" Score="5" ViewCount="" Body="&lt;p&gt;My interpretation is that you want to know the relation between $y$ and $x$&#xD;&#xA;so that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\left( \dfrac{1+x^{2}}{1-x^{2}}\right) ^{2}=\dfrac{1}{1-y^{2}}.$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My detailed computation is as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\dfrac{\left( 1+x^{2}\right) ^{2}}{\left( 1-x^{2}\right) ^{2}}= \dfrac{x^{4}+2x^{2}+1}{x^{4}-2x^{2}+1}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\left( \dfrac{1+x^{2}}{1-x^{2}}\right) ^{2}=\dfrac{1}{1-y^{2}}\Leftrightarrow &#xD;&#xA;\dfrac{x^{4}+2x^{2}+1}{x^{4}-2x^{2}+1}=\dfrac{1}{1-y^{2}}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Leftrightarrow \left( x^{4}+2x^{2}+1\right) \left( 1-y^{2}\right)&#xD;&#xA;=x^{4}-2x^{2}+1$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Expanding&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\left( x^{4}+2x^{2}+1\right) \left( 1-y^{2}\right)&#xD;&#xA;=2x^{2}-y^{2}+x^{4}-2x^{2}y^{2}-x^{4}y^{2}+1$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;you get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$2x^{2}-y^{2}+x^{4}-2x^{2}y^{2}-x^{4}y^{2}+1=x^{4}-2x^{2}+1$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Leftrightarrow 4x^{2}-y^{2}-2x^{2}y^{2}-x^{4}y^{2}=0\qquad\text{the same as in the question}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Leftrightarrow (1+2x^{2}+x^{4})y^{2}=4x^{2}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Leftrightarrow (1+x^{2})^{2}y^{2}=4x^{2}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Leftrightarrow (1+x^{2})y=\pm 2x$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Taking the positive root, we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$y+x^{2}y=2x$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and finally&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x^{2}y=2x-y$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Added: Or&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Leftrightarrow 4x^{2}-y^{2}-2x^{2}y^{2}-x^{4}y^{2}=0\qquad\text{the same as in the question}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Leftrightarrow (1+2x^{2}+x^{4})y^{2}=4x^{2}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Leftrightarrow (1+x^{2})^{2}y^{2}=4x^{2}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Taking the negative root gives&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(1+x^{2})y=-2x$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Leftrightarrow y+x^{2}y=-2x$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and finally&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x^{2}y=-2x-y$&lt;/p&gt;&#xA;" OwnerUserId="752" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2010-11-21T15:30:51.493" LastActivityDate="2010-11-21T15:30:51.493" CommentCount="6" />
  <row Id="11222" PostTypeId="2" ParentId="11214" CreationDate="2010-11-21T14:50:31.750" Score="2" ViewCount="" Body="&lt;p&gt;This is a very interesting question. I am acquainted only very little with these matters. Hopefully other people can fill you in better. There is indeed some analogy of the type you are seeking. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Fourier series of a periodic function captures all the information for a function(signal) over a fixed period. When you determine the Fourier coefficients, you compute suitable integrals over a fixed period and you capture all the frequency information for a certain period. Fourier transforms capture frequency domain information for non-periodic signals as well. If you look at the Fourier transform of a function, you would in general be at a loss to guess the function behavior at a particular point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, note that Taylor Series is giving information about a function specifically at a point. Taylor series captures very very &lt;em&gt;local&lt;/em&gt; information.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems there is such a tradeoff always happening. If you want more information on the frequency domain, you have to sacrifice information on the time domain, and vice versa. On the one extreme we have the Taylor series and the other the Fourier theory. It seems there are a range of transforms called the &lt;a href=&quot;http://en.wikipedia.org/wiki/Wavelet_transform&quot; rel=&quot;nofollow&quot;&gt;wavelet transforms&lt;/a&gt; which are in between these two extremes and captures a bit of both frequency domain and time domain information without completely sacrificing one of these. It seems for such transforms for precisely formulating such a tradeoff, there is a certain &lt;a href=&quot;http://en.wikipedia.org/wiki/Uncertainly_principle&quot; rel=&quot;nofollow&quot;&gt;uncertainty principle&lt;/a&gt; which is a precise theorem in Harmonic analysis.&lt;/p&gt;&#xA;" OwnerUserId="1119" LastActivityDate="2010-11-21T14:50:31.750" CommentCount="1" />
  <row Id="11223" PostTypeId="1" CreationDate="2010-11-21T14:57:55.093" Score="3" ViewCount="286" Body="&lt;p&gt;There are 1000 light bulbs and 1000 tutors. All light bulbs are off. Tutor 1 goes flipping light bulb 1,2,3,4... tutor 2 then flips 2,4,6,8...tutor 3 then 3,6,9...etc until all 1000 tutors have done this.  What is the status of light bulb 25? 93? 576?  Is there a general solution to find out the status of any light bulb?  How many light bulbs are on after all 1000 tutors have gone by?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Not homework but a UK university interview question. Please help.&lt;/p&gt;&#xA;" LastActivityDate="2011-03-27T09:09:49.220" Title="A tedious puzzle (but not homework)" Tags="&lt;sequences-and-series&gt;&lt;algebra&gt;&lt;puzzle&gt;" AnswerCount="6" CommentCount="3" FavoriteCount="1" />
  <row Id="11224" PostTypeId="2" ParentId="11223" CreationDate="2010-11-21T15:02:24.790" Score="8" ViewCount="" Body="&lt;p&gt;Hint: Bulb number $N$ will be flipped by all tutors with number $m$ such that $m$ divides $N$ and no others.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For e.g. light bulb number $36$ will be flipped by tutors numbered $1,2,3,4,6,9,12,18,36$, so $9$ times. Since every bulb was initially off, an odd number of switches will turn it on, which is what will happen with number $36$.&lt;/p&gt;&#xA;" OwnerUserId="3431" LastEditorUserId="3431" LastEditorDisplayName="" LastEditDate="2010-11-21T15:26:29.957" LastActivityDate="2010-11-21T15:26:29.957" />
  <row Id="11225" PostTypeId="2" ParentId="11220" CreationDate="2010-11-21T15:05:14.313" Score="3" ViewCount="" Body="&lt;p&gt;Outside of a sufficiently large compact set, $\|g\|_q &amp;lt;\epsilon$. Use Holder naively, and on that portion the integral is bounded above by $(\|f_n\|_p + \|f\|_p)\epsilon$. On the inside of the compact set $\|g\|_1 \leq C \|g\|_q$ where $C$ is given by the volume of the compact set. Then you can replace $|f_n - f|$ by $\epsilon$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that this requires $\|f_n\|_p$ be uniformly bounded, which would be true if you additionally assume $f_n \to f$ in $L^p$. If you don't assume that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider over $\mathbb{R}$, take $f_n = \frac1n \chi_{[2^n,2^{n+1}]}$, where $\chi$ denotes the characteristic function. $f_n \to 0$ uniformly as a sequence of functions (i.e. $f_n\to f$ in $L^\infty$). For this sequence $\|f_n\|_p = \frac1n \cdot 2^{n/p} \nearrow \infty$ (unless $p = \infty$). (Note that since it is not a bounded sequence, you cannot assert that it has a weakly converging subsequence.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now take $g$ be the function such that $g(x) = 2^{-n/(q-\delta)}$ if $2^n &amp;lt; |x| \leq 2^{n+1}$. By construction it is a function in $L^q$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But $|f_n - f| |g| = \frac1n \cdot 2^{-n / (q - \delta)} \chi_{[2^n,2^{n+1}]}$. If you integrate it, you get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \frac1n \cdot 2^{n( 1 - \frac{1}{q - \delta}) }$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So unless $q = 1$ and $p = \infty$, you can always choose $\delta$ sufficiently small such that the exponent in the integrated expression is positive, and provide a counterexample to what you want proved. (And of course, in the case where $p = \infty$, you can just replace $|f-f_n|$ by $\epsilon$ and be done with it.  &lt;/p&gt;&#xA;" OwnerUserId="1543" LastActivityDate="2010-11-21T15:05:14.313" CommentCount="4" />
  <row Id="11226" PostTypeId="2" ParentId="11204" CreationDate="2010-11-21T15:11:55.077" Score="1" ViewCount="" Body="&lt;p&gt;This is far too long to be a comment, so it's an answer instead.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't think anyone usually gives $C^\infty(\Omega)$ another name. Now I presume you've already shown the space to be &lt;a href=&quot;http://en.wikipedia.org/wiki/Barrelled_space&quot; rel=&quot;nofollow&quot;&gt;barreled&lt;/a&gt;, so we just need $X \subset C^\infty(\Omega)$ compact iff $X$ is closed and bounded.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First note that your sufficient family of seminorms is too large. Because $\Omega$ is $\sigma$-compact, we can thin out the family by taking a countable family of compact sets $K_j$ that exhaust $\Omega$, i.e. $K_j \subsetneq K_{j+1}$ and $\cup_j K_j = \Omega$, so &#xA;$$ \Vert f \Vert_{i,j} = \max_{|\alpha|\leq i} \; \sup_{x \in K_j} |\partial^\alpha f(x)|$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is a countable family of seminorms that generate the same topology. From these we can define a metric on $C^\infty(\Omega)$ as follows&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ d(f,g) = \sum_{i=0}^\infty\; \sup_{j \in \mathbb{N}} \frac{\Vert f-g \Vert_{i,j}}{1 + \Vert f-g \Vert_{i,j}} 2^{-i}.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is easy to verify that this metric generates the same topology as the family of seminorms and that the metric is complete and translation-invariant. Thus $C^\infty(\Omega)$ is a Fréchet space and we know that the compact subsets of complete metric spaces are precisely those that are closed and totally bounded. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the strategy is clear. First show that $X$ bounded w.r.t. the seminorms implies $X$ bounded w.r.t. the metric. Then show that $X$ bounded w.r.t. the metric implies that $X$ is totally bounded.&lt;/p&gt;&#xA;" OwnerUserId="1269" LastEditorUserId="1269" LastEditorDisplayName="" LastEditDate="2010-11-21T15:19:42.647" LastActivityDate="2010-11-21T15:19:42.647" CommentCount="1" />
  <row Id="11227" PostTypeId="1" AcceptedAnswerId="11289" CreationDate="2010-11-21T15:25:38.430" Score="1" ViewCount="158" Body="&lt;p&gt;Let $\mathbb{Z}_p$ be te ring of p-adic integers and let $T$,$T'$ be two free $\mathbb{Z}_p$-module with a continuous action of $G_{\mathbb{Q}_p}$ (the absolute Galois group of $\mathbb{Q}_p$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is not hard to see that if $T$ and $T&amp;#39;$ are of rank one, then there is a non zero $G_{\mathbb{Q}_p}$-equivariant map $T \to T&amp;#39; \otimes \mathbb{Q}_p / \mathbb{Z}_p$ if and only if $T \simeq T&amp;#39;$ mod $p$ (just write down explicitely the characters).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this true in a more general case, namely when $T&amp;#39;$ and $T&amp;#39;$ mod $p$ are irreducible and $T$ is any $\mathbb{Z}_p$-representation ?&lt;/p&gt;&#xA;" OwnerUserId="3781" LastActivityDate="2010-11-22T16:29:04.707" Title="torsion representation" Tags="&lt;number-theory&gt;&lt;representation-theory&gt;" AnswerCount="1" CommentCount="0" />
  <row Id="11228" PostTypeId="2" ParentId="11210" CreationDate="2010-11-21T15:28:01.337" Score="2" ViewCount="" Body="&lt;p&gt;You can formalize what you're doing by showing that $-1 \leq a_{n+1} &amp;lt; 4$ whenever $a_n \geq -1$. So once you get an $n$ such that $a_n \geq -1$ you're done.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If some $a_n &amp;lt; -3$, your recursion gives $a_{n+1} &amp;gt; -1$ and you're in the above situation. The only remaining option is that $-3 &amp;lt; a_n &amp;lt; -1$ for all $n$... but then the sequence is still bounded. So you're done.&lt;/p&gt;&#xA;" OwnerUserId="3035" LastEditorUserId="3035" LastEditorDisplayName="" LastEditDate="2010-11-21T15:54:16.287" LastActivityDate="2010-11-21T15:54:16.287" CommentCount="2" />
  <row Id="11229" PostTypeId="2" ParentId="11210" CreationDate="2010-11-21T15:32:09.963" Score="0" ViewCount="" Body="&lt;p&gt;Your last argument is sound.  If you know that the terms are bounded over $[0,\infty)$, (or any interval, for that matter) that iteration doesn't take it out of that interval, and that all starting values eventually enter that interval, the terms are bounded.  As AkiRoss says, prepending terms cannot change the bounded nature, it only changes the bound.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-11-21T15:32:09.963" />
  <row Id="11230" PostTypeId="2" ParentId="154" CreationDate="2010-11-21T15:59:05.797" Score="1" ViewCount="" Body="&lt;p&gt;It may help to think of negative numbers as something other than &quot;negative.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The concept that helped me was to think of -1 as the opposite of 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the context was distance, for example, and I had a value of 1, then I went one unit in a direction.  If, on the other hand, I had a value of -1, then I'm traveling in the opposite direction of 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In calculus and series work, it will become more clear that this is a terminology problem most people have.  So whether you're dealing with space-time or voltages or something else, the negative symbol is just a relative thing, not an absolute value.  In other words, don't think of -5 as negative 5 but as something opposite of something else... &lt;/p&gt;&#xA;" OwnerUserId="3783" LastActivityDate="2010-11-21T15:59:05.797" CommunityOwnedDate="2011-04-15T20:45:29.080" CommentCount="1" />
  <row Id="11231" PostTypeId="1" AcceptedAnswerId="11316" CreationDate="2010-11-21T16:03:13.343" Score="6" ViewCount="245" Body="&lt;p&gt;Let's say you have a single-lane road (single in this case meaning single lane in each direction, or what you could also call two-lane). Let's say you have a random number of vehicles on a given 10-mile stretch of road. The speed limit is 45 mph, but some people go faster and some go slower.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any math that describes the likelihood that faster vehicles will overtake slower vehicles at &quot;choke points&quot; (points where oncoming traffic makes passing impossible)? (Assume a straight, flat road where safe passing is always allowed.) EDIT: Choke points are points where oncoming cars make it impossible to pass without the overtaking vehicle having to slow down and wait for the oncoming car (or cars) to pass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously, the answer has to be some kind of curve, because if there is only one vehicle on the road the probability is 0 that a choke point will be encountered, and if the road is completely filled with vehicles in both directions (or even one) the probability is 1 that a choke point will be encountered at any given time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sorry if this is a stupid question. But it seems like there should be a way to express this mathematically.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT 2: Limiting the scope of the problem&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks to Mike Spivey, and I hope this helps narrow the problem so that may be in some way answerable. If not, I'll just declare defeat and retire from the field. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's say that we have two cars going in one direction: a slow vehicle (Vehicle A) traveling at 30mph and a faster vehicle (Vehicle B) traveling at 60mph. At the start of the problem the A is somewhere between mile 2 and mile 4 on the 10-mile road, and B is at mile 0. Additionally, there will be four oncoming vehicles randomly distributed, and these vehicles are randomly traveling at 30 or 60mph. The overtaking vehicle requires .5 miles of &quot;clear&quot; space to pass safely if the oncoming vehicle is traveling at 60mph, and .25 if it is traveling at 30mph.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume further that the car lengths do not matter, but that a &quot;choke point&quot; (or bottleneck, if you prefer) is not reached until B gets within .05 miles of the rear bumper of A, at which point B has to pass A or decelerate. We will call the latter condition a &quot;deceleration event,&quot; (DE) and I would like to be able to calculate the probability that at least one DE will occur for Vehicle A along the 10-mile route.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps this clarification is still insufficient to render the problem solvable, but I appreciate anyone taking the time to consider it.&lt;/p&gt;&#xA;" OwnerUserId="3782" LastEditorUserId="3782" LastEditorDisplayName="" LastEditDate="2010-11-22T13:58:19.363" LastActivityDate="2010-11-22T13:58:19.363" Title="Probability in single-lane traffic flow: What are the odds of &quot;choke points&quot; being encountered?" Tags="&lt;probability&gt;&lt;mathematical-modeling&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="11232" PostTypeId="1" AcceptedAnswerId="11281" CreationDate="2010-11-21T16:14:14.313" Score="12" ViewCount="170" Body="&lt;p&gt;Let $f: \mathbb{R}^2 \rightarrow \mathbb{R}^2$ be bijective map with following properties:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) $f|_{\mathbb{Q}^2}=id$;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Image of any line under map $f$ is again a line.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it right that $f=id$?&lt;/p&gt;&#xA;" OwnerUserId="3784" LastActivityDate="2010-11-21T21:23:42.827" Title="Almost identical map" Tags="&lt;geometry&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="3" />
  <row Id="11233" PostTypeId="1" CreationDate="2010-11-21T16:21:31.667" Score="0" ViewCount="562" Body="&lt;p&gt;For a 2D Gaussian distribution with&lt;br&gt;&#xA;$$ \mu = \begin{pmatrix} \mu_x \\ \mu_y \end{pmatrix}, \quad \Sigma = \begin{pmatrix} \sigma_x^2 &amp;amp; \rho \sigma_x \sigma_y \\ \rho \sigma_x \sigma_y &amp;amp; \sigma_y^2 \end{pmatrix},&#xD;&#xA;$$&#xD;&#xA;its probability density function is&#xA;$$&#xD;&#xA;    f(x,y) = \frac{1}{2 \pi \sigma_x \sigma_y \sqrt{1-\rho^2}} \exp\left( -\frac{1}{2(1-\rho^2)}\left[ \frac{(x-\mu_x)^2}{\sigma_x^2} + \frac{(y-\mu_y)^2}{\sigma_y^2} - \frac{2\rho(x-\mu_x)(y-\mu_y)}{\sigma_x \sigma_y} \right] \right), &#xD;&#xA;$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was wondering if there is also a similarly clean formula for 3D Gaussian distribution density? What is it? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks and regards!&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;EDIT:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I ask is after taking the inverse of the covariance matrix, if the density has a clean form just as in 2D case?&lt;/p&gt;&#xA;" OwnerUserId="1281" LastEditorUserId="1281" LastEditorDisplayName="" LastEditDate="2010-11-21T17:10:38.580" LastActivityDate="2010-11-21T17:10:38.580" Title="density of 3D Gaussian distribution" Tags="&lt;probability&gt;" AnswerCount="1" />
  <row Id="11235" PostTypeId="2" ParentId="11218" CreationDate="2010-11-21T16:36:14.017" Score="2" ViewCount="" Body="&lt;p&gt;If $|x|&amp;gt;1$ then $\frac{1}{|x|}&amp;lt;1$, so:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ln\left(\frac{x+1}{x-1}\right)=\ln\left(\frac{x(1+\frac{1}{x})}{x(1-\frac{1}{x})}\right)=\ln\left(\frac{1+\frac{1}{x}}{1-\frac{1}{x}}\right)=\ln(1+t)-\ln(1-t)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With $t=\displaystyle\frac{1}{x}&amp;lt;1$, then $|t|&amp;lt;1$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Expanding :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ln(1+t)=t-\frac{t^{^2}}{2}+\frac{t^{^3}}{3}-\frac{t^{^4}}{4}+...$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ln(1-t)=-t-\frac{t^{^2}}{2}-\frac{t^{^3}}{3}-\frac{t^{^4}}{4}-...$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ln(1+t)-\ln(1-t)=\left(t-\frac{t^{^2}}{2}+\frac{t^{^3}}{3}-\frac{t^{^4}}{4}+...\right)-\left(-t-\frac{t^{^2}}{2}-\frac{t^{^3}}{3}-\frac{t^{^4}}{4}-...\right)=$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$2\left(t+\frac{t^{^3}}{3}+\frac{t^{5}}{5}...\right)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now $t=\displaystyle\frac{1}{x}$;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ln\left(\frac{x+1}{x-1}\right)=2\left(\frac{1}{x}+\frac{1}{3x^{^3}}+\frac{1}{5x^{^5}}+\frac{1}{7x^{^7}}...\right)$$&lt;/p&gt;&#xA;" OwnerUserId="3421" LastEditorUserId="3421" LastEditorDisplayName="" LastEditDate="2010-11-22T01:19:06.663" LastActivityDate="2010-11-22T01:19:06.663" />
  <row Id="11236" PostTypeId="1" CreationDate="2010-11-21T16:49:02.163" Score="0" ViewCount="277" Body="&lt;p&gt;Hi&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\lim\limits_{x\rightarrow 0 } \frac{x}{ln x}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This was wrong, I got a big red wrong! Why doesn't L'Hopital work on this one? &#xA;The problem is that ln is not defined for 0. It needs to be rewritten?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Thanks to everyone helping me out with my homework, due anxiety I'm not able attend the class workshops.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: I did get the correct limit ($0$), but that was a coincidence.  &lt;/p&gt;&#xA;" OwnerUserId="2230" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-21T21:28:11.830" LastActivityDate="2010-11-21T21:28:11.830" Title="Another limit task, x over ln x, L'Hopital won't do it?" Tags="&lt;calculus&gt;&lt;homework&gt;" AnswerCount="4" CommentCount="2" />
  <row Id="11237" PostTypeId="2" ParentId="11177" CreationDate="2010-11-21T16:50:56.887" Score="1" ViewCount="" Body="&lt;p&gt;I've looked at many set theory books and I think the best one for a beginner is one called Classic Set Theory by Goldrei.  It goes through loads of examples and is designed for self-study.&lt;/p&gt;&#xA;" OwnerUserId="3785" LastActivityDate="2010-11-21T16:50:56.887" />
  <row Id="11238" PostTypeId="2" ParentId="11223" CreationDate="2010-11-21T16:51:44.733" Score="1" ViewCount="" Body="&lt;p&gt;You need to factorise each lightbulb's number. If it has an odd number of divisors then it is ON. If it has an even number of divisors it is OFF. I'm sorry I don't have a general n-th rule though.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To answer the question, 25 is ON, 93 is OFF and 576 is ON.&lt;/p&gt;&#xA;" OwnerUserId="3746" LastActivityDate="2010-11-21T16:51:44.733" CommentCount="1" />
  <row Id="11239" PostTypeId="2" ParentId="11236" CreationDate="2010-11-21T16:55:58.533" Score="6" ViewCount="" Body="&lt;p&gt;L'Hopital's rule is best forgotten about. Of course as $x\to0^+$,&#xA;$\log x\to-\infty$ and so $1/\log x\to0$. &lt;em&gt;A fortiori&lt;/em&gt;&#xA;$x/\log x\to 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A better problem is to find $\lim_{x\to0^+}x\log x$.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-11-21T16:55:58.533" CommentCount="3" />
  <row Id="11240" PostTypeId="2" ParentId="11217" CreationDate="2010-11-21T17:00:56.377" Score="1" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\rm\quad\ 0 \ \ = \ \ (y^2-1)\ (1+x^2)^2 + (1 - x^2)^2$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\quad\quad\quad\quad\quad\quad\ = \ \ y^2\:(1+x^2)^2 - 4\:x^2$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\quad\quad\quad\quad\quad\quad\ =\ \ (y\:(1+x^2)-2\:x)\ \ (y\:(1+x^2)+2\:x)$&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-11-21T17:07:56.117" LastActivityDate="2010-11-21T17:07:56.117" />
  <row Id="11241" PostTypeId="2" ParentId="11236" CreationDate="2010-11-21T17:02:58.813" Score="1" ViewCount="" Body="&lt;p&gt;$\frac{0}{\infty}$  is not indeterminate form, is zero, then $\lim\limits_{ x \rightarrow 0^+ }\frac{x}{lnx}=0$ but is a right limit. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But L'Hôpital's rule applies..&lt;/p&gt;&#xA;" OwnerUserId="3421" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-21T21:26:44.843" LastActivityDate="2010-11-21T21:26:44.843" CommentCount="0" />
  <row Id="11242" PostTypeId="2" ParentId="11223" CreationDate="2010-11-21T17:03:22.567" Score="3" ViewCount="" Body="&lt;p&gt;Each number that is a square will be ON.&lt;/p&gt;&#xA;" OwnerUserId="3567" LastActivityDate="2010-11-21T17:03:22.567" />
  <row Id="11243" PostTypeId="2" ParentId="11233" CreationDate="2010-11-21T17:05:18.977" Score="4" ViewCount="" Body="&lt;p&gt;There is a standard, general formula for the density of the joint normal (or multivariate normal) distrubution of dimension $n$, provided that the ($n \times n$) covariance matrix $\Sigma$ is non-singular (see, e.g., &lt;a href=&quot;http://planetmath.org/encyclopedia/JointNormalDistribution.html&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; or &lt;a href=&quot;http://en.wikipedia.org/wiki/Multivariate_normal_distribution&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt;). In particular, you can apply for $n=3$. When the covariance matrix is singular, the distribution is expressed in terms of the characteristic function.&lt;/p&gt;&#xA;" OwnerUserId="2810" LastActivityDate="2010-11-21T17:05:18.977" CommentCount="14" />
  <row Id="11244" PostTypeId="1" AcceptedAnswerId="11258" CreationDate="2010-11-21T17:20:09.923" Score="4" ViewCount="188" Body="&lt;p&gt;Let $\hat{\mathbb{C}} = \mathbb{C} \cup \{ \infty \}$ denote the &lt;a href=&quot;http://en.wikipedia.org/wiki/Riemann_sphere&quot; rel=&quot;nofollow&quot;&gt;Riemann Sphere&lt;/a&gt;. While reading the wikipedia article about it I found a passage that said that &lt;a href=&quot;http://en.wikipedia.org/wiki/Riemann_sphere#Rational_functions&quot; rel=&quot;nofollow&quot;&gt;every rational function on the complex plane can be extended to a continuous function on the Riemann Sphere&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The particular construction is as follows, let $R(z) = \frac{f(z)}{g(z)} \in \mathbb{C}(z)$ be a rational function. Lets assume for simplicity that $f(z)$ and $g(z)$ share no common factor. Then for any point $a \in \mathbb{C}$ such that $g(a) = 0$ but $f(a) \neq 0$ we define $R(a) = \infty$. Also we define $R(\infty) := \lim_{z \to \infty} R(z)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So to quote wikipedia, with this definitions $R(z)$ becomes a continuous function from the Riemann Sphere to itself. The problem for me is that the article doesn't add any details as to how one may go about showing that in fact $R(z)$ is continuous. So my question is exactly that, to make it simple, if I have for instance $R(z) = \frac{z-1}{z+1}$ or even $R(z) = \frac{1}{z}$, how do I show that it is a continuous function on $\hat{\mathbb{C}}$? I think I can build up from an easy example such as this (assumming this is easy).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, I'm a little bit confused about how to interpret this continuity, how should I see the continuity on the Riemann Sphere?, does it involve an argument with stereographic projection?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I added in the Riemann Surface tag just in case they are involved, which I'm not sure. Thank you very much in advance.&lt;/p&gt;&#xA;" OwnerUserId="900" LastActivityDate="2011-05-04T19:45:37.283" Title="Continuity of Rational Functions on the Riemann Sphere $\hat{\mathbb{C}}$" Tags="&lt;complex-analysis&gt;&lt;riemann-surface&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="11246" PostTypeId="1" CreationDate="2010-11-21T17:28:23.137" Score="10" ViewCount="610" Body="&lt;p&gt;How can we prove the following trigonometric identity?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle \tan(3\pi/11) + 4\sin(2\pi/11) =\sqrt{11}$$&lt;/p&gt;&#xA;" OwnerUserId="3786" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-11-21T22:17:17.057" LastActivityDate="2011-05-13T23:29:57.280" Title="How to prove that: $\tan(3\pi/11) + 4\sin(2\pi/11) = \sqrt{11}$" Tags="&lt;trigonometry&gt;" AnswerCount="4" CommentCount="0" FavoriteCount="3" />
  <row Id="11247" PostTypeId="2" ParentId="11194" CreationDate="2010-11-21T17:34:09.720" Score="2" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\quad$ Solve it the same way you would solve $\rm\ f(x)\ =\ 1 + 2 \int f(x)\ dx\ $ but instead of $\rm\ d/dx\ $ apply $\rm\ \Delta_n\: f(n) = f(n+1) - f(n)\ $ to eliminate (invert) the $\rm\: \sum = \Delta^{-1}\:,\ $ i.e. employ the the &lt;strong&gt;fundamental theorem of difference calculus&lt;/strong&gt;.&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-11-21T17:34:09.720" />
  <row Id="11249" PostTypeId="2" ParentId="11246" CreationDate="2010-11-21T17:41:59.247" Score="5" ViewCount="" Body="&lt;p&gt;Hi--&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can find the solution in this page:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://natto.2ch.net/math/kako/1002/10029/1002903143.html&quot; rel=&quot;nofollow&quot;&gt;http://natto.2ch.net/math/kako/1002/10029/1002903143.html&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Translation of the page into  &lt;em&gt;English.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I = tan (3π/11) +4 sin (2π/11)&lt;br&gt;&#xA;and t = 3π/11 &lt;/p&gt;&#xA;&#xA;&lt;p&gt;　11t = 3π&lt;br&gt;&#xA;　⇔ 6t = 3π-5t&lt;br&gt;&#xA;　⇒ $\sin (6t) = \sin (3π-5t)$ taking sin of both sides&lt;br&gt;&#xA;　⇔ $2\sin (3t) \cos (3t) = \sin(5t)$ double angle formula&lt;br&gt;&#xA;  ⇔ $[3\sin(t)-4 \sin^3 (t)] [4 \cos^3 (t)-3\cos(t)] = 16 \sin^5(t) -20 \sin^3(t) +5 \sin(t)$&lt;br&gt;&#xA;　⇔ $[3-4 \sin^2 t ] [4 \cos^3 t -3\cos t] = 16 \sin^4 t - 20 \sin^2 t  +5$ dividing by  $\sin t ≠ 0$&lt;br&gt;&#xA;　⇔ $32x ^ 5-16x ^ 4-32x ^ 3 +12 x ^ 2 +6 t-1 = 0$, where $\sin^2 t  = 1 - \cos^2 t$, $x = \cos t$    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus $x = \cos (3π/11)$ is a solution of &#xA;$ 32x ^ 5-16x ^ 4-32x ^ 3 +12 x ^ 2 +6 t-1 = 0 $ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $(2π/11) = [1 - (9 / 11)] π = (π-3t)$, so&lt;br&gt;&#xA;$I = \tan (3π/11) +4 \sin (2π/11)$&lt;br&gt;&#xA;$　= \tan t +4 sin (π-3t)$&lt;br&gt;&#xA;$　= \tan t +4 \sin (3t)$&lt;br&gt;&#xA;$　= (\sin t / \cos t) +4 [3\sin t-4 \sin^3 t ]$&lt;br&gt;&#xA;$　= (\sin t / \cos t) [16 \cos^3 t- 4 \cos t +1]$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$I ^ 2 = (\sin t / \cos t) ^ 2 [16 \cos^3 t -4 \cos t +1]^2$&lt;br&gt;&#xA;$　= [(1 - \cos^2 t) / \cos^2 t] [16 \cos^3 t -4 \cos t +1] ^ 2$&lt;br&gt;&#xA;$　= [(1-x^2) (16x^3-4x +1)^2]/x^2$, where $x = \cos t$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Molecule {(1-x ^ 2) (16x ^ 3-4x +1) ^ 2} a &#xA;{32x ^ 5-16x ^ 4-32x ^ 3 +12 x ^ 2 +6 t-1} is divided by &#xA;← 2 11x ^ quotient remainder omitted &lt;/p&gt;&#xA;" LastEditorUserId="2532" LastEditorDisplayName="" LastEditDate="2011-04-27T19:44:11.310" LastActivityDate="2011-04-27T19:44:11.310" />
  <row Id="11250" PostTypeId="1" CreationDate="2010-11-21T17:43:53.780" Score="1" ViewCount="88" Body="&lt;p&gt;I was wondering if f(x)=O(x^(c+a)) for all a&gt;0 then is it necessarily true that f(x)=O((x^c)logx)?&#xA;I suspect it's not true but want to know why.  (I know the converse is true.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any help is much appreciated,&#xA;Thank you.&lt;/p&gt;&#xA;" OwnerUserId="3785" LastActivityDate="2010-11-21T19:50:41.800" Title="If f is big O of x^c+a for all a&gt;0 then is f big O of (x^c)logx?" Tags="&lt;number-theory&gt;&lt;real-analysis&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="11251" PostTypeId="2" ParentId="11193" CreationDate="2010-11-21T17:46:36.320" Score="0" ViewCount="" Body="&lt;p&gt;Suppose you want to visualize an $8$-dimensional space. One possible approach(in fact the only one I know) is to first visualize an $n$-dimensional space in the form of $\mathbb R^n$, and then let $n = 8$.&lt;/p&gt;&#xA;" OwnerUserId="1119" LastActivityDate="2010-11-21T17:46:36.320" CommunityOwnedDate="2010-11-21T17:46:36.320" CommentCount="3" />
  <row Id="11252" PostTypeId="2" ParentId="11236" CreationDate="2010-11-21T17:48:27.080" Score="2" ViewCount="" Body="&lt;p&gt;As others have mentioned, you don't need L'Hospital's rule for this simple example. However, in fact, there is a general form of the rule that does apply to this case, and it does come in handy in some circumstances - so it is worth knowing. For an exposition see e.g.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A. E. Taylor, $\ $ L'Hospital's Rule&lt;br&gt;&#xA;Amer. Math. Monthly, Vol. 59, No. 1 (Jan., 1952), pp. 20-24.&lt;br&gt;&#xA;&lt;a href=&quot;http://www.jstor.org/stable/2307183&quot; rel=&quot;nofollow&quot;&gt;http://www.jstor.org/stable/2307183&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A. M. Ostrowski, $\ $ Note on the Bernoulli-L'Hospital Rule&lt;br&gt;&#xA;Amer. Math. Monthly, Vol. 83, No. 4 (Apr., 1976), pp. 239-242.&lt;br&gt;&#xA;&lt;a href=&quot;http://www.jstor.org/stable/2318210&quot; rel=&quot;nofollow&quot;&gt;http://www.jstor.org/stable/2318210&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-11-21T17:48:27.080" CommentCount="4" />
  <row Id="11254" PostTypeId="2" ParentId="11206" CreationDate="2010-11-21T17:50:39.583" Score="0" ViewCount="" Body="&lt;p&gt;You seem to be asking about an ellipse whose axes are aligned with the $x$ and $y$ directions. Note that an ellipse which is rotated by an arbitrary angle is still an ellipse. But you cannot uniquely determine such an ellipse from the information given.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If two points $(x_1, y_1)$ and $(x_2, y_2)$ pass through an ellipse centred at $(x_0, y_0)$ with &lt;em&gt;semi-axes&lt;/em&gt; along $x$ and $y$ of length $r_x$ and $r_y$ respectively, then the points $(x_1/r_x, y_1/r_y)$ and $(x_2/r_x, y_2/r_y)$ pass through a circle of unit radius with centre $(x_0/r_x, y_0/r_y)$. So if you can find the centre of this circle, you can find the centre of the corresponding ellipse.&lt;/p&gt;&#xA;" OwnerUserId="856" LastActivityDate="2010-11-21T17:50:39.583" />
  <row Id="11256" PostTypeId="2" ParentId="11250" CreationDate="2010-11-21T17:53:52.177" Score="7" ViewCount="" Body="&lt;p&gt;Consider $f(x) = x^c (\log x)^b$, where $b&amp;gt;1$. This function is $O(x^{(c+a)})$ but it is not $O(x^c \log x)$. &lt;/p&gt;&#xA;" OwnerUserId="2641" LastEditorUserId="2641" LastEditorDisplayName="" LastEditDate="2010-11-21T18:00:50.357" LastActivityDate="2010-11-21T18:00:50.357" CommentCount="1" />
  <row Id="11257" PostTypeId="2" ParentId="11194" CreationDate="2010-11-21T17:55:28.443" Score="1" ViewCount="" Body="&lt;p&gt;Using a spreadsheet, I note that $T(n)=3^{(n-1)}$  This is easily verified by induction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$T(1)=1=3^0$.&lt;br&gt;&#xA;Then if it is true up to $n$, $$T(n+1)=1+2\sum_{i=0}^{n-1}3^i=1+2\frac{3^n-1}{3-1}$$&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-11-21T17:55:28.443" />
  <row Id="11258" PostTypeId="2" ParentId="11244" CreationDate="2010-11-21T17:56:59.753" Score="3" ViewCount="" Body="&lt;p&gt;One way of doing this is to use the sequential definition of continuity... this works because of the way the Riemann sphere has a metric (distance function) defined on it, inherited from the regular Euclidean distance function in 3-d. In this viewpoint, the statement that $\lim_{z \rightarrow a} f(z) = \infty$ is equivalent to saying $\lim_{z \rightarrow a} |f(z)| = \infty$. In the case that $a$ itself is infinity, then $\lim_{z \rightarrow \infty} f(z) = \infty$ becomes $\lim_{|z| \rightarrow \infty } |f(z)| = \infty$. In other words, for every $N$ there's an $M$ such that if $|z| &amp;gt; M$ then $|f(z)| &amp;gt; N$. Similarly, for some finite $z_0$ the statement $\lim_{z \rightarrow \infty} f(z) = z_0$ means $\lim_{|z| \rightarrow \infty } f(z) = z_0$; for every $\epsilon &amp;gt; 0$ there's an $N$ such that $|z| &amp;gt; N$ implies $|f(z) - z_0| &amp;lt; \epsilon$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this way many statements like the ones you're trying to prove become pretty routine.&lt;/p&gt;&#xA;" OwnerUserId="3035" LastEditorUserId="3035" LastEditorDisplayName="" LastEditDate="2010-11-21T18:03:04.010" LastActivityDate="2010-11-21T18:03:04.010" CommentCount="4" />
  <row Id="11259" PostTypeId="1" AcceptedAnswerId="11265" CreationDate="2010-11-21T17:59:01.543" Score="1" ViewCount="111" Body="&lt;p&gt;Hi there,&#xA;I have two questions and I hope it is alright to ask them both at the same time. I'm currently trying to solve an online-exercise about differential calculus and the Taylor series and I'm having some trouble there.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Let $f: (a,b) \rightarrow \mathbb{R}$ be differentiable at the point $x_0$. Which of the following statements is true?&lt;/p&gt;&#xA;  &#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;$f(x) = f(x_0) + O(x-x_0) \;\; (x \rightarrow x_0)$&#xD;&lt;/li&gt;&#xA;  &lt;li&gt;$f(x) = f(x_0) + f&amp;#39;(x_0)(x-x_0) + O((x-x_0)^2) \;\; (x \rightarrow x_0)$&#xD;&lt;/li&gt;&#xA;  &lt;li&gt;$f(x) = f(x_0) + f&amp;#39;(x_0)(x-x_0) + o((x-x_0)) \;\; (x \rightarrow x_0)$&#xD;&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am pretty sure the third statement is true, because that's basically our definition of the Taylor series. However, I'm not quite sure about the other two statements, though I believe either both or none of them is true. When proving the Fundamental Theorem of Algebra, we once &quot;replaced&quot; $o(r^k)$ by $O(r^{k+1})$ which already then I found hard to understand. Can one always do this? In that case, all three statements would be correct...&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Let $f(x) = e^{\frac{-1}{|x|}} \cdot \cos(x^{-1})$ for $x \neq 0$ and $0$ for $x = 0$. $f$ has a local maximum or minimum at $x=0$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Again, I am not sure at all. I plotted the function and came to the conclusion that it has a minimum at $x=0$. However, we usually have a minimum when $f&amp;#39;(x_0) = 0$ and $f^{(n)} (x_0) &amp;gt; 0$, where $n$ is an even number. This function is arbitrarily often differentiable and I believe at none of the derivatives $f^{(n)} (x_0) \neq 0$...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd be delighted if someone could help me out here.&lt;/p&gt;&#xA;" OwnerUserId="3787" LastActivityDate="2010-11-21T18:12:11.697" Title="Landau symbols and a question about the turning point of an arbitrarily often differentiable function" Tags="&lt;analysis&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="11260" PostTypeId="2" ParentId="11250" CreationDate="2010-11-21T18:00:21.500" Score="0" ViewCount="" Body="&lt;p&gt;Consider $f(x) = (\log x)^2$.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-21T18:00:21.500" />
  <row Id="11261" PostTypeId="1" AcceptedAnswerId="11266" CreationDate="2010-11-21T18:04:44.143" Score="5" ViewCount="224" Body="&lt;p&gt;What is your recommended symbolic computation program/software for free and commercial respectively?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are its strength and weakness? For example, efficiency, comprehensiveness, etc&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="1281" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-21T23:23:16.343" LastActivityDate="2011-03-16T03:50:17.520" CommunityOwnedDate="2010-11-22T07:36:53.763" Title="symbolic computation program/software" Tags="&lt;math-software&gt;&lt;symbolic-computation&gt;" AnswerCount="5" CommentCount="8" FavoriteCount="1" />
  <row Id="11262" PostTypeId="2" ParentId="11259" CreationDate="2010-11-21T18:05:30.197" Score="1" ViewCount="" Body="&lt;p&gt;For your second question, there are points arbitrarily close to zero, such as $\frac{1}{2k\pi}$ for k a natural number, where f(x)&gt;0 and also points $\frac{1}{(2k+1)\pi}$where f(x)&amp;lt;0.  So zero is neither a maximum nor a minimum.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-11-21T18:05:30.197" />
  <row Id="11263" PostTypeId="2" ParentId="11261" CreationDate="2010-11-21T18:06:51.427" Score="2" ViewCount="" Body="&lt;p&gt;Both Mathematica and Maple are excellent. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems they have added symbolic computation toolbox for scilab (which is free). Since you are linux user, you might want to explore it. I have never personally used it though (I have used scilab itself and it does a good job of being a lilliputian Matlab)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://packages.ubuntu.com/lucid/scilab-scimax-doc&quot; rel=&quot;nofollow&quot;&gt;http://packages.ubuntu.com/lucid/scilab-scimax-doc&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3431" LastEditorUserId="3431" LastEditorDisplayName="" LastEditDate="2010-11-21T18:40:30.137" LastActivityDate="2010-11-21T18:40:30.137" CommunityOwnedDate="2010-11-22T07:36:53.763" CommentCount="5" />
  <row Id="11264" PostTypeId="2" ParentId="11259" CreationDate="2010-11-21T18:07:58.033" Score="1" ViewCount="" Body="&lt;p&gt;For your first question, the third statement is the usual definition of the derivative. For the second statement, take the absolute value function and integrate it to get a function $f(x)$; check what happens at $x_0=0$.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-21T18:07:58.033" />
  <row Id="11265" PostTypeId="2" ParentId="11259" CreationDate="2010-11-21T18:12:11.697" Score="1" ViewCount="" Body="&lt;p&gt;For your first question, the first statement is true; it follows from the third.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second statement is not always true; it essentially requires the function to be twice differentiable.  Try a function that is only once differentiable at $x_0$ and see what happens.&lt;/p&gt;&#xA;" OwnerUserId="822" LastActivityDate="2010-11-21T18:12:11.697" CommentCount="3" />
  <row Id="11266" PostTypeId="2" ParentId="11261" CreationDate="2010-11-21T18:25:14.267" Score="7" ViewCount="" Body="&lt;p&gt;I recommend  sageMath  &lt;a href=&quot;http://www.sagemath.org/&quot; rel=&quot;nofollow&quot;&gt;http://www.sagemath.org/&lt;/a&gt;, also Mathematica and Maple&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An easy to use  is Maxima  &lt;a href=&quot;http://maxima.sourceforge.net/&quot; rel=&quot;nofollow&quot;&gt;http://maxima.sourceforge.net/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can  to work with sage here: &lt;a href=&quot;http://www.sagenb.org/&quot; rel=&quot;nofollow&quot;&gt;http://www.sagenb.org/&lt;/a&gt; , not is necesary installing !!!&lt;/p&gt;&#xA;" OwnerUserId="3421" LastEditorUserId="3421" LastEditorDisplayName="" LastEditDate="2010-11-21T19:53:14.183" LastActivityDate="2010-11-21T19:53:14.183" CommunityOwnedDate="2010-11-22T07:36:53.763" />
  <row Id="11267" PostTypeId="1" CreationDate="2010-11-21T18:43:33.330" Score="1" ViewCount="156" Body="&lt;p&gt;John Von Neumann once said to Felix Smith, &quot;Young man, in mathematics you don't understand things. You just get used to them.&quot; This was a response to Smith's fear about the method of characteristics. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Did he mean that with experience and practice, one obtains understanding? &lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2010-11-21T18:43:33.330" Title="What are some interpretations of Von Neumann's quote?" Tags="&lt;soft-question&gt;" CommentCount="6" />
  <row Id="11268" PostTypeId="1" AcceptedAnswerId="11274" CreationDate="2010-11-21T19:01:08.457" Score="5" ViewCount="125" Body="&lt;p&gt;Let $X$, $Y$ be pointed CW complexes, $Y$ connected and $f:X\to Y$ a mapping.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does the assertion '$\Sigma f:\Sigma X\to\Sigma Y$ is a homotopy equivalence' imply that $f$ is a homotopy equivalence? '$\Sigma$' is the reduced suspension.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If not, is it true with some additional hypotheses on $Y$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Addition:&lt;/em&gt;&#xA;Does the assertion '$\Omega f:\Omega X\to\Omega Y$ is a homotopy equivalence' imply that $f$ is a homotopy equivalence? '$\Omega$' is the loopspace.&lt;/p&gt;&#xA;" OwnerUserId="3789" LastEditorUserId="3789" LastEditorDisplayName="" LastEditDate="2010-11-21T20:32:13.053" LastActivityDate="2010-11-21T21:46:05.177" Title="Is a map a homotopy equivalence if its suspension is so?" Tags="&lt;topology&gt;&lt;algebraic-topology&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="11269" PostTypeId="2" ParentId="11261" CreationDate="2010-11-21T19:11:33.280" Score="3" ViewCount="" Body="&lt;p&gt;Another very popular free program for symbolic computations is &lt;a href=&quot;http://pari.math.u-bordeaux.fr/&quot; rel=&quot;nofollow&quot;&gt;PARI/GP&lt;/a&gt;. It is aimed somehow at computations in number theory, but its functionality is not only restricted to this. You can read a little bit about it also in the &lt;a href=&quot;http://en.wikipedia.org/wiki/PARI/GP&quot; rel=&quot;nofollow&quot;&gt;wikipedia article&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="900" LastActivityDate="2010-11-21T19:11:33.280" CommunityOwnedDate="2010-11-22T07:36:53.763" />
  <row Id="11270" PostTypeId="2" ParentId="11177" CreationDate="2010-11-21T19:39:06.460" Score="5" ViewCount="" Body="&lt;p&gt;I am going to go out on a limb and recommend a more elementary book than (I think) any of the ones others have mentioned.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I claim that as a pure mathematician who is not a set theorist, all the set theory I have ever needed to know I learned from Irving Kaplansky's &lt;em&gt;Set Theory and Metric Spaces&lt;/em&gt;.  (And, you know, I also enjoyed the part about metric spaces).  Kaplansky spent most of his career at the University of Chicago.  Although he had left for MSRI by the time I got there in the mid 1990's, nevertheless his text was still used for the one undergraduate set theory course they offered there.  (Not that I actually took that course, but I digress...)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact I think that if you work through this book carefully -- it's beautifully written and reads easily, but is not always as innocuous as it appears -- you will actually come out with more set theory than the average pure mathematician knows.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Apologies if you actually do need or want to know some more serious stuff: there's nothing about, say, cofinalities in there, let alone forcing and whatever else comes later on.  But maybe this answer will be appropriate for someone else, if not for you.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Added&lt;/b&gt;: I suppose I might as well mention my own lecture notes, available online &lt;a href=&quot;http://www.math.uga.edu/~pete/expositions.html&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; (scroll down to Set Theory).  I think it is fair to say that these are a digest version of Kaplansky's book, even though they were for the most part not written with that book in hand.  [However, last week David Speyer emailed me to kindly point out that I had completely screwed up (not his words!) one of the proofs.  He also suggested the correct fix, but I didn't feel sanguine about it until I went back to Kaplansky to see how he did it.]  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The description &lt;em&gt;All the set theory I have ever needed to know&lt;/em&gt; on the main page is not meant to be offensive to set theorists (and I hope it isn't) but rather an honest admission: here is the little bit of material that goes a very long way indeed.  Note especially the word &lt;em&gt;need&lt;/em&gt;: this is not to say that these 40 pages contain all the set theory I &lt;em&gt;want&lt;/em&gt; to know.  For instance, I own Cohen's book on forcing and the Continuum Hypothesis, and I would certainly like to know how that stuff goes...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[Come to think of it: I would be highly amused and interested to read 40 pages of notes entitled &lt;em&gt;All the number theory I have ever needed to know&lt;/em&gt; written by one of the several eminent set theorists / logicians who frequent this site and MO.  What would make the cut?]&lt;/p&gt;&#xA;" OwnerUserId="299" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2010-12-27T12:43:26.587" LastActivityDate="2010-12-27T12:43:26.587" />
  <row Id="11271" PostTypeId="2" ParentId="11261" CreationDate="2010-11-21T20:12:38.713" Score="1" ViewCount="" Body="&lt;p&gt;I have been using Mathematica since fall 2007. I absolutely love it. It has good palettes included that help you typeset your equations more easily. It also has nice display options for giving presentations. However, if you are just wanting to typeset a paper, LaTeX is probably your best bet. TeXmaker is a very good free mac OS software for LaTeX.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maple is another very popular computational software; however, I have not used it.&lt;/p&gt;&#xA;" OwnerUserId="3791" LastActivityDate="2010-11-21T20:12:38.713" CommunityOwnedDate="2010-11-22T07:36:53.763" />
  <row Id="11272" PostTypeId="2" ParentId="11194" CreationDate="2010-11-21T20:13:37.233" Score="2" ViewCount="" Body="&lt;p&gt;You can also use generating functions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;if $$\displaystyle f(x) = \sum_{n=1}^{\infty} T(n) x^n$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle \frac{2f(x)}{1-x} = \sum_{n=1}^{\infty} (2\sum_{k=1}^{n} T(k)) x^n$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i.e.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle \frac{2f(x)}{1-x} = \sum_{n=1}^{\infty} (3T(n) - 1) x^n = 3f(x) - \frac{1}{1-x} + 1$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This gives us&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle f(x) = \frac{x}{1-3x} = \sum_{n=1}^{\infty} 3^{n-1} x^n$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle T(n) = 3^{n-1}$$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-11-21T20:13:37.233" />
  <row Id="11273" PostTypeId="2" ParentId="11194" CreationDate="2010-11-21T20:16:20.090" Score="0" ViewCount="" Body="&lt;p&gt;Take a look at Kelley and Peterson's textbook [1]. They provide a very good discussion of difference equations in this text. I believe you can relate the information in this book to answer your question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You will find further discussion regarding the answers posted here by Moron and Ross Millikan.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me know if you still cannot figure it out!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[1] Kelley, W. &amp;amp; Peterson, A. (2001). Difference Equations: An Introduction with Applications (2nd Ed.). San Diego, CA: Academic Press.&lt;/p&gt;&#xA;" OwnerUserId="3791" LastActivityDate="2010-11-21T20:16:20.090" />
  <row Id="11274" PostTypeId="2" ParentId="11268" CreationDate="2010-11-21T20:18:47.157" Score="6" ViewCount="" Body="&lt;p&gt;I believe the answer to your first question is no. Let $X$ be any connected acyclic CW-complex with non-trivial fundamental group, for example the space constructed as example 2.38 in Hatcher. Such a space has the property that $H_i(X) = 0$ for $i&amp;gt;0$ and $H_0(X) = \mathbb{Z}$, but $\pi_1(X) \neq 0$. (In particular $\pi_1(X)$ must be perfect). Consider the projection map $f: X \to pt$. By looking $\pi_1$, $f$ cannot be a homotopy equivalence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, $\Sigma f: \Sigma X \to \Sigma pt$ is a homotopy equivalence. To see this, note that suspension increases the connectivity, which implies that both spaces are simply connected. Hence the homology Whitehead theorem applies, which says that a map between simply connected CW-complexes is a homotopy equivalence if and only if it induces isomorphisms on all homology groups. Using the suspension axiom in homology, we see that all $H_i(\Sigma X)$ and $H_i(\Sigma pt)$ for all $i&amp;gt;0$ are zero and for $i=0$ are $\mathbb{Z}$. It is then easy to check that $\Sigma f$ is an isomorphism in all degrees.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;edit: in your addition, I think the answer is yes, if we replace homotopy equivalence by weak homotopy equivalence. The Whitehead theorem says that $f: X \to Y$ is homotopy equivalence if and only if induces an isomorphism on all $\pi_i$. Because $\Omega X$ and $\Omega Y$ have the homotopy type of CW-complexes, we can replace them by CW-complexes with the price of replacing homotopy equivalence with weak homotopy equivalence. Now note that $Map_+(S^n,\Omega X) \cong Map_+(S^{n+1},X)$ and similarly $Map_+(S^n,\Omega Y) \cong Map_+(S^{n+1},Y)$. Under this isomorphism $(\Omega f)_*$ corresponds to $f_*$.&lt;/p&gt;&#xA;" OwnerUserId="3790" LastEditorUserId="3790" LastEditorDisplayName="" LastEditDate="2010-11-21T21:46:05.177" LastActivityDate="2010-11-21T21:46:05.177" />
  <row Id="11275" PostTypeId="2" ParentId="11150" CreationDate="2010-11-21T20:23:48.617" Score="1" ViewCount="" Body="&lt;p&gt;Take a look at WolframMathWorld's [1] discussion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See if this gives you any clarification.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[1] Weisstein, Eric W. &quot;Indeterminate.&quot; From MathWorld--A Wolfram Web Resource. &lt;a href=&quot;http://mathworld.wolfram.com/Indeterminate.html&quot; rel=&quot;nofollow&quot;&gt;http://mathworld.wolfram.com/Indeterminate.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3791" LastActivityDate="2010-11-21T20:23:48.617" />
  <row Id="11276" PostTypeId="2" ParentId="11236" CreationDate="2010-11-21T20:27:33.590" Score="2" ViewCount="" Body="&lt;p&gt;The limit, as written, does not exist because the function is not defined on an open neighborhood of the limit point: the function is not defined on $(-\infty,0)$, so you cannot take the limit as $x\to 0$. That would be a full answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If, however, the limit is taken only from the right, then the other answers you have received are correct: the limit is not an indeterminate form, it is equal to $0$. But you cannot get that by 'evaluating', because $\ln(x)$ is not defined at $0$.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-11-21T20:27:33.590" CommentCount="1" />
  <row Id="11277" PostTypeId="1" AcceptedAnswerId="11278" CreationDate="2010-11-21T20:32:11.483" Score="2" ViewCount="185" Body="&lt;p&gt;Is there an easy way to see that the $n$-dimensional Fourier transform of $1/|x|^a$ is equal to $1/|x|^{n-a}$ (up to multiplicative constant) where $x$ is an $n$ dimensional vector (assuming that $ 0 &amp;lt; a &amp;lt; n $) ?&lt;/p&gt;&#xA;" LastActivityDate="2010-11-21T21:45:12.383" Title="how to calculate fourier transform of a power of radial function" Tags="&lt;fourier-analysis&gt;" AnswerCount="1" />
  <row Id="11278" PostTypeId="2" ParentId="11277" CreationDate="2010-11-21T20:45:50.943" Score="3" ViewCount="" Body="&lt;p&gt;Yes... you use the way the definition of Fourier transform changes under rotations and dilations. Namely, recall that&#xA;$$\hat{f}(\xi) = \int_{R^n} f(x) e^{-2\pi i x \cdot \xi} \,dx$$&#xD;&#xA;So that if $f(x) = |x|^{-a}$ this becomes&#xA;$$\hat{f}(\xi) = \int_{R^n} |x|^{-a} e^{-2\pi i x \cdot \xi} \,dx$$&#xD;&#xA;Note this is an improper integral, so you have to be careful here with how one interprets the above. Next, note that if $R$ is a rotation, then $Rx \cdot \xi = x \cdot R^{-1}{\xi}$. So changing variables in the above, turning $x$ into $Rx$, and observing that $|x| = |Rx|$, leads to &#xA;$$\hat{f}(\xi) = \int_{R^n} |x|^{-a} e^{-2\pi i x \cdot R^{-1}\xi} \,dx$$&#xD;&#xA;$$ = \hat{f}(R^{-1}\xi)$$&#xD;&#xA;Since $R$ is any rotation this means that $\hat{f}$ is radial.&#xA;Next, change variables $x \rightarrow tx$ for some $t &amp;gt; 0$ in the above definition of $\hat{f}$. This time you get &#xA;$$\hat{f}(\xi) = \int_{R^n} |tx|^{-a} e^{-2\pi i tx \cdot \xi} t^n \,dx$$&#xD;&#xA;$$ = |t|^{n-a}\int_{R^n} |x|^{-a} e^{-2\pi i x \cdot t\xi} \,dx$$&#xD;&#xA;So $\hat{f}(\xi) = |t|^{n-a} \hat{f}(t\xi)$, or equivalently $\hat{f}(t\xi) = &#xD;&#xA;|t|^{a-n}\hat{f}(\xi)$. This means on each radial line, $\hat{f(t\xi)}$ is equal to $C|t|^{a-n}$. And since $\hat{f}(\xi)$ is a radial function, $C$ is independent of which radial line you're on. This gives what you're looking for.&lt;/p&gt;&#xA;" OwnerUserId="3035" LastEditorUserId="3035" LastEditorDisplayName="" LastEditDate="2010-11-21T21:45:12.383" LastActivityDate="2010-11-21T21:45:12.383" CommentCount="3" />
  <row Id="11279" PostTypeId="2" ParentId="11212" CreationDate="2010-11-21T20:52:49.513" Score="0" ViewCount="" Body="&lt;p&gt;If you have an interest in engineering optics or linear systems, try &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0471292885&quot; rel=&quot;nofollow&quot;&gt;Gaskill's book&lt;/a&gt;.  It's good to start from.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0521709792&quot; rel=&quot;nofollow&quot;&gt;David Kammler's book&lt;/a&gt; is also a a good introduction.  The first part explains the connections between Fourier Series, Fourier Transform, DTFT, DFT very clearly which other books often gloss over.&lt;/p&gt;&#xA;" LastActivityDate="2010-11-21T20:52:49.513" />
  <row Id="11280" PostTypeId="1" AcceptedAnswerId="11287" CreationDate="2010-11-21T20:54:45.377" Score="2" ViewCount="112" Body="&lt;p&gt;I'm reading Serre's &quot;Lie algebra and Lie groups&quot; now, and I found a description of Lie algebra of an algebraic matrix group, given for a free algebra (free as a module) $k&amp;#39;$ with a basis $\lbrace 1, \varepsilon\rbrace$, $\varepsilon^2 = 0$ as all $X \in M(n,k)$ such that $1 + \varepsilon X \in G(k&amp;#39;)$, where $G(k&amp;#39;)$ is a set of zeros of a number of polinomials.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$k&amp;#39;$ struck me as being extremely close to how smooth infinitesimals are defined in &lt;a href=&quot;http://arxiv.org/abs/0805.3307&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/0805.3307&lt;/a&gt; (a link I found on Wiki)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm a grad student who recently switched from applied math, so I know very little about algebra, differential geometry etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you please clarify how is such a free algebra related to smooth infinitesimals, and when were the latter invented, and how close is synthetic differential geometry to this construction? I currently don't know much about synthetic differential geometry, but I know it uses smooth infinitesimals to define differential, tangent space etc. differently from how they are usually defined.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sorry if the questions are not clear enough :) I want to dive into the interesting subject, I just need a push in the right direction. It amazes me how the example fits the concept of a Lie algebra as being an 'infinitesimal Lie group' when you think of $\varepsilon$ as an infinitesimal, but I don't know what to make out of it, or even if I am right in my intuition.&lt;/p&gt;&#xA;" OwnerUserId="3793" LastEditorUserId="3793" LastEditorDisplayName="" LastEditDate="2010-11-21T21:09:27.877" LastActivityDate="2010-11-21T22:05:46.853" Title="Synthetic differential geometry" Tags="&lt;abstract-algebra&gt;&lt;lie-algebra&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="11281" PostTypeId="2" ParentId="11232" CreationDate="2010-11-21T21:23:42.827" Score="5" ViewCount="" Body="&lt;p&gt;(Posting this as CW so Alex can accept the answer.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Trutheality re-asked the question on MathOverflow, and it turns out the answer is given by what is known as the &quot;Fundamental Theorem of Affine Geometry&quot;. See &lt;a href=&quot;http://mathoverflow.net/questions/46854/continuity-in-terms-of-lines&quot; rel=&quot;nofollow&quot;&gt;http://mathoverflow.net/questions/46854/continuity-in-terms-of-lines&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1543" LastActivityDate="2010-11-21T21:23:42.827" CommunityOwnedDate="2010-11-21T21:23:42.827" CommentCount="2" />
  <row Id="11282" PostTypeId="2" ParentId="11193" CreationDate="2010-11-21T21:33:55.047" Score="1" ViewCount="" Body="&lt;p&gt;The most useful way to think of the $n$-cube is to imagine the set of configurations of $n$-points in the interval. For a more explicit recipe, wiggle your index finger; its set of configurations describes an interval. Wiggle two fingers: now you have a square. Wiggle $4$-fingers and pretend that the motions are independent (or ask an Indian percussionists to demonstrate); the set of configurations describes a $4$-cube. The configuration space method is useful if you want to decompose a (hyper)cube into simplicies. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A beautiful rendition of an $n$-cube is obtained by projecting the unit coordinate vectors to the $n$-th roots of unity. As for &quot;how to&quot; see the MO link above.&lt;/p&gt;&#xA;" OwnerUserId="722" LastActivityDate="2010-11-21T21:33:55.047" CommunityOwnedDate="2010-11-21T21:33:55.047" />
  <row Id="11283" PostTypeId="1" AcceptedAnswerId="11285" CreationDate="2010-11-21T21:35:01.053" Score="1" ViewCount="184" Body="&lt;p&gt;I am being more stupid than usual: embarrassingly so. Let $G$ be a group. Let $\phi:G \rightarrow G$ be an automorphism with inverse $\psi$. Is it true (and why is it true) that $\psi(x)=\phi(x^{-1})$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Right now, I got myself in a gumption trap and have not thought my way out of it.&lt;/p&gt;&#xA;" OwnerUserId="722" LastActivityDate="2010-11-22T00:25:22.773" Title="Inverse of an automorphism" Tags="&lt;group-theory&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="11285" PostTypeId="2" ParentId="11283" CreationDate="2010-11-21T21:44:39.000" Score="6" ViewCount="" Body="&lt;p&gt;This is not true in general. For instance consider the case of the identity $I_{G}:G \rightarrow G $. It is its own inverse, nevertheless it isn't true that $I_{G}(x) = I_{G}(x^{-1})$ because that would imply that $x = x^{-1}$ for every $x \in G$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another counterexample along the same lines. If $G$ is an abelian group then you have available the automorphism $\phi : G \rightarrow G$ given by $\phi (x) = x^{-1}$. Then in this case also $\phi$ is its own inverse and the same argument as before applies.&lt;/p&gt;&#xA;" OwnerUserId="900" LastEditorUserId="900" LastEditorDisplayName="" LastEditDate="2010-11-21T22:05:30.567" LastActivityDate="2010-11-21T22:05:30.567" CommentCount="2" />
  <row Id="11286" PostTypeId="2" ParentId="11283" CreationDate="2010-11-21T21:48:51.587" Score="1" ViewCount="" Body="&lt;p&gt;$\phi(x^{-1})=\phi(x)^{-1}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But $\phi(x)^{-1}= \phi^{-1}(x)$, not is necesarily true .&lt;/p&gt;&#xA;" OwnerUserId="3421" LastActivityDate="2010-11-21T21:48:51.587" />
  <row Id="11287" PostTypeId="2" ParentId="11280" CreationDate="2010-11-21T22:05:46.853" Score="2" ViewCount="" Body="&lt;p&gt;Your intuition is right. This $\varepsilon^2 = 0$ stuff is basically an algebraic way to extract differential information, but it's certainly not as sophisticated as the full edifice of synthetic differential geometry because for polynomials everything is trivial. The basic observation here is that $f(a + \varepsilon b) = f(a) + b f&amp;#39;(a) \varepsilon$ and from here everything else follows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Recall that if $f_1, ... f_k$ are a family of polynomials, then a $k$-homomorphism $k[x_1, ... x_n]/(f_1, ... f_k) \to k$ is the same thing as a $k$-point $(p_1, ... p_n)$ on the variety $f_1 = ... = f_k = 0$.  The appropriate generalization here is that a $k$-homomorphism $k[x_1, ... x_n]/(f_1, ... f_k) \to k[\varepsilon]/\varepsilon^2$ is the same thing as a &quot;generalized point&quot; $(p_1 + \varepsilon q_1, ... p_n + \varepsilon q_n)$ on the variety.  If you write down what it means for $f_1, ... f_k$ to vanish on this point, the condition you get is that $p = (p_1, ... p_n)$ must be a point in the ordinary sense and $(q_1, ... q_n)$ must be orthogonal to the gradients of all the $f_i$; but this is equivalent to being a &lt;strong&gt;tangent vector&lt;/strong&gt; at $p$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is one way to define the &lt;a href=&quot;http://en.wikipedia.org/wiki/Zariski_tangent_space&quot; rel=&quot;nofollow&quot;&gt;Zariski tangent space&lt;/a&gt; at a point of a variety, and from here the algebraic definition of the Lie algebra follows.  Again, since we only have to deal with polynomials, we don't need anything close to the full power of synthetic differential geometry, which is a much more recent development.  &lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-11-21T22:05:46.853" />
  <row Id="11288" PostTypeId="2" ParentId="11246" CreationDate="2010-11-21T22:16:05.513" Score="17" ViewCount="" Body="&lt;p&gt;This is a famous problem! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A proof, which I got from just googling, appears as a solution Problem 218 in the College Mathematics Journal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Snapshot:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/PYq30.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You should be able to find a couple of different proofs more and references here: &lt;a href=&quot;http://arxiv.org/PS_cache/arxiv/pdf/0709/0709.3755v1.pdf&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/PS_cache/arxiv/pdf/0709/0709.3755v1.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-11-21T22:16:05.513" CommentCount="0" />
  <row Id="11289" PostTypeId="2" ParentId="11227" CreationDate="2010-11-21T22:30:29.227" Score="2" ViewCount="" Body="&lt;p&gt;Consider the exact sequence of trivial $G$-modules:&#xA;$$0 \rightarrow \mathbf{Z}/p \rightarrow&#xD;&#xA;\mathbf{Q}_p/\mathbf{Z}_p \rightarrow \mathbf{Q}_p/\mathbf{Z}_p \rightarrow 0,$$&#xD;&#xA;(where the second map is multiplication by $p$) and tensor with $T&amp;#39;$. Since $T&amp;#39;$ is free over $\mathbf{Z}_p$, it is flat, and so the sequence remains exact.&#xA;Thus $T&amp;#39;/p$ injects into $T&amp;#39; \otimes \mathbf{Q}_p/\mathbf{Z}_p$. In particular,&#xA;any non-trivial map $T/p \rightarrow T&amp;#39;/p$ extends to a non-trivial map&#xA;$$T \rightarrow T/p \rightarrow T&amp;#39;/p \rightarrow T&amp;#39; \otimes \mathbf{Q}_p/\mathbf{Z}_p.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, for the &quot;if&quot; part, you only need the hypothesis that $T&amp;#39;$ is flat and there is a non-zero $G$-equivariant map from $T/p$ to $T&amp;#39;/p$ (for any $G$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conversely, any map $T \rightarrow T&amp;#39; \otimes \mathbf{Q}_p/\mathbf{Z}_p$ factors through $T/p^n$ for some $n$, and hence as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$T \rightarrow T/p^n \rightarrow (T&amp;#39; \otimes \mathbf{Q}_p/\mathbf{Z}_p)[p^n]&#xD;&#xA;= T&amp;#39;/p^n.$$&#xD;&#xA;If $T&amp;#39;/p$ is irreducible, then the composition factors of the RHS are all of the form $T&amp;#39;/p$, and hence there is a surjective map $T/p \rightarrow T&amp;#39;/p$, which must be an isomorphism if $T/p$ is also irreducible.&lt;/p&gt;&#xA;" LastActivityDate="2010-11-22T16:29:04.707" CommentCount="2" />
  <row Id="11290" PostTypeId="2" ParentId="11145" CreationDate="2010-11-21T22:41:36.917" Score="1" ViewCount="" Body="&lt;p&gt;Here's an idea, although I don't have time to try it: draw lines at angles of multiples of $\frac{\pi}{4}$ from the origin to both of the unit circles you're investigating and compute the curvature at the corresponding intersection points.  If you can show that the curvatures don't match up, the corresponding unit circles can't be related by a scaling and a rotation.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-11-21T22:41:36.917" CommentCount="4" />
  <row Id="11291" PostTypeId="2" ParentId="11193" CreationDate="2010-11-21T22:51:33.827" Score="0" ViewCount="" Body="&lt;p&gt;Whenever I think about $S^{2n-1}$, I draw $n$ complex planes and a unit disk in each one.  $S^{2n-1}$ is the set of ordered $n$-tuples of points, one from each disk, so that the sum of the squares of their absolute values is one; you can rotate each &quot;coordinate&quot; in a circle in its own plane independently of the others, but if you move it &quot;outwards&quot; or &quot;inwards,&quot; the others have to change in magnitude as well.  From there, it's only a small leap to $\mathbb{CP}^n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For complex functions, which live in 4 dimensions, one idea is to graph either the real part or the magnitude as a function in 3-space, and then use color to represent the imaginary part or the argument.  Or you could just do several different graphs, &lt;a href=&quot;http://mathworld.wolfram.com/ExponentialFunction.html&quot; rel=&quot;nofollow&quot;&gt;like Wolfram MathWorld does&lt;/a&gt;.  This breaks down for things sitting in 4-space that aren't graphs of functions, because you could wind up with two points sharing x-, y-, and z-coordinates but with different &quot;colors.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe people who study knotted surfaces like to use &quot;movies,&quot; each &quot;frame&quot; of which is a 2-d cross section of the surface.  The advantage of this is that the Reidemeister moves for knots generalize to &quot;movie moves&quot; for knotted surfaces.  I don't know that much about this, though.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, my algebraic topology professor told me that he just thinks of CW-complexes in terms of their cell structure.  I asked him to draw $\mathbb{CP}^n$ and he just wrote $1,0,1,0,\dotsc$ in an upwards column, each number representing the number of cells in that dimension.  Obviously, you'd need to take attaching maps into account as well in the general case.  But this lets you just read off homology, etc.  Thinking like this is really why fields like algebraic topology exist: it's easier to visualize the invariants than the actual objects.  I've yet to talk to a geometer about this, though, and it would be interesting to hear how you think of, say, 3-manifolds when it's no longer &quot;up to homotopy equivalence.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the answer to your question is: depends on what you want to visualize!&lt;/p&gt;&#xA;" OwnerUserId="2215" LastActivityDate="2010-11-21T22:51:33.827" CommunityOwnedDate="2010-11-21T22:51:33.827" />
  <row Id="11292" PostTypeId="2" ParentId="11085" CreationDate="2010-11-21T23:30:37.467" Score="-1" ViewCount="" Body="&lt;p&gt;If you use a 60 degree coordinate system, the height of a regular tet is 1. It's volume is 1. It's area is 4.&lt;/p&gt;&#xA;" OwnerUserId="3797" LastActivityDate="2010-11-21T23:30:37.467" CommentCount="3" />
  <row Id="11293" PostTypeId="1" AcceptedAnswerId="11299" CreationDate="2010-11-21T23:42:20.600" Score="0" ViewCount="81" Body="&lt;p&gt;This isn't homework.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Prove or dis-prove the statement: If $R$ is a regular language and both $L$ \ $R$ and $L \cup R$ are context-free, then $L$ is context-free.&lt;/p&gt;&#xA;" OwnerUserId="3051" LastActivityDate="2010-11-22T00:20:03.480" Title="context free languages closure properties" Tags="&lt;computer-science&gt;" AnswerCount="1" />
  <row Id="11294" PostTypeId="1" AcceptedAnswerId="11296" CreationDate="2010-11-21T23:45:02.283" Score="1" ViewCount="267" Body="&lt;p&gt;Let $f(x)= \displaystyle \sum \limits_{n=1}^\infty \frac{\sin(nx)}{n^3}.$ Show that $f(x)$ is differentiable and that the derivative $f&amp;#39;(x)$ is continuous.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In class we solved a similar problem, and I think we had to show that both $f(x)$ and $f&amp;#39;(x)$ converge uniformly, but I am not really sure &lt;em&gt;why&lt;/em&gt; that is what we have to show. I think we also showed that the partial sums are continuous.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would really appreciate it if someone could solve this problem and maybe explain a little the theory behind it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="2467" LastActivityDate="2010-11-22T00:15:14.367" Title="Convergence of Series of Functions" Tags="&lt;real-analysis&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="11295" PostTypeId="2" ParentId="11294" CreationDate="2010-11-21T23:53:47.643" Score="0" ViewCount="" Body="&lt;p&gt;Well, to see that $f(x)$ *converges uniformly* please apply the &lt;em&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Weierstrass_M-test&quot; rel=&quot;nofollow&quot;&gt;Weierstrass M-Test&lt;/a&gt;&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, i am aware of the fact that term by term differentiation and integration is possible if the series is uniformly convergent. But i don't know much about this, i don't want to answer that, but i am sure someone here will help you out regarding this.&lt;/p&gt;&#xA;" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-22T00:00:39.197" LastActivityDate="2010-11-22T00:00:39.197" CommentCount="3" />
  <row Id="11296" PostTypeId="2" ParentId="11294" CreationDate="2010-11-22T00:02:32.597" Score="1" ViewCount="" Body="&lt;p&gt;To show uniform convergence of the original series and the series of derivatives you can use the facts that $\sin$ and $\cos$ are bounded and that $\sum_{n=1}^\infty \frac{1}{n^3}$ and $\sum_{n=1}^\infty \frac{1}{n^2}$ converge.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The uniform limit of continuous functions is continuous.  To see that $f$ is differentiable with derivative equal to the series of derivatives, you can apply an integral convergence theorem.  E.g., uniform convergence allows you to pass limits through integrals, and (using the fundamental theorem of calculus) $f(x)=\sum_{n=1}^\infty \int_0^x \frac{\cos(nt)}{n^2}dt$.  You can use the fundamental theorem of calculus again to finish, also using the fact that the series of derivatives represents a continuous function by the first sentence of this paragraph.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I'm intentionally leaving some details for you to fill in.)&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-11-22T00:02:32.597" CommentCount="4" />
  <row Id="11298" PostTypeId="2" ParentId="11294" CreationDate="2010-11-22T00:15:14.367" Score="1" ViewCount="" Body="&lt;p&gt;You had to show that both $(f_n)$ and $(f_n&amp;#39;)$ converge uniformly because you want to apply a theorem like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Theorem 10.7&lt;/strong&gt; (from &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0521497566&quot; rel=&quot;nofollow&quot;&gt;&quot;Real Analysis&quot;&lt;/a&gt; by N.L. Carothers):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose that $(f_n)$ is a sequence of real-valuled functions, each having a continuous derivative on $[a,b]$, and suppose that the sequence of derivatives $(f_n&amp;#39;)$ converges uniformly to a function $g$ on $[a,b]$. If $(f_n(x_0))$ converges at any point $x_0$ in $[a,b]$, then in fact, $(f_n)$ converges uniformly to a differentiable function $f$ on $[a,b]$. Moreover $f&amp;#39; = g$. That is, $(f_n&amp;#39;)$ converges uniformly to $f&amp;#39;$ on $[a,b]$.&lt;/p&gt;&#xA;" OwnerUserId="1269" LastActivityDate="2010-11-22T00:15:14.367" />
  <row Id="11299" PostTypeId="2" ParentId="11293" CreationDate="2010-11-22T00:20:03.480" Score="2" ViewCount="" Body="&lt;p&gt;Take $R = A^*$, where $A$ is your alphabet.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-22T00:20:03.480" />
  <row Id="11300" PostTypeId="2" ParentId="11283" CreationDate="2010-11-22T00:25:22.773" Score="1" ViewCount="" Body="&lt;p&gt;After brushing off some cobwebs, I thought I might see for which groups the identity can hold.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Claim&lt;/em&gt;: Let $G$ be a finite group for which there exists $\phi \in \mathrm{Aut}(G)$ for which $\phi(x)=\phi^{-1}(x^{-1})$ for all $x \in G$.  Then $G$ is an &lt;a href=&quot;http://en.wikipedia.org/wiki/Elementary_abelian_group#Automorphism_group&quot; rel=&quot;nofollow&quot;&gt;elementary abelian 2-group&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Proof&lt;/em&gt;:  If $\phi(x)=\phi^{-1}(x^{-1})$ for all $x \in G$, then $\phi(\phi(x))=\phi^{-1}(\phi^{-1}(x^{-1}))$ and so on.  Hence $x=x^{-1}$ and $x^2=\mathrm{id}$ for all $x \in G$, and so $G$ must be an elementary abelian 2-group. $\square$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, a converse of the above is true, since each elementary abelian 2-group admits a the identity automorphism $\phi$, which satisfies the identity $\phi(x)=\phi^{-1}(x^{-1})$ for all $x \in G$.  But, as we will see, not all automorphisms of elementary abelian 2-groups satisfy the identity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Claim&lt;/em&gt;:  Let $G$ be a finite group for which each $\phi \in \mathrm{Aut}(G)$ satisfies $\phi(x)=\phi^{-1}(x^{-1})$ for all $x \in G$.  Then $G \cong \mathbb{Z}_2$ or $G$ is the trivial group.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Proof&lt;/em&gt;:  By the previous claim, $G$ must be an elementary abelian 2-group.  Applying $\phi$ to both sides of the identity $\phi(x)=\phi^{-1}(x^{-1})$, we find that $\phi(\phi(x))=x^{-1}=x$ for all $x \in G$.  Hence $\phi^2$ is the identity automorphism for all $\phi \in \mathrm{Aut}(G)$.  Therefore, every automorphism of $G$ must not contain a 3-cycle (in fact t-cycle where $t \geq 3$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The elementary abelian 2-group of order 4 admits an automorphism that contains a 3-cycle.  Consequently, we can prove by induction that elementary abelian 2-group of order $2^a$ for all $a \geq 2$ each admit an automorphism that contains a 3-cycle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The claim is true for the trivial group and $\mathbb{Z}_2$ by inspection. $\square$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="139" LastActivityDate="2010-11-22T00:25:22.773" CommentCount="1" />
  <row Id="11301" PostTypeId="1" AcceptedAnswerId="11309" CreationDate="2010-11-22T00:49:45.567" Score="3" ViewCount="203" Body="&lt;p&gt;Gaussian integers are the set:&#xA;$$\mathbb{Z}[\imath] =\{a+b\imath : a,b \in\mathbb{Z} \}$$&#xD;&#xA;With norm: &#xA;$$\mathrm{N}(a+b\imath)=a^{2}+b^{2}.$$&#xD;&#xA;It satisfies $\mathrm{N}(\alpha\cdot \beta )=\mathrm{N}(\alpha)\cdot\mathrm{N}( \beta )$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The units of $\mathbb{Z}[\imath]$ are  precisely: $1,-1,\imath,-\imath$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Result: If $p=4k+1$ ( p : prime), there are $x&amp;lt;p$  such that $p/(x^{2}+1)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$Theorem$: If $p$ is a prime with&#xA;  $p=4k+1$, then $p$ is an sum of two&#xA;  squares.&#xA;  &lt;em&gt;Proof.&lt;/em&gt;&#xA;  If $p=4k+1$ there are $x&amp;lt;p$ such that $p/(x^{2}+1)$, then  $p/(x+\imath)(x-\imath)$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Note that, if $p/(x+\imath)$, there are $(a+b\imath)$ such that:&#xA;$$(x+\imath)=p(a+b\imath)=pa+pb\imath.$$  &#xA;This implies that $x=pa$, this is imposible as $x\lt p$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore $p\nmid(x+\imath)$, also $p\nmid(x-\imath)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then $p$ is not an Gaussian prime, so $p = \alpha \beta$ with $\mathrm{N}(\alpha)\gt 1$ and $\mathrm{N}(\beta)\gt 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $ \alpha=a+b\imath $ and $\beta=c+d\imath$, we get:&#xA;$$\mathrm{N}(p)=\mathrm{N}(\alpha \beta)=\mathrm{N}\alpha\cdot\mathrm{N}\beta,$$&#xD;&#xA;which implies:&#xA;$$p^{2}=(a^{2}+b^{2})(c^{2}+d^{2}).$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the last equation is an integer, so $(a^{2}+b^{2})|p^{2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By this last equation $(a^{2}+b^{2}),(c^{2}+d^{2})\neq 1,$ so therefore:&#xA;$$p=a^{2}+b^{2}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;**&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;It's beautiful!  Does anyone have other&#xA;  proofs  ?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;**&lt;/p&gt;&#xA;" OwnerUserId="3421" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-22T02:36:58.293" LastActivityDate="2010-11-22T02:42:19.567" Title="Fermat's Christmas theorem on sums of two squares with Gaussian integers" Tags="&lt;number-theory&gt;&lt;prime-numbers&gt;" AnswerCount="4" CommentCount="7" FavoriteCount="1" />
  <row Id="11302" PostTypeId="2" ParentId="11301" CreationDate="2010-11-22T00:55:01.790" Score="0" ViewCount="" Body="&lt;p&gt;This &lt;a href=&quot;http://en.wikipedia.org/wiki/Proofs_of_Fermat%27s_theorem_on_sums_of_two_squares#Lagrange.27s_proof_through_quadratic_forms&quot; rel=&quot;nofollow&quot;&gt;wiki&lt;/a&gt; page has the other proofs which I know of. Other novel proofs are welcome as always.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastActivityDate="2010-11-22T00:55:01.790" />
  <row Id="11303" PostTypeId="2" ParentId="11301" CreationDate="2010-11-22T01:09:03.303" Score="1" ViewCount="" Body="&lt;p&gt;There is a proof by &lt;a href=&quot;http://eprints.maths.ox.ac.uk/677/&quot; rel=&quot;nofollow&quot;&gt;Roger Heath-Brown&lt;/a&gt; and another by &lt;a href=&quot;http://www.jstor.org/pss/2323282&quot; rel=&quot;nofollow&quot;&gt;John Ewell&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Heath-Brown's proof was later adapted into a &quot;one-sentence&quot; proof by Zagier. Zagier's proof is available at the wikipedia link given in Sivaram's answer.&lt;/p&gt;&#xA;" OwnerUserId="1119" LastActivityDate="2010-11-22T01:09:03.303" CommentCount="1" />
  <row Id="11304" PostTypeId="1" AcceptedAnswerId="11318" CreationDate="2010-11-22T01:29:11.273" Score="6" ViewCount="174" Body="&lt;p&gt;Background: Let $k$ be an even integer. The Eisenstein series are defined by $$E_{k} =  1 - \frac{2k}{B_{k}}\sum_{n=1}^{\infty} \sigma_{k-1}(n)q^{n}$$ where &#xA;$$\sigma_{k-1}(n)= \sum\limits_{d \mid n,\;d \geq 1} d^{k-1}$$ and $B_{k}$ is the $k$-th Bernoulli number. If $k \gt 2$ then $E_{k}$ is a modular form of weight $k$. In the case when $k = 2$, $E_{k}$ is not a modular form. Let $P = E_{2}$. Thus $P = 1 - 24\sum_{n = 1}^{\infty} \sigma_{1}(n)q^{n}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $\theta$ be the differential operator defined by $\theta = q \frac{d}{dq}$. If $f$ is a modular form of weight $k$, define $\Delta f = 12 \theta f - kPf$. (Note that the symbol $\circ$ means composition.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let &#xA;$$\alpha =  \left( \begin{array}{cc}&#xD;&#xA;a &amp;amp; b  \\&#xD;&#xA;c &amp;amp; d \\&#xD;&#xA;\end{array} \right) \in \textrm{SL}_{2}(\mathbf{R}).$$&#xD;&#xA;Define &#xA;$$\alpha \tau = \frac{a \tau + b}{c \tau + d}.$$&#xD;&#xA;Define $f \circ [a]_{k}(\tau) = f(\alpha \tau)(c \tau + d)^{-k}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lang claims (as a lemma on page 161 of his &quot;Introduction to Modular Forms&quot;) that &#xA;$\Delta(f \circ [\alpha]_{k}) = (\Delta f) \circ [\alpha]_{k + 2}$. My question is why is this true?&lt;/p&gt;&#xA;" OwnerUserId="4457" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-22T03:03:16.653" LastActivityDate="2010-11-22T06:18:27.670" Title="why does a certain formula in Lang's book on modular forms hold?" Tags="&lt;number-theory&gt;&lt;algebraic-number-theory&gt;&lt;modular-forms&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" />
  <row Id="11305" PostTypeId="1" AcceptedAnswerId="11308" CreationDate="2010-11-22T01:33:31.977" Score="2" ViewCount="67" Body="&lt;p&gt;Hello, I'm working through a little proof of this statement, and here is an excerpt of the problem I'm following:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Suppose $p \equiv 1\pmod{4}$. Show that if $x^2-py^2=1$ the $x$ is odd and $y$ is even. Suppose that ${x_0}^2-p{y_0}^2=1$ with $y_0&amp;gt;0$, $y_0$ minimal. Show that $gcd(x_0+1,x_0-1)=2$. Deduce that one of two cases arises: Case 1: $x_0-1=2pu^2$, $x_0+1=2v^2$. Case 2: $x_0-1=2u^2$, $x_0+1=2pv^2$. Show that in Case 1, $v^2-pu^2=1$ with $|u|&amp;lt;y_0$ a contradiction to the minimality of $y_o$. show that in Case 2, $u^2-pv^2=-1$. Conclude that if $p\equiv 1\pmod{4}$ then the equation $x^2-py^2=-1$ has an integral solution.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I note that $0$ and $1$ are the only squares modulo $4$, thus $x^2$ and $y^2$ can only take values $0,1$ modulo $4$. Then mod 4, $x^2-py^2=1$ reduces to $x^2-y^2=1$, and this can only occur when $x^2\equiv 1$, $y^2\equiv 0$, and thus $x$ is odd, and $y$ is even. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore, since $x_0$ is then odd, $x_0-1$ and $x_0+1$ are both even, so their gcd is at least $2$. Also, any common divisor must divide $(x_0+1)-(x_0-1)=2$, and so the gcd is 2. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I'm not sure why the two cases must arise, as I'm not sure what $u,v$ are. My guess is they are coprime factors of $y_0$? I'm thinking this because since $y_0$ is even, $y_0=2uv$ for some factors $u,v$ of $y_0$. Then $(x_0-1)(x_0+1)=p{y_0}^2=4pu^2v^2=(2v^2)(2pu^2)$? Is this the right train of thought? I'm not sure why it should follow that $(x_0+1),(x_0-1)$ must each be equal to one of $(2u^2),(2pv^2)$, and not some other arrangement. Is it because this is the only way to group the factors such that their products have gcd $2$? If so, I don't see why $u$ or $v$ would be coprime to $p$. Thanks for any explanation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After this, I believe I can show the remaining steps of the problem.&lt;/p&gt;&#xA;" OwnerUserId="1583" LastEditorUserId="1583" LastEditorDisplayName="" LastEditDate="2010-11-22T01:50:42.360" LastActivityDate="2010-11-22T02:06:47.187" Title="Small question in showing if $p\equiv 1\pmod{4}$, then $x^2-py^2=-1$ has integral solution." Tags="&lt;number-theory&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="11306" PostTypeId="1" CreationDate="2010-11-22T01:48:42.800" Score="5" ViewCount="109" Body="&lt;h1&gt;Motivation&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;Suppose that $u \in \mathbb{R}^d$ is a unit-norm vector, $\|u\| = 1$, $a, b, c$ are some positive constants and $\xi \in [0,1]$ is another constant (usually chosen close to 1). I am interested in solving the following problem &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;\sup_{v \in \mathbb{R}^d}\ (\xi + (1-\xi)\|v\|_1^2)\left(\sqrt{&#xD;&#xA;\frac{a \langle u, v \rangle^2 + b}{\xi + (1-\xi)\|v\|_1^2}} - c&#xD;&#xA;\right)&#xD;&#xA;$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;subject to $\|v\| = 1$. &lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Question&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;While any suggestions on how to find an optimal $v$ are welcome, I am specifically interested in the following question. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How to find a vector $v$ that maximizes $\langle v, u \rangle$ and satisfies $\|v\|_2 = 1$ and $\|v\|_1 = x$? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A related question (that may be an easier one): given a vector $v_1$ that maximizes $\langle v_1, u \rangle$ and satisfies $\|v_1\|_2 = 1$ and $\|v_1\|_1 = x$ and another vector $v_2$ that maximizes $\langle v_2, u \rangle$ and satisfies $\|v_2\|_2 = 1$ and $\|v_2\|_1 = x + \delta$, how to find the change $\langle v_1, u \rangle - \langle v_2, u \rangle$ as a function of $\delta$?&lt;/p&gt;&#xA;" OwnerUserId="2384" LastActivityDate="2010-11-22T06:14:24.600" Title="How to compute the change in the angle between two unit norm vectors as the $\ell_1$ norm of one vector changes?" Tags="&lt;geometry&gt;&lt;linear-algebra&gt;&lt;optimization&gt;&lt;euclidean-geometry&gt;" AnswerCount="0" CommentCount="4" />
  <row Id="11307" PostTypeId="1" CreationDate="2010-11-22T02:02:57.120" Score="4" ViewCount="311" Body="&lt;p&gt;If someone could give answers and explain, it would be greatly appreciated. Help required studying for a final.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One hundred tickets, numbered 1,2,3,…,100, are sold to 100 different people for a drawing.&#xA;Four different prizes are awarded, including a grand prize (a trip to Tahiti). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A) How many ways are there to award the prizes?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;B) How many ways are there to award the prizes if the person holding ticket 47 wins the grand prize?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;C) How many ways are there to award the prizes if the person holding ticket 47 wins one of the prizes?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;D) How many ways are there to award the prizes if the person holding ticket 47 does not win a prize?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;E) How many ways are there to award the prizes if the people holding tickets 19 and 47 both win prizes?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;F) How many ways are there to award the prizes if the people holding tickets 19, 47, 73, and 97 all win prizes?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;G) How many ways are there to award the prizes if none of the people holding tickets 19, 47, 73, and 97 wins a prize?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;H) How many ways are there to award the prizes if the grand prize winner is a person holding ticket 18, 47, 73, or 97?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I) How many ways are there to award the prizes if the people holding tickets 19 and 47 win prizes, but the people holding tickets 73 and 97 do not win prizes?&lt;/p&gt;&#xA;" OwnerUserId="3800" LastEditorUserId="2778" LastEditorDisplayName="" LastEditDate="2010-11-22T07:28:25.833" LastActivityDate="2010-11-22T07:28:25.833" Title="How do I do Combinatorics / Counting?" Tags="&lt;combinatorics&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="3" />
  <row Id="11308" PostTypeId="2" ParentId="11305" CreationDate="2010-11-22T02:06:47.187" Score="3" ViewCount="" Body="&lt;p&gt;Firstly, since $(x_0+1)(x_0-1)=x_0^2-1 = py_0^2$, exactly one of $x_0+1$ and $x_0-1$ is divisible by $p$. If you take the common factor of 2 out of both of them and that factor of $p$ out of one of them, then you are left with two &lt;strong&gt;co-prime&lt;/strong&gt; integers whose product is a square, therefore each must be a square by unique factorisation. The co-primality condition is crucial for that step.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So $u$ and $v$ are simply new variables that are introduced to express the fact that either both $\frac{x_0+1}{2p}$ and $\frac{x_0-1}{2}$ or both $\frac{x_0+1}{2}$ and $\frac{x_0-1}{2p}$ are squares.&lt;/p&gt;&#xA;" OwnerUserId="3212" LastActivityDate="2010-11-22T02:06:47.187" CommentCount="2" />
  <row Id="11309" PostTypeId="2" ParentId="11301" CreationDate="2010-11-22T02:09:04.490" Score="1" ViewCount="" Body="&lt;p&gt;This is best viewed from a slightly more general perspective as follows. In any $\rm UFD$, if $\rm\ a\ $ is not prime, i.e.  $\rm\ a\:|\:bc\ $ but $\rm\ a\nmid b,\ a\nmid c\ $ then $\rm\ \gcd(a,b)\ $ is a &lt;em&gt;proper&lt;/em&gt; factor of $\rm\:a\:$. Moreover this gcd can be computed  when a $\rm UFD$ has a constructive Euclidean algorithm, as does $\rm \mathbb Z[i]\:$. Therefore this yields a &lt;em&gt;constructive&lt;/em&gt; proof that nonprime nonunits are reducible in Euclidean domains (i.e. the nontrivial half of the &lt;a href=&quot;http://math.stackexchange.com/questions/4268/alternate-definition-of-prime-number/4399#4399&quot;&gt;equivalence&lt;/a&gt; of irreducible and prime elements in $\rm UFDs$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Applying this above &lt;a href=&quot;http://math.stackexchange.com/questions/5877/efficiently-finding-two-squares-which-sum-to-a-prime/5895#5895&quot;&gt;we deduce&lt;/a&gt; that $\rm\ gcd(p,x-i)\ = a + bi $ is a proper factor of $\rm\:p\:$, so it must have norm a proper factor of $\rm\ N(p) = p^2\ $, i.e. it must have norm $\rm\:p\:$. Therefore $\rm\ p = a^2 + b^2\ $ as desired. This leads to an elegant efficient few-line Euclidean algorithm to compute a representation of a prime $\rm\ p = 4k+1\ $ as a sum of two squares - see my &lt;a href=&quot;http://math.stackexchange.com/questions/5877/efficiently-finding-two-squares-which-sum-to-a-prime/6159#6159&quot;&gt;post here&lt;/a&gt; for an implementation.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-11-22T02:16:12.907" LastActivityDate="2010-11-22T02:16:12.907" />
  <row Id="11311" PostTypeId="2" ParentId="11193" CreationDate="2010-11-22T02:27:06.600" Score="0" ViewCount="" Body="&lt;p&gt;If you want to visualize four dimensional space time, then imagine three dimensional space except each point has a clock; this is just three space where we associate a number with each point.  When the fourth dimension is curved, then time passes at different rates for these clocks.&lt;/p&gt;&#xA;" OwnerUserId="1417" LastActivityDate="2010-11-22T02:27:06.600" CommunityOwnedDate="2010-11-22T02:27:06.600" />
  <row Id="11312" PostTypeId="2" ParentId="11307" CreationDate="2010-11-22T02:33:02.213" Score="17" ViewCount="" Body="&lt;p&gt;The basic rules of counting are:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;Sum rule:&lt;/em&gt; If one event can occur in $n$ different ways, and another, independent, event can occur in $m$ different ways, then the number of different ways in which either one or the other event can occur is $n+m$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;Product rule:&lt;/em&gt; If one event can occur in $n$ different ways, and another, independent, event can occur in $m$ different ways, then the number of different ways in which &lt;em&gt;both&lt;/em&gt; evens can occur is $nm$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;When making choices, you have two basic kinds: if the order in which you make the choices matters, they are called &lt;strong&gt;permutations.&lt;/strong&gt; (For example, choosing the Cesar salad as an appetizer and the shrimp cocktail as the main course is different from choosing the shrimp cocktail as an appetizer and the Cesar salad as a main course). If the order in which you make the choices does not matter, you have &lt;strong&gt;combinations.&lt;/strong&gt; (For example, when choosing teams, it doesn't matter if Bill is chosen first to join team A and Clara is chosen second also for team A, or if Clara is chosen first to join team A and Bill is chosen second also for team A; all that matters is that both Bill and Clara are in team A). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Combining these, you get some basic formulas:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;The number of ways in which you can make $n$ choices, with $m$ options for each, allowing repetitions but where the order of the choice matters (permutations with repetitions), is $m^n$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The number of ways in which you can make $n$ choices, with $m$ possibilities, where the order matters but with repetitions not allowed (permutations without repetitions) is $m(m-1)(m-2)\cdots(m-n+1) = \frac{m!}{(m-n)!}$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The number of ways in which you can make $n$ choices out of $m$ options, if the order does not matter and repetitions are not allowed, is $\binom{m}{n} = \frac{n!}{n!(m-n)!}$ (called &quot;$m$-choose-$n$&quot;, because you are choosing $n$ out of $m$). These are &quot;combinations&quot;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The number of ways in which you can make $n$ choices, with $m$ options, if the order does not matter and repetitions &lt;em&gt;are&lt;/em&gt; allowed is $\binom{m+n-1}{n}$. (Combinations with repetitions).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;You probably knew all that, but still...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That said:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A. You need to choose 4 tickets, out of 100 possibilities; you are not allowed to choose the same ticket twice (no repetitions). Since the prizes are all different, the order in which you make the choices &lt;em&gt;matters&lt;/em&gt; (draw for the 4th prize, then the 3rd prize, then the 2nd prize, then the Grand Prize; or in whichever order you want). So, which formula above applies, and what is the answer?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;B. If ticket 47 will win the Grand Prize, you still need to assign the remaining three prizes. They have to be awarded to people &lt;em&gt;other&lt;/em&gt; than ticket 47, no repetitions, but order still matters. How many options do you have, and how many choices to do you need to make? Order matters, no repetitions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;C. This is similar to the above, but &lt;em&gt;this&lt;/em&gt; time, ticket 47 may win the Grand Prize, the second, the third, or the fourth place prize. Count each of the four outcomes separately, then apply the &lt;em&gt;Sum rule&lt;/em&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;D. Well, if you are going to exclude ticket 47, you have to choose from among the remaining 99 tickets. Order still matters, repetitions are not allowed. So the only difference is the number of options you have.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;E. Well, you have two prizes awarded, and two more to award. First award the other two prizes (i.e., count how many ways you can do that). Then decide which prizes go to tickets 19 and 47: you have four prizes you can award, so you need to choose two prizes (to give to 19 and 47); order matters (first prize chosen goes to 19, second to 47), no repetitions. Count &lt;em&gt;that&lt;/em&gt;. In total, you need to (i) pick to other winners; pick them in order, so first winner gets the top prize not awarded to 19 or 47, second gets the lower prize not awarded); and (ii) pick which prizes to give to 19 and to 47. You need &lt;em&gt;both&lt;/em&gt; things to happen, so you should then use the &lt;em&gt;Product rule.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;F. Now you know who the four winners are; you just need to figure out which prizes they get. Count how many ways you can distribute the prizes among the four winners.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;G. Similar to D, but now you have to exclude four people instead of one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;H. First decide who wins the Grand Prize. Then pick the other three winners. Then use the &lt;em&gt;Product rule.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I. This is a combination of the ideas from E and D. Use the same method as in E to figure out how many ways there are to award prizes to 19 and 47; then figure out how many ways there are to assign the remaining two prizes to the remaining people if you exclude 73 and 97. Then combine the two answers using an appropriate rule mentioned above. &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-22T06:21:04.260" LastActivityDate="2010-11-22T06:21:04.260" CommentCount="2" />
  <row Id="11313" PostTypeId="2" ParentId="11307" CreationDate="2010-11-22T02:35:24.787" Score="0" ViewCount="" Body="&lt;p&gt;Try to solve the questions one award at a time, for each award asking yourself how many people could receive this award? Note that this approach may lead to branching, question C for instance you would get 100*99*98 for the first 3 awards, but the number of persons who may receive the final award is either 1 or 97 depending on whether ticket 47 has already been awarded.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some of the questions may be simpler to answer if you initially consider what awards a specific person or persons may have, and thereafter how the rest may be distributed.&lt;/p&gt;&#xA;" OwnerUserId="3584" LastActivityDate="2010-11-22T02:35:24.787" />
  <row Id="11314" PostTypeId="2" ParentId="11301" CreationDate="2010-11-22T02:42:19.567" Score="1" ViewCount="" Body="&lt;p&gt;There is a nice proof using &lt;a href=&quot;http://en.wikipedia.org/wiki/Minkowskis_theorem&quot; rel=&quot;nofollow&quot;&gt;Minkowski's theorem&lt;/a&gt; which also proves the four-square theorem.  It is Proof #2 &lt;a href=&quot;http://www.artofproblemsolving.com/Forum/blog.php?u=14052&amp;amp;b=10589&amp;amp;&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-11-22T02:42:19.567" />
  <row Id="11315" PostTypeId="1" CreationDate="2010-11-22T03:52:14.470" Score="6" ViewCount="190" Body="&lt;p&gt;Suppose you are given a question that goes like this:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Consider three planets that revolve&#xA;  around a star on separate circular&#xA;  orbits sharing the same orbital plane.&#xA;  The first planet, A, takes 385 days to&#xA;  complete its orbit. The second planet,&#xA;  B, takes 814 days, while the third&#xA;  planet, C, takes 1672 days. Suppose,&#xA;  at one moment in time, the three&#xA;  planets and their star are in one&#xA;  straight line among each other (i.e.&#xA;  they are collinear). Calculate how&#xA;  many days the star and planets will be&#xA;  collinear again.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It looks like a simple question. Obvious enough, there are at least two trivial solutions for this kind of question. The first is the linear product of all the figures: 385 x 814 x 1672 = 523988080 days, while the second is their Least Common Multiple (LCM): LCM(385, 814, 1672) = 2165240 days. But surprisingly, it just takes $\frac{2165240}{39}$ days for the planets to be collinear again, albeit not at the same location. The question does not restrict the answer to be an integer, after all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My &quot;Orbital Collinearity Theorem&quot; :P&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By trial and error, I found out that the number of days for the planets to become collinear again can be calculated by the formula:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\dfrac{LCM(d_1, d_2, ... , d_n)}{M}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where LCM(…) denotes Least Common Multiple and M is the largest integer for which the following modular congruence holds:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{d_1}{G} \equiv \frac{d_2}{G} \equiv ... \equiv \frac{d_n}{G} \pmod{M}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where G is the Greatest Common Divisor of the numbers of days the planets take to complete their respective orbits ($d_1$, $d_2$, ... , $d_n$). In other words, M is the largest integer that can give the same remainder when it divides each of $\frac{d_1}{G}$, $\frac{d_2}{G}$, ... , $\frac{d_n}{G}$. If there is no such M that is larger than 1 (since all integers will give remainder of 0 when divided by 1), the answer will be the LCM, which is very much expected.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For 385, 814 and 1672, the value of M is 39:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$GCD(385, 814, 1672) = 11$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{385}{11}=35$ and $35 \bmod 39 = 35$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{814}{11}=74$ and $74 \bmod 39 = 35$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{1672}{11}=152$ and $152 \bmod 39 = 35$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don’t really have the proof/working since it is trial and error, but so far after testing with many samples the formula seems to hold for each case. So let’s check if after $\frac{2165240}{39}$ days, the three planets are collinear or not:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Planet A: if it takes 385 days to complete its orbit, after $\frac{2165240}{39}$ days, it has traveled $\frac{2165240}{39} \cdot \frac{1}{385}=144 \frac{3080}{15015} = 144 \frac{8}{39}$ cycles&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Planet B: if it takes 814 days to complete its orbit, after $\frac{2165240}{39}$ days, it has traveled $\frac{2165240}{39} \cdot \frac{1}{814}=68 \frac{6512}{31746} = 68 \frac{8}{39}$ cycles&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Planet C: if it takes 1672 days to complete its orbit, after $\frac{2165240}{39}$ days, it has traveled $\frac{2165240}{39} \cdot \frac{1}{1672}=33 \frac{13376}{65208} = 33 \frac{8}{39}$ cycles&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;It should be obvious that the numbers of cycles that the three planets have gone have the same fractional part, which is $\frac{8}{39}$. Meaning that they are all $\frac{8}{39}$ cycle into their current cycle (34th, 69th and 145th). Hence they all form the same angle from their last collinear position, thus they are collinear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Second sample: minute and hour hands&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The minute and hour hands overlap at&#xA;  12 o’clock. At what time will the&#xA;  hands overlap again?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;By considering the hands as planets revolving around the clock center, we can use the formula to solve it. First consider that minute hand takes 60 minutes to complete a cycle, while hour hand takes 720 minutes. LCM(60,720) = 720 and GCD(60, 720) = 60. That will give us M = 11. Thus it will take $\frac{720}{11} = 65.454545454545454545454545454545$ minutes for them to overlap again. And translating that into time gives us 1:05 and 27.27272727272 seconds. Correct. Moreover, this sample is an example of why we must divide the numbers by their GCD to get the M, a step that was not significant with the three planets question; because if we didn’t, M would be 60, giving us 12 minute as the answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;So, my real question is: Is there official name and proof for this formula/theorem?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;p/s: forgive the inappropriate tags since I don't have enough points to add new tags and there are no suitable tags among the existing ones.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;EDIT:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;OK, I've worked out the proof half-way but I still can't connect it to the last piece. I'm appending the half-proof here with hope that any of you guys can help me complete it:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $L = LCM(d_1, d_2, ... , d_n)$. Then it's a given that L is the least number of days for the planets to be collinear again at the same position (i.e. all of them have revolves $c_1, c_2, ... , c_n$ integral number of cycles around the star). Proof by contradiction: suppose there exists another number $K &amp;lt; L$ for which the planets are collinear at the same position, then L must be a common multiple of $d_1, d_2, ... , d_n$, but $L$ is the least of their common multiples, hence contradiction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yet, we still have a possibility that L could be the M-th time the planets are collinear regardless of position. If such $M &amp;gt; 1$ exists, then the number of cycles that the planets have undergone after L days, $c_1, c_2, ... , c_n$, should have the same remainder if divided by M. Therefore, the following modular congruence must hold:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{L}{d_1} \equiv \frac{L}{d_2} \equiv ... \equiv \frac{L}{d_n} \pmod{M}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's where I am now. I'm trying to connect the above congruence to:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{d_1}{G} \equiv \frac{d_2}{G} \equiv ... \equiv \frac{d_n}{G} \pmod{M}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's trivial to do so with $d_1$ and $d_2$ only because $LCM(d_1,d_2)GCD(d_1,d_2) = d_1d_2$ but the relationship between LCM and GCD becomes less simplistic for more than two numbers. The simplest I could find is $\displaystyle a_1 ... a_n = GCD(a_1, a_2, ... a_n) LCM \left( \frac{a_1 ... a_n}{a_1}, \frac{a_1 ... a_n}{a_2}, ... \frac{a_1 ... a_n}{a_n} \right)$ (&lt;a href=&quot;http://math.stackexchange.com/questions/1579/n-ary-version-of-gcda-blcma-b-ab/1581#1581&quot;&gt;from another question in Math SE&lt;/a&gt;) but it cannot be used directly in the formula because the GCD and LCM terms are different.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone help?&lt;/p&gt;&#xA;" OwnerUserId="3803" LastEditorUserId="3803" LastEditorDisplayName="" LastEditDate="2010-11-23T01:31:53.753" LastActivityDate="2010-11-23T01:31:53.753" Title="Is there formal name and proof for this formula/theorem (for now I'm calling it Orbital Collinearity Theorem)?" Tags="&lt;number-theory&gt;&lt;modular-arithmetic&gt;" AnswerCount="1" FavoriteCount="2" />
  <row Id="11316" PostTypeId="2" ParentId="11231" CreationDate="2010-11-22T04:06:05.287" Score="9" ViewCount="" Body="&lt;p&gt;Your question is an interesting one, but I think the reason you haven't received an answer to it yet is that it's ill-constrained.  That's not a criticism but a technical term.  Basically it means that the problem isn't well-defined enough for there to be a clear, distinct answer.  (This is what the second part of Rahul Narain's comment is getting at.)  To borrow from the Math Overflow FAQ: &quot;The site works best for well-defined questions: math questions that actually have a specific answer.&quot;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instead, your question would be great in my mathematical modeling course.  I would pose it and then ask the students what assumptions would need to be made in order to get a well-defined mathematical model, or what aspects of the problem would be parameters to the model that we could vary and then see how the solution changes (such as those Rahul Narain mentions), or what existing mathematical tools we could use to help understand traffic behavior (such as queuing theory, as Yuval Filmus mentions).  An answer to the question would then require some student taking this on as an extended assignment.  He or she would have to make and justify assumptions, create a model (or two or three, which might require some programming), testing it (or them), refine the model(s), vary parameters, and then interpret the output from the model(s).  There probably wouldn't be a single answer but a range of answers along the lines of &quot;Based on these assumptions our model says this.&quot;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I've actually just described the process of mathematical modeling.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another aspect of your question is that traffic problems are hard.  They frequently show up as problems in the &lt;a href=&quot;http://www.comap.com/undergraduate/contests/mcm/&quot; rel=&quot;nofollow&quot;&gt;Mathematical Contest in Modeling&lt;/a&gt; (see, for example, the &lt;a href=&quot;http://www.comap.com/undergraduate/contests/mcm/contests/2009/problems/&quot; rel=&quot;nofollow&quot;&gt;2009&lt;/a&gt; and &lt;a href=&quot;http://www.comap.com/undergraduate/contests/mcm/contests/2005/problems/&quot; rel=&quot;nofollow&quot;&gt;2005&lt;/a&gt; contests).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Traffic problems are an active area of research, too, because there's a lot we don't understand about the way traffic behaves.  There's not even a consensus about the best way to go about modeling traffic.  &lt;a href=&quot;http://math.mit.edu/projects/traffic/&quot; rel=&quot;nofollow&quot;&gt;Fluid flow models&lt;/a&gt; seem to be the most popular, but some people use &lt;a href=&quot;http://www.rci.rutgers.edu/~gursoy/papers/ifacconf-Stochastic%20Traffic%20Flow%20Models-v2.pdf&quot; rel=&quot;nofollow&quot;&gt;queuing&lt;/a&gt;, and I've also seen &lt;a href=&quot;http://link.aip.org/link/?JCNDDM/3/021206/1&quot; rel=&quot;nofollow&quot;&gt;discrete dynamical system&lt;/a&gt; and even &lt;a href=&quot;http://sjsu.rudyrucker.com/~han.jiang/paper/&quot; rel=&quot;nofollow&quot;&gt;cellular automata&lt;/a&gt; models.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're interested in learning more about research on traffic, you could also check out &lt;a href=&quot;http://www.physorg.com/news117283969.html&quot; rel=&quot;nofollow&quot;&gt;this article&lt;/a&gt; (&quot;Traffic jam mystery solved by mathematicians&quot;) or the book &lt;em&gt;&lt;a href=&quot;http://books.google.com/books?hl=en&amp;amp;lr=&amp;amp;id=U-Aj1nvo3sgC&amp;amp;oi=fnd&amp;amp;pg=PA1&amp;amp;dq=mathematical+theories+of+traffic+flow&amp;amp;ots=etyrVdoQko&amp;amp;sig=xiApBvgh8ijvc8HP036Ezg3HZrQ#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;Mathematical Theories of Traffic Flow&lt;/a&gt;&lt;/em&gt;. &lt;/p&gt;&#xA;" OwnerUserId="2370" LastActivityDate="2010-11-22T04:06:05.287" CommentCount="4" />
  <row Id="11317" PostTypeId="2" ParentId="11315" CreationDate="2010-11-22T04:32:23.403" Score="1" ViewCount="" Body="&lt;p&gt;Here's how you'd start working on it. Suppose $t$ is a time in which planets $i$ and $j$ are collinear. Then $t/d_i - t/d_j$ is some integer $n$. Thus $$t = \frac{nd_id_j}{d_i-d_j}.$$ Therefore you're looking for the smallest $t$ in $$\bigcap_{i &amp;lt; j} \frac{d_id_j}{d_i-d_j} \mathbb{Z}_+.$$ We can take the common denominator and find the answer (up to sign) $$\frac{LCM(\{d_id_j\prod_{(l,k)\neq (i,j)} (d_k-d_l) : i&amp;lt;j\})}{\prod_{k &amp;lt; l} (d_k-d_l)}.$$ Of course, that form is not as explicit as yours.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Here is the explicit result (up to sign) for $n=3$, where the orbits are denoted $x,y,z$: $$\frac{LCM(xy(z-x)(z-y),xz(y-x)(y-z),yz(x-y)(x-z))}{(x-y)(x-z)(y-z)}.$$&lt;/p&gt;&#xA;" OwnerUserId="1277" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2010-11-22T08:40:32.490" LastActivityDate="2010-11-22T08:40:32.490" CommentCount="4" />
  <row Id="11318" PostTypeId="2" ParentId="11304" CreationDate="2010-11-22T04:56:32.030" Score="6" ViewCount="" Body="&lt;p&gt;Note to begin with that the symbol $\circ$ does &lt;em&gt;not&lt;/em&gt; mean composition; rather, it means the operation given by the formula further on in your question, namely:&#xA;$$f\circ[\alpha]_k(\tau) = (c\tau + d)^k f(\alpha \tau).$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now to say that $f$ is a modular form of weight $k$ is to say that&#xA;$$f\circ[\alpha]_k(\tau) = f(\tau),$$&#xD;&#xA;for all $\alpha \in SL_2(\mathbb Z),$&#xD;&#xA;or equivalently,&#xA;that $$f(\alpha \tau) = (c\tau + d)^k f(\tau).$$&#xD;&#xA;Differentiating both sides with respect to $\tau,$ &#xA;we find that&#xA;$$f&amp;#39;(\alpha \tau) (c\tau + d)^{-2} = (c\tau + d)^k f&amp;#39;(\tau) + c k (c\tau + d)^{k-1}f(\tau).$$&#xD;&#xA;Since $\theta = \dfrac{1}{2\pi i} \dfrac{d}{d\tau},$&#xD;&#xA;we may rewrite this as&#xA;$$(c\tau + d)^{-(k+2)}(\theta f)(\alpha\tau) = (\theta f)(\tau)  + \dfrac{c k}{2 \pi i(c\tau + d)} f(\tau),$$&#xD;&#xA;or equivalently,&#xA;that $$(\theta f)\circ [\alpha]_{k+2}(\tau) = (\theta f)(\tau) &#xD;&#xA;+ \dfrac{c k}{c \tau + d} f(\tau).$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, &#xA;$$(k P f)\circ[\alpha]_{k+2}(\tau) = k (P\circ[\alpha]_2)(f\circ [\alpha]_k)(\tau),$$&#xD;&#xA;and so the formula you ask about is equivalent to the formula&#xA;$$P\circ [\alpha]_2(\tau) = P(\tau) + \dfrac{12 c}{2 \pi i(c\tau + d)}.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a standard formula for $P$ (which shows that although $P$ is not&#xA;a modular form of weight 2, it is not so far off).&#xA;One way to obtain it is to consider the weight 12 cuspform&#xA;$\Delta = q\prod_{n = 1}^{\infty}(1 - q^n)^{24}.$&#xD;&#xA;(Sorry for the conflict in notation, but $\Delta$ is the standard notation here.&#xA;The operator that you have labelled $\Delta$ is often denoted $\delta$, perhaps &#xA;to avoid precisely this notational conflict.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A computation (using the product formula for $\Delta$) shows that $\theta(\log \Delta) = P.$  &#xA;Now writing down the modularity equation $\Delta(\alpha \tau) =&#xD;&#xA;(c\tau + d)^{12} \Delta$ and applying $\theta$ to the log of each side,&#xA;we find that indeed $P$ satisfies the required functional equation,&#xA;and thus the formula you asked about holds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an aside, note that one can also reverse this last argument: suppose that we know that the formula&#xA;in the question is true.  Applying it to the modular form&#xA;$\Delta$, we find that $\theta \Delta - P \Delta$ is a cuspform of weight $14$, but there are no non-zero such cuspforms. Thus we find that&#xA;$\theta\Delta = P\Delta,$ or equivalently,&#xA;that $\theta(\log \Delta) = P.$  If you look in Serre's &lt;em&gt;Course in arithmetic&lt;/em&gt;,&#xA;you will see that this is how he proves the product formula for $\Delta$:&#xA;he verifies the functional equation for $P$ directly (in the special case&#xA;when $\alpha = \begin{pmatrix}0 &amp;amp; 1 \\ - 1 &amp;amp; 0\end{pmatrix}$; but since&#xA;the functional equation is obvious for $\begin{pmatrix} 1 &amp;amp; 1 \\ 0 &amp;amp; 1 \end{pmatrix},$ and since these two matrices together generate &#xA;$PSL_2({\mathbb Z})$, verifying it for this single choice of $\alpha$ is enough), and then&#xA;essentially applies the argument I just gave to deduce the product formula for &#xA;$\Delta$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Incidentally, I highly recommend Serre's book; as an introduction to modular forms it is both shorter and clearer than Lang's.  If you then want to go further with the theory, there are other, better sources.  (For example, after&#xA;reading Serre's book you could try directly reading Atkin and Lehner's paper on newforms.)&lt;/p&gt;&#xA;" OwnerUserId="221" LastEditorUserId="221" LastEditorDisplayName="" LastEditDate="2010-11-22T06:18:27.670" LastActivityDate="2010-11-22T06:18:27.670" CommentCount="1" />
  <row Id="11319" PostTypeId="1" AcceptedAnswerId="11324" CreationDate="2010-11-22T05:26:02.240" Score="2" ViewCount="120" Body="&lt;p&gt;The question :&#xA;Prove that :$$ \text{ if } y = 2x^2 - 1,\text{ then } \biggl[ \frac{1}{y} + \frac{1}{3y^3} + \frac{1}{5y^5}+ \cdots \biggr]$$  is equal to $$\frac{1}{2} \biggl[ \frac{1}{x^2} + \frac{1}{2x^4} + \frac{1}{3x^6} + \cdots \biggr]$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here I have modified the question, in the actual question (from my paper) there were four other options given,my approach was to reduce the first expression to $\frac{1}{2} \ln \biggl( \frac{y+1}{y-1} \biggr) $  and then trying to check for each options to find the match, now is there any other approach for this one ? Since it took me sometime for checking each options :( and the desired answer is at the last option!&lt;/p&gt;&#xA;" OwnerUserId="2109" LastActivityDate="2010-11-22T05:46:32.290" Title="Any tricky method to solve this one ?" Tags="&lt;algebra-precalculus&gt;&lt;logarithms&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" />
  <row Id="11320" PostTypeId="1" AcceptedAnswerId="11323" CreationDate="2010-11-22T05:35:39.357" Score="5" ViewCount="302" Body="&lt;p&gt;This question might turn out to be really trivial.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f$ is a one-to-one function from the interval $[0,1]$ to $\mathbb{R}$. Is it necessary that $\exists q \in \mathbb{Q}$ such that $f(x) = q$ for some $x \in [0,1]$ i.e. is it necessary that the image of $f$ contains a rational number?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I came across this question when I was browsing through some website.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think this is false. But I am unable to come up with a counter example.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2010-11-22T06:40:09.337" LastActivityDate="2010-11-22T06:40:09.337" Title="One-to-one function from the interval $[0,1]$ to $\mathbb{R}\setminus\mathbb{Q}$?" Tags="&lt;real-analysis&gt;&lt;functions&gt;" AnswerCount="2" CommentCount="5" />
  <row Id="11321" PostTypeId="1" AcceptedAnswerId="11326" CreationDate="2010-11-22T05:38:38.167" Score="3" ViewCount="179" Body="&lt;p&gt;I'm interested because I want to show that $x^2-34y^2\equiv -1\pmod{m}$ has solutions for all integers $m$. I started by using the following reasoning:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $3\nmid m$, then $gcd(m,3)=1$. Then there exists a multiplicative inverse $\bar{3}$ modulo $m$. I note that $5^2-34=-(3^2)$, and thus $\bar{3}^2(5^2-34)\equiv (\bar{3}\cdot 5)^2-34(\bar{3}^2)\equiv -(\bar{3})^2(3^2) \equiv -1\pmod{4}$. And thus $(\bar{3}\cdot 5, \bar{3})$ is a solution modulo $m$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, if $5\nmid m$, then $(m,5)=1$. Then since $3^2-34=-(5^2)$, then I also have $\bar{5}^2(3^2-34)\equiv (\bar{5}\cdot 3)^2-34(\bar{5}^2)\equiv -(\bar{5})^2(5^2)\equiv -1\pmod{m}$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So for any $m$ not divisible by $3$ or $5$, there exists a solution. Then for $m$ such that $3|m$ and $5|m$, then $m$ has prime factorization $m=3^a5^b{p_1}^{q_1}\cdots {p_r}^{q_r}$. This would give the system of congruences&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x^2-34y^2 \equiv -1 \pmod{3^a},&#xD;&#xA;x^2-34y^2 \equiv -1 \pmod{5^b},&#xD;&#xA;x^2-34y^2 \equiv -1 \pmod{{p_i}^{q_i}}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then $5\nmid 3^a$, $3\nmid 5^b$, and $3\nmid {p_i}^{q_i}$ and $5\nmid {p_i}^{q_i}$, so each of the congruences has a solution. Does the Chinese Remainder Theorem then imply that there is a solution modulo $m$? I know that it holds for polynomials in one variable $x$, and that the number of solutions is the product of the number of solutions for each prime power modulus. Would the same result hold now that there are two variables in the polynomial? I haven't found any proofs to support or contradict the result. Thanks!&lt;/p&gt;&#xA;" OwnerUserId="1583" LastEditorUserId="1583" LastEditorDisplayName="" LastEditDate="2010-11-22T05:55:31.483" LastActivityDate="2010-11-22T06:04:20.537" Title="How does the Chinese Remainder Theorem behave with polynomials with multiple variables?" Tags="&lt;number-theory&gt;&lt;modular-arithmetic&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="11322" PostTypeId="2" ParentId="11320" CreationDate="2010-11-22T05:40:11.580" Score="3" ViewCount="" Body="&lt;p&gt;It is obviously false since $[0,1]$ and $\mathbb{R}\setminus\mathbb{Q}$ have the same cardinality, so there exists a function $f: [0,1] \to \mathbb{R}\setminus\mathbb{Q}$ that is bijective. Do codomain expansion and you get an injective map from the unit interval to $\mathbb{R}$ that has no rationals in its range.&lt;/p&gt;&#xA;" OwnerUserId="1269" LastActivityDate="2010-11-22T05:40:11.580" CommentCount="1" />
  <row Id="11323" PostTypeId="2" ParentId="11320" CreationDate="2010-11-22T05:43:54.097" Score="12" ViewCount="" Body="&lt;p&gt;If you want an explicit counterexample, define $f:[0,1]\to\mathbb{R}$ by $f(x)=x$ if $x$ is irrational, $f(x)=\sqrt2+x$ if $x$ is rational.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2010-11-22T05:56:41.603" LastActivityDate="2010-11-22T05:56:41.603" CommentCount="2" />
  <row Id="11324" PostTypeId="2" ParentId="11319" CreationDate="2010-11-22T05:46:32.290" Score="3" ViewCount="" Body="&lt;p&gt;One you simplify to $\frac{1}{2}\log(\frac{y+1}{y-1})$, plug in $y = 2x^2 -1$ and simplify it to get $-\frac{1}{2} \log(1-\frac{1}{x^2})$. Now expand to get the desired answer.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastActivityDate="2010-11-22T05:46:32.290" />
  <row Id="11326" PostTypeId="2" ParentId="11321" CreationDate="2010-11-22T05:57:58.547" Score="4" ViewCount="" Body="&lt;p&gt;The Chinese Remainder Theorem says that if the modulii $m_1,\ldots,m_k$ are pairwise relatively prime, and $a_1,\ldots,a_k$ are arbitrary, then there is a solution $x$ to the congruence system&#xA;\begin{align*}&#xA;x &amp;amp;\equiv a_1 &amp;amp;\pmod{m_1}\\&#xA;x &amp;amp;\equiv a_2 &amp;amp;\pmod{m_2}\\&#xA;&amp;amp;\vdots\\&#xA;x &amp;amp;\equiv a_k &amp;amp;\pmod{m_k}&#xA;\end{align*}&#xA;and the solution is unique modulo $m_1\cdots m_k$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, for each $i$ you find a value $r_i$ and a value $s_i$ such that $r_i^2 - 34s_i^2 \equiv -1 \pmod{p_i^{q_i}}$, and values $r$ and $s$ for the congruence modulo $3^a$, and values $r&amp;#39;$ and $s&amp;#39;$ for the congruence modulo $5^b$. Apply the Chinese Remainder Theorem to the $r_i$, $r$, and $r&amp;#39;$ (with appropriate modulii) to get a single value of $x$ that is congruent to what you want for each congruence. Do the same with the $s_i$, $s$, and $s&amp;#39;$ to get a value for $y$.  The single value of $x$ and $y$ is the solution you want.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice that you are not applying the Chinese Remainder Theorem to a 2-variable polynomial (or to a one-variable polynomial other than $p(x)=x$, for that matter): rather, you are finding the values you want for each variable for each prime power, and then you are using the Chinese Remainder Theorem to find a single value that will have the appropriate remainders for each of the variables separately.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-22T06:04:20.537" LastActivityDate="2010-11-22T06:04:20.537" CommentCount="2" />
  <row Id="11329" PostTypeId="1" AcceptedAnswerId="11333" CreationDate="2010-11-22T07:45:18.040" Score="2" ViewCount="74" Body="&lt;p&gt;I've got a multiplicative-with-noise model  $F(x,y)=S(x)*R(y)*D(x,y)+N$, where $S(x)$ and $R(y)$ are unknown functions, $D(x,y)$ is a distance function, that is, a function that depends only on $|x-y|$ and decreases quickly when distance increases. $N$ is an uncorrelated &quot;small random noise&quot; function. All of the functions except noise are positive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have $F(x,y)$ sampled for almost any not-very-distant pair of discrete $(x,y)$, that is, for $(x,y): |x-y| \leq D_{max}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd like to decompose $F(x,y)$ to obtain the &quot;form&quot; of $S(x), R(y), D(x,y)$. How can I do it? My only idea is to calculate averages $S_{est}(x_i)=average(F(x,y), x=x_i)$, $R_{est}(y_j)=average(F(x,y),y=y_j)$ etc, assume $N=0$, then produce an estimated $F_{est}=S_{est}(x)*R_{est}(y)*D_{est}(x,y)$. I dont know what to do next.&lt;/p&gt;&#xA;" OwnerUserId="3806" LastActivityDate="2010-11-22T08:38:18.530" Title="How to decompose a result of a multiplication?" Tags="&lt;approximation&gt;" AnswerCount="1" />
  <row Id="11330" PostTypeId="1" AcceptedAnswerId="11334" CreationDate="2010-11-22T07:59:53.717" Score="5" ViewCount="259" Body="&lt;p&gt;Is there a proof in the spirit of Euclid to prove Dirichlet's theorem on primes in arithmetic progression? (By the spirit of Euclid, I mean assuming finite number of primes we try to construct another number which has a prime factor which falls in the same equivalence class as the other primes but the number is not divisible by any of the primes we considered in the initial list.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am aware of the proof using $L$ functions but I am curious to know if Euclid's &quot;simple&quot; idea can be extended to all other cases as well. I tried googling but am unable to find a proof other than the ones relying on $L$ functions. If it is not possible, to what cases can Euclid's idea be extended to?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any other proofs are welcome as well.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastActivityDate="2010-11-22T09:27:41.657" Title="Dirichlet's theorem on primes in arithmetic progression" Tags="&lt;number-theory&gt;&lt;prime-numbers&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="11331" PostTypeId="2" ParentId="9807" CreationDate="2010-11-22T08:11:11.890" Score="2" ViewCount="" Body="&lt;p&gt;I got the same answer as the book: 20,077,200&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And here's the equation: 27 * 26 * 26 * 11 * 10 * 10&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Simply put, since the license plate can have 2 or 3 letters and 2 or 3 numbers, you can consider the first letter and first number to have extra one choice: the 'null' letter/number!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the first letter can be A to Z or null: 27 choices&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second and third letters can be A to Z: 26 choices each&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first number can be 0 to 9 or null: 11 choices&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second and third letters can be 0 to 9: 10 choices each&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Multiply them together to get 20077200.&lt;/p&gt;&#xA;" OwnerUserId="3803" LastActivityDate="2010-11-22T08:11:11.890" />
  <row Id="11332" PostTypeId="2" ParentId="11330" CreationDate="2010-11-22T08:30:28.940" Score="1" ViewCount="" Body="&lt;p&gt;There are similar proofs for primes of the forms $4n+1$ and $4n+3$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose $p_1,\ldots,p_k$ are all primes of the form $4n+1$. Consider $$P = (p_1 \cdots p_k)^2 + 2^2.$$ Since $P$ is a sum of two squares but not a square (or twice a square) itself, it must have a prime factor $q$ of the type $4n+1$ (by Fermat's theorem on numbers representable as sums of two squares). Since $P \mod p_i = 4 \mod p_i \neq 0$, $q$ is a &lt;i&gt;new&lt;/i&gt; prime of the form $4n+1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose $p_1,\ldots,p_k$ are all primes of the form $4n+3$. Consider $P = p_1 \cdots p_k + C$, where $C$ is chosen to be either $2$ or $4$ so that $P$ itself is of the form $4n+3$. Since primes of the form $4n+1$ never multiply to a number of the form $4n+3$, $P$ must have a prime number $q$ of the type $4n+3$. As before, $q$ must be new.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-22T08:30:28.940" CommentCount="1" />
  <row Id="11333" PostTypeId="2" ParentId="11329" CreationDate="2010-11-22T08:38:18.530" Score="3" ViewCount="" Body="&lt;p&gt;The standard way would be to minimize an error function, perhaps least-min-squares $$\mathbb{E} [F(x,y) - S(x)R(y)D(|x-y|)]^2,$$ where $S,R,D$ are all variables. You can also add a regularization term reflecting your belief that $D(d)$ is small for large $d$, e.g. something like $\sum c_d D(d)^2$ for some chosen weights $c_d \rightarrow \infty$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now that you have a clear goal, you can try to find an explicit solution by finding the minimum analytically. Alternatively, use any standard iterative minimizing technique in the literature, for example gradient descent, to solve the problem in practice.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-22T08:38:18.530" CommentCount="4" />
  <row Id="11334" PostTypeId="2" ParentId="11330" CreationDate="2010-11-22T08:43:21.320" Score="11" ViewCount="" Body="&lt;p&gt;As Jonas says, Keith Conrad's paper will tell you that the answer is essentially no.  He defines a certain notion of &quot;Euclidean&quot; proof coming from writing down a polynomial with certain properties and gives two classic results which say that these proofs exist for primes in arithmetic progression $a \bmod n$ if and only if $a^2 \equiv 1 \bmod n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first case these proofs can't handle is primes congruent to $2 \bmod 5$ (equivalently, primes congruent to $3 \bmod 5$).  The basic problem is that there is no way to force a positive integer to have factors congruent to $2 \bmod 5$ and not congruent to $3 \bmod 5$ (or the other way around) solely by controlling its residue class $\bmod 5$.  In more sophisticated language, $2$ and $3$ lie in all the same subgroups of $(\mathbb{Z}/5\mathbb{Z})^{\ast}$.  &lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-11-22T08:43:21.320" CommentCount="1" />
  <row Id="11335" PostTypeId="1" CreationDate="2010-11-22T08:46:26.950" Score="2" ViewCount="92" Body="&lt;p&gt;Suppose that a in the plane a given conic curve is compelled to pass through two fixed points of that plane.&#xA;What are the curves covered by a fixed point of the conic, its center (for an ellipse), its focus, etc ?&#xA;(I apologize for the bad english ...)  &lt;/p&gt;&#xA;" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2011-04-29T16:44:36.020" LastActivityDate="2011-04-29T16:44:36.020" Title="Curve of a fixed point of a conic compelled to pass through 2 points" Tags="&lt;geometry&gt;&lt;euclidean-geometry&gt;&lt;analytic-geometry&gt;&lt;plane-curves&gt;" CommentCount="5" />
  <row Id="11337" PostTypeId="2" ParentId="11330" CreationDate="2010-11-22T09:27:41.657" Score="5" ViewCount="" Body="&lt;p&gt;A couple more references:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ram Murty and Nithum Thain,&#xA;&lt;a href=&quot;http://www.mast.queensu.ca/~murty/murty-thain2.pdf&quot; rel=&quot;nofollow&quot;&gt;Prime numbers in certain arithmetic progressions&lt;/a&gt;,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Paul Pollack, &#xA;&lt;a href=&quot;http://www.math.uiuc.edu/~pppollac/impossibility.pdf&quot; rel=&quot;nofollow&quot;&gt;Hypothesis H and an impossiblity theorem of Ram Murty&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-11-22T09:27:41.657" />
  <row Id="11338" PostTypeId="2" ParentId="11210" CreationDate="2010-11-22T10:31:59.510" Score="3" ViewCount="" Body="&lt;p&gt;Write $a_n={2+b_n \over 1-b_n}$. Then $b_{n+1}={3\over5} b_n$. This implies $\lim_{n\to\infty} b_n=0$ and $\lim_{n\to\infty} a_n=2$, unless $b_n=1$ for some $n\ge0$. This is the case if $b_0=(5/3)^k$ for some $k\ge0$. In addition there is the fixed point $a_0=a_1=\ldots=-1$ which is not covered by this &quot;Ansatz&quot;.&lt;/p&gt;&#xA;" OwnerUserId="1303" LastActivityDate="2010-11-22T10:31:59.510" CommentCount="2" />
  <row Id="11339" PostTypeId="1" CreationDate="2010-11-22T10:44:39.157" Score="0" ViewCount="122" Body="&lt;p&gt;Suppose $T_1, T_2 : \ell^2(Z_N)-&amp;gt; \ell^2(Z_N)$ are translation invariant linear transformation. How to prove that the composition $T_2 \circ T_1$ is translation invariant?&#xA;Recall the translation operator $R_k$ defined by &#xA;$(R_k z)(n)=z(n-k) $&#xD;&#xA;Also recall definition:&#xA;Let $T : \ell^2(Z_N)-&amp;gt; \ell^2(Z_N)$ be a linear transformation. $T$ is translation invariant if &#xA;$T(R_k z)=R_k T(z)$&lt;/p&gt;&#xA;" OwnerUserId="2723" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-22T15:28:30.147" LastActivityDate="2010-12-22T18:44:26.177" Title="How to prove that the composition $T_2 \circ T_1$ is translation invariant?" Tags="&lt;linear-algebra&gt;&lt;homework&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="11340" PostTypeId="1" CreationDate="2010-11-22T11:35:01.250" Score="2" ViewCount="135" Body="&lt;p&gt;I ask this in a partly recreational, and partly research-related spirit, and I realize my problem might be ill-posed, so any suggestions for clarification might go a long way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Succinctly, my problem can be stated as&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Find the period of a &quot;black box&quot;&#xA;  pseudorandom number generator&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;but I suppose the following formulation might be more attractive:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most programs for playing MP3 files (and MP3-playing devices as well) possess a &quot;shuffle&quot; feature that allows the player to play a list of songs in essentially &quot;random&quot; order.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you run a music player set to shuffle long enough, you will note that sometimes the time interval between the instances your favorite song in the playlist can be short, or can be long.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, for a number of reasons, you can't take a look at what pseudorandom number generator your music player is using; you are however interested in how frequently your favorite song in the playlist would be played.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, one now asks,&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Can you determine how often your favorite song in the playlist would be played based on the time intervals between the different instances your song is played?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;How long should your playlist be? Obviously, you can't conclude much if you only have two songs in your playlist...&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;How long should you leave your music player playing to answer question number 1?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I should probably make the drastic assumption that all the songs in the playlist have the same length, e.g. 3 minutes or so.&lt;/p&gt;&#xA;" OwnerUserId="498" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-22T11:41:58.763" LastActivityDate="2010-11-23T01:06:46.327" Title="Determining the period of a music player's &quot;shuffle&quot; feature." Tags="&lt;random&gt;" AnswerCount="1" CommentCount="9" />
  <row Id="11341" PostTypeId="1" CreationDate="2010-11-22T11:41:21.037" Score="0" ViewCount="253" Body="&lt;p&gt;How to prove that convolution is associative, that is &#xA;$(x*y)*z = x*(y*z)$, for $x,y,z \in l^2(Z_N)$. Either directly from the definition of convolution( Definition of convolution: For $z,w \in l^2(Z_N)$, the convolution $z*w \in l^2(Z_N)$ is the vector with components &#xA;$z*w(m) = \sum_{n=0}^{N-1} z(m-n)w(n)$ ) or using following lemma:Suppose $z,w \in l^2(Z_N)$. Then for each m, $(z*w)\hat{}(m) = \hat{z(m)}\hat{w(m)}$( where ^ is discrete fourier transform $\sum_{n=0}^{N-1} z(n)e^{-2 \pi i m n/N}$) and also Fourier inversion formula: $z(n) = 1/N \sum_{m=0}^{N-1} \hat{z(m)} e^{2 \pi i m n/N}$, for $n=0,1,\dots, N-1$.&#xA;In my solution I have used lemma above and Fourier inversion formula as following:&#xA;\begin{align*}&#xA;(x*y)*z(n) &amp;amp;= \frac{1}{N} \sum_{m=0}^{N-1} ((x*y)*z)\widehat{(m)}e^{2 \pi i m n/N}\\ &#xA;&amp;amp;= \frac{1}{N} \sum_{m=0}^{N-1} (x*y)\widehat{(m)}*\widehat{z(m)}e^{2 \pi i m n/N}\\&#xA;&amp;amp;= \frac{1}{N} \sum_{m=0}^{N-1} \widehat{x(m)}\widehat{y(m)}\widehat{z(m)}e^{2 \pi i m n/N} \\&#xA;&amp;amp;= &#xA;\frac{1}{N} \sum_{m=0}^{N-1} \widehat{x(m)}(y*z)\widehat{(m)}e^{2 \pi i m n/N}\\&#xA;&amp;amp;= \frac{1}{N} \sum_{m=0}^{N-1} (x*(y*z))\widehat{(m)}e^{2 \pi i m n/N}\\&#xA;&amp;amp;= x*(y*z).&#xA;\end{align*}&lt;/p&gt;&#xA;" OwnerUserId="2723" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-22T15:33:10.703" LastActivityDate="2010-11-22T15:33:10.703" Title="How to prove that convolution is associative?" Tags="&lt;linear-algebra&gt;&lt;homework&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="11342" PostTypeId="1" CreationDate="2010-11-22T12:15:12.087" Score="0" ViewCount="88" Body="&lt;p&gt;Suppose $b_1, b_2 \in l^2(Z_N)$. How to prove that the composition $T_{b_2} \circ T_{b_1} $ of the convolution operators $T_{b_2}$ and $T_{b_1}$ is the convolution operator $T_b$ with $b=b_2 \cdot b_1$? Hint use previous question where I asked: How to prove that convolution is associative? and also &#xA;$b_2 \cdot b_1(m) = \sum_{n=0}^{N-1} b_2(m-n)b_1(n)$.&lt;/p&gt;&#xA;" OwnerUserId="2723" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-22T15:27:17.357" LastActivityDate="2010-12-23T14:44:26.140" Title="How to prove that the composition $T_{b_2} \circ T_{b_1} $ of the convolution operators $T_{b_2}$ and $T_{b_1}$ is the convolution operator $T_b$ ?" Tags="&lt;linear-algebra&gt;&lt;homework&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="11343" PostTypeId="1" CreationDate="2010-11-22T12:36:51.950" Score="1" ViewCount="180" Body="&lt;p&gt;$$\sum_{i=1}^n \frac1{4i-1}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know I have to integrate the function but from what to find  lower and upper bound. &lt;/p&gt;&#xA;" OwnerUserId="3741" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2010-11-22T16:30:19.150" LastActivityDate="2010-11-23T01:18:05.417" Title="how to find bounds of the series $\sum_{i=1}^n \frac1{4i-1}$" Tags="&lt;calculus&gt;&lt;sequences-and-series&gt;" AnswerCount="3" />
  <row Id="11344" PostTypeId="2" ParentId="11343" CreationDate="2010-11-22T13:01:01.287" Score="1" ViewCount="" Body="&lt;p&gt;$\sum_{i=1}^n \frac{1}{4i - 1} = \frac{1}{3} + \sum_{i=2}^n \frac{1}{4i-1}.$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now apply the usual technique for obtaining a lower bound on the sum on the right.&lt;/p&gt;&#xA;" OwnerUserId="247" LastActivityDate="2010-11-22T13:01:01.287" CommentCount="2" />
  <row Id="11345" PostTypeId="2" ParentId="10693" CreationDate="2010-11-22T13:08:27.933" Score="1" ViewCount="" Body="&lt;p&gt;In my previous answer, I have shown that definition (1) implies definition (2). Now I prove the converse implication.&#xA;Suppose that the condition in definition (2) holds. That is, $\mu_n ((a,b))$ converges for any $a,b \in D$, $a &amp;lt; b$, where $D$ is a dense subset of $\mathbb{R}$. We are now going to specify the limit. For any $a,b \in D$, $a &amp;lt; b$, define&#xA;$$&#xD;&#xA;\mu ((a,b)) = \mathop {\lim }\limits_{n \to \infty } \mu _n ((a,b)).&#xD;&#xA;$$&#xD;&#xA;Since $\mu_n$ is a sequence of subprobability measures (s.p.m.), $\mu ((a,b)) \in [0,1]$.&#xA;Next, fix any $a,b \in {\mathbb R}$, $a &amp;lt; b$. If $a_m$ is a decreasing sequence converging to $a$, and $b_m$ is an increasing sequence converging to $b$, $a_m,b_m \in D$, then, by the last definition, $\mu ((a_m,b_m))$ is a monotone increasing sequence, bounded from above by $1$. Hence, this sequence converges, and we can define&#xA;$$&#xD;&#xA;\mu ((a,b)) = \mathop {\lim }\limits_{m \to \infty } \mu ((a_m ,b_m )) \in [0,1].&#xD;&#xA;$$&#xD;&#xA;This actually determines a s.p.m. $\mu$ on $\mathbb{R}$. Now, define $\tilde D = \lbrace x \in {\mathbb R}:\mu (\lbrace x \rbrace ) = 0 \rbrace$ and proceed similarly to my previous answer, based on &#xA;$$&#xD;&#xA;\mu _n ((a&amp;#39;,b&amp;#39;)) \le \mu _n ((a,b]) \le \mu _n ((a&amp;#39;&amp;#39;,b&amp;#39;&amp;#39;)),&#xD;&#xA;$$&#xD;&#xA;to conclude that for any $a &amp;lt; b$ in (the dense set) $\tilde D$, $\mu _n ((a,b]) \to \mu ((a,b])$.&lt;/p&gt;&#xA;" OwnerUserId="2810" LastActivityDate="2010-11-22T13:08:27.933" />
  <row Id="11346" PostTypeId="1" AcceptedAnswerId="17394" CreationDate="2010-11-22T13:15:22.943" Score="7" ViewCount="483" Body="&lt;p&gt;Given two vectors $u, v \in \mathbb{R}^d$ represented the spherical coordinates is there a simple formula to compute the angle between the two vectors? Without loss of generality, we can assume that the vectors $u$ and $v$ have unit norm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not sure that the following notation is standard, but I assume that the vector $u$ is defined with $\rho = 1$ and the angular components $\theta_1, \ldots, \theta_{d-1}$. We can obtain the Euclidean components as follows: $$u_{x_1} = \cos \theta_1$$ $$u_{x_2} = \sin \theta_1 \cos \theta_2$$ $$\ldots$$ $$u_{x_{d-1}} = \sin \theta_1 \ldots \sin \theta_{d-2} \cos \theta_{d-1}$$ $$u_{x_d} = \sin \theta_1 \ldots \sin \theta_{d-2} \sin \theta_{d-1}.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One way to find the angle is to represent the two vectors in the Euclidean coordinates and compute arccos of the dot product. Is there a simpler way?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As pointed out in comments, is there a generalization of the Haversine formula?&lt;/p&gt;&#xA;" OwnerUserId="2384" LastEditorUserId="2384" LastEditorDisplayName="" LastEditDate="2010-11-23T10:02:27.780" LastActivityDate="2011-04-15T15:18:20.030" Title="How to compute the angle between two vectors expressed in the spherical coordinates?" Tags="&lt;geometry&gt;" AnswerCount="3" CommentCount="12" FavoriteCount="3" />
  <row Id="11347" PostTypeId="2" ParentId="11212" CreationDate="2010-11-22T13:28:11.667" Score="0" ViewCount="" Body="&lt;p&gt;This link is really, really good!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fourier Transform in one day&#xA;&lt;a href=&quot;http://www.dspdimension.com/admin/dft-a-pied/&quot; rel=&quot;nofollow&quot;&gt;http://www.dspdimension.com/admin/dft-a-pied/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3783" LastActivityDate="2010-11-22T13:28:11.667" />
  <row Id="11348" PostTypeId="1" CreationDate="2010-11-22T13:29:14.397" Score="0" ViewCount="59" Body="&lt;p&gt;If you have a non-causal function (i.e. the Hodrick-Prescott filter) and you want to turn this into a causal filter that resembles it as closely as possible, how would you do this? Train a  Multilayer feedforward network? Try to find a FIR filter with .. coefficients? Fuzzy systems? A RBF network?&lt;/p&gt;&#xA;" LastActivityDate="2010-11-22T13:29:14.397" Title="turn non-causal spline into causal filter" Tags="&lt;statistics&gt;" />
  <row Id="11349" PostTypeId="1" CreationDate="2010-11-22T14:04:24.587" Score="0" ViewCount="72" Body="&lt;p&gt;I was recently studying some literature on chaos theory and non-linear equations . where in various ciphers were created using non- linear equations like Lorenz equation . Are the data generated from these equations truly random , is it true that the cipher created using non -linear equations are much stronger than AES and other ciphers used currently.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS: I am very naive in maths and I apologies , if this sounds ridiculous. &lt;/p&gt;&#xA;" OwnerUserId="3814" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-22T14:49:08.100" LastActivityDate="2010-11-22T17:52:06.023" Title="Are the values generated by non-linear equations truly random ?" Tags="&lt;random&gt;&lt;dynamical-systems&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="11350" PostTypeId="2" ParentId="11341" CreationDate="2010-11-22T14:16:44.083" Score="3" ViewCount="" Body="&lt;p&gt;This (very standard) computation is done in more generality on Proposition 46, p. 41 of &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://math.uga.edu/~pete/integral.pdf&quot; rel=&quot;nofollow&quot;&gt;http://math.uga.edu/~pete/integral.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The OP is asking for the special case in which the algebra $R$ is $\mathbb{C}$ and the semigroup $M$ is the (group) $(\mathbb{Z}/N\mathbb{Z},+)$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course use of the language of algebras and semigroups is overkill for this particular special case.  But if you look at the three lines of computation in the middle of the page, you see that the point is just to expand each of $(x*y)*z$ and $x*(y*z)$ into a triple sum and then notice that these two triple sums are identical.  In other words, &lt;em&gt;just do it&lt;/em&gt;: it sounds harder than it actually is.&lt;/p&gt;&#xA;" OwnerUserId="299" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2010-11-22T14:37:33.027" LastActivityDate="2010-11-22T14:37:33.027" CommentCount="3" />
  <row Id="11351" PostTypeId="2" ParentId="11349" CreationDate="2010-11-22T14:17:08.323" Score="1" ViewCount="" Body="&lt;p&gt;&quot;Random&quot; is usually a poorly defined term, so it is impossible to give a clear answer.  Many times what is called random behavior in a dynamical system is more appropriately called chaotic.  If you plot the output values (however you take them) you notice that certain values are much more common than others, just from one value you can't predict the next.  Usually for random numbers the first of many requirements is that all numbers are equally likely.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you speak of the strength of an encryption algorithm, what you are usually asking is &quot;given certain data (read about cryptology attacks to see what sorts of data are considered), can you learn something about the key or about the plaintext that was encrypted.  Encryption algorithms are (should be) carefully designed on this point.  Generally speaking you need to scramble the bits in well-defined ways.  Dynamical systems are not designed with this in mind and may well have vulnerabilities when used for encryption.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-11-22T14:17:08.323" CommentCount="2" />
  <row Id="11352" PostTypeId="1" AcceptedAnswerId="11354" CreationDate="2010-11-22T14:36:58.423" Score="1" ViewCount="315" Body="&lt;p&gt;I'm looking for a program or utility to help construct DFA/NFA/PDA etc. and the like. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The main reason I'm interested in this, is because I want to be able to move the various states around without breaking their component connections, or having to redraw all the connections to their new place (otherwise I'd be doing this by hand! :P )&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please let me know if there's a better SE site for this.&lt;/p&gt;&#xA;" OwnerUserId="1760" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2011-03-21T05:32:11.220" LastActivityDate="2011-03-21T05:32:11.220" Title="Program for working with DFA/NFA/PDA ?" Tags="&lt;computer-science&gt;&lt;automata&gt;" AnswerCount="5" CommentCount="9" />
  <row Id="11353" PostTypeId="2" ParentId="11352" CreationDate="2010-11-22T15:01:00.460" Score="4" ViewCount="" Body="&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;If you want to construct and test and play with your automata, you might find &lt;a href=&quot;http://www.cs.duke.edu/csed/jflap/&quot; rel=&quot;nofollow&quot;&gt;JFlap&lt;/a&gt; usefull.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If you want to produce images and are familiar with LaTeX then TikZ is great (&lt;a href=&quot;http://www.texample.net/tikz/examples/feature/automata-and-petri-nets/&quot; rel=&quot;nofollow&quot;&gt;TikZ automata examples&lt;/a&gt;)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If you want to produce images maybe a vector drawing tool like &lt;a href=&quot;http://inkscape.org/download/&quot; rel=&quot;nofollow&quot; title=&quot;Inkscape&quot;&gt;Inkscape&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="571" LastActivityDate="2010-11-22T15:01:00.460" CommentCount="1" />
  <row Id="11354" PostTypeId="2" ParentId="11352" CreationDate="2010-11-22T15:06:30.580" Score="1" ViewCount="" Body="&lt;p&gt;JFlap is a pretty useful tool. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have also used Grapher which may be downloaded from here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.ii.pw.edu.pl/~ratajcza/csm/grapher/grapher.html&quot; rel=&quot;nofollow&quot;&gt;http://www.ii.pw.edu.pl/~ratajcza/csm/grapher/grapher.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3431" LastActivityDate="2010-11-22T15:06:30.580" CommentCount="1" />
  <row Id="11355" PostTypeId="2" ParentId="11341" CreationDate="2010-11-22T15:08:27.850" Score="0" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt;  $\ $ Put $\rm\ \hat f = \sum_{i=0}^{N-1} \:f(i)\: X^i\ $ and note $\rm\ \hat f \:(\hat g\: \hat h) \equiv (\hat f\: \hat g)\:\hat h\ \ (mod\:\ X^N - 1)\ $&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-11-22T15:08:27.850" />
  <row Id="11356" PostTypeId="1" AcceptedAnswerId="11364" CreationDate="2010-11-22T15:16:36.017" Score="0" ViewCount="85" Body="&lt;p&gt;I want to show that if $X$ and $Y$ are the two loss variables such that $X\leq Y$, then $\text{VaR}_\delta(X)\leq\text{VaR}_\delta(Y)$.  &lt;/p&gt;&#xA;" OwnerUserId="626" LastEditorUserId="1154" LastEditorDisplayName="" LastEditDate="2010-11-22T15:56:58.743" LastActivityDate="2010-11-22T17:10:27.743" Title="proof that value at risk VaR is monotonic" Tags="&lt;probability&gt;&lt;finance&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="11357" PostTypeId="2" ParentId="11352" CreationDate="2010-11-22T15:44:24.527" Score="1" ViewCount="" Body="&lt;p&gt;Have a look into the &lt;a href=&quot;http://en.wikipedia.org/wiki/DOT_language&quot; rel=&quot;nofollow&quot;&gt;dot file format&lt;/a&gt;. It is a quasi-standard for graph descriptions. Once you have written down your graph, you can generate an SVG from it using e.g. &lt;a href=&quot;http://www.graphviz.org/&quot; rel=&quot;nofollow&quot;&gt;graphviz&lt;/a&gt; and edit the result in &lt;a href=&quot;http://inkscape.org/&quot; rel=&quot;nofollow&quot;&gt;Inkscape&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="3330" LastActivityDate="2010-11-22T15:44:24.527" />
  <row Id="11358" PostTypeId="2" ParentId="10972" CreationDate="2010-11-22T16:02:56.867" Score="7" ViewCount="" Body="&lt;p&gt;The constant $C$ given in Moron's answer, as suspected, is exactly&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\gamma + \log \frac{4}{\pi},$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which, together with Moron's answer, nicely rounds off the second part of this question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $ \text{sech} x = 2(e^{-x} – e^{-3x} + e^{-5x} + \cdots ) \quad (1)$ we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_0^1 \frac{\tanh x}{x} dx =&#xD;&#xA;2\int_0^1 \frac{\sinh x}{x}(e^{-x} – e^{-3x} + e^{-5x} + \cdots ) dx$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$2\int_0^1 \frac{\sinh x}{x} e^{-x} dx = - Ei(-2) + \gamma + \log 2$$&#xD;&#xA;$$2\int_0^1 \frac{\sinh x}{x} e^{-3x} dx = - Ei(-4) + Ei(-2) + \log 2$$&#xD;&#xA;$$2\int_0^1 \frac{\sinh x}{x} e^{-5x} dx = - Ei(-6) + Ei(-4) + \log (3/2)$$&#xD;&#xA;$$2\int_0^1 \frac{\sinh x}{x} e^{-7x} dx = - Ei(-8) + Ei(-6) + \log (4/3)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and so on, where $Ei(x)$ is the &lt;a href=&quot;http://mathworld.wolfram.com/ExponentialIntegral.html&quot; rel=&quot;nofollow&quot;&gt;exponential integral.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, interchanging the order of summation, summing and using &lt;a href=&quot;http://en.wikipedia.org/wiki/Wallis_product&quot; rel=&quot;nofollow&quot;&gt;Wallis's product&lt;/a&gt;&#xA;we obtain&#xA;$$\int_0^1 \frac{\tanh x}{x} dx = \gamma + \log \frac{4}{\pi}&#xD;&#xA;-2Ei(-2)+2Ei(-4)-2Ei(-6) + \cdots. \quad (2)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using $(1)$ for $\text{sech} x$ we also have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_0^1 \frac{2}{x(e^{2/x}+1) } dx = 2 \int_1^\infty \frac{1}{x(e^{2x}+1)} dx$$&#xD;&#xA;$$=  \int_1^\infty \frac{\text{sech} x}{x} e^{-x} dx&#xD;&#xA;= 2  \int_1^\infty \frac{e^{-2x}}{x} - \frac{e^{-4x}}{x} + \frac{e^{-6x}}{x} - \cdots dx $$&#xD;&#xA;$$=  -2Ei(-2)+2Ei(-4)-2Ei(-6) + \cdots.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And so the result follows from $(2).$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="1301" LastEditorUserId="1301" LastEditorDisplayName="" LastEditDate="2010-11-22T18:06:27.977" LastActivityDate="2010-11-22T18:06:27.977" CommentCount="1" />
  <row Id="11359" PostTypeId="1" CreationDate="2010-11-22T16:36:20.750" Score="5" ViewCount="146" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is known that for any locally compact Hausdorff space X, we can define a Hausdorff one-point compactification containing X.&#xA;In the case of the (differentiable) manifold $\mathbb R^n$ this one-point compactification turns out to be (homeomorphic to) $\mathbb S^n$, which is again a (differentiable) manifold.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This leads to the following question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What does the picture look like in the general case for compactifications of an arbitrary manifold $M$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Although the one-point compactification of $M$ is not a manifold in general (e.g. $\mathbb R^n - 0$); is it possible to view every manifold as an open (dense?) subset of a compact manifold by taking some other kind of compactification?&#xA;In the differentiable case? In the $C^0$-case?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I had thought along the following lines at first:&#xA;By the Whitney embedding theorem, every manifold $M$ can be thought of as a closed submanifold of $\mathbb R^n$ for some $n$. And by embedding $\mathbb R^n$ into $\mathbb S^n$, we can think of $M$ as an embedded submanifold of a compact manifold. &#xA;But I guess taking the closure of $M$ in $\mathbb S^n$ will not in general leave us with a manifold anymore (?), so this does not answer my question...&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Has this been looked into?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for any thoughts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;S.L.&lt;/p&gt;&#xA;" OwnerUserId="3208" LastActivityDate="2010-11-22T16:36:20.750" Title="Compactification of Manifolds" Tags="&lt;topology&gt;&lt;differential-topology&gt;" CommentCount="3" />
  <row Id="11360" PostTypeId="2" ParentId="11352" CreationDate="2010-11-22T16:53:24.157" Score="1" ViewCount="" Body="&lt;p&gt;If I were attempting to draw a DFA or NFA, I would probably use &lt;a href=&quot;http://www.graphviz.org/&quot; rel=&quot;nofollow&quot;&gt;Graphviz&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1004" LastActivityDate="2010-11-22T16:53:24.157" />
  <row Id="11361" PostTypeId="2" ParentId="11352" CreationDate="2010-11-22T16:58:34.243" Score="1" ViewCount="" Body="&lt;p&gt;If you already have access to Mathematica, you can also use it to make images of DFAs.  The recent versions &lt;a href=&quot;http://reference.wolfram.com/mathematica/tutorial/GraphDrawingIntroduction.html&quot; rel=&quot;nofollow&quot;&gt;have the ability to draw graphs&lt;/a&gt;. &lt;/p&gt;&#xA;" OwnerUserId="1004" LastActivityDate="2010-11-22T16:58:34.243" CommentCount="2" />
  <row Id="11362" PostTypeId="1" AcceptedAnswerId="11427" CreationDate="2010-11-22T17:02:10.780" Score="1" ViewCount="66" Body="&lt;p&gt;I've come across a problem with the size of coefficients of a series expansion.  Here's an example of an expression that I expand into a power series:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle\frac{1-e^{i3n\cdot t}}{1-e^{i\cdot n\cdot t}}$ where $n \in \mathbb{N}$ and $t \in \mathbb{R}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is that I'd like to work with small coefficient values, but the values increase exponentially with $n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One question that seems natural to ask in this case is, &quot;Can we find another series with smaller coefficients?&quot;.  Let me explain a little.  I'm fiddling with an integration technique for specific forms of integrals, and I use the series expansions to help with the integration.  I can fairly easily work with more terms, but it's hard to work with larger values of coefficients.  This may seem a bit contrary to many mathematicians' experiences.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I'm getting at is that I'd like to find alternatives to power series expansions.  In other words, is there literature on series expansions other than power series expansions?  I've seen series like Dirchlet series and whatnot, but I'm specifically interested in finding series expansions for functions of exponentials.  And, of course, I'm hoping that I can find ones whose terms are smaller than the corresponding power series expansions. &lt;/p&gt;&#xA;" OwnerUserId="2626" LastEditorUserId="2626" LastEditorDisplayName="" LastEditDate="2010-11-22T17:16:39.560" LastActivityDate="2010-11-23T01:47:35.287" Title="References to Information on Alternative Series Representations" Tags="&lt;sequences-and-series&gt;" AnswerCount="1" />
  <row Id="11363" PostTypeId="1" AcceptedAnswerId="11372" CreationDate="2010-11-22T17:08:13.750" Score="27" ViewCount="500" Body="&lt;p&gt;Let $\mu\left(n\right)$ be the Mobius function. Let $\phi\left(n\right)$ be Euler's totient function. Let $\sigma\left(n\right)$ be the sum of divisors and $\tau\left(n\right)$ be the number of divisors functions. I am curious to know whether or not the system:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\mu\left(n\right)=a$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\phi\left(n\right)=b$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sigma\left(n\right)=c$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\tau\left(n\right)=d$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;has at most one solution. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Motivation: I remember a number theory assignment I had where we were given particular values for each of these functions and asked to recover the original number. I can't for the life of me remember how (or if) I managed to solve this problem. I tried to work out a general proof, but couldn't. I also wrote a loop in maple to check for counterexamples, but haven't found any yet. I feel like this is something I should know, but probably have forgotten the relevant facts to approaching this problem.&lt;/p&gt;&#xA;" OwnerUserId="249" LastActivityDate="2010-11-22T18:45:54.480" Title="Do the Mobius function, totient function, sum of divisors and number of divisors uniquely specify a number?" Tags="&lt;number-theory&gt;" AnswerCount="2" CommentCount="7" FavoriteCount="5" />
  <row Id="11364" PostTypeId="2" ParentId="11356" CreationDate="2010-11-22T17:10:27.743" Score="2" ViewCount="" Body="&lt;p&gt;From the link given by whuber, we have &#xA;$$&#xD;&#xA;{\rm VaR}_\alpha  (L) = \inf \lbrace l \in \mathbb{R}:{\rm P}(L &amp;gt; l) \le 1 - \alpha \rbrace. &#xD;&#xA;$$&#xD;&#xA;If $X$ and $Y$ are random variables such that $X \leq Y$, then obviously ${\rm P}(X &amp;gt; l) \leq {\rm P}(Y &amp;gt; l)$ for any $l \in \mathbb{R}$.&#xA;So, if $l$ belongs to the set $\lbrace l \in \mathbb{R}:{\rm P}(Y &amp;gt; l) \le 1 - \alpha \rbrace$, then $l$ belongs also to the set $\lbrace l \in \mathbb{R}:{\rm P}(X &amp;gt; l) \le 1 - \alpha \rbrace$. Hence, in particular,&#xA;$$&#xD;&#xA;\inf \lbrace l \in \mathbb{R}:{\rm P}(X &amp;gt; l) \le 1 - \alpha \rbrace \leq  &#xD;&#xA;\inf \lbrace l \in \mathbb{R}:{\rm P}(Y &amp;gt; l) \le 1 - \alpha \rbrace,&#xD;&#xA;$$&#xD;&#xA;that is ${\rm VaR}_\alpha  (X) \leq {\rm VaR}_\alpha  (Y)$. Of course, nothing changes if the $\leq$ in $X \leq Y$ refers to stochastic dominance.&lt;/p&gt;&#xA;" OwnerUserId="2810" LastActivityDate="2010-11-22T17:10:27.743" CommentCount="1" />
  <row Id="11365" PostTypeId="1" CreationDate="2010-11-22T17:21:12.550" Score="0" ViewCount="60" Body="&lt;p&gt;I'm searching for a statistical method to determine if a player is cheating in an online game. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given a number  of positive points and a number of negative points per player (score) and given $n$ players ($n &amp;lt;= 64$). Each player can cheat independent of each other. I used standard deviation, which works great on a single player but fails miserably as soon as several players are cheating at the same time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any useful algorithm or formula for this problem?&lt;/p&gt;&#xA;" OwnerUserId="3234" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-11-22T18:02:13.687" LastActivityDate="2010-11-22T21:55:22.927" Title="Method to reliably determine abnormal statistical values" Tags="&lt;probability&gt;&lt;statistics&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="11366" PostTypeId="2" ParentId="11349" CreationDate="2010-11-22T17:52:06.023" Score="0" ViewCount="" Body="&lt;p&gt;The random (chaotic) result from x&amp;lt;-4x(1-x) (which is truly chaotic if x starts as almost any irrational number between 0 and 1) actually ends up reflecting the final bit of rounding error when using any precision floating point representation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is still pretty random, but not worth using for any cryptographic uses.&lt;/p&gt;&#xA;" OwnerUserId="820" LastActivityDate="2010-11-22T17:52:06.023" />
  <row Id="11367" PostTypeId="2" ParentId="11339" CreationDate="2010-11-22T18:14:24.647" Score="2" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Some advices:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, try to break down the problem into pieces you are familiar with and understand - if you feel uncomfortable with some part, then you should study that again and again.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you attend lectures, try to study the chapter before the lecture.   &lt;/p&gt;&#xA;" OwnerUserId="1154" LastActivityDate="2010-11-22T18:14:24.647" />
  <row Id="11368" PostTypeId="2" ParentId="11363" CreationDate="2010-11-22T18:21:37.903" Score="9" ViewCount="" Body="&lt;p&gt;First, congratulations on asking a number theory question which is quite unlike any I have seen before.  It's certainly more fun to read about questions like this rather than computing $a^b$ modulo $n$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regarding the assignment you got: one can certainly find particular values of $a$, $b$, $c$ and $d$ such that there is exactly one $n$ solving the equations.  For instance, if $p$ is a prime number then taking $a = -1$, $b = p-1$, $c = p+1$, $d = 2$ has the unique &#xA;solution $n = p$: $\tau(n) = 2$ forces $n$ to be prime and thus $\varphi(n) = n-1$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Although I am a number theorist, I don't have an expert opinion on the general question.  (I'm not really &quot;the right kind&quot; of number theorist to answer this question.  You might ask Carl Pomerance, for instance.)  I suppose the more reasonable guess is that there are values of $a,b,c,d$ which yield more than one solution.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Probably the first thing to do is what you're doing: search for counterexamples by computer.  When you get tired of doing that, let us know how far you looked (and, of course, whether you've found any).  The second thing to do is to come up with a heuristic on how often one might expect multiple solutions to be found...&lt;/p&gt;&#xA;" OwnerUserId="299" LastActivityDate="2010-11-22T18:21:37.903" CommentCount="4" />
  <row Id="11369" PostTypeId="2" ParentId="8854" CreationDate="2010-11-22T18:21:39.680" Score="1" ViewCount="" Body="&lt;p&gt;I would be extremely surprised if there existed a translation or a LaTeX version. I have a searchable pdf of it though. Annoying Megaupload link: &lt;a href=&quot;http://www.megaupload.com/?d=F6GBYUBU&quot; rel=&quot;nofollow&quot;&gt;http://www.megaupload.com/?d=F6GBYUBU&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="677" LastActivityDate="2010-11-22T18:21:39.680" />
  <row Id="11370" PostTypeId="2" ParentId="6568" CreationDate="2010-11-22T18:27:26.227" Score="1" ViewCount="" Body="&lt;p&gt;We could also write set difference in terms of intersection and complement as : $A \cap B = A - (A - B)$. (I do also prefer $A - B$ to $A \setminus B$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This expression comes handy when you are dealing with cardinality.&lt;/p&gt;&#xA;" OwnerUserId="2109" LastEditorUserId="2109" LastEditorDisplayName="" LastEditDate="2010-11-22T18:34:27.377" LastActivityDate="2010-11-22T18:34:27.377" />
  <row Id="11371" PostTypeId="1" AcceptedAnswerId="11397" CreationDate="2010-11-22T18:30:19.813" Score="0" ViewCount="96" Body="&lt;p&gt;Let $s = (\frac{1}{\sqrt{d}}, \ldots, \frac{1}{\sqrt{d}})$ and $u \in \mathbb{R}^d$ be two distinct unit norm vectors in the first orthant. Consider moving along the great circle defined by $s$ and $u$ (in the direction from $s$ to $u$) until the first intersection of the great circle and one of the following hyperplanes $\{x_1 = 0, x_2 = 0, \ldots, x_d = 0\}$ is reached. Let $v$ denote that intersection. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it possible to explicitly give an equation for the point $v$ or at least the angle between $u$ and $v$?&lt;/p&gt;&#xA;" OwnerUserId="2384" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-22T19:12:31.710" LastActivityDate="2010-11-22T22:33:35.407" Title="How to find the intersection between the great circle and a hyperplane?" Tags="&lt;geometry&gt;&lt;linear-algebra&gt;&lt;spherical-geometry&gt;" AnswerCount="1" CommentCount="0" />
  <row Id="11372" PostTypeId="2" ParentId="11363" CreationDate="2010-11-22T18:31:14.103" Score="21" ViewCount="" Body="&lt;p&gt;The answer is No. The smallest counterexamples I could find are {1836, 1824), {5236, 4960}, {5742, 5112}, {6764, 6368}, {9180, 9120} and {9724, 9184}. I think those are all the pairs in which both numbers are less than 10,000. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, both $n=1836$ and $n=1824$ satisfy $\mu(n)=0$, $\varphi(n)=576$, $\sigma(n)=5040$ and $\tau(n)=24$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: here's the code of the program I used in GAP.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;vec := function(n)&#xA; return [MoebiusMu(n), Phi(n), Sigma(n), Tau(n)];&#xA;end;&#xA;&#xA;dic:=NewDictionary([1,2,3,4], true);&#xA;&#xA;for i in [2..10000] do&#xA; v:=vec(i);&#xA; if (LookupDictionary(dic, v) &amp;lt;&amp;gt; fail) then Print(i,&quot; &amp;lt;=&amp;gt; &quot;, LookupDictionary(dic, v), &quot;\n&quot;); fi;&#xA; AddDictionary(dic, v, i);&#xA;od;&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="308" LastEditorUserId="308" LastEditorDisplayName="" LastEditDate="2010-11-22T18:45:54.480" LastActivityDate="2010-11-22T18:45:54.480" CommentCount="3" />
  <row Id="11373" PostTypeId="1" AcceptedAnswerId="11375" CreationDate="2010-11-22T18:56:18.640" Score="2" ViewCount="198" Body="&lt;p&gt;I just want to make sure that I understand these two objects correctly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $R$ be a ring. A subset $R&amp;#39;\subseteq R$ is a subring provided $R&amp;#39;$ is an additive subgroup that is closed with respect to multiplication. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A subset $I\subseteq R$ is an ideal provided $I$ is an additive subgroup such that $rI\subseteq I$ and $Ir\subseteq I$ for all $r\in R$ (where $rI$ and $Ir$ are defined in the &quot;obvious&quot; way). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, in my notes, for the definition of a subring, I didn't specify what I meant by &quot;with respect to multiplication.&quot; Looking back now, I assume I meant left multiplication. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So then, if this is correct, an ideal is a subring that is closed with respect to both left and right multiplication. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this correct?&lt;/p&gt;&#xA;" OwnerUserId="2313" LastActivityDate="2010-11-25T07:35:13.650" Title="Relationship between subrings and ideals..." Tags="&lt;abstract-algebra&gt;&lt;ring-theory&gt;" AnswerCount="4" CommentCount="3" />
  <row Id="11374" PostTypeId="1" AcceptedAnswerId="11380" CreationDate="2010-11-22T18:57:41.380" Score="2" ViewCount="65" Body="&lt;p&gt;Suppose you have two options for making a payment:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(A) Pay $90 \%$ of the purchase price two months after the date of the sale.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(B) Deduct $X \%$ off the purchase price and pay cash on the date of sale.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Determine $X$ so that a customer is indifferent between the two options when valuing them using an effective annual interest rate of $8 \%$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $P$ be the purchase price. So option (B) is: $P \left(1- \frac{X}{100} \right)$. Now option (A) would be $0.9P(1.08)^{1/6}$ since we want to know what $0.9P$ would be worth in $2$ months? Then set $P \left(1- \frac{X}{100} \right) = 0.9P(1.08)^{1/6}$ to find $X$? &lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2010-11-22T21:13:21.997" Title="Payments and Interest Rates" Tags="&lt;finance&gt;" AnswerCount="1" />
  <row Id="11375" PostTypeId="2" ParentId="11373" CreationDate="2010-11-22T19:03:37.757" Score="4" ViewCount="" Body="&lt;p&gt;A left-ideal is an abelian subgroup (under addition) of the ring which is closed under left-multiplication by elements of the ring, and not just elements in the ideal. In the commutative case we identify left and right multiplication and then an ideal is a subring of the ring. However, that is not enough. Since the subring requires closure under multiplication of elements contained in itself while a left or right ideal requires closure under corresponding multiplication by elements of the ring. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, consider the polynomial ring $R=k[x]$ in one variable over a field $k$. Then $R$ is a commutative ring with identity. $k$ is a subring of $R$ being a field. However, any ideal that contains any element of $k$ is actually the whole ring (since if it contains an element of $k$, it contains $1$ by multiplying it by it's inverse and hence contain every element of $R$). So being an ideal is a stronger requirement.&lt;/p&gt;&#xA;" OwnerUserId="3431" LastEditorUserId="3431" LastEditorDisplayName="" LastEditDate="2010-11-22T19:46:19.123" LastActivityDate="2010-11-22T19:46:19.123" CommentCount="2" />
  <row Id="11376" PostTypeId="2" ParentId="3006" CreationDate="2010-11-22T19:12:28.373" Score="0" ViewCount="" Body="&lt;p&gt;&lt;em&gt;A particle on a trigonometric hyperbola has acceleration described by the same hyperbola.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is, differentiation is idempotent on the hyperbolic cosine. Since $e^{x} = (\cosh x)^{\prime} + \cosh x$, the claim follows immediately.&lt;/p&gt;&#xA;" OwnerUserId="2720" LastEditorUserId="2720" LastEditorDisplayName="" LastEditDate="2010-11-22T19:27:45.473" LastActivityDate="2010-11-22T19:27:45.473" CommentCount="1" />
  <row Id="11377" PostTypeId="2" ParentId="11373" CreationDate="2010-11-22T19:15:46.010" Score="2" ViewCount="" Body="&lt;p&gt;In the case of the multiplication operation in a (sub)ring the phrase 'with respect to multiplication' doesn't refer to any 'left multiplication': In a ring there is just one multiplication which may be or not be commutative. Being closed with respect to multiplication means that if $a$ and $b$ belong to $R&amp;#39;$, then also $ab$&amp;#160;(and $ba$ by exchanging the roles of $a$ and $b$) belong to $R&amp;#39;$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the non-commutative situation we often distinguish the products $ab$ and $ba$ of two elements $a$ and $b$&amp;#160;by saying that we 'multiply $b$ on left/right by $a$'.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ideal is more than a subring in the sense that we require that products stay in the ideal even when multiplying by elements that don't belong to the ideal.&lt;/p&gt;&#xA;" OwnerUserId="3776" LastActivityDate="2010-11-22T19:15:46.010" CommentCount="1" />
  <row Id="11378" PostTypeId="2" ParentId="11373" CreationDate="2010-11-22T19:27:31.020" Score="5" ViewCount="" Body="&lt;p&gt;Remember from groups and vector spaces the notions of &quot;sub-group&quot; and of &quot;sub-space&quot;: you must have a &#xA;&lt;strong&gt;subset&lt;/strong&gt; which is also &quot;closed&quot; with respect to the operations. That is, if you take elements &lt;em&gt;from the subset&lt;/em&gt;, and you subject them to the operations you have, you should end up with results that are also in your subset.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With groups, that means that if a subset $S$ of a group $G$ is going to be a group, then it must be the case that whenever $s,t\in S$, then $st\in S$ (&quot;closed under multiplication&quot;; if you take two things in $S$ and multiply them, you don't need to go &quot;outside of $S$&quot; to find the answer, it is all already inside $S$; no need to go outside); and if $s\in S$, then you need $s^{-1}\in S$ (&quot;closed under inverses&quot;); and you need $e\in S$ (closed under the operation that gives the identity element).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With vector spaces, if a subset $W$ of a vector space $V$ is going to be a subspace, you need that $\mathbf{0}\in W$ (closed under the operation that gives the identity element); that if $w_1,w_1\in W$, then $w_1+w_2\in W$ (closed under vector addition); and that if $w\in W$ and $\alpha\in F$, then $\alpha w\in W$ (closed under scalar multiplication).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same is true for rings: if you have a ring $R$, then in order for a subset $S$ to be a &lt;strong&gt;subring&lt;/strong&gt; you need it to be (i) closed under addition; (ii) closed under additive inverses; (iii) contain the identity element of the sum; and (iv) closed under multiplication (if $a,b\in S$, then $ab\in S$). Saying &quot;is a subgroup&quot; is the same as saying (i), (ii), and (iii). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ideals, however, are a bit more than simply subrings. They play exactly the analogous role to rings as normal subgroups play to groups. You may recall that a normal subgroup $N$ of a group $G$ is a &quot;subgroup-with-something-extra&quot;: not only is it &quot;closed&quot; under the three usual operations of the group (identity element, addition, and inverses), it must also be &quot;closed&quot; under a bunch of other operations (conjugation).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly with ideals: not only must they be &quot;closed&quot; under the four usual operations of the ring (additive identity element, addition, additive inverses, and multiplication), it must also be &quot;closed&quot; under a bunch of other operations (that are like the scalar multiplications in the vector space case): left multiplication by any element of $R$, and right multiplication by any element of $R$. Just like &quot;normal subgroup&quot; is &quot;subgroup with something extra&quot;, likewise ideals are &quot;subrings with something extra.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem with saying &quot;an ideal is a subring that is closed under left and right multiplication&quot; is that saying &quot;closed under left and right multiplication&quot; &lt;em&gt;only&lt;/em&gt; says that if $a,b\in I$, then $ab\in I$; you really want to add the coda of &quot;...left and right multiplication &lt;strong&gt;by elements of the ring.&lt;/strong&gt;&quot; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Examples abound. Take $R=\mathbb{R}[x]$, the polynomials with coefficients in the real numbers, and consider $\mathbb{Z}[x]$, the polynomials with integer coefficients. The latter is a subggroup of $R$, and if $p(x),q(x)\in\mathbb{Z}[x]$, then the product $p(x)q(x)$ is also in $\mathbb{Z}[x]$. So $\mathbb{Z}[x]$ is a subring of $R$. It is not, however, an &lt;em&gt;ideal&lt;/em&gt; of $R$, because $x\in\mathbb{Z}[x]$, and $\pi\in R$, but their product $\pi x\notin\mathbb{Z}[x]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or take $S=\mathbb{R}[x^2]$, the polynomials with real coefficients in which only even powers of $x$ occur. This is a subring of $R$ but not an ideal, because $x^2 \in S$, $x\in R$, but $x(x^2)\notin S$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, $S=\{p(x)\in R\mid p(0)=0\}$, the collection of all polynomials with constant term equal to $0$, is both a subring and an ideal: if $p(x)\in S$ and $q(x)\in R$, then $q(x)p(x)\in S$ since $q(0)p(0)=0$ (and $p(x)q(x)=q(x)p(x)$). That is, $q(x)S\subseteq S$ and $Sq(x)\subseteq S$. &lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-11-22T19:27:31.020" />
  <row Id="11379" PostTypeId="1" CreationDate="2010-11-22T19:58:21.487" Score="0" ViewCount="122" Body="&lt;p&gt;How to prove that a power series in more than one variable is invertible?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this still true if the variables are noncommutative?&lt;/p&gt;&#xA;" LastActivityDate="2010-11-22T21:06:02.253" Title="Inverting a power series" Tags="&lt;power-series&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="11380" PostTypeId="2" ParentId="11374" CreationDate="2010-11-22T20:02:10.430" Score="2" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Edited.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Not quite. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's how it works: let $P$ be the price. Under option (A), you take the money $P$ you would have paid, put it in the bank generating interest, and at the end of the two months you pay $0.9P$, keeping the difference between what you have at the end of the two months and the price you pay. So under option (A), at the end of two months you have both the item plus $P(1.08)^{1/6}-0.9P$ in cash.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Under option (B), you pay $(1- X/100)P$ now, keeping $(\frac{X}{100})P$ of the price, which you can then put in the bank for two months generating interest. So under option (B), at the end of the two months, you have both the item and $\left(\frac{X}{100}P\right)(1.08)^{1/6}$ in cash. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are trying to find the value of $X$ that will make both options equal. (Note that the value $P$ does not really matter, because it will cancel out.) So the quantities you need to set equal to one another are not exactly what you have, but something slightly different.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added.&lt;/strong&gt;&#xA;Equivalently, as you note in the comments, we may instead want to compare the present value of option (A) with option (B) (rather than comparing how much money you will end up with if you set aside $P$ under options (A) and (B), two months down the line). The present value of option (A) is the amount of money $Y$ you would need to put in the bank at $8\%$ interest so that at the end of two months you have $0.9P$ to pay for the item; that is, $Y(1.08)^{1/6} = 0.9P$, or $Y = \frac{0.9P}{(1.08)^{1/6}}$. So you want to find $x$ for which the current expenditure $(1-x)P$ equals the current value of option (A), $(1-x)P = \frac{0.9P}{(1.08)^{1/6}}$. Again, $P$ drops out, and we get the formula you have in the comments,&#xA;$$ 1 - x = \frac{0.9}{(1.08)^{1/6}}.$$&#xD;&#xA;This gives $x = \frac{X}{100} = 1 - \frac{0.9}{(1.08)^{1/6}}$, or a little over $11$% discount.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you proceed as I described originally, and fix the mistake I originally made, you get exactly the same formula and the same answer. &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-22T21:13:21.997" LastActivityDate="2010-11-22T21:13:21.997" CommentCount="4" />
  <row Id="11381" PostTypeId="2" ParentId="11379" CreationDate="2010-11-22T20:06:43.407" Score="2" ViewCount="" Body="&lt;p&gt;(I am answering with respect to the commutative case.) This is not completely true. A power series with a nonzero constant term over a field is invertible. The power series ring in finitely many variables $k[[x_1,...,x_n]]$ over a field is a local ring and the ideal generated by all the variables is the unique maximal ideal. By an elementary result on local rings, every element outside the unique maximal ideal is invertible. So if a power series has a nonzero constant term, it is not in the unique maximal ideal and hence invertible. If the constant term is zero, it is in this maximal ideal and hence not invertible (this ideal is proper).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: You only need the ring of coefficients to be local (and not necessarily a field) for the formal power series ring to be local. In this case the maximal ideal is $m+(x_1,...,x_n)$ where $m$ is the maximal ideal of the coefficient ring $R$. So in this case a power series is invertible iff the constant term is outside the maximal ideal of the coefficient ring.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not sure if this is true in the noncommutative case.&lt;/p&gt;&#xA;" OwnerUserId="3431" LastEditorUserId="3431" LastEditorDisplayName="" LastEditDate="2010-11-22T21:06:02.253" LastActivityDate="2010-11-22T21:06:02.253" CommentCount="2" />
  <row Id="11382" PostTypeId="1" CreationDate="2010-11-22T20:23:14.117" Score="34" ViewCount="2334" Body="&lt;p&gt;I'm hoping that this isn't such a basic question that it gets completely laughed off the site, but why would I want to multiply two polynomials together? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I flipped through some algebra books and have googled around a bit, and whenever they introduce polynomial multiplication they just say 'Suppose you have two polynomials you wish to multiply', or sometimes it's just as simple as 'find the product'. I even looked for some example story problems, hoping that might let me in on the secret, but no dice. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand that a polynomial is a basically a set of numbers (or, if you'd rather, a mapping of one set of numbers to another), or, in another way of thinking about it, two polynomials are functions, and the product of the two functions is a new function that lets you apply the function once, provided you were planning on applying the original functions to the number and then multiplying the result together. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Elementary multiplication can be described as 'add X to itself Y times', where Y is a nice integer number of times. When Y is not a whole number, it doesn't seem to make as much sense. &lt;/p&gt;&#xA;" OwnerUserId="3818" LastActivityDate="2011-01-19T18:00:47.290" Title="Why would I want to multiply two polynomials?" Tags="&lt;algebra-precalculus&gt;" AnswerCount="14" CommentCount="13" FavoriteCount="11" />
  <row Id="11383" PostTypeId="2" ParentId="11379" CreationDate="2010-11-22T20:32:00.143" Score="0" ViewCount="" Body="&lt;p&gt;Timothy Wagner's answer extends to the case of power series over a field with in noncommuting variables with nonzero constant terms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the power series with zero constant terms is both a maximal left ideal and a maximal right ideal. The same argument as in the commutative case gives left and right inverses, and by associativity both must be equal.&lt;/p&gt;&#xA;" OwnerUserId="1119" LastActivityDate="2010-11-22T20:32:00.143" />
  <row Id="11384" PostTypeId="1" CreationDate="2010-11-22T20:42:15.287" Score="1" ViewCount="78" Body="&lt;p&gt;What is the order of the set of distinct (up to similarity) nxn matrices over $\mathbb{R}$ with determinant equal to some non-zero scalar... say 6? (eg. countable, uncountable etc.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The set of matrices over $\mathbb{R}$ is uncountable. So is the order of the set consisting of classes of matrices with the same determinant. In each of those classes we have further subsets, the equivalence classes formed by grouping similar matrices. Each equivalence class of similar matrices represents one linear transformation expressed in terms of all possible bases of $\mathbb{R}^n$, so the size of each equivalence class is also uncountable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I'm not certian about is the size of the set of &quot;different&quot; transformations that have the same determinant. Is it uncountable too?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What can I put it in a correspondence with to show this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Apologies if this is poorly worded-- let me know if there is a better way to ask this question.&lt;/p&gt;&#xA;" OwnerUserId="2563" LastActivityDate="2010-11-22T22:07:04.793" Title="What is the order of the set of distinct (up to similarity) nxn matrices over R?" Tags="&lt;linear-algebra&gt;&lt;elementary-set-theory&gt;&lt;cardinals&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="11385" PostTypeId="2" ParentId="11384" CreationDate="2010-11-22T20:47:06.030" Score="3" ViewCount="" Body="&lt;p&gt;It has the same cardinality as $\mathbb{R}$ if $n\gt 1$.  Consider diagonal matrices with entries $(x,6/x,1,1,\ldots,1)$, $x\neq0$.  Two such are similar only in a situation where $\{x,6/x\}=\{y,6/y\}$, hence the conclusion on cardinality.  (Of course $6$ is not special.)&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-11-22T20:47:06.030" CommentCount="3" />
  <row Id="11386" PostTypeId="2" ParentId="11382" CreationDate="2010-11-22T20:47:27.863" Score="1" ViewCount="" Body="&lt;p&gt;If you were working with a &lt;a href=&quot;http://en.wikipedia.org/wiki/Polynomial_ring&quot; rel=&quot;nofollow&quot;&gt;polynomial ring&lt;/a&gt;, you'd definitely want to be able to multiply polynomials together. This can allow you to use power series to solve recurrences with generating functions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Polynomial rings are also used in some cryptographic systems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a few years, you'll take calculus, and then you'll need to multiply polynomials in order to take derivatives of rational functions, to give one example.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The algebra you learn now will be very useful to your mathematics education later on. You want to get this stuff down cold now, so that later on you can focus on more complicated subjects.&lt;/p&gt;&#xA;" OwnerUserId="2994" LastActivityDate="2010-11-22T20:47:27.863" CommentCount="4" />
  <row Id="11388" PostTypeId="2" ParentId="11382" CreationDate="2010-11-22T20:53:56.683" Score="40" ViewCount="" Body="&lt;p&gt;You have two questions, the explicit one about why you would want to multiply polynomials, and an implicit one in your final paragraph about what multiplication by a non-integer might mean or why we would care to multiply by a non-integer in the first place. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To address the last one first: once you have multiplication by integers, multiplication by fractions will very quickly rear its head. What does multiplying by &quot;one and a half&quot; mean, if multiplying by 2 means &quot;add to itself&quot;, etc? Well, imagine you have a chocolate bar, those that are made up of smaller squares. You can imagine breaking the bar in half, and then figuring out what three times that half will be; that will be multiplying by &quot;three halves&quot; (a.k.a one and a half). You are really multiplying by an integer, after suitably modifying $X$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, if you need to multiply $X$ by a fraction, $\frac{p}{q}$, imagine dividing $X$ into $q$ equal parts, and then multiplying such a $q$th part of $X$ by $p$ in the sense you have above. That is the same as &quot;multiplying by $\frac{p}{q}$&quot;. So multiplying by a fraction &lt;em&gt;is&lt;/em&gt; like &quot;abbreviated addition&quot;: it means &quot;break up into $q$ equal parts, and then add a $q$th part repeatedly $p$ times.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So at least, multiplying by fractions makes just as much &quot;natural sense&quot; as multiplying by integers does.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why bother with numbers other than fractions then? Well, in one sense you don't have to: you can try to stick to fractions and nothing more complicated than that, and you can go very far. But as the Greeks discovered a long time ago, you also run into very big walls very quickly. For instance, if you draw a square which is $1$ foot long on each side, and you try to measure how long its diagonal is (say, for construction purposes), then it turns out that the diagonal is &lt;em&gt;not&lt;/em&gt; a number that can be expressed as a fraction; it is an &lt;a href=&quot;http://en.wikipedia.org/wiki/Irrational_number&quot; rel=&quot;nofollow&quot;&gt;irrational number&lt;/a&gt;. So very soon you end up having to consider numbers that are &lt;em&gt;not&lt;/em&gt; fractions, and if they are lying around sooner or later you are going to have to multiply them to compute stuff. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you end up having to find some way of multiplying irrationals as well, even though they no longer seem to fit with that same &quot;natural&quot; meaning they had back when we started with integers. One solution is that every irrational can be approximated by a suitable sequence of fractions (think about computing the decimals one at a time; every time you stop, what you have so far as a rational; for example, $\sqrt{2} = 1.4142\ldots$, and you get that $1.4 = \frac{14}{10}$, $1.41=\frac{141}{100}$, $1.414=\frac{1414}{1000}$, etc.) We know what it means to multiply $X$ by each of those fractions in a sensible way, so we say that multiplying $X$ by $\sqrt{2}$ is the number you get by doing the successive multiplications, just like $\sqrt{2}$ is the number you get by doing the successive fractional approximations. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This no longer makes sense as &quot;abbreviated addition&quot;, but it turns out that it is very, very necessary and very, very useful, in order to make sense of things and be able to compute things that we need to be able to compute (areas, productivity, interest, etc). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for multiplying polynomials...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One answer: multiplying functions lets you construct more complicated functions out of simpler ones. Or more to the point, it lets you express more complicated functions in terms of simpler ones. This is particularly important if you want to perform complex computations, as then you my be able to &quot;get away&quot; with performing much simpler computations and then multiplying the results, rather than do the really complicated expression instead.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, say you have a single polynomial like $p(x) = x^2-7x+10$. If you realize that $p(x)$ is the result of multiplying the simpler polynomial $x-2$ by the (also simpler) polynomial $x-5$, then whenever you need to evaluate $p(x)$ at a number, say $17$, instead of having to square $17$, then multiply it by $7$, subtract &lt;em&gt;that&lt;/em&gt; form the square you computed, and then adding $10$ (three multiplications and two additions/subtractions), you can instead take $17$, subtract $2$ to get $15$; then take $17$, subtract $5$ to get $12$; and then multiply $15$ by $12$ (one multiplication and two additions/subtractions), because $x^2-7x+10 = (x-2)(x-5)$, so $(17)^2 - 7(17) + 10 = (17-2)(17-5)$. Much simpler to do.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another: it is usually very hard to find a value $x$ for which the result of doing some complex series of operations will be a desired quantity, $d$. For example, you want to know how much money to put in the bank so that, at the end of five months at a particular interest rate, you will have exactly the amount of money you need to buy that new wide-screen TV. This involves solving equations. Many natural equations can be written down in the form $p(x)=c$ where $p(x)$ is a polynomial expression in the unknown quantity $x$, and $c$ is the desired value. Solving such equations can be dificult in general. If you don't know the quadratic formula, then figuring out the values of $x$ for which the polynomial above $x^2-7x+10$ is equal to zero can be pretty difficult. Or think about something like $x^4 + x^3 - 120x^2 - 121x = 121$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, figuring out when a product is equal to $0$ is very easy, because the only way for a product to be zero is if one of the two factors is equal to zero. So if you take the equation above and you write it as $x^4+x^3-120x^2-121x-121 =0$, then you are trying to find when a certain polynomial is equal to $0$. If you can write $q(x)=x^4+x^3-120x^2-121x-121$ as a &lt;em&gt;product&lt;/em&gt;, $q(x) = p(x)r(x)$, then you have that $q(x)=0$ if and only if either $p(x)=0$ or $r(x)=0$. With some luck, $p$ and $r$ will be &quot;easier&quot; than $q$, so you can solve them. (In the above case, $q(x) = (x^2-121)(x^2+x+1)=(x-11)(x+11)(x^2+x+1)$, so the only way you can get $q(x)=0$ is if $x=11$ or $x=-11$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, this is one way to figure out the quadratic formula (did you ever wonder where it came from?). Why are the solutions to $ax^2 + bx+c = 0$ given by $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$? You can factor out $a$ and get $a(x^2 + Bx + C) = 0$, with $B=\frac{b}{a}$ and $C=\frac{c}{a}$. For this to be zero, you need $x^2+Bx+C=0$. Now, imagine you could write it as a product, &#xA;$$x^2 + Bx+C = (x-r_1)(x-r_2).$$&#xD;&#xA;What would $r_1$ and $r_2$ be? If you know how to multiply polynomials, you get that $(x-r_1)(x-r_2)=x^2 - (r_1+r_2)x + r_1r_2$, so you need $r_1r_2=C$ and $r_1+r_2 = -B$. Squaring the latter you get $(r_1+r_2)^2 = B^2$; but&#xA;$(r_1+r_2)^2 = r_1^2 +2r_1r_2 + r_2^2$. On the other hand, &#xA;$$(r_1-r_2)^2 = r_1^2 - 2r_1r_2 + r_2^2 = (r_1^2+2r_1r_2+r_2^2) - 4r_1r_2 = B^2 - 4C.$$&#xD;&#xA;So $(r_1-r_2)^2 =B^2-4C$. Taking square roots, you have that $r_1-r_2 = \pm \sqrt{B^2-4C}$. And you already know that $r_1+r_2 = -B$. Adding them you get&#xA;$$2r_1 = -B\pm\sqrt{B^2-4C}\qquad\text{or}\qquad r_1 = \frac{-B\pm\sqrt{B^2-4C}}{2}$$&#xD;&#xA;and taking the difference between $r_1+r_2 = -B$ and $r_1-r_2 = \pm\sqrt{B^2-4C}$ you get&#xA;$$2r_2 = -B\mp\sqrt{B^2-4C}\qquad\text{or}\qquad r_2 = \frac{-B\mp\sqrt{B^2-4C}}{2}.$$&#xD;&#xA;So you get that $r_1 = \frac{-B+\sqrt{B^2-4C}}{2}$ and $r_2 = \frac{-B-\sqrt{B^2-4C}}{2}$, and plugging in $B=\frac{b}{a}$ and $C=\frac{c}{a}$ gives the usual quadratic formula. No way to find it without knowing how to multiply polynomials!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you get to Calculus (&lt;em&gt;added:&lt;/em&gt; I'm assuming you will &quot;get to Calculus&quot; because you tagged the question as being (algebra-precalculus), so presumably you are taking a course labeled as 'precalculus'; but this may not be the case. If you are not going to &quot;get to Calculus&quot;, then this paragraph will not tell you anything useful), you will find that there is a particular operation (differentiation, taking derivatives) which is &lt;em&gt;very&lt;/em&gt; useful and &lt;em&gt;very&lt;/em&gt; important. It tells you how fast a certain quantity is changing, and it can be used to find all sorts of useful things, like what production level will maximize profit in a factory, how big a dose of medicine and how often you should give to a patient based on how fast they metabolize it, and many other applications. Computing derivatives from first principles with an arbitrary function is pretty labor-intensive; but by reconizing a function as being &quot;made up&quot; (through sums, products, quotients, and compositions) of other, simpler, functions, makes it a very straighforward and easy job. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But in order to be able to recognize that a function is a product of two other functions, you first need to know how to multiply two functions together. Polynomials are one case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another situation occurs when the polynomials are measuring different things, and their product is somehow meaningful; maybe one polynomial gives you the length and the other polynomial gives you the width of a certain figure? Their product will be the area, which may be something you need to compute.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And more generally, you can think of polynomials as &quot;abbreviations&quot; for more complicated operations that you are doing with numbers, just like you are thinking of multiplication as &quot;abbreviated addition&quot;. In that case, multiplying the two polynomials represents another complicated operation that you need to express in terms of the two simpler ones (addition and multiplication). &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-24T04:17:12.877" LastActivityDate="2010-11-24T04:17:12.877" CommentCount="4" />
  <row Id="11389" PostTypeId="1" CreationDate="2010-11-22T20:57:48.267" Score="0" ViewCount="47" Body="&lt;p&gt;Let ( L, ≤ ) be a lattice, x, y, z ∈ L&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am unable to understand why  ( x ∧ z ) ∧ ( y ∧ z ) = z ∧ ( y ∧ x ) is mentioned as fact in  Y.N. Singh's &quot;Mathematical Foundation of Computer Science&quot; - Pg 157&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could I prove this somehow, or is it an axiom?&lt;/p&gt;&#xA;" OwnerUserId="3740" LastActivityDate="2010-11-22T21:37:10.783" Title="In a lattice, ( x ∧ z ) ∧ ( y ∧ z ) = z ∧ ( y ∧ x )" Tags="&lt;order-theory&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="11390" PostTypeId="2" ParentId="11382" CreationDate="2010-11-22T21:08:40.630" Score="2" ViewCount="" Body="&lt;p&gt;When you take calculus, you will need to factor a polynomial p as a product of two polynomials a and b.  If you know how polynomial multiplication works, then finding factorizations is easier.  Learn how to multiply now so that you can factor easily later. :)&lt;/p&gt;&#xA;" OwnerUserId="3331" LastActivityDate="2010-11-22T21:08:40.630" CommentCount="4" />
  <row Id="11391" PostTypeId="1" AcceptedAnswerId="11392" CreationDate="2010-11-22T21:08:54.667" Score="0" ViewCount="127" Body="&lt;p&gt;It is well known that $\text{Ext}(\mathbb Z_p,\mathbb Z)$ is the trivial group, because $\mathbb Z_p$ is projective; this seems to be in contradiction with the Exercise 1.1 in Hilton - Stammbach, pag. 88: the two sequences&#xA;$$&#xD;&#xA;0 \to \mathbb Z \xrightarrow{\mu} \mathbb Z \xrightarrow{\epsilon}\mathbb Z_3 \to 0&#xD;&#xA;$$&#xD;&#xA;and&#xA;$$&#xD;&#xA;0 \to \mathbb Z \xrightarrow{\mu} \mathbb Z \xrightarrow{\epsilon&amp;#39;}\mathbb Z_3 \to 0&#xD;&#xA;$$&#xD;&#xA;where $\mu$ is the multiplication by 3 in $\mathbb Z$ and $\epsilon(n)=n\pmod{3}$, $\epsilon&amp;#39;(n)=n+1\pmod{3}$ are not equivalent, and then represent different elements in the set of extensions of $\mathbb Z_3$ and $\mathbb Z$... Where am I wrong?&lt;/p&gt;&#xA;" OwnerUserId="685" LastActivityDate="2010-11-22T21:33:40.643" Title="Computing $\text{Ext}(\mathbb Z_p,\mathbb Z)$" Tags="&lt;commutative-algebra&gt;&lt;homological-algebra&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="11392" PostTypeId="2" ParentId="11391" CreationDate="2010-11-22T21:18:16.347" Score="2" ViewCount="" Body="&lt;p&gt;Well, $\mathbb Z_p=\mathbb Z/p\mathbb Z$ certainly isn't projective as a&#xA;$\mathbb Z$-module.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added&lt;/strong&gt;&#xA;Now $\mathrm{Ext}(\mathbb Z/p\mathbb Z,\mathbb Z)$ is cyclic of&#xA;order $p$. The zero element corresponds to the split extension of&#xA;$\mathbb Z$ by $\mathbb Z_p$. The other elements correspond to extensions&#xA;$$0\to \mathbb Z\to \mathbb Z\to \mathbb Z/p\mathbb Z\to0.\qquad\qquad\qquad(*)$$&#xD;&#xA;In this exact sequence the surjection can be taken to map $1\in \mathbb Z$&#xD;&#xA;to any given nonzero $a\in\mathbb Z/p\mathbb Z$. Thus there are&#xA;$p-1$ non-isomorphic extensions looking like $(*)$.&lt;/p&gt;&#xA;" OwnerUserId="226" LastEditorUserId="226" LastEditorDisplayName="" LastEditDate="2010-11-22T21:33:40.643" LastActivityDate="2010-11-22T21:33:40.643" CommentCount="1" />
  <row Id="11393" PostTypeId="1" AcceptedAnswerId="11409" CreationDate="2010-11-22T21:21:12.863" Score="7" ViewCount="373" Body="&lt;p&gt;Suppose I can calculate the extraordinary cohomology encoded in &lt;em&gt;topological&lt;/em&gt; $K$-groups of a topological space $X$ with CW structure. What information does this give me about $C^{*}$-algebras associated with $X$? What is the algebraic analogue of topological suspension or the algebraic version of Bott Periodicity?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Bigger Question&lt;/strong&gt;: More generally, what is the &lt;em&gt;deep&lt;/em&gt; connection between topological $K$-theory and algebraic $K$-theory?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Motivation&lt;/strong&gt;: I've computed the reduced topological $K$-groups of a wedge sum of $m$ $n$-spheres. They are given by&#xA;\begin{eqnarray}&#xA;\tilde{K}^{p}_{\text{top}} \left(\bigvee^{m} S^{n} \right) \cong&lt;br&gt;&#xA;\mathbb{Z}^{m} \quad \text{if} \quad p + n \quad \text{is even}&#xA;\end{eqnarray} &#xA;and trivial otherwise. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question&lt;/strong&gt;: Is the positive integer $m$ the dimension of an algebra associated with the wedge sum? If so, is there an intuitive description of this algebra?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="2720" LastEditorUserId="2720" LastEditorDisplayName="" LastEditDate="2011-06-03T17:10:57.783" LastActivityDate="2011-06-03T17:10:57.783" Title="Topological vs. Algebraic $K$-Theory" Tags="&lt;algebraic-topology&gt;&lt;cohomology&gt;&lt;topological-k-theory&gt;&lt;algebraic-k-theory&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="3" />
  <row Id="11394" PostTypeId="2" ParentId="11389" CreationDate="2010-11-22T21:37:10.783" Score="2" ViewCount="" Body="&lt;p&gt;There are at least two ways to do this:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;As Robin Chapman suggests, show that each side of the equation is just the infimum of the set $\{x,y,z\}$&#xD;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Prove that $\wedge$ is an associative and commutative operation, and then use this to reduce the left side to $x \wedge y \wedge (z \wedge z)$ which is $x \wedge y \wedge z$ because $z \wedge z = z$. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="630" LastActivityDate="2010-11-22T21:37:10.783" CommentCount="2" />
  <row Id="11395" PostTypeId="1" CreationDate="2010-11-22T21:50:01.963" Score="0" ViewCount="108" Body="&lt;p&gt;The upper-half plane $\mathbb H$ carries a hyperbolic metric and the geodesics are semicircles with base on the real line. We consider oriented geodesics. Let $x \in \mathbb H$ and let $v$ be a unit tangent vector at $v$. How to prove the following statement:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;There exists a unique geodesic $\gamma$ on $\mathbb H$ such that $\gamma^\prime(0) = v$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-12-23T00:46:15.253" LastActivityDate="2010-12-23T00:46:15.253" Title="Geodesic on half-plane determined by tangent vector" Tags="&lt;differential-geometry&gt;&lt;hyperbolic-geometry&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="11396" PostTypeId="2" ParentId="11365" CreationDate="2010-11-22T21:55:22.927" Score="3" ViewCount="" Body="&lt;p&gt;Your question needs some more information:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How is their score generated (what kind of game is it)? What should your non-cheating data look like? &#xA;How do people cheat? How will their score be different (in a statistical sense) when they are not cheating?&#xA;Do you know roughly the proportion that are cheating? Or is that something you also want to find out?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would also look at outlier detection algorithms: wikipedia looks useful on this topic (&lt;a href=&quot;http://en.wikipedia.org/wiki/Outlier#Identifying_outliers&quot; rel=&quot;nofollow&quot;&gt;link&lt;/a&gt;). Using a &lt;a href=&quot;http://en.wikipedia.org/wiki/Q-Q_plot&quot; rel=&quot;nofollow&quot;&gt;Q-Q plot&lt;/a&gt; on your data may also be useful if your non-cheating data should be approximately Normally distributed; points that are significantly above the line might be cheaters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You may also try posting this on the stats.stackexchange.com since this is really more of a statistics question.&lt;/p&gt;&#xA;" OwnerUserId="3480" LastActivityDate="2010-11-22T21:55:22.927" />
  <row Id="11397" PostTypeId="2" ParentId="11371" CreationDate="2010-11-22T21:58:51.227" Score="1" ViewCount="" Body="&lt;p&gt;$s$ and $u$ span a 2-dimensional plane, and the great circle that contains them is precisely this plane intersected with the unit sphere.  Let $p_i$ be the point of intersection of the great circle and the hyperplane $\{x_i = 0\}$.  Then you know $p_i$ is of the form $as + bu$, and you can solve the equations&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$as_i + bu_i = 0$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\lVert as + bu \rVert = 1$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where, $s_i$ and $u_i$ indicate the $i^{th}$ components of $s$ and $u$, respectively.  This gives you the (typically) two points where the the great circle crosses the hyperplane, and by finding the $i$ whose solution has all nonnegative coordinates, you should get your answer.&lt;/p&gt;&#xA;" OwnerUserId="3823" LastEditorUserId="3823" LastEditorDisplayName="" LastEditDate="2010-11-22T22:33:35.407" LastActivityDate="2010-11-22T22:33:35.407" />
  <row Id="11398" PostTypeId="1" CreationDate="2010-11-22T22:05:20.470" Score="19" ViewCount="345" Body="&lt;p&gt;This question is a follow up to &lt;a href=&quot;http://math.stackexchange.com/questions/11363/do-the-mobius-function-totient-function-sum-of-divisors-and-number-of-divisors&quot;&gt;this&lt;/a&gt; excellent mathematics stackexchange question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $\mu(n)$ be the Möbius function, $\phi(n)$ Euler's totient function, $\sigma(n)$ the sum of divisors function and $\tau(n)$ the number of divisors function. Define the set $S_N,$ for a natural number $N,$ by &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$S_N = \lbrace (m,n) \in \mathbb{N} \times \mathbb{N} \mid m \ne n, \,&#xD;&#xA;\mu(m)=\mu(n), \, \phi(m)=\phi(n),$$&#xD;&#xA;$$\sigma(m)=\sigma(n), \, \tau(m)=\tau(n) \textrm{ and }&#xD;&#xA;\text{max} \lbrace m,n \rbrace \le N \rbrace .$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How large is the set $ S_N $ ? &lt;/p&gt;&#xA;" OwnerUserId="1301" LastActivityDate="2010-11-22T22:05:20.470" Title="The order of the number of integer pairs satisfying certain arithmetical function relationships." Tags="&lt;number-theory&gt;" CommentCount="2" FavoriteCount="3" />
  <row Id="11400" PostTypeId="2" ParentId="872" CreationDate="2010-11-22T22:08:07.980" Score="0" ViewCount="" Body="&lt;p&gt;The basic concept of the Hodge dual has been around since Grassmann -- it is essentially Grassmann's complement. The treatments of the Hodge dual you can find on the web are usually sketchy and unmotivated. My notes on vector spaces give a more thorough treatment than most of what you can find on the web; just google &quot;Vector Spaces, Vector Algebras, and Vector Geometries&quot;. --Richard     &lt;/p&gt;&#xA;" OwnerUserId="3824" LastActivityDate="2010-11-22T22:08:07.980" />
  <row Id="11401" PostTypeId="2" ParentId="11382" CreationDate="2010-11-22T22:10:21.410" Score="10" ViewCount="" Body="&lt;p&gt;Suppose you want to make error-correcting codes (to do things like communicating with spacecraft, manufacturing CD's, and designing computer memory systems). Encoding and decoding BCH codes involves multiplying polynomials.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, nowadays the best error correcting codes are turbo codes or LPDC codes, which aren't based on polynomials. But for nearly 40 years many digital communications systems were based on multiplying polynomials.&lt;/p&gt;&#xA;" OwnerUserId="2501" LastActivityDate="2010-11-22T22:10:21.410" CommentCount="2" />
  <row Id="11402" PostTypeId="1" AcceptedAnswerId="11404" CreationDate="2010-11-22T22:22:24.053" Score="2" ViewCount="206" Body="&lt;p&gt;In a few examples i noted that the existence of k-regular graph on n vertices is :&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;True , for k or n even.&lt;/li&gt;&#xA;&lt;li&gt;False , for k and n odd . But we can find a graph with n-1 vertices with&#xA;degree k and one vertex with degree  k-1. There doesn't exists a k-regular graph for &#xA;k and n odd  because k=deg(G) = 2*|E(G)| / |V(G)|&#xA;|E(G)| = k*n/2 , and |E(G)|= m is not a natural number if n and k is odd.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Any proof idea ??&lt;/p&gt;&#xA;" OwnerUserId="3825" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2010-11-23T00:54:25.427" LastActivityDate="2010-11-23T00:54:25.427" Title="Existence of k-regular graph" Tags="&lt;graph-theory&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="11403" PostTypeId="2" ParentId="11343" CreationDate="2010-11-22T22:25:03.013" Score="1" ViewCount="" Body="&lt;p&gt;It sounds like you may be asking for a representation of $$\sum_{i=1}^n \frac{1}{4i-1}$$ as a Riemann sum estimate of two different integrals, one of which underestimates the true value of the integral (giving you an upper bound on your sum), and the other of which overestimates the true value of the integral (giving you a lower bound on your sum).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The right-hand Riemann sum approximation of $$\int_0^n \frac{1}{4x-1} dx$$ with $\Delta x = 1$ is $\sum_{i=1}^n \frac{1}{4i-1}$ and will underestimate the integral.  However, the integrand is undefined at $x = 1/4$, and so you're better off looking at $$\frac{1}{3} + \sum_{i=2}^n \frac{1}{4i-1}$$ as an underestimate of $$\frac{1}{3} + \int_1^n \frac{1}{4x-1} dx.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, the left-hand Riemann sum approximation of $$\int_0^n \frac{1}{4(x+1)-1} dx$$ with $\Delta x = 1$ is $\sum_{i=1}^n \frac{1}{4i-1}$ and will overestimate the integral.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Drawing the Riemann sum approximations and the graphs of the functions will help to see this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Putting it all together, we have &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_0^n \frac{1}{4x-3} dx &amp;lt; \sum_{i=1}^n \frac{1}{4i-1} \leq \frac{1}{3} + \int_1^n \frac{1}{4x-1} dx,$$&#xD;&#xA;with equality holding in the second case only when $n=1$.&lt;/p&gt;&#xA;" OwnerUserId="2370" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-11-22T22:48:04.687" LastActivityDate="2010-11-22T22:48:04.687" />
  <row Id="11404" PostTypeId="2" ParentId="11402" CreationDate="2010-11-22T22:30:28.427" Score="2" ViewCount="" Body="&lt;p&gt;There is a theorem (&lt;a href=&quot;http://mathworld.wolfram.com/GraphicSequence.html&quot; rel=&quot;nofollow&quot;&gt;Erdos-Gallai&lt;/a&gt;) on degree sequences:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle d_i$ is a degree sequence of some graph if and only if&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle \sum_{i=1}^{m} d_i \leq m(m-1) + \sum_{i=m+1}^{n} \min \{d_i, m\} \ \  \text{for} \ \ m \in \{1,2, \dots, n\}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle \sum_{i=1}^{n} d_i$ is even.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It should be easy to verify for the case when $\displaystyle d_i = k \ \  \forall i \in \{1,2, \dots, n\}$ (cumbersome verification at the end of the answer).&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The case $\displaystyle k=n-1$, we trivially know the existence of a regular graph ($\displaystyle K_n$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose $\displaystyle k \lt n-1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now for $\displaystyle 1 \le m \le k$ we have that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle m(m-1) + (k-m)m + (n-k)k -mk = nk - k^2 - m \ge nk-k^2 - k = k(n-k-1) &amp;gt; 0$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $\displaystyle  m \gt k$ we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle m(m-1) + (n-m)k - mk = m^2 - m(2k+1) + nk = (m-(2k+1)/2)^2 +nk - ((2k+1)/2)^2$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $k \lt n-1$ we have that $\displaystyle 4nk \gt 4k^2 + 4k$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i.e. $4nk \ge 4k^2 + 4k + 1 = (2k+1)^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence &#xA;$\displaystyle m(m-1) + (n-m)k - mk \ge 0$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus if $\displaystyle kn$ is even, there exists a $k$-regular graph on n vertices.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other part I leave to you.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-11-22T23:11:52.450" LastActivityDate="2010-11-22T23:11:52.450" CommentCount="2" />
  <row Id="11405" PostTypeId="2" ParentId="227" CreationDate="2010-11-22T22:32:04.050" Score="0" ViewCount="" Body="&lt;p&gt;Of interest to anyone who finds this page, a new stack exchange Economics site is nascent and needs your commitment if you can lend it: &lt;a href=&quot;http://area51.stackexchange.com/proposals/1618/economics?referrer=iym-fVnycBaB379vI_74Bg2&quot;&gt;http://area51.stackexchange.com/proposals/1618/economics?referrer=iym-fVnycBaB379vI_74Bg2&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3746" LastActivityDate="2010-11-22T22:32:04.050" CommunityOwnedDate="2010-11-22T22:32:04.050" />
  <row Id="11406" PostTypeId="2" ParentId="11395" CreationDate="2010-11-22T22:48:35.610" Score="1" ViewCount="" Body="&lt;p&gt;Here is an attempt at solution without getting into the intricacies of Riemannian geometry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let a geodesic for the upper-half plane have the center $(a,0)$ and radius $r$. Thus it has the equation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(x-a)^2 + y^2 = r^2.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Specifying a unit tangent vector at a point is a choice of a direction there, which might be given by a directed line passing through that point. Your requirement translates to the assertion that the slope of the circle at this point agrees with that of the directed line corresponding to the unit vector. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have two unknowns in this situation, viz., $a$ and $r$. Partial differentiation of the circle equation to get the $x$ and $y$ components of slope would give two linear equations, for which there would be a unique solution.&lt;/p&gt;&#xA;" OwnerUserId="1119" LastActivityDate="2010-11-22T22:48:35.610" />
  <row Id="11408" PostTypeId="2" ParentId="8854" CreationDate="2010-11-22T22:53:10.733" Score="1" ViewCount="" Body="&lt;p&gt;There is no translation in my knowledge. But there is a later book by Katz and Mazur on &quot;Arithmetic Moduli of Elliptic Curves&quot;, which is in English and which ought to a good amount of the things in the paper of Deligne and Rapoport.&lt;/p&gt;&#xA;" OwnerUserId="1119" LastActivityDate="2010-11-22T22:53:10.733" CommentCount="1" />
  <row Id="11409" PostTypeId="2" ParentId="11393" CreationDate="2010-11-22T23:05:01.803" Score="9" ViewCount="" Body="&lt;p&gt;I cannot fully answer your questions, but hopefully I can help orient you.  I know a little about C*-algebra K-theory, and less about topological K-theory, so take this with a grain of salt.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The $K_0$ group of a (complex) C*-algebra $A$ is defined as the Grothendieck group of a commutative monoid of equivalence classes of projections in the matrix algebras over $A$.  The $K_1$ group of $A$ can be defined as a group of equivalence classes of unitary elements of matrix algebras over $A$.  (The subscripts are used instead of superscripts because the corresponding functors are covariant rather than contravariant.)  To define higher $K_n$ groups, &quot;suspensions&quot; are used.  In this setting, the suspension $SA$ of $A$ is defined to be $C_0(0,1)\otimes A\cong C_0((0,1),A)$, which can be thought of as the C*-algebra of continuous $A$-valued functions on the unit interval that vanish at $0$ and $1$.  When $n$ is greater than $1$, $K_n(A)$ is defined to be $K_{n-1}(SA)$.  It turns out that this is consistent with the $n=1$ case, i.e., $K_1(A)$ is isomorphic to $K_0(SA)$.  Bott periodicity says that $K_{n+2}(A)$ is isomorphic to $K_n(A)$ for all $n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the commutative case, there is a locally compact Hausdorff space $X$ such that $A$ is isomorphic to $C_0(X)$, the algebra of complex-valued continuous functions on $X$ vanishing at infinity.  The homeomorphism class of $X$ is determined by $A$, and the isomorphism class of $A$ is determined by $X$.  One can consider either the operator K-theory of $C_0(X)$ or the topological K-theory of $X$.  It turns out that the groups are isomorphic: $K_n(C_0(X))\cong K^n(X)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The K-groups don't tell you anything literally about the dimension of the C*-algebra.  For example, the infinite dimensional C*-algebra $C(S^2)$ of continuous complex-valued functions on the $2$-sphere has K-groups $K_0(C(S^2))=\mathbb{Z}^2$ and $K_1(C(S^2))=0$.  These are the same as the K-groups of the algebra $\mathbb{C}\oplus\mathbb{C}$ (which is the algebra of continuous complex-valued functions on a $2$-point discrete space).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a terminological aside, usually a distinction is made between &lt;a href=&quot;http://en.wikipedia.org/wiki/Algebraic_K-theory&quot; rel=&quot;nofollow&quot;&gt;algebraic K-theory&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/Operator_K-theory&quot; rel=&quot;nofollow&quot;&gt;K-theory of operator algebras&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are some references.  For topological K-theory, see &lt;a href=&quot;http://books.google.com/books?id=RfArvjhOG44C&amp;amp;dq=atiyah+k-theory&amp;amp;source=gbs_navlinks_s&quot; rel=&quot;nofollow&quot;&gt;Atiyah&lt;/a&gt; and &lt;a href=&quot;http://books.google.com/books?id=fJzlLkHBYTYC&amp;amp;dq=karoubi+k-theory&amp;amp;source=gbs_navlinks_s&quot; rel=&quot;nofollow&quot;&gt;Karoubi&lt;/a&gt;.  For operator K-theory, in increasing order of difficulty and assumed background, see &lt;a href=&quot;http://books.google.com/books?id=EVOjQgAACAAJ&amp;amp;dq=k-theory%20wegge-olsen&amp;amp;source=gbs_book_other_versions&quot; rel=&quot;nofollow&quot;&gt;Wegge-Olsen&lt;/a&gt;, &lt;a href=&quot;http://books.google.com/books?id=SMiB8VIB5UIC&amp;amp;dq=rordam+k-theory&amp;amp;source=gbs_navlinks_s&quot; rel=&quot;nofollow&quot;&gt;Rørdam et al.&lt;/a&gt;, and &lt;a href=&quot;http://books.google.com/books?id=214a1Wri63QC&amp;amp;dq=blackadar+k-theory&amp;amp;source=gbs_navlinks_s&quot; rel=&quot;nofollow&quot;&gt;Blackadar&lt;/a&gt;.  Blackadar's book is a wonderful reference and the most comprehensive, while the other two serve as good introductions.  (Rørdam et al. is the one I've studied the most.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some insight on the relationship between the two theories can be gained in light of &lt;a href=&quot;http://en.wikipedia.org/wiki/Swan%27s_theorem&quot; rel=&quot;nofollow&quot;&gt;Swan's theorem&lt;/a&gt; on the equivalence between finitely generated projective modules over $C(X)$ and vector bundles over $X$.  Roughly speaking, since projective modules are direct summands of free modules, they correspond to idempotents in the endomorphism rings of free modules, and in the C*-algebra case this corresponds to the projections in matrix algebras mentioned above.  Each of the operator K-theory books mentioned above includes at least some material on the relationship to topological K-theory.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-11-22T23:05:01.803" CommentCount="1" />
  <row Id="11410" PostTypeId="2" ParentId="11382" CreationDate="2010-11-22T23:14:42.253" Score="5" ViewCount="" Body="&lt;p&gt;When solving real-world problems with simple algebra, it is not uncommon to have polynomials scattered throughout the equation. Given that 1/(3x+4)=2x-1, which isn't an extraordinarily complex equation, the first thing you might want to do is multiply both sides by 3x+4, which means you will need to be able to deal with (2x-1)*(3x+4).&lt;/p&gt;&#xA;" OwnerUserId="1995" LastActivityDate="2010-11-22T23:14:42.253" />
  <row Id="11411" PostTypeId="1" AcceptedAnswerId="11413" CreationDate="2010-11-22T23:25:46.047" Score="4" ViewCount="227" Body="&lt;p&gt;I was re-reading &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0486661695&quot; rel=&quot;nofollow&quot;&gt;this old book&lt;/a&gt; of mine; and I noticed that in defining the rules of differential forms, it &quot;makes sense&quot; that we have the rule $dx \wedge dx=0$ because if $dx$ is infinitesimal, then to first order approximations we can ignore powers of $dx$.  Similarly, the definition for the exterior derivative $d$, of a differential form $\omega=Adx+Bdy+Cdz$, $d\omega=\frac{dA}{dx}dx + \frac{dB}{dy}dy + \frac{dC}{dz}dz $ &quot;makes sense&quot; because it feels like we are just multiplying the top and bottom by the differentials $dx,dy,$ and $dz$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But it is practically a miracle that by introducing the simple anti-symmetrical commutation relations for differential forms, and applying very elementary operations, we can arrive at all the results of vector calculus such as gradient and cross product, among a large amount of other well known results.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this particular book, the authors motivate the anti-symmetry condition by properties of determinants and Jacobian's for change of variables in integration.  But I was wondering if there are other ways to think about why differential forms should commute anti-symmetrically which might provide some more intuition on just why this &quot;miracle&quot; works.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="1417" LastActivityDate="2010-11-23T00:17:41.593" Title="Intuition behind $dx \wedge dy=-dy \wedge dx$." Tags="&lt;differential-forms&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" />
  <row Id="11412" PostTypeId="1" AcceptedAnswerId="11423" CreationDate="2010-11-23T00:10:38.773" Score="9" ViewCount="294" Body="&lt;p&gt;Let $B$ be a Stone space (compact, Hausdorff, and totally disconnected).  Then I am basically certain (because of Stone's representation theorem) that if $a, b \in B$ are two distinct points in $B$, then $B$ can be written as a disjoint union $U \cup V$ of open sets where $a \in U, b \in V$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I can't seem to prove this directly.  The proof should be fairly straightforward, so I am sure I'm missing something obvious.  (As an exercise for myself, I'm trying to prove Stone's representation theorem, and I need this as a lemma.)&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-11-23T16:25:46.667" Title="Any two points in a Stone space can be disconnected by clopen sets" Tags="&lt;topology&gt;&lt;general-topology&gt;&lt;boolean-algebra&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="11413" PostTypeId="2" ParentId="11411" CreationDate="2010-11-23T00:11:44.077" Score="5" ViewCount="" Body="&lt;p&gt;One way of looking at the antisymmetric relation is a consequence of $dx∧dx=0$ (which feels intuitive to you). Applied to $(dx+dy)∧(dx+dy)=0$, we get $(dx∧dx)+(dx∧dy)+(dy∧dx)+(dy∧dy)=0$. So, $(dx∧dy)+(dy∧dx)=0$, so $(dx∧dy)=-(dy∧dx)$&lt;/p&gt;&#xA;" OwnerUserId="3431" LastActivityDate="2010-11-23T00:11:44.077" CommentCount="2" />
  <row Id="11414" PostTypeId="2" ParentId="11411" CreationDate="2010-11-23T00:17:41.593" Score="4" ViewCount="" Body="&lt;p&gt;I like the motivation given by Jack Lee's book Introduction to Smooth Manifolds.  Roughly, we want to capture volume by the exterior algebra: say $\omega$ is a tensor that we want to apply to $n$ vectors to get the $n$-dimensional volume of the parallelogram they form. In the case $n=2$ for example, we should have $\omega(X,X) = 0$ since we get a line and not a 2-d region (so 0 area).  Now by linearity, $\omega(X,X) = 0$ forces $X$ to be alternating (as in Timothy's answer).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The algebra of forms is the algebra of alternating tensors.&lt;/p&gt;&#xA;" OwnerUserId="9" LastActivityDate="2010-11-23T00:17:41.593" />
  <row Id="11415" PostTypeId="1" AcceptedAnswerId="11416" CreationDate="2010-11-23T00:23:01.947" Score="1" ViewCount="202" Body="&lt;p&gt;The chance to throw a 6 with one die is 1/6&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And 6 times 1/6 = 1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, if I throw with 6 dice, the chance to throw at least 1 six should be 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But when I throw 6 dice, I sometimes don't throw any 6 at all..&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How come?&lt;/p&gt;&#xA;" OwnerUserId="3535" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-23T01:05:14.870" LastActivityDate="2011-06-03T01:55:26.973" Title="chance on throwing a six with 6 dice" Tags="&lt;probability&gt;" AnswerCount="4" CommentCount="1" />
  <row Id="11416" PostTypeId="2" ParentId="11415" CreationDate="2010-11-23T00:29:31.463" Score="7" ViewCount="" Body="&lt;p&gt;With $6$ dice there are $6^6$ possible outcomes. Of these, $5^6$ don't lead to six on any of the dice. So the number of outcomes that lead to at least one six are $6^6-5^6$, so the probability of at least one six is $\frac{6^6-5^6}{6^6} \approx 0.6651$&lt;/p&gt;&#xA;" OwnerUserId="3431" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-06-03T01:55:26.973" LastActivityDate="2011-06-03T01:55:26.973" CommentCount="3" />
  <row Id="11418" PostTypeId="2" ParentId="11415" CreationDate="2010-11-23T00:43:14.107" Score="4" ViewCount="" Body="&lt;p&gt;The &lt;i&gt;expected&lt;/i&gt; number of dice showing $6$ is $1$ when throwing six dice. The &lt;i&gt;probability&lt;/i&gt; to see at least one $6$ is $1 - (5/6)^6$, as explained in Timothy's answer.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-23T00:43:14.107" />
  <row Id="11419" PostTypeId="2" ParentId="11415" CreationDate="2010-11-23T00:44:02.953" Score="1" ViewCount="" Body="&lt;p&gt;If you flip a coin, there's a 50-50 chance you get heads. If you flip two coins there's a $3/4$ chance you get at least one head, since there's a $1/2 \times 1/2 = 1/4 $ chance they're both tails. So similarly, there's a $5/6$ chance each die is not $6$. The chance they're all not $6$ is $(5/6) \times (5/6) \times (5/6) \times (5/6) \times (5/6) \times (5/6) = (5/6)^6$. So the chance that at least one is a six is $1 - (5/6)^6$. &lt;/p&gt;&#xA;" OwnerUserId="3035" LastActivityDate="2010-11-23T00:44:02.953" />
  <row Id="11420" PostTypeId="2" ParentId="11415" CreationDate="2010-11-23T00:51:37.973" Score="4" ViewCount="" Body="&lt;p&gt;The probability of the union of independent events (unlike disjoint events) is not the sum of the individual probabilities. If $E_i$ denotes the event &quot;$6$ is obtained on the $i$th throw&quot;, then $E_i$ are independent events, and it does not hold ${\rm P}(E_1  \cup  \cdots  \cup E_6 ) = {\rm P}(E_1 ) +  \cdots  + {\rm P}(E_6 )$. The left-hand side probability can be found as follows.&#xA;The complement of the event $E_1  \cup  \cdots  \cup E_6$ is $E_1^c  \cap  \cdots  \cap E_6^c $, where $E_i^c$ is the event &quot;$6$ is not obtained on the $i$th throw&quot;. Hence, &#xA;$$&#xD;&#xA;{\rm P}(E_1  \cup  \cdots  \cup E_6 ) = 1 - {\rm P}(E_1^c  \cap  \cdots  \cap E_6^c ).&#xD;&#xA;$$&#xD;&#xA;Now, the probability of the intersection of independent events is the product of their individual probabilities.&#xA;So, ${\rm P}(E_1^c  \cap  \cdots  \cap E_6^c ) = {\rm P}(E_1^c) \cdots {\rm P}(E_6^c) = (5/6)^6$.&#xA;Hence, ${\rm P}(E_1  \cup  \cdots  \cup E_6 ) = 1 - (5/6)^6$.&lt;/p&gt;&#xA;" OwnerUserId="2810" LastActivityDate="2010-11-23T00:51:37.973" CommentCount="2" />
  <row Id="11421" PostTypeId="2" ParentId="11402" CreationDate="2010-11-23T00:54:03.867" Score="1" ViewCount="" Body="&lt;p&gt;At the outset, you should assume that $k &amp;lt; n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $n = 2m$ is even, construct a graph with vertex set $$ \{ X_i : i \in \mathbb{Z}_m \} \cup \{ Y_i : i \in \mathbb{Z}_m \}, $$ where $\mathbb{Z}_m$ is the integers modulo $m$. Connect $X_i$ to $Y_j$ if $j-i \in \{1,\ldots,k\}$, where subtraction is done modulo $m$. Each $X_i$ is connected to $Y_{i+1},\ldots,Y_{i+k}$, and each $Y_j$ is connected to $Y_{j-1},\ldots,Y_{j-k}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $k = 2l$ is even, construct a graph with vertex set $$\{X_i : i \in \mathbb{Z}_n\}.$$ Connect $X_i$ and $X_j$ if $i \neq j$ and $i-j \in \{-l,\ldots,l\}$. This relation is symmetric (since $j-i = -(i-j)$), and every $X_i$ is connected to $X_{i-l},\ldots,X_{i+l}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you mentioned, when both $n,k$ are odd, we have $2e = nk$ where $e$ is the number of edges (this formula is obtained by counting all endpoints of all edges), which is a contradiction.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-23T00:54:03.867" CommentCount="4" />
  <row Id="11422" PostTypeId="2" ParentId="8814" CreationDate="2010-11-23T00:59:40.413" Score="2" ViewCount="" Body="&lt;p&gt;Let $f$ be a symbol with the property that $f^n = n!$. Consider $d_n$, the number of ways of putting $n$ letters in $n$ envelopes so that no letter gets to the right person (aka derangements). Many people initially think that $d_n = (n-1)! = f^{n-1}$ (the first object has $n-1$ legal locations, the second $n-2$, ...). The correct answer isn't that different actually:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$d_n = (f-1)^n$. &lt;/p&gt;&#xA;" OwnerUserId="308" LastActivityDate="2010-11-23T00:59:40.413" CommunityOwnedDate="2010-11-23T00:59:40.413" />
  <row Id="11423" PostTypeId="2" ParentId="11412" CreationDate="2010-11-23T01:00:31.443" Score="7" ViewCount="" Body="&lt;p&gt;Before proving the fact you want, we'll need the notion of quasicomponents and some basic propositions about it. In these terms, you are just asking if distinct points are in distinct quasicomponents. Let $X$ be a topological space. Given $x,y \in X$ , define $x \sim y$ if $X$ cannot be written as a disjoint union of open sets $U$ and $V$ containing $x$ and $y$, respectively. It is straightforward to verify that $\sim$ is an equivalence relation. The equivalence classes are called the quasicomponents of $X$. It's easy to prove that the quasicomponent of a point $x$ is the intersection of all closed-open subsets of $X$ containing $x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In every topological space, the component $C$ of a point $x$ is contained in the quasicomponent $Q$ of the point $x$. In fact, if $F$ is a closed-open set containing $x$, then $X = F \ \cup (X-F \ )$ is a separation of $X$. Since $C \cap F \ne \emptyset$, it follows that $C \subseteq F$. Thus we have $C \subseteq Q$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll prove now, based on Engelking's proof in the book 'General Topology', that in every compact Hausdorff space, components and quasicomponents coincide. Let $C$ and $Q$ as above. We just need to prove that the quasicomponent $Q$ is connected. Then will follow that $ Q = C$. Suppose that $Q = X_1 \cup X_2$, where $X_1, X_2$ are two disjoint closed subsets of the space $Q$. We have that $X_1$ and $X_2$ are closed in $X$, since $Q$ is closed in $X$. By normality of compact Hausdorff spaces, there exist disjoint open subsets $U,V$ of $X$ containing $X_1, X_2$, respectively. Hence, we have $Q \subseteq U \cup V$ and, by compactness, there exist closed-open sets $F_1, \ldots, F_k$ such that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$Q \subseteq \bigcap_{i=1}^k F_i \subseteq U \cup V.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$F = \bigcap_{i=1}^k F_i$ is clearly closed-open. Since $ \overline{U \cap F} \subseteq \overline{U} \cap F = \overline{U} \cap (U \cup V) \cap F = U \cap F$, the intersection $U \cap F$ is also closed-open. As $x \in U \cap F$, we have $Q \subseteq U \cap F$ and $X_2 \subseteq Q \subseteq U \cap F \subseteq U$. It follows that $X_2 \subseteq U \cap V = \emptyset$, which shows that the set $Q$ is connected.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now it's simple to prove your statement. Let $B$ be a Stone Space. Since $B$ is compact Hausdorff, quasicomponents coincide with components. $B$ is totally disconnected, so the quasicomponent of a point $a \in B$ is $\{a \}$. If $b \in B$ is a different point, then $b$ isn't in the quasicomponent of $a$. Thus there exist disjoint open sets $U,V$ containing $a,b$, respectively, such that $B = U \cup V$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added&lt;/strong&gt; This was motivated by Pete's answer. Actually, to prove that a locally compact Hausdorff totally disconnected space $X$ is zero-dimensional, we just need what I've proved for compact Hausdorff spaces. Indeed, let $x \in X$ and $U$ an open set containing $x$. Since $X$ is regular, there is an open set $V$ containing $x$ such that $\overline{V}$ is compact and $\overline{V} \subseteq U$. Also, $\overline{V}$ is a totally disconnected space. Using the fact that quasicomponents and components coincide in compact Hausdorff spaces, we have that the quasicomponent of $x$ in $\overline{V}$ is $\{x\}$. Now, the compactness of $\overline{V}$ guarantees that there are closed-open sets $F_1, \ldots, F_k$ such that $x \in F_1 \cap \ldots \cap F_k \subseteq V$. Let $F$ be $\bigcap_{i=1}^k F_i$. Then $F \ $ is an open set in $V$ and since $V \ $ is open in $X$, $F \ $ is open in $X$. Also, $F \ $ is compact, since each $F_i$ is a closed subset of the compact space $\overline{V}$, and then $F \ $ is closed in $X$. We conclude that for each open set $U$ containing $x$, there is a closed-open set $F \ $ such that $x \in F \subseteq U$. Thus $X$ is zero-dimensional. &lt;/p&gt;&#xA;" OwnerUserId="329" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2010-11-23T16:25:46.667" LastActivityDate="2010-11-23T16:25:46.667" CommentCount="4" />
  <row Id="11424" PostTypeId="2" ParentId="11340" CreationDate="2010-11-23T01:06:46.327" Score="2" ViewCount="" Body="&lt;p&gt;Here's a practical procedure, which assumes that the Shuffler is cyclic. Construct some hash function $H$ from $k$-tuples of songs to $n$ bits (the parameters $k,n$ will be decided upon later). Store the value of $H$ on the first $k$ songs, and look for a matching value of $H$ on a future $k$-tuple. If $k$ is large enough, then the expected number of occurrences of this $k$-tuple in the entire cycle is $&amp;lt;1$, and so its likely that a repeat means that you've reached the end of the cycle. If $n$ is big enough (larger than $\sqrt{N}$, where $N$ is the period), then it's probable that matching hashes means matching $k$-tuples.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to increase your certainty, you can store a much larger prefix, and then use it whenever the hash value matches. In fact, you can probably use several hash functions this way to optimize the procedure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you can't assume that the Shuffler is cyclic, but rather that from some point on it will cycle, you can store all the hashes in a hash table (along with their time) and wait for a hash that appears twice. If you're allowed to read the input twice, you can use any of the Pollard rho algorithms, as hardmath commented.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-23T01:06:46.327" CommentCount="1" />
  <row Id="11425" PostTypeId="2" ParentId="11343" CreationDate="2010-11-23T01:18:05.417" Score="0" ViewCount="" Body="&lt;p&gt;You can write $$\frac{1}{4i-1} = \frac{1}{4i} + \frac{1}{4i(4i-1)}.$$ Therefore $$\sum_{i=1}^n \frac{1}{4i-1} = \frac{1}{4} \sum_{i=1}^n \frac{1}{i} + \sum_{i=1}^n \frac{1}{4i(4i-1)}.$$ The second term is bounded (by comparison with the convergent infinite series $\sum_{i=1}^\infty 1/i^2$), and the first term is familiar.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-23T01:18:05.417" />
  <row Id="11426" PostTypeId="1" CreationDate="2010-11-23T01:43:55.140" Score="0" ViewCount="290" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Show that the term ZZ where Z is λz.λx. x(z z x) satisfies the requirement for fixed point combinators that ZZM =β M(ZZM).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2010-11-23T04:43:05.273" LastActivityDate="2011-03-09T04:39:43.063" Title="Proving a combinator is a fixed point" Tags="&lt;lambda-calculus&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="11427" PostTypeId="2" ParentId="11362" CreationDate="2010-11-23T01:47:35.287" Score="1" ViewCount="" Body="&lt;p&gt;I'm not entirely sure how helpful this suggestion (this was supposed to be a comment, but it got very long) will ultimately be, but you might be interested in knowing that there exists a generalization of the usual Taylor series, called the (Lagrange-)Bürmann series.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Briefly, given a function $f(x)$ to be expanded, a &quot;basis function&quot; $g(x)$, an expansion point $a$, and the assumption that $g^{\prime}(x)$ is nonzero over some interval containing $a$, such that $g(x)-g(a)$ is monotonic as $x$ increases over said interval, the Bürmann series of $f(x)$ with basis function $g(x)$ and expansion point $x=a$ reads as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f(x)=f(a)+\sum_{k=1}^{\infty} \frac{\beta_k(a)}{k!}(g(x)-g(a))^k$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where the $\beta_k(x)$ satisfy the recursion&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\beta_0(x)=f(x),\qquad \beta_k(x)=\frac1{g^{\prime}(x)}\frac{\mathrm d}{\mathrm dx}\beta_{k-1}(x)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The series may or may not converge, of course, so the analysis of convergence (which &lt;em&gt;highly&lt;/em&gt; depends on the nature of your $f(x)$ and $g(x)$) is still your burden.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's not terribly popular, probably because it's difficult to choose $g(x)$ such that your expansion of $f(x)$ is meaningful, but it's worth a shot.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-11-23T01:47:35.287" CommentCount="4" />
  <row Id="11428" PostTypeId="1" AcceptedAnswerId="11432" CreationDate="2010-11-23T02:01:30.070" Score="2" ViewCount="252" Body="&lt;p&gt;How to solve  $y&amp;#39;&amp;#39;-\displaystyle\frac{y&amp;#39;}{x}=4x^{2}y$ ?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that the solution of this equation is: $y = e^{x^{2}}$, but I cannot resolve.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First I thought that $z=y&amp;#39;$ could be, but was not.&lt;/p&gt;&#xA;" OwnerUserId="3421" LastEditorUserId="3421" LastEditorDisplayName="" LastEditDate="2010-11-23T04:05:58.730" LastActivityDate="2010-11-23T04:05:58.730" Title="How to solve this differential equation? $y''-\frac{y'}{x}=4x^{2}y$" Tags="&lt;calculus&gt;&lt;differential-equations&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="11429" PostTypeId="1" AcceptedAnswerId="15583" CreationDate="2010-11-23T02:16:19.473" Score="0" ViewCount="129" Body="&lt;p&gt;A loan for $50,000$ has level payments to be made at the end of each year for 10 years at an annual rate of $9 \%$. (a) Find the balance at the end of $3$ years. (b) Find the interest paid in the third payment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I think this involves using amortization. That is, one pays the interest due first and then the principal afterwards. In other words, one covers the &quot;creeping costs&quot; first, and then pays the full amount. So this is a immediate annuity. So the interest due would be $50000(1.09)^{10}$?&lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2011-04-29T10:40:31.643" Title="Loan Repayment and Annuities" Tags="&lt;finance&gt;" AnswerCount="4" />
  <row Id="11431" PostTypeId="1" AcceptedAnswerId="11442" CreationDate="2010-11-23T02:17:14.730" Score="3" ViewCount="239" Body="&lt;p&gt;Today I was thinking about composition of functions.  It has nice properties, its always associative, there is an identity, and if we restrict to bijective functions then we have an inverse.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But then I thought about commutativity.  My first intuition was that bijective self maps of a space should commute but then I saw some counter-examples. The symmetric group is only abelian if n&amp;lt;=2 so clearly there needs to be more restrictions on functions than bijectivity for them to commute.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only examples I could think of were boring things like multiplying by a constant or maximal tori of groups like O(n) (maybe less boring).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question: In a euclidean space, what are (edit) some nice characterizations of sets of functions that commute? What about in a more general space? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bonus: Is this notion of commutativity important anywhere in analysis?&lt;/p&gt;&#xA;" OwnerUserId="565" LastEditorUserId="565" LastEditorDisplayName="" LastEditDate="2010-11-23T04:07:44.667" LastActivityDate="2010-11-23T04:07:44.667" Title="When functions commute under composition" Tags="&lt;functions&gt;" AnswerCount="2" CommentCount="7" FavoriteCount="1" />
  <row Id="11432" PostTypeId="2" ParentId="11428" CreationDate="2010-11-23T02:17:31.750" Score="7" ViewCount="" Body="&lt;p&gt;Note that we have $xy&amp;#39;&amp;#39; - y&amp;#39; = 4x^3y$. The LHS in some sense is dimensional consistent and looks something similar to a quotient rule provided we divide by $x^2$. So dividing by $x^2$ and doing algebraic manipulations we get $(\frac{1}{x}y&amp;#39;)&amp;#39; = 4xy$. Now this looks familiar to some extent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rewriting, we get $(\frac{y&amp;#39;}{2x})&amp;#39; = 2xy$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let $\frac{y&amp;#39;}{2x} = z(x)$. Plug this in and simplify to get $z&amp;#39; = \frac{y&amp;#39;}{z}y$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Replace $2x$ by $\frac{y'}{z}$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we have $z^2 = y^2 + c$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus we have now converted a second order differential equation in terms of first order differential equation, viz,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{1}{2x}\frac{dy}{dx} = \pm \sqrt{y^2 + c}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $c$ is a constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(You could plug this in and check that this satisfies the second order differential equation.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We now need other conditions (boundary/ initial conditions) to completely solve the problem i.e. to eliminate $c$ and other constant which will arise after solving the first order differential equation to get $y(x) = \exp(x^2)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Note that taking $c =0 $ we get a simple ode and the solution to which is $y(x) = y(0) \exp(\pm x^2)$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\textbf{EDIT:}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first order ODE can be solved as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\textbf{CASE 1:}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $c &amp;gt; 0$, then let $c = a^2$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rearranging, we get &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{dy}{\sqrt{y^2 + a^2}} = \pm d(x^2)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$y = a \tan(\theta)$, we get $dy = a \sec^2(\theta) d\theta$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, the ode now becomes,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sec(\theta) d\theta = \pm d(x^2)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$d(log(\sec(\theta) + \tan(\theta))) = \pm d(x^2)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$log(\sec(\theta) + \tan(\theta)) = \pm (x^2 + k)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sec(\theta) + \tan(\theta) = \exp(\pm (x^2 + k))$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Substitute for $\theta$ in terms of $y$ to get,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{y}{a} \pm \sqrt{1+(\frac{y}{a})^2} = K \exp(\pm x^2)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\textbf{CASE 2:}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $c &amp;gt; 0$, then let $c = -a^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rearranging, we get &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{dy}{\sqrt{y^2 - a^2}} = \pm d(x^2)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$y = a \sec(\theta)$, we get $dy = a \sec(\theta) \tan(\theta) d\theta$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, the ode now becomes,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sec(\theta) d\theta = \pm d(x^2)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$d(log(\sec(\theta) + \tan(\theta))) = \pm d(x^2)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$log(\sec(\theta) + \tan(\theta)) = \pm (x^2 + k)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sec(\theta) + \tan(\theta) = \exp(\pm (x^2 + k))$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Substitute for $\theta$ in terms of $y$ to get,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{y}{a} \pm \sqrt{(\frac{y}{a})^2 - 1} = K \exp(\pm x^2)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\textbf{CASE 3:}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $c = 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The equation, we have now is $\frac{dy}{dx} = \pm 2xy$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solving, we get $y(x) = K \exp(\pm x^2)$.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2010-11-23T02:58:11.263" LastActivityDate="2010-11-23T02:58:11.263" CommentCount="1" />
  <row Id="11433" PostTypeId="2" ParentId="11429" CreationDate="2010-11-23T02:31:31.470" Score="0" ViewCount="" Body="&lt;p&gt;You need to determine the payment so you come out even at the end of 10 years.  There should be a formula in your text for that.  If I recall correctly, it looks like payment = interest * principal/(1-(1+i)^-t), but I don't know what symbols you use for the terms.  Then you are right, as each payment is made you deduct the interest that has accrued in the past year, then apply the rest to the principal.  The interest due in the third year will be somewhat less than 9% of 50000 because the principal will have been reduced by the first two year's payments.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-11-23T02:31:31.470" />
  <row Id="11434" PostTypeId="1" CreationDate="2010-11-23T02:37:24.847" Score="0" ViewCount="248" Body="&lt;p&gt;Hello, I need help studying for an exam.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Give a recursive form (including bases) for the following functions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;f(n) = 5 + (-1)^n&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;f(n) = n(n+3)&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lastly,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;S(0) = 3&lt;/p&gt;&#xA;&#xA;&lt;p&gt;S(1) = 10&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Derive the closed form for this recursion and prove it true by induction.&lt;/p&gt;&#xA;" OwnerUserId="3800" LastActivityDate="2010-11-27T18:02:32.883" Title="How do I do Recursion?" Tags="&lt;discrete-mathematics&gt;" AnswerCount="4" CommentCount="3" />
  <row Id="11435" PostTypeId="1" AcceptedAnswerId="11481" CreationDate="2010-11-23T02:55:16.287" Score="4" ViewCount="146" Body="&lt;p&gt;&lt;strong&gt;Question&lt;/strong&gt;: What is the correct notion of a &lt;em&gt;product&lt;/em&gt; of integral (or rational) polytopes which induces a factorization of its Ehrhart (quasi-)polynomial into two primitive Ehrhart (quasi-)polynomials corresponding to its constituent polytopes, viz., $L_{P \times Q}(t) = L_{P}(t) L_{Q}(t)$? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(&lt;em&gt;Motivation&lt;/em&gt;) Given two closed integral polytopes $P$ and $Q$ each with vertices at $\{ \mathbf{0} , b_{1} \mathbf{e}_{1}, \dots, b_{n} \mathbf{e}_{n} \}$ and $\{ \mathbf{0} , d_{1} \mathbf{e}_{1}, \dots, d_{m} \mathbf{e}_{m} \}$, respectively, where $n, b_{i}, m, d_{j} \in \mathbb{N}$, define the integral polytope $R$ with vertices at $\{ \mathbf{0}, b_{1} \mathbf{e}_{1}, \dots, b_{n} \mathbf{e}_{n}, d_{1} \mathbf{e}_{n+1}, \dots, d_{m} \mathbf{e}_{n+m} \}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The above construction cannot be the sought after product $P \times Q$. Suppose $P$ and $Q$ are defined by $b_{1} = b_{2} = d_{1} = d_{2} = 2$. It is easy to show that $L_{P}(1) = L_{Q}(1) = 6$. Define $R$ as above with vertices of $P$ and $Q$. It is true that $L_{R}(1) = 15 \neq 6^{2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question&lt;/strong&gt;: What is $R$ in terms of $P$ and $Q$? Is it special in some way?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="2720" LastEditorUserId="2720" LastEditorDisplayName="" LastEditDate="2010-11-23T04:41:18.210" LastActivityDate="2010-11-23T16:55:25.710" Title="Product of Polytopes" Tags="&lt;number-theory&gt;&lt;combinatorics&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="2" />
  <row Id="11436" PostTypeId="2" ParentId="11428" CreationDate="2010-11-23T02:57:05.413" Score="5" ViewCount="" Body="&lt;p&gt;$\rm\quad\quad\quad \ y&amp;#39;\ =\ f\ y $  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\Rightarrow\ y&amp;#39;&amp;#39;\ =\ f\:&amp;#39;\ y + f^{\:2}\ y $  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\displaystyle\Rightarrow\ y&amp;#39;&amp;#39;\ = \frac{f\:&amp;#39;}f\ y&amp;#39; + f^{\:2}\ y $&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So $\rm\quad\displaystyle y&amp;#39;&amp;#39;\ = \ \frac{1}x\ y&amp;#39; + 4x^2\ y\ \ \Rightarrow\ \ f\ =\ \pm 2\:x$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and $\rm\ y&amp;#39;\ =\ \pm 2\:x\ y\ \ \Rightarrow\ \ y\ =\ c\ e^{\pm x^2}$&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-11-23T03:02:41.553" LastActivityDate="2010-11-23T03:02:41.553" CommentCount="6" />
  <row Id="11438" PostTypeId="1" AcceptedAnswerId="11445" CreationDate="2010-11-23T03:04:25.443" Score="2" ViewCount="147" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am trying to compute the Kolmogorov distance between two univariate gaussian distributions $\mathcal{N}(0,n)$ and $\mathcal{N}(0,2n)$ for large $n$.  I have a feeling this should be simple but whatever I have tried so far doesn't work.  Could anyone give me some hints?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the Kolmogorov distance between two distributions $P$ and $Q$ I mean:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle max_t | \Pr[P&amp;gt;t] - \Pr[Q&amp;gt;t] | $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks,&lt;/p&gt;&#xA;" OwnerUserId="3831" LastActivityDate="2010-11-23T03:38:16.203" Title="Kolmogorov distance between univariate gaussians" Tags="&lt;real-analysis&gt;&lt;probability&gt;" AnswerCount="1" CommentCount="0" />
  <row Id="11439" PostTypeId="2" ParentId="11434" CreationDate="2010-11-23T03:22:31.793" Score="3" ViewCount="" Body="&lt;p&gt;Recursion refers to a method where you figure out the &quot;next&quot; value in terms of previous values. For example, to solve the &lt;a href=&quot;http://math.stackexchange.com/questions/2650/how-to-prove-the-optimal-towers-of-hanoi-strategy&quot;&gt;Towers of Hanoi&lt;/a&gt; puzzle, you can explain the solution recursively: to move $n$ disks, first move the top $n-1$ disks to the auxiliary rod, then move the bottom disk to the target rod, then move the other $n-1$ disks to the target rod. The &quot;move the top $n-1$ disks&quot; parts of the instruction are recursive, because you need to know how to solve the $n-1$ case. And you solve that case by solving the $n-2$ case. And you solve &lt;em&gt;that&lt;/em&gt; case by solving the $n-3$ case. And you ... etc. Eventually you need to touch bottom, which you do at the $n=1$ case, the base.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, let's think about $f(n)=5+(-1)^n$. Can you express $f(n+1)$ in terms of $f(n)$? If so, then this expression plus the value of $f(0)$ will give you the recursive forms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To guide you, let me do a different example. Consider $f(n) = 2^{n}-1$. How do we express it recursively? First, I try to see if I can express $f(n+1)$ in terms of $f(n)$:&#xA;$$f(n+1) = 2^{n+1}-1 = 2(2^n) - 1 = 2\Bigl((2^n - 1)+1\Bigr) -1 = 2(f(n)+1)-1 = 2f(n)+1.$$&#xD;&#xA;(Why did I do that? Because I need to &quot;find&quot; the expression $2^n-1$ somehow &lt;em&gt;in&lt;/em&gt; the expression $2^{n+1}-1$, so that I can replace it with $f(n)$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This, together with the fact that $f(0)=2^0-1 = 0$ gives me the recursive form:&#xA;\begin{align*}&#xA;f(0) &amp;amp;= 0;\\&#xA;f(n+1) &amp;amp;= 2f(n) + 1.&#xA;\end{align*}&#xA;Try something similar with your two $f(n)$s.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for your &quot;lastly&quot;, I have no idea what you have. Giving just the first two values does not tell you what a &quot;closed form&quot; for $S$ will be (here, &quot;closed form&quot; is the opposite process: if you know the recursive formula, you want to find a formula of the form $S(n) = $blah, where blah depends only on $n$ and not on previous values of $S$). &lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-11-23T03:22:31.793" />
  <row Id="11440" PostTypeId="2" ParentId="11434" CreationDate="2010-11-23T03:23:38.097" Score="2" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\rm\ \ (-1)^{n+1}/(-1)^n = -1,\ $ so substituting $\rm\ (-1)^n\ =\ f(n)-5\ \Rightarrow\ \ldots$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the second, note that $\rm\ f(n+1) - f(n)\ $ reduces the degree of polynomial $\rm\ f(n)\:$.&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-11-23T03:23:38.097" />
  <row Id="11441" PostTypeId="1" CreationDate="2010-11-23T03:30:13.690" Score="2" ViewCount="129" Body="&lt;p&gt;I'd like to know about the best numerical methods for approximating an integral.  Unfortunately, I want to know about a fairly general case, so I cannot give a lot of information.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Essentially, I have an integral that is full of exponential functions and little else.  I therefore assume as given, although possibly incorrectly, that the power series expansion of the integrand can be integrated to give an exact value to the integral.  In other words, we can take the power series expansion of the integrand (the function inside the integral), and integrate it to obtain the same value as integrating the integral itself gives.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I then &quot;define&quot; the value $n$, saying that we need at least $n$ terms of the power series expansion to integrate to the wanted degree of accuracy/precision.  I also &quot;define&quot; $m$, such that by numerical methods, the integral must be evaluated at at least $m$ points to obtain the same degree of accuracy/precision.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hoping I've stated the integral and defined some restraints well, I have some curiosities.  I'm wondering how well numerical methods can hope to achieve these requirements.  In other words, how many calculations, and how much memory, etc. is needed to approximate this integral?  For example, if we know that we have to evaluate the integral at $m$ points, does it take $O(m)$ calculations to evaluate this integral, or perhaps $O(m \log(m))$?  Similarly, if we know that the power series expansion requires $n$ terms to get the job done, how much computing power do we need?  $O(n)$?  $O(n \log(n))$?  Some other value?  I know that this is an extremely general case, but I'm curious as to how much computing requirements various methods take?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm asking because I want to compare a method that I'm experimenting with.  I doubt that I can do as well as the best methods, but I would like to know what the best method is so I can eventually use it.&lt;/p&gt;&#xA;" OwnerUserId="2626" LastActivityDate="2010-11-23T03:30:13.690" Title="Numerical Methods: Approximating an Integral with Exponentials" Tags="&lt;calculus&gt;&lt;sequences-and-series&gt;&lt;intuition&gt;&lt;numerical-methods&gt;&lt;numerical-optimization&gt;" CommentCount="4" />
  <row Id="11442" PostTypeId="2" ParentId="11431" CreationDate="2010-11-23T03:32:23.577" Score="9" ViewCount="" Body="&lt;p&gt;A &lt;a href=&quot;http://www.ams.org/journals/tran/1923-025-03/S0002-9947-1923-1501252-3/S0002-9947-1923-1501252-3.pdf&quot; rel=&quot;nofollow&quot;&gt;classic result of Ritt&lt;/a&gt; shows that polynomials that commute under composition must be, up to a linear homeomorphism, either both powers of $x$, both iterates of the same polynomial, or both  Chebychev polynomials. Actually Ritt proved a more general rational function case - follow the link. His work was motivated by work of Julia and Fatou's work on Julia sets of rational functions, e.g. see &lt;a href=&quot;http://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.47.8482&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; for a modern presentation.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-11-23T03:45:15.153" LastActivityDate="2010-11-23T03:45:15.153" CommentCount="3" />
  <row Id="11443" PostTypeId="1" CreationDate="2010-11-23T03:36:51.807" Score="17" ViewCount="339" Body="&lt;p&gt;Let $n$ be a positive integer. Suppose we have an equilateral polygon in the Euclidean plane with the property that all angles except possibly two consecutive ones are an integral multiple of $\pi/n$, then all angles are an integral multiple of $\pi/n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This problem is #28 on page 61 in these notes restated here for convenience:&#xA;&lt;a href=&quot;http://websites.math.leidenuniv.nl/algebra/ant.pdf&quot; rel=&quot;nofollow&quot;&gt;http://websites.math.leidenuniv.nl/algebra/ant.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have seen a number-theoretic proof of this. I was wondering if there are any geometric (or at least non number-theoretic) proofs of this result.&lt;/p&gt;&#xA;" OwnerUserId="3431" LastEditorUserId="3431" LastEditorDisplayName="" LastEditDate="2010-12-12T01:02:54.250" LastActivityDate="2011-05-16T13:13:48.740" Title="Equilateral polygon in a plane" Tags="&lt;geometry&gt;&lt;algebraic-number-theory&gt;" AnswerCount="1" CommentCount="6" FavoriteCount="1" />
  <row Id="11444" PostTypeId="2" ParentId="11431" CreationDate="2010-11-23T03:37:47.033" Score="4" ViewCount="" Body="&lt;p&gt;According to Wikipedia, a set of diagonalizable matrices commute if and only if they are simultaneously diagonalizable. There is a far-reaching generalization, namely the &lt;a href=&quot;http://en.wikipedia.org/wiki/Gelfand_representation#Statement_of_the_commutative_Gelfand-Naimark_theorem&quot; rel=&quot;nofollow&quot;&gt;Gelfand representation theorem&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Gelfand representation theorem for commutative $C^*$ algebras represents every commutative $C^*$ algebra as an algebra of functions with pointwise multiplication; the domain of the latter algebra is the spectrum of the former algebra.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-23T03:37:47.033" CommentCount="2" />
  <row Id="11445" PostTypeId="2" ParentId="11438" CreationDate="2010-11-23T03:38:16.203" Score="2" ViewCount="" Body="&lt;p&gt;Because the Gaussians have the same mean and Gaussians are symmetric about the mean, the expression $\Pr[P&amp;gt;t] - \Pr[Q&amp;gt;t]$ should have a single maximum and a single minimum whose absolute values are the same.  So you could just solve $\max_t (\Pr[P&amp;gt;t] - \Pr[Q&amp;gt;t])$.  To do that, I would express $\Pr[P&amp;gt;t]$ as &#xA;$$\int_t^{\infty} \frac{1}{\sqrt{2\pi n}} e^{-x^2/(2n)} dx,$$&#xD;&#xA;and $\Pr[Q&amp;gt;t]$ similarly.  Then take the derivative of the differences with respect to $t$ via the Fundamental Theorem of Calculus, set it to $0$, and solve for $t$.  Since you don't want a full answer, I'll stop there. &lt;/p&gt;&#xA;" OwnerUserId="2370" LastActivityDate="2010-11-23T03:38:16.203" CommentCount="3" />
  <row Id="11446" PostTypeId="2" ParentId="11426" CreationDate="2010-11-23T03:42:29.170" Score="1" ViewCount="" Body="&lt;p&gt;Hint: Expand ZZM using the definition of Z &lt;i&gt;once&lt;/i&gt; (for the outer Z).&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-23T03:42:29.170" />
  <row Id="11447" PostTypeId="2" ParentId="11382" CreationDate="2010-11-23T04:16:16.673" Score="2" ViewCount="" Body="&lt;p&gt;In terms of simply the math, it is essentially a skill you need to learn. Mathematics becomes (a long with many other sciences) more &lt;em&gt;abstract&lt;/em&gt; as you go up in rank, so to speak. You can think of multiplying and adding polynomials as the next logical step after learning how to multiply decimals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a natural progression, that is both historic and customary to progress from the most concrete to the more abstract. In grade school, the first mathematical concept you learned is numbers, shortly thereafter you were introduced to how you could &lt;em&gt;combine&lt;/em&gt; two numbers to form a third one (namely through addition and multiplication). After being introduced to natural numbers, the concept of combining two numbers was expanded to something called fractions, which also introduced the concept of division (the opposite of multiplication, if you will). After getting comfortable with fractions, you were introduced to decimals and &lt;em&gt;irrational&lt;/em&gt; numbers, and the rules for combining and manipulating those numbers. At this point, mathematical education turns to more abstract concepts, the idea of &lt;em&gt;functions&lt;/em&gt; and &lt;em&gt;variables&lt;/em&gt;. You spend plenty of time (depending on how far you go in math, maybe the rest of your mathematical career) studying functions, how they behave, how to manipulate them. It turns out that polynomials, are some of the simplest functions you can think of, so it is only natural that you start with them. Following the trend as before, after understanding what functions (specifically polynomials) &lt;em&gt;are&lt;/em&gt;, a natural next step is considering how you can &lt;em&gt;combine&lt;/em&gt; those polynomials to make new polynomials. Later on in math, you will be introduced with even more ways of combining and manipulating functions to create new functions. This is where you are right now, you are learning to manipulate and combined some (if you think about it) incredibly abstract concept, you are learning to manipulate an object that took thousands of years of human thought to finalize. Of course, in reality this is just a small stepping stone to the greener pastures ahead, assuming you go on in mathematics you will encounter some of the most mind boggling abstract concepts (some of the negatively voted answers to this question will give you just a small hint of whats out there). You will find that the very simple concepts of addition and multiplication (and their respective inverses) will go a long way in the study of mathematics, but they are stripped of all their cuddly original meaning, to become the most beefed up behemoths of abstract mathematical thought.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, enough with the theoretical &quot;its the way it should be&quot;. Lets move on to practicalities. Applications of numbers, are easy enough to see, addition,multiplication, subtraction, and division of numbers is an everyday thing. Considering how to apply polynomials to your everyday life (especially why you would ever think of combining polynomials) requires a little greater stretch of the mind (or a little more mathematics :D ). Without appealing to calculus (which is certainly an easy cop-out).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Polynomials, Describe a large variety of every day processes and events. The most common being the quadratic polynomial (this is due to a very deep physical fact, but I promised to avoid calculus). The free fall trajectory of an object in a gravitational field is described (neglecting friction) by a quadratic polynomial. Even in simple physical problems you can see the multiplication of monomials and quadratics. For example, solving for when a shot cannon hits the ground is merely the root of the polynomial, conversely, if you know where the canon hit the ground you can multiply to monomials $(x-r)$ where $r$ is the root, to get the trajectory. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we deviate a little from the math you have been exposed to, there are more abstract polynomials called &lt;a href=&quot;http://en.wikipedia.org/wiki/Laguerre_polynomials&quot; rel=&quot;nofollow&quot;&gt;Laguerre Polynomials&lt;/a&gt; that are present in the solution to the hydrogen atom. So if you (for some reason) wanted to calculate the shape of a 3D orbital of the hydrogen atom, you would need to multiply a lot of polynomials. Related polynomials play important roles in other problems in quantum mechanics and classic physics (Legendre Polynomials for example). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hopefully this clears some of it up, unfortunately in a large sense it is really a stepping stone, into the world of higher mathematics. They are important on their own, but for reasons that... also become much more obvious later on. &lt;/p&gt;&#xA;" OwnerUserId="1906" LastActivityDate="2010-11-23T04:16:16.673" CommentCount="3" />
  <row Id="11448" PostTypeId="1" AcceptedAnswerId="11449" CreationDate="2010-11-23T04:22:22.423" Score="7" ViewCount="478" Body="&lt;p&gt;Let &lt;em&gt;n&lt;/em&gt; and &lt;em&gt;k&lt;/em&gt; be integers with 1 $\leq$ k $\leq$ n. Show that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{k=1}^n {n \choose k}{n \choose k-1} = \frac12{2n+2 \choose n+1} - {2n \choose n}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was told this is supposed to use a combinatorial proof and while I'm not that comfortable with that, many similar proofs use a mathematical proof that shows equality. Any guidance would be much appreciated.&lt;/p&gt;&#xA;" OwnerUserId="1880" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2011-05-11T18:29:23.767" LastActivityDate="2011-05-11T18:44:35.423" Title="Combinatorial Proof of Binomial Coefficient Summation" Tags="&lt;combinatorics&gt;&lt;binomial-coefficients&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="11449" PostTypeId="2" ParentId="11448" CreationDate="2010-11-23T04:30:10.307" Score="10" ViewCount="" Body="&lt;p&gt;The Combinatorial Proof goes as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider three boxes $A$,$B$ and $C$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Box $A$ contains $n$ balls, Box $B$ contains $n$ balls, and Box $C$ contains $2$ balls.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We want to choose $n+1$ balls out of these three boxes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The total number of ways is $C(2n+2,n+1)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This can be split as follows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You choose $2$ balls from box $C$ and the remaining $n-1$ balls from boxes $A$ and $B$ or you choose $1$ ball from box $C$ and the remaining $n$ balls from boxes $A$ and $B$ or you choose all the $n+1$ balls from boxes $A$ and $B$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\textbf{Case 1:}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose you choose $2$ balls from box $C$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is only one way of choosing these $2$ balls from the box $C$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We now need the number of ways of choosing $n-1$ balls from boxes $A$ and $B$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Say if you choose $k$ balls from box $A$, you need to choose $n-k-1$ balls from box $B$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Number of ways of choosing $k$ balls from box $A$ and $n-k-1$ balls from box $B$ is given by $C(n,k)C(n,n-k-1)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now summing over all possible values of $k$ from $0$ to $n-1$ we get,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Total number of ways is $\displaystyle \sum_{k=0}^{n-1} C(n,k)C(n,n-k-1)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, $C(n,r) = C(n,n-r)$. (For a combinatorial argument of this see the bottom of the post).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, Total number of ways for the current case is $1 \times \displaystyle \sum_{k=0}^{n-1} C(n,k)C(n,k+1) = \displaystyle \sum_{k=1}^{n} C(n,k-1)C(n,k)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\textbf{Case 2:}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose you choose $1$ ball from the box $C$. You need to now choose the remaining $n$ balls from boxes $A$ and $B$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Number of ways of choosing $1$ ball from box $C$ is given by $C(2,1) = 2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are $2n$ balls in total in boxes $A$ and $B$ and hence the total number of ways of choosing $n$ balls from boxes $A$ and $B$ is $C(2n,n)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, the total number of ways for the current case is $2 \times C(2n,n)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\textbf{Case 3:}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose you choose all the $n+1$ balls from boxes $A$ and $B$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Number of ways of choosing no balls from box $C$ is $1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We now need to count the number of ways of choosing $n+1$ balls from boxes $A$ and $B$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is similar to Case 1 and we get the total number of ways for this case is $1 \times \displaystyle \sum_{k=1}^{n} C(n,k)C(n,n+1-k) = \displaystyle \sum_{k=1}^{n} C(n,k)C(n,k-1)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Note: Here $k$ goes from $1$ to $n$ since we need to choose atleast $1$ from one of the boxes and we can choose atmost $n$ from each box).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Also Note: You could directly state that Case 1 and Case 3 should be the same since $C(2n,n-1)=C(2n,n+1)$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Combine all the three cases to get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$2 \times \displaystyle \sum_{k=1}^{n} C(n,k)C(n,k-1) + 2 \times C(2n,n) = C(2n+2,n+1)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now do the desired algebraic manipulation to get the result.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\textbf{APPENDIX:}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The combinatorial argument of $C(n,r) = C(n,n-r)$ is that the number of ways of selecting $r$ out of $n$ things is the same as the number of ways of discarding $n-r$ out of $n$ things.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="6179" LastEditorDisplayName="" LastEditDate="2011-05-11T18:44:35.423" LastActivityDate="2011-05-11T18:44:35.423" CommentCount="6" />
  <row Id="11450" PostTypeId="2" ParentId="11448" CreationDate="2010-11-23T04:32:02.817" Score="4" ViewCount="" Body="&lt;p&gt;Hint: Use $\displaystyle {n \choose k} = {n \choose n-k}$ and try to count something which equals $\displaystyle {n \choose n-k}{n \choose k-1}$&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-11-23T04:32:02.817" />
  <row Id="11451" PostTypeId="1" AcceptedAnswerId="11466" CreationDate="2010-11-23T04:33:50.287" Score="2" ViewCount="412" Body="&lt;p&gt;Hello, I need help studying for an exam. How do I find &quot;Big Oh,&quot; &quot;Big Omega,&quot; and &quot;Big Theta&quot;?&#xA;How do I combine this with Induction?&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Prove that $f(x)$ is $O(x^n)$, where&#xA;$$f(x) = a_n x^n + a_{n-1} x^{n-1} + a_{n-2} x^{n-2} + \cdots + a_2 x^2 + a_1 x + a_0$$&#xD;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Find &quot;Big O&quot;: (Usually log in Computer Science is assumed to be base 2)&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$(2n+\log(n!)+3)(n^2+3\log n)(2^n+2)$&#xD;&lt;/li&gt;&#xA;&lt;li&gt;$(3n+1)\log(n^5+7)+n!\log(n+\log n)$&#xD;&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;For each function, circle the sets that $f(n)$, $g(n)$ and $h(n)$ belong to:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$f(n) = n\log(n!) + 100$: $O(g(n))$, $\Theta(g(n))$, $\Omega(g(n))$&#xD;&lt;/li&gt;&#xA;&lt;li&gt;$g(n) = (n \log n)^2 + n^2$:&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$O(n^3)$, $\Theta(n^3)$, $\Omega(n^3)$&#xD;&lt;/li&gt;&#xA;&lt;li&gt;$O(n^2)$, $\Theta(n^2)$, $\Omega(n^2)$&#xD;&lt;/li&gt;&#xA;&lt;li&gt;$O(n \log n)$, $\Theta(n \log n)$, $\Omega(n \log n)$&#xD;&lt;/li&gt;&#xA;&lt;/ol&gt;&lt;/li&gt;&#xA;&lt;li&gt;$h(n) = n(1000 + 2^n) + n!$:&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$O(2^n)$, $\Theta(2^n)$, $\Omega(2^n)$&#xD;&lt;/li&gt;&#xA;&lt;li&gt;$O(n!)$, $\Theta(n!)$, $\Omega(n!)$&#xD;&lt;/li&gt;&#xA;&lt;li&gt;$O(n^n)$, $\Theta(n^n)$, $\Omega(n^n)$&lt;/li&gt;&#xA;&lt;/ol&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="3800" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2010-11-23T04:57:16.560" LastActivityDate="2010-11-23T05:56:40.463" Title="How do I find &quot;Big Oh,&quot; &quot;Big Omega,&quot; and &quot;Big Theta&quot;?" Tags="&lt;discrete-mathematics&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" />
  <row Id="11452" PostTypeId="1" AcceptedAnswerId="11462" CreationDate="2010-11-23T04:36:04.340" Score="0" ViewCount="104" Body="&lt;p&gt;I have a formula which is G-force = velocity*omega/9.8. Omega is the angular velocity.&#xA;I've seen on the internet that G force is actually acceleration/9.8. I'm confused as to which formula is correct. For simulating the motion of particle taking a turn, would omega simply be velocity divided by radius of turn? Assuming Cartesian coordinates.&lt;/p&gt;&#xA;" OwnerUserId="3737" LastActivityDate="2010-11-23T05:40:21.483" Title="Calculation of G force" Tags="&lt;computer-science&gt;&lt;physics&gt;&lt;simulation&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="11453" PostTypeId="1" AcceptedAnswerId="11461" CreationDate="2010-11-23T04:37:59.397" Score="4" ViewCount="177" Body="&lt;p&gt;I found this puzzle posted on a student website I frequent, but no one including the poster nor I could solve it. So I'm posting this puzzle here with hope that some of math whiz in here could illuminate on how to approach it:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;There is an alien on the surface of a&#xA;  spherical planet. The alien can run at a top&#xA;  speed of $u$. An alien hunter is hunting&#xA;  the alien in his spaceship which can&#xA;  fly at a top speed of $v$. Once the&#xA;  hunter sees the alien, he fires, and&#xA;  the alien dies. Show that the alien&#xA;  will always die if $v &amp;gt; 10u$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Note that there's no information given about the planet radius nor the altitude at which the spaceship hovers so I'm suspecting that the solution somehow ignores them or they get canceled out somewhere in the calculation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;p/s: again I repeat that I do not know of the solution myself&lt;/p&gt;&#xA;" OwnerUserId="3803" LastActivityDate="2010-11-23T05:28:01.250" Title="Alien vs Alien Hunter Puzzle" Tags="&lt;geometry&gt;&lt;puzzle&gt;&lt;proof-strategy&gt;" AnswerCount="2" CommentCount="7" />
  <row Id="11454" PostTypeId="1" CreationDate="2010-11-23T04:40:57.877" Score="0" ViewCount="194" Body="&lt;p&gt;Help would be appreciated. The notes are poor on the subject, and im clueless. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Verify the following equalities:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A) SIII=βI, where S is λxyz.(xz)(yz) and I is λx.x&lt;/p&gt;&#xA;&#xA;&lt;p&gt;B) twice (twice) f x= β f(f(f(f x))), where twice is λfx.f(f x)&lt;/p&gt;&#xA;" LastActivityDate="2011-01-22T06:46:45.197" Title="lambda calculus, equalities" Tags="&lt;lambda-calculus&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="11455" PostTypeId="2" ParentId="11453" CreationDate="2010-11-23T04:49:11.133" Score="2" ViewCount="" Body="&lt;p&gt;There's a similar puzzle &lt;a href=&quot;http://domino.research.ibm.com/Comm/wwwr_ponder.nsf/challenges/May2001.html&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; (with solution).&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-23T04:49:11.133" CommentCount="2" />
  <row Id="11456" PostTypeId="2" ParentId="11454" CreationDate="2010-11-23T04:58:34.577" Score="1" ViewCount="" Body="&lt;p&gt;Try substituting the arguments in the definitions of the combinators.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2010-11-23T06:15:58.457" LastActivityDate="2010-11-23T06:15:58.457" CommentCount="4" />
  <row Id="11457" PostTypeId="2" ParentId="11412" CreationDate="2010-11-23T05:04:11.173" Score="4" ViewCount="" Body="&lt;p&gt;The point of this answer is mostly to provide some terminology with which to express the distinction here.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A topological space is &lt;strong&gt;totally disconnected&lt;/strong&gt; if the only nonempty connected subsets are the singleton sets.  Equivalently, the connected components are all singleton sets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A topological space is &lt;strong&gt;zero-dimensional&lt;/strong&gt; if it admits a base of clopen [i.e., both open and closed] sets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A topological space is &lt;strong&gt;separated&lt;/strong&gt; (more traditionally: $T_1$) if all the singleton sets are closed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First easy observation: A totally disconnected space is separated.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Proof: Otherwise the closure of a point would give a larger connected component.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second easy observation: In a separated zero-dimensional space $X$, for any two distinct points $x_1$, $x_2$, there exists a separation $X = U_1 \coprod U_2$ with $x_i \in U_i$. &#xA;(In the terminology of Nuno's answer -- which is relatively standard if not well-known -- the conclusion is that the quasi-components are singleton sets.) In particular, $X$ is totally disconnected. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Proof: Let $x_1$, $x_2$ be distinct points in $X$.  Since $X$ is separated, $N = X \setminus \{x_1\}$ is open.  By definition of a base, there exists a clopen set $U_1$ with $x \in U_1 \subset N$ and then $U_1$, $U_2 = X \setminus U_1$ is the desired separation.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By contrast, we have the following&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nontrivial result: A locally compact Hausdorff space is zero-dimensional iff it is totally disconnected.  For instance &lt;a href=&quot;http://en.wikipedia.org/wiki/Zero-dimensional_space&quot; rel=&quot;nofollow&quot;&gt;this wikipedia article&lt;/a&gt; gives a reference.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that Nuno's answer (which should be the accepted one, IMO) gives a proof of this in the compact case.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps a cleaner approach is to &quot;redefine&quot; a Stone space to be a space which is compact, Hausdorff and zero-dimensional.  Then when you are given a space that someone says is a Stone space, the first thing you'll do is look for a base of clopen sets, which tends to be a good idea anyway.  For instance, given any profinite space (i.e., an inverse limit of finite discrete spaces, yet another term for the same class of spaces!) it is very easy to do this.  &lt;/p&gt;&#xA;" OwnerUserId="299" LastActivityDate="2010-11-23T05:04:11.173" CommentCount="3" />
  <row Id="11458" PostTypeId="1" AcceptedAnswerId="11459" CreationDate="2010-11-23T05:08:23.137" Score="0" ViewCount="121" Body="&lt;p&gt;How to proceed in this problem ?&lt;/p&gt;&#xA;" OwnerUserId="2109" LastEditorUserId="2109" LastEditorDisplayName="" LastEditDate="2010-11-23T05:14:45.273" LastActivityDate="2010-11-23T06:27:53.110" Title="If $ x = \log_{12} 27 \text {,then what is the the value of } \log_6 16 $ ?" Tags="&lt;algebra-precalculus&gt;&lt;logarithms&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="11459" PostTypeId="2" ParentId="11458" CreationDate="2010-11-23T05:15:09.143" Score="5" ViewCount="" Body="&lt;p&gt;$x = \log_{12}(27) = 3 \log_{12}(3)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{1}{x} = \frac{1}{3} \frac{1}{\log_{12}(3)} = \frac{1}{3} \log_{3}(12) = \frac{1}{3} \log_{3}(3 \times 4) = \frac{1}{3} (1 + 2 \log_{3}(2))$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Simplifying, we get $\log_{3}(2) = \frac{3}{2x} - \frac{1}{2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $y = \log_{6}(16) = 4 \log_{6}(2) = 4 \frac{1}{\log_{2}(6)} = \frac{4}{1 + \log_{2}(3)}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now make use of the fact that $\log_{3}(2) = \frac{1}{\log_{2}(3)}$ to get $y$ in terms of $x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I get $y = 4 (\frac{3-x}{3+x})$&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2010-11-23T06:25:40.647" LastActivityDate="2010-11-23T06:25:40.647" CommentCount="1" />
  <row Id="11460" PostTypeId="1" CreationDate="2010-11-23T05:22:43.470" Score="0" ViewCount="42" Body="&lt;p&gt;What does it mean full 3D per-point anisotropic covariance estimates of a matrix?&lt;/p&gt;&#xA;" OwnerUserId="3836" LastActivityDate="2010-11-23T05:22:43.470" Title="Anisotropic covariance estimates" Tags="&lt;linear-algebra&gt;" CommentCount="1" />
  <row Id="11461" PostTypeId="2" ParentId="11453" CreationDate="2010-11-23T05:28:01.250" Score="3" ViewCount="" Body="&lt;p&gt;Let us measure distance in planet radius and time such that the alien can make one radian per unit time on the surface.  If I fly along the surface at altitude h, I can see a swath of width $2 \arccos (\frac{1}{1+h})$ and can fly around the equator in $\frac{2 \pi (1+h)}{10}$, so as long as $2 \arccos (\frac{1}{1+h})-\frac{2 \pi (1+h)}{10} \gt 0$ I can make a spiral starting at one pole and push the alien in front of me down to the other pole.  At h of about 0.55, this only requires my speed to be about 5.6 instead of 10.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-11-23T05:28:01.250" CommentCount="5" />
  <row Id="11462" PostTypeId="2" ParentId="11452" CreationDate="2010-11-23T05:40:21.483" Score="0" ViewCount="" Body="&lt;p&gt;For circular motion, the two expressions are equivalent.  Yes, in circular motion omega is velocity/radius.  Acceleration in circular motion is velocity^2/radius.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-11-23T05:40:21.483" />
  <row Id="11463" PostTypeId="2" ParentId="11458" CreationDate="2010-11-23T05:45:31.477" Score="1" ViewCount="" Body="&lt;p&gt;Put $\rm\ \ell\: n\ =\ \log_6 n\:.\ \rm\ 27\ =\ 12^x\ \Rightarrow\ \ell\: 27\ =\ x\:(\ell\:6 + \ell\: 2)\ = x\:(1 + \ell\:2)\:.\ $ Now solve this for $\rm\:\ell\:2\:$ and plug the result into $\rm \ell\:16\ =\ 4\ \ell\: 2\ =\ \ldots$ &lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-11-23T06:27:53.110" LastActivityDate="2010-11-23T06:27:53.110" CommentCount="6" />
  <row Id="11464" PostTypeId="1" AcceptedAnswerId="11549" CreationDate="2010-11-23T05:52:25.493" Score="3" ViewCount="194" Body="&lt;p&gt;Hello all,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given $$1\cdot 2^1 + 2\cdot 2^2 + 3\cdot 2^3 + 4\cdot 2^4 + \cdots + d \cdot 2^d = \sum_{r=1}^d r \cdot 2^r,$$&#xD;&#xA;how can we infer to the following solution? $$2 (d-1) \cdot 2^d + 2. $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you&lt;/p&gt;&#xA;" OwnerUserId="3838" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2010-11-23T05:58:44.127" LastActivityDate="2010-11-23T18:42:09.403" Title="How to compute the following formula" Tags="&lt;discrete-mathematics&gt;" AnswerCount="7" FavoriteCount="1" />
  <row Id="11465" PostTypeId="1" CreationDate="2010-11-23T05:56:22.870" Score="8" ViewCount="139" Body="&lt;p&gt;I am an undergraduate learning about gauge theory and I have been tasked with working through the two examples given on pages 65 and 66 of &quot;&lt;a href=&quot;http://www.jstor.org/pss/1971013&quot; rel=&quot;nofollow&quot;&gt;Characteristic forms and geometric invariants&lt;/a&gt;&quot; by Chern and Simon. I will recount the examples and my progress at a solution. For ease here is the relevant text:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;em&gt;Example 1.&lt;/em&gt; Let $M = \mathbb{R}P^3 = SO(3)$ together with the standard metric of constant curvature 1. Let $E_1, E_2, E_3$ be an orthonormal basis of left invariant fields on $M$, oriented positively. Then it is easily seen that $\nabla_{E_1}E_2 = E_3, \nabla_{E_1}E_3 = - E_2, \text{ and } \nabla_{E_2}E_3 = E_1$. Let $\chi : M \rightarrow F(M)$ be the cross-section determined by this frame.&#xA;  $$\Phi(SO(3)) = \frac{1}{2}.$$&#xD;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;em&gt;Example 2.&lt;/em&gt; Again let $M = SO(3)$, but this time with left invariant metric $g_{\lambda}$, with respect to which $\lambda E_1, E_2, E_3$ is an orthonormal frame. Direct calculation shows&#xA;  $$\Phi(SO(3),g_{\lambda}) = \frac{2\lambda^2 - 1}{2\lambda^4}.$$  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For each of these examples I am expected to calculate&lt;br&gt;&#xA;$$\Phi(M) = \int_{\chi(M)} \frac{1}{2} TP_1(\theta)$$&#xD;&#xA;which lies in $\mathbb{R}/\mathbb{Z}$. Previously in the paper they give an explicit formulation of $TP_1(\theta)$ in terms of the &quot;component&quot; forms of the connection $\theta$ and its curvature $\Omega$,&#xA;$$TP_1(\theta) = \frac{1}{4\pi^2}\left( \theta_{12}\wedge\theta_{13}\wedge\theta_{23} + \theta_{12}\wedge\Omega_{12} + \theta_{13}\wedge\Omega_{13} + \theta_{23}\wedge\Omega_{23}\right).$$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have verified this formula for myself given the information in the paper. Using the structural equation $\Omega = d\theta + \theta\wedge\theta$ I am able to reduce the expression for $TP_1(\theta)$ to&#xA;$$TP_1(\theta) = \frac{-1}{2\pi^2}\left( \theta_{12}\wedge\theta_{13}\wedge\theta_{23} \right).$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't believe I have assumed anything about the structure of $M$ during that reduction so I believe it should hold for both examples. I continue by claiming that since $E_1, E_2, E_3 \in so(3)$, the Lie algebra of $SO(3)$ I should be able to compute $\theta$ by considering&#xA;$$\nabla_{E_i}E_j := (\nabla E_j)(E_i) = \sum_k E_k \otimes \theta^{k}{}_{ij}(E_i)$$&#xD;&#xA;and comparing it with the given derivatives.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example one this yielded for me $\theta_{12} = E^3, \theta_{13} = -E^2, \theta_{23} = E^1$ where $E^i$ are the 1-forms dual to the basis $E_i$. Then I think that $\chi^*$ should act trivially on $TP_1(\theta)$ as it is a horizontal form in $\Lambda^*(T^*F(M))$. Therefore I find that $\chi^*(TP_1(\theta)) = \frac{1}{2\pi^2}\omega$, where $\omega$ is the volume form of $M$, and when integrated this yields the correct answer of $\frac{1}{2}$ for the first example.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, my approach fails completely for the second example. I assume that the set $\lambda E_1, E_2, E_3$ obeys the same derivate relationships as given in the first example, but this does not seem to give me enough factors of $\lambda$. I suspect that I am not handling the computation of the $\theta_{ij}$ forms or the application of $\chi^*$ correctly, however I am uncertain what my exact issue is. Is there a fundamental flaw in my understanding? I am hoping someone with more experience can point me in the right direction.&lt;/p&gt;&#xA;" OwnerUserId="2887" LastActivityDate="2010-11-23T05:56:22.870" Title="Computing the Chern-Simons invariant of SO(3)" Tags="&lt;homework&gt;&lt;differential-geometry&gt;" CommentCount="1" />
  <row Id="11466" PostTypeId="2" ParentId="11451" CreationDate="2010-11-23T05:56:40.463" Score="3" ViewCount="" Body="&lt;p&gt;For the first part, since $x^k \leq x^n$ for $k \leq n$ (whenever $x \geq 1$), the polynomial is at most $$(|a_n| + \cdots + |a_0|)x^n.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the second part, the following facts are useful:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$\log n! = O(n\log n)$.&lt;/li&gt;&#xA;&lt;li&gt;If $g(n) = o(f(n))$ then $f(n) + g(n) = O(f(n))$. For example, $2^n + 2 = O(2^n)$. The notation $g(n) = o(f(n))$ means that $\lim_{n\rightarrow\infty} \frac{g(n)}{f(n)} = 0$, i.e. $g$ grows much slower than $f$.&lt;/li&gt;&#xA;&lt;li&gt;If $f_1(n) = O(g_1(n))$ and $f_2(n)) = O(g_2(n))$ then $f_1(n)f_2(n) = O(g_1(n)g_2(n))$.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;For the third part, the following facts are useful:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$n! = \Theta(\sqrt{n}(n/e)^n)$ (Stirling's approximation).&lt;/li&gt;&#xA;&lt;li&gt;$\log n = O(n^{\epsilon})$ for every $\epsilon &amp;gt; 0$.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-23T05:56:40.463" />
  <row Id="11467" PostTypeId="2" ParentId="11464" CreationDate="2010-11-23T06:06:14.420" Score="8" ViewCount="" Body="&lt;p&gt;There are at least five ways:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Use plain induction.&lt;/li&gt;&#xA;&lt;li&gt;Differentiate the identity for the sum of a geometric series.&lt;/li&gt;&#xA;&lt;li&gt;Find some combinatorial interpretation for both sides and provide a bijection.&lt;/li&gt;&#xA;&lt;li&gt;Write as a sum of geometric series, sum each individual series, and sum the resulting geometric series.&lt;/li&gt;&#xA;&lt;li&gt;Divide both sides by $2^d$ and slightly modify the sum so that it has some probabilistic interpretation. Use known properties of the relevant random variable.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-23T06:06:14.420" />
  <row Id="11468" PostTypeId="1" AcceptedAnswerId="11475" CreationDate="2010-11-23T06:07:37.113" Score="4" ViewCount="243" Body="&lt;p&gt;Need help studying for an exam.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Practice Question:&#xA;If there are 200 students in the library, how many ways are there for them to be split among the floors of the library if there are 6 floors?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint: The students can not be told apart (they are indistinguishable).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer must bein terms of P(n,r), C(n,r), powers, or combinations of these. The answers do not have to be calculated.&lt;/p&gt;&#xA;" OwnerUserId="3800" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2010-12-21T01:42:54.787" LastActivityDate="2010-12-21T03:00:26.217" Title="If there are 200 students in the library, how many ways are there for them to be split among the floors of the library if there are 6 floors?" Tags="&lt;combinatorics&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="1" />
  <row Id="11470" PostTypeId="2" ParentId="11468" CreationDate="2010-11-23T06:12:03.060" Score="3" ViewCount="" Body="&lt;p&gt;Since the students are indistinguishable, we can number them from $1$ to $200$, and can assume that students $a_i$ to $a_{i+1}-1$ get assigned to floor $i$, where $a_1 = 1$ and $a_7 = 201$. Since $a_1,a_7$ are fixed, we have to choose $a_2,\ldots,a_6$. Note that $1 \leq a_2 \leq \cdots \leq a_6 \leq 200$, and this is the only restriction on them.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-23T06:12:03.060" />
  <row Id="11471" PostTypeId="2" ParentId="11464" CreationDate="2010-11-23T06:12:49.290" Score="5" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\displaystyle\rm\ \ \ r\ x^r\ =\ x \frac{d}{dx} (x^r)\:.\ $ Apply this to $\rm\ \sum_{r=1}^d x^r\ $ then put $\rm\ x = 2\:$.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-11-23T06:28:40.533" LastActivityDate="2010-11-23T06:28:40.533" />
  <row Id="11472" PostTypeId="2" ParentId="11145" CreationDate="2010-11-23T06:15:24.477" Score="8" ViewCount="" Body="&lt;p&gt;You seem to have solved this on your own with &quot;bare hands&quot; but I wanted to point out some connections that this problem has with slightly more general concepts from Banach space theory.  It is a little sad that nobody else has pointed these things out yet, but not surprising; Banach space theory does not seem to be very popular on the mathematical Internet.  (I suspect if you had posted this problem on Mathoverflow, a few people would vote to close it as &quot;too specialized&quot; out of their own ignorance, and then gone off to write answers to much simpler, much more specific problems in algebraic geometry.)  In what follows I write $\ell^p$ for what you have written $\ell^n_p$; suppressing the dependence on $n$ since it can clearly be determined from the dimension.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One way to distinguish these spaces is with the modulus of convexity.  For $X$ a Banach space, the modulus of convexity of $X$ is the function $\delta_X: (0,2] \to [0,1]$ given by&#xA;$$&#xD;&#xA;\delta_X(\epsilon) = \inf \{1 - \frac{1}{2} \|x+y\|: x, y \in X, \|x\|=\|y\|=1, \|x-y\| \geq \epsilon\}.&#xD;&#xA;$$&#xD;&#xA;(There is not total agreement on this; you will see the data encoded slightly differently in the literature.  For example what I have just written is what some people would call $\delta_X(\epsilon/2)$, and they would define it only for $\epsilon \in (0,1]$.  Keep your eyes open.)  It is evident from the definition that the modulus of convexity is invariant under isometric isomorphism, that is, if $X$ and $Y$ are Banach spaces and there is an isometric isomorphism $f: X \to Y$, then $\delta_X = \delta_Y$ as functions on $(0,2]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To compute the modulus of convexity for $\ell^p$, go and prove Hanner's inequality for $\ell^p$.  For $p \in [1,2]$ this is the statement&#xA;$$&#xD;&#xA;\|x+y\|_p^p + \|x-y\|_p^p \geq (\|x\|_p + \|y\|_p)^p + | \|x\|_p - \|y\|_p|^p, \qquad x, y \in \ell^p,&#xD;&#xA;$$&#xD;&#xA;and for $p \in (2, \infty)$ the inequality reverses.  There are various nice proofs of Hanner's inequality in the literature--- Hanner's original proof is pretty straightforward, although maybe not the simplest.  My advice would be to search Math Reviews for a proof that is to your liking.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A bit of work, that I omit, shows that Hanner's inequality completely determines the modulus of convexity for $\ell^p$, in that you can write down an actual formula for it.  And once you have done that, you see that different $p$s give you different moduli of convexity.  (For $p \geq 2$ it is something like $1 - (1 - (\epsilon/2)^p)^{1/p}$; for $p \in [1,2)$ the formula is more simply written with $\delta$ as an &quot;implicit&quot; function of $\epsilon$.)  The details are in Hanner's original paper, &quot;On the uniform convexity of $L^p$ and $\ell^p$&quot;, if you don't want to supply them yourself.  (To recover $p$ from the formula in the $[2, \infty)$ case, you can look at the value at $\epsilon = 1$, for example, and use calculus to show that this is strictly decreasing in $p$.  Or you could look at the limit of the second $\epsilon$-derivative as $\epsilon$ approaches $0$; I think it's a constant multiple of $(p-1)$.  In any case this distinguishes all the $\ell^p$ with $p \in [2, \infty)$ from one another, and doing the calculation for $p \in [1,2)$ you can finish the job in a similar way.)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you read up on the modulus of convexity you see it has a very geometric interpretation (it quantifies in some sense &quot;how convex&quot; the unit ball of $X$ is) and contains many of the same ideas you were probably getting at with an ad hoc approach.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This problem is also solved in Banach's textbook (Theory of linear operators), I think, in another way.  I don't have the book on me at the moment but the idea is that any sequence $(x_k)_{k=1}^{\infty}$ in $\ell^p$ that converges to $0$ has a subsequence $(x_{k_j})_{j=1}^{\infty}$ with the property that the function $n \mapsto \|\sum_{j=1}^n x_{k_j}\|$ is $O(n^{1/p})$, and if you think about this long enough, this can be used to distinguish the $\ell^p$ from one another.  There is a more modern or high tech approach that abstracts this idea to an invariant defined more generally in Banach spaces (e.g. you can distinguish the $\ell^p$ using the indices of &quot;Rademacher type&quot; and &quot;Rademacher cotype&quot;).&lt;/p&gt;&#xA;" OwnerUserId="3840" LastActivityDate="2010-11-23T06:15:24.477" CommentCount="5" />
  <row Id="11473" PostTypeId="2" ParentId="11464" CreationDate="2010-11-23T06:37:00.943" Score="5" ViewCount="" Body="&lt;p&gt;Perhaps a sixth way...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle S = \sum_{r=1}^{d} r\cdot 2^r$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle 2S = \sum_{r=1}^{d} r\cdot 2^{r+1} = \sum_{r=2}^{d+1} (r-1)2^{r}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle 2S -S = d\cdot 2^{d+1} - \sum_{r=1}^{d} 2^r = d\cdot 2^{d+1} - 2^{d+1} +2  = (d-1)2^{d+1} + 2$$&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-11-23T06:37:00.943" CommentCount="1" />
  <row Id="11474" PostTypeId="1" AcceptedAnswerId="12178" CreationDate="2010-11-23T06:44:49.313" Score="1" ViewCount="212" Body="&lt;p&gt;I recall reading somewhere that the supremum of a family of topologies on a set is simply the topology generated by the union of all topologies in the family. My question is, what does &quot;generated&quot; mean in this case? For instance, I know when dealing with filter bases, the generated filter is the set of all subsets containing some set in the filter base. But for topologies, how do we know what subsets to include in the generated topology? I would like to be able to visualize what kind of subsets are contained in this supremum. Thanks for any insight.&lt;/p&gt;&#xA;" OwnerUserId="1583" LastActivityDate="2010-11-28T04:42:50.983" Title="What exactly is the topology generated by the union of a family of topologies?" Tags="&lt;topology&gt;&lt;general-topology&gt;" AnswerCount="3" CommentCount="3" />
  <row Id="11475" PostTypeId="2" ParentId="11468" CreationDate="2010-11-23T06:53:21.640" Score="14" ViewCount="" Body="&lt;p&gt;Note that if they are distinguishable then the number of ways is given by $6^{200}$ since each of the 200 students have $6$ choices of floors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, we are given that the students are indistinguishable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, we are essentially interested in solving $a_1 + a_2 + a_3 + a_4 + a_5 + a_6 = 200$, where $a_i$ denotes the number of students in the $i^{th}$ floor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The constraints are $0 \leq a_i \leq 200$, $\forall i \in \{1,2,3,4,5,6\}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We will in fact look at a general version of this problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We want to find the total number of natural number solutions for the following equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle \sum_{i=1}^{n} a_i = N$, where $a_i \in \mathbb{N}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The method is as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider $N$ sticks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$| | | | | | | | ... | | |$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We want to do partition these $N$ sticks into $n$ parts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This can be done if we draw $n-1$ long vertical lines in between these $N$ sticks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The number of gaps between these $N$ sticks is $N-1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the total number of ways of drawing these $n-1$ long vertical lines in between these $N$ sticks is $C(N-1,n-1)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the number of natural number solutions for $\displaystyle \sum_{i=1}^{n} a_i = N$ is $C(N-1,n-1)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we are interested in the number of non-negative integer solutions, all we need to do is replace $a_i = b_i - 1$ and count the number of natural number solutions for the resulting equation in $b_i$'s.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i.e. $\displaystyle \sum_{i=1}^{n} (b_i - 1) = N$ i.e. $\displaystyle \sum_{i=1}^{n} b_i = N + n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the number of non-negative integer solutions to $\displaystyle \sum_{i=1}^{n} a_i = N$ is given by $C(N+n-1,n-1)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, for the current problem assuming that some of the floors can be empty, the answer is $C(200+5,5) = C(205,5) = 2872408791$.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2010-11-23T18:41:02.233" LastActivityDate="2010-11-23T18:41:02.233" CommentCount="1" />
  <row Id="11476" PostTypeId="2" ParentId="11464" CreationDate="2010-11-23T07:13:28.263" Score="2" ViewCount="" Body="&lt;p&gt;Denote the solution by $\rm\:S(d)\:,$ and put $\rm\ s(d) = S(d)/2 - 1   = (d-1)\:2^d\:.\ $ It is the unique solution of the recurrence $\rm\ s(2) = 4,\ \ s(d+1)/s(d) = 2d/(d-1)\ \:$ i.e. $\rm\ (d-1)\ s(d+1)\ =\ 2d\ s(d)\:.\: $ It suffices to verify that $\rm\ sum/2 - 1$ satisfies the same recurrence - a simple calculation (probably essentially the same calculation that is in Moron's answer, but I haven't checked that).&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-11-23T17:53:58.640" LastActivityDate="2010-11-23T17:53:58.640" />
  <row Id="11477" PostTypeId="1" AcceptedAnswerId="11479" CreationDate="2010-11-23T07:15:38.417" Score="7" ViewCount="383" Body="&lt;p&gt;Question: Let $F_n$ the sequence of Fibonacci numbers, given by $F_0 = 0, F_1 = 1$ and $F_n = F_{n-1} + F_{n-2}$ for $n \geq 2$. Show for $n, m \in \mathbb{N}$: $$F_{n+m} = F_{n-1}F_m + F_n F_{m+1}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My (very limited) attempt so far: after creating a small list of the values $F_0=0, F_1=1, F_2=1, F_3=2, F_4=3, F_5=5, F_6=8, F_7=13, F_8=21, F_9=34, F_{10}=55$ i can see that yes it does seem to work for instance $F_{6+3}=F_5 F_3 +F_6 F_4 = 10 +24 = 34 = F_9$. However, I really don't know where to begin as showing that this must hold in general terms. Should I be looking to use limits? Or perhaps induction? What is the best way to solve this?&lt;/p&gt;&#xA;" OwnerUserId="3711" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-11-23T17:21:23.490" LastActivityDate="2010-12-24T22:01:27.810" Title="Showing that an equation holds true with a Fibonacci sequence: $F_{n+m} = F_{n-1}F_m + F_n F_{m+1}$" Tags="&lt;discrete-mathematics&gt;&lt;fibonacci-number&gt;" AnswerCount="8" CommentCount="3" FavoriteCount="1" />
  <row Id="11478" PostTypeId="2" ParentId="11477" CreationDate="2010-11-23T07:19:19.893" Score="1" ViewCount="" Body="&lt;p&gt;One simple way is to use the formula for the $n^{th}$ Fibonacci number, viz, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$F_n = \frac{\phi^n - (1-\phi)^n}{\sqrt{5}}$ where $\phi$ is the golden ratio. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\phi = \frac{1 + \sqrt{5}}{2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or another equally simple way is to use induction on $n$ and then on $m$ or just using induction on one of these might turn out to suffice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It will be interesting to see a direct proof...&lt;/p&gt;&#xA;" OwnerUserId="3302" LastActivityDate="2010-11-23T07:19:19.893" CommentCount="5" />
  <row Id="11479" PostTypeId="2" ParentId="11477" CreationDate="2010-11-23T07:28:31.730" Score="2" ViewCount="" Body="&lt;p&gt;Fix $m \in \mathbb{N}$. We shall use induction on $n$. For $n=1$, the &lt;strong&gt;RHS&lt;/strong&gt; of the equation becomes $$F_{m-1}F_{1} + F_{m}F_{2} = F_{m-1} + F_{m}$$, which is equal to $F_{m+1}$. When $n=2$, the equation is also true.( I hope you can prove this!).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now assume, that the result is true for $k=3,4, \cdots , n$. We want to show that the result is true for $k=n+1$. $$ \text{For} \ k=n-1 \ \text{we have} \quad F_{m+n-1} = F_{m-1}F_{n-1} + F_{m}f_{n}$$ and $$ \text{For} \ k = n \ \text{we have} \quad F_{m+n}=F_{m-1}F_{n} + F_{m}F_{n+1}$$ Adding both the sides you will get $$F_{m+n-1} + F_{m+n} = F_{m+n-1} = F_{m-1}F_{n+1} + F_{m}F_{n+2}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Oh, i reversed the notations. This is for proving $$F_{m+n} = F_{m-1}F_{n} + F_{m}F_{n+1}$$&lt;/p&gt;&#xA;" LastActivityDate="2010-11-23T07:28:31.730" CommentCount="1" />
  <row Id="11480" PostTypeId="2" ParentId="11477" CreationDate="2010-11-23T07:30:49.987" Score="1" ViewCount="" Body="&lt;p&gt;These can almost always be solved by induction. (In general sequences that are defined by recursion go often well together with induction.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For this formula we can do the following: Induct on $m$. If $m=1$, then&#xA;$F_{n+m} = F_{n}F_{m-1} + F_{n-1}F_m = F_{n}F_{0} + F_{n-1}F_1 = F_n + F_{n-1} = F_{n+1}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose then that the result holds for $m$ and smaller. Now&#xA;$$F_{n+m+1} = F_{n+m} + F_{n+m-1} = F_{n}F_{m-1} + F_{n-1}F_m + F_{n}F_{m-2} + F_{n-1}F_{m-1}$$&#xD;&#xA;$$= F_{n}(F_{m-1} + F_{m-2}) + F_{n-1}(F_m + F_{m-1}) = F_{n}F_{m} + F_{n-1}F_{m+1}.$$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="3776" LastActivityDate="2010-11-23T07:30:49.987" />
  <row Id="11481" PostTypeId="2" ParentId="11435" CreationDate="2010-11-23T07:32:21.323" Score="4" ViewCount="" Body="&lt;p&gt;Isn't this obvious (or have I missed something?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The appropriate product of rational polytopes $P\subseteq\mathbb R^m$&#xD;&#xA;and $Q\subseteq\mathbb R^n$ is surely their Cartesian product&#xA;$P\times Q\subseteq\mathbb R^{m+n}$. For a positive&#xA;integer $t$, the integer points in $t(P\times Q)$ are those of the form&#xA;$(a,b)$ where $a\in tP$ and $b\in tQ$, so there are $L_P(t)L_Q(t)$&#xD;&#xA;of them.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-11-23T07:32:21.323" CommentCount="2" />
  <row Id="11482" PostTypeId="2" ParentId="11477" CreationDate="2010-11-23T07:35:51.993" Score="4" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\ \ $ If you put the Fibonacci recurrence into matrix form then the result is obvious, viz.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ M^n\ :=\ \left(\begin{array}{ccc} &#xD;&#xA;1 &amp;amp; 1 \\\&#xD;&#xA;1 &amp;amp; 0 \end{array}\right)^n\ =\ \left(\begin{array}{ccc}&#xD;&#xA;F_{n+1} &amp;amp; F_n \\\&#xD;&#xA;F_n &amp;amp; F_{n-1} \end{array}\right) $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now compute $\ M^{n+m}\ =\ M^n \ M^m$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-11-23T07:46:22.153" LastActivityDate="2010-11-23T07:46:22.153" CommentCount="3" />
  <row Id="11483" PostTypeId="2" ParentId="11477" CreationDate="2010-11-23T07:37:09.553" Score="5" ViewCount="" Body="&lt;p&gt;If pressed, I'd nuke with Binet's formula myself, but here's another&#xA;approach. By an easy induction, if&#xA;$$A=\pmatrix{0&amp;amp;1\\\\ 1&amp;amp;1}$$&#xD;&#xA;then&#xA;$$A^n=\pmatrix{F_{n-1}&amp;amp;F_n\\\\ F_n&amp;amp;F_{n+1}}.$$&#xD;&#xA;Comparing the top right entry in the equation $A^m A^n=A^{m+n}$&#xD;&#xA;gives&#xA;$$F_{m-1}F_n+F_m F_{n+1}=F_{m+n}.$$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-11-23T07:37:09.553" />
  <row Id="11484" PostTypeId="2" ParentId="11477" CreationDate="2010-11-23T07:44:40.313" Score="3" ViewCount="" Body="&lt;p&gt;Try proving this statement:&#xA;Claim: If $f(n) = f(n-1)+f(n-2)$, then $f(n) =  F_n f_1 + F_{n-1} f_0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now &quot;fix&quot; $m$ and think of $F_{n+m}$ as a linear recurrence in $n$ with initial conditions $F_{m+1}$ and $F_m$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then your claim will follow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the way, there is a short and clean proof of Claim, but you should know it uses the fact $F_{-1} = 1$.(Something you can easily verify if you did not already know).&lt;/p&gt;&#xA;" OwnerUserId="899" LastActivityDate="2010-11-23T07:44:40.313" />
  <row Id="11485" PostTypeId="2" ParentId="11145" CreationDate="2010-11-23T07:55:49.753" Score="0" ViewCount="" Body="&lt;p&gt;An elementary approach involving the third mixed partial derivatives shows that such an isometry in the form given in the question is only possible with $p=q=2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume $1&amp;lt; p,\ q &amp;lt; \infty$ and $1/p + 1/q = 1$. Assume also that  there exist an isometry of the given form, i.e., a normalized rotation of $\pi/4$ between $(\mathbb R^2,\|\ \|_p)$ and $(\mathbb R^2,\|\ \|_q)$.   Then the corresponding matrix would be $A = \lambda&amp;#39;\left(\matrix{1 &amp;amp; 1\\1 &amp;amp; -1}\right)$ for some scalar $\lambda&amp;#39;$. It follows that &#xA;$$|x|^p + |y|^p = \lambda &amp;#39;&amp;#39;(|x + y|^q + |x - y|^q)^{p/q}$$  &#xA;for some constant $\lambda&amp;#39;&amp;#39;$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we compute&lt;sup&gt;1&lt;/sup&gt; the mixed third derivative $\frac {\partial^3}{\partial x \partial y^2}$ of both sides at the point $(1,0)$, after simplifications we obtain&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$0 = (-2+q) (-1+q)+\left(-1+\frac{p}{q}\right) (-1+q) q.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This plus the assumptions on $p$ and $q$ gives us that the unique solution is $p=2$ and $q =2$. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;1&lt;/sup&gt; with the help of Mathematica: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;D[((x + y)^q + (x - y)^q)^(p/q), x, y, y] /. {x -&amp;gt; 1, y -&amp;gt; 0}&lt;/code&gt;&lt;/p&gt;&#xA;" OwnerUserId="145" LastEditorUserId="145" LastEditorDisplayName="" LastEditDate="2010-11-23T10:38:59.083" LastActivityDate="2010-11-23T10:38:59.083" />
  <row Id="11486" PostTypeId="2" ParentId="11464" CreationDate="2010-11-23T07:59:39.253" Score="4" ViewCount="" Body="&lt;p&gt;This is an arithmetic-geometric progression. I remember doing these&#xA;at school, I think even before O-levels. The typical A-G progression&#xA;has the form&#xA;$$ab,ar(b+d),ar^2(b+2d),\ldots,a r^n(b+nd)\ldots$$&#xD;&#xA;(it's the pointwise product of an arithmetic progression and a geometric&#xA;progression). To obtain the sum of the first $N$ terms, one uses the same&#xA;trick as for summing $N$ terms of a GP. Let&#xA;$$S=\sum_{k=0}^{N-1} ar^k(b+kd).$$&#xD;&#xA;Then&#xA;$$rS=\sum_{k=0}^{N-1} ar^{k+1}(b+kd)=\sum_{k=1}^N ar^k(b+kd-d).$$&#xD;&#xA;Hence&#xA;$$(1-r)S=ab-ar^N(b+(N-1)d)+\sum_{k=1}^{N-1}adr^k.$$&#xD;&#xA;The last sum is that of a GP, which one already knows about...&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-11-23T07:59:39.253" CommentCount="4" />
  <row Id="11487" PostTypeId="1" CreationDate="2010-11-23T08:33:57.043" Score="1" ViewCount="154" Body="&lt;p&gt;Y= X+N where X,N are normally distributed with zero mean &lt;/p&gt;&#xA;&#xA;&lt;p&gt;i know that Y is also normally distributed with zero mean and sum of variances &lt;/p&gt;&#xA;&#xA;&lt;p&gt;i don't know how to get f(X|y) in order to calculate E(X|y)&lt;/p&gt;&#xA;" OwnerUserId="3490" LastActivityDate="2010-11-23T21:08:32.720" Title="Y= X+N what is E(X|Y=y)" Tags="&lt;probability&gt;" AnswerCount="4" CommentCount="1" />
  <row Id="11488" PostTypeId="2" ParentId="11487" CreationDate="2010-11-23T09:01:20.477" Score="1" ViewCount="" Body="&lt;p&gt;\begin{equation}&#xA;p(x|y) = \frac{p(y|x)p(x)}{p(y)} &#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You know $p(y)$ and $p(x)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;p(y|x) = \frac{1}{\sqrt{2 \pi \sigma_N^2}} e^{\frac{-(y-x)^2}{2\sigma_N^2}}&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using these expressions, you can easily get both $p(x|y)$ and $E(X|Y=y)$.&lt;/p&gt;&#xA;" OwnerUserId="2641" LastActivityDate="2010-11-23T09:01:20.477" />
  <row Id="11489" PostTypeId="2" ParentId="11382" CreationDate="2010-11-23T09:07:59.393" Score="4" ViewCount="" Body="&lt;p&gt;It's possible to write down the set of matches between two strings as the result of multiplying two polynomials constructed from the strings. This turns out to be very useful: indeed, fast algorithms for string matching exploit the fast Fourier transform method used to multiply polynomials. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To see how this works, think of a binary string as representing coefficients of a polynomial. Then matches between the strings correspond to nonzero coefficients of the product of the polynomial from one string and the reverse of the polynomial from the second. Here, the &quot;multiplication&quot; is over the AND-OR field, rather than the usual integer or GF(2) field. &lt;/p&gt;&#xA;" OwnerUserId="865" LastActivityDate="2010-11-23T09:07:59.393" CommentCount="2" />
  <row Id="11490" PostTypeId="2" ParentId="11487" CreationDate="2010-11-23T09:11:11.540" Score="2" ViewCount="" Body="&lt;p&gt;You can find ${\rm E}(X|Y=y)$ immediately as follows.&#xA;If $X \sim {\rm N}(0,\sigma_1^2)$ and $N \sim {\rm N}(0,\sigma_2^2)$&#xD;&#xA;(and $X$ and $N$ are independent), then&#xA;$$&#xD;&#xA;{\rm E}(X|X + N) = \frac{{\sigma _1^2 }}{{\sigma _1^2  + \sigma _2^2 }}(X + N),&#xD;&#xA;$$&#xD;&#xA;so, in particular,&#xA;$$&#xD;&#xA;{\rm E}(X|X + N = y) = \frac{{\sigma _1^2 }}{{\sigma _1^2  + \sigma _2^2 }}y.&#xD;&#xA;$$&#xD;&#xA;I will (probably) justify this (i.e., the first formula) later on today.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The point is that, in some respect, the normal distribution does not play a special role in this result.&#xA;For example, if $X$ and $N$ are independent Poisson random variables with means $\lambda_1$ and $\lambda_2$, respectively, then&#xA;$$&#xD;&#xA;{\rm E}(X|X + N) = \frac{{\lambda _1 }}{{\lambda_1  + \lambda_2 }}(X+N).&#xD;&#xA;$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is common to the normal distribution and the Poisson distribution in this context is that both are infinitely divisible. More details later on.&lt;/p&gt;&#xA;" OwnerUserId="2810" LastEditorUserId="2810" LastEditorDisplayName="" LastEditDate="2010-11-23T15:06:55.690" LastActivityDate="2010-11-23T15:06:55.690" CommentCount="2" />
  <row Id="11491" PostTypeId="2" ParentId="11342" CreationDate="2010-11-23T09:18:10.753" Score="1" ViewCount="" Body="&lt;p&gt;$T_{b_2} \circ T_{b_1} = T_{b_2}(T_{b_1}(z))= T_{b_2}(b_1*z)=b_2*(b_1*z)=(b_2*b_1)*z = b*z$.&lt;/p&gt;&#xA;" OwnerUserId="2723" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-23T14:43:51.860" LastActivityDate="2010-11-23T14:43:51.860" />
  <row Id="11492" PostTypeId="1" CreationDate="2010-11-23T09:43:15.927" Score="3" ViewCount="227" Body="&lt;p&gt;I would like to show that the quantity: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$-2\sigma\left(rx^{2}+y^{2}+b\left(z-r\right)^{2}-br^{2}\right)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is negative on the surface:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$rx^{2}+\sigma y^{2}+\sigma\left(z-2r\right)^{2}=C$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for some sufficiently large value of $C$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was not able to massage the first quantity any more in order to make it look like the second. I also considered a change of coordinates, but had no luck.&#xA;$\sigma, b, r$  are positive parameters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a step in exercise 9.2.2 from Strogatz Nonlinear Dynamics and Chaos.&lt;/p&gt;&#xA;" OwnerUserId="2256" LastActivityDate="2010-12-23T21:23:13.623" Title="Create a trapping region for Lorenz Attractor" Tags="&lt;geometry&gt;&lt;dynamical-systems&gt;&lt;vector-calculus&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="11493" PostTypeId="2" ParentId="11382" CreationDate="2010-11-23T09:57:25.633" Score="2" ViewCount="" Body="&lt;p&gt;Multiplication of polynomials has a very nice combinatorial interpetation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take a simple example: $1+x=x^0+x^1$. If you perform an experiment once, that experiment can either succeed or fail. You can fail the experiment in only one way in one performance, or you can succeed in one way. This is precisely encoded in the coefficients of the polynomial.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Look now at the product: $(1+x)\times(1+x)=1\times x^0+2\times x^1+1\times x^2$. The coefficients now encode in how many ways you can fail twice, how many ways you can succeed once and fail once and how many ways you can succeed twice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, I guess you get the trick now, $(1+x)^n$ will have encoded in its $k$th coefficient in how many ways you can succeed $k$ times but fail $n-k$ times.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This can be generalized to all kinds of polynomials.&lt;/p&gt;&#xA;" OwnerUserId="3567" LastActivityDate="2010-11-23T09:57:25.633" />
  <row Id="11494" PostTypeId="2" ParentId="11382" CreationDate="2010-11-23T10:24:48.220" Score="1" ViewCount="" Body="&lt;p&gt;Apart from the mundane reasons, e.g. &quot;you need it to do certain computations&quot; there are also some more technical reasons:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;The Stone-Weierstrass theorem: &lt;em&gt;Let $X$ be a compact metric space and let $A$ be a $\mathbb{C}$-sub&lt;a href=&quot;http://en.wikipedia.org/wiki/Algebra_over_a_field&quot; rel=&quot;nofollow&quot;&gt;algebra&lt;/a&gt; of $C(X,\mathbb{C})$. If $A$ separates points in $X$ and vanishes at no point of $X$ and is self-conjugate, then $A$ is dense in $C(X,\mathbb{C})$.&lt;/em&gt; &lt;br/&gt;&#xA;The polynomial functions on $[a,b]$ with complex coefficients are the prototypical example of such a subalgebra. Of course verifying that it is an algebra and has the stated properties is an application of multiplication of polynomials.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Orthogonal polynomials. These are somewhat important in the theory of special functions. For example in &lt;a href=&quot;http://books.google.com/books?id=nMm13WXpLt8C&amp;amp;pg=PA240&quot; rel=&quot;nofollow&quot;&gt;[Andrews, Askey and Roy]&lt;/a&gt; there is an entire chapter devoted to families of orthogonal polynomials.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1269" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-23T10:35:22.103" LastActivityDate="2010-11-23T10:35:22.103" CommentCount="2" />
  <row Id="11495" PostTypeId="1" CreationDate="2010-11-23T10:37:34.663" Score="0" ViewCount="287" Body="&lt;p&gt;Hi to all, &#xA;I'm in Italian guy almost in my 30. I have a regular job as a programmer. When in school I've never been much interested nor good in math and even at the university I've been studying languages, hence something more on the literary side that on the scientific one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When in high school I've had mostly literary classes, so, again, my current knowledge of algebra, geometry and arithmetic is quite basic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've always been interested in programming, though, up to making it my profession. With my growing interest in math, I'm tinkering with the idea of even getting a Computer Science or Computer Engineering degree at the university. Both rely heavily on maths and I would have to study it by myself, as I cannot afford following classes (as said before, I have a full-time job), starting from the basis that I don't have.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Before attending to it, I want to start learning math by myself, to gain the necessary base knowledge I don't currently have. Do you think it is possible to successfully study mathematics with such a short initial knowledge and not having my full day to do it? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any general advice or book / site to get me started with the &quot;preparation&quot; study? What are the basic areas I should concentrate on to be prepared to the first year math exams?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you in advance. &lt;/p&gt;&#xA;" OwnerUserId="3847" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-23T10:40:50.583" LastActivityDate="2010-11-25T14:18:04.083" CommunityOwnedDate="2010-11-25T12:19:27.163" Title="Preparing for first year CS" Tags="&lt;soft-question&gt;&lt;self-learning&gt;" AnswerCount="4" CommentCount="1" FavoriteCount="1" />
  <row Id="11496" PostTypeId="2" ParentId="11495" CreationDate="2010-11-23T10:55:51.707" Score="1" ViewCount="" Body="&lt;p&gt;Your first course with programming in it will probably use some functional programming language such as an ML variant, Scheme, Haskell, a LISP, etc. As I understand it, universities start with that for two reasons, (i) it works well for teaching (this is debatable, though), and (ii) whatever misconceptions and terrible habits the neophytes bring to the table from unguided tinkering with e.g. Visual Basic will be almost &lt;em&gt;entirely useless&lt;/em&gt;, thus ensuring that you also have a good chance of beating some bad habits out of them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first heavy math class in a CS education (other than the typically required first course in mathematics) is probably algorithms. Curiously the math requirements for an algorithms course are rather low in the &quot;skills&quot; department, but high in the &quot;mathematical maturity&quot; department. That first course in math will teach you the method of induction and - equally important - the ability to reason mathematically.&lt;/p&gt;&#xA;" OwnerUserId="1269" LastActivityDate="2010-11-23T10:55:51.707" CommunityOwnedDate="2010-11-25T12:19:27.163" CommentCount="1" />
  <row Id="11497" PostTypeId="2" ParentId="11382" CreationDate="2010-11-23T10:58:55.453" Score="2" ViewCount="" Body="&lt;p&gt;I think a better metaphor for multiplication comes from physics.  If your speed s is constant over some time t, then your distance traveled is just s times t.  Whatever fraction of time you travel, you will expect to have traveled some distance.  So even fractional multiplication makes some sense.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the regular world, it's very common to use a polynomials (often linear ones) to approximate things that are changing.  For instance, if you are a grocer and you are trying to predict how many tomatoes you will sell tomorrow, it's not a bad approach to guess it will be the same as yesterday.  It's also not a bad approach to guess it might be the average of the last two days.  Or, you might come up with a more complicated formula involving the last 50 days.  All of these attempts to guess will have an error.  This is the point of studying these things, to do a better job and make smaller errors.  (For reasons, I can't go into, the squares and cubes of values can sometimes help make approximations more accurate.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My point is that any quantity is a potential polynomial.  You might approximate your customers per day as a polynomial of the number of customers of the last two days.  You might approximate the value of sales per customer as another polynomial.  Well, if you wanted to know your potential income, customers times sales, you might multiply them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I use polynomials all the time.  For instance, when I was helping a friend lose weight, I approximated their weight loss behavior using a polynomial.  It was fairly accurate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, keep with polynomials, they were worth it in the long run!&lt;/p&gt;&#xA;" OwnerUserId="1004" LastActivityDate="2010-11-23T10:58:55.453" />
  <row Id="11498" PostTypeId="1" CreationDate="2010-11-23T11:06:48.587" Score="0" ViewCount="164" Body="&lt;p&gt;I'm really embarrassed to ask this - my apologies. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a piece of software I'm writing, I am using the following: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$Y = X+\frac{X}{10}N$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's all fine - but how do I calculate X when I know Y and N? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know it's really basic, but for some reason I get stuck half-way through moving the equation around. &lt;/p&gt;&#xA;" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-23T11:25:49.097" LastActivityDate="2010-11-23T15:18:45.993" Title="Basic algebra gone out the window" Tags="&lt;algebra-precalculus&gt;" AnswerCount="1" CommentCount="7" />
  <row Id="11499" PostTypeId="1" AcceptedAnswerId="11512" CreationDate="2010-11-23T11:26:45.163" Score="1" ViewCount="307" Body="&lt;h2&gt;Edit:&lt;/h2&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The answers I seem to be getting here are that this is not possible, but I have come across a formulae for subdividing an icosahedron (which I dont have to hand, but goes something like this)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;V = 10 * u² + 2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;E = 30 * u²&lt;/p&gt;&#xA;&#xA;&lt;p&gt;F = 20 * u² &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where U = the frequency of subdivision&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I may have it the wrong way around but you get the picture) this allows us to calculate the amount of vertecies, edges and faces for each subdivision.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using this forumlea (or the correct one anyway) you can incrediment the frequency untill the number of faces reach 1000. It would appear in theory this will allow me to imitate a sphere with 1000 congruent polygon's (faces) by constructing a subdivided icosahedron because using V-E+F=2 I get 502-1500+1000=2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone confirm if the faces will wrap around to make a full icosahedron without any missing faces?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Know the answer? reply and you will most likley get the vote.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;complete icosahedron with 20,40,80,120,160,200,240,280,320...1000 faces = true?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Hi, Does anyone know if it possible to reprisent a sphere with 1000 equal surfaces?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I say equal surfaces, I am talking in terms of area. However it would also be preferable if the shape of these surfaces match.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, I am looking at subdivided icosahedron's, all surfaces are equal triangles in terms of shape and area, but to produce a nice sphere it would seem the number of surfaces would need to be 20,80,320 etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also I have come across zonohedron's - Does this allow any amount of surface? I can see that the surfaces are different shapes (but i should hope they have the same area)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know this may sound abit confusing, but any feedback will be much appreachiated&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks, Chris&lt;/p&gt;&#xA;" OwnerUserId="3849" LastEditorUserId="3849" LastEditorDisplayName="" LastEditDate="2010-12-03T14:14:24.483" LastActivityDate="2011-02-23T10:16:42.830" Title="Possible to imitate a sphere with 1000 congruent polygons?" Tags="&lt;geometry&gt;" AnswerCount="3" CommentCount="14" />
  <row Id="11500" PostTypeId="2" ParentId="5713" CreationDate="2010-11-23T12:00:56.430" Score="0" ViewCount="" Body="&lt;p&gt;I think the question requires that the language must be finite. So, it's quite hard.&lt;/p&gt;&#xA;" LastActivityDate="2010-11-23T12:00:56.430" />
  <row Id="11501" PostTypeId="1" CreationDate="2010-11-23T12:09:32.897" Score="0" ViewCount="164" Body="&lt;p&gt;I'm trying to teach myself mathematics. I only started last year with a previous knowledge of basic arithmetic. I often make a detour into regions featuring problems that are a bit too deep for me. I'm often too curious to wait until I was able to catch up on the necessary bedding. I hope you'll excuse me :-)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.wolframalpha.com/input/?i=f%28x%29+%3D+f%28x-1%29%2ab&quot; rel=&quot;nofollow&quot;&gt;Wolfram|Alpha says&lt;/a&gt;: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f(x) = f(x-1)*b \to f(x) = c_1 b^{x-1}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is how I would compute the original recursive algorithm(?):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f(x) = f(x-1)*b$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align} \\&#xD;&#xA;f(4) &amp;amp;= f(4-1)*b \\&#xD;&#xA;&amp;amp;= f(f(3-1)*b)*b \\&#xD;&#xA;&amp;amp;= f(f(f(2-1)*b)*b)*b \\&#xD;&#xA;&amp;amp;= f(f(f(f(1-1)*b)*b)*b)*b \\&#xD;&#xA;&amp;amp;= \qquad ... &#xD;&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But &lt;strong&gt;how do I compute the recurrence relation $f(x) = c_1 b^{x-1}$&lt;/strong&gt;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would compute the Fibonacci numbers like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xD;&#xA;F_n &amp;amp;= F_{n-1} + F_{n-2}\\&#xD;&#xA;    F_0 &amp;amp;= 0\\&#xD;&#xA;    F_1 &amp;amp;= 1\\&#xD;&#xA;&#xD;&#xA;   F_5 &amp;amp;= F_{5-1} + F_{5-2}\\&#xD;&#xA;       &amp;amp;= F_4 + F_3\\&#xD;&#xA;       &amp;amp;= (F_3 + F_2) + (F_2 + F_1)\\&#xD;&#xA;       &amp;amp;= ((F_2 + F_1) + (F_1 + F_0)) + ((F_1 + F_0) + F_1)\\&#xD;&#xA;       &amp;amp;= (((F_1 + F_0) + F_1) + (F_1 + F_0)) + ((F_1 + F_0) + F_1) \\&#xD;&#xA;       &amp;amp;= 5&#xD;&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm also not sure about the difference between a recursive algorithm and its recurrence relation. I'd be grateful if someone could enlighten me or simply link me up. &lt;/p&gt;&#xA;" OwnerUserId="3850" LastEditorUserId="9754" LastEditorDisplayName="" LastEditDate="2011-06-03T21:13:20.143" LastActivityDate="2011-06-03T21:13:20.143" Title="$f(x) = f(x-1)*b  \to  f(x) = c_1 b^{x-1}$" Tags="&lt;algorithms&gt;&lt;recurrence-relations&gt;&lt;wolframalpha&gt;&lt;recursive-algorithms&gt;" AnswerCount="2" />
  <row Id="11502" PostTypeId="1" AcceptedAnswerId="11506" CreationDate="2010-11-23T12:15:30.550" Score="3" ViewCount="127" Body="&lt;p&gt;Is there any &quot;algorithm&quot; or steps to follow to get a formula from a table of values.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using this values:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;X                Result&#xA;1                3&#xA;2                5&#xA;3                7&#xA;4                9&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I'd like to obtain:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Result = 2X+1&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe using excel?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit 2&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Additional info:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is not going to be always a polynomial and it may have several parameters (I think 2).&lt;/p&gt;&#xA;" OwnerUserId="3314" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-23T14:30:15.317" LastActivityDate="2010-11-23T14:30:15.317" Title="Find formula from values" Tags="&lt;functions&gt;&lt;approximation&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="1" />
  <row Id="11503" PostTypeId="2" ParentId="11501" CreationDate="2010-11-23T12:19:24.133" Score="2" ViewCount="" Body="&lt;p&gt;I believe that your way to compute $f(n)$ is not correct, as you have outlined in the question. Observe that $f(4) = b * f(3) = b^2 * f(2)$ and not $f(4) = b * f(f(2)*b)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That said, you have $$f(n) = b * f(n-1) = b^2 * f(n-2) = \ldots = b^{n-1} * f(1).$$&#xD;&#xA;Say that your initial condition states that $f(1) = c_1$, you can conclude that $f(n) = c_1 * b^{n-1}$.&lt;/p&gt;&#xA;" OwnerUserId="2384" LastActivityDate="2010-11-23T12:19:24.133" />
  <row Id="11504" PostTypeId="2" ParentId="11502" CreationDate="2010-11-23T12:24:55.393" Score="1" ViewCount="" Body="&lt;p&gt;One way to obtain the desired relationship is to use the &lt;a href=&quot;http://en.wikipedia.org/wiki/Regression_analysis&quot; rel=&quot;nofollow&quot;&gt;regression analysis&lt;/a&gt;. In the case that you know what form of the relationship you are expecting, say you know that $f(x) = a * x + b$ it is very simple to find the parameters $a$ and $b$ (see for &lt;a href=&quot;http://en.wikipedia.org/wiki/Linear_regression&quot; rel=&quot;nofollow&quot;&gt;example&lt;/a&gt;). In the case that you don't know what form your formula may take, it is common to use the &lt;a href=&quot;http://en.wikipedia.org/wiki/Nonparametric_regression&quot; rel=&quot;nofollow&quot;&gt;nonparametric regression&lt;/a&gt; techniques.&lt;/p&gt;&#xA;" OwnerUserId="2384" LastActivityDate="2010-11-23T12:24:55.393" />
  <row Id="11505" PostTypeId="2" ParentId="11502" CreationDate="2010-11-23T12:38:40.513" Score="2" ViewCount="" Body="&lt;p&gt;An easier method (assuming that the values were generated by a polynomial) would be to note that successive divided differences $\frac{y_{i+1}-y_i}{x_{i+1}-x_i}$ are constant and equal to 2; thus your function is of the form $y=2x+c$. The constant $c$ is then determined by replacing both $x$ and $y$ with appropriate values, and then solving for $c$.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-11-23T12:38:40.513" CommentCount="1" />
  <row Id="11506" PostTypeId="2" ParentId="11502" CreationDate="2010-11-23T13:07:28.313" Score="3" ViewCount="" Body="&lt;p&gt;The best tool for doing this is that very, very impressive piece of software:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://ccsl.mae.cornell.edu/eureqa&quot; rel=&quot;nofollow&quot;&gt;http://ccsl.mae.cornell.edu/eureqa&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; For your abovementioned very easy example, even WA will find the right formula:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.wolframalpha.com/input/?i=3,+5,+7,+9&quot; rel=&quot;nofollow&quot;&gt;http://www.wolframalpha.com/input/?i=3,+5,+7,+9&lt;/a&gt;,...&lt;/p&gt;&#xA;" OwnerUserId="346" LastActivityDate="2010-11-23T13:07:28.313" CommentCount="4" />
  <row Id="11508" PostTypeId="2" ParentId="10392" CreationDate="2010-11-23T13:51:31.893" Score="0" ViewCount="" Body="&lt;p&gt;You can also use &lt;a href=&quot;http://en.wikipedia.org/wiki/Disjunctive_normal_form&quot; rel=&quot;nofollow&quot;&gt;disjunctive normal form&lt;/a&gt; or conjunctive Normal Form(no link because of spam prevention mechanism)&#xA;for finding the right formula. So for the first answe&lt;/p&gt;&#xA;" LastActivityDate="2010-11-23T13:51:31.893" />
  <row Id="11509" PostTypeId="1" AcceptedAnswerId="11516" CreationDate="2010-11-23T13:54:10.520" Score="2" ViewCount="218" Body="&lt;p&gt;Hi all, I've got a question concerning how to proof the Pythagoras theorem using the following assumption:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x$ is perpendicular to $y$ (if and only if) $||x+y||^2 = ||x||^2 + ||y||^2$,&#xA;where $x$ and $y$ are vectors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a basic understanding of linear algebra however I'm a beginner with this. The question provides hints how to prove the above mentioned equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Namely that I should use the properties of the dot product and the definition of the norm of a vector.&#xA;Those being symmetry, scaling and distributivity as the dot product properties and the norm of a vector being the squared root of the dot product between the same vector.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was thinking about using the fact that if a vector is perpendicular to another vector the dot product between those vectors should be 0. But that is not provided as a hint so I'm not sure. I know the under lying thought behind it is the cosine rule for vectors, that being:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x\cdot y = ||x||\,||y|| \cos(\theta)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the angle between the two vectors is perpendicular you should use $\cos(\pi/2)$ which is $0$ and $||x||\cdot 0 = 0$ and $||y||\cdot 0 = 0$ with the vectors not necessarily being $0$. Thus $x\cdot y = 0$.&#xA;How would I apply this to the equation I first mentioned to prove the Pythagoras theorem?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a few more thoughts on how I could prove this but I'm not sure if they're correct. &#xA;I hope someone could point me in the right direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance, Rope.&lt;/p&gt;&#xA;" OwnerUserId="3854" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-23T15:40:26.223" LastActivityDate="2010-12-05T23:07:00.273" Title="How to prove the Pythagoras theorem using vectors" Tags="&lt;linear-algebra&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="1" />
  <row Id="11511" PostTypeId="2" ParentId="11509" CreationDate="2010-11-23T14:05:44.590" Score="0" ViewCount="" Body="&lt;p&gt;If I understood the question correctly, you have to prove that&#xA;$$\|x+y\|^2 = \|x\|^2 + \|y\|^2$$&#xD;&#xA;if and only if $x$ and $y$ are perpendicular. That is, as you noted, $x \cdot y = 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint: You can use the following formulas:&#xA;$$\|x\|^2 = x \cdot x$$&#xD;&#xA;$$(x+y) \cdot z = x \cdot z + y \cdot z$$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="3776" LastActivityDate="2010-11-23T14:05:44.590" CommentCount="4" />
  <row Id="11512" PostTypeId="2" ParentId="11499" CreationDate="2010-11-23T14:11:11.433" Score="6" ViewCount="" Body="&lt;p&gt;If you want the faces congruent, it will not work.  If you try to  use congruent polygons you cannot satisfy Euler's formula.  This says if you use pentagons and hexagons there must be 12 pentagons.  The &lt;a href=&quot;http://en.wikipedia.org/wiki/Truncated_icosahedron&quot; rel=&quot;nofollow&quot;&gt;truncated icosahedron&lt;/a&gt;, which has 12 pentagonal and 20 hexagonal faces is quite close to a sphere and there are only two kinds of faces.  I thnk you could subdivide the hexagons into 7 and pentagons into one pentagon and five hexagons to get 200 hexagons and 12 pentagons, but I don't think the hexagons are all the same shape any more.  Another subdivision has 1460 hexagons and 12 pentagons.  I think there are more in between, but a quick web search didn't turn up any good references.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-11-23T14:11:11.433" CommentCount="2" />
  <row Id="11513" PostTypeId="2" ParentId="11464" CreationDate="2010-11-23T14:16:56.950" Score="1" ViewCount="" Body="&lt;p&gt;Have a look at Kelley and Peterson's textbook[1]. I think this book can give you a little more understanding of where everything comes from. It talks about difference calculus, which is the discrete analogue of continuous calculus. In difference calculus, we have the delta and sum operators where we have the derivative and integral respectively in continuous calculus.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take a look at it and let me know if it helps.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[1] Kelley, W. &amp;amp; Peterson, A. (2001). Difference Equations: An Introduction with Applications (2nd Ed.). San Diego, CA: Academic Press.&lt;/p&gt;&#xA;" OwnerUserId="3791" LastActivityDate="2010-11-23T14:16:56.950" />
  <row Id="11514" PostTypeId="2" ParentId="11492" CreationDate="2010-11-23T14:18:50.737" Score="0" ViewCount="" Body="&lt;p&gt;Just simplifying:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$-2\sigma\left(rx^{2}+y^{2}+b\left(z-r\right)^{2}-br^{2}\right) &amp;lt; 0$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$=&amp;gt; rx^{2}+y^{2}+b\left(z-r\right)^{2}-br^{2} &amp;gt; 0 ... \sigma &amp;gt; 0$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$=&amp;gt; rx^{2}+y^{2}+bz^{2}-2brz+br^{2}-br^{2} &amp;gt; 0 $&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$=&amp;gt; rx^{2}+y^{2}+bz^{2}-2brz &amp;gt; 0 $&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$=&amp;gt; rx^{2}+y^{2}+bz^{2} &amp;gt; 2brz $&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$=&amp;gt; z&amp;lt;=0$ or $z&amp;gt;=2r$ for any $C$ or&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ rx^{2}+y^{2} &amp;gt; 2brz - bz^{2}, 0&amp;lt;z&amp;lt;2r $&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now,&#xA;$rx^{2}+\sigma y^{2}+\sigma\left(z-2r\right)^{2}=C$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$=&amp;gt; rx^{2}+y^{2}+(\sigma-1) y^{2}+\sigma\left(z-2r\right)^{2}=C$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$=&amp;gt; rx^{2}+y^{2}= C -(\sigma-1) y^{2}-\sigma\left(z-2r\right)^{2}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the original expression is negative when $z$ is outside $(0,2r)$ or&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ 2brz - bz^{2} &amp;lt; C -(\sigma-1) y^{2}-\sigma\left(z-2r\right)^{2}, 0&amp;lt;z&amp;lt;2r $&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$=&amp;gt; bz(2r - z) &amp;lt; C -(\sigma-1) y^{2}-\sigma\left(z-2r\right)^{2}, 0&amp;lt;z&amp;lt;2r $&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$=&amp;gt; 0 &amp;lt; C -(\sigma-1) y^{2}-\sigma\left(z-2r\right)^{2}+bz(z - 2r), 0&amp;lt;z&amp;lt;2r $&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$=&amp;gt; 0 &amp;lt; C -(\sigma-1) y^{2}-\sigma\left(z(1-b/2)-2r\right)^{2}+(bz/2)^{2}, 0&amp;lt;z&amp;lt;2r $ ... completing the square&lt;/p&gt;&#xA;" OwnerUserId="820" LastEditorUserId="820" LastEditorDisplayName="" LastEditDate="2010-11-23T14:33:11.407" LastActivityDate="2010-11-23T14:33:11.407" />
  <row Id="11515" PostTypeId="1" AcceptedAnswerId="11532" CreationDate="2010-11-23T14:22:48.797" Score="2" ViewCount="311" Body="&lt;p&gt;Let's say that we have three points: $p = (x_p,y_p)$, $q = (x_q,y_q)$ and $a = (x_a,y_a)$. How can i find point $b$ which is reflection of $a$ across a line drawn through $p$ and $q$? I know it's simple to calculate, when we have $p$, $q$ etc. But I want to do this in my program, and I'm not sure, how to compute this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;OK, I've found solution by myself (but answers in this topic really helped me).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose, that we have a line $Ax+By+C=0$, and $A^2+B^2 \not= 0$. $M (a,b)$ reflection across the line is point: $M&amp;#39; (\frac{aB^2-aA^2-2bAB-2AC}{A^2+B^2}, \frac{bA^2-bB^2-2aAB-2BC}{A^2+B^2})$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In my case, we don't have line, but only 2 points. How we can find $A,B,C$? It's simple:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's say, that $p=(p_x,p_y)$ and $q = (q_x,q_y)$. Line equation is: $(y-p_y)(q_x-p_x) - (q_y-p_y)(x-p_x) = 0$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After some calculations we have: $y(q_x-p_x) - x(q_y-p_y) - (p_y(q_x-p_x)+p_x(p_y-q_y)) = 0$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So:&#xA;$A = p_y-q_y$, $B = q_x-p_x$ and $C = -p_y(q_x-p_x)-p_x(p_y-q_y)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's all.&lt;/p&gt;&#xA;" OwnerUserId="3856" LastEditorUserId="3856" LastEditorDisplayName="" LastEditDate="2010-11-23T21:42:03.613" LastActivityDate="2010-11-23T21:42:03.613" Title="Point reflection across a line" Tags="&lt;reflection&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="11516" PostTypeId="2" ParentId="11509" CreationDate="2010-11-23T14:28:38.993" Score="2" ViewCount="" Body="&lt;p&gt;$x,y$ are perpendicular if and only if $x\cdot y=0$. Now, $||x+y||^2=(x+y)\cdot (x+y)=(x\cdot x)+(x\cdot y)+(y\cdot x)+(y\cdot y)$. The middle two terms are zero if and only if $x,y$ are perpendiculat. So, $||x+y||^2=(x\cdot x)+(y\cdot y)=||x||^2+||y||^2$ if and only if $x,y$ are perpendicular.&lt;/p&gt;&#xA;" OwnerUserId="3431" LastActivityDate="2010-11-23T14:28:38.993" CommentCount="1" />
  <row Id="11517" PostTypeId="2" ParentId="11515" CreationDate="2010-11-23T14:31:50.813" Score="2" ViewCount="" Body="&lt;p&gt;You find the equation of the line through p and q, one version of which is $$\frac{y-y_p}{x-x_p}=\frac{y_q-y_p}{x_q-x_p}$$  The slope is the RHS.  So the slope of the line through a and perpendicular to the line through p and q is $$\frac{-(x_q-x_p)}{y_q-y_p}$$  and the line is $$y-y_a=\frac{-(x_q-x_p)}{y_q-y_p}(x-x_a)$$  Calculate the distance from a to the line through p and q and extend that far again.  I think there is a formula for this, but don't know it, so I would have to solve the two equations for x and y to get the intersection.  Note that there is a divide by zero concern.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-11-23T14:31:50.813" CommentCount="5" />
  <row Id="11518" PostTypeId="1" AcceptedAnswerId="11530" CreationDate="2010-11-23T14:42:02.587" Score="2" ViewCount="182" Body="&lt;p&gt;How to prove that $\mathcal{O}_{\sqrt[3]{3}}$ is an euclidean domain? I heard that one should prove the following but why it is enough?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For any $ a,b,c\in\mathbb{R}$, prove that there are $ x,y,z\in\mathbb{R}$ such that $ x-a,y-b,z-c\in\mathbb{Z}$ and that&#xA;$$-1\leq x^3+3y^3+9z^3-9xyz\leq 1.$$&lt;/p&gt;&#xA;" OwnerUserId="2856" LastActivityDate="2010-11-23T16:47:21.970" Title="Euclidean domain" Tags="&lt;inequality&gt;&lt;algebraic-number-theory&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="11519" PostTypeId="2" ParentId="11474" CreationDate="2010-11-23T14:44:17.990" Score="2" ViewCount="" Body="&lt;p&gt;The topology generated by any family of subsets $F$ of a set $X$ is the intersection of all topologies on $X$ containing $F$; $F$ then forms a &lt;a href=&quot;http://en.wikipedia.org/wiki/Subbase&quot; rel=&quot;nofollow&quot;&gt;subbase&lt;/a&gt; for the resulting topology.  In practice this means that you take all finite intersections of the elements of $F$ to get a &lt;a href=&quot;http://en.wikipedia.org/wiki/Base_(topology)&quot; rel=&quot;nofollow&quot;&gt;base&lt;/a&gt; $F&amp;#39;$, then take arbitrary unions of the elements of $F&amp;#39;$.  &lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-11-23T14:44:17.990" />
  <row Id="11520" PostTypeId="2" ParentId="11499" CreationDate="2010-11-23T14:46:28.063" Score="3" ViewCount="" Body="&lt;p&gt;Draw 1000 meridians enclosing an angle of 0.36 degrees between two successive ones. Or, if you prefer 1000 congruent triangles: Draw an equator and 500 meridians enclosing an angle of 0.72 degrees between two successive ones.&lt;/p&gt;&#xA;" OwnerUserId="1303" LastEditorUserId="1303" LastEditorDisplayName="" LastEditDate="2010-12-03T19:41:07.227" LastActivityDate="2010-12-03T19:41:07.227" CommentCount="1" />
  <row Id="11521" PostTypeId="2" ParentId="11487" CreationDate="2010-11-23T14:53:25.453" Score="2" ViewCount="" Body="&lt;p&gt;For some later purposes, I give here the complete solution based on svenkatr's approach (thus also verifying my previous answer). In what follows, I use  a different (somewhat more natural) notation, so don't get confused.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose that $X$ and $Y$ are independent ${\rm N}(0,\sigma_1^2)$ and ${\rm N}(0,\sigma_2^2)$ variables, respectively.&#xA;We can find ${\rm E}(X|X+Y=z)$ as follows.&#xA;$$&#xD;&#xA;{\rm E}(X|X + Y = z) = \int_{ - \infty }^\infty  {xf_{X|X + Y} (x|z)\,{\rm d}x} = \int_{ - \infty }^\infty  {x\frac{{f_X (x)f_{X + Y|X} (z|x)}}{{f_{X + Y} (z)}}\, {\rm d}x}&#xD;&#xA;$$&#xD;&#xA;(the notation should be clear from the context). Noting that $f_{X+Y|X}(\cdot|x)$ is the ${\rm N}(x,\sigma_2^2)$ density function, some algebra shows that the right-hand side integral is equal to&#xA;$$&#xD;&#xA;\int_{ - \infty }^\infty  {x\sqrt {\frac{{\sigma _1^2  + \sigma _2^2 }}{{\sigma _1^2 \sigma _2^2}}} \frac{1}{{\sqrt {2\pi } }}\exp \bigg\{  - \frac{{[x - z \sigma _1^2 /(\sigma _1^2  + \sigma _2^2 )]^2 }}{{2[\sigma _1^2 \sigma _2^2 /(\sigma _1^2  + \sigma _2^2 )]}}} \bigg \} \,{\rm d}x.&#xD;&#xA;$$&#xD;&#xA;Now, from $\int_{ - \infty }^\infty  {x\frac{1}{{\sqrt {2\pi \sigma ^2 } }}{\rm e}^{ - (x - \mu )^2 /(2\sigma ^2 )} \,{\rm d}x}  = \mu $ (expectation of a ${\rm N}(\mu,\sigma^2)$ random variable), we can find that&#xA;$$&#xD;&#xA;{\rm E}(X|X + Y = z) = \frac{{\sigma _1^2 }}{{\sigma _1^2  + \sigma _2^2 }}z.&#xD;&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="2810" LastActivityDate="2010-11-23T14:53:25.453" />
  <row Id="11522" PostTypeId="2" ParentId="11498" CreationDate="2010-11-23T15:18:45.993" Score="1" ViewCount="" Body="&lt;p&gt;If $Y=X+\frac{X}{10}N=X\left(1+\frac{N}{10}\right)$, then $X=\frac{Y}{1+\frac{N}{10}}$ .&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-11-23T15:18:45.993" CommunityOwnedDate="2010-11-23T15:18:45.993" />
  <row Id="11523" PostTypeId="1" CreationDate="2010-11-23T15:25:17.580" Score="2" ViewCount="67" Body="&lt;p&gt;Hello friends,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let X be a smooth projective variety with canonical class K.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let a be defined to be the maximum dimension of the image of X under the rational map induced by the linear system |nK| as n ranges over all positive integers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let b be defined to be the smallest positive integer c such that the sequence (dim(H^0 (X, nK))/n^c) is bounded (as n gets large).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second definition I guess doesnt include the case with kodaira dimension smaller than 0 (ie -1 or -infinity depending on how you like to define it), so we're only talking about non-negative kodaira dimension.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Will someone please tell me why a=b?&lt;/p&gt;&#xA;" OwnerUserId="3860" LastActivityDate="2011-05-15T21:16:58.857" Title="Different definitions of Kodaira dimension" Tags="&lt;algebraic-geometry&gt;" AnswerCount="1" />
  <row Id="11524" PostTypeId="1" AcceptedAnswerId="11526" CreationDate="2010-11-23T15:28:43.410" Score="4" ViewCount="157" Body="&lt;p&gt;I'm asking here because no textbook or website that I know of gives a definition of the above mentioned term.&#xA;Since there's no obvious way (that I can think of) to define a normal subloop, I don't see how the definition of a simple group can be modified for the case of loops.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So - &lt;strong&gt;what is the definition of a simple loop?&lt;/strong&gt; (And what is the motivation behind the definition?)&lt;/p&gt;&#xA;" OwnerUserId="766" LastActivityDate="2011-04-12T04:25:44.270" Title="What is a simple loop?" Tags="&lt;abstract-algebra&gt;&lt;definition&gt;&lt;loops&gt;&lt;quasigroups&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="1" />
  <row Id="11525" PostTypeId="2" ParentId="11509" CreationDate="2010-11-23T15:32:46.947" Score="2" ViewCount="" Body="&lt;p&gt;The definition of $||x||$ for vectors is:&#xA;$$||x|| = \sqrt{x\cdot x}.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, you have that&#xA;\begin{align*}&#xA;||x+y||^2 &amp;amp;= (x+y)\cdot(x+y)  &amp;amp;\text{(by definition)}\\&#xA;&amp;amp;= x\cdot x + x\cdot y + y\cdot x + y\cdot y&#xA;&amp;amp;\text{(by distributivity)}\\&#xA;&amp;amp;= x\cdot x + y\cdot y + 2(x\cdot y) &amp;amp;\text{(by symmetry)}\\&#xA;&amp;amp;= ||x||^2 + ||y||^2 + 2(x\cdot y) &amp;amp;\text{(by definition)}\\&#xA;&amp;amp;= ||x||^2 + ||y||^2 + 2||x||\,||y||\cos(\theta),&#xA;\end{align*}&#xA;where $\theta$ is the angle between $x$ and $y$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This holds in any case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, $||x+y||^2 = ||x||^2 + ||y||^2$ if and only if $2||x||\,||y||\cos(\theta)=0$. One possibility is $||x||=0$; another is $||y||=0$; and the final one is $\cos(\theta)=0$. So the equality holds if and only if one of the following happens:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$||x||=0$;&lt;/li&gt;&#xA;&lt;li&gt;$||y||=0$; or&lt;/li&gt;&#xA;&lt;li&gt;$\cos(\theta)=0$ where $\theta$ is the angle between $x$ and $y$, $x\neq 0$, $y\neq 0$.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-05T23:07:00.273" LastActivityDate="2010-12-05T23:07:00.273" CommentCount="2" />
  <row Id="11526" PostTypeId="2" ParentId="11524" CreationDate="2010-11-23T15:55:52.660" Score="6" ViewCount="" Body="&lt;p&gt;A subloop $H$ of a loop $L$ is a normal subloop if $H$ is the kernel of a homomorphism from $L$ onto some loop $K$, and $L$ is a simple loop if it has no nontrivial normal subloops; that is, the only homomorphic images of $L$ have order $1$ or are isomorphic to $L$. (This is Qiaochu's answer above).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I took the definition out of the review  (MR0977475 (90b:20057)) of&#xA;Bannai, Eiichi; Song, Sung-Yell, &lt;em&gt;The character tables of Paige's simple Moufang loops and their relationship to the character tables of ${\rm PSL}(2,q)$&lt;/em&gt;.&#xA;Proc. London Math. Soc. (3) &lt;strong&gt;58&lt;/strong&gt; (1989), no. 2, 209–236,  &lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-11-23T15:55:52.660" />
  <row Id="11527" PostTypeId="2" ParentId="11477" CreationDate="2010-11-23T16:01:59.470" Score="6" ViewCount="" Body="&lt;p&gt;With Fibonacci numbers usually there are &lt;em&gt;multiple&lt;/em&gt; ways of proving identities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One way (which is one of my favourites) to prove your identity is the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the following problem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A person climbs up $\displaystyle n$ steps, by taking either one step, or two steps at a time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The total number of ways the person can climb up all the $\displaystyle n$ steps is $\displaystyle F_{n+1}$ (Why?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now consider climbing $\displaystyle m+n-1$ steps and split into the cases when the person lands on step $\displaystyle n$ and the cases when the person lands on step $\displaystyle n-1$ and takes two steps at that point (and so does not land on step $\displaystyle n$ in those cases). These two cases cover all possibilities, and so we have:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle F_{m+n} = F_{n+1}F_{m} + F_{n}F_{m-1}$$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-11-23T16:01:59.470" CommentCount="5" />
  <row Id="11528" PostTypeId="1" AcceptedAnswerId="11545" CreationDate="2010-11-23T16:12:44.790" Score="1" ViewCount="187" Body="&lt;p&gt;What is the next term of the following sequences ? &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$ 3,8,13,24,41, \cdots $ &#xA;       &lt;strong&gt;OPTIONS :&lt;/strong&gt; $\{ 70,75,80,85 \}$&#xD;&lt;/li&gt;&#xA;&lt;li&gt;$ 4,23,60,121, \cdots $  &lt;strong&gt;OPTIONS :&lt;/strong&gt; $\{212,221,241,242\}$&#xD;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This two question is taken from a test paper of mine,I am no able to figure out the correct-answer.&lt;/p&gt;&#xA;" OwnerUserId="2109" LastEditorUserId="1543" LastEditorDisplayName="" LastEditDate="2011-05-03T10:11:39.540" LastActivityDate="2011-05-03T10:11:39.540" Title="Find the missing term :" Tags="&lt;sequences-and-series&gt;" AnswerCount="5" CommentCount="6" FavoriteCount="1" />
  <row Id="11529" PostTypeId="2" ParentId="11501" CreationDate="2010-11-23T16:13:04.837" Score="-1" ViewCount="" Body="&lt;p&gt;A &lt;i&gt;recurrence relation&lt;/i&gt; defines a function depending on its values on points of lower &quot;rank&quot;. Since you want to be able to calculate the function, you also need some bases cases that are just given (like axioms). For example, the recurrence relation defining the Fibonacci numbers defines $F_n$ in terms of $F_{n-1},F_{n-2}$; the base cases $F_1=1,F_0=0$ are not defined in terms of previous elements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can have more complicated definitions of &quot;rank&quot;. Here are two examples. One can define the binomial coefficients using Pascal's identity $$\binom{n}{k} = \binom{n-1}{k-1} + \binom{n-1}{k}$$ with appropriate bases cases. Written out, that would be $$C(n,k) = C(n-1,k-1) + C(n-1,k), \quad C(0,k) = 1_{k=0}.$$ So in that cases $(n-1,\cdot)$ is of lower rank than $(n,\cdot)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another example is the determinant - it's a function on $n \times n$ matrices (arrays of numbers) which is defined as the single value in case $n=1$, and using some formula involving $(n-1)\times(n-1)$ matrices otherwise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, for the definition to work, the notation of rank must be a &lt;i&gt;well ordering&lt;/i&gt;.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;A recursive algorithm is a procedure which calls itself. For example, you can use the following pseudocode to calculate the Fibonacci numbers:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;if ((n == 0) || (n==1)) return n;&#xA;else return F(n-1) + F(n-2);&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Every recurrence relation can be computed this way. In this particular case, the algorithm for calculating $F_n$ runs in exponential time (in other words, it's very inefficient), and there are better ways to calculate $F_n$; for example, you could calculate the values iteratively:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;F[0] = 0; F[1] = 1;&#xA;for (int i = 2; i &amp;lt;= n; i++) F[i] = F[i-1] + F[i-2];&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;There are even better algorithms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conversely, recursive algorithms have other uses, for example the best algorithms for sorting a list of numbers are recursive and use the &quot;divide and conquer&quot; paradigm.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-23T16:13:04.837" />
  <row Id="11530" PostTypeId="2" ParentId="11518" CreationDate="2010-11-23T16:16:49.640" Score="2" ViewCount="" Body="&lt;p&gt;Note that the ring of integers of $\mathbb{Q}(\sqrt[3]{3})$ is $\mathbb{Z}[\sqrt[3]{3}]$ with basis $1,\sqrt[3]{3},\sqrt[3]{9}$ over $\mathbb{Z}$. You want to show that the norm, defined by&#xA;\begin{eqnarray*}&#xA;N(a+b\sqrt[3]{3} + c\sqrt[3]{9}) &amp;amp; = &amp;amp; (a+b\sqrt[3]{3} + c\sqrt[3]{9})(a+\zeta_3b\sqrt[3]{3} + \zeta_3^2c\sqrt[3]{9})(a+\zeta_3^2b\sqrt[3]{3} + \zeta_3c\sqrt[3]{9})\\&#xA;&amp;amp; = &amp;amp; a^3 + 3b^3 + 9c^3 - 9abc&#xA;\end{eqnarray*}&#xA;gives in fact a Euclidean norm upon taking absolute values, where $\zeta_3$ is a fixed primitive cube root of unity. Now, the first (and maybe the main) thing you might wonder about is where I pulled this norm out. For that you have to know a little bit of Galois theory. The basic idea is that from the perspective of $\mathbb{Q}$, the elements $\sqrt[3]{3}$ and $\zeta_3\sqrt[3]{3}$ are indistinguishable: they are both just &lt;strong&gt;some&lt;/strong&gt; roots of the polynomial $x^3-3$ and can be thought of as mirror images of each other. Essentially, the factors that I am multiplying are like &quot;mirror images&quot; of my given element of $\mathbb{Z}[\sqrt[3]{3}]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the strategy is exactly the same as in the case of some quadratic rings like $\mathbb{Z}[\sqrt{2}]$: given $\alpha = u+v\sqrt[3]{3} + w\sqrt[3]{9}$ and&#xA;$\beta = f+g\sqrt[3]{3} + h\sqrt[3]{9}\in \mathbb{Z}[\sqrt[3]{3}]$, you want to show that there exist $p$ and $r$ in that ring such that&#xA;\begin{eqnarray*}\alpha = p\beta + r,&#xA;\end{eqnarray*}&#xA;where either $r=0$ or $N(r)&amp;lt;N(\beta)$. Dividing by $\beta$, this becomes equivalent to showing that given any $\alpha/\beta=a+b\sqrt[3]{3} + c\sqrt[3]{9}$ in the field of fraction $\mathbb{Q}(\sqrt[3]{3})$ of $\mathbb{Z}[\sqrt[3]{3}]$, there exists $r/\beta\in \mathbb{Q}(\sqrt[3]{3})$ such that the difference is in the ring of integers itself (since the difference is $p$) and $|N(r/\beta)|&amp;lt;1$. That is the statement you posted.&lt;/p&gt;&#xA;" OwnerUserId="3212" LastActivityDate="2010-11-23T16:16:49.640" CommentCount="1" />
  <row Id="11531" PostTypeId="2" ParentId="11528" CreationDate="2010-11-23T16:16:59.267" Score="1" ViewCount="" Body="&lt;ol&gt;&#xA;&lt;li&gt;3+8+2=13,8+13+3=24,13+24+4=41,24+41+5=70&lt;/li&gt;&#xA;&lt;li&gt;I don't know&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;PS: you should use the tag &quot;homework&quot;&lt;/p&gt;&#xA;" OwnerUserId="3631" LastActivityDate="2010-11-23T16:16:59.267" CommentCount="3" />
  <row Id="11532" PostTypeId="2" ParentId="11515" CreationDate="2010-11-23T16:18:22.383" Score="5" ViewCount="" Body="&lt;p&gt;It is useful to think about this by using vectors. Suppose that your points are $p,q,a \in \mathbb{R^2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then $x = p + (q-p)t$ is a point on the line $pq$. We wish to find the projection of $a$&amp;#160;on the line. To do this we require&#xA;$\|x - a\|^2 = \|x\|^2 + \|a\|^2 - 2 (x \cdot a)$ to be minimal. That is we have to minimize&#xA;$$\|p + (q-p)t\|^2 + \|a\|^2 - 2 (p + (q-p) t) \cdot a)$$&#xD;&#xA;w.r.t $t$. To do this we write this as&#xA;$$\|p\|^2 + \|q-p\|^2 t^2 + 2 t p \cdot (q-p) + \|a\|^2 - 2(p \cdot a) - 2 t (q-p) \cdot a.$$&#xD;&#xA;This is a quadratic in $t$ with minimum at the tip of the parabola:&#xA;$$t = \frac{-2 p \cdot (q-p) + 2 (q-p) \cdot a}{2\|q-p\|^2} = \frac{(q-p) \cdot (a-p)}{\|q-p\|^2}.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus the projection is given by&#xA;$$x = p + (q-p) \frac{(q-p) \cdot (a-p)}{\|q-p\|^2}$$ and the reflection is then just $x + (x-a) = 2x-a$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This method doesn't have problems with infinite slopes.&lt;/p&gt;&#xA;" OwnerUserId="3776" LastActivityDate="2010-11-23T16:18:22.383" CommentCount="5" />
  <row Id="11533" PostTypeId="2" ParentId="11435" CreationDate="2010-11-23T16:21:22.647" Score="2" ViewCount="" Body="&lt;p&gt;The (Motivation) description appears to be similar to a direct sum (a.k.a. tegum product) as defined in Section 1.2 of this paper:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://people.reed.edu/~davidp/homepage/seniors/mcmahan.pdf&quot; rel=&quot;nofollow&quot;&gt;http://people.reed.edu/~davidp/homepage/seniors/mcmahan.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The difference is that for a direct sum, 0 is in the relative interior of both of the factor polytopes, rather than a common vertex.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same article answers your question about what the nm vertices of a Cartesian product are.&lt;/p&gt;&#xA;" OwnerUserId="1782" LastEditorUserId="1782" LastEditorDisplayName="" LastEditDate="2010-11-23T16:55:25.710" LastActivityDate="2010-11-23T16:55:25.710" />
  <row Id="11534" PostTypeId="2" ParentId="11528" CreationDate="2010-11-23T16:21:28.437" Score="4" ViewCount="" Body="&lt;p&gt;For part (2), try adding $4$ to the numbers.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-23T16:21:28.437" CommentCount="2" />
  <row Id="11535" PostTypeId="2" ParentId="11518" CreationDate="2010-11-23T16:40:32.590" Score="2" ViewCount="" Body="&lt;p&gt;As Alex mentioned, the proof that you have in mind amounts to showing that your cubic field is Euclidean with respect to the norm. In fact it is known that there are only three pure cubic fields $\rm\ Q(\sqrt[3] m)\ $ that are norm Euclidean, viz. $\rm\ \mathbb Q(\sqrt[3] 2),\ \mathbb Q(\sqrt[3] 3),\ \mathbb Q(\sqrt[3] {10})\:.\ $ You can find the proofs in Cioffari: &lt;a href=&quot;http://www.ams.org/journals/mcom/1979-33-145/S0025-5718-1979-0514835-5/S0025-5718-1979-0514835-5.pdf&quot; rel=&quot;nofollow&quot;&gt;The Euclidean Condition in Pure Cubic and Complex Quartic Fields.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-11-23T16:47:21.970" LastActivityDate="2010-11-23T16:47:21.970" CommentCount="3" />
  <row Id="11536" PostTypeId="2" ParentId="10468" CreationDate="2010-11-23T16:46:43.467" Score="0" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://tutorteddy.com/site/&quot; rel=&quot;nofollow&quot;&gt;http://tutorteddy.com/site/&lt;/a&gt;&lt;/p&gt;&#xA;" LastActivityDate="2010-11-23T16:46:43.467" />
  <row Id="11537" PostTypeId="1" AcceptedAnswerId="11546" CreationDate="2010-11-23T17:02:38.837" Score="2" ViewCount="134" Body="&lt;p&gt;According to &lt;a href=&quot;http://en.wikipedia.org/wiki/Discontinuous_linear_map#Axiom_of_choice&quot; rel=&quot;nofollow&quot;&gt;wikipedia&lt;/a&gt;, you need the Axiom of Choice to find a discontinuous map between two Banach spaces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does this procedure also apply for Banach algebras yielding a discontinuous multiplicative linear map?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or, is there some obstruction, ensuring that every algebraic homomorphism between two Banach algebras is continuous? (I know that this is true for $*$-homomorphisms between C*-algebras.)&lt;/p&gt;&#xA;" OwnerUserId="367" LastEditorUserId="367" LastEditorDisplayName="" LastEditDate="2010-11-23T21:04:26.220" LastActivityDate="2010-11-23T21:04:26.220" Title="Is there an algebraic homomorphism between two Banach algebras which is not continuous?" Tags="&lt;set-theory&gt;&lt;functional-analysis&gt;" AnswerCount="1" />
  <row Id="11538" PostTypeId="1" AcceptedAnswerId="11540" CreationDate="2010-11-23T17:04:54.977" Score="5" ViewCount="353" Body="&lt;p&gt;I understand the formal definition of uniform continuity of a function,&#xA;and how it is different from standard continuity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is: Is there an intuitive way to classify a function&#xA;on $\mathbb{R}$ as being&#xA;uniformly continuous, just like there is for regular continuity?&#xA;I mean, that for a &quot;nice&quot; function $f:\mathbb{R} \to \mathbb{R}$,&#xA;it is usually easy to tell if it is continuous on an interval by&#xA;looking at or thinking of the graph of the function on the interval,&#xA;or on all of $\mathbb{R}$. Can I do the same for uniform continuity?&#xA;Can I tell that a function is uniformly continuous just by what it&#xA;looks like? Ideally, this intuition would fit with the Heine-Cantor theorem for&#xA;compact sets on $\mathbb{R}$.&lt;/p&gt;&#xA;" OwnerUserId="2029" LastActivityDate="2010-11-23T21:04:02.507" Title="Intuition for uniform continuity of a function on $\mathbb{R}$" Tags="&lt;real-analysis&gt;" AnswerCount="3" FavoriteCount="2" />
  <row Id="11539" PostTypeId="1" AcceptedAnswerId="11544" CreationDate="2010-11-23T17:06:00.450" Score="0" ViewCount="187" Body="&lt;p&gt;Consider a function f:[0,1] -&gt; R  with f(0)=0&lt;br&gt;&#xA;having a finite derivative at each x in (0,1)&lt;br&gt;&#xA;Prove that if f' is an increasing function (at least on the interval (0,1))&lt;br&gt;&#xA;then h(x)=f(x)/x is also increasing&lt;/p&gt;&#xA;" OwnerUserId="3333" LastActivityDate="2010-11-23T17:42:57.830" Title="Prove an increasing derivative implies an increasing function" Tags="&lt;real-analysis&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="11540" PostTypeId="2" ParentId="11538" CreationDate="2010-11-23T17:14:16.323" Score="9" ViewCount="" Body="&lt;p&gt;I like to think of the following fact: a $C^1$ function on $\mathbb{R}$ with bounded derivative is uniformly continuous.  So in order for a function not to be uniformly continuous, there have to be places where its graph is &quot;arbitrarily steep&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because of the theorem that any continuous function on a compact set is uniformly continuous, this can only happen if you make the function steeper and steeper as you go off to $\infty$, either unboundedly (like $x^2$) or boundedly (like $\sin(x^2)$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: As pointed out, the converse is false: a function with unbounded derivative can still be uniformly continuous.  $\sin(x^4)/(1+x^2)$ is such an example, I believe.  So this may not be great intuition after all.&lt;/p&gt;&#xA;" OwnerUserId="822" LastEditorUserId="822" LastEditorDisplayName="" LastEditDate="2010-11-23T21:04:02.507" LastActivityDate="2010-11-23T21:04:02.507" CommentCount="6" />
  <row Id="11541" PostTypeId="2" ParentId="11538" CreationDate="2010-11-23T17:14:43.090" Score="7" ViewCount="" Body="&lt;p&gt;One possible intuition for uniform continuity is that a function shouldn't oscillate arbitrarily wildly. If the function is differentiable e.g. on an open interval, then the derivative should be bounded in absolute value. That's something that you can see by looking at the graph. For example if you consider the function $sin(1/x)$ on the open interval $(0,1)$, then just by looking at it, you will realise that it's not uniformly continuous, since it's slope becomes arbitrarily steep as you approach $0$ from the right.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The way this intuition ties in with the Heine-Cantor theorem is that on compact sets, bounded functions attain their extrema. If your function is differentiable, then this is in particular true for its derivative, so the slope of the function cannot get arbitrarily steep on a compact interval.&lt;/p&gt;&#xA;" OwnerUserId="3212" LastEditorUserId="3212" LastEditorDisplayName="" LastEditDate="2010-11-23T17:58:47.390" LastActivityDate="2010-11-23T17:58:47.390" CommentCount="2" />
  <row Id="11542" PostTypeId="1" AcceptedAnswerId="11572" CreationDate="2010-11-23T17:15:28.323" Score="6" ViewCount="195" Body="&lt;p&gt;I'm trying to understand a proof in &quot;Combinatorial Group Theory&quot; by Lyndon &amp;amp; Schupp, page 26, Lemma 4.7.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's my problem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$F$ is a free group with $X$ as basis. $x \in X$. $\alpha \in IA(F)$ - meaning it's an automorphism of $F$ which induces the identity automorphism on $F/[F,F]$. $k \in F_n$ - a term in the descending central series for F. $x\alpha$ and $xk$ are conjugates.&#xA;In addition: for some $N\geq1$, $\alpha^N \in JA(F)$ - the group of inner automorphisms of F.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Elementary Fact&quot; used in the proof: $\alpha \in IA(F)$ and $k \in F_n$ implies $k \alpha = k$ (modulo $F_{n+1}$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Right after mentioning this &quot;elementary fact&quot; they say it implies: $x\alpha^N = xk^N$ (modulo $F_{n+1}$). Why is that?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;: A little clarification: $x\alpha$ is the image of $x$ under the automorphism $\alpha$, following the notation in this book. $xk$ is just $x$ times $k$, multiplication in the group $F$.&lt;/p&gt;&#xA;" OwnerUserId="3533" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-28T04:44:43.387" LastActivityDate="2010-11-28T21:36:56.230" Title="Combinatorial Group Theory / Lyndon &amp; Schupp, Lemma 4.7. Automorphisms of free groups." Tags="&lt;group-theory&gt;" AnswerCount="1" CommentCount="12" />
  <row Id="11543" PostTypeId="2" ParentId="11538" CreationDate="2010-11-23T17:20:06.753" Score="4" ViewCount="" Body="&lt;p&gt;This is slightly tricky because your intuitive picture for &quot;continous&quot; probably more closely matches &quot;uniformly continuous.&quot;  That is if you can &lt;em&gt;actually&lt;/em&gt; draw a graph without lifting your pencil then it's uniformly continuous.  In order to get something continuous but not uniformly continuous you have to do something that you can't actually draw like going off to infinity on an open interval or oscillating really wildly (as explained in the other two answers).&lt;/p&gt;&#xA;" OwnerUserId="8" LastActivityDate="2010-11-23T17:20:06.753" />
  <row Id="11544" PostTypeId="2" ParentId="11539" CreationDate="2010-11-23T17:23:06.327" Score="0" ViewCount="" Body="&lt;p&gt;Hint: show that $h&amp;#39;(x)$ is positive on $(0,1)$, using the mean value theorem applied to $f(x)-f(0)$.&lt;/p&gt;&#xA;" OwnerUserId="2810" LastActivityDate="2010-11-23T17:23:06.327" CommentCount="3" />
  <row Id="11545" PostTypeId="2" ParentId="11528" CreationDate="2010-11-23T17:32:16.657" Score="3" ViewCount="" Body="&lt;p&gt;For the second one, take a look at the Online Encyclopedia of Integer Sequences[1]. Sequence #A030653 is what you are wanting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(You can look at [1] for more extensive information including sources, but I will tell a bit about it here.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $a(n)$ be the n-th number of the sequence for $n \in \{1,2,3...\}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then we have that $a(n) = n^{3} + 3n^{2} + 3n - 3$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Examples:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a(1) = 1 + 3(1) + 3(1) - 3 = 4$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a(2) = 8 + 3(4) + 3(2) -3 = 8 + 12 + 6 - 3 = 23$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the next number in the sequence, we need to calculate $a(5)$. Using the formula above, we get,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a(5) = 125 + 3(25) + 3(5) - 3 = 125 + 75 + 15 -3 = 212$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So 212 is the next number in the sequence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[1] &lt;a href=&quot;http://oeis.org/A030653&quot; rel=&quot;nofollow&quot;&gt;http://oeis.org/A030653&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3791" LastActivityDate="2010-11-23T17:32:16.657" CommentCount="2" />
  <row Id="11546" PostTypeId="2" ParentId="11537" CreationDate="2010-11-23T17:42:54.460" Score="6" ViewCount="" Body="&lt;p&gt;Rasmus:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;There are&quot; examples of discontinuous homomorphisms between Banach algebras. However, the quotes are there because the question is &lt;em&gt;independent&lt;/em&gt; of the usual axioms of set theory. I quote from the introduction to W. Hugh Woodin, &quot;A discontinuous homomorphism from $C(X)$ without CH&quot;, J. London Math. Soc. (2) 48 (1993), no. 2, 299-315:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Suppose that $X$ is an infinite compact Hausdorff space and let $C(X)$ be the algebra of continuous real-valued functions on $X$. Then $C(X)$ is a commutative Banach algebra relative to the supnorm: ${}\|f\|= \sup\{f(p)\mid p\in X\}$. A well-known question of I. Kaplansky posed around 1947 asks whether every algebra homomorphism of $C(X)$ into a Banach algebra $B$ is necessarily continuous. &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;There is a discontinuous homomorphism of $C(X)$ if and only if there is a discontinuous homomorphism of $C(X, {\mathbb C})$, the $C^*$-algebra of continuous complex-valued functions on $X$. We prefer to deal with the real case; some of the references adopt the complex view. The question is now known to be independent of the axioms of set theory, ZFC. H. G. Dales &lt;a href=&quot;http://en.wikipedia.org/wiki/Martin%27s_axiom&quot; rel=&quot;nofollow&quot;&gt;1&lt;/a&gt; and J. Esterle [3] independently constructed discontinuous homomorphisms of $C(X)$ for any infinite space $X$ assuming the &#xA;  Continuum Hypothesis, CH. About the same time R. Solovay [4] proved that it is relatively consistent with ZFC that every homomorphism of $C(X)$ for any space X is necessarily continuous. Solovay's result was improved [5] fairly soon thereafter, to obtain the relative consistency with ZFC + &lt;a href=&quot;http://en.wikipedia.org/wiki/Martin%27s_axiom&quot; rel=&quot;nofollow&quot;&gt;Martin's Axiom&lt;/a&gt; (ZFC + MA) that every homomorphism of $C(X)$ for any space $X$ is continuous. We refer the reader to &lt;a href=&quot;http://en.wikipedia.org/wiki/Proper_forcing_axiom&quot; rel=&quot;nofollow&quot;&gt;2&lt;/a&gt; for an exposition of the latter result concerning MA, historical points and related results. &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;After these results several questions remained. This paper is concerned with the question of whether the existence of a discontinuous homomorphism of $C(X)$ is possible given the failure of the Continuum Hypothesis.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The references listed above are:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;H. G. DALES, &quot;A discontinuous homomorphism from $C(X)$&quot; Amer. J. Math. 101 (1979) 647-734. &lt;/li&gt;&#xA;&lt;li&gt;H. G. DALES and H. WOODIN, An introduction to independence for analysts, London Mathematical Society Lecture Notes 115 (University Press, Cambridge, 1987). &lt;/li&gt;&#xA;&lt;li&gt;J. R. ESTERLE, &quot;Injection de semi-groupes divisibles dans des algebres de convolution et construction d'homomorphismes discontinus de $C(K)$&quot; Proc. London Math. Soc. (3) 36 (1978) 59-85. &lt;/li&gt;&#xA;&lt;li&gt;R. SOLOVAY, personal communication, unpublished. &lt;/li&gt;&#xA;&lt;li&gt;H. WOODIN, Set theory and discontinuous homomorphisms from Banach algebras, Ph.D. Thesis, University of California at Berkeley, 1984. &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I believe Solovay's original construction is simply superseded by Hugh's result, as described in his thesis. In any case, reference 2 is particularly good to learn about this fascinating topic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Shortly after, in Stevo Todorcevic, Partition Problems in Topology (Contemporary Mathematics), American Mathematical Society (January 1989), it is shown that automatic continuity of homomorphisms between Banach algebras is a theorem of ZFC + PFA. PFA is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Proper_forcing_axiom&quot; rel=&quot;nofollow&quot;&gt;proper forcing axiom&lt;/a&gt;, a strengthening of Martin's axiom.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is also a much more recent reference: H. Garth Dales, Banach Algebras and Automatic Continuity (London Mathematical Society Monographs New Series), Oxford University Press, USA (May 17, 2001).&lt;/p&gt;&#xA;" OwnerUserId="462" LastActivityDate="2010-11-23T17:42:54.460" />
  <row Id="11547" PostTypeId="2" ParentId="11539" CreationDate="2010-11-23T17:42:57.830" Score="0" ViewCount="" Body="&lt;p&gt;This seems to be a straightforward application of Cauchy's Mean value theorem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Cauchy%27s_mean_value_theorem#Cauchy.27s_mean_value_theorem&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Cauchy%27s_mean_value_theorem#Cauchy.27s_mean_value_theorem&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2641" LastActivityDate="2010-11-23T17:42:57.830" />
  <row Id="11548" PostTypeId="2" ParentId="11528" CreationDate="2010-11-23T17:48:34.737" Score="1" ViewCount="" Body="&lt;p&gt;To add to Gerry's answer...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the first sequence, you can use a recurrence relation to find the answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can say that,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$y(n+2) = y(n+1) + y(n) + n - 1$, $y(1) = 3$, and $y(2) = 8$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Examples:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$y(3) = y(2) + y(1) + 3 - 1 = 8 + 3 + 3 - 1 = 13$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$y(4) = y(3) + y(2) + 4 - 1 = 13 + 8 + 4 - 1 = 24$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we are trying to find the next number in the sequence which is $y(6)$. So we have,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$y(6) = y(5) + y(4) + 6 - 1 = 41 + 24 + 6 - 1 = 70$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, 70 should be the next number in the sequence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I get a chance to find a formula for the sequence that does not involve recursion, I will post it here.&lt;/p&gt;&#xA;" OwnerUserId="3791" LastActivityDate="2010-11-23T17:48:34.737" />
  <row Id="11549" PostTypeId="2" ParentId="11464" CreationDate="2010-11-23T18:13:11.567" Score="1" ViewCount="" Body="&lt;p&gt;As noted above, observe that&#xA;\begin{eqnarray}&#xA;\sum_{r = 1}^{d} x^{r} = \frac{x(x^{d} - 1)}{x - 1}.&#xA;\end{eqnarray}&#xA;Differentiating both sides and multiplying by $x$, we find&#xA;\begin{eqnarray}&#xA;\sum_{r = 1}^{d} r x^{r} = \frac{dx^{d + 2} - x^{d+1}(d+1) + x}{(x - 1)^{2}}.&#xA;\end{eqnarray}&#xA;Substituting $x = 2$, &#xA;\begin{eqnarray}&#xA;\sum_{r = 1}^{d} r 2^{r} =  d2^{d + 2} - (d+1) 2^{d+1} + 2 = (d - 1) 2^{d+1} + 2.&#xA;\end{eqnarray}&lt;/p&gt;&#xA;" OwnerUserId="2720" LastEditorUserId="2720" LastEditorDisplayName="" LastEditDate="2010-11-23T18:42:09.403" LastActivityDate="2010-11-23T18:42:09.403" CommentCount="1" />
  <row Id="11550" PostTypeId="2" ParentId="10392" CreationDate="2010-11-23T18:21:44.007" Score="1" ViewCount="" Body="&lt;p&gt;Pick out the rows where a &lt;em&gt;t&lt;/em&gt; appears in the rightmost column, and write down a disjunctive normal form.  In your example, there are only two rows with a &lt;em&gt;t&lt;/em&gt; and your expression will have two terms:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ (X \cdot \bar{Y} \cdot Z)  + (\bar{X} \cdot \bar{Y} \cdot Z) $&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now you have a logical formula for your truth table. You can stop there, or you can use the laws of &lt;a href=&quot;http://en.wikipedia.org/wiki/Boolean_algebra_%28logic%29&quot; rel=&quot;nofollow&quot;&gt;boolean algebra&lt;/a&gt; to get a simpler expression. In this case, you can use the distributive law:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ (X \cdot \bar{Y} \cdot Z)  + (\bar{X} \cdot \bar{Y} \cdot Z) = (X + \bar{X})\cdot (\bar{Y} \cdot Z) = 1 \cdot (\bar{Y} \cdot Z) = \bar{Y} \cdot Z$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="130" LastActivityDate="2010-11-23T18:21:44.007" />
  <row Id="11551" PostTypeId="1" CreationDate="2010-11-23T18:24:11.590" Score="16" ViewCount="767" Body="&lt;p&gt;Why in a Hausdorff sequentially compact space the size of the closure of a countable subset is less or equal than $c$ ? I can see why this is true when the space if first countable but we are not assuming so.&lt;/p&gt;&#xA;" OwnerUserId="4219" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2010-11-23T22:57:26.453" LastActivityDate="2011-02-08T19:48:36.590" Title="Size of the closure of a set" Tags="&lt;topology&gt;&lt;general-topology&gt;" AnswerCount="3" CommentCount="10" FavoriteCount="4" />
  <row Id="11552" PostTypeId="1" AcceptedAnswerId="11558" CreationDate="2010-11-23T18:34:15.410" Score="1" ViewCount="247" Body="&lt;p&gt;$f\colon (a,b) \to \mathbb{R}$ is continuous, with finite derivatives everywhere in $(a,b)$, except maybe at $c$.&lt;br&gt;&#xA;If &#xA;$$\lim_{x\to c}f&amp;#39;(x) = B,$$&#xD;&#xA;show that $f&amp;#39;(c)$ exists and equals $B$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure where to start on this. I've tried using the definitions of continuity and f' but it isn't working out  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well I started out with &#xA;$$\lim_{x\to c}\frac{f(x)-f(c)}{x-c} = f&amp;#39;(c)$$&#xD;&#xA;is equivalent to saying&lt;br&gt;&#xA;given $\epsilon\gt 0$ there exists $\delta\gt 0$ such that&lt;br&gt;&#xA;$$|x-c| \lt \delta \Longrightarrow \left|\frac{f(x)-f(c)}{x-c} - f&amp;#39;(c)\right| \lt\epsilon.$$  &#xA;I was trying to prove $f&amp;#39;(c)$ must exist since $f$ is continuous but I feel like I'm assuming what I'm trying to prove. &lt;/p&gt;&#xA;" OwnerUserId="3333" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-23T19:32:50.280" LastActivityDate="2010-11-24T00:42:04.950" Title="Prove a derivative exists" Tags="&lt;real-analysis&gt;&lt;homework&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="11553" PostTypeId="2" ParentId="11552" CreationDate="2010-11-23T18:41:57.230" Score="0" ViewCount="" Body="&lt;p&gt;So what can go wrong?  We can certainly define a function g(x)=something nice except when x=c, something different at c.  Then $\lim_{x\rightarrow c} g(x) \neq g(c)$ Can you rule out this behavior in $f&amp;#39;(x)$?  Or maybe $f&amp;#39;(x)$ has a step at $c$.  Why can't it?&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-11-23T18:41:57.230" CommentCount="1" />
  <row Id="11554" PostTypeId="2" ParentId="11528" CreationDate="2010-11-23T18:46:16.560" Score="1" ViewCount="" Body="&lt;p&gt;I am adding lots of answers here, sorry. I think since they are giving different information, they are best put separately.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have not had experience with this, you may be interested in how people derive formulas for these problems. Difference calculus is a nice method to use. Kelley and Peterson's textbook [1] is a very good resource for this (I tend to recommend this on a lot of different questions on here.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, you can take a look at the notes I took in my difference equations course for this type of problem [2].&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have any additional questions, I'd be happy to try to help you. Feel free to email me, my contact information is on my website ( &lt;a href=&quot;http://www.tylerclark12.com/blog/?page_id=11&quot; rel=&quot;nofollow&quot;&gt;http://www.tylerclark12.com/blog/?page_id=11&lt;/a&gt; ). Good luck with your exam! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;[1] Kelley, W. &amp;amp; Peterson, A. (2001). Difference Equations: An Introduction with Applications (2nd Ed.). San Diego, CA: Academic Press.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[2] &lt;a href=&quot;http://www.tylerclark12.com/blog/wp-content/uploads/2010/10/Math_542_9-24.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.tylerclark12.com/blog/wp-content/uploads/2010/10/Math_542_9-24.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3791" LastActivityDate="2010-11-23T18:46:16.560" />
  <row Id="11555" PostTypeId="2" ParentId="11382" CreationDate="2010-11-23T18:48:31.940" Score="6" ViewCount="" Body="&lt;p&gt;While some of the other answers look good, there are two particular related uses for polynomial multiplication which I didn't see while skimming.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One is related to probabilities. We can use a polynomial to represent the possible outcomes of a random process (so long as there are finitely many), where there is a variable for each possible outcome, and the coefficient of that variable is the probability of it occurring. (We'll just go with linear polynomials for now.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So for example, a toss of a fair coin might be represented as 1/2 H + 1/2 T. An unfair coin might be 2/3 H + 1/3 T. Someone's plays in the game of rock/paper/scissors might be summarised as 3/5 rock + 1/5 paper + 1/5 scissors. (Dude likes rock.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, what would it mean to multiply two such polynomials? Well, supposing that the random events are independent of each other (that is, the outcome of one doesn't affect the probabilities involved in the other), the product of two such polynomials tells us the likelihood of various pairs of outcomes. If we flip that unbalanced coin, and have that guy throw rock/paper/scissors, the combined possibilities are&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2/3 H + 1/3 T) (3/5 rock + 1/5 paper + 1/5 scissors) = 2/5 H rock + 1/5 T rock + 2/15 H paper + 1/15 T paper + 2/15 H scissors + 1/15 T scissors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we can read off that, for instance, the combined probability of getting a head on the coin and having the guy throw scissors is the coefficient of H scissors, which is 2/15.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So that's a decent way to organise information about probabilities in a way that lets us combine multiple independent random processes. Raising a polynomial to the power of n in this setting gives us the probabilities when the process is repeated n times, assuming again that the outcome of the previous trials doesn't affect the next one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, let's flip the unbalanced coin three times in a row:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2/3 H + 1/3 T)^3 = 8/27 H^3 + 4/9 H^2 T + 2/9 H T^2 + 1/27 T^3&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This tells us that the probability that we get two heads and one tail is 4/9, and so on. (If we didn't allow H and T to commute with each other, we would get the probability of each possible sequence.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's also worth noting that it's quite effective to use polynomials in keeping track of proportions when forming mixtures of things, which are then mixed together (say, in chemistry), using substitution of one variable representing a mixture with the polynomial representing its contents. Multiplication doesn't immediately have such a useful interpretation there though.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another thing we can do is to use the coefficients of polynomials to count the number of things of a given type. If we have a (finite) bunch of things with integer &quot;sizes&quot; or &quot;weights&quot;, and we want to record the number of things with a given weight, we can use the coefficient of x^n in a polynomial to record the number of objects with weight n. The product of two such polynomials will then count the number of ways to form a pair of objects with total weight n, for each n.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, suppose we want to know the number of ways to put change for n cents into a vending machine using 5 coins, for each possible n. (And we'll assume that the coins have values {1, 5, 10, 25, 100, 200}). We can use the polynomial c(x) = (x^1 + x^5 + x^10 + x^25 + x^100 + x^200) to represent our set of coins. Sequences of 5 coins are then represented by (c(x))^5. This is a polynomial with a fairly large number of terms (representing the fact that there are lots of ways to choose a sequence of 5 coins in order, specifically, 6^5 = 7776, and moreso the fact that there are many different possible sums that those coins could have).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This polynomial is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;x^5 + 5 x^9 + 10 x^13 + 5 x^14 + 10 x^17 + 20 x^18 + 5 x^21 + &#xA; 30 x^22 + 10 x^23 + x^25 + 20 x^26 + 30 x^27 + 5 x^29 + 5 x^30 + &#xA; 30 x^31 + 10 x^32 + 20 x^33 + 10 x^35 + 20 x^36 + 30 x^37 + &#xA; 20 x^38 + 10 x^40 + 25 x^41 + 60 x^42 + 10 x^45 + 60 x^46 + &#xA; 30 x^47 + 21 x^50 + 60 x^51 + 10 x^53 + 30 x^55 + 20 x^56 + &#xA; 30 x^57 + 20 x^60 + 30 x^61 + 30 x^62 + 15 x^65 + 60 x^66 + &#xA; 30 x^70 + 30 x^71 + 30 x^75 + 10 x^77 + 10 x^80 + 20 x^81 + &#xA; 10 x^85 + 20 x^86 + 20 x^90 + 10 x^95 + 5 x^101 + 5 x^104 + &#xA; 5 x^105 + 20 x^108 + 5 x^110 + 30 x^112 + 20 x^113 + 20 x^116 + &#xA; 60 x^117 + 5 x^120 + 60 x^121 + 30 x^122 + 21 x^125 + 60 x^126 + &#xA; 20 x^128 + 30 x^130 + 20 x^131 + 60 x^132 + 20 x^135 + 60 x^136 + &#xA; 60 x^137 + 25 x^140 + 120 x^141 + 60 x^145 + 60 x^146 + 60 x^150 + &#xA; 30 x^152 + 20 x^155 + 60 x^156 + 30 x^160 + 60 x^161 + 60 x^165 + &#xA; 30 x^170 + 20 x^176 + 20 x^180 + 20 x^185 + 5 x^200 + 10 x^203 + &#xA; 5 x^204 + 30 x^207 + 20 x^208 + 30 x^211 + 60 x^212 + 20 x^213 + &#xA; 10 x^215 + 80 x^216 + 60 x^217 + 35 x^220 + 90 x^221 + 30 x^222 + &#xA; 50 x^225 + 60 x^226 + 30 x^227 + 20 x^228 + 40 x^230 + 80 x^231 + &#xA; 60 x^232 + 50 x^235 + 120 x^236 + 60 x^237 + 85 x^240 + 120 x^241 + &#xA; 90 x^245 + 60 x^246 + 60 x^250 + 30 x^251 + 30 x^252 + 50 x^255 + &#xA; 60 x^256 + 60 x^260 + 60 x^261 + 60 x^265 + 30 x^270 + 10 x^275 + &#xA; 20 x^276 + 20 x^280 + 20 x^285 + 5 x^300 + 10 x^302 + 20 x^303 + &#xA; 20 x^306 + 60 x^307 + 10 x^310 + 80 x^311 + 60 x^312 + 40 x^315 + &#xA; 120 x^316 + 70 x^320 + 60 x^321 + 60 x^325 + 20 x^326 + 60 x^327 + &#xA; 40 x^330 + 120 x^331 + 80 x^335 + 120 x^336 + 120 x^340 + 60 x^345 + &#xA; 10 x^350 + 60 x^351 + 60 x^355 + 60 x^360 + 20 x^375 + 5 x^401 + &#xA; 30 x^402 + 10 x^403 + 5 x^405 + 60 x^406 + 30 x^407 + 35 x^410 + &#xA; 90 x^411 + 30 x^412 + 70 x^415 + 60 x^416 + 60 x^420 + 30 x^421 + &#xA; 35 x^425 + 60 x^426 + 30 x^427 + 70 x^430 + 60 x^431 + 90 x^435 + &#xA; 60 x^436 + 60 x^440 + 30 x^445 + 30 x^450 + 30 x^451 + 30 x^455 + &#xA; 30 x^460 + 10 x^475 + x^500 + 20 x^501 + 30 x^502 + 20 x^505 + &#xA; 60 x^506 + 50 x^510 + 60 x^511 + 60 x^515 + 30 x^520 + 20 x^525 + &#xA; 60 x^526 + 60 x^530 + 60 x^535 + 30 x^550 + 5 x^600 + 30 x^601 + &#xA; 10 x^602 + 30 x^605 + 20 x^606 + 40 x^610 + 20 x^611 + 20 x^615 + &#xA; 10 x^620 + 30 x^625 + 20 x^626 + 20 x^630 + 20 x^635 + 10 x^650 + &#xA; 10 x^700 + 20 x^701 + 20 x^705 + 20 x^710 + 20 x^725 + 10 x^800 + &#xA; 5 x^801 + 5 x^805 + 5 x^810 + 5 x^825 + 5 x^900 + x^1000&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I just used my computer to calculate it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This tells us, for instance, that there are 120 distinct ways to put coins in a vending machine to make $1.41, since the coefficient of x^141 is 120. There is of course, only one way to make change for 10 dollars (which is to put in all 2 dollar coins), so the coefficient of x^1000 is 1. Note also that if we were to evaluate this polynomial at x = 1, we would get the total number of possible sequences, which is 7776. It's also maybe interesting to see which terms are missing from this polynomial (have a coefficient of 0), telling us that there's no way to make change for that amount using 5 coins.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, both of these approaches generalise to power series (which are like polynomials, but which have infinite sequences of terms), and the effectiveness of the approach really becomes more apparent in that more general setting. This technique is called generating series (or the somewhat unfortunate, but perhaps more common name &quot;generating functions&quot;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, I think that gives some impression of some of the variety of things that one can do with polynomials. (I'll just leave that pun there for the algebraic geometers. ;)&lt;/p&gt;&#xA;" OwnerUserId="3869" LastEditorUserId="3869" LastEditorDisplayName="" LastEditDate="2010-11-23T19:05:30.990" LastActivityDate="2010-11-23T19:05:30.990" />
  <row Id="11556" PostTypeId="2" ParentId="11551" CreationDate="2010-11-23T18:49:41.970" Score="0" ViewCount="" Body="&lt;p&gt;It seems like you don't need all the conditions, so maybe I am missing something.  There are only continuum many sequences of points from a countable set.  So the limit points are at most continuum many.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-11-23T18:49:41.970" CommentCount="5" />
  <row Id="11557" PostTypeId="2" ParentId="11382" CreationDate="2010-11-23T18:52:26.927" Score="7" ViewCount="" Body="&lt;p&gt;This is really several questions.  In increasing order of depth:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Why multiply polynomials?  Because they are functions, and multiplication is a natural or desirable operation on functions.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Why multiply functions?   Because the values of the functions are integers, or rational numbers, or real numbers, or objects of some other more complicated kind that can be multiplied (such as matrices or rotations).  If multiplication is natural for these numbers or number-like objects, it is natural for functions whose values are such objects.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Why multiply integers, rational numbers, real numbers, etc?  Here is where explanation becomes difficult.  The natural operation is &lt;em&gt;not&lt;/em&gt; multiplication but the more structured operation of &quot;tensor multiplication&quot; that retains information about the factors that are multiplied.  One can represent an Area as a product of type Length x Length that remembers its two-dimensionality, instead of a one-dimensional numerical value of that product that has forgotten its origins.  In the same way, for integer multiplication it is most direct to multiply 5 and 6 by drawing 5x6 as a rectangular 2-dimensional array of dots instead of the &quot;numerical evaluation&quot; of that array as a one-dimensional string of 30 dots. The latter is less natural in that it requires a method of enumerating the dots in the grid and there is no preferred order in which to count them.  This is also reflected in the ability to naturally multiply 5 gadgets by 6 gizmos, without there being any given ordering of the gadgets or gizmos.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Polynomials, especially polynomials in several variables ($x +3x^2y + 2 z^{10} x$ and such), have more inner structure than numbers and thus can reflect -- in fact, they can be defined and derived from -- the tensor structure of multiplication.   So it is not that polynomials exist and we &lt;em&gt;might&lt;/em&gt; want to multiply them; it is that multiplication of numbers (or of finite sets) naturally involves more than just numerical information and polynomials are an enhancement of numbers that more directly embody that information &lt;em&gt;so that&lt;/em&gt; the multiplication can be performed in a way that retains more of the inner structure.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A hint of this is that integers in base 10 are values of polynomials at $x = 10$ (and those polynomials can be thought of as a &quot;liberation&quot; or &quot;upgrading&quot; of the numbers).  Multiplication of integers can sometimes be seen to replicate the patterns in the coefficients when the polynomials are multiplied for general $x$, e.g., compare powers of $102$ and $x^2 + 0x + 2$.  Later there are things like generating functions and convolutions that directly exploit polynomials as carriers of information that is enriched compared to using numbers alone.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(This is glossing over some technical questions about commutativity. Also, areas should be sums of LxL products, and one should explain the role of sums as well as products.  But these are details that do not affect the main point.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In applications, multiplication represents interaction or correlation between different effects.  Situations where there are several independent processes isolated from each other and contributing to some outcome lead to sums  of functions of the different variables, such as $f(x) + g(y) + h(z)$, the terms in the sum accounting for the separate effects.  Situations where different parts of the mechanism can interact with each other in producing the outcome, or where there is correlation between different effects (e.g., one intensifies or suppresses the other), almost always involve multiplication when expressed mathematically.   For example, if you count the number of handshakes in a group of $n$ people the answer will be of order $n^2$, and if you count the number of possible 3-person interactions in this group it will be of order $n^3$.  Nonlinearities reflect the ability to organize the people into pairs, triples etc, and this again reflects the fact that the natural multiplication of two finite sets $A$ and $B$ is the set of ordered pairs $(a,b)$ of elements, one from each set, and the same for triples and higher numbers of sets. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A somewhat faddish term for related ideas is (de- or re-)&lt;em&gt;categorification&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="467" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-11-23T23:55:32.883" LastActivityDate="2010-11-23T23:55:32.883" />
  <row Id="11558" PostTypeId="2" ParentId="11552" CreationDate="2010-11-23T19:38:26.030" Score="1" ViewCount="" Body="&lt;p&gt;You are right that you are kind of assuming what you want, because you cannot even write down $f&amp;#39;(c)$ until you prove it exists; this can be fixed if you replace $f&amp;#39;(c)$ by a specific limit, and the specific limit you want is probably $B$. But even so ran into a bit of an alley, so let me help you out.&#xA;Here is the beginning of half a proof, following the Hint of Robin Chapman:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider first the limit as $x\to c^+$. For any $h\gt 0$, the function $f(x)$ is continuous on $[c,c+h]$, and is differentiable on $(c,c+h)$, so by the Mean Value Theorem there exists a point $d_h$ (which depends on $h$) such that &#xA;$$\frac{f(c+h)-f(c)}{h} = f&amp;#39;(d_h).$$&#xD;&#xA;Therefore, we have that&#xA;\begin{align*}&#xA;\lim_{x\to c^+}\frac{f(x)-f(c)}{x-c} &amp;amp;= \lim_{h\to 0^+}\frac{f(c+h)-f(c)}{h}\&#xA;&amp;amp;=\lim_{h\to 0^+}f'(d_h).&#xA;\end{align*}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we've managed to turn the (one-sided) limit that will tell us whether $f$ is differentiable at $c$ into a limit about $f&amp;#39;(x)$. Luckily, we have information about limits of $f&amp;#39;(x)$. So perhaps we can leverage that information into an actual value for this limit? If so, perhaps we can try doing the same thing for the limit from the left (as $x\to c^-$), so that the two together shows us that &#xA;$$\lim_{x\to c}\frac{f(x)-f(c)}{x-c}$$&#xD;&#xA;exists; and will tell us what the value of $f&amp;#39;(c)$ must be in that case.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-24T00:42:04.950" LastActivityDate="2010-11-24T00:42:04.950" />
  <row Id="11559" PostTypeId="2" ParentId="11382" CreationDate="2010-11-23T19:38:55.593" Score="5" ViewCount="" Body="&lt;p&gt;Why not?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;" OwnerUserId="3757" LastActivityDate="2010-11-23T19:38:55.593" CommentCount="2" />
  <row Id="11560" PostTypeId="1" AcceptedAnswerId="11564" CreationDate="2010-11-23T19:45:37.543" Score="1" ViewCount="939" Body="&lt;p&gt;I have 2 matrices and have been trying to multiply them but to no avail. Then I found this &lt;a href=&quot;http://easycalculation.com/matrix/matrix-multiplication.php&quot; rel=&quot;nofollow&quot;&gt;online&lt;/a&gt; site and trying feeding it the values but yet no success.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;- R' . T&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;is what i would like to do but R is a 3x3 (3 rows, 3 cols) matrix and T is a 1 x 3 matrix (1 row, 3 cols).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any idea how to go ahead with this?&lt;/p&gt;&#xA;" OwnerUserId="3866" LastActivityDate="2010-11-23T20:06:23.857" Title="How to multiply a 3x3 matrix with a 1x3 matrix?" Tags="&lt;matrices&gt;" AnswerCount="4" FavoriteCount="1" />
  <row Id="11561" PostTypeId="2" ParentId="11199" CreationDate="2010-11-23T19:47:54.120" Score="1" ViewCount="" Body="&lt;p&gt;Mathematica apparently does not have a nice way of looking at the sum as it is given,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$S_n = \sum_{k=0}^n \frac{n-k+1}{(n+1) (n+2) (n-k+2)}.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, when you calculate this sum as a definite integral, for both the equation above and the one you derived, ie&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{k=0}^{n=p} \frac{n^{\underline{k}}}{(n+2)^{\underline{k}}(n-k+2)^2}.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where p is the number you enter to make it a definite sum, you get the same calculation as when using the original equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I initially used the FactorialPower[n,k] command and then I thought that maybe the domain for k was causing issues, so I rewrote the falling factorials using the gamma function to see if that would help. If you are not familiar with how to do this,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ n^{\underline{k}} = \frac{ \Gamma (n+1)}{ \Gamma (n-k+1)}$$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Either way I write the falling factorial, I still come up with the equation you are getting when I use an indefinite sum, which is,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$S_n = \frac{7}{6(n+1)(n+2)}$$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you know, this is in fact not correct. I am not sure why it works for the definite sum, but not the indefinite sum. I will keep thinking about this and let you know what I come up with, if I come up with anything.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you find an answer, please let us know here. Sorry I cannot be of more help.&lt;/p&gt;&#xA;" OwnerUserId="3791" LastActivityDate="2010-11-23T19:47:54.120" CommentCount="1" />
  <row Id="11563" PostTypeId="2" ParentId="11560" CreationDate="2010-11-23T19:49:51.883" Score="2" ViewCount="" Body="&lt;p&gt;The usual matrix multiplication is only defined for multiplying an $m\times n$ matrix with an $n\times R$ matrix. So the number of columns of the first matrix must be equal to the number of rows of the second for matrix multiplication to be defined. This is not satisfied by the matrices you have. So you cannot multiply them in that order. You can, however, multiply $T$ with $R$ i.e. the product $TR$ is defined but not $RT$.&lt;/p&gt;&#xA;" OwnerUserId="3431" LastActivityDate="2010-11-23T19:49:51.883" CommentCount="2" />
  <row Id="11564" PostTypeId="2" ParentId="11560" CreationDate="2010-11-23T19:50:56.560" Score="4" ViewCount="" Body="&lt;p&gt;You can only multiply an $n\times m$ matrix by a $p\times q$ matrix, in that order, if $m=p$. That is, only if the number of columns of the first factor is equal to the number of rows of the second factor. So you &lt;em&gt;cannot&lt;/em&gt; multiply a $3\times 3$ matrix by a $1\times 3$ matrix; on the other hand, you &lt;em&gt;can&lt;/em&gt; multiply a $1\times 3$ matrix by a $3\times 3$ matrix (so you can do $TR$); or you can multiply the $3\times 3$ matrix by the &lt;em&gt;transpose&lt;/em&gt; of $T$, which would be $3\times 1$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You &lt;em&gt;can&lt;/em&gt; try extending $T$ into a $3\times 3$ by adding two rows of zeros (that is what the site you linked to does), but then you are not really multiplying $R$ by $T$. All you are doing is multiplying the first column of $R$ by the first entry of $T$, the second column of $R$ by the second entry of $T$, and the third column of $R$ by the third entry of $T$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the real question is: &lt;em&gt;where&lt;/em&gt; did the matrices come from and &lt;em&gt;why&lt;/em&gt; are you trying to multiply them in that order? Perhaps it turns out you shouldn't be doing that product at all...&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-11-23T19:50:56.560" CommentCount="9" />
  <row Id="11565" PostTypeId="1" AcceptedAnswerId="11571" CreationDate="2010-11-23T19:58:31.467" Score="1" ViewCount="72" Body="&lt;p&gt;This is initially a funny question, because I've found this on old notes but I do not find/recover my own derivation... But then the question is more general.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Q1:     &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I considered the function&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ f(x) = - \frac {2x^2+3x}{(x+1)^2}  $ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I expressed this by a powerseries $ f_1(x) = -3x + 4x^2 - 5x^3 + 6x^4 - \ldots $ &#xA;and stated without the derivation that this is also&#xA;$  f_2(x) = \frac {-2}{1} -\frac {-1}{x} + 0 - \frac  {1} {x^3} + \frac {2}{x^4} - \ldots + \ldots $ &lt;br&gt;&#xA;and - well: hell, - don't see it now how I did it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What was interesting to me was, that after looking for the fixpoints $ x_0=0, x_{1,2} =-2 $ the range of convergence in the expression by $f_1$ is obviously $ |x|&amp;lt;1 $ limited to the unit-interval but in that by $f_2$ it is infinity and $ |x|&amp;gt;1 $ .      &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Q2:&lt;br&gt;&#xA;I would like to be able to translate also other powerseries into an $f_2$-type-expression. (I remember to have read a remark of &lt;em&gt;&quot;expanding a powerseries at infinity&quot;&lt;/em&gt; but have never seen an explanation of this - so this might be irrelevant for this case?) So: what is the technique to do this given a function in terms of a usual powerseries, for instance for the geometric series $ g(x)=1+x+x^2+ \ldots $ or some series $ h(x) = K + a*x + b*x^2 + c*x^3 + \ldots $ ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[edit: minus-sign in f(x) was missing, one numerator in f2 was wrong]&lt;/p&gt;&#xA;" OwnerUserId="1714" LastEditorUserId="1714" LastEditorDisplayName="" LastEditDate="2010-11-23T21:12:30.750" LastActivityDate="2010-11-23T21:12:30.750" Title="conversion of a powerseries $-3x+4x^2-5x^3+\ldots $ into $ -2+\frac 1 x - 0 - \frac 1 {x^3} + \ldots $" Tags="&lt;functions&gt;&lt;functional-equations&gt;&lt;taylor-expansion&gt;" AnswerCount="1" />
  <row Id="11566" PostTypeId="2" ParentId="11560" CreationDate="2010-11-23T19:59:26.810" Score="2" ViewCount="" Body="&lt;p&gt;Suppose you have two matrices A and B. Let A be a 3x3 matrix and B be a 1x3 matrix.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As stated above, you can only multiply matrix C (mxn) by matrix D (pxq) if n=p. That is, the number of columns in C = the number of rows in D.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We see that for your problem, we have the following information,&#xA;$m=n=q=3$ and $p=1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus we have that $ n \neq p$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you insist on multiplying these two matrices together, you could always take the transpose of matrix B, denoted $B^{T}$. (For information about transposed matrices, see [1].)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$B^{T}$ would be a 3x1 matrix. So you could in fact do $AB^{T}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For more on matrix multiplication in general, see [2]. I hope this is helpful!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[1] Weisstein, Eric W. &quot;Transpose.&quot; From MathWorld--A Wolfram Web Resource. &lt;a href=&quot;http://mathworld.wolfram.com/Transpose.html&quot; rel=&quot;nofollow&quot;&gt;http://mathworld.wolfram.com/Transpose.html&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;[2] Weisstein, Eric W. &quot;Matrix Multiplication.&quot; From MathWorld--A Wolfram Web Resource. &lt;a href=&quot;http://mathworld.wolfram.com/MatrixMultiplication.html&quot; rel=&quot;nofollow&quot;&gt;http://mathworld.wolfram.com/MatrixMultiplication.html&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="3791" LastActivityDate="2010-11-23T19:59:26.810" />
  <row Id="11567" PostTypeId="1" CreationDate="2010-11-23T20:03:55.613" Score="5" ViewCount="311" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Possible Duplicate:&lt;/strong&gt;&lt;br&gt;&#xA;  &lt;a href=&quot;http://math.stackexchange.com/questions/7473/number-theory-proving-question&quot;&gt;Number theory proving question?&lt;/a&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;p&gt;Dear friends,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since b, x, y, z, ... integers greater than 1, how can we prove that &#xA;gcd (b ^ x - 1, b ^ y - 1, b ^ z - 1 ,...)= [b ^ gcd (x, y, z, .. .)] - 1?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Paulo Argolo&lt;/p&gt;&#xA;" OwnerUserId="2938" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2010-11-24T04:42:38.897" LastActivityDate="2011-03-16T19:42:16.707" ClosedDate="2010-11-25T20:07:55.940" Title="$\gcd(b^x - 1, b^y - 1, b^ z- 1,...) = b^{\gcd(x, y, z,...)} -1$" Tags="&lt;number-theory&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="2" />
  <row Id="11568" PostTypeId="2" ParentId="11560" CreationDate="2010-11-23T20:06:23.857" Score="0" ViewCount="" Body="&lt;p&gt;From what I can see, perhaps you have the R and Xw matrix switched around. If you had Xw*R, then I think it would be okay. Did you derive this formula or did you find it somewhere?&lt;/p&gt;&#xA;" OwnerUserId="3791" LastActivityDate="2010-11-23T20:06:23.857" CommentCount="1" />
  <row Id="11570" PostTypeId="2" ParentId="11567" CreationDate="2010-11-23T20:07:58.660" Score="10" ViewCount="" Body="&lt;p&gt;It suffices to prove it for two terms, that is, $\gcd(a^n - 1, a^m - 1) = a^{\gcd(n,m)} - 1$.  The basic idea is that we can use the Euclidean algorithm on the exponents, as follows: if $n &amp;gt; m$, then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\gcd(a^n - 1, a^m - 1) = \gcd(a^n - 1, a^n - a^{n-m}) = \gcd(a^{n-m} - 1, a^m - 1).$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we can keep subtracting one exponent from the other until we get $\gcd(n, m)$ as desired.  Another way to look at this computation is to write $d = \gcd(a^n - 1, a^m - 1)$ and note that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$a^n \equiv 1 \bmod d, a^m \equiv 1 \bmod d \Rightarrow a^{nx+my} \equiv 1 \bmod d$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;from which it readily follows, as before, that $a^{\gcd(n,m)} \equiv 1 \bmod d$, so $d$ dividess $a^{\gcd(n,m)} - 1$.  On the other hand, $a^{\gcd(n, m)} - 1$ also divides $d$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What's really nice about this result is that it holds both for particular values of $a$ and also for $a$ as a &lt;strong&gt;variable&lt;/strong&gt;, e.g. in a polynomial ring with indeterminate $a$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can readily deduce several seemingly nontrivial results from this; for example, the sequence defined by $a_0 = 2, a_n = 2^{a_{n-1}} - 1$ is a sequence of pairwise relatively prime integers, from which it follows that there are infinitely many primes.  By working only slightly harder you can deduce that in fact there are infinitely many primes congruent to $1 \bmod p$ for any prime $p$.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-11-23T20:07:58.660" />
  <row Id="11571" PostTypeId="2" ParentId="11565" CreationDate="2010-11-23T20:11:41.853" Score="3" ViewCount="" Body="&lt;p&gt;Divide the numerator and denominator of $f(x)$ by $x^2$ and set $y=1/x$ then expand for $y$ and you have your expansion at infinity.&lt;/p&gt;&#xA;" OwnerUserId="1301" LastActivityDate="2010-11-23T20:11:41.853" CommentCount="2" />
  <row Id="11572" PostTypeId="2" ParentId="11542" CreationDate="2010-11-23T20:14:59.527" Score="5" ViewCount="" Body="&lt;p&gt;For the first assertion, I will work with $F_n$ being the $n$th term of the lower central series; but the argument should be about the same for any central series.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We show that if $\alpha$ is the identity on $F/[F,F]$, then it induces the identity on $F_n/F_{n+1}$ for all $n$. The proof is by induction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The assertion for $n=1$ is the original assumption. Assume now that the result holds for $k$, so that if $w\in F_k$, then $w\alpha\equiv w\pmod{F_{k+1}}$. We want to show that the result holds for $k+1$. The generators of $F_{k+1}$ are contained in the set $[F,F_{k+1}]$, so let $u\in F$ and $w\in F_k$; if we prove the result for $[u,w]$, then it will follow for $F_{k+1}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We will use the following commutator identity: in any group, for all $x,y,z,t$, we have:&#xA;$$[xy,zt] = [x,t]^y[y,t][x,z]^{yt}[y,z]^t,$$&#xD;&#xA;where $a^b = b^{-1}ab$ is conjugation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, since $u\alpha\equiv u\pmod{F_2}$, then $u\alpha=uc$ with $c\in F_2$. And since $w\alpha\equiv w\pmod{F_{k+1}}$, then $w\alpha = wd$ with $d\in F_{k+1}$. Therefore,&#xA;$$[u,w]\alpha = [u\alpha,w\alpha]= [uc,wd]&#xD;&#xA;= [u,d]^{c}[c,d][u,w]^{cd}[c,w]^{d} \equiv [u,w]\pmod{F_{k+2}},$$&#xD;&#xA;because $[u,d]\in[F_1,F_{k+1}]$, $[c,d]\in[F_2,F_{k+1}]$, $[c,w]\in[F_2,F_k]$, and $[u,w]^{cw}=[u,w][[u,w],cw]\equiv [u,w]\pmod_{F_{k+2}}$. &#xA;Therefore, $\alpha$ acts as the identity on $F_{k+1}/F_{k+2}$, as desired.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Alternatively, the basic commutators of weight $k$ give a basis for the free abelian group $F_{k}/F_{k+1}$, and it is an easy matter to check that if $\alpha$ induces the identity on $F/F_{2}$, then it maps a basic commutator of weight $c$ to the product of itself times commutators of weight greater than $c$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm still working on the second assertion, but for instance, for $n=1$, we have that $x\alpha= xk$ for some $k\in F_2$, hence $x\alpha\equiv xk\pmod{F_3}$, and since $k\alpha\equiv k\pmod{F_3}$, you get that $x\alpha^2 \equiv x\alpha k \equiv x k^2\pmod{F_3}$, and so inductively $x\alpha^N \equiv x k^{N}\pmod{F_3}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added:&lt;/strong&gt; I double-checked the book on my way out of the office and to the gym, and I noticed that the book did not have &quot; $x\alpha^N\equiv xk^N\pmod{F_{n+1}}$, but only $x\alpha^N\equiv xk^N$. I can show that under the given assumptions, $x\alpha^N$ is &lt;em&gt;conjugate&lt;/em&gt;, modulo $F_{n+1}$, to $xk^N$, and that should be enough for the purposes of the argument, since the next step of the proof is to note that since $\alpha^N\in JA(F)$, then this means that $uxu^{-1}\equiv xk^N\pmod{F_{n+1}}$ for some $u$, and this will also hold if all you have is that $x\alpha^N$ is conjugate to $xk^N$ modulo $F_{n+1}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To show this weaker claim, I proceed by induction on $m$. We know that $x\alpha$ is conjugate to $xk$, hence $x\alpha \equiv (xk)^u\pmod{F_{n+1}}$ for some $u$. Assume that $x\alpha^m\equiv (xk^m)^v\pmod{F_{n+1}}$ for some $v\in F$, where $a^b = b^{-1}ab$ is just conjugation. Applying $\alpha$ on both sides, we have $x\alpha^{m+1}$ on the left hand side, and &#xA;\begin{align*}&#xA;(xk^m)^v\alpha &amp;amp;= ((x\alpha)(k^m\alpha))^{v\alpha}\\&#xA;&amp;amp; \equiv \bigl((xk)^u k^m\bigr)^{v\alpha}\pmod{F_{n+1}}\\&#xA;&amp;amp;\equiv \bigl( (xk^{m+1})^u\bigr)^{v\alpha}\pmod{F_{n+1}} &amp;amp;\text{(as $k$ is central in $F_/F_{n+1}$)}\\&#xA;&amp;amp;\equiv (xk^{m+1})^{u(v\alpha)}\pmod{F_{n+1}}.&#xA;\end{align*}&#xA;Therefore, $x\alpha^{m+1}\equiv (xk^{m+1})^w\pmod{F_{n+1}}$ for some $w\in F$; that is, $x\alpha^{m+1}$ is conjugate, modulo $F_{n+1}$, to $xk^{m+1}$. In particular, $x\alpha^N$ is conjugate, modulo $F_{n+1}$ to $xk^{N}$, and since $x$ is conjugate to $x\alpha^N$, then $x$ is conjugate to $xk^N$ modulo $F_{n+1}$. Now I think the argument given in the book can be continued from this point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;hr/&gt;&#xA;&lt;hr/&gt;&#xA;So, continuing on with the proof of the proposition. We know that $x\alpha^N$ is conjugate to $xk^N$ modulo $F_{n+1}$, and that $\alpha^N$ is equal to conjugation by some element. So there exists $u$ such that $x\alpha^N = u^{-1}xu = x[x,u]$ is equal, modulo $F_{n+1}$, to $xk^N$, and therefore $[x,u]=k^N \pmod{F_{n+1}}$ by cancelling $x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, $F_n/F_{n+1}$ is a finitely generated free abelian group generated by the basic commutators; $k\in F_n$, so we can write $k$ as a product of basic commutators, and the same is true for $[x,u]$. Because $[x,u]$ is an $N$th power, commutator calculus tells us that in fact $u$ must be an $N$th power, so that in fact we can find a $v$ such that $[x,v]\equiv k \pmod{F_{n+1}}$ (essentially, every commutator that shows up is an $N$th power, so dividing the exponent by $N$ we can construct an appropriate $v$). So now we have $[x,v]\equiv k\pmod{F_{n+1}}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now: $x\alpha$ is conjugate to $xk$, so $v^{-1}x\alpha v$ is conjugate to $vxkv^{-1}$ modulo $F_{n+1}$. But $vxkv^{-1} \equiv vxv^{-1}k\pmod{F_{n+1}}$ (since $k$ is central in $F/F_{n+1}$), so &#xA;$$vxkv^{-1} \equiv vxv^{-1}k \equiv x[x,v^{-1}]k \equiv x[x,v]^{-1}k \equiv xk^{-1}k\pmod{F_{n+1}}$$&#xD;&#xA;(the next to last congruence by using commutator identities, since we are in the lowest term of the lower central series of $F/F_{n+1}$). In summary, $x\alpha$ is conjugate to $x$ modulo $F_{n+1}$. That means that there exists some $k&amp;#39;\in F_{n+1}$ and some $w\in F$ such that $w^{-1}(x\alpha)w\equiv x \pmod{F_{n+1}}$. But this means that $w^{-1}(x\alpha)w =xk&amp;#39;$ for some $k&amp;#39;\in F_{n+1}$, which is precisely what we wanted to prove to complete the inductive step.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-28T21:36:56.230" LastActivityDate="2010-11-28T21:36:56.230" CommentCount="1" />
  <row Id="11573" PostTypeId="1" AcceptedAnswerId="11575" CreationDate="2010-11-23T20:19:33.050" Score="0" ViewCount="148" Body="&lt;p&gt;Let $f\colon A\to B$, and $g\colon B\to C$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Prove: If $g\circ f$ is onto and $g$ is 1-1, then $f$ is onto.&lt;/p&gt;&#xA;&#xA;&lt;hr/&gt;&#xA;&#xA;&lt;p&gt;Here is all I can think of:&#xA;$g\circ f\colon A\to C$, and for all $x\in B$ there is a unique $y\in C$.   &lt;/p&gt;&#xA;" OwnerUserId="2587" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-23T20:25:58.047" LastActivityDate="2010-11-23T20:35:49.577" Title="Prove: If $g\circ f$ is onto and $g$ is 1-1, then $f$ is onto" Tags="&lt;functions&gt;" AnswerCount="2" />
  <row Id="11574" PostTypeId="1" CreationDate="2010-11-23T20:23:15.320" Score="0" ViewCount="107" Body="&lt;p&gt;This is just a simple doubt I wanted to clear. I'm calculating the Hamming and Euclidean distance between two columns in a matrix, where each column is storing time information in seconds of when a particular event occurred. When I calculate the Hamming and Euclidean distance between the two columns, I get back a certain value. What is the units of measure for that value? Is it in seconds? &lt;/p&gt;&#xA;" OwnerUserId="2747" LastActivityDate="2010-11-23T20:29:26.867" Title="Units of Hamming, and Euclidean distance" Tags="&lt;statistics&gt;&lt;discrete-mathematics&gt;&lt;recreational-mathematics&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="11575" PostTypeId="2" ParentId="11573" CreationDate="2010-11-23T20:28:17.560" Score="3" ViewCount="" Body="&lt;p&gt;Given an element of $B$ we need to find an element in $A$ which maps to it under $f$. Given $b\in B$, there is an element $c\in C$ such that $g(b)=c$. Since $g\circ f$ is onto, there exists an element $a\in A$ such that $g(f(a))=c$. Since $g$ is $1-1$ $f(a)=b$.&lt;/p&gt;&#xA;" OwnerUserId="3431" LastActivityDate="2010-11-23T20:28:17.560" />
  <row Id="11576" PostTypeId="2" ParentId="11573" CreationDate="2010-11-23T20:29:09.400" Score="2" ViewCount="" Body="&lt;p&gt;Since $g\circ f$ is onto, then for every $c\in C$ there exists (at least one) $a\in A$ such that $g\circ f(a) = c$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, since $f(a)=b\in B$, and $g(b)=c$, then $g$ is onto.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So $g$ is onto, and $g$ is also 1-1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So $g$ is ......&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you fill in the blank, use that to get that $f$ is onto. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Added:&lt;/em&gt; This is not the most direct way of doing it; Timothy Wagner's direct argument is more streamlined. I did it this way, because through long use,  &quot;if a $g\circ f$ is onto, then $g$ is onto&quot; and &quot;if $g\circ f$ is 1-1 then $f$ is 1-1&quot; are so ingrained that they immediately jump out, just like if I see &quot;$2\times 7$&quot; I immediately think &quot;$14$&quot;. &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-23T20:35:49.577" LastActivityDate="2010-11-23T20:35:49.577" CommentCount="1" />
  <row Id="11577" PostTypeId="2" ParentId="11574" CreationDate="2010-11-23T20:29:26.867" Score="2" ViewCount="" Body="&lt;p&gt;The Euclidean distance has the same units as the positions are measured in.  The Hamming distance has no units.  Think of changing units, say from meters to centimeters.  The numeric value of the Euclidean distance will multiply by 100, but the Hamming distance will not change.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-11-23T20:29:26.867" />
  <row Id="11578" PostTypeId="1" CreationDate="2010-11-23T20:34:19.207" Score="0" ViewCount="85" Body="&lt;p&gt;Let's say I have an &quot;S&quot; shaped curve made up of 50 points.  I am only given the data points, and there is no simple or obvious equation for each point on the curve.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So given only the data points, how would I be able to calculate the slope of the curve at a particular data point?&lt;/p&gt;&#xA;" OwnerUserId="3874" LastActivityDate="2010-11-23T21:03:18.360" Title="Calculating The Slope Of A Line At A Specific Point" Tags="&lt;calculus&gt;" AnswerCount="3" />
  <row Id="11579" PostTypeId="1" AcceptedAnswerId="11583" CreationDate="2010-11-23T20:35:05.960" Score="1" ViewCount="57" Body="&lt;p&gt;Suppose $f\colon [a,b] \to \mathbb{R}$ is continuous and has a finite derivative $f&amp;#39;$ everywhere on $(a,b)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $f(a)=f(b)=0$ prove that for every $y\in\mathbb{R}$ there is some $c$ in $(a,b)$ with $f&amp;#39;(c) = yf(c)$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We were given a hint to apply Rolle's theorem to $h(x)f(x)$ for an $h$ that depends on $y$ but I don't understand how to apply Rolle's theorem here. &lt;/p&gt;&#xA;" OwnerUserId="3333" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-23T20:50:23.867" LastActivityDate="2010-11-23T20:50:23.867" Title="For every $y\in \mathbb{R}$ there is some $c$ in $(a,b)$ with $f'(c) = yf(c)$" Tags="&lt;calculus&gt;&lt;real-analysis&gt;&lt;homework&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="11580" PostTypeId="2" ParentId="11578" CreationDate="2010-11-23T20:39:49.293" Score="1" ViewCount="" Body="&lt;p&gt;Estimate the tangent by taking the secants you can compute; you the tangent will lie somewhere between the two. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is: you cannot compute the slope exactly, but you can estimate it from the information you have.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have immediately preceding and immediately succeeding points, $t_{i-1}$ and $t_{i+1}$, as well as value $f(t_{i-1})$, $f(t_i)$, and $f(t_{i+1})$. The secant between $(t_{i-1},f(t_{i-1}))$, and $(t_i,f(t_i))$ has a certain slope $L_1$; the secant between $(t_i,f(t_i))$ and $(t_{i+1},f(t_{i+1}))$ has a certain slope $L_2$. Unless the graph is very nasty, the tangent at $(t_i,f(t_i))$ should have a slope that is somewhere between $L_1$ and $L_2$.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-11-23T20:39:49.293" />
  <row Id="11581" PostTypeId="2" ParentId="11578" CreationDate="2010-11-23T20:40:23.367" Score="0" ViewCount="" Body="&lt;p&gt;The simplest is just to take one of the neighboring points and find the slope of the line through those two.  This approach is sensitive to measurement error in the points, particularly if they are measured at x values that are close together.  Another approach is to fit some functional form (like an arctangent or logistic function is your data is S shaped) through the data points.  This is smoothing, so will be less sensitive to measurement error (particularly if it has fewer parameters than you have measurements), but is sensitive to modeling error.  There is a discussion in chapters 3 and 15 of &lt;a href=&quot;http://www.nr.com/&quot; rel=&quot;nofollow&quot;&gt;Numerical Recipes&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-11-23T20:40:23.367" />
  <row Id="11582" PostTypeId="2" ParentId="11578" CreationDate="2010-11-23T20:42:18.177" Score="0" ViewCount="" Body="&lt;p&gt;Cubic Splines! You fit cubic spline through the points, and then take slope from the cubic spline definition. Note that if the data have any noise, you need to smooth it first.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;links&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://mathworld.wolfram.com/CubicSpline.html&quot; rel=&quot;nofollow&quot;&gt;Cubic Splines - Wolfram&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Spline_%28mathematics%29&quot; rel=&quot;nofollow&quot;&gt;Splines - Wikipedia&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.nrbook.com/a/bookf90pdf.php&quot; rel=&quot;nofollow&quot;&gt;Interpolation - Numerical Recipies Online&lt;/a&gt; - Open charter B3&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Once the spline is defined, for each segment with curve coordinates ($x(\alpha)$,$y(\alpha)$) the slope is $$s=\frac{\mathrm{d}y/\mathrm{d}\alpha}{\mathrm{d}x/\mathrm{d}\alpha}$$ evaluated at the desired location parameter $\alpha=0\ldots1$.&lt;/p&gt;&#xA;" OwnerUserId="3301" LastEditorUserId="3301" LastEditorDisplayName="" LastEditDate="2010-11-23T21:03:18.360" LastActivityDate="2010-11-23T21:03:18.360" />
  <row Id="11583" PostTypeId="2" ParentId="11579" CreationDate="2010-11-23T20:43:14.997" Score="2" ViewCount="" Body="&lt;p&gt;$\textbf{HINT:}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given $y \in \mathbb{R}$, consider the function $g_y(x)$ as below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$g_y(x) = e^{-yx} f(x)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Apply Rolle's to $g_y(x)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(In general, if you find equations where the derivative is proportional to the function as in this case where $f&amp;#39;(c) \propto f(c)$, the detective in you must suspect immediately that the exponential function must be there somewhere in the picture)&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2010-11-23T20:50:01.293" LastActivityDate="2010-11-23T20:50:01.293" />
  <row Id="11584" PostTypeId="2" ParentId="11579" CreationDate="2010-11-23T20:49:45.753" Score="2" ViewCount="" Body="&lt;p&gt;Suppose you have a function $h(x)$ which is differentiable, and you consider the function $\mathcal{F}(x) = h(x)f(x)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $h$ and $f$ are both continuous on $[a,b]$, so is $\mathcal{F}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $h$ and $f$ are both differentiable on $(a,b)$, so is $\mathcal{F}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $f(a)=f(b)=0$, then $\mathcal{F}(a)=h(a)f(a)=0$ and $\mathcal{F}(b)=h(b)f(b)=0$, regardless of what $h(x)$ is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, $\mathcal{F}$ will &lt;em&gt;also&lt;/em&gt; satisfy Rolle's Theorem, and so you know that there exists $c\in (a,b)$ such that $\mathcal{F}&amp;#39;(c)=0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, what is $\mathcal{F}&amp;#39;(c)$? Well,&#xA;$$\mathcal{F}&amp;#39;(x) = h&amp;#39;(x)f(x) + h(x)f&amp;#39;(x).$$&#xD;&#xA;So $\mathcal{F}&amp;#39;(c) = h&amp;#39;(c)f(c) + h(c)f&amp;#39;(c)$. You want $f&amp;#39;(c)=yf(c)$, so you want&#xA;$$ 0 = h&amp;#39;(c)f(c) + h(c)f&amp;#39;(c) = h&amp;#39;(c)f(c) + h(c)yf(c) = f(c)\Bigl(h&amp;#39;(c)+yh(c)\Bigr).$$&#xD;&#xA;So either $f(c)=0$, or else $h&amp;#39;(c)+yh(c)=0$. You don't want $f(c)=0$ unless $y=0$, so in general you just want $h&amp;#39;(c)+yh(c)=0$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How about trying to find some function $h(x)$ such that $h&amp;#39;(x) = -yh(x)$ for all $x$? Then use that for $\mathcal{F}$. Obviously, your choice of $h$ will depend on $y$.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-11-23T20:49:45.753" CommentCount="2" />
  <row Id="11585" PostTypeId="1" AcceptedAnswerId="11589" CreationDate="2010-11-23T21:07:21.840" Score="3" ViewCount="150" Body="&lt;p&gt;How  to show :&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If $A \subseteq \mathbb{I}$ is not&#xA;  countable, then $A$ is not closed set&#xA;  in $\mathbb{R}$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I think that if $A$ is closed, then $A^{c}$ is open thus: $A^{c}$ is an union of open intervals, but...&lt;/p&gt;&#xA;" OwnerUserId="3421" LastEditorUserId="3421" LastEditorDisplayName="" LastEditDate="2010-11-23T21:11:13.493" LastActivityDate="2010-11-23T22:13:49.767" Title="Question about irrational numbers" Tags="&lt;real-analysis&gt;&lt;general-topology&gt;" AnswerCount="3" CommentCount="8" FavoriteCount="1" />
  <row Id="11586" PostTypeId="2" ParentId="11487" CreationDate="2010-11-23T21:08:32.720" Score="1" ViewCount="" Body="&lt;p&gt;What follows is too long to be a comment. It is also very useful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose first that $X_i$, $i=1,2$, are independent ${\rm N}(0,\sigma_i^2)$ variables. Their characteristic functions are ${\rm E}[{\rm e}^{{\rm i}zX_i} ] = {\rm e}^{\sigma _i^2 ( - z^2 /2)}$. Suppose now that $X_i$, $i=1,2$, are independent Poisson variables, with means $\lambda_i$. Their characteristic functions are ${\rm E}[{\rm e}^{{\rm i}zX_i} ] = {\rm e}^{\lambda _i ({\rm e}^{{\rm i}z}  - 1)}$. For the normal distribution case&#xA;${\rm E}(X_1 | X_1 + X_2) = \frac{{\sigma _1^2 }}{{\sigma _1^2  + \sigma _2^2 }}(X_1  + X_2 )$, and for the Poisson case&#xA;${\rm E}(X_1 | X_1 + X_2) = \frac{{\lambda_1 }}{{\lambda_1  + \lambda_2 }}(X_1  + X_2 )$ (the later result is easy to prove). Finally, consider independent rv's $X_i$, $i=1,2$, having finite expectations and characteristic functions of the form ${\rm E}[{\rm e}^{{\rm i}zX_i} ] = {\rm e}^{c _i \psi(z)}$, $c_i &amp;gt; 0$. What would you expect ${\rm E}(X_1 | X_1 + X_2)$ to be, based on the previous observations? Though very informal, this is very useful, since the conclusion can be applied  to a vast number of rv's, e.g. gamma rv's (specifically, the conclusion applies to all integrable infinitely divisible rv's).&lt;/p&gt;&#xA;" OwnerUserId="2810" LastActivityDate="2010-11-23T21:08:32.720" />
  <row Id="11587" PostTypeId="1" AcceptedAnswerId="11593" CreationDate="2010-11-23T21:10:14.273" Score="2" ViewCount="119" Body="&lt;p&gt;I am trying to upper bound $\sum_{l=1}^\infty (1 - n! 2^{-ln} {2^l \choose n})$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried to use the simple ${2^l \choose n} \ge (\frac{2^l}{n})^n$, but then $l$ disappears and it diverges. Any ideas?&lt;/p&gt;&#xA;" OwnerUserId="2649" LastActivityDate="2010-11-23T22:55:56.877" Title="Upper bound binomial sum" Tags="&lt;inequality&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" />
  <row Id="11588" PostTypeId="2" ParentId="11188" CreationDate="2010-11-23T21:11:35.103" Score="1" ViewCount="" Body="&lt;p&gt;With the identity, $ x \leq y \Leftrightarrow x = x \wedge y$, the implication is straightforward to prove.  That is, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xD;&#xA;x \leq y \text{ and } x \leq z &amp;amp; \Leftrightarrow x = x \wedge y \text{ and } x = x \wedge z \\&#xD;&#xA;&amp;amp; \Leftrightarrow x \wedge x = (x\wedge y) \wedge (x\wedge z)\\&#xD;&#xA;&amp;amp; \Leftrightarrow x = x \wedge (y \wedge z)\\&#xD;&#xA;&amp;amp; \Leftrightarrow x \leq y \wedge z. &#xD;&#xA;\end{align}$$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="3877" LastActivityDate="2010-11-23T21:11:35.103" />
  <row Id="11589" PostTypeId="2" ParentId="11585" CreationDate="2010-11-23T21:23:53.143" Score="4" ViewCount="" Body="&lt;p&gt;The statement is not true.  There exist perfect&lt;sup&gt;1&lt;/sup&gt; (hence closed) subsets of $\mathbb R$ that doesn't intersect $\mathbb Q$, and &lt;a href=&quot;http://pirate.shu.edu/~wachsmut/ira/topo/proofs/pfctuncb.html&quot; rel=&quot;nofollow&quot;&gt;perfect sets are not countable&lt;/a&gt;.  See: &lt;a href=&quot;http://math.stackexchange.com/questions/1064/perfect-set-without-rationals&quot;&gt;Perfect set without rationals&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;1&lt;/sup&gt;  A set $S$ is perfect if it is closed and every point of $S$ is an accumulation point of $S$. &lt;/p&gt;&#xA;" OwnerUserId="145" LastEditorUserId="145" LastEditorDisplayName="" LastEditDate="2010-11-23T21:39:37.690" LastActivityDate="2010-11-23T21:39:37.690" />
  <row Id="11590" PostTypeId="1" AcceptedAnswerId="11591" CreationDate="2010-11-23T21:38:59.967" Score="1" ViewCount="67" Body="&lt;p&gt;Given $c \in \mathbb{R}$ and $m,n \in \mathbb{N}$, is it possible to combine the inequations&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$n &amp;lt; m$&#xD;&lt;/li&gt;&#xA;&lt;li&gt;$\frac{8}{c} &amp;lt; m$&#xD;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;into one inequation?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I argued that due $n &amp;lt; m$ and $\frac{8}{c} &amp;lt; m$, there has to be one $n$ between $m$ and $\frac{8}{c}$, being $n = \left\lceil \frac{8}{c} \right\rceil$, so that $m &amp;gt; n &amp;gt; \frac{8}{c}$, but I'm not sure if that is correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Context&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\forall c \in R^{+} \exists n \in \mathbb{N}: \forall m \in \mathbb{N} \land m &amp;gt; n: |f(m)|&amp;lt; c*|g(m)) \implies f \in \mathcal{o}(g)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, this is about the landau-notation of $f(x)$ raises slower than $g(x)$, the task is to show that a specific case ($f(x) = 8x$ and $g(x) = x^2$) complies to this definition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Note:&lt;/strong&gt; I already solved it until I encounter both inequations together and want to show that there is an $n$ for all cases.&lt;/p&gt;&#xA;" OwnerUserId="3880" LastEditorUserId="3880" LastEditorDisplayName="" LastEditDate="2010-11-23T21:55:08.433" LastActivityDate="2010-11-23T22:41:28.817" Title="Combining two inequations" Tags="&lt;homework&gt;&lt;inequality&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="11591" PostTypeId="2" ParentId="11590" CreationDate="2010-11-23T21:47:00.620" Score="1" ViewCount="" Body="&lt;p&gt;Nope. What if $m=2$, $c=7$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's really no way of finding out the what the relationship b/w $n$ and $8/c$ is without additional information.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, the task of finding an $n$ that satisfies certain properties is very different from the task of determining whether $n$ satisfies certain properties. (It's the difference between &quot;there exists&quot; and &quot;for all&quot;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Edit RE: Your edits:&lt;/b&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I understand you correctly, what you actually need is to find an $n$ such that $n&amp;lt;m \Rightarrow \frac8c&amp;lt;m$. Then $n = \lfloor \frac8c \rfloor$ should satisfy that.&lt;/p&gt;&#xA;" OwnerUserId="3638" LastEditorUserId="3638" LastEditorDisplayName="" LastEditDate="2010-11-23T22:32:24.797" LastActivityDate="2010-11-23T22:32:24.797" />
  <row Id="11592" PostTypeId="1" AcceptedAnswerId="11989" CreationDate="2010-11-23T21:52:36.407" Score="1" ViewCount="193" Body="&lt;p&gt;Let $a,b,c,x,y,z\in\mathbb{R}$. Prove that $$ \left(\frac{ax+by+cz}{x-y}\right)^2+\left(\frac{ay+bz+cx}{y-z}\right)^2+\left(\frac{az+bx+cy}{z-x}\right)^2\geq(c-a)^2+(c-b)^2$$&lt;/p&gt;&#xA;" OwnerUserId="2856" LastActivityDate="2010-11-26T20:28:41.563" Title="Second degree polynomial inequality" Tags="&lt;polynomials&gt;&lt;inequality&gt;" AnswerCount="1" CommentCount="13" />
  <row Id="11593" PostTypeId="2" ParentId="11587" CreationDate="2010-11-23T22:03:26.360" Score="3" ViewCount="" Body="&lt;p&gt;We have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \frac{n!}{2^{kn}} { 2^k \choose n } =&#xD;&#xA;\left( 1 - \frac{1}{2^k} \right) \left( 1 - \frac{2}{2^k} \right) \cdots&#xD;&#xA;\left( 1 - \frac{n-1}{2^k} \right) \ge&#xD;&#xA;1 - \frac{n(n-1)}{2^{k+1}}.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And so&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{n(n-1)}{2^{k+1}} \ge 1 - \frac{n!}{2^{kn}} { 2^k \choose n }.$$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="1301" LastActivityDate="2010-11-23T22:03:26.360" CommentCount="1" />
  <row Id="11594" PostTypeId="2" ParentId="11585" CreationDate="2010-11-23T22:09:52.533" Score="4" ViewCount="" Body="&lt;p&gt;As another counterexample, enumerate the rationals as $\{q_k\}_{k=1}^\infty$, and let $U = \bigcup_{k=1}^\infty (q_k - 2^{-(k+2)}, q_k + 2^{-(k+2)})$.  $U$ is open and contains all the rationals, so its complement $U^c$ is closed and contains none of them.  But $U$ is a union of intervals of lengths $2^{-(k+1)}$, so its Lebesgue measure (i.e. length) is at most $\sum_{k=1}^\infty 2^{-(k+1)} = 1/2$.  So $U^c$ has measure at least $1/2$ and in particular must be uncountable.  You can modify this construction to make the  measure of $U^c$ as close to $1$ as desired.&lt;/p&gt;&#xA;" OwnerUserId="822" LastActivityDate="2010-11-23T22:09:52.533" />
  <row Id="11595" PostTypeId="2" ParentId="11585" CreationDate="2010-11-23T22:13:49.767" Score="1" ViewCount="" Body="&lt;p&gt;As AgCl pointed out (and François G. Dorais, in the answer AgCl linked to), the statement is false. (And Nate also pointed out, as I was writing my answer.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would even say that the point is that ${\mathbb R}\setminus{\mathbb Q}$ is &lt;em&gt;homeomorphic&lt;/em&gt; to the set ${\mathbb N}^{\mathbb N}$ (with the product topology of discrete ${\mathbb N}$). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, obviously, ${\mathbb N}^{\mathbb N}$ contains lots of perfect subsets, such as $\{1,2\}^{\mathbb N}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[A really pretty example of such a homeomorphism in the first chapter of the book &quot;Descriptive Set Theory and Forcing: how to prove theorems about Borel sets the hard way&quot; by Arnold Miller (which you can download from his webpage). The usual construction (which is also pretty) is to note that a sequence $(a_1,a_2,\dots)$ can be identified with the continued fraction of an irrational in $(0,1)$, and a bit of thought shows that this map is indeed a homeomorphism.]&lt;/p&gt;&#xA;" OwnerUserId="462" LastActivityDate="2010-11-23T22:13:49.767" />
  <row Id="11596" PostTypeId="1" CreationDate="2010-11-23T22:23:29.743" Score="0" ViewCount="57" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the univariate central limit theorem, the Berry-Esseen theorem gives a quantitative bound on the rate of convergence of distributions to the Normal distribution under Kolmogorov distance:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Berry%E2%80%93Esseen_theorem&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Berry%E2%80%93Esseen_theorem&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are similar statements known for the multivariate version of the central limit theorem, that use some standard distance measure?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Central_limit_theorem#Multidimensional_central_limit_theorem&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Central_limit_theorem#Multidimensional_central_limit_theorem&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks,&lt;/p&gt;&#xA;" OwnerUserId="3831" LastActivityDate="2010-11-23T22:23:29.743" Title="Quantitative bounds for multivariate central limit theorem" Tags="&lt;real-analysis&gt;&lt;probability-theory&gt;&lt;reference-request&gt;" CommentCount="1" />
  <row Id="11597" PostTypeId="2" ParentId="11590" CreationDate="2010-11-23T22:33:20.130" Score="0" ViewCount="" Body="&lt;p&gt;So, you need to show that for every $c&amp;gt;0 \in \mathbb{R}$ there exists $n \in \mathbb{R}^+$ such that $|f(x)|&amp;lt;c|g(x)|$ for all $x&amp;gt;n$. Given a $c&amp;gt;0$, let $n=\frac{8}{c}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then...&#xA;$x&amp;gt;n \implies |f(x)|&amp;lt;c|g(x)|$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="2596" LastEditorUserId="2596" LastEditorDisplayName="" LastEditDate="2010-11-23T22:41:28.817" LastActivityDate="2010-11-23T22:41:28.817" />
  <row Id="11598" PostTypeId="1" AcceptedAnswerId="11607" CreationDate="2010-11-23T22:46:53.097" Score="0" ViewCount="97" Body="&lt;p&gt;I've been playing around with this equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle\int_{-\pi}^\pi{\displaystyle\frac{1-e^{3it}}{1-e^{it}}dt}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now it seems to me that we can (possibly) split the integral into four seperate integrals; one for each quadrant of the complex plane.  Doing this, we can determine the complex part if we know the real part, and vice versa.  So I therefore ponder that it might be possible to simplify the exponential expressions into expressions of variables.  In other words, I wonder if we can rewrite the $e^{it}$s as $u$s.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm extremely interested in the potential of a variable substitution, but maybe I'm missing the obvious.  I can't tell you exactly what I'm looking for, except for information concerning the possibility of this method.  What is known?  If there is a known way, I'd really enjoy seeing it.  It would make many series expansions that I'm thinking about much easier, I'm almost certain.  If no methods are known, I'd appreciate it very much if someone could take the time to elaborate on all of the reasons why this method is not known.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm very interested in going into detail on this matter, because it seems to decide issues concerning a great amount of potential.&lt;/p&gt;&#xA;" OwnerUserId="2626" LastActivityDate="2010-11-23T23:35:41.940" Title="Complex integrals and the possibility of avoiding parametric equations" Tags="&lt;complex-analysis&gt;&lt;intuition&gt;&lt;complex-variables&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="11599" PostTypeId="2" ParentId="11477" CreationDate="2010-11-23T22:54:08.843" Score="4" ViewCount="" Body="&lt;p&gt;There are several good answers already, but I thought I would add the following derivation because it is one of the few uses I know for the sum property of permanents; namely,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $A$, $B$, and $C$ are matrices with identical entries except that one row (column) of $C$, say the $k^{th}$, is the sum of the $k^{th}$ rows (columns) of $A$ and $B$, then $\text{ per } A + \text{ per } B = \text{per } C$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Start with the matrices $\begin{bmatrix} F_n &amp;amp; F_{n-1} \\ F_0 &amp;amp; F_1 \end{bmatrix}$ and $\begin{bmatrix} F_n &amp;amp; F_{n-1} \\ F_1 &amp;amp; F_2 \end{bmatrix}$.  Since $F_0 = 0$ and $F_1 = F_2 = 1$, they have permanents $F_n$ and $F_n + F_{n-1} = F_{n+1}$, respectively.  Applying the Fibonacci recurrence and the permanent sum property,  we have $\text{ per } \begin{bmatrix} F_n &amp;amp; F_{n-1} \\ F_2 &amp;amp; F_3 \end{bmatrix} = F_{n+2}$.  By continuing to construct new matrices whose second rows are the sums of the second rows of the previous two matrices, this process continues until we have $F_n F_{m+1} + F_{n-1}F_m = \text{ per} \begin{bmatrix} F_n &amp;amp; F_{n-1} \\ F_m &amp;amp; F_{m+1} \end{bmatrix} = F_{n+m}.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For more on this approach (but with determinants), see this paper I wrote a few years ago: &quot;&lt;a href=&quot;http://math.pugetsound.edu/~mspivey/FibDetermPageProofs.pdf&quot; rel=&quot;nofollow&quot;&gt;Fibonacci Identities via the Determinant Sum Property&lt;/a&gt;,&quot; &lt;em&gt;The College Mathematics Journal&lt;/em&gt;, 37 (4): 286-289, 2006. &lt;/p&gt;&#xA;" OwnerUserId="2370" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-12-24T22:01:27.810" LastActivityDate="2010-12-24T22:01:27.810" CommentCount="1" />
  <row Id="11600" PostTypeId="2" ParentId="11587" CreationDate="2010-11-23T22:55:56.877" Score="0" ViewCount="" Body="&lt;p&gt;I was working on this and noticed an interesting thing. Do not know if it is obvious. I wanted to try and improve the bound Derek has (which is a quadratic $n$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we reformulate the problem as Derek has (since for this we need $2^k \geq n$, so we forget the original problem and think the problem given is as follows)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i.e. Let $S(n) = \displaystyle \sum_{k=1}^{\infty} (1 - (1-\frac{1}{2^k})(1-\frac{2}{2^k}) \cdots (1-\frac{n-1}{2^k}))$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We see that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$S(1) = 0$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$S(2) = 1$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$S(3) = \frac{7}{3} \approx 2.3333$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$S(4) = \frac{67}{21} \approx 3.1904$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$S(5) = \frac{407}{105} \approx 3.8762$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$S(6) = \frac{4789}{1085} \approx 4.4138$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$S(7) = \frac{5289}{1085} \approx 4.8747$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$S(8) = \frac{726093}{137795} \approx 5.2694$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$S(9) = \frac{118399669}{21082635} \approx 5.61598$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We see that the growth is slow as $n$ increases. My question is if this converges to some limit or it diverges. I have been working on it for the past couple of hours but am unable to come to a conclusion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the question is what is $\displaystyle \lim_{n \rightarrow \infty} S(n)$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it converges, can we find the limit?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it diverges, how fast does it diverge?&lt;/p&gt;&#xA;" OwnerUserId="3302" LastActivityDate="2010-11-23T22:55:56.877" CommentCount="2" />
  <row Id="11601" PostTypeId="1" CreationDate="2010-11-23T23:00:49.960" Score="4" ViewCount="359" Body="&lt;p&gt;From its definition a combination $(^n_k)$, is the number of distinct subsets of size &lt;em&gt;k&lt;/em&gt; from a set of &lt;em&gt;n&lt;/em&gt; elements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is clearly an integer, however I was curious as to why the equation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{n!}{k!(n-k)!}$ always evaluates to an integer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So far I figured:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$n!$, is clearly divisible by $k!$, and $(n-k)!$, individually, but I could not seem to make the jump to proof that that $n!$ is divisible by their product.&lt;/p&gt;&#xA;" OwnerUserId="3881" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-11-23T23:03:28.677" LastActivityDate="2010-11-24T15:35:06.263" Title="Proof that a Combination is an integer" Tags="&lt;binomial-coefficients&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="2" />
  <row Id="11602" PostTypeId="1" AcceptedAnswerId="11662" CreationDate="2010-11-23T23:02:25.277" Score="4" ViewCount="272" Body="&lt;p&gt;This is a follow up of &lt;a href=&quot;http://math.stackexchange.com/questions/11587/upper-bound-binomial-sum&quot;&gt;Upper bound binomial sum&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was working on the problem in the above thread and noticed an interesting thing. I wanted to try and improve the bound Derek had (which was a quadratic in $n$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we reformulate the problem as Derek has (since for this we need $2^k \geq n$, so we forget the original problem and think the problem given is as follows)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i.e. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $S(n) = \displaystyle \sum_{k=1}^{\infty} \left(1 - \prod_{j=1}^{n-1}\left(1-\frac{j}{2^k}\right)\right)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We see that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$S(1) = 0$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$S(2) = 1$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$S(3) = \frac{7}{3} \approx 2.3333$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$S(4) = \frac{67}{21} \approx 3.1904$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$S(5) = \frac{407}{105} \approx 3.8762$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$S(6) = \frac{4789}{1085} \approx 4.4138$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$S(7) = \frac{5289}{1085} \approx 4.8747$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$S(8) = \frac{726093}{137795} \approx 5.2694$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$S(9) = \frac{118399669}{21082635} \approx 5.61598$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$S(10) = \frac{9120486643}{1539032355} \approx 5.92612$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$S(11) = \frac{105065594573}{16929355905} \approx 6.20612$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$S(12) = \frac{31986101239583}{4950627362505} \approx 6.4610$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We see that the growth is slow as $n$ increases. My question is if this converges to some limit or it diverges. I have been working on it for the past couple of hours but am unable to come to a conclusion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the question is what is $\displaystyle \lim_{n \rightarrow \infty} S(n)$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it converges, can we find the limit?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it diverges, how fast does it diverge?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;$\textbf{EDIT:}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;So Mike has proved that $S(n)$ in fact diverges.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The conjecture is now that $\lim_{n \rightarrow \infty} (2 \log_{2}(n) - S(n)) = \alpha$ where $\alpha \approx 0.667$.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Look at &lt;a href=&quot;http://math.stackexchange.com/questions/11726/limit-of-sn-displaystyle-sum-k1-infty-left1-prod-j1n-1-le&quot;&gt;Limit of $S(n) = \sum_{k=1}^{\infty} \left(1 - \prod_{j=1}^{n-1}\left(1-\frac{j}{2^k}\right)\right)$ - Part II&lt;/a&gt; for further discussions.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2010-11-24T22:55:14.717" LastActivityDate="2010-11-24T22:55:14.717" Title="Limit of $S(n) = \sum_{k=1}^{\infty} \left(1 - \prod_{j=1}^{n-1}\left(1-\frac{j}{2^k}\right)\right)$" Tags="&lt;real-analysis&gt;&lt;sequences-and-series&gt;" AnswerCount="2" CommentCount="9" />
  <row Id="11603" PostTypeId="2" ParentId="11601" CreationDate="2010-11-23T23:08:13.087" Score="5" ViewCount="" Body="&lt;p&gt;As Jonas mentioned, it counts something so it has to be a natural number. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another way is to notice that product of $m$ consecutive natural numbers is divisible by $m!$.(Prove this!)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if we write $n! = n(n-1)(n-2) \cdots (k+1) \times (k!)$, we find that $k!$ divides $k!$ and &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$n(n-1)(n-2) \cdots (k+1)$ is a product of $(n-k)$ consecutive natural numbers and hence $(n-k)!$ divides it.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastActivityDate="2010-11-23T23:08:13.087" CommentCount="3" />
  <row Id="11604" PostTypeId="2" ParentId="11601" CreationDate="2010-11-23T23:14:00.720" Score="6" ViewCount="" Body="&lt;p&gt;Well, one noncombinatorial way is to induct on $n$ using Pascal's triangle; that is, using the fact that ${n \choose k} = {n-1 \choose k - 1} + {n-1 \choose k}$ (easy to verify directly) and that each ${n - 1 \choose 0}$ is just $1$.&lt;/p&gt;&#xA;" OwnerUserId="3035" LastActivityDate="2010-11-23T23:14:00.720" />
  <row Id="11605" PostTypeId="2" ParentId="11602" CreationDate="2010-11-23T23:21:28.080" Score="4" ViewCount="" Body="&lt;p&gt;If $(x-1)(x-2)\cdots(x-(n-1)) = \sum_{j=0}^{n-1} a_j x^j$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(As Mike points out, the $a_j$ are nothing but &lt;a href=&quot;http://en.wikipedia.org/wiki/Stirling_numbers_of_the_first_kind#Definition&quot; rel=&quot;nofollow&quot;&gt;Stirling Numbers of the first kind&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then the $k^{th}$ term is $$\displaystyle - \sum_{j=0}^{n-2} \frac{a_j}{2^{(n-1-j)k}}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;by setting $\displaystyle x=2^k$ in the above polynomial and dividing by an appropriate power of $2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And so the limit is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$S(n) = \displaystyle - \sum_{j=0}^{n-2} \frac{a_j}{2^{n-1-j}-1}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, for $\displaystyle n=5$, we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle (x-1)(x-2)(x-3)(x-4) = x^4 - 10x^3 + 35x^2 - 50x + 24$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle S(5) = 10/1 - 35/3 + 50/7 - 24/15 = 407/105$.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-11-23T23:47:28.730" LastActivityDate="2010-11-23T23:47:28.730" CommentCount="6" />
  <row Id="11606" PostTypeId="2" ParentId="11601" CreationDate="2010-11-23T23:25:25.263" Score="7" ViewCount="" Body="&lt;p&gt;See my &lt;a href=&quot;http://math.stackexchange.com/questions/2158/division-of-factorials/2192#2192&quot;&gt;post here&lt;/a&gt; for a simple purely arithmetical proof that every binomial coefficient is an integer. The proof shows how to rewrite any binomial coefficient fraction as a product of fractions whose denominators are all coprime to any given prime $\rm\:p\:$. This implies that no primes divide the denominator (when written in lowest terms), so the fraction is an integer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The key property that lies at the heart of this proof is that, among all products of $\rm\: n\:$ consecutive integers, $\rm\ n!\ $ has the least possible power of $\rm\:p\:$ dividing it - for all primes $\rm\:p\:$. Thus $\rm\ n!\ $ divides every product of $\rm\:n\:$ consecutive integers, since it has a smaller power of every prime divisor. Therefore&#xA;$$\rm\displaystyle\quad\quad {m \choose n}\ =\ \frac{m!/(m-n)!}{n!}\ =\ \frac{m\:(m-1)\:\cdots\:(m-n+1)}{n\:(n-1)\quad\quad\:\cdots\:\quad\quad 1\quad\quad}\ \in\ \mathbb Z$$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-11-24T15:35:06.263" LastActivityDate="2010-11-24T15:35:06.263" />
  <row Id="11607" PostTypeId="2" ParentId="11598" CreationDate="2010-11-23T23:27:17.077" Score="5" ViewCount="" Body="&lt;p&gt;I don't really see why you don't want to just divide and then integrate the three terms... but if you insist on not using parameterizations: you can rewrite your integral as&#xA;$$-i\int_{-\pi}^{\pi} {1 - e^{3it} \over 1 - e^{it}} ie^{it} {1 \over e^{it}}\,dt$$&#xD;&#xA;Then you do a complex variables version of the $u$-substitution you are trying. You let $z = e^{it}$ and $dz = i e^{it} dt$. Your integral becomes &#xA;$$-i\int_{|z| = 1} {1 - z^3 \over z(1 - z)}\,dz$$&#xD;&#xA; Dividing $(1 - z)$ into $(1 - z^3)$ turns this into&#xA;$$-i \int_{|z| = 1} ({1 \over z} + 1 + z) \,dz$$&#xD;&#xA;Using the residue theorem this immediately evaluates into $-i 2\pi i = 2\pi$.&lt;/p&gt;&#xA;" OwnerUserId="3035" LastEditorUserId="3035" LastEditorDisplayName="" LastEditDate="2010-11-23T23:35:41.940" LastActivityDate="2010-11-23T23:35:41.940" CommentCount="5" />
  <row Id="11608" PostTypeId="1" AcceptedAnswerId="11615" CreationDate="2010-11-23T23:29:46.517" Score="1" ViewCount="111" Body="&lt;p&gt;I am currently a mathematics graduate student at Western Kentucky University in Bowling Green, KY. I am looking for some kind of summer opportunity to participate in during summer 2011.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anyone have any suggestions of good opportunities or a good list of opportunities?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would really appreciate it!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am more interested in pure mathematics than I am in applied mathematics. My favorite areas are set theory, complex variables/analysis, topology, difference equations, difference(discrete) calculus, and time scale calculus.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am really looking to broaden my horizons though with new and interesting topics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Naturally, I am going to speak with my advisor and professors, but I would like to get insight from others as well.&lt;/p&gt;&#xA;" OwnerUserId="3791" LastActivityDate="2010-11-23T23:54:01.360" Title="Mathematics Graduate Student Summer Opportunities" Tags="&lt;education&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="2" />
  <row Id="11609" PostTypeId="1" AcceptedAnswerId="11623" CreationDate="2010-11-23T23:40:13.187" Score="1" ViewCount="214" Body="&lt;p&gt;Question: Let $v = (1,1,0,1) \in \mathbb{R}^{4}$ Let $$V:= \{f: \mathbb{R}^{4} \rightarrow \mathbb{R}^{2} \mbox{ linear } | f(v) = 0\}.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(a) Show that $V$ forms a vectorspace.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(b) Determine dim$(V)$. (Tip: Describe the elements in $V$ by extending $v$ to a basis of $\mathbb{R}^{4}$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My attempt: for (a): ok i know that a vectorspace is defined through those 8 axioms, since it says &quot;linear&quot; i know that $f(v+u) = f(v)+f(u)$ and $\lambda f(v) = f(\lambda v)$, however i don't see how to proceed. what more should i be inferring from the given information?&#xA;for (b) if i understand correctly, i can extend $v=(1,1,0,1)$ to a basis of $\mathbb{R}^{4}$ with $u=(0,1,0,0), w=(0,0,1,0), x=(0,0,0,1)$ correct? of course i wouldn't know what to do with that.. in the problems i've encountered so far it seems that dim$(V)$ has always been the amount of elements in a basis, however i'm doubting that it's so simple as dim$(V)=4$ just because dim$(\mathbb{R}^{4})=4$... i notice that $f(v)=0$, perhaps i could do something with the sum of the kernel and the image? as always, any help is greatly appreciated!&lt;/p&gt;&#xA;" OwnerUserId="3711" LastActivityDate="2010-11-24T21:27:35.580" Title="Show that this is a vector space and determine the dimension" Tags="&lt;linear-algebra&gt;&lt;vector-space&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="11610" PostTypeId="1" CreationDate="2010-11-23T23:41:35.090" Score="0" ViewCount="63" Body="&lt;p&gt;Dear friends,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The length of any curve that goes from A to B (other than the line segment AB) is greater than the length of that segment. This statement may be a theorem? Or, necessarily, is an axiom?&#xA;It seems to me that any attempt to build a theorem eventually turning into begging.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you very much!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Paulo Argolo&lt;/p&gt;&#xA;" OwnerUserId="2938" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-24T02:47:35.123" LastActivityDate="2010-11-24T03:03:10.427" Title="Lower curve from A to B: axiom or theorem?" Tags="&lt;geometry&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="11611" PostTypeId="1" CreationDate="2010-11-23T23:43:40.127" Score="1" ViewCount="335" Body="&lt;p&gt;I was trying to wrap my brain around this but could not think of anything. I am very poor at maths but trying to learn of it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit:&#xA;Exact question is&#xA;If x^2+x-6 is a composite function f(g(x)); then figure out f(x) and g(x) ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To me information looks incomplete to figure out such details.&lt;/p&gt;&#xA;" OwnerUserId="3882" LastEditorUserId="3882" LastEditorDisplayName="" LastEditDate="2010-11-23T23:55:33.267" LastActivityDate="2011-01-24T13:13:04.340" Title="How to decompose function like x^2+x-6 ?" Tags="&lt;polynomials&gt;" AnswerCount="4" CommentCount="10" />
  <row Id="11612" PostTypeId="2" ParentId="11610" CreationDate="2010-11-23T23:44:46.427" Score="0" ViewCount="" Body="&lt;p&gt;This is a (simple) theorem.  See &lt;a href=&quot;http://en.wikipedia.org/wiki/Arc_length#Definition&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Arc_length#Definition&lt;/a&gt;.&lt;/p&gt;&#xA;" LastActivityDate="2010-11-23T23:44:46.427" CommentCount="4" />
  <row Id="11615" PostTypeId="2" ParentId="11608" CreationDate="2010-11-23T23:54:01.360" Score="3" ViewCount="" Body="&lt;p&gt;Robert Krasny from University of Michigan, has an excellent resource page&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.math.lsa.umich.edu/~krasny/fellowships_etc.html&quot; rel=&quot;nofollow&quot;&gt;http://www.math.lsa.umich.edu/~krasny/fellowships_etc.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is also a good list here&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.phds.org/graduate-school/preparing/summer-internships-2/&quot; rel=&quot;nofollow&quot;&gt;http://www.phds.org/graduate-school/preparing/summer-internships-2/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3431" LastActivityDate="2010-11-23T23:54:01.360" CommentCount="1" />
  <row Id="11616" PostTypeId="2" ParentId="11611" CreationDate="2010-11-23T23:56:43.687" Score="1" ViewCount="" Body="&lt;p&gt;We always have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(x+a)&lt;em&gt;(x+b) = (x&lt;/em&gt;(x+b))+(a*(x+b)) = (x*x)+(x*b)+(a*x)+(a*b) = (x^2)+(b*x)+(a*x)+(a*b)&#xA;(x^2)+(b*x)+(a*x)+(a*b) = (x^2)+(a*x)+(b*x)+(a*b) = (x^2)+((a+b)*x)+(a*b)&#xA;(x+a)*(x+b) = (x^2)+((a+b)*x)+(a*b)&#xA;&lt;br&gt;&lt;br&gt;&lt;br&gt;&#xA;Converting your equation to that form gives&lt;/p&gt;&#xA;&#xA;&lt;p&gt;x^2+x-6 = (x^2)+(1*x)+(-6)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where we want&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(x^2)+(1*x)+(-6) = (x^2)+((a+b)*x)+(a*b)&#xA;&lt;br&gt;&lt;br&gt;&lt;br&gt;&#xA;So, we're looking for numbers whose sum is 1 and whose product is -6.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a+b = 1  and  a*b = -6&#xA;&lt;br&gt;&lt;br&gt;&lt;br&gt;&#xA;That should be enough for you to figure it out, but in case it's not,&#xA;&lt;br&gt;&#xA;if you just want the answer you can put&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;n vf rdhny gb artngvir gjb naq o vf rdhny gb cbfvgvir guerr&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;into rot13.com&lt;/p&gt;&#xA;" LastActivityDate="2010-11-23T23:56:43.687" CommentCount="2" />
  <row Id="11618" PostTypeId="1" AcceptedAnswerId="11621" CreationDate="2010-11-24T01:20:51.570" Score="1" ViewCount="232" Body="&lt;p&gt;Prove the following: $\displaystyle a^{n}-b^{n} = (a-b) \sum\limits_{k=0}^{n-1} a^{k}b^{n-1-k}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So one could use induction on $n$? Could one also use trichotomy or some type of combinatorial argument? &lt;/p&gt;&#xA;" OwnerUserId="2548" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2010-11-24T04:38:48.420" LastActivityDate="2010-11-24T15:29:24.757" Title="Algebraic Identity $a^{n}-b^{n} = (a-b) \sum\limits_{k=0}^{n-1} a^{k}b^{n-1-k}$" Tags="&lt;algebra-precalculus&gt;" AnswerCount="7" FavoriteCount="1" />
  <row Id="11619" PostTypeId="2" ParentId="11618" CreationDate="2010-11-24T01:22:32.263" Score="2" ViewCount="" Body="&lt;p&gt;Can we build a combinatorial argument along these lines?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Say if we have say $n$ students to be allotted in $a$ rooms with $b$ of the $a$ room being non air-conditioned. (Assume the students are distinguishable so we can order the student as $1,2,3,...,n$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The total number of ways is $a^n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose all students are allotted to the $b$ rooms, the number of ways is $b^n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the first $n-1$ students are allotted to the $b$ rooms, and the final dude in some other room, the number of possible ways is $b^{n-1} \times (a-b)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now if the first $n-2$ students are allotted to the $b$ rooms, and the remaining two students are now left. If the $(n-1)^{th}$ student chooses from the $b$ rooms then we are back to the earlier case. So the $(n-1)^{th}$ student needs to choose from the remaining $(a-b)$ rooms. Now the $n^{th}$ student can choose from any of the $a$ rooms. The number of possible ways is $b^{n-2} \times (a-b) \times a$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, if the first $n-k$ students are allotted to the $b$ rooms, and the remaining $k$ students are now left. If the $(n-k+1)^{th}$ student chooses from the $b$ rooms then we are back to the previous case. So the $(n-k+1)^{th}$ student needs to choose from the remaining $(a-b)$ rooms. Now the students from $(n-k+2)$ to $n$ can choose from any of the $a$ rooms. The number of possible ways is $b^{n-k} \times (a-b) \times a^{k-1}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the total number of ways is $b^n + \displaystyle \sum_{k=1}^n b^{n-k} \times (a-b) \times a^{k-1}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both the counting must add up and hence we get $a^n = b^n + \displaystyle \sum_{k=1}^n b^{n-k} \times (a-b) \times a^{k-1}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could use geometric series to conclude the result as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The right hand side is $(a-b) b^{n-1} \displaystyle \sum_{k=0}^{n-1} (\frac{a}{b})^k = (a-b) b^{n-1} \frac{((\frac{a}{b})^n - 1)}{(\frac{a}{b})-1} = (a-b) \frac{a^n - b^n}{a-b} = a^n - b^n$&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2010-11-24T01:50:08.750" LastActivityDate="2010-11-24T01:50:08.750" CommentCount="2" />
  <row Id="11620" PostTypeId="2" ParentId="11618" CreationDate="2010-11-24T01:24:38.537" Score="0" ViewCount="" Body="&lt;p&gt;Yes, you could use induction on n.  I don't see an easy trichotomy or combinatorial argument.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-11-24T01:24:38.537" />
  <row Id="11621" PostTypeId="2" ParentId="11618" CreationDate="2010-11-24T01:36:10.593" Score="7" ViewCount="" Body="&lt;p&gt;I have no idea what you mean by &quot;use trichotomy,&quot; but here is the combinatorial argument.  $a^n$ counts the number of words of length $n$ on the alphabet $\{ 1, 2, ... a \}$ and $b^n$ counts the number of words of length $n$ on the alphabet $\{ 1, 2, ... b \}$.  Assume $a &amp;gt; b$.  Then $a^n - b^n$ counts the number of words of length $n$ on the alphabet $\{ 1, 2, ... a \}$ such that at least one letter is greater than $b$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given such a word, suppose the last letter greater than $b$ occurs at position $k+1$.  Then there are $a - b$ choices for this letter, $a^k$ choices for the letters before this letter, and $b^{n-k-1}$ choices for the letters after this letter.  Thus there are $(a - b) a^k b^{n-k-1}$ such words, and summing over all $k$ gives&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$a^n - b^n = (a - b) \sum_{k=0}^{n-1} a^k b^{n-k-1}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;as desired.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-11-24T01:36:10.593" CommentCount="3" />
  <row Id="11622" PostTypeId="1" AcceptedAnswerId="11625" CreationDate="2010-11-24T01:48:44.867" Score="2" ViewCount="62" Body="&lt;p&gt;This is a follow up to a question I asked earlier. I had previously asked that if a space was regular if and only if any closed set $Z$ was the intersection of all open sets containing $Z$, but the backwards implication turned out to be untrue. I'm now curious to see if this new bi-implication is fully true. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My definition of regular here is that for any $p\in X$ and any closed set $Q$ not containing $p$, there exist open sets $U,V$ such that $p\in U$, $Q\subseteq V$, and $U\cap V=\emptyset$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I assume $X$ is regular. I think it is clear that $$Z\subseteq\bigcap_{\text{closed}\ N\in\mathscr{N}_Z}N$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the other containment, I suppose that $p\not\in Z$. Then since $X$ is regular, there exist open sets $U,V$ such that $p\in U$, $Z\subseteq V$, and $U\cap V=\emptyset$. Then $U\subseteq V^c\subseteq Z^c$, and thus $U^c\supseteq V\supseteq Z$. So $U^c$ is a closed neighborhood of $Z$ which does not contain $p$. Hence $p\not\in\cap\mathcal{N}$, where $\mathcal{N}$ is the family of closed neighborhood of $Z$, and the equality follows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I feel I have a hole in my reasoning to show the other implication. I suppose now that $Z=\cap\mathcal{N}$, and take some $p\not\in Z$, and so $p\not\in\cap\mathcal{N}$. Anyway, $p\in Z^c$, which is an open neighborhood of $p$. I then take some closed neighborhood $Q$ of $p$ such that $Q\subseteq Z^c$. I then let $U={Q}^{\circ}$, and $V=Q^c$. So both are are open and disjoint. Also, $p\in U$, and $V=Q^c\supseteq (Z^c)^c=Z$, so $X$ is regular. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I don't think I'm able to make the assumption that there is some closed neighborhood of $p$ contained in the open neighborhood $Z^c$. Is there some way to use the fact that $p\not\in\cap\mathcal{N}$ to show this is indeed true? Thanks. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: In case it is not entirely clear, my definition of a neighborhood $N$ of a set $Z$ is that $Z$ is contained in the interior of $N$, so $N$ is a neighborhood of each point $Z$. Then my definition of a neighborhood $M$ of a point $p$ is that $M$ contains an open set containing $p$. $\mathscr{N}_Z$ is my notation for the set of all neighborhoods of $Z$, and $\mathcal{N}$ is a shortcut way of writing the set of $\text{closed}\ N\in\mathscr{N}_Z$.&lt;/p&gt;&#xA;" OwnerUserId="1583" LastEditorUserId="1583" LastEditorDisplayName="" LastEditDate="2010-11-24T02:08:36.840" LastActivityDate="2010-11-24T02:09:45.777" Title="A topological space is regular if and only if any closed set $Z$ is the intersection of its closed neighborhoods?" Tags="&lt;topology&gt;" AnswerCount="1" CommentCount="0" />
  <row Id="11623" PostTypeId="2" ParentId="11609" CreationDate="2010-11-24T02:04:00.557" Score="8" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Re-rewritten&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems, from your comments, that you aren't even very comfortable with the idea of the set $V$ as a vector space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As it happens, it is better to start from a bit further back, and consider the set of &lt;strong&gt;all&lt;/strong&gt; linear transformations from $\mathbb{R}^4\to\mathbb{R}^2$, which I'm going to call $\mathbf{V}$; that is,&#xA;$$\mathbf{V}=\left\{ T\colon\mathbb{R}^4\to\mathbb{R}^2\mid \text{ $T$ is linear}\right\}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We want to make $\mathbf{V}$ into a vector space. So the &lt;em&gt;elements&lt;/em&gt; of $\mathbf{V}$ are going to be our &quot;vectors&quot;; the fact that these vectors are actually functions when they are at home is immaterial: when they are at their &quot;job&quot; in $\mathbf{V}$, they are &quot;vectors&quot;. In order for this to be a vector space, we need to have a way of adding these vectors, and of multiplying them by scalars so that the operations satisfy the axioms of a vector space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How are we going to add these vectors? Remember from Calculus that if we have two functions $f\colon \mathbb{R}\to\mathbb{R}$ and $g\colon\mathbb{R}\to\mathbb{R}$, we can make a new function called the &quot;sum of $f$ and $g$&quot;. This function is called &quot;$f+g$&quot;. In order to tell you what function &quot;$f+g$&quot; is, I need to tell you how to evaluate it at any real number $r$. So here's the rule: To find the value of $f+g$ at $r$, first find the value of $f$ at $r$, then find the value of $g$ at $r$, and then add the two results. That is, $$(f+g)(r) = f(r)+g(r).$$ &#xA;Now, there are two &lt;code&gt;+&lt;/code&gt; symbols in that equation, but they mean different things: the first &lt;code&gt;+&lt;/code&gt; sign is the symbol that represents the sum of two functions; the second &lt;code&gt;+&lt;/code&gt; is the usual sum sign that represents the sum of two real numbers (notice that both $f(r)$ and $g(r)$ are real numbers). So, for example, if $f(x) = 2x-1$ and $g(x)=x^3$, then to find the value of $(f+g)(7)$, we first find $f(7)$ (which is $13$), then we find the value of $g(7)$ (which is $343$), and then we add them ($13+343=356$, so $(f+g)(7) = 356$). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We are going to do the same thing with the elements of $\mathbf{V}$. The elements of $\mathbf{V}$ are functions from $\mathbb{R}^4$ to $\mathbb{R}^2$. As it happens, we know how to &quot;add&quot; two elements of $\mathbb{R}^2$ (because this is a vector space). That means that we can play the same game with functions $T\colon\mathbb{R}^4\to\mathbb{R}^2$ and $U\colon \mathbb{R}^4\to\mathbb{R}^2$ that we played in calculus: we are going to define a new function, which I'm going to call $T\oplus U$ (I'm using a different symbol so we are clear that this is a &lt;em&gt;new&lt;/em&gt; kind of addition which I am defining), which is going to be a function from $\mathbb{R}^4$ to $\mathbb{R}^2$. In order to tell you what this function is, I need to tell you how you evaluate it at any element of $\mathbb{R}^4$. Okay, here goes: given any $\mathbf{v}\in\mathbb{R}^4$, first find out what $T(\mathbf{v})$ is; then find out what $U(\mathbf{v})$ is. Both of these are elements of $\mathbb{R}^2$, so we can add them. The result of adding them is what $(T\oplus U)(\mathbf{v})$ is. That is, if &lt;code&gt;+&lt;/code&gt; represents the sum of vectors in $\mathbb{R}^2$, then&#xA;$$ (T\oplus U)(\mathbf{v}) = T(\mathbf{v}) + U(\mathbf{v}).$$&#xD;&#xA;(I put the parentheses around $T\oplus U$ so that we are clear that it is a single entity with a long name which we are evaluating at $\mathbf{v}$; that way we don't think that we are doing $U(\mathbf{v})$ first, and then $\oplus$-ing something called $T$ to &lt;em&gt;that&lt;/em&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, this makes sense for &lt;em&gt;any&lt;/em&gt; two functions from $\mathbb{R}^4$ to $\mathbb{R}^2$, whether they are linear or not. For instance, if $T\colon\mathbb{R}^4\to\mathbb{R}^2$ is the function $T(a,b,c,d) = (a+b,cd)$, and $U\colon\mathbb{R}^4\to\mathbb{R}^2$ is the function $U(a,b,c,d) = (a,d)$, then the value of $T\oplus U$ at $(1,2,3,4)$ is the result of adding $T(1,2,3,4)=(3,12)$ and $U(1,2,3,4)=(1,4)$; so $(T\oplus U)(1,2,3,4)=(3,12)+(1,4) = (4,16)$. If $T$ and $U$ are &lt;em&gt;any&lt;/em&gt; functions from $\mathbb{R}^4$ to $\mathbb{R}^2$, then $T\oplus U$ is &lt;em&gt;also&lt;/em&gt; a function from $\mathbb{R}^4$ to $\mathbb{R}^2$. (As in Calculus, though, if $T$ and $U$ happen to have &quot;nice&quot; properties, then there is a good chance that these properties will be inherited by $T\oplus U$). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This kind of operation is called &lt;strong&gt;pointwise addition&lt;/strong&gt;: whenever you have functions from the same domain to the same codomain, and you know how to &quot;add&quot; elements of the codomain, you can define a &quot;sum of functions&quot; $f+g$ by saying &quot;first figure out the value of $f$, then figure out the value of $g$, and then add the results together.&quot; It's &quot;pointwise&quot; because the way we compute the sum is by first figuring out the values at &lt;em&gt;each point&lt;/em&gt;, and then adding those values. Notice that the domain $\mathbb{R}^4$ is not playing any role here; you could replace it with any set and still be able to define the sum of functions, because the sum is taking place in the &lt;em&gt;image&lt;/em&gt;, not in the domain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can play the same game with scalar multiplication. If $T\colon\mathbb{R}^4\to\mathbb{R}^2$ is &lt;em&gt;any&lt;/em&gt; function, and $\alpha$ is any real number, then we can remember that we know how to &quot;multiply&quot; any element of $\mathbb{R}^2$ by $\alpha$; so we can define a &quot;pointwise multiplication by $\alpha$&quot; as well. I'm going to define a new function, called $\alpha\odot T\colon\mathbb{R}^4\to\mathbb{R}^2$. The rule for evaluating $\alpha\odot T$ is: to find the value at $\mathbf{v}$, first evaluate $T$ at $\mathbf{v}$, and then multiply the result by $\alpha$. That is,&#xA;$$(\alpha\odot T)(\mathbf{v}) = \alpha\cdot T(\mathbf{v}).$$&#xD;&#xA;Again, I put the parenthesis around $\alpha\odot T$ to remind us that it is a single entity with a long name. The $\cdot$ on the right hand side is the usual scalar multiplication of $\mathbb{R}^2$. For instance, going back to my $T\colon\mathbb{R}^4\to\mathbb{R}^2$ from the previous paragraph, $T(a,b,c,d)=(a+b,cd)$, if $\alpha=\frac{3}{2}$ then $\frac{3}{2}\odot T$ at $(1,2,3,4)$ is computed by first doing $T(1,2,3,4)=(3,12)$, and then doing $\frac{3}{2}(3,12) = (\frac{9}{2},18)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again, this definition makes sense for &lt;em&gt;any&lt;/em&gt; function into $\mathbb{R}^2$, whether linear or not, whether from $\mathbb{R}^4$ or not. This one is called &lt;strong&gt;pointwise (scalar) multiplication&lt;/strong&gt;, because again the &quot;multiplication&quot; is happening after we figure out the value &quot;at each point&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, now let's go back and look at the set $\mathbf{V}$,&#xA;$$\mathbf{V}= \bigl\{ T\colon\mathbb{R}^4\to\mathbb{R}^2\ \bigm|\ \text{$T$ is linear}\bigr\}.$$&#xA;This is a set of functions, and if we take two functions $T$ and $U$ from $\mathbf{V}$ then we can &quot;add&quot; them using $\oplus$ to get a function $T\oplus U\colon\mathbb{R}^4\to\mathbb{R}^2$, and if we take $\alpha\in\mathbb{R}$ and $T\in\mathbf{V}$, we can &quot;scalar multiply them&quot; and get a function $\alpha\odot T\colon\mathbb{R}^4\to\mathbb{R}^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I claim that if $T$ and $U$ are in $\mathbf{V}$ (that is, if in addition to just being functions they are also &lt;em&gt;linear&lt;/em&gt; functions), then $T\oplus U$ is also going to be in $\mathbf{V}$ (it's already a function from $\mathbb{R}^4$ to $\mathbb{R}^2$, so we will just need to check if it is linear). And that if $T\in\mathbf{V}$ and $\alpha\in\mathbb{R}$, then $\alpha\odot T$ is &lt;em&gt;also&lt;/em&gt; going to be in $\mathbf{V}$. That is, $\oplus$ will give me a way of adding elements of $\mathbf{V}$ and get elements of $\mathbf{V}$, and $\odot$ will give me a way to multiply elements of $\mathbf{V}$ by scalars to get elements of $\mathbf{V}$. (This is the &lt;em&gt;first&lt;/em&gt; requirement for a set to be a vector space: it must have some kind of &quot;vector addition&quot; and some kind of &quot;scalar multiplication&quot;). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, suppose that you start by taking $T$ and $U$ in $\mathbf{V}$. In order for $T\oplus U$ to also belong to the club $\mathbf{V}$, $T\oplus U$ must satisfy the membership requirements: it must be a function, it must map $\mathbb{R}^4$ to $\mathbb{R}^2$, and it must be linear. We already know it is a function that maps $\mathbb{R}^4$ to $\mathbb{R}^2$, so we just need to check that it is linear. Remember that each of $T$ and $U$ are themselves linear. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How does one check that &lt;em&gt;a&lt;/em&gt; function $\mathfrak{f}\colon\mathbb{R}^4\to\mathbb{R}^2$ is linear? We check that if $u,w\in\mathbb{R}^4$ are any vectors, then $\mathfrak{f}(u+w) = \mathfrak{f}(u)+\mathfrak{f}(w)$ (the first &lt;code&gt;+&lt;/code&gt; is the sum in $\mathbb{R}^4$; the second is the sum in $\mathbb{R}^2$; the equality will take place in $\mathbb{R}^2$); and we check that for any $u\in\mathbb{R}^4$ and any real number $\alpha$, $\mathfrak{f}(\alpha u) = \alpha\mathfrak{f}(u)$ (first scalar multiplication is in $\mathbb{R}^4$; the second is in $\mathbb{R}^2$; and the equality is taking place in $\mathbb{R}^2$). So, we do this for our function $T\oplus U$: let $u,w\in\mathbb{R}^4$. Then:&#xA;\begin{align*}&#xA;(T\oplus U)(u+w) &amp;amp;= T(u+w) + U(u+w) &amp;amp;\text{(by definition of $(T\oplus U)$)}\\&#xA;&amp;amp;= \bigl(T(u) + T(w)\bigr) + \bigl(U(u) + U(w)\bigr) &amp;amp;\text{(because $T$ and $U$ are each linear)}\\&#xA;&amp;amp;= \bigl(T(u) + U(u)\bigr) + \bigl(T(w) + U(w)\bigr) &amp;amp;\text{(because addition in $\mathbb{R}^2$}\\&#xA;&amp;amp;&amp;amp;\text{ is associative and commutative)}\\&#xA;&amp;amp;= (T\oplus U)(u) + (T\oplus U)(w). &amp;amp;\text{(by definition of $(T\oplus U)$)}&#xA;\end{align*}&#xA;And if $u\in\mathbb{R}^4$ and $\alpha\in\mathbb{R}$, then&#xA;\begin{align*}&#xA;(T\oplus U)(\alpha u) &amp;amp;= T(\alpha u) + U(\alpha u) &amp;amp;\text{(by definition of $(T\oplus U)$)}\\&#xA;&amp;amp;= \alpha T(u) + \alpha U(u) &amp;amp;\text{(because $T$ and $U$ are each linear)}\\&#xA;&amp;amp;= \alpha(T(u) + U(u)) &amp;amp;\text{(because scalar multiplication distributes over sum in $\mathbb{R}^2$)}\\&#xA;&amp;amp;=\alpha(T\oplus U)(u).&amp;amp;\text{(by definition of $T\oplus U$)}&#xA;\end{align*}&#xA;So, $T\oplus U$ is indeed a linear function from $\mathbb{R}^4$ to $\mathbb{R}^2$, and so by rights belongs to $\mathbf{V}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Likewise, to check that if $T\in\mathbf{V}$ and $\gamma\in\mathbb{R}$ then $\gamma \odot T$ is linear (I'm using $\gamma$ so we don't confuse it with the $\alpha$ that we need to check linearity) let $u,w\in\mathbb{R}^4$; then:&#xA;\begin{align*}&#xA;(\gamma\odot T)(u+w) &amp;amp;= \gamma T(u+w) &amp;amp;\text{(by definition of $\gamma\odot T$)}\\&#xA;&amp;amp;= \gamma\bigl(T(u) + T(w)\bigr) &amp;amp;\text{(because $T$ is linear)}\\&#xA;&amp;amp;= \gamma T(u) + \gamma T(w) &amp;amp;\text{(because scalar multiplication distributes in $\mathbb{R}^2$)}\\&#xA;&amp;amp;= (\gamma\odot T)(u) + (\gamma\odot T)(w). &amp;amp;\text{(by definition of $\gamma\odot T$)}&#xA;\end{align*}&#xA;So $(\gamma\odot T)(u+w) = (\gamma\odot T)(u) + (\gamma\odot T)(w)$. And, if $u\in \mathbb{R}^4$ and $\alpha\in\mathbb{R}$, then:&#xA;\begin{align*}&#xA;(\gamma\odot T)(\alpha u) &amp;amp;= \gamma T(\alpha u) &amp;amp;\text{(by definition of $\gamma\odot T$)}\\&#xA;&amp;amp;= \gamma\left(\alpha T(u)\right) &amp;amp;\text{(because $T$ is linear)}\\&#xA;&amp;amp;= (\gamma \alpha)T(u) &amp;amp;\text{(scalar multiplication is associative in $\mathbb{R}^2$)}\\&#xA;&amp;amp;= (\alpha\gamma)T(u) &amp;amp;\text{(real multiplication is commutative)}\\&#xA;&amp;amp;= \alpha\left(\gamma T(u)\right) &amp;amp;\text{(scalar multiplication is associative in $\mathbb{R}^2$)}\\&#xA;&amp;amp; = \alpha (\gamma\odot T)(u). &amp;amp;\text{(by definition of $\gamma\odot T$)}&#xA;\end{align*}&#xA;So $(\gamma\odot T)(\alpha u) = \alpha(\gamma\odot T)(u)$. Together with the additivity, this tells us that $\gamma\odot T$ is linear, and therefore is, by all rights, a member of $\mathbf{V}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So... now we have a set $\mathbf{V}$, and operations $\oplus$ and $\odot$ on $\mathbf{V}$. To check that this makes it into a vector space, we need to check the axioms. That is, we need to check that:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;If $T$ and $U$ are in $\mathbf{V}$, then $T\oplus U = U\oplus T$.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;But what does it mean for $T\oplus U$ to equal $U\oplus T$? Well, they are both functions, so we need to check that they have the same domain and the exact same value at every point of the domain; that is, we need to check that for every $u\in\mathbb{R}^4$, $(T\oplus U)(u) = (U\oplus T)(u)$.&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If $S$, $T$, and $U$ are in $\mathbf{V}$, then $(S\oplus T)\oplus U = S\oplus(T\oplus U)$.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;That is, the two functions, $(S\oplus T)\oplus U$ and $S\oplus(T\oplus U)$ have the same domain, and for every $u$ in the domain they have the same value at $u$.&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;There is a function $O$ in $\mathbf{V}$ such that for all functions $T\in \mathbf{V}$, $O\oplus T = T$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;For each $T\in\mathbf{V}$ there is a function $U\in\mathbf{V}$ such that $T\oplus U = O$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;For each $T\in\mathbf{V}$, $1\odot T = T$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;For each pair of real numbers $\alpha$ and $\beta$ and every $T\in\mathbf{V}$, $\alpha\odot(\beta \odot T) = (\alpha\beta)\odot T$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;For each $\alpha\in\mathbb{R}$ and every $T$ and $U$ in $\mathbf{V}$, $\alpha\odot(T\oplus U) = (\alpha\odot T) \oplus (\alpha\odot U)$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;For every $\alpha$ and $\beta$ in $\mathbb{R}$ and every $T\in\mathbf{V}$, $(\alpha+\beta)\odot T = (\alpha\odot T)\oplus (\beta\odot T)$.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Again, both $(\alpha+\beta)\odot T$ and $(\alpha\odot T)\oplus(\beta\odot T)$ are functions, so to check if they are equal you need to check that they have the same domain and the same value at each point of the domain.&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;If all of these conditions are satisfied, then by all rights we must acknowledge that $\mathbf{V}$, with the operations $\oplus$ for &quot;vector addition&quot; and $\odot$ as &quot;scalar multiplication&quot; &lt;strong&gt;is&lt;/strong&gt; a vector space. The &quot;vectors&quot; of this vector space are linear transformations when they are relaxing at home, but at their job in $\mathbf{V}$ they are &quot;vectors&quot; (because the elements of a vector space are called &quot;vectors&quot;). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that you have several different meanings of &quot;vector&quot; at play: there's the &quot;vectors&quot; that are elements of $\mathbf{V}$ (who, in reality, are linear maps from $\mathbb{R}^4$ to $\mathbb{R}^2$). You also have the &quot;vectors&quot; that are elements of $\mathbb{R}^4$ (at home, they are $4$-tuples of real numbers); and you have the elements of $\mathbb{R}^2$ (who are &quot;really&quot; ordered pairs of real numbers). You need to keep them straight, but it is not hard to do that if you just keep your wits about you and simpy &lt;em&gt;refuse&lt;/em&gt; to get confused.  &lt;/p&gt;&#xA;&#xA;&lt;hr/&gt;&#xA;&#xA;&lt;p&gt;So, hopefully now we know that $\mathbf{V}$ actually &lt;em&gt;is&lt;/em&gt; a vector space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Our set $V$ is a &lt;em&gt;subset&lt;/em&gt; of $\mathbf{V}$.  In order to show that $V$ is itself a vector space (with the same operations that we have for $\mathbf{V}$, we are in the situation of checking if something is a &lt;em&gt;subspace&lt;/em&gt;: because $\oplus$ is associative and commutative for &lt;em&gt;all&lt;/em&gt; functions in $\mathbf{V}$, then it is also so in $V$; similarly with axioms 5 through 8 for $\odot$ and $\oplus$. Really, the only sticking points for $V$ to be a vector space are four things: what I call the &quot;hidden axioms&quot; of a vector space, and axioms 3 and 4. Explicitly:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;We know that if we take $T$ and $U$ in $\mathbf{V}$, then $T\oplus U$ is also in $\mathbf{V}$. But for $V$ to be a vector space, we need to make sure that if we take $T$ and $U$ in $V$, then $T\oplus U$ is not just going to be in $\mathbf{V}$, but actually in $V$ (we know the sum is in the same &quot;state&quot;, but we actually  need to make sure it lives in the same &lt;em&gt;city&lt;/em&gt;). This is one of the &quot;hidden axioms&quot; of vector spaces, because it is not listed in the usual list; it is implicit when we talk about addition of vectors being an &quot;operation&quot;. So we need to check that if $T$ and $U$ are in $V$ (and not merely in $\mathbf{V}$), then $T\oplus U$ is &lt;em&gt;also&lt;/em&gt; in $V$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Likewise, we know that if $T$ is in $\mathbf{V}$ and $\alpha\in\mathbb{R}$, then $\alpha\odot T$ is going to be in $\mathbf{V}$. So if we start with $T\in V$, then certainly $\alpha\odot T$ is going to be in $\mathbf{V}$, but we actually need it to be in the more exclusive neighborhood of $V$. So we need to check &lt;em&gt;that&lt;/em&gt;: if $T\in V$ and $\alpha\in\mathbb{R}$, we need to make sure that $\alpha\odot T$ is &lt;em&gt;also&lt;/em&gt; in $V$ (and not merely in $\mathbf{V}$, which we know is the case).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Axiom 3 is not immediate either: we know that there is a vector $O$ in $\mathbf{V}$ such that for every $T\in\mathbf{V}$, $O\oplus T = T$. In particular, if we take a $T\in V$, then we wil also have $O\oplus T=T$. But the problem is that for $V$ to be a vector space, we need this special vector $O$ to actually be in $V$, and not somewhere outside of $V$. So we need to check that this happens.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Axiom 4 is also not immediate: if $T\in V$, then since $T$ is also in $\mathbf{V}$ we know there is a $U$ in $\mathbf{V}$ such that $T\oplus U = O$. But for $V$ to be a vector space, we need to make sure that $U$ is &lt;em&gt;also&lt;/em&gt; in $V$, and not just in $\mathbf{V}$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;But if we can check these four new items, then we will have that $V$ is also a vector space. (In fact, there are shortcuts, which you may or may not have seen if you have already studied &quot;subspaces&quot;; you are really trying to show that $V$ is a subspace of $\mathbf{V}$, so if you know the shortcuts, you can use them here, keeping in mind that &quot;vectors&quot; means &quot;elements of $V$&quot; (or $\mathbf{V}$), etc.). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, how do we check that if $T$ and $U$ are in $V$, then $T\oplus U$ is in $V$? Well, we know $T\oplus U$ is linear, and maps $\mathbb{R}^4$ to $\mathbb{R}^2$ (because we know that $T\oplus U$ is in $\mathbf{V}$). The only extra membership requirement to be in $V$ is that you &lt;em&gt;must&lt;/em&gt; have that $(T\oplus U)(v) = 0$ (that is, that $(T\oplus U)(1,1,0,1) = (0,0)$).  Since we are assuming that $T$ and $U$ are &lt;em&gt;each&lt;/em&gt; in $V$, that means that $T(v)=0$ and that $U(v)=0$. But that means that&#xA;\begin{align*}&#xA;(T\oplus U)(v) &amp;amp;= T(v) + U(v) &amp;amp;\text{(by definition of $T\oplus U$)}\\&#xA;&amp;amp;= 0 + 0 &amp;amp;\text{(since $T$ and $U$ are each in $V$)}\\&#xA;&amp;amp;= 0. &amp;amp;\text{(well... duh)}&#xA;\end{align*}&#xA;So... guess what? $T\oplus U$ satisfies all membership requirements for belonging to $V$, so it is in $V$! Great!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then we need to check the rest of the conditions similarly.&lt;/p&gt;&#xA;&#xA;&lt;hr/&gt;&#xA;&#xA;&lt;p&gt;What about (b)? There are a couple of ways; following the suggestion, remember that the elements of $V$ are &lt;em&gt;really&lt;/em&gt; linear transformations. To &quot;know&quot; what a linear transformation is, it is enough to know what it does to a basis, any basis. So, let's start by taking the vector $v$, and extending $\{v\}$ to a basis for $\mathbb{R}^4$; say $\beta=[v, v_2, v_3, v_4]$. In order for me to give you an element $f$ of $\mathbf{V}$, it is enough for me to tell you what $f(v)$, $f(v_2)$, $f(v_3)$, and $f(v_4)$ are. From that information, and knowing that it is a linear map, you can figure out who $f$ is. And, what is more, any values I give you will give you an element of $\mathbf{V}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, what do I need to give you in order to describe an element of $V$? I still need to tell you what $f(v)$, $f(v_2)$, $f(v_3)$ and $f(v_4)$ are, but as it happens, in order for $f$ to be in $V$ we need $f(v)=0$. So I just need to tell you what happens to $v_2$, $v_3$, and $v_4$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might be tempted to think this means the dimension of $V$ is $3$, because you have three degrees of freedom, but in fact this is not the case: each choice of $f(v_2)$, $f(v_3)$ and $f(v_4)$ is a &lt;em&gt;vector&lt;/em&gt; in $\mathbb{R}^2$; so you actually have to specify &lt;em&gt;two&lt;/em&gt; things for &lt;em&gt;each&lt;/em&gt; of them. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This should suggest what the answer &lt;em&gt;should&lt;/em&gt; be; you still have to actually prove it, though.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-24T21:27:35.580" LastActivityDate="2010-11-24T21:27:35.580" CommentCount="20" />
  <row Id="11624" PostTypeId="1" CreationDate="2010-11-24T02:08:54.810" Score="5" ViewCount="158" Body="&lt;p&gt;I'm trying to classify (at least as fully as possible) the values of $t$ in $(0,\pi/2)$ for which the following equation has a solution for some natural number, $n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac1{n}\sin(nt)=\frac1{n+2}\sin((n+2)t)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anyone know of any results that will help me?&lt;/p&gt;&#xA;" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-24T02:35:37.193" LastActivityDate="2010-11-24T04:38:51.990" Title="When is $\frac1{n}\sin(nt)=\frac1{n+2}\sin((n+2)t)$?" Tags="&lt;number-theory&gt;&lt;trigonometry&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="11625" PostTypeId="2" ParentId="11622" CreationDate="2010-11-24T02:09:45.777" Score="5" ViewCount="" Body="&lt;p&gt;I agree that you don't know that there is a closed neighborhood of $p$ contained in $Z^c$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, this perhaps will work: Since $p$ is not in the intersection $\cap \mathcal{N}$, there is a closed set $V$ with $Z\subseteq V^{\circ}$ and with $p\notin V$. Since $V$ is closed, then $V^c$ is an open set containing $p$; and the interior of $V$ is an open set containing $Z$. Of course, $V^c$ is disjoint from $V^{\circ}$, so don't $V^{\circ}$ and $V^c$ give you open sets, disjoint, one containing $Z$ and the other containing the point $p\notin Z$?&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-11-24T02:09:45.777" CommentCount="3" />
  <row Id="11626" PostTypeId="1" CreationDate="2010-11-24T02:38:16.783" Score="4" ViewCount="575" Body="&lt;p&gt;I have a few weeks off from school soon, and I was hoping to self-study a bit of algebra. I don't think this question has been asked on here before, but does anyone have any suggestions for algebra books which are particularly suited to self-study? &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Here is some background and motivation, if it's helpful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm currently a junior in high school, but I have some familiarity with groups, rings, polynomials and fields, as I went through Fraleigh's book. I liked the writing style of that book, but I felt many of the problems were not helpful. I would like to get better at algebra with some other book, particularly one that's aimed at someone who already has a little familiarity with the objects, but is by no means completely comfortable with them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of my longer-term goals is to learn a bit of algebraic number theory, as right now I've only been studying elementary number theory. That's why I feel it would be good to learn more algebra first. (I'd also be interested in suggestions of other good subjects to learn prior to tackling algebraic number theory.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At first I was considering using one of Lang's books, since they seem to be pretty universal, but I've heard rumors that they are very terse and not good for learning by oneself. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any suggestions would be very appreciated, thank you!&lt;/p&gt;&#xA;" OwnerUserId="1583" LastActivityDate="2010-11-24T04:33:54.587" CommunityOwnedDate="2010-11-24T06:01:08.860" Title="Good books for self-studying algebra?" Tags="&lt;big-list&gt;&lt;books&gt;&lt;self-learning&gt;" AnswerCount="3" CommentCount="4" FavoriteCount="9" />
  <row Id="11627" PostTypeId="2" ParentId="11626" CreationDate="2010-11-24T02:52:10.173" Score="4" ViewCount="" Body="&lt;p&gt;For Algebra you can look at these books:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;Topics in Algebra&lt;/em&gt; by I.N. Herstein&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;Abstract Algebra&lt;/em&gt; by Dummit and Foote&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;Algebra&lt;/em&gt; by Michael Artin&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;Algebra&lt;/em&gt; by T.Hungerford (Springer)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;Lectures in Abstract Algebra&lt;/em&gt; by N.Jacobson (Has 3 volumes!)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;Algebra&lt;/em&gt; by Anthony Knapp. (2 Volumes.)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;My feeling of &lt;em&gt;Herstein&lt;/em&gt; is it has lot of problems which are challenging. For theory part i would like to use &lt;em&gt;Dummit and Foote&lt;/em&gt;. Artin's Algebra is very well written and contains a lot of Linear Algebra. Anthony Knapps treatment of Algebra is very comprehensive, and contains a lot of Algebra. Since your aim is to read &lt;em&gt;Algebraic Number Theory&lt;/em&gt; you might want to learn some &lt;em&gt;Galois theory&lt;/em&gt; also for which there many good books like:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;Lectures in Galois theory&lt;/em&gt; by Emil Artin&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;Field theory and its Classical problems&lt;/em&gt; by Charles Hadlock.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;Galois theory&lt;/em&gt; by J.Rotmain (Springer.)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" LastActivityDate="2010-11-24T02:59:32.057" CommunityOwnedDate="2010-11-24T06:01:08.860" CommentCount="2" />
  <row Id="11628" PostTypeId="2" ParentId="11611" CreationDate="2010-11-24T02:58:33.177" Score="0" ViewCount="" Body="&lt;p&gt;The 'simplest' possibility is f(x) := x-3 and g(x) := x^2+x-3.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is the more natural of the 2 pairs of integer polynomials of degree at most 2 that minimize the sum of squares of their coefficients while composing to give x^2+x-6.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and apparently it's not this easy to get back the account I created on another computer.&lt;/p&gt;&#xA;" OwnerUserId="3888" LastActivityDate="2010-11-24T02:58:33.177" CommentCount="0" />
  <row Id="11629" PostTypeId="2" ParentId="11610" CreationDate="2010-11-24T03:03:10.427" Score="1" ViewCount="" Body="&lt;p&gt;It's a consequence of the so-called triangle inequality.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-24T03:03:10.427" />
  <row Id="11630" PostTypeId="2" ParentId="11624" CreationDate="2010-11-24T03:11:22.560" Score="0" ViewCount="" Body="&lt;p&gt;EDIT: This solution is wrong! The error is highlighted below (thanks for Zaricuse and Rahul).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is no solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given $t$, define a function $$f(n) = \frac{\sin(nt)}{n}.$$&#xD;&#xA;Its derivative is $$f&amp;#39;(n) = \frac{\cos(nt)nt - \sin(nt)}{n^2}.$$&#xD;&#xA;If $f(n) = f(n+2)$ then the derivative must have a zero somewhere in between, say $f&amp;#39;(m) = 0$. That implies $$\tan(mt) = mt.$$ However, for $x \in (0,\pi)$ we have $\tan(x) &amp;gt; x$ (consider the Taylor expansion of $\tan$, for example). &lt;b&gt;The unfortunate error is that although $t &amp;lt; \pi$, $x = mt$ is not necessarily less than $\pi$.&lt;/b&gt;&lt;/p&gt;&#xA;" OwnerUserId="1277" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2010-11-24T03:58:33.910" LastActivityDate="2010-11-24T03:58:33.910" CommentCount="3" />
  <row Id="11631" PostTypeId="2" ParentId="11618" CreationDate="2010-11-24T03:15:20.887" Score="2" ViewCount="" Body="&lt;p&gt;EDIT&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof by induction&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$n=1$ is valid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Supose valid by n, then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$a^{n+1}-b^{n+1}=a(a^{n})+b(b^{n})$$, using the hipotesis :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$a(a^{n})+b(b^{n})=a\left[b^{n}+(b-a)\displaystyle\sum\limits_{k=0}^{n-1} a^{k}b^{n-1-k}\right] + b\left[a^{n}-(b-a)\displaystyle\sum\limits_{k=0}^{n-1} a^{k}b^{n-1-k}\right]=$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left[ab^{n}+(b-a)a\displaystyle\sum\limits_{k=0}^{n-1} a^{k}b^{n-1-k}\right] + \left[a^{n}b-(b-a)b\displaystyle\sum\limits_{k=0}^{n-1} a^{k}b^{n-1-k}\right]=$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left[ab^{n}+(b-a)\displaystyle\sum\limits_{k=0}^{n-1} a^{k+1}b^{n-1-k}\right] + \left[a^{n}b-(b-a)\displaystyle\sum\limits_{k=0}^{n-1} a^{k}b^{n-k}\right]=$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left[ab^{n}+(b-a)\displaystyle\sum\limits_{k=0}^{n-1} a^{k+1}b^{n-1-k}+(b-a)b^{n}-(b-a)b^{n}\right] +$$&#xD;&#xA;$$ \left[a^{n}b-(b-a)\displaystyle\sum\limits_{k=0}^{n-1} a^{k}b^{n-k}-(b-a)a^{n}+(b-a)a^{n}\right]=$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left[(b-a)[\displaystyle\sum\limits_{k=0}^{n-1} a^{k+1}b^{n-1-k}+b^{n}]+b^{n+1}\right] +\left[(b-a)[\displaystyle\sum\limits_{k=0}^{n-1} a^{k}b^{n-k}+a^{n}]-a^{n+1}\right] +$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left[(b-a)\displaystyle\sum\limits_{k=0}^{n} a^{k}b^{n-k}+b^{n+1}\right] +\left[(b-a)\displaystyle\sum\limits_{k=0}^{n} a^{k}b^{n-k}-a^{n+1}\right] =$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$-a^{n+1}+b^{n+1}+2\left[(b-a)\displaystyle\sum\limits_{k=0}^{n} a^{k}b^{n-k}\right] =$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus:&#xA;$$a^{n+1}-b^{n+1}=-a^{n+1}+b^{n+1}+2\left[(b-a)\displaystyle\sum\limits_{k=0}^{n} a^{k}b^{n-k}\right] $$, then &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$2(a^{n+1}-b^{n+1})=2\left[(b-a)\displaystyle\sum\limits_{k=0}^{n} a^{k}b^{n-k}\right]$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;thus:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$a^{n+1}-b^{n+1}=\left[(b-a)\displaystyle\sum\limits_{k=0}^{n} a^{k}b^{n-k}\right]$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So $n+1$ is valid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Complete the proof&lt;/p&gt;&#xA;" OwnerUserId="3421" LastEditorUserId="3421" LastEditorDisplayName="" LastEditDate="2010-11-24T03:59:39.820" LastActivityDate="2010-11-24T03:59:39.820" CommentCount="1" />
  <row Id="11632" PostTypeId="2" ParentId="11618" CreationDate="2010-11-24T03:27:39.387" Score="4" ViewCount="" Body="&lt;p&gt;Someone should mention the &quot;polynomial multiplication&quot; or &quot;telescoping&quot; proof, which may be viewed as a variant of the &quot;geometric series&quot; method.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align*} (a-b)\sum_{k=0}^{n-1} a^k b^{n-1-k} &amp;amp;= \sum_{k=0}^{n-1} a^{k+1} b^{n-1-k} - \sum_{k=0}^{n-1} a^k b^{n-k} \\ &amp;amp;= \sum_{k=1}^n a^k b^{n-k} - \sum_{k=0}^{n-1} a^k b^{n-k} = a^n - b^n. \end{align*}$$&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-24T03:27:39.387" />
  <row Id="11633" PostTypeId="2" ParentId="11624" CreationDate="2010-11-24T03:32:53.173" Score="3" ViewCount="" Body="&lt;p&gt;I don't know if there is any closed form for the solutions, but by replacing $\sin\big((n+2)t\big)$ with $\sin nt \cos 2t + \cos nt \sin 2t$ and rearranging, your problem is equivalent to finding the solutions of $$\tan nt = \frac{n \sin 2t}{n + 2 - n \cos 2t}.$$ For any $n &amp;gt; 0$, the RHS is bounded, while the LHS crosses from $-\infty$ to $\infty$ exactly $2n$ times over a $2 \pi$ interval. Some experimentation suggests that there are indeed exactly $2n$ solutions; possibly you could bound the slopes of the two expressions to make this rigorous. This would also let you find bounds for the locations of the solutions, and easily find them numerically through a bisection procedure.&lt;/p&gt;&#xA;" OwnerUserId="856" LastEditorUserId="856" LastEditorDisplayName="" LastEditDate="2010-11-24T04:03:06.170" LastActivityDate="2010-11-24T04:03:06.170" CommentCount="5" />
  <row Id="11634" PostTypeId="1" AcceptedAnswerId="11649" CreationDate="2010-11-24T03:38:12.787" Score="1" ViewCount="299" Body="&lt;p&gt;Assume $f$ has a finite derivative and $|f&amp;#39;(x)| \leq y &amp;lt; 1$ for all $x \in (a,b)$ &lt;br&gt;&#xA;$f$ is continuous and $a \leq f(x) \leq b$ for all $x \in [a,b]$. Prove $f$ has a unique fixed point in $[a,b]$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So far I have for every c in (a,b) |f'(c)| ≤ y&#xA;=&gt; lim x-&gt;c |f(x) - f(c)|/|x-c| ≤ y&#xA;=&gt; lim x-&gt;c |f(x) - f(c)| ≤ y lim x-&gt;c |x-c|&lt;br&gt;&#xA;Would that be the definition of a contractive map in R?&lt;br&gt;&#xA;Therefore by Banach Fixed Point Theorem, f has a unique fixed point.&lt;br&gt;&#xA;Can I prove Banach's theorem using the mean value theorem?&lt;/p&gt;&#xA;" OwnerUserId="3333" LastActivityDate="2010-11-24T04:37:01.133" Title="Prove Fixed Point Theorem using the Mean Value Theorem" Tags="&lt;real-analysis&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="11635" PostTypeId="2" ParentId="11626" CreationDate="2010-11-24T03:39:27.930" Score="0" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/1568814542&quot; rel=&quot;nofollow&quot;&gt;A Survey of Modern Algebra&lt;/a&gt; by Birkhoff and Maclane.&lt;/p&gt;&#xA;" OwnerUserId="1417" LastActivityDate="2010-11-24T03:39:27.930" CommunityOwnedDate="2010-11-24T06:01:08.860" CommentCount="1" />
  <row Id="11636" PostTypeId="2" ParentId="11567" CreationDate="2010-11-24T03:39:59.053" Score="2" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\ $ The simple equivalences given below demonstrate that both sides of your equation have the same common divisors $\rm\:d\:,\ $ therefore they have the same &lt;em&gt;greatest&lt;/em&gt; common divisor. $\ \ $ &lt;strong&gt;QED&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\quad\quad mod\ d:\ \ b^M,\:b^N\equiv 1\iff ord(b)\:|\:M,N\iff ord(b)\:|\:gcd(M,N)\iff b^{gcd(M,N)}\equiv 1\ $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;NOTE&lt;/strong&gt; $\ $ The conceptual structure at the heart of this simple proof is the ubiquitous &lt;strong&gt;order ideal.&lt;/strong&gt; See my &lt;a href=&quot;http://math.stackexchange.com/questions/4713/if-the-order-divides-a-prime-p-then-the-order-is-p-or-1/4715#4715&quot;&gt;post here&lt;/a&gt; for more on this and the more familiar additive form of a &lt;strong&gt;denominator ideal.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More generally $\rm\ gcd(f(m), f(n))\ =\ f(gcd(m,n))\ \ \ if\ \ \ f(n)\ \equiv\ f(n-m)\ \ (mod\ f(m)),\ \ \ f(0)\ =\ 0\:.\ $ See my &lt;a href=&quot;http://math.stackexchange.com/questions/7473/number-theory-proving-question/7561#7561&quot;&gt;post here&lt;/a&gt; for a simple inductive proof.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact there is a q-analog: the result also holds true for polynomials $\rm\ \ f(n)\ =\ (x^n-1)/(x-1)\:,\ $ and $\rm\ x\to 1\ $ yields the integer case (Bezout identity) - see my &lt;a href=&quot;http://math.stackexchange.com/questions/7473/number-theory-proving-question/7479#7479&quot;&gt;post here&lt;/a&gt; for the simple proof.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2011-03-16T19:42:16.707" LastActivityDate="2011-03-16T19:42:16.707" />
  <row Id="11639" PostTypeId="1" AcceptedAnswerId="11641" CreationDate="2010-11-24T03:48:54.650" Score="1" ViewCount="356" Body="&lt;p&gt;Is there a fundamental relationship between circuits and cycles in an arbitrary graph? Are they mutually independent objects?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="5664" LastActivityDate="2010-11-24T15:55:15.157" Title="Eulerian Circuits and Hamiltonian Cycles" Tags="&lt;graph-theory&gt;" AnswerCount="3" />
  <row Id="11640" PostTypeId="1" AcceptedAnswerId="11644" CreationDate="2010-11-24T03:50:07.473" Score="2" ViewCount="138" Body="&lt;p&gt;As is well known, if $x,y$ are natural numbers then their GCD $(x,y)$ has a representation $(x,y) = ax + by$ where $a,b$ are natural. Now let's prove that if $(x,y)=1$ then $(xy,x+y) = 1$. Starting from $ax+by=1$, $$1 = (ax+by)^2 = a^2x^2 + b^2y^2 + 2abxy = (a^2x+b^2y)(x+y) - (a-b)^2xy.$$ The reverse implication is trivial in this method. The proof can be generalized to show that if $(x,y,z)=1$ then $(xyz,xy+xz+yz,x+y+z)=1$ and vice versa.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;i&gt;Can this idea be molded into a reasonably strong formal proof system for some fragment of number theory?&lt;/i&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We're competing with divisibility proofs, which we'd like to forbid somehow. For example, the original statement can be proven as follows. Suppose $p$ is a prime dividing $(xy,x+y)$. Since $p|xy$ and $p$ is prime, either $p|x$ or $p|y$. From $p|x+y$ we get that $p|(x,y)$.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-24T04:13:52.040" Title="A proof system for GCD?" Tags="&lt;number-theory&gt;&lt;logic&gt;" AnswerCount="1" />
  <row Id="11641" PostTypeId="2" ParentId="11639" CreationDate="2010-11-24T03:53:40.013" Score="3" ViewCount="" Body="&lt;p&gt;A cycle with two chords has a Hamiltonian cycle but no Eulerian circuit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A lemniscate ($\infty$, two cycles pasted at a vertex) has an Eulerian circuit but no Hamiltonian cycle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A cycle has both a Hamiltonian cycle and an Eulerian circuit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A star with at least 3 edges has neither a Hamiltonian cycle nor an Eulerian circuit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wikipedia describes the graphs which have Eulerian circuits; Hamiltonian cycles are much more complicated, and in particular it is very probable that there's no simple characterization of graphs that have them (unless P=NP).&lt;/p&gt;&#xA;" OwnerUserId="1277" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2010-11-24T04:09:21.113" LastActivityDate="2010-11-24T04:09:21.113" CommentCount="5" />
  <row Id="11642" PostTypeId="2" ParentId="11634" CreationDate="2010-11-24T03:55:46.253" Score="3" ViewCount="" Body="&lt;p&gt;To show that there is some fixed point, consider $g(x) = f(x)-x$. Then $g(a)\geq 0$ and $g(b) \leq 0$. To show that it is unique, use the mean-value theorem for the two purported fixed points.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-24T03:55:46.253" CommentCount="4" />
  <row Id="11643" PostTypeId="2" ParentId="11639" CreationDate="2010-11-24T04:08:14.900" Score="0" ViewCount="" Body="&lt;p&gt;By definition a &lt;em&gt;path graph&lt;/em&gt; cannot have an Eulerian circuit or a Hamiltonian cycle. A &lt;em&gt;loop&lt;/em&gt; graph (consisting of one edge and one vertex) has both an Eulerian circuit and a Hamiltonian cycle. As above, there are examples where a graph might have one but not the other. The answer to your question is that there is &lt;em&gt;no&lt;/em&gt; fundamental relationship between the two.&lt;/p&gt;&#xA;" OwnerUserId="2720" LastEditorUserId="2720" LastEditorDisplayName="" LastEditDate="2010-11-24T04:46:12.090" LastActivityDate="2010-11-24T04:46:12.090" />
  <row Id="11644" PostTypeId="2" ParentId="11640" CreationDate="2010-11-24T04:08:33.573" Score="4" ViewCount="" Body="&lt;p&gt;First, work with ideals, which provides much more power and transparency. For example, in terms of ideals your first proof is simply:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\quad (x,y) \ \supset\ (xy,\ x+y)\ \supset\ (x,y)^2\ \ $ therefore $\rm\ \ (xy,\ x+y)\ =\ 1\ \iff\ (x,y)\ =\ 1$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice how employing ideals has eliminated obfuscatory information such as the extraneous coefficients $\ a,\:b\ $ in the original proof. That done, the innate structure becomes much clearer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second, there are various generalizations of the Grobner basis algorithm over Euclidean domains. Whether or not they will suffice for your application is hard to say without knowing further details. &lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-11-24T04:13:52.040" LastActivityDate="2010-11-24T04:13:52.040" CommentCount="2" />
  <row Id="11645" PostTypeId="1" AcceptedAnswerId="11652" CreationDate="2010-11-24T04:10:55.083" Score="1" ViewCount="143" Body="&lt;p&gt;I was wondering if we have a finite group $G$ with subgroups $H$ and $K$.  With $X:=G/H$.&#xA;Is there a simple way of finding $X^K$ and when will $X^K=\emptyset$? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also does this have any baring on when $G/H≅ G/K$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(This is the first question I've ever asked here so apologies for errors etc.)&lt;/p&gt;&#xA;" OwnerUserId="3890" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-24T04:33:19.707" LastActivityDate="2010-11-25T01:14:06.170" Title="Fixed points of finite group." Tags="&lt;group-theory&gt;" AnswerCount="1" CommentCount="9" />
  <row Id="11646" PostTypeId="2" ParentId="11624" CreationDate="2010-11-24T04:20:09.017" Score="2" ViewCount="" Body="&lt;p&gt;EDIT: See J.M.'s comment to Rahul's answer: it's better to use the &quot;other&quot; Chebyshev polynomials.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Square your identity: $$(n+2)^2 \sin^2 nt  = n^2 \sin^2 (n+2)t.$$ Convert from sines to cosines: $$(n+2)^2 + n^2 \cos^2 (n+2)t = n^2 + (n+2)^2 \cos^2 nt.$$ Now let $x = \cos t$ to get $$(n+2)^2 + n^2 T_{n+2}(x)^2 = n^2 + (n+2)^2 T_n(x)^2,$$ where the $T_m$ are the Chebyshev polynomials. We get a polynomial equation of degree $2(n+2)$ in $x$ which you can solve to find all solutions (you will need to check that the signs of the sines match).&lt;/p&gt;&#xA;" OwnerUserId="1277" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2010-11-24T04:38:51.990" LastActivityDate="2010-11-24T04:38:51.990" CommentCount="1" />
  <row Id="11648" PostTypeId="2" ParentId="11626" CreationDate="2010-11-24T04:33:54.587" Score="4" ViewCount="" Body="&lt;p&gt;I learned algebra (before taking Artin's course) by self-studying Nathan Jacobson's &lt;em&gt;Basic Algebra I and II&lt;/em&gt;. This is a beautiful algebra textbook. However, it may help to supplement it with other presentations since it is a bit too concise in some places. A good choice for a supplement is Karlheinz Spindler's &lt;em&gt;Abstract algebra with applications in two volumes&lt;/em&gt;. It has many interesting well-chosen motivational examples.&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-11-24T04:33:54.587" CommunityOwnedDate="2010-11-24T06:01:08.860" CommentCount="1" />
  <row Id="11649" PostTypeId="2" ParentId="11634" CreationDate="2010-11-24T04:37:01.133" Score="2" ViewCount="" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;$\lim_{x\to c} |f(x) - f(c)|\leq y\ \lim_{x\to c} |x-c|$&#xD;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Would that be the definition of a contractive map in R?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No, this is just the statement that $f$ is continuous at $c$, because the right-hand side is $0$.  The fixed point theorem will apply, but to show that $f$ is contractive you will want to use the mean value theorem.  Suppose that $a\leq z\lt x\leq b$.  By the mean value theorem, there is a $c$ in $(z,x)$ such that $f&amp;#39;(c)=\frac{f(x)-f(z)}{x-z}$.  Apply absolute values, rearrange, and use the hypothesis on the derivative to conclude that $f$ is contractive.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-11-24T04:37:01.133" CommentCount="1" />
  <row Id="11650" PostTypeId="1" AcceptedAnswerId="11667" CreationDate="2010-11-24T04:42:09.430" Score="9" ViewCount="294" Body="&lt;p&gt;Quote from Don Zagier (Mathematicians: An Outer View of the Inner World):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;    I like explicit, hands-on formulas. To me they have a beauty of their own. They can be deep or not. As an example, imagine you have a series of numbers such that if you add 1 to any number you will get the product of its left and right neighbors. Then this series will repeat itself at every fifth step! For instance, if you start with 3, 4 then the sequence continues: 3, 4, 5/3, 2/3, 1, 3, 4, 5/3, etc. The difference between a mathematician and a nonmathematician is not just being able to discover something like this, but to care about it and to be curious about why it's true, what it means, and what other things in mathematics it might be connected with. In this particular case, the statement itself turns out to be connected with a myriad of deep topics in advanced mathematics: hyperbolic geometry, algebraic K-theory, the Schrodinger equation of quantum mechanics, and certain models of quantum field theory. I find this kind of connection between very elementary and very deep mathematics overwhelmingly beautiful.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The recursion described here is a(n+1)=(1+a(n))/a(n-1) with a(0)=1, a(1)=3.&#xA;It's fairly straightforward to show that for every positive a(0),a(1) the sequence repeats after 5 steps (and a(0)=a(1)=golden ratio is a fixed point). One can also think of this recursion as the planar map f(x,y)=((1+x)/y,x), and every point in the plane is a period-5 point of this map.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what is the connection with the deep topics listed in the quote? &lt;/p&gt;&#xA;" OwnerUserId="3891" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-11-24T17:37:13.890" LastActivityDate="2010-12-21T10:16:22.013" Title="What is the connection of the sequence 3, 4, 5/3, 2/3, 1 with deep topics?" Tags="&lt;sequences-and-series&gt;&lt;hyperbolic-geometry&gt;&lt;algebraic-k-theory&gt;" AnswerCount="3" CommentCount="0" />
  <row Id="11651" PostTypeId="1" AcceptedAnswerId="11655" CreationDate="2010-11-24T04:56:57.320" Score="1" ViewCount="123" Body="&lt;p&gt;What would be an example of a matrix that would not have a $LDL^T$ decomposition? This is a trivial case but I was thinking of a zero matrix which would result in L being an identity matrix and D would be a zero matrix. When you compute $LDL^T$ it still gives you the original zero matrix so I don't believe this is correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What about a singular matrix?&lt;/p&gt;&#xA;" OwnerUserId="1880" LastEditorUserId="1880" LastEditorDisplayName="" LastEditDate="2010-11-24T05:03:09.293" LastActivityDate="2010-11-24T05:20:43.807" Title="Symmetric Matrix without $LDL^T$ Decomposition" Tags="&lt;linear-algebra&gt;&lt;matrices&gt;" AnswerCount="1" CommentCount="12" />
  <row Id="11652" PostTypeId="2" ParentId="11645" CreationDate="2010-11-24T05:00:03.433" Score="4" ViewCount="" Body="&lt;p&gt;I'm assuming that $G/H$ represents the left cosets of $H$ in $G$, that is the sets of the form $gH$ with $g\in G$, with the action of $G$ on $X$ being given by $x\cdot(gH) = xgH$ (left multiplication), and that $X^K$ is the set of fixed points of $K$; that is, all $gH\in X$ such that for all $k\in K$, $k(gH)=gH$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $kgH = gH$, then $g^{-1}kg\in H$. Thus, $gH$ is fixed by all $k\in K$ if and only if $g^{-1}Kg\subseteq H$. So&#xA;$$X^K = \{ gH\in G/H \mid g^{-1}Kg\subseteq H\}.$$&#xD;&#xA;Note that this is well defined: if $gH = g&amp;#39;H$, then $g&amp;#39;^{-1}g\in H$, so if $g^{-1}Kg\subseteq H$, then &#xA;$$g&amp;#39;^{-1}Kg&amp;#39; = (g&amp;#39;^{-1}g)(g^{-1}Kg)(g&amp;#39;^{-1}g)^{-1}\subseteq g&amp;#39;^{-1}gH(g&amp;#39;^{-1}g)^{-1}=H.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That gives an easy way to find $X^K$ (by looking at the conjugates of $K$) and to figure out if $X^K=\emptyset$ (if no conjugate of $K$ is contained in $H$; e.g., if the order of $K$ is larger than the order of $H$; if $H$ is normal and $K$ is not contained in $H$; many other criteria).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, when are the $G$-sets $G/H$ and $G/K$ isomorphic? First, you must have $[G:H]=[G:K]$. Also, there must be a bijection $\psi\colon G/H\to G/K$ such that for all $x,g\in G$, $\psi(gxH) = g\psi(xH)$. In particular, for all $h\in H$ we must have $\psi(H) = \psi(hH) = h\psi(H)$. If $\psi(H) = xK$, then that means that $HxK = xK$, or that $x^{-1}Hx\subseteq K$. In particular, we need a conjugate of $H$ to be contained in $K$. Since $G$ is finite and $H$ and $K$ must have the same order, that means that $H$ is a conjugate of $K$. If $x^{-1}Hx=K$, then for any coset $gH$ we must have $\psi(gH) = g\psi(H) =gxK$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conversely, if $H$ is a conjugate of $K$, $x^{-1}Hx = K$, then let $\psi\colon G/H\to G/K$ be the map that sends the coset $gH$ to the coset $gxK$. First, I claim this is well defined: if $yH=zH$, we need to show that $yxK = zxK$. Since $z^{-1}y\in H$, then $x^{-1}z^{-1}yx\in x^{-1}Hx = K$, so $yxK=zxK$, as claimed. If $gH\in G/H$ and $a\in G$, then $\psi(a\cdot gH) = \psi(agH) = agxK = a\cdot gxK = a\psi(gH)$, so $\psi$ is a $G$-set homomorphism. The map $G/K\to G/H$ given by $gK\mapsto gx^{-1}K$ is also a $G$-set homomorphism (as $xKx^{-1}=H$), and is the inverse of the previous map, so $\psi$ is an isomorphism. Thus, the $G$-set $G/H$ is isomorphic to the $G$-set $G/K$ if and only if $H$ and $K$ are conjugate. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added:&lt;/strong&gt; It is interesting to note that the definition of $\psi$ above &lt;em&gt;does&lt;/em&gt; depend on the choice of $x$; if $x$ and $y$ are two distinct elements such that $xHx^{-1}=yHy^{-1}=K$, then the maps $\psi(gH) = gxK$ and $\phi(gH) = gyK$ are the same map if and only if $y^{-1}x\in K$, that is, if and only if $y$ and $x$ are in the same coset of $K$. So you get one map for each coset of $K$ that intersects the set of elements that conjugate $H$ to $K$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note, however, that this does not give you any light into the situation in which $H$ and $K$ are normal, and you are interested in knowing whether the quotient groups $G/H$ and $G/K$ are isomorphic &lt;em&gt;as groups&lt;/em&gt;: obviously, if $H$ and $K$ are normal, then they are conjugate if and only if they are equal, but it is trivial to find examples of two distinct normal sugroups of a finite group $G$ that give you isomorphic quotients. For example, take any two distinct nontrivial proper subgrups of the Klein $4$-group.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-25T01:14:06.170" LastActivityDate="2010-11-25T01:14:06.170" CommentCount="2" />
  <row Id="11653" PostTypeId="2" ParentId="11650" CreationDate="2010-11-24T05:04:38.730" Score="1" ViewCount="" Body="&lt;p&gt;Google &quot;cluster algebra&quot;, e.g. see this paper &lt;a href=&quot;http://arxiv.org/PS_cache/math/pdf/0602/0602408v5.pdf&quot; rel=&quot;nofollow&quot;&gt;Combinatorial interpretations for rank-two cluster algebras of affine type.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-11-24T05:04:38.730" CommentCount="9" />
  <row Id="11654" PostTypeId="1" AcceptedAnswerId="11716" CreationDate="2010-11-24T05:17:50.057" Score="1" ViewCount="218" Body="&lt;p&gt;Let $f(x,y)$, with $0\le x,y\le 1$, be integrable in $y$ for each $x$. Further suppose $\partial f(x,y)/\partial x$ is a bounded function of $(x,y)$. Show that $\partial f(x,y)/\partial x$ is a measurable function of $y$ for each $x$ and&#xA;$$\frac{d}{dx}\int_0^1{f(x,y)\,dy} = \int_0^1{\frac{\partial}{\partial x}f(x,y)\,dy}\,.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wish I had some work to really show. At this point all I've done is written down each side of the desired equation in terms of limits, and I know ultimately this problem is actually about passing limits inside integrals... But we've done almost no examples of this sort of thing (we literally just finished the limiting theorems today: Fatou, MCT, DCT, etc), and I'm not sure where to begin.&lt;/p&gt;&#xA;" OwnerUserId="2313" LastActivityDate="2010-11-26T04:27:09.650" Title="Passing the derivative inside the integral" Tags="&lt;real-analysis&gt;&lt;homework&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="11655" PostTypeId="2" ParentId="11651" CreationDate="2010-11-24T05:20:43.807" Score="5" ViewCount="" Body="&lt;p&gt;So that this does not remain unanswered:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{pmatrix}0&amp;amp;1\\1&amp;amp;0\end{pmatrix}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is the simplest matrix that does not possess an $\mathbf L\mathbf D\mathbf L^T$ decomposition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, any symmetric matrix whose leading submatrix (the submatrix formed by the first few rows and columns) is singular will not possess such a decomposition.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-11-24T05:20:43.807" CommentCount="1" />
  <row Id="11656" PostTypeId="2" ParentId="11654" CreationDate="2010-11-24T05:46:34.500" Score="2" ViewCount="" Body="&lt;p&gt;You can write the limits of difference quotients as limits of sequences of difference quotients.  The measurability of the partial derivative follows from the fact that a pointwise limit of a sequence of measurable functions is measurable.  The left-hand side of the equation you want to prove is a limit of integrals of difference quotients, and the right-hand side is an integral of limits of difference quotients, so you want to justify passing a limit through an integral.  You can do this because the sequence of difference quotients is uniformly bounded, by the mean value theorem and the hypothesis that $\frac{\partial f}{\partial x}$ is bounded. &lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-11-24T05:46:34.500" />
  <row Id="11657" PostTypeId="1" AcceptedAnswerId="11660" CreationDate="2010-11-24T05:58:32.797" Score="0" ViewCount="171" Body="&lt;p&gt;For&lt;br&gt;&#xA;$$f(x,y) = \begin{cases}&#xD;&#xA;\frac{xy}{x^2+y^2} &amp;amp;\text{if }(x,y)\neq (0,0);\\&#xD;&#xA;0 &amp;amp;\text{if }(x,y)=(0,0).\end{cases}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm trying to prove that $\frac{\partial f}{\partial x}$, the partial derivative with respect to $x$, exists.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Taking the partial derivative freehand I get&lt;br&gt;&#xA;$$\frac{y(x^2+y^2)-xy(2x)}{(x^2+y^2)^2}=  &#xD;&#xA;\frac{y(y^2-x^2)}{(x^2+y^2)^2}.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But using the limits I get&#xA;$$\lim_{x\to c}\frac{f(x,y)-f(c,y)}{x-c} = y(y^2-c^2).$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure what I'm doing wrong...&lt;/p&gt;&#xA;" OwnerUserId="3333" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-24T06:08:30.007" LastActivityDate="2010-11-24T06:29:27.237" Title="Partial Derivative proof" Tags="&lt;real-analysis&gt;&lt;homework&gt;" AnswerCount="2" CommentCount="6" />
  <row Id="11658" PostTypeId="2" ParentId="11657" CreationDate="2010-11-24T06:12:34.177" Score="0" ViewCount="" Body="&lt;p&gt;You have an error in the calculation of the limit (see Arturo's remark about the denominator). When you correct it, you'll get the expression you calculated &lt;i&gt;freehand&lt;/i&gt;, with $c$ substituted for $x$.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-24T06:12:34.177" />
  <row Id="11659" PostTypeId="1" AcceptedAnswerId="11664" CreationDate="2010-11-24T06:15:14.050" Score="1" ViewCount="55" Body="&lt;p&gt;How many elements are in the set:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\{(x,y,z) \mid x+y+z = k \text{ and } x,y,z \in \{0,\dotsc,n\}\}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can see that when $k \le n$, it's $\binom{k+2}{2}$.&#xA;And, by symmetry, the value for $k$ is the same as that for $3n-k$.&#xA;So, I've figured out the two pyramids, and I'm just missing the octahedron in the middle.&lt;/p&gt;&#xA;" OwnerUserId="774" LastActivityDate="2010-11-24T06:41:14.063" Title="Number of nodes on diagonal planes of a cube" Tags="&lt;combinatorics&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="0" />
  <row Id="11660" PostTypeId="2" ParentId="11657" CreationDate="2010-11-24T06:16:37.003" Score="1" ViewCount="" Body="&lt;p&gt;I have no idea what you did with the limit. For $y\neq 0$ or $c\neq 0$, we have:&#xA;\begin{align*}&#xA;\lim{x\to c}\frac{f(x,y)-f(c,y)}{x-c} &amp;amp;= \lim_{x\to c}\frac{\frac{xy}{x^2+y^2} - \frac{cy}{c^2+y^2}}{x-c}\\&#xA;&amp;amp;= \lim_{x\to c}\frac{xy(c^2+y^2)-cy(x^2+y^2)}{(c-y)(x^2+y^2)(c^2+y^2)}\\&#xA;&amp;amp;= \lim_{x\to c}\frac{(x-c)(y^3-xyc)}{(x-c)(x^2+y^2)(c^2+y^2)}\\&#xA;&amp;amp;= \lim{x\to c}\frac{y^3-xyc}{(x^2+y^2)(c^2+y^2)}\\&#xA;&amp;amp;= \frac{y(y^2-c^2)}{(c^2+y^2)^2}.&#xA;\end{align*}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, how did you get what you claim you got?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Added.&lt;/em&gt; For the partial at $(x,y)=(0,0)$, you have:&#xA;$$\lim_{x\to 0}\frac{f(x,0)-f(0,0)}{x-0} = \lim_{x\to 0}\frac{0 - 0}{x} = 0.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the limits exist, the partial derivative exists at each point.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-24T06:29:27.237" LastActivityDate="2010-11-24T06:29:27.237" CommentCount="2" />
  <row Id="11661" PostTypeId="1" AcceptedAnswerId="11663" CreationDate="2010-11-24T06:30:08.983" Score="0" ViewCount="129" Body="&lt;p&gt;How can I prove that $a^x$ &amp;lt; $a^{x + \Delta x}$, where $\Delta x &amp;gt; 0$, $a$ is a constant, and $a &amp;gt; 1$ (in my case, $a=2$)?&#xA;I don't want to use a graph, of course.&lt;/p&gt;&#xA;" OwnerUserId="2736" LastEditorUserId="1543" LastEditorDisplayName="" LastEditDate="2011-05-03T10:11:19.057" LastActivityDate="2011-05-03T10:11:19.057" Title="Proof that $a^x$ &lt; $a^{x + \Delta x}$" Tags="&lt;inequality&gt;&lt;exponentiation&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="0" />
  <row Id="11662" PostTypeId="2" ParentId="11602" CreationDate="2010-11-24T06:36:42.197" Score="5" ViewCount="" Body="&lt;p&gt;$S(n)$ diverges at a rate at least as large as $\log_2 n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose $n &amp;gt; 2^k$.  Then, for some $1 \leq j \leq n-1$, $j = 2^k$.  Thus&#xA;$$\prod_{j=1}^{n-1}\left(1-\frac{j}{2^k}\right) = 0.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore,&#xA;$$S(n) = \sum_{k=1}^{\infty} \left(1 - \prod_{j=1}^{n-1}\left(1-\frac{j}{2^k}\right)\right) &amp;gt; \sum_{k=1}^{\log_2 (n-1)} \left(1 - \prod_{j=1}^{n-1}\left(1-\frac{j}{2^k}\right)\right) = \sum_{k=1}^{\log_2 (n-1)} 1 = \lfloor \log_2 (n-1) \rfloor.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;HR&gt;&#xA;As far as an upper bound, the following graph is of $2 \log_2 n - S(n)$ for $n \leq 300$.  I conjecture that there exists some $\alpha \approx \frac{2}{3}$ such that $S(n) \leq 2 \log_2 n - \alpha$ for $n \geq 2$ and that $\lim_{n \to \infty} (2 \log_2 n - S(n)) = \alpha$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/2OH3g.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(More numerical evidence: The value of $2 \log_2 n - S(n)$, is, for $n = 1000$, $2000$, and $3000$, respectively, $0.667734$, $0.667494$, and $0.667413$.)&lt;/p&gt;&#xA;" OwnerUserId="2370" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-11-24T06:56:50.807" LastActivityDate="2010-11-24T06:56:50.807" CommentCount="9" />
  <row Id="11663" PostTypeId="2" ParentId="11661" CreationDate="2010-11-24T06:38:40.500" Score="3" ViewCount="" Body="&lt;p&gt;It all depends on what you know. Do you know that if $a\gt 1$ and $b\gt 0$ then $a^b\gt 1$? If so, $2^{x+\Delta x}-2^x = 2^x(2^{\Delta x}-1)$ is a product of positive numbers.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-11-24T06:38:40.500" />
  <row Id="11664" PostTypeId="2" ParentId="11659" CreationDate="2010-11-24T06:41:14.063" Score="3" ViewCount="" Body="&lt;p&gt;When $n &amp;lt; k &amp;lt; 2n$, the number of points for which $x,y,z \in \mathbb N$ is $\binom{k+2}{2}$. Of these, $\binom{k-n+1}{2}$ have $x &amp;gt; n$ (as you can see by letting $x = x'+n+1$), and similarly for $y$ and $z$. These are the only exceptions, because when $k &amp;lt; 2n$, more than one of the variables cannot exceed $n$. So the number of points with $x,y,z \le n$ is $\binom{k+2}{2} - 3\binom{k-n+1}{2}$. This looks somewhat asymmetrical, but it's actually equal to $(k-n)(2n-k) + \binom{n+2}{2}$. I like writing it in this form because then it's obvious that it satisfies the &quot;boundary conditions&quot;.&lt;/p&gt;&#xA;" OwnerUserId="856" LastActivityDate="2010-11-24T06:41:14.063" />
  <row Id="11665" PostTypeId="1" AcceptedAnswerId="12982" CreationDate="2010-11-24T07:02:41.940" Score="2" ViewCount="206" Body="&lt;p&gt;Let $x$ be a positive rational number of the form $\displaystyle x = \sum\limits_{k=1}^{n}  \frac{a_k}{k!}$ where each $a_k$ is a nonnegative integer with $a_k \leq k-1$ for $k \geq 2$ and $a_n &amp;gt;0$. Prove that $a_1 = [x]$, $a_k = [k!x]-k[(k-1)!x]$ for $k = 2, \dots, n$, and that $n$ is the smallest integer such that $n!x$ is an integer. Conversely, show that every positive rational number can be expressed in this form in one and only one way. Note that $[x]$ is the greatest integer function. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I think there are two parts to this: (i) an inductive proof and (ii) a proof by contradiction. Would this be the correct &quot;high level&quot; approach to this problem?  &lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2010-12-04T15:34:05.293" Title="Rational Numbers and Uniqueness" Tags="&lt;real-analysis&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" />
  <row Id="11666" PostTypeId="2" ParentId="5627" CreationDate="2010-11-24T07:19:41.617" Score="1" ViewCount="" Body="&lt;p&gt;For a treatment of these basic facts about what happens to injectivity and surjectivity under composition, see Section 2.5 of &lt;a href=&quot;http://www.math.uga.edu/~pete/3200relationsfunctions.pdf&quot; rel=&quot;nofollow&quot;&gt;these lecture notes&lt;/a&gt; from a &quot;transition to upper level mathematics&quot; course that I taught twice at UGA.  Note that elsewhere in the course there were exercises asking students to prove that certain other statements which are &lt;em&gt;not&lt;/em&gt; listed here are not always true: e.g. if $g \circ f$ is surjective, then $g$ must be surjective (as is in the notes) but $f$ need not be.  Similarly, if $g \circ f$ is injective, then $f$ must be injective (...) but $g$ need not be.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given how fundamental and ubiquitous these results are, it's surprising that they are not always covered in standard texts.  (I guess the student is supposed to figure them out for herself, but in my experience most actual undergraduates do much better if they are exposed to these concepts in the context of an actual formal course.)  &lt;/p&gt;&#xA;" OwnerUserId="299" LastActivityDate="2010-11-24T07:19:41.617" />
  <row Id="11667" PostTypeId="2" ParentId="11650" CreationDate="2010-11-24T07:37:05.133" Score="7" ViewCount="" Body="&lt;p&gt;Hyperbolic geometry / algebraic K-theory must mean that $(1+x)/y$ and its iterates appear in the functional equations of one of the dilogarithm functions (Abel's relation or similar).  There is a 5-term functional equation with some cyclic symmetry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Models of QFT may refer to the &quot;pentagon relation&quot; from integrable models.  This again reflects the order 5 property.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[added:]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This does appear to be what Zagier was pointing to.  Back in the 1990's,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ADE functional dilogarithm identities and integrable models&#xA;F. Gliozzi,  R. Tateo&#xA;&lt;a href=&quot;http://xxx.lanl.gov/abs/hep-th/9411203&quot; rel=&quot;nofollow&quot;&gt;http://xxx.lanl.gov/abs/hep-th/9411203&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;formulated a functional equation generalizing some identities for the Rogers dilogarithm evaluated at roots of unity.  The latter identities were known to correspond to torsion in $K$-theory of fields (see papers of Frenkel and Szenes on dilogarithm identities, also early and mid-90's, on arxiv, including proofs of Gliozzi-Tateo formulae and some comments on $K_3(\mathbb{C})$).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On page 4-5 of the paper it is pointed out that the 5-fold periodicity comes from an inspection of the cyclic symmetry of the Abel 5-term functional equation of the Rogers dilogarithm.  If the terms are numbered correctly then $(1+x)/y$ is the generator of this symmetry.  The periodicity of this transformation was probably observed about 200 years ago.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cluster algebras appeared later and provided the technology to prove (among many other things) more general periodicity conjectures from integrable models of QFT, going back to earlier papers of Zamolodchikov.  However, the connection of this one period-5 transformation to the dilogarithm is probably easiest to see in the Gliozzi/Tateo paper above.  K-theory and hyperbolic geometry have no known direct connection to cluster algebras, though the latter are related to triangulations of polygons and this is sometimes connected to hyperbolic 2- and 3-dimensional geometry.&lt;/p&gt;&#xA;" OwnerUserId="467" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-11-24T17:33:48.167" LastActivityDate="2010-11-24T17:33:48.167" />
  <row Id="11668" PostTypeId="1" CreationDate="2010-11-24T08:39:02.637" Score="1" ViewCount="325" Body="&lt;p&gt;As far as I am concerned, probability &lt;strong&gt;distribution&lt;/strong&gt; function is for discrete random variables while probability &lt;strong&gt;density&lt;/strong&gt; function is for continuous random variables. To find the probability value of continuous random variable, we have to take the total area under the function which differ from discrete random variable, where we can take the probability value directly from the function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But here's my confusion,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's say Z=X+Y. X is discrete and can take in value of -1 and 1. Y is Gaussian random variable. So,Z is Gaussian random variable with probability &lt;strong&gt;density&lt;/strong&gt; function of p(Z)=p(Z|X=1)+p(Z|X=-1).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However when we try to find p(X=1|Z), it is equals to [p(Z|X=1)p(X=1)]/p(Z).&#xA;My questions, how can we time p(Z|X=1) and p(X=1) since former is probability &lt;strong&gt;density&lt;/strong&gt; function while the latter is probability &lt;strong&gt;distribution&lt;/strong&gt; function? What's more, p(Z) is also a probability &lt;strong&gt;density&lt;/strong&gt; function. In the end, what is p(X=1|Z), a probability &lt;strong&gt;density&lt;/strong&gt; or &lt;strong&gt;distribution&lt;/strong&gt; function?&lt;/p&gt;&#xA;" LastActivityDate="2010-11-28T06:14:09.763" Title="Confusion between probability distribution function and probability density function" Tags="&lt;probability&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="11669" PostTypeId="1" AcceptedAnswerId="11671" CreationDate="2010-11-24T09:24:05.267" Score="121" ViewCount="27125" Body="&lt;p&gt;The division of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Chromatic_scale&quot; rel=&quot;nofollow&quot;&gt;chromatic scale&lt;/a&gt; in 7 natural notes (white keys in a piano) and 5 &lt;a href=&quot;http://en.wikipedia.org/wiki/Accidental_%28music%29&quot; rel=&quot;nofollow&quot;&gt;accidental&lt;/a&gt; ones (black) seems a bit arbitrary to me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Apparently, adjacent notes in a piano (including white or black) are always separated by a semitone. Why the distinction, then? Why not just have scales with 12 notes? (apparently there's a musical scale called &lt;a href=&quot;http://en.wikipedia.org/wiki/Swara&quot; rel=&quot;nofollow&quot;&gt;Swara&lt;/a&gt; that does just that)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've asked several musician friends, but they lack the math preparation for giving me a valid answer. &quot;Notes are like that because they are like that&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need some mathematician with musical knowledge (or a musician with mathematical knowledge) to help me out with this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mathematically, is there any difference between white and black notes, or do we make the distinction just for historical reasons?&lt;/p&gt;&#xA;" OwnerUserId="3893" LastEditorUserId="10557" LastEditorDisplayName="" LastEditDate="2011-05-15T02:30:22.653" LastActivityDate="2011-05-15T02:30:22.653" Title="Mathematical difference between white and black notes in a piano" Tags="&lt;music-theory&gt;" AnswerCount="20" CommentCount="11" FavoriteCount="65" />
  <row Id="11670" PostTypeId="2" ParentId="11668" CreationDate="2010-11-24T09:25:37.827" Score="1" ViewCount="" Body="&lt;p&gt;You have assumed $X$ to be discrete, so $P(X=1|Z)$ had better be discrete or something really weird is going on. Just because you know $Z$ doesn't still change that $X$ still &quot;lives on&quot; a countable set.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It will all make a lot more sense after you've had measure theory and you see that it's all really the same thing:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the continuous case, the probability that you get an outcome in $B \subset \mathbb{R}$ is $\int_B p(x) \; \mathrm{d}m(x)$, where $p$ is your density and $m$ is the Lebesgue measure. &quot;Of course&quot; this can be rewritten into $\mu(B)$ for some probability measure $\mu$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the discrete case, the probability that you get an element in $A \subset \mathbb{N}$ is exactly $\sum_{a \in \mathbb{N}} p(a)$, but this can be rewritten as an integral with respect to the counting measure $c$: $\int_\mathbb{N} p(a) \; \mathrm{d}c(a)$. And this too can be rewritten into a probability measure: $\nu(A)$, where $\nu = \sum_{a \in \mathbb{N}} p(a) \delta_a$, $p$ being your distribution and $\delta_i$ the Dirac measure at $i$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Everything is &quot;just&quot; integrals with respect to some measure. And finally a word of warning: one shouldn't forget that the statement: &quot;Probability theory is just measure theory in the special case of total measure one&quot; is a gross oversimplification.&lt;/p&gt;&#xA;" OwnerUserId="1269" LastActivityDate="2010-11-24T09:25:37.827" CommentCount="4" />
  <row Id="11671" PostTypeId="2" ParentId="11669" CreationDate="2010-11-24T09:49:53.367" Score="143" ViewCount="" Body="&lt;p&gt;The first thing you have to understand is that notes are not uniquely defined.  Everything depends on what tuning you use.  I'll assume we're talking about &lt;a href=&quot;http://en.wikipedia.org/wiki/Equal_temperament&quot; rel=&quot;nofollow&quot;&gt;equal temperament&lt;/a&gt; here.  In equal temperament, a half-step is the same as a frequency ratio of $\sqrt[12]{2}$; that way, twelve half-steps makes up an octave.  Why twelve?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the end of the day, what we want out of our musical frequencies are nice ratios of small lintegers.  For example, a perfect fifth is supposed to correspond to a frequency ratio of $3 : 2$, or $1.5 : 1$, but in equal temperament it doesn't; instead, it corresponds to a ratio of $2^{ \frac{7}{12} } : 1 \approx 1.498 : 1$.  As you can see, this is not a fifth; however, it is quite close.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, a perfect fourth is supposed to correspond to a frequency ratio of $4 : 3$, or $1.333... : 1$, but in equal temperament it corresponds to a ratio of $2^{ \frac{5}{12} } : 1 \approx 1.335 : 1$.  Again, this is not a perfect fourth, but is quite close.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And so on.  What's going on here is a massively convenient mathematical coincidence: several of the powers of $\sqrt[12]{2}$ happen to be good approximations to ratios of small integers, and there are enough of these to play Western music.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's how this coincidence works.  You get the white keys from $C$ using (part of) the circle of fifths.  Start with $C$ and go up a fifth to get $G$, then $D$, then $A$, then $E$, then $B$.  Then go down a fifth to get $F$.  These are the &quot;neighbors&quot; of $C$ in the circle of fifths.  You get the black keys from here using the rest of the circle of fifths.  After you've gone up a &quot;perfect&quot; perfect fifth twelve times, you get a frequency ratio of $3^{12} : 2^{12} \approx 129.7 : 1$.  This happens to be rather close to $2^7 : 1$, or seven octaves!  And if we replace $3 : 2$ by $2^{ \frac{7}{12} } : 1$, then we get exactly seven octaves.  In other words, the reason you can afford to identify these intervals is because $3^{12}$ happens to be rather close to $2^{19}$.  Said another way, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\log_2 3 \approx \frac{19}{12}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;happens to be a good rational approximation, and this is the main basis of equal temperament.  (The other main coincidence here is that $\log_2 \frac{5}{4} \approx \frac{4}{12}$; this is what allows us to squeeze major thirds into equal temperament as well.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is a fundamental fact of music that $\log_2 3$ is irrational, so it is impossible for any kind of equal temperament to have &quot;perfect&quot; perfect fifths regardless of how many notes you use.  However, you can write down good rational approximations by looking at the &lt;a href=&quot;http://en.wikipedia.org/wiki/Continued_fraction&quot; rel=&quot;nofollow&quot;&gt;continued fraction&lt;/a&gt; of $\log_2 3$ and writing down convergents, and these will correspond to equal-tempered scales with more notes.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, you can use other types of temperament, such as &lt;a href=&quot;http://en.wikipedia.org/wiki/Well_temperament&quot; rel=&quot;nofollow&quot;&gt;well temperament&lt;/a&gt;; if you stick to $12$ notes (which not everybody does!), you will be forced to make some intervals sound better and some intervals sound worse.  In particular, if you don't use equal temperament then &lt;strong&gt;different keys sound different.&lt;/strong&gt;  This is a major reason many Western composers composed in different keys; during their time, this actually made a difference.  As a result when you're playing certain sufficiently old pieces you aren't actually playing them as they were intended to be heard - you're using the wrong tuning.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt;  I suppose it is also good to say something about why we care about frequency ratios which are ratios of small integers.  This has to do with the physics of sound, and I'm not particularly knowledgeable here, but this is my understanding of the situation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You probably know that sound is a wave.  More precisely, sound is a &lt;a href=&quot;http://en.wikipedia.org/wiki/Longitudinal_wave&quot; rel=&quot;nofollow&quot;&gt;longitudinal wave&lt;/a&gt; carried by air molecules.  You might think that there is a simple equation for the sound created by a single note, perhaps $\sin 2\pi f t$ if the corresponding tone has frequency $f$.  Actually this only occurs for tones which are produced electronically; any tone you produce in nature carries with it &lt;a href=&quot;http://en.wikipedia.org/wiki/Overtone&quot; rel=&quot;nofollow&quot;&gt;overtones&lt;/a&gt; and has a Fourier series&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum \left( a_n \sin 2 \pi n f t + b_n \cos 2 \pi n f t \right)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where the coefficients $a_n, b_n$ determine the &lt;a href=&quot;http://en.wikipedia.org/wiki/Timbre&quot; rel=&quot;nofollow&quot;&gt;timbre&lt;/a&gt; of the sound; this is why different instruments sound different even when they play the same notes, and has to do with the physics of vibration, which I don't understand too well.  So any tone which you hear at frequency $f$ almost certainly also has components at frequency $2f, 3f, 4f, ...$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you play two notes of frequencies $f, f&amp;#39;$ together, then the resulting sound corresponds to what you get when you add their Fourier series.  Now it's not hard to see that if $\frac{f}{f&amp;#39;}$ is a ratio of small integers, then many (but not all) of the overtones will match in frequency with each other; the result sounds a more complex note with certain overtones.  Otherwise, you get dissonance as you hear both types of overtones simultaneously and their frequencies will be similar, but not similar enough.  &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt;  You should probably check out &lt;a href=&quot;http://www.maths.abdn.ac.uk/~bensondj/html/maths-music.html&quot; rel=&quot;nofollow&quot;&gt;the book Rahul Narain recommended&lt;/a&gt; in the comments for the full story.  There was a lot I didn't know, and I'm only in the introduction!&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-11-26T04:07:28.117" LastActivityDate="2010-11-26T04:07:28.117" CommentCount="17" />
  <row Id="11672" PostTypeId="2" ParentId="11668" CreationDate="2010-11-24T10:04:37.307" Score="1" ViewCount="" Body="&lt;p&gt;Let $X$ be random variable over $\mathbb{Z}$ with probability weights $p_k = Pr[X=k]$. Many discrete scenario can be expressed in these terms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The probability distribution function is by virtue of definition $F(x) = Pr[X \leq x] = \sum\limits_{k=-\infty}^{\lfloor x \rfloor}Pr[X=k] = \sum\limits_{k=-\infty}^{\lfloor x \rfloor} p_k$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A density function corresponds to the weights $p_k$ while the definition of distribution function remains the same in the continuous case, only that $Pr[X \leq x]$ has to be calculated by an integral instead of a sum. See kahen's answer for details.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So no, you should not exchange distribution function and density. They do different things.&lt;/p&gt;&#xA;" OwnerUserId="3330" LastActivityDate="2010-11-24T10:04:37.307" CommentCount="4" />
  <row Id="11673" PostTypeId="1" CreationDate="2010-11-24T10:08:28.270" Score="0" ViewCount="107" Body="&lt;p&gt;The following result must be well known, but it is instructive to give it here in view of the recent posts &lt;a href=&quot;http://math.stackexchange.com/questions/11487/y-xn-what-is-exyy&quot;&gt;Y= X+N what is E(X|Y=y)&lt;/a&gt; and &lt;a href=&quot;http://mathoverflow.net/questions/47168/ex-1-x-1-x-2-where-x-i-are-integrable-independent-infinitely-divisib/47204#47204&quot; rel=&quot;nofollow&quot;&gt;http://mathoverflow.net/questions/47168/ex-1-x-1-x-2-where-x-i-are-integrable-independent-infinitely-divisib/47204#47204&lt;/a&gt;. This is also an interesting exercise in its own right.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose that $X_1$ and $X_2$ are independent ${\rm Gamma}(c_i,\lambda)$ rv's, meaning that $X_i$, $i=1,2$, has density function $f_{X_i } (x) = \lambda ^{c_i } {\rm e}^{ - \lambda x} x^{c_i  - 1} /\Gamma (c_i )$, $x &amp;gt; 0$ ($c_i$ are positive constants, $\Gamma$ is the gamma function). Show that&#xA;$$&#xD;&#xA;{\rm E}(X_1 | X_1 + X_2 = z) = \frac{{c_1 }}{{c_1  + c_2 }} z.&#xD;&#xA;$$ &lt;/p&gt;&#xA;" OwnerUserId="2810" LastActivityDate="2010-11-27T04:39:49.707" Title="$E(X_1 | X_1 + X_2)$ for independent gamma random variables" Tags="&lt;probability&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" />
  <row Id="11674" PostTypeId="1" CreationDate="2010-11-24T10:31:14.427" Score="5" ViewCount="79" Body="&lt;p&gt;For Riemann zeta function and $L$-functions of number fields, the width of critical strip is $1$. For $L$-functions of modular forms of weight $k$, the width of the critical strip is $k$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is there a variation in the width of the critical strip for various $L$-functions? Is there a conceptual explanation or an underlying heuristics? &lt;/p&gt;&#xA;" LastActivityDate="2010-11-24T11:00:24.293" Title="Why is width of critical strip what it is?" Tags="&lt;analytic-number-theory&gt;" AnswerCount="1" FavoriteCount="2" />
  <row Id="11675" PostTypeId="1" CreationDate="2010-11-24T10:42:43.470" Score="5" ViewCount="182" Body="&lt;p&gt;For proving the quadratic reciprocity, Gauss sums are very useful. However this seems an ad-hoc construction. Is this useful in a wider context? What are some other uses for Gauss sums?&lt;/p&gt;&#xA;" LastActivityDate="2010-11-24T17:03:56.313" Title="Applications of Gauss sums" Tags="&lt;number-theory&gt;" AnswerCount="4" FavoriteCount="5" />
  <row Id="11676" PostTypeId="2" ParentId="11669" CreationDate="2010-11-24T10:45:47.540" Score="3" ViewCount="" Body="&lt;p&gt;I'm no musician, but as far as I know, audio waves are felt only then &quot;round/sound&quot;, iff they repeat faster than a specific frequency. That frequency is probably that of our brain waves: being awake and in a non-meditating state, that is faster than like 18 or more Hz; neither can you shiver faster nor hear lower frequencies than your brain waves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Audio waves have a length of $$\mathrm{lcm}\{m,n\}&amp;#183;2\pi$$ if they are of the shape $$a_1&amp;#183;\sin(m&amp;#183;2\pi&amp;#183;t+s_1)+a_2&amp;#183;\sin(n&amp;#183;2\pi&amp;#183;t+s_2)$$. The notes double their frequency each octave; therefore they have a logarithmic scale and not a linear. Good violin and harp players can play all fitting (&quot;sound&quot; sounding) combination of frequencies, but instruments with keys lack the variety.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Qiaochu Yuan did answer faster than me, while I was on the phone. Seems to be more complete than I could have answered. I have nothing to add.)&lt;/p&gt;&#xA;" OwnerUserId="3226" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-26T03:20:39.270" LastActivityDate="2010-11-26T03:20:39.270" />
  <row Id="11677" PostTypeId="2" ParentId="11675" CreationDate="2010-11-24T10:55:36.000" Score="5" ViewCount="" Body="&lt;p&gt;Not just &lt;strong&gt;quadratic&lt;/strong&gt; reciprocity, one can use them to prove higher&#xA;reciprocity laws: see Ireland and Rosen's &lt;em&gt;A Classical Introduction&#xA;to Modern Number Theory&lt;/em&gt;. They also turn up in the functional&#xA;equation for Dirichlet L-functions (and are massively generalized in the&#xA;topic of root numbers).&lt;/p&gt;&#xA;" OwnerUserId="226" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-24T11:15:31.403" LastActivityDate="2010-11-24T11:15:31.403" />
  <row Id="11678" PostTypeId="2" ParentId="11674" CreationDate="2010-11-24T11:00:24.293" Score="6" ViewCount="" Body="&lt;p&gt;Roughly speaking the critical strip is where the $L$-function&#xA;is hard to compute/understand. Its right edge is the boundary&#xA;of the region where the Dirichlet series/Euler product converges&#xA;nicely (absolutely, locally uniformly etc). Its left edge&#xA;is the image of the right edge under the functional equation&#xA;(to compute it to the left of the critical strip you can use&#xA;the functional equation to reduce it to comuting a nice series/product).&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-11-24T11:00:24.293" />
  <row Id="11679" PostTypeId="1" CreationDate="2010-11-24T11:11:59.440" Score="0" ViewCount="76" Body="&lt;p&gt;The context is the definition of Hecke Größencharakter:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Hecke_character&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Hecke_character&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is supposed to generalize the Dirichlet $L$-series for number fields. Dirichlet characters are characters of the multiplicative groups of $\mathbb Z/p\mathbb Z$. An appropriate generalization would be instead to consider characters of the multiplicative group of $cal O_K/\cal P$ where $\cal P$ is a prime ideal in the ring of integers of a number field $K$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But Hecke Größencharakter goes to more trouble than this. It brings in ideles and such for a more complicated generalization. Why is this necessary?&lt;/p&gt;&#xA;" LastActivityDate="2010-11-24T11:19:51.820" Title="Motivation for Hecke characters" Tags="&lt;analytic-number-theory&gt;" AnswerCount="1" />
  <row Id="11680" PostTypeId="2" ParentId="11675" CreationDate="2010-11-24T11:17:28.050" Score="2" ViewCount="" Body="&lt;p&gt;It can be used to prove a certain theorem called the &lt;a href=&quot;http://en.wikipedia.org/wiki/Stickelberger_theorem&quot; rel=&quot;nofollow&quot;&gt;Stickelberger's theorem&lt;/a&gt;, the likes of which can in turn be used for proving things like &lt;a href=&quot;http://en.wikipedia.org/wiki/Catalan_conjecture&quot; rel=&quot;nofollow&quot;&gt;Catalan's Conjecture&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1119" LastActivityDate="2010-11-24T11:17:28.050" />
  <row Id="11681" PostTypeId="2" ParentId="11679" CreationDate="2010-11-24T11:19:51.820" Score="1" ViewCount="" Body="&lt;p&gt;One reason for the greater generality is that Hecke&#xA;characters do more than describe Abelian extensions of number fields&#xA;(essentially Dirichlet characters describe Abelian extenions&#xA;of $\mathbb Q$). For instance the L-function of an elliptic&#xA;curve with complex multiplication is the L-function of a Hecke&#xA;character of infinite order.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-11-24T11:19:51.820" />
  <row Id="11682" PostTypeId="1" CreationDate="2010-11-24T11:28:18.193" Score="0" ViewCount="77" Body="&lt;p&gt;Let X be a topological space and A be an abelian group. Give A the discrete topology. For any open set U of X, Let $\cal A(U)$ be the group of all continuous aps of U into A. Thus with the usual restriction maps we obtain a sheaf $\cal A$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, why for every connected open set U, $\cal A(U)\cong A$ for all U?&#xA; And, what is the sheafifcation of the presheaf $U \mapsto A$?&lt;/p&gt;&#xA;" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-24T17:22:55.067" LastActivityDate="2010-11-24T17:22:55.067" Title="the sheafification of a constant presheaf" Tags="&lt;algebraic-geometry&gt;" AnswerCount="1" />
  <row Id="11683" PostTypeId="1" CreationDate="2010-11-24T11:46:31.063" Score="0" ViewCount="107" Body="&lt;p&gt;I need to solve this equation. Please help me!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\dfrac{-\mathrm d^2u}{\mathrm dt^2}+\dfrac{u}{LC}=\dfrac{V}{LC},\qquad L,C,V,I=\text{constant}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and $u(0)=0$, $\left. \dfrac{\mathrm du}{\mathrm dt}\right\vert_{t=0}=\dfrac{I}{C}$&#xD;&lt;/p&gt;&#xA;" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2010-11-24T16:08:24.397" LastActivityDate="2010-11-24T16:08:24.397" Title="Solution for the equation $\frac{-\mathrm d^2u}{\mathrm dt^2}+\frac{u}{LC}=\frac{V}{LC}$" Tags="&lt;differential-equations&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="11684" PostTypeId="2" ParentId="11618" CreationDate="2010-11-24T11:47:18.227" Score="2" ViewCount="" Body="&lt;p&gt;You can apply &lt;a href=&quot;http://en.wikipedia.org/wiki/Ruffini%27s_rule&quot; rel=&quot;nofollow&quot;&gt;Ruffini's rule&lt;/a&gt;. Here is a copy from my Algebra text book (Compêndio de Álgebra, VI, by Sebastião e Silva and Silva Paulo) where the following formula is obtained:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x^n-a^n\equiv (x-a)(x^{n-1}+ax^{n-2}+a^2x^{n-3}+\cdots +a^{n-2}x+a^{n-1}).$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/HtIOy.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Translation: The Ruffini's rule can be used to find the quotient of $x^n-a^n$ by $x-a$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Figure)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, if $n$ is a natural number, we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x^n-a^n\equiv (x-a)(x^{n-1}+ax^{n-2}+a^2x^{n-3}+\cdots +a^{n-2}x+a^{n-1})$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="752" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2010-11-24T15:29:24.757" LastActivityDate="2010-11-24T15:29:24.757" CommentCount="3" />
  <row Id="11685" PostTypeId="2" ParentId="4619" CreationDate="2010-11-24T12:01:14.003" Score="0" ViewCount="" Body="&lt;p&gt;An example of a decidable, NP-Hard but not NP-Complete problem is the Towers of Hanoi problem. Intuitively, the T.of H. problem is not in NP because the exploration of all possible configurations cannot be done in polynomial time, on a non-deterministic machine.&lt;/p&gt;&#xA;" LastActivityDate="2010-11-24T12:01:14.003" CommentCount="1" />
  <row Id="11686" PostTypeId="1" AcceptedAnswerId="11691" CreationDate="2010-11-24T12:57:27.540" Score="2" ViewCount="305" Body="&lt;p&gt;Is there a rational number (with denominator not greater than 200) between 15/106 and 16/113? &lt;/p&gt;&#xA;" OwnerUserId="3901" LastActivityDate="2011-01-08T12:14:04.437" Title="Is there a rational number (with denominator not greater than 200) between 15/106 and 16/113?" Tags="&lt;number-theory&gt;&lt;prime-numbers&gt;&lt;divisibility&gt;" AnswerCount="5" CommentCount="3" />
  <row Id="11687" PostTypeId="2" ParentId="11682" CreationDate="2010-11-24T13:09:44.737" Score="2" ViewCount="" Body="&lt;p&gt;Some hints:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For your first question: $A$ has the discrete topology, so elements of $A$ are open and closed. Then, what properties does the preimage of a single element under a continuous map $U \to A$ have? Recall the definition of connectedness. What does this imply for continuous maps $U \to A$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For your second question, write out explicitly the definition of the sheafification functor for your example and see that the result you will get is exactly the sheaf you described in the first part of your question.&lt;/p&gt;&#xA;" OwnerUserId="1766" LastActivityDate="2010-11-24T13:09:44.737" />
  <row Id="11688" PostTypeId="2" ParentId="11686" CreationDate="2010-11-24T13:44:16.563" Score="6" ViewCount="" Body="&lt;p&gt;A brute-force search in &lt;em&gt;Mathematica&lt;/em&gt;, using&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Select[Flatten[Table[m/n, {n, 2, 200}, {m, 1, n - 1}]], (15/106 &amp;lt; # &amp;lt; 16/113) &amp;amp;]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;indicates that there is no rational number satisfying your requirement.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-11-24T13:44:16.563" CommentCount="5" />
  <row Id="11689" PostTypeId="2" ParentId="11669" CreationDate="2010-11-24T13:44:53.840" Score="6" ViewCount="" Body="&lt;p&gt;Just to let you see that other tunings are possible and thus other keyboards:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.kylegann.com/tuning.html&quot; rel=&quot;nofollow&quot;&gt;http://www.kylegann.com/tuning.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3567" LastActivityDate="2010-11-24T13:44:53.840" />
  <row Id="11690" PostTypeId="2" ParentId="11686" CreationDate="2010-11-24T13:45:22.360" Score="10" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\ \ $Google &quot;farey sequence&quot; and &quot;mediant&quot; and / or see my many sci.math posts on these topics, e.g. see &lt;a href=&quot;http://groups.google.com/groups?selm=y8zpsy7ixpf.fsf%40nestle.csail.mit.edu&quot; rel=&quot;nofollow&quot;&gt;this post&lt;/a&gt; which explains the lucid lattice geometric viewpoint, and see &lt;a href=&quot;http://groups.google.com/groups?selm=y8zwv4xtxj3.fsf%40nestle.ai.mit.edu&quot; rel=&quot;nofollow&quot;&gt;this post&lt;/a&gt; on how to apply this to perform Farey mediant binary searching to find the simplest fraction(s) between two fractions. See also my &lt;a href=&quot;http://groups.google.com/groups/search?q=group%3a%2Amath%2A+author%3Adubuque+%28farey+OR+mediant%29&quot; rel=&quot;nofollow&quot;&gt;other posts&lt;/a&gt; on these topics, which includes the fascinating connection with &lt;a href=&quot;http://en.wikipedia.org/wiki/Pick%27s_theorem&quot; rel=&quot;nofollow&quot;&gt;Pick's theorem&lt;/a&gt; on the area of lattice polygons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;NOTE&lt;/strong&gt; $\ \ $ If someone here is handy with graphing programs and/or Java you'd be doing a great service if you could draw/animate the mediant lattice construction depicted in ascii art in my first post. Or perhaps someone has done this elsewhere? That yields a &quot;proof without words&quot; that succinctly captures the essence of the matter.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-11-24T15:05:04.850" LastActivityDate="2010-11-24T15:05:04.850" CommentCount="1" />
  <row Id="11691" PostTypeId="2" ParentId="11686" CreationDate="2010-11-24T13:47:18.153" Score="11" ViewCount="" Body="&lt;p&gt;The sandwich ${15\over 106}&amp;lt;{p\over q}&amp;lt; {16\over 113}$ implies $15 q&amp;lt;106p$ and $16q&amp;gt;113p$, so necessarily $q&amp;gt;7p$. Write $q=7p+ q_1$ with $q_1&amp;gt;0$. It follows that $15q_1&amp;lt;p$ and $16 q_1&amp;gt;p$, so necessarily $p= 15 q_1+p_1$ with $0&amp;lt;p_1&amp;lt;q_1$. The optimal choice is $p_1=1$, $q_1=2$ from which we get $p=31$, $q=219&amp;gt;200$.&lt;/p&gt;&#xA;" OwnerUserId="1303" LastActivityDate="2010-11-24T13:47:18.153" CommentCount="2" />
  <row Id="11692" PostTypeId="2" ParentId="11675" CreationDate="2010-11-24T13:53:01.273" Score="8" ViewCount="" Body="&lt;p&gt;Gauss sums are not an ad-hoc construction!  I know two ways to motivate the definition, one of which requires that you know a little Galois theory and the other which is totally mysterious to me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is the Galois-theoretic explanation.  Let $\zeta_p$ be a primitive $p^{th}$ root of unity, for $p$ prime.  The &lt;a href=&quot;http://en.wikipedia.org/wiki/Cyclotomic_field&quot; rel=&quot;nofollow&quot;&gt;cyclotomic field&lt;/a&gt; $\mathbb{Q}(\zeta_p)$ is &lt;a href=&quot;http://en.wikipedia.org/wiki/Galois_extension&quot; rel=&quot;nofollow&quot;&gt;Galois&lt;/a&gt;, so one can define its &lt;a href=&quot;http://en.wikipedia.org/wiki/Galois_group&quot; rel=&quot;nofollow&quot;&gt;Galois group&lt;/a&gt;, the group of all field automorphisms which preserve $\mathbb{Q}$.  Such an automorphism is determined by what it does to $\zeta_p$, and it must send $\zeta_p$ to another primitive $p^{th}$ root of unity.  It follows that the Galois group $G = \text{Gal}(\mathbb{Q}(\zeta_p)/\mathbb{Q})$ is isomorphic to $(\mathbb{Z}/p\mathbb{Z})^{\times}$, which is cyclic of order $p-1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now suppose $p$ is odd.  As a cyclic group of even order, $G$ has a unique subgroup $H$ of index two given precisely by the multiplicative group of quadratic residues $\bmod p$, so by the &lt;a href=&quot;http://en.wikipedia.org/wiki/Fundamental_theorem_of_Galois_theory&quot; rel=&quot;nofollow&quot;&gt;fundamental theorem of Galois theory&lt;/a&gt; the fixed field $\mathbb{Q}(\zeta_p)^H$ is the unique quadratic subextension of $\mathbb{Q}(\zeta_p)$.  And it's not hard to see that this unique quadratic subextension must be generated by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{\sigma \in H} \sigma(\zeta_p) = \sum_{a \text{ is a QR}} \zeta_p^a = \frac{1}{2} \left( \sum_{a=1}^{p-1} \zeta_p^{a^2} \right)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which you will of course recognize as a Gauss sum!  So the Gauss sum generates a quadratic subextension, and any of various methods will tell you that this subextension is precisely $\mathbb{Q}(\sqrt{p^{\ast}})$ where $p^{\ast} = (-1)^{ \frac{p-1}{2} } p$.  (This does not actually require any computation: if you know enough algebraic number theory, it follows from a consideration of which primes ramify in cyclotomic extensions.)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The totally mysterious explanation is that Gauss sums naturally appear when you start thinking about the &lt;a href=&quot;http://en.wikipedia.org/wiki/Discrete_Fourier_transform&quot; rel=&quot;nofollow&quot;&gt;discrete Fourier transform&lt;/a&gt;.  For example, the trace of the DFT matrix is a Gauss sum.  But more mysteriously, Gauss sums are eigenfunctions of the DFT in a certain sense.  (I sketch how this works &lt;a href=&quot;http://qchu.wordpress.com/2010/01/11/some-quadratic-reciprocity/&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.)  There is a sort of mysterious connection here to the Gaussian distribution, which is an eigenfunction of the continuous Fourier transform; see &lt;a href=&quot;http://mathoverflow.net/questions/11053/whats-the-relationship-between-gauss-sums-and-the-normal-distribution&quot; rel=&quot;nofollow&quot;&gt;this MO question&lt;/a&gt;.  Again, I don't know what to make of this.  There is a book by Berg called &lt;a href=&quot;http://books.google.com/books?id=fxVZybOHF8EC&amp;amp;pg=PR17&amp;amp;lpg=PR17&amp;amp;dq=fourier-theoretic+proof+of+quadratic+reciprocity&amp;amp;source=bl&amp;amp;ots=o4aiY9ferP&amp;amp;sig=Wj7JvvAdomtfCwy-UAFfgOKYPWs&amp;amp;hl=en&amp;amp;ei=ARntTKCBF6GShAf_6eTNDA&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=1&amp;amp;ved=0CBIQ6AEwAA&quot; rel=&quot;nofollow&quot;&gt;The Fourier-analytic proof of quadratic reciprocity&lt;/a&gt; and it may or may not be about this construction.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-11-24T13:53:01.273" />
  <row Id="11693" PostTypeId="1" CreationDate="2010-11-24T13:57:56.773" Score="10" ViewCount="146" Body="&lt;p&gt;The Manin-Drinfeld theorem asserts that for a modular curve $X_0(N)$ and Jacobian $J_0(N)$ with the former being embdedded in the latter under the map that takes $i\infty$ to $0$, the cusps are torsion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The proof Manin-Drinfeld theorem seems to use Hecke operators and thus seems to be valid only for congruence subgroups of $PSL_2(\mathbb Z)$. Is there possibly a finite-index subgroup of $PSL_2(\mathbb Z)$ such that the statement of Manin-Drinfeld theorem is still true for the quotient? &lt;/p&gt;&#xA;" OwnerUserId="3903" LastActivityDate="2010-11-27T06:58:21.037" Title="Generality of the Manin-Drinfel'd theorem" Tags="&lt;number-theory&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="2" />
  <row Id="11694" PostTypeId="2" ParentId="11669" CreationDate="2010-11-24T14:02:58.830" Score="2" ViewCount="" Body="&lt;p&gt;If you only had a repetitive series of keys on your piano it would be a bit difficult to visually get some reference points. I think this is the main reason wh&lt;/p&gt;&#xA;" OwnerUserId="3904" LastActivityDate="2010-11-24T14:02:58.830" />
  <row Id="11695" PostTypeId="2" ParentId="11618" CreationDate="2010-11-24T14:05:00.467" Score="-2" ViewCount="" Body="&lt;p&gt;If you know the sum of a geometric sequence, then set $x=a/b$ and conclude&#xA;$&#xD;&#xA;x^n-1 = (x-1)(1+x+x^2+\cdots+x^{n-1})&#xD;&#xA;$. Now multiply by $b^n$. (If $b=0$, then the identity is obvious.)&lt;/p&gt;&#xA;" OwnerUserId="589" LastActivityDate="2010-11-24T14:05:00.467" CommentCount="1" />
  <row Id="11696" PostTypeId="1" CreationDate="2010-11-24T14:15:01.887" Score="0" ViewCount="247" Body="&lt;p&gt;The Coordinates of two points are A(-2,6) and B(9,3).  Find the coordinates of the point C on the x-axis such that AC = BC&lt;/p&gt;&#xA;" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-24T14:20:22.367" LastActivityDate="2010-11-24T15:52:54.900" Title="Find the coordinates of a point equidistant from two given points" Tags="&lt;analytic-geometry&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="11697" PostTypeId="2" ParentId="11673" CreationDate="2010-11-24T14:34:07.723" Score="1" ViewCount="" Body="&lt;p&gt;First of all, we note that $X_1 + X_2 \sim {\rm Gamma}(c_1 + c_2 , \lambda)$.&#xA;Given that $X_1 + X_2 = z$, $X_1$ cannot exceed $z$. We can thus find ${\rm E}(X_1|X_1+X_2=z)$ as follows.&#xA;$$&#xD;&#xA;{\rm E}(X_1|X_1 + X_2 = z) = \int_{0}^z  {xf_{X_1|X_1 + X_2} (x|z)\,{\rm d}x} = \int_{0}^z  {x\frac{{f_{X_1} (x)f_{X_1 + X_2|X_1} (z|x)}}{{f_{X_1 + X_2} (z)}}\, {\rm d}x}. &#xD;&#xA;$$&#xD;&#xA;Noting that $f_{X_1 + X_2|X_1} (z|x) = f_{X_2} (z-x)$, it follows after some algebra and a change of variable that the right-hand side integral is equal to&#xA;$$&#xD;&#xA;z\frac{{\Gamma (c_1  + c_2 )}}{{\Gamma (c_1 )\Gamma (c_2 )}}\int_0^1 {x^{c_1 } (1 - x)^{c_2  - 1} \,{\rm d}x}  = z\frac{{\Gamma (c_1  + c_2 )}}{{\Gamma (c_1 )\Gamma (c_2 )}}{\rm B}(c_1  + 1,c_2 ),&#xD;&#xA;$$&#xD;&#xA;where ${\rm B}(a,b) = \int_0^1 {t^{a - 1} (1 - t)^{b - 1} \,{\rm d}t}$ is the beta function. Finally, from the alternative form ${\rm B}(a,b) = \frac{{\Gamma (a)\Gamma (b)}}{{\Gamma (a + b)}}$ followed by $\Gamma (p+1) = p \Gamma (p)$, we obtain&#xA;$$&#xD;&#xA;{\rm E}(X_1|X_1 + X_2 = z) = \frac{{c_1 }}{{c_1  + c_2 }} z.&#xD;&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="2810" LastActivityDate="2010-11-24T14:34:07.723" />
  <row Id="11698" PostTypeId="1" AcceptedAnswerId="29042" CreationDate="2010-11-24T14:44:30.650" Score="1" ViewCount="145" Body="&lt;p&gt;If I have an arc (which comes as part of an ellipse), can I represent it (or at least closely approximate) by cubic Bezier curve? And if yes, how can I calculate control points for that Bezier curve?&lt;/p&gt;&#xA;" OwnerUserId="3777" LastEditorUserId="3777" LastEditorDisplayName="" LastEditDate="2010-11-24T15:26:48.757" LastActivityDate="2011-03-25T17:53:49.987" Title="How elliptic arc can be represented by cubic Bezier curve?" Tags="&lt;geometry&gt;&lt;bezier-curve&gt;&lt;elliptic-curves&gt;&lt;conic-sections&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="11699" PostTypeId="2" ParentId="11683" CreationDate="2010-11-24T14:45:22.737" Score="0" ViewCount="" Body="&lt;p&gt;Hint:  What functions do you know whose second derivative is proportional to the function?  There should be a two-dimensional solution space to the homogeneous equation (where the RHS is set to zero)&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-11-24T14:45:22.737" />
  <row Id="11700" PostTypeId="2" ParentId="11683" CreationDate="2010-11-24T14:49:52.833" Score="2" ViewCount="" Body="&lt;p&gt;Here are some hints. It's a standard form of 2nd order ordinary differential equation. You first find a &quot;particular solution&quot; $u_p$ to your equation (not including your conditions on $u(0)$ and ${du \over dt}(0)$). In this case it's just a constant solution. Then you find the general solution to the homogeneous equation $-{d^2 u \over dt} + {1 \over LC} u = 0$. This will be of the form $a u_1(t) + b u_2(t)$ for the right two functions, which will be certain exponentials here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then any solution to $-{d^2 u \over dt} + {1 \over LC} u = {V \over LC}$ will be of the form $a u_1(t) + b u_2(t) + u_p(t)$. Plug in the conditions on $u(0)$ and ${du \over dt}(0)$ to figure which $a$ and $b$ correspond to your situation. Voila, you're done.&lt;/p&gt;&#xA;" OwnerUserId="3035" LastActivityDate="2010-11-24T14:49:52.833" />
  <row Id="11701" PostTypeId="2" ParentId="7570" CreationDate="2010-11-24T14:50:17.427" Score="0" ViewCount="" Body="&lt;p&gt;If they commute with each other we can interchange the limits........&lt;/p&gt;&#xA;" LastActivityDate="2010-11-24T14:50:17.427" />
  <row Id="11702" PostTypeId="2" ParentId="7570" CreationDate="2010-11-24T14:55:56.357" Score="1" ViewCount="" Body="&lt;p&gt;If the total sum converges absolutely, you can reorder the terms however you want.  Can you prove that?&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-11-24T14:55:56.357" />
  <row Id="11703" PostTypeId="1" AcceptedAnswerId="11704" CreationDate="2010-11-24T14:56:28.337" Score="0" ViewCount="349" Body="&lt;p&gt;Proof to the formula &#xA;$$1+x+x^2+x^3+\cdots+x^n = \frac{x^{n+1}-1}{x-1}.$$&lt;/p&gt;&#xA;" OwnerUserId="3907" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2011-04-07T14:19:32.473" LastActivityDate="2011-04-07T14:19:32.473" Title="proof of the formula $1+x+x^2+x^3+...+x^n =\frac{x^{n+1}-1}{x-1}$" Tags="&lt;algebra&gt;" AnswerCount="7" CommentCount="4" />
  <row Id="11704" PostTypeId="2" ParentId="11703" CreationDate="2010-11-24T15:02:54.680" Score="5" ViewCount="" Body="&lt;p&gt;Let $S=1+x+x^2+...+x^n$. Then, $xS=x+x^2+...+x^{n+1}=1+x+x^2+...+x^n+(x^{n+1}-1)=S+x^{n+1}-1$. So, $xS-S=x^{n+1}-1$. So, $S=\frac{x^{n+1}-1}{x-1}$. (The exponent of the $x$ in the numerator of the RHS should be $n+1$ not $n$).&lt;/p&gt;&#xA;" OwnerUserId="3431" LastActivityDate="2010-11-24T15:02:54.680" />
  <row Id="11705" PostTypeId="2" ParentId="11703" CreationDate="2010-11-24T15:10:23.360" Score="4" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\ \ $ The sum $\rm\:S\:$ is &quot;almost&quot; preserved by a shift symmetry $\rm\ S \to x\:S$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Examine the discrepancy $\rm\ x\:S - S\:.\ \ $ It's just the finite case of &lt;a href=&quot;http://en.wikipedia.org/wiki/Hilbert%27s%5Fparadox%5Fof%5Fthe%5FGrand%5FHotel&quot; rel=&quot;nofollow&quot;&gt;Hilbert's infinite Hotel&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-11-24T15:15:42.807" LastActivityDate="2010-11-24T15:15:42.807" />
  <row Id="11706" PostTypeId="1" CreationDate="2010-11-24T15:29:47.680" Score="3" ViewCount="193" Body="&lt;p&gt;In English (I think this is universal anyway) we use the 1s, 10s, and 100s in a cycle. One, one thousand, one million; twenty two, twenty two thousand, twenty two million; one hundred and forty six, one hundred and forty six thousand, one hundred and forty six million, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Who decided to cycle every (3x+1)th digit?&lt;/p&gt;&#xA;" OwnerUserId="3911" LastEditorUserId="2720" LastEditorDisplayName="" LastEditDate="2010-11-24T15:53:01.650" LastActivityDate="2010-11-24T17:37:15.740" Title="Why do our number start over at million, billion, etc." Tags="&lt;soft-question&gt;&lt;math-history&gt;&lt;counting&gt;" AnswerCount="2" CommentCount="8" FavoriteCount="1" />
  <row Id="11707" PostTypeId="2" ParentId="11703" CreationDate="2010-11-24T15:31:06.573" Score="1" ViewCount="" Body="&lt;p&gt;It equals (x^(n+1)-1)/(x-1), not what you wrote.&lt;/p&gt;&#xA;" LastActivityDate="2010-11-24T15:31:06.573" CommentCount="1" />
  <row Id="11708" PostTypeId="2" ParentId="11696" CreationDate="2010-11-24T15:52:54.900" Score="3" ViewCount="" Body="&lt;p&gt;A general approach: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given two points $A$ and $B$, the line that is perpendicular to the segment $AB$ and goes through its midpoint is &lt;em&gt;precisely&lt;/em&gt; the set of points that are equidistant from $A$ and $B$.&lt;/p&gt;&#xA;" OwnerUserId="462" LastActivityDate="2010-11-24T15:52:54.900" CommentCount="3" />
  <row Id="11709" PostTypeId="2" ParentId="11686" CreationDate="2010-11-24T15:54:17.200" Score="5" ViewCount="" Body="&lt;p&gt;Any rational number between 15/106 and 16/113 is closer to their mean than either of them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course this mean 3391/(2*106*113) has denominator larger than 200.  However if we examine the best approximations to this mean by rational numbers of increasing denominators, we will find if there are any before the denominator 200 is reached.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We say a rational number $p/q$ is a &lt;strong&gt;best rational approximation&lt;/strong&gt; to real number $x$ iff no rational number with denominator of smaller absolute value is closer to x.  Without loss of generality we restrict discussion to the case p,q,x &gt; 0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sequence of such best rational approximations can be derived from &lt;a href=&quot;http://en.wikipedia.org/wiki/Continued_fraction&quot; rel=&quot;nofollow&quot;&gt; the convergents of its continued fraction expansion&lt;/a&gt;.  The convergents are themselves entries in this sequence, and the only other entries are &quot;interpolants&quot; between the convergents in the following sense.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose $p_1/q_1$ and $p_2/q_2$ are a consecutive pair of convergents to $x$ (in the terminology of continued fractions).  The next convergent $p_3/q_3$ will by definition have the form $(p_1 + d*p_2)/(q_1 + d*q_2)$ where $d$ is a positive integer unless already $x$ = $p_2/q_2$.  For integers $a = 1,...,d-1$ the &quot;semiconvergents&quot; $(p_1 + a*p_2)/(q_1 + a*q_2)$ are also best rational approximations, and having denominators between $q_2$ and $q_3$, they furnish the entries in the sequence of best rational approximations between $p_2/q_2$ and $p_3/q_3$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We thus begin by &lt;a href=&quot;http://www.maths.surrey.ac.uk/hosted-sites/R.Knott/Fibonacci/cfCALC.html&quot; rel=&quot;nofollow&quot;&gt; forming the convergents&lt;/a&gt; to 3391/(2*106*113) until denominators exceed 200:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;0/1&lt;br&gt;&#xA;1/7&lt;br&gt;&#xA;15/106&lt;br&gt;&#xA;31/219&lt;br&gt;&#xA;...  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice that the only semiconvergent between the last two convergents is exactly:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;16/113 = (1+15)/(7+106)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This proves there is no rational number between 15/106 and 16/113 with denominator less than the next entry in the sequence 31/219, whose denominator exceeds 200.&lt;/p&gt;&#xA;" OwnerUserId="3111" LastActivityDate="2010-11-24T15:54:17.200" CommentCount="1" />
  <row Id="11710" PostTypeId="2" ParentId="11639" CreationDate="2010-11-24T15:55:15.157" Score="1" ViewCount="" Body="&lt;p&gt;Even for planar 3-connected graphs, which are the vertex-edge graphs that arise from convex 3-dimensional polyhedra, one can have all four possibilities: non-HC and non-Eulerian graphs, non-HC and Eulerian graphs; HC and Non-Eulerian graphs; HC and and Euler graphs, where HC means has a Hamiltonian circuit, and Eulerian means has an Eulerian circuit. (It is well known that a connected graph has an Eulerian circuit if and only if it is even-valent.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, for planar and 3-connected graphs it is of considerable interest to look at operations on the graphs that preserve the property of having an HC or Eulerian circuit or special kinds of Eulerian circuits. For example if G is plane, 3-connected and has an HC when does the dual of G have an HC? When does the medial graph for G have an HC, etc.? A famous problem which remains open is known as Barnette's conjecture: Does every 3-valent, bipartite, plane, 3-connected graph have an HC? It is also unkown if all the planar 3-connected 3-valent graphs with exactly 12 pentagons and some number of hexagonal faces have HC's.  (There graphs are known as the fullerenes.)&lt;/p&gt;&#xA;" OwnerUserId="1369" LastActivityDate="2010-11-24T15:55:15.157" />
  <row Id="11711" PostTypeId="2" ParentId="11703" CreationDate="2010-11-24T16:20:53.090" Score="2" ViewCount="" Body="&lt;p&gt;Observe that&#xA;\begin{eqnarray}&#xA;x^{n+1} - 1 = x^{n+1} + (x^{n} - x^{n}) + \cdots + (x - x) - 1 = (x^{n} + x^{n-1} + \cdots + x + 1)(x - 1).&#xA;\end{eqnarray}&lt;/p&gt;&#xA;" OwnerUserId="2720" LastActivityDate="2010-11-24T16:20:53.090" />
  <row Id="11714" PostTypeId="1" AcceptedAnswerId="11727" CreationDate="2010-11-24T16:59:56.897" Score="2" ViewCount="138" Body="&lt;p&gt;Factoring a rational function as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{(x-a)(x-b)\cdots(x-z)}{(x-A)(x-B)\cdots(x-Z)}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clearly displays the x-intercepts (roots) and vertical asymptotes (poles) of a rational function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, dividing the denominator D into the numerator N to get N = DQ+R and writing N/D as Q + R/D clearly displays the end-behavior (horizontal or even diagonal asymptotes and so forth).&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Is there some similar factorization that displays the local minimum and maximums of the real-valued real-input rational function?&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In other words, if the local extrema occur at a,b,…,z then the expression for the function has the a,b,…,z explicit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only example I can think of is &quot;completing the square&quot; exhibits the vertex of a parabola.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; After a few hours of work, I've also managed to easily express a cubic in terms of its two vertices (I had no idea cubics were so symmetrical).  I'm hoping similar ideas will work for any polynomial.&lt;/p&gt;&#xA;" OwnerUserId="583" LastEditorUserId="583" LastEditorDisplayName="" LastEditDate="2010-11-25T03:50:53.090" LastActivityDate="2010-11-25T03:50:53.090" Title="Exhibiting local extrema of rational functions" Tags="&lt;algebra&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="11715" PostTypeId="2" ParentId="11675" CreationDate="2010-11-24T17:03:56.313" Score="2" ViewCount="" Body="&lt;p&gt;They are also used to describe something called the Talbot Effect:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.phy.bris.ac.uk/people/berry_mv/research.html&quot; rel=&quot;nofollow&quot;&gt;http://www.phy.bris.ac.uk/people/berry_mv/research.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;look at #8 in the list. I attended a seminar by Mike Berry about 12 years ago where he claimed that the Talbot Effect was a physical manifestation of Gauss Sums.&lt;/p&gt;&#xA;" OwnerUserId="3919" LastActivityDate="2010-11-24T17:03:56.313" />
  <row Id="11716" PostTypeId="2" ParentId="11654" CreationDate="2010-11-24T17:14:51.317" Score="2" ViewCount="" Body="&lt;p&gt;Write $\frac{d}{dx}\int_0^1 f(x,y)dy$ as $$\lim_{h\to 0}\int_0^1 \frac{f(x+h,y)-f(x,y)}{h}dy.$$(You can do that because integration is linear.) If $$|\frac{\partial f}{\partial x}|\le M,$$ then by mean valued theorem&#xA;$$|\frac{f(x+h,y)-f(x,y)}{h}|\le M.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now apply Bounded Convergence Theorem (a weaker form of Lebesgue convergence theorem in this case.)&lt;/p&gt;&#xA;" OwnerUserId="3249" LastEditorUserId="3249" LastEditorDisplayName="" LastEditDate="2010-11-26T04:27:09.650" LastActivityDate="2010-11-26T04:27:09.650" CommentCount="3" />
  <row Id="11717" PostTypeId="2" ParentId="11010" CreationDate="2010-11-24T17:20:15.197" Score="2" ViewCount="" Body="&lt;p&gt;To amplify Chandru1's response, it is important to note the difference in how we check that a map is linear and how we check that a map is &lt;em&gt;not&lt;/em&gt; linear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to show that a map $\varphi$ is linear, we need to show that &lt;strong&gt;for every&lt;/strong&gt; $\mathbf{x}$ and &lt;strong&gt;every&lt;/strong&gt; $\mathbf{y}$ $\varphi(\mathbf{x}+\mathbf{y})=\varphi(\mathbf{x})+\varphi(\mathbf{y})$; and that &lt;strong&gt;for every&lt;/strong&gt; scalar $\alpha$ and &lt;strong&gt;every&lt;/strong&gt; $\mathbf{x}$, we have $\varphi(\alpha\mathbf{x})=\alpha\varphi(\mathbf{x})$. This is often done at the level of &quot;general formulas&quot;, checking that both expressions in each case lead to &lt;em&gt;identical&lt;/em&gt; formulas. Thus, for example, to check that you function (b) is a linear transformation, we verify that both $\varphi(f+g)$ and $\varphi(f)+\varphi(g)$ lead to the same expression:&#xA;$$\varphi(f+g) = (f+g)(1) = f(1)+g(1);\qquad \varphi(f)+\varphi(g) = f(1)+g(1);$$&#xD;&#xA;and likewise with $\varphi(af)$ and $a\varphi(f)$:&#xA;$$\varphi(af) = (af)(1) = a\cdot f(1);\qquad a\varphi(f) = a\cdot f(1).$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;However&lt;/strong&gt;, what you write in your comment for (a) is &lt;em&gt;not&lt;/em&gt; a good proof that the map $\varphi(x) = \bigl(1-x, (1-x)^2\bigr)$ is not linear. When you write something like:&#xA;$$ \varphi(x+y) = \bigl( 1-x-y, (1-x-y)^2\bigr) \neq (2 - x -y, (1-x)^2+(1-y)^2\bigr) = \varphi(x)+\varphi(y)$$&#xD;&#xA;you are actually asserting that the two expressions as &lt;strong&gt;never equal&lt;/strong&gt;. But in many instances, students will write this even when the expressions may sometimes be equal, yet think this constitutes a proof that the map is not linear. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is &lt;em&gt;not&lt;/em&gt; the case. You must either explain why the two expressions &lt;em&gt;actually&lt;/em&gt; are never equal (this happens in this case, because $1-x-y$ is never equal to $2-x-y$), or you must exhibit specific values of $x$ and $y$ for which the two expressions evaluate to different values. It is &lt;em&gt;not&lt;/em&gt; enough for the formulas to &quot;look different&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to show that the function is not linear, you must show that there exist &lt;strong&gt;specific&lt;/strong&gt; $x$ and $y$ such that $\varphi(x+y)$ is not equal to $\varphi(x)+\varphi(y)$, or that there exist a &lt;strong&gt;specific&lt;/strong&gt; $\alpha$ and a &lt;strong&gt;specific&lt;/strong&gt; $x$ such that $\varphi(\alpha x)\neq \alpha\varphi(x)$. That is, the verification that the equality does not always hold is &lt;em&gt;not&lt;/em&gt; done at the level of &quot;the formulas don't look the same&quot;, but rather by saying: &lt;em&gt;&quot;here are specific values, and, look, they don't match&quot;&lt;/em&gt;. So to show that (a) is &lt;em&gt;not&lt;/em&gt; linear, rather than working out the expressions and seeing that they don't &lt;em&gt;seem&lt;/em&gt; to be identical (just because two things don't &lt;em&gt;look&lt;/em&gt; identical doesn't mean they don't always take the same value; &quot;$\sin^2(x)+\cos^2(x)$&quot; looks very different from &quot;$1$&quot;, but they always take the same value), you should find &lt;em&gt;specific&lt;/em&gt; values of $x$ and $y$ for which the answers don't come out the same. For instance, you could show that if $x=y=1$, then&#xA;$$\varphi(1+1) = \varphi(2) = \bigl( 1-2, (1-2)^2\bigr) = (-1,1),$$&#xD;&#xA;but &#xA;$$\varphi(1)+\varphi(1) = \bigl( 1-1, (1-1)^2\bigr) + \bigl(1-1, (1-1)^2\bigr) = (0,0)+(0,0)=(0,0),$$&#xD;&#xA;and since $(-1,1)\neq (0,0)$, then $\varphi(1+1)\neq\varphi(1)+\varphi(1)$. &lt;em&gt;This&lt;/em&gt; shows the map is not linear, not the fact that the formulas for $\varphi(x+y)$ and for $\varphi(x)+\varphi(y)$ don't seem to be identical.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Think about it this way: to show that not every student in your class has red hair, it is not enough to argue that it would be highly unlikely that everyone in a big class would have such an uncommon color of hair, that it never happens, that it doesn't look that way when one takes a quick look, etc. The &lt;em&gt;sure&lt;/em&gt; way of showing that not every student in your class has red hair is to stand someone up and &lt;em&gt;show&lt;/em&gt; that he/she does not have red hair, and then you're absolutely done.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same is true for (d).&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-24T17:38:46.187" LastActivityDate="2010-11-24T17:38:46.187" CommentCount="2" />
  <row Id="11718" PostTypeId="2" ParentId="11706" CreationDate="2010-11-24T17:25:37.780" Score="3" ViewCount="" Body="&lt;p&gt;This is not universal. In Spanish, for example, you need a thousand thousands to get a million. And &quot;billion&quot; is not, as in the United States, a thousand million; no, &quot;billion&quot; means &quot;a million millions&quot;; then trillion means &quot;a billion billions&quot;; quatrillions are &quot;a trillion trillion&quot;, etc.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-24T17:35:01.840" LastActivityDate="2010-11-24T17:35:01.840" CommunityOwnedDate="2010-11-24T17:25:43.877" CommentCount="1" />
  <row Id="11719" PostTypeId="1" CreationDate="2010-11-24T17:26:11.820" Score="1" ViewCount="46" Body="&lt;p&gt;For my understanding of modular forms, I have &quot;properly&quot; read the last chapter of Serre's book on arithmetic. I have not &quot;properly&quot; read the setup of Hecke operators for congruence subgroups.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was intending to read the paper of Atkin and Lehner for filling this lacuna, as well as for understanding the proof of Atkin-Lehnher theorem. For some reason I could not relate very well with the exposition in the book of Diamond &amp;amp; Shurman or the book of Shimura.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would be most grateful if somebody can point out some good reference for the Atkin-Lehner theorem other than these.&lt;/p&gt;&#xA;" OwnerUserId="3903" LastActivityDate="2010-11-24T17:26:11.820" Title="Reference for Atkin-Lehner theorem" Tags="&lt;number-theory&gt;&lt;modular-forms&gt;" CommentCount="1" />
  <row Id="11720" PostTypeId="1" CreationDate="2010-11-24T17:30:40.493" Score="2" ViewCount="145" Body="&lt;p&gt;What is a good reference for the story of congruences such as &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle \tau(n) \equiv \sigma(11)(n) \mod\ 691$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with a conceptual explanation with connections to étale cohomology, etc?&lt;/p&gt;&#xA;" LastEditorUserId="8246" LastEditorDisplayName="" LastEditDate="2011-04-26T20:05:57.960" LastActivityDate="2011-04-26T20:05:57.960" Title="Ramanujan congruences and étale cohomology" Tags="&lt;number-theory&gt;&lt;algebraic-geometry&gt;&lt;modular-forms&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="11721" PostTypeId="2" ParentId="11706" CreationDate="2010-11-24T17:37:15.740" Score="3" ViewCount="" Body="&lt;p&gt;Warning: I am not a linguist, so my answer is to be taken as an educated guess, no more.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is essentially historical and the development of this system is told at  &lt;a href=&quot;http://en.wikipedia.org/wiki/Long_and_short_scales&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Long_and_short_scales&lt;/a&gt; . As said above, it is not universal, even in other base-10 languages : East Asia and India have different systems. Even in western scientific tradition, Archimedes used a myriad (10⁴) based system for big numbers and French use the long scale, a 10⁶ based system. The choice of a system grouping the digits by 3 is therefore an historical accident, and we could as well have ended with a different grouping.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the need of very large number seems to only exists for some technical use (science, accounting, religion) and do not emerge naturally in the language evolution. Therefore, the systems for big numbers are constructed by the specialist in question and do not blend well with the other numeral. In French, for example, the word &quot;mille&quot; (thousand) is grammatically a number (a cardinal numeral) but &quot;million&quot; is a simple noun. Hence the &quot;need&quot; for a &quot;restart&quot;. I guess it's similar for other languages.&lt;/p&gt;&#xA;" OwnerUserId="3918" LastActivityDate="2010-11-24T17:37:15.740" />
  <row Id="11722" PostTypeId="2" ParentId="11720" CreationDate="2010-11-24T17:43:05.237" Score="3" ViewCount="" Body="&lt;p&gt;The paper of Swinnerton-Dyer in Lecture Notes in Math 350, as well as Serre's Bourbaki talk on the same topic. &lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2010-11-24T17:43:05.237" />
  <row Id="11723" PostTypeId="2" ParentId="11703" CreationDate="2010-11-24T17:50:23.837" Score="0" ViewCount="" Body="&lt;p&gt;For a more mechanical proof, you could use induction. The proof then boils down to finding a common denominator:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{x^{n+1}-1}{x-1} + x^{n+1} = \frac{x^{n+1}-1+(x-1)x^{n+1}}{x-1} = \frac{x^{n+2}-1}{x-1}$&lt;/p&gt;&#xA;" OwnerUserId="3060" LastActivityDate="2010-11-24T17:50:23.837" />
  <row Id="11724" PostTypeId="1" CreationDate="2010-11-24T17:50:43.870" Score="8" ViewCount="293" Body="&lt;p&gt;What is the compelling need for introducing a theory of $p$-adic integration?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do the existing theories of $p$-adic integration use some kind of analogues of Lebesgue measures? That is, do we put a Lebesgue measure on $p$-adic spaces, and just integrate real or complex valued functions on $p$-adic spaces, or is something more possible like integrating $p$-adic valued functions on $p$-adic spaces? What is the machinery used?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then again, does the integration on spaces like $\mathbb C_p$ give something more than the usual integration in real analysis? I mean, the integration of complex valued functions of complex variables, or more precisely holomorphic functions, is much a much more interesting topic than measure theory. Is a similar analogue true in $p$-adic cases?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have also seen mentioned that Grothendieck's cohomology theories like etale cohomology, crystalline cohomology etc., fit into such $p$-adic integration theories. What could possibly be the connection?&lt;/p&gt;&#xA;" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-11-25T18:43:01.407" LastActivityDate="2010-11-26T05:33:52.650" Title="Theories of $p$-adic integration" Tags="&lt;number-theory&gt;&lt;algebraic-geometry&gt;&lt;arithmetic-geometry&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" />
  <row Id="11726" PostTypeId="1" AcceptedAnswerId="12020" CreationDate="2010-11-24T17:56:34.630" Score="4" ViewCount="327" Body="&lt;p&gt;This is a follow up of &lt;a href=&quot;http://math.stackexchange.com/questions/11602/limit-of-sn-displaystyle-sum-k1-infty-left1-prod-j1n-1-le&quot;&gt;Limit of $S(n) = \sum_{k=1}^{\infty} \left(1 - \prod_{j=1}^{n-1}\left(1-\frac{j}{2^k}\right)\right)$&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More details can be found in the above thread.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $S(n) = \displaystyle \sum_{k=1}^{\infty} \left(1 - \prod_{j=1}^{n-1}\left(1-\frac{j}{2^k}\right)\right)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mike has proved that $S(n)$ in fact diverges at-least faster than $\log_2(\lfloor n-1 \rfloor)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now based on what Mike has worked this conjectures arises:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle \lim_{n \rightarrow \infty} (2 \log_{2}(n) - S(n)) = \alpha$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, can $\alpha$ be expressed in terms of other familiar constants. $\frac{\pi \gamma}{e}$ seems to be a close guess.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The numerical evidence seem to suggest they are true.  For example, we have the following graph of $2 \log_2 n - S(n)$ for $n \leq 300$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/E5byB.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(More numerical evidence: The value of $2 \log_2 n - S(n)$, is, for $n = 1000$, $2000$, and $3000$, respectively, $0.667734$, $0.667494$, and $0.667413$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An alternative expression for $S(n)$ was worked out by Moron in the previously-mentioned question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$S(n) = - \sum_{k=1}^{n-1} \frac{s(n,k)}{2^{n-k}-1},$$ &#xA;where $s(n,k)$ is a &lt;a href=&quot;http://en.wikipedia.org/wiki/Stirling_numbers_of_the_first_kind&quot; rel=&quot;nofollow&quot;&gt;Stirling number of the first kind&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-11-27T05:10:59.820" LastActivityDate="2010-11-27T21:14:25.560" Title="Limit of $S(n) = \sum_{k=1}^{\infty} \left(1 - \prod_{j=1}^{n-1}\left(1-\frac{j}{2^k}\right)\right)$ - Part II" Tags="&lt;number-theory&gt;&lt;real-analysis&gt;&lt;combinatorics&gt;&lt;sequences-and-series&gt;&lt;asymptotics&gt;" AnswerCount="3" CommentCount="12" FavoriteCount="1" />
  <row Id="11727" PostTypeId="2" ParentId="11714" CreationDate="2010-11-24T18:08:55.120" Score="3" ViewCount="" Body="&lt;p&gt;I don't think there is a nice answer for the general case of your question. However, there is nice answer to a closely related question - see the review below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;MR1752251 (2001c:11035) 11D25 (11D41 11G05 11G30)&lt;br&gt;&#xA;Buchholz, Ralph H.; MacDougall, James A.(5-NEWC)&lt;br&gt;&#xA;&lt;strong&gt;When Newton met Diophantus: a study of rational-derived polynomials&lt;br&gt;&#xA;and their extension to quadratic fields.&lt;/strong&gt;&lt;br&gt;&#xA;J. Number Theory 81 (2000), no. 2, 210–233.&lt;br&gt;&#xA;&lt;a href=&quot;http://dx.doi.org/10.1006/jnth.1999.2473&quot; rel=&quot;nofollow&quot;&gt;http://dx.doi.org/10.1006/jnth.1999.2473&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is an interesting paper, which surveys the problem of determining the set $D(n)$ of all &quot;$k$-derived'' univariate polynomials of degree $n$ (where a polynomial $f \in k[x]$ is $k$-derived if $f$ and each of its successive derivatives has all roots in the ground field $k$). Define two polynomials $f_1,f_2\in k[x]$ to be equivalent if $f_1(x)=r f_2(s x+t)$ for $r,s,t\in k$, $r,s \neq 0$. Then up to equivalence, the following is known about $\mathbb Q$-derived polynomials: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$D(1)=\{x\};\quad D(2)=\{x^2,x(x-1)\};$$&#xD;&#xA;$$D(3)=\{x^3\}\cup\bigg\{x(x-1)(x-a)\ :\ a=\frac{w(w-2)}{w^2-1},w\in \mathbb Q\bigg\};$$&#xD;&#xA;$$ D(4)\supseteq \{x^4\}\cup\bigg\{x^2(x-1)(x-a)\ :\ a=\frac{9(2w+z-12)(w+z)}{(z-w-18)(8w+z)},&#xD;&#xA; (w,z)\in E(\mathbb Q), E\colon z^2=w(w-6)(w+18)\bigg\};$$&#xD;&#xA;$$ D(n)\supseteq \{x^n, x^{n-1}(x-1)\}\ {\rm for}\ n\geq 5.$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The authors prove that determining $D(n)$ in general devolves into two conjectures: (1) that no quartic with four distinct roots is $\mathbb Q$-derived; (2) that no quintic of type $x^3(x-a)(x-b)$, $a\neq b,\ a,b\neq0$, is $\mathbb Q$-derived. The first conjecture can be solved by determining all rational points on a hyperelliptic surface of degree 10. The second conjecture can be solved by determining all rational points on a curve of genus 2 (E. V. Flynn [&quot;On $\mathbb Q$-derived polynomials'', Preprint; per revr.] has now proved this second conjecture). The authors also discuss briefly the situation of $K$-derived polynomials for quadratic extensions $K$ of $\mathbb Q$; there is, for example, the quartic $y^2=x^2(x-1)(x-\frac{37-20\sqrt{3}}{13})$ which is a ${\mathbb Q}(\sqrt{3})$-derived polynomial.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reviewed by Andrew Bremner&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-11-24T18:08:55.120" CommentCount="2" />
  <row Id="11728" PostTypeId="1" AcceptedAnswerId="11733" CreationDate="2010-11-24T18:12:55.553" Score="0" ViewCount="56" Body="&lt;p&gt;An annuity immediate has quarterly payments of $1000$ for $25$ years at a rate of $6 \%$ converted quarterly. Find its present value. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The effective interest rate $i$ is $0.06/4$. So let $v = \frac{1}{1+i}$. Then wouldn't the present value be $1000 \left(\frac{1-v^{25}}{i} \right)$? &lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2010-11-24T19:25:27.380" Title="Annuity immediate" Tags="&lt;finance&gt;" AnswerCount="3" />
  <row Id="11729" PostTypeId="2" ParentId="11720" CreationDate="2010-11-24T18:16:12.040" Score="1" ViewCount="" Body="&lt;p&gt;The paper of Deligne, Formes modulaires et représentations $l$-adiques, available at &lt;a href=&quot;http://www.numdam.org/numdam-bin/fitem?id=SB_1968-1969__11__139_0&quot; rel=&quot;nofollow&quot;&gt;numdam&lt;/a&gt;, has these things.&lt;/p&gt;&#xA;" OwnerUserId="1119" LastEditorUserId="1119" LastEditorDisplayName="" LastEditDate="2010-11-24T18:26:24.980" LastActivityDate="2010-11-24T18:26:24.980" />
  <row Id="11731" PostTypeId="1" CreationDate="2010-11-24T18:24:52.840" Score="2" ViewCount="65" Body="&lt;p&gt;In various descriptions of the moduli space of elliptic curves with level structures, such as the description of $X_0(N)$ being defined over $Spec\ \mathbb{Z}[1/N]$, the primitive $N^{th}$ roots of unity pop up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is somewhere a description available as to how this connection arises? I mean, somewhere manageable, without wading through the heavy volumes of Deligne &amp;amp; Rapoport or Katz &amp;amp; Mazur?&lt;/p&gt;&#xA;" LastActivityDate="2010-11-24T18:24:52.840" Title="Primitive nth roots of unity and moduli of elliptic curves" Tags="&lt;arithmetic-geometry&gt;" />
  <row Id="11732" PostTypeId="2" ParentId="11728" CreationDate="2010-11-24T18:32:19.490" Score="1" ViewCount="" Body="&lt;p&gt;For starters, there are 100 periods, not 25.  If you just reduce each payment to present value, what is the value of the nth payment?  Can you then sum this from 1 to 100?&lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2010-11-24T19:25:27.380" LastActivityDate="2010-11-24T19:25:27.380" />
  <row Id="11733" PostTypeId="2" ParentId="11728" CreationDate="2010-11-24T18:36:55.193" Score="1" ViewCount="" Body="&lt;p&gt;The present value of an income stream with $n$ equally spaced payments, with an interest of $r$ at each payment, each payment of $C$, is obtained by computing the present value of each payment and adding it up. Here, you have $r=0.06/4 = 0.015$, as you note, so you get&#xA;$$1000 + \frac{1000}{1+r} + \frac{1000}{(1+r)^2} + \cdots + \frac{1000}{(1+r)^{100}},$$&#xD;&#xA;since there are 100 payments (one payment per quarter, 25 years, beginning now).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is just&#xA;$$1000\left(1+\frac{1}{1+r} + \left(\frac{1}{1+r}\right)^2 + \cdots + \left(\frac{1}{1+r}\right)^{100}\right)$$&#xD;&#xA;which is&#xA;$$1000\left( \frac{1 - \frac{1}{(1+r)^{101}}}{1-\frac{1}{1+r}}\right).$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you only get 1000 payments (no payment at the end of the last quarter 25 years from now) then the sum only goes up to exponent 99; adjust accordingly.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-24T19:23:21.733" LastActivityDate="2010-11-24T19:23:21.733" />
  <row Id="11735" PostTypeId="1" AcceptedAnswerId="11738" CreationDate="2010-11-24T18:56:11.987" Score="7" ViewCount="235" Body="&lt;p&gt;Hi there,&#xA;I want to find the volume of a torus with a given thickness and a given radius.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let r be the radius of a circle with its midpoint at $M(0|b)$ ($b \geq r$). Now I want to rotate this circle about the x-axis, that is to say about a circular path which has the length $2 \pi \cdot |b|$. So I thought I'd simply integrate:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$V = \int\limits_0^{2 \pi \cdot |b|} \pi r^2 dz = 2 \pi \cdot r^2 \cdot |b|$, which turns out to be the correct result.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I don't find it trivial that the volume of this torus is the same as the volume of a cylinder with the corresponding height. I read the article on Wikipedia about the torus and it said that this was due to Cavalieri's theorem, which to my mind doesn't really have a lot to do with the torus vs. the cylinder...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there some easy way to prove that a torus has the same volume as a cylinder with the height equal to the torus' perimeter?&lt;/p&gt;&#xA;" OwnerUserId="3787" LastActivityDate="2010-11-24T20:37:55.747" Title="How to prove that a torus has the same volume as a cylinder (with the height equal to the torus' perimeter)" Tags="&lt;geometry&gt;" AnswerCount="3" CommentCount="3" />
  <row Id="11736" PostTypeId="2" ParentId="11728" CreationDate="2010-11-24T19:19:42.583" Score="1" ViewCount="" Body="&lt;p&gt;If the first payment is received at the end of the first quarter, then the present value is $$1000\left(\frac{1-v^{100}}{i}\right).$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the first payment is received at the beginning of the first quarter (this seems to be the case, since you stated &quot;immediate&quot;), then the present value is $$1000\left(\frac{1+i-v^{99}}{i}\right),$$&#xD;&#xA;assuming a total of 100 payments are made.&lt;/p&gt;&#xA;" OwnerUserId="3249" LastActivityDate="2010-11-24T19:19:42.583" />
  <row Id="11737" PostTypeId="2" ParentId="11735" CreationDate="2010-11-24T19:34:49.757" Score="-1" ViewCount="" Body="&lt;p&gt;Perhaps I don't understand the question, but if you have a cylinder, you can imagine bending it so that the top and bottom are connected. Then you have a torus.&lt;/p&gt;&#xA;" OwnerUserId="2954" LastActivityDate="2010-11-24T19:34:49.757" CommentCount="3" />
  <row Id="11738" PostTypeId="2" ParentId="11735" CreationDate="2010-11-24T19:47:09.163" Score="14" ViewCount="" Body="&lt;p&gt;(i) Slice the torus into a million near-disks.&lt;br&gt;&#xA;(ii) Rotate every second disk through 180 degrees.&lt;br&gt;&#xA;(iii) Stick them all together again.&lt;br&gt;&#xA;You get a near-cylinder whose height is nearly 2πb. Now let a million tend to infinity.&lt;/p&gt;&#xA;" OwnerUserId="1508" LastActivityDate="2010-11-24T19:47:09.163" CommentCount="1" />
  <row Id="11739" PostTypeId="1" AcceptedAnswerId="11741" CreationDate="2010-11-24T20:02:47.903" Score="3" ViewCount="154" Body="&lt;p&gt;If $\ln(1-x+x^2)  = a_1x+a_2x^2 + \cdots \text{ then } a_3+a_6+a_9+a_{12} + \cdots = $ ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My approach is to write $1-x+x^2 = \frac{1+x^3}{1+x}$ then expanding the respective logarithms,I got a series (of coefficient) which is nothing but $\frac{2}{3}\ln 2$.But this approach took some time for me (I can't solved it during the test but after the test I solved it)... Any other quick method?&lt;/p&gt;&#xA;" OwnerUserId="2109" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-11-24T20:31:48.770" LastActivityDate="2010-11-24T23:24:26.027" Title="Summing up the series $a_{3k}$ where $\log(1-x+x^2) = \sum a_k x^k$" Tags="&lt;calculus&gt;&lt;logarithms&gt;&lt;power-series&gt;" AnswerCount="3" />
  <row Id="11740" PostTypeId="2" ParentId="11726" CreationDate="2010-11-24T20:07:33.100" Score="-4" ViewCount="" Body="&lt;p&gt;We have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$S(n) - \beta \log(n) = \displaystyle S(n) - \sum_{k=1}^{\infty} \beta \frac{x^k}{k} = \displaystyle \sum_{k=1}^{\infty} \left(1 - \prod_{j=1}^{n-1}\left(1-\frac{j}{2^k}\right) - \beta \frac{x^k}{k}\right)$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $x = \frac{n-1}{n}$. If in (i) we can guarantee the interchange of the limit and the summation we can conclude&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle \lim_{n \rightarrow \infty} S(n) - \beta \log(n) = \lim_{n \rightarrow \infty} \left[  \sum_{k=1}^{\infty} \left(1 - \prod_{j=1}^{n-1}\left(1-\frac{j}{2^k}\right) - \beta \frac{x^k}{k}\right) \right] = \sum_{k=1}^{\infty}  \left(1 - \lim_{n \rightarrow \infty}\prod_{j=1}^{n-1}\left(1-\frac{j}{2^k}\right) - \beta \lim_{n \rightarrow \infty}\frac{x^k}{k}\right)= \sum_{k=1}^{\infty} 1 - \frac{\beta}{k}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which diverges.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am, however, still thinking about whether the interchange is legitimate, I think it should be... perhaps by the &lt;a href=&quot;http://en.wikipedia.org/wiki/Dominated_convergence_theorem&quot; rel=&quot;nofollow&quot;&gt;Dominated convergence theorem&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Possible Completion of Proof of Divergence:&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;We can show that the interchange of limits holds by the &lt;a href=&quot;http://en.wikipedia.org/wiki/Dominated_convergence_theorem&quot; rel=&quot;nofollow&quot;&gt;Dominated convergence theorem&lt;/a&gt; if $\displaystyle \sum_{k=1}^{\infty} \left|1 - \prod_{j=1}^{n-1}\left(1-\frac{j}{2^k}\right) - \beta \frac{x^k}{k}\right| &amp;lt; \infty$ does converge absolutely. But then the interchange of limits shown above shows that (i) diverges. &lt;/p&gt;&#xA;" OwnerUserId="1077" LastEditorUserId="1077" LastEditorDisplayName="" LastEditDate="2010-11-24T21:06:41.467" LastActivityDate="2010-11-24T21:06:41.467" CommentCount="4" />
  <row Id="11741" PostTypeId="2" ParentId="11739" CreationDate="2010-11-24T20:13:20.440" Score="13" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Lemma:&lt;/strong&gt; Let $f(x) = \sum a_n x^n$ be a power series.  Then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum a_{3n} x^{3n} = \frac{f(x) + f(\omega x) + f(\omega^2 x)}{3}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\omega = e^{ \frac{2\pi i}{3} }$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Proof.&lt;/em&gt;  Ignoring convergence it suffices to prove this for a single term, and then it boils down to the identity&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{1 + \omega^n + \omega^{2n}}{3} = \begin{cases} 1 \text{ if } 3 | n \\\ 0 \text{ otherwise} \end{cases}.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a special case of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Discrete_Fourier_transform&quot; rel=&quot;nofollow&quot;&gt;discrete Fourier transform&lt;/a&gt;.  Applying the lemma, we readily obtain that the desired sum is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\ln 1 + \ln (1 - \omega + \omega^2) + \ln (1 - \omega^2 + \omega)}{3} = \frac{1}{3} \ln (2 \omega^2 \cdot 2 \omega) = \frac{2}{3} \ln 2$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where we use the fact that $1 + \omega + \omega^2 = 0$.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-11-24T20:13:20.440" CommentCount="8" />
  <row Id="11744" PostTypeId="2" ParentId="11735" CreationDate="2010-11-24T20:37:55.747" Score="3" ViewCount="" Body="&lt;p&gt;To use Cavalieri's theorem, lay the torus and cylinder on a table and slice them with planes parallel to the table. Then it suffices to show that the torus slices (annuli) have the same area as the cylinder slices (rectangles).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At height $h$ (measured from the centre of the torus or cylinder), the annulus has inner and outer radii $$R_\pm = |b| \pm \sqrt{r^2-h^2},$$ so its area is $$A_1 = \pi R_+^2 - \pi R_-^2 = \pi(R_+ + R_-)(R_+ - R_-) = \pi \cdot 2|b| \cdot 2\sqrt{r^2-h^2}.$$ The rectangle has width $2\sqrt{r^2-h^2}$ and length $2\pi |b|$, so its area is $$A_2 = 2\sqrt{r^2-h^2} \cdot 2\pi|b|.$$ The areas are equal, so we're done!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I prefer Pappus's centroid theorem though...&lt;/p&gt;&#xA;" OwnerUserId="3747" LastActivityDate="2010-11-24T20:37:55.747" />
  <row Id="11746" PostTypeId="2" ParentId="11739" CreationDate="2010-11-24T21:23:20.390" Score="1" ViewCount="" Body="&lt;p&gt;Generally one can employ evaluations at roots of unity to take arbitrary such &lt;strong&gt;multisections&lt;/strong&gt; of power series. See my &lt;a href=&quot;http://math.stackexchange.com/questions/3510/how-to-prove-eulers-formula-expi-t-costi-sint/3514#3514&quot;&gt;post here&lt;/a&gt; and the links there for discussion and examples. Generally, if we let $\rm\:w\:$ denote a primitive $\rm n$'th root of unity, then the $\rm m$'th $\rm n$-section of a power series selects the linear progression of $\rm\: m + k\:n\:$ indexed terms, viz. if $\rm\ f\:(X)\ =\ a_0 + a_1 X + a_2 X^2 +\: \cdots\ $ then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\quad\quad\quad\quad a_{m}\ X^m + a_{m+n}\ X^{m+n} + a_{m+2n}\ X^{m+2n}\ +\: \cdots $&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\quad\displaystyle\ =\ \frac{1}n\: (\:f\:(X) + f\:(Xw)\ w^{-m} + f\:(X\:w^2)\ w^{-2m} +\:\cdots\: \:+\ f\:(X\:w^{n-1})\ w^{(1-n)\:m}\:) $&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See Riordan's classic textbook &quot;Combinatorial Identities&quot; for many applications of  multisections.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-11-24T23:24:26.027" LastActivityDate="2010-11-24T23:24:26.027" />
  <row Id="11747" PostTypeId="1" AcceptedAnswerId="11752" CreationDate="2010-11-24T21:37:08.870" Score="1" ViewCount="370" Body="&lt;p&gt;How can I prove that dividing an irrational number by an irrational number (besides himself) &lt;strong&gt;can&lt;/strong&gt; result in an integer?&lt;/p&gt;&#xA;" OwnerUserId="1445" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2011-01-02T04:29:41.227" LastActivityDate="2011-01-02T04:29:41.227" Title="Proof that dividing irrational number by an irrational number can result in an integer?" Tags="&lt;real-number-computation&gt;&lt;irrational-numbers&gt;" AnswerCount="9" CommentCount="10" />
  <row Id="11748" PostTypeId="2" ParentId="3510" CreationDate="2010-11-24T21:40:12.020" Score="0" ViewCount="" Body="&lt;p&gt;An easy way out to answer the question is when we try to extend the definition of exponential to complex plane in a &quot;nice&quot; way (read &quot;nice&quot; as holomorphic) we then end up with this definition. And as you are probably aware there is only at most one such extension. So if we want to define our exponential in the complex plane such that the exponential function is holomorphic, and it matches with the conventional exponential function on the real line, we end up with $e^{it} = \cos(t) + i \sin(t)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;All the other answers I think are circular in some sense. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, how do we know that we can expand $e^{i x}$ as a power series. All we know is that we can expand $e^{x}$ as a power series when $x$ is real. We do not know apriori that the expansion can be carried forward if $x$ happens to be complex.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2010-11-24T21:54:47.170" LastActivityDate="2010-11-24T21:54:47.170" CommentCount="1" />
  <row Id="11749" PostTypeId="2" ParentId="11747" CreationDate="2010-11-24T21:40:40.650" Score="5" ViewCount="" Body="&lt;p&gt;By giving an example. For instance, $2\sqrt{2}$ divided by $\sqrt{2}$? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Note well:&lt;/em&gt; To prove that something &lt;strong&gt;can&lt;/strong&gt; happen, all you need to do is exhibit a single instance of it happening. To prove that something &lt;strong&gt;must&lt;/strong&gt; happen you need to show that it happens always. So to show that dividing an irrational number by an irrational number other than itself &lt;strong&gt;can&lt;/strong&gt; be an integer, all you need to do is cook up a single example of it happening. If you wanted to prove that, say, a rational divided by rational other than zero &lt;strong&gt;must&lt;/strong&gt; be a rational, an example would not suffice; you would need to show that for &lt;em&gt;any&lt;/em&gt; rational $q$ and &lt;em&gt;any&lt;/em&gt; rational $r\neq 0$, you would have $\frac{q}{r}$ a rational. &lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-11-24T21:40:40.650" />
  <row Id="11750" PostTypeId="2" ParentId="11747" CreationDate="2010-11-24T21:42:03.377" Score="4" ViewCount="" Body="&lt;p&gt;A trivial example would be $2\pi/\pi=2$. In fact every example would be of this form (one irrational would be an integer multiple of the other), for if $a,b$ are distinct irrational numbers such that $\frac{a}{b}=c$ where $c$ is an integer, then, $a=bc$&lt;/p&gt;&#xA;" OwnerUserId="3431" LastEditorUserId="3431" LastEditorDisplayName="" LastEditDate="2010-11-24T21:47:07.847" LastActivityDate="2010-11-24T21:47:07.847" />
  <row Id="11751" PostTypeId="2" ParentId="11747" CreationDate="2010-11-24T21:42:35.700" Score="0" ViewCount="" Body="&lt;p&gt;$\frac{nr}{r}$, where $n \in \mathbb{Z} \backslash \{0\}$ and $r \in \mathbb{R} \backslash \mathbb{Q}$&lt;/p&gt;&#xA;" OwnerUserId="3302" LastActivityDate="2010-11-24T21:42:35.700" />
  <row Id="11752" PostTypeId="2" ParentId="11747" CreationDate="2010-11-24T21:42:36.153" Score="4" ViewCount="" Body="&lt;p&gt;An example would suffice to prove this claim. If you don't want to divide an irrational number by itself, how about $2\sqrt{2}/\sqrt{2}$. You should be able to prove that if a number is irrational, then twice that number is also irrational.&lt;/p&gt;&#xA;" OwnerUserId="576" LastActivityDate="2010-11-24T21:42:36.153" />
  <row Id="11753" PostTypeId="2" ParentId="11747" CreationDate="2010-11-24T21:42:45.263" Score="1" ViewCount="" Body="&lt;p&gt;@Arturo Magidin: Did you read the &quot;besides himself&quot; part?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;@Alon: How about $\frac{2 \pi}{\pi}$ or something similar?&lt;/p&gt;&#xA;" OwnerUserId="3787" LastActivityDate="2010-11-24T21:42:45.263" CommentCount="1" />
  <row Id="11754" PostTypeId="2" ParentId="11747" CreationDate="2010-11-24T21:44:35.793" Score="2" ViewCount="" Body="&lt;p&gt;Start by assuming that you have a case where it &lt;em&gt;does&lt;/em&gt; result in an integer, and see what that says about the irrational numbers.  From there you should be able to give examples.  That is, if $x/y = n$ for some integer $n$, what does that say about the relationship between $x$ and $y$?  Choosing a specific integer $n$ and a specific $y$ should lead you to an example.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(While writing this, many answers were posted.  I'm adding this anyway to give a naive approach to use if you don't already see how it will come out.)&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-11-24T21:44:35.793" />
  <row Id="11755" PostTypeId="2" ParentId="11739" CreationDate="2010-11-24T21:51:11.717" Score="3" ViewCount="" Body="&lt;p&gt;Qiaochu Yuan gives an excellent exposition of the general method. Sometimes in specific cases you can get the answer via some hands-on calculations as well, and expanding the logs as you mentioned can actually be done pretty quickly: using that $1 - x + x^2 = {1 + x^3 \over 1 + x}$ one gets that $\ln(1 - x + x^2) = \ln(1 + x^3) - \ln(1 + x)$. You now can use the fact that the sum of $a_3 + a_6 + ...$ is obtained by subtracting the corresponding sums for $\ln(1 + x^3)$ and $\ln(1 + x)$. Since the power series for $\ln(1 + x^3)$ only contains powers of $x^3$, the contribution to $a_3 + a_6 + ...$ coming from that term is what you get from plugging in $x = 1$, namely $\ln(2)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The power series of $\ln(1 + x)$ may be written as $-\sum_{n &amp;gt; 0} {(-x)^n \over n}$.&#xA;Taking every third term gives $-\sum_{n &amp;gt; 0}  {(-x)^{3n} \over 3n} = -{1 \over 3}\sum_{n &amp;gt; 0} {(-x^3)^n \over n}$. Note this series is again the series of $\ln(1 + x)$, but applied to $x^3$ in place of $x$. Plugging in $x = 1$ gives ${1 \over 3}\ln(2)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the answer you want is $\ln(2) - {1 \over 3}\ln(2) = {2 \over 3}\ln(2)$.&lt;/p&gt;&#xA;" OwnerUserId="3035" LastEditorUserId="3035" LastEditorDisplayName="" LastEditDate="2010-11-24T22:03:06.470" LastActivityDate="2010-11-24T22:03:06.470" />
  <row Id="11756" PostTypeId="2" ParentId="11747" CreationDate="2010-11-24T21:51:13.367" Score="1" ViewCount="" Body="&lt;p&gt;If x is an irrational then x/-x works and is the canonical example that satisfies the OP's rules&lt;/p&gt;&#xA;" OwnerUserId="3878" LastActivityDate="2010-11-24T21:51:13.367" />
  <row Id="11757" PostTypeId="1" CreationDate="2010-11-24T22:31:20.513" Score="3" ViewCount="203" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The series for $\log (1+x)$ is convergent(by estimating radius of convergence) only with $|x| &amp;lt;1$. Then how is it still true for $x = 1$?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;How is it still apparently true when $x$ is a roots of unity other than $-1$? Such as in the answers to this question: &lt;a href=&quot;http://math.stackexchange.com/questions/11739&quot;&gt;Summing up the series $a_{3k}$ where $\log(1-x+x^2) = \sum a_k x^k$&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" LastActivityDate="2010-11-24T22:53:04.547" Title="Convergence for log 2" Tags="&lt;real-analysis&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="11758" PostTypeId="2" ParentId="11757" CreationDate="2010-11-24T22:36:12.127" Score="1" ViewCount="" Body="&lt;p&gt;Look up alternating series test or generalized alternating series test (Also known as Dirichlet's test as George pointed out).&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2010-11-24T22:53:04.547" LastActivityDate="2010-11-24T22:53:04.547" CommentCount="2" />
  <row Id="11759" PostTypeId="2" ParentId="11757" CreationDate="2010-11-24T22:38:21.207" Score="13" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Abel%27s_theorem&quot; rel=&quot;nofollow&quot;&gt;Abel's theorem&lt;/a&gt; is the key to why you can say that the limit as you approach the boundary comes out to the same as evaluating the series at the boundary.  Sivaram's answer addresses why the series converges at those boundary points.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Explicitly, Abel's theorem implies that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\lim_{r\nearrow 1}\log(1+zr)=\sum_{n=1}^\infty\frac{(-1)^{n+1}}{n}z^n$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for each complex number $z$ with $|z|=1$ such that the series converges.  By continuity of the logarithm, this will be $\log(1+z)$ in such cases.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2010-11-24T22:51:44.813" LastActivityDate="2010-11-24T22:51:44.813" />
  <row Id="11760" PostTypeId="2" ParentId="11747" CreationDate="2010-11-24T22:46:54.420" Score="3" ViewCount="" Body="&lt;p&gt;The basic idea works much more generally: it amounts to how a subgroup acts on its complement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm If\:\ \ a,b \in A \subset B\ \ \: then\ \ ab \in  A \iff b \in A\ \ $ if  $\rm\ \ a^{-1}\in A,\ $ and $\rm\ A,\:B\ $ are monoids.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus $\rm\ \ b\not\in A\ \ \Rightarrow\ \ ab\not\in A\ \ and\ \ (ab)/b = a\in A\ $ as you desire.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your case is simply $\rm\ \ A = \mathbb Q,\ \ B = \mathbb R,\ \ a\in \mathbb Z$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The key property above is essentially this &quot;complementary&quot; view of a subgroup:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;THEOREM&lt;/strong&gt;  $\ \ $ A nonempty subset $\rm\:A\:$ of abelian group $\rm\:B\:$ &#xA;comprises a subgroup $\rm\iff\ A\ \bar A\ = \bar A\ $ where  $\rm\: \bar A\:$ is the complement of $\rm\:A\:$ in $\rm\:B$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instances of this are ubiquitous in concrete number systems, e.g. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;&lt;pre&gt;&lt;code&gt;                                 transcendental &#xA;     algebraic * nonalgebraic  =  nonalgebraic  if  nonzero &#xA;      rational * irrrational   =   irrational   if  nonzero &#xA;          real *   nonreal     =    nonreal     if  nonzero &#xA;&#xA;         even  +     odd       =      odd          additive example&#xA;       integer + noninteger    =   noninteger&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-11-24T23:16:32.743" LastActivityDate="2010-11-24T23:16:32.743" />
  <row Id="11761" PostTypeId="1" CreationDate="2010-11-24T23:29:25.130" Score="4" ViewCount="176" Body="&lt;p&gt;From the wikipedia page:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Motive_(algebraic_geometry&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Motive_(algebraic_geometry&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;it appears that the category of Mixed motives $MM(k)$ over a field $k$ is still conjectural; but there is a good derived category $DMM(k)$ already constructed for this. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is a good reference for this construction, and why this derived category is the &quot;suitable&quot; one.&lt;/p&gt;&#xA;" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-11-25T10:24:07.307" LastActivityDate="2010-11-26T05:50:07.007" Title="Status of mixed motives" Tags="&lt;number-theory&gt;&lt;algebraic-geometry&gt;&lt;homological-algebra&gt;" AnswerCount="2" />
  <row Id="11762" PostTypeId="2" ParentId="5231" CreationDate="2010-11-24T23:35:50.863" Score="0" ViewCount="" Body="&lt;p&gt;At the moment, I went with a somewhat simple solution that allows me to draw a plane/face with 4 vertices with arbitrary rotations, in 2D:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's how it works:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;PVector[] unRotateVerts(PVector[] verts,PVector n){&#xA;  //get the angle between the face4 normal and Y&#xA;  angle = PVector.angleBetween(n,y);//acos(n.dot(y));&#xA;  //get the axis of rotation, by getting the perpendicular&#xA;  axis = n.cross(y);&#xA;  axis.normalize();&#xA;  //clone vertices&#xA;  int vl = verts.length;&#xA;  PVector[] clone = new PVector[vl];&#xA;  for(int i = 0; i&amp;lt;vl;i++) clone[i] = verts[i].get();&#xA;  //make a rotation matrix and rotate it&#xA;  PMatrix3D rMat = new PMatrix3D();&#xA;  rMat.rotate(angle,axis.x,axis.y,axis.z);&#xA;  //multiply each vertex with the rotation matrix&#xA;  PVector[] dst = new PVector[vl];&#xA;  for(int i = 0; i&amp;lt;vl;i++) {&#xA;    dst[i] = new PVector();&#xA;    rMat.mult(clone[i],dst[i]);&#xA;  }&#xA;  return dst;&#xA;}&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This works perfectly for aligning with XZ, but rotations on the Z axis are still there. Any hints on how to remove that would be handy.&lt;/p&gt;&#xA;" OwnerUserId="304" LastActivityDate="2010-11-24T23:35:50.863" />
  <row Id="11763" PostTypeId="1" CreationDate="2010-11-24T23:50:56.167" Score="6" ViewCount="248" Body="&lt;p&gt;It is a bit easier to understand the homology $H_1(X, \mathbb Z)$ for various compact surfaces in analogy with handles and so on. There seems to be a nice intuitive picture with handles, holes, etc to think of the first homology group, and similar heuristics for higher homology groups.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But almost all axiomatic treatment of homology groups uses instead the relative homology. But it is not so intuitively clear how to visualize the relative homology groups.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are some intuitive crutches for dealing with these relative homology groups, particularly for surfaces?&lt;/p&gt;&#xA;" LastActivityDate="2010-12-27T17:28:46.127" Title="Meaning of relative homology" Tags="&lt;algebraic-topology&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="1" />
  <row Id="11764" PostTypeId="1" CreationDate="2010-11-24T23:55:37.283" Score="0" ViewCount="96" Body="&lt;p&gt;Sheaf cohomology is a well-studied topic with a lot of references available. For example Hartshorne's book. But for a certain paper I am reading now, I have to understand sheaf homology.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could someone tell me some references? This is just sheaf homology in the topological sense; not in algebraic geometry.&lt;/p&gt;&#xA;" LastActivityDate="2010-11-25T20:42:07.207" Title="References for sheaf homology" Tags="&lt;algebraic-topology&gt;" AnswerCount="2" />
  <row Id="11765" PostTypeId="1" CreationDate="2010-11-25T00:04:45.040" Score="2" ViewCount="74" Body="&lt;p&gt;Consider the power series ring in noncommutative associative formal variables $X, Y$ over the field $\mathbb C$. Let $P$ be a power series. Define the exponential $e^P$ and logarithm $\log (1+P)$ using the usual series expansions. The exponential maps from a neighborhood of $0$ to a neighborhood of $1$ and the logarithm goes the other way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $P$ be such a power series with zero constant term. How to show that the logarithm and exponential maps defined above are inverses of each other?&lt;/p&gt;&#xA;" LastActivityDate="2010-11-25T00:17:01.087" Title="Exponential and Logarithm in noncommutative setting" Tags="&lt;abstract-algebra&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="11766" PostTypeId="2" ParentId="11765" CreationDate="2010-11-25T00:17:01.087" Score="1" ViewCount="" Body="&lt;p&gt;Powers of $P$ commute so you need not worry about the non-commutative underlying variables. Expand $e^{\log(1+P)}$ in powers of $P$ and calculate the power of $P^t$ for each $t$; you'll get $1+P$. That's something like Moebius inversion. You can also do it the opposite way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, you don't even need to calculate it, since we already know the answer by substitution for the corresponding &quot;real&quot; power series, which do converge in some neighborhood, and the answers must match.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-25T00:17:01.087" />
  <row Id="11767" PostTypeId="2" ParentId="11747" CreationDate="2010-11-25T00:24:36.527" Score="2" ViewCount="" Body="&lt;p&gt;An example from Euler together with the proof of the irrationality of $\pi^2$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\dfrac{\pi^2}{\zeta(2)}=6,$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\zeta(2)=\displaystyle\sum_{k=1}^{\infty}\dfrac{1}{k^2}.$$&lt;/p&gt;&#xA;" OwnerUserId="752" LastActivityDate="2010-11-25T00:24:36.527" />
  <row Id="11768" PostTypeId="1" CreationDate="2010-11-25T00:27:07.227" Score="1" ViewCount="114" Body="&lt;p&gt;I was wondering about the generalization of the concept of convolution from the familiar one on real spaces and how many properties still remain. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For convolution on Lebesgue-integrable real-valued functions defined on a group equipped with measure, does each of the following properties holds? If not, what extra requirements needed on the group, measure and/or functions in order that it can become true:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) Commutativity: $f * g = g * f \,$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) Associativity: $f * (g * h) = (f * g) * h \,$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(3) Linearity (This, I am more sure, holds, but I might be wrong):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Distributivity:    $f * (g + h) = (f * g) + (f * h) \,$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Associativity with scalar multiplication: $  a (f * g) = (a f) * g = f * (a g) \, $&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(4) Time invariant: $f(t+\tau)*g(t)=h(t+\tau), \forall t, \tau$ in the group.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(5) the reverse of (3) and (4), i.e.: Any linear and time-invariant operation on the set of Lebesgue-integrable real-valued functions must be the convolution with some Lebesgue-integrable real-valued function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks and regards!&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Edit:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The definition for the convolution on a group with measure is defined according to &lt;a href=&quot;http://en.wikipedia.org/wiki/Convolution#Convolutions_on_groups&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If G is a suitable group endowed with&#xA;  a measure λ, and if f and g are real&#xA;  or complex valued integrable functions&#xA;  on G, then we can define their&#xA;  convolution by&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;$(f * g)(x) = \int_G f(y) g(-y+x)\,d\lambda(y). \,$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="1281" LastEditorUserId="1543" LastEditorDisplayName="" LastEditDate="2011-04-03T17:59:20.390" LastActivityDate="2011-04-03T17:59:20.390" Title="Convolution on group with measure" Tags="&lt;group-theory&gt;&lt;harmonic-analysis&gt;&lt;convolution&gt;&lt;classical-analysis&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="11769" PostTypeId="1" AcceptedAnswerId="11776" CreationDate="2010-11-25T00:44:33.407" Score="7" ViewCount="121" Body="&lt;p&gt;Let $M$ be a smooth connected manifold. Is it always possible to find a connected dense open subset $U$ of $M$ which is diffeomorphic to an open subset of &lt;strong&gt;R&lt;/strong&gt;$^n$? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we don't require $U$ to be connected, the answer is yes: it is enough to construct a countable collection of disjoint open &quot;affines&quot; whose union is dense, and this is not terribly difficult.&lt;/p&gt;&#xA;" OwnerUserId="546" LastEditorUserId="546" LastEditorDisplayName="" LastEditDate="2010-11-25T01:16:20.663" LastActivityDate="2010-11-25T02:19:45.387" Title="Do all manifolds have a densely defined chart?" Tags="&lt;topology&gt;&lt;differential-topology&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="1" />
  <row Id="11770" PostTypeId="1" AcceptedAnswerId="11772" CreationDate="2010-11-25T00:56:44.640" Score="0" ViewCount="174" Body="&lt;p&gt;I'm trying to solve a differential equation problem that has to do with barometric pressure at a certain altitude.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rate at which barometric pressure decreases with altitude is proportional to the pressure at that altitude. If the altitude in feet, then the constant of proportionality is 3.7 x 10&lt;sup&gt;5&lt;/sup&gt;. The barometric pressure at sea level is 29.92 inches of mercury.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How would I calculate the barometric pressure, in inches of mercury, at any altitude using the information above?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The differential equation that I came up with looks like this:&lt;br /&gt;&#xA;Let $P=$ Barometric Pressure&lt;br /&gt;&#xA;Let $a=$ Altitude&lt;br /&gt;&#xA;$\dfrac{dP}{da}=-k(29.92-P)$ &lt;br /&gt;&#xA;But when I solve it, it doesn't look like it's the right answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for your help!&lt;/p&gt;&#xA;" OwnerUserId="3946" LastActivityDate="2010-11-25T01:02:45.637" Title="Differential Equations: Calculating Barometric Pressure at an Altitude" Tags="&lt;calculus&gt;&lt;differential-equations&gt;" AnswerCount="1" />
  <row Id="11771" PostTypeId="1" AcceptedAnswerId="11774" CreationDate="2010-11-25T01:01:01.740" Score="3" ViewCount="112" Body="&lt;p&gt;I have a question about the set of functions from a set to another set. I am wondering about the degenerate cases. Suppose $X^Y$ denotes the set of functions from a set $Y$ to a set $X$, why is $|Y^{\emptyset}|=1$ but $|\emptyset^Y|=0$ where $Y\neq \emptyset$?&lt;/p&gt;&#xA;" OwnerUserId="3947" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-25T01:03:23.953" LastActivityDate="2010-11-25T01:23:23.637" Title="Why is $|Y^{\emptyset}|=1$ but $|\emptyset^Y|=0$ where $Y\neq \emptyset$" Tags="&lt;elementary-set-theory&gt;" AnswerCount="2" />
  <row Id="11772" PostTypeId="2" ParentId="11770" CreationDate="2010-11-25T01:02:45.637" Score="2" ViewCount="" Body="&lt;p&gt;Pressure decreases exponentially with altitude and thats what you will get provided you fix your signs in the equation. Your equation will look like:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{dP}{da} = -kP$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $k$ is some positive constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solve this to get $P(a) = P(0) e^{-ka}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and at $a = 0$, $P(0) = 29.92$. Make sure to keep track of all your units.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastActivityDate="2010-11-25T01:02:45.637" CommentCount="1" />
  <row Id="11773" PostTypeId="2" ParentId="11771" CreationDate="2010-11-25T01:05:21.173" Score="4" ViewCount="" Body="&lt;p&gt;Because the &lt;a href=&quot;http://en.wikipedia.org/wiki/Empty_function&quot; rel=&quot;nofollow&quot;&gt;empty function&lt;/a&gt; is the unique function from the empty set to an arbitrary set $Y$, while if $Y\neq\emptyset$, then there exists $y\in Y$, but there's no place in $\emptyset$ for a function to send $y$.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-11-25T01:05:21.173" CommentCount="2" />
  <row Id="11774" PostTypeId="2" ParentId="11771" CreationDate="2010-11-25T01:09:10.903" Score="6" ViewCount="" Body="&lt;p&gt;The definition of $A^B$ is &quot;the set of all functions with domain $B$ and codomain $A$&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A function $f$ from $B$ to $A$ is a set of ordered pairs such that:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;If $(x,y)\in f$, then $x\in B$ and $y\in A$.&lt;/li&gt;&#xA;&lt;li&gt;For every $b\in B$ there exists $a\in A$ such that $(b,a)\in f$.&lt;/li&gt;&#xA;&lt;li&gt;If $(b,a)$ and $(b,a&amp;#39;)$ are in $f$, then $a=a&amp;#39;$.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Now, what happens if $B=\emptyset$? Well, then there can be no pair in $f$, because you cannot have $x\in B$. But notice that in that case, 2 is satisfied &quot;by vacuity&quot; (if it were false, you would be able to exhibit a $b\in\emptyset$ for which there is no $a\in A$ with $(b,a)\in f$; but there &lt;em&gt;are&lt;/em&gt; no $b\in\emptyset$, so you cannot make such an exhibition; the statement is true because the premise, &quot;$b\in\emptyset$&quot;, can never hold). Likewise 3 holds by vacuity. So it turns out that if we take $f=\emptyset$, then $f$ satisfies 1, 2, and 3, and therefore it is by all rights a &quot;function from $\emptyset$ to $A$&quot;.  But this is the only possible function from $\emptyset$ to $A$, because only the empty set works.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By contrast, if $A=\emptyset$, but $B\neq\emptyset$, then no set $f$ can satisfy both 1 and 2, so no set can be a function from $B$ to $A$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That means that $Y^{\emptyset}$ always contains exactly one element, namely the &quot;empty function&quot;, $\emptyset$. But if $Y\neq\emptyset$, then $\emptyset^Y$ contains &lt;em&gt;no&lt;/em&gt; elements; that is, it is empty. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, since $Y^{\emptyset}$ has exactly one element, $|Y^{\emptyset}|=1$ regardless of what $Y$ is. But if $Y\neq\emptyset$, then $\emptyset^{Y}$ is empty, so $|\emptyset^{Y}| = 0$. &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-25T01:15:45.263" LastActivityDate="2010-11-25T01:15:45.263" CommentCount="2" />
  <row Id="11776" PostTypeId="2" ParentId="11769" CreationDate="2010-11-25T01:41:19.397" Score="4" ViewCount="" Body="&lt;p&gt;If the manifold is compact, the answer is yes.  The geometric idea is to put a small balloon in the manifold, and to inflate it.  Since the manifold is compact, you eventually fill the manifold, so you're left with the interior of the balloon (an open ball) with identifications on the boundary, so it's dense. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can make this precise by putting a Riemann metric on the manifold and using the Hopf-Rinow Theorem. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The relation between my answer and Jason's comment would be to take a handle decomposition of the manifold and to consider a maximal tree in the dual 1-skeleton as giving a &quot;tree-like ball&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;edit: responding to Qiaochu's comment, I believe the answer is affirmative for connected, non-compact manifolds as well.  The technique of the proof has to be adapted some.  Step 1: construct a proper Morse function on $M$, meaning a Morse function $f : M \to [0,\infty)$ such that it has only one local minimum -- which is the global minimum, $0$, and demand that $f^{-1}[0,c]$ is connected and compact for all $c\geq 0$.  Step 2: you construct the charts on $f^{-1}[c_i,c_{i+1}]$ where $c_1&amp;lt;c_2&amp;lt;\cdots$ are an increasing sequence $\lim_{n\to\infty} c_n =\infty$ which are regular values of $f$.  Step 3: paste the charts together along little arcs connecting them, pairwise. &lt;/p&gt;&#xA;" OwnerUserId="642" LastEditorUserId="642" LastEditorDisplayName="" LastEditDate="2010-11-25T02:19:45.387" LastActivityDate="2010-11-25T02:19:45.387" CommentCount="1" />
  <row Id="11777" PostTypeId="2" ParentId="11769" CreationDate="2010-11-25T02:06:48.710" Score="1" ViewCount="" Body="&lt;p&gt;Depending on what you consider a manifold, the &lt;a href=&quot;http://en.wikipedia.org/wiki/Long_line_%28topology%29&quot; rel=&quot;nofollow&quot;&gt;long line&lt;/a&gt; may be a counterexample.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And for a non-connected manifold, surely the answer is no?  Take your favorite smooth manifold, and take a disjoint union of more than $\mathfrak{c}$ copies of it.  Again, unless your definition of &quot;manifold&quot; rules this out (by assuming separability, etc).&lt;/p&gt;&#xA;" OwnerUserId="822" LastActivityDate="2010-11-25T02:06:48.710" CommentCount="2" />
  <row Id="11778" PostTypeId="1" CreationDate="2010-11-25T02:54:13.690" Score="0" ViewCount="198" Body="&lt;p&gt;I hope that someone's up for the challenge; I'm attempting to solve this via computer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;\int_{-\pi}^\pi{\displaystyle \frac{e^{i\cdot a\cdot t}(e^{i\cdot b\cdot t}-1)(e^{i\cdot c \cdot t}-1)}{(e^{i\cdot t}-1)(e^{i\cdot d \cdot t}+1)(e^{i\cdot f \cdot t}-1)} \dots dt}&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to know if it's possible to break this up into simpler subproblems.  You can use just about anything to do this, but there is one restriction.  In computer science terms, I want this to be in $P$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me try to explain.  I don't want the work I have to do grow exponentially.  For instance, if I break up the integral into two integrals, I don't want to double the amount of work I have to do.  I don't want to dramatically increase the amount of work I have to do by breaking things apart, I want to keep things fairly fast.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;  I'd prefer to see integration techniques used for this.&lt;/p&gt;&#xA;" OwnerUserId="2626" LastEditorUserId="2626" LastEditorDisplayName="" LastEditDate="2010-11-25T03:23:14.230" LastActivityDate="2011-01-28T00:02:22.903" Title="Insidious exponential integral" Tags="&lt;intuition&gt;&lt;integral&gt;" AnswerCount="1" CommentCount="9" />
  <row Id="11779" PostTypeId="2" ParentId="11669" CreationDate="2010-11-25T02:55:59.880" Score="2" ViewCount="" Body="&lt;p&gt;I agree with the theory that the distinction between the notes is used for visual aid and reference points. In addition to that, it was meant to be treated as a vertically rising instrument as if you were to go up a ladder of sorts and those accidentals ( in the case of C, the black notes) are the grips to reach to the next level. As we would refer to them as leading notes. There is also another reason why there are that many notes. Almost all scales are a variation of the major scale or aeolian mode. This scale is designed to have a certain number of tones and semitones to give it the feel of a major scale. If there were too many tones or too many semitones it would not be the same because it will produce too much dissonance or invariably consonance. That is why there is a standard tuning for pianos i.e. A440. If the interval in vibration were to be changed it would not be the same because if the vibrations aren't in sync, the resonation will be totally off. That is why there can be only soo many notes on a piano and make sense to the human ear. Other tunings are possible but the same effect is made that the intervals are kept in a strict way to keep harmony. So, getting back to your question Mathematically, yes there is a reason for that specific order of white keys and black keys. Most of its relation deals with the mode theory of 12 notes and the circle of fifths where if you were to expand the notes on the piano it will form a perfect circle in diminished chords of C as the cardinal points. If you were to go in fifths in a clockwise direction the circle will be C g d A e b F#/Gb d#/db Ab Eb Bb F where when compressed into one octave it turns out with 7 natural notes and 5 accidentals&lt;/p&gt;&#xA;" OwnerUserId="3954" LastActivityDate="2010-11-25T02:55:59.880" />
  <row Id="11780" PostTypeId="1" CreationDate="2010-11-25T03:06:04.110" Score="7" ViewCount="112" Body="&lt;p&gt;I have the following problem. There is a binary relation $\succeq$ defined on a separable metric space $X$, and $\succeq$ is total, transitive, and closed (as a subset of $X\times X$, in the product metric). There exists a set $Y\subseteq X$ that bounds the elements of $X$ in order (that is, for each $x \in X$ one can find $y_{1},y_{2} \in Y$ such that $y_{1}\succeq x\succeq y_{2}$). The set $X$ is itself convex, and the order $\succeq$ is affine when restricted to $Y$ (you may interpret the restriction of $\succeq$ to $Y$ as a convex subset of $Y \times Y$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The equivalence class $[x]$ the element $x$ belongs to intersects the set $Y$. Informally, I want to construct a measurable selection from that intersection: a (Borel) measurable function that maps $x$ to an element $y$ of $Y$ such that $y\in [x]$. When the set $X$ is compact I can construct it easily. In general, for $X$ separable, I am not able to find one. I came across some measurable selection theorems, but they all work for Polish spaces. I also found a theorem in Kechris's book (Classical Descriptive Set Theory) that could be useful, but it also works for Polish spaces only.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;My question is&lt;/em&gt;: Where can I find (measurable) selection theorems from equivalence classes? Is there any special theory for it (or counterexamples)? (Sorry for the vague question, but I am not quite acquainted with the literature.)&lt;/p&gt;&#xA;" OwnerUserId="3959" LastActivityDate="2011-04-17T13:42:08.387" Title="Measurable selection from equivalence classes" Tags="&lt;reference-request&gt;" AnswerCount="1" CommentCount="0" />
  <row Id="11781" PostTypeId="2" ParentId="11495" CreationDate="2010-11-25T03:31:23.667" Score="0" ViewCount="" Body="&lt;p&gt;Get a solid foundation in problem solving. Many CS educations are actually not that math heavy but often rely on what is generally know as a &quot;certain mathematical maturity&quot;. Luckily quite a few good books exists about this subject, but none more important (and accssible/enjoyable) than: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.amazon.com/How-Solve-Aspect-Mathematical-ebook/dp/B001R23FWA/&quot; rel=&quot;nofollow&quot;&gt;How to solve it&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you stay clear of the graphics/optimization (and image processing) curses then you might get a graduate degree without doing any Linear Algebra, however you should properly pick up a basic book on discrete mathematics. Plus if you plan to do any programming lanugage theory you should read a book about mathematical proofs (In which case you should pick &lt;a href=&quot;http://www.amazon.com/How-Prove-Structured-Approach-ebook/dp/B000VHZRU0/&quot; rel=&quot;nofollow&quot;&gt;How to prove it&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And last, the only way to learn math is to do math, and in the context of CS you can get lots of good training by starting on &lt;a href=&quot;http://projecteuler.net/&quot; rel=&quot;nofollow&quot;&gt;project Euler&lt;/a&gt;. Finally top it off with reading a good algorithm book like &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/1848000693&quot; rel=&quot;nofollow&quot;&gt;The Algorithm Design Manual&lt;/a&gt; by&#xA;Steven S. Skiena.&lt;/p&gt;&#xA;" OwnerUserId="3493" LastActivityDate="2010-11-25T03:31:23.667" CommunityOwnedDate="2010-11-25T12:19:27.163" />
  <row Id="11782" PostTypeId="1" AcceptedAnswerId="11783" CreationDate="2010-11-25T03:35:12.920" Score="0" ViewCount="66" Body="&lt;p&gt;Let $\{r_1, r_2, r_3,\ldots \}$ be an enumeration of the rationals in $[0,1]$, and let $(a_k)$ be a sequence such that $\sum_{k=1}^{\infty}{a_k}$ converges absolutely. Set $f_k(x) = a_k\,|x-r_k|^{-1/2}$. I'd like to show that $$\sum_{k=1}^{\infty}{\int_{[0,1]}{|f_k(x)|\,dx}}&amp;lt;\infty\,.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've rewritten the sum as $$\sum{|a_k|\,\int_0^1{\dfrac{1}{|x-r_k|^{1/2}}\,dx}}\,.$$ For each $k$, I broke up the integral into two integrals to account for the absolute value. After integrating, I got $2(\sqrt{r_k}+\sqrt{1-r_k})$. (If anyone is willing to double check that... that'd be awesome.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But now how do I show that $$2\sum{|a_k|\,(\sqrt{r_k}+\sqrt{1-r_k})}$$ converges? That's where I'm stuck.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: You know, I typed all this stuff up, and then as soon as I submitted, I realized that $\sqrt{r_k}$ and $\sqrt{1-r_k}$ are bounded by $1$. So $\sum{|a_k|\,(\sqrt{r_k}+\sqrt{1-r_k})}\le \sum{2|a_k|}$, and we know that $\sum{|a_k|}$ is convergent. It's a simple comparison. (Someone please tell me I'm correct, haha.)&lt;/p&gt;&#xA;" OwnerUserId="2313" LastActivityDate="2010-11-25T03:52:11.700" Title="Show a series of integrals is convergent" Tags="&lt;real-analysis&gt;&lt;homework&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="11783" PostTypeId="2" ParentId="11782" CreationDate="2010-11-25T03:40:13.207" Score="1" ViewCount="" Body="&lt;p&gt;$\textbf{HINT:}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that $\sqrt{r_k} + \sqrt{1-r_k} \leq \sqrt{2}$, $\forall r_k \in \mathbb{Q} \cap [0,1]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Prove the above by Arithmetic mean - Geometric Mean Inequality&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2010-11-25T03:52:11.700" LastActivityDate="2010-11-25T03:52:11.700" />
  <row Id="11784" PostTypeId="1" AcceptedAnswerId="11800" CreationDate="2010-11-25T03:45:49.537" Score="2" ViewCount="172" Body="&lt;p&gt;So I am studying for my calculus midterm and I had the following question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $f_{x}(a, b)$ and $f_{y}(a,b)$ both exist then $f$ is differentiable at $(a,b)$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I answered true, but the answer is false. Why is that? I haven't done real analysis yet (dropped it this term) so I won't know what your talking about if its too advanced. If you attempt to use epsilons and deltas then please try to explain it as simple as possible. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="287" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-25T20:00:07.407" LastActivityDate="2010-11-25T23:14:23.170" Title="Differentiable function - Easy Calc." Tags="&lt;calculus&gt;" AnswerCount="5" CommentCount="1" />
  <row Id="11785" PostTypeId="2" ParentId="11784" CreationDate="2010-11-25T03:50:14.870" Score="1" ViewCount="" Body="&lt;p&gt;Think about when can you say a function is differentiable?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remember that $f_x$ and $f_y$ gives you the partial derivatives in just 2 directions namely $X$ axis and $Y$ axis respectively.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastActivityDate="2010-11-25T03:50:14.870" />
  <row Id="11786" PostTypeId="2" ParentId="11784" CreationDate="2010-11-25T03:52:24.927" Score="4" ViewCount="" Body="&lt;p&gt;$f=\frac{xy}{x^2+y^2}$ when $(x,y)\neq (0,0)$ and $f(0,0)=0$ is a standard counterexample for this. $f_x(0,0)=f_y(0,0)=0$, but $f$ is not differentiable (or even continuous at the origin).&lt;/p&gt;&#xA;" OwnerUserId="3431" LastActivityDate="2010-11-25T03:52:24.927" />
  <row Id="11787" PostTypeId="1" AcceptedAnswerId="11794" CreationDate="2010-11-25T04:00:45.723" Score="4" ViewCount="128" Body="&lt;p&gt;Why is it that cubics are separable over fields that are not of characteristic $2$ or $3$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the starting point for some a discussion of the Galois group of a cubic, but I seem to be stuck right off the bat. &lt;/p&gt;&#xA;" OwnerUserId="3656" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-25T04:19:48.467" LastActivityDate="2010-11-25T07:39:52.907" Title="Why are cubics separable over fields that are not characteristic 2 or 3" Tags="&lt;abstract-algebra&gt;&lt;polynomials&gt;&lt;galois-theory&gt;" AnswerCount="2" />
  <row Id="11788" PostTypeId="2" ParentId="11495" CreationDate="2010-11-25T04:01:30.817" Score="0" ViewCount="" Body="&lt;p&gt;I'm currently in CS, so i could give you some titles of the books i have used over the years. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Discrete mathematics with applications by Susanna S.Epp&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Calculus by Larson, Hostetler, Edwards ISBN:  0-618-14918-X I like this book the best for calc. I have like 5 calc books so...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also You will probably be learning some sort of assembly language. I can't remember if i even had a book for that, but we use MIPS.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We had to learn java for first year programming language, which almost any book from amazon should do. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For C, once you learn java you should pick up C pretty quickly. I really don't even read this book. The class i am taking now is Haskell, ruby and prolog in one semester. So, haskell would be a nice little language to learn. Its pretty cool.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But i would definitely learn Calc and discrete math or statistics. discrete math is somewhat similar to stats so it would help.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if a place to start right now is Calc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also i recommend CLEPing some of your general lower division course. This will save you money. Any REA book or even &quot;5 steps to A 5&quot; by mcgraw hill. While it will say AP what subject you take i found it useful for passing. I studied probably 2 weeks for one class i have never taken in my life and passed. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are also plenty of resources out there like MIT online and Stanford online. They have free courses online.&lt;/p&gt;&#xA;" OwnerUserId="3961" LastActivityDate="2010-11-25T04:01:30.817" CommunityOwnedDate="2010-11-25T12:19:27.163" />
  <row Id="11789" PostTypeId="2" ParentId="11044" CreationDate="2010-11-25T04:06:19.677" Score="0" ViewCount="" Body="&lt;p&gt;So I'll give the proof I'm used to.  It relies on what (I've been led to believe) Brendan McKay coined the &quot;switching principle&quot;:  Suppose you have a bipartite multi-graph, with parts $A$ and $B$.  Suppose every vertex in $A$ has degree $c$ and every vertex in $B$ has degree $d$.  Then $Ac=Bd$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We say a Latin rectangle is &lt;em&gt;normalised&lt;/em&gt; if the first row is $(1,2,\ldots,n)$.  Let $K_{k,n}$ be the number of $k \times n$ normalised Latin rectangles.  By permuting the columns of a $k \times n$ Latin rectangle, we generate $n!$ distinct Latin rectangles of which exactly one is normalised.  Hence $L_{k,n}=n! K_{k,n}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We will now prove the identity $K_{k,n}=\frac{(n-1)!}{(n-k)!} R_{k,n}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We take A to be the set of all &lt;em&gt;normalised&lt;/em&gt; $k \times n$ Latin rectangles and take B to be the set of all &lt;em&gt;reduced&lt;/em&gt; $k \times n$ Latin rectangles.  [Actually, B is a subset of A, so this is strictly speaking a bit suspect, but this can be overcome this problem by labelling (I won't do this here, as it's a bit messy).]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $G$ be the group of permutations of $\{1,2,\ldots,n\}$ that fix 1.  Given any reduced Latin rectangle $L \in B$ and $\alpha \in G$ we can construct a normalised Latin rectangle by: starting from $L$, permute the columns by $\alpha$, then permute the symbols by $\alpha$.  Call this Latin square $L_\alpha$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We construct the bipartite multigraph by: for every $L \in B$ add an edge from $L$ to $L_\alpha$ for all $\alpha \in G$.  Since $L_\alpha$ might equal $L_\beta$, the graph might not be simple.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Every vertex in $B$ has degree $(n-1)!$.  So it is sufficient to show that every vertex in $A$ has degree $(n-k)!$.  There is an edge between $L=(l_{ij}) \in A$ and some vertex in $B$ for each permutation $\alpha \in G$ satisfying $\alpha(i)=l_{i1}$ for all $1 \leq i \leq k$.  There are $(n-k)!$ such permutations, and hence every vertex in $A$ has degree $(n-k)!$.&lt;/p&gt;&#xA;" OwnerUserId="139" LastActivityDate="2010-11-25T04:06:19.677" />
  <row Id="11790" PostTypeId="1" CreationDate="2010-11-25T04:10:10.633" Score="0" ViewCount="77" Body="&lt;p&gt;Can elliptic arc (defined as part of an ellipse, with extent not bigger than 90 degrees) be represented by quadratic Bezier curve?&lt;/p&gt;&#xA;" OwnerUserId="3777" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-25T04:33:59.260" LastActivityDate="2010-11-25T04:33:59.260" Title="Can elliptic arc be represented by quadratic Bezier curve?" Tags="&lt;geometry&gt;&lt;bezier-curve&gt;" CommentCount="4" />
  <row Id="11792" PostTypeId="1" CreationDate="2010-11-25T04:26:48.053" Score="4" ViewCount="212" Body="&lt;p&gt;I went through this thread on &lt;a href=&quot;http://math.stackexchange.com/questions/329/best-ever-book-on-number-theory&quot;&gt;this question&lt;/a&gt; asking about a text on number theory, but my question is slightly different. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was looking for a short and sweet book on elementary number theory. I don't need it to be a definitive treatise on number theory. I just need to have book which covers enough material to advance to algebraic number theory. I have looked at the books by Niven-Zuckerman-Montgomery and the Jones' and the one by Burton. All of them were very readable and I would like to read at least one of them at some point. For now, I would just like a book (say 100 pages long) which can get me acquainted with most of the material needed for algebraic number theory. Basically there is a course on algebraic number theory that I wish to take which begins in a month's time and I have not done enough elementary number theory. &lt;/p&gt;&#xA;" OwnerUserId="3947" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-25T04:47:23.330" LastActivityDate="2010-11-25T09:47:12.820" Title="Best introductory text on elementary number theory that is concise" Tags="&lt;elementary-number-theory&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="11794" PostTypeId="2" ParentId="11787" CreationDate="2010-11-25T04:32:33.070" Score="3" ViewCount="" Body="&lt;p&gt;A polynomial has multiple roots if and only if it is not relatively prime to its formal derivative. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose that $f(x) = x^n + a_{n-1}x^{n-1}+\cdots+a_0$ is an irreducible polynomial of degree $n\gt 0$ over a field $\mathbf{F}$. Then $f&amp;#39;(x) = nx^{n-1}+(n-1)a_{n-1}x^{n-1}+\cdots + a_1$. Because $f(x)$ does not have any proper factors in $\mathbf{F}[x]$, then the only possible common factors of $f(x)$ and $f&amp;#39;(x)$ are, up to units, $1$ and $f(x)$. But if $f&amp;#39;(x)\neq 0$, then it is of degree strictly smaller than $f(x)$, so it cannot be divisible by $f(x)$. So &lt;em&gt;either&lt;/em&gt; $f&amp;#39;(x)=0$, or else $\gcd(f,f&amp;#39;)=1$. But the only way in which we can have $f&amp;#39;(x)=0$ is if $n=0$ in $\mathbf{F}$; that is, the characteristic of $\mathbf{F}$ divides $n$. (In fact, more is true: $f(x)$ must be expresssible as $g(x^p)$, where $g(x)$ is a polynomial and $p$ is the characteristic of $\mathbf{F}$, but that is not needed here). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, a polynomial is separable if and only if its &lt;em&gt;irreducible factors&lt;/em&gt; are separable. Take a cubic polynomial $f(x) = x^3 + ax^2+bx+c$. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;If $f(x)$ is irreducible, then the only way it can fail to be separable is if $\mathrm{char}(\mathbf{F})|3$, that is, if the characteristic of $\mathbf{F}$ is $3$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If $f(x)$ is the product of an irreducible quadratic and a linear polynomial, $f(x) = \ell(x)q(x)$, then $\ell(x)$ is certainly separable, so the only way for $f(x)$ to fail to be separable is for $q(x)$ to not be separable; this is a monic polynomial of degree $2$, so for $q(x)$ to &lt;em&gt;not&lt;/em&gt; be separable we need $\mathrm{char}(\mathbf{F})$ to divide $2$; that is, $\mathbf{F}$ to be of characteristic $2$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If $f(x)$ is a product of three linear polynomials (possibly repeated), then $f(x)$ is separable, because each irreducible factor is linear and hence separable.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So, a &lt;em&gt;necessary&lt;/em&gt; condition for a cubic polynomial to &lt;strong&gt;not&lt;/strong&gt; be separable is for the characteristic of the field of definition to be either $2$ or $3$. In consequence, if $f(x)$ is a cubic over a field of characteristic not equal to $2$ or to $3$, then it &lt;em&gt;must&lt;/em&gt; be separable.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-11-25T04:32:33.070" CommentCount="2" />
  <row Id="11795" PostTypeId="2" ParentId="11784" CreationDate="2010-11-25T04:45:25.623" Score="3" ViewCount="" Body="&lt;p&gt;Need the partial derivatives exist and are $continuous$ in a open  ball around $(a, b)$. This is sufficient. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;&#xA;You can check : &lt;a href=&quot;http://en.wikipedia.org/wiki/Partial_derivative&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Partial_derivative&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3421" LastEditorUserId="3421" LastEditorDisplayName="" LastEditDate="2010-11-25T13:22:23.633" LastActivityDate="2010-11-25T13:22:23.633" />
  <row Id="11796" PostTypeId="2" ParentId="7090" CreationDate="2010-11-25T04:45:47.753" Score="4" ViewCount="" Body="&lt;p&gt;A complete residue system modulo $n$ is a set of $n$ integers containing a representative element from each congruence class modulo $n$ (see &lt;a href=&quot;http://en.wikipedia.org/wiki/Modular_arithmetic&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt; for more details).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A commonly used complete residue system is $\{0,1,\ldots,n-1\}$ where it is easy to see that any two elements are incongruent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The claim states that if gcd(a,n)=1, then $\{0,a,2a,\ldots,(n-1)a\}$ is a complete residue system modulo $n$.  Since there are exactly $n$ elements in this set, in order to prove the claim, it is sufficient to show that $ka \not\equiv ra \pmod n$ whenever $0 \leq k&amp;lt;r \leq n-1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Step 1: $ka \equiv ra \pmod n$ implies $k \equiv r \pmod n$.  While true, I think it should be remarked that we may &quot;divide&quot; by $a$ only because it is comprime to $n$.  In fact, it's not really division, it's multiplication by an integer $b$ such that $ab \equiv 1 \pmod n$ (which can be shown to exist using the &lt;a href=&quot;http://en.wikipedia.org/wiki/Extended_Euclidean_algorithm&quot; rel=&quot;nofollow&quot;&gt;Extended Euclidean Algorithm&lt;/a&gt;).  So $ka \equiv ra \pmod n$ implies $kab \equiv rab \pmod n$ implies $k \equiv r \pmod n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Step 2: $k \equiv r \pmod n$ implies $n$ divides $k-r$.  There's a typo in the OP's statement.  This is by the definition of congruent modulo n.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Step 3: $n$ divides $k-r$ implies $k=r$ because $0 \leq k&amp;lt;r \leq n-1$.  There's some information left out here in the OP's statement.  In order to have $n$ dividing $k-r$ while both $k$ and $r$ are between $0$ and $n-1$, we must have $k=r$ (note that 0 is divisible by n).&lt;/p&gt;&#xA;" OwnerUserId="139" LastActivityDate="2010-11-25T04:45:47.753" />
  <row Id="11797" PostTypeId="2" ParentId="11792" CreationDate="2010-11-25T04:46:26.843" Score="3" ViewCount="" Body="&lt;p&gt;The thing about algebraic number theory is that, although the problems you will be looking at come from &quot;classical&quot; number theory, the techniques owe a lot more to abstract algebra than to classical number theory. You would be better served making sure that you have a solid understanding of commutative rings, field extensions, Galois Theory, and the basics of modules over commutative rings. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In that respect, the book by Ireland and Rosen that was mentioned in that thread, &lt;em&gt;A Classical Introduction to Modern Number Theory&lt;/em&gt;, may be a good one, since it is leading you in &lt;em&gt;that&lt;/em&gt; direction in any case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I took Algebraic Number Theory (which I absolutely loved; it was a graduate two-course sequence), I had not had any number theory beyond the very elementary congruence stuff; I had never seen anything like Quadratic Reciprocity, or Fermat's Christmas Theorem, or anything like that. I did not feel hobbled by this lack. But I certainly had to lean heavily on the commutative algebra and Galois Theory I knew! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It was only &lt;em&gt;later&lt;/em&gt; that I sat in on an upper-division number theory course (based on the Ireland-Rosen book mentioned above), and that I read through Gauss's &lt;em&gt;Disquisitiones&lt;/em&gt; (on my own) and other number theory books. Then again, I'm not a number theorist, so perhaps I am in fact deeply handicapped by this but I don't know it.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-25T04:52:17.443" LastActivityDate="2010-11-25T04:52:17.443" CommentCount="3" />
  <row Id="11798" PostTypeId="2" ParentId="6575" CreationDate="2010-11-25T05:02:56.693" Score="1" ViewCount="" Body="&lt;p&gt;There's a lot of similar things that are called &quot;trees&quot; in graph theory.  I interpret this question to mean full binary trees (see &lt;a href=&quot;http://en.wikipedia.org/wiki/Binary_tree&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;) -- that is, each non-leaf node has a &quot;left&quot; and a &quot;right&quot; descendant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Hint&lt;/em&gt;:  How can you modify a full binary tree with n+1 nodes to a full binary tree with n nodes?  Well, you could try deleting a leaf-node, but then you will end up with a non-leaf node with only one descendant (which isn't a full binary tree).  So this isn't going to work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So how can you ensure that a full binary tree (with fewer nodes) is constructed?&lt;/p&gt;&#xA;" OwnerUserId="139" LastActivityDate="2010-11-25T05:02:56.693" />
  <row Id="11799" PostTypeId="2" ParentId="6575" CreationDate="2010-11-25T05:16:57.633" Score="0" ViewCount="" Body="&lt;p&gt;How do you define a leaf?  The idea of structural induction would be to show that whenever you add a new node to an existing tree, the difference between leaves and nodes without right siblings stays constant.  Without the leaf definition I can't help further.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-11-25T05:16:57.633" />
  <row Id="11800" PostTypeId="2" ParentId="11784" CreationDate="2010-11-25T05:24:19.337" Score="3" ViewCount="" Body="&lt;p&gt;Others have given you some of the standard examples that you can verify are indeed examples. Let me try to give you some intuition of &lt;em&gt;why&lt;/em&gt; the partials are not enough.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Intuitively, a function $f\colon\mathbb{R}^2\to\mathbb{R}$ is differentiable at a point $(x_0,y_0)$ if you can find a plane that is tangent to the graph of $f$ at the point $(x_0,y_0,f(x_0,y_0))$. Here, &quot;tangent&quot; is interpreted the same way that a line is said to be tangent to a curve when dealing with $g\colon\mathbb{R}\to\mathbb{R}$: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the one-variable case, a line $y=mx+b$ is tangent to the graph of $y=g(x)$ at the point $c$ if and only $mc+b=g(c)$ (they go through the same point) &lt;em&gt;and&lt;/em&gt; the approximation given by the tangent as $x$ approaches $c$ goes to $0$ &#xA;&lt;em&gt;faster&lt;/em&gt; than $x$ approaches $c$; that is, &#xA;$$\lim_{x\to c}\frac{g(x)-(mx+b)}{x-c}=0.$$&#xD;&#xA;&lt;em&gt;Note:&lt;/em&gt; To see that this certainly will happen for the tangent, remember that the tangent will have $m=g&amp;#39;(c)$ and $b=g(c) - g&amp;#39;(c)c$, so the expression in the limit becomes&#xA;$$\frac{g(x) - (g&amp;#39;(c)x + g(c)-g&amp;#39;(c)c)}{x-c} = \frac{g(x) - g(c) - g&amp;#39;(c)(x-c)}{x-c} = \frac{g(x)-g(c)}{x-c} - g&amp;#39;(c).$$&#xD;&#xA;But in fact, one can &lt;em&gt;define&lt;/em&gt; the tangent as the (unique) line with this property.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We do the same thing with two variables, but with planes instead of lines. Differentiability at $(x_0,y_0)$ for $f(x,y)$ means that there is a plane $z=ax+by+c$, with $ax_0 + by_0 + c = f(x_0,y_0)$, and such that &#xA;$$\frac{f(x,y) - (ax + by + c)}{||(x,y)-(x_0,y_0)||}$$&#xD;&#xA;goes to $0$ as $(x,y)\to (x_0,y_0)$: the plane approaches the graph, in &lt;em&gt;every&lt;/em&gt; direction, faster than the arguments are approaching the point. (Here, $||\cdot||$ is the usual norm in $\mathbb{R}^2$: $||(a,b)-(c,d)||=\sqrt{(a-c)^2+(b-d)^2}$). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But this has to happen along &lt;em&gt;all&lt;/em&gt; directions of approach to $(x_0,y_0)$; just like with limits of two variables, approaching along vertical lines alone or along horizontal lines alone is not enough (even approaching along &lt;em&gt;lines&lt;/em&gt; is not enough; it really has to be along &lt;em&gt;all&lt;/em&gt; directions). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;By contrast, the partial derivative with respect to $x$ only tell you that if you &quot;slice&quot; the graph with a plane parallel to the XZ-plane and you look at the &quot;shadow&quot; on that plane (which will look just like the graph of a one-variable function), then you can approach the bit of graph with an appropriate line; that is, you can approach the graph of $z=f(x,y)$ (which is a three-dimensional thing) with a &lt;em&gt;line&lt;/em&gt; parallel to the XZ plane; and the partial derivative with respect to $y$ tells you that you can approach the graph with a &lt;em&gt;line&lt;/em&gt; parallel to the YZ-plane. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Certainly, &lt;em&gt;if&lt;/em&gt; you can approach the graph with a plane appropriately, then you can also approach it with lines (just make sure the lines are contained in the plane). But being able to approach it in the two coordinate directions alone is not enough to make sure you can actually get a whole plane close to the graph in &lt;em&gt;all&lt;/em&gt; directions. That is why knowing that the partial derivatives exist does not tell you that the function is differentiable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One nice thing, as mentioned by Bryan Yocks, is that if if the partial derivatives, viewed as functions of two variables, not only exist but are also &lt;em&gt;continuous&lt;/em&gt;, then you can leverage this extra information into showing you can find a plane. Intuitively, what happens here is that the lines do give you enough information because you know that points close to $(x_0,y_0)$ will have similar lines that are very close to the ones you have, and so you can picture in your head perhaps that as you draw all these lines, &quot;locally&quot; it's going to look very much like a plane because the change in either of the two directions is very small. But if all you know is that the partial derivatives exist, this is not enough for the reasons described above. &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-25T05:46:08.157" LastActivityDate="2010-11-25T05:46:08.157" />
  <row Id="11801" PostTypeId="2" ParentId="11787" CreationDate="2010-11-25T05:28:06.573" Score="4" ViewCount="" Body="&lt;p&gt;You can actually do this &quot;by hand&quot; too. Our goal is to show that a monic irreducible polynomial $f(x)$ of degree $2$ or $3$ over $F$ splits into distinct linear factors in an algebraic closure $E$, as long as the characteristic of the $F$ is not $2$ or $3$. Suppose the irreducible $f(x)$ does not split into linear factors in $E$; we will get a contradiction. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, if the polynomial is of the form $(x - a)^3 = x^3 - 3ax^2 + 3a^2x + a^3$, then the $x^2$ coefficient $-3a$ must be in $F$. So $a$ is in $F$ too, using that $F$ is not of characteristic $3$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The remaining case is that the polynomial is of the form $(x - a)^2(x - b)$.&#xA;For this we use the fact that for any polynomial $p(x)$, ${p&amp;#39;(x) \over p(x)} = \sum {1 \over x - r}$, where the sum is taken over the roots of $p(x)$. Applying this to $f(x)$ gives that ${f&amp;#39;(x) \over f(x)} = {2 \over (x - a)} + {1 \over x - b}$.  Note that $f(x) = x^3 - (b + 2a)x^2 + ...$, so that $b + 2a \in F$. Hence ${b + 2a \over 3} \in F$ too. Substituting this value of $x$ into ${f&amp;#39;(x) \over f(x)}$ gives gives ${9 \over 2(b - a)}$, which must be in $F$. Thus $b - a\in F$; note we use the characteristic is not $2$ or $3$ here. So $b = {2 \over 3}(b - a) + {1 \over 3}(b + 2a)$ is in $F$ as well, again contradicting the irreducibility of $f(x)$.&lt;/p&gt;&#xA;" OwnerUserId="3035" LastEditorUserId="3035" LastEditorDisplayName="" LastEditDate="2010-11-25T07:39:52.907" LastActivityDate="2010-11-25T07:39:52.907" />
  <row Id="11802" PostTypeId="1" AcceptedAnswerId="11805" CreationDate="2010-11-25T05:34:07.760" Score="0" ViewCount="157" Body="&lt;p&gt;I've never had to do this before, so I'm not really sure how to do it. These problems also don't even really relate to what the subject of the book is as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given: $f(px+(1-p)y)\le pf(x)+(1-p)f(y)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the function&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f(x)\left\{ \begin{array}{cc}&#xD;&#xA;x &amp;amp; x\le1\\&#xD;&#xA;2-x &amp;amp; x&amp;gt;1\end{array}\right\} $&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Show that $f(x)$ is convex on $-\infty &amp;lt; x \le 1$ and convex on $1 &amp;lt; x &amp;lt; \infty$ but not convex on $-\infty &amp;lt; x &amp;lt; \infty$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another problem using the same convex definition:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose $f(x)$ is defined on the interval I. If $x_{1}&amp;lt;x_{2}&amp;lt;x_{3}$ are in I and $f(x_{1})&amp;lt;f(x_{2})$ and $f(x_{3})&amp;lt;f(x_{2})$, then show that $f(x)$ is not convex on I.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: Im not sure how to use the definition. Thats kind of where i am stuck. I'm sure it's easy, but like i said i've never seen this before, the book has nothing to do with this, it's just randomly thrown in here. There are no examples for this either.&lt;/p&gt;&#xA;" OwnerUserId="3961" LastEditorUserId="3961" LastEditorDisplayName="" LastEditDate="2010-11-26T22:22:25.193" LastActivityDate="2010-11-26T22:22:25.193" Title="show that function is not convex" Tags="&lt;convex-analysis&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="11803" PostTypeId="2" ParentId="11802" CreationDate="2010-11-25T05:39:42.907" Score="0" ViewCount="" Body="&lt;p&gt;You just need to apply the definition for these two problems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the first problem,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider two points $x,y$ such that $x,y \leq 1$. Argue why the definition of convexity is true in this case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, consider two points $x,y$ such that $x,y &amp;gt; 1$. Argue why the definition of convexity is true in this case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now consider $x &amp;lt; 1$ and $y &amp;gt; 1$ and argue why the definition is violated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the second problem, your statement should read &quot;If $x_1 &amp;lt; x_2 &amp;lt; x_3$ are in $I$ and $\ldots$&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Use the same idea as before (argue from the definition)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that $x_2$ can be written as a weighted linear combination of $x_1$ and $x_3$.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2010-11-25T05:45:13.480" LastActivityDate="2010-11-25T05:45:13.480" />
  <row Id="11804" PostTypeId="2" ParentId="11792" CreationDate="2010-11-25T05:56:55.420" Score="0" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://books.google.com/books?id=ktCZ2MvgN3MC&amp;amp;printsec=frontcover&amp;amp;dq=sierpinski&amp;amp;hl=en&amp;amp;ei=e_rtTK2HNcSAlAeC1rWtDA&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=1&amp;amp;ved=0CCMQ6AEwAA#v=onepage&amp;amp;q=sierpinski&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;Elementary Theory of Numbers&lt;/a&gt; by Sierpinski.&lt;/p&gt;&#xA;" OwnerUserId="1417" LastActivityDate="2010-11-25T05:56:55.420" CommentCount="3" />
  <row Id="11805" PostTypeId="2" ParentId="11802" CreationDate="2010-11-25T05:57:21.693" Score="2" ViewCount="" Body="&lt;p&gt;If you look at the definition geometrically (on $\mathbb{R}$) it states that a function is convex if each point of the graph $(x, f(x))$ is below any line that adjoins graph points $(a, f(a))$ and $(b, f(b))$ where $a&amp;lt;x&amp;lt;b$. In other words, for any ordered triple $a&amp;lt;x&amp;lt;b$ &#xA;the point $(x,f(x))$ is below the lint through $(a,f(a))$ and $(b,f(b))$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now think about the graph of your function - look at $x&amp;lt;1$... look at $x&amp;gt;1$... look at $x=1$...&lt;/p&gt;&#xA;" OwnerUserId="1154" LastActivityDate="2010-11-25T05:57:21.693" />
  <row Id="11806" PostTypeId="2" ParentId="11669" CreationDate="2010-11-25T06:10:02.810" Score="3" ViewCount="" Body="&lt;p&gt;Check out &lt;a href=&quot;http://arxiv.org/abs/0711.1873&quot; rel=&quot;nofollow&quot;&gt;this paper&lt;/a&gt; which is about the regular 12-gon and music theory.  It will help you answer this question, as well as many others that are similar to it.&lt;/p&gt;&#xA;" OwnerUserId="1417" LastActivityDate="2010-11-25T06:10:02.810" />
  <row Id="11807" PostTypeId="1" AcceptedAnswerId="11816" CreationDate="2010-11-25T06:12:32.357" Score="1" ViewCount="116" Body="&lt;p&gt;I was trying to solve this congruence equation: $2ax+b\equiv 0 \pmod 5$. Here, $b$ is any integer and $a$ is an integer not divisible by $5$. Does a solution always exist? I don't understand why they have set the condition above on $a$. As should be clear from my other question on elementary number theory, I am not very good at it, and I have no idea where to begin when solving congruences. Any help would be much appreciated.&lt;/p&gt;&#xA;" OwnerUserId="3947" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-25T06:32:29.103" LastActivityDate="2010-11-25T07:32:58.683" Title="Solution of a congruence $2ax+b\equiv 0 \pmod 5$" Tags="&lt;elementary-number-theory&gt;" AnswerCount="3" />
  <row Id="11808" PostTypeId="2" ParentId="11807" CreationDate="2010-11-25T06:25:02.020" Score="2" ViewCount="" Body="&lt;p&gt;If $\ a\equiv 0\ $ then $\ b\equiv 0\:,\ $ so it wouldn't be solvable for $b\not\equiv 0$. Conversely, if $\ a\not\equiv 0 \ $ then $\ 1/a \equiv a^3\ $ since $\ a^4\equiv 1 \ $ by Fermat's little theorem, so you can cancel the leading coefficient and solve for $x$.&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-11-25T06:25:02.020" />
  <row Id="11809" PostTypeId="1" AcceptedAnswerId="11813" CreationDate="2010-11-25T06:25:19.633" Score="2" ViewCount="130" Body="&lt;p&gt;Question: Let $U$ denote a subset of $\mathbb{R}$. We say that $U$ is an open set when for every $x \in U$ there is an $\epsilon &amp;gt; 0$ for which $U_{\epsilon}(x) = (x - \epsilon, x +\epsilon)$ is contained in $U$. Show:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(i) For $a,b \in \mathbb{R}$ the interval $(a,b)$ is open, but the interval $[a,b]$ is not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(ii) If $U,W \subset \mathbb{R}$ are open, so are $U \cup W$ and $U \cap W$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(iii) If $(U_{i})_{i \in I}$ is a set of open subsets from $\mathbb{R}$, $\cup_{i \in I}U_{i}$ is also open.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(iv) If $(U_{i})_{i \in I}$ is a set of open subsets from $\mathbb{R}$, $\cap_{i \in I}U_{i}$ is not necessarily open.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My attempt:&#xA;I think that with (i) i can conceptually understand everything:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\forall x \in (a,b): a&amp;lt; x - \frac{|a-x|}{2} &amp;lt; x &amp;lt; x + \frac{|b-x|}{2} &amp;lt; b$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $x&amp;gt; \frac{|b-a|}{2}$ let $\epsilon = \frac{|b-x|}{2}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $x&amp;lt; \frac{|b-a|}{2}$ let $\epsilon = \frac{|a-x|}{2}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Rightarrow \forall x \in (a,b) \exists \epsilon &amp;gt; 0 : U_{\epsilon}(x) = (x - \epsilon, x +\epsilon) \Rightarrow$ the interval (a,b) is open.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;max$[a,b] = b \Rightarrow \forall x \in [a,b]: b \geq x$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $\epsilon &amp;gt; 0, b \ngeq b+ \epsilon \Rightarrow$ there exists an $x$ for which  $U_{\epsilon}(x) = (x - \epsilon, x +\epsilon)$ is not contained in $U \Rightarrow$ the interval is not open&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With (ii) and (iii) i think i understand, but i am not sure how to start stating that they are true mathematically. to be sure,  $\cup_{i \in I}U_{i}$ denotes several unions in a row, correct? i'm even reading a proof which goes: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given $v \in U \cap W$, there exists an open ball $B$ of radius $r$ centered at $v$ contained in $U$, and there exists an open ball $B&amp;#39;$ of radius $r&amp;#39;$ centered at $v$ contained in $W$. Let $\delta =$ min$(r, r')$. Then the open ball of radius $\delta$ centered at $v$ is contained in $U \cap W$, which is therefore open. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately i'm having trouble interpreting this and am wondering if there is another way to put it or approach it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With (iv) i am not even sure intuitively why it should be true. i can't think of a situation where it wouldn't be true, i guess it has to do with whether the sets are finite or not, but i don't see how this would make a difference (especially given that openness is apparently preserved with infinite unions).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If anyone can help provide suggestions, corrections, hints, etc i appreciate it, thanks!&lt;/p&gt;&#xA;" OwnerUserId="3711" LastEditorUserId="640" LastEditorDisplayName="" LastEditDate="2010-11-25T14:25:45.493" LastActivityDate="2010-11-25T19:31:35.010" Title="Proving statements related to open sets" Tags="&lt;analysis&gt;&lt;general-topology&gt;" AnswerCount="1" />
  <row Id="11810" PostTypeId="1" AcceptedAnswerId="11833" CreationDate="2010-11-25T06:27:07.237" Score="7" ViewCount="199" Body="&lt;p&gt;This question arises from a Fourier transform class I took about a year back.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The poisson summation formula is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle \sum_{n= - \infty}^{\infty} f(n) = \displaystyle \sum_{k= - \infty}^{\infty} \hat{f}(k)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\hat{f}$ is the Fourier Transform of $f(x)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is interesting since this is true for all $f(x)$ for which we can define Fourier transform.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a nice (probably physical) interpretation for this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am wondering if this characterizes some property which is invariant, some sort of conservation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, if we consider Parseval's theorem, one interpretation of it is that the total energy across all time is the total energy across all of its frequency components.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, from a mathematical standpoint what does this mean? Is this a manifestation of some property of integers?&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2010-11-25T08:41:44.670" LastActivityDate="2010-11-25T11:34:33.910" Title="Interpretation of Poisson Summation Formula" Tags="&lt;number-theory&gt;&lt;fourier-analysis&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" />
  <row Id="11811" PostTypeId="2" ParentId="11807" CreationDate="2010-11-25T06:41:52.683" Score="0" ViewCount="" Body="&lt;p&gt;Solving the congruence in this particular example means, we need to get solution for $x$ as $x \equiv m (\bmod 5)$, where $m$ depends on $a$ and $b$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that when $5|a$, we have $2ax \equiv 0 (\bmod 5)$. Hence, for arbitrary integer $b$, $2ax + b \equiv b (\bmod 5)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, if we want a solution to exist for $2ax + b \equiv 0 (\bmod 5)$, then this forces $b$ to be $0 (\bmod 5)$, which is not necessarily true since $b$ is an arbitrary integer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are getting started on Number Theory, then you could argue by cases. This will be a good exercise when you try to generalize this to other numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Case 1:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider $a \equiv 1 (\bmod 5)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then $ax \equiv x (\bmod 5)$, $2ax \equiv 2x (\bmod 5)$ and hence $2ax + b \equiv 2x+b (\bmod 5)$. And now we want this to be $0 (\bmod 5)$. So we need $2x + b \equiv 0 (\bmod 5)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Multiply the equation by $3$ and we get $6x + 3b \equiv 0 (\bmod 5)$. Now observe that $5x \equiv 0 (\bmod 5)$. Hence, we get $x + 3b \equiv 0 (\bmod 5)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(The reason you multiply by $3$ is $3 \times 2 \equiv 1 (\bmod 5)$. So $3$ is the inverse of $2$, when you are working in $(\bmod 5)$ setting. Also note that this gives you the reason why you couldn't do when $5|a$, since $5$ doesn't have an inverse in the $(\bmod 5)$ setting.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the solution is $x \equiv -3b (\bmod 5)$, when $a \equiv 1 (\bmod 5)$.Proceed via the same idea for other cases as well viz $a \equiv 2 (\bmod 5)$, $a \equiv 3 (\bmod 5)$, $a \equiv 4 (\bmod 5)$.You could then generalize this once you observe what is happening. Note that here since $5$ is a prime all the other elements will have inverses. Think what will happen when $5$ is replaced by '$n$' where '$n$' is a composite number.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2010-11-25T07:13:16.883" LastActivityDate="2010-11-25T07:13:16.883" />
  <row Id="11813" PostTypeId="2" ParentId="11809" CreationDate="2010-11-25T06:57:37.330" Score="4" ViewCount="" Body="&lt;p&gt;Intuitively, a set $U$ is open if every point in the set can be completely surrounded by points that are all in the set; here, &quot;surrounded&quot; means that an entire little interval centered at $x$ is completely contained in $U$, which means, in a sense, that $x$ is &lt;em&gt;really&lt;/em&gt; inside $U$. A set is open if every point in the set is &lt;em&gt;really&lt;/em&gt; &quot;inside&quot; the set. Compare with $[a,b]$ (which you show is not open), where one might say that although $b$ &#xA;&lt;em&gt;is&lt;/em&gt; in the set, it is &quot;barely&quot; in the set: move just a bit to the right, and you fall outside $[a,b]$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That said, on to your work:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your idea for the first part of (i) is okay, but the way you describe $\varepsilon$ is not. For example, say $x=1000$, $a=999.9$, and $b=1000.5$. Then $|b-a|/2 = .3$, and $x\gt |b-a|/2$, so you take $\varepsilon = 0.3$. But $(x-0.3,x+0.3)$ is not contained in $(a,b)$, because $x-0.3 = 999.7\lt a$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you &lt;em&gt;really&lt;/em&gt; want to say is that $\varepsilon=\min(|x-b|/2, |x-a|/2)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the second part, you've got it right, but there is no need to say &quot;there exists an $x$ ...&quot;, just say that for every $\varepsilon\gt 0$, $(b-\varepsilon,b+\varepsilon)$ is not contained in $[a,b]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For (ii); not quite: $\cup_{i\in I} U_i$ is not &quot;several unions in a row&quot;, because the set may not something that you can do &quot;one after another&quot;; maybe there is one for every real number! No, $\cup_{i\in I}U_i$ just means &quot;the union of all the $U_i$&quot;; this is the collection of all points that are in at least one of the $U_i$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think you mean $U\cap W$ when you said $V\cap V$. But you've go the idea right for the intersection. Conceptually: you know that you can find an open interval around $x$ that &quot;works&quot; for $U$, and one that works for $W$; to find one that works for &lt;em&gt;both&lt;/em&gt;, just take the smaller one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for (iv), it's &lt;em&gt;not&lt;/em&gt; true: you are being asked to show it is not true. Can you think of a way to get a closed interval as an intersection of open ones?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added&lt;/strong&gt;: Okay, here's a hint for (iv): for every positive integer $n$, $\frac{1}{n}$ is stricly larger than $0$; but the infinimum of $\{ \frac{1}{n}\mid n\in\mathbb{Z}_{\gt 0}\}$ is &lt;em&gt;not&lt;/em&gt; strictly larger than $0$.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-25T19:31:35.010" LastActivityDate="2010-11-25T19:31:35.010" CommentCount="2" />
  <row Id="11814" PostTypeId="2" ParentId="11223" CreationDate="2010-11-25T07:20:10.310" Score="0" ViewCount="" Body="&lt;p&gt;You simply need to look at the number of factors that a given number has.  This equals the number of tutors that have &quot;changed&quot; that light bulb.  For instance, 7 only has one and itself as a factor, so only 2 tutors affect bulb 7; tutors 1 and 7 (the bulb goes from off, to on then off where it remains).  169 on the other hand, has three factors; 1,13, and 169, so this bulb is affected by the tutors numbered 1,13 and 169 (from off to on to off back to on).  &lt;/p&gt;&#xA;" OwnerUserId="3971" LastActivityDate="2010-11-25T07:20:10.310" />
  <row Id="11815" PostTypeId="2" ParentId="10073" CreationDate="2010-11-25T07:27:23.627" Score="0" ViewCount="" Body="&lt;p&gt;I'm Not too sure how to answer the first part of your question.  For the second part, a complex variable function can be thought of as a real variable function whose domain and range are both subsets of R^2.  This is usually different then those functions you study in multivariable calculus: functions from R^n to R.  One noticible difference is that for functions of a single complex variable, there is a total derivaitve which exists if the cauchy riemaan differential equations hold.  Real functions on R^2 would have two partial derivatives instead.&lt;/p&gt;&#xA;" OwnerUserId="3971" LastActivityDate="2010-11-25T07:27:23.627" />
  <row Id="11816" PostTypeId="2" ParentId="11807" CreationDate="2010-11-25T07:32:58.683" Score="2" ViewCount="" Body="&lt;p&gt;Perhaps you've come across the following theorem in your studies?&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If $gcd(a,m)=1$, then there is an $x$ such that $ax\equiv 1\pmod{m}$, and any two such $x$ are congruent $\pmod{m}$. If $gcd(a,m)&amp;gt;1$, then there is no such $x$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Essentially, this means an integer $a$ has a multiplicative inverse modulo $m$ if and only if it $a$ is relatively prime to the modulus $m$. This nice thing about primes is that if an integer $a$ is not divisible by a prime $p$, then it is relatively prime to that prime, as the only other common factor can be $1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So since $5\nmid a$, you know $gcd(a,5)=1$, and so there is some $x$, let's call it $\bar{a}$, such that $\bar{a}a\equiv 1\pmod{5}$. This allows you to solve for $x$ in this linear congruence. Note &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$2ax+b\equiv 0\Rightarrow 2ax\equiv -b\Rightarrow \bar{2}2ax\equiv -\bar{2}b\Rightarrow \bar{a}ax\equiv -\bar{a}\bar{2}b\Rightarrow x\equiv -\bar{2}\bar{a}b\equiv -3\bar{a}b\pmod{5}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $gcd(2,5)=1$, you also know that $2$ has a multiplicative inverse modulo $5$, namely $3$. This is the same reason why there is such a condition on $a$.&lt;/p&gt;&#xA;" OwnerUserId="1583" LastActivityDate="2010-11-25T07:32:58.683" CommentCount="1" />
  <row Id="11817" PostTypeId="2" ParentId="11373" CreationDate="2010-11-25T07:35:13.650" Score="1" ViewCount="" Body="&lt;p&gt;In noncommutative rings, the terms &quot;left ideal&quot; and &quot;right ideal&quot; are used indicate if you are multiplying from the left or right.  The term &quot;ideal&quot; usualy means that it is a left and right ideal at the same time.  In commutative ring theory, which is probably what you are studying, this notion of left and right is meaningless since ab and ba are always the same.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Simply put, an ideal is a very special type of subring, with the added property that if a is in the ideal and r is in the ring, r times a is in the ideal (even if r is not in the ideal).  Subrings don't have this extra property in general.&lt;/p&gt;&#xA;" OwnerUserId="3971" LastActivityDate="2010-11-25T07:35:13.650" />
  <row Id="11818" PostTypeId="1" AcceptedAnswerId="11819" CreationDate="2010-11-25T08:03:48.497" Score="2" ViewCount="143" Body="&lt;p&gt;In the situation of having a high entropy random number generator, that generates numbers in the range of 0 and 2,147,000,000.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If i have a list of 1,000,000 integer values, what are the chances that a random number will already be on my list?&lt;/p&gt;&#xA;" OwnerUserId="3472" LastActivityDate="2010-11-25T17:32:29.240" Title="Probability of Random number repeating" Tags="&lt;probability&gt;&lt;random&gt;" AnswerCount="2" />
  <row Id="11819" PostTypeId="2" ParentId="11818" CreationDate="2010-11-25T08:10:49.267" Score="4" ViewCount="" Body="&lt;p&gt;One minus the probability that they are all different:&#xA;$$1 - \left( 1- \frac{1}{n} \right)&#xD;&#xA;\left( 1- \frac{2}{n} \right)&#xD;&#xA;\cdots \left( 1- \frac{k-1}{n} \right),$$&#xD;&#xA;where $n=2,147,000,001$ (since you include $0$) and $k=1,000,000.$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See the &lt;a href=&quot;http://en.wikipedia.org/wiki/Birthday_problem&quot; rel=&quot;nofollow&quot;&gt;birthday problem&lt;/a&gt; for more information.&lt;/p&gt;&#xA;" OwnerUserId="1301" LastActivityDate="2010-11-25T08:10:49.267" CommentCount="0" />
  <row Id="11820" PostTypeId="2" ParentId="11669" CreationDate="2010-11-25T09:21:21.250" Score="31" ViewCount="" Body="&lt;p&gt;The first answer is great, so I'll try to approach the question from another angle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, there are several different scales, and different cultures use different ones. It depends on the mathematics of the instruments as much as on cultural factors. Our scale has a very long history that can be traced to the ancient Greeks and Pythagoras in particular. They noticed (by hearing) that stringed instruments could produce different notes by adjusting the length of the string, and that some combinations sounded better.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Greeks had a lot of interest in mathemathics, and it seemed &quot;right&quot; for them to search for &quot;perfect&quot; combinations - perfect meaning that they should be expressed in terms of fractions of small integer numbers. They noticed that if you double or halve the string length, you get the same note (the concept of an octave); other fractions, such as 2/3, 3/4, also produced &quot;harmonic&quot; combinations. That's also the reason why some combinations sound better, as it can be explained by physics. When you combine several sine waves, you hear several different notes that are the result of the interference betwen the original waves. Some combinations sound better while others produce what we call &quot;disonance&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, in theory, you can start from an arbitrary frequency (or note) and build a scale of &quot;harmonic&quot; notes using these ratios (I'm using quotes because the term harmonic has a very specific meaning in music, and I'm talking in broad and imprecise terms). The major and minor scales of Western music can be approximately derived from this scheme. Both scales (major and minor) have 7 notes. The white keys in the piano correspond to the major scale, starting from the C note.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if you get the C note and use the &quot;perfect&quot; fractions, you'll get the &quot;true&quot; C major scale. And that's where the fun begins. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you take any note in the C major scale, you can treat that note as the start of another scale. Take for instance the fifth of C (it's the G), and build a new major scale, now starting from G instead of C. You'll get another seven notes. Some of them are also on teh scale of C; others are very close, but not exactly equal; and some fall in the middle of the notes in the scale of C. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you repeat this exercise with all notes, you'll end up building 12 different scales. The problem is that the interval is not regular, and there are some imprecisions. You need to retune the instrument if you want to have the perfect scale.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The concept of &quot;chromatic&quot; scale (with 12 notes, equaly spaced) was invented to solve this &quot;problem&quot;. The chromatic scale is a mathematical approximation, that is close enough for MOST people (but not all). People with &quot;perfect&quot; ear can listen the imperfections. In the chromatic scale, notes are evenly space using the twelfth root of two. It's a geometric progression, that matches with good precision all possible major and minor scales. The invention of the chromatic scale allows players to play music in arbitrary scales without retuning the instrument - you only need to ajust the scale by &quot;offsetting&quot; a fixed number of positions, or semitones, from the base one of the original scale.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All in all, that's just convention, and a bit of luck. The white keys are an &quot;historical accident&quot;, being the keys of the major scale of C. The other ones are needed to allow for transposition. Also bear in mind that (1) the keys need to have a minimum width to allow for a single finger, and (2) if you didn't have the black keys, the octave would be too wide for &quot;normal&quot; hands to play. So the scheme with a few intermediate keys is needed anyway, and the chromatic scale that we use is at least as good (or better) as any other possible scale.&lt;/p&gt;&#xA;" OwnerUserId="3981" LastActivityDate="2010-11-25T09:21:21.250" CommentCount="1" />
  <row Id="11821" PostTypeId="2" ParentId="3510" CreationDate="2010-11-25T09:26:35.567" Score="-2" ViewCount="" Body="&lt;p&gt;Ok, here is &quot;an engineer's proof&quot;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why not simply differentiate the Euler formula $e^{ix}=\cos x+i\sin x$ twice by $x$:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$i^2e^{ix}=-cos x-i\sin x$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $i^2=-1$, the proof is ended.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;{Sorry, i am not true mathematician :)}&lt;/p&gt;&#xA;" OwnerUserId="2323" LastActivityDate="2010-11-25T09:26:35.567" CommentCount="8" />
  <row Id="11822" PostTypeId="2" ParentId="11611" CreationDate="2010-11-25T09:27:29.497" Score="0" ViewCount="" Body="&lt;p&gt;The context of this question suggests that $f=f(x)$ and $g=g(x)$ are going to be polynomials (with either real, rational or integer coefficients).  If this is the case then...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Hint&lt;/em&gt;: While there's going to be an infinite number of functions $f,g$ for which $f(g(x))=x^2+x-6$, there will only be two situations when this can occur (consider the degrees of $f$ and $g$, and how that effects $f(g(x))$).  These can be classified by setting $f$ and $g$ as arbitrary polynomials of the appropriate degrees, composing them, then equating coefficients with $f(g(x))=x^2+x-6$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $f$ and $g$ are not restricted to polynomials, there's going to be all sorts of functions $f$ and $g$ for which $f(g(x))=x^2+x-6$.  For example: Let $A=\left(\begin{matrix}-2 &amp;amp; 1 \\ 4 &amp;amp; 1 \end{matrix}\right)$.  Then the characteristic polynomial $\det(A-xI)=(-2-x)(1-x)-4=x^2+x-6$.&lt;/p&gt;&#xA;" OwnerUserId="139" LastActivityDate="2010-11-25T09:27:29.497" />
  <row Id="11823" PostTypeId="2" ParentId="11792" CreationDate="2010-11-25T09:47:12.820" Score="0" ViewCount="" Body="&lt;p&gt;The book of Harold Davenport, &lt;strong&gt;The Higher Arithmetic: An introduction to the theory of numbers&lt;/strong&gt;,  fits your bill. &lt;/p&gt;&#xA;" OwnerUserId="1119" LastActivityDate="2010-11-25T09:47:12.820" />
  <row Id="11824" PostTypeId="2" ParentId="11810" CreationDate="2010-11-25T09:55:26.433" Score="1" ViewCount="" Body="&lt;p&gt;If you want analogies from electrical engineering, one possible way to understand the consequences of the Poisson summation is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Nyquist%E2%80%93Shannon_sampling_theorem&quot; rel=&quot;nofollow&quot;&gt;Nyquist sampling theorem&lt;/a&gt;. It says that periodic sampling of a signal is enough to capture all the signals whose frequency is less than or equal to half the sampling rate.&lt;/p&gt;&#xA;" OwnerUserId="1119" LastActivityDate="2010-11-25T09:55:26.433" CommentCount="2" />
  <row Id="11825" PostTypeId="1" CreationDate="2010-11-25T09:56:46.607" Score="0" ViewCount="91" Body="&lt;p&gt;Let $\{e_0, e_1, \dots, e_{N-1} \}$ be the Euclidian basis for $l^2(Z_N)$, and let $\{F_0, F_1, \dots, F_{N-1} \}$ be the fourier basis &#xA;i. How to show that $\hat{e}_m(k) = e^{-2 \pi mk/N}$ for all k? Notice that $\hat{e}_m$ is very nearly ( up to a reflection and a normalization ) an element of the Fourier basis.&#xA;ii. How to show that $\hat{F}_m = e_m$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First some definitions $F_m=N^{-1/2} E_m = \frac{1}{N}e^{2 \pi i m n/N}$ and&#xA;$\hat{z}(m) = \sum_{n=0}^{N-1} z(n) e^{-2 \pi i m n/N}$&#xD;&#xA;i.How to show that $\hat{e}_m(k) = e^{-2 \pi mk/N}$?&#xA;(my) Solution:&#xA;$\hat{e}_m(k) = \sum_{n=0}^{N-1} e_m(k) e^{-2 \pi i m k/N} = e^{-2 \pi i m k/N}$&#xD;&#xA;ii.  How to show that $\hat{F}_m = e_m$? &#xA;(my solution:&#xA;$\hat{F}_m = \sum_{n=0}^{N-1} \frac{1}{N}e^{2 \pi i m n/N} e^{-2 \pi i m n/N} = \sum_{n=0}^{N-1} \frac{1}{N} = ? $&#xD;&#xA;Any comments on my solutions.&lt;/p&gt;&#xA;" OwnerUserId="2723" LastEditorUserId="2723" LastEditorDisplayName="" LastEditDate="2010-11-25T10:43:38.750" LastActivityDate="2010-11-25T10:43:38.750" Title="How to show that $\hat{e}_m(k) = e^{-2 \pi mk/N}$ for all k?" Tags="&lt;homework&gt;&lt;algebra&gt;&lt;linear-algebra&gt;" AnswerCount="0" CommentCount="1" />
  <row Id="11826" PostTypeId="2" ParentId="11768" CreationDate="2010-11-25T10:02:20.810" Score="1" ViewCount="" Body="&lt;p&gt;The proofs of all these assertions in the case when your group is $\mathbb R$, can be found in any elementary book defining convolution. These same proofs will work in the case of arbitrary groups.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, take the case of associativity. A proof for the case of real numbers is available at &lt;a href=&quot;http://planetmath.org/encyclopedia/AssociativityOfConvolution.html&quot; rel=&quot;nofollow&quot;&gt;planetmath&lt;/a&gt;. The same proof technique carries over to the general case. All you need additionally is to formally state two theorems: 1. Fubini's theorem on integration wrt product measures, and 2. translation invariance of the measure. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here note that you always have to use a Haar measure.&lt;/p&gt;&#xA;" OwnerUserId="1119" LastActivityDate="2010-11-25T10:02:20.810" />
  <row Id="11828" PostTypeId="2" ParentId="11693" CreationDate="2010-11-25T10:07:54.377" Score="2" ViewCount="" Body="&lt;p&gt;I like your question! I wish I could give a full answer, but I'm not familiar enough with the Manin-Drinfeld theorem.  After I finish a couple weeks of work, I hope I can return and think about it in detail.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would guess the answer is yes, and I have some candidates you could try.  The Fermat curves can be identified with compactified quotients $\Gamma\backslash\mathcal{H}$ for finite index subgroups $\Gamma$ in the following way:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The modular function $\lambda(z) = \frac{\theta_2^4(q)}{\theta_3^4(q)}=16 q^{1/2} -128q +704q^{3/2} + \ldots$ &#xA;is a Hauptmodul of the genus 0 congruence subgroup $\Gamma(2)$.  (I.e., it is invariant under the action &#xA;of $\Gamma(2)$ by fractional linear transformation, and parametrizes the genus 0 modular surface $X_{\Gamma(2)} = \Gamma(2) \backslash \mathcal{H} \cup \{0,1,i_\infty\}$.)   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We define Hauptmoduln $x:= \sqrt[n]{\lambda}$ and $y:= \sqrt[n]{1-\lambda}$, which determine finite index  genus 0 subgroups $\Gamma_x$ and $\Gamma_y$. Then $\Gamma:= \Gamma_x \cap \Gamma_y$ is subgroup with compactified quotient $X_{\Gamma}: x^n +  y^n = 1$, and $\Gamma$ is a noncongruence subgroup for $n \neq 1,2,4,8$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This construction is from a survey paper by Ling Long.  We can use essentially the same idea to obtain quotient curves of the Fermat curves: If we set $x:= -\sqrt[5]{\lambda}$ and $y:= \sqrt{1-\lambda}$, we obtain a noncongruence subgroup $\Gamma$ with model $y^2=x^5+1$, which is a nice genus 2 curve, and its Jacobian has CM.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you can verify what happens in this case, I would be interested to hear about it.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Added later:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I just did a search and found a relevant paper: &quot;The Manin—Drinfeld theorem and Ramanujan sums&quot; by V. Kumar Murty and Dinakar Ramakrishnan.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They give the same construction of the Fermat curves as I gave above, attributing it to Fricke and Klein.  They also cite a result of Rohrlich which establishes that the cusps of $\Gamma$ map to torsion points of the Jacobian.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the answer is yes; the noncongruence groups $\Gamma$ corresponding to the Fermat curves also satisfy the statement of the Manin-Drinfeld theorem.&lt;/p&gt;&#xA;" OwnerUserId="1461" LastEditorUserId="1461" LastEditorDisplayName="" LastEditDate="2010-11-27T06:58:21.037" LastActivityDate="2010-11-27T06:58:21.037" CommentCount="1" />
  <row Id="11829" PostTypeId="2" ParentId="11761" CreationDate="2010-11-25T10:16:59.820" Score="6" ViewCount="" Body="&lt;p&gt;The Wikipedia references are a reasonable starting point in the sense that some of them contain Voevodsky's construction(s) and related expositions.  Beyond that the literature on motives is enormous and difficult and relies on many ideas that are not apparent from the papers themselves -- not to mention unpublished letters and oral tradition found in a few major mathematical centers.  The state of documentation is far better in the internet era but this is still a highly technical field that is difficult to penetrate by casual reading.  In one respect the main breakthrough made by Voevodsky is that he created a theory with a framework that can be learned (e.g., by topologists) knowing scheme theory and a bounded number of other things, whereas before that motives were considered a pinnacle of inaccessible mystery, or a foundation-less heuristic used informally.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for what is &quot;suitable&quot;, Voevodsky proved the universality of motivic cohomology as defined in his approach; in the 2-category of homology theories satisfying some axioms, he constructs an initial object (similar to characterizing singular homology as the universal model of Eilenberg-Steenrod axioms).  Also, he builds a motivic homotopy theory with analogies to the topological theory, and not &quot;only&quot;(!!) the motivic homology and cohomology.  There are other respects in which his approach is better motivated (so to speak) or conceptual compared to other approaches.  (Grothendieck's approach is also conceptual but assumes Standard Conjectures and the work on mixed motives was not published.)   So although nobody knows the ultimate construction of mixed motives, Voevodsky's approach is the leading contender at present.&lt;/p&gt;&#xA;" OwnerUserId="467" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-11-25T18:34:26.653" LastActivityDate="2010-11-25T18:34:26.653" />
  <row Id="11830" PostTypeId="2" ParentId="11724" CreationDate="2010-11-25T10:50:29.990" Score="2" ViewCount="" Body="&lt;p&gt;I must say I do not know anything much about it; but the following is the general idea. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that in the case of cohomology theories for smooth projective varieties over $\mathbb C$, there are various canonical isomorphisms between de Rham cohomology, Betti cohomology, etale cohomology, etc.. For instance the isomorphism between de Rham cohomology and Betti cohomology involves integrating differential forms over homology classes. Actually the situation is a bit more intricate; the de Rham cohomology decomposes into various other spaces via the Hodge decomposition. This whole stuff is all done over the complex numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the $p$-adic case, an analogous theory is built, called the $p$-adic Hodge theory. Some details are available at the &lt;a href=&quot;http://en.wikipedia.org/wiki/P-adic_Hodge_theory&quot; rel=&quot;nofollow&quot;&gt;wikipedia page&lt;/a&gt;. There is an isomorphism between algebraic de Rham cohomology and $p$-adic etale cohomology, as per the Hodge-Tate conjecture. There is another conjectural functor from the algebraic de Rham cohomology to $p$-adic etale cohomology, which is called Grothendieck's mysterious functor. A theory for this was constructed by Jean-Marc Fontaine and it was applied in various situations such as the study of $p$-adic Galois representations. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The theories of $p$-adic integration are useful in connection with such investigations. Pierre Colmez has a theory and so does Robert Coleman. Coleman's homepage has a &lt;a href=&quot;http://math.berkeley.edu/~coleman/fontaine.html&quot; rel=&quot;nofollow&quot;&gt;course&lt;/a&gt; on this material. Also see Perrin-Riou's Asterisque volume [3]:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[3]: Fonctions L p-adiques des représentations p-adiques, Perrin-Riou, B.&lt;/p&gt;&#xA;" OwnerUserId="1119" LastActivityDate="2010-11-25T10:50:29.990" />
  <row Id="11831" PostTypeId="1" CreationDate="2010-11-25T11:23:56.250" Score="0" ViewCount="126" Body="&lt;p&gt;Need some help on following induction problem:&#xA;$\frac{1}{1*2} + \frac{1}{2*3} + \frac{1}{3*4}+...+\frac{1}{n(n+1)} = \frac{n}{n+1}$&#xD;&lt;/p&gt;&#xA;" LastActivityDate="2010-11-25T14:02:49.233" Title="proof by induction: n/(n+1)" Tags="&lt;induction&gt;" AnswerCount="3" />
  <row Id="11832" PostTypeId="2" ParentId="11831" CreationDate="2010-11-25T11:31:38.847" Score="3" ViewCount="" Body="&lt;p&gt;Hint:&#xA;$$\frac{1}{k(k+1)}= \frac{1}{k} - \frac{1}{k+1}.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint2:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{n}{n+1} = 1 - \frac{1}{n+1}.$$&lt;/p&gt;&#xA;" OwnerUserId="1301" LastActivityDate="2010-11-25T11:31:38.847" />
  <row Id="11833" PostTypeId="2" ParentId="11810" CreationDate="2010-11-25T11:34:33.910" Score="1" ViewCount="" Body="&lt;p&gt;The formula is not about the integers in the sense that it doesn't involve their multiplicative structure; rather, it's about how the integers sit inside the reals as a discrete subgroup.  You can see this in the way the formula generalizes to $\mathbb{Z}^n$ sitting inside $\mathbb{R}^n$ or more generally in how it generalizes to the &lt;a href=&quot;http://en.wikipedia.org/wiki/Selberg_trace_formula&quot; rel=&quot;nofollow&quot;&gt;Selberg trace formula&lt;/a&gt;.  The trace formula makes it clear that Poisson summation is a representation-theoretic fact; indeed one of its special cases is Frobenius reciprocity.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-11-25T11:34:33.910" />
  <row Id="11834" PostTypeId="1" AcceptedAnswerId="12298" CreationDate="2010-11-25T11:34:48.933" Score="4" ViewCount="343" Body="&lt;p&gt;For a common basis, I'll state basic definitions of a category and the relation type I'm thinking of. They're here for quick clarity, not precision, so feel free to revise for an answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Category:&lt;br&gt;&#xA;A collection of objects, a collection of arrows each with a source and target among said objects, identity arrows, and a composition operator satisfying an associative law over arrows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reflexive relation on AxA:&lt;br&gt;&#xA;A collection of objects, a labelled multi-set of pairs of said objects satisfying the reflexive law, and the standard relation combinator operation (which satisfies the associative law).&lt;br&gt;&#xA;Note that labelled multi-set relations are commonly expressed and utilized in discrete math as multi-graphs. I collapse graphs and relations here for unified axiomatic treatment. This is not to treat, e.g., paths over graphs as different from relation composition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What, exactly, separates these two? They appear essentially identical in definition and meaning to me. Relations immediately appear more general, given that categories only analogize to a certain restricted class of relations.&lt;br&gt;&#xA;I note some notational difference, such as how categories name each arrow; I don't see how that would change mathematical power. Similarly with the explicit treatment of composition. While such contrivances can come in handy for certain proofs or explorations, I don't see how it justifies treating the two as separate branches rather than syntactic shims on identical concepts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[EDITS: fixed associativity statement, extended relations to multi-set representation with graph analogy]&lt;/p&gt;&#xA;" OwnerUserId="3976" LastEditorUserId="3976" LastEditorDisplayName="" LastEditDate="2010-11-29T00:03:05.563" LastActivityDate="2010-11-29T07:51:23.107" Title="What is the difference between Categories and Relations?" Tags="&lt;category-theory&gt;&lt;relations&gt;" AnswerCount="2" CommentCount="8" FavoriteCount="2" />
  <row Id="11835" PostTypeId="2" ParentId="11831" CreationDate="2010-11-25T11:36:13.693" Score="2" ViewCount="" Body="&lt;p&gt;$\textbf{HINT:}$ Actually this is the answer in itself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{n}{n+1} + \frac{1}{(n+1)(n+2)} = \frac{n+1}{n+2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is all that you will need when you do induction.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastActivityDate="2010-11-25T11:36:13.693" />
  <row Id="11836" PostTypeId="1" AcceptedAnswerId="11841" CreationDate="2010-11-25T11:46:10.217" Score="2" ViewCount="302" Body="&lt;p&gt;Is the essential difference between one-variable calculus and multi-variable calculus exterior differential or something else?&lt;/p&gt;&#xA;" OwnerUserId="1382" LastActivityDate="2010-11-25T21:03:45.707" Title="What is the essential difference between one-variable calculus and multi-variable calculus?" Tags="&lt;calculus&gt;" AnswerCount="4" CommentCount="2" FavoriteCount="1" />
  <row Id="11838" PostTypeId="2" ParentId="11836" CreationDate="2010-11-25T12:22:24.783" Score="3" ViewCount="" Body="&lt;p&gt;It seems to me that an important difference is that while in one-variable calculus one only deals with one derivative, in multi-variable calculus there are infinitely many derivatives, the &lt;a href=&quot;http://en.wikipedia.org/wiki/Directional_derivative&quot; rel=&quot;nofollow&quot;&gt;directional derivatives&lt;/a&gt;, a particular case of which are the partial derivatives. There are still the total derivatives for functions whose variables depend on another variable. Though all these derivatives are a generalization of the derivative of a single variable function.&lt;/p&gt;&#xA;" OwnerUserId="752" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2010-11-25T20:42:59.967" LastActivityDate="2010-11-25T20:42:59.967" />
  <row Id="11839" PostTypeId="2" ParentId="18" CreationDate="2010-11-25T12:34:05.153" Score="0" ViewCount="" Body="&lt;p&gt;Where,&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;  a = transverse radius = semi-major axis (for ellipse/oblate spheroid);&#xA;  b = conjugate radius = semi-minor axis (&quot;     &quot;     &quot;);&#xA; oe = angular eccentricity = arccos(b/a);&#xA;  f = flattening = (a-b)/a = 1 - b/a) = 1 - cos(oe) = 2*sin(.5*oe)^2;&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;then b = a*cos(oe) = a*(1-f).&lt;/p&gt;&#xA;" LastActivityDate="2010-11-25T12:34:05.153" />
  <row Id="11841" PostTypeId="2" ParentId="11836" CreationDate="2010-11-25T13:12:14.430" Score="4" ViewCount="" Body="&lt;p&gt;There's essentially no difference in the &lt;em&gt;concept&lt;/em&gt; of the derivative. It is still the best linear approximation to your function at each point. I.e. $f$ differentiable at $x_0$ still means that&#xA;$$ f(x_0+h) = f(x_0) + f&amp;#39;(x_0)x_0 + \varphi(h), $$&#xD;&#xA;for $h$ in some nbhd. of $0$ and $\varphi$ a continuous function that tends to $0$ at $0$ faster than linear. Except now $f&amp;#39;(x_0)$ is an $m \times n$ matrix called the &lt;a href=&quot;http://en.wikipedia.org/wiki/Jacobian_matrix&quot; rel=&quot;nofollow&quot;&gt;Jacobian matrix&lt;/a&gt; and much more complicated to work with. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first new thing we get when we go to higher dimensions is the notion of directional derivative, i.e. &quot;how much does $f$ change in the direction of $v$?&quot; We actually already have that on $\mathbb{R}$, it's just that there is only one direction (actually two, but it just comes down to a sign change), so it's never really looked at that way. The notion of directional derivative is exactly what you want when you want to generalize further to smooth manifolds, except you have to be a bit clever since you don't have an ambient space in which to have tangent vectors and instead use &lt;a href=&quot;http://en.wikipedia.org/wiki/Derivation_%28abstract_algebra%29&quot; rel=&quot;nofollow&quot;&gt;derivations&lt;/a&gt; (of e.g. smooth functions).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Integrals in multiple variables are much more complicated that the usual Riemann integral. Even when the functions are continuous. As demonstrated by &lt;a href=&quot;http://en.wikipedia.org/wiki/Fubini%27s_theorem&quot; rel=&quot;nofollow&quot;&gt;Fubini's theorem&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Tangentially calculus-related (really it's more analysis, but those are related anyway): The fact that $\mathbb{R}$ has an ordering allows one to define things like the &lt;a href=&quot;http://en.wikipedia.org/wiki/Henstock%E2%80%93Kurzweil_integral&quot; rel=&quot;nofollow&quot;&gt;Henstock–Kurzweil integral&lt;/a&gt;. Such an extension of the Lebesgue integral is (AFAIK) not possible in $\mathbb{R}^n$ for $n &amp;gt; 1$.&lt;/p&gt;&#xA;" OwnerUserId="1269" LastActivityDate="2010-11-25T13:12:14.430" />
  <row Id="11842" PostTypeId="2" ParentId="11831" CreationDate="2010-11-25T14:02:49.233" Score="7" ViewCount="" Body="&lt;p&gt;Every question of the form: prove by induction that&#xA;$$\sum_{k=1}^n f(k)=g(n)$$&#xD;&#xA;can be done by verifying two facts about the functions&#xA;$f$ and $g$:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$f(1)=g(1)$&#xD;&#xA;and&lt;/li&gt;&#xA;&lt;li&gt;$g(n+1)-g(n)=f(n+1)$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-11-25T14:02:49.233" />
  <row Id="11843" PostTypeId="1" AcceptedAnswerId="12753" CreationDate="2010-11-25T14:05:24.377" Score="4" ViewCount="223" Body="&lt;p&gt;Hi all!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lets imagine that we have some experiments. Each experiment may result in one of the outcomes: A, B, C. So we have probability distributions for each experiment $P_A, P_B, P_C$ which is context-dependent, e.g.:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$Context_1 \Rightarrow \{P_A^1, P_B^1, P_C^1\},$ experimental outcome is A&lt;/li&gt;&#xA;&lt;li&gt;$Context_2 \Rightarrow \{P_A^2, P_B^2, P_C^2\},$ experimental outcome is B&lt;/li&gt;&#xA;&lt;li&gt;$Context_3 \Rightarrow \{P_A^3, P_B^3, P_C^3\},$ experimental outcome is A&lt;/li&gt;&#xA;&lt;li&gt;$Context_4 \Rightarrow \{P_A^4, P_B^4, P_C^4\},$ experimental outcome is C&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;These probabilities are calculated by some function $F:Context\rightarrow \{P_A, P_B, P_C\}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to estimate an absolute trust rate of this function. In other words, I want to be able to say &quot;we can trust this function on 86%&quot; like we do when we deal with Pearson's chi-square test.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any suggestions?&lt;/p&gt;&#xA;" OwnerUserId="4003" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-28T15:01:38.453" LastActivityDate="2010-12-02T13:46:47.453" Title="Chi-square analog for context-dependent distributions" Tags="&lt;probability&gt;&lt;statistics&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="1" />
  <row Id="11844" PostTypeId="2" ParentId="11495" CreationDate="2010-11-25T14:18:04.083" Score="0" ViewCount="" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Do you think it is possible to&#xA;  successfully study mathematics with&#xA;  such a short initial knowledge and not&#xA;  having my full day to do it?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That depends on your university course and how faster you can learn and get skills, but you will like to devote at least 4-5 hours day, and will want to do as much exercises as you can.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;When in high school I've had mostly&#xA;  literary classes, so, again, my&#xA;  current knowledge of algebra, geometry&#xA;  and arithmetic is quite basic.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You should start with that basic stuff, they are the foundations for everything that comes next and doing that way will prevent you to go back very often while you are learning the college's math. Once you get in touch with Algebra, Calculus and Discrete Math you will see how there is sinergy about programming and doing math that makes the hours worthy besides&#xA;the CS degree.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to give you an advice from my personal experience being a math student but like CS more ;) : &lt;/p&gt;&#xA;&#xA;&lt;p&gt;half efforts don't produce half results, you must do as more exercises as you can practicing&#xA;both theory and plumping work (when you get to lineral algebra you might remember my words),&#xA;don't let that the amount of work to-do grow bigger every week, be sure to understand the theorems (that is, dont left the desk every time you get stuck, happens to everyone). After you dive in for a couple of weeks you will start noticing the difference... you will become more analytic and insightful.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Any general advice or book / site to&#xA;  get me started with the &quot;preparation&quot;&#xA;  study? What are the basic areas I&#xA;  should concentrate on to be prepared&#xA;  to the first year math exams?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Consult your courses specifications, they are many books out there, first read the ones your professor recommends... &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Vi auguro buona fortuna (i wish you good luck right?)&lt;/p&gt;&#xA;" LastActivityDate="2010-11-25T14:18:04.083" CommunityOwnedDate="2010-11-25T14:18:04.083" />
  <row Id="11845" PostTypeId="2" ParentId="11669" CreationDate="2010-11-25T14:25:00.390" Score="11" ViewCount="" Body="&lt;p&gt;Note also that many cultures use a &lt;a href=&quot;http://en.wikipedia.org/wiki/Pentatonic_scale&quot; rel=&quot;nofollow&quot;&gt;pentatonic scale&lt;/a&gt;. This would correspond to playing only the notes CDEGA. As explained in Qiaochu's answer, we want notes that are in small rational intervals, and particularly notes that are in small rational intervals from the tonic. Exactly which set of notes is chosen varies from culture to culture, with Western music using the 7 white keys, but many other cultures only using the 5 pentatonics.&lt;/p&gt;&#xA;" OwnerUserId="448" LastActivityDate="2010-11-25T14:25:00.390" CommentCount="2" />
  <row Id="11846" PostTypeId="1" CreationDate="2010-11-25T14:39:18.680" Score="1" ViewCount="182" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am learning about discrete probability distributions and found 2 definitions for Geometric Distributions from &lt;a href=&quot;http://en.wikipedia.org/wiki/Geometric_distribution&quot; rel=&quot;nofollow&quot;&gt;wikipedia&lt;/a&gt;:&#xA;1. The probability distribution of the number X of Bernoulli trials needed to get one success, supported on the set { 1, 2, 3, ...}&#xA;2. The probability distribution of the number Y = X − 1 of failures before the first success, supported on the set { 0, 1, 2, 3, ... }&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems like a subtle difference, but i'm having trouble wrapping my head around when i would use which? Any insights appreciated.&lt;/p&gt;&#xA;" OwnerUserId="4007" LastActivityDate="2010-11-27T05:28:44.690" Title="Which geometric distribution to use?" Tags="&lt;probability&gt;&lt;statistics&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="0" />
  <row Id="11847" PostTypeId="2" ParentId="11836" CreationDate="2010-11-25T14:42:02.847" Score="1" ViewCount="" Body="&lt;p&gt;The question is too wide - the answer depends on who you ask.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Essential in analysis is to look at limits. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) In one real variable there are two ways/directions to narrow a point. &#xA;In two variables there are infinitely lines through a point, hence there are infinitely many directions to narrow a point and moreover you might get different results on different other kind of curves. (I think this is what Américo Tavares meant by infinitely many derivatives - directional derivatives). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) Integration over a 2-dimensional set might be done in at least three ways - a double integral or two iterated integrations -  which if the conditions of Fubini-Tonelli's theorem is fulfilled has the same result.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(3) The fundamental theorem of calculus is a bit harder - Stoke's theorem. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(4) Partial differential equations are, generally speaking, harder than ordinary differential equations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(5) Also, several variables makes one variable easier, as an example you will learn an easy way to calculate &#xA;$$\int_{-\infty}^\infty e^{-x^2}dx$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(6) ...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(7) ...&lt;/p&gt;&#xA;" OwnerUserId="1154" LastActivityDate="2010-11-25T14:42:02.847" />
  <row Id="11848" PostTypeId="2" ParentId="11834" CreationDate="2010-11-25T14:50:32.157" Score="11" ViewCount="" Body="&lt;p&gt;Qiaochu is right: categories are much more general. Let me provide a concrete example which, while in a sense trivial, may illustrate just how much more you could get from a category than you can from a reflexive and &lt;em&gt;transitive&lt;/em&gt; relation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the positive integers $\mathbb P$. The usual total order $\leqslant$ is a perfectly good reflexive, transitive relation on this set. We can think of this as a category which consists of the identity map $\mathrm{id}_n$ on each integer &lt;em&gt;n&lt;/em&gt;&amp;nbsp;&amp;isin;&amp;nbsp;$\mathbb P$, and the composition of the successor maps $\sigma_n: n \to n+1$. The result is essentially an &lt;em&gt;almost&lt;/em&gt; acyclic directed graph (all circuits are stationary walks on a single vertex).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let me describe to you a different category on the positive integers: the category of injective linear maps on finite-dimensional real vector spaces &amp;mdash; which is to say, the category whose objects are the positive integers, and whose arrows are &lt;em&gt;m&lt;/em&gt;&amp;nbsp;&amp;times;&amp;nbsp;&lt;em&gt;n&lt;/em&gt; matrices over $\mathbb R$, with rank &lt;em&gt;n&lt;/em&gt;, for &lt;em&gt;m&lt;/em&gt;&amp;nbsp;&amp;ge;&amp;nbsp;&lt;em&gt;n&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;em&gt;&lt;strong&gt;An aside&lt;/em&gt;&lt;/strong&gt;. &quot;What!?&quot; you may exclaim; &quot;How can you talk about matrices having domains and codomains consisting of positive integers?&quot; Well, the shape of each matrix defines what sorts of other matrices it can be composed with, and any action of a matrix &lt;em&gt;M&lt;/em&gt; on a vector &lt;strong&gt;v&lt;/strong&gt;&amp;nbsp;&amp;isin;&amp;nbsp;$\mathbb R^n$ could be viewed as composing the &lt;em&gt;m&lt;/em&gt;&amp;nbsp;&amp;times;&amp;nbsp;&lt;em&gt;n&lt;/em&gt; matrix &lt;em&gt;M&lt;/em&gt; with the &lt;em&gt;n&lt;/em&gt;&amp;nbsp;&amp;times;&amp;nbsp;1 matrix &lt;strong&gt;v&lt;/strong&gt; to obtain an &lt;em&gt;m&lt;/em&gt;&amp;nbsp;&amp;times;&amp;nbsp;1 matrix &lt;em&gt;M&lt;/em&gt;&lt;strong&gt;v&lt;/strong&gt;. The integers stand in for the dimension of the vector space $\mathbb R^n$ or $\mathbb R^m$ which are the &quot;real&quot; domain and codomain of the matrix; and because we can substitute the linear action of operators on $\mathbb R^n$ with the effect of composing two linear operators (one of which happens to have a domain of $\mathbb R^1$), we may ignore the structure inside of the vector spaces and refer to them just by their dimensions.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As with the directed graph generated by the total order &amp;le;, there exist arrows &lt;em&gt;m&lt;/em&gt;&amp;nbsp;&amp;rarr;&amp;nbsp;&lt;em&gt;n&lt;/em&gt; for any &lt;em&gt;m&lt;/em&gt;&amp;nbsp;&amp;ge;&amp;nbsp;&lt;em&gt;n&lt;/em&gt; (by definition). All of the maps from &lt;em&gt;n&lt;/em&gt; to &lt;em&gt;m&lt;/em&gt; for &lt;em&gt;m&lt;/em&gt;&amp;nbsp;&amp;gt;&amp;nbsp;&lt;em&gt;n&lt;/em&gt;+1 can be obtained by composing maps &lt;em&gt;n&lt;/em&gt;&amp;nbsp;&amp;rarr;&amp;nbsp;&lt;em&gt;n&lt;/em&gt;+1 and &lt;em&gt;n&lt;/em&gt;+1&amp;nbsp;&amp;rarr;&amp;nbsp;&lt;em&gt;m&lt;/em&gt;; and furthermore, all maps &lt;em&gt;n&lt;/em&gt;&amp;nbsp;&amp;rarr;&amp;nbsp;&lt;em&gt;n&lt;/em&gt;+1 can be obtained by composing &lt;em&gt;any single designated&lt;/em&gt; map $\sigma_n: n \to n+1$ with a suitable map $\tau: n \to n$. So this category has a number of things in common with the one defined by the total order &amp;le;. But there are important differences, owing in part to the fact that there are infinitely maps &lt;em&gt;n&lt;/em&gt;&amp;nbsp;&amp;rarr;&amp;nbsp;&lt;em&gt;n&lt;/em&gt; for each &lt;em&gt;n&lt;/em&gt;. Because of this, this second category has much more structure than a binary relation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, it becomes meaningful and interesting to talk about &lt;em&gt;coproducts&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;em&gt;&lt;strong&gt;Definition.&lt;/em&gt;&lt;/strong&gt; Let &lt;em&gt;A,B&lt;/em&gt; be two objects. A &lt;em&gt;coproduct&lt;/em&gt; of &lt;em&gt;A&lt;/em&gt; and &lt;em&gt;B&lt;/em&gt; is an object (&lt;em&gt;A&lt;/em&gt;&amp;nbsp;$\amalg$&amp;nbsp;&lt;em&gt;B&lt;/em&gt;), together with two maps &lt;em&gt;i&lt;sub&gt;A&lt;/sub&gt;&lt;/em&gt;&amp;nbsp;:&amp;nbsp;&lt;em&gt;A&lt;/em&gt;&amp;nbsp;&amp;rarr;&amp;nbsp;(&lt;em&gt;A&lt;/em&gt;&amp;nbsp;$\amalg$&amp;nbsp;&lt;em&gt;B&lt;/em&gt;) and &lt;em&gt;i&lt;sub&gt;B&lt;/sub&gt;&lt;/em&gt;&amp;nbsp;:&amp;nbsp;&lt;em&gt;B&lt;/em&gt;&amp;nbsp;&amp;rarr;&amp;nbsp;(&lt;em&gt;A&lt;/em&gt;&amp;nbsp;$\amalg$&amp;nbsp;&lt;em&gt;B&lt;/em&gt;) such that, for any other object &lt;em&gt;X&lt;/em&gt; and maps &lt;em&gt;f&lt;/em&gt;&amp;nbsp;:&amp;nbsp;&lt;em&gt;A&lt;/em&gt;&amp;nbsp;&amp;rarr;&amp;nbsp;&lt;em&gt;X&lt;/em&gt; and &lt;em&gt;g&lt;/em&gt;&amp;nbsp;:&amp;nbsp;&lt;em&gt;B&lt;/em&gt;&amp;nbsp;&amp;rarr;&amp;nbsp;&lt;em&gt;X&lt;/em&gt;, there exists a unique map (&lt;em&gt;f&lt;/em&gt;&amp;nbsp;|&amp;nbsp;&lt;em&gt;g&lt;/em&gt;)&amp;nbsp;:&amp;nbsp;(&lt;em&gt;A&lt;/em&gt;&amp;nbsp;$\amalg$&amp;nbsp;&lt;em&gt;B&lt;/em&gt;)&amp;nbsp;&amp;rarr;&amp;nbsp;&lt;em&gt;X&lt;/em&gt; such that &lt;em&gt;f&lt;/em&gt;&amp;nbsp;=&amp;nbsp;(&lt;em&gt;f&lt;/em&gt;&amp;nbsp;|&amp;nbsp;&lt;em&gt;g&lt;/em&gt;)&amp;nbsp;$\circ$&amp;nbsp;&lt;em&gt;i&lt;sub&gt;A&lt;/sub&gt;&lt;/em&gt; and &lt;em&gt;g&lt;/em&gt;&amp;nbsp;=&amp;nbsp;(&lt;em&gt;f&lt;/em&gt;&amp;nbsp;|&amp;nbsp;&lt;em&gt;g&lt;/em&gt;)&amp;nbsp;$\circ$&amp;nbsp;&lt;em&gt;i&lt;sub&gt;B&lt;/sub&gt;&lt;/em&gt;&amp;nbsp;. That is, in the following diagram, any two directed walks between a common pair of points represent the same map:&lt;br&gt;&lt;br&gt;&#xA;  $\begin{matrix} \qquad\! A \!\!&amp;amp;\!\! \xrightarrow{\textstyle \;i_A\;} \!\!&amp;amp;\!\! (A \amalg B) \!\!&amp;amp;\!\! \xleftarrow{\textstyle \;i_B\;} \!\!&amp;amp;\!\! B \!\qquad \\&#xD;&#xA;\mathrm{id_A}\Bigg\updownarrow &amp;amp; &amp;amp; \Bigg\downarrow(f|g)\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!&amp;amp; &amp;amp;\Bigg\updownarrow\mathrm{id_B} \\&#xD;&#xA;\qquad\! A \!\!&amp;amp;\!\! \xrightarrow[\textstyle \quad f \quad] \!\!\!\!\!\!\!\!\!\!&amp;amp;\!\! X \!\!&amp;amp;\!\!\!\!\!\!\!\!\!\! \xleftarrow[\textstyle \quad g \quad] \!\!&amp;amp;\!\! B\!\qquad \end{matrix}$&#xD;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;In the case of the partial order &amp;le;, the coproduct of &lt;em&gt;A&lt;/em&gt; and &lt;em&gt;B&lt;/em&gt; is just &amp;thinsp;&lt;b&gt;max&lt;/b&gt;&amp;thinsp;{&amp;thinsp;&lt;em&gt;A, B&lt;/em&gt;&amp;thinsp;}&amp;thinsp;, which you should verify for yourself.&#xA;&#xA;&lt;li&gt;In the category of matrices we described above, the coproduct of &lt;em&gt;A&lt;/em&gt; and &lt;em&gt;B&lt;/em&gt; is not the maximum of &lt;em&gt;A&lt;/em&gt; and &lt;em&gt;B&lt;/em&gt;, but is instead the integer &lt;em&gt;A&lt;/em&gt;+&lt;em&gt;B&lt;/em&gt;, corresponding to splitting a vector of dimension &lt;em&gt;A&lt;/em&gt;+&lt;em&gt;B&lt;/em&gt; into a block of size &lt;em&gt;A&lt;/em&gt; and a block of size &lt;em&gt;B&lt;/em&gt;.&lt;br&gt;&lt;br&gt;&#xA;&lt;ul&gt;&lt;li&gt; The &quot;inclusion&quot; map $i_A$ would correspond to mapping a vector in &lt;b&gt;v&lt;/b&gt;&amp;nbsp;&amp;isin;&amp;nbsp;$\mathbb R^A$ into the first &lt;em&gt;A&lt;/em&gt; coefficients of a vector in $\mathbb R^{A+B}$, with the others set to zero;&#xA;&lt;li&gt; The &quot;inclusion&quot; map $i_B$ would be similar, mapping vectors in $\mathbb R^B$ into the final &lt;em&gt;B&lt;/em&gt; coefficients.&#xA;&lt;li&gt; For functions &lt;em&gt;f&lt;/em&gt;:&amp;nbsp;&lt;em&gt;A&lt;/em&gt;&amp;nbsp;&amp;rarr;&amp;nbsp;&lt;em&gt;A&lt;/em&gt;+&lt;em&gt;B&lt;/em&gt; and &lt;em&gt;g&lt;/em&gt;:&amp;nbsp;&lt;em&gt;B&lt;/em&gt;&amp;nbsp;&amp;rarr;&amp;nbsp;&lt;em&gt;A&lt;/em&gt;+&lt;em&gt;B&lt;/em&gt;, the map (&lt;em&gt;f&lt;/em&gt;&amp;thinsp;|&amp;thinsp;&lt;em&gt;g&lt;/em&gt;) corresponds to performing the map &lt;em&gt;f&lt;/em&gt; to the first block and &lt;em&gt;g&lt;/em&gt; to the second block of the vector.&lt;/ul&gt;&#xA;&lt;br&gt;Of course, there's more than one way to decompose $\mathbb R^{A+B}$ into blocks; we could have chosen a different definition for the maps &lt;em&gt;i&lt;sub&gt;A&lt;/sub&gt;&lt;/em&gt; and &lt;em&gt;i&lt;sub&gt;B&lt;/sub&gt;&lt;/em&gt;, by having &lt;em&gt;i&lt;sub&gt;B&lt;/sub&gt;&lt;/em&gt; map vectors in $\mathbb R^B$ into the first &lt;em&gt;B&lt;/em&gt; coefficients, and &lt;em&gt;i&lt;sub&gt;A&lt;/sub&gt;&lt;/em&gt; map vectors into the final &lt;em&gt;A&lt;/em&gt; coefficients. This way of embedding $\mathbb R^A$ and $\mathbb R^B$ into the same space at the same time is equivalent to the one we describe above; in fact we can show that there is a unique isomorphism between these two ways of constructing (&lt;em&gt;A&lt;/em&gt;&amp;nbsp;$\amalg$&amp;nbsp;&lt;em&gt;B&lt;/em&gt;). So, the coproduct (&lt;em&gt;A&lt;/em&gt;&amp;nbsp;$\amalg$&amp;nbsp;&lt;em&gt;B&lt;/em&gt;) is dependent on the choice of maps &lt;em&gt;i&lt;sub&gt;A&lt;/sub&gt;&lt;/em&gt; and &lt;em&gt;i&lt;sub&gt;B&lt;/sub&gt;&lt;/em&gt; &amp;mdash; but not in any way which is really important. &#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Even though the first category generated by &amp;le; is what you get when you forget the matrix structure of the second category &amp;mdash; by replacing each infinite collection of arrows in each case with a single arrow between the same points &amp;mdash; that second category of injective linear operators is substantially different from the simple one described by &amp;le;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is just scratching the surface of one fingernail of category theory; but hopefully it convinces you that categories and reflexive, transitive binary relations are not trivially equivalent.&lt;/p&gt;&#xA;" OwnerUserId="439" LastEditorUserId="439" LastEditorDisplayName="" LastEditDate="2010-11-25T21:00:27.800" LastActivityDate="2010-11-25T21:00:27.800" CommentCount="5" />
  <row Id="11849" PostTypeId="1" CreationDate="2010-11-25T15:23:39.890" Score="2" ViewCount="112" Body="&lt;p&gt;I am looking for a proof that this inequality:&#xA;$$(W_1+W_2+\cdots+W_n)^a \leq  W_1^a +\cdots +  W_n^a$$ &#xA;is valid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a power function $f(W)=W^a$ where $a$ is a real number, constant but usually $0\lt a\lt 1$. The $W_i$ are real positive numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then I want to check if $\left(\sum W_i\right)^a \leq \sum(W_i^a)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example for $n=2$ : $(x+y)^a \leq x^a + y^a$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="4011" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-25T23:36:50.523" LastActivityDate="2010-11-25T23:36:50.523" Title="$(W_1+W_2+\cdots+W_n)^a \leq W_1^a +\cdots + W_n^a$ for $n$ integer, $n\geq 2$, $W\gt 0$ and $a$ constant, real, $0\lt a\lt 1$" Tags="&lt;inequality&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="11850" PostTypeId="2" ParentId="11849" CreationDate="2010-11-25T15:48:29.343" Score="1" ViewCount="" Body="&lt;p&gt;To show that $(w_1+w_2)^a\leq w_1^a+w_2^a$ whenever $w_1$ and $w_2$ are positive, we can divide by $w_1^a$ on both sides and rewrite $\frac{w_2}{w_1}$ as $t$ to get the equivalent statement $(1+t)^a\leq 1+ t^a$ when $t\geq 0$.  Notice that the two sides agree when $t$ goes to zero, and the derivative of the right-hand side with respect to $t$ is always bigger than that of the left-hand side (because $0\lt a\lt 1$), so the inequality follows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The case of general $n$ follows by induction, using the $n=2$ case for the inductive step. &lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2010-11-25T15:53:40.880" LastActivityDate="2010-11-25T15:53:40.880" />
  <row Id="11851" PostTypeId="2" ParentId="11849" CreationDate="2010-11-25T16:04:55.173" Score="4" ViewCount="" Body="&lt;p&gt;This is equivalent to proving $(\frac{W_1}{W_1+W_2+...+W_n})^a+(\frac{W_2}{W_1+W_2+...+W_n})^a+...+(\frac{W_n}{W_1+W_2+...+W_n})^a\geq 1$. Write $a_i=\frac{W_1}{W_1+W_2+...+W_n}$. Then, $a_1+a_2+...+a_n=1$, $0&amp;lt; a_1,a_2,...,a_n,a &amp;lt; 1$. So, $a_i^a &amp;gt; a_i$. So, $a_1^a+...+a_n^a&amp;gt; a_1+a_2+...+a_n=1$. QED. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: I think the inequality should be strict since $a&amp;lt;1$.&lt;/p&gt;&#xA;" OwnerUserId="3431" LastEditorUserId="3431" LastEditorDisplayName="" LastEditDate="2010-11-25T21:41:55.580" LastActivityDate="2010-11-25T21:41:55.580" CommentCount="4" />
  <row Id="11852" PostTypeId="1" CreationDate="2010-11-25T16:37:51.623" Score="0" ViewCount="99" Body="&lt;p&gt;Let f be a permutation on n letters. I want to count the number of k-long increasing subsequences quickly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First I use divide and conquer to split the problem into finding increasing subsequences in pairs of numbers. The resulting 'tree' has height logn. I'm wondering how I can combine the subproblems in an efficient way. I thought about prefix sums and using mergesort, but I can't see how to effectively implement them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One idea to save time when combining two subproblems is to compare elements at the end of your increasing subsequences in your first subproblem with the values in your second subproblem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My goal is to get to O(knlogn) time but this doesn't do it.&lt;/p&gt;&#xA;" LastActivityDate="2010-11-25T16:46:35.393" Title="computing the number of increasing k-long subsequences" Tags="&lt;algorithms&gt;&lt;permutations&gt;" AnswerCount="0" CommentCount="3" />
  <row Id="11854" PostTypeId="2" ParentId="3048" CreationDate="2010-11-25T17:19:43.410" Score="1" ViewCount="" Body="&lt;p&gt;It seems not to have yet been mentioned here that there is no universal agreement on the meaning of countable.  &quot;Countably infinite&quot; is unambiguous, but some authors use &quot;countable&quot; to mean countably infinite, while many (perhaps most) use countable to mean finite or countably infinite, as the other answers indicate.  When authors use countable to refer only to sets in bijection with $\mathbb{N}$, they often end up using the phrase &quot;at most countable&quot;.  This is seen for example in Rudin's analysis texts.   Springer's online encyclopedia also &lt;a href=&quot;http://eom.springer.de/C/c026710.htm&quot; rel=&quot;nofollow&quot;&gt;defines&lt;/a&gt; countable to mean countably infinite. &lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-11-25T17:19:43.410" />
  <row Id="11855" PostTypeId="2" ParentId="11818" CreationDate="2010-11-25T17:32:29.240" Score="0" ViewCount="" Body="&lt;p&gt;If the list is drawn with replacement, the chance that a given number (the new draw) is not in it is $$(1-\frac{1}{2147000001})^{1000000}$$  The birthday problem only comes up if you ask for the chance of a collision anywhere in the 1000001 numbers.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-11-25T17:32:29.240" />
  <row Id="11856" PostTypeId="1" AcceptedAnswerId="12092" CreationDate="2010-11-25T17:40:15.383" Score="3" ViewCount="140" Body="&lt;p&gt;I think the cone (or what is also called the &quot;half cone&quot;) is a differential manifold but not a smooth manifold. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone help me understand this the nuts and bolts way? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How explicitly can I write down the differential structure on the cone?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also what is the natural metric on it? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Is that what is called the ``Sasakian Metric&quot;? If yes then can one kindly explain that or give an expository reference on this?)&lt;/p&gt;&#xA;" OwnerUserId="965" LastActivityDate="2010-11-27T18:13:57.443" Title="Differential and Riemannian structure on the cone" Tags="&lt;differential-geometry&gt;" AnswerCount="2" CommentCount="0" />
  <row Id="11857" PostTypeId="2" ParentId="11849" CreationDate="2010-11-25T17:48:03.253" Score="0" ViewCount="" Body="&lt;p&gt;See my answer to &lt;a href=&quot;http://math.stackexchange.com/questions/10993/please-explain-inequality-xp-yp-leq-x-yp&quot;&gt;Please explain inequality $|x^p-y^p|\le |x-y|^p$&lt;/a&gt; -&#xA;the inequality you ask for follows by induction.&lt;/p&gt;&#xA;" OwnerUserId="1154" LastActivityDate="2010-11-25T17:48:03.253" />
  <row Id="11858" PostTypeId="2" ParentId="11669" CreationDate="2010-11-25T17:57:36.350" Score="4" ViewCount="" Body="&lt;p&gt;Start at F and go up a fifth (to C).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/xEMSY.jpg&quot; alt=&quot;piano&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(In a keyboard with 12-key octaves, that's 7 steps.) Repeat that process (through the circle of fifths). You'll hit all the white keys and then all the black keys  -- F, C, G, D, A, E, B, F#, C#, G#, D#, A# -- note, these keys are usually represented with flats). So it turns out if you're splitting tones on 3 : 2 (fifth) or 4 : 3 (fourth), the least common multiple is twelve. In practice, the 3 : 2 is similar enough that it gives a sort of 'secure' or content feeling. The 4 : 3 gives a slightly edgier feeling but one which is somewhat counterposed perfectly against this secure feeling. So a fourth + a fifth will give you an octave. So why we want all twelve keys is that we're saying that we want the fifth (dominant) and the fourth (subdominant) to come together and make a whole. This is sort of mirrored with the fact that the first 7/12 of the circle of fifths form the basis and the second 5/12 are 'overlayed' on top (with black keys).&lt;/p&gt;&#xA;" OwnerUserId="4022" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-26T03:16:40.100" LastActivityDate="2010-11-26T03:16:40.100" />
  <row Id="11859" PostTypeId="1" AcceptedAnswerId="11860" CreationDate="2010-11-25T18:00:35.497" Score="3" ViewCount="164" Body="&lt;p&gt;I have seen this claim on Wikipedia and other places. Which branch of mathematics does this result come from?&lt;/p&gt;&#xA;" OwnerUserId="4025" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-25T23:06:21.997" LastActivityDate="2010-11-25T23:06:21.997" Title="Why can't erf be expressed in terms of elementary functions?" Tags="&lt;special-functions&gt;" AnswerCount="3" />
  <row Id="11860" PostTypeId="2" ParentId="11859" CreationDate="2010-11-25T18:04:38.467" Score="3" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Differential_Galois_theory&quot; rel=&quot;nofollow&quot;&gt;Differential Galois theory.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2937" LastActivityDate="2010-11-25T18:04:38.467" />
  <row Id="11861" PostTypeId="2" ParentId="11859" CreationDate="2010-11-25T18:06:45.210" Score="1" ViewCount="" Body="&lt;p&gt;I am not aware of any easy proof of this. Here is a proof that $\int e^{x^2}$ cannot be expressed in terms of elementary functions:&#xA;&lt;a href=&quot;http://math.hunter.cuny.edu/ksda/papers/rick_09_02.pdf&quot; rel=&quot;nofollow&quot;&gt;http://math.hunter.cuny.edu/ksda/papers/rick_09_02.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3431" LastActivityDate="2010-11-25T18:06:45.210" />
  <row Id="11862" PostTypeId="2" ParentId="11859" CreationDate="2010-11-25T18:07:15.880" Score="2" ViewCount="" Body="&lt;p&gt;This is a consequence of the Liouville-Ritt theory of integration in finite terms. You can find a brief sketch and references in my &lt;a href=&quot;http://math.stackexchange.com/questions/155/how-can-you-prove-that-a-function-has-no-closed-form-integral/5723#5723&quot;&gt;post here.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-11-25T18:07:15.880" />
  <row Id="11863" PostTypeId="2" ParentId="11856" CreationDate="2010-11-25T18:10:35.103" Score="1" ViewCount="" Body="&lt;p&gt;The differentiable structure of the cone can be viewed as the same as the one of $S^1\times\mathbb R$, or $\mathbb{R}^2 - \{0\}$ . That is, you construct a homeomorphism between $\mathbb{R}^2 - \{0\}$ and the cone (as subsets of $\mathbb{R}^3$) and say that this is a diffeomorphism, in the most intuitive way: $(x,y)\mapsto (x,y,\sqrt{x^2 + y^2})$. With this imposition, the differentiable structure of $\mathbb{R}^2 - \{0\}$ is carried to the cone.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the &quot;natural&quot; metric is the one induced from $\mathbb{R}^3$: $\tilde g = i^*g$, where $i$ is the inclusion map from the cone to $\mathbb{R}^3$ and $g$ is the Euclidean metric of $\mathbb{R}^3$. &lt;/p&gt;&#xA;" OwnerUserId="1178" LastActivityDate="2010-11-25T18:10:35.103" CommentCount="2" />
  <row Id="11864" PostTypeId="1" AcceptedAnswerId="11893" CreationDate="2010-11-25T18:30:18.393" Score="4" ViewCount="187" Body="&lt;p&gt;Let $A$ be the infinitesimal generator of a $C_0$-semigroup $(S(t))_{t \geq 0}$. Now, for every $x_0 \in X$ the map $t \mapsto S(t) x_0$ is a mild solution of&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}\label{eq:1}&#xA;\dot{x} = Ax, \quad x(0) = x_0. \qquad \text{(*)}&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, a continuous function $x: [0, \infty) \to X$ is called a mild solution of $\text{(*)}$ if $\int_0^t x(s) \, ds \in D(A)$ where $D(A)$ is the domain of $A$, $x(0) = x_0$ and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x(t) - x(0) = A \int_0^t x(\tau) \, d\tau \text{ for all $t \geq 0$}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, I have a proof of this but it uses Hille's theorem, but it is quite involved (needs a few tricks) and Hille's theorem is not elementary, does someone know an elementary proof of the uniqueness?&lt;/p&gt;&#xA;" OwnerUserId="1120" LastEditorUserId="1120" LastEditorDisplayName="" LastEditDate="2010-12-04T23:51:13.470" LastActivityDate="2010-12-04T23:51:13.470" Title="Uniqueness mild solution of $\dot{x} = A x$." Tags="&lt;differential-equations&gt;&lt;semigroup&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="11865" PostTypeId="1" AcceptedAnswerId="11900" CreationDate="2010-11-25T18:33:11.130" Score="4" ViewCount="289" Body="&lt;p&gt;&lt;a href=&quot;http://www-history.mcs.st-and.ac.uk/Biographies/Khayyam.html&quot; rel=&quot;nofollow&quot;&gt;Omar Khayyam&lt;/a&gt; is known for his significant progress in solving cubic polynomial equations. For example, his biography on &lt;a href=&quot;http:///www-history.mcs.st-andrews.ac.uk&quot; rel=&quot;nofollow&quot;&gt;www-history.mcs.st-andrews.ac.uk&lt;/a&gt; says&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;(...) This problem in turn led Khayyam to&#xA;  solve the cubic equation x^3 + 200x =&#xA;  20x^2 + 2000 and he found a positive&#xA;  root of this cubic by considering the&#xA;  intersection of a rectangular&#xA;  hyperbola and a circle. &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;(...) Indeed Khayyam did produce such a work, the Treatise &#xA;  on Demonstration of Problems of&#xA;  Algebra  which contained a complete&#xA;  classification of cubic equations with&#xA;  geometric solutions found by means of&#xA;  intersecting conic sections.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But I still can't see the big picture of those days.  I'm possibly omitting something about the idea of geometric solutions of algebraic equations, but why were they trying hard to find intersections of conic sections, and building large classification schemes for it?. If the idea was to get a numerical value out of these constructions by measuring lengths on paper,  they could just have prepared a careful template for the function $y = x^3$, and then solved all the cubic equations by intersecting it with a parabola, like in the figure below for the mentioned equation.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would appreciate answers that would clarify my confusion.  Was it that they did not conceive $y=x^3$ as a curve, if they were interested in getting a numerical value? Or was it a conceptual challenge to show that all cubic equations can be represented as an intersection of two conic sections? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/E4U2q.png&quot; alt=&quot;plot&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="145" LastEditorUserId="145" LastEditorDisplayName="" LastEditDate="2010-11-25T18:51:20.547" LastActivityDate="2010-11-26T03:58:11.750" Title="Khayyam's work on cubic equations" Tags="&lt;math-history&gt;" AnswerCount="2" CommentCount="7" FavoriteCount="1" />
  <row Id="11866" PostTypeId="2" ParentId="11669" CreationDate="2010-11-25T18:43:35.760" Score="3" ViewCount="" Body="&lt;p&gt;Other responses do a good job of explaining the 12-note chromatic scale. From those 12 tones, if one starts to build a series of tones starting on a single note and going up the circle of fifths, there are two natural stopping points where you have a complete-sounding scale that spans the octaves and has relatively equal spacing between the notes with no gaps: five notes, which gives whole-step and minor-third intervals; and seven notes, which gives whole-step and half-step intervals. These two scales (pentatonic and heptatonic) correspond to the spacing of the black keys and white keys on the keyboard.  They are mirror images of each other around the circle of fifths. So the two colors of notes are not &quot;different,&quot; but rather a natural division into two symmetrical sclaes built from going opposite directions around the circle of fifths.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the standard tuning system C is &quot;privileged&quot; because it is (essentially) the note where we start building the circle of fifths to create these two scales.&lt;/p&gt;&#xA;" OwnerUserId="4030" LastActivityDate="2010-11-25T18:43:35.760" />
  <row Id="11867" PostTypeId="2" ParentId="6647" CreationDate="2010-11-25T18:48:17.030" Score="1" ViewCount="" Body="&lt;p&gt;Here is a good writeup about implementing Stirling's approximation, along with a reference implementation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://threebrothers.org/brendan/blog/stirlings-approximation-formula-clojure/&quot; rel=&quot;nofollow&quot;&gt;http://threebrothers.org/brendan/blog/stirlings-approximation-formula-clojure/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4032" LastActivityDate="2010-11-25T18:48:17.030" />
  <row Id="11868" PostTypeId="2" ParentId="11865" CreationDate="2010-11-25T18:57:25.520" Score="6" ViewCount="" Body="&lt;p&gt;I don't think they had the idea of y=x^3 as a curve. For one thing there was no analytic geometry until Descartes. With coordinates some problems become much simpler. &lt;/p&gt;&#xA;" OwnerUserId="771" LastActivityDate="2010-11-25T18:57:25.520" CommentCount="4" />
  <row Id="11869" PostTypeId="2" ParentId="6546" CreationDate="2010-11-25T19:24:01.080" Score="2" ViewCount="" Body="&lt;p&gt;Have you tried taking the semi-direct product of two subgroups (one needs to be a normal subgroup).  The order of G is the product of the orders of the subgroups.  One subgroup needs to be normal.  If the other is not, then the semidirect product is usually (if not always) non-abelian&lt;/p&gt;&#xA;" OwnerUserId="3971" LastActivityDate="2010-11-25T19:24:01.080" />
  <row Id="11870" PostTypeId="2" ParentId="11764" CreationDate="2010-11-25T19:28:34.587" Score="1" ViewCount="" Body="&lt;p&gt;Here is a reference -first result in google:&#xA;&lt;a href=&quot;http://projecteuclid.org/DPubS?service=UI&amp;amp;version=1.0&amp;amp;verb=Display&amp;amp;handle=euclid.ijm/1256060422&quot; rel=&quot;nofollow&quot;&gt;http://projecteuclid.org/DPubS?service=UI&amp;amp;version=1.0&amp;amp;verb=Display&amp;amp;handle=euclid.ijm/1256060422&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The author (who now works in a bank) needs his construction of sheaf homology to prove a criterion for the vanishing of certain integrals of differential forms, though the paper treats general topological spaces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The overarching goal seems interesting, as the names of Langlands, Igusa, Denef are mentioned, but is beyond me, unhelpfully.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway that looks like very nice mathematics. There are actually 2 constructions of sheaf homology. Both rely on triangulations, the first proceeds defining modules of sections on simplices and restriction maps to subsimplices and extending by linearity to complexes. The second, used to prove Poincaré duality/isomorphism with (usual) sheaf cohomology, proceeds defining sheaves &quot;on simplicial complexes&quot; and taking the homology of the chain complex of global sections of these.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The 2 constructions are isomorphic, and I guess they could be isomorphic to a more direct construction which only uses open covers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the paper you are referring to?&lt;/p&gt;&#xA;" OwnerUserId="3407" LastActivityDate="2010-11-25T19:28:34.587" />
  <row Id="11871" PostTypeId="1" AcceptedAnswerId="11877" CreationDate="2010-11-25T19:39:48.103" Score="4" ViewCount="435" Body="&lt;p&gt;Given $N$ random iid variables, $X_1, \ldots, X_N$, with a uniform probability distribution on $[0, 1)$ what is the distribution of $\displaystyle \max_{i = 1 \ldots N}(X_i)$?&lt;/p&gt;&#xA;" OwnerUserId="958" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2010-11-26T01:29:11.853" LastActivityDate="2010-11-26T01:29:11.853" Title="Probability distribution of the maximum of random variables" Tags="&lt;probability&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="11872" PostTypeId="1" CreationDate="2010-11-25T19:43:38.947" Score="0" ViewCount="287" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Possible Duplicate:&lt;/strong&gt;&lt;br&gt;&#xA;  &lt;a href=&quot;http://math.stackexchange.com/questions/4467/a1-2-is-either-an-integer-or-an-irrational-number&quot;&gt;$a^{1/2}$ is either an integer or an irrational number.&lt;/a&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;p&gt;Will every $n^{th}$ root of $2$ be an irrational number? If yes, how can I prove that?&lt;/p&gt;&#xA;" OwnerUserId="1445" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2010-11-26T01:08:29.040" LastActivityDate="2010-11-26T01:08:29.040" ClosedDate="2010-11-26T20:05:27.770" Title="Is $n^{th}$ root of 2 an irrational number?" Tags="&lt;roots&gt;" AnswerCount="2" CommentCount="7" FavoriteCount="1" />
  <row Id="11873" PostTypeId="2" ParentId="11669" CreationDate="2010-11-25T19:46:22.057" Score="16" ViewCount="" Body="&lt;p&gt;The math of frequency relationships here is sound (pun intended) but they don't help explain the white vs black key piano layout.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's the historical imperitive thay led to this layout for &quot;Western Music&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First consider the major triad: root + third + fifth notes of the &quot;diatonic scale&quot;.&#xA;They follow the harmonic series:&#xA;1 - root&#xA;2 - octave (doubling of root frequency)&#xA;3 - fifth (triple of the root - 3:2 relationship to the octave)&#xA;4 - double octave (4x)&#xA;5 - 10th (double octave of a third)&#xA;6 - octave fifth&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These are notes a static length of tubing can produce by blowing into it: the bugle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Combinations of these notes create frequencies that make choirs sound heaven-ly.&#xA;The frequencies align and blend into pure complex vibraations that are the sum and&#xA;the differencies (harmonic overtones) of these relationships.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Choirs can tune themselves dynamically to create these frequency alignments that&#xA;are percieved as being perfectly consonant. Upbeat western music focus on the 3 major&#xA;chords found in the diatonic scale:&#xA;1+3+5 root major chord - white keys C - E - G&#xA;4+6+1 4th chord - white keys F - A - C&#xA;5+7+2 5th chord - whote keys G - B - D&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The basics of western folk music are the 1 - 4 - 5 sequences of chords.&#xA;Learn C, F and G on a guitar and you can play the bulk of the classic Country &#xA;song book.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Put the notes of these chords into a scale and you get that row of 7 white keys:&#xA;C - D - E - F - G - A - B (repeat until you can't hear it).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, they western scale is based upon frequency relationships that make combinations&#xA;of notes &quot;ring&quot; in consonance in it's purest form... like the Gregorian Chants of&#xA;the Roman Church.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, a basic &quot;western keyboard&quot; could be made from just these 7 notes repeated across the frequency spectrum. Look at the layout of a Greek Lyre (a harp) and that's what you&#xA;will find. A sequence following the diatonic scale which sounds pleasant if you just strum across the strings due to the tuning of even multiples (adjusted by octaves).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;OK... now adding the black keys is a compromise of tuning specific notes so that you can build these 1+3+5 chords from any starting point and thus play a song adjusted up or down to any starting point. The piano will never achieve that sonic mathematical glimpse into the &quot;music of the spheres&quot; that the self-adjusting choir can to make a chord mathematically perfect in alignment but it's the &quot;keyboard&quot; for the modern composer... the effective &quot;musical qwerty&quot; that a composer or a pianist begins to visualize chord &quot;shapes&quot; as hand positions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With a lot of practice a pianist can pre-visualize sound in terms of finger and hand movements much like a solid touch typist starts to set words and sentences as a sequence of movements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The addition of the black keys was called a &quot;Well Tempered&quot; tuning and Bach was one of the first composers to create whole bodies of compositions that worked through the Major and Minor keys of the 12 scales that you noticed intially when inspecting the keyboard.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you look into other musical cultures you will find a different approaches to standardizing sound relationships that do not focus on the 1 - 4 - 5 chords.&#xA;This music to a culturally trained western ear is less predicatable in nature and&#xA;that lack of predictablility can make the music frustrating or exciting... music &quot;speaks&quot; to us in terms of pure sensory inputs that can move, excite, bore or confuse us.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the piano keyboard is designed to be the perfect delivery system for an individual to produce the range of complexity that western music has achieved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The modern keyboard synthesizers are now able to produce the full range of the western orchestra in terms of &quot;instruments&quot; and I'm hoping someone create one that micro-adjusts notes based upon the surrounding context... shifting a note up or down slightly from the &quot;well tempered&quot; compromise to the pitch that makes a chord &quot;ring&quot; and&#xA;produce the upper harmonic overtones that make a great orchestra truly &quot;heavenly&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe it's already been done.&lt;/p&gt;&#xA;" OwnerUserId="4040" LastActivityDate="2010-11-25T19:46:22.057" CommentCount="2" />
  <row Id="11874" PostTypeId="1" AcceptedAnswerId="11876" CreationDate="2010-11-25T19:48:04.957" Score="4" ViewCount="281" Body="&lt;p&gt;What does this expression mean?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f\in C^2[a,b]$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More specifically, I don't know what $C$ means.&lt;/p&gt;&#xA;" OwnerUserId="2637" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2010-11-26T05:17:06.330" LastActivityDate="2010-11-26T05:17:06.330" Title="What does &quot;$f\in C^2[a,b]$&quot; mean?" Tags="&lt;notation&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="11875" PostTypeId="2" ParentId="11872" CreationDate="2010-11-25T19:50:16.513" Score="9" ViewCount="" Body="&lt;p&gt;Yes. In fact, for every integer $k$ and every $n\gt 1$, the $n$th root of $k$ is either an integer or irrational.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One way to prove it is to use exactly the same idea as for proving the square root of $2$ is irrational: suppose $\sqrt[n]{k} =\frac{p}{q}$, with $p$ and $q$ integers, relatively prime. Then $q^nk = p^n$. Now think about the prime factorizations: every prime that divides $q$ must divide $p$, but $p$ and $q$ are relatively prime, so $q=1$. That means that you must have $k=p^n$ with $p$ an integer. That is, the only way for the $n$th root of $k$ to be a rational is if $k$ is an $n$th power of an integer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or you can use the Rational Root Test: an $n$th root of $k$ is a root of the polynomial $x^n - k$. But a rational root of a polynomial with integer coefficients that is written in lowest term $\frac{p}{q}$ must have denominator $q$ that divides the leading coefficient and numerator $q$ that divides the constant coefficient. So any rational root of $x^n-k$ must be an integer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Getting this back to your question, since $2$ is not an $n$th power of an integer for any $n\gt 0$, $\sqrt[n]{2}$ is not a rational for any integer $n\gt 0$. &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-25T20:00:33.820" LastActivityDate="2010-11-25T20:00:33.820" />
  <row Id="11876" PostTypeId="2" ParentId="11874" CreationDate="2010-11-25T19:51:54.963" Score="5" ViewCount="" Body="&lt;p&gt;$f : [a,b] \rightarrow \mathbb{R}$ is a function that is twice differentiable with each derivative continuous.  That is, $f&amp;#39;$ and $f&amp;#39;&amp;#39;$ both exist and are both continuous.&lt;/p&gt;&#xA;" OwnerUserId="3537" LastEditorUserId="3537" LastEditorDisplayName="" LastEditDate="2010-11-26T01:19:25.570" LastActivityDate="2010-11-26T01:19:25.570" CommentCount="5" />
  <row Id="11877" PostTypeId="2" ParentId="11871" CreationDate="2010-11-25T19:51:58.687" Score="5" ViewCount="" Body="&lt;p&gt;Hint: $\max \lbrace X_1,...,X_N \rbrace \leq x$ if and only if $X_i \leq x$ for all $i=1,...,N$.&lt;/p&gt;&#xA;" OwnerUserId="2810" LastActivityDate="2010-11-25T19:51:58.687" />
  <row Id="11878" PostTypeId="1" AcceptedAnswerId="11901" CreationDate="2010-11-25T19:53:03.613" Score="2" ViewCount="141" Body="&lt;p&gt;Dear MSE-community,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When we consider the Taylor Series expansion of $f(x)=b^x$ for some $b \in \mathbb{R}$, we see that $$b^x = 1 + \sum_{n=1}^{\infty}\frac{(\log(b))^n}{n!}x^n.$$ We can substitute $x$ for $b^x$ to find that $$ b^{b^{x}} = 1 + \sum_{n=1}^{\infty}\frac{(\log(b))^n}{n!}b^{xn}.$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, let's say that we want to find a function that accurately describes how we should raise $b$ to $b$'th power $x$ times. We say that $b^{(b)}_x$ says that $b$ should be raised to itself $x$ times, for some $x\in\mathbb{R}$. We can generalize the previous examples, and write $$b^{(b)}_x = 1 + \sum_{n=1}^{\infty}\frac{(\log(b))^n}{n!}b^{(b)n}_{x-1}.\qquad \text{(1)}$$ (Please note that $b^{(b)n}_{x}$ is &lt;em&gt;not&lt;/em&gt; equal to $b^{(bn)}_{x}$.) Using this equation we can also state that $$b^{(b)}_{x-1} = 1 + \sum_{n=1}^{\infty}\frac{(\log(b))^n}{n!}b^{(b)n}_{x-2}.\qquad\text{(2)}$$&#xD;&#xA;When we substitute (2) in (1), we find that $$ b^{(b)}_x = 1 + \sum_{n=1}^{\infty}\frac{(\log(b))^n}{n!}\Big( 1 + \sum_{n=1}^{\infty}\frac{(\log(b))^n}{n!}b^{(b)n}_{x-2} \Big).$$&#xD;&#xA;From now on, we write $ \sum_{n=1}^{\infty}\frac{(\log(b))^n}{n!} = A$, because then the next equations look more clear. &#xA;We could go on substituting in this manner $*$, until we arive at $b^{(b)}_{(x-(x-1))}=b$. We then see that the equality $$ b^{(b)}_x = 1 + \sum A \Big( 1 + \sum A \Big(\cdots \Big(1 + \sum_{n=1}^{\infty} \frac{(\log(b^b))^n}{n!}\Big)\Big)\Big) $$&#xD;&#xA;holds when we iterate $f(x) = 1 +  \sum_{n=1}^{\infty}\frac{(\log(b))^n}{n!}$ an $x-2$ amount of times, for real $x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question 1:&lt;/strong&gt; Is this possible? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question 2:&lt;/strong&gt; If so, how is this done? How do you describe the formula that precisely defines how $f(x)$ looks like after being iterated $x-2$ times? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question 3:&lt;/strong&gt; If such a formula is found, would it imply that a nice way of finding the fourth hyper operator (or &quot;&lt;a href=&quot;http://en.wikipedia.org/wiki/Tetration&quot; rel=&quot;nofollow&quot;&gt;tetration&lt;/a&gt;&quot;) is found for real numbers?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Max&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$*$EDIT: When we proceed in this manner when $x$ is not an integer, we will not find $x=1$. We should be able to iterate the function a real amount of times to find $x=1$.&lt;/p&gt;&#xA;" OwnerUserId="1055" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2010-11-26T06:26:46.490" LastActivityDate="2010-11-26T06:26:46.490" Title="Could real iterates of the Taylor Series expansion of $b^x$ help to find a way to define tetration?" Tags="&lt;real-analysis&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="11879" PostTypeId="2" ParentId="11874" CreationDate="2010-11-25T19:55:45.337" Score="3" ViewCount="" Body="&lt;p&gt;$f:[a,b]\to \mathbb{R}$ or $\to \mathbb{C}$. And both the first  and second derivatives of $f$ are continuous.&lt;/p&gt;&#xA;" OwnerUserId="3249" LastActivityDate="2010-11-25T19:55:45.337" />
  <row Id="11880" PostTypeId="2" ParentId="8040" CreationDate="2010-11-25T20:03:39.290" Score="0" ViewCount="" Body="&lt;p&gt;It works because the Leibniz's notation for derivatives dx/dt means two things that are both true:&#xA; (1) a name for the derivative x'(t);&#xA; (2) the fraction or quotient of differentials: numerator dx, denominator dt (arbitrary number); so dx=x'(t)dt.&lt;/p&gt;&#xA;" LastActivityDate="2010-11-25T20:03:39.290" />
  <row Id="11881" PostTypeId="2" ParentId="11669" CreationDate="2010-11-25T20:12:28.107" Score="1" ViewCount="" Body="&lt;p&gt;If you really want to know all about it then you should read 'On the sensation of tone' by Helmholz.&#xA;&lt;a href=&quot;http://store.doverpublications.com/0486607534.html&quot; rel=&quot;nofollow&quot;&gt;http://store.doverpublications.com/0486607534.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4045" LastActivityDate="2010-11-25T20:12:28.107" CommentCount="2" />
  <row Id="11882" PostTypeId="1" AcceptedAnswerId="11884" CreationDate="2010-11-25T20:12:29.370" Score="3" ViewCount="233" Body="&lt;p&gt;Prove that $\lim\limits_{x \to 2} \frac{x^{2}-2x+9}{x+1}$ using an epsilon delta proof.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I have most of the work done.  I choose $\delta = min{\frac{1}{2}, y}$,&lt;br&gt;&#xA;$f(x)$ factors out to $\frac{|x-3||x-2|}{|x+1|}$  &#xA;But $|x-3| \lt \frac{3}{2}$ for $\delta = \frac{1}{2}$ and also $|x+1| &amp;gt; 5/2$ (I'll spare you the details).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure how to choose my y here. If I take $\lim\limits_{x \to 2} \frac{x^{2}-2x+9}{x+1}$ &amp;lt; $(3/5) \delta$   How do I choose my epsilon here (replace y with this) to satisfy this properly?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="2533" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-25T20:13:57.857" LastActivityDate="2010-11-26T04:57:02.610" Title="$\epsilon$-$\delta$ limit proof, $\lim_{x \to 2} \frac{x^{2}-2x+9}{x+1}$" Tags="&lt;calculus&gt;&lt;analysis&gt;&lt;real-analysis&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="11884" PostTypeId="2" ParentId="11882" CreationDate="2010-11-25T20:26:34.630" Score="10" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;First:&lt;/strong&gt; You &lt;strong&gt;don't&lt;/strong&gt; choose $\epsilon$! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In an $\epsilon$-$\delta$ proof, you begin by taking an &lt;em&gt;arbitrary&lt;/em&gt; $\epsilon$; you then need to show that it is possible to choose a $\delta$ that &quot;works&quot; for &lt;em&gt;that&lt;/em&gt; $\epsilon$. So you are going at things &lt;em&gt;exactly&lt;/em&gt; backwards, by first trying to work with $\delta$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Second:&lt;/strong&gt; You cannot &quot;prove $\lim\limits_{x\to 2}\frac{x^2-2x+9}{x+1}$&quot; using $\epsilon$-$\delta$ proofs. What you &lt;em&gt;can&lt;/em&gt; try to prove is that the limit is equal to &lt;em&gt;something&lt;/em&gt; (in this case, to $3$). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Third:&lt;/strong&gt; $f(x)$ does not factor as $\frac{|x-2||x-3|}{|x+1|}$; what factors that way is $|f(x)-3|$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So. Let's start from scratch, shall we?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An $\epsilon$-$\delta$ proof that $\lim\limits_{x\to 2}\frac{x^2-2x+9}{x+1}=3$ would work as follows: given some arbitrary $\epsilon\gt 0$, we must find a $\delta\gt 0$ (which may depend on $\epsilon$), with the property that if $0\lt |x-2|\lt \delta$, then we will have $\left|\frac{x^2-2x+9}{x+1} - 3\right|\lt \epsilon$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, let $\epsilon\gt 0$ be given. You want to make sure that&#xA;$$\left|\frac{x^2-2x+9}{x+1} - 3\right| = \left|\frac{x^2-2x+9-3x-3}{x+1}\right| = \left|\frac{x^2-5x+6}{x+1}\right| = \left|\frac{(x-3)(x-2)}{x+1}\right|$$&#xD;&#xA;is small, by making sure that $|x-2|$ is small (that is, that $0\lt |x-2|\lt\delta$ for some chosen $\delta$). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clearly, you also want to make sure that $\delta\lt \frac{|x+1|}{|x-3|}\epsilon$, because then we will have:&#xA;$$\left|\frac{x^2-x+0}{x+1}-3\right| = \frac{|x-3||x-2|}{|x+1|} \lt \delta\frac{|x-3|}{|x+1|} \lt \frac{|x+1|\epsilon}{|x-3|}\frac{|x-3|}{|x+1|} = \epsilon.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can make sure that $\frac{1}{|x+1|}$ is no larger than some constant by making sure that $x$ is close enough to $2$. You can do the same thing for $|x-3|$. So &lt;strong&gt;then&lt;/strong&gt; you can pick a $\delta$ that is simultaneously small enough to ensure that $\frac{1}{|x+1|}$ is smaller than some $C$, that $|x-3|$ is smaller than some $D$, &lt;em&gt;and&lt;/em&gt; that $|x-2|$ is smaller than $CD\epsilon$. Try that.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-26T04:57:02.610" LastActivityDate="2010-11-26T04:57:02.610" CommentCount="5" />
  <row Id="11885" PostTypeId="2" ParentId="11872" CreationDate="2010-11-25T20:26:36.580" Score="2" ViewCount="" Body="&lt;p&gt;If $\rm\ \sqrt[n] c = a/b,\ \ gcd(a,b) = 1\ $ then $\rm\ c\ b^n = a^n\ \Rightarrow \ b\:|\: a^n\:.\: $ But $\rm\ gcd(a,b) = 1\ \Rightarrow\ gcd(a^n,b) = 1\ $ by &lt;a href=&quot;http://en.wikipedia.org/wiki/Euclid%27s_lemma&quot; rel=&quot;nofollow&quot;&gt;Euclid's Lemma&lt;/a&gt;. Thus $\rm b = 1\ $ hence $\rm\:\ c \  =\ a^n\:.\ $ In particular $\rm\ c = 2\ \Rightarrow\ n = 1\:.$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are many possible variations on such irrationality proofs, e.g. using the &lt;a href=&quot;http://en.wikipedia.org/wiki/Rational_root_theorem&quot; rel=&quot;nofollow&quot;&gt;Rational Root Test&lt;/a&gt; or directly using Unique Factorization of integers, or using the principality of &lt;a href=&quot;http://math.stackexchange.com/questions/4713/if-the-order-divides-a-prime-p-then-the-order-is-p-or-1/4715#4715&quot;&gt;denominator ideals&lt;/a&gt;. Perhaps the most elegant is to employ Dedekind's notion of the &lt;a href=&quot;http://math.stackexchange.com/questions/2883/show-that-3p2q2-implies-3p-and-3q/2941#2941&quot;&gt;&lt;strong&gt;conductor  ideal&lt;/strong&gt;&lt;/a&gt; - which yields a one-line proof that a  $\rm PID$ is integrally closed, i.e. satisfies the monic case of the rational root test.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-11-25T20:42:04.907" LastActivityDate="2010-11-25T20:42:04.907" />
  <row Id="11886" PostTypeId="2" ParentId="11669" CreationDate="2010-11-25T20:36:27.457" Score="5" ViewCount="" Body="&lt;p&gt;To add to this thread, you can understand why certain notes sound good/bad together by looking at trig sum/product formulas, e.g.:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;cos(a) + cos(b) = 2 * cos(a - b) * cos(a + b)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What this means is that when you add two tones/frequencies 'a' and 'b', it is equivalent to taking one wave of frequency 'a + b' and modulating its amplitude with another of frequency 'a - b'. The frequency 'a + b' will be a faster vibration, and the frequency 'a - b' will be a slower vibration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the two original frequencies are close (e.g. A = 440Hz and A# = 466Hz), the 'a - b' component will be heard as an unpleasant low frequency beating (here, 26Hz).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the two original frequencies are integer ratios of each other (e.g. 3/2, 4/3) as in chords, then the resulting 'a + b' and 'a - b' frequencies will also be integer ratios of each other. The resulting wave will be simple and sounds harmonious. This is why integer ratios of notes are so important in music.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It helps to plot sums of sines graphically to see this in action.&lt;/p&gt;&#xA;" OwnerUserId="4048" LastActivityDate="2010-11-25T20:36:27.457" CommentCount="1" />
  <row Id="11887" PostTypeId="2" ParentId="11882" CreationDate="2010-11-25T20:36:47.203" Score="0" ViewCount="" Body="&lt;p&gt;I'm going to go out on a limb and guess that you're trying to show the limit is 3 and that $f(x) = {x^2 - 2x + 9 \over x + 1} - 3$. I suggest trying to translate what you've done into the fact that  $|{x^2 - 2x + 9 \over x + 1} - 3| &amp;lt; {3 \over 5}|x - 2|$ whenever $|x - 2| &amp;lt; {1 \over 2}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means that if you choose any $\epsilon &amp;lt; {1 \over 2}$, then you have that $|{x^2 - 2x + 9 \over x + 1} - 3| &amp;lt; {3 \over 5}\epsilon$ whenever $|x - 2| &amp;lt; \epsilon$. So, given $\epsilon$, the natural choice for $\delta$ is ${3 \over 5}\epsilon$. (you got the $\delta$ and $\epsilon$ reversed.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now verify that the &quot;for every $\epsilon$ there is a $\delta$&quot; definition is satisfied in this way. &lt;/p&gt;&#xA;" OwnerUserId="3035" LastActivityDate="2010-11-25T20:36:47.203" />
  <row Id="11888" PostTypeId="2" ParentId="11764" CreationDate="2010-11-25T20:42:07.207" Score="1" ViewCount="" Body="&lt;p&gt;See Glen  Bredon's book &lt;em&gt;Sheaf Theory&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2010-11-25T20:42:07.207" />
  <row Id="11889" PostTypeId="2" ParentId="11836" CreationDate="2010-11-25T21:03:45.707" Score="3" ViewCount="" Body="&lt;p&gt;The topology of $R^n$ is much more complicated than the topology of $R$. For example, a simply connected subset of $R$ is just an interval. A simply connected subset in $R^n$ can be very complicated.  This matters because simple connectivity enters as a hypothesis in several theorems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The boundary of an interval in $R$ is simply a pair of points (or 1 or 0 points if the interval is unbounded).  But the boundary of an open set in $R^n$ can be a manifold or something more complicated.  Since the fundamental theorem of calculus requires integrating over a boundary, the theorem is more complicated in several variables. Integration over the boundary of an interval in $R$ is simply evaluating a function at two points.  Integration over a manifold is more subtle and requires a large amount of machinery to do formally.&lt;/p&gt;&#xA;" OwnerUserId="247" LastActivityDate="2010-11-25T21:03:45.707" />
  <row Id="11890" PostTypeId="2" ParentId="11669" CreationDate="2010-11-25T21:25:24.103" Score="4" ViewCount="" Body="&lt;p&gt;The &quot;circle of fifths&quot; is a by product of the preference for diatonic scales.&#xA;If you layout the chromatic (12-tone) scale without the white and raised black arrangement you'd use the same logic to describe a &quot;circle of sevenths&quot; (counting&#xA;upo semitones from C to G).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the arrangement makes solid sense when applied to the human hand. We need to be able to span interesting distances with the &quot;octave&quot; interval be very useful for most pianists&#xA;as a basic required for anyone older that 10-12. Some pianists can span 10ths with relative ease but they are in the minority. The piano music of Rachmaninoff is riddled with these massive but musically sonorus intervals. The are the major third expanded to&#xA;the pure natural interval (10 keys apart) of the &quot;bugle&quot; overtone series.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can reach the 10th's on the white to white instances but the black to white (Bb to D for example) are beyond me. And doing them quickly and accurately is the mark of true mastery of the instrument... it's like being able to dunk: genetics help and no amount of effort can help a small handed pianist.&lt;/p&gt;&#xA;" OwnerUserId="4040" LastActivityDate="2010-11-25T21:25:24.103" />
  <row Id="11892" PostTypeId="1" CreationDate="2010-11-25T21:55:07.347" Score="3" ViewCount="134" Body="&lt;p&gt;Suppose that $f:U\subset\mathbb{C}\to\mathbb{C}$, where $U$ is a region in the complex plane, is a holomorphic function. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $c\in\mathbb{R}$ is a regular value for $\text{Re}(f(z))$ then it follows from implicit function theorem that &#xA;$\text{Re}(f(z))^{-1}(c)$ is at least locally a differentiable curve in the plane. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt; Question: &lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1- If $c$ is a regular value is any connected component of $\text{Re}(f(z))^{-1}(c)$ a global differentiable curve ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2-If $c$ is not a regular value and $\text{Re}(f(z))^{-1}(c)$ have at least one cluster point is this set locally a curve ?&lt;/p&gt;&#xA;" OwnerUserId="633" LastActivityDate="2010-11-28T06:50:18.003" Title="What Re(f(z))=c is if f is a holomorphic function ?" Tags="&lt;complex-analysis&gt;&lt;differential-geometry&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="0" />
  <row Id="11893" PostTypeId="2" ParentId="11864" CreationDate="2010-11-25T22:14:50.810" Score="1" ViewCount="" Body="&lt;p&gt;You just need some simple properties of the Banach space-valued integral, namely linearity (the integral commutes with continuous linear operators) substitution for translations and the fundamental theorem of calculus, which says: $\frac{1}{h}\int_0 ^h f(t) d t \to f(0)$ (in the Banach space) for $h \to 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So after writing out all definitions (and using linearity and the semigroup properties), your job is to show that for any $t&amp;gt;0$ we have $$\frac{1}{h}\int_0 ^t (S(\tau+h) x_0 - S(\tau)x_0) d \tau \to S(t)x_0 - x_0$$ for $h\to 0$ from above. We can assume $t&amp;gt;h$. Then split the first integral into $\int_0 ^{t-h}+\int_{t-h} ^t$ and the second one into $\int_0 ^h+ \int_h ^t$. By translation the integrals $\int_0 ^{t-h}$ and $\int_h ^t$ cancel, and you just have to apply the fundamental theorem of calculus.&lt;/p&gt;&#xA;" OwnerUserId="1609" LastActivityDate="2010-11-25T22:14:50.810" CommentCount="6" />
  <row Id="11894" PostTypeId="2" ParentId="4170" CreationDate="2010-11-25T22:44:52.867" Score="0" ViewCount="" Body="&lt;p&gt;For my work in this area, I refer to :&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Richard Epstein &quot;Classical Mathematical Logic&quot;&lt;/li&gt;&#xA;&lt;li&gt;Wolfgang Rautenberg &quot;A Concise Introduction to Mathematical Logic&quot;&lt;/li&gt;&#xA;&lt;li&gt;Jon Barwise &quot;Handbook of Mathematical Logic&quot;&lt;/li&gt;&#xA;&lt;li&gt;Jean Heijenoort &quot;From Frege to Godel&quot;&lt;/li&gt;&#xA;&lt;li&gt;We Li &quot;Mathematical Logic&quot;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Rautenberg has a lot of examples, exercise, but is very heavy going (at least for me).  Epstein is fairly recent and very well laid out. Barwise is the most comprehensive for when you need to deep dive.&lt;/p&gt;&#xA;" OwnerUserId="4057" LastActivityDate="2010-11-25T22:44:52.867" CommunityOwnedDate="2010-11-25T22:44:52.867" />
  <row Id="11895" PostTypeId="2" ParentId="4170" CreationDate="2010-11-25T23:03:32.873" Score="0" ViewCount="" Body="&lt;p&gt;Yuri Manin, A course in mathematical logic.&lt;/p&gt;&#xA;" OwnerUserId="1119" LastActivityDate="2010-11-25T23:03:32.873" CommunityOwnedDate="2010-11-25T23:03:32.873" />
  <row Id="11897" PostTypeId="1" CreationDate="2010-11-25T23:07:42.280" Score="21" ViewCount="931" Body="&lt;p&gt;Many years ago a lecturer of mine had a photocopy of a page from a book containing a math problem ( I think it was a simple quadradic equation ) that was stated/solved in Cuneiform, Arabic, Latin scripts and Finally in modern math notation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have contacted my lecturer but he has no idea where it was from, nor I have been able to find it using google books searches etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anyone know where to find it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you in advance&lt;/p&gt;&#xA;" OwnerUserId="3936" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-25T23:11:59.090" LastActivityDate="2011-04-18T08:16:55.700" Title="Image of a math problem that was stated in Cuneiform, Arabic, Latin and Finally in modern math notation" Tags="&lt;reference-request&gt;&lt;math-history&gt;" AnswerCount="2" CommentCount="6" FavoriteCount="4" />
  <row Id="11898" PostTypeId="2" ParentId="11784" CreationDate="2010-11-25T23:14:23.170" Score="0" ViewCount="" Body="&lt;p&gt;An even easier example than Timothy's  is $f(x,y)=1$ when $x=0$ or $y=0$, and $f(x,y)=0$ for all other points. Then $f_x(0,0)=0, f_y(0,0)=0$ but $f$ is not differentiable (or even continuous at the origin).&lt;/p&gt;&#xA;" OwnerUserId="3249" LastActivityDate="2010-11-25T23:14:23.170" />
  <row Id="11899" PostTypeId="1" AcceptedAnswerId="11913" CreationDate="2010-11-25T23:17:48.850" Score="5" ViewCount="231" Body="&lt;p&gt;Hello alltogether&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At university, we are currently introduced in various methods of proving that a series converges. For example the ComparisonTest, the RatioTest or the RootTest. However we aren't told of how to calculate the point of convergence. The only series of which we know the point of convergence is the geometric series. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe this to be a non trivial task in general, but there have to be at least some methods to calculate the point of convergence for simple series, haven't there?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you could point me in the direction of some of these methods I would very much appreciate it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance&lt;/p&gt;&#xA;" OwnerUserId="3188" LastEditorUserId="3188" LastEditorDisplayName="" LastEditDate="2010-11-25T23:26:47.783" LastActivityDate="2010-11-26T19:08:24.190" Title="Convergence of Series" Tags="&lt;analysis&gt;&lt;sequences-and-series&gt;&lt;convergence&gt;" AnswerCount="4" CommentCount="6" FavoriteCount="1" />
  <row Id="11900" PostTypeId="2" ParentId="11865" CreationDate="2010-11-25T23:27:02.570" Score="5" ViewCount="" Body="&lt;p&gt;There's a brief note in &lt;a href=&quot;http://books.google.com/books?id=LLEZQC74gVcC&amp;amp;pg=PA157&quot; rel=&quot;nofollow&quot;&gt;this book&lt;/a&gt; on how Khayyam bumped into having to solve a cubic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll only make the note that you should remember the context of the time: there was no concept of negative, much less complex, solutions. Corresponding to our current Cartesian system, Khayyam only looked at intersections in the first quadrant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another note should be made that the curves of the time were constructed with geometric tools (straightedge, compass, and a bunch of other contraptions), and $y=x^3$ isn't really a sort of curve that easily lends itself to such a construction (but is now easily constructed thanks to our current knowledge of coordinate geometry).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://books.google.com/books?id=EIdtVPeD7GcC&amp;amp;pg=PA107&quot; rel=&quot;nofollow&quot;&gt;Here&lt;/a&gt; is a more explicit mention of the hyperbola-circle intersection problem Khayyam studied and was mentioned in the OP.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://books.google.com/books?id=ucQXDt2WRJgC&amp;amp;pg=PA89&quot; rel=&quot;nofollow&quot;&gt;Here&lt;/a&gt; is a (more or less) complete table of all the intersection cases Khayyam studied. (The book has an appendix containing a (translated) section of Khayyam's work.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://books.google.com/books?id=V_B5SIGTnnoC&amp;amp;pg=PA92&quot; rel=&quot;nofollow&quot;&gt;Here&lt;/a&gt; is yet another reference.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I'll keep updating this answer as I comb through more books; watch this space! As an aside, it's funny that my attempts to look for answers to &lt;a href=&quot;http://math.stackexchange.com/questions/11897&quot;&gt;this question&lt;/a&gt; are leading me to references for this question!)&lt;/p&gt;&#xA;" OwnerUserId="498" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-26T03:58:11.750" LastActivityDate="2010-11-26T03:58:11.750" CommentCount="2" />
  <row Id="11901" PostTypeId="2" ParentId="11878" CreationDate="2010-11-25T23:31:46.023" Score="2" ViewCount="" Body="&lt;p&gt;The process you describe can be better expressed in terms of the Bell-/Carleman-matrix (see &lt;a href=&quot;http://en.wikipedia.org/wiki/Carleman_matrix&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt; )  associated to function $f(x) = b^x$ . I got myself used to the following notation:&lt;br&gt;&#xA;let $V(x)$ denote a rowvector of infinite dimension of consecutive powers of an argument x $V(x)=[1,x,x^2,x^3, \dots]$ then let &lt;em&gt;B&lt;/em&gt; denote the infinite matrix which performs $ V(x) * B = V(b^x) $ . The columns of &lt;em&gt;B&lt;/em&gt; contain the coefficients for the powerseries for the consecutive powers of $(b^x)^0, (b^x)^1 , (b^x)^2, \ldots $. &lt;em&gt;B&lt;/em&gt; is then the (transposed) Carleman-matrix.    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What your formula and iteration constructs can simply be expressed by the notion of powers of &lt;em&gt;B&lt;/em&gt;; the h-fold nested infinite sums (your &lt;em&gt;A&lt;/em&gt;) are captured by the formal matrixproducts $ V(x)*B*B*\ldots*B = V(x)*B^h = V(b^(b^(b^\ldots (b^x)))  $ . Unfortunately the fractional part of iteration must then be expressed by a fractional power of &lt;em&gt;B&lt;/em&gt; --- which is not trivial.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But that is only one problem. We have already the problem of powers of &lt;em&gt;B&lt;/em&gt; (or nested summation of your example): the convergence using matrices of finite size gets lost after few iterations and it is a special art to find closed forms of arbitrarily precise computable expressions for the powerseries only for third or fourth iteration. I've tried this different ways and could not get significant improvement of the convergence behaviour as long as I used that nesting which you describe above for more than some exotic bases b near 1.    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But there are ways to express the coefficients of the powerseries in finite expressions/sums including &lt;em&gt;exp/log&lt;/em&gt; which we can assume to be available always in arbitrary precision. This can be found by triangular decomposition of &lt;em&gt;B&lt;/em&gt; and a modified way to arrive at the required powers. (See for a not very explanative technical description &lt;a href=&quot;http://go.helms-net.de/math/tetdocs/TTetrationExactEntries_short.htm&quot; rel=&quot;nofollow&quot;&gt;exact entries&lt;/a&gt;, remark: there might be better descriptions...)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But all this covers only the part of integer height (integer number of iterations). And for this we do not really need that formal powerseries/matrixpowers approximated by finitely truncated matrices: we have precise exponentiation for each base at hand. The crux is the part of the fractional height/iteration count. Here there are some different approaches available; one of it is the approach to compute fractional powers of &lt;em&gt;B&lt;/em&gt; by diagonalization to get the coefficients for a formal powerseries which represents the fractional iteration of $b^x$ However - to have series with real coefficients we must restrict ourselves to bases b between $exp(-exp(1))\ldots exp(exp(-1)) $ &lt;/p&gt;&#xA;" OwnerUserId="1714" LastEditorUserId="1714" LastEditorDisplayName="" LastEditDate="2010-11-25T23:40:34.727" LastActivityDate="2010-11-25T23:40:34.727" CommentCount="1" />
  <row Id="11902" PostTypeId="2" ParentId="11899" CreationDate="2010-11-25T23:38:27.723" Score="1" ViewCount="" Body="&lt;p&gt;&quot;there have to be at least some methods to calculate the point of convergence for simple series, haven't there?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are sure a given sequence of numbers converges, you have to just keep iterating everything manually, unless you can think of a clever trick.  I suggest you study how Euler proved that $\zeta(2)=\frac{\pi^2}{6}$, that will be very instructive for you because it will demonstrate how clever you have to sometimes be in order to solve this type of problem.&lt;/p&gt;&#xA;" OwnerUserId="1417" LastActivityDate="2010-11-25T23:38:27.723" CommentCount="1" />
  <row Id="11903" PostTypeId="1" CreationDate="2010-11-25T23:42:05.523" Score="2" ViewCount="164" Body="&lt;p&gt;In &lt;a href=&quot;http://math.stackexchange.com/questions/11849/w1w2-wna-w1a-wna-for-n-integer-2-w0-and-aconstant-real-0&quot;&gt;a previous post&lt;/a&gt;, the following inequality has been proven&#xA;$${\left( {\sum\limits_{i = 1}^n {{W_i}} } \right)^a} \le \sum\limits_{i = 1}^n {{W_i}^a}$$&#xD;&#xA;where $W_i\gt 0$, $0\lt a\lt 1$.&#xA;I guess it is more correct to say that this is always greater, and it is valid for $a\gt 0$ not just $0\lt a \lt 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am trying to see if one can generalize it to something like&#xA;$$f\left( \sum\limits_{i = 1}^n W_i  \right) \le \sum\limits_{i = 1}^n {f({W_i})}$$ &#xA;where ${W_i}\gt 0$ ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Under what circumstances and functions can this be true ? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It has been proven for the power function $f(x)=kx^a$ where $a,k\gt 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there any other cases? Do you think there is any basic inequality to prove this?&lt;/p&gt;&#xA;" OwnerUserId="4011" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-25T23:59:29.310" LastActivityDate="2010-11-26T11:41:21.750" Title="$f\left(\sum X_i\right) \leq \sum f(X_i)$, where $X_i\gt 0$; for what functions is this true ?" Tags="&lt;functions&gt;&lt;inequality&gt;" AnswerCount="2" CommentCount="0" />
  <row Id="11905" PostTypeId="2" ParentId="11899" CreationDate="2010-11-25T23:52:20.970" Score="3" ViewCount="" Body="&lt;p&gt;Hi!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Right after the geometric series, the easiest method I know of to calculate the sum of a series is the property of derivation term-by-term of convergent power series. I don't know if you've already covered it in the lessons, so I'll try to give you an idea of how it works:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You know that $\sum x^n=\frac{1}{1-x}$ if $x&amp;lt;1$; now, you will see that you can derive a series like the one before with respect to $x$, and you can do it term-by-term so that you obtain (by differentiating both sides of the previous identity): $\frac{d}{dx}\sum x^n=\sum \frac{d}{dx}x^n=\sum nx^{n-1}=\frac{1}{(1-x)^2}$.&lt;/p&gt;&#xA;" OwnerUserId="4054" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-26T00:04:00.927" LastActivityDate="2010-11-26T00:04:00.927" CommentCount="1" />
  <row Id="11906" PostTypeId="2" ParentId="11903" CreationDate="2010-11-26T00:00:08.793" Score="2" ViewCount="" Body="&lt;p&gt;Many inequalities similar to this one come from &lt;a href=&quot;http://en.wikipedia.org/wiki/Jensen%27s_inequality&quot; rel=&quot;nofollow&quot;&gt;Jensen's Inequality&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="4054" LastEditorUserId="4054" LastEditorDisplayName="" LastEditDate="2010-11-26T11:41:21.750" LastActivityDate="2010-11-26T11:41:21.750" CommentCount="5" />
  <row Id="11907" PostTypeId="2" ParentId="11899" CreationDate="2010-11-26T00:02:26.013" Score="5" ViewCount="" Body="&lt;p&gt;There are some convergent series whose exact sum can be evaluated. For instance:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle\sum_{n=1}^{\infty}\dfrac{n}{(n+1)!}.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Observing that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\dfrac{n}{(n+1)!}=\dfrac{(n+1)-1}{(n+1)!}=\dfrac{1}{n!}-\dfrac{1}{(n+1)!}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the series &lt;a href=&quot;http://en.wikipedia.org/wiki/Telescoping_series&quot; rel=&quot;nofollow&quot;&gt;telescopes&lt;/a&gt;. Therefore,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle\sum_{n=1}^{\infty}\dfrac{n}{(n+1)!}=1-\underset{n\rightarrow\infty}{\lim}\dfrac{1}{(n+1)!}=1-0=1.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But in general it is a difficult task to find the exact sum, as you said.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Added: Unlike $\displaystyle\sum_{n=1}^{\infty }\frac{1}{n^{2}}$ whose sum equals $\dfrac{\pi ^{2}}{6}$, the series $\displaystyle\sum_{n=1}^{\infty }\dfrac{1}{n^{3}}$, although convergent, nobody knows a closed form in terms of other mathematical constants. This sum is therefore a mathematical constant in itself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Added 2: Another example that uses integration and differentiation of the geometric series $$\sum_{n=1}^{\infty }x^{n}=\frac{x}{1-x},\qquad\left\vert x\right\vert &amp;lt;1$$  is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{n=1}^{\infty }\frac{n}{(n+1)e^{n}}=e\left( \log \frac{e-1}{e}+\frac{1%&#xD;&#xA;}{e-1}\right) .$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Added 3: From&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{n=1}^{\infty }x^{n}=\frac{x}{1-x}\qquad\left\vert x\right\vert &amp;lt;1,$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;we get for $\left\vert x\right\vert &amp;lt;1$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int \sum_{n=1}^{\infty }x^{n}dx=\sum_{n=1}^{\infty }\frac{1}{n+1}%&#xD;&#xA;x^{n+1}=\int \frac{x}{1-x}dx=-x-\log \left\vert 1-x\right\vert .$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{n=1}^{\infty }\frac{1}{n+1}x^{n}=-1-\frac{1}{x}\log \left\vert&#xD;&#xA;1-x\right\vert $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now if we diferentiate, we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{n=1}^{\infty }\frac{n}{n+1}x^{n-1}=\frac{1}{x^{2}}\log \left\vert&#xD;&#xA;1-x\right\vert -\frac{1}{x^{2}-x},$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or equivalently&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{n=1}^{\infty }\frac{n}{n+1}x^{n}=\frac{1}{x}\log \left\vert&#xD;&#xA;1-x\right\vert -\frac{1}{x-1}\qquad\left\vert x\right\vert &amp;lt;1.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally for $x=e^{-1}$, we obtain&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{n=1}^{\infty }\frac{n}{(n+1)e^{n}}=e\left( \log \frac{e-1}{e}+\frac{1%&#xD;&#xA;}{e-1}\right) .$$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="752" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2010-11-26T19:08:24.190" LastActivityDate="2010-11-26T19:08:24.190" CommentCount="3" />
  <row Id="11908" PostTypeId="1" AcceptedAnswerId="11910" CreationDate="2010-11-26T00:07:50.230" Score="2" ViewCount="329" Body="&lt;p&gt;I am working through an economics &lt;a href=&quot;http://faculty.wcas.northwestern.edu/~lchrist/papers/aeragency.pdf&quot; rel=&quot;nofollow&quot;&gt;paper&lt;/a&gt; and I need to take the derivative of the following function:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$h\left(\overline{\omega}\right) = \int^{\infty}_{\overline{\omega}} \omega \Phi \left(d\omega\right)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even though I don't understand it well, I can do the derivative for the case&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$g\left(\overline{\omega}\right) = \int^{\overline{\omega}}_{0} \omega \Phi \left(d\omega\right)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where the derivative is simply&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$g&amp;#39;\left(\overline{\omega}\right) = \overline{\omega} \phi \left(\overline{\omega}\right)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But for $h\left(\overline{\omega}\right)$ where the upper bound is $\infty$ I really have no idea of what to do. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone help me? Any explanation or even a pointer to where I can learn those things would be greatly appreciated.&lt;/p&gt;&#xA;" OwnerUserId="85" LastActivityDate="2010-11-26T00:27:34.907" Title="Derivative of an expectation (using integrals)" Tags="&lt;calculus&gt;&lt;integral&gt;&lt;probability&gt;" AnswerCount="3" CommentCount="3" />
  <row Id="11909" PostTypeId="2" ParentId="11908" CreationDate="2010-11-26T00:17:09.643" Score="3" ViewCount="" Body="&lt;p&gt;Since $$h(\overline{\omega})=-\int_{\infty}^{\overline{\omega}}\omega\Phi(d\omega),$$&#xD;&#xA;$h&amp;#39;(\overline{\omega})=-\overline{\omega}\phi(\overline{\omega})$.&lt;/p&gt;&#xA;" OwnerUserId="3249" LastActivityDate="2010-11-26T00:17:09.643" CommentCount="1" />
  <row Id="11910" PostTypeId="2" ParentId="11908" CreationDate="2010-11-26T00:17:43.187" Score="4" ViewCount="" Body="&lt;p&gt;If the integral is sufficiently nice by which I mean it doesn't blow up to infinity, we can write $h(\bar{\omega}) = \int_{l}^{\infty} \omega d(\Phi(w)) - \int_{l}^{\bar{\omega}} \omega d(\Phi(w))$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $l$ is some constant number. It could be $0$ or $\infty$ based on your problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the first integral is just a constant, and the second integral is similar to $g(\bar{\omega})$&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2010-11-26T00:27:34.907" LastActivityDate="2010-11-26T00:27:34.907" CommentCount="10" />
  <row Id="11911" PostTypeId="2" ParentId="11908" CreationDate="2010-11-26T00:18:29.647" Score="4" ViewCount="" Body="&lt;p&gt;Note that $h(\bar \omega ) + g(\bar \omega )$ is constant, and use the result for $g(\bar \omega )$.&lt;/p&gt;&#xA;" OwnerUserId="2810" LastActivityDate="2010-11-26T00:18:29.647" />
  <row Id="11912" PostTypeId="2" ParentId="11903" CreationDate="2010-11-26T00:27:33.893" Score="3" ViewCount="" Body="&lt;p&gt;The condition $$f\left( {\sum\limits_{i = 1}^n {{W_i}} } \right) \le \sum\limits_{i = 1}^n {f({W_i})} $$ where ${W_i}$&gt;0 is equivalent to $$f\left( {\sum\limits_{i = 1}^2 {{W_i}} } \right) \le \sum\limits_{i = 1}^2 {f({W_i})} $$ where ${W_i}$&gt;0. These are subadditive functions. See &lt;a href=&quot;http://en.wikipedia.org/wiki/Subadditivity&quot; rel=&quot;nofollow&quot;&gt; Subadditivity&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3249" LastActivityDate="2010-11-26T00:27:33.893" CommentCount="1" />
  <row Id="11913" PostTypeId="2" ParentId="11899" CreationDate="2010-11-26T01:03:30.250" Score="6" ViewCount="" Body="&lt;p&gt;The short answer is that there are no general methods and we use whatever's available.  The first thing you can do is learn the &lt;a href=&quot;http://en.wikipedia.org/wiki/Taylor_series&quot; rel=&quot;nofollow&quot;&gt;Taylor series&lt;/a&gt; of common functions; then evaluation of the Taylor series at a point where the series converges to the function corresponds to evaluation of the function.  For example, $e^x = \sum \frac{x^n}{n!}$, hence $e = \sum \frac{1}{n!}$.  Less trivially, $\ln (1 + x) = \sum \frac{(-1)^{n-1} x^n}{n}$, hence&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ln(2) = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} \pm ....$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The more Taylor series you know, the more series you can evaluate &quot;explicitly&quot; in some sense.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Beyond that, life gets difficult.  Mastering the use of Taylor series is already highly nontrivial - especially recognizing when the method is applicable - so worry about learning how to do that properly first.  &lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-11-26T01:03:30.250" CommentCount="2" />
  <row Id="11916" PostTypeId="1" AcceptedAnswerId="11919" CreationDate="2010-11-26T02:45:05.160" Score="1" ViewCount="119" Body="&lt;p&gt;Let $$p_n(x)e^{-x^2}$$ be the $n$th derivative of $$e^{-x^2}.$$&#xD;&#xA;Find a formula for $p_n(x)$. We have $p_1(x)=-2x, p_2(x)=4x^2-2$, etc. But what is the general formula for $p_n$?&lt;/p&gt;&#xA;" OwnerUserId="3249" LastActivityDate="2010-11-26T02:57:47.390" Title="Higher derivatives of an exponential function" Tags="&lt;calculus&gt;&lt;analysis&gt;&lt;classical-analysis&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="11917" PostTypeId="2" ParentId="11669" CreationDate="2010-11-26T02:48:20.080" Score="13" ViewCount="" Body="&lt;p&gt;The math in this thread is awesome, but I'm not sure it addresses the original question about the &quot;difference between white and black notes&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other responses in this thread provide enough math to understand that each octave can be more-or-less naturally divided into twelve semitones. The Western music tradition further evolved to be based around what's called the &quot;diatonic scale&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A musical scale is a sequence of pitches within one octave; scales can be defined by the number of semitones between each successive note.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, the Whole Tone Scale consists entirely of whole tones; it has six distinct pitches, each of which is two semitones higher than the last. So you might represent it with the string '222222' — that is, take a note, then the note 2 semitones higher, then the note 2 semitones higher, etc., until the last &quot;2&quot; takes you to the note an octave above where you started.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Diatonic Scale that Western music is based around could likewise be represented by the string '2212221'.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you start with a C on a keyboard and go up, you'll see that the white keys conform to that pattern of semitones. That, generally, is why the black keys are in that particular pattern.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, you can start a scale on any pitch, not just C. That's why the &quot;same&quot; diatonic scale in a different key will involve a unique set of sharps and flats.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the Diatonic Scale can also be represented by '2212221' shifted to the left or right any number of times. For example, '2122212', '1222122', etc. are also Diatonic; these are called the &quot;modes&quot; of the Diatonic scale. Each Diatonic mode can be played on only the white keys of the piano by starting on a different pitch.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2212221 is called the Ionian mode (this is also generically called the Major scale), and can be played on the white keys starting with C.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2122212 is the Dorian mode and can be played on the white keys starting with D.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1222122 is the Phrygian mode, starting on E.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2221221 is the Lydian mode, starting on F.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2212212 is the Mixolydian mode, starting on G.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2122122 is the Aeolian mode (the Minor scale), starting on A.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1221222 is the (awesome) Locrian mode, starting on B.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each mode has its own unique &quot;sound&quot;, which (in my opinion, at least) derives precisely from the different placement of the semitones within each scale.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And of course there are scrillions of non-Diatonic scales that have nothing whatsoever to do with how the modern keyboard came to be.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT to add a shorter, less implicit answer: The white keys alone can be used to play the set of diatonic scales listed above; the black keys are &quot;different&quot; because they are the remaining chromatic pitches not used in that set of diatonic scales.&lt;/p&gt;&#xA;" OwnerUserId="4067" LastEditorUserId="4067" LastEditorDisplayName="" LastEditDate="2010-11-27T22:21:51.630" LastActivityDate="2010-11-27T22:21:51.630" CommentCount="3" />
  <row Id="11918" PostTypeId="2" ParentId="11916" CreationDate="2010-11-26T02:49:28.713" Score="1" ViewCount="" Body="&lt;p&gt;$p_n$ satisfies the recurrence, $p_{n+1}(x) = p_n&amp;#39;(x) - 2xp_n(x)$ with $p_0(x) = 1$. &lt;s&gt; This looks to me like it might give you some Tchebyshev polynomial. Something like $p_n(x) = 2 (-1)^n T_n(x)$ &lt;/s&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt;&#xA;I was wrong with the Tchebyshev polynomial thing. Sorry.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2010-11-26T02:54:44.400" LastActivityDate="2010-11-26T02:54:44.400" />
  <row Id="11919" PostTypeId="2" ParentId="11916" CreationDate="2010-11-26T02:57:47.390" Score="7" ViewCount="" Body="&lt;p&gt;That depends on what you mean by &quot;general formula.&quot;  These are (up to some normalization) the &lt;a href=&quot;http://en.wikipedia.org/wiki/Hermite_polynomials&quot; rel=&quot;nofollow&quot;&gt;Hermite polynomials&lt;/a&gt;.  They satisfy a nice recurrence and have a nice generating function.  You could torture some kind of general formula out of the generating function but I really don't see the point.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-11-26T02:57:47.390" CommentCount="4" />
  <row Id="11920" PostTypeId="1" AcceptedAnswerId="11921" CreationDate="2010-11-26T03:06:35.913" Score="3" ViewCount="112" Body="&lt;p&gt;I am reminded of this question that appeared in a regional Physics Olympiad I had appeared. I was wondering if there is a &quot;mathematical&quot; way of doing it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you start from a point $A$ at midnight along a specified path down a mountain and reach a specified point $B$ exactly 24 hours later. At this point you reverse your direction of travel and travel along the same path to point $A$ and reach $A$ exactly 24 hours later. At any point of time during these two days, your velocity can be positive, negative or zero (and of course less than $c$). Prove that there exists at least one point along the path where you were at the same time on day 1 and day 2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It &quot;seems&quot; like an application of intermediate value theorem for some appropriately defined function, but I am not sure (though I will tag it as calculus for the moment). Any ideas?&lt;/p&gt;&#xA;" OwnerUserId="3947" LastActivityDate="2010-11-26T04:00:22.167" Title="Traveling up and down a mountain" Tags="&lt;calculus&gt;" AnswerCount="1" />
  <row Id="11921" PostTypeId="2" ParentId="11920" CreationDate="2010-11-26T03:10:45.550" Score="8" ViewCount="" Body="&lt;p&gt;This a classic!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A &quot;layman&quot; proof is to consider you and your ghost from the previous day.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you turn back, let your ghost of the previous day start from point $A$. At some point, you and ghost must meet!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In stricter mathematical terms:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have two continuous functions, which measure the distance from point $A$ with time (in days say). Let the distance from $A$ to $B$ be 1km.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have that $f(0) = 0, f(1) = 1, g(0) = 1, g(1) = 0$. $f$ is the function for day 1 and $g$ is the function for day 2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now $h(x) = f(x) - g(x)$ is continuous with $h(0) &amp;lt; 0$ and $h(1) &amp;gt; 0$ and so must be $0$ at some point in between, using the intermediate value theorem.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-11-26T04:00:22.167" LastActivityDate="2010-11-26T04:00:22.167" CommentCount="2" />
  <row Id="11922" PostTypeId="1" CreationDate="2010-11-26T03:52:09.197" Score="1" ViewCount="111" Body="&lt;p&gt;Let $A$ and $B$ be finite-dimensional Hermitian matrices, and let $v$ $:=$ $max${$||A||,||B||$}.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Show that $||(e^{-iAt/m}e^{-iBt/m})^m -e^{-i(A+B)t}||\leq\epsilon$ provided $m = \Omega(v^2t^2/\epsilon)$.&lt;/p&gt;&#xA;" LastActivityDate="2010-11-26T07:42:41.853" Title="Lie Product Formula" Tags="&lt;linear-algebra&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="11923" PostTypeId="1" AcceptedAnswerId="11926" CreationDate="2010-11-26T04:15:14.580" Score="17" ViewCount="1844" Body="&lt;p&gt;Can anyone recommend a good self-contained reference for completion of rationals to get reals using Cauchy sequences?&lt;/p&gt;&#xA;" OwnerUserId="3947" LastActivityDate="2011-03-22T23:19:36.977" Title="Completion of rational numbers via Cauchy sequences" Tags="&lt;real-analysis&gt;&lt;reference-request&gt;" AnswerCount="5" FavoriteCount="13" />
  <row Id="11924" PostTypeId="1" AcceptedAnswerId="11927" CreationDate="2010-11-26T04:37:25.723" Score="2" ViewCount="108" Body="&lt;p&gt;I made the following observation following a suggestion from a friend. This question is about the Microsoft calculator program implemented on Windows. Since I don't have access to any other calculator at the moment, I cannot confirm whether this is a general feature. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you enter an integer in the calculator program, say $N$, hit the minus button followed by the same integer again, you get zero as expected.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if you input $N^2$, then hit square root button, you get $N$, then subtract $N$, you get a very small number, but not zero. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This has been my observation for several integers that I have tried. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, with $N=5$, you get $1.232\times 10^{-24}$ in the final step.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, I understand computers use finite precision (and hence the square root function must be working under finite precision). However, I don't understand why this shows up in the final step only? Is it only because this is a small number in comparison to any integer, but when you get $0$, you can no longer make that approximation?&lt;/p&gt;&#xA;" OwnerUserId="3947" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-26T04:52:48.210" LastActivityDate="2010-11-26T05:53:23.320" Title="Rounding trouble in a calculator's square root function" Tags="&lt;algorithms&gt;&lt;computer-science&gt;" AnswerCount="2" CommentCount="6" />
  <row Id="11925" PostTypeId="1" AcceptedAnswerId="11928" CreationDate="2010-11-26T04:43:11.550" Score="2" ViewCount="122" Body="&lt;p&gt;Let $f:A \rightarrow B$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The complements indicated below are taken within $A$ or $B$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need to prove that&#xA;$f(S)^c \subseteq f(S^c)$, $\forall S \subseteq A$, if and only if $f$ is surjective. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I need to prove $f(A) = B$ right?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I prove this?&lt;/p&gt;&#xA;" OwnerUserId="2587" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-26T05:40:50.090" LastActivityDate="2010-11-26T06:00:15.420" Title="$f(S)^c \subseteq f(S^c)$, $\forall S\subseteq A$, if and only if $f\colon A\to B$ is surjective." Tags="&lt;elementary-set-theory&gt;" AnswerCount="3" CommentCount="12" FavoriteCount="1" />
  <row Id="11926" PostTypeId="2" ParentId="11923" CreationDate="2010-11-26T04:50:49.867" Score="5" ViewCount="" Body="&lt;p&gt;Here is a link that came up in a Google search.  You can also search for &quot;construction of real numbers&quot;.  Since this is standard material, you can find it in many books on analysis.    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://math.mit.edu/~tkemp/18.024/R.pdf&quot; rel=&quot;nofollow&quot;&gt;math.mit.edu/~tkemp/18.024/R.pdf&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="145" LastEditorUserId="145" LastEditorDisplayName="" LastEditDate="2010-11-26T06:54:27.670" LastActivityDate="2010-11-26T06:54:27.670" CommentCount="4" />
  <row Id="11927" PostTypeId="2" ParentId="11924" CreationDate="2010-11-26T04:52:33.207" Score="7" ViewCount="" Body="&lt;p&gt;You'll find your answer and much more in the articles below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://docs.sun.com/source/806-3568/ncg_goldberg.html&quot; rel=&quot;nofollow&quot;&gt;What Every Computer Scientist Should Know About Floating-Point Arithmetic&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also see this article on &lt;a href=&quot;http://www.rskey.org/~mwsebastian/miscprj/forensics.htm&quot; rel=&quot;nofollow&quot;&gt;Calculator Forensics&lt;/a&gt;, esp. &quot;Hints and Tricks For Obtaining Guard Digits&quot;.&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-11-26T04:52:33.207" CommentCount="1" />
  <row Id="11928" PostTypeId="2" ParentId="11925" CreationDate="2010-11-26T05:04:40.917" Score="3" ViewCount="" Body="&lt;p&gt;You want to prove &lt;em&gt;two&lt;/em&gt; things:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;If&lt;/em&gt; the complement of the image is always contained in the image of the complement, &lt;em&gt;then&lt;/em&gt; $f(A)=B$; and&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;If&lt;/em&gt; $f(A)=B$, then for every subset $S$ of $A$ you have that the complement of the image is contained in the image of the complement.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;So, &quot;I have to prove that $f(A)=B$&quot; is only true for half of the problem at issue.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As to how you prove 1 above, well, since the containment holds for &lt;em&gt;all&lt;/em&gt; subsets $S$ of $A$, perhaps you can pick a &lt;em&gt;particular&lt;/em&gt; subset $S$ of $A$ in some clever way that will tell you that $f(A)$ must be equal to $B$. (&lt;strong&gt;HINT:&lt;/strong&gt; $f(A)=B$ if and only if $f(A)^c$ is... )&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For 2, you have to &lt;strong&gt;assume&lt;/strong&gt; that $f(A)=B$ (that $f$ is surjective). Then you need to take an &lt;em&gt;arbitrary&lt;/em&gt; subset $S$ of $A$. Then to test whether $f(S)^c\subseteq f(S^c)$ holds, you need to take an arbitrary $b\in f(S)^c$, and show that it must lie in $f(S^c)$; that is, that there exists some $a\notin S$ such that $b=f(a)$. Now go to it.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-26T05:32:17.450" LastActivityDate="2010-11-26T05:32:17.450" />
  <row Id="11929" PostTypeId="2" ParentId="11925" CreationDate="2010-11-26T05:15:49.450" Score="2" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\rm\ (\Rightarrow)\ $ Put $\rm S\ =\ \ldots\ \ (\Leftarrow)\ $ Consider $\rm\  {\overline {f(S)}} \cap (f(S) \cup f(\overline S))$&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-11-26T05:15:49.450" CommentCount="3" />
  <row Id="11930" PostTypeId="2" ParentId="11724" CreationDate="2010-11-26T05:33:52.650" Score="2" ViewCount="" Body="&lt;p&gt;I would normally take $p$-adic integration to mean &quot;integration of $p$-adic valued functions&quot; or &quot;integration of differential forms with some kind of $p$-adic valued functions as coefficients&quot;, where the integration is also taking place over some kind of $p$-adic space or manifold.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason for wanting such theories are various.  One reason is indicated in George S.'s answer: there are &lt;em&gt;known&lt;/em&gt; analogues of classical Hodge theory, known as $p$-adic Hodge theory, whose proofs however are not analytic, but rather proceed via arithmetic geometry.  One would like to have more analytic ways of thinking about them, and this is one goal of Robert Coleman's theory.  (In a recent volume of Asterisque, namely vol. 331, Coleman and Iovita have an article, &lt;a href=&quot;http://math.berkeley.edu/~coleman/Hidden/HiddenStr.pdf&quot; rel=&quot;nofollow&quot;&gt;&quot;Hidden structures on semistable curves&quot;&lt;/a&gt;, related to this problem.)  (Note also that $p$-adic Hodge theory relates $p$-adic etale cohomology to crystalline cohomology, which gives on answer to your question of how $p$-adic integration might be related to those topics.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another reason is that many integral formulas (involving usual archimedean integrals) appear in the theory of classical $L$-functions attached&#xA;to automorphic forms, and one would like, at least in certain contexts, to be able to write down $p$-adic analogues so as to construct $p$-adic $L$-functions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for what machinery is used: in the theory of $p$-adic $L$-functions and related contexts in Iwasawa theory, often nothing more is used than basic computations with Riemann sums.  In the material related to $p$-adic Hodge theory, much more substantial theoretical foundations are used: tools from arithemtic geometry, rigid analysis, possibly Berkovic spaces, and related topics.  &lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2010-11-26T05:33:52.650" />
  <row Id="11931" PostTypeId="2" ParentId="11925" CreationDate="2010-11-26T05:39:11.903" Score="1" ViewCount="" Body="&lt;p&gt;You need to prove more than that $f(A) = B$. You need to show that if $f$ is surjective, then $(f(S))^c \subset f(S^c)$ for every possible subset $S$ of $A$, and that if $f$ is not surjective, then there is some $S$ for which $(f(S))^c$ is not a subset of  $f(S^c)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Largish hint for the first part: If $f$ is surjective then $f(S) \cup f(S^c)$ is all of $B$, regardless of what $S$ is. Use this to prove the contrapositive of the first part.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint for the second part: Pick a special subset $S$ of $A$ and show it works for that $S$. &lt;/p&gt;&#xA;" OwnerUserId="3035" LastEditorUserId="3035" LastEditorDisplayName="" LastEditDate="2010-11-26T06:00:15.420" LastActivityDate="2010-11-26T06:00:15.420" CommentCount="6" />
  <row Id="11932" PostTypeId="2" ParentId="11669" CreationDate="2010-11-26T05:42:04.033" Score="9" ViewCount="" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Mathematically, is there any difference between white and black notes, or do we make the distinction just for historical reasons?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There is no mathematical difference between the white and black notes. Adjacent notes on modern piano keyboards are typically tuned 1/12 of an octave apart. Quiaochu explains this most completely, but what it boils down to is that there is no difference.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We haven't always and don't today always use equal temperament on keyboard instruments but even then the difference between white and black notes would be arbitrary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The distinction is for historical and cultural reasons. There is a cool picture here showing Nicholas Farber's Organ (1361), which used an 8 + 4 layout rather than the modern 7 + 5 layout we see today. &lt;a href=&quot;http://en.wikipedia.org/wiki/Musical_keyboard#Size_and_historical_variation&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Musical_keyboard#Size_and_historical_variation&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are examples of instruments in use today that use a chromatic keyboard with no differentiation between the &quot;white&quot; and the &quot;black&quot; notes. See the Bayan and the Bandoneon accordion type instruments.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the New England Conservatory in the classroom where they teach a class on quarter tones, they keep two pianos tuned a quarter tone off from each other. In that case, a full 24-note chromatic quarter tone octave must be played alternating notes on the two pianos.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is only the beginning of this particular rabbit hole.&lt;/p&gt;&#xA;" OwnerUserId="4076" LastActivityDate="2010-11-26T05:42:04.033" CommentCount="1" />
  <row Id="11933" PostTypeId="2" ParentId="11761" CreationDate="2010-11-26T05:50:07.007" Score="5" ViewCount="" Body="&lt;p&gt;As a supplement to T..'s nice answer, let me write something that is more of a cultural remark than a direct answer to the question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many people learnt about the yoga of mixed motives by reading &lt;a href=&quot;http://www.math.ias.edu/files/deligne/GaloisGroups.pdf&quot; rel=&quot;nofollow&quot;&gt;Deligne's article&lt;/a&gt;&#xA;on the thrice punctured sphere.  (This was certainly the case for me; and a hat-tip to &lt;a href=&quot;http://quomodocumque.wordpress.com/2009/07/24/le-groupe-fondamental-de-la-droite-projective-moins-trois-points-is-now-online/&quot; rel=&quot;nofollow&quot;&gt;Quomodocumque&lt;/a&gt; for providing the link.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This article helps explain what one should expect from the category of mixed motives, and thus helps one recognize the (or perhaps a) correct construction when one sees it. &lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2010-11-26T05:50:07.007" CommentCount="0" />
  <row Id="11934" PostTypeId="2" ParentId="11924" CreationDate="2010-11-26T05:53:23.320" Score="0" ViewCount="" Body="&lt;p&gt;J.M., since when is 24 decimal digits accuracy considered a bad approximation?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Original poster, they could show sqrt(5)**2 as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;5.000000000000000000000001232&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but since the display probably has fewer display digits than this, they chop it to &quot;5&quot; as that is the closest representation in the allowed number of digits.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;BTW, I just tried this in &quot;calc&quot; in windows 7 and the 5-sqrt(5)**2 is 1.7599e-37&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want exact math, you will need to use a symbolic math package.&lt;/p&gt;&#xA;" LastActivityDate="2010-11-26T05:53:23.320" CommentCount="2" />
  <row Id="11935" PostTypeId="2" ParentId="11923" CreationDate="2010-11-26T06:03:21.543" Score="3" ViewCount="" Body="&lt;p&gt;Pages &lt;a href=&quot;http://books.google.com/books?id=3-nrPB7BQKMC&amp;amp;lpg=PP1&amp;amp;pg=PA142#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;142&lt;/a&gt;-154 of Stoll's &lt;em&gt;Set theory and logic&lt;/em&gt; looks like a good self-contained treatment.  The chapter starts on &lt;a href=&quot;http://books.google.com/books?id=3-nrPB7BQKMC&amp;amp;lpg=PP1&amp;amp;pg=PA130#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;page 130&lt;/a&gt; and is called &quot;The extension of the natural numbers to the real numbers.&quot;&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-11-26T06:03:21.543" />
  <row Id="11936" PostTypeId="1" CreationDate="2010-11-26T06:16:29.437" Score="1" ViewCount="99" Body="&lt;p&gt;Let Q= {q_1,...,q_n} be a set of permutations of some finite set X.  Let &#xA;Q' be {q in Q^* | q=1 in the symmetric group on X}.  Is Q' a regular language?&lt;/p&gt;&#xA;" LastActivityDate="2010-11-27T04:22:51.837" Title="regular languages and permutations" Tags="&lt;group-theory&gt;" AnswerCount="2" />
  <row Id="11937" PostTypeId="2" ParentId="11936" CreationDate="2010-11-26T06:20:37.763" Score="3" ViewCount="" Body="&lt;p&gt;Construct a graph $\Gamma$ having as vertices the elements of the symmetric group $S(X)$, and for each $g\in S(X)$ and each $i\in\{1,\dots,n\}$ an arrow from the vertex $g$ to the vertex $q_ig$ labeled $q_i$. Upgrade $\Gamma$ into a deterministic automatic so that the initial state is the identity element $e$ of $S(X)$, and the only accepted state is $e$ itself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the language accepted by the automaton $\Gamma$?&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2010-11-26T06:20:37.763" />
  <row Id="11938" PostTypeId="2" ParentId="11846" CreationDate="2010-11-26T06:25:36.740" Score="1" ViewCount="" Body="&lt;p&gt;When you would use which really depends on the random variable you're interested in.  If you're interested in the number of trials needed to obtain the first success, use the first kind of geometric distribution.  If you're interested in the number of successes you are able to achieve before the first failure occurs then use the second kind of geometric distribution.  (Inverting the interpretation of the second version like this also requires you to redefine the success probability as $1-p$ and the failure probability as $p$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the Wikipedia page says, &quot;Which of these one calls 'the' geometric distribution is a matter of convention and convenience.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I find the two different versions confusing myself.  When I read somewhere that $X$ has a &quot;geometric distribution,&quot; the writer isn't always careful to specify which one is meant, and then I have to spend time figuring out which one it is.  This confusion also spreads to the &lt;a href=&quot;http://en.wikipedia.org/wiki/Negative_binomial_distribution&quot; rel=&quot;nofollow&quot;&gt;negative binomial distribution&lt;/a&gt;, which is a generalization of the geometric distribution.  The &lt;a href=&quot;http://en.wikipedia.org/wiki/Exponential_distribution&quot; rel=&quot;nofollow&quot;&gt;exponential distribution&lt;/a&gt;, which is a continuous version of the geometric distribution, and the &lt;a href=&quot;http://en.wikipedia.org/wiki/Gamma_distribution&quot; rel=&quot;nofollow&quot;&gt;gamma distribution&lt;/a&gt; (a generalization of the exponential), have more than one definition, too.  It's too bad that these weren't standardized with one definition for each, but part of the reason they weren't is that the different versions are useful in different scenarios. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In response to Josh Guffin's comment: Yes, in many contexts it is easy to figure out which geometric distribution a writer is referring to.  However, in some it is not.   &lt;a href=&quot;http://en.wikipedia.org/wiki/Moment-generating_function&quot; rel=&quot;nofollow&quot;&gt;Wikipedia's page on moment-generating functions&lt;/a&gt; is a classic example.  The table there gives the mgf's for the negative binomial (and thus geometric), exponential, and gamma distributions, but it doesn't specify which convention for each one is being used.&lt;/p&gt;&#xA;" OwnerUserId="2370" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-11-27T05:28:44.690" LastActivityDate="2010-11-27T05:28:44.690" CommentCount="1" />
  <row Id="11939" PostTypeId="1" AcceptedAnswerId="11959" CreationDate="2010-11-26T06:51:20.807" Score="2" ViewCount="290" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's consider a 2x2 linear system:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A &lt;strong&gt;u&lt;/strong&gt; = &lt;strong&gt;b&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The solution will still be the same even after we interchange the rows in A and B. I know this to be true because algebraically, we will get the same set of equations before and after the row interchange.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the vectors in columns of A and B are different. So how can the system still have the same solution as before the row interchange?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you.&lt;/p&gt;&#xA;" OwnerUserId="2973" LastActivityDate="2010-11-26T10:03:08.690" Title="Elementary Row Operations - Interchange a Matrix's rows" Tags="&lt;linear-algebra&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="11940" PostTypeId="2" ParentId="9934" CreationDate="2010-11-26T07:01:58.430" Score="3" ViewCount="" Body="&lt;p&gt;Just thought, I would make this obvious remark, extending Arturo's answer: since, even in the infinite case, any subgroup of finite index in $G$ must be normal for $G$ to satisfy the requirement, it follows that for any $H$ of finite index in $G$, any subgroup of the quotient $G/H$ will also correspond to a subgroup of $G$ of finite index. In particular, for any $H$ of finite index, $H$ must be normal and the quotient $G/H$ must be cyclic by Arturo's argument.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A class of such groups considerably extending that of cyclic groups are the pro-cyclic group, i.e. inverse limits of cyclic ones. Examples include $\mathbb{Z}_p$ or any product $\prod_p \mathbb{Z}_p$ over distinct primes $p$. In particular, $\hat{\mathbb{Z}}$ is another example. In fact, Arturo's argument shows that any pro-finite group satisfying the above condition must be pro-cyclic.&lt;/p&gt;&#xA;" OwnerUserId="3212" LastActivityDate="2010-11-26T07:01:58.430" CommentCount="3" />
  <row Id="11941" PostTypeId="1" AcceptedAnswerId="11950" CreationDate="2010-11-26T07:02:12.060" Score="2" ViewCount="259" Body="&lt;p&gt;I was assisting a TA for an introductory calculus class with the following limit,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\lim_{x \rightarrow \infty} \left(\frac{20x}{20x+4}\right)^{8x}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and I came to simple solution which involved evaluating the &quot;reciprocal&quot; limit&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\lim_{z \rightarrow 0} \left(\frac{1}{1+\frac{z}{5}}\right)^{8/z}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;by using the Taylor expansion of $\log(1+z)$ around $z=0$. However, the TA claims that the students have not learned about series expansions so that would not be a valid solution for the course. I tried applying L'Hopital's rule, which I was told the class did cover, but I was unsuccessful. As a note I will mention that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\lim_{x \rightarrow \infty} \left(\frac{20x}{20x+4}\right)^{8x} = e^{-8/5}.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any ideas for a solution to this problem using only knowledge from a first quarter (or semester) calculus course which hasn't covered series expansions?&lt;/p&gt;&#xA;" OwnerUserId="2887" LastActivityDate="2010-12-08T02:58:40.887" Title="Evaluation of the limit, $\lim_{x\rightarrow\infty} \left(\frac{20x}{20x+4}\right)^{8x}$, using only elementary methods." Tags="&lt;calculus&gt;&lt;homework&gt;" AnswerCount="4" CommentCount="2" FavoriteCount="1" />
  <row Id="11942" PostTypeId="2" ParentId="11939" CreationDate="2010-11-26T07:02:28.527" Score="3" ViewCount="" Body="&lt;p&gt;Rearranging the rows of $A$ and $b$ corresponds to premultiplying them with a &lt;a href=&quot;http://en.wikipedia.org/wiki/Permutation_matrix&quot; rel=&quot;nofollow&quot;&gt;permutation matrix&lt;/a&gt; $P$. It's not hard to see that the new system $PAu = Pb$ has the same solution as the original system. This holds for linear systems of all sizes, not just $2\times 2$ ones.&lt;/p&gt;&#xA;" OwnerUserId="856" LastActivityDate="2010-11-26T07:02:28.527" />
  <row Id="11944" PostTypeId="2" ParentId="11941" CreationDate="2010-11-26T07:08:32.163" Score="3" ViewCount="" Body="&lt;p&gt;Hint: Consider taking logarithms before applying L'Hospital's rule.&lt;/p&gt;&#xA;" OwnerUserId="3431" LastActivityDate="2010-11-26T07:08:32.163" />
  <row Id="11945" PostTypeId="1" AcceptedAnswerId="11969" CreationDate="2010-11-26T07:08:40.347" Score="7" ViewCount="546" Body="&lt;p&gt;Where does this sequence converge?&#xA;$\sqrt{7},\sqrt{7+\sqrt{7}},\sqrt{7+\sqrt{7+\sqrt{7}}}$,...&lt;/p&gt;&#xA;" OwnerUserId="4082" LastEditorUserId="4082" LastEditorDisplayName="" LastEditDate="2010-11-26T17:59:40.423" LastActivityDate="2010-11-27T00:03:34.820" Title="Limit of $\sqrt{7+\sqrt{7+\sqrt{7+\cdots}}}$" Tags="&lt;real-analysis&gt;" AnswerCount="3" />
  <row Id="11946" PostTypeId="2" ParentId="11941" CreationDate="2010-11-26T07:09:13.630" Score="3" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT:&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align*}&#xA;\lim_{x \to \infty} \biggl(\frac{20x}{20x+4}\biggr)^{8x} &amp;amp;=\lim_{x \to \infty} \biggl( 1 - \frac{4}{20x+4}\biggr)^{8x} \\ &amp;amp;=\lim_{x \to \infty} \biggl(1 - \frac{4}{20x+4}\biggr)^{\frac{20x+4}{4} \cdot \frac{8x}{20x+4} \cdot 4}&amp;amp;\end{align*}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now use the fact that $$\lim_{n \to \infty} \biggl(1 + \frac{1}{n}\biggr)^{n} =e$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;ADDED:&lt;/strong&gt; After simplyfing you should get this. $$e^{\displaystyle \lim_{x \to \infty} - \frac{8x}{20x+4} \cdot 4} = e^{-8/5}$$  &lt;/p&gt;&#xA;" LastActivityDate="2010-12-08T02:58:40.887" />
  <row Id="11947" PostTypeId="2" ParentId="11945" CreationDate="2010-11-26T07:11:20.163" Score="3" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT:&lt;/strong&gt; Suppose the sequence converges to $x$. Then solve for $x = \sqrt{7+x}$&lt;/p&gt;&#xA;" LastActivityDate="2010-11-26T07:11:20.163" />
  <row Id="11948" PostTypeId="1" CreationDate="2010-11-26T07:12:40.850" Score="1" ViewCount="207" Body="&lt;p&gt;I have read a result on computing the highest power of a prime that divides $n!$. I was wondering if there are any results on how to compute the highest power of a prime dividing $f(x)$, where $f$ is a function on the real numbers that outputs integer values. &lt;/p&gt;&#xA;" OwnerUserId="3947" LastActivityDate="2010-11-29T19:30:19.347" Title="The highest power of a prime that divides $f(x)$" Tags="&lt;number-theory&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="11949" PostTypeId="2" ParentId="11939" CreationDate="2010-11-26T07:14:08.910" Score="4" ViewCount="" Body="&lt;p&gt;Perhaps I don't fully understand your question. So suppose you have a $2\times 2$ linear system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{bmatrix}a &amp;amp; b\\c &amp;amp; d \end{bmatrix}\begin{bmatrix}u_1\\ u_2\end{bmatrix}=\begin{bmatrix} e\\ f\end{bmatrix}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This gives you the augmented matrix&#xA;$$\begin{bmatrix}a &amp;amp; b &amp;amp; e\\ c &amp;amp; d &amp;amp; f\end{bmatrix}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This gives the set of equations &#xA;$$au_1+bu_2=e\quad\text{and}\quad cu_1+du_2=f.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you interchange the rows, you get the augmented matrix&#xA;$$\begin{bmatrix}c &amp;amp; d &amp;amp; f\\ a &amp;amp; b &amp;amp; e\end{bmatrix}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which indeed has different columns. But this gives the equations &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$cu_1+du_2=f\quad\text{and}\quad au_1+bu_2=e$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;just the other way around as before. The entries of $\textbf{u}$ haven't been switched, if that was your concern.&lt;/p&gt;&#xA;" OwnerUserId="1583" LastActivityDate="2010-11-26T07:14:08.910" />
  <row Id="11950" PostTypeId="2" ParentId="11941" CreationDate="2010-11-26T07:16:19.770" Score="5" ViewCount="" Body="&lt;p&gt;If they know the definition of $e$ as&#xA;$$\lim_{n\rightarrow \infty} \left(1+\frac{1}{n}\right)^n,$$&#xD;&#xA;then set $\alpha=\lim_{x\rightarrow\infty}\left(\frac{20x}{20x+4}\right)^{8x}$&#xD;&#xA;and note that&#xA;$$1/\alpha^5 = \lim_{x\rightarrow\infty}\left(1 + \frac{1}{5x}\right)^{40x}=&#xD;&#xA;\left(\lim_{x\rightarrow\infty}\left(1 + \frac{1}{5x}\right)^{5x}\right)^8 = e^8.$$&lt;/p&gt;&#xA;" OwnerUserId="3212" LastActivityDate="2010-11-26T07:16:19.770" CommentCount="2" />
  <row Id="11952" PostTypeId="2" ParentId="11945" CreationDate="2010-11-26T07:18:02.840" Score="14" ViewCount="" Body="&lt;p&gt;Hint: It morally converges to $\sqrt{7+\sqrt{7+\sqrt{7+.....}}}$. Call this limit $l$. Then, since the nested series of square roots extends indefinitely, $l=\sqrt{7+l}$&lt;/p&gt;&#xA;" OwnerUserId="3431" LastActivityDate="2010-11-26T07:18:02.840" CommentCount="4" />
  <row Id="11953" PostTypeId="1" CreationDate="2010-11-26T07:24:23.833" Score="1" ViewCount="204" Body="&lt;p&gt;I have the following constraint functions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$g_{i_{min}} \leq y_{i+1}-y_{i} \leq g_{i_{max}}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$y_{i_{max}}-y_{i} \geq h_{i}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$v_{i_{min}} \leq \Biggl[\frac{(y_{i+1}-y_{i})^{3} (h_{i}+a)^{4}}{(h_{i}+b)^{4}}\Biggr] \leq v_{i_{max}}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$Q_{i_{min}} \leq \Biggl[\frac{(y_{i+1}-y_{i})^{3} (h_{i}+a)^{10}}{(h_{i}+b)^{4}}\Biggr]$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $i=1,2,3$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The variables for minimization are $y_i$ and $h_i$. &lt;strong&gt;Note that $y_i$ are continuous and $h_i$ are discrete.&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;All $g_{i_{min}}, g_{i_{max}}, Q_{i_{min}},v_{i_{min}},v_{i_{max}}, y_{i_{max}},a,b$ are constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are &lt;em&gt;two&lt;/em&gt; objectives functions to minimize against:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f_{1}=\sum_{i=1}^3 h_i  \quad\quad \text{(1)}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f_{2}=-\sum_{i=1}^4 y_i  \quad\quad \text{(2)}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The idea is that $(1)$ should always be minimized first, and in the case of a draw, minimize $(2)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any idea how to tackle this question?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On a second thought, I think for my application, I can combine the two objective functions into a single function. &lt;strong&gt;So my question how is is there anyway to minimize a set of nonlinear constraint equations, which some of the variables are discrete, and some are continuous?&lt;/strong&gt; &lt;/p&gt;&#xA;" OwnerUserId="550" LastEditorUserId="550" LastEditorDisplayName="" LastEditDate="2010-11-30T15:34:41.643" LastActivityDate="2010-12-30T15:44:30.017" Title="Optimizing Nonlinear Constraint Equations with Discrete Variables and Multiple Objective Functions" Tags="&lt;nonlinear-optimization&gt;&lt;integer-programming&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="11955" PostTypeId="2" ParentId="11922" CreationDate="2010-11-26T07:42:41.853" Score="1" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT:&lt;/strong&gt; Consider the product of power series $\exp(A/m)\exp(B/m) = I + \frac{A}{m} + \frac{B}{m} + O\left(\frac{1}{m^2}\right)$. Then use this to estimate the logarithm for sufficiently large $m$, then exponentiate and raise to the $m$-th power.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is from the proof in section 2.4 of &quot;Lie Groups, Lie Algebras, and Representations&quot; by Brian C. Hall.&lt;/p&gt;&#xA;" OwnerUserId="2887" LastActivityDate="2010-11-26T07:42:41.853" CommentCount="1" />
  <row Id="11956" PostTypeId="2" ParentId="11941" CreationDate="2010-11-26T08:15:46.117" Score="2" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Always start by simplifying&#xA;$$\left(\frac{20x}{20x+4}\right)^{8x}=\left(\frac{20x+4}{20x}\right)^{-8x}=\left(1+\frac{1}{5x}\right)^{-8x}$$&#xD;&#xA;perhaps a substitution helps... &lt;/p&gt;&#xA;" OwnerUserId="1154" LastActivityDate="2010-11-26T08:15:46.117" />
  <row Id="11957" PostTypeId="2" ParentId="11948" CreationDate="2010-11-26T08:26:57.417" Score="5" ViewCount="" Body="&lt;p&gt;Finding divisors of combinatorial numbers can range from trivial to extremely difficult.  I'm going to explain a technique that I used numerous times to find divisors of combinatorial numbers, particularly the number of &lt;a href=&quot;http://en.wikipedia.org/wiki/Latin_square&quot; rel=&quot;nofollow&quot;&gt;Latin squares&lt;/a&gt; and Latin rectangles. (check out my website for numerous papers including this tactic).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose you have a set $C$ and want to find $|C| \pmod \mu$ for some $\mu$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Step 1&lt;/strong&gt;: Identify a group $G$ such that (a) $G$ acts on $C$ and (b) $|G|$ divisible by $\mu$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $B=\{x \in C:$ there exist a non-trivial $g \in G$ such that $gx=x\}$, that is, $B$ is the set of objects in $C$ that admit non-trivial automorphisms in $G$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that $G$ acts on $C \setminus B$, and moreover, the orbit of any $x \in C \setminus B$ is of cardinality $|G|$ (by the &lt;a href=&quot;http://en.wikipedia.org/wiki/Group_action&quot; rel=&quot;nofollow&quot;&gt;Orbit-Stabiliser Theorem&lt;/a&gt;, when all the stabilisers are trivial).  We can therefore deduce $|G|$ divides $|C \setminus B|$, and hence $|C| \equiv |B| \pmod {|G|}$ (and hence $|C| \equiv |B| \pmod \mu$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most of the time, $x \in C$ will not admit a non-trivial automorphism in $G$.  This means, that if we want to compute $|C| \pmod \mu$, we can instead find $|B|$, which already results in a significant computational saving.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sometimes, however, this approach is still inappropriate (e.g. $|B|$ could be massive).  But not all is lost.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Step 2&lt;/strong&gt;: Identify a set $A$ such that:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$B \subseteq A \subseteq C$,&lt;/li&gt;&#xA;&lt;li&gt;$G$ acts on $A$,&lt;/li&gt;&#xA;&lt;li&gt;members of $A$ have some property (such as a certain substructure) that can be used to find a divisor $\nu$ of $|A|$ (in the case of Latin squares, the substructure is a subsquare, which we can replace by any other subsquare, inducing an equivalence relation on $A$, thus giving a divisor of $|A|$).&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Then the same argument as above implies $|C| \equiv |A| \pmod {|G|}$ and so $\gcd(|A|,|G|)$ divides $|C|$.  Since we know $\nu$ divides $|A|$ and $\mu$ divides $|G|$, we can deduce that $\gcd(\mu,\nu)$ divides $|C|$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I depict this concept below.  The area shaded gray represents $C$, and is partitioned into orbits by $G$.  The blue orbits have cardinality $|G|$ and the red orbits have cardinality strictly less than $|G|$.  We &quot;break a bit off&quot; and call it $A$.  If we can find a divisor of $|A|$, then we can use it to find a divisor of $|C|$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/3zg3S.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="139" LastActivityDate="2010-11-26T08:26:57.417" />
  <row Id="11958" PostTypeId="1" CreationDate="2010-11-26T09:15:52.073" Score="4" ViewCount="112" Body="&lt;p&gt;I start from a simple example. I through a (possibly unfair) coin 20 times. I got &quot;eagle&quot; and &quot;tail&quot; 15 and 5 times, respectively. Now I need to estimate probabilities for eagle and tail. For eagle I get 15/20 = 0.75 and for the tail I get 0.25.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I know that the probability that I got are not accurate. It is still possible that I have a fair coin (with probability for eagle equal to 0.5) and I get more eagles in my experiment just by chance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I want to estimate probabilities of probabilities. In other words I want to know how likely it is that probability for eagle is 0.5 (or 0.3, or 0.9). I can solve this problem, but I would like to know if there is a name for this problem. I an also interested in the generalization of the problem for the case of more than two events (not just &quot;eagle&quot; and &quot;tail&quot;).&lt;/p&gt;&#xA;" OwnerUserId="4088" LastActivityDate="2010-11-26T17:26:53.003" Title="What is the standard method to estimate probabilities of events based on observations?" Tags="&lt;probability&gt;&lt;statistics&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="11959" PostTypeId="2" ParentId="11939" CreationDate="2010-11-26T09:51:42.353" Score="3" ViewCount="" Body="&lt;p&gt;Let us consider a $2 \times 2$ example. We will then extend this higher dimensions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $$A = \begin{bmatrix}A_{11} &amp;amp; A_{12}\\A_{21} &amp;amp; A_{22} \end{bmatrix}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$b = \begin{bmatrix}b_1 \\b_2 \end{bmatrix}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you now want to solve $Ax_1 = b$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x_1$ is given by $A^{-1}b$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now you swap the two rows of $A$ and $b$. Call them $\tilde{A}$ and $\tilde{b}$ respectively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\tilde{A} = \begin{bmatrix}A_{21} &amp;amp; A_{22}\\A_{11} &amp;amp; A_{12} \end{bmatrix}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\tilde{b} = \begin{bmatrix}b_2 \\b_1 \end{bmatrix}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now how do we relate $\tilde{A}$ and $A$ and similarly $\tilde{b}$ and $b$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The relation is given by a Permutation matrix $P$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\tilde{A} = P A$ and $\tilde{b} = P b$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The matrix $P$ is given by:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\tilde{P} = \begin{bmatrix}0 &amp;amp; 1\\1 &amp;amp; 0 \end{bmatrix}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Check that $\tilde{A} = P A$ and $\tilde{b} = P b$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we look at solving the system $\tilde{A}x_2 = \tilde{b}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Substitute for $\tilde{A}$ and $\tilde{b}$ in terms of $A$ and $b$ respectively to get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$PAx_2 = Pb$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the important thing to note is that $P^2 = I$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This can be verified algebraically or by a simple argument by seeing that $P^2$ swaps and swaps again which reverts back to the original giving $I$ or the other way of looking is $P^2$ is $P$ applied to $P$ which swaps the two rows of $P$ giving back the identity matrix.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So from $P^2 = I$, we get $P^{-1} = P$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we have $PAx_2 = Pb$ and premultiplying by $P^{-1}$ gives $Ax_2 = b$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we have $Ax_1 = b$ and $Ax_2 = b$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And if we assume $A$ is invertible this gives us a unique solution and hence we get $x_1 = x_2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or the other way to look at is to write $x_2 = \tilde{A}^{-1} \tilde b = (PA)^{-1}Pb = A^{-1} P^{-1} P b = A^{-1} I b = A^{-1} b$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All you need to observe in the above step is that the matrix $P$ is invertible and hence the matrix $(PA)$ is also invertible (since $A$ is assumed to be invertible and that $(PA)^{-1} = A^{-1}P^{-1}$ and matrix multiplication is associative.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same argument with permutation matrix holds true for a $n \times n$ system as well.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2010-11-26T10:03:08.690" LastActivityDate="2010-11-26T10:03:08.690" />
  <row Id="11960" PostTypeId="2" ParentId="11958" CreationDate="2010-11-26T11:17:48.863" Score="1" ViewCount="" Body="&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Maximum_likelihood&quot; rel=&quot;nofollow&quot;&gt;maximum likelihood estimate&lt;/a&gt; is a typical method for estimating the probability distribution given some outcomes.  To describe it simply, you assume that your probability distribution you are sampling from is actually one from some set $\Omega$, and you are trying to find which one fits best the data (i.e. which is most likely to be correct).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your case, you could reasonably assume that the probability distribution $\theta_\pi$ for which &quot;heads&quot; has probability $\pi$ and &quot;eagle&quot; has probability $1-\pi$.  So the question is:  Which $\pi \in [0,1]$ is most likely to by the actual distribution, given the observed outcomes?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A description of how to answer the question is given on the linked Wikipedia page under &quot;Discrete distribution, finite parameter space&quot;.&lt;/p&gt;&#xA;" OwnerUserId="139" LastActivityDate="2010-11-26T11:17:48.863" />
  <row Id="11961" PostTypeId="1" CreationDate="2010-11-26T12:16:55.353" Score="0" ViewCount="89" Body="&lt;p&gt;Why are consistently ordered matrices important?  Are there any applications where they are turn up?&lt;/p&gt;&#xA;" OwnerUserId="1794" LastActivityDate="2010-11-26T12:16:55.353" Title="Consistently ordered matrices." Tags="&lt;matrices&gt;" CommentCount="3" />
  <row Id="11962" PostTypeId="1" AcceptedAnswerId="12318" CreationDate="2010-11-26T12:38:38.577" Score="1" ViewCount="108" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope this question is not off-topic, but I'm currently discovering the existence of mixing processes, which attempt to express the dependence between random variables using miscellaneous coefficients ($\alpha$, $\beta$, $\phi$ and $\rho$ are the ones I've met so far).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there any references you would advise that survey them, perhaps along with more or less concrete examples of situations where these various coefficients can be used?&lt;/p&gt;&#xA;" OwnerUserId="689" LastActivityDate="2010-11-29T09:12:29.183" Title="Mixing processes" Tags="&lt;probability-theory&gt;&lt;stochastic-processes&gt;" AnswerCount="1" />
  <row Id="11963" PostTypeId="2" ParentId="11669" CreationDate="2010-11-26T12:42:33.087" Score="1" ViewCount="" Body="&lt;p&gt;One day I shall do a serious study of this!  there is truth in all these answers, the white notes give us our do-re-mi (major scale) starting on C, this scale has a mixture of tones and semitones, and dictate where the black notes should go and how many we need.  The re-tuning to equal temperament is a fudge, and if you were to analyse a tuned keyboard, not all semitones are equally spaced.  Other intervals are also compromised, so a major third in one key may have notes further apart than one in a different key.  Composers have long been aware of this, and aware that the key they select for a composition can make a significant difference to the &quot;mood&quot; (that is after you have selected major or minor).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Classical Indian Music uses a system of scales (ragas).  There are several hundred of these, and they will be fitted to specific moods, times of day, types of occasion etc. These are not random variations from any Western scale, and have nothing to do with the keyboards we commonly use.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Our keyboard system is just for keyboards - a string instrument may not play exactly the same pitch as a piano for a given note (unless it is an open string), because they will tend to use something closer to the original pythagorean scale.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS I am a working musician with a bachelors degree in maths!&lt;/p&gt;&#xA;" LastActivityDate="2010-11-26T12:42:33.087" />
  <row Id="11964" PostTypeId="1" CreationDate="2010-11-26T13:02:17.070" Score="1" ViewCount="176" Body="&lt;p&gt;Came across this math problem while programming something:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given positive $y$, $a$, and $b$, find an integer $x$, $a\lt x\lt b$, so that $y\bmod x$ (the remainder of $y$ when divided by $x$) is positive and less than or equal to $100$,&#xA;$$0 \lt y\bmod x \leq 100.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know sometimes it's not solvable depending on $y$ and the range. I'm not really sure where to begin because I wasn't very good at modular algebra.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I really don't want to have to try every possible number for $x$ between $a$ and $b$ because the range I'm using it for has over 25 million in length.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: Sorry the question wasn't very clear. $y$, $a$, and $b$ are given. And I'm trying to find an $x$ that satisfies those conditions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit 2: I've realize $y \bmod x$ need to be less than or equal to $100$ and also over $0$.&lt;/p&gt;&#xA;" OwnerUserId="4093" LastEditorUserId="4093" LastEditorDisplayName="" LastEditDate="2010-11-27T07:31:17.580" LastActivityDate="2010-11-29T10:45:22.450" Title="Solving for x in $0 &lt; y \bmod x \le 100$ where x is in the range (a, b)" Tags="&lt;inequality&gt;&lt;modular-arithmetic&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="11965" PostTypeId="1" AcceptedAnswerId="11966" CreationDate="2010-11-26T15:40:46.563" Score="2" ViewCount="86" Body="&lt;p&gt;This is my first post here. Apologies if there are any misformats or mis-tagging.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also if the post doesn't belong here, sorry, and I will delete it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I derived this formula from a probability calculation. How would I go about finding the maximum value for the following?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{T}{S}\frac{S!}{(S-D)!}\frac{(T-D)!}{T!}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$T&amp;gt;S&amp;gt;D$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and T, S, and D are all integers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you.&lt;/p&gt;&#xA;" OwnerUserId="4102" LastEditorUserId="4102" LastEditorDisplayName="" LastEditDate="2010-11-26T15:48:45.963" LastActivityDate="2010-11-26T15:50:55.353" Title="Maximum value of probability" Tags="&lt;probability&gt;" AnswerCount="1" CommentCount="0" />
  <row Id="11966" PostTypeId="2" ParentId="11965" CreationDate="2010-11-26T15:50:55.353" Score="3" ViewCount="" Body="&lt;p&gt;Hint: We can rewrite this as (by multiplying and dividing by $(D-1)!$) as $\frac{{S-1}\choose {D-1}}{{T-1}\choose {D-1}}$.&lt;/p&gt;&#xA;" OwnerUserId="3431" LastActivityDate="2010-11-26T15:50:55.353" CommentCount="1" />
  <row Id="11967" PostTypeId="2" ParentId="11726" CreationDate="2010-11-26T16:05:30.503" Score="4" ViewCount="" Body="&lt;p&gt;Here's an argument that pushes the lower bound for $S(n)$ closer to a factor of $2$ times&#xA;$\log_2 n.$ More precisely, we show&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ S(n) \ge \left( 2 - \frac{1}{e} \right) \left( \lfloor \log_2 n \rfloor – 1 \right). $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $ a= \lfloor \log_2 n \rfloor $ and write&#xA;$ f(n,x) = 1 - \prod_{j=1}^{n-1} ( 1 – jx),$&#xD;&#xA;and so&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ S(n) = \sum_{k=1}^\infty \left( 1 - \prod_{j=1}^{n-1} \left( 1 - \frac{j}{2^k} \right) \right)$$&#xD;&#xA;$$ = \sum_{k=1}^\infty f(n, \frac{1}{2^k} ) = &#xD;&#xA;a-1 + \sum_{k=a}^\infty f(n, \frac{1}{2^k} ),$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;since for each $k \le a-1$ $\exists$ $j=2^k$ in $\prod_{j=1}^{n-1} ( 1 - j/2^k ),$ and thus this product is $0.$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So neglecting terms with $k \ge 2a-1,$ we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$S(n) \ge a-1 + \sum_{k=a}^{2a-2} f(n, \frac{1}{2^k} ). \qquad (1)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the AM-GM for $x \le 1/(m-1)$ we have for $m \ge 2$ and $2/m(m-1) \le x \le 1/(m-1)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(1-x)(1-2x) \cdots (1-(m-1)x) \le&#xD;&#xA;\left( 1 - \frac{mx}{2} \right)^{m-1} \le&#xD;&#xA;\left( 1 - \frac{1}{m-1} \right)^{m-1} &amp;lt; \frac{1}{e}. $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus for $k=a, a+1,\ldots, 2a-2$ we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ f(n, \frac{1}{2^k} ) \ge 1 - \frac{1}{e}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and substituting this into $(1)$ the result follows.&lt;/p&gt;&#xA;" OwnerUserId="1301" LastActivityDate="2010-11-26T16:05:30.503" CommentCount="1" />
  <row Id="11968" PostTypeId="1" AcceptedAnswerId="11972" CreationDate="2010-11-26T16:18:56.453" Score="4" ViewCount="107" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Can a finite group G have a cyclic Sylow 3-subgroup of order 9, such that the intersection of the Sylow 3-subgroups of G has order exactly 3, without having non-identity normal subgroups of order coprime to 3?&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In classifying the finite groups with cyclic Sylow 3-subgroups of order 9, it seems reasonable to split into cases based on the size of the 3-core (the intersection of the Sylow 3-subgroups) mod the 3'-core (the largest normal subgroup of order coprime to 3).  3-Cores of sizes 1 and 9 are easy to handle, but size 3 is devolving into a number of cases none of which seem to work, but for no systematic reason.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there some systematic reason it cannot occur, or is there an example I've overlooked?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; The original left out the important condition on normal subgroups of order coprime to 3, without which the classification is intractable.  Alex's answer shows how to use such normal subgroups to get fairly arbitrary behavior (for any type of Sylow).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit 2:&lt;/strong&gt; If the perfect residuum X of the group is nontrivial, it contains Ω(P) of the cyclic Sylow p-subgroup P, and so O&lt;sub&gt;p&lt;/sub&gt;(G) ≤ O&lt;sub&gt;p&lt;/sub&gt;(X).  In this case, O&lt;sub&gt;p&lt;/sub&gt;(X) = 1.  Hence the group is solvable, and Fit(G) = O&lt;sub&gt;p&lt;/sub&gt;(G), so G/Fit(G) ≤ Aut(p) = p−1 has the wrong order.&lt;/p&gt;&#xA;" OwnerUserId="583" LastEditorUserId="583" LastEditorDisplayName="" LastEditDate="2010-11-26T17:53:13.760" LastActivityDate="2010-11-26T17:53:13.760" Title="Cyclic sylow subgroup of order 9 with 3-core of order 3" Tags="&lt;group-theory&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="11969" PostTypeId="2" ParentId="11945" CreationDate="2010-11-26T16:21:11.883" Score="17" ViewCount="" Body="&lt;p&gt;For a proof of convergence,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Define the sequence as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle x_{0} = 0$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle x_{n+1} =\sqrt{7 + x_n}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that $\displaystyle x_n \geq 0 \ \ \forall n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice that $\displaystyle x^2 - x - 7 = (x-a)(x-b)$ where $\displaystyle a \lt 0$ and $\displaystyle b \gt 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We claim the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i) $\displaystyle x_n \lt b \Longrightarrow x_{n+1} \lt b$ &lt;br/&gt;&#xA;ii) $\displaystyle x_n \lt b \Longrightarrow x_{n+1} \gt x_n$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a proof of i)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle x_n \lt b = b^2 - 7$ and so $x_n +7  \lt b^2$ and thus by taking square roots $x_{n+1} \lt b$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a proof of ii) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle (x_{n+1})^2 - (x_n)^2 = -(x^2_n - x_n -7) = -(x_n-a)(x_n-b) \gt 0$ if $x_n \lt b$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus $\displaystyle \{x_{n}\}$ is monotonically increasing and bounded above and so is convergent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By setting $L = \sqrt{7+L}$, we can easily see that the limit is $\displaystyle b = \dfrac{1 + \sqrt{29}}{2}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;In fact, we can show that the convergence is linear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle \dfrac{b-x_{n+1}}{b-x_n} = \dfrac{b^2 - (7+x_n)}{(b+\sqrt{7+x_n})(b-x_n)} = \dfrac{1}{b + x_{n+1}}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus $\displaystyle \lim_{n\to \infty}  \dfrac{b-x_{n+1}}{b-x_n} = \dfrac{1}{2b}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can also show something a bit stronger:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $\displaystyle t_n = b - x_n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The we have shown above that $\displaystyle t_n \gt 0$ and $\displaystyle t_n \lt b^2$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle b - t_{n+1} = \sqrt{7 + b - t_n} = \sqrt{b^2 - t_n}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dividing by $\displaystyle b$ throughout we get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle 1 - \dfrac{t_{n+1}}{b} = \sqrt{1 - \dfrac{t_n}{b^2}}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using $\displaystyle 1 - \dfrac{x}{2} \gt \sqrt{1-x} \gt 1 - x \ \ 0 \lt x \lt 1$ we have that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle 1 - \dfrac{t_n}{2b^2} \geq 1 - \dfrac{t_{n+1}}{b} \geq 1 - \dfrac{t_n}{b^2}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And so &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle  \dfrac{t_n}{2b}  \leq t_{n+1} \leq \dfrac{t_n}{b}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This gives us that $\displaystyle  b - \dfrac{b}{b^n} \leq x_n \leq b - \dfrac{b}{(2b)^n}$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-11-27T00:03:34.820" LastActivityDate="2010-11-27T00:03:34.820" />
  <row Id="11970" PostTypeId="2" ParentId="11964" CreationDate="2010-11-26T16:28:43.027" Score="1" ViewCount="" Body="&lt;p&gt;I'm not sure I understood your question. Let's suppose your question is the following : &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $100=c&amp;lt;a&amp;lt;b&amp;lt;y$ be 4 given integers, how to find $x$ such that&#xA;$$&#xD;&#xA;a\le x \le b \text{ and }&#xD;&#xA;y \mod x \le c&#xD;&#xA;$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a tentative algorithm, which is probably not optimal, but (slightly) better that trying all the possible values for $x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, try the first possibility $x=a$ and perform an euclidean division to find $q$ and $r$ such that $y=qa+r$.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;if $0&amp;lt;r\le c$, $x=a$ and you're done&lt;/li&gt;&#xA;&lt;li&gt;otherwise, divide $r$ by $q$ to find $k,l$ such that $r=kq+l$.&#xA;&lt;ul&gt;&#xA;&lt;li&gt;if $0&amp;lt;l\le c$, $x=a+k$ is your solution if it is in $[a,b]$. If not, try $b$. If $b$ is not a solution either, then tere is no solution.&lt;/li&gt;&#xA;&lt;li&gt;otherwise, increase $a$ by $k+1$ and start again from the beginning.&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Of course, you stop repeating the above as soon as $a$ becomes bigger than $b$, which would mean that ther is no solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit to take into account the $0 &amp;lt; y \mod c$ condition&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't think this condition change the search much, at least if you use the above algorithm. Hoxever, if you have a $z$ such that $y \mod z =0$, you can write $y=mz$, which might be helpful. For example, if $m\le c$, $x=z+1$ will be a solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit to correct for the $a+k&amp;gt;b$ case&lt;/strong&gt;&#xA;  &lt;strong&gt;Edit&lt;/strong&gt;&#xA;To be more clear, the algorithm above find $k,q,l$ such that $y=(a+k)q+l$, and tries to get $l$ as small as possible (without trying too much)&lt;/p&gt;&#xA;" OwnerUserId="3918" LastEditorUserId="3918" LastEditorDisplayName="" LastEditDate="2010-11-29T10:45:22.450" LastActivityDate="2010-11-29T10:45:22.450" CommentCount="10" />
  <row Id="11971" PostTypeId="1" AcceptedAnswerId="11976" CreationDate="2010-11-26T16:40:07.103" Score="3" ViewCount="271" Body="&lt;p&gt;I am learning group theory, and while learning automorphisms, I came across conjugation as an example in many textbooks. Though the definition itself, (and when considering the case of abelian groups), it seems pretty innocent, I have to admit that I have no intuition about whats happening, or why even such a map is important. Did anyone feel this way when you learnt it? Any examples/references that you have found useful when you learnt it?&lt;/p&gt;&#xA;" OwnerUserId="1441" LastActivityDate="2011-04-16T17:14:46.527" Title="Intuition behind conjugation in group theory" Tags="&lt;group-theory&gt;" AnswerCount="5" CommentCount="1" FavoriteCount="1" />
  <row Id="11972" PostTypeId="2" ParentId="11968" CreationDate="2010-11-26T16:52:22.483" Score="3" ViewCount="" Body="&lt;p&gt;Such groups are easy to construct: let $C_9$ act on something through a $C_3$-quotient. E.g. construct the semi-direct product $C_7\rtimes C_9$ such that $C_9$ acts on $C_7$ through the unique automorphism of order 3. Then, the Sylow 3-subgroups are not normal, so there is more than one of them, but there is a unique $C_3$, which is the centre of $G$.&lt;/p&gt;&#xA;" OwnerUserId="3212" LastActivityDate="2010-11-26T16:52:22.483" CommentCount="3" />
  <row Id="11973" PostTypeId="2" ParentId="11971" CreationDate="2010-11-26T16:57:07.053" Score="1" ViewCount="" Body="&lt;p&gt;My intuition goes back to permutations.  Namely, conjugating everything by some group element can be identified with some permutation of the group.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the importance of conjugation, one easy answer is that it gives insight into the commutative portions of the group.  If $x$ is closed under conjugation by  $g$, that is $x = gxg^{-1}$, then we have $xg = gx$, so the elements commute.  The set of elements that are closed under conjugation by all group elements forms the &lt;em&gt;center subgroup&lt;/em&gt;, while any subgroup that is closed under conjugation by all group elements is a &lt;em&gt;normal subgroup&lt;/em&gt;.  These are fundamental to understanding the structure of a given group.&lt;/p&gt;&#xA;" OwnerUserId="3537" LastActivityDate="2010-11-26T16:57:07.053" CommentCount="1" />
  <row Id="11974" PostTypeId="2" ParentId="11971" CreationDate="2010-11-26T16:59:04.650" Score="2" ViewCount="" Body="&lt;p&gt;There are several reasons:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Conjugation is a way of measuring, how commutative or otherwise your group is. Namely $aba^{-1} = b\Leftrightarrow ab=ba$. Since abelian groups are much easier to understand than arbitrary ones, this measure is important.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Conjugation is more or less the only known way of obtaining automorphisms of any given group without knowing anything further about the group (of course, these automorphisms can all turn out to be trivial if your group is abelian).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;You have to bear in mind that normal subgroups and the action of conjugation were first introduced by Galois, because for a subgroup to be normal was exactly the criterion that he needed for an automorphism of a field to give an automorphism of a subfield upon restriction. This comment might be slightly over your head if you are only learning groups at the moment, but bear in mind that historically, the notion of normal subgroups arose in the context of interpreting groups as symmetries of field extensions, when Galois developed a theory to decide whether a given polynomial was solvable by using elementary operations of arithmetic and roots.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="3212" LastActivityDate="2010-11-26T16:59:04.650" />
  <row Id="11975" PostTypeId="2" ParentId="11669" CreationDate="2010-11-26T17:07:45.710" Score="1" ViewCount="" Body="&lt;p&gt;A somewhat grapical representation of what Carlos Ribeiro was talking about.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;If you take any note in the C major scale, you can treat that note as the start of another scale. Take for instance the fifth of C (it's the G), and build a new major scale, now starting from G instead of C. You'll get another seven notes. Some of them are also on teh scale of C; others are very close, but not exactly equal; and some fall in the middle of the notes in the scale of C. &quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note the semi-tone interval EF and BC on the C scale.&#xA;When trying to reproduce the same scale starting at D, we run into a problem.&#xA;Alphabetically, the third note should be a F, but F is a semitone too low for that spot. In order to maintain the same sounding scale, we need to introduce a NEW note, called F#.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;C - D - E F - G - A - B C  (C scale)&lt;/li&gt;&#xA;&lt;li&gt;D - E - F#G - A - B - C#D  (D scale)&lt;/li&gt;&#xA;&lt;li&gt;E - F#-G#A - B - C#- D#E  (E scale)&lt;/li&gt;&#xA;&lt;li&gt;F - G - AA#- C - D - E F  (F scale)  &lt;/li&gt;&#xA;&lt;li&gt;etc.&lt;br&gt;&#xA;Note that in actual writing of music the A-A# would be written as A-Bb so that the 'A' line of the staff wouldn't be ambiguous.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" LastActivityDate="2010-11-26T17:07:45.710" />
  <row Id="11976" PostTypeId="2" ParentId="11971" CreationDate="2010-11-26T17:12:52.693" Score="10" ViewCount="" Body="&lt;p&gt;One way to think about conjugation is as a generalization of changing coordinates when rewriting matrices or, from a physical point of view, as the process of &quot;seeing a group from a different perspective.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's make this precise.  Let $G$ be a group and let $X$ be a set on which $G$ acts faithfully.  For example, if $G$ is the Euclidean group of isometries of the plane, $X$ could be... well, the plane.  Let's say we have a reasonably concrete description of what a particular element $g \in G$ does to $X$.  For example, if $G, X$ are as above, then perhaps $g$ is a rotation counterclockwise by $\theta$ centered at the origin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given that description, what does the element $hgh^{-1}$ do?  Well, this is actually quite simple: in the description of $g$, replace all the elements $x \in X$ that occur by the corresponding elements $hx \in X$.  For example, if $h$ is a translation by a vector $v$, then $hgh^{-1}$ is a rotation counterclockwise by $\theta$ centered at $v$ instead of the origin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, what conjugation does in terms of group actions (and it is always a good idea to think of groups in terms of their actions) is it corresponds to a &quot;change of coordinates&quot; on the underlying set $X$.  This is a basic reason conjugation and group theory are important in understanding Newtonian mechanics (where $G$ is the Galilean group) or relativity (where $G$ is the Lorentz group) or really much of modern physics in general: in physics groups are always studied because of their actions and we always want our concepts and equations to be invariant under these actions.  (For example, mass and charge are invariant concepts.  The mass or charge of an object doesn't change when you rotate or translate it.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This gives a very intuitive definition of a normal subgroup: it's a subgroup that &quot;looks the same from every perspective.&quot;  For example, the subgroup of translations in the Euclidean group is always normal because the description &quot;$g$ is a translation&quot; is the same from every perspective (that is, it's invariant under conjugation).  It's a good exercise to look at some groups you're familiar with and see if you can identify which subgroups are normal based on this principle.  (This principle underlies the importance of normal subgroups in Galois theory as well.)&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-11-26T17:18:41.670" LastActivityDate="2010-11-26T17:18:41.670" CommentCount="4" />
  <row Id="11977" PostTypeId="2" ParentId="11971" CreationDate="2010-11-26T17:13:43.613" Score="3" ViewCount="" Body="&lt;p&gt;One example of conjugation you probably already know from linear algebra is &lt;strong&gt;change of base&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The linear transformation A that takes (x,y) to (3x−y,2x) is reasonably nice and can be represented by the matrix $$A = \begin{pmatrix}3&amp;amp;2\\-1&amp;amp;0\end{pmatrix}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the basis u=2x−y, v=x−y, Au = 2Ax − Ay = 2(3x−y) − 2x = 4x − 2y = 2u and Av = Ax−Ay = (3x−y) - 2x = x−y = v.  So in this basis, the linear transformation is incredibly nice and can be represented by the matrix $$\tilde A = \begin{pmatrix}2&amp;amp;0\\0&amp;amp;1\end{pmatrix}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we take the matrix $$B = \begin{pmatrix} 2&amp;amp;1\\-1&amp;amp;-1\end{pmatrix}$$&#xD;&#xA;then $B^{-1} A B = \tilde A$ expresses the matrix in the new basis defined by B, (u,v) = (2x-1y,1x-1y).  Note that Bx = u, and By = v.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In some sense this is what conjugation does in general: it changes the basis on which your group acts.  If you conjugate by g, it changes the way G acts by converting the old basis to the a new basis already acted on by g.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A good example to work through to see this conjugation in a new context is with &lt;strong&gt;permutations&lt;/strong&gt;.  If a permutation A takes 1,2,3 to 3,2,1 and the permutation B takes 1,2,3 to 1,3,2, then the conjugated element $B^{-1} A B$ takes 1,3,2 to 2,3,1.  In other words, instead of acting on 1,2,3 it acts on 1,3,2.  In cycle notation, you can write this as A=(1,3) and B=(2,3) and then A^B = (1,2).  The &quot;3&quot; in (1,3) has been replaced by &quot;2&quot;, since B replaces 3s by 2s.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Try to think of groups as acting on something.  Either let them be matrices moving vectors around, or permutations moving arbitrary things around.  Conjugation just lets you &lt;strong&gt;relabel&lt;/strong&gt; the things you are moving around.&lt;/p&gt;&#xA;" OwnerUserId="583" LastActivityDate="2010-11-26T17:13:43.613" />
  <row Id="11978" PostTypeId="2" ParentId="11958" CreationDate="2010-11-26T17:26:53.003" Score="1" ViewCount="" Body="&lt;p&gt;We want to find $p,0\le p\le 1,$ such that $${20\choose 15}p^{15}(1-p)^5$$ is maximum.&#xA;By simple calculus, this $p$ turns out to be 15/20, as expected.&lt;/p&gt;&#xA;" OwnerUserId="3249" LastActivityDate="2010-11-26T17:26:53.003" />
  <row Id="11979" PostTypeId="1" CreationDate="2010-11-26T17:30:09.023" Score="-3" ViewCount="271" Body="&lt;p&gt;How many moves can a horse make on a chessboard?&lt;/p&gt;&#xA;" OwnerUserId="2779" LastEditorUserId="2779" LastEditorDisplayName="" LastEditDate="2010-11-26T17:50:38.587" LastActivityDate="2010-11-27T00:12:39.823" Title="How many moves can a horse make on a chessboard?" Tags="&lt;homework&gt;&lt;graph-theory&gt;" AnswerCount="2" CommentCount="10" FavoriteCount="0" />
  <row Id="11980" PostTypeId="1" AcceptedAnswerId="11984" CreationDate="2010-11-26T17:39:46.090" Score="2" ViewCount="142" Body="&lt;p&gt;In other words, if X = Y in distribution, is it true that EX = EY?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think this must be true, but I've tried to prove it a few times and I always get stuck.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance for any hints or reference.&lt;/p&gt;&#xA;" LastActivityDate="2010-11-26T23:10:20.193" Title="Does equality in distribution imply equality of expected value?" Tags="&lt;probability&gt;" AnswerCount="1" />
  <row Id="11981" PostTypeId="2" ParentId="11979" CreationDate="2010-11-26T17:42:11.340" Score="-1" ViewCount="" Body="&lt;p&gt;That piece can move in an L-shape. So potential, it can move in 1 of 4 directions (unless it's restricted by one edges of the board).&lt;/p&gt;&#xA;" OwnerUserId="4007" LastActivityDate="2010-11-26T17:42:11.340" CommentCount="2" />
  <row Id="11983" PostTypeId="1" CreationDate="2010-11-26T17:44:47.037" Score="3" ViewCount="208" Body="&lt;p&gt;How would one go about showing how many solutions the following congruence has?&#xA;$$x^2 + y^2 \equiv 23 \pmod{93}.$$&lt;/p&gt;&#xA;" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-26T19:16:00.140" LastActivityDate="2010-11-26T22:22:32.653" Title="Quadratic Congruence and Sum of Two Squares" Tags="&lt;number-theory&gt;&lt;number&gt;&lt;quadratic-reciprocity&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="3" />
  <row Id="11984" PostTypeId="2" ParentId="11980" CreationDate="2010-11-26T17:45:13.637" Score="1" ViewCount="" Body="&lt;p&gt;Yes, this is true.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The expectation value depends only on the measure $P^X$ on $\mathbb R$ induced by the random variable $X$, and $X$ and $Y$ are, by definition, equal in distribution if and only if $P^X=P^Y$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can find this here: &lt;a href=&quot;http://en.wikipedia.org/wiki/Expectation_value&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Expectation_value&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="367" LastEditorUserId="367" LastEditorDisplayName="" LastEditDate="2010-11-26T23:10:20.193" LastActivityDate="2010-11-26T23:10:20.193" CommentCount="1" />
  <row Id="11985" PostTypeId="2" ParentId="11983" CreationDate="2010-11-26T18:16:01.507" Score="0" ViewCount="" Body="&lt;p&gt;This is a start:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You would like to know for which y it holds that 23 - y^2 is a square modulo 93. As 93 is 31*3, you can find these y by finding the squares of the form 23 - y^2 modulo 3 and modulo 31. Can you go on from there? &lt;/p&gt;&#xA;" LastActivityDate="2010-11-26T18:16:01.507" CommentCount="3" />
  <row Id="11986" PostTypeId="1" CreationDate="2010-11-26T19:17:07.363" Score="2" ViewCount="178" Body="&lt;p&gt;My homework question:  Show that all zeros of $p(z)=z^4 + 6z + 3$ lie in the circle of radius $2$ centered at the origin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know $p(z)$ has a zero-count of $4$ by using the  Fundamental Theorem of Algebra. Then using the Local Representation Theorem the $\int \frac{n}{z+a} = 4(2 \pi i)$. I am assuming $a=0$ since we are centered at the origin. I apologize for my lack of math-type.  What does $= 8 \pi i$ mean?  Am I going around the unit circle $4$ times?  Or is it even relevant to my final answer.  Which I am assuming is finding the coordinates to the $4$ singularities. I have always looked for my singularities in the values that make the denominator zero, but in this question my denominator is $z$.  $z=0$ doesn't seem right.  So the question is, am I suppose to factor the polynomial $z^4 + 6z + 3$ to find the zeros?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2010-11-26T20:39:59.130" LastActivityDate="2010-11-26T20:39:59.130" Title="Guidance on a Complex Analysis question" Tags="&lt;homework&gt;&lt;complex-analysis&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" />
  <row Id="11987" PostTypeId="2" ParentId="11986" CreationDate="2010-11-26T19:25:15.167" Score="4" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Hint:&lt;/strong&gt; This kind of questions are usually handled using &lt;a href=&quot;http://en.wikipedia.org/wiki/Rouch%C3%A9%27s_theorem&quot; rel=&quot;nofollow&quot;&gt;Rouche's Theorem&lt;/a&gt;. I suggest you look it up in the wikipedia article, where you can see an example of its usage. Also here's an &lt;a href=&quot;http://books.google.co.cr/books?id=wDsrQTPUaSsC&amp;amp;pg=PA77&amp;amp;dq=rouche%27s+theorem&amp;amp;hl=es&amp;amp;ei=YAvwTNmTOIX7lwej4JH9DA&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=2&amp;amp;ved=0CCoQ6AEwAQ#v=onepage&amp;amp;q=rouche%27s%20theorem&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;example&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The key is choosing wisely another function $f(z)$ with which to compare in the inequality in Rouche's theorem and such that you can easily decide how many zeroes does $f(z)$ have inside the region you are considering, which in your case is the circle $|z| &amp;lt; 2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;About your other question, you don't need to factor the polynomial in order to answer this.&lt;/p&gt;&#xA;" OwnerUserId="900" LastEditorUserId="900" LastEditorDisplayName="" LastEditDate="2010-11-26T20:06:27.580" LastActivityDate="2010-11-26T20:06:27.580" />
  <row Id="11988" PostTypeId="2" ParentId="11983" CreationDate="2010-11-26T19:26:11.877" Score="4" ViewCount="" Body="&lt;p&gt;If you are working modulo a prime $p$, then you can obtain the result using the Legendre symbol or by simple trial and error if it is small. For instance, working modulo $3$, as CJost suggests, you are trying to solve $x^2 + y^2 \equiv 23 \equiv 2 \pmod{3}$. then you want to solve $x^2 \equiv 2-y^2\pmod{3}$. Since $y^2$ can only be congruent to $0$ or to $1$ modulo $3$, you only have solutions when $x^2\equiv y^2\equiv 1\pmod{3}$. That gives you four solutions modulo $3$: $(1,1)$, $(1,-1)$, $(-1,1)$, and $(-1,-1)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You need to do something similar for $31$: for each possibly value of $y^2$, decide if $23-y^2$ is a square modulo $31$; if it is, that gives you two possible values for $x$ and two possible values for $y$ (unless one of the two needs to be $0$ modulo $23$). But remember that you are working &lt;em&gt;modulo&lt;/em&gt; $31$; for example, with $y=5$, there is no problem with the fact that $23-y^2 = -2$, because you are trying to solve $x^2\equiv -2\pmod{31}$, and this is the same as $x^2\equiv 29 \pmod{31}$; the negative number is not a problem. And of course, to decide if $x^2\equiv 29\pmod{31}$ has a solution, you can use the &lt;a href=&quot;http://en.wikipedia.org/wiki/Legendre_symbol&quot; rel=&quot;nofollow&quot;&gt;Legendre symbol&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/Quadratic_reciprocity&quot; rel=&quot;nofollow&quot;&gt;quadratic reciprocity&lt;/a&gt; to compute it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Added:&lt;/em&gt; As it happens, $23$ is &lt;em&gt;not&lt;/em&gt; a square modulo $31$, so any solutions to $x^2+y^2\equiv 23\pmod{31}$ will necessarily have both $x$ and $y$ not congruent to zero modulo $31$; so you don't need to worry about that when you are counting solutions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, suppose you find a solution $x=a$ and $y=b$ modulo $31$. How do you get solutions modulo $93$? You take one solution modulo $3$, say $(1,1)$, and the solution modulo $31$, say $(a,b)$, and you use the Chinese Remainder Theorem: you can find an integer $x$ such that $x\equiv 1 \pmod{3}$ and $x\equiv a\pmod{31}$; then find an integer $y$ such that $y\equiv 1\pmod{3}$ and $y\equiv b\pmod{31}$. Then the pair $(x,y)$ will be a solution modulo $93$ that corresponds to the solution $(1,1)$ modulo $3$ and the solution $(a,b)$ modulo $31$. So each choice of a solution modulo $3$ and a solution modulo $31$ will give you a solution modulo $93$.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-26T22:22:32.653" LastActivityDate="2010-11-26T22:22:32.653" />
  <row Id="11989" PostTypeId="2" ParentId="11592" CreationDate="2010-11-26T20:28:41.563" Score="2" ViewCount="" Body="&lt;p&gt;With help from Maple, I got &#xA;$$\left(\frac{ax+by+cz}{x-y}\right)^2+\left(\frac{ay+bz+cx}{y-z}\right)^2+\left(\frac{az+bx+cy}{z-x}\right)^2-(c-a)^2-(c-b)^2$$&#xD;&#xA;equal to&#xA;$$\frac{(c(x^3+y^3+z^3)+(a-c)(x^2y+y^2z+z^2x)+(b-c)(x^2z+y^2x+z^2y)-3(a+b-c)xyz)^2}{(x-y)^2(y-z)^2(x-z)^2}$$ which of course is $\ge 0$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But with no help from a computer algebra, how would one prove this? &lt;/p&gt;&#xA;" OwnerUserId="3249" LastActivityDate="2010-11-26T20:28:41.563" />
  <row Id="11991" PostTypeId="1" AcceptedAnswerId="11993" CreationDate="2010-11-26T20:57:09.110" Score="2" ViewCount="194" Body="&lt;p&gt;Are &lt;a href=&quot;http://en.wikipedia.org/wiki/Order_theory&quot; rel=&quot;nofollow&quot;&gt;order&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/Binary_relation&quot; rel=&quot;nofollow&quot;&gt;binary relation&lt;/a&gt; the same thing, or order is just some special binary relation? If it is the latter, what kinds of properties distinguish an order from a general binary relation?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks and regards!&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Sorry for the confusion already caused. What I thought was that a partial order is just some special order, and there are other orders that are not partial orders, such as &lt;a href=&quot;http://en.wikipedia.org/wiki/Preorder&quot; rel=&quot;nofollow&quot;&gt;preorder&lt;/a&gt;, &lt;a href=&quot;http://en.wikipedia.org/wiki/Directed_set&quot; rel=&quot;nofollow&quot;&gt;the order on a directed set&lt;/a&gt;. So I was wondering how an order was defined then, and I saw there was no definition for an order in the Wikipedia page for order theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe  I was not lazy or could not read, what I understood was different from yours, and I did not realize that when I first posted my question. So sorry about that!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the end, if possible, I would like to know if there is a concept that can include partial order, preorder and the &quot;order&quot; on a directed set, but more special than a binary relation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks! &lt;/p&gt;&#xA;" OwnerUserId="1281" LastEditorUserId="1281" LastEditorDisplayName="" LastEditDate="2010-11-26T22:33:24.433" LastActivityDate="2010-11-26T22:33:24.433" Title="order and binary relation" Tags="&lt;elementary-set-theory&gt;&lt;order-theory&gt;" AnswerCount="1" CommentCount="8" />
  <row Id="11992" PostTypeId="2" ParentId="11429" CreationDate="2010-11-26T21:00:44.720" Score="0" ViewCount="" Body="&lt;p&gt;I am not sure if this is a homework. So I will give you the answers and you figure out yourself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $i=0.09, v=(1+i)^{-1}=0.917431192$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Answer for (a): $$50000\frac{1-v^7}{1-v^{10}}$$&#xD;&#xA;Answer for (b): $$50000\times 0.09\times\frac{1-v^8}{1-v^{10}}$$&lt;/p&gt;&#xA;" OwnerUserId="3249" LastActivityDate="2010-11-26T21:00:44.720" />
  <row Id="11993" PostTypeId="2" ParentId="11991" CreationDate="2010-11-26T21:01:56.217" Score="5" ViewCount="" Body="&lt;p&gt;Orders are a special kind of binary relation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A binary relation $R$ on a set $A$ is just a collection of ordered pairs $R\subseteq A\times A$, with absolutely no conditions on $R$. So long as it is a subset, it is a binary relation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A (partial) order on $A$, by contrast, is a subset $P\subseteq A\times A$ which satisfies three conditions:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The relation is &lt;strong&gt;reflexive&lt;/strong&gt;: For every $a\in A$, $(a,a)\in P$.&lt;/li&gt;&#xA;&lt;li&gt;The relation is &lt;strong&gt;anti-symmetric&lt;/strong&gt;: for every $a,b\in A$, if $(a,b)\in P$ and $(b,a)\in P$, then $a=b$.&lt;/li&gt;&#xA;&lt;li&gt;The relation is &lt;strong&gt;transitive&lt;/strong&gt;: for every $a,b,c\in A$, if $(a,b)\in P$ and $(b,c)\in P$, then $(a,c)\in P$. &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The partial order is a &lt;strong&gt;total&lt;/strong&gt; or linear order if, in addition, for every $a,b\in A$, either $(a,b)\in P$ or $(b,a)\in P$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So an order on $A$ is a special kind of binary relation, because it must satisfy the three properties above, whereas a &quot;general binary relation&quot; doesn't have to satisfy anything other than being a collection of ordered pairs of elements of the set.&lt;/p&gt;&#xA;&#xA;&lt;hr/&gt;&#xA;&#xA;&lt;p&gt;Okay, addressing your additions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most basic notion is &quot;partial order&quot;. Every other kind of order is a &quot;partial order with extra conditions&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The one exception is a &lt;strong&gt;pre-order&lt;/strong&gt;, because, as the name indicates, a pre-order is something which is not yet an order (again, &quot;order&quot; is just a short way of saying &quot;partial order&quot;), but may become one (when it grows up, so to speak).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Pre-orders&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A &lt;strong&gt;pre-order&lt;/strong&gt; is a binary relation which is reflexive and transitive, but is not necessarily anti-symmetric. An example of this is the relation &quot;divides&quot; in the integers: define $a\preceq b$, with $a$ and $b$ integers, if and only if $a$ divides $b$. This is reflexive (every integer divides itself), and transitive (if $a$ divides $b$ and $b$ divides $c$, then $a$ divides $c$), but it is not anti-symmetric, because if $a$ divides $b$ and $b$ divides $a$, the best you can conclude is that $|a|=|b|$; for instance, $2$ and $-2$ are &lt;em&gt;different&lt;/em&gt;, but $2\preceq -2$ and $-2\preceq 2$. So this relation is &lt;em&gt;not&lt;/em&gt; a partial order because it lacks anti-symmetry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One way to solve this lack is to use the relation to define something which &lt;em&gt;is&lt;/em&gt; a partial order on a closely related set. We define an &lt;a href=&quot;http://en.wikipedia.org/wiki/Equivalence_relation&quot; rel=&quot;nofollow&quot;&gt;equivalence relation&lt;/a&gt; on the integers by saying $a\sim b$ if and only if $a\preceq b$ and $b\preceq a$ (in this case, if and only if $|a|=|b|$). Then we can consider the &lt;a href=&quot;http://en.wikipedia.org/wiki/Equivalence_relation#Quotient_set&quot; rel=&quot;nofollow&quot;&gt;quotient set&lt;/a&gt;, and define the relation $\leq$ on the quotient set by $[a]\leq [b]$ if and only if $a\preceq b$ (here, $[a]$ is the equivalence class of $a$ under $\sim$, etc). Then one can show that this is a partial order on the quotient set; it is patently closely related to the original pre-order $\preceq$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's why relations that are reflexive and transitive are called &lt;strong&gt;pre-orders&lt;/strong&gt;: you only need one further step to get to a partial order, though not usually on the set you started with.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Other kinds of orders&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The basic notion, as I said above, is &quot;partial order&quot;. There are lots of extra conditions one can put on a partially ordered set to make it a &quot;nicer&quot; kind of ordered set. These extra conditions define special kinds of orders. A random sampling off the top of my head (I'm using $\leq$ to denote whatever the partial order relation in question is; $a\leq b$ means $(a,b)$ is in the relation that defines the order):&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Total&lt;/strong&gt; or &lt;strong&gt;linear orders&lt;/strong&gt;. In a partial order, it is possible for two elements to not be comparable at all, that is, you have neither $a\leq b$ nor $b\leq a$. For example, if you take a set $X$, you can define a partial order on $\mathcal{P}(X)$, the power set of $X$ (the set of all subsets of $X$) by saying that $A\leq B$ if and only if $A\subseteq B$ (in fact, this is in a sense the most basic kind of &quot;partial order&quot;). If $X$ has more than one element, though, then you have elements of $\mathcal{P}(X)$ that cannot be compared: take $x,y\in X$, $x\neq y$; then $\{x\}\not\leq \{y\}$ and $\{y\}\not\leq\{x\}$. This in contrast to the order relations we are used to in the natural numbers, real numbers, etc., where any two elements can be compared: one is bigger than the other. So we add a condition: the partially ordered set $A$ is &lt;strong&gt;totally ordered&lt;/strong&gt; or &lt;strong&gt;linearly ordered&lt;/strong&gt; if the partial order &lt;em&gt;also&lt;/em&gt; satisfies the condition that for every $a,b\in A$, either $a\leq b$ or $b\leq a$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Well order.&lt;/strong&gt; This is a very strong generalization of total orders. In a total order, any subset with finitely many elements has a smallest element. A partially ordered set $A$ is said to be &lt;strong&gt;well-ordered&lt;/strong&gt; if &lt;em&gt;any&lt;/em&gt; nonempty subset $B$ of $A$ has a smallest element. For instance, the natural numbers with their usual order is well-ordered, but the real numbers with their usual order are not (the positive reals have no least element). A well-order must be a total order, but the converse is not true in general.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Lattice order.&lt;/strong&gt; A lattice order is a partial order where, even though it is not true that any two elements are comparable, it is nevertheless the case that any two elements have a &lt;em&gt;least upper bound&lt;/em&gt; (given $a,b\in A$, there exists a $c\in A$ such that $a\leq c$, $b\leq c$, and for every $d\in A$, if $a\leq d$ and $b\leq d$, then $c\leq d$), and a &lt;em&gt;greatest lower bound&lt;/em&gt; (the dual concept; just reverse all the inequalities in the definition above). The typical example is going back to the partially ordered set of subsets of a given $X$ under inclusion; the least upper bound of $A$ and $B$ is $A\cup B$, and the greatest lower bound is $A\cap B$. You can weaken the definition to require only least upper bounds (&lt;em&gt;upper semi-lattice&lt;/em&gt;) or just greatest upper bounds (&lt;em&gt;lower semi-lattices&lt;/em&gt;). Or you can strengthen it to require least upper bound and/or greatest lower bounds for any nonempty subset (&lt;em&gt;complete lattices&lt;/em&gt;); you can also require the existence of a smallest element and/or a greatest elements, or the existence of &quot;complements&quot; (if $0$ is the smallest element and $1$ is the greatest element, then a complement of $a\in A$ is an element $b\in A$ such that the least upper bound of $\{a,b\}$ is $1$ and the greatest lower bound of $\{a,b\}$ is $0$; in the case of $\mathcal{P}(X)$, the complement is the usual set-complement relative to $X$).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Directed sets.&lt;/strong&gt; This is a kind of compromise between partially ordered sets and linear sets, in one direction: a partially ordered set $A$ is &lt;strong&gt;directed&lt;/strong&gt; if and only if for every $a,b\in A$ there exists a $c\in A$ such that $a\leq c$ and $b\leq c$ (any two elements have a common upper bound). the dual notion, where any two elements have a common lower bound, is called an &lt;strong&gt;inversely directed sets.&lt;/strong&gt; They play an important role in many areas of mathematics to construct special kinds of limits; when the index set is a directed system, you get something called a &lt;em&gt;directed limit&lt;/em&gt;; when the index set is an inversely directed set, you get something called an &quot;inverse limit&quot;. The &lt;a href=&quot;http://en.wikipedia.org/wiki/P-adic_number&quot; rel=&quot;nofollow&quot;&gt;$p$-adic integers&lt;/a&gt; are an example of an inverse limit.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Note:&lt;/em&gt; These are just a few of the ones I know, and they are likely a very limited sample of the ones that people who actually work with orders know. I don't do research anywhere near these topics, I just use orders all the time, like most mathematicians.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Strict orders&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In addition, there is a &quot;sister&quot; notion to partial orders, called &quot;strict orders&quot;. A binary relation $\prec$ is a &lt;em&gt;strict order&lt;/em&gt; if and only if it is:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Anti-reflexive: for every $a\in A$, $a\not\prec a$.&lt;/li&gt;&#xA;&lt;li&gt;Asymmetric: for all $a,b\in A$, if $a\prec b$, then $b\not\prec a$. &lt;/li&gt;&#xA;&lt;li&gt;Transitive: for every $a,b,c\in A$, if $a\prec b$ and $b\prec c$, then $a\prec c$. &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;However, strict and partial orders are actually closely connected. If you have a partial order $\preceq$, then you can define a strict order $\prec$ by &#xA;$$a\prec b \Longleftrightarrow a\preceq b\text{ and }a\neq b.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conversely, if you have a strict order $\prec$, then you can use it to define a partial order $\preceq$ by&#xA;$$a\preceq b \Longleftrightarrow a\prec b\text{ or } a=b.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you can take either notion as your &quot;original&quot; notion, and define the other one in terms of it. These days, it is more common to take &quot;partial order&quot; as the original, and define strict order in terms of it. &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-26T22:00:31.167" LastActivityDate="2010-11-26T22:00:31.167" CommentCount="13" />
  <row Id="11994" PostTypeId="2" ParentId="250" CreationDate="2010-11-26T21:43:39.813" Score="3" ViewCount="" Body="&lt;p&gt;The Fold-and-Cut Theorem is pretty unintuitive. &lt;a href=&quot;http://erikdemaine.org/foldcut/&quot; rel=&quot;nofollow&quot;&gt;http://erikdemaine.org/foldcut/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="677" LastActivityDate="2010-11-26T21:43:39.813" CommunityOwnedDate="2010-11-26T21:43:39.813" />
  <row Id="11995" PostTypeId="1" AcceptedAnswerId="11996" CreationDate="2010-11-26T22:48:19.783" Score="2" ViewCount="284" Body="&lt;p&gt;Hi, first question!&#xA;Equations it's on Latex Format, I'm still trying to understand how MathJax works.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given the following integral:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\int_0^{+\infty } \frac{1}{x \sqrt{x}} \, dx$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm pretty sure that does not converge.&#xA;And if the integral was a indefinite one, also I'm almost sure the result is  $\ln(\sqrt x)$ using substitution.&#xA;If I am correct, why wolfram alpha says that the result is $- 2/ \sqrt x$ instead of $\ln(\sqrt x)$ ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://mathway.com/&quot; rel=&quot;nofollow&quot;&gt;MathWay&lt;/a&gt; shows the correct results for both definite and indefinite integrals.&#xA;I'm new to any mathematical soft like wolfram alfa /mathematica, and my idea is to test my pen and paper results with a software that checks my results.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any hints will be greatly appreciated.&#xA;Thanks&lt;/p&gt;&#xA;" OwnerUserId="4122" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2010-11-26T23:25:09.550" LastActivityDate="2010-11-27T00:58:05.090" Title="Comparing integral resolutions using Wolfram Alpha / Mathematica" Tags="&lt;calculus&gt;&lt;math-software&gt;&lt;mathematica&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="11996" PostTypeId="2" ParentId="11995" CreationDate="2010-11-26T22:56:22.283" Score="3" ViewCount="" Body="&lt;p&gt;The indefinite integral is $$\int \frac{1}{x\sqrt{x}} \, dx = \int \frac{1}{x^{^{3/2}}}\, dx = \frac{-2}{\sqrt{x}} + C$$ I suppose you made a mistake when integrating it by making a wrong substitution $u = \sqrt{x}$. Probably that's how you got the $\ln{\sqrt{x}}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;About the convergence issue, you're right, the integral is divergent as you can see by taking the corresponding limit at $0$ of the indefinite integral $\frac{-2}{\sqrt{x}}$.&lt;/p&gt;&#xA;" OwnerUserId="900" LastEditorUserId="900" LastEditorDisplayName="" LastEditDate="2010-11-26T23:03:15.240" LastActivityDate="2010-11-26T23:03:15.240" CommentCount="3" />
  <row Id="11997" PostTypeId="1" AcceptedAnswerId="11999" CreationDate="2010-11-26T23:20:10.337" Score="0" ViewCount="85" Body="&lt;p&gt;Given: $f(px+(1-p)y)\le pf(x)+(1-p)f(y)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose $f(x)$ is defined on the interval I. If $x_{1}&amp;lt;x_{2}&amp;lt;x_{3}$ are in I and $f(x_{1})&amp;lt;f(x_{2})$ and $f(x_{3})&amp;lt;f(x_{2})$, then show that $f(x)$ is not convex on I.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Not sure where to begin on this one&lt;/p&gt;&#xA;" OwnerUserId="3961" LastActivityDate="2010-11-26T23:37:40.517" Title="convex with generic function" Tags="&lt;convex-analysis&gt;" AnswerCount="2" />
  <row Id="11999" PostTypeId="2" ParentId="11997" CreationDate="2010-11-26T23:27:54.853" Score="1" ViewCount="" Body="&lt;p&gt;First express $x_2$ as a weighted linear combination of $x_1$ and $x_3$ i.e. $x_2 = tx_1 + (1-t)x_3$ such that $0 &amp;lt; t &amp;lt;1$. This is always possible since $x_2$ lies on the line segment connecting $x_1$ and $x_3$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now if $f(x)$ is convex in the interval $[x_1,x_2]$, then $f(x_2) = f(tx_1 + (1-t)x_3) &amp;lt; tf(x_1) + (1-t)f(x_3)$. But we are given $f(x_1) &amp;lt; f(x_2)$ and $f(x_3) &amp;lt; f(x_2)$. And hence $tf(x_1) + (1-t)f(x_3) &amp;lt; t f(x_2) + (1-t) f(x_2) = f(x_2)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, we get $tf(x_1) + (1-t)f(x_3) &amp;lt; f(x_2)$ which contradicts the fact that $f$ is convex on the interval.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastActivityDate="2010-11-26T23:27:54.853" />
  <row Id="12000" PostTypeId="2" ParentId="11997" CreationDate="2010-11-26T23:29:45.743" Score="0" ViewCount="" Body="&lt;p&gt;What you say is &quot;given&quot; is actually the definition of convex, which you want to show does &lt;em&gt;not&lt;/em&gt; hold for any $f$ with the property given subsequently.  The convexity property says that the function value at a point between $x_1$ and $x_3$ has to be less than or equal to a certain weighted average of the values $f(x_1)$ and $f(x_3)$.  This is impossible if the in-between value is larger than the endpoint values.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-11-26T23:29:45.743" />
  <row Id="12001" PostTypeId="2" ParentId="11979" CreationDate="2010-11-26T23:44:19.787" Score="2" ViewCount="" Body="&lt;p&gt;Perhaps you want to know the total number of moves that a single knight could theoretically make, where (since this is a graph theory homework question) I'll offer the following hint:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Hint&lt;/em&gt;:  You could construct a graph (the &lt;a href=&quot;http://en.wikipedia.org/wiki/Knight%27s_graph&quot; rel=&quot;nofollow&quot;&gt;knight's graph&lt;/a&gt;) where edges represent the moves of the knight.  The total number of moves can be deduced from a graph-theoretical property of this graph.&lt;/p&gt;&#xA;" OwnerUserId="139" LastEditorUserId="139" LastEditorDisplayName="" LastEditDate="2010-11-27T00:12:39.823" LastActivityDate="2010-11-27T00:12:39.823" CommentCount="1" />
  <row Id="12002" PostTypeId="1" CreationDate="2010-11-27T00:17:31.380" Score="0" ViewCount="180" Body="&lt;p&gt;Given a problem, we are asked to use a statistical method to come up with a conclusion. e.g. XYZ drug decreases heart diseases. (null hypothesis)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After searching on internet, I got a list of these tests:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Analysis of variance (ANOVA),&#xA;Chi-square test,&#xA;Correlation,&#xA;Factor analysis,&#xA;Mann–Whitney U,&#xA;Mean square weighted deviation (MSWD),&#xA;Pearson product-moment correlation coefficient,&#xA;Regression analysis,&#xA;Spearman's rank correlation coefficient,&#xA;Student's t-test,&#xA;Time series analysis&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I presume all methods above can be applied. So I pick up student's t-test as my choice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How could I know whether it is a maximum likelihood estimation which is unbiased, consistent and efficient? What I think is that we should know the Mean and Variance of the estimate to conclude. But it is nowhere in the notes I found online. =/&lt;/p&gt;&#xA;" LastEditorUserId="2641" LastEditorDisplayName="" LastEditDate="2010-11-27T05:19:17.803" LastActivityDate="2010-11-27T05:19:17.803" Title="Maximum likelihood estimation," Tags="&lt;probability-theory&gt;&lt;estimation-theory&gt;" CommentCount="7" />
  <row Id="12003" PostTypeId="1" AcceptedAnswerId="12004" CreationDate="2010-11-27T00:23:49.990" Score="5" ViewCount="417" Body="&lt;p&gt;This is probably a bit of a boring and ridiculously noob question, so apologies in advance, but:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is everything in maths essentially built on adding, subtracting, multiplying and dividing?&lt;/p&gt;&#xA;" OwnerUserId="3229" LastEditorUserId="3229" LastEditorDisplayName="" LastEditDate="2010-12-01T22:34:49.730" LastActivityDate="2010-12-02T02:16:21.710" Title="Is everything in mathematics built on add, subtract, multiply and divide?" Tags="&lt;soft-question&gt;" AnswerCount="2" CommentCount="11" FavoriteCount="1" />
  <row Id="12004" PostTypeId="2" ParentId="12003" CreationDate="2010-11-27T00:38:48.287" Score="28" ViewCount="" Body="&lt;p&gt;Not really. For one thing, before you can &quot;add&quot;, &quot;subtract&quot;, &quot;multiply&quot;, and &quot;divide&quot;, you have to have &lt;strong&gt;something&lt;/strong&gt; that you are &quot;adding&quot;, &quot;subtracting&quot;, &quot;multiplying&quot;, or &quot;dividing.&quot; So that &quot;something&quot; has to precede these operations. And for another, there are many notions that cannot really be expressed in these terms. To take a familiar one if you've taken Calculus, the idea of &quot;limit&quot; is not one that can be expressed in terms of adding, subtracting, multiplying, or dividing, but rather is expressed in terms of &quot;approaching&quot;, or of being &quot;arbitrarily close to&quot;.  (This is one reason why Newton and Leibnitz are so rightly hailed as revolutionaries: the ideas of Calculus were very much unlike what mathematics had been up to that point.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, one has to start &lt;em&gt;somewhere&lt;/em&gt;. Historically, mathematics started from different starting points in different cultures; counting was certainly one of the firsts things that comes up. But some cultures then realized that they need some more general ideas that went beyond simply counting (with its addition, multiplication, etc). Greeks settled on geometry as their underlying structure, and in fact translated everything about numbers into statements about geometry: lines, points, circles, squares, etc. Other cultures did not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These days we usually take the route of &lt;a href=&quot;http://en.wikipedia.org/wiki/Axiomatic_theory&quot; rel=&quot;nofollow&quot;&gt;axiomatic theories&lt;/a&gt;, where we have some &quot;primitive notions&quot; that are not defined, and some rules about how we can talk about them and their properties, and then we try to build everything on top of &lt;em&gt;that&lt;/em&gt;. One &lt;em&gt;can&lt;/em&gt; start from the natural numbers (the &quot;counting numbers&quot;) using, for instance, &lt;a href=&quot;http://en.wikipedia.org/wiki/Peano_axioms&quot; rel=&quot;nofollow&quot;&gt;the Peano axioms&lt;/a&gt;, and start building from there. Then you are actually starting &lt;em&gt;before&lt;/em&gt; you get to &quot;addition&quot;, &quot;multiplication&quot;, &quot;subtraction&quot;, and &quot;division&quot;, with only the notion of &quot;number&quot;, &quot;next one&quot;, and induction. Or we can have other starting points.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, &lt;em&gt;today&lt;/em&gt;, there is usually one of two starting points: Set Theory, and Category Theory. Both presume that we agree on the basics of logic (the meaning of things like &quot;and&quot;, &quot;or&quot;, &quot;if... then...&quot;, and &quot;not&quot;, the meaning of &quot;is equal to&quot;, and things like that); and the &lt;a href=&quot;http://en.wikipedia.org/wiki/Rules_of_inference&quot; rel=&quot;nofollow&quot;&gt;rules of inference&lt;/a&gt;, which are rules that let you go from some statements others; for instance, the rule that says that if you know that A implies B, and you also know that A is true, then you know that B is true.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In Set Theory, you start with the idea of &quot;set&quot; and &quot;element of&quot;; these ideas are not actually defined, you just give some axioms that describe things you can say about them. From these axioms one can &lt;em&gt;define&lt;/em&gt; numbers, addition, multiplication, etc. And also build &lt;em&gt;other&lt;/em&gt; notions that are orthogonal to natural numbers and their usual operations. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In Category Theory, you start with the idea of &quot;objects&quot; and &quot;maps&quot; between objects (again, not defined; you only get some axioms that tell you how you can talk about them); then you can use these notions to construct everything else in terms of them. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, rather, in modern mathematics everything is either &quot;sets all the way down&quot;, or everything is &quot;categories all the way down&quot;. If there is one common set of presuppositions, it's just &lt;a href=&quot;http://en.wikipedia.org/wiki/First_order_logic&quot; rel=&quot;nofollow&quot;&gt;basic logic&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-27T00:44:05.333" LastActivityDate="2010-11-27T00:44:05.333" />
  <row Id="12005" PostTypeId="2" ParentId="11995" CreationDate="2010-11-27T00:58:05.090" Score="3" ViewCount="" Body="&lt;p&gt;If you are at the point of computing integrals, then you are probably already comfortable with computing derivatives.  The way to check whether a potential antiderivative is correct is to take its derivative.  If you take the derivative of $f(x)=\ln(\sqrt x)$ (either by first simplifying to $\frac{1}{2}\ln(x)$ or using the chain rule) you get $f&amp;#39;(x)=\frac{1}{2x}$. If you take the derivative of $g(x)=-\frac{2}{\sqrt x}=-2x^{-1/2}$, you get $g&amp;#39;(x)=x^{-3/2}=\frac{1}{\sqrt{x}^3}=\frac{1}{x\sqrt{x}}$.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-11-27T00:58:05.090" />
  <row Id="12006" PostTypeId="1" CreationDate="2010-11-27T01:32:19.110" Score="1" ViewCount="134" Body="&lt;p&gt;I remember reading in RPF's biographical book, &quot;Surely you're joking Mr. Feynman&quot; that he used to have timed contest about writing down the biggest number using standard symbols. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instead of the time restriction, I was wondering what if we restrict the number of characters, to say $N=10$. What is the largest number we can write using $N$ characters using numbers and standard functions known to most mathematicians? Of course I think the real difficulty would be in proving the said number is the largest. What is the largest number you can think of with $N=10$?&lt;/p&gt;&#xA;" OwnerUserId="3947" LastActivityDate="2010-11-27T05:09:27.390" Title="Largest number that can be written using N characters" Tags="&lt;soft-question&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="12007" PostTypeId="1" AcceptedAnswerId="12029" CreationDate="2010-11-27T01:35:20.923" Score="10" ViewCount="169" Body="&lt;p&gt;I'm trying to reconstruct a problem I saw in the Monthly, years ago. Perhaps it'll look familiar to someone.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the integer lattice in the plane, we grow a tree in the following natural way: Initially the tree is just the origin. At each step, we find the set of lattice points that are neighbors (distance 1) to precisely one vertex of our tree, and add them (simultaneously) to the tree.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus on day 0 the tree is $\{(0,0)\}$; on day 1 it contains $\{(0,0), (1,0), (-1,0),(0,1),(0,-1)\}$; on day 2 it contains those vertices along with $(2,0),(-2,0),(0,2)$ and $(0,-2)$ (note that $(1,1)$ is not added because it has two neighbors already in the tree), and on day 3 we add 12 new vertices. It looks like a pretty familiar fractal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The thing I'm not sure of is what exactly was asked of that tree... Possible candidates include its asymptotic density, some sort of simple formula to determine which lattice points ultimately make it into the tree and the # of vertices added on day $n$. There are lots of interesting questions and I'm happy to try and solve them but I prefer to work on the ones that were actually posed! &lt;/p&gt;&#xA;" OwnerUserId="308" LastEditorUserId="308" LastEditorDisplayName="" LastEditDate="2011-04-06T06:26:02.027" LastActivityDate="2011-04-06T06:26:02.027" Title="Reconstructing a Monthly problem: tree growth on the 2D integer lattice" Tags="&lt;number-theory&gt;&lt;reference-request&gt;&lt;fractals&gt;&lt;discrete-geometry&gt;" AnswerCount="1" />
  <row Id="12008" PostTypeId="1" CreationDate="2010-11-27T01:40:47.643" Score="1" ViewCount="109" Body="&lt;p&gt;I'm trying to understand the proof for Lemma 4.8 in Lyndon &amp;amp; Schupp Combinatorial Group Theory, page 26, proposition 4.8.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the very end of this proof, we have: $N \unlhd F$ where $F$ is a finitely generated free group. $[F:N]=p$, $p$ prime. $K \unlhd N$, $K \unlhd F $. $u,v \in N$ such that their images in $N/K$ are not conjugate.&#xA;It is then claimed that the images of $u,v$ in $F/K$ are not conjugate as well. Why is that? &lt;/p&gt;&#xA;" OwnerUserId="3533" LastActivityDate="2010-11-28T22:38:23.330" Title="Simplified Version: $N \unlhd F$, $a,b \in N$. $[F:N]=p$, prime. $a \sim b$ in $F$ (conjugates). Are $a \sim b$ in $N$? (more inside)" Tags="&lt;group-theory&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="12009" PostTypeId="2" ParentId="11923" CreationDate="2010-11-27T02:40:20.727" Score="33" ViewCount="" Body="&lt;h3&gt;Update 1&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;This &lt;em&gt;should&lt;/em&gt; be the final edit. Lots of typos have been corrected and the presentation in the section on existence has been greatly improved. It should be very close to a form that can be used as a basis for a project for students.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Update 2&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Even more typos corrected. Added example of non-archimedean ordered field. Fleshed out a few bits more. &lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Update 3&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Fixed up the proof sketch in the last exercise. Also I keep finding these small annoying errors... I'm sorry that I keep up bumping this thread, I just hope that you don't mind too much.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Introduction&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;I'll do one better: I'll write you one (mostly because I actually have already written one - consider this sort of a Cliff's notes version with most of the details left as exercises). Recall that a first course in analysis typically asserts that there exists a field with an order and the least upper bound property, and you then develop the theory of sequences and series from there. Thus our goal should be to prove existence and uniqueness (up to some isomorphism notion) of such an object.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Uniqueness&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Definition 1.&lt;/strong&gt; An &lt;em&gt;ordered field&lt;/em&gt; is a field, $K$ along with a total order $\leq$ on $K$ such that &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$\forall x,y,z \in K : x \leq y \implies x+z \leq y+z$ (Transitivity).&lt;/li&gt;&#xA;&lt;li&gt;$\forall x,y \in K : 0 \leq x \land 0 \leq y \implies 0 \leq xy$ (Positivity of product).&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Moreover, define the &lt;em&gt;absolute value&lt;/em&gt; function $\vert \cdot \vert: K \to K$ by &#xA;$$\vert x \vert = \begin{cases} &#xD;&#xA;    x  &amp;amp; \text{if } 0 \leq x \\\\ &#xD;&#xA;    -x &amp;amp; \text{otherwise.}&#xD;&#xA;\end{cases} $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Exercise 1.&lt;/strong&gt; There is an alternate definition of an ordered field: $K$ is said to be ordered if there is a subset $P$ of $K$ - called the &lt;em&gt;positive elements of $K$&lt;/em&gt; - such that &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$K$ is the disjoint union of $P$, $-P$ and $\lbrace 0 \rbrace$.&lt;/li&gt;&#xA;&lt;li&gt;$x,y \in P \implies x+y \in P$ and $xy \in P$.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Make formal the colloquial statement &quot;The two definitions of ordered field are equivalent&quot; and prove said equivalence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that we could just as well require that the inequality in Definition 1 be strict since $x \lt y \lor x = y \iff x \leq y$ allows us to pass between strict and non-strict orderings as we please.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Definition 2.&lt;/strong&gt; Let $K$ and $L$ be ordered fields. An order-preserving ($f$ is order-preserving means that $x\leq  y \implies f(x)\leq f(y)$) ringhomomorphism is called an &lt;em&gt;embedding&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that any embedding must be injective since the kernel of a ringhomomorphism is and ideal and $0 \lt 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proposition 3.&lt;/strong&gt; If $K$ is an ordered field, then $K$ has a subfield isomorphic to $\mathbb{Q}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Exercise 2.&lt;/strong&gt; Prove Proposition 3 by showing that $\varphi: \mathbb{N} \to K$ defined by &#xA;$$ \varphi(n) = \underbrace{1+\dotsb+1}_{n \text{ } 1\text{s}} $$&#xD;&#xA;extends to an embedding of $\mathbb{Q}$ in $K$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Definition 4.&lt;/strong&gt; An ordered field $K$ is called an &lt;em&gt;archimedean field&lt;/em&gt; if for all nonzero $x$ in $K$ there exists a natural number $n$ such that $n|x| \gt 1$. Or equivalently: there exists a natural number $n$ such that $n \gt |x|$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proposition 5.&lt;/strong&gt; Let $K$ be an ordered field. If $K$ has the least upper bound property, then $K$ is archimedean.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Exercise 3.&lt;/strong&gt; Show the contrapositive of Proposition 5 by considering the set&#xA;$$ I = \lbrace x \in K \;\vert\; 0 \lt x \land \forall n \in \mathbb{N} : nx &amp;lt; 1 \rbrace. $$&#xD;&#xA;Show that $I$ is nonempty, $1$ is an upper bound for $I$ and $I$ has no least upper bound [Hint: assume that $y$ is the least upper bound and split in to cases (i) for $y \in I$ look at $2y$, and (ii) for $y \notin I$ look at $y/2$].&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a bit of an abuse of language, the elements of $I$ can be said to be &quot;infinitely small&quot;. For an example of an ordered field which is not archimedean, consider $\mathbb{Q}(X)$, the set of &lt;a href=&quot;http://en.wikipedia.org/wiki/Rational_function&quot; rel=&quot;nofollow&quot;&gt;rational functions&lt;/a&gt; in one variable over $\mathbb{Q}$ (see [&lt;a href=&quot;http://en.wikipedia.org/wiki/Ordered_field#Examples_of_ordered_fields&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;] for how the order is defined) - $0 \lt \frac{1}{X} \lt q$ for all positive rationals $q$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now it's time to consider the order topology on ordered and archimedean fields, i.e. the topology generated by the open intervals $(a,b)$ defined in the obvious way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Exercise 4.&lt;/strong&gt; Show that the open intervals in an ordered field $K$ are a basis for a topology on $K$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proposition 6.&lt;/strong&gt; $\mathbb{Q}$ is dense in any archimedean field $K$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Sketch of proof.&lt;/em&gt; Since the open intervals are a basis, we just need to verify that any open interval contains a rational. Define the ceiling function $\lceil \cdot \rceil : K \to \mathbb{Z}$ by &#xA;$$ \lceil x \rceil = \min \lbrace n \in \mathbb{Z} \;\vert\; n \geq x\rbrace, $$&#xD;&#xA;which is well-defined since $K$ is archimedean.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let $x \in K$ and a $K \owns \varepsilon \gt 0$ be given. Since $K$ is archimedean we can find $n \in \mathbb{N}$ such that $n\varepsilon &amp;gt; 1$. Now we have:&#xA;$$ 0 \leq \left\lvert \frac{\lceil nx \rceil}{n} - x \right\rvert &#xD;&#xA;     \leq \left\lvert \frac{\lceil nx \rceil - nx}{n} \right\rvert&#xD;&#xA;     \leq \frac{1}{n} \lt \varepsilon&#xD;&#xA;$$&#xD;&#xA;Hence $\frac{\lceil nx \rceil}{n} \in (x-\varepsilon,x+\varepsilon)$ from which that result follows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Corollary 7.&lt;/strong&gt; $\mathbb{Q}$ is densely ordered in any archimedean field, i.e. if $K$ is archimedean, $x,y \in K$, and $x \lt y$, then there exists a $q \in \mathbb{Q}$ such that $x \lt q \lt y$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proposition 8.&lt;/strong&gt; Any archimedean field is 2nd-countable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Exercise 5.&lt;/strong&gt; Prove Proposition 8 [Hint: Consider the set of intervals with rational endpoints. Show that it is a basis for a topology. Argue that it is coarser than the standard topology, and then show that any basis set for the standard topology (i.e. an interval) contains an interval with rational endpoints]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proposition 9.&lt;/strong&gt; Any archimedean field $K$ is a topological field, i.e. $K \times K \owns (x,y) \mapsto x+y \in K$, $K^\times \times K^\times \owns (x,y) \mapsto xy \in K^\times$ and $K^\times \owns x \mapsto x^{-1} \in K^\times$ are continuous.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Exercise 6.&lt;/strong&gt; Prove Proposition 9 [Hint: Consider sequences $x_n$ and $y_n$ that converge to $x$ and $y$ respectively, then use that convergent sequences are bounded].&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that proposition 9 in fact holds for ordered fields too. You just have to consider nets instead.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Definition 10.&lt;/strong&gt; Let $G$ be a topological group. A sequence $(x_n)$ in $G$ is said to be a &lt;em&gt;Cauchy sequence&lt;/em&gt; if for every neighbourhood $U$ of $e$ exists a $N$ such that $m,n \geq N \implies x_nx_m^{-1} \in U$. We shall denote the set of Cauchy sequences in $G$ by $G^\mathbb{N}_c$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that this definition too can be used to define Cauchy nets in topological groups, but we shall not need that. Observe that the intervals with rational endpoints gives us a countable basis at $0$ in any archimedean field, so we see that for archimedean fields, the definition boils down to this:&#xA;$$ \forall \varepsilon \in \mathbb{Q}^+ \exists N : n,m \geq N \implies |x_n - x_m| &amp;lt; \varepsilon. $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proposition 11.&lt;/strong&gt; Let $K$ be an archimedean field and $(x_n)$ a sequence in $K$. The following hold:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;If $(x_n)$ is convergent, then $(x_n)$ is a Cauchy sequence, and any subsequence of $(x_n)$ converges to the same limit.&lt;/li&gt;&#xA;&lt;li&gt;If $(x_n)$ is a Cauchy sequence, then it is bounded.&lt;/li&gt;&#xA;&lt;li&gt;If $(x_n)$ is bounded, then $(x_n)$ has a Cauchy subsequence (&quot;pre-Bolzano-Weierstraß&quot;)&lt;/li&gt;&#xA;&lt;li&gt;If $(x_n)$ is a Cauchy sequence then any subsequence is a Cauchy sequence. Moreover if $(x_n)$ has a convergent subsequence, then $(x_n)$ converges to the same limit.&lt;/li&gt;&#xA;&lt;li&gt;If $(x_n)$ is a Cauchy sequence and $(x_{k_n})$ is a subsequence, then $\lim_{n \to \infty} |x_n - x_{k_n}| = 0$.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Exercise 7.&lt;/strong&gt; Prove Proposition 11.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proposition 12.&lt;/strong&gt; Let $K$ be an archimedean field. TFAE:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$K$ has the least upper bound property.&lt;/li&gt;&#xA;&lt;li&gt;Any Cauchy sequence in $K$ is convergent (&quot;$K$ is complete&quot;).&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Exercise 8.&lt;/strong&gt; Flesh out the following sketch of a proof of Proposition 12: &lt;br/&gt;&#xA;$1 \implies 2$: Let $(x_n)$ be a Cauchy sequence in $K$. If $(x_n)$ is eventually constant, then it is obviously convergent, so assume that it is not so, and consider the set &#xA;$$ X = \lbrace x \in K \;\vert\; \exists N \in \mathbb{N} : n \geq N \implies x \lt x_n \rbrace. $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$X$ is nonempty and has an upper bound (why?) and thus $\sup X$ exists, and we claim that $x_n \to \sup X$. Given an $\varepsilon &amp;gt; 0$ find $N$ such that $|x_n - x_m| &amp;lt; \frac{\varepsilon}{2}$, and consider the interval $(x_N-\frac{\varepsilon}{2}, x_N + \frac{\varepsilon}{2})$. There are only finitely many elements of the sequence outside of this interval, so $x_N - \frac{\varepsilon}{2} \in X$. Moreover, $x + \frac{\varepsilon}{2}$ is an upper bound for $X$. The triangle inequality (note: prove the triangle inequality for archimedean fields) now gives &#xA;$$ |x_n - \sup X| \leq |x_n - x_N| + |x_N - \sup X| \lt \varepsilon. $$&#xD;&#xA;$2 \implies 1$: Let $A \subset K$ be nonempty and bounded above by $y_0 \in K$. Let $x_0$ be some non-upper bound for $A$ e.g. $a-1$ for some $a \in A$, and recursively define a pair of sequences $(x_n)$ and $(y_n)$ over $\mathbb{N}_0$ by&#xA;$$ \begin{align}&#xD;&#xA;y_{n+1} &amp;amp; = \begin{cases}&#xD;&#xA;    \tfrac{y_n + x_n}{2} &#xD;&#xA;        &amp;amp; \text{if } \tfrac{y_n+x_n}{2} \text{ is an upper bound for } A, \\\\&#xD;&#xA;    y_n &amp;amp; \text{otherwise}&#xD;&#xA;\end{cases} \\\\&#xD;&#xA;x_{n+1} &amp;amp; = \begin{cases}&#xD;&#xA;    \tfrac{y_n+x_n}{2}&#xD;&#xA;        &amp;amp; \text{if } \tfrac{y_n+x_n}{2} \text{ is a non-upper bound for } A, \\\\&#xD;&#xA;    x_n &amp;amp; \text{otherwise.}&#xD;&#xA;\end{cases}&#xD;&#xA;\end{align} $$&#xD;&#xA;By induction we have that $(x_n)$ is increasing, $(y_n)$ is decreasing, and  for all $i,j$: $x_i \leq y_j$. Likewise by induction we get that for all $i$: $0 \leq y_i - x_i \leq 2^{-i}(y_0 - x_0)$. It is now easy to show that $(y_n)$ is a Cauchy sequence. Assume that $n \geq m$ and we have:&#xA;$$ \begin{align}&#xD;&#xA;y_m-y_n &amp;amp; = y_m - y_{m+1} + y_{m+1} - \dotsb - y_{n-1} + y_{n-1} - y_n \\\\&#xD;&#xA;        &amp;amp; \leq (2^{-m-1} + 2^{-m-2} + 2^{-n})(y_0 - x_0) \\\\&#xD;&#xA;        &amp;amp; \leq 2^{-m}(y_0 - x_0)&#xD;&#xA;\end{align} $$&#xD;&#xA;from which it is easy to conclude that $(y_n)$ is Cauchy. Similarly $(x_n)$ is shown to be Cauchy. Thus the two sequences are convergent, and their limits must be the same since $y_n - x_n \leq 2^{-n}(y_0 - x_0)$ -- denote the limit by $s$. Finally observe that for all $a \in A$, $a \leq y_n$, so $a \leq s$ (why?) i.e. $s$ is an upper bound for $A$. Moreover if $u$ is an upper bound for $A$ then $x_n \leq u$ for all $n$, so $s \leq u$ (why?). Hence $s$ is the least upper bound for $A$. QED.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proposition 12.&lt;/strong&gt; Let $K$ and $L$ be archimedean fields. If $f: K \to L$ and $g: L \to K$ are embeddings, then $K$ and $L$ are isomorphic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Proof.&lt;/em&gt; Let $g \circ f = \phi: K \to K$. By definition $\phi$ is an embedding. Assume that $\phi$ is not the identity. Then there exists an $x$ for which $\phi(x) \neq x$. Find a $q \in \mathbb{Q}$ between $x$ and $\phi(x)$ and observe that $\phi$ has to be the identity on $\mathbb{Q}$, but it can't preserve the ordering of $x$ and $q$. This gives a contradiction, and thus $\phi$ must be the identity. The proof that $f \circ g$ is the identity is similar. QED.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Lemma 13.&lt;/strong&gt; Let $G$ and $H$ be topological groups and let $G&amp;#39;$ be a dense subgroup of $G$. If $\phi: G \to H$ is continuous and $\phi|_{G&amp;#39;}$ is a homomorphism, then $\phi$ is a homomorphism.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Proof.&lt;/em&gt; Let $x,y \in G$ and assume that $\phi(xy) \neq \phi(x)\phi(y)$. Since $G&amp;#39;$ is dense there exists nets $(x_\alpha)$ and $(y_\alpha)$ in $G&amp;#39;$ converging to $x$ and $y$ respectively. Since $\phi$ is a homomorphism on $G&amp;#39;$ we have for all $\alpha$ that $\phi(x_\alpha y_\alpha) = \phi(x_\alpha)\phi(y_\alpha)$, and since $\phi$ is continuous and $\cdot$ is continuous we have that $\phi(xy) = \phi(x)\phi(y)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Remark.&lt;/em&gt; This is not a vacuous use of reductio ad absurdum, since a direct proof would have to work with arbitrary nets converging to $x$ and $y$. Of course this lemma immediately generalizes to also apply to rings in an obvious manner.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Theorem 14.&lt;/strong&gt; (Embedding theorem for archimedean fields). Let $K$ and $L$ be archimedean fields where $L$ is complete. Then there exists an embedding $\phi: K \to L$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Proof.&lt;/em&gt; Define $\phi: K \to L$ by $\phi(x) = \sup \lbrace q \in \mathbb{Q} \;\vert\; q \lt x \rbrace$. It is easy to show that $\phi$ is order-preserving, so the only hard part is showing that it's a homomorphism. We wish to use the preceding lemma, so first we note that it's obvious that $\phi$'s restriction to $\mathbb{Q}$ is the identity, so the only hard part is verifying that it's continuous. Consider a $k \in K$ and let $V$ be a neighbourhood of $\phi(k)$. Find rationals $p$ and $q$ such that $\phi(k) \in (p,q) \subset V$. We claim that $\phi((p,q)) \subset (p,q)$. Let $k&amp;#39; \in (p,q) \subset K$ and find $q&amp;#39; \in \mathbb{Q}$ such that $p \lt k&amp;#39; \lt q&amp;#39; \lt q$. It's easy to see that $p \lt \phi(k&amp;#39;) \leq q&amp;#39; \lt q$, so $\phi(k&amp;#39;) \in (p,q)$, and we conclude that $\phi$ is continuous. QED.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Theorem 15.&lt;/strong&gt; (Uniqueness of complete ordered field) There exists up to order-preserving isomorphism one and only one complete ordered field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Proof.&lt;/em&gt; Let $K$ and $L$ be complete ordered fields. Theorem 14 gives embeddings $\phi: K \to L$ and $\varphi: L \to K$. Proposition 12 gives that $K$ and $L$ are isomorphic. QED.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Existence&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Much more has been written on this subject than uniqueness, so I'll write a lot less here. Recall that $\mathbb{Q}$ under addition is a topological group, so we use the previously established notation for its set of Cauchy sequences: $\mathbb{Q}^\mathbb{N}_c$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proposition 16.&lt;/strong&gt; $\mathbb{Q}^\mathbb{N}_c$ is a commutative unital ring under coordinatewise addition and multiplication.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Exercise 8.&lt;/strong&gt; Prove Proposition 16.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proposition 17.&lt;/strong&gt; The set of rational Cauchy sequences that converge to zero, i.e.&#xA;$$ \mathfrak{z} = \left\lbrace (z_n) \in \mathbb{Q}^\mathbb{N}_c \;\bigg\vert\; z_n \to 0 \right\rbrace,$$&#xD;&#xA;is a maximal ideal in $\mathbb{Q}^\mathbb{N}_c$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Exercise 9.&lt;/strong&gt; Prove Proposition 17. [Hint: Consider an ideal $\mathfrak{a} \supset \mathfrak{z}$ with a $(q_n) \in \mathfrak{a}\setminus\mathfrak{z}$. Show that there exists an $N$ such that $n \geq N \implies q_n \neq 0$ and consider the sequence $(z_n) = (1-q_1,\dotsc,1-q_N,0,0,\dotsc)$. Argue that $(q_n + z_n) \in \mathfrak{a}$ and conclude that $\mathfrak{a} = \mathbb{Q}^\mathbb{N}_c$.]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus $\mathfrak{z}$ is a maximal ideal and hence the quotient, which we shall denote by $R$, must be a field. From here on out the rest is sketchy as there are a lot of details to check and it's really rather boring.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We adopt the convention that we write sequences with &lt;strong&gt;bold&lt;/strong&gt;, e.g. $(x_n) = {\bf x}$ and use the notation $[{\bf x}]$ for the equivalence class containing ${\bf x}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Define $\lt$ on $R$ by $[{\bf p}] \lt [{\bf q}] \iff \exists (p_n) \in [{\bf p}], (q_n) \in [{\bf q}], \alpha \in \mathbb{Q}^+ : p_n \lt q_n + \alpha$ eventually. In particular note that the existence of the positive $\alpha$ cannot be dropped; e.g. look at $(1/n)$, $(0)$ and $(-1/n)$. Without the requirement that such a positive $\alpha$ exists, you'd thus have $0 \lt 0$, which is clearly rubbish.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Exercise 10.&lt;/strong&gt; Verify that if $(p_n) \in [{\bf p}]$ and $(q_n) \in [{\bf q}]$ are witnesses that $[{\bf p}] \lt [{\bf q}]$, then &lt;em&gt;any&lt;/em&gt; pair of representatives will do just as well, i.e. if $(p&amp;#39;_n) \in [{\bf p}]$ and $(q&amp;#39;_n) \in [{\bf q}]$, then $(p&amp;#39;_n)$ and $(q&amp;#39;_n)$ are witnesses as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Exercise 11.&lt;/strong&gt; Show that $\lt$ is a strict total order, i.e. $\lt$ is irreflexive, transitive and total [Hint (for totality): You need to show that exactly one of $[{\bf x}] \lt [{\bf y}]$, $[{\bf x}] = [{\bf y}]$ and $[{\bf y}] \lt [{\bf x}]$ holds, so first show that at least one holds by assuming that none of them holds and derive a contradiction, and then show that at most one holds likewise by contradiction and using that $[{\bf x}] \nless [{\bf y}] \iff [{\bf y}] \leq [{\bf x}]$. You'll probably need to pass to &quot;nicer&quot; subsequences many times, so Proposition 11(5) will come in handy].&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And sit back and reap the reward with the following easy &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Exercise 12.&lt;/strong&gt; Show that $R$ is an ordered field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Exercise 13.&lt;/strong&gt; Show that $R$ is archimedean.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Exercise 14.&lt;/strong&gt; Fill out the details of the following sketch of a proof that any Cauchy sequence in $R$ is convergent: &lt;br/&gt;&#xA;&lt;em&gt;Proof.&lt;/em&gt; Starting with an arbitrary Cauchy sequence $([{\bf q}]^i)$ of equivalence classes, Proposition 11(4) allows us to assume WLOG that for all $i,j$ with $j \geq i$ that $\left\lvert[{\bf q}]^i - [{\bf q}]^j]\right\rvert \lt 4^{-i}$. Note that convergence means showing some inequalities hold, so we only need to show those for &lt;em&gt;some&lt;/em&gt; representatives, so we are free to pick nice ones. &#xA;Pick representatives, $(q^i_n)$, such that $|q^i_n - q^i_m| \lt 4^{-n}$.&#xA;Take ${\bf x} = (q^1_1,q^2_2,\dotsc)$. This is obviously a Cauchy sequence (why?), and we claim that $[{\bf q}]^i \to [{\bf x}]$. Convince yourself that this is the same as this: &#xA;$$\forall \varepsilon \in \mathbb{Q}^+ \exists K,N : k \geq K, n \geq N \implies \left\lvert q^k_n - q^n_n \right\rvert \lt \varepsilon. $$&#xD;&#xA;Note that the requirement of the existence of the positive $\alpha$ dropped out since the above has to hold for every $\varepsilon &amp;gt; 0$. So let $\varepsilon \gt 0$ be given. We use the standard &quot;trick&quot; of inserting a clever element into the inequality: $|q^k_n - q^n_n| \leq |q^k_n - q^k_k| + |q^k_k - q^n_n|$ and use that the both the sequence and the representatives have that &quot;$4^{-m}$&quot;-property to conclude that if we choose $N,K$ such that $N=K$ and $2^{-N} \lt \varepsilon$ then RHS is eventually smaller than $\varepsilon$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;CONGRATULATIONS! You're done and you now have a fully certified driver's license for the real numbers!&lt;/p&gt;&#xA;" OwnerUserId="1269" LastEditorUserId="1269" LastEditorDisplayName="" LastEditDate="2010-11-28T03:08:15.393" LastActivityDate="2010-11-28T03:08:15.393" CommentCount="12" />
  <row Id="12010" PostTypeId="2" ParentId="12008" CreationDate="2010-11-27T02:46:56.587" Score="2" ViewCount="" Body="&lt;p&gt;I don't have the book handy, but in your setup, a,b could easily be non-conjugate in N; similarly u,v could easily be conjugate in F/K.  Take F/K to be nonabelian of order 6, N/K to be index 2 and order 3.  None of its distinct elements are conjugate in N/K, but its non-identity elements are all conjugate in F/K.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, something you haven't mentioned so far is critical for their proof.&lt;/p&gt;&#xA;" OwnerUserId="583" LastActivityDate="2010-11-27T02:46:56.587" CommentCount="0" />
  <row Id="12011" PostTypeId="2" ParentId="12008" CreationDate="2010-11-27T02:47:12.480" Score="2" ViewCount="" Body="&lt;p&gt;As you have stated it, I think the result is not true. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a counterexample: for odd prime $p$, take the nonabelian group of order $p^3$ and exponent $p$, &#xA;$$G = \langle a,b\,\bigm|\, a^p = b^p = [a,b]^p = [a,b,a]=[a,b,b]=1\rangle.$$&#xD;&#xA;Let $F(x,y,z)$ be the free group on three generators. Consider $\varphi\colon F\to G$ given by $x\mapsto a$, $y\mapsto b$, and $z\mapsto [a,b]$. Let $K$ be the kernel of $\varphi$, and let $N$ be the preimage under $\varphi$ of the subgroup $\langle b,[a,b]\rangle$. Then $N\triangleleft F$, $K\triangleleft N$, $K\triangleleft F$, and $[F:N] = [G:\langle b,[a,b]\rangle] = p$. Note that since $N/K$ is abelian, if $u,v\in N$ are conjugate in $N/K$, then they are equivalent modulo $K$. Now consider $y$ and $yz$ in $N$; viewing $N/K$ as being $\langle b,[a,b]\rangle$, you have the elements $b$ and $b[a,b]$, which are not equal and hence not conjugate in $N/K$. However, $y^x = y[x,y]$. But $y[x,y] \equiv yz \pmod{K}$, since $z$ maps to $[a,b]$, and so does $[x,y]$. So $y$ and $yz$ &#xA;&lt;em&gt;are&lt;/em&gt; conjugate in $F/K$, even though they are not conjugate in $N/K$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can cook up a similar example with the quaternion group or the dihedral group with $p=2$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So either the lemma is incorrect, or else your simplification has eradicated some crucial bit of information.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-11-27T02:47:12.480" />
  <row Id="12013" PostTypeId="2" ParentId="275" CreationDate="2010-11-27T03:14:15.540" Score="3" ViewCount="" Body="&lt;p&gt;I think Mathematics : A very short Introduction by Timothy Gowers is a very interesting read. As such, it's not something you &quot;learn&quot; from, but it makes for a very short and sweet introduction to someone who is just curious about mathematics. (Also its also a great read for people who actually do work in mathematics, because that's where I lift my examples from, when I explain to my family and friends what I am doing!) ;)&lt;/p&gt;&#xA;" OwnerUserId="1441" LastActivityDate="2010-11-27T03:14:15.540" CommunityOwnedDate="2010-11-27T03:14:15.540" />
  <row Id="12015" PostTypeId="2" ParentId="275" CreationDate="2010-11-27T03:45:13.880" Score="1" ViewCount="" Body="&lt;p&gt;I'm surprised nobody has mentioned &lt;strong&gt;Mathematics and the Imagination&lt;/strong&gt; by Edward Kasner and James Newman; originally published in 1940, still &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0486417034&quot; rel=&quot;nofollow&quot;&gt;available&lt;/a&gt; in an edition from Dover. Among other things, the book that introduced the world to the name &quot;googol&quot; for $10^{100}$. It's a classic, and I've never heard anything bad about it. The book is meant for non-mathematicians.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I second the recommendation of Martin Gardner's columns as a follow-up. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A more recent addition to this genre is &lt;strong&gt;The Calculus Diaries: How math can help you lose weight, win in Vegas, and survive zombie apocalypse&lt;/strong&gt; by Jennifer Ouellette; it was reviewed favorably in NPR's &quot;Science Friday&quot;; written by a non-mathematician who never got through Calculus in school, also for non-mathematicians. I've heard some minor criticisms of the style, but otherwise generally positive reviews.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-11-27T03:45:13.880" CommunityOwnedDate="2010-11-27T03:45:13.880" />
  <row Id="12016" PostTypeId="1" CreationDate="2010-11-27T03:48:58.057" Score="4" ViewCount="122" Body="&lt;p&gt;I have just begun delving into p-adic number theory. I was wondering, given a poynomial $f(x)$ with integer coefficients, what does it mean when we say, $f(x)$ has a root in $\mathbb{Z}_2$, for instance.&lt;/p&gt;&#xA;" OwnerUserId="3947" LastActivityDate="2010-11-27T03:48:58.057" Title="What does it mean for a polynomial with integer coefficients to have a root in p-adic integers?" Tags="&lt;number-theory&gt;" CommentCount="2" />
  <row Id="12017" PostTypeId="1" CreationDate="2010-11-27T03:50:45.647" Score="3" ViewCount="100" Body="&lt;p&gt;The group $S_n \times S_n \times S_n$ acts on the set of &lt;a href=&quot;http://en.wikipedia.org/wiki/Latin_square&quot; rel=&quot;nofollow&quot;&gt;Latin squares&lt;/a&gt; $L$ of order $n$, with $\theta:=(\alpha,\beta,\gamma)$, starting from $L$, permuting the rows by $\alpha$, permuting the columns by $\beta$ and permuting the symbols by $\gamma$.  Each $\theta \in S_n \times S_n \times S_n$ is called an &lt;em&gt;isotopism&lt;/em&gt;.  In some instances $\theta(L)=L$, whence we call $\theta$ an &lt;em&gt;autotopism&lt;/em&gt; of $L$.  The set of autotopisms of a Latin square forms a group under composition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[Side note: we reserve the name automorphism for autotopisms of the form $(\alpha,\alpha,\alpha)$ to correspond with the notions in group theory and quasigroup theory]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From Brendan McKay's &lt;a href=&quot;http://cs.anu.edu.au/~bdm/data/latin.html&quot; rel=&quot;nofollow&quot;&gt;data&lt;/a&gt;, we can deduce that no Latin square of order 7 has an autotopism group of order 7.  [There are autotopisms of order 7, but no autotopism &lt;em&gt;group&lt;/em&gt; has order 7.]&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What's a clever proof of this observation that doesn't look through all the non-isotopic Latin squares?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Alexander Hulpke, Petteri Kaski, Patric R. J. Östergård, &lt;a href=&quot;http://arxiv.org/abs/0909.3402&quot; rel=&quot;nofollow&quot;&gt;The number of Latin squares of order 11&lt;/a&gt;, claims that there is a Latin square of order 11 that admits an autotopism group of order $11$.&lt;/p&gt;&#xA;" OwnerUserId="139" LastActivityDate="2010-11-27T03:50:45.647" Title="Why is there no Latin square of order 7 with an autotopism group of order 7?" Tags="&lt;combinatorics&gt;&lt;latin-square&gt;" CommentCount="2" FavoriteCount="1" />
  <row Id="12018" PostTypeId="2" ParentId="11673" CreationDate="2010-11-27T04:22:37.850" Score="2" ViewCount="" Body="&lt;p&gt;another way to get this result is to first recall that $S := X_1+X_2$ and $R := X_2/X_1$ are independent. [this is well-known and can be proved by the usual method of derived distributions for two random variables. the mapping $(X_1,X_2) \to (R,S)$ is 1-1 and easily inverted. this is a standard example or exercise in many intro probability and math stat books - such as  &lt;a href=&quot;http://books.google.com/books?id=L95tVBSOgpUC&amp;amp;pg=PA377&amp;amp;dq=functions+of+two+gamma+variables&amp;amp;hl=en&amp;amp;ei=PXvwTJDpEIT68Abu3N2HDA&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=35&amp;amp;ved=0CNYBEOgBMCI#v=onepage&amp;amp;q=functions%20of%20two%20gamma%20variables&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;then, noting that $\frac{X_1}{X_1+X_2} = \frac{1}{1+R}$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ {\mathrm E}\{ X_1 | S\} = {\mathrm E}\{ \frac{S}{1+R} | S\} = S\kern1pt {\mathrm E}\{ \frac{1}{1+R} | S\} = S\kern1pt{\mathrm E}\frac{1}{1+R}. \kern10pt (1)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the last equality in (1) follows from independence. also, [on repeating the argument - or just taking unconditional expectations in (1)],&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ c_1 = {\mathrm E}X_1 = {\mathrm E}\frac{S}{1+R} = {\mathrm E}\frac{1}{1+R}\kern2pt {\mathrm E}S = {\mathrm E}\frac{1}{1+R}(c_1+c_2),$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\kern84pt {\mathrm E}\frac{1}{1+R} = \frac{c_1}{c_1+c_2}, \kern84pt (2)$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;which can be plugged into (1) to get the result.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[btw, (2) says that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$${\mathrm E}\frac{X_1}{X_1+X_2} = \frac{{\mathrm E}X_1}{{\mathrm E}(X_1+X_2)},$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;a relation which, in general, is a fond wish of students in an intro prob or stats course - and perhaps the instructors too.]&lt;/p&gt;&#xA;" OwnerUserId="1547" LastEditorUserId="1547" LastEditorDisplayName="" LastEditDate="2010-11-27T04:39:49.707" LastActivityDate="2010-11-27T04:39:49.707" />
  <row Id="12019" PostTypeId="2" ParentId="11936" CreationDate="2010-11-27T04:22:51.837" Score="0" ViewCount="" Body="&lt;p&gt;Thanks Mariano!&lt;/p&gt;&#xA;" LastActivityDate="2010-11-27T04:22:51.837" />
  <row Id="12020" PostTypeId="2" ParentId="11726" CreationDate="2010-11-27T05:05:05.037" Score="4" ViewCount="" Body="&lt;p&gt;Here's a proof.  The value of $\alpha$ is the value of the infinite sum $$\sum_{m= -\infty}^{\infty} (e^{-2^{-m-1}} - [m \geq 1]),$$  where $[m \geq 1]$ is 1 if $m \geq 1$ and 0 otherwise.  Mathematica gives this value (to 6 decimal places) as $0.667253$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The full argument in all its rigor is too long to post on this site, so I'm only going to give an extended outline.  There are a couple of strange claims in here, but bear with me.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;HR&gt;&#xA;&lt;strong&gt;Part 1&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From my &lt;a href=&quot;http://math.stackexchange.com/questions/11602/limit-of-sn-sum-k1-infty-left1-prod-j1n-1-left1-fracj/11662#11662&quot;&gt;previous post&lt;/a&gt; we know that $1 - \prod_{j=1}^{n-1} \left(1 - \frac{j}{2^k}\right) = 1$ when $k \leq \lfloor \log_2 (n-1) \rfloor$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $p = 2 \log_2 n - \lfloor 2 \log_2 n \rfloor$.    Thus  $2 \log_2 n - S(n)$ is  $$2 \log_2 n - \lfloor 2 \log_2 n \rfloor + \sum_{k= \lfloor \log_2 (n-1) \rfloor +1}^{\lfloor 2 \log_2 n \rfloor} \prod_{j=1}^{n-1} \left(1 - \frac{j}{2^k}\right) + \sum_{k= \lfloor 2 \log_2 n \rfloor + 1}^{\infty} \left(\prod_{j=1}^{n-1} \left(1 - \frac{j}{2^k}\right) - 1\right)$$&#xD;&#xA;$$= p + \sum_{k= \lfloor \log_2 (n-1) \rfloor +1}^{2 \log_2 n - p} \prod_{j=1}^{n-1} \left(1 - \frac{j}{2^k}\right) + \sum_{k= 2 \log_2 n - p + 1}^{\infty} \left(\prod_{j=1}^{n-1} \left(1 - \frac{j}{2^k}\right) - 1\right) .$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the expression $$\prod_{j=1}^{n-1} \left(1 - \frac{j}{2^k}\right)$$ is very close to 0 when $k &amp;lt; 2 \log_2 n$ and very close to 1 when $k &amp;gt; 2 \log_2 n$.  So most of the contribution to $2 \log_2 n - S(n)$ occurs when $k$ is close to $2 \log_2 n$.  The next step, then, is to reindex with $m = k - \lfloor 2\log_2 n \rfloor$.  Now we basically have &#xA;$$2 \log_2 n - S(n) = p + \sum_{m= - \log_2 n}^0 \prod_{j=1}^{n-1} \left(1 - \frac{j}{2^{m-p} n^2}\right) + \sum_{m= 1}^{\infty} \left(\prod_{j=1}^{n-1} \left(1 - \frac{j}{2^{m-p} n^2}\right) - 1\right).$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;HR&gt;&#xA;&lt;strong&gt;Part 2&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next, we need a good approximation to $\prod_{j=1}^{n-1} \left(1 - \frac{j}{2^{m-p} n^2}\right)$.  It turns out that $e^{-2^{-m+p-1}}$ is an excellent approximation (which surprises me some - despite the fact that I have verified it numerically - as it is independent of $n$).  To see this, rewrite $\prod_{j=1}^{n-1} \left(1 - \frac{j}{2^{m-p} n^2}\right)$ as $$\exp \left( \sum_{j=1}^{n-1} \ln\left(1 - \frac{j}{2^{m-p} n^2}\right)\right)$$ and then expand the $\log$ expression with the Maclaurin series for $\ln (1+x)$.  The first term in the expansion dominates when $m$ is positive or constant, and we get&#xA;$$\prod_{j=1}^{n-1} \left(1 - \frac{j}{2^{m-p} n^2}\right) = \exp \left(\sum_{j=1}^{n-1} \left(- \frac{j}{2^{m-p} n^2} \right) + O\left(\frac{j^2}{4^{m-p} n^4}\right) \right)$$&#xD;&#xA;$$=\exp \left( - \frac{1}{2^{m-p+1}} + O\left(\frac{1}{2^m n}\right) \right) = \exp \left( - \frac{1}{2^{m-p+1}}\right) + O\left(\frac{1}{2^m n}\right)$$&#xD;&#xA;Thus $$\sum_{m= 1}^{\infty} \left(\prod_{j=1}^{n-1} \left(1 - \frac{j}{2^{m-p} n^2}\right) - 1\right) = \sum_{m= 1}^{\infty} \left(\exp \left( - \frac{1}{2^{m-p+1}} \right) - 1\right) + O\left(\frac{1}{n}\right).$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When $m$ is negative and not constant, things are a little trickier, as the higher-order Maclaurin series terms make $\exp \left( - \frac{1}{2^{m-p+1} } \right)$ not a good relative approximation to the product.  However, all the terms in the Maclaurin series are negative, so truncating after the first term does yield an upper bound.  In addition, $\exp \left( - \frac{1}{2^{m-p+1}} \right)$ goes to zero extremely fast as $m \to -\infty$.  (For example, if $m = - \log_2 (\log n)$ (which heads to $-\infty$ very slowly), we still have $\exp \left( - \frac{1}{2^{m-p+1}} \right) = \frac{1}{n^{2^{p-1}}}$.)  Thus $\exp \left( - \frac{1}{2^{m-p+1}} \right)$ is still an excellent absolute approximation for the product when $m$ is negative.  Since there are only $\log_2 n$ negative terms in the sum we are considering, we have &#xA;$$\sum_{m= - \log_2 n}^0 \prod_{j=1}^{n-1} \left(1 - \frac{j}{2^{m-p} n^2}\right) = \sum_{m = - \log_2 n}^0 \exp \left( - \frac{1}{2^{m-p+1}} \right) + E(n),$$&#xD;&#xA;where $E(n) \to 0$ as $n \to \infty$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;HR&gt;&#xA;&lt;strong&gt;Part 3&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, let $$f(p) = \sum_{m= - \infty}^0  e^{-2^{-m+p-1}} + \sum_{m= 1}^{\infty} (e^{-2^{-m+p-1}} - 1).$$&#xD;&#xA;The function $f$ is linear in $p$!  (Despite having verified this numerically and despite the argument below, this still seems weird to me!)  The slope is $-1$.  To see this, differentiate to get $$f&amp;#39;(p) = - \ln 2 \sum_{m= - \infty}^{\infty} 2^{-m+p-1} e^{-2^{-m+p-1}}.$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, apply the &lt;a href=&quot;http://en.wikipedia.org/wiki/Euler-Maclaurin_formula&quot; rel=&quot;nofollow&quot;&gt;Euler-Maclaurin summation formula&lt;/a&gt;.  Because we have the product of two exponentials, $f^{(k)}(p)$ will have the expression $2^{-m+p-1} e^{-2^{-m+p-1}}$ in every term.  As $m \to \infty$, $2^{-m+p-1} \to 0$ and $e^{-2^{-m+p-1}} \to 1$.  As $m \to -\infty$, $2^{-m+p-1} \to \infty$ and $e^{-2^{-m+p-1}} \to 0$, but the latter expression dominates.  Thus  $f^{(k)}(p) \to 0$ as $m \to \infty$ and as $m \to -\infty$.  Thus the Euler-Maclaurin formula says that&lt;br&gt;&#xA;$$f&amp;#39;(p) = - \ln 2 \int_{-\infty}^{\infty} 2^{-m+p-1} e^{-2^{-m+p-1}} dp = \left. e^{-2^{-m+p-1}}\right|_{-\infty}^{\infty} = -1.$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, $$\lim_{n \to \infty} \left(2 \log_2 n - S(n)\right) = p + \sum_{m= - \infty}^0  e^{-2^{-m+p-1}} + \sum_{m= 1}^{\infty} (e^{-2^{-m+p-1}} - 1) $$&#xD;&#xA;$$= p - p + \sum_{m= - \infty}^0  e^{-2^{-m-1}} + \sum_{m= 1}^{\infty} (e^{-2^{-m-1}} - 1) = \sum_{m= - \infty}^0  e^{-2^{-m-1}} + \sum_{m= 1}^{\infty} (e^{-2^{-m-1}} - 1).$$&#xD;&#xA;Again, this value is approximately $0.667253$.&lt;/p&gt;&#xA;" OwnerUserId="2370" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-11-27T21:14:25.560" LastActivityDate="2010-11-27T21:14:25.560" CommentCount="8" />
  <row Id="12021" PostTypeId="2" ParentId="12006" CreationDate="2010-11-27T05:09:27.390" Score="2" ViewCount="" Body="&lt;p&gt;Well, just for fun I thought I'd try to answer the question with the operations restricted to:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;nullary operations $\{0,1,2,\ldots,9\}$ and&lt;/li&gt;&#xA;&lt;li&gt;binary operation $*$ (i.e. multiplication).&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;It turns out that&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;9999999999&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;is the greatest in this case.  The closest runner-up is:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;99999*9999 = 999890001&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;We can make some simplifying assumptions:  (a) all 10 slots are to be used (otherwise, we just add in another 9 at the end) and (b) any of $0,1,2\ldots,8$ is better off being replaced by a $9$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This exercise didn't seem all that satisfying really, there's so many sequences that result in syntax errors.  E.g.:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;*999*99999&#xA;999999999*&#xA;999999**99&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;In my opinion, it gets even less satisfying if you keep going.  E.g.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Is 1/0 allowed?&lt;/li&gt;&#xA;&lt;li&gt;Does exponentation require a character?  E.g. &lt;code&gt;2^6&lt;/code&gt; vs. $2^6$.&lt;/li&gt;&#xA;&lt;li&gt;Do you require suitable bracketing?  When?  &lt;code&gt;1/2^{-6}&lt;/code&gt; vs. &lt;code&gt;1/2^-6&lt;/code&gt;.  What about &lt;code&gt;9^9^9&lt;/code&gt;?&lt;/li&gt;&#xA;&lt;li&gt;How many characters does $\log$ require?&lt;/li&gt;&#xA;&lt;li&gt;The result varies on human factors -- e.g. what number system is used (base 10?), how many characters it takes to write a function, etc.&lt;/li&gt;&#xA;&lt;li&gt;And, of course, there's the inevitable question of self-referential sequences (which might be difficult to arrange for 10 characters).&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="139" LastActivityDate="2010-11-27T05:09:27.390" CommentCount="1" />
  <row Id="12022" PostTypeId="1" AcceptedAnswerId="12024" CreationDate="2010-11-27T05:09:36.953" Score="1" ViewCount="165" Body="&lt;p&gt;The problem is &quot;Calculating $\oint_{L} \frac{xdy - ydx}{x^2 + y^2}$, where L is a smooth, simple closed, and postively oriented curve that does not pass through the orgin&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is my solution:&lt;br&gt;&#xA;Let $$P(x,y) = \frac{-y}{x^2 + y^2}, Q(x,y) = \frac{x}{x^2 + y^2}$$&#xD;&#xA;Get $$\frac{\partial{P}}{\partial{y}} = \frac{y^2 - x^2}{(x^2 + y^2)^2} = \frac{\partial{Q}}{\partial{x}}$$&#xD;&#xA;Acorrding to the Green formula: &#xA;$$\oint_{L} \frac{xdy - ydx}{x^2 + y^2} = \iint (\frac{\partial{Q}}{\partial{x}} - \frac{\partial{P}}{\partial{y}})dxdy = 0$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What's wrong with my solution?&lt;/p&gt;&#xA;" OwnerUserId="1382" LastEditorUserId="1382" LastEditorDisplayName="" LastEditDate="2010-11-27T05:50:47.697" LastActivityDate="2011-01-30T21:03:33.763" Title="Calculating $\oint_{L} \frac{xdy - ydx}{x^2 + y^2}$" Tags="&lt;calculus&gt;" AnswerCount="5" CommentCount="3" FavoriteCount="1" />
  <row Id="12023" PostTypeId="2" ParentId="12022" CreationDate="2010-11-27T05:12:47.750" Score="2" ViewCount="" Body="&lt;p&gt;Hint: What does the hypothesis of &lt;a href=&quot;http://en.wikipedia.org/wiki/Green%27s_theorem&quot; rel=&quot;nofollow&quot;&gt;Green's theorem&lt;/a&gt; say about the requirements on functions $P$ and $Q$?&lt;/p&gt;&#xA;" OwnerUserId="3431" LastActivityDate="2010-11-27T05:12:47.750" />
  <row Id="12024" PostTypeId="2" ParentId="12022" CreationDate="2010-11-27T05:16:45.993" Score="1" ViewCount="" Body="&lt;p&gt;That is valid as long as $L$ is a simple closed curve and the origin is not &quot;inside&quot; $L$, i.e., in the bounded region determined by $L$.  For example, this would work if $L$ is a unit circle centered at $(2,0)$, but not if it is the unit circle centered at the origin, and not if it is not a simple closed curve.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(As Timothy Wagner already pointed out before I finished writing, you need to check the hypotheses on the theorem you invoked to see when it works.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $L$ is closed, your integral should be $2\pi$ times the number of times $L$ &lt;a href=&quot;http://en.wikipedia.org/wiki/Winding_number#Differential_geometry&quot; rel=&quot;nofollow&quot;&gt;winds&lt;/a&gt; around the origin.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2010-11-27T05:25:47.780" LastActivityDate="2010-11-27T05:25:47.780" CommentCount="10" />
  <row Id="12025" PostTypeId="2" ParentId="12022" CreationDate="2010-11-27T05:21:03.107" Score="0" ViewCount="" Body="&lt;p&gt;The answer should be $2\pi$ if the origin lies &quot;inside&quot; $L$, otherwise it should be 0.&lt;/p&gt;&#xA;" OwnerUserId="3249" LastEditorUserId="3249" LastEditorDisplayName="" LastEditDate="2010-11-27T06:01:41.497" LastActivityDate="2010-11-27T06:01:41.497" CommentCount="2" />
  <row Id="12026" PostTypeId="1" AcceptedAnswerId="12033" CreationDate="2010-11-27T06:00:20.420" Score="1" ViewCount="366" Body="&lt;p&gt;In &lt;a href=&quot;http://math.stackexchange.com/questions/12022/calculating-oint-l-fracxdy-ydxx2-y2&quot;&gt;this&lt;/a&gt; problem, I know that the hypothsis of Green theorem must ensure that the simple closed curve is smooth, but what is smooth? Could you give a definition and an intuitive explaination?&lt;/p&gt;&#xA;" OwnerUserId="1382" LastActivityDate="2011-01-30T23:32:09.720" Title="What does smooth curve mean?" Tags="&lt;calculus&gt;" AnswerCount="4" CommentCount="5" />
  <row Id="12027" PostTypeId="2" ParentId="12026" CreationDate="2010-11-27T06:04:52.053" Score="-1" ViewCount="" Body="&lt;p&gt;Smooth means differentiable (at least once). In other words, smooth means continuous and without &quot;corners&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: changed &quot;edges&quot; to &quot;corners&quot; after J.M.&lt;/p&gt;&#xA;" OwnerUserId="3891" LastEditorUserId="3891" LastEditorDisplayName="" LastEditDate="2010-11-27T06:12:40.937" LastActivityDate="2010-11-27T06:12:40.937" CommentCount="4" />
  <row Id="12028" PostTypeId="2" ParentId="12022" CreationDate="2010-11-27T06:05:29.370" Score="0" ViewCount="" Body="&lt;p&gt;One way of understanding why Green's theorem doesn't necessarily work in this context is to view Green's theorem as two separate equations added together: $\oint_{L} P dx = \int\int-{\partial P \over \partial y} dy dx$ and $\oint_{L} Q dy = \int\int{\partial Q \over \partial x} dx dy$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically, Green's theorem holds because you can prove each of these equations by integrating by parts. In the first equation you integrate by parts in $y$ for a fixed $x$ and then integrate the result in $x$. In the second, you integrate by parts in $x$ for a fixed $y$ and then integrate the result in $y$. Since in your case both $-{\partial P \over \partial y}$ and  ${\partial Q \over \partial x}$ blow up at the origin, if the origin is contained in the interior of your curve these integrations by parts won't work... and you'll get the wrong answer. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(The curve is assumed to be simple and closed in the above.)&lt;/p&gt;&#xA;" OwnerUserId="3035" LastActivityDate="2010-11-27T06:05:29.370" />
  <row Id="12029" PostTypeId="2" ParentId="12007" CreationDate="2010-11-27T06:06:41.177" Score="9" ViewCount="" Body="&lt;p&gt;I think I got it. It seems to be problem 10360, originally published in 1994 (volume 101, issue 1, page 76), proposed by Richard Stanley. I'm looking at &lt;a href=&quot;http://www.jstor.org/pss/2589004&quot; rel=&quot;nofollow&quot;&gt;the JSTOR page of the solution&lt;/a&gt; (by &lt;a href=&quot;http://math.stackexchange.com/users/226/robin-chapman&quot;&gt;Robin Chapman&lt;/a&gt;), which appeared in October 1998, volume 105, no 8, pages 769-771). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is the statement of the problem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $L$ be the &lt;em&gt;integer lattice&lt;/em&gt; in $\mathbb{R}^d$, i.e., the $L$ is the set of points $(x_1,x_2,\ldots,x_d)$ with all $x_j\in\mathbb{Z}$. Consider $L$ as a graph by declaring two lattice points to be adjacent if the distance between them is $1$. Define a sequence $S_0$, $S_1,\ldots$ of subsets of $L$ inductively as follows:&#xA;\begin{align*}&#xA;S_0 &amp;amp;= \Bigl\{ (0,0,\ldots,0)\Bigr\}\\&#xA;S_{n} &amp;amp;= \Bigl\{ P\in L-\mathop{\cup}\limits_{0\leq k\lt n}S_k\ \Bigm|\  \text{$P$ is adjacent to exactly one element of $\mathop{\cup}\limits_{0\leq k\lt n}S_k$}\Bigr\}.&#xA;\end{align*}&#xA;Let $S$ be the subgraph of $L$ whose vertices are $\cup S_n$. Thus, $P\in S$ is adjacent to $P&amp;#39;\in S$ if the distance between $P$ and $P&amp;#39;$ is $1$.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Find a simple condition for a point of $L$ to belong to $S$.&lt;/li&gt;&#xA;&lt;li&gt;For $P\in S$, find a simple rule to determine $i$ such that $P\in S_i$.&lt;/li&gt;&#xA;&lt;li&gt;How many elements are in $S_i$?&lt;/li&gt;&#xA;&lt;li&gt;How many $P\in S_i$ are adjacent to no points in $S_{i+1}$?&lt;/li&gt;&#xA;&lt;li&gt;Show that $S$ is a tree.&lt;/li&gt;&#xA;&lt;li&gt;Investigate the (vertex) density of $S$ in $L$, and compare it to the largest density of a subset of $L$ for which the induced subgraph is a tree.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-28T23:54:13.190" LastActivityDate="2010-11-28T23:54:13.190" CommentCount="3" />
  <row Id="12030" PostTypeId="1" AcceptedAnswerId="12037" CreationDate="2010-11-27T06:10:58.757" Score="1" ViewCount="135" Body="&lt;p&gt;&lt;a href=&quot;http://math.stackexchange.com/questions/12022/calculating-oint-l-fracxdy-ydxx2-y2&quot;&gt;The orginal problem&lt;/a&gt; is &quot;Calculating $\oint_{L} \frac{xdy - ydx}{x^2 + y^2}$, where L is a smooth, simple closed, and postively oriented curve that does not pass through the orgin&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But what if I modify the hypothesis and allow non-simple closed curve? I mean is there something like green formula that allows us calculuating &quot;non-simple closed curve integral&quot;? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;:&lt;br&gt;&#xA;It seems to me that this integral should also resemble the simple closed one. Because common interior line integral should be canceled only remaining the outter curve. Am I right?&lt;/p&gt;&#xA;" OwnerUserId="1382" LastEditorUserId="1382" LastEditorDisplayName="" LastEditDate="2010-11-27T07:54:22.260" LastActivityDate="2010-11-27T09:14:57.473" Title="Calculating $\oint_{L} \frac{xdy - ydx}{x^2 + y^2}$ PartII" Tags="&lt;calculus&gt;&lt;vector-calculus&gt;" AnswerCount="4" />
  <row Id="12032" PostTypeId="1" AcceptedAnswerId="12036" CreationDate="2010-11-27T06:16:00.823" Score="2" ViewCount="239" Body="&lt;p&gt;Was trying to calculate $$\int_{0}^{\infty}e^{-x}\ln x dx=-\gamma$$ and I found this question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to analyze $$\int\frac{e^{-x}}{x}dx$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With $u=\displaystyle\frac{1}{x} \Rightarrow du = \displaystyle\frac{-1}{x^{2}} dx  $,  and $dv=e^{-x} \Rightarrow v=-e^{-x}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int\frac{e^{-x}}{x}dx = \displaystyle\frac{1}{x}\cdot-e^{-x}-\int-e^{-x}\cdot\displaystyle\frac{-1}{x^{2}} dx = -\displaystyle\frac{e^{-x}}{x}-\int \displaystyle\frac{e^{-x}}{x^{2}} dx$$   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Integrating from same form get:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int\frac{e^{-x}}{x}dx = -\displaystyle\frac{e^{-x}}{x} + \displaystyle\frac{e^{-x}}{x^{2}} + 2\int\frac{e^{-x}}{x^{3}}dx$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These calculations are correct?, and more is  valid say : &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int\frac{e^{-x}}{x}dx = \displaystyle\sum\limits_{n=0}^\infty (-1)^{n+1}n!\frac{e^{-x}}{x^{n+1}}\ ?$$  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$\bf{EDIT}$: This serie  helps me to&#xA;  calculate it ? :&#xA;  $$\int_{0}^{\infty}e^{-x}\ln xdx=-\gamma$$ I don't know  how to turn&#xA;  this serie in something armonic. If&#xA;  not is this way to calculate that this&#xA;  integral converges to $-\gamma$, which&#xA;  is the form ?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="3421" LastEditorUserId="3421" LastEditorDisplayName="" LastEditDate="2010-11-28T20:20:08.303" LastActivityDate="2010-11-28T20:20:08.303" Title="These calculations are correct ? About $\int\frac{e^{-x}}{x}dx$ $ \bf{[EDITED]}$:" Tags="&lt;calculus&gt;&lt;integral&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" />
  <row Id="12033" PostTypeId="2" ParentId="12026" CreationDate="2010-11-27T06:17:36.133" Score="2" ViewCount="" Body="&lt;p&gt;There are many ways you can characterize the smoothness of a curve.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Typically, people use the notation $C^{(n)}(\Omega)$ where $n \in \mathbb{N}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So when we say $f(x) \in C^{(n)}(\Omega)$, we mean that $f(x)$ has $n$ derivatives in the entire domain ($\Omega$ denotes the domain of the function) and the $n^{th}$ derivative of $f(x)$ is continuous i.e. $f^{n}(x)$ is continuous.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also by convention, if $f(x)$ is just continuous, then we say $f(x) \in C^{(0)}(\Omega)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, $f(x) \in C^{(\infty)}$ if the function is differentiable any number of times. For instance, $e^{x} \in C^{(\infty)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An example to illustrate is to consider the following function $f: \mathbb{R} \rightarrow \mathbb{R}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f(x) = 0, x \leq 0$$ $$f(x) = x^2, x&amp;gt;0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This function is in $C^{(1)}(\mathbb{R})$ but not in $C^{(2)}(\mathbb{R})$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, when the domain of the function is the largest set over which the function definition makes sense, then we omit $\Omega$ and write that $f \in C^{(n)}$ the domain being understood as the largest set over which the function definition makes sense.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, note the containment $C^{(n)} \subseteq C^{(m)}$ whenever $n&amp;gt;m$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In case of Green's theorem when we apply the formula&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\oint_c (Ldx + Mdy) = \int \int_D (\frac{\partial M}{\partial x} - \frac{\partial L}{\partial y}) dx dy$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;we need $L,M \in C^1{(\Omega)}$ where $\Omega$ is a domain containing the curve and the interior of the curve viz $D$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The simple closed curve $C$ should be piecewise smooth or more generally the curve $C$ should be in $C^{(0)}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We say that the curve $C$ is piecewise smooth curve when the two conditions below are satisfied:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(i) $C \in C^{(0)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(ii) The domain over which the curve is defined can be partitioned into disjoint subsets such that the curve is in $C^{(\infty)}$ (or sufficiently smooth i.e. the curve is in $C^{(n)}$ for some $n$ till which we are interested) over each of these subsets&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2011-01-30T22:40:24.567" LastActivityDate="2011-01-30T22:40:24.567" CommentCount="8" />
  <row Id="12034" PostTypeId="1" AcceptedAnswerId="12041" CreationDate="2010-11-27T06:20:08.473" Score="0" ViewCount="171" Body="&lt;p&gt;What are the conditions for existence of fourier series expansion of a function $f\colon\mathbb{R}\to\mathbb{R}$&lt;/p&gt;&#xA;" OwnerUserId="2987" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-27T06:22:13.937" LastActivityDate="2010-11-27T07:11:42.837" Title="What are the conditions for existence of fourier series expansion of a function $f\colon\mathbb{R}\to\mathbb{R}$" Tags="&lt;fourier-series&gt;" AnswerCount="1" CommentCount="16" />
  <row Id="12035" PostTypeId="2" ParentId="12026" CreationDate="2010-11-27T06:20:11.717" Score="0" ViewCount="" Body="&lt;p&gt;Consider the following curve in the plane, $(x(t),y(t))$, this curve is called smooth if the functions $x(t)$ and $y(t)$ are smooth, which simply means that for all $N$, the derivatives $\frac{d^Nx}{dt^N}$ and $\frac{d^Ny}{dt^N}$ exist.&lt;/p&gt;&#xA;" OwnerUserId="1417" LastActivityDate="2010-11-27T06:20:11.717" />
  <row Id="12036" PostTypeId="2" ParentId="12032" CreationDate="2010-11-27T06:24:13.777" Score="2" ViewCount="" Body="&lt;p&gt;The series diverges, but converges to your integral asymptotically: If you add up the first $n$ terms the ratio of the error to the $n$th term goes to zero as $x$ goes to infinity&lt;/p&gt;&#xA;" OwnerUserId="3035" LastActivityDate="2010-11-27T06:24:13.777" CommentCount="1" />
  <row Id="12037" PostTypeId="2" ParentId="12030" CreationDate="2010-11-27T06:25:30.037" Score="2" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT:&lt;/strong&gt; Suppose $L$ intersects with it self at finitely many points. Then, if we consider a parametrization of $L$, we deduce $L=L_1\cup L_2\cup\cdots L_n$ where $L_k$ is simple. &#xA;Now consider&#xA;$$ \oint_L f = \sum_k\oint_{L_k} f$$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="1154" LastActivityDate="2010-11-27T06:25:30.037" />
  <row Id="12038" PostTypeId="1" CreationDate="2010-11-27T06:44:35.637" Score="0" ViewCount="49" Body="&lt;p&gt;For the simpler flexagons, the notion of a face is obvious. However, there are flexagons where it's possible to mix faces. In those cases, how is a face defined?&lt;/p&gt;&#xA;" LastActivityDate="2010-11-27T06:44:35.637" Title="What is a face of a flexagon?" Tags="&lt;topology&gt;" />
  <row Id="12039" PostTypeId="2" ParentId="11434" CreationDate="2010-11-27T06:49:18.007" Score="0" ViewCount="" Body="&lt;p&gt;Finding a recurrence for the first equation should be obvious (just look at the first few values).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the second function $f(n) = n(n+3)$, here's an approach that seems natural to me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You want to express $f(n)$ in terms of $f(n-1)$, $f(n-2)$, etc.  So start with $f(n)=a f(n-1) + b f(n-2)$ and equate coefficients.  That is, we have \[n^2+3n=a(n^2+n-2)+b(n^2-n-2)\] so&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$a+b=1$,&lt;/li&gt;&#xA;&lt;li&gt;$a-b=3$,&lt;/li&gt;&#xA;&lt;li&gt;$-2a-2b=0$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;We observe that this system of linear equations has no solutions (the first and third equations are inconsistent).  So, we now consider a deeper recurrence: Let $f(n)=a f(n-1) + b f(n-2) + c f(n-3)$, now solve the resultant system of linear equations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[Note: in both cases, remember to describe sufficiently many initial values.]&lt;/p&gt;&#xA;" OwnerUserId="139" LastActivityDate="2010-11-27T06:49:18.007" />
  <row Id="12040" PostTypeId="2" ParentId="8814" CreationDate="2010-11-27T06:58:45.720" Score="3" ViewCount="" Body="&lt;p&gt;\[\sqrt{n}^{\log n}=n^{\log \sqrt{n}}\]&lt;/p&gt;&#xA;" OwnerUserId="139" LastActivityDate="2010-11-27T06:58:45.720" CommunityOwnedDate="2010-11-27T06:58:45.720" CommentCount="3" />
  <row Id="12041" PostTypeId="2" ParentId="12034" CreationDate="2010-11-27T07:11:42.837" Score="6" ViewCount="" Body="&lt;p&gt;If $f\in L^1_\text{loc}(\mathbb{R})$, then on an interval $I=(a,b)$ we can define &#xA;$$\hat{f}(n)=\frac{1}{b-a}\int_a^b f(x)e^{-2\pi inx/(b-a)}dx.$$&#xD;&#xA;However, in order for the formal Fourier series $$S[f](x)=\sum_{-\infty}^{\infty} \hat{f}(n)e^{2\pi inx/(b-a)}$$ to converge we need more conditions on $f$. &#xA;Kolmogorov proved in 1925 that there is there is $f\in L^1(0,2\pi)$ such that $S[f]$ diverges almost everywhere. In 1966 Carleson proved that $S[f]$ converges almost everywhere provided $f\in L^2(0,2\pi)$.&lt;/p&gt;&#xA;" OwnerUserId="1154" LastActivityDate="2010-11-27T07:11:42.837" CommentCount="6" />
  <row Id="12042" PostTypeId="1" AcceptedAnswerId="12055" CreationDate="2010-11-27T07:30:50.853" Score="4" ViewCount="196" Body="&lt;p&gt;Let $L(n)=\lfloor(1+\sqrt{5})^n\rfloor$. What kind of a linear recurrence is satisfied by $L(n)$? I have no idea how to go about this, because of the presence of the greatest integer function. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please feel free to retag it as I kept getting an error on every tag I thought was appropriate.&lt;/p&gt;&#xA;" OwnerUserId="3947" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-27T07:36:08.510" LastActivityDate="2010-11-27T11:33:53.923" Title="Recurrence relation satisfied by $\lfloor(1+\sqrt{5})^n\rfloor$" Tags="&lt;discrete-mathematics&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" />
  <row Id="12043" PostTypeId="2" ParentId="12030" CreationDate="2010-11-27T07:42:01.740" Score="1" ViewCount="" Body="&lt;p&gt;You should parameterize your curve using polar coordinates:&#xA;write the typical point $(x,y)=(r(t)\cos\theta(t),r(t)\sin\theta(t))$&#xD;&#xA;where $r$ and $\theta$ are functions of $t$ on an interval $[a,b]$.&#xA;Also note that as the curve &quot;closes up&quot; then $r(b)=r(a)$&#xD;&#xA;and $\theta(b)=\theta(a)+2m\pi$ for some integer $m$.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-11-27T07:42:01.740" />
  <row Id="12044" PostTypeId="2" ParentId="12042" CreationDate="2010-11-27T07:46:54.087" Score="0" ViewCount="" Body="&lt;p&gt;Here is a possible approach. If $x=1+\sqrt 5$, notice that $x^2=4+2x$, and you could use this to simplify higher powers: $(a+bx)x=4b+(a+2b)x$ leads to recurrence relations for $a_n$ and $b_n$ in the expression $x^n=a_n+b_nx$.  If you determine simple expressions for $a_n$ and $b_n$ and thus $x^n$, you can then find $[x^n]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You may also want to study the closely related &lt;a href=&quot;http://en.wikipedia.org/wiki/Fibonacci_number#Relation_to_the_golden_ratio&quot; rel=&quot;nofollow&quot;&gt;closed-form expression for the Fibonacci sequence&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[Edit: In the original version I made a mistake in the equation for $x^2$.]&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2010-11-27T08:04:19.437" LastActivityDate="2010-11-27T08:04:19.437" />
  <row Id="12045" PostTypeId="2" ParentId="12030" CreationDate="2010-11-27T07:50:00.240" Score="1" ViewCount="" Body="&lt;p&gt;First you need to prove that $\oint_{L} \frac{xdy - ydx}{x^2 + y^2}$ is the same as calculating the integral over a circle of radius '$r$' such that the circle of radius '$r$' lies completely within the domain enclosed by the curve $L$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you prove that the problem now becomes simple. All you need is to evaluate the integral over a circle. This can be done by parameterizing the circle as $x = r \cos(\theta)$ and $y = r \sin(\theta)$. Note that $dx = - r \sin(\theta)d\theta$ and $dy = r cos(\theta)d\theta$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, $x^2 + y^2 = r^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The integral then becomes&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \int_{0}^{2 \pi} \frac{r\cos(\theta) \times r \cos(\theta) d\theta - r\sin(\theta) \times -r \sin(\theta) d\theta}{r^2} = \int_{0}^{2 \pi} d\theta =2\pi$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, we get $$\oint_{L} \frac{xdy - ydx}{x^2 + y^2} = 2\pi$$&lt;/p&gt;&#xA;" OwnerUserId="3302" LastActivityDate="2010-11-27T07:50:00.240" />
  <row Id="12046" PostTypeId="2" ParentId="11953" CreationDate="2010-11-27T07:56:11.597" Score="1" ViewCount="" Body="&lt;p&gt;This is not a complete answer. I'm assuming that you also have non-negativity constraints on all variables.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After looking at your question more carefully, I rearranged it to the following constraints:&#xA;$$\begin{align}&#xD;&#xA;y_{i+1} - y_i &amp;amp; \in [g_{i_{\min}}, g_{i_{\max}}], \\&#xD;&#xA;y_i &amp;amp; \le y_{i_{\max}} - h_i, \\&#xD;&#xA;y_{i+1} - y_i &amp;amp; \in \left(1 + \frac{c}{h_i+b}\right)^{4/3}[w_{i_{\min}}, w_{i_{\max}}], \\&#xD;&#xA;y_{i+1} - y_i &amp;amp; \ge R_{i_{\min}} \left(1 + \frac{c}{h_i+b}\right)^{4/3}\frac{1}{(h_i+a)^2},\end{align}$$&#xD;&#xA;where $c = b - a$, $w_{i_{*}} = v_{i_{*}}^{1/3}$, and $R_{i_{\min}} = Q_{i_{\min}}^{1/3}$. Hopefully the abuse of notation in the third equation is understandable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now all the $y_i$ are on the left side of the constraints and the $h_i$ are on the right. This form should be easier to reason about. For example, I was initially concerned that the optimal $f_1$ might always be $0$. Note that all the right-hand sides get larger as the $h_i$ get smaller, so a solution with all $h_i = 0$ would only fail to be feasible if one of the lower bounds for $y_{i+1} - y_i$ became too large, i.e. incompatible with one of the upper bounds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, since all the constraints except one involve only the &lt;em&gt;differences&lt;/em&gt; between the $y_i$, you can decrease all $y_i$ by the same amount and still have a feasible solution, as long as none of them fall below zero. So when you get an optimal solution for $f_1$, you can subtract $\min\{y_j\}$ from $y_i$ and get the solution that minimizes $f_2$ for fixed $h_i$. That's not the same thing as only fixing $\sum h_i$, but it's something...&lt;/p&gt;&#xA;" OwnerUserId="856" LastActivityDate="2010-11-27T07:56:11.597" CommentCount="2" />
  <row Id="12047" PostTypeId="1" AcceptedAnswerId="12056" CreationDate="2010-11-27T08:05:39.530" Score="3" ViewCount="616" Body="&lt;p&gt;How to prove the following trigonometric identities ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) If $\displaystyle \tan (\alpha) \cdot \tan(\beta) = 1 \text{ then } \alpha + \beta = \frac{\pi}{2}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried to prove it by using the the formula for $\tan(\alpha + \beta)$ but ain't it valid only when $\alpha + \beta \neq \frac{\pi}{2}$ ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) $\displaystyle\sec\theta + \tan \theta = \frac{1}{ \sec\theta - \tan \theta}, \theta \neq (2n+1)\frac{\pi}{2}, n \in \mathbb{Z} $&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For this one I tried substituting them with the sides of the triangle, but not successful to the final result.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These are not my homework, I am trying to learn maths almost on my own, so ...&lt;/p&gt;&#xA;" OwnerUserId="3744" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2010-11-27T13:27:04.257" LastActivityDate="2011-04-06T22:38:08.683" Title="Proving basic/standard trigonometric identities" Tags="&lt;trigonometry&gt;" AnswerCount="4" CommentCount="7" />
  <row Id="12048" PostTypeId="2" ParentId="12047" CreationDate="2010-11-27T08:26:25.987" Score="1" ViewCount="" Body="&lt;p&gt;1) tan(a)tan(b) = 1 is equivilant to sin(a)sin(b) = Cos(a)cos(b) **using the fact that tan is sine divided by cosine.  Next use &#xA;                    cos(a+b) = cos(a)cos(b) -sin(a)sin(b)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which implies      sin(a)sin(b)  = cos(a)cos(b) -cos(a+b)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Use this on the left hand side of ** to get &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;                              cos(a+b) = 0&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;which implies a+b = Pi/2 or -Pi/2&lt;/p&gt;&#xA;" OwnerUserId="3971" LastActivityDate="2010-11-27T08:26:25.987" CommentCount="5" />
  <row Id="12050" PostTypeId="2" ParentId="12030" CreationDate="2010-11-27T09:14:57.473" Score="0" ViewCount="" Body="&lt;p&gt;In the punctuated plane $\dot{\mathbb R}^2$ we have the arg-&quot;function&quot; taking values in $\mathbb R/(2\pi\mathbb Z)$. Its gradient is a well defined vector field, namely $\nabla$arg$=(-{y\over x^2+y^2}, {x\over x^2+y^2})$. It follows that  for any nice curve $L$, closed or not, simple or not (but not going through the origin), your integral sums up the ongoing argument increases/decreases along $L$. If $L$ is closed then these arg-increases/decreases sum up to $2\pi$ times the winding number of $L$ around the origin.&lt;/p&gt;&#xA;" OwnerUserId="1303" LastActivityDate="2010-11-27T09:14:57.473" />
  <row Id="12051" PostTypeId="2" ParentId="12047" CreationDate="2010-11-27T10:29:19.570" Score="2" ViewCount="" Body="&lt;p&gt;For (2)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sec \theta + \tan \theta = \frac{1}{\cos \theta} + \frac{\sin \theta}{\cos \theta}= \frac{1 + \sin \theta}{\cos \theta}$$ $$ = \frac{(1 + \sin \theta)\cdot (1 - \sin \theta)}{\cos \theta \cdot(1 - \sin \theta)}\text{ [Multiplying both sides by } (1 - \sin \theta)\text{]} $$&#xD;&#xA;$$ = \frac{\cos^2 \theta}{ \cos \theta - \sin \theta \cdot \cos \theta} = \frac{1}{\sec \theta - \tan \theta} \text{ (Q.E.D) }$$&lt;/p&gt;&#xA;" OwnerUserId="2109" LastEditorUserId="2109" LastEditorDisplayName="" LastEditDate="2010-11-27T10:42:26.570" LastActivityDate="2010-11-27T10:42:26.570" />
  <row Id="12052" PostTypeId="1" AcceptedAnswerId="12053" CreationDate="2010-11-27T10:38:02.127" Score="5" ViewCount="146" Body="&lt;p&gt;Hey,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given the following series: $\sum_{n=1}^{\infty} (\sqrt[3]{n+1} - \sqrt[3]{n-1})^{\alpha }$ &#xA;where $\alpha \in \mathbb{R}$. Does the series converge or diverge?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Attempts to solve the problem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) $\lim_{n\to\infty} (\sqrt[3]{n+1} - \sqrt[3]{n-1})^{\alpha } = 0$ - not helpful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Used the formula $a^{3} - b^{3} = (a-b)(a^{2} + ab + b^{2})$ - not helpful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) The ration test is not helpful either.&lt;/p&gt;&#xA;" OwnerUserId="3637" LastActivityDate="2010-11-27T10:57:16.480" Title="Series converge or diverge" Tags="&lt;calculus&gt;&lt;homework&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="12053" PostTypeId="2" ParentId="12052" CreationDate="2010-11-27T10:51:22.867" Score="5" ViewCount="" Body="&lt;p&gt;The formula $a^{3} - b^{3} = (a-b)(a^{2} + ab + b^{2})$ actually can be quite helpful.  Note that it implies that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sqrt[3]{n+1}-\sqrt[3]{n-1}=\frac{2}{(n+1)^{2/3}+(n^2-1)^{1/3}+(n-1)^{2/3}}.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For sufficiently large $n$ ($n\gt2$ will do for sure, and you can get tighter bounds by going further), that denominator is squeezed between $\frac{3}{2}n^{2/3}$ and $6n^{2/3}$, and therefore the convergence question is the same as for the series $\sum_{n=1}^\infty n^{-2\alpha/3}$.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2010-11-27T10:57:16.480" LastActivityDate="2010-11-27T10:57:16.480" CommentCount="2" />
  <row Id="12054" PostTypeId="1" AcceptedAnswerId="13755" CreationDate="2010-11-27T11:13:22.527" Score="5" ViewCount="344" Body="&lt;p&gt;Hey there,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using Lagrange's remainder, I have to prove that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\log(1+x) = \sum\limits_{n=1}^\infty (-1)^{n+1} \cdot \frac{x^n}{n}, \; \forall |x| &amp;lt; 1$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not quite sure how to do this. I started with the Taylor series for $x_0 = 0$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f(x_0) = \sum\limits_{n=0}^\infty \frac{f^{(n)}(x_0)}{n!} \cdot x^n + r_n$, where $r_n$ is the remainder. Then, I used induction to prove that the n-th derivative of $\log(1+x)$ can be written as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f^{(n)} = (-1)^{n+1} \cdot \frac{(n-1)!}{(1+x)^n}, \forall n \in \mathbb{N}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I plugged this formula into the Taylor series for $\log(1+x)$ and ended up with:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f(x_0) = \sum\limits_{n=1}^\infty (-1)^{n+1} \cdot \frac{x^n}{n} + r_n$, which already looked quite promising.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the formula which I have to prove doesn't have that remainder $r_n$, I tried to show that $\lim_{n \to \infty} r_n = 0$, using Lagrange's remainder formula (for $x_0 = 0$ and $|x| &amp;lt; 1$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So now I basically showed that the formula was valid for $x \to x_0 = 0$. I also showed that the radius of convergence of this power series is $r = 1$, that is to say the power series converges $\forall |x| &amp;lt; 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is bugging me, is the fact, that to my opinion, the formula is only valid for $x \to 0$. I mean sure, the radius of convergence is 1, but does this actually tell me that the formula is valid within $(-1,1)$? I've never done something like this before, thus the insecurity. I'd be delighted, if someone could help me out and tell me, whether the things I've shown are already sufficient or whether I still need to prove something.&lt;/p&gt;&#xA;" OwnerUserId="3787" LastActivityDate="2010-12-10T17:00:14.863" Title="How to use the Lagrange's remainder to prove that log(1+x) = sum(...)?" Tags="&lt;analysis&gt;&lt;taylor-expansion&gt;&lt;power-series&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="12055" PostTypeId="2" ParentId="12042" CreationDate="2010-11-27T11:33:53.923" Score="3" ViewCount="" Body="&lt;p&gt;Do you have any reason to suspect $L(n)$ should satisfy a linear recurrence?  Here's a way you can prove that it doesn't satisfy a linear recurrence of depth 2 (which can be generalised to any depth).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Step 1&lt;/em&gt;: Compute $L(n)$ for small $n$:  3, 10, 33, 109, 354, 1148, 3716, ...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Step 2&lt;/em&gt;: Assume $b L(n-2)+a L(n-1)=L(n)$ for some $a,b$.  Using the data from Step 1 we obtain the system of linear equations:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$3b+10a=33$,&lt;/li&gt;&#xA;&lt;li&gt;$10b+33a=109$,&lt;/li&gt;&#xA;&lt;li&gt;$33b+109a=354$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In fact, we can keep going forever adding equations $109b+354a=1148$, and so on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Step 3&lt;/em&gt;: Solve the system of linear equations (or get your computer to do it for you (using e.g. &lt;a href=&quot;http://www.wolframalpha.com/input/?i=3b%2B10a%3D33,+10b%2B33a%3D109,+33b%2B109a%3D354&quot; rel=&quot;nofollow&quot;&gt;WolframAlpha&lt;/a&gt;)).  In this case there are no solutions, so $L(n)$ does not satisfy a linear recurrence of depth 2.  If you feel that the starting point shouldn't be $L(1)$, you can use the same argument starting later in the sequence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming I coded things correctly, I have checked that $L(n)$ doesn't satisfy a linear recurrence of depth 10 (or less).  [It's probably a good idea to check this yourself if you end up relying on this result.]  I also attempted to find a linear recurrence with polynomial coefficients for $L(n)$ to a limited extent (see &lt;a href=&quot;http://www.math.upenn.edu/~wilf/AeqB.html&quot; rel=&quot;nofollow&quot;&gt;A=B&lt;/a&gt; about Sister Celine's Technique for more info).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, if you're allowed to use an auxiliary function, then let $s(1)=1+\sqrt{5}$ and for $n \geq 2$ let $s(n)=(1+\sqrt{5}) \cdot s(n-1)$.  Then $L(n)=\lfloor s(n) \rfloor$ for all $n \geq 1$.&lt;/p&gt;&#xA;" OwnerUserId="139" LastActivityDate="2010-11-27T11:33:53.923" CommentCount="2" />
  <row Id="12056" PostTypeId="2" ParentId="12047" CreationDate="2010-11-27T11:35:04.217" Score="3" ViewCount="" Body="&lt;p&gt;Added (and corrected twice):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\tan \alpha \tan \beta =1\Leftrightarrow \dfrac{\sin \alpha }{\cos \alpha }%&#xD;&#xA;\dfrac{\sin \beta }{\cos \beta }=1$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Multiplying by $\cos\alpha\cos\beta\ne 0$, gives&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sin \alpha \sin \beta -\cos \alpha\cos \beta =0 \Leftrightarrow \cos (\alpha +\beta )=0$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is equivalent to &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\alpha +\beta =\dfrac{\pi }{2}+n\pi,\qquad (\ast)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;to which we still have to add the condition written above ($\cos\alpha\cos\beta\ne 0$), which means the constraint &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\alpha,\beta\ne\dfrac{\pi}{2}+n\pi,\qquad (\ast\ast)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $n$ is an integer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: In the original equation $\tan \alpha \tan \beta =1$, neither $\alpha$ nor $\beta$ can be zero.  The combined conditions $(\ast)$ and $(\ast\ast)$ assures that.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The identity $$\sec \theta +\tan \theta =\dfrac{1}{\sec \theta -\tan \theta }\qquad \theta \neq (2n+1)\dfrac{\pi}{2}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is equivalent to $$\sin ^{2}\theta +\cos ^{2}\theta =1.$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Indeed, if &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\theta \neq (2n+1)\dfrac{\pi }{2}\Leftrightarrow \sin \theta \neq \pm 1 \Leftrightarrow \dfrac{\pm1}{\cos \theta }-\dfrac{\sin \theta }{\cos \theta }\neq 0\Leftrightarrow \pm\sec \theta -\tan \theta \neq 0,$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sec \theta +\tan \theta =\dfrac{1}{\sec \theta -\tan \theta }\Leftrightarrow \left( \sec \theta +\tan \theta \right) \left( \sec \theta&#xD;&#xA;-\tan \theta \right) =1$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Leftrightarrow \sec ^{2}\theta -\tan ^{2}\theta =1\Leftrightarrow \dfrac{1}{\cos ^{2}\theta }-\dfrac{\sin ^{2}\theta }{\cos^{2}\theta }=1\Leftrightarrow 1-\sin ^{2}\theta=\cos ^{2}\theta$$ $$\Leftrightarrow \sin ^{2}\theta +\cos ^{2}\theta =1.$$&lt;/p&gt;&#xA;" OwnerUserId="752" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2010-12-01T15:06:28.400" LastActivityDate="2010-12-01T15:06:28.400" CommentCount="5" />
  <row Id="12057" PostTypeId="1" AcceptedAnswerId="12059" CreationDate="2010-11-27T11:45:24.857" Score="0" ViewCount="166" Body="&lt;p&gt;Let $p,q$ belong to $\mathbb{N}$ and are relatively prime to each other. If $\alpha,\beta$ belong to $\mathbb{N}$, are also relatively prime to each other,then are $(p\beta+q\alpha)$ and $q\beta$ always relatively prime ?&lt;/p&gt;&#xA;" OwnerUserId="2987" LastEditorUserId="2987" LastEditorDisplayName="" LastEditDate="2010-11-27T12:29:03.593" LastActivityDate="2010-12-28T02:15:51.310" Title="a basic question on relatively prime numbers" Tags="&lt;number-theory&gt;" AnswerCount="3" />
  <row Id="12058" PostTypeId="1" CreationDate="2010-11-27T11:49:37.903" Score="1" ViewCount="46" Body="&lt;p&gt;Hi everyone,&#xA;I have a problem of commutative algebra. Take $A$ an excellent integrally closed noetherian ring with field of quotients $K$, $I\subset A$ a radical ideal such that $A$ is $I$-adically complete. Consider $G=Spec(A)\times \mathbb{G}_m^r$ and $Y\subset G(K)$ a subgroup isomorphic to $\mathbb{Z}^r$. Let $X$ the set of character of $\mathbb{G}_m^r$ and consider an homomorphism $f: Y\rightarrow X$ such that $X_{f(y)}(z)=X_{f(z)}(y)$ and $X_{f(y)}(y)\in I$ for every $y,z\in Y$ where $X_{a}$ is the element of $\Gamma(G,O_G)$ corresponding to $a\in X$.&#xA;The question is the following: why there exists only a finite set of minimal primes $p\subset A$ such that $ord_p X_{f(y)}(z)\neq 0$ for all $y,z\in Y$?&lt;/p&gt;&#xA;" LastActivityDate="2010-11-27T11:49:37.903" Title="existence of finitely many non zero valuation" Tags="&lt;algebraic-geometry&gt;" />
  <row Id="12059" PostTypeId="2" ParentId="12057" CreationDate="2010-11-27T11:51:38.610" Score="7" ViewCount="" Body="&lt;p&gt;The answer is clearly no: take $p=\alpha$, $q=\beta$. Then $p\beta+q\alpha=2pq$ and $q\beta=pq$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If however $\beta$ is co-prime to $q$, then the answer is equally clearly yes, since no divisor of $q$ divides $p\beta$ and no divisor of $\beta$ divides $q\alpha$.&lt;/p&gt;&#xA;" OwnerUserId="3212" LastActivityDate="2010-11-27T11:51:38.610" CommentCount="0" />
  <row Id="12060" PostTypeId="2" ParentId="12057" CreationDate="2010-11-27T11:55:00.463" Score="2" ViewCount="" Body="&lt;p&gt;No, e.g. $p=3,q=5,\alpha=3,\beta=5$ is a counterexample.&lt;/p&gt;&#xA;" OwnerUserId="3249" LastActivityDate="2010-11-27T11:55:00.463" />
  <row Id="12061" PostTypeId="2" ParentId="12054" CreationDate="2010-11-27T11:56:16.257" Score="6" ViewCount="" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;$f(x_0) = \sum\limits_{n=1}^\infty (-1)^{n+1} \cdot \frac{x^n}{n} + r_n$&#xD;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That should say&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f(x)=\sum_{n=1}^k (-1)^{n+1} \cdot \frac{x^n}{n} + r_k(x),$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $r_k$ is the error term of the $k^\text{th}$ partial sum.  You want to use estimates to show that the error term goes to $0$ as $k$ goes to $\infty$, which will justify convergence of the series to $f(x)=\log(1+x)$.  &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Edit: I've struck through part of my answer that relied on a wrong estimate of the derivatives, as pointed out by Robert Pollack.  With the missing $k!$ term, the estimate only works on $[-\frac{1}{2},1)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Added: To make this answer a little more useful, I decided to look up a correct method.  Spivak in his book &lt;em&gt;Calculus&lt;/em&gt; (3rd Edition, page 423) uses the formula &#xA;$$\frac{1}{1+t}=1-t+t^2-\cdots+(-1)^{n-1}t^{n-1}+\frac{(-1)^nt^n}{1+t}$$&#xD;&#xA;in order to write the remainder as $r_n(x)=(-1)^n\int_0^x\frac{t^n}{1+t}dt$.  The estimate $\int_0^x\frac{t^n}{t+1}dt\leq\int_0^xt^ndt=\frac{x^{n+1}}{n+1}$ holds when $x\geq0$, and the harder estimate &#xA;$\left|\int_0^x\frac{t^n}{1+t}\right|\leq\frac{|x|^{n+1}}{(1+x)(n+1)}$, when $-1\lt x\leq0$,&#xA;is given as Problem 11 on page 430.  Combining these, you can show that the sequence of remainders converges uniformly to $0$ on $[-r,1]$ for each $r\in(0,1)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strike&gt;Lagrange's form of the error term can be used to do this.  The estimates, which follow from Taylor's theorem, are also found on &lt;a href=&quot;http://en.wikipedia.org/wiki/Taylor%27s_theorem#Estimates_of_the_remainder&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;.  In this case, if $0\lt r\lt 1$, then $|f^{k+1}(x)|\leq \frac{1}{(1-r)^{k+1}}$ whenever $x\geq-r$, so you have the estimate $|r_k(x)|\leq \frac{r^{k+1}}{(1-r)^{k+1}}\frac{1}{(k+1)!}$ for all $x$ in $(-r,r)$, which you can show goes to $0$ (because (k+1)! grows faster than the exponential function $\left(\frac{r}{(1-r)}\right)^{k+1}$), thus showing that the series converges uniformly to $\log(1+x)$ on $(-r,r)$.  Since $r$ was arbitrary, this shows that the series converges on $(-1,1)$, and the convergence is uniform on compact subintervals.&lt;/strike&gt;&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2010-12-10T04:41:03.077" LastActivityDate="2010-12-10T04:41:03.077" CommunityOwnedDate="2010-12-10T04:41:03.077" CommentCount="6" />
  <row Id="12062" PostTypeId="1" CreationDate="2010-11-27T12:24:07.040" Score="1" ViewCount="162" Body="&lt;p&gt;With help from Maple, I got &#xA;$$\left(\frac{ax+by+cz}{x-y}\right)^2+\left(\frac{ay+bz+cx}{y-z}\right)^2+\left(\frac{az+bx+cy}{z-x}\right)^2-(c-a)^2-(c-b)^2$$&#xD;&#xA;equal to&#xA;$$\frac{(c(x^3+y^3+z^3)+(a-c)(x^2y+y^2z+z^2x)+(b-c)(x^2z+y^2x+z^2y)-3(a+b-c)xyz)^2}{(x-y)^2(y-z)^2(x-z)^2}$$ which of course is $\ge 0$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But with no help from a computer algebra, how would one prove:$$\left(\frac{ax+by+cz}{x-y}\right)^2+\left(\frac{ay+bz+cx}{y-z}\right)^2+\left(\frac{az+bx+cy}{z-x}\right)^2\ge (c-a)^2+(c-b)^2 ?$$&lt;/p&gt;&#xA;" OwnerUserId="3249" LastActivityDate="2010-11-27T17:20:52.033" Title="How to prove this inequality without use of computers?" Tags="&lt;inequality&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="1" />
  <row Id="12063" PostTypeId="1" CreationDate="2010-11-27T12:41:52.823" Score="3" ViewCount="91" Body="&lt;p&gt;Recently, I consider a problem: Let $E/F$ is a Galois extension of number field, denote the idele group of $E$ (resp $F$) by $I_E$ (resp $I_F$). There is a homomorphism induced by norm map $N_{E/F}$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$N: I_E/E^{\ast} \to I_F/F^{\ast}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there is a finite abel extension $K/E$ ,from Global Class Field Theory,there is an isomorphism called Artin map:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$A_K/E: I_E/E^{\ast} \text{Nm}(I_K) \to \text{Gal}(L/K)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there is a subset $U$ of $I_E/E^{\ast}$ s.t $A_K/E(U) = 1$, where $1$ is the identity element of $\text{Gal}(K/E)$, So i really want know is there a special relationship between $U$ and the set $\text{Ker}(I_E/E{\ast} \to I_F/F^{\ast})$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="4138" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-11-27T12:51:42.113" LastActivityDate="2010-11-27T12:51:42.113" Title="A subtle relationship from class field theory" Tags="&lt;number-theory&gt;&lt;algebraic-number-theory&gt;" CommentCount="1" FavoriteCount="1" />
  <row Id="12065" PostTypeId="1" CreationDate="2010-11-27T14:08:20.037" Score="4" ViewCount="315" Body="&lt;p&gt;How can we prove that the product of n consecutive integers is divisible by n factorial?&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Note: In this &lt;a href=&quot;http://math.stackexchange.com/questions/12067/the-product-of-n-consecutive-integers-is-divisible-by-n-without-using-the-prope&quot;&gt;subsequent question&lt;/a&gt; and the comments here the OP has clarified that he seeks a proof that &quot;does not use the properties of binomial coefficients&quot;. Please post answers in said newer thread so that this incorrectly-posed question may be closed as a duplicate.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2938" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-11-28T00:12:14.287" LastActivityDate="2010-11-28T00:12:14.287" Title="The product of n consecutive integers is divisible by n factorial" Tags="&lt;number-theory&gt;" AnswerCount="4" CommentCount="12" FavoriteCount="1" />
  <row Id="12066" PostTypeId="2" ParentId="12065" CreationDate="2010-11-27T14:18:45.863" Score="5" ViewCount="" Body="&lt;p&gt;This is almost immediate from the fact that the binomial coefficient $$\binom{k+n}{k}$$ is an integer. Just write the product $(k+1) \cdots (k+n)$ accordingly and you'll have your answer.&lt;/p&gt;&#xA;" OwnerUserId="900" LastActivityDate="2010-11-27T14:18:45.863" />
  <row Id="12067" PostTypeId="1" CreationDate="2010-11-27T14:33:38.090" Score="0" ViewCount="345" Body="&lt;p&gt;How can we prove, without using the properties of binomial coefficients, the product of n consecutive integers is divisible by n factorial?&lt;/p&gt;&#xA;" OwnerUserId="2938" LastActivityDate="2010-11-27T20:59:43.410" Title="The product of n consecutive integers is divisible by n! (without using the properties of binomial coefficients)" Tags="&lt;number-theory&gt;" AnswerCount="3" CommentCount="17" FavoriteCount="1" />
  <row Id="12068" PostTypeId="1" AcceptedAnswerId="12080" CreationDate="2010-11-27T14:49:08.863" Score="3" ViewCount="217" Body="&lt;p&gt;If $x_i$ is exponentially distributed $(i=1,...,n)$ with parameter $\lambda$ and $x_i$'s are mutually independent, what is the expectation of&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$E\left\{ \left(\sum_{i=1}^n {x_i} \right)^2 \right\}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;in terms of $n$ and $\lambda$ and possibly other constants?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Note:&lt;/strong&gt; This question has gotten a statistical answer on &lt;a href=&quot;http://stats.stackexchange.com/q/4959/2148&quot;&gt;http://stats.stackexchange.com/q/4959/2148&lt;/a&gt;. The readers would take a look at it too.&lt;/p&gt;&#xA;" OwnerUserId="4051" LastEditorUserId="4051" LastEditorDisplayName="" LastEditDate="2011-03-05T18:05:09.870" LastActivityDate="2011-03-05T18:05:09.870" Title="expectation of $E\left\{ \left(\sum_{i=1}^n {x_i} \right)^2 \right\}$" Tags="&lt;statistics&gt;&lt;probability&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="12069" PostTypeId="1" AcceptedAnswerId="12071" CreationDate="2010-11-27T14:58:54.910" Score="1" ViewCount="114" Body="&lt;p&gt;For example, if $f(x) = \sin x$ and $x$ is uniformly distributed on $[0, \pi]$, how is the equation found that satisfies the probability distribution function of $f(x)$? I imagine the distribution function will be greater when the derivative of $f(x)$ is closer to zero, but this is just a guess.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I apologize if this question is vague or not advanced enough, but I can't find the answer anywhere.&lt;/p&gt;&#xA;" OwnerUserId="227" LastActivityDate="2010-12-29T18:03:44.410" Title="Given a function $f(x)$ where $x$ is uniformly distributed between $a$ and $b$, how do I find the probability density function of $f$?" Tags="&lt;probability&gt;&lt;statistics&gt;" AnswerCount="2" />
  <row Id="12070" PostTypeId="2" ParentId="12065" CreationDate="2010-11-27T15:13:55.950" Score="0" ViewCount="" Body="&lt;p&gt;The identity below shows that the problem is equivalent to the fact that binomial coefficients are integral - for which various proofs are known, e.g. using their recursion, or their well-known combinatorial  interpretation, or their minimality in terms of prime divisors - see this &lt;a href=&quot;http://math.stackexchange.com/questions/11601/proof-that-a-combination-is-an-integer&quot;&gt;prior question&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\rm\displaystyle\quad\quad {m \choose n}\ =\ \frac{m!/(m-n)!}{n!}\ =\ \frac{m\:(m-1)\:\cdots\:(m-n+1)}{n\:(n-1)\quad\quad\:\cdots\:\quad\quad 1\quad\quad}$$&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-11-27T15:13:55.950" />
  <row Id="12071" PostTypeId="2" ParentId="12069" CreationDate="2010-11-27T15:30:32.713" Score="2" ViewCount="" Body="&lt;p&gt;Note that $\sin(x)$ increases from $x = 0$ to $x = {\pi \over 2}$, then decreases from ${\pi \over 2}$ to $\pi$, in a way symmetric about ${\pi \over 2}$. So for a given $0 \leq \alpha \leq 1$, the $x \in [0,\pi]$ for which $\sin(x) \leq \alpha$ consists of two segments, $[0,\beta]$ and $[\pi - \beta, \pi]$, where $\beta$ is the number for which $\sin(\beta) = \alpha$. In other words $\beta = \arcsin(\alpha)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $x$ is uniformly distributed on $[0,\pi]$, the probability $x$ is in $[0,\beta]$ is ${\beta \over \pi}$, and the probability $x$ is in $[\pi - \beta, \pi]$ is also ${\beta \over \pi}$. So the chance that $x$ is in one of these two segments is $2{\beta \over \pi}$. This means the probability $\sin(x) \leq \alpha$ is $2{\beta \over \pi}$, or ${2 \over \pi} \arcsin(\alpha)$. Thus this gives the distribution function of $\sin(x)$. The density function is obtained by differentiating with respect to $\alpha$; the result is ${2 \over \pi \sqrt{1 - \alpha^2}}$.&lt;/p&gt;&#xA;" OwnerUserId="3035" LastActivityDate="2010-11-27T15:30:32.713" CommentCount="1" />
  <row Id="12072" PostTypeId="2" ParentId="12067" CreationDate="2010-11-27T15:33:11.357" Score="0" ViewCount="" Body="&lt;p&gt;Since this statement is equivalent to the fact that binomial coefficients are integral it is a bit tricky to make precise the notion of a proof that does not &quot;use properties of binomial coefficients&quot;. I will interpret this proviso to mean that the proof is not essentially combinatorial, i.e. it does not employ properties that are immediate from the combinatorial interpretation of binomial coefficients.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In my &lt;a href=&quot;http://math.stackexchange.com/questions/2158/division-of-factorials/2192#2192&quot;&gt;post here&lt;/a&gt; you'll find a purely arithmetical proof that $\rm\: n!\: $ divides the product of $\rm\:n\:$ consecutive integers. The proof shows how to rewrite the associated fraction as a product of fractions whose denominators are all coprime to any given prime $\rm\:p\:$. This implies that no primes divide the denominator (when written in lowest terms), hence the fraction is an integer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The key property that lies at the heart of this proof is that, among all products of $\rm\: n\:$ consecutive integers, $\rm\ n!\ $ has the least possible power of $\rm\:p\:$ dividing it - for all primes $\rm\:p\:$. Thus $\rm\ n!\ $ divides every product of $\rm\:n\:$ consecutive integers, since it has a smaller power of every prime divisor. &lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-11-27T15:45:52.353" LastActivityDate="2010-11-27T15:45:52.353" CommentCount="1" />
  <row Id="12073" PostTypeId="2" ParentId="12065" CreationDate="2010-11-27T15:37:57.733" Score="3" ViewCount="" Body="&lt;p&gt;You might be interested in this blog post of Timothy Gowers:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://gowers.wordpress.com/2010/09/18/are-these-the-same-proof/&quot; rel=&quot;nofollow&quot;&gt;http://gowers.wordpress.com/2010/09/18/are-these-the-same-proof/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3736" LastActivityDate="2010-11-27T15:37:57.733" CommunityOwnedDate="2010-11-27T15:37:57.733" CommentCount="1" />
  <row Id="12074" PostTypeId="2" ParentId="11669" CreationDate="2010-11-27T16:03:00.670" Score="2" ViewCount="" Body="&lt;p&gt;I think in a tiny tiny nutshell... the reason for the 12-division is because a very practical solution for Western music, and the layout of black/white &quot;evolved&quot; into this form because it lacked re-engineering.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is no particularly &quot;mathemagical&quot; thing about it. In other words... square two: it's an arbitrary choice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're looking for something that uses 12-divided octave as a practical solution and is engineered for facility, check out the layout of the Russian Bayan (accordion). It's pretty awesome.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for something that is engineered for facility but does not divide the octave into 12 parts, your common fretless string instruments are good examples.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again, all I've said has been mentioned above. Just beware of the overtly &quot;mathemagical&quot; ones, they don't say much about the music but rather put it in a fancy straightjacket.&lt;/p&gt;&#xA;" OwnerUserId="4142" LastActivityDate="2010-11-27T16:03:00.670" />
  <row Id="12076" PostTypeId="2" ParentId="12067" CreationDate="2010-11-27T16:10:22.777" Score="3" ViewCount="" Body="&lt;p&gt;There is an alternate proof at the following link which uses the result about the highest power of a prime dividing a factorial (scroll down for the proof)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://2000clicks.com/MathHelp/BasicFactorialConsecutiveIntegerProducts.aspx&quot; rel=&quot;nofollow&quot;&gt;http://2000clicks.com/MathHelp/BasicFactorialConsecutiveIntegerProducts.aspx&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3431" LastActivityDate="2010-11-27T16:10:22.777" CommentCount="1" />
  <row Id="12077" PostTypeId="1" AcceptedAnswerId="12079" CreationDate="2010-11-27T16:27:24.993" Score="1" ViewCount="105" Body="&lt;p&gt;Let $(X,d)$ be a metric space and $S \subset X$. Show that $d_S(x):=\text{inf}\{d(x,s): s \in S\}=0 \Leftrightarrow x \in \overline S .$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notes: $\overline S$ is the closure of S. Maybe you can use that a closed set is also closed for sequences in the set? I think the difficult part is when S is open, otherwise its trivial as the closure would be equal to S.&lt;/p&gt;&#xA;" OwnerUserId="3123" LastActivityDate="2010-11-27T16:39:37.637" Title="Closure of a subset in a metric space" Tags="&lt;topology&gt;&lt;metric-spaces&gt;" AnswerCount="2" CommentCount="5" />
  <row Id="12078" PostTypeId="2" ParentId="12077" CreationDate="2010-11-27T16:36:58.123" Score="1" ViewCount="" Body="&lt;p&gt;If $x \in \overline{S}$, then there is a sequence $(s_n)$ in $S$ converging to $x$ (e.g. take $s_n$ to be some element in $S \cap B(x;\tfrac{1}{n})$). Then $d_S(x) \leq \inf\lbrace d(x,s_n) \;\vert\; n \in \mathbb{N}\rbrace = 0$ (why?).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the other directtion, we prove the contrapositive: If $x \notin \overline{S}$, then there is some ball around $x$ disjoint from $S$. This gives that the infimum in $d_S(x)$ must be greater than $0$ (why?).&lt;/p&gt;&#xA;" OwnerUserId="1269" LastActivityDate="2010-11-27T16:36:58.123" CommentCount="2" />
  <row Id="12079" PostTypeId="2" ParentId="12077" CreationDate="2010-11-27T16:39:37.637" Score="4" ViewCount="" Body="&lt;p&gt;If $x\notin \overline{S}$, then there exists $r&amp;gt;0$ such that $B_r(x)\cap S=\phi$. It follows that $d(x,s)\ge r&amp;gt;0$ for all $s\in S$ and hence $d_S(x)&amp;gt;0$. &#xA;And these steps can be reversed, i.e. the steps above are actually &quot;if and only if&quot;. So you get the proof.&lt;/p&gt;&#xA;" OwnerUserId="3249" LastActivityDate="2010-11-27T16:39:37.637" CommentCount="2" />
  <row Id="12080" PostTypeId="2" ParentId="12068" CreationDate="2010-11-27T16:43:33.123" Score="5" ViewCount="" Body="&lt;p&gt;In what follows, $x_j$ are assumed to be independent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ E[x_0^2] = \int_0^{\infty} \lambda t^2 e^{-\lambda t}  dt = \frac{2}{\lambda^2} $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ E[x_0 x_1] = \int_0^{\infty} \int_0^{\infty} \lambda^2 t_0 t_1 e^{-\lambda t_0} e^{-\lambda t_1 } dt_0 dt_1 = \frac{1}{\lambda^2} $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ E[ (\sum_{j=0}^{n-1} x_j)^2 ] = n E[ x_0^2] + n(n-1) E[x_0 x_1] = \frac{ n^2 + n }{ \lambda^2 } $$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="4143" LastEditorUserId="4143" LastEditorDisplayName="" LastEditDate="2010-11-27T16:57:14.777" LastActivityDate="2010-11-27T16:57:14.777" CommentCount="3" />
  <row Id="12081" PostTypeId="1" CreationDate="2010-11-27T16:59:19.983" Score="1" ViewCount="90" Body="&lt;p&gt;In my website, users can either &quot;like&quot; or &quot;dislike&quot; a posted comment. I want to put a link to sort comments by liking such that the most liked ones becomes on top. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course I cannot just sort by the number of likes only. I have to subtract the dislikes. But what if the difference (likes - dislikes) are the same. e.g. 10 - 8 = 2 and 5 - 3 = 2. I think in this case, the comment of 10 likes and 8 dislikes has to come before the 5 likes and 3 dislikes comment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So is there an equation that you feed it the number of likes and number of dislikes and then it gives you a meaningful rating number that I can sort with?&lt;/p&gt;&#xA;" OwnerUserId="4145" LastActivityDate="2010-11-27T23:50:19.010" Title="Equation to calculate best liked comments" Tags="&lt;sorting&gt;" AnswerCount="2" CommentCount="6" />
  <row Id="12082" PostTypeId="1" CreationDate="2010-11-27T17:00:27.580" Score="8" ViewCount="176" Body="&lt;p&gt;If there is a field $F$ that is a field reduction of the real numbers, that is $F(a)=\mathbb{R}$ for some $a$, let's also denote this $F=\mathbb{R}(\setminus a)$, then given $x \in \mathbb{R}$ is there a general method to determine whether $x$ is in $F$ or $x$ is in $\mathbb{R}\setminus F$ ?&lt;/p&gt;&#xA;" OwnerUserId="4144" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-28T04:20:59.577" LastActivityDate="2010-11-28T04:20:59.577" Title="Field reductions" Tags="&lt;abstract-algebra&gt;&lt;field-theory&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="2" />
  <row Id="12083" PostTypeId="2" ParentId="10937" CreationDate="2010-11-27T17:01:22.133" Score="1" ViewCount="" Body="&lt;p&gt;I guess your question is: Why is $E(X)=\int_0^\infty 1-F(x) dx$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer is as follows: By definition $$E(X)=\int_0^\infty tf(t) dt.$$&#xD;&#xA;Now \begin{eqnarray*} &amp;amp; &amp;amp; \int_0^\infty tf(t) dt\&#xA;&amp;amp;=&amp;amp;\int_0^\infty f(t)\int_0^t 1 dx dt\&#xA;&amp;amp;=&amp;amp;\int_0^\infty\int_0^t f(t) dx dt\&#xA;&amp;amp;=&amp;amp;\int_0^\infty\int_x^\infty f(t) dt dx \&#xA;&amp;amp;=&amp;amp;\int_0^\infty 1-F(x) dx.&#xA;\end{eqnarray*}&lt;/p&gt;&#xA;" OwnerUserId="3249" LastActivityDate="2010-11-27T17:01:22.133" />
  <row Id="12084" PostTypeId="1" AcceptedAnswerId="12086" CreationDate="2010-11-27T17:17:57.340" Score="1" ViewCount="134" Body="&lt;p&gt;The empty set is a member of $P({a,b}) \times P({p,q})$. True or false?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My first instinct was false, since the empty set is a member of each power set individually, but when multiplied together, you get {0,0}, which I'm not sure represents the empty set. But my counter argument is that the empty set is a member of the power set of anything, right?&lt;/p&gt;&#xA;" OwnerUserId="2985" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2010-11-27T21:30:59.580" LastActivityDate="2010-11-27T21:30:59.580" Title="Empty set and power sets" Tags="&lt;elementary-set-theory&gt;" AnswerCount="1" />
  <row Id="12085" PostTypeId="2" ParentId="12062" CreationDate="2010-11-27T17:20:52.033" Score="3" ViewCount="" Body="&lt;p&gt;This problem may be amenable to the &quot;Purkiss Principle&quot;. I'll leave it as an exercise for you to determine if it can be applied here. But even if not you should read the following beautiful article on it by Wm. Waterhouse &lt;a href=&quot;http://mathdl.maa.org/mathDL/22/?pa=content&amp;amp;sa=viewDocument&amp;amp;nodeId=2941&quot; rel=&quot;nofollow&quot;&gt;Do Symmetric Problems Have Symmetric Solutions?&lt;/a&gt; I recall thinking that this was one of the most beautiful Monthly articles that I ever read as an undergraduate. Apparently others felt similarly since it won a prestigious Lester R. Ford award for expository excellence.&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-11-27T17:20:52.033" CommentCount="1" />
  <row Id="12086" PostTypeId="2" ParentId="12084" CreationDate="2010-11-27T17:21:16.683" Score="9" ViewCount="" Body="&lt;p&gt;Your first instinct was right: $P(a,b) \times P(p,q)$ contains $(\emptyset, \emptyset)$, but not $\emptyset$. $(\emptyset, \emptyset)$ is not the empty set, so $P(a,b) \times P(p,q)$ does not contain the empty set.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And yes, every powerset contains the empty set, but $P(a,b) \times P(p,q)$ is not a powerset, it's the cartesian product of two powersets.&lt;/p&gt;&#xA;" OwnerUserId="11" LastActivityDate="2010-11-27T17:21:16.683" />
  <row Id="12087" PostTypeId="2" ParentId="12082" CreationDate="2010-11-27T17:37:59.340" Score="8" ViewCount="" Body="&lt;p&gt;In fact there are no nontrivial &quot;field reductions&quot; of $\mathbb{R}$: if $\mathbb{R} = F(a)$, then $a \in F$ and $F = \mathbb{R}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Case 1: $a$ is algebraic over $F$, hence $F(a) = F[a]$, so $d = [F[a]:F]$ is finite.  Then $[\mathbb{C}:F] = 2d$, i.e., &quot;the&quot; algebraic closure of $F$ has finite degree over $F$.  &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Theorem: Let $K$ be a field and $\overline{K}$ any algebraic closure.  If $[\overline{K}:K] = d$ is finite, then $d = 1$ or $d = 2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proof: This is essentially the Grand Artin-Schreier Theorem (see e.g. Section 12.5 of &lt;a href=&quot;http://math.uga.edu/~pete/FieldTheory.pdf&quot; rel=&quot;nofollow&quot;&gt;http://math.uga.edu/~pete/FieldTheory.pdf&lt;/a&gt; for a proof of that.)  Namely, Artin-Schreier says that if $\overline{K}/K$ is a finite &lt;strong&gt;Galois&lt;/strong&gt; extension, then the degree is either $1$ or $2$.  Certainly $\overline{K}/K$ is normal.  And for the purposes of this question we are in characteristic zero, so everything is separable.  Therefore $\overline{K}/K$ is Galois.  (But here is a proof of the separability in the general case: if $\overline{K}/K$ is not separable, then there is a nontrivial subextension $L$ such that $[\overline{K}:L] = p^n$ and $\overline{K} = L(a^{p^{-n}})$.  But this is impossible: the polynomial $t^p - a$ is irreducible over $L$ iff all of the polynomials $t^{p^n} - a$ are irreducible over $L$.  So $\overline{K}/K$ is separable.)&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Thus we must have $d = 1$, i.e., $F = \mathbb{R}$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Case 2: $a$ is transcendental over $F$.  But then the field $F(a)$ is isomorphic to the rational function field $F(t)$.  Such a field cannot be isomorphic to $\mathbb{R}$, because it admits finite extensions of degree $n$ for all $n \in \mathbb{Z}^+$, e.g. &#xA;$F(t^{\frac{1}{n}})$.  &lt;/p&gt;&#xA;" OwnerUserId="299" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2010-11-27T18:03:03.437" LastActivityDate="2010-11-27T18:03:03.437" CommentCount="18" />
  <row Id="12088" PostTypeId="1" AcceptedAnswerId="12089" CreationDate="2010-11-27T17:52:30.617" Score="2" ViewCount="152" Body="&lt;p&gt;No subset of the real numbers is countable. True or false.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In looking at the &lt;a href=&quot;http://en.wikipedia.org/wiki/Real_number#Uses&quot; rel=&quot;nofollow&quot;&gt;wikipedia article&lt;/a&gt; for real numbers, I'm not really clear on the answer to this. They use the words computable and countable, and I'm not sure of the difference. Also, I dont think I fully understand the term countable. The &lt;a href=&quot;http://en.wikipedia.org/wiki/Countable_set&quot; rel=&quot;nofollow&quot;&gt;definition&lt;/a&gt; states that a set is countable if it has the same cardinality as some subset of the natural numbers, but I'm not sure what that means. &lt;/p&gt;&#xA;" OwnerUserId="2985" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2010-11-27T21:30:29.440" LastActivityDate="2010-11-27T21:30:29.440" Title="Real numbers and countability" Tags="&lt;elementary-set-theory&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" />
  <row Id="12089" PostTypeId="2" ParentId="12088" CreationDate="2010-11-27T17:54:52.020" Score="4" ViewCount="" Body="&lt;p&gt;A set is said to be countable if there is bijection between a  subset of natural numbers (or even integers) and that set.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Countable_set&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Countable_set&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cantor's diagonalization argument shows that the set of reals is uncountable. So, there are uncountably many subsets of real numbers that are countable. For example the set consisting of any single real number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Computability has nothing to do with counting the number of elements in a set. This is related to the idea of computable functions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Computable_function&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Computable_function&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A real number is computable if there is finite algorithm that can yield it's digits to any desired level of accuracy.&lt;/p&gt;&#xA;" OwnerUserId="3431" LastEditorUserId="3431" LastEditorDisplayName="" LastEditDate="2010-11-27T18:29:27.190" LastActivityDate="2010-11-27T18:29:27.190" CommentCount="4" />
  <row Id="12090" PostTypeId="2" ParentId="11434" CreationDate="2010-11-27T18:02:32.883" Score="0" ViewCount="" Body="&lt;p&gt;I'd look at it this way:&lt;br&gt;&#xA;$ f(n)  -f(n-1) = (n^2 + 3n) - (n^2+n-2) = +2n + 2 $ &lt;br&gt;&#xA;$ f(n-1)-f(n-2) = (n^2 + n-2) - (n^2-n-2)= +2n  $     &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So by the first differences the square terms vanish and by the second order differences also the linear terms vanish, so only a constant term remains if second order difference is used, so we can leave this as    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ (f(n)-f(n-1)) - (f(n-1)-f(n-2)) = 1*f(n) - 2*f(n-1) + 1*f(n-2)  = 2 $&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and finally get by this&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ f(n) = 2*f(n-1)-f(n-2) + 2 $ &lt;/p&gt;&#xA;" OwnerUserId="1714" LastActivityDate="2010-11-27T18:02:32.883" />
  <row Id="12091" PostTypeId="1" CreationDate="2010-11-27T18:11:17.953" Score="3" ViewCount="192" Body="&lt;p&gt;well i was wondering if, there s a way in which one could compare these two theories on the ground that which one will provide more accurate answers to a given problem. &#xA;Fuzzy logic always focuses on analysing between 0 and 1 and logic only dolls down to 0 and 1 , so does that imply that fuzzy logic provides us with more insight into a given problem or creates more parameters than necessary and compounds the problem ?&lt;/p&gt;&#xA;" OwnerUserId="3814" LastEditorUserId="3731" LastEditorDisplayName="" LastEditDate="2010-12-20T23:37:32.260" LastActivityDate="2010-12-20T23:37:32.260" Title="Fuzzy Logic vs Logic" Tags="&lt;logic&gt;&lt;fuzzy-logic&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="12092" PostTypeId="2" ParentId="11856" CreationDate="2010-11-27T18:13:57.443" Score="1" ViewCount="" Body="&lt;p&gt;There seems to be fair bit of confusion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First of, there are different notions of &quot;cone&quot;. The standard one refers to a cone on a topological space (see &lt;a href=&quot;http://en.wikipedia.org/wiki/Cone_%28topology%29&quot; rel=&quot;nofollow&quot;&gt;wiki article&lt;/a&gt;), and is defined to be  $CX = (X \times [0,1])/(X \times \{0\})$. This can, in particular, be applied to manifolds. What one gets is a topological space which, in general, is not a manifold, but can, in some special cases, be given a structure of a smooth manifold (see for example the discussion &lt;a href=&quot;http://math.stackexchange.com/questions/7175/topology-of-a-cone-of-rp2&quot;&gt;here&lt;/a&gt;). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if one removes the &quot;cone point&quot; $X\times \{0\}$, what remains is just the product  $X \times (0,1]$, which, as any product of a manifold and a manifold with boundary, is a manifold with boundary. The boundary is $X\times \{1\}$, which can be identified with $X$. This is the kind of cone one talks about in &lt;a href=&quot;http://en.wikipedia.org/wiki/Sasakian_manifold&quot; rel=&quot;nofollow&quot;&gt;Sasakian geometry&lt;/a&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a completely separate issue, but the distinction between smooth and C^1 manifolds is largely nonexistent, due to a theorem of &lt;a href=&quot;http://en.wikipedia.org/wiki/Differential_structure#Existence_and_uniqueness_theorems&quot; rel=&quot;nofollow&quot;&gt;Whitney&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for Sasaki manifolds, those are very special types of Riemannian manifolds. As the wiki referenced above explains, one way to define when a given Riemannian manifold $X$ is Sasaki, is to use the Riemannian metric on $X$ to put a metric  on the cone, and see if the resulting metric makes the cone &lt;a href=&quot;http://en.wikipedia.org/wiki/K%C3%A4hler_manifold&quot; rel=&quot;nofollow&quot;&gt;Kahler&lt;/a&gt;. (The article uses $X \times (0, \infty)$, which is diffeomorphic to the interior of the manifold we have, so this amounts to the same thing.) Usually it will not (Kahler metrics are quite special), but sometimes it will. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is done this way because Kahler manifolds are more well known - they appear as projective or affine complex varieties. The meaning of this construction is that Sasaki manifolds are boundaries of (non-compact) Kahler manifolds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, with these things sorted out, one can actually ask how does one put the Riemannian metric on the cone $X \times (0, \infty)$. The formula is in the wiki article - it calls it the &quot;cone metric&quot;. If it is unclear what it means, please ask - here as a comment, or elsewhere.  &lt;/p&gt;&#xA;" OwnerUserId="2633" LastActivityDate="2010-11-27T18:13:57.443" CommentCount="6" />
  <row Id="12093" PostTypeId="1" AcceptedAnswerId="12094" CreationDate="2010-11-27T18:33:02.860" Score="2" ViewCount="103" Body="&lt;p&gt;Imagine a two-dimensional grid consisting of 20 points along the x-axis and&#xA;10 points along the y-axis. Suppose the origin (0,0) is in the bottom-left corner and the&#xA;point (20,10) is the top-right corner. A path on the grid consists of a series of moves in&#xA;which each move is either one unit to the right or one unit up. Diagonal moves are not&#xA;allowed. How many diﬀerent ways are there to construct a path starting at (0,0) and&#xA;ending at (20,10)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm a little stuck on this one. I feel I'm headed towards the right direction, but I'm not sure if I'm doing this right. For every move, there are 2 possible choices. If we want to get to (20,10), then there are 200 points from the origin to this point. And I think order matters here, so we would use permutations, so I come up with 200 P 2, which is 39,800.&lt;/p&gt;&#xA;" OwnerUserId="2985" LastEditorUserId="1016" LastEditorDisplayName="" LastEditDate="2010-11-27T18:35:26.780" LastActivityDate="2010-11-27T23:16:53.997" Title="Counting number of moves on a grid" Tags="&lt;combinatorics&gt;&lt;discrete-mathematics&gt;&lt;permutations&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" />
  <row Id="12094" PostTypeId="2" ParentId="12093" CreationDate="2010-11-27T18:36:29.087" Score="5" ViewCount="" Body="&lt;p&gt;You have 30 moves to make, 10 of which must be up and 20 of which must be right.  How many ways to pick which are the 10?&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-11-27T18:36:29.087" CommentCount="6" />
  <row Id="12095" PostTypeId="2" ParentId="12081" CreationDate="2010-11-27T18:49:49.670" Score="1" ViewCount="" Body="&lt;p&gt;You have to define what criterion will say that a pair $(L_1,D_1)$ for comment 1 is better than $(L_2,D_2)$ for comment 2.  One way is just to subtract, so $(L_1,D_1) \geq (L_2,D_2)$ if $L_1-D_1 \geq L_2-D_2$.  But on stackexchange there are many more upvotes than downvotes, so maybe you want $(L_1,D_1) \geq (L_2,D_2)$ if $L_1-10*D_1 \geq L_2-10*D_2$ or some such.  Maybe you want to compare on $\frac{L-D}{L+D}$.  There are many choices, and you need to consider your audience and their behavior to select one.  Any such function will map (L,D) to some number, which you can then sort.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-11-27T18:49:49.670" />
  <row Id="12096" PostTypeId="2" ParentId="12093" CreationDate="2010-11-27T19:06:30.277" Score="3" ViewCount="" Body="&lt;p&gt;There are two ways of doing this. One is Ross Millikan's: you will make ten &quot;up&quot; moves, and 20 &quot;right&quot; moves; the only question is which order you make them in. Imagine placing the &quot;right&quot; moves on a row; now you need to decide where to do the &quot;up&quot; moves: you do so by inserting them &quot;in between&quot; (or before, or after) the &quot;right&quot; moves. So you need to choose ten places to put &quot;up&quot; moves: there are 21 locations for them (nineteeen in between the &quot;right&quot; moves, one before all of them, one after), and you are allowed to choose the same location more than once.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a combinations-with-repetitions: the formula is $\binom{n+r-1}{r}$, where you have $n$ possibilities, and must make $r$ choices with repetitions allowed. In this case, $n=21$, $r=10$, so you get $\binom{30}{10}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is another way of doing it, which is more graphical. I'll do it with a 4 by 3 array so you see how it works. You have this array:&#xA;$$\begin{array}{cccc}&#xD;&#xA;\cdot &amp;amp; \cdot &amp;amp; \cdot &amp;amp; \cdot\\&#xD;&#xA;\cdot &amp;amp; \cdot &amp;amp; \cdot &amp;amp; \cdot\\&#xD;&#xA;\cdot &amp;amp; \cdot &amp;amp; \cdot &amp;amp; \cdot&#xD;&#xA;\end{array}$$&#xD;&#xA;Now, you start at the bottom left, so there is only one way to get there; we put a $1$ next to it.&#xA;$$\begin{array}{llll}&#xD;&#xA;\cdot &amp;amp; \cdot &amp;amp; \cdot &amp;amp; \cdot\\&#xD;&#xA;\cdot &amp;amp; \cdot &amp;amp; \cdot &amp;amp; \cdot\\&#xD;&#xA;\cdot\;1&amp;amp; \cdot &amp;amp; \cdot &amp;amp; \cdot&#xD;&#xA;\end{array}$$&#xD;&#xA;Then, you can go either up or right; there is only one way to get to those points (via the first move); we put a $1$ next to them:&#xA;$$\begin{array}{llll}&#xD;&#xA;\cdot &amp;amp; \cdot &amp;amp; \cdot &amp;amp; \cdot\\&#xD;&#xA;\cdot\;1 &amp;amp; \cdot &amp;amp; \cdot &amp;amp; \cdot\\&#xD;&#xA;\cdot\;1&amp;amp; \cdot\;1 &amp;amp; \cdot &amp;amp; \cdot&#xD;&#xA;\end{array}$$&#xD;&#xA;Now: to get to $(1,1)$, you can either get to it from $(1,0)$ or from $(0,1)$; since there is only one way to get to each of those, there are two ways to get to $(1,1)$. On the other hand, only one way to get to $(2,0)$ or to $(0,2)$:&#xA;$$\begin{array}{llll}&#xD;&#xA;\cdot\;1 &amp;amp; \cdot &amp;amp; \cdot &amp;amp; \cdot\\&#xD;&#xA;\cdot\;1 &amp;amp; \cdot\;2 &amp;amp; \cdot &amp;amp; \cdot\\&#xD;&#xA;\cdot\;1&amp;amp; \cdot\;1 &amp;amp; \cdot\;1 &amp;amp; \cdot&#xD;&#xA;\end{array}$$&#xD;&#xA;Next: to get to $(1,2)$, you can arrive either from $(0,2)$ (one way of being there), or from $(1,1)$ (two ways of getting &lt;em&gt;there&lt;/em&gt;); so in total, three ways. Likewise, you have three ways to get to $(2,1)$, because you can either go up from $(2,0)$, and there is only one way to do all of that, or you can go right from $(1,1)$ (and there are two ways of doing that, corresponding to the two ways there are to get &lt;em&gt;to&lt;/em&gt; $(1,1)$; so we have:&#xA;$$\begin{array}{llll}&#xD;&#xA;\cdot \;1 &amp;amp;\cdot\;3 &amp;amp; \cdot &amp;amp; \cdot\\&#xD;&#xA;\cdot\;1 &amp;amp; \cdot\;2 &amp;amp; \cdot\;3 &amp;amp; \cdot\\&#xD;&#xA;\cdot\;1&amp;amp; \cdot\;1 &amp;amp; \cdot\;1 &amp;amp; \cdot&#xD;&#xA;\end{array}$$&#xD;&#xA;Continuing this way, we get:&#xA;$$\begin{array}{llll}&#xD;&#xA;\cdot\;1 &amp;amp; \cdot\;3 &amp;amp; \cdot\;6 &amp;amp; \cdot\;10\\&#xD;&#xA;\cdot\;1 &amp;amp; \cdot\;2 &amp;amp; \cdot\;3 &amp;amp; \cdot\;4\\&#xD;&#xA;\cdot\;1 &amp;amp; \cdot\;1 &amp;amp; \cdot\;1 &amp;amp; \cdot\;1&#xD;&#xA;\end{array}$$&#xD;&#xA;So there are $10$ ways to get to the top right corner in the 4 by 3 case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You may even recognize that these numbers are just Pascal's triangle lying on its side! Well, there is a combinatorial formula for the entries of Pascal's triangle: the $r$th entry in the $m$th row corresponds to the coefficient of $a^{m-r}b^{r-1}$ in the binomial expansion of $(a+b)^{m-1}$, so it equals $\binom{m-1}{r-1}$. To figure out the entry that corresponds to the top right corner, note that you go &quot;down&quot; one row for each position on the $x$-axis, and another one for each step up. So here we have gone to the 4th row on the horizontal steps, and then to the 6th row on the by going up twice. Each step up is a move &quot;right&quot; on the row. So with a 4 by 3, we are in row $4+(3-1)=6$, and we are in position $3$ of that row. According to the formula above, that corresponds to $\binom{4+2-1}{3-1}=\binom{5}{2}$. This corresponds to the need to make $3$ moves right and two moves up, so we need choose where to place the two up moves among the three right moves; there are four places to put them in (before the three, after the three, or in the two spaces in between), so the formula I gave above gives this answer as well.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-27T23:16:53.997" LastActivityDate="2010-11-27T23:16:53.997" CommentCount="1" />
  <row Id="12097" PostTypeId="1" AcceptedAnswerId="12272" CreationDate="2010-11-27T19:08:33.473" Score="5" ViewCount="274" Body="&lt;p&gt;Original post on Mathoverflow &lt;a href=&quot;http://mathoverflow.net/questions/47487/probability-of-the-maximum-levy-stable-random-variable-in-a-list-being-greater&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given a list of identical and independently distributed &lt;a href=&quot;http://en.wikipedia.org/wiki/Stable_distribution&quot; rel=&quot;nofollow&quot;&gt;Levy Stable&lt;/a&gt; random variables, $(X_0, X_1, \dots, X_{n-1})$, what is the is the probability that the maximum exceeds the sum of the rest?  i.e.:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ M = \text{Max}(X_0, X_1, \dots, X_{n-1}) $$&#xD;&#xA;$$ \text{Pr}( M &amp;gt; \sum_{j=0}^{n-1} X_j - M ) $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where, in &lt;a href=&quot;http://academic2.american.edu/~jpnolan/stable/chap1.pdf&quot; rel=&quot;nofollow&quot;&gt;Nolan&lt;/a&gt;'s notation, $X_j \in S(\alpha, \beta=1, \gamma, \delta=0 ; 0)$, where $\alpha$ is the critical exponent, $\beta$ is the skew, $\gamma$ is the scale parameter and $\delta$ is the shift.  For simplicity, I have taken the skew parameter, $\beta$, to be 1 (maximally skewed to the right) and $\delta=0$ so everything has its mode centered in an interval near 0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From numerical simulations, it appears that for the region of $0 &amp;lt; \alpha &amp;lt; 1$, the probability converges to a constant, irregardless of $n$ or $\gamma$.  Below is a plot of this region for $n=500$, $0&amp;lt; \alpha &amp;lt; 1$, where each point represents the result of 10,000 random draws.  The graph looks exactly the same for $n=100, 200, 300$ and $400$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/qX5W7.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $1 &amp;lt; \alpha &amp;lt; 2$ it appears to go as $O(1/n^{\alpha - 1})$ (maybe?) irregardless of $n$ or $\gamma$.  Below is a plot of the probability for $\alpha \in (1.125, 1.3125)$ as a function of $n$.  Note that it is a log-log plot and I have provided the graphs $1/x^{.125}$ and $1/x^{.3125}$ for reference.  It's hard to tell from the graph unless you line them up, but the fit for each is a bit off, and it appears as if the (log-log) slope of the actual data is steeper than my guess for each.  Each point represents 10,000 iterations.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/niTeg.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $\alpha=1$ it's not clear (to me) what's going on, but it appears to be a decreasing function dependent on $n$ and $\gamma$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have tried making a heuristic argument to the in the form of:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\text{Pr}( M &amp;gt; \sum_{j=0}^{n-1} X_j - M) \le n \text{Pr}( X_0 - \sum_{j=1}^{n-1} X_j &amp;gt; 0 )$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then using formula's provided by Nolan (pg. 27) for the parameters of the implicit r.v. $ U = X_0 - \sum_{j=1}^{n-1} X_j$ combined with the tail approximation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \text{Pr}( X &amp;gt; x ) \sim \gamma^{\alpha} c_{\alpha} ( 1 + \beta ) x^{-\alpha} $$&#xD;&#xA;$$ c_{\alpha} = \sin( \pi \alpha / 2) \Gamma(\alpha) / \pi $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but this leaves me nervous and a bit unsatisfied.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just for comparison, if $X_j$ were taken to be uniform r.v.'s on the unit interval, this function would decrease exponentially quickly.  I imagine similar results hold were the $X_j$'s Gaussian, though any clarification on that point would be appreciated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Getting closed form solutions for this is probably out of the question, as there isn't even a closed form solution for the pdf of Levy-Stable random variables, but getting bounds on what the probability is would be helpful.  I would appreciate any help with regards to how to analyze these types of questions in general such as general methods or references to other work in this area.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If this problem is elementary, I would greatly appreciate any reference to a textbook, tutorial or paper that would help me solve problems of this sort.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;UPDATE&lt;/strong&gt;:  George Lowther and Shai Covo have answered this question below.  I just wanted to give a few more pictures that compare their answers to some of the numerical experiments that I did.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Below is the probability of the maximum element being larger than the rest for a list size of $n=100$ as a function of $\alpha$, $\alpha \in (0,1)$.  Each point represents 10,000 simulations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/khCj3.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Below are two graphs for two values of $\alpha \in \{1.53125, 1.875\}$.  Both have the function $ (2/\pi) \sin(\pi \alpha / 2) \Gamma(\alpha) n (( \tan(\pi \alpha/2) (n^{1/\alpha} - n))^{-\alpha} $ with different prescalars in front of them to get them to line up ( $1/4$ and $1/37$, respectively) superimposed for reference.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/6Bt49.png&quot; alt=&quot;alt text&quot;&gt;&#xA;&lt;img src=&quot;http://i.imgur.com/7OBsH.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As George Lowther correctly pointed out, for the relatively small $n$ being considered here, the effect of the extra $n^{1/\alpha}$ term (when $1 &amp;lt; \alpha &amp;lt; 2$) is non-negligible and this is why my original reference plots did not line up with the results of the simulations.  Once the full approximation is put in, the fit is much better.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I get around to it, I will try and post some more pictures for the case when $\alpha=1$ as a function of $n$ and $\gamma$.&lt;/p&gt;&#xA;" OwnerUserId="4143" LastEditorUserId="4143" LastEditorDisplayName="" LastEditDate="2010-12-01T07:36:23.203" LastActivityDate="2010-12-01T07:36:23.203" Title="Probability of the maximum (Levy Stable) random variable in a list being greater than the sum of the rest?" Tags="&lt;probability&gt;&lt;analysis&gt;&lt;reference-request&gt;" AnswerCount="4" CommentCount="13" FavoriteCount="1" />
  <row Id="12098" PostTypeId="1" AcceptedAnswerId="12105" CreationDate="2010-11-27T19:09:33.090" Score="3" ViewCount="504" Body="&lt;p&gt;It's not really typical math question,today while studying graphs I suddenly got inquisitive about whether there exiat any function that could possible to draw a heart-shaped graph.Out of sheer curiousity I clicked on Google, which took me to this &lt;a href=&quot;http://mathworld.wolfram.com/HeartCurve.html&quot; rel=&quot;nofollow&quot;&gt;page&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The page seems informative, and I am glad to learn certain new things! Now I am interested in drawing them by my own by using mathematica.So my question is it possible to draw them in mathematica if yes please show me how.&lt;/p&gt;&#xA;" OwnerUserId="2109" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-28T01:17:21.810" LastActivityDate="2010-11-28T03:15:14.563" Title="Drawing heart in mathematica" Tags="&lt;geometry&gt;&lt;plane-curves&gt;&lt;mathematica&gt;" AnswerCount="4" CommentCount="2" FavoriteCount="2" />
  <row Id="12099" PostTypeId="2" ParentId="12098" CreationDate="2010-11-27T19:14:38.520" Score="1" ViewCount="" Body="&lt;p&gt;This is really about plotting polar plots, parametric plots and implicitly defined functions in Mathematica.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the info on how to draw polar plots&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://mathworld.wolfram.com/PolarPlot.html&quot; rel=&quot;nofollow&quot;&gt;http://mathworld.wolfram.com/PolarPlot.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Parametric plots&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://reference.wolfram.com/mathematica/ref/ParametricPlot.html&quot; rel=&quot;nofollow&quot;&gt;http://reference.wolfram.com/mathematica/ref/ParametricPlot.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This provides info on implicit plots&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://grosz.math.txstate.edu/~dhaz/prob_sets/LTs09cal1lab8.pdf&quot; rel=&quot;nofollow&quot;&gt;http://grosz.math.txstate.edu/~dhaz/prob_sets/LTs09cal1lab8.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3431" LastEditorUserId="3431" LastEditorDisplayName="" LastEditDate="2010-11-27T19:30:45.427" LastActivityDate="2010-11-27T19:30:45.427" CommentCount="2" />
  <row Id="12100" PostTypeId="2" ParentId="12088" CreationDate="2010-11-27T19:19:12.403" Score="3" ViewCount="" Body="&lt;p&gt;It is false, because you can always pick a countable subset of your choice (e.g. {1, 2}). A set is countable if it can be put into one-to-one correspondence with the set of Natural numbers or some subset of Natural numbers. Since Natural numbers are contained in Real numbers, you can always pick a subset (like the one mentioned above) that would be countable.&lt;/p&gt;&#xA;" OwnerUserId="3731" LastEditorUserId="3731" LastEditorDisplayName="" LastEditDate="2010-11-27T19:37:02.873" LastActivityDate="2010-11-27T19:37:02.873" />
  <row Id="12101" PostTypeId="2" ParentId="11964" CreationDate="2010-11-27T19:25:40.980" Score="0" ViewCount="" Body="&lt;p&gt;Your question is equivalent to asking whether any element of {y-1, y-2, $\ldots$, y-100} has a divisor in the range (a,b).  So if you can factor all the 100 numbers, you can check that.  Unfortunately, factoring is pretty hard and then you have the NP complete knapsack problem to solve in looking through all the factors for one that is in range.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-11-27T19:25:40.980" />
  <row Id="12102" PostTypeId="1" AcceptedAnswerId="12103" CreationDate="2010-11-27T19:40:09.830" Score="1" ViewCount="149" Body="&lt;p&gt;Suppose a graph with 12 vertices is colored with exactly 5 colors. By the pigeonhole principle, each color appears on at least two vertices. True or false?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The correct answer is false, but I assumed it to be true. Why is this so?&lt;/p&gt;&#xA;" OwnerUserId="2985" LastActivityDate="2010-11-27T19:45:10.213" Title="Pigeonhole principle question" Tags="&lt;number-theory&gt;&lt;counting&gt;&lt;pigeonhole-principle&gt;" AnswerCount="1" CommentCount="14" />
  <row Id="12103" PostTypeId="2" ParentId="12102" CreationDate="2010-11-27T19:45:10.213" Score="3" ViewCount="" Body="&lt;p&gt;The Pigeonhole Principle implies that there is &lt;em&gt;at least one&lt;/em&gt; color which appears on at least $2$ vertices, not that &lt;em&gt;each&lt;/em&gt; color appears.  Is it possible to color $12$ vertices with five colors in such a way that one of the colors is used only on a single vertex?&lt;/p&gt;&#xA;" OwnerUserId="299" LastActivityDate="2010-11-27T19:45:10.213" />
  <row Id="12105" PostTypeId="2" ParentId="12098" CreationDate="2010-11-27T20:00:55.967" Score="6" ViewCount="" Body="&lt;p&gt;For the fifth function in the link you &lt;a href=&quot;http://mathworld.wolfram.com/HeartCurve.html&quot; rel=&quot;nofollow&quot;&gt;mentioned&lt;/a&gt; (which I thought it was the most similar to a heart):&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;PolarPlot[(Sin[t]Sqrt[Abs[Cos[t]]])/(Sin[t]+7/5)-2Sin[t]+2, {t, 0, 10}]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Similarly, using W|A:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/pZEwC.gif&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="2569" LastActivityDate="2010-11-27T20:00:55.967" CommentCount="5" />
  <row Id="12106" PostTypeId="1" AcceptedAnswerId="12108" CreationDate="2010-11-27T20:02:46.873" Score="2" ViewCount="90" Body="&lt;p&gt;My TI-83 calculator doesnt allow me to do this, but using Windows calculator, I can compute the factorial of say 5.8. What does this mean and how does it work?&lt;/p&gt;&#xA;" OwnerUserId="2985" LastActivityDate="2010-11-27T20:05:56.630" Title="Factorial of a non-integer number" Tags="&lt;number-theory&gt;&lt;factorial&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="12107" PostTypeId="1" AcceptedAnswerId="12115" CreationDate="2010-11-27T20:04:11.850" Score="6" ViewCount="338" Body="&lt;p&gt;This is probably simple, but I'm solving a practice problem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\lim_{n \to \infty}\frac{1}{n}\left( \cos{\frac{\pi}{n}} + \cos{\frac{2\pi}{n}} + \ldots +\cos{\frac{n\pi}{n}} \right)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I recognize this as the Riemann sum from 0 to $\pi$ on $\cos{x}$, i.e. I think its the integral&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\int_0^\pi{ \cos{x}dx }$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is 0, but the book I'm using says it should be&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ \frac{1}{\pi}\int_0^\pi{ \cos{x}dx }$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Still 0 anyway, but where did the $\frac{1}{\pi}$ in front come from?&lt;/p&gt;&#xA;" OwnerUserId="1069" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-27T23:28:55.760" LastActivityDate="2010-11-28T14:40:10.017" Title="Is $\lim\limits_{n \to \infty}\frac{1}{n}\left( \cos{\frac{\pi}{n}} + \cos{\frac{2\pi}{n}} + \ldots + \cos{\frac{n\pi}{n}} \right)$ a Riemann sum?" Tags="&lt;calculus&gt;&lt;integral&gt;" AnswerCount="5" CommentCount="3" FavoriteCount="1" />
  <row Id="12108" PostTypeId="2" ParentId="12106" CreationDate="2010-11-27T20:05:56.630" Score="10" ViewCount="" Body="&lt;p&gt;Perhaps you're looking for the Gamma function. It restricts to the usual factorial on natural numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Gamma_function&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Gamma_function&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3431" LastActivityDate="2010-11-27T20:05:56.630" CommentCount="1" />
  <row Id="12110" PostTypeId="1" AcceptedAnswerId="12111" CreationDate="2010-11-27T20:24:55.517" Score="4" ViewCount="181" Body="&lt;p&gt;I accidentally typed &lt;code&gt;4 ** 5&lt;/code&gt; into Google and received &lt;code&gt;1,024&lt;/code&gt; instead of &lt;code&gt;20&lt;/code&gt;. But &lt;code&gt;4^5&lt;/code&gt; is also &lt;code&gt;1024&lt;/code&gt;...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does the double multiplication sign also mean the power sign? Why two symbols?&lt;/p&gt;&#xA;" OwnerUserId="3143" LastActivityDate="2010-11-28T03:26:01.440" Title="What is the double multiplication sign for?" Tags="&lt;powers&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="12111" PostTypeId="2" ParentId="12110" CreationDate="2010-11-27T20:25:21.527" Score="8" ViewCount="" Body="&lt;p&gt;Exponentiation in some software. So a**b means a^b.&lt;/p&gt;&#xA;" OwnerUserId="1120" LastActivityDate="2010-11-27T20:25:21.527" />
  <row Id="12112" PostTypeId="2" ParentId="12107" CreationDate="2010-11-27T20:25:46.763" Score="1" ViewCount="" Body="&lt;p&gt;The sum is also the real part of&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{1}{n}\left(e^{i\frac{\pi}{n}}+e^{i\frac{2\pi}{n}}+\ldots+e^{i\frac{n\pi}{n}}\right) \; .$$&lt;/p&gt;&#xA;" OwnerUserId="3567" LastActivityDate="2010-11-27T20:25:46.763" CommentCount="2" />
  <row Id="12113" PostTypeId="1" AcceptedAnswerId="12117" CreationDate="2010-11-27T20:36:19.150" Score="1" ViewCount="127" Body="&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/UkWn5.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The answer for this turns out to be only irreflexive. However, how is this not transitive? The definition I have for transitive states &quot;whenever there is a&#xA;path from x to y then there must be a direct arrow from x to y&quot;. So for the above graph, if there exists a path from one point to another, then there should be a direct arrow. Well that appears to be the case for the above image, no? Theres a path from C to B, and there is a direct arrow. There is no path from C to E, so no arrow needed. No path from B to C, B to D, so no arrows needed here. There is a path from Z to A and there is also a direct arrow, and same for A to Z. Am I misunderstanding this?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Transitive_relation#Properties_of_transitivity&quot; rel=&quot;nofollow&quot;&gt;Wikipedia article for transitive relation&lt;/a&gt; says:&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;For a transitive relation the&#xA;  following are equivalent:&lt;/p&gt;&#xA;  &#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;Irreflexivity  &lt;/li&gt;&#xA;  &lt;li&gt;Asymmetry   &lt;/li&gt;&#xA;  &lt;li&gt;Being a strict partial order&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Does this mean that if a relation is transitive, then it is also irreflexive? If thats the case, then shouldn't the above graph then be transitive?&lt;/p&gt;&#xA;" OwnerUserId="2985" LastActivityDate="2010-11-27T20:49:07.193" Title="Transitive and reflexive graph" Tags="&lt;graph-theory&gt;&lt;relations&gt;" AnswerCount="1" />
  <row Id="12114" PostTypeId="2" ParentId="12067" CreationDate="2010-11-27T20:36:59.297" Score="3" ViewCount="" Body="&lt;p&gt;You could argue by induction on $n$. The result is obvious if $n=1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the inductive step, assume that $(n-1)!$ divides any product of $n-1$ consecutive integers. Now consider products of $n$ consecutive integers. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Say your product is $P(k)=(k+1)(k+2)\dots(k+n)$. First, you may assume $k\ge0$: If one of the factors $k+j$ is 0, then $P(k)=0$ and obviously $n!$ divides it. If $k+n=-t-1&amp;lt;0$, this is the same as $$(-1)^n(t+1)(t+2)\dots(t+n)=(-1)^nP(t),$$ and $t\ge0$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now argue by induction on $k$. The result clearly holds if $k=0$, since $P(0)=n!$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose $n!|P(k)$. Then $P(k+1)=(k+2)(k+3)\dots(k+n+1)$. Split the last factor as $(k+1)+n$. We have $$P(k+1)=[(k+2)\dots(k+n)](k+1)+[(k+2)\dots(k+n)]n.$$ The first summand is $P(k)$, which is divisible by $n!$ by the inductive assumption on $k$. The second summand is $n$ times a product of $n-1$ consecutive integers, thus $n$ times a multiple of $(n-1)!$, by the inductive assumption on $n$. Clearly, $n$ times a multiple of $(n-1)!$ is a multiple of $n!$, and the sum of two multiples of $n!$ is a multiple of $n!$, so $n!|P(k+1)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We are done by induction.&lt;/p&gt;&#xA;" OwnerUserId="462" LastEditorUserId="462" LastEditorDisplayName="" LastEditDate="2010-11-27T20:59:43.410" LastActivityDate="2010-11-27T20:59:43.410" />
  <row Id="12115" PostTypeId="2" ParentId="12107" CreationDate="2010-11-27T20:39:24.593" Score="6" ViewCount="" Body="&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Riemann_sum#Definition&quot; rel=&quot;nofollow&quot;&gt;Riemann sum&lt;/a&gt; is given by $$S=\sum_{i=1}^{n}f(y_{i})(x_{i}-x_{i-1})$$ where $x_{i-1}\leq y_{i}\leq x_{i}$. If you choose $f(y_{i})=\cos(y_{i})$, $y_{i}=\frac{i \pi}{n}$ and $x_{i}=\frac{i \pi}{n}$ you get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_{0}^{\pi} \cos(x)dx=\lim_{n\rightarrow{\infty}}\sum_{i=1}^{n}\cos\left(\frac{i \pi}{n}\right)(x_{i}-x_{i-1})$$ which is just the sum you have with an extra factor $\pi$. Therefore&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{1}{\pi}\int_{0}^{\pi} \cos(x)dx=\lim_{n\to\infty}\frac{1}{n}\left( \cos{\frac{\pi}{n}} + \cos{\frac{2\pi}{n}} + \ldots \cos{\frac{n\pi}{n}} \right)$$&lt;/p&gt;&#xA;" OwnerUserId="2569" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-27T23:26:22.783" LastActivityDate="2010-11-27T23:26:22.783" CommentCount="7" />
  <row Id="12116" PostTypeId="2" ParentId="12107" CreationDate="2010-11-27T20:43:09.740" Score="2" ViewCount="" Body="&lt;p&gt;Assume that $x_{k-1}\leq c_{k}\leq x_{k}$, $x_{0}=0,x_{n}=\pi $, and that the interval of integration is divided into $n$ sub-intervals of equal width. Under these circumstances $x_{k}-x_{k-1}=\dfrac{\pi }{n}$ and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{eqnarray*}\lim_{n\rightarrow \infty }\frac{1}{n}\sum_{k=1}^{n}\cos \frac{k\pi }{n}&amp;amp;=&amp;amp;\lim_{n\rightarrow \infty }\sum_{k=1}^{n}\cos \left( c_{k}\right) \frac{%&#xD;&#xA;x_{k}-x_{k-1}}{\pi }\\&#xD;&#xA;&amp;amp;=&amp;amp;\lim_{n\rightarrow \infty }\sum_{k=1}^{n}\left(\frac{1}{\pi }\cos \left(&#xD;&#xA;c_{k}\right)\right) (x_{k}-x_{k-1})\\&#xD;&#xA;&amp;amp;=&amp;amp;\int_{0}^{\pi }\frac{1}{\pi }\cos x\ dx\end{eqnarray*}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sum $\displaystyle\sum_{k=1}^{n}\left(\dfrac{1}{\pi }\cos \left(&#xD;&#xA;c_{k}\right)\right) (x_{k}-x_{k-1})$ is a &lt;a href=&quot;http://en.wikipedia.org/wiki/Riemann_sum#Definition&quot; rel=&quot;nofollow&quot;&gt;Riemann Sum&lt;/a&gt; of the function $f(x)=\dfrac{1}{\pi }\cos x$ in the interval $[0,\pi]$, but &lt;em&gt;not&lt;/em&gt; of the function $\cos x$.&lt;/p&gt;&#xA;" OwnerUserId="752" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2010-11-28T14:40:10.017" LastActivityDate="2010-11-28T14:40:10.017" />
  <row Id="12117" PostTypeId="2" ParentId="12113" CreationDate="2010-11-27T20:43:19.557" Score="2" ViewCount="" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;However, how is this not transitive?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There is an edge from Z to A and one from A to Z, but no edge from Z to itself or from A to itself.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Does this mean that if a relation is transitive, then it is also irreflexive?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No, it means that if a transitive relation is irreflexive, it is also asymmetric and a strict partial order (and if it is asymmetric, it is also irreflexive etc.).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if the graph was transitive, its irreflexivity would imply that it's also antisymmetric, but since the graph is not transitive, this does not apply.&lt;/p&gt;&#xA;" OwnerUserId="11" LastEditorUserId="11" LastEditorDisplayName="" LastEditDate="2010-11-27T20:49:07.193" LastActivityDate="2010-11-27T20:49:07.193" CommentCount="3" />
  <row Id="12118" PostTypeId="1" CreationDate="2010-11-27T20:55:38.597" Score="2" ViewCount="107" Body="&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Exterior_algebra&quot; rel=&quot;nofollow&quot;&gt;grassmann algebra&lt;/a&gt; $G$ is the algebra over a field $\mathbb{F}$ generated by the variables $e_i$ such that $e_i^2=0$ and $e_i e_j = -e_j e_i$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm looking for some references on algebras $G\otimes A$ (In this case the grassmann algebra is over $\mathbb{C}$) where A is the group algebra $\mathbb {C} \mathbb{Z}_n$ generated by $U_t$ and $e_i U_t = \zeta U_t e_i$ where $\zeta$ is a $n$-th primitive root of unity. Or in other words, I want to attach another variable $t$ to the grassman algebra such that $t^n=1$ and $e_i t = \zeta t e_i $.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason I ask this, is that I came across algebras that have the same polynomial identities as the ones above, so I want to see what can I say about them from only the identities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think now that these algebras are Matrix algebras with coefficients in $\mathbb{Q}(\zeta)G$ such that each diagonal is multiplied by a (single) element from G. something like&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$&#xD;&#xA;\left(\begin{array}{cccc}&#xD;&#xA;g_{0} &amp;amp; g_{1} &amp;amp; g_{2} &amp;amp; g_{3}\\&#xD;&#xA;g_{3} &amp;amp; g_{0} &amp;amp; g_{1} &amp;amp; g_{2}\\&#xD;&#xA;g_{2} &amp;amp; g_{3} &amp;amp; g_{0} &amp;amp; g_{1}\\&#xD;&#xA;g_{1} &amp;amp; g_{2} &amp;amp; g_{3} &amp;amp; g_{0}\end{array}\right)&#xD;&#xA;$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $g_0, g_2 \in G_0$ (even) and $g_1, g_3 \in G_1$ (odd). The $t$ variable is the diagonal matrix with coefficients $1,i,-1,-i$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="2125" LastEditorUserId="2125" LastEditorDisplayName="" LastEditDate="2010-11-28T06:43:32.953" LastActivityDate="2010-11-28T06:43:32.953" Title="Grassmann Algebras" Tags="&lt;algebra&gt;&lt;exterior-algebra&gt;" CommentCount="2" />
  <row Id="12119" PostTypeId="2" ParentId="12097" CreationDate="2010-11-27T20:57:04.353" Score="0" ViewCount="" Body="&lt;p&gt;So $M=\mbox{Max}(X_0,\ldots,X_{n-1})$ then you know that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathbb{P}(M\geq x)=1-F(x)^n$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with $F(x)$ the Levy-stable cumulative distribution. Now, you also have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathbb{P}(M\geq \sum X_i) = \int p(M\geq \sum X_i | \sum X_i = x) p(x) dx$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $p(x)$ is the density function for Levy-stable distributions. (Since by definition Levy-stable distributions are stable under summation of random Levy-stable variables.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So it seems to me you have in principle all the ingredients to try some asymptotics, even if you don't have an explicit formula for all the distributions. Maybe you can try with some for which the formula exists, like Cauchy and normal distributions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sorry for the sloppiness, just trying to sketch an approach for now.&lt;/p&gt;&#xA;" OwnerUserId="3567" LastEditorUserId="3567" LastEditorDisplayName="" LastEditDate="2010-11-27T21:12:39.637" LastActivityDate="2010-11-27T21:12:39.637" />
  <row Id="12120" PostTypeId="1" AcceptedAnswerId="12127" CreationDate="2010-11-27T21:07:29.847" Score="0" ViewCount="73" Body="&lt;p&gt;There's one basilar math thing that keeps bugging me: the fact that a really simple 2D geometrical figure (like a circle) might not be a function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know what the definition of a function is. A circle is not a function (of one variable) because it would associate two values of the codomain to a single value of the domain. But this doesn't help my intuitivity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It sounds terribly weird that a given curve (say a sinusoidal) is a function only unless you rotate it by 45 or more degrees...&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Is there any simple way (a concept similar to that of how most people imagine a function: a curve on a graph) to represent 2D geometrical figures like a circle (or a rotated sinusoidal, or whatever)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only one I can think of is using a function of two or more variables, but this sounds pretty dirty to me: why should I use a function in three dimensions just to see its shadow on two dimensions?&lt;br&gt;&#xA;Besides if I think of the function as a real object (in our real, 3D space), I cannot help thinking that it's not a 2D circle, it's a 3D weird object which can be seen as a circle when rotated in a particular way (just like the &lt;a href=&quot;http://en.wikipedia.org/wiki/Penrose_stairs&quot; rel=&quot;nofollow&quot;&gt;Penrose stairs&lt;/a&gt; look &lt;em&gt;possible&lt;/em&gt; when rotated in a special way).&lt;/p&gt;&#xA;" OwnerUserId="3846" LastActivityDate="2010-11-27T21:42:29.897" Title="From geometrical figures to function" Tags="&lt;geometry&gt;&lt;functions&gt;&lt;plane-curves&gt;" AnswerCount="1" CommentCount="6" />
  <row Id="12121" PostTypeId="2" ParentId="12065" CreationDate="2010-11-27T21:16:00.180" Score="0" ViewCount="" Body="&lt;p&gt;Let us prove that $m^{(k)}=m(m+1)...(m+k-1)$ is divided by $k!$ for all integer $m$.&#xA;Induction by $k$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$k=1$: Every integer $m$ is divided by $1$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$k\to k+1$: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;induction by $m$: $m=0$: $0^{k+1}=0$&#xD;&#xA;is divided by $(k+1)!$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$m\to m+1$: $(m+1)^{(k+1)}=((m+1)^{(k+1)}-m^{(k+1)})+m^{(k+1)}=$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$=(m+1)(m+2)...(m+k+1)-m(m+1)...(m+k)+m^{(k+1)}=$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(k+1)(m+1)...(m+k)+m^{(k+1)}=(k+1)(m+1)^{(k)}+m^{(k+1)}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and first term is divided by $(k+1)\cdot k!=(k+1)!$ because of induction by k and the second term is divided by $(k+1)!$ because of induction by $m$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the same works for $m\to m-1$&#xD;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update&lt;/strong&gt; Oops, essentially the same proof found in the thread mentioned in this &lt;a href=&quot;http://math.stackexchange.com/questions/12065/the-product-of-n-consecutive-integers-is-divisible-by-n-factorial/12073#12073&quot;&gt;answer&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="425" LastEditorUserId="425" LastEditorDisplayName="" LastEditDate="2010-11-27T21:21:26.043" LastActivityDate="2010-11-27T21:21:26.043" CommunityOwnedDate="2010-11-27T21:21:26.043" />
  <row Id="12122" PostTypeId="1" AcceptedAnswerId="12124" CreationDate="2010-11-27T21:16:24.347" Score="1" ViewCount="107" Body="&lt;p&gt;Question: Let $V$ be a $K$-Vectorspace and $f: V \rightarrow V$ be linear. It holds that $f \circ f = f$. Show that $V = \mbox{ker}(f) \oplus \mbox{im}(f)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My attempt: So i guess that the $\oplus$ denotes a direct sum which means i have to show that (i) $V = \mbox{ker}(f) + \mbox{im}(f)$ and (ii) $\mbox{ker}(f) \cap \mbox{im}(f) = \{0\}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried to do (ii) first: Let $v \in \mbox{im}(f) \cap \mbox{ker}(f)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Rightarrow \exists u: f(u)=v \wedge f(v) = 0$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(can i put a &quot;Rightarrow&quot; here?) $(f \circ f)(u)=f(f(u)) = f(v) = 0$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for (i) i am having difficulty with an approach to showing that $V = \mbox{ker}(f) + \mbox{im}(f)$. Should I even be trying to do this in the first place? if so, any advice as to how?&lt;/p&gt;&#xA;" OwnerUserId="3711" LastActivityDate="2010-11-27T21:30:47.993" Title="Show that $V = \mbox{ker}(f) \oplus \mbox{im}(f)$ for a linear map with $f \circ f = f$" Tags="&lt;linear-algebra&gt;" AnswerCount="3" />
  <row Id="12123" PostTypeId="2" ParentId="12122" CreationDate="2010-11-27T21:25:55.747" Score="0" ViewCount="" Body="&lt;p&gt;$v = (v - f(v)) + f(v)$&lt;/p&gt;&#xA;" OwnerUserId="3035" LastActivityDate="2010-11-27T21:25:55.747" />
  <row Id="12124" PostTypeId="2" ParentId="12122" CreationDate="2010-11-27T21:28:30.790" Score="4" ViewCount="" Body="&lt;p&gt;Let's see (ii) first: (Yes, your use of the implication symbol $\Rightarrow$ is appropriate.) You want to use that $f\circ f=f$, which you do not mention explicitly. The point is: In (ii) you need to show that the $v$ you are considering is $0$. You have: $f\circ f=f$, so $v=f(u)=(f\circ f)(u)$ etc, as you did. This gives $v=0$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For (i), let $v$ be any element of $V$. What can we say about $v-f(v)$? Let's call it $w$. We have $f(w)=f(v-f(v))=f(v)-f(f(v))=f(v)-f(v)=0$. Here, I used that $f$ is linear and (again) that $f\circ f=f$. This shows that $w\in{\rm ker}(f)$. So, $v=(v-f(v))+f(v)=w+f(v)$ is the sum of an element of the kernel (namely, $w$) and one of the image (namely, $f(v)$), and that is precisely what you needed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(A meta-question here is how would one think of considering $v-f(v)$. Ok, let's try to work backwards. Suppose we already know that $v=a+b$ with $a$ in the kernel and $b$ in the range. Then $f(v)=f(a)+f(b)=f(b)$, since $a$ is in the kernel. Also, $f(v)=f(f(v))$, so it is reasonable, as a first approach, to let $b=f(v)$, and see if we can directly prove that $a=v-b$ is in the kernel. And this is precisely what we did.)&lt;/p&gt;&#xA;" OwnerUserId="462" LastActivityDate="2010-11-27T21:28:30.790" CommentCount="2" />
  <row Id="12125" PostTypeId="2" ParentId="12122" CreationDate="2010-11-27T21:30:47.993" Score="0" ViewCount="" Body="&lt;p&gt;Hi.  For 1) let v be an arbitrary element in the intersection of ker f and image f.  This second statement, as you have pointed out, means that v=f(u) for some u, and the first statement, means that f(v) = 0.  So 0=f(v)=f(f(u)) which is f(u) (or v) by assumption.  This shows v=0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For 2) I believe you can do this with nothing more than set theory.  Just prove V is a subset of ker(f) + im(f) and then prove the reverse inclusion (this second inclusion is trivial since ker(f) and im(f) are clearly both subsets of V)&lt;/p&gt;&#xA;" OwnerUserId="3971" LastActivityDate="2010-11-27T21:30:47.993" />
  <row Id="12126" PostTypeId="1" AcceptedAnswerId="12151" CreationDate="2010-11-27T21:40:04.383" Score="11" ViewCount="202" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Lennart_Carleson&quot; rel=&quot;nofollow&quot;&gt;Lennart Carleson&lt;/a&gt; proved Luzin's conjecture that &lt;a href=&quot;http://en.wikipedia.org/wiki/Carleson%27s_theorem&quot; rel=&quot;nofollow&quot;&gt;the Fourier series of each $f\in L^2(0,2\pi)$ converges almost everywhere&lt;/a&gt;. Also, &lt;a href=&quot;http://en.wikipedia.org/wiki/Richard_Allen_Hunt&quot; rel=&quot;nofollow&quot;&gt;Richard Hunt&lt;/a&gt; extended the result to &lt;a href=&quot;http://en.wikipedia.org/wiki/Carleson%27s_theorem#CITEREFHunt1968&quot; rel=&quot;nofollow&quot;&gt;$L^p$ ($p&gt;1$)&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some time ago I tried to read Carleson's paper, but I would say it is fairly hard to assimilate. &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Is there an easier proof? Can someone point out of the core or give an outline?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;What did Hunt do? Can someone give an outline of that proof?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="1154" LastActivityDate="2010-11-28T00:52:20.577" Title="Accessible proof of Carleson's $L^2$ theorem" Tags="&lt;fourier-analysis&gt;&lt;fourier-series&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="4" />
  <row Id="12127" PostTypeId="2" ParentId="12120" CreationDate="2010-11-27T21:42:29.897" Score="2" ViewCount="" Body="&lt;p&gt;Some possibilities:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First of all, you might also want to switch to &lt;a href=&quot;http://en.wikipedia.org/wiki/Polar_coordinates&quot; rel=&quot;nofollow&quot;&gt;polar coordinates&lt;/a&gt;, in which points are defined through angle $\varphi$ and radius $r$ rather than $x$/$y$ coordinates. For example, in a polar coordinate system, a circle (let's take the unit circle) $K$ actually &lt;strong&gt;is&lt;/strong&gt; &lt;em&gt;a function&lt;/em&gt; (now of form $r(\varphi)$ rather than $y(x)$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the radius is constant, we end up with&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$K: r(\varphi) = 1$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But most figures still aren't functions in polar coordinates, so we might have to take a more general approach: A curve&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A curve is a function that produces &lt;em&gt;coordinates&lt;/em&gt; rather than a single value from some parameter, i.e. $\mathbb{R} \to \mathbb{R}^2$ in our case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $K$ be our unit circle again - now parameterized by some angle $\varphi$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$K(\varphi) = (\cos \varphi, \sin \varphi)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At least, this should work for most figures. The most general form though is simply an equation that the coordinates have to satisfy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$K: x^2+y^2 = 1$$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="917" LastActivityDate="2010-11-27T21:42:29.897" />
  <row Id="12128" PostTypeId="1" CreationDate="2010-11-27T21:44:52.540" Score="2" ViewCount="115" Body="&lt;p&gt;Hey,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given the series: $\sum_{n=0}^{\infty}(-1)^{n}(\sqrt[n]{n} - 1)^{n}$. Does the series converge?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Attempt to solution (might be incorrect):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(\sqrt[n]{n} - 1)^{n}&amp;gt; (1+\frac{1}{n})^{n}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(1+\frac{1}{n})^{n} \to e \Rightarrow (\sqrt[n]{n} - 1)^{n}$ lower-bounded by $e$. Based on Leibniz Criterion the sequence $\{A_n\}$ (in our case, $(\sqrt[n]{n} - 1)^{n}$) is monotone decreasing, but its limit is not $0$ at infinite $\Rightarrow$ series diverge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it enough to say that since the sequence is lower-bounded, the limit of it at infinite is not $0$, or should I actually calculate the limit of the sequence? &lt;/p&gt;&#xA;" OwnerUserId="3637" LastActivityDate="2010-11-27T23:32:58.540" Title="Leibniz criterion for alternating series" Tags="&lt;calculus&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="12129" PostTypeId="2" ParentId="12107" CreationDate="2010-11-27T21:49:36.213" Score="3" ViewCount="" Body="&lt;p&gt;$$\sum_{k=0}^n \cos\left(\frac{k\pi}{n}\right) = 0$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the expression equals to $-1/n$, and the limit is trivially equal to zero.&lt;/p&gt;&#xA;" OwnerUserId="3406" LastActivityDate="2010-11-27T21:49:36.213" />
  <row Id="12130" PostTypeId="2" ParentId="12128" CreationDate="2010-11-27T22:06:42.410" Score="2" ViewCount="" Body="&lt;p&gt;Your lower bound for $(n^{1/n} -1)^n$ is not correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint: To show that&#xA;$(n^{1/n} -1)^n \rightarrow 0$ write $n^{1/n} = 1 + r_n$ and use the binomial theorem&#xA;to show that for $n \ge 2$ we have $r_n &amp;lt; \sqrt{2/(n-1)}.$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint2: When you expand $(1+r_n)^n$ the important term in the expansion&#xA;is the one in ${r_n}^2 .$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this way you can show that the series converges absolutely, you don't really need the Leibniz criterion although they will do the job. &lt;/p&gt;&#xA;" OwnerUserId="1301" LastEditorUserId="1301" LastEditorDisplayName="" LastEditDate="2010-11-27T22:24:09.040" LastActivityDate="2010-11-27T22:24:09.040" />
  <row Id="12131" PostTypeId="2" ParentId="12128" CreationDate="2010-11-27T22:15:01.340" Score="1" ViewCount="" Body="&lt;p&gt;Write your series as $\sum_{n = 1}^\infty (-1)^n a_n$ where $a_n = (n^{1/n} - 1)^n$. Note that $(a_n)$ is monotonically decreasing to zero and all terms are positive, hence by the alternating series test (or Leibniz' test) the series converges.&lt;/p&gt;&#xA;" OwnerUserId="1120" LastActivityDate="2010-11-27T22:15:01.340" />
  <row Id="12132" PostTypeId="1" AcceptedAnswerId="12148" CreationDate="2010-11-27T22:15:37.003" Score="4" ViewCount="212" Body="&lt;p&gt;Hi, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question 1&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lets consider a 3x3 matrix, A.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The inverse of this matrix can be found by finding the adjoint of the matrix and dividing it by its determinant. Can someone please explain why this process work and how it can be visualized?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that the determinant can be thought of as the volume spanned by the column vectors and that the adjoint of matrix A is the transpose of A's cofactors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question 2&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's consider a 3x3 matrix, A.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The cofactors of this matrix can be found by the cross product of each of its column vectors. So what does the cofactor matrix tell us? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you.&lt;/p&gt;&#xA;" OwnerUserId="2973" LastActivityDate="2010-11-28T00:13:09.543" Title="Finding the inverse of a Matrix using determinants" Tags="&lt;linear-algebra&gt;&lt;matrices&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" />
  <row Id="12134" PostTypeId="1" AcceptedAnswerId="12136" CreationDate="2010-11-27T22:18:15.423" Score="2" ViewCount="125" Body="&lt;p&gt;I've recently read a article 'Theories of computational complexity'. In the article it states that with Rice's theorem one can prove that, for example, {$i∈[\mathbb N:\phi_i$ is a constant function}, is not recursive, but I can't see how. I already know the theorem, but I don't really get how I can use it. Thank you for your answer!&lt;/p&gt;&#xA;" LastActivityDate="2010-11-27T22:35:40.373" Title="Rice’s theorem explanation" Tags="&lt;logic&gt;" AnswerCount="1" />
  <row Id="12135" PostTypeId="2" ParentId="11669" CreationDate="2010-11-27T22:28:27.287" Score="12" ViewCount="" Body="&lt;p&gt;The answers given are pretty good from a musical, mathematical, and socialogical / historical reason.  But they miss the fundamental reason why there are 12 notes in a western scale (or 5 notes in an eastern pentatonic, etc.), and why it's those particular 12 notes (or 5).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Qiaochu almost nailed it by pointing out that we like notes which are simple integer ratios.  But why?  The fundamental reason stems from the physics of common early instruments -- flutes (including the human voice) and plucked strings -- and from the physics of the tympanum in the ear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As Qiaochu noted, sound is not composed of a single sine wave frequency but rather a sum of many sine waves.  The &quot;note&quot; we hear is the frequency of the primary (loudest) wave coming from these instruments.  But frequencies exist in that wave as well, albeit largely masked by the primary.  These are known informally as harmonics or overtones.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first several harmonics of flutes and plucked strings are similar and very straightforward:  If the primary is normalized to frequency 1, then the second loudest harmonic is typically 1/2 (an octave above), the third is usually 1/3 (an octave and a fifth above), the fourth is usually 1/4 (two octaves), the fifth is usually 1/5 (two octaves and a major third), and the sixth is usually 1/6 (two octaves and a fifth).  If the primary note is C1, these translate roughly into C2, G3, C4, E4, and G4.  If the harmonics continued in this way -- and they don't always -- various other notes appear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This matters because if you want to play TWO instruments together, you'd like their harmonics to coincide even if they're playing different notes.  Otherwise the excess of harmonics sounds bad to the ear.  In the worst case, very close but not entirely overlapping harmonics create &quot;beats&quot; -- seeming alternating loud and soft periods of time -- which are irritating to listen to and tough on the ear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To get harmonics to coincide in multiple instruments or even successive notes, you have to pick notes for them to play where their harmonics have a strong overlap.  For example, this is also why the major fourth is useful even though it doesn't often appear early.  It's because if one instrument is playing C, if the other instrument is playing major fourth but &lt;em&gt;lower&lt;/em&gt; by an octave, they'll overlap nicely.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe these note selections (guaranteeing harmonics in harmony, so to speak) influenced the evolution of scale choices -- especially the pentatonic, that is, the black notes), and the division of the octave into 12 pieces. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One early instrument which is totally out of whack from this is the bell.  Bells and gongs can be tuned to have a variety of harmonics, but the most common ones -- foundry bells -- have a very loud, unusual third harmonic: minor third or E flat.  It is so loud and incongruous that they sound terrible, even disturbing, when played along with strings, flutes, voices, etc.  In fact, entire musical pieces have to be written specially for carillons (large multibell instruments) in order to guarantee proper overlap of harmonics.  Generally this means that the entire piece has to be written in fully diminished chords.  Major chords sound among the worst because of the clash between the major third in the chord and the minor third coming from the root's loud third harmonic.&lt;/p&gt;&#xA;" OwnerUserId="4151" LastActivityDate="2010-11-27T22:28:27.287" CommentCount="2" />
  <row Id="12136" PostTypeId="2" ParentId="12134" CreationDate="2010-11-27T22:35:40.373" Score="3" ViewCount="" Body="&lt;p&gt;Rice's theorem says that for any subset $F$ of the class $T$ of partial computable functions, the set $\{i\mid \phi_i\in F\}$ is recursive iff $F=\emptyset$ or $F=T$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $F$ be the set of partial computable functions that are constant on their domain. $F$ is certainly non-empty, since the function that always takes the value 0 is certainly computable. $F$ is also not the whole class $T$, since the function $\phi$ with $\phi(0)=0$, $\phi(n)=1$ for $n&amp;gt;0$ is certainly computable and not constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It follows that the set you are interested in is not recursive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If your confusion lies in how to translate the usual statement of Rice's theorem (in terms of decision problems) into the version I wrote, simply remember that formally a decision problem is a set $D\subseteq{\mathbb N}$, and the problem is decidable iff $D$ is recursive. Of course, this is usually stated by saying that there is a Turing machine that, given $n$, decides whether $n\in D$ or not. But this is certainly the same as saying that the machine computes the characteristic function of $D$, which is (the definition of) being recursive.&lt;/p&gt;&#xA;" OwnerUserId="462" LastActivityDate="2010-11-27T22:35:40.373" CommentCount="1" />
  <row Id="12137" PostTypeId="1" AcceptedAnswerId="12138" CreationDate="2010-11-27T22:49:27.500" Score="1" ViewCount="84" Body="&lt;p&gt;State in words the negation of the following sentence: For every martian M, if M is green, then M is tall and ticklish. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I got the right answer to this, give or take a few words, but this is a question of form more than anything. After converting this statement to symbols and negating everything, I come up with: $\exists M(P \wedge (\neg \text{Tall } \vee \neg\text{ Ticklish})$ and so in word format that would be: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;There exists a martian such that it is&#xA;  green and not tall or not ticklish.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;However the really correct answer is: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;There is a martian M such that M is&#xA;  green but M is not tall or M is not&#xA;  ticklish.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The difference between these two is a 'but' and an 'and'. Does this mean anything mathematically? Is my version correct?&lt;/p&gt;&#xA;" OwnerUserId="2985" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-27T23:51:59.617" LastActivityDate="2010-11-27T23:51:59.617" Title="Negating a statement" Tags="&lt;propositional-calculus&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="12138" PostTypeId="2" ParentId="12137" CreationDate="2010-11-27T22:52:07.410" Score="4" ViewCount="" Body="&lt;p&gt;Your answer is &quot;really correct&quot;. It is perhaps more usual in English to use &quot;but&quot; rather than &quot;and&quot; in a sentence as the one in your example. Formally (i.e., mathematically), there is no difference.&lt;/p&gt;&#xA;" OwnerUserId="462" LastActivityDate="2010-11-27T22:52:07.410" CommentCount="1" />
  <row Id="12139" PostTypeId="1" AcceptedAnswerId="12143" CreationDate="2010-11-27T23:29:20.897" Score="4" ViewCount="209" Body="&lt;p&gt;Consider a non-empty set A containing n objects. How many relations on A are both symmetric and reflexive?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer to this is $2^p$ where $p=$ $n \choose 2$. However, I dont understand why this is so. Can anyone explain this?&lt;/p&gt;&#xA;" OwnerUserId="2985" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-27T23:54:00.673" LastActivityDate="2010-11-28T03:42:11.043" Title="Number of relations that are both symmetric and reflexive" Tags="&lt;combinatorics&gt;&lt;discrete-mathematics&gt;&lt;relations&gt;" AnswerCount="4" CommentCount="1" />
  <row Id="12140" PostTypeId="2" ParentId="12128" CreationDate="2010-11-27T23:32:58.540" Score="2" ViewCount="" Body="&lt;p&gt;The alternatingness of the series is something of a red herring as the series converges absolutely. By the root test, to show this it suffices to show that $\lim_{n \rightarrow \infty} |n^{1 \over n} - 1| = 0$. In other words, it suffices to show that $\lim_{n \rightarrow \infty} n^{1 \over n} = 1$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are a few ways to show this limit is in fact $1$. One way is to note that $\ln (n^{1 \over n}) = {\ln(n) \over n}$, and the latter is seen to go to zero as $n$ goes to infinity using L'Hopital's rule. Since the natural log of the $n$th term of the sequence goes to zero, the $n$th term of the sequence goes to $e^0 = 1$. &lt;/p&gt;&#xA;" OwnerUserId="3035" LastActivityDate="2010-11-27T23:32:58.540" />
  <row Id="12142" PostTypeId="2" ParentId="12139" CreationDate="2010-11-27T23:45:04.277" Score="5" ViewCount="" Body="&lt;p&gt;Being reflexive means that $(x,x)\in R$ for all $x\in A$. Being symmetric means that $(x,y)\in R$ implies that $(y,x)\in R$ as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Begin by listing $A$ as $A=\{a_1,\dots,a_n\}$. Then let $B$ be the set $$\{(a_i,a_j)\mid 1\le i&amp;lt;j\le n\}.$$ Note that if $x\ne y$ are elements of $A$, then either $(x,y)\in B$ or $(y,x)\in B$ but not both. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $S$ be any subset of $B$. Let $$R_S=S\cup\{(y,x)\mid (x,y)\in S\}\cup\{(x,x)\mid x\in A\}.$$ Then $R_S$ is a symmetric and reflexive relation on $A$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that there are $2^{|B|}$ subsets of $B$, and that if $S\ne S&amp;#39;$ are subsets of $B$, then $R_S\ne R_{S&amp;#39;}$. Also, note that $|B|=\binom{n}2$. (If the last equality is not clear, note that $$B=\{(a_1,a_j)\mid j&amp;gt;1\}\cup\{(a_2,a_j)\mid j&amp;gt;2\}\cup\dots$$ so $|B|=(n-1)+(n-2)+\dots+1$, and it is well-known that the last sum equals $n(n-1)/2=\binom n2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This shows that the number of symmetric, reflexive relations on $A$ is at least $2^p$ with $p=\binom n2$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To see the equality, it is enough to check that any such relation $R$ is $R_S$ for some $S\subseteq B$. But, given $R$, let $S=\{(a_i,a_j)\in R\mid i&amp;lt;j\}$. This is a subset of $B$, and it is easy to check that $R=R_S$. &lt;/p&gt;&#xA;" OwnerUserId="462" LastActivityDate="2010-11-27T23:45:04.277" />
  <row Id="12143" PostTypeId="2" ParentId="12139" CreationDate="2010-11-27T23:45:20.950" Score="9" ViewCount="" Body="&lt;p&gt;To be reflexive, it must include all pairs $(a,a)$ with $a\in A$. To be symmetric, whenever it includes a pair $(a,b)$, it must include the pair $(b,a)$. So it amounts to choosing which $2$-element subsets from $A$ will correspond to associated pairs.  If you pick a subset $\{a,b\}$ with two elements, it corresponds to adding both $(a,b)$ and $(b,a)$ to your relation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How many $2$-element subsets does $A$ have? Since $A$ has $n$ elements, it has exactly $\binom{n}{2}$ subsets of size $2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So now you want to pick a collection of subsets of $2$-elements. There are $\binom{n}{2}$ of them, and you can either pick or not pick each of them. So you have $2^{\binom{n}{2}}$ ways of picking the pairs of distinct elements that will be related. &lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-11-27T23:45:20.950" />
  <row Id="12144" PostTypeId="2" ParentId="12081" CreationDate="2010-11-27T23:50:19.010" Score="0" ViewCount="" Body="&lt;p&gt;You can try $$f(L,D) = L - D + \frac{1}{D+2}.$$ This would sort first according to $L - D$ (ascending) and then according to $D$ (descending).&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-27T23:50:19.010" CommentCount="2" />
  <row Id="12145" PostTypeId="2" ParentId="12139" CreationDate="2010-11-27T23:52:59.200" Score="3" ViewCount="" Body="&lt;p&gt;Hi!&#xA;Maybe you can see it like this: a relation $R$ on $A$ is a subset of $A\times A$, and it is symmetric if and only if $(x,y)\in R \implies (y,x)\in R$, moreover, if the relation is reflexive, then $(x,x)\in R$ for all $x\in A$. Then you can determine uniquely such a relation by saying which subsets of two distinct elements of $A$ &quot;belong&quot; to $R$, in the sense that $\{x,y\}\in R \iff (x,y),(y,x)\in R$. Now, you know that the number of subsets with two distinct elements of $A$ is $\binom{n}{2}$, and the number of subset of a set with $p$ elements is $2^p$.&#xA;I'm sorry if i was too obscure. &lt;/p&gt;&#xA;" OwnerUserId="4054" LastActivityDate="2010-11-27T23:52:59.200" CommentCount="1" />
  <row Id="12147" PostTypeId="2" ParentId="12139" CreationDate="2010-11-28T00:09:09.953" Score="1" ViewCount="" Body="&lt;p&gt;There is only one way to make the relation reflexive -- all ordered pairs $(x,x), x\in A$ must be in the relation. So the number of reflexive symmetric relations on $A$ is the same as the number of ways of adding symmetric pairs $(a,b),(b,a)$, where $a\neq b$ into the relation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $S$ be a subset of $2^A$ consisting of subsets of 2 elements. Then $S$ gives rise to exactly one reflexive symmetric relation on $A$. For example, if $A=\lbrace 1,2,3,4\rbrace$, then an example of $S$ is $\lbrace \lbrace 1,2\rbrace, \lbrace 1,4\rbrace, \lbrace 2,3\rbrace\rbrace$.   The relation induced by $S$  is $$\lbrace (1,2), (2,1), (1,4), (4,1), (2,3), (3,2)\rbrace$$ plus all $(x,x), x\in A$. &#xA;Conversely, every reflexive symmetric relation on $A$ arises in this way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since there are $p={n\choose 2}$ subsets of 2 elements, there are $2^p$  such $S$'s. The answer to your question is therefore $2^p$. &lt;/p&gt;&#xA;" OwnerUserId="3249" LastEditorUserId="3249" LastEditorDisplayName="" LastEditDate="2010-11-28T03:42:11.043" LastActivityDate="2010-11-28T03:42:11.043" CommentCount="2" />
  <row Id="12148" PostTypeId="2" ParentId="12132" CreationDate="2010-11-28T00:13:09.543" Score="1" ViewCount="" Body="&lt;p&gt;There is some explanation in the Geometric Interpretation &lt;a href=&quot;http://en.wikipedia.org/wiki/Cramer%27s_rule&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;, but I don't really like it too much.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think it's easier to think this way: The cross product of $v$ and $w$ is a vector $x$ orthogonal to the two given ones, whose length is twice the area of the triangle. Hence the volume of the tetrahedron on $u, v, w$ is one sixth of the the dot product of $x$ with the remaining vector $u$ (one should be careful with signs, but it works out). The dot product of $x$ with $v$ or $w$ is zero. Similar things are true about the cross product of $u$ and $w$ dotted with $v$ (and $u$ and $w$) and the cross product of $u$ and $v$ dotted with $w$ (and $u$ and $v$).  What all of this implies is that the matrix of cross product vectors (transposed) multiplied with the matrix of $u$, $v$, $w$ is diagonal, with all diagonal entries equal to six times the volume of the tetrahedron, i.e. the determinant of the $u$, $v$, $w$ matrix. The result follows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Higher dimensional versions are fairly similar, with volumes of $n$ and $n-1$ simplices.&lt;/p&gt;&#xA;" OwnerUserId="2633" LastActivityDate="2010-11-28T00:13:09.543" />
  <row Id="12149" PostTypeId="1" CreationDate="2010-11-28T00:22:26.130" Score="4" ViewCount="85" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Mitchell%27s_embedding_theorem&quot; rel=&quot;nofollow&quot;&gt;Freyd-Mitchell's embedding theorem&lt;/a&gt; states that: if A is a small abelian category, then there exists a ring R and a full, faithful and exact functor F: A → R-Mod.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is quite the theorem and has several useful applications (it allows one to do diagram chasing in abstract abelian categories, etc.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have been asked to state and prove the theorem in class (a homological algebra course). However, by reading the texts I was recommended, I'm about to give in:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Freyd's Abelian Categories says that the text, excepting the exercises, tries to be a geodesic leading to the theorem. If you take out the exercises, probably the text is 120 pages long. Impossible to do in 2:30 hours. To give you an idea, the course I'm taking is based in Rotman's &quot;An Introduction to Homological Algebra&quot; which works in R-Mod...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mitchell's Theory of Categories is very hard to read, and also to prove the theorem you have tons of definitions and propositions and lemmas to prove.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Weibel's An Introduction to Homological Algebra redirects me to Swan, The Theory of Sheaves, a book which is unavailable in my university's library. I've leafed through Swan's Algebraic K-Theory: the theorem is proved, but it is also long, hard and painful to read, and assumes a lot of knowledge I don't have (I had never seen a weakly effaceable functor, or a Serre subcategory; and it certainly is not well known to me that the category of additive functors from a small abelian category to the category of abelian groups is well-powered, right complete, and has injective envelopes!)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm starting to believe it's an impossible task. But maybe there are more modern proofs which require less heavy machinery and technicalities?&lt;/p&gt;&#xA;" OwnerUserId="2614" LastActivityDate="2010-11-28T00:22:26.130" Title="Freyd-Mitchell's embedding theorem" Tags="&lt;reference-request&gt;" CommentCount="6" FavoriteCount="1" />
  <row Id="12150" PostTypeId="1" CreationDate="2010-11-28T00:37:21.787" Score="2" ViewCount="99" Body="&lt;p&gt;On my mid-term exam tonight, I had a problem which went something like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which algebraic structure is $(M,*)$, where $M= \left \{ \begin{bmatrix}&#xD;&#xA;1 &amp;amp; 0\\ &#xD;&#xA;0 &amp;amp; 1&#xD;&#xA;\end{bmatrix} \right., \begin{bmatrix}&#xD;&#xA;0 &amp;amp; 1\\ &#xD;&#xA;-1 &amp;amp;0 &#xD;&#xA;\end{bmatrix},\begin{bmatrix}&#xD;&#xA;-1 &amp;amp; 0\\ &#xD;&#xA;0 &amp;amp; -1&#xD;&#xA;\end{bmatrix},\begin{bmatrix}&#xD;&#xA;0 &amp;amp;-1 \\ &#xD;&#xA;1 &amp;amp; 0&#xD;&#xA;\end{bmatrix}\left.  \right \}  $   and * is matrix multiplication.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There must be some smart way of doing it, but I couldn't figure it out. In the end I made a table with all elements of M and relationship between them and then just read structure's properties. To me this just feels like excessive application of brute force.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I was trying to prove that operation is closed, I tried with something like $x=\begin{bmatrix}&#xD;&#xA;a &amp;amp; b\\ &#xD;&#xA;c &amp;amp; d&#xD;&#xA;\end{bmatrix}, y= \begin{bmatrix}&#xD;&#xA;p &amp;amp; q\\ &#xD;&#xA; s&amp;amp; t&#xD;&#xA;\end{bmatrix}$ and multiply that and then prove that result is part of the set, but I run out of ideas on how to prove that result is part of the set.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I had basically same problem for inverse element and commutativity too.&lt;/p&gt;&#xA;" OwnerUserId="348" LastActivityDate="2010-11-28T01:39:12.027" Title="Looking for a smart way to find out what algebraic structure this is..." Tags="&lt;linear-algebra&gt;" AnswerCount="2" CommentCount="6" />
  <row Id="12151" PostTypeId="2" ParentId="12126" CreationDate="2010-11-28T00:52:20.577" Score="13" ViewCount="" Body="&lt;p&gt;A few years after Carleson's proof Fefferman came up with a shorter proof of the $L^2$ and $L^p$ results. Later, in the context of their work in multilinear harmonic analysis Lacey and Thiele came up with a quite short and easy to understand proof for the $L^2$ theorem which is to some extent a descendant of Fefferman's proof. It's only 10 pages long and can be found in&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lacey, Michael; Thiele, Christoph (2000), &quot;A proof of boundedness of the Carleson operator&quot;, Mathematical Research Letters 7 (4): 361–370.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(By the way, this paper has an amusing Mathscinet review which begins with &quot;This is one of the greatest papers written in Fourier analysis.&quot;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They also wrote an expository article describing this and a number of related results:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lacey, Michael T. (2004), &quot;Carleson's theorem: proof, complements, variations&quot;, Publicacions Matemàtiques 48 (2): 251–307&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They also put an expanded version of this last paper on the arxiv &lt;a href=&quot;http://arxiv.org/abs/math/0307008v4&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/math/0307008v4&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3035" LastActivityDate="2010-11-28T00:52:20.577" CommentCount="4" />
  <row Id="12152" PostTypeId="2" ParentId="12150" CreationDate="2010-11-28T01:08:31.190" Score="2" ViewCount="" Body="&lt;p&gt;Max has got the right idea in the comments.  If you look at what these matrices do to points in the plane the answer should be clear.  &lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-11-28T01:08:31.190" />
  <row Id="12154" PostTypeId="1" AcceptedAnswerId="12155" CreationDate="2010-11-28T01:23:45.460" Score="2" ViewCount="74" Body="&lt;p&gt;If any edge from a given spanning tree T0 is contained in some minimum spanning tree T*, does this imply that T0 is also a minimum spanning tree ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Right now, I'm trying to draw on paper some graphs to prove that it doesn't. Please correct me if it does, or help me find an example if it doesn't.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S. smb please add MST or spanning-trees tag.&lt;/p&gt;&#xA;" OwnerUserId="2963" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2010-11-28T01:32:00.020" LastActivityDate="2010-11-28T01:34:09.270" Title="Fast question about minimum spanning trees" Tags="&lt;graph-theory&gt;&lt;trees&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="12155" PostTypeId="2" ParentId="12154" CreationDate="2010-11-28T01:34:09.270" Score="4" ViewCount="" Body="&lt;p&gt;Consider a triangle with edge weights $2,2,1$.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-28T01:34:09.270" CommentCount="1" />
  <row Id="12156" PostTypeId="2" ParentId="12150" CreationDate="2010-11-28T01:39:12.027" Score="0" ViewCount="" Body="&lt;p&gt;Another option is to take one of the matrices and apply it several times. Say you take the second matrix and apply it to $(x,y)$. Magically, you will immediately recognize the other matrices (in order), so you've got a cyclic group of order $4$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As Max and Qiaochu have commented, there's a geometric interpretation (think what geometric transformation keeps the origin and has order $4$).&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-28T01:39:12.027" CommentCount="2" />
  <row Id="12157" PostTypeId="2" ParentId="12098" CreationDate="2010-11-28T02:18:04.533" Score="6" ViewCount="" Body="&lt;p&gt;You can plot the &lt;a href=&quot;http://www.wolframalpha.com/entities/surfaces/taubin%27s_heart_surface/cr/rn/yc/&quot; rel=&quot;nofollow&quot;&gt;Taubin's heart surface&lt;/a&gt; using ContourPlot3D&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;ContourPlot3D[(2 x^2 + y^2 + z^2 - 1)^3 - (1/10) x^2 z^3 - y^2 z^3 == 0, {x, -1.5, 1.5}, {y, -1.5, 1.5}, {z, -1.5, 1.5}, Mesh -&amp;gt; None, ContourStyle -&amp;gt; Directive[Red, Opacity[0.8]]]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I would post a image but it won't let me because I don't have reputation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/lXM95.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1938" LastEditorUserId="1938" LastEditorDisplayName="" LastEditDate="2010-11-28T03:15:14.563" LastActivityDate="2010-11-28T03:15:14.563" CommentCount="1" />
  <row Id="12158" PostTypeId="2" ParentId="12098" CreationDate="2010-11-28T02:49:36.433" Score="3" ViewCount="" Body="&lt;p&gt;Consider the map $T \colon \mathbb R^2 \rightarrow \mathbb R^2, \ (x,y) \mapsto (x, y+ \sqrt{|x|})$.  With a little examination, you can see that this will define a warping on the plane that will map the unit circle to a heart shaped curve: &#xA;&lt;img src=&quot;http://i.imgur.com/qpKuu.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if you know that a parametrization for the circle is $(\cos(t),\ \sin(t)),\  t\in [-\pi,\pi]$, then the parametrization for its heart-shaped image would be $(\cos(t),\ \sin(t) + \sqrt{|\cos(t)|}),\  t\in [-\pi,\pi]$. You can plot the curve with the following Mathematica code:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;ParametricPlot[{Cos[t], Sin[t] + Sqrt[Abs[Cos[t]]]}, {t, -Pi, Pi}]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="145" LastActivityDate="2010-11-28T02:49:36.433" />
  <row Id="12159" PostTypeId="2" ParentId="12107" CreationDate="2010-11-28T03:21:44.597" Score="0" ViewCount="" Body="&lt;p&gt;The key to this last assertion is the simple fact that $\cos(\pi - x) = -cos(x)$.  Said symmetry can be observed directly from the definition of the cosine function via the unit circle.&lt;/p&gt;&#xA;" OwnerUserId="4154" LastActivityDate="2010-11-28T03:21:44.597" />
  <row Id="12160" PostTypeId="1" AcceptedAnswerId="12270" CreationDate="2010-11-28T03:25:57.893" Score="8" ViewCount="422" Body="&lt;p&gt;I was wondering if the following properties of the Legendre polynomials are true in general. They hold for the first ten or fifteen polynomials.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1)Are the roots always simple (ie. multiplicity 1)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Except for low-degree cases, the roots can't be calculated exactly,only approximated.(unlike Chebyshev polynomials) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) Are roots of the entire family of Legendre Polynomials dense in the interval [0,1] (ie, its not possible to find a subinterval, no matter how small, that doesn't contain at least one root of one polynomial).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If anyone knows of an article/text that proves any of the above, please let me know.  The definition of these polynomials can be found on Wikipedia&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Legendre_polynomials&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Legendre_polynomials&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3971" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-28T13:13:05.537" LastActivityDate="2011-01-08T09:44:03.503" Title="Roots of Legendre Polynomial" Tags="&lt;real-analysis&gt;&lt;polynomials&gt;&lt;special-functions&gt;&lt;roots&gt;" AnswerCount="5" CommentCount="12" />
  <row Id="12161" PostTypeId="2" ParentId="12110" CreationDate="2010-11-28T03:26:01.440" Score="2" ViewCount="" Body="&lt;p&gt;Caveat: The symbol ^ in C family languages is bitwise xor.  This is true in Python, too.  The ** symbol is also seen in Python.  &lt;/p&gt;&#xA;" OwnerUserId="4154" LastActivityDate="2010-11-28T03:26:01.440" />
  <row Id="12162" PostTypeId="1" CreationDate="2010-11-28T03:39:58.390" Score="2" ViewCount="88" Body="&lt;p&gt;In an attempt to model the cassette tape spooling action, I'm hoping to solve this problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Required feed speed 5cm/second.&#xA;Two Spindles with starting radius of &#xA;S1: 0.8cm (empty spool) &#xA;S2: 1.8cm (full spool)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The spooled tape thickness is unknown but can be assumed to be around 15micrometer if needed.&#xA;The time to complete the spooling from one spool to the other spool is required to be 45 minutes.&lt;/p&gt;&#xA;" OwnerUserId="4153" LastActivityDate="2010-11-28T05:13:57.377" Title="How do I calculate the instantaneous angular speed of a spindle given a spooled radius and required feed speed?" Tags="&lt;mathematical-modeling&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="12163" PostTypeId="1" CreationDate="2010-11-28T03:42:31.793" Score="4" ViewCount="184" Body="&lt;p&gt;How can we find $\displaystyle \lim_{y\to{b}}\frac{y-b}{\ln{y}-\ln{b}}$ without using:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(a) L'Hôpital's rule, (b) the limit $\displaystyle \lim_{h \to 0}\frac{e^h-1}{h} = 1$, and (c) the fact that $\displaystyle \frac{d}{dx}\left(e^x\right) = e^x$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason for the conditions is that with this limit I'm trying to prove (c), and I've done so with (b) and I gather it would be circular to use (a). So that's that. Also, I would appreciate if you could share one or more ways of proving that the derivative of $e^x$ is $e^x$. Thanks a lot for your time.  &lt;/p&gt;&#xA;" OwnerUserId="3207" LastActivityDate="2010-11-28T10:17:29.837" Title="Calculate $\displaystyle \lim_{y\to{b}}\frac{y-b}{\ln{y}-\ln{b}}$" Tags="&lt;calculus&gt;&lt;limit&gt;" AnswerCount="4" CommentCount="5" FavoriteCount="3" />
  <row Id="12164" PostTypeId="2" ParentId="12163" CreationDate="2010-11-28T03:47:04.190" Score="0" ViewCount="" Body="&lt;p&gt;Say $y = b(1+\epsilon)$. Then $\ln y - \ln b \approx \epsilon$ whereas $y - b = b\epsilon$. So it all boils down to showing $\ln (1+\epsilon) \approx \epsilon$.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-28T03:47:04.190" CommentCount="1" />
  <row Id="12165" PostTypeId="2" ParentId="11474" CreationDate="2010-11-28T03:49:10.973" Score="0" ViewCount="" Body="&lt;p&gt;You can also do this.  The intersection of any family of topologies on a set is itself a topology.  The discrete topology (all subsets open) is a topology for any set.  So, take the set of all topologies containing the family of topologies (remember: the discrete one is here) and you will get a topology.  It will be the smallest topology containing all topologies in the family.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is integral to the notion of weak topology.  The first answer has a second correct characterization of this topology.  See the book {\it Topology} by Wilansky.  &lt;/p&gt;&#xA;" OwnerUserId="4154" LastActivityDate="2010-11-28T03:49:10.973" CommentCount="2" />
  <row Id="12166" PostTypeId="1" CreationDate="2010-11-28T03:50:11.767" Score="2" ViewCount="150" Body="&lt;p&gt;&quot;When you draw a circle in a plane of radius 1 you can perfectly surround it with 6 other circles of the same radius.&quot; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;BUT when you draw a circle in a plane of radius 1 and try to perfectly surround the central circle with 7 circles you have to change the radius of the surround circles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;how can i find the radius of the surround circles if i want too use more that 6 cirles?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ex :&#xA;7 circles = radius 0.4&lt;/p&gt;&#xA;&#xA;&lt;p&gt;8 circles = radius 0.2&lt;/p&gt;&#xA;" OwnerUserId="4155" LastEditorUserId="139" LastEditorDisplayName="" LastEditDate="2010-11-28T04:33:18.257" LastActivityDate="2010-11-28T04:33:18.257" Title="Numbers of circles around a circle" Tags="&lt;geometry&gt;&lt;circle&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="12167" PostTypeId="1" AcceptedAnswerId="12168" CreationDate="2010-11-28T03:59:04.013" Score="0" ViewCount="233" Body="&lt;p&gt;The set of rationals $\mathbb{Q}$ has the same cardinality as&#xA;the set of integers $\mathbb{Z}$. True or false?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This was a question on an old exam for our class. The correct answer is true. However, I did some additional reading and came across Cantor's transfinite numbers. In the book I'm reading, it says that &quot;there are more real numbers (which include rational and irrational numbers) than there are integers&quot;. So can it also be said that there are more rational numbers than integers? And so can we say that the above statement is false?&lt;/p&gt;&#xA;" OwnerUserId="2985" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-28T04:00:34.500" LastActivityDate="2010-12-22T21:21:46.343" Title="The set of rationals has the same cardinality as the set of integers" Tags="&lt;elementary-set-theory&gt;" AnswerCount="3" CommentCount="6" />
  <row Id="12168" PostTypeId="2" ParentId="12167" CreationDate="2010-11-28T04:02:11.240" Score="2" ViewCount="" Body="&lt;p&gt;No. The statement is still true.&#xA;The cardinality of the natural number set is the same as the cardinality of the rational number set. In fact, this cardinality is the first transfinite number denoted by $\aleph_0$ i.e. $|\mathbb{N}| = |\mathbb{Q}| = \aleph_0$. By first I mean the &quot;smallest&quot; infinity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The cardinality of the set of real numbers is typically denoted by $\mathfrak{c}$. We have $\mathfrak{c}  &amp;gt; \aleph_0$, since we can set up a bijection from $\mathbb{R}$ to the power set of the natural numbers and by Cantor's theorem, for any set $X$, we have $|X| &amp;lt; |2^{X}|$. So we have $|\mathbb{R}| = |2^{\mathbb{N}}| &amp;gt; |\mathbb{N}|$. So what this essentially says is that &quot;there are more real numbers (which include rational and irrational numbers) than there are integers&quot; in some sense.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/The_Continuum_Hypothesis&quot; rel=&quot;nofollow&quot;&gt;continuum hypothesis&lt;/a&gt; states that &quot;there is no set whose cardinality is strictly between that of the natural numbers and that of the real numbers&quot; which essentially means real numbers form the second &quot;smallest&quot; infinity.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2010-11-28T05:11:34.897" LastActivityDate="2010-11-28T05:11:34.897" CommentCount="7" />
  <row Id="12169" PostTypeId="2" ParentId="9703" CreationDate="2010-11-28T04:03:09.460" Score="1" ViewCount="" Body="&lt;p&gt;Here is a very prosaic answer. If you go into a grocery store with an order, you expect the cash register to read 0 before any items are rung up.  This is because $0 + x = x$ for any number $x$.  It is the fact that zero is the neutral element of addition that makes it a valid start for the register; starting there will not cheat you on your order or give you any unfair advantage.  To wit: an empty sum is 0.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Likewise, suppose we are multiplying numbers in a multiplying machine.  What is neutral there?  We know $1*x = x$ for any number $x$.  Since an empty product, like an empty sum should be neutral and not affect the product of the numbers coming in after it, the multiplying machine should be set at 1 before it starts work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now $2^0$ is an empty product so $2^0 = 1$.  Said identity should be true for any base.&lt;/p&gt;&#xA;" OwnerUserId="4154" LastActivityDate="2010-11-28T04:03:09.460" />
  <row Id="12170" PostTypeId="2" ParentId="12167" CreationDate="2010-11-28T04:05:02.967" Score="6" ViewCount="" Body="&lt;p&gt;You can say whatever you want. However, if you want to be &lt;em&gt;correct&lt;/em&gt;, and you are talking about cardinalities, then you cannot say that there are more rationals than integers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason we &lt;em&gt;can&lt;/em&gt; say that there are more reals than rational numbers is because they actually have &lt;em&gt;greater cardinality&lt;/em&gt;, not just because the set of reals properly contains the natural numbers. While we can find a bijection between $\mathbb{N}$ and $\mathbb{Q}$, &lt;em&gt;no&lt;/em&gt; function from $\mathbb{N}$ to $\mathbb{R}$ is surjective (and, naturally, no function from $\mathbb{Q}$ to $\mathbb{R}$ can be surjective either). &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-28T04:11:10.290" LastActivityDate="2010-11-28T04:11:10.290" />
  <row Id="12171" PostTypeId="2" ParentId="12163" CreationDate="2010-11-28T04:07:35.833" Score="4" ViewCount="" Body="&lt;p&gt;If $e^x=\sum_{n\geq0}x^n/n!$, then you can show easily that $e^{x+y}=e^xe^y$. Then $$\frac{e^{x+h}-e^x}{h} = \frac{e^h-1}{h}e^x,$$ and to conclude that $\frac{d}{dx}e^x=e^x$ we need only then show that $$\frac{e^h-1}{h}\to1$$ if $h\to0$. Using your definition, that is easy.&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2010-11-28T04:07:35.833" CommentCount="6" />
  <row Id="12172" PostTypeId="1" AcceptedAnswerId="12206" CreationDate="2010-11-28T04:08:17.420" Score="1" ViewCount="171" Body="&lt;p&gt;Does there exist a nonzero linear transformation $T$ such that&#xA;$$\alpha^t T \alpha = 0\text{ for all $\alpha\in\mathbb{R}^n$,}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where ${\alpha}^{t}$ denotes the transpose of the matrix. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks! =D&lt;/p&gt;&#xA;" OwnerUserId="647" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-28T04:12:59.900" LastActivityDate="2010-11-28T13:06:50.410" Title="Does there exist a nonzero linear transformation $T$ such that ${\alpha}^{T} T \alpha = 0$ for all $\alpha \in \mathbb{R}^n$?" Tags="&lt;linear-algebra&gt;" AnswerCount="4" CommentCount="2" />
  <row Id="12173" PostTypeId="2" ParentId="12172" CreationDate="2010-11-28T04:13:32.813" Score="2" ViewCount="" Body="&lt;p&gt;Sure. If $n=2$, let $T$ be a $90^\circ$ rotation. A suitable modification of this works for all $n\ge 2$: Simply rotate the first two coordinates $90^\circ$, and map the rest to 0. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $n=1$, there is an $r\in{\mathbb R}$ such that $T\alpha=r\alpha$ for all $\alpha$, and $\alpha^T T\alpha$ is simply $r\alpha^2$, so if $n=1$, $T$ must be 0.&lt;/p&gt;&#xA;" OwnerUserId="462" LastActivityDate="2010-11-28T04:13:32.813" />
  <row Id="12174" PostTypeId="2" ParentId="12172" CreationDate="2010-11-28T04:15:38.770" Score="2" ViewCount="" Body="&lt;p&gt;If $n$ is even then the block matrix  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$T = \begin{pmatrix} 0 &amp;amp; I \\ -I &amp;amp; 0 \end{pmatrix}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;works. For odd $n$ the matrix $T$ may be adjusted by simply including $\mathbb{R}^n$ in $\mathbb{R}^{n-1}$ by sending one component to zero. Of course this fails for $n=1$, and as noted in other answers $T$ must be zero for $n=1$.&lt;/p&gt;&#xA;" OwnerUserId="2887" LastEditorUserId="2887" LastEditorDisplayName="" LastEditDate="2010-11-28T04:19:29.717" LastActivityDate="2010-11-28T04:19:29.717" />
  <row Id="12175" PostTypeId="2" ParentId="12172" CreationDate="2010-11-28T04:15:46.870" Score="4" ViewCount="" Body="&lt;p&gt;Presumably, your vectors are column vectors and $T$ is $n\times n$, so that $\alpha^t T\alpha$ is a $1\times 1$ matrix (i.e., a scalar). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Multiplying $v^t$ by $w$ amounts to taking the dot product of $v$ with $w$. Since $v\cdot w=0$ if and only if $v$ and $w$ are orthogonal, the question is equivalent to asking if you can find a linear transformation that sends every vector to a vector orthogonal to itself, and does not send all vectors to zero. In two dimensions, this is trivial. If you can do it in two dimensions, you can do it in any larger number (think about it). &lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-11-28T04:15:46.870" />
  <row Id="12176" PostTypeId="2" ParentId="12166" CreationDate="2010-11-28T04:32:42.077" Score="7" ViewCount="" Body="&lt;p&gt;Imagine there are $n \geq 3$ circles surrounding your unit circle.  Then the situation would look like:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/M9dm8.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whence $\cos(x)=\frac{r}{r+1}$.  The angle $x$ is half of the &lt;a href=&quot;http://www.mathsisfun.com/geometry/interior-angles-polygons.html&quot; rel=&quot;nofollow&quot;&gt;interior angle&lt;/a&gt; of the corresponding regular polygon, so $x=\frac{n-2}{2n} \cdot 180^{\circ}$.  You can then solve for $r=\frac{\cos(x)}{1-\cos(x)}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;when $n=4$ we have $r=\frac{\cos(45^{\circ})}{1-\cos(45^{\circ})}=1+\sqrt{2}=2.41421...$.&lt;/li&gt;&#xA;&lt;li&gt;when $n=6$ we have $r=\frac{\cos(60^{\circ})}{1-\cos(60^{\circ})}=1$.&lt;/li&gt;&#xA;&lt;li&gt;when $n=8$ we have $r=\frac{\cos(67.5^{\circ})}{1-\cos(67.5^{\circ})}=0.619914...$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="139" LastActivityDate="2010-11-28T04:32:42.077" />
  <row Id="12177" PostTypeId="2" ParentId="12163" CreationDate="2010-11-28T04:38:45.723" Score="5" ViewCount="" Body="&lt;p&gt;Let us look at the limit in the title and the definition of $\log x$  (please excuse me for using $\log$ notation instead of $\ln$) as in your comment,&#xA;$$\log x =\int_1^x \frac{1}{t}dt$$&#xD;&#xA;By the fundamental theorem of calculus we have &#xA;$$\frac{d}{dx}\log x = \frac{1}{x}$$&#xD;&#xA;on the other hand &#xA;$$f&amp;#39;(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}=\lim_{y\to x}\frac{f(y)-f(x)}{y-x}$$&#xD;&#xA;hence &#xA;$$\lim_{y\to x}\frac{y-x}{\log y-\log x}=\lim_{y\to x}\frac{1}{\frac{\log y-\log x}{y-x}}=\frac{1}{1/x}=x$$&#xD;&#xA;where in the last step we used the quotient rule &#xA;$$\lim_{y\to a}\,g(y)=A\, \text{ and }\,\lim_{y\to a}\,h(y)=B\ne0\text{ implies }\lim_{y\to a}\,\frac{g(y)}{h(y)}=\frac{A}{B}$$&#xD;&#xA;with $g(y)=1$.&lt;/p&gt;&#xA;" OwnerUserId="1154" LastActivityDate="2010-11-28T04:38:45.723" />
  <row Id="12178" PostTypeId="2" ParentId="11474" CreationDate="2010-11-28T04:42:50.983" Score="1" ViewCount="" Body="&lt;p&gt;The following short piece of lecture notes is almost entirely devoted to an answer of your question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://math.uga.edu/~pete/TopSection4.pdf&quot; rel=&quot;nofollow&quot;&gt;http://math.uga.edu/~pete/TopSection4.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, for reasons that now escape me, I decided to break the file here, so the important remark that the statements &quot;$\tau$ is the topology generated by the family of sets $\mathcal{F}$&quot; and &quot;The family of sets $\mathcal{F}$ is a subbase for the topology $\tau$&quot; are equivalent does not appear until the beginning of the next little piece of notes:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://math.uga.edu/~pete/TopSection5.pdf&quot; rel=&quot;nofollow&quot;&gt;http://math.uga.edu/~pete/TopSection5.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="299" LastActivityDate="2010-11-28T04:42:50.983" CommentCount="2" />
  <row Id="12179" PostTypeId="1" CreationDate="2010-11-28T04:46:09.903" Score="3" ViewCount="300" Body="&lt;p&gt;How does one diagonalize a large sparse symmetric matrix to get the eigenvalues and the eigenvectors?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is the matrix could be very large (though it is sparse), at most 2500*2500. Is there a good algorithm to do that and most importantly, I can implement it into my own code. Thanks a lot!&lt;/p&gt;&#xA;" OwnerUserId="2419" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-28T14:36:05.683" LastActivityDate="2010-11-28T14:36:05.683" Title="how to diagonalize a large sparse symmetric matrix, to get the eigenvalues and eigenvectors" Tags="&lt;linear-algebra&gt;&lt;algorithms&gt;&lt;matrices&gt;&lt;numerical-methods&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="1" />
  <row Id="12180" PostTypeId="1" AcceptedAnswerId="12184" CreationDate="2010-11-28T05:02:05.237" Score="0" ViewCount="218" Body="&lt;p&gt;A function $f(x)$ is defined $\forall x \in \mathbb{Q}$ and $ x \in (0,1)$ as $f(x) = q $ where $x = (p/q)$ in simplest form. Find $\sup_{\tau} \quad \min_{a} \quad f(a) + f(\tau-a)$    where $ a \in \mathbb{Q}$ and $ a \in (0,1)$ and $ \tau \in (\alpha,\beta)$ where   $ 0&amp;lt; \alpha &amp;lt; \beta &amp;lt;2 $ and $ \alpha,\beta \in \mathbb{Q} $. $\sup$ and $\min$ are taken over only those $x$ where $f(x)$ is defined. Is the supremum infinity ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The supremum may or may not depend on $\alpha$ and $\beta$...this has not been ruled out.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT 2&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I will try something !&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider $\tau = l/m$ where $l,m \in \mathbb{N}$ in simplest form and $\tau \in \mathbb{Q}\cap(0,2)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;what is minimum possible value of $q$ where $\tau-a = p/q $ written in simplest form&#xA;such that $a \in \mathbb{Q} \cap (0,1)$ and $(\tau-a) \in (0,1)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As Yuval Filmus suggested in his answer...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;one way to choose $a$ is that it has same denominator (in simplest form) as that of $\tau$ and numerator $n$ such that $l-n$ is relatively prime to $m$, (If such a choice is possible under given conditions), then $f(a)+f(\tau-a)$ becomes $2m$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The question is whether such a choice leading to $f(a)+f(\tau-a) = 2m$ is the minimum possible value ? If not I request you to give a counter example.&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it is, then the supremum (over $\tau$ varying in any arbitrary open interval contained in $(0,2)$) is $\infty$ as we can increase $m$ arbitrarily by varying $\tau$ in any arbitrary open interval $(\alpha,\beta)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT 3&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;small change in Question...instead of&lt;/strong&gt; $0&amp;lt;\alpha&amp;lt;\beta&amp;lt;2$ we have $-1&amp;lt;\alpha&amp;lt;\beta&amp;lt;1$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;this change is not made with any intention of bringing out a certain solution.Also its influence on the solution is hoped to be very small.Inconvenience is regretted.&lt;/strong&gt; &lt;/p&gt;&#xA;" OwnerUserId="2987" LastEditorUserId="2987" LastEditorDisplayName="" LastEditDate="2010-11-28T17:58:43.037" LastActivityDate="2010-11-28T21:26:39.943" Title="A question on Rational numbers in simplest form." Tags="&lt;number-theory&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="12181" PostTypeId="1" CreationDate="2010-11-28T05:02:21.190" Score="2" ViewCount="154" Body="&lt;p&gt;I propose the following restriction on universal instantiation: UI may not be used to introduce new variables. The variable specified should be an &quot;old&quot; variable, i.e. it must already have been introduced by either an active premise or by existential instantiation (E-elimination). (An &lt;em&gt;active&lt;/em&gt; premise is one that has not been closed-off or discharged by what I call a &lt;em&gt;conclusion&lt;/em&gt; statement.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The primary benefit of this restriction is to eliminate the need to consider dependencies among variables that are created with existential instantiation. Such considerations never seem to come up in &quot;real mathematics.&quot; The restriction I propose seems to be implicit even in the most rigorous mathematical proofs, if not in standard formal logic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With this restriction in place, dependencies among variables need not be considered because the only variables on which you could then make universal generalizations are those introduced in a premise that has subsequently been closed-off or discharged by what I call a conclusion statement. The conclusion statement and subsequent statements that are derived from it and also refer to these variables can always be universally generalized regardless of the presence of any other variables that were introduced by E-elimination.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have implemented this restriction in my proof checker/editor (available free at &lt;a href=&quot;http://www.dcproof.com&quot; rel=&quot;nofollow&quot;&gt;http://www.dcproof.com&lt;/a&gt;). Play around with it. Try to &quot;break it&quot; if you can. Work through the brief tutorial to learn the system. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another benefit, from a pedagogical perspective, has been the elimination from my program of several what must have been confusing warning messages about dependencies among variables. It also simplifies the requirements for universal generalizations (A-introduction). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Following is a summary of the rules of inference I use in my system that are relevant to the handling of free variables:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Free variables may be introduced only by means of a premise (assumption) or by existential specification (E-elimination).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Existential specification (E-elimination) allows an unused free variable to be specified for any active, existentially quantified statement.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Universal specification (A-elimination) allows any free variable introduced by an active premise or any algebraic expression in such free variables to be specified for any active, universally quantified statement.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Existential generalization (E-introduction) may be applied to any free variable or any algebraic expression in one or more free-variables that is found in any active statement.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Universal generalizations (A-introduction) may be applied to any free variable that (a) is found in an active statement, and (b) is &lt;em&gt;not&lt;/em&gt; referred to by any active premise, and (c) was &lt;em&gt;not&lt;/em&gt; introduced by existential specification (E-elimination). (There are no considerations given for any dependencies among variables.)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Free variables introduced after a premise statement and before the corresponding conclusion statement may &lt;em&gt;not&lt;/em&gt; appear in that conclusion statement.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;In my program, I use color-coding for free variables. Green indicates that a Universal Generalization is possible. Red indicates that a Universal Generalization is &lt;em&gt;not&lt;/em&gt; possible. When first introduced, all free variables are red. A free variable that was introduced by a premise is changed to green when that premise is closed-off (or discharged) by a conclusion statement. It is effectively reintroduced, changing back to red if it is referred to in subsequent premise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your comments would be appreciated.&lt;/p&gt;&#xA;" OwnerUserId="3515" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-11-29T17:44:51.867" LastActivityDate="2011-02-10T16:21:11.857" Title="Proposed Restriction on Universal Instantiation (Natural Deduction)" Tags="&lt;logic&gt;&lt;proof-theory&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="12182" PostTypeId="2" ParentId="12181" CreationDate="2010-11-28T05:11:26.617" Score="1" ViewCount="" Body="&lt;p&gt;You can always add some tautology as an assumption (say $x \rightarrow x$) and then later discharge it, so it seems like your restriction is valid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Check &quot;Structural Proof Theory&quot;, a book by von Plato and Negri, which discusses such issues; it also sponsors a proof editor PESCA by Aarne Ranta.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-28T05:11:26.617" CommentCount="5" />
  <row Id="12183" PostTypeId="2" ParentId="12162" CreationDate="2010-11-28T05:13:57.377" Score="2" ViewCount="" Body="&lt;p&gt;Suppose that the feed spindle has outer radius $r_f(t)$ and the intake spindle has outer radius $r_i(t)$, similarly let the spindle's angular velocity be given by $\omega_f(t)$ and $\omega_i(t)$. Note that both $r_f$ and $r_i$ lie in the interval $[r_0, R_0]$, where $r_0= .8$ cm and $R_0 = 1.8$ cm.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only constraint is that the tape must pass between the two spindles at a constant rate of $s = 5$ cm/s. For the tape to remain taught under this condition the tangential velocity of each spindle must be $s$. Thus,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$s = r_f(t)\cdot\omega_f(t) = r_i(t)\cdot\omega_i(t).$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\omega(t) = \frac{s}{r(t)}$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;for either spool. So now we simply need to find a description of the outer radius of the spindle. Consider that the tape has thickness $a \ll r_0$ and width $w$, thus the volume of tape transfered per unit time is  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \frac{dV}{dt} = \pm a s w$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and since $V = \pi w r(t)^2$ then we also have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \frac{dV}{dt} = 2\pi r(t) \dot{r}(t) w$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and then by equating these expressions and solving the IVP for the feeding spindle yields,  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$r_f(t) = \sqrt{R_0^2 - \frac{as}{\pi} t}.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, the IVP for the intake spindle yields&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$r_i(t) = \sqrt{r_0^2 + \frac{as}{\pi} t}.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From those expressions it is easy to see that the proper speeds to drive the spindles are approximately&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\omega_f(t) = \left(\sqrt{\left(\frac{R_0}{s}\right)^2 - \frac{a}{s\pi} t}\right)^{-1}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\omega_i(t) = \left(\sqrt{\left(\frac{r_0}{s}\right)^2 + \frac{a}{s\pi} t}\right)^{-1}.$$&lt;/p&gt;&#xA;" OwnerUserId="2887" LastActivityDate="2010-11-28T05:13:57.377" CommentCount="3" />
  <row Id="12184" PostTypeId="2" ParentId="12180" CreationDate="2010-11-28T05:15:03.290" Score="2" ViewCount="" Body="&lt;p&gt;EDIT: This is a consolidated solution which previously appeared in parts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lemma 1. For each natural $M$ and real $\alpha &amp;lt; \beta$ there's a rational number in $(\alpha,\beta)$ whose reduced form has a denominator larger than $M$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proof. Choose some power of two $2^n &amp;gt; \max\left(M, \frac{2}{\beta-\alpha}\right)$. Since $2^{-n} &amp;lt; \frac{\beta-\alpha}{2}$, there are at least two fractions of the form $x/2^n$ in $(\alpha,\beta)$. One of them has odd $x$ and so is in reduced form.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lemma 2. If $p_1/q_1 + p_2/q_2 = P/Q$, all rationals in reduced form, then $(q_1+q_2)^2 \geq 4Q$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proof. Taking a common denominator, $\frac{p_1 q_2 + p_2 q_1}{q_1q_2} = \frac{P}{Q}$ and so $Q|q_1q_2$ (since $P/Q$ is reduced). In particular $Q \leq q_1q_2 \leq \left(\frac{q_1+q_2}{2}\right)^2$, where the last inequality is the arithmetic-geometric mean inequality.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Theorem. For every real $M$ and reals $\alpha &amp;lt; \beta$ there is a rational $r \in (\alpha,\beta)$ such that $\min_{a \in \mathbb{Q}} f(a) + f(r-a) &amp;gt; M$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proof. Using lemma 1, take a rational $r \in (\alpha,\beta)$ with denominator $Q &amp;gt; M^2/4$. Using lemma 2, we conclude that for every rational $a$, $f(a) + f(r-a) \geq 2\sqrt{Q} = M$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Corollary. $$\sup_r \min_a f(a) + f(r-a) = \infty.$$&lt;/p&gt;&#xA;" OwnerUserId="1277" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2010-11-28T21:26:39.943" LastActivityDate="2010-11-28T21:26:39.943" CommentCount="15" />
  <row Id="12185" PostTypeId="2" ParentId="9934" CreationDate="2010-11-28T05:31:50.503" Score="3" ViewCount="" Body="&lt;p&gt;Let $G$ be a group.  The &lt;em&gt;canonical residually finite quotient&lt;/em&gt; of $G$ is $R(G)=G/K$ where $K$ is the intersection of all the finite-index subgroups of $G$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Lemma:&lt;/strong&gt;  &lt;em&gt;If $G$ is finitely generated (update)&lt;/em&gt; then $G$ has at most one subgroup of each index if and only if $R(G)$ is cyclic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Proof:&lt;/em&gt; First, note that $R(G)$ is residually finite.  If every finite quotient of $R(G)$ is cyclic then $R(G)$ is residually cyclic, and it follows that $R(G)$ is abelian.  So $R(G)$ has a non-cyclic finite quotient unless $R(G)$ is cyclic.  Therefore, if $R(G)$ is not cyclic then $R(G)$, and hence $G$, has a finite non-cyclic quotient, and hence, by Artuto's answer, has a two distinct finite-index subgroups of the same index.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conversely, suppose that $R(G)$ is cyclic.  Every finite-index subgroup of $G$ contains $K$, so the quotient map $G\to R(G)$ maps finite-index subgroups to finite-index subgroups bijectively and preserves the index.  Therefore, if $R(G)$ is cyclic then $G$ has at most one subgroup of each index.  &lt;em&gt;QED&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe that it is an open question whether or not there is an algorithm to determine whether a fp group has a proper finite-index subgroup, ie whether or not $R(G)$ is non-trivial.  So it may be open whether or not it is possible to determine if $R(G)$ is cyclic, too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Note:&lt;/strong&gt; Earlier, I forgot to mention that I had implicitly assumed that $G$ is finitely generated.  This assumption is clearly necessary; otherwise the additive group of the rationals is a counterexample.  If $G$ is not finitely generated, then the same argument shows that if $G$ has at most one subgroup of each finite index then $R(G)$ is &lt;em&gt;residually&lt;/em&gt; cyclic.  But it's not clear to me that the converse of this statement is true.   So I'll finish with a question:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If $G$ is residually cyclic, does $G$ have at most one subgroup of each finite index?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2287" LastEditorUserId="2287" LastEditorDisplayName="" LastEditDate="2010-11-28T15:01:03.837" LastActivityDate="2010-11-28T15:01:03.837" CommentCount="1" />
  <row Id="12186" PostTypeId="1" CreationDate="2010-11-28T05:32:49.567" Score="1" ViewCount="191" Body="&lt;p&gt;I posted this on &lt;a href=&quot;http://gamedev.stackexchange.com/q/6009/3637&quot;&gt;gamedev.stackexchange.com&lt;/a&gt; earlier&lt;s&gt;, but I didn't get any answers... and only a few views&lt;/s&gt;.&#xA;EDIT: I've got two answers there (as of now), and I'm investigating them now.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I find out the distance traveled by a Bezier Curve?&#xA;For example, the distance traveled by a Linear Bezier Curve is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$distance = \sqrt{(x_1 - x_0)^2 + (y_1 - y_0)^2}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But what of a Quadratic, Cubic, Nth-Degree Bezier Curve?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My goal is to figure out the &quot;resolution&quot; ($resolution = 1-t$) beforehand, so I don't have to waste time checking if the next point is touching the previous point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From &lt;a href=&quot;http://en.wikipedia.org/wiki/B%C3%A9zier_curve#Generalization&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/B%C3%A9zier_curve#Generalization&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$\mathbf{B}(t) = \sum_{i=0}^n {n\choose i}(1-t)^{n-i}t^i\mathbf{P}_i$&#xD;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2736" LastEditorUserId="2736" LastEditorDisplayName="" LastEditDate="2010-11-28T21:35:12.987" LastActivityDate="2010-11-28T21:35:12.987" Title="Arc Length of Bezier Curves" Tags="&lt;bezier-curve&gt;" CommentCount="7" FavoriteCount="1" />
  <row Id="12188" PostTypeId="2" ParentId="11668" CreationDate="2010-11-28T06:14:09.763" Score="2" ViewCount="" Body="&lt;p&gt;the real complication here is the &lt;em&gt;joint&lt;/em&gt; probability diatribution of $X$ and $Y$. in introductory courses, one learns about the joint pdf of two continuous random variables, or the joint pmf of two discrete random variables. the mixed case you mention, where one variable is continuous and the other is discrete is not commonly mentioned in introductory texts - altho some manage to sneak it in without calling attention to just what they are dealing with.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a common example is this: suppose $X$ has the binomial distribution bin($n,u$), where $u$ is the success probability. each day a new $X$ is observed. however, the value of $u$ changes from day to day in a random way: it is selected each time at random uniformly from the interval (0,1). because the success probability $u$ is also random, we should denote this random quantity [before it is selected] by $U$. [in bayesian contexts, this uniform distribution for $U$ is referred to as a 'prior pdf' for $U$. however, unlike the bayesian context, which focuses on $U$, our focus here is on $X$.]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so we have here a mixed pair of random variables $(X,U)$, where one is discrete and one is continuous [as in the OP's question.] here the [mixed] joint probability distribution of $(X,U)$ is specified by two distributions: the [marginal] pdf of $U$: $f_U(u) = 1$ for $0 &amp;lt; u &amp;lt; 1$ [and is $0$ elsewhere] -  and what is really the conditional distribution of $X$ given $U$, for which the conditional pmf is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f_{X|U}(x|u) = P(X=x|U=u) = P(bin(n,u) = x) = {n\choose x} u^x(1-u)^{n-x},\quad x = 0,1,\cdots, n.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;it turns out that the joint [mixed] pmf-pdf of $(X,U)$ can be written as the product of the marginal pdf of $U$ and the conditional pmf of $X|U$, analogous to what one does in the case where both variables are discrete, or both are continuous. [the justification for this involves what kahen alludes to in his answer - but it goes beyond the scope of this reply.]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so we can write the mixed joint pmf-pdf of $(X,U)$ as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ f_{X,U}(x,u) = f_U(u)f_{X|U}(x|u) = {n\choose x} u^x(1-u)^{n-x},\quad x = 0,1,\cdots, n,\quad 0&amp;lt;u&amp;lt;1.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[here $f_{X,U}(x,u)$ is understood to vanish for other values of $(x,u)$.]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;this mixed pmf-pdf can be treated similarly to the way one manipulates a joint pmf or pdf [to get marginal and conditional distributions, for example] - except that one integrates out $u$ to get the marginal pmf of $X$ and sums out $x$ to get the marginal pdf of $U$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so for example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f_X(x) = \int_0^1 f_{X,U}(x,u)du  =\int_0^1 {n\choose x} u^x(1-u)^{n-x}du = \frac{1}{n+1},\quad x = 0,1,\dots,n.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;thus $X$ is marginally uniformly distributed on the set of outcomes $\{ 0,1,\dots, n\} $.  [this means that over a large number of days, all of the possible values of $X$ (from $0$ to $n$) will turn up about equally often. given the way $U$ varies from day to day, this seems plausible.]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\large \mathbf [$ the evaluation of the integral is straightforward on noting that it is a beta function integral, so that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_0^1 u^x(1-u)^{n-x}du = Beta(x+1, n-x+1) = \frac{x!(n-x)!}{(n+1)!}. {\mathbf ]}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;one can similarly get the conditional pdf of $U|X$ as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f_{U|X}(u|x) = \frac{f_{X,U}(x,u)}{f_X(x)} = (n+1){n\choose x}u^x(1-u)^{n-x},\quad 0&amp;lt;u&amp;lt;1.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;this means that looking only at the days when $X=x$, conditionally, $U$ has a beta distribution with parameters $(x+1, n-x+1)$. this result is what interests bayesians in their contexts - altho their scenario would usually involve considering the value of $X$ only for day 1, and their interpretation of $f_{U|X}(u|x)$ isn different from that given here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the bottom line is that in working with a mixed pmf-pdf, each variable retains its discrete or continuous nature whether one considers marginal, joint or conditional distributions, and the joint mixed pmf-pdf  can be manipulated as in the pure [= unmixed] cases, remembering to sum or integrate as is appropriate. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;this reply does not address the specific problem posed in the OP, but it can be handled by the same methods. [only the imputs are a bit different: there one starts with [the joint pmf-pdf of] $(X,Y)$ and proceeds to $(X,Z)$. starting from the mixed pmf-pdf of $(X,Y)$, it is straightforward to get the conditional pmf of $Z|X$ and then the joint mixed pmf-pdf of $(X,Z)$.] &lt;/p&gt;&#xA;" OwnerUserId="1547" LastActivityDate="2010-11-28T06:14:09.763" />
  <row Id="12189" PostTypeId="1" AcceptedAnswerId="12191" CreationDate="2010-11-28T06:43:58.010" Score="3" ViewCount="65" Body="&lt;p&gt;As a homework assignment, I've been given a particular subring of $M_2(\mathbb{Q})$, and asked to list all the ideals... For reference, $S$ = { set of matrices in $M_2(\mathbb{Q})$ with bottom left entry $0$ } is the subring in question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't really see how to get a handle on this. Is there any sort of algorithmic way of doing this, or...? Where do I start?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: I posted this over at Ask an Algebraist originally: &lt;a href=&quot;http://at.yorku.ca/cgi-bin/bbqa?forum=ask_an_algebraist;task=show_msg;msg=2257&quot; rel=&quot;nofollow&quot;&gt;http://at.yorku.ca/cgi-bin/bbqa?forum=ask_an_algebraist;task=show_msg;msg=2257&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2313" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-28T12:12:45.637" LastActivityDate="2010-11-28T12:12:45.637" Title="Ideals of subrings of $M_2(\mathbb{Q})$" Tags="&lt;homework&gt;&lt;abstract-algebra&gt;&lt;ring-theory&gt;" AnswerCount="1" />
  <row Id="12190" PostTypeId="2" ParentId="11892" CreationDate="2010-11-28T06:50:18.003" Score="1" ViewCount="" Body="&lt;p&gt;(1) To use your notation, if $c \in \mathbb{R}$ is a regular value, then the level set $Re(f(z))^{-1}(c)$ will be a 1-dimensional embedded submanifold of $U \subset \mathbb{R}^2$.  Therefore, every connected component will be a connected 1-manifold.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, I'm not sure what exactly you mean by &quot;global differentiable curve.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you mean &quot;something of the form $f(t) = (x(t), y(t))$,&quot; then it follows from the comments in &lt;a href=&quot;http://math.stackexchange.com/questions/4699/which-smooth-1-manifolds-can-be-represented-by-a-single-smooth-parametrization&quot;&gt;this question&lt;/a&gt; that yes, every connected component can be put in that form.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you mean &quot;something of the form $y = f(x)$ or $x = f(y)$&quot; then the answer is (I think) no.  For example, consider $f(z) = \log z$ on $U = R^2 - \{x \geq 0, y = 0\}$, i.e. the plane with the non-negative x-axis deleted.  Then $Re(f(z)) = \log(\sqrt{x^2+y^2})$, so the level set $Re(f(z))^{-1}(1)$ is the circle $x^2 + y^2 = e^2$ minus the point $(e,0)$.  So, it doesn't seem like you'd be able to represent this curve by a single function $x = f(y)$ or $y = f(x)$.&lt;/p&gt;&#xA;" OwnerUserId="640" LastActivityDate="2010-11-28T06:50:18.003" CommentCount="3" />
  <row Id="12191" PostTypeId="2" ParentId="12189" CreationDate="2010-11-28T06:50:24.407" Score="7" ViewCount="" Body="&lt;p&gt;Start by looking at ideals generated by one element - choose a matrix and see what happens when you multiply it from the left and right, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Every ideal contains an ideal generated by one element, so now you need to see if you can extend some of the ideals you got to larger ideals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One more thing is that ideals in your ring are also vector spaces over the Rationals, because the ring &quot;contains&quot; $\mathbb{Q}$ as the scalar matrices, so you know that once you have an ideal of dimension 2 it is maximal and cannot be extended (since the entire vector space is of dimension 3).&lt;/p&gt;&#xA;" OwnerUserId="2125" LastActivityDate="2010-11-28T06:50:24.407" />
  <row Id="12192" PostTypeId="1" CreationDate="2010-11-28T06:57:48.860" Score="3" ViewCount="74" Body="&lt;p&gt;I would appreciate some help with factoring a trivariate polynomial.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The polynomial in question is&#xA;$p(x,y,z)=a_1 x^7+a_2 x^5*y+a_3 x^3*y^2+a_4 x*y^3+a_5 x^4*z+a_6 x^2*y*z+a_7 y^2*z+a_8 x*z^2$, where the coefficients are integers. I would like to assign values to these coefficients such that $p$ can be factored.&#xA;For example, one possible factorization might be&#xA;$p(x,y,z)=(x^3+b_1 x*y+b_2 z)(b_3 x^4+b_4 x^2*y+b_5 x*z+b_6 y^2)$.&#xA;One can expand this factorization and equate the coefficients of the expansion with those of the above equation and hope that the system of equations can be solved for integers b's.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can one find other possible factorizations? Is there an easy way?&lt;/p&gt;&#xA;" OwnerUserId="3891" LastActivityDate="2010-11-28T06:57:48.860" Title="Factoring a trivariate polynomial" Tags="&lt;algebra&gt;&lt;diophantine-equations&gt;" FavoriteCount="1" />
  <row Id="12193" PostTypeId="1" AcceptedAnswerId="12267" CreationDate="2010-11-28T07:40:27.607" Score="3" ViewCount="191" Body="&lt;p&gt;This is a follow-up to the question &#xA;&lt;a href=&quot;http://math.stackexchange.com/questions/12082/field-reductions&quot;&gt;on field reductions&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(EDIT: Originally this question used the notation $\{\mathbb{R}(\setminus a)\}$, but now uses $\mathbb{R}(\setminus a)$ instead.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There isn't a unique largest subfield of $\mathbb{R}$ not containing $a$,&#xA;so let $\mathbb{R}(\setminus a)$ denote the set of maximal subfields of $\mathbb{R}$ not containing $a$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: The question &quot;What is the cardinality of $\mathbb{R}(\setminus a)$ ?&quot; has now been asked on &lt;a href=&quot;http://mathoverflow.net/questions/47810/set-of-maximal-subfields-not-containing-particular-elements&quot; rel=&quot;nofollow&quot;&gt;Mathoverflow&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the previous question if $F\in \mathbb{R}(\setminus a)$ then $F$ is uncountable, but what is the size of $\mathbb{R}\setminus F$ the complement of $F$ in $\mathbb{R}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each $F \in \mathbb{R}(\setminus a)$ should be arrived at by removing elements of $\mathbb{R}$, but if you start with $\mathbb{R}$ and remove $a$ and then remove more elements until you get a field then surely you arrive at a single field, so where do the multiple maximal fields come from ? - It must be because you get different fields depending on the order in which elements are removed from $\mathbb{R}$, in which case what is the most natural order to remove elements to get a definition of a single canonical subfield $\mathbb{R}(\setminus a)$ of $\mathbb{R}$ that doesn't contain $a$, that we could call 'the' field reduction of $\mathbb{R}$ by $a$ ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Given Arturo Magidin's comments and answer, if we abandon the idea of a distinguished element in the abstract/in general, then since the notation $\{\mathbb{R}(\setminus a)\}$ looks like a set containing a single element called $\mathbb{R}(\setminus a)$, it makes more sense to drop the brackets and just write $\mathbb{R}(\setminus a)$ on the understanding that this is a set of field reductions i.e. a set of subfields not a single subfield.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From Pete L. Clark's answer to &lt;a href=&quot;http://math.stackexchange.com/questions/12082/field-reductions&quot;&gt;the previous question&lt;/a&gt;, if $F(a)=\mathbb{R}$ then $F=\mathbb{R}$, so let's define the notion of a super-extension of $F$ by $a$, $F(a..)=R$ being the sequence of field extensions to get to $\mathbb{R}$ from $F\in\mathbb{R}(\setminus a)$. More generally, let $@K(\setminus a)$ be an element of $K(\setminus a)$ then $@K(\setminus a)(a..)=K$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We could also reduce $K$ by a set $A$, with $K(\setminus A)$ being the set of maximal subfields that don't contain any elements from $A$, and again write the super-extension $@K(\setminus A)(A..)=K$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $\mathbb{A}&amp;#39;$ be the set of non-rational algebraic numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the cardinality of $\mathbb{R}(\setminus \mathbb{A}&amp;#39;)$ ?&lt;/p&gt;&#xA;" OwnerUserId="4157" LastEditorUserId="4157" LastEditorDisplayName="" LastEditDate="2010-11-30T17:01:02.337" LastActivityDate="2010-11-30T17:01:02.337" Title="Field reductions. part two" Tags="&lt;abstract-algebra&gt;&lt;field-theory&gt;" AnswerCount="1" CommentCount="10" />
  <row Id="12194" PostTypeId="1" CreationDate="2010-11-28T08:30:24.823" Score="5" ViewCount="492" Body="&lt;p&gt;Ibid the topic title.&lt;/p&gt;&#xA;" OwnerUserId="4158" LastEditorUserId="900" LastEditorDisplayName="" LastEditDate="2010-11-29T02:22:57.623" LastActivityDate="2010-11-29T02:22:57.623" Title="I know that &quot;x exp(x) = e&quot; means x = 1, but how do you solve for it?" Tags="&lt;calculus&gt;&lt;algebra-precalculus&gt;" AnswerCount="4" CommentCount="5" />
  <row Id="12195" PostTypeId="2" ParentId="12194" CreationDate="2010-11-28T08:42:52.680" Score="17" ViewCount="" Body="&lt;p&gt;If you want to prove the proposition that if $x e^x = e$ then $x = 1$, one way to do it is to just show that (1) $x = 1$ is indeed a solution of $x e^x = e$ by plugging the value in, and that (2) $x = 1$ is the &lt;em&gt;only&lt;/em&gt; solution, for which you can argue that $x e^x$ is an increasing function on $x \ge 0$, and the equation clearly has no solution for $x &amp;lt; 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If that makes it sounds like you have to know the answer beforehand before you can solve for it, well, that's not far from the truth. In general, if the right-hand side is an arbitrary real number $y$ instead of $e$, then you cannot solve $x e^x = y$ for $x$ in terms of elementary functions. One in fact &lt;em&gt;defines&lt;/em&gt; the &lt;a href=&quot;http://en.wikipedia.org/wiki/Lambert_W_function&quot; rel=&quot;nofollow&quot;&gt;Lambert W function&lt;/a&gt; as precisely the solution of this equation.&lt;/p&gt;&#xA;" OwnerUserId="856" LastEditorUserId="856" LastEditorDisplayName="" LastEditDate="2010-11-28T09:07:58.680" LastActivityDate="2010-11-28T09:07:58.680" CommentCount="3" />
  <row Id="12196" PostTypeId="2" ParentId="11091" CreationDate="2010-11-28T09:20:19.337" Score="2" ViewCount="" Body="&lt;p&gt;I &lt;a href=&quot;http://en.wikipedia.org/w/index.php?title=Lights_Out_%28game%29&amp;amp;oldid=399287365&quot; rel=&quot;nofollow&quot;&gt;tried&lt;/a&gt; (and &lt;a href=&quot;http://en.wikipedia.org/w/index.php?title=Lights_Out_%28game%29&amp;amp;oldid=399483159&quot; rel=&quot;nofollow&quot;&gt;tried again&lt;/a&gt;) to improve the Wikipedia article but the edits were promptly reverted as if well-known solutions of Lights Out are &quot;original research&quot;, despite providing several citations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If any Stack Exchange members are Wikipedians who can assist in restoring the article, it would be greatly appreciated. Also, I would direct this request more appropriately if I could, but don't see a way to contact John Smith directly. Feel free to delete it.&lt;/p&gt;&#xA;" OwnerUserId="1174" LastEditorUserId="1174" LastEditorDisplayName="" LastEditDate="2010-11-29T08:10:25.650" LastActivityDate="2010-11-29T08:10:25.650" CommentCount="2" />
  <row Id="12197" PostTypeId="1" AcceptedAnswerId="12226" CreationDate="2010-11-28T09:29:22.530" Score="7" ViewCount="154" Body="&lt;p&gt;My friend claims that one could reduce the number of states of a given turning  machine by somehow blowing up the tape alphabet. He does not have any algorithm though. He only has the intuition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I say it's not possible. Else one could arbitrarily keep decreasing the states via the same algorithm and arrive at some constant sized machine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Who is right?&lt;/p&gt;&#xA;" OwnerUserId="813" LastActivityDate="2010-11-28T17:53:33.493" Title="Can we reduce the number of states of a Turing Machine?" Tags="&lt;algorithms&gt;&lt;computer-science&gt;&lt;computability&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="12198" PostTypeId="2" ParentId="12163" CreationDate="2010-11-28T10:17:29.837" Score="3" ViewCount="" Body="&lt;p&gt;We can convert your $\lim $ into the inverse of the derivative of $f(y)=\ln y&#xD;&#xA;$, evaluated at $y=b$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\underset{y\rightarrow b}{\lim }\dfrac{y-b}{\ln y-\ln b}=\dfrac{1}{\underset{&#xD;&#xA;y\rightarrow b}{\lim }\dfrac{\ln y-\ln b}{y-b}}=\dfrac{1}{\dfrac{d}{dy}\left.&#xD;&#xA;\ln y\right\vert _{y=b}}=\dfrac{1}{\left. \dfrac{1}{y}\right\vert _{y=b}}=%&#xD;&#xA;\dfrac{1}{\dfrac{1}{b}}=b.$$&lt;/p&gt;&#xA;" OwnerUserId="752" LastActivityDate="2010-11-28T10:17:29.837" />
  <row Id="12199" PostTypeId="2" ParentId="12197" CreationDate="2010-11-28T10:21:09.927" Score="0" ViewCount="" Body="&lt;p&gt;I think this in the same vein as creating a compression algorithm that will compress any given file, i.e. than we can compress the output again and again, until we reach a single bit that will represent all possible files. Yet, compression algorithms do exist, and they do compress &lt;strong&gt;some&lt;/strong&gt; files.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, even it the number of states of a given Turing machine is reducible, it does not mean that all Turing machines are reducible, since that would mean that all Turing machines are just different interpretations of one and the same one-state machine.&lt;/p&gt;&#xA;" OwnerUserId="4160" LastEditorUserId="4160" LastEditorDisplayName="" LastEditDate="2010-11-28T10:41:52.147" LastActivityDate="2010-11-28T10:41:52.147" CommentCount="2" />
  <row Id="12201" PostTypeId="2" ParentId="12179" CreationDate="2010-11-28T12:05:01.273" Score="3" ViewCount="" Body="&lt;p&gt;$2500 \times 2500$ is a &lt;strong&gt;small&lt;/strong&gt; matrix by current standards. The standard eig command of matlab should be able to handle this size with ease. Iterative sparse matrix eigensolvers like those implemented in ARPACK, or SLEPc will become more preferable if the matrix is much larger.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, if you want to implement an eigensolver into your own code, just use the LAPACK library that comes with very well developed routines for such purpose. Matlab also ultimately invokes LAPACK routines for doing most of its numerical linear algebra.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Semi-related note: the matrix need not be explicitly available for the large sparse solvers, because they usually just depend on being able to compute $A*x$ and $A&amp;#39;*x$.&lt;/p&gt;&#xA;" OwnerUserId="1709" LastEditorUserId="1709" LastEditorDisplayName="" LastEditDate="2010-11-28T12:15:30.517" LastActivityDate="2010-11-28T12:15:30.517" CommentCount="4" />
  <row Id="12202" PostTypeId="1" AcceptedAnswerId="12207" CreationDate="2010-11-28T12:10:29.180" Score="1" ViewCount="180" Body="&lt;p&gt;Define $A$ as a nonempty set, $\mathcal{B}:=\{f: A \rightarrow \mathbb{R}: f(A) \text{is bounded} \} ,d_\infty:=\text{sup}\{|f(x)-g(x)|:x \in A\}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For which $A$ is $\overline {B_1(0)} \subset \mathcal{B}$ compact?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notes: ${B_1(0)}$ is the unit sphere. I already proved that $(\mathcal{B},d_\infty)$ is a metric space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My thoughts: I investigated a bit on this and found some proofs that the unit sphere in a banach space is compact when the banach space is finite dimensional. But I don't want to use this here, I don't even know if $(\mathcal{B},d_\infty)$ is a banach space. All proofs we did who depend on the dimension of a metric space is that the intersections of compact subsets is compact and that the finite union of compact subsets is compact. Maybe we can use this?&lt;/p&gt;&#xA;" OwnerUserId="3123" LastEditorUserId="3123" LastEditorDisplayName="" LastEditDate="2010-11-28T13:55:34.680" LastActivityDate="2010-11-28T13:55:34.680" Title="Unit sphere compactness in a metric space" Tags="&lt;functions&gt;&lt;metric-spaces&gt;" AnswerCount="1" CommentCount="6" />
  <row Id="12203" PostTypeId="1" AcceptedAnswerId="12210" CreationDate="2010-11-28T12:40:48.570" Score="1" ViewCount="178" Body="&lt;p&gt;Hello!&#xA;I want to ask, how can I simplify this formula ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ e^{-i0.5t}+e^{i0.5t} $&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that it can be simplify to $\cos(0.5t)$, but I don't know how :/&lt;/p&gt;&#xA;" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2010-11-28T15:27:20.403" LastActivityDate="2011-01-13T09:12:11.213" Title="Simplify formula $e^{-i0.5t}+e^{i0.5t}$" Tags="&lt;complex-numbers&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="12204" PostTypeId="2" ParentId="12203" CreationDate="2010-11-28T12:47:20.613" Score="3" ViewCount="" Body="&lt;p&gt;Use $e^{ia}  = \cos a + i\sin a$, and find that the answer is slightly different from what you wrote.&lt;/p&gt;&#xA;" OwnerUserId="2810" LastActivityDate="2010-11-28T12:47:20.613" />
  <row Id="12205" PostTypeId="1" CreationDate="2010-11-28T12:53:14.977" Score="5" ViewCount="130" Body="&lt;p&gt;I wonder if someone could shed some light in the following question&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $(x,y)$ denote the greatest common divisor of $x$ and $y$, and let $x_1,y_1,x_2,y_2$ be integers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is  the following statement true?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $(x_1,y_1)=(x_2,y_2)=(x_1^2+y_1^2,x_2^2+y_2^2)=1$, then $$(x_1x_2\pm y_1y_2, x_1y_2\mp x_2y_1)=1$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If not, what further hypotheses are necessary to guarantee the claim?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance,&#xA;Guillermo&lt;/p&gt;&#xA;" LastActivityDate="2010-11-29T11:11:11.917" Title="About the greatest common divisor of certain expressions..." Tags="&lt;number-theory&gt;" AnswerCount="2" FavoriteCount="2" />
  <row Id="12206" PostTypeId="2" ParentId="12172" CreationDate="2010-11-28T13:06:50.410" Score="11" ViewCount="" Body="&lt;p&gt;The property holds if and only if $T$ is skew-symmetric (meaning that $T^t=-T$). (The other answers so far are special cases of this.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proof: Let $x=\alpha^t T \alpha$. Since $x$ is a $1\times 1$ matrix, it equals its own transpose. Thus $2x=x+x^t=\alpha^t T \alpha+(\alpha^t T \alpha)^t =\alpha^t (T+T^t)\alpha$, which is a quadratic form in $\alpha$ that corresponds to the symmetric matrix $T+T^t$; thus it is zero for all $\alpha$ iff $T+T^t=0$.&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2010-11-28T13:06:50.410" />
  <row Id="12207" PostTypeId="2" ParentId="12202" CreationDate="2010-11-28T13:50:09.807" Score="1" ViewCount="" Body="&lt;p&gt;The result you mention has a converse: the closed unit ball in a Banach space is compact &lt;strong&gt;if and only if&lt;/strong&gt; the Banach space is finite-dimensional.  That should suggest to you what is true here.  &lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-11-28T13:50:09.807" CommentCount="2" />
  <row Id="12209" PostTypeId="2" ParentId="12160" CreationDate="2010-11-28T14:18:33.437" Score="3" ViewCount="" Body="&lt;p&gt;I'll answer question 1 only for now, but I might edit this to address the others later.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One should note that corresponding to any set of orthogonal polynomials, there exists a symmetric tridiagonal matrix, called a &lt;strong&gt;Jacobi matrix&lt;/strong&gt;, whose characteristic polynomial is the monic (leading coefficient is 1) version of the set of orthogonal polynomials considered. To use the Legendre polynomials as an explicit example, we first note that the monic Legendre polynomials satisfy the following two-term recurrence relation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\hat{P}_{n+1}(x)=x \hat{P}_n(x)-\frac{n^2}{4 n^2-1}\hat{P}_{n-1}(x)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\hat{P}_n(x)=\frac{(n!)^2 2^n}{(2n)!}P_n(x)$ is the monic Legendre polynomial.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From this, we can derive an explicit expression for the corresponding Jacobi matrix (here I give the 5-by-5 case):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{pmatrix}0&amp;amp;\frac{1}{\sqrt{3}}&amp;amp;0&amp;amp;0&amp;amp;0\\\frac{1}{\sqrt{3}}&amp;amp;0&amp;amp;\frac{2}{\sqrt{15}}&amp;amp;0&amp;amp;0\\0&amp;amp;\frac{2}{\sqrt{15}}&amp;amp;0&amp;amp;\frac{3}{\sqrt{35}}&amp;amp;0\\0&amp;amp;0&amp;amp;\frac{3}{\sqrt{35}}&amp;amp;0&amp;amp;\frac{4}{\sqrt{63}}\\0&amp;amp;0&amp;amp;0&amp;amp;\frac{4}{\sqrt{63}}&amp;amp;0\end{pmatrix}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(the general pattern is that you have $\frac{n}{\sqrt{4 n^2-1}}$ in the $(n,n+1)$ and $(n+1,n)$ positions, and 0 elsewhere.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We now note that $\frac{n}{\sqrt{4 n^2-1}}$ can never be 0, and then use the fact that if a symmetric tridiagonal matrix has no zeroes in its sub- or superdiagonal, then all its eigenvalues have multiplicity 1. (A proof of this fact can be found in Beresford Parlett's &lt;a href=&quot;http://books.google.com/books?id=mWin91cLndsC&amp;amp;pg=PA134&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;The Symmetric Eigenvalue Problem&lt;/em&gt;&lt;/a&gt;.) Thus, all the roots of the Legendre polynomial are simple roots.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A more conventional proof of this fact is in page 27 of Theodore Chihara's &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0677041500&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;An Introduction to Orthogonal Polynomials&lt;/em&gt;&lt;/a&gt;. Briefly, the argument is that $P_n(x)$ changes sign at least once within $[-1,1]$ (and thus has at least one zero of odd multiplicity within the support interval) since&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_{-1}^1 P_n(u)\mathrm du=0$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the polynomial&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$P_n(x)\prod_{j=1}^k(x-\xi_j)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where the $\xi_j$ are the distinct zeroes of odd multiplicity within $[-1,1]$, should be greater than or equal to zero within $[-1,1]$, and thus its integral over $[-1,1]$ should be greater than zero. However, since&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_{-1}^1 P_n(u) u^k\mathrm du=0\qquad\text{if}\qquad k &amp;lt; n$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;we have a contradiction, and thus all the roots of the Legendre polynomial are simple (and within the support interval $[-1,1]$).&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-11-28T14:18:33.437" CommentCount="1" />
  <row Id="12210" PostTypeId="2" ParentId="12203" CreationDate="2010-11-28T14:27:32.040" Score="2" ViewCount="" Body="&lt;p&gt;By the &lt;a href=&quot;http://en.wikipedia.org/wiki/Euler%27s_formula&quot; rel=&quot;nofollow&quot;&gt;Euler's formula&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$e^{ix}=\cos x+i\sin x$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for $x=-0.5t$ , we obtain&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$e^{-0.5t}=e^{i\left( -0.5t\right) }=\cos \left( -0.5t\right) +i\sin \left(&#xD;&#xA;-0.5t\right) $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and for $x=0.5t$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$e^{0.5t}=\cos \left( 0.5t\right) +i\sin \left( 0.5t\right) .$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $\cos$ is an &lt;a href=&quot;http://en.wikipedia.org/wiki/Even_function#Even_functions&quot; rel=&quot;nofollow&quot;&gt;even function&lt;/a&gt; and $\sin$ an &lt;a href=&quot;http://en.wikipedia.org/wiki/Even_function#Even_functions&quot; rel=&quot;nofollow&quot;&gt;odd function&lt;/a&gt;, we obtain&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{eqnarray*}&#xD;&#xA;e^{-0.5t}+e^{0.5t} &amp;amp;=&amp;amp;\cos \left( -0.5t\right) +i\sin \left( -0.5t\right)&#xD;&#xA;+\cos \left( 0.5t\right) +i\sin \left( 0.5t\right)  \\&#xD;&#xA;&amp;amp;=&amp;amp;\left( \cos \left( -0.5t\right) +\cos \left( 0.5t\right) \right) +i\left(&#xD;&#xA;\sin \left( -0.5t\right) +\sin \left( 0.5t\right) \right)  \\&#xD;&#xA;&amp;amp;=&amp;amp;\left( \cos \left( 0.5t\right) +\cos \left( 0.5t\right) \right) +i\left(&#xD;&#xA;-\sin \left( 0.5t\right) +\sin \left( 0.5t\right) \right)  \\&#xD;&#xA;&amp;amp;=&amp;amp;2\cos \left( 0.5t\right) +i\cdot 0 \\&#xD;&#xA;&amp;amp;=&amp;amp;2\cos \left( 0.5t\right) &#xD;&#xA;\end{eqnarray*}$$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="752" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2010-11-28T14:34:24.890" LastActivityDate="2010-11-28T14:34:24.890" CommentCount="2" />
  <row Id="12211" PostTypeId="2" ParentId="12180" CreationDate="2010-11-28T15:10:24.373" Score="1" ViewCount="" Body="&lt;p&gt;For a given $\tau=\frac{1}{q}, q$ prime it seems that $a$ should be (perhaps among other choices) $\frac{\tau}{2}$ with $\sup_{\tau} \quad \min_{a} \quad f(a) + f(\tau-a)=2q$.  Can you prove this?  If so, then the sup over $\tau$ is infinity as you can find $\tau$ with arbitrarily large denominator.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2010-11-28T15:29:24.163" LastActivityDate="2010-11-28T15:29:24.163" CommentCount="2" />
  <row Id="12212" PostTypeId="2" ParentId="8224" CreationDate="2010-11-28T15:22:04.847" Score="0" ViewCount="" Body="&lt;p&gt;We have (using ab for those who favor I and II, n for those who favor none, etc)&#xA;a+b+c+2ab+2ac+2bc+3abc=?&#xA;a+b+c+2ab+2ab+2ac=?&#xA;a+b+c+ab+ac+bc+abc=?&#xA;a+b+c+ab+ac+bc=?&#xA;and so?&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-11-28T15:22:04.847" />
  <row Id="12213" PostTypeId="1" AcceptedAnswerId="12214" CreationDate="2010-11-28T15:32:13.077" Score="4" ViewCount="157" Body="&lt;p&gt;Let C be a collection of subsets of $\mathbb{N}$ such that $A,B\in C \Rightarrow A \not\subseteq B$. Can C be uncountable?&lt;/p&gt;&#xA;" OwnerUserId="2934" LastActivityDate="2010-11-28T16:22:32.067" Title="Can a collection of subsets of N such that no one set contains another be uncountable?" Tags="&lt;elementary-set-theory&gt;" AnswerCount="2" />
  <row Id="12214" PostTypeId="2" ParentId="12213" CreationDate="2010-11-28T15:45:08.570" Score="8" ViewCount="" Body="&lt;p&gt;Yes. Consider the collections of subsets which contain exactly one of $2i$ and $2i+1$, for all $i$.&lt;/p&gt;&#xA;" OwnerUserId="2501" LastActivityDate="2010-11-28T15:45:08.570" CommentCount="2" />
  <row Id="12215" PostTypeId="2" ParentId="8224" CreationDate="2010-11-28T16:16:04.023" Score="1" ViewCount="" Body="&lt;p&gt;The simplest way is probably to draw Venn diagrams and stuff, but here's how you do it using the inclusion-exclusion principle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For any three sets $A,B,C$, we have $$|A \cup B \cup C| = S_1 - S_2 + S_3,$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $S_1 = |A|+|B|+|C|$, $S_2 = |A \cap B| + |B \cap C| + |C \cap A|$, and $S_3 = |A \cap B \cap C|$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Letting $A,B,C$ denote the obvious sets, you are given the values of $|A \cup B \cup C|$, $S_1$ and $S_3$ so you can solve for $S_2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now you are asked to find $x = |(A \cap B) \cup (B \cap C) \cup (C \cap A)|$, which we can expand with the same formula (using the fact that $(A \cap B) \cap (B \cap C) = A \cap B \cap C$ etc.) as $$x = |A \cap B| + |B \cap C| + |C \cap A| - 3|A \cap B \cap C| + 3|A \cap B \cap C| = S_2 - 2S_3.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you do this right, and assuming I've done it right, the answer should be 17%.&lt;/p&gt;&#xA;" OwnerUserId="3747" LastActivityDate="2010-11-28T16:16:04.023" />
  <row Id="12216" PostTypeId="2" ParentId="12205" CreationDate="2010-11-28T16:16:43.400" Score="6" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\ \: $ Interpret it in Gaussian integers $\rm\: \mathbb Z[i]\:.\: $ Define $\rm\ a + b\ i\ $ to be primitive if it has no rational prime divisor, i.e. $\rm\ (a,b) = 1\:.\:$ Then the statement says that the product of two primitive Gaussian integers is primitive if they have coprime norms. Put $\rm\ \ \alpha = x_1+y_1\ i,\ $ $\rm\ \beta = x_2+y_2\ i\:.\ $ Suppose $\rm\ p\:|\:\alpha\beta\ $ for a prime $\rm\: p\in \mathbb Z\:.\:$ If $\rm \: p\:$ is prime in $\rm\: \mathbb Z[i]\: $ then $\rm\ p\:|\:\alpha\ $ or $\rm\ p\:|\:\beta\ $ contra $\ldots\ $  Else $\rm\: p = \pi\pi&amp;#39;\: $ for a prime $\rm\: \pi \in \mathbb Z[i]\:$ hence $\rm\ \pi\pi&amp;#39;|\:\alpha\beta\ \Rightarrow\ \ldots$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively, if you are not familiar with Gaussian integer arithmetic then you may employ the &lt;a href=&quot;http://en.wikipedia.org/wiki/Brahmagupta%E2%80%93Fibonacci_identity&quot; rel=&quot;nofollow&quot;&gt;Brahmagupta–Fibonacci identity&lt;/a&gt; for composition of squares. It is a consequence of the fact that the norm is &lt;em&gt;multiplicative:&lt;/em&gt; $\rm\ \ \ N(\alpha)\ N(\beta)\ =\ N(\alpha\beta)\ $ i.e. $\rm\ \alpha\alpha&amp;#39;\beta\beta&amp;#39; = (\alpha\beta) (\alpha\beta)&amp;#39;\: $ which, rationally, is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\rm (x_1^2 + y_1^2)\ (x_2^2 + y_2^2)\ =\ (x_1x_2\pm y_1y_2)^2 + ( x_1y_2\mp x_2y_1)^2 $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The upper $\pm$ signs are from $\rm\ N(\alpha)\ N(\beta&amp;#39;)\ = \ N(\alpha\beta&amp;#39;)\:$.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-11-28T17:43:56.817" LastActivityDate="2010-11-28T17:43:56.817" CommentCount="1" />
  <row Id="12217" PostTypeId="2" ParentId="12213" CreationDate="2010-11-28T16:22:32.067" Score="4" ViewCount="" Body="&lt;p&gt;Here is another one, perhaps due to Donald J Newman.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For each real number $x$ consider an (infinite) sequence of rationals $R_x = \{r_{x1}, r_{x2}, \cdots \}$ which converges to $x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $x \neq y$, $R_x$ and $R_y$ have at most finite intersection.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus the set $S = \{R_x : x \in \mathbb{R}\}$ is an uncountable collection of infinite subsets of a countable set, none of which is a subset of other. In fact, any two members of the collection have only a finite intersection.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-11-28T16:22:32.067" />
  <row Id="12218" PostTypeId="2" ParentId="12194" CreationDate="2010-11-28T16:27:39.797" Score="11" ViewCount="" Body="&lt;p&gt;Another way is thinking of this as asking where do the graphs of the functions given by $f(x) = e^{x-1}$ and $g(x) = \frac{1}{x}$ intersect. This is because $$ xe^{x} = e \Leftrightarrow e^{x-1} = \frac{1}{x} \quad \text{for $x \neq 0$}$$ Then the answer you have would correspond to the only intersection point $(1, 1)$. So if you have no idea of where to start you could just draw the corresponding graphs and that could give you an idea of what is happening.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/upxAx.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then the graphs maybe would suggest to you the following line of reasoning. If $0 &amp;lt; x &amp;lt; 1$ then $$ \frac{1}{x} &amp;gt; 1 \quad \text{but} \quad e^{x-1} &amp;lt; e^{0} = 1$$ so their graphs can't meet in the interval $]0, 1[$. Then if $x &amp;gt; 1$ we have $$\frac{1}{x} &amp;lt; 1 \quad \text{but} \quad e^{x-1} &amp;gt; e^{0} = 1$$ so again there's no intersection point in the interval $]1, \infty[$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This rules out all the possible solutions apart from $x = 1$ which you already knew and which could be guessed by looking at the graphs.&lt;/p&gt;&#xA;" OwnerUserId="900" LastEditorUserId="900" LastEditorDisplayName="" LastEditDate="2010-11-28T18:21:15.210" LastActivityDate="2010-11-28T18:21:15.210" CommentCount="4" />
  <row Id="12219" PostTypeId="1" AcceptedAnswerId="12228" CreationDate="2010-11-28T16:33:02.907" Score="4" ViewCount="96" Body="&lt;p&gt;Let $T$ be a bounded linear operator on $L^2(\mathbb R)$. So, let us now assume that $T$ commutes with the translations $\tau_x$. How do I now show that $T$ is given by a convolution with respect to a distribution?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the way, I know I can probably find the proof somewhere in one of Stein's books, but I would like to prove it myself without knowing what it should be but I'm struggling a bit. So I would like some hints. Especially I would like a method of deriving the result without knowing what it should be. If that is not possible, an intuitive argument why it should be true would be nice as well.&lt;/p&gt;&#xA;" OwnerUserId="1120" LastEditorUserId="1120" LastEditorDisplayName="" LastEditDate="2010-11-28T16:40:04.340" LastActivityDate="2010-11-28T19:01:19.687" Title="Operators commuting with translations." Tags="&lt;harmonic-analysis&gt;&lt;distribution-theory&gt;" AnswerCount="2" />
  <row Id="12220" PostTypeId="1" AcceptedAnswerId="12756" CreationDate="2010-11-28T16:52:56.750" Score="2" ViewCount="193" Body="&lt;p&gt;Hi all, I have the following situation:&#xA;I have a hermitian Matrix $A$ that satisfies some symmetries which I can express via &#xA;$AS = SA$ for a unitary matrix $S$. Now I am interested in the eigenvectors of $A$, but I want that these eigenvectors also respect my symmetries ( compare to Bloch waves in physics where the eigenvectors are chosen to reflect the translational invariance of the lattice ).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $A$ has degenerate (repeated) eigenvalues, the standard numerical techniques will return some arbitrary (yet orthonormal) eigenvectors spanning the eigenspace. I, however, want to obtain unique results and thus want to make use of the symmetries. How can I do this numerically? I know that commuting matrices can be diagonalized simultaneously - in theory. But I don't know how to do it practically. Would I have to diagonalize one of them, apply the unitary transform thus obtained to the other one, arriving at a block diagonal form where I then have to diagonalize each block separately? Or is there something more elegant I can do?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: The matrix is dense, but quite small (12x12 to 18x18).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The symmetry would be something like translation symmetry:&#xA;$\begin{pmatrix} 0 &amp;amp; 0 &amp;amp; 0 &amp;amp; -1 \\ 1 &amp;amp; 0 &amp;amp; 0 &amp;amp; 0 \\ 0 &amp;amp; 1 &amp;amp; 0 &amp;amp; 0 \\ 0  &amp;amp; 0 &amp;amp; 1 &amp;amp; 0 \end{pmatrix}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where each entry is a 3x3-block in the case of a 12x12 matrix, which looks something like&#xA;$\begin{pmatrix} a &amp;amp; b &amp;amp; 0 &amp;amp; -b\\ b&amp;amp;a&amp;amp;b&amp;amp;0\\ 0&amp;amp;b&amp;amp;a&amp;amp;b\\ -b&amp;amp;0&amp;amp;b&amp;amp;a\end{pmatrix}$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="4171" LastEditorUserId="4171" LastEditorDisplayName="" LastEditDate="2010-11-28T17:09:58.957" LastActivityDate="2010-12-02T15:45:40.987" Title="Symmetrize eigenvectors of degenerate (repeated) eigenvalue" Tags="&lt;linear-algebra&gt;&lt;numerical-methods&gt;" AnswerCount="1" CommentCount="6" />
  <row Id="12221" PostTypeId="2" ParentId="11843" CreationDate="2010-11-28T17:05:27.957" Score="2" ViewCount="" Body="&lt;p&gt;Your function is really &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$F: Context \mapsto \mathbb{P}(X|Context)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or in other words you just want to determine the probability of a particular outcome, conditional on a certain context. The question is, what are your data? Do you know about the different contexts? Or do you only dispose of the outcomes? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, since $$\mathbb{P}(X|Context)=\frac{\mathbb{P}(X \cap Context)}{\mathbb{P}(Context)} \; ,$$ I think any test you'd want to devise for this function would boil down to the Chi-squared test anyway.&lt;/p&gt;&#xA;" OwnerUserId="3567" LastActivityDate="2010-11-28T17:05:27.957" CommunityOwnedDate="2010-11-28T17:05:27.957" />
  <row Id="12222" PostTypeId="2" ParentId="12219" CreationDate="2010-11-28T17:09:14.560" Score="2" ViewCount="" Body="&lt;p&gt;Maybe you should look at $F \circ T \circ F^{-1}$ where $F$ denotes Fourier transform, so that $\tau_x$ becomes multiplication by $t \mapsto e^{itx}$ and convolution becomes multiplication by something as well.&lt;/p&gt;&#xA;" OwnerUserId="2660" LastActivityDate="2010-11-28T17:09:14.560" />
  <row Id="12223" PostTypeId="1" CreationDate="2010-11-28T17:11:35.790" Score="4" ViewCount="230" Body="&lt;p&gt;Hello,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to &lt;a href=&quot;http://en.wikipedia.org/wiki/Logical_connective#Order_of_precedence&quot; rel=&quot;nofollow&quot;&gt;the precedence of logical connectives&lt;/a&gt;, operator $\rightarrow$ gets higher precedence than $\leftrightarrow$ operator. But what about associativity of $\rightarrow$ operator? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I could not locate any book/webpage that mentions about associativity of logical operators in discrete mathematics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please also cite the reference (book/reliable webpage) that you use to answer my question (as I'm planning to add this to wikipedia page about 'logical connectives').&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS: I got this question when I saw this problem:&#xA;Check if following compound  proposition is tautology or not:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \mathrm{p} \leftrightarrow (\mathrm{q} \wedge \mathrm{r}) \rightarrow \neg\mathrm{r} \rightarrow \neg\mathrm{p}$$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="4170" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-29T05:44:58.277" LastActivityDate="2010-11-29T05:44:58.277" Title="Associativity of logical connectives" Tags="&lt;logic&gt;&lt;propositional-calculus&gt;" AnswerCount="3" CommentCount="4" />
  <row Id="12225" PostTypeId="1" AcceptedAnswerId="12241" CreationDate="2010-11-28T17:42:28.737" Score="2" ViewCount="110" Body="&lt;p&gt;For a given matrix $A$ the question &quot;is there a vector (or are there vectors)&#xA;whose direction remain the same when it is multiplied with the matrix&quot;&#xA;leads to the well-known eigenvalue problem $A\mathbf{v}=\lambda\mathbf{v}$,&#xA;where $\mathbf{v}$ is an eigenvector of the matrix. Since a vector is fully&#xA;described by its direction and amplitude, one can also ask the other obvious question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a given matrix $A$ is there a vector (or are there vectors) whose&#xA;amplitude remains the same when it is multiplied with the matrix? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This would&#xA;amount to finding &quot;eigenrotations&quot; that satisfy&#xA;$\left\Vert A\mathbf{r}\right\Vert =\left\Vert \mathbf{r}\right\Vert .$&#xD;&#xA;The eigenrotations are thus characterized by&#xA;$\mathbf{r}^{T}\left(  A^{T}A-I\right)  \mathbf{r}=0,$&#xD;&#xA;and under some conditions on the coefficients of $A$ there exist such vectors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Question: In what contexts do &quot;eigenrotations&quot; come up? (other than a mathexchange question)&lt;/p&gt;&#xA;" OwnerUserId="3891" LastActivityDate="2010-11-28T19:17:54.067" Title="&quot;Eigenrotations&quot; of a matrix" Tags="&lt;linear-algebra&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="12226" PostTypeId="2" ParentId="12197" CreationDate="2010-11-28T17:53:33.493" Score="6" ViewCount="" Body="&lt;p&gt;Yes it is possible to reduce the number of states a Turing machine uses to decide a problem $X$ by increasing the number of symbols.&#xA;However it gets very tricky when you are close to the &lt;em&gt;minimum number&lt;/em&gt; of states/symbols needed to solve  $X$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a nice survey paper about the efforts to find the minimum number of states and symbols Turing machines need for universality. &#xA;&lt;a href=&quot;http://portal.acm.org/citation.cfm?id=1498068&quot; rel=&quot;nofollow&quot;&gt;http://portal.acm.org/citation.cfm?id=1498068&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;The complexity of small universal Turing machines: A survey&quot;&#xA;Damien Woods and Turlough Neary,&#xA;Theoretical Computer Science archive&#xA;Volume 410 Issue 4-5, February, 2009&lt;/p&gt;&#xA;" OwnerUserId="571" LastActivityDate="2010-11-28T17:53:33.493" />
  <row Id="12227" PostTypeId="1" AcceptedAnswerId="12233" CreationDate="2010-11-28T18:11:01.550" Score="1" ViewCount="162" Body="&lt;p&gt;I'm trying to wrap my head around this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a \equiv b^{-1}\pmod{y}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x \equiv c^{-1} \pmod{y}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know all values but $x$ and $y$. Is it possible to solve this problem for $x$?&#xA;Can anyone point out a good resource on this kind of problems?&lt;/p&gt;&#xA;" OwnerUserId="4166" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-28T19:09:56.787" LastActivityDate="2010-11-29T00:19:09.960" Title="Understanding congruence equations" Tags="&lt;number-theory&gt;" AnswerCount="2" />
  <row Id="12228" PostTypeId="2" ParentId="12219" CreationDate="2010-11-28T18:14:37.663" Score="1" ViewCount="" Body="&lt;p&gt;Some suggestions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even though the delta function is not in $L^2$, heuristically what does your condition on $T$ say when applied to the delta function? Now, given the answer to the previous question and the fact that every function is an average of translated delta functions, what does one expect $Tf$ to be for an arbitrary $L^2$ function $f(x)$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can also do this on the Fourier transform side.. i.e. consider $G = F \circ T \circ F^{-1}$ as Plop suggested and then look at how $G$ behaves on a given $\delta(x - a)$. Then again use the idea that an $L^2$ function is an average of translated delta functions.&lt;/p&gt;&#xA;" OwnerUserId="3035" LastActivityDate="2010-11-28T18:14:37.663" CommentCount="5" />
  <row Id="12229" PostTypeId="2" ParentId="12194" CreationDate="2010-11-28T18:17:24.903" Score="2" ViewCount="" Body="&lt;p&gt;Assuming that we are only looking for real solutions, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$xe^x = e.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that $e^x$ is always positive and so no solution for this equation is possible for $x&amp;lt;0$ or $x=0$.  As x increases, both x and $e^x$ are strictly increasing.  Therefore, there is only one value for which $xe^x = e$.  We already know that solution to be $x=1$.&lt;/p&gt;&#xA;" OwnerUserId="1004" LastEditorUserId="1004" LastEditorDisplayName="" LastEditDate="2010-11-29T00:33:55.103" LastActivityDate="2010-11-29T00:33:55.103" />
  <row Id="12233" PostTypeId="2" ParentId="12227" CreationDate="2010-11-28T18:36:38.487" Score="3" ViewCount="" Body="&lt;p&gt;If $b,c$  have an inverse in $\pmod y$ then is valid :$$ ab\equiv 1 \pmod y $$ and $$ xc\equiv 1 \pmod y $$&#xD;&#xA;&#xD;&#xA;From this you get:  $$ ab-1 =  k_{1}y  $$  $$ xc-1 =  k_{2}y  $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(For some  $k_{1}$ and $k_{2}$  integers )&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you can find $x$ and $y$ from this ecuations system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$Edit:$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $y=5$, the elements with inverse are: $2,3$, also $b=2$ and $b^{-1}=3$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Know that (with $a=8$ ):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ 8 \equiv 3 \pmod 5 $, is say , $ a \equiv b^{-1} \pmod y $,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that also:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ 16 =8 \cdot 2\equiv 1 \pmod 5 $, is say , $ ab \equiv 1 \pmod y $ &lt;/p&gt;&#xA;" OwnerUserId="3421" LastEditorUserId="3421" LastEditorDisplayName="" LastEditDate="2010-11-29T00:19:09.960" LastActivityDate="2010-11-29T00:19:09.960" CommentCount="4" />
  <row Id="12234" PostTypeId="2" ParentId="12227" CreationDate="2010-11-28T18:39:25.890" Score="0" ViewCount="" Body="&lt;p&gt;$\rm x\ =\ (c^{-1}\ mod\ y)\ +\ y\ \mathbb Z\ \ $ where $\rm\ \ y\:|\: ab-1,\ \ gcd(y,bc)=1$&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-11-28T18:45:17.410" LastActivityDate="2010-11-28T18:45:17.410" />
  <row Id="12235" PostTypeId="2" ParentId="12223" CreationDate="2010-11-28T18:39:43.923" Score="2" ViewCount="" Body="&lt;p&gt;Presumably $p \Rightarrow q \Rightarrow r$ means $(p \Rightarrow q) \Rightarrow r$, since the other option $p \Rightarrow (q \Rightarrow r)$ is equivalent to $(p \land q) \Rightarrow r$. That's similar to the laws of powering $p^{q^r} = p^{\left(q^r\right)}$ since $\left(p^q\right)^r = p^{qr}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Best practice is just &lt;i&gt;not&lt;/i&gt; to put the reader into a state of ambiguity. Some mathematical writers dislike punctuation, but that's not an excuse for the rest of us to take advantage of it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $\land$ and $\lor$ are associative, there's no ambiguity there, and it makes sense not to put parentheses. Sometimes one even thinks of an expression like $p \land q \land r$ as a conjunction $\land(p,q,r)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The latter convention is the case in proof complexity when one is measuring the depth of a formula. If a first-order &lt;i&gt;quantified&lt;/i&gt; formula is translated to propositional logic, then $\forall$ will correspond to a conjunction, and $\exists$ to a disjunction; it makes sense not to penalize the formula for the size of the underlying universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An even more daring &quot;convention&quot; is having infinite conjunctions/disjunctions - this wouldn't make sense at all for $\Rightarrow$. So your textbook authors are at fault (even though having the $\Leftrightarrow$ bind weakly is a reasonable convention if not overused).&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-28T18:39:43.923" CommentCount="4" />
  <row Id="12237" PostTypeId="1" AcceptedAnswerId="12249" CreationDate="2010-11-28T18:46:26.933" Score="1" ViewCount="132" Body="&lt;p&gt;For a given group of data, I have both a mean and variance (already calculated). Looking over the raw spread of data, and its numerous outliers, I expect that variance will be high.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is -- now that I have my variance calculated, what do I need to compare it to to determine if the variance is significant (i.e., indicative of highly skewed data)?&lt;/p&gt;&#xA;" OwnerUserId="1760" LastActivityDate="2010-11-28T21:02:08.393" Title="How do I determine if my variance is significant?" Tags="&lt;statistics&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="12238" PostTypeId="1" CreationDate="2010-11-28T18:49:35.293" Score="1" ViewCount="85" Body="&lt;p&gt;I have skimmed this video that I found on mathoverflow:&#xA;&lt;a href=&quot;http://tube.sfu-kras.ru/video/407?playlist=397&quot; rel=&quot;nofollow&quot;&gt;http://tube.sfu-kras.ru/video/407?playlist=397&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At about 15:05 the lecturer wrote down an equality&#xA;$\sum F(m_1, \ldots, m_m)z^{m_1}\ldots z^{m_m} = \frac{1}{det(zI - A)}$, where F is given in terms of coefficients of the $m$-by-$m$ matrix $A$. The definition of $F$ doesn't yet make much sense to me, but that's beyond the point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is, how is this equality possible? the right hand side obviously has poles—the eigenvalues of $A$, but the left hand side is just a polinomial, it can't have poles! Is there something I'm missing?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: well, it was actually&#xA;$\sum F(m_1, \ldots, m_m) z_1^{m_1}\ldots z_m^{m_m} = \frac{1}{det(z - A)}$.&#xA;Here I have abused the notation slightly: the z means actually the diagonalization of the corresponding vector. However, when you take $z_1 = \ldots = z_m$, the question still stands.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The domain is real, as I understand from the context of the lecture :), so the use of the term pole isn't really correct, I just forgot the right word in English :)&lt;/p&gt;&#xA;" OwnerUserId="3793" LastEditorUserId="3793" LastEditorDisplayName="" LastEditDate="2010-11-28T19:33:11.947" LastActivityDate="2010-11-28T19:33:11.947" Title="A relation between permanents and determinants" Tags="&lt;real-analysis&gt;&lt;algebra&gt;" CommentCount="8" />
  <row Id="12239" PostTypeId="1" AcceptedAnswerId="12259" CreationDate="2010-11-28T18:51:44.143" Score="3" ViewCount="83" Body="&lt;p&gt;I know this is a trivial problem but am stuck with this thing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$A \in \mathbb{R}^{m \times m}$ is a symmetric positive definite matrix and $C \in \mathbb{R}^{m \times p}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do I show that the matrix $$B = \begin{bmatrix} A &amp;amp; C \\ C^T &amp;amp; 0 \end{bmatrix}$$ is indefinite?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can show that there exists a vector $v$ such that $v^TBv &amp;gt; 0$.&#xA;(This is trivial. For instance, $v = [x , 0]^T$ where $x \in \mathbb{R}^{1 \times m}$ and $0 \in \mathbb{R}^{1 \times p}$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now how do I find a vector such that $v^TBv &amp;lt; 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I do not need the answer. A clue/hint is welcome.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can assume that $p \leq m$ and the matrix $C$ is full rank that is to say that all the constraints are linearly independent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2010-11-28T22:54:36.977" LastActivityDate="2010-11-29T07:25:28.463" Title="Indefiniteness of KKT system" Tags="&lt;linear-algebra&gt;&lt;homework&gt;&lt;optimization&gt;" AnswerCount="2" />
  <row Id="12240" PostTypeId="1" AcceptedAnswerId="12242" CreationDate="2010-11-28T18:59:17.207" Score="3" ViewCount="104" Body="&lt;p&gt;Hey there,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $P$ and $Q$ be complex polynomials with $deg(P) &amp;lt; deg(Q)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $Q(z) = (z-z_1)^{k_1} (z-z_2)^{k_2}...(z-z_m)^{k_m}, z_i \in \mathbb{C} \text{ and } k_i \in \mathbb{N}$ be a complete decomposition. Then, I have to prove that there exists a unique exposition of the following form:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{P(z)}{Q(z)} = \sum\limits_{i=1}^m \sum\limits_{j=1}^{k_i} \frac{a_{ij}}{(z-z_i)^j}, a_{ij} \in \mathbb{C}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I started with the proof of the uniqueness of this exposition and succeeded. However, I don't feel like I am doing well now, while trying to prove its existence. Basically, I want to do two inductions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First I want to do an induction over $deg(Q)$ with $deg(P) = 0$, then I want to go on with an induction over $deg(P)$ with an arbitrary $deg(Q)$. The start of the first induction is easy, but I can't get to an end at the induction step. I have something like:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{P}{Q} = \frac{P}{(z-z*)Q&amp;#39;} = \frac{1}{z-z*} \frac{P}{Q&amp;#39;}$, where $Q&amp;#39;$ is a polynomial with $deg(Q&amp;#39;) = n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What do I do next? I know that a unique exposition exists for the latter fraction and if $z* = z_i$ for some $i$, it's proven I guess. But what if this $z*$ is a completely new complex number?&lt;/p&gt;&#xA;" OwnerUserId="3787" LastActivityDate="2010-11-28T19:55:03.160" Title="Stuck at the proof of the existence of the partial fraction expansion" Tags="&lt;analysis&gt;" AnswerCount="2" />
  <row Id="12241" PostTypeId="2" ParentId="12225" CreationDate="2010-11-28T19:17:54.067" Score="5" ViewCount="" Body="&lt;p&gt;Allow me to generalize the question a bit. It's a very natural operation to multiply a matrix by a scalar. Most properties of matrices, including the eigenvectors and eigenvalues, are either not affected by scalar multiplication or are multiplied by the same scalar. But $\lVert Ax \rVert = \lVert x \rVert$ does not remain true for the same vector $x$. So it's more natural to look at all the sets $\lVert Ax \rVert = c \lVert x \rVert$ as a function of a scalar $c$. These are just the level sets of $f(x) := \lVert Ax \rVert / \lVert x \rVert$. Of course, unlike the eigendecomposition, where you get a discrete set of (at most) $n$ vectors out of it, you will generically have infinitely many solutions for any $c$. Since all nonzero scalar multiples of $x$ have the same value of $f(x)$, we can without loss of generality restrict to the surface of the unit sphere $\lVert x \rVert = 1$. Then we are talking about solutions to the equations $\lVert x \rVert = 1$, $\lVert Ax \rVert = c$. Clearly, the solutions form a $(n-2)$-dimensional manifold.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is the only context in which I have seen such a construction. I think it's called &lt;a href=&quot;http://en.wikipedia.org/wiki/Poinsot%27s_ellipsoid&quot; rel=&quot;nofollow&quot;&gt;Poinsot's ellipsoid&lt;/a&gt;. Consider a rigid body with moment of inertia $I$ tossed into the air with some angular velocity $\omega$. Since no external forces are acting on the body, its angular momentum $L = I\omega$ is conserved, and so is its rotational kinetic energy $T = \frac{1}{2}\omega^T I\omega$. Take a reference frame attached to the body. Then $I$ becomes a constant matrix. $L$ is no longer constant, but its magnitude still must be. So we get two equations: $\lVert L \rVert = \mathrm{const}$, and $\frac{1}{2} L^T I^{-1} L = \mathrm{const}$, which has almost the same form as your equations. This blog post called &quot;&lt;a href=&quot;http://gmunu.wordpress.com/2008/06/29/on-tossing-stuff-into-the-air/&quot; rel=&quot;nofollow&quot;&gt;On tossing stuff into the air&lt;/a&gt;&quot; has some nice pictures, as well as discussion about how this tells you why things wobble if you spin them about the intermediate moment of inertia, and what happens when you toss a bag of peas into the air.&lt;/p&gt;&#xA;" OwnerUserId="856" LastActivityDate="2010-11-28T19:17:54.067" />
  <row Id="12242" PostTypeId="2" ParentId="12240" CreationDate="2010-11-28T19:27:48.717" Score="5" ViewCount="" Body="&lt;p&gt;By the inductive hypothesis, $\frac{P}{Q&amp;#39;}$ has a partial fraction decomposition.  Show that by multiplying this decomposition by $\frac{1}{z - z^{\ast}}$ you can write down a partial fraction decomposition for $\frac{P}{Q}$.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-11-28T19:27:48.717" />
  <row Id="12243" PostTypeId="2" ParentId="2923" CreationDate="2010-11-28T19:38:16.860" Score="0" ViewCount="" Body="&lt;p&gt;Hello, I don't have Billingsley's book here, but I guess what he says is that uniform integrability and convergence in distribution imply convergence &lt;strong&gt;of the means&lt;/strong&gt; (and not &lt;strong&gt;in mean&lt;/strong&gt;), i.e. E[X_n] converges to E[X]. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One simple (though not truely elementary) way to see it is to use Skorokhod's theorem stating that convergence in distribution is equivalent to almost-sure convergence of copies of the random variables in some abstract probability state. Then we just apply the usual Vitali theorem about uniform integrability for these copies. It gives convergence in mean of the copies of X_n to the copy of X, and therefore convergence of the mean of (the copy of) X_n to that of (the copy of) X. I put parentheses since the mean only depends on the law and hence is the same for X_n or a copy of it. This is the result. &lt;/p&gt;&#xA;" LastActivityDate="2010-11-28T19:38:16.860" />
  <row Id="12244" PostTypeId="2" ParentId="12240" CreationDate="2010-11-28T19:49:58.297" Score="1" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\rm\quad\displaystyle 1 = gcd(F,G)\ \ \Rightarrow\ \ 1 = AF+BG\ \ \Rightarrow\ \ \frac{1}{FG}\ = \ \frac{A}G + \frac{B}F$&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-11-28T19:55:03.160" LastActivityDate="2010-11-28T19:55:03.160" />
  <row Id="12245" PostTypeId="2" ParentId="12194" CreationDate="2010-11-28T20:02:37.150" Score="4" ViewCount="" Body="&lt;p&gt;With logarithms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Firstly  $x&amp;gt;0$, Then :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$xe^{x}=e \Rightarrow \ln (xe^{x})=\ln e \Rightarrow \ln x + x = 1 $.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $x&amp;gt;1$ then $ \ln x &amp;gt;0$, thus $\ln x +x &amp;gt; 1+0 =1$  (contradiction)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $x&amp;lt;1$ then $ \ln x &amp;lt; 0$, thus $\ln x +x &amp;lt; 1+0 =1$  (contradiction)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus $x=1$&lt;/p&gt;&#xA;" OwnerUserId="3421" LastEditorUserId="3421" LastEditorDisplayName="" LastEditDate="2010-11-29T02:02:20.783" LastActivityDate="2010-11-29T02:02:20.783" />
  <row Id="12246" PostTypeId="2" ParentId="10092" CreationDate="2010-11-28T20:09:30.240" Score="1" ViewCount="" Body="&lt;p&gt;Here is something even more spectacular. Let $\epsilon &amp;gt;0$.  Let $z_n$ be an enumeration of all points in $R^d$ with all rational coordinates. We know this sequence's range is dense in $R^d$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;About each $z_n$ choose an open ball of small radius $B_n$ whose volume is less than $\epsilon/2^n$.  Now write&#xA;$$U_\epsilon = \bigcup_{n=1}^\infty B_n.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a union of open balls and is therefore open.  Its total Lebesgue measure (volume) is less than $\epsilon$. It is an open dense subset of $R^d$.  An open dense subset of $R^d$, can be very small indeed.  &lt;/p&gt;&#xA;" OwnerUserId="4154" LastActivityDate="2010-11-28T20:09:30.240" />
  <row Id="12247" PostTypeId="1" AcceptedAnswerId="12248" CreationDate="2010-11-28T20:13:22.163" Score="2" ViewCount="137" Body="&lt;p&gt;If a cyclic subgroup has a proper subgroup of infinite order, how many finite subgroups does it have?  Explain.  Any help appreciated!  Thank you.&lt;/p&gt;&#xA;" OwnerUserId="4179" LastActivityDate="2010-11-28T20:38:40.020" Title="Abstract Algebra - Cyclic Subgroups with a Proper Subgroup of Infinite Order" Tags="&lt;abstract-algebra&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="12248" PostTypeId="2" ParentId="12247" CreationDate="2010-11-28T20:38:40.020" Score="5" ViewCount="" Body="&lt;p&gt;It's easy to see that a cyclic infinite group is isomorphic to the set of integers $\mathbb{Z}$: being cyclic, it is generated my one element, $x$ say, and being infinite, no power of $x$ is equal to the identity. Consequently it is formed by the integer powers of $x$, and not two of them are equal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer to your question consequently is that an infinite cyclic group cannot have a nontrivial finite subgroup.&lt;/p&gt;&#xA;" OwnerUserId="2137" LastActivityDate="2010-11-28T20:38:40.020" />
  <row Id="12249" PostTypeId="2" ParentId="12237" CreationDate="2010-11-28T20:49:21.240" Score="0" ViewCount="" Body="&lt;p&gt;Taken from &lt;a href=&quot;http://forumserver.twoplustwo.com/15/poker-theory/how-does-standard-deviation-relate-poker-867847/&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; post of mine from a poker forum.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;You need to somehow compute the standard deviation for your distribution, and then count how many standard deviations away from the expected value your actual result is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's say we are flipping coins, and each time it lands heads you give me a dollar, and each time it's tails I give you a dollar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then let's keep track of how much money I have made or lost. On average I expect to make nothing, but in practice I will most likely lose or win a finite amount. In fact if we flip the coin $N$ times, then I am most likely to win or lose somewhere around the $\sqrt{N}$ dollars. Why is this true? It's actually easy to show.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's call my profit/loss $M$. Then $M$ starts at zero, and the average change in $M$ after one or more flips is zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is the clever trick: let's consider the average change in $M^2$. Well after one flip it's quite clear that this will be 1, since +1 and -1 squared are both equal to 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice that after one flip the average profit/loss is zero, but in reality I either gained or lost a dollar. $M^2$ however is 1, so you can see it sort of measures the fluctuations in my profit/loss.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Call $M(N)$ my profit/loss after N flips. Then what is $M(N+1)^2$? Since I can either gain or lose 1 dollar, we have either&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$M(N+1) = M(N) + 1$ or $M(N+1) = M(N) - 1$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;lets square these two expressions and take the average, which I write as $&amp;lt;M(N+1)^2 &amp;gt;$. If you do this, which is quite easy, you will find that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$&amp;lt;M(N+1)^2&amp;gt;$ = $&amp;lt;M(N)&amp;gt;^2 + 1$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you think about it, this implies that after $N$ flips, the average of $M^2$ will be $N$; therefore the square root of the average of $M^2$ is $\sqrt{N}$, which is just a fancy way of saying the SD for flipping a coin N times is $\sqrt{N}$.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;This shows that the SD for flipping a coin (1:1 odds) is $\sqrt{N}$, and it's simple to modify the above argument to find the SD for weighted coins.  Therefore, besides telling us how good/bad you are running, the magnitude of the SD also tells you, in poker, how well on average you are getting your money in.  In other words, SD can be used to measure what average chance you have to win the pot when you put all your chips in with cards left to come.&lt;/p&gt;&#xA;" OwnerUserId="1417" LastEditorUserId="1417" LastEditorDisplayName="" LastEditDate="2010-11-28T21:02:08.393" LastActivityDate="2010-11-28T21:02:08.393" />
  <row Id="12250" PostTypeId="1" AcceptedAnswerId="12265" CreationDate="2010-11-28T21:07:49.127" Score="0" ViewCount="227" Body="&lt;p&gt;Euler's formula for connected planar graphs (i.e. a single connected component) states that $v-e+f=2$. State the generalization of Euler's formula for planar graphs with $k$ connected components (where $k\geq1$). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The correct answer is $v-e+f=1+k$, but I'm not understanding the reasoning behind it. Anyone care to share some insight?&lt;/p&gt;&#xA;" OwnerUserId="2985" LastActivityDate="2010-11-28T23:33:57.980" Title="Euler's formula for connected planar graphs" Tags="&lt;graph-theory&gt;" AnswerCount="2" CommentCount="5" />
  <row Id="12251" PostTypeId="2" ParentId="12250" CreationDate="2010-11-28T21:33:25.040" Score="2" ViewCount="" Body="&lt;p&gt;The usual proof of Euler's formula works by first triangulating the graphs, then removing triangles one by one until you reach a single triangle; all these respect the Euler characteristic $v-e+f$. The proof is completed by calculating $v-e+f$ for a triangle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can do the same here - the first phase is the same, and in the end you will be left with $k$ triangles. Then all you have to do is calculate $v-e+f$ for $k$ unconnected triangles. Since $v=e$ for any collection of triangles, the result is the same as the number of faces, which is $1+k$: one face per triangle, and an extra outer face.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-28T21:33:25.040" />
  <row Id="12252" PostTypeId="2" ParentId="8875" CreationDate="2010-11-28T21:53:51.233" Score="1" ViewCount="" Body="&lt;p&gt;First, one has $$e^{ir}-1=O(|r|) $$ for all $r\in \mathbb{R}.$ This means that there exist constant (independent of $r$) $A&amp;gt;0$ such that $$|e^{ir}-1|\le A|r| \mbox{ for all } r\in \mathbb{R} $$ (In fact, you can take $A=1$, as $|e^{ir}-1|$ is the length of the chord from 1 to $e^{ir}$ and $r$ is the corresponding arc length on the unit circle.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let us prove your question in the case $X $ is contiuous. (The discrete case can be proved similarly.) Suppose $f(x)$ is the probability density function of $X$.  Then&#xA;$$ \phi(t)=E(e^{itX})=\int_{-\infty}^\infty f(x)e^{itx} dx$$ and $\phi(0)=\int_{-\infty}^\infty f(x) dx =1$. So by linearity of integral&#xA;$$\phi&amp;#39;(0)=\lim_{t\to 0}\frac{\phi(t)-\phi(0)}{t}=\lim_{t\to 0}\int_{-\infty}^\infty f(x)\frac{e^{itx}-1}{t} dx.$$ Since&#xA;$$\left|f(x)\frac{e^{itx}-1}{t}\right|\le f(x) A \frac{|tx|}{|t|}=A f(x)|x|,$$ and $$\lim_{t\to 0}\frac{e^{itx}-1}{t}=\frac{d}{dt} e^{itx}|_{t=0}=ix e^0=ix,$$&#xD;&#xA;it follows from DCT that if $E(|X|)&amp;lt;\infty$, then $$\phi&amp;#39;(0)= \int_{-\infty}^\infty f(x)ix dx=i E(X).$$&lt;/p&gt;&#xA;" OwnerUserId="3249" LastEditorUserId="3249" LastEditorDisplayName="" LastEditDate="2010-11-28T22:46:44.740" LastActivityDate="2010-11-28T22:46:44.740" />
  <row Id="12253" PostTypeId="2" ParentId="12223" CreationDate="2010-11-28T22:03:41.640" Score="3" ViewCount="" Body="&lt;p&gt;Some logical operators are associative: both $\wedge$ and $\vee$ are associative, as a simple check of truth tables verifies. Likewise, the biconditional $\leftrightarrow$ is associative. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the implication $\rightarrow$ is &lt;em&gt;not&lt;/em&gt; associative. Compare $(p\rightarrow q)\rightarrow r$ and $p\rightarrow(q\rightarrow r)$. If $p$ and $r$ are true and $q$ is false, then $(p\rightarrow q)$ is false, so $(p\rightarrow q)\rightarrow r$ is true. But under the same conditions $q\rightarrow r$ is true, and therefore $p\rightarrow(q\rightarrow r)$ is true. So $(p\rightarrow q)\rightarrow r$ and $p\rightarrow(q\rightarrow r)$ cannot be equivalent; that is, $\rightarrow$ is not associative.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-11-28T22:03:41.640" CommentCount="4" />
  <row Id="12254" PostTypeId="2" ParentId="12097" CreationDate="2010-11-28T22:06:46.883" Score="0" ViewCount="" Body="&lt;p&gt;Some rough idea.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I consider only the special case where the $X_i$ are positive rv's such that, by self-similarity, $X_1 + \cdots + X_{n-1}$ is equal in distribution to $(n-1)^{1/\alpha}X_1$ where $\alpha$ is the associated index. Then, it seems useful to start (similarly to what you did above) with&#xA;$$&#xD;&#xA;P = n{\rm P}(X_0  &amp;gt; X_1  +  \cdots  + X_{n-1}),&#xD;&#xA;$$&#xD;&#xA;where $P$ is the probability in the title. Then, recalling the self-similarity argument, a straightforward application of the law of total probability (conditioning on $X_0$) yields&#xA;$$&#xD;&#xA;P = n\int F (s/(n - 1)^{1/\alpha } )f(s){\rm d}s,&#xD;&#xA;$$&#xD;&#xA;where $F$ and $f$ are the common distribution and density functions of the $X_i$, respectively. See if you can continue from here.&lt;/p&gt;&#xA;" OwnerUserId="2810" LastEditorUserId="2810" LastEditorDisplayName="" LastEditDate="2010-11-28T23:08:02.063" LastActivityDate="2010-11-28T23:08:02.063" CommentCount="2" />
  <row Id="12255" PostTypeId="1" AcceptedAnswerId="12257" CreationDate="2010-11-28T22:08:43.243" Score="4" ViewCount="148" Body="&lt;p&gt;I'm trying to help my son study 8th grade math (in Texas).  He's having trouble with the &quot;Nth rule&quot;.  Is anyone familiar with this terminology?  I think that it may be related to the Nth term in a series, but I also think that it has a more specific meaning.&lt;/p&gt;&#xA;" OwnerUserId="4181" LastActivityDate="2010-11-28T22:23:02.583" Title="Terminology - Nth rule" Tags="&lt;sequences-and-series&gt;&lt;terminology&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="12256" PostTypeId="1" AcceptedAnswerId="12260" CreationDate="2010-11-28T22:14:58.773" Score="3" ViewCount="248" Body="&lt;p&gt;What's the largest regular tetrahedron (side length x) you can fit inside a sphere with a radius of one?&lt;/p&gt;&#xA;" OwnerUserId="3746" LastActivityDate="2010-11-29T00:42:21.850" Title="Tetrahedron inside a sphere" Tags="&lt;geometry&gt;" AnswerCount="2" CommentCount="7" FavoriteCount="1" />
  <row Id="12257" PostTypeId="2" ParentId="12255" CreationDate="2010-11-28T22:15:32.140" Score="9" ViewCount="" Body="&lt;p&gt;A quick google search suggests that &quot;the $n$th rule&quot; means finding the formula for the $n$th term of a given sequence of numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, given $1$, $3$, $5$, $7,\ldots$, the expected &quot;nth rule&quot; is probably &quot;$2n-1$&quot;, since the 1st term is $2(1)-1$, the 2nd term is $2(2)-1$, the 3rd term is $2(3)-1$, etc. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;What follows is a bit of a diatribe completely irrelevant to your question:&lt;/em&gt; It is an utterly silly request, as noted long ago by Carl Linderholm in his book, &lt;strong&gt;Mathematics made difficult&lt;/strong&gt;. Any answer that gives the right values at the given points is valid; and if you pick your favorite number for the &quot;next&quot; number, you can find a formula that will give &lt;em&gt;those&lt;/em&gt; using Lagrange interpolation; for example, if you have the sequence $1$, $2$, $4$, $8$, $16,\ldots$, then the &quot;next term&quot; is $31$, either through Lagrange interpolation, or by counting the number of sectors that a circle is divided into if you place $n$ points on its circumference and connect them with straight lines. Both show much deeper mathematical understanding than &quot;$2^{n-1}$&quot;. The problem is not really  &quot;What is the formula for the $n$th term of this sequence?&quot; the question is &quot;of all possible formulas that would give these values for $n=1,2,\ldots$, which is the one that the person who is grading this test is thinking of?&quot;&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-28T22:23:02.583" LastActivityDate="2010-11-28T22:23:02.583" CommentCount="1" />
  <row Id="12258" PostTypeId="2" ParentId="12255" CreationDate="2010-11-28T22:16:23.460" Score="4" ViewCount="" Body="&lt;p&gt;What they mean is this: Suppose I give you a sequence like&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2,4,6,8,...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then I ask you: How can I obtain the n-th term in this sequence?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And you'll say: It's just $2n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, for sequences of the form $an + b$, it is quite easy to obtain the rule: Just look how the element increases in each step, this gives you $a$. Then see what the first element is and choose $b$ so that it's obtained for $n = 1$.&lt;/p&gt;&#xA;" OwnerUserId="4171" LastActivityDate="2010-11-28T22:16:23.460" />
  <row Id="12259" PostTypeId="2" ParentId="12239" CreationDate="2010-11-28T22:28:49.753" Score="1" ViewCount="" Body="&lt;p&gt;Ok. So I figured it out. But I think my argument is really convoluted. I would appreciate if someone could come up with a simpler argument.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Without loss of generality, we can assume that the matrix $C$ is of full-rank $p$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First let $B = \begin{bmatrix} A &amp;amp; C \\ C^T &amp;amp; 0 \end{bmatrix}$, where $A \in \mathbb{R}^{m \times m}$ and $C \in \mathbb{R}^{m \times p}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first observation is that all the eigenvalues of this matrix have to be real. This is so since the matrix $B$ is symmetric. Also, the matrix $B$ has an eigenvalue decomposition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clearly, $\exists x$ such that $x^TBx &amp;gt; 0$. For instance, $x = [x_1 , 0]^T$ where $x_1 \in \mathbb{R}^{1 \times m}$ and $x_1 \neq 0$ and $0 \in \mathbb{R}^{1 \times p}$. Then $x^TBx =x_1^TAx_1 &amp;gt; 0$ since $A$ is positive definite.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if we consider $x=[0,x_2]^T$ where $x_2 \in \mathbb{R}^{1 \times p}$ and $x_2 \neq 0$ and $0 \in \mathbb{R}^{1 \times m}$, then $x^TBx = 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So $B$ is not positive definite. And since $B$ is symmetric, there is an eigenvalue decomposition for $B$. And hence $\exists \lambda$ such that $\lambda \leq 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, we need to prove that $\exists \lambda$ such that $\lambda &amp;lt; 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, we recognize that the matrix $B$ is full-rank.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This can be seen from the fact that the null space of the matrix $B$ is trivial.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $ z = \begin{bmatrix} x \\ y \end{bmatrix}$ such that $Bz = 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then we have $Ax + Cy = 0$ and $C^Tx = 0$. So we have $x = -A^{-1}Cy$ and hence we get $C^TA^{-1}Cy = 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The matrix $C^TA^{-1}C$ is positive definite. This is because $C$ is full rank and is a skinny matrix. So whenever $x \neq 0$, $x^* = Cx \neq 0$ and hence $x^TC^TA^{-1}Cx = {x^*}^TA^{-1}x^* &amp;gt; 0$ (Since $A^{-1}$ is also positive definite).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, $y=0$ is the only possible solution which implies $x=0$ and hence $z=0$ is the only vector in the null space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the matrix $B$ is of full-rank.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, none of the eigenvalues are zero i.e. $\forall \lambda$, $\lambda \neq 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, from previous argument, we have that $\exists \lambda$ such that $\lambda \leq 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, we can conclude that $\exists \lambda$ such that $\lambda &amp;lt; 0$. So if I choose the (or) a eigenvector $v$ corresponding to this $\lambda$, then I have $v^TBv = - \lambda ||v||_2^2 &amp;lt; 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, the matrix $B$ is Indefinite.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2010-11-28T22:59:06.410" LastActivityDate="2010-11-28T22:59:06.410" CommentCount="4" />
  <row Id="12260" PostTypeId="2" ParentId="12256" CreationDate="2010-11-28T22:36:20.223" Score="5" ViewCount="" Body="&lt;p&gt;We are clearly looking for the regular tetrahedron inscribed in a sphere of radius 1 (i.e. with all its vertices lying on the sphere). The neat trick with regular tetrahedra is to inscribe them in a cube.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wikipedia has a picture of the two regular tetrahedra you can find in a cube: &lt;a href=&quot;http://en.wikipedia.org/wiki/File:CubeAndStel.gif&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/File:CubeAndStel.gif&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The cube inscribed in a unit sphere has side length $\frac{2}{\sqrt 3}$, so the regular tetrahedron has side length $x = \sqrt{2} \frac{2}{\sqrt 3} = \sqrt{\frac{8}{3}}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is another example of this idea: &lt;a href=&quot;http://math.stackexchange.com/questions/11085/height-of-a-tetrahedron/11090#11090&quot;&gt;Height of a tetrahedron&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3747" LastActivityDate="2010-11-28T22:36:20.223" />
  <row Id="12262" PostTypeId="1" CreationDate="2010-11-28T23:04:34.433" Score="2" ViewCount="111" Body="&lt;p&gt;At Wikipedia's &lt;a href=&quot;http://en.wikipedia.org/wiki/Erlangen_program#Abstract_returns_from_the_Erlangen_program&quot; rel=&quot;nofollow&quot;&gt;Erlangen program&lt;/a&gt; I read that &quot;quite often, it appears there are two or more distinct geometries with isomorphic automorphism groups&quot;. Some examples are given.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;But what are examples of geometries nevertheless &lt;em&gt;being&lt;/em&gt;&#xA;  uniquely defined (up to isomorphism) &#xA;  by the isomorphism class of their&#xA;  automorphism group?&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Are there examples of non-Euclidean&#xA;  geometries with an automorphism group&#xA;  isomorphic to the Euclidean group?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1792" LastEditorUserId="1792" LastEditorDisplayName="" LastEditDate="2010-11-28T23:47:03.413" LastActivityDate="2010-11-28T23:47:03.413" Title="Euclidean geometry and the Euclidean group" Tags="&lt;geometry&gt;&lt;euclidean-geometry&gt;&lt;lie-groups&gt;" CommentCount="3" />
  <row Id="12263" PostTypeId="1" AcceptedAnswerId="12333" CreationDate="2010-11-28T23:23:38.753" Score="0" ViewCount="109" Body="&lt;p&gt;I'm attempting to understand how B-Spline derivatives work. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To do this, i've generated some data in MATLAB to plot sin(x) over 0 - 2*pi. I then use the splinetool to fit an order 2 spline to the data which it fits perfectly. However, when I look at the 1st derivative of this sine wave data it's max is 0.017 and it's min is -0.017.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't understand why the derivative of this data doesn't give exactly cos(x) with a max and min of 1, -1. Any insights please?&lt;/p&gt;&#xA;" OwnerUserId="1588" LastActivityDate="2010-11-29T11:08:09.980" Title="B-Spline Derivatives" Tags="&lt;calculus&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="12264" PostTypeId="2" ParentId="12256" CreationDate="2010-11-28T23:25:37.967" Score="2" ViewCount="" Body="&lt;p&gt;You are looking at the regular tetrahedron inscribed in a sphere of radius 1. Denote the center of the sphere by O, and the vertices by A, B, C and D.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fact: In the regular tetrahedron, the altitude from A is cut by O in 3:1 ratio (Note: In an equilateral triangle the analogous ratio is 2:1).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proof:  The four vectors from O to the vertices sum to zero (the sum vector is invariant under rotations preserving the tetrahedron, so must be zero). Pick a vertex A. The projections of the other three vectors OB, OC and OD onto the line OA are all equal and, as their sum must cancel OA, each of these projections is equal to 1/3 of OA.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Denote the base of the altitude by H. Then we just showed OH=1/3, AH=4/3. Now by Pythagoras theorem, $HB^2=BO^2-OH^2=1^2-(1/3)^2=8/9$. Finally, by Pythagoras again, $AB^2=AH^2+HB^2=(4/3)^2+8/9=24/9$, so $AB=\sqrt(8/3)$.&lt;/p&gt;&#xA;" OwnerUserId="2633" LastEditorUserId="2633" LastEditorDisplayName="" LastEditDate="2010-11-29T00:42:21.850" LastActivityDate="2010-11-29T00:42:21.850" />
  <row Id="12265" PostTypeId="2" ParentId="12250" CreationDate="2010-11-28T23:33:57.980" Score="3" ViewCount="" Body="&lt;p&gt;For many approaches to Euler's formula look at:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.ics.uci.edu/~eppstein/junkyard/euler/&quot; rel=&quot;nofollow&quot;&gt;http://www.ics.uci.edu/~eppstein/junkyard/euler/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1369" LastActivityDate="2010-11-28T23:33:57.980" />
  <row Id="12266" PostTypeId="1" CreationDate="2010-11-28T23:38:29.083" Score="1" ViewCount="50" Body="&lt;p&gt;Hecke operators are things that act on modular forms and give rise to a lot of interesting arithmetical results:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Hecke_operator&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Hecke_operator&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand on the wikpedia page for Hecke algebras, which should naively be an algebra of Hecke operators, the term seems to acquire very different meanings, such as deformations of group algebra of a Coxeter group.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the connection between the two?&lt;/p&gt;&#xA;" LastActivityDate="2010-11-28T23:38:29.083" Title="Connection between Hecke operators and Hecke algebras" Tags="&lt;modular-forms&gt;&lt;representation-theory&gt;" CommentCount="2" />
  <row Id="12267" PostTypeId="2" ParentId="12193" CreationDate="2010-11-28T23:45:33.107" Score="2" ViewCount="" Body="&lt;p&gt;I'll place something as an answer in the interest of not having questions with no answers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know what the cardinality of $\{\mathbb{R}(\setminus a)\}$ is, but I would be surprised if it is not uncountable. I'll see if I can think about it some in the next couple of days.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for your final paragraph, the different fields come not from the order in which you would remove elements, but rather on the &lt;em&gt;choices&lt;/em&gt; you make as to what to remove at each (of infinitely many) steps. As I noted in the comments, think of the similar process whereby we look for subgroups of a given group that are maximal among those that do not contain a given element. If we start with the Klein $4$-group, $C_2\times C_2$, and remove the element $a=(1,1)$, then we get two different choices of what other element to remove in order to get a group: we can remove &lt;em&gt;either&lt;/em&gt; $(1,0)$ &#xA;&lt;em&gt;or&lt;/em&gt; &#xA;$(0,1)$; each of the choices will lead to a different subgroup. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, consider for instance the case where $a=\sqrt{2}$; then as noted in your previous question, a maximal subfield of $\mathbb{R}$ that does not contain $a$ may contain either $\sqrt{3}$ or $\sqrt{6}$, but not both, and you get one field that is maximal and &lt;em&gt;does&lt;/em&gt; contain $\sqrt{3}$ (simple application of Zorn's Lemma), and one that is maxiimal and &lt;em&gt;does&lt;/em&gt; contain $\sqrt{6}$ (ditto); so with your process you can either remove $\sqrt{3}$ at some point, or remove $\sqrt{6}$ at some point, and these choices lead to different maximal subfields.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The very word &quot;choices&quot; suggests how the Axiom of Choice is playing a role in this &quot;top-down&quot; construction of elements of $\{\mathbb{R}(\setminus a)\}$ (in the &quot;bottoms-up&quot; approach, it plays a role via Zorn's Lemma). In order to be able to obtain something one could call &lt;strong&gt;the&lt;/strong&gt; &quot;field reduction&quot;, we would need some procedure or algorithm that determines &lt;em&gt;which&lt;/em&gt; choice we make at any stage. One way would be by endowing $\mathbb{R}$ with a well-ordering, and then at each step we can consider the set of all elements we can remove; if it is empty, we are done; if it is not empty, we remove the least element of the set. But, of course, well-ordering $\mathbb{R}$ is a bit of a problem in and of itself: there is no consistently agreed-upon well-ordering of $\mathbb{R}$ (and in some cases, not even an explicit one). We could try to pick a particular model in which there are explicit well-orderings (as mentioned by Adres Caicedo in the comments, such models exist), &lt;em&gt;and&lt;/em&gt; a particular well-ordering in there, and &lt;em&gt;then&lt;/em&gt; define &quot;the&quot; field reduction, but that seems to be as much of a problem in and of itself: which model, and which order, and why? Perhaps there are other mechanisms whereby one could try to establish a distinguished element in $\{\mathbb{R}(\setminus a)\}$ (or perhaps, a distinguished element for specific instances of $a$), but it seems to me to be a hopeless task in the abstract/in general. &#xA;&lt;hr/&gt;&#xA;&lt;em&gt;Added and edited:&lt;/em&gt; Recall that any field extension $K/k$ can be broken up as $K/F/k$, where $K/F$ is algebraic and $F/k$ is a purely transcendental extension. In particular, if $T$ is a transcendence basis for $K$ over $k$, then $k(T)$ will be a maximal subfield of $K$ that contains $k$ and does not contain any elements of $K$ that are algebraic over $k$ but not in $k$. So for any field extension $K/k$, if we let&#xA;\begin{equation*}&#xA;\mathbb{A}_{K/k} =\bigl\{ a \in K\ \bigm|\ \mbox{$a$ is algebraic over $k$, $a\notin k$}\bigr\},&#xA;\end{equation*}&#xA;then &#xA;$$K(\setminus \mathbb{A}_{K/k}) = \{ k(T)\ |\ \mbox{$T$ is a transcendence basis for $K$ over $k$}\}.$$&#xA;If $K$ is algebraic over $k$, then you just get $k$, of course; if $K$ is purely transcendental over $k$, then $\mathbb{A}_{K/k}$ is empty, so you get $K$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I had originally written that the set would always be singleton, but this is not the case. If $T$ is any transcendence basis for $\mathbb{R}$ over $\mathbb{Q}$, you can modify an element of $T$ by multiplying it by a non-rational algebraic number and you get a slightly different subfield. For example, fix $t\in T$ and replace $t$ with $\sqrt{2}t$ to obtain a new set $T&amp;#39;$. This is still a transcendence basis for $\mathbb{R}$ over $\mathbb{Q}$; but whereas $\sqrt{2}t\in \mathbb{Q}(T&amp;#39;)$, it does not lie in $\mathbb{Q}(T)$. If it did, then we would immediately get $\sqrt{2}\in\mathbb{Q}(T)$, contradicting the fact that $\mathbb{Q}(T)$ is purely transcendental over $\mathbb{Q}$. You can do the same thing taking arbitrary subsets of $T$ and multiplying them through by a nonrational algebraic number. Since $|T|=\mathfrak{c}$, this gives at least $2^{\mathfrak{c}}$ distinct elements in $\mathbb{R}(\setminus\mathbb{A}&amp;#39;)$, and since that is the number of &lt;em&gt;subsets&lt;/em&gt; of $\mathbb{R}$, it follows that the cardinality of $\mathbb{R}(\setminus\mathbb{A}&amp;#39;)$ is $2^\mathfrak{c}$. &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-29T18:16:40.417" LastActivityDate="2010-11-29T18:16:40.417" CommentCount="9" />
  <row Id="12268" PostTypeId="1" CreationDate="2010-11-29T00:03:20.383" Score="1" ViewCount="125" Body="&lt;p&gt;Problem:&#xA;Let $\omega\in\Omega^r(M^n)$ suppose that $\int_\sum \omega = 0$ for every oriented smooth manifold $\sum \subseteq M^n$ that is diffeomorphic to $S^r$. Show that $d\omega = 0$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proof:&#xA;Assume $d\omega \neq 0$. Then there exists $v_1, \ldots, v_{r+1}\in T_pM$ such that $d\omega_p(v_1, \ldots, v_{r+1}) \neq 0$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$D^{r+1}\subseteq \mathbb{R}^{r+1}$ a smooth submanifold of $\mathbb{R}^{n}$ with boundary $S^r$. Let $(h,U)$ be a chart around $p$ such that $D^{r+1}$ (with &lt;em&gt;some&lt;/em&gt; radius) is mapped to $N = h^{-1}(D^{r+1})$ around $p$. Then $N$ is a smooth submanifold of $M^n$ with boundary equal to $\partial N =  h^{-1}(S^r)$ (diffeomorphic to $S^r$). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;By definition of the integral and Stokes' theorem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\int_{\partial N} \omega = \int_N d\omega = \int_{D^{r+1}}(h^{-1})^* (d\omega)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let $\alpha = (h^{-1})^*(d\omega)$. Then $\alpha = f(x)dx_1\wedge\ldots\wedge dx_{r+1}$(topform in $\mathbb{R}^{r+1}$). Since $f(x) \neq 0$, it has to be different from zero on a small domain. Assume that $f(x) &amp;gt; 0$. Then &#xA;$\int_{D^{r+1}}\alpha = \int_{D^{r+1}}f(x)d\mu_{r+1} &amp;gt; 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Contradiction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-- I feel my idea is correct, but I'm not fully sure this is a full good proof. Could this have been done easier? I'm grateful for any feedback.&lt;/p&gt;&#xA;" OwnerUserId="2900" LastActivityDate="2010-11-29T00:03:20.383" Title="Diffeomorphisms and Stokes' theorem" Tags="&lt;differential-topology&gt;" CommentCount="3" />
  <row Id="12269" PostTypeId="1" AcceptedAnswerId="12693" CreationDate="2010-11-29T00:03:37.653" Score="6" ViewCount="275" Body="&lt;p&gt;Recall that a &lt;strong&gt;model category&lt;/strong&gt; is a complete and cocomplete category with classes of morphisms called &lt;em&gt;cofibrations, fibrations,&lt;/em&gt; and &lt;em&gt;weak equivalences.&lt;/em&gt; These are closed under composition and satisfy certain axioms, such as lifting properties. Furthermore, according to Mark Hovey (but this apparently varies in the literature), any morphism must admit a functorial factorization into the composite of a cofibration and a trivial fibration (i.e. a fibration which is a weak equivalence), as well as a composite of a trivial cofibration and a fibration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The standard examples of model categories  are simplicial sets and chain complexes. Yet the words &quot;fibration&quot; and &quot;cofibration&quot; suggest not category theory but the topological homotopy lifting property. A morphism of topological spaces $f: X \to Y$ is called a (Hurewicz) fibration if whenever $p: T \times [0,1]  \to Y$ is a map and $\widetilde{p}: T \to X$ is a map lifting $p|_{T \times \{0\}}$, $\widetilde{p}$ can be extended to a lifting of $p$. Cofibrations can be defined by an analogous homotopy extension property, and for Hausdorff spaces this is equivalent to being a deformation retract of a suitable neighborhood (defined as $\{x: u(x)&amp;lt;1\}$ where $u$ is a suitable continuous function).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, although this is not explicitly stated anywhere I see, I take it that topological spaces with the usual notions of fibrations and cofibrations (with weak equivalences the homotopy equivalences) do indeed form a model category. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; Am I right in assuming this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The axioms are slightly tricky to check in general. Topological spaces admit limits and colimits. I believe that the lifting properties of cofibrations with respect to trivial fibrations (or trivial cofibrations with respect to fibrations) follows directly from the definition of a cofibration (namely, if you can extend something up to homotopy, you can do it exactly for a cofibration). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm getting slightly confused on the functorial factorizations. Let $f: X \to Y$ be a morphism. We need a &lt;em&gt;canonical&lt;/em&gt; way of factoring this (actually, two canonical ways). &lt;del&gt;One way is to use the inclusion of $X$ in the mapping cylinder $M_f$ (which is a cofibration because $(M_f, X)$ is checked to be an NDR-pair).&#xA;Moreover, $M_f \to Y$ is a homotopy equivalence. &#xA;I don't see why this is a fibration though (it's true that the fibers have the same homotopy type at least).&lt;/del&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Correction: As Aaron observes in the comments, there are easy counterexamples for when the map from the mapping cylinder is not a fibration. This means that another approach is needed to construct the functorial factorization of a map into a cofibration and a trivial fibration?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could someone clarify?&lt;/p&gt;&#xA;" OwnerUserId="536" LastEditorUserId="536" LastEditorDisplayName="" LastEditDate="2010-11-29T02:59:24.847" LastActivityDate="2010-12-02T03:18:40.370" Title="Why is Top a model category?" Tags="&lt;algebraic-topology&gt;&lt;category-theory&gt;&lt;homotopy-theory&gt;" AnswerCount="2" CommentCount="8" />
  <row Id="12270" PostTypeId="2" ParentId="12160" CreationDate="2010-11-29T00:12:26.997" Score="4" ViewCount="" Body="&lt;p&gt;To resolve the second question, note first that the Legendre polynomials are odd functions for odd order (0 then is one root of the polynomial), and even functions for even order. Thus, with regards to solubility in terms of radicals, you should be able to derive (possibly complicated!) radical expressions at least up until $P_9(x)$. To use that as an example, note that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{P_9(\sqrt{x})}{\sqrt{x}}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is a quartic; thus, one can use the quartic formula to derive explicit expressions for its roots, and then you can easily derive the roots of $P_9(x)$ .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$P_{10}(x)$ is where your trouble starts. If we take a look at the polynomial&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$P_{10}(\sqrt{x})$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;we have a quintic to contend with. I'll skip the relatively tedious details, but you can verify that its Galois group is not a solvable group, and thus the solution cannot be expressed in terms of radicals (you can use theta or hypergeometric functions, though).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, not much hope in the symbolic front. In the &lt;em&gt;numeric&lt;/em&gt; front, things are much easier. The slickest way of getting accurate values of the roots of the Legendre polynomial is to use the Jacobi matrix in my previous answer. Since there exist stable and efficient algorithms (e.g. QR algorithm or divide-and-conquer) for the symmetric eigenproblem (in &lt;a href=&quot;http://www.netlib.org/lapack/&quot; rel=&quot;nofollow&quot;&gt;LAPACK&lt;/a&gt;, for instance), and things can be set such that only eigenvalues are returned, you have a good way of generating good approximate values of Legendre polynomial roots. (In the context of Gaussian quadrature, where the roots of orthogonal polynomials play a pivotal role, the scheme is referred to as the &lt;a href=&quot;http://www.ams.org/journals/mcom/1969-23-106/S0025-5718-69-99647-1/S0025-5718-69-99647-1.pdf&quot; rel=&quot;nofollow&quot;&gt;Golub-Welsch algorithm&lt;/a&gt;.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively, as I mentioned in the comments, there exist asymptotic approximations for the roots, which can then be subsequently polished with a few applications of Newton-Raphson. One such asymptotic approximation is due to Francesco Tricomi. Letting $\xi_{n,k}$ be the $k$-th root of $P_n(x)$, ordered in decreasing order, we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\xi_{n,k}\approx\left(1-\frac1{8n^2}+\frac1{8n^3}\right)\cos\left(\pi\frac{4k-1}{4n+2}\right)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and $O(n^{-4})$ and further terms are omitted. Other asymptotic approximations due to Luigi Gatteschi use roots of Bessel functions, but I won't say more about those.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-11-29T00:12:26.997" />
  <row Id="12271" PostTypeId="2" ParentId="7096" CreationDate="2010-11-29T01:28:25.750" Score="0" ViewCount="" Body="&lt;p&gt;I cannot answer your question but if we take $E$ to be the &lt;a href=&quot;http://en.wikipedia.org/wiki/Topologist%27s_sine_curve&quot; rel=&quot;nofollow&quot;&gt;topologist's sine curve&lt;/a&gt; and its projection to $B=[0,1]$ we have a topological (i.e. continuous) family of points (compact subspaces, analogs of the compact complex submanifolds) which however is not proper as a map of topological spaces, since $E=\pi^{-1}(B)$ is not compact while $B$ is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We may obtain similar nonproper examples of topological families with compact fibers modifying the fibers or the base.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here the base is a manifold but not the total space/family, and I do not see immediately such counterexamples with families of compact topological manifolds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Actually I do not understand how Ehresmann's theorem is used in your reasonings, in particular whether you obtain that $\pi$ is proper from it.&lt;/p&gt;&#xA;" OwnerUserId="3407" LastActivityDate="2010-11-29T01:28:25.750" CommentCount="1" />
  <row Id="12272" PostTypeId="2" ParentId="12097" CreationDate="2010-11-29T01:57:26.827" Score="3" ViewCount="" Body="&lt;p&gt;What you are asking for is a special case of the joint distribution of the maximum and sum of a sequence of random variables. The fact that the variables are stable simplifies this and allows us to compute the joint distribution in the limit as n goes to infinity. This limit can be described in terms of the joint distribution of the value of a stable &lt;a href=&quot;http://en.wikipedia.org/wiki/L%C3%A9vy_process&quot; rel=&quot;nofollow&quot;&gt;L&amp;eacute;vy process&lt;/a&gt; at time $t=1$ and its maximum jump size over $t\le1$. L&amp;eacute;vy processes are right continuous with left limits (&lt;a href=&quot;http://en.wikipedia.org/wiki/C%C3%A0dl%C3%A0g&quot; rel=&quot;nofollow&quot;&gt;cadlag&lt;/a&gt;) and have stationary independent increments. Stable L&amp;eacute;vy processes have increments with stationary distributions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $X_i$ each have the $\alpha$-stable distribution denoted by $S(\alpha,\beta,\gamma,\delta)=S(\alpha,\beta,\gamma,\delta;0)$ in the text you link, then the sum $S=\sum_{i=0}^{n-1}X_i$ has the $S(\alpha,\beta,n^{1/\alpha}\gamma,\delta_n)$ distribution where&#xA;$$&#xD;&#xA;\delta_n=\begin{cases}&#xD;&#xA;n\delta + \gamma\beta\tan\frac{\pi\alpha}{2}(n^{1/\alpha}-n),&amp;amp;\textrm{if }\alpha\not=1,\\&#xD;&#xA;n\delta +\gamma\beta\frac{2}{\pi}n\log n,&amp;amp;\textrm{if }\alpha=1.&#xD;&#xA;\end{cases}\qquad\qquad{\rm(1)}&#xD;&#xA;$$&#xD;&#xA;This is equation (1.9) on page 20 of the linked notes. It will be helpful to rescale the sum to remove the dependence on n. The random variable $Y=n^{-1/\alpha}(S-\delta_n)$ has the $S(\alpha,\beta,\gamma,0)$ distribution (by Proposition 1.16).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, to relate this to L&amp;eacute;vy processes. Let $Z_t$ be the L&amp;eacute;vy process with $Z_1\sim Y$. We can take $X_i=n^{1/\alpha}(Z_{(i+1)/n}-Z_{i/n})+\delta_n/n$, which will be independent with the required $S(\alpha,\beta,\gamma,\delta)$ distribution. In fact, in this case we have $Z_1=Y$ and $S=n^{1/\alpha}Z_1+\delta_n$.&#xA;Writing $M=\max_iX_i$ and $\Delta Z^*_t$ for the maximum of $\Delta Z_s=Z(s)-Z(s-)$ over $s\le t$ gives $n^{-1/\alpha}(M-\delta_n/n)\to\Delta Z^*_1$ as n goes to infinity.&#xA;We can then compute the probability &lt;em&gt;P&lt;/em&gt; that you are asking for to leading order as &lt;em&gt;n&lt;/em&gt; becomes large,&#xA;$$&#xD;&#xA;\begin{align}&#xD;&#xA;P&amp;amp;\equiv\mathbb{P}\left(M &amp;gt; S-M\right)\\&#xD;&#xA;&amp;amp;=\mathbb{P}\left(2n^{-1/\alpha}(M-\delta_n/n) &amp;gt; n^{-1/\alpha}(S-2\delta_n/n)\right)\\&#xD;&#xA;&amp;amp;\sim\mathbb{P}\left(2\Delta Z^*_1 &amp;gt; Z_1+n^{-1/\alpha}\delta_n(1-2/n)\right)&#xD;&#xA;\end{align}&#xD;&#xA;$$&#xD;&#xA;So,&#xA;$$&#xD;&#xA;P\sim\mathbb{P}\left(2\Delta Z^*_1 - Z_1 &amp;gt; n^{-1/\alpha}\delta_n\right)&#xD;&#xA;$$&#xD;&#xA;and the leading term in the asymptotics for &lt;em&gt;P&lt;/em&gt; are determined by the distribution of $2\Delta Z^*_1-Z_1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $\alpha &amp;lt; 1$ it can be seen from (1) that $n^{-1/\alpha}\delta_n\to\gamma\beta\tan\frac{\pi\alpha}{2}$ as &lt;em&gt;n&lt;/em&gt; goes to infinity, so we get a finite and positive limit&#xA;$$&#xD;&#xA;P\to\mathbb{P}\left(2\Delta Z^*_1-Z_1 &amp;gt; \gamma\beta\tan\frac{\pi\alpha}{2}\right)&#xD;&#xA;$$&#xD;&#xA;as $n\to\infty$. The tangent has only appeared in the expression here because of the particular parameterization that we are using. It is a bit simpler if we rewrite this in terms of $\tilde Z_t=Z_t+\gamma\beta t\tan\frac{\pi\alpha}{2}$. Then, for the $\beta=1$ case you are using, $\tilde Z_t$ has support $[0,\infty)$ (Lemma 1.10 in the linked text). This means that $\tilde Z$ is a &lt;em&gt;stable subordinator&lt;/em&gt; and the required probability is&#xA;$$&#xD;&#xA;P\to\mathbb{P}\left(2\Delta\tilde Z^*_1 &amp;gt; \tilde Z_1\right)=\frac{\sin(\pi\alpha)}{\pi\alpha}.&#xD;&#xA;$$&#xD;&#xA;[I've been a bit sneaky here, and substituted in the expression that Shai Covo gave in &lt;a href=&quot;http://math.stackexchange.com/questions/12097/probability-of-the-maximum-levy-stable-random-variable-in-a-list-being-greater/12347#12347&quot;&gt;his answer&lt;/a&gt; for $\mathbb{P}(2\Delta\tilde Z^*_1 &amp;gt; \tilde Z_1)$ and simplified a bit using &lt;a href=&quot;http://en.wikipedia.org/w/index.php?title=Reflection_formula&amp;amp;oldid=358060311&quot; rel=&quot;nofollow&quot;&gt;Euler's reflection formula&lt;/a&gt;.]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, if $\alpha &amp;gt; 1$ then (1) gives $n^{-1/\alpha}\delta_n\sim n^{1-1/\alpha}(\delta-\gamma\beta\tan\frac{\pi\alpha}{2})$ and, if $\alpha=1$, then $n^{-1/\alpha}\delta_n\sim\gamma\beta\frac{2}{\pi}\log n$. In this case, P tends to zero at a rate dependent on the tail of the distribution function of $2\Delta Z^*_1-Z_1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is also possible to use the &lt;a href=&quot;http://en.wikipedia.org/w/index.php?title=L%C3%A9vy_process&amp;amp;oldid=396531725#L.C3.A9vy.E2.80.93Khintchine_representation&quot; rel=&quot;nofollow&quot;&gt;L&amp;eacute;vy-Khintchine representation&lt;/a&gt; to compute the tail of the distribution function of $2\Delta Z^*_1-Z_1$. The jumps of a L&amp;eacute;vy process are described by a L&amp;eacute;vy measure $\nu$ on the real numbers, so that the jumps of size belonging to a set $A$ occur according to a &lt;a href=&quot;http://en.wikipedia.org/w/index.php?title=Poisson_process&amp;amp;oldid=391609118&quot; rel=&quot;nofollow&quot;&gt;Poisson process&lt;/a&gt; of rate $\nu(A)$. An $\alpha$-stable process has L&amp;eacute;vy measure $d\nu(x)=c\vert x\vert^{-\alpha-1}\,dx$, where $c$ only depends on the sign of $x$. We can back out the value of $c$ by calculating the characteristic function of $Z_1$ using the L&amp;eacute;vy-Khintchine formula (essentially the Fourier transform of $\nu$, see &lt;a href=&quot;http://en.wikipedia.org/w/index.php?title=L%C3%A9vy_process&amp;amp;oldid=396531725#L.C3.A9vy.E2.80.93Khintchine_representation&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;, which uses W for the L&amp;eacute;vy measure) and comparing with expression (1.4) from your text. Going through this computation gives&#xA;$$&#xD;&#xA;d\nu(x)=\pi^{-1}\alpha\gamma^\alpha\sin\frac{\pi\alpha}{2}\Gamma(\alpha)(1+\beta{\rm sign}(x))\vert x\vert^{-\alpha-1}\,dx.&#xD;&#xA;$$&#xD;&#xA;The number of jumps of size greater than $x$ is Poisson distributed of rate $\nu((x,\infty))$ giving,&#xA;$$&#xD;&#xA;\begin{align}\mathbb{P}(\Delta Z^*_1&amp;gt; x) &amp;amp;= 1-e^{-\nu((x,\infty))}\\&#xD;&#xA;&amp;amp;\sim\nu((x,\infty))\\&#xD;&#xA;&amp;amp;=\pi^{-1}\gamma^\alpha\sin\frac{\pi\alpha}{2}\Gamma(\alpha)(1+\beta)x^{-\alpha}.&#xD;&#xA;\end{align}\qquad{\rm(2)}&#xD;&#xA;$$&#xD;&#xA;Comparing with your linked text (Theorem 1.12), this is exactly the same as the tail of $Z_1$! That is, $\mathbb{P}(\Delta Z^*_1 &amp;gt; x)\sim\mathbb{P}(Z_1 &amp;gt; x)$. This means that the tail of the distribution is entirely due to the occasional big jump of the L&amp;eacute;vy process. This does actually correspond to experience. I have plotted Cauchy process sample paths before, and you do get some paths with a single big jump drowning out all other behaviour. We can also see why this should be the case. The density of the jump distribution is sub-exponential, proportional to $\vert x\vert^{-\alpha-1}$. So one big jump of size $x$ has a much bigger probability than &lt;em&gt;n&lt;/em&gt; small jumps of size $x/n$. It also suggests that, in your calculation of $X_i$, the samples contributing to the probability come mainly from the case where all of them are reasonably close to the mean except for one extreme (positive or negative) $X_i$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, from the argument above, the leading asymptotic of the probability &lt;em&gt;P&lt;/em&gt; comes from the case where $\Delta Z_1^*$ is very large compared to $Z_1-\Delta Z_1^*$, or when the largest negative jump $\Delta(-Z)^*_1$ is large compared to $Z_1+\Delta(-Z)^*_1$. This gives&#xA;$$&#xD;&#xA;\begin{align}&#xD;&#xA;\mathbb{P}(2\Delta Z^*_1-Z_1 &amp;gt; x)&#xD;&#xA;&amp;amp;\sim\mathbb{P}(\Delta Z^*_1 &amp;gt; x)&#xD;&#xA;+\mathbb{P}(\Delta (-Z)^*_1 &amp;gt; x)\\&#xD;&#xA;&amp;amp;\sim 2\pi^{-1}\gamma^{\alpha}\sin\frac{\pi\alpha}{2}\Gamma(\alpha)x^{-\alpha},&#xD;&#xA;\end{align}&#xD;&#xA;$$&#xD;&#xA;where I substituted in (2). Plugging in the asymptotic form for $x=n^{-1/\alpha}\delta_n$ calculated above gives&#xA;$$&#xD;&#xA;P\sim2\pi^{-1}\gamma^{\alpha}\sin\frac{\pi\alpha}{2}\Gamma(\alpha)\left(\delta-\gamma\beta\tan\frac{\pi\alpha}{2}\right)^{-\alpha}n^{1-\alpha}&#xD;&#xA;$$&#xD;&#xA;for $\alpha &amp;gt; 1$. This confirms your $O(1/n^{\alpha-1})$ guess from the numerical simulation! Also, using $x=n^{-1/\alpha}\delta_n\sim\gamma\beta\frac{2}{\pi}\log n$ for $\alpha=1$ gives &#xA;$$&#xD;&#xA;P\sim\frac{1}{\beta\log n}.&#xD;&#xA;$$&#xD;&#xA;Hopefully this is all correct, but I can't rule out making an arithmetic slip above. &lt;strong&gt;Edit&lt;/strong&gt;: The limit above for $\alpha &amp;gt; 1$ does not look very good. It appears to be slightly above the plot in the original post. However, even for n=1000, it should not be very close to the asymptotic limit described. Replacing the asymptotic limit for $\delta_n$ by its exact form effective replaces the $n^{1-\alpha}$ term for &lt;em&gt;P&lt;/em&gt; by $(n^{1-1/\alpha}-1)^{-\alpha}$. For n=1000 this approximately doubles the calculated number so, the more accurate value for $\delta_n$ moves you further away from the plot. Something has gone wrong somewhere...&lt;/p&gt;&#xA;" OwnerUserId="1321" LastEditorUserId="1321" LastEditorDisplayName="" LastEditDate="2010-11-30T01:04:58.173" LastActivityDate="2010-11-30T01:04:58.173" CommentCount="20" />
  <row Id="12273" PostTypeId="1" AcceptedAnswerId="12277" CreationDate="2010-11-29T02:01:34.797" Score="5" ViewCount="215" Body="&lt;p&gt;This is some sort of standard puzzle, which many do it using trial and error or brute-force method. The question goes like this, given the numbers, 11,13,31,33,42,44,46 what is choices of numbers (a number can be chosen more than once) so that sum adds up to 100.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a formula like way or approach to this other than brute force only?&lt;/p&gt;&#xA;" OwnerUserId="5516" LastEditorUserId="5516" LastEditorDisplayName="" LastEditDate="2010-11-29T05:06:15.197" LastActivityDate="2010-11-29T17:53:34.207" Title="Minimum multi-subset sum to a target." Tags="&lt;combinatorics&gt;" AnswerCount="2" CommentCount="5" />
  <row Id="12274" PostTypeId="1" CreationDate="2010-11-29T02:05:25.773" Score="2" ViewCount="43" Body="&lt;p&gt;Are there any interesting examples of second order stationary processes on ${\mathcal R}^2$ or ${\mathcal R}^3$ that are not isotropic? The book I am looking at has no examples.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Update : I asked this &lt;a href=&quot;http://mathoverflow.net/questions/47684/stationary-non-isotropic-spatial-stochastic-processes&quot; rel=&quot;nofollow&quot;&gt;question&lt;/a&gt; in mathoverflow, apparently such examples are not easy to come by.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Update: Processes with &lt;a href=&quot;http://books.google.com/books?id=ZvoibTTS9QwC&amp;amp;pg=PA113&amp;amp;dq=anisotropic+variograms&amp;amp;hl=en&amp;amp;ei=fvbzTMm-I8WqlAeP4aX0DA&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=1&amp;amp;ved=0CCcQ6AEwAA#v=onepage&amp;amp;q=anisotropic%2520variograms&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;anisotropic variograms&lt;/a&gt; are examples of non-isotropic stationary processes.&lt;/p&gt;&#xA;" OwnerUserId="404" LastEditorUserId="404" LastEditorDisplayName="" LastEditDate="2010-11-29T18:56:46.977" LastActivityDate="2010-11-29T18:56:46.977" Title="stationary non-isotropic spatial stochastic processes" Tags="&lt;probability-theory&gt;&lt;statistics&gt;" />
  <row Id="12275" PostTypeId="1" AcceptedAnswerId="12280" CreationDate="2010-11-29T02:06:10.020" Score="11" ViewCount="145" Body="&lt;p&gt;Could you prove or disprove the following statement? &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Let &#xA;  $f\colon[0,1]^2\rightarrow \mathbb R$ &#xA;  be a continuous function.  Then&#xA;  there are continuous functions &#xA;  $g,\ h\colon [0,1]\rightarrow \mathbb R$ and&#xA;  $\Phi\colon \mathbb R \to \mathbb R$&#xD;&#xA;  such that  $$ f(x,y) = \Phi(g(x) + h(y)).$$&#xD;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(This problem popped up in my mind while I was thinking about &lt;a href=&quot;http://mathoverflow.net/questions/47557/&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; related one on MO.  I couldn't find an easy proof or a disproof. This version is much weaker than the one asked at MO, since $g$ and $h$ do depend on $f$ here.)&lt;/p&gt;&#xA;" OwnerUserId="145" LastActivityDate="2010-11-29T04:47:34.353" Title="Writing $f(x,y)$ as $\Phi(g(x) + h(y))$" Tags="&lt;real-analysis&gt;" AnswerCount="2" FavoriteCount="2" />
  <row Id="12276" PostTypeId="1" AcceptedAnswerId="12286" CreationDate="2010-11-29T02:07:17.977" Score="20" ViewCount="370" Body="&lt;p&gt;Green's Theorem has the form: &#xA;$$\oint P(x,y)dx = - \iint \frac{\partial P}{\partial x}dxdy , \oint Q(x,y)dy = \iint \frac{\partial Q}{\partial y}dxdy $$&#xD;&#xA;The Cauchy-Riemann equations have the following form:(Assuming $z = P(x,y) + iQ(x,y)$)&#xA;$$\frac{\partial P}{\partial x} = \frac{\partial Q}{\partial y}, \frac{\partial P}{\partial y} = - \frac{\partial Q}{\partial x}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any connection between this two equations?&lt;/p&gt;&#xA;" OwnerUserId="1382" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-29T02:58:50.557" LastActivityDate="2011-01-30T20:32:01.167" Title="Is there any connection between Green's Theorem and the Cauchy-Riemann equations?" Tags="&lt;calculus&gt;&lt;complex-analysis&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="6" />
  <row Id="12277" PostTypeId="2" ParentId="12273" CreationDate="2010-11-29T02:17:33.820" Score="5" ViewCount="" Body="&lt;p&gt;SUBSET-SUM is NP-complete, so in general it's difficult to tell even if there is a choice of numbers that adds up to a given target.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Moron has raised the valid point that numbers can be reused.&#xA;We therefore replace each number $x$ with a sequence $x,2x,4x,\ldots,2^kx,\ldots$, stopping before reaching a number larger than our target.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's carefully analyze the blowup in size. Suppose the original numbers $x_i$ had size $l_i$, and the target $T$ had size $L$. We replace each number $x_i$ with $L$ numbers of size at most $L$, and so the new size is $$\sum_i L^2 + L \leq L^2 (\sum_i x_i + L),$$ so the blowup is at most cubic, and in particular polynomial. &lt;i&gt;So the problem is still NP-hard (it is trivially in NP):&lt;/i&gt; no, since the reduction is in the wrong direction... (thanks, Moron, for noticing)&lt;/p&gt;&#xA;" OwnerUserId="1277" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2010-11-29T16:41:42.380" LastActivityDate="2010-11-29T16:41:42.380" CommentCount="5" />
  <row Id="12278" PostTypeId="2" ParentId="12276" CreationDate="2010-11-29T02:17:39.937" Score="7" ViewCount="" Body="&lt;p&gt;If you consider $P(x,y)$ and $Q(x,y)$ as  the part real and imaginary of an analitic function you can get the connection.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then $P(x,y)$ and $Q(x,y$) verify the Cauchy-Riemann equations, thus :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \frac{\partial P}{\partial x} = \frac{\partial Q}{\partial y} ,  \frac{\partial P}{\partial x} = -\frac{\partial Q}{\partial y} $$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Depending of oriented get that the integral is Zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This satisfies the Cauchy's integral theorem that an analytic function  on a closed curve is zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can see it &lt;a href=&quot;http://en.wikipedia.org/wiki/Cauchy%27s_integral_theorem&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;, where the proof of Cauchy's integral theorem  uses Green's Theorem .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A little deeper you can see, Complex Analysis by Lars Ahlfors, section 4.6 page 144.&lt;/p&gt;&#xA;" OwnerUserId="3421" LastEditorUserId="3421" LastEditorDisplayName="" LastEditDate="2010-11-29T03:26:53.690" LastActivityDate="2010-11-29T03:26:53.690" />
  <row Id="12279" PostTypeId="2" ParentId="12275" CreationDate="2010-11-29T02:57:21.143" Score="1" ViewCount="" Body="&lt;p&gt;Suppose $f$ is such that $f(x,0)$ and $f(0,y)$ are monotonic in $x$ and $y$ respectively over $[0,1]$. Since $f(x,0) = \Phi(g(x) + h(0))$, it must be the case that $g$ is monotonic over $[0,1]$. Similarly, $h$ is monotonic over $[0,1]$. So $g(x) + h(y)$ is the sum of monotonic functions. It's not hard to show that for any $(x,y)$ in the interior of $[0,1]^2$, there must be a point on the boundary with the same value of $g(x) + h(y)$, and therefore the same value of $\Phi(g(x) + h(y))$. Since it's easy to construct a continuous $f$ which satisfies the first assumption yet takes a larger range of values in the interior than it does on the boundary (for any $f$ that doesn't, try adding $M x(1-x)y(1-y)$ for large enough $M$), this contradicts the conjecture.&lt;/p&gt;&#xA;" OwnerUserId="856" LastEditorUserId="856" LastEditorDisplayName="" LastEditDate="2010-11-29T03:15:50.177" LastActivityDate="2010-11-29T03:15:50.177" CommentCount="2" />
  <row Id="12280" PostTypeId="2" ParentId="12275" CreationDate="2010-11-29T03:05:07.337" Score="8" ViewCount="" Body="&lt;p&gt;The statement is not true.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To make the counterexample simpler, I will take $f:[-1,1]^2 \rightarrow \mathbb{R}$.  Let &lt;em&gt;f(x,y)=xy&lt;/em&gt;.  So we have &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;---0+++&#xA;---0+++&#xA;0000000&#xA;+++0---&#xA;+++0---&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Consider the image of &lt;em&gt;xy=0&lt;/em&gt; in &lt;em&gt;g(x)+h(y)&lt;/em&gt;.  Since it's a connected, compact set, the image is connected and compact, so an interval [m,n], with $\Phi([m,n])=0$.  We may assume that $\Phi(n+\epsilon)&amp;gt;0$ and $\Phi(m-\epsilon)&amp;lt;0$.  Let &lt;em&gt;A=g(1)&lt;/em&gt;, &lt;em&gt;a=g(-1)&lt;/em&gt;, &lt;em&gt;B=h(1)&lt;/em&gt;, &lt;em&gt;b=h(-1)&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then we have $A+B&amp;gt;n, \; a+b&amp;gt;n, \;  A+b&amp;lt;m, \; a+B&amp;lt;m$, which leads to a contradiction.&lt;/p&gt;&#xA;" OwnerUserId="1461" LastEditorUserId="1461" LastEditorDisplayName="" LastEditDate="2010-11-29T04:47:34.353" LastActivityDate="2010-11-29T04:47:34.353" CommentCount="2" />
  <row Id="12281" PostTypeId="1" AcceptedAnswerId="12282" CreationDate="2010-11-29T03:13:36.400" Score="3" ViewCount="62" Body="&lt;p&gt;In propositional logic, for example:&#xA;$$\neg p \vee q.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $p$ is true at the outset, does that mean it must be considered false when comparing with q in the disjunction?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S. I am unsure about tags for this question.&lt;/p&gt;&#xA;" OwnerUserId="3540" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-29T03:23:09.577" LastActivityDate="2010-11-29T05:03:01.170" Title="Are standalone propositions affected by negation operators?" Tags="&lt;logic&gt;&lt;propositional-calculus&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="12282" PostTypeId="2" ParentId="12281" CreationDate="2010-11-29T03:17:14.653" Score="4" ViewCount="" Body="&lt;p&gt;If $p$ is true, then $\neg p$ is false. To evaluate $\neg p \vee q$, you must evaluate $\neg p$ and you must evaluate $q$. If either $\neg p$ is true or $q$ is true, then $\neg p\vee q$ is true. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, you really need to figure out $(\neg p)\vee q$, performing first the operation inside the parentheses, then the disjunction. &lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-11-29T03:17:14.653" />
  <row Id="12283" PostTypeId="2" ParentId="12281" CreationDate="2010-11-29T03:17:16.960" Score="4" ViewCount="" Body="&lt;p&gt;If $p$ is true, then $\lnot p \lor q \Leftrightarrow q$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, $p$ and $\lnot p$ have the opposite value: if one is true then the other is false, and vice versa.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can think of $p$ as some proposition, say &quot;today is Sunday&quot;. Then $\lnot p$ stands for &quot;today is &lt;i&gt;not&lt;/i&gt; Sunday&quot;.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-29T03:17:16.960" />
  <row Id="12284" PostTypeId="2" ParentId="12273" CreationDate="2010-11-29T03:21:59.810" Score="4" ViewCount="" Body="&lt;p&gt;Assuming that the numbers are positive integers,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is closely related to the &lt;a href=&quot;http://mathworld.wolfram.com/CoinProblem.html&quot; rel=&quot;nofollow&quot;&gt;Frobenius Coin Problem&lt;/a&gt; which says that there is a maximum number $\displaystyle F$ (called the Frobenius number) which is &lt;em&gt;not&lt;/em&gt; representable. It is NP-Hard to find out the Frobenius number when there are at least $\displaystyle 3$ numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a formula like approach to determine if such a representation is possible or not, you can use generating functions, which can be used to give a pseudo polynomial time algorithm, polynomial in size $\displaystyle W = n_1 + n_2 + \cdots + n_k$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the numbers are $\displaystyle n_1, n_2, \dots, n_k$ and you need to see if they can be summed to $\displaystyle S$ then the number of ways it can be done is the coefficient of $\displaystyle x^S$ in&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle (1+x^{n_1} + x^{2n_1} + x^{3n_1} + \cdots )(1+ x^{n_2} + x^{2n_2} +  x^{3n_2} + \cdots ) \cdots (1 + x^{n_k} + x^{2n_k} + x^{3n_k} + \cdots )$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle  = \dfrac{1}{(1-x^{n_1})(1-x^{n_2}) \cdots (1-x^{n_k})}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using partial fractions this can be written as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle \sum_{j=1}^{m} \dfrac{C_j}{c_j - x}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\displaystyle C_j$ and $\displaystyle c_j$ are appropriate complex numbers and $\displaystyle m \le n_1 + n_2 + \cdots + n_k$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The coefficient of $\displaystyle x^S$ is thus given by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle \sum_{j=1}^{m} \dfrac{C_j}{c_j^{S+1}}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which you need to check is zero or not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, this might require quite precise floating point operations and does not actually tell you what numbers to choose.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-11-29T17:53:34.207" LastActivityDate="2010-11-29T17:53:34.207" CommentCount="0" />
  <row Id="12285" PostTypeId="1" CreationDate="2010-11-29T03:24:32.057" Score="10" ViewCount="191" Body="&lt;p&gt;Why is the following true?  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If $(X,d)$ is a compact metric space and $f: X \rightarrow X$ is non-expansive (i.e $d(f(x),f(y)) \leq d(x,y)$) and surjective then $f$ is an isometry.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4219" LastEditorUserId="4219" LastEditorDisplayName="" LastEditDate="2010-11-29T03:36:07.740" LastActivityDate="2010-11-29T11:20:42.060" Title="Isometry in compact metric spaces" Tags="&lt;metric-spaces&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="2" />
  <row Id="12286" PostTypeId="2" ParentId="12276" CreationDate="2010-11-29T03:40:57.813" Score="17" ViewCount="" Body="&lt;p&gt;Absolutely yes!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In Section 7 of &lt;a href=&quot;http://math.uga.edu/~pete/handouteight.pdf&quot; rel=&quot;nofollow&quot;&gt;these notes on Green's Theorem&lt;/a&gt;, I explain how Green's Theorem plus the Cauchy-Riemann equations immediately yields the Cauchy Integral Theorem.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Probably most serious students of mathematics realize this on their own at some point, but it is surprising how few standard texts make this connection.  In (especially American) undergraduate texts on Subject X, there is a distressing tendency to politely ignore the existence of Subject Y, even when any student of Subject X will almost surely have already have studied / be concurrently studying / soon be studying Subject Y.&lt;/p&gt;&#xA;" OwnerUserId="299" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2011-01-30T20:32:01.167" LastActivityDate="2011-01-30T20:32:01.167" CommentCount="2" />
  <row Id="12287" PostTypeId="1" AcceptedAnswerId="12290" CreationDate="2010-11-29T04:23:53.613" Score="3" ViewCount="255" Body="&lt;p&gt;It is my question when I was in Senior High School. Up to now, I have no idea about the correct explanation. I just accept it by faith :D&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here  is the question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/HuwRD.gif&quot; alt=&quot;tangent line diagram&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the symbol $\Delta x \to 0$ does not mean $\Delta x =0$, how can the points $A$ and $B$ get overlapped which in turn, how can the line joining $A$ and $B$ become $C$, the tangent line to the curve $y=f(x)$ at point $A$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you in advance.&lt;/p&gt;&#xA;" OwnerUserId="3901" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-29T04:55:35.967" LastActivityDate="2010-11-29T15:37:20.493" Title="Approaching to zero, but not equal to zero, then why do the points get overlapped?" Tags="&lt;calculus&gt;&lt;analysis&gt;&lt;functions&gt;&lt;limit&gt;&lt;differential&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="1" />
  <row Id="12288" PostTypeId="2" ParentId="12287" CreationDate="2010-11-29T04:34:59.907" Score="7" ViewCount="" Body="&lt;p&gt;$A$ and $B$ do not literally become overlapped, and the line $C$ is not obtained literally as a line joining points $A$ and $B$ on the curve.  The diagram is designed to motivate the plausibility of the fact that as $B$ gets &quot;sufficiently close&quot; (but not equal) to $A$, the line connecting $A$ and $B$ gets &quot;arbitrarily close&quot; to the tangent line $C$ to the curve at the point $A$.  This is consistent with the idea that $\Delta x$ &quot;approaches $0$&quot;, but is never equal to $0$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;More precisely, the slope of the line $C$ is the derivative of $f$ at $x_0$, $f&amp;#39;(x_0)$, defined by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f&amp;#39;(x_0)=\lim_{\Delta x\to 0}\frac{f(x_0+\Delta x)-f(x_0)}{\Delta x}.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More informally again, this &lt;a href=&quot;http://en.wikipedia.org/wiki/Limit_of_a_function&quot; rel=&quot;nofollow&quot;&gt;limit&lt;/a&gt; means that you can get &quot;arbitrarily close&quot; to $f&amp;#39;(x_0)$ by taking $\Delta x$ &quot;sufficiently small&quot; (but not $0$) in the difference quotient $\frac{f(x_0+\Delta x)-f(x_0)}{\Delta x}$, which is the slope of the line connecting $A$ to $B$ in the diagram.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2010-11-29T05:25:53.573" LastActivityDate="2010-11-29T05:25:53.573" CommentCount="4" />
  <row Id="12289" PostTypeId="2" ParentId="12285" CreationDate="2010-11-29T04:37:03.290" Score="5" ViewCount="" Body="&lt;p&gt;In fact more is true. For a compact, metric space, $f$ locally non-expansive $\implies$ $f$ local isometry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is theorem 4.2 in the paper linked below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://projecteuclid.org/DPubS/Repository/1.0/Disseminate?view=body&amp;amp;id=pdf_1&amp;amp;handle=euclid.pjm/1102970464&quot; rel=&quot;nofollow&quot;&gt;http://projecteuclid.org/DPubS/Repository/1.0/Disseminate?view=body&amp;amp;id=pdf_1&amp;amp;handle=euclid.pjm/1102970464&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3431" LastActivityDate="2010-11-29T04:37:03.290" CommentCount="1" />
  <row Id="12290" PostTypeId="2" ParentId="12287" CreationDate="2010-11-29T04:51:14.567" Score="14" ViewCount="" Body="&lt;p&gt;Basically, what you have put your finger on is the original difficulty with &lt;a href=&quot;http://en.wikipedia.org/wiki/Infinitesimals&quot; rel=&quot;nofollow&quot;&gt;infinitesimals&lt;/a&gt;, and the basis on which Bishop Berkeley made some of his famous objections (mind you, Berkeley didn't really care about the logical foundations of Calculus; he was engaged in a theological debate at the time&lt;sup&gt;1&lt;/sup&gt;): if infinitesimals are not zero, then what you have is not a tangent but a secant, because the line touches the curve at two distinct points; but if the two points are the same, then they don't define a tangent because you cannot determine a line with a single point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer is that when we talk about limits, we are talking about what the quantities are &lt;em&gt;approaching&lt;/em&gt;, not what the quantities &lt;em&gt;are&lt;/em&gt;. The point $B$ never &quot;gets overlapped&quot; with $A$, it just &lt;em&gt;approaches&lt;/em&gt; $A$; the line between $A$ and $B$ never &quot;becomes&quot; the tangent (which in your diagram is $C$), but its slope &lt;em&gt;approaches&lt;/em&gt; the slop of $c$. These &quot;approaches&quot; have a very precise meaning (made formal by Weierstrass).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, what would it mean to say that the values of a function $g(x)$ &quot;approach&quot; the value $L$ as $x$ approaches $a$? It means that you can make all the values of $g(x)$ be arbitrarily close to $L$ provided that $x$ is &quot;close enough&quot; to $a$. If you specify a narrow horizontal band around  the value $y=L$, then by specifying a narrow vertical band around the value $a$ we can guarantee that the graph of $y=g(x)$ for values of $x$ inside the narrow vertical band will necessarily lie inside the horizontal line.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason this is sensible is that if the values of $g(x)$ approach some other number $M$ as well, then by specifying a band around $L$ which is smaller than the distance from $M$ to $L$ you will always run into problems: the graph of $y=g(x)$ will always end up with parts outside this band, because $M$ is outside the band and you are also approaching $M$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Formally, we say it like this: the limit of $g(x)$ as $x$ approaches $a$ is $L$, which is written:&#xA;$$\lim_{x\to a}g(x) = L$$&#xD;&#xA;if and only if for every $\epsilon\gt 0$ there exists a $\delta\gt 0$ such that if $0\lt |x-a|\lt \delta$, then $|f(x)-L|\lt \epsilon$; $\epsilon$ is how wide the horizontal band around $L$ is, $\delta$ is how thin you need to make the vertical band around $a$ needs to be to make sure the graph is completely inside the rectangle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What the picture suggests is that as $\Delta x$ approaches $0$, the slope of the line joining $A$ and $B$  &lt;em&gt;should&lt;/em&gt; approach the slope of the tangent at $a$; however, the line joining $A$ and $B$ never actually &quot;becomes&quot; the tangent $C$, and the point $B$ never actually &quot;becomes&quot; the point $A$. They are just approaching.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, does the line joining $A$ and $B$ have a slope that &lt;em&gt;really&lt;/em&gt; approaches the slope of the tangent? The key is that we can characterize the tangent algebraically: the tangent is the unique line that affords the best approximation to $y=f(x)$ near the point $x_0$, where by &quot;best approximation&quot; we mean one in which the relative error approaches zero. That is: if we take a line $mx+b$ that goes through $(x_0,f(x_0))$, then we can ask for the &quot;relative error&quot; in approximating using $mx+b$ instead of $f(x)$:&#xA;$$\frac{f(x) - (mx+b)}{x-x_0}.$$&#xD;&#xA;The tangent is the unique line through $(x_0,f(x_0))$ for which the relative error approaches $0$ as $x$ approaches $x_0$ (if such a line exists at all). Since $y=mx+b$ goes through $(x_0,f(x_0))$ if and only if $f(x_0) = mx_0 + b$, then $b= f(x_0)-mx_0$; so we have that the relative error will be&#xA;$$\frac{f(x) - mx - f(x_0)+mx_0}{x-x_0} = \frac{f(x)-f(x_0) - m(x-x_0)}{x-x_0} = \frac{f(x)-f(x_0)}{x-x_0} - m.$$&#xD;&#xA;So if we want the relative error to approach $0$, then we need &#xA;$\frac{f(x)-f(x_0)}{x-x_0}$&#xD;&#xA;to approach $m$; that is, the slope of the tangent &lt;em&gt;must&lt;/em&gt; be whatever quantity the numbers&#xA;$$\frac{f(x)-f(x_0)}{x-x_0}$$&#xD;&#xA;are approaching as $x$ approaches $x_0$ (if they are approaching any particular number). Setting $\Delta x = x-x_0$, so that $x = x_0+\Delta x$, the above fraction is the same as&#xA;$$\frac{f(x_0 + \Delta x) - f(x_0)}{\Delta x}.$$&#xD;&#xA;And $x$ approaches $x_0$ if and only if $\Delta x$ approaches $0$. So the slope of the tangent needs to be whatever quantity this fraction approaches as $\Delta x$ approaches $0$; that is, the slope of the tangent at $x_0$ is $m$ if and only if &#xA;$$ \lim_{\Delta x \to 0} \frac{f(x_0+\Delta x)-f(x_0)}{\Delta x} = m.$$&#xD;&#xA;Turning this assertion into a picture gives precisely the picture you have.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;hr/&gt;&#xA;&lt;sup&gt;1&lt;/sup&gt; In case anyone is interested: many people viewed Newtonian mechanics and its consequent &quot;clockwork universe&quot; as a direct attack on the Christian notion of a deity that was directly involved in and modifying his creation, and also on the notion of free will. Some deists were in fact using the new physics as evidence in favor of the deist God, who created the universe, set it motion, but does not actively participate in it, in contrast to the theistic deity. By attacking the mathematical foundation of the new physics, Berkeley was defending the notions of free will and of the active deity. If you read Augustus de Morgan's &lt;em&gt;A Budget of Paradoxes&lt;/em&gt;, he reviews many pamphlets and booklets written during those years which attack Newton and Calculus because they view the latter as an attack on religion. The &quot;morality&quot; arguments raised against Calculus are eerily similar (when not downright &lt;em&gt;identical&lt;/em&gt;) to those raised against Darwin and the Theory of Evolution in later days. &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-29T15:37:20.493" LastActivityDate="2010-11-29T15:37:20.493" CommentCount="5" />
  <row Id="12291" PostTypeId="1" AcceptedAnswerId="12535" CreationDate="2010-11-29T04:58:08.367" Score="6" ViewCount="91" Body="&lt;p&gt;Given a Noetherian, local ring $(R,m)$, can we always find a system of parameters whose images in the cotangent space $m/m^2$ are linearly independent. We can do this in the regular case, by just choosing a basis for the cotangent space and looking at their preimages in $m$ under the canonical map. By Nakayama's lemma these generate $m$ and hence form a system of parameters. Can we do this for any general Noetherian, local ring?&lt;/p&gt;&#xA;" OwnerUserId="3947" LastActivityDate="2010-11-30T21:29:09.977" Title="System of parameters which have linear independent images in the cotangent space" Tags="&lt;algebraic-geometry&gt;&lt;commutative-algebra&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="12292" PostTypeId="2" ParentId="12281" CreationDate="2010-11-29T05:03:01.170" Score="1" ViewCount="" Body="&lt;p&gt;I'm not sure I understand your question, but this may help.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Truth Table for ~p v q:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;~ p v q&lt;br&gt;&#xA;F T T T&lt;br&gt;&#xA;F T F F&lt;br&gt;&#xA;T F T T&lt;br&gt;&#xA;T F T F&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If p is true, and ~p v q is true (first line only), then q is true.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that ~p v q is logically equivalent to p =&gt; q. &lt;/p&gt;&#xA;" OwnerUserId="3515" LastActivityDate="2010-11-29T05:03:01.170" />
  <row Id="12293" PostTypeId="2" ParentId="12287" CreationDate="2010-11-29T05:22:33.170" Score="0" ViewCount="" Body="&lt;p&gt;You have already got good answers. However, I would just want to add a suggestion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Take a look at the diagram and picture other points $B_1,B_2,\ldots$ closer and closer to $A$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Think of the lines $L_1,L_2,\ldots$ joining $A$ with and the points $B_1,B_2,\ldots$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you see how the lines approach the tangent line $C$?&lt;/p&gt;&#xA;" OwnerUserId="1154" LastActivityDate="2010-11-29T05:22:33.170" />
  <row Id="12294" PostTypeId="2" ParentId="12223" CreationDate="2010-11-29T05:31:29.347" Score="1" ViewCount="" Body="&lt;p&gt;Counter example: Suppose p, q, r are all false. Then (p=&gt;q)=&gt;r is false, and p=&gt;(q=&gt;r) is true.&lt;/p&gt;&#xA;" OwnerUserId="3515" LastActivityDate="2010-11-29T05:31:29.347" CommentCount="3" />
  <row Id="12295" PostTypeId="1" AcceptedAnswerId="12297" CreationDate="2010-11-29T05:48:06.060" Score="1" ViewCount="67" Body="&lt;p&gt;I encountered a problem which is asks for the first negative term in the expansion of $(1+x)^{\frac{27}{5}}$, where $x$ is always positive,the solution suggested in my module is like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle \text{ For the first negative term: } n-r+1 \lt 0 \text{ or, } r \gt 6\frac{2}{5}$, since $n = \frac{27}{5}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, the 8th term is the first negative term.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But my problem is that I am not getting the rationale behind this approach.Also,if you are aware of any other method for the same problem you may like to post too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS:Inquisitive to verify the same using Mathematica,I tried &lt;code&gt;Expand[(1 + x)^(27/5)]&lt;/code&gt; but I guess this is not the correct function.&lt;/p&gt;&#xA;" OwnerUserId="2109" LastEditorUserId="2109" LastEditorDisplayName="" LastEditDate="2010-11-29T05:53:25.327" LastActivityDate="2010-11-29T06:07:27.920" Title="First non-negative term in an Expansion" Tags="&lt;algebra-precalculus&gt;&lt;algebra&gt;" AnswerCount="1" />
  <row Id="12296" PostTypeId="1" CreationDate="2010-11-29T05:48:19.867" Score="0" ViewCount="91" Body="&lt;p&gt;Let H be a subgroup of G and V the trivial representation of C[H]. Prove that we have an isomorphism of C[G]-modules between $\mathrm{Ind}_{H}^{G}(V)$ and C[G/H].&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for your help.&lt;/p&gt;&#xA;" LastActivityDate="2010-11-29T06:23:52.073" Title="Induced representation" Tags="&lt;representation-theory&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="12297" PostTypeId="2" ParentId="12295" CreationDate="2010-11-29T05:52:47.003" Score="2" ViewCount="" Body="&lt;p&gt;Hint: Use generalized binomial theorem, when does a negative factor appear in the binomial coefficient. (I am asssuming $x$ is positive here and $n,r$ have their usual meanings).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;${n\choose r}=\frac{(n)(n-1)...(n-r+1)}{r!}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Binomial_theorem#Newton.27s_generalized_binomial_theorem&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Binomial_theorem#Newton.27s_generalized_binomial_theorem&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: $r,x$ are positive here. So the only way we can get a negative term is having a negative coefficient and hence only by having a negative numerator of the coefficient. Since the numerator is a product of real numbers it first becomes negative when the last factor is negative (since the first factor is positive and the value of the factors keeps decreasing by one successively). So this happens when $n-r+1&amp;lt;0$. Set $n=\frac{27}{5}$, so $\frac{27}{5}-r+1&amp;lt;0\implies \frac{32}{5}-r&amp;lt;0 \implies r&amp;gt;\frac{32}{5}=6\frac{2}{5}$&lt;/p&gt;&#xA;" OwnerUserId="3431" LastEditorUserId="3431" LastEditorDisplayName="" LastEditDate="2010-11-29T06:07:27.920" LastActivityDate="2010-11-29T06:07:27.920" CommentCount="5" />
  <row Id="12298" PostTypeId="2" ParentId="11834" CreationDate="2010-11-29T06:11:21.257" Score="6" ViewCount="" Body="&lt;p&gt;A directed multigraph can be thought of as a one-dimensional simplicial complex: the vertices are points, and the edges are 1-simplices.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given a category, we can form its nerve, and the 1-skeleton will essentially be a multi-graph.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The observation of the OP is that these two constructions are roughly inverse to each other.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The relationship between simplicial sets and categories is an important one,&#xA;and is at the basis of the theory of higher categories and its relationship to homotopy theory.  Thus the OP's observation and question shouldn't be dismissed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, I'm not sure that I agree with the sentiment of the OP's final non-parenthetical sentence:&#xA;&quot;I don't see how it justifies treating the two as separate branches rather than syntactic shims on identical concepts.&quot;   Certainly (multi-)graph theorists and (higher-)category/homotopy theorists both make itensive studies of 1-dimensional&#xA;simplicial sets,&#xA;but the questions they ask and the constructions they are concerned with are typically extremely different.   If they do find common ground, that's great, but they needn't be obliged to seek it.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most working mathematicians using categories are, furthermore, not thinking in a homotopical/simplicial way at all about the structures involved, but are just taking advantage of the very useful language and range of concepts that category theory provides.  These are concepts that (as far as I know) are not used at all by those studying multi-graphs and related structures, and so I don't see any reason that people should be obliged or even encouraged to translate (what to them are) familiar category-theoretic notions into a different terminological framework of relations/multi-graphs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[Added: Ryan Budney's comment above makes essentially the same point in a pithier fashion.]&lt;/p&gt;&#xA;" OwnerUserId="221" LastEditorUserId="221" LastEditorDisplayName="" LastEditDate="2010-11-29T07:51:23.107" LastActivityDate="2010-11-29T07:51:23.107" CommentCount="1" />
  <row Id="12299" PostTypeId="1" AcceptedAnswerId="12303" CreationDate="2010-11-29T06:20:14.720" Score="2" ViewCount="127" Body="&lt;p&gt;Hi, my Calculus II class is nearing the end of the quarter and we've just started differential equations to get ready for Calculus III. In my homework, I came upon these problems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of the problems was:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Find the general solution to the differential equation &#xA;$$\frac{dy}{dt} = t^3 + 2t^2 - 8t.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The teacher just said to integrate. So I did. Then in question 8a it gives the differential equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{dy}{dt} = y^3 + 2y^2 - 8y.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and asks &quot;Why can't we find a solution like we did to the previous problem? My guess was: &quot;In 7 we were integrating with respect to t. Since this equation is the highest order derivative, we can't solve it like # 7&quot;. Although, I have no confidence in that answer and I'm not sure it makes total sense even to me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, part 8B. asks: Show that the constant function $y(t) = 0$ is a solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've done a problem like this before, except that it wasn't a constant function. This problem seems like a question that asks: &quot;show that every member of the family of functions $y = (\ln x + C)/x$ is a solution to the differential equation (some diff. equation)&quot; except it seems a little bit different.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any hints on how I can solve this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you.&lt;/p&gt;&#xA;" OwnerUserId="3514" LastEditorUserId="3514" LastEditorDisplayName="" LastEditDate="2010-11-29T06:31:55.743" LastActivityDate="2010-11-29T06:52:08.630" Title="Help solving a differential equation" Tags="&lt;homework&gt;" AnswerCount="3" CommentCount="0" />
  <row Id="12300" PostTypeId="2" ParentId="12296" CreationDate="2010-11-29T06:23:52.073" Score="1" ViewCount="" Body="&lt;p&gt;For instance, one of the standard descriptions of the induced module $\operatorname{Ind}_H^G(V)$ is as a direct sum $\bigoplus_{s \in S} s V$ of copies of $V$ indexed by a set $S$ of coset representatives of $G/H$ with a certain natural $G$-action.  See for instance Lemma 2.1 &lt;a href=&quot;http://math.berkeley.edu/~serganov/math252/notes4.pdf&quot; rel=&quot;nofollow&quot;&gt;in these notes&lt;/a&gt;.  (I have no particular connection to these course notes; they just came up as one of the first hits upon googling &quot;induced module&quot;.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you are trying to prove follows almost immediately from this description of the induced module.  Indeed, in the aforementioned lecture notes, your exercise appears as Example 1 on the following page (with no further explanation).  If you can't see how to deduce this, you should probably book up (and possibly ask further questions) on the definition of an induced module.  &lt;/p&gt;&#xA;" OwnerUserId="299" LastActivityDate="2010-11-29T06:23:52.073" />
  <row Id="12301" PostTypeId="2" ParentId="12299" CreationDate="2010-11-29T06:30:20.560" Score="3" ViewCount="" Body="&lt;p&gt;The difference is that the main variable is $y$ and you separate the variables to integrate and obtain the solution&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint 1:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first equation is very easy, so :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{dy}{dt}=t^3 + 2t^2 - 8t \Rightarrow dy = (t^3 + 2t^2 - 8t) dt \Rightarrow $$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$  \int dy = \int (t^3 + 2t^2 - 8t) dt $$   ( which is the left side ? )&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint2:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the other.&#xA;$$\frac{dy}{dt}=y^3 + 2y^2 - 8y \Rightarrow  \frac{1}{y^3 + 2y^2 - 8y}dy =  dt $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then proceed integrating.&lt;/p&gt;&#xA;" OwnerUserId="3421" LastEditorUserId="3421" LastEditorDisplayName="" LastEditDate="2010-11-29T06:52:08.630" LastActivityDate="2010-11-29T06:52:08.630" />
  <row Id="12302" PostTypeId="1" CreationDate="2010-11-29T06:30:21.690" Score="2" ViewCount="144" Body="&lt;p&gt;Find the value of:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ 1+ \biggl(\frac{1}{10}\biggr)^2 + \frac{1 \cdot 3}{1 \cdot 2} \biggl(\frac{1}{10}\biggr)^4 + \frac{1\cdot 3 \cdot 5}{1 \cdot 2 \cdot 3} \biggl(\frac{1}{10}\biggr)^6 + \cdots  $$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="2109" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-29T06:35:56.060" LastActivityDate="2010-11-29T07:17:31.593" Title="Suggest a tricky method for this problem" Tags="&lt;sequences-and-series&gt;" AnswerCount="3" />
  <row Id="12303" PostTypeId="2" ParentId="12299" CreationDate="2010-11-29T06:34:44.813" Score="3" ViewCount="" Body="&lt;p&gt;Parts of problems that say &quot;show that [explicitly given function] is a solution&quot; can be solved by simply &quot;plugging in&quot;.  E.g., if you're supposed to show that $y(t)=\frac{1}{1-t}$ is a solution to the differential equation $dy/dt=y^2$, then you should compute $dy/dt=\frac{1}{(1-t)^2}$, compute $y^2=\frac{1}{(1-t)^2}$, and &quot;plug them in&quot; to check that the differential equation is satisfied for this choice of $y$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You're correct that the difference in the second problem is that you no longer have the derivative with respect to $t$ expressed as a function of $t$.  When the equation involves both a function $y$ and its derivative (or derivatives), it can generally be more difficult to solve.  In this case, a hint I'll give you is that $dy/dt=1/(dt/dy)$ (e.g., see &lt;a href=&quot;http://en.wikipedia.org/wiki/Derivative_rule_for_inverses&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;). You can first turn things around and think of $t$ as a function of $y$.  You can find $t(y)$ using similar methods to the first problem (but with a more complicated function), and then find the inverse function to get $y(t)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A somewhat more general method of solving differential equations that applies in this case is &lt;a href=&quot;http://en.wikipedia.org/wiki/Separation_of_variables&quot; rel=&quot;nofollow&quot;&gt;separation of variables&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-11-29T06:34:44.813" />
  <row Id="12304" PostTypeId="2" ParentId="12299" CreationDate="2010-11-29T06:34:50.583" Score="1" ViewCount="" Body="&lt;p&gt;In the first question, you are given the derivative in terms of the variable. But in the second question, you are given an expression for the derivative that involves &lt;em&gt;the function&lt;/em&gt;. For instance, it would be one thing if you were told $\frac{dy}{dx} = x$ (which would mean that $y = \frac{1}{2}x^2 + C$), and a completely different thing if you are told $\frac{dy}{dx}=y$ (this tells you that the function $y$ is equal to its derivative; which means that $y=Ae^x$ for some constant $A$). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Actually, we &lt;em&gt;can&lt;/em&gt; solve the second differential equation; but we don't solve it by simple integration, just like we don't solve the equation $y&amp;#39; = y$ by integrating. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Showing that $y(t)=0$ is a solution to the second equation is just a question of plugging in $0$ for $y$ and verifying that you get a solution by verifying that the resulting equality is true.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You never said what question 8a was, through; were you asked to &lt;em&gt;solve&lt;/em&gt; the differential equation, or just to say what is the difference between it and the previous one?&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-11-29T06:34:50.583" />
  <row Id="12305" PostTypeId="2" ParentId="12302" CreationDate="2010-11-29T06:50:49.783" Score="6" ViewCount="" Body="&lt;p&gt;Hint: What is the generalized binomial expansion of $(1-2(\frac{1}{10})^2)^{-\frac{1}{2}}$&lt;/p&gt;&#xA;" OwnerUserId="3431" LastActivityDate="2010-11-29T06:50:49.783" CommentCount="3" />
  <row Id="12306" PostTypeId="2" ParentId="12302" CreationDate="2010-11-29T06:50:51.537" Score="5" ViewCount="" Body="&lt;p&gt;Note that we can rewrite the series as &#xA;$$ \displaystyle\sum\limits_{n=0}^\infty \frac{(2n)!}{2^n(n!)^2} \cdot \bigg(\frac{1}{10}\bigg)^{2n}$$&#xD;&#xA;which is exactly&#xA;$$ \displaystyle\sum\limits_{n=0}^\infty \frac{(2n)!}{2^n(n!)^2} \cdot x^{2n}$$&#xD;&#xA;evaluated at $x = \frac{1}{10}$. It is easy to see that this sum has radius of convergence $\frac{1}{\sqrt{2}}$, and so converges absolutely for $x = \frac{1}{10}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also note that this sum is exactly the Taylor series for $\frac{1}{\sqrt{1-2x^2}}$ centered at $x = 0$. Thus&#xA;$$ \displaystyle\sum\limits_{n=0}^\infty \frac{(2n)!}{2^n(n!)^2} \cdot \bigg(\frac{1}{10}\bigg)^{2n} = \frac{10}{7\sqrt{2}}$$&lt;/p&gt;&#xA;" OwnerUserId="1016" LastActivityDate="2010-11-29T06:50:51.537" />
  <row Id="12307" PostTypeId="1" CreationDate="2010-11-29T06:56:47.140" Score="2" ViewCount="355" Body="&lt;p&gt;I know from an online calculator &lt;a href=&quot;http://www.numberempire.com/derivatives.php&quot; rel=&quot;nofollow&quot;&gt;http://www.numberempire.com/derivatives.php&lt;/a&gt; that &#xA;$\lim_{x \to \infty} \left ( \frac{x+2}{x} \right )^{x}=e^{2}$.  How do you calculate this step by step?&lt;/p&gt;&#xA;" OwnerUserId="4192" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2010-11-29T11:09:38.797" LastActivityDate="2010-11-29T18:12:27.953" Title="How to calculate $\displaystyle\lim_{x \to \infty} \left ( \frac{x+2}{x} \right )^{x}=e^{2}$" Tags="&lt;calculus&gt;" AnswerCount="5" CommentCount="2" FavoriteCount="1" />
  <row Id="12308" PostTypeId="2" ParentId="12307" CreationDate="2010-11-29T07:03:24.933" Score="7" ViewCount="" Body="&lt;p&gt;This is more or less by definition of $e$, depending on which definition you use. Do you know the definition $e=\lim_{n\rightarrow\infty} (1+\frac{1}{n})^n$? Then set $x=2n$ and you are done.&lt;/p&gt;&#xA;" OwnerUserId="3212" LastActivityDate="2010-11-29T07:03:24.933" CommentCount="3" />
  <row Id="12309" PostTypeId="2" ParentId="12307" CreationDate="2010-11-29T07:04:26.867" Score="2" ViewCount="" Body="&lt;p&gt;Rewriting the function whose limit is taken $$\left(1 + \frac{2}{x}\right)^x,$$ we immediately recognize the definition of $e^2$.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-29T07:04:26.867" />
  <row Id="12310" PostTypeId="1" CreationDate="2010-11-29T07:12:34.383" Score="1" ViewCount="65" Body="&lt;p&gt;Three planes &lt;em&gt;A&lt;/em&gt;, &lt;em&gt;B&lt;/em&gt; and &lt;em&gt;C&lt;/em&gt; intersect at point P.  The dihedral angle between &lt;em&gt;A&lt;/em&gt; and &lt;em&gt;B&lt;/em&gt; is $\theta$ and the dihedral angle between &lt;em&gt;B&lt;/em&gt; and &lt;em&gt;C&lt;/em&gt; is $\psi$.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Solve for the dihedral angle between by &lt;em&gt;A&lt;/em&gt; and &lt;em&gt;C&lt;/em&gt;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Planes &lt;em&gt;A&lt;/em&gt; and &lt;em&gt;B&lt;/em&gt; intersect plane &lt;em&gt;C&lt;/em&gt; in two lines that form angle QPR.  That is, point Q lies in both &lt;em&gt;A&lt;/em&gt; and &lt;em&gt;C&lt;/em&gt;, and point R lies in both &lt;em&gt;B&lt;/em&gt; and &lt;em&gt;C&lt;/em&gt;. Solve for the angle of QPR in terms of $\theta$ and $\psi$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Does this fall under trigonometry? If not, what?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="4158" LastActivityDate="2010-11-29T09:04:12.357" Title="Solve for the third dihedral angle in an intersection of three planes." Tags="&lt;trigonometry&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="12311" PostTypeId="2" ParentId="12302" CreationDate="2010-11-29T07:17:31.593" Score="2" ViewCount="" Body="&lt;p&gt;This can also be done using Wallis' product.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance see a similar problem: &lt;a href=&quot;http://math.stackexchange.com/questions/3763/summing-the-series-frac12n1-frac12-cdot-frac12n3-cdots/3766#3766&quot;&gt;Summing the series $ \frac{1}{2n+1} + \frac{1}{2} \cdot \frac{1}{2n+3} + \cdots \ \text{ad inf}$ &lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://math.stackexchange.com/questions/3502/summing-the-series-1k-frac2k2k1-a2k1&quot;&gt;Summing the series $(-1)^k \frac{(2k)!!}{(2k+1)!!} a^{2k+1}$&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Use&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle \dfrac{2}{\pi} \int_{0}^{\pi/2} \sin^{2n} x \ \text{dx} = \dfrac{1 \cdot 3 \cdot 5 \cdots (2n-1)}{2 \cdot 4 \cdot 6 \cdots 2n} =  \dfrac{1}{2^n} \dfrac{1 \cdot 3 \cdot 5 \cdots (2n-1)}{1 \cdot 2 \cdot 3 \cdots n}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle \dfrac{2}{\pi} \int_{0}^{\pi/2} \left(\dfrac{2\sin^2 x}{100}\right)^n \ \text{dx} = \dfrac{1 \cdot 3 \cdot 5 \cdots (2n-1)}{1 \cdot 2 \cdot 3 \cdots n} \dfrac{1}{10^{2n}}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus your sum is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle  \sum_{n=0}^{\infty} \dfrac{2}{\pi} \int_{0}^{\pi/2} \left(\dfrac{2\sin^2 x}{100}\right)^n \ \text{dx} =  \dfrac{2}{\pi} \int_{0}^{\pi/2} \sum_{n=0}^{\infty} \left(\dfrac{2\sin^2 x}{100}\right)^n  \ \text{dx}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle = \dfrac{2}{\pi}\int_{0}^{\pi/2} \dfrac{100}{100 - 2\sin^2x} \ \text{dx} = \dfrac{2}{\pi} \dfrac{5\pi}{7\sqrt{2}} = \dfrac{10}{7\sqrt{2}}$$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-11-29T07:17:31.593" CommentCount="2" />
  <row Id="12312" PostTypeId="2" ParentId="12307" CreationDate="2010-11-29T07:17:38.487" Score="1" ViewCount="" Body="&lt;p&gt;Find a step-by-step derivation here (and not only for this special problem):&#xA;&lt;a href=&quot;http://www.wolframalpha.com/input/?i=limit+((x%2B2)/x&quot; rel=&quot;nofollow&quot;&gt;http://www.wolframalpha.com/input/?i=limit+((x%2B2)/x&lt;/a&gt;)^x+x-%3Eoo&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(if the link doesn't work copy and paste the whole line)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;...or go to &lt;a href=&quot;http://www.wolframalpha.com&quot; rel=&quot;nofollow&quot;&gt;http://www.wolframalpha.com&lt;/a&gt; directly and type:&lt;br&gt;&#xA;&quot;limit ((x+2)/x)^x x-&gt;oo&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Click on &quot;Show steps&quot; - Done! ;-)&lt;/p&gt;&#xA;" OwnerUserId="346" LastActivityDate="2010-11-29T07:17:38.487" />
  <row Id="12313" PostTypeId="2" ParentId="12239" CreationDate="2010-11-29T07:25:28.463" Score="2" ViewCount="" Body="&lt;p&gt;Hint for a less convoluted argument: Unless $C$ is zero (in which case the statement is false), there are vectors $x$ and $y$ such that $x^T C y$ is a nonzero $1\times 1$ matrix. Now consider $v=[x,ky]^T$ for large $k$ (positive or negative, depending on the sign of $x^T C y$).&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2010-11-29T07:25:28.463" />
  <row Id="12314" PostTypeId="1" AcceptedAnswerId="12315" CreationDate="2010-11-29T07:53:59.113" Score="7" ViewCount="153" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could someone give me an example of a (smooth) $n$-manifold $(n=2, 3)$ which cannot be embedded (or immersed) in $\mathbb R^4$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;S. L.&lt;/p&gt;&#xA;" OwnerUserId="3208" LastActivityDate="2010-11-29T09:19:58.180" Title="Examples of manifolds that cannot be embedded in $\mathbb R^4$" Tags="&lt;topology&gt;&lt;differential-topology&gt;" AnswerCount="2" />
  <row Id="12315" PostTypeId="2" ParentId="12314" CreationDate="2010-11-29T08:02:02.410" Score="13" ViewCount="" Body="&lt;p&gt;All 2-manifolds embed in $\mathbb R^4$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $3$-manifolds, $\mathbb RP^3$ doesn't embed in $\mathbb R^4$.  An interesting aspect for 3-manifolds in $\mathbb R^4$ is the subject fractures into tame topological embeddings and smooth/PL embeddings.  For example, Poincare Dodecahedral space has a tame topological embedding in $\mathbb R^4$ but not a smooth or PL-embedding. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want more examples, check out this pre-print: &lt;a href=&quot;http://front.math.ucdavis.edu/0810.2346&quot; rel=&quot;nofollow&quot;&gt;http://front.math.ucdavis.edu/0810.2346&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="642" LastActivityDate="2010-11-29T08:02:02.410" CommentCount="1" />
  <row Id="12316" PostTypeId="1" AcceptedAnswerId="12365" CreationDate="2010-11-29T08:47:53.323" Score="1" ViewCount="141" Body="&lt;p&gt;In Linear Algebra, when computing an inner product $&amp;lt;x,y&amp;gt; = y^*Wx$, what is the name of the matrix W?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it doesn't have a name, where can I find a practical explanation of how to construct it for a particular problem or space?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a text on the subject that explains this simply? A note on my background, I am approaching this from the perspective of an engineering student and not that of a mathematician; I don't have an understanding of the finer points of topology or differential geometry (not yet at least :) ).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe that this is the same matrix used in vector calculus to perform a change of variables. As in, we perform our change of variables, then multiply the new expression by $\frac{det(J(W))}{det(J(V))}$ where $W$ is this magic matrix in the new space, $V$ is this magic matrix the old space, and J(X) is the Jacobian operator.  Am I correct?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strike&gt;Also, is this related to one of the the matrices that comes up in Singular Value Decomposition (namely the diagonal matrix containing the singular values)?&lt;/strike&gt;&lt;/p&gt;&#xA;" OwnerUserId="4194" LastEditorUserId="4194" LastEditorDisplayName="" LastEditDate="2010-11-29T18:18:01.140" LastActivityDate="2010-11-29T18:18:01.140" Title="What is the name of the matrix used to weight an inner product?" Tags="&lt;linear-algebra&gt;&lt;differential-geometry&gt;&lt;inner-product-space&gt;" AnswerCount="1" CommentCount="6" FavoriteCount="1" />
  <row Id="12317" PostTypeId="2" ParentId="12310" CreationDate="2010-11-29T09:04:12.357" Score="1" ViewCount="" Body="&lt;p&gt;Question 1: The third angle is not uniquely determined by $\theta$ and $\psi$. If $a$, $b$ and $c$ are the unit normal vectors to the planes $A$, $B$ and $C$, then the problem is equivalent to determining $a \cdot c$ from $a \cdot b$ and $c \cdot b$, which cannot be done. (Think of $b$ being fixed; then giving values of $a \cdot b$ and $c \cdot b$ constrains $a$ and $c$ to lie on certain cones around $b$, but the positions of $a$ and $c$ on these cones are arbitrary.) &lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2010-11-29T09:04:12.357" />
  <row Id="12318" PostTypeId="2" ParentId="11962" CreationDate="2010-11-29T09:12:29.183" Score="1" ViewCount="" Body="&lt;p&gt;I found the following references by Richard C. Bradley:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;http://www.emis.de/journals/PS/images/ps-2005-37baf3.pdf&quot; rel=&quot;nofollow&quot;&gt;&quot;Basic Properties of Strong Mixing Conditions. A Survey and Some Open Questions&quot;&lt;/a&gt;;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;http://www.kendrickpress.com/MixingConditions.html&quot; rel=&quot;nofollow&quot;&gt;&quot;Introduction to Strong Mixing Conditions, Volumes 1-3&quot;&lt;/a&gt;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I've only been able to have a look at the first one; not very intuitive (to me, at least), but still a great overview.&lt;/p&gt;&#xA;" OwnerUserId="689" LastActivityDate="2010-11-29T09:12:29.183" />
  <row Id="12319" PostTypeId="2" ParentId="12314" CreationDate="2010-11-29T09:19:58.180" Score="7" ViewCount="" Body="&lt;p&gt;No compact nonorientable $(n-1)$-manifold embeds in $\mathbb R^n$:&#xA;this follows from the Alexander duality theorem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Immersibility is a harder problem....&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-11-29T09:19:58.180" />
  <row Id="12320" PostTypeId="1" AcceptedAnswerId="12321" CreationDate="2010-11-29T09:30:07.420" Score="3" ViewCount="267" Body="&lt;p&gt;I am looking for a correct proof of this statement: If $G$ is a group such that $G/Z(G)$ is cyclic, then $G$ is commutative.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proof: $G/Z(G)$ is isomorphic to $\operatorname{Inn}(G)$ and is cyclic, and then for every $a$ and $b$ in $G$ the inner isomorphisms $\gamma_a$ and $\gamma_b$ satisfy $\gamma_a \gamma_b = \gamma_{ab} = \gamma_{ba} = \gamma_b \gamma_a$, and therefore for every $a,b \in G$, $ab = ba$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is that proof complete, or am I missing something? Thanks a lot for the help.&lt;/p&gt;&#xA;" OwnerUserId="4195" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2010-11-29T21:38:36.243" LastActivityDate="2010-11-29T21:38:36.243" Title="Proof that if group $G/Z(G)$ is cyclic, then $G$ is commutative." Tags="&lt;group-theory&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="12321" PostTypeId="2" ParentId="12320" CreationDate="2010-11-29T09:44:29.367" Score="3" ViewCount="" Body="&lt;p&gt;A problem with your attempt is that you seem to want to use only the fact that the inner automorphism group is abelian, but this does not suffice.  There are nonabelian groups $G$ such that $G/Z(G)$ is abelian, like the group of symmetries of the square.  Thus, it does not follow from $\gamma_{ab}=\gamma_{ba}$ that $ab=ba$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you need to use the hypothesis that $G/Z(G)$ has a single generator.  Robin Chapman pointed out to you &lt;a href=&quot;http://mathoverflow.net/questions/47650/proof-that-if-group-g-zg-is-cyclic-then-g-is-commutative-closed&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; what you can conclude from this, so I might as well quote him: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Each element of $G$ has the form $a^nz$ where $a$ is fixed and $z\in Z(G)$...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-11-29T09:44:29.367" CommunityOwnedDate="2010-11-29T09:44:29.367" CommentCount="1" />
  <row Id="12322" PostTypeId="1" AcceptedAnswerId="12324" CreationDate="2010-11-29T10:04:31.390" Score="4" ViewCount="333" Body="&lt;p&gt;How to prove DeMorgan's Law?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$A - (B \cup C) = (A - B) \cap (A - C)$$&#xD;&#xA;$$A - (B \cap C) = (A - B) \cup (A - C)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;:&lt;br&gt;&#xA;Here is what I have tried so far:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Considering the first equation, assuming $x \in A - (B \cup C)$ then $x \in A$ and $x \not\in B$ and $x \not\in C$, while the right hand means ($x \in A$ and $x \not\in B$) or ($x \in A $ and $x \not\in C$) which is the same as $x \in A$ and $x \not\in B$ and $x \not\in C$. So the two set is the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I do not know whether this is sufficient for a proof. Am I wrong?&lt;/p&gt;&#xA;" OwnerUserId="1382" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-29T15:55:21.217" LastActivityDate="2010-11-29T16:02:17.670" Title="How to prove DeMorgan's law?" Tags="&lt;elementary-set-theory&gt;" AnswerCount="2" CommentCount="5" />
  <row Id="12323" PostTypeId="1" AcceptedAnswerId="12357" CreationDate="2010-11-29T10:08:56.793" Score="5" ViewCount="276" Body="&lt;p&gt;A lot of textbooks said it was hard for human to accept zero when it was first introduced.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How could it be? It seems to me as natural as postive integer which represent there is no elements at all.&lt;/p&gt;&#xA;" OwnerUserId="1382" LastActivityDate="2010-11-30T19:25:41.630" Title="What's the hard part of zero?" Tags="&lt;soft-question&gt;" AnswerCount="4" CommentCount="1" FavoriteCount="1" />
  <row Id="12324" PostTypeId="2" ParentId="12322" CreationDate="2010-11-29T10:10:30.407" Score="5" ViewCount="" Body="&lt;p&gt;I'd &quot;reduce&quot; it to logic&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x\in A-(B\cup C) \iff x\in A \text{ and not } (x\in B \text{ or } x\in C)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;thus by DeMorgan in logic (which can be proved with truth table)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x\in A \text{ and } x\notin B \text{ and } x\notin C$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus $(x \in A  \text{ and } x \notin B) \text{ and } (x \in A  \text{ and } x \notin C)$. Thus&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x\in (A- B)\cap (A- C)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and vice versa.&lt;/p&gt;&#xA;" OwnerUserId="52" LastEditorUserId="52" LastEditorDisplayName="" LastEditDate="2010-11-29T16:02:17.670" LastActivityDate="2010-11-29T16:02:17.670" />
  <row Id="12325" PostTypeId="2" ParentId="12322" CreationDate="2010-11-29T10:12:26.040" Score="3" ViewCount="" Body="&lt;p&gt;Maybe you're familiar with the fact that if $A\subseteq B$ and $B\subseteq A$, then $A=B$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since you're viewing $A,B,C$ as sets, you can prove these by showing the set on the left of $=$ is a subset of the set on the right of $=$, and vice versa.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, suppose $x\in A-(B\cup C)$. So $x\in A$, but $x\not\in B\cup C$. In particular, $x\not\in B$, and $x\not\in C$. It follows that $x\in A-B$ and $x\in A-C$, that is, $x\in (A-B)\cap (A-C)$. Thus $A-(B\cup C)\subseteq (A-B)\cap (A-C)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Try using a similar method to show the reverse containment, and you'll get your result. This works for the second law also.&lt;/p&gt;&#xA;" OwnerUserId="1583" LastActivityDate="2010-11-29T10:12:26.040" />
  <row Id="12326" PostTypeId="2" ParentId="12323" CreationDate="2010-11-29T10:18:55.953" Score="0" ViewCount="" Body="&lt;p&gt;You can't count out zero 'things'; I would imagine that that was the big issue.  Same goes for the negative side of $\mathbb{Z}$.&lt;/p&gt;&#xA;" OwnerUserId="4194" LastActivityDate="2010-11-29T10:18:55.953" CommentCount="5" />
  <row Id="12327" PostTypeId="2" ParentId="12323" CreationDate="2010-11-29T10:27:42.093" Score="12" ViewCount="" Body="&lt;p&gt;Of course it seems natural to you; you grew up in the modern world, where everyone accepts zero.  More importantly, people now accept the abstract concept of numbers and are capable of divorcing them from the things that they represent.  This is a sophisticated point of view.  From a more naive point of view, a number is a property of a collection of objects: when I say there are $2$ apples somewhere, that is a property of a collection of apples.  If there are no apples, then what is there to &quot;hold&quot; the corresponding property?  So instead of saying there are $0$ apples, people said there are no apples.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In computer science terms, &lt;code&gt;apples.num()&lt;/code&gt; is not defined if there is no &lt;code&gt;apples&lt;/code&gt; variable!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You also have to understand that &quot;human&quot; in your statement means (from my understanding) &quot;Western mathematicians.&quot;  Indian mathematicians had no trouble with zero.  &lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-11-29T10:27:42.093" CommentCount="5" />
  <row Id="12328" PostTypeId="1" AcceptedAnswerId="12367" CreationDate="2010-11-29T10:45:34.587" Score="2" ViewCount="325" Body="&lt;p&gt;I stumbled across this paragraph in a paper:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Hence, user b cannot decrypt C&#xA;  directly. But using e and d , user b&#xA;  can quickly factor N.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;How is  it possible to speedup the prime factorization when knowing e (public key) and d (private key)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;For clarification:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;RSA provides us with these equations:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$n = pq$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\phi = (p-1)(q-1)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$gcd(e, \phi) = 1$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$de = 1\pmod{\phi}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to determine $p$ and $q$ an attacker has to factor n which is not feasible. However the paper stated that it is easy to reconstruct $p$ and $q$ when a person knows both (his) private and public keys.&lt;/p&gt;&#xA;" OwnerUserId="4166" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-29T12:39:06.533" LastActivityDate="2010-11-29T17:23:39.080" Title="RSA: Fast factorization of N if d and e are know" Tags="&lt;number-theory&gt;&lt;cryptography&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="12329" PostTypeId="1" AcceptedAnswerId="12343" CreationDate="2010-11-29T10:45:56.917" Score="4" ViewCount="148" Body="&lt;p&gt;How possible methods for finding groups of given order we have?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sylow theorems - and next? We know something about center of group, about possible orders of these elements and subgroups. But what is most effective way, to find all non-isomorphic groups of given order? Is there any other strong &quot;weapon&quot; for this (like Sylow theorems)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, I took groups order 15.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Commutative group is only $\mathbb{Z}_{15}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;About non-commutative we know, they have 1 Syllow 3-subgroup and 1 Syllow 5-subgroup. And next, that $M_3 \cup M_5 = G$ (if $M_3$, resp. $M_5$ is Syllow 3-subgroup, resp. Syllow 5-subgroup).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$M_3$ and $M_5$ are single generated, than we can choose elements $f$ and $g$ satysfaing $&amp;lt;f&amp;gt;=M_3$ and $&amp;lt;g&amp;gt;=M_5$. Because there are characterictis, we get these equations: $f^{-1}gf=g^m$, where $m \in \{1,2,3,4\}$ and $g^{-1}fg=f^n$, where $n \in \{1,2\}$. $m$ and $n$ must supplying $|g|=|g^m|$ and $|f|=|f^n|$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But there is end of my way, i don't know how continue... Can anyone help? (Sorry for bad English)&lt;/p&gt;&#xA;" OwnerUserId="4195" LastActivityDate="2010-11-29T13:16:34.893" Title="How to find finite groups" Tags="&lt;group-theory&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="12330" PostTypeId="2" ParentId="12307" CreationDate="2010-11-29T11:00:10.287" Score="5" ViewCount="" Body="&lt;p&gt;Observe that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left( \dfrac{x+2}{x}\right) ^{x}=\left( \left( 1+\dfrac{1}{\frac{x}{2}}%&#xD;&#xA;\right) ^{x/2}\right) ^{2}.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, by the definition of $e$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$e=\lim_{n\rightarrow \infty }\left( 1+\dfrac{1}{n}\right) ^{n}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{eqnarray*}&#xD;&#xA;\lim_{x\rightarrow \infty }\left( \dfrac{x+2}{x}\right) ^{x}&#xD;&#xA;&amp;amp;=&amp;amp;\lim_{x\rightarrow \infty }\left( \left( 1+\frac{1}{x/2}\right) ^{x/2}\right) ^{2} \\&#xD;&#xA;&amp;amp;=&amp;amp;\left( \lim_{x\rightarrow \infty }\left( 1+\frac{1}{x/2}\right) ^{%&#xD;&#xA;x/2}\right) ^{2} \\&#xD;&#xA;&amp;amp;=&amp;amp;\left( \lim_{y\rightarrow \infty }\left( 1+\frac{1}{y}\right) ^{y}\right)&#xD;&#xA;^{2} \\&#xD;&#xA;&amp;amp;=&amp;amp;e^{2}&#xD;&#xA;\end{eqnarray*}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;For instance, by the same argument, for $k\in\mathbb{N}$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\lim_{x\rightarrow \infty }\left( \dfrac{x+k}{x}\right) ^{x}=\left( \lim_{y\rightarrow \infty }\left( 1+\frac{1}{y}\right) ^{y}\right)&#xD;&#xA;^{k}=e^{k}$$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="752" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2010-11-29T18:12:27.953" LastActivityDate="2010-11-29T18:12:27.953" />
  <row Id="12331" PostTypeId="1" AcceptedAnswerId="12372" CreationDate="2010-11-29T11:03:17.620" Score="1" ViewCount="152" Body="&lt;p&gt;Here is what my textbook told me:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming the formula for surface $\Sigma$ is $$F(x,y,z) = 0$$ &#xA;Suppose $X_0 = (x_0, y_0, z_0)$ is a point on the surface $\Sigma$ and we assuming F(x,y,z) is differentiable and $$\mathbf{J}F(X_0) = (\frac{\partial F(X_0)}{\partial x}, \frac{\partial F(X_0)}{\partial y}, \frac{\partial F(X_0)}{\partial z}) \neq 0$$ &#xA;Draw a line $\Gamma$ in the surface $\Sigma$ passing through the point $X_0$, assuming the equations for $\Sigma$ is $$x = x(t), y = y(t), z = z(t)$$  $t = t_0$ correspond to the point $X_0$ and $x&amp;#39;(t_0), y&amp;#39;(t_0), z&amp;#39;(t_0)$ does not all vanish. Because of the line $\Gamma$ is on the surface $\Sigma$, so $$F(x(t), y(t), z(t)) = 0$$ So $$ \frac{dF}{dt}\mid_{t=t_0} = {F_x}&amp;#39;(X_0)x&amp;#39;(t_0) + {F_y}&amp;#39;(X_0)y&amp;#39;(t_0) + {F_z}&amp;#39;(X_0)z&amp;#39;(t_0) = 0 $$ So $$ ({F_x}&amp;#39;(X_0), {F_y}&amp;#39;(X_0), {F_z}&amp;#39;(X_0))\cdot(x&amp;#39;(t_0), y&amp;#39;(t_0), z&amp;#39;(t_0)) = 0 $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know the vector $\mathbf{T} = (x&amp;#39;(t_0), y&amp;#39;(t_0), z&amp;#39;(t_0))$ is the tangent vector for the line $\Gamma$ on the point $X_0$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My questions are&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Why the vector $\mathbf{T}$ is the tangent vector for line $\Gamma$ at point $X_0$?&lt;/li&gt;&#xA;&lt;li&gt;Why the $\mathbf{J}F(X_0)$ should not equal to zero? What if it is zero?&lt;/li&gt;&#xA;&lt;li&gt;Why $x&amp;#39;(t_0), y&amp;#39;(t_0), z&amp;#39;(t_0)$ should not all vanish? What if all vanish?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="1382" LastEditorUserId="1382" LastEditorDisplayName="" LastEditDate="2010-11-29T11:14:31.350" LastActivityDate="2010-11-29T18:26:05.883" Title="Help understanding the definition of tangent vector or tangent plane?" Tags="&lt;calculus&gt;&lt;geometry&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="12332" PostTypeId="1" CreationDate="2010-11-29T11:03:39.490" Score="2" ViewCount="114" Body="&lt;p&gt;For $n &amp;gt; 2$, is it possible to embed $\underbrace{S^1 \times \cdots \times S^1}_{n\text{ times}}$ into $\mathbb R^{n+1}$?&lt;/p&gt;&#xA;" LastEditorUserId="856" LastEditorDisplayName="" LastEditDate="2010-11-29T11:16:24.500" LastActivityDate="2010-11-29T18:14:35.043" Title="Embedding torus in Euclidean space" Tags="&lt;algebraic-topology&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="12333" PostTypeId="2" ParentId="12263" CreationDate="2010-11-29T11:08:09.980" Score="1" ViewCount="" Body="&lt;p&gt;A B-spline of order 2 is a continuous piecewise-linear function that interpolates between the assigned data points (&quot;knots&quot;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the slopes of such a function vary between +0.017 and -0.017 on the interval $[0,2\pi]$, then by the Mean Value Theorem there cannot be more than $0.034\pi$ or about 0.1 between the minimum and maximum function values.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So either there are no data points near the minimum and maximum of $sin(x)$ on this interval or else some error has been made in computation (perhaps a mistake in scaling the x-axis that stretches everything out).&lt;/p&gt;&#xA;" OwnerUserId="3111" LastActivityDate="2010-11-29T11:08:09.980" CommentCount="1" />
  <row Id="12334" PostTypeId="2" ParentId="12205" CreationDate="2010-11-29T11:11:11.917" Score="0" ViewCount="" Body="&lt;p&gt;@Bill: Thanks, I see that $\pi\pi&amp;#39;|\alpha\beta$ implies that $p^2|N(\alpha)N(\beta)$. Since $(N(\alpha),N(\beta))=1$ we have that $p^2$ divides $N(\alpha)$ or $N(\beta)$, say $p^2|N(\alpha)$. Thus, $p$ must divide $\alpha$, a contradiction.&lt;/p&gt;&#xA;" LastActivityDate="2010-11-29T11:11:11.917" />
  <row Id="12335" PostTypeId="2" ParentId="12285" CreationDate="2010-11-29T11:20:42.060" Score="3" ViewCount="" Body="&lt;p&gt;Consider the map $f\times f: X\times X \mapsto X\times X$. It is continuous (because f is)  and surjective (because f is). Consider the orbit of $(x,y)$ under $f$, i.e. the sequence $p_n=(f^n(x), f^n(y))$. Because $X\times X$  is compact, it has a convergent subsequence, hence for any $\epsilon$ (sorry, I hoped to do no epsilons, but can't), there is an $N_0, k&amp;gt;0$, such that $f^{N_0}(x)$ is $\epsilon$ close to $f^{k+N_0}(x)$ and $f^{N_0}(y)$ is epsilon close to $f^{k+N}(y)$, hence $d(f^{k+N_0}(x), f^{k+N_0}(y))$ is $2\epsilon$ close to $d(f^{N_0}(x), f^{N_0}(y))$. Then because $f$ does not expand distances, induction shows the same is true for all $N \geq N_0$. But we want more. We want this to be true for some fixed $N$ (and varying $k$) for all $(x,y)$. That is achievable by compactness as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As we showed (replacing $\epsilon$ by $\epsilon/2$) we have $d(f^{k+N_0}(x), f^{k+N_0}(y))$ is $\epsilon$ close to $d(f^{N_0}(x), f^{N_0}(y))$. Because $f$ does not expand distances (and triangle inequality), this means that for any $(v,w)$ with $d(v,x)&amp;lt;\epsilon/4$ and $d(w,y)&amp;lt;\epsilon/4$ the we have $d(f^{k+N_0}(v), f^{k+N_0}(w))$ is $2\epsilon$ close to $d(f^{N_0}(v), f^{N_0}(w))$. Now cover $X$ with $\epsilon/4$ balls and pick a finite subcover. For any pair of centers $(x_i,y_i)$ of the balls we have some $N_0$ which gives $\epsilon$-closeness, and hence gives $2\epsilon$ closeness on the ball, hence the maximum $M$ of these $N_0$'s gives $2\epsilon$ closeness everywhere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we can prove $f$ is an isometry. To get contradiction, assume you have $(x,y)$ with $d(f(x), f(y))&amp;lt; d(x,y)$, hence for some $\epsilon$ also $d(f(x), f(y))&amp;lt; d(x,y)-2\epsilon$. Then for any $k&amp;gt;0$, we have $d(f^k(x), f^k(y))&amp;lt; d(x,y)-2\epsilon$.  Finally we use surjectivity. Take $(a,b)$ with $f^M(a)=x, f^M(b)=y$. Then $d(f^{M+k}(a), f^{M+k} (b))= d(f^k(x), f^k(y))$ is $2\epsilon$ close to $d(f^M (a), f^M(b))=d(x,y)$ for some $k$, contradiction.&lt;/p&gt;&#xA;" OwnerUserId="2633" LastActivityDate="2010-11-29T11:20:42.060" />
  <row Id="12336" PostTypeId="1" CreationDate="2010-11-29T11:26:59.793" Score="0" ViewCount="108" Body="&lt;p&gt;Hi all,&#xA;I have a uniform grid of square cells and a point on that grid.  I want to calculate the intersection of cells on the grid, with a circular area centered on the point, with a radius R.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anyone know of an efficient way of doing this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" LastActivityDate="2010-11-29T11:26:59.793" Title="Simple method for detecting grid intersection with circle" Tags="&lt;geometry&gt;" CommentCount="1" />
  <row Id="12337" PostTypeId="1" CreationDate="2010-11-29T11:36:17.353" Score="2" ViewCount="78" Body="&lt;p&gt;Follow-up to &lt;a href=&quot;http://math.stackexchange.com/questions/12193/field-reductions-part-two&quot;&gt;Field reductions. part two&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a field $K$, an element $a \in K$, and $K(\setminus a)$ the set of field reductions (maximal subfields of $K$ not containing $a$) are there any interesting equivalence classes on $K(\setminus a)$ or binary operations, or distance functions, or ...? Real numbers are represented by infinite decimals. How would you represent elements of $\mathbb{R}(\setminus a)$ ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Terry Tao's &lt;a href=&quot;http://terrytao.wordpress.com/2010/11/27/nonstandard-analysis-as-a-completion-of-standard-analysis/&quot; rel=&quot;nofollow&quot;&gt;latest blog post&lt;/a&gt; is about the concept of extending and completing things. What about going the other way and reducing things ? So moving on from fields to general spaces:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $X$ be a space, i.e. a set with some added structure&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $a \in X$, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $X(\setminus a)$ be the set of maximal subspaces of $X$ not containing $a$, called the $a$-reduced spaces of $X$ or space reductions of $X$ by $a$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What examples are there in mathematics that have been previously studied that could be described as a maximal subspace not containing $a$ ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $@X(\setminus a)$ denote an element of $X(\setminus a)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a subspace $S$ of $X$, let $S(a..)$ denote the extension of $S$ to $X$, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;so then $@X(\setminus a)(a..)=X$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there some category-theoretic framework that can be used here ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Morphisms from $X(\setminus x_1)$ to $X(\setminus x_2)$ or $Y(\setminus y_1)$ or $Z(\setminus z_1)$ ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you extend a field by adding a new element and then also adding all its arithmetic combinations with the existing elements, then this is like adding an idea to a body of knowledge or piece of information to your list and then combining it in all possible ways with all existing items to get an extended list, like in Sudoku or Logic problems or ... , so information is like a universal algebra with lots of operations of different arities, where not only the elements can be added to, but also the operations. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you remove a piece of information from a list and remove all the information that depended on it then there seems to be only one way of doing that so the analogy between information reduction and field reduction breaks down, but the analogy can be repaired by removing not only what depends on it, but also what leads to it, and there would be a choice because removing one thing that leads to it would mean not needing to remove another thing. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Reverse_mathematics&quot; rel=&quot;nofollow&quot;&gt;Reverse mathemtaics&lt;/a&gt;: take a set of axioms of a theory and adjoin all true statements, now take a statement that you are interested in finding the necessary axioms for, remove that statement and anything that follows from it or leads to it, then you'd get maximal subtheories for which that statement was independent of. If any of the axioms are not in a subtheory then the statement depended on that axiom (Probably. Correct me if I'm wrong.) &lt;/p&gt;&#xA;" OwnerUserId="4157" LastEditorUserId="4157" LastEditorDisplayName="" LastEditDate="2010-11-29T14:19:17.853" LastActivityDate="2010-11-29T14:19:17.853" Title="Field reductions. part three" Tags="&lt;abstract-algebra&gt;&lt;category-theory&gt;&lt;field-theory&gt;" CommentCount="1" />
  <row Id="12338" PostTypeId="2" ParentId="12332" CreationDate="2010-11-29T11:40:08.027" Score="2" ViewCount="" Body="&lt;p&gt;Let $e_0, \ldots, e_n$ be the standard basis of $\mathbb {R}^{n+1}$. Take $\epsilon$ small. Consider the vector $v_1$ of length 1 in the span of  $e_0, e_1$. Then the  vector $v_2$ of length $\epsilon$ in the span of  $v_1, e_2$, and in general the vector $v_i$  of length $\epsilon^{i-1}$ in the span of  $v_{i-1}, e_i$. Now consider the vector $w=v_1+\ldots v_n$  For small $\epsilon$ the set of $w$'s is a torus embedded $\mathbb {R}^{n+1}$ (any $\epsilon &amp;lt; 1$ will do, actually).&lt;/p&gt;&#xA;" OwnerUserId="2633" LastEditorUserId="2633" LastEditorDisplayName="" LastEditDate="2010-11-29T18:14:35.043" LastActivityDate="2010-11-29T18:14:35.043" />
  <row Id="12340" PostTypeId="1" AcceptedAnswerId="12341" CreationDate="2010-11-29T11:59:15.350" Score="10" ViewCount="288" Body="&lt;p&gt;What is the limit of the series 1/(2n)! for n in [0, infinity) ?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've ground out the sum of the 1st 1000 terms to 1000 digits using Python,&#xA;(see &lt;a href=&quot;http://tutoree7.pastebin.com/pv0i1XQb&quot; rel=&quot;nofollow&quot;&gt;http://tutoree7.pastebin.com/pv0i1XQb&lt;/a&gt; ), but how would a mathematician calculate the limit? And what is it? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;No, this isn't homework. I'm 73. Just curious. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="1964" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2010-11-29T15:10:38.763" LastActivityDate="2010-11-29T15:10:38.763" Title="What is limit of $\sum_{n=0}^{\infty}\frac{1}{(2n)!} $ ?" Tags="&lt;sequences-and-series&gt;&lt;limit&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" />
  <row Id="12341" PostTypeId="2" ParentId="12340" CreationDate="2010-11-29T12:12:36.387" Score="21" ViewCount="" Body="&lt;p&gt;It's half the sum of $e^1=\sum 1/n!$ and $e^{-1}=\sum (-1)^{n}/n!$ (or $\cosh 1$, in other words).&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2010-11-29T12:12:36.387" CommentCount="13" />
  <row Id="12343" PostTypeId="2" ParentId="12329" CreationDate="2010-11-29T13:04:44.323" Score="4" ViewCount="" Body="&lt;p&gt;If you don't like semi-direct products, then you will not get very far in your endeavour. I recommend that you play around with them until you start liking them. The reason is that the single most powerful result in trying to classify all finite groups of given order is the so-called Schur-Zassenhaus theorem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Schur-Zassenhaus Theorem&lt;/strong&gt;: Let $N$ be a normal subgroup of a finite group $G$ and suppose that the order of $N$ is co-prime to its index. Then $G$ is a semidirect product $G=N\rtimes H$ for a subgroup $H$ of $G$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is this useful? I will show you, how to finish your project for $n=15$ essentially in one line: as you have noted, there is one Sylow 5-subgroup in $G$, so it's normal. Since its index is 3, Schur-Zassenhaus applies and tells us that $G=C_5\rtimes C_3$. But the automorphism group of $C_5$ is cyclic of order 4, so there are no non-trivial maps from $C_3$ to the automorphism group of $C_5$ and $G$ must in fact be a direct product.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Exercise: classify all finite groups of order 20 using this approach.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S. One more thing: don't ever take unions of subgroups! What you wanted to say is that $G=\langle M_3,M_5\rangle$, i.e. is generated by (all possible products of elements of) the subgroups.&lt;/p&gt;&#xA;" OwnerUserId="3212" LastActivityDate="2010-11-29T13:04:44.323" />
  <row Id="12344" PostTypeId="2" ParentId="12307" CreationDate="2010-11-29T13:05:54.377" Score="2" ViewCount="" Body="&lt;p&gt;For ease rewrite as &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;\lim_{n\rightarrow \infty}\left(1+\frac{2}{x}\right)^x.&#xD;&#xA;$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First we compute &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;\lim_{n\to\infty}\ln\left(\left(1+\frac{2}{x}\right)^x\right).&#xD;&#xA;$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using laws of logarithms we get &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;\lim_{n\to\infty}x\ln\left(1+\frac{2}{x}\right)  =\lim_{n\to\infty}\frac{\ln\left(1+\frac{2}{x}\right)}{\frac{1}{x}}.&#xD;&#xA;$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We are now in a position to apply L'Hopital's Rule.  Taking derivatives gives&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;\lim_{n\to\infty} \frac{\left(\frac{1}{1+\frac{2}{x}}\right)\left(-\frac{2}{x^2}\right)}{-\frac{1}{x^2}}=\lim_{n\to\infty}\frac{2}{1+\frac{2}{x}}=2.&#xD;&#xA;$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, your limit is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;\lim_{n\to\infty}e^{\ln\left(\left(1+\frac{2}{x}\right)^x\right)}=e^2&#xD;&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="2838" LastActivityDate="2010-11-29T13:05:54.377" />
  <row Id="12345" PostTypeId="2" ParentId="12329" CreationDate="2010-11-29T13:16:34.893" Score="2" ViewCount="" Body="&lt;p&gt;Here's how to finish without assuming the Sylow 5-subgroup is normal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Assume by way of contradiction:) If $n=2$, so $g^{-1} f g = ff$, then: $$g^{-2} f g^2 = g^{-1}( g^{-1} f g ) g = g^{-1}( ff )g = g^{-1}( fg g^{-1}f)g = (g^{-1}fg) (g^{-1}fg) = (ff) (ff) = f^4.$$ Every time you apply $g^{-1}( * )g$, you double the number of $f$, so $g^{-i} f g^i = f^{(2^i)}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we use $g^5 = 1$.  If you apply $g^{-1}( * )g$ five times, you get $g^{-5} f g^5 = 1 f 1 = f$ on the one hand, but $f^{(2^5)} = f^{32} = f^2$ on the other, using the fact that 2 ≡ 32 mod 3.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can $f = f^2$?  It does not, since $f$ has order 3.  Assuming $n=2$ gives a contradiction, so $n=1$, and $g^{-1} f g = f$.  Multiply on the left by $g$ to get $f g = g f$.  The group must be abelian, since all of its generators commute.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Now try this with &quot;5&quot; replaced by &quot;2&quot; and you will still get the Sylow 3-subgroup is normal, but the Sylow 2-subgroup need not be normal.  The same argument will get you that both n=1 (cyclic group of order six), and n=2 (dihedral group of order six) are possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More precisely, you'll check that $g^{-2} f g^2 = f^{(2^2)} = f^4 = f$ just as it should be.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Congratulations, you have just analyzed some semi-direct products using automorphisms!  With cyclic Sylow subgroups it makes a lot of sense.  We didn't even need to call &quot;n&quot; an automorphism, nor did we need to call M&lt;sub&gt;3&lt;/sub&gt;∪M&lt;sub&gt;5&lt;/sub&gt; a semi-direct product.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the subgroups are not Sylow subgroups, then it is harder to check (1) that they are normal or (2) that M∩N = 1, but there are more general techniques for this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the subgroups are not cyclic, then you cannot just use one &quot;n&quot; for the Sylow 3-subgroup.  You get a matrix of &quot;m&quot;s and &quot;n&quot;s.  Not just any matrix will do, but the ones that do work are called automorphisms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, the general case of semi-direct products is similar to what you are now doing.  Don't give up.&lt;/p&gt;&#xA;" OwnerUserId="583" LastActivityDate="2010-11-29T13:16:34.893" />
  <row Id="12347" PostTypeId="2" ParentId="12097" CreationDate="2010-11-29T13:53:37.097" Score="2" ViewCount="" Body="&lt;p&gt;My previous answer was useless. The new answer below completes George's answer for the  $0 &amp;lt; \alpha &amp;lt; 1$ case, illustrated in the first graph above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For this case, George provided the asymptotic approximation $P_n \sim {\rm P}(2 \Delta Z^*_1 - Z_1 &amp;gt; 0)$ as $n \to \infty$, where $P_n$ is the probability the OP asked for, $Z = \lbrace Z_t : t \geq 0 \rbrace$ is a strictly stable L\'evy process of index $\alpha$, and $\Delta Z^*_1$ is the largest jump of $Z$ in the time interval $[0,1]$. It is not easy to obtain &#xA;${\rm P}(2 \Delta Z^*_1 - Z_1 &amp;gt; 0)$ in closed form, but indeed possible, and in fact the following more general result is known:&#xA;$$&#xD;&#xA;{\rm P}\bigg(\frac{{\Delta Z_1^* }}{{Z_1 }} &amp;gt; \frac{1}{{1 + y}}\bigg) = \frac{{y^\alpha  }}{{\Gamma (1 - \alpha )\Gamma (1 + \alpha )}}, \;\; y \in [0,1],&#xD;&#xA;$$&#xD;&#xA;where $\Gamma$ is the gamma function. Letting $y = 1$ thus gives $P_n \sim 1/[\Gamma(1 - \alpha) \Gamma(1 + \alpha)]$. &#xA;This asymptotic expression agrees very well with the graph above, as I confirmed by checking various values of $\alpha \in (0,1)$. (For example, $\alpha = 1/2$ gives $P_n \sim 2/\pi \approx 0.6366$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A generalization. Instead of just considering the probability ${\rm P}(M &amp;gt; \sum\nolimits_{i = 0}^{n - 1} {X_i }  - M)$, we can consider the probability ${\rm P}(M &amp;gt; y^{-1}[\sum\nolimits_{i = 0}^{n - 1} {X_i }  - M])$, $y \in (0,1]$. In view of George's answer, this should correspond to the formula given above for ${\rm P}(\Delta Z_1^* / Z_1 &amp;gt; 1/(1+y))$.&#xA;This can be further generalized, in various directions, based on known results from the literature (three useful references in this context are indicated somewhere in the comments above/below).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As George observed, the term $\Gamma(1 - \alpha) \Gamma(1 + \alpha)$ can be simplified to $(\pi\alpha) / \sin(\pi\alpha)$. The former expression corresponds to $[\Gamma(1 - \alpha)]^k \Gamma(1 + k \alpha)$, which is incorporated in the explicit formula for ${\rm P}(\Delta Z_1^* / Z_1 &amp;gt; 1/(1+y))$, $y &amp;gt; 0$, of the form&#xA;$\sum\nolimits_{k = 1}^{\left\lceil y \right\rceil } {a_k \varphi _k (y)} $. The function $\varphi _k (y)$ is some $(k-1)$-dimensional integral, hence that formula is computationally very expensive already for moderate values of $y$.&lt;/p&gt;&#xA;" OwnerUserId="2810" LastEditorUserId="2810" LastEditorDisplayName="" LastEditDate="2010-11-29T21:19:24.743" LastActivityDate="2010-11-29T21:19:24.743" />
  <row Id="12348" PostTypeId="1" AcceptedAnswerId="12351" CreationDate="2010-11-29T14:07:57.407" Score="2" ViewCount="331" Body="&lt;p&gt;folks,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i am new to this forum and not a math expert. so please bear with me if am asking silly questions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;the question is &quot;probability of getting 50 heads from tossing a coin 100 times&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so the answer for this is, i guess, (100 C 50)*(2 ^-100) [[ fixed typo ]]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so all am trying to get is easier way to calculate 100 C 50. or another approach to the parent problem only&lt;/p&gt;&#xA;&#xA;&lt;p&gt;thanks all, appreciate that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;raj&lt;/p&gt;&#xA;" OwnerUserId="4205" LastEditorUserId="4205" LastEditorDisplayName="" LastEditDate="2010-11-30T07:49:13.857" LastActivityDate="2010-11-30T12:15:03.573" Title="probability of getting 50 heads from tossing a coin 100 times" Tags="&lt;probability&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="12349" PostTypeId="2" ParentId="12331" CreationDate="2010-11-29T14:14:09.347" Score="3" ViewCount="" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The coordinates $x(t)$ and so on tell you how to trace out a curve, as $t$ changes. The vector &lt;strong&gt;T&lt;/strong&gt; describes the change in the $x$, $y$ and $z$-coordinates as $t$ changes infinitesimally. Surely, that's exactly what the tangent vector to a curve should do. More algebraically, the tangent to the curve at $X_0$ is defined as&#xA;$$\lim_{\delta\rightarrow 0} \frac{(x(t_0),y(t_0)z(t_0)) - (x(t_0+\delta),y(t_0+\delta)z(t_0+\delta))}{\delta},$$&#xD;&#xA;whenever the limit is defined.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;As J.M. wrote, in that case you don't have a well-defined tangent. Again, algebraically, your calculation has shown that &lt;strong&gt;J&lt;/strong&gt;$F(X_0)$ is normal to tangents to all curves on the surface passing through $X_0$, so it &lt;em&gt;should&lt;/em&gt; be the normal vector to the surface at that point. But if the surface has a singularity, then there is no well-defined normal vector.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;As remarked above, the derivatives $x&amp;#39;(t_0)$ and so on describe, how quickly the coordinates change as you trace out your curve. In pictorial terms, you can imagine walking down the curve as the parameter $t$ slowly increases. If all these derivatives vanish, it means that as $t$ approaches $t_0$, your walk is slowing down to a grinding halt. That means that either you have chosen a bad parametrisation, i.e. you are staying in one place for some range of $t$, which you might as well leave out then, or you are approaching a singularity, which is bad news if you are computing tangents.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="3212" LastActivityDate="2010-11-29T14:14:09.347" />
  <row Id="12350" PostTypeId="2" ParentId="12348" CreationDate="2010-11-29T14:16:05.217" Score="4" ViewCount="" Body="&lt;p&gt;See &lt;a href=&quot;http://www.johndcook.com/blog/2008/04/24/how-to-calculate-binomial-probabilities/&quot; rel=&quot;nofollow&quot;&gt;How to calculate binomial probabilities&lt;/a&gt; for how to calculate these probabilities while avoiding numerical overflow or underflow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, I believe you meant to type 2^-100 instead of 2^-10 in your question.&lt;/p&gt;&#xA;" OwnerUserId="247" LastActivityDate="2010-11-29T14:16:05.217" CommentCount="1" />
  <row Id="12351" PostTypeId="2" ParentId="12348" CreationDate="2010-11-29T14:31:55.533" Score="4" ViewCount="" Body="&lt;p&gt;The coefficient ${2n \choose n}$ for $n$ large can be approximated well (using Stirling's formula) by ${2n \choose n} \approx 4^n / \sqrt{n \pi} $. In particular, letting $n=50$ gives ${100 \choose 50} \approx 4^{50}/ \sqrt {50 \pi}$.&lt;/p&gt;&#xA;" OwnerUserId="2810" LastActivityDate="2010-11-29T14:31:55.533" />
  <row Id="12352" PostTypeId="1" AcceptedAnswerId="12355" CreationDate="2010-11-29T14:35:05.700" Score="0" ViewCount="157" Body="&lt;p&gt;how to calculate the sides and hypotenuse length of the &lt;strong&gt;right triangle&lt;/strong&gt; if I know&#xA;the bigger side = 60 &#xA;one angle = 60 second angle = 30 (3rd = 90)&lt;/p&gt;&#xA;" OwnerUserId="699" LastEditorUserId="699" LastEditorDisplayName="" LastEditDate="2010-11-29T14:47:22.153" LastActivityDate="2010-11-29T15:38:50.647" Title="calculate sides of the right triangle if I know 1 side and all the angles" Tags="&lt;geometry&gt;" AnswerCount="4" />
  <row Id="12353" PostTypeId="2" ParentId="12352" CreationDate="2010-11-29T14:40:57.320" Score="3" ViewCount="" Body="&lt;p&gt;You could use the identities &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$sin(\theta)=\frac{oposite-side}{hypotenuse} $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$cos(\theta)=\frac{adyacent-side}{hypotenuse} $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact you have :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sin(60)=\frac{60}{hypotenuse} \Rightarrow hypotenuse = \frac{120}{\sin(60)}=\frac{120}{\sqrt{3}}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\cos(60)=\frac{adyacent-side}{hypotenuse} \Rightarrow adyacent-side = \cos(60)\cdot hypotenuse =\frac{60}{\sqrt{3}} $$&lt;/p&gt;&#xA;" OwnerUserId="3421" LastEditorUserId="3421" LastEditorDisplayName="" LastEditDate="2010-11-29T14:53:27.953" LastActivityDate="2010-11-29T14:53:27.953" CommentCount="2" />
  <row Id="12354" PostTypeId="2" ParentId="12352" CreationDate="2010-11-29T14:41:23.213" Score="1" ViewCount="" Body="&lt;p&gt;tan(60) = side1/side2 = 60/side2 =&gt; side2 = 60/tan(60) = 60/sqrt(3) = 20sqrt(3)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;sin(60) = side1/hypotenuse =&gt; hypotenuse = 60/sin(60)=120/sqrt(3) = 40sqrt(3)&lt;/p&gt;&#xA;" OwnerUserId="1030" LastActivityDate="2010-11-29T14:41:23.213" CommentCount="1" />
  <row Id="12355" PostTypeId="2" ParentId="12352" CreationDate="2010-11-29T14:41:45.183" Score="2" ViewCount="" Body="&lt;p&gt;Some hints: what is the third angle? Draw a picture. Can you spot some symmetry?&#xA;If not, then decode the following hint with &lt;a href=&quot;http://www.rot13.com/index.php&quot; rel=&quot;nofollow&quot;&gt;http://www.rot13.com/index.php&lt;/a&gt;.&#xA;But please, spend some time trying first.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ersyrpg gur gevnatyr va gur evtug natyr. Jung vf fb fcrpvny va gur erfhygvat gevnatyr? Gel znxvat hfr bs gur rkgen flzzrgel.&lt;/p&gt;&#xA;" OwnerUserId="3212" LastActivityDate="2010-11-29T14:41:45.183" CommentCount="4" />
  <row Id="12356" PostTypeId="1" AcceptedAnswerId="12359" CreationDate="2010-11-29T14:48:41.560" Score="1" ViewCount="154" Body="&lt;p&gt;What are the necessary and sufficient conditions on a vector field $F$ for the divergence $\nabla\cdot F$ to exist at a given point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In &lt;a href=&quot;http://en.wikipedia.org/wiki/Divergence&quot; rel=&quot;nofollow&quot;&gt;Divergence&lt;/a&gt; in the second line under the heading &quot;&lt;strong&gt;Application in Cartesian coordinates&lt;/strong&gt;&quot;, why is it assumed that $\vec{F}$ to be a continuously differentiable vector field ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT 2&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ideally one would expect each component of $F$ to be differentiable at a given point $\vec{a}$ no matter through which continuous contour you traverse the point $\vec{a}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT 3&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or is it that each component of $F$ to be differentiable and the derivative being continuous at a given point $\vec{a}$ no matter through which continuous contour you traverse the point $\vec{a}$. &lt;/p&gt;&#xA;" OwnerUserId="2987" LastEditorUserId="2987" LastEditorDisplayName="" LastEditDate="2010-11-29T17:06:39.950" LastActivityDate="2010-11-29T17:06:39.950" Title="Conditions for existence of divergence of a vector field." Tags="&lt;vector-calculus&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" />
  <row Id="12357" PostTypeId="2" ParentId="12323" CreationDate="2010-11-29T15:27:21.570" Score="3" ViewCount="" Body="&lt;p&gt;The problem was that math started out of necessity.  Shepherds needed to count how many sheep they own, parents needed to count how many children they had, etc.  For most simple mathematics, zero is simply a place-holder.  Who needs to count zero sheep?  Even the most basic forms of counting--using your fingers--who has zero fingers?  Zero is also pretty closely tied to the concept of negative numbers as well, which are a very non-natural inclusion for mathematics that simply evolve out of necessity.  What shepherd in his right mind has -1 sheep.  Wrapping your head around the idea of zero is easy once the idea is conveyed, but somewhat non-intuitive to arbitrarily include.  Even worse, once you do include zero, how do you deal with all the problems that come along with it such as division by zero.  A great history on the story of how we came to understand zero can be found in the book: &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0140296476&quot; rel=&quot;nofollow&quot;&gt;Zero: The Biography of a Dangerous Idea&lt;/a&gt; by Charles Seife&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/vXuTw.jpg&quot; alt=&quot;Book Cover&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="2470" LastEditorUserId="2470" LastEditorDisplayName="" LastEditDate="2010-11-30T19:25:41.630" LastActivityDate="2010-11-30T19:25:41.630" />
  <row Id="12358" PostTypeId="2" ParentId="12352" CreationDate="2010-11-29T15:38:50.647" Score="2" ViewCount="" Body="&lt;p&gt;Well, you don't even need to assume that the triangle is a right triangle. Simply use the &lt;a href=&quot;http://en.wikipedia.org/wiki/Law_of_sines&quot; rel=&quot;nofollow&quot;&gt;Law of sines&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{a}{\sin A} \,=\, \frac{b}{\sin B} \,=\, \frac{c}{\sin C} .$$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="1119" LastActivityDate="2010-11-29T15:38:50.647" />
  <row Id="12359" PostTypeId="2" ParentId="12356" CreationDate="2010-11-29T15:48:54.700" Score="1" ViewCount="" Body="&lt;p&gt;Perhaps the point is that existence of partial derivatives is not enough for a function of several variables to be &lt;strong&gt;differentiable&lt;/strong&gt; in the multivariable sense, or even necessarily continuous: see &lt;a href=&quot;http://mathoverflow.net/questions/19303/good-example-of-a-non-continuous-function-all-of-whose-partial-derivatives-exist&quot; rel=&quot;nofollow&quot;&gt;this MO question&lt;/a&gt; for a standard example of a function of two variables all of whose partial (and even all directional) derivatives exist but is not even continuous at the origin.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;[&lt;b&gt;Edit&lt;/b&gt;: the following paragraph was added after reading Qiaochu's comment.]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is also possible for all the partial derivatives in a given coordinate system to exist -- e.g. $\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z}$ -- but for directional derivatives in other directions not to exist.  (For instance, let $f: \mathbb{R}^3 \rightarrow \mathbb{R}$ be the function which is $0$ at $(x,y,z)$ if at least two of the variables are zero or if all of $x$,$y$,$z$ have rational coordinates and which is $1$ at all other points.)  Here the divergence will be defined with respect to the standard coordinate system but not with respect to a different coordinate system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, in general a function which has a continuous derivative (or partial derivatives) is much better behaved than a merely differentiable function.  Inserting the hypothesis that a function is $C^1$ -- unless you really need to be considering weaker hypotheses than that for a specific application -- is generally a prudent practice.&lt;/p&gt;&#xA;" OwnerUserId="299" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2010-11-29T16:03:25.687" LastActivityDate="2010-11-29T16:03:25.687" CommentCount="6" />
  <row Id="12360" PostTypeId="2" ParentId="12160" CreationDate="2010-11-29T16:07:17.367" Score="1" ViewCount="" Body="&lt;p&gt;The Abramowitz–Stegun Handbook of Mathematical Functions claims on page 787 that all the roots are simple: &lt;a href=&quot;http://convertit.com/Go/ConvertIt/Reference/AMS55.ASP?Res=150&amp;amp;Page=787&quot; rel=&quot;nofollow&quot;&gt;http://convertit.com/Go/ConvertIt/Reference/AMS55.ASP?Res=150&amp;amp;Page=787&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2307" LastActivityDate="2010-11-29T16:07:17.367" />
  <row Id="12361" PostTypeId="2" ParentId="12160" CreationDate="2010-11-29T16:35:11.370" Score="4" ViewCount="" Body="&lt;p&gt;I think there is a simpler proof that the roots are simple.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Legendre polynomial $\displaystyle P_n(x)$ satisfies the differential equation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle (1-x^2) y&amp;#39;&amp;#39; -2x y&amp;#39; + n(n+1) y = 0$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that, we scale the polynomials so that $\displaystyle P_n(1) = 1$, so if $\displaystyle \alpha$ is a root, then $\displaystyle \alpha \neq 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose $\displaystyle \alpha$ is a root of multiplicity $\displaystyle \gt 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then we must have that $P_n(\alpha) = P_n&amp;#39;(\alpha) = 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The above equation implies that $P_n&amp;#39;&amp;#39;(\alpha) = 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By induction, by differentiating the above equation we can show that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle (1-x^2) y^{(k+2)&amp;#39;} - f_k(x) y^{(k+1)&amp;#39;} + g_k(x) y^{k&amp;#39;} = 0$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\displaystyle y^{l&amp;#39;}$ is the $\displaystyle l^{th}$ derivate of $\displaystyle y$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $\displaystyle \alpha \neq 1$ we see that $\displaystyle P_n^{k&amp;#39;}(\alpha) = 0 \ \ \forall k$ and thus the only root of $\displaystyle P_n(x)$ is $\displaystyle \alpha$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe it is easy to show that $\displaystyle P_n(x) \neq C(x-\alpha)^n$.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-11-29T16:35:11.370" />
  <row Id="12362" PostTypeId="1" CreationDate="2010-11-29T16:42:23.533" Score="0" ViewCount="280" Body="&lt;p&gt;What is the square root of $-4$ (negative 4)?&lt;/p&gt;&#xA;" OwnerUserId="4211" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-29T17:10:37.350" LastActivityDate="2010-12-11T04:54:58.200" Title="what is the square root of $-4$?" Tags="&lt;complex-numbers&gt;" AnswerCount="5" CommentCount="2" />
  <row Id="12363" PostTypeId="2" ParentId="12362" CreationDate="2010-11-29T16:43:19.227" Score="1" ViewCount="" Body="&lt;p&gt;Plus or minus 2i&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-11-29T16:43:19.227" />
  <row Id="12364" PostTypeId="2" ParentId="12362" CreationDate="2010-11-29T16:44:07.667" Score="2" ViewCount="" Body="&lt;p&gt;-4 has no real square roots, since $x^2 \geq 0$.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-29T16:44:07.667" CommentCount="4" />
  <row Id="12365" PostTypeId="2" ParentId="12316" CreationDate="2010-11-29T16:44:46.227" Score="1" ViewCount="" Body="&lt;p&gt;I think you will need to give more details on what is motivating your search for $W$ before you'll get a satisfactory answer. I will make a guess that you have a vector space that has a preferred basis and you are seeking an inner product which renders this basis orthonormal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know of a name for $W$; it is the matrix of the inner product. It should be (symmetric) positive definite.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose we have $d$ linearly independent vectors $\{x_i\}$ in $\mathbb{R}^d$ that are a convenient basis for our work, except that they aren't orthonormal. We may as well assume they are normalized with respect to the usual inner product on $\mathbb{R}^d$. Make a matrix $X$ whose columns are these vectors. Then $X$ is the matrix of a change of basis. Specifically, $X^{-1}y=x$ gives us $x$ as the representation of $y$ in the basis $\{x_i\}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now consider $x^Tx$. That is the inner product we want. It is the inner product which renders our basis orthonormal. Since $x^Tx = y^TX^{-T}X^{-1}y$, we can set $W = X^{-T}X^{-1}$ and we have the desired inner product without having to explicitly perform a change of basis. (Here $X^{-T}$ is the inverse of the transpose of $X$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One context where I have dealt with this is when $X$ was the complete basis of eigenvectors of a linear operator, and I wanted an inner product for which the operator was self-adjoint.&lt;/p&gt;&#xA;" OwnerUserId="1728" LastActivityDate="2010-11-29T16:44:46.227" CommentCount="3" />
  <row Id="12366" PostTypeId="2" ParentId="12362" CreationDate="2010-11-29T16:45:52.063" Score="4" ViewCount="" Body="&lt;p&gt;no it's not -2. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The square root is a number $x$ such that its square is the original number. Depending on the number system you are looking at there might be no square root, for example in the integers or reals. Over the complex numbers there is a square root which other posters have given as 2i or -2i&lt;/p&gt;&#xA;" OwnerUserId="2809" LastEditorUserId="2809" LastEditorDisplayName="" LastEditDate="2010-12-11T04:30:17.237" LastActivityDate="2010-12-11T04:30:17.237" />
  <row Id="12367" PostTypeId="2" ParentId="12328" CreationDate="2010-11-29T17:17:20.133" Score="4" ViewCount="" Body="&lt;p&gt;It is not difficult to prove that one can factor $\rm N$ in polynomial time given any multiple of $\rm \phi(N)$ (here $\rm \:de-1\:$).$\ $ See, for example, Gary Miller: &lt;a href=&quot;http://www.cs.cmu.edu/~glmiller/Publications/Papers/Mi76.pdf&quot; rel=&quot;nofollow&quot;&gt;Riemann's hypothesis and tests for primality. 1976&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;NOTE&lt;/strong&gt; $\ $ This fact was well-known to the discoverers of RSA. Indeed they mention it explicitly in section IX of the &lt;a href=&quot;http://people.csail.mit.edu/rivest/Rsapaper.pdf&quot; rel=&quot;nofollow&quot;&gt;original 1978 paper on RSA&lt;/a&gt;, which is quite readable.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-11-29T17:23:39.080" LastActivityDate="2010-11-29T17:23:39.080" />
  <row Id="12368" PostTypeId="1" AcceptedAnswerId="12371" CreationDate="2010-11-29T17:19:56.447" Score="2" ViewCount="245" Body="&lt;p&gt;Given $x(t) = f(t) \cdot g(t)$, what is the Fourier transform of $x(t)$?  If possible, please explain your answer.&lt;/p&gt;&#xA;" OwnerUserId="4212" LastEditorUserId="5363" LastEditorDisplayName="" LastEditDate="2011-04-17T20:30:06.627" LastActivityDate="2011-04-17T20:30:06.627" Title="What is the Fourier transform of the product of two functions?" Tags="&lt;fourier-analysis&gt;" AnswerCount="2" CommentCount="8" />
  <row Id="12369" PostTypeId="1" AcceptedAnswerId="12406" CreationDate="2010-11-29T17:23:10.080" Score="2" ViewCount="109" Body="&lt;p&gt;I have a couple questions that I'd like answered:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1)  How do I evaluate this integral?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2)  Is it possible to do a substitution?  What about $u=e^{i \cdot t}$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(3)  What if 5 and 3 are replaced by other constant integers?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(4)  What should be the most effective method to handle question (3), using the fewest calculations (like in numerical methods)?  In other words, what is the most effecient way to handle the general case?  (Note that it doesn't have to be a numerical method; I just want something fast!)&lt;/p&gt;&#xA;" OwnerUserId="2626" LastEditorUserId="2626" LastEditorDisplayName="" LastEditDate="2010-11-29T17:34:56.600" LastActivityDate="2010-11-29T23:21:38.443" Title="Help With $\int_{-\pi}^\pi{\cos{(y e^{i\cdot 5\cdot t} + z e^{i\cdot 3\cdot t})}dt}$" Tags="&lt;calculus&gt;&lt;integral&gt;&lt;numerical-methods&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" />
  <row Id="12370" PostTypeId="2" ParentId="12368" CreationDate="2010-11-29T17:41:52.147" Score="2" ViewCount="" Body="&lt;p&gt;Transforms such as Fourier transform or Laplace transform, takes a product of two functions to the convolution of the integral transforms, and vice versa.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is called the &lt;a href=&quot;http://en.wikipedia.org/wiki/Convolution_theorem&quot; rel=&quot;nofollow&quot;&gt;Convolution Theorem&lt;/a&gt;, and is available with proof at &lt;a href=&quot;http://en.wikipedia.org/wiki/Convolution_theorem&quot; rel=&quot;nofollow&quot;&gt;wikipedia&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1119" LastActivityDate="2010-11-29T17:41:52.147" />
  <row Id="12371" PostTypeId="2" ParentId="12368" CreationDate="2010-11-29T17:43:13.563" Score="2" ViewCount="" Body="&lt;p&gt;Fourier transform of a product is the convolution of the corresponding transforms. For details on conditions on the functions refer links below&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Fourier_transform&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Fourier_transform&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Fourier_transform#Convolution_theorem&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Fourier_transform#Convolution_theorem&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3431" LastEditorUserId="3431" LastEditorDisplayName="" LastEditDate="2010-11-29T17:55:23.220" LastActivityDate="2010-11-29T17:55:23.220" CommentCount="2" />
  <row Id="12372" PostTypeId="2" ParentId="12331" CreationDate="2010-11-29T18:26:05.883" Score="3" ViewCount="" Body="&lt;p&gt;This is intended as a complement to Alex Bartel's answer. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1)  Consider the 2D case given by $x = x(t)$, $y = y(t)$.  Here is a graph of a parametrized function with $(x(t_0),y(t_0))$ (in red) and $(x(t_0 + \delta), y(t_0 + \delta))$ (in yellow) marked on it.  The vector $(x(t_0 + \delta), y(t_0 + \delta)) - (x(t_0),y(t_0))$ (from the numerator of the derivative definition) is turquoise, sits almost on top of the graph of the function, and thus is very close to being tangent to the graph of the function at $(x(t_0), y(t_0))$.  Dividing by $\delta$ only changes the length of that vector, not its direction.  Now imagine letting $\delta \to 0$ and watching the yellow and turquoise vectors change.  Graphically, the turquoise vector is getting closer and closer to being a tangent vector at $t_0$.  In the limit, it should actually be a tangent vector.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/l8N24.png&quot; alt=&quot;alt text&quot;&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) (from another perspective)  If $(x(t), y(t), z(t))$ describes the position of some object moving through space, then $(x&amp;#39;(t), y&amp;#39;(t), z&amp;#39;(t))$ is its velocity.  Since the direction of the velocity vector gives the direction the object is moving at an instant in time, $(x&amp;#39;(t_0), y&amp;#39;(t_0), z&amp;#39;(t_0))$ will be tangent to the graph of the position at $t_0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(3) Here's an example in the 2D case where $x&amp;#39;(t)$ and $y&amp;#39;(t)$ both vanish.  The parametrized function is $x(t) = 1+t^3$ and $y(t) = t^2$, with $x&amp;#39;(0) = y&amp;#39;(0) = 0$.  Graphically, we have a cusp, and so there is no well-defined tangent vector.  This is not the only situation that can occur (see Alex's answer and J.M.'s comment), but it does show why an assumption like this is needed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/2u7Wm.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="2370" LastActivityDate="2010-11-29T18:26:05.883" CommentCount="3" />
  <row Id="12373" PostTypeId="2" ParentId="12362" CreationDate="2010-11-29T18:49:53.527" Score="4" ViewCount="" Body="&lt;p&gt;There is no solution to this in the real numbers. The square of any number is greater or equal to zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In complex numbers however, we have &lt;em&gt;defined&lt;/em&gt; something called $i$ with $i \cdot i = -1$ or equivalently $i = \sqrt{-1}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x^2 = -4 = 4\cdot(-1)$$&#xD;&#xA;$$x = \pm \sqrt{4}\cdot\sqrt{-1} = \pm2i$$&lt;/p&gt;&#xA;" OwnerUserId="917" LastActivityDate="2010-11-29T18:49:53.527" />
  <row Id="12374" PostTypeId="1" CreationDate="2010-11-29T19:09:07.240" Score="2" ViewCount="182" Body="&lt;p&gt;I have a system that is modeled by the following differential equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;db/dt = j(t)&lt;em&gt;ha + k&lt;/em&gt;(Ta-b(t))&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where db/dt is the rate of temperature change, j(t) is an input, ha, k, Ta are all constants, and b(t) is an output.  Note that this is newtonian cooling with a heating input, j(t)*ha.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to find the transfer function is the laplace domain, B(s)/J(s).&#xA;Taking the laplacian of the equation of interest, assuming all IC's are 0, yields:&#xA;s*B(s)+k*B(s)-(k*Ta)/s = ha*J(s)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I can't figure out is the term k*Ta/s is not a function of J(s) or B(s) so I can not get a transfer function of purely B(s)/J(s).  Does anyone know how to solve this equation or a better way to find the transfer function relating the input to the output?&lt;/p&gt;&#xA;" LastActivityDate="2010-11-30T21:01:14.707" Title="Transfer Function, Newtonian Cooling" Tags="&lt;differential-equations&gt;&lt;control-theory&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="12376" PostTypeId="1" CreationDate="2010-11-29T19:20:56.743" Score="0" ViewCount="112" Body="&lt;p&gt;I have data for points of a 3D grid. The points of the grid are generated from three nonorthogonal vectors: each grid point has coordinates $\mathbf{q}_{ijk} = i \mathbf{a} + j \mathbf{b} + k \mathbf{c}$, where $\mathbf{a}$, $\mathbf{b}$ and $\mathbf{c}$ are nonorthogonal, noncoplanar vectors. (In cristallography, it's called a &lt;a href=&quot;http://en.wikipedia.org/wiki/Triclinic&quot; rel=&quot;nofollow&quot;&gt;triclinic system&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, my question is: how would you adapt the marching tetrahedra (or marching cubes, if easier) to this case? Has this been treated already somewhere, is there software for that (I haven't found any).&lt;/p&gt;&#xA;" OwnerUserId="3406" LastActivityDate="2010-11-29T22:43:41.683" Title="Marching tetrahedra (or cubes) on a non-orthogonal 3D grid" Tags="&lt;algorithms&gt;&lt;marching-cubes&gt;&lt;marching-tetrahedra&gt;" AnswerCount="1" />
  <row Id="12377" PostTypeId="1" CreationDate="2010-11-29T19:20:58.407" Score="0" ViewCount="81" Body="&lt;p&gt;Suppose I is an index set (not necessarily countable). Say we have a family of subsets $\{C_{i}: i \in I\}$ and suppose that this family is a cover of a space $X$, that is, for each $x$ in $X$ we have $x \in C_{j}$ for some $j \in I$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can we say the following? for each $x \in X$ there exists some natural number $n$ such that $x \in C_{i_{n}}$ where $i \in I$ ? or do we require countability of $I$ in order to guarantee this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me state the question more precisely. I'm trying to understand the proof of the following fact: Let $X$ be a $T_{1}$ space with locally-finite basis. Then $X$ must be discrete. Proof of the author: say $B$ is a locally finite basis. Now pick $x \in B$ then by definition there is some element $B_{1}$ of $B$ such that $x \in B_{1}$. Using recursion and the assumption of $T_{1}$ we can find $B_{n}$ in B such that $x$ in all $B_{n}$ and for all n we have the inclusion $B_{n+1} \subset B_{n}$. Thus $B$ cannot be locally finite, contradiction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Isn't the above proof assuming I is countable? Can you please ellaborate more?&lt;/p&gt;&#xA;" OwnerUserId="4219" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-29T20:18:34.903" LastActivityDate="2010-11-29T20:42:42.567" Title="Doubt (index sets)" Tags="&lt;topology&gt;&lt;elementary-set-theory&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="12379" PostTypeId="2" ParentId="12377" CreationDate="2010-11-29T19:29:13.577" Score="1" ViewCount="" Body="&lt;p&gt;We cannot say it. It depends on the specific family you consider. This is possible even if $I$ is uncountable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, say $X$ has size $\aleph_1$, the first uncountable cardinal. Let $I=X$, and suppose $i\in C_i$ and $X\setminus C_i$ is &lt;em&gt;countable&lt;/em&gt; for each $i\in X$. A natural example is to let $X=\omega_1$, the first uncountable ordinal, and $C_i=[i,\infty)$ with the usual order of ordinals. Then Only countably many $C_i$ are needed to cover $X$, so we can pick for each $x$ an $i_n$ as you require.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if $I=X$ and $C_i=\{i\}$ for each $i$, then of course all $C_i$ are needed. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that we can easily put a topology in each case so the $C_i$ are open sets, so that additional condition would not change the answer.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(So, to &lt;em&gt;guarantee&lt;/em&gt; that it is always possible to pick such $i_n$, we need $I$ to be countable, or some topological condition if we want the $C_i$ to be open, such as $X$ being Lindelöf, which is a weakening of being compact.)&lt;/p&gt;&#xA;" OwnerUserId="462" LastActivityDate="2010-11-29T19:29:13.577" CommentCount="2" />
  <row Id="12380" PostTypeId="2" ParentId="11948" CreationDate="2010-11-29T19:30:19.347" Score="0" ViewCount="" Body="&lt;p&gt;This brought to mind a conjecture I made a couple of years ago which involved a generalisation of the factorial function.&#xA;The integer n! is the product of x taken over all integers from 1 to n and as you know Legendre (and also I think de Polignac) proved the highest power of a prime p diving n! is given by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(floor(n/p)) + (floor(n/p^2)) + (floor(n/p^3)) + . . ..&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Define the arithmetical function&lt;/p&gt;&#xA;&#xA;&lt;p&gt;F(n) := Product {1&amp;lt;=x&amp;lt;=n,1&amp;lt;=y&amp;lt;=n} GCD(x,y).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here GCD(x,y) is the greatest common divisor of x and y. Clearly, we can consider this function as a generalisation of n! = Product {1&amp;lt;=x&amp;lt;=n} GCD(x). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sequence F(n) begins [1,1,2,6,96,480,414720,..], which I submitted as A092287 in the OEIS. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There I conjectured that the highest power of a prime p dividing F(n) is given by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(floor(n/p))^2 + (floor(n/p^2))^2 + (floor(n/p^3))^2 + . . ..&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So F(n) is, in some sense, the next step up from n!.&#xA;Continuing further in this manner, consider products of the GCD of triples of integers:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;G(n) := Product {1&amp;lt;=x&amp;lt;=n,1&amp;lt;=y&amp;lt;=n,1&amp;lt;=z&amp;lt;=n} GCD(x,y,z).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See A129454 in the OEIS.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Calculation suggests that the highest power of a prime p diving G(n) is given by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(floor(n/p))^3 + (floor(n/p^2))^3 + (floor(n/p^3))^3 + . . ..&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we may have an infinite sequence n!, F(n), G(n), ... of arithmetical functions answering your question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One can also use the least common multiple function LCM instead of the GCD function in the above to get similar though more complicated conjectural factorisation results. See A090494 in Sloane for an example. &#xA;Finally, take a look at A092143 and A129365 for yet more conjectural factorisations which have a similar feel to Legendre's result.&lt;/p&gt;&#xA;" LastActivityDate="2010-11-29T19:30:19.347" />
  <row Id="12381" PostTypeId="1" CreationDate="2010-11-29T19:37:40.500" Score="3" ViewCount="98" Body="&lt;p&gt;Dear All,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's say that $f(x)=f^{1}(x)$ and that $f(f(x))=f^{2}(x)$. Moreover, $f^{n}(x)$ is the n-th iterate of $f(x)$, for $n \in \mathbb{N}$. I'm curious about extending iteration to larger number sets. For $n \in \mathbb{R}$, there's the concept &lt;a href=&quot;http://en.wikipedia.org/wiki/Flow_(mathematics)&quot; rel=&quot;nofollow&quot;&gt;flow&lt;/a&gt; (I think?). I don't understand the Wikipedia-article on this subject very well, though. I was hoping for some nice, concrete examples of iterated functions extended to the real or even complex numbers with which I might understand things better. If we take $f(x) = x^2 +3$, for example, what would $f^{\sqrt(2)}(x)$ be? Or, even more ambitiously, say that $g(x)=e^x$ How do we find $g^{\pi^2 + 3i}(x)$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Max Muller&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Editorial to the moderators: perhaps this should be CW?&lt;/p&gt;&#xA;" OwnerUserId="1055" LastEditorUserId="1055" LastEditorDisplayName="" LastEditDate="2010-11-30T13:35:07.610" LastActivityDate="2010-11-30T13:35:07.610" CommunityOwnedDate="2010-12-17T11:21:07.453" Title="Explicit examples of functions with flow?" Tags="&lt;big-list&gt;&lt;functional-analysis&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="12382" PostTypeId="1" AcceptedAnswerId="12398" CreationDate="2010-11-29T19:43:46.313" Score="2" ViewCount="135" Body="&lt;p&gt;Given a topological space (with nice enough conditions, maybe Hausdorff, compactly generated, or CW complex, I'm not sure) $X$ and a subspace $A\subset X$, is it true that $H^n(X,A)\cong H^n(X\backslash A)$ whenever there is an open set containing $A$ which can be retracted to $A$?  Can this be shown easily using the Mayer-Vietoris Sequence, or something else similar to that? (I'm basically assuming simplicial or singular cohomology here, with integer coefficients.)&lt;/p&gt;&#xA;" OwnerUserId="3595" LastActivityDate="2010-11-29T21:22:23.360" Title="Relative Cohomology Isomorphic to Cohomology of Quotient" Tags="&lt;topology&gt;&lt;algebraic-topology&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="12383" PostTypeId="1" AcceptedAnswerId="12385" CreationDate="2010-11-29T19:46:45.943" Score="6" ViewCount="298" Body="&lt;p&gt;Question: (a) Let $f: V \rightarrow W$ with $ V,W \simeq \mathbb{R}^{3}$ given by: $$f(x_1, x_2, x_3) = (x_1 - x_3, 2x_1 -5x_2 -x_3, x_2 + x_3).$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Determine the matrix of $f$ relative to the basis $\{(0,2,1),(-1,1,1),(2,-1,1)\}$ of $V$ and $\{(-1,-1,0),(1,-1,2),(0,2,0)\}$ of $W$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(b) Let $n \in \mathbb{N}$ and $U_n$ the vector space of real polynomials of degree $\leq n$. The linear map $f: U_n \rightarrow U_n$ is given by $f(p) = p&amp;#39;$. Determine the matrix of $f$ relative to the basis $\{1,t,t^{2},...,t^{n}\}$ of $U_n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My attempt so far: &#xA;(a): First relative to the bases of $W$ I found the coordinates of an arbitrary vector: $\left( \begin{array}{r} a \\ b \\ c \end{array} \right) = x \left( \begin{array}{r} -1 \\ -1 \\ 0 \end{array} \right) + y \left( \begin{array}{r} 1 \\ -1 \\ 2 \end{array} \right) + z \left( \begin{array}{c} 0 \\ 2 \\ 0 \end{array} \right)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\begin{array}{l} a = -x + y \\ b = - x - y + 2z \\ c = 2y \end{array}$ or $\begin{array}{l} x = -a + \frac{1}{2}c \\ z = -\frac{1}{2}a + \frac{1}{2}b + \frac{1}{2}c \\ y = \frac{1}{2}c \end{array}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At this point I believe I have the linear combinations of the given basis in $W$ for an arbitrary vector, so next I take the vectors from $V$ and send them to $W$ using the given function:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\begin{array}{l} f(v_1) = f(0,2,1) = (-1,-11,3) = (1 + \frac{3}{2})w_1 + \frac{3}{2}w_2 + (\frac{1}{2} - \frac{11}{2} + \frac{3}{2})w_3 \\ f(v_2) = f(-1,1,1) = (-2,-8,2) = (2+1)w_1 + w_2 + (1 - 4 +1)w_3 \\ f(v_3) = f(2,-1,1) = (1,8,0) = w_1 + (-\frac{1}{2} + 4)w_3 \end{array}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or $\left( \begin{array}{rrc} \frac{5}{2} &amp;amp; 3 &amp;amp; 1 \\ \frac{3}{2} &amp;amp; 1 &amp;amp; 0 \\ -\frac{7}{2} &amp;amp; -2 &amp;amp; \frac{7}{2}\end{array} \right)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Was I taking the correct steps? I didn't really do anything differently based on the fact that $V,W$ were isometric... Is there a particular significance or interpretation for the resulting matrix?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(b): Not really sure here... &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f(p) = p&amp;#39;$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;would it make sense to write something like:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f(1,t,t^{2},\dots, t^{n}) = (0,1,2t, \dots, nt^{n-1})$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and if a basis for $(1,t,t^{2},\dots, t^{n})$ would be $A = \left( \begin{array}{ccccc} 1 &amp;amp; 0 &amp;amp; 0 &amp;amp; \cdots &amp;amp; 0 \\ 0 &amp;amp; 1 &amp;amp; 0 &amp;amp; \cdots &amp;amp; 0 \\ 0 &amp;amp; 0 &amp;amp; 1 &amp;amp; \cdots &amp;amp; 0 \\ \vdots &amp;amp; \vdots &amp;amp; \vdots &amp;amp; \ddots &amp;amp; \vdots \\0 &amp;amp; \cdots &amp;amp; \cdots &amp;amp; 0 &amp;amp; 1 \end{array} \right)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;could i write: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$A&amp;#39; =  \left( \begin{array}{ccccc} 0 &amp;amp; 0 &amp;amp; 0 &amp;amp; \cdots &amp;amp; 0 \\ 1 &amp;amp; 0 &amp;amp; 0 &amp;amp; \cdots &amp;amp; 0 \\ 0 &amp;amp; 1 &amp;amp; 0 &amp;amp; \cdots &amp;amp; 0 \\ \vdots &amp;amp; \vdots &amp;amp; \vdots &amp;amp; \ddots &amp;amp; \vdots \\0 &amp;amp; 0 &amp;amp; 0 &amp;amp; 1 &amp;amp; 0 \end{array} \right)$?&lt;/p&gt;&#xA;" OwnerUserId="3711" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-29T20:14:02.773" LastActivityDate="2010-11-29T21:01:20.560" Title="Determine the matrix relative to a given basis" Tags="&lt;linear-algebra&gt;&lt;matrices&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="12384" PostTypeId="1" AcceptedAnswerId="12388" CreationDate="2010-11-29T20:01:46.603" Score="-6" ViewCount="306" Body="&lt;p&gt;&lt;strong&gt;This question has been edited in view of comments received.Hope it is better now.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the set of all points on the circumference of a circle and the set of all points inside a circle. what is difference in the measure of these two sets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is &lt;a href=&quot;http://en.wikipedia.org/wiki/Bertrand_paradox_%28probability%29&quot; rel=&quot;nofollow&quot;&gt;Bertrand&lt;/a&gt; paradox related to measure theory ? Can bertrand paradox be resolved with the notion of a measure of a set. If so please suggest some good references/books for measure theory. &lt;/p&gt;&#xA;" OwnerUserId="2987" LastEditorUserId="2987" LastEditorDisplayName="" LastEditDate="2010-12-09T08:42:59.643" LastActivityDate="2010-12-09T08:42:59.643" Title="A soft Question Suggestions for good introductory books" Tags="&lt;measure-theory&gt;&lt;paradox&gt;" AnswerCount="2" CommentCount="11" />
  <row Id="12385" PostTypeId="2" ParentId="12383" CreationDate="2010-11-29T20:02:26.830" Score="2" ViewCount="" Body="&lt;p&gt;Yes, your solution to (a) is correct assuming your computations for how to express the elements is correct. Though you don't need to find how to write an &lt;em&gt;arbitrary&lt;/em&gt; vector in terms of the basis of $W$, you only need to find out how to write $f(0,2,1)$, $f(-1,1,1)$, and $f(2,-1,1)$ (the images of the vectors in the given basis for $V$) in terms of the given basis of $W$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;No, you never need to use the fact that $V$ and $W$ are isomorphic (we don't consider a metric here, so &quot;isometric&quot; is not appropriate here). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your answer for (b), on the other hand, is not well done (or correct).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The basis for your vector space is $\{1,t,\ldots,t^n\}$. To find the matrix representation of $f$ relative to this basis, you need to find the image of each basis vector in the domain, and express it in terms of the basis vectors of the range. But remember: the &lt;em&gt;vectors&lt;/em&gt; in $U_n$ are not tuples, they are &lt;em&gt;polynomials&lt;/em&gt;. So $(1,t,\ldots,t^n)$ is not an element of $U_n$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, you would have to find each of $f(1)$, $f(t)$, $f(t^2),\ldots, f(t^n)$, and then express them in terms of the basis of the range, which happens to again be $\{1,t,t^2,\ldots,t^n\}$. So for instance, $f(t^2) = 2t$, so&#xA;$$f(t^2) = 0\cdot 1 + 2\cdot t + 0\cdot t^2 + \cdots + 0\cdot t^n.$$&#xD;&#xA;That means that the third column of the matrix (the one corresponding to the third basis vector, which is $t^2$) will be the transpose of $(0,2,0,\ldots,0)$, that is&#xA;$$\left(\begin{array}{c} 0\\2\\0\\ \vdots \\ 0\end{array}\right).$$&#xD;&#xA;Etc. &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-29T21:01:20.560" LastActivityDate="2010-11-29T21:01:20.560" CommentCount="2" />
  <row Id="12386" PostTypeId="2" ParentId="12384" CreationDate="2010-11-29T20:04:47.433" Score="2" ViewCount="" Body="&lt;p&gt;Here's one:&#xA;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0387952241&quot; rel=&quot;nofollow&quot;&gt; Beginning Functional Analysis by Karen Saxe &lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4216" LastActivityDate="2010-11-29T20:04:47.433" />
  <row Id="12387" PostTypeId="2" ParentId="12381" CreationDate="2010-11-29T20:07:39.473" Score="4" ViewCount="" Body="&lt;p&gt;I thought the wikipedia article is pretty straightforward. Define a function $\phi: \mathbb{R}^2 \to \mathbb{R}: (x,t) \mapsto \phi(x,t)$. Now, you want the second parameter $t$ to be interpreted as the number of times you have applied the function to $x$, in a way. To formalize this, you introduce the following rule on $\phi$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\phi(\phi(x,t),s)=\phi(x,t+s)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and this for all $x,t$ and $s$. In particular, you see that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\phi(\phi(x,t),t)=\phi(x,2t)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and more generally if we define $\phi_t:\mathbb{R}\to\mathbb{R}:x\mapsto\phi(x,t)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\phi_t^n(x)=\phi(x,nt)=\phi_{nt}(x)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is exactly the behaviour you would like to have.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Determining a flow $\phi(x,t)$ starting from the condition that $\phi(x,1)=f(x)$ is not an easy task however and can not be done for any arbitrary function I think. I remember another post related to the question. I think it was a question about the Vieta product.&lt;/p&gt;&#xA;" OwnerUserId="3567" LastActivityDate="2010-11-29T20:07:39.473" CommunityOwnedDate="2010-12-17T11:21:07.453" CommentCount="3" />
  <row Id="12388" PostTypeId="2" ParentId="12384" CreationDate="2010-11-29T20:08:55.213" Score="4" ViewCount="" Body="&lt;p&gt;Lang's book is a fairly comprehensive treatise on Differential Geometry which will introduce you to Riemannian manifold and other notions you mention in first paragraph&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://books.google.com/books?id=VfxGB5nYv1MC&quot; rel=&quot;nofollow&quot;&gt;http://books.google.com/books?id=VfxGB5nYv1MC&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Measure theory can be picked up from Paul Halmos' book&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0387900888&quot; rel=&quot;nofollow&quot;&gt;http://www.amazon.com/dp/0387900888&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: Following Rajesh D's comment below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I liked Munkres book on point set Topology. It also gives a flavor of Algebraic Topology towards the end. Though I have heard mixed opinions from others about this. But it can be read with almost zero background in any sort of math. It builds up from basic set theory (this was incidentally the first mathematics text I ever read).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0131816292&quot; rel=&quot;nofollow&quot;&gt;http://www.amazon.com/dp/0131816292&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Janich book on basic topology is wonderful and reads like a novel. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0387908927&quot; rel=&quot;nofollow&quot;&gt;http://www.amazon.com/dp/0387908927&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3431" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-30T01:36:29.813" LastActivityDate="2010-11-30T01:36:29.813" CommentCount="6" />
  <row Id="12389" PostTypeId="2" ParentId="12381" CreationDate="2010-11-29T20:11:55.010" Score="2" ViewCount="" Body="&lt;p&gt;Max, besides the existence of a lot of literature about this subject I've written two very basic/introductive articles about it. Perhaps the better is the shorter (and newer) one, I deal with the iteration of $ f(x) = 1/(1+x) $ The idea stems originally from a consideration of continued fractions in a general form $ x; 1/(1+x); 1/(1+1/(1+x));....$ . On the other hand, it is simple and the iteration of the function can be approached from different viewpoints, so the approaches are crosschecked mutually.&#xA;See : &lt;a href=&quot;http://go.helms-net.de/math/tetdocs/FracIterAltGeom.htm&quot; rel=&quot;nofollow&quot;&gt;fractional functional iteration of f(x)=1/(1+x) &lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;additional comment: the difficulties of defining a continuous iteration of $f(x) = e^x$ compared with that of a polynomial $f(x)=x^2 + 3$ are so eminent that putting them into the same question requires a much too wide area for discussion in a MO-thread[Edit: Oh sorry, we're not in MO, I was jumping between tabs in my browser, sorry]. However in the last years a lot of material, including scans of historical articles, was put online and can be read. Possibly I can provide useful links if you ask more specific questions. Also you might browse the tetration-forum at &lt;a href=&quot;http://math.eretrandre.org/tetrationforum/index.php&quot; rel=&quot;nofollow&quot;&gt;Tetration-forum&lt;/a&gt; There is also an online reference-database. &lt;a href=&quot;http://math.eretrandre.org/tetrationforum/showthread.php?tid=365&quot; rel=&quot;nofollow&quot;&gt;see tet-forum&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[Update] i constructed a function f(x,h) where f(x,0)=x, f(x,1)=x^2+3 and f(x,h) the h'th iterate using fixpoint-shift and matrix-diagonalization. Unfortunately this gives complex powerseries for fractional iterates. The construction gives the following complex plot for x=1,h=0..2 . note that for real, noninteger heights h the curve extends into the complex plane! So I think, there should be a better solution...    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/g8FDe.png&quot; alt=&quot;provisorial construction of f(x,h)&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1714" LastEditorUserId="1714" LastEditorDisplayName="" LastEditDate="2010-11-29T23:06:47.277" LastActivityDate="2010-11-29T23:06:47.277" CommunityOwnedDate="2010-12-17T11:21:07.453" CommentCount="3" />
  <row Id="12390" PostTypeId="2" ParentId="12383" CreationDate="2010-11-29T20:18:10.467" Score="2" ViewCount="" Body="&lt;p&gt;Here's a suggestion for seeing the bigger picture in your first question:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The key idea below is that the column vectors of a matrix $M$ are the images of the standard ordered basis when the matrix $M$ is viewed as a transformation&#xA;$M: \mathbb{R}^n \to \mathbb{R}^n$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;It is easy to find a matrix of f with respect to the standard ordered basis of $\mathbb{R}^3$: it is given by a matrix with columns $\{1, 2, 0\}$, $\{0, -5, 1\}$, and $\{-1, -1, 1\}$.  Call this matrix $C$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The matrix which transforms the given ordered basis of $V$ to the standard basis of $\mathbb{R}^3$ is given by writing the basis vectors as columns of this matrix. Similarly, this can be done for the basis of $W$.  Let's call these matrices $A$ and $B$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;You desire a matrix which represents $f$.  Interpreting the matrices above as transformations, consider the following diagram:&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;$$\begin{array}{ccc}&#xD;&#xA;V &amp;amp; \to &amp;amp; \mathbb{R}^3\\&#xD;&#xA;\downarrow &amp;amp; \qquad &amp;amp; \downarrow\\&#xD;&#xA;W &amp;amp; \to &amp;amp; \mathbb{R}^3&#xD;&#xA;\end{array}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where the horizontal maps are $A$ and $B$ and the right vertical map is $C$.&#xA;Your desired matrix is the left vertical arrow.  It is now clear that it is given by the matrix $B^{-1} C A$.&lt;/p&gt;&#xA;" OwnerUserId="4009" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-29T20:26:18.437" LastActivityDate="2010-11-29T20:26:18.437" CommentCount="1" />
  <row Id="12391" PostTypeId="2" ParentId="12381" CreationDate="2010-11-29T20:19:22.710" Score="2" ViewCount="" Body="&lt;p&gt;Since you ask for explicit examples, the solution to any time-invariant ordinary differential equation as a function of time and the initial value of the variable is a flow. For example, $dx/dt = \lambda x$ with initial value $x(0) = x_0$ has the solution $f(t,x_0) = e^{\lambda t} x_0$, which in your notation can be interpreted as $f^t(x) = e^{\lambda t} x$. This is a somewhat boring example; the interesting ones happen in more than one dimension. Rotation, for example: let $f^t(x)$ be the rotation of $x$ about a fixed axis by an angle $\omega t$. I'll leave it to you to think of more complicated ODEs.&lt;/p&gt;&#xA;" OwnerUserId="856" LastActivityDate="2010-11-29T20:19:22.710" CommunityOwnedDate="2010-12-17T11:21:07.453" CommentCount="0" />
  <row Id="12392" PostTypeId="2" ParentId="12377" CreationDate="2010-11-29T20:42:42.567" Score="1" ViewCount="" Body="&lt;p&gt;You are misunderstanding the author's argument.  Suppose $x \in B_1$.  By local finiteness, $B_1$ only contains finitely many elements of $X$, say $n$.  If $y \in B_1$ is any element other than $x$, then by the $T_1$ assumption $B_1 - \{ y \} = B_2$ is also an open set containing $x$.  Continuing in this way we obtain a sequence $B_1 \supset B_2 \supset ... $ of strictly decreasing open sets containing $x$ which stabilizes after at most $n$ steps; this gives an open set $B_n = \{ x \}$, hence the topology is discrete.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-11-29T20:42:42.567" />
  <row Id="12393" PostTypeId="1" CreationDate="2010-11-29T20:45:09.880" Score="1" ViewCount="144" Body="&lt;p&gt;I'm looking for a graphic that shows the isomorphism theorems illustrated using a finite vector space. (that is a finite dimensional vector space over a finite field.) Something that shows nicely how the cosets &quot;collapse&quot; in to single elements or how cosets at different &quot;levels&quot; are matched up to form the isomorphisms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have tried searching but I keep finding arrow diagrams.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternately, suggestions for good examples that don't take too long to draw would be appreciated.   &lt;/p&gt;&#xA;" OwnerUserId="2563" LastActivityDate="2011-05-26T05:02:07.653" Title="Pictorial representations of the isomorphism theorems" Tags="&lt;linear-algebra&gt;&lt;abstract-algebra&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="12394" PostTypeId="1" AcceptedAnswerId="12409" CreationDate="2010-11-29T20:46:47.580" Score="2" ViewCount="269" Body="&lt;p&gt;I have a radius, &lt;code&gt;R&lt;/code&gt;, for an aircraft traveling at velocity, &lt;code&gt;V&lt;/code&gt;. If we start at point, &lt;code&gt;(X,Y)&lt;/code&gt;, what is the position of the point at time, &lt;code&gt;t&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example:&lt;br&gt;&#xA;The aircraft is at point &lt;code&gt;(0,0)&lt;/code&gt; and traveling at &lt;code&gt;250&lt;/code&gt; knots and initiates a turn with a bank angle, phi, of &lt;code&gt;5&lt;/code&gt; degrees. Assume that the aircraft can instantaneously rotate to the five degree bank. The equation for the turn radius, R where g is the acceleration due to gravity (9.81) is:&#xA;\begin{equation}&#xA;\text{R} = \frac{V^2}{\text{g} \tan{\phi}}&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For this example, &lt;code&gt;R = 10.4&lt;/code&gt; nautical miles. Where is the aircraft at &lt;code&gt;t = 2&lt;/code&gt; if the aircraft is traveling at a heading of &lt;code&gt;90&lt;/code&gt; degrees (straight along the y axis)?&lt;/p&gt;&#xA;" OwnerUserId="4222" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2010-11-30T14:22:07.523" LastActivityDate="2011-04-19T18:02:15.470" Title="Given a radius and velocity calculate position of an aircraft banking to make a turn" Tags="&lt;geometry&gt;&lt;homework&gt;&lt;physics&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="1" />
  <row Id="12395" PostTypeId="2" ParentId="12394" CreationDate="2010-11-29T20:51:01.497" Score="1" ViewCount="" Body="&lt;p&gt;I don't know what you mean by R.  If you start at the origin, $v_x=v\cos(\theta)$ and $v_y=v\sin(\theta)$, $x=v_xt$ and $y=v_yt$.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-11-29T20:51:01.497" CommentCount="4" />
  <row Id="12396" PostTypeId="2" ParentId="1901" CreationDate="2010-11-29T20:51:56.813" Score="1" ViewCount="" Body="&lt;p&gt;The following (famous) question is perhaps in the spirit of the OP's question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $M$ and $N$ are closed aspherical $n$-manifolds with isomorphic fundamental groups, are $M$ and $N$ homeomorphic?  (This is known as the Borel Conjecture.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Scott Carter's example illustrates the need for the assumption that the manifolds are closed.  But the question above suggests that perhaps the fundamental group is sufficient to characterize the topological type of a closed aspherical manifold.&lt;/p&gt;&#xA;" OwnerUserId="4009" LastActivityDate="2010-11-29T20:51:56.813" />
  <row Id="12397" PostTypeId="1" AcceptedAnswerId="16108" CreationDate="2010-11-29T21:07:00.270" Score="1" ViewCount="137" Body="&lt;p&gt;For a given radius, theta, and position, can you calculate a Bezier curve?&lt;/p&gt;&#xA;" OwnerUserId="4222" LastActivityDate="2011-01-02T09:44:32.633" Title="Given a radius, calculate Bezier curve" Tags="&lt;geometry&gt;&lt;bezier-curve&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="12398" PostTypeId="2" ParentId="12382" CreationDate="2010-11-29T21:22:23.360" Score="5" ViewCount="" Body="&lt;p&gt;It should say reduced cohomology of $X/A$, I believe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, it is always true that the reduced cohomology of $X \cup CA$ (for $CA$ the cone) is isomorphic to the relative cohomology. To see this,  one removes a small contractible piece near the top. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The exact sequence shows that $\widetilde{H}^*(X \cup CA) \simeq \widetilde{H}^*(X \cup CA, C A)$ because the cone is contractible. Then, excise the top of the cone. The pair then becomes homotopy equivalent to the pair $(X, A)$. So this proves the claim.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The next fact that one needs to show is that $X/A$ has the same cohomology as $X \cup CA$. This is true if the inclusion $A \to X$ is a cofibration. In general, it is a fact that if $B \subset Y$ is a cofibration, and $B$ is contractible, then $Y$ and $Y/B$  have the same homotopy type. (Apply this with $Y = X \cup &#xD;&#xA;CA, B = CA$ to see that $X \cup CA$ and $X/A$ have the same homotopy type under cofibration conditions.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To see this, one takes a contracting homotopy $B \times I \to B$ of the identity to a point, and extends it to a homotopy $\phi: Y \times I \to Y$ by the cofibration property. This sends $B \times \{1\}$ into a point. So $\phi(., 1)$ is a map $Y/B \to Y$. The composite of $\phi(.,1)$ and $Y \to Y/B$ is homotopic to $1_{Y/B}$ by $\phi$ (since $\phi(.,t)$ always sends $B$ into itself).&lt;/p&gt;&#xA;" OwnerUserId="536" LastActivityDate="2010-11-29T21:22:23.360" CommentCount="1" />
  <row Id="12399" PostTypeId="2" ParentId="12394" CreationDate="2010-11-29T22:18:39.520" Score="2" ViewCount="" Body="&lt;p&gt;There is still not enough information to answer the question.  The initial speed is specified, but not the direction.  Given the direction of travel, you can find the center of the circle.  Then (as in my last comment) the angular velocity is V/R and you can apply the same trig functions.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-11-29T22:18:39.520" CommentCount="3" />
  <row Id="12400" PostTypeId="1" CreationDate="2010-11-29T22:23:35.473" Score="4" ViewCount="96" Body="&lt;p&gt;Let $F_n$ be an $n$-generator free group with a free basis $x_1,\ldots,x_n.$ Is it true that the stabilizer of $x_1$ in $\mathrm{Aut}(F_n)$ is generated by all left and right Nielsen moves $\lambda_{ij}$ and $\rho_{ij}$ such that $i \ne 1$ and by the element of order two $\epsilon_n$ such that $\epsilon_n(x_n)=x_n^{-1}$ while other elements of the basis remain fixed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $i \ne j$ and $1 \le i,j \le n.$ The left Nielsen move $\lambda_{ij}$ takes $x_i$ to $x_j x_i$ and the right Nielsen move $\rho_{ij}$ takes $x_i$ to $x_i x_j;$ both $\lambda_{ij}$ and $\rho_{ij}$ fix all $x_k$ with $k \ne i.$ &lt;/p&gt;&#xA;" LastActivityDate="2010-11-29T23:29:20.220" Title="Stab(x_1) in Aut( F(x_1,...x_n ) )" Tags="&lt;group-theory&gt;" CommentCount="3" />
  <row Id="12401" PostTypeId="2" ParentId="12376" CreationDate="2010-11-29T22:43:41.683" Score="1" ViewCount="" Body="&lt;p&gt;Nothing in Marching Cubes (or Marching Tetrahedra, AFAIK) actually requires that the grid be cubical or even orthogonal; the algorithms are inherently topological (or arguably, combinatorial) in nature.  The heart of Marching Cubes is really just a table matching the $2^8$ combinations of positive/negative function values at the cube corners to a set of topological triangulations that are consistent with the 'simplest' isosurface of a function taking on those values (e.g., when only one vertex is positive we have a single triangle, with vertices along the three edges connecting the positive vertex to its neighbors), and those triangulations are invariant under linear transformations - which is all your skewed grid represents.&lt;/p&gt;&#xA;" OwnerUserId="785" LastActivityDate="2010-11-29T22:43:41.683" CommentCount="1" />
  <row Id="12402" PostTypeId="1" CreationDate="2010-11-29T22:46:59.990" Score="3" ViewCount="89" Body="&lt;p&gt;Problem: The series $\sum_{n=1}^{\infty} a_n$ diverge and positive. What can be said on the series $\sum_{n=1}^{\infty} \frac{a_n}{1+n^{2}a_n}$ ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Approach:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;First, I tried separating it into two cases: $a_n \to \infty$, $a_n \to A$ (where $A$ is some const value). &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Tried the ratio test and got nowhere with that.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I think that the right approach is to check each case separately, but my hunch tells me that there is a workaround.&lt;/p&gt;&#xA;" OwnerUserId="3637" LastActivityDate="2010-11-29T22:52:19.520" Title="Question about Infinite Series" Tags="&lt;calculus&gt;&lt;sequences-and-series&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" />
  <row Id="12403" PostTypeId="2" ParentId="4362" CreationDate="2010-11-29T22:48:46.400" Score="3" ViewCount="" Body="&lt;p&gt;Noam Elkies gives an example &lt;a href=&quot;http://www.math.harvard.edu/~elkies/Misc/sol8.html&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1770" LastActivityDate="2010-11-29T22:48:46.400" />
  <row Id="12404" PostTypeId="2" ParentId="12402" CreationDate="2010-11-29T22:52:07.147" Score="7" ViewCount="" Body="&lt;p&gt;Hint: Rewrite as $$\sum_{n=1}^{\infty} \frac{1}{\frac{1}{a_n} + n^2}.$$&lt;/p&gt;&#xA;" OwnerUserId="2370" LastActivityDate="2010-11-29T22:52:07.147" CommentCount="3" />
  <row Id="12405" PostTypeId="2" ParentId="12402" CreationDate="2010-11-29T22:52:19.520" Score="2" ViewCount="" Body="&lt;p&gt;Divide the numerator and denominator by $a_n$, and compare what you get with $1/n^2$.&lt;/p&gt;&#xA;" OwnerUserId="2810" LastActivityDate="2010-11-29T22:52:19.520" />
  <row Id="12406" PostTypeId="2" ParentId="12369" CreationDate="2010-11-29T23:11:25.843" Score="3" ViewCount="" Body="&lt;p&gt;This is another contour integral. Replace $y$ by $a$ and $z$ by $b$ to simplify notation. Then your integral is&#xA;$$-i \int_{-\pi}^{\pi}{\cos(ae^{5it} + be^{3it}) \over e^{it}} ie^{it}\,dt$$&#xD;&#xA;$$= -i \int_{|z| = 1}{ \cos(az^5 + bz^3) \over z}\,dz$$&#xD;&#xA;By the residue theorem this integral is $2\pi$ times the residue of $\cos(az^5 + bz^3)$ at $z = 0$. Since $\cos(az^5 + bz^3) = 1$ when $z = 0$, this residue is $1$. So your overall integral is $2\pi$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that it doesn't matter what $a$ or $b$ are, and that $3$ and $5$ can be replaced by any nonnegative integers.&lt;/p&gt;&#xA;" OwnerUserId="3035" LastEditorUserId="3035" LastEditorDisplayName="" LastEditDate="2010-11-29T23:21:38.443" LastActivityDate="2010-11-29T23:21:38.443" CommentCount="6" />
  <row Id="12407" PostTypeId="1" CreationDate="2010-11-29T23:34:12.947" Score="0" ViewCount="91" Body="&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/bBDqg.png&quot; alt=&quot;N=2762&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;N=2762&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've been exploring a data set that seems to give rise to this kind of qqplot rather frequently. Would you say this is one population with a different than normal population? Or are two populations confounding the normal distribution?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It used matplotlib and scipy.stats:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;      (x,y), (slope, inter, cor) = stats.probplot(data, dist='norm')&#xA;&#xA;      osmf = x.take([0, -1])  # endpoints&#xA;      osrf = slope * osmf + inter&#xA;      self.ax.plot(x, y, ',', osmf, osrf, '-', **self.kwargs)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="4225" LastActivityDate="2010-11-29T23:34:12.947" Title="Does this qqplot suggest a distribution?" Tags="&lt;statistics&gt;" CommentCount="2" />
  <row Id="12408" PostTypeId="1" AcceptedAnswerId="12411" CreationDate="2010-11-29T23:40:13.343" Score="2" ViewCount="86" Body="&lt;p&gt;A &lt;a href=&quot;http://en.wikipedia.org/wiki/Regular_sequence_%28algebra%29&quot; rel=&quot;nofollow&quot;&gt;regular sequence&lt;/a&gt; is an $n$-fold collection $\{r_1, \cdots, r_n\} \subset R$ of elements of a ring $R$ such that for any $2 \leq i \leq n$, $r_i$ is not a zero divisor of the quotient ring&#xA;$$ \frac R {\langle r_1, r_2, \cdots, r_{i-1} \rangle}.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Does the order of the $r_i$'s matter? That is, is any permutation of a regular sequence regular?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="3008" LastActivityDate="2010-11-29T23:59:15.177" Title="Is a regular sequence ordered?" Tags="&lt;commutative-algebra&gt;" AnswerCount="2" />
  <row Id="12409" PostTypeId="2" ParentId="12394" CreationDate="2010-11-29T23:42:45.413" Score="3" ViewCount="" Body="&lt;p&gt;See &lt;a href=&quot;http://en.wikipedia.org/wiki/Bank_angle#Banked_turn_in_aeronautics&quot; rel=&quot;nofollow&quot;&gt;banked turn in  aeronautics&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;In a motion along a circular path centered at $(0,0)$ (see figure below) with constant angular velocity $\omega$ the position of a point $P(x,y)$ (in &lt;a href=&quot;http://en.wikipedia.org/wiki/Cylindrical_coordinate_system&quot; rel=&quot;nofollow&quot;&gt;Cartesian coordinates&lt;/a&gt; and $P(r,\varphi)$ in &lt;a href=&quot;http://en.wikipedia.org/wiki/Cylindrical_coordinate_system&quot; rel=&quot;nofollow&quot;&gt;cylindrical coordinates&lt;/a&gt;) starting at $(r,0)$ at $t=0$ is given by the law of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Circular_motion&quot; rel=&quot;nofollow&quot;&gt;uniform circular motion&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x=r\cos(\omega t)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$y=r\sin(\omega t),$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $r$ is the radius of the circumference and $\omega=\dfrac{d\varphi}{dt}$. The magnitude of the tangential velocity is $v=\omega r$ and the radial velocity is zero. Thus $\omega=\dfrac{v}{r}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x=r\cos\left(\frac{vt}{r}\right)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$y=r\sin\left(\frac{vt}{r}\right)$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your case $r=R$, $v=V$ (it is assumed that the aircraft mantains the speed $V$) and the centre of the circumference is $(X,Y)=(-R,0)$. If you make the change of coordinates $X=x-R$ and $Y=y$, you get the following $X,Y$ coordinates as a function of time $t$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$X=R\left(\cos\left(\frac{V}{R}t\right)-1\right)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$Y=R\sin\left(\frac{V}{R}t\right)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this $XY$-coordinate system the motion starts at $(X,Y)=(0,0)$ ($t=0$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I assumed that the direction of the turn is to the left. If it is to the right, then one has a symmetric motion with respect to the $y-$axis:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$X=R\left(-\cos\left(\frac{V}{R}t\right)+1\right)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$Y=R\sin\left(\frac{V}{R}t\right).$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/S7zj4.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The numerical result is obtained for $t=2$ and the other given data. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Added&lt;/em&gt;: For a different starting position $\left( X_{0},Y_{0}\right) $ at $t=0$ we have to incorporate these values in the motion equations. Integrating $%&#xD;&#xA;\omega =\dfrac{d\varphi }{dt}$, we obtain $\varphi =\omega t+C$, where $C$ is&#xA;a constant. Then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x=r\cos \left( \omega t+C\right) ,y=r\sin \left( \omega t+C\right) $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x_{0}=r\cos \left( C\right) ,y_{0}=r\sin \left( C\right) $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\dfrac{y_{0}}{x_{0}}=\dfrac{\sin \left( C\right) }{\cos \left( C\right) }%&#xD;&#xA;=\tan \left( C\right) ,C=\arctan \left( \dfrac{y_{0}}{x_{0}}\right) $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x=r\cos \left( \omega t+\arctan \left( \dfrac{y_{0}}{x_{0}}\right) \right)&#xD;&#xA;,y=r\sin \left( \omega t+\arctan \left( \dfrac{y_{0}}{x_{0}}\right) \right) .$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$X=x-r=r\cos \left( \omega t+\arctan \left( \dfrac{y_{0}}{x_{0}}\right)&#xD;&#xA;\right) -r,Y=y=r\sin \left( \omega t+\arctan \left( \dfrac{y_{0}}{x_{0}}%&#xD;&#xA;\right) \right) $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;we get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$X_{0}=x_{0}-r,Y_{0}=y_{0},\dfrac{y_{0}}{x_{0}}=\dfrac{Y_{0}}{X_{0}+r}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$r=R,\omega =\dfrac{v}{r}=\dfrac{V}{R},\dfrac{y_{0}}{x_{0}}=\dfrac{Y_{0}}{X_{0}+R}.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, we obtain the equations of the motion of the aircraft in the $X,Y$-plane:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$X=x-R=R\left( \cos \left(\frac{V}{R}t+\arctan \left(\frac{Y_{0}}{X_{0}+R} \right) \right) -1\right) $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$Y=y=R\sin \left(\frac{V}{R}t+\arctan \left(\frac{Y_{0}}{X_{0}+R}\right) \right),$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\left( X_{0},Y_{0}\right) $ is the position of the aircraft at $t=0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the direction of the turn is to the right, then the motion is symmetric with respect to the $y-$axis:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$X=-x+R=R\left( -\cos \left(\frac{V}{R}t+\arctan \left(\frac{Y_{0}}{X_{0}+R} \right) \right) +1\right) $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$Y=y=R\sin \left(\frac{V}{R}t+\arctan \left(\frac{Y_{0}}{X_{0}+R}\right) \right).$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Note&lt;/em&gt;: I changed the notation; in the question the position of the aircraft&#xA;at $t=0$ is $\left( X,Y\right) $.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The velocity vector is $\overrightarrow{v}=v\overrightarrow{e}_{\varphi }$ and the acceleration vector $\overrightarrow{a}=-\dfrac{v^{2}}{r}\overrightarrow{e}_{r}$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="752" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2010-11-30T12:49:11.563" LastActivityDate="2010-11-30T12:49:11.563" CommentCount="2" />
  <row Id="12410" PostTypeId="1" AcceptedAnswerId="12420" CreationDate="2010-11-29T23:43:27.350" Score="4" ViewCount="149" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider a particle starting at the the origin and moving along the positive real line at a constant speed of 1.  Suppose there is a counter which clicks at random time intervals following the &lt;a href=&quot;http://en.wikipedia.org/wiki/Exponential_distribution&quot; rel=&quot;nofollow&quot;&gt;exponential distribution&lt;/a&gt; with parameter $\lambda$ and whenever the counter clicks, the position $x &amp;gt; 0$ of the particle at that time instantaneously changes to the position $x/2$.  We wish to calculate the expected average speed of the particle.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't really have any idea of how to go about solving this.  Here are a couple of related problems which seem even more difficult to me:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Modify the puzzle so that when the counter clicks, the particle moves from $x$ to a point chosen uniformly at random from $[0,x]$.&lt;/li&gt;&#xA;&lt;li&gt;The particle starts moving as above but whenever the counter clicks, its speed increases by 1 (the initial speed was 1).  What is the expected time when the particle hits the position 1?  What is the expected speed when the particle hits the position 1?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;This is not a homework problem.  Any solutions, hints, thoughts will be appreciated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks,&lt;/p&gt;&#xA;" OwnerUserId="3831" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-11-29T23:53:19.333" LastActivityDate="2010-12-01T06:14:38.493" Title="Particle moving at constant speed with Poisson setbacks" Tags="&lt;puzzle&gt;&lt;probability&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="0" />
  <row Id="12411" PostTypeId="2" ParentId="12408" CreationDate="2010-11-29T23:44:55.880" Score="4" ViewCount="" Body="&lt;p&gt;Not in general. A standard example is $R=k[x,y,z]$, where $k$ is a field. Then, $x,y(1-x),z(1-x)$ is regular but $y(1-x),z(1-x),x$ is not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the bright side, if $R$ is Noetherian, local then every permutation of a regular sequence is regular.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact more is true. We can extend this notion to modules over rings analogously. Then, if $M$ is a finitely generated module over a Noetherian, local ring, then every permutation of a regular sequence is regular.&lt;/p&gt;&#xA;" OwnerUserId="3431" LastEditorUserId="3431" LastEditorDisplayName="" LastEditDate="2010-11-29T23:59:15.177" LastActivityDate="2010-11-29T23:59:15.177" CommentCount="2" />
  <row Id="12412" PostTypeId="2" ParentId="12408" CreationDate="2010-11-29T23:45:50.257" Score="2" ViewCount="" Body="&lt;p&gt;In general, no. If $R$ is a local noetherian ring, then yes, though. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A counterexample is if $R = k[x,y,z]/(x-1)z$ over a field $k$, and the elements $x, (x-1)y$. This is a regular sequence even though $(x-1)y$ is a zerodivisor (so the swapped sequence is not regular).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is true under local hypotheses by the Krull intersection theorem; see&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://amathew.wordpress.com/2010/10/30/the-notion-of-a-regular-sequence/&quot; rel=&quot;nofollow&quot;&gt;http://amathew.wordpress.com/2010/10/30/the-notion-of-a-regular-sequence/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="536" LastActivityDate="2010-11-29T23:45:50.257" />
  <row Id="12413" PostTypeId="1" AcceptedAnswerId="12415" CreationDate="2010-11-29T23:59:40.840" Score="3" ViewCount="109" Body="&lt;p&gt;Currently taking a logic class and trying to understand this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have two set A and B. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both sets are empty sets. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is set A a subset of the complement of set B?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume the context is the universal set.&lt;/p&gt;&#xA;" LastEditorUserId="462" LastEditorDisplayName="" LastEditDate="2010-11-30T00:09:57.003" LastActivityDate="2010-11-30T00:09:57.003" Title="Is there and empty set in the complement of an empty set?" Tags="&lt;elementary-set-theory&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="12414" PostTypeId="2" ParentId="12413" CreationDate="2010-11-30T00:06:03.700" Score="1" ViewCount="" Body="&lt;p&gt;The empty set is a subset of any other set.&lt;/p&gt;&#xA;" OwnerUserId="3947" LastActivityDate="2010-11-30T00:06:03.700" />
  <row Id="12415" PostTypeId="2" ParentId="12413" CreationDate="2010-11-30T00:09:35.100" Score="7" ViewCount="" Body="&lt;p&gt;The answer is yes. But there are several comments that need to be made: &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;There is &lt;em&gt;only one&lt;/em&gt; empty set. So it is better to say that $A=B=\emptyset$ rather than saying that &quot;both $A$ and $B$ are empty sets&quot;, as the latter erroneously suggests that there is more than one. This is because two sets are equal precisely if they have the same elements. So any two empty sets are equal, since they have precisely the same elements (namely, none).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I assume by &quot;context&quot; you mean that the complement of $B$ is computed with respect to the &quot;universal set.&quot; In the standard system of set theory, there can be no &quot;universal set&quot;, as assuming its existence leads to problems (Russell paradox). [Though, yes, it is usual to talk of a &quot;universal set&quot; as a way to delineate what objects we are interested in.]&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The reason why $A$ is contained in the complement of $B$ is that $A$ (being the empty set) is a subset of &lt;em&gt;any&lt;/em&gt; set. This is because we define &quot;$A$ is contained in $C$&quot; to mean that any element of $A$ is also an element of $C$. Now, since nothing is an element of $A$, this condition is satisfied in this case (one typically says that it is satisfied &lt;em&gt;vacuously&lt;/em&gt;.)&lt;/p&gt;&#xA;" OwnerUserId="462" LastActivityDate="2010-11-30T00:09:35.100" />
  <row Id="12416" PostTypeId="1" AcceptedAnswerId="12417" CreationDate="2010-11-30T00:13:04.817" Score="4" ViewCount="127" Body="&lt;p&gt;This is a follow up to a &lt;a href=&quot;http://math.stackexchange.com/questions/12042/recurrence-relation-satisfied-by-lfloor1-sqrt5n-rfloor&quot;&gt;question&lt;/a&gt; I had asked earlier about a linear recurrence relationship satsified by $\lfloor(1+\sqrt{5})^n\rfloor$. I messed up there, and I actually meant to ask about $L(n)=\lfloor(1+\sqrt{3})^n\rfloor$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Following Douglas' suggestion I have determined that the values (at least the first 1000) satisfy the following recurrence: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$L(2n+5)=8L(2n+3)-4L(2n+1)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is how do I prove something like this. I can prove the recurrence for the values inside the floor function, but floor function in general does not commute with addition and multiplication. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Explicitly, it's easy to show&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(1+\sqrt{3})^{2n+5}=8(1+\sqrt{3})^{2n+3}-4(1+\sqrt{3})^{2n+1}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but I am not sure how to prove the recurrence from here.&lt;/p&gt;&#xA;" OwnerUserId="3947" LastActivityDate="2010-11-30T00:18:19.133" Title="Recurrence relation satisfied by $\lfloor(1+\sqrt{3})^n\rfloor$" Tags="&lt;discrete-mathematics&gt;" AnswerCount="1" />
  <row Id="12417" PostTypeId="2" ParentId="12416" CreationDate="2010-11-30T00:18:19.133" Score="8" ViewCount="" Body="&lt;p&gt;Prove the same relation for $1-\sqrt 3$, and check that the resulting powers are all smaller than 1, all negative, and when added to the same powers of $1+\sqrt 3$ you end up with an integer: For all $k=1,2,\dots$, $$(1+\sqrt 3)^k+(1-\sqrt 3)^k $$ is an integer. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are several ways of checking this fact. For example, by induction. Or by using the Binomial theorem.&lt;/p&gt;&#xA;" OwnerUserId="462" LastActivityDate="2010-11-30T00:18:19.133" CommentCount="4" />
  <row Id="12418" PostTypeId="1" CreationDate="2010-11-30T00:23:22.993" Score="0" ViewCount="91" Body="&lt;p&gt;$$\sum_{n=1}^{\infty}\frac{1}{2^n3^{(n-1)}}$$&#xD;&lt;/p&gt;&#xA;" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2010-11-30T00:30:15.337" LastActivityDate="2010-11-30T00:38:20.147" Title="whether the following series converge or diverge" Tags="&lt;calculus&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="12419" PostTypeId="2" ParentId="12418" CreationDate="2010-11-30T00:29:03.877" Score="4" ViewCount="" Body="&lt;p&gt;Your series can be written as $$\sum_{n=1}^\infty \frac {3}{6^n}.$$&#xD;&#xA;It is a geometric series with $a=3/6=1/2, r=1/6$. &lt;/p&gt;&#xA;" OwnerUserId="3249" LastActivityDate="2010-11-30T00:29:03.877" />
  <row Id="12420" PostTypeId="2" ParentId="12410" CreationDate="2010-11-30T00:31:24.137" Score="3" ViewCount="" Body="&lt;p&gt;Sketch of solution: Let $X_t$ denote the location of the particle at time $t$, and set $f(t) = {\rm E}(X_t)$. Consider the time $t + \Delta t$, $\Delta t \approx 0+$. With probability of about $1 - \lambda \Delta t$ the particle continues to position $X_t + \Delta t$, whereas with probability of about &#xA;$\lambda \Delta t$ it moves to position of about $X_t/2$. This leads straightforwardly to an elementary differential equation in terms of $f(t)$, from which you obtain $f(t)$. The expected average speed is then $f(t)/t$. &lt;/p&gt;&#xA;" OwnerUserId="2810" LastEditorUserId="2810" LastEditorDisplayName="" LastEditDate="2010-11-30T00:48:41.180" LastActivityDate="2010-11-30T00:48:41.180" CommentCount="7" />
  <row Id="12421" PostTypeId="2" ParentId="12418" CreationDate="2010-11-30T00:38:20.147" Score="1" ViewCount="" Body="&lt;p&gt;You could use the &lt;a href=&quot;http://en.wikipedia.org/wiki/Comparison_test&quot; rel=&quot;nofollow&quot;&gt;comparison test&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\[\sum_{n=1}^\infty \frac{1}{2^n 3^{n-1}}&amp;lt;\sum_{n=1}^\infty \frac{1}{2^n}=1.\]&lt;/p&gt;&#xA;" OwnerUserId="139" LastActivityDate="2010-11-30T00:38:20.147" />
  <row Id="12422" PostTypeId="2" ParentId="12374" CreationDate="2010-11-30T00:42:29.967" Score="1" ViewCount="" Body="&lt;p&gt;When you take the Laplace transform and solve for B(s), you should consider &lt;em&gt;both&lt;/em&gt; nonhomogeneous terms in the input:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;B(s) = \frac{1}{s+k} \left (h_a J(s) + \frac{kT_a}{s} \right ).&#xD;&#xA;$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The transfer function is just first factor,  $H(s) = \displaystyle \frac{1}{s+k}$, and the solution $b(t)$ is given by the convolution of $h(t) = e^{-kt}$ with the forcing term $h_a j(t) + k T_a$.&lt;/p&gt;&#xA;" OwnerUserId="4229" LastActivityDate="2010-11-30T00:42:29.967" />
  <row Id="12423" PostTypeId="1" CreationDate="2010-11-30T00:44:48.953" Score="8" ViewCount="383" Body="&lt;p&gt;I am in the process of proving that if a space curve (in R^3) has infinite length and the curvature tends towards 0 as the natural parameter s tends to infinity, the curve must be unbounded - i.e. not contained in any sphere of finite radius. This seems correct intuitively, but I have no guarantee it is correct, unless I am missing something obvious. One way to prove my hunch, I have deduced, is to use a lemma that any curve contained in the open unit ball with curvature always less than one must have a finite upper bound on its length (possibly 2π, but it could be greater for all I know).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How might one go about proving such an upper bound exists, or if it exists? It might also be nice to know what the bound specifically is, too. I've thought it might be possible to pose this as a variational problem - maximizing length - and then reducing it into a simpler problem, but that appears to be hellishly complicated. Thoughts?&lt;/p&gt;&#xA;" OwnerUserId="3122" LastActivityDate="2011-02-05T20:17:46.387" Title="Supremum length of space curves contained in the open unit ball having always less than unity curvature" Tags="&lt;differential-geometry&gt;&lt;calculus-of-variations&gt;" AnswerCount="3" CommentCount="8" />
  <row Id="12424" PostTypeId="2" ParentId="12091" CreationDate="2010-11-30T00:51:32.490" Score="2" ViewCount="" Body="&lt;p&gt;This is the type of questions to which the answer should be obvious: It depends on what you want to do. If you are interested in practical things like decisions about some outcomes of an experiment or even electrical circuits, then fuzzy logic may be the choice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, it is a generalization of two-valued boolean logic, which is just a small (and mathematically not that interesting) part of logic. For many applications, you do not even explicitly use truth values.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;logic vs. fuzzy logic&quot; is the wrong question.&lt;/p&gt;&#xA;" OwnerUserId="1858" LastActivityDate="2010-11-30T00:51:32.490" />
  <row Id="12425" PostTypeId="1" CreationDate="2010-11-30T01:09:08.080" Score="1" ViewCount="147" Body="&lt;p&gt;{An} is a sequence of events which may NOT be independent. Also A is any event such that P(A) greater than zero. Again the question says nothing about whether {An} is an independent sequence and also it says nothing about whether A or{An} are independent (we assume no independence).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;we are asked to prove P[An i.o.]=1 iff ∑ P(A ∩ An) = ∞. (Sum goes fron n=1 to infinity).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is what I have so far. I think if we assume that ∑ P(A ∩ An) = ∞,  That means &#xA;P[(A ∩ An) i.o.]=1, that is the    lim sup (A ∩ An) =1. If I am not wrong,lim sup (A ∩ An) is a subset of  lim sup (An).Therefore P(lim sup (An)) must be greater than or equal to &#xA;P (lim sup (A ∩ An)) and we have proved that P[An i.o.]= P( lim sup An ) = 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we assume P[An i.o.]= P( lim sup An ) = 1, how can we prove this implies ∑ P(A ∩ An) = ∞ ? Once again please remember we cannot use independence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;can someone enlighten me on this one please :)&lt;/p&gt;&#xA;" LastActivityDate="2010-11-30T01:29:02.453" Title="Problem Related To Borel Cantelli Lemma and independence" Tags="&lt;probability-theory&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="12426" PostTypeId="1" CreationDate="2010-11-30T01:13:44.897" Score="1" ViewCount="191" Body="&lt;p&gt;Hi!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am an ESL learner and most of the time I am able to understand English but In my courses, we don't study mathematics, and I think that it is a mistake or wrong.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1-What are really the best sources (book or books) to study and review completely high school mathematics?&#xA;-titles&#xA;-authors&#xA;-edition&#xA;-publishers&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2-What are really the best sources (book or books) to study and review completely college mathematics?&#xA;-titles&#xA;-authors&#xA;-edition&#xA;-publishers&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3-What are really the best sources (book or books) to study and review completely university mathematics; the basic courses before to study a specialty in a university for example. I ignore how matematics courses work in an American, Canadian or English university.&#xA;-titles&#xA;-authors&#xA;-edition&#xA;-publishers&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4-What is or are the best calculator or calculators? I mean software applications used to calculate or the usual electronic device used in university courses? I know DreamCalc but are there others? I know what is spreadsheet but is there a better thing that I can use? I know of the IrisPen to develop solutions to math problems using a computer but is there a better thing to use? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I sincerely thank you to develop extensively comments and suggest really good books, products, software applications and os on.&lt;/p&gt;&#xA;" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-30T01:26:19.480" LastActivityDate="2010-12-01T01:39:58.107" Title="I would like to buy references in order to review and study mathematics." Tags="&lt;soft-question&gt;&lt;education&gt;" AnswerCount="2" CommentCount="6" />
  <row Id="12427" PostTypeId="2" ParentId="12425" CreationDate="2010-11-30T01:29:02.453" Score="1" ViewCount="" Body="&lt;p&gt;This is not correct as stated.  Suppose $B$ is an event with probability $1/2$, and let $A$ and all the $A_n$ be equal to $B$.  Then $P(A_n \text{i.o.}) = P(B) = 1/2$ but $\sum_{n=1}^\infty P(A \cap A_n) = \sum_{n=1}^\infty \frac{1}{2} = \infty$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your sentence beginning &quot;I think if we assume&quot; is not true in general.  It is the so-called second Borel-Cantelli lemma, which requires independence.&lt;/p&gt;&#xA;" OwnerUserId="822" LastActivityDate="2010-11-30T01:29:02.453" CommentCount="2" />
  <row Id="12428" PostTypeId="1" CreationDate="2010-11-30T02:02:10.550" Score="0" ViewCount="251" Body="&lt;p&gt;Prove that any isomorphism between two cyclic groups always maps every generator to a generator.&lt;/p&gt;&#xA;" OwnerUserId="4179" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-30T04:22:59.433" LastActivityDate="2010-11-30T04:22:59.433" Title="Isomorphism, Cyclic Groups, and Generators" Tags="&lt;abstract-algebra&gt;&lt;group-theory&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="12429" PostTypeId="2" ParentId="12428" CreationDate="2010-11-30T02:05:18.543" Score="1" ViewCount="" Body="&lt;p&gt;Step 1: Show that any homomorphism from a cyclic group is determined by its image on a generator of the group.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Step 2: Show that the image of the homomorphism is generated by the image of that generator.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Step 3: What can you say about the kernel of such a homomorphism? What powers of the generator does it send to the identity?&lt;/p&gt;&#xA;" OwnerUserId="3212" LastActivityDate="2010-11-30T02:05:18.543" />
  <row Id="12430" PostTypeId="2" ParentId="12428" CreationDate="2010-11-30T02:26:20.640" Score="1" ViewCount="" Body="&lt;p&gt;Suppose you have two cyclic groups $G$ and $G&amp;#39;$ and some isomorphism $\phi\colon G\to G&amp;#39;$. Since $\phi$ is surjective, for any $g&amp;#39;\in G&amp;#39;$ there exists some $g\in G$ such that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$g&amp;#39;=\phi(g).$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Try expressing $g$ and/or $g&amp;#39;$ in terms of the generators, and see what happens from there. The fact that $\phi$ is a homomorphism is particularly useful here.&lt;/p&gt;&#xA;" OwnerUserId="1583" LastActivityDate="2010-11-30T02:26:20.640" />
  <row Id="12431" PostTypeId="1" AcceptedAnswerId="12503" CreationDate="2010-11-30T02:37:58.180" Score="0" ViewCount="198" Body="&lt;p&gt;I have deleted the content of the original post.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The following exercise is inspired by answers to the questions &lt;a href=&quot;http://math.stackexchange.com/questions/11487/y-xn-what-is-exyy&quot;&gt;this&lt;/a&gt; and &lt;a href=&quot;http://math.stackexchange.com/questions/11673/ex-1-x-1-x-2-for-independent-gamma-random-variables&quot;&gt;this&lt;/a&gt;, but no knowledge of probability theory is required. The solution is straightforward assuming familiarity with the relevant theorems/properties. The proof is supposed to be informal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $\hat \varphi (\omega)$ denote the Fourier transform of a function $\varphi$, and $*$ convolution.&#xA;Express $\hat g(\omega)$ in terms of $\hat f(\omega)$ and a constant $a \in (0,1)$, so that&#xA;$$&#xD;&#xA;(f_2  * g)(t) = at(f * g)(t),&#xD;&#xA;$$ &#xA;where $f_2 (x) = xf(x)$. &lt;/p&gt;&#xA;" OwnerUserId="2810" LastEditorUserId="2810" LastEditorDisplayName="" LastEditDate="2010-11-30T14:14:50.950" LastActivityDate="2010-12-01T09:08:45.633" Title="A convolution identity (revision of the question &quot;Is this convolution identity known?&quot;)" Tags="&lt;calculus&gt;&lt;real-analysis&gt;&lt;convolution&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="12433" PostTypeId="1" AcceptedAnswerId="12450" CreationDate="2010-11-30T02:58:10.820" Score="6" ViewCount="393" Body="&lt;p&gt;Hi,&#xA;Can someone please guide me to a way by which I can solve the following problem.&#xA;There is a die and 2 players. Rolling stops as soon as some exceeds 100(not including 100 itself). Hence you have the following choices: 101, 102, 103, 104, 105, 106. &#xA;Which should I choose given first choice. &#xA;I'm thinking Markov chains, but is there a simpler way? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: I wrote dice instead of die. There is just one die being rolled&lt;/p&gt;&#xA;" OwnerUserId="2988" LastEditorUserId="2988" LastEditorDisplayName="" LastEditDate="2010-11-30T03:49:15.467" LastActivityDate="2010-12-01T01:26:14.597" Title="Probability of dice sum just greater than 100" Tags="&lt;probability&gt;&lt;game-theory&gt;" AnswerCount="4" CommentCount="4" FavoriteCount="3" />
  <row Id="12434" PostTypeId="1" AcceptedAnswerId="12441" CreationDate="2010-11-30T03:04:30.347" Score="2" ViewCount="145" Body="&lt;p&gt;If $(x_n)$ and $(y_n)$ are positive real sequences such that $(y_n)$ is bounded and $\lim\frac{(x_n)}{(y_n)} = 0$, prove that $\lim(x_n) = 0$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So since $(y_n)$ is bounded, there exists a real number $M$ such that for all n natural numbers, $(y_n) &amp;lt; |M|$  &#xA;So by the limit definition :&lt;br&gt;&#xA;$\lim|\frac{(x_n)}{(y_n)}-0| = \lim\frac{(x_n)}{(y_n)} &amp;lt; \epsilon$&#xD;&#xA;So $|(x_n)| &amp;lt; \epsilon(y_n)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I need to find a natural number $N$ such that for all $n &amp;gt; N$, $|(x_n)| &amp;lt; \epsilon$ is this correct which would prove this statement.  However I'm having trouble picking something that will work here?  Am I on the right track? I have to be very careful also about anything I use to quote the theorem since this is an introduction to real analysis class.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="2533" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-30T04:55:13.807" LastActivityDate="2010-11-30T04:55:13.807" Title="$(y_n)$ bounded sequence and $\lim\frac{(x_n)}{(y_n)}=0$ show $\lim(x_n) = 0$" Tags="&lt;real-analysis&gt;&lt;homework&gt;&lt;sequences-and-series&gt;" AnswerCount="2" />
  <row Id="12435" PostTypeId="2" ParentId="12423" CreationDate="2010-11-30T03:04:47.087" Score="1" ViewCount="" Body="&lt;p&gt;Here is something along these lines, coming from estimates using elementary calculus, but it is much weaker than the lemma you want to prove.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $\gamma$ be a regular $C^2$ curve in $\mathbb{R}^3$ parametrized with respect to arclength $s$ (I will assume for simplicity that $s$ starts at $0$).  If the curvature $\|\gamma&amp;#39;&amp;#39;(s)\|$ is always less than $K$ and $\gamma$ is contained in a ball of radius $R\leq\frac{1}{4K}$, then the length of $\gamma$ is less than $\frac{1}{K}(1-\sqrt{1-4KR})$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To see this, write $\gamma(s)=\gamma(0)+s\gamma&amp;#39;(0)+\int_0^s(s-t)\gamma&amp;#39;&amp;#39;(t)dt$ (derived as in this &lt;a href=&quot;http://en.wikipedia.org/wiki/Taylor%27s_theorem#Integral_version&quot; rel=&quot;nofollow&quot;&gt;Wikipedia article&lt;/a&gt;).  Moving $s\gamma&amp;#39;(0)$ to the left and $\gamma(s)$ to the right and applying the triangle inequality, &#xA;$$&#xD;&#xA;\begin{align}&#xD;&#xA;s &amp;amp;\leq \|\gamma(s)-\gamma(0)\|+\|\int_0^s(s-t)\gamma&amp;#39;&amp;#39;(t)dt\| \\&#xD;&#xA; &amp;amp;\leq 2R +\int_0^s\|(s-t)\gamma&amp;#39;&amp;#39;(t)\|dt \\&#xD;&#xA; &amp;amp;\lt 2R +K\int_0^s(s-t)dt \\&#xD;&#xA; &amp;amp;=2R+\frac{K}{2}s^2.&#xD;&#xA;\end{align}&#xD;&#xA;$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For this inequality to always hold, $s$ must remain smaller than the smallest root of $\frac{K}{2}x^2-x+2R$.  Thus,  $s\lt\frac{1}{K}(1-\sqrt{1-4KR})$ as claimed.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could still apply this to prove the original result as stated, but hopefully someone comes along with a stronger result.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2010-11-30T03:14:06.033" LastActivityDate="2010-11-30T03:14:06.033" CommentCount="2" />
  <row Id="12437" PostTypeId="2" ParentId="12433" CreationDate="2010-11-30T03:24:06.690" Score="2" ViewCount="" Body="&lt;p&gt;Maybe I didn't understand the question, but it seems clear, considering $\lbrace 95 + j + k:j = 0, \ldots ,5;k = 1, \ldots 6 \rbrace$, that you should choose $101$. Here, $95+j$ corresponds to the sum just before exceeding $100$ for the first time, and $95+j+k$ to the sum when exceeding $100$ for the first time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me elaborate a little on my argument above. First, you can easily show by induction on $j$ that if the current sum is $100 - j$, $j=0,\ldots,4$, then there is no strictly better choice than $101$ (the base case $j=0$ is trivial). Then consider the case where the current sum is $95$, to conclude that $101$ is strictly the best choice.&lt;/p&gt;&#xA;" OwnerUserId="2810" LastEditorUserId="2810" LastEditorDisplayName="" LastEditDate="2010-11-30T08:30:07.790" LastActivityDate="2010-11-30T08:30:07.790" CommentCount="8" />
  <row Id="12438" PostTypeId="2" ParentId="12434" CreationDate="2010-11-30T03:26:38.350" Score="3" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Hint&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $|y_{n}|\leq M$ then $1 \leq \frac{M}{|y_{n}|} $&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$| x_{n}| \leq |x_{n}| \cdot \frac{M}{|y_{n}|} = |\frac{x_{n}}{y_{n}}| \cdot M &amp;lt; \varepsilon_{0} \cdot M = \frac{\varepsilon}{M} \cdot M = \varepsilon$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $|\frac{x_{n}}{y_{n}}| &amp;lt; \varepsilon_{0}$ *&lt;em&gt;for hypotesis&lt;/em&gt;* and choosing $\varepsilon_{0} = \frac{\varepsilon}{M}$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="3421" LastEditorUserId="3421" LastEditorDisplayName="" LastEditDate="2010-11-30T04:13:11.950" LastActivityDate="2010-11-30T04:13:11.950" CommentCount="8" />
  <row Id="12439" PostTypeId="1" AcceptedAnswerId="12452" CreationDate="2010-11-30T03:41:14.323" Score="6" ViewCount="115" Body="&lt;p&gt;Let $H=\left\{ z\in\mathbb{C}\mid\Im\left(z\right)&amp;gt;0\right\}$  be the upper-half Poincare plane. Let $GL\left(2,\mathbb{R}\right)$ be the general linear group, $Z\left(GL\left(2,\mathbb{R}\right)\right)$ be the center of the general linear group and $O\left(2,\mathbb{R}\right)$ be the orthogonal subgroup of $GL\left(2,\mathbb{R}\right)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What does it mean to say $H=GL\left(2,\mathbb{R}\right)/\left(Z\left(GL\left(2,\mathbb{R}\right)\right)\cdot O\left(2,\mathbb{R}\right)\right)$? The left-hand side is a metric space and the right hand side is a set of cosets of $GL\left(2,\mathbb{R}\right)$. So I'm confused about what it means to write that they are equal or to say &quot;the upper half plane is...&quot; It seems like this would be the group of orientation preserving isometries of H, but I still find the terminology confusing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've been trying to figure out what this could possibly mean, but my searches on the internet have not been fruitful. I've also looked at 2 sources on standard modular groups but they make no mention of this fact. An explanation or reference would be greatly appreciated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Motivation: I am reading a paper titled &quot;On Modular Functions in characteristic p&quot; by Wen-Ch'ing Winnie Li which can be found at &lt;a href=&quot;http://www.jstor.org/stable/1997973&quot; rel=&quot;nofollow&quot;&gt;http://www.jstor.org/stable/1997973&lt;/a&gt;. The claim appears on page 3 of the pdf (page 232 of the journal). It is also stated on the wikipedia page: &lt;a href=&quot;http://en.wikipedia.org/wiki/Poincar%C3%A9_half-plane_model&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Poincar%C3%A9_half-plane_model&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="249" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-30T04:24:22.840" LastActivityDate="2010-11-30T05:26:58.993" Title="What does $H=GL(2,\mathbb{R})/(Z(GL(2,\mathbb{R}))\cdot O(2,\mathbb{R}))$ mean?" Tags="&lt;geometry&gt;&lt;group-theory&gt;&lt;reference-request&gt;" AnswerCount="2" />
  <row Id="12440" PostTypeId="2" ParentId="12439" CreationDate="2010-11-30T04:03:59.797" Score="1" ViewCount="" Body="&lt;p&gt;H is not a group, but rather a space with a group acting on it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If K ≤ G is a subgroup, then G/K is not usually a group, but rather a set of cosets on which G acts.  G/K is only a group if K is normal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your situation, the subgroup is not normal, so you only get a group action.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One keyword form the wikipedia article is &quot;isotropy subgroup&quot;.  If G acts transitively on a space H, then the set K of elements of G that fix some particular point of H forms a subgroup (the &quot;isotropy subgroup&quot;), and the action of G on H is isomorphic to the action of G on the cosets in G/K.&lt;/p&gt;&#xA;" OwnerUserId="583" LastActivityDate="2010-11-30T04:03:59.797" CommentCount="5" />
  <row Id="12441" PostTypeId="2" ParentId="12434" CreationDate="2010-11-30T04:07:25.533" Score="6" ViewCount="" Body="&lt;p&gt;I would approach this by contradiction: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the sequence does not converge to 0, then there is a subsequence that is bounded away from 0. So we may fix $a&amp;gt;0$ such that $|x_n|&amp;gt;a$ for infinitely many values of $n$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the sequence of $y_n$ is bounded, say by $M$, we have $|y_n|&amp;lt;M$ for all $n$, so if $n$ is one of the indices in our subsequence, we have $|x_n/y_n|&amp;gt;a/M$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This inequality holds for infinitely many $n$, and we have found a subsequence of $(x_n/y_n)$ that does not converge to 0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does this make sense? &lt;/p&gt;&#xA;" OwnerUserId="462" LastActivityDate="2010-11-30T04:07:25.533" CommentCount="5" />
  <row Id="12442" PostTypeId="1" CreationDate="2010-11-30T04:08:33.717" Score="8" ViewCount="213" Body="&lt;p&gt;I am trying to get a grasp on what a representation is, and a professor gave me a simple example of representing the group Z_12 as the twelve roots of unity, or corresponding 2x2 matrices. Now I am wondering how GL(1,C) and GL(2,R) are related, since the elements of both groups are automorphisms of the complex numbers. GL(C), the group of automorphisms of C, is (to my understanding) isomorphic to both GL(1,C) and GL(2,R) since the complex numbers are a two-dimensional vector space over R. But it doesn't seem like these two groups are isomorphic to each other. &lt;/p&gt;&#xA;" OwnerUserId="4238" LastActivityDate="2010-11-30T22:29:31.530" Title="How is GL(1,C) related to GL(2,R)?" Tags="&lt;group-theory&gt;&lt;lie-groups&gt;&lt;representation-theory&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="1" />
  <row Id="12444" PostTypeId="1" CreationDate="2010-11-30T04:16:18.093" Score="0" ViewCount="50" Body="&lt;p&gt;I posted this question in MO (Mathoverflow) and didn't get an answer, so I try my luck here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $C$ be a nonempty bounded closed convex subset, say the unit ball, of $\ell^2(\mathbb{N})$. Let $T: C\to 2^C$ be a map such that $T(x)$ is nonempty closed for each $x$, and that $$D(Tx,Ty)\le \|x-y\|$$ for all $x,y\in C$, where $D$ is the Hausdorff metric defined by $$D(A,B)=\inf\lbrace r&amp;gt;0: N_r(A)\supset B, N_r(B)\supset A\rbrace,$$ $N_r(S) =\lbrace x\in C: d(x,S)\lt r\rbrace$ being the $r$-neighborhood of $S$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is whether $T$ has a fixed point, i.e. a point $x\in C$ such that $x\in Tx$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer is yes if $T$ is a contraction, i.e. replacing the $\le \|x-y\|$ by $\le \lambda \|x-y\|$ for some $0\le\lambda&amp;lt;1$ in the definition; or if $Tx$ is compact for each $x$.&lt;/p&gt;&#xA;" OwnerUserId="3249" LastActivityDate="2010-11-30T04:16:18.093" Title="Multi-valued nonexpansive maps" Tags="&lt;geometry&gt;&lt;analysis&gt;&lt;functional-analysis&gt;&lt;metric&gt;" />
  <row Id="12445" PostTypeId="2" ParentId="12442" CreationDate="2010-11-30T04:22:46.940" Score="7" ViewCount="" Body="&lt;p&gt;No, they're not isomorphic. $GL(1,\mathbb{C})$ injects into $GL(2, \mathbb{R})$ but the latter consists of the set of &lt;em&gt;real&lt;/em&gt; automorphisms of $\mathbb{R}^2$, while the former consists of the set of all &lt;em&gt;complex&lt;/em&gt; automorphisms of $\mathbb{C}$. For instance, an element of $GL(1, \mathbb{C})$ is the composite of a scaling (dilation) by a positive real number and a rotation, so is $\mathbb{R}_{&amp;gt;0} \times SO(2) = \mathbb{R}_{&amp;gt;0} \times S^1$. However, $GL(2, \mathbb{R})$ includes more complicated things (e.g. reflections, things which even if scaled do not become orthogonal transformations).&lt;/p&gt;&#xA;" OwnerUserId="536" LastEditorUserId="536" LastEditorDisplayName="" LastEditDate="2010-11-30T12:52:26.117" LastActivityDate="2010-11-30T12:52:26.117" CommentCount="6" />
  <row Id="12446" PostTypeId="2" ParentId="12442" CreationDate="2010-11-30T04:23:20.933" Score="4" ViewCount="" Body="&lt;p&gt;$GL(1,\mathbb{C})$ can be thought of as the group of automorphisms of $\mathbb{C}$ as a complex vector space.  Each such automorphism is given by multiplication by a nonzero complex number. On the other hand, $GL(2,\mathbb{R})$ can be thought of as the group of automorphisms of $\mathbb{C}$ as a &lt;em&gt;real&lt;/em&gt; vector space, and hence it is larger.  In terms of matrices, $GL(1,\mathbb{C})$ corresponds to all $2$-by-$2$ real matrices of the form &#xA;$\begin{bmatrix}&#xD;&#xA;a &amp;amp; -b \\&#xD;&#xA;b &amp;amp; a&#xD;&#xA;\end{bmatrix}$ with $a^2+b^2\gt0$, which is a proper subgroup of the group $GL(2,\mathbb{R})$ of all $2$-by-$2$ real invertible matrices.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-11-30T04:23:20.933" CommentCount="3" />
  <row Id="12447" PostTypeId="1" AcceptedAnswerId="12448" CreationDate="2010-11-30T04:30:35.970" Score="4" ViewCount="54" Body="&lt;p&gt;Hello, I started reading about uniform spaces in Bourbaki, and the closure operation has been defined as such:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\bar{A}=\bigcap_{V\in\mathcal{U}}V(A)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $\mathcal{U}$ is the uniform structure of the space, and $V(A)$ is the set of all left-relatives of $A$. One thing I don't understand is why $\bar{A}=\bar{\bar{A}}$, as is necessary for a closure operation. In particular, I don't see why $\bar{\bar{A}}\subseteq\bar{A}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After playing around with it for a while, I have this. Take some $s\in\bar{\bar{A}}$. So $s\in V(\bar{A})$ for all $V\in\mathcal{U}$. So taking any $V$, there is some $r\in\bar{A}$ such that $(s,r)\in V$. But $r\in V(A)$, so $(r,t)\in V$ for some $t\in A$. Is there some why to show that $(s,t)\in V$, to conclude that $s\in V(A)$? At best I can see that $(s,t)\in V\circ V$. &lt;/p&gt;&#xA;" OwnerUserId="1583" LastActivityDate="2010-11-30T04:43:45.740" Title="Closure Operation on a Uniform Topology" Tags="&lt;topology&gt;" AnswerCount="1" />
  <row Id="12448" PostTypeId="2" ParentId="12447" CreationDate="2010-11-30T04:36:49.727" Score="4" ViewCount="" Body="&lt;p&gt;For any $V\in {\cal U}$ choose $U\in {\cal U}$ such that $U\circ U\subset V$. Then you proceed with your proof above with $U$ (instead of $V$).&lt;/p&gt;&#xA;" OwnerUserId="3249" LastEditorUserId="3249" LastEditorDisplayName="" LastEditDate="2010-11-30T04:43:45.740" LastActivityDate="2010-11-30T04:43:45.740" CommentCount="1" />
  <row Id="12449" PostTypeId="2" ParentId="12426" CreationDate="2010-11-30T04:38:00.173" Score="1" ViewCount="" Body="&lt;p&gt;Hi&#xA;1) &quot;High School&quot; Math really varies from region to region, but if you aim is the review for University, you should get any book on analytic geometry, or &quot;college&quot; algebra.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2,3) Calculus and Linear algebra make up the bulk of first year University Math.  Any old edition of &quot;James Stewart Calculus&quot; should suit you fine.  I think David Lay's &quot;Linear algebra and it's applications&quot; is also pretty good.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4) Most University math is done without a calculator, but some students purchase the computer software &quot;Maple&quot; as an aid.  Maple is easier to learn than other programs like Mathamatica, and you shouldn't need anything more advanced than this for intro college/university&lt;/p&gt;&#xA;" OwnerUserId="3971" LastActivityDate="2010-11-30T04:38:00.173" CommentCount="1" />
  <row Id="12450" PostTypeId="2" ParentId="12433" CreationDate="2010-11-30T04:40:46.490" Score="5" ViewCount="" Body="&lt;p&gt;My attempt: a combination of my comment and Shai's comment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A partition means a partition of n using only 1,2,3,4,5,6.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Number of ways to get to 101: Number of partitions of 101.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Number of ways to get to 102: partitions of 96 + partitions of 97... + partitions of 100. Since we can have 96+6, 97+5, ...100+2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Number of ways to get to 106: Partitions of 100. Since we can get to 106 only by adding to 100 then rolling 6.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A partition of n will be the nth coefficient x in $\prod_{k=0}^6 \frac{1}{1-x^k}$ using the geometric series. But an easier observation is that $a_n = a_{n-1}+a_{n-2}+a_{n-3}+a_{n-4}+a_{n-5}+a_{n-6}$. Using this &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Number of ways to get to 101 is  $a_{101}=a_{100}+a_{99}+a_{98}+a_{97}+a_{96}+a_{95}$&#xD;&#xA;Number of ways to get to 102 is $a_{100}+a_{99}+a_{98}+a_{97}+a_{96}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Number of ways to get to 106 is $a_{100}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $a_n &amp;gt; 0$ we see that 101 will occur the most frequently.&lt;/p&gt;&#xA;" OwnerUserId="565" LastActivityDate="2010-11-30T04:40:46.490" />
  <row Id="12451" PostTypeId="1" CreationDate="2010-11-30T05:25:30.740" Score="3" ViewCount="526" Body="&lt;p&gt;Can someone please share references (websites or books) where I can find problems related with Fubini's theorem and applications of Radon-Nikodym theorem? I have googled yes and don't find many problems. What are the &quot;typical&quot; problems (if there are any) related with these topics? [Yes, exam is coming soon so I don't know what to expect and don't have access to midterms from previous years].&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you&lt;/p&gt;&#xA;" OwnerUserId="4240" LastActivityDate="2011-01-06T17:32:20.957" Title="Typical applications of Fubini's theorem and Radon-Nikodym" Tags="&lt;measure-theory&gt;" AnswerCount="3" CommentCount="0" />
  <row Id="12452" PostTypeId="2" ParentId="12439" CreationDate="2010-11-30T05:26:58.993" Score="6" ViewCount="" Body="&lt;p&gt;My comments were getting too long, so I will post this as an answer: the isomorphism referred to is an isomorphism of $G$-sets. This works in great generality: let $H$ be a set, let $G$ act on $H$ transitively. Pick an arbitrary point $z\in H$ and let $K=\text{Stab}_G(z)$ be the point stabiliser in $G$. Note that $K$ is not normal in general, since the group $gKg^{-1}$ stabilises the point $g(z)$ (my action is on the left). So, $K$ is normal if and only if it acts trivially on $H$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nevertheless, the set of cosets $G/K$ is always a $G$-set, i.e. a set with an action of $G$:&#xA;$$g: hK\mapsto (gh)K$$&#xD;&#xA;for all $g\in G$ and $hK\in G/K$. This is the usual coset action. Now, check that the map&#xA;$$\phi:G/K\rightarrow H,\;gK\mapsto g(z)$$&#xD;&#xA;is a bijection of $G$-sets, i.e. a bijection of sets that respects the $G$-action.&lt;/p&gt;&#xA;" OwnerUserId="3212" LastActivityDate="2010-11-30T05:26:58.993" CommentCount="2" />
  <row Id="12453" PostTypeId="1" AcceptedAnswerId="12474" CreationDate="2010-11-30T06:38:39.760" Score="10" ViewCount="216" Body="&lt;p&gt;I heard that using some relatively basic differential geometry, you can show that the only spheres which are Lie groups are $S^0$, $S^1$, and $S^3$.  My friend who told me this thought that it involved de Rham cohomology, but I don't really know anything about the cohomology of Lie groups so this doesn't help me much.  Presumably there are some pretty strict conditions we can get from talking about invariant differential forms -- if you can tell me anything about this it will be a much-appreciated bonus :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(A necessary condition for a manifold to be a Lie group is that is must be parallelizable, since any Lie group is parallelized (?) by the left-invariant vector fields generated by a basis of the Lie algebra.  Which happens to mean, by some pretty fancy tricks, that the only spheres that even have a chance are the ones listed above plus $S^7$.  The usual parallelization of this last one comes from viewing it as the set of unit octonions, which don't form a group since their multiplication isn't associative; of course this doesn't immediately preclude $S^7$ from admitting the structure of a Lie group.  Whatever.  I'd like to avoid having to appeal to this whole parallelizability business, if possible.)&lt;/p&gt;&#xA;" OwnerUserId="2159" LastEditorUserId="2159" LastEditorDisplayName="" LastEditDate="2010-11-30T08:45:21.293" LastActivityDate="2010-11-30T14:02:13.187" Title="Is there an easy way to show which spheres can be Lie groups?" Tags="&lt;differential-geometry&gt;&lt;differential-topology&gt;&lt;lie-groups&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="2" />
  <row Id="12454" PostTypeId="1" AcceptedAnswerId="12455" CreationDate="2010-11-30T06:53:01.767" Score="1" ViewCount="145" Body="&lt;p&gt;On a 2D plane, how to construct a graph with 7 vertexes and 21 edges? I tried various combination but couldn't seem to draw that kind of contrived graph on a paper.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But my understanding is that &lt;a href=&quot;http://mathoverflow.net/questions/30759/in-a-graph-is-it-always-possible-to-construct-a-set-of-cycle-bases-with-each-an/30767#30767&quot; rel=&quot;nofollow&quot;&gt;it is possible&lt;/a&gt;. So anyone can help me with it?&lt;/p&gt;&#xA;" OwnerUserId="550" LastActivityDate="2010-11-30T20:23:48.337" Title="Construct a Graph with 7 Vertex, 21 Edges on 2D Plane" Tags="&lt;graph-theory&gt;" AnswerCount="2" CommentCount="8" FavoriteCount="1" />
  <row Id="12455" PostTypeId="2" ParentId="12454" CreationDate="2010-11-30T07:20:01.227" Score="5" ViewCount="" Body="&lt;p&gt;My guess is you're after $K_7$ drawn on the torus; see &lt;a href=&quot;http://www.amotlpaa.org/math/k7torus.html&quot; rel=&quot;nofollow&quot;&gt;http://www.amotlpaa.org/math/k7torus.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To answer you question as to why $K_5$ is not planar:  If a complete graph is planar, then for every $K_3$ subgraph, either every vertex (that's not part of the $K_3$) is inside the $K_3$ or outside the $K_3$ (otherwise there's a crossing edge from outside to inside the $K_3$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, if we attempt to draw $K_5$ vertex-by-vertex, then we first draw a triangle $K_3$.  We can place the 4-th vertex either inside or outside of the triangle.  In either case, the drawing obtained will look like:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/vSihQ.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, wherever you put the 5-th vertex, you will form some $K_3$ subgraph for which the other two vertices are not both inside and not both outside.&lt;/p&gt;&#xA;" OwnerUserId="139" LastEditorUserId="139" LastEditorDisplayName="" LastEditDate="2010-11-30T07:42:24.270" LastActivityDate="2010-11-30T07:42:24.270" CommentCount="2" />
  <row Id="12456" PostTypeId="2" ParentId="12451" CreationDate="2010-11-30T07:25:49.127" Score="1" ViewCount="" Body="&lt;p&gt;Radon-Nikodym is used to prove the existence of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Conditional_expectation#Formal_definition&quot; rel=&quot;nofollow&quot;&gt;conditional expectation&lt;/a&gt; in probability theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fubini's theorem is, among other things, a very useful device to compute integrals over product spaces.&lt;/p&gt;&#xA;" OwnerUserId="367" LastEditorUserId="367" LastEditorDisplayName="" LastEditDate="2010-11-30T08:05:58.640" LastActivityDate="2010-11-30T08:05:58.640" CommentCount="3" />
  <row Id="12457" PostTypeId="2" ParentId="12433" CreationDate="2010-11-30T07:43:40.307" Score="1" ViewCount="" Body="&lt;p&gt;101 is the most probable stop state.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The operation &quot;replace the final toss of the dice by the one that gives a sum of 101&quot; is a probability-conserving transformation on the set of possible games.  It does not cover the whole set (i.e. the image of the transformation is a subset) of possible games stopping at 101, such as the ones with a transition from 95 to 101 by a roll of 6.  This shows that all the terminal states higher than 101 have a smaller probability than that of finishing at 101. &lt;/p&gt;&#xA;" OwnerUserId="467" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-11-30T22:26:17.523" LastActivityDate="2010-11-30T22:26:17.523" />
  <row Id="12458" PostTypeId="1" CreationDate="2010-11-30T07:50:47.630" Score="-2" ViewCount="99" Body="&lt;p&gt;Suppose $f(n)$ is $2n$.&#xA;Suppose $g_i(n)$ is $n+i$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, we have a recurrence given by $a_{2n} = c_0 a_n + c_1 a_{n+1} + \dots c_m a_{n+m}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) What is known about this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) What type of recurrence is this?  (My guess is that this is still a linear recurrence, since it is based linearly on parts of the series.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(3) Where are there references and information on recurrences such as this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(4) What, if anything, remains unknown about this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What if we simplify this to $a_{2n} = a_n + n$.  Then what is known?&lt;/p&gt;&#xA;" OwnerUserId="2626" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2011-06-03T20:03:22.670" LastActivityDate="2011-06-03T20:03:22.670" Title="Concerning $a_{f(n)} = \displaystyle\sum_{i=0}^m{c_i a_{g_i(n)}}$" Tags="&lt;sequences-and-series&gt;&lt;intuition&gt;&lt;recurrence-relations&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="12459" PostTypeId="1" AcceptedAnswerId="12491" CreationDate="2010-11-30T08:01:01.063" Score="3" ViewCount="70" Body="&lt;p&gt;I have a network, with $V$ vertex, $E$ edges and $C$ cycles, which has &quot;current&quot; flowing in and out from the network, as shown below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://docs.google.com/drawings/pub?id=1Micr1JiaNxGw3FTF-vJOR3h94IjNPjq1mQqEBS7QkGY&amp;amp;w=960&amp;amp;h=720&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each edge has a current value associated to it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The equations governs the current flow in the network are the so-called &quot;current conservation theorem&quot;, which means that all the current flowing into a node is equal to the current flowing outside from it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, in the above network, such equations can be written as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$Q_{in}=e1+e2    \quad\quad \text{(at v1)}$$  &#xA;$$e3+e13=e7+e5+e6   \quad\quad \text{(at v4)}$$&#xD;&#xA;$$e8+e10=e11+Q_{out2} \quad\quad \text{(at v8)}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;at every vertex. the node location and the value of $Q_{outi}$ are given beforehand, but only the location of $Q_{in}$ are given, the value of $Q_{in}$ is not given. $e_i$ is not known as well.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The interesting thing to note is that from the above sets of equation, one can deduce that, for any kinds of graph system, all the incoming external sources &lt;strong&gt;must&lt;/strong&gt; equal to the outgoing external sources, i.e.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$Q_{in}=Q_{out1}+Q_{out2}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this way one can deduce the value of $Q_{in}$ quite easily.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another point to note is that in a $V$ vertexes network with $E$ edges, the total number independent equations one can form from the above requirements are $V-1$. This is because if you manipulate the above equations for any kinds of graph, you will always find that there are only $V-1$ independent equations.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus the number of current value that one can arbitrary set is $E-(V-1)$. Recall that $C=E-V+1$, one can conclude that &lt;em&gt;the number of cycles corresponds to the number of indeterminate variables&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words this is an &lt;a href=&quot;http://en.wikipedia.org/wiki/Indeterminate_system&quot; rel=&quot;nofollow&quot;&gt;indeterminate system&lt;/a&gt; with $E$ variables but $V-1$ independent equations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is, out of all the $e1, e2, e3$ and so on, how to choose which $e_{i}$ for arbitrary value assignment ( so that the system will be reduced to a &lt;em&gt;unique system&lt;/em&gt;, with $V-1$ variables and $V-1$ independent equations)? My hunch is that I will use all the edges that are not inside the &lt;a href=&quot;http://en.wikipedia.org/wiki/Spanning_tree&quot; rel=&quot;nofollow&quot;&gt;spanning tree&lt;/a&gt; of the graph and set their current value arbitrarily. But I can't prove ( or disprove) this. Any ideas?&lt;/p&gt;&#xA;" OwnerUserId="550" LastEditorUserId="550" LastEditorDisplayName="" LastEditDate="2010-11-30T15:14:09.423" LastActivityDate="2010-12-02T10:36:30.170" Title="How to Set the Redundant Current Value in a Network?" Tags="&lt;linear-algebra&gt;&lt;graph-theory&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="0" />
  <row Id="12460" PostTypeId="1" AcceptedAnswerId="12461" CreationDate="2010-11-30T08:10:41.627" Score="3" ViewCount="130" Body="&lt;p&gt;Is Knopp's &lt;em&gt;Theory and Application of Infinite Series&lt;/em&gt; out of date? It's looks terrific to me, but the Dover edition I bought new maybe a year ago: &lt;a href=&quot;http://preview.tinyurl.com/2eprqps&quot; rel=&quot;nofollow&quot;&gt;http://preview.tinyurl.com/2eprqps&lt;/a&gt; seems to be the same as an edition published in 1951 and may go back as far as 1921. 60 or 90 years is a lot of math years. How about it? Does my book leave out some important developments? Is it old-fashioned in some other ways?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've seen this question: &lt;a href=&quot;http://math.stackexchange.com/questions/6845/what-is-the-current-state-of-the-art-in-methods-of-summing-exotic-series&quot;&gt;what is the current state of the art in methods of summing &quot;exotic&quot; series?&lt;/a&gt; but it doesn't have a full answer yet.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="1964" LastActivityDate="2010-11-30T08:14:56.833" Title="Is Knopp's &quot;Theory and Application of Infinite Series&quot; out of date?" Tags="&lt;sequences-and-series&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" />
  <row Id="12461" PostTypeId="2" ParentId="12460" CreationDate="2010-11-30T08:14:56.833" Score="0" ViewCount="" Body="&lt;p&gt;Konrad Knopp's books are nice to read and the results still holds. I remember some strange &quot;hunt&quot; for a &quot;convergence border&quot; in that book, which is a bit absurd to me..&lt;/p&gt;&#xA;" OwnerUserId="1154" LastActivityDate="2010-11-30T08:14:56.833" CommentCount="1" />
  <row Id="12462" PostTypeId="1" AcceptedAnswerId="12497" CreationDate="2010-11-30T08:25:43.217" Score="6" ViewCount="454" Body="&lt;p&gt;I've never actually done a delta-epsilon proof, so I thought I'd try my hand at one. I decided to try it out for $f(x)=1/x$. If I understand correctly from the wikipedia article, I want to show for any $\epsilon&amp;gt;0$, there exists a $\delta&amp;gt;0$ such that if $|x-c|&amp;lt;\delta$, then $|f(x)-f(c)|&amp;lt;\epsilon$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, I noticed that I want something like $|f(x)-f(c)|=|1/x-1/c|=\frac{|x-c|}{|xc|}&amp;lt;\epsilon$. So $|x-c|&amp;lt;|xc|\epsilon$, which looks similar to the fact that I want $|x-c|&amp;lt;\delta$. However, I've also heard that one is never supposed to let $\delta$ depend on $x$. Is this the right direction? How would I use this information to find a corresponding $\delta$ for each $\epsilon$? Thanks!&lt;/p&gt;&#xA;" OwnerUserId="1583" LastActivityDate="2010-11-30T20:23:11.010" Title="Is $f(x)=1/x$ continuous on $(0,\infty)$?" Tags="&lt;calculus&gt;" AnswerCount="3" CommentCount="6" FavoriteCount="1" />
  <row Id="12463" PostTypeId="1" AcceptedAnswerId="12467" CreationDate="2010-11-30T08:30:39.873" Score="3" ViewCount="61" Body="&lt;p&gt;Let $Z$ be a random variable with finite mean and consider the function $F: R \rightarrow R$ defined by $$ F(x) = E[\max(x+Z,0)]$$ How to prove that $F(x)$ is a continuous function of $x$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is not homework. A textbook I am reading states that $F(x)$ is continuous without providing any explanation, and I can't seem to fill the gap. I would be happy with a hint. &lt;/p&gt;&#xA;" OwnerUserId="4244" LastEditorUserId="4244" LastEditorDisplayName="" LastEditDate="2010-11-30T08:50:19.473" LastActivityDate="2010-11-30T09:50:57.533" Title="Continuity of $E[\max(x+Z,0)]$ as a function of $x$" Tags="&lt;probability&gt;" AnswerCount="1" />
  <row Id="12464" PostTypeId="2" ParentId="12462" CreationDate="2010-11-30T08:45:37.620" Score="2" ViewCount="" Body="&lt;p&gt;Hi, your proof looks ok.  Delta can depend on epsilon and on c in the definition of &quot;continuity&quot;.  A function is called uniformly continuous if you can proof that given epsion, the required value of delta depends on epsilon but NOT c.  1/x is NOT uniformly continuous on (0,1) which is why you can't get around the fact that delta depends on c.  1/x would be uniformly continous on other intervals however (intervals where f' is bounded).&lt;/p&gt;&#xA;" OwnerUserId="3971" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-30T15:04:15.263" LastActivityDate="2010-11-30T15:04:15.263" CommentCount="5" />
  <row Id="12465" PostTypeId="1" AcceptedAnswerId="12471" CreationDate="2010-11-30T08:46:25.597" Score="0" ViewCount="52" Body="&lt;p&gt;This is a slightly different question than the one asked &lt;a href=&quot;http://math.stackexchange.com/questions/1340/in-a-graph-is-it-always-possible-to-construct-a-set-of-cycle-basis-with-each-an&quot;&gt;here&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand that for non-planar graph there are edges in the graph which has more than 2 cycles sharing on it. But for &lt;a href=&quot;http://en.wikipedia.org/wiki/Planar_graph&quot; rel=&quot;nofollow&quot;&gt;planar graph&lt;/a&gt;, can we prove that &lt;em&gt;it is always possible to find a complete set of cycle basis that each edge is shared by at most 2 cycles?&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="550" LastActivityDate="2010-11-30T11:26:52.927" Title="For a Planar Graph, is it always possible to construct a set of cycle basis, with each and every edge Is shared by at most 2 cycle bases?" Tags="&lt;graph-theory&gt;" AnswerCount="1" />
  <row Id="12466" PostTypeId="2" ParentId="12462" CreationDate="2010-11-30T08:51:19.753" Score="2" ViewCount="" Body="&lt;p&gt;The statement in your first paragraph would be the definition of uniform continuity (I interpret what you wrote as &quot;for all $\epsilon$, there exists a $\delta$ such that for all $x$...&quot;). For a related question, see &lt;a href=&quot;http://math.stackexchange.com/questions/11538/intuition-for-uniform-continuity-of-a-function-on-mathbbr&quot;&gt;Intuition for uniform continuity of a function on $\mathbb{R}$&lt;/a&gt;. Uniform continuity implies continuity, but is strictly stronger, and in fact, the function at hand is continuous on $(0,\infty)$, but not uniformly continuous.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The statement of continuity would be that for all $\epsilon$ and all $x$ in the range, there exists a $\delta$ such that... So $\delta$ depends on $\epsilon$ *and* on $x$. Intuitively: the greater the slope at $x$, the closer you will have to get to $x$ for the function values to be close. &lt;/p&gt;&#xA;" OwnerUserId="3212" LastEditorUserId="3212" LastEditorDisplayName="" LastEditDate="2010-11-30T08:56:26.097" LastActivityDate="2010-11-30T08:56:26.097" CommentCount="6" />
  <row Id="12467" PostTypeId="2" ParentId="12463" CreationDate="2010-11-30T08:53:23.590" Score="3" ViewCount="" Body="&lt;p&gt;If $\lvert x-y\rvert\leq\delta$, then $\lvert (x+Z)-(y+Z)\rvert\leq\delta$, and also $$\lvert \mathrm{max}(x+Z,0)-\mathrm{max}(y+Z,0)\rvert\leq\delta,$$&#xD;&#xA;right? (Notice that $\mathrm{max}(\bullet+Z,0)$ is 1-Lipschitz.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence we also have $\lvert F(x)-F(y)\rvert\leq\delta$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;s&gt;Remark: In order for $\mathrm{max}(x+Z,0)$ to be well-defined, we should assume that $Z$ is bounded, I guess.&lt;/s&gt; (According to the comments this is not neccessary.)&lt;/p&gt;&#xA;" OwnerUserId="367" LastEditorUserId="367" LastEditorDisplayName="" LastEditDate="2010-11-30T09:50:57.533" LastActivityDate="2010-11-30T09:50:57.533" CommentCount="3" />
  <row Id="12468" PostTypeId="2" ParentId="12458" CreationDate="2010-11-30T09:17:09.390" Score="0" ViewCount="" Body="&lt;p&gt;To ensure that the equations define a nonzero, finite dimensional space of sequences, some condition on the functions $f$ and $g_i$ is needed.  If they are too fast-growing the sequence will be underconstrained and if they are slow-growing it will be overdetermined.  To avoid excessive generality such as a trivial encoding of the $3x+1$ problem (or, I think, the halting problem) as a statement about a recurrence of this type, the functions should be of an easy to analyze kind, such as polynomial or linear.  In that case asymptotic analysis should be possible but the details will depend on the individual recurrence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(edit: I see the question has changed to include specific functions $f(n)=2n$ and $g_i=n+i$.  In that case it should be true that $a_n$ has relatively smooth asymptotics, such as $n^u (\log n)^b$, once it is well-defined, and this could be substituted in the recurrence to get the approximation $a_{2n} = Ca_n$ where $C = \Sigma c_i$, so that $u$ can be calculated and the procedure &quot;bootstrapped&quot; to get more precise information.)&lt;/p&gt;&#xA;" OwnerUserId="467" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-11-30T09:36:29.490" LastActivityDate="2010-11-30T09:36:29.490" CommentCount="2" />
  <row Id="12469" PostTypeId="1" CreationDate="2010-11-30T10:28:15.360" Score="0" ViewCount="59" Body="&lt;p&gt;Question: (a) Determine each of the absolute conditions of the linear map $f: \mathbb{R} \rightarrow \mathbb{R}:$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$f(x) = |x|$&#xD;&lt;/li&gt;&#xA;&lt;li&gt;$f(x) = \frac{\pi}{2}$&#xD;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;(b) Provide both an example of a well conditioned and an ill conditioned function evaluation (both the function $f$ and the argument $x$). Do not use any of the examples from (a).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My attempt: so although this seems really simple I'm unfortunately having trouble with exactly what to do. The given definition of absolute condition $\kappa_{abs}$ is the smallest number for which: $|f(x_0) - f(x)| \leq \kappa_{abs}|x_0 - x| + o(|x_0 - x|)$. However in the one example I have it seems that the last term with $o$ is ignored (I assume it is too small...). I was thinking that perhaps for (a) 2. I could write something like: $|f(x_0) - f(x)| = |\frac{\tilde{\pi}}{2} - \frac{\pi}{2}|= \frac{1}{2}|\tilde{\pi}-\pi| \Rightarrow \kappa_{abs} = \frac{1}{2}$ ? With 1. (a) I am don't know which sort of answer would be expected...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For (b) I am guessing it must be measured in terms of relative condition, right? Since it appears that the rule is condition(not specified which) greater than 1 is ill conditioned. I know that subtraction of almost equal numbers does not do well. I thought a way to guarantee that the function would always subtract two such numbers would to be to let $f(x) = x - \frac{99x}{100}$ using the formula for relative condition of subtraction($\kappa = \frac{|x|+|y|}{|x-y|}$) I would have $$\frac{|x| + |\frac{99x}{100}|}{|x| - |\frac{99x}{100}|} = 199 &amp;gt; 1$$ $\Rightarrow$ This function is ill-conditioned ? For the well conditioned function I have the theorem: for $x,y &amp;gt; 0$ $$\frac{|(x+y) - (\tilde{x} + \tilde{y})|}{|x+y|} \leq 1\epsilon \Rightarrow \kappa = 1$$ but I don't understand how to use that here since it seems self explanatory that a positive number divided by a larger positive number is less than 1. Can I simply substitue any positive number for $y$ and let my function be $f(x) = x^{2} +10$ ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As if it weren't obvious I am very lost with this and any help, tips, advice, etc is as usual greatly appreciated!&lt;/p&gt;&#xA;" OwnerUserId="3711" LastEditorUserId="3711" LastEditorDisplayName="" LastEditDate="2010-11-30T12:32:21.680" LastActivityDate="2010-11-30T12:32:21.680" Title="Determine absolute conditions of mappings, provide examples of ill/well conditioned functions" Tags="&lt;numerical-methods&gt;" CommentCount="5" />
  <row Id="12470" PostTypeId="2" ParentId="12348" CreationDate="2010-11-30T10:52:22.457" Score="0" ViewCount="" Body="&lt;p&gt;The numbers in question here, of course, can be computed exactly.  For example, using &lt;a href=&quot;http://gmplib.org/&quot; rel=&quot;nofollow&quot;&gt;bignum&lt;/a&gt; or &lt;a href=&quot;http://www.gap-system.org/&quot; rel=&quot;nofollow&quot;&gt;GAP&lt;/a&gt; (or even &lt;a href=&quot;http://www.wolframalpha.com/&quot; rel=&quot;nofollow&quot;&gt;WolframAlpha&lt;/a&gt; -- the exactly link won't work on here, but I'm sure you can type in &quot;(100 choose 50)*2^(-100)&quot; yourself).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On my home computer, in GAP, it took less than a millisecond.  To make it more interesting, I also computed ${100000 \choose 50000} \cdot 2^{-100000}$, which took a bit more than 17 seconds.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;gap&amp;gt; Binomial(100,50)/2^100;&#xA;12611418068195524166851562157/158456325028528675187087900672&#xA;gap&amp;gt; time;&#xA;0&#xA;gap&amp;gt; Binomial(100000,50000)/2^100000;&#xA;&amp;lt;&amp;lt;an integer too large to be printed&amp;gt;&amp;gt;/&amp;lt;&amp;lt;an integer too large to be printed&amp;gt;&amp;gt;&#xA;gap&amp;gt; time;&#xA;17266&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;In fact, provided the coin has probability 1/2, the probability will always have a terminating decimal expansion (since binomial coefficients are integers, and 2 divides 10).  Here it is in this case:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;0.0795892373871787614981270502421704614029315404247333213573478705171737601631321012973785400390625&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="139" LastEditorUserId="139" LastEditorDisplayName="" LastEditDate="2010-11-30T12:15:03.573" LastActivityDate="2010-11-30T12:15:03.573" CommentCount="0" />
  <row Id="12471" PostTypeId="2" ParentId="12465" CreationDate="2010-11-30T11:26:52.927" Score="1" ViewCount="" Body="&lt;p&gt;This is of course true, with the basis cycles being boundaries of the bounded faces (after killing off all pairs {edge e, reverse of edge e}). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are you looking for a rigorous proof? It should follow from the argument establishing the fact that there are such things as faces etc (see for example chapter 4 &lt;a href=&quot;http://diestel-graph-theory.com/basic.html&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;). How much of relevant topology are you willing to assume? &lt;/p&gt;&#xA;" OwnerUserId="2633" LastActivityDate="2010-11-30T11:26:52.927" CommentCount="4" />
  <row Id="12472" PostTypeId="1" CreationDate="2010-11-30T11:47:54.887" Score="1" ViewCount="303" Body="&lt;p&gt;Can this expression be simplified?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ \sum_{i=0}^n (k^i)/i!  $&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT:&#xA;This expression I have got as the number of possible ways to select n objects or less from k different infinite objects (you can select as many as you can from any type)...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe it must be equal to the number of possible ways to select n objects from k+1 different infinite objects, where the extra +1 is a dummy object, selecting one of this type means letting an empty space in the final selection (i.e. selecting n-1 not n)...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so this must be equal to:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ (k+1)^n/n! $&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why that is not right??&lt;/p&gt;&#xA;" OwnerUserId="896" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2011-02-16T15:25:30.890" LastActivityDate="2011-02-17T09:02:48.603" Title="Can $\sum_{i=0}^n (k^i)/i!$ be simplified?" Tags="&lt;combinatorics&gt;&lt;sequences-and-series&gt;" AnswerCount="3" CommentCount="10" />
  <row Id="12474" PostTypeId="2" ParentId="12453" CreationDate="2010-11-30T12:11:06.713" Score="12" ViewCount="" Body="&lt;p&gt;Here is the sketch of the proof.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Start with a compact connected Lie group G.  Let's break into 2 cases - either $G$ is abelian or not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $G$ is abelian, then one can easily show the Lie algebra is abelian, i.e., $[x,y]=0$ for any $x$ and $y$ in $\mathfrak{g}$.  Since $\mathbb{R}^n$ is simply connected and has the same Lie algebra as $G$, it must be the universal cover of $G$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, if $G$ is a sphere, it's $S^1$, since all the others are simply connected, and hence are their own universal covers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next, we move onto the case where $G$ is nonabelian.  For $x,y,$ and $z$ in the lie algebra, consider the map $t(x,y,z) = \langle [x,y], z\rangle$.  This map is clearly multilinear.  It obviously changes sign if we swap $x$ and $y$.  What's a bit more surprising is that it changes sign if we swap $y$ and $z$ or $x$ and $z$.  Said another way, $t$ is a 3 form!  I believe $t$ is called the Cartan 3-form.  Since $G$ is nonabelian, there are some $x$ and $y$ with $[x,y]\neq 0$.  Then $t(x,y,[x,y]) = ||[x,y]||^2 \neq 0$ so $t$ is not the 0 form.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next, use left translation on $G$ to move $t$ around: define $t$ at the point $g\in G$ to be $L_{g^{-1}}^*t$, where $L_{g^{-1}}:G\rightarrow G$ is given by $L_{g^{-1}}(h) = g^{-1}h$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This differential 3-form is automatically left invariant from the way you've defiend it everywhere.  It takes a bit more work (but is not too hard) to show that it's also right invariant as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next one argues that a biinvariant form is automatically closed.  This means $t$ defines an element in the 3rd de Rham cohomology of $G$.  It must be nonzero, for if $ds = t$, then we may assume wlog that $s$ is biinvariant in which case $ds = 0 = t$, but $t$ is not $0$ as we argued above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, for a nonabelian Lie group, $H^3_{deRham}(G)\neq 0$.  But this is isomporphic to singluar homology.  Hence, for a sphere to have a nonabelian Lie group stucture, it must satisfy $H^3(S^n)\neq 0$.  This tells you $n=3$.&lt;/p&gt;&#xA;" OwnerUserId="331" LastEditorUserId="331" LastEditorDisplayName="" LastEditDate="2010-11-30T14:02:13.187" LastActivityDate="2010-11-30T14:02:13.187" CommentCount="17" />
  <row Id="12476" PostTypeId="1" AcceptedAnswerId="12482" CreationDate="2010-11-30T12:57:59.400" Score="3" ViewCount="90" Body="&lt;p&gt;This is just a completely random question of the top of my head.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $\Omega$ be the set of bijections $\phi:\mathbb{N} \cup \{0\} \rightarrow \mathbb{Q}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can imagine any $\phi \in \Omega$ to be equivalent to a wacky &quot;staircase&quot;, where for $x \in \mathbb{R}^+$ we have $x \mapsto \phi(\lfloor x \rfloor)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we walk along this staircase, the biggest step we need to take is of size $\sup_{n \geq 0} |\phi(n+1)-\phi(n)|$.  Suppose we want to make the steps as small as possible.  How small can we make them?  That is, what is \[\inf_{\phi \in \Omega} \sup_{n \geq 0} |\phi(n+1)-\phi(n)|\ \ ?\]&lt;/p&gt;&#xA;" OwnerUserId="139" LastActivityDate="2010-11-30T14:01:49.270" Title="Which bijection from the natural numbers to the rationals has the smallest steps?" Tags="&lt;real-analysis&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="12477" PostTypeId="1" AcceptedAnswerId="12483" CreationDate="2010-11-30T13:12:27.553" Score="4" ViewCount="151" Body="&lt;p&gt;Many introductory books on vector spaces mention that the scalars need not be reals, and might even have sections discussing complex vector spaces or vector spaces over the integers mod 2.  I have never seen any such book mention that all of the theory goes through as well if one restricts the scalars to be just rational numbers.  Perhaps this is because there is a dearth of interesting problems about such vector spaces accessible at this level that couldn't simply be discussed in the context of real scalars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wonder if there is an interesting introductory-level problem or topic about vector spaces that would be most naturally conducted by allowing rational number scalars.  Does anyone know of such, perhaps one with a number-theoretic aspect?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(By introductory:  I envision a first course on linear algebra, including non-math majors.  They would be seeing vector spaces (and that level of abstraction) for the first time.  Perhaps they would be seeing matrix multiplication for the first time.  Usually, in my experience, such courses primarily use the real numbers as scalars.)&lt;/p&gt;&#xA;" OwnerUserId="4253" LastEditorUserId="4253" LastEditorDisplayName="" LastEditDate="2010-11-30T15:24:26.533" LastActivityDate="2010-11-30T16:43:27.057" Title="Does anyone know an interesting introductory topic involving vector spaces over the rationals" Tags="&lt;linear-algebra&gt;" AnswerCount="6" CommentCount="2" FavoriteCount="1" />
  <row Id="12478" PostTypeId="2" ParentId="12477" CreationDate="2010-11-30T13:26:22.097" Score="4" ViewCount="" Body="&lt;p&gt;An abelian group is torsionfree and divisible iff it has the (unique) structure of a vector space over $\mathbb{Q}$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using this observation the classification of vector spaces over a field in terms of their dimension gives rise to a classification of torsionfree divisible abelian groups.  In particular, since if $V$ is a vector space over a field $K$ with $\operatorname{dim} V&amp;gt; \# K$ we have $\operatorname{dim} V = \# V$, we find that any two uncountable torsionfree divisible abelian groups of the same cardinality are isomorphic.  In particular $\mathbb{R}$ and $\mathbb{C}$ are isomorphic as abelian groups.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[&lt;b&gt;Added&lt;/b&gt;: In model-theoretic terms, we have established that the elementary theory of torsionfree divisible abelian groups is &lt;strong&gt;uncountably categorical&lt;/strong&gt;.  Since it has no finite models, by Vaught's Test this theory is complete.  See &lt;a href=&quot;http://www.math.uga.edu/~pete/modeltheory2010Chapter4.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.math.uga.edu/~pete/modeltheory2010Chapter4.pdf&lt;/a&gt;, especially Theorem 6.  (This is one of the most inexpensive examples of a complete theory that I know.)  Now I feel honorbound to advise you &lt;em&gt;not&lt;/em&gt; to mention this in a linear algebra class!]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whether one would or could want to introduce any of this in a linear algebra class is highly dubious (a graduate-level linear algebra class, perhaps).  However, I can't help but recall that Paul Halmos was very fond of the conclusion in the last paragraph and apparently liked to include as a problem on his exams: &quot;Can $\mathbb{R}$ be given the &#xA;structure of a vector space over $\mathbb{C}$?&quot;  He describes this in his autobiography. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Although I am a number theorist, no number-theoretic applications are springing to mind.  A little voice in my head is saying &quot;Hilbert 90&quot;, but that's probably not a direction you want to go in either.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You may well know this already, but there are lots of cool combinatorial problems which -- in a highly non-obvious way! -- can be solved using linear algebra over finite fields, especially $\mathbb{F}_2$.  I highly recommend L. Babai's manuscript &quot;Linear algebra methods in combinatorics&quot;.&lt;/p&gt;&#xA;" OwnerUserId="299" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2010-11-30T13:39:45.843" LastActivityDate="2010-11-30T13:39:45.843" CommentCount="1" />
  <row Id="12479" PostTypeId="2" ParentId="12477" CreationDate="2010-11-30T13:32:47.147" Score="2" ViewCount="" Body="&lt;p&gt;The question is kind of general, but here is an attempt at an answer:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;In the general introductory books on vector spaces, one usually tries to develop as much of the theory as possible without specifying the field of scalars at all. This has a good reason: the main advantage in talking about finite dimensional vector spaces rather than tuples of numbers is that one gets rid of baggage, as it were, and leaves only the essential parts of a theory. The same reasoning suggests to not specify the field. This way, the relevant structure becomes visible and is not obstructed by unnecessary details. There is no need to specify the field, if you want to define vectors, linear maps, bases, matrices, change of basis, determinants etc. In the usual order of teaching these topics, the first place when it becomes necessary to specify the field is when one starts talking about inner products, because to define non-degeneracy, one needs a notion of positivity. To sum up point one: the reason that books rarely say &quot;by the way, this works also for the rationals&quot; is that most books state and prove as much as possible by starting &quot;Let $V$ be a vector space over a field $F$...&quot; They don't feel that they have to remind the reader that $\mathbb{Q}$ is one such field. Indeed, there are soooo many others out there.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;There are however areas neighbouring linear algebra where the field of scalars becomes important and in many of them, $\mathbb{Q}$ is one of the fields that receive special attention. One example is: classical representation theory is usually first developed over $\mathbb{C}$. But when all the necessary theory is in place, one starts wondering about fields of definition. Questions like &quot;which representations are defined over $\mathbb{Q}$&quot; and &quot;how many irreducible rational representations does a finite group have&quot; are very natural and very non-trivial.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Since your thoughts are already going in that direction, I will also mention, that some theories such as number theory are even more interested in free $\mathbb{Z}$-modules than in vector spaces. Those are even more special (and difficult) than rational vector spaces. Such beasts arise naturally in algebraic number theory. For example Galois groups of finite extensions of number fields naturally act on many free $\mathbb{Z}$-modules, such as rings of integers, units therein (modulo torsion) and many more. Determining the structure of these actions is a very active area of research, but you will have difficulty introducing that in an early undergraduate course. Sometimes, determining the integral structure is so difficult that one is forced to tensor with $\mathbb{Q}$ and turn these modules into rational vector spaces, which brings us back to where you started from.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;So to summarise all this: the very basic theory of linear algebra tries to be oblivious to the field itself. The more advanced theory is very much interested in the field of definition. $\mathbb{C}$ tends to be the easiest to understand, then comes $\mathbb{R}$, the rationals are much more difficult than these two. So I guess, the reason that you don't see $\mathbb{Q}$ often in introductory undergraduate books is that, as soon as the field actually matters, $\mathbb{Q}$ is a difficult one to work with.&lt;/p&gt;&#xA;" OwnerUserId="3212" LastEditorUserId="3212" LastEditorDisplayName="" LastEditDate="2010-11-30T13:50:42.007" LastActivityDate="2010-11-30T13:50:42.007" CommentCount="6" />
  <row Id="12480" PostTypeId="2" ParentId="12476" CreationDate="2010-11-30T13:58:12.100" Score="4" ViewCount="" Body="&lt;p&gt;If $\phi$ is a bijection, so is $r \phi$ for any $r \in \mathbb{Q}$, so if the supremum is ever finite then the infimum you ask for is zero.  And it's not hard to describe a bijection with bounded step size:  number the rationals in $[-1, 1]$ with denominator at most $2$ from left to right, then weave back and number the rationals in $[-2, 1]$ with denominator at most $4$ that you haven't already, then weave back and number the rationals in $[-2, 2]$ with denominator at most $8$ that you haven't already... and so forth.  The step size here, I think, is always bounded by $\frac{1}{2}$.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-11-30T13:58:12.100" />
  <row Id="12481" PostTypeId="2" ParentId="12477" CreationDate="2010-11-30T14:01:19.037" Score="3" ViewCount="" Body="&lt;p&gt;One nontrivial example of vector spaces over $\mathbb{Q}$ being used to solve a problem is the construction of the Dehn invariant in the resolution of &lt;a href=&quot;http://en.wikipedia.org/wiki/Hilbert%27s_third_problem&quot; rel=&quot;nofollow&quot;&gt;Hilbert's third problem&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt;  After reading your comments on Alex Bartel's answer, I guess this is not really &quot;introductory level&quot; as there are two nontrivial $\mathbb{Q}$-vector spaces involved and one has to tensor them... &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think if you want to impress on your students that stuff like the rank-nullity theorem goes through over an arbitrary field, you should be looking for combinatorial applications over $\mathbb{F}_2$.  If your students can digest the definitions, these are very elegant examples of just how general the theory is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, much of the theory covered in a standard textbook - eigenvalues, eigenvectors - does &lt;em&gt;not&lt;/em&gt; go through over an arbitrary field.  Perhaps this will all just be unnecessarily confusing.&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-11-30T14:15:36.087" LastActivityDate="2010-11-30T14:15:36.087" CommentCount="4" />
  <row Id="12482" PostTypeId="2" ParentId="12476" CreationDate="2010-11-30T14:01:49.270" Score="3" ViewCount="" Body="&lt;p&gt;I would claim the limit is zero.  First, here is a version with the step under 2.  Partition $\mathbb{R}^+$ into sets $(m,m+1]$ for $m\in\mathbb{N}$.  There are countably many rationals in each interval, so order them in type $\omega$.  Now let $\phi(0)$ be the first item in (0,1], $\phi(1)$ the first item in (1,2], $\phi(2)$ the second item in (0,1], $\phi(3)$ the second item in (1,2], $\phi(4)$ the first item in (2,3], etc..  We step up through the intervals, going one higher each run, then back down to (0,1], and start up again.  No step is longer than twice an interval.  Now we can shrink the intervals as small as we wish.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-11-30T14:01:49.270" CommentCount="3" />
  <row Id="12483" PostTypeId="2" ParentId="12477" CreationDate="2010-11-30T14:09:28.543" Score="4" ViewCount="" Body="&lt;p&gt;One standard example of using $\mathbb{Q}$-vector spaces is to construct discontinuous additive functions $f: \mathbb{R} \to \mathbb{R}$. The approach is to choose a $\mathbb{Q}$-basis for $\mathbb{R}$. For each $x \in \mathbb{R}$, define $f(x)$ to be the sum of the coefficients in this basis. This is linear but discontinuous (it maps into $\mathbb{Q}$).&lt;/p&gt;&#xA;" OwnerUserId="536" LastEditorUserId="536" LastEditorDisplayName="" LastEditDate="2010-11-30T14:59:16.517" LastActivityDate="2010-11-30T14:59:16.517" CommentCount="2" />
  <row Id="12484" PostTypeId="2" ParentId="12472" CreationDate="2010-11-30T14:30:47.830" Score="1" ViewCount="" Body="&lt;p&gt;In response to the edited question, it is not right.  If we just let n=2, $1+k+\frac{k^2}{2}\neq \frac{(k+1)^2}{2}$.  The difference is $\frac{1}{2}$&lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2010-11-30T15:13:14.800" LastActivityDate="2010-11-30T15:13:14.800" CommentCount="1" />
  <row Id="12485" PostTypeId="1" AcceptedAnswerId="12490" CreationDate="2010-11-30T14:31:25.957" Score="4" ViewCount="172" Body="&lt;p&gt;I've run across the recurrence $a_{n+1} = (a_n)^2 + 1$ in the past.  Unfortunately, the referrence escapes me.  However, my impression was that recurrences involving the product of previous terms (such as $a_{n+1} = (a_n)(a_{n-1})$) are difficult to solve.  I'm wondering what is known for this very general problem.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1)  Is there a known way to solve recurrences involving the product of previous terms?  Or, what is known about these?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2)  What are these recurrences called?  Do they have a general name (such as non-linear recurrences)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(3)  Where can I find more literature on the subject?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(4)  Who are some experts that have dealt with this?&lt;/p&gt;&#xA;" OwnerUserId="2626" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2011-06-03T20:03:05.090" LastActivityDate="2011-06-03T20:03:05.090" Title="What's known about recurrences involving $(a_n)^2$?" Tags="&lt;sequences-and-series&gt;&lt;recurrence-relations&gt;" AnswerCount="4" CommentCount="4" />
  <row Id="12486" PostTypeId="1" AcceptedAnswerId="12496" CreationDate="2010-11-30T14:31:53.077" Score="7" ViewCount="190" Body="&lt;p&gt;A general problem for the Calculus of Variations asks us to minimize the value of a functional $A[f]$, where $f$ is usually a differentiable function defined on $\mathbb{R}^n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What if, however, the domain of $A$ is not actually all differentiable functions. Suppose there is a constraint equation on $f$, such as (for example):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$L[f] = \int_{-1}^1 \sqrt{1 + f&amp;#39;(x)^2} dx = \pi$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and we want to minimize over functions satisfying the above and the property that $f(-1)=f(1)=0$ the functional&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$A[f] = \int_{-1}^1 f(x) dx $&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This sort of problem seems to me to be very similar to the problem in multivariate calculus of minimizing a function $f(x)$ with respect to a constraint equation $g(x) = 0$. In this case we are trying to minimize a functional $A[f]$ with respect to a functional constraint equation $L[f] = \pi$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the former, one can use Lagrange multipliers to reduce the problem to that of solving a system of equations. Is there such a technique for the variational version?&lt;/p&gt;&#xA;" OwnerUserId="1089" LastEditorUserId="1089" LastEditorDisplayName="" LastEditDate="2010-11-30T15:25:50.587" LastActivityDate="2010-11-30T15:25:50.587" Title="Calculus of Variations and Lagrange Multipliers" Tags="&lt;calculus&gt;&lt;real-analysis&gt;&lt;calculus-of-variations&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="12487" PostTypeId="1" CreationDate="2010-11-30T14:34:47.747" Score="1" ViewCount="90" Body="&lt;p&gt;There are presumably three ways of characterizing the abstract Euclidean space $E^n$ that are quite different in spirit:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;axiomatically (with axioms concerning dimension)&lt;/li&gt;&#xA;&lt;li&gt;by the abstract Euclidean group $E(n)$ (as its symmetry group, determining $E^n$ uniquely)&lt;/li&gt;&#xA;&lt;li&gt;by presupposing a metric and requiring &lt;em&gt;that the space is a maximal one with respect to the property&lt;/em&gt; that the $(n+1)$-dimensional Cayley-Menger determinant vanishes for all $(n+2)$-tuples of points &lt;em&gt;and does not vanish for all $k$-tuples of points &quot;in general position&quot; for $k &amp;lt; n+2$&lt;/em&gt;.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;[Having tried to &quot;rescue&quot; 3 by adding &lt;em&gt;conditions in italics&lt;/em&gt;, due to Robin's comment.]&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Question 1:&lt;/strong&gt; Is it correct, actually, that $E^n$ is uniquely&#xA;  determined via 2 and 3?&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;Question 2:&lt;/strong&gt; What are other ways of&#xA;  characterizing $E^n$ that are quite different in spirit?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1792" LastEditorUserId="1792" LastEditorDisplayName="" LastEditDate="2010-11-30T15:23:43.617" LastActivityDate="2010-11-30T15:23:43.617" Title="Characterizations of Euclidean space" Tags="&lt;geometry&gt;&lt;euclidean-geometry&gt;" CommentCount="1" />
  <row Id="12488" PostTypeId="1" AcceptedAnswerId="13069" CreationDate="2010-11-30T14:37:48.863" Score="4" ViewCount="221" Body="&lt;p&gt;In a planar graph $G$, one can easily find all the cycle basis by first finding the spanning tree ( &lt;em&gt;any&lt;/em&gt; spanning tree would do), and then use the remaining edge to complete cycles. Given Vertex $V$, edge $E$, there are $C=E-V+1$ number of cycles, and there are $C$ number of edges that are inside the graph, but not inside the spanning tree.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, &lt;a href=&quot;http://math.stackexchange.com/questions/12465/for-a-planar-graph-is-it-always-possible-to-construct-a-set-of-cycle-basis-with&quot;&gt;there always exists a set of cycle basis such that each and every edge inside the $G$ is shared by at most 2 cycles&lt;/a&gt;. My question is, is there any algorithm that allows me to find such a set of cycle basis? The above procedure I outlined only guarantees to find a set of cycle basis, but doesn't guarantee that all the edges in the cycle basis is shared by at most two cycles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Note: Coordinates for each vertex are &lt;em&gt;not&lt;/em&gt; known, even though we do know that the graph &lt;em&gt;must&lt;/em&gt; be planar.&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;del&gt;Note: All the edges are of the same weight.&lt;/del&gt; (deleted for not relevant)&lt;/p&gt;&#xA;" OwnerUserId="550" LastEditorUserId="550" LastEditorDisplayName="" LastEditDate="2010-12-01T12:24:58.440" LastActivityDate="2010-12-05T07:10:16.833" Title="For a Planar Graph, Find the Algorithm that Constructs A Cycle Basis, with each Edge Shared by At Most 2 Cycles" Tags="&lt;graph-theory&gt;" AnswerCount="2" CommentCount="13" />
  <row Id="12489" PostTypeId="1" AcceptedAnswerId="12494" CreationDate="2010-11-30T14:44:04.150" Score="2" ViewCount="155" Body="&lt;p&gt;(Pun definitely intended.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dear MSE-Community,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I were to choose one artist that has made interesting works of art not only because of their beauty but also because of their connections to Mathematics, I would choose Escher. His works have always intrigued me. Some of his paintings are mind-boggling when looked at for a long time, but I have the feeling that it can be described accurately with mathematics. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's compare some drawings. In the first and second drawing ((1),(2)), the artist chooses to depict some simple geometrical objects in one- and two point perspective respectively. Although the objects seem to float in space, both the images look &quot;all right&quot; to me. These drawing are, however, though not 'Eschers', illusory too. The brain somehow creates 3D space from 2D space, but that's more of a biological issue.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One Point Perspective&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/ilLt5.gif&quot; alt=&quot;alt text&quot;&gt;  (1)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Two Point Perspective&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/7HxIg.gif&quot; alt=&quot;alt text&quot;&gt;  (2)&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Sub-question 1: How does one describe these seemingly &quot;sound&quot; drawings mathematically? How do (1) and (2) compare to one another?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now, lets get to to part I find most interesting, Escher's etchings, prints and lithographs. When I look at the following pictures:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-- M.C. Escher 1960 lithograph Ascending and Descending&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/ViM3c.jpg&quot; alt=&quot;alt text&quot;&gt;  (3)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-- M.C. Escher 1953 Relativity&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/FeUTO.gif&quot; alt=&quot;alt text&quot;&gt;  (4)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I recognize that these are two different types of paradoxes because Escher plays with perspective in two different ways. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Sub-question 2: How could the difference between these and other visual paradoxes of artists (mostly Escher, but I guess there are a lot more artists that mimic and extend his style) be formalized with the aid of mathematics?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Thanks,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Max Muller&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S. I'm sorry these images are all of different sizes and some are too large. I'm a bit in a hurry so I didn't make them equally large. &lt;/p&gt;&#xA;" OwnerUserId="1055" LastEditorUserId="1055" LastEditorDisplayName="" LastEditDate="2010-11-30T15:06:25.473" LastActivityDate="2010-11-30T15:09:25.460" Title="Can we put the mathematics of paradoxes in visual art into perspective?" Tags="&lt;visualization&gt;" AnswerCount="1" CommentCount="13" FavoriteCount="1" />
  <row Id="12490" PostTypeId="2" ParentId="12485" CreationDate="2010-11-30T14:47:34.863" Score="11" ViewCount="" Body="&lt;p&gt;$a_{n+1} = a_n a_{n-1}$ happens to be quite easy to solve: if $b_n = \log a_n$ then $b_n$ satisfies the Fibonacci recurrence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only recurrences of the form $a_{n+1} = a_n^2 + c$ that have anything like a closed form solution occur when $c = 0$ (obvious) and when $c = -2$ (where the solution is given by $a_n = 2 \cos 2^n \theta$ where $\theta$ is determined by $a_0$).  For all other values of $c$ you get very complicated behavior, and the situation for more general nonlinear recurrences is even worse.  People study questions like these in &lt;a href=&quot;http://en.wikipedia.org/wiki/Dynamical_system&quot; rel=&quot;nofollow&quot;&gt;dynamical systems&lt;/a&gt; and related areas; one keyword you might use to start your exploration is &lt;a href=&quot;http://en.wikipedia.org/wiki/Julia_set&quot; rel=&quot;nofollow&quot;&gt;Julia set&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Like many other classes of sufficiently general problems in mathematics, the problem is hopelessly complicated in general and people study interesting special cases.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-11-30T14:47:34.863" />
  <row Id="12491" PostTypeId="2" ParentId="12459" CreationDate="2010-11-30T14:53:22.897" Score="2" ViewCount="" Body="&lt;p&gt;Update: Upon a bit of reflection I decided that the original answer is unnecesarily long. What is really being computed is the homology of the graph. Let $C(V)$ be the vector space (over real numbers, since the flows are real) freely generated by the vertices, and $C(E)$ the vector space freely generated by the edges. We have the map $\delta: C(E) \mapsto C(v)$ which assigns to a flow (thought of as a weighted set of edges) the tuple of net flows through vertices. The question is what is it's kernel, which is of course the $H_1$ of the graph an is in fact exactly the subspace of cycles in $C(E)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(See  the last paragraph for interpretation of what follows in this language.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Original answer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact you can certainly take any spanning tree $T$. The point is that any edge in the complement of $T$ can be completed to a cycle using edges in $T$ in a unique way, and the resulting cycles form a basis of the cycle space. This is your ambiguity. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The general statement is: Flows on the graph with prescribed sinks and sources are in bijection with the linear maps on the cycle space. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The argument is as follows: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We need two facts:&#xA;Fact 1) If the flow around any cycle is zero, then the flow actually comes from a potential - that is a function $f$ on vertices, such that flow on edge e is the difference of the value of $f$ on its head and its tail. This $f$ is uniquely determined by the flow up to an overall scalar (assuming the graph is connected; start at any vertex and any value of $f$ for it and fololow edges assigning the unique possible values for $f$ as you go; the flow around cycle being zero gurantees that you never assign contradictory values). We will show that this potential is also uniquely determined, up to overall scalar, by the net flow in each vertex, and as a consequence the corresponding flow on edges is uniquely defined.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fact 2) If the net flow at each vertex is zero (no sources or sinks) then the flow is uniquely determined by a function on the cycle space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proof of Fact 1. Consider the map $Fl$ sending a potential $f$ to the set of net flows at each vertex. $Fl$ is linear, and constant potentials are in the kernel. I claim that this is the whole kernel. Indeed, if there is a vertex with a maximal potential, it will have a negative net flow from it, unless all its neighbours have the same potential. As the graph is connected, the only potentials with no flow at any vertex are constant ones. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$Fl$ is a map between vector spaces of the same dimension. As the kernel of $Fl$ is one dimensional, so is the cokernel.  Note that $Fl$ lands in the subset of all tuples of net flows with the zero sum (in the summ of all net flows each edge is counted twice, once with plus and once with minus). Hence $Fl$ hits any net flow tuple with total flow zero, and any such tuple of net flows comes from unique potential, up to an overall scalar, as wanted. End of proof of Fact 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proof of Fact 2: Given any linear function $G$ on the cycle space, we can define the flow $E=F(G)$ on each edge $e$ to be the sum of $G(c)$ for cycles $c$ with $e$ in $c$. This of course lands in the flows with zero net flow at each vertex (this is clear by considering what happens to a dual function to any cycle $c$ and the fact that the assignment $F$ is linear). Observe that there is a basis $c_k$ of the cycle space and edges $e_k$ with $e_k$ in $c_k$ and no $c_l$ for $l \neq k$ (the edges in the complement of the spanning tree will do), so $F$ is injective. Finally, dimension count shows $F$ is an isomorphism. End of proof of Fact 2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now if we fix a collection of sources and sinks, this determines a unique flow $E$ which is zero on cycles by Fact 1. The difference of any flow $E&amp;#39;$ with these sinks and cycles and $E$ has zero net flow on each vertex, and hence corresponds to a linear function on the cycle space by Fact 2. The end.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Of original answer.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Comment: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What the long argument above does is as folows: One can view flows as cochains instead of chains. That is, since $C(E)$ comes with a basis, it has a prefered isomorphism $I_e$ to $\operatorname{Hom}(C(E), \mathbb{R})$. Similarly there is a prefered isomorphism $I_v$ from $C(v)$ to $\operatorname{Hom}(C(V), \mathbb{R})$. The map $\delta$ has the dual $\delta ^*:\operatorname{Hom}(C(V), \mathbb{R}) \mapsto \operatorname{Hom}(C(E), \mathbb{R})$ sending a potential (thought of as a map from vertices to $\mathbb{R}$) to its flow (there should be a (non-commutative) diagram here). Fact 1 above is that the composition $\delta \cdot I_c \cdot \delta^*$ is an isomorphism from $\operatorname{Hom}(C(V), \mathbb{R})/\operatorname{constants}$ to $\operatorname{Im} \delta$. Fact 2 is that the kernel of $\delta$ is the cycle space (or its dual). All of this presumably in some standard book. &lt;/p&gt;&#xA;" OwnerUserId="2633" LastEditorUserId="2633" LastEditorDisplayName="" LastEditDate="2010-12-02T10:36:30.170" LastActivityDate="2010-12-02T10:36:30.170" CommentCount="7" />
  <row Id="12492" PostTypeId="2" ParentId="12485" CreationDate="2010-11-30T14:57:34.720" Score="2" ViewCount="" Body="&lt;p&gt;If we have a sequence $a_{n+1} = f(a_n, ..., a_{n-k})$ where $f$ is a quadratic polynomial in $k$ variables, then we can use this recurrence relation to write down a &quot;simplified&quot; recurrence relation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a_{n+1} = g(a_{n-k})$ where $g$ is a polynomial of even degree.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Define a new sequence $b_j = a_{(k+1) \cdot j}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then $b_{j+1} = a_{kj + k + j + 1} = g(a_{(k+1)j}) = g(b_j)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice that if $a_n$ converges, then so does $b_n$ and $\lim a_n = \lim b_n$. Denote $\lim b_n$ by $b$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As $g$ is continuous,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\lim_{n \rightarrow \infty} b_n = \lim_{n \rightarrow \infty} g(b_{n+1}) = g( \lim_{n \rightarrow \infty} b_n )$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, $g(b) = b$, so the limit of the sequence $a_n$ must be one of the fixed points of $g$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To find the fixed points of $g(x)$, one must find the roots of the polynomial $g(x) - x$ (of which there are only finitely many), and to solve this problem one can use a variety of methods, like for example Newton's method.&lt;/p&gt;&#xA;" OwnerUserId="1089" LastEditorUserId="1089" LastEditorDisplayName="" LastEditDate="2010-11-30T15:21:54.147" LastActivityDate="2010-11-30T15:21:54.147" CommentCount="2" />
  <row Id="12493" PostTypeId="2" ParentId="12485" CreationDate="2010-11-30T14:59:23.063" Score="2" ViewCount="" Body="&lt;p&gt;Sometime ago I considered the sequence of fractions defined by $a_1 = 1/2$, $a_{n+1} = (a_n^2 + 1)/2$, in some probabilistic context; see &lt;a href=&quot;http://oeis.org/A167424&quot; rel=&quot;nofollow&quot;&gt;OEIS seqence A167424 &lt;/a&gt;. This sequence is closely related to the one defined by $a_1 = 1/2$, $a_{n+1}=a_n-a_n^2$, see &lt;a href=&quot;http://oeis.org/A076628&quot; rel=&quot;nofollow&quot;&gt;OEIS sequence A076628&lt;/a&gt;, whose asymptotics as $n \to \infty$ was analyzed &lt;a href=&quot;http://www.math.niu.edu/~rusin/known-math/99/somos&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2810" LastEditorUserId="2810" LastEditorDisplayName="" LastEditDate="2010-12-16T01:24:43.107" LastActivityDate="2010-12-16T01:24:43.107" />
  <row Id="12494" PostTypeId="2" ParentId="12489" CreationDate="2010-11-30T15:09:25.460" Score="3" ViewCount="" Body="&lt;p&gt;Question 1: this is &lt;a href=&quot;http://en.wikipedia.org/wiki/Projective_geometry&quot; rel=&quot;nofollow&quot;&gt;projective geometry&lt;/a&gt;.  In this context the horizon line is known as the line at infinity, and the drawings illustrate that &quot;parallel lines&quot; in the projective plane meet at the line at infinity while non-parallel lines meet at a finite point and intersect the line at infinity in different points.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Question 2: I don't know that you have to do anything sophisticated to describe what is going on in Ascending and Descending.  The drawing suggests the existence of four heights $h_1, h_2, h_3, h_4$ (the heights of the corners) such that $h_1 &amp;gt; h_2 &amp;gt; h_3 &amp;gt; h_4 &amp;gt; h_1$, and this is a contradiction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And what is paradoxical about Relativity?  As far as I can tell, that room is buildable.  Maybe I don't understand what you are trying to ask here.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-11-30T15:09:25.460" CommentCount="2" />
  <row Id="12496" PostTypeId="2" ParentId="12486" CreationDate="2010-11-30T15:18:39.097" Score="7" ViewCount="" Body="&lt;p&gt;Yes. For the general theorem, see &lt;a href=&quot;http://en.wikipedia.org/wiki/Lagrange_multipliers_on_Banach_spaces&quot; rel=&quot;nofollow&quot;&gt;this Wikipedia page&lt;/a&gt;. In the particular case which you consider, you can actually consider the minimization problem&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \int_{-1}^1 f(x) + \lambda (\sqrt{ 1 + f&amp;#39;(x)^2} - \pi) dx $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which leads to the Euler-Lagrange equation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$  \left( \frac{f&amp;#39;}{\sqrt{1 + (f&amp;#39;)^2}}\right)&amp;#39; = \lambda^{-1} $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which leads to a somewhat unappealing looking Lagrangian with $f&amp;#39;&amp;#39;$ in the denominator when you plug $\lambda$ back in. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(This is, of course, not completely unexpected, as your functional $A$ is not bounded below on the set $L[f] = \pi$. To see this, it suffice to note that $L[f + c] = L[f]$ by definition, while $A[f + c] = A[f] + c$. So for a decreasing sequence of $c\searrow -\infty$, $A[f + c]$ decreases without bound.)&lt;/p&gt;&#xA;" OwnerUserId="1543" LastActivityDate="2010-11-30T15:18:39.097" CommentCount="2" />
  <row Id="12497" PostTypeId="2" ParentId="12462" CreationDate="2010-11-30T15:21:00.873" Score="10" ViewCount="" Body="&lt;p&gt;There have been some confusing comments regarding dependence on $x$ or on $c$, so let me try to put it all together.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are correct that $\delta$ should not depend on $x$. However, when one is proving that $f(x)$ is continuous &lt;em&gt;at $c$&lt;/em&gt;, then $\delta$ is allowed to depend on both $\epsilon$  &lt;em&gt;and&lt;/em&gt; $c$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remember the definition: $f(x)$ is continuous &lt;em&gt;at $c$&lt;/em&gt; if and only for every $\epsilon\gt 0$ there exists a $\delta\gt 0$ such that for all $x$, if $|x-c|\lt \delta$, then $|f(x)-f(c)|\lt \epsilon$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice how the existence of $\delta$ is mentions &lt;em&gt;before&lt;/em&gt; $x$ ever comes into the picture? That's an indication that $\delta$ cannot depend on $x$. On the other hand, both $f(x)$, $\epsilon$, and $c$ occur &lt;em&gt;before&lt;/em&gt; $\delta$, which means that, absent any indication to the contrary, $\delta$ is allowed to depend on $f(x)$ (obviously), on $\epsilon$, &lt;em&gt;and&lt;/em&gt; on $c$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So here, you cannot pick $\delta=|xc|\epsilon$, because that would make $\delta$ depend on $x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The way to get around it is to get rid of the dependence on $x$. The key here is that since we are trying to make sure everything works if $x$ is &quot;close enough&quot; to $c$, then we will also have that $|x|$ will be very close to $|c|$. So we should be able to to control that division by $x$ in the expression $\frac{|x-c|}{|xc|}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How? Well, if a particular $\delta_0$ works, then any smaller one will work as well. So we can always shrink $\delta$ a bit more if needed. Wo the first thing we can note is that we can always require that $\delta$ be smaller than both $1$ and than $\frac{c}{2}$; that is, we will require $\delta\lt\min\{1,\frac{c}{2}\}$. Why $1$? Because then I know that $c-1\lt x \lt c+1$; if $c-1\gt 0$, then this means that $\frac{1}{c+1}\lt \frac{1}{x} \lt \frac{1}{c-1}$, so we can &quot;control&quot; the value of $\frac{1}{x}$. Why less than $\frac{c}{2}$? Just in case $c-1\lt 0$. So let $\mu=\min\{1,\frac{c}{2}\}$. Then we can conclude that $\frac{1}{x}\lt \frac{1}{c-\mu}$. (We could get away with simply putting $\delta\lt\frac{c}{2}$; restricting it to less than $1$ is a common practice, though, which is why I put it here).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, by requiring that $\delta\lt\min\{1,\frac{c}{2}\}$, we guarantee that $\frac{1}{|x|}\lt \frac{1}{c-\mu}$ (remember that we are working on $(0,\infty)$). What do we gain by this? Well, look:&#xA;$$|f(x)-f(c)| = \left|\frac{1}{x}-\frac{1}{c}\right| = \left|\frac{c-x}{xc}\right| = |x-c|\frac{1}{c}\cdot\frac{1}{x} \leq |x-c|\frac{1}{c(c-\mu)}.$$&#xD;&#xA;So if we &lt;em&gt;also&lt;/em&gt; ask that $\delta\lt c(c-\mu)\epsilon$, then we have:&#xA;$$|f(x)-f(c)|\leq |x-c|\frac{1}{c(c-\mu)} \lt \frac{\delta}{c(c-\mu)} \lt \frac{c(c-\mu)\epsilon}{c(c-\mu)} = \epsilon$$&#xD;&#xA;which is what we want!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, in summary, what do we need? We need to make sure that $\delta$ is:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Less than $1$;&lt;/li&gt;&#xA;&lt;li&gt;Less than $\frac{c}{2}$; (both of these to ensure $\frac{1}{x}\lt\frac{1}{c-\mu}$);&lt;/li&gt;&#xA;&lt;li&gt;Less than $c(c-\mu)\epsilon$, where $\mu=\min\{1,\frac{c}{2}\}$. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So, for instance, we can just let $\delta = \frac{1}{2}\min(1,\frac{c}{2},c(c-\mu)\epsilon)$, where $\mu=\min\{1,\frac{c}{2}\}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, if you can let your $\delta$ depend only on $f(x)$ and on $\epsilon$ but not $c$, then we say $f(x)$ is &lt;em&gt;uniformly continuous&lt;/em&gt;. This is a stronger condition than continuity, and often very useful. $\frac{1}{x}$ is not uniformly continuous on $(0,\infty)$, however (though it is on $[a,\infty)$ for any $a\gt 0$). &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-30T20:23:11.010" LastActivityDate="2010-11-30T20:23:11.010" CommentCount="4" />
  <row Id="12498" PostTypeId="2" ParentId="12485" CreationDate="2010-11-30T15:21:05.617" Score="2" ViewCount="" Body="&lt;p&gt;You may want to take a look at Kelley and Peterson's textbook [1]. The problem you are looking at here is a recurrence relation as you have stated. You also hear them called difference equations. If you are familiar with differential equations, these difference equations are the discrete analog of differential equations [2].&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Kelley and Peterson go very in depth with solving different types of difference equations. They give a nice overview of several different methods.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope this helps you! If you have any other questions, feel free to email me or contact me through my blog &lt;a href=&quot;http://www.tylerclark12.com/blog&quot; rel=&quot;nofollow&quot;&gt;http://www.tylerclark12.com/blog&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[1] Kelley, W. &amp;amp; Peterson, A. (2001). Difference Equations: An Introduction with Applications (2nd Ed.). San Diego, CA: Academic Press.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[2] Weisstein, Eric W. &quot;Recurrence Equation.&quot; From MathWorld--A Wolfram Web Resource. &lt;a href=&quot;http://mathworld.wolfram.com/RecurrenceEquation.html&quot; rel=&quot;nofollow&quot;&gt;http://mathworld.wolfram.com/RecurrenceEquation.html&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="3791" LastActivityDate="2010-11-30T15:21:05.617" />
  <row Id="12500" PostTypeId="2" ParentId="12477" CreationDate="2010-11-30T15:30:10.837" Score="0" ViewCount="" Body="&lt;p&gt;I just thought of one that is at about the level I envisioned.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let V be a two-dimensional Q-subspace of the Q-vector space R.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(a)  Show that if V is closed under multiplication of real numbers, and if the nonzero vectors in V are closed under division of real numbers, then V has a basis of the form {1, \alpha} where alpha is a root of a quadratic polynomial whose coefficients are integers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(b)  Show that under the same hypotheses as in part (a), V has a basis of the form {1, \sqrt{N}} where N is a positive integer. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(c)  Show that if V has a basis of the form {1,\sqrt{N}} where N is a positive integer, then V is closed under multiplication of real numbers and the nonzero vectors of V are closed under division of real numbers.&lt;/p&gt;&#xA;" OwnerUserId="4253" LastActivityDate="2010-11-30T15:30:10.837" />
  <row Id="12501" PostTypeId="1" AcceptedAnswerId="12517" CreationDate="2010-11-30T16:18:42.870" Score="-1" ViewCount="138" Body="&lt;p&gt;For a constant, &lt;code&gt;N&lt;/code&gt;, what value of &lt;code&gt;x&lt;/code&gt; will maximize the cosine (or any trig) function?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;1 = \cos{(Nx)}&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am looking for the exact form, not the approximation because,&#xA;\begin{equation}&#xA;\frac{\arccos{(1)}}{N} = x = 0&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, &lt;a href=&quot;http://www.wolframalpha.com/input/?i=1+%3D+cos%2819.013+%2a+x%29&quot; rel=&quot;nofollow&quot;&gt;WolframAlpha.com&lt;/a&gt; states that if &lt;code&gt;N = 19.013&lt;/code&gt;, then,&#xA;\begin{equation}&#xA;x = \frac{2000 \pi n}{19013} , n \text{ } \varepsilon \text{ } \text{set of integers}&#xA;\end{equation}&#xA;How was that solution calculated?&lt;/p&gt;&#xA;" OwnerUserId="4222" LastActivityDate="2010-11-30T18:55:06.283" Title="Calculate exact solution to maximize cosine" Tags="&lt;homework&gt;&lt;trigonometry&gt;" AnswerCount="3" CommentCount="10" />
  <row Id="12503" PostTypeId="2" ParentId="12431" CreationDate="2010-11-30T16:28:40.380" Score="2" ViewCount="" Body="&lt;p&gt;I may be messing up my usage of Fourier transforms and convolutions with distributions, but I think this can be solved with a few identities (namely how transforms work on products and convolutions, and how convolutions work with derivatives and Dirac deltas). Given the starting point $(xf) * g = a t (f * g)$, take the (unitary, ordinary frequency) transform to obtain&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(\hat{x}*\hat{f} ) \hat{g} = a\hat{x} * (\hat{f} \hat{g}) \implies (\delta&amp;#39;*\hat{f} ) \hat{g} = a\delta&amp;#39; * (\hat{f} \hat{g}) \implies \hat{f}&amp;#39; \hat{g} = a (\hat{f}\hat{g})&amp;#39;$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you foil and rewrite you get the simple differential equation $a \hat{g} &amp;#39; + (a-1) (\hat{f}&amp;#39; / \hat{f}) \hat{g} =0 $, which is easy to solve and comes out with $\hat{g} = \hat{f}^{1/a - 1}$ (I hope I did that all correctly!).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[EDIT - thanks to below comment.]&lt;/p&gt;&#xA;" OwnerUserId="3122" LastEditorUserId="3122" LastEditorDisplayName="" LastEditDate="2010-11-30T16:45:30.553" LastActivityDate="2010-11-30T16:45:30.553" CommentCount="2" />
  <row Id="12504" PostTypeId="2" ParentId="275" CreationDate="2010-11-30T16:39:01.727" Score="0" ViewCount="" Body="&lt;p&gt;I rather enjoyed Professor Stewart's book [1]. Take a look at it; I hope you enjoy it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have blogged about a selection from his book, you can view it at [2]. This is just one of the many different mathematical concepts covered in the book. It is more of a fun book than lots of theory. It will get you to think.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[1] Stewart, I. (2009). Professor Stewart’s Hoard of Mathematical Treasures. New York: Basic Books.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[2] &lt;a href=&quot;http://www.tylerclark12.com/blog/?p=159&quot; rel=&quot;nofollow&quot;&gt;http://www.tylerclark12.com/blog/?p=159&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3791" LastActivityDate="2010-11-30T16:39:01.727" CommunityOwnedDate="2010-11-30T16:39:01.727" />
  <row Id="12505" PostTypeId="2" ParentId="12477" CreationDate="2010-11-30T16:43:27.057" Score="4" ViewCount="" Body="&lt;p&gt;Continuing Akhil's answer, let's prove a theorem of Dehn: if a rectangle is tiled by squares, the the ratio of the lengths of its sides is rational.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose to the contrary that the sides of the rectangle $x,y$ are &lt;i&gt;not&lt;/i&gt; rationally dependent. Then we can find some linear homomorphism $f\colon \mathbb{R} \rightarrow \mathbb{Q}$ such that $f(x) = 1$ and $f(y) = -1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We define the $f$-area $A(R)$ of a rectangle $R$ with edge lengths $h,v$ to be $f(h)f(v)$. If a rectangle $R$ is tiled by rectangles $R_i$ &lt;i&gt;forming a grid&lt;/i&gt;, then from linearity it immediately follows that $$A(R) = \sum_i A(R_i).$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Denote the big rectangle $R$ and the squares $S_i$. Take your tiling and extend all the lines to form a grid inside the rectangle. Denote the grid rectangles by $G_j$. Then $$A(R) = \sum_j A(G_j) = \sum_i A(S_i).$$ Since a square has both sides equal, $A(S_i) \geq 0$. On the other hand, by construction $A(R) &amp;lt; 0$. This contradiction shows that the two sides of the big rectangle are, in fact, rationally dependent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instead of taking a linear mapping from $\mathbb{R}$ to $\mathbb{Q}$ we could take a linear mapping from a smaller, finite dimensional domain by only considering the lengths in the grid $G_j$ - then everything becomes beginner's linear algebra.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-30T16:43:27.057" />
  <row Id="12506" PostTypeId="2" ParentId="275" CreationDate="2010-11-30T17:06:10.613" Score="1" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://books.google.com/books?id=Oac_PgAACAAJ&amp;amp;dq=how+to+solve+it&amp;amp;hl=en&amp;amp;ei=0S71TPb-BIX7lweo3viUBg&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=2&amp;amp;ved=0CD4Q6AEwAQ&quot; rel=&quot;nofollow&quot;&gt;How to solve it&lt;/a&gt; by G. Polya.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://books.google.com/books?id=NNQEAAAAYAAJ&amp;amp;printsec=frontcover&amp;amp;dq=value+of+science&amp;amp;hl=en&amp;amp;ei=LS_1TO6XG8SAlAfYwvCbBg&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=1&amp;amp;ved=0CCsQ6AEwAA#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;The Value of Science&lt;/a&gt; by H. Poincare.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://books.google.com/books?id=T43Cmu_EaZAC&amp;amp;printsec=frontcover&amp;amp;dq=symmetry&amp;amp;hl=en&amp;amp;ei=-i_1TO6qKcGqlAf90sDQBQ&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=1&amp;amp;ved=0CDUQ6AEwAA#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;Symmetry&lt;/a&gt; by H. Weyl&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://books.google.com/books?id=R6E0AAAAMAAJ&amp;amp;printsec=frontcover&amp;amp;dq=flatland&amp;amp;hl=en&amp;amp;ei=VjD1TKnfAoOdlgeuh-jiBQ&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=2&amp;amp;ved=0CC0Q6AEwAQ#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;Flatland&lt;/a&gt; by E. Abbott&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://books.google.com/books?id=vslEPgAACAAJ&amp;amp;dq=chaos&amp;amp;hl=en&amp;amp;ei=hi_1TIyJBIOClAevkJXTBQ&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=1&amp;amp;ved=0CCwQ6AEwAA&quot; rel=&quot;nofollow&quot;&gt;Chaos: Making of a New Science&lt;/a&gt; by J. Gleick&lt;/p&gt;&#xA;" OwnerUserId="1417" LastEditorUserId="1417" LastEditorDisplayName="" LastEditDate="2010-11-30T17:13:39.373" LastActivityDate="2010-11-30T17:13:39.373" CommunityOwnedDate="2010-11-30T17:06:10.613" CommentCount="1" />
  <row Id="12507" PostTypeId="1" AcceptedAnswerId="12510" CreationDate="2010-11-30T17:18:30.947" Score="10" ViewCount="246" Body="&lt;p&gt;I've been wondering about the following question:&#xA;Suppose that $P$ is a polynomial of degree $n$&amp;#160;with complex coefficients. Assume that $a_0, a_1, \dots, a_n \in \mathbb{C}$ are distinct. Are the polynomials $$P(x + a_0), P(x + a_1), \dots, P(x + a_n)$$ linearly independent?&lt;/p&gt;&#xA;" OwnerUserId="3776" LastActivityDate="2011-02-16T17:46:37.887" Title="Are translations of a polynomial linearly independent?" Tags="&lt;linear-algebra&gt;&lt;polynomials&gt;" AnswerCount="4" CommentCount="2" FavoriteCount="1" />
  <row Id="12508" PostTypeId="2" ParentId="275" CreationDate="2010-11-30T17:30:51.137" Score="0" ViewCount="" Body="&lt;p&gt;Here's what you're looking for. It's an amazing back, and you'll be blown away by all the stuff mentioned in here:&#xA;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/1402757964&quot; rel=&quot;nofollow&quot;&gt;http://www.amazon.com/Math-Book-Pythagoras-Milestones-Mathematics/dp/1402757964&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2985" LastActivityDate="2010-11-30T17:30:51.137" CommunityOwnedDate="2010-11-30T17:30:51.137" />
  <row Id="12510" PostTypeId="2" ParentId="12507" CreationDate="2010-11-30T17:33:50.993" Score="7" ViewCount="" Body="&lt;p&gt;Consider the matrix of coefficients, where each row corresponds to an evaluation at a point, and each column to a power of $x$. Any linear dependency among the columns amounts to some degree~$n$ polynomial in the point, which is supposed to have $n+1$ roots. Since this can't happen, the polynomials are independent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example (almost Vandermonde) - $P(x) = x^2$: $$\begin{pmatrix} 1 &amp;amp; 2a &amp;amp; a^2 \\ 1 &amp;amp; 2b &amp;amp; b^2 \\ 1 &amp;amp; 2c &amp;amp; c^2 \end{pmatrix}$$ A linear dependency of the columns is a quadratic polynomial having three different roots $a,b,c$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Just to clarify what happens in the general case. Here's the matrix corresponding to $P(x) = x^2 + \alpha x + \beta$: $$\begin{pmatrix} 1 &amp;amp; 2a + \alpha &amp;amp; a^2 + \alpha a + \beta \\ 1 &amp;amp; 2b + \alpha &amp;amp; b^2 + \alpha b + \beta \\ 1 &amp;amp; 2c + \alpha &amp;amp; c^2 + \alpha c + \beta \end{pmatrix}$$&#xD;&#xA;Since the degree of the polynomial in column $t$ is $t$ (if we start counting from zero), it's easy to see that the any non-zero combination of columns results in a non-zero polynomial (consider the rightmost non-zero column).&lt;/p&gt;&#xA;" OwnerUserId="1277" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2010-11-30T18:27:05.197" LastActivityDate="2010-11-30T18:27:05.197" CommentCount="3" />
  <row Id="12511" PostTypeId="2" ParentId="12501" CreationDate="2010-11-30T17:53:34.637" Score="0" ViewCount="" Body="&lt;p&gt;We know that   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;1 = \cos{(2\pi)}&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, for a given &lt;code&gt;N&lt;/code&gt; to &quot;maximize&quot; cosine:&lt;br&gt;&#xA;\begin{aligned}&#xA;2\pi = Nx \&#xA;\end{aligned}&#xA;\begin{aligned}&#xA;x = \frac{2\pi}{N}&#xA;\end{aligned}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the above equation:&#xA;\begin{aligned}&#xA;x &amp;amp;= \frac{2\pi}{19.013} \&#xA;\end{aligned}&#xA;\begin{aligned}&#xA;x &amp;amp;\approx 0.33046 \&#xA;\end{aligned}&lt;/p&gt;&#xA;" OwnerUserId="4222" LastEditorUserId="4222" LastEditorDisplayName="" LastEditDate="2010-11-30T18:02:03.180" LastActivityDate="2010-11-30T18:02:03.180" CommentCount="1" />
  <row Id="12512" PostTypeId="1" CreationDate="2010-11-30T17:56:21.013" Score="3" ViewCount="189" Body="&lt;p&gt;Is it even worth the theorem below?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For every positive integer $n$, there is a real number $r$, &#xA;and $\frac{1}{12n+1} \lt r \lt\frac{1}{12n}$, such that:&#xA;$$ n! = \sqrt{2n\pi}\left(\frac{n}{e}\right)^n e^r.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I saw this statement on some sites, but got no further details.&#xA;I think the statement refers to an exact value of $n!$, not an approximation.&lt;/p&gt;&#xA;" OwnerUserId="2938" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-30T18:27:00.250" LastActivityDate="2010-11-30T18:44:19.250" Title="$n! =\sqrt{2n\pi}\left(\frac{n}{e}\right)^n e^r$?" Tags="&lt;number-theory&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" />
  <row Id="12513" PostTypeId="2" ParentId="12512" CreationDate="2010-11-30T17:59:17.463" Score="13" ViewCount="" Body="&lt;p&gt;It is the strirling's aproximation  &lt;a href=&quot;http://en.wikipedia.org/wiki/Stirling%27s_approximation&quot; rel=&quot;nofollow&quot;&gt;See article&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$n!\approx \left(\frac{n}{e}\right)^{n}\sqrt{2\pi n}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The error estimates are very interesting:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$n!=\left(\frac{n}{e}\right)^{n}\sqrt{2\pi n} \cdot e^{r_{n}}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With $\frac{1}{12n+1}&amp;lt;r_{n}&amp;lt;\frac{1}{12n}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the strirling's aproximation with remainder &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; This aproximation is very useful, for example in the calculation of limits of sequences&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\sqrt[n]{n!}}{n} \approx \frac{\sqrt[n]{\left(\frac{n}{e}\right)^{n}\sqrt{2\pi n}}}{n} = \frac{\left(\frac{n}{e}\right)\sqrt[n]{\sqrt{2\pi n}}}{n} = \frac{\sqrt[2n]{2\pi n}}{e} \longrightarrow \frac{1}{e}, ( n\longrightarrow \infty ) $$   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the fact $\sqrt[n]{n} \longrightarrow 1,  ( n\longrightarrow \infty )$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="3421" LastEditorUserId="3421" LastEditorDisplayName="" LastEditDate="2010-11-30T18:44:19.250" LastActivityDate="2010-11-30T18:44:19.250" />
  <row Id="12514" PostTypeId="2" ParentId="12512" CreationDate="2010-11-30T17:59:28.430" Score="2" ViewCount="" Body="&lt;p&gt;I think you are referring to the Stirling's approximation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Stirlings_approximation&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Stirlings_approximation&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3431" LastActivityDate="2010-11-30T17:59:28.430" />
  <row Id="12515" PostTypeId="2" ParentId="12507" CreationDate="2010-11-30T18:14:23.440" Score="0" ViewCount="" Body="&lt;p&gt;If you write $P(x) = \sum_{j=0}^n p_j x^j$ and set $ \sum_{i=0}^n c_i P(x+a_i) = 0$, the full expression is written $\sum_{i=0}^n \sum_{k=0}^n \sum_{j=0}^k c_i p_k {k \choose j} a_i^{k-j} x^j$. If we find the coefficient of each power $x_j$ as a matrix times the vector $(c_0, ..., c_n)$, and then setting the system equal to zero, we can deduce the determinant of the matrix is zero. The matrix is $A_{ij} = \sum_{k=j}^n p_k {k \choose j} a_i^{k-j}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Setting the determinant equal to zero gives a polynomial in all $a_i$. I don't know if this polynomial is simpler than the matrix hints at, but it is certainly possible to find coefficients $a_i$ such that the polynomials are linearly independent, though for &quot;most&quot; values it won't be the case, as the solution set must be of at least one dimension less than $C^{n+1}$ (seen by fixing all $a_1, ..., a_n$ arbitrarily and thereby determining a finite set of possible $a_0$).&lt;/p&gt;&#xA;" OwnerUserId="3122" LastActivityDate="2010-11-30T18:14:23.440" CommentCount="2" />
  <row Id="12516" PostTypeId="2" ParentId="12501" CreationDate="2010-11-30T18:52:23.657" Score="0" ViewCount="" Body="&lt;p&gt;$x=0$ is always a solution, since $\cos(N\times0)=\cos0=1$ . &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want the set of all solutions, use the fact that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\cos\theta=1 \Leftrightarrow \theta=2k\pi \text{ with } k\in\mathbb{Z}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Replacing $\theta$ by $Nx$ above gives you the set of solutions :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ 1=\cos{Nx}\Leftrightarrow x\in \left\{\frac{2k\pi}N\right\}_{k\in\mathbb{Z}}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the solution given by Wolfram Alpha for $N=19.013=\frac{19013}{1000}$.&lt;/p&gt;&#xA;" OwnerUserId="3918" LastActivityDate="2010-11-30T18:52:23.657" />
  <row Id="12517" PostTypeId="2" ParentId="12501" CreationDate="2010-11-30T18:55:06.283" Score="1" ViewCount="" Body="&lt;p&gt;Cosine takes it's maximum when the argument is $2k\pi$, where $k$ is any integer. Therefore &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{aligned}2k\pi = Nx \\\\&#xD;&#xA;&#xD;&#xA;x=\frac{2k\pi}{N}\end{aligned}$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus for $N=19.013$&#xD;&#xA;$$x= \frac{2k\pi}{19.013} = \frac{2000k\pi}{19013}$$.&lt;/p&gt;&#xA;" OwnerUserId="4216" LastActivityDate="2010-11-30T18:55:06.283" />
  <row Id="12518" PostTypeId="2" ParentId="12323" CreationDate="2010-11-30T19:13:18.450" Score="1" ViewCount="" Body="&lt;p&gt;When did 'none' come to be a number? My guess is that it was when merchants or scribes first codified symbolic addition and subtraction to calculate quantities of food and supplies on hand. Intuitively, they must have known that 'none' would be the result of subtracting 100 bushels of wheat from same. And that adding 100 bushels to 'none' would give you a measurable quantity. Philosophers may have quibbled about whether 'none' was a real and true number, but I'm sure it was all too real to the merchants and scribes of antiquity.&lt;/p&gt;&#xA;" OwnerUserId="3515" LastActivityDate="2010-11-30T19:13:18.450" />
  <row Id="12519" PostTypeId="2" ParentId="12507" CreationDate="2010-11-30T19:27:52.670" Score="0" ViewCount="" Body="&lt;p&gt;Here a fancy proof by induction that uses differentiation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Case $n=0$.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A polynomial of degree $0$ has the form $P(x)=c_0$ with $c_0\neq 0$ which is clearly linearly independent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Case $n=k+1$.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume that there is a linear dependence&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\lambda_0 P(x+a_0) + \lambda_1 P(x+a_1) + \dots + \lambda_n P(x+a_n) = 0.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since this relation holds for all $x$, we can differentiate this relation. After differentiating $n$ times, we obtain a relation for the $n$-th derivative&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\lambda_0 P^{(n)}(x+a_0) + \lambda_1 P^{(n)}(x+a_1) + \dots + \lambda_n P^{(n)}(x+a_n) = 0.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But since $P$ has degree $n$, we know that $P^{(n)}(x+a_i)$ is equal to a constant $c\neq 0$, which implies&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \lambda_0 + \lambda_1 + \dots + \lambda_n = 0 .$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By expressing $\lambda_0$ in terms of the other coefficients, we obtain a relation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \lambda_1 (P(x+a_1) - P(x+a_0)) + \dots + \lambda_n (P(x+a_n) - P(x+a_0)) = 0$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;of polynomials of degree $n-1$. By induction, we conclude that all the $\lambda_i$ must vanish.&lt;/p&gt;&#xA;" OwnerUserId="434" LastActivityDate="2010-11-30T19:27:52.670" CommentCount="2" />
  <row Id="12520" PostTypeId="1" AcceptedAnswerId="12529" CreationDate="2010-11-30T19:28:21.500" Score="1" ViewCount="262" Body="&lt;p&gt;Can we call the &lt;a href=&quot;http://mathworld.wolfram.com/FloorFunction.html&quot; rel=&quot;nofollow&quot;&gt;greatest integer function&lt;/a&gt; as a periodic function with no fundamental period or is it just non-periodic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please explain your answer.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;To my understanding, if we consider $f(x) = [x]$ now, $f(3) = [3] = 3$ and $f(3+0.5) = [3.5] = 3$ So, can't we say that it is periodic (&lt;strong&gt;A constant function is periodic with no fundamental period&lt;/strong&gt;) ? But the problem is to derive the fundamental period.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt; After checking out some aswer I am quite inquisitive to know is it really necessary to have a fundamental period to call a function periodic? However,If you go by my book it is not.&lt;/p&gt;&#xA;" OwnerUserId="2109" LastEditorUserId="2109" LastEditorDisplayName="" LastEditDate="2010-11-30T19:48:38.933" LastActivityDate="2010-12-01T16:47:46.237" Title="Doubt on greatest integer function" Tags="&lt;functions&gt;" AnswerCount="6" CommentCount="3" />
  <row Id="12521" PostTypeId="2" ParentId="12520" CreationDate="2010-11-30T19:30:34.430" Score="1" ViewCount="" Body="&lt;p&gt;A periodic functions repeats it's output values after regular intervals. That is there exists $T$ such that $f(x+T)=f(x)$ for all $x$. The greatest integer function does not satisfy this equation for any $T$. So it is not periodic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Periodic_function&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Periodic_function&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fundamental period is just the smallest (positive) such $T$ (if one exists). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sawtooth function which measures the deviation from the greatest integer function (also called floor function) is periodic with period 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Sawtooth_function&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Sawtooth_function&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3431" LastEditorUserId="3431" LastEditorDisplayName="" LastEditDate="2010-11-30T19:52:12.423" LastActivityDate="2010-11-30T19:52:12.423" />
  <row Id="12522" PostTypeId="1" AcceptedAnswerId="12527" CreationDate="2010-11-30T19:35:38.783" Score="3" ViewCount="106" Body="&lt;p&gt;What is the value of n $\in \mathbb{Z} $ for which the function $\displaystyle f(x) = \frac{\sin nx} { \sin \biggl( \frac{x}{n} \biggr) } \text { has } 4\pi $ as period?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also could it be possible to solve this if we need $x\pi$ as period ?I am interested in learning the general approach for this particular type of the problem.&lt;/p&gt;&#xA;" OwnerUserId="2109" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2010-11-30T19:54:53.840" LastActivityDate="2010-12-02T12:03:03.493" Title="Period of a function" Tags="&lt;trigonometry&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="12523" PostTypeId="2" ParentId="7096" CreationDate="2010-11-30T19:36:19.993" Score="2" ViewCount="" Body="&lt;p&gt;I think the following argument answers your question. It reminds me a lot of the exercises in the basic topology course I took during my undergrad. Fun times.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Recall that $\pi$ is a submersion, so for any $x$ in $E$ there exists a neighborhood of the form $U \times V$, such that $\pi$ identifies with the projection $U \times V \to V$. Suppose we only know that $E_t := \pi^{-1}(t)$ is compact for any $t$ in $B$, and let's show that $\pi$ is proper.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $K \subset B$ be compact and set $L = \pi^{-1}(K)$. Let $(\mathcal U_\alpha)$ be an covering of $L$ by open subsets of $E$. We will show that there exists a finite covering of $E$ by open subsets of $(\mathcal U_\alpha)$, and thus by sets of the original covering.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take $t \in K$. Find neighborhoods $\mathcal U_1, \ldots \mathcal U_{n(t)}$ that cover $E_t$. For any point $x \in E_t$, find neighborhoods $U_{t,x} \times V_{t,x} \subset \mathcal U_{j(x)}$ for some $j(x)$ (the $\mathcal U_{j(x)}$ being one of the above), such that the map $\pi$ identifies with the projection $U_{t,x} \times V_{t,x} \to V_{t,x}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The compactness of $E_t$ gives finitely many such $U_{t,x_\nu} \times V_{t,x_\nu}$ which cover $E_t$ and such that there is a $V_t \subset B$ contained in the image of any $V_{t,x_\nu}$ by $\pi$ (the intersection of the finitely many such $V$ contains $t$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now: $K$ is compact, so it is covered by finitely many of those $V_t$. The corresponding finite collection $(U_{t,x_\nu} \times V_{t,x_\nu})$ covers $L$, and consists of subsets of elements of $(\mathcal U_\alpha)$. Thus finitely many elements of $(\mathcal U_\alpha)$ cover $L$.&lt;/p&gt;&#xA;" OwnerUserId="3225" LastActivityDate="2010-11-30T19:36:19.993" CommentCount="2" />
  <row Id="12524" PostTypeId="2" ParentId="12520" CreationDate="2010-11-30T19:37:28.103" Score="1" ViewCount="" Body="&lt;p&gt;I am inclined to say that the greatest integer function (floor function) is not periodic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mathworld [1] tells us that, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A function f(x) is said to be periodic (or, when emphasizing the presence of a single period instead of multiple periods, singly periodic) with period p if $f(x)=f(x+np)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With the floor function, we can see that $\lfloor x+n \rfloor = \lfloor x \rfloor + n$ for $ x \not \in \mathbb{Z}$ and $n \in \mathbb{Z}$. This would give us reason to believe that we cannot find a period $p$ such that $f(x)=f(x+np)$ is satisfied for the greatest integer function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[1] Weisstein, Eric W. &quot;Periodic Function.&quot; From MathWorld--A Wolfram Web Resource. &lt;a href=&quot;http://mathworld.wolfram.com/PeriodicFunction.html&quot; rel=&quot;nofollow&quot;&gt;http://mathworld.wolfram.com/PeriodicFunction.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think I know what you are saying about the constant function now though. I suppose that if you restrict the domain of the function, then you could say that the greatest integer function is periodic. Suppose that we just want to look at the greatest integer function on the interval [0,1). Then we can say that the definition of periodicity is satisfied. However on the entire domain of the greatest integer function, we do not have this property.&lt;/p&gt;&#xA;" OwnerUserId="3791" LastEditorUserId="3791" LastEditorDisplayName="" LastEditDate="2010-11-30T19:49:38.823" LastActivityDate="2010-11-30T19:49:38.823" CommentCount="6" />
  <row Id="12525" PostTypeId="2" ParentId="12520" CreationDate="2010-11-30T19:40:43.717" Score="0" ViewCount="" Body="&lt;p&gt;Even if you choose your period $T$ arbitrarily small, you can't satisfy $f(x+T) = f(x)$ for the arguments that are closer to any integer than T, so yo can't say it's a function with no fundamental period, it is non-periodic.&lt;/p&gt;&#xA;" OwnerUserId="4216" LastActivityDate="2010-11-30T19:40:43.717" />
  <row Id="12526" PostTypeId="1" AcceptedAnswerId="14774" CreationDate="2010-11-30T19:47:04.550" Score="6" ViewCount="187" Body="&lt;p&gt;Let $B$ be the unit ball of $\ell^2(\mathbb{N})$, i.e. $B=\lbrace x\in \ell^2(\mathbb{N}): \|x\|\le 1\rbrace.$ For each $x=(x_1,x_2,\cdots)\in B$, let &#xA;$$f(x)=(1-\|x\|,x_1,x_2,\cdots).$$ Define $T:B\to 2^B$ by $$T(x)=B(f(x),r(x))\cap B, \mbox{ where }r(x)=\frac{1}{2}(\|x-f(x)\|).$$&#xD;&#xA;Is it true that $$D(Tx,Ty)\le \|x-y\| \mbox{ for all }x,y\in B?$$&#xD;&#xA;Here $B(y,r)$ denotes the closed ball with radius $r$ centered at $y$, and $D$ is the Hausdorff metric defined by $$D(A,B)=\inf\lbrace r&amp;gt;0: N_r(A)\supset B, &#xD;&#xA;N_r(B)\supset A\rbrace,$$ $N_r(S) =\lbrace x\in C: d(x,S)\lt r\rbrace$ being the &#xA;$r$-neighborhood of $S$. &lt;/p&gt;&#xA;" OwnerUserId="3249" LastEditorUserId="3249" LastEditorDisplayName="" LastEditDate="2010-11-30T20:25:59.820" LastActivityDate="2010-12-18T17:38:16.397" Title="A map on the unit ball" Tags="&lt;geometry&gt;&lt;functional-analysis&gt;&lt;metric-spaces&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="0" />
  <row Id="12527" PostTypeId="2" ParentId="12522" CreationDate="2010-11-30T19:49:01.723" Score="4" ViewCount="" Body="&lt;p&gt;With $n=2$, $\sin(2x)$ has period $\pi$ and $\sin(x/2)$ has period $4\pi$ so their ratio must have a period of $4\pi$ since the latter period is an integral multiple of the former.&lt;/p&gt;&#xA;" OwnerUserId="3431" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-02T12:03:03.493" LastActivityDate="2010-12-02T12:03:03.493" CommentCount="9" />
  <row Id="12528" PostTypeId="2" ParentId="12522" CreationDate="2010-11-30T19:53:28.817" Score="5" ViewCount="" Body="&lt;p&gt;You want $$\frac{\sin n(x + 4\pi)}{\sin \frac{x + 4\pi}{n}} = \frac{\sin nx}{\sin \frac{x}{n}}.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is equivalent with&#xA;$$\sin \frac{x + 4\pi}{n} = \sin \frac{x}{n}.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore $\frac{x}{n} = \frac{x + 4\pi}{n} + 2k\pi$ or $\frac{x}{n} = \pi - \frac{x + 4\pi}{n} + 2k\pi$ for some $k \in \mathbb{Z}$. In the first case $x = x + 4 \pi + 2k \pi n$ and thus $n = \frac{4}{2k}$. In the second case $x = \pi n - x - 4\pi + 2k\pi n$ and thus $n = \frac{2x + 4\pi}{2k\pi + \pi}$, which is impossible since this should hold for every $x \in \mathbb{R}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus $n = \pm 1$ or $\pm 2$.&lt;/p&gt;&#xA;" OwnerUserId="3776" LastActivityDate="2010-11-30T19:53:28.817" />
  <row Id="12529" PostTypeId="2" ParentId="12520" CreationDate="2010-11-30T19:54:23.467" Score="11" ViewCount="" Body="&lt;p&gt;A function $f(x)$ is &quot;periodic with period $c$&quot;, with $c\gt 0$, if $f(x+c) = f(x)$ for all values of $x$. In particular, $f(x+mc) = f(x)$ for all integers $m$, by using induction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A function is periodic if it is periodic with period $c$ for some $c\neq 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Every periodic function has lots of periods: if $c$ is a period, then so is $mc$ for any positive integer $m$, at the very least; it may have others. So we may be interested in knowing what is the &quot;quickest&quot; that the function starts repeating. For instance, $\tan x$ certainly repeats every $2\pi$, since $\tan(x+2\pi) = \tan(x)$ for all $x$ (either both are defined and equal, or both are undefined); and by the comment above, it will repeat every $2m\pi$ for any integer $m$. But it actually repeats &lt;em&gt;more often&lt;/em&gt; than every $2\pi$: we know that $\tan(x+\pi) = \tan(x)$ for all $x$. So altough it is true that $\tan(x)$ is periodic with period $2\pi$, it is &lt;em&gt;also&lt;/em&gt; true that $\tan(x)$ is periodic with period $\pi$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we would like to have something more than just a &quot;period&quot;; we would like to be able to talk about the &lt;em&gt;smallest&lt;/em&gt; possible period. That's what the &quot;fundamental period&quot; is trying to capture:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We say that a periodic function $f(x)$ has &quot;fundamental period&quot; $k$ if and only if $k$ is the smallest positive number such that $f(x)$ is periodic with period $k$. That means that we require:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$f$ to be periodic with period $k$: so $f(x) = f(x+k)$ for all $x$;&lt;/li&gt;&#xA;&lt;li&gt;$k$ to be positive; and&lt;/li&gt;&#xA;&lt;li&gt;if $c$ is a positive constant such that $f(x+c) = f(x)$ for all $x$, then $k\leq c$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So, in general, we would take the set of all positive periods of the function $f$, and look for its minimum (smallest element). The one possible problem is that not every set of positive real numbers has a smallest element, so it may be possible, at least in principle, that a function is periodic but has no fundamental period, because it &lt;em&gt;does&lt;/em&gt; have periods, but it doesn't have one that is the &lt;em&gt;smallest&lt;/em&gt; period.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The constant functions, $f(x) = a$ for all $x$, are an example of this pathology: they are periodic, because they satisfy the defintion of being periodic. In order to show that they satisfy the definition, we need to show that there is at least one number $c\gt 0$ such that $f(x)=f(x+c)$ for all $x$. Well, we can take $c=1$, because $f(x) = a = f(x+1)$ for all $x$. So the constant function is certainly periodic. But in fact, for &lt;em&gt;every&lt;/em&gt; positive number $c$, $f(x+c) = a =f(x)$. That means that if you take &lt;em&gt;any&lt;/em&gt; $c\gt 0$, then you can rightly say that &quot;$f(x)$ is periodic with period $c$&quot;. So if we look at the set of all periods,&#xA;$$P=\{ c \in \mathbb{R}\mid c\gt 0\text{ and $f(x)$ is periodic of period $c$}\}$$&#xA;then $P=(0,\infty)$. Since the &quot;fundamental period&quot; is supposed to be the &lt;em&gt;smallest&lt;/em&gt; element of $P$, but $P$ does &lt;em&gt;not&lt;/em&gt; have a smallest element, then the constant function does &lt;em&gt;not&lt;/em&gt; have a fundamental period. But it &lt;em&gt;is&lt;/em&gt; periodic (we just saw it satisfies the definition).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of the greatest integer function, $f$ is not periodic at all. Again, &quot;$f$ is periodic&quot; means &quot;there exists $c\gt 0$ such that $f(x+c)=f(x)$ for all $x$&quot;. So the negation of &quot;$f$ is periodic&quot; is &quot;for every $c\gt 0$, there exists at least one $x$ such that $f(x)\neq f(x+c)$.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to show that the greatest integer function $f(x)=\lfloor x\rfloor$ is &lt;strong&gt;not&lt;/strong&gt; periodic, we need to show that given any $c\gt 0$ (think of it as a candidate for a period), there is at least one point $x$ for which $f(x)\neq f(x+c)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, let $c\gt 0$ be given. I'm going to come up with some point $x$ (which will depend on $c$) with the property that $\lfloor x\rfloor\neq \lfloor x+c\rfloor$. The idea is simply to take a value of $x$ that is before some integer, but once you add $c$ to it you go &lt;em&gt;above&lt;/em&gt; that integer. Then the greatest integer function will give two different values. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The simplest example to take is $x=-\frac{c}{2}$; this number is negative, so $f(-\frac{c}{2}) = \lfloor -\frac{c}{2}\rfloor\lt 0$; on the other hand, $x+c = -\frac{c}{2}+c = \frac{c}{2}$ is positive, so $f(x+c) = \lfloor \frac{c}{2}\rfloor \geq 0$. Since $f(x)$ is strictly less than $0$ and $f(x+c)$ is at least $0$, $f(x)\neq f(x+c)$. So $f(x)$ is not periodic with period $c$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But $c$ was an &lt;em&gt;arbitrary&lt;/em&gt; positive number. That means that we cannot say &quot;$f(x)$ is periodic with period $c$&quot; for &lt;em&gt;any&lt;/em&gt; positive number, and that means, by definition, that $f(x) = \lfloor x\rfloor$ is &lt;strong&gt;not&lt;/strong&gt; periodic.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-30T20:28:01.800" LastActivityDate="2010-11-30T20:28:01.800" CommentCount="5" />
  <row Id="12530" PostTypeId="2" ParentId="12454" CreationDate="2010-11-30T19:59:04.300" Score="0" ViewCount="" Body="&lt;p&gt;Although the software below does not allow one to check for planarity you might find it useful in seeing drawings of graphs (not typically drawn in the plane) with a small number of vertices and specified number of edges:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.gfredericks.com/sandbox/graphs/browse&quot; rel=&quot;nofollow&quot;&gt;http://www.gfredericks.com/sandbox/graphs/browse&lt;/a&gt;  &lt;/p&gt;&#xA;" OwnerUserId="1369" LastEditorUserId="1369" LastEditorDisplayName="" LastEditDate="2010-11-30T20:23:48.337" LastActivityDate="2010-11-30T20:23:48.337" />
  <row Id="12531" PostTypeId="1" AcceptedAnswerId="12532" CreationDate="2010-11-30T20:32:25.087" Score="7" ViewCount="181" Body="&lt;p&gt;Question: Let $V \subset M(n,n,\mathbb{R})$ be the set of all symmetric, real $(n \times n)$ matrices, that is $a_{ij} = a_{ji}$ for all $i,j$. Show that $V$ is a subspace of $M(n,n,\mathbb{R})$ and calculate dim$(V)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My attempt so far:&#xA;First part: To show that $V$ is a subspace I need to show:&#xA; (a) $ 0 \in V$ and &#xA; (b) $\forall A,B \in V:&#xD;&#xA; 	(i) A + B \in V&#xD;&#xA; 	(ii) \lambda A \in V$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For (a) I would say: Let $a_{ij} \in 0$(this should represent a zero matrix, is that how to write it?) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a_{ij} = 0 = a_{ji} \Rightarrow 0 \in V$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For (b) I am actually confused since I would first think: both a $(2 \times 2)$ matrix and a $(3 \times 3)$ matrix belong to $V$ but addition of matrices of different size is undefined $\Rightarrow$ $V$ is not closed under addition $\Rightarrow$ $V$ is not a subspace of $M(n,n,\mathbb{R})$... what am I missing here? (To start I don't really understand the notation $M(n,n,\mathbb{R})$... what exactly does the $\mathbb{R}$ represent there?).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Disregarding my confusion I would still try to show (b), but my mathematical notation is still lacking competence... Is the following somewhat clear? Would anyone ever use &quot;$\in$&quot; to denote &quot;is an element of matrix&quot;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(i)Let $a_{ij},a_{ji} \in A$ and $b_{ij}, b_{ji} \in B$. Let $A,B \in V$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Rightarrow a_{ij} = a_{ji}, b_{ij} = b_{ji}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$A + B = C \Rightarrow c_{ij} = (a_{ij}+b_{ij}) = (a_{ji} + b_{ij}) = (a_{ij} + b_{ji}) = c_{ji} = (a_{ji} + b_{ji})$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Rightarrow C \in V$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(ii) Let $A\in V, \lambda \in \mathbb{R}$. Let $a_{ij},a_{ji} \in A$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Rightarrow a_{ij} = a_{ji}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\lambda \cdot A = A&amp;#39;$ with $\lambda a_{ij} = \lambda a_{ji} \Rightarrow A&amp;#39; \in V$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second part: I feel that I understand the answer... For an $(n \times n)$ matrix, the diagonal length $ = n$ and these are the elements which have no counterpart and are not critical to the symmetry. When these elements are subtracted from the total$(n^{2})$, half of the remainder can be independently selected and the other half will follow as a result. Therefore I think it makes sense to write that dim$(V) = n + \frac{n^{2}-n}{2}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this correct? If so, given the context of the exercise, how could I make my answer more acceptable?&lt;/p&gt;&#xA;" OwnerUserId="3711" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-01T11:01:04.557" LastActivityDate="2010-12-01T11:01:04.557" Title="Show that the set of all symmetric, real matrices is a subspace, determine the dimension" Tags="&lt;linear-algebra&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="12532" PostTypeId="2" ParentId="12531" CreationDate="2010-11-30T20:44:48.927" Score="7" ViewCount="" Body="&lt;p&gt;To write that the matrix is the zero matrix, you should write &quot;let $a_{ij}=0$ for all $i$ and $j$&quot;, not &quot;$a_{ij}\in 0$&quot;. (Nothing is an element of $0$). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For (b): No, notice that the $n$ is fixed. You are only considering matrices that are symmetric of a &lt;em&gt;fixed size&lt;/em&gt;. If $n=2$, then you only consider $2\times 2$ matrices; if $n=3$, then you only consider $3\times 3$ matrices. You never consider &lt;em&gt;both&lt;/em&gt; $2\times 2$ and $3\times 3$ matrices at the same time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$M(n,n,\mathbb{R})$ means:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Matrices  (that's the &quot;$M$&quot;);&lt;/li&gt;&#xA;&lt;li&gt;with $n$ rows (that's the first $n$);&lt;/li&gt;&#xA;&lt;li&gt;with $n$ columns (that's the second $n$);&lt;/li&gt;&#xA;&lt;li&gt;and each entry is a real number (that is the $\mathbb{R}$).&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So $M(2,3,\mathbb{Z})$ would mean &quot;matrices with 2 rows and 3 columns each, and each entry is an integer.&quot; $M(7,2,\mathbb{C})$ means &quot;matrices with 7 rows and 2 columns, and every entry is a complex number. Etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The way you want to say (b) is: Let $A=(a_{ij})$ and $B=(b_{ij})$ (that is, let's call the entries of $A$ &quot;$a_{ij}$&quot;, and let's call the entries of $B$ &quot;$b_{ij}$&quot;). Because $A$ is symmetric, we know that for all $i$ and $j$, $a_{ij}=a_{ji}$; and since $B$ is symmetric we know that for all $i$ and $j$, $b_{ij}=b_{ji}$. Now, let $C=A+B$. If we call the $(i,j)$th entry of $C$ &quot;$c_{ij}$&quot;, then you want to show that $c_{ij}=c_{ji}$ for all $i$ and $j$. How do you do that? You use the fact that you can express $c_{ij}$ in terms of the entries of $A$ and of $B$, and that $A$ and $B$ are symmetric, pretty much how you did; it's just a matter of writing it clearly. Same with the scalar multiplication.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your argument for the second part is essentially correct. To make it water tight, you should produce a basis of the appropriate size. Each element would be used to determine a particular entry as you describe them. &lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-11-30T20:44:48.927" CommentCount="0" />
  <row Id="12533" PostTypeId="2" ParentId="12374" CreationDate="2010-11-30T21:01:14.707" Score="0" ViewCount="" Body="&lt;p&gt;Thanks cch.  Another approach a coworker recommended is to create a new time dependent term that includes all the constants:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;c(t) = (ha/k)*j(t)+Ta&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that ha, k, and Ta are all constant so c(t) and j(t) have no dynamics between themselves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then the differential equation simplifies to:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;db/dt = k*(c(t)-b(t))&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And the laplacian becomes:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;sB(s) + kB(s) = kC(s)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And the transfer function becomes:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;B(s)/C(s) = k/(s+k)&lt;/p&gt;&#xA;" LastActivityDate="2010-11-30T21:01:14.707" />
  <row Id="12534" PostTypeId="1" CreationDate="2010-11-30T21:21:20.580" Score="2" ViewCount="107" Body="&lt;p&gt;Given an angle $\theta$ (in radians), write $\theta = 2\pi t + \phi$, where $\phi$ is between $0$ and $2\pi$, $0\leq \phi\lt 2\pi$, and $t$ is an integer. Call $k=2\pi t$ the number of &quot;full circles&quot; or &quot;full turns&quot; corresponding to the angle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $\theta = \frac{13\pi}{4}$, &#xA;am I right that k=$2\pi$?&lt;/p&gt;&#xA;" OwnerUserId="4263" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-11-30T22:21:10.897" LastActivityDate="2010-11-30T22:21:10.897" Title="Basic trigonometry" Tags="&lt;homework&gt;&lt;trigonometry&gt;" AnswerCount="1" CommentCount="6" />
  <row Id="12535" PostTypeId="2" ParentId="12291" CreationDate="2010-11-30T21:29:09.977" Score="2" ViewCount="" Body="&lt;p&gt;By induction let's just worry about picking one parameter element $x$. We need $x$ to be in $m$, but  outside all the minimal primes of $R$ and $m^2$. This smells exactly like &lt;a href=&quot;http://commalg.wiki-site.com/index.php/Prime_avoidance_lemma&quot; rel=&quot;nofollow&quot;&gt;Prime Avoidance&lt;/a&gt; (note that they don't have to be all primes)! Now replace $R$ by $R/(x)$ and repeat.&lt;/p&gt;&#xA;" OwnerUserId="223" LastActivityDate="2010-11-30T21:29:09.977" />
  <row Id="12536" PostTypeId="1" CreationDate="2010-11-30T21:47:11.727" Score="1" ViewCount="544" Body="&lt;p&gt;I heard that they are difficult. Is this true? Are they like the qualifying exams in grad school? For example, is the probability exam and the financial math exam comparable to qualifying exams (e.g. needs months and months of study)?&lt;/p&gt;&#xA;" LastActivityDate="2011-01-18T18:03:19.497" Title="Are the actuarial exams hard?" Tags="&lt;soft-question&gt;" AnswerCount="4" CommentCount="6" />
  <row Id="12537" PostTypeId="1" AcceptedAnswerId="12538" CreationDate="2010-11-30T21:48:54.800" Score="4" ViewCount="98" Body="&lt;p&gt;Let C be a category of sets, which has objects all sets and arrows all functions, with usual identity functions and the usual composition of functions. For any set S, the assignment s-s for all s in S describes the identity function. If S is a subset of Y, the assignment also describes the inclusion function from S to Y. These functions are different unless S is equal to Y. My question is why these functions are different since they have the same domain and assignment.&lt;/p&gt;&#xA;" OwnerUserId="2177" LastActivityDate="2010-11-30T22:10:06.230" Title="Category of sets" Tags="&lt;category-theory&gt;" AnswerCount="2" />
  <row Id="12538" PostTypeId="2" ParentId="12537" CreationDate="2010-11-30T21:52:09.857" Score="4" ViewCount="" Body="&lt;p&gt;In Categories, we have two operations called &lt;em&gt;Domain&lt;/em&gt; and &lt;em&gt;Codomain&lt;/em&gt; from the collection of all arrows to the collection of all objects. Each arrow $f$ belongs to the collection $C(\mathrm{dom}f,\mathrm{codom} f)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in the Category of Sets, we actually are interested not just in the domain and the assignment of each function (which is what the arrows are), but also in the codomain. That is, you want to think of a function in the category $\mathcal{S}et$ as a triple, consisting of the domain, the codomain, and the actual set of pairs that make up the function (the assignment), $(A,B,f)$ for a function $f\colon A\to B$. Then the operator &lt;em&gt;Domain&lt;/em&gt; will just give you the first component, $A$; the operator &lt;em&gt;Codomain&lt;/em&gt; will give you the second component $B$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here, two functions are equal if and only if they have the same domain, the same assignment, &lt;em&gt;and the same codomain&lt;/em&gt;. The codomain is important because it is an important attribute of arrows in categories. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the inclusion function from $S$ to $Y$ has domain $S$, codomain $Y$, and rule $s\mapsto s$. The identity function from $S$ to itself has domain $S$, codomain $S$, and rule $s\mapsto s$. So the first function corresponds to&#xA;$$\Bigl(S, Y, \{(s,s)\in S\times S\}\Bigr)$$&#xD;&#xA;while the second function is&#xA;$$\Bigl(S,S,\{(s,s)\in S\times S\}\Bigr).$$&#xD;&#xA;If the two arrows are equal, then the value of &lt;em&gt;Domain&lt;/em&gt; and of &lt;em&gt;Codomain&lt;/em&gt; must be the same in both; that is, we must have $S=Y$. And of course, if $S=Y$, then they are the same function.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-30T22:10:06.230" LastActivityDate="2010-11-30T22:10:06.230" />
  <row Id="12539" PostTypeId="2" ParentId="12537" CreationDate="2010-11-30T21:53:00.773" Score="2" ViewCount="" Body="&lt;p&gt;The two functions have different codomains. Any map $S\to S$ lies in $Mor(S,S)$ while any map $S\to Y$ lies in $Mor(S,Y)$.&lt;/p&gt;&#xA;" OwnerUserId="3431" LastActivityDate="2010-11-30T21:53:00.773" />
  <row Id="12540" PostTypeId="2" ParentId="12534" CreationDate="2010-11-30T21:53:28.863" Score="3" ViewCount="" Body="&lt;p&gt;I'm guessing that the homework problem is something like &quot;find an angle between 0 and $2\pi$ that's equal to $13\pi/4$&quot;.  That is, for what $k$ such that $k$ is a multiple of $2\pi$ do we have &#xA;$$ {13\pi\over 4} = k + {5\pi\over 4}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer is indeed $k=2\pi$.  The first equation in the original question should just have &quot;$k$&quot; in place of &quot;($2\pi$)&quot;.&lt;/p&gt;&#xA;" OwnerUserId="2682" LastActivityDate="2010-11-30T21:53:28.863" />
  <row Id="12541" PostTypeId="2" ParentId="12536" CreationDate="2010-11-30T22:20:53.110" Score="1" ViewCount="" Body="&lt;p&gt;If you are referring to the actuarial exams in the United States, obviously different people have different reactions, but they are designed to be hard in the sense that they are designed so that students will study for months to pass them. The pass rate for the first exam (exam P) is about 40 percent, and the pass rate only slowly increases with the higher exams. Most actuaries fail exams along the way, and it's not uncommon for an actuary to have one exam in particular he/she gets stuck on and has to retake several times. And some give up altogether. &lt;/p&gt;&#xA;" OwnerUserId="3035" LastActivityDate="2010-11-30T22:20:53.110" />
  <row Id="12542" PostTypeId="2" ParentId="12536" CreationDate="2010-11-30T22:27:09.490" Score="1" ViewCount="" Body="&lt;p&gt;Personal experience: I took the first two of the &lt;a href=&quot;http://www.soa.org/&quot; rel=&quot;nofollow&quot;&gt;SOA&lt;/a&gt;/ASA/CAS exams (which is, if you are in the United States, the same as what you called the Exam P-Probability and Exam FM-Financial Mathematics) when I was finishing high school and starting college. Compared to graduate school (in mathematics) qualifying exams, they are much, much easier. (Though I heard that the latter exams in the CAS or ASA requirement series get quite a bit harder.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The probability exam, modulo some terminology/jargon issue, is actually not beyond the grasp of high-school level pre-calculus. You just need to have some experience dealing with probability and statistics. I think it is more of a test on problem solving skills than anything else. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The financial mathematics one has quite a bit more specialist knowledge involved. I remember having to spend several months studying for it just because there were many concepts which were new to me. At the time I managed to pass it by mentally associating the various words with the mathematical concepts/formulae they involved; in hindsight it was definitely the wrong way to study. Anyway, the study material from the SOA is actually quite comprehensive for the first two exams. If you spend a few months going through them you shouldn't have too much trouble.&lt;/p&gt;&#xA;" OwnerUserId="1543" LastActivityDate="2010-11-30T22:27:09.490" CommentCount="4" />
  <row Id="12543" PostTypeId="2" ParentId="12442" CreationDate="2010-11-30T22:29:31.530" Score="0" ViewCount="" Body="&lt;p&gt;I'd like to supplement the above answers with two important examples.  Consider a &lt;em&gt;hyperbolic&lt;/em&gt; transformation $H(x, y) = (2x, y/2)$ and a &lt;em&gt;shear&lt;/em&gt; $S(x,y) = (x+y, y)$, both elements of $GL(2,R)$.  If you draw how these act on points of $R^2$ you will see that these distort angles.  On the other hand, elements of $GL(1,C)$ always preserve angles (and thus so do holomorphic maps, away from critical points).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is, the extra algebraic structure preserved by $GL(1,C)$ translates to the geometric property of preserving angles and &quot;handedness&quot;.&lt;/p&gt;&#xA;" OwnerUserId="1307" LastActivityDate="2010-11-30T22:29:31.530" />
  <row Id="12544" PostTypeId="1" CreationDate="2010-11-30T22:35:36.933" Score="6" ViewCount="302" Body="&lt;p&gt;Can n! be a perfect square when n is an integer greater than 1?&lt;/p&gt;&#xA;" OwnerUserId="2938" LastEditorUserId="1778" LastEditorDisplayName="" LastEditDate="2010-12-04T06:36:18.557" LastActivityDate="2010-12-04T06:36:18.557" Title="Can n! be a perfect square when n is an integer greater than 1?" Tags="&lt;number-theory&gt;&lt;diophantine-equations&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="1" />
  <row Id="12545" PostTypeId="2" ParentId="12544" CreationDate="2010-11-30T22:37:22.353" Score="4" ViewCount="" Body="&lt;p&gt;There is a prime between n/2 and n, if I am not mistaken. &lt;/p&gt;&#xA;" OwnerUserId="223" LastActivityDate="2010-11-30T22:37:22.353" />
  <row Id="12546" PostTypeId="2" ParentId="12544" CreationDate="2010-11-30T22:56:09.077" Score="11" ViewCount="" Body="&lt;p&gt;Assume, $n\geq 4$. By Bertrand's postulate there is a prime, let's call it $p$ such that $\frac{n}{2}&amp;lt;p&amp;lt;n$ . Suppose, $p^2$ divides $n$. Then, there should be another number $m$ such that $p&amp;lt;m\leq n$ such that  $p$ divides $m$. So, $\frac{m}{p}\geq 2$, then, $m\geq 2p &amp;gt; n$. This is a contradiction.  So, $p$ divides $n!$ but $p^2$ does not. So $n!$ is not a perfect square.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Bertrand_postulate&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Bertrand_postulate&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That leaves two more cases. We check directly, $2!=2$ and $3!=6$ are not perfect squares.&lt;/p&gt;&#xA;" OwnerUserId="3431" LastEditorUserId="3431" LastEditorDisplayName="" LastEditDate="2010-11-30T23:09:44.190" LastActivityDate="2010-11-30T23:09:44.190" CommentCount="2" />
  <row Id="12547" PostTypeId="1" AcceptedAnswerId="12569" CreationDate="2010-11-30T22:57:19.120" Score="15" ViewCount="447" Body="&lt;p&gt;Please excuse the naive question.  I have had two classes now in which this theorem was taught and proven, but I have only ever seen a single (indirect?) application involving the quantum harmonic oscillator.  Even if this is not the strongest spectral theorem, it still seems useful enough that there should be many nice examples illustrating its utility.  So... what are some of those examples?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I couldn't readily find any nice examples looking through a few functional analysis textbooks, either.  Maybe I have the wrong books.)&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-12-01T03:35:44.840" Title="What is the spectral theorem for compact self-adjoint operators on a Hilbert space actually for?" Tags="&lt;big-list&gt;&lt;functional-analysis&gt;&lt;examples&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="3" />
  <row Id="12548" PostTypeId="1" AcceptedAnswerId="12549" CreationDate="2010-11-30T23:11:41.863" Score="2" ViewCount="111" Body="&lt;p&gt;I'm not quite sure where such a concept fits. Suppose $X$ is a topological space. I know then that the diagonal $\Delta=\{(x,x)\ | x\in X\}$, so $\Delta\subseteq X\times X$. What then would a neighborhood of the diagonal be comprised of? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;By this I mean, suppose $E\in\mathscr{N}_\Delta$, where $\mathscr{N}_\Delta$ is the filter of all neighborhoods of the diagonal. Is $E\in\mathscr{P}(X)$, or is $E\in\mathscr{P}(X\times X)$? What is the criterion for a subset to be in a neighborhood of $\Delta$? My guess is that $E\in\mathscr{P}(X\times X)$, and if this is correct, we would say&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(x,x&amp;#39;)\in E\quad\text{if and only if...}?$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="1583" LastEditorUserId="1583" LastEditorDisplayName="" LastEditDate="2010-11-30T23:19:01.417" LastActivityDate="2010-11-30T23:23:19.850" Title="What exactly is a neighborhood of the diagonal on a topological space?" Tags="&lt;topology&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="12549" PostTypeId="2" ParentId="12548" CreationDate="2010-11-30T23:18:17.513" Score="4" ViewCount="" Body="&lt;p&gt;A neighborhood of the diagonal $A$ is a subset of $X\times X$ containing an open set $E$ in $X\times X$ with the product topology such that $\Delta\subset E$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The set of all neighborhoods of the diagonal is a subset of $\mathcal{P}(X\times X)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might be interested in checking &lt;a href=&quot;http://en.wikipedia.org/wiki/Uniform_space&quot; rel=&quot;nofollow&quot;&gt;uniform spaces&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2614" LastEditorUserId="2614" LastEditorDisplayName="" LastEditDate="2010-11-30T23:23:19.850" LastActivityDate="2010-11-30T23:23:19.850" CommentCount="1" />
  <row Id="12550" PostTypeId="1" CreationDate="2010-12-01T00:08:39.127" Score="2" ViewCount="157" Body="&lt;p&gt;If topology is called rubber-sheet geometry, would it be accurate to describe the &quot;cut and shuffle&quot; topic of &quot;piecewise isometries&quot; as broken glass geometry ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Isometry sounds more geometrical than topological. What would be a good name for &quot;piecewise topology&quot; ?  If you take a topological space and break it up and put it back together in a random way, not necessarily in one piece, then what could you say about the topology of the resulting jumble ?&lt;/p&gt;&#xA;" OwnerUserId="4157" LastActivityDate="2010-12-01T01:41:34.937" Title="Broken glass geometry" Tags="&lt;geometry&gt;&lt;topology&gt;&lt;terminology&gt;&lt;dynamical-systems&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="12551" PostTypeId="2" ParentId="12547" CreationDate="2010-12-01T00:14:38.257" Score="10" ViewCount="" Body="&lt;p&gt;Maybe a definition of functional calculus? Using quantum mechanics notation, if $A$ is self-adjoint and compact, then $A = \sum \lambda_k |k\rangle\langle k|$, which means that for $f:\mathbb{R}\to\mathbb{R}$ we can define $f(A) = \sum f(\lambda_k) |k\rangle\langle k|$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This allows us to give a simple demonstration of Stone's theorem for such operators: that $\exp itA$ is a strongly continuous one-parameter unitary group on your Hilbert space. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It also gives very simple motivation for the construction of Green's functions and resolvent operators.  &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Besides the usual quantum harmonic oscillator, a similar construction can be used to give the decomposition of $L^2$ via eigenfunctions of the Laplacian on a compact manifold. This naturally leads to the Hodge decomposition, which I'm told is generally considered to be somewhat useful :-) &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;That on a compact manifold, the inverse of the Laplacian is a compact operator means that, discarding the harmonic functions, the Laplacian has a lowest eigenvalue. This fact (and that self-adjointness allows it to be diagonalized) allows you to define the Zeta-function &lt;a href=&quot;http://en.wikipedia.org/wiki/Functional_determinant&quot; rel=&quot;nofollow&quot;&gt;determinant&lt;/a&gt; of the Laplacian using some analytic continuation tricks. On 2-dimensional closed surfaces, this is an interesting invariant with nice geometric properties. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Note that in the case of non-compact domains, the inverse Laplacian is no longer a compact operator, and the Laplacian has a continuous spectrum. So the summation in the zeta-function determinant no longer makes any sense...)&lt;/p&gt;&#xA;" OwnerUserId="1543" LastActivityDate="2010-12-01T00:14:38.257" />
  <row Id="12552" PostTypeId="2" ParentId="12550" CreationDate="2010-12-01T00:41:31.763" Score="2" ViewCount="" Body="&lt;p&gt;If the pieces are of given shapes, then this is essentially a jigsaw puzzle, perhaps in a higher dimension than 2.  A mathematical term for problems of this type is &quot;dissection&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the pieces don't have a specific form and don't have to be put back together &quot;in one piece&quot;, I don't think much can be proved about the &quot;resulting jumble&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Banach%E2%80%93Tarski_paradox&quot; rel=&quot;nofollow&quot;&gt; Banach-Tarski paradox&lt;/a&gt; gives an example of what is possible, at least if the Axiom of Choice is allowed.  This is a result that proves a ball (solid sphere) can be partitioned into a finite number of subsets that can be recombined to form two balls, each congruent to the original.  (The subsets are not measurable.  It is not a true paradox, but it is so called because it shows the Axiom of Choice has implications that defy common intuition.)&lt;/p&gt;&#xA;" OwnerUserId="3111" LastActivityDate="2010-12-01T00:41:31.763" CommentCount="1" />
  <row Id="12553" PostTypeId="2" ParentId="12433" CreationDate="2010-12-01T01:03:33.780" Score="1" ViewCount="" Body="&lt;p&gt;It's a nice problem. The chance of hitting $101$ first is surprisingly large. &#xA; Let $a(x)$ be the chance of hitting $101$ first, starting at $x$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We solve recursively by setting $a(106)=a(105)=a(104)=a(103)=a(102)=0$ and $a(101)=1$. Then, for $x$ from $100$ to $1$, put $a(x)={1\over 6}\sum_{j=1}^6 a(x+j)$. By the renewal theorem, $a(x)$ converges to $1/\mu$, where $\mu=(1+2+3+4+5+6)/6=7/2$. &#xA;The value $a(1)$ is very close to this limit.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, the whole hitting distribution is approximately $[6/21,5/21,4/21,3/21,2/21,1/21]$.&lt;/p&gt;&#xA;" OwnerUserId="940" LastEditorUserId="940" LastEditorDisplayName="" LastEditDate="2010-12-01T01:26:14.597" LastActivityDate="2010-12-01T01:26:14.597" CommunityOwnedDate="2010-12-01T01:03:33.780" CommentCount="1" />
  <row Id="12554" PostTypeId="2" ParentId="12536" CreationDate="2010-12-01T01:09:52.810" Score="2" ViewCount="" Body="&lt;p&gt;I have taken the US SOA exams up through M and taken a qualifying exam at a US graduate school.  They are very different kinds of exams and so are a different kind of hard.  The mathematics involved in an actuarial exam isn't particularly advanced, but there is a lot of terminology to learn and concepts that you wouldn't spend much time on in math degree.  You shouldn't consider them trivial.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My understanding of the later exams is they get closer to quals in character.  They are several hour written exams and they serve a similar purpose (making sure you have the required knowledge of the discipline).  In particular, there are a set of exams in the US called the EA exams that require memorization and application of a large chunk of IRS regulations.  I'm glad I decided to go to grad school before having to take those.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The nice thing about actuarial exams (and I think this is relatively universal) is that once you get a job you will get paid study time and financial rewards for passing them.   &lt;/p&gt;&#xA;" OwnerUserId="4266" LastActivityDate="2010-12-01T01:09:52.810" />
  <row Id="12555" PostTypeId="1" AcceptedAnswerId="12558" CreationDate="2010-12-01T01:13:36.377" Score="22" ViewCount="661" Body="&lt;p&gt;Sorry if this is a really naive question, but in my reading of a lot of textbooks and articles, there is a lot of mention of how many decimals we know of a certain number today, such as $\pi$ or $e$. An excerpt from my textbook:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In 1748, Leonard Euler used the sum of the infinite series of $e$ (mentioned in the book in a section about Taylor Series) to find the value of $e$ to 23 digits. In 2003, Shigeru Kondo, again using the series, computed $e$ to 50 billion decimals places&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My question is &lt;em&gt;why&lt;/em&gt; does it matter how many decimals we know? Isn't this just a huge waste of time? What could we ever do with so many decimal places? And, if $e$ can be represented as a sum of infinite series of $1/n!$, can't we just plug that into a computer that just loops the same equation but increasing $n$ every iteration, and find as many decimals of $e$ as we like?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Once again, I realize this may be an ignorant/naive question, but I've always been curious about this)&lt;/p&gt;&#xA;" OwnerUserId="2985" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-01T02:56:16.027" LastActivityDate="2010-12-01T03:51:45.757" Title="$e$ to 50 billion decimal places" Tags="&lt;calculus&gt;&lt;soft-question&gt;&lt;taylor-expansion&gt;" AnswerCount="3" CommentCount="14" FavoriteCount="4" />
  <row Id="12556" PostTypeId="2" ParentId="12555" CreationDate="2010-12-01T01:23:32.527" Score="7" ViewCount="" Body="&lt;p&gt;The answer to your last question is no, for a certain value of &quot;no.&quot;  The problem with your idea is that as $n$ gets bigger it gets harder to calculate $n!$, so it gets harder to tell what the digits of $\frac{1}{n!}$ are.  In other words, if you actually tried to carry out your plan, you would quickly find that it is computationally infeasible.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So instead one has to resort to a smarter algorithm.  Thus being able to compute constants to high precision is a measure of how smart our algorithms are.  If your algorithm is smarter than mine, the most concrete way to prove that is to use it to compute more digits in a reasonable amount of time than I can.  So while it's probably true that nobody is actually going to use these digits for anything (except possibly to verify certain conjectures), the fact that we know them is a measure of our knowledge, both about algorithms and about $e$.  (It is also a measure of how good our hardware is, but I'm trying to emphasize the math here.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, here is an &lt;a href=&quot;http://www.numberworld.org/misc_runs/pi-5t/announce_en.html&quot; rel=&quot;nofollow&quot;&gt;announcement from Kondo&lt;/a&gt; about a more recent version of this computation (for $\pi$) and here are, among other things, his &lt;a href=&quot;http://www.numberworld.org/misc_runs/pi-5t/details.html&quot; rel=&quot;nofollow&quot;&gt;reasons for doing it&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-12-01T01:23:32.527" />
  <row Id="12558" PostTypeId="2" ParentId="12555" CreationDate="2010-12-01T01:26:32.460" Score="21" ViewCount="" Body="&lt;p&gt;There are a couple of reasons why people compute $\pi$ and $e$ to so many digits.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One is simply that it is a way to test hardware. The values are known (or can be known), and you want to see if your hardware is fast and accurate with these computations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another is that there are actually a lot of questions about the decimal expansions of $\pi$ and $e$ (and other numbers) for which we simply don't know the answer yet. For example: is $\pi$  &lt;em&gt;normal in base 10&lt;/em&gt;? That is, will a particular sequence of digits occur in its decimal expansion in about the frequency you expect? More precisely, given a specific sequence of $n$ digits, will&#xA;$$\frac{\text{number of times that the specific sequence occurs in the first $m$ digits of $\pi$}}{m}$$&#xA;approach $1/10^n$ as $m\to\infty$ (there are $10^n$ different sequences of $n$ digits, so this is what you would expect if the sequence was completely random)? While the answer cannot be known simply by computing digits of $\pi$, knowing more and more digits helps us see whether it seems at least plausible or not, at least early on. We can also perform other tests of randomness to see whether the digits of $\pi$ (or $e$) seem to pass it or not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As to your question about $e$, yes, in principle we can do that. But the process gets more and more complex as $n$ gets larger. Normally, a computer only knows how to represent numbers that are no larger than a certain quantity (depending on the number of bytes it uses to represent numbers), so you need to &quot;explain&quot; to your computer how to handle big numbers like $n!$ for large $n$. The storage space needed to perform the computations also gets larger and larger and larger. And that idea only works if you start from a known point. So while theoretically we could compute $e$ to as many digits as we want, in practice if we want the computations to finish sometime before the Sun runs out of hydrogen we can't really go that far. There are other algorithms known to compute decimals of $\pi$ or $e$ that are faster, or that don't require you to know the previous digits to figure out the next one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And that leads to yet another thing that people who are computing so many digits of $\pi$ and $e$ may be doing: testing algorithms for large number computations, for floating point accuracy, or for parallel computing (we are not very good at parallel computing, and trying to figure out how to do it effectively is a Very Big Deal; coming up with ways to do computations such as &quot;compute $\pi$ to the $n$th millionth digit&quot; are ways to test ideas for doing parallel computing). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That leads to a third interest in the computations: coming up with mathematical ideas that can zero in on desired digits of the decimal expansion quickly; not because we are particularly interested in the digits of $\pi$,  necessarily, but because we often &lt;em&gt;are&lt;/em&gt; interested in particular digits of &lt;em&gt;other&lt;/em&gt; numbers. $\pi$ and $e$ are just very convenient benchmarks to test things with.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why did Euler compute the numbers? Because he wanted to show off some of the uses for Taylor series; up to that time, the methods for approximating the value of $\pi$ were much more onerous. Ludolph van Ceulen famously used the methods of Archimedes (inscribing and circumscribing polygons in circles) to find the value of $\pi$ to 16 decimals (and after many more years of effort, to 25 places); this was in the late 16th century, before it was known that $\pi$ was transcendental (or even before the notion of being transcendental was very clear), so trying to compute $\pi$ precisely actually had a point. He was so proud of the accomplishment (given all the hard work it had entailed) that he had the value of $\pi$ he computed put into his tombstone. Euler showed that Taylor series could be used to obtain the same results with a lot less effort and with more precision.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-01T03:51:45.757" LastActivityDate="2010-12-01T03:51:45.757" CommentCount="5" />
  <row Id="12559" PostTypeId="2" ParentId="12426" CreationDate="2010-12-01T01:39:58.107" Score="0" ViewCount="" Body="&lt;p&gt;At first when I had the idea to ask this question, I tought about books such as Schaum's Outline of College Mathematics and other books of the same kind for high school and university mathematics courses; then I found out this Website and I had the idea to demand to &quot;math experts.&quot; I knew Mathematica, this great math software application, but I am also happy that you talked about Maple that also really seems to be a great tool too. Thank you to tell me that Maple is less complex. At least, it seems that they developed portals and resources for students. It &quot;sounds&quot; great. In practice it is sometimes &quot;another game.&quot; In my opinion, I suppose that someone who buys the IrisPen to develop math problems on a computer and on paper, and that is also enable to use an application such as Maple 14 Academic or another version is really in a good position to be good at solving math problems and tasks of the like.      &lt;/p&gt;&#xA;" LastActivityDate="2010-12-01T01:39:58.107" />
  <row Id="12560" PostTypeId="2" ParentId="12550" CreationDate="2010-12-01T01:41:34.937" Score="3" ViewCount="" Body="&lt;p&gt;You might be interested in reading about &lt;a href=&quot;http://en.wikipedia.org/wiki/Hilbert%27s_third_problem&quot; rel=&quot;nofollow&quot;&gt;Hilbert's Third Problem&lt;/a&gt;. Hilbert asked the question: Given two polytopes of the same volumes, can we cut one up into polytopal pieces and reassemble it to form the other. The answer is NO in dimensions $\geq 3$; there are many additional subtle obstacles. &lt;/p&gt;&#xA;" OwnerUserId="448" LastActivityDate="2010-12-01T01:41:34.937" />
  <row Id="12561" PostTypeId="1" AcceptedAnswerId="12567" CreationDate="2010-12-01T02:01:46.293" Score="2" ViewCount="123" Body="&lt;p&gt;Theorem: If a function f is analytis throughout a simply connected domain D, then for every closed contour C lying in D, the integral is o along the contour.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a theorem from the text I am using without complete proof. The problem is that when the closed contour has an infinite number of self-intersection points. Who can give me a rigorous proof or a reference about it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="2177" LastActivityDate="2010-12-01T03:07:47.390" Title="Simply connected domains" Tags="&lt;complex-analysis&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="12562" PostTypeId="2" ParentId="12555" CreationDate="2010-12-01T02:08:46.537" Score="15" ViewCount="" Body="&lt;p&gt;I'd just like to give two quotes from the book &lt;a href=&quot;http://www.siam.org/books/100digitchallenge/&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;The SIAM 100-Digit Challenge: A Study in High-Accuracy Numerical Computing&lt;/em&gt;&lt;/a&gt; that might help explain motivation. Here is the one from their chapter on computing constants to 10,000 digits:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;While such an exercise might seem frivolous, the fact is we learned a&#xA;  lot from the continual refinement of our algorithms to work efficiently at&#xA;  ultrahigh precision. The reward is a deeper understanding of the theory,&#xA;  and often a better algorithm for low-precision cases.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and here is something from the foreword written by David Bailey, one of the pioneers of &lt;a href=&quot;http://www.experimentalmath.info/&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;experimental mathematics&lt;/em&gt;&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Some may question why anyone would  care about such prodigious precision,&#xA;  when in the “real” physical world, hardly any quantities are known to an&#xA;  accuracy beyond about 12 decimal digits. For instance, a value of π&#xA;  correct to 20 decimal digits would suffice to calculate the circumference&#xA;  of a circle around the sun at the orbit of the earth to within the width&#xA;  of an atom. So why should anyone care about finding any answers to 10,000&#xA;  digit accuracy?&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;In fact, recent work in experimental mathematics has provided an important&#xA;  venue where numerical results are needed to very high numerical&#xA;  precision, in some cases to thousands of decimal digit accuracy. In&#xA;  particular, precision of this scale is often required when applying integer&#xA;  relation algorithms to discover new mathematical identities. An integer&#xA;  relation algorithm is an algorithm that, given $n$ real numbers&#xA;  ($x_i,\quad 1\leq i\leq n$), in the form of high-precision floating-point&#xA;  numerical values, produces $n$ integers, not all zero, such that&#xA;  $a_1x_1+a_2x_2+\cdots+a_n x_n=0$.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The best known example of this sort is the discovery in 1995 of a new formula&#xA;  for π:&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$$\pi=\sum_{k=0}^{\infty}\frac1{16^k}\left(\frac{4}{8k+1}-\frac{2}{8k+4}-\frac1{8k+5}-\frac1{8k+6}\right)$$&#xD;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;This formula was found by a computer program implementing the PSLQ integer&#xA;  relation algorithm, using (in this case) a numerical precision of&#xA;  approximately 200 digits. This computation also required, as an input&#xA;  real vector, more than 25 mathematical constants, each computed to 200-digit&#xA;  accuracy. The mathematical significance of this particular formula is that it &#xA;  permits one to directly calculate binary or hexadecimal digits of &#xA;  π beginning at any arbitrary position, using an algorithm that is very simple,&#xA;  requires almost no memory, and does not require multiple-precision arithmetic.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-12-01T02:08:46.537" />
  <row Id="12563" PostTypeId="1" CreationDate="2010-12-01T02:13:14.107" Score="5" ViewCount="198" Body="&lt;p&gt;I want to solve equations like this (mod $2^n$):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{array}{rcrcrcr} 3x&amp;amp;+&amp;amp;4y&amp;amp;+&amp;amp;13z&amp;amp;=&amp;amp;3&amp;amp;\pmod{16} \\ x&amp;amp;+&amp;amp;5y&amp;amp;+&amp;amp;3z&amp;amp;=&amp;amp;5&amp;amp;\pmod{16} \\ 4x&amp;amp;+&amp;amp;7y&amp;amp;+&amp;amp;11z&amp;amp;=&amp;amp;12&amp;amp;\pmod{16}\end{array}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since we are working over a ring and not a field, Gaussian elimination doesn't work. So how can I still solve these types of equations?&lt;/p&gt;&#xA;" OwnerUserId="3163" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-01T17:02:29.487" LastActivityDate="2010-12-01T17:02:29.487" Title="Solving systems of linear equations over a finite ring" Tags="&lt;number-theory&gt;&lt;linear-algebra&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="12564" PostTypeId="2" ParentId="12563" CreationDate="2010-12-01T02:23:23.043" Score="2" ViewCount="" Body="&lt;p&gt;You basically do Gaussian elimination as usual, although you can get stuck if at some point all coefficients of a variable are, say, even. This just means that you'll have two solutions to that variable. In general, you'll pick the row in which the coefficient has the least power of $2$.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-12-01T02:23:23.043" />
  <row Id="12565" PostTypeId="2" ParentId="12563" CreationDate="2010-12-01T02:27:34.803" Score="5" ViewCount="" Body="&lt;p&gt;You can still use Gaussian elimination as long as you don't &quot;divide&quot; by things that are not relatively prime to the modulus. In this case, you can &quot;divide&quot; by any odd number, and perform all the usual computations. In this case, you can perform Gaussian pretty well:&#xA;\begin{align*}&#xA;\left(\begin{array}{ccc|c}&#xA;3 &amp;amp; 4 &amp;amp; 13 &amp;amp; 3\\&#xA;1 &amp;amp; 5 &amp;amp; 3 &amp;amp; 5\\&#xA;4 &amp;amp; 7 &amp;amp; 11 &amp;amp; 12&#xA;\end{array}\right) &amp;amp;\rightarrow&#xA;\left(\begin{array}{ccc|c}&#xA;1 &amp;amp; 5 &amp;amp; 3 &amp;amp; 5\\&#xA;3 &amp;amp; 4 &amp;amp; 13 &amp;amp; 3\\&#xA;4 &amp;amp; 8 &amp;amp; 11 &amp;amp; 12&#xA;\end{array}\right) &amp;amp;&amp;amp; \rightarrow&#xA;\left(\begin{array}{ccr|c}&#xA;1 &amp;amp; 5 &amp;amp; 3 &amp;amp; 5\\&#xA;0 &amp;amp; 5 &amp;amp; 4 &amp;amp; 4\\&#xA;0 &amp;amp; 4 &amp;amp; -1 &amp;amp; 8&#xA;\end{array}\right)\\&#xA;&amp;amp;\rightarrow&#xA;\left(\begin{array}{ccr|r}&#xA;1 &amp;amp; 5 &amp;amp; 3 &amp;amp; 5\\&#xA;0 &amp;amp; 1 &amp;amp; 5 &amp;amp; -4\\&#xA;0 &amp;amp; 4 &amp;amp; -1 &amp;amp; 8&#xA;\end{array}\right)&#xA;&amp;amp;&amp;amp;\rightarrow&#xA;\left(\begin{array}{ccr|r}&#xA;1 &amp;amp; 5 &amp;amp; 3 &amp;amp; 5\\&#xA;0 &amp;amp; 1 &amp;amp; 5 &amp;amp; -4\\&#xA;0 &amp;amp; 0 &amp;amp; 11 &amp;amp; 8&#xA;\end{array}\right).&#xA;\end{align*}&#xA;So here you get that $11z\equiv 8 \pmod{16}$. Since $11^{-1} \equiv 3\pmod{16}$, this means $z \equiv 24 \equiv 8\pmod{16}$. Then you can backsubstitute and solve. (Assuming I didn't make any mistakes with my modular arithmetic, anyway...)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are unlucky enough to get a congruence in which all the coefficients are even, then you can divide through by $2$ and get a congruence modulo $8$ (instead of $16$); that will lead to two solutions modulo $16$ (if you get $x\equiv 4 \pmod{8}$, that means $x\equiv 4 \pmod{16}$ or $x\equiv 8+4=12\pmod{16}$, for instance).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically, so long as you are careful, you can &lt;em&gt;certainly&lt;/em&gt; do Gaussian elimination. You can even do it over more general rings, through in that case you have to other restrictions on what you can or cannot conclude.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-12-01T02:27:34.803" />
  <row Id="12566" PostTypeId="1" AcceptedAnswerId="12588" CreationDate="2010-12-01T02:39:10.637" Score="3" ViewCount="323" Body="&lt;p&gt;How to prove this: $$\sum\limits_{p \leq x} \frac{1}{\sqrt{p}} \geq \frac{1}{2}\log{x} -\log{\log{x}}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From &lt;em&gt;Apostol's&lt;/em&gt; number theory text i know that $$\sum\limits_{p \leq x} \frac{1}{p} = \log{\log{x}} + A + \mathcal{O}\Bigl(\frac{1}{\log{x}}\Bigr)$$ But how can i use this to prove my claim. &lt;/p&gt;&#xA;" LastActivityDate="2010-12-02T16:46:37.057" Title="Proving $\sum\limits_{p \leq x} \frac{1}{\sqrt{p}} \geq \frac{1}{2}\log{x} -\log{\log{x}}$" Tags="&lt;analytic-number-theory&gt;" AnswerCount="3" CommentCount="4" FavoriteCount="1" />
  <row Id="12567" PostTypeId="2" ParentId="12561" CreationDate="2010-12-01T03:07:47.390" Score="2" ViewCount="" Body="&lt;p&gt;You can find a proof in &lt;a href=&quot;http://books.google.com/books?id=9LtfZr1snG0C&amp;amp;lpg=PP1&amp;amp;dq=conway%20complex&amp;amp;pg=PA87#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;Section IV.6&lt;/a&gt; of J.B. Conway's &lt;em&gt;Functions of one complex variable&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-12-01T03:07:47.390" CommunityOwnedDate="2010-12-01T03:07:47.390" />
  <row Id="12568" PostTypeId="1" AcceptedAnswerId="12581" CreationDate="2010-12-01T03:13:53.077" Score="3" ViewCount="83" Body="&lt;p&gt;Suppose I take $n$ sets $\{S_i\}$ where each $S_i$ is some subset of integers. Let $P$ indicate a set of integers obtainable by drawing exactly one factor from each $\{S_i\}$ and multiplying them together. What conditions on these sets do I need to ensure that&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$P$ forms a contiguous interval&lt;/li&gt;&#xA;&lt;li&gt;Each element of $P$ has a unique expansion of this form&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;This is probably a well known problem, so any keyword pointers are welcome&lt;/p&gt;&#xA;" OwnerUserId="998" LastEditorUserId="998" LastEditorDisplayName="" LastEditDate="2010-12-01T07:57:14.490" LastActivityDate="2010-12-01T07:57:14.490" Title="Generalizing decimal expansions" Tags="&lt;number-theory&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" />
  <row Id="12569" PostTypeId="2" ParentId="12547" CreationDate="2010-12-01T03:35:44.840" Score="12" ViewCount="" Body="&lt;p&gt;What is the spectral theorem for compact operators good for?  Here are some examples.  (I am ignoring the self-adjoint aspects, since they don't really play&#xA;a role in the theorem.  And it is valid for more general spaces than Hilbert spaces too, so I will also ignore that part, in the sense that I won't pay too much attention to whether my examples deal with Hilbert spaces on the nose, rather than some variant.)&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Proving the Peter--Weyl theorem.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Proving the Hodge decomposition for cohomology of manifolds (using the&#xA;fact that the inverse to the Laplacian is compact); Willie noted this example in his answer too.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Proving the finiteness of cohomology of coherent sheaves on compact complex analytic manifolds.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;In its $p$-adic version, the theory of compact operators is basic to the theory of $p$-adic automorphic forms: e.g. in the construction of so-called eigenvarieties parameterizing $p$-adic families of automorphic Hecke eigenforms of finite slope.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;It is also a basic tool in more classical problems, such as the theory of integral equations.  (It is in this context that the theory was first developed; see Dieudonne's book on the history of functional analysis for a very nice account of the historical development of the theory.)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="221" LastActivityDate="2010-12-01T03:35:44.840" CommentCount="8" />
  <row Id="12570" PostTypeId="1" CreationDate="2010-12-01T04:06:38.983" Score="1" ViewCount="53" Body="&lt;p&gt;Just curious. You pay electronically, and you get a string of numbers. My question, what is the likelihood for a person to type in the same serial number to say, top up their mobile credit? What is the range of probability electronic payment companies are using?Then, does electronic payment in China have longer serial number?&lt;/p&gt;&#xA;" LastActivityDate="2010-12-01T04:35:37.950" Title="What probability methodology is electronic payment using?" Tags="&lt;probability&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="12571" PostTypeId="1" CreationDate="2010-12-01T04:10:02.817" Score="4" ViewCount="107" Body="&lt;p&gt;If $K$ is a field complete with respect to a non-archimedean absolute value, then the $n$th term test (checking whether the $n$th term of a series goes to zero) is sufficient to check convergence of a series in $K$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was wondering if we have any results that guarantee, under additional hypothesis, whether a series in $\mathbb{R}$ converges by just making sure that the $n$th term goes to zero. An example would be the alternating series test. Are there others?&lt;/p&gt;&#xA;" OwnerUserId="3947" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-01T05:17:25.667" LastActivityDate="2010-12-01T05:17:25.667" Title="When is the $n$th term sufficient to guarantee convergence of a series" Tags="&lt;real-analysis&gt;" CommentCount="2" />
  <row Id="12572" PostTypeId="1" AcceptedAnswerId="12579" CreationDate="2010-12-01T04:13:39.087" Score="3" ViewCount="201" Body="&lt;p&gt;Every Fourier transform formula that I know of has a corresponding Fourier series analog, except the multiplication formula&#xA;$$\int_{-\infty}^\infty f(x)\hat{g}(x) dx=\int_{-\infty}^\infty \hat{f}(y)g(y) dy.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So my question is whether a Fourier series analog for this formula exists.&lt;/p&gt;&#xA;" OwnerUserId="3249" LastEditorUserId="3249" LastEditorDisplayName="" LastEditDate="2010-12-01T04:22:05.970" LastActivityDate="2010-12-01T16:27:58.560" Title="Fourier series analog of a formula in Fourier transform" Tags="&lt;fourier-analysis&gt;&lt;fourier-series&gt;" AnswerCount="3" />
  <row Id="12573" PostTypeId="1" CreationDate="2010-12-01T04:28:04.810" Score="3" ViewCount="94" Body="&lt;p&gt;Let $f$ be Lebesgue measurable in $[0,1]$ and assume $f$ takes finitely many values. Assuming $f(x) - f(y)$ is Lebesgue measurable in $[0,1] \times [0,1]$ show that $f$ is integrable over $[0,1]$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Stuck for a while with this one. (Not homework, just practice)&lt;/p&gt;&#xA;" OwnerUserId="4240" LastActivityDate="2010-12-01T04:56:34.397" Title="Integration over the unit square" Tags="&lt;measure-theory&gt;" AnswerCount="1" />
  <row Id="12574" PostTypeId="2" ParentId="12570" CreationDate="2010-12-01T04:35:37.950" Score="1" ViewCount="" Body="&lt;p&gt;Credit card transactions generally involve an input of much more information that just the credit card number (usually there is name, address, security code, card type as well. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even assuming just the card number is required, assuming it is 16 digits (15 with AMEX), since each digit can be anything from $0-9$, there are $10$ possibilities for each place, and hence $10^{16}$ possible card numbers. The current population of the world is of the order of $10^{10}$. Even if everyone in the world owns a card issued by the same company, the chances of someone just &quot;guessing&quot; someone else's number are pretty slim (of the order of $10^{-6}$). Of course as I said above there are more security measures in place and most merchants generally don't allow multiple attempts to enter card info beyond a certain threshold. &lt;/p&gt;&#xA;" OwnerUserId="3431" LastActivityDate="2010-12-01T04:35:37.950" />
  <row Id="12575" PostTypeId="2" ParentId="12573" CreationDate="2010-12-01T04:49:27.463" Score="4" ViewCount="" Body="&lt;p&gt;The fact that $f(x)-f(y)$ is Lebesgue measurable implies that it is defined almost everywhere on the square. For this to be possible, $f$ cannot be $+\infty$ or $-\infty$ on a set $S$ of measure greater than 0 (otherwise $f(x)-f(y)$ will be undefined on the set $S\times S$ of positive measure). Since $f$ also takes on only finitely values, it is integrable.&lt;/p&gt;&#xA;" OwnerUserId="3249" LastEditorUserId="3249" LastEditorDisplayName="" LastEditDate="2010-12-01T04:56:34.397" LastActivityDate="2010-12-01T04:56:34.397" CommentCount="0" />
  <row Id="12576" PostTypeId="1" CreationDate="2010-12-01T05:16:16.157" Score="3" ViewCount="256" Body="&lt;p&gt;Hello everyone.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wanted to ask these questions here, feeling that this place place is without doubt is better than anywhere else, feeling hesitant and shy to ask such question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For several reasons, the environment which I live in doesn't encourage the learning of natural sciences, it permits it but doesn't put efforts in teaching it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a result I missed a great deal of exciting things as I went to college which show me the several weak points I developed through the years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Being taught by other people who came from different locations, environments cultures.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I could see the huge contrast.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Feeling the pain of such experience, I decided to relearn everything form scratch.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Beginning from basics of math.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now that I'm studying in college Computer Information Systems, I have some college level Math courses (Calc 1, Calc 2, Discrete Math, Linear algebra and some other courses like QA). And I am delaying them to prepare myself to accomplish basic courses like algebra1, algebra2, Geometry, College algebra,Trignometry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I think about it. I found that I don't have the time (though I enjoy them+ feeling pain that I missed learning them properly) to do like all these algebra courses in a short time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Is it ok just to take a quick review algebra 1,2 using (cliffs notes) then study this college algebra book &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0618825150&quot; rel=&quot;nofollow&quot;&gt; College Algebra and Trigonometry&lt;/a&gt;  or should I finish algebra 1 then jump into this book, as I find algebra one some how very easy!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I actually want to shorten the time to focus on geometry, so I guess learning geometry prepares me for trignometry and learning algebra 1 (at least) prepares me for college algebra which leads to higher mathematics. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Thank you for your time.&lt;/p&gt;&#xA;" OwnerUserId="4272" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-01T05:21:13.573" LastActivityDate="2011-04-21T10:35:19.010" Title="about math in general." Tags="&lt;soft-question&gt;&lt;education&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="1" />
  <row Id="12577" PostTypeId="1" AcceptedAnswerId="12578" CreationDate="2010-12-01T05:36:04.520" Score="1" ViewCount="107" Body="&lt;p&gt;Let $\rho : S_n \to \mathrm{GL}(\mathbb{C}^n)$, where $\rho(\sigma)(x_1, \ldots, x_n) = (x_{\sigma^{-1}(1)}, \ldots, x_{\sigma^{-1}(n)})$.  How can you prove that $W = \{ (x_1, \ldots, x_n) : x_1 + \cdots + x_n = 0 \} \subset \mathbb{C}^n$ is an irreducible representation?&lt;/p&gt;&#xA;" OwnerUserId="314" LastActivityDate="2010-12-01T10:24:39.140" Title="Representation is irreducible" Tags="&lt;representation-theory&gt;" AnswerCount="2" />
  <row Id="12578" PostTypeId="2" ParentId="12577" CreationDate="2010-12-01T05:41:21.107" Score="5" ViewCount="" Body="&lt;p&gt;Use the trace formula for $\rho$ to get that $\rho$ can be decomposed into two irreducible representations.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-12-01T05:41:21.107" />
  <row Id="12579" PostTypeId="2" ParentId="12572" CreationDate="2010-12-01T05:44:06.397" Score="8" ViewCount="" Body="&lt;p&gt;Your formula is another way of writing &lt;a href=&quot;http://en.wikipedia.org/wiki/Fourier_transform#The_Plancherel_theorem_and_Parseval.27s_theorem&quot; rel=&quot;nofollow&quot;&gt;Parseval's theorem&lt;/a&gt;, namely that Fourier transform preserves inner products:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_{-\infty}^{\infty} f \overline{g} = \int_{-\infty}^{\infty} \hat{f}\overline{\hat{g}} .$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(You should combine this with the identities $\hat{\hat{g}}(x) = g(-x)$&#xD;&#xA;and $\overline{\hat{g}}(y) = \hat{\overline{g}}(-y).$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is also a Fourier series version of Parseval's theorem: namely&#xA;if $f$ (with period 1) has Fourier series $a_n$ and $g$ (with period 1)&#xA;has Fourier series $b_n$, then&#xA;$$\int_0^1 f \overline{g} = \sum_{n = -\infty}^{\infty} a_n \overline{b}_n.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I don't see how to rewrite this in a way analogous to your formula,&#xA;just because (unlike in the Fourier transform context, where the funtion $f$&#xD;&#xA;and the Fourier transform $\hat{g}$ are both functions on the real line) a periodic function and its Fourier series are different beasts, and so it's not clear (to me) how one could mix them together in an integral or something similar.&lt;/p&gt;&#xA;" OwnerUserId="221" LastEditorUserId="221" LastEditorDisplayName="" LastEditDate="2010-12-01T05:51:55.020" LastActivityDate="2010-12-01T05:51:55.020" />
  <row Id="12580" PostTypeId="2" ParentId="12576" CreationDate="2010-12-01T05:46:19.343" Score="2" ViewCount="" Body="&lt;p&gt;You will need some geometry for calculus.  You should be able to teach yourself everything you need to know in terms of geometry within 1 month max, I would say if you work hard 2-3 weeks is a realistic timeframe.&#xA;You really need a solid foundation in algebra if you need to take calc 1 and calc 2 though.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would recommend taking some sort of algebra class but not more than one.  Alternatively, if you are disciplined, you can self-teach yourself, but this is not quite as good in my opinion.  I'm in my 2nd year of math/computer science joint program and my math classes are making my computer science courses a breeze.  In my (biased) opinion a solid foundation in math will REALLY help you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Best of luck.&lt;/p&gt;&#xA;" OwnerUserId="2533" LastActivityDate="2010-12-01T05:46:19.343" CommentCount="1" />
  <row Id="12581" PostTypeId="2" ParentId="12568" CreationDate="2010-12-01T05:48:58.480" Score="1" ViewCount="" Body="&lt;p&gt;Let's assume that all the integers are in fact at least $1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $\max(P) = \prod_i s_i$. If $\max(P) \neq \min(P)$ then $\max(P)-1$ must be representable. Now $(s_1 - 1) \prod_{i&amp;gt;1} s_i = \prod_i s_i - \prod_{i&amp;gt;1} s_i$, so there must be an index $I$ such that $S_i = \{1\}$ for $i \neq I$. $S_I$ itself is of course of the form $\{1,\ldots,\max(P)\}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Summarizing, we have found two solutions: $|S_i|=1$ for all $i$ (the sets are otherwise arbitrary), and $S_i = \{1\}$ apart for one set which is an interval.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the sets are infinite (i.e. $\max(P) = \infty$) then for each prime $p$ take the set $S_p = \{p^k : k \geq 0\}$. There are other solutions: you can decompose the set of primes arbitrarily as the disjoint union $\bigcup P_i$ and then take as $S_i$ all numbers whose prime factors are in $P_i$. All these use the convention that you only select finitely many numbers which are not $1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Open question: are these the only solutions?&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-12-01T05:48:58.480" CommentCount="0" />
  <row Id="12582" PostTypeId="2" ParentId="12566" CreationDate="2010-12-01T05:56:28.040" Score="1" ViewCount="" Body="&lt;p&gt;You can try to break up the sum according to some exponential scale in $p$, and then use the previous estimate to sum up every step of the ladder, multiplying the result by $\sqrt{p_{\min}}$ where $p_{\min}$ is the lowest prime in your ladder (you should use a smarter factor since you'll be summing each $1/p$ several times, and you want the sum of the relevant factors to be $\sqrt{p_{\min}}$).&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-12-01T05:56:28.040" />
  <row Id="12583" PostTypeId="2" ParentId="12410" CreationDate="2010-12-01T06:14:38.493" Score="3" ViewCount="" Body="&lt;p&gt;Just for the record, I like Shai Covo's answer better.  But the OP asked me to post my solution as well, so here it is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $X_t$ be the position of the object at time $t$.  Given $N$ clicks in $[0,t]$, let $\tau_1, \tau_2, \ldots \tau_N$ be the times of those clicks.  Let $T_i$ be the $i$th interarrival time, so that $T_1 = \tau_1$, $T_{N+1} = t - \tau_N$, and $T_i = \tau_i - \tau_{i-1}$ otherwise.  Thus $t = \sum_{i=1}^{N+1} T_i$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By properties of the exponential distribution, $E[T_i|N] = E[T_j|N]$ for all $i, j$.  Thus $$t = \sum_{i=1}^{N+1} E[T_i|N] = E[T_i|N] (N+1) \Rightarrow E[T_i|N] = \frac{t}{N+1}.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $N=0$, then $X_t = T_1$.  If $N = 1$, $X_t = \frac{1}{2}T_1 + T_2$.  If $N = 2$, $X_t = \frac{1}{4}T_1 + \frac{1}{2}T_2 + T_3$, and, in general, $X_t = \sum_{i=0}^N \frac{T_{N+1-i}}{2^i} $.  Thus &#xA;$$E[X_t|N] = \sum_{i=0}^N \frac{E[T_{N+1-i}|N]}{2^i}  = \frac{t}{N+1}\left(2 - \frac{1}{2^N}\right).$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $E[X_t] = E[E[X_t|N]]$, we just have to calculate $E\left[\frac{1}{N+1}\right]$ and $E\left[\frac{1}{(N+1)2^N}\right]$.  Since $N$ is Poisson$(\lambda t)$, we have&#xA;$$E\left[\frac{1}{N+1}\right] = \sum_{n=0}^{\infty} \frac{(\lambda t)^{n} e^{-\lambda t}}{(n+1) n!} = e^{-\lambda t} \sum_{n=0}^{\infty} \frac{(\lambda t)^{n} }{(n+1)!} = \frac{e^{-\lambda t}}{\lambda t} \sum_{n=0}^{\infty} \frac{(\lambda t)^{n+1} }{(n+1)!} $$&#xD;&#xA;$$= \frac{e^{-\lambda t}}{\lambda t} \left(\sum_{n=0}^{\infty} \frac{(\lambda t)^n }{n!} - 1 \right) = \frac{e^{-\lambda t}}{\lambda t} \left(e^{\lambda t} - 1 \right) = \frac{1}{\lambda t} - \frac{e^{-\lambda t}}{\lambda t}.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, &#xA;$$E\left[\frac{1}{(N+1)2^N}\right] = \frac{2e^{-\lambda t}}{\lambda t} \sum_{n=0}^{\infty} \frac{(\frac{\lambda t}{2})^{n+1} }{(n+1)!} = \frac{2e^{-\lambda t}}{\lambda t} \left(e^{\lambda t/2} - 1 \right) = \frac{2e^{-\lambda t/2}}{\lambda t} - \frac{2e^{-\lambda t}}{\lambda t}.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$E[X_t] = E\left[\frac{2t}{N+1}\right] - E\left[\frac{t}{(N+1)2^N}\right] = \frac{2}{\lambda}\left(1 - e^{-\lambda t/2}\right),$$&#xD;&#xA;which is exactly what you get if you solve the differential equation in Shai Covo's answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the expected average speed (velocity, actually, since the average speed is technically 1) is &#xA;$$\frac{2}{\lambda t}\left(1 - e^{-\lambda t/2}\right).$$&lt;/p&gt;&#xA;" OwnerUserId="2370" LastActivityDate="2010-12-01T06:14:38.493" />
  <row Id="12584" PostTypeId="1" AcceptedAnswerId="12585" CreationDate="2010-12-01T06:24:32.777" Score="5" ViewCount="175" Body="&lt;p&gt;A &lt;a href=&quot;http://en.wikipedia.org/wiki/Doubly_stochastic_matrix&quot; rel=&quot;nofollow&quot;&gt;doubly-stochastic matrix&lt;/a&gt; is an $n\times n$ matrix $P$ such that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle\sum_{i=1}^n{p_{ij}}=1$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle\sum_{j=1}^n{p_{ij}}=1$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $p_{ij}\ge 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can someone suggest an algorithm for generating these matrices uniform randomly?&lt;/p&gt;&#xA;" OwnerUserId="1004" LastActivityDate="2010-12-01T21:03:49.730" Title="How do I Generate Doubly-Stochastic Matrices Uniform Randomly?" Tags="&lt;probability&gt;&lt;combinatorics&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="12585" PostTypeId="2" ParentId="12584" CreationDate="2010-12-01T06:36:25.930" Score="5" ViewCount="" Body="&lt;p&gt;What we want is to generate a bistochastic matrix according to the Haar measure, which is the unique distribution which is invariant to multiplication by bistochastic matrices from both sides.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The standard algorithm is to take some iid matrix (each entry is chosen iid from some distribution over non-negative numbers) and then repeatedly make it row-stochastic and column-stochastic - this is like projecting the matrix to the linear subspace of stochastic matrices. The process converges pretty fast, and we get some random bistochastic matrix, even though not according to the wanted distribution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://chaos.if.uj.edu.pl/~karol/pdf/CSBZ09.pdf&quot; rel=&quot;nofollow&quot;&gt;A recent paper&lt;/a&gt; discusses these issues in the context of estimating the measure of all bistochastic matrices (an open problem).&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-12-01T06:36:25.930" CommentCount="1" />
  <row Id="12586" PostTypeId="2" ParentId="12584" CreationDate="2010-12-01T06:36:55.790" Score="3" ViewCount="" Body="&lt;p&gt;Suppose you want a random rational point in the Birkhoff polytope (set of double stochastic matrices), with denominators dividing $N$.  This is equivalent to a random contingency table with all row and column sums equal to $N$.  A reversible Markov chain exists on the set of tables and is rapidly mixing.  This can be used for approximate sampling by following the chain for a large number of steps.  The limit distribution of the chain is close to uniform because away from the boundary of the polytope, all vertices of the chain have the same degree.  Exact sampling from the limit distribution of the chain is possible, but because the limit distribution is not quite uniform this is not the same as obtaining an exact sample from the uniform distribution on contingency tables.  The latter is a hard unsolved problem as far as I know. &lt;/p&gt;&#xA;" OwnerUserId="467" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-12-01T21:03:49.730" LastActivityDate="2010-12-01T21:03:49.730" CommentCount="2" />
  <row Id="12587" PostTypeId="1" CreationDate="2010-12-01T06:47:37.353" Score="9" ViewCount="242" Body="&lt;p&gt;I have a homework problem in my textbook that has stumped me so far. There is a similar one to it that has not been assigned and has an answer in the back of the textbook. It reads: How many ways are there for 8 men and 5 women to stand in a line so that no two women stand next to each other?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer is 609638400, but no matter what I try I cannot reach that number. I have tried doing 2(8!5!/3!)*(8!/5!) since each woman must be paired with a man in order to prevent two women getting near each other, but of course it's the wrong answer. What am I doing wrong here?&lt;/p&gt;&#xA;" OwnerUserId="3071" LastEditorUserId="2109" LastEditorDisplayName="" LastEditDate="2010-12-01T07:11:15.033" LastActivityDate="2010-12-01T07:20:40.270" Title="Homework related: Permutations and Combinations question." Tags="&lt;homework&gt;&lt;combinatorics&gt;&lt;permutations&gt;" AnswerCount="1" />
  <row Id="12588" PostTypeId="2" ParentId="12566" CreationDate="2010-12-01T07:01:53.993" Score="10" ViewCount="" Body="&lt;p&gt;We can use your second estimate&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum\limits_{p \leq x} \frac{1}{p} = \log{\log{x}} + A + R(x)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\displaystyle R(x) = \mathcal{O}\Bigl(\frac{1}{\log{x}}\Bigr)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;to show the asymptotic estimate:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum\limits_{p \leq x} \frac{1}{\sqrt{p}} = \mathcal{\theta}\left(\frac{\sqrt{x}}{\log x}\right)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact something stronger is known, that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle R(x) = \mathcal{O}\left(\frac{1}{\log^2{x}}\right)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and I believe this gives us&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum\limits_{p \leq x} \frac{1}{\sqrt{p}} = \frac{2\sqrt{x}}{\log x} + \mathcal{O}\left(\frac{\sqrt{x}}{\log ^2x}\right)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For this, we use &lt;a href=&quot;http://books.google.com/books?id=iJx1lP9ZcIkC&amp;amp;pg=PA25&quot; rel=&quot;nofollow&quot;&gt;Abel's Identity&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Define $\displaystyle A(x) = \sum\limits_{n \leq x} \frac{a(n)}{n}$, where $\displaystyle a(n) = 1$ iff $\displaystyle n$ is prime and $\displaystyle 0$ otherwise. The above estimate tells us that $\displaystyle A(x) = \log\log x + A + R(x)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Setting $\displaystyle f(t) = \sqrt{t}$ and using Abel's Identity we obtain&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle \sum\limits_{3 \le p \le x} \frac{1}{\sqrt{p}} + C = A(x) \sqrt{x} - \int_{3}^{x} \frac{A(t)}{2\sqrt{t}} \ \text{d}t$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle \int_{3}^{x} \frac{A(t)}{2\sqrt{t}} \ \text{d}t = \int_{3}^{x} \frac{\log \log t + A + R(t)}{2\sqrt{t}} \ \text{d}t$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle  = C^{\prime} + A\sqrt{x} + \sqrt{x} \log \log x - \mathrm{Ei}\left(\frac{\log x}{2}\right) + \mathcal{O}\left(\frac{\sqrt{x}}{\log^2 x}\right)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;using&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle \int \frac{\log \log x }{\sqrt{x}} = 2\sqrt{x} \log \log x - 2\mathrm{Ei}\left(\frac{\log x}{2}\right) + K$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle \int \frac{R(x)}{\sqrt{x}} = \mathcal{O}\left(\int \frac1{\log^2 x \sqrt{x}}\right) = \mathcal{O}\left(\frac{\sqrt{x}}{\log^2 x}\right) $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\displaystyle \mathrm{Ei}(x)$ is the exponential integral.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(All gotten using Wolfram Alpha).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $\displaystyle \mathrm{Ei}(x) = \frac{e^x}{x} + \mathcal{O}\left(\frac{e^x}{x^2}\right)$ (again using Wolfram Alpha), we get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle \sum\limits_{3 \le p \le x} \frac{1}{\sqrt{p}} = C + R(x) \sqrt{x} + \frac{2\sqrt{x}}{\log{x}} + \mathcal{O}\left(\frac{\sqrt{x}}{\log^2 x}\right) = \frac{2\sqrt{x}}{\log x} + \mathcal{O}\left(\frac{\sqrt{x}}{\log^2 x}\right) $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Hopefully, I got the computations right).&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I will leave my earlier answer here (which is more relevant to the question as asked)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;An elementary lower bound&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is one elementary way (which does not use your second estimate or the prime number theorem) which proves something slightly stronger.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the fact the each integer is uniquely expressible as the product of a square and a square free integer, we have that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle \sum_{j=1}^{n} \dfrac1{\sqrt{j}} \leq \prod_{p \le n} \left(1 + \dfrac1{\sqrt{p}}\right) \sum_{k=1}^{n} \dfrac1{k}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now $\displaystyle e^x \gt 1 + x$, we have that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle \sum_{j=1}^{n} \dfrac1{\sqrt{j}} \leq \left(\prod_{p \le n} e^{\dfrac1{\sqrt{p}}}\right)\left(\sum_{k=1}^{n} \dfrac1{k}\right)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We use the estimate that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle \sum_{j=1}^{n} \dfrac{1}{\sqrt{j}} \ge 2\sqrt{n-1}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle \sum_{k=1}^{n} \dfrac1{j} \le \log (n+2)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;to get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle 2\sqrt{n-1} \leq \log(n+2) \prod_{p \le n} e^{\dfrac1{\sqrt{p}}}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Taking logs, we see that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle \dfrac{1}{2}\log (n-1) - \log \log(n+2)  + \log 2 \le \sum_{p \le n} \dfrac1{\sqrt{p}}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $\displaystyle \log \log (n+2) - \log \log n \to 0 \ \text{as} \ n \to \infty$, $\displaystyle \log (n) - \log (n-1) \to 0 \ \text{as} \ n \to \infty$ and $\displaystyle \log 2 \gt 0$, you just need to verify you inequality for finite number of $\displaystyle n$.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-12-02T16:46:37.057" LastActivityDate="2010-12-02T16:46:37.057" CommentCount="4" />
  <row Id="12590" PostTypeId="2" ParentId="12587" CreationDate="2010-12-01T07:04:51.107" Score="16" ViewCount="" Body="&lt;p&gt;For this type of problem first consider the position of the men, then the position of the women.In general first consider the the group with more people.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How many possible ways are there to arrange eight men in a row? It will be $ ^8P_8 = 8! = 40320$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now as &lt;code&gt;no two women stand next to each other&lt;/code&gt;,we can imagine the situation as &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;&lt;pre&gt;&lt;code&gt;                   * M * M * M * M * M * M * M * M *&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hence, we need to find how many ways we can arrange $5$ women in the $9$ possible (as shown above) places,this is actually $ ^9P_5 = 9*8*7*6*5 = 15120$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now applying the fundamental law of counting (precisely &lt;a href=&quot;http://en.wikipedia.org/wiki/Rule_of_product&quot; rel=&quot;nofollow&quot;&gt;product rule&lt;/a&gt;), total number of possible arrangements satisfying both constraints is: $15120 * 40320 = 609638400$ which is your required/desired answer. &lt;/p&gt;&#xA;" OwnerUserId="2109" LastEditorUserId="2109" LastEditorDisplayName="" LastEditDate="2010-12-01T07:20:40.270" LastActivityDate="2010-12-01T07:20:40.270" CommentCount="3" />
  <row Id="12591" PostTypeId="1" CreationDate="2010-12-01T07:10:41.960" Score="1" ViewCount="134" Body="&lt;p&gt;Suppose $\frac{dx}{dt}=ax+b$ and then assume that $a=c+g$ where $g$ is a Wiener process. &lt;/p&gt;&#xA;" OwnerUserId="4274" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-12-02T01:04:38.047" LastActivityDate="2011-01-15T21:54:08.013" Title="How can I obtain from a differential equation a stochastic version?" Tags="&lt;differential-equations&gt;&lt;stochastic-processes&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="1" />
  <row Id="12592" PostTypeId="2" ParentId="12566" CreationDate="2010-12-01T07:28:01.853" Score="10" ViewCount="" Body="&lt;p&gt;I wasn't sure that you wanted a proof that used that fact from Apostol.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One easy method &lt;em&gt;not&lt;/em&gt; using the result you quote from Apostol is as follows:&#xA;$$&#xD;&#xA;\sum_{p\le x} \frac{1}{\sqrt{p}} &amp;gt; \sum_{p \le x} \frac{1}{\sqrt{x}} = \frac{\pi(x)}{\sqrt{x}} &amp;gt; c \frac{\sqrt{x}}{\log{x}}&#xD;&#xA;$$&#xD;&#xA;where $c$ is a constant you can get from Chebyshev or &lt;a href=&quot;http://projecteuclid.org/DPubS?service=UI&amp;amp;version=1.0&amp;amp;verb=Display&amp;amp;handle=euclid.ijm/1255631807&quot; rel=&quot;nofollow&quot;&gt;Rosser and Schoenfeld&lt;/a&gt;, and maybe do a little computation to take care of small $x$, and you've got it.  This is a much better lower bound than what you are trying to prove.&lt;/p&gt;&#xA;" OwnerUserId="2937" LastActivityDate="2010-12-01T07:28:01.853" CommentCount="1" />
  <row Id="12593" PostTypeId="1" CreationDate="2010-12-01T07:52:34.953" Score="1" ViewCount="124" Body="&lt;p&gt;Is there a &quot;SETI at Home&quot; style initiative for calculating constants with distributed computing?&lt;/p&gt;&#xA;" OwnerUserId="4277" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-12-01T08:16:06.370" LastActivityDate="2010-12-31T09:44:30.140" Title="Distributed software for calculating $\pi$ or $e$" Tags="&lt;math-software&gt;&lt;pi&gt;" AnswerCount="1" />
  <row Id="12594" PostTypeId="1" AcceptedAnswerId="12595" CreationDate="2010-12-01T07:57:13.950" Score="2" ViewCount="88" Body="&lt;p&gt;For what values of $x = (x_0 x_1 ... x_n)$ the following inequality holds for all positive values of the vector $a = (a_0 a_1 ... a_n)$: $a x^T \ge 0$?&lt;/p&gt;&#xA;" OwnerUserId="1041" LastEditorUserId="1041" LastEditorDisplayName="" LastEditDate="2010-12-01T08:56:35.547" LastActivityDate="2010-12-01T08:56:35.547" Title="Simple-looking vector inequality ..." Tags="&lt;inequality&gt;&lt;vector-space&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" />
  <row Id="12595" PostTypeId="2" ParentId="12594" CreationDate="2010-12-01T08:08:59.437" Score="4" ViewCount="" Body="&lt;p&gt;The condition is that $x_i\geq 0$ for all $i$.  It is clear that $ax^T\geq0$ if this holds.  On the other hand, if $x_i\lt0$ for some $i$, then choose $a$ such that $a_i=1$ and $|a_jx_j|\lt\frac{|x_i|}{n}$ for all $j\neq i$.  Then $ax^T=x_i+(a_0x_0+\cdots+a_{i-1}x_{i-1}+a_{i+1}x_{i+1}+\cdots+a_nx_n)&#xD;&#xA;\lt x_i +n\frac{|x_i|}{n}=0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or, if it was assumed that $a$ is fixed beforehand, then your inequality would describe a &lt;a href=&quot;http://en.wikipedia.org/wiki/Half-space&quot; rel=&quot;nofollow&quot;&gt;half-space&lt;/a&gt; whose boundary is the hyperplane described by $ax^T=0$, which would include those vectors $x$ such that $x_i\geq0$ for all $i$.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2010-12-01T08:14:20.887" LastActivityDate="2010-12-01T08:14:20.887" CommentCount="1" />
  <row Id="12596" PostTypeId="1" AcceptedAnswerId="12606" CreationDate="2010-12-01T08:15:22.203" Score="2" ViewCount="257" Body="&lt;p&gt;While learning &lt;em&gt;topology&lt;/em&gt; one learns about compact set. The standard definition is:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;A set $X$ is said to be compact &lt;em&gt;if open cover has a finite subcover.&lt;/em&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Since $[0,1]$ is compact, if we take a open cover for this we should be able to get a finite subcover. I know, that $(0,1)$ is not compact, so there must exists some open cover for $(0,1)$ which doesn't admit any finite subcover. But how does one prove this fact?&lt;/p&gt;&#xA;" LastActivityDate="2010-12-01T12:03:50.010" Title="Finite sub cover for $(0,1)$." Tags="&lt;analysis&gt;&lt;topology&gt;" AnswerCount="4" CommentCount="1" />
  <row Id="12598" PostTypeId="2" ParentId="12596" CreationDate="2010-12-01T08:20:33.367" Score="15" ViewCount="" Body="&lt;p&gt;Isn't this clear from $(0,1) = (0,1/2) \cup (0,2/3) \cup (0,3/4) \cup  \cdots$?&lt;/p&gt;&#xA;" OwnerUserId="2810" LastActivityDate="2010-12-01T08:20:33.367" />
  <row Id="12599" PostTypeId="2" ParentId="12596" CreationDate="2010-12-01T08:26:08.650" Score="3" ViewCount="" Body="&lt;p&gt;Let $A$ be a non-closed subset of a metric space, and let $x\in\overline A\setminus A$.  Then $A$ is covered by open sets given by the complements of closed balls of positive radius centered at $x$, but not by any finite subcollection.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-12-01T08:26:08.650" />
  <row Id="12600" PostTypeId="1" AcceptedAnswerId="12602" CreationDate="2010-12-01T08:43:07.687" Score="3" ViewCount="77" Body="&lt;p&gt;I have a plane defined through a point P and 2 3d-vectors Xdir and yDir.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wish to convert coordinates of points on this plane between local 2d-parametric and world 3d coordinate systems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know the conversion from 2d Parametric to 3d is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;C(u, v) = P + u*XDir + v*YDir &lt;/p&gt;&#xA;&#xA;&lt;p&gt;however i have been unable to find a way for the inverse case &lt;/p&gt;&#xA;&#xA;&lt;p&gt;C'(x,y,z) which should give me the parameters u and v for any point x,y,z in the plane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How does this conversion work?&lt;/p&gt;&#xA;" OwnerUserId="1357" LastActivityDate="2010-12-01T09:25:08.900" Title="Given a plane and a 3d-Vertex, what are U and V?" Tags="&lt;geometry&gt;" AnswerCount="1" />
  <row Id="12601" PostTypeId="1" CreationDate="2010-12-01T08:46:19.227" Score="1" ViewCount="95" Body="&lt;p&gt;I am reading by myself this book &lt;a href=&quot;http://tinyurl.com/37z4bbt&quot; rel=&quot;nofollow&quot;&gt;http://tinyurl.com/37z4bbt&lt;/a&gt;. But to be honest, I have several problems to fully understand some part of the text. Maybe because I have not yet solid knowledge or I still need to learn more about the subject, hence asked a little help&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To be exact, is this part of the construction concerning to Algebraic Blow-Up that I do not understand like trying...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I quote part of the building in which I do not understand how to deal with some concepts&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Consider the canonical map from $\mathbb{C}^2$ to the projective line $\mathbb{CP}^1$ that associates with each point $(x,y)$ different from the origin, the line {$(tx, ty) : t \in \mathbb{C}$} passing through this point. The graph of this map is a complex 2-dimensional surface in the complex 3-dimensional manifold (the Cartesian product) $\mathbb{C}^2$ × $\mathbb{CP}^1$, &lt;strong&gt;which is not closed&lt;/strong&gt;. To obtain the closure, one has to add the exceptional curve $E={0} \times \mathbb{CP^{1}} \subset \mathbb{C^{2}} \times \mathbb{CP^{1}}$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;then, I do not understand is how to handle this graph. If, as a &lt;strong&gt;topological space&lt;/strong&gt; or as &lt;strong&gt;complex manifold&lt;/strong&gt;. Well, in the sense of complex manifold, say it is not closed, it means that the manifold is not compact without boundary. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am really a little confused&lt;/p&gt;&#xA;" OwnerUserId="4278" LastActivityDate="2011-05-19T11:59:07.550" Title="Blow-up. Graph of canonical map: is a complex manifold no closed or a topological space no closed" Tags="&lt;topology&gt;&lt;differential-equations&gt;&lt;algebraic-topology&gt;&lt;analytic&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="12602" PostTypeId="2" ParentId="12600" CreationDate="2010-12-01T08:57:52.457" Score="3" ViewCount="" Body="&lt;p&gt;The inverse of the 3-by-3 matrix $A$ whose columns are given by $X$, $Y$, and the cross product $X\times Y$ changes basis from $\{X,Y,X\times Y\}$ to the standard basis.  Given $(x,y,z)^T=P+uX+vY$ in the plane, subtracting $P$ and multiplying $A^{-1}$ by the resulting column vector gives the column vector $(u,v,0)^T$.  So one way to write the map would be $\pi(A^{-1}((x,y,z)^T-P))$, where $\pi:\mathbb{R}^3\to\mathbb{R}^2$ is projection onto the first two coordinates.  To write this more explicitly, you could use the matrix $\left(\begin{matrix}&#xD;&#xA;1 &amp;amp; 0 &amp;amp; 0 \\&#xD;&#xA;0 &amp;amp; 1 &amp;amp; 0&#xD;&#xA;\end{matrix}\right)&#xD;&#xA;$&#xD;&#xA;of $\pi$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The cross product isn't really necessary, it just seemed like the most straightforward way to complete $\{X,Y\}$ to a basis for $\mathbb{R}^3$.  Any vector $Z$ not in the span of $\{X,Y\}$ would do.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2010-12-01T09:25:08.900" LastActivityDate="2010-12-01T09:25:08.900" />
  <row Id="12603" PostTypeId="2" ParentId="12431" CreationDate="2010-12-01T09:08:45.633" Score="0" ViewCount="" Body="&lt;p&gt;My answer is similar to chroma's answer, only that I get &#xA;$$&#xD;&#xA;(xf) * g = at(f * g) \Rightarrow \hat f&amp;#39;\hat g = a(\hat f \hat g)&amp;#39; &#xD;&#xA;$$&#xD;&#xA;straight from the following standard property (combined, of course, with the convolution theorem):&#xA;if $\psi = x \varphi$, then $\hat \psi = i \hat \varphi&amp;#39;$. &lt;/p&gt;&#xA;" OwnerUserId="2810" LastActivityDate="2010-12-01T09:08:45.633" />
  <row Id="12604" PostTypeId="1" CreationDate="2010-12-01T09:30:33.900" Score="3" ViewCount="114" Body="&lt;p&gt;Hi I have a problem with quadratic splines, I am supposed to find $&#xD;&#xA;S_1 $ and $S_2$ that interpolates the following points $S(-1)=0$, $S(0)=1$, $S(1)=2$, and at the same time we want to find $S$ such that  $\int_{-1}^1 \! (S(x))^2 \, \mathrm{d}x$  is minimal. The answer is on the form&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$S_1(x)=a_1 x^2 +b_1x +c_1$ on $[-1,0]$ and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$S_2(x)=a_2 x^2 +b_2x +c_2$ on $[0,1]$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;my answer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I use the data points and find that $a_1=-a_2$, $b_1=b_2$ and $c_1=c_2=1$, but I have no idea how to use minimize $\int_{-1}^1 \! (S(x))^2 \, \mathrm{d}x$&#xD;&#xA;, can I divide it up?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\min \int_{-1}^1 \! (S(x))^2 \, \mathrm{d}x=\min (\int_{-1}^0 \! (S_1(x))^2 \, \mathrm{d}x + \int_0^1 \! (S_2(x))^2 \, \mathrm{d}x )$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know I should get an expression and probably set the derivative to zero but I just don't know how to attack the minimizing integral since the function has two parts. Help greatly appreciated.&lt;/p&gt;&#xA;" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-02T11:52:46.080" LastActivityDate="2010-12-02T11:58:20.910" Title="Quadratic splines, minimizing integral" Tags="&lt;numerical-methods&gt;&lt;approximation&gt;" AnswerCount="1" />
  <row Id="12605" PostTypeId="2" ParentId="12593" CreationDate="2010-12-01T09:41:25.023" Score="1" ViewCount="" Body="&lt;p&gt;There was &lt;a href=&quot;http://oldweb.cecm.sfu.ca/projects/pihex/index.html&quot; rel=&quot;nofollow&quot; title=&quot;Pi Hex&quot;&gt;PiHex&lt;/a&gt; which has now ended. I am not aware of any others - someone may correct me&lt;/p&gt;&#xA;" OwnerUserId="2219" LastActivityDate="2010-12-01T09:41:25.023" />
  <row Id="12606" PostTypeId="2" ParentId="12596" CreationDate="2010-12-01T09:43:37.913" Score="4" ViewCount="" Body="&lt;p&gt;To prove that a set $X$ isn't compact is generally easy. You only need to give an example of a cover, that doesn't admit a finite subcover. To expand on Shai's answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\mathcal U := \{(0,n/(n+1): n \in \mathbb N\}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and observe that if you take any $U_1,U_2,\dots U_n \in \mathcal U$, then &#xA;there exists some k such that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(0,k/(k+1))=\bigcup_{i=1}^n U_1.$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If this isn't obvious, note that for any two elements in $\mathcal U$ that one is contained the other. So if you take a finite number of them, then one of them will contain the rest. Furthermore since $(0,k/(k+1))$ does not contain $(0,1)$,&#xA;$\mathcal U$ does not have a finite subcover, so X is not compact. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I guess you should also show that $\mathcal U$ is a cover; I'll leave it as an exercise to the reader.&lt;/p&gt;&#xA;" OwnerUserId="1930" LastActivityDate="2010-12-01T09:43:37.913" />
  <row Id="12607" PostTypeId="2" ParentId="12604" CreationDate="2010-12-01T10:03:43.013" Score="7" ViewCount="" Body="&lt;p&gt;Here's a hint: since you need your piecewise expression to be a &lt;em&gt;spline&lt;/em&gt;, one thing you should have is the imposition of a derivative condition at the point $x=0$; namely, $S_1^{\prime}(0)=S_2^{\prime}(0)$. (This is apart from the obvious equations you get from substituting appropriate values of $x$ and $y$ into the expressions for $S_1$ and $S_2$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus far, you should have five equations in six unknowns; you can eliminate some unknowns, and be able to express some of them in terms of just one of the unknowns, such that both $S_1$ and $S_2$ depend on a single parameter. You had the right idea to split up the integral into integrals over $[-1,0]$ and $[0,1]$ ; if you do that, you should end up with a quadratic in a single variable, whose minimum is trivial to find.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Shorter version of the hint: you should be able to express $a_1$ (and thus $a_2$ as well) in terms of either $b_1$ or $b_2$. Picking $b_1$ as the variable of interest, and substituting into your integral, you should have an expression of the form&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$u(b_1)^2+v\cdot b_1+w$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $u$, $v$, and $w$ are certain constants. If $u$ is positive, then $-\frac{v}{2u}$ is the minimum value of $b_1$ you seek. Substitute into all your other relations, and you should now have the desired spline.&lt;/p&gt;&#xA;" OwnerUserId="498" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-02T11:58:20.910" LastActivityDate="2010-12-02T11:58:20.910" CommentCount="1" />
  <row Id="12608" PostTypeId="1" AcceptedAnswerId="12640" CreationDate="2010-12-01T10:22:34.880" Score="0" ViewCount="68" Body="&lt;p&gt;Let $S$ be a collection of disjoint sets. Let a binary operation '*' defined $\forall x,y$ each belonging to two different sets or a same set in $S$ with the property that $z=x*y$ belongs to some set in $S$ . Let $\forall A_1,A_2 \in S$ a  binary operation $ \otimes$ is defined between $A_1,A_2$ as $B=A_1 \otimes A_2$ as the collection of all $z=(x*y)$ where $x\in A_1,y\in A_2$. Now the set $S$ is closed under the binary operation $\otimes$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to study the properties of such a set $S$ under the operation $\otimes$.My question is what is the subject in mathematics which deals with such a situation ?&lt;/p&gt;&#xA;" OwnerUserId="2987" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-01T10:45:40.170" LastActivityDate="2010-12-01T17:07:43.413" Title="a reference request to study properties of a set which is closed under a binary operation." Tags="&lt;reference-request&gt;&lt;elementary-set-theory&gt;" AnswerCount="1" />
  <row Id="12609" PostTypeId="2" ParentId="12577" CreationDate="2010-12-01T10:24:39.140" Score="4" ViewCount="" Body="&lt;p&gt;The first time I was assigned this exercise I was explicitly told not to use character theory.  There is a straightforward direct proof (showing directly that the orbit of any nonzero vector is all of $W$) and I implore you to try to find it.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-12-01T10:24:39.140" CommentCount="2" />
  <row Id="12610" PostTypeId="1" AcceptedAnswerId="12612" CreationDate="2010-12-01T11:18:05.400" Score="1" ViewCount="99" Body="&lt;p&gt;In a 2-d plane, let $A$ be the set of all points inside a circle $C$ and $P$ be the set of all points on the perimeter of the circle $C$. Is there any generalized measure in set theory which distinguishes between two such sets ?&lt;/p&gt;&#xA;" OwnerUserId="2987" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-01T11:20:14.460" LastActivityDate="2010-12-01T14:22:23.760" Title="a question about a measurable property of a set" Tags="&lt;elementary-set-theory&gt;" AnswerCount="2" />
  <row Id="12611" PostTypeId="2" ParentId="12520" CreationDate="2010-12-01T11:45:26.763" Score="2" ViewCount="" Body="&lt;p&gt;Though the floor function is not periodic, it is &lt;em&gt;quasiperiodic&lt;/em&gt;; that is, it is a function that satisfies the functional equation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f(x+\alpha)=f(x)+ux+v$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the specific case of the floor function, you have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\lfloor x+1\rfloor=\lfloor x\rfloor+1$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so 1 can be thought of as the floor function's quasiperiod.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's a rather trivial example of a quasiperiodic function, though.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-12-01T11:45:26.763" />
  <row Id="12612" PostTypeId="2" ParentId="12610" CreationDate="2010-12-01T11:50:21.670" Score="5" ViewCount="" Body="&lt;p&gt;Yes there is. It's called the &lt;a href=&quot;http://en.wikipedia.org/wiki/Hausdorff_measure&quot; rel=&quot;nofollow&quot;&gt;Hausdorff measure&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1269" LastActivityDate="2010-12-01T11:50:21.670" />
  <row Id="12613" PostTypeId="2" ParentId="12596" CreationDate="2010-12-01T12:03:50.010" Score="1" ViewCount="" Body="&lt;p&gt;Alternatively, use the fact that every compact subspace of a Hausdorff space is closed. Since (0,1) is not closed, it cannot be compact. (this is theorem 26.3 in Munkres)&lt;/p&gt;&#xA;" OwnerUserId="4284" LastActivityDate="2010-12-01T12:03:50.010" />
  <row Id="12614" PostTypeId="1" CreationDate="2010-12-01T12:34:49.837" Score="1" ViewCount="116" Body="&lt;p&gt;Suppose a &lt;a href=&quot;http://en.wikipedia.org/wiki/Parabolic_mirror&quot; rel=&quot;nofollow&quot;&gt;parabolic mirror&lt;/a&gt; of a telescope is large enough (or equivalently that the camera &lt;a href=&quot;http://en.wikipedia.org/wiki/Shutter_speed&quot; rel=&quot;nofollow&quot;&gt;exposure time&lt;/a&gt; is short enough) that you need to take into account the transit time of the light signal from hitting the mirror to reaching the focal point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What shape should the mirror be to have the same effect as a parabolic mirror that doesn't take into account light transit time ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Added after comments:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose you want to view surface details on an exoplanet with an array of space mirrors spanning millions or even billions of km, also redirecting photons towards a detector at a focal point. The property to preserve is photons from an event on the exoplanet arriving at the focal point at the same time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the transit time from hitting the mirror to reaching the focus is long enough then given that objects aren't ideal or at infinity, then any discrepancy at all in arrival times is magnified by the huge transit-time so if the object being viewed changes fast it can change significantly during the discrepancy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In talking about shutter speed I was thinking about viewing an object whose properties change very quickly (say clouds moving on a giant exoplanet). If there is a discrepancy in the time it takes for photons leaving the object to reach the focal point and the object changes during that time then the image would be blurred (temporally aberrated).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As Ross Millikan points out distances to the focus point should be the same, so I was misunderstanding things, however looking at Wikipedia exact focus only exists for point sources otherwise there is &lt;a href=&quot;http://en.wikipedia.org/wiki/Coma_%28optics%29&quot; rel=&quot;nofollow&quot;&gt;coma aberration&lt;/a&gt;. The article also says there is something called bestform or aplantic lenses to minimise this, but you can't apply a lense to an array of mirrors spanning billions of km, so the original question stands but with a changed understanding.&lt;/p&gt;&#xA;" OwnerUserId="4157" LastEditorUserId="4157" LastEditorDisplayName="" LastEditDate="2010-12-01T15:50:52.493" LastActivityDate="2010-12-01T15:50:52.493" Title="Temporal aberration. Shape of telescope mirror large enough that light transit time is significant" Tags="&lt;geometry&gt;&lt;physics&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="0" />
  <row Id="12615" PostTypeId="1" AcceptedAnswerId="12616" CreationDate="2010-12-01T12:42:11.133" Score="3" ViewCount="186" Body="&lt;p&gt;I know that there was a guy that could get 100 decimal digits of $\pi$ before computers were able to get thousands.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How did the guy do that?&lt;/p&gt;&#xA;" OwnerUserId="3314" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-12-01T16:37:55.730" LastActivityDate="2010-12-01T16:37:55.730" Title="Get $\pi$ decimals manually" Tags="&lt;numerical-methods&gt;&lt;pi&gt;" AnswerCount="3" CommentCount="3" />
  <row Id="12616" PostTypeId="2" ParentId="12615" CreationDate="2010-12-01T12:48:11.177" Score="5" ViewCount="" Body="&lt;p&gt;According to Wikipedia, &lt;a href=&quot;http://en.wikipedia.org/wiki/John_Machin&quot; rel=&quot;nofollow&quot;&gt;John Machin&lt;/a&gt; combined the formula &#xA;$$\frac{\pi}{4}=4\cot^{-1}5-\cot^{-1}239$$&#xD;&#xA;with the Taylor series expansion for the inverse tangent in order to compute $\pi$ to 100 decimal places.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A previous record was due to &lt;a href=&quot;http://en.wikipedia.org/wiki/Abraham_Sharp&quot; rel=&quot;nofollow&quot;&gt;Abraham Sharp&lt;/a&gt; who used an arcsine series to find 72 decimal digits. &lt;/p&gt;&#xA;" OwnerUserId="723" LastEditorUserId="723" LastEditorDisplayName="" LastEditDate="2010-12-01T13:02:06.933" LastActivityDate="2010-12-01T13:02:06.933" />
  <row Id="12617" PostTypeId="2" ParentId="12572" CreationDate="2010-12-01T13:00:13.157" Score="1" ViewCount="" Body="&lt;p&gt;Given an orthogonal function system $\phi_{n}(x)$ ($n=1,2,\dots$) with norm&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left\vert \left\vert \phi _{n}\right\vert \right\vert =\sqrt{\left( \phi_{n}\cdot \overline{\phi }_{n}\right) }=\displaystyle\sqrt{\displaystyle\int_{a}^{b}\phi _{n}\left( x\right) \,\overline{\phi }_{n}\left( x\right) \;dx}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and two functions $f(x)$ e $g(x)$ represented by the Fourier series&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f(x)\sim\displaystyle\sum_{n\ge 1} c_{n}\phi_{n}(x)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$g(x)\sim\displaystyle\sum_{n\ge 1} d_{n}\phi_{n}(x)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with coefficients $c_n,d_n$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$c_n=\dfrac{\left ( f\cdot \overline{\phi }_{n}\right)}{\left\vert \left\vert \phi _{n}\right\vert \right\vert ^{2}}=\dfrac{\displaystyle\int_{a}^{b}f\left( x\right)\overline{\phi }_{n}\left( x\right) \;dx}{\left\vert \left\vert \phi _{n}\right\vert \right\vert ^{2}}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$d_n=\dfrac{\left ( f\cdot \overline{\phi }_{n}\right)}{\left\vert \left\vert \phi _{n}\right\vert \right\vert ^{2}}=\dfrac{\displaystyle\int_{a}^{b}g\left( x\right)\overline{\phi }_{n}\left( x\right) \;dx}{\left\vert \left\vert \phi _{n}\right\vert \right\vert ^{2}},$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and if&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle\int_{a}^{b}\left\vert f(x)-\displaystyle\sum_{n=1}^{\infty}c_{n}\phi_{n}(x)\right\vert ^{2}dx=0$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle\int_{a}^{b}\left\vert g(x)-\displaystyle\sum_{n=1}^{\infty}d_{n}\phi_{n}(x)\right\vert ^{2}dx=0,$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;then the following equality holds &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle\int_{a}^{b}f(x)\overline{g}_{n}(x)\; dx=\displaystyle\sum_{n\ge 1} c_{n}\overline{d}_{n}||\phi_n||^2,$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a particular case of which, for $g(x)=f(x)$, is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle\int_{a}^{b}|f(x)|^2\; dx=\displaystyle\sum_{n\ge 1} |c_{n}|^2||\phi_n||^{2}.$$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="752" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2010-12-01T15:24:57.847" LastActivityDate="2010-12-01T15:24:57.847" />
  <row Id="12618" PostTypeId="2" ParentId="12615" CreationDate="2010-12-01T13:09:51.073" Score="3" ViewCount="" Body="&lt;p&gt;This page on the &lt;a href=&quot;http://www-history.mcs.st-and.ac.uk/history/HistTopics/Pi_chronology.html&quot; rel=&quot;nofollow&quot;&gt;chronology of pi&lt;/a&gt; contains many useful notes on how the pre-computer era calculations of $\pi$ were performed.&lt;/p&gt;&#xA;" OwnerUserId="1301" LastActivityDate="2010-12-01T13:09:51.073" />
  <row Id="12619" PostTypeId="1" CreationDate="2010-12-01T13:14:59.247" Score="1" ViewCount="101" Body="&lt;p&gt;Hello!&#xA;I am looking for a website that contains a collection of functions, more than Matlab contains, to download I use from Matlab. &#xA;The extension should be *.m or the code should appear.&#xA;Thank you! &lt;/p&gt;&#xA;" OwnerUserId="4255" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-12-01T16:41:17.973" LastActivityDate="2010-12-01T16:41:17.973" Title="Matlab: Website with a lot of functions to download" Tags="&lt;functions&gt;&lt;math-software&gt;&lt;matlab&gt;" CommentCount="2" />
  <row Id="12620" PostTypeId="1" CreationDate="2010-12-01T13:21:00.520" Score="2" ViewCount="87" Body="&lt;p&gt;The &lt;a href=&quot;http://stackoverflow.com/questions/4324490/gaussian-distribution-hash-tables&quot;&gt;original question&lt;/a&gt; I posted on StackOverflow. I think it's more mathematically inclined so I posted it here again. In terms of math.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's a class of students. Each students has a score between 0 and 300 (gaussian, with some known average and standard deviation). I need to find classes of marks (like 0-100, 101-125, 126-140, 141-160, 161-175, 176-200, 201-300) such that:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&lt;li&gt;The number of classes is minimum&lt;/li&gt;&lt;li&gt;The number of students in each class is minimum&lt;/li&gt;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;How do I go about doing this? Is this a standard problem? Also, is it possible to prove that only this set of classes will have the above property.&lt;/p&gt;&#xA;" OwnerUserId="4285" LastActivityDate="2010-12-01T14:07:28.767" Title="Gaussian Distribution + Hash Tables" Tags="&lt;probability&gt;&lt;normal-distribution&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="12621" PostTypeId="1" AcceptedAnswerId="12624" CreationDate="2010-12-01T13:25:39.363" Score="2" ViewCount="94" Body="&lt;p&gt;Let us assume that we are given a closed, orientable 2D manifold embedded in $R^3$, and lets call it $M$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think it is clear that in a coordinate neighborhood $(U, \phi)$ it is possible to at each point define a unit normal vector $N_p$ by looking at the tangent space $T_pM$ as a subspace of $T_pR^3$ and considering the 1D vector space $T_pM^\perp$. However, there are always two choices of $N_p$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Intuitively, an orientation should should define $N_p$ uniquely, but I am not sure how to use that $M$ is oriented to do this. If we have an orientation 2-form $\Omega$ on $M$, how can we use this to figure out which of the two possible $N_p$'s is correct?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore, since we have $N_p$ defined at each point in $U$, we have a vector field on $U$. I was not sure how to show that this is a smooth vector field, but I believe that once you've done this, you can somehow use a partition of unity to extend $N$ to the entire manifold. But, I think that there is some subtlety even to this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I mean, for a non-orientable manifold such as a mobius strip, it seems to me that you can define locally a smooth unit-normal vector field, but when you try to extend it to the whole manifold it is no longer a continuous vector field, so somehow the orientation must play a role in the extension, and I am not sure how.&lt;/p&gt;&#xA;" OwnerUserId="1089" LastActivityDate="2010-12-01T14:45:04.013" Title="Unit Normal Field on a 2D manifold embedded in R^3" Tags="&lt;calculus&gt;&lt;topology&gt;&lt;differential-geometry&gt;" AnswerCount="1" />
  <row Id="12622" PostTypeId="2" ParentId="12620" CreationDate="2010-12-01T13:47:15.827" Score="1" ViewCount="" Body="&lt;p&gt;You question is a bit underdefined, but in general what you're looking for is index (order?) statistics. For example, if the number of classes is to be $2$, you need to cut them at the median. If it's $3$, you cut them at the $1/3$ and $2/3$ order statistics, and so on. The classes will be entirely evenly-spread.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The former paragraph assumed the classes can depend on the data. If the classes are not supposed to depend only on the distribution, then you just need to find the index statistics for the distribution. For example, the median of a Gaussian is equal to its mean, and for the other points you can consult a table (there is no closed form). You can then estimate how evenly spread the scores are going to be, e.g. the spread in the L2 spread is given by a $\chi^2$ distribution, whose mean you can look up in Wikipedia.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-12-01T13:47:15.827" />
  <row Id="12623" PostTypeId="2" ParentId="12620" CreationDate="2010-12-01T13:47:23.690" Score="1" ViewCount="" Body="&lt;p&gt;To have the minimum number of classes, take a class from 0 to 300.  To have the minimum number of students per class, take each different score to be a class.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More seriously, you generally need to define the number of classes you want, and whether you want all the classes to be about a given size.  You might want this when giving grades, say 50% A's, 40% B's and 10% C's.  For other applications you are just trying to group &quot;things that are close enough together that we can treat them the same.&quot;  If the distribution is a true Gaussian, with no noise, there is no nice answer.  But with a reasonably small sample, you can just plot a histogram, look for scores that are relatively rare, and make the cut there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In more than one dimension it gets harder.  You might look at &lt;a href=&quot;http://en.wikipedia.org/wiki/Cluster_analysis&quot; rel=&quot;nofollow&quot;&gt;cluster analysis&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-12-01T13:47:23.690" />
  <row Id="12624" PostTypeId="2" ParentId="12621" CreationDate="2010-12-01T14:04:51.550" Score="2" ViewCount="" Body="&lt;p&gt;There is a standard unit volume form $\omega = dx\wedge dy\wedge dz$&#xD;&#xA;on $\wedge R^3$. Contracting with repect to a unit vector $v$ gives&#xA;a 2-form $\omega&amp;#39;$. Pulling back to your surface if $v$ is normal&#xA;to $M$ at $p$ then $\omega&amp;#39;$ pulls back to a multiple of $\Omega$&#xD;&#xA;at $p$. If this is a positive multiple then $v$ and $\Omega$ are&#xA;compatibly oriented.&lt;/p&gt;&#xA;" OwnerUserId="226" LastEditorUserId="226" LastEditorDisplayName="" LastEditDate="2010-12-01T14:45:04.013" LastActivityDate="2010-12-01T14:45:04.013" CommentCount="0" />
  <row Id="12625" PostTypeId="2" ParentId="12620" CreationDate="2010-12-01T14:07:28.767" Score="2" ViewCount="" Body="&lt;p&gt;Statisticians would usually refer to your &quot;classes&quot; as &quot;bins&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your requirements for &quot;minimum number of bind&quot; and &quot;minimum size of the bins&quot; are in opposition, and you would have to clarify how the conflict should be resolved.  For instance, if one arrangement has 10 bins with 5 marks each, and another has 5 bins with 10 marks each, which is better?  Or, one arrangement has 10 bins with 10 marks each, and another has 10 bins total, 9 with 9 marks each, and the last one with 19 marks.  You have to specify a rule for how to identify which of any two arrangements is better.  Only then can you ask how to achieve the best one, and whether it is unique.&lt;/p&gt;&#xA;" OwnerUserId="822" LastActivityDate="2010-12-01T14:07:28.767" CommentCount="1" />
  <row Id="12626" PostTypeId="1" AcceptedAnswerId="12643" CreationDate="2010-12-01T14:12:55.940" Score="4" ViewCount="242" Body="&lt;p&gt;Let $L,K$ be number fields and $L|K$ a galois extension.&#xA;Let $(0)\neq Q$ a prime ideal in $\mathcal O_L$ (=ring of integers in $L$) and $P=Q \cap \mathcal O_K$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$Z_Q $ denotes the decomposition field of $Q$ and $T_Q$ denotes the inertia field of $Q$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now put $Q&amp;#39; :=Q\cap Z_Q$ and $Q&amp;#39;&amp;#39; :=Q\cap T_Q$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How does one prove, that $e(Q|Q&amp;#39;&amp;#39;)=e(Q|P)$ and $f(Q&amp;#39;&amp;#39;|Q&amp;#39;)=f(Q|P)$, if $e$ denotes the ramification index and $f$ the inertia degree?&lt;/p&gt;&#xA;" OwnerUserId="4287" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-01T17:09:36.673" LastActivityDate="2010-12-01T22:02:45.747" Title="Ramification index and inertia degree" Tags="&lt;algebraic-number-theory&gt;" AnswerCount="1" />
  <row Id="12627" PostTypeId="2" ParentId="12614" CreationDate="2010-12-01T14:18:38.163" Score="1" ViewCount="" Body="&lt;p&gt;As Willie Wong points out, this is not an issue for a parabolic mirror.  In fact, another way to think about optical focus is that it makes constructive interference at the image point.  This means that all light rays coming from the subject take the same travel time to reach the focus or else the interference wouldn't be constructive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that shutter speed doesn't matter unless you are thinking about exposing different parts of the sensor at different times (like the old 35mm curtain cameras) and the subject is moving.  Focus is focus for any shutter speed.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-12-01T14:18:38.163" CommentCount="6" />
  <row Id="12628" PostTypeId="2" ParentId="12610" CreationDate="2010-12-01T14:22:23.760" Score="1" ViewCount="" Body="&lt;p&gt;The area of A is positive, while the area of P is zero.  Did you mean to distinguish between the interior of the circle and the closure of that set (i. e. the union of the perimeter and the interior)?&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-12-01T14:22:23.760" />
  <row Id="12629" PostTypeId="1" AcceptedAnswerId="12632" CreationDate="2010-12-01T15:22:00.353" Score="10" ViewCount="325" Body="&lt;p&gt;Can you partition an infinite set, into an infinite number of infinite sets?&lt;/p&gt;&#xA;" OwnerUserId="3317" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-12-01T16:21:38.633" LastActivityDate="2011-01-09T06:16:48.997" Title="Partitioning an infinite set" Tags="&lt;elementary-set-theory&gt;&lt;infinity&gt;" AnswerCount="4" CommentCount="4" FavoriteCount="3" />
  <row Id="12630" PostTypeId="2" ParentId="12629" CreationDate="2010-12-01T15:27:18.297" Score="7" ViewCount="" Body="&lt;p&gt;Certainly.  Look at the &lt;a href=&quot;http://en.wikipedia.org/wiki/Cantor_pairing_function#Cantor_pairing_function&quot; rel=&quot;nofollow&quot;&gt;Cantor pairing function&lt;/a&gt; that shows a correspondence between the naturals and pairs of naturals.  You have $f:\mathbb{N}\mapsto \mathbb{N\times N}$ which  is a byjection.  All sets of the form (x,y) for a given x are infinite.  So the sets of n that go to (x,y) for a given x are infinite.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On second thought, this is too complicated.  $(x,y) \in \mathbb{N \times N}$ is infinite.  The sets of the form $(x,1)$ are infinite, as are $(x,2)$, as are $\ldots$.  So we have divided an infinite set into an infinite number of infinite sets.  The pairing function just projects this onto $\mathbb{N}$.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2010-12-01T15:37:21.730" LastActivityDate="2010-12-01T15:37:21.730" />
  <row Id="12631" PostTypeId="2" ParentId="12615" CreationDate="2010-12-01T15:33:12.300" Score="1" ViewCount="" Body="&lt;p&gt;See also &lt;a href=&quot;http://www.springer.com/mathematics/history+of+mathematics/book/978-0-387-20571-7&quot; rel=&quot;nofollow&quot;&gt;Pi: A Source Book&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="589" LastActivityDate="2010-12-01T15:33:12.300" />
  <row Id="12632" PostTypeId="2" ParentId="12629" CreationDate="2010-12-01T15:33:25.900" Score="15" ViewCount="" Body="&lt;p&gt;Yes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, note that it's enough to do it for countably infinite sets.  For if $X$ is uncountable and $Y$ is a countably infinite subset, then $Z = X-Y$ is also infinite.  Then, if we divide $Y$ into infintely many infinite sets, we've divided $X$ into infinitely many infinite sets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, let's consider a countable set $Y$.  Of course, we may as well label the elements of $Y$ as $0,1,2,...$.  In short, I'm going to break $\mathbb{N}$ into infinitely many infinite sets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First note that there are infinitely many prime numbers.  Further, if $p$ and $q$ are prime numbers, and if $p^a = q^b$, then we must have $p=q$ and $a=b$.  This follows from unique factorization of numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, for each prime $p$, consider the set $Y_p = \{ p^a$ with $a&amp;gt;0\in\mathbb{N}\}$.  Then, e.g., $Y_2 = \{2,4,8,16,32,64,128,...\}$ and $Y_3 =\{3,9,27,81,243,...\}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clearly, each $Y_p$ is infinite.  Further, if $Y_p$ and $Y_q$ have anything in common, then by what we said before, we must have that $p=q$.  In other words, for different primes $p$ and $q$, the sets $Y_p$ and $Y_q$ have no overlap.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since there are infinitely many primes, there are infinitely many $Y_p$ with no overlap.  Finally, let $Z$ be all the naturals which are NOT powers of a prime number.  $Z$ is also infinite since, for example, it contains $2*3, 2*3^2, 2*3^3, 2*3^4,...$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, we've dividied $\mathbb{N}$ into infinitely many infinite sets:  $Z$ and each of the $Y_p$.&lt;/p&gt;&#xA;" OwnerUserId="331" LastActivityDate="2010-12-01T15:33:25.900" />
  <row Id="12633" PostTypeId="2" ParentId="12629" CreationDate="2010-12-01T15:54:49.077" Score="6" ViewCount="" Body="&lt;p&gt;For another example with a somewhat different flavour, chop the real line up at integers, dividing it into countable union of unit intervals, each of which has uncountably many points:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \mathbb{R} = \bigcup_{n \in \mathbb{Z}} [n,n+1)$$&lt;/p&gt;&#xA;" OwnerUserId="2439" LastActivityDate="2010-12-01T15:54:49.077" />
  <row Id="12635" PostTypeId="1" AcceptedAnswerId="12658" CreationDate="2010-12-01T15:59:04.203" Score="5" ViewCount="103" Body="&lt;p&gt;Update: Steve points out in comments that many direct products of residually cyclic groups will be counterexamples.  Indeed, I should have spotted this: every finitely generated abelian group is residually cyclic!  I should have asked about &lt;strong&gt;fully&lt;/strong&gt; residually cyclic groups, ie groups G such that for any finite subset $X\subseteq G\smallsetminus 1$, there is a homomorphism from G to a cyclic group that doesn't kill any elements of $X$.  I suspect the question is now quite easy, though I haven't had a chance to think about it yet.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;This question is motivated by &lt;a href=&quot;http://math.stackexchange.com/questions/9934/groups-having-at-most-one-subgroup-of-any-given-finite-index&quot;&gt;this recent question&lt;/a&gt;, which asked for a characterisation of groups with at most one subgroup of each finite index.  Arturo Magidin's answer showed that every such finite group is cyclic, but the questioner, Louis Burkill, added in comments that he is really interested in the infinite case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In my answer to that question, I argue that a &lt;em&gt;finitely generated&lt;/em&gt; group has at most one subgroup of each finite index if and only if its canonical residually finite quotient, R(G), is cyclic. The `finitely generated' hypothesis is necessary - otherwise the additive group of the rationals provides a counterexample.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the general case, one can reduce from the case of G to R(G) as before (this gets rid of pathological examples like infinite simple groups), and the same argument shows that if R(G) has at most one subgroup of each finite index then R(G) is &lt;em&gt;fully&lt;/em&gt; residually cyclic, in particular abelian.  But the converse is not clear to me.  Hence the question in the title, which I'll reiterate here, with some extra hypotheses that should indicate where the difficulty lies.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If G is an infinitely generated, &lt;strong&gt;fully&lt;/strong&gt; residually cyclic (in particular, abelian) group, must G have at most one subgroup of each finite index?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2287" LastEditorUserId="2287" LastEditorDisplayName="" LastEditDate="2010-12-01T18:34:22.190" LastActivityDate="2010-12-01T23:28:26.750" Title="If G is fully residually cyclic, does G have at most one subgroup of each finite index?" Tags="&lt;group-theory&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="12636" PostTypeId="2" ParentId="12572" CreationDate="2010-12-01T16:27:58.560" Score="1" ViewCount="" Body="&lt;p&gt;There is such a formula, but $f$ and $g$ live on different domains: Let $f: t\mapsto f(t)$ be periodic with period $1$ and $g: k\mapsto g(k)$ be a function defined on $\mathbb Z$. Both these functions have Fourier transforms defined by&#xA;$$\hat f(k)=\int_0^1 f(t) e^{-2\pi i k t} dt, \ \ \ \hat g(t)=\sum_{k=-\infty}^\infty g(k)e^{-2\pi i k t}.$$ Put $h(t):= \overline{\hat g(t)}$; this is a function of period 1. Then $$h(t)= \sum_{k=-\infty}^\infty \overline{g(k)}e^{2\pi i k t},$$ so the Fourier coefficients of $h$ are given by $\hat h(k)=\overline{g(k)}$.  Therefore, by Parseval's theorem for Fourier series one has $$\int_0^1 f(t)\hat g(t)dt = \int_0^1 f(t)\overline{h(t)}dt=\sum_{k=-\infty}^\infty \hat f(k)\overline{\hat h(k)}=\sum_{k=-\infty}^\infty \hat f(k) g(k).$$&lt;/p&gt;&#xA;" OwnerUserId="1303" LastActivityDate="2010-12-01T16:27:58.560" CommentCount="0" />
  <row Id="12637" PostTypeId="1" CreationDate="2010-12-01T16:29:31.807" Score="4" ViewCount="190" Body="&lt;p&gt;How do we show the following?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $X$ be a topological space and let $x \in X$. Show that if $x$ has a countable neighborhood basis in $X$ then $x$ has a countable neighborhood basis in $\beta X$. Here $\beta X$ denotes the Stone-Cech compactification of $X$.&lt;/p&gt;&#xA;" OwnerUserId="4240" LastActivityDate="2011-02-10T17:06:32.033" Title="Stone-Cech compactification problem" Tags="&lt;topology&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="12638" PostTypeId="2" ParentId="12520" CreationDate="2010-12-01T16:47:46.237" Score="2" ViewCount="" Body="&lt;p&gt;The answers so far do not catch the fact that the floor function has indeed some built in periodicity: The function $f(x):=\lfloor x\rfloor - x$ is in fact periodic and so can be developped into a Fourier series. &quot;For all practical purposes&quot; one has&#xA;$$\lfloor x\rfloor =x-{1\over2} +{1\over\pi}\sum_{k=1}^\infty{\sin(2\pi k t)\over k},$$&#xD;&#xA;but here the right hand side for $t\in\mathbb Z$ does not produce the correct value.&lt;/p&gt;&#xA;" OwnerUserId="1303" LastActivityDate="2010-12-01T16:47:46.237" />
  <row Id="12639" PostTypeId="1" AcceptedAnswerId="12649" CreationDate="2010-12-01T17:05:13.137" Score="4" ViewCount="259" Body="&lt;p&gt;My understanding of integration-by-parts is a little shaky.  In particular, I'm not totally certain that I understand how to properly calculate the limits of integration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, the formula I have is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\int_{v_1}^{v_2}{u dv} = (u_2 v_2 - u_1 v_1) - \int_{u_1}^{u_2}{v du}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd like to see how to calculate $u_1$ and $u_2$, preferably in a complete example (that solves a definite integral.)  I'm really interested in an example where the limits of integration change; i.e. $u_1$ and $u_2$ are different than $v_1$ and $v_2$, if possible.&lt;/p&gt;&#xA;" OwnerUserId="2626" LastEditorUserId="2626" LastEditorDisplayName="" LastEditDate="2010-12-01T17:44:36.893" LastActivityDate="2010-12-01T18:54:32.033" Title="Understanding Limits of Integration in Integration-by-Parts" Tags="&lt;calculus&gt;&lt;integral&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="12640" PostTypeId="2" ParentId="12608" CreationDate="2010-12-01T17:07:43.413" Score="1" ViewCount="" Body="&lt;p&gt;Sets with a binary operation that is not assumed to be associative are called &lt;a href=&quot;http://en.wikipedia.org/wiki/Magma_%28algebra%29&quot; rel=&quot;nofollow&quot;&gt;Magmas&lt;/a&gt; (they are sometimes also called &quot;quasigroups&quot;, especially in older literature, but that term has a different meaning in category Theory). Your set $S$ with operation $\otimes$ is a magma. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the operation happens to be associative, then you have &lt;a href=&quot;http://en.wikipedia.org/wiki/Semigroup&quot; rel=&quot;nofollow&quot;&gt;semigroup&lt;/a&gt;, though that will depend, in this case, on the binary operation *. &lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-12-01T17:07:43.413" CommentCount="1" />
  <row Id="12641" PostTypeId="1" AcceptedAnswerId="12650" CreationDate="2010-12-01T17:26:03.100" Score="1" ViewCount="174" Body="&lt;p&gt;Can you show me how to prove this::&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$66^{11 k+66} + 11^{101 k+55}\equiv \left\{\begin{array}{ll}&#xA;22 \pmod{165}&amp;amp; \text{if $k$ is odd,}\\&#xA;77 \pmod{165} &amp;amp; \text{if $k$ is even.}&#xA;\end{array}\right.$$&lt;/p&gt;&#xA;" OwnerUserId="4290" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2010-12-01T17:55:23.933" LastActivityDate="2010-12-01T18:17:27.710" Title="hard problem of modular arithmetic" Tags="&lt;number-theory&gt;" AnswerCount="3" />
  <row Id="12642" PostTypeId="2" ParentId="12639" CreationDate="2010-12-01T17:31:22.747" Score="0" ViewCount="" Body="&lt;p&gt;Okay.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_1^2 \ln x \, dx = [x \ln x]_{x = 1}^2 - \int_1^2 1 \, dx = 2 \ln 2 - 1$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A prototype example. Where $u = \ln x$ and $v = x$.&lt;/p&gt;&#xA;" OwnerUserId="1120" LastActivityDate="2010-12-01T17:31:22.747" CommentCount="3" />
  <row Id="12643" PostTypeId="2" ParentId="12626" CreationDate="2010-12-01T17:42:30.580" Score="2" ViewCount="" Body="&lt;p&gt;This is essentially the argument in Daniel Marcus's &lt;em&gt;Number Fields&lt;/em&gt; (which I very highly recommend, especially for its exercises), Theorem 28 on page 100.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $G=\mathrm{Gal}(L/K)$ be the Galois group. The decomposition group of $Q$ is&#xA;$$D = \{ \sigma\in G\mid \sigma Q = Q\}$$&#xD;&#xA;and the inertia group of $Q$ is&#xA;$$E = \{\sigma\in G\mid \sigma(a)\equiv a\pmod{Q}\text{ for all $a\in \mathcal{O}_L$}\}.$$&#xA;Then $Z_Q$ is defined to be the fixed field of $D$, and $T_Q$ is the fixed field of $E$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As usual, let $[L:K]=n = efr$, where $e$ is the ramification degree and $f$ is the inertia degree.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, I claim that $[Z_Q:K]=r$. Indeed, by the Fundamental theorem of Galois Theory, $[Z_Q:K] = [G:D]$. If $\tau\in G$, then every element of the coset $\tau D$ maps $Q$ to $\tau Q$; moreover, if $\tau Q=\rho Q$, then $\rho^{-1}\tau \in T$. So we have a one-to-one correspondence between the cosets of $D$ in $G$, and the primes over $P$ of the form $\tau Q$ with $\tau\in G$.  Since $L$ is Galois over $K$, the action of $G$ is transitive on the primes lying above $P$, and there are $r$ of them. So the index $[G:D]$ equals $r$, hence $[Z_Q:K]=r$, as claimed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next, we show that $e(Q&amp;#39;|Q)=f(Q&amp;#39;|Q)=1$. Notice that $Q$ is the only prime of $\mathcal{O}_L$ that lies over $Q&amp;#39;$: because $\mathrm{Gal}(L/Z_Q)=D$ by the Fundamental Theorem of Galois Theory, and $D$ acts transitively on the primes of $\mathcal{O}_L$ lying over $Q&amp;#39;$; but every element of $D$ maps $Q$ to itself, so $Q$ is the only prime lying over $Q&amp;#39;$. Since $[L:Z_Q]=e(Q|Q&amp;#39;)f(Q|Q&amp;#39;)r(Q|Q&amp;#39;) = e(Q|Q&amp;#39;)f(Q|Q&amp;#39;)$, and since $erf=[L:K]=[L:Z_Q][Z_Q:K]=[L:Z_Q]r$, then $[L:Z_Q]=ef$. And since $e(Q|Q&amp;#39;)\leq e$ and $f(Q|Q&amp;#39;)\leq f$, it follows that we must have $e(Q|Q&amp;#39;)=e$, $f(Q|Q&amp;#39;)=f$. And since $e=e(Q|P) = e(Q|Q&amp;#39;)e(Q&amp;#39;|P)$, and $f=f(Q|P) = f(Q|Q&amp;#39;)f(Q&amp;#39;|P)$, then $e(Q&amp;#39;|P)=f(Q&amp;#39;|P)=1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, since $1 = e(Q&amp;#39;|P) = e(Q&amp;#39;|Q&amp;#39;&amp;#39;)e(Q&amp;#39;&amp;#39;|P)$, then $e(Q&amp;#39;|Q&amp;#39;&amp;#39;)=1$. Thus, $e(Q|Q&amp;#39;&amp;#39;) = e(Q|Q&amp;#39;)e(Q&amp;#39;|Q&amp;#39;&amp;#39;) = e = e(Q|P)$, which proves the first equality.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, we show that $f(Q|Q&amp;#39;&amp;#39;) = 1$, or equivalently that $\mathcal{O}_{T_Q}/Q&amp;#39;&amp;#39;$ is equal to $\mathcal{O}_L/Q$. For this it suffices to show the corresponding Galois group is trivial.  If we can establish this, then we will have &#xA;$f = f(Q|Q&amp;#39;) = f(Q|Q&amp;#39;&amp;#39;)f(Q&amp;#39;&amp;#39;|Q&amp;#39;) = f(Q&amp;#39;&amp;#39;|Q&amp;#39;)$, which will give the second equality you want.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To show that $\mathcal{O}_L/Q$ is the trivial extension of $\mathcal{O}_{T_Q}/Q&amp;#39;&amp;#39;$, it is enough to show that for every $\overline{a}\in\mathcal{O}_L/Q$, the polynomial $(x-\overline{a})^m$ lies in $\mathcal{O}_{T_Q}/Q&amp;#39;&amp;#39;[x]$ for some positive $m$. If this is the case, then every element of the Galois group must send $\overline{a}$ to itself. Pick any preimage $a\in \mathcal{O}_L$ of $\overline{a}$. Then the polynomial&#xA;$$\prod_{\sigma\in E}(x - \sigma(a))$$&#xD;&#xA;has coefficients in the fixed field of $E$, that is, in $T_Q$; reduce modulo $Q&amp;#39;&amp;#39;$ to get a polynomial with coefficients in $\mathcal{O}_{T_Q}/Q&amp;#39;&amp;#39;$; since $\sigma(a)\equiv a \pmod{Q}$ for all $\sigma\in E$, the reduced polynomial is of the form $(x-\overline{a})^m$, with $m=|E|$. This proves that every element of $\mathcal{O}_L/Q$ is fixed by every element of the Galois group, so the extension is trivial, hence the inertia degree is $1$. This proves the second equality, as outlined above.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-01T22:02:45.747" LastActivityDate="2010-12-01T22:02:45.747" CommentCount="2" />
  <row Id="12644" PostTypeId="1" AcceptedAnswerId="13587" CreationDate="2010-12-01T17:45:53.383" Score="8" ViewCount="227" Body="&lt;p&gt;A student came to me showing a question from his exam in basic group theory, in which they are asked to prove that $\mathbb{C}^*$ modulo the subgroup of roots of unity is isomorphic to $\mathbb{R}^+$ (in both cases we mean the multiplicative groups).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now this seems to me to be a simple error in the question. I believe they meant to ask to prove that $\mathbb{C}^*$ modulo &lt;strong&gt;all&lt;/strong&gt; the elements of absolute value 1 is isomorphic to $\mathbb{R}^+$ which is very easy to prove (take the homomorphism mapping $z$ to $|z|$ and use the first homomorphism theorem). However, I couldn't prove the claim about the roots of unity is wrong; is there an easy way to show this?&lt;/p&gt;&#xA;" OwnerUserId="1818" LastActivityDate="2010-12-09T00:35:28.517" Title="Is $\mathbb{C}^*$ modulo the roots of unity isomorphic to $\mathbb{R}^+$?" Tags="&lt;group-theory&gt;&lt;complex-numbers&gt;&lt;counter-examples&gt;" AnswerCount="1" CommentCount="7" FavoriteCount="1" />
  <row Id="12645" PostTypeId="1" AcceptedAnswerId="12654" CreationDate="2010-12-01T17:46:56.190" Score="1" ViewCount="109" Body="&lt;p&gt;Hello everyone,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm currently trying to do some Fourier transformations, or at least trying to understand them. The only thing I'm worried about is the complex part of the function. All I have is some basic, self thought, understanding about complex numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as I'm concerned, a complex number exists of a real part and a complex part. In the following DFT it seems it has no real part. Is that correct?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/EUcLT.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think we can simplify it to the following, since the others are just modifiers. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$e^{-i}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But for some reason, im not getting the good value's. Perhaps i'm doing something wrong, but im not quite sure. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I just realised this may be a bit vague, but my questions are&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Does the function F(k,l) return a complex number, with no real part.&lt;/li&gt;&#xA;&lt;li&gt;Is there a special rule about $e^{-i}$&#xD;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit2:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Okay, I'm sorry for asking these basic questions, this is all way beyond the math I've learned at school xD. But I understand most of the DFT now. Only there's one thing I don't understand and that:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;What does the $F(k,l)$ part of the function define. &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="3605" LastEditorUserId="3605" LastEditorDisplayName="" LastEditDate="2010-12-03T08:02:36.057" LastActivityDate="2010-12-03T08:02:36.057" Title="Getting the value of a Fourier Transform, problem with the complex part" Tags="&lt;fourier-analysis&gt;&lt;complex-numbers&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="12646" PostTypeId="1" AcceptedAnswerId="12681" CreationDate="2010-12-01T17:46:57.720" Score="3" ViewCount="293" Body="&lt;p&gt;Has anyone ever come across a good cheat sheet for a list of definitions for the various algebraic structures out there, i.e. groups, fields, rings etc. Every time I come across the name of some structure, I have to look it up on wikipedia just to be sure I'm thinking of the right one, figured it would be cool to print out a cheat sheet and hang it on a wall nearby. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The table on the &lt;a href=&quot;http://en.wikipedia.org/wiki/Algebraic_structure&quot; rel=&quot;nofollow&quot;&gt;algebraic structure&lt;/a&gt; article on wikipedia is almost what I want, however it's a bit cryptic and lacks some structures, e.g. that of a vector space or a module.&lt;/p&gt;&#xA;" OwnerUserId="4291" LastActivityDate="2010-12-02T04:59:23.770" Title="Algebraic structure cheat sheet anyone ?" Tags="&lt;abstract-algebra&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="4" />
  <row Id="12647" PostTypeId="1" AcceptedAnswerId="12653" CreationDate="2010-12-01T17:51:31.473" Score="1" ViewCount="104" Body="&lt;p&gt;Hello,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;suppose, $f(x,y)$ is a bounded continuous function on $\mathbb{R}^2$. Consider&#xA;$$\lim_{ y \rightarrow y&amp;#39; }\&amp;gt; \sup_{x \in \mathbb{R}} f(x,y).$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In how far can you switch suprema and limits, or which possible term comes closest to switching these two? Ideally, there would be something like&#xA;$$\sup_{x \in \mathbb{R}}\&amp;gt; \lim_{ y \rightarrow y&amp;#39; } f(x,y).$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Best&lt;/p&gt;&#xA;" OwnerUserId="3557" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-01T18:03:30.413" LastActivityDate="2010-12-01T19:37:22.027" Title="Switching Limits and Suprema" Tags="&lt;analysis&gt;" AnswerCount="1" />
  <row Id="12648" PostTypeId="1" AcceptedAnswerId="12699" CreationDate="2010-12-01T17:53:21.500" Score="4" ViewCount="427" Body="&lt;p&gt;I have an idea for solving the knapsack problem, but it looks too good to be true. I would like someone to explain potential problems with this approach. I'll give an example: I want to find a subset of {2,7,11} which has elements that sum to 13. Here is the algorithm for solving this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In binary notation 2=0010, 7=0111, 11=1011, 13=1101. Suppose that 2x+7y+11z=13, where x,y,z are {0,1}. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then y+z=1 (mod 2), since the last bits of 7, 11, and 13 are 1 and the last bit of 2 is 0. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And (y+z-1)+2*(x+y+z)=0 (mod 4), since the second to last bits of 2, 7, and 11 are 1 and the second to last bit of 13 is 0. (The y+z-1 is the carry bit from before.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And ((y+z-1)+2*(x+y+z)-0)+4y=4 (mod 8), since the third to last bits of 7 and 13 are 1 and the third to last bit of 2 and 11 are 0. (The (y+z-1)+2*(x+y+z)-0 is the carry bit from before.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And finally ((y+z-1)+2*(x+y+z)-0)+4y-4)+8z=8 (mod 16), since the first bits of 11 and 13 are 1 and the first bit of 2 and 7 are 0. (The ((y+z-1)+2*(x+y+z)-0)+4y-4 is the carry bit from before.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we have four linear equations over a finite ring, if we change each one to mod 16.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;8y+8z=8 (mod 16)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;8x+12y+12z=4 (mod 16)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4x+14y+6z=10 (mod 16)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2x+7y+11z=13 (mod 16)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As we can see, a solution is x=1, y=0, z=1, as one would expect, since the set {2,11} has elements that sum to 13. We could have gotten this through Gaussian elimination (according to answers to my last question on this site).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By request, here is the general algorithm that I just wrote (which can be induced from the above example): We want to solve a_1*x_1 + ... a_n*x_n = b, where each a_j and b is an integer and each x_j is in {0,1}. Putting each a_j and b in binary, we have a_j=(a_{mj} ... a_{1j}) and b=(b_m ... b_1), for suitable m. For instance, a_j=5 in binary is (a_{4j} ... a_{1j})=(0 1 0 1) - highest bit on the left side, lowest on the right side.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is the simple algorithm for constructing a new matrix A=(a_{ij}) from this original (a_{ij}) and a new b=(b_i) from this original b:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;`Let b_1 = 2^{m-1}*b_1;&#xA;For j=1 to n &#xA; Let a_{1j} = 2^{m-1}*a_{1j};&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For i=2 to m {&#xA; Let b_i = 2^{m-1}*b_i + b_{i-1}/2;&#xA; For j=1 to n&#xA;  Let a_{ij} = 2^{m-1}*a_{ij} + a_{(i-1),j}/2;&#xA;}`&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next solve the matrix equation Ax=b (mod 2^m), for the new A=(a_{ij}) and b=(b_i). If there is a solution to the knapsack problem, there should be a solution for x in {0,1} in Ax=b (mod 2^m) and vice versa (since the last row is just the original a_j and b). I'm curious to see if it works for large m and n. Anyone want to give it a shot?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What's wrong with this approach? It shouldn't work, but I don't know why.&lt;/p&gt;&#xA;" OwnerUserId="3163" LastEditorUserId="3163" LastEditorDisplayName="" LastEditDate="2010-12-02T02:10:03.117" LastActivityDate="2011-04-17T18:52:11.607" Title="knapsack algorithm that looks too good to be true" Tags="&lt;linear-algebra&gt;&lt;algorithms&gt;&lt;computer-science&gt;" AnswerCount="3" CommentCount="13" FavoriteCount="2" />
  <row Id="12649" PostTypeId="2" ParentId="12639" CreationDate="2010-12-01T17:53:22.890" Score="8" ViewCount="" Body="&lt;p&gt;A more precise notation is this one&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_{x_{1}}^{x_{2}}u(x)v^{\prime }(x)dx=\left(&#xD;&#xA;u(x_{2})v(x_{2})-u(x_{1})v(x_{2})\right) -\int_{x_{1}}^{x_{2}}u^{\prime&#xD;&#xA;}(x)v(x)dx$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is derived from the derivative rule for the product&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(u(x)v(x))^{\prime }=u^{\prime }(x)v(x)+u(x)v^{\prime }(x)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$u(x)v^{\prime }(x)=(u(x)v(x))^{\prime }-u^{\prime }(x)v(x).$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{eqnarray*}&#xD;&#xA;\int_{x_{1}}^{x_{2}}u(x)v^{\prime }(x)dx&#xD;&#xA;&amp;amp;=&amp;amp;\int_{x_{1}}^{x_{2}}(u(x)v(x))^{\prime }dx-\int_{x_{1}}^{x_{2}}u^{\prime&#xD;&#xA;}(x)v(x)dx \\&#xD;&#xA;&amp;amp;=&amp;amp;\left. (u(x)v(x))\right\vert&#xD;&#xA;_{x=x_{1}}^{x=x_{2}}-\int_{x_{1}}^{x_{2}}u(x)v(x)dx \\&#xD;&#xA;&amp;amp;=&amp;amp;\left( u(x_{2})v(x_{2})-u(x_{1})v(x_{2})\right)&#xD;&#xA;-\int_{x_{1}}^{x_{2}}u^{\prime }(x)v(x)dx.&#xD;&#xA;\end{eqnarray*}.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you write $dv=v^{\prime }(x)dx$ and $du=u^{\prime }(x)dx$, you get your&#xA;formula but with $u,v$ as a function of $x$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_{v_{1}(x)}^{v_{2}(x)}u(x)dv=\left(&#xD;&#xA;u(x_{2})v(x_{2})-u(x_{1})v(x_{2})\right)&#xD;&#xA;-\int_{u_{1}(x)}^{u_{2}(x)}v(x)du$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example: Assume you want to evaluate $\int_{x_{1}}^{x_{2}}\log&#xD;&#xA;xdx=\int_{x_{1}}^{x_{2}}1\cdot \log xdx$. You can choose $v^{\prime }(x)=1$&#xD;&#xA;and $u(x)=\log x$. Then $v(x)=x$ (omitting the constant of integration) and &#xA;$u^{\prime }(x)=\frac{1}{x}$. Hence&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{eqnarray*}&#xD;&#xA;\int_{x_{1}}^{x_{2}}\log xdx &amp;amp;=&amp;amp;\int_{x_{1}}^{x_{2}}1\cdot \log xdx \\&#xD;&#xA;&amp;amp;=&amp;amp;\left( \log x_{2}\cdot x_{2}-\log x_{1}\cdot x_{1}\right)&#xD;&#xA;-\int_{x_{1}}^{x_{2}}\frac{1}{x}\cdot xdx \\&#xD;&#xA;&amp;amp;=&amp;amp;\left( \log x_{2}\cdot x_{2}-\log x_{1}\cdot x_{1}\right)&#xD;&#xA;-\int_{x_{1}}^{x_{2}}dx \\&#xD;&#xA;&amp;amp;=&amp;amp;\left( \log x_{2}\cdot x_{2}-\log x_{1}\cdot x_{1}\right) -\left(&#xD;&#xA;x_{2}-x_{1}\right) &#xD;&#xA;\end{eqnarray*}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The same example with your formula:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$u=\log x,v=x,dv=dx,v=x,du=\frac{1}{x}dx$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$u_{2}=\log x_{2},u_{1}=\log x_{1},v_{2}=x_{2},v_{1}=x_{1}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{eqnarray*}&#xD;&#xA;\int_{v_{1}}^{v_{2}}udv &amp;amp;=&amp;amp;\left( u_{2}v_{2}-u_{1}v_{2}\right)&#xD;&#xA;-\int_{u_{1}}^{u_{2}}vdu \\&#xD;&#xA;\int_{x_{1}}^{x_{2}}\log xdx &amp;amp;=&amp;amp;\left( \log x_{2}\cdot x_{2}-\log x_{1}\cdot&#xD;&#xA;x_{1}\right) -\int_{\log x_{1}}^{\log x_{2}}xdu \\&#xD;&#xA;&amp;amp;=&amp;amp;\left( \log x_{2}\cdot x_{2}-\log x_{1}\cdot x_{1}\right)&#xD;&#xA;-\int_{x_{1}}^{x_{2}}x\cdot \frac{1}{x}dx \\&#xD;&#xA;&amp;amp;=&amp;amp;\left( \log x_{2}\cdot x_{2}-\log x_{1}\cdot x_{1}\right) -\left(&#xD;&#xA;x_{2}-x_{1}\right). &#xD;&#xA;\end{eqnarray*}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: The limits of integration, although different in terms of $u(x),v(x)$, when expressed in terms of the &lt;em&gt;same&lt;/em&gt; variable $x$ of functions $u(x),v(x)$ are the &lt;em&gt;same&lt;/em&gt; in both sides.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a strategy on how to chose the $u$ and $v$ terms see &lt;a href=&quot;http://math.stackexchange.com/questions/10146/intuition-behind-the-ilate-rule&quot;&gt;this question&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="752" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2010-12-01T18:54:32.033" LastActivityDate="2010-12-01T18:54:32.033" />
  <row Id="12650" PostTypeId="2" ParentId="12641" CreationDate="2010-12-01T17:53:28.990" Score="2" ViewCount="" Body="&lt;p&gt;It's  $\rm \equiv 0\ (mod\ 11)\:,\: $ and $\rm\: mod\ 15\ $ it's $\rm\ 6^n \equiv 6\ $ plus $\rm\ (-4)^n \equiv -4,\:1,-4\:, 1\:, \ldots\ $ equals $\rm\ 2, 7, 2, 7\:\ldots$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence it's $\rm\ 22,77,22,77\:\ldots\ (mod\ 11*15)$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-12-01T18:17:27.710" LastActivityDate="2010-12-01T18:17:27.710" />
  <row Id="12651" PostTypeId="2" ParentId="12641" CreationDate="2010-12-01T17:54:13.747" Score="7" ViewCount="" Body="&lt;p&gt;First, $66^2 \equiv 66 \pmod{165}$. Therefore $66^{11k+66} \equiv 66 \pmod{165}$. Second, $11^3 \equiv 11 \pmod{165}$. Therefore for even $n &amp;gt; 0$, $11^n \equiv 121 \pmod{165}$, whereas for odd $n &amp;gt; 0$, $11^n \equiv 11 \pmod{165}$. If $n=101k+55$ then $n$ is odd iff $k$ is even. Thus for odd $k$ we get $66+121 \equiv 22 \pmod{165}$, and for even $k$ we get $66+11 \equiv 77 \pmod{165}$.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-12-01T17:54:13.747" CommentCount="1" />
  <row Id="12652" PostTypeId="2" ParentId="12641" CreationDate="2010-12-01T17:54:38.723" Score="1" ViewCount="" Body="&lt;p&gt;Since $165=3\times 5\times 11$, work modulo each of $3$, $5$, and $11$ separately and then combine them using the Chinese Remainder Theorem.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-01T18:00:15.907" LastActivityDate="2010-12-01T18:00:15.907" />
  <row Id="12653" PostTypeId="2" ParentId="12647" CreationDate="2010-12-01T18:09:55.803" Score="3" ViewCount="" Body="&lt;p&gt;You cannot in general. To take a very simple example, consider $f(x,y) = \sin(xy)$, and $y&amp;#39;=0$. Then we have:&#xA;$$\lim_{y\to 0}\&amp;gt;\sup_{x\in\mathbb{R}}\sin(xy) = 1.$$&#xD;&#xA;(Remember that if we take the limit as $y\to 0$, then we do not consider $y=0$, so $y\neq 0$ in $\sin(xy)$). &#xA;But&#xA;$$\sup_{x\in\mathbb{R}}\&amp;gt;\lim_{y\to 0}\sin(xy) = \sup_{x\in\mathbb{R}}\ 0 = 0.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if the limit of $f(x,y)$ as $y\to y&amp;#39;$ exists for every $x$, and the limit of the suprema exist, then you get one inequality:&#xA;$$\sup_{x\in\mathbb{R}}\&amp;gt;\lim_{y\to y&amp;#39;}f(x,y) \leq \lim_{y\to y&amp;#39;}\&amp;gt;\sup_{x\in\mathbb{R}} f(x,y).$$&#xD;&#xA;To see this, note that for each fixed $x_0\in\mathbb{R}$, $f(x_0,y) \leq \sup\limits_{x\in\mathbb{R}}f(x,y)$, so taking limits we have&#xA;$$\lim_{y\to y&amp;#39;} f(x_0,y) \leq \lim_{y\to y&amp;#39;}\&amp;gt;\sup_{x\in\mathbb{R}}f(x,y).$$&#xD;&#xA;Since this holds for each $x_0$, the supremum also satisfies the inequality, so&#xA;$$\sup_{x\in\mathbb{R}}\&amp;gt;\lim_{y\to y&amp;#39;}f(x,y) \leq \lim_{y\to y&amp;#39;}\&amp;gt;\sup_{x\in\mathbb{R}}f(x,y).$$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-01T19:37:22.027" LastActivityDate="2010-12-01T19:37:22.027" CommentCount="10" />
  <row Id="12654" PostTypeId="2" ParentId="12645" CreationDate="2010-12-01T18:20:56.897" Score="4" ViewCount="" Body="&lt;p&gt;Your sum has both real and imaginary parts. Because&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$e^{ix}= \cos(x) + i \sin(x) \; .$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is $f(a,b)$ in your formula a real number? Then the transform has real and imaginary components in general. (For certain choices of $f(a,b)$, you can have purely real numbers.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you mean $e^{-i}$, then use the formula I provided you and you should be able to compute it.&lt;/p&gt;&#xA;" OwnerUserId="3567" LastActivityDate="2010-12-01T18:20:56.897" CommentCount="8" />
  <row Id="12655" PostTypeId="1" AcceptedAnswerId="12660" CreationDate="2010-12-01T19:13:38.683" Score="2" ViewCount="167" Body="&lt;p&gt;Just a quick question about the geometry of Hilbert spaces from an intuitive standpoint.  Maybe just assuming we're working with $L^2$ would simplify the situation. Basically, in something like $\mathbb{R}^2$ we have the situation that $\cos(\theta)=\frac{\langle a,b\rangle}{\vert a\vert\cdot\vert b\vert}$, and the idea of an angle between vectors is very meaningful, geometrically.  We can easily extend this idea to $\mathbb{R}^n$ because when we talk about the angle between two vectors, we mean we are choosing the plane that both of them lie in, and picking the vector in there.  But what does this really mean in a Hilbert space like $L^2$? I have a good intuition about functions, and about geometry (topology) separately, but not really the &quot;geometry of functions&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, there may be no visualization of this in $L^2$, and I'm not asking for one, but is there any sense to doing geometry (i.e. actual polygons, things like that) in a space like $L^2$?  Also, what sort of applications do ideas like this have in functional analysis?  Are we ever interested in ideas like &quot;planes&quot; of functions, polygons, surfaces, solids, etc.? What do we really mean by angles, projections, normal vectors?  And do these sorts of things ever have any sort of interesting relationships?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm primarily asking for an intuitive idea here.  It's easy to just do the math, prove theorems about inner products, norms, etc. Maybe geometry gives some clues or intuitive ideas when doing functional analysis?&lt;/p&gt;&#xA;" OwnerUserId="3595" LastActivityDate="2010-12-02T04:23:52.583" Title="Soft Question Hilbert Space Geometry" Tags="&lt;soft-question&gt;&lt;functional-analysis&gt;&lt;hilbert-spaces&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="2" />
  <row Id="12656" PostTypeId="1" CreationDate="2010-12-01T19:16:14.860" Score="1" ViewCount="102" Body="&lt;p&gt;First, let me start by saying, before you spend too much time on this, that it is still an open question, so I am just trying to gather some ideas and views on the problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $(\Omega ,\mathcal{A},\mathbb{P)}$ be a probability space and $(B_{t})_{t\geq 0}$ a &lt;em&gt;Brownian motion&lt;/em&gt; defined on this space and starting at $0$. Let us introduce the &lt;em&gt;local time&lt;/em&gt; at $0$, denoted by $(L_{t})_{t\geq 0}$.&#xA;A simple consequence of Tanaka formula and Levy characterisation of Brownian motion is that :&#xA;$$&#xD;&#xA;\tilde{B}_{t}=\left| B_{t}\right| -L_{t}=\int_{0}^{t}\mathrm{sgn}(B_{s})\mathrm{d}B_{s}&#xD;&#xA;$$&#xA;is a Brownian motion.  In other words, the &lt;em&gt;Levy transform&lt;/em&gt; : &#xA;\begin{align*}&#xA;T:\mathbb{W}&amp;amp; \longrightarrow \mathbb{W} \\&#xA;B&amp;amp; \longmapsto \tilde{B}&#xA;\end{align*}&#xA;defined almost everywhere on the Wiener space $\mathbb{W}=\mathcal{C}&#xD;&#xA;\mathbb{(R}_{+},\mathbb{R}\mathbb{)}$ leaves invariant the Wiener measure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $\mathcal{F}_{t}=\sigma (B_{s},s\leq t)$ denote the natural filtration of the Brownian motion $B$, completed by the &lt;em&gt;negligible&lt;/em&gt; $\sigma$-field $\mathcal{N}$, $\mathcal{F}_{t}^{(1)}=\sigma (\tilde{B}_{s},s\leq t)$ and similarly $\mathcal{F}_{t}^{(n)}$ the natural filtration of $T^{n}(B)$. Then one can wonder if the asymptotic filtration is trivial, i.e. if : &#xA;\begin{equation*}&#xA;\forall t\geq 0,\quad\bigcap_{n\geq 0}\mathcal{F}_{t}^{(n)}=\mathcal{N}.&#xA;\end{equation*}&#xA;In an &lt;em&gt;ergodic theory&lt;/em&gt; context this would mean that the transformation $T$ is &lt;em&gt;exact&lt;/em&gt; and would imply that $T$ is &lt;em&gt;ergodic&lt;/em&gt;. There is an indication that it is as the discrete case is. But the many attempts in the continuous case have been unsuccessful, although we are making steps in the right direction, for example the fact that orbits are dense is known.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See the book by Revuz and Yor, and articles by Dubins, Smorodinsky, Malriq, and others.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="2905" LastEditorUserId="6179" LastEditorDisplayName="" LastEditDate="2011-03-20T16:10:52.267" LastActivityDate="2011-03-20T16:10:52.267" Title="Is the Levy Transform Ergodic?" Tags="&lt;probability-theory&gt;&lt;measure-theory&gt;&lt;stochastic-processes&gt;" CommentCount="5" />
  <row Id="12657" PostTypeId="1" AcceptedAnswerId="12666" CreationDate="2010-12-01T19:18:21.753" Score="7" ViewCount="159" Body="&lt;p&gt;This is a problem that a professor proposed for the highschool mathematical olympiad in Costa Rica that we haven't been able to solve. Therefore it cannot be asked since we don't have a solution yet in general.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $\mathcal{F}_{k}$ for a fixed $k \in \mathbb{N}$   be the family of subsets $A_i \subset \mathbb{N}$ that satisfy the following conditions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) The cardinality of $A_i$ is $k$ for every  index $i$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) For every $A_i$ it holds that given any two different subsets of two elements $ \{ x, y \} \neq \{ z, w \}$, the absolute value of the differences between the elements of each subset are different&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ |x - y| \neq |z - w|$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we define a function $f: \mathcal{F}_k \rightarrow \mathbb{N}$ given by $$f(A_i) =  \max{A_i}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is to find the minimum of the image of $f$, that is to find&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\min{f(\mathcal{F}_k)}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, we know that for $k = 4$ the answer is $\min{f(\mathcal{F}_4)} = 7$ and for $k = 3 $ the answer is $\min{f(\mathcal{F}_3)} = 4$ but we don't have a general pattern for the solution, basically these were done by &quot;brute force&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We would very much appreciate any help with this problem. Thanks a lot in advance.&lt;/p&gt;&#xA;" OwnerUserId="900" LastEditorUserId="900" LastEditorDisplayName="" LastEditDate="2010-12-01T20:54:27.800" LastActivityDate="2010-12-02T12:08:20.180" Title="Finding the least element of the greatest elements of certain subsets of the natural numbers" Tags="&lt;number-theory&gt;&lt;combinatorics&gt;&lt;counting&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="1" />
  <row Id="12658" PostTypeId="2" ParentId="12635" CreationDate="2010-12-01T19:32:34.033" Score="2" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;The odds are against us.  Even &quot;fully&quot; is not enough.&lt;/strong&gt;&#xA;With the &quot;fully&quot; hypothesis, probably it is easy:  If G has two subgroups H,K of finite index n, then their intersection has finite index.  Take X to be a set of (non-identity) coset reps of G/H∩K, then there is a quotient where no element non-identity element of G/H∩K is sent to the identity &lt;s&gt;(and WLOG, G/H∩K is that quotient)&lt;/s&gt;, but since G is fully residually cyclic, that quotient is cyclic, and &lt;s&gt;the two subgroups must be identical by the lattice homomorphism theorem&lt;/s&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;s&gt;In other words, you seem to have exactly answered your own question!&lt;/s&gt; :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At least K4 is no longer a counterexample.&lt;/p&gt;&#xA;" OwnerUserId="583" LastEditorUserId="583" LastEditorDisplayName="" LastEditDate="2010-12-01T23:28:26.750" LastActivityDate="2010-12-01T23:28:26.750" CommentCount="6" />
  <row Id="12659" PostTypeId="2" ParentId="5819" CreationDate="2010-12-01T19:35:26.107" Score="1" ViewCount="" Body="&lt;p&gt;As an addition: the separating closed disjoint sets part is often called normality (a space is normal if it satisfies this), so $T_4$ is normal &lt;strong&gt;plus&lt;/strong&gt; $T_1$, and similarly for regular: $T_3$ is regular &lt;strong&gt;and&lt;/strong&gt; $T_1$. Spaces that have no disjoint non-empty closed sets (besides the trivial topology, we have examples like $\mathbf{N}$ with the topology generated by the sets of the form $U(n) = \{ k : k \ge n \}$ e.g.) trivially satisfy normality. $T_4$ is to avoid these pathologies: the extra $T_1$ ensures that at least all finite sets are closed, so we have some &quot;relevant&quot; closed sets to apply normality to...&lt;/p&gt;&#xA;" OwnerUserId="4280" LastActivityDate="2010-12-01T19:35:26.107" />
  <row Id="12660" PostTypeId="2" ParentId="12655" CreationDate="2010-12-01T19:45:09.070" Score="3" ViewCount="" Body="&lt;p&gt;I think in infinite-dimensional spaces one should interpret the angle as a measure of how correlated two functions are.  The intuition manifests itself most clearly when the functions are random variables of mean zero; then their inner product is precisely their covariance, which is zero when they are independent (that is, uncorrelated).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Certainly one of the most important thing about Hilbert spaces is that there is a good notion of projection, and things one would expect to intuitively be true about projection (for example, that the projection of a vector onto a subspace is the part of the subspace closest to the vector) are actually true because it turns out the inner product makes everything work.  So in that sense, one can sometimes reliably use finite-dimensional intuition in the infinite-dimensional case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps one can also draw intuition from quantum mechanics, where the inner product is a &quot;probability amplitude.&quot;  I don't know enough about quantum mechanics to really elaborate on this point, though.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-12-01T19:45:09.070" CommentCount="1" />
  <row Id="12661" PostTypeId="1" AcceptedAnswerId="12672" CreationDate="2010-12-01T20:10:21.987" Score="0" ViewCount="103" Body="&lt;p&gt;When ever I find definition of the quantum plane it says $A_q^2 = C&amp;lt;x,y&amp;gt;/I$, where $I = C&amp;lt;xy-qyx&amp;gt;$. What I want to know is whether they mean the unital free algebra or just the free algebra. In brief, is the quantum plane unital?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moreover, when people write $A_q^N$, they mean the free (unital) algebra with $N$ generaterators, where every generator just commutes with every other generator? &lt;/p&gt;&#xA;" OwnerUserId="2099" LastActivityDate="2010-12-01T21:37:22.393" Title="Definition of the Quantum Plane" Tags="&lt;algebra&gt;&lt;quantum-groups&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="12662" PostTypeId="1" AcceptedAnswerId="12663" CreationDate="2010-12-01T20:10:37.217" Score="4" ViewCount="152" Body="&lt;p&gt;Prove or disprove:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $\rho : \mathbb{N} \rightarrow \mathbb{N}$ injective. Let $(a_{n})_{n \in \mathbb{N}}$ be a sequence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(i) If $\displaystyle\sum\limits_{n=1}^\infty a_{n}$ absolutely converges then $\displaystyle\sum\limits_{n=1}^\infty a_{\rho(n)}$ also converges absolutely.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(ii) If $\displaystyle\sum\limits_{n=1}^\infty a_{n}$ converges then $\displaystyle\sum\limits_{n=1}^\infty a_{\rho(n)}$ also converges.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So my understanding is that a series converges if the infinite sum of the series is the limit? It converges absolutely if $\displaystyle\sum\limits_{n=1}^\infty |a_{n}|$ converges. The way I am reading the $a_{\rho(n)}$ 's is that they denote some sort of permutation or rearrangement of the original series. Aside from that though, I am absolutely lost when it comes to approaching this problem... In general am I looking for an $\epsilon &amp;gt;0$ which is greater than $a_{\rho(1)}+ \dots +a_{\rho(n)}$ ? If so, how do I begin trying to find it?&lt;/p&gt;&#xA;" OwnerUserId="3711" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-02T11:06:02.013" LastActivityDate="2010-12-02T19:48:05.987" Title="Proving or disproving statements related to convergent and absolutely convergent series" Tags="&lt;analysis&gt;&lt;sequences-and-series&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="1" />
  <row Id="12663" PostTypeId="2" ParentId="12662" CreationDate="2010-12-01T20:30:19.370" Score="7" ViewCount="" Body="&lt;p&gt;For (i):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose that $\sum a_n$ converges to $a \in \mathbb{R}$. So if $\epsilon &amp;gt; 0$ there exists N such that if $n, l &amp;gt; N$ and $s_n = a_1 + \ldots + a_n$ then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$|a - s_n| &amp;lt; \epsilon \text{    and    } \sum_{k = N + 1}^l |a_n| &amp;lt; \epsilon.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let $M \in \mathbb N$ such that the terms $a_1, \ldots, a_N$ are contained as sum elements in $t_m = a_{\rho(1)} + \ldots + a_{\rho(M)}$. So now we have that for $m \geq M$ that $t_m - s_n$ is a sum of finitely many terms $a_l$ with $l &amp;gt; N$. So, for some $l &amp;gt; N$ we have that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$|t_m - s_n| \leq \sum_{k = N + 1}^l |a_n| &amp;lt; \epsilon.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So for $m \geq M$ we have:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$|t_m - a| \leq |t_m - s_n| + |s_n - a| &amp;lt; \epsilon + \epsilon = 2\epsilon.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the rearrangement converges.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(ii) is not true. Let $\sum_n (-1)^n/n$ then take the $\rho$ which maps the integers injectively to the even integers.&lt;/p&gt;&#xA;" OwnerUserId="1120" LastActivityDate="2010-12-01T20:30:19.370" />
  <row Id="12664" PostTypeId="1" AcceptedAnswerId="12665" CreationDate="2010-12-01T20:31:42.973" Score="6" ViewCount="182" Body="&lt;p&gt;Let $G$ be a group and $G&amp;#39;$ its commutator subgroup. Let $\pi: G\to G/G&amp;#39;$ be the natural projection. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Statement 1: $G/G&amp;#39;$ is the largest Abelian quotient of $G$ in the sense that if $H\unlhd G$ and $G/H$ is Abelian, then $G&amp;#39;\le H$. Conversely, if $G&amp;#39;\le H$, then $H\unlhd G$ and $G/H$ is Abelian.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Statement 2: If $\varphi:G\to A$ is any homomorphism of $G$ into an Abelian group $A$, then $\varphi$ factors through $G&amp;#39;$; i.e., $G&amp;#39;\le \ker{\varphi}$ and there is a homomorphism $\hat{\varphi}:G/G&amp;#39;\to A$ such that $\varphi(g) = (\hat{\varphi}\circ \pi)(g)$. (That is, we have a fancy commutative diagram.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is from Dummit and Foote, p.169, Proposition 7. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The proof of (1) is very straightforward. However, the authors claim that (1) is a restatement of (2) in terms of homomorphisms. Can anyone explain this? Because it is not clear to me. Also, if I wanted to prove (2) outright, what should the map $\hat{\varphi}$ be? My first thought was defining it as $\hat{\varphi}(aG&amp;#39;)= \varphi(a)$, but I don't think this works.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="2313" LastActivityDate="2010-12-01T22:00:09.133" Title="These two group theory statements are &quot;the same&quot;?" Tags="&lt;abstract-algebra&gt;&lt;group-theory&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="12665" PostTypeId="2" ParentId="12664" CreationDate="2010-12-01T20:41:53.207" Score="6" ViewCount="" Body="&lt;p&gt;Your idea for (2) exactly works.  Any two representatives $a$, $b$ with $aG&amp;#39; = bG&amp;#39;$ will be related by an element $g&amp;#39; \in G&amp;#39;$ with $ag&amp;#39; = b$.  Then, because $G&amp;#39;$ is a subset of the kernel of $\varphi$, $$\varphi(b) = \varphi(ag&amp;#39;) = \varphi(a)\varphi(g&amp;#39;) = \varphi(a).$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This really is a restatement of the first isomorphism theorem for groups to a &quot;first homomorphism theorem for groups&quot;, so to speak.  The isomorphism theorem says that map $\theta: G \to A$ with kernel $K$ will factor through the quotient map $G \to G/K$ --- but really for any subgroup $N$ of $K$ with $N$ normal in $G$ will give a map $G \to G / N$ by the same recipe described above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your specific case, when the target is abelian, the kernel necessarily contains the commutator subgroup $G&amp;#39;$ of $G$, and so you use the &quot;first homomorphism theorem&quot; to get your map $\hat\varphi: G / G&amp;#39; \to A$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That the first isomorphism theorem can be weakened in this way means that, whereas the first isomorphism theorem alone relates surjective maps off $G$ to normal subgroups of $G$, the weakening says that the inclusions $N_1 \subseteq N_2$ of normal subgroups of $G$ is also reflected on the level of homomorphisms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope this helps!&lt;/p&gt;&#xA;" OwnerUserId="2574" LastActivityDate="2010-12-01T20:41:53.207" CommentCount="1" />
  <row Id="12666" PostTypeId="2" ParentId="12657" CreationDate="2010-12-01T20:43:53.347" Score="9" ViewCount="" Body="&lt;p&gt;Here are some more values, starting from $\mathcal{F}_1$: $$1,2,4,7,12,18,26,35,45,56,73,86,107,128,152,178,200,217,247,284,334,357,373,426,481,493,\ldots$$&#xD;&#xA;You're trying to find the shortest &lt;a href=&quot;http://en.wikipedia.org/wiki/Golomb_ruler&quot; rel=&quot;nofollow&quot;&gt;Golomb ruler&lt;/a&gt; of given size. I believe that the exact optimal values are unknown in general.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: This is &lt;a href=&quot;http://oeis.org/A003022&quot; rel=&quot;nofollow&quot;&gt;A003022&lt;/a&gt; (in which all numbers are $1$ less, e.g. $(0,)1,3,6,\ldots$).&lt;/p&gt;&#xA;" OwnerUserId="1277" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-02T12:08:20.180" LastActivityDate="2010-12-02T12:08:20.180" CommentCount="2" />
  <row Id="12667" PostTypeId="2" ParentId="12664" CreationDate="2010-12-01T20:44:21.803" Score="8" ViewCount="" Body="&lt;p&gt;By the Homomorphism Theorem, any homomorphism $f\colon G\to K$ factors through $G/\mathrm{ker}f$, meaning that there is a map $\hat{f}\colon G/\mathrm{ker}f \to K$ such that $ f = \hat{f}\pi$. The map is indeed $\hat{f}(g\,\mathrm{ker}f) = f(g)$. This applies to the specific case given in Statement 2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Edit:&lt;/em&gt; In fact, the full Statement 1 is not equivalent to Statement 2, in the sense that if you replace $G&amp;#39;$ with an arbitrary subgroup $M$ of $G$ in both statements, then Statement 1 characterizes $G&amp;#39;$, but Statement 2 does not. That is, if you have&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Statement 1': If $H\triangleleft G$ and $G/H$ is abelian, then $M\subseteq H$; and if $M\subseteq H$, then $H\triangleleft G$ and $G/H$ is abelian.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Statement 2': If $\varphi\colon G\to A$ is any homomorphism of $G$ into an abelian group $A$, then $\varphi$ factors through $M$; that is, $M\subseteq \ker\varphi$ and there is a homomorphism $\hat{\varphi}\colon G/M\to A$ such that $\varphi(g) = \hat{\varphi}\circ\pi(g)$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The only subgroup $M$ of $G$ that satisfies Statement 1' is $M=G&amp;#39;$. However, any subgroup of $G&amp;#39;$ that is normal in $G$ will satisfy Statement 2'. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, Statement 2 is equivalent to the first clause of Statement 1, namely that if $H\triangleleft G$ and $G/H$ is abelian, then $G&amp;#39;\subseteq H$, plus the implicit assertion that $G&amp;#39;$ itself is normal in $G$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming the first clause of Statement 1 plus the fact that $G&amp;#39;\triangleleft G$, if $\varphi\colon G\to A$ is a homomorphism, then by the Homomorphism Theorem, letting $H=\ker\varphi$, then $G/H$ is (isomorphic to) a subgroup of $A$, hence abelian, so we must have $G&amp;#39;\subseteq H = \mathrm{ker}\varphi$; this is Statement 2 (with the final clause of 2 given by the homomorphism theorem as above).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming Statement 2, (which implicitly asserts that $G&amp;#39;$ is normal) suppose that $H$ is a normal subgroup of $G$ such that $G/H$ is abelian. Then considering $\pi\colon G\to G/H$ and applying 2, you conclude that $G&amp;#39;\subseteq \mathrm{ker}\pi = H$. And normality of $G&amp;#39;$ follows from the statement of 2, which requires it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is, they aren't quite equivalent, because Statement 1 has &lt;em&gt;another&lt;/em&gt; clause, namely the &quot;Conversely...&quot; clause, which is not a consequence of assuming Statement 2. But the first part of Statement 1 (plus &quot;$G'\triangleleft G$&quot;) &lt;em&gt;is&lt;/em&gt; equivalent to Statement 2. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Added earlier:&lt;/em&gt; To see that the two are not quite equivalent as stated, let me give you an example of a subgroup $M$ of $G$ that satisfies Statement 2' but not Statement 1': consider the case of $G=S_4$; then $G&amp;#39; = A_4$. Now let $M = \{ 1, (12)(34), (13)(24), (14)(23)\}$. Then $M\triangleleft G$, and the statement in 2 holds for $M$: given any homomorphism $f\colon G\to A$ with $A$ abelian, the map $f$ factors through $G/M$ and there exists a homomorphism $\hat{f}\colon G/M\to A$ such that $f=\hat{f}\pi$. However, $M$ is not the commutator subgroup of $G$.  What is missing in Statement 2 for it to be a true equivalent of Statement 1 is some statement that corresponds to the assertion that $G/G&amp;#39;$ is &lt;em&gt;itself&lt;/em&gt; abelian, which is what follows from the &quot;Conversely...&quot; clause in Statement 1. One way to do it is to simply state that $G/G&amp;#39;$ is itself abelian. Another is to consider the intersection of all kernels of all homomorphisms into abelian groups, and say that $G&amp;#39;$ must be equal to that intersection.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-01T22:00:09.133" LastActivityDate="2010-12-01T22:00:09.133" CommentCount="2" />
  <row Id="12668" PostTypeId="2" ParentId="12648" CreationDate="2010-12-01T20:48:28.740" Score="1" ViewCount="" Body="&lt;p&gt;When you solve your equations mod 16, you get x, y, and z mod 16. Not mod 2. So all you have is integers x, y, and z between 0 and 15 inclusive such that 2x+7y+11z=13.&lt;/p&gt;&#xA;" OwnerUserId="1508" LastActivityDate="2010-12-01T20:48:28.740" />
  <row Id="12669" PostTypeId="2" ParentId="12662" CreationDate="2010-12-01T21:28:05.090" Score="6" ViewCount="" Body="&lt;p&gt;I just want to give an alternative proof for (i).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To say that $\rho : \mathbb{N} \rightarrow \mathbb{N}$ is injective means that for every $n \in \mathbb{N}$, $\rho(n) \in \mathbb{N}$, and if $n \neq m$, then $\rho(n) \neq \rho(m)$. Define $S^*_n = |a_1| + |a_2| + \cdots + |a_n|$, and $\tilde S^*_n = |a_{\rho(1)}| + |a_{\rho(2)}| + \cdots + |a_{\rho(n)}|$. Obviously, since $\rho$ is injective, $\tilde S^*_n \leq |a_1| + |a_2| + |a_3| + \cdots$. But the right-hand side exists as a finite nonnegative number (since $\displaystyle\sum\nolimits_{n=1}^\infty a_{n}$ absolutely converges, or, equivalently, $S^*_n$ converges). So, $\tilde S^*_n$ is a monotone increasing sequence, bounded from above by $\sum\nolimits_{n = 1}^\infty  {|a_n |}$. Hence, $\tilde S^*_n$ converges to a finite nonnegative number $\tilde S^* \leq \sum\nolimits_{n = 1}^\infty  {|a_n |}$. That is, $\displaystyle\sum\nolimits_{n=1}^\infty a_{\rho(n)}$ converges absolutely to $\tilde S^*$.&lt;/p&gt;&#xA;" OwnerUserId="2810" LastActivityDate="2010-12-01T21:28:05.090" />
  <row Id="12670" PostTypeId="1" AcceptedAnswerId="12673" CreationDate="2010-12-01T21:35:26.410" Score="2" ViewCount="135" Body="&lt;p&gt;Show that: $\sum _{k=n}^{\infty } \frac{1}{k!}\leq \frac{2}{n!}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am clueless here, I tried to multiply both sides with $n!$ but it doesn't make things better. I know that the left one converges against $e$ for $n=0$, but I better don't want to use its numerical value.&lt;/p&gt;&#xA;" OwnerUserId="3123" LastActivityDate="2010-12-01T21:42:59.960" Title="Series inequality" Tags="&lt;homework&gt;&lt;sequences-and-series&gt;&lt;inequality&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="12671" PostTypeId="2" ParentId="12670" CreationDate="2010-12-01T21:36:07.697" Score="4" ViewCount="" Body="&lt;p&gt;Estimate it using a geometric series.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-12-01T21:36:07.697" CommentCount="2" />
  <row Id="12672" PostTypeId="2" ParentId="12661" CreationDate="2010-12-01T21:37:22.393" Score="0" ViewCount="" Body="&lt;p&gt;Kassel, p. 3 ($k$ is the ground field):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;An algebra is a ring together with a ring map $\eta_A : k \to A$ whose image is contained in the center of $A$. ... A morphism of algebras or an algebra morphism_ is a ring map $f : A \to B$ such that $f \circ \eta_A = \eta_B.$ (1.1)  As a consequence of (1.1), $f$ preserves the &lt;strong&gt;units&lt;/strong&gt;, i.e., we have $f(1) = 1$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There is also a definition of free algebras on p. 7 where he explicitly states that they have a unit.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-12-01T21:37:22.393" CommentCount="4" />
  <row Id="12673" PostTypeId="2" ParentId="12670" CreationDate="2010-12-01T21:42:59.960" Score="5" ViewCount="" Body="&lt;p&gt;Hint:&#xA;$$\frac{1}{(n+1)!} + \frac{1}{(n+2)!} + \cdots $$&#xD;&#xA;$$= \frac{1}{n!} \left( \frac{1}{(n+1)} + \frac{1}{(n+1)(n+2)} + \cdots \right) $$&#xD;&#xA;$$ &amp;lt; \frac{1}{n!} \times \text{some geometric series}$$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="1301" LastActivityDate="2010-12-01T21:42:59.960" CommentCount="2" />
  <row Id="12674" PostTypeId="1" CreationDate="2010-12-01T21:58:29.103" Score="0" ViewCount="138" Body="&lt;p&gt;Say I have a numeric sequence A and a set of sequences B that vary with time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suspect that there is a relationship between one or more of the B sequences and sequence A,  that changes in Bn are largely or wholly caused by changes in sequence A.  However there is an unknown time delay between changes in A and their effect on each of the B sequences (they are each out of phase by varying amounts)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am looking for a means of finding the most closely correlating B to A regardless of the time delay.  What options are available to me? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;** EDIT **&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The crux of the problem here is that I have millions of B sequences to test, and there are approx 2 million data points within the lag window that I would like to test over.  Working out a correllation for each B for each possible lag scenario is just going to be too computationally expensive (especially as in reality there will be a more dynamic relationship than just lag between A and B, so I will be looking to test variations of relationships as well).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what I am looking for is a means of taking the lag out of calculation.&lt;/p&gt;&#xA;" LastActivityDate="2011-05-31T21:32:59.520" Title="Correlation between out of phase signals" Tags="&lt;analysis&gt;" AnswerCount="3" CommentCount="4" FavoriteCount="1" />
  <row Id="12675" PostTypeId="1" CreationDate="2010-12-01T22:06:59.783" Score="3" ViewCount="87" Body="&lt;p&gt;&lt;strong&gt;The context:&lt;/strong&gt;&#xA;I'm working on geocoded documents. This is, documents have  latitude and longitude attributes, as well as some other geo attributes such as an address. Right now, I'm performing text search operations on the indexed docs and showing their geolocation on a map. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The problem:&lt;/strong&gt;&#xA;Most of search criterias include a location expressed by a name. But this name may appear in the document in fields other than those that express location. This usally gives irrelevan results on the map. &#xA;In other cases, a doc within a resultset may be wrongly geocoded.&#xA;I both cases, the consequence is that the map appears with some relevant results geographically grouped and a few irrelevant results scattered far away.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The required solution:&lt;/strong&gt;&#xA;I'm stuck trying to find an algorithm that processes the latitude/longitude of each doc in the results to determine which points are grouped and discard those that are not grouped.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any ideas? Thanks in advance!&lt;/p&gt;&#xA;" LastActivityDate="2011-05-02T19:44:02.110" Title="Algorithm for discarding geographically disgregate points" Tags="&lt;algorithms&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="12676" PostTypeId="1" AcceptedAnswerId="12682" CreationDate="2010-12-01T22:13:15.593" Score="2" ViewCount="91" Body="&lt;p&gt;What exactly is &quot;Approximation Theory&quot;? If I read the wikipedia-article I doesn't get much clearer. Why are &quot;pure&quot; mathematicians interested in it? I see a lot of people that do harmonic analysis also do approximation theory.&lt;/p&gt;&#xA;" OwnerUserId="1120" LastEditorUserId="1120" LastEditorDisplayName="" LastEditDate="2010-12-01T22:27:58.087" LastActivityDate="2010-12-01T23:47:25.097" Title="Approximation Theory." Tags="&lt;reference-request&gt;&lt;approximation-theory&gt;" AnswerCount="1" />
  <row Id="12678" PostTypeId="2" ParentId="12591" CreationDate="2010-12-01T22:22:35.347" Score="1" ViewCount="" Body="&lt;p&gt;I think that 'Stochastic Differential Equations' by Bernt Oksendal is a good book about SDE.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According your equation, lets rewrite it:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$dx = ax\,dt +b\,dt$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and add noise&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$dx = (cx+b)\,dt + xg\,dt$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $g$ is a Wiener process, then $g\,dt$ is a Brownian motion, so your SDE equation is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$dX_t(\omega) = (cX_t(\omega)+b)\,dt + X_t(\omega)dB_t(\omega)$$&lt;/p&gt;&#xA;" OwnerUserId="2016" LastActivityDate="2010-12-01T22:22:35.347" CommentCount="4" />
  <row Id="12679" PostTypeId="2" ParentId="12674" CreationDate="2010-12-01T22:30:01.530" Score="1" ViewCount="" Body="&lt;p&gt;Will depend a bit on what kind of sequences you have, but assuming you are talking about discrete sequences, let's say $A=(A(t))_{t=-\infty}^{\infty}=(\ldots,A(-2),A(-1),A(0),A(1),A(2),\ldots)$ and the same for a sequence $B$. If your sequence doesn't run indefinitely, just put past values and future values equal to zero from a certain time on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then you can look at the following quantity:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ C(A,B;t_0,T,\tau)=\sum_{t=t_0}^{t_0+T} [A(t) - \bar{A}(t_0,T)][B(t-\tau)-\bar{B}(t_0-\tau,T)] $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \bar{A}(t_0,T) = \frac{1}{T+1} \sum_{t=t_0}^{t_0+T} A(t) $$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;is a moving average over a time window $T+1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The quantity $C(A,B;t_0,T,\tau)$ then measures correlation between signals $A$ and $B$. You can normalize it by dividing with the square roots of the autocorrelations of $A$ and $B$, that is $C(A,A;t_0,T,0)$ and $C(B,B;t_0-\tau,T,0)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me take a periodic sequence as an example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$A= (\ldots,0,1,2,0,1,2,0,1,2,0,1,2,\ldots)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the pattern $(0,1,2)$ repeats itself indefinitely.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The time average over a window of 6 time units is $\bar{A}=(0+1+2+0+1+2)/6=1$ and this regardless of where I chose to start the sum. (I chose 6 on purpose, this independence need not be generally true.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The autocorrelation of A with itself is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ C(A,A;0,6,\tau)=\sum_{t=0}^{6} [A(t) - 1][A(t-\tau)-1] $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $\tau=0$, this sum will be $$C(A,A;0,6,0)= [0-1]^2 + [1-1]^2 + [2-1]^2 + [0-1]^2 + [1-1]^2 + [2-1]^2 = 4 \; .$$ In fact, if $\tau$ is any multiple of 3, you should get the same response, since the function is periodic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $\tau=1$, this sum will be $$C(A,A;0,6,1)= 2\left([0-1][2-1] + [1-1][0-1] + [2-1][1-1]\right) = -2 \; .$$ Which in absolute value is smaller than the previous result, meaning that the correlation is less strong, plus since the sign is negative, it can be interpreted as being closer to anti-correlation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And that's how you'll be able to read patterns from these correlation functions.&lt;/p&gt;&#xA;" OwnerUserId="3567" LastEditorUserId="3567" LastEditorDisplayName="" LastEditDate="2010-12-01T23:53:39.087" LastActivityDate="2010-12-01T23:53:39.087" CommentCount="4" />
  <row Id="12680" PostTypeId="1" AcceptedAnswerId="12707" CreationDate="2010-12-01T22:40:52.470" Score="1" ViewCount="137" Body="&lt;p&gt;$S$ is a collection of disjoint sets. $(S,\cdot)$ is a commutative monoid and $(S,*)$ is a commutative semigroup. The identity element $e$ of $(S,\cdot)$ is the zero element of $(S,*)$.&#xA;The monoid $(S,\cdot)$ does not have a zero element. The binary operator $&amp;#39;*&amp;#39;$ is distributive over the binary operator $&amp;#39;\cdot&amp;#39;$. What can be said about such an algebraic structure ? What could be its usefulness ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a third binary operator $\otimes$ with which $(S,\otimes)$ is a commutative semigroup and the operator $&amp;#39;\otimes&amp;#39;$ is distributive over $&amp;#39;\cdot&amp;#39;$. $(S,\otimes)$ does not have a zero (absorption) element.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Motivation behind the question :&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I had something and wanted to check where it fits in a formalism and it happened to turn out like this. I have a naive question, why should i even bother about such a formalism like semigroup,semiring etc., how is it useful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS: I thought it is OK/appropriate to edit a question to add to ask more on the same subject. Please let me know if it is not OK.I can make it as another question.&lt;/p&gt;&#xA;" OwnerUserId="2987" LastEditorUserId="2987" LastEditorDisplayName="" LastEditDate="2010-12-02T16:52:20.410" LastActivityDate="2010-12-02T16:52:20.410" Title="On a set which is a monoid and a semigroup under two different binary operators with some properties." Tags="&lt;abstract-algebra&gt;&lt;group-theory&gt;" AnswerCount="1" CommentCount="6" />
  <row Id="12681" PostTypeId="2" ParentId="12646" CreationDate="2010-12-01T23:06:24.683" Score="4" ViewCount="" Body="&lt;p&gt;This might not be as comprehensive as you wish, but here is one. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.glump.net/content/abstract_algebra_cheat/index.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.glump.net/content/abstract_algebra_cheat/index.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are a few others here if you wish&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://math-blog.com/2008/09/20/13-useful-math-cheat-sheets/&quot; rel=&quot;nofollow&quot;&gt;http://math-blog.com/2008/09/20/13-useful-math-cheat-sheets/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I still reckon there is no substitute to doing it on your own. You can cater it to your needs and typeset it as beautifully as you want in tex&lt;/p&gt;&#xA;" OwnerUserId="3431" LastActivityDate="2010-12-01T23:06:24.683" CommentCount="1" />
  <row Id="12682" PostTypeId="2" ParentId="12676" CreationDate="2010-12-01T23:47:25.097" Score="5" ViewCount="" Body="&lt;p&gt;Approximation theory includes many subject areas of analysis, but the common idea is how well a target in a topological space (often a metric space) can be approximated by the points of a narrower subspace.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some examples will illustrate the breadth of this topic.  Given a real number x of a certain kind (e.g. algebraic), what are the rational numbers of bounded denominator that best approximate x, using the least absolute value of their difference as the objective?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given a real function $f$ on $[0,1]$ of a certain kind (e.g. twice continuously differentiable), what are the polynomials of bounded degree that best approximate $f$?  There are a variety of objectives that might be used, such as minimizing the square integral of the difference or minimizing the maximum difference.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Problems involving function approximation can be extended to higher dimensions, and the scope of approximating candidates can be varied endlessly (splines, trigonometric series, rational functions, etc.) and subject to many different restrictions (monotonicity, analyticity, symmetry, etc.).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a strong sense all of analysis uses approximation theory.&lt;/p&gt;&#xA;" OwnerUserId="3111" LastActivityDate="2010-12-01T23:47:25.097" CommentCount="2" />
  <row Id="12683" PostTypeId="1" CreationDate="2010-12-01T23:53:59.913" Score="0" ViewCount="53" Body="&lt;p&gt;I'm currently looking at the equation for a multiclass classifier of the form &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum\limits_{i=1}^{N_I} \sum\limits_{k=1,\atop k \neq y_i}^{N_K} L(1+ \mathbf{w_k}\cdot\mathbf{x_i}-\mathbf{w_{y_i}}\cdot\mathbf{x_i})$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where &lt;br/&gt;&#xA;$X$ is a $N_K \times N_F$ data matrix&lt;br/&gt;&#xA;$y$ is a vector of class labels  &lt;br/&gt;&#xA;$W$ is an $N_K \times N_I$ matrix where each corresponds to the weights for the hyperplane splitting one class from the rest &lt;br/&gt;&#xA;$L$ is some arbitrary loss function returning a real number eg square loss, absolute, hinge etc&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've been told that an answer for the above equation looks like&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\partial_{w_k} = \sum\limits_{i=1 \atop y_i \neq k}^{N_I} \ L&amp;#39;(1+ \mathbf{w_k}\cdot\mathbf{x_i}-\mathbf{w_{y_i}}\cdot\mathbf{x_i})\cdot\mathbf{x_i}&#xD;&#xA;&#xD;&#xA;-&#xD;&#xA;\sum\limits_{i=1 \atop y_i =k}^{N_I} \sum\limits_{l=1,\atop l \neq k}^{N_K} L&amp;#39;(1+ \mathbf{w_k}\cdot\mathbf{x_i}-\mathbf{w_{y_i}}\cdot\mathbf{x_i})\cdot\mathbf{x_i}$$&#xD;&#xA;Where $L&amp;#39;$ is the derivative of the loss function&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This answer doesn't make sense to me. Where does the second summation ($\sum\limits_{i=1 \atop y_i =k}^{N_I} \sum\limits_{l=1,\atop l \neq k}^{N_K} )$ come from? After all, in the initial equation, when $i$ is such that $y_i=k$, we skip over it. How then does this case have a large impact on the partial derivative? When I've attempted to do this myself, I've gotten the first part of the derivative but not the second.&lt;/p&gt;&#xA;" OwnerUserId="4286" LastEditorUserId="4286" LastEditorDisplayName="" LastEditDate="2010-12-03T01:04:17.290" LastActivityDate="2010-12-03T01:04:17.290" Title="Partial derivative of a multiclass linear classifier" Tags="&lt;calculus&gt;" CommentCount="2" />
  <row Id="12684" PostTypeId="1" CreationDate="2010-12-02T00:21:21.883" Score="0" ViewCount="103" Body="&lt;p&gt;To construct a multivariate B-spline, we simply take the Kronecker tensor product between the univariate basis functions constructed for each individual dimension.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I'd like to know is how do you construct a B-spline derivative for a multivariate function? Is it the same method of using the tensor product, but with the univariate B-spline derivatives? &lt;/p&gt;&#xA;" OwnerUserId="1588" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-02T00:45:18.820" LastActivityDate="2010-12-02T00:45:18.820" Title="Multivariate B-Spline Derivatives" Tags="&lt;approximation&gt;" CommentCount="5" />
  <row Id="12685" PostTypeId="2" ParentId="12646" CreationDate="2010-12-02T01:11:17.840" Score="3" ViewCount="" Body="&lt;p&gt;What might be an even better question is How can I remember all of the different common algebraic structures? How do I keep track of all of the &quot;different&quot; definitions?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I use quotes because most of the definitions are very similar.&#xA;It is important that we understand homomorphisms : whenever someone says homomorphism they mean it preserves all of the structure that it can! (unit, product, sum, everything in sight)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Firstly two rules that you always have whenever you could have them (for the most part) :&#xA;1) associative law : whenever there is an operation it is associative, this is a rule that only involves one operation. Again, almost every algebraic structure has every binary operation being associative... except for non-associative algebras.&#xA;2) distributive law : whenever there are two operations there is the distributive law. I am pretty sure this one is always there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most basic :&#xA;Monoids (some people call them different things, and there are more basic notions, but i dont think they are actively studied in algebra. In fact, I dont think monoids are actively studied, this definition is handy though for later purposes) : You have one operation and a unit identity element for that operation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Commutative monoid : like above but the operation is commutative.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Groups : a monoid with inverses FOR EVERYTHING!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Abelian Group : a commutative monoid with inverses FOR EVERYTHING!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From this point on everything is an abelian group plus some extra stuff.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Two operations :&#xA;these come in one of two flavors: another operation, or the action of something external on the guy you are looking at.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;another operation (usually we think of one of the opeerations as additive and the other as multiplicative, this is common and don't let it scare or worry you) :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rings with unit : these are monoid objects in the category of abelian groups... whatever that means. The right way to think about that last sentence is that there is some structure sitting on top of the abelian group structure, and they play nicely together (the distributive law! multiplication by a fixed element is a homomorphism of abelian groups). (after we understand the rules for rings with units we can think about rings without units... if we even want to worry about them now. Some people take ring to mean ring with unit, it really depends on the book)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Commutative Ring with unit : like a commutative monoid in the category of abelian groups right? yep... whatever that means. Mostly just look at the above. Looking a tthe commutative versions of things is usually an easy transition to make.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Division algebra : This is like a group object in the category of abelian groups except there can't be an inverse for zero, or rather the additive identity. So $R$ is a division algebra if $x \in R\setminus 0$ has an inverse for all $a \in R$. Note here $R$ must have a multiplicative identity. Essentially $R$ is a ring and $R\setminus 0$ is a group.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Field: Like a division algebra except commutative, so $R$ is a ring and $R\setminus 0$ is an abelian group.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;External operation : &lt;/p&gt;&#xA;&#xA;&lt;p&gt;All of these are Modules in one form or another. Always.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Module: an abelian group with an action of a ring on it so that the ring action satisfies some distributivity condition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Vector space: A module over a field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ideal: A submodule of $R$. Note that if $R$ is a ring than we can think of it as acting on itself via multiplication, this is the sense in which an ideal is a submodule.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Later I may come back and add examples if you like.&#xA;Let me know if I left something out that you want me to add. I strongly think the above intuitive definition is far better than memorizing the equations one needs. If you learn something about commutative diagrams then you oly need the diagrams, and there are only really 2 or 3 of those. Then you will see it really is always the same concept.&lt;/p&gt;&#xA;" OwnerUserId="627" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-02T01:16:03.197" LastActivityDate="2010-12-02T01:16:03.197" CommentCount="2" />
  <row Id="12686" PostTypeId="2" ParentId="12269" CreationDate="2010-12-02T01:52:29.923" Score="2" ViewCount="" Body="&lt;p&gt;As Aaron observes in the comments, this is in Strom's &quot;The homotopy category is a homotopy category.&quot;&lt;/p&gt;&#xA;" OwnerUserId="536" LastActivityDate="2010-12-02T01:52:29.923" CommunityOwnedDate="2010-12-02T01:52:29.923" />
  <row Id="12687" PostTypeId="6" CreationDate="2010-12-02T02:12:13.200" Score="0" ViewCount="0" Body="&lt;p&gt;I have been an active member of math.SE since the public beta, and before that I was an active member of MathOverflow; I have participated in meta on both sites.  While MathOverflow is a uniquely valuable resource to the research mathematics community, math.SE is a valuable resource to the wider community of those interested in mathematics, and for that reason I am committed to seeing it continue to grow and to run smoothly. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I check math.SE frequently, so it would be relatively easy for me to carry out janitorial duties such as closing spam questions.  I am particularly interested in merging tags; they are quite cluttered at the moment.  I won't take a particularly active role in closing questions other than those which obviously don't belong.  I just want to make it easier for interesting questions to be found and answered.  &lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-12-02T02:12:13.200" LastActivityDate="2010-12-02T02:12:13.200" CommunityOwnedDate="2010-12-02T02:12:13.200" CommentCount="6" />
  <row Id="12688" PostTypeId="2" ParentId="12003" CreationDate="2010-12-02T02:16:21.710" Score="1" ViewCount="" Body="&lt;p&gt;This is an interesting rhetorical question. On the one hand, many theoretical mathematical concepts transcend arithmetic. However, computer algebra systems (such as Mathematica) are able to express answers to those sorts of questions using a CPU that performs a limited range of simple computations. So the answer is a matter of opinion. Or there is no answer?&lt;/p&gt;&#xA;" LastActivityDate="2010-12-02T02:16:21.710" />
  <row Id="12689" PostTypeId="1" AcceptedAnswerId="12717" CreationDate="2010-12-02T02:32:32.990" Score="4" ViewCount="135" Body="&lt;p&gt;A little help here. Exercise 21, Ch. 2 from &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0471257087&quot; rel=&quot;nofollow&quot;&gt;Feller's book&lt;/a&gt; reads &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In a town a $n+1$ inhabitants, a person tells a rumor to a second person, who in turn repeats it to a third person, etc. At each step, the recipient of the rumor is chosen at random from the $n$ people available. Find the probability that the rumor will be told $r$ times without: a) returning to the originator, b) being repeated to any person. Do the same problem when at each step the rumor is told by one person to a gathering of $N$ randomly chosen people. (The first question is the special case N=1).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I already did a) and b) for the first description of the problem and a) for the case when the rumor is spreading through a gathering of $N$ people, however, my solution for b) in this second case is not correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I reasoned in the following way: In a first instance, $n$ people to receive the rumor, however, it's needed to spread such rumor through a group of $N$ people, therefore, there are $\displaystyle n \choose N$ ways to choose those gatherings. Once one of these people is chosen, he/she can choose from another gathering of $N$ people, taking care of not choosing someone who already know the rumor, which is, there are $\displaystyle n-1 \choose N$, and so on, until we reach the $r$ step in this process. Therefore, the probability I get is: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\displaystyle {n \choose N} {n-1 \choose N} {n-2 \choose N} ... {n-r+1 \choose N}}{\displaystyle {n \choose N}^{r}}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to the book, the solution must be $\displaystyle \frac{(n)_{Nr}}{(n_{N})^{r}}$ (which is not equivalent to the first expression).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I will appreciate any help.&lt;/p&gt;&#xA;" OwnerUserId="2569" LastEditorUserId="2569" LastEditorDisplayName="" LastEditDate="2010-12-02T03:51:20.727" LastActivityDate="2010-12-02T05:04:38.650" Title="Probability on spreading of rumors" Tags="&lt;probability&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="12690" PostTypeId="1" CreationDate="2010-12-02T02:50:17.483" Score="3" ViewCount="111" Body="&lt;p&gt;I understand Lie groups are defined by the structure constants associated with the lie brackets, which are treated as commutators in quantum mechanics, but i dont know of a math theory related to group theory to define or use an anti commutator. If Lie groups theory uses the commutator, what theory uses the anti commutator?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finite groups (not Lie groups, which are continuous), can be specified by structure equations analogous to a Lie bracket, but more general, of which commutation or anti commutation relations are just one of infinite possibilities. Is there such variety of possible structure equations in continuous groups too?&lt;/p&gt;&#xA;" LastActivityDate="2010-12-02T23:06:19.323" Title="What is the formal mathematical theory behind the concept of the anti commutator used to quantize fermions?" Tags="&lt;lie-groups&gt;" AnswerCount="2" />
  <row Id="12691" PostTypeId="2" ParentId="12488" CreationDate="2010-12-02T02:51:53.617" Score="3" ViewCount="" Body="&lt;p&gt;I agree with all the commenters who say that you should just find a planar embedding. However, I happened to stumble across a description that might make you happy:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Let $G$ be a &lt;a href=&quot;http://en.wikipedia.org/wiki/Connectivity_%28graph_theory%29&quot; rel=&quot;nofollow&quot;&gt;three-connected&lt;/a&gt;&#xA;  planar graph and let $C$ be a cycle.&#xA;  Let $G/C$ be the graph formed by&#xA;  contracting $C$ down to a point. Then&#xA;  $C$ is a face of the planar graph if&#xA;  and only if $G/C$ is &lt;a href=&quot;http://en.wikipedia.org/wiki/Connectivity_%28graph_theory%29&quot; rel=&quot;nofollow&quot;&gt;two-connected&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In particular, this lemma is useful in proving that a three-connected graph can only be planar in one way, a result of Whitney.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But testing this for every cycle is much less efficient than just finding the planar embedding.&lt;/p&gt;&#xA;" OwnerUserId="448" LastActivityDate="2010-12-02T02:51:53.617" />
  <row Id="12692" PostTypeId="2" ParentId="12690" CreationDate="2010-12-02T03:07:29.580" Score="5" ViewCount="" Body="&lt;p&gt;If Lie algebras are (in light of the Poincare-Birkhoff-Witt theorem) a complete axiomatization of the antisymmetric multiplication $AB-BA$ in an associative algebra, Jordan algebras are an almost-complete axiomatization of the symmetric multiplication $(AB+BA)/2$. ( &lt;a href=&quot;http://en.wikipedia.org/wiki/Jordan_algebra&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Jordan_algebra&lt;/a&gt; ).  They are the closest thing known to an intrinsic structure related to anticommutators.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unlike Lie algebras, there are exceptional Jordan algebras that do not come from the multiplication in associative algebras, and Jordan algebras are not known to be infinitesimal objects for groups or any other structure.  So they do not arise as often outside of (or in) quantum mechanics and the initial hopes for a Lie-like theory have not been realized.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you mean not the anti-commutator but the supercommutator, $AB \pm BA$ with the sign depending on parity of $A$ and $B$ or their consituents (e.g., negative for bosons and positive for fermions) there is a theory of Lie superalgebras and Lie supergroups. There are new phenomena in the classification, such as continuous families of nonisomorphic simple objects.  The theory can be phrased as the &quot;Lie group theory in the category of super-vector spaces&quot;, which in turn is a special case of Lie theory (or group theory, or algebraic geometry) in a tensor category.  So in principle there are many possible theories of commutator-like objects, but I don't know if any have been found to be interesting besides the usual theory, its super-version, and analogues in characteristic $p$.  &lt;/p&gt;&#xA;" OwnerUserId="467" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-12-02T19:20:50.783" LastActivityDate="2010-12-02T19:20:50.783" />
  <row Id="12693" PostTypeId="2" ParentId="12269" CreationDate="2010-12-02T03:18:40.370" Score="4" ViewCount="" Body="&lt;p&gt;This is in many places. I think that the whole point of model categories is that people saw you could get a lot of mileage out of the adjunction between Top and sSet (geometric realization and singular simplices). Just how good is the combinatorial model for topological spaces? The answer is that it is a Quillen equivalence (whatever that means). But for that sentence to exist we need to talk about model categories. I would say that Top is one of the most basic examples, before simplicial sets and chain complexes. For one thing the model structure on sSet requires use of the model structure on Top (the standard definition of a weak equivalence of simplicial sets is that its geometric realization is a weak equivalence in Top). The model structure on chain complexes sort of came out of homological algebra, and i don't know how early it was realized that this model category stuff shed light on the homological algebra that was going on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When thinking about this stuff I find the model theoretic approach helps me understand what is going on in top. Also, the weak equivalences are the maps that induce isos in homotopy for all choices of basepoints.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ps Peter May has a write up of a verification of the model category axioms for Top under Misc notes on his website. (There is tons of great stuff on there)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;pps Dwyer and Spalinski is a great resource as well &lt;a href=&quot;http://www.nd.edu/~wgd/&quot; rel=&quot;nofollow&quot;&gt;http://www.nd.edu/~wgd/&lt;/a&gt; number 75&lt;/p&gt;&#xA;" OwnerUserId="627" LastActivityDate="2010-12-02T03:18:40.370" CommentCount="2" />
  <row Id="12694" PostTypeId="1" CreationDate="2010-12-02T03:27:32.737" Score="1" ViewCount="178" Body="&lt;p&gt;So i've been stuck on this problem for about an hour. I can't figure out how to do it, and help would be absolutely amazing. This is what's given:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$0 + 1 = 1$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\forall x (x + 0 = x)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\forall x \forall y [x + (y+1)= (x+y)+1]$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$[0 + 1 = 1 + 0 \wedge \forall x(x + 1 = 1 + x \rightarrow (x+1)+1 = 1+(x+1)))] \rightarrow  \forall x (x+1=1+x)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And with these premises, I need to get $\forall x (x+1=1+x)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Help would be great. I keep getting closeish, only to find I'm actually doing it totally wrong.&lt;/p&gt;&#xA;" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-02T04:11:21.460" LastActivityDate="2010-12-02T17:29:35.757" Title="Formal logic to prove x+1 = 1 +x" Tags="&lt;logic&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="12695" PostTypeId="2" ParentId="10512" CreationDate="2010-12-02T03:32:04.433" Score="1" ViewCount="" Body="&lt;p&gt;This is really a comment on ryan'sanswer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have to disagree with Ryan. Group cohomology was in its early stages before Eilenberg and Maclane came along. There are awful and ugly formulations of just $H^1$ and $H^2$ that lead me to believe that they must have been formulated before E&amp;amp;M did their work. I am thinking of factor sets and cocycle conditions that people wrote down a long time ago. I think this is a matter of history though.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: I just checked wikipedia &lt;a href=&quot;http://en.wikipedia.org/wiki/Group_cohomology&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Group_cohomology&lt;/a&gt; the history section near the end. It seems to agree with what I said above.&lt;/p&gt;&#xA;" OwnerUserId="627" LastEditorUserId="627" LastEditorDisplayName="" LastEditDate="2010-12-02T03:49:56.953" LastActivityDate="2010-12-02T03:49:56.953" />
  <row Id="12696" PostTypeId="2" ParentId="12655" CreationDate="2010-12-02T03:43:26.097" Score="0" ViewCount="" Body="&lt;p&gt;Let us say we are dealing with real Hilbert space $X=L^2[0,1]$. Any two functions $f,g\in X$ are contained in the plane $P$ spanned by $f,g$. This plane is isometrically isomorphic to the Euclidean space $\mathbb{R}^2$; this means that $P$ is spanned by two orthonormal vectors $e_1, e_2$ (any two orthonormal vectors), and the linear map $T$ that maps $e_1$ to $(1,0)$ and $e_2$ to $(0,1)$  preserves distance and inner product (and hence also angle). The angle between $f$ and $g$ is precisely the angle between $Tf$ and $Tg$ in $\mathbb{R}^2$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $f, g$ are linearly independent, $e_1,e_2$ can be obtained from them by Gram-Schmidt process, i.e.&#xA;$$ e_1= \frac{f}{\|f\|},  e_2=\frac{g-(e_1,g)e_1}{\|g-(e_1,g)e_1\|}.$$&lt;/p&gt;&#xA;" OwnerUserId="3249" LastEditorUserId="3249" LastEditorDisplayName="" LastEditDate="2010-12-02T04:23:52.583" LastActivityDate="2010-12-02T04:23:52.583" />
  <row Id="12697" PostTypeId="2" ParentId="12648" CreationDate="2010-12-02T03:54:24.893" Score="2" ViewCount="" Body="&lt;p&gt;I'm not sure how one sees that finding solutions to the linear system over {0,1} can be accomplished in polynomial time.  Reducing the matrix to row-echelon form is not sufficient for finding such solutions.  Suppose that the reduced system has a parametrized solution set, as happens in your example.  Is there any way, short of plugging in 0 and 1 for the free parameters in all possible ways and then checking the values of the remaining variables, of finding {0,1} solutions?  Maybe, but I don't think this is immediately clear.  It seems possible that there might be many free parameters, and that there would then be a combinatorial explosion in the number of cases that needed to be checked in order to find {0,1} solutions.&lt;/p&gt;&#xA;" OwnerUserId="3736" LastActivityDate="2010-12-02T03:54:24.893" />
  <row Id="12698" PostTypeId="1" AcceptedAnswerId="12701" CreationDate="2010-12-02T04:01:40.807" Score="6" ViewCount="253" Body="&lt;p&gt;I want to why $\mathbb{Z}[x]/(1-x,p)$ is isomorphic to $\mathbb{Z_{p}}$ where p is a prime integer?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's what I have so far, but I am unsure if I am correct.   Every $f\in \mathbb{Z}[x]$ can be written as $(1-x)q+ r$ where $q\in \mathbb{Z}[x]$ and $r$ is in $\mathbb{Z}$.   Does It follows that there are $p$ cosets of $(1-x,p)$  ( namely 0+(1-x,p), 1+(1-x,p),2+(1-x,p)  etc ..) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That would imply $\mathbb{Z}[x]/(1-x,p)$ is isomorphic to $\mathbb{Z_{p}}$&lt;/p&gt;&#xA;" OwnerUserId="832" LastActivityDate="2010-12-02T17:49:30.520" Title="Why is $\mathbb{Z}[x]/(1-x,p)$ isomorphic to $\mathbb{Z_{p}}$ where p is a prime integer." Tags="&lt;abstract-algebra&gt;" AnswerCount="5" CommentCount="2" FavoriteCount="2" />
  <row Id="12699" PostTypeId="2" ParentId="12648" CreationDate="2010-12-02T04:04:41.053" Score="7" ViewCount="" Body="&lt;p&gt;The problem is that the equations generated are all linearly dependent on the original equation, so finding solutions of the matrix is just as hard as finding solutions to just the original equation. It works in the given example only because the matrix is so small that trial-and-error makes a solution apparent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To use your example, let $S = \{2,7,11\}$ and $N = 13$. We want to find a subset of $S$ whose elements sum to $N$. Looking at your generated solutions modulo $2^n$, notice that&#xA;$$2x + 7y + 11z \equiv 13 \pmod{16} \, \Rightarrow \, 4x + 14y + 6z \equiv 10 \pmod{16}$$&#xA;$$4x + 14y + 6z \equiv 10 \pmod{16} \, \Rightarrow \, 8x + 12y + 12z \equiv 4 \pmod{16}$$&#xA;$$8x + 12y + 12z \equiv 4 \pmod{16} \, \Rightarrow \, 8y + 8z \equiv 8 \pmod{16}$$&#xA;Since the reverse implications are not true in general, &quot;simplifying&quot; the equation in this manner does no good, as solving a simpler equation has no effect than to rule out possibilities (and as the problem scales it becomes infeasible to keep track of all remaining possibilities).&lt;/p&gt;&#xA;" OwnerUserId="1016" LastEditorUserId="1016" LastEditorDisplayName="" LastEditDate="2011-04-17T18:52:11.607" LastActivityDate="2011-04-17T18:52:11.607" CommentCount="1" />
  <row Id="12700" PostTypeId="2" ParentId="12698" CreationDate="2010-12-02T04:07:37.480" Score="0" ViewCount="" Body="&lt;p&gt;Remember that finite fields are isomorphic if and only if they have the same number of elements.&lt;/p&gt;&#xA;" OwnerUserId="3971" LastActivityDate="2010-12-02T04:07:37.480" CommentCount="1" />
  <row Id="12701" PostTypeId="2" ParentId="12698" CreationDate="2010-12-02T04:08:41.387" Score="6" ViewCount="" Body="&lt;p&gt;Basically, yes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is a bit easier if you consider it in steps, though. First, consider $\mathbb{Z}[x]/(p)$, which is isomorphic to $\mathbb{Z}_p[x]$; then consider $\mathbb{Z}_p[x]/(1-x)$, which is isomorphic to $\mathbb{Z}_p$ under the map &quot;evaluate at 1&quot; (or using the division algorithm, like you do above).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or you can do it the other way. First consider the quotient $\mathbb{Z}[x]/(x-1)$, which is isomorphic to $\mathbb{Z}$ under evaluation at $x=1$; then mod out by $(p)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(What you are doing if you go this route is using the homomorphism theorems; the ideal $(p + (x-1))$ of $\mathbb{Z}[x]/(x-1)$ corresponds to the ideal $(p,x-1)$ of $\mathbb{Z}[x]$, and $(R/I)/(J/I)\cong R/J$, where $I$ and $J$ are ideals of $R$ with $I\subseteq J$). &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-02T04:16:30.017" LastActivityDate="2010-12-02T04:16:30.017" />
  <row Id="12702" PostTypeId="2" ParentId="12698" CreationDate="2010-12-02T04:10:45.877" Score="2" ViewCount="" Body="&lt;p&gt;Define a map $\mathbb{Z}[x]\to \mathbb{Z}_p$ that sends $f(x)$ to the equivalence class of $f(1)$ modulo $p$. This is a homomorphism of rings. Since, every $f(x)$ can be written as $(x-1)q+r$ as you did, this map is surjective (just look at the images of the constant functions $0,1,...,p-1$), and the kernel is precisely $(x-1,p)$. This gives the required isomorphism by the first isomorphism theorem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Counting cosets only gives you an abstract isomorphism, while this gives you a canonical one.&lt;/p&gt;&#xA;" OwnerUserId="3431" LastActivityDate="2010-12-02T04:10:45.877" />
  <row Id="12703" PostTypeId="2" ParentId="12694" CreationDate="2010-12-02T04:12:53.720" Score="5" ViewCount="" Body="&lt;p&gt;The last axiom is basically your induction statement: it says that if you can prove that $x+1 = 1+x$ for $x=0$, and you can prove that if it holds for $x$ then it holds for $x+1$, then you can conclude that the statement $x+1=1+x$ holds for all $x$. So you want to use your first and second axioms, as well as your third axiom, to show that the premises of the last axiom hold. That will give the conclusion you want.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-12-02T04:12:53.720" />
  <row Id="12704" PostTypeId="1" AcceptedAnswerId="12712" CreationDate="2010-12-02T04:30:10.350" Score="3" ViewCount="86" Body="&lt;p&gt;Hello,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume I have linear equation system in $n$ variables and the solution is $x_1=x_2=\cdots = x_n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What properties must the matrix of the system have, or how do I prove that the system has such a solution?&lt;/p&gt;&#xA;" OwnerUserId="4311" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-30T06:47:33.773" LastActivityDate="2011-01-20T12:59:37.787" Title="matrix properties if all the solutions are the same" Tags="&lt;linear-algebra&gt;&lt;matrices&gt;" AnswerCount="2" />
  <row Id="12705" PostTypeId="1" AcceptedAnswerId="12710" CreationDate="2010-12-02T04:32:06.397" Score="3" ViewCount="74" Body="&lt;p&gt;What does $\mathbb{Z}[\omega]$ usually mean when $\omega$ is a primitive root of unity?&lt;/p&gt;&#xA;" OwnerUserId="832" LastActivityDate="2010-12-02T04:39:41.703" Title="Definition of $\mathbb{Z}[\omega]$ where $\omega$ is a primitive root of unity." Tags="&lt;abstract-algebra&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="12706" PostTypeId="2" ParentId="12698" CreationDate="2010-12-02T04:32:23.993" Score="2" ViewCount="" Body="&lt;p&gt;Just to add something to the nice answers you already have. In fact this is a special case of a more general proposition. It can be shown that all the maximal ideals $\mathfrak{m}$ of the polynomial ring $\mathbb{Z}[x]$ have the form $\mathfrak{m} = (g(x), p)$ where $p \in \mathbb{Z}$ is a prime number and $g(x) \in \mathbb{Z}[x]$ satisfies that its reduction modulo $p$ is an irreducible element $\bar{g}(x)$ in the polynomial ring $\mathbb{Z}_{p}[x]$. As a consequence of this you get that the quotient ring $ \mathbb{Z}[x] / \mathfrak{m} $, which in fact is a field since $\mathfrak{m}$ is maximal, is a finite algebraic extension field of $\mathbb{Z}_p$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can find this theorem for instance as &lt;strong&gt;proposition 1.5&lt;/strong&gt; in page 22 of Miles Reid's &lt;a href=&quot;http://books.google.co.cr/books?id=mUL1us0mRrAC&amp;amp;pg=PA22&amp;amp;dq=undergraduate+commutative+algebra+Worked+examples&amp;amp;hl=es&amp;amp;ei=6B_3TN6YIYKB8gbjnszjBQ&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=1&amp;amp;ved=0CCoQ6AEwAA#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;Undergraduate Commutative Algebra&lt;/a&gt;. Another place where you may want to look is &lt;strong&gt;Section II.4.3&lt;/strong&gt; in page 84 of Eisenbud and Harris' &lt;a href=&quot;http://books.google.co.cr/books?id=BpphspzsasEC&amp;amp;printsec=frontcover&amp;amp;dq=The+Geometry+of+Schemes&amp;amp;hl=es&amp;amp;ei=5az3TPLGEYG0lQf9nIyhAg&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=1&amp;amp;ved=0CCUQ6AEwAA#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;The Geometry of Schemes&lt;/a&gt;, where this is discussed more generally in the context of the prime spectrum $\text{Spec} \, \mathbb{Z}[x]$ and you can see a nice picture of it and how its &quot;points&quot; correspond to the points in $\text{Spec} \,  \mathbb{Z}$.&lt;/p&gt;&#xA;" OwnerUserId="900" LastEditorUserId="900" LastEditorDisplayName="" LastEditDate="2010-12-02T14:29:23.610" LastActivityDate="2010-12-02T14:29:23.610" />
  <row Id="12707" PostTypeId="2" ParentId="12680" CreationDate="2010-12-02T04:33:59.763" Score="1" ViewCount="" Body="&lt;p&gt;I think what you are describing is a &lt;a href=&quot;http://en.wikipedia.org/wiki/Semiring&quot; rel=&quot;nofollow&quot;&gt;semirings without identity&lt;/a&gt;: a semiring would be aset $R$ with two operations, $+$ and $\cdot$, such that $(R,+)$ is a commutative monoid, $(R,\cdot)$ is a commutative monoid, $\cdot$ distributes over $+$, and the neutral element relative to $+$ is a zero element relative to $\cdot$. The only difference between this and your structure is the existence of an identity for the operation *. Some authors allow semirings to not have a multiplicative identity, which would be exactly what you have.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-12-02T04:33:59.763" />
  <row Id="12708" PostTypeId="1" AcceptedAnswerId="12714" CreationDate="2010-12-02T04:34:21.510" Score="2" ViewCount="94" Body="&lt;p&gt;The idea of a read-only right-moving Turing Machine already exists [1].&#xA;Is there a relaxed version of this where an input-reading head only moves &#xA;right to read inputs as needed? There may be other heads doing the computation&#xA;on work tapes; the movement of these heads is unrestricted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course the right-moving head could simply copy the contents to another unrestricted TM tape and start that; I'm trying to study TM configurations where the TM works as much as possible before requesting another input character.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm assuming this machine is Turing complete; correct me if I'm wrong.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;References with your answer would be helpful. Thanks&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[1] &lt;a href=&quot;http://en.wikipedia.org/wiki/Read-only_right_moving_Turing_Machines&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Read-only_right_moving_Turing_Machines&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1884" LastEditorUserId="1884" LastEditorDisplayName="" LastEditDate="2010-12-02T05:16:04.220" LastActivityDate="2010-12-02T05:16:04.220" Title="Right-moving Turing Machines" Tags="&lt;computer-science&gt;" AnswerCount="1" />
  <row Id="12709" PostTypeId="2" ParentId="12694" CreationDate="2010-12-02T04:36:31.470" Score="1" ViewCount="" Body="&lt;p&gt;Step 1: Prove 0+1=1+0&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Apply premise #2 for x=1, and substitute into premise #1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Step 2: Prove Ax [x+1=1+x -&gt; x+1+1=1+(x+1)]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose k+1=1+k (premise)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Apply premise #3 for x=1, y=k. Substitute from last premise. Apply symmetry of = to get k+1+1=1+(k+1).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Generalize. Apply premise #4 to get required result.&lt;/p&gt;&#xA;" OwnerUserId="3515" LastEditorUserId="3515" LastEditorDisplayName="" LastEditDate="2010-12-02T17:29:35.757" LastActivityDate="2010-12-02T17:29:35.757" />
  <row Id="12710" PostTypeId="2" ParentId="12705" CreationDate="2010-12-02T04:36:42.240" Score="4" ViewCount="" Body="&lt;p&gt;$\mathbb{Z}[\omega]$ is the ring generated by $\mathbb{Z}$ and $\omega$ (inside, say $\mathbb{C}$). You can think of this as a ring whose elements are polynomials in $\omega$ with coefficients in $\mathbb{Z}$. Since, $\omega^n=1$ for the appropriate $n$, you need to only consider polynomials of degree less than $n$. Then the usual addition and multiplication of polynomials give you the ring structure.&lt;/p&gt;&#xA;" OwnerUserId="3431" LastActivityDate="2010-12-02T04:36:42.240" CommentCount="1" />
  <row Id="12711" PostTypeId="2" ParentId="12705" CreationDate="2010-12-02T04:39:41.703" Score="1" ViewCount="" Body="&lt;p&gt;When $\omega$ is any algebraic element of $\mathbb{C}$ satisfying a monic polynomial with integral coefficients, $\mathbb{Z}[\omega]$ denotes the ring $\{a_0+a_1\omega+\ldots+a_{n-1}\omega^{n-1}: a_i\in \mathbb{Z}, n=\text{deg }\omega\}$, where the degree of $\omega$ is the degree of the minimal polynomial of $\omega$ over $\mathbb{Q}$. When $\omega$ is a primitive $n$-th root of unity, this ring happens to be the ring of integers of the cyclotomic field $\mathbb{Q}(\mu_n)$, but for general algebraic $\omega$, $\mathbb{Z}[\omega]$ is a proper finite index subring of the ring of integers of $\mathbb{Q}[\omega]$.&lt;/p&gt;&#xA;" OwnerUserId="3212" LastActivityDate="2010-12-02T04:39:41.703" CommentCount="2" />
  <row Id="12712" PostTypeId="2" ParentId="12704" CreationDate="2010-12-02T04:49:24.830" Score="6" ViewCount="" Body="&lt;p&gt;I assume that you have an $n\times n$ linear system $Ax=0$, whose solutions are precisely $x_1=x_2=\cdots=x_n$. &#xA;This means that the null space is the linear span of the vector $e=(1,1,\cdots,1)$. So the row space is the orthogonal complement of $e$. That means the sum of each row of $A$ must be 0, and that $n-1$ of these rows must be linearly independent.&lt;/p&gt;&#xA;" OwnerUserId="3249" LastActivityDate="2010-12-02T04:49:24.830" />
  <row Id="12713" PostTypeId="2" ParentId="12698" CreationDate="2010-12-02T04:51:39.257" Score="1" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;LEMMA&lt;/strong&gt; $\rm\quad S\ :=\ R[x]/(x-a,b)\ \cong\ R/b\ \ $ for $\rm\ a,b\in R\ $ any ring. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof&lt;/strong&gt; $\ \ $ In $\rm\:S\ $ we have $\rm\ x = a\ $ hence $\rm\ f(x) = f(a)\:.\ $ Therefore the natural  map of $\rm R $ into $\rm S$ is onto, with kernel $\rm K \supset b\:R\:.\ $  If $\rm\ c\in K\ $ then $\rm\ c = (x-a)\ f(x) + b\ g(x)\ \Rightarrow\ c\in b\:R\ \ $  via $\ $ eval $\: $ at $\rm\ x = a\:.$ Therefore $\rm\ \  \ K = b\:R\ \ $ so $\rm\: \ S \cong R/K =  R/b\:.$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-12-02T17:49:30.520" LastActivityDate="2010-12-02T17:49:30.520" />
  <row Id="12714" PostTypeId="2" ParentId="12708" CreationDate="2010-12-02T04:53:48.047" Score="2" ViewCount="" Body="&lt;p&gt;You could copy the input to a work tape and go on from there.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-12-02T04:53:48.047" />
  <row Id="12715" PostTypeId="1" AcceptedAnswerId="12730" CreationDate="2010-12-02T04:55:20.510" Score="1" ViewCount="71" Body="&lt;p&gt;Suppose I have a function $f \in L^2(R^n)$ with compact support, and a mollifier $\phi(x) \in C^\infty(R^n)$ with compact support s.t. $\int \phi(x) dx = \int \phi_\epsilon(x) = 1$ and $lim_{\epsilon \to 0} \phi_\epsilon(x) = \delta(x)$ where $\phi_{\epsilon}(x) = \epsilon^{-n} \phi(x/\epsilon)$ and $\delta(x)$ denotes the dirac delta. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $f_\epsilon = f * \phi_\epsilon$, where $*$ denotes convolution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, is it possible to pass the limit of the integral of $f_\epsilon$ inside and say that  $\lim_{\epsilon \to 0} \int f * \phi_\epsilon = \int \lim_{\epsilon \to 0} f *\phi_\epsilon$? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And if it is not, what type of stronger condition would I need to be able to do that? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I thought about using the Dominated Convergence Theorem, but I wasn't sure how to come up with a dominating function that works for every $\epsilon$.   &lt;/p&gt;&#xA;" OwnerUserId="1736" LastActivityDate="2010-12-02T06:10:10.757" Title="Integrating mollifiers and their limits" Tags="&lt;real-analysis&gt;" AnswerCount="1" />
  <row Id="12716" PostTypeId="2" ParentId="12646" CreationDate="2010-12-02T04:59:23.770" Score="3" ViewCount="" Body="&lt;p&gt;Here's a different suggestion on how to remember all the different algebraic structures, one that doesn't require memorization of the list of axioms: think of a &quot;canonical&quot; example for each type, an example which doesn't fit into any of the stronger structures.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-12-02T04:59:23.770" />
  <row Id="12717" PostTypeId="2" ParentId="12689" CreationDate="2010-12-02T05:04:38.650" Score="4" ViewCount="" Body="&lt;p&gt;Liberalkid is right. Using his suggestion, you get $$\frac{\binom{n}{N}\binom{n-N}{N}\cdots\binom{n-(r-1)N}{N}}{\binom{n}{N}^r} = \frac{n_N (n-N)_N \cdots (n-(r-1)N)_N}{(n_N)^r} = \frac{n_{Nr}}{(n_N)^r}.$$ In the first step you cancel $N!$ from each side $r$ times.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-12-02T05:04:38.650" CommentCount="3" />
  <row Id="12718" PostTypeId="2" ParentId="12675" CreationDate="2010-12-02T05:09:43.920" Score="1" ViewCount="" Body="&lt;p&gt;You can look, for each point, at the maximum distance among the $k$ nearest neighbors, and throw the point away if the maximum is too big (perhaps, relative to some average score on the specific map). You can determine $k$ by hand.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the way, these points are known as &lt;i&gt;outliers&lt;/i&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-12-02T05:09:43.920" />
  <row Id="12719" PostTypeId="1" CreationDate="2010-12-02T05:15:48.240" Score="1" ViewCount="61" Body="&lt;p&gt;Earlier I had posted question about Borel cantelli Lemma. Thanks to Nate and Shai for your kind response. Please see my profile for the earlier questions. As you can see I do not understand the lemma well. Here are some confusions I have. I hope someone will be kind enough to help me understand the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it true that P ([An i.o.]∩ A) = P [( A ∩ An) i.o.]. If so, how do we prove this? or is it trivial? And assuming that we know that P[An i.o.] =1 , can we establish that the sum {P(A ∩ An)} diverges (meaning = infinity) ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks &lt;/p&gt;&#xA;" LastActivityDate="2010-12-02T06:09:28.087" Title="Borell Cantelli lemma question part 2" Tags="&lt;probability-theory&gt;" AnswerCount="1" />
  <row Id="12720" PostTypeId="1" AcceptedAnswerId="12721" CreationDate="2010-12-02T05:40:02.223" Score="1" ViewCount="130" Body="&lt;p&gt;I like to understand how things work. It's the only way I'll remember the rule. If anyone has a way of breaking this down into logical steps to get from one to the other... it would be greatly appreciated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;why is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$X^n\cdot X^m = X^{n+m}$ ?&lt;/p&gt;&#xA;" OwnerUserId="4313" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-02T05:42:25.683" LastActivityDate="2010-12-02T05:49:46.937" Title="why is $X^n\cdot X^m = X^{n+m}$" Tags="&lt;algebra&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="12721" PostTypeId="2" ParentId="12720" CreationDate="2010-12-02T05:43:19.613" Score="6" ViewCount="" Body="&lt;p&gt;$X^n$ is $X$ multiplied by itself $n$ times. $X^m$ is $X$ multiplied by itself $m$ times. When you multiply these together, you'll get $X$ multiplied by itself $n+m$ times, which is $X^{n+m}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's an example with $n=2$ and $m=3$: $$X^2 \cdot X^3 = (X \cdot X) \cdot (X \cdot X \cdot X) = X^5.$$&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-12-02T05:43:19.613" CommentCount="1" />
  <row Id="12722" PostTypeId="2" ParentId="12720" CreationDate="2010-12-02T05:49:46.937" Score="6" ViewCount="" Body="&lt;p&gt;If $n$ and $m$ are positive integers, then remember that $X^n$ is shorthand for &#xA;$$ \underbrace{X\cdot X\cdots X}_{\text{$n$ factors}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So $X^n\cdot X^m$ is:&#xA;$$X^n \cdot X^m = \underbrace{(X\cdot X\cdots X)}_{\text{$n$ factors}} \cdot \underbrace{(X\cdot X \cdots X)}_{\text{$m$ factors}} = \underbrace{X \cdot X\cdots X}_{\text{$n+m$ factors}} = X^{n+m}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For negative integers, remember that if $n\gt 0$, then $X^{-n} = \frac{1}{X^n}$. So we have that &#xA;$$X^{-n}\cdot X^{-m} = \frac{1}{X^n}\cdot \frac{1}{X^m} = \frac{1}{X^n\cdot X^m} = \frac{1}{X^{n+m}} = X^{-(n+m)} = X^{-n-m}.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From there, using the other exponent rules, you can also get that it works for exponents that are fractions; and then once you define arbitrary exponents, for any exponent.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-12-02T05:49:46.937" />
  <row Id="12723" PostTypeId="1" AcceptedAnswerId="12727" CreationDate="2010-12-02T05:53:21.107" Score="2" ViewCount="109" Body="&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/xLuHb.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For part b, I'm getting a little stuck. So I'm trying to show that if $(x,y,r)R(a,b,s)$ and $(a,b,s)R(x,y,r)$, then $(x,y,r)=(a,b,s)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And so $(x,y,r)R(a,b,s)$ implies $\sqrt{(x-a)^2+(y-b)^2} \leq s-r$&#xD;&#xA;and $(a,b,s)R(x,y,r)$ implies $\sqrt{(a-x)^2+(b-y)^2} \leq r-s$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After some algebra, here's what I get:&#xA;For the first one, &#xA;$x^2+y^2-s^2-2xa-2yb+2sr+b^2+a^2 \leq r^2$&#xD;&#xA;and for the second one, &#xA;$x^2+y^2-s^2-2ax-2by+2sr+b^2+a^2 \leq r^2$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and so they come out to be the same, but in an example from the notes, I think I should have came up with something like: $a \leq c$ and $c \leq a$, so $c=a$, and I didnt get anything of that form, so I'm wondering if my steps above are also correct. Have I proven anything, or if not, what have I done/assumed wrong?&lt;/p&gt;&#xA;" OwnerUserId="2985" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-02T05:56:03.483" LastActivityDate="2010-12-02T06:34:56.087" Title="Proving a relation is antisymmetric" Tags="&lt;relations&gt;" AnswerCount="3" CommentCount="3" />
  <row Id="12724" PostTypeId="2" ParentId="12723" CreationDate="2010-12-02T05:56:42.343" Score="3" ViewCount="" Body="&lt;p&gt;Your two square roots are the same, so  $r=s$ follows from the antisymmetry of $\leq$&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-12-02T05:56:42.343" />
  <row Id="12725" PostTypeId="2" ParentId="12723" CreationDate="2010-12-02T05:58:41.727" Score="2" ViewCount="" Body="&lt;p&gt;The relation is saying that the first circle is completely included inside the second circle. If two circles are completely included in one another, they must be the same circle.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2010-12-02T06:10:08.557" LastActivityDate="2010-12-02T06:10:08.557" CommentCount="5" />
  <row Id="12726" PostTypeId="2" ParentId="7966" CreationDate="2010-12-02T06:06:41.990" Score="3" ViewCount="" Body="&lt;p&gt;If you know a bit of algebraic topology Hatcher's book mentioned above is great. I thought it was a little slow going, so instead I prefer Mosher and Tangora, which is now a Dover book!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think there are some great exercises that really made me feel comfortable with Spectral sequences. The first Two SS I would look at are Bockstein and Serre. First look at the description/construction/definition of $sq^1$ in MT. It IS the first differential in the mod two bockstein SS. so now you can compute integral cohomology from mod two cohomology once you know how $sq^1$ acts (this also shed light on $sq^1$ for me). To do these computations you should adjoin some indeterminate and use adams grading. then $d_1$ goes up one and to the left one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now you can use the serre SS and various simple fibre sequences to compute the cohomology of all the complex and real projective spaces WITH their ring structure! (first compute it for the infinite cases then restrict). This is all spelled out in MT (maybe not the part about restricting to get the finite dimensional projective spaces). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then compute the cohomology of $\Omega S^n$ for every n, with ring structure!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then prove the Thom Isomorphism Thm with a relative form of the Serre SS.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, when you are looking at the bockstein SS draw one of boardman's unrolled exact couples and look at what is going on. And dont use their construction of the steenrod operations, there are better places imo.&lt;/p&gt;&#xA;" OwnerUserId="627" LastActivityDate="2010-12-02T06:06:41.990" CommentCount="2" />
  <row Id="12727" PostTypeId="2" ParentId="12723" CreationDate="2010-12-02T06:07:50.847" Score="3" ViewCount="" Body="&lt;p&gt;Part (a) would give you the intuition. But the first thing to notice here (even before you notice that the square roots have the same value) is that since the square root is always nonnegative, the first condition gives you that $s-r\geq 0$ and the second gives you that $r-s\geq 0$. Those two together tell you &lt;em&gt;what&lt;/em&gt; about $s$ and $r$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you have that, you know something about the square root; and since the quantity inside is a sum of squares, the only way that &lt;em&gt;something&lt;/em&gt; can happen is if both squares are -&lt;strong&gt;fill in the blank&lt;/strong&gt;-. Which tells you something about $x$, $y$, $a$, and $b$. &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-02T06:34:56.087" LastActivityDate="2010-12-02T06:34:56.087" CommentCount="11" />
  <row Id="12728" PostTypeId="1" CreationDate="2010-12-02T06:08:11.007" Score="7" ViewCount="149" Body="&lt;p&gt;Hello all this is my first math question, I hope this is the right place to post.&#xA;Anyway, this is basic stuff but I'm a bit confused here, here's the problem's legend:&#xA;&quot;Find the equation of the plane that passes through the intersection line ($l$) of the planes $2x-y-z=2, x+y-3z+4=0$ and has a distance of 2 from the origin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ok, So here's what I've done, the family of planes that pases through $l$ is $$2x-y-z-2+k(x+y-3z+4)=0$$ We work the equation so it looks more like a planes general form: $$Ax+By+Cz+D=0$$ giving us this:$$(2+k)x+(k-1)y+(-3k-1)z +2(2k-1)=$$&#xA;where$$A=(2+k)$$ $$B=(k-1)$$ $$C=(-1-3k)$$ $$D=2(2k-1)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem states that it's distance to the origin is 2 and the normal form of a plane($x\cos{\alpha}+y\cos{\beta}+z\cos{\gamma}-p=0$) where $p$ is the distance of the plane to the origin. We know that to transform a planes equation from the general form to the normal form we must divide each term by $$r=\pm \sqrt{A^2+B^2+C^2}$$&#xD;&#xA;So $p=\frac{D}{r}$ and we have this:$$2=\frac{2(2k-1)}{\pm \sqrt{(2+k)^2+(k-1)^2+(-1-3k)^2}}$$&#xA;And here is where my doubt is, what sign should $r$ have positive or negative, the following theorem says:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;li&gt;if $D \neq 0$, $r$ is opposite sign of $D$&lt;/li&gt;&#xA;&lt;li&gt;if $D=0$ and $C \neq 0$ $r$ is equal sign of $C$&lt;/li&gt;&#xA;&lt;li&gt;if $D=0$ and $C=0$ and $B \neq 0$ $r$ is equal sign of $B$&lt;/li&gt;&#xA;&lt;li&gt;if $D=0$ and $C=0$ and $B=0$ and $A \neq 0$ $r$ is equal sign of $A$&lt;/li&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since we don't know the value of k and $D(k)=2(2k-1)$ and to have $D$ we must find k but k is determined by the sign of D(what a mess hehe) how do we chose $r$'s sign?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any help is really appreciated.&lt;/p&gt;&#xA;" OwnerUserId="4312" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-02T10:57:55.980" LastActivityDate="2010-12-02T16:07:19.603" Title="Analytic Geometry Doubt" Tags="&lt;geometry&gt;&lt;homework&gt;" AnswerCount="2" CommentCount="5" />
  <row Id="12729" PostTypeId="2" ParentId="12719" CreationDate="2010-12-02T06:09:28.087" Score="2" ViewCount="" Body="&lt;p&gt;The events &quot;$A_n$ for infinitely many $n$; and $A$&quot; and &quot;$A_n \land A$ for infinitely many $n$&quot; are one and the same since $\land$ distributes over $\lor$: $$ A \land \bigwedge_{n \geq 0} \bigvee_{m \geq n} A_m \Leftrightarrow \bigwedge_{n \geq 0} A \land \bigvee_{m \geq n} A_m \Leftrightarrow \bigwedge_{n \geq 0} \bigvee_{m \geq n} A \land A_m.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for your second question, if $A$ never happens then $P(A \land A_n) = 0$.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-12-02T06:09:28.087" />
  <row Id="12730" PostTypeId="2" ParentId="12715" CreationDate="2010-12-02T06:10:10.757" Score="3" ViewCount="" Body="&lt;p&gt;Note that &#xA;$$\int f \ast \phi_{\epsilon}(x)\,dx= \int(\int f(x - y)\phi_{\epsilon}(y)\,dy)\,dx$$&#xD;&#xA;So by applying Fubini's theorem and integrating with respect to $y$ first, $\int f \ast \phi_{\epsilon}(x)\,dx = \int f \int \phi_{\epsilon} = \int f$. So to show your equality, it suffices to show that $\lim_{\epsilon \rightarrow 0} f \ast \phi_{\epsilon}(x) = f(x)$ for almost all $x$. It's actually true at all Lebesgue points of $f(x)$ which is a standard fact about these mollifiers. So this gives you what you want.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another way: Since your functions are compactly supported you can use that $\int |f -  f \ast \phi_{\epsilon}| \leq C (\int |f -  f \ast \phi_{\epsilon}|^2)^{1 \over 2}$. (This follows by Schwartz's inequality for example). Then you can use Plancherel's theorem on the right hand side and get that&#xA;$$\int |f -  f \ast \phi_{\epsilon}|^2 = \int |\hat{f} - \hat{\phi_{\epsilon}}\hat{f}|^2$$&#xD;&#xA;Since $\phi_{\epsilon}$ are mollifiers, $\hat{\phi_{\epsilon}}(x) = \hat{\phi}(\epsilon x)$ converges to 1 for all $x$ as $\epsilon$ goes to zero. So $\hat{f} - \hat{\phi_{\epsilon}}\hat{f}$ converges to zero. Since $|\hat{f} - \hat{\phi_{\epsilon}}\hat{f}|^2$ has the dominating function $|2\hat{f}|^2$, one can now apply the dominated convergence theorem to the right-hand side above and get that zero is the limit. As a result, the left-hand side must also converge to zero. So since $\int |f -  f \ast \phi_{\epsilon}| \leq C (\int |f -  f \ast \phi_{\epsilon}|^2)^{1 \over 2}$, one also has that $\int |f -  f \ast \phi_{\epsilon}|$ goes to zero. So the same is true for $|\int f - \int f \ast \phi_{\epsilon}| = |\int (f - f \ast \phi_{\epsilon})| \leq \int |f -  f \ast \phi_{\epsilon}|$, which is what you are looking for.&lt;/p&gt;&#xA;" OwnerUserId="3035" LastActivityDate="2010-12-02T06:10:10.757" CommentCount="1" />
  <row Id="12731" PostTypeId="1" CreationDate="2010-12-02T06:10:55.977" Score="10" ViewCount="495" Body="&lt;p&gt;Given a number of 3in squares and 2in squares, how many of each are needed to get a total area of 35 in^2?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Through quick trial and error (the method they wanted I believe) you find that you need 3 3in squares and 2 2in squares, but I got to thinking on how to solve this exactly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have 2 unknowns and the following info:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4x + 9y = 35&lt;/p&gt;&#xA;&#xA;&lt;p&gt;x &gt;= 0, y &gt;= 0, x and y are both integers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It also follows then that x &amp;lt;= 8 and y &amp;lt;= 3&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure how to use the inequalities or the integer only info to form a direct 2nd equation in order to solve the system of equations.  How would you do this without trial and error?&lt;/p&gt;&#xA;" OwnerUserId="4315" LastActivityDate="2010-12-02T07:07:45.390" Title="My Daughter's 4th grade math question got me thinking" Tags="&lt;algebra&gt;&lt;inequality&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="12732" PostTypeId="2" ParentId="12728" CreationDate="2010-12-02T06:23:04.383" Score="1" ViewCount="" Body="&lt;p&gt;Correct me if I'm wrong here, but if two planes are not co-planar (and these are not), then the intersection between the plane will form a single line.  That line will be whatever distance it happens to be from the origin.  I'm assuming that by distance from origin, they mean closest point from origin to the line.  But anyways, that distance from the line seems just extra.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then again, I'm probably missing the point here.  All I can think of when seeing this problem is forming normals for the 2 planes and forming the cross product of those vectors to get the intersecting vector.  Been doing computer graphics for too long :)&lt;/p&gt;&#xA;" OwnerUserId="4315" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2010-12-02T16:07:19.603" LastActivityDate="2010-12-02T16:07:19.603" CommentCount="1" />
  <row Id="12733" PostTypeId="2" ParentId="12731" CreationDate="2010-12-02T06:29:25.640" Score="10" ViewCount="" Body="&lt;p&gt;A quick way to see the answer is to convert both sides of the equation mod 4. So the left hand side is y  (mod 4) (b/c 4=0, 9=1 mod 4), and the right hand side is 3 (mod 4). So y=3 mod 4. Since $y\le 3$ as you observed, the only solution (if there is any ) is $y=3$. Then you check that $4x+27=35$ and hence $x=2$. &lt;/p&gt;&#xA;" OwnerUserId="3249" LastActivityDate="2010-12-02T06:29:25.640" CommentCount="4" />
  <row Id="12734" PostTypeId="1" AcceptedAnswerId="12737" CreationDate="2010-12-02T06:37:40.700" Score="2" ViewCount="235" Body="&lt;p&gt;I have read &lt;a href=&quot;http://dx.doi.org/10.1016%2F0020-0190%2879%2990002-4&quot; rel=&quot;nofollow&quot;&gt;this paper&lt;/a&gt; and the wikipedia summary &lt;a href=&quot;http://en.wikipedia.org/wiki/2-satisfiability&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; and even &lt;a href=&quot;http://docs.google.com/viewer?a=v&amp;amp;q=cache%3aDwWbM7Q3o1EJ%3awww.soe.ucsc.edu/classes/cmps132/Winter05/hw/hw8sols.pdf+2sat+in+p&amp;amp;hl=en&amp;amp;gl=us&amp;amp;pid=bl&amp;amp;srcid=ADGEESjTJ-300yEMXJh2u4othjtyh57czm-IkIBugnUlIZiWBNFLo82JfweSmUHniqQYVtU23zWxRQuuwXVM0XIwoAOPmuZpRqutv8YXt4vkB-RnJT777wC-4nuT1zSlquia3m0lZ7P5&amp;amp;sig=AHIEtbRZa564hezVFCgcggCp2rDREeEiGg&quot; rel=&quot;nofollow&quot;&gt;this homework problem&lt;/a&gt;, but I do not understand how 2SAT is in P. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is my concern: the idea is that we start by setting a random variable. Then we set any implied variables. Repeat, and if we eventually reach a contradiction, then the problem has no solution. For example, that homework solutions thing says:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Begin with an arbitrary vertex α such that (α, ¬α) is not an edge in the graph. [Fill out implied variables]. Repeating this process for every unassinged literal in I creates a&#xA;  satisfying assignment.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I do not think this is always works. Take $(x\vee y)\wedge(\bar{x}\vee\bar{x})$. Set x = true. Now the second clause doesn't work, and we've reached a contradiction. But this formula does have a solution: $\bar{x}\wedge y$. Yet if we look at the implication for the second clause, it is $x\implies x$, so it's not as though we didn't set an implied variable. There is no rule $x\implies\bar{x}$ either, so it doesn't contradict that constraint.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.google.com/url?sa=t&amp;amp;source=web&amp;amp;cd=5&amp;amp;ved=0CDAQFjAE&amp;amp;url=http%3A%2F%2Fwww.cs.tau.ac.il%2F~safra%2FComplexity%2F2SAT.ppt&amp;amp;rct=j&amp;amp;q=2sat%20in%20p&amp;amp;ei=zy73TJedO8P38AaqsemxBw&amp;amp;usg=AFQjCNGvmrsStd0iVzF5g8FEEvQkhIJRrw&amp;amp;cad=rja&quot; rel=&quot;nofollow&quot;&gt;This link&lt;/a&gt; says to &quot;rewind&quot; once if you reach a contradiction, but I don't see why it's not possible to reach contradictions any number of times even if the proposition is satisfiable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What am I missing here?&lt;/p&gt;&#xA;" OwnerUserId="2549" LastEditorUserId="2549" LastEditorDisplayName="" LastEditDate="2010-12-02T14:01:12.943" LastActivityDate="2010-12-02T14:01:12.943" Title="Help understanding the proof that 2SAT is in P" Tags="&lt;algorithms&gt;&lt;computer-science&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="12735" PostTypeId="6" CreationDate="2010-12-02T06:45:55.713" Score="0" ViewCount="0" Body="&lt;p&gt;I've been a user of math.SE and MO since the inception of both, as well as the&#xA;meta forums on each. I would take a relatively light approach to moderation:&#xA;that is, let the community as a whole close questions (except for obvious spam&#xA;or duplicates), and allow generally free discussions on meta. On the other hand, I'd be certainly be interested in &lt;em&gt;discussing&lt;/em&gt; policies (about, say, homework) with the community. I would of course be willing to intervene when discussions degenerate into flamewars, but on the whole, I would aim to facilitate a more open community.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think it a useful and worthwhile project to create an active mathematical Q/A&#xA;at a more general level than MathOverflow and would be happy to serve if elected. Outside of here, I've participated in various aspects of the internet mathematical community (e.g. blogs, MathOverflow, and for a brief while nLab); it appears to hold promise to me, as I have found involvement in it to be productive and enlightening. &lt;/p&gt;&#xA;" OwnerUserId="536" LastEditorUserId="536" LastEditorDisplayName="" LastEditDate="2010-12-02T06:57:02.543" LastActivityDate="2010-12-02T06:57:02.543" CommunityOwnedDate="2010-12-02T06:45:55.713" CommentCount="2" />
  <row Id="12736" PostTypeId="2" ParentId="12728" CreationDate="2010-12-02T06:48:02.467" Score="1" ViewCount="" Body="&lt;p&gt;First of all, your equation should be $$(k+1)x+(k-1)y-\cdots.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the distance formula is  $$2=\frac{|2(2k-1)|}{\cdots}$$ (no $\pm$). Square both sides (note that square of $|\text{anything}|$ is $\text{anything}^2$, then solve a quadratic equation. You will get two $k$'s. Substitute back to see which one is correct. In this case, it turns out that both $k$'s are correct.&lt;/p&gt;&#xA;" OwnerUserId="3249" LastEditorUserId="3249" LastEditorDisplayName="" LastEditDate="2010-12-02T16:00:01.310" LastActivityDate="2010-12-02T16:00:01.310" CommentCount="2" />
  <row Id="12737" PostTypeId="2" ParentId="12734" CreationDate="2010-12-02T06:50:50.143" Score="3" ViewCount="" Body="&lt;p&gt;The graph in the paragraph you're quoting is the transitive closure of the original graph. Your algorithm can be adapted as follows: choose an arbitrary literal $\alpha$ and make all the necessary assignments; if you reach $\lnot \alpha$ (or reach both $\beta$ and $\lnot \beta$ for some other variable $\beta$), then $\alpha$ must be false, so try again from $\lnot \alpha$; if you reach $\alpha$ (or reach both $\beta$ and $\lnot \beta$ for some other variable $\beta$), then the instance is unsatisfiable. Continue until you set the values of all variables.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-12-02T06:50:50.143" CommentCount="5" />
  <row Id="12738" PostTypeId="2" ParentId="154" CreationDate="2010-12-02T06:55:08.583" Score="1" ViewCount="" Body="&lt;p&gt;I'm of the idea that complex numbres exist in the same way real number exists, we can be ware of real number every time because they're involved in our daily operations. Lets talk about &lt;strong&gt;phasors&lt;/strong&gt;, they aren't more that convenient ways to express &lt;em&gt;senoidal equations&lt;/em&gt;, based on the &lt;strong&gt;Eulers Identity&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;e^jT = Cos(T) + jSin(T) &amp;lt;- Eulers identity&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because this is easir to express and solve &lt;em&gt;differential equations&lt;/em&gt; that involves senoidal functions in terms of exponential notation, then they become the prefered manner for &lt;em&gt;analysis of electrical circuits of alternating current (CA)&lt;/em&gt;. Some quantities sucha as impedance (Z=V/I) where V,I are phasors, have no sense without the complex numbers. In this application, physically the complex part is associated with &lt;em&gt;reactive elements&lt;/em&gt; (inductors and capacitors) while the real part is associated with &lt;em&gt;resistive elements&lt;/em&gt; (resistors). Also other kinds of transformation such as Laplace transformation (for solving linear differential equations) are based on complex numbers. I have had some &lt;a href=&quot;http://www.monografias.com/trabajos-pdf/principios-fisicos-matematicos-analisis-sistemas/principios-fisicos-matematicos-analisis-sistemas.pdf&quot; rel=&quot;nofollow&quot;&gt;work&lt;/a&gt; about these kind of &lt;em&gt;notations and transformations&lt;/em&gt; and their application in Electrical Enginnering.&lt;/p&gt;&#xA;" OwnerUserId="4314" LastActivityDate="2010-12-02T06:55:08.583" CommunityOwnedDate="2011-04-15T20:45:29.080" />
  <row Id="12739" PostTypeId="2" ParentId="12731" CreationDate="2010-12-02T07:00:42.280" Score="7" ViewCount="" Body="&lt;p&gt;There is an algorithmic way to solve this which works when you have two types of squares.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;if $\displaystyle \text{gcd}(a,b) = 1$, then for any integer $c$ the linear diophantine equation $\displaystyle ax + by = c$ has an infinite number of solution, with integer $\displaystyle x,y$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact if $\displaystyle x_0, y_0$ are such that $\displaystyle  a x_0 - b y_0 = 1$, then all the solutions of $\displaystyle ax + by = c$ are given by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle x = -tb + cx_0$, $\displaystyle y = ta - cy_0$, where $\displaystyle t$ is an arbitrary integer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle x_0 , y_0$ can be found using the &lt;a href=&quot;http://en.wikipedia.org/wiki/Extended_Euclidean_algorithm&quot; rel=&quot;nofollow&quot;&gt;Extended Euclidean Algorithm&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since you also need $\displaystyle x \ge 0$ and $\displaystyle y \ge 0$ you must pick a $\displaystyle t$ such that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle c x_0 \ge tb$ and $ta \ge cy_0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there is no such $\displaystyle t$, then you do not have a solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your case, $\displaystyle a= 9, b= 4$, we need a solution of $\displaystyle ax + by = 35$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can easily see that $\displaystyle x_0 = 1, y_0 = 2$ gives us $\displaystyle a x_0 - by_0 = 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus we need to find a $\displaystyle t$ such that $ 35 \ge t\times 4$ and $ t\times 9 \ge 35\times 2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i.e.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle 35/4 \ge t \ge 35\times 2/9$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i.e.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle  8.75 \ge t \ge 7.77\dots$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus $t = 8$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This gives us $\displaystyle x = cx_0 - tb  = 3$, $\displaystyle y = ta- cy_0 = 2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Note: I have swapped your x and y).&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-12-02T07:07:45.390" LastActivityDate="2010-12-02T07:07:45.390" />
  <row Id="12741" PostTypeId="1" AcceptedAnswerId="12742" CreationDate="2010-12-02T09:38:43.603" Score="5" ViewCount="92" Body="&lt;p&gt;How to find the least positive root of the equation $\cos 3x + \sin 5x = 0$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My approach so far is to represent $\sin 5x$ as $\cos \biggl(\frac{\pi}{2} - 5x\biggr)$ then the whole equation reduces to $$2\cos \biggl(\frac{\pi}{4} - x\biggr)\cdot \cos \biggl(\frac{\pi}{4} - 4x\biggr) = 0$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From here we can write:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\biggl(\frac{\pi}{4} - x\biggr) = n\pi + \frac{\pi}{2} , n \in \mathbb{Z}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\biggl(\frac{\pi}{4} - 4x\biggr) = n\pi + \frac{\pi}{2} , n \in \mathbb{Z}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now there can be infinitely many solutions for this, what I am not getting how to compute the minimum among them? And what about if I am asked to find the maximum?&lt;/p&gt;&#xA;" OwnerUserId="3744" LastEditorUserId="3744" LastEditorDisplayName="" LastEditDate="2010-12-02T10:10:34.063" LastActivityDate="2010-12-02T10:17:59.270" Title="Finding the least positive root" Tags="&lt;trigonometry&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="12742" PostTypeId="2" ParentId="12741" CreationDate="2010-12-02T09:51:55.717" Score="4" ViewCount="" Body="&lt;p&gt;You've already done the difficult part! Now just find all the solutions of $\cos(\pi/4-x)=0$ and $\cos(\pi/4-4x)=0$, and pick the smallest positive one.&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2010-12-02T09:51:55.717" CommentCount="9" />
  <row Id="12743" PostTypeId="2" ParentId="12741" CreationDate="2010-12-02T10:00:39.527" Score="0" ViewCount="" Body="&lt;p&gt;EDIT: So apparently you want a full solution... &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A product of two real numbers is $0$ precisely when a factor is $0$, so you must have that &#xA;$$ \begin{align}&#xD;&#xA;&amp;amp; \cos\left(\tfrac{\pi}{4} - x\right) = 0 \qquad\;\; (1) \text{, or} \\\\&#xD;&#xA;&amp;amp; \cos\left(\tfrac{\pi}{4} - 4x\right) = 0 \qquad (2).&#xD;&#xA;\end{align} $$&#xD;&#xA;Now we use that $\cos a = 0 \iff a \in \tfrac{\pi}{2} + \pi\mathbb{Z}$:&#xA;$$ \begin{align}&#xD;&#xA;\cos \left(\tfrac{\pi}{4} - x\right) = 0 &#xD;&#xA;  &amp;amp; \iff \tfrac{\pi}{4} - x \in \tfrac{\pi}{2} + \pi\mathbb{Z} \\\\&#xD;&#xA;  &amp;amp; \iff x \in \tfrac{3\pi}{4} + \pi \mathbb{Z},&#xD;&#xA;\end{align} $$&#xD;&#xA;the smallest positive $x$ satisfying this clearly being $\tfrac{3\pi}{4}$.&#xA;$$ \begin{align}&#xD;&#xA;\cos \left(\tfrac{\pi}{4} - 4x\right) = 0 &#xD;&#xA;  &amp;amp; \iff \tfrac{\pi}{4} - 4x \in \tfrac{\pi}{2} + \pi\mathbb{Z} \\\\&#xD;&#xA;  &amp;amp; \iff 4x \in \tfrac{3\pi}{4} + \pi \mathbb{Z} \\\\&#xD;&#xA;  &amp;amp; \iff x \in \tfrac{3\pi}{16} + \tfrac{\pi}{4}\mathbb{Z},&#xD;&#xA;\end{align} $$&#xD;&#xA;the smallest positive $x$ satisfying this clearly being $\tfrac{3\pi}{16}$. The problem is now reduced to picking the smallest of &lt;em&gt;two&lt;/em&gt; real numbers.&lt;/p&gt;&#xA;" OwnerUserId="1269" LastEditorUserId="1269" LastEditorDisplayName="" LastEditDate="2010-12-02T10:17:59.270" LastActivityDate="2010-12-02T10:17:59.270" CommentCount="3" />
  <row Id="12744" PostTypeId="7" CreationDate="2010-12-02T11:09:03.700" Score="0" ViewCount="0" Body="&lt;p&gt;&#xA;    On Stack Exchange, we believe the core moderators should come &lt;i&gt;from&lt;/i&gt; the community, and be elected by the community&#xA;    itself through popular vote. We hold regular elections to determine who these community moderators will be.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&#xA;    See &lt;a href=&quot;http://blog.stackoverflow.com/2009/05/a-theory-of-moderation/&quot; rel=&quot;nofollow&quot;&gt;a theory of moderation&lt;/a&gt; for the typical roles and abilities of a moderator. Once elected, moderators may hold the position as long as they wish, unless they become inactive or exhibit gross misbehavior.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&#xA;    Community moderators are accorded the highest level of privilege and trust on our community, and should themselves be exemplars of&#xA;    positive behavior within the community. Our general criteria for moderators is as follows:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;    &lt;li&gt;patient and fair&lt;/li&gt;&#xA;    &lt;li&gt;leads by example&lt;/li&gt;&#xA;    &lt;li&gt;shows respect for their fellow community members in their actions and words&lt;/li&gt;&#xA;    &lt;li&gt;open to some light but firm moderation to keep the community on track and resolve (hopefully) uncommon disputes and exceptions&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&#xA;    Every election has three phases:&#xA;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;    &lt;li&gt;Nomination&lt;/li&gt;&#xA;    &lt;li&gt;Primary&lt;/li&gt;&#xA;    &lt;li&gt;Election&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;p&gt;&#xA;Depending on the number of nominees that enter, and the number of moderator positions to be filled, in some circumstances the election may skip the Primary phase and proceed directly to the Election phase.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&#xA;    Please participate in the moderator elections by voting, and perhaps even by nominating yourself to be a community moderator!&#xA;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(You may pose questions for the nominees over the &lt;a href=&quot;http://meta.math.stackexchange.com/questions/1254/list-of-questions-for-candidates-in-2010-moderator-election&quot;&gt;meta thread&lt;/a&gt;)&lt;/p&gt;&#xA;" OwnerUserId="-1" LastEditorUserId="153" LastEditorDisplayName="" LastEditDate="2010-12-09T00:16:21.587" LastActivityDate="2010-12-09T00:16:21.587" />
  <row Id="12745" PostTypeId="1" AcceptedAnswerId="12747" CreationDate="2010-12-02T11:18:02.913" Score="3" ViewCount="346" Body="&lt;p&gt;I'm reading a paper on fluid dynamics and it references a unit vector between two particles i and j. I'm not clear what it means by a unit vector in this instance. How do I calculate the unit vector between the two particles?&lt;/p&gt;&#xA;" OwnerUserId="4318" LastActivityDate="2010-12-02T12:28:48.240" Title="How do you calculate the unit vector between two points?" Tags="&lt;geometry&gt;&lt;vector-space&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" />
  <row Id="12746" PostTypeId="2" ParentId="12745" CreationDate="2010-12-02T11:24:50.750" Score="3" ViewCount="" Body="&lt;p&gt;If particle $i$'s position is described by a position vector $\vec{r}_i$ and particle $j$'s position is described by a position vector $\vec{r}_j$, then you can define the position of $j$ relative to $i$ as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\vec{r}_{ji}= \vec{r}_j-\vec{r}_i$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if you divide this vector by its length: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\vec{r}_{ji}}{\|\vec{r}_{ji}\|}=\frac{\vec{r}_j-\vec{r}_i}{\|\vec{r}_j-\vec{r}_i\|}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;you get a vector with unit length and aligned along the direction of the line through particles $i$ and $j$, pointing towards $j$.&lt;/p&gt;&#xA;" OwnerUserId="3567" LastActivityDate="2010-12-02T11:24:50.750" CommentCount="1" />
  <row Id="12747" PostTypeId="2" ParentId="12745" CreationDate="2010-12-02T11:26:37.180" Score="4" ViewCount="" Body="&lt;p&gt;Two particles i, j are located in some reference frame at vectorial positions $\vec{r}_i$ and $\vec{r}_j$. Therefore the vector which starts at the position of i and ends at j, is just the difference $\vec{r}_j-\vec{r}_i$; its modulus $||\vec{r}_j-\vec{r}_i||$ is the distance between the particles so one can construct the unit vector in that direction (from i to j) by just&#xA;$$\vec{u}_{ij}=\frac{1}{||\vec{r}_j-\vec{r}_i||}(\vec{r}_j-\vec{r}_i)$$&#xD;&#xA;Indeed this is a unit vector for its a multiple of the original with unit modulus since $||\vec{u}_{ij}||=\left|\frac{1}{||\vec{r}_j-\vec{r}_i||}\right|\cdot||\vec{r}_j-\vec{r}_i||=1$ using the property $||\lambda\cdot\vec{v}||=|\lambda|\cdot ||\vec{v}||$.&lt;/p&gt;&#xA;" OwnerUserId="4058" LastEditorUserId="4058" LastEditorDisplayName="" LastEditDate="2010-12-02T12:28:48.240" LastActivityDate="2010-12-02T12:28:48.240" />
  <row Id="12748" PostTypeId="2" ParentId="8349" CreationDate="2010-12-02T12:18:02.770" Score="4" ViewCount="" Body="&lt;p&gt;If someone gives you a random large number, the last thing you want to do is perform a deterministic primality test -- it's very likely to be composite.  Firstly, you should perform:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;trial division and&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Fermat_primality_test&quot; rel=&quot;nofollow&quot;&gt;Fermat's test&lt;/a&gt;.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;These can be performed using PFGW (from &lt;a href=&quot;http://tech.groups.yahoo.com/group/primeform/&quot; rel=&quot;nofollow&quot;&gt;http://tech.groups.yahoo.com/group/primeform/&lt;/a&gt;).  You might even want to perform a stronger pseudo-primality test.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once these tests have been passed, then you should be somewhat confident you have a prime.  If you're happy with a probable prime, then you can stop.  Otherwise, you'll need a proof.  [Curiously, if sufficiently many probabilistic primality tests have been performed, the probability that the number is composite can be less than the probability of a random error occurring in a &quot;proof&quot; of its primality.]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If your number has no particularly special form you can exploit, then your best hope is &lt;a href=&quot;http://en.wikipedia.org/wiki/Elliptic_curve_primality_proving&quot; rel=&quot;nofollow&quot;&gt;elliptic curve primality proving&lt;/a&gt; (see &lt;a href=&quot;http://primes.utm.edu/top20/page.php?id=27&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; for ECPP records).  You could use &lt;a href=&quot;http://www.ellipsa.eu/&quot; rel=&quot;nofollow&quot;&gt;Primo&lt;/a&gt; or &lt;a href=&quot;http://www.alpertron.com.ar/ECM.HTM&quot; rel=&quot;nofollow&quot;&gt;Alpertron&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if your candidate prime happens to have a special form, then there are various deterministic primality proofs you could use.  For example, the &lt;a href=&quot;http://en.wikipedia.org/wiki/Lucas%E2%80%93Lehmer_primality_test&quot; rel=&quot;nofollow&quot;&gt;Lucas-Lehmer test&lt;/a&gt; (cf. &lt;a href=&quot;http://www.mersenne.org/&quot; rel=&quot;nofollow&quot;&gt;GIMPS&lt;/a&gt;), Proth's test (http://en.wikipedia.org/wiki/Proth's_theorem), &lt;a href=&quot;http://en.wikipedia.org/wiki/Lucas%E2%80%93Lehmer%E2%80%93Riesel_test&quot; rel=&quot;nofollow&quot;&gt;Lucas-Lehmer-Riesel test&lt;/a&gt;.  In fact, these tests are great if you're the type of person who aspires to finding a &lt;a href=&quot;http://primes.utm.edu/primes/home.php&quot; rel=&quot;nofollow&quot;&gt;top-5000 prime&lt;/a&gt; (something I used to do: see &lt;a href=&quot;http://primes.utm.edu/bios/page.php?id=776&quot; rel=&quot;nofollow&quot;&gt;my profile&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again, before using these tests, you should perform trial division.  If you're looking for primes that can be proved by the latter two tests, you will probably find &lt;a href=&quot;http://primes.utm.edu/programs/NewPGen/&quot; rel=&quot;nofollow&quot;&gt;NewPGen&lt;/a&gt; invaluable.&lt;/p&gt;&#xA;" OwnerUserId="139" LastActivityDate="2010-12-02T12:18:02.770" CommentCount="2" />
  <row Id="12749" PostTypeId="1" AcceptedAnswerId="12751" CreationDate="2010-12-02T13:16:39.020" Score="6" ViewCount="220" Body="&lt;p&gt;I can't use any of the convergence tests I learned because I have no information on $f(x)$, in particular I don't know if it's continuous or positive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only thing I could think of was that if $\displaystyle \int_{1}^{\infty}f(x)dx$ was absolutely convergent, then $|f(x)\sin x| \leq |f(x)|$ would imply by the comparison test that $\displaystyle \int_{1}^{\infty}f(x)\sin xdx$ converges.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if I want to find a counter-example I have to pick $f(x)$ so that $\displaystyle \int_{1}^{\infty}f(x)dx$ conditionally converges, but I can't think of one.&lt;/p&gt;&#xA;" OwnerUserId="3237" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-02T14:36:19.560" LastActivityDate="2010-12-02T14:36:19.560" Title="if $\int_1^{\infty}f(x)\mathrm dx$ converges, must $\int_1^{\infty}f(x)\sin x\mathrm dx$ converge?" Tags="&lt;calculus&gt;&lt;homework&gt;&lt;integral&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="12750" PostTypeId="1" CreationDate="2010-12-02T13:19:42.263" Score="4" ViewCount="103" Body="&lt;p&gt;Can we alter &lt;a href=&quot;http://en.wikipedia.org/wiki/Hilbert%27s_axioms&quot; rel=&quot;nofollow&quot;&gt;Hilbert's axioms&lt;/a&gt; to have $\mathbb{Q}^3$ as a unique model?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The critical axioms seem to be the &lt;a href=&quot;http://en.wikipedia.org/wiki/Hilbert%27s_axioms#IV._Congruence&quot; rel=&quot;nofollow&quot;&gt;congruence axioms&lt;/a&gt; IV.1 and IV.4, and presumably the &lt;a href=&quot;http://en.wikipedia.org/wiki/Hilbert%27s_axioms#V._Continuity&quot; rel=&quot;nofollow&quot;&gt;line completeness axiom&lt;/a&gt; V.2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But how are they to be modified? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;IV.1 might be replaced by requiring that there are counter-examples (irrationality of $\sqrt{2}$) and appropriately relaxing &quot;congruent&quot; to &quot;almost congruent&quot; (= &quot;arbitrarily close to congruence&quot;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But what about line completeness then, since it might be possible to add irrational points to $\mathbb{Q}^3$ such that the modified axioms still hold?&lt;/p&gt;&#xA;" OwnerUserId="1792" LastActivityDate="2010-12-02T13:19:42.263" Title="Can we alter Hilbert's axioms to have $\mathbb{Q}^3$ as a unique model?" Tags="&lt;geometry&gt;&lt;euclidean-geometry&gt;" CommentCount="3" />
  <row Id="12751" PostTypeId="2" ParentId="12749" CreationDate="2010-12-02T13:24:24.397" Score="15" ViewCount="" Body="&lt;p&gt;Consider $f(x)=\sin(x) / x$.&lt;/p&gt;&#xA;" OwnerUserId="2810" LastActivityDate="2010-12-02T13:24:24.397" CommentCount="5" />
  <row Id="12752" PostTypeId="1" AcceptedAnswerId="12754" CreationDate="2010-12-02T13:40:19.077" Score="6" ViewCount="193" Body="&lt;p&gt;It is proved in &lt;em&gt;Advanced Calculus&lt;/em&gt; by Angus Taylor, § 20.8, that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\log n!=\log \left( \left( \frac{n}{e}\right) ^{n}\sqrt{2\pi n}\right)&#xD;&#xA;+r_{n},$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$r_{n}=\sum_{k=1}^{\infty }S_{k}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$S_{k}=\sum_{p=n+1}^{\infty }\frac{k}{2(k+1)(k+2)p^{k+1}}.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This formula for $r_{n}$ provides a method for finding the estimate&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{1}{12\left( n+1\right) }&amp;lt;r_{n}&amp;lt;\frac{1}{12\left( n-1\right) }.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Indeed &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{1}{k\left( n+1\right) ^{k}}=\sum_{p=n+1}^{\infty }\int_{p}^{p+1}\frac{%&#xD;&#xA;1}{x^{k+1}}\mathrm dx&amp;lt;\sum_{p=n+1}^{\infty }\frac{1}{p^{k+1}}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{p=n+1}^{\infty }\frac{1}{p^{k+1}}&amp;lt;\sum_{p=n+1}^{\infty }\int_{p-1}^{p}%&#xD;&#xA;\frac{1}{x^{k+1}}\mathrm dx=\frac{1}{kn^{k}},$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and so&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{1}{2(k+1)(k+2)\left( n+1\right) ^{k}}&amp;lt;S_{k}&amp;lt;\frac{1}{2(k+1)(k+2)n^{k}}&#xD;&#xA;$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{1}{12\left( n+1\right) }&amp;lt;\sum_{k=1}^{\infty }\frac{1}{%&#xD;&#xA;2(k+1)(k+2)\left( n+1\right) ^{k}}&amp;lt;r_{n}&amp;lt;\sum_{k=1}^{\infty }\frac{1}{%&#xD;&#xA;2(k+1)(k+2)n^{k}}&amp;lt;\frac{1}{12\left( n-1\right) }.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added&lt;/strong&gt;:The &lt;a href=&quot;http://en.wikipedia.org/wiki/Stirling%27s_approximation#Speed_of_convergence_and_error_estimates&quot; rel=&quot;nofollow&quot;&gt;Wikipedia article&lt;/a&gt; pointed to in the comment gives an approximation based on the Euler-MacLaurin formula in terms of the Bernoulli numbers and states that it can also be obtained by repeated integration by parts. The Stirling series that appears in the same article is different from the one above. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question&lt;/strong&gt; (edited): Are there better estimates of $r_{n}$ based on different methods?&lt;/p&gt;&#xA;" OwnerUserId="752" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2010-12-04T17:53:48.850" LastActivityDate="2010-12-04T17:53:48.850" Title="Estimates for Stirling's formula remainder" Tags="&lt;calculus&gt;&lt;approximation&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="12753" PostTypeId="2" ParentId="11843" CreationDate="2010-12-02T13:46:47.453" Score="1" ViewCount="" Body="&lt;p&gt;If your predictions are repetitive I think you can do this.  For example, if you always predict 50% chance of one choice, 30% for a second and 20% for the third (even if they are permuted on a particular play) you can do a standard chi-square test.  Did you in fact get 50% right on the one you claim?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If your predictions vary continuously, you can still pick the highest probability and compute the observed number of times it is right compared to the prediction and the variance using the variance on each prediction is p(1-p).  Then you can compare with the null hypothesis of 1/3, 1/3, 1/3&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure how to understand the statement &quot;we can trust this function on 86%&quot;.  I think the chi-square test can say &quot;it is 86% probably that my predictions are better than the null hypothesis&quot;, but that doesn't tell you that the predictions are right.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-12-02T13:46:47.453" />
  <row Id="12754" PostTypeId="2" ParentId="12752" CreationDate="2010-12-02T14:40:17.450" Score="8" ViewCount="" Body="&lt;p&gt;Here's a paper &lt;a href=&quot;http://biblio.ugent.be/input/download?func=downloadFile&amp;amp;fileOId=451038&quot; rel=&quot;nofollow&quot;&gt;Stirling's Series Made Easy&lt;/a&gt; by Chris Impens that states in Corollary 3&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \frac{1}{12n} - \frac{1}{360n^3} &amp;lt; r_n &amp;lt; \frac{1}{12n}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and (the interesting part) that the constant $1/360$ cannot be improved. I confess to not having&#xA;read the paper closely.&lt;/p&gt;&#xA;" OwnerUserId="1301" LastActivityDate="2010-12-02T14:40:17.450" />
  <row Id="12755" PostTypeId="1" AcceptedAnswerId="12758" CreationDate="2010-12-02T15:41:43.920" Score="2" ViewCount="148" Body="&lt;p&gt;Set $S$ is a collection of disjoint sets each having cardinality of that of $\mathbb{R}$.The cardinality of set $S$ is also that of $\mathbb{R}$. $F$ is the set which is union of all the elements of $S$. Is the cardinality of set $F$ is equal to that of $\mathbb{R}$ ?&lt;/p&gt;&#xA;" OwnerUserId="2987" LastActivityDate="2010-12-02T18:35:48.717" Title="on cardinality of a set." Tags="&lt;elementary-set-theory&gt;" AnswerCount="2" />
  <row Id="12756" PostTypeId="2" ParentId="12220" CreationDate="2010-12-02T15:45:40.987" Score="0" ViewCount="" Body="&lt;p&gt;I think I solved it: If matrices $A$ and $S$ commute, i.e. $AS = SA$, then I eigenspaces of one matrix are invariant under the other matrix, because if $Ax = \lambda x$, then &#xA;$A Sx = SAx = S\lambda x$, so if $x$ is eigenvector of $A$ with eigenvalue $\lambda$, then so is $S\lambda x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This in mind, I choose one of the matrices and diagonalize it, say I choose $S$. Then I get&#xA;$S = T D T^\top$ with $T$ being unitary and $D$ being diagonal. If I then use $T$ to transform $A$, I get &#xA;$A&amp;#39; = T A T^\top$ is &lt;em&gt;block diagonal&lt;/em&gt;, because of the invariance mentioned earlier. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can then diagonalize each block of $A$ on its own, because that way, the resulting eigenvectors of $A$ will remain eigenvectors of $S$. That way, I have simultaneously diagonalized $A$. The nice thing is that this can help make the resulting eigenvectors unique, because now eigenvectors are characterized not only by their $A$-matrix eigenvalue but also by their $S$-matrix eigenvalue.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If that doesn't suffice, I could search for more symmetries that all simultaneously commute with each other. An example from quantum mechanics is the hydrogen atom, where one find that the Hamiltonian commutes with the modulus of the angular momentum and its z-component, which ultimately gives rise to the three quantum numbers n, l and m.&lt;/p&gt;&#xA;" OwnerUserId="4171" LastActivityDate="2010-12-02T15:45:40.987" />
  <row Id="12757" PostTypeId="2" ParentId="12755" CreationDate="2010-12-02T15:48:53.870" Score="6" ViewCount="" Body="&lt;p&gt;You are asking whether $\mathbb{R} \times \mathbb{R}$ has the same cardinality as $\mathbb{R}$.  The answer is yes, thanks to the fact that $|\mathbb{R}| = 2^{\aleph_0}$ and $2^{\aleph_0} \times 2^{\aleph_0} = 2^{\aleph_0 + \aleph_0} = 2^{\aleph_0}$.  More explicitly, $\mathbb{R}$ has the same cardinality as the set $S = \{ 0, 1 \}^{\mathbb{N}}$ of binary sequences, and there is an obvious bijection $S \times S \to S$ given by &quot;interweaving&quot; sequences: that is, sending&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(a_1, a_2, a_3, ...) \times (b_1, b_2, b_3, ...) \to (a_1, b_1, a_2, b_2, a_3, b_3, ...).$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, the statement that $A \times A$ has the same cardinality as $A$, for $A$ infinite, is true for all &lt;a href=&quot;http://en.wikipedia.org/wiki/Aleph_number&quot; rel=&quot;nofollow&quot;&gt;alephs&lt;/a&gt;.  I think I have been told that whether this statement is true for all infinite sets is equivalent to AC.  &lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-12-02T15:48:53.870" CommentCount="3" />
  <row Id="12758" PostTypeId="2" ParentId="12755" CreationDate="2010-12-02T15:49:22.963" Score="10" ViewCount="" Body="&lt;p&gt;Yes. The cardinality of $F$ is $\sum\limits_{x\in S}|x| = \sum\limits_{x\in S}\mathfrak{c} = \mathfrak{c}\mathfrak{c}=\mathfrak{c}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In cardinal arithmetic, if $\kappa$ and $\lambda$ are nonzero cardinals, and at least one is infinite, you have&#xA;$$\kappa+\lambda = \kappa\lambda = \max\{\kappa,\lambda\}.$$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-02T18:35:48.717" LastActivityDate="2010-12-02T18:35:48.717" CommentCount="2" />
  <row Id="12759" PostTypeId="1" AcceptedAnswerId="12761" CreationDate="2010-12-02T16:17:04.223" Score="0" ViewCount="150" Body="&lt;p&gt;Assume that Y has a beta distribution with parameters a and b. Find the density function of U = 1 - Y.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know how to do then when they give the density function of Y, but i'm confused here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="3049" LastActivityDate="2010-12-02T16:46:09.100" Title="density function of U when Y has a beta distribution" Tags="&lt;probability&gt;&lt;homework&gt;&lt;statistics&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="12760" PostTypeId="2" ParentId="10564" CreationDate="2010-12-02T16:23:32.170" Score="0" ViewCount="" Body="&lt;p&gt;Your answer is mathematically correct, however I figured out another way to solve this equation, which leads to a simpler result.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I applied the technique for 2D projective geometry, which is described &lt;a href=&quot;http://books.google.de/books?id=si3R3Pfa98QC&amp;amp;pg=PA35&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; to the 1D case and it works out fine.&lt;/p&gt;&#xA;" OwnerUserId="3573" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-02T16:25:39.533" LastActivityDate="2010-12-02T16:25:39.533" />
  <row Id="12761" PostTypeId="2" ParentId="12759" CreationDate="2010-12-02T16:46:09.100" Score="3" ViewCount="" Body="&lt;p&gt;First, for the density function of a beta$(\alpha,\beta)$ random variable, see &lt;a href=&quot;http://en.wikipedia.org/wiki/Beta_distribution&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Beta_distribution&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if $Y$ is beta distributed, then it takes values in $(0,1)$. Hence, $U = 1 - Y$ also takes values in $(0,1)$.&#xA;In order to find the density of $U$, it is useful to find first its distribution function, and then differentiate it.&#xA;The distribution function of $U$ at $x \in (0,1)$ is the tail distribution function of $Y$ at $1-x$. By taking complement, you get the distribution function of $U$ expressed in terms of that of $Y$. Differentiating it, you get the density of $U$ expressed in terms of that of $Y$. You should find out that $U$ and $Y$ are very closely related.&lt;/p&gt;&#xA;" OwnerUserId="2810" LastActivityDate="2010-12-02T16:46:09.100" CommentCount="3" />
  <row Id="12762" PostTypeId="6" CreationDate="2010-12-02T16:57:08.190" Score="0" ViewCount="0" Body="&lt;p&gt;I am an active member of both math.stackexchange.com and&#xA;mathoverflow.net. So far mathoverflow has been a valuable&#xA;source of research-level questions and answers, and I see&#xA;math.se becoming the same at a more general level.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to see m.se as a site for well-focused mathematical&#xA;questions and I would take a robust line against spammers,&#xA;trolls, flamers and so on. So far the site has attracted few&#xA;of these individuals, but with increasing success, the danger&#xA;is that more would come. As a former regular on sci.math, over the years I&#xA;watched the newsgroup degenerate into a bear pit&#xA;filled with flame wars and crankery. I wish to maintain the high&#xA;signal-to-noise ratio presently here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Were I elected, I would regard my appointment as for a period&#xA;of not more than one year, and I would resign or seek re-election&#xA;before or at the end of this time.&lt;/p&gt;&#xA;" OwnerUserId="226" LastEditorUserId="226" LastEditorDisplayName="" LastEditDate="2010-12-02T16:57:08.190" LastActivityDate="2010-12-02T16:57:08.190" CommunityOwnedDate="2010-12-02T16:57:08.190" CommentCount="1" />
  <row Id="12763" PostTypeId="1" AcceptedAnswerId="12794" CreationDate="2010-12-02T17:01:18.817" Score="3" ViewCount="242" Body="&lt;p&gt;In the following picture, the author of the &lt;em&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0201128195&quot; rel=&quot;nofollow&quot;&gt;Field and Wave Electromagnetics&lt;/a&gt;&lt;/em&gt; shows the geometrical meaning of the direction of the gradient. That is, only by following the direction of the normal vector to the curve at that pointer could the rate of change be the maximum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/zpSIu.jpg&quot; alt=&quot;alt text&quot;&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But what about the geometrical interpretation of the &lt;strong&gt;magnitude&lt;/strong&gt; of gradient &lt;strong&gt;generally&lt;/strong&gt; or maybe is there a &lt;strong&gt;geometrical interpreation&lt;/strong&gt; of the magnitude of graident?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;thanks.&lt;/p&gt;&#xA;" OwnerUserId="1382" LastEditorUserId="1382" LastEditorDisplayName="" LastEditDate="2010-12-02T17:25:09.053" LastActivityDate="2010-12-03T16:58:11.923" Title="What's the geometrical interpretation of the magnitude of gradient generally?" Tags="&lt;calculus&gt;&lt;vector-analysis&gt;" AnswerCount="4" FavoriteCount="1" />
  <row Id="12764" PostTypeId="2" ParentId="12763" CreationDate="2010-12-02T17:06:33.880" Score="2" ViewCount="" Body="&lt;p&gt;If you look at the set of points satisfying f(x) = c, the gradient of f is normal to the surface and points in the direction of greatest increase of f. The magnitude of the gradient is proportional to the rate of increase.&lt;/p&gt;&#xA;" OwnerUserId="247" LastEditorUserId="247" LastEditorDisplayName="" LastEditDate="2010-12-02T20:22:23.850" LastActivityDate="2010-12-02T20:22:23.850" CommentCount="4" />
  <row Id="12765" PostTypeId="2" ParentId="12763" CreationDate="2010-12-02T17:13:03.340" Score="1" ViewCount="" Body="&lt;p&gt;If you're standing on a point on a hill, the gradient gives the direction of the steepest ascent from that location immediately outward. The magnitude of the gradient gives how steep that ascent is, by elementary slope.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you work out the algebra, the rate of change in a function $f$ in the direction of the unit vector $\eta$ is $\nabla f \cdot \eta$ (This is simply the chain rule.). This quantity reaches its maximum when $\eta$ shares the same direction as $\nabla f$ (to maximize the cosine of the angle that's part of the dot product formula). Thus the magnitude of $\nabla f$ gives the rate of change in that direction, as well.&lt;/p&gt;&#xA;" OwnerUserId="3122" LastEditorUserId="3122" LastEditorDisplayName="" LastEditDate="2010-12-02T17:18:32.663" LastActivityDate="2010-12-02T17:18:32.663" CommentCount="1" />
  <row Id="12766" PostTypeId="1" AcceptedAnswerId="12797" CreationDate="2010-12-02T17:20:03.920" Score="2" ViewCount="124" Body="&lt;p&gt;I have another question about correctness of my proof. We know, that $N$ and $G/N$ are p-groups, than $G$ is p-group.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proof: $|N|=p^m$ and thus $|G|=p^m q$. Suppose that $(p^m,q)=1$, than in $G$ exists element of order $k$, such that $k | q$ and $(k,p^m)=1$ - suppose it's $b \in G$. Element $bN \in G/N$ is also of order $k$, and thus $k | \left| G/N \right|$ which contradicts with prerequisite, that $G/N$ is p-group.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's correct? It's here simply way to prove that? Thanks.&lt;/p&gt;&#xA;" OwnerUserId="4195" LastActivityDate="2010-12-02T19:40:50.273" Title="Proof that if group $N$ and $G/N$ are p-groups, than $G$ is p-group." Tags="&lt;group-theory&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="12767" PostTypeId="1" AcceptedAnswerId="12774" CreationDate="2010-12-02T17:27:22.437" Score="3" ViewCount="77" Body="&lt;p&gt;My question is similar to &lt;a href=&quot;http://math.stackexchange.com/questions/9627/cohomology-of-a-tensor-product-of-sheaves&quot;&gt;this&lt;/a&gt;, but not identical.  I believe the following to be true, but I'd like a reference.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given (quasicoherent?) sheaves of $\mathcal O_X$ modules $E$ and $F$ on a projective variety $X$, &#xA;$$&#xD;&#xA;H^n(X, E\otimes F) \cong \bigoplus_{p+q=n} H^p(X, E) \otimes H^q(X, F).&#xD;&#xA;$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Is this true, and what is a good reference or counterexample? If true, is quasicoherent necessary?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="3008" LastActivityDate="2010-12-02T18:57:28.790" Title="Cohomological decomposition of tensor sheaves?" Tags="&lt;algebraic-geometry&gt;&lt;reference-request&gt;&lt;quasicoherent-sheaves&gt;" AnswerCount="2" />
  <row Id="12768" PostTypeId="1" AcceptedAnswerId="12950" CreationDate="2010-12-02T17:29:26.363" Score="6" ViewCount="332" Body="&lt;p&gt;I'm reading &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/059680279X&quot; rel=&quot;nofollow&quot;&gt;High Performance JavaScript&lt;/a&gt;, and I think the graphs in one chapter are just plain wrong. &lt;a href=&quot;http://books.google.com/books?id=ED6ph4WEIoQC&amp;amp;pg=PA41&quot; rel=&quot;nofollow&quot;&gt;Here is one on Google Books&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The y axis is &quot;Times faster&quot;, and it runs from -1.5 to +4.0. Now, I would have thought that &quot;1 times faster&quot; means &quot;no faster&quot;, &quot;2 times faster&quot; means &quot;twice as fast&quot;, and &quot;0.5 times faster&quot; means &quot;half as fast&quot;/&quot;twice as slow&quot;. Have they just got completely confused in that graph, or is it me?&#xA;&lt;img src=&quot;http://i.imgur.com/q48tl.jpg&quot; alt=&quot;screenshot&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4331" LastEditorUserId="4331" LastEditorDisplayName="" LastEditDate="2010-12-02T17:39:23.700" LastActivityDate="2010-12-03T22:01:03.900" Title="What does -1.13 times faster mean?" Tags="&lt;statistics&gt;" AnswerCount="6" CommentCount="9" />
  <row Id="12769" PostTypeId="1" AcceptedAnswerId="12773" CreationDate="2010-12-02T17:30:53.573" Score="1" ViewCount="77" Body="&lt;p&gt;Let $a_n$ be a sequence in $\mathbb{R}$, $\phi: \mathbb{N} \rightarrow \mathbb{N}$ a monotonically increasing function with $\phi (0)=0$. Show that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sum _{n=0}^{\infty } a_n$ converges $\Rightarrow  \sum _{i=0}^{\infty } {(\sum _{j=\phi (i)}^{\phi (i+1)-1 }a_{j})}$ converges.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My notes: Can you use that ${(\sum _{j=\phi (i)}^{\phi (i+1)-1 }a_{j})}$ has to be a cauchy sequence and can be as small as you want? Certainly it will be smaller than $a_n$ for all $n$ bigger than some $N_e$&lt;/p&gt;&#xA;" OwnerUserId="3123" LastActivityDate="2010-12-02T17:47:08.323" Title="Convergence of Sequences" Tags="&lt;real-analysis&gt;&lt;sequences-and-series&gt;&lt;convergence&gt;" AnswerCount="2" />
  <row Id="12770" PostTypeId="1" AcceptedAnswerId="12783" CreationDate="2010-12-02T17:31:50.120" Score="2" ViewCount="120" Body="&lt;p&gt;I'm given the following exercise:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\iint\limits_D \exp(x^{2}+y^{2})dA$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And I dont even know where to start, any chance someone could give me a hint?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;D is a half circle, given by:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$9\le x^{2}+y^{2}\le 16, y\ge 0$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wheres A, I don't know what it is, its not given in the exercise. - Thats partly whats bugging me.&lt;/p&gt;&#xA;" LastActivityDate="2010-12-02T19:59:23.273" Title="Multiple integral Issue" Tags="&lt;integral&gt;" AnswerCount="3" CommentCount="10" />
  <row Id="12771" PostTypeId="2" ParentId="12768" CreationDate="2010-12-02T17:36:21.893" Score="1" ViewCount="" Body="&lt;p&gt;I would say $-1.13$ faster means 1.13 slower. For example $-1.13$ faster than 160 mph is $-180.8$ mph, i.e. 180.8 going in the opposite direction.&lt;/p&gt;&#xA;" OwnerUserId="3249" LastActivityDate="2010-12-02T17:36:21.893" CommentCount="5" />
  <row Id="12772" PostTypeId="2" ParentId="12769" CreationDate="2010-12-02T17:44:59.620" Score="2" ViewCount="" Body="&lt;p&gt;When you take a fixed N you get &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sum _{i=0} ^{N} \sum _{j=\phi(i)} ^{\phi(i+1)-1} a_j = \sum_{j=0}^{j=\phi(N+1)-1} a_j$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so basically the new sequence (meaning the partial summations) is a subsequence of the original one.&lt;/p&gt;&#xA;" OwnerUserId="2125" LastActivityDate="2010-12-02T17:44:59.620" CommentCount="2" />
  <row Id="12773" PostTypeId="2" ParentId="12769" CreationDate="2010-12-02T17:47:08.323" Score="1" ViewCount="" Body="&lt;p&gt;I suppose you meant $\phi$ is strictly increasing, otherwise the inner sum could be an empty sum. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $b_i=\sum_{j=\phi(i)}^{\phi(i+1)-1} a_j$. For any $\epsilon &amp;gt;0$, you can find $N$ such that $$|a_j+\cdots +a_{j+k}|&amp;lt;\epsilon$$ for all $j\ge N, k&amp;gt;0$. Let $I$ be such that $\phi(I)\ge N$. Then for all $i\ge I$ you have &#xA;$$ |b_i+\cdots +b_{i+k}|&amp;lt;\epsilon $$&#xD;&#xA;for all $k&amp;gt;0$. This proves that your series is Cauchy and hence converges.&lt;/p&gt;&#xA;" OwnerUserId="3249" LastActivityDate="2010-12-02T17:47:08.323" />
  <row Id="12774" PostTypeId="2" ParentId="12767" CreationDate="2010-12-02T17:53:09.903" Score="5" ViewCount="" Body="&lt;p&gt;No, it is false, consider e.g. $E=F=\mathcal{O}_X$ on a curve $X$ of genus one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Here is something that is true though: let $p$ and $q$ be the two projection maps $X \times X \to X$. Then $H^\bullet(X\times X,p^\ast E \otimes q^\ast F) \cong H^\bullet(X,E) \otimes H^\bullet(X,F)$. There are lots of basic variants of this theorem and all of them are called the Kunneth formula.)&lt;/p&gt;&#xA;" OwnerUserId="677" LastActivityDate="2010-12-02T17:53:09.903" CommentCount="4" />
  <row Id="12775" PostTypeId="1" AcceptedAnswerId="12778" CreationDate="2010-12-02T17:56:08.750" Score="3" ViewCount="200" Body="&lt;p&gt;I'm looking for an example of a non-commutative ring, $R$, with identity s.t $R$ does not contain a non-trival 2 sided ideal and $R$ is not a division ring&lt;/p&gt;&#xA;" OwnerUserId="832" LastEditorUserId="832" LastEditorDisplayName="" LastEditDate="2010-12-02T21:00:30.620" LastActivityDate="2010-12-02T21:00:30.620" Title="Example of a non-commutative rings with identity that do not contain non-trival ideals and are not division rings" Tags="&lt;abstract-algebra&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="12776" PostTypeId="1" AcceptedAnswerId="12782" CreationDate="2010-12-02T17:59:21.677" Score="1" ViewCount="97" Body="&lt;p&gt;-5 = 2x + 5&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to &quot;re-start&quot; my mathematical education.  I am needing it more and more at work.  A co-worker asked me to break that down (he couldn't remember either).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where should I start?  What books?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="4333" LastEditorUserId="1120" LastEditorDisplayName="" LastEditDate="2010-12-02T18:07:04.320" LastActivityDate="2010-12-02T18:07:04.320" Title="20 years since I was in high school. How do I break this down?" Tags="&lt;algebra-precalculus&gt;" AnswerCount="3" />
  <row Id="12777" PostTypeId="2" ParentId="12770" CreationDate="2010-12-02T17:59:43.527" Score="0" ViewCount="" Body="&lt;p&gt;The A in $dA$ refers to the infinitesimal thing you're integrating over. It just means that instead of a line, you integrate over a surface. A for Area.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for a hint to start: Since you are dealing with circles, polar coordinates are a natural choice. I am sure these coordinates have been discussed in your lecture before. Note that $x^2 + y^2 = r^2$, where $r$ is the distance of the point from the coordinate origin. Also note that $dA = r dr d\phi$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, you can characterize the area $D$ by simple intervals for your polar coordinates: r will go from $3$ to $4$ and $\phi$ will go from $0$ to $\pi$.&lt;/p&gt;&#xA;" OwnerUserId="4171" LastActivityDate="2010-12-02T17:59:43.527" CommentCount="2" />
  <row Id="12778" PostTypeId="2" ParentId="12775" CreationDate="2010-12-02T18:00:12.253" Score="8" ViewCount="" Body="&lt;p&gt;If &quot;ideal&quot; means &quot;two-sided ideal,&quot; then the &lt;a href=&quot;http://en.wikipedia.org/wiki/Weyl_algebra&quot; rel=&quot;nofollow&quot;&gt;Weyl algebra&lt;/a&gt; $\mathbb{C} \langle x, y \rangle / (xy - yx - 1)$ is an example.  So is $\mathcal{M}_n(R)$ for any division ring $R$ and $n \ge 2$.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-12-02T18:00:12.253" />
  <row Id="12779" PostTypeId="2" ParentId="12775" CreationDate="2010-12-02T18:00:44.797" Score="7" ViewCount="" Body="&lt;p&gt;If you mean two-sided ideals, you are looking for simple rings (that are not division rings):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Simple_ring&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Simple_ring&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;E.g. as in the wikipedia article, is the ring of matrices (of a certain size) over a field. Clearly this is not a division ring since not every matrix is invertible (matrices with zero determinants are not invertible).&lt;/p&gt;&#xA;" OwnerUserId="3431" LastEditorUserId="3431" LastEditorDisplayName="" LastEditDate="2010-12-02T19:25:27.297" LastActivityDate="2010-12-02T19:25:27.297" />
  <row Id="12780" PostTypeId="2" ParentId="12776" CreationDate="2010-12-02T18:05:36.787" Score="2" ViewCount="" Body="&lt;p&gt;Hint: In general for such one variable problems, where you need to find the value of the variable from a given single equation, you collect terms in the variable on one side of the equation and terms not containing the variable on the other side. Then divide the equation by the coefficient of the variable ($x$ in this case) to solve for it.&lt;/p&gt;&#xA;" OwnerUserId="3431" LastActivityDate="2010-12-02T18:05:36.787" />
  <row Id="12781" PostTypeId="2" ParentId="12776" CreationDate="2010-12-02T18:06:20.140" Score="0" ViewCount="" Body="&lt;p&gt;This is elementary algebra, there are Schaum's books on that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, you solve for $x$ like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$-5 - 5 = 2x + 5 - 5$ (substract 5 from each side), this gives $-10 = 2x$. Now divide both sides by $2$ to get $-10/2 = 2x/2$ so $x = -5$.&lt;/p&gt;&#xA;" OwnerUserId="1120" LastActivityDate="2010-12-02T18:06:20.140" />
  <row Id="12782" PostTypeId="2" ParentId="12776" CreationDate="2010-12-02T18:06:42.183" Score="11" ViewCount="" Body="&lt;p&gt;This is an equation, because it says &quot;Left hand side = Right hand side&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Think ow an equation as a scale in perfect balance. Now you have to rearrange everything, without destroying the balance, so that the $x$ will appear at one side alone. Because then you know what $x$ is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, if you take away the same amount from &lt;em&gt;both&lt;/em&gt; sides, the scale will remain in balance. So, let us take away $5$ from both sides, and we end up with &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$-10 = 2x$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if the scale is in balance now, it will remain in balance when we make each side half as big, because if two things are equal, then their halves are equal as well. This leaves us with&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$-5 = x$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The scale is in balance, on the left side we have a $-5$ and on the right side we have $x$. Therefore, $x$ has to be $-5$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, to re-start your education: The problem above is elementary algebra, so I'd pick up with an algebra book that starts from the beginning. Something like &quot;Math for Dummies&quot; might be a nice idea as they are fun to read.&lt;/p&gt;&#xA;" OwnerUserId="4171" LastActivityDate="2010-12-02T18:06:42.183" CommentCount="1" />
  <row Id="12783" PostTypeId="2" ParentId="12770" CreationDate="2010-12-02T18:09:52.880" Score="5" ViewCount="" Body="&lt;p&gt;The notation $\underset{D}{\displaystyle\int \int }f(x,y)dA$ has the same meaning as $\underset{D}{\displaystyle\int \int }f(x,y)dxdy$. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;To evaluate an integral in a different coordinate system one has to find the absolute value of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Jacobian_matrix_and_determinant&quot; rel=&quot;nofollow&quot;&gt;Jacobian determinant&lt;/a&gt; of the transformation. Using polar coordinates, we can transform the initial integral &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\underset{D}{\int \int }e^{x^{2}+y^{2}}dA=\underset{D}{\int \int }e^{x^{2}+y^{2}}dxdy$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;into this one&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_{0}^{\pi }\left( \int_{3}^{4}e^{r^{2}}rdr\right) d\theta ,$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where the conversion factor $r$ is the Jacobian determinant (see Example 3 on the Wikipedia article), and observe that $r^{2}=x^{2}+y^{2}$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{eqnarray*}&#xD;&#xA;\underset{D}{\int \int }e^{x^{2}+y^{2}}dA &amp;amp;=&amp;amp;\underset{D}{\int \int }%&#xD;&#xA;e^{x^{2}+y^{2}}dxdy \\&#xD;&#xA;&amp;amp;=&amp;amp;\int_{0}^{\pi }\left( \int_{3}^{4}e^{r^{2}}rdr\right) d\theta  \\&#xD;&#xA;&amp;amp;=&amp;amp;\int_{0}^{\pi }\left[ \frac{1}{2}e^{r^{2}}\right] _{3}^{4}d\theta  \\&#xD;&#xA;&amp;amp;=&amp;amp;\int_{0}^{\pi }\frac{1}{2}(e^{16}-e^{9})d\theta  \\&#xD;&#xA;&amp;amp;=&amp;amp;\frac{\pi }{2}(e^{16}-e^{9}).&#xD;&#xA;\end{eqnarray*}$$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="752" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2010-12-02T19:59:23.273" LastActivityDate="2010-12-02T19:59:23.273" />
  <row Id="12785" PostTypeId="2" ParentId="12768" CreationDate="2010-12-02T18:21:32.430" Score="0" ViewCount="" Body="&lt;p&gt;I suppose Skilldrick is right, and $-x$ times faster is $x$ times slower or $1/x$ times faster. For example, if A takes 2 seconds and B is $2$ times faster, B takes 1 second; if C is $-2$ times faster, it takes 4 seconds. And A is $0$ times faster than itself.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-12-02T18:21:32.430" />
  <row Id="12786" PostTypeId="2" ParentId="12770" CreationDate="2010-12-02T18:31:31.963" Score="0" ViewCount="" Body="&lt;p&gt;Some confusion is coming from not understanding what the $dA$ represents, so here's a link on &lt;a href=&quot;http://keep2.sjfc.edu/faculty/kgreen/vector/Block3/jacob/node4.html&quot; rel=&quot;nofollow&quot;&gt;the area element in polar coordinates&lt;/a&gt; that hopefully will help clear up this point.&lt;/p&gt;&#xA;" OwnerUserId="1301" LastActivityDate="2010-12-02T18:31:31.963" />
  <row Id="12787" PostTypeId="1" CreationDate="2010-12-02T18:37:21.577" Score="4" ViewCount="258" Body="&lt;p&gt;My son is taking algebra and I'm a little rusty. Not using a calculator or the internet, how would you find the roots of 2x&lt;sup&gt;4&lt;/sup&gt; + 3x&lt;sup&gt;3&lt;/sup&gt; - 11x&lt;sup&gt;2&lt;/sup&gt; - 9x + 15 = 0. Please list step by step. Thanks, Brian&lt;/p&gt;&#xA;" OwnerUserId="4338" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-03T02:39:10.197" LastActivityDate="2010-12-03T02:39:10.197" Title="Finding roots of fourth degree polynomial" Tags="&lt;algebra-precalculus&gt;&lt;polynomials&gt;" AnswerCount="2" />
  <row Id="12788" PostTypeId="2" ParentId="9092" CreationDate="2010-12-02T18:41:42.373" Score="0" ViewCount="" Body="&lt;p&gt;First, what's $\lim_{a \to 0^+} \int_1^\infty ae^{-ax} \: dx$?  Second, can you show that your integral is not so far from $\int_1^\infty ae^{-ax} \: dx$?&lt;/p&gt;&#xA;" OwnerUserId="173" LastActivityDate="2010-12-02T18:41:42.373" />
  <row Id="12789" PostTypeId="2" ParentId="12787" CreationDate="2010-12-02T18:43:51.280" Score="7" ViewCount="" Body="&lt;p&gt;You can try first finding the rational roots using the &lt;a href=&quot;http://en.wikipedia.org/wiki/Rational_root_theorem&quot; rel=&quot;nofollow&quot;&gt;rational root theorem&lt;/a&gt; in combination with the &lt;a href=&quot;http://en.wikipedia.org/wiki/Factor_theorem&quot; rel=&quot;nofollow&quot;&gt;factor theorem&lt;/a&gt; in order to reduce the degree of the polynomial until you get to a quadratic, which can be solved by means of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Quadratic_equation#Quadratic_formula&quot; rel=&quot;nofollow&quot;&gt;quadratic formula&lt;/a&gt; or by &lt;a href=&quot;http://en.wikipedia.org/wiki/Completing_the_square&quot; rel=&quot;nofollow&quot;&gt;completing the square&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, to complement a little bit on Moron's answer, the first solution he found $x = 1$ can be guessed by using the rational root theorem since the theorem tells you to look for rational solutions only in the set of fractions $\frac{\pm a}{b}$ where $a, b \in \mathbb{Z}$ are integers such that $a$ divides $15$ and $b$ divides $2$. Thus you list all divisors of $15$, which are $\pm 1, \pm 3, \pm 5, \pm 15$ and the divisors of $2$ are $\pm 1, \pm 2$. then your list of possible rational roots would be $\pm \frac{1}{1}, \pm \frac{1}{2}, \pm \frac{3}{1}, \pm \frac{3}{2}, \pm \frac{5}{1}, \pm \frac{5}{2}, \pm \frac{15}{1}, \pm \frac{15}{2}$, and you have to start proving to see if any of those is a root of your polynomial.&lt;/p&gt;&#xA;" OwnerUserId="900" LastEditorUserId="900" LastEditorDisplayName="" LastEditDate="2010-12-02T18:58:34.990" LastActivityDate="2010-12-02T18:58:34.990" />
  <row Id="12790" PostTypeId="2" ParentId="12787" CreationDate="2010-12-02T18:47:35.440" Score="10" ViewCount="" Body="&lt;p&gt;Guessing one root sometimes opens up the whole equation for you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First notice that $\displaystyle x=1$ gives 0. So $\displaystyle x-1$ is a factor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next, rewrite as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle 2x^4 - 2x^3 + 5x^3 - 5x^2 -6x^2 + 6x - 15x + 15$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is to try and get $x-1$ as a factor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This gives us&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle 2x^3(x-1) + 5x^2(x-1) - 6x(x-1) - 15(x-1) = (x-1)(2x^3 + 5x^2 - 6x - 15)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now notice that $\displaystyle 2x^3 - 6x = 2x(x^2-3)$ and $5x^2 - 15 = 5(x^2 - 3)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle (x-1)(2x^3 + 5x^2 - 6x - 15) = (x-1)(2x(x^2 - 3) + 5(x^2 - 3)) = (x-1)(x^2-3)(2x+5)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and so the roots are $\displaystyle 1, \pm\sqrt{3}, -\frac{5}{2}$&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-12-02T18:47:35.440" CommentCount="2" />
  <row Id="12791" PostTypeId="2" ParentId="12767" CreationDate="2010-12-02T18:57:28.790" Score="2" ViewCount="" Body="&lt;p&gt;It is false.  Take $E = \mathcal O_{\mathbb P^n}(2)$ and $F= \mathcal O_{\mathbb P^n}(-2)$.  Then,&#xA;$$&#xD;&#xA;H^0(\mathbb P^n, \mathcal O_{\mathbb P^n}) = &#xD;&#xA;H^0(\mathbb P^n, E\otimes F) \ne &#xD;&#xA;H^0(\mathbb P^n, \mathcal O_{\mathbb P^n}(2)) \otimes&#xD;&#xA;H^0(\mathbb P^n, \mathcal O_{\mathbb P^n}(-2)).&#xD;&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="3009" LastActivityDate="2010-12-02T18:57:28.790" />
  <row Id="12794" PostTypeId="2" ParentId="12763" CreationDate="2010-12-02T19:16:23.437" Score="0" ViewCount="" Body="&lt;p&gt;Here's a geometric way of thinking that might be helpful: Consider a family of level surfaces $f(x,y,z)=C$ for some evenly spaced values of $C$ (where the spacing should be fairly small). These level surfaces will lie closely stacked in space near points where $|\nabla f|$ is large, and farther apart near points where $|\nabla f|$ is small.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(The two-dimensional counterpart is curves of constant elevation on a map; they are densely packed where the slope of the terrain is steep.)&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2010-12-02T19:16:23.437" />
  <row Id="12795" PostTypeId="2" ParentId="12768" CreationDate="2010-12-02T19:24:37.093" Score="1" ViewCount="" Body="&lt;p&gt;Suppose browser $A$ finishes loading a given page in five seconds, while browser $B$ completes in half the time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How many times faster is $B$ than $A$? Well, twice, of course. But what about the inverse? How many &quot;times faster&quot; is $B$ than $A$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One way to answer this is to assume that &quot;$-x$ times faster than&quot; just means &quot;$1/x$ times the speed of&quot;. The intuitive problem with this is determining how much faster something is moving compared to something moving slower than it. It's just a really awkward way of putting it, but basically we can look at the number of &quot;times faster&quot; as a negative &lt;em&gt;or&lt;/em&gt; a fractional value.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They're both more or less valid interpretations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now if we ask the question, how many times slower than $B$ is $A$, we get the original answer -- it's twice as slow. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So how many times faster is it? It depends on which interpretation we're using. It's correct to say that it is -2 times &quot;as fast&quot; since it is twice as slow; but also (and more intuitively) correct to say the speed of $A$ is 0.5 times the speed of $B$.&lt;/p&gt;&#xA;" OwnerUserId="4300" LastEditorUserId="4300" LastEditorDisplayName="" LastEditDate="2010-12-02T23:20:47.490" LastActivityDate="2010-12-02T23:20:47.490" />
  <row Id="12796" PostTypeId="1" CreationDate="2010-12-02T19:31:09.090" Score="1" ViewCount="58" Body="&lt;p&gt;What is the probability of getting 3 heads in a row? Would it be 1/8?&lt;/p&gt;&#xA;" LastActivityDate="2010-12-02T19:36:45.130" Title="probability of heads" Tags="&lt;probability&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="12797" PostTypeId="2" ParentId="12766" CreationDate="2010-12-02T19:33:05.680" Score="4" ViewCount="" Body="&lt;p&gt;There is a slight mistake in your argument: a priori, you don't know if the order $bN$ in $G/N$ is exactly $k$, but you do know that it is of order &lt;em&gt;dividing&lt;/em&gt; $k$, hence relatively prime to $p$. In order to show that the order cannot be $1$, you need to use the fact that $N$ itself is a $p$-group (did you notice that you never used it?) so that $g\notin N$. Once you know that, &lt;em&gt;then&lt;/em&gt; your contradiction would follow. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As to a simpler way, it depends on your definition of $p$-group! If to you a $p$-group is a group whose order is a power of $p$, then the very simplest way is simply to remember that for any group $G$ and any subgroups $H\subseteq K\subseteq G$, you have $[G:H]=[G:K][K:H]$ (cardinal multiplication in the case of infinite indices if necessary). So $|G|=[G:1]=[G:N][N:1] = |G/N||N|$ is a power of $p$, hence $G$ is a $p$-group.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, there are meanings of $p$-group: some authors define $G$ to be a $p$-group if for every $g\in G$ there exists $k\gt 0$ such that $g^{p^k}=1$ (that is, the order of every element is a power of $p$). Lagrange's and Cauchy's Theorems tell you that for finite groups, the two definitions coincide. But  there are infinite groups that satisfy the second meaning (but obviously not the first); the &lt;a href=&quot;http://en.wikipedia.org/wiki/Pr%C3%BCfer_group&quot; rel=&quot;nofollow&quot;&gt;Prüfer group for $p$&lt;/a&gt; for example. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do we prove the statement under this definition, for possibly infinite groups (it's also true)? Suppose $N$ is a $p$-group and normal in $G$, and $G/N$ is a $p$-group. Let $g\in G$. We want to show that the order of $g$ is some power of $p$. Look at $gN$ in $G/N$: since $G/N$ is a $p$-group by hypothesis, then there exists $k\gt 0$ such that $(gN)^{p^k} = eN$; that is, $g^{p^k}N = eN$. That means that $g^{p^k}\in N$. Now use the fact that $N$ is itself a $p$-group (under this definition) to deduce that the order of $g$ must be a power of $p$.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-02T19:40:50.273" LastActivityDate="2010-12-02T19:40:50.273" />
  <row Id="12798" PostTypeId="2" ParentId="12796" CreationDate="2010-12-02T19:36:05.940" Score="4" ViewCount="" Body="&lt;p&gt;If you are assuming 3 coin tosses of a fair coin, then there are $2\times 2\times 2=8$ outcomes. Three consective heads can occur only one way. So, yes, the probability is $\frac{1}{8}$&lt;/p&gt;&#xA;" OwnerUserId="3431" LastActivityDate="2010-12-02T19:36:05.940" CommentCount="3" />
  <row Id="12799" PostTypeId="2" ParentId="12796" CreationDate="2010-12-02T19:36:45.130" Score="1" ViewCount="" Body="&lt;p&gt;Assuming a fair coin, how many sequences of three throws are there? Are they equally likely? How many of those are &quot;3 heads in a row&quot;? &lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-12-02T19:36:45.130" />
  <row Id="12800" PostTypeId="1" AcceptedAnswerId="12807" CreationDate="2010-12-02T19:48:03.197" Score="6" ViewCount="239" Body="&lt;p&gt;What are some motivating consequences of the axiom of choice (or its omission)? I know that weak forms of choice are sometimes required for interesting results like Banach-Tarski; what are some important consequences of a strong formulation of the axiom of choice? &lt;/p&gt;&#xA;" OwnerUserId="4300" LastEditorUserId="4300" LastEditorDisplayName="" LastEditDate="2010-12-02T19:54:35.620" LastActivityDate="2010-12-02T21:29:50.693" Title="Motivating implications of the axiom of choice?" Tags="&lt;set-theory&gt;&lt;teaching&gt;&lt;axioms&gt;&lt;axiom-of-choice&gt;" AnswerCount="9" CommentCount="1" FavoriteCount="1" />
  <row Id="12801" PostTypeId="2" ParentId="12662" CreationDate="2010-12-02T19:48:05.987" Score="0" ViewCount="" Body="&lt;p&gt;Here is a third alternative to statement (i).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $A=\sum_{n=1}^\infty |a_n|$ and consider the sequence of partial sums $S_N=\sum_{n=1}^N |a_{\rho(n)}|$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) Since $\rho$ is injective $S_N\le A$ for all $N$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) Note that $S_N\le S_{N+1}$ for all $N$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;By (1) $S_N$ has a finite least upper bound (that is $\sup S_N&amp;lt;\infty$), and by (2) the sequence grows to that bound.&lt;/p&gt;&#xA;" OwnerUserId="1154" LastActivityDate="2010-12-02T19:48:05.987" />
  <row Id="12802" PostTypeId="2" ParentId="12800" CreationDate="2010-12-02T19:49:26.703" Score="6" ViewCount="" Body="&lt;p&gt;Axiom of Choice $\iff$ A non-empty product of non-empty sets is non-empty.&lt;/p&gt;&#xA;" OwnerUserId="1269" LastActivityDate="2010-12-02T19:49:26.703" CommentCount="4" />
  <row Id="12803" PostTypeId="1" AcceptedAnswerId="12812" CreationDate="2010-12-02T19:52:16.873" Score="2" ViewCount="148" Body="&lt;p&gt;According to Wikipedia, a (non-commutative) ring R is local if and only if there do not exist two proper (principal) (left) ideals I1, I2 such that R = I1 + I2. It is easy to see that local rings have this property, but I'm having trouble showing the other way: if there do not exist two proper co-maximal (principal) (left) ideals then the ring is local.&lt;/p&gt;&#xA;" OwnerUserId="4341" LastEditorUserId="4341" LastEditorDisplayName="" LastEditDate="2010-12-02T21:43:39.270" LastActivityDate="2010-12-02T21:43:39.270" Title="Local rings characterization" Tags="&lt;abstract-algebra&gt;" AnswerCount="1" />
  <row Id="12804" PostTypeId="2" ParentId="12800" CreationDate="2010-12-02T19:53:42.660" Score="3" ViewCount="" Body="&lt;p&gt;A consequence that motivates me is: If $I$ is some non-empty class and $X_i$ is a non-empty class $i\in I$, then $$\prod_{i\in I}X_i\ne\emptyset$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An other consequence that motivates me is: If $I$ is some non-empty set and $X_i$ is a compact topological space $i\in I$, then $\prod_{i\in I}X_i\ne\emptyset$ is compact.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course these are known equivalences of the Axiom of choice.&lt;/p&gt;&#xA;" OwnerUserId="1154" LastActivityDate="2010-12-02T19:53:42.660" CommentCount="1" />
  <row Id="12805" PostTypeId="2" ParentId="12800" CreationDate="2010-12-02T19:54:16.073" Score="5" ViewCount="" Body="&lt;p&gt;The Axiom of choice is required to prove for example that every ring has a maximal ideal (or a minimal prime), or the existence of an algebraic closure.&lt;/p&gt;&#xA;" OwnerUserId="4284" LastActivityDate="2010-12-02T19:54:16.073" CommentCount="1" />
  <row Id="12806" PostTypeId="2" ParentId="12800" CreationDate="2010-12-02T19:54:38.803" Score="3" ViewCount="" Body="&lt;p&gt;To construct left- and right inverses of functions where the domain is not well-ordered we need the axiom of choice. Also, we can always get a well-ordering (this is equivalent to the axiom of choice) so that way we also could obtain this result (then you take for the many inverses the smallest one).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, it is important in functional analysis, for example Hahn-Banach depends on it. The axiom of choice implies Hahn-Banach but not the other way around.&lt;/p&gt;&#xA;" OwnerUserId="1120" LastActivityDate="2010-12-02T19:54:38.803" />
  <row Id="12807" PostTypeId="2" ParentId="12800" CreationDate="2010-12-02T19:55:09.400" Score="7" ViewCount="" Body="&lt;p&gt;Each of the following is equivalent to the Axiom of Choice:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Every vector space (over any field) has a basis.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Every surjection has a right inverse.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Zorn's Lemma.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The first is &lt;em&gt;extremely&lt;/em&gt; important and useful. The third is used all the time, especially in algebra, also very important and useful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could write an entire book on important consequences (and equivalents) of the Axiom of Choice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, any publisher worth his salt would reject it, since both have already been written:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Rubin, Herman; Rubin, Jean E.&#xA;&lt;strong&gt;Equivalents of the axiom of choice.&lt;/strong&gt; North-Holland Publishing Co., Amsterdam 1963 xxiii+134 pp.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Rubin, Herman; Rubin, Jean E.&#xA;&lt;strong&gt;Equivalents of the axiom of choice. II.&lt;/strong&gt;&#xA;Studies in Logic and the Foundations of Mathematics, 116. North-Holland Publishing Co., Amsterdam, 1985. xxviii+322 pp. ISBN: 0-444-87708-8 &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Howard, Paul; Rubin, Jean E.&#xA;&lt;strong&gt;Consequences of the axiom of choice.&lt;/strong&gt; Mathematical Surveys and Monographs, 59. American Mathematical Society, Providence, RI, 1998. viii+432 pp. ISBN: 0-8218-0977-6 &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;These books are probably not the best place to start, though; the first book is okay, listing some of the most important equivalents as they were known before Cohen's work, but at least the last is pretty difficult to slough through.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want a good introduction to the Axiom of Choice and some idea of its uses, Horst Herrlich's &lt;strong&gt;Axiom of Choice&lt;/strong&gt;, Lecture Notes in Mathematics v. 1876, Springer-Verlag (2006) ISBN: 3-540-30989-6 is pretty good, discussing some of the bad things that happen if you don't have AC, some of the bad things that happen if you &lt;em&gt;do&lt;/em&gt; have AC, and some alternative axioms that contradict AC but lead to very nice theorems. &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-02T20:17:44.290" LastActivityDate="2010-12-02T20:17:44.290" CommentCount="3" />
  <row Id="12808" PostTypeId="2" ParentId="12800" CreationDate="2010-12-02T20:01:36.133" Score="3" ViewCount="" Body="&lt;p&gt;There Lebesgue measure is non-trivial - there are non-measurable sets in $\mathbb{R}$.&lt;/p&gt;&#xA;" OwnerUserId="1154" LastActivityDate="2010-12-02T20:01:36.133" CommentCount="2" />
  <row Id="12809" PostTypeId="1" AcceptedAnswerId="12814" CreationDate="2010-12-02T20:02:26.253" Score="2" ViewCount="155" Body="&lt;p&gt;This argument appeared in a proof in a paper (I am paraphrasing), and I am not sure why it is true. It should be rather simple.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $(\Omega,P)$ be some sample space. Let $X$ be a random variable between $0$ and $1$. Let $U \subseteq \Omega$ and let&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$U^\prime = ${$ \omega \colon X(\omega) \ge E[X \mid U] - 1/m$}.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then $P(U^\prime) \ge P(U)/m$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can't see exactly why this is true.&lt;/p&gt;&#xA;" OwnerUserId="7076" LastActivityDate="2010-12-05T05:08:55.283" Title="a simple argument?" Tags="&lt;probability&gt;&lt;statistics&gt;" AnswerCount="3" CommentCount="3" />
  <row Id="12810" PostTypeId="2" ParentId="12800" CreationDate="2010-12-02T20:03:52.293" Score="2" ViewCount="" Body="&lt;p&gt;Adding a 2nd answer in order to keep things to one example per answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There was some time ago a question on MO about whether &lt;a href=&quot;http://mathoverflow.net/questions/45928/does-arzela-ascoli-require-choice&quot; rel=&quot;nofollow&quot;&gt;Arzelà-Ascoli requires choice&lt;/a&gt;, and the answer turns out to be that it indeed does require choice - though not the full strength of AC.&lt;/p&gt;&#xA;" OwnerUserId="1269" LastActivityDate="2010-12-02T20:03:52.293" />
  <row Id="12811" PostTypeId="2" ParentId="12763" CreationDate="2010-12-02T20:10:23.287" Score="0" ViewCount="" Body="&lt;p&gt;If you are at a point ${\bf p}$ in the domain of $f$ then the rate of change of $f$ when starting from ${\bf p}$ to nearby points depends on the direction you take. The gradient $\nabla f({\bf p})$ is a vector attached at the point ${\bf p}$; it points into the direction of maximal positive rate of change of $f$. The actual rate of change in this direction is $|\nabla f({\bf p})|$,  and in the opposite direction it is $-|\nabla f({\bf p})|$. For an arbitrary unit vector ${\bf e}$ attached at ${\bf p}$ the $directional\ derivative$ of $f$ in direction ${\bf e}$ is given by the scalar product $\nabla f({\bf p})\bullet {\bf e}$.&lt;/p&gt;&#xA;" OwnerUserId="1303" LastActivityDate="2010-12-02T20:10:23.287" />
  <row Id="12812" PostTypeId="2" ParentId="12803" CreationDate="2010-12-02T20:16:31.983" Score="0" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\ $ If $\rm\ R \ne 0\ $ is nonlocal then there exists an $\rm\: r\:$ such that both $\rm\: r\:$ and $\rm\: 1-r\:$ are nonunits. Choose maximal ideals containing them.&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-12-02T20:16:31.983" CommentCount="5" />
  <row Id="12813" PostTypeId="2" ParentId="12800" CreationDate="2010-12-02T20:18:41.263" Score="4" ViewCount="" Body="&lt;p&gt;The Hahn-Banach theorem and Tychonoff's theorem are two major ways in which the axiom of choice gets used in analysis (I am echoing other answers here because these are really important).  In addition to &lt;a href=&quot;http://en.wikipedia.org/wiki/Zorns_lemma&quot; rel=&quot;nofollow&quot;&gt;Zorn's lemma&lt;/a&gt;, the third &quot;classic&quot; theorem known to be equivalent to AC is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Well-ordering_theorem&quot; rel=&quot;nofollow&quot;&gt;well-ordering theorem&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a somewhat surprising one I was recently made aware of: you cannot prove that a countable union of countable sets is countable without some form of countable choice.  The problem is that in order for any of the standard proofs to go through, you need to choose, for each countable set in the union, a bijection of it with $\mathbb{N}$, and there is no way to do this canonically in general.  (Fortunately, for most countable sets we encounter in practice we &lt;em&gt;can&lt;/em&gt; define explicit bijections, so there's usually no problem.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A weak form of choice, the &lt;a href=&quot;http://en.wikipedia.org/wiki/Boolean_prime_ideal_theorem&quot; rel=&quot;nofollow&quot;&gt;Boolean prime ideal theorem&lt;/a&gt; (or equivalently the ultrafilter lemma), also has many useful consequences.  For example, it is equivalent (I think) to Tychonoff's theorem for compact Hausdorff spaces.  It is equivalent to &lt;a href=&quot;http://en.wikipedia.org/wiki/G%C3%B6del%27s_completeness_theorem&quot; rel=&quot;nofollow&quot;&gt;Gödel's completeness theorem&lt;/a&gt;, as well as to the &lt;a href=&quot;http://en.wikipedia.org/wiki/Compactness_theorem&quot; rel=&quot;nofollow&quot;&gt;compactness theorem&lt;/a&gt;.  It allows you to construct &lt;a href=&quot;http://en.wikipedia.org/wiki/Ultraproduct&quot; rel=&quot;nofollow&quot;&gt;ultraproducts&lt;/a&gt;, which has a whole genre of uses such as, for example, defining &lt;a href=&quot;http://en.wikipedia.org/wiki/Non-standard_analysis&quot; rel=&quot;nofollow&quot;&gt;nonstandard models&lt;/a&gt; of $\mathbb{R}$ or more exotic things.  Terence Tao has written several good blog posts, including &lt;a href=&quot;http://terrytao.wordpress.com/2007/06/25/ultrafilters-nonstandard-analysis-and-epsilon-management/&quot; rel=&quot;nofollow&quot;&gt;this one&lt;/a&gt;, on these and related subjects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might be interested in browsing &lt;a href=&quot;http://consequences.emich.edu/CONSEQ.HTM&quot; rel=&quot;nofollow&quot;&gt;http://consequences.emich.edu/CONSEQ.HTM&lt;/a&gt; for more consequences of AC.  There are really quite a lot.&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-12-02T20:24:27.280" LastActivityDate="2010-12-02T20:24:27.280" CommentCount="1" />
  <row Id="12814" PostTypeId="2" ParentId="12809" CreationDate="2010-12-02T20:21:52.057" Score="2" ViewCount="" Body="&lt;p&gt;We can assume that $U = \Omega$. So you want to show $$\Pr[X \geq E[X] - \alpha] \geq \alpha,$$ where $\alpha = 1/m$. If not, then $$E[X] &amp;lt; \alpha + (1-\alpha)(E[X] - \alpha) = (1-\alpha) E[X] + \alpha^2,$$ so that $\alpha E[X] &amp;lt; \alpha^2$ and $E[X] &amp;lt; \alpha$. But in that case, the inequality trivially holds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Some more details. Suppose that $\Pr[X &amp;gt; E[X] - \alpha] = p &amp;lt; \alpha$. Given $p$, the distribution maximizing $E[X]$ under this constraint is $$\Pr[X=E[X]-\alpha]=1-p,\quad \Pr[X=1]=p.$$ Its expectation is $$p + (1-p)(E[X] - \alpha)= E[X] - \alpha + p(1 - E[X] + \alpha).$$ Since $p &amp;lt; \alpha$ and $1 - E[X] + \alpha &amp;gt; 0$, we get that $$E[X] &amp;lt; E[X] - \alpha + \alpha(1 - E[X] + \alpha).$$ Rearranging, $$\alpha (1 + E[X]) &amp;lt; \alpha (1 + \alpha)$$ and so $E[X] &amp;lt; \alpha$, in which case the inequality trivially holds.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2010-12-03T05:12:09.257" LastActivityDate="2010-12-03T05:12:09.257" CommentCount="5" />
  <row Id="12815" PostTypeId="1" AcceptedAnswerId="12819" CreationDate="2010-12-02T20:43:23.167" Score="8" ViewCount="500" Body="&lt;p&gt;Starting with the famous Basel problem, Euler evaluated the Riemann zeta function for all even positive integers and the result is a compact expression involving Bernoulli numbers. However, the evaluation of the zeta function at odd positive integers (in terms of getting a closed form sum) is still open. There has been some progress in the form of Apery's theorem and other results such as &quot;infinitely many of $\zeta(2n+1)$ are irrational&quot; or &quot;at least one of $\zeta(5),\zeta(7),\zeta(9)$ or $\zeta(11)$ is irrational&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Question(s): Is there a high level understanding for this disparity between even and odd integers? Is it a case of there being a simple expression for $\zeta(3)$ that is out there waiting for an ingenious attack like Euler did with $\zeta(2)$? Or is the belief that such a closed form summation is unlikely? Where do the many many proofs powerful enough to evaluate $\zeta(2n)$ stumble when it comes to evaluating $\zeta(2n+1)$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Motivation: The Basel problem and Euler's solution are my all-time favorites for the sheer surprise factor and ingenuity of proof (what do $\pi$ and $\frac{sin(x)}{x}$ have to do with $\zeta(2)$??). However, I currently lack the more advanced analytical tools to appreciate the deeper results of this area. I have wondered for a while about the questions above and Internet search hasn't helped much. I would greatly appreciate any answers/references. Thanks.&lt;/p&gt;&#xA;" OwnerUserId="618" LastActivityDate="2011-01-04T13:40:59.233" Title="Riemann zeta function at odd positive integers" Tags="&lt;analytic-number-theory&gt;&lt;riemann-zeta&gt;" AnswerCount="4" CommentCount="1" FavoriteCount="4" />
  <row Id="12816" PostTypeId="2" ParentId="12768" CreationDate="2010-12-02T20:43:25.520" Score="1" ViewCount="" Body="&lt;p&gt;There are many abuses like this in the world.  I have seen things on sale that were 400% OFF!!!  Would they give me 3x the original price if I took it?  Maybe -1.13 times faster means that you send them 113% as many bits as you received in the base case.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-12-02T20:43:25.520" />
  <row Id="12817" PostTypeId="1" CreationDate="2010-12-02T20:49:30.647" Score="2" ViewCount="178" Body="&lt;p&gt;Given the following formula:&#xA;$$\bigg[\forall x P(x,x) \wedge \forall x \forall y \forall z\bigg( P(x,y) \wedge  P(y,z) \Rightarrow P(x,z)\bigg) \wedge \forall x \forall y (P(x,y) \vee P(y,x)) \bigg] \Longrightarrow \exists y \forall x P(y,x)&#xD;&#xA;$$ &#xA;It is known that this formula is always true if x,y,z are in any domain with finite size, but not always true in infinite case!!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe that the first part can be proved by induction! If the domain contains only one point x then formula is true. Suppose its true for a domain of size k, how to prove the k+1 case?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the second part I need to find a counterexample for such P in some infinite domain, say the integers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any help?!&lt;/p&gt;&#xA;" OwnerUserId="4343" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-12-07T07:56:37.257" LastActivityDate="2010-12-07T07:56:37.257" Title="Model Theory-logic" Tags="&lt;logic&gt;&lt;model-theory&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="12818" PostTypeId="2" ParentId="12817" CreationDate="2010-12-02T20:52:50.333" Score="4" ViewCount="" Body="&lt;p&gt;The premises tell you that $P$ is reflexive and transitive, as well as total; this means that it is a total pre-order on the domain. The conclusion is that the pre-order has a smallest element.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How would you go about doing it by induction? Well, suppose you know the result holds in any domain with fewer than $k$ elements, and your domain has $k$ elements. Pick $x_0\in D$ ($D$ the domain), and let $D&amp;#39;=D\setminus\{x_0\}$, and let $P&amp;#39;=P\cap D&amp;#39;\times D&amp;#39;$. If you can prove that $P&amp;#39;$ satisfies the premises of the formula relative to $D&amp;#39;$ (given that $P$ satisfies them relative to $D$), then you'll be able to conclude that there is a $y\in D&amp;#39;$ such that $(y,x)\in P&amp;#39;$ for every $x\in D&amp;#39;$. Now, how can $y$ fail to be an element with the right property if you now consider it in $D$? And if that happens, can you find an element that &lt;em&gt;will&lt;/em&gt; work?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As to a counterexample, given my first paragraph, does something suggest itself?&lt;/p&gt;&#xA;&#xA;&lt;hr/&gt;&#xA;&#xA;&lt;p&gt;Okay, you are asking for some more help on getting the induction going.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $n$ be the number of elements in the model $D$. We want to prove the formula holds for $D$. We do so by induction on $n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The base case is $n=1$. To establish the proposition, we are allowed to assume the premise holds: that for all $x\in D$ we have $P(x,x)$, that for all $x$, $y$, and $z$ in $D$, if $P(x,y)$ and $P(y,z)$ both hold, then $P(x,z)$ holds; and that for all $x$ and $y$ in $D$, either $P(x,y)$ holds or $P(y,x)$ holds. We want to show that there exists a $y\in D$ such that for all $x\in D$, $P(y,x)$ holds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $D = \{x_1\}$ has a single element, what can $y$ be? It must be $x_1$. So you need to show that for every $x$ in $D$, $P(x_1,x)$ holds. Surely this is easy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Okay, so now assume that you know the proposition is true in &lt;em&gt;any&lt;/em&gt; domain with $k$ elements, and assume that $D=\{x_1,x_2,\ldots,x_k,x_{k+1}\}$ has $k+1$ elements. We are also allowed to assume that $P(x,x)$ holds for all $x\in D$, that for all $x,y,z$ in $D$, if $P(x,y)$ and $P(y,z)$ holds, then $P(x,z)$ holds; and that for all $x$ and $y$ in $D$, either $P(x,y)$ holds or $P(y,x)$ holds. We want to show that there is a $y\in D$ such that for all $x\in D$, $P(y,x)$ holds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider $D&amp;#39;=D-\{x_{k+1}\}$, and let $P&amp;#39;$ be a binary relation on $D&amp;#39;$ defined by &#xA;$$P'(x,y)\text{ in $D'$} \Leftrightarrow P(x,y) \text{in $D$.}$$&#xA;Verify that $P&amp;#39;$ satisfies the premise of the proposition in $D&amp;#39;$: for all $x\in D&amp;#39;$, $P&amp;#39;(x,x)$ holds; if $P&amp;#39;(x,y)$ and $P&amp;#39;(y,z)$ hold, then $P&amp;#39;(x,z)$ hold; and for all $x$ and $y$ in $D&amp;#39;$, either $P&amp;#39;(x,y)$ holds or $P&amp;#39;(y,x)$ holds. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the induction hypothesis, the formula is &lt;em&gt;true&lt;/em&gt; in $D&amp;#39;$; so if the &lt;em&gt;premise&lt;/em&gt; is also true in $D&amp;#39;$, then the conclusion is true in $D&amp;#39;$. So there is a $y_0\in D&amp;#39;$ such that for all $x\in D&amp;#39;$ you have $P&amp;#39;(y_0,x)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, go back to $D$; $y_0$ is in $D$, and for every $x\in D$, if $x\in D&amp;#39;$ (that is, if $x\neq x_{k+1}$) then $P(y_0,x)$ holds. Now, what happens to $x_{k+1}$? Well, by our premises, either $P(y_0,x_{k+1})$ holds, or else $P(x_{k+1},y_0)$ holds because $y_0$ and $x_{k+1}$ are both in $D$. Think about what each of those two things would tell you in terms of trying to find a $y\in D$ such that $P(y,x)$ holds for all $x$ in $D$.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-03T04:56:10.807" LastActivityDate="2010-12-03T04:56:10.807" CommentCount="4" />
  <row Id="12819" PostTypeId="2" ParentId="12815" CreationDate="2010-12-02T21:00:02.087" Score="12" ViewCount="" Body="&lt;p&gt;The zeta function is defined as a sum over the positive integers, but as far as actually evaluating it, it turns out to be more natural to think of it as a sum over all nonzero integers; thus we should really be thinking about $\sum_{n \neq 0} \frac{1}{n^k}$.  For $k$ even this is just $2 \zeta(k)$ and there are various ways to evaluate this more symmetric sum, e.g. by writing down a meromorphic function with the right poles, or a Fourier series with the right coefficients, etc.  But for $k$ odd this is equal to zero, since terms cancel with their negatives!  Written in this way, the zeta function at even integers reveals its alter ego as an &lt;a href=&quot;http://en.wikipedia.org/wiki/Eisenstein_series&quot; rel=&quot;nofollow&quot;&gt;Eisenstein series&lt;/a&gt; in one dimension.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This cancellation phenomenon occurs in Euler's classic &quot;proof,&quot; since the infinite product for $\sin z$ that he uses has zeroes at all integer multiples of $\pi$, not just the positive ones.  It also occurs in the general proof that proceeds by considering the generating function $\frac{z}{e^z - 1}$ for the &lt;a href=&quot;http://en.wikipedia.org/wiki/Bernoulli_number&quot; rel=&quot;nofollow&quot;&gt;Bernoulli numbers&lt;/a&gt;.  As you might know, the closed form of $\zeta(2k)$ involves Bernoulli numbers, and again $\frac{z}{e^z - 1}$ has poles at $2 \pi i n$ for all nonzero integers $n$, not just the positive ones.  I describe how this works in slightly more detail &lt;a href=&quot;http://math.stackexchange.com/questions/8337/different-methods-to-compute-sum-n1-infty-frac1n2/8373#8373&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another way to think about the difference between the even and odd cases is that one can think of the even cases as $L^2$ norms of appropriate Fourier series; this is precisely how a standard proof of the evaluation of $\zeta(2)$ works.  But for the odd cases we don't get an $L^2$ norm; instead we get a mysterious inner product.  &lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-12-02T21:05:33.360" LastActivityDate="2010-12-02T21:05:33.360" CommentCount="2" />
  <row Id="12820" PostTypeId="1" AcceptedAnswerId="12861" CreationDate="2010-12-02T21:08:26.180" Score="0" ViewCount="106" Body="&lt;p&gt;Motivated by my ignorance &lt;a href=&quot;http://math.stackexchange.com/questions/12767/cohomological-decomposition-of-tensor-sheaves/12774#12774&quot;&gt;here&lt;/a&gt;, if $X$ is a projective toric variety, is &#xA;$$H^m(X, \mathcal O_X) \cong &#xD;&#xA;\begin{cases}&#xD;&#xA;0 &amp;amp; m &amp;gt; 0 \\&#xD;&#xA;\mathbb C &amp;amp; m = 1&#xD;&#xA;\end{cases}&#xD;&#xA;$$&#xD;&#xA;as for $\mathbb P^n$?&lt;/p&gt;&#xA;" OwnerUserId="3008" LastEditorUserId="3008" LastEditorDisplayName="" LastEditDate="2010-12-02T23:10:42.287" LastActivityDate="2010-12-03T03:31:10.913" Title="Cohomology of $\mathcal O_X$ for toric varieties" Tags="&lt;algebraic-geometry&gt;&lt;cohomology&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="12821" PostTypeId="1" CreationDate="2010-12-02T21:11:20.800" Score="1" ViewCount="85" Body="&lt;p&gt;If $F_{1}$ is a subfield of $F_{2}$, is $GL(N,F_{1})$ contained in $GL(N,F_{2})$?   I am particular interested in the case of when $F_{1} =\mathbb{Q}(\omega)$ and $F_{2} = \mathbb{C}$.   Here $\omega$ is a complex root of unity.&lt;/p&gt;&#xA;" OwnerUserId="832" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-03T04:57:45.000" LastActivityDate="2010-12-03T04:57:45.000" Title="If $F_{1}$ is a subfield of $F_{2}$, is $GL(N,F_{1})$ contained in $GL(N,F_{2})$?" Tags="&lt;abstract-algebra&gt;" CommentCount="9" />
  <row Id="12822" PostTypeId="1" AcceptedAnswerId="12827" CreationDate="2010-12-02T21:11:34.823" Score="2" ViewCount="228" Body="&lt;p&gt;I am trying to better understand how unique factorization of algebraic integers in an algebraic number ring implies that the class number of that number ring is 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am asking for some examples of this to get me started.  &lt;/p&gt;&#xA;" OwnerUserId="2212" LastActivityDate="2010-12-02T23:22:17.380" Title="Examples of Class Group" Tags="&lt;number-theory&gt;" AnswerCount="3" />
  <row Id="12823" PostTypeId="1" AcceptedAnswerId="12830" CreationDate="2010-12-02T21:19:13.233" Score="0" ViewCount="139" Body="&lt;p&gt;Hi. I need to re-arrange the following equation (see link JPG below please, sorry I'm not familiar with how to write an equation on here and as I'm a newbie it won't let me post images) to solve for $d$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$L = 4\pi C_x d^2.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you.&lt;/p&gt;&#xA;" OwnerUserId="3686" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-02T21:37:44.287" LastActivityDate="2010-12-02T22:32:09.323" Title="Re-arranging a complex equation" Tags="&lt;algebra-precalculus&gt;" AnswerCount="1" CommentCount="7" />
  <row Id="12824" PostTypeId="1" AcceptedAnswerId="12835" CreationDate="2010-12-02T21:22:21.943" Score="7" ViewCount="261" Body="&lt;p&gt;So one way to define a group presentation is to say, well, let's generate the free group with some number of generators, and then quotient by saying certain elements (relators) cancel just as $aa^{-1}$ does (and other things, you have to take the normal subgroup generated by the elements, but the basic idea works, I think). That is, we have for some words $R_1, R_2, \ldots$ the relations $R_1 = R_2 = \ldots = e$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A simple example is the cyclic group of order n: one presentation is $\langle g \mid g^n \rangle$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is: can and do we do this for rings? I'm imagining you would do something like take the &quot;free ring&quot; consisting of all sums of products of generators, and perhaps instead of having all the relators equal to the trivial group they would equal the zero ideal, because then if you wanted some word $R$ to equal the multiplicative identity you would just say $R - 1 = 0$. (In particular, the comments on the vaguely related question that got me thinking about this is &lt;a href=&quot;http://math.stackexchange.com/questions/12698/why-is-mathbbzx-1-x-p-isomorphic-to-mathbbz-p-where-p-is-a-prime&quot;&gt;here&lt;/a&gt;: could you formulate a rigorous proof by &quot;dividing by the relations&quot;?)&lt;/p&gt;&#xA;" OwnerUserId="2954" LastActivityDate="2010-12-30T23:47:04.843" Title="Do people study &quot;ring presentations&quot;? Is this a dumb question?" Tags="&lt;reference-request&gt;&lt;ring-theory&gt;" AnswerCount="2" CommentCount="7" />
  <row Id="12825" PostTypeId="1" CreationDate="2010-12-02T21:25:50.507" Score="8" ViewCount="171" Body="&lt;p&gt;My research has brought me to the following, very general problem.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given a fixed, but arbitrary, natural number, $\displaystyle v$, consider the following family of polynomials: The $\displaystyle (n-1)^{th}$ derivative of &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle (1-x^2)^{v+n} \ \ \forall n \in \mathbb{N} $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to prove (or disprove) that the roots of this entire family of polynomials forms a dense subset of the interval $\displaystyle [0,1]$ for any value of $\displaystyle v$ (I am not interested in roots outside the interval $\displaystyle [0,1]$).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, given any subinterval, $\displaystyle [a,b]$,no mater how small, at least one of these polynomials has at least one root in the interval $\displaystyle [a,b]$ (for any fixed value of $\displaystyle v$).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I realize my question is very general and will happily accept any partial solutions.&lt;/p&gt;&#xA;" OwnerUserId="3971" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-12-02T22:23:01.593" LastActivityDate="2011-04-19T17:55:09.233" Title="density of roots of a family of polynomials: $(1-x^2)^{v+n}$" Tags="&lt;real-analysis&gt;&lt;polynomials&gt;&lt;special-functions&gt;&lt;roots&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="2" />
  <row Id="12826" PostTypeId="2" ParentId="12800" CreationDate="2010-12-02T21:29:50.693" Score="1" ViewCount="" Body="&lt;p&gt;I think that one of the most important implications of the axiom of choice is actually the equivalence of continuity between the Cauchy definition of $\epsilon$-$\delta$ and the Heine definition using sequences.&lt;/p&gt;&#xA;" OwnerUserId="622" LastActivityDate="2010-12-02T21:29:50.693" />
  <row Id="12827" PostTypeId="2" ParentId="12822" CreationDate="2010-12-02T21:30:20.677" Score="4" ViewCount="" Body="&lt;p&gt;I'm not sure what you mean by giving you examples... the abstract argument seems simple enough to me:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Every principal ideal that is generated by a prime element is a prime ideal: suppose $ab\in (p)$; then $p|ab$, and since $p$ is prime, $p|a$ or $p|b$, so $a\in (p)$ or $b\in (p)$. Conversely, a nonzero principal ideal is prime if and only if the generator is a prime element.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose $\mathfrak{M}$ is a maximal ideal of the number ring; let $a\in\mathfrak{M}$. Then $a = p_1\cdots p_k$ for some primes $p_i$ by unique factorization into primes, hence $(a) = (p_1\cdots p_k) = (p_1)(p_2)\cdots(p_k)\subseteq \mathfrak{M}$. Since $\mathfrak{M}$ is maximal, it is prime, so there exists $i$ such that $(p_i)\subseteq \mathfrak{M}$. But since $p_i$ is a prime element, then $(p_i)$ is a prime ideal, and in a number ring every nonzero prime ideal is maximal; hence $\mathfrak{M}=(p_i)$. Therefore, every maximal ideal is principal; in particular, every nonzero prime ideal is principal, since nonzero prime ideals are maximal. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let $\mathfrak{I}$ be any ideal; if $\mathfrak{I}=(0)$, then it is principal. Otherwise, $\mathfrak{I}$ is a product of prime ideals by the Fundamental Theorem of Dedekind Domains, so $\mathfrak{I}=(p_1)\cdots (p_k)$ with $p_i$ a prime element (since primes are maximal, which are principal, and therefore generated by prime elements); but $(p_1)\cdots(p_k) = (p_1\cdots p_k)$ is principal, so $\mathfrak{I}$ is principal. Thus, if you have unique factorization, then every ideal is principal, and therefore the class number is $1$: given any two nonzero ideals $\mathfrak{I}$ and $\mathfrak{J}$, we can find elements $a$ and $b$ such that $\mathfrak{I}=(a)$ and $\mathfrak{J}=(b)$, and therefore we have that $b\mathfrak{I}=a\mathfrak{J}$, hence $\mathfrak{I}\sim \mathfrak{J}$ in the ideal class group. Since this holds for any two nonzero ideals, the ideal class group consists of exactly one class. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;hr/&gt;&#xA;The key really is that you have unique factorization into prime ideals and that nonzero primes are maximal; without that, you would be out of luck. One problem with the simplest examples ($\mathbb{Z}[i]$, $\mathbb{Z}[\sqrt{-2}]$, for instance) is that are also Euclidean domains, so that's really how you get the unique factorization (as a further consequence, a prior one already being that it is a PID). I think the smallest imaginary quadratic which is a PID but not Euclidean is $\mathbb{Z}[(1+\sqrt{-19})/2]$, and I'm not sure I would want to try to work out explicitly the argument above with specific elements of that ring.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-02T21:35:27.703" LastActivityDate="2010-12-02T21:35:27.703" CommentCount="1" />
  <row Id="12829" PostTypeId="2" ParentId="12822" CreationDate="2010-12-02T21:55:54.123" Score="3" ViewCount="" Body="&lt;p&gt;$\rm PID\:$s are &lt;em&gt;precisely&lt;/em&gt; the $\rm UFD\:$s having dimension $\le 1\ $ (i.e. every nonzero prime ideal is maximal). Therefore $\rm UFD $ Dedekind domains are $\rm PID\:,$ so they have trivial class group. Here's the key result:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;THEOREM&lt;/strong&gt; $\rm\ \ \ TFAE\ $ for a $\rm UFD\ D$ &lt;br/&gt;&#xA;$1)\ $ prime ideals are maximal if nonzero&lt;br&gt;&#xA;$2)\ $ prime ideals are principal&lt;br&gt;&#xA;$3)\ $ maximal ideals are principal&lt;br&gt;&#xA;$4)\ \rm\ gcd(a,b) = 1\ \Rightarrow\ (a,b) = 1$ &lt;br/&gt;&#xA;$5)\ $ $\rm D$ is Bezout&lt;br&gt;&#xA;$6)\ $ $\rm D$ is a $\rm PID$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof&lt;/strong&gt; $\ $ (sketch of $1 \Rightarrow 2 \Rightarrow 3 \Rightarrow 4 \Rightarrow 5 \Rightarrow 6 \Rightarrow 1$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$1\Rightarrow 2)$ $\rm\ \ P\supset (p)\ \Rightarrow\ P = (p)$ &lt;br/&gt;&#xA;$2\Rightarrow 3)$ $\ \: $ Clear.&lt;br&gt;&#xA;$3\Rightarrow 4)$ $\ \ \rm (a,b) \subsetneq P = (p)\ $ so $\rm\ (a,b) = 1$ &lt;br/&gt;&#xA;$4\Rightarrow 5)$ $\ \ \rm c = \gcd(a,b)\ \Rightarrow\ (a,b) = c\ (a/c,b/c) = (c)$ &lt;br/&gt;&#xA;$5\Rightarrow 6)$ $\ \ \rm 0 \ne I \subset D\:$ Bezout is generated by an elt with the least number of prime factors&lt;br&gt;&#xA;$6\Rightarrow 1)$ $\ \ \rm P \supset (p),\ a \not\in (p)\ \Rightarrow\ (a,p) = (1)\ \Rightarrow\ P = (p)$&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-12-02T22:44:15.793" LastActivityDate="2010-12-02T22:44:15.793" CommentCount="0" />
  <row Id="12830" PostTypeId="2" ParentId="12823" CreationDate="2010-12-02T22:32:09.323" Score="1" ViewCount="" Body="&lt;p&gt;As Arturo said&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$L=4 \pi C_x d^2$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\implies d^2 = \frac{L}{4 \pi C_x}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\implies d= \pm\sqrt{\frac{L}{4 \pi C_x}}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your equation looks like it is physical, so the negative square root may not be relevant&lt;/p&gt;&#xA;" OwnerUserId="2219" LastActivityDate="2010-12-02T22:32:09.323" />
  <row Id="12831" PostTypeId="2" ParentId="12822" CreationDate="2010-12-02T22:54:29.837" Score="2" ViewCount="" Body="&lt;p&gt;This is actually a frequently used (e.g. in the theory of divisors) fact of commutative algebra. A noetherian domain $A$ is a UFD if and only if every prime ideal of height one (by Krull's principal ideal theorem, this is the same thing as being minimal over a nonzero element) is principal (which corresponds, with a little work, to the statement that the &lt;a href=&quot;http://en.wikipedia.org/wiki/Weil_divisor&quot; rel=&quot;nofollow&quot;&gt;Weil divisor class group&lt;/a&gt; of a normal ring is trivial iff it is factorial). In the case of a Dedekind domain (of which a ring of integers in a number field is a paradigm example), this means that every prime ideal is principal (as $(0)$ obviously is).&#xA;In a Dedekind domain, every ideal is a product of prime ideals. So if every prime ideal is principal, so is every ideal. This is the statement that the class number is one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So this answers your question by reducing it to another result. You can find the proof and discussion of this result as Theorem 18.6 in &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://people.fas.harvard.edu/~amathew/CAnotes.pdf&quot; rel=&quot;nofollow&quot;&gt;http://people.fas.harvard.edu/~amathew/CAnotes.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="536" LastEditorUserId="536" LastEditorDisplayName="" LastEditDate="2010-12-02T23:22:17.380" LastActivityDate="2010-12-02T23:22:17.380" CommentCount="10" />
  <row Id="12832" PostTypeId="2" ParentId="12690" CreationDate="2010-12-02T23:00:54.493" Score="0" ViewCount="" Body="&lt;p&gt;The algebraic structure corresponding most naturally to the anticommutator is that of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Clifford_algebra&quot; rel=&quot;nofollow&quot;&gt;Clifford algebras&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One can construct a Clifford algebra by defining $n$ generating elements $\mathbf{e}_j$ with $j \in \{1,\ldots,n\}$ such that they satisfy the conditions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \mathbf{e}_i \mathbf{e}_j + \mathbf{e}_j \mathbf{e}_i = 2 \delta_{ij} \; ,$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\delta_{ij}$ is the Kronecker delta. From these elements you can form products of elements of the form&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathbf{e}_{i_1}\mathbf{e}_{i_2}\ldots\mathbf{e}_{i_p} \; ,$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for $p$ varying from $0$ to $n$. There's a total of $2^n$ products that can be formed and from these, taking linear combination you can generate the rest of the algebra.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can check that Pauli matrices and quaternions form Clifford algebras.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if you take a Clifford algebra with $n$ generating elements, and construct the elements $M_{ij}=\frac{1}{2}\mathbf{e}_{i}\mathbf{e}_{j}$, you can show that they form the basis of the Lie algebra $so(n)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another construct related to fermions, and which is important in the path integral representation of fermions, are the &lt;a href=&quot;http://en.wikipedia.org/wiki/Grassmann_number&quot; rel=&quot;nofollow&quot;&gt;Grassman numbers&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="3567" LastEditorUserId="3567" LastEditorDisplayName="" LastEditDate="2010-12-02T23:06:19.323" LastActivityDate="2010-12-02T23:06:19.323" CommentCount="3" />
  <row Id="12833" PostTypeId="1" CreationDate="2010-12-02T23:06:40.527" Score="7" ViewCount="146" Body="&lt;p&gt;If $X$ is a locally compact and metrizable space such that its Alexandroff compactification is &lt;em&gt;not&lt;/em&gt; first countable. Does this imply that no other compactification of $X$ can be first countable? Why?&lt;/p&gt;&#xA;" LastActivityDate="2011-02-09T21:24:43.430" Title="Alexandroff compactification question" Tags="&lt;topology&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" />
  <row Id="12834" PostTypeId="2" ParentId="7516" CreationDate="2010-12-02T23:07:35.427" Score="0" ViewCount="" Body="&lt;p&gt;An excellent book is also &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0444520775&quot; rel=&quot;nofollow&quot;&gt;http://www.amazon.com/Lectures-Curry-Howard-Isomorphism-Foundations-Mathematics/dp/0444520775&lt;/a&gt;.  This book is more focused on the $\lambda$-Calculus but it has some really excellent sections on general intuitionist logic, including Heyting Algebras, Hilbert Proofs, the (Gentzen's) sequent calculus and things like that.  &lt;/p&gt;&#xA;" OwnerUserId="3595" LastActivityDate="2010-12-02T23:07:35.427" />
  <row Id="12835" PostTypeId="2" ParentId="12824" CreationDate="2010-12-02T23:16:05.387" Score="11" ViewCount="" Body="&lt;p&gt;You certainly can: the constructions of the tensor algebra, symmetric algebra, and exterior algebra can all be performed using generators and relations (though are perhaps usually thought of conceptually via  a universal property). The operation of localization is also an example of this: one formally adjoins &quot;denominators&quot; satisfying the requisite relations (localization of $A$ at a multiplicative set $S$ can be thought of adjoining variables $x_s, s \in S$ with the relations $s x_s = 1$). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The notion of a ring having a &quot;finite presentation&quot; over another ring $R$ (i.e. be  the quotient of a polynomial ring $R$ by a finitely generated ideal--that is, by finitely many generators and relations) is of importance in algebraic geometry, as many properties proved in, say, the finite type case over noetherian schemes extend here (e.g. I believe Chevalley's theorem of the constructibility of the image of a constructible set under a morphism of schemes is true under finite presentation hypotheses). I am not sufficiently qualified to say much more on this topic, but you might find BCnrd's answer &lt;a href=&quot;http://mathoverflow.net/questions/1788/does-finitely-presented-mean-always-finitely-presented-answered-yes&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; enlightening.&lt;/p&gt;&#xA;" OwnerUserId="536" LastActivityDate="2010-12-02T23:16:05.387" CommentCount="1" />
  <row Id="12836" PostTypeId="2" ParentId="12768" CreationDate="2010-12-02T23:28:16.910" Score="0" ViewCount="" Body="&lt;p&gt;Strictly speaking, $-1.13$ times faster would mean that the program is doing things in reverse order. And Opera would completely stand still. Since that can't be what they really mean, it must be an abuse of language as has already been pointed out. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could it be though that the original scale was logarithmic and that the person then just interpreted the numbers incorrectly? Which would mean that the graph is correct, but the person who added the 'times faster' doesn't get the concept of logarithm?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, $-1.13$ should be $a^{-1.13}$ in some suitable base, probably base $10$, $2$ or $e$?&lt;/p&gt;&#xA;" OwnerUserId="3567" LastActivityDate="2010-12-02T23:28:16.910" CommentCount="2" />
  <row Id="12837" PostTypeId="1" AcceptedAnswerId="12840" CreationDate="2010-12-02T23:42:49.027" Score="7" ViewCount="186" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;There are ten people in a room:&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Person 1 knows 9 people. Person 2&#xA;  knows 8 people. etc. Person 9 knows 1&#xA;  person.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;How many people does person 10 know?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What is the answer and what is the principle behind this question?&lt;/p&gt;&#xA;" OwnerUserId="4350" LastEditorUserId="4350" LastEditorDisplayName="" LastEditDate="2010-12-02T23:51:36.443" LastActivityDate="2010-12-03T00:01:17.413" Title="Ten people in a room" Tags="&lt;logic&gt;&lt;puzzle&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="12838" PostTypeId="1" CreationDate="2010-12-02T23:50:53.137" Score="2" ViewCount="84" Body="&lt;p&gt;(Assume a unit speed parametrization)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Prove that an admissible curve $c:(a,b) \to \mathbb R^3$ (hence $c^{\prime} \times c^{\prime\prime}$ is never zero) with a zero torsion is contained in a plane.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand why this should be true, if there is no torsion element to the curve then $c$ can't twist into the third dimension.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\mathbf T = c^{\prime}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\mathbf N = c^{\prime\prime}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\mathbf B = \mathbf T \times \mathbf N$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\mathbf T^{\prime} = \kappa \mathbf N$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\mathbf N^{\prime} = -\kappa \mathbf T + \tau\mathbf B$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\mathbf B^{\prime} = -\tau\mathbf N$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if the torsion $\tau$ is zero, $\mathbf B^{\prime} = \mathbf 0$, therefore $\mathbf B$ is a constant not equal to zero. I'm not sure where to go from here. If the normal vector of $\mathbf T$ and $\mathbf N$ is a constant, does that mean no matter what point you are at on the curve $\mathbf B$ points in the same direction?&lt;/p&gt;&#xA;" OwnerUserId="3333" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-03T00:33:05.113" LastActivityDate="2010-12-03T02:46:16.250" Title="Admissible curve contained in a plane" Tags="&lt;homework&gt;&lt;differential-geometry&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="12839" PostTypeId="1" AcceptedAnswerId="12841" CreationDate="2010-12-02T23:57:41.700" Score="2" ViewCount="194" Body="&lt;p&gt;The following is a multiplication problem in traditional base ten notation. Each letter represents a different digit. What digit does each letter represent? How did you get your foot in the door? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/cXC4Y.png&quot; alt=&quot;i hate this course&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4351" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-03T05:14:15.310" LastActivityDate="2010-12-03T05:14:15.310" Title="Algorithm (cryptarithm) question, yes it is homework, but I can't get it" Tags="&lt;homework&gt;&lt;arithmetic&gt;&lt;cryptarithm&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="12840" PostTypeId="2" ParentId="12837" CreationDate="2010-12-03T00:01:17.413" Score="6" ViewCount="" Body="&lt;p&gt;You can think of this as a problem in &lt;a href=&quot;http://en.wikipedia.org/wiki/Graph_theory&quot; rel=&quot;nofollow&quot;&gt;graph theory&lt;/a&gt;.  You have a graph $G$ on ten vertices (the ten people in the room ${ 1, 2, 3, ... 10 }$) where two people are connected by an edge if and only if they know each other.  The &lt;strong&gt;degree&lt;/strong&gt; or &lt;strong&gt;valence&lt;/strong&gt; of a vertex $v \in G$, denoted $d_v$, is the number of edges connected to it (the number of people someone knows).  You are given that $d_1 = 9, d_2 = 8, ... d_9 = 1$ and you want to know $d_{10}$ (so the problem gives you information here: it tells you that $d_{10}$ is uniquely determined by these conditions).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's one way to solve it, in steps.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Person $1$ must know everyone else; that is, he must be connected to all of the other vertices (including person $10$).  In particular, he must be the only person that person $9$ knows.  &lt;/li&gt;&#xA;&lt;li&gt;Person $2$ must know everyone else except one person, and since person $9$ only knows person $1$, person $2$ must know everyone else except person $9$ (including person $10$).  In particular, he must be the only other person that person $8$ knows.&lt;/li&gt;&#xA;&lt;li&gt;Person $3$ must know everyone else except two people, and since persons $8$ and $9$ have all their friends accounted for, person $3$ must know everyone else except persons $8$ and $9$ (including person $10$).  In particular, he must be the only other person that person $7$ knows.&lt;/li&gt;&#xA;&lt;li&gt;... and so forth.  By &lt;a href=&quot;http://en.wikipedia.org/wiki/Mathematical_induction&quot; rel=&quot;nofollow&quot;&gt;induction&lt;/a&gt; we conclude that person $10$ knows persons $1, 2, 3, 4, 5$ but not persons $6, 7, 8, 9$.  So $d_{10} = 5$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-12-03T00:01:17.413" CommentCount="2" />
  <row Id="12841" PostTypeId="2" ParentId="12839" CreationDate="2010-12-03T00:03:01.467" Score="2" ViewCount="" Body="&lt;p&gt;For the first equation $X=1,Y=0$ works. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This should be the only solution since, $XY\cdot YX= XY$ implies $XY=0$ or if $XY\neq 0$ then, $YX=1$. If $XY=0$ then $X=Y=0$ (remember $X,Y$ are digits here), so they are not distinct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the second part, $Y=0$, so the only way to get a third digit in the final solution, would be to have $X+Z=1+Z$ be a two digit sum. This can only happen if $Z=9$, so $V=0$ and $W=1$&lt;/p&gt;&#xA;" OwnerUserId="3431" LastEditorUserId="3431" LastEditorDisplayName="" LastEditDate="2010-12-03T00:15:39.067" LastActivityDate="2010-12-03T00:15:39.067" CommentCount="3" />
  <row Id="12842" PostTypeId="1" AcceptedAnswerId="12843" CreationDate="2010-12-03T00:03:31.640" Score="4" ViewCount="168" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;A newspaper launches a competition. It said that readers should submit one number between 1 and 1000. A £2000 prize would be awarded to the person that got the closest to 2/3 of the mean of all the entries. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What would a smart entry be? What is the underlying principle behind this type of question. I know it involves game theory but I do not fully understand this type of problem yet. Can anyone draw the game theory tables involved / explain the Nash Equilibria?&lt;/p&gt;&#xA;" OwnerUserId="4350" LastActivityDate="2010-12-03T00:34:24.143" Title="Newspaper competition" Tags="&lt;puzzle&gt;&lt;game-theory&gt;&lt;nash-equilibrium&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" />
  <row Id="12843" PostTypeId="2" ParentId="12842" CreationDate="2010-12-03T00:17:20.233" Score="12" ViewCount="" Body="&lt;p&gt;The only Nash equilibrium is that everybody submits $1$.  To see this, consider any other set of strategies for all the players.  Suppose each player $i, 1 \le i \le n$ submits numbers with an expected value of $p_i$ and that at least one of the $p_i$ is greater than $1$.  Then the expected value of the mean of the entries is $\frac{p_1 + ... + p_n}{n}$, and there must be at least one player $j$ for which this mean is less than or equal to their number.  But then the expected value of $\frac{2}{3}$ of the mean must be strictly less than $p_j$, so player $j$ has an incentive to switch to a lower number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a funny example.  Although the Nash equilibrium suggests that playing $1$ should be an optimal strategy, in practice you will always lose when you do this.  The issue, I think, is that it only makes sense to adopt the Nash equilibrium strategy if you expect everybody to be playing optimally, but in practice most people will not think this game through enough to actually find the Nash equilibrium strategy.  I've played this game a few times, and what seems to be the case in practice is that to most players you can associate a number $n$ such that the number they pick is $\left( \frac{2}{3} \right)^n$ times the maximum.  I like to think of $n$ as being a very rough measure of how far the players think ahead, since one can think of $n$ as being the number of times they recurse through the following argument:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Since the maximum you can play is $1000$, the maximum the average could be is $\frac{2}{3} \cdot 1000$, so I should never guess more than that.&lt;/li&gt;&#xA;&lt;li&gt;But since nobody should guess more than $\frac{2}{3} \cdot 1000$, the maximum the average could be is $\left( \frac{2}{3} \right)^2 1000$, so I should never guess more than that.&lt;/li&gt;&#xA;&lt;li&gt;But...&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;and so forth.  If the people you're playing with are smart enough, $n$ will actually measure something much more complicated: each player will be trying to guess how far everyone else will think ahead, and then guessing how far everyone else will guess how far everyone else thinks ahead, and then... this phenomenon, in my opinion, is much more interesting than Nash equilibria; I think it is related to the idea of &lt;a href=&quot;http://en.wikipedia.org/wiki/Common_knowledge_(logic)&quot; rel=&quot;nofollow&quot;&gt;common knowledge&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(The last time I played this game, the possible numbers were $0$ through $100$.  I and one other person submitted $100$ just to throw off the results.  The actual average ended up being something like $19$.  So here $n \approx 9.7$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, here's a cute observation.  In general the existence of Nash equilibria can be proven using the &lt;a href=&quot;http://en.wikipedia.org/wiki/Nash_equilibrium#Alternate_proof_using_the_Brouwer_fixed_point_theorem&quot; rel=&quot;nofollow&quot;&gt;Brouwer fixed point theorem&lt;/a&gt;.  The second argument I gave above establishes the existence (and uniqueness) of the Nash equilibrium using essentially the &lt;a href=&quot;http://en.wikipedia.org/wiki/Banach_fixed_point_theorem&quot; rel=&quot;nofollow&quot;&gt;Banach fixed point theorem&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-12-03T00:34:24.143" LastActivityDate="2010-12-03T00:34:24.143" CommentCount="3" />
  <row Id="12845" PostTypeId="2" ParentId="12842" CreationDate="2010-12-03T00:26:04.687" Score="2" ViewCount="" Body="&lt;p&gt;When solving problems like this the first thing to do is (almost) always to try and reduce the problem to a simpler problem; in this case, for example, pretending the number of allowed numbers is 10, and not 1000, is a good first thing to do.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You may also simplify the problem by allowing everyone to know the current mean value. That way you'll see that everybody has an incentive to submit lower and lower numbers - resulting in Qiaochu's answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(as a &quot;by the way&quot;, I'll recommend G. Polya's book &quot;How to solve it&quot;, which talks about general problem solving techniques.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: This problem reminds me of a &quot;paradox&quot; I read about yesterday:&#xA;&quot;A prisoner is awating death sentence, and is told he will be executed withing this week (it's Monday now), but on a day he's not expecting it. The prisoner reasons that then he can't be executed on Sunday (because if he lives then, that must be the day), and the same reasoning tells us that he will not be executed on Saturday either, and so on. The prisoner concludes that he wont be executed after all. (without expecting it, he is executed on Monday)&quot;&lt;/p&gt;&#xA;" OwnerUserId="4284" LastActivityDate="2010-12-03T00:26:04.687" CommentCount="1" />
  <row Id="12846" PostTypeId="1" AcceptedAnswerId="12847" CreationDate="2010-12-03T01:20:36.677" Score="2" ViewCount="209" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have this question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a \equiv 12^{772} \pmod{71}$, when $0 \leq a &amp;lt; 71$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and I am having troubles getting it started.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{772}{71}$ is $10$ with a remainder of $62$;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do I do this question?&lt;/p&gt;&#xA;" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-03T01:34:47.233" LastActivityDate="2010-12-03T18:47:18.977" Title="Using Fermat's little theorem" Tags="&lt;number-theory&gt;&lt;modular-arithmetic&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="12847" PostTypeId="2" ParentId="12846" CreationDate="2010-12-03T01:24:50.010" Score="3" ViewCount="" Body="&lt;p&gt;$12^{772}=12^{71\cdot 10+62}\equiv 12^{10+62}=12^{71+1}\equiv 12\cdot 12=144\equiv 2$ mod $71$. So $a=2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternately,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$12^{772}=12^{70\cdot 11+2}\equiv 12^2=144\equiv 2$ mod $71$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first one uses the form $a^p\equiv a$ mod $p$ and the second uses the form $a^{p-1}\equiv 1$ mod $p$ if $gcd(a,p)=1$ where $p$ is prime.&lt;/p&gt;&#xA;" OwnerUserId="3431" LastEditorUserId="3431" LastEditorDisplayName="" LastEditDate="2010-12-03T03:28:40.430" LastActivityDate="2010-12-03T03:28:40.430" />
  <row Id="12848" PostTypeId="1" AcceptedAnswerId="12857" CreationDate="2010-12-03T01:29:35.727" Score="6" ViewCount="203" Body="&lt;p&gt;Is there an infinite set of strings whose Kolmogorov complexities are computable?&lt;/p&gt;&#xA;" OwnerUserId="4353" LastEditorUserId="630" LastEditorDisplayName="" LastEditDate="2010-12-03T02:58:30.117" LastActivityDate="2010-12-03T02:58:30.117" Title="Is there an infinite set of strings whose Kolmogorov complexities are computable?" Tags="&lt;computability&gt;&lt;kolmogorov-complexity&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="2" />
  <row Id="12849" PostTypeId="2" ParentId="12839" CreationDate="2010-12-03T01:50:00.883" Score="3" ViewCount="" Body="&lt;p&gt;The image changes the problem and turns this into a classic 'long multiplication' cryptarithm problem.  For starters, we know (for the same reasons given above) that X=1, because multiplying X by XY gives XY again.  The second line tells us that Y*XY = YZ; in particular, since Y*X = Y, we know that Y*Y can't introduce a 'carry' digit to pollute the 1-by-2 multiplication.  This means that Y is either 2 or 3, leaving Z to be either 4 or 9 respectively (i.e., '2*12 = 24' or '3*13 = 39' - notice how 4*14 = 56, and the carry digit I mentioned before turns the tens digit into a 5, one more than the Y that it needs to be).  From here it's just trial and error, performing the rest of the calculation and checking to see whether either one gives duplicate digits.  For XY=12, we get XY*YX = 12*21 = 252 - but notice that this would be written as 'YVY', since we're told that each letter corresponds to a unique digit; we couldn't have W and Y both be 2.  However, with XY=13, we get XY*YX = 13*31 = 403, so ultimately we get the solution: X=1, Y=3, Z=9, W=4, V=0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(ETA: The multiplication can be confusing if you haven't seen it before - this is the old-style 'long multiplication' way of doing products, which I seem to recall isn't taught as much any more.  Each line between the two horizontal rules corresponds to the result of multiplying one of the lower digits by the entire upper number, and then those lines are all added together to produce the sum beneath the lower horizontal rule.)&lt;/p&gt;&#xA;" OwnerUserId="785" LastActivityDate="2010-12-03T01:50:00.883" />
  <row Id="12850" PostTypeId="1" AcceptedAnswerId="12854" CreationDate="2010-12-03T01:53:56.307" Score="3" ViewCount="90" Body="&lt;p&gt;Assume $f:(a,b) \to \mathbb R^3$ is an admissible unit speed curve (hence $f^{\prime} \times f^{\prime\prime}$ is never zero) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $f$ lies on the sphere with center $a$ and radius $r$ prove that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f = a - (1/\kappa) \mathbf N - (1/\kappa)^{\prime} (1/\tau) \mathbf B$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With $|f^{\prime}| = 1$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\mathbf T = f^{\prime}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\mathbf N = f^{\prime\prime}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\mathbf B = \mathbf T \times \mathbf N$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\mathbf T^{\prime} = \kappa \mathbf N$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\mathbf N^{\prime} = -\kappa \mathbf T + \tau\mathbf B$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\mathbf B^{\prime} = -\tau\mathbf N$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've been using the hint that since $f$ lies on the sphere then  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(f-a) \cdot (f-a) = r^2$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And trying to differentiate it (three times) to get what I want but I'm not getting very far. I start with  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(f-a) \cdot (f-a) = r^2$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Differentiate&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$2\mathbf T \cdot (f-a) = 0$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Differentiate&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$2 + 2\kappa\mathbf N \cdot (f-a) = 0$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Differentiate&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\mathbf N \cdot \mathbf T + (-\kappa\mathbf T + \tau\mathbf B + 2(\kappa^{\prime}/\kappa)\mathbf N) \cdot (f-a) = 0$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then I'm not sure where to go.&lt;/p&gt;&#xA;" OwnerUserId="3333" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-03T01:57:57.877" LastActivityDate="2010-12-03T02:33:01.223" Title="Prove using an admissible unit speed curve and a Frenet frame" Tags="&lt;homework&gt;&lt;differential-geometry&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="12851" PostTypeId="2" ParentId="12846" CreationDate="2010-12-03T02:00:38.827" Score="2" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\rm\ \ \ X^N \equiv X\ \ \Rightarrow\ \ X^{A + B\ N + C\ N^2 + \cdots} \equiv\ X^{A + B + C + \cdots}\ (mod\ N)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;NOTE&lt;/strong&gt; $\ $ This is nothing but &quot;casting out orders&quot; - a cyclic group analog of &lt;a href=&quot;http://en.wikipedia.org/wiki/Casting_out_nines&quot; rel=&quot;nofollow&quot;&gt;&lt;strong&gt;casting out nines&lt;/strong&gt;.&lt;/a&gt; Namely, $\rm\ X^{N-1} = 1\ \Rightarrow\ X^N = X\ $ so we can map $\rm\ N\to 1$ when calculating integral powers of $\rm\:X\:$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When $\rm\:X\:$ has order dividing $\:9\:,\:$ then it amounts precisely to casting out nines from its exponents.&#xA;For example, $\rm\ mod\ 27:\ (-2)^9 \equiv 1\ \Rightarrow\ (-2)^{721} \equiv -2\ \ $ since $\rm\ \ 721\equiv 7+2+1\equiv 1\ \:(mod\ 9)$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-12-03T18:47:18.977" LastActivityDate="2010-12-03T18:47:18.977" />
  <row Id="12852" PostTypeId="1" AcceptedAnswerId="12853" CreationDate="2010-12-03T02:20:36.133" Score="2" ViewCount="133" Body="&lt;p&gt;Hi, I'm not very expert in math, but I have a curiosity that could help me many times ..&#xA;Is there any way to find a function, even just similar, from a set of values?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;example:&#xA;I have some values (x, y):&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;{(200, 11), (72, 5), (36,3), (28,3), (18,2), (12,2), ...}&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;what is the function that can return these values and all others who follow this trend on a graph, more easily and better possible?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;thank you, hello!&lt;/p&gt;&#xA;" OwnerUserId="4356" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-03T02:28:18.387" LastActivityDate="2010-12-03T02:31:43.187" Title="Find a function from values" Tags="&lt;functions&gt;" AnswerCount="1" CommentCount="6" />
  <row Id="12853" PostTypeId="2" ParentId="12852" CreationDate="2010-12-03T02:25:37.267" Score="3" ViewCount="" Body="&lt;p&gt;Assuming those values came from a polynomial, what you want to do is &lt;em&gt;polynomial interpolation&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case, since you have six points, the degree of the polynomial is at most five, since six coefficients are needed to uniquely determine a quintic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wolfram Alpha is able to determine interpolating polynomials from given data, e.g. &lt;a href=&quot;http://tinyurl.com/35kldba&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="498" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-03T02:31:43.187" LastActivityDate="2010-12-03T02:31:43.187" CommentCount="9" />
  <row Id="12854" PostTypeId="2" ParentId="12850" CreationDate="2010-12-03T02:33:01.223" Score="4" ViewCount="" Body="&lt;p&gt;For each $t\in(a,b)$, $\{T(t),N(t),B(t)\}$ forms an orthonormal basis for $\mathbb{R}^3$, so you can express $f(t)-a$ in terms of this basis as $f(t)-a=c_1(t)T(t)+c_2(t)N(t)+c_3(t)B(t)$, and the coefficients are obtained by taking the dot products with the corresponding vectors.  Thus the result of differentiating once shows that $c_1(t)\equiv 0$.  Solving for $(f-a)\cdot N$ after differentiating twice shows that $c_2(t)=-1/\kappa(t)$.  After differentiating three times, notice that $N$ and $T$ are orthogonal, and use what you already know from the previous steps to solve for $(f-a)\cdot B$ and obtain $c_3(t)$.   You will want to use the fact that $(1/\kappa)&amp;#39;=-\kappa&amp;#39;/\kappa^2$ to simplify the final coefficient to the given form.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also note that it is not correct that $f&amp;#39;&amp;#39;=N$.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-12-03T02:33:01.223" />
  <row Id="12855" PostTypeId="2" ParentId="12838" CreationDate="2010-12-03T02:46:16.250" Score="1" ViewCount="" Body="&lt;p&gt;Show that if $t_0$ is in $(a,b)$, then for all $t\in(a,b)$, $c(t)-c(t_0)$ lies in the plane through the origin orthogonal to (the constant vector) $B$ by showing that the function $f(t)=(c(t)-c(t_0))\cdot B$ is identically zero. &lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-12-03T02:46:16.250" />
  <row Id="12856" PostTypeId="1" CreationDate="2010-12-03T02:53:26.077" Score="6" ViewCount="217" Body="&lt;p&gt;I was just reading a proof of the dimension theorem in Steven Roman's Advanced Linear Algebra. In addressing the cases of infinite bases, Roman proceeds to show that if $\mathcal{B}$ and $\mathcal{C}$ are bases of a space $V$, then $|\mathcal{B}|\leq |\mathcal{C}|$, working up to an application of the BSC-Theorem. Anyway, he uses the string&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$|\mathcal{B}|\leq\aleph_0|\mathcal{C}|=|\mathcal{C}|.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sorry if it's an elementary question, but why does the equality follow? Here $\mathcal{C}$ is any infinite basis. Is it definition? I tried looking up multiplication of ordinals, but didn't find anything useful. Thanks.&lt;/p&gt;&#xA;" LastActivityDate="2010-12-03T08:43:45.570" Title="Multiplying Cardinal Numbers" Tags="&lt;linear-algebra&gt;&lt;set-theory&gt;" AnswerCount="3" />
  <row Id="12857" PostTypeId="2" ParentId="12848" CreationDate="2010-12-03T02:54:37.263" Score="9" ViewCount="" Body="&lt;p&gt;I think you are asking this: is there an infinite r.e. set of pairs $(\sigma,n)$ where $\sigma \in 2^{&amp;lt;\omega}$ is a string of Kolmogorov complexity $n$. The answer to that is no. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a contradiction, assume such a list is r.e. - then there arbitrarily long strings in it, and thus strings of arbitrarily high Kolmogorov complexity. Define a function $P$ that takes input $\tau \in 2^{&amp;lt;\omega}$ and does the following. First, it effectively enumerates that list until it finds a pair $(\sigma, n)$ where $n &amp;gt; |\tau|$. Then it prints out $\sigma$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The assumptions we have made ensure that $P$ is a total computable function.  Therefore, applying Kleene's recursion theorem to $P$ gives a program $e_0$ that, when run with no input, computes $P(e_0)$.  Thus the output of program $e_0$, run with no input, is a string of Kolmogorov complexity larger than $|e_0|$, which is impossible. &lt;/p&gt;&#xA;" OwnerUserId="630" LastActivityDate="2010-12-03T02:54:37.263" CommentCount="3" />
  <row Id="12859" PostTypeId="2" ParentId="12856" CreationDate="2010-12-03T03:25:17.113" Score="6" ViewCount="" Body="&lt;p&gt;No, it's not a definition. A theorem of Tarski says that if $A$ is an infinite set, then $|A| = |A \times A|$. An acessible proof of this can be found in Hungerford's &quot;Algebra&quot;. So using it and the Schröder-Bernstein theorem, it's easy to prove that if $A$ is an infinite set and $B$ is a non-empty set such that $|B| \le |A|$, then $|A||B| = |A|$.&lt;/p&gt;&#xA;" OwnerUserId="329" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2010-12-03T03:38:06.880" LastActivityDate="2010-12-03T03:38:06.880" CommentCount="2" />
  <row Id="12860" PostTypeId="2" ParentId="12856" CreationDate="2010-12-03T03:25:54.303" Score="6" ViewCount="" Body="&lt;p&gt;By hypothesis $\mathcal C$ is infinite so $\aleph_0|\mathcal{C}|= \max(\aleph_0,|\mathcal C|)  = |\mathcal{C}|$ by basic &lt;a href=&quot;http://en.wikipedia.org/wiki/Cardinal_number&quot; rel=&quot;nofollow&quot;&gt;cardinal arithmetic.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-12-03T03:25:54.303" CommentCount="1" />
  <row Id="12861" PostTypeId="2" ParentId="12820" CreationDate="2010-12-03T03:31:10.913" Score="3" ViewCount="" Body="&lt;p&gt;Yes, this is true, at least for varieties over the complex numbers $\mathbb{C}$.  Indeed, a toric variety over an algebraically closed field is rational (i.e., birational to projective space).  In characteristic zero, rational connectedness is a birational invariant, so toric varieties are rationally connected.  Finally, any rationally connected variety is $\mathcal{O}$-acyclic, which is the name for the conclusion that you want.  See e.g. &lt;a href=&quot;http://www.aimath.org/WWN/qptsurface2/articles/html/14a/&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; for this last implication.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The conclusion might well hold more generally; I am not an expert in these matters.  You may want to ask your question on MathOverflow if you are not satisfied with this answer.&lt;/p&gt;&#xA;" OwnerUserId="299" LastActivityDate="2010-12-03T03:31:10.913" CommentCount="1" />
  <row Id="12862" PostTypeId="1" AcceptedAnswerId="12866" CreationDate="2010-12-03T04:53:31.777" Score="4" ViewCount="258" Body="&lt;p&gt;Suppose I have a function f that is infinitely differentiable on some interval I.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I construct a Taylor series P for it, using some point a in I, does f(x) = P(x) for all x in I?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm confused as to whether Taylor series approximate (or equal - I'm not about that either) functions on one point, or on an interval.&lt;/p&gt;&#xA;" OwnerUserId="181" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-12-03T06:19:46.927" LastActivityDate="2010-12-03T08:31:15.613" Title="On what interval does a Taylor series approximate (or equal?) its function?" Tags="&lt;calculus&gt;&lt;taylor-expansion&gt;" AnswerCount="4" />
  <row Id="12863" PostTypeId="1" AcceptedAnswerId="12868" CreationDate="2010-12-03T04:54:44.397" Score="10" ViewCount="274" Body="&lt;p&gt;On an infinite grid of ideal one-ohm resistors, what's the equivalant resistance between two nodes a knights move away?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://imgs.xkcd.com/comics/nerd_sniping.png&quot; alt=&quot;I first saw this problem on the Google Labs Aptitude Test.  A professor and I filled a blackboard without getting anywhere.  Have fun.&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;&lt;em&gt;(please fix the tags, I didn't really know where to put it)&lt;/em&gt;&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="4360" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-12-03T05:13:24.073" LastActivityDate="2010-12-03T05:13:24.073" Title="What is the resistance between two points a knights move away on a infinite grid of 1-ohm resistors" Tags="&lt;physics&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="4" />
  <row Id="12864" PostTypeId="2" ParentId="12862" CreationDate="2010-12-03T05:01:56.893" Score="1" ViewCount="" Body="&lt;p&gt;The (truncated) Taylor expansion agrees with the function only at the point of expansion; otherwise, the approximation gradually becomes worse the further you get away from your point of expansion. How bad it gets depends on the behavior of the coefficients.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, if you're asking if a Taylor expansion and the function where it came from are the same: in most cases, yes, but there are exceptions like $\exp\left(-\frac1{x^2}\right)$ ...&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-12-03T05:01:56.893" CommentCount="4" />
  <row Id="12865" PostTypeId="2" ParentId="12815" CreationDate="2010-12-03T05:03:07.680" Score="6" ViewCount="" Body="&lt;p&gt;One expects that the numbers $\zeta(2n+1)$ are algebraically independent of one another, and of $\pi$, and so one should think of them as ``new'' numbers; you can't expect any closed form expression in terms of powers of $\pi$, say.  Unfortunately, this conjecture seems very much out of reach at the moment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For an explanation of why people believe this conjecture, one can see for example&#xA;&lt;a href=&quot;http://mathoverflow.net/questions/30659/establishing-zeta3-as-a-definite-integral-and-its-computation/30698#30698&quot; rel=&quot;nofollow&quot;&gt;this answer&lt;/a&gt; to a related mathoverflow question.  &lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2010-12-03T05:03:07.680" CommentCount="1" />
  <row Id="12866" PostTypeId="2" ParentId="12862" CreationDate="2010-12-03T05:03:40.717" Score="10" ViewCount="" Body="&lt;p&gt;Let $f$ be an infinitely differentiable function, and let $T(x) = \sum_{n=0}^{\infty} a_n x^n$ be its Taylor series (say at $x = 0$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) The Taylor series of $f$ need not converge at any point other than $x = 0$.  Indeed, by a famous Theorem of Borel, for any sequence $\{a_n\}_{n=0}^{\infty}$ of real numbers whatsoever, there exists an infinitely differentiable function with Taylor series equal to $\sum_{n=0}^{\infty} a_n x_n$.  If the $a_n$ grow too rapidly -- e.g. $a_n = n!$ -- then there will be convergence only at $x = 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Even if the Taylor series has a positive radius of convergence $R$, there is no guarantee that $f(x)$ and $T(x)$ must be equal on $(-R,R)$.  A function with this property is said to be &lt;strong&gt;analytic&lt;/strong&gt; at $x = 0$.  (It should be mentioned that most of the familiar functions one encounters in calculus are analytic.)  Perhaps the simplest example of a non-analytic function is $f(x) = e^{\frac{-1}{x^2}}$ (and $f(0) = 0$), for which $f$ and all of its derivatives vanish at $x = 0$, so $T(x) = 0$.  But $f(x)$ is clearly positive at any $x \neq 0$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Added&lt;/b&gt;: To show that a function is equal to its Taylor series in some interval $I$, one has to show, for each $x$ in $I$, that the remainder function $R_n(x) = f(x) - T_n(x) = f(x) - \sum_{k=0}^n \frac{f^{(k)}(0)}{k!} x^n$ approaches $0$ as $n$ approaches infinity.  The most useful basic tool for this is Taylor's Formula for the Remainder.  The issue comes down to having a good understanding of the growth of the derivatives at $0$ as $n$ increases.  For instance, if there exists a fixed $M$ such that &#xA;$|f^{(n)}(a)| \leq M$ for all $a \in I$, then Taylor's Formula immediately implies that &#xA;$R_n(x) \rightarrow 0$ for all $x \in I$.  This is the case for instance for $f(x) = \cos x, \sin x, e^x$ (in the last case we need to assume that the interval $I$ is bounded, although we can take it to be bounded as large as we want, so the eventual conclusion will be that $e^x$ is equal to its Taylor series on the entire real line).  In general this can be a hard problem, for instance because the conclusion need not be true!  Books on advanced calculus / elementary real analysis will have some worked examples.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;More Added&lt;/b&gt;: Gosh, for a few minutes there I forgot about the thousand pages or so of lecture notes I have online!  In particular see &lt;a href=&quot;http://math.uga.edu/~pete/243functions2.pdf&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; for some further details on applying Taylor's Formula with Remainder.  However this discussion is at a somewhat higher level than strictly necessary (it came from a second semester undergraduate real analysis course).  As it happens, starting in January I'll be teaching a sophomore-level course on sequences and series, so in the fullness of time I might have more detailed notes.  Anyway, I'm sure that a little googling will find plenty of better notes on this topic...(In fact if you flip towards the back of a good calculus book, you'll certainly find this material there.)&lt;/p&gt;&#xA;" OwnerUserId="299" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2010-12-03T05:50:12.883" LastActivityDate="2010-12-03T05:50:12.883" CommentCount="4" />
  <row Id="12867" PostTypeId="2" ParentId="12862" CreationDate="2010-12-03T05:06:08.683" Score="2" ViewCount="" Body="&lt;p&gt;No, in fact the Taylor series may not even converge to the function. Consider&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f(x) = \left\{&#xD;&#xA;&#xD;&#xA; \begin{array}{ll}&#xD;&#xA;  f(x)=e^{-1/x^2}  &amp;amp; \mbox{if } x \ne 0 \\&#xD;&#xA;  f(x)=0 &amp;amp; \mbox{if } x = 0&#xD;&#xA; \end{array}&#xD;&#xA;\right.&#xD;&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="2219" LastEditorUserId="2219" LastEditorDisplayName="" LastEditDate="2010-12-03T05:07:32.177" LastActivityDate="2010-12-03T05:07:32.177" />
  <row Id="12868" PostTypeId="2" ParentId="12863" CreationDate="2010-12-03T05:07:26.817" Score="9" ViewCount="" Body="&lt;p&gt;After Google asked this in their aptitude test, this has become a famous problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can find a nice discussion and a more general solution here: &lt;a href=&quot;http://www.mathpages.com/home/kmath668/kmath668.htm&quot; rel=&quot;nofollow&quot;&gt;http://www.mathpages.com/home/kmath668/kmath668.htm&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe the answer for your case is $\displaystyle \dfrac{4}{\pi} - \dfrac{1}{2}$.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-12-03T05:07:26.817" CommentCount="2" />
  <row Id="12869" PostTypeId="2" ParentId="12809" CreationDate="2010-12-03T05:18:43.110" Score="0" ViewCount="" Body="&lt;p&gt;looks like chebyshev inequality.&lt;/p&gt;&#xA;" LastActivityDate="2010-12-03T05:18:43.110" CommentCount="3" />
  <row Id="12870" PostTypeId="1" AcceptedAnswerId="12945" CreationDate="2010-12-03T05:34:39.887" Score="0" ViewCount="127" Body="&lt;p&gt;I have been interested in learning formal math formally enough so that I could write a proof assistant using some simple parsing tools, and explain to someone else with little math knowledge how to write one. Naturally, it would be a very clunky and tedious assistant, but this is the standard I have set for myself for &quot;really understanding&quot; a formal system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd like to do this with a typed calculus, but you work with what you can get. The most formal explanations out there seem to be for the untyped lambda calculus. However, I haven't found a good formal explanation of how one detects/avoids problems with bound/unbound variables and problems with the same variable being bound in overlapping scopes. It seems like these kinds of problems should not just be avoided but somehow simply the result of an incorrect usage of the formal rules for constructing expressions. Does anyone know of a good way to understand these problems?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I asked a related question &lt;a href=&quot;http://math.stackexchange.com/questions/10181/resources-for-learning-formal-math&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="3480" LastActivityDate="2010-12-03T21:00:18.173" Title="formal rules for avoiding bound/unbound variable problems in lambda calculus" Tags="&lt;logic&gt;&lt;reference-request&gt;&lt;formal-systems&gt;&lt;type-theory&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="12871" PostTypeId="1" AcceptedAnswerId="12891" CreationDate="2010-12-03T06:06:27.330" Score="3" ViewCount="223" Body="&lt;p&gt;The usual statement of the Lebesgue Decomposition Theorem says that given two $\sigma$-finite measures $\mu$ and $\nu$ on a measure space, we can decompose $\nu = \nu_1 + \nu_2$, where $\nu_1$ is absolutely continuous with respect to $\mu$ and $\nu_2$ and $\mu$ are mutually singular. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wikipedia (&lt;a href=&quot;http://en.wikipedia.org/wiki/Lebesgue%27s_decomposition_theorem&quot; rel=&quot;nofollow&quot; title=&quot;Lebesgue Decomposition Theorem&quot;&gt;link text&lt;/a&gt;) says that there is a &quot;refinement&quot; of this result, where the singular part $\nu_2$ can be further decomposed into a discrete measure and a singular continuous measure. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand what a discrete measure is, but what exactly is the definition of a singular continuous measure? I was also wondering if anyone knew of a reference for this refined result, since I haven't been able to find it anywhere. &lt;/p&gt;&#xA;" OwnerUserId="1736" LastActivityDate="2010-12-03T08:45:52.027" Title="Lebesgue Decomposition Theorem" Tags="&lt;real-analysis&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" />
  <row Id="12872" PostTypeId="1" AcceptedAnswerId="12880" CreationDate="2010-12-03T06:12:19.150" Score="24" ViewCount="743" Body="&lt;p&gt;I understand that $\pi$ and e are transcendental and that these are not simple facts. I mean, I have been told that these results are deep and difficult, and I am happy to believe them. I am curious what types of techniques are used and just how difficult of a problem it is. Would this result be a reasonable capstone to any course? (any course that isn't essentially &quot;how to prove...&quot;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another part of this is the following observation: as time passes deep results become easier to understand or rather assimilate into ones body of knowledge and some problems are just hard. I am wondering if people feel like this result is something that a grad student could spend some leisure time and understand, or if it really is something only graspable by &quot;experts&quot; (meaning people in the appropriate field and not a general mathematical audience). How specialized are the techniques used for the problem at hand? Have they been used to prove different results? are the techniques drastically different for e and $\pi$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope this isn't too soft of a question. I was talking with my roommate, also a math grad student, and it came up. I said that it was a classically difficult result, but then wondered if that was right, so here I am.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: I don't want a proof or a sketch of one, but maybe a heuristic as to why new techniques were needed or explaining the troubles one has when being naive or using early methods to attack the problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: As Matt E pointed out I should also ask: what are the old techniques?&#xA;Also, did I even tag the question correctly? These seem to be the areas I would put them in, but I don't know the anything about this stuff.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the above it also isnt clear that I am wondering if the proof of this result is gotten by some clever new trick or by lots of hard had work that people couldn't/didn't do before? essentially, is it all elbow grease, or some clever new machinery or something completely different?&lt;/p&gt;&#xA;" OwnerUserId="627" LastEditorUserId="627" LastEditorDisplayName="" LastEditDate="2010-12-03T06:38:13.070" LastActivityDate="2011-05-24T21:45:48.737" Title="How hard is the proof of $\pi$ or e being transcendental?" Tags="&lt;analysis&gt;&lt;soft-question&gt;&lt;math-history&gt;&lt;analytic-number-theory&gt;" AnswerCount="5" CommentCount="2" FavoriteCount="7" />
  <row Id="12875" PostTypeId="2" ParentId="12872" CreationDate="2010-12-03T06:30:54.970" Score="9" ViewCount="" Body="&lt;p&gt;As mentioned above, it can be easily proven that e is transcendental. One can then combine it with $e^{iπ}+1=0$ (Euler's equation). Lindemann (I think) proved that if $e^x+1=0$, then $x$ has to be transcendental. And since $i$π is transcendental, ($i$ is obviously algebraic), π has to be transcendental, too. So, this way the proof seems rather easy, but there are much more complicated ways to prove it (as far as I know), which I will leave for others to elaborate on.&lt;/p&gt;&#xA;" OwnerUserId="3731" LastEditorUserId="3731" LastEditorDisplayName="" LastEditDate="2010-12-23T04:50:52.643" LastActivityDate="2010-12-23T04:50:52.643" CommentCount="10" />
  <row Id="12876" PostTypeId="1" AcceptedAnswerId="12877" CreationDate="2010-12-03T06:51:43.297" Score="2" ViewCount="111" Body="&lt;p&gt;Does anybody know where I can find a general form, in terms of n, of the sum $\left(\sum\limits_{i=1}^n a_i\right)^n$.  What I mean is, there appears to be some sort of pattern, if you look at $\left(\sum\limits_{i=1}^2 a_i\right)^2$ you get $a_1^2+2a_1a_2+a_2^2$ and if you let $n=3$ then you get $a_1^3+a_2^3+a_3^3+3a_1^2(a_2+a_3)+3a_2^2(a_1+a_3)+3a_3^2(a_2+a_3)+6a_1a_2a_3$.  I'm working with something right now where I'd like to subtract out the cubed terms and have a general formula left over for the remaining terms which I can write simply in terms of $n$.  Sorry about the lack of clarity, I hope this helps.&lt;/p&gt;&#xA;" OwnerUserId="3595" LastEditorUserId="3595" LastEditorDisplayName="" LastEditDate="2010-12-03T07:03:19.853" LastActivityDate="2010-12-03T07:03:19.853" Title="General form for $(\sum\limits_{i=1}^n a_i)^n$" Tags="&lt;sequences-and-series&gt;&lt;reference-request&gt;&lt;arithmetic&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="12877" PostTypeId="2" ParentId="12876" CreationDate="2010-12-03T06:54:57.637" Score="6" ViewCount="" Body="&lt;p&gt;It isn't really clear what you are asking.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps this will help: &lt;a href=&quot;http://en.wikipedia.org/wiki/Multinomial_theorem&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Multinomial_theorem&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-12-03T06:54:57.637" CommentCount="2" />
  <row Id="12878" PostTypeId="2" ParentId="12870" CreationDate="2010-12-03T06:55:47.583" Score="1" ViewCount="" Body="&lt;p&gt;A great reference for $\lambda$-calculus and formal logic in general, which treats the subject highly rigorously is Sorensen and Urzyczyn's &quot;Lectures on the Curry-Howard Isomorphism.&quot;  I'm doing a little reading on the subject myself, but not enough to really answer your question.&lt;/p&gt;&#xA;" OwnerUserId="3595" LastActivityDate="2010-12-03T06:55:47.583" />
  <row Id="12879" PostTypeId="1" AcceptedAnswerId="12883" CreationDate="2010-12-03T07:06:16.723" Score="1" ViewCount="160" Body="&lt;p&gt;Hi, this step in the proof is confusing me:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_1^\infty {\frac{4^{n}}{3^{n-1}}}\qquad \longrightarrow \qquad\sum_1^\infty 4\left(\frac43\right)^{n-1}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;please explain how/why this happened?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;cheers,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;gregg&lt;/p&gt;&#xA;" OwnerUserId="4362" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-03T14:10:27.563" LastActivityDate="2010-12-03T14:10:27.563" Title="sequence /series convergence of 2^2n 3^(1-n)" Tags="&lt;sequences-and-series&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="12880" PostTypeId="2" ParentId="12872" CreationDate="2010-12-03T07:14:07.567" Score="20" ViewCount="" Body="&lt;p&gt;This is an answer elaborating on my comment above, and largely addressing the&#xA;edits to the original question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't think there were any &quot;old techniques&quot; before Hermite proved&#xA;transcendence of $e$ in the early 1870s.  As far as I know, the only transcendental numbers known before then were Liouville's interesting but&#xA;somewhat artificial examples from early that century.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The subject of transcendence, and the related subject of Diophantine approximation (i.e. approximating irrational numbers, especially irrational algebraic numbers, by rationals), is relatively new.  Liouville proved the first results showing that it is not so easy to approximate an irrational algebraic number by rational ones, and used this to construct his transcendental numbers, which he could recognize as being transcendental because they are too well approximated by rational numbers.  (&quot;Too well&quot; and &quot;not so easy&quot; here refer to the following problem: if you try to approximate $\alpha$ by the rational number $p/q$, can you get within a distance of $O(q^{-n})$ for some given $n$ as you let the denominator $q$ get arbitrarily large. (The larger $n$ is, the smaller $q^{-n}$ is, and so the better the rate of approximation.) Liouville showed, using the pigeon hole principal more-or-less, that if $\alpha$ is algebraic of degree $d$ then you can't do better than $n = d$.)&#xA;But this left open the problem of showing that various given numbers (like $e$ and $\pi$) are transcendental.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you like, here is one way to think of the problem: if you want to show that $\alpha$ is transcendental, then&#xA;you want to show that $f(\alpha) \neq 0$ for any non-zero polynomial $f$&#xD;&#xA;with rational coefficients.   The difficulty is that there will be lots of polynomials with real coefficients that have $\alpha$ as a root, and any one of them can be approximated as closely as you like by an $f$ with rational coefficients, so we &lt;em&gt;can&lt;/em&gt; find (lots of!) $f$ with rational coefficients such that $f(\alpha)$ is as close to zero as we like.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you have to find some way to pin down the difference between $f(\alpha)$ being zero and $f(\alpha)$ being very close to zero.  This is not so easy!&#xA;(For example, computationally, you can't tell the difference between $0$ and any real number that is smaller than your computational accuracy can recognize.)&#xA;And now one sees why Diophantine approximation ideas of the type mentioned above&#xA;are relevant.  They are related to quantifying how close we can make $f(\alpha)$ to zero while bounding the denominators of the rational numbers involved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is not coincidence that bounding the denominators is relevant: morally this is an attempt to pass from working over $\mathbb Q$ to working over $\mathbb Z$.&#xA;Why do we want to do this?  Well, as I already noted, it's pretty hard to tell the difference between $\mathbb Q$ and $\mathbb R$, since the former is dense in the latter, but we &lt;em&gt;can&lt;/em&gt; tell the difference between $\mathbb Z$ and $\mathbb R$, since the former is discrete: a non-zero integer is some definite positive distance (i.e. at least 1) away from $0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The preceding remarks are somewhat philosophical, and they reflect my (limited) experience of thinking about these kinds of questions.  If you look at the proofs in the link above, it may not be obvious that they are relevant, but I believe that they in fact do have some relevance: e.g. you will see that the arguments reduce to considering integer rather than rational polynomials, and that growth considerations play a key role.  Another thing you will see is that certain auxiliary polynomials enter the proof, and a key fact about them is that they have a high order of vanishing at their zeroes.  The appearance of auxiliary polynomials, often with a high order of vanishing, is ubiquitous in this theory.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One more (somewhat cultural) remark:&#xA;Roth's theorem, for which he got the Fields medal, is the ultimate strengthening of Liouville's theorem: he shows that if $\alpha$ is irrational algebraic, then one can't do better than $O(q^{-2})$ in the problem of Diophantine approximation discussed above.  The proof involves (among other things) constructions with auxiliary polynomials.  So my impression is that Liouville, Hermite, and Lindemann (and there are probably other names that should be here) invented a new subject, namely Diophantine approximation and transcendence theory, whose modern methods are an outgrowth of the methods that they introduced.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S.  Reading the first few pages of Baker's book &lt;a href=&quot;http://books.google.com/books?id=SmsCqiQMvvgC&amp;amp;printsec=frontcover&quot; rel=&quot;nofollow&quot;&gt;Transcendental Number Theory&lt;/a&gt;&#xA;(Google Books link) might help.&lt;/p&gt;&#xA;" OwnerUserId="221" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-03T11:21:37.447" LastActivityDate="2010-12-03T11:21:37.447" CommentCount="2" />
  <row Id="12881" PostTypeId="1" AcceptedAnswerId="12884" CreationDate="2010-12-03T07:26:04.077" Score="3" ViewCount="86" Body="&lt;p&gt;I'm trying to figure out whether my hardware function for mapping operating system pages to DDR memory controllers is &lt;a href=&quot;http://en.wikipedia.org/wiki/Injective_function&quot; rel=&quot;nofollow&quot;&gt;injective&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$F(x) = a x \pmod{b},\text{ where $x$ in $[0,b)$, $a$ and $b$ are co-prime}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My hunch is that it is injective and I think it has to do with the &lt;a href=&quot;http://en.wikipedia.org/wiki/Chinese_remainder_theorem&quot; rel=&quot;nofollow&quot;&gt;Chinese Remainder Theorem&lt;/a&gt;, though it may be some other theorem.  Also, is &quot;$a,b$ are co-prime&quot; sufficient?  Or does $a$ also need to be greater than $b$?&lt;/p&gt;&#xA;" OwnerUserId="4363" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-03T15:53:46.107" LastActivityDate="2010-12-03T15:53:46.107" Title="When $a$ and $b$ are co-prime, is $F(x) = a x \mod b$, $x$ in $[0,b)$ , $a$ and $b$ co-prime, an injective function?" Tags="&lt;prime-numbers&gt;&lt;modular-arithmetic&gt;" AnswerCount="2" />
  <row Id="12882" PostTypeId="2" ParentId="12881" CreationDate="2010-12-03T07:32:24.750" Score="3" ViewCount="" Body="&lt;p&gt;It is injective, your hunch is correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $\displaystyle ax = ay \mod b$ then $\displaystyle a(x-y) = 0 \mod b$. Since $\displaystyle a$ and $\displaystyle b$ are co-prime, $\displaystyle x-y= 0 \mod b$ and hence $\displaystyle x = y$, as $\displaystyle x,y \in [0,b)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It does not matter whether $\displaystyle a \gt b$ or not.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-12-03T07:32:24.750" />
  <row Id="12883" PostTypeId="2" ParentId="12879" CreationDate="2010-12-03T07:38:59.017" Score="7" ViewCount="" Body="&lt;p&gt;The $n^\text{th}$ term was rewritten by pulling out a factor of $4$ from the numerator.  Maybe seeing a couple of extra steps will help:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{4^n}{3^{n-1}}=\frac{4\cdot4^{n-1}}{3^{n-1}}=4\frac{4^{n-1}}{3^{n-1}}=4\left(\frac{4}{3}\right)^{n-1}$$&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-12-03T07:38:59.017" CommunityOwnedDate="2010-12-03T07:38:59.017" CommentCount="4" />
  <row Id="12884" PostTypeId="2" ParentId="12881" CreationDate="2010-12-03T07:40:28.533" Score="2" ViewCount="" Body="&lt;p&gt;Yes, $\rm\ gcd(a,b)=1,\ \ b\:|\:ax\ \Rightarrow\ b\:|\:x\ $ by &lt;a href=&quot;http://en.wikipedia.org/wiki/Euclid%27s_lemma&quot; rel=&quot;nofollow&quot;&gt;Euclid's Lemma&lt;/a&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Said functionally: $\rm\ ax\equiv 0\ \Rightarrow\ x\equiv 0\ \ (mod\ b)\ $ so $\rm\ x\to ax\ $ is injective.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More intuitively, by &lt;a href=&quot;http://en.wikipedia.org/wiki/B%C3%A9zout%27s_identity&quot; rel=&quot;nofollow&quot;&gt;Bezout's identity&lt;/a&gt;, $\rm\ gcd(a,b)=1\ \Rightarrow\ a^{-1}\ $ exists $\rm\ (mod\ b)\:.\: $ Therefore one can multiply $\rm \ ax\equiv ay\ $ by $\rm\:a^{-1}\:$ to deduce $\rm\ x\equiv y\:$.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-12-03T07:56:37.950" LastActivityDate="2010-12-03T07:56:37.950" />
  <row Id="12885" PostTypeId="1" AcceptedAnswerId="12886" CreationDate="2010-12-03T07:43:12.430" Score="3" ViewCount="185" Body="&lt;p&gt;I am not getting this theorem:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;em&gt;&lt;strong&gt;Angle between chord AB and tangent at A is the same as subtended by segment AB at any point on the circumference.&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;How to prove this theorem?&lt;/p&gt;&#xA;" OwnerUserId="2109" LastActivityDate="2010-12-03T12:59:55.380" Title="Angle between chord and tangent" Tags="&lt;geometry&gt;" AnswerCount="3" />
  <row Id="12886" PostTypeId="2" ParentId="12885" CreationDate="2010-12-03T07:50:38.160" Score="5" ViewCount="" Body="&lt;p&gt;Use the fact that if $O$ is the center of circle, and $C$ is foot of the perpendicular from $O$ to $AB$, then the angle subtended at any point on the circumference is half $\angle{AOB}$ which is $\angle{AOC}$ and try showing that the angle between chord $AB$ and tangent at $A$ is the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/Eowxw.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note, you do have to be careful here, there are two different angles subtended by points on the circumference, corresponding to the two different arcs which $AB$ forms. In fact those two angles add up to $180^{\circ}$.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-12-03T08:03:33.340" LastActivityDate="2010-12-03T08:03:33.340" CommentCount="4" />
  <row Id="12887" PostTypeId="2" ParentId="12885" CreationDate="2010-12-03T07:54:33.553" Score="2" ViewCount="" Body="&lt;p&gt;It is well known that the angle subtended by segment $AB$ at any point on the circumference is constant on either of the arcs $AB$. Since the sum of those angles is $\pi$, it suffices to prove this for the point $C$&amp;#160;that is opposite to point $A$. Then it is well known that the angle $ABC$ is a right angle and also the segment $CA$ meets the tangent at $A$ in a right angle. Therefore $CAB = \frac{\pi}{2} - BCA$ and your angle is then just $\frac{\pi}{2} - CAB = BCA$.&lt;/p&gt;&#xA;" OwnerUserId="3776" LastActivityDate="2010-12-03T07:54:33.553" />
  <row Id="12888" PostTypeId="1" CreationDate="2010-12-03T08:13:24.933" Score="0" ViewCount="86" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Possible Duplicate:&lt;/strong&gt;&lt;br&gt;&#xA;  &lt;a href=&quot;http://math.stackexchange.com/questions/12848/is-there-an-infinite-set-of-strings-whose-kolmogorov-complexities-are-computable&quot;&gt;Is there an infinite set of strings whose Kolmogorov complexities are computable?&lt;/a&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;p&gt;I &lt;a href=&quot;http://math.stackexchange.com/questions/12848/&quot;&gt;asked a similar question&lt;/a&gt; but it turned out to be ambiguous. I am attempting to clarify what I meant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there an infinite, recursively enumerable set of pairs $(\sigma,n)$ where $\sigma \in S$ is a string of Kolmogorov complexity $n$ where $S$ is some infinite set of strings each over the same finite alphabet for some finite alphabet?&lt;/p&gt;&#xA;" OwnerUserId="4353" LastActivityDate="2010-12-03T08:13:24.933" ClosedDate="2010-12-17T11:23:13.160" Title="Is Kolmogorov complexity recursively enumerable for some infinite set of strings?" Tags="&lt;computability&gt;&lt;kolmogorov-complexity&gt;" CommentCount="6" />
  <row Id="12889" PostTypeId="2" ParentId="12862" CreationDate="2010-12-03T08:31:15.613" Score="2" ViewCount="" Body="&lt;p&gt;Since you speak about intervals (on the real line), perhaps it should also be mentioned that the &quot;natural habitat&quot; for power series is really the complex plane; computing a power series involves only +, -, *, /, and limits, which are well defined operations on complex numbers. And for so-called &quot;complex analytic&quot; (or &quot;holomorphic&quot;) functions, which includes most functions that you encounter in calculus, it is a fact that the Taylor series at any point in the complex plane is convergent (equal to the function) in a circle around that point. The size of this circle is such that it exactly reaches out to the nearest singularity of the function. (Circles in the complex plane are the counterparts to intervals on the real line in this context.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A simple example is $f(x)=1/(1+x^2)$. If you just look at the graph of this function (for $x$ real) it looks perfectly nice, and there seems to be no reason why the Taylor series at $x=0$ only manages to converge to the function in the interval $(-1,1)$. But if you think of $x$ as a complex variable, it's clear that there are singularities (division by zero) at the points $x=\pm i$, which lie at distance one from the origin, and that explains why the Taylor series converges inside the circle with radius one. (Note that the intersection of this disk with the real axis is just the interval $(-1,1)$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same thing goes for $g(x)=\arctan x$. It's derivative is $f(x)$ above, and where the derivative has singularitites, the function has too. So the Taylor series for $g(x)$ at $x=0$ is also convergent inside the unit circle.&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2010-12-03T08:31:15.613" CommentCount="1" />
  <row Id="12890" PostTypeId="2" ParentId="12856" CreationDate="2010-12-03T08:43:45.570" Score="3" ViewCount="" Body="&lt;p&gt;Seeing how the essential question was answered, I want to stress something else in your post which needs to be pointed out:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is true that cardinals (namely, &lt;a href=&quot;http://en.wikipedia.org/wiki/Aleph_number&quot; rel=&quot;nofollow&quot;&gt;Aleph numbers&lt;/a&gt;) are usually treated as ordinals, however the multiplications and addition of cardinals and ordinals are very different, and most of all - exponentiation is different as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For ordinals $\alpha$ and $\beta$ we define the sum to be:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\alpha + 0 = \alpha$&#xD;&lt;/li&gt;&#xA;&lt;li&gt;$\alpha + (\beta + 1) = (\alpha + \beta) + 1$ (where $+1$ is the successor ordinal)&lt;/li&gt;&#xA;&lt;li&gt;$\alpha + \beta$ for a limit ordinal $\beta$ is the limit of $\alpha+\gamma$ for $\gamma&amp;lt;\beta$&#xD;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;One can notice that it is usually non-commutative as $2+\omega = \sup\{2+n\colon n&amp;lt;\omega\} = \omega \neq \omega+2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Ordinal_arithmetic#Multiplication&quot; rel=&quot;nofollow&quot;&gt;ordinal multiplication&lt;/a&gt; is defined in a similar way, as well &lt;a href=&quot;http://en.wikipedia.org/wiki/Ordinal_arithmetic#Exponentiation&quot; rel=&quot;nofollow&quot;&gt;exponentiation&lt;/a&gt;. (Namely a simple rule for zero, and successor and a limit for limits) and an interesting result is that $\omega^\omega$ is &lt;strong&gt;countable&lt;/strong&gt; when dealing with ordinal exponentiation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In contrast, if $\lambda$ and $\mu$ are cardinals then $\lambda + \mu = \mu + \lambda = \lambda \cdot \mu = \mu \cdot \lambda = \max \{\lambda, \mu\}$, and the exponentiation is defined as $\lambda^\mu = |\{f | f\colon\mu\to\lambda\}|$ - that is the cardinality of the collection of functions from $\mu$ into $\lambda$.&#xA;For further information and definitions you can see &lt;a href=&quot;http://en.wikipedia.org/wiki/Cardinal_number#Cardinal_arithmetic&quot; rel=&quot;nofollow&quot;&gt;this wikipedia link&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So once you were dealing with the cardinality of the basis you were looking for cardinal arithmetics and not ordinal arithmetics. Which are two different things.&lt;/p&gt;&#xA;" OwnerUserId="622" LastActivityDate="2010-12-03T08:43:45.570" />
  <row Id="12891" PostTypeId="2" ParentId="12871" CreationDate="2010-12-03T08:45:52.027" Score="4" ViewCount="" Body="&lt;p&gt;That decomposition is commonly encountered in probability theory. In a more general setting, suppose that $\rho$ is a $\sigma$-finite measure on $\mathcal{B}(\mathbb{R}^n)$. Then there is a unique decomposition $\rho = \rho_{ac} + \rho_d + \rho_{cs}$, such that: 1) $\rho_{ac}$ is absolutely continuous, that is, $\rho_{ac}$ is zero on sets of Lebesgue measure zero; 2) $\rho_d$ is discrete, that is, $\rho_d$ is zero on the complement of some countable set $C$;  3) $\rho_{cs}$ is continuous singular, that is, $\rho_{cs}$ is zero at every point $x \in \mathbb{R}^n$ (= continuous measure), and is zero on the complement of some set $B$ of Lebesgue measure zero (= singular measure).&lt;/p&gt;&#xA;" OwnerUserId="2810" LastActivityDate="2010-12-03T08:45:52.027" CommentCount="1" />
  <row Id="12892" PostTypeId="2" ParentId="12872" CreationDate="2010-12-03T10:02:46.547" Score="5" ViewCount="" Body="&lt;p&gt;For the proofs, you would of course need to know the necessary notions from mathematical analysis, covered in a good course, for instance the equivalent of what is in W. Rudin's principles of mathematical analysis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming you are comfortable with the analytic notions, then proving the transcendence of $e$ is quite easy. It can be done using the series expansion for $e$. This is done in the introductory algebra book of I. N. Herstein, &quot;Topics in algebra&quot;. The transcendence of $pi$ is harder. But it is still within the grasp of a beginning graduate student, though the proof is usually not included in the curriculum. It is given for instance in the algebra book of N. Jacobson, &quot;Basic Algebra&quot;, Vol 1.. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A proof sketch of the Lindemann-Weierstrass theorem is in fact given in &lt;a href=&quot;http://en.wikipedia.org/wiki/Lindemann%E2%80%93Weierstrass_theorem&quot; rel=&quot;nofollow&quot;&gt;wikipedia&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the answer to your question of how difficult the theorems are, the answer is that they are accessible to beginning graduate students, and undergraduates who are willing to put in the extra labor.&lt;/p&gt;&#xA;" OwnerUserId="1119" LastActivityDate="2010-12-03T10:02:46.547" CommentCount="2" />
  <row Id="12893" PostTypeId="2" ParentId="12872" CreationDate="2010-12-03T10:38:51.663" Score="9" ViewCount="" Body="&lt;p&gt;This paper by Ivan Niven,&#xA;&lt;a href=&quot;http://www.jstor.org/stable/2302515&quot; rel=&quot;nofollow&quot;&gt;The transcendence of π&lt;/a&gt;&#xA;[&lt;em&gt;Amer. Math. Monthly&lt;/em&gt; 46, (1939). 469–471], is short and readable. His book &lt;a href=&quot;https://www.maa.org/ebusppro/Bookstore/ProductDetail/tabid/170/Default.aspx?ProductId=947&quot; rel=&quot;nofollow&quot;&gt;Irrational numbers&lt;/a&gt; contains this result and much more and is highly recommended.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See also &lt;a href=&quot;http://mathoverflow.net/questions/34055/transcendence-of-pi&quot; rel=&quot;nofollow&quot;&gt;Transcendence of PI&lt;/a&gt; in MO.&lt;/p&gt;&#xA;" OwnerUserId="589" LastEditorUserId="589" LastEditorDisplayName="" LastEditDate="2011-05-19T12:45:36.560" LastActivityDate="2011-05-19T12:45:36.560" CommentCount="1" />
  <row Id="12894" PostTypeId="1" AcceptedAnswerId="12901" CreationDate="2010-12-03T11:21:39.693" Score="2" ViewCount="115" Body="&lt;p&gt;Hello,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;in functional analysis, you encounter the terms 'adjoint' and 'formal adjoint'.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What does 'formal' in that case mean? It Sounds like a hint that 'formal adjoints' lack a certain property to make them a 'true' adjoint.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have nowhere found a definition, and would be eager to know.&lt;/p&gt;&#xA;" OwnerUserId="3557" LastActivityDate="2010-12-03T12:39:51.327" Title="Distinction between 'adjoint' and 'formal adjoint'" Tags="&lt;terminology&gt;&lt;functional-analysis&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" />
  <row Id="12895" PostTypeId="1" AcceptedAnswerId="12899" CreationDate="2010-12-03T11:26:27.880" Score="8" ViewCount="233" Body="&lt;p&gt;And what about the other direction? If $(u_{n})$ is increasing, what about $(a_{n})$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm &lt;em&gt;guessing&lt;/em&gt; the former is true since we know that if $(a_{n})$ converges, $(u_{n})$ converges to the same limit. That tells me that around infinity the sequences behave roughly the same. I tried proof by induction but got stuck.&lt;/p&gt;&#xA;" OwnerUserId="3237" LastActivityDate="2010-12-03T18:56:08.390" Title="If $(a_{n})$ is increasing, is $u_{n}=\frac{a_{1}+\cdots+a_{n}}{n}$ increasing as well?" Tags="&lt;calculus&gt;&lt;homework&gt;&lt;sequences-and-series&gt;" AnswerCount="2" CommentCount="5" FavoriteCount="0" />
  <row Id="12896" PostTypeId="1" CreationDate="2010-12-03T11:51:18.423" Score="1" ViewCount="121" Body="&lt;p&gt;Guys is my logic correct? If Earth travels around the Sun in 29.78 km/s and I have to calculate how long will it take to travel 1km I did following:&lt;br&gt;&#xA;1s / 29.78 = 0.03358 (to 5 d.p)&lt;br&gt;&#xA;I think it's ok but I would like to get confirmation on this.&lt;br&gt;&#xA;Thank you.&lt;/p&gt;&#xA;" OwnerUserId="4366" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-03T12:01:16.990" LastActivityDate="2010-12-03T12:01:16.990" Title="Earth and Sun - fast question" Tags="&lt;arithmetic&gt;" CommentCount="4" />
  <row Id="12897" PostTypeId="1" AcceptedAnswerId="12898" CreationDate="2010-12-03T11:52:56.943" Score="8" ViewCount="350" Body="&lt;p&gt;How do i show that $f_{1}(x)=1$, $f_{2}(x)=e^{x}$ and $f_{3}(x)=\sin{x}$ are linearly independent, as elements of the vector space, of continuous functions $\mathcal{C}[0,1]$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So for showing these elements are linearly independent, one needs to show that if $$ a_{1} \cdot 1 + a_{2} \cdot e^{x} + a_{3} \cdot \sin{x}=0$$ then from this we should conclude that $a_{1}=a_{2}=a_{3}=0$. But i am not being able to deduce this. &lt;/p&gt;&#xA;" LastActivityDate="2010-12-04T21:19:58.240" Title="Showing $1,e^{x}$ and $\sin{x}$ are linearly independent in $\mathcal{C}[0,1]$" Tags="&lt;linear-algebra&gt;&lt;analysis&gt;&lt;vector-space&gt;" AnswerCount="6" CommentCount="3" />
  <row Id="12898" PostTypeId="2" ParentId="12897" CreationDate="2010-12-03T12:01:10.767" Score="15" ViewCount="" Body="&lt;p&gt;If you differentiate your equation with respect to $x$ four times, you get that $a_2 e^x+a_3 \sin x = 0$, from which it follows that $a_1=0$. Now, letting $x=0$, you see that $a_2 e^x=0$, so $a_2=0$, and it follows readily that $a_3=0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With problems like this, always remember that you are dealing with functions. That means that you can treat them as functions as well: you can differentiate them and set their values equal to something.&lt;/p&gt;&#xA;" OwnerUserId="4284" LastActivityDate="2010-12-03T12:01:10.767" CommentCount="1" />
  <row Id="12899" PostTypeId="2" ParentId="12895" CreationDate="2010-12-03T12:06:34.003" Score="17" ViewCount="" Body="&lt;p&gt;For the first question, we have for increasing $a_n$:&#xA;\begin{align}&#xA;u_{n+1} - u_n &#xA;    &amp;amp;= \frac{a_1+\dotsb+a_{n+1}}{n+1} - \frac{a_1+\dotsb+a_n}{n} \\&#xA;    &amp;amp;= \frac{n(a_1+\dotsb+a_{n+1})}{n(n+1)} - \frac{(n+1)(a_1+\dotsb+a_n)}{n(n+1)} \\&#xA;    &amp;amp;= \frac{na_{n+1} - \sum_{k=1}^n a_k}{n(n+1)} \\&#xA;    &amp;amp;= \frac{a_{n+1}}{n+1} - \frac{u_n}{n+1} \\&#xA;    &amp;amp;\geq \frac{a_{n+1}}{n+1} - \frac{a_n}{n+1} \gt 0,&#xA;\end{align}&#xA;where we have used that $u_n \leq a_n$ (the mean is at most the largest of the elements we're averaging over).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regarding whether $u_n$ increasing $\implies a_n$ increasing, this seems to be false. For example we can take $a_1 = 0$, $a_2 = 1$, and for $n&amp;gt;2$: $a_n = \frac{1}{2}(u_{n-1}+a_{n-1})$. For $n \geq 2$ we have $a_{n+1} \lt a_n$ and $u_{n+1} \gt u_n$.&lt;/p&gt;&#xA;" OwnerUserId="1269" LastEditorUserId="1269" LastEditorDisplayName="" LastEditDate="2010-12-03T12:37:26.057" LastActivityDate="2010-12-03T12:37:26.057" CommentCount="3" />
  <row Id="12900" PostTypeId="2" ParentId="12894" CreationDate="2010-12-03T12:18:05.543" Score="1" ViewCount="" Body="&lt;p&gt;Is it this thing you're looking for?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Differential_operator#Formal_adjoint_in_one_variable&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Differential_operator#Formal_adjoint_in_one_variable&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2010-12-03T12:18:05.543" />
  <row Id="12901" PostTypeId="2" ParentId="12894" CreationDate="2010-12-03T12:32:17.310" Score="8" ViewCount="" Body="&lt;p&gt;Are you talking about differential operators on functions over a domain in $\mathbb{R}^d$? (This is the context in which the phrase &quot;formal adjoint&quot; usually comes up.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The idea is that working with, say, smooth functions with compact support, we have the integration by parts formula&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \int Du\cdot v ~dx + \int u\cdot Dv~ dx = 0 $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if a linear partial differential operator is defined as $P = \sum A_\alpha D^\alpha$ where $\alpha$ are multi-indices, you can write $P&amp;#39;$ as a linear partial differential operator $P&amp;#39;\phi = \sum (-1)^{|\alpha|} D^\alpha(A_\alpha \phi)$ and generalize the integration by parts formula&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \int Pu \cdot v~dx = \int u \cdot P&amp;#39;v~ dx $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which looks, in &lt;em&gt;form&lt;/em&gt;, suspiciously like the adjoint with respect to the $L^2$ inner product. That is, writing $\langle,\rangle$ for the $L^2$ inner product of real valued functions, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \langle Pu,v\rangle = \langle u, P&amp;#39;v\rangle $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason that we call this a formal adjoint is because, technically, to take an adjoint (in the Hilbert space sense, there is also a different notion for Banach spaces) of an operator, you need to specify which Hilbert space you are working over. In the case of the formal adjoint, it is left unspecified: indeed, the formula only really hold for sufficiently smooth function decaying sufficiently fast at infinity, and not in general for arbitrary functions $u,v\in L^2$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general for differential operators, the operator itself will not be bounded on an $L^2$ Hilbert space, and so the operator is only densely defined on your Hilbert space. Therefore the adjoint can only be defined on another subset of the Hilbert space, the domain of the adjoint. (In the most general cases, the domain of the adjoint can be a much, much smaller set [even finite dimensional], so does not make much sense as an operator on the original Hilbert space. For differential operators, the adjoint is still densely defined using the density of $C^\infty_0$ in $L^2$.) (Note that also if the spactial domain has a boundary, the integration by parts formula picks up a boundary term in general, so you pick up a further problem with the notion of adjoints, related to the fact that $C^\infty_0(\Omega)$ is not dense in the Sobolev space $W^{1,2}(\Omega)$ when $\Omega$ has boundary.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While the word &quot;formal&quot; is, I think, not mentioned explicitly, a lot of the problems that can arise when you deal with unbounded operators are discussed in chapter 8 of Reed-Simon, &quot;Methods of mathematical physics&quot;. &lt;/p&gt;&#xA;" OwnerUserId="1543" LastEditorUserId="1543" LastEditorDisplayName="" LastEditDate="2010-12-03T12:39:51.327" LastActivityDate="2010-12-03T12:39:51.327" />
  <row Id="12902" PostTypeId="2" ParentId="12885" CreationDate="2010-12-03T12:54:24.840" Score="2" ViewCount="" Body="&lt;p&gt;As a diagram of the other common method of proof of the Alternate Segment Theorem has not yet been posted, here it is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/9XvQV.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;AB is the diameter and so $ \angle BCA $ is a right angle. Hence $y = \pi/2 - \angle BAC = x,$&#xD;&#xA;and we just note that the angle subtended by $AC$ is constant in this segment.&lt;/p&gt;&#xA;" OwnerUserId="1301" LastEditorUserId="1301" LastEditorDisplayName="" LastEditDate="2010-12-03T12:59:55.380" LastActivityDate="2010-12-03T12:59:55.380" />
  <row Id="12904" PostTypeId="2" ParentId="12897" CreationDate="2010-12-03T13:11:58.180" Score="5" ViewCount="" Body="&lt;p&gt;You may calculate the &lt;a href=&quot;http://en.wikipedia.org/wiki/Wronskian&quot; rel=&quot;nofollow&quot;&gt;Wronskian&lt;/a&gt; of $f_1(x)$, $f_2(x)$, $f_3(x)$ and check that it doesn't vanish on $[0,1]$. This implies that the functions are linearly independent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;A caveat.&lt;/strong&gt; If the Wronskian of a set of functions is equal to zero identically, it doesn't&#xA;follow in general that the functions are linearly dependent.&lt;/p&gt;&#xA;" OwnerUserId="723" LastActivityDate="2010-12-03T13:11:58.180" />
  <row Id="12905" PostTypeId="1" CreationDate="2010-12-03T13:25:56.560" Score="0" ViewCount="41" Body="&lt;p&gt;So I've got just the update step of the k-means algorithm running for a single cluster in a program.  I'm really only calculating a single centroid given a set number of observations.  So what does this make the running time?  Is it O(n)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/aqT6I.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, what if I add a single computation to the end of this update step.  Would this be added to the running time or would it be omitted as its not the leading order?  Resources to learn this would be cool too...&lt;/p&gt;&#xA;" OwnerUserId="3105" LastActivityDate="2010-12-03T13:25:56.560" Title="What is the time complexity of the update step of the K-Means Algorithm?" Tags="&lt;algorithms&gt;" CommentCount="1" />
  <row Id="12906" PostTypeId="1" AcceptedAnswerId="12907" CreationDate="2010-12-03T13:44:56.300" Score="54" ViewCount="3004" Body="&lt;p&gt;What is wrong with this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/GU8wd.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;a href=&quot;http://www.bestwtf.com/2010/11/explaining-why-pi-is-4.html&quot; rel=&quot;nofollow&quot;&gt;SOURCE&lt;/a&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="705" LastActivityDate="2011-05-30T17:07:31.027" Title="Is value of $\pi$ = 4 ?" Tags="&lt;pi&gt;" AnswerCount="11" CommentCount="11" FavoriteCount="25" />
  <row Id="12907" PostTypeId="2" ParentId="12906" CreationDate="2010-12-03T13:56:11.760" Score="43" ViewCount="" Body="&lt;p&gt;This question is usually posed as the length of the diagonal of a unit square.  You start going from one corner to the opposite one following the perimeter and observe the length is 2, then take shorter and shorter stairsteps and the length is 2 but your path approaches the diagonal.  So $\sqrt{2}=2$  In both cases, you are approaching the area but not the path length.  You can make this more rigorous by breaking into increments and following the proof of the Riemann sum.  The difference in area between the two curves goes nicely to zero, but the difference in arc length stays constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit:  making the square more explicit.  Imagine dividing the diagonal into n segments and a stairstep approximation.  Each triangle is $(\frac{1}{n},\frac{1}{n},\frac{\sqrt{2}}{n})$.  So the area between the stairsteps and the diagonal is $n \frac{1}{n^2}$ which converges to 0.  The path length is $n \frac{2}{n}$, which converges even more nicely to 2.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2010-12-03T19:06:58.983" LastActivityDate="2010-12-03T19:06:58.983" CommentCount="6" />
  <row Id="12909" PostTypeId="1" AcceptedAnswerId="12911" CreationDate="2010-12-03T14:01:25.873" Score="8" ViewCount="323" Body="&lt;p&gt;I'm interested in the potential of such a technique.  I got the idea from Moron's answer to &lt;a href=&quot;http://math.stackexchange.com/questions/9402/9409#9409&quot;&gt;this question&lt;/a&gt;, which uses the technique of differentiation under the integral.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, I'd like to consider this integral:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_{-\pi}^\pi \cos{(y(1-e^{i\cdot n \cdot t}))}\mathrm dt$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd like to differentiate with respect to y.  This will give the integral:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_{-\pi}^\pi -(1-e^{i\cdot n \cdot t})(\sin{(y(1-e^{i\cdot n \cdot t}))}\mathrm dt$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;...If I'm correct.  Anyways, I'm interested in obtaining the results to this second integral, using this technique.  So I'm wondering if solving the first integral can help give results for the second integral.  I'm thinking of setting $y=1$ in the second integral.  This should eliminate $y$ from the result, and give me the integral involving $x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The trouble is, I'm not sure I can use the technique of differentiation under the integral.  I want to know how I can apply this technique to the integrals above.  Any pointers are appreciated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, for what values of $y$ is this valid?&lt;/p&gt;&#xA;" OwnerUserId="2626" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-03T14:05:41.890" LastActivityDate="2011-03-13T15:48:10.303" Title="Will moving differentiation from inside, to outside an integral, change the result?" Tags="&lt;calculus&gt;&lt;integral&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="5" />
  <row Id="12910" PostTypeId="2" ParentId="12897" CreationDate="2010-12-03T14:13:12.053" Score="3" ViewCount="" Body="&lt;p&gt;Following up on Fredrik Meyer's post, you can just pick points to evaluate the functions at and show that the resulting equations require $a_1, a_2$, and $a_3$ are 0.  So at 0 you have $a_1+a_2=0$.  At 1 you have $a_1+a_2\exp (1) +a_3\sin (1)=0$.  At $\frac{\pi}{4}$ you have $a_1+a_2\exp(\frac{\pi}{4}) + a_3\frac{\sqrt{2}}{2}=0$.  These will show that $a_1, a_2$, and $a_3$ are 0.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-12-03T14:13:12.053" />
  <row Id="12911" PostTypeId="2" ParentId="12909" CreationDate="2010-12-03T14:14:49.530" Score="12" ViewCount="" Body="&lt;p&gt;Wikipedia doesn't seem to have a precise statement of this theorem.  Here's a very general statement.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Theorem (Differentiation under the integral sign):&lt;/strong&gt; Let $U$ be an open subset of $\mathbb{R}$ and let $E$ be a measure space (which you can freely take to be any open subset of $\mathbb{R}^n$ if you want).  Let $f : U \times E \to \mathbb{R}$ have the following properties:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$x \mapsto f(t, x)$ is integrable for all $t$,&lt;/li&gt;&#xA;&lt;li&gt;$t \mapsto f(t, x)$ is differentiable for all $x$,&lt;/li&gt;&#xA;&lt;li&gt;for some integrable function $g$, for all $x \in E$, and for all $t \in U$, &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;$$\left| \frac{\partial f}{\partial t}(t, x) \right| \le g(x).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then the function $x \mapsto \frac{\partial f}{\partial t}(t, x)$ is integrable for all $t$.  Moreover, the function $F : U \to \mathbb{R}$ defined by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$F(t) = \int_E f(t, x) \mu(dx)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is differentiable, and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$F&amp;#39;(t) = \int_E \frac{\partial f}{\partial t}(t, x) \mu(dx).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In practice the only condition that isn't easily satisfiable is the third one.  It is satisfied in this case, so you're fine (for all $y$).&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2011-03-13T15:48:10.303" LastActivityDate="2011-03-13T15:48:10.303" CommentCount="2" />
  <row Id="12912" PostTypeId="1" AcceptedAnswerId="12915" CreationDate="2010-12-03T14:20:02.480" Score="3" ViewCount="180" Body="&lt;p&gt;I run a website that generates &lt;a href=&quot;http://www.degreedays.net/&quot; rel=&quot;nofollow&quot;&gt;degree days&lt;/a&gt;, a specialist form of weather data used for calculations relating to building energy consumption.  Without going into unnecessary detail, temperature is a function of time, and degree days are essentially the integral of that function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the moment our site calculates the data using an approximation method based on daily average, maximum, and minimum temperatures.  But I'm working on improving that method by using finer-grained temperature measurements.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;These finer-grained temperature measurements are taken throughout each day, and the recording interval can be anything from 1 minute to several hours - it depends on the weather station making the recordings.  For most weather stations, the readings are fairly regular, but they're not completely regular.  A weather station might typically record the temperature every half hour, but there will often be extra readings or missing readings, or readings taken at less regular intervals for certain periods.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Initially I've been using the trapezoidal method to numerically integrate the function of temperature against time.  It's working pretty well, but I'm wondering if I might be able to improve it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not a mathematician, and my understanding of numerical integration is only very basic.  I understand that Simpson's 1/3 rule and Simpson's 3/8 rule typically work better than the trapezoidal rule when numerically integrating mathematical functions.  But real-world temperature readings don't follow an exact mathematical function.  Also I understand that Simpson's rules require equal intervals, which my temperature readings don't consistently have.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wonder if it might make sense to use Simpson's rules to integrate stretches of temperature readings that have 2 or more consecutive time intervals of equal length, and use the trapezoidal rule for stretches of irregular readings.  But then I see &lt;a href=&quot;http://www.emis.de/journals/JIPAM/images/031_02_JIPAM/031_02.pdf&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; (a paper that I don't pretend to understand properly) that the trapezoidal rule can often work better than Simpson's rule for various classes of &quot;rougher&quot; functions.  I would guess that outside air-temperature variation would be classed as pretty rough - the temperature jumps up and down throughout the day for all sorts of reasons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I could probably come up with some way to estimate the effectiveness of various methods, but it's tricky because there's no &quot;right answer&quot; to compare figures against.  So I'm trying to figure out what method would make most sense from a theoretical standpoint.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do you think the trapezoidal rule is likely to be the best approach for me?  Or are there other approaches that might make more sense?&lt;/p&gt;&#xA;" OwnerUserId="4367" LastEditorUserId="4367" LastEditorDisplayName="" LastEditDate="2010-12-03T14:28:14.690" LastActivityDate="2010-12-13T15:39:03.940" Title="Numerical Integration of Irregular Temperature Readings" Tags="&lt;integral&gt;&lt;numerical-methods&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" />
  <row Id="12913" PostTypeId="1" AcceptedAnswerId="12918" CreationDate="2010-12-03T14:38:43.360" Score="1" ViewCount="126" Body="&lt;p&gt;If I have matrices $\mathbf A$ and $\mathbf B$ then what is $\mathbf x$ to minimize $\|\mathbf A-\mathbf x\mathbf B\|_F$?&lt;/p&gt;&#xA;" OwnerUserId="4370" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-03T14:46:33.037" LastActivityDate="2011-05-03T02:26:32.410" Title="Minimize Frobenius norm" Tags="&lt;linear-algebra&gt;" AnswerCount="2" CommentCount="9" />
  <row Id="12914" PostTypeId="1" AcceptedAnswerId="17442" CreationDate="2010-12-03T14:39:38.547" Score="3" ViewCount="192" Body="&lt;p&gt;Let $n$ be composite. I'm trying to figure out if the set $H$ of $a$ such that&#xA;1) $a$ is relatively prime to $n$ and &#xA;2) the Miller-Rabin test fails to show compositeness of $n$ with $a$&#xD;&#xA;is a subgroup of the multiplicative group mod $n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My instinct is no: I am trying to show that the set is not closed. But then again we won't get any factors of $n$ by multiplication because of condition 1).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Update: I've run some experiments and it seems that the number of strong liars relatively prime to n divides the order of the multiplicative group. I wanted to get a contradiction with Lagrange's theorem.&lt;/p&gt;&#xA;" OwnerUserId="2298" LastEditorUserId="2298" LastEditorDisplayName="" LastEditDate="2010-12-03T18:57:51.663" LastActivityDate="2011-01-14T00:22:24.930" Title="Miller-Rabin Primality Testing failure and a subgroup" Tags="&lt;number-theory&gt;&lt;prime-numbers&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="12915" PostTypeId="2" ParentId="12912" CreationDate="2010-12-03T14:44:29.250" Score="4" ViewCount="" Body="&lt;p&gt;Since your temperature readings aren't equispaced, you can't directly apply Simpson's rule; the approach equivalent to this is to construct the parabola that interpolates three consecutive points (i.e., across two panels), and integrate that. The problem with this approach, of course, is that you need to have an odd number of data points (even number of panels) to do this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can use the trapezoidal rule, of course, across each panel, but a probably better idea might be to construct a &lt;a href=&quot;http://en.wikipedia.org/wiki/Cubic_Hermite_spline&quot; rel=&quot;nofollow&quot;&gt;cubic Hermite interpolant&lt;/a&gt; for each panel, and then integrate that. An obvious problem is that four conditions are needed to uniquely determine a cubic for each panel (two points and two derivative values), but estimates of derivative values can be constructed from the data such that the piecewise interpolant is &lt;em&gt;locally monotonic&lt;/em&gt;; briefly, an piecewise interpolant is locally monotonic if there are no spurious inflection points or extrema within a panel. One approach to estimating derivative values for monotonic interpolation, due to &lt;a href=&quot;http://dx.doi.org/10.1137/0717021&quot; rel=&quot;nofollow&quot;&gt;Fritsch and Carlson&lt;/a&gt;, is implemented in FORTRAN as &lt;a href=&quot;http://www.netlib.org/slatec/pchip/&quot; rel=&quot;nofollow&quot;&gt;the &lt;code&gt;pchip&lt;/code&gt; package&lt;/a&gt;, and in MATLAB as &lt;a href=&quot;http://www.mathworks.com/help/techdoc/ref/pchip.html&quot; rel=&quot;nofollow&quot;&gt;the function &lt;code&gt;pchip&lt;/code&gt;&lt;/a&gt;. A more modern approach, and one which may be better is some situations, is due to &lt;a href=&quot;http://adsabs.harvard.edu/full/1990A%26A...239..443S&quot; rel=&quot;nofollow&quot;&gt;Steffen&lt;/a&gt;. You may have to experiment which of the trapezoidal rule, Fritsch-Carlson, or Steffen would be best for integrating your data.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Let me detail the way one uses cubic Hermite interpolation for integrating data:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each panel is bounded by two points, $(x_i,y_i)$ and $(x_{i+1},y_{i+1})$. The trapezoidal rule consists of constructing the line joining these two points (&lt;em&gt;linear interpolation&lt;/em&gt;):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f_i(x)=\frac{x_{i+1}-x}{x_{i+1}-x_i}y_i+\frac{x-x_i}{x_{i+1}-x_i}y_{i+1}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and integrating that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_{x_i}^{x_{i+1}}f_i(x)\mathrm dx=\frac{x_{i+1}-x_i}{2}(y_i+y_{i+1})$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Integrating with cubic Hermite interpolation can be considered as a further &quot;improvement&quot; of the trapezoidal rule; briefly, in addition to points, one has derivative values $y_i^{\prime}$ and $y_{i+1}^{\prime}$, from which one constructs the cubic&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$g_i(x)=y_i+y_i^{\prime}(x-x_i)+c_i(x-x_i)^2+d_i(x-x_i)^3$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle c_i=\frac{3\frac{y_{i+1}-y_i}{x_{i+1}-x_i}-2y_i^{\prime}-y_{i+1}^{\prime}}{x_{i+1}-x_i}$ and $\displaystyle d_i=\frac{y_i^{\prime}+y_{i+1}^{\prime}-2\frac{y_{i+1}-y_i}{x_{i+1}-x_i}}{(x_{i+1}-x_i)^2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Integrating $g_i(x)$ might look slightly complicated, however, there is a nice expression for the integral:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_{x_i}^{x_{i+1}}g_i(x)\mathrm dx=\frac{x_{i+1}-x_i}{6}\left(y_i+4g_i\left(\frac{x_{i+1}+x_i}{2}\right)+y_{i+1}\right)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;whose verification I'll leave up to you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll just add the note that in the case of equispaced data, integration with a piecewise cubic Hermite interpolant is equivalent to integration with the trapezoidal rule plus corrections at the beginning and end.&lt;/p&gt;&#xA;" OwnerUserId="498" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-13T15:39:03.940" LastActivityDate="2010-12-13T15:39:03.940" CommentCount="4" />
  <row Id="12916" PostTypeId="2" ParentId="12913" CreationDate="2010-12-03T15:28:11.300" Score="3" ViewCount="" Body="&lt;p&gt;The problem is equivalent to minimizing $\|A - xB\|_F^2$, which is in turn:&#xA;$\min_{x} \sum_{i, j}(A_{ij}-xB_{ij})^2$&#xD;&#xA;This is a simple quadratic in $x$, whose minimizer is obtained by setting the derivative to 0.&lt;/p&gt;&#xA;" OwnerUserId="1527" LastActivityDate="2010-12-03T15:28:11.300" CommentCount="0" />
  <row Id="12917" PostTypeId="2" ParentId="12906" CreationDate="2010-12-03T15:32:35.210" Score="27" ViewCount="" Body="&lt;p&gt;This problem illustrates the fact that two functions can be very close: $|f(x)-g(x)|&amp;lt;\epsilon$&#xD;&#xA;for all $x\in [0,1]$, but their derivatives can still be far apart, $|f&amp;#39;(x)-g&amp;#39;(x)|&amp;gt;c$ for some&#xA;constant $c&amp;gt;0$. &#xA;In our case, let $x=a(t),y=b(t),0\le t\le 1$ and $x=c(t),y=d(t), 0\le t\le 1$ be the &#xA;parametrizations of the two curves. By smoothing the corners, we may assume that both&#xA;are smooth. $$ \|(a(t),b(t))\|\approx \|(c(t),d(t))\|$$ does not imply&#xA;$$  \|(a&amp;#39;(t),b&amp;#39;(t))\|\approx \|(c&amp;#39;(t),d&amp;#39;(t))\|$$&#xD;&#xA;Therefore $\int_0^1 \|(a&amp;#39;(t),b&amp;#39;(t))\| dt$ need not be close to $\int_0^1 \|(c&amp;#39;(t),d&amp;#39;(t))\| dt.$&#xD;&#xA;Here $\|(x,y)\|$ denotes $\sqrt{x^2+y^2}$.&lt;/p&gt;&#xA;" OwnerUserId="3249" LastEditorUserId="3249" LastEditorDisplayName="" LastEditDate="2010-12-03T19:18:13.137" LastActivityDate="2010-12-03T19:18:13.137" CommentCount="3" />
  <row Id="12918" PostTypeId="2" ParentId="12913" CreationDate="2010-12-03T15:35:06.470" Score="3" ViewCount="" Body="&lt;p&gt;The Frobenius norm is the regular vector Euclidean norm when $A$ and $B$ are viewed as $n^2$-dimensional vectors. So the $xB$ minimizing $||A - xB||_F$ would be the vector projection of $A$ on $B$, given by $(A \cdot {B \over ||B||_F}) B$, where $\cdot$ is the $n^2$ dimensional dot product. Hence $x$ is given by $A \cdot {B \over ||B||_F}$. &lt;/p&gt;&#xA;" OwnerUserId="3035" LastActivityDate="2010-12-03T15:35:06.470" />
  <row Id="12919" PostTypeId="2" ParentId="12906" CreationDate="2010-12-03T15:40:10.990" Score="15" ViewCount="" Body="&lt;p&gt;The pithy expression for this &quot;paradox&quot; is as follows: let $x_n(t)$ be a sequence of parameterized curves which converges uniformly to a limit curve $x(t)$.  Then it need not be the case that the arclengths of $x_n(t)$ approach the arclength of $x(t)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[&lt;b&gt;Added after seeing TCL's answer&lt;/b&gt;: it is also true that uniform convergence of a sequence of functions does not imply convergence of their derivatives.  See Section 3 &lt;a href=&quot;http://math.uga.edu/~pete/243functions1.pdf&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; for some discussion of this.  As TCL points out, since arclength elements are computed using derivatives, the observation about derivatives may be in some sense more fundamental.  In other words, I think I like TCL's answer better than mine.]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As Ross Millikan points out, this is more familiarly shown by approximating the hypotenuse of a right triangle by a staircase pattern of horizontal and vertical line segments.  I still remember being a senior in high school and having a friend (whom I had had no prior mathematical interactions with) show this to me.  I definitely remember thinking that it was not paradoxical but certainly surprising.  (And I have mathematically respected this person ever since, even though I haven't seen her since I was a teenager.)  &lt;/p&gt;&#xA;" OwnerUserId="299" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2010-12-03T15:48:09.107" LastActivityDate="2010-12-03T15:48:09.107" />
  <row Id="12920" PostTypeId="2" ParentId="12906" CreationDate="2010-12-03T15:57:41.907" Score="15" ViewCount="" Body="&lt;p&gt;Hilarious! Of course, the circumference is &lt;em&gt;not&lt;/em&gt; approximated by the sum of lengths of the lines constructed as shown, but by the sum of the hypotenuses of each of the right-angle triangles formed around the edge of the circle (forming a polygon with vertices on the circle).&lt;/p&gt;&#xA;" OwnerUserId="3515" LastEditorUserId="3515" LastEditorDisplayName="" LastEditDate="2010-12-03T18:05:55.073" LastActivityDate="2010-12-03T18:05:55.073" />
  <row Id="12921" PostTypeId="1" CreationDate="2010-12-03T16:00:11.567" Score="-1" ViewCount="150" Body="&lt;p&gt;What is an R-trivial language? What is an R-trivial monoid?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Context: Formal languages. Afaik, R-trivial languages is a subset of the starfree languages.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I mostly have background in formal languages and automata theory but not so much with the syntactic monoid characterization. So it would be nice to give a basic definition with maybe a small example of such a language.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;(In order to support multiple QA-sites because I don't want to have any QA-site stay behind and to have that question also represented there, I have also posted this question on these other sites: &lt;a href=&quot;http://stackoverflow.com/questions/4346608/formal-languages-what-does-r-trivial-mean&quot;&gt;stackoverflow.com&lt;/a&gt;, &lt;a href=&quot;http://cstheory.stackexchange.com/questions/3503/formal-languages-what-does-r-trivial-mean&quot;&gt;cstheory.stackexchange.com&lt;/a&gt;, &lt;a href=&quot;http://mathoverflow.net/questions/48181/formal-languages-what-does-r-trivial-mean&quot; rel=&quot;nofollow&quot;&gt;mathoverflow.net&lt;/a&gt;. In general I am against cross-posting but in this case, as they all have the same goal to be a complete reference of questions in the specific area, having the question cross posted is the best thing you can do.)&lt;/p&gt;&#xA;" OwnerUserId="4371" LastEditorUserId="4371" LastEditorDisplayName="" LastEditDate="2010-12-03T20:09:38.077" LastActivityDate="2010-12-04T17:05:22.223" Title="formal languages: what does R-trivial mean?" Tags="&lt;monoid&gt;" AnswerCount="1" CommentCount="7" />
  <row Id="12922" PostTypeId="1" CreationDate="2010-12-03T16:18:53.813" Score="3" ViewCount="101" Body="&lt;p&gt;I've been attempting to find out more research and resources on generating random math problems based on performance and calculated difficulty. There's a few places that I'm just now stumbling upon, but I figured I'd ask the community what they recommend reading up on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.faqs.org/patents/app/20090017427&quot; rel=&quot;nofollow&quot;&gt;http://www.faqs.org/patents/app/20090017427&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's an article here on an application based on research at MIT&#xA;&lt;a href=&quot;http://www.mathscore.com/math/mathProblems.html&quot; rel=&quot;nofollow&quot;&gt;http://www.mathscore.com/math/mathProblems.html&lt;/a&gt;&#xA;I'm wondering where specifically that research resides or what I can learn from it and others.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My goal is to first understand how probability, priority queues and anything else plays a role (if at all) in generating varying math problems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any resources and recommended reading would be most useful if you can provide that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks :)&lt;/p&gt;&#xA;" OwnerUserId="4372" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-12-04T05:12:51.430" LastActivityDate="2010-12-04T05:12:51.430" Title="A method for generating random math problems based on performance" Tags="&lt;probability-theory&gt;&lt;statistics&gt;" AnswerCount="1" CommentCount="0" />
  <row Id="12924" PostTypeId="2" ParentId="12897" CreationDate="2010-12-03T16:23:05.820" Score="2" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\rm\ \ y(0) = a+b,\ \ y&amp;#39;(0) = b+c,\ \ y&amp;#39;&amp;#39;(0) = b\ $ and they must all be $\:0\:$ if $\rm\ y(x) = 0 $&#xD;&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-12-03T17:53:54.223" LastActivityDate="2010-12-03T17:53:54.223" />
  <row Id="12925" PostTypeId="1" AcceptedAnswerId="12980" CreationDate="2010-12-03T16:38:15.683" Score="2" ViewCount="186" Body="&lt;p&gt;There are 2 improper points: $0, 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found that $\displaystyle \int_{\frac{1}{2}}^{1}\frac{\ln^\alpha(1+x)}{x^\beta(1-x)^\alpha}\mathrm dx$ exists for $\alpha &amp;lt; 1$ (by using the limit comparison test, with $g(x)=\frac{1}{(1-x)^\alpha}$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I'm having trouble picking $g(x)$ for the other interval. If $\alpha \lt 0$ then $\ln^\alpha(1+x) \underset{x\to 0}{\to} \infty$, and everything I try gives me $\frac{f(x)}{g(x)}\to 0$ or $\frac{f(x)}{g(x)}\to \infty$ which doesn't give me all possible solutions.&lt;/p&gt;&#xA;" OwnerUserId="3237" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-04T11:48:51.367" LastActivityDate="2010-12-04T11:48:51.367" Title="For which values of $\alpha, \beta$, $\int_0^1 \frac{\ln^\alpha(1+x)}{x^\beta(1-x)^\alpha}\mathrm dx$ exists?" Tags="&lt;calculus&gt;&lt;homework&gt;&lt;improper-integrals&gt;" AnswerCount="1" CommentCount="11" FavoriteCount="1" />
  <row Id="12926" PostTypeId="2" ParentId="12906" CreationDate="2010-12-03T16:38:51.283" Score="10" ViewCount="" Body="&lt;p&gt;What is wrong with this? Fundamentally, that you have jumped in without a definition of the length of a arc.&lt;/p&gt;&#xA;" OwnerUserId="1312" LastActivityDate="2010-12-03T16:38:51.283" />
  <row Id="12927" PostTypeId="1" AcceptedAnswerId="12933" CreationDate="2010-12-03T16:43:54.613" Score="1" ViewCount="116" Body="&lt;p&gt;The problem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&lt;strong&gt;Three poles standing at the points A,B and C subtend angles $\alpha$, $\beta$ and $\gamma$ respectively,at the circumcentre of the triangle ABC.If the heights of these poles are in arithmetic progression; then show that $\cot \alpha$, $\cot \beta$ and $\cot \gamma $ are in harmonic progression.&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now,what I could not understand is subtending of the angle part,precisely &lt;strong&gt;how a point subtend angle at another point?&lt;/strong&gt; So,what I am looking for a proper explanation of the problem statement with a figure,since it's troubling me from sometime.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS:I am not looking for the solution (as of now) or any hints regarding the solution,just a clear explanation will be appreciated.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;My solution using &lt;a href=&quot;http://math.stackexchange.com/users/1102/moron&quot;&gt;Moron&lt;/a&gt;'s &lt;a href=&quot;http://math.stackexchange.com/questions/12927/help-to-understanding-the-problem/12933#12933&quot;&gt;interpretation&lt;/a&gt;,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $a$ $b$ and $c$ are the length of three sides of the poles and $O$ be the circumcentre then, $$ \cot \alpha = \frac{OA}{a}$$ $$ \cot \beta = \frac{OB}{b}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \cot \gamma = \frac{OC}{c}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As O is the circumcentre,$OA = OB = OC = k $(say) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again, $a$ $b$ and $c$ are in arithmetic progression,hence,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$2 \cdot \frac{k}{\cot \alpha} = \frac{k}{\cot \beta} + \frac{k}{\cot \gamma}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Canceling $k$ from both sides,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$2 \cdot \frac{1}{\cot \alpha} = \frac{1}{\cot \beta} + \frac{1}{\cot \gamma}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, $\cot \alpha$, $\cot \beta$ and $\cot \gamma $ are in harmonic progression. (&lt;strong&gt;QED&lt;/strong&gt;)&lt;/p&gt;&#xA;" OwnerUserId="2109" LastEditorUserId="2109" LastEditorDisplayName="" LastEditDate="2010-12-04T07:37:28.083" LastActivityDate="2010-12-04T07:37:28.083" Title="Help to understanding the problem" Tags="&lt;geometry&gt;&lt;trigonometry&gt;" AnswerCount="2" />
  <row Id="12928" PostTypeId="2" ParentId="12927" CreationDate="2010-12-03T16:47:35.803" Score="0" ViewCount="" Body="&lt;p&gt;I would read it that the poles have a diameter&gt;0.  The pole at A has diameter so the angle seen from the circumcenter is $\alpha$.  See if that works.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-12-03T16:47:35.803" CommentCount="1" />
  <row Id="12929" PostTypeId="1" CreationDate="2010-12-03T16:52:44.297" Score="7" ViewCount="132" Body="&lt;p&gt;Let $\displaystyle AD$, $\displaystyle BE$, $\displaystyle CF$ be 3 cevians concurrent at $\displaystyle P$ inside the triangle $\displaystyle ABC$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Prove or disprove that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle \dfrac{AD}{AP} + \dfrac{BE}{BP} + \dfrac{CF}{CP} \ge \dfrac{9}{2}$$&lt;/p&gt;&#xA;" OwnerUserId="3538" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-12-04T07:35:51.297" LastActivityDate="2010-12-04T20:23:30.670" Title="An inequality on Cevians" Tags="&lt;geometry&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="12930" PostTypeId="2" ParentId="12675" CreationDate="2010-12-03T16:56:17.963" Score="0" ViewCount="" Body="&lt;p&gt;It sounds like you are marking all the pizza joints in an area on Google maps, or something similar. More like a programming problem than a math problem. I have two notions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, look under the covers of a Hash Table for ideas. Not my field, but I believe they have ways to group a set in one pass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or, second, use a statistical sampling approach. Find the median and quartiles of the lat/long of a subset and make your decisions based on that. A little research plus some experimentation will give an idea about how big the subset has to be. &lt;/p&gt;&#xA;" OwnerUserId="1312" LastActivityDate="2010-12-03T16:56:17.963" />
  <row Id="12931" PostTypeId="1" CreationDate="2010-12-03T17:07:01.760" Score="3" ViewCount="178" Body="&lt;p&gt;One can write out an integral whose solution gives the number of solutions to the NP-Complete Number Partition Problem and I'm wondering if anyone has an suggestions or ideas on who to solve or approximate this integral analytically or numerically.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given $a_0, a_1, \dots, a_{n-1} \in \mathbb{Z}$, the Number Partition Problem asks for a partition of the $a_k$'s such that $\sum_{k=0}^{n-1} \sigma_k a_k = 0$, where $\sigma_k \in  \{  -1,1 \} $.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the following Integral:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ I(a_0, a_1, \dots, a_{n-1}) = \frac{1}{2 \pi} \int_{-\pi}^{\pi} \prod_{k=0}^{n-1} (e^{ i a_k \theta } + e^{ -i a_k \theta } ) d\theta $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$I(\dots)$ will count the number of solutions for a given instance of $(a_0, a_1, \dots, a_{n-1})$.  Solving this integral is worse than NP-Hard (it's #P) so asking for a general solution is out of the question.  But can one do any sort of approximation, either analytically or numerically?  If you choose the $a_k$'s with some distribution, say uniform on some interval, can you exploit that randomness to help you approximate this integral?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any ideas would be appreciated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;note:  This has been studied by Borgs et all for the &lt;a href=&quot;http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.89.9577&amp;amp;rep=rep1&amp;amp;type=pdf&quot; rel=&quot;nofollow&quot;&gt;NPP Phase Transition&lt;/a&gt; and that's where I first saw this integral representation of the Number Partition Problem, but their analysis relies on approximating the family of instances given a uniform distribution on the $a_k$'s rather than trying to solve a particular instance, as I'm trying to do above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One can write out an integral whose solution gives the number of solutions to the NP-Complete Number Partition Problem and I'm wondering if anyone has an suggestions or ideas on who to solve or approximate this integral analytically or numerically.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given $a_0, a_1, \dots, a_{n-1} \in \mathbb{Z}$, the Number Partition Problem asks for a partition of the $a_k$'s such that $\sum_{k=0}^{n-1} \sigma_k a_k = 0$, where $\sigma_k \in \{ -1,1 \} $.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the following Integral:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ I(a_0, a_1, \dots, a_{n-1}) = \frac{1}{2 \pi} \int_{-\pi}^{\pi} \prod_{k=0}^{n-1} (e^{ i a_k \theta } + e^{ -i a_k \theta } ) d\theta $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$I(\dots)$ will count the number of solutions for a given instance of $(a_0, a_1, \dots, a_{n-1})$.  Solving this integral is worse than NP-Hard (it's #P) so asking for a general solution is out of the question.  But can one do any sort of approximation, either analytically or numerically?  If you choose the $a_k$'s with some distribution, say uniform on some interval, can you exploit that randomness to help you approximate this integral?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any ideas would be appreciated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This has been studied by Borgs et all for the &lt;a href=&quot;http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.89.9577&amp;amp;rep=rep1&amp;amp;type=pdf&quot; rel=&quot;nofollow&quot;&gt;NPP Phase Transition&lt;/a&gt; and that's where I first saw this integral representation of the Number Partition Problem, but their analysis relies on approximating the family of instances given a uniform distribution on the $a_k$'s rather than trying to solve a particular instance, as I'm trying to do above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;NOTE&lt;/strong&gt;: This is a crosspost, and I have asked this on &lt;a href=&quot;http://mathoverflow.net/questions/38253/approximating-an-integral-representation-of-the-number-partition-problem&quot; rel=&quot;nofollow&quot;&gt;mathoverflow&lt;/a&gt;.  I have posted it here in hopes of getting more answers. &lt;/p&gt;&#xA;" OwnerUserId="4143" LastEditorUserId="4143" LastEditorDisplayName="" LastEditDate="2011-04-23T09:15:06.457" LastActivityDate="2011-04-23T09:15:06.457" Title="Approximating an integral representation of the Number Partition Problem" Tags="&lt;calculus&gt;&lt;analysis&gt;&lt;integral&gt;&lt;numerical-methods&gt;" AnswerCount="1" CommentCount="8" />
  <row Id="12932" PostTypeId="1" AcceptedAnswerId="12934" CreationDate="2010-12-03T17:16:01.230" Score="1" ViewCount="142" Body="&lt;p&gt;This is a pretty simple question, I'm afraid I have been out of math-land for too long!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have two currencies, the U.S. dollar and the rupee.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;$1 us dollar = .015 rupees&lt;/code&gt; (just an example, probably no where near the actual conversion).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's say I have $100,000$ rupees, how do I figure out the new value of my rupees if the value raises to be equal to the U.S. dollar ($1 U.S. dollar = 1 rupee)&lt;/p&gt;&#xA;" OwnerUserId="4374" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-03T18:00:58.833" LastActivityDate="2010-12-03T18:00:58.833" Title="(really) Simple percentage calculation with currency conversion" Tags="&lt;algebra-precalculus&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="12933" PostTypeId="2" ParentId="12927" CreationDate="2010-12-03T17:24:51.063" Score="1" ViewCount="" Body="&lt;p&gt;My guess is the poles are of (different) heights $h_A$, $h_B$, $h_C$ and the angle is from the foot of pole to the circumcentre of $\triangle ABC$ to the top of the pole.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-12-03T17:24:51.063" CommentCount="11" />
  <row Id="12934" PostTypeId="2" ParentId="12932" CreationDate="2010-12-03T17:30:47.873" Score="2" ViewCount="" Body="&lt;p&gt;Multiply the number of rupees by the value in dollars of one rupee.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the problem is that your conversion &quot;goes the other way&quot; (that is, you have something like 1 US dollar = 0.75 rupees, then &quot;solve&quot; for how many dollars equals one rupee first:&#xA;\begin{align*}&#xA;1\text{ dollars} &amp;amp;= 0.75\text{ rupees}\\&#xA;\frac{1}{0.75} \text{ dollars} &amp;amp;= \frac{0.75}{0.75} \text{ rupees}\\&#xA;\frac{4}{3}\text{ dollars} &amp;amp;= 1\text{ rupees.}&#xA;\end{align*}&#xA;and then you can just multiply the number of rupees by $\frac{4}{3}$ to get the number of dollars.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-12-03T17:30:47.873" CommentCount="1" />
  <row Id="12937" PostTypeId="1" AcceptedAnswerId="12946" CreationDate="2010-12-03T18:47:31.257" Score="1" ViewCount="55" Body="&lt;p&gt;Let&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$R(t) = u + ct - \sum_{k=1}^{X(t)}Z_{k}, t\geq 0 $,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;be a risk process, where $u&amp;gt; 0$ is the initial capital of the insurance company and $c&amp;gt; 0$ is a premium rate. We know that the number of claims received by the insurance company is in accordance with a homogenous Poisson Process $X(t), t\geq 0$, with parameter (intensity) $\lambda &amp;gt; 0$. Assume that the amounts $Z_k, k = 1, 2, ...$ of successive claims are independent random variables with common probability density function $f_{Z_k}(z), z \geq 0$. Suppose that random variables $Z_{k}, k = 1, 2, ...$ are independent of $X(t)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would first like to be able to compute the ranges of $q, q \geq 0$ such that the function $f(z)=q\alpha \exp (-\alpha z) + (1 - q)\beta \exp (-\beta z); \beta &amp;gt; \alpha &amp;gt; 0,$ may be taken as $f_{Z_k}(z), z \geq 0$ in the risk process.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, I would like to be able to compute $E[R(t)]$, and $Var[R(t)]$ in terms of the parameters $u, c, \lambda, \alpha, \beta$ and $q$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone help? I'm pretty stuck with this problem.&lt;/p&gt;&#xA;" OwnerUserId="2976" LastActivityDate="2010-12-03T21:30:51.537" Title="Risk process in Insurance" Tags="&lt;probability&gt;&lt;statistics&gt;&lt;applications&gt;" AnswerCount="1" />
  <row Id="12938" PostTypeId="1" CreationDate="2010-12-03T18:51:18.770" Score="1" ViewCount="159" Body="&lt;p&gt;I have a question concerning Gödel's second incompleteness theorem.&lt;br&gt;&#xA;Let $\sharp \psi$ be the Gödel number of a sentence $\psi\in L_A$. $\mathcal{N}$ is the standard model of Peano arithmetic,$\mathcal{P}$.&#xA;Using Gödel's second incompleteness theorem,one can show that $\mathcal{N}\models \psi\rightarrow\exists bPf_\mathcal{P}(\underline{\sharp\psi},b)$ is not true. ($Pf_\mathcal{P}(a, b)$ means that $b$ is the Gödel number of proof from $\mathcal{P}$ of the $L_A$ formula with Gödel number $a$.)  My question is how can I understand $Con_\mathcal{P}$, in this case and more generally?&#xA;Thank you for your answer.&lt;/p&gt;&#xA;" OwnerUserId="4377" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-04T13:11:07.507" LastActivityDate="2010-12-04T13:11:07.507" Title="Gödel's second incompleteness theorem" Tags="&lt;logic&gt;" CommentCount="1" />
  <row Id="12939" PostTypeId="2" ParentId="12895" CreationDate="2010-12-03T18:56:08.390" Score="3" ViewCount="" Body="&lt;p&gt;Here is a &quot;proof by physics&quot;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The average of $a_1, \dots, a_n$ is the center of mass of $n$ point masses, each with unit mass, having positions (along the $x$-axis) given by their values, i.e. $a_i$ has mass 1 and position $x = a_i$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given $(n+1)$ unit point masses, $a_1 \leq \dots \leq a_{n+1}$, let $u_n$ be the center of mass of the first $n$ points.  The center of mass $u_{n+1}$ can be computed by replacing the first $n$ points by a mass of $n$ with position $u_{n}$.  Since $a_{n+1} \geq a_n \geq u_n$, the addition of $a_{n+1}$ to the system shifts the center of mass to the right.  Hence, $u_{n+1} \geq u_n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, if $n$ points of unit mass are distributed evenly along the $x$-axis, it is easy to see that the addition of a unit mass at a position to the right of $u_n$ but to the left of $a_n$ will shift the center of the mass to the right (even though the position of the additional mass is to the left of $a_n$).  Hence, the converse is false.&lt;/p&gt;&#xA;" OwnerUserId="4009" LastActivityDate="2010-12-03T18:56:08.390" />
  <row Id="12941" PostTypeId="1" AcceptedAnswerId="12942" CreationDate="2010-12-03T19:25:22.180" Score="5" ViewCount="146" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;You have two empty bags and 100 bread&#xA;  rings. The goal is to put all your&#xA;  rings into the bags in such a way that one&#xA;  of the bags will contain exactly&#xA;  twice as many rings then the other.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;You can't eat, break and lose the rings.&#xA;  All of them should go into the bags.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Let me know if my explanation is not clear. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Actually, I know the answer but I'd like to share it with you. IMHO, it is pretty nice.&lt;/p&gt;&#xA;" OwnerUserId="3761" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-12-03T19:33:04.053" LastActivityDate="2010-12-03T19:33:04.053" Title="Nice puzzle: 100 bread rings and two bags" Tags="&lt;puzzle&gt;" AnswerCount="1" CommentCount="6" />
  <row Id="12942" PostTypeId="2" ParentId="12941" CreationDate="2010-12-03T19:30:35.103" Score="7" ViewCount="" Body="&lt;p&gt;Put half in one bag, and put half in the other bag. Then put one bag inside the other. &lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-12-03T19:30:35.103" CommentCount="4" />
  <row Id="12944" PostTypeId="1" AcceptedAnswerId="12988" CreationDate="2010-12-03T20:54:35.600" Score="2" ViewCount="332" Body="&lt;p&gt;what is the product of delta function with itself ? what is the dot product with itself ?&lt;/p&gt;&#xA;" OwnerUserId="2987" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-04T15:15:40.067" LastActivityDate="2010-12-04T22:00:27.020" Title="what is product of delta function with itself ?" Tags="&lt;distribution-theory&gt;" AnswerCount="2" CommentCount="15" />
  <row Id="12945" PostTypeId="2" ParentId="12870" CreationDate="2010-12-03T21:00:18.173" Score="2" ViewCount="" Body="&lt;p&gt;There are a number of ways of dealing with free/bound variables. The simplest (and least user friendly) is &lt;a href=&quot;http://en.wikipedia.org/wiki/De_Bruijn_indices&quot; rel=&quot;nofollow&quot;&gt;de Bruijn Indices&lt;/a&gt;. In this system, a bound variable is named based on how &quot;far&quot; it is from the binder. For example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\lambda x. \lambda y.~ x~y$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\lambda . \lambda .~ 1~0$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;in de Bruijn notation. This makes substitution of closed terms straight-forward. Free variables are represented by indices &quot;beyond&quot; the number of binders. For example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\lambda x. x~a$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\lambda. 0~1$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since there is only one binder, $1$ is free (note that this is indexed from 0). For the full details, look at the wikipedia page linked above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A great resource for this and other basic questions about lambda calculi is Benjamin Pierce's &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0262162091&quot; rel=&quot;nofollow&quot;&gt;Types and Programming Languages&lt;/a&gt;. Very well written and readable. It's the closest we have to a &quot;standard&quot; Programming Languages book.&lt;/p&gt;&#xA;" OwnerUserId="4388" LastActivityDate="2010-12-03T21:00:18.173" />
  <row Id="12946" PostTypeId="2" ParentId="12937" CreationDate="2010-12-03T21:30:51.537" Score="1" ViewCount="" Body="&lt;p&gt;I'll start with some suggestions that should work.  If you're still stuck after trying them, just comment under my answer, and I'll give more detail.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;For the range of $q$, it sounds like you want to find the values of $q$ that make $f(z)$ a legitimate probability density function.  If so, you want to find the range of $q$ values for which $f(z) \geq 0$ and $\int_0^{\infty} f(z) dz = 1$.  That should be doable.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;For $E[R(t)]$, use the fact that $E[R(t)] = E[E[R(t)|X(t)]]$.  Knowing the distribution of $X(t)$ should make this doable as well.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;For $Var[R(t)]$, try $Var[R(t)] = E[Var[R(t)|X(t)]] + Var[E[R(t)|X(t)]]$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="2370" LastActivityDate="2010-12-03T21:30:51.537" />
  <row Id="12947" PostTypeId="1" CreationDate="2010-12-03T21:33:11.577" Score="2" ViewCount="93" Body="&lt;p&gt;Hello,&#xA;where can I find a proof that truncated octahedron tiles Euclidean 3D space?&lt;/p&gt;&#xA;" OwnerUserId="4389" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-03T23:59:12.943" LastActivityDate="2011-04-20T23:45:25.393" Title="Truncated octahedron tiles 3D space. Proof?" Tags="&lt;geometry&gt;&lt;tiling&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="12949" PostTypeId="1" AcceptedAnswerId="12956" CreationDate="2010-12-03T21:39:57.937" Score="0" ViewCount="185" Body="&lt;p&gt;Let $$F(x)=\sum_{n=1}^\infty \frac{(nx)}{n^2},$$ where $(x)=x$ for $x\in (-1/2,1/2]$ and $(x)$ is continued to $\mathbb{R}$ by periodicity, that is $(x+1)=(x)$. Prove that $F$ is discontinuous whenever $x=m/2n$, where $m,n\in \mathbb{Z}$ with $m$ odd and $n\neq 0$. This does not seem obvious to me.&lt;/p&gt;&#xA;" OwnerUserId="3249" LastActivityDate="2010-12-04T04:11:59.337" Title="An integrable function with many discontinuities" Tags="&lt;real-analysis&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="12950" PostTypeId="2" ParentId="12768" CreationDate="2010-12-03T22:01:03.900" Score="4" ViewCount="" Body="&lt;p&gt;sorry for the confusion, but you got it absolutely right&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2 times faster means two times as fast :) &#xA;A takes 100ms&#xA;B takes 200ms&#xA;so A is two times faster than B&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This particular graph is probably the most confusing taken out of the flow of the previous ones in the chapter. This one shows an optimization that most often you shouldn't bother. Especially when there are cases of stuff being 100 times faster. And especially when it's not consistent among browsers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The point was not to worry about micro-optimizations (unless it's critical to you and you've done all else). So 20% faster (or 1.2 times faster) is probably not worth it most of the time when there are other optimization that will make something 10 times faster&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Best,&#xA;Stoyan&lt;/p&gt;&#xA;" OwnerUserId="4390" LastActivityDate="2010-12-03T22:01:03.900" CommentCount="1" />
  <row Id="12951" PostTypeId="1" CreationDate="2010-12-03T22:09:42.717" Score="1" ViewCount="91" Body="&lt;p&gt;Any idea how to proof that when 3D space is tiled with truncated octahedra, all vertices can be colored black and white such that no 2 vertices sharing the same color are adjacent?&lt;/p&gt;&#xA;" OwnerUserId="4389" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-04T00:36:19.090" LastActivityDate="2010-12-04T00:36:19.090" Title="Truncated octahedron is bipartite. Proof?" Tags="&lt;geometry&gt;&lt;tiling&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="12952" PostTypeId="1" CreationDate="2010-12-03T22:12:54.927" Score="0" ViewCount="53" Body="&lt;p&gt;I need to know what are these beasts, but I can't get my hands on Gray &amp;amp; Hervella paper :) Is there some other paper that describes them and is available for free on the Internet?&lt;/p&gt;&#xA;" OwnerUserId="3793" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-03T23:44:28.627" LastActivityDate="2010-12-03T23:44:28.627" Title="Gray-Hervella classes" Tags="&lt;geometry&gt;&lt;reference-request&gt;" CommentCount="3" />
  <row Id="12953" PostTypeId="2" ParentId="7090" CreationDate="2010-12-03T22:45:09.197" Score="0" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\rm\ \ a\:$ invertible $\rm\:(mod\ n)\ \ \Rightarrow\ \ a\ \{0,1,2,\ldots,n-1\}\ \equiv\ \{0,1,2,\ldots, n-1\}\ \ (mod\ n)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i.e. the map $\rm\ x \to a\:x\ $ is a bijection $\rm\ (mod\ n)\ $ so it simply permutes the congruence classes.&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-12-03T22:45:09.197" />
  <row Id="12954" PostTypeId="2" ParentId="12951" CreationDate="2010-12-03T22:55:56.250" Score="1" ViewCount="" Body="&lt;p&gt;The graph of a truncated octahedron is bipartite but there are space fillers with regular polygons as faces which are not bipartite.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://mathworld.wolfram.com/Space-FillingPolyhedron.html&quot; rel=&quot;nofollow&quot;&gt;http://mathworld.wolfram.com/Space-FillingPolyhedron.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1369" LastActivityDate="2010-12-03T22:55:56.250" CommentCount="3" />
  <row Id="12956" PostTypeId="2" ParentId="12949" CreationDate="2010-12-03T23:41:34.770" Score="1" ViewCount="" Body="&lt;p&gt;Note that $\lim_{x \rightarrow a}(x) = a$ except when $a$ is an odd multiple of ${1 \over 2}$, in which case $\lim_{x \rightarrow a^-}(x) - \lim_{x \rightarrow a^+}(x) = 1$. Another way of saying this is that $\lim_{x \rightarrow a^-}(x) - \lim_{x \rightarrow a^+}(x) = 0$ unless $a$ is an odd multiple of ${1 \over 2}$ in which case it is 1. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So for the term $f_n(x) = {(nx) \over n^2}$, $\lim_{x \rightarrow a^-}f_n(x) - \lim_{x \rightarrow a^+}f_n(x) = 0$ unless $x = {m \over 2n}$ for some odd integer $m$, in which case it is ${1 \over n^2}$. So for your arbitrary ${m \over 2n}$ with $m$ odd, it will incur a jump of ${1 \over n^2}$ due to $f_n(x)$, and other terms can only increase or keep the same the size of the jump. For any $\epsilon &amp;gt; 0$, if enough terms have been added, the absolute value of the sum of the remaining terms will be at most $\epsilon$, so that no matter how these remaining terms add up there will still be a jump of at least ${1 \over n^2} - \epsilon$ at ${m \over 2n}$. Letting $\epsilon \rightarrow 0$ this means the infinite sum has a jump of at least  ${1 \over n^2}$ at ${m \over 2n}$. (By jump I now mean the $\liminf$ from the left minus the $\limsup$ from the right.) In particular it's discontinuous there.&lt;/p&gt;&#xA;" OwnerUserId="3035" LastEditorUserId="3035" LastEditorDisplayName="" LastEditDate="2010-12-04T01:43:42.233" LastActivityDate="2010-12-04T01:43:42.233" CommentCount="2" />
  <row Id="12957" PostTypeId="1" AcceptedAnswerId="12976" CreationDate="2010-12-04T00:13:39.370" Score="0" ViewCount="95" Body="&lt;p&gt;How do you find out how much a put contract would cost?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The premium for a 99.00 Eurodollar futures put option is 0.55 and the&#xA;  spot price is 98.51&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So I have the intrinsic value which is &lt;code&gt;$1225&lt;/code&gt;, but I am not sure how to compute the time value and how much that put contract would cost. Still trying to understand options.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="3961" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-04T00:29:28.070" LastActivityDate="2010-12-04T07:27:52.770" Title="Stock/Futures put options" Tags="&lt;finance&gt;" AnswerCount="1" CommentCount="0" />
  <row Id="12958" PostTypeId="1" CreationDate="2010-12-04T00:17:33.350" Score="3" ViewCount="135" Body="&lt;p&gt;Both answers to this question seem equally reasonable to me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the answer is positive, I have no idea what the construction of such a space would look like....&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the answer is negative, I assume one would try to subdivide the cells somehow... but I don't really know how that would go.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I guess this came up because I was trying to think of an example of a CW-complex that wasn't homeomorphic to a Delta-complex... and figured the easiest way to make such a thing would be to make one that is not homotopic. This, of course, doesn't seem much easier to build, but at least easier to prove once you're done building.&lt;/p&gt;&#xA;" OwnerUserId="423" LastActivityDate="2010-12-04T00:17:33.350" Title="Is every CW complex homotopic to a Delta-Complex?" Tags="&lt;topology&gt;&lt;algebraic-topology&gt;" CommentCount="2" />
  <row Id="12959" PostTypeId="1" AcceptedAnswerId="13005" CreationDate="2010-12-04T00:21:22.370" Score="6" ViewCount="157" Body="&lt;p&gt;Let $\{B(t), t \in \mathbb{R} \}$ be a two sided brownian motion defined as&#xA;$$&#xD;&#xA;B(t) = \begin{cases} B_1(t),\quad t &amp;gt;0 \\&#xD;&#xA;0, \quad t = 0 \\&#xD;&#xA;B_2(-t), \quad t &amp;lt; 0 \end{cases}&#xD;&#xA;$$&#xD;&#xA;where $B_1$ and $B_2$ are independent standard Brownian motions on $\mathbb{R}^+$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fix $x_0 &amp;gt; 0$ and let $x_k = B(x_{k-1})$ for $k=1,2,\dots$.  What can we say about $\lim_{k\rightarrow \infty} x_k$?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it converges, then I imagine the limit would have to be to $0$ a.s., since the limit $y$ would satisfy $B(y) = y$ a.s..  But I don't know how to show this sequence converges.  Any ideas? (this isn't homework, just a problem my friend and I thought up)&lt;/p&gt;&#xA;" OwnerUserId="2515" LastEditorUserId="2515" LastEditorDisplayName="" LastEditDate="2010-12-04T19:19:00.593" LastActivityDate="2010-12-04T19:19:00.593" Title="Question about Brownian motion" Tags="&lt;probability&gt;&lt;stochastic-processes&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="12960" PostTypeId="1" AcceptedAnswerId="13017" CreationDate="2010-12-04T01:53:24.480" Score="12" ViewCount="141" Body="&lt;p&gt;&lt;a href=&quot;http://books.google.com/books?id=ZlY7gZacogsC&amp;amp;pg=PA172&quot; rel=&quot;nofollow&quot;&gt;Exercise 7.11 of  Kurzweil–Stellmacher&lt;/a&gt; asks me to:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Prove that the centralizer of an involution is nonsolvable inside any perfect group with generalized quaternion Sylow 2-subgroups of order at least 16.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I would like to oblige, but I'm having trouble getting much of a handle on the centralizer itself, rather than just its action.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The center T=⟨t⟩ of a quaternion group is weakly closed, and so the normalizer X=N&lt;sub&gt;G&lt;/sub&gt;(T)=C&lt;sub&gt;G&lt;/sub&gt;(t) of T in that perfect group G controls G-fusion within any Sylow 2-subgroup P that happens to contain T.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only fusion possible on P in a perfect group is the &quot;full&quot; fusion where all (both classes of) quaternion subgroups of order 8 are acted on by their full automorphism group (and otherwise no fusion beyond that induced by P itself).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I know X contains at least two copies of SL(2,3) and X/Z contains at least two copies of S4.  Since G is perfect, P = [P,X] ≤ [X,X].  I think the copies of SL(2,3) pairwise intersect in a particular cyclic subgroup of P of order 4.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More precisely: If Q is a quaternion subgroup of P of order 8, then Y=N&lt;sub&gt;X&lt;/sub&gt;(Q) contains SL(2,3) and Y/Z contains S4.  [Y,Q] ≥ Q.  P is generated by its [T:Q] quaternion subgroups of order 8, so also by its [Y,Q]s.  The intersection of any pair of distinct Qs is the order 4 cyclic subgroup of the (unique) maximal subgroup that happens to be cyclic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So it is a special configuration that has to fit lots of solvable groups inside, but I don't see any particular reason the result could not be solvable.  The closest thing to nonsolvable I get is that the derived length of N has to be at least 3.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I get roughly the same results for a quaternion Sylow of order 8, except one only has the single SL(2,3).  I have not found a perfect G with solvable X in this case, but of course I can if I only require the same fusion pattern, since SL(2,3) and SL(2,5) have the same 2-fusion.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;More:&lt;/strong&gt; With a little reminder from reading the work of Bender's students, I can now do this using Brauer–Suzuki or Z*, but this does not seem much like a homework solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The truth seems to be X⋅O(G)=G, and every non-trivial quotient of a perfect group is perfect, so the homomorphic image G/O(G) of X is perfect.  Since G has Sylow 2-subgroups, G/O(G) is non-identity, so X has a non-identity perfect quotient, so X is nonsolvable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure how to show X⋅O(G)=G without using tools significantly beyond chapter 7.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Smaller question:&lt;/strong&gt; I vaguely recall Suzuki proved that if G is a perfect group with no non-identity normal subgroups of odd order and with quaternion Sylow 2-subgroups, then each coset of C&lt;sub&gt;G&lt;/sub&gt;(t) contains exactly one conjugate of t, where t is an involution.  In other words, if the product of two involutions centralizes t, then that product is the identity.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Can someone prove this with ideas from K–S up to chapter 7?&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is hard for me to check since in reality such a group has exactly one involution t and G = C&lt;sub&gt;G&lt;/sub&gt;(t), but that comes later in my solution.&lt;/p&gt;&#xA;" OwnerUserId="583" LastEditorUserId="583" LastEditorDisplayName="" LastEditDate="2010-12-04T18:22:12.760" LastActivityDate="2010-12-05T12:16:24.170" Title="Involution centralizer of perfect group with quaternion Sylow 2-subgroups" Tags="&lt;group-theory&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="2" />
  <row Id="12961" PostTypeId="1" AcceptedAnswerId="12971" CreationDate="2010-12-04T04:06:03.290" Score="3" ViewCount="90" Body="&lt;p&gt;Got stuck on this one:&#xA;Show that there is a constant $C&amp;gt;0$ such that $$\left|\int_0^x \frac{\sin (N+1/2)t}{\sin t/2} dt \right|\le C$$ for all&#xA;$x\in [-\pi,\pi]$ and integer $N\ge 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I thought this should follow from $\int_0^\infty \frac{\sin t}{t}dt&amp;lt;\infty$, but somehow&#xA;don't get the connection.&lt;/p&gt;&#xA;" OwnerUserId="3249" LastEditorUserId="3249" LastEditorDisplayName="" LastEditDate="2010-12-04T04:53:43.513" LastActivityDate="2010-12-04T05:58:23.280" Title="Boundedness of an integral" Tags="&lt;analysis&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="12962" PostTypeId="2" ParentId="12949" CreationDate="2010-12-04T04:11:59.337" Score="1" ViewCount="" Body="&lt;p&gt;A direct computation shows that &#xA;$$F(x+)=F(x)-\frac{1}{2n^2}\sum_{k=0}^{\infty}\frac{1}{(2k+1)^2}=F(x)-\frac{\pi^2}{16n^2},$$&#xD;&#xA;$$F(x-)=F(x)+\frac{1}{2n^2}\sum_{k=0}^{\infty}\frac{1}{(2k+1)^2}=F(x)+\frac{\pi^2}{16n^2},$$ &#xA;when $x=m/2n$, with $m$ and $2n$ being relatively prime. (The infinite sum is easily derived  from the &lt;a href=&quot;http://en.wikipedia.org/wiki/Basel_problem&quot; rel=&quot;nofollow&quot;&gt;Euler result&lt;/a&gt;, of course.)&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The function $F(x)$ was suggested by Riemann in his thesis as an example of a (Riemann)-integrable function which has a point of discontinuity in every interval (see &#xA;&lt;a href=&quot;http://books.google.com/books?id=QdrCXs4L4v4C&amp;amp;printsec=frontcover&amp;amp;dq=Gesammelte+Mathematische+Werke+und+++Wissenschaftlicher+Nachlass&amp;amp;cd=2#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;&quot;Gesammelte Mathematische Werke und Wissenschaftlicher Nachlass&quot;&lt;/em&gt;&lt;/a&gt;, Springer -Verlag 1990, pp. 259-296).&lt;/p&gt;&#xA;" OwnerUserId="723" LastActivityDate="2010-12-04T04:11:59.337" CommentCount="2" />
  <row Id="12963" PostTypeId="2" ParentId="12961" CreationDate="2010-12-04T04:13:12.267" Score="0" ViewCount="" Body="&lt;p&gt;If you expand the numerator you get a sum of two terms.  One doesn't depend on t, one is proportional to cot(t/2)&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-12-04T04:13:12.267" CommentCount="4" />
  <row Id="12965" PostTypeId="1" AcceptedAnswerId="12972" CreationDate="2010-12-04T04:35:53.060" Score="1" ViewCount="132" Body="&lt;p&gt;I am trying to see why $\frac{\sin(nx)}{1+\vert x\vert^2}$ has no convergent subsequence in $L^1(\mathbb{R})$.  This is an &quot;optional&quot; homework problem from my disorganized analysis professor which is worth no points, and which I wouldn't do anyway, because I'm terrible at analysis and have no clue how to deal with this.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To me it seems clear though that I want to be looking at the integral $\int\limits_\mathbb{R} \frac{\vert\sin(n_kx)-\sin(n_jx)\vert}{1+\vert(x)\vert^2}$ and finding some kind of constant, maybe in terms of when $n_k\neq n_j$ based on vague memories of undergrad PDE.  Any hints?&lt;/p&gt;&#xA;" OwnerUserId="3595" LastActivityDate="2010-12-04T06:05:19.033" Title="Convergence of $\frac{\sin(nx)}{1+\vert x\vert^2}$" Tags="&lt;real-analysis&gt;&lt;functional-analysis&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="12967" PostTypeId="2" ParentId="12922" CreationDate="2010-12-04T05:07:49.920" Score="3" ViewCount="" Body="&lt;p&gt;The only stated relation to MIT at the second link is &quot;developed by MIT graduates&quot;. That or equivalent words appear in a few places on the web site, but I see no claim that the program or patents are related to research conducted at any university. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wikipedia has a page on adaptive testing: &lt;a href=&quot;http://en.wikipedia.org/wiki/Computerized_adaptive_testing&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Computerized_adaptive_testing&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The ideas are similar to the statistical protocols used in medical trials, that allow for early stopping in case of unexpectedly good or bad results.&lt;/p&gt;&#xA;" OwnerUserId="467" LastActivityDate="2010-12-04T05:07:49.920" CommentCount="1" />
  <row Id="12968" PostTypeId="1" AcceptedAnswerId="12969" CreationDate="2010-12-04T05:22:56.020" Score="14" ViewCount="378" Body="&lt;p&gt;I was wondering if there are examples of results in mathematics that were first proven using axiom of choice and later someone found a proof of the result without using the axiom of choice.&lt;/p&gt;&#xA;" OwnerUserId="3947" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-04T06:12:11.653" LastActivityDate="2010-12-04T18:40:50.923" Title="Axiom of choice - to use or not to use" Tags="&lt;reference-request&gt;&lt;axiom-of-choice&gt;" AnswerCount="4" CommentCount="5" FavoriteCount="4" />
  <row Id="12969" PostTypeId="2" ParentId="12968" CreationDate="2010-12-04T05:33:25.243" Score="14" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Cantor%E2%80%93Bernstein%E2%80%93Schroeder_theorem&quot; rel=&quot;nofollow&quot;&gt;Cantor-Schroder-Berstein&lt;/a&gt; theorem was &lt;a href=&quot;http://en.wikipedia.org/wiki/Cantor%E2%80%93Bernstein%E2%80%93Schroeder_theorem#Original_proof&quot; rel=&quot;nofollow&quot;&gt;first proved&lt;/a&gt; by Cantor using the axiom of choice.  Proofs without using it came later. &lt;/p&gt;&#xA;" OwnerUserId="145" LastActivityDate="2010-12-04T05:33:25.243" CommentCount="3" />
  <row Id="12970" PostTypeId="2" ParentId="12961" CreationDate="2010-12-04T05:46:30.377" Score="2" ViewCount="" Body="&lt;p&gt;Consider: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sin (Nt + \frac12 t) = \frac1{2i} \left[ \exp (i Nt + \frac{i}2 t) - \exp (-Nt - \frac{i}2 t)\right] = \sin \frac{t}2  \sum_{m = -N}^{N} \exp (imt) $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So your integrand reduces to &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ 1 + 2\sum_{m = 1}^N \cos(mt) $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, $\int_0^x \cos(mt) dt = \frac{\sin (mx)}{m}$. To estimate $\sum_1^N \frac{\sin(mx)}{m}$ you can now use your fact that $\int_1^\infty \frac{\sin t}{t} dt &amp;lt; \infty$.  &lt;/p&gt;&#xA;" OwnerUserId="1543" LastActivityDate="2010-12-04T05:46:30.377" CommentCount="2" />
  <row Id="12971" PostTypeId="2" ParentId="12961" CreationDate="2010-12-04T05:58:23.280" Score="3" ViewCount="" Body="&lt;p&gt;This is a classic integral from summing Fourier series. I saw the following argument in Katznelson's book I believe. By symmetry we can assume $x &amp;gt; 0$. We divide the integral into $0$ to ${1 \over N}$ and ${1 \over N}$ to $x$ parts. (The second term will be $0$ if $x$ is small enough). Since $|\sin((N + {1 \over 2})t)| \leq (N + {1 \over 2})t$ and $\sin({t \over 2}) \geq C&amp;#39;t$ the integrand is bounded by $CN$ and the first term is at most $C$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the second term, you can integrate by parts, integrating the $\sin((N + {1 \over 2})t)$ and differentiating the ${1 \over \sin({t \over 2})}$. You get a factor of ${1 \over N + {1 \over 2}}$ from the integration, and the resulting integrand is now bounded by $C{1 \over Nt^2}$. Taking absolute values and integrating from ${1 \over N}$ to $x$ again gives a bound of $C$. The left endpoint term in this integration by parts is bounded by $C{1 \over Nt}$ at $t = {1 \over N}$, so once again you just get a $C$. The right-endpoint term is even smaller. Thus you're done.&lt;/p&gt;&#xA;" OwnerUserId="3035" LastActivityDate="2010-12-04T05:58:23.280" />
  <row Id="12972" PostTypeId="2" ParentId="12965" CreationDate="2010-12-04T06:05:19.033" Score="6" ViewCount="" Body="&lt;p&gt;This is homework-like so I'll just give some hints. Show that if ${\sin(n_kx) \over 1 + x^2}$ converges in $L^1$ to some $f(x)$, then $\int_R{\sin(n_kx) \over 1 + x^2} g(x)$ converges to $\int_R f(x)g(x)$ for any bounded $g(x)$. Show this contradicts the Riemann-Lebesgue lemma in some way. Note you have to consider the case where $f(x) = 0$ too.&lt;/p&gt;&#xA;" OwnerUserId="3035" LastActivityDate="2010-12-04T06:05:19.033" CommentCount="1" />
  <row Id="12973" PostTypeId="2" ParentId="12968" CreationDate="2010-12-04T06:47:56.037" Score="6" ViewCount="" Body="&lt;p&gt;This  article is an example of sorts where Doyle and Conway prove without using AC that for any two sets $A,B$ such that there is a bijection between the sets $3\times A$ and $3\times B$, where $3=\{0,1,2\}$, there is a bijection between $A$ and $B$. &lt;a href=&quot;http://www.math.dartmouth.edu/~doyle/docs/three/three.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.math.dartmouth.edu/~doyle/docs/three/three.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This should be relatively easy to prove using well ordering principle which is equivalent to axiom of choice.&lt;/p&gt;&#xA;" OwnerUserId="3431" LastActivityDate="2010-12-04T06:47:56.037" />
  <row Id="12974" PostTypeId="1" AcceptedAnswerId="12977" CreationDate="2010-12-04T06:50:37.313" Score="0" ViewCount="50" Body="&lt;p&gt;So, I'm playing a game where there is a certain percent value that changes over time.  I look at it occasionally, it seems to be approximately linear in time.  However, the change is slow enough that I want something better than simple linear regression.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Problem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f(x) := ax+b$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$g(x) := \mathrm{round}(f(x))$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given pairs $\{(x_1,g(x_1)),\dots,(x_n,g(x_n))\}$, estimate $a$ and $b$.&lt;/p&gt;&#xA;" OwnerUserId="3888" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-04T09:02:29.130" LastActivityDate="2010-12-04T09:02:29.130" Title="Estimating the equation of a line from its rounded values" Tags="&lt;approximation&gt;&lt;regression&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="12975" PostTypeId="2" ParentId="12929" CreationDate="2010-12-04T07:06:42.020" Score="6" ViewCount="" Body="&lt;p&gt;The inequality is true!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/ffgYO.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It can be shown that (see proof at the end of the answer)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle \frac{PD}{AD} + \frac{PE}{BE} + \frac{PF}{CF} = 1$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that this implies that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle \frac{AP}{AD} + \frac{BP}{BE} + \frac{CP}{CF}  = 2$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;as $\displaystyle 1 - \frac{PD}{AD} = \frac{AP}{AD}$ etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we have the inequality (easily shown using $\text{AM} \ge \text{GM}$) that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle (a_1 + a_2 + a_3)(\frac{1}{a_1} + \frac{1}{a_2} + \frac{1}{a_3}) \ge 9$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This shows that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle (\frac{AP}{AD} + \frac{BP}{BE} + \frac{CP}{CF})(\frac{AD}{AP} + \frac{BE}{BP} + \frac{CF}{CP}) \ge 9$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and so&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle 2(\frac{AD}{AP} + \frac{BE}{BP} + \frac{CF}{CP}) \ge 9$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i.e.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle \frac{AD}{AP} + \frac{BE}{BP} + \frac{CF}{CP} \ge \frac{9}{2}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the equality occurs only when $\displaystyle \frac{AP}{AD} = \frac{BP}{BE} = \frac{CP}{CF} = \frac{2}{3}$, which implies that $\displaystyle P$ is the centroid.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let us try showing that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle \frac{PD}{AD} + \frac{PE}{BE} + \frac{PF}{CF} = 1$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the figure (repeated from above for convenience).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/ffgYO.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note, if you are worried about acute triangle vs obtuse etc, a simple affine transformation will do to transform the triangle into an equilateral triangle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let X be the foot of perpendicular from A to BC and Y be the foot of the perpendicular from P to BC.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle \triangle AXD$ and $\triangle PYD$ are similar and thus&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle \frac{PY}{AX} = \frac{PD}{AD}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now $\displaystyle \frac{PY}{AX} = \frac{|\triangle PBC|}{|\triangle ABC|}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\displaystyle |\triangle MNO|$ is the area of $\displaystyle \triangle MNO$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus $\displaystyle \frac{PD}{AD} = \frac{|\triangle PBC|}{|\triangle ABC|}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle \frac{PE}{BE} = \frac{|\triangle PAC|}{|\triangle ABC|}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle \frac{PF}{CF} = \frac{|\triangle PAB|}{|\triangle ABC|}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Adding gives us&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle \frac{PD}{AD} + \frac{PE}{BE} + \frac{PF}{CF} = \frac{|\triangle PAB| + |\triangle PAC| + |\triangle PBC|}{|\triangle ABC|} = \frac{|\triangle ABC|}{|\triangle ABC|} = 1$$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-12-04T20:23:30.670" LastActivityDate="2010-12-04T20:23:30.670" CommentCount="3" />
  <row Id="12976" PostTypeId="2" ParentId="12957" CreationDate="2010-12-04T07:27:52.770" Score="1" ViewCount="" Body="&lt;p&gt;Skipping your example, for a &lt;em&gt;European&lt;/em&gt; put option on a future contract you can apply the formula known as &lt;a href=&quot;http://en.wikipedia.org/wiki/Black_model&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;Black -76&lt;/em&gt;&lt;/a&gt;. For &lt;em&gt;American&lt;/em&gt; or &lt;em&gt;Bermudan&lt;/em&gt; options on futures this Black -76 is also adequate because if you exercise early than you would stand with the future which is delivered at expiry anyway - of course the difference is that it might be easier to sell the future..   &lt;/p&gt;&#xA;" OwnerUserId="1154" LastActivityDate="2010-12-04T07:27:52.770" CommentCount="2" />
  <row Id="12977" PostTypeId="2" ParentId="12974" CreationDate="2010-12-04T07:58:04.087" Score="1" ViewCount="" Body="&lt;p&gt;Given pairs $(x_i,y_i)$, you can try to find $a,b$ that minimize the $L^\infty$ distance from $y_i$ to $ax_i+b$ using LP, which is $\max_i |y_i-ax_i-b|$; if $y_i$ indeed depends linearly on $x_i$ up to rounding, then there's a choice of $a,b$ with $L^\infty$ distance of less than $1/2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There may be a simpler way to solve this particular optimization problem, but an LP would work, especially with your small dataset. But if your model is wrong, you might get garbage because of the unforgiving $L^\infty$ metric. Minimizing the $L^1$ metric can also be formulated as an LP, and minimizing the $L^2$ metric is the well-known least-min-squares, for which there is an explicit solution.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-12-04T07:58:04.087" CommentCount="1" />
  <row Id="12980" PostTypeId="2" ParentId="12925" CreationDate="2010-12-04T09:49:44.480" Score="0" ViewCount="" Body="&lt;p&gt;We can pick $g(x)=\frac{1}{x^(\beta-\alpha)}$ and then $\frac{f(x)}{g(x)}\to 1$ as $x\to 0$. By the LCT we get $\int_{0}^{\frac{1}{2}}\frac{\ln^\alpha(1+x)}{x^\beta(1-x)^\alpha}$ exists iif $\beta-\alpha \lt 1$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So overall $\int_{0}^{1}\frac{\ln^\alpha(1+x)}{x^\beta(1-x)^\alpha}$ exists iif $\beta-1 \lt \alpha \lt 1$.&lt;/p&gt;&#xA;" OwnerUserId="3237" LastActivityDate="2010-12-04T09:49:44.480" />
  <row Id="12982" PostTypeId="2" ParentId="11665" CreationDate="2010-12-04T10:38:21.023" Score="4" ViewCount="" Body="&lt;p&gt;Since $a_k \le k-1$ for $k \ge 2$ we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \lfloor x \rfloor = \left\lfloor \sum_{k=1}^n \frac{a_k}{k!} \right\rfloor \le a_1&#xD;&#xA;+ \left\lfloor \sum_{k=2}^n \frac{k-1}{k!} \right\rfloor = a_1 $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;as the latter term is $0$ since $ \sum_{k=2}^n \frac{k-1}{k!} = \sum_{k=2}^n \lbrace \frac{1}{(k-1)!} - \frac{1}{k!} \rbrace&#xD;&#xA;= 1 – 1/n! &amp;lt; 1.$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To show $ a_m = \lfloor m! x \rfloor – m \lfloor (m-1)! x \rfloor $ for $m=2,3,\ldots,n$ we note that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \lfloor m! x \rfloor = \left\lfloor m! \sum_{k=1}^n \frac{a_k}{k!} \right\rfloor = &#xD;&#xA;A_m + \left\lfloor m! \sum_{k=m+1}^n \frac{a_k}{k!} \right\rfloor, $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $A_m$ is an integer given by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$A_m = m! \frac{a_1}{1!} + m! \frac{a_2}{2!} + \cdots + m! \frac{a_m}{m!}.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ m! \sum_{k=m+1}^n \frac{a_k}{k!} \le m! \sum_{k=m+1}^n \frac{k-1}{k!}&#xD;&#xA;= m! \sum_{k=m+1}^n \left\lbrace \frac{1}{(k-1)!} - \frac{1}{k!} \right\rbrace $$&#xD;&#xA;$$ = m! \left( \frac{1}{m!} - \frac{1}{n!} \right) = 1 - \frac{m!}{n!} &amp;lt; 1 $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and hence&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \left\lfloor m! \sum_{k=m+1}^n \frac{a_k}{k!} \right\rfloor = 0.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also for $ m &amp;gt; 1 $&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ m \lfloor (m-1)! x \rfloor = m \left\lfloor (m-1)! \sum_{k=1}^n \frac{a_k}{k!} \right\rfloor $$&#xD;&#xA;$$=m \left\lbrace  (m-1)! \frac{a_1}{1!} + (m-1)! \frac{a_2}{2!} + \cdots + (m-1)! &#xD;&#xA;\frac{a_{m -1}}{(m-1)!} \right\rbrace + m \left\lfloor \sum_{k=m}^n \frac{a_k}{k!} \right\rfloor $$&#xD;&#xA;$$= \left( A_m - a_m \right) + m \left\lfloor \sum_{k=m}^n \frac{a_k}{k!} \right\rfloor&#xD;&#xA;=  A_m – a_m . $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;since&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \sum_{k=m}^n \frac{a_k}{k!}  \le \sum_{k=m}^n \frac{k-1}{k!}&#xD;&#xA;= \frac{1}{(m-1)!} - \frac{1}{n!} &amp;lt; 1 .$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which proves that for $ m &amp;gt; 1 $&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \lfloor m! x \rfloor -  m \lfloor (m-1)! x \rfloor = A_m – (A_m – a_m) = a_m . \qquad (1)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To prove that $n$ is the smallest integer such that $n! x $ is an integer, suppose that&#xA;$ (n-1)! x $  is an integer. Then by $(1)$ we have for $n &amp;gt; 1$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ a_{n-1} = (n-1)! x – n! x &amp;lt; 0 $$&#xD;&#xA;contradicting the fact that $a_k \ge 0 .$ And so $ (n-1)! x $ cannot be an integer, and if $ m! x $ is an integer for any $ m &amp;lt; n $ we have $(n-1)(n-2) \cdots m! x = (n-1)! x $ is an integer, another contradiction. Hence $ m! x $ cannot be an integer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To show that every positive rational number $x$ can be expressed in this form, let $ x = p/q, $ where $ gcd(p,q)=1 \textrm{ and } p,q \in \mathbb{N} $ and let $ n $ be the smallest integer such that $ n! p/q $ is an integer. Define&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \begin{align*}&#xD;&#xA;a_1 &amp;amp;= \left\lfloor \frac{p}{q} \right\rfloor&#xD;&#xA;\\  a_m &amp;amp;= \left\lfloor m! \frac{p}{q} \right\rfloor -  m \left\lfloor (m-1)!  \frac{p}{q} \right\rfloor&#xD;&#xA;\quad \textrm{ for } m &amp;gt; 1. \quad (2)&#xD;&#xA;\end{align*} $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We note that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \left\lfloor (n-1)!  \frac{p}{q} \right\rfloor &amp;lt; (n-1)!  \frac{p}{q} , $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;since $ (n-1)! p/q $ is not an integer, and so $a_n &amp;gt; 0 .$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, since $ (m-1)! p/q $ is not an integer, for $ m \le n $ we can write&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ (m-1)!  \frac{p}{q} = N_m + r_m \quad \textrm{ where } 0 &amp;lt; r_m &amp;lt; 1 $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and $N_m$ is an integer. Hence for $m=2,3,\ldots,n$ from $(2)$ we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ a_m = \lfloor mN_m + mr_m \rfloor - mN_m = \lfloor m r_m \rfloor \le m-1.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note also that the $a_m$ are non-negative. Now assume that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \frac{p}{q} = \sum_{k=1}^n \frac{a_k}{k!} \quad (3)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;then as the $a_k$ satisfy the conditions that each is a non-negative integer with $a_k \le k-1 $ for $k=2,3\ldots,n$ they are uniquely determined by $(1)$ and $a_1 = \lfloor p/q \rfloor .$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It only remains to prove $(3).$ To show this we note that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \begin{align*} &#xD;&#xA;\sum_{k=1}^n \frac{a_k}{k!} &amp;amp;= \left\lfloor \frac{p}{q} \right\rfloor +  \sum_{k=2}^n \frac{a_k}{k!}&#xD;&#xA;\\ &amp;amp;= \left\lfloor \frac{p}{q} \right\rfloor +&#xD;&#xA;\sum_{k=2}^n \left\lbrace&#xD;&#xA;\frac{ \left\lfloor k! \frac{p}{q} \right\rfloor -  k \left\lfloor (k-1)!  \frac{p}{q} \right\rfloor }{k!}&#xD;&#xA;\right\rbrace&#xD;&#xA;\\ &amp;amp;= \left\lfloor \frac{p}{q} \right\rfloor + &#xD;&#xA;\sum_{k=2}^n \frac{ \left\lfloor k! \frac{p}{q} \right\rfloor }{k!} - &#xD;&#xA;\sum_{k=2}^n \frac{ \left\lfloor (k-1)! \frac{p}{q} \right\rfloor }{(k-1)!}&#xD;&#xA;\\ &amp;amp;= \frac {n! p/q}{n!} = \frac{p}{q},&#xD;&#xA;\end{align*} $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;since $n! p/q$ is an integer and all the terms cancel, except the last. This completes the proof.&lt;/p&gt;&#xA;" OwnerUserId="1301" LastEditorUserId="1301" LastEditorDisplayName="" LastEditDate="2010-12-04T15:34:05.293" LastActivityDate="2010-12-04T15:34:05.293" CommentCount="2" />
  <row Id="12983" PostTypeId="1" CreationDate="2010-12-04T11:43:06.430" Score="7" ViewCount="216" Body="&lt;p&gt;I'm studying statistics, and would like to better understand the Central Limit Theorem. The proof I found on Wikipedia requires some previous knowledge I do not currently possess.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a quick intuitive explanation you can give as to why this theorem is correct?&lt;/p&gt;&#xA;" OwnerUserId="4401" LastActivityDate="2010-12-04T20:21:32.580" Title="Intuition about the Central Limit Theorem" Tags="&lt;statistics&gt;" AnswerCount="3" CommentCount="4" FavoriteCount="2" />
  <row Id="12984" PostTypeId="2" ParentId="8004" CreationDate="2010-12-04T12:09:59.680" Score="3" ViewCount="" Body="&lt;p&gt;Your proof of (2) is incorrect. The quantifiers are wrong. &#xA;You have shown that for any $N$ (neighbourhood of $x$, in the remainder): $N$ intersects $E$ or $N$ intersects $F$.&#xA;But you have to show (for any $N$: $N$ intersects $E$) or (for any $N$, $N$ intersects $F$), i.e. that you intersect the &lt;strong&gt;same&lt;/strong&gt; set for any $N$. &#xA;So you have to argue by contradiction: suppose $x \in \overline{E \cup F}$, but $x \notin \overline{E}$ &lt;strong&gt;and&lt;/strong&gt; $x \notin \overline{F}$. The latter 2 imply that there is a neighbourhood $N_1$ of $x$ that misses $E$ and a neighbourhood $N_2$ of $x$ that misses $F$. But then $N_1 \cap N_2$ is a neighbourhood of $x$ as well and this misses $E \cup F$, contradicting that $x \in \overline{E \cup F}$. Done.&lt;/p&gt;&#xA;" OwnerUserId="4280" LastActivityDate="2010-12-04T12:09:59.680" CommentCount="1" />
  <row Id="12985" PostTypeId="2" ParentId="12983" CreationDate="2010-12-04T13:03:27.457" Score="6" ViewCount="" Body="&lt;p&gt;I don't think you should expect any short, snappy answers because I think this is a very deep question.  Here is a guess at a conceptual explanation, which I can't quite flesh out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Our starting point is something called the &lt;a href=&quot;http://en.wikipedia.org/wiki/Principle_of_maximum_entropy&quot; rel=&quot;nofollow&quot;&gt;principle of maximum entropy&lt;/a&gt;, which says that in any situation where you're trying to assign a probability distribution to some events, you should choose the distribution with maximum entropy which is consistent with your knowledge.  For example, if you don't know anything and there are $n$ events, then the maximum entropy distribution is the uniform one where each event occurs with probability $\frac{1}{n}$.  There are lots more examples in &lt;a href=&quot;http://www.math.uconn.edu/~kconrad/blurbs/analysis/entropypost.pdf&quot; rel=&quot;nofollow&quot;&gt;this expository paper&lt;/a&gt; by Keith Conrad.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now take a bunch of independent identically distributed random variables $X_i$ with mean $\mu$ and variance $\sigma^2$.  You know exactly what the mean of $\frac{X_1 + ... + X_n}{n}$ is; it's $\mu$ by linearity of expectation.  Variance is also linear, at least on independent variables (this is a probabilistic form of the Pythagorean theorem), hence&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\text{Var}(X_1 + ... + X_n) = \text{Var}(X_1) + ... + \text{Var}(X_n) = n \sigma^2$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but since variance scales quadratically, the variance of $\frac{X_1 + ... + X_n}{n}$ is actually $\frac{\sigma^2}{n}$; in other words, it goes to zero!  This is a simple way to convince yourself of the (weak) &lt;a href=&quot;http://en.wikipedia.org/wiki/Law_of_large_numbers&quot; rel=&quot;nofollow&quot;&gt;law of large numbers&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we can convince ourselves that (under the assumptions of finite mean and variance) the average of a bunch of iid random variables tends to its mean.  If we want to study &lt;em&gt;how&lt;/em&gt; it tends to its mean, we need to instead consider $\frac{(X_1 - \mu) + ... + (X_n - \mu)}{\sqrt{n}}$, which has mean $0$ and variance $\sigma^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose we suspected, for one reason or another, that this tended to some fixed limiting distribution in terms of $\sigma^2$ alone.  We might be led to this conclusion by seeing this behavior for several particular distributions, for example.  Given that, it follows that we don't know anything about this limiting distribution except its mean and variance.  So we should choose the distribution of maximum entropy with a fixed mean and variance.  And this is &lt;strong&gt;precisely&lt;/strong&gt; the corresponding normal distribution!  Intuitively, each iid random variable is like a particle moving randomly, and adding up the contributions of all of the random particles adds &quot;heat,&quot; or &quot;entropy,&quot; to your system.  (I think this is why the normal distribution shows up in the description of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Heat_kernel&quot; rel=&quot;nofollow&quot;&gt;heat kernel&lt;/a&gt;, but don't quote me on this.)  In information-theoretic terms, the more iid random variables you sum, the less information you have about the result.  &lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-12-04T14:20:18.843" LastActivityDate="2010-12-04T14:20:18.843" />
  <row Id="12986" PostTypeId="1" AcceptedAnswerId="12987" CreationDate="2010-12-04T14:11:34.690" Score="7" ViewCount="283" Body="&lt;p&gt;Find examples of two series $\sum a_n$ and $\sum b_n$ both of which diverge but for which $\sum \min(a_n, b_n)$ converges. To make it more challenging, produce examples where $a_n$ and $b_n$ are positive and decreasing.&lt;/p&gt;&#xA;" OwnerUserId="4403" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-04T14:19:51.253" LastActivityDate="2010-12-04T14:43:48.013" Title="find examples of two series $\sum a_n$ and $\sum b_n$ both of which diverge but for which $\sum \min(a_n, b_n)$ converges" Tags="&lt;analysis&gt;&lt;sequences-and-series&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" />
  <row Id="12987" PostTypeId="2" ParentId="12986" CreationDate="2010-12-04T14:29:45.943" Score="13" ViewCount="" Body="&lt;p&gt;Here is a hint for the first part: you can make min$(a_n,b_n)=0$ for all $n$ (it doesn't get more convergent than that), while the two series $\sum a_n$ and $\sum b_n$ are very far from converging. Once you solve this one, let me know and I will give you a hint for the challenging part. It will build on ideas from the first one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit:&#xA;Nice solutions to the first part! Now for the second: of course the idea will have to be the same: the sequences will have to somehow alternate in which one is lower at any given point. But think about it: if they alternate at each step, like before, then we will have that $\sum_n a_{2n}$ converges and also $a_{2n+1} &amp;lt; a_{2n}$, so we will have that $\sum_n a_n &amp;lt; 2\sum_n a_{2n}$ will converge. That's not good. If you think about this issue for a while, you will realise that the intervals at which one sequence dives below the other one must get longer and longer for the two to diverge. Now, can you make your previous idea work?&lt;/p&gt;&#xA;" OwnerUserId="3212" LastEditorUserId="3212" LastEditorDisplayName="" LastEditDate="2010-12-04T14:43:48.013" LastActivityDate="2010-12-04T14:43:48.013" CommentCount="4" />
  <row Id="12988" PostTypeId="2" ParentId="12944" CreationDate="2010-12-04T14:34:34.800" Score="8" ViewCount="" Body="&lt;p&gt;A distribution is actually a linear functional on the space of compactly supported infinitely differentiable functions (the so called &quot;test functions&quot;). A function $f$ is compactly supported if $\overline{\{x : f(x) \neq 0\}}$ is compact (the overline denotes the closure).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The $\delta$-distribution is a linear functional such that for all $\phi \in C_c^\infty(\mathbb{R}^n)$ we have that $\langle \delta, \phi \rangle = \phi(0)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you want to compute the product of distributions the problem is that you don't have a property which you would really like to have, that is associativity. So for distributions $\alpha$, $\beta$ and $\gamma$ we usually have that $(\alpha \cdot \beta) \cdot \gamma \neq \alpha \cdot (\beta \cdot \gamma)$. &lt;a href=&quot;http://en.wikipedia.org/wiki/Distribution_%28mathematics%29%23Problem_of_multiplication&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt; gives an example. However this does not really turn out to be a problem in applications. What we do have is convolution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When we want to do convolution we prefer a smaller class of distributions (for example because on the smaller class the Fourier transform of a distribution in this class is again a distribution in this class). This actually has a rougher class of test-functions, as test functions here we take the Schwartz functions, that are the smooth functions of which the function itself and all its derivatives are rapidly decreasing. $f$ is said to be rapidly decreasing if there are constants $M_n$ such that $|f(x)| \leq M_N |x|^{-N}$ as $x \to \infty$ for $N = 1,2,3,\ldots$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To begin defining the convolution we first define what the convolution of Schwartz-function is with a tempered distribution. Let $f$ be our tempered distribution, then we can show that the following definition actually makes sense:&#xA;$$\langle \phi * f, \psi \rangle := \langle f, \tilde{\phi} * \psi \rangle$$&#xD;&#xA;where $\tilde{\phi}(x) = \phi(-x)$. Note that the RHS is well-defined. Convolution is a nice thing, we can see that if we start with a tempered distribution and convolute it with a test function, the result will be smooth.&#xA;Now, $L_1 * L_2$ is the unique distribution $L$ with the property that $L * \phi = L_1 * (L_2 * \phi)$. We can show that this is commutative.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fine, now note that $\delta * \phi(x) = \phi(x - y)|_{y = 0} = \phi(x)$. So we see that $\delta * \delta = \delta$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want me to comment on the dot product of distributions, you first would have to explain what you mean with that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So far for this short digression on distributions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;: Okay, you want to compute $\delta^2$. Let $\phi_n$ be an approximation to the identity and let it converge to $\delta$ in the sense of distributions, but $\phi_n^2$ does not converge at all since the integral against a test function that does not vanish at the origin blows up as $n \to \infty$.&lt;/p&gt;&#xA;" OwnerUserId="1120" LastEditorUserId="1120" LastEditorDisplayName="" LastEditDate="2010-12-04T15:53:08.383" LastActivityDate="2010-12-04T15:53:08.383" />
  <row Id="12989" PostTypeId="1" AcceptedAnswerId="12990" CreationDate="2010-12-04T14:48:57.280" Score="7" ViewCount="193" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am wondering how much a smooth function may be non-analytic, because in proofs, whilst there non-analytic smooth functions, it would suffice if a smooth function were analytic on only a &quot;small set&quot;. More exactly:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $U \in \mathbb R^n$ be open, $C^\infty = C^\infty(U,\mathbb R)$. A smooth function in $C^\infty$ is analytic in $a \in U$, iff there exists $\epsilon &amp;gt; 0$, s.t. the function is equal to its own Taylor series in $B_\epsilon(a)$. There exist smooth functions that are non-analytic, i.e. there exists $f \in C^\infty, b \in U, \epsilon &amp;gt; 0$ s.t. the function is not its taylor series at $x$ in $B_\epsilon (b)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $A$ be the union of all $\epsilon$-Balls in $U$ where $f$ is analytic. By definition, $A$ is open. It's complement $C = A^c$is the closed set of points where $f$ is non-analytic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does $C$ have an interior?&lt;/p&gt;&#xA;" OwnerUserId="3557" LastActivityDate="2011-03-17T14:59:19.950" Title="A smooth function's domain of being non-analytic" Tags="&lt;real-analysis&gt;&lt;analysis&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" />
  <row Id="12990" PostTypeId="2" ParentId="12989" CreationDate="2010-12-04T15:09:18.907" Score="14" ViewCount="" Body="&lt;p&gt;Yes, that can happen. The canonical example (as far as I know) is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Fabius_function&quot; rel=&quot;nofollow&quot;&gt;Fabius function&lt;/a&gt; which is smooth and &lt;em&gt;nowhere&lt;/em&gt; analytic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;hr&gt;&#xA;I never really liked this answer because it's hard — if not outright impossible — to find references online on how to prove the nonanalyticity of $Fb$. So here is a short exposition on a different example adapted from the discussion archived in &lt;a href=&quot;http://www.math.niu.edu/~rusin/known-math/99/nowhere_analy&quot; rel=&quot;nofollow&quot;&gt;this text file&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h1&gt;Introduction&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;Students often see&#xA;$$ f(x) = \begin{cases}&#xD;&#xA;    \exp(-\tfrac{1}{x}) &amp;amp; \text{for } x &amp;gt; 0 \\\\&#xD;&#xA;    0                  &amp;amp; \text{for } x \leq 0&#xD;&#xA;\end{cases}$$&#xA;as an example of a smooth function that's not analytic at $0$, but this as well as the other usual examples makes it very easy to intuit that smooth functions are “mostly analytic”, i.e. everywhere analytic except possibly at some isolated points. And given that this is the only example that one typically sees this is in fact a reasonable thing to believe. But nonetheless there are plenty of examples out there — for example the following: &lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Example&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;Define $F: \mathbb{R} \to \mathbb{C}$ by&#xA;$$ F(x) := \sum_{n=0}^\infty \frac{\exp(i2^nx)}{n!}. $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Theorem:&lt;/strong&gt; $F$ and $\Re F$, the real part of $F$, are smooth nowhere analytic functions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Proof:&lt;/em&gt; Computing the derivatives of $F$ we get&#xA;$$ F^{(k)}(x) = \sum_{n=0}^\infty \frac{\exp(i2^nx)(i2^n)^k}{n!}, $$&#xA;which is uniformly convergent everywhere, so it's continuous and there are no issues with moving the differentiation in under the summation. In other words $F$ and $\Re F$ are smooth. Furthermore we have that &#xA;$$ F^{(k)}(0) = \sum_{n=0}^\infty \frac{(i2^n)^k}{n!} = i^k \exp(2^k). $$&#xA;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Cauchy-Hadamard_theorem&quot; rel=&quot;nofollow&quot;&gt;Cauchy-Hadamard formula&lt;/a&gt; for the radius of convergence, $r$, gives at $x=0$ with the usual conventions about infinities that &#xA;$$ \frac{1}{r} = \limsup_{k \to \infty} \left\lvert F^{(k)}(0)\right\rvert^{1/k} = \infty, $$&#xA;and so $r = 0$. The same will be true for $\Re f$ as the even-indexed parts of $\Re F$ have the same magnitude as those of $F$. $F$ is $2\pi$-periodic, so the same is true for every $x$ which is an integer multiple of $2\pi$. Moreover we see that if we throw away the first $p$ terms, we get a function with period $\omega = \frac{2\pi}{2^p}$ and with points of non-analyticity at all integer multiples of $\omega$. Hence $F$ and $\Re F$ must also be nonanalytic at these points and we conclude that the set of points where the two functions are nonanalytic is dense in $\mathbb{R}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally observe that if a function is analytic in a point it must be analytic on a neighbourhood of that point, thus there are no possible open sets where $F$ and $\Re F$ can be analytic, i.e. they are smooth nowhere analytic functions. $\;\square$&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Good Questions to Ponder&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;Suppose $\Omega \subseteq \mathbb{R}^n$ is open.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there an $f \in C^\infty(\mathbb{R}^n)$ such that $\Omega$ is its locus of analyticity?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are the functions that are analytic at some point of the first category in $C^\infty(\Omega)$?&lt;/p&gt;&#xA;" OwnerUserId="1269" LastEditorUserId="1269" LastEditorDisplayName="" LastEditDate="2011-03-17T14:59:19.950" LastActivityDate="2011-03-17T14:59:19.950" CommentCount="2" />
  <row Id="12991" PostTypeId="1" CreationDate="2010-12-04T15:28:27.717" Score="1" ViewCount="183" Body="&lt;p&gt;Is Robinson Arithmetic complete in the sense of Gödels completeness theorem? And is Robinson Arithmetic incomplete in the sense of Gödels first incompleteness theorem?&#xA;If RA is both it would be a good example to explain the different notions of &quot;completeness&quot; used in the two theorems mentioned. &lt;/p&gt;&#xA;" LastActivityDate="2010-12-05T11:33:39.327" Title="Is Robinson Arithmetic complete and not-complete?" Tags="&lt;logic&gt;&lt;incompleteness&gt;" AnswerCount="1" />
  <row Id="12992" PostTypeId="1" AcceptedAnswerId="12993" CreationDate="2010-12-04T15:52:01.147" Score="0" ViewCount="173" Body="&lt;p&gt;If I have 16 different symbols, howmany of these symbols can I combine to create no more than 16.777.216 unique combiations?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Eg. I have all symbols of the Hexadecimal language (16).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Howmany combinations can I create with X amount of symbols to not have more than 16.777.216 unique combinations maximum?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The position of each symbol is of importance.&lt;/p&gt;&#xA;" OwnerUserId="1771" LastEditorUserId="1543" LastEditorDisplayName="" LastEditDate="2010-12-04T15:54:18.193" LastActivityDate="2010-12-04T15:55:20.860" Title="If I have 16 different symbols, howmany of these symbols can I combine to create no more than 16.777.216 unique combiations?" Tags="&lt;combinatorics&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="12993" PostTypeId="2" ParentId="12992" CreationDate="2010-12-04T15:55:20.860" Score="2" ViewCount="" Body="&lt;p&gt;Six, because $16^6 = 16.777.216$.&lt;/p&gt;&#xA;" OwnerUserId="506" LastActivityDate="2010-12-04T15:55:20.860" CommentCount="6" />
  <row Id="12994" PostTypeId="2" ParentId="12991" CreationDate="2010-12-04T15:58:53.587" Score="4" ViewCount="" Body="&lt;p&gt;Gödel's completeness theorem isn't about theories but about the logic used in the language of a theory, namely about the connection between the syntax and semantics of that logic. The logic used to describe Robinson's arithmetic (first order logic) is complete and at the same time RA is not complete as a theory. The same applies to first order Peano Arithmetic. The logic used in the language of PA is complete but PA is not a complete theory. Same thing applies to ZFC. Generally, every incomplete theory described in first order logic has that property.&lt;/p&gt;&#xA;" OwnerUserId="1772" LastEditorUserId="1772" LastEditorDisplayName="" LastEditDate="2010-12-05T11:33:39.327" LastActivityDate="2010-12-05T11:33:39.327" CommentCount="8" />
  <row Id="12996" PostTypeId="1" CreationDate="2010-12-04T16:33:14.373" Score="3" ViewCount="120" Body="&lt;p&gt;Let $R$ a root system and $\Delta$ be a simple system of roots of a Lie algebra $\mathfrak g$, $\Delta&amp;#39;\subset \Delta$ and $R(\Delta&amp;#39;)=R\cap \mathbb Z(\Delta&amp;#39;)$. Define &#xA;$$p(\Delta&amp;#39;)=\mathfrak h \bigoplus_{\alpha \in R(\Delta&amp;#39;)} \mathfrak g_{\alpha}  \bigoplus_{\alpha \in R^+ \setminus R^+(\Delta&amp;#39;)}\mathfrak g_{\alpha}$$ the parabolic subalgebra associated to $\Delta&amp;#39;$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $\alpha$ is a simple root in $R^+(\Delta)\setminus R^+ (\Delta&amp;#39;)$, then $\beta(h_\alpha)=0$ for all $\beta$ in $R(\Delta&amp;#39;)$???&lt;/p&gt;&#xA;" OwnerUserId="2764" LastEditorUserId="2764" LastEditorDisplayName="" LastEditDate="2010-12-04T19:42:11.300" LastActivityDate="2010-12-28T19:35:11.343" Title="Parabolic subalgebra" Tags="&lt;representation-theory&gt;" AnswerCount="1" />
  <row Id="12997" PostTypeId="2" ParentId="12921" CreationDate="2010-12-04T17:05:22.223" Score="0" ViewCount="" Body="&lt;p&gt;The best answer has been given &lt;a href=&quot;http://cstheory.stackexchange.com/questions/3503/formal-languages-what-does-r-trivial-mean/3515#3515&quot;&gt;here&lt;/a&gt; by &lt;a href=&quot;http://cstheory.stackexchange.com/users/508/michael-blondin&quot;&gt;Michael Blondin&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A semigroup $S$ is $R\text{-trivial}$ iff $a \: R \: b \Rightarrow a = b$ for all $a, b \in S$ where $R$ is &lt;a href=&quot;http://en.wikipedia.org/wiki/Green%27s_relations#The_L.2C_R.2C_and_J_relations&quot; rel=&quot;nofollow&quot;&gt;Green's relation&lt;/a&gt; $a \: R \: b \Leftrightarrow aS^1 = bS^1$. The set of $R\text{-trivial}$ monoids forms a variety which can be ultimately defined by the equations $(xy)^n x = (xy)^n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A language is $R\text{-trivial}$ if its &lt;a href=&quot;http://en.wikipedia.org/wiki/Syntactic_monoid&quot; rel=&quot;nofollow&quot;&gt;syntactic monoid&lt;/a&gt; is $R\text{-trivial}$. This variety of languages is alternatively defined as the set of all languages that can be written as a disjoint union  of languages of the form $A_0^* a_1 A_1^* a_2 \ldots a_n A_n^*$ where $n \geq 0$, $a_1, \ldots, a_n \in A$, $A_i \subseteq A \setminus \{a_{i+1}\}$ for $0 \leq i \leq n-1$. Another definition given in &lt;a href=&quot;http://www.liafa.jussieu.fr/~jep/Resumes/Varietes.html&quot; rel=&quot;nofollow&quot;&gt;[Pin]&lt;/a&gt;, which I'm not familiar with, uses the so-called &lt;em&gt;automates extensifs&lt;/em&gt; (&quot;extensive automata&quot;). You can find more results about those languages in &lt;a href=&quot;http://www.liafa.jussieu.fr/~jep/Resumes/Varietes.html&quot; rel=&quot;nofollow&quot;&gt;[Pin]&lt;/a&gt;. There is an English version of this book, I've never read it but I'm pretty sure that you can find the same content.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the sake of completeness, here is an example of a $R\text{-trivial}$ language: $\{b\}^* a \{a,c\}^* b \{a\}^* b \{a,b,c\}^* \cup \{d\}^* \cup abcd$. You can build other examples with the previous definitions. Note that all of the properties of varieties of languages hold for $R\text{-trivial}$ languages, therefore they are closed under union, intersection and complementation. It should help to build more complicated languages.&lt;/p&gt;&#xA;" OwnerUserId="4371" LastActivityDate="2010-12-04T17:05:22.223" CommentCount="2" />
  <row Id="12998" PostTypeId="2" ParentId="12968" CreationDate="2010-12-04T17:16:06.933" Score="1" ViewCount="" Body="&lt;p&gt;Things come full circle:  Constructive proofs of the Hilbert Basis theorem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.andrew.cmu.edu/user/avigad/Students/hertz.pdf&quot; rel=&quot;nofollow&quot;&gt;Click with all of your justice&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="126" LastActivityDate="2010-12-04T17:16:06.933" CommentCount="3" />
  <row Id="12999" PostTypeId="1" AcceptedAnswerId="13002" CreationDate="2010-12-04T17:23:18.677" Score="2" ViewCount="179" Body="&lt;p&gt;When you multiply two power series' that have difference intervals of convergence, what is the product's interval of convergence?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ex.  f(x) converges for |x|&amp;lt;6, g(x) converges for |x|&amp;lt;4.  What does f(x)g(x) converge for?&lt;/p&gt;&#xA;" OwnerUserId="3691" LastActivityDate="2010-12-04T18:08:14.847" Title="Multiplication of Power Series and their convergences" Tags="&lt;sequences-and-series&gt;&lt;convergence&gt;" AnswerCount="2" />
  <row Id="13000" PostTypeId="1" CreationDate="2010-12-04T17:25:53.403" Score="3" ViewCount="65" Body="&lt;p&gt;The 4-associahedron is the 4-dimensional version of Mac Lane's pentagon diagram.  If you look at &lt;a href=&quot;http://math.ucr.edu/home/baez/trimble/tetracategories.html&quot; rel=&quot;nofollow&quot;&gt;Trimble's notes on tetracategories&lt;/a&gt;, you can see the obvious difficulty in working with such a diagram (using it to verify things like coherence etc).  Are there any techniques and/or approaches to working with  such unwieldy higher-dimensional diagrams?  &lt;/p&gt;&#xA;" OwnerUserId="126" LastActivityDate="2010-12-04T17:25:53.403" Title="Practical approaches to working with nonplanar commutative diagrams?" Tags="&lt;category-theory&gt;" />
  <row Id="13001" PostTypeId="2" ParentId="12999" CreationDate="2010-12-04T17:30:59.047" Score="0" ViewCount="" Body="&lt;p&gt;The power series for a function $f:\mathbb{C}\to\mathbb{C}$ converges for all $z$ within the largest open disk such that $f$ is analytic within the disk (i.e. no poles, discontinuities, etc.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example:&#xA;$f(z)=\frac{1}{1+z^2}$. The function is analytic within $|z| &amp;lt; 1$, but has poles at $\pm i$. Its power series converges within the same disk. Multiplying with for example $g(z)=e^z$ (everywhere analytic), yields $h(z)=f(z)g(z)$ with radius of convergence $1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strike&gt;(actually, if $R,S$ is the radius of convergence for $f,g$ respectively, then the radius of convergence for $fg$ is $\min(R,S)$.)&lt;/strike&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: I lived in a complex world, see Pete L. Clark's comment below.&lt;/p&gt;&#xA;" OwnerUserId="4284" LastEditorUserId="4284" LastEditorDisplayName="" LastEditDate="2010-12-04T18:08:14.847" LastActivityDate="2010-12-04T18:08:14.847" CommentCount="4" />
  <row Id="13002" PostTypeId="2" ParentId="12999" CreationDate="2010-12-04T17:31:33.167" Score="5" ViewCount="" Body="&lt;p&gt;Assuming both power series have the same center, the radius of convergence of the product is at least the minimum of the radii of the convergence of the two individual series. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;See Proposition 2 in &lt;a href=&quot;http://math.stackexchange.com/users/299/pete-l-clark&quot;&gt;Pete Clark'&lt;/a&gt;s notes here&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://math.uga.edu/~pete/243series7.pdf&quot; rel=&quot;nofollow&quot;&gt;http://math.uga.edu/~pete/243series7.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3431" LastActivityDate="2010-12-04T17:31:33.167" CommentCount="1" />
  <row Id="13003" PostTypeId="1" AcceptedAnswerId="13089" CreationDate="2010-12-04T17:48:13.253" Score="4" ViewCount="227" Body="&lt;p&gt;Consider a parallelogram which is neither a rhombus nor a rectangle. It is well-known that such shapes do not have an axis of symmetry.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Is there a simple proof for this?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I prefer a proof that I can give to a kid (~12 year old), but more involved proofs will do as well.&lt;/p&gt;&#xA;" OwnerUserId="2420" LastActivityDate="2010-12-05T09:55:37.517" Title="Parallelograms &amp; Axis of Symmetry" Tags="&lt;geometry&gt;&lt;symmetric-groups&gt;" AnswerCount="4" CommentCount="2" />
  <row Id="13004" PostTypeId="1" AcceptedAnswerId="13007" CreationDate="2010-12-04T17:50:41.633" Score="3" ViewCount="97" Body="&lt;p&gt;Are there any advanced topics--preferably at the research frontier--which can nevertheless be explained accurately, if not efficiently, using very down-to-earth ideas which would be accessible to most college students, not necessarily even math majors? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a prototypical example of what I mean, I offer the book &lt;em&gt;QED&lt;/em&gt; by Feynman, which explains the ideas of quantum electrodynamics without even using the word &quot;complex number.&quot; In mathematics, the book &lt;em&gt;Naive Lie Theory&lt;/em&gt; by John Stillwell comes close to what I'm looking for. &lt;em&gt;Visual Complex Analysis&lt;/em&gt; by Tristan Needham is a pseudo-example, the main issue being that its subject matter is very standard undergraduate material.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, any such explanation cannot be a fully faithful representation of the original work, but ideally it would still be possible to convey the essence of the methods without introducing any real inaccuracies.&lt;/p&gt;&#xA;" OwnerUserId="251" LastActivityDate="2010-12-04T18:19:05.367" Title="Examples of advanced results and ideas explained in a down-to-earth way" Tags="&lt;intuition&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="13005" PostTypeId="2" ParentId="12959" CreationDate="2010-12-04T17:52:30.063" Score="3" ViewCount="" Body="&lt;p&gt;For any fixed $x_0$, there is positive probability that the sequence starting from $x_0$ fails to converge.   To see why, take, say, $x_0 = 15$.  There is positive probability that $50 &amp;lt; B_t &amp;lt; 60$ for $10 &amp;lt; t &amp;lt; 15$ and $10 &amp;lt; B_t &amp;lt; 20$ for $50 &amp;lt; t &amp;lt; 60$.  On this event the sequence oscillates between the intervals $(10,20)$ and $(50,60)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, almost surely there exist infinitely many $t$ in any interval $(0,\epsilon)$ with $B_t = t$, so even when the sequence does converge the limit need not be 0.  For a proof, note that $P(B_t &amp;gt; t) = P(N &amp;gt; \sqrt{t}) \ge 1/4$ for sufficiently small $t$ (here $N$ is a standard normal random variable).  So for any sequence $t_n$ decreasing to $0$, we have $P(B_{t_n} &amp;gt; t_n \text{ i.o.}) \ge 1/4$; by the Blumenthal 0-1 law, $P(B_{t_n} &amp;gt; t_n \text{ i.o.}) =1$.  However, we also have $P(B_{t_n} &amp;lt; t_n) \ge P(B_{t_n} &amp;lt; 0) = 1/2$ so by a similar argument $P(B_{t_n} &amp;lt; t_n \text{ i.o.}) =1$.  The result follows by continuity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One could ask some other questions:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;For a fixed $x_0$, what is the probability that the sequence starting from $x_0$ converges?  My guess is $0$ but I don't see a proof offhand.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Consider the (random) set $C$ of $x$ such that the sequence starting from $x$ converges.  What is the Lebesgue measure of $C$?  My guess is that $m(C) = 0$ a.s. but again no proof.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; Another interesting fact is that almost surely, for every starting point $x_0$, the sequence $x_k$ is bounded, and hence has a convergent subsequence.  Let $M_r = \sup_{t \in [-r,r]} |B_t|$.  By the strong law of large numbers, $B_t/t \to 0$ a.s. as $t \to \pm \infty$, and it follows that $M_r / r \to 0$ a.s. as $r \to \infty$. In particular, a.s. there exists $r &amp;gt; x_0$ with $M_r &amp;lt; r$, and then $|x_k| \le M_r$ for all $k \ge 1$.&lt;/p&gt;&#xA;" OwnerUserId="822" LastEditorUserId="822" LastEditorDisplayName="" LastEditDate="2010-12-04T18:17:45.897" LastActivityDate="2010-12-04T18:17:45.897" CommentCount="1" />
  <row Id="13007" PostTypeId="2" ParentId="13004" CreationDate="2010-12-04T18:19:05.367" Score="5" ViewCount="" Body="&lt;p&gt;I can think of a couple of books that may be close to what you claim to be looking for. One is &lt;a href=&quot;http://books.google.co.cr/books?id=h7oUr88S7JQC&amp;amp;printsec=frontcover&amp;amp;dq=fearless+symmetry&amp;amp;hl=es&amp;amp;ei=_4H6TLKHFoL88AaU7sWCCw&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=1&amp;amp;ved=0CCUQ6AEwAA#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;Fearless Symmetry&lt;/a&gt; by Avner Ash and Robert Gross. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;They try to &quot;explain&quot; (without entering in the details of course) the general ideas behind Wiles' proof of Fermat's Last Theorem. So they begin in chapter 1 with an intuitive explanation of the purpose and aim of the idea of a representation and follow their way through elliptic curves, reciprocity laws and even Galois Representations. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is certainly a book that &quot;explains&quot; very advanced stuff, but that is aimed at a non-specialist audience. It is supposed to be a popular math book, but I'm not sure how accurate that is since it is not that easy to digest if you don't have at least some undergraduate mathematics under your belt. At least that's how I see it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other book I have in mind is &lt;a href=&quot;http://books.google.co.cr/books?id=T_o0CnMZecMC&amp;amp;printsec=frontcover&amp;amp;dq=visual+group+theory&amp;amp;hl=es&amp;amp;ei=LIH6TM_bGYL58AaUy7zJCw&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=1&amp;amp;ved=0CCUQ6AEwAA#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;Visual Group Theory&lt;/a&gt;, but this is closer to the spirit of Visual Complex Analysis, I suppose, and as in that case it explains standard undergraduate mathematics.&lt;/p&gt;&#xA;" OwnerUserId="900" LastActivityDate="2010-12-04T18:19:05.367" CommentCount="1" />
  <row Id="13008" PostTypeId="2" ParentId="12968" CreationDate="2010-12-04T18:26:21.143" Score="7" ViewCount="" Body="&lt;p&gt;The issue is complicated by the fact that many (most?) &quot;mainstream mathematicians&quot; are not really keeping track of whether they are using some form of AC in their arguments.  Moreover, they often do not write their proofs in such a way which makes clear either whether they are using AC or whether this use of AC is really necessary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is an example from my own experience. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Theorem (Luther Claborn): For every abelian group $A$, there exists a Dedekind domain $R$ such that the ideal class group of $R$ is isomorphic to $A$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know of three proofs of this theorem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Claborn's original 1966 proof:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://math.uga.edu/~pete/claborn66.pdf&quot; rel=&quot;nofollow&quot;&gt;http://math.uga.edu/~pete/claborn66.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) C. Leedham-Green's 1972 proof: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://math.uga.edu/~pete/leedhamgreen.pdf&quot; rel=&quot;nofollow&quot;&gt;http://math.uga.edu/~pete/leedhamgreen.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) My 2008 proof:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://math.uga.edu/~pete/ellipticded.pdf&quot; rel=&quot;nofollow&quot;&gt;http://math.uga.edu/~pete/ellipticded.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of these, I am sorry to say that the only one I am truly familiar with is my own.  My proof uses transfinite induction.  To be quite honest, I was pretty tickled by this: it was maybe the fourth time in my life that I had written down a transfinite induction argument and the first time that it was for any serious purpose.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, I soon heard from the eminent mathematician Bjorn Poonen (then at Berkeley) that he had been thinking about this very result because a student of Eisenbud's had recently given a seminar talk on Leedham-Green's proof (which I would describe as &quot;clever and complicated&quot;).  The proof I had given was much more up his own alley, and he immediately found a way to recast / improve it.  Among other things, his recasting completely avoided use of the Axiom of Choice.  (But I hasten to add that there was more to it than that.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We speculated briefly on the possibility that this modified argument was the &lt;em&gt;first&lt;/em&gt; proof of Claborn's theorem to avoid AC.  I asked whether Leedham-Green's proof used AC and he told me that indeed it was also a transfinite induction argument.  So then I looked back at Claborn's paper and asked my student about it.  We were able to point to some part of Claborn's argument (which, if you look at the paper, you'll see is rather terse: I wish the paper were more than four pages long!) which we thought used AC.  But I don't remember what that was, and I am only moderately confident that we were right: if you told me otherwise, I wouldn't be too surprised.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And now I am afraid to have to tell you that I didn't include in my paper any hint that Poonen's modification would give an AC-less proof (and possibly the first such proof!) &#xA;of Claborn's theorem.  I felt that Poonen's argument was nice enough so that it deserved further exploration in its own right: I wanted to use it -- and still do -- to give a complete characterization of all abelian groups which can be the group of rational points on some elliptic curve over some field.  So in essence I have been keeping to myself the fact that this theorem of Claborn (which lots of people know about) can be proved independently of AC.  Sorry!&lt;/p&gt;&#xA;" OwnerUserId="299" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2010-12-04T18:40:50.923" LastActivityDate="2010-12-04T18:40:50.923" />
  <row Id="13009" PostTypeId="1" AcceptedAnswerId="13011" CreationDate="2010-12-04T18:28:19.487" Score="3" ViewCount="92" Body="&lt;p&gt;What order is the field: $\mathbb{F}_p(\zeta_{p-1}^{1/n})$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\mathbb{F}_p^{\times}$ is size $p-1$ and cyclic (with a generator we shall call $\zeta_{p-1}$). Naively, it seems that by taking the $n^{th}$ root of $\zeta_{p-1}$ we get a group of units of size $C_{n(p-1)}$. This, of course, doesn't make sense. What should be true is that $C_{n(p-1)}$ is some subgroup of the group of units of the $\mathbb{F}_{p^r}$ we get.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Question&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;What is the order of the field $\mathbb{F}_p[x]/(m_{\zeta_{p-1}^{1/n}})$ where $m_{\zeta_{p-1}^{1/n}}$ is the minimal polynomial of $\zeta_{p-1}^{1/n}$ (which, unless I'm mistaken, is the same as the splitting field of $x^n-\zeta_{p-1}$)? Does it depend on whether $p|n$ or not?&lt;/p&gt;&#xA;" OwnerUserId="4410" LastEditorUserId="4410" LastEditorDisplayName="" LastEditDate="2010-12-04T18:41:16.477" LastActivityDate="2010-12-04T18:45:57.403" Title="What field is $\mathbb{F}_p(\zeta_{p-1}^{1/n})$" Tags="&lt;finite-fields&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="13010" PostTypeId="1" AcceptedAnswerId="13012" CreationDate="2010-12-04T18:33:24.827" Score="3" ViewCount="220" Body="&lt;p&gt;So I read the answers in &lt;a href=&quot;http://math.stackexchange.com/questions/713/how-to-visualize-a-rank-2-tensor&quot;&gt;this&lt;/a&gt; question, and don't feel that much closer to an answer about how tensors as multi-linear maps and tensors as &quot;multi-dimensional&quot; matrices are truly related.  For instance, it seems that a (1,1)-tensor should be able to be realized as a 2 by 2 matrix, as something that looks like $(x_1,x_2)\otimes(y_1,y_2)$ and as a multi-linear map from $\mathbb{R}^2\times(\mathbb{R}^2)^\ast$ to $\mathbb{R}$.  Is there a formal isomorphism between these three ideas?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My motivation for asking this is that I'm trying to work with deRham cohomology without just doing the mechanics of it.  I'd like to actually know what's going on, and so I'm trying to see &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) How is $\mathrm{Alt}^p(\mathbb{R}^n)$ actually the same as $\wedge^p(\mathbb{R}^n)$ where the former is the space of alternating p-linear maps and the latter is the algebraically defined quotient of $\mathbb{R}^n\otimes\ldots\otimes\mathbb{R}^n$, the tensor product of $p$ copies of $\mathbb{R}^n$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) How is $\Omega^p(\mathbb{R}^n)$ actually the dual of p-vector fields (if in fact it is...)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="3595" LastActivityDate="2010-12-04T19:37:59.113" Title="Tensors as matrices vs. Tensors as multi-linear maps" Tags="&lt;soft-question&gt;&lt;algebra&gt;&lt;differential-geometry&gt;&lt;cohomology&gt;&lt;tensors&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" />
  <row Id="13011" PostTypeId="2" ParentId="13009" CreationDate="2010-12-04T18:45:57.403" Score="1" ViewCount="" Body="&lt;p&gt;Every nonzero element of a finite field is a root of unity. Roots of unity behave very nicely, since the other roots of unity of the same order are just powers of the first.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance i is a primitive 4th root of unity, and the other one, -i, is just a power i,-1,-i,1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The roots of unity form these nice cyclic subgroups of the group of units of a field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want one primitive k'th root of unity, you get all of them for free, so we are just interested in the smallest finite field of characteristic p with a primitive k'th root of unity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now you get into trouble if p divides k: a finite field of characteristic p has size p^m, so its group of units has order p^m-1, and so k has to divide p^m-1.  Such numbers are coprime to p.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your question, you'll want to assume n is coprime to p, otherwise you won't get very good roots.  For instance, when p=2, the 8th roots of unity are exactly {1}; none of them are primitive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Groups of units of finite fields are very well behaved: they are all cyclic.  So to find a primitive k'th root of unity, you just want to find the smallest m such that k divides p^m-1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The field obtained by adjoining a primitive k=n(p-1)st root of unity is the field of size p&lt;sup&gt;m&lt;/sup&gt;, where m is the order of p in the group of units of the ring &lt;strong&gt;Z&lt;/strong&gt; / k &lt;strong&gt;Z&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Note that k has to be coprime to p in order for a primitive k'th root of unity to exist.  Just divide k by p until it is coprime if you want to adjoin the &quot;closest to primitive&quot; root.&lt;/p&gt;&#xA;" OwnerUserId="583" LastActivityDate="2010-12-04T18:45:57.403" />
  <row Id="13012" PostTypeId="2" ParentId="13010" CreationDate="2010-12-04T18:49:16.793" Score="5" ViewCount="" Body="&lt;p&gt;0)  Let $V$ be a (finite-dimensional) real vector space.  A $(1, 1)$-tensor over $V$ is any of the following equivalent objects:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;A linear transformation $V \to V$,&lt;/li&gt;&#xA;&lt;li&gt;An element of $V^{\ast} \otimes V$,&lt;/li&gt;&#xA;&lt;li&gt;A linear map $V \otimes V^{\ast} \to \mathbb{R}$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The isomorphism between these three pictures comes from the definition of the dual and the universal property of the tensor product.  A linear transformation is an object such that, when you feed it an element of $V$, it spits out an element of $V$.  So if you feed it an element of $V$ and an element of $V^{\ast}$, then by the dual pairing (tensor contraction) you get an element of $\mathbb{R}$.  Explicitly, if $f : V \to V$ is a linear transformation, then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\langle f(v), w^{\ast} \rangle : V \times V^{\ast} \to \mathbb{R}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is bilinear (where $\langle \cdot, \cdot \rangle$ denotes the dual pairing), and every bilinear map can be uniquely written this way.  Then one uses the universal property of the tensor product.  That gives you the identification with the third picture.  The identification with the second picture comes from the fact that dual distributes over tensor product (which again comes down to tensor contraction) and the fact that $V^{\ast \ast} \simeq V$.  Alternately, again by tensor contraction, there is a natural bilinear map $V \times (V^{\ast} \otimes V) \to V$ which identifies an element of $V^{\ast} \otimes V$ with a linear transformation $V \to V$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You get the &quot;explicit&quot; version of all these pictures by taking a basis $e_1, ... e_n$ of $V$ and the corresponding dual basis $e_1^{\ast}, ... e_n^{\ast}$ of $V^{\ast}$.  These define bases of every space of $(m, n)$-tensor (just multiply).  In particular the space of $(1, 1)$-tensors has basis $e_i^{\ast} \otimes e_j, 1 \le i, j \le n$.  Writing a $(1, 1)$-tensor in this basis corresponds exactly to writing a linear transformation as a square matrix.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1)  It's not.  If by $\text{Alt}^p(V)$ you mean the space of alternating $p$-linear maps $V^p \to \mathbb{R}$, this is naturally isomorphic to $\Lambda^p (V^{\ast})$ where $V^{\ast}$ is the dual space to $V$ (the space of linear maps $V \to \mathbb{R}$).  You can't identify $V$ with its dual without, say, an inner product.  To see the isomorphism, $(V^{\ast})^{\otimes p}$ can naturally be identified with the space of all $p$-linear maps $V^p \to \mathbb{R}$, more or less by definition of the dual space and tensor product, and taking the subspace of alternating maps in one picture corresponds to taking the quotient in the other.  You can make this really explicit by writing down a basis for both spaces if you want.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2)  This all comes down to tensor contraction.  Locally a $p$-form is the same thing as an alternating $p$-linear map $V^p \to \mathbb{R}$ where $V$ is the tangent space, and by the universal property of the exterior power this is the same thing as a linear map $\Lambda^p V \to \mathbb{R}$.  And a $p$-vector is the same as an element of $\Lambda^p V$.  So the natural dual pairing (tensor contraction) comes into play.&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-12-04T19:08:55.333" LastActivityDate="2010-12-04T19:08:55.333" />
  <row Id="13013" PostTypeId="1" AcceptedAnswerId="13014" CreationDate="2010-12-04T18:49:34.687" Score="1" ViewCount="121" Body="&lt;p&gt;So, I guess divide it into six sections representing the difference between the numbers, so the first is $0\rightarrow n$, the second through fifth are $1\rightarrow n$ and the sixth is $0\rightarrow n$ again, and that becomes a generating function of $((1+x+x^2+\cdots+x^n)^2)((x^2+\cdots+x^n)^4)$. Does this make sense, and what coefficient represents what value of $n$?&lt;/p&gt;&#xA;" OwnerUserId="4411" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2010-12-04T19:08:53.213" LastActivityDate="2010-12-07T14:57:15.787" Title="Generating function for modeling the number of ways to select five integers from 1 to n where no two are consecutive." Tags="&lt;combinatorics&gt;&lt;generating-functions&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="13014" PostTypeId="2" ParentId="13013" CreationDate="2010-12-04T18:58:28.560" Score="0" ViewCount="" Body="&lt;p&gt;Let $f_{k,n}$ be the number of ways to select $k$ integers in $\{ 1, 2, ... n \}$ such that no two are consecutive.  There are two ways to do this: either the largest one is not $n$, and then you choose $k$ integers in $\{ 1, 2, ... n-1 \}$, or the largest one is $n$, and then you choose $k-1$ integers in $\{ 1, 2, ... n-2 \}$.  This gives&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f_{k,n} = f_{k,n-1} + f_{k-1, n-2}.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The generating function you want is $\sum_{n \ge 0} f_{5,n} x^n$, but I think it will be easier to find the bivariate generating function $\sum_{n,k \ge 0} f_{k, n} y^k x^n$ first.  Do you know how to do this from a recursion?&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-12-04T18:58:28.560" CommentCount="3" />
  <row Id="13015" PostTypeId="2" ParentId="356" CreationDate="2010-12-04T19:02:05.233" Score="0" ViewCount="" Body="&lt;p&gt;The Mythical Man-Month more for software engineering but absolutely one of the best books ever written about programing IMO. &lt;/p&gt;&#xA;" LastActivityDate="2010-12-04T19:02:05.233" CommunityOwnedDate="2010-12-04T19:02:05.233" />
  <row Id="13016" PostTypeId="2" ParentId="12983" CreationDate="2010-12-04T19:04:58.737" Score="1" ViewCount="" Body="&lt;p&gt;There's an almost-formal argument using cumulants. Given a random variable $X$, define its moment generating function $$M(X) = E[e^{tX}].$$ It's called the moment generating function since opening the Taylor series of the exponential, we get $$M(X) = 1 + E[X]t + \frac{1}{2}E[X]^2t^2 + \cdots.$$ The moment generating function is useful because of its relation to convolution of two &lt;i&gt;independent&lt;/i&gt; random variables: $$M(X+Y) = E[e^{t(X+Y)}] = E[e^{tX}e^{tY}] = E[e^{tX}]E[e^{tY}] = M(X)M(Y).$$ One proof of the CLT takes the route of the mgf, but we would like to replace the multiplication by a addition since we only really know how to handle sums. So we define the cumulant generating function $$K(X) = \log M(X).$$ We can calculate the first few coefficients (which are called cumulant) by substituting into the (formal) power series of $\log(1+x) = x - x^2/2 + \cdots$: $$K(X) = \log (1+E[X]t+E[X^2]t^2/2 + \cdots) = E[X]t + E[X^2]t^2/2 - (E[X]^2t^2 + E[X]E[X^2]t^3 + E[X^2]t^4/4)/2 + \cdots = E[X]t + V[X]t^2/2 + \cdots.$$ Also, if $X$ and $Y$ are independent then $$K[X+Y] = K[X]+K[Y].$$ Now suppose $X_1,\ldots,X_n$ are iid variables distributed like $X$ with zero expectation. Then $$K[X_1+\cdots+X_n] = nK[X] = \frac{1}{2}nV[X]t^2 + \frac{1}{6}nK_3(X)t^3 + \cdots,$$ where $K_m(X)$ are just the (normalized) coefficients of the cgf, i.e. the cumulants (they are normalized by $1/m!$). If we scale this sum down by $\sqrt{n}$, then the second cumulant becomes $V[X]$ (i.e. the variance is the same), but the rest of the cumulants $K_m$ for $m \geq 3$ get multiplied by $n^{1-m/2} \rightarrow 0$, so in the limit they disappear, and the cumulant of the limit is just $$K\left[\frac{X_1+\cdots+X_n}{\sqrt{n}}\right] = \frac{1}{2}V[X]t^2.$$ Therefore there is one 'domain of attraction' for distributions, which must be the normal distribution with zero mean and variance $V[X]$; it can be calculated directly from this representation. The same idea can be used to analyze the case where the variables are independent but not identically distributed. The main step missing to make this proof formal is reasoning about the limit distribution from the limit cgf; this is the Levy continuity lemma, which shows that the 'inverse Fourier transform' is continuous.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Had we taken the route of mgf's, we would have had to use the identity $(1+1/n)^n \rightarrow e^n$ somewhere, but otherwise the argument would be much the same.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-12-04T19:04:58.737" />
  <row Id="13017" PostTypeId="2" ParentId="12960" CreationDate="2010-12-04T19:06:04.510" Score="5" ViewCount="" Body="&lt;p&gt;Let $P$ be generated by $x$ of order $2^n &amp;gt; 4$ and $y$ of order $4$, and let $z = x^{2^{n-2}}$ be a power of $x$ of order 4. Then the two classes of $Q_8$ in $P$ are represented by $\langle z,y \rangle$ and $\langle z, xy \rangle$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From what you say, you know that both of these subgroups are normalized in $X$ by subgroups isomorphic to ${\rm SL}(2,3)$. That means that $z$ is conjugate in $X$ to both $y$ and to $xy$. I don't think that's possible in a solvable group $X$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume that $X$ is solvable, so it has a chief series with elementary abelian factors. So $z \in M \setminus N$ for one of these factors $M/N$. If $z$ is conjugate to both $y$ and $xy$, then we have $y, xy \in M \setminus N$. But then $x \in M$ and so $x^2 \in N$ and hence $z \in N$, contradiction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ADDED: For your smaller question if, for conjugates $u$ and $v$ of $t$, $x=uv$ is nontrivial and centralizes $t$, then $x$ has odd order and both $t$ and $u$ lie in the normalizer $N$ of $\langle x \rangle$ and hence are conjugate in $N$, but that is impossible because $t$ centralizes and $u$ inverts $x$.&lt;/p&gt;&#xA;" OwnerUserId="2820" LastEditorUserId="2820" LastEditorDisplayName="" LastEditDate="2010-12-05T12:16:24.170" LastActivityDate="2010-12-05T12:16:24.170" CommentCount="2" />
  <row Id="13019" PostTypeId="1" AcceptedAnswerId="24072" CreationDate="2010-12-04T19:08:13.823" Score="2" ViewCount="231" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Show that $\Gamma(y) = \int_0^{\infty}{e^{-x}x^{y-1}\,dx}$ is finite for $y&amp;gt;0$ both as an improper Riemann integral and as a Lebesgue integral. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Show $\Gamma&amp;#39;(y) = \int_0^{\infty}{e^{-x}x^{y-1}\ln{x}\,dx}$ for $y&amp;gt;0$. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;For one: I've tried simply integrating it as an improper Riemann integral, but you always end up with another integral of the &quot;same type&quot; (which is how you eventually show $\Gamma(y+1)=y\Gamma(y)$ ). How do I get around this? As for the Lebesgue integral, I think it'd be easiest to compare the integrand to a larger function whose integral converges, but I haven't come up with a good candidate. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For two: Fix $y_0&amp;gt;0$. Write $$\Gamma&amp;#39;(y) = \lim_{y\to y_0}{\int_0^{\infty}{\frac{e^{-x}x^{y-1}-e^{-x}x^{y_0-1}}{y-y_0}\,dx}}\,.$$ By the MVT, there exists $\eta$ between $y$ and $y_0$ such that the above limit is equal to $$\lim_{y\to y_0}{\int_0^{\infty}{e^{-x}x^{\eta-1}\ln{x}\,dx}}\,.$$ But now I'm not sure what to do. This is similar to a previous question I posted; for that problem we knew the derivative of the original integrand was bounded, so we applied the bounded convergence theorem. Would it be enough to prove that the derivative of my integrand is bounded, and apply BCT? &lt;/p&gt;&#xA;" OwnerUserId="2313" LastActivityDate="2011-02-27T20:27:35.513" Title="Some questions about the gamma function" Tags="&lt;real-analysis&gt;&lt;homework&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" />
  <row Id="13020" PostTypeId="2" ParentId="13010" CreationDate="2010-12-04T19:37:59.113" Score="0" ViewCount="" Body="&lt;p&gt;This interpretation makes judicious use of duality.  Recall that $Hom_R(A\otimes B, C)\cong Hom_R(A,Hom_R(B,C))$ (the proof of this fact is extremely simple.  Please give it a try.).  With finite dimensional vector spaces, we have a canonical isomorphism $V\cong V^{**}=Hom_R(Hom_R(V,R),R)$.  Then to give a map $A\to B$, this is the same as giving a map $A\to B^{**}$ by composition, and by the earlier note about the hom-tensor adjunction, this is identical to giving a map $A\otimes B^*\to R$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;With regards to your two questions, they fall out from these elementary observations.  I urge you to try to work them out.&lt;/p&gt;&#xA;" OwnerUserId="126" LastActivityDate="2010-12-04T19:37:59.113" CommentCount="1" />
  <row Id="13022" PostTypeId="2" ParentId="13003" CreationDate="2010-12-04T20:02:32.490" Score="4" ViewCount="" Body="&lt;p&gt;Perhaps this will be convincing?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) The axis of symmetry intersects one of the sides at point which is not the vertex.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case, by considering the perpendicular to the axis at the point of intersection of the axis and the side, we see that the axis has to be perpendicular to the side and intersects it at the midpoint. In which case it is not an axis of symmetry as it won't intersect the opposite side at the midpoint.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) The axis of symmetry intersects a side at a vertex. Then by 1) it also intersects the opposite side at the vertex, and hence has to be a diagonal, in which case, it is not an axis of symmetry either.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-12-04T20:02:32.490" CommentCount="1" />
  <row Id="13023" PostTypeId="2" ParentId="12983" CreationDate="2010-12-04T20:21:32.580" Score="1" ViewCount="" Body="&lt;p&gt;Working out a few simple examples might help. This would indeed show you that the theorem works in special cases. Thus it would go a long way towards convincing oneself of the validity of the central limit theorem. The central limit theorem first appeared in the work of Abraham de Moivre, in which he proved that the normal distribution to approximate the distribution of the number of heads resulting from many tosses of a fair coin. Later Laplace showed that the same for the binomial distribution, approximating it with the normal distribution. I suggest that you work out these two simpler cases to get a feeling of how the approximation happens. All the necessary background for doing this yourself is available in the book of Hoel, Port and Stone.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you find the theorem hard to understand, it might make you feel better to hear that probabilists needed a long time to properly formulate and understand the theorem. It was only done in the twentieth century by Lyapunov.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are oriented towards practical applications, then getting used to some topics of your choice, for instance noise analysis in communication theory, might help you in convincing yourself of the truth of the central limit theorem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The best way to understand the central limit theorem would of course be to take a course in probability theory. An introductory course usually ends with a proof of this theorem. And if you take a course, you would see other interesting theorems such as the weak and strong laws of large numbers, and this would put the central limit theorem in better perspective. Even after all this, you might still need to contemplate a bit to really absorb the theorem. The proof I have seen is using some &quot;characteristic functions&quot; and a sort of &quot;Fourier transform&quot;. I have to regretfully confess that I didn't fully understand it when I took the course. I never had to take up probability theory later; but if the occasion arises, I intend to go all properly through the proof and understand the machinery.&lt;/p&gt;&#xA;" OwnerUserId="1119" LastActivityDate="2010-12-04T20:21:32.580" />
  <row Id="13024" PostTypeId="2" ParentId="13003" CreationDate="2010-12-04T20:22:43.750" Score="1" ViewCount="" Body="&lt;p&gt;Let's prove the converse: if a parallelogram $ABCD$ has a symmetry axis, it's either a rectangle or a rhombus.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clearly the image of a vertex is a vertex.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If some of the vertices (say $A$) is mapped to itself, than there must be another vertex mapped to itself (because otherwise the rest of vertices must be split into pairs by mapping). This another vertex must be $C$, because if it's adjacent to $A$, than the symmetry axis is one of the edges, which is impossible. So, $AC$ is the axis, $B$ maps to $D$, thus $BD$ is perpendicular to $AC$, therefore the parallelogram $ABCD$ is a rhombus.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the case when every vertex is mapped to some other vertex. If $A$ maps to $C$, than $B$ must map to $D$, so the axis should be a line perpendicular to both $AC$ and $BD$, which is impossible since $AC$ and $BD$ are parallel. So, $A$ maps to an adjacent vertex (let it be $B$), $C$ maps to $D$, so the axis connects the midpoints of $AB$ and $CD$ and is perpendicular to them. This implies that $AB$ is perpendicular to $CD$.&lt;/p&gt;&#xA;" OwnerUserId="2977" LastEditorUserId="2977" LastEditorDisplayName="" LastEditDate="2010-12-04T20:28:15.070" LastActivityDate="2010-12-04T20:28:15.070" />
  <row Id="13025" PostTypeId="2" ParentId="12897" CreationDate="2010-12-04T20:30:45.060" Score="2" ViewCount="" Body="&lt;p&gt;You have $a + b \sum_{k\geq 0} \frac{x^k}{k!} + c \sum_{k\geq 0} (-1)^{k} \frac{x^{2k+1}}{(2k+1)!} = 0$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All the coefficients must be 0 above, so, for the constant term we have $a+b = 0$ and comparing coefficients of $x$ we have $b+c=0$, comparing coefficient of $x^2$ we have $b=0$, from which it follows $a=b=c=0.$&lt;/p&gt;&#xA;" OwnerUserId="404" LastActivityDate="2010-12-04T20:30:45.060" />
  <row Id="13026" PostTypeId="1" AcceptedAnswerId="13048" CreationDate="2010-12-04T20:56:59.197" Score="2" ViewCount="182" Body="&lt;p&gt;OK, so I was told in &lt;a href=&quot;http://cstheory.stackexchange.com/questions/3516/are-there-problems-that-cant-be-expressed-as-languages&quot;&gt;CSTheory&lt;/a&gt; that I should be asking here. So my question is the following:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I've taken my first course on Language Theory and we saw the &lt;a href=&quot;http://www.cs.lmu.edu/~ray/notes/languagetheory/&quot; rel=&quot;nofollow&quot;&gt;&quot;standard&quot;&lt;/a&gt; classification of languages. We saw that many problems can be expressed as a membership question of a corresponding language. I'm wondering if there are any problems for which no language can be specified.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1965" LastActivityDate="2011-03-15T18:12:46.570" Title="Are there problems that can't be expressed as languages?" Tags="&lt;computability&gt;" AnswerCount="3" CommentCount="3" />
  <row Id="13027" PostTypeId="2" ParentId="12897" CreationDate="2010-12-04T21:04:35.023" Score="0" ViewCount="" Body="&lt;p&gt;You need to show that there are no $\alpha$, $\beta$ such that $\alpha e^x + \beta \sin x = \mathrm{const} \neq 0$ on $[0, 1]$. To do that, observe that $e^x$ is strictly convex and $\sin x$ is strictly concave on $[0, 1]$ and both are strictly increasing. It is easy to show that convexness and concaveness are invariant in respect to multiplication by a positive number, and they swap under multiplication by a negative number, so if $\alpha &amp;gt; 0$ and $\beta &amp;gt; 0$ or both are less than zero, then the linear combination is not constant (otherwise the summands would be either both strictly convex or both strictly concave). If $\alpha &amp;gt; 0$ and $\beta &amp;lt; 0$, then one summand is strictly increasing while the other is strictly decreasing, which raises the same problem. The last case to check is when either $\alpha$ or $\beta$ is zero, but it is then obvious.&lt;/p&gt;&#xA;" OwnerUserId="3793" LastEditorUserId="3793" LastEditorDisplayName="" LastEditDate="2010-12-04T21:19:58.240" LastActivityDate="2010-12-04T21:19:58.240" />
  <row Id="13028" PostTypeId="1" CreationDate="2010-12-04T21:05:17.920" Score="2" ViewCount="199" Body="&lt;p&gt;Let T1 and T2 be two theories having the same set of symbols. Assume that any interpretation of T1 is a model of T1 if and only if it is not a model of T2. Then:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;T1 and T2 are finitely axiomatizable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(i.e. there are finite sets of sentences A1 and A2 such that, for any sentence S:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;T1 proves S if and only if A1 proves S, and T2 proves S if and only if A2 proves S).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;/The proof will be by contradiction; assume T1 or T2 are not finitely axiomatizable, then .....?/&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any one have any idea of how to prove this argument!&lt;/p&gt;&#xA;" OwnerUserId="4413" LastActivityDate="2010-12-05T12:26:08.600" Title="Finitely axiomatizable theories" Tags="&lt;logic&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="13029" PostTypeId="1" CreationDate="2010-12-04T21:22:55.387" Score="2" ViewCount="145" Body="&lt;p&gt;How to show that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \displaystyle (a_1 + a_2 + a_3)(\frac{1}{a_1} + \frac{1}{a_2} + \frac{1}{a_3}) \ge 9 $$&#xD;&lt;/p&gt;&#xA;" LastActivityDate="2010-12-05T08:22:48.990" Title="AM-GM application" Tags="&lt;algebra-precalculus&gt;" AnswerCount="4" CommentCount="1" />
  <row Id="13031" PostTypeId="2" ParentId="13029" CreationDate="2010-12-04T21:31:10.213" Score="1" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT:&lt;/strong&gt; The function   $f(x,y,z)=(x+y+z)(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})$ has the absolute minimum in $(1,1,1)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle\frac{\partial f}{\partial x}=\frac{1}{y}+\frac{1}{z}-\frac{y+z}{x^2}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle\frac{\partial f}{\partial y}=\frac{1}{x}+\frac{1}{z}-\frac{x+z}{y^2}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle\frac{\partial f}{\partial x}=\frac{1}{x}+\frac{1}{y}-\frac{x+y}{z^2}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remember that you must resolve:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left\{&#xD;&#xA;    \begin{array}{ll}&#xD;&#xA;      \displaystyle\frac{\partial f}{\partial x}=0  \\&#xD;&#xA;      \displaystyle\frac{\partial f}{\partial y}=0  \\&#xD;&#xA;      \displaystyle\frac{\partial f}{\partial z}=0 &#xD;&#xA;    \end{array}&#xD;&#xA;  \right.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The solution of this system is: (1,1,1) (critical point).&lt;/p&gt;&#xA;" OwnerUserId="3421" LastEditorUserId="3421" LastEditorDisplayName="" LastEditDate="2010-12-04T21:54:40.327" LastActivityDate="2010-12-04T21:54:40.327" CommentCount="4" />
  <row Id="13033" PostTypeId="2" ParentId="13029" CreationDate="2010-12-04T21:41:48.563" Score="1" ViewCount="" Body="&lt;p&gt;One way, although not very thoughtfull, is to simply expand the brackets.  You will get a sum of 9 terms on the LHS.  3 of these terms will be 1's and the other 6 will be fractions involving the three &quot;a&quot; terms.  Get a common denominator for the pair of fractions a/b and b/a (three such pairs) then you can use the following to show each of these three fractions is at least 2, producing a total of at least 9.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Inequality_of_arithmetic_and_geometric_means&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Inequality_of_arithmetic_and_geometric_means&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sorry, you can only use this inequality if the numbers are non-negative.&lt;/p&gt;&#xA;" OwnerUserId="3971" LastEditorUserId="3971" LastEditorDisplayName="" LastEditDate="2010-12-04T21:49:12.443" LastActivityDate="2010-12-04T21:49:12.443" />
  <row Id="13034" PostTypeId="2" ParentId="13029" CreationDate="2010-12-04T21:44:52.820" Score="6" ViewCount="" Body="&lt;p&gt;Hint: &lt;a href=&quot;http://en.wikipedia.org/wiki/Cauchy%E2%80%93Schwarz_inequality&quot; rel=&quot;nofollow&quot;&gt;Cauchy-Schwarz&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint2: Two applications of the AM-GM will also solve it: one is $(x+y+z)/3 \ge \cdots $ and the other is ... (try to fill in the details).&lt;/p&gt;&#xA;" OwnerUserId="1301" LastEditorUserId="1301" LastEditorDisplayName="" LastEditDate="2010-12-04T22:08:04.297" LastActivityDate="2010-12-04T22:08:04.297" />
  <row Id="13035" PostTypeId="1" AcceptedAnswerId="13041" CreationDate="2010-12-04T21:46:39.317" Score="4" ViewCount="182" Body="&lt;p&gt;Hey there,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have to calculate the surface area of the solid of revolution which is produced from rotating $f: (-1,1) \rightarrow \mathbb{R}$, $f(x) = 1-x^2$ about the $x$-axis. I do know there is a formula:&#xA;$$S=2 \pi \int_{a}^b f(x)   \sqrt{1+f&amp;#39;(x)^2}\, \mathrm dx$$&#xD;&#xA;Which will work very well. However, I am not very comfortable with the integral&#xA;$$\int_{-1}^1 (1-x^2)\sqrt{1+4x^2}\, \mathrm dx$$&#xD;&#xA;which I would have to calculate in order to get to the surface area (I tried to substitute $x=\frac{1}{2} \sinh(u)$, but it did not work out too well). Thus, I had the idea to apply &lt;a href=&quot;http://en.wikipedia.org/wiki/Pappus%27s_centroid_theorem&quot; rel=&quot;nofollow&quot;&gt;Pappus' centroid theorem&lt;/a&gt;. I first found the centroid of the area between the parabola and the x-axis to be at $y=\frac{2}{5}$, hence the surface area of the solid of revolution would be:&#xA;$$S = 2 \pi \frac{2}{5} \int_{-1}^1 \sqrt{1+4x^2}\, \mathrm dx$$&#xD;&#xA;But this leads me to a different result than I should get (I calculated the value of the first integral with the help of wolframalpha, it's about ~11...).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What did I do wrong? My best guess is that I misunderstood Pappus' centroid theorem, but what's the mistake? How can I fix it?&lt;/p&gt;&#xA;" OwnerUserId="3787" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-05T00:38:17.257" LastActivityDate="2010-12-05T00:38:17.257" Title="Calculate the surface area of a solid of revolution" Tags="&lt;calculus&gt;" AnswerCount="3" />
  <row Id="13036" PostTypeId="2" ParentId="12944" CreationDate="2010-12-04T22:00:27.020" Score="0" ViewCount="" Body="&lt;p&gt;Here is a heuristic to suggest that it will be difficult to define the square of the delta function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Fourier transform has the property that it takes the convolution of two functions to the product of their Fourier transforms, and vice versa, ie, it takes the product of two functions to their convolution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remember that the Fourier transform of the delta function is the constant function($=1$). Now suppose that $\delta^2$ exists. Then its Fourier transform would be the convolution of two constant functions. Such a convolution would shoot to infinity at every point. Even the theory of distributions can't handle such kind of stuff. So how would you imagine the inverse Fourier transform of this? How would it make sense?&lt;/p&gt;&#xA;" OwnerUserId="1119" LastActivityDate="2010-12-04T22:00:27.020" />
  <row Id="13037" PostTypeId="2" ParentId="13026" CreationDate="2010-12-04T22:14:17.260" Score="1" ViewCount="" Body="&lt;p&gt;Let $P$ be a property of objects of type $X$, i.e. of $x \in X$. Then $P$ can be expressed as membership in $$\{ x \in X : P(x) \}.$$&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-12-04T22:14:17.260" CommentCount="6" />
  <row Id="13038" PostTypeId="2" ParentId="13035" CreationDate="2010-12-04T22:17:48.007" Score="1" ViewCount="" Body="&lt;p&gt;I'm not too sure, but if I recall correctly, one way you can make this a little easier is to note that $g(x)=(1-x^2)\sqrt{1+4x^2}$ is even $g(-x) = g(x)$ and its range is from $-a$ to $a$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you can do $2\int_0^a g(x)\, dx$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;btw: Can anyone show me a link on how to type out formulas so that they appear in mathematical format?&lt;/p&gt;&#xA;" OwnerUserId="3514" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-04T22:24:56.963" LastActivityDate="2010-12-04T22:24:56.963" CommentCount="2" />
  <row Id="13039" PostTypeId="2" ParentId="13035" CreationDate="2010-12-04T22:24:46.670" Score="4" ViewCount="" Body="&lt;p&gt;If you want to try the original integral again, you can use the trig substitution x = 0.5 tan(t).  This will give you an integral involving tan and sec which can be worked out.&lt;/p&gt;&#xA;" OwnerUserId="3971" LastActivityDate="2010-12-04T22:24:46.670" CommentCount="1" />
  <row Id="13041" PostTypeId="2" ParentId="13035" CreationDate="2010-12-04T22:37:10.217" Score="5" ViewCount="" Body="&lt;p&gt;You did misinterpret Pappus' Theorem.  You used the geometric centroid of the &lt;em&gt;region&lt;/em&gt; between $1-x^2$ and the $x$-axis, whereas Pappus' Theorem wants you to use the geometric centroid of the &lt;em&gt;curve&lt;/em&gt; $1-x^2$.  The geometric centroid of the latter is not $\frac{2}{5}$ but (by definition)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\int_{-1}^1 (1-x^2) \sqrt{1+4x^2} dx}{\int_{-1}^1 \sqrt{1+4x^2} dx} \approx 0.59002.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, multiplying this by $2\pi$ times the arc length just gives you the integral you started with.  So it doesn't appear that Pappus' Theorem is an easier route to take.  You could also try switching to an integral in $dy$, but I doubt that will be any better.  I would try Joe's suggestion on your first integral.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For more on finding geometric centroids of curves, see &lt;a href=&quot;http://www.maplesoft.com/applications/view.aspx?SID=32536&amp;amp;view=html&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2370" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-12-04T22:42:16.967" LastActivityDate="2010-12-04T22:42:16.967" />
  <row Id="13043" PostTypeId="1" AcceptedAnswerId="13184" CreationDate="2010-12-04T22:52:00.983" Score="2" ViewCount="152" Body="&lt;p&gt;X|N,P ~ Binomial(N, P)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;N ~ Poisson(11)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P ~ Beta(2,3)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the moment generating function for X?&lt;/p&gt;&#xA;" OwnerUserId="4416" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-12-06T00:04:43.037" LastActivityDate="2010-12-06T05:04:58.393" Title="binomial-Poisson/beta hierarchy" Tags="&lt;probability&gt;&lt;statistics&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="0" />
  <row Id="13044" PostTypeId="1" AcceptedAnswerId="13046" CreationDate="2010-12-04T22:54:43.247" Score="9" ViewCount="188" Body="&lt;p&gt;I'm going over a chapter by chapter review for my calculus final and discovered this problem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$y=\int_{\sqrt{x}}^{x^3}\sqrt{t}\sin{t}\;\mathrm dt$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They split it up so that it became:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$-\int_1^{\sqrt{x}}\sqrt{t}\sin{t}\;\mathrm dt + \int_1^{x^3}\sqrt{t}\sin{t}\;\mathrm dt $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why 1? How was that determined?&lt;/p&gt;&#xA;" OwnerUserId="3514" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-05T00:34:57.230" LastActivityDate="2010-12-05T01:52:42.163" Title="Ranges and the Fundamental Theorem of Calculus 1" Tags="&lt;calculus&gt;&lt;homework&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="13045" PostTypeId="2" ParentId="13044" CreationDate="2010-12-04T23:07:19.263" Score="5" ViewCount="" Body="&lt;p&gt;You can take any lower limit, it doesn't need to be 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\int_b^c f(x) dx = \int_a^c f(x) dx - \int_a^b f(x) dx$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just think of it as the area under the curve.&lt;/p&gt;&#xA;" OwnerUserId="3787" LastActivityDate="2010-12-04T23:07:19.263" />
  <row Id="13046" PostTypeId="2" ParentId="13044" CreationDate="2010-12-04T23:18:38.440" Score="13" ViewCount="" Body="&lt;p&gt;It doesn't have to be $1$; all it needs to be is in the domain of the function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is going on is that you have the Fundamental Theorem of Calculus, which tells you that if $f(x)$ is continuous on an interval containing $a$, then the function $F(x)$ defined on that interval by:&#xA;$$F(x) = \int_a^x f(t)\,dt$$&#xD;&#xA;is differentiable, and in fact $F&amp;#39;(x) = f(x)$ for all $x$ in the interval. But this requires (i) the lower limit to be constant; and (ii) the upper limit to be the variable with respect to which you are taking the derivative.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the first question is: what does one do if you have the variable in the &lt;em&gt;lower&lt;/em&gt; limit instead of the upper limit? Well, that only requires you to use the property of the integral that says&#xA;$$\int_a^b f(t)\,dt = -\int_b^a f(t)\,dt,$$&#xD;&#xA;and the fact that $(-G(x))&amp;#39; = -G&amp;#39;(x)$ for any $G$. So if &#xA;$$F_2(x) = \int_x^a f(t)\,dt$$&#xD;&#xA;then&#xA;$$F_2&amp;#39;(x) = \frac{d}{dx}\int_x^a f(t)\,dt = \frac{d}{dx}\left(-\int_a^x f(t)\,dt\right) = -\frac{d}{dx}\int_a^x f(t)\,dt = -f(x),$$&#xD;&#xA;with the last equality holding by the FTC.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next, what if the upper limit is not $x$, but a function of $x$, say&#xA;$$F_3(x) = \int_a^{g(x)} f(t)\,dt\ ?$$&#xD;&#xA;Then we use the Chain Rule:&#xA;$$\frac{dF_3}{dx} = \frac{dF_3}{dg}\frac{dg}{dx} = &#xD;&#xA;\left(\frac{d}{dg}\int_a^{g(x)}f(t)\,dt\right)\frac{dg}{dx} = f(g(x))g&amp;#39;(x),$$&#xD;&#xA;with the last equality again by the FTC.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What if the lower limit is the function and the upper limit the constant? We combine the two &quot;tricks&quot; above to get the derivative.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And finally, what if both upper and lower limit are functions? Then we use the property of the integrals mentioned by Huy:&#xA;$$\int_a^b f(t)\,dt = \int_a^cf(t)\,dt + \int_c^b f(t)\,dt$$&#xD;&#xA;for any $c$ such that $f(x)$ is defined and continuous on an interval containing $a$, $c$, and $b$. In the case at hand, they pick $1$ simply because it's an easy point; you can use any point on $[0,\infty)$ (cannot be a negative point because you have $\sqrt{t}$ in the integrand). Pick your favorite $a$ on $[0,\infty)$, and you have&#xA;$$F(X) = \int_{\sqrt{x}}^{x^2}f(t)\,dt = \int_{\sqrt{x}}^a f(t)\,dt + \int_a^{x^2}f(t)\,dt = -\int_a^{\sqrt{x}}f(t)\,dt + \int_a^{x^2}f(t)\,dt$$&#xD;&#xA;and each of the integrals on the right is an integral we know how to do using the FTC. So if we let the first integral be some $G(X)$ and the second be $H(X)$, then $F(X)=G(X)+H(X)$, so $F&amp;#39;(X) = G&amp;#39;(X)+H&amp;#39;(X)$, and we can find $G&amp;#39;(X)$ and $H&amp;#39;(X)$ using the methods outlined above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They did it with $a=1$, but you could do it with $a=\pi$, or $a=4$, or any nonnegative $a$.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-05T01:52:42.163" LastActivityDate="2010-12-05T01:52:42.163" CommentCount="1" />
  <row Id="13047" PostTypeId="2" ParentId="13028" CreationDate="2010-12-04T23:24:24.127" Score="4" ViewCount="" Body="&lt;p&gt;There's a result using compactness that says that if two sets of sentences, $T_1,T_2$ have no common model then there exists a sentence $\phi$ such that $T_1\models\phi$ and $T_2\models\lnot\phi$. To see this use the contrapositive: If for every sentence $\phi$ we have $T_1\models\phi$ has as a consequence that $T_2\cup\{\phi\}$ is satisfiable, then for every sentence $\phi$ that is a conjunction of sentences of $T_1$ we have $T_2\cup\{\phi\}$ is satisfiable and through compactness $T_1$ and $T_2$ share a model.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let $\phi$ be such that $T_1\models\phi$ and $T_2\models\lnot\phi$. Then $\phi$ is an axiomatization of $T_1$ and $\lnot\phi$ is an axiomatization of $T_2$. This is because, if a model satisfies $\phi$ then it can't satisfy $T_2$ and therefore satisfies $T_1$ and vice versa.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's the answer I came up with, but maybe there's a more direct way to it.&lt;/p&gt;&#xA;" OwnerUserId="1772" LastActivityDate="2010-12-04T23:24:24.127" CommentCount="1" />
  <row Id="13048" PostTypeId="2" ParentId="13026" CreationDate="2010-12-04T23:39:38.327" Score="6" ViewCount="" Body="&lt;p&gt;There are many types of &quot;problems&quot;.  Not all types of problems correspond to languages, but some do.  Here are some relevant examples.  &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;A &lt;em&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Decision_problem&quot; rel=&quot;nofollow&quot;&gt;decision problem&lt;/a&gt;&lt;/em&gt; is the problem of deciding membership in some set. Usually in CS the set is a set of natural numbers or strings. Any such problem can be represented as a language in the most general sense. In the notes you linked, they only consider languages on sets of strings over a finite alphabet; these languages correspond to a certain type of decision problem, which is related to the usual reducibilities such as Turing reducibility and P-time reducibility. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;A &lt;em&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Function_problem&quot; rel=&quot;nofollow&quot;&gt;function problem&lt;/a&gt;&lt;/em&gt; is a problem that asks you to be able to compute a specific function. For example, given an integer, produce an explicit prime factorization. These problems do not directly correspond to languages, but each function problem can be converted into the related decision problem of the graph of the function. When the underlying set is countable, this conversion preserves Turing degree.  &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;A &lt;em&gt;mass problem&lt;/em&gt; is a more general type of problem that is represented by a subset of $\omega^\omega$. The &quot;problem&quot; is to produce some element of the set. These problems correspond to Medvedev reducibility and Muchnik reducibility.  They do not correspond to decision problems for sets of strings on finite alphabets. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="630" LastActivityDate="2010-12-04T23:39:38.327" CommentCount="1" />
  <row Id="13049" PostTypeId="2" ParentId="13003" CreationDate="2010-12-05T01:29:32.397" Score="1" ViewCount="" Body="&lt;p&gt;I would start by pointing out that parallelogram has two points farther apart than any other pair-two of the vertices.  Any symmetry must leave them in place or exchange them.  To leave them in place, the axis has to go through them.  To exchange them, the axis has to be the perpendicular bisector of the segment between them.  By inspection, neither is a symmetry.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-12-05T01:29:32.397" />
  <row Id="13050" PostTypeId="1" AcceptedAnswerId="13052" CreationDate="2010-12-05T02:16:37.627" Score="10" ViewCount="584" Body="&lt;p&gt;I was reading the following article on &lt;a href=&quot;http://en.wikipedia.org/wiki/Ultrafinitism&quot; rel=&quot;nofollow&quot;&gt;Ultrafinitism&lt;/a&gt;, and it mentions that one of the reasons ultrafinitists believe that &lt;strong&gt;N&lt;/strong&gt; is not infinite is because the floor of $e^{e^{e^{79}}}$ is not computable. I was wondering if that's the case because of technological limitations, or whether there is another reason we cannot find a floor of this number.&lt;/p&gt;&#xA;" OwnerUserId="3731" LastActivityDate="2010-12-05T12:53:08.707" Title="$e^{e^{e^{79}}}$ and ultrafinitism" Tags="&lt;finitism&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="6" />
  <row Id="13052" PostTypeId="2" ParentId="13050" CreationDate="2010-12-05T02:37:36.223" Score="17" ViewCount="" Body="&lt;p&gt;In the formal meaning of &quot;&lt;a href=&quot;http://en.wikipedia.org/wiki/Computable_number&quot; rel=&quot;nofollow&quot;&gt;computable&lt;/a&gt;&quot; the floor of that number is indeed computable. This is to say that a patient immortal human with access to unlimited paper and pencil could, in principle, work out the answer. (Here I assume, for technical reasons, that the number in question is not an integer - I assume someone who knows enough number theory will be able to cite a result that implies this.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The article linked makes the weaker claim that the value has not yet been calculated, which seems likely to me. The issue they are concerned about is that humans are not immortal and that our supply of paper is very limited. If the number of decimal digits in the value is too large, it would be impossible to actually represent it in any physical way within our universe. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, I think it is more accurate to say that ultrafinitists don't accept that the set of all natural numbers is a coherent entity - not that they think it is finite. However, as the article you linked alludes, it is very difficult to find a coherent but non-arbitrary way to say what natural numbers are without accepting that there are an infinite number of them.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&lt;strong&gt;Addendum&lt;/em&gt;&lt;/strong&gt; Here is why I am worried whether $e^{e^{e^{79}}}$ is an integer. It's certainly correct that no matter what, the floor of that number is an integer and is therefore computable. That part of my argument is fine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, if $e^{e^{e^{79}}}$ is not an integer, then I can tell you a specific algorithm to use to compute it. Namely, compute better and better upper and lower bounds until they fall strictly between two consecutive integers (which they must, since their limit is not an integer) and then pick the smaller of those two integers.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $e^{e^{e^{79}}}$ is an integer, then that algorithm won't work, because it will never stop. But if we knew that $e^{e^{e^{79}}}$ was an integer then we could take better and better upper and lower approximations until they straddle a single integer, and then pick that. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the reason that I am interested whether the number is an integer is that, beyond merely knowing that the floor is an integer, I'd like to know which algorithm could be used to compute it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In any case, I don't think that the point of the example was to pick a number that is not known to be integer or known to not be an integer. The point of the example should be to pick a number which is simply too large to represent physically.  I was hoping that someone would have a quick answer that confirms $e^{e^{e^{79}}}$ is not an integer, so I could edit my response with that info. But the non-integer property seems more difficult than I thought. &lt;/p&gt;&#xA;" OwnerUserId="630" LastEditorUserId="630" LastEditorDisplayName="" LastEditDate="2010-12-05T12:53:08.707" LastActivityDate="2010-12-05T12:53:08.707" CommentCount="6" />
  <row Id="13053" PostTypeId="1" AcceptedAnswerId="14407" CreationDate="2010-12-05T02:46:48.630" Score="3" ViewCount="282" Body="&lt;p&gt;The following is a plot comparing &lt;code&gt;Exp[Derivative[1,0][Zeta][0,x]+1/2Log[2 Pi]]&lt;/code&gt; and &lt;code&gt;Gamma[x]&lt;/code&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/Adlxh.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In theory the blue and the red graphs should be identical, giving the Gamma function. But in reality we have the blue plot (the plot of &lt;code&gt;Exp[Derivative[1,0][Zeta][0,x]+1/2Log[2 Pi]]&lt;/code&gt;) representing the absolute value of the Gamma function instead. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is: is this my fault (bad formula); and if so, how then do I correct it? Or it is a &lt;em&gt;Mathematica&lt;/em&gt; bug?&lt;/p&gt;&#xA;" OwnerUserId="2513" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-14T17:22:56.877" LastActivityDate="2010-12-15T11:34:42.273" Title="Is it problem of Mathematica or my own?" Tags="&lt;math-software&gt;&lt;special-functions&gt;&lt;gamma-function&gt;&lt;riemann-zeta&gt;" AnswerCount="2" CommentCount="13" />
  <row Id="13054" PostTypeId="1" CreationDate="2010-12-05T02:47:36.093" Score="47" ViewCount="1575" Body="&lt;p&gt;In &lt;a href=&quot;http://math.stackexchange.com/questions/13050/eee79-and-ultrafinitism/&quot;&gt;this question&lt;/a&gt;, I needed to assume in my answer that $e^{e^{e^{79}}}$ is not an integer. Is there some standard result in number theory that applies to situations like this? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Much later addendum:&lt;/strong&gt; when I asked this, I didn't expect it to be an open question. If it is open, if someone could give an answer to that effect, I would accept it simply to close off the question. If some author has stated something similar as an open question, that would be helpful for me to know; I'm not a number theorist and my knowledge of the field is not very deep. &lt;/p&gt;&#xA;" OwnerUserId="630" LastEditorUserId="630" LastEditorDisplayName="" LastEditDate="2011-03-20T12:01:23.143" LastActivityDate="2011-03-20T16:08:17.500" Title="How to show $e^{e^{e^{79}}}$ is not an integer" Tags="&lt;number-theory&gt;&lt;transcendence-theory&gt;" AnswerCount="1" CommentCount="13" FavoriteCount="18" />
  <row Id="13055" PostTypeId="1" CreationDate="2010-12-05T03:08:06.427" Score="5" ViewCount="208" Body="&lt;p&gt;In some math job advertisments, it shows &quot;the deadeline is...&quot;, but to surprised some schools review applications before that, does that mean we submit application as earlier as possible could be in a better position? (because I normally do that when date is close to the deadline). (I just want to make sure I understand things correctly). Thanks!&lt;/p&gt;&#xA;" OwnerUserId="1944" LastActivityDate="2010-12-07T07:10:53.883" Title="deadline in math jobs application" Tags="&lt;soft-question&gt;" AnswerCount="2" />
  <row Id="13056" PostTypeId="2" ParentId="13055" CreationDate="2010-12-05T03:21:50.113" Score="10" ViewCount="" Body="&lt;p&gt;I attended a panel talk about this recently and the uniform opinion of the panelists was that one should submit job applications as soon as possible regardless of the deadline. Some schools start reviewing applications much earlier than the deadline (this includes my university). Of course, as long as you send in your application before the deadline, it will be looked at. The reason the panelists mentioned for early applications is that review committees tend to swamped just before and after the deadline. So if you get in your application early, there is more chance that you will get a more through look and if you are a candidate who might otherwise be on the fringe, this might work in your favor. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, here is a useful article in the AMS: (page 1024 answers your question)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.ams.org/notices/200609/fea-lewis.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.ams.org/notices/200609/fea-lewis.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am quoting the relevant part for convenience:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;When Do I Apply for Jobs?&#xA;  Most deadlines are between December 1 and January&#xA;  15. For nonpostdoctoral positions, it is very&#xA;  helpful to send in your application by mid-&#xA;  December even if the deadline is much later, especially&#xA;  if you will be at the Joint Mathematics&#xA;  Meetings in January. For schools that conduct interviews&#xA;  at the JMM, you are better off if you have&#xA;  already applied and the schools have had the&#xA;  chance to look over your application. In mid-December,&#xA;  you may want to email the schools and remind&#xA;  them that you will be at the JMM and would&#xA;  be happy to meet with them there. If you are giving&#xA;  a talk, this also lets you send them the title, time,&#xA;  and location if you didn’t know those earlier. A&#xA;  short polite email, especially if you mention&#xA;  specifics about the individual school, may lead directly&#xA;  to an interview and is unlikely to bother anyone.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="3431" LastEditorUserId="3431" LastEditorDisplayName="" LastEditDate="2010-12-05T04:39:54.773" LastActivityDate="2010-12-05T04:39:54.773" CommentCount="1" />
  <row Id="13058" PostTypeId="1" AcceptedAnswerId="13081" CreationDate="2010-12-05T04:13:49.767" Score="2" ViewCount="77" Body="&lt;p&gt;Let $$D_N(x)=\frac{\sin [(N+(1/2))t]}{\sin (t/2)}$$ be the Dirichlet kernel. Let $x(N)$ be the number in $0&amp;lt;x&amp;lt;\pi/N$ such that $D_N(x)=1$. Is $$\left|\int_{x(N)}^{\pi/N} D_N(t)\mathrm dt \right|=O\left(\frac1{N}\right)$$ true?&#xA;This question arises from my attempt to give a rigorous proof of Gibbs phenomenon. In fact, I only need that the limit is 0 as $N\to\infty$.&lt;/p&gt;&#xA;" OwnerUserId="3249" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-05T05:36:58.337" LastActivityDate="2010-12-05T09:06:44.097" Title="A limit related to the Gibbs phenomenon" Tags="&lt;analysis&gt;&lt;fourier-analysis&gt;&lt;fourier-series&gt;" AnswerCount="1" />
  <row Id="13060" PostTypeId="2" ParentId="12809" CreationDate="2010-12-05T05:08:55.283" Score="0" ViewCount="" Body="&lt;p&gt;We want to show that &#xA;$$&#xD;&#xA;{\rm P}[X \ge {\rm E}(X|U) - 1/m] \ge {\rm P}(U)/m.&#xD;&#xA;$$&#xD;&#xA;This is trivially not true if $0 &amp;lt; m &amp;lt; 1$, for if $U = \Omega$, the right-hand side is greater than $1$. So, assume that $m \geq 1$, and ${\rm P}(U)&amp;gt;0$. Further, if ${\rm E}(X|U) - 1/m \leq 0$, then the inequality is trivial; hence we assume that&#xA;${\rm E}(X|U) - 1/m &amp;gt; 0$. Let $P_{X|U}$ denote the conditional distribution of $X$ given $U$, so that&#xA;$$&#xD;&#xA;P_{X|U} {(A)} = {\rm P}(X \in A | U) = \frac{{{\rm P}([X \in A],U)}}{{{\rm P}(U)}}.&#xD;&#xA;$$&#xD;&#xA;Then,&#xA;$$&#xD;&#xA;{\rm E}(X|U) = \int_{[0,1]} {xP_{X|U} ({\rm d}x)}. &#xD;&#xA;$$&#xD;&#xA;Next, decompose $[0,1]$ into the union of the intervals $I_1 = \lbrace 0 \leq x &amp;lt; {\rm E}(X|U) - 1/m \rbrace$ and $I_2 = \lbrace {\rm E}(X|U) - 1/m \leq x \leq 1 \rbrace$. On the one hand,&#xA;$$&#xD;&#xA;\int_{I_1 } {xP_{X|U} ({\rm d}x)}  \le  [{\rm E}(X|U) - 1/m]P_{X|U} (I_1 ) \le {\rm E}(X|U) - 1/m,&#xD;&#xA;$$ &#xA;and on the other hand, &#xA;$$&#xD;&#xA;\int_{I_2 } {xP_{X|U} ({\rm d}x)} \leq  \int_{I_2 } {P_{X|U} ({\rm d}x)} = P_{X|U} {(I_2)} \leq \frac{{{\rm P}(X \in I_2 )}}{{{\rm P}(U)}}.&#xD;&#xA;$$&#xD;&#xA;Combining it all, we get&#xA;$$&#xD;&#xA;{\rm E}(X|U) \leq {\rm E}(X|U) - 1/m + {\rm P}(X \in I_2)/{\rm P}(U).&#xD;&#xA;$$&#xD;&#xA;That is ${\rm P}(X \in I_2) \geq {\rm P}(U)/m$, which is exactly what we want.&lt;/p&gt;&#xA;" OwnerUserId="2810" LastActivityDate="2010-12-05T05:08:55.283" />
  <row Id="13062" PostTypeId="1" AcceptedAnswerId="13066" CreationDate="2010-12-05T06:29:28.217" Score="2" ViewCount="254" Body="&lt;p&gt;Let $G$ be a finitely generated group and $H$ a subgroup of $G$. If the index of $H$ in $G$ is finite, show that $H$ is also finitely generated.&lt;/p&gt;&#xA;" OwnerUserId="4426" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-05T09:08:17.637" LastActivityDate="2010-12-06T18:05:35.107" Title="subgroups of finitely generated groups with a finite index." Tags="&lt;homework&gt;&lt;group-theory&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="13063" PostTypeId="1" CreationDate="2010-12-05T06:47:33.047" Score="1" ViewCount="45" Body="&lt;p&gt;I know a little bit of probability; but I do not know anything from dynamical systems. I want to understand the notion of Lyapunov exponent for a certain sequence of random variables arising out of a dynamical system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the wikipedia article on Lyapunov exponent is not giving any suitable references. I would be most grateful if someone can give some helpful pointers.&lt;/p&gt;&#xA;" LastActivityDate="2010-12-05T07:59:01.827" Title="Lyapunov exponent reference" Tags="&lt;reference-request&gt;&lt;dynamical-systems&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="13065" PostTypeId="1" AcceptedAnswerId="13070" CreationDate="2010-12-05T06:56:27.863" Score="1" ViewCount="38" Body="&lt;p&gt;I don't understand an equation I am reading in my notes: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose, $||$ is a nonarchimedean absolute value on a field $K$ complete wrt this absolute value. Suppose, $|a_0|&amp;gt;|a_i|$ for all $i&amp;gt;0$. Then, $|a_0+a_1+a_2+...|=|a_0|$. I understand that $|a_0+a_1+a_2+...|\leq max\{|a_i|:i\geq 0\}=|a_0|$ by ultrametric inequality. I am not sure how to get the reverse inequality.&lt;/p&gt;&#xA;" OwnerUserId="3947" LastEditorUserId="3947" LastEditorDisplayName="" LastEditDate="2010-12-05T07:17:06.090" LastActivityDate="2010-12-05T07:17:06.090" Title="A question regarding non archimedean absolute values" Tags="&lt;number-theory&gt;" AnswerCount="1" />
  <row Id="13066" PostTypeId="2" ParentId="13062" CreationDate="2010-12-05T06:58:59.920" Score="2" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Hint:&lt;/strong&gt; Suppose $G$ has generators $g_1, \ldots, g_n$. We can assume that the inverse of each generator is a generator. Now let $Ht_1, \ldots, Ht_m$ be all right cosets, with $t_1 = 1$. For all $i,j$, there is $h_{ij} \in H$ with $t_i g_j = h_{ij} t_{{k}_{ij}}$, for some $t_{{k}_{ij}}$. It's not hard to prove that $H$ is generated by all the $h_{ij}$.&lt;/p&gt;&#xA;" OwnerUserId="329" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2010-12-05T07:31:51.330" LastActivityDate="2010-12-05T07:31:51.330" CommentCount="4" />
  <row Id="13067" PostTypeId="1" CreationDate="2010-12-05T07:02:17.123" Score="6" ViewCount="226" Body="&lt;p&gt;When I read some issues here I see from time to time incorrect references to the field &lt;em&gt;special functions&lt;/em&gt;, it might e.g. be a discussion around Dirac's $\delta$-function which is tagged (special-functions) or a discussion around some function reminding of the Weierstrass no where differentiable continuous function. These examples make me think - what would classify a special function? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A vague bad definition could be &quot;A function is a special function if it has some resemblance to some Hypergeometric function&quot; or &quot;A function is a special function if it fits into the &lt;a href=&quot;http://en.wikipedia.org/wiki/Bateman_manuscript_project&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;Bateman manuscript project&lt;/em&gt;&lt;/a&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To me the &lt;a href=&quot;http://en.wikipedia.org/wiki/Gamma_function&quot; rel=&quot;nofollow&quot;&gt;Gamma function&lt;/a&gt; and the &lt;a href=&quot;http://en.wikipedia.org/wiki/Riemann_zeta_function&quot; rel=&quot;nofollow&quot;&gt;Zeta function&lt;/a&gt; are definitively special functions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, I have worked on &lt;a href=&quot;http://en.wikipedia.org/wiki/Legendre_function&quot; rel=&quot;nofollow&quot;&gt;Legendre functions&lt;/a&gt; $P_\lambda$ and $Q_\lambda$ of the first and second kind , which I would call special functions, but not individually however (except perhaps $P_{-1/2}$). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would not say that elementary functions (such as trigonometric functions and the exponential function) are special functions - but I am not totally convinced about this..&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I do not agree with &lt;a href=&quot;http://en.wikipedia.org/wiki/Special_functions&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;, it says: &#xA;&quot;Special functions are particular mathematical functions which have more or less established names and notations due to their importance in mathematical analysis, functional analysis, physics, or other applications.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is no general formal definition, but the &lt;a href=&quot;http://en.wikipedia.org/wiki/List_of_mathematical_functions&quot; rel=&quot;nofollow&quot;&gt;list of mathematical functions&lt;/a&gt; contains functions which are commonly accepted as special. In particular, elementary functions are also considered as special functions.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, looking at the Wikipedia list (linked above) the indicator function, step functions, the absolute value function and the sign are special functions -- this sounds very wrong to me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what is a special function and what should be under the (special-function) tag?&lt;/p&gt;&#xA;" OwnerUserId="1154" LastActivityDate="2010-12-06T06:19:27.813" Title="What is a special function?" Tags="&lt;special-functions&gt;&lt;definition&gt;" AnswerCount="2" CommentCount="8" FavoriteCount="2" />
  <row Id="13068" PostTypeId="1" AcceptedAnswerId="13085" CreationDate="2010-12-05T07:07:58.413" Score="2" ViewCount="236" Body="&lt;p&gt;Evaluate the integral as a power series:&lt;br&gt;&#xA;$\int(x^{11}(\tan^{-1}(x^2))\mathrm dx)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have been using Abel's theorem to do this (and the fact that the function is differentiable and integrable on it's interval of convergence in this case)&#xA;From what I can tell, an approach to this question would be to find a closed form similar to $\frac1{1-x}$ by taking the derivative of $\tan^{-1}(x)$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;so let $f(x) = \int(x^{11}(\tan^{-1}(x^2))\mathrm dx)$&lt;br&gt;&#xA;then let $t = x^2, g(t) = f&amp;#39;(x^2)$ &#xA;then divide by $x^{11/2}$,&lt;br&gt;&#xA;$\frac{g(t)}{t^{11/2}} = \tan^{-1}(t)$&lt;br&gt;&#xA;then take the derivative of both sides so&lt;br&gt;&#xA;derivative of $\frac{g(t)}{t^{11/2}} = \frac1{t^2+1}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I am stuck here relating this to $\frac1{1-x}$.  Maybe I am looking at this the wrong way.  Any help would be appreciated. Thanks&lt;/p&gt;&#xA;" OwnerUserId="2533" LastEditorUserId="2533" LastEditorDisplayName="" LastEditDate="2011-03-30T19:15:20.647" LastActivityDate="2011-03-30T19:15:20.647" Title="Evaluate the integral as a power series $\int(x^{11}(\tan^{-1}(x^2))\mathrm dx)$" Tags="&lt;calculus&gt;&lt;homework&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="13069" PostTypeId="2" ParentId="12488" CreationDate="2010-12-05T07:10:16.833" Score="3" ViewCount="" Body="&lt;p&gt;To suppliment Moron's comment: yes, using plane embedding algorithm will do. One of such algorithm is &lt;a href=&quot;http://jgaa.info/accepted/2004/BoyerMyrvold2004.8.3.pdf&quot; rel=&quot;nofollow&quot;&gt;Boyer and Myrvold&lt;/a&gt;. &lt;/p&gt;&#xA;" OwnerUserId="550" LastActivityDate="2010-12-05T07:10:16.833" />
  <row Id="13070" PostTypeId="2" ParentId="13065" CreationDate="2010-12-05T07:14:33.407" Score="2" ViewCount="" Body="&lt;p&gt;Let $a = \sum a_i$. You can apply the argument to $a_0 = a- (a_1 + \dots + a_n)$. We have then that &#xA;$$ |a_0| \leq \max(|a|, |a_1 + \dots a_n|) = |a|$$&#xD;&#xA;because $|a_0|$ is bigger than $|a_1 + \dots + a_n|$, and consequently the second term cannot contribute to the maximum. This implies that $|a| \geq |a_0|$, which is the reverse inequality you wanted.&lt;/p&gt;&#xA;" OwnerUserId="536" LastActivityDate="2010-12-05T07:14:33.407" CommentCount="3" />
  <row Id="13072" PostTypeId="1" CreationDate="2010-12-05T07:52:18.373" Score="0" ViewCount="44" Body="&lt;p&gt;Let $R$ be a unitary ring, $\mathfrak J$ an ideal of $R$ such that ($\mathfrak J+G=\{a+g\mid a\in\mathfrak J,\ g\in G\}$, and $R^\times$ is obviously the group of unity of $R$.) $\mathfrak J+G\le R^\times$ for any $G\le R^\times$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The group $G$ is said to be $\mathfrak J-$ &lt;em&gt;seminal&lt;/em&gt; iff $\mathfrak J+G=R^\times$. Let now $\Sigma_R(\mathfrak J)$ be the set of all $\mathfrak J-$seminal subgroups of $R^\times$, and let $\sigma_R(\mathfrak J)$ denote the intersection of all elements in $\Sigma_R(\mathfrak J)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The set $\sigma_R(\mathfrak J)$ is the &lt;em&gt;seed&lt;/em&gt; of $R$ &lt;em&gt;rooted&lt;/em&gt; at the ideal $\mathfrak J$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider now the set &#xA;$$&#xD;&#xA;N_Z(R)=\mathcal Z(R)\cap \mathcal N(R)&#xD;&#xA;$$ &#xA;intersection of the center of $(R,\cdot)$ with the set of all nilpotent elements (which coincides with $\bigcap \text{Spec}(R)$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The $N_Z(R)-$seminal ideals of $R$ are said to be its &lt;em&gt;nilseminal&lt;/em&gt; ideals.&#xA; The seed of $R$ rooted at $N_Z(R)$ is said to be the &lt;em&gt;nilseed&lt;/em&gt; of $R$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The ring $R$ is said to be &lt;em&gt;nilseminal&lt;/em&gt; iff its nilseed is an element of $\Sigma_R(N_Z(R))$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the key step is to show that in fact &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$N_Z(R)+G\le R^\times$ for any $G\le R^\times$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Other questions (those marked with ``true'' have been solved) which involve more general rings are&lt;/p&gt;&#xA;&#xA;&lt;p&gt;true A) If $n$ is a squarefree integer, then $\mathbb Z/n\mathbb Z$ is nilseminal;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;true B) If $n=p^\alpha$ for some $\alpha\in\mathbb N$, then $\mathbb Z/n\mathbb Z$ is nilseminal;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Let $R$ and $S$ be isomorphic rings via $\phi\colon R\to S$. Prove or disprove that for any ideal $\mathfrak J$ of $R$, a subgroup $G\le R^\times$ is $\mathfrak J-$seminal iff $\phi(G)$ is $\phi(\mathfrak J)-$seminal.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Can a ring $R$ with nilpotent elements of any positive order be nilseminal? Can a nilseminal ring have nilpotents of any positive order?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Let $S$ be a unitary subring of a ring $R$. Prove or disprove that $R$ is nilseminal only if $S$ is.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Prove or disprove that ``being nilseminal'' is a local property, i.e. $R$ is nilseminal iff $R_\mathfrak p$ is nilseminal for any $\mathfrak p\in \text{Spec}(R)$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="685" LastEditorUserId="685" LastEditorDisplayName="" LastEditDate="2010-12-07T11:41:44.787" LastActivityDate="2010-12-07T11:41:44.787" Title="Properties of (nil)seminal rings" Tags="&lt;abstract-algebra&gt;&lt;ring-theory&gt;" CommentCount="1" />
  <row Id="13073" PostTypeId="2" ParentId="13063" CreationDate="2010-12-05T07:59:01.827" Score="1" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://books.google.com/books?id=FIYHiBLWCJMC&amp;amp;printsec=frontcover&amp;amp;dq=nonlinear+dynamics&amp;amp;hl=en&amp;amp;ei=2kX7TKeKFISClAeVs5GlDA&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=1&amp;amp;sqi=2&amp;amp;ved=0CCoQ6AEwAA#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;Non Linear Dynamics and Chaos&lt;/a&gt; by Strogatz, page 322.  I read this a long time ago, and the whole book is worth reading in my opinion.&lt;/p&gt;&#xA;" OwnerUserId="1417" LastActivityDate="2010-12-05T07:59:01.827" />
  <row Id="13074" PostTypeId="2" ParentId="13068" CreationDate="2010-12-05T07:59:49.040" Score="2" ViewCount="" Body="&lt;p&gt;$\frac{1}{1+x^2} = \frac{1}{1-(-x^2)}$.&lt;/p&gt;&#xA;" OwnerUserId="3787" LastActivityDate="2010-12-05T07:59:49.040" CommentCount="1" />
  <row Id="13075" PostTypeId="1" AcceptedAnswerId="13079" CreationDate="2010-12-05T08:12:02.450" Score="1" ViewCount="91" Body="&lt;p&gt;Hi i am getting confused with this diagram &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://img35.imageshack.us/img35/8175/picture1654x310.jpg&quot; rel=&quot;nofollow&quot;&gt;http://img35.imageshack.us/img35/8175/picture1654x310.jpg&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The bold line is a vector. I haven't worked much with vectors, but it looks similar to a linear graph (polynomial degree 1)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would i be correct in saying these are the differences: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When working with linear graphs (polynomial degree 1) we deal with multiple x values and multiple y values. But vectors are different to linear graphs because we only deal with the length of the line (and direction) and the length of the projected lines?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and that a linear graph (polynomial degree 1) represents multiple times. But a vector only represents an instant in time (but you can have multiple vectors for each time).&lt;/p&gt;&#xA;" OwnerUserId="4429" LastActivityDate="2010-12-05T09:03:51.810" Title="What are the differences between linear graphs (polynomial degree 1) and vectors" Tags="&lt;vector-analysis&gt;" AnswerCount="2" />
  <row Id="13076" PostTypeId="2" ParentId="13075" CreationDate="2010-12-05T08:18:35.780" Score="1" ViewCount="" Body="&lt;p&gt;The vector you are referring here is a Euclidean vector&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Euclidean_vector&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Euclidean_vector&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A degree 1 polynomial is a straight line&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Line_(geometry&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Line_(geometry&lt;/a&gt;)&lt;/p&gt;&#xA;" OwnerUserId="3431" LastActivityDate="2010-12-05T08:18:35.780" />
  <row Id="13077" PostTypeId="2" ParentId="13029" CreationDate="2010-12-05T08:22:48.990" Score="1" ViewCount="" Body="&lt;p&gt;One more way is to note that it's just the AM-HM:&#xA;$$\frac{a_1 + a_2 + a_3}{3} \ge \frac{3}{\frac{1}{a_1} + \frac{1}{a_2} + \frac{1}{a_2}}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general it's useful to remember the power mean inequality: Suppose that $\alpha \ge \beta$. Then&#xA;$$\left(\frac{a_1^\alpha + \dots + a_n^\alpha}{n}\right)^{1/\alpha} \ge \left(\frac{a_1^\beta + \dots + a_n^\beta}{n}\right)^{1/\beta}.$$ The inequality works also in the case that $\alpha$&amp;#160;or $\beta$&amp;#160;is zero, when we take that to mean the geometric mean.&lt;/p&gt;&#xA;" OwnerUserId="3776" LastActivityDate="2010-12-05T08:22:48.990" CommentCount="2" />
  <row Id="13079" PostTypeId="2" ParentId="13075" CreationDate="2010-12-05T08:32:35.367" Score="1" ViewCount="" Body="&lt;p&gt;The graphs of degree one polynomial functions are the same as non-vertical lines. Vectors are not lines; as you indicate, they consist of direction and (finite) length (and, depending on context, a base point).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a close relationship between the two.  A line is determined by taking a point in the plane and then choosing the direction to travel from that point.  This direction can be specified by a vector.  So instead of thinking of a line as the graph of a degree one polynomial function you could think of it as the set of all points $p+tv$, where $p$ is point in the plane, $v$ is a nonzero vector, and $t$ ranges over the real numbers.  The parameter $t$ tells you how much you travel in the $v$ direction (technically, divided by the length of $v$), with the understanding that when $t$ is negative you are going $|t|$ much in the opposite direction.  This is actually more general than thinking of graphs of functions, because you could also describe vertical lines by choosing $v$ to point straight up or down. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To summarize: A vector and a starting point determine a line, and a line determines a direction vector up to multiplication by a nonzero real number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To be concrete, consider $f(x)=2x+3$.  The point $p=(0,3)$ is on the graph, and the slope $2$ determines the direction.  This direction can be described by the vector $v$ with $x$ component $1$ and $y$ component $2$.  We can write $v=(1,2)$ to summarize this information. Here going $t$ in the $v$ direction from $p$ takes you to the point $p+tv=(0,3)+(t,2t)=(t,2t+3)$, which is the same as the point $(t,f(t))$ on the graph of $f$.  Notice that the slope of the line is the ratio of the $y$ component to the $x$ component of the vector.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, say we take the point $p=(3,-1)$ and the vector $v=(-3,6)$.  The line through $p$ in the $v$ direction consists of all of the points $p+tv=(3,-1)+(-3t,6t)=(3-3t,6t-1)$ as $t$ ranges over the real numbers.  To determine the corresponding function, just find the slope-intercept form of the equation of the line through the point $(3,-1)$ with slope $\frac{6}{-3}=-2$, namely $y=-2x+5$, so that $f(x)=-2x+5$ is the desired function.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2010-12-05T09:03:51.810" LastActivityDate="2010-12-05T09:03:51.810" />
  <row Id="13080" PostTypeId="1" AcceptedAnswerId="13101" CreationDate="2010-12-05T08:44:20.740" Score="1" ViewCount="89" Body="&lt;p&gt;I have a sequence of  points:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;(x1, y1), (x2, y2), (x3, y3)....&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I would like to know is it exponential sequence and how to get expression of it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/0PHR0.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4430" LastEditorUserId="4430" LastEditorDisplayName="" LastEditDate="2010-12-05T09:26:11.593" LastActivityDate="2010-12-05T12:07:10.690" Title="How to detect exponent" Tags="&lt;sequences-and-series&gt;" AnswerCount="2" CommentCount="7" />
  <row Id="13081" PostTypeId="2" ParentId="13058" CreationDate="2010-12-05T09:06:44.097" Score="2" ViewCount="" Body="&lt;p&gt;Looks true to me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle x(N) = \dfrac{\pi}{N+1}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now $\displaystyle D_n(x) = 1 + 2\sum_{k=1}^{n} \cos(kx)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and $\displaystyle |\sin x - \sin y| \leq |x-y|$ (easily seen using Mean Value Theorem)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle \left|\int_{\pi/(N+1)}^{\pi/N} D_N(t) \ \text{dt}\right| = \left|\int_{\pi/(N+1)}^{\pi/N} 1 + 2\sum_{k=1}^{N} \cos(kx)  \ \text{dt}\right|$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle = \frac{\pi}{N(N+1)} + 2 \sum_{k=1}^{N} \mathcal{O}(\frac{1}{N^2})  = \mathcal{O}(\frac{1}{N})$$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-12-05T09:06:44.097" CommentCount="1" />
  <row Id="13085" PostTypeId="2" ParentId="13068" CreationDate="2010-12-05T09:15:16.733" Score="6" ViewCount="" Body="&lt;p&gt;Recall that&#xA;$$\tan^{-1}x=\sum_{n=0}^\infty(-1)^n\frac{x^{2n+1}}{2n+1}.$$&#xD;&#xA;Use this to find a power series expansion for the integrand,&#xA;and integrate it termwise.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-12-05T09:15:16.733" CommentCount="2" />
  <row Id="13087" PostTypeId="2" ParentId="13080" CreationDate="2010-12-05T09:24:38.310" Score="4" ViewCount="" Body="&lt;p&gt;Set $Y=\ln y$. If the points $(x_k,y_k)$ lie on the curve $y=a e^{bx}$, then the points $(x_k,Y_k)$ lie on the straight line $Y=(\ln a) + bx$.&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2010-12-05T09:24:38.310" CommentCount="2" />
  <row Id="13088" PostTypeId="1" AcceptedAnswerId="17787" CreationDate="2010-12-05T09:36:08.783" Score="3" ViewCount="119" Body="&lt;p&gt;Hi, I need some help with my homework in probability.&#xA;I need to prove that if&#xA;$X(n) =$ the number of different coupons that the collector has in time $n$&#xD;&#xA;then $X(n)$ represents a Markov chain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I proved that &#xA;$$P(X(m+1)=j)= \left\{\begin{array}{ll}&#xA;\frac{X(m)}{n}&amp;amp;\text{if $j=X(m)$}\\&#xA;1-\frac{X(m)}{n}&amp;amp;\text{if $j=X(m)+1$}\\&#xA;0 &amp;amp;\text{otherwise}&#xA;\end{array}\right.$$&#xA;Now I need to show from there that $P(X(m+1=j)|X(m)=Km,\ldots ,X(0)=K0)=P(X(m+1)=j|X(m)=Km)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;thanks for the help.&#xA;benny.&lt;/p&gt;&#xA;" OwnerUserId="3755" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-12-05T21:59:31.023" LastActivityDate="2011-01-16T22:58:10.583" Title="coupon collector and Markov chains" Tags="&lt;homework&gt;&lt;probability-theory&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="13089" PostTypeId="2" ParentId="13003" CreationDate="2010-12-05T09:43:15.270" Score="2" ViewCount="" Body="&lt;p&gt;If a symmetry pairs two vertices, then the angles at those vertices must be congruent. A symmetry that pairs &lt;em&gt;adjacent&lt;/em&gt; vertices of a parallelogram requires that the figure be a rectangle: adjacent angles are supplementary; if they're also congruent, then they must be right angles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Having ruled out rectangles, your proposed symmetry must pair each vertex (a) with its non-adjacent counterpart, or (b) with itself. In (a), the axis would be the perpendicular bisector of the diagonal between the paired vertices; in (b) the axis would contain the vertex. If both (a) and (b) occur, then we have a vertex (from (b)) on the perpendicular bisector of the diagonal joining two non-adjacent others (from (a)); this implies that we have two congruent adjacent edges, which in turn implies that the figure is a rhombus.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To avoid both rectangles and rhombi, we must have that the axis provides only-(a) symmetry, or only-(b) symmetry. An axis of only-(a) symmetry must be the perpendicular bisector of both diagonals, which (as perpendiculars of that axis, and &lt;em&gt;lying in the same plane of that axis&lt;/em&gt;) must therefore lie on the same line: the vertices of the figure are collinear. An axis of only-(b) symmetry contains all the vertices, making them collinear (but in a different arrangement than required by (a)).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you disallow &quot;degenerate&quot; parallelograms with all vertices collinear, then you have your impossibility argument. However, I prefer to recognize degenerate figures as legitimate, and I try to avoid stigmatizing them whenever possible. (After all, they come in handy as intermediate steps in transforming one figure smoothly into another. Besides, it's not like the &quot;Parallelogram Law of Vector Addition&quot; becomes invalid when the vectors involved are linearly dependent.) So, I would go so far as to observe that an axis of only-(a) symmetry implies the existence of a (separate) axis of only-(b) symmetry, and vice versa; that is, we've determined a third class of parallelograms --along with rectangles and rhombi-- that have &lt;em&gt;two&lt;/em&gt; axes of symmetry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Consideration of parallelograms whose vertices coincide is left to the reader.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: An only-(a) axial symmetry can also be realized with an axis perpendicular to the plane of the parallelogram (which need not be degenerate), but you've ruled this out.&lt;/p&gt;&#xA;" OwnerUserId="409" LastEditorUserId="409" LastEditorDisplayName="" LastEditDate="2010-12-05T09:55:37.517" LastActivityDate="2010-12-05T09:55:37.517" />
  <row Id="13090" PostTypeId="1" CreationDate="2010-12-05T09:43:55.817" Score="2" ViewCount="140" Body="&lt;p&gt;I need help this question&lt;/p&gt;&#xA;&#xA;&lt;p&gt;prove that every finite ring has the Invariant Dimension Property(IDP).( Assume $1_R \neq 0_R$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is what I know I should do.  Let $X$ and $Y$ be two sets such that the free module with basis $X$ is isomorphic to the free module with basis $Y$. Next, I have to show that $|X| = |Y|$. This is where I'm stuck.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="4433" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-05T09:48:26.370" LastActivityDate="2010-12-05T10:20:07.750" Title="Invariant dimension property" Tags="&lt;homework&gt;&lt;abstract-algebra&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="13091" PostTypeId="1" CreationDate="2010-12-05T09:47:10.480" Score="4" ViewCount="97" Body="&lt;p&gt;Hello all, this is me again, I have another problem which I haven't been able to solve, the &#xA;legend goes like this:&lt;br&gt;&#xA;Find the equation of the plane that contains the points $P_1(1,0,-1)$, $P_2(2,0,2)$ and forms a 60&lt;sup&gt;0&lt;/sup&gt; ($\theta$) angle with the other plane $2x-2y+z+6=0$ (Two Solutions).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's what I've done:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since we already know two points, P&lt;sub&gt;1&lt;/sub&gt;(1,0,-1) and P&lt;sub&gt;2&lt;/sub&gt;(2,0,2) they both satisfy a planes equation:&lt;br&gt;&#xA;$$Ax+By+Cz+D=0$$&#xD;&#xA;So, we substitute $x,y,z$ respectively for each point, giving use these two equations&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$A(1)+B(0)+C(-1)+D=0$&lt;/li&gt; &#xA;&lt;li&gt;$A(2)+B(0)+C(2)+D=0$&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Simplifying we get:  &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt; $A-C+D=0$&lt;/li&gt; &#xA;&lt;li&gt; $2A+2C+D=0$&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Now, according to the following theorem the angle that is formed by the normals of to planes is the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\cos{\theta} = \pm \frac{AA&amp;#39;+BB&amp;#39;+CC&amp;#39;}{\sqrt{A^2+B^2+C^2} \sqrt{(A&amp;#39;)^2+(B&amp;#39;)^2+(C&amp;#39;)^2}}$$ Where $[A,B,C]$ Are the director numbers of the normal in the first plane and $[A&amp;#39;,B&amp;#39;,C&amp;#39;]$ Are the director numbers of the normal in the second plane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, given those and taking the known angle of 60&lt;sup&gt;o&lt;/sup&gt;, we substitute the director numbers of $2x-2y+z+6-0:[A&amp;#39;=2,B&amp;#39;=-2,C&amp;#39;=1]$ and the angle in the last formula, giving us:&#xA;$$\cos{60^\circ} = \pm \frac{A(2)-B(-2)+C(1)}{\sqrt{A^2+B^2+C^2} \sqrt{(2)^2+(-2)^2+(1)^2}}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Simplifying, we get(positive because $\theta$ is acute):&#xA;$$\frac{1}{2}=\frac{2A-2B+C}{3 \sqrt{A^2+B^2+C^2}}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another simplification finally for:&#xA;$$3(A^2+B^2+C^2)=(4A-4B+2C)^2$$&#xD;&#xA;Now, we have 3 equations:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt; $A-C+D=0$&lt;/li&gt; &#xA;&lt;li&gt; $2A+2C+D=0$&lt;/li&gt;&#xA;&lt;li&gt;$3(A^2+B^2+C^2)=(4A-4B+2C)^2$&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;If we take 1) and 2) and write them as $D(A,C)$ we get:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt; &#xA;&lt;li&gt;$D=C-A$ &#xA;&lt;li&gt;$D=-2A-2C$&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;From which we can obtain A in terms of C&#xA;$$-2A-2C=C-A$$&#xD;&#xA;$$A=-3C$$&#xD;&#xA;Now we substitute back in 1):&#xA;$$(-3C)-C+D=0$$&#xD;&#xA;$$C=\frac{1}{4}D$$&#xD;&#xA;Since I want every director number in terms of D we substitute again but in A since A was in function of C.&#xA;$$A=-3C$$&#xD;&#xA;$$A=-3(\frac{1}{4}D)$$&#xD;&#xA;$$A=\frac{-3}{4}D$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, we have found $A(D)$, $C(D)$ we have yet to find $B(D)$ I'll explain why I think and want everything in terms of D, if we take the general form of a planes equation $Ax+By+Cz+D=0$ and substitute $A$ and $C$ with our known values we get:&#xA;$$(\frac{-3}{4}D)x+By+(\frac{1}{4}D)z+D=0$$&#xD;&#xA;if we divide by $D \neq 0$ and then multiply by 4 to eliminate fractions:&#xA;$$\frac{-3}{4}x+\frac{B}{D}y+\frac{1}{4}z+1=0$$&#xD;&#xA;$$-3x+\frac{4B}{D}y+z+4=0$$&#xD;&#xA;From here it's clear that if we find B in terms of D and B is linear we have the planes equation, so, in 3) $3(A^2+B^2+C^2)=(4A-4B+2C)^2$ we substitute A and C:&#xA;$$3((\frac{-3}{4}D)^2+B^2+(\frac{1}{4}D)^2)=(4(\frac{-3}{4})-4B+2(\frac{1}{4}D))^2$$&#xD;&#xA;$$\vdots$$&#xD;&#xA;$$104B^2+80DB+35D^2=0$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is were I get stuck, I get imaginary parts, since I have to use the quadratic formula, so any help is really appreciated, the problem states that there are two solutions, but this is what I have thinked of if anyone can supply any solution I'll be very gratefull.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regards.. Tristian&lt;/p&gt;&#xA;" OwnerUserId="4312" LastEditorUserId="4312" LastEditorDisplayName="" LastEditDate="2010-12-07T07:34:55.500" LastActivityDate="2010-12-07T08:35:25.010" Title="Analytic Geometry | Two Planes and a Angle | Two Solutions" Tags="&lt;geometry&gt;&lt;homework&gt;&lt;euclidean-geometry&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="13092" PostTypeId="2" ParentId="13090" CreationDate="2010-12-05T09:55:12.417" Score="1" ViewCount="" Body="&lt;p&gt;Just count the number of elements. Since $R$ is finite, you can say exactly how many elements a free modulo of rank $k$ has. For finite $k$, that will do it. In the infinite case, you'll need to do a bit of cardinal work.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-12-05T09:55:12.417" />
  <row Id="13094" PostTypeId="1" AcceptedAnswerId="13095" CreationDate="2010-12-05T09:58:12.893" Score="9" ViewCount="622" Body="&lt;p&gt;There is a formula given in my module :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \displaystyle\sqrt[n]{a^n} = a \text{ if n is odd } $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \displaystyle\sqrt[n]{a^n} = |a| \text{ if n is even } $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't really understand the differences between them,kindly explain with an example.&lt;/p&gt;&#xA;" OwnerUserId="3744" LastEditorUserId="2937" LastEditorDisplayName="" LastEditDate="2011-04-20T22:56:54.420" LastActivityDate="2011-04-20T22:56:54.420" Title="Significance of $\displaystyle\sqrt[n]{a^n} $?!" Tags="&lt;algebra-precalculus&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="13095" PostTypeId="2" ParentId="13094" CreationDate="2010-12-05T10:01:07.087" Score="18" ViewCount="" Body="&lt;p&gt;$\sqrt{(-1)^2} =\sqrt{1} = 1 = |-1|$ because in order to make the square root an unambiguous operation, we agree that &lt;strong&gt;the&lt;/strong&gt; square root of a nonnegative number $x$ is always &lt;strong&gt;the&lt;/strong&gt; (unique) nonnegative number $r$ such that $r^2=x$.  But with cubic roots there is no problem: $\sqrt[3]{-1}=-1$, because &lt;em&gt;every&lt;/em&gt; real number has a &lt;em&gt;unique&lt;/em&gt; cubic root.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same is true with 4th, 6th, 8th, 10th, etc. powers, since $a^n = (-a)^n$, and the 4th, 6th, 8th, 10th, etc roots are defined to be the unique nonnegative real number that &quot;works&quot;, so that they are unambiguous.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is, there are two numbers which when squared will given you the value $2^2$: both $2$ and $-2$. There are two number that when taken to the fourth power will give you $(-6)^4$: both $-6$ and $6$. And so on. Generally, both $a$ and $-a$ will, when raised to an even $n$th power, give the same answer: $a^n = (-a)^n$. And we agree that a square root (fourth root, sixth root, etc) will always be the nonnegative answer, so the $n$th root of $a^n$ will be $|a|$ when $n$ is even. (Don't let the big $-$ in &quot;$-a$&quot; fool you; that does not mean that $-a$ is negative, it just means the additive inverse of whatever $a$ is; if $a$ is positive, then $-a$ is negative, but if $a$ is negative, say $a=-3$, then $-a$ is positive, $-a = -(-3) = 3$. Repeat after me: the proper way to pronounce &quot;-a&quot; is not &quot;&lt;em&gt;negative a&lt;/em&gt;&quot;, the proper pronunciation is &quot;&lt;em&gt;minus a&lt;/em&gt;&quot;). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if $n$ is odd, then every number has a unique $n$th root. In particular, the only number that when cubed gives $2^3$ is $2$; the only number which, when raised to the fifth power, gives $(-6)^5$, is $-6$. There is no longer the problem that both $6$ and $-6$ are possible answers, so we can simply say that the cubic root of $(-2)^3$ is $-2$, the fifth root of $7^5$ is $7$, etc.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-06T00:09:02.257" LastActivityDate="2010-12-06T00:09:02.257" CommentCount="8" />
  <row Id="13096" PostTypeId="2" ParentId="13091" CreationDate="2010-12-05T10:04:02.950" Score="3" ViewCount="" Body="&lt;p&gt;The normal vector to the sought plane is orthogonal to the&#xA;line $P_1 P_2$ and at angle $\pi/3$ to the normal to your given plane.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-12-05T10:04:02.950" CommentCount="6" />
  <row Id="13097" PostTypeId="1" CreationDate="2010-12-05T10:19:38.933" Score="1" ViewCount="98" Body="&lt;p&gt;I have two sets $M$ and $H$. $M$ is an arbitrary string of length $k$ and $H$ is an  string of length $p$. Both are constructed from a charset of length $r$. And $p&amp;lt;k$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hash function $f(m)=h$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand that $r^p$ hashes are possible and that $r^k-r^p$ collisions occur for the complete set $M$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But how can I predict the overall probability for finding any element of $M$ that correctly maps to a given element of $h$?&lt;/p&gt;&#xA;" OwnerUserId="4166" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-05T16:52:04.937" LastActivityDate="2010-12-05T16:53:46.627" Title="Understanding birthday attack probabilities" Tags="&lt;probability&gt;&lt;cryptography&gt;" AnswerCount="1" />
  <row Id="13098" PostTypeId="2" ParentId="13090" CreationDate="2010-12-05T10:20:07.750" Score="1" ViewCount="" Body="&lt;p&gt;As $R$ is finite, it is Artinian, so the modules $R^m$ and $R^n$&#xD;&#xA;have finite length. You can now use the Jordan-Holder theorem&#xA;to deduce that $R^m\cong R^n$ implies $m=n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or you can just count :-)&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-12-05T10:20:07.750" />
  <row Id="13099" PostTypeId="2" ParentId="2611" CreationDate="2010-12-05T10:28:01.937" Score="5" ViewCount="" Body="&lt;p&gt;It should intuitively feel like it's not (theoretically) possible -- the cost to play and the average win are determined by a bookie (or a casino, etc.), whereas the probability of winning is determined by the game itself.  Given two games, one with a probability of winning $p_1$ and the other with a probability of winning $p_2$, the bookie can adjust the returns on bets so as the cost of play and average win remain constant.  For example:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Consider a tossing coin game, the player bets 1 dollar, if the coin is &quot;heads&quot; it will return 2 dollars and if &quot;tales&quot; then there is no return.  So the expected win is +1-1=0.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Now consider a die rolling game, where the player bets 1 dollar, if the die rolls 6, then 6 dollars are returned, otherwise there is no return.  Here the expected win is +5-1-1-1-1-1=0.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In both cases the cost to play is 1 dollar and in both cases the expected win is 0, but the probability of winning is different (1/2 vs. 1/6).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In practice, however, you might be able to infer an approximate probability of winning based on past experiences of the bookie, familiarity with the game being played, etc.&lt;/p&gt;&#xA;" OwnerUserId="139" LastActivityDate="2010-12-05T10:28:01.937" />
  <row Id="13100" PostTypeId="2" ParentId="13062" CreationDate="2010-12-05T11:57:38.973" Score="1" ViewCount="" Body="&lt;p&gt;Well, the standard argument is as follows. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let &#xA;$$&#xD;&#xA;g \mapsto [g] \qquad (g \in G)&#xD;&#xA;$$&#xD;&#xA;be a function which is constant on all right cosets of $H,$ and we require&#xA;$$&#xD;&#xA;[e]=e.&#xD;&#xA;$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is easy to see that&#xA;$$&#xD;&#xA;u [u]^{-1} \in H, \quad [[u]]=[u], \quad [[u]v]=[uv] \qquad (u,v \in G)  \qquad \qquad (*)&#xD;&#xA;$$&#xD;&#xA;Now let&#xA;$$&#xD;&#xA;S = \{ [g] : g \in G\}&#xD;&#xA;$$&#xD;&#xA;and $Y=Y^{-1}$ be a symmetric generating set of $G.$ Then the set&#xA;$$&#xD;&#xA;\{ s y [sy]^{-1} : s \in S, y \in Y\}&#xD;&#xA;$$&#xD;&#xA;is a generating set of $H$ (a &lt;i&gt;finite&lt;/i&gt; one, if both $S$ and $Y$ are finite $\iff$ the index of $H$ in $G$ is finite and $G$ is finitely generated).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For suppose that a product $y_1 \ldots y_r$ is in $H$ where $y_k \in Y$ ($k=1,\ldots,r$). Let, for example's sake, $r=3.$ Then&#xA;$$&#xD;&#xA;y_1 y_2 y_3 = y_1 [y_1]^{-1} \cdot [y_1] y_2 [[y_1] y_2]^{-1} \cdot [[y_1] y_2] y_3&#xD;&#xA;[[[y_1] y_2] y_3]^{-1} \qquad \qquad (**)&#xD;&#xA;$$&#xD;&#xA;where in the right hand side we have a product of elements of $H$ by (*), since&#xA;$$&#xD;&#xA;[[[y_1] y_2] y_3]=[y_1 y_2 y_3]=e;&#xD;&#xA;$$&#xD;&#xA;the same $(*)$ also simplifies the right hand side of $(**)$ as&#xA;$$&#xD;&#xA;y_1 [y_1]^{-1} \cdot [y_1] y_2 [y_1 y_2]^{-1} \cdot [y_1 y_2] y_3 [y_1 y_2 y_3]^{-1}=y_1 y_2 y_3 [y_1 y_2 y_3]^{-1}=y_1 y_2 y_3&#xD;&#xA;$$ &#xA;Now the induction step in general must be easy. &lt;/p&gt;&#xA;" OwnerUserId="1100" LastEditorUserId="1100" LastEditorDisplayName="" LastEditDate="2010-12-06T18:05:35.107" LastActivityDate="2010-12-06T18:05:35.107" CommentCount="1" />
  <row Id="13101" PostTypeId="2" ParentId="13080" CreationDate="2010-12-05T12:07:10.690" Score="4" ViewCount="" Body="&lt;p&gt;I suppose it's time to be putting my dough where my pie hole is. This'll be a tad long, and I hope your collective eyes don't glaze over too much!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The way I'd approach the problem of fitting $y=ab^x$ to a given set of points contaminated with error is to do a &lt;em&gt;weighted&lt;/em&gt; linear regression, similar to Hans's proposal, but with a tiny wrinkle that is required for reasons similar to what I mentioned in this &lt;a href=&quot;http://math.stackexchange.com/questions/3625/3636#3636&quot;&gt;answer&lt;/a&gt; to a related question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Our linearization here takes the form&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ln\;y=\ln\;a+x\ln\;b$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, since the supplied $y_i$ are in fact $\hat{y}_i\pm\sigma_i$ where the $\hat{y}_i$ are the &quot;true values&quot; obscured by the measurement errors $\sigma_i$ , any application of linearization should take into account that the errors as well as the &quot;true values&quot; are being transformed (in this case, by a logarithm).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We now use the fact that if the $y_i$ are transformed by a function $f(y)$, the $\sigma_i$ are transformed according to $f&amp;#39;(y_i)\sigma_i$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From this, the objective function that needs to be minimized for a weighted linear fit is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f(a,b)=\sum_i{y_i^2\;(\ln\;y_i-\ln\;a-x_i\ln\;b)^2}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and the recipe for getting the required $a$ and $b$ now reads as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$m=\sum_i y_i^2$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\bar{x}=\frac{\displaystyle \sum_i y_i^2 x_i}{m}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$t=\sum_i y_i^2 (x_i-\bar{x})^2$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$b=\exp\left(\frac{\displaystyle \sum_i y_i^2\ln\;y_i (x_i-\bar{x})}{t}\right)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$a=\exp\left(\frac{\displaystyle \sum_i y_i^2\ln\;y_i}{m}-b\bar{x}\right)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It turns out that to do a nonlinear fit, we don't even need a provisional value for the linear parameter $a$; we can set things up so that the nonlinear expression we have to minimize only involves $b$. We can then use the $b$ returned by the above formula as a seed, have a secant or Newton-Raphson iteration polish that provisional $b$, and use the polished $b$ to get a proper $a$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Skipping the details (which can be adapted from the derivation I gave in the answer I linked to), the nonlinear equation that has to be solved for $b$ is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left(\sum_i y_i b^{x_i}\right)\left(\sum_i x_i b^{2x_i-1}\right)-\left(\sum_i x_i y_i b^{x_i-1}\right)\left(\sum_i b^{2x_i}\right)=0$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;from which&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$a=\left(\sum_i y_i b^{x_i}\right) / \left(\sum_i b^{2x_i}\right)$$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-12-05T12:07:10.690" CommentCount="4" />
  <row Id="13102" PostTypeId="2" ParentId="13028" CreationDate="2010-12-05T12:26:08.600" Score="4" ViewCount="" Body="&lt;p&gt;The union $T_1\cup T_2$ has no models, and so by the Compactness theorem there is a finite subtheory with no models. This amounts to finite $A_1\subset T_1$ and $A_2\subset T_2$ such that $A_1\cup A_2$ has no models. Any model $M$ of $A_1$ is therefore not a model of $A_2$ and so $M$ is not a model of $T_2$ and hence by your assumption it is a model of $T_1$. So $A_1\vdash T_1$ and similarly $A_2\vdash T_2$, so both theories are finitely axiomatizable. &lt;/p&gt;&#xA;" OwnerUserId="413" LastActivityDate="2010-12-05T12:26:08.600" />
  <row Id="13103" PostTypeId="2" ParentId="13094" CreationDate="2010-12-05T12:27:56.003" Score="-3" ViewCount="" Body="&lt;p&gt;The comments for this answer have more value than the answer I posted. I am removing the contents of my answer but leaving the responses in tact ( that is why I am not deleting this post, if there is an alternative method please let me know). Regarads&lt;/p&gt;&#xA;" OwnerUserId="3936" LastEditorUserId="3936" LastEditorDisplayName="" LastEditDate="2011-04-20T22:25:06.313" LastActivityDate="2011-04-20T22:25:06.313" CommentCount="27" />
  <row Id="13104" PostTypeId="1" AcceptedAnswerId="13105" CreationDate="2010-12-05T13:11:52.473" Score="2" ViewCount="107" Body="&lt;p&gt;&lt;em&gt;&lt;strong&gt;If the $5$-th term of the G.P is its middle term and its value is 2,then find the product of all terms of this GP?&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My approach:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $a_5$ is the middle term, so the GP contains 9 terms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$P_9 = a\cdot ar\cdot ar^2 \cdot ar^3 \cdot 2\cdot ar^5 \cdots ar^8$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which leaves us $P_9 = 2\cdot (ar^4)^8$ ... I am not getting how to proceed further? Is this approach even correct?&lt;/p&gt;&#xA;" OwnerUserId="2109" LastActivityDate="2010-12-05T13:19:34.233" Title="Product of a geometric series from it's middle term" Tags="&lt;algebra-precalculus&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="13105" PostTypeId="2" ParentId="13104" CreationDate="2010-12-05T13:19:34.233" Score="4" ViewCount="" Body="&lt;p&gt;The fourth term is $2/r$ and the sixth term is $2r$.  What happens when you multiply these?  What about the third and seventh terms?&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-12-05T13:19:34.233" CommentCount="2" />
  <row Id="13106" PostTypeId="2" ParentId="13067" CreationDate="2010-12-05T14:07:13.197" Score="3" ViewCount="" Body="&lt;p&gt;The term &quot;special function&quot; has a historical connection with solutions of ordinary differential equations, often second order differential equations that arise from a separation of variables treatment of second order linear partial differential equations with constant coefficients.  This connection is mentioned briefly &lt;a href=&quot;http://en.wikipedia.org/wiki/Special_functions&quot; rel=&quot;nofollow&quot;&gt; in the Wikipedia article&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://www.sagemath.org/doc/reference/sage/functions/special.html&quot; rel=&quot;nofollow&quot;&gt; Sage documentation of special functions&lt;/a&gt;, for example, lists the ODE satisfied by the function together with remarks about boundary conditions.  Stephen Wolfram gave a nice historical review in his talk &lt;a href=&quot;http://www.stephenwolfram.com/publications/recent/specialfunctions/&quot; rel=&quot;nofollow&quot;&gt; The History and Future of Special Functions&lt;/a&gt;, in which he says, &quot;Most continuous special functions are in effect defined implicitly, usually from differential equations.&quot;  Naturally he is not shy about what he perceives as Mathematica's contribution to this history, developing automated searches for accurate numerical methods that proceed directly from the underlying definition by differential equations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;He is not alone in exploring the automatic generation of algorithms for special functions using symbolic computation.  Focusing on special functions that are the solutions of linear ordinary differential equations with polynomial coefficients, &lt;a href=&quot;http://algo.inria.fr/meunier/articles/MeSa03.pdf&quot; rel=&quot;nofollow&quot;&gt; Meunier and Salvy (2003)&lt;/a&gt; describe the design of their website &lt;a href=&quot;http://algo.inria.fr/esf&quot; rel=&quot;nofollow&quot;&gt; The Encyclopedia of Special Functions (ESF)&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The elementary transcendental functions fit easily into this scheme as special special functions, since they solve differential equations too.  For example, the exponential function solves $y&amp;#39; = y$ and sine and cosine, with different boundary conditions, solve the constant coefficient ODE $y&amp;#39;&amp;#39; = -y$.&lt;/p&gt;&#xA;" OwnerUserId="3111" LastActivityDate="2010-12-05T14:07:13.197" CommentCount="1" />
  <row Id="13107" PostTypeId="1" AcceptedAnswerId="13108" CreationDate="2010-12-05T14:09:01.353" Score="3" ViewCount="109" Body="&lt;p&gt;&lt;em&gt;&lt;strong&gt;Prove that if the system of equations $x+2ay + az = 0$,$x+3by+bz = 0$ ,$x+4cy+cz = 0$ has a non-zero solution then a,b,c are in harmonic progression.&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am looking for a suitable approach for this problem.&lt;/p&gt;&#xA;" OwnerUserId="2109" LastActivityDate="2010-12-05T14:54:35.277" Title="Prove in harmonic progression" Tags="&lt;algebra-precalculus&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="13108" PostTypeId="2" ParentId="13107" CreationDate="2010-12-05T14:29:41.690" Score="6" ViewCount="" Body="&lt;p&gt;Write these equations in determinants form and use the following operations:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;$C_{2} \to C_{2} -2C_{3}$&#xD;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$R_{3} \to R_{3} - R_{2}$ and $R_{2} \to R_{2}-R_{1}$&#xD;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;When these two operations are done, then the resulting determinant should be:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left|\begin{array}{lll} 1 &amp;amp; 0 &amp;amp; a \\ 0 &amp;amp; b &amp;amp; b-a \\ 0 &amp;amp; 2c-b &amp;amp; c-b \end{array}\right| = b(c-b) - (b-a)(2c-b) =0 $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which will give $$ \frac{2}{b} = \frac{1}{a} + \frac{1}{c}$$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="9413" LastEditorUserId="9413" LastEditorDisplayName="" LastActivityDate="2010-12-05T14:29:41.690" />
  <row Id="13109" PostTypeId="1" CreationDate="2010-12-05T14:35:14.010" Score="1" ViewCount="195" Body="&lt;p&gt;I have some troubles proving soundness for $\forall$ Introduction. Let $T_1,...,T_n\vdash Q$. $\forall$ Intro gives that $T_1,...,T_n\vdash \forall x Q[x/t]$. Here, t is a constant symbol that does not occur in ${T_1,...,T_n}$, x is a variable which does not occur in Q. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand it intuitively, but have some problems with tarskis' truth definitions. Thank you for your answer.&lt;/p&gt;&#xA;" LastActivityDate="2010-12-09T08:22:30.440" Title="Soundness for $\forall$Introduction" Tags="&lt;logic&gt;" AnswerCount="2" />
  <row Id="13110" PostTypeId="2" ParentId="13107" CreationDate="2010-12-05T14:54:35.277" Score="4" ViewCount="" Body="&lt;p&gt;OK, Chandru1 solved it, so I'm gonna explain the way I would have approached it. (Which is actually the same approach up to ordering my equations differently.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1/ A homogeneous system of equations has a non-zero solution if the determinant is zero. Therefore look at this equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left|\begin{array}{ccc}&#xD;&#xA;1 &amp;amp; 2a &amp;amp; a \\&#xD;&#xA;1 &amp;amp; 3b &amp;amp; b \\&#xD;&#xA;1 &amp;amp; 4c &amp;amp; c &#xD;&#xA;\end{array}\right| = 0$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2/ Since we want to check for a harmonic progression, we need to consider the inverses of $a,b$ and $c$. If one of these coefficients is zero but not the others, there can't be a harmonic progression, but you can also show that in these cases, the system of equations can't have a non-zero solution, so they are ruled out anyway. If two or more of these coefficients are zero, the system is underdetermined and there is an infinity of solutions. So, it seems that there is a small loophole here in the question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, assuming that all coefficients are different from zero, we can rewrite the determinant as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left|\begin{array}{ccc}&#xD;&#xA;1/a &amp;amp; 2 &amp;amp; 1 \\&#xD;&#xA;1/b &amp;amp; 3 &amp;amp; 1 \\&#xD;&#xA;1/c &amp;amp; 4 &amp;amp; 1 &#xD;&#xA;\end{array}\right| = 0$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Subtracting the second row from the first and third row, you get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left|\begin{array}{ccc}&#xD;&#xA;1/a-1/b &amp;amp; -1 &amp;amp; 0 \\&#xD;&#xA;1/b &amp;amp; 3 &amp;amp; 1 \\&#xD;&#xA;1/c-1/b &amp;amp; 1 &amp;amp; 0 &#xD;&#xA;\end{array}\right| = 0$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And this can be developed with respect to the third column to get as an end result:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{1}{a}+\frac{1}{c}=\frac{2}{b}$$&lt;/p&gt;&#xA;" OwnerUserId="3567" LastActivityDate="2010-12-05T14:54:35.277" />
  <row Id="13111" PostTypeId="1" AcceptedAnswerId="13224" CreationDate="2010-12-05T16:21:32.023" Score="4" ViewCount="353" Body="&lt;p&gt;So Ive been trying to prove that there's no simple group of order 300. This is what I did and I was wondering if it was enough. Let |$G$|=2$^2$ . 3 . 5$^2$. Suppose $G$ is simple. Then there would be 6 Sylow 5-groups, one of which will have an index of 6. But then |$G$|=300 does not divide 6! which leads to a contradiction. So the number of Sylow 5-groups is 1 and $\exists$ a proper normal Sylow 5-subgroup in $G$. Hence $G$ is not simple.  &lt;/p&gt;&#xA;" OwnerUserId="4426" LastEditorUserId="8575" LastEditorDisplayName="" LastEditDate="2011-05-19T09:09:18.987" LastActivityDate="2011-05-19T09:09:18.987" Title="No simple group of order 300" Tags="&lt;homework&gt;&lt;group-theory&gt;&lt;simple-groups&gt;" AnswerCount="3" CommentCount="5" FavoriteCount="1" />
  <row Id="13112" PostTypeId="2" ParentId="13097" CreationDate="2010-12-05T16:43:36.843" Score="1" ViewCount="" Body="&lt;p&gt;If the hash function is well constructed, the elements of $M$ will be equally distributed across the hash values.  So each element of $M$ will have $r^{-p}$ chance of mapping to a given element of $H$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this model, I'm not sure how you are counting collisions.  Each element of $H$ will receive $r^{(k-p)}$ of the strings in $M$.  If your count is to go through all the strings in $M$ and count one for each that hashes to an already used value, you are correct.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2010-12-05T16:53:46.627" LastActivityDate="2010-12-05T16:53:46.627" />
  <row Id="13114" PostTypeId="1" CreationDate="2010-12-05T17:16:29.470" Score="1" ViewCount="144" Body="&lt;p&gt;Let $\mathbf G$ be a given $m \times n$ matrix, $\mathbf y$ a given $m \times 1$ column vector and $\mathbf x$ an unknown $n \times 1$ column vector such that $\mathbf x \ge 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) How do you find $\displaystyle \min_{\mathbf x} (\mathbf y - \mathbf G\mathbf x)^T(\mathbf y - \mathbf G\mathbf x)$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Whether there is a solution of that problem similar to the simple pseudo-inverse $x=(G^TG)^{−1}G^Ty$ used for solving the least-squares problem?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) And what if we have $n \times n$ matrix $\mathbf C$ and the constraint $\mathbf C\mathbf x \ge 0$ instead of $\mathbf x \ge 0$? &lt;/p&gt;&#xA;" OwnerUserId="1041" LastEditorUserId="1041" LastEditorDisplayName="" LastEditDate="2010-12-08T10:38:29.843" LastActivityDate="2010-12-08T10:38:29.843" Title="Finding $\min_{\mathbf x} (\mathbf y - \mathbf G\mathbf x)^T(\mathbf y - \mathbf G\mathbf x)$" Tags="&lt;linear-algebra&gt;&lt;optimization&gt;&lt;vector-space&gt;&lt;vector-calculus&gt;" AnswerCount="1" CommentCount="8" FavoriteCount="2" />
  <row Id="13115" PostTypeId="1" CreationDate="2010-12-05T17:17:37.503" Score="0" ViewCount="86" Body="&lt;p&gt;Consider a discrete or a continuous time dynamical system. Show that the complement of a forward invariant set is backward invariant, and vice versa. &lt;/p&gt;&#xA;" LastActivityDate="2010-12-05T17:17:37.503" Title="Complement of forward invariant set" Tags="&lt;dynamical-systems&gt;" CommentCount="1" />
  <row Id="13116" PostTypeId="1" AcceptedAnswerId="13123" CreationDate="2010-12-05T17:18:56.507" Score="4" ViewCount="115" Body="&lt;p&gt;Let $p(n)$ be a polynomial of degree $a$. Start of with plunging in arguments from zero and go up one integer at the time. Go on until you have come at an integer argument $n$ of which $p(n)$'s value is not prime and count the number of distinct primes your polynomial has generated. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Question: what is the maximum number of distinct primes a polynomial of degree $a$ can generate by the process described above? Furthermore, what is the general form of such a polynomial $p(n)$?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This question was inspired by &lt;a href=&quot;http://mathworld.wolfram.com/Prime-GeneratingPolynomial.html&quot; rel=&quot;nofollow&quot;&gt;this article&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Max&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[Please note that your polynomial does not need to generate consecutive primes, only primes at consecutive positive integer arguments.]&lt;/p&gt;&#xA;" OwnerUserId="1055" LastEditorUserId="1055" LastEditorDisplayName="" LastEditDate="2010-12-05T17:48:22.160" LastActivityDate="2010-12-05T18:42:09.023" Title="What is the maximum number of primes generated consecutively generated by a polynomial of degree $a$ ?" Tags="&lt;number-theory&gt;&lt;prime-numbers&gt;&lt;polynomials&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="13118" PostTypeId="2" ParentId="13114" CreationDate="2010-12-05T17:37:16.823" Score="2" ViewCount="" Body="&lt;p&gt;You can start by writing the expression as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;F(x) = (y^T - x^T G^T)(y - Gx) = y^T y - y^T Gx - x^TG^Ty + x^TG^TGx &#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can then differentiate wrt $x$ and set the expression to zero. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;F'(x) = -2G^Ty + 2G^TG x = 0&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you get $x = (G^TG)^{-1}G^T y$.&lt;/p&gt;&#xA;" OwnerUserId="2641" LastActivityDate="2010-12-05T17:37:16.823" CommentCount="2" />
  <row Id="13119" PostTypeId="1" CreationDate="2010-12-05T17:39:47.893" Score="1" ViewCount="152" Body="&lt;p&gt;Is there a way to verify whether a non-empty intersection exists between two simple polygons (not necessarily convex) using the Chazelle's simplicity test ?&lt;/p&gt;&#xA;" OwnerUserId="4444" LastActivityDate="2011-01-04T19:44:35.003" Title="Using Chazelle's simplicity test to verify simple polygons intersection" Tags="&lt;computational-geometry&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="13120" PostTypeId="1" CreationDate="2010-12-05T17:40:28.237" Score="1" ViewCount="128" Body="&lt;p&gt;Louis de Branges  had a paper on his &lt;a href=&quot;http://www.math.purdue.edu/~branges/site//Papers&quot; rel=&quot;nofollow&quot;&gt;homepage&lt;/a&gt; claiming a solution for the &lt;a href=&quot;http://en.wikipedia.org/wiki/Invariant_subspace_problem&quot; rel=&quot;nofollow&quot;&gt;Invariant Subspace Problem&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I don't see that paper anymore, though he still has a &quot;proof&quot; of  Riemann Hypothesis on his website.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anyone know whether he actually admitted his proof of the Invariant Subspace Problem was wrong? Or someone pointed that out?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;His proposed proof of the RH has been on his website for many years now, and yet no one has pointed out an error.&lt;/p&gt;&#xA;" OwnerUserId="3249" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-05T22:11:44.603" LastActivityDate="2010-12-06T08:12:08.410" Title="Invariant Subspace Problem" Tags="&lt;soft-question&gt;&lt;reference-request&gt;&lt;functional-analysis&gt;&lt;open-problem&gt;" AnswerCount="0" CommentCount="3" FavoriteCount="1" />
  <row Id="13121" PostTypeId="2" ParentId="13116" CreationDate="2010-12-05T17:50:02.830" Score="2" ViewCount="" Body="&lt;p&gt;Here is result by Rabinowitsch for quadratic polynomials.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$n^2+n+A$ is prime for $n=0,1,2,...,A-2$ if and only if $d=1-4A$ is squarefree and the class number of $\mathbb{Q}[\sqrt{d}]$ is $1$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;See this article for details.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://matwbn.icm.edu.pl/ksiazki/aa/aa89/aa8911.pdf&quot; rel=&quot;nofollow&quot;&gt;http://matwbn.icm.edu.pl/ksiazki/aa/aa89/aa8911.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also here is a list of imaginary quadratic fields with class number $1$&#xD;&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/List_of_number_fields_with_class_number_one#Imaginary_quadratic_fields&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/List_of_number_fields_with_class_number_one#Imaginary_quadratic_fields&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are many other articles about prime generating (quadratic) polynomials that you can google.&lt;/p&gt;&#xA;" OwnerUserId="3431" LastActivityDate="2010-12-05T17:50:02.830" CommentCount="3" />
  <row Id="13122" PostTypeId="1" CreationDate="2010-12-05T17:57:55.257" Score="1" ViewCount="188" Body="&lt;p&gt;For any positive integer $N$, the binomial$(N!,p)$ distribution has the following property: for any $1 \leq n \leq N$,&#xA;there exist i.i.d. random variables $X_1,\ldots,X_n$ such that $X_1 + \cdots + X_n \sim {\rm binomial}(N!,p)$ (specifically, we take $X_1,\ldots,X_n$ to be i.i.d. binomial$(N!/n,p)$ rv's). It may be interesting to consider the following question: given $N \geq 3$, arbitrary but fixed, is there a continuous bounded distribution $\mu = \mu_N$ having the same property? (I stress: continuous and bounded.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Well, it turns out this is a really trivial problem, but worth remembering...; see my answer below.&lt;/p&gt;&#xA;" OwnerUserId="2810" LastEditorUserId="2810" LastEditorDisplayName="" LastEditDate="2010-12-06T15:33:14.530" LastActivityDate="2010-12-06T15:33:54.187" Title="A continuous analogue of the binomial distribution" Tags="&lt;probability&gt;" AnswerCount="2" CommentCount="5" />
  <row Id="13123" PostTypeId="2" ParentId="13116" CreationDate="2010-12-05T18:00:02.710" Score="6" ViewCount="" Body="&lt;p&gt;The Green-Tao Theorem states that there are arbitrarily long arithmetic progressions of primes; that is, sequences of primes of the form&#xA;$$ b , b+a, b+2a, b+3a,... ,b+na $$&#xD;&#xA;Since such a progression will be the first $n$ values of the polynomial $ax+b$, this implies that even for degree 1, there is no upper bound to how many primes in a row a polynomial can generate.&lt;/p&gt;&#xA;" OwnerUserId="1062" LastActivityDate="2010-12-05T18:00:02.710" CommentCount="6" />
  <row Id="13124" PostTypeId="1" AcceptedAnswerId="13130" CreationDate="2010-12-05T18:19:28.223" Score="4" ViewCount="89" Body="&lt;p&gt;I have heard that given two sheaves $A$ and $B$ on a variety, one can identify elements of $Ext^d(A,B)$ with complexes of sheaves&#xA;$$0\to B \to C_1 \to \cdots \to C_d \to A \to 0.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My questions are, &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;How do I see that this is true?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If I have obtained an element of $Ext^n$ by some other method, can I explicitly construct the $C_j$ sheaves and the differentials?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am sure this is well-known, so I'm marking it also as &quot;reference-request&quot;.&lt;/p&gt;&#xA;" OwnerUserId="3008" LastActivityDate="2011-03-07T02:44:08.523" Title="Ext &amp; Complexes" Tags="&lt;reference-request&gt;&lt;homological-algebra&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="13125" PostTypeId="1" AcceptedAnswerId="13133" CreationDate="2010-12-05T18:31:22.570" Score="7" ViewCount="201" Body="&lt;p&gt;On $\mathbb{R}^n$ and $p\ge 1$ the $p$-norm is defined as $$\|x\|_p=\left ( \sum _{j=1} ^n |x_j| ^p \right ) ^{1/p}$$&#xD;&#xA;and there is the $\infty$-norm which is $\|x\|_\infty=\max _j |x_j|$. It's called the $\infty$ norm because it is the limit of $\|\cdot\|_p$ for $p\to \infty$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we can use the definition above for $p&amp;lt;1$ as well and define a $p$-&quot;norm&quot; for these $p$. The triangle inequality is not satisfied, but I will use the term &quot;norm&quot; nonetheless. For $p\to 0$ the limit of $\|x\|_p$ is obviously $\infty$ if there are at least two nonzero entries in $x$, but if we use the following modified definition&#xA;$$\|x\|_p=\left ( \frac{1}{n} \sum _{j=1} ^n |x_j| ^p \right ) ^{1/p}$$&#xD;&#xA;then this should have a limit for $p\to 0$, which should be called 0-norm. What is this limit?&lt;/p&gt;&#xA;" OwnerUserId="1609" LastActivityDate="2010-12-05T19:05:01.900" Title="What is the 0-norm?" Tags="&lt;real-analysis&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="13127" PostTypeId="2" ParentId="13116" CreationDate="2010-12-05T18:42:09.023" Score="1" ViewCount="" Body="&lt;p&gt;Here is a related fact which might also be of interest.  There exists a polynomial in 26 variables with the property that, if you plug in integers for all 26 variables, and the output is a positive number (it will always be an integer), then the output is prime.  The polynomial can be found here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Formula_for_primes&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Formula_for_primes&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore, every prime occurs as an output of this polynomial.  This is a very indirect way to generate primes, because there is no easy way to know if a given input will give a positive output.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is apparently another polynomial in 10 variables that does this, but Wikipedia doesn't explicitly write it.&lt;/p&gt;&#xA;" OwnerUserId="1062" LastActivityDate="2010-12-05T18:42:09.023" />
  <row Id="13128" PostTypeId="1" CreationDate="2010-12-05T18:45:15.730" Score="0" ViewCount="56" Body="&lt;p&gt;Okay, I am in no way a mathematics student but I happened to be writing a program on a map with coordinates of locations displayed on the map. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My problem is that the points could be scattered on the map with an indefinite boundary and I need to find a boundary (rectangle) around these points and thus I need to get a formula to form an algorithm so that the program will find the optimum boundary itself. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope this is understandable and thanks in anticipation of any response.&lt;/p&gt;&#xA;" LastActivityDate="2010-12-05T23:45:08.293" Title="optimum rectangle around a group of coordinates (longitude and latitude)" Tags="&lt;polar-coordinates&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="13129" PostTypeId="1" AcceptedAnswerId="13167" CreationDate="2010-12-05T18:47:36.463" Score="3" ViewCount="222" Body="&lt;p&gt;Let $K$ be a field and let $K(x)$ be the field of rational functions in $x$ whose coefficients are in $K$. Let $\theta(x)$ $\in$ Aut($K(x))$ such that $\theta|_K = id_K$.  Show that $\theta(x)$ =$\frac{ax+b}{cx+d}$, with $ad$ $\neq$ $bc$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is my attempt.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $\theta(x)$ = $\frac{f}{g}$, $f$, $g$ $\in $K$[$x$] $, with gcd($f, $g)=1. let $h$ $\in$ $K(x)$. Then  $h(\frac{f}{g})$ = $x$ and $\frac{f(x)}{g(x)}$ $\neq$ $\frac{f(y)}{g(y)}$ if $x\neq y$ . Suppose deg $f$ &gt; 1 or deg $g$ &gt; 1, then the equation $g(x)b=f(x)$, $b\in K$ ,  will have more than one solution for $x$. Hence $f$ and $g$ have degrees at most 1 and thus $\theta(x)$ =$\frac{ax+b}{cx+d}$ with inverse function $\theta(x)$ =$\frac{dx-b}{a-cx}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I'd like to know is if Ive approached the problem in the right manner. Please any input will be very much appreciated. Thanks.&lt;/p&gt;&#xA;" OwnerUserId="4426" LastActivityDate="2010-12-06T16:32:44.347" Title="Automorphism of the Field of rational functions." Tags="&lt;homework&gt;" AnswerCount="1" FavoriteCount="2" />
  <row Id="13130" PostTypeId="2" ParentId="13124" CreationDate="2010-12-05T18:48:54.713" Score="4" ViewCount="" Body="&lt;p&gt;For modules, Weibel discusses this in &quot;Introduction to homological algebra.&quot; Section 3.4 deals with d=1 case and in Vista 3.4.6 is about the general case. He gives no proof for d&gt;1 and refers to Bourbaki &quot;Algebre homologique&quot; 7.5 and Maclane &quot;Homology&quot; pp82-87.&lt;/p&gt;&#xA;" OwnerUserId="2633" LastEditorUserId="2633" LastEditorDisplayName="" LastEditDate="2010-12-05T18:55:40.853" LastActivityDate="2010-12-05T18:55:40.853" CommentCount="2" />
  <row Id="13131" PostTypeId="1" CreationDate="2010-12-05T18:58:45.990" Score="5" ViewCount="119" Body="&lt;p&gt;Prove that for any finite sequence of decimal digits, there exists an $n$ such that the decimal expansion of $2^n$ begins with these digits.&lt;/p&gt;&#xA;" LastActivityDate="2010-12-05T19:06:57.797" Title="Starting digits of 2^n" Tags="&lt;homework&gt;&lt;dynamical-systems&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="13132" PostTypeId="2" ParentId="13122" CreationDate="2010-12-05T19:04:21.770" Score="0" ViewCount="" Body="&lt;p&gt;OK, I think I've got one for you, but I have to admit it's a bit of a cheat. If you replace independence in your requirements by &lt;a href=&quot;http://en.wikipedia.org/wiki/Free_probability&quot; rel=&quot;nofollow&quot;&gt;free independence&lt;/a&gt; then there exists a continuous bounded and infinitely divisible distribution, namely the &lt;a href=&quot;http://en.wikipedia.org/wiki/Wigner_semicircle_distribution&quot; rel=&quot;nofollow&quot;&gt;Wigner semi-circle distribution&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's the only example I can think of for now.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also give this &lt;a href=&quot;http://books.google.be/books?id=NUyG98QZEl8C&amp;amp;printsec=frontcover&amp;amp;dq=free+probability+denes+petz&amp;amp;source=bl&amp;amp;ots=YAwO-jcpk7&amp;amp;sig=KRweScXQiDuUJCAhcFNxuxkFJSg&amp;amp;hl=nl&amp;amp;ei=7eL7TM2sJcqZ8QPX7bWHDA&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=4&amp;amp;ved=0CDMQ6AEwAw#v=onepage&amp;amp;q=free%20probability%20denes%20petz&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;reference&lt;/a&gt; for details.&lt;/p&gt;&#xA;" OwnerUserId="3567" LastActivityDate="2010-12-05T19:04:21.770" CommentCount="2" />
  <row Id="13133" PostTypeId="2" ParentId="13125" CreationDate="2010-12-05T19:05:01.900" Score="10" ViewCount="" Body="&lt;p&gt;When $p$ is small, $$x^p = \exp(p \log x) \approx 1 + p\log x.$$ Therefore $$\frac{1}{n} \sum_{j=1}^n x_j^p \approx 1 + p\frac{1}{n} \sum_{j=1}^n \log x_j = 1 + p \log \sqrt[n]{\prod_{i=1}^n x_j}.$$ On the other hand, we have $$(1+py)^{1/p} \longrightarrow \exp(y),$$ and so we easily get that the norm approaches the geometric mean, as Raskolnikov commented.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-12-05T19:05:01.900" />
  <row Id="13134" PostTypeId="2" ParentId="13131" CreationDate="2010-12-05T19:06:57.797" Score="8" ViewCount="" Body="&lt;p&gt;Take $\log_{10} (2^n) = n \log_{10} 2$, note that $\log_{10} 2$ is irrational, and use the equidistribution theorem (or prove what you want directly using the pigeonhole principle).&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-12-05T19:06:57.797" CommentCount="1" />
  <row Id="13135" PostTypeId="1" AcceptedAnswerId="13136" CreationDate="2010-12-05T19:07:54.323" Score="1" ViewCount="120" Body="&lt;p&gt;Let $f\colon A\to A$; prove that if $f\circ f = \mathrm{id}_A$ then $f$ is a bijection.&lt;/p&gt;&#xA;" OwnerUserId="2587" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-05T22:02:04.847" LastActivityDate="2010-12-06T00:00:22.393" Title="Let $f\colon A\to A$; prove that if $f\circ f = \mathrm{id}_A$ then $f$ is a bijection" Tags="&lt;elementary-set-theory&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="13136" PostTypeId="2" ParentId="13135" CreationDate="2010-12-05T19:13:45.250" Score="3" ViewCount="" Body="&lt;p&gt;Hints: Injectivity: Suppose $f(x)=f(y)$, what can you say about $f(f(x))$ and $f(f(y))$.&#xA;Surjectivity: Given $a\in A$, there exists $b\in A$ such that $f(b)=a$ since, $f(f(a))=id(a)=a$ so b=?&lt;/p&gt;&#xA;" OwnerUserId="3431" LastActivityDate="2010-12-05T19:13:45.250" CommentCount="7" />
  <row Id="13137" PostTypeId="1" AcceptedAnswerId="14611" CreationDate="2010-12-05T19:24:07.973" Score="5" ViewCount="162" Body="&lt;p&gt;If $A$ is a subset of $\mathbb{R}$ with Lebesgue measure strictly greater than $0$, does it follow then that are there $a$ and $b$ such that the measure of $[a,b]\cap A$ is $b-a$? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you.&lt;/p&gt;&#xA;" OwnerUserId="2934" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-05T22:03:41.930" LastActivityDate="2010-12-17T02:27:25.063" Title="If A is a subset of R with Lebesgue measure &gt; 0 then are there a,b such that the measure of $[a,b]\cap A$ is b-a?" Tags="&lt;measure-theory&gt;" AnswerCount="2" CommentCount="6" />
  <row Id="13138" PostTypeId="2" ParentId="13128" CreationDate="2010-12-05T19:26:54.550" Score="2" ViewCount="" Body="&lt;p&gt;If the sides of the rectangle must be aligned with the edges of the map, you can just take the maximum and minimum coordinates in each direction.  If you allow arbitrary orientation of the rectangle you have a one dimensional minimization problem.  You can compute the coordinates of each point as a function of rotation angle, again take the maximum and minimum in each dimension, calculate the perimeter, and minimize over the rotation angle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are not devoted to a rectangle, maybe you would prefer the &lt;a href=&quot;http://en.wikipedia.org/wiki/Convex_hull&quot; rel=&quot;nofollow&quot;&gt;convex hull&lt;/a&gt; of your points.  There are efficient algorithms to find it and it doesn't need the minimization.  It also has a smaller perimeter yet.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2010-12-05T23:45:08.293" LastActivityDate="2010-12-05T23:45:08.293" />
  <row Id="13139" PostTypeId="1" CreationDate="2010-12-05T19:27:51.217" Score="1" ViewCount="109" Body="&lt;p&gt;Hi, I have a school assignment to do, but I have no idea, where to start. I hope you can help. Here it is: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have a finite sequence $A = (a_1, a_2,\ldots, a_n)$, length of $A$ is $n$, elements of $A$ are natural numbers. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $S(i,j)$ be partial sum of this sequence from $a_i$ to $a_j$, where $1\leq i\leq j\leq n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Formally written as $S(i,j) = a_i+a_{i+1}+\cdots+a_j$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can call sequence $A$ &quot;nice&quot;, when all partial sums $S(i,j)$ are different from each other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Formally written - For every $1\leq i\leq j\leq n$ and $1\leq k\leq l\leq n$ it holds that &quot;$S(i,j)=S(k,l)$ implies $i=k$ and $j=l$&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;E.g. $(1,3,2)$ is &quot;nice&quot; and $(1,4,2,3)$ is not &quot;nice&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have following tasks:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Prove that for every natural $n$ a &quot;nice&quot; sequence exists.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The main question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the smallest possible value of the greatest element of a &quot;nice&quot; sequence $A$ of length $n$? Or in other words - for every $n$ find the smallest natural number $p$ for which a &quot;nice&quot; sequence $A$ of length $n$, in which no element is larger than $p$, exists.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint: search for upper and lower bounds on $p$ in the form of appropriate functions of $n$.&lt;/p&gt;&#xA;" OwnerUserId="4447" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2010-12-05T20:30:21.133" LastActivityDate="2010-12-05T20:30:21.133" Title="Series partial sums" Tags="&lt;sequences-and-series&gt;" CommentCount="2" />
  <row Id="13140" PostTypeId="2" ParentId="13119" CreationDate="2010-12-05T19:31:01.860" Score="2" ViewCount="" Body="&lt;p&gt;Independent of Chazelle, a discussion of the case of intersecting convex polygons can be found at:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.iro.umontreal.ca/~plante/compGeom/algorithm.html&quot; rel=&quot;nofollow&quot;&gt;http://www.iro.umontreal.ca/~plante/compGeom/algorithm.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.springerlink.com/content/q744325k7350m237/&quot; rel=&quot;nofollow&quot;&gt;http://www.springerlink.com/content/q744325k7350m237/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1369" LastActivityDate="2010-12-05T19:31:01.860" CommentCount="1" />
  <row Id="13141" PostTypeId="1" AcceptedAnswerId="13144" CreationDate="2010-12-05T20:08:56.147" Score="1" ViewCount="41" Body="&lt;p&gt;&lt;strong&gt;Problem:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;If the sum of the first $p$ terms of an arithmetic progression is $q$ and the sum of the first $q$ terms is $p$, then find the sum of $p+q$ terms.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the problem we can write (considering $a$ is the first term and $d$ is the common difference):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{p}{2}\cdot \biggl[2a + (p-1)d \biggr] = q \qquad \cdots (1)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{q}{2}\cdot \biggl[2a + (q-1)d \biggr] = p \qquad \cdots (2)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now in my module it is given that from these we can write $\displaystyle d= \frac{-2(p+q)}{pq}$; I am not getting how we can get that value of $d$ ?!&lt;/p&gt;&#xA;" OwnerUserId="2109" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2010-12-05T20:25:11.777" LastActivityDate="2010-12-05T20:25:11.777" Title="A small doubt on calculation" Tags="&lt;algebra-precalculus&gt;" AnswerCount="2" />
  <row Id="13142" PostTypeId="1" AcceptedAnswerId="13143" CreationDate="2010-12-05T20:15:38.940" Score="0" ViewCount="62" Body="&lt;p&gt;&lt;strong&gt;Problem:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $a_1,a_2,a_3 \cdots a_n$ are in HP then find the value of $ a_1 \cdot a_2 +  a_2 \cdot a_3 +   a_3 \cdot a_4 + \cdots +  a_{n-1} \cdot a_n$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;My initial approach,using the property of HP that $ \frac{1}{a_1},  \frac{1}{a_2}, \frac{1}{a_3},\cdots ,\frac{1}{a_n}$ is in AP, I am getting this form:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{a_1-a_2}{a_1 \cdot a_2} = \frac{a_2-a_3}{a_2 \cdot a_3} = \cdots = \frac{a_{n-1}-a_n}{a_{n-1} \cdot a_n}= d$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How to proceed next?&lt;/p&gt;&#xA;" OwnerUserId="2109" LastEditorUserId="2109" LastEditorDisplayName="" LastEditDate="2010-12-05T20:34:29.087" LastActivityDate="2010-12-05T20:34:29.087" Title="Suggest a tricky method" Tags="&lt;algebra-precalculus&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="13143" PostTypeId="2" ParentId="13142" CreationDate="2010-12-05T20:20:57.893" Score="1" ViewCount="" Body="&lt;p&gt;Write $a_i a_{i+1}$ as a difference of two fractions to get a telescopic sum.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-12-05T20:20:57.893" />
  <row Id="13144" PostTypeId="2" ParentId="13141" CreationDate="2010-12-05T20:23:32.937" Score="3" ViewCount="" Body="&lt;p&gt;$q \times (1) - p \times (2)$ will give you $d$ and then you can work out $a.$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="1301" LastActivityDate="2010-12-05T20:23:32.937" />
  <row Id="13145" PostTypeId="2" ParentId="13141" CreationDate="2010-12-05T20:24:04.617" Score="0" ViewCount="" Body="&lt;p&gt;You get two equations in two unknowns $a,d$; solve them to find $d$. Since we only want to find $d$, we can equate the coefficient of $a$ by multiplying the first equation by $q$, the second by $p$, and subtracting to get a linear equation for $d$.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-12-05T20:24:04.617" />
  <row Id="13146" PostTypeId="1" CreationDate="2010-12-05T20:47:09.183" Score="1" ViewCount="125" Body="&lt;p&gt;Define $h\colon \mathbb{Z}/\sim ~16 \to \mathbb{Z}/\sim 24$ by $h([a]16) = [3a]24$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a. Prove $h$ is well defined.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;b. Compute $h(a)$ where $a = \{[0]16, [3]16, [6]16\}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;c. Compute $h^{-1}([10]24)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is the following correct:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a. $h$ is well defined since each for all $a$ $|h([a]16)|\equiv 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;b. $\{[0]24, [9]24, [18]24\}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;c. $\emptyset$&lt;/p&gt;&#xA;" OwnerUserId="2587" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-05T22:04:51.303" LastActivityDate="2011-03-05T23:39:34.143" Title="Define $h\colon \mathbb{ Z}/\sim 16 \to \mathbb{Z}/\sim 24$ by $h([a]16) = [3a]24$" Tags="&lt;elementary-set-theory&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="13147" PostTypeId="2" ParentId="13146" CreationDate="2010-12-05T21:04:42.233" Score="1" ViewCount="" Body="&lt;p&gt;Your answers to (b) and (c) are correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For (a), you need to prove that if a=b mod 16, i.e. $a-b=16n $ for some integer $n$, then $3a=3b$ mod 24, i.e. $3a-3b=24m$ for some integer $m$. Can you do that?&lt;/p&gt;&#xA;" OwnerUserId="3249" LastActivityDate="2010-12-05T21:04:42.233" />
  <row Id="13148" PostTypeId="1" CreationDate="2010-12-05T21:07:44.233" Score="0" ViewCount="164" Body="&lt;p&gt;How do you show that the convex hull of a given set of points S, always has the minimum perimeter ? By perimeter i mean the length of the boundary of the hull&lt;/p&gt;&#xA;" OwnerUserId="4444" LastActivityDate="2010-12-06T00:01:07.710" Title="Convex hull has the smallest perimeter" Tags="&lt;algorithms&gt;&lt;convex-analysis&gt;&lt;computational-geometry&gt;&lt;convex-sets&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="13149" PostTypeId="1" CreationDate="2010-12-05T21:21:43.067" Score="2" ViewCount="52" Body="&lt;p&gt;The rough question is: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;How do the finite subgroups of SL(2,R/I) vary as I varies, where R is a nice local ring?&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;SL(2,A) = { [a,b;c,d] : ad-bc=1, a,b,c,d in A } is the group of a 2&amp;times;2 matrices with entries from A and determinant 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Two cases seem basic and approachable but already very interesting:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;R the 2-adic integers, and R/I = &lt;strong&gt;Z&lt;/strong&gt;/2&lt;sup&gt;n&lt;/sup&gt;&lt;strong&gt;Z&lt;/strong&gt; with n varying, and&lt;/li&gt;&#xA;&lt;li&gt;R the power series over the field with 2 elements, and R/I = &lt;strong&gt;Z&lt;/strong&gt;[x]/(2,x&lt;sup&gt;n&lt;/sup&gt;) with n varying.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In each case G = SL(2,R/I) is finite and solvable, with G/O&lt;sub&gt;2&lt;/sub&gt;(G) ≅ SL(2,2) ≅ S&lt;sub&gt;3&lt;/sub&gt; being a nice familiar group.  In other words, nothing radically strange can happen in any one particular group.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I am having quite a bit of trouble understanding how the subgroups vary with n.  Call the first sequence of groups G(2,n) and the second sequence of groups G(x,n).&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Which groups G(2,n) and G(x,n) contain Q&lt;sub&gt;8&lt;/sub&gt;, a quaternion group of order 8?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Q8 seems to be contained in G(x,2n+1) and not contained in G(x,2n) or G(x,1) or any G(2,n).&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Which dihedral groups are contained in each G(2,n) and G(x,n)?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The G(2,n+1) seem to be bereft of dihedrals, but G(2,1) has one, and G(x,n+1) seem to have the full crew.  Why does the &quot;2&quot; destroy dihedrals while the &quot;x&quot; invites more? On the abelian side:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What is the 2-rank (maximum n such that G contains the direct product of n cyclic groups of order 2) of each G(2,n) and G(x,n)?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Again the pattern seems a bit wobbly, especially if you count how many subgroups of each rank there are (these do not form monotonic sequences).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any answer to the bold question should be able to address the strange (conjectured) answers to the smaller questions, but I'm also just fine with some ad hoc answers to the small ones as these groups are currently a complete mystery.  I think larger primes (or residue fields) would help with the very restricted nature of the subgroups available for testing, but at the beginning p=2 seems to help.  More independent nilpotents has not struck me as counter-intuitive yet.  Also in my limited experiments, &lt;strong&gt;GL&lt;/strong&gt; seems to be a lot less strange, but feel free to share observations about it as well.&lt;/p&gt;&#xA;" OwnerUserId="583" LastEditorUserId="583" LastEditorDisplayName="" LastEditDate="2010-12-05T21:29:04.137" LastActivityDate="2010-12-05T21:29:04.137" Title="Subgroups of SL(2,R/I) for R local as I varies" Tags="&lt;group-theory&gt;" CommentCount="1" />
  <row Id="13150" PostTypeId="1" AcceptedAnswerId="13165" CreationDate="2010-12-05T21:24:32.117" Score="1" ViewCount="434" Body="&lt;p&gt;Hi, how can i extract rotation, scale values from 2d transformation matrix? i mean a have a 2d transformation&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;matrix = [1, 0, 0, 1, 0, 0]&#xA;&#xA;matrix.rotate(45 / 180 * PI)&#xA;matrix.scale(3, 4)&#xA;matrix.translate(50, 100)&#xA;matrix.rotate(30 / 180 * PI)&#xA;matrix.scale(-2, 4)&#xA;now my matrix have values [a, b, c, d, tx, ty]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;lets forget about the processes above and imagine that we have only the values a, b, c, d, tx, ty&lt;/p&gt;&#xA;&#xA;&lt;p&gt;how can i find total rotation and scale values via a, b, c, d, tx, ty&lt;/p&gt;&#xA;&#xA;&lt;p&gt;sorry for my english&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks your advance&lt;/p&gt;&#xA;" OwnerUserId="4452" LastActivityDate="2011-06-04T11:00:04.003" Title="extracting rotation, scale values from 2d transformation matrix" Tags="&lt;matrices&gt;&lt;transformation&gt;" AnswerCount="3" />
  <row Id="13151" PostTypeId="1" AcceptedAnswerId="13173" CreationDate="2010-12-05T21:29:10.530" Score="2" ViewCount="101" Body="&lt;p&gt;Question: Calculate the rank of the following matrices:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$A = \left( \begin{array}{cc} 1 &amp;amp; n \\ n &amp;amp; 1 \end{array} \right), n \in \mathbb{Z}$ and $B = \left( \begin{array}{ccc} 1 &amp;amp; x &amp;amp; x^{2} \\ 1 &amp;amp; y &amp;amp; y^{2} \\ 1 &amp;amp; z &amp;amp; z^{2} \end{array} \right)$,  $x,y,z \in \mathbb{R}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the way I understand rank($A$), is the number of pivots in an echelon form of $A$. To put $A$ into echelon form I would subtract $n$ times the first row from the second row: $A \sim \left( \begin{array}{cc} 1 &amp;amp; n \\ n &amp;amp; 1 \end{array} \right) \sim \left( \begin{array}{cc} 1 &amp;amp; n \\ 0 &amp;amp; 1 - n^{2} \end{array} \right) \Rightarrow $rank$(A) = 2$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;With $B$ I would have done pretty much the same thing, subtracting row 1 from both row 2 and row 3: $B \sim \left( \begin{array}{ccc} 1 &amp;amp; x &amp;amp; x^{2} \\ 1 &amp;amp; y &amp;amp; y^{2} \\ 1 &amp;amp; z &amp;amp; z^{2} \end{array} \right) \sim \left( \begin{array}{ccc} 1 &amp;amp; x &amp;amp; x^{2} \\ 0 &amp;amp; y - x &amp;amp; y^{2} - x^{2} \\ 0 &amp;amp; z - x &amp;amp; z^{2} - x^{2} \end{array} \right)$ (at this point I could multiply row 2 by $-(\frac{z-x}{y-x})$ and add it to row 3 which ends up being a long polynomial....) However, with both parts, I am pretty confident that it is not so simple and that I am missing the point of this exercise. Could somebody please help point me in the right direction?&lt;/p&gt;&#xA;" OwnerUserId="3711" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-05T22:54:09.620" LastActivityDate="2010-12-05T23:36:11.253" Title="Calculate the rank of the following matrices" Tags="&lt;linear-algebra&gt;&lt;matrices&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="13152" PostTypeId="1" AcceptedAnswerId="13158" CreationDate="2010-12-05T21:33:05.617" Score="2" ViewCount="120" Body="&lt;p&gt;Prove that a vector perpendicular to the line $Ax+By+C=0$ in the $xy$-plane is $&amp;lt;A, B&amp;gt;$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So intuitively this is clear, since we can always find a vector normal to a line by looking at the coefficients of $x,y,z$ So in this case $&amp;lt;A, B, 0&amp;gt;$ or $&amp;lt;A, B&amp;gt;$ in 2d space is clearly a normal vector of the line.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried using this approach but got stuck :  take two points on the line, find a vector parallel to the line, take the dot product with $&amp;lt;A, B&amp;gt;$ and show this is $= 0$&lt;br&gt;&#xA;So&lt;br&gt;&#xA;Let $P_1 = Ax_1 + By_1 = -C$    be a point on the line and&lt;br&gt;&#xA;Let $P_2 = Ax_2 + By_2 = -C$    be another point on the line.&lt;br&gt;&#xA;(let this denote the vector) Then $P_1P_2 = &amp;lt;Ax_2 - Ax_1, By_2 - By_1&amp;gt;$   is parallel to the line.&lt;br&gt;&#xA;So  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$&amp;lt;A, B&amp;gt; \cdot P_1P_2 = A^2x_2-A^2x_1+B^2y_n-B^2y_1$&lt;br&gt;&#xA;                     $= A^2x_2 + B^2y_n - (A^2x_1+B^2y_1)$&lt;br&gt;&#xA;But I wasn't able to get this to equal 0 using the fact that $Ax + By = -C$&lt;br&gt;&#xA;Any help is appreciated!&lt;br&gt;&#xA;Thanks.  &lt;/p&gt;&#xA;" OwnerUserId="2533" LastActivityDate="2010-12-05T22:44:05.747" Title="Prove that a vector perpendicular to the line $Ax+By+C=0$ in the $xy$-plane is $&lt;A, B&gt;$" Tags="&lt;calculus&gt;&lt;linear-algebra&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="13153" PostTypeId="1" AcceptedAnswerId="13161" CreationDate="2010-12-05T21:44:23.017" Score="1" ViewCount="137" Body="&lt;p&gt;The problem was to find $\int \tan^5(x)\, \sec^7(x)\, dx$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The solution the book got was different from mine, but I'm sure I did the right steps. Since the solutions are completely different, I have a feeling they might be the same thing, but I'm not sure how I would know this. Here's what the book got:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sec^{11}(x)/11-2\sec^9(x)/9+\sec^7(x)/7+C$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's what I got:&#xA;$\tan^6(x)/6 + \tan^{11}(x)/11 + C$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: So my answer is wrong then, so let me list my steps see if you can catch my mistake&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\int \tan^5(x)\sec^7(x)\, dx$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\int \tan^4x \tan x \sec^5x \sec^2x\, dx$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\int \tan^4 x\tan x \ (1+\tan^5x)\,\sec^2x\,dx$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$u=\tan x \ du=\sec^2x$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\int u^4(u)(1+u^5)\,du$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\int u^5(1+u^5)\,du$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\int u^5+u^{10}\,du$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$=\tan^6(x)/6 + \tan^{11}(x)/11 + C$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="2985" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-05T23:47:24.810" LastActivityDate="2010-12-06T21:45:42.803" Title="Trigonometric integral question" Tags="&lt;calculus&gt;&lt;integral&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="13154" PostTypeId="1" AcceptedAnswerId="13204" CreationDate="2010-12-05T21:53:38.613" Score="0" ViewCount="49" Body="&lt;p&gt;Hello,&#xA;how do I found the resolvent of this formula?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(A \vee C) \wedge ( \neg A \vee B) \wedge C$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it as easy as take every clause with the same statements?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(A \vee C)$ and $( \neg A \vee B)$ and put out the contradiction (in this case $A$ and $\neg A$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could you recommend me some source for resolution method? except &lt;a href=&quot;http://en.wikipedia.org/wiki/Resolution_%28logic%29&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;, because there is too easy examples :-)&lt;/p&gt;&#xA;" OwnerUserId="4449" LastActivityDate="2010-12-06T05:22:35.313" Title="The resolution method" Tags="&lt;soft-question&gt;&lt;logic&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="13155" PostTypeId="1" AcceptedAnswerId="13164" CreationDate="2010-12-05T21:53:43.930" Score="11" ViewCount="296" Body="&lt;p&gt;In most undergraduate differential geometry courses -- I am thinking of do Carmo's &quot;Differential Geometry of Curves and Surfaces&quot; -- the topic of study is curves and surfaces in $\mathbb{R}^3$.  However, the definition of &quot;curve&quot; and &quot;surface&quot; are usually presented in very different ways.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A curve is defined simply as a differentiable map $\gamma\colon I \to \mathbb{R}^3$, where $I \subset \mathbb{R}$ is an interval.  Of course, some authors prefer to define a curve as the image of such a map, and others require piecewise-differentiability, but the general concept is the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, surfaces are essentially defined as 2-manifolds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, in graduate courses on manifolds -- I am thinking of John Lee's &quot;Introduction to Smooth Manifolds&quot; -- one talks about curves $\gamma\colon I \to M$ in a manifold, and can do line integrals over such curves, but talks separately about embedded/immersed 1-dimensional submanifolds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question, then, is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Why make (parametrized) curves the object of study rather than 1-manifolds?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Earlier, I asked a &lt;a href=&quot;http://math.stackexchange.com/questions/4699/which-smooth-1-manifolds-can-be-represented-by-a-single-smooth-parametrization&quot;&gt;question&lt;/a&gt; that was perhaps meant to hint at this one, though I didn't say so explicitly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ultimately, I would simply like to say &quot;curves are 1-manifolds and surfaces are 2-manifolds,&quot; and am looking for reasons why this is correct/incorrect or at least a good/bad idea.  (So, yes, I'm looking for a standard definition of &quot;curve.&quot;)&lt;/p&gt;&#xA;" OwnerUserId="640" LastActivityDate="2010-12-06T13:34:16.697" Title="Why study &quot;curves&quot; instead of 1-manifolds?" Tags="&lt;soft-question&gt;&lt;differential-geometry&gt;&lt;differential-topology&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" />
  <row Id="13156" PostTypeId="1" CreationDate="2010-12-05T21:55:55.790" Score="0" ViewCount="140" Body="&lt;p&gt;How is a complex number a group under addition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How is a complex number a group under multiplication(without zero)&lt;/p&gt;&#xA;" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2010-12-06T00:54:41.647" LastActivityDate="2010-12-06T00:54:41.647" Title="How does the set of complex numbers (resp., nonzero complex numbers) form a group under addition (resp., multiplication)?" Tags="&lt;group-theory&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="13157" PostTypeId="1" AcceptedAnswerId="13160" CreationDate="2010-12-05T21:56:17.250" Score="1" ViewCount="98" Body="&lt;p&gt;A friend and I were doing some math together when we had to integrate 2(100 + t).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I just multiplied it out and integrated 200 + 2t, which should be 200t + t^2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;He did u-substitution and got (100 + t)^2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I take the derivative of both of ours, I seem to get the same thing. But (100 + t)^2 = 10,000 + 200t + t^2, which is not equal to just 200t + t^2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is going on here?&lt;/p&gt;&#xA;" OwnerUserId="3143" LastActivityDate="2010-12-05T22:39:46.440" Title="Integrating 2(100+t)" Tags="&lt;calculus&gt;&lt;integral&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="13158" PostTypeId="2" ParentId="13152" CreationDate="2010-12-05T21:56:32.920" Score="1" ViewCount="" Body="&lt;p&gt;The only problem you have is that your expression for the vector $\overrightarrow{P_1P_2}$ is slightly off. As you say, let $P_1 = (x_1,y_1)$ and $P_2 = (x_2,y_2)$ be points on the line. Then what are the coordinates of the vector $\overrightarrow{P_1P_2}$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you calculate that, then notice that the coordinates of $P_1$ satisfy the equation $Ax_1 + By_1 = -C$, and similarly for $P_2$. Then calculate the scalar product of $(A,B)$ with the vector $\overrightarrow{P_1P_2}$, and interpret what comes out.&lt;/p&gt;&#xA;" OwnerUserId="3225" LastActivityDate="2010-12-05T21:56:32.920" CommentCount="1" />
  <row Id="13159" PostTypeId="2" ParentId="13152" CreationDate="2010-12-05T22:00:28.517" Score="1" ViewCount="" Body="&lt;p&gt;vector is equals $&amp;lt;x_2 - x_1, y_2 - y_1&amp;gt;$, not $&amp;lt;Ax_2 - Ax_1, By_2 - By_1&amp;gt;$ then $&amp;lt;A,B&amp;gt; \cdot P_1P_2 = A\cdot x_2-A\cdot x_1+B\cdot y_2-B\cdot y_1$&lt;/p&gt;&#xA;" OwnerUserId="4454" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2010-12-05T22:44:05.747" LastActivityDate="2010-12-05T22:44:05.747" />
  <row Id="13160" PostTypeId="2" ParentId="13157" CreationDate="2010-12-05T22:02:07.500" Score="4" ViewCount="" Body="&lt;p&gt;The difference is just the constant of integration.  Both your solutions are correct, as is any one of the form C+200t+t^2 for any constant C.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-12-05T22:02:07.500" />
  <row Id="13161" PostTypeId="2" ParentId="13153" CreationDate="2010-12-05T22:02:24.687" Score="3" ViewCount="" Body="&lt;p&gt;If they differ by a constant then their difference should be constant. You can check that this is not so by setting $x=0$ and $x=\pi.$&lt;/p&gt;&#xA;" OwnerUserId="1301" LastActivityDate="2010-12-05T22:02:24.687" CommentCount="7" />
  <row Id="13162" PostTypeId="2" ParentId="13157" CreationDate="2010-12-05T22:03:18.187" Score="1" ViewCount="" Body="&lt;p&gt;When you integrate indefinitely, you must add an arbitrary constant. So in essence, your answers are the same, up to addition of a constant (10,000 in this case). In other words, his constant will be 10,000 less than yours.&lt;/p&gt;&#xA;" OwnerUserId="2029" LastActivityDate="2010-12-05T22:03:18.187" />
  <row Id="13163" PostTypeId="2" ParentId="13156" CreationDate="2010-12-05T22:04:26.297" Score="5" ViewCount="" Body="&lt;p&gt;A complex number is not a group under addition.  The set of all complex numbers is a group under addition.  Just look at the definition of a group and see that you can verify the axioms.  Similarly for the set of complex numbers without zero and multiplication.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-12-05T22:04:26.297" />
  <row Id="13164" PostTypeId="2" ParentId="13155" CreationDate="2010-12-05T22:17:03.660" Score="12" ViewCount="" Body="&lt;p&gt;Essentially because the connected 1-manifolds are (&quot;up to...&quot;) $(0,1)$ and $S^1$, so the notion of curve captures all of the possibilities of 1-manifolds sitting in higher-dimensional manifolds. In other words, the situation for 1-dimensional manifolds is so simple that it really makes no sense to use the full machinery of embeddings and immersions to talk about them other than just checking that the definitions of embedding and immersion are compatible with the definition of curve.&lt;/p&gt;&#xA;" OwnerUserId="1269" LastActivityDate="2010-12-05T22:17:03.660" />
  <row Id="13165" PostTypeId="2" ParentId="13150" CreationDate="2010-12-05T22:18:44.373" Score="2" ViewCount="" Body="&lt;p&gt;Esentially you need to solve the following&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left[\begin{array}{ccc}&#xD;&#xA;\mathrm{a} &amp;amp; \mathrm{b} &amp;amp; \mathrm{ty}\\&#xD;&#xA;\mathrm{c} &amp;amp; \mathrm{d} &amp;amp; \mathrm{ty}\end{array}\right]=\left[\begin{array}{ccc}&#xD;&#xA;s_{x}\cos\psi &amp;amp; -s_{x}\sin\psi &amp;amp; x_{c}\\&#xD;&#xA;s_{y}\sin\psi &amp;amp; s_{y}\cos\psi &amp;amp; y_{c}\end{array}\right]$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $s_x$, $s_y$ are the scalings, $x_c$, $y_c$ is the translation and $\psi$ is the rotation angle. The results I get are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x_{c}=\mathrm{tx}$$&#xD;&#xA;  $$y_{c}=\mathrm{ty}$$&#xD;&#xA;  $$s_{x}=\mathrm{sign(a)\,}\sqrt{\mathrm{a}^{2}+\mathrm{b}^{2}}$$&#xD;&#xA;  $$s_{y}=\mathrm{sign(d)\,}\sqrt{\mathrm{c}^{2}+\mathrm{d}^{2}}$$&#xD;&#xA;  $$\tan\psi=-\frac{\mathrm{b}}{\mathrm{a}}=\frac{\mathrm{c}}{\mathrm{d}}$$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="3301" LastEditorUserId="3301" LastEditorDisplayName="" LastEditDate="2010-12-05T22:30:43.847" LastActivityDate="2010-12-05T22:30:43.847" CommentCount="3" />
  <row Id="13166" PostTypeId="1" AcceptedAnswerId="13185" CreationDate="2010-12-05T22:29:10.250" Score="3" ViewCount="383" Body="&lt;p&gt;Sometimes when you want to solve an equation you can just use algebra and rearrange it then you are done. But sometimes no amount of algebra can ever prove the equation, and then you need an idea, here is what I mean by the idea:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;parity (or modular arithmetic for higher numbers)&lt;/li&gt;&#xA;&lt;li&gt;complex integers.&lt;/li&gt;&#xA;&lt;li&gt;the irrational number&lt;/li&gt;&#xA;&lt;li&gt;a special function&lt;/li&gt;&#xA;&lt;li&gt;a strange curve&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;A lot of equations can be proved impossible if the left hand side is odd and the right hand side is even, sometimes I had a pair of equations that turned out to be easy if you read it as a real and imaginary equation, I read to solve Pell's equation you need a fraction close to the square roots of $d$ and to solve $ax + by = 1$ you have use the greatest common divisor function. To solve Fermat's Last theorem you need elliptic curves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have two questions, Where can I find more examples of this (an equation and then what it is really - which lets you prove it)? and how do you find out what is behind the equation aka. what is the idea (because I only see the + - * ^ and variables..)?&lt;/p&gt;&#xA;" OwnerUserId="4290" LastEditorUserId="4290" LastEditorDisplayName="" LastEditDate="2010-12-06T00:27:05.840" LastActivityDate="2010-12-07T01:40:06.353" Title="What are Diophantine equations REALLY?" Tags="&lt;number-theory&gt;&lt;soft-question&gt;&lt;diophantine-equations&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" />
  <row Id="13167" PostTypeId="2" ParentId="13129" CreationDate="2010-12-05T22:36:59.673" Score="2" ViewCount="" Body="&lt;p&gt;It is unclear to me on what grounds you claim that $g(x)b=f(x)$ will have more than one solution for $x$. Even assuming that $K$ were algebraically closed, how do you know that you don't have one solution which is a multiple solution? For a very simple example, consider the case of $K=\mathbf{F}_2$, the field with two elements, and take $g(x) = x^2+x+1$ and $f(x)=x$. Then $g(x)0 = f(x)$ has one solution ($x=0$) and $g(x)1 = f(x)$ has one solution ($x=1$), so it does not follow merely from the degrees of $f$ and $g$: there is more work to be done. I don't know for sure if you can push it through, but you are correct that the first step is showing that the degrees have to be at most one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know two proofs: a direct way and a clever way. The direct way is to start as you do, with $\theta(x)=\frac{f(x)}{g(x)}$. Then consider an $\alpha = p(x)/q(x)$. If you write $f(x) = a_nx^n+\cdots +a_0$ and $g(x)=b_nx^n+\cdots+b_0$, with $a_n$ and $b_n$ not both zero, and you write $p(x) = c_mx^m+\cdots+c_0$, $q(x) = d_mx^m+\cdots+d_0$, with $c_m$ and $d_m$ both nonzero, then you can think about the degrees of the numerator and denominator of $\theta(\alpha)$. Try to think about what you would need for the degrees to drop after applying $\theta$ in order to conclude information about what can map &lt;strong&gt;to&lt;/strong&gt; $x$ under $\theta$. From there what you want will follow. This proof basically amounts to working things through in the obvious manner until you get to where you want to go.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second proof I know is the &quot;clever proof&quot;. Again, take $\theta(x)=\frac{f(x)}{g(x)}$ as you do. Note that $\theta(x)\notin K$. Show that $\theta(x)$ is transcendental over $K$ by showing that $x$ is algebraic over $K(\theta(x))$, and find its irreducible polynomial. Then use the fact that $K(\theta(x))$ is isomorphic to $K(x)$ to deduce that the irreducible polynomial must be of degree $1$. The irreducible polynomial turns out to be closely related to $f$ and $g$, of course. (I also like the fact that Gauss's Lemma comes into play to show that the obvious polynomial satisfied by $\theta(x)$ is in fact irreducible in the field you are looking at).&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-06T16:32:44.347" LastActivityDate="2010-12-06T16:32:44.347" CommentCount="0" />
  <row Id="13168" PostTypeId="2" ParentId="13157" CreationDate="2010-12-05T22:39:46.440" Score="1" ViewCount="" Body="&lt;p&gt;Failure to recognize that an indefinite integral is only defined up to a constant can lead to strange results:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}\int \frac{1}{x} dx &amp;amp; = \frac{1}{x} x - \int x \frac{(-1)}{x^2} dx \\ &#xD;&#xA;\int \frac{1}{x} dx &amp;amp; = 1 + \int \frac{1}{x} dx \end{align}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or, subtracting the integral from both sides, we obtain $0=1$. As troll face would say:&quot;Math will go bankrupt!&quot;.&lt;/p&gt;&#xA;" OwnerUserId="3567" LastActivityDate="2010-12-05T22:39:46.440" />
  <row Id="13169" PostTypeId="1" AcceptedAnswerId="13175" CreationDate="2010-12-05T22:41:48.630" Score="8" ViewCount="212" Body="&lt;p&gt;Let $X = \textrm{Spec} \mathbf{C}[[x,y]]/(y^{2} - x^{3} - x^{2})$. I would like to describe X set-theoretically. My questions are: Can one explicitly say what the elements in X are? Is it possible to interpret them geometrically? And is X irreducible? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not really sure where to get started. Any help would be much appreciated. &lt;/p&gt;&#xA;" OwnerUserId="4457" LastActivityDate="2010-12-05T23:37:16.537" Title="What is Spec $\mathbf{C}[[x,y]]/(y^{2} - x^{3} - x^{2})$?" Tags="&lt;homework&gt;&lt;algebraic-geometry&gt;&lt;commutative-algebra&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="13170" PostTypeId="2" ParentId="13122" CreationDate="2010-12-05T22:49:00.013" Score="1" ViewCount="" Body="&lt;p&gt;EDIT: NEW ANSWER&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is convenient to express the solution in terms of convolutions: for any distribution $\mu$, positive integer $N$, and $1 \leq n \leq N$, we have&#xA;$$&#xD;&#xA;\mu ^{*N!}  = \big(\mu ^{*(N!/n)} \big)^{*n}.&#xD;&#xA;$$ &#xA;So simple, yet instructive...&lt;/p&gt;&#xA;" OwnerUserId="2810" LastEditorUserId="2810" LastEditorDisplayName="" LastEditDate="2010-12-06T15:33:54.187" LastActivityDate="2010-12-06T15:33:54.187" CommentCount="3" />
  <row Id="13171" PostTypeId="2" ParentId="13166" CreationDate="2010-12-05T22:52:37.047" Score="3" ViewCount="" Body="&lt;p&gt;It sounds like what you mean is how can I find the motivation for studying a Diophantine equation by looking at the equation.  In the case of Fermat's Last Theorem, it is quite a natural equation to study.  I would disagree that it is really about elliptic curves, but rather say that it is about the natural numbers.  It so happens that elliptic curves gave the tools to solve the problem.  Similarly for Pell's equation, you can just wonder about solutions to $x^2-2y^2=1$, find several by hand, and find the recurrence that generates them without knowing about continued fractions and the expansion of $\sqrt{2}$.  Knowing about continued fractions and rings where you add a square root of some element suggest new areas of study, which has been quite productive.  But I don't know how to discern that from the equation itself.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-12-05T22:52:37.047" CommentCount="7" />
  <row Id="13172" PostTypeId="2" ParentId="13151" CreationDate="2010-12-05T22:56:33.190" Score="1" ViewCount="" Body="&lt;p&gt;For your first matrix, the rank could be 1 if $n=1$ or $n=-1$ (because there would only be one pivot column).  For you second example, the rank could be 1,2,or 3 depending on x,y, and z.  For instance, if $x=y=z$ there are only non-zero entries in the first row of the reduced matrix.  You may want to look at the invertible matrix theorem to help you with this second example.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.ams.sunysb.edu/~yan2000/ams210_f2005/TheInvertibleMatrixTheorem.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.ams.sunysb.edu/~yan2000/ams210_f2005/TheInvertibleMatrixTheorem.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, a square matrix has &quot;full rank&quot; iff it is invertible.  This makes your first question trivial.  For the second one, think about the values of $x,y,z$ that make the matrix singular, then classify these as rank 1 or 2.  Any combination of $x,y,z$ making the matrix invertible implies the resulting matrix has rank 3.&lt;/p&gt;&#xA;" OwnerUserId="3971" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-05T23:36:11.253" LastActivityDate="2010-12-05T23:36:11.253" />
  <row Id="13173" PostTypeId="2" ParentId="13151" CreationDate="2010-12-05T23:01:27.443" Score="5" ViewCount="" Body="&lt;p&gt;You seem to be assuming that because &quot;$1-n^2$&quot; doesn't look like $0$, then it cannot be zero. That is a common, but often fatal, mistake.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remember that $n$ stands for &lt;em&gt;some&lt;/em&gt; integer. Once you get to&#xA;$$A = \left(\begin{array}{cc}&#xD;&#xA;1 &amp;amp; n\\&#xD;&#xA;0 &amp;amp; 1-n^2&#xD;&#xA;\end{array}\right),$$&#xD;&#xA;you cannot just jump to saying there are two pivots: your next step would be to divide the second row by $1-n^2$ to make the second pivot, but whenever you divide by something, that little voice in your head should be whispering in your ear: &quot;Wait! Are you sure you are not dividing by zero?&quot; (remember, if you divide by zero, &lt;em&gt;the universe explodes!&lt;/em&gt;). And the thing is, you &lt;strong&gt;aren't&lt;/strong&gt; sure you are not dividing by zero. It depends on what $n$ is! So, your answer should be that it will be rank $2$  &lt;em&gt;if&lt;/em&gt; $1-n^2\neq 0$, and rank $1$  &lt;em&gt;if&lt;/em&gt; $1-n^2 = 0$. But you don't want the person who is grading/reading to have to figure out &lt;em&gt;when&lt;/em&gt; that will happen. You want them to be able to glance at the original matrix, and then be able to immediately say (correctly) &quot;Rank is 1&quot; or &quot;Rank is 2&quot;. So you should express the conditions in terms of $n$ alone, not in terms of some computation involving $n$. So your final answer should be something like &quot;$\mathrm{rank}(A)=2$ if $n=\text{someting}$, and $\mathrm{rank}(A)=1$ if $n=\text{something else}$.&quot; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same thing happens with the second matrix: in order to be able to multiply by $-(\frac{z-x}{y-x})$, that little voice in your head will whisper &quot;Wait! are you sure you are not dividing by zero?&quot;, which leads you to consider what happens when $y-x=0$. But more: even if you are sure that $y-x\neq 0$, that meddlesome little voice should be whispering &quot;Wait! Are you sure you are not &lt;em&gt;multiplying&lt;/em&gt; the row by zero?&quot; (because, remember, multiplying a row by zero is &lt;em&gt;not&lt;/em&gt; an elementary row operation). (And be careful: if you don't pay attention to that voice, it's going to start yelling instead of whispering...) So that means that you &lt;em&gt;also&lt;/em&gt; need to worry about what happens when $z-x=0$. The answer on the rank of $B$, then, will depend on how $x$, $y$, and $z$ relate, and so your solution should reflect that.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2010-12-05T23:16:28.437" LastActivityDate="2010-12-05T23:16:28.437" CommentCount="2" />
  <row Id="13174" PostTypeId="1" AcceptedAnswerId="34047" CreationDate="2010-12-05T23:23:32.597" Score="10" ViewCount="233" Body="&lt;p&gt;I have the following system of equations where the $m$'s are known but $a, b, c, x, y, z$ are unknown. How does one go about solving this system? All the usual linear algebra tricks I know don't apply and I don't want to do it through tedious substitutions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{aligned}&#xA;a + b + c &amp;amp;= m_0 \\&#xA;ax + by + cz &amp;amp;= m_1 \\&#xA;ax^2 + by^2 + cz^2 &amp;amp;= m_2 \\&#xA;ax^3 + by^3 + cz^3 &amp;amp;= m_3 \\&#xA;ax^4 + by^4 + cz^4 &amp;amp;= m_4 \\&#xA;ax^5 + by^5 + cz^5  &amp;amp;= m_5&#xA;\end{aligned}&lt;/p&gt;&#xA;" OwnerUserId="1427" LastEditorUserId="1427" LastEditorDisplayName="" LastEditDate="2010-12-06T02:58:13.257" LastActivityDate="2011-04-25T10:08:58.493" Title="Solving a peculiar system of equations" Tags="&lt;algebra&gt;" AnswerCount="5" CommentCount="6" FavoriteCount="3" />
  <row Id="13175" PostTypeId="2" ParentId="13169" CreationDate="2010-12-05T23:37:16.537" Score="16" ViewCount="" Body="&lt;p&gt;Geometrically you are considering germ of singular node at the origin, that is at the singularity. The node is  $  N = \textrm{Spec} ( \mathbf{C}[x,y]/(y^{2} - x^{3} - x^{2})) $  and is irreducible affine curve, because polynomial $y^{2} - x^{3} - x^{2}$ is irreducible in polynomial ring  $\mathbf{C}[x,y]$. However your germ $X$ is reducible: this is because polynomial $y^{2} - x^{3} - x^{2}$ become reducible in formal powers series ring : $\; y^{2} - x^{3} - x^{2}=(y-x\sqrt{1+x})(y+x\sqrt{1+x})\in \mathbf{C}[[x,y]]$&#xD;&#xA; where $\sqrt{1+x}= 1+\frac{1}{2}x+...$ can be developed by Newton binomial in &#xA;$\mathbf{C}[[x,y]]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in scheme sense $X$ contains three points:  origin (= singularity )and two generic points of two irreducible components of $N$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is very intersting because node remains irreducible in every neighbourhood of origin in affine plane  $\textrm{Spec} ( \mathbf{C}[x,y])$ but by going to formal series you obligate curve to split in two components. So intuition should be that going to formal series is strong form of localization.&lt;/p&gt;&#xA;" OwnerUserId="2872" LastActivityDate="2010-12-05T23:37:16.537" CommentCount="1" />
  <row Id="13176" PostTypeId="1" AcceptedAnswerId="13179" CreationDate="2010-12-05T23:42:22.073" Score="2" ViewCount="433" Body="&lt;p&gt;Given the line $x+2=\frac{y+4}{2}=\frac{z-5}{-2}$&lt;br&gt;&#xA;I want to find the closest point on this line to $(1,1,1)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suppose the details here don't matter but in general how is this done?  We need a vector perpendicular to the line but that also reaches $(1,1,1)$.&lt;/p&gt;&#xA;" OwnerUserId="2533" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2010-12-06T00:00:13.110" LastActivityDate="2010-12-06T00:19:47.157" Title="How to find a point on a line closest to another given point?" Tags="&lt;calculus&gt;&lt;geometry&gt;&lt;euclidean-geometry&gt;" AnswerCount="4" CommentCount="2" />
  <row Id="13177" PostTypeId="2" ParentId="13153" CreationDate="2010-12-05T23:44:58.857" Score="0" ViewCount="" Body="&lt;p&gt;Instead of working with $\tan $ ,$\sec $ , use their $\sin \text {and} \cos $ definitions in their place, the problem will be reduced to reverse application of product rule. way easier than to play around with than  $\tan $ and $\sec $ .&lt;/p&gt;&#xA;" OwnerUserId="3936" LastActivityDate="2010-12-05T23:44:58.857" />
  <row Id="13178" PostTypeId="2" ParentId="13176" CreationDate="2010-12-05T23:49:12.113" Score="2" ViewCount="" Body="&lt;p&gt;Using multi-variable calculus you would write a formula for distance from the point (1,1,1) to the arbitrary point (x,y,z) (this second points needs to satisfy your equation since it lies on the line).  You would then apply the multi-variable optimization procedure:  Find the critical points of this functions (those points where both partial derivatives equal zero or those points where either partial derivative is undefined) and examine these to find the minimum value for distance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you do want to use linear algebra, a vector from the point (x,y,z) (on the  line) to the point (1,1,1) is $&amp;lt;1-x,1-y,1-z&amp;gt;$  Now take a directional vector from your line and compute the dot product of this with $&amp;lt;1-x,1-y,1-z&amp;gt;$.  As you said, this dot product should be zero so solve for x,y, and z.&lt;/p&gt;&#xA;" OwnerUserId="3971" LastActivityDate="2010-12-05T23:49:12.113" CommentCount="1" />
  <row Id="13179" PostTypeId="2" ParentId="13176" CreationDate="2010-12-05T23:59:36.993" Score="7" ViewCount="" Body="&lt;p&gt;One point on the line is (-2,-4,5) and another is (0,0,1) (found by setting the equation to 0, then to 2).  So the general point is (-2t, -4t,1+4t).  You want the dot product of the vector from a point on the line to (1,1,1) and a vector along the line to be zero.  So&#xA;$$(-2,-4,4)\cdot(-2t-1,-4t-1,4t)=0$$ &#xA; $$4t+2+16t+4+16t=0$$&#xD;&#xA; $$t=\frac{-1}{6}$$&#xD;&#xA;And the point is $$(\frac{1}{3},\frac{2}{3},\frac{1}{3})$$&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-12-05T23:59:36.993" />
  <row Id="13180" PostTypeId="2" ParentId="13135" CreationDate="2010-12-06T00:00:22.393" Score="2" ViewCount="" Body="&lt;p&gt;This is a job for the &lt;strong&gt;Green and Brown Fact&lt;/strong&gt;, namely Theorem 8 on page 14 of &lt;a href=&quot;http://www.math.uga.edu/~pete/3200relationsfunctions.pdf&quot; rel=&quot;nofollow&quot;&gt;these notes&lt;/a&gt;.  (The name comes from the colors of the chalk I used to draw a box around it that day in class.)  If you have a composition of functions $g \circ f$, this tells you that if the composite function is injective, $f$ is injective, and if the composite function is surjective, $g$ is surjective.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question asked by the OP fits so nicely into this framework that I feel embarrassed not to have discussed it either time I taught the course.  Next time I will.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ooh, another good exercise: what if $f \circ f \circ f = \operatorname{Id}_A$, and so forth.&lt;/p&gt;&#xA;" OwnerUserId="299" LastActivityDate="2010-12-06T00:00:22.393" />
  <row Id="13181" PostTypeId="2" ParentId="13148" CreationDate="2010-12-06T00:01:07.710" Score="4" ViewCount="" Body="&lt;p&gt;Take an arbitrary simple polygon $\pi$ including every point of $S$, and take some segment $PQ$ of the convex hull. Extend it into a line $\ell$. Form a new polygon $\pi&amp;#39;$ by locating the &quot;left&quot;most and &quot;right&quot;most intersection points $L,R$ of $\ell$ with $\pi$, adding new vertices to $\pi$ at $L,R$, and replacing the part of $\pi$ between $L$ and $R$ by the single line segment $LR$. The resulting polygon $\pi&amp;#39;$ is a simple polygon including all points of $S$, and the triangle inequality implies that its perimeter is smaller or equal to the perimeter of $\pi$.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-12-06T00:01:07.710" CommentCount="1" />
  <row Id="13182" PostTypeId="2" ParentId="13176" CreationDate="2010-12-06T00:01:28.890" Score="1" ViewCount="" Body="&lt;p&gt;First express $y$ and $z$ in terms of $x$: $y=y(x)$, $z=z(x)$. You want to find the minimum of $f(x) = (x-1)^2 + (y(x)-1)^2 + (z(x)-1)^2$. Noting that $f(x)$ is a parabola, you should find easily the (single) $x$ that minimizes $f(x)$ (global minimum). Then, you get $y(x)$ and $z(x)$ by substitution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Explicitly, $y(x)=2x$ and $z(x)=1-2x$. This leads to $f(x)=9x^2-6x+2$, whose minimum is obtained at $x=1/3$. The solution is then $(x,y,z)=(1/3,2/3,1/3)$, as already given in two previous answers.&lt;/p&gt;&#xA;" OwnerUserId="2810" LastEditorUserId="2810" LastEditorDisplayName="" LastEditDate="2010-12-06T00:19:47.157" LastActivityDate="2010-12-06T00:19:47.157" />
  <row Id="13183" PostTypeId="2" ParentId="13176" CreationDate="2010-12-06T00:03:07.703" Score="2" ViewCount="" Body="&lt;p&gt;One method is to parametrize the line. So&#xA;$$\vec{r}(t) = (t-2)\hat{\imath} + (2t-4)\hat{\jmath} + (-2t+5)\hat{k}$$&#xD;&#xA;We can then define the real-valued function $d(t)$ representing the distance from the line to $(1,1,1)$ for any value of $t$. Hence&#xA;$$d(t) = \sqrt{(t-3)^2 + (2t-5)^2 + (-2t+4)^2} = \sqrt{9t^2 - 42t + 40}$$&#xD;&#xA;To find the minimum, it's now only a single variable optimization problem. So by standard procedure, look at $d&amp;#39;(t)$&#xD;&#xA;$$d&amp;#39;(t) = \frac{9t - 21}{\sqrt{9t^2 - 42t + 40}}$$&#xD;&#xA;The critical point occurs at $t = \frac{7}{3}$, so the closest point is $\vec{r}(\frac{2}{3}) = (\frac{1}{3},\frac{2}{3},\frac{1}{3})$.&lt;/p&gt;&#xA;" OwnerUserId="1016" LastEditorUserId="1016" LastEditorDisplayName="" LastEditDate="2010-12-06T00:08:13.737" LastActivityDate="2010-12-06T00:08:13.737" />
  <row Id="13184" PostTypeId="2" ParentId="13043" CreationDate="2010-12-06T00:04:27.217" Score="1" ViewCount="" Body="&lt;p&gt;You've got the right start by conditioning on $N$ and $P$ to get $E[e^{tX}] = E[(Pe^t + 1-P)^N]$.  Now, evaluate the second expression via&#xA;$$E[(Pe^t + 1-P)^N] = E[E[(Pe^t + 1-P)^N|P]].$$&#xD;&#xA;Since $N$ is Poisson, you will be able to get a closed-form answer for the inner expectation.  The only remaining variable will then be $P$.  But since $P$ is beta with small integer parameters the outer expectation will involve an integral of a low-degree polynomial times an exponential.  That can be evaluated via integration by parts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;HR&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Update with more details, after comments below.  This problem is more complicated than I had realized at first!) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Applying the calculations described above, we have, where $\lambda = 11$,&#xA;$$E[(Pe^t + 1-P)^N|P] = e^{\lambda P(e^t-1)}.$$&#xD;&#xA;Then $E[e^{\lambda P(e^t-1)}]$ must be done piecewise, depending on the value of $t$.  If $t = 0$, we have&#xA;$$E[e^{\lambda P(e^t-1)}] = E[1] = 1.$$&#xD;&#xA;Otherwise, $E[e^{\lambda P(e^t-1)}]$ is a much more complicated expression with, as the OP indicates, $(e^t-1)^4$ in the denominator:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$E[e^{\lambda P(e^t-1)}] = \frac{12 \left(6 - 6 e^{\lambda (e^t-1)} + 2 \lambda (2 + e^{\lambda (e^t-1)}) (e^t-1) + &#xD;&#xA;   \lambda^2 (e^t-1)^2\right)}{\lambda^4 (e^t-1)^4}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because the limit of this latter expression as $t \to 0$ is $1$, $E[e^{\lambda P(e^t-1)}]$ is still continuous at $0$, despite the fact that it has to be calculated piecewise.  It also turns out to be differentiable at $t = 0$ as well, and so one can still find moments with it.  For example, to find $E[X]$ we differentiate the expression for $E[e^{\lambda P(e^t-1)}]$, $t \neq 0$, once with respect to $t$ and then take the limit as $t \to 0$.  This gives $E[X] = 4.4$, which is the same result one would obtain by calculating $E[X]$ directly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Incidentally, a similar situation occurs with the continuous uniform distribution on $[a,b]$.  Its mgf is, for $t \neq 0$, $$\frac{e^{tb}-e^{ta}}{t(b-a)}.$$  The moments can be found by successively differentiating this expression and then taking the limit as $t \to 0$ rather than by substituting $0$ for $t$.  See, for example, MathWorld's article on the &lt;a href=&quot;http://mathworld.wolfram.com/UniformDistribution.html&quot; rel=&quot;nofollow&quot;&gt;uniform distribution&lt;/a&gt;. &lt;/p&gt;&#xA;" OwnerUserId="2370" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-12-06T05:04:58.393" LastActivityDate="2010-12-06T05:04:58.393" CommentCount="4" />
  <row Id="13185" PostTypeId="2" ParentId="13166" CreationDate="2010-12-06T00:40:46.630" Score="7" ViewCount="" Body="&lt;p&gt;I interpret the question as &quot;how do I know how to attack a Diophantine problem&quot;. The question is in fact not easy and it is this skill that algebraic number theorists hone. Moreover, the methods that you listed are where the story was about 300 years ago. Today, there are more sophisticated and more general techniques available. If you are more interested in the motivation, you should have a look at &lt;a href=&quot;http://mathoverflow.net/questions/42406/why-certain-diophantine-equations-are-interesting-and-others-are-not&quot; rel=&quot;nofollow&quot;&gt;this MO question&lt;/a&gt;, which somewhat goes in your direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Before I start listing the modern methods, let me directly address your questions &quot;what is behind a given equation&quot; and &quot;how do I know what to try?&quot;, starting with the latter: you don't. After a while you develop some intuition, but still, the basic way to find an approach that works is to try them all out in the order of decreasing likelihood of success. It is this likelihood that you can estimate better and better, as you become more proficient, but you will never know for sure, until you try an approach. As for the question &quot;what is behind a Diophantine equation&quot;, there is no good way of making sense of this question at present. Some people will view the equation as describing a geometric object (see last paragraph of this post), some people will look at it from a &quot;modular angle&quot; (see penultimate paragraph). But at the end of the day, when you are interested in integral or rational solutions, the equation is just that: a Diophantine equation. If you must categorise equations, then to categorise them according to the geometry and topology of the set of complex solutions is probably the most sensible thing to do (see last paragraph).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are two or three rather broad themes in modern research, where modern means everything from the last 150-odd years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, a classical method that you have hinted at but that has much more potential is that often, to understand integer solutions, you are forced to work in a bigger ring. You have listed square roots of integers, but the technique is more general than that. To master it, you need to learn some classical algebraic number theory, as it was developed at the end of the 19th - beginning of the 20th century and that's also where I would recommend you to start reading. Have a look at an introductory book into algebraic number theory, such as &lt;a href=&quot;http://books.google.co.uk/books?id=PIibasv45boC&amp;amp;dq=algebraic+number+theory+books&amp;amp;printsec=frontcover&amp;amp;source=in&amp;amp;hl=en&amp;amp;ei=riz8TKfEEZCosAPB69X3DQ&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=12&amp;amp;sqi=2&amp;amp;ved=0CGgQ6AEwCw#v=onepage&amp;amp;q=algebraic%20number%20theory%20books&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;the book by Ian Stewart&lt;/a&gt;, which I personally quite like.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another broad theme is the one successfully used by Ribet, Frey, Wiles and several others along the way to prove Fermat's Last Theorem. It is nowadays subsumed under the mysterious bracketing term &quot;modularity&quot;. To start understanding, what this is about, you first need to learn about modular forms and elliptic curves. The basic idea is that the Shimura-Tanyiama-Weil conjecture, which was the actual result Wiles proved, relates two seemingly unrelated objects: rational elliptic curves and modular forms. This is extremely useful, because modular forms are extremely well-behaved. The &quot;modularity&quot;-idea of solving Diophantine equations then is to construct an elliptic curve out of a putative solution to your given equation that has such strange properties that it cannot possibly be modular. That would then contradict Wiles's theorem, so there cannot be such a solution. The places to start reading about elliptic curves and modular forms are (after you have completely read an introductory book on algebraic number theory and done all the exercises) &lt;a href=&quot;http://books.google.com/books?id=Z90CA_EUCCkC&amp;amp;lpg=PR1&amp;amp;ots=3K9dkyWh06&amp;amp;dq=Silverman%20elliptic%20curves&amp;amp;pg=PR1#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;Silverman&lt;/a&gt; - the classic on elliptic curves, and maybe &lt;a href=&quot;http://www.springer.com/mathematics/numbers/book/978-0-387-23229-4&quot; rel=&quot;nofollow&quot;&gt;the book by Diamond and Schurman&lt;/a&gt; for modular forms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, a very broad theme is that often, the geometry or the topology of the complex solutions of the equation controls its arithmetic (i.e. Diophantine) behaviour. It is difficult to point to one place where to learn about this, but elliptic curves are definitely the right point to start. I think, once you have read a book on algebraic number theory and one on elliptic curves, you should just come back here and ask this question again with your new background.&lt;/p&gt;&#xA;" OwnerUserId="3212" LastEditorUserId="3212" LastEditorDisplayName="" LastEditDate="2010-12-07T01:40:06.353" LastActivityDate="2010-12-07T01:40:06.353" CommentCount="2" />
  <row Id="13186" PostTypeId="1" AcceptedAnswerId="13189" CreationDate="2010-12-06T00:51:34.697" Score="5" ViewCount="242" Body="&lt;p&gt;I was working through a textbook on topology and I came across a problem I couldn't solve.&lt;/p&gt;&#xA;&#xA; &lt;p&gt;1) It is known that &lt;b&gt;if a space is T1, it is countably compact if and only if every &lt;i&gt;countable&lt;/i&gt; open cover has a &lt;i&gt;finite&lt;/i&gt; subcover.&lt;/b&gt; (See below for the definition of countably compact, which might be different than the conventional definition.)&lt;/p&gt; &#xA;&#xA;&lt;p&gt;2) It is also true that &lt;b&gt;if a space is T1, it is countably compact if and only if every &lt;i&gt;infinite&lt;/i&gt; open cover has a &lt;i&gt;proper&lt;/i&gt; subcover.&lt;/b&gt; (why?)&lt;/p&gt; &#xA;&#xA;&lt;p&gt;Intuitively, both properties seem to talk about how open covers can be removed of unnecessary elements and still work as a cover, under conditions where points are sufficiently close together. However, I cannot figure out a proof for the second statement. Because the cover may contain uncountably many sets, it is very hard to deal with.&lt;/p&gt; &#xA;&#xA;&lt;p&gt;This question appears in &quot;Elements of Point Set Topology&quot; by John D. Baum as exercise 3.33. The question and related hint can be viewed &lt;a href=&quot;http://books.google.com/books?id=hlC6nK0tFBEC&amp;printsec=frontcover&amp;dq=elements+of+point+set+topology&amp;hl=en&amp;ei=KkH8TOrdGYKclgeUpIiZBQ&amp;sa=X&amp;oi=book_result&amp;ct=result&amp;resnum=1&amp;ved=0CCwQ6AEwAA#v=onepage&amp;q&amp;f=false&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Terminology used in this text:&lt;/b&gt;&lt;/p&gt; &#xA;&#xA;&lt;p&gt;A &lt;b&gt;T1 space&lt;/b&gt; is a topological space such that, if x is an element of the space, the set {x} is closed.&lt;/p&gt; &#xA;&#xA;&lt;p&gt;A &lt;b&gt;countably compact space&lt;/b&gt; is a space such that every infinite subset of the space has a limit point in the space. I am under the impression that other texts refer to this property as &lt;b&gt;limit point compactness&lt;/b&gt;.&lt;/p&gt; &#xA;&#xA;&lt;p&gt;An &lt;b&gt;infinite open cover&lt;/b&gt; is a collection of infinitely many open sets which cover the space.&lt;/p&gt; &#xA;" OwnerUserId="4460" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-06T02:13:54.223" LastActivityDate="2010-12-06T23:10:01.890" Title="A challenging question about T1 spaces and countable compactness" Tags="&lt;general-topology&gt;" AnswerCount="1" CommentCount="0" />
  <row Id="13187" PostTypeId="1" CreationDate="2010-12-06T01:43:10.033" Score="0" ViewCount="125" Body="&lt;p&gt;On a practice final exam (for a Computer Security course), I am given the following equation to solve, but I have no idea how to solve it:&#xA;$$324x \mod{121} = 1.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any direction?&lt;/p&gt;&#xA;" OwnerUserId="4461" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-06T03:03:06.080" LastActivityDate="2010-12-06T06:50:43.517" Title="Newbie to modular arithmetic needs help" Tags="&lt;algebra&gt;&lt;modular-arithmetic&gt;" AnswerCount="2" />
  <row Id="13188" PostTypeId="2" ParentId="13187" CreationDate="2010-12-06T01:48:18.857" Score="3" ViewCount="" Body="&lt;p&gt;You need the &lt;a href=&quot;http://en.wikipedia.org/wiki/Euclidean_algorithm&quot; rel=&quot;nofollow&quot;&gt;Extended Euclidean Algorithm&lt;/a&gt; which solves the problem of finding modular inverses.  &lt;a href=&quot;http://en.wikipedia.org/wiki/B%C3%A9zout%27s_identity&quot; rel=&quot;nofollow&quot;&gt;Bezout's identity&lt;/a&gt; guarantees that there is a solution.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-12-06T01:48:18.857" />
  <row Id="13189" PostTypeId="2" ParentId="13186" CreationDate="2010-12-06T02:17:48.520" Score="7" ViewCount="" Body="&lt;p&gt;We shall show that if $X$ is a $T_1$ space, then it is countably compact if and only if every infinite open cover has a proper subcover. The key idea is to proceed by contrapositive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;($\Rightarrow$ Needs $T_1$) Suppose that $ \ \mathcal{U} \ $ is an infinite open cover of $X$ with no proper subcover. Then for each $U \in \mathcal{U}$, there is a point $p_U \in U$ that doesn't belong to any other member of $ \ \mathcal{U} \ $. The set $A = \{p_U : U \in \mathcal{U} \}$ is infinite and doesn't have a limit point. Indeed, if $x \in X$, then there is $U \in \mathcal{U}$ containing $x$ and $ U \cap A = \{ p_U \}$. If $x = p_U$, then it isn't a limit point of $A$. If $x \ne p_U$, then, using $T_1$, $U- \{ p_U \}$ is open and $(U- \{ p_U \}) \cap A =\emptyset$. So $x$ isn't a limit point of $A$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;George Lowther gave an example showing that the $T_1$ hypothesis is fundamental. Since there are many comments, I'll reproduce it here:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;Consider the example of the real numbers where the open sets are unions of intervals $[n,a)$ for integer $n$ and real $a&amp;gt;n$. This is $T_0$ but not $T_1$. It is also countably compact, but the infinite cover $\mathcal{U}=\{[n,n+1)\colon n\in\mathbb{Z}\}$ has no proper subcovers. So, $T_1$ is needed.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A direct approach to prove this implication was suggested by Carl Mummert. Let $ \ \mathcal{U} \ $ be an infinite open cover of $X$ and $ \ \mathcal{U}_0 $ be a countably infinite subset of $ \ \mathcal{U} $. Now consider $ \ \mathcal{U}_1 = \mathcal{U} - \mathcal{U}_0$ and let $V \ $ be the union of all sets in $ \ \mathcal{U}_1$. Then $ \ \mathcal{U}_0 \cup \{ V \ \}$ is a countable open cover of $X$ and follows from $(1)$ that it has a finite subcover $U_1, \ldots, U_n$. Now the set consisting of all $U_i$, with possible exception of $V$, adjoined with the sets $ U \in \mathcal{U}_1$ is a proper subcover of $ \ \mathcal{U}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;($\Leftarrow$ Doesn't need $T_1$) Now, suppose that $X$ isn't countably compact. Then there is an infinite set $A$ with no limit points. Thus $A$ is closed. By the definition of limit point, for each $x \in A$, there is an open set $U_x$ containing $x$ such that $ U_x \cap A = \{ x \} $. Consider $ \ \mathcal{U} \ $ the set of all $U_x$ plus, if necessary, $X-A$. Then $ \ \mathcal{U} \ $ is an infinite cover of $X$ with no proper subcovers. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S. Thanks Carl Mummert, George Lowther and Mark for your efforts to clarify and improve this answer.&lt;/p&gt;&#xA;" OwnerUserId="329" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2010-12-06T23:10:01.890" LastActivityDate="2010-12-06T23:10:01.890" CommentCount="40" />
  <row Id="13190" PostTypeId="2" ParentId="13187" CreationDate="2010-12-06T02:22:43.773" Score="2" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\ $ One can easily solve it mod $11$ then lift the solution to one mod $11^2$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\ \ mod\ 11:\ \ \ \ x = 1/(324) = 1/(-6) = -2\ \ $  so $\rm\ x = -2 + 11\ n$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\ \ mod\ 121:\ \ \: 1 = 324\ x = (-6 + 11\cdot 30)\ (-2 + 11\ n)\ \ =\ 55\ n - 43\ \ \Rightarrow\ \ 55\ n = 44$   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dividing by $11$ we deduce $\rm\ 5\ n = 4\ \:(mod\ 11)\ $ so $\rm\ n =\ \ldots\ \: $ The method generalizes, e.g.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\quad a\ x\ =\ 1 + b\ m\ \ \Rightarrow\ \ a\ x\ (2-a\ x)\ =\ (1 + b\ m)\ (1-b\ m) = 1 - (b\ m)^2\equiv 1\:\ (mod\ m^2) $  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;so $\rm\ a^{-1}\equiv x\:\ (mod\ m)\ $ lifts to $\rm\ a^{-1}\equiv x\ (2-a\ x)\ \ (mod\ m^2)\:.\:$&#xD;&#xA;This method of lifting solutions is a special case of a very powerful generalization of Newton's method known as &lt;a href=&quot;http://en.wikipedia.org/wiki/Hensel%27s_lemma&quot; rel=&quot;nofollow&quot;&gt;Hensel's Lemma.&lt;/a&gt;  &lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-12-06T06:50:43.517" LastActivityDate="2010-12-06T06:50:43.517" />
  <row Id="13191" PostTypeId="1" AcceptedAnswerId="13192" CreationDate="2010-12-06T02:38:08.510" Score="0" ViewCount="108" Body="&lt;p&gt;How many residue classes satisfy the congruence $x^3 \equiv 3 \pmod{21}$?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't understand what this question is asking me to do.&lt;br&gt;&#xA;Can someone simplify the question for me, thanks.&lt;/p&gt;&#xA;" OwnerUserId="2588" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-06T03:28:26.797" LastActivityDate="2010-12-06T04:27:49.063" Title="How many residue classes satisfy the congruence $x^3 \equiv 3 \pmod{21}$?" Tags="&lt;modular-arithmetic&gt;" AnswerCount="3" CommentCount="4" />
  <row Id="13192" PostTypeId="2" ParentId="13191" CreationDate="2010-12-06T02:48:31.450" Score="7" ViewCount="" Body="&lt;p&gt;The question is asking you to check which of the 21 residue classes of integers modulo $21$ (to wit, the class of $0$, the class of $1$, the class of $2$, etc) are solutions to $x^3\equiv 3\pmod{21}$. If nothing else occurs to you, you can certainly plug and chug and figure out which ones are solutions and which ones are not.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-06T03:07:26.107" LastActivityDate="2010-12-06T03:07:26.107" />
  <row Id="13193" PostTypeId="2" ParentId="13191" CreationDate="2010-12-06T02:49:30.527" Score="1" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\rm\ \ mod\ 7:\ \ x^3 = 3\ \Rightarrow\ x^6 = \ldots\ $ contra a well-known &quot;little&quot; theorem.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-12-06T03:03:37.717" LastActivityDate="2010-12-06T03:03:37.717" />
  <row Id="13194" PostTypeId="2" ParentId="13191" CreationDate="2010-12-06T02:59:13.417" Score="3" ViewCount="" Body="&lt;p&gt;Just a little something to add to Arturo's answer. Note that any solution to $x^3\equiv 3 \pmod{21}$ will also be a solution to $x^3\equiv 3\pmod{3}$ and $x^3\equiv 3\pmod{7}$. So instead of checking all $21$ congruence classes, you can begin by checking the congruence classes modulo the prime powers of $21$. If one happens to have no solution, then modulo $21$ there should be no solution either. It should save you some time, as there are fewer classes to check.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is an application of the following theorem.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Let $f(x)$ be a fixed polynomial with integral coefficients, and for any positive interger $m$, let $N(m)$ denote the number of solutions of the congruence $f(x)\equiv 0\pmod{m}$. If $m=m_1m_2$, where $gcd(m_1,m_2)=1$, then $N(m)=N(m_1)N(m_2)$. If $m=\prod p^\alpha$ is the canonical factorization of $m$, then $N(m)=\prod N(p^\alpha)$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1583" LastEditorUserId="1583" LastEditorDisplayName="" LastEditDate="2010-12-06T04:27:49.063" LastActivityDate="2010-12-06T04:27:49.063" CommentCount="7" />
  <row Id="13195" PostTypeId="2" ParentId="13174" CreationDate="2010-12-06T03:22:30.797" Score="5" ViewCount="" Body="&lt;p&gt;Look at the first three equations as a system of linear equations determining $a$, $b$ and $c$ in terms of $x$, $y$, $z$, $m_0$, $m_1$, $m_2$; the matrix of coefficients is a van der Monde matrix for the triple $x$, $y$, $z$, so it not hard to use Cramer's rule to solve it when no two of them are equal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now replace the values you got for $a$, $b$ and $c$ in the second group of three equations. Magically, after simplifying and canceling things common to numerators and denominators, it is a system of &lt;em&gt;polynomial&lt;/em&gt; equations in $x$, $y$ and $z$. Moreover, the three left hand sides are symmetric polynomials in $x$, $y$ and $z$, so after expressing those left hand sides in terms of the elementary symmetric polynomials on $x$, $y$ and $z$, you end up with a not too difficult polynomial system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solve it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now you have the values of $x+y+z$, $xy+xz+yz$ and $xyz$. Solving a cubic equation now gives you $x$, $y$ and $z$ and, from the first part, also $a$, $b$ and $c$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;NB:&lt;/strong&gt; Following these steps is easily doable in &lt;em&gt;Mathematica&lt;/em&gt; and you get an answer. The answer is simply too huge, and I doubt it can be massaged into something sensible---and I hence doubt there is a sensible way to solve the system of equations by hand.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Mma&lt;/em&gt;, on the other hand, did not solve the system as it is originally given in the time I gave it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;NB2:&lt;/strong&gt; For comparison, the system $$\left\{\begin{aligned}a + b = m_0\\ax+by=m_1\\ax^2+by^2=m_2\\ax^3+by^3=m_3\end{aligned}\right.$$ has as solution, according to &lt;em&gt;Mma&lt;/em&gt; (and this one it does compute it by itself):&#xA;$$\begin{aligned}&#xD;&#xA; &amp;amp;a= \frac{-2 m_1^3+3 m_0 m_2 m_1+m_0 \left(\sqrt{4 m_3 m_1^3-3&#xD;&#xA;   m_2^2 m_1^2-6 m_0 m_2 m_3 m_1+4 m_0 m_2^3+m_0^2&#xD;&#xA;   m_3^2}-m_0 m_3\right)}{2 \sqrt{4 m_3 m_1^3-3 m_2^2&#xD;&#xA;   m_1^2-6 m_0 m_2 m_3 m_1+m_0 \left(4 m_2^3+m_0&#xD;&#xA;   m_3^2\right)}} \\&#xD;&#xA; &amp;amp;b = \frac{2 m_1^3-3 m_0 m_2 m_1+m_0 \left(m_0 m_3+\sqrt{4 m_3&#xD;&#xA;   m_1^3-3 m_2^2 m_1^2-6 m_0 m_2 m_3 m_1+4 m_0 m_2^3+m_0^2&#xD;&#xA;   m_3^2}\right)}{2 \sqrt{4 m_3 m_1^3-3 m_2^2 m_1^2-6 m_0&#xD;&#xA;   m_2 m_3 m_1+m_0 \left(4 m_2^3+m_0 m_3^2\right)}} \\&#xD;&#xA; &amp;amp;x = \frac{-m_1 m_2+m_0 m_3+\sqrt{\left(m_1 m_2-m_0&#xD;&#xA;   m_3\right){}^2+4 \left(m_1^2-m_0 m_2\right) \left(m_1&#xD;&#xA;   m_3-m_2^2\right)}}{2 \left(m_0 m_2-m_1^2\right)} \\&#xD;&#xA; &amp;amp;y = \frac{m_1 m_2-m_0 m_3+\sqrt{4 m_3 m_1^3-3 m_2^2 m_1^2-6 m_0&#xD;&#xA;   m_2 m_3 m_1+4 m_0 m_2^3+m_0^2 m_3^2}}{2 m_1^2-2 m_0 m_2}&#xD;&#xA;\end{aligned}&#xD;&#xA;$$ and&#xA;$$\begin{aligned}&#xD;&#xA; &amp;amp;a=\frac{2 m_1^3-3 m_0 m_2 m_1+m_0 \left(m_0 m_3+\sqrt{4 m_3&#xD;&#xA;   m_1^3-3 m_2^2 m_1^2-6 m_0 m_2 m_3 m_1+4 m_0 m_2^3+m_0^2&#xD;&#xA;   m_3^2}\right)}{2 \sqrt{4 m_3 m_1^3-3 m_2^2 m_1^2-6 m_0&#xD;&#xA;   m_2 m_3 m_1+m_0 \left(4 m_2^3+m_0 m_3^2\right)}} \\&#xD;&#xA; &amp;amp;b=\frac{-2 m_1^3+3 m_0 m_2 m_1+m_0 \left(\sqrt{4 m_3 m_1^3-3&#xD;&#xA;   m_2^2 m_1^2-6 m_0 m_2 m_3 m_1+4 m_0 m_2^3+m_0^2&#xD;&#xA;   m_3^2}-m_0 m_3\right)}{2 \sqrt{4 m_3 m_1^3-3 m_2^2&#xD;&#xA;   m_1^2-6 m_0 m_2 m_3 m_1+m_0 \left(4 m_2^3+m_0&#xD;&#xA;   m_3^2\right)}} \\&#xD;&#xA; &amp;amp;x=-\frac{m_1 m_2-m_0 m_3+\sqrt{\left(m_1 m_2-m_0&#xD;&#xA;   m_3\right){}^2+4 \left(m_1^2-m_0 m_2\right) \left(m_1&#xD;&#xA;   m_3-m_2^2\right)}}{2 \left(m_0 m_2-m_1^2\right)} \\&#xD;&#xA; &amp;amp;y=-\frac{-m_1 m_2+m_0 m_3+\sqrt{4 m_3 m_1^3-3 m_2^2 m_1^2-6&#xD;&#xA;   m_0 m_2 m_3 m_1+4 m_0 m_2^3+m_0^2 m_3^2}}{2&#xD;&#xA;   \left(m_1^2-m_0 m_2\right)}&#xD;&#xA;\end{aligned}&#xD;&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="274" LastEditorUserId="274" LastEditorDisplayName="" LastEditDate="2010-12-06T03:43:36.293" LastActivityDate="2010-12-06T03:43:36.293" />
  <row Id="13196" PostTypeId="1" AcceptedAnswerId="13199" CreationDate="2010-12-06T03:47:10.210" Score="2" ViewCount="176" Body="&lt;p&gt;Given that $f\colon [-1,1] \to \mathbb{R}$ is a continuous function such that $ \int_{-1}^{1} f(t) \ dt =1$, how do I evaluate the limit of this integral: $$\lim_{n \to \infty} \int_{-1}^{1} f(t) \cos^{2}{nt} \,dt$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I did was to write $\cos^{2}{nt} = \frac{1+\cos{2nt}}{2}$ and substitute it in the integral so that I can make use of the given hypothesis of $\int_{-1}^{1} f(t) \ dt =1$. So the integral becomes, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align*}&#xA;\int_{-1}^{1} f(t)\cos^{2}{nt} \ dt = \int_{-1}^{1} f(t) \biggl[\frac{1+\cos{2nt}}{2}\biggr] \ dt  &amp;amp;  \\  \hspace{3cm} =&#xA;\frac{1}{2}\int_{-1}^{1}f(t) \ dt + \int_{-1}^{1} \frac{f(t)\cos{2nt}}{2} \ dt \end{align*}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I don't really know how I can evaluate the second integral and also I can't realize as to why that &lt;em&gt;integral condition&lt;/em&gt; on $f$ has been assumed. Moreover without assuming that condition on $f$ is it possible to evaluate this integral? If yes, then what would the answer be?&lt;/p&gt;&#xA;" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-06T03:59:09.770" LastActivityDate="2010-12-06T06:26:10.663" Title="Evaluating the integral $\mathop{\lim}\limits_{n \to \infty} \int_{-1}^{1} f(t)\cos^{2}(nt) \ dt $" Tags="&lt;real-analysis&gt;&lt;analysis&gt;" AnswerCount="2" />
  <row Id="13197" PostTypeId="2" ParentId="13174" CreationDate="2010-12-06T03:54:11.680" Score="10" ViewCount="" Body="&lt;p&gt;Let $\displaystyle x,y,z$ be roots of $\displaystyle t^3 + pt^2 + qt + r = 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i.e. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle x^3 + px^2 + qx + r = 0$ &amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;   -----   (1)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle y^3 + py^2 + qy + r = 0$&amp;nbsp;&amp;nbsp;&amp;nbsp; -----  (2)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle z^3 + pz^2 + qz + r = 0$&amp;nbsp;&amp;nbsp;&amp;nbsp; ----- (3)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Multiply the (1) by $\displaystyle a$, (2) by $\displaystyle b$ and (3) by $\displaystyle c$ and adding gives&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle m_3 + p m_2 + q m_1 + rm_0 = 0$&amp;nbsp;&amp;nbsp;&amp;nbsp; ----- (4)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Multiply the (1) by $\displaystyle ax$, (2) by $\displaystyle by$ and (3) by $\displaystyle cz$ and adding gives&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle m_4 + pm_3 + qm_2 + rm_1 = 0$&amp;nbsp;&amp;nbsp;&amp;nbsp; ----- (5)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Multiply the (1) by $\displaystyle ax^2$, (2) by $\displaystyle by^2$ and (3) by $\displaystyle cz^2$ and adding gives&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle m_5 + pm_4 + qm_3 + rm_2 = 0$&amp;nbsp;&amp;nbsp;&amp;nbsp; ----- (6)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now (4), (5), (6) is a set of 3 linear equations in 3 variables ($\displaystyle p,q,r$) and can be solved easily.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This give us the cubic which $\displaystyle x,y,z$ satisfy (because we can find out that $\displaystyle p,q,r$ are) which can be solved using &lt;a href=&quot;http://en.wikipedia.org/wiki/Cubic_function#Cardano.27s_method&quot; rel=&quot;nofollow&quot;&gt;Cardano's Method&lt;/a&gt;, or more simply by the &lt;a href=&quot;http://en.wikipedia.org/wiki/Cubic_function#Trigonometric_.28and_hyperbolic.29_method&quot; rel=&quot;nofollow&quot;&gt;Trigonometric and Hyperbolic Method&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you know $\displaystyle x,y,z$ you can solve for $\displaystyle a,b,c$, as those become just linear equations.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-12-06T07:23:11.153" LastActivityDate="2010-12-06T07:23:11.153" CommentCount="5" />
  <row Id="13198" PostTypeId="1" AcceptedAnswerId="13200" CreationDate="2010-12-06T04:01:47.673" Score="2" ViewCount="164" Body="&lt;p&gt;Hello all,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is probably one of the more embarrassing moments of my life, but I will soldier on. Having not dealt with inequalities for quite some time, I was rather befuddled by a question from a friend of mine regarding solving for two variables given a pair of inequalities. The problem goes like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given $x + y \lt 11$ and $x \gt 6$, what is the lower bound of $x - y$ ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm fairly sure she told me the answer was $4$, but I've exhausted my bag of tricks trying to get there. If these were equalities, it would be fairly easy, of course, but would someone be so kind as to help nudge me in the right direction?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks a lot in advance.&lt;/p&gt;&#xA;" OwnerUserId="1720" LastEditorUserId="1720" LastEditorDisplayName="" LastEditDate="2010-12-06T06:44:31.230" LastActivityDate="2010-12-06T06:44:31.230" Title="Solving for two unknowns in an inequality" Tags="&lt;algebra&gt;&lt;inequality&gt;" AnswerCount="2" CommentCount="10" />
  <row Id="13199" PostTypeId="2" ParentId="13196" CreationDate="2010-12-06T04:04:39.920" Score="10" ViewCount="" Body="&lt;p&gt;Define $g(t):[-\pi,\pi]\to \mathbb{R}$ such that $g(t)=f(t)$ for $t\in [-1,1]$ and $g(t)=0$ elsewhere. Then $g$ is integrable and by Lebesgue-Riemann lemma  $$\lim_{n\to\infty}\int_{-\pi}^\pi  g(t)\cos 2nt dt=0.$$ But this is just the same as $$\lim_{n\to\infty}\int_{-1}^1  f(t)\cos 2nt dt=0.$$ &lt;/p&gt;&#xA;" OwnerUserId="3249" LastActivityDate="2010-12-06T04:04:39.920" />
  <row Id="13200" PostTypeId="2" ParentId="13198" CreationDate="2010-12-06T04:15:00.920" Score="7" ViewCount="" Body="&lt;p&gt;Are $x$ and $y$ required to be &lt;em&gt;integers&lt;/em&gt;? Because if they are not, then the answer cannot be $4$: take $x=6.5$ and $y=4$. Then $x+y = 10.5\lt 11$, but $x-y = 2.5 \lt 4$. In fact, taking $x=6+\epsilon$ and $y=5-(\epsilon+\delta)$ with $\epsilon\gt 0$ and $\delta\gt 0$ arbitrary but positive, and this gives you $x+y = 11-\delta\lt 11$, but $x-y = 1+2\epsilon+\delta$, so we can make $x-y$ as close to $1$ as we like (without it ever being equal to $1$); on the other hand, since $x\gt 6$ and $x+y\lt 11$, then $y\lt 11-x \lt 11-6 = 5$, so $x-y \gt 6-5 = 1$. Thus, the difference is always greater than $1$; but since we can make it as close to $1$ as we want, the best possible lower bound for $x-y$ is $1$ (without it actually being achieved).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if $x$ and $y$ are required to be integers, then the smallest value of $x-y$ will occur when $x$ is as small as possible and $y$ as large as it possibly can be given that. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Added:&lt;/em&gt; To see why, first suppose that $x+y\lt 11$ with $x\gt 6$, but $y$ is not as large as it can be; that is, that $x+(y+1)$ is &lt;em&gt;also&lt;/em&gt; less than $11$. Setting $y&amp;#39;=y+1$, you have that $x-y&amp;#39; = x-y-1 \lt x-y$, so you get a smaller value with $y&amp;#39;$. That means that in order to find the smallest possible value of $x-y$ for a given $x$, you may assume $y$ is as large as it possibly can be. Once you have that, you can check that larger values of $x$ will give you &lt;em&gt;larger&lt;/em&gt; differences, so you also want $x$ to be as small as it possibly can be.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-06T04:37:30.830" LastActivityDate="2010-12-06T04:37:30.830" CommentCount="1" />
  <row Id="13201" PostTypeId="2" ParentId="13198" CreationDate="2010-12-06T04:15:23.557" Score="2" ViewCount="" Body="&lt;p&gt;The answer is $x-y&amp;gt;1$. What you can do is to graph the set $A$: $x+y&amp;lt;11, x&amp;gt;6$ and note that $x-y=1$ passes through the intersection of the two lines $x+y=11$ and $x=6$. The set $A$ is to the right of $x-y=1$, which is $x-y&amp;gt;1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $x,y$ are integers, then the smallest integer $k$ such that the set $x-y\ge k$ includes all the points in $A$ is 4 ( $(7,3)$ lies on the line $x-y=4$). This is clear from the graph.&lt;/p&gt;&#xA;" OwnerUserId="3249" LastEditorUserId="3249" LastEditorDisplayName="" LastEditDate="2010-12-06T06:00:18.477" LastActivityDate="2010-12-06T06:00:18.477" CommentCount="2" />
  <row Id="13202" PostTypeId="1" AcceptedAnswerId="13203" CreationDate="2010-12-06T05:08:25.753" Score="0" ViewCount="31" Body="&lt;p&gt;Alright, so I'm still trying to make my players movements relative to the camera. I got a good answer &lt;a href=&quot;http://stackoverflow.com/questions/4321382?tab=votes#tab-top&quot;&gt;here&lt;/a&gt; But he ends it with: $s(Jx \widehat{\mathbf{x}} + Jy \widehat{\mathbf{y}}$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So how can I turn that equation into an $x$ and $y$?&lt;/p&gt;&#xA;" OwnerUserId="4464" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-06T05:10:37.467" LastActivityDate="2010-12-06T05:21:49.807" Title="Splitting this into a x and y point?" Tags="&lt;geometry&gt;&lt;trigonometry&gt;" AnswerCount="1" CommentCount="0" />
  <row Id="13203" PostTypeId="2" ParentId="13202" CreationDate="2010-12-06T05:21:49.807" Score="1" ViewCount="" Body="&lt;p&gt;If you read the answer you reference, $\widehat{\mathbf{x}}$ is a unit vector perpendicular to the vector from the camera to the player and $\widehat{\mathbf{y}}$ is a unit vector pointing towards the player from the camera.  Each has x and y coordinates in your coordinate plane.  $Jx$ is the x coordinate of the joystick position with a range depending on how you read it out.  Maybe it goes from 0 to 10 degrees.  Similarly for $Jy$.  So you form a new vector by multiplying and adding, then scale it by a constant $s$ that represents how fast the player should move for a given motion of the joystick.  This gives you a vector that you add to the current player position to get the new player position.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-12-06T05:21:49.807" CommentCount="2" />
  <row Id="13204" PostTypeId="2" ParentId="13154" CreationDate="2010-12-06T05:22:35.313" Score="1" ViewCount="" Body="&lt;p&gt;Yes, you're going to find clauses with contradictory literals and apply the resolution rule to those. So you're going to resolve the first two clauses in your example.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are a couple of pages you can look at:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.ai.mit.edu/courses/6.825/fall02/pdf/6.825-lecture-07.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.ai.mit.edu/courses/6.825/fall02/pdf/6.825-lecture-07.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://logic.stanford.edu/classes/cs157/2005/notes/chap05.pdf&quot; rel=&quot;nofollow&quot;&gt;http://logic.stanford.edu/classes/cs157/2005/notes/chap05.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1884" LastActivityDate="2010-12-06T05:22:35.313" CommentCount="1" />
  <row Id="13205" PostTypeId="1" CreationDate="2010-12-06T05:24:11.497" Score="4" ViewCount="372" Body="&lt;p&gt;I would like to know why $\mathbb{C}[x,y]$ is not isomorphic to $\mathbb{C}[x] \otimes _{\mathbb{Z}} \mathbb{C}[y]$ as rings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you! 1&lt;/p&gt;&#xA;" OwnerUserId="4465" LastEditorUserId="4465" LastEditorDisplayName="" LastEditDate="2010-12-06T06:03:04.490" LastActivityDate="2010-12-06T13:56:26.000" Title="Why is $\mathbb{C}[x,y]$ not isomorphic to $\mathbb{C}[x] \otimes _{\mathbb{Z}} \mathbb{C}[y]$ as rings?" Tags="&lt;abstract-algebra&gt;&lt;algebraic-geometry&gt;&lt;commutative-algebra&gt;" AnswerCount="3" CommentCount="7" FavoriteCount="2" />
  <row Id="13206" PostTypeId="2" ParentId="13205" CreationDate="2010-12-06T05:35:24.417" Score="5" ViewCount="" Body="&lt;p&gt;Because $\mathbb C \otimes_{\mathbb Z} \mathbb C$ is not isomorphic to&#xA;$\mathbb C$.  (There is a natural map from the first to the second, with an&#xA;absolutely enormous kernel.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Added: In light of the discussion below, it might help to add the remark that&#xA;this kernel is full of non-trivial zero-divisors (see e.g. Robin's answer).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One way to think of this is that $\mathbb C \otimes_{\mathbb Z} \mathbb C = \mathbb C \otimes_{\mathbb Q} \mathbb C$ contains $\overline{\mathbb Q}\otimes_{\mathbb Q} \overline{\mathbb Q}$, and this latter ring already is full of zero divisors.  (Robin's answer gives one explicit example.  More generally,&#xA;$\overline{\mathbb Q}$ is the union of all finite Galois extensions $L$ of $\mathbb Q$ lying in $\mathbb C$, and for any such $L$, the tensor product&#xA;$L \otimes_{\mathbb Q} L$ is isomorphic to a product of $[L:\mathbb Q]$ copies of $L$.)&lt;/p&gt;&#xA;" OwnerUserId="221" LastEditorUserId="221" LastEditorDisplayName="" LastEditDate="2010-12-06T13:56:26.000" LastActivityDate="2010-12-06T13:56:26.000" CommentCount="5" />
  <row Id="13207" PostTypeId="1" CreationDate="2010-12-06T05:37:50.593" Score="1" ViewCount="368" Body="&lt;p&gt;Find the complete integral of partial differential equation&lt;br&gt;&#xA;&lt;code&gt;z^2 = pqxy&lt;/code&gt; ?&lt;br&gt;&#xA;I have solved this equation till auxiliary equation:   &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;dp/(-pqy+2pz) = dq/(-pqx+2qz) = dz/(2pqxy) = dx/(qxy) = dy/(pxy)  &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;But I have unable to find value of p and q.&lt;br&gt;&#xA;&lt;strong&gt;EDIT:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;p = ∂z/∂x&#xA;q = ∂z/∂y&#xA;r = ∂²z/∂x²  = ∂p/∂x&#xA;s = ∂²z/∂x∂y = ∂p/∂y or ∂q/∂x&#xA;t = ∂²z/∂y²  = ∂q/∂y&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="4463" LastEditorUserId="4463" LastEditorDisplayName="" LastEditDate="2010-12-06T15:50:43.167" LastActivityDate="2011-01-05T20:44:39.410" Title="Charpit's Method" Tags="&lt;differential-equations&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="13208" PostTypeId="2" ParentId="13205" CreationDate="2010-12-06T05:54:49.707" Score="4" ViewCount="" Body="&lt;p&gt;Note that the two are not even isomorphic as $\mathbb{C}$-vector spaces, since the LHS has countable dimension, while for the RHS $\{1\otimes z|\;\bar{z}\in \mathbb{C}/\mathbb{Z}\}$ is an uncountable set that is linearly independent over $\mathbb{C}$.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Edit: another way to see that the rings are not isomorphic, which is just an addition to Matt's answer, is e.g. to note that there is a different number of elements of finite multiplicative order $n$ for any $n&amp;gt;2$. E.g. on the LHS, an element must clearly be a scalar to be of finite multiplicative order, so the only elements of order 4 are $i$ and $-i$. On the other hand, on the RHS, $i\otimes 1$, $-i\otimes 1$, $1\otimes i$ and $-1\otimes i = 1\otimes -i$ are all distinct elements of order 4. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To prove that two elements of the tensor product are really distinct (this applies to any tensor products), you have to construct a $\mathbb{Z}$-bilinear map from $\mathbb{C}[x]\times\mathbb{C}[y]$ to some other $\mathbb{Z}$-module that takes the two corresponding elements (in this case ($i$,1) and (1,$i$), say) to different images. To understand, why that is a necessary and sufficient condition for them to be distinct in the tensor product, go back to the definition of tensor product via a universal property.&lt;/p&gt;&#xA;" OwnerUserId="3212" LastEditorUserId="3212" LastEditorDisplayName="" LastEditDate="2010-12-06T07:35:49.417" LastActivityDate="2010-12-06T07:35:49.417" CommentCount="6" />
  <row Id="13209" PostTypeId="2" ParentId="13109" CreationDate="2010-12-06T05:55:11.647" Score="1" ViewCount="" Body="&lt;p&gt;Note that $x$ cannot occur free in any of the $T_{i}$ in order to apply the Generalization Theorem (namely: $\forall$ Introduction).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A quick look at my old copy of [1] revealed the following:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Enderton proves the Generalization Theorem purely syntactically by induction on the logical axioms, formulas in which $x$ does not occur, and formulas derivable by modus ponens. (In Enderton's system, modus ponens is the only actual rule of inference; all the rest is lumped into lots of logical axioms.)&lt;/li&gt;&#xA;&lt;li&gt;He then proves Soundness entirely separately from the Generalization Theorem with a similar inductive argument with a lemma added to assure the validity of the logical axioms which surprisingly occupies the bulk of the proof.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;[1] &lt;a href=&quot;http://www.math.ucla.edu/~hbe/amil/&quot; rel=&quot;nofollow&quot;&gt;http://www.math.ucla.edu/~hbe/amil/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1884" LastEditorUserId="1884" LastEditorDisplayName="" LastEditDate="2010-12-06T06:10:37.450" LastActivityDate="2010-12-06T06:10:37.450" />
  <row Id="13210" PostTypeId="2" ParentId="13062" CreationDate="2010-12-06T06:07:30.803" Score="5" ViewCount="" Body="&lt;p&gt;Here's the topological argument.  The fact that $G$ is finitely generated means that $G=\pi_1(K)$ for $K$ a CW-complex with finite 1-skeleton.  Let $\widehat{K}$ be the covering space corresponding to $H$.  Then $H=\pi_1(\widehat{K})$, and $\widehat{K}$ also has finite 1-skeleton, so $H$ is finitely generated.&lt;/p&gt;&#xA;" OwnerUserId="2287" LastActivityDate="2010-12-06T06:07:30.803" CommentCount="2" />
  <row Id="13211" PostTypeId="2" ParentId="13067" CreationDate="2010-12-06T06:19:27.813" Score="0" ViewCount="" Body="&lt;p&gt;Maybe it would better to say that &quot;&lt;em&gt;Special functions&lt;/em&gt; does not refer to any particular function, it is merely a mathematical field devoted to the study of some particular explicit function or class of functions in mathematical analysis.&quot;&lt;/p&gt;&#xA;" OwnerUserId="1154" LastActivityDate="2010-12-06T06:19:27.813" />
  <row Id="13212" PostTypeId="2" ParentId="13196" CreationDate="2010-12-06T06:26:10.663" Score="1" ViewCount="" Body="&lt;p&gt;\begin{equation}&#xA;\int_{-1}^1 f(t) \cos^2 nt dt = 1 - \int_{-1}^1 f(t) \sin^2 nt dt &#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can use $\sin^2 nt = \cos^2(nt + \pi/2)$ and some manipulation to show that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;\lim_{n \to \infty} \int_{-1}^1 f(t) \cos^2 nt dt = \lim_{n \to \infty} \int_{-1}^1 f(t) \sin^2 nt dt&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Combining these, you get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;\lim_{n \to \infty} \int_{-1}^1 f(t) \cos^2 nt dt = \frac{1}{2}&#xA;\end{equation}&lt;/p&gt;&#xA;" OwnerUserId="2641" LastActivityDate="2010-12-06T06:26:10.663" />
  <row Id="13213" PostTypeId="1" AcceptedAnswerId="13214" CreationDate="2010-12-06T07:07:14.383" Score="2" ViewCount="72" Body="&lt;p&gt;&lt;strong&gt;If $f(x)$ is a function satisfying $ \displaystyle f(x+y) = f(x) \cdot f(y)  \text { for all } x,y \in \mathbb{N} \text{ such that } f(1) = 3 \text { and } $ $ \sum_{x=1}^{n} f(x) = 120 $, then find the value of $n$.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How to approach this one?&lt;/p&gt;&#xA;" OwnerUserId="2109" LastActivityDate="2010-12-06T07:12:31.667" Title="Function Problem" Tags="&lt;algebra-precalculus&gt;" AnswerCount="1" />
  <row Id="13214" PostTypeId="2" ParentId="13213" CreationDate="2010-12-06T07:12:31.667" Score="7" ViewCount="" Body="&lt;p&gt;Hint: Knowing the value of $f(1)$, you can get the value of $f(2)=f(1+1)$ from the defining equation. Knowing these two, you can get value of $f(3)=f(1+2)$ and so on.&lt;/p&gt;&#xA;" OwnerUserId="3431" LastActivityDate="2010-12-06T07:12:31.667" CommentCount="2" />
  <row Id="13215" PostTypeId="1" CreationDate="2010-12-06T07:59:47.883" Score="1" ViewCount="106" Body="&lt;p&gt;Let $(X,M,\mu)$ be a measure space. Why are the following statements equivalent?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i) There exists a sequence of pairwise disjoint measurable sets $\{A_{n}\}$ in $M$ , each of finite measure, such that $\mu(A) = \sum_{n=1}^{\infty} \mu(A \cap A_{n})$ for every measurable set $A \in M$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ii) $\mu$ is a countable sum of pairwise mutually singular finite measures.&lt;/p&gt;&#xA;" OwnerUserId="4467" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-07T19:28:52.773" LastActivityDate="2010-12-07T19:28:52.773" Title="Equivalence of two properties of a measure space" Tags="&lt;measure-theory&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="13217" PostTypeId="2" ParentId="13215" CreationDate="2010-12-06T08:23:09.347" Score="1" ViewCount="" Body="&lt;p&gt;(ii) $\implies$ (i)&#xA;Suppose $\mu = \sum_{n=1}^\infty \mu_n$, where $\mu_i \perp \mu_j$ for all $i \neq j$.  By definition of &quot;mutually singular,&quot; there exist pairwise disjoint $A_n \in M$ such that $\mu_n(A) = \mu_n(A \cap A_n)$ for all $A \in M$.  Therefore, $\mu(A_k) = \sum_{n=1}^\infty \mu_n(A_k) = \mu_k(A_k) &amp;lt; \infty$, so that each $A_n$ has finite $\mu$-measure.  Moreover, $\mu(A) = \sum_{n=1}^\infty \mu_n(A_n) = \sum_{n=1}^\infty \mu_n(A \cap A_n) = \sum_{n=1}^\infty \mu(A \cap A_n)$, where in the last step we used the fact that $\mu(A \cap A_n) = \mu_n(A \cap A_n)$ for all $n \in N$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(i) $\implies$ (ii)&#xA;Let $\mu_n$ be a measure concentrated on $A_n \in M$, normalized so that $\mu_n(A_n) = 1$.  Then $\mu(A) = \sum_{n=1}^\infty \mu(A\cap A_n) = \sum_{n=1}^\infty \mu_n(A)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(EDIT following Nate's comment: We can define $\mu_n(A) = \mu(A \cap A_n)$.)&lt;/p&gt;&#xA;" OwnerUserId="640" LastEditorUserId="640" LastEditorDisplayName="" LastEditDate="2010-12-06T18:42:38.047" LastActivityDate="2010-12-06T18:42:38.047" CommentCount="2" />
  <row Id="13218" PostTypeId="2" ParentId="13174" CreationDate="2010-12-06T08:29:30.680" Score="4" ViewCount="" Body="&lt;p&gt;@Moron's solution in insightful, but I'll present the (tedious, but thoughtlessly mechanical) methodology I tend to use to approach such problems: &lt;a href=&quot;http://mathworld.wolfram.com/Resultant.html&quot; rel=&quot;nofollow&quot;&gt;resultants&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The resultant of two polynomials is a polynomial with one variable eliminated. Therefore, you can proceed to solve a polynomial system in roughly the same way you would with a linear system (in theory).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, convert your equations to polynomials, $p_i$, so that the &lt;em&gt;roots&lt;/em&gt; of these polynomials are the solutions to your equations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{eqnarray}&#xD;&#xA;p_0 &amp;amp;=&amp;amp; a+b+c-m_0\\&#xD;&#xA;p_1 &amp;amp;=&amp;amp; a x + b y + c z - m_1\\&#xD;&#xA;p_2 &amp;amp;=&amp;amp; a x^2 + b y^2 + c z^2 - m_2\\&#xD;&#xA;p_3 &amp;amp;=&amp;amp; a x^3 + b y^3 + c z^3 - m_3\\&#xD;&#xA;p_4 &amp;amp;=&amp;amp; a x^4 + b y^4 + c z^4 - m_4\\&#xD;&#xA;p_5 &amp;amp;=&amp;amp; a x^5 + b y^5 + c z^5 - m_5&#xD;&#xA;\end{eqnarray}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pick your favorite one (say, $p_0$), and use it to compute a bunch of resultants (using &lt;em&gt;Mathematica&lt;/em&gt;'s ${\tt Resultant[]}$ function) that eliminate one variable (say, $a$):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{eqnarray}&#xD;&#xA;r_0 &amp;amp;=&amp;amp; {\tt Resultant[} p_0, p_1, a {\tt ]} = -m_1 - b x - c x + m_0 x + b y + c z\\&#xD;&#xA;r_1 &amp;amp;=&amp;amp; {\tt Resultant[} p_0, p_2, a {\tt ]} = -m_2 - b x^2 - c x^2 + m_0 x^2 + b y^2 + c z^2 \\&#xD;&#xA;r_2 &amp;amp;=&amp;amp; {\tt Resultant[} p_0, p_3, a {\tt ]} = -m_3 - b x^3 - c x^3 + m_0 x^3 + b y^3 + c z^3\\&#xD;&#xA;r_3 &amp;amp;=&amp;amp; {\tt Resultant[} p_0, p_4, a {\tt ]} = -m_4 - b x^4 - c x^4 + m_0 x^4 + b y^4 + c z^4\\&#xD;&#xA;r_4 &amp;amp;=&amp;amp; {\tt Resultant[} p_0, p_5, a {\tt ]} = -m_5 - b x^5 - c x^5 + m_0 x^5 + b y^5 + c z^5\\&#xD;&#xA;\end{eqnarray}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Of course, the above amounts to simply solving $p_0 = 0$ for $a$ and substituting into the other $p_i$, since the system is linear in $a$. The same will be true when eliminating $b$ and $c$, so using resultants here is a little bit of overkill, but the process is mechanical enough to be about as easy to (not-)think about as, say, Gaussian elimination.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, we use, say, $r_0$ to compute resultants that eliminate $b$, and so forth. For your special system, the polynomials at each step happen to have common factors that give rise to possible solutions; you might need to consider those separately, but they tend to represent &quot;special case&quot; scenarios. In the following, I base the next resultant step on the uncommon factors (indicated with upper-case variables).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{eqnarray}&#xD;&#xA;s_i &amp;amp;=&amp;amp; {\tt Resultant[} r_0, r_i, b {\tt ]} = S_i \cdot (x-y), \hspace{0.2in}i = 1, 2, 3, 4\\&#xD;&#xA;t_i &amp;amp;=&amp;amp; {\tt Resultant[} S_0, S_i, c {\tt ]} = T_i \cdot (y-z)(z-x), \hspace{0.2in}i = 1, 2, 3 \\&#xD;&#xA;u_i &amp;amp;=&amp;amp; {\tt Resultant[} T_0, T_i, x {\tt ]} = U_i \cdot (m_2-m_1 y - m_1 z + m_0 y z ), \hspace{0.2in}i = 1, 2 \\&#xD;&#xA;v_1 &amp;amp;=&amp;amp; {\tt Resultant[} U_0, U_1, y {\tt ]}&#xD;&#xA;\end{eqnarray}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The final polynomial, $v_1$, has just one variable, $z$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{eqnarray}v_1 &amp;amp;=&amp;amp; \left(m_2^2 - m_1 m_3 - m_1 m_2 z + m_0 m_3 z + m_1^2 z^2 - m_0 m_2 z^2\right)^4 \\ &#xD;&#xA;&amp;amp;&amp;amp; \left(-m_3^3 + 2 m_2 m_3 m_4 - m_1 m_4^2 - m_2^2 m_5 + m_1 m_3 m_5 \right.\\&#xD;&#xA;&amp;amp;&amp;amp;\left. + z\left(m_2 m_3^2 - m_2^2 m_4 - m_1 m_3 m_4 + m_0 m_4^2 + m_1 m_2 m_5 - m_0 m_3 m_5 \right) \right. \\&#xD;&#xA;&amp;amp;&amp;amp;\left. +z^2\left(- m_2^2 m_3 + m_1 m_3^2 + m_1 m_2 m_4 - m_0 m_3 m_4 - m_1^2 m_5 + m_0 m_2 m_5\right) \right. \\&#xD;&#xA;&amp;amp;&amp;amp;\left. + z^3\left( m_2^3 - 2 m_1 m_2 m_3 + m_0 m_3^2 + &#xD;&#xA;   m_1^2 m_4 - m_0 m_2 m_4\right)\right)^2\end{eqnarray}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first factor is another &quot;special case&quot;, so that $z$, in general, should be a root of the cubic polynomial in the second factor. To get $y$, and $x$, and the rest, you could back-substitute into polynomials from earlier in the resultant chain (just as you would with a linear system), or you could just compute resultant chains that eliminate every variable except $y$, then every variable except $x$, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It turns out that (as one might expect) your particular system is such that the &quot;final&quot; polynomials for $x$, $y$, and $z$ are equivalent; that is, $x$, $y$, and $z$ are roots of the cubic in the second factor of $v_1$ above. (This is, in fact, @Moron's cubic.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll note that resultants in general are computationally expensive (especially in symbolic form). Very often, the process can bog down completely in &lt;em&gt;Mathematica&lt;/em&gt;, even for fairly modest systems, and/or the final polynomial can be of enormous degree. Sometimes, a bit of finesse in the order of elimination helps. My first attempt at finding (a chain of resultants that eliminated everything but) $a$ got boggy, but eliminating in the order $b$, $c$, $z$, $y$, $x$ gave me a cubic equation (and some &quot;special case&quot; factors) ... which is &lt;em&gt;huge&lt;/em&gt;, so I won't present it here. A more-practical approach here might be to find $x$, $y$, and $z$ from the cubic polynomial, substitute these values back into three of the $p_i$, and solve the linear system for $a$, $b$, and $c$.&lt;/p&gt;&#xA;" OwnerUserId="409" LastActivityDate="2010-12-06T08:29:30.680" />
  <row Id="13219" PostTypeId="1" AcceptedAnswerId="13227" CreationDate="2010-12-06T08:47:45.277" Score="3" ViewCount="112" Body="&lt;p&gt;From Statistical Inference Second Edition (George Casella, Roger L. Berger)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;My telephone rings 12 times each week, the calls being randomly distributed among the 7 days. What is the probability that I get a least one call each day?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer is .2285, but I don't know how they got it. My reasoning was as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are 12 calls and thus 13 places to put &quot;day dividers&quot; to produce possible distributions of calls. There should be 6 day dividers for the week. One possibility:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;_1|2_3|4_5_6|7|8||9_10_11_12&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;(1 on Monday, 2 on Tuesday, 3 on Wednesday, 1 on Thursday, 1 on Friday, 0 on Saturday, 4 on Sunday)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are &lt;code&gt;13^6&lt;/code&gt; possible distributions (using this method). In order to satisfy 1 call/day, dividers can't be at the beginning or end, nor can they be on top of one another (signifying a day with no calls). This means there are:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;11*10*9*8*7*6&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Valid distributions and a probability of: 0.069. Where am I going wrong?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(P.S. This isn't homework as I'm not in school.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: I don't think each &quot;distribution&quot; is equally likely. That's probably my error. But I still don't know how to get to the correct answer :)&lt;/p&gt;&#xA;" OwnerUserId="4468" LastEditorUserId="4468" LastEditorDisplayName="" LastEditDate="2010-12-06T08:53:31.400" LastActivityDate="2010-12-06T11:51:31.283" Title="Statistical Inference Question" Tags="&lt;probability&gt;&lt;statistics&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="13220" PostTypeId="1" AcceptedAnswerId="13222" CreationDate="2010-12-06T09:03:55.007" Score="1" ViewCount="156" Body="&lt;p&gt;I have a vector A and a point P. My problem is how to find out if point is on the left or right side of vector looking from the point of origin of a vector in direction of it.&lt;/p&gt;&#xA;" OwnerUserId="3563" LastActivityDate="2010-12-06T12:01:45.943" Title="On which side of vector the point is" Tags="&lt;euclidean-geometry&gt;&lt;vector-analysis&gt;" AnswerCount="2" CommentCount="5" />
  <row Id="13222" PostTypeId="2" ParentId="13220" CreationDate="2010-12-06T09:58:44.330" Score="3" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://mathoverflow.net/questions/44096/detecting-whether-directed-cycle-is-clockwise-or-counterclockwise/44098#44098&quot; rel=&quot;nofollow&quot;&gt;This answer&lt;/a&gt; on MO should do it for you.&lt;/p&gt;&#xA;" OwnerUserId="3212" LastActivityDate="2010-12-06T09:58:44.330" />
  <row Id="13224" PostTypeId="2" ParentId="13111" CreationDate="2010-12-06T11:05:53.543" Score="4" ViewCount="" Body="&lt;p&gt;OK, the following results lead to a solution to this and similar problems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Theorem. Let $G$ be a finite simple group with a subgroup $H$ of index $n&amp;gt;1$. Then $n \ge 5$, and $|G|$ divides $n!/2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proof. Let $\phi: G \rightarrow S_n$ be the permutation representation of $G$ acting by (left or right depending on whether you use left or right group actions) multiplication on the set of (left or right) cosets of $H$ in $G$. Then  $G/{\rm Ker}(\phi) \cong {\rm im}(\phi) \le S_n$. Since $n&amp;gt;1$ and ${\rm im}(\phi)$ is transitive, $|{\rm im}(\phi)| &amp;gt; 1$ and so $G$ simple implies ${\rm Ker}(\phi) = 1$, and hence $G \cong {\rm im}(\phi)$. Now $S_n$ is solvable for $n &amp;lt; 5$, so we must have $n \ge 5$. Furthermore, we must have ${\rm im}(\phi) \le A_n$, since otherwise ${\rm im}(\phi) \cap A_n$ would be a normal subgroup of ${\rm im}(\phi)$ of index 2, and so $G$ would not be simple.&#xA;Hence $|G|$ divides $|A_n| = n!/2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Corollary. Let $G$ be a finite simple group and $n = |{\rm Syl}_p(G)|$ for some prime $p$ dividing $|G|$. Then  $n \ge 5$ and $|G|$ divides $n!/2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proof. Let $P \in {\rm Syl}_p(G)$. We cannot have $n=1$ because then  $P$ would be normal in $G$. Now apply the theorem to the subgroup $N_G(P)$ of index $n$ in $G$.&lt;/p&gt;&#xA;" OwnerUserId="2820" LastActivityDate="2010-12-06T11:05:53.543" CommentCount="1" />
  <row Id="13225" PostTypeId="2" ParentId="13205" CreationDate="2010-12-06T11:17:46.260" Score="6" ViewCount="" Body="&lt;p&gt;To show that $1\otimes 1+i\otimes i$ is nonzero&#xA;in $\mathbb{C}[X]\otimes_{\mathbb{Z}}\mathbb{C}[X]$ note that it maps&#xA;to $1\otimes 1+i\otimes i$ in&#xA;$\mathbb{C}[X]\otimes_{\mathbb{R}}\mathbb{C}[X]$.&#xA;This is a tensor product over a field, so a basis as an&#xA;$\mathbb{R}$-vector space is got by tensoring togther bases&#xA;on each side. Now as the set of elements of the forms $X^n$&#xD;&#xA;and $iX^n$ are bases of $\mathbb{C}[X]$ over $\mathbb{R}$&#xD;&#xA;then $1\otimes 1$ and $i\otimes i$ are linearly independent&#xA;over $\mathbb{R}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As&#xA;$$(1\otimes 1+i\otimes i)(1\otimes 1-i\otimes i)=0$$&#xD;&#xA;then $\mathbb{C}[X]\otimes_{\mathbb{Z}}\mathbb{C}[X]$&#xD;&#xA;isn't an integral domain, unlike $\mathbb{C}[X]\otimes_{\mathbb{C}}\mathbb{C}[X]$.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-12-06T11:17:46.260" />
  <row Id="13226" PostTypeId="2" ParentId="13220" CreationDate="2010-12-06T11:33:13.707" Score="0" ViewCount="" Body="&lt;p&gt;Let $B$ be the vector from the end point of $A$ to $P$, then observe the sign of the coefficient of $A \wedge B$ in a chosen basis. The exact sign is up to $O(2)/SO(2)$, but you only need to distinguish between the two cases, so it's ok :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;UPD: Now that I think about it, letting $B$ be the vector from the starting point of $A$ to $P$ is even neater.&lt;/p&gt;&#xA;" OwnerUserId="3793" LastEditorUserId="3793" LastEditorDisplayName="" LastEditDate="2010-12-06T12:01:45.943" LastActivityDate="2010-12-06T12:01:45.943" />
  <row Id="13227" PostTypeId="2" ParentId="13219" CreationDate="2010-12-06T11:51:31.283" Score="2" ViewCount="" Body="&lt;p&gt;Here is an easy approach to solve the problem. Consider the tuples $(n_1,n_2,\ldots,n_7)$ such that $n_i$ are positive integers satisfying $n_1 + n_2 + \cdots + n_7 = 12$. Further, assume for the moment that $n_1 \geq n_2 \geq \cdots \geq n_7$.&#xA;You should find that there are only seven such tuples. Take, for example, the tuple $(4,3,1,1,1,1,1)$. The corresponding probability is, according to the multinomial distribution with parameters $n=12$ and $p_1 = p_2 = \cdots = p_7 = 1/7$, &#xA;$$&#xD;&#xA;\frac{{12!}}{{4!3!1!1!1!1!1!}}\bigg(\frac{1}{7}\bigg)^{4}\bigg(\frac{1}{7}\bigg)^{3}\bigg(\frac{1}{7}\bigg)^{1}\bigg(\frac{1}{7}\bigg)^{1}\bigg(\frac{1}{7}\bigg)^{1}\bigg(\frac{1}{7}\bigg)^{1}\bigg(\frac{1}{7}\bigg)^{1} = &#xD;&#xA;\frac{{12!}}{{4!3!}}\bigg(\frac{1}{7}\bigg)^{12}.&#xD;&#xA;$$&#xD;&#xA;Now, if we don't assume that $n_1 \geq n_2 \geq \cdots \geq n_7$, then you should multiply the above probability by ${7 \choose 2}2$. Doing the same for the other tuples, you obtain seven probabilities. Their sum is the probability you are looking for. I have done the calculation and the result is $\approx 0.228452440447$.&lt;/p&gt;&#xA;" OwnerUserId="2810" LastActivityDate="2010-12-06T11:51:31.283" CommentCount="1" />
  <row Id="13228" PostTypeId="1" CreationDate="2010-12-06T12:30:24.297" Score="0" ViewCount="70" Body="&lt;p&gt;Had asked this in the Physics forum, and they told me to ask it here.&#xA;I've used a &lt;a href=&quot;http://en.wikipedia.org/wiki/Kalman_filter&quot; rel=&quot;nofollow&quot;&gt;Kalman filter&lt;/a&gt; with a constant velocity motion model, where the $X$ matrix is [$x$ $v_x$ $y$ $v_y$]'.&#xA;But for the Nearly Coordinated Turn model, the $X$ matrix is [$x$ $v_x$ $y$ $v_y$ $\omega$]'. &lt;em&gt;($\omega$ is angular velocity)&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've tried an NCT model from &lt;a href=&quot;http://books.google.co.in/books?id=xz9nQ4wdXG4C&amp;amp;printsec=frontcover&amp;amp;dq=estimation+with+applications+to+tracking+and+navigation&amp;amp;source=bl&amp;amp;ots=Vn_Vt3d1u0&amp;amp;sig=MWNdwWa8V53k2wN-8bg8PO8wtxo&amp;amp;hl=en&amp;amp;ei=m3j8TJWqOcjJrAe5qqz-Bw&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=1&amp;amp;ved=0CBwQ6AEwAA#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; book, which says that the omega value has to be estimated with a separate equation containing white noise acceleration. I want to know how this could be possible, as filtering is done to eliminate noise as much as possible. Then how can adding noise to the angular velocity help, because wouldn't that make the predictions randomly turn left and right even if the target is turning left constantly? (If I'm asking the question in the wrong forum, kindly redirect me. I'll delete this question)&lt;/p&gt;&#xA;" OwnerUserId="3737" LastActivityDate="2010-12-06T12:30:24.297" Title="How is a Kalman filter with a Nearly Coordinated Turn model meant to work?" Tags="&lt;algorithms&gt;&lt;taylor-expansion&gt;&lt;matlab&gt;" FavoriteCount="1" />
  <row Id="13229" PostTypeId="1" CreationDate="2010-12-06T12:30:37.850" Score="0" ViewCount="103" Body="&lt;p&gt;Hello,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am computing the integral of some discrete data, in the form:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;t  m^3/s&#xA;0  0.75116&#xA;30 0.55621&#xA;...&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;However, the results for the smallest possible step (120) are not good at all:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Trapezoids rule&#xA;S: 0.111159 (step 120)&#xA;S': 0.111177 (step 60)&#xA;S'': 0.111188 (step 30)&#xA;------------------&#xA;Convergence Quotient ~= 1.73291 (should be 4&#xA;Error ~= 3.44458e-006 (meaningless for CQ != 4)&#xA;&#xA;&#xA;Simpson's method&#xA;------------------&#xA;h: 120&#xA;------------------&#xA;S: 0.111003 (step 120)&#xA;S': 0.111183 (step 60)&#xA;S'': 0.111191 (step 30)&#xA;------------------&#xA;Convergence quotient ~= 23.0815 (should be 16)&#xA;Error ~= 5.20611e-007 (meaningless for CQ != 16)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Samples range from 0 to 86400 in increments of 30.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why does this happen? Floating point imprecisions? Something else I am missing?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="4470" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-07T15:04:00.683" LastActivityDate="2010-12-07T15:04:00.683" Title="Trapezoid/Simpson's method for discrete data gives unreliable results" Tags="&lt;integral&gt;&lt;numerical-methods&gt;" CommentCount="1" FavoriteCount="1" />
  <row Id="13231" PostTypeId="1" AcceptedAnswerId="13232" CreationDate="2010-12-06T12:40:53.987" Score="0" ViewCount="188" Body="&lt;p&gt;What does D = max[0; M(x)] mean?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;M(x) is a function.&lt;/p&gt;&#xA;" OwnerUserId="3532" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-07T00:04:59.307" LastActivityDate="2010-12-07T00:04:59.307" Title="What does max[] mean?" Tags="&lt;notation&gt;" AnswerCount="2" />
  <row Id="13232" PostTypeId="2" ParentId="13231" CreationDate="2010-12-06T12:52:36.967" Score="3" ViewCount="" Body="&lt;p&gt;Taking the maximal number amongst the parameters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\max\{x_1,x_2\} = \cases{x_1, \text{if }x_1 &amp;gt; x_2\\x_2, \text{otherwise}}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can define like that the maximum of any finitely many elements. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the parameters are an infinite set of values, then it is implied that one of them is maximal (namely that there is a greatest one, unlike the set $\{-\frac{1}{n} | n\in\mathbb{N}\}$ where there is no greatest element)&lt;/p&gt;&#xA;" OwnerUserId="622" LastActivityDate="2010-12-06T12:52:36.967" CommentCount="1" />
  <row Id="13233" PostTypeId="1" AcceptedAnswerId="13239" CreationDate="2010-12-06T13:23:19.617" Score="1" ViewCount="70" Body="&lt;p&gt;Can somebody prove that when we add 2 bit integers and if there is an overflow then the result always will be lesser than the 2 operands used??&lt;/p&gt;&#xA;" OwnerUserId="1353" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-06T14:01:52.920" LastActivityDate="2010-12-06T15:31:47.027" Title="detecting 2 n bits integer addition overflow" Tags="&lt;arithmetic&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="13234" PostTypeId="2" ParentId="13155" CreationDate="2010-12-06T13:34:16.697" Score="2" ViewCount="" Body="&lt;p&gt;Also, in agreement to what kahen said:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a regular parametrization, you can always find a metric that is constant along the curve leading to zero intrinsic curvature. This is also a reason why it might be somehow not very enlightening to discuss curves as one dimensional manifolds in its own right.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Curves, on the other hand, can have extrinsic curvature which can for some body moving along a trajectory be interpreted as the acting force.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Greets&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Robert&lt;/p&gt;&#xA;" OwnerUserId="2671" LastActivityDate="2010-12-06T13:34:16.697" />
  <row Id="13235" PostTypeId="2" ParentId="13231" CreationDate="2010-12-06T13:35:33.057" Score="4" ViewCount="" Body="&lt;p&gt;In this case, I can also think of $D$ as a function in $x$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$D(x) = max(0, M(x)) = \begin{cases} M(x), &amp;amp;M(x) &amp;gt;0 \\ 0, &amp;amp;else\end{cases}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The result is essentially $M$ &quot;cut off&quot; at $0$.&lt;/p&gt;&#xA;" OwnerUserId="3330" LastActivityDate="2010-12-06T13:35:33.057" CommentCount="3" />
  <row Id="13236" PostTypeId="2" ParentId="13109" CreationDate="2010-12-06T13:38:39.067" Score="0" ViewCount="" Body="&lt;p&gt;Hi&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do I understand your query correctly that you want to universally abstract from constant $t$, to a variable $x$. Actually I would write $Q[x/t]$ instead of $Q[t/x]$, since you are replacing $t$ by $x$, right? Also I rather prefer writing $Q[c/x]$ instead $Q[t/x]$ to indicate dealing with a constant c and not with a term t.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think it can be shown that your rule is a combination of a lemma and another rule. If you can show soundness of the lemma and the other rule you are done.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The lemma is the lifting lemma which holds for a great deal of logics. It can be formulated for variables and it can be formulated for constants. The version for &#xA;constants reads:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$T \vdash Q$            c/t does not cause name clash&lt;br&gt;&#xA;---------------- (Lifting)&lt;br&gt;&#xA;$T[c/t] \vdash Q[c/t]$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The above lemma can be proved via induction on the proof. The other rule is&#xA;now abstract of a variable:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$T \vdash Q$            x not in T&lt;br&gt;&#xA;---------------  (Right Forall)&lt;br&gt;&#xA;$T \vdash \forall x Q$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now how can we combine the two to arrive at your result? We simply need to observe&#xA;that when c does not occur in T, then T[c/t]=T. Therefore we have:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$T \vdash Q$        c not in T&lt;br&gt;&#xA;-----------  (Lifting)  c/x does not cause name clash&lt;br&gt;&#xA;$T \vdash Q[c/x]$&lt;br&gt;&#xA;-------------------- (Right Forall) x not in T&lt;br&gt;&#xA;$T \vdash \forall x Q[c/x]$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you can see an additional condition sneaks in. We do not only have to have c not&#xA;in $T$, but also $x$ not in $T$ and $c/x$ does not cause a name clash. I am not &#xA;sure whether the later two conditions correspond to your condition that $x$ may not &#xA;occur in $Q$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Should $x$ not occur in $Q$ mean not occur free or not occur bound? Normally non &#xA;occurence refers to free occurence. If it occurs bound we might observe a name &#xA;clash and generate unsoundness. Need to find a counter example still.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Best Regards&lt;/p&gt;&#xA;" OwnerUserId="4414" LastEditorUserId="4414" LastEditorDisplayName="" LastEditDate="2010-12-09T08:22:30.440" LastActivityDate="2010-12-09T08:22:30.440" />
  <row Id="13237" PostTypeId="1" AcceptedAnswerId="13240" CreationDate="2010-12-06T13:40:40.000" Score="0" ViewCount="85" Body="&lt;p&gt;Hey!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a mathematical function such that;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;f(3, 5)   = 3&#xA;f(10, 2)  = 2&#xA;f(14, 15) = 14&#xA;f(9, 9)   = 9&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;It would be even more cool if there's a function that takes three (3) parameters, but that one could be solved by using recursive functionality;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;f( f(3, 5), 4) = 3&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="4477" LastActivityDate="2010-12-07T15:31:56.650" Title="Arithmetic function to return lowest in-parameter" Tags="&lt;functions&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="13239" PostTypeId="2" ParentId="13233" CreationDate="2010-12-06T13:45:08.993" Score="3" ViewCount="" Body="&lt;p&gt;Suppose $a$ and $b$&amp;#160;are the numbers you are adding. Suppose that $0 \le a,b &amp;lt; 2^n$. Then we must have that $a+b - 2^n &amp;lt; a$ and $a+b - 2^n &amp;lt; b$. Hence if overflow occurs the result is smaller than $a$ or $b$.&lt;/p&gt;&#xA;" OwnerUserId="3776" LastActivityDate="2010-12-06T13:45:08.993" />
  <row Id="13240" PostTypeId="2" ParentId="13237" CreationDate="2010-12-06T13:46:36.730" Score="6" ViewCount="" Body="&lt;p&gt;$$f(x,y)=\frac{x+y-|x-y|}{2}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Oscar gave a nice interpretation of the above formula in &lt;a href=&quot;http://math.stackexchange.com/questions/13253&quot;&gt;his follow-up question&lt;/a&gt;, but I'll give a dumb derivation here for completeness.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Making use of &lt;a href=&quot;http://mathworld.wolfram.com/IversonBracket.html&quot; rel=&quot;nofollow&quot;&gt;Iversonian brackets&lt;/a&gt;, we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\min(x,y)=x[y \geq x]+y[y &amp;lt; x]$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and since $[\neg p]=1-[p]$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\min(x,y)=x[y \geq x]+y(1-[y \geq x])=y-(y-x)[y-x \geq 0]$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, there is the identity&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{u+|u|}{2}=u[u \geq 0]$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and so we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\min(x,y)=y+\frac{x-y-|x-y|}{2}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which simplifies to the desired expression.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The extension to more than two arguments is no longer as compact, though, since one now has to contend with products of Iversonian brackets ($[p \land q]=[p]\cdot[q]$).&lt;/p&gt;&#xA;" OwnerUserId="498" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-07T15:31:56.650" LastActivityDate="2010-12-07T15:31:56.650" CommentCount="3" />
  <row Id="13241" PostTypeId="1" AcceptedAnswerId="13243" CreationDate="2010-12-06T13:47:32.250" Score="0" ViewCount="245" Body="&lt;p&gt;We know that the set of continuous functions and the set of smooth functions in $\mathcal{L}^1$ or $\mathcal{L}^2$ are dense in $\mathcal{L}^1$ and $\mathcal{L}^2$. Is the set of $\mathcal{C}^k$ functions dense in $\mathcal{L}^1$ or $\mathcal{L}^2$ for any $k\in\mathbb{N}$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is the set $\mathcal{C}^k\setminus \mathcal{C}^{\infty}$ in $\mathcal{L}^1$ or $\mathcal{L}^2$ dense in $\mathcal{L}^1$ or $\mathcal{L}^2$. How would one prove this one way or the other?&lt;/p&gt;&#xA;" OwnerUserId="2987" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-06T15:18:47.637" LastActivityDate="2011-03-30T06:33:44.833" Title="Are the $\mathcal{C}^k$ functions dense in either $\mathcal{L}^2$ or $\mathcal{L}^1$ ?" Tags="&lt;real-analysis&gt;" AnswerCount="3" CommentCount="8" />
  <row Id="13242" PostTypeId="1" AcceptedAnswerId="33813" CreationDate="2010-12-06T14:03:19.453" Score="0" ViewCount="82" Body="&lt;p&gt;I recently made a model to solve a conflict problem in sports.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The model is based on Bayesian Games: a complete definition can be found &lt;a href=&quot;http://en.wikipedia.org/wiki/Bayesian_game&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The thing is that I am now looking to compute the equilibria of the system, but I don't know any algorithm to do so.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do you know any?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which language do you think I should use, I was a priori going to use R.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anybody has any experience with this kind of problem?&lt;/p&gt;&#xA;" OwnerUserId="3354" LastActivityDate="2011-04-19T09:40:10.137" Title="Computing Bayesian Nash Equilibrium" Tags="&lt;game-theory&gt;" AnswerCount="1" />
  <row Id="13243" PostTypeId="2" ParentId="13241" CreationDate="2010-12-06T14:23:55.523" Score="2" ViewCount="" Body="&lt;p&gt;Here is the idea behind the proof.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Step functions are dense in $L^p$ ($p=1,2$). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Given a step function $T$ and $\varepsilon&amp;gt;0$ there is a $g\in C^k$ &#xA;$$\|T-g\|_{L^p}\lt\varepsilon$$&#xD;&#xA;($p=1,2$). &lt;/p&gt;&#xA;" OwnerUserId="1154" LastActivityDate="2010-12-06T14:23:55.523" CommentCount="9" />
  <row Id="13244" PostTypeId="1" CreationDate="2010-12-06T14:29:56.613" Score="3" ViewCount="121" Body="&lt;p&gt;Let $G$ and $H$ be two possibly directed, non necessarily simple, vertex-labelled graphs with respective adjacency matrices $A_G$ and $A_H$ and $V(G)=V(H)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) What is the name of the graph $M$ with adjacency matrix $A_M=A_HA_G$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) In the unlikely event that I am the first to think about using that operation, which symbols should I NOT use to denote it in order to avoid confusion with &lt;a href=&quot;http://en.wikipedia.org/wiki/Graph_product&quot; rel=&quot;nofollow&quot;&gt;other graph products&lt;/a&gt;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;UPDATE:&lt;/strong&gt; I have now asked this question on &lt;a href=&quot;http://mathoverflow.net/questions/49233/name-of-an-operation-on-graphs&quot; rel=&quot;nofollow&quot;&gt;MathOverflow&lt;/a&gt; as well.&lt;/p&gt;&#xA;" OwnerUserId="689" LastEditorUserId="689" LastEditorDisplayName="" LastEditDate="2010-12-13T14:18:04.337" LastActivityDate="2010-12-13T14:18:04.337" Title="Name of an operation on graphs" Tags="&lt;reference-request&gt;&lt;graph-theory&gt;&lt;terminology&gt;" AnswerCount="3" CommentCount="5" />
  <row Id="13245" PostTypeId="1" AcceptedAnswerId="13385" CreationDate="2010-12-06T14:32:13.693" Score="2" ViewCount="290" Body="&lt;p&gt;I'm having trouble understanding this proof:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Lemma:&lt;/strong&gt; Let $\lambda$ be the Lebesgue measure restricted to the Borel sets of $[0,1]$. Then there exists a decreasing sequence $(E_n)_{n\ge0}$ of subsets of $[0,1]$ such that each of them is $\lambda$-thick and $\bigcap E_n = \emptyset$.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;Proof:&lt;/strong&gt; Let $E_n = [0,1]\cap\{f\ge n\}$, where $f$ is a noncontinuous solution $f:\mathbb{R}\longrightarrow\mathbb{R}$ of $f(x+y) = f(x) + f(y)$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A continuous solution of that functional equation is a line through the origin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let ~ be the equivalence relation on $\mathbb{R}-\{0\}$ such that $x$ ~ $y$ $\leftrightarrow$ $\exists q\in\mathbb{Q}$ s.t. $x = qy$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $A$ be a set of representatives from each equivalence class. Let $g:A\longrightarrow\mathbb{R}$ be a non-constant &quot;assignment of slopes&quot;. Let $f:\mathbb{R} - \{0\}\longrightarrow\mathbb{R}$ s.t. $x \mapsto x\cdot g(y)$ where $x$~$y\in A$, let $f(0) = 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To put it roughly, a noncontinuous solution is a stack of lines through the origin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strike&gt;a) Is this description of a noncontinuous solution correct?&lt;/strike&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;No it's not! As TCL helped me realize, this construction does satisfy the desired functional equation only for x and y in the same equivalence class. I need to require something more from g.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;b) Why does the proof work? I fail to see why are the $E_n$ &lt;strike&gt;measurable, or why are they &lt;/strike&gt;$\lambda$-thick.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They don't need to be measurable! In fact, the concept of thickness is introduced to get around measurability issues.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's construct $f$ this way: let H be an Hamel basis of $\mathbb{R}$ over $\mathbb{Q}$, and $a\in H$. Let $g(x) = x$ for $x\in$ H$- {a}$ and $g(a) = 0$. Let's extend this function to $f:\mathbb{R}\longrightarrow\mathbb{R}$ by linearity using the Hamel basis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using this $f$ we have that $E_0 = [0,1]$, $E_1 = \{1\}$, which clearly isn't $\lambda$-thick. &lt;strike&gt;Where's the flaw?&lt;/strike&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The flaw is that $E_1 \ne \{1\}$ because... Hamel bases don't work that way :)&#xA;See the comments for details.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I give up. I will ask my professor about this.&lt;/p&gt;&#xA;" OwnerUserId="4471" LastEditorUserId="4471" LastEditorDisplayName="" LastEditDate="2010-12-09T09:41:07.940" LastActivityDate="2010-12-10T15:17:26.020" Title="Noncontinuous solutions of f(x+y) = f(x) + f(y), their use in a measure theory proof." Tags="&lt;homework&gt;&lt;measure-theory&gt;" AnswerCount="1" CommentCount="11" />
  <row Id="13246" PostTypeId="2" ParentId="13111" CreationDate="2010-12-06T14:46:02.463" Score="0" ViewCount="" Body="&lt;p&gt;Assume $G$ is simple.  Then the existence of 6 Sylow 5-groups implies $G$ embeds in $S_{6}$ (let $G$ act on the Sylow 5-subgroups by conjugation and use the assumption that $G$ is simple).   But 300 does not divide 6 factorial.   So $G$ is not simple.  &lt;/p&gt;&#xA;" OwnerUserId="832" LastActivityDate="2010-12-06T14:46:02.463" CommentCount="1" />
  <row Id="13247" PostTypeId="2" ParentId="13241" CreationDate="2010-12-06T14:49:10.050" Score="2" ViewCount="" Body="&lt;p&gt;One basic way to see that smooth functions are dense in any reasonable function space is to use &quot;&lt;a href=&quot;http://en.wikipedia.org/wiki/Nascent_delta_function#nascent_delta_function&quot; rel=&quot;nofollow&quot;&gt;approximations to the identity&lt;/a&gt;.&quot; The point is that convolving with the delta function is the identity operator. However, the delta function can be approximated by smooth functions (take a smooth function supported in a small ball of total integral one). If $\phi$ is such a smooth function (which approximates the $\delta$ function), and $f$ is any function, then $\phi \ast f$ is probably close to $f$ in the metric you are interested in (this is always true in $L^p$, at least). But $\phi \ast f$ is smooth since to differentiate this, just differentiate $\phi$. (It is a nice property of convolution that convolving anything with a smooth function gives a smooth function---&lt;em&gt;if&lt;/em&gt; the integrals all converge nicely.)&lt;/p&gt;&#xA;" OwnerUserId="536" LastActivityDate="2010-12-06T14:49:10.050" CommentCount="1" />
  <row Id="13248" PostTypeId="2" ParentId="13174" CreationDate="2010-12-06T15:04:40.140" Score="1" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\ $ Exploit the innate symmetry of the problem - here power sums as symmetric functions.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The power sum $\rm\ \ \Sigma_n = a\ x^n + b\ y^n + c\ z^n\ \ $ satisfies the recursion&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\rm \Sigma_{n+3} - (x+y+z)\ \Sigma_{n+2} + (xy+yz+zx)\ \Sigma_{n+1} - xyz\ \Sigma_n\ =\ 0 $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $\rm\ n = 0,1,2\ $ the above yields a linear system for $\rm\: \ x+y+z,\ \ xy+yz+zx,\ \ xyz\ ,\:$ which are the coefficients of the cubic whose roots are $\rm\ x, y, z$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;REMARK&lt;/strong&gt; $\ $ As I mentioned in an &lt;a href=&quot;http://math.stackexchange.com/questions/6965/system-of-equations/6990#6990&quot;&gt;answer&lt;/a&gt; to a similar question, readers familiar with the  theory of &lt;strong&gt;Lucas-Lehmer sequences&lt;/strong&gt; may recognize this recursion as one of many useful identities satisfied by sums of powers.  Such identities arise in many diverse contexts. Here is an interesting example: Capdegelle's work on &lt;strong&gt;FLT&lt;/strong&gt; (Fermat's Last Theorem) employed the power sum recursion $\rm\ F_{n+3} + S_2\: F_{n+2} + S_1\: F_{n+1} + S_0\: F_n = 0\ $ where $\rm\ F_n = X^n + Y^n - Z^n\ $ is the Fermat polynomial and the polynomials $\rm\ S_k\ $ are the elementary symmetric polynomials $\rm\ S_0 = -XYZ,$ $\rm\ S_1 = XY + XZ + YZ,\ $ $\rm S_2 = -(X+Y+Z)\:.\ $ &lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-12-06T15:17:48.337" LastActivityDate="2010-12-06T15:17:48.337" />
  <row Id="13249" PostTypeId="2" ParentId="13233" CreationDate="2010-12-06T15:31:47.027" Score="0" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\rm\ \ \ a+b-c\ &amp;lt;\ a,b\ \iff\ a,b\ &amp;lt;\ c$&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-12-06T15:31:47.027" />
  <row Id="13250" PostTypeId="1" CreationDate="2010-12-06T15:56:00.750" Score="4" ViewCount="197" Body="&lt;p&gt;We know that $$e^{i \pi} = -1 .$$ We can transform all of the components of this identity into (generalized) continued fractions. When we start of with $\pi$, we see that $$  \Big(3+ \mathbf{K}_{i=1}^{\infty} \frac{(\sqrt{a_{i-1}}+2)^2}{6}\Big)^{e \cdot i} = -1 $$ When we take the $\frac{1}{i}$'th power, we see that &#xA;$$ \Big(3 + \mathbf{K}_{i=1}^{\infty} \frac{(\sqrt{a_{i-1}}+2)^2}{6}\Big)^{e} = (-1)^{-i} ,$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In which $a_1=1^2,a_2=3^2$. We could re-state this equality by writing $e, -1$ and the root the latter as &#xA;continued fractions, but I think the question is quite clear without doing so (also: other &lt;a href=&quot;http://en.wikipedia.org/wiki/Generalized_continued_fraction#Pi&quot; rel=&quot;nofollow&quot;&gt;GCF's of pi&lt;/a&gt;). What information about (generalized) continued fractions and calculating with them can we extract from Euler's identity?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Motivation: I was hoping that, if we could find a good explanation for this, we would have more insight in how continued fractions behave under arithmetic operations. With that insight, perhaps we could find ways to combine the continued fractions of certain constants and finding the values of some constants, like Apery's Constant and Catalan's, of which we &lt;em&gt;do&lt;/em&gt; know the continued fraction representations but few more.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Max&lt;/p&gt;&#xA;" OwnerUserId="1055" LastEditorUserId="1055" LastEditorDisplayName="" LastEditDate="2010-12-06T16:55:06.563" LastActivityDate="2010-12-06T16:55:06.563" Title="What can Euler's identity teach us about (generalised) continued fractions?" Tags="&lt;number-theory&gt;&lt;continued-fractions&gt;" CommentCount="8" FavoriteCount="2" />
  <row Id="13251" PostTypeId="1" AcceptedAnswerId="13252" CreationDate="2010-12-06T16:21:33.083" Score="2" ViewCount="46" Body="&lt;p&gt;I'm currently reading Artificial Intelligence, by Russel &amp;amp; Norvig. They state that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A) A sentence is valid if it is true in all models&lt;/p&gt;&#xA;&#xA;&lt;p&gt;B) The deduction theorem: &quot;For any sentences $\alpha$ and $\beta$, $\alpha \models \beta$ if and only if ($\alpha \implies \beta$) is valid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My mental blockage consists of the fact that implication is False if $\alpha = T$ and $\beta = F$ -- thus resulting in one model that is not true. But, in order for ($\alpha \implies \beta$) to be valid, it has to be true for &lt;em&gt;all&lt;/em&gt; models.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is it I'm missing?&lt;/p&gt;&#xA;" OwnerUserId="4478" LastActivityDate="2010-12-06T16:24:58.600" Title="The deduction theorem according to AIMA" Tags="&lt;logic&gt;" AnswerCount="1" />
  <row Id="13252" PostTypeId="2" ParentId="13251" CreationDate="2010-12-06T16:24:58.600" Score="2" ViewCount="" Body="&lt;p&gt;In order for $\alpha\Rightarrow\beta$ to be valid, it must hold in all models; for $\alpha\Rightarrow\beta$ to not be valid, there must be a model where it is false. If there is a model where it is false, then there is a model in which $\alpha$ is true but $\beta$ is false, which means that $\alpha\models\beta$ does not hold. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remember: you are proving an implication. You are trying to prove that &lt;strong&gt;if&lt;/strong&gt; $\alpha\models\beta$, &lt;strong&gt;then&lt;/strong&gt; $\alpha\Longrightarrow \beta$ is valid. You are not trying to prove that $\alpha\Longrightarrow\beta$ is valid in all cases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Of course, you also need to prove the converse: &lt;strong&gt;if&lt;/strong&gt; $\alpha\Longrightarrow\beta$ is valid, so that it holds in all models, &lt;strong&gt;then&lt;/strong&gt; you need to show that $\alpha\models\beta$ holds). &lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-12-06T16:24:58.600" CommentCount="5" />
  <row Id="13253" PostTypeId="1" CreationDate="2010-12-06T16:35:46.163" Score="14" ViewCount="385" Body="&lt;p&gt;As we saw &lt;a href=&quot;http://math.stackexchange.com/questions/13237/13240#13240&quot;&gt;here&lt;/a&gt;, the minimum of two quantities can be written using elementary functions and the absolute value function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\min(a,b)=\frac{a+b}{2} - \frac{|a-b|}{2}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's even a nice intuitive explanation to go along with this: If we go to the point half way between two numbers, then going down by half their difference will take us to the smaller one. So my question is: &quot;Is there a similar formula for three numbers?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously $\min(a,\min(b,c))$ will work, but this gives us the expression:&#xA;$$\frac{a+\left(\frac{b+c}{2} - \frac{|b-c|}{2}\right)}{2} - \frac{\left|a-\left(\frac{b+c}{2} - \frac{|b-c|}{2}\right)\right|}{2},$$&#xD;&#xA;which isn't intuitively the minimum of three numbers, and isn't even symmetrical in the variables, even though its output is. Is there some nicer way of expressing this function?&lt;/p&gt;&#xA;" OwnerUserId="1149" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-07T15:12:43.280" LastActivityDate="2010-12-07T15:12:43.280" Title="Nice expression for minimum of three variables?" Tags="&lt;functions&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="4" />
  <row Id="13254" PostTypeId="2" ParentId="13253" CreationDate="2010-12-06T16:46:41.853" Score="18" ViewCount="" Body="&lt;p&gt;This probably isn't what you were thinking of, but if $a_1, ... a_n$ are non-negative, then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\min(a_1, ... a_n) = \lim_{k \to -\infty} \sqrt[k]{a_1^k + ... + a_n^k}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;whereas&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\max(a_1, ... a_n) = \lim_{k \to \infty} \sqrt[k]{a_1^k + ... + a_n^k}.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are several applications of these identities (at least the second one), e.g. to functional analysis.  They are also related to the way in which tropical arithmetic arises as a &quot;limit&quot; of ordinary arithmetic.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-12-06T16:46:41.853" />
  <row Id="13255" PostTypeId="2" ParentId="13244" CreationDate="2010-12-06T16:48:22.057" Score="1" ViewCount="" Body="&lt;p&gt;I believe this is known as &lt;i&gt;Graph Squaring&lt;/i&gt;. See also &lt;a href=&quot;http://mathworld.wolfram.com/GraphPower.html&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-12-06T16:48:22.057" CommentCount="1" />
  <row Id="13256" PostTypeId="2" ParentId="13253" CreationDate="2010-12-06T16:54:22.103" Score="12" ViewCount="" Body="&lt;p&gt;If you want a symmetric expression, you can take $\frac13 (\min(a,\min(b,c))+\min(b,\min(a,c))+\min(c,\min(a,b)))$. But if you rewrite it using that absolute-value trick, I still don't think it gives something that's &quot;intuitively the minimum&quot;, I'm afraid...&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2010-12-06T16:54:22.103" />
  <row Id="13257" PostTypeId="1" CreationDate="2010-12-06T17:07:08.530" Score="3" ViewCount="164" Body="&lt;p&gt;Hi all. I am stuck in trying to solve the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given two points $(x_{1}; y_{1})$ and $(x_{2}; y_{2})$, to determine the parameters $a$ and $b$ in the equation:&lt;br&gt;&#xA;$$y=\frac{e^{a+bx} - e^{a}}{1+e^{a+bx}}.$$&lt;br&gt;&#xA;In other words, I have two unknowns and two set of points, so there's for sure a solution, but how can the following system of equation be rewritten explicitly on $a$ and $b$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left\{\begin{matrix}&#xD;&#xA;y_{1}&amp;amp;=\frac{e^{a+bx_{1}} - e^{a}}{1+e^{a+bx_{1}}}\\ &#xD;&#xA;y_{2}&amp;amp;=\frac{e^{a+bx_{2}} - e^{a}}{1+e^{a+bx_{2}}}&#xD;&#xA;\end{matrix}\right.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Should I implement an equation solver instead? In which case, do you have any suggestion on the method?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks very much, I hope it's not a too stupid question :)&lt;/p&gt;&#xA;" OwnerUserId="2916" LastEditorUserId="2916" LastEditorDisplayName="" LastEditDate="2010-12-06T19:33:51.187" LastActivityDate="2010-12-06T20:07:48.427" Title="Solving a &quot;logistic-like&quot; function with two unknowns, two data points." Tags="&lt;functions&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="13258" PostTypeId="2" ParentId="13244" CreationDate="2010-12-06T17:09:53.357" Score="1" ViewCount="" Body="&lt;p&gt;As described in Brualdi and Cvetkovic's &lt;a href=&quot;http://books.google.com/books?id=pwx6t8QfZU8C&amp;amp;printsec=frontcover&amp;amp;dq=combinatorial+approach+to+matrix+theory&quot; rel=&quot;nofollow&quot;&gt;Combinatorial Approach to Matrix Theory&lt;/a&gt;,  rather than associating a directed graph in the usual way to a matrix, it's more natural to associate the &lt;strong&gt;König digraph&lt;/strong&gt; for the purposes of thinking about matrix multiplication.  For an $m \times n$ non-negative integer matrix $A$, the Konig digraph is a bipartite digraph with vertex classes $X$ (with $n$ elements) and $Y$ (with $m$ elements) such that the number of edges from $x \in X$ to $y \in Y$ is $A_{yx}$.  Composition of matrices then corresponds to &quot;plugging&quot; one digraph into another, and whatever purpose you want to multiply adjacency matrices for, I think it will be more natural (or at least equivalent) phrased in this language instead.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I've described is a category whose objects are the non-negative integers $[n], n \ge 0$ and where a morphism from $[n]$ to $[m]$ is a collection of arrows from an $n$-element set to an $m$-element.  This category is a certain refinement of the category of finite sets and &lt;a href=&quot;http://en.wikipedia.org/wiki/Binary_relation&quot; rel=&quot;nofollow&quot;&gt;relations&lt;/a&gt; and is a very natural setting for certain combinatorial arguments (as well as for, as the book says, a combinatorial approach to matrix theory).  I don't know if it has a name, though.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-12-06T17:09:53.357" />
  <row Id="13259" PostTypeId="1" CreationDate="2010-12-06T17:17:11.367" Score="1" ViewCount="94" Body="&lt;p&gt;Hi, I hope you can help me with this excersise:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lets have finite sequence of characters 0 and 1. We are doing following algorithm:&#xA;&lt;br&gt;If the sequence starts with 0, we add 00 to the end.&#xA;&lt;br&gt;If the sequence starts with 1, we add 1101 to the end.&#xA;&lt;br&gt;Then we remove first three characters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Prove, that if the sequence repeats in cycles, the cycles are of even length.&lt;/b&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proof:&lt;br&gt;&#xA;We are changing the length of the sequence, doing &lt;i&gt;n&lt;/i&gt; steps, by &lt;i&gt;n&lt;/i&gt;. To get to the starting state of the loop, the length has to be changed doing &lt;i&gt;n&lt;/i&gt; steps, by -&lt;i&gt;n&lt;/i&gt;. The total number of steps is &lt;i&gt;2n&lt;/i&gt;, which is even.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We assume, that if &lt;i&gt;n = n+1&lt;/i&gt;, the number of steps is still even.&lt;br&gt;&#xA;We are changing the length of the sequence, doing &lt;i&gt;n+1&lt;/i&gt; steps, by &lt;i&gt;n+1&lt;/i&gt;. To get to the starting state of the loop, the length has to be changed doing &lt;i&gt;n+1&lt;/i&gt; steps, by -&lt;i&gt;(n+1)&lt;/i&gt;. The total number of steps is &lt;i&gt;2n + 2&lt;/i&gt;, which is even.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are some examples:&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to create a 2 step loop, we need a sequence starting which some time comes to state &lt;br&gt;0xx1xx or &lt;br&gt;1xx0x. &lt;br&gt;Both these constructions will generate the other one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To create a 6 step loop, we need &lt;br&gt;1xx1xx1xx0xx0 or &lt;br&gt;1xx1xx0xx0xx0x or &lt;br&gt;1xx0xx0xx0xx1xx or &lt;br&gt;0xx0xx0xx1xx1xx1 or &lt;br&gt;0xx0xx1xx1xx1xx or &lt;br&gt;0xx1xx1xx1xx0x. &lt;br&gt;They generate the other one in that order.&lt;/p&gt;&#xA;" OwnerUserId="4447" LastEditorUserId="4447" LastEditorDisplayName="" LastEditDate="2010-12-06T20:55:05.813" LastActivityDate="2010-12-06T20:55:05.813" Title="Cycling sequence of 0 and 1 - is cycle length even?" Tags="&lt;sequences-and-series&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="13260" PostTypeId="1" AcceptedAnswerId="13271" CreationDate="2010-12-06T17:41:30.083" Score="7" ViewCount="210" Body="&lt;p&gt;I'm interested in methods of computing the homology groups and Euler characteristics of the classical Lie groups ($GL(n,\mathbb{R}), SL(n,\mathbb{R})$, etc.). (But I'd be interested in techniques which are more generally applicable to arbitrary Lie groups as well.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's the current state of my knowledge on this issue with some explicit questions interspersed:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;A compact Lie group of positive dimension admits a nowhere vanishing vector field and hence has Euler characteristic zero.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I remember reading in the past that a connected Lie group deformation retracts onto its maximal compact subgroup, which by the foregoing has Euler characteristic zero. Given this, we can say that a connected Lie group is either contractible or has Euler characteristic zero. However, my knowledge of Lie theory is (very) weak at best, so I am not fully comfortable accepting this fact. Is there a simple proof? (I feel that there should be but I'm just missing it.)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The Euler characteristic respects fiber bundle structures: if $F \rightarrow E \rightarrow B$ is a fiber bundle with total space $E$, base $B$, and fiber $F$, then $\chi(E) = \chi(F)\cdot\chi(B)$. Some suitable conditions are needed; I think $B$ path-connected suffices, but I'm not 100% certain. I can think of one explicit example of this: Identify $S^3$ with $SU(2)$ and implement the Hopf fibration $S^1 \rightarrow S^3 \rightarrow S^2$. The subgroup $U(1)$ is realized as $S^1$ and the quotient $SU(2)/U(1)$ is realized as $S^2$. So we have a fibration $SU(2) \rightarrow SU(2)/U(1)$ with fiber $U(1)$; hence $\chi(SU(2)) = \chi(SU(2)/U(1)) \cdot \chi(U(1)) = 0$. But this is already known since $SU(2)$ is compact.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The most direct approach to computing both the homology and the Euler characteristics would possibly be to find an explicit cellular decomposition of these groups. Again, my knowledge of basic Lie theory is failing me here: I don't see how to find such decompositions of the matrix groups.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Surely there are other methods of computing the homology which would exploit the Lie group structure, but I don't know of any.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="251" LastActivityDate="2010-12-06T18:58:10.187" Title="Homology and Euler characteristics of the classical Lie groups" Tags="&lt;algebraic-topology&gt;&lt;lie-groups&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="13261" PostTypeId="1" CreationDate="2010-12-06T17:47:01.107" Score="0" ViewCount="162" Body="&lt;p&gt;I'm doing a raytracing exercise. I have a vector representing the normal of a surface at an intersection point, and a vector of the ray to the surface. How can I determine what the reflection will be?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the below image, I have &lt;code&gt;d&lt;/code&gt; and &lt;code&gt;n&lt;/code&gt;. How can I get &lt;code&gt;r&lt;/code&gt;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://imgur.com/9BHM2.png&quot; alt=&quot;Vector d is the ray; n is the normal; t is the refraction; r is the reflection&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="2755" LastActivityDate="2010-12-06T23:44:55.710" Title="How to get a reflection vector?" Tags="&lt;linear-algebra&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="13262" PostTypeId="2" ParentId="13244" CreationDate="2010-12-06T17:51:15.623" Score="1" ViewCount="" Body="&lt;p&gt;Maybe it would be helpful... from position that $M$ is a hypergraph, value of matrix element $A_M(i,j)$ seems to be equal to the number of edges from vertex i to vertex j with length 2 (minus edges from j to i). But I'm not sure, if $A_M$ is adjacency matrix for hypergraph $M$... Especially after googling &lt;a href=&quot;http://books.google.ru/books?id=DWZc9Ti_vBwC&amp;amp;pg=PA312&amp;amp;lpg=PA312&amp;amp;dq=hypergraph+adjacency+matrix+definition&amp;amp;source=bl&amp;amp;ots=2R8o1vC0Z3&amp;amp;sig=h-2ubx2ytHCYCatkQ8YlG12mWtA&amp;amp;hl=ru&amp;amp;ei=8RX9TN_eH4eXOp_nwdQK&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=4&amp;amp;sqi=2&amp;amp;ved=0CDcQ6AEwAw#v=onepage&amp;amp;q=hypergraph%20adjacency%20matrix%20definition&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;&quot;hypergraph adjacency matrices&quot;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4454" LastActivityDate="2010-12-06T17:51:15.623" />
  <row Id="13263" PostTypeId="2" ParentId="13261" CreationDate="2010-12-06T17:54:33.690" Score="1" ViewCount="" Body="&lt;p&gt;$$r = d - 2 (d \cdot n) n$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $d \cdot n$ is the dot product, and&#xA;$n$ must be normalized.&lt;/p&gt;&#xA;" OwnerUserId="4241" LastActivityDate="2010-12-06T17:54:33.690" CommentCount="2" />
  <row Id="13264" PostTypeId="2" ParentId="13260" CreationDate="2010-12-06T17:58:26.030" Score="3" ViewCount="" Body="&lt;p&gt;The explicit cellular decomposition you're looking for is called &lt;a href=&quot;http://en.wikipedia.org/wiki/Bruhat_decomposition&quot; rel=&quot;nofollow&quot;&gt;Bruhat decomposition&lt;/a&gt;.  Counting the number of cells in each dimension automatically gives you the homology of classical groups over $\mathbb{C}$ (since then there are only cells in even dimensions), but I think over $\mathbb{R}$ you have to work harder (although I think you do get the Euler characteristic, in any case).&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-12-06T17:58:26.030" />
  <row Id="13265" PostTypeId="2" ParentId="13259" CreationDate="2010-12-06T17:59:53.647" Score="2" ViewCount="" Body="&lt;p&gt;Assuming you are talking about the sequence of bitstrings repeating in cycles, and the period of that,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint: You either increase the length of the bitstring by 1 or reduce it by 1.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-12-06T17:59:53.647" CommentCount="15" />
  <row Id="13266" PostTypeId="2" ParentId="13261" CreationDate="2010-12-06T18:04:09.657" Score="4" ViewCount="" Body="&lt;p&gt;Let $\hat{n} = {n \over \|n\|}$. Then $\hat{n}$ is the vector of magnitude one in the same direction as $n$. The projection of $d$ in the $n$ direction is given by $\mathrm{proj}_{n}d = (d \cdot \hat{n})\hat{n}$, and the projection of $d$ in the orthogonal direction is therefore given by $d - (d \cdot \hat{n})\hat{n}$. Note that $-r$ has the same projection onto $d$, with its orthogonal projection given by $-1$ times that of $d$. Thus we have &#xA;$$d = (d \cdot \hat{n})\hat{n} + [d - (d \cdot \hat{n})\hat{n}]$$&#xD;&#xA;$$-r = (d \cdot \hat{n})\hat{n} - [d - (d \cdot \hat{n})\hat{n}]$$&#xD;&#xA;The later equation is exactly&#xA;$$r = -(d \cdot \hat{n})\hat{n} + [d - (d \cdot \hat{n})\hat{n}]$$&#xD;&#xA;Hence one can get $r$ from $d$ via&#xA;$$r = d - 2(d \cdot \hat{n})\hat{n}$$&#xD;&#xA;Stated in terms of $n$ itself, this becomes&#xA;$$r = d - {2 d \cdot n\over \|n\|^2}n$$&lt;/p&gt;&#xA;" OwnerUserId="3035" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-06T23:44:55.710" LastActivityDate="2010-12-06T23:44:55.710" />
  <row Id="13268" PostTypeId="1" CreationDate="2010-12-06T18:24:39.900" Score="3" ViewCount="139" Body="&lt;p&gt;Let $A$ and $B$ be two symmetric matrices of the same size, and let $\alpha_k$ and $\beta_k$ be the $k$th largest eigenvalues of $A$ and $B$ respectively.  What can we say about $|\beta_k - \alpha_k|$ in terms of $||B - A||_2$?  To be clearer, I want the tightest possible upper bound on $|\beta_k - \alpha_k|$ using only the stated assumptions, and computed as a function of $||B - A||_2$.&lt;/p&gt;&#xA;" OwnerUserId="4370" LastEditorUserId="4370" LastEditorDisplayName="" LastEditDate="2010-12-06T18:42:06.197" LastActivityDate="2010-12-07T18:32:12.300" Title="eigenvalue error" Tags="&lt;linear-algebra&gt;" AnswerCount="2" CommentCount="13" />
  <row Id="13269" PostTypeId="2" ParentId="1676" CreationDate="2010-12-06T18:26:48.483" Score="1" ViewCount="" Body="&lt;p&gt;The picture on Mumford's Red book for the arithmetic surface $Spec\ \mathbb Z[x]$ is very enlightening. The book itself is a fantastic read.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://matrix.cmi.ua.ac.be/DATA2/mumforddrawing.jpg&quot; rel=&quot;nofollow&quot;&gt;Lieven le Bruyn&lt;/a&gt; has a copy of that picture. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The book ought to be read. What is equally enlightening is the functorial point of view emphasized by Grothendieck, that schemes are representing their functor of points and thus can be thought of as the functor instead. This is an easy consequence of Yoneda lemma. A delightful book with this point of view is Mumford, &quot;Lectures on curves on an algebraic surface&quot;.&lt;/p&gt;&#xA;" OwnerUserId="1119" LastActivityDate="2010-12-06T18:26:48.483" CommentCount="2" />
  <row Id="13270" PostTypeId="1" AcceptedAnswerId="13281" CreationDate="2010-12-06T18:29:07.980" Score="4" ViewCount="244" Body="&lt;p&gt;Since I started university, almost every exam I've had in mathematics have ended up as failed mostly due to my lack of using proper &quot;syntax&quot; on paper, written by hand. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where can I learn the standard way of writing mathematics, beginning with the basics? Does this make sense? Is there a standard &quot;uniform&quot; syntax, or does it depend on locale - country from country?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt; With syntax I mean the way you write down your logic on paper - for instance, how you should put up a solution for display; what format to use, when to use deduction/implication arrows, how your answer should be represented etc. When to tell whether you are you allowed to skip mental steps and so on?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Very simple example:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;x(x + 3)(x + 7) =                     &#xA;= x(x^2 + 10x + 21) =                              &#xA;= x^3 + 10x^2 + 21x         &#xA;Answer: x^3 + 10x^2 + 21x&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;vs&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;x(x + 3)(x + 7) = x^3 + 10x^2 + 21x&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;and so on... Hope this makes it a bit more clear?&lt;/p&gt;&#xA;" OwnerUserId="470" LastEditorUserId="470" LastEditorDisplayName="" LastEditDate="2010-12-06T19:29:34.293" LastActivityDate="2011-05-05T06:04:17.777" Title="Where to learn the syntax of mathematics?" Tags="&lt;learning&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="13271" PostTypeId="2" ParentId="13260" CreationDate="2010-12-06T18:58:10.187" Score="5" ViewCount="" Body="&lt;p&gt;So we know that $GL_n(\mathbb{R})$ and $SL_n(\mathbb{R})$ are homotopy equivalent to $O_n$&#xD;&#xA;and $SO_n$ respectively (Similar things are true for working over $\mathbb{C}$). Now you can work inductively using a Serre SS. Look at the fibration coming from the inclusion of one classical group into the next higher one, if you set it up right the homogenous space will be a sphere. This should get most of the work done for you, and it doesn't require lie theory. I don't think it will generalize though to exceptional lie groups though, you need to understand the homogenous spaces in order for this to work. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is another way using the Serre SS: Use the fact that we know what $H^*(BG)$ is for $G$ unitary or orthoganl, special or not (with integer or mod 2 coefficients depending on whether or not it is $U$ or $O$). We know these rings because they give us chern and stiefel-whitney classes respectively (I believe the effect of looking a the special groups is just to kill $c_1$ or $w_1$). Also, the dimension of the vector space the group is acting on tells you the index of the top characteristic class. Now use our good old friend the loop path space fibration: $\Omega BG \to PBG \to BG$ and remember that $\Omega BG$ is homotopy equivalent to $G$ and $PBG$ is contractible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both of these approaches give you the ring structure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then use a Bockstein SS to get the integral cohomology for the stuff with mod two coefficients ($sq^1$ is the differential!).&#xA;Once you get the cohomology you can compute the euler characteristic by taking the alternating sums of the ranks. &lt;/p&gt;&#xA;" OwnerUserId="627" LastActivityDate="2010-12-06T18:58:10.187" CommentCount="3" />
  <row Id="13272" PostTypeId="1" AcceptedAnswerId="13304" CreationDate="2010-12-06T19:04:18.277" Score="0" ViewCount="124" Body="&lt;p&gt;How would I create a rotation matrix that rotates X by a, Y by b, and Z by c?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need to formulas, unless you're using the ardor3d api's functions/methods.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Matrix is set up like this&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;xx, xy, xz,&#xA;yx, yy, yz,&#xA;zx, zy, zz&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;A Quaternion is fine too.&lt;/p&gt;&#xA;" OwnerUserId="4464" LastActivityDate="2010-12-06T22:22:23.850" Title="How would I create a rotation matrix that rotates X by a, Y by b, and Z by c?" Tags="&lt;geometry&gt;&lt;matrices&gt;&lt;quaternions&gt;" AnswerCount="2" />
  <row Id="13273" PostTypeId="6" CreationDate="2010-12-06T19:05:01.087" Score="0" ViewCount="0" Body="&lt;p&gt;Let me introduce myself first. My name is Willie Wong, and for what it is worth, I am a mathematician. And (at the risk of the nomination statements reading like broken records) I have also been a long-time participant in MathOverflow, have been lurking on Math.SE since public beta, and have been involved with TeX.SE since its inception. So I have some familiarity with the type of community the StackExchange platform is trying to forge. Compared to the other nominees, my name may be much less familiar to you, but I assure you this is not due to lack of activity: I do visit the website every working day, so I pledge to be usually available for moderating duties. My (relatively) low reputation score should be taken as a reflection of the large number of very capable people providing answers on this website, and of my personal belief in suggesting improvements to existing answers, instead of posting duplicate ones. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If elected as a moderator, I promise to moderate in moderation :) That is, with the exception of extreme cases of abuse (spammers, trolls, or offensive posts or comments), I will in general defer to the concensus of the community on what is or isn't appropriate, and I ask voters to also help set this website's scope and nature &lt;a href=&quot;http://meta.math.stackexchange.com/questions/1222/please-help-construct-moderation-policies-for-our-newcoming-moderators&quot;&gt;in Meta&lt;/a&gt;. I also promise to not involve myself in personal conflicts against other users, and be generally courteous in tone when dealing with whatever situations that may arise (unfortunately, the necessity of using the written form on the internet sometimes lead to unintended interpretations of words; I will try my best in my communications to avoid this). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As to why I want to be a moderator: I was rather late coming to this website, and initially had some reservations about its purpose. I was afraid that it may be overrun with cranks or homework problems. After lurking for a while during public beta, I've been won over by the consistent level and focus of the discussions that take place here, and I want to do my best to help maintain this standard at Math.SE. &lt;/p&gt;&#xA;" OwnerUserId="1543" LastEditorUserId="1543" LastEditorDisplayName="" LastEditDate="2010-12-06T19:05:01.087" LastActivityDate="2010-12-06T19:05:01.087" CommunityOwnedDate="2010-12-06T19:05:01.087" />
  <row Id="13274" PostTypeId="2" ParentId="13272" CreationDate="2010-12-06T19:09:29.097" Score="3" ViewCount="" Body="&lt;p&gt;I think that 3d rotation is more complicated than this; rotating around each of the axes separately can give different cumulative results depending on the order in which you choose to do the rotations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;see also&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Rotation_matrix#Three_dimensions&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Rotation_matrix#Three_dimensions&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4370" LastActivityDate="2010-12-06T19:09:29.097" CommentCount="2" />
  <row Id="13275" PostTypeId="2" ParentId="13270" CreationDate="2010-12-06T19:12:45.037" Score="2" ViewCount="" Body="&lt;p&gt;Read your textbook and then use the syntax in your textbook when you take the exam.&lt;/p&gt;&#xA;" OwnerUserId="4370" LastActivityDate="2010-12-06T19:12:45.037" />
  <row Id="13276" PostTypeId="1" AcceptedAnswerId="13279" CreationDate="2010-12-06T19:14:09.470" Score="2" ViewCount="184" Body="&lt;p&gt;I recently learned about finding density functions of functions of random variables using the cumulative distribution function.  For example, computing the density function for the difference of two uniform random variables on [0,1] (as in Grinstead &amp;amp; Snell's free online book).  I gave myself the following problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Say $X$ and $Y$ are independent random variables with exponential density $f(t) = a e^{-at}$. I'm trying to compute the density function for $Z =  X-Y$, but I'm getting nonsense. Since $z&amp;lt;0$ is ok and the density function will go like $e^{-az}$, trying to check that $\int_{-\infty}^\infty f_{X-Y}(z) dz$ diverges!  What is going wrong? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: The calculations are as follows.  If $z &amp;gt; 0$, then &#xA;        $$&#xD;&#xA;		\begin{split}&#xD;&#xA;		  P(Z &amp;lt; z) &amp;amp;= \int_0^\infty dy \int_0^{y+z} dx a^2&#xD;&#xA;		  e^{-a(x+y)} \\&#xD;&#xA;		  &amp;amp;= 1 - \frac 1 2 e^{-a z}&#xD;&#xA;		\end{split}&#xD;&#xA;		$$&#xD;&#xA;I found this by thinking about the (x,y) points satisfying $Y &amp;gt; X - z$, giving a region $y &amp;gt; 0, y+z &amp;gt; x &amp;gt; 0$.  If $ z &amp;lt; 0$, &#xA;$$ \begin{split}&#xD;&#xA;\displaystyle P(Z &amp;lt; z) &amp;amp;= \int_z^\infty dy \int_0^{y+z} dx a^2 e^{-a(x+y)} \\&#xD;&#xA;&amp;amp;= -\frac 1 2 e^{-3 a z } + e^{-a z}.\end{split}$$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="3008" LastEditorUserId="3008" LastEditorDisplayName="" LastEditDate="2010-12-06T20:59:28.930" LastActivityDate="2010-12-06T20:59:28.930" Title="Density of a function of random variables." Tags="&lt;probability&gt;" AnswerCount="2" CommentCount="5" />
  <row Id="13277" PostTypeId="2" ParentId="13270" CreationDate="2010-12-06T19:14:21.047" Score="1" ViewCount="" Body="&lt;p&gt;It is not clear what you mean with &quot;syntax&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming you by &quot;syntax&quot; mean your way of reasoning, i.e. your &quot;path&quot; via logical connectives to the result, my best tip is to carefully read all proofs in your math book. Not only read through them, but understand them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You must be able to prove things correctly, i.e. using only valid deductions. They way they do it in your math book is not a &quot;local&quot; thing, but &quot;global&quot; (no pun!). As long as your reasoning is understandable by anyone else reading it, it should be fine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you could be more spesific with what you mean by &quot;syntax&quot;, I'm sure somebody could give a clearer answer.&lt;/p&gt;&#xA;" OwnerUserId="4284" LastActivityDate="2010-12-06T19:14:21.047" />
  <row Id="13278" PostTypeId="2" ParentId="13207" CreationDate="2010-12-06T19:18:06.587" Score="2" ViewCount="" Body="&lt;p&gt;If the equation is $z^2=xy\frac{\partial z}{\partial x}\frac{\partial z}{\partial y}$, I would be tempted to see the symmetry in $x$ and $y$ and try solutions of the form $z=(xy)^n$.  What happens then?&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-12-06T19:18:06.587" />
  <row Id="13279" PostTypeId="2" ParentId="13276" CreationDate="2010-12-06T19:20:29.300" Score="1" ViewCount="" Body="&lt;p&gt;It is a laplace distibution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Laplace_distribution#Relation_to_the_exponential_distribution&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Laplace_distribution#Relation_to_the_exponential_distribution&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4370" LastActivityDate="2010-12-06T19:20:29.300" />
  <row Id="13280" PostTypeId="2" ParentId="13276" CreationDate="2010-12-06T19:24:36.093" Score="0" ViewCount="" Body="&lt;p&gt;In general, you are actually interested in &lt;a href=&quot;http://en.wikipedia.org/wiki/Cross-correlation&quot; rel=&quot;nofollow&quot;&gt;Cross-correlation&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's consider your problem in the setting of cross-correlation. We'll begin by considering the distribution function of $X-Y$ for $t&amp;gt;0$ (this is enough, since $X-Y$ is a symmetric random variable, hence having a symmetric density function).&#xA;Then, by the law of total probability (conditioning on $Y$), we have&#xA;$$&#xD;&#xA;{\rm P}(X - Y \le t) = \int_0^\infty  {{\rm P}(X - Y \le t|Y = \tau )f(\tau ){\rm d}\tau } = &#xD;&#xA;\int_0^\infty  {{\rm P}(X \le t + \tau )f(\tau ){\rm d}\tau }.  &#xD;&#xA;$$&#xD;&#xA;Of course, you can now calculate the right-hand side integral directly, and then differentiate with respect to $t$, but it is instructive to first differentiate with respect to $t$. Thus, we get&#xA;$$&#xD;&#xA;f_{X - Y} (t) = \frac{{\rm d}}{{{\rm d}t}}\int_0^\infty  {F(t + \tau )f(\tau ){\rm d}\tau }  = \int_0^\infty  {f(t + \tau )f(\tau ){\rm d}\tau } ,&#xD;&#xA;$$&#xD;&#xA;where $F$ denotes the distribution function of $X$ (and $Y$). According to Wikipedia's link given above, this can be denoted as &#xA;$f_{X-Y}(t) = (f \star f)(t)$ (read &quot;$f_{X-Y}(t)$ is the cross-correlation of $f$ with itself&quot;). Finally, from $f(u)= a {\rm e}^{-a u}$, $u &amp;gt; 0$, we find $f_{X-Y}(t) = (a/2){\rm e}^{-at}$, $t &amp;gt; 0$; then, by the aforementioned symmetry, $f_{X-Y}(t) = (a/2){\rm e}^{ - a|t|}$, for all $t \in \mathbb{R}$.&lt;/p&gt;&#xA;" OwnerUserId="2810" LastEditorUserId="2810" LastEditorDisplayName="" LastEditDate="2010-12-06T20:33:52.693" LastActivityDate="2010-12-06T20:33:52.693" />
  <row Id="13281" PostTypeId="2" ParentId="13270" CreationDate="2010-12-06T19:27:27.310" Score="10" ViewCount="" Body="&lt;p&gt;Learning to write mathematics properly is much like learning to write any language properly, only more so because such a premium is placed on logical organization and clarity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mathematics, both the part that is written in words and the part that is written in symbols, is attempting to convey meaning. When you write and read symbols, think about what they &lt;strong&gt;say&lt;/strong&gt; when you read them out loud. Every time you see the symbol $=$, remember that the symbol has a pronunciation when you read it, and it says &quot;is equal to&quot;. So if you write things like&#xA;$$2x = 4 = \frac{4}{2}=2,$$&#xA;(which I see far too often on exams) then you are saying &quot;twice $x$ is equal to four, which is equal to four halves, which is equal to $2$&quot;, which of course is false and liable to cost you points, even though you probably don't &lt;em&gt;actually&lt;/em&gt; think that $4$ and $\frac{4}{2}$ are equal. Remember, first and foremost, that every symbol has a meaning and a pronunciation. Unless you recognize that, you won't be able to get very far. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So first you need to be clear on what you want to say, and then make sure that what you wrote actually conveys that meaning and not other meanings. If you can do that, even if it is with &quot;plain English&quot;, you will have gotten over more than half the problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That said, mathematics is also a technical language with a number of conventions and jargon. The very best thing you can do to become familiar with, and good at using those conventions, is to read a lot of mathematics, with an eye towards &lt;em&gt;understanding&lt;/em&gt; what is written and how the language helps that understanding (just like doing a lot of reading is one of the best ways to improve one's spelling).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A close second is to read books that are meant to help introduce you to proofs and logical arguments, usually with subtitles like &quot;first course in advanced mathematics&quot; or &quot;introduction ot abstract/advanced mathematics&quot;. Find out what the &quot;Intro to proofs&quot; course is at your school, and look at the textbook they use. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One important thing is not to simply try to mimic the language you see: that will result in the mathematical equivalent of saying &lt;em&gt;&quot;Buenos días. Yo quiero estación de tren ser, por favor&quot;&lt;/em&gt;  when trying to ask for directions to the train station (&quot;Good morning. I want be train station, please.&quot;) You want to keep an eye on the meaning that the words are conveying, and how that particular choice of words (and even the &lt;em&gt;order&lt;/em&gt; of the words) matters. Notice, for instance, that saying &quot;For every $x$ there is a $y$&quot; is &lt;em&gt;not&lt;/em&gt; the same thing as saying &quot;There is a $y$ such that for every $x$...&quot;, even though they may seem very similar when thought of in English.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So: always think first about what you are trying to say and make sure you say &lt;em&gt;that&lt;/em&gt;. Read what you've written, pronouncing every symbol to yourself to make sure you aren't saying that you &quot;want to be train station&quot;. And read your books and professors' notes to see how the language works and become familiar with it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added:&lt;/strong&gt; It seems I rather badly misinterpret the true thrust of your question (despite the fact that you seem to have &quot;accepted&quot; my answer). In so far as what steps to add or what steps to skip, as has been pointed out, it depends on your level. I would not object to a student in my graduate abstract algebra class going from $x(x+1)(x+2)$ directly to $x^3+3x^2+2x$ (or even not writing out the computations before using it!) but I would definitely request a precalculus student to write out how he got to that final answer. If you have to stop and think carefully about what the answer &lt;em&gt;is&lt;/em&gt;, then you should not skip the step and write it down. For a particular class, look at the textbook and which steps it works out explicitly and which steps it skips. Look at what steps your professor works out explicitly and which steps he skips. You'll want to not stray too far from them as far as skipping more steps (though of course you can always skip &lt;em&gt;fewer&lt;/em&gt; steps if you are unsure about a calculation/argument).&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2011-05-05T06:04:17.777" LastActivityDate="2011-05-05T06:04:17.777" CommentCount="3" />
  <row Id="13282" PostTypeId="1" AcceptedAnswerId="13311" CreationDate="2010-12-06T19:32:13.700" Score="7" ViewCount="875" Body="&lt;p&gt;Hello,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm trying to create a program, that will decompose a matrix using the Cholesky decomposition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The decomposition itself isn't a difficult algorithm, but a matrix, to be eligible for Cholesky decomposition, must be symmetric and positive-definite. Checking whether a matrix is symmetric is easy, but the positive part proves to be more complex.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've read about the Sylvester's criterion, but that leads to determinants, and based on what I found on the web, those are quite extensive and hard on computers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a nutshell - is there something I might be missing? Due to the fact the the matrix is square or something like that, is there possibly a simpler way to determine whether it's positive?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regards,&#xA;Paul&lt;/p&gt;&#xA;" OwnerUserId="3543" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-06T23:56:43.870" LastActivityDate="2010-12-27T22:55:51.737" Title="Determining whether a symmetric matrix is positive-definite (algorithm)" Tags="&lt;linear-algebra&gt;&lt;algorithms&gt;&lt;matrices&gt;&lt;numerical-methods&gt;" AnswerCount="4" CommentCount="5" />
  <row Id="13286" PostTypeId="2" ParentId="13282" CreationDate="2010-12-06T19:42:35.407" Score="3" ViewCount="" Body="&lt;p&gt;I don't think there is a simpler way than computing a decomposition or determinant unless your matrix has a special form.  For example if it is a sample covariance matrix then it is positive semidefinite by construction.&lt;/p&gt;&#xA;" OwnerUserId="4370" LastActivityDate="2010-12-06T19:42:35.407" CommentCount="1" />
  <row Id="13289" PostTypeId="2" ParentId="13282" CreationDate="2010-12-06T19:57:27.890" Score="0" ViewCount="" Body="&lt;p&gt;Perhaps someone has an iterative approach or some other approximate approach, but it looks like to be sure, one has to calculate determinants and apply &lt;a href=&quot;http://en.wikipedia.org/wiki/Sylvester%27s_criterion&quot; rel=&quot;nofollow&quot;&gt;Sylvester's criterion&lt;/a&gt; (as you pointed out).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sylvester's states that iff, for all $k&amp;lt;n$, the $\det(A_k) &amp;gt; 0$, where $A_k$ is the $k$'th principle minor, then the matrix is positive definite.  The only deterministic, efficient, algorithm to calculate determinants that I know of is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Bareiss_algorithm&quot; rel=&quot;nofollow&quot;&gt;Bareiss algorithm&lt;/a&gt;, for which you can see &lt;a href=&quot;http://www.jstor.org/stable/2004533&quot; rel=&quot;nofollow&quot;&gt;Bareiss's original paper&lt;/a&gt; or Chee Yap's excellent reference &lt;a href=&quot;http://cs.nyu.edu/~yap/book/alge/ftpSite/l10.ps.gz&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Bareiss algorithm, in it's intermediate steps, ends up calculating the determinant of the $k$ minors of the matrix for all $k&amp;lt;n$, so you get a simple test for positive definiteness should you end up implementing the Bareiss algorithm yourself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd be interested to know if there are any iterative methods that would give you some confidence of a matrices positive definiteness.&lt;/p&gt;&#xA;" OwnerUserId="4143" LastActivityDate="2010-12-06T19:57:27.890" CommentCount="1" />
  <row Id="13290" PostTypeId="2" ParentId="13257" CreationDate="2010-12-06T20:07:48.427" Score="4" ViewCount="" Body="&lt;p&gt;I don't know how to solve exactly (or even if it's possible) but maybe this will help: Let $A = e^a$ and $B = e^b$. It suffices to find $A$ and $B$ and then take logarithms. Your two equations can be written as&#xA;$$y_1 = {A(B^{x_1} - 1) \over 1 + AB^{x_1}}$$&#xD;&#xA;$$y_2 = {A(B^{x_2} - 1) \over 1 + AB^{x_2}}$$&#xD;&#xA;You can solve each of these equations for $A$ via some algebra:&#xA;$$A = {y_1 \over B^{x_1}(1 - y_1) - 1}$$&#xD;&#xA;$$A = {y_2 \over B^{x_2}(1 - y_2) - 1}$$&#xD;&#xA;So the reciprocals of the right-hand sides are the same (the original equation ensures that $y_1$ and $y_2$ are nonzero):&#xA;$${B^{x_1}(1 - y_1) - 1 \over y_1} = { B^{x_2}(1 - y_2) - 1 \over y_2}$$&#xD;&#xA;In other words,&#xA;$${1 - y_1 \over y_1}B^{x_1} - {1 - y_2 \over y_2}B^{x_2} = {1 \over y_1} - {1 \over y_2}$$&#xD;&#xA;This doesn't have a general solution you can write out (for example if $x_1$ and $x_2$ are large positive integers it's a polynomial equation of large degree). But it shouldn't be too hard to deal with numerically. Once you know what $B$ is (or have a good enough approximation) you can plug back in above to get $A$.&lt;/p&gt;&#xA;" OwnerUserId="3035" LastActivityDate="2010-12-06T20:07:48.427" CommentCount="4" />
  <row Id="13291" PostTypeId="1" CreationDate="2010-12-06T20:10:50.300" Score="2" ViewCount="137" Body="&lt;p&gt;In logic theory, if we have a finite domain A of k elements, then we can construct only finitely many structures (finite structures) each of which has A as domain. I think the number of structures will be $2^{k}$. How to prove it - if my claim is right? &lt;/p&gt;&#xA;" OwnerUserId="4481" LastEditorUserId="4481" LastEditorDisplayName="" LastEditDate="2010-12-06T23:49:42.587" LastActivityDate="2010-12-06T23:49:42.587" Title="In Logic Theory...." Tags="&lt;logic&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="13292" PostTypeId="2" ParentId="13291" CreationDate="2010-12-06T20:18:00.417" Score="-1" ViewCount="" Body="&lt;p&gt;2^k is the number of subsets of a set with k elements&lt;/p&gt;&#xA;" OwnerUserId="4370" LastActivityDate="2010-12-06T20:18:00.417" />
  <row Id="13293" PostTypeId="1" AcceptedAnswerId="13295" CreationDate="2010-12-06T20:27:15.833" Score="4" ViewCount="118" Body="&lt;p&gt;I have some kind of &quot;homework&quot; question. I have the following theorem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Theorem&lt;/strong&gt; (Transfinite Recursion over the class of of ordinals $\mathbf{ON}$: Let $\mathbf{V}$ be the class of all sets. If $\mathbf{F} : \mathbf{V} \to \mathbf{V}$ then there exists a unique $\mathbf{G} : \mathbf{ON} \to \mathbf{V}$ such that for all $\alpha$ $\mathbf{G}(\alpha) = \mathbf{F}(\mathbf{G}|_{\alpha}))$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I want to show that if $\hat{x} = \{y : y &amp;lt; x\}$ and if $\mathbf{F} : \mathbf{V} \to \mathbf{V}$ and $(X, &amp;lt;)$ is a well-ordering then there exists a function $f$ with domain $X$ such that $f(x) = \mathbf{F}(f|_{\hat{x}}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I let $0 = \text{min}(X)$, and then I note that I want $f(0) = \mathbf{F(\emptyset})$. I continue like this and I note that $f(1) = \mathbf{F}(f(0))$ and so on. Is this correct? If so, can someone give me a hint how I apply the above theorem to say that this construction works?&lt;/p&gt;&#xA;" OwnerUserId="1120" LastEditorUserId="622" LastEditorDisplayName="" LastEditDate="2010-12-07T04:24:46.697" LastActivityDate="2010-12-07T04:24:46.697" Title="Recursion on ordinals." Tags="&lt;homework&gt;&lt;set-theory&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="13294" PostTypeId="1" AcceptedAnswerId="13299" CreationDate="2010-12-06T20:41:31.527" Score="0" ViewCount="76" Body="&lt;p&gt;How would I go about it?&lt;/p&gt;&#xA;" OwnerUserId="4464" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-06T22:40:09.427" LastActivityDate="2010-12-06T22:40:09.427" Title="figuring out the x,y, and z rotation of a right triangle?" Tags="&lt;geometry&gt;&lt;trigonometry&gt;" AnswerCount="1" CommentCount="6" />
  <row Id="13295" PostTypeId="2" ParentId="13293" CreationDate="2010-12-06T20:48:42.050" Score="3" ViewCount="" Body="&lt;p&gt;If you can use the transfinite recursion theorem you stated, why not simply apply it to $F$, and then use an isomorphism between $X$ and some ordinal to transfer the function $G$ from $ON$ to $X$?  It seems like you are trying to reprove the transfinite recursion theorem rather than just applying it. &lt;/p&gt;&#xA;" OwnerUserId="630" LastActivityDate="2010-12-06T20:48:42.050" />
  <row Id="13296" PostTypeId="1" CreationDate="2010-12-06T20:53:59.247" Score="-1" ViewCount="109" Body="&lt;p&gt;Let $\lambda$ denote the Lebesgue measure and let $C$ be a collection of intervals and let $\sigma(C)$ denote the sigma-algebra generated by the collection $C$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is the total variation $v$ of the Lebesgue measure $\lambda$ always absolutely continuous with respect $\lambda$?&lt;/p&gt;&#xA;" OwnerUserId="4482" LastActivityDate="2010-12-06T22:13:14.000" Title="Total variation of a measure absolutely continuous?" Tags="&lt;measure-theory&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="13298" PostTypeId="2" ParentId="13153" CreationDate="2010-12-06T21:45:42.803" Score="0" ViewCount="" Body="&lt;p&gt;Here's a hint: calculate the derivative of $\sec^n(x)$ for integer n &gt; 0, then think how you can use this , and the fact that $\tan^2(x) = 1 + \sec^2(x)$, to calculate the integral.&lt;/p&gt;&#xA;" OwnerUserId="1470" LastActivityDate="2010-12-06T21:45:42.803" />
  <row Id="13299" PostTypeId="2" ParentId="13294" CreationDate="2010-12-06T21:50:09.790" Score="0" ViewCount="" Body="&lt;p&gt;So you have the co-ordinates $P_i = (x_i, y_i, z_i), \ \ i=1,2,3$ of the vertices and you want to find the angles of the triangle formed by $P_1, P_2, P_3$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let the angle at $P_1$ be $\theta_1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle \vec{a} = (x_3 -x_1, y_3 - y_1, z_3 - z_1)$ &lt;br/&gt;&#xA;$\displaystyle \vec{b} =  (x_2 - x_1, y_2 - y_1, z_2  -z_1)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The we have that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle |\vec{a}| |\vec{b} | \cos \theta_1 = \vec{a} \cdot \vec{b}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $|\vec{v}|$ is the length of $\vec{v}$ and $\cdot $ is the dot product.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, you can find the other angles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For more information, see here: &lt;a href=&quot;http://www.mvps.org/directx/articles/math/dot/index.htm&quot; rel=&quot;nofollow&quot;&gt;Understanding the Dot Product&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-12-06T22:16:45.867" LastActivityDate="2010-12-06T22:16:45.867" CommentCount="1" />
  <row Id="13300" PostTypeId="1" AcceptedAnswerId="13305" CreationDate="2010-12-06T21:55:54.887" Score="5" ViewCount="325" Body="&lt;p&gt;I have a function $S(x)$ which is bounded above and below as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f(x) + C_1 + \mathcal{O}(g(x)) &amp;lt; S(x) &amp;lt; f(x) + C_2 + \mathcal{O}(g(x))$ as $x \rightarrow \infty$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can I conclude that $$S(x) = f(x) + C + \mathcal{O}(g(x))$$ as $x \rightarrow \infty$? Can anyone give a short proof for this fact?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(To clear up the confusion, I am stating the problem below. I am wondering if what I did was right.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Prove that as $x \rightarrow \infty$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle \sum_{n \leq x} \frac{\log n}{n} = \frac{(\log x)^2}{2} + C + \mathcal{O}(\frac{\log(x)}{x})$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $C$ is a constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is how I did it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Approximate the summation as an integral $\int \frac{\log(x)}{x} dx$ from above and from below. (The usual way of coming up with bounds for $p$ series). Then we will get the lower bound as $\frac{\log([x])^2}{2} + C_1$ and the upper bound as $\frac{\log([x])^2}{2} + C_2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, $\log([x]) = \log(x) + \log(\frac{[x]}{x}) = \log(x) + \log(1-\frac{\{x\}}{x}) = \log(x) + \mathcal{O}(\frac{\{x\}}{x}) = \log(x) + \mathcal{O}(\frac{1}{x})$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Plugging this into the above I get the the question I asked.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I did this when I was asked on a timed exam since I could not think of other ways to get to the answer. Other answers and suggestions welcome as always.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-12-07T06:04:14.213" LastActivityDate="2010-12-07T15:54:08.410" Title="Big $\mathcal{O}$ Notation question while estimating $\sum \frac{\log n}{n}$" Tags="&lt;asymptotics&gt;" AnswerCount="4" CommentCount="4" FavoriteCount="1" />
  <row Id="13301" PostTypeId="2" ParentId="13300" CreationDate="2010-12-06T22:04:29.627" Score="2" ViewCount="" Body="&lt;p&gt;I don't think so because $g(x)$ might be too small.  Take $f(x)=0, C_1=-1.5, C_2=1.5, g(x)=\exp(-x)$.  Then doesn't $S(x)=\sin(x)$ violate it?&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-12-06T22:04:29.627" CommentCount="4" />
  <row Id="13302" PostTypeId="2" ParentId="13296" CreationDate="2010-12-06T22:13:14.000" Score="1" ViewCount="" Body="&lt;p&gt;What do you mean? The Lebesgue measure is a positive measure. But anyway we can show (quite directly) that if $\mu$ is a measure and $\nu$ is a positive measure on $(X, \Omega)$ then $\mu \ll \nu$ is equivalent with $|\mu| \ll \nu$.&lt;/p&gt;&#xA;" OwnerUserId="1120" LastActivityDate="2010-12-06T22:13:14.000" />
  <row Id="13303" PostTypeId="2" ParentId="13300" CreationDate="2010-12-06T22:19:58.703" Score="0" ViewCount="" Body="&lt;p&gt;If $a&amp;lt;x&amp;lt;b$, then $|x|\le \max \lbrace |a|,|b|\rbrace$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, if $C_1=C_2=C$,  what you have there implies &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ |S(x)-f(x)-C|\le K |g(x)| $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for some $K&amp;gt;0$, i.e. $S(x)=f(x)+C+O(g(x))$.&lt;/p&gt;&#xA;" OwnerUserId="3249" LastActivityDate="2010-12-06T22:19:58.703" CommentCount="3" />
  <row Id="13304" PostTypeId="2" ParentId="13272" CreationDate="2010-12-06T22:22:23.850" Score="3" ViewCount="" Body="&lt;p&gt;The rotation matrices around the x, y, and z axes, respectively, are&#xA;$$R_x(\theta) = \begin{pmatrix} 1 &amp;amp; 0 &amp;amp; 0 \\ 0 &amp;amp; \cos \theta &amp;amp; -\sin \theta \\ 0 &amp;amp; \sin \theta &amp;amp; \cos \theta \end{pmatrix}$$&#xD;&#xA;$$R_y(\phi) = \begin{pmatrix} \cos \phi &amp;amp; 0 &amp;amp; \sin \phi \\ 0 &amp;amp; 1 &amp;amp; 0 \\ - \sin \phi &amp;amp; 0 &amp;amp; \cos \phi \end{pmatrix}$$&#xD;&#xA;$$R_z(\psi) = \begin{pmatrix} \cos \psi &amp;amp; - \sin \psi &amp;amp; 0 \\ \sin \psi &amp;amp; \cos \psi &amp;amp; 0 \\ 0 &amp;amp; 0 &amp;amp; 1 \end{pmatrix}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to rotate in the order specified in your comment on mathcast's answer, then you want&#xA;$$R_z(\psi) \cdot R_y(\phi) \cdot R_x(\theta) = $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{pmatrix} \cos \phi \cos \psi &amp;amp; \cos \psi \sin \theta \sin \phi - \cos \theta \sin \psi &amp;amp; \cos \theta \cos \psi \sin \phi + \sin \theta \sin \psi \\ \cos \phi \sin \psi &amp;amp; \cos \theta \cos \psi + \sin \theta \sin \phi \sin \psi &amp;amp; \cos \theta \sin \phi \sin \psi - \cos \psi \sin \theta \\ - \sin \phi &amp;amp; \cos \phi \sin \theta &amp;amp; \cos \theta \cos \phi \end{pmatrix}$$&lt;/p&gt;&#xA;" OwnerUserId="1016" LastActivityDate="2010-12-06T22:22:23.850" CommentCount="3" />
  <row Id="13305" PostTypeId="2" ParentId="13300" CreationDate="2010-12-06T22:41:16.147" Score="8" ViewCount="" Body="&lt;p&gt;A misconception needs to be cleared up here.  Recall that $f(x) = O(g(x))$ means that there exists a constant $C$ such that $|f(x)| \le C |g(x)|$ in a neighborhood of whatever value of $x$ you're interested in.  It is an abuse of notation that we use the equality symbol here at all; $O(g(x))$ is not a function, so it cannot be &quot;equal&quot; to $f(x)$ in the usual sense.  See the discussion at &lt;a href=&quot;http://en.wikipedia.org/wiki/Big_O_notation#Equals_sign&quot; rel=&quot;nofollow&quot;&gt;the Wikipedia article&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Based on this definition, people sometimes say that $f(x) = h(x) + O(g(x))$ if $f(x) - h(x) = O(g(x))$.  This is an additional abuse of notation, but it generally does not create problems and can be extremely convenient.  I guess you could write $f(x) &amp;lt; h(x) + O(g(x))$ if you really wanted to, but this is redundant, since the inequality is already built into the definition of big-O notation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I have no idea what $f(x) &amp;gt; h(x) + O(g(x))$ is supposed to mean.  If you are trying to say that there is a constant $C$ such that $|f(x) - h(x)| &amp;gt; C |g(x)|$, the correct way to write this is using big-Omega, not big-O, as $f(x) = h(x) + \Omega(g(x))$.&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-12-06T22:46:42.087" LastActivityDate="2010-12-06T22:46:42.087" CommentCount="4" />
  <row Id="13306" PostTypeId="2" ParentId="4781" CreationDate="2010-12-06T22:55:46.200" Score="2" ViewCount="" Body="&lt;p&gt;Your solution is almost correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the definition of RL, we require that the algorithm halt with probability $1$ on any input. As you note, your algorithm runs forever on &quot;No&quot; instances. To fix this, we'll take advantage of the fact that we're allowed a small chance of error. In particular, we're allowed to sometimes reject &quot;Yes&quot; instances, as long as we never accept a &quot;No&quot; instance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The basic idea is to keep a counter of how many iterations we've used so far. If a path exists, we should find it within $2N^N$ iterations (with probability at least $3/4$). So if we've gone that long without finding a valid path, we'll reject the graph. Note that we'll never accept a bad graph this way, and we'll rarely reject a good graph.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can't naively count to $N^N$, however, since this would require $N\lg N$ space. Instead, we'll probabilistically &quot;count&quot; that high.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After each iteration, we'll flip $N\lg N + 2$ coins. If every coin lands heads, then we stop (and reject the graph). If at least one coin lands tails, then we begin another iteration. This can be done in log-space, since we only need to count how many flips have occurred.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The probability of getting all heads is $1/2^{N\lg N + 2}=1/4N^N$, so with probability at least $3/4$, this won't happen until at least $2N^N$ iterations have elapsed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Taking a union bound over both ways the algorithm could fail, we see that everything works with probability at least $1/2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, note that this halts with probability $1$, since we'll flip all heads eventually.&lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2010-12-06T22:55:46.200" />
  <row Id="13307" PostTypeId="2" ParentId="13291" CreationDate="2010-12-06T23:01:57.043" Score="3" ViewCount="" Body="&lt;p&gt;Janice : As Carl pointed out, the answer depends on the signature (language) one uses. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, suppose the language consists of a single unary relation symbol $\dot R$. Any structure in this language has the form $(A,R)$, where $R\subseteq A$. If we fix $A$ with $|A|=k$, there are precisely $2^k$ structures of this form (one for each subset of $A$). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Carl's comment was: Suppose your language consists of a single unary function symbol $\dot f$. Any structure of this language has the form $(A,f)$ where $f:A\to A$. The number of functions from a finite set $A$ to a finite set $B$ is $|B|^{|A|}$, so there are $k^k$ such functions $f$, and therefore we have $k^k$ models in this case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, a language consists of: &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;For each positive integer $n$, some $n$-ary function symbols $\dot f$. Each of them will be interpreted as a function $f:A^n\to A$. There are $k^{k^n}$ possible interpretations, if $A$ is fixed and $|A|=k$.&lt;/li&gt;&#xA;&lt;li&gt;For each positive integer $n$, some $n$-ary relation symbols $\dot R$. Each of them will be interpreted as a set $R\subseteq A^n$. There are $2^{k^n}$ many possible interpretations.&lt;/li&gt;&#xA;&lt;li&gt;Some constant symbols $\dot c$. Each of them will be interpreted as an element of $A$, and there are $k$ possible interpretations.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;We can interpret different symbols independently of one another, and we understand that order matters, so $(A,R,S)$ and $(A,S,R)$ are interpreted as different structures if $R\ne S$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Say that $\rho_n,\lambda_n,\tau$ are cardinals (they may be $0$, they may be infinite), and that our language has precisely $\rho_n$ $n$-ary function symbols, precisely $\lambda_n$ $n$-ary relation symbols (for each positive integer $n$), and precisely $\tau$ constant symbols. Then the number of structures in this language that have domain $A$ with $|A|=k$ finite is precisely $$ \prod_{n\ge 1}(k^{k^n})^{\rho_n}\cdot\prod_{n\ge 1}(2^{k^n})^{\lambda_n}\cdot k^\tau, $$ where (as usual) we understand that empty products equal 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here, the products are cardinals in the set theoretic sense. If you are only interested in finite languages, this product gives the right answer, when we interpret products as numbers the usual way.&lt;/p&gt;&#xA;" OwnerUserId="462" LastActivityDate="2010-12-06T23:01:57.043" CommentCount="3" />
  <row Id="13308" PostTypeId="1" CreationDate="2010-12-06T23:24:39.220" Score="5" ViewCount="154" Body="&lt;p&gt;If I have the Keplerian elements for two orbits, how do I compute the shortest distance between these two orbits in 3D space? Is there any analytical expression to compute that?&lt;/p&gt;&#xA;" OwnerUserId="4486" LastActivityDate="2011-03-07T00:39:33.933" Title="Analytical Expression to find the Shortest Distance between Two Ellipses?" Tags="&lt;geometry&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="13309" PostTypeId="1" AcceptedAnswerId="13313" CreationDate="2010-12-06T23:29:35.730" Score="0" ViewCount="81" Body="&lt;p&gt;Hello&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It all about maths I don't understand, how can I solve this exercise, someone say that is a 2 dimensional problem, but I can not figure out for myself, I already understand what to do, but I don't know why is the formula, can someone explain this exercise for a dummy? &lt;a href=&quot;http://www.codeforces.com/contest/1/problem/A&quot; rel=&quot;nofollow&quot;&gt;http://www.codeforces.com/contest/1/problem/A&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;why the solution of this problem is ((m+a-1)/a)*((n+a-1)/a)?&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Help really appreciated&lt;/p&gt;&#xA;" OwnerUserId="4487" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2010-12-06T23:57:45.640" LastActivityDate="2010-12-06T23:57:45.640" Title="Covering a plaza with square flagstones" Tags="&lt;algebra-precalculus&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="13310" PostTypeId="2" ParentId="13308" CreationDate="2010-12-06T23:33:28.907" Score="1" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://www.mathworks.com/matlabcentral/fileexchange/26329-determine-the-distance-between-two-ellipses-in-3d&quot; rel=&quot;nofollow&quot;&gt;http://www.mathworks.com/matlabcentral/fileexchange/26329-determine-the-distance-between-two-ellipses-in-3d&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4488" LastActivityDate="2010-12-06T23:33:28.907" CommentCount="1" />
  <row Id="13311" PostTypeId="2" ParentId="13282" CreationDate="2010-12-06T23:34:22.823" Score="6" ViewCount="" Body="&lt;p&gt;Mathcast had it; in fact, in practical work, one uses the Cholesky decomposition $\mathbf G\mathbf G^T$ for &lt;strong&gt;efficiently testing&lt;/strong&gt; if a symmetric matrix is positive definite. The only change you need to make to turn your decomposition program into a check for positive definiteness is to insert a check before taking the required square roots that the quantity to be rooted is positive. If it is zero, you have a positive semidefinite matrix; if not, then your symmetric matrix isn't positive (semi)definite. (Programming-wise, it should be easy to throw an exception within a loop! If your language has no way to break a loop, however, you have my pity.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively, one uses the $\mathbf L\mathbf D\mathbf L^T$ decomposition here (an equivalent approach in the sense that $\mathbf G=\mathbf L\sqrt{\mathbf D}$); if any nonpositive entries show up in $\mathbf D$, then your matrix is not positive definite. Note that one could set things up that the loop for computing the decomposition is broken once a negative element of $\mathbf D$ is encountered, before the decomposition is finished!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In any event, I don't understand why people are shying away from using Cholesky here; the statement is &quot;a matrix is positive definite &lt;strong&gt;if and only if&lt;/strong&gt; it possesses a Cholesky decomposition&quot;. It's a biconditional; exploit it! It's exceedingly cheaper than successively checking minors or eigendecomposing, FWIW.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-12-06T23:34:22.823" CommentCount="1" />
  <row Id="13312" PostTypeId="2" ParentId="13309" CreationDate="2010-12-06T23:38:53.223" Score="0" ViewCount="" Body="&lt;p&gt;The answer is (the area to cover) / (the area per flagstone) + (penalty for having to use whole flagstones).  Note that your answer is almost equal to (m*n)/(a*a) so it looks about right.&lt;/p&gt;&#xA;" OwnerUserId="4488" LastActivityDate="2010-12-06T23:38:53.223" CommentCount="10" />
  <row Id="13313" PostTypeId="2" ParentId="13309" CreationDate="2010-12-06T23:57:01.380" Score="1" ViewCount="" Body="&lt;p&gt;So break it into one dimensional problems.  How many flagstones of length a does it take to cover a path of m meters?  If you can trust a and m to be integers, m/a is correct if a divides m.  Otherwise you need floor (m/a) +1.  One way to combine these is floor((m+a-1)/a).  Try it with some small numbers to see how it works.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For two dimensions, just multiply two one dimensional problems.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-12-06T23:57:01.380" CommentCount="1" />
  <row Id="13314" PostTypeId="2" ParentId="13300" CreationDate="2010-12-07T00:41:42.893" Score="6" ViewCount="" Body="&lt;p&gt;Since you wanted a different proof approach, you can try using &lt;a href=&quot;http://books.google.com/books?id=iJx1lP9ZcIkC&amp;amp;pg=PA25&quot; rel=&quot;nofollow&quot;&gt;Abel's Identity&lt;/a&gt;, which has turned out to be quite useful in analytic number theory. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance see this: &lt;a href=&quot;http://math.stackexchange.com/questions/12566/proving-sum-limits-p-leq-x-frac1-sqrtp-geq-frac12-logx-log/12588#12588&quot;&gt;An estimate for sum of reciprocals of square roots of prime numbers&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To apply this to your question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since we know that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle \sum_{1 \le n \le x} \frac1{n} = \log x + \gamma + R(x)$, where $\displaystyle R(x) = \mathcal{O}\left(\frac1{x}\right)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using Abel's identity we have that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle \sum_{1 \le n \le x} \frac{\log n}{n} = (\log x+ \gamma + R(x))\log x - \int_{1}^{x} \frac{\log t+ \gamma + R(t)}{t} \ \text dt$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i.e.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle \sum_{1 \le n \le x} \frac{\log n}{n} = \frac{\log^2 x}{2} + R(x)\log x - \int_{1}^{x} \frac{R(t)}{t} \ \text dt$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $\displaystyle R(x) = \mathcal{O}\left(\frac1{x}\right)$ we have that $\displaystyle \int_{1}^{\infty} \frac{R(t)}{t} \ \text dt = \eta$ exists. We also have that $\displaystyle R(x) \log x \to 0 \ \text{as} \ x \to \infty$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus we have that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle \sum_{1 \le n \le x} \frac{\log n}{n} = \frac{\log^2 x}{2} -\eta + \mathcal{O}\left(\frac{\log x}{x}\right) \ \ \text{as} \ \ x \to \infty$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another useful approach is to try using the &lt;a href=&quot;http://en.wikipedia.org/wiki/Euler%E2%80%93Maclaurin_formula&quot; rel=&quot;nofollow&quot;&gt;Euler-Maclaurin Summation formula&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-07T15:54:08.410" LastActivityDate="2010-12-07T15:54:08.410" CommentCount="4" />
  <row Id="13315" PostTypeId="1" AcceptedAnswerId="13316" CreationDate="2010-12-07T01:10:55.193" Score="1" ViewCount="65" Body="&lt;p&gt;I'm given a number (n), an interval (i), and the number of steps (s).  I want to distribute these numbers over that criteria as best as possible.  The upper bounds would be n (which is critical) and lower bounds is greater than or equal to 0.75 * n (but divisible by the interval).  So, if n=100, i=5, and s=6, you'd have [100, 95, 90, 85, 80, 75].  This one is easy because 100*.75 is &amp;lt;= s*i.  How would you do n=25, i=2.5, and s=6?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This was a problem I came across while programming and was wondering if there was some mathematical concept that fit this problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, I had no idea what to tag this under.&lt;/p&gt;&#xA;" OwnerUserId="4489" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2010-12-07T01:21:00.950" LastActivityDate="2010-12-07T01:21:00.950" Title="How do I distribute these numbers?" Tags="&lt;algebra-precalculus&gt;&lt;algebra&gt;" AnswerCount="2" />
  <row Id="13316" PostTypeId="2" ParentId="13315" CreationDate="2010-12-07T01:18:38.987" Score="1" ViewCount="" Body="&lt;p&gt;If I understand you, the ideal situation is to have your set be $[n, n-i, n-2i, \ldots,n-(s-1)i]$, which has s members.  To keep this within the top $0.75n$, you need $(s-1)i\leq .25n$, which is just satisfied in your first example.  In your second it is not, so you need to decide what to give up.  You could forget about the 0.75 criterion and keep the interval, giving $[25, 22.5, \ldots,12.5]$  Or you could maintain the 0.75 criterion and reduce the step size, giving an interval of 1.25.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-12-07T01:18:38.987" CommentCount="4" />
  <row Id="13317" PostTypeId="2" ParentId="13315" CreationDate="2010-12-07T01:20:08.017" Score="0" ViewCount="" Body="&lt;p&gt;in R:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;seq(from=25, by=-2.5, length=6)&lt;/p&gt;&#xA;" OwnerUserId="4488" LastActivityDate="2010-12-07T01:20:08.017" />
  <row Id="13318" PostTypeId="1" CreationDate="2010-12-07T01:49:43.813" Score="-1" ViewCount="148" Body="&lt;p&gt;Alright, so I have a triangle that is part of a mesh (3d model) and I need to get the rotation of it's x,y and z axis. Did I say all that right?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;rotation of y axis&#xA;    (Math.toDegrees(Math.atan(min.getYf()/max.getYf())&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How to get min and max:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Vector3 min = new Vector3((float)Math.min((float) Math.min(v3[s][0].getXf(), v3[s][1].getXf()), v3[s][2].getXf()),&#xA;                    (float)Math.min((float)Math.min(v3[s][0].getYf(), v3[s][1].getYf()), v3[s][2].getYf()),&#xA;                    (float)Math.min((float)Math.min(v3[s][0].getZf(), v3[s][1].getZf()), v3[s][2].getZf()));&#xA;&#xA;            Vector3 max = new Vector3((float) Math.max((float)Math.max(v3[s][0].getXf(), v3[s][1].getXf()), v3[s][2].getXf()),&#xA;                    (float)Math.max((float)Math.max(v3[s][0].getYf(), v3[s][1].getYf()), v3[s][2].getYf()),&#xA;                    (float)Math.max((float)Math.max(v3[s][0].getZf(), v3[s][1].getZf()), v3[s][2].getZf()));&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="4464" LastEditorUserId="4464" LastEditorDisplayName="" LastEditDate="2010-12-07T02:10:46.330" LastActivityDate="2010-12-07T02:10:46.330" Title="I am so confused!" Tags="&lt;geometry&gt;&lt;trigonometry&gt;" AnswerCount="1" CommentCount="9" />
  <row Id="13319" PostTypeId="1" AcceptedAnswerId="13321" CreationDate="2010-12-07T01:55:21.970" Score="2" ViewCount="280" Body="&lt;p&gt;Okay, so almost 3 months into my abstract algebra, we just started rings. I have a few questions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A &quot;trivial ring&quot; is a ring with only one element. So $R={0}$ is a trivial ring. Understandable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then a definition states: Let $R$ be a ring. If there is an element $x \in R$ s.t. for all $a \in R$ you have $a * x = a = x * a$ then R is called a &quot;ring with identity&quot;. The notation is $1_{R}$. I also know it is possible to have a right identity and not a left identity (in the case of a 2 x 2 matrix). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then we have a definition for an integral domain. An integral domain is a commutative ring $R$ with identity $1_{R} \neq 0_{R}$. Okay so this is where I am confused now. What does this mean? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it saying that the identity element can not be the zero element? And the identity element could be anything right? For integers its 1. So basically it states that if you have a TRIVIAL RING and if the element (since there can only be one) is not the zero element then its an ID? &lt;/p&gt;&#xA;" OwnerUserId="287" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-12-07T16:58:46.180" LastActivityDate="2010-12-07T17:17:19.043" Title="What is an Integral Domain?" Tags="&lt;abstract-algebra&gt;" AnswerCount="4" CommentCount="3" />
  <row Id="13320" PostTypeId="2" ParentId="13318" CreationDate="2010-12-07T01:59:59.993" Score="0" ViewCount="" Body="&lt;p&gt;You need some kind of reference.  Where did the triangle start and what kind of transformation was applied to it?  Was it only rotated or was it also translated?&lt;/p&gt;&#xA;" OwnerUserId="4488" LastActivityDate="2010-12-07T01:59:59.993" CommentCount="3" />
  <row Id="13321" PostTypeId="2" ParentId="13319" CreationDate="2010-12-07T02:01:45.670" Score="4" ViewCount="" Body="&lt;p&gt;An integral domain is a commutative ring with unit $1\neq 0$ such that if $ab=0$ then either $a=0$ or $b=0$.  The idea that $1\neq 0$ means that the multiplicative unit, the element $x$ such that $xa=a$ for all $a\in R$ is not the same element as the additive unit, the element $y$ such that $a+y=a$ for all $a\in R$.  We denote the multiplicative unit by 1 and the additive unit by 0, since they are similar to the same thing in $\mathbb{Z}$.  The second part says that if $a,b\neq 0$, then $ab\neq 0$.  For instance $\mathbb{Z}/6$ has two elements $[2]$ and $[3]$ such that $[2]\cdot [3]=[0]$.  So, it is not a integral domain.&lt;/p&gt;&#xA;" OwnerUserId="2838" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-07T02:12:42.227" LastActivityDate="2010-12-07T02:12:42.227" CommentCount="20" />
  <row Id="13322" PostTypeId="2" ParentId="13268" CreationDate="2010-12-07T02:09:57.953" Score="0" ViewCount="" Body="&lt;p&gt;I am too untrusted to add a comment so i'm spamming this as an answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;J.M: I think you are considering the converse of the question.  If you have two very different-looking matrices it is possible that they have the same spectrum.  But if you have two similar-looking matrices, then is it possible that their spectra can be very different?&lt;/p&gt;&#xA;" OwnerUserId="4488" LastActivityDate="2010-12-07T02:09:57.953" CommentCount="8" />
  <row Id="13323" PostTypeId="2" ParentId="13319" CreationDate="2010-12-07T02:10:27.727" Score="6" ViewCount="" Body="&lt;p&gt;There is only one trivial ring up to isomorphism.  It has one element $x$ which satisfies $x + x = x \cdot x = x$.  That means that $x$ is both the additive identity and the multiplicative identity; symbolically, $x = 1_R = 0_R$.  This is an if-and-only-if, too: if a ring $R$ has the property that $1_R = 0_R$, then every element is equal to $0_R$, since $x \cdot 1_R = x = x \cdot 0_R = 0_R$.  So a ring is the trivial ring if and only if $1_R = 0_R$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the definition of an integral domain, we require that the ring is nontrivial.  There are several good reasons for this, but they are sort of hard to motivate at the level of a first course in abstract algebra.  (But as Joe Johnson points out, this is not the main part of the definition of an integral domain.)&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-12-07T02:10:27.727" CommentCount="1" />
  <row Id="13324" PostTypeId="1" CreationDate="2010-12-07T03:08:11.680" Score="0" ViewCount="137" Body="&lt;p&gt;This might be an easy question, but I haven't been able to up come up with a solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The image of the map $$f : \mathbb{R} \to \mathbb{R}^2, a \mapsto (\frac{2a}{a^2+1}, \frac{a^2-1}{a^2+1})$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is the unit circle take away the north pole. $f$ extends to a function $$g: \mathbb{C} \backslash \{i, -i \} \to \mathbb{C}^2. $$  Can anything be said about the image of $g$?&lt;/p&gt;&#xA;" OwnerUserId="1725" LastActivityDate="2011-02-05T05:17:42.270" Title="A stereographic projection related question" Tags="&lt;complex-analysis&gt;&lt;functions&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="13325" PostTypeId="1" AcceptedAnswerId="13350" CreationDate="2010-12-07T03:21:09.970" Score="-2" ViewCount="115" Body="&lt;p&gt;Alright, so I got two vectors that each contain a x y and z. We'll call them a and b.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;atan2(double y, double x)&#xA;Converts rectangular coordinates (x, y) to polar (r, theta).&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I create a new point in the middle of the two points.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Vector3 middle = new Vector3(//order x,y,z&#xA;        (max.getX() - min.getX())/2,&#xA;        (max.getY() - min.getY())/2,&#xA;        (max.getZ() - min.getZ())/2&#xA;        );&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;To get the y angle I use the formula &#xA;&lt;code&gt;Math.toDegrees(Math.atan2(middle.getX(), middle.getZ()))&lt;/code&gt;&lt;/p&gt;&#xA;" OwnerUserId="4464" LastActivityDate="2010-12-07T08:21:31.780" Title="getting the x,y, and z angle of two points?" Tags="&lt;geometry&gt;&lt;trigonometry&gt;&lt;vector-space&gt;" AnswerCount="1" CommentCount="0" />
  <row Id="13326" PostTypeId="2" ParentId="13324" CreationDate="2010-12-07T03:24:52.883" Score="1" ViewCount="" Body="&lt;p&gt;Note that although $a$ is complex, is valid :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left(\frac{2a}{a^2+1}\right)^2+\left(\frac{a^2-1}{a^2+1}\right)^2= \frac{4a^2}{(a^2+1)^2}+\frac{(a^2-1)^2}{(a^2+1)}=$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{4a^2+a^4-2a^2+1}{(a^2+1)^2}=\frac{(a^4+2a^2+1)}{(a^2+1)^2}=\frac{(a^2+1)^2}{(a^2+1)^2}=1$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus is also an circle&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is say, the points of the set $\{g(a)\in \mathbb{C}^2 :a\in \mathbb{C}/ \{\imath,-\imath\}\}$ meet the above.&lt;/p&gt;&#xA;" OwnerUserId="3421" LastEditorUserId="3421" LastEditorDisplayName="" LastEditDate="2010-12-07T03:50:37.113" LastActivityDate="2010-12-07T03:50:37.113" CommentCount="1" />
  <row Id="13327" PostTypeId="2" ParentId="13324" CreationDate="2010-12-07T03:28:28.323" Score="0" ViewCount="" Body="&lt;p&gt;It is the unit circle. Substituting $\tan t$ for $a$ and simplifying, you get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ x=\sin 2t, y=-\cos 2t $$ which is the unit circle.&lt;/p&gt;&#xA;" OwnerUserId="3249" LastEditorUserId="3249" LastEditorDisplayName="" LastEditDate="2010-12-07T03:53:10.150" LastActivityDate="2010-12-07T03:53:10.150" CommentCount="4" />
  <row Id="13328" PostTypeId="2" ParentId="13319" CreationDate="2010-12-07T03:29:46.763" Score="3" ViewCount="" Body="&lt;p&gt;An integral domain is a ring with no zero divisors, i.e. $\rm\ xy = 0\ \Rightarrow\ x=0\ \ or\ \ y=0\:.\:$ Additionally it is a widespread &lt;em&gt;convention&lt;/em&gt; to disallow as a domain the trivial one-element ring (or, equivalently, the ring with $\: 1 = 0\:$). It is the nonexistence of zero-divisors that is the important hypothesis in the definition. The exclusion of the trivial ring is merely a convention that proves convenient in many contexts (see my &lt;a href=&quot;http://math.stackexchange.com/questions/120/is-1-a-prime-number/5735#5735&quot;&gt;post here&lt;/a&gt; for some reasons why the convention proves convenient).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One may think of a domain as a ring-theoretic analog of a field, since a ring is a domain iff it is a subring of some field. Indeed, a field clearly has no zero-divisors so ditto for all of its subrings. Conversely, any domain may be enlarged to its field of fractions by adjoining an inverse for every nonzero element. Although domains needn't contain inverses for every $\rm\ c\ne 0\:,\:$ they do retain a trace of this property. Namely, nonzero elements $\rm\:c\:$ are cancellable: $\rm\ c\ x = c\ y\ \Rightarrow\ x = y\:.\ $ This enables transfer of many (but not all) properties of fields to domains. Here is one useful example: $\ $ a polynomial $\rm\ f(x)\in D[x]\ $ has at most $\rm\ deg\ f\ $ roots in the ring $\rm\:D\ $ iff $\rm\ D\:$ is a domain. For the simple proof see my &lt;a href=&quot;http://math.stackexchange.com/questions/4597/simple-explanation-and-examples-of-the-miller-rabin-primality-test/4604#4604&quot;&gt;post here&lt;/a&gt;, where I illustrate it constructively in $\rm\ \mathbb Z/m\ $ by showing that, given any $\rm\:f(x)\:$ with more roots than its degree, we can quickly compute a nontrivial factor of $\rm\:m\:$ via a $\rm\:gcd\:$. The quadratic case of this result is at the heart of many integer factorization algorithms, which attempt to factor $\rm\:m\:$ by searching for a nontrivial square root in $\rm\: \mathbb Z/m\:,\:$ e.g. a square root of $1$ that is not $\:\pm 1$.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-12-07T17:17:19.043" LastActivityDate="2010-12-07T17:17:19.043" CommentCount="1" />
  <row Id="13330" PostTypeId="1" AcceptedAnswerId="13331" CreationDate="2010-12-07T03:46:45.337" Score="-1" ViewCount="604" Body="&lt;p&gt;how does 1+1 = 3?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've seen it around and I mean the obvious response is that its just wrong but people say stuff like there's just some special way to look at it or something. any ideas?&lt;/p&gt;&#xA;" OwnerUserId="1141" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-07T05:04:06.223" LastActivityDate="2010-12-07T10:02:44.423" ClosedDate="2010-12-07T12:02:46.113" Title="I'm totally confused about this addition problem" Tags="&lt;addition&gt;" AnswerCount="4" CommentCount="7" />
  <row Id="13331" PostTypeId="2" ParentId="13330" CreationDate="2010-12-07T04:04:52.693" Score="2" ViewCount="" Body="&lt;p&gt;It follows directly from the elementary theorem 2=1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $a=b$ be integers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a = b$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Multiply both sides by $a$: $a^2 = ab$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Subtract $b^2$ from both sides: $a^2 -b^2 = ab - b^2$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Factor: $(a-b)(a+b) = b(a-b)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cancel $(a-b)$ on both sides: $a+b = b$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Use the fact $a=b$: $2b = b$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Divide both sides by $b$: $2 = 1$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now that we have $2=1$, we get $3 = 1 + 1$ as a corollary. And an elementary proof of the Riemann Hypothesis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S. $a-b = 0$, so canceling is the same as division by $0$.&lt;/p&gt;&#xA;" OwnerUserId="4388" LastActivityDate="2010-12-07T04:04:52.693" CommentCount="3" />
  <row Id="13332" PostTypeId="1" CreationDate="2010-12-07T04:08:53.677" Score="2" ViewCount="134" Body="&lt;p&gt;It is a local constant sheaf, but why the local system on topological space $X$ is one-to-one correspondes to $\tilde{X}\times_G V$, where $G$is fundamental group of $X$ and $\tilde{X}$ is the universal space of $X$, and $V$ is a $G$-module,  I am wondering why $\tilde{X}\times_G V$ is a local system (local constant sheaf)?&lt;/p&gt;&#xA;" OwnerUserId="1944" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-07T05:34:04.393" LastActivityDate="2010-12-07T05:45:05.957" Title="What local system really is" Tags="&lt;algebraic-topology&gt;&lt;coherent-sheaves&gt;" AnswerCount="1" CommentCount="0" />
  <row Id="13333" PostTypeId="2" ParentId="13330" CreationDate="2010-12-07T04:14:31.683" Score="3" ViewCount="" Body="&lt;p&gt;I have no idea which people are saying $1+1 = 3$ and why they are saying it.  Certainly, without any additional explanation this sounds like a simply false statement: in standard arithmetic, $1+1 = 2$, which does not equal $3$!  (It is not even the case that $1+ 1 = 3$ &lt;em&gt;modulo n&lt;/em&gt; for any $n &amp;gt; 1$, and considering numbers modulo $1$ is, literally, trivial: I believe there was another recent question about that.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But there &lt;em&gt;are&lt;/em&gt; real-world phenomena which are somewhat reasonably described as $1 +1 = 3$.  For instance, suppose I am living on a desert island and I have fashioned a very crude scale out of coconuts, bamboo, crab exoskeleta, and so forth.  My scale doesn't have decimals on it: it says that the weight of something is some whole number of pounds.  Now it is entirely possible that I can have two rocks, each of which, according to the scale, weighs one pound, but when I put both rocks on the scale, it tells me that together they weigh three pounds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course this is an instance of rounding error: e.g. maybe the true weight of each rock is something like $1.35$ pounds -- which gets rounded to $1$ -- so the true weight of the two rocks is something like $2.7$ pounds -- which gets rounded to $3$.  I have explained this in the context of a somewhat unlikely scenario, but rounding errors are something that one has to take very seriously in applied mathematics and the sciences.&lt;/p&gt;&#xA;" OwnerUserId="299" LastActivityDate="2010-12-07T04:14:31.683" />
  <row Id="13334" PostTypeId="1" AcceptedAnswerId="13338" CreationDate="2010-12-07T05:04:19.207" Score="3" ViewCount="126" Body="&lt;p&gt;I am reading Joseph Bernstein's notes on D-modules which are available online &lt;a href=&quot;http://www.math.columbia.edu/~khovanov/resources/Bernstein-dmod.pdf&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In section 1 (page 1 of the pdf) Bernstein writes &quot;By Serre's theorem this condition is local.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was wondering to what theorem Bernstein is referring and wondering where I could find a statement and proof of this theorem.&lt;/p&gt;&#xA;" OwnerUserId="249" LastActivityDate="2010-12-07T17:35:15.013" Title="What theorem of Serre's is being referenced?" Tags="&lt;reference-request&gt;&lt;d-modules&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="13335" PostTypeId="2" ParentId="13330" CreationDate="2010-12-07T05:04:26.910" Score="0" ViewCount="" Body="&lt;p&gt;More generally, if somewhere in our reasoning we make a mistake (by dividing by zero, assuming that $a^2 = b^2 \implies a = b$, assuming that $\lnot P \implies \lnot Q$ follows from $P \implies Q$), we are able to prove anything.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is, $P \implies Q$ is always true if $P$ is false to begin with. Wikipedia calls this being &quot;&lt;a href=&quot;http://en.wikipedia.org/wiki/Vacuous_truth&quot; rel=&quot;nofollow&quot;&gt;vacuously true&lt;/a&gt;&quot;, although I think the term is usually used when the hypothesis is not satisfied but the result is still true, as in the example given of all cell phones being turned off in a room simply because there are no cell phones.&lt;/p&gt;&#xA;" OwnerUserId="2954" LastActivityDate="2010-12-07T05:04:26.910" />
  <row Id="13336" PostTypeId="1" AcceptedAnswerId="13337" CreationDate="2010-12-07T05:06:18.217" Score="1" ViewCount="153" Body="&lt;p&gt;Let $R$ be a non-zero commutative ring. Prove that the ideal $(x)$ of $R[x]$ is prime if and only if $R$ is an integral domain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm working on the left-to-right direction right now. I know that $R[x]/(x)$ is an integral domain since $(x)$ is prime. So I want to fix $r\in R$ and use the evaluation map $e_r:R[x]\to R$ given by $f(x)\mapsto f(r)$, a surjective ring homomorphism, and apply the first isomorphism theorem. But I'm having trouble with the kernel of $e_r$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;First of all, is it even true that $\ker{(e_r)}=(x)$? If so, any hints you could drop would be great. Thanks! &lt;/p&gt;&#xA;" OwnerUserId="2313" LastActivityDate="2010-12-07T06:17:51.950" Title="Understanding the kernel of the evaluation map" Tags="&lt;abstract-algebra&gt;&lt;ring-theory&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="13337" PostTypeId="2" ParentId="13336" CreationDate="2010-12-07T05:09:39.603" Score="4" ViewCount="" Body="&lt;p&gt;No, the kernel of $e_r$ consists of all polynomials that are zero at $r$: remember, $\mathrm{ker}(e_r) = \{ f(x)\in R[x]\mid e_r(f(x)) = f(r) = 0\}$. It will contain $(x-r)$ (assuming your ring has an identity), but in general rings it may be all sorts of things (&lt;em&gt;edit&lt;/em&gt;: as Bill Dubuque points out, the Factor Theorem will be on hand in any ring with $1$, but in rings without $1$ weird things may occur). But that does not really matter here, because you don't want to look at an &lt;em&gt;arbitrary&lt;/em&gt; $r$ in $R$. You want to pick a particular $r$ that gives you the kernel you want. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And for you to get $(x)$ as the kernel, you should look at $e_0$, evaluation at $0$. Trivially, $(x)$ is contained in the kernel; showing that every element of the kernel is in $(x)$ should be straightforward.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-07T05:42:54.867" LastActivityDate="2010-12-07T05:42:54.867" CommentCount="5" />
  <row Id="13338" PostTypeId="2" ParentId="13334" CreationDate="2010-12-07T05:12:07.993" Score="3" ViewCount="" Body="&lt;p&gt;I believe the first theorem in article 1.2 &lt;a href=&quot;http://books.google.com/books?id=ZhzXJHUgcRUC&amp;amp;pg=PA28&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3431" LastEditorUserId="3431" LastEditorDisplayName="" LastEditDate="2010-12-07T17:35:15.013" LastActivityDate="2010-12-07T17:35:15.013" CommentCount="1" />
  <row Id="13340" PostTypeId="2" ParentId="6381" CreationDate="2010-12-07T05:35:16.680" Score="1" ViewCount="" Body="&lt;p&gt;If you compute homology with twisted coefficients, where the coefficient system involves vector spaces of dimension $d$, then the Euler characteristic of the resulting homology spaces (i.e. the alternating sum of the $H_i$ with twisted coefficients) is equal to $d$ times the Euler characteristic of the space computed via homology with trivial coefficients.&lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2010-12-07T05:35:16.680" />
  <row Id="13341" PostTypeId="1" AcceptedAnswerId="13364" CreationDate="2010-12-07T05:44:43.540" Score="5" ViewCount="187" Body="&lt;p&gt;I have a helix in parametric equations that wraps around the Z axis and a point in space. I want to determine the shortest distance between this helix and the point, how would i go about doing that?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've tried using Pythagorean theorem to get the distance and then taking the derivative of the distance function to find the zeros but I can't seem to get an explicit equation for T and I'm stuck at that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I apologize for the tags, not sure how to tag it and I cant create new ones either)&lt;/p&gt;&#xA;" OwnerUserId="3507" LastActivityDate="2010-12-07T13:37:14.433" Title="Shortest distance between a point and a helix" Tags="&lt;geometry&gt;" AnswerCount="1" CommentCount="6" />
  <row Id="13342" PostTypeId="2" ParentId="13332" CreationDate="2010-12-07T05:45:05.957" Score="2" ViewCount="" Body="&lt;p&gt;The group $G$ acts properly discontinuously on $\tilde{X}$, and so if $x$ is any&#xA;point of $\tilde{X}$, it admits a neighbourhood $U$ s.t. that $U g$ is disjoint&#xA;from $U$ if $g \in G$ is non-trivial.   Thus the natural map from&#xA;$U$ to $\tilde{X}/G = X$ is an embedding.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus the natural map from $U \times V$ to $\tilde{X}\times_G V$ is also an&#xA;embedding, and so $\tilde{X}\times_G V$ is locally constant (i.e. locally a&#xA;product).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;More detailed remarks:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;We should equip $V$ with its discrete topology&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The object $\tilde{X}\times_G V$ is not itself actually a sheaf, but is rather the espace etale of a sheaf.  To get the actual sheaf we consider the natural projection $\tilde{X}\times_G V \to \tilde{X}/G = X$, and form the associated sheaf of sections.  Over the open set $U \hookrightarrow X,$ this restricts to the sheaf of sections&#xA;to the projection $U\times V \to U$, whose sections are precisely the constant sheaf on $U$ attached to the vector space $V$.  (Here is where we see that it is important to equip $V$ with the discrete topology.)  Thus our original sheaf&#xA;of sections is locally constant, as claimed.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="221" LastActivityDate="2010-12-07T05:45:05.957" CommentCount="2" />
  <row Id="13343" PostTypeId="2" ParentId="13336" CreationDate="2010-12-07T06:05:47.833" Score="2" ViewCount="" Body="&lt;p&gt;You are on the right track by applying: $\rm\ A/I\ $ is a domain iff $\rm\ I\ $ is prime. For $\rm\ A = R[x],\ \ I = (x)\ $ the problem reduces to showing that $\rm R[x]/(x)\cong R\:.\:$ As you surmise, this is easily achieved by exhibiting an evaluation map with kernel $\rm\ (x)\:,\:$ i.e. an evaluation which maps $\rm\ x\to \ \ldots$?   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Presumably you already know the existence (and uniqueness) of such evaluation maps for polynomial rings. In fact this &lt;a href=&quot;http://en.wikipedia.org/wiki/Universal_property&quot; rel=&quot;nofollow&quot;&gt;universal mapping property&lt;/a&gt; characterizes polynomial rings. It can be employed to give slick conceptual proofs of results about polynomial rings. For a simple example see &lt;a href=&quot;http://math.stackexchange.com/questions/2514/why-cant-the-polynomial-ring-be-a-field/2523#2523&quot;&gt;this proof&lt;/a&gt; that $\rm\:x\:$ is not invertible in $\rm\: R[x]\:.$ The importance of such universal characterizations will become clearer when you study universal algebra and category theory.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-12-07T06:17:51.950" LastActivityDate="2010-12-07T06:17:51.950" />
  <row Id="13344" PostTypeId="1" AcceptedAnswerId="13346" CreationDate="2010-12-07T06:12:05.373" Score="10" ViewCount="553" Body="&lt;p&gt;I am looking for a short proof that $$\int_0^\infty \left(\frac{\sin x}{x}\right)^2 dx=\frac{\pi}{2}.$$&#xD;&#xA;What do you think?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is kind of amazing that $$\int_0^\infty \frac{\sin x}{x} dx$$ is also $\frac{\pi}{2}.$ Many proofs of this latter one are already in this &lt;a href=&quot;http://math.stackexchange.com/questions/5248/solving-the-integral-int-0-infty-frac-sinxx-dx-frac-pi2/8350#8350&quot;&gt;post&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3249" LastEditorUserId="3249" LastEditorDisplayName="" LastEditDate="2010-12-07T06:45:17.673" LastActivityDate="2010-12-07T20:21:55.740" Title="Proof for an integral involving sinc function" Tags="&lt;analysis&gt;&lt;integral&gt;" AnswerCount="6" CommentCount="2" FavoriteCount="4" />
  <row Id="13345" PostTypeId="1" CreationDate="2010-12-07T06:20:56.720" Score="1" ViewCount="135" Body="&lt;p&gt;Let $\displaystyle X$ be a Polish space. Let $\displaystyle d,d&amp;#39;$ be two complete compatible metrics. Is $\displaystyle 1:(X,d)\to(X,d&amp;#39;)$ uniformly continuous.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I really want to know is: can we unambiguously speak of $\displaystyle f:X\to K$ from one Polish space to a compact Polish space as being uniformly continuous? Clearly it won't depend on the compatible metric on $\displaystyle K$, and clearly if the above question is answered in the positive, then the answer will be „yes“.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So far I have done the following. Suppose $\displaystyle 1:(X,d)\to(X,d&amp;#39;)$ were not uniformly cts. Then there would be an $\displaystyle \epsilon\in\mathbb{R}^{+}$ for which $\displaystyle (\forall{\delta\in\mathbb{R}^{+}})(\exists{x\in X})\ \  \mathcal{B}^{d}_{&amp;lt;\delta}(x)\nsubseteq\mathcal{B}^{d&amp;#39;}_{&amp;lt;\epsilon}(x)$.&#xA;For each $\displaystyle \delta\in\mathbb{R}^{+}$, let $\displaystyle S_{\delta}:=\{x\mid \mathcal{B}^{d}_{&amp;lt;\delta}(x)\nsubseteq\mathcal{B}^{d&amp;#39;}_{&amp;lt;\epsilon}(x)\}$. Then for all $\displaystyle \delta&amp;lt;\delta&amp;#39;$ in $\displaystyle \mathbb{R}^{+}$, we would have $\displaystyle S_{\delta}\neq\emptyset$ and $\displaystyle S_{\delta}\subseteq S_{\delta&amp;#39;}$. We also see that $\displaystyle \overline{S_{\delta}}\subseteq\bigcup_{\delta&amp;#39;&amp;gt;\delta}S_{\delta&amp;#39;}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So were $\displaystyle \bigcap_{n\in\omega}\overline{S_{\delta_{n}}}\neq\emptyset$ where $\displaystyle (\delta_{n})_{n\in\omega}\searrow 0$ in $\displaystyle \mathbb{R}$, then it would be so that $\displaystyle \bigcap_{\delta\in\mathbb{R}}S_{\delta}\neq\emptyset$. But then there would be an $\displaystyle x$ and a sequence $\displaystyle (x_{n}\in\mathcal{B}^{d}_{&amp;lt;\delta_{n}}(x)\setminus\mathcal{B}^{d&amp;#39;}_{&amp;lt;\epsilon}(x))_{n\in\omega}$ converging to $\displaystyle x$ in the space $\displaystyle (X,d)$ and not converging to $\displaystyle x$ in $\displaystyle (X,d&amp;#39;)$, which would contradict that $\displaystyle d,d&amp;#39;$ are compatible for $\displaystyle X$. Thus it would be that the countable intersection $\displaystyle \bigcap_{n\in\omega}\overline{S_{\delta_{n}}}=\emptyset$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I can show, this müsse not be the case, then I would be done.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However (1) there might be a completely different way to proving this; (2) the question might have a negative answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any ideas would be great.&lt;/p&gt;&#xA;" OwnerUserId="4493" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-12-08T09:06:05.273" LastActivityDate="2010-12-08T15:18:18.507" Title="Is uniform continuity well defined on Polish spaces?" Tags="&lt;real-analysis&gt;&lt;general-topology&gt;" AnswerCount="4" CommentCount="2" />
  <row Id="13346" PostTypeId="2" ParentId="13344" CreationDate="2010-12-07T06:26:47.087" Score="10" ViewCount="" Body="&lt;p&gt;Let $f(x)=\max\{0,1-|x|\}$. It is easy to calculate the Fourier transform&#xA;$$\hat{f}(\xi)=\int_{-\infty}^{\infty}f(x)e^{-ix\xi}dx=\left(\frac{\sin(\xi/2)}{\xi/2}\right)^2.$$&#xD;&#xA;Taking the inverse Fourier transform, we get &#xA;$$\int_{-\infty}^{\infty}\left(\frac{\sin(\xi/2)}{\xi/2}\right)^2e^{ix\xi}d\xi=2\pi f(x),$$ &#xA;and the result follows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second integral can be computed in a similar way. Just take $f(x)=\chi_{[-1,1]}(x)$ (the indicator function of the interval $[-1,1]$).&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit.&lt;/strong&gt; It might be interesting to note that there are analogous formulas for the sinc &#xA;sums&#xA;$$\sum_{n=1}^{\infty}\frac{\sin n}{n}=\sum_{n=1}^{\infty}\left(\frac{\sin n}{n}\right)^2=&#xD;&#xA;\frac{\pi}{2}-\frac{1}{2}.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I learned about this from the note &lt;em&gt;&quot;Surprising Sinc Sums and Integrals&quot;&lt;/em&gt; by Baillie, Borwein, and Borwein (can be found through a quick web search).&lt;/p&gt;&#xA;" OwnerUserId="723" LastEditorUserId="723" LastEditorDisplayName="" LastEditDate="2010-12-07T09:03:27.610" LastActivityDate="2010-12-07T09:03:27.610" CommentCount="6" />
  <row Id="13347" PostTypeId="2" ParentId="13345" CreationDate="2010-12-07T06:47:18.793" Score="1" ViewCount="" Body="&lt;p&gt;Hmm, yes that is. Also $X=[0,\infty)$ with $d(x,y)=|y-x|$ $d&amp;#39;(x,y)=|y^{2}-x^{2}|$ would work too. Thanks for that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Okay then, scrapping that too general a question\ldots can my other concern be addressed, &lt;em&gt;i.e.&lt;/em&gt; can we unambiguously call a map from a Polish space to a compact Polish space uniformly continuous without referring to the metrics?&lt;/p&gt;&#xA;" OwnerUserId="4494" LastActivityDate="2010-12-07T06:47:18.793" CommentCount="1" />
  <row Id="13348" PostTypeId="2" ParentId="13344" CreationDate="2010-12-07T06:59:17.263" Score="10" ViewCount="" Body="&lt;p&gt;Well, it's not hard to reduce this integral to $\displaystyle \int_0^{\infty} {\sin(x) \over x}\,dx$: Just integrate by parts in $\displaystyle \int_0^{\infty} {\sin^2(x) \over x^2}\,dx$, integrating $\displaystyle {1 \over x^2}$ and differentiating $\displaystyle \sin^2(x)$. You're left with $\displaystyle \int_0^{\infty} {\sin(2x) \over x}\,dx$ which reduces to the $\displaystyle \int_0^{\infty} {\sin(x) \over x}\,dx$ integral after changing variables from $\displaystyle x$ to $\displaystyle 2x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So any elementary proof that $\displaystyle \int_0^{\infty} {\sin(x) \over x}\,dx = {\pi \over 2}$ is effectively also an elementary proof that $\displaystyle \int_0^{\infty} {\sin^2(x) \over x^2}\,dx$ is also $\displaystyle {\pi \over 2}$.&lt;/p&gt;&#xA;" OwnerUserId="3035" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-12-07T07:05:46.230" LastActivityDate="2010-12-07T07:05:46.230" />
  <row Id="13349" PostTypeId="2" ParentId="13055" CreationDate="2010-12-07T07:10:53.883" Score="0" ViewCount="" Body="&lt;p&gt;My only experience is applying for research postdocs last year, and a lot of schools started looking at applications shortly after Thanksgiving. Some of my friends had offers the first week of December. So I would say apply by mid-November to be safe. This might only apply to research postdocs though.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't like the blanket advice &quot;apply as early as possible&quot; because for many people, including me, the late summer and early fall of application season is a very productive time, and you might get a lot of research done. If you're hot on the trail of something good it's better to finish it off than apply early, especially if you're a grad student with not so many results. Of course if you're at a natural stopping point in your research (e.g. because you're essentially stuck) then it's better not to procrastinate and get the applications in early. Professors have access to your file on mathjobs as soon as you apply, and maybe someone in the department where you're applying is bored and browsing...&lt;/p&gt;&#xA;" OwnerUserId="4495" LastActivityDate="2010-12-07T07:10:53.883" />
  <row Id="13350" PostTypeId="2" ParentId="13325" CreationDate="2010-12-07T08:00:58.700" Score="0" ViewCount="" Body="&lt;p&gt;if you have two points, we can state the following let $P_1(x_1,y_1,z_1)$ and $P_2(x_2,y_2,z_2)$ be your two points and $\alpha, \beta, \gamma$, the angles to the X,Y,Z axes respectively; if we want to find the direction (or the angles two the X,Y,Z axes) of the segment formed by the two points $\overline{P_1 P_2}$ these are obtained from these formulas:&#xA;$$\cos{\alpha}=\frac{\Delta x}{d},\cos{\beta}=\frac{\Delta y}{d},\cos{\gamma}=\frac{\Delta z}{d}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where d is the distance between the two points which is given by the following formula:&#xA;$$d=\sqrt{(\Delta x)^2+(\Delta y)^2+(\Delta z)^2}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you've found the values of each cosine, you can use their inverse trigonometric function to find each angle, we would have this:&#xA;$$\alpha = \arccos{\frac{\Delta x}{d}}, \beta = \arccos{\frac{\Delta y}{d}}, \gamma = \arccos{\frac{\Delta z}{d}}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively, if you are looking for th angles that each point forms with the three axes, let $O(0,0,0)$ be the origin and use it to find the direction of the two segments $\overline{O P_1}$ and $\overline{O P_2}$ as explained above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope that this is what your looking for.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regards Tristian.&lt;/p&gt;&#xA;" OwnerUserId="4312" LastEditorUserId="4312" LastEditorDisplayName="" LastEditDate="2010-12-07T08:21:31.780" LastActivityDate="2010-12-07T08:21:31.780" CommentCount="2" />
  <row Id="13351" PostTypeId="2" ParentId="13344" CreationDate="2010-12-07T08:17:45.853" Score="4" ViewCount="" Body="&lt;p&gt;More generally, there is a result due to Wolstenholme (I can't find a link) that says&#xA;$$ \int_0^\infty \left( \frac {\sin x}{x} \right)^n dx = &#xD;&#xA;\frac{1}{(n-1)!} \frac{\pi}{2^n} \left\lbrace&#xD;&#xA;n^{n-1} - { n \choose 1 } (n-2)^{n-1} + { n \choose 2 } (n-4)^{n-1} - \cdots&#xD;&#xA;\right\rbrace .$$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="1301" LastActivityDate="2010-12-07T08:17:45.853" CommentCount="2" />
  <row Id="13352" PostTypeId="2" ParentId="13091" CreationDate="2010-12-07T08:35:25.010" Score="0" ViewCount="" Body="&lt;p&gt;To flesh out Robin's answer a little: The thing that you really want to find is the unit normal vector $n=(A,B,C)^t$ to the sought plane. Once you have that, it's easy to find the equation of the plane. This vector has to satisfy three conditions:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Length one: $A^2+B^2+C^2=1$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Orthogonal to the vector $\overrightarrow{P_1 P_2} = (1,0,3)^t$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;60 degrees to the normal vector of the other plane. If we normalize that vector too, this condition becomes $\frac13 (2,-2,1)^t \cdot (A,B,C) = \cos 60^{\circ}$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Conditions 2 and 3 give a linear system with two equations and three unknowns $A$, $B$, $C$; solve it and you will get a family of solutions depending on one parameter $t$. Condition 1 then gives a quadratic eqation for determining the two possible values of $t$.&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2010-12-07T08:35:25.010" />
  <row Id="13353" PostTypeId="1" AcceptedAnswerId="13354" CreationDate="2010-12-07T08:42:26.443" Score="1" ViewCount="73" Body="&lt;p&gt;I have some problems understanding a question in a math class test for 10 year olds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Someone pays in 200€ to a bank. The interest rate is 3.75%. Some time later, he looks at the account balance and he sees that there are 235€ now. How much time passed?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I would calculate this using logarithms, but the interesting part here is that they don't even know square roots, not to mention logarithms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any easy way to calculate this without logarithms that I overlooked?&lt;/p&gt;&#xA;" OwnerUserId="4497" LastActivityDate="2010-12-07T09:05:01.600" Title="Interest of interest calculations without logarithm" Tags="&lt;finance&gt;&lt;logarithms&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="13354" PostTypeId="2" ParentId="13353" CreationDate="2010-12-07T09:05:01.600" Score="4" ViewCount="" Body="&lt;p&gt;It depends on whether the interest is simple or compound.  If the interest is simple, you look at the fact the the 35C gain is a gain of 17.5%.  Then, 17.5% divided by 3.75% is 4 and two third years (assuming the interest is 3.75% per year).  If the interest is compound, I see no other way then using logarithms producing an answer of about 4.38 years.  If this is meant to be an easy problem, they are probably assuming simple interest since 4 and two third years is not a very complicated answer.&lt;/p&gt;&#xA;" OwnerUserId="3971" LastActivityDate="2010-12-07T09:05:01.600" CommentCount="2" />
  <row Id="13355" PostTypeId="2" ParentId="5804" CreationDate="2010-12-07T09:15:17.707" Score="1" ViewCount="" Body="&lt;p&gt;I think you will find the proof for this in Mallat 1989, 'Multiresolution approximations and wavelet orthonormal bases of L^2'. Theorem 1 (in particular Equations (23), (36)) is what you are after. It is not trivial, longer to prove than I immediately thought. Perhaps there is a very fast proof but I can't think of it now.&lt;/p&gt;&#xA;" OwnerUserId="4322" LastEditorUserId="4322" LastEditorDisplayName="" LastEditDate="2010-12-09T09:58:54.617" LastActivityDate="2010-12-09T09:58:54.617" CommentCount="6" />
  <row Id="13356" PostTypeId="2" ParentId="13344" CreationDate="2010-12-07T09:33:04.820" Score="2" ViewCount="" Body="&lt;p&gt;Apply Parseval-Plancherel to $\chi_{[-1,1]}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we consider the Fourier transform as given by &#xA;$$f\mapsto \hat{f}(\xi)=\int_{-\infty}^\infty f(x)e^{-2\pi i x\xi}dx$$&#xD;&#xA;then &#xA;$$\int_{-\infty}^\infty |f(x)|^{2}dx=\int_{-\infty}^\infty |\hat{f}(\xi)|^{2}d\xi$$&#xD;&#xA;for $f\in L^2(\mathbb{R})$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $f(x)=\chi_{[-1,1]}(x)$, the characteristic function on $I=[-1,1]$ (that is $f(x)=1$ for $x\in I$ and $f(x)=0$ otherwise), we have&#xA;$$\hat{f}(x)=\int_{-\infty}^\infty f(x)e^{-2\pi i x\xi}dx =\int_{-1}^1 e^{-2\pi i x\xi}dx=\frac{e^{-2\pi i \xi}-e^{2\pi i \xi}}{{-2\pi i \xi}}=\frac{\sin 2\pi\xi}{\pi\xi}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence &#xA;$$\int_{-\infty}^\infty \left(\frac{\sin 2\pi\xi}{\pi\xi}\right)^2d\xi=\int_{-1}^1dx = 2$$&#xD;&#xA;by a change of variables, $y=2\pi \xi$,  and using symmetry we arrive at&#xA;$$2=\int_{-\infty}^\infty \left(\frac{2\sin y}{y}\right)^2\frac{dy}{2\pi}=\frac{8}{2\pi}\int_{0}^\infty \left(\frac{\sin y}{y}\right)^2 dy$$&#xD;&#xA;or &#xA;$$\int_{0}^\infty \left(\frac{\sin y}{y}\right)^2 dy=\frac{\pi}{2}$$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="1154" LastEditorUserId="1154" LastEditorDisplayName="" LastEditDate="2010-12-07T10:57:26.293" LastActivityDate="2010-12-07T10:57:26.293" />
  <row Id="13357" PostTypeId="1" CreationDate="2010-12-07T09:42:09.163" Score="0" ViewCount="122" Body="&lt;p&gt;What does it mean to say that an algebra is complete with respect to a filtration? Thanks!&lt;/p&gt;&#xA;" OwnerUserId="1944" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-09T03:10:02.727" LastActivityDate="2010-12-09T03:38:19.087" Title="What do we mean by an algebra is complete with respect to a filtration" Tags="&lt;algebra&gt;" AnswerCount="2" />
  <row Id="13358" PostTypeId="1" AcceptedAnswerId="13361" CreationDate="2010-12-07T09:50:08.880" Score="2" ViewCount="134" Body="&lt;p&gt;The question is how to calculate for the general problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The specific problem is this:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Given a door code of $n$ numbers on a 10 digit panel, how many combinations do you have to try?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The catch is that the mechanism that senses whether the correct code has been entered has no concept of the start and end of the code. So the sequence&#xA;123456 is actually testing 3 unique door codes - my particular case uses a 4 digit code.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is, given this modification how many combinations are there given a code length of $n$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In answer to the comment:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yes, I am looking for the shortest sequence of digits that tests all combinations.&lt;/p&gt;&#xA;" OwnerUserId="4499" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-07T16:08:38.100" LastActivityDate="2010-12-07T16:10:20.663" Title="Combinations: security code with no defined start and finish" Tags="&lt;combinatorics&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="13359" PostTypeId="2" ParentId="13357" CreationDate="2010-12-07T09:55:49.700" Score="3" ViewCount="" Body="&lt;p&gt;A filtration induces a topology on the algebra, by providing a basis of open neighbourhoods of 0. The algebra is then complete if the associated topological space is complete. Concretely, if the filtration is called ${\mathfrak{F}}^i$, indexed by natural numbers, where the subalgebras are decreasing as the index increases, say, then you can define a sequence $a_i$ in the algebra to converge if, after possibly replacing it by some translate, it satisfies that for any $i$ there exists an $N\in\mathbb{N}$ such that $a_n\in {\mathfrak{F}}^i$ for all $n\geq N$. You can define Cauchy sequences analogously and the algebra is complete if every Cauchy sequence converges.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Local rings are a prominent example, where the filtration is given by powers of the maximal ideal.&lt;/p&gt;&#xA;" OwnerUserId="3212" LastActivityDate="2010-12-07T09:55:49.700" />
  <row Id="13360" PostTypeId="2" ParentId="13330" CreationDate="2010-12-07T10:02:44.423" Score="1" ViewCount="" Body="&lt;p&gt;It's actually much simpler than that. It's not a maths problem though.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;OK, brace yourselves, I'm gonna tell you about the little bees and the pretty flowers...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More seriously, 1+1=3 is a colloquial expression for saying that when a man and a woman get together, they usually are soon 3 in the family.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;No math there. More like a riddle, a word game.&lt;/p&gt;&#xA;" OwnerUserId="3567" LastActivityDate="2010-12-07T10:02:44.423" CommentCount="1" />
  <row Id="13361" PostTypeId="2" ParentId="13358" CreationDate="2010-12-07T10:09:22.997" Score="5" ViewCount="" Body="&lt;p&gt;Consider a directed graph whose nodes are all possible strings of $n-1$ digits, and with 10 edges labelled 0, ..., 9 going out from each node; the edge $x$ from node $abc\dots de$ leads to $bc\dots dex$, and should be thought of as &quot;testing the code $abc \dots dex$&quot;. The first $n-1$ digits that you enter put you in an initial state, and for each subsequent digit you follow the corresponding edge to another state.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your example, you start in state 123, and move through states 234, 345, 456.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the in- and out-degree of each node is equal (=10), you can find an &lt;a href=&quot;http://en.wikipedia.org/wiki/Euler_path&quot; rel=&quot;nofollow&quot;&gt;Eulerian circuit&lt;/a&gt; that passes every edge exactly once; that is, there is a sequence of length $(n-1)+10^n$ which tests all the $10^n$ codes in the shortest way imaginable (each code is tried exactly once).&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2010-12-07T10:09:22.997" CommentCount="5" />
  <row Id="13362" PostTypeId="2" ParentId="13344" CreationDate="2010-12-07T10:12:57.343" Score="3" ViewCount="" Body="&lt;p&gt;From squaring the identity&#xA;$$\frac{\sin nx}{\sin x}=\frac{e^{inx}-e^{-inx}}{e^{ix}-e^{-ix}}&#xD;&#xA;=\sum_{k=0}^{n-1}e^{(2k-n+1)ix}$$&#xD;&#xA;and integrating we get&#xA;$$n\pi=\int_{-\pi/2}^{\pi/2}\frac{\sin^2 nx}{\sin^2 x}\,dx.$$&#xD;&#xA;Let&#xA;$$I_n=\int_{-\pi/2}^{\pi/2}\frac{\sin^2 nx}{nx^2}\,dx&#xD;&#xA;=\int_{-n\pi/2}^{n\pi/2}\frac{\sin^2y}{y^2}\,dy.$$&#xD;&#xA;Then&#xA;$$\pi-I_n=\frac{1}{n}&#xD;&#xA;\int_{-\pi/2}^{\pi/2}\sin^2nx(\csc^2x-x^{-2})\,dx.$$&#xD;&#xA;and so&#xA;$$|\pi-I_n|\le\frac{1}{n}\int_{-\pi/2}^{\pi/2}|\csc^2x-x^{-2}|\,dx&#xD;&#xA;=O(1/n)$$&#xD;&#xA;as $x\mapsto\csc^2x-x^{-2}$ extends to a continuous&#xA;function on $[-\pi/2,\pi/2]$.&#xA;Hence $I_n\to\pi$ as $n\to\infty$ and&#xA;$$\pi=\int_\infty^\infty\frac{\sin^2y}{y^2}\,dy.$$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="226" LastEditorUserId="226" LastEditorDisplayName="" LastEditDate="2010-12-07T20:21:55.740" LastActivityDate="2010-12-07T20:21:55.740" CommentCount="0" />
  <row Id="13363" PostTypeId="2" ParentId="13344" CreationDate="2010-12-07T10:28:33.243" Score="3" ViewCount="" Body="&lt;p&gt;Integrating by parts:&#xA;$$V_n=\int_{n\pi}^{(n+1)\pi} \frac{\sin^2x}{x^2} \mathrm dx = U_{2n} + U_{2n+1}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where:&#xA;$$U_n=\int_{n\pi}^{(n+1)\pi} \frac{\sin x}{x} \mathrm dx$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus:&#xA;$$\sum U_n = \sum V_n$$&lt;/p&gt;&#xA;" OwnerUserId="2380" LastActivityDate="2010-12-07T10:28:33.243" />
  <row Id="13364" PostTypeId="2" ParentId="13341" CreationDate="2010-12-07T10:34:22.757" Score="7" ViewCount="" Body="&lt;p&gt;Let the helix be given by $(\cos t, \sin t, ht)$ (after scaling). If $P$ is your point $(a,b,c)$, and $Q = (\cos t, \sin t, ht)$ is the nearest point on the helix, then $PQ$ is perpendicular to the tangent at $Q$, which is just $(-\sin t, \cos t, h)$:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$-(\cos t - a)\sin t + (\sin t - b)\cos t + (ht - c)h = 0  $&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This simplifies to $A \sin(t+B) + Ct + D = 0$ for some constants $A,B,C,D,$ as Moron said. But then you have to solve this numerically. There will be more than one solution in general, but (as Jonas Kibelbek pointed out in the comments) you only need to check the solutions with $z$-coordinate in the interval $[c-\pi h, c+\pi h)$.&lt;/p&gt;&#xA;" OwnerUserId="1508" LastEditorUserId="1508" LastEditorDisplayName="" LastEditDate="2010-12-07T13:37:14.433" LastActivityDate="2010-12-07T13:37:14.433" CommentCount="7" />
  <row Id="13365" PostTypeId="2" ParentId="13319" CreationDate="2010-12-07T10:58:05.480" Score="0" ViewCount="" Body="&lt;p&gt;Some examples might be in place:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;INTEGRAL DOMAINS:&#xA;$\mathbb{Z},\mathbb{Z}/(p) \mathbb{Q}, \mathbb{R}, k[x], \mathbb{Z}[x], \mathbb{Z}[x_1,x_2,\ldots]$&#xD;&#xA;(where the latter denotes a polynomial ring with countably many indeterminates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;NON-INTEGRAL DOMAINS: (i.e. every ring in which you can multiply a non-zero number to get zero)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\mathbb{Z}/(6), k[x]/(x^2)$ ($x \times x = x^2=0$ and $x \neq 0$)&lt;/p&gt;&#xA;" OwnerUserId="4284" LastActivityDate="2010-12-07T10:58:05.480" />
  <row Id="13366" PostTypeId="2" ParentId="13150" CreationDate="2010-12-07T11:02:02.000" Score="0" ViewCount="" Body="&lt;p&gt;Is there a way to get the sign of the calculated angle? Right now the calculation gives me an angle of 90° no matter wether I rotate clockwise or counterclockwise.&lt;/p&gt;&#xA;" OwnerUserId="4500" LastActivityDate="2010-12-07T11:02:02.000" />
  <row Id="13367" PostTypeId="1" AcceptedAnswerId="13374" CreationDate="2010-12-07T11:20:27.727" Score="2" ViewCount="105" Body="&lt;p&gt;&lt;strong&gt;Problem:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&lt;strong&gt;In a school competition,the probability of hitting the target bu Dick is $\frac{1}{2}$,by Betty is $\frac{1}{3}$ and by Joe is $\frac{3}{5}$.If all of them fire independently,calculate the probability that the target will be hit.&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This general approach for solving is to &lt;strong&gt;find the complement of the probability that the target would not be hit&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I am not getting is that if the there events are mutually exclusive/independent then why not summing the three probabilities doesn't gives the correct answer?!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please explain.&lt;/p&gt;&#xA;" OwnerUserId="2109" LastActivityDate="2010-12-07T13:17:44.490" Title="Doubts on Mutually exclusive and Independent events" Tags="&lt;probability&gt;&lt;probability-theory&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="13368" PostTypeId="1" AcceptedAnswerId="13369" CreationDate="2010-12-07T11:24:06.050" Score="2" ViewCount="73" Body="&lt;p&gt;Repeating for my exam in commutive algebra.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let G be a topological abelian group, i.e. such that the mappings $+:G\times G \to G$ and $-:G\to G$ are continous. Then we have the following Lemma:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let H be the intersection of all neighborhoods of $0$ in $G$. Then $H$ is a subgroup.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The proof in the books is the following one-liner: &quot;follows from continuity of the group operations&quot;. (this is from &quot;Introduction to Commutative Algebra&quot; by Atiyah-MacDonald)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I must admit that I don't really see how that &quot;follows&quot;. If there is an easy explanation aimed at someone who has not encountered topological groups in any extent, I'd be happy to read it.&lt;/p&gt;&#xA;" OwnerUserId="4284" LastActivityDate="2010-12-07T16:20:21.050" Title="Intersection of neighborhoods of 0. Subgroup?" Tags="&lt;group-theory&gt;&lt;commutative-algebra&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="13369" PostTypeId="2" ParentId="13368" CreationDate="2010-12-07T11:42:51.490" Score="5" ViewCount="" Body="&lt;p&gt;If $U$ is a neighbourhood of $0$ then so is $-U=\{-x:x\in U\}$.&#xA;This shows that if $x\in H$ then $-x\in H$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To show that $H$ is closed under addition, use the fact that&#xA;if $U$ is a neighbourhood of $0$ then there is another&#xA;neighbourhood $V$ of $0$ with $V+V\subseteq U$. The existence&#xA;of $V$ follows from the continuity of addition at $(0,0)$.&lt;/p&gt;&#xA;" OwnerUserId="226" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-07T16:20:21.050" LastActivityDate="2010-12-07T16:20:21.050" CommentCount="2" />
  <row Id="13370" PostTypeId="1" AcceptedAnswerId="13372" CreationDate="2010-12-07T11:54:27.837" Score="1" ViewCount="118" Body="&lt;p&gt;Hello,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've read the formula written on the title on a mathematics' book and it doesn't seem correct for me: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;For the first part of the formula [A ∨ (B∧C)] I have the following possible values:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;A; B and C; A and B and C (the Or is not exclusive)&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;For the second part of the formula [(A ∨ B) ∧ (A ∨ C)] I have the following values:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;A and A (A); A and C; B and A; B and C; ...; A and B and A and C (A and B and C)&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I can have for the second part of the formula &lt;em&gt;A and B; A and C&lt;/em&gt; which I can't obtain with the first part of the formula.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I'm mistaken can somebody please tell me how and give some examples.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;thanks,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bruno&lt;/p&gt;&#xA;" OwnerUserId="4475" LastActivityDate="2010-12-07T12:04:52.897" Title="Is the formula A∨ (B∧C) = (A ∨ B) ∧ (A ∨ C) correct ?" Tags="&lt;logic&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="13371" PostTypeId="2" ParentId="13370" CreationDate="2010-12-07T11:58:41.747" Score="4" ViewCount="" Body="&lt;p&gt;Draw a Venn diagram and think of $A,B,C$ as sets. (consisting of True,False). Then you'll see that the formula is correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or you could draw a truth table.&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Truth_table&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Truth_table&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="4284" LastActivityDate="2010-12-07T11:58:41.747" />
  <row Id="13372" PostTypeId="2" ParentId="13370" CreationDate="2010-12-07T12:04:52.897" Score="3" ViewCount="" Body="&lt;p&gt;Suppose you have $A \vee (B \wedge C)$.  Then you either have $A$ or $B \wedge C$.  If you have $A$, then you have $A \vee B$ and $A \vee C$.  If, on the other hand, you have $B \wedge C$, then you have both $B$ and $C$, in which case you have both $A \vee B$ and $A \vee C$.  So $A \vee (B \wedge C) \Rightarrow (A \vee B) \wedge (A \vee C)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Going in the other direction, suppose you have $(A \vee B) \wedge (A \vee C)$.  So you have both $A \vee B$ and $A \vee C$.  Either you have $A$, or you don't.  If you have $A$, then you have $A \vee (B \wedge C)$.  If, on the other hand, you don't have $A$, then the fact that you &lt;em&gt;do&lt;/em&gt; have both $A \vee B$ and $A \vee C$ implies that you must have both $B$ and $C$.  Therefore you have $B \wedge C$, and so you have $A \vee (B \wedge C)$, i.e. $(A \vee B) \wedge (A \vee C) \Rightarrow A \vee (B \wedge C)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we've established that $A \vee (B \wedge C) \Rightarrow (A \vee B) \wedge (A \vee C)$ and $(A \vee B) \wedge (A \vee C) \Rightarrow A \vee (B \wedge C)$.  Therefore $A \vee (B \wedge C) = (A \vee B) \wedge (A \vee C)$.&lt;/p&gt;&#xA;" OwnerUserId="506" LastActivityDate="2010-12-07T12:04:52.897" />
  <row Id="13373" PostTypeId="1" AcceptedAnswerId="13839" CreationDate="2010-12-07T12:21:44.253" Score="19" ViewCount="347" Body="&lt;p&gt;A friend of mine introduced me to the following question:&#xA;Does there exist a smooth function $f: \mathbb{R} \to \mathbb{R}$ ($f \in C^{\infty}$), such that $f$ maps rationals to rationals and irrationals to irrationals and is nonlinear?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;He has been able to prove that such a polynomial (with degree at least 2) doesn't exist.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem has been asked before at least at &lt;a href=&quot;http://www.artofproblemsolving.com&quot; rel=&quot;nofollow&quot;&gt;http://www.artofproblemsolving.com&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="3776" LastEditorUserId="3776" LastEditorDisplayName="" LastEditDate="2010-12-07T13:52:47.810" LastActivityDate="2010-12-10T19:26:51.563" Title="Smooth functions for which $f(x)$ is rational if and only if $x$ is rational" Tags="&lt;number-theory&gt;&lt;real-analysis&gt;" AnswerCount="1" CommentCount="10" FavoriteCount="6" />
  <row Id="13374" PostTypeId="2" ParentId="13367" CreationDate="2010-12-07T13:01:43.207" Score="6" ViewCount="" Body="&lt;p&gt;&quot;Mutually exclusive&quot; and &quot;independent&quot; mean different things.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Two events are &lt;strong&gt;mutually exclusive&lt;/strong&gt; if they can't both happen. For example, &quot;my first name is Steve&quot; and &quot;my first name is Fred&quot; are mutually exclusive. When events are mutually exclusive, you are allowed to add their probabilities to get the probability that one of them occurs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Independent&lt;/strong&gt; events are events where finding out about one doesn't change the probability of the other. Finding out that &quot;it is raining&quot; doesn't tell you anything about &quot;my car is red&quot; so those events are independent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the question the events &quot;Dick hits the target&quot;, &quot;Betty hits the target&quot; and &quot;Joe hits the target&quot; are independent, but are not mutually exclusive. For instance, Betty and Dick could both hit the target. Since they are not mutually exclusive adding the probabilities will not give the correct answer.&lt;/p&gt;&#xA;" OwnerUserId="1149" LastActivityDate="2010-12-07T13:01:43.207" CommentCount="1" />
  <row Id="13375" PostTypeId="1" AcceptedAnswerId="13383" CreationDate="2010-12-07T13:57:23.453" Score="0" ViewCount="305" Body="&lt;p&gt;I had many interaction with young girls when I was a private math teacher. Most of them have an assumption that learning mathematics is not necessary for them, because they just want to be a house wife in the future.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have some suggestions but I will not say here now (only to avoid interfering your idea).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have the same experience, could you tell me what is your suggestion to them? &lt;/p&gt;&#xA;" OwnerUserId="3901" LastEditorUserId="3901" LastEditorDisplayName="" LastEditDate="2010-12-07T14:07:04.553" LastActivityDate="2010-12-07T15:26:43.423" ClosedDate="2010-12-07T16:56:50.923" Title="How to suggest young girls who think of mathematics as an unnecessary thing if they will be a house wife in the future?" Tags="&lt;soft-question&gt;&lt;teaching&gt;" AnswerCount="3" CommentCount="11" />
  <row Id="13376" PostTypeId="1" AcceptedAnswerId="13390" CreationDate="2010-12-07T14:18:02.027" Score="12" ViewCount="457" Body="&lt;p&gt;(&lt;strong&gt;Disclaimer:&lt;/strong&gt; I'm an engineer)&lt;br&gt;&#xA;Hi everybody, I found this “riddle” posted on the internet:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/kLBmn.jpg&quot; alt=&quot;Portal Riddle&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's meant as a joke, but I do think it deserves an answer :)&lt;br&gt;&#xA;A bit of background: the orange and blue ellipses are a citation from the videogame Portal. They are &quot;portals&quot; connected so that everything that goes into one of them comes out of the other, mantaining his momentum (this last part is the foundation for the game). For further info, you can also &lt;a href=&quot;http://www.youtube.com/watch?v=TluRVBhmf8w&quot; rel=&quot;nofollow&quot;&gt;watch the trailer&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So.. is it a circle, a straight line or what else?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(also, feel free to retag)&lt;/p&gt;&#xA;" OwnerUserId="894" LastActivityDate="2011-05-13T06:29:39.820" Title="Have I made a straight line, or a circle?" Tags="&lt;geometry&gt;&lt;circle&gt;" AnswerCount="4" CommentCount="8" />
  <row Id="13377" PostTypeId="2" ParentId="13375" CreationDate="2010-12-07T14:21:09.420" Score="2" ViewCount="" Body="&lt;p&gt;Arguably the most important role of a housewife is the raising of children. This is a difficult, often thankless job but no job is more important in the entire world. The more education a mother has, the better prepared she is to help her children experience the world and understand how it works. Mathematics in particular is so very important to quantifying and comparing the relationships that we see all around us, in nature, in populations, and even in the kitchen!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Note&lt;/strong&gt;: I am not saying that girls should feel obligated to be housewives. My response has the assumption, as stated in the questions, that the girls plan on being housewives.&lt;/p&gt;&#xA;" OwnerUserId="4503" LastEditorUserId="4503" LastEditorDisplayName="" LastEditDate="2010-12-07T14:28:54.653" LastActivityDate="2010-12-07T14:28:54.653" CommentCount="2" />
  <row Id="13378" PostTypeId="2" ParentId="13376" CreationDate="2010-12-07T14:25:03.627" Score="6" ViewCount="" Body="&lt;p&gt;One way to capture the roundness of a circle is to embed it in a plane and measure distance between two points using the ambient plane. A different way to measure distance in a circle is to pretend you are an ant restricted to move within the circle. Then you measure the distance based on how far you walk in the circle. This is a &quot;flat&quot; circle. It is isometric to the circle depicted by the rope in your picture. (An ant walking on that rope couldn't tell any intrinsic difference between the rope and the circle in the plane.)&lt;/p&gt;&#xA;" OwnerUserId="2941" LastActivityDate="2010-12-07T14:25:03.627" />
  <row Id="13379" PostTypeId="2" ParentId="13375" CreationDate="2010-12-07T14:27:18.540" Score="3" ViewCount="" Body="&lt;p&gt;I don't believe k-12 mathematics curriculum is optimized from the point of view of helping students understand the nature and usefulness of mathematics when the students are not likely to go on in math and science. I think girls are especially disadvantaged by current curriculum because there are societal factors that already seem to discourage girls from pursing mathematics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A very short look at my views here can be found in this paper about public perceptions of mathematics:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.education.umd.edu/MathEd/conference/vbook/public-perceptions_Malkevitch.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.education.umd.edu/MathEd/conference/vbook/public-perceptions_Malkevitch.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1369" LastActivityDate="2010-12-07T14:27:18.540" />
  <row Id="13381" PostTypeId="2" ParentId="13013" CreationDate="2010-12-07T14:57:15.787" Score="1" ViewCount="" Body="&lt;p&gt;I don't understand why this question is asked in the language of generating functions.  The easy way to solve the problem is to take any subset $\{a,b,c,d,e\}$ of $\{1,\dots,n-4\}$, and choose the subset $\{a,b+1,c+2,d+3,e+4\}$.  In particular, there are $\binom{n-4}{5}$  ways to do this.  We can recover the generating function using basic generating function knowledge:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{n\geq 0} \binom{n-4}{5}x^n =\sum_{n\geq 0} \binom{n-9+5}{5}x^n=x^9\sum_{n\geq 0} \binom{n-9+5}{5}x^{n-9}$$&#xD;&#xA;Changing indices and recognizing the form of the generating function, we have &#xA;$$=x^9\sum_{N\geq -9} \binom{N+5}{5}x^N=\frac{x^9}{(1-x)^6}.$$&lt;/p&gt;&#xA;" OwnerUserId="4504" LastActivityDate="2010-12-07T14:57:15.787" />
  <row Id="13382" PostTypeId="2" ParentId="13345" CreationDate="2010-12-07T15:24:16.563" Score="0" ViewCount="" Body="&lt;p&gt;Yeah i worked it out on the train. It's not the case. Consider $X=[0,\infty)$ and $d:(x,y)\mapsto|y-x|$, $d&amp;#39;:(x,y)\mapsto|\sqrt{y}-\sqrt{x}|$ and consider the map $f:X\to[0,1]$ a periodic continuous function like $sin$. Then $f:(X,d)\to[0,1]$ is uniformly cts, but $f:(X,d&amp;#39;)\to[0,1]$ is not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ultimately I'm looking for this: let $X$ be a polish space, then does it embed continuously to a unique &lt;em&gt;polish&lt;/em&gt; compactification, universal in the sense that every continuous map from $X$ to another compact polish space lifts uniquely to a map from the compactification. Doesn't seem likely. I can't see that the Stone-Cech compactification is polish (it's at least separable compact hausdorff, but that actually does not imply second countable).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I can do, is take $(X,d)$ where $d$ is a compatible complete metric, and embed this via a uniformly continuous map to a compact polish space, universal in the sense that every uniformly cts map from $(X,d)$ to another compact polish space lifts uniquely to a map from the compactification.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What that compactification is might depend on $d$. Though it would be nice if it didn't, so that we could talk about „the“ compactification of a polish space.&lt;/p&gt;&#xA;" OwnerUserId="4508" LastActivityDate="2010-12-07T15:24:16.563" />
  <row Id="13383" PostTypeId="2" ParentId="13375" CreationDate="2010-12-07T15:26:43.423" Score="11" ViewCount="" Body="&lt;p&gt;The most important thing you need to do for these girls, more important than teaching them any particular mathematical topic, is to inspire them to have a broader vision of what they might attain in their life. Sure, being a housewife is possible, but also they might aim to be a scientist, doctor, philosopher or engineer; they could be an astronaut or a musical performer or a computer programmer. Why should a young girl decide at a young age on such a narrow and impoverishing plan for their life?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I find it shocking that any young girl would assert such a plan for their life (to say nothing of &quot;most of them&quot; as you mentioned). These girls need urgent direction.&lt;/p&gt;&#xA;" OwnerUserId="413" LastActivityDate="2010-12-07T15:26:43.423" CommunityOwnedDate="2010-12-07T15:26:43.423" CommentCount="5" />
  <row Id="13384" PostTypeId="1" CreationDate="2010-12-07T15:35:59.840" Score="2" ViewCount="58" Body="&lt;p&gt;Let $(X,d),(Y,d)$ be polish spaces and $Y$ compact. Consider the spaces $C_{u}(X,Y)$ of uniformly continuous functions, and $C(X,Y)$ of continuous functions. We can endow these with the uniform-metric.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can $C(X,Y)$ be generated from $C_{u}(X,Y)$ using only point-wise limits of $\omega$-sequences? &lt;em&gt;I.e.&lt;/em&gt; Define $\mathcal{B}_{0}:=C_{u}(X,Y)$, and $\mathcal{B}_{\alpha}:=$ the set of point-wise limits of a sequence $(f_{n})_{n\in\omega}$ of functions $f_{n}\in\bigcup_{\xi&amp;lt;\alpha}\mathcal{B}_{\xi}$. Then this question asks whether $C(X,Y)\subseteq\mathcal{B}_{\omega_{1}}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course if it's true, then we get that all borel functions from $X$ to $Y$ can be generated from the uniformly continuous functions using only point-wise limits.&lt;/p&gt;&#xA;" OwnerUserId="4508" LastActivityDate="2010-12-07T15:35:59.840" Title="generating continuous functions from uniformly continuous functions" Tags="&lt;real-analysis&gt;&lt;general-topology&gt;" />
  <row Id="13385" PostTypeId="2" ParentId="13245" CreationDate="2010-12-07T15:38:00.553" Score="1" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;: I think my previous answer can be adapted to the case we are interested in.&#xA;So let $E_n=\{ f \geq n\}$.&#xA;Let $m$ be a positive integer.&#xA;If $s = \sup_{F \subset [-m,m] \setminus E_n} \lambda(F)$ (the sets $F$ are taken measurable here), then there is a measurable set $F \subset [-m,m] \setminus E_n$ such that $\lambda(F)=s$ (take a sequence and a union).&#xA;We are going to prove that for any real M, for almost all $x \in F$, $f(x)&amp;lt;-M$, that is, there is a measurable $A(M) \subset F$ of zero measure such that $f(x)&amp;lt;-M$ for all $x \in F \setminus A(M)$.&#xA;For this, it is enough to find, for every positive integer $k$, $A_k(M) \subset F$ such that $f(x)&amp;lt;-M$ for all $x \in F \setminus A_k(M)$ and $\lambda(A_k(M)) \leq 1/k$.&#xA;Since $f$ is not continous and $\mathbb{Q}$-linear, there exists $a_k \in \mathbb{R}$ such that $|a_k| \lt 1/(2k)$ and $f(a_k)&amp;lt;-n-M$.&#xA;Then $a_k+F \subset [-m-1/(2k), m + 1/(2k)] \setminus E_n$, so $\lambda((a_k+F) \cap [-m,m] \setminus F) = 0$ and consequently, $\lambda(F \setminus (a_k+F)) \leq 1/k$, so we can set $A_k(M) = F \setminus (a_k+F)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now since $f(x)&amp;lt;-M$ for all $x \in F \setminus A(M)$, $F = \cup_{M \in \mathbb{N}} A(M)$ has measure $0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have shown that $E_n$ is $\lambda$-thick in $[-m,m]$ for all $m$, so $E_n$ is $\lambda$-thick in $\mathbb{R}$, and this is stronger than $E_n$ being $\lambda$-thick in $[0,1]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Old answer&lt;/strong&gt;:&#xA;This is not exactly what you want, but here is a similar proof of your lemma.&#xA;Define $E_n = \{ |f| \geq n \}$. Then $E_n$ is $\lambda$-thick (in $\mathbb{R}$).&#xA;We need to show that if $F$ is measurable and does not intersect $E_n$, then $\lambda(F)=0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume otherwise, and let $m$ be big enough so that $F \cap [-m,m]$ has positive measure.&#xA;Since $f$ is not continuous and $\mathbb{Q}$-linear, $f_{|[0,1]}$ is not bounded, so there is a sequence $(a_k)_{k \geq 0}$ of elements of $[0,1]$ such that $|f(a_{k+1})| &amp;gt; |f(a_k)|+2n$.&#xA;Then $\cup_{k \geq 0} \left( a_k + (F \cap [-m,m]) \right)$ is included in $[-m,m+1]$ and its measure is equal to $\sum_{k \geq 0} \lambda (F \cap [-m,m]) = + \infty$, this is because the $a_k + F$ are disjoint (look at the images by $f$), so we get a contradiction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So $E_n$ is $\lambda$-thick, and $E_n \cap [0,1]$ is $\lambda$-thick in $[0,1]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe there is a way to deduce your statement from this one.&lt;/p&gt;&#xA;" OwnerUserId="2660" LastEditorUserId="2660" LastEditorDisplayName="" LastEditDate="2010-12-10T15:17:26.020" LastActivityDate="2010-12-10T15:17:26.020" CommentCount="1" />
  <row Id="13386" PostTypeId="1" AcceptedAnswerId="13391" CreationDate="2010-12-07T15:48:55.947" Score="9" ViewCount="312" Body="&lt;p&gt;I read that any continuous function can be represented as a sum of convex and concave function, meaning for all $f(x)$, $f(x) = g(x) + h(x)$ where $g$ is convex and $h$ is concave.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There could be infinitely many decompositions of that sort.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyone knows where I can see a proof or this, or knows of a proof of this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="7076" LastEditorUserId="3431" LastEditorDisplayName="" LastEditDate="2010-12-07T16:38:22.260" LastActivityDate="2010-12-07T17:32:27.657" Title="can any continuous function be represented as a sum of convex and concave function?" Tags="&lt;functions&gt;&lt;convex-analysis&gt;" AnswerCount="6" CommentCount="3" FavoriteCount="1" />
  <row Id="13387" PostTypeId="1" AcceptedAnswerId="13444" CreationDate="2010-12-07T16:01:31.837" Score="3" ViewCount="241" Body="&lt;p&gt;What is the best program to factor large arbitrary-form integers on a single computer, or on a few disjointed computers? &quot;Best&quot; is obviously subjective, but what do you recommend?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm working on a project to factor general-form numbers that I know are composite, with a scale of 100 - 1000 digits. I have a few computers I can use to process in parallel, but they're nowhere close to a cluster - and they certainly don't have the horsepower / memory needed for something like GNFS.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've looked around a little, but what I've found can be classified as:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Prove primality of general-form numbers &lt;em&gt;(mine = known composites)&lt;/em&gt;&lt;/li&gt;&#xA;&lt;li&gt;Factor special-form numbers &lt;em&gt;(mine = general-form numbers)&lt;/em&gt;&lt;/li&gt;&#xA;&lt;li&gt;Factor general-form numbers on a huge cluster &lt;em&gt;(no cluster)&lt;/em&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I'm missing the last one, &quot;Factor general-form numbers on one or two sneakernet systems&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Final caveats: I'd prefer something free, but I'd be willing to spring for Maple / Mathematica / whatnot if it's my best option. Also I'd prefer something that's already built (binary package), but I can compile from source if that's a better option.&lt;/p&gt;&#xA;" OwnerUserId="4507" LastActivityDate="2010-12-10T18:36:05.197" Title="Factoring large integers without a cluster" Tags="&lt;soft-question&gt;&lt;prime-numbers&gt;&lt;math-software&gt;" AnswerCount="1" CommentCount="7" />
  <row Id="13388" PostTypeId="2" ParentId="13386" CreationDate="2010-12-07T16:07:23.797" Score="-2" ViewCount="" Body="&lt;p&gt;what would this look like for sin(x)&lt;/p&gt;&#xA;" OwnerUserId="4370" LastActivityDate="2010-12-07T16:07:23.797" CommentCount="2" />
  <row Id="13390" PostTypeId="2" ParentId="13376" CreationDate="2010-12-07T16:12:27.923" Score="23" ViewCount="" Body="&lt;p&gt;Though I (slightly) disagree with Nate's characterization of geometers in his comments, I think there is actually some deep notions worth discussing contained in it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The heart of the matter is, &lt;strong&gt;how do you define a circle&lt;/strong&gt;, and &lt;strong&gt;how do you define a line&lt;/strong&gt;? And in the apparent paradox of the original question, the resolution is that there are (at least) two different ways of distinguishing circles versus lines, which, when taken together, allows for an object that is simultaneously a line in one definition but a circle in the other. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the &lt;strong&gt;topological&lt;/strong&gt; category, the circle and the line are the two examples of connected smooth one-dimensional manifolds. Since we are dealing with topology, everything is allowed to be stretchy/flabby, so the only difference we really care about between the circle and the line is that the latter is simply connected, while the former is not. More precisely, you can (trivially) find a closed loop in the circle which can not be continuously deformed to a point, while any closed loop on the line must &quot;backtrack&quot; sufficiently that it can be continuously shrunk down to a point. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A second description of the circle and the line comes from the &lt;strong&gt;geometrical&lt;/strong&gt; category. (Here I'll just discuss the distinction between lines and non-lines.) A line in geometry is, intuitively, the straightest possible curve, which we take to mean &quot;a curve that locally minimizes the distance between two points&quot;. It is an interesting special case of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Cartan-Hadamard_theorem&quot; rel=&quot;nofollow&quot;&gt;Cartan-Hadamard theorem&lt;/a&gt; that, in a simply connected manifold of non-positve curvature (in particular the usual flat Euclidean space), rays (geodesics) emanating from the same point in two different directions will diverge forever and never intersect. (In fact, the divergence of geodesic rays is a characterization of non-positive curvature; also compare this to the case of positive curvature on a sphere, where any two great circles intersect at exactly two points.) So since the world in which we live our usual, everyday lives is more-or-less flat, our common intuition is that a line must be non-self-intersecting. Of course, if you consider traveling on the oblate spheroid that is the earth, you may come to a somewhat different conclusion. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The situation of &lt;em&gt;Portal&lt;/em&gt; is precisely at the level of breaking the hypothesis of Cartan-Hadamard theorem in topology. As stated above, the Cartan-Hadamard theorem requires the space to be simply connected. By allowing the Portal, which is an identification two distinct subsets of usual Euclidean space, you pick up a non-trivial topology. And therefore Cartan-Hadamard theorem can fail. Hence your rope is allowed to be simultaneously a &lt;strong&gt;line&lt;/strong&gt; in the sense that it is a straightest possible curve and a &lt;strong&gt;circle&lt;/strong&gt; in the sense that it is a curve that returns to its starting point, and intrinsically has non-trivial topology. &lt;/p&gt;&#xA;" OwnerUserId="1543" LastActivityDate="2010-12-07T16:12:27.923" CommentCount="2" />
  <row Id="13391" PostTypeId="2" ParentId="13386" CreationDate="2010-12-07T16:22:50.100" Score="9" ViewCount="" Body="&lt;p&gt;If $f$ is $C^2$ (meaning the derivatives $f&amp;#39;$ and $f&amp;#39;&amp;#39;$ are defined and continuous), and if $f$ is defined on a closed interval like $[0,1]$, then there is an easy solution. Since $f&amp;#39;&amp;#39;$ is continuous, it has a minimal value $m$. Choose $c$ positive and greater than $m$. Then $f(x)+c x^2/2$ is convex, by the double derivative test, and $-c x^2$ is concave. Of course, $[f(x)+c x^2/2] + [-cx^2/2] = f(x)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It shouldn't be too much harder to do a general continuous function, but I don't see it right now.&lt;/p&gt;&#xA;" OwnerUserId="448" LastActivityDate="2010-12-07T16:22:50.100" CommentCount="2" />
  <row Id="13392" PostTypeId="1" AcceptedAnswerId="13400" CreationDate="2010-12-07T16:48:26.870" Score="2" ViewCount="119" Body="&lt;p&gt;Lets say I take a computation that involves only addition and multiplication:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;(a+b)*(c+d)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;which can be done in many other ways, eg.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;a*(c+d) + b*(c+d)&#xA;a*c + a*d + b*c + b*d&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;In terms of additions and multiplications the number of operations required for each of the three examples shown are (2,1) (3,2) (3,4) respectively. Clearly, if the goal is to reduce the total number of operations the first is superior. Is there a way, given an arbitrary expression to find the computation order that requires the least number of operations? &lt;/p&gt;&#xA;" OwnerUserId="196" LastActivityDate="2010-12-07T18:10:40.200" Title="Reduce the number of operations on a simple expression" Tags="&lt;computer-science&gt;&lt;symbolic-computation&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="13393" PostTypeId="2" ParentId="13386" CreationDate="2010-12-07T16:50:09.327" Score="2" ViewCount="" Body="&lt;p&gt;Counterexample:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f(x)=x^2sin(\frac{1}{x^2})$      (with f(0)=0)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;this is a continuous function, whose gradient is unbounded as x goes to 0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;g'(x) and h'(x) are monotonic, and hence bounded on [-1,1], so the gradient of g(x)+h(x) is bounded as x goes to 0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: Looking at it again, this doesn't look fantastically rigorous, but I think it holds up.&lt;/p&gt;&#xA;" OwnerUserId="1149" LastActivityDate="2010-12-07T16:50:09.327" />
  <row Id="13394" PostTypeId="2" ParentId="5226" CreationDate="2010-12-07T16:53:38.637" Score="1" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://mathoverflow.net/questions/48411/are-there-any-abelian-2-groups-with-complete-holomorphs-other-than-c2-2-and-c&quot; rel=&quot;nofollow&quot;&gt;Derek Holt helped me&lt;/a&gt; figure this out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose V is not just normal, but characteristic in A.  Then Out(A) has a normal series with factors N, S, H with:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;N ≤ Out(G) consisting of those automorphisms that take V to an isomorphic G-module&lt;/li&gt;&#xA;&lt;li&gt;S a quotient of the group Aut&lt;sub&gt;G&lt;/sub&gt;(V) of G-module automorphisms of V by the normal subgroup of automorphisms induced by the center of G&lt;/li&gt;&#xA;&lt;li&gt;H = H&lt;sup&gt;1&lt;/sup&gt;(G,V), the first cohomology group&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;For A to be complete, Out(A) = 1, so N=S=H=1.  In particular,&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;No (non-identity) outer automorphism can take V to an isomorphic G-module&lt;/li&gt;&#xA;&lt;li&gt;V has to be multiplicity-free and split, and the center of G needs to include all the (block) scalar matrices&lt;/li&gt;&#xA;&lt;li&gt;The first cohomology group has to vanish&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Assuming then that V is irreducible and characteristic, then the second condition is just that V is absolutely irreducible and G contains the scalar matrices Z(GL(V)).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If V is not characteristic (say if G is very small and unipotent) then this analysis fails, but I think if G starts out close to being complete, V is likely to be characteristic.  Certainly if V is irreducible.&lt;/p&gt;&#xA;" OwnerUserId="583" LastActivityDate="2010-12-07T16:53:38.637" CommentCount="1" />
  <row Id="13395" PostTypeId="2" ParentId="13386" CreationDate="2010-12-07T16:55:38.293" Score="18" ViewCount="" Body="&lt;p&gt;You need more conditions than even absolute continuity. A characterisation of continuous, convex functions defined on open intervals is that they must be the indefinite integral of a monotonically non-decreasing function. (See &lt;a href=&quot;http://eom.springer.de/c/c026240.htm&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; for example.) In particular, this means that continuous, convex functions are absolutely continuous. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if you take any function that is continuous and not absolutely continuous (Cantor's stair case comes to mind), it cannot be decomposed as a sum of a convex and a concave function. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What's more important is that a convex function, by &lt;a href=&quot;http://www.math.sc.edu/~howard/Notes/alex.pdf&quot; rel=&quot;nofollow&quot;&gt;Alexandrov's theorem&lt;/a&gt;, must have a second derivative almost everywhere. Therefore your initial function must be even better than just absolutely continuous, it needs to also admit almost everywhere second derivatives. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And even assuming almost everywhere second derivatives is not enough, if you take the function $x \sin(1/x)$  which is analytic away from the origin, you cannot represent it as the difference of two convex functions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, there is still a gap between twice almost everywhere differentiable and David Speyer's everywhere twice-continuously differentiable condition. I am not entirely sure where the correct boundary lies, or if there is a correct boundary using just classical differentiability notions. &lt;/p&gt;&#xA;" OwnerUserId="1543" LastActivityDate="2010-12-07T16:55:38.293" CommentCount="3" />
  <row Id="13396" PostTypeId="1" AcceptedAnswerId="13465" CreationDate="2010-12-07T17:08:09.030" Score="2" ViewCount="462" Body="&lt;p&gt;So I have this problem to solve...&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Let X denote the number of paint defects found in a square yard section of a car body painted by a robot.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These data are obtained:&#xA;8, 5, 0, 10, 0, 3, 1, 12, 2, 7, 9, 6&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Assume that X has a Poisson distribution with parameter λs.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a) Find an unbiased estimate for λs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;b) Find an unbiased estimate for the average number of flaws per square yard.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;c) Find an unbiased estimate for the average number of flaws per square foot.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I have no idea where to begin. I mean, how do I even find ANY unbiased estimate? The textbook is worthless imo and I can't find any good readings on the web either... please help.&lt;/p&gt;&#xA;" OwnerUserId="713" LastActivityDate="2010-12-08T01:14:45.323" Title="Find an unbiased estimate for λs (Poisson distribution)" Tags="&lt;probability&gt;&lt;statistics&gt;&lt;discrete-mathematics&gt;" AnswerCount="1" CommentCount="8" FavoriteCount="1" />
  <row Id="13397" PostTypeId="2" ParentId="13386" CreationDate="2010-12-07T17:17:53.003" Score="2" ViewCount="" Body="&lt;p&gt;A $C^1$ function $f$ on a compact interval is convex (or is it concave?)&#xA;if and only if $f&amp;#39;$ is increasing and is concave (or is it convex?)&#xA;if and only if $f&amp;#39;$ is decreasing. If a function on a compact interval&#xA;is a sum of a convex and a concave function, each of which is $C^1$, then&#xA;$f&amp;#39;$ is the sum of an increasing and a decreasing function and so&#xA;has bounded variation. There are continuous functions which are not&#xA;of bounded variation on a compact interval. Integrating one gives a&#xA;$C^1$ function $f$ which is not the sum of a $C^1$ convex and&#xA;a $C^1$ concave function. Perhaps it could be the sum of some&#xA;more general convex and concave functions, but I wouldn't bet on it :-)&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-12-07T17:17:53.003" CommentCount="3" />
  <row Id="13398" PostTypeId="2" ParentId="13386" CreationDate="2010-12-07T17:32:27.657" Score="0" ViewCount="" Body="&lt;p&gt;Assume $f$ to be differentiable. We need to write $f&amp;#39;$ as the sum of nondecreasing and a nonincreasing functions $g$ and $h$. This is possible if and only if $f&amp;#39;$ has bounded variation: for every $a&amp;lt;b$, there exists $M$ such that for any integer $n$ and any $a=x_1 \lt \ldots \lt x_{n+1} = b$, $\sum_{i=1}^n |f(x_{i+1})-f(x_i)| \leq M$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case we define, for $x \geq 0$, $g(x)=\mathrm{sup}_{0=x_1 \lt \ldots \lt x_{n+1}=x} \sum_{i=1}^n \left(f(x_{i+1})-f(x_i) \right)^+$ where $y^+ = y$ if $y \geq 0$ and $0$ otherwise. Define $g$ similarly on $]-\infty,0]$, and $h=f-g$. It is easy to check that $g$ is nondecreasing, $h$ is nonincreasing, and their primitives give the result.&lt;/p&gt;&#xA;" OwnerUserId="2660" LastActivityDate="2010-12-07T17:32:27.657" />
  <row Id="13399" PostTypeId="2" ParentId="13268" CreationDate="2010-12-07T18:03:42.073" Score="3" ViewCount="" Body="&lt;p&gt;I am slightly unsure about the following approach, so you should double check if that is correct. I'll use $\|\cdot\|_2$ to denote the spectral norm as above, and $\|\cdot \|_F$ for the Frobenius norm. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is well known that for $n\times n$ matrices,  $\|\cdot \|_F \leq \sqrt{n} \|\cdot\|_2$ is sharp from just the definition of the norms. Now, I claim that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ |\beta_k - \alpha_k| \leq \|B - A\|_F $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is also sharp. Fix the eigenvalues $\beta_k$ and $\alpha_k$. Consider orthogonal transformations of the symmetric matrix $B$ by $Q$. It suffices to compute the minimum&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \inf_{Q\in O(n)} \| QBQ^T - A\|_F $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But using the definition of the Frobenius norm as an inner product, you see that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \| QBQ^T - A\|_F^2 = \mathop{Tr}\left[ (QBQ^T - A)(QBQ^T - A) \right] = \|B\|_F^2 + \|A\|_F^2 - 2 A\cdot_F (QBQ^T)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's an exercise to see that $A\cdot_F (QBQ^T)$ is maximized when all the $k$th eigenspace of $A$ lines up with that of $QBQ^T$ (some sort of Cauchy-Schwarz plus re-arrangement inequality). When the eigen-spaces line up, you have that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \| QBQ^T - A\|_F^2 = \sum_{k = 1}^n |\beta_k - \alpha_k|^2 $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and so the desired inequality holds, and is attained with the eigenvalues only differ in one position. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, the composition of these two inequalities is not automatically sharp, since the bound of $|\beta_k-\alpha_k|$ by $\|B-A\|_F$ is only sharp with $B-A$ has rank 1, while the first bound can be sharpened $\|C\|_F \leq \sqrt{\mathop{rank}(C)} \|C\|_2$. But it is clear that the most general sharp bound should be&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ |\beta_k-\alpha_k| \leq s\|B-A\|_2$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with $1 \leq s \leq \sqrt{n}$. &lt;/p&gt;&#xA;" OwnerUserId="1543" LastEditorUserId="1543" LastEditorDisplayName="" LastEditDate="2010-12-07T18:32:12.300" LastActivityDate="2010-12-07T18:32:12.300" CommentCount="4" />
  <row Id="13400" PostTypeId="2" ParentId="13392" CreationDate="2010-12-07T18:10:40.200" Score="4" ViewCount="" Body="&lt;p&gt;Here's a closely related question which is easier to answer.  Suppose you only care about the number of multiplications.  (This is reasonable in many situations where addition is cheap but multiplication is expensive, so you want to minimize the number of multiplications you're going to do.)  So suppose someone gives you an expression like a*c + a*d + b*c + b*d and you want to rearrange using as few multiplications as possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case what you're doing is computing the &quot;rank&quot; of a tensor.  (Warning: &quot;rank&quot; of a tensor can mean two totally different things, this is the rank that's analogous to rank of a matrix, not the rank that tells you which kind of tensor you're looking at.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, suppose your expression only ever involves products of two things, then this becomes computing the rank of a matrix.  In your example, a*c + a*d + b*c + b*d, you're looking at a 4x4 matrix (since there are 4 variables: a,b,c,d) which is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{pmatrix}0&amp;amp;0&amp;amp;1&amp;amp;1\\ 0&amp;amp;0&amp;amp;1&amp;amp;1 \\ 0&amp;amp;0&amp;amp;0&amp;amp;0 \\ 0&amp;amp;0&amp;amp;0&amp;amp;0 \end{pmatrix}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This matrix has rank 1 as is easily seen using row-reduction, hence it can be written using only one multiplication.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know whether there are good algorithms for computing ranks of higher tensors the way there are for matrices.&lt;/p&gt;&#xA;" OwnerUserId="8" LastActivityDate="2010-12-07T18:10:40.200" CommentCount="3" />
  <row Id="13401" PostTypeId="2" ParentId="13345" CreationDate="2010-12-07T18:19:14.177" Score="2" ViewCount="" Body="&lt;p&gt;Dear Raj,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is an answer to the question in one of your answers: is the Stone-Cech Compactification of a Polish space necessarily Polish?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer is a resounding no: not even close, in some sense.  Let $X$ be a countably &#xA;infinite set equipped with the discrete metric (and thus the discrete topology).  Its Stone-Cech compactification is the space of ultrafilters on $X$, which has cardinality &#xA;$2^{2^{|X|}}$ (&lt;b&gt;added&lt;/b&gt;: e.g., &lt;a href=&quot;http://planetmath.org/encyclopedia/NumberOfUltrafilters.html&quot; rel=&quot;nofollow&quot;&gt;see here&lt;/a&gt;). Evidently this space is separable: $X$ itself is a countable dense subspace.  Now, if you consult &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Separable_space#Cardinality&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Separable_space#Cardinality&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;you will see that a first countable separable space can have at most continuum cardinality.  (Disclosure: I wrote this part of the article.)  So the Stone-Cech &#xA;compactification is not first countable, so is not even close to being metrizable.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not an expert on Polish spaces, but I think you just do not have the right structure here to define a reasonable notion of uniform continuity.&lt;/p&gt;&#xA;" OwnerUserId="299" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2010-12-08T15:18:18.507" LastActivityDate="2010-12-08T15:18:18.507" />
  <row Id="13402" PostTypeId="1" CreationDate="2010-12-07T18:23:32.777" Score="1" ViewCount="130" Body="&lt;p&gt;Alright, so I got two points in 3d space, so they have a x,y, and z.&#xA;Now if the line's y - which I get like so:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Vector3 v = new Vector3();&#xA;v = a.subtract(b, v);&#xA;v.normalizeLocal();&#xA;&#xA;float cosine = (float) v.dot(v);&#xA;float angle = (float) Math.toDegrees(Math.acos( cosine ));&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Now I have a third point (the c point), which is inside the line of a and b.&#xA;now I need to know how to get point a and b to 0 and 1, than I need to find out where c is on the line. my goal is to have something like c.y = b.y - (a.y * c.x and c.z's point on the line (which will be between 0 and 1))&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So how do I do this?&lt;/p&gt;&#xA;" OwnerUserId="4464" LastActivityDate="2011-01-07T02:44:39.043" Title="Getting the % that a point is on a line" Tags="&lt;geometry&gt;&lt;trigonometry&gt;&lt;vector-space&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="13403" PostTypeId="1" CreationDate="2010-12-07T18:39:28.703" Score="2" ViewCount="210" Body="&lt;p&gt;Please help me, I can't find the theorem anywhere that states something like:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;For a bounded set $U\subset\mathbb R$ there exists a non-descreasing sequence $(a_n)_{n\in \mathbb N}$ with $a_n \in U$ with $\lim_{n\to\infty}a_n=\sup U$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Thank you!&lt;/p&gt;&#xA;" OwnerUserId="4514" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-08T04:46:03.943" LastActivityDate="2010-12-08T04:46:03.943" Title="Sequence converging to the supremum" Tags="&lt;sequences-and-series&gt;" AnswerCount="2" CommentCount="13" />
  <row Id="13404" PostTypeId="1" CreationDate="2010-12-07T18:41:28.003" Score="1" ViewCount="399" Body="&lt;p&gt;I have a camera looking at a computer monitor from varying angles. Since the camera is a grid of pixels, I can define the bounds of the monitor in the camera image as:&#xA;&lt;img src=&quot;http://i.imgur.com/MGwUh.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope that makes sense. What I want to do is come up with an algorithm to translate points within this shape to this:&#xA;&lt;img src=&quot;http://i.imgur.com/niJYY.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have points within the same domain as ABCD, as determined from the camera, but I need to draw these points in the domain of the monitor's resolution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does that makes sense? Any ideas?&lt;/p&gt;&#xA;" OwnerUserId="4515" LastActivityDate="2010-12-10T04:15:35.923" Title="Mapping Irregular Quadrilateral to a Rectangle" Tags="&lt;geometry&gt;" AnswerCount="2" CommentCount="5" FavoriteCount="1" />
  <row Id="13405" PostTypeId="2" ParentId="13403" CreationDate="2010-12-07T18:48:03.087" Score="6" ViewCount="" Body="&lt;p&gt;You can prove the theorem (under the assumption that $\sup U$ exists) directly from the definition of supremum. For each $n$ there is some point in $U$ within $1/n$ of the supremum. Use the axiom of choice to choose a sequence $(a_n)$ so that for each $n$, $a_n$ is within $1/n$ of the supremum. Then prove that this sequence converges to the supremum. &lt;/p&gt;&#xA;" OwnerUserId="630" LastActivityDate="2010-12-07T18:48:03.087" CommentCount="3" />
  <row Id="13406" PostTypeId="2" ParentId="13403" CreationDate="2010-12-07T18:49:10.693" Score="4" ViewCount="" Body="&lt;p&gt;Because it follows directly from the definition of supremum. You can create the sequence $\{a_n\}$ easily:&#xA;Let $M = \sup U $.&#xA;Assume that $M \notin U$ (otherwise just let $a_i = M \, \forall i \in \mathbb{N}$).&#xA;Fix $a_0 \in U$.&#xA;Choose $a_i \in (\frac{M + a_{i-1}}{2}, M) \subset U$ for $i = 1, 2, \dots$&#xD;&#xA;Such $a_i$ must exist since $M$ is the supremum. Otherwise $\frac{M + a_{i-1}}{2} &amp;lt; M$ would be a lower upper bound for $U$.&lt;/p&gt;&#xA;" OwnerUserId="1016" LastEditorUserId="1016" LastEditorDisplayName="" LastEditDate="2010-12-07T18:55:39.300" LastActivityDate="2010-12-07T18:55:39.300" CommentCount="1" />
  <row Id="13407" PostTypeId="2" ParentId="8946" CreationDate="2010-12-07T18:50:51.227" Score="1" ViewCount="" Body="&lt;p&gt;Try this approach:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the partial derivative with respect to $\xi$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\partial_{\xi}\phi(t,\xi)=i\mathbb{E}\left[X_t e^{i\xi X_t}\right] \; .$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the partial derivative with respect to $t$, I'll take the differential but only varying t, so as to connect with the SDE:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$d_t\phi(t,\xi)=\mathbb{E}\left[e^{i\xi (X_t+dX_t)} - e^{i\xi X_t}\right]=\mathbb{E}\left[(i\xi dX_t-\frac{\xi^2}{2}dX_t^2)e^{i\xi X_t}\right] \; .$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ dX_t = (X_t - \mu) dt + \sigma \sqrt{X_t}dW_t $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ dX_t^2 = \sigma^2 X_t dt $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This implies&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$d_t\phi(t,\xi)=i\xi \mathbb{E}\left[dX_t e^{i\xi X_t}\right]-\frac{\xi^2}{2}\mathbb{E}\left[dX_t^2 e^{i\xi X_t}\right] $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and thus&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$d_t\phi(t,\xi)=i\xi \mathbb{E}\left[(X_t-\mu) e^{i\xi X_t}\right]dt+i\xi \mathbb{E}\left[\sqrt{X_t} dW_t e^{i\xi X_t}\right]-\frac{\sigma^2 \xi^2}{2}\mathbb{E}\left[X_t e^{i\xi X_t}\right]dt $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the middle term contains $dW_t$ as a consequence, taking its expectation gives zero. You can now recognize the other terms as containing $\phi$ or $\partial_{\xi}\phi$, so you've got yourself an ordinary PDE for $\phi$. That should be the aim of the computation.&lt;/p&gt;&#xA;" OwnerUserId="3567" LastEditorUserId="3567" LastEditorDisplayName="" LastEditDate="2010-12-07T21:10:49.700" LastActivityDate="2010-12-07T21:10:49.700" CommentCount="12" />
  <row Id="13408" PostTypeId="2" ParentId="13215" CreationDate="2010-12-07T19:19:27.720" Score="0" ViewCount="" Body="&lt;p&gt;@Jesse: Thanks a lot! I will read carefully this.&lt;/p&gt;&#xA;" OwnerUserId="4482" LastActivityDate="2010-12-07T19:19:27.720" />
  <row Id="13409" PostTypeId="2" ParentId="13404" CreationDate="2010-12-07T19:33:07.270" Score="3" ViewCount="" Body="&lt;p&gt;In general there is no affine transformation that maps an arbitrary quadrangle onto a rectangle. But there is (exactly one) projective transformation $T$ that maps a given quadrangle $(A, B, C, D)$ in the projective plane onto a given quadrangle $(A&amp;#39;, B&amp;#39;, C&amp;#39; D&amp;#39;)$ in the same or another projective plane. This $T$ is ${\it collinear}$, i.e., it maps lines to lines. To do the calculations you have to introduce homogeneous coordinates $(x,y,z)$ such that $D=(0,0,1)$, $C=(1,0,1)$, $A=(0,1,1)$, $B=(1,1,1)$ and similarly for $A&amp;#39;$, $B&amp;#39;$, $C&amp;#39;$, $D&amp;#39;$. With respect to these coordinates the map $T$ is linear and its matrix is the identity matrix.&lt;/p&gt;&#xA;" OwnerUserId="1303" LastActivityDate="2010-12-07T19:33:07.270" CommentCount="2" />
  <row Id="13410" PostTypeId="1" CreationDate="2010-12-07T19:48:45.853" Score="0" ViewCount="70" Body="&lt;p&gt;Hi!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried to show the following estimation and am not sure, if it holds:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Given $U\subseteq\mathbb R$ and $f:\mathbb R\times \mathbb R\to\mathbb R$. Its derivative with respect to the second argument exists and is denoted by $f_b(a,b)$:&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$\displaystyle\frac{d}{db}\left(\sup_{a\in U}f(a,b)\right)\ge \inf_{a\in U}f_b(a,b)$&#xD;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h2&gt;My proof:&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;There exists a sequence $a_n$ in $U$ such that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle\lim_{n\to\infty}f(a_n,b)=\sup_{a\in U}f(a,b)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\forall n\in\mathbb N$ the following holds:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sup_{a\in U}f(a,b+h)-\sup_{a\in U}f(a,b)\ge f(a_n,b+h)-\sup_{a\in U}f(a,b)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the differentiability of $f$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$=f(a_n,b)+h f_b(a_n,b)-\sup_{a\in U}f(a,b)+\mathcal O(h^2)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is always greater than&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle\ge f(a_n,b)+h\;\inf_{a\in U} f_b(a,b)-\sup_{a\in U}f(a,b)+\mathcal O(h^2)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As this holds for an arbitrary $n$ and thus $f(a_n,n)$ comes arbitrarily close to $\sup_{a\in U}f(a,b)$ it holds:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sup_{a\in U}f(a,b+h)-\sup_{a\in U}f(a,b)\ge h\;\inf_{a\in U} f_b(a,b)+\mathcal O(h^2)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Deviding by $h$ and performing the limit yields the assertion.&lt;/p&gt;&#xA;" OwnerUserId="4514" LastActivityDate="2010-12-07T19:48:45.853" Title="Estimate of the derivative of a function that is supremum of a function of two variables" Tags="&lt;real-analysis&gt;&lt;optimization&gt;" CommentCount="2" />
  <row Id="13411" PostTypeId="1" AcceptedAnswerId="13419" CreationDate="2010-12-07T19:52:16.753" Score="2" ViewCount="168" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Given a continuous function $f(x)$ and&#xA;  $\forall a\gt 0, |\int_{0}^{a}f(x)dx| \leq M$, show&#xA;  $\int_{0}^{\infty}f(x^2)dx$ exists.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I tried substituting $t=x^2$ which gave me $\displaystyle \int_{0}^{\infty}f(x^2)dx=\frac{1}{2}\int_{0}^{\infty}\frac{f(t)}{\sqrt t}dt$ but I can't see how to use the fact that $f(x)$ has a bounded indefinite integral in $[0, \infty)$. Simply doing $\frac{1}{2}|\int_{0}^{\infty}\frac{f(t)}{\sqrt t}dt|\leq \frac{M}{2}|\int_{0}^{\infty}\frac{1}{\sqrt t}dt|$ doesn't really help because $\int \frac{1}{\sqrt t}$ doesn't converge in $[0, \infty)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After substituting I tried integration by parts (using $u = \frac{1}{\sqrt t}, v' = f(t) \implies v = F(t)$) but that seems to complicate things further.&lt;/p&gt;&#xA;" OwnerUserId="3237" LastEditorUserId="3237" LastEditorDisplayName="" LastEditDate="2010-12-07T20:04:31.737" LastActivityDate="2010-12-07T20:44:38.420" Title="Show $\int_{0}^{\infty}f(x^2)dx$ exists if $f(x)$ is continuous and $\forall a\gt 0, |\int_{0}^{a}f(x)dx| \leq M$" Tags="&lt;calculus&gt;&lt;homework&gt;&lt;integral&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="13412" PostTypeId="1" AcceptedAnswerId="13413" CreationDate="2010-12-07T20:03:57.340" Score="7" ViewCount="240" Body="&lt;p&gt;Given the following expression:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(7y-1)(y-7) \le 0$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To me this inequality implies $y \le 7$ and $y \le \frac{1}{7}$ but the correct expression (from my module) happens to be $\frac{1}{7} \le y \le 7$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where exactly I am wrong?&lt;/p&gt;&#xA;" OwnerUserId="3744" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2011-03-08T09:24:16.263" LastActivityDate="2011-03-08T09:24:16.263" Title="An elementary doubt on Inequalities" Tags="&lt;algebra-precalculus&gt;" AnswerCount="4" CommentCount="9" />
  <row Id="13413" PostTypeId="2" ParentId="13412" CreationDate="2010-12-07T20:07:48.903" Score="7" ViewCount="" Body="&lt;p&gt;Hint: If the product of two numbers is nonpostive, one of them must be nonpositive and one of them must be nonnegative.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: $(7y-1)(y-7)\leq 0$. So either, $7y-1 \leq 0$ and $y-7 \geq 0$ or $7y-1 \geq 0$ and $y-7 \leq 0$. Now, can you eliminate one of these two cases?&lt;/p&gt;&#xA;" OwnerUserId="3431" LastEditorUserId="3431" LastEditorDisplayName="" LastEditDate="2010-12-07T20:27:49.470" LastActivityDate="2010-12-07T20:27:49.470" CommentCount="7" />
  <row Id="13414" PostTypeId="1" AcceptedAnswerId="13449" CreationDate="2010-12-07T20:19:17.900" Score="12" ViewCount="319" Body="&lt;p&gt;I want to find a nice simple expression for the definite integral&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_0^\infty \frac{x^2\,dx}{(x^2-a^2)^2 + x^2}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, I can numerically compute this integral, and it seems to converge to $\pi/2$ for all real values of $a$. Is this integral actually always equal to $\pi/2$? How can I show this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, why does Wolfram Alpha give me something that appears to depend on $a$? Is there a good reason it doesn't eliminate $a$?&lt;/p&gt;&#xA;" OwnerUserId="4519" LastActivityDate="2011-01-10T14:18:49.067" Title="Is this definite integral really independent of a parameter? How can it be shown?" Tags="&lt;integral&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="1" />
  <row Id="13415" PostTypeId="2" ParentId="6323" CreationDate="2010-12-07T20:26:39.753" Score="0" ViewCount="" Body="&lt;p&gt;Sometimes the quotient space is badly behaved. For instance when a Lie group acts smoothly on a manifold, the quotient space does not carry, in general, the structure of a smooth manifold. Then the equivariant (co)homology construction can help out. (This is of course not that much of an issue if you're considering singular homology, but for Morse homology there is a HUGE difference...)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also it is possible to &quot;resolve&quot; the singularities of a group action and compute a certain (co)homology on the resolved space. The homology obtained in this way is equivalent to equivariant (co)homology, see &quot;equivariant cohomology and resolution&quot; &lt;a href=&quot;http://www-math.mit.edu/~rbm/paper.html&quot; rel=&quot;nofollow&quot;&gt;http://www-math.mit.edu/~rbm/paper.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4520" LastEditorUserId="4520" LastEditorDisplayName="" LastEditDate="2010-12-08T15:45:21.453" LastActivityDate="2010-12-08T15:45:21.453" />
  <row Id="13416" PostTypeId="2" ParentId="4894" CreationDate="2010-12-07T20:33:03.647" Score="0" ViewCount="" Body="&lt;p&gt;There is a simple &quot;philosophical&quot; answer: Symplectic forms allow you to only measure two-dimensional quantities, not one-dimensional ones (you can measure area infinitesimally). That's the basic difference with Riemannian geometry, where you can measure length infinitesimally.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore symplectic geometry pops up in connection with the cotangent bundle. For instance if you have a (scalar) first order differential operator on a manifold, then the principal symbol transforms like a covector (when changing coordinates). Note that this does not require a metric.&lt;/p&gt;&#xA;" OwnerUserId="4520" LastActivityDate="2010-12-07T20:33:03.647" />
  <row Id="13417" PostTypeId="2" ParentId="9655" CreationDate="2010-12-07T20:37:25.857" Score="1" ViewCount="" Body="&lt;p&gt;Just minimize the energy of the path in the homotopy class of paths connecting p and q.&lt;/p&gt;&#xA;" OwnerUserId="4520" LastActivityDate="2010-12-07T20:37:25.857" CommentCount="2" />
  <row Id="13418" PostTypeId="2" ParentId="8808" CreationDate="2010-12-07T20:40:29.613" Score="1" ViewCount="" Body="&lt;p&gt;Well, my answer to 1. is that you can convienently do standard analysis in a chart. Did you ever try working on more complicated (singular) objects? It's a pain in the neck! (2. and 3. have already been answered)&lt;/p&gt;&#xA;" OwnerUserId="4520" LastActivityDate="2010-12-07T20:40:29.613" />
  <row Id="13419" PostTypeId="2" ParentId="13411" CreationDate="2010-12-07T20:44:38.420" Score="2" ViewCount="" Body="&lt;p&gt;If $f$ is only assumed continuous on $(0,\infty)$, then $\int_0^\infty  {f(x^2 )dx}$ might diverge (consider $f$ such that $f(x)=1/\sqrt{x}$ near $0+$). So, assume that $f$ is continuous on $[0,\infty)$. Then, obviously $\int_0^1 {f(x^2 )dx}$ exists (since $f$ is uniformly bounded on $[0,1]$). So, we only have to prove that $\int_1^\infty  {f(x^2 )dx} $ exists. Now, after substituting and performing integration by parts as you did, it suffices to show, using your notation, that $F(t)/\sqrt{t} \to 0$ as $t \to \infty$, and that $\int_1^\infty  {\frac{{F(t)}}{{t^{3/2} }}} dt$ exists. This is clearly true... &lt;/p&gt;&#xA;" OwnerUserId="2810" LastActivityDate="2010-12-07T20:44:38.420" CommentCount="1" />
  <row Id="13420" PostTypeId="1" CreationDate="2010-12-07T20:50:09.967" Score="0" ViewCount="96" Body="&lt;p&gt;Let R be a commutative ring and I and J be two ideals of R, how to show that IJ= the intersection of I and J if R=I+J?&lt;/p&gt;&#xA;" OwnerUserId="4521" LastEditorUserId="4521" LastEditorDisplayName="" LastEditDate="2010-12-07T21:02:34.433" LastActivityDate="2010-12-07T21:12:51.520" Title="A homework problem about ideal" Tags="&lt;ring-theory&gt;" AnswerCount="1" />
  <row Id="13421" PostTypeId="1" AcceptedAnswerId="13425" CreationDate="2010-12-07T20:53:56.297" Score="1" ViewCount="223" Body="&lt;p&gt;By definition, the weak law states that for a specified large $n$, the average  is likely to be near $\mu$. Thus, it leaves open the possibility that $|\bar{X_n}-\mu| \gt \eta$ happens an infinite number of times, although at infrequent intervals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The strong law shows that this almost surely will not occur. In particular, it implies that with probability 1, we have that for any $\eta &amp;gt; 0$ the inequality $|\bar{X_n}-\mu| \lt \eta$ holds for all large enough $n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now my question is application of these laws. How do I know which distribution satisfies the strong law vs the weak law. For example, consider a distribution $X_n$ be iid with finite variances and zero means. Does the mean $\frac{\sum_{k=1}^{n} X_k}{n}$ converge to $0$ almost surely (strong law of large numbers) or only in probability (weak law of large numbers)?&lt;/p&gt;&#xA;" OwnerUserId="957" LastActivityDate="2010-12-07T22:19:22.923" Title="application of strong vs weak law of large numbers" Tags="&lt;probability-theory&gt;&lt;measure-theory&gt;&lt;convergence&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" />
  <row Id="13422" PostTypeId="2" ParentId="13420" CreationDate="2010-12-07T20:54:54.353" Score="2" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\ $ One direction is trivial. The other direction requires further hypotheses, e.g. if $\rm\ I + J = 1\ $ then start from the fact that $\rm\ (I+J)\ \ (I\cap J) \ \subseteq\ I\:J\ $ follows by applying the distributive law. It's an ideal form of the integer law $\rm gcd(i,j)\ lcm(i,j) = i\:j\ \ $ so $\rm\ \ gcd(i,j)=1\ \Rightarrow\ lcm(i,j) = i\:j\:.$&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-12-07T21:12:51.520" LastActivityDate="2010-12-07T21:12:51.520" />
  <row Id="13423" PostTypeId="2" ParentId="13412" CreationDate="2010-12-07T21:06:20.157" Score="5" ViewCount="" Body="&lt;p&gt;Another way is to analyze as a quadratic function $f(y)=7y^2-50y+7$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that you want solve : $f(y)=7y^2-50y+7 \leq 0$, whose graph is: &lt;img src=&quot;http://i.imgur.com/ophP5.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The intersections with the x-axis are : $x=\frac{1}{7}$ and $x=7$, this graph we say that $f(y) \leq 0$ in $[\frac{1}{7},7]$, this  wanted to see.&lt;/p&gt;&#xA;" OwnerUserId="3421" LastEditorUserId="3421" LastEditorDisplayName="" LastEditDate="2010-12-08T02:06:32.777" LastActivityDate="2010-12-08T02:06:32.777" />
  <row Id="13424" PostTypeId="1" AcceptedAnswerId="13428" CreationDate="2010-12-07T21:06:25.647" Score="2" ViewCount="169" Body="&lt;p&gt;How would you expand a moment generating function with a negative fractional power as a power series??&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What method would you be able to use?&lt;/p&gt;&#xA;" OwnerUserId="4522" LastEditorUserId="4522" LastEditorDisplayName="" LastEditDate="2010-12-14T19:37:22.140" LastActivityDate="2010-12-14T19:37:22.140" Title="Expansion of a function with a negative fractional power." Tags="&lt;statistics&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="13425" PostTypeId="2" ParentId="13421" CreationDate="2010-12-07T21:11:44.767" Score="1" ViewCount="" Body="&lt;p&gt;If $X_1,X_2,\ldots$ is a sequence of i.i.d. random variables with finite mean $\mu$ (in your example, $\mu = 0$),&#xA;then by the strong law of large numbers, $\frac{{\sum\nolimits_{i = 1}^n {X_i } }}{n}$ converges to $\mu$ almost surely. In particular, $\frac{{\sum\nolimits_{i = 1}^n {X_i } }}{n}$ converges to $\mu$ in probability. So, you actually don't have to assume finite variance. &lt;/p&gt;&#xA;" OwnerUserId="2810" LastActivityDate="2010-12-07T21:11:44.767" CommentCount="5" />
  <row Id="13426" PostTypeId="2" ParentId="13424" CreationDate="2010-12-07T21:11:47.043" Score="1" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Binomial_series&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Binomial_series&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1508" LastActivityDate="2010-12-07T21:11:47.043" />
  <row Id="13428" PostTypeId="2" ParentId="13424" CreationDate="2010-12-07T21:23:14.487" Score="2" ViewCount="" Body="&lt;p&gt;See &lt;a href=&quot;http://mathworld.wolfram.com/GammaDistribution.html&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt;, in particular equations (17)-(24). This is the moment-generating function of a gamma distribution, and so having the moments you are actually done.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Specifically, take $\theta = 2$ and $\alpha = n/2$. Then, the moments $\mu&amp;#39;_r$ are given by &#xA;$$&#xD;&#xA;\mu&amp;#39;_r = \frac{{2^r \Gamma (n/2 + r)}}{{\Gamma (n/2)}} = n(n+2)(n+4) \cdots (n+2r-2),&#xD;&#xA;$$&#xD;&#xA;where for the last equality we have used $\Gamma(p+1) = p \Gamma(p)$.&#xA;Now, the moment-generating function can be expanded as&#xA;$$&#xD;&#xA;M(t) = 1 + \mu&amp;#39; _1 t + \frac{1}{{2!}}\mu&amp;#39; _2 t^2  + \frac{1}{{3!}}\mu&amp;#39; _3 t^3  +  \cdots,&#xD;&#xA;$$&#xD;&#xA;giving us&#xA;$$&#xD;&#xA;\frac{1}{{(1 - 2t)^{n/2} }} = 1 + nt + \frac{{n(n + 2)}}{{2!}}t^2  + \frac{{n(n + 2)(n + 4)}}{{3!}}t^3  +  \cdots &#xD;&#xA;$$&#xD;&#xA;(for $|t|&amp;lt;1/2$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT 2&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For completeness, let us show how to find the moments $\mu&amp;#39;_r$. Your moment-generating function corresponds to a gamma distribution with density&#xA;$$&#xD;&#xA;f(x)= \frac{{x^{\alpha  - 1} e^{ - x/\theta } }}{{\theta ^\alpha  \Gamma (\alpha )}}, \;\; x&amp;gt;0,&#xD;&#xA;$$&#xD;&#xA;where $\theta = 2$ and $\alpha = n/2$. So,&#xA;$$&#xD;&#xA;\mu&amp;#39;_r = \frac{1}{{\theta ^\alpha  \Gamma (\alpha )}}\int_0^\infty  {x^{r + \alpha  - 1} e^{ - x/\theta } \,{\rm d}x} = &#xD;&#xA;\frac{{\theta ^r }}{{\Gamma (\alpha )}}\int_0^\infty  {x^{r + \alpha  - 1} e^{ - x} \,{\rm d}x} = \theta ^r \frac{{\Gamma (r + \alpha )}}{{\Gamma (\alpha )}}.&#xD;&#xA;$$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="2810" LastEditorUserId="2810" LastEditorDisplayName="" LastEditDate="2010-12-08T17:16:26.277" LastActivityDate="2010-12-08T17:16:26.277" CommentCount="1" />
  <row Id="13430" PostTypeId="1" AcceptedAnswerId="13432" CreationDate="2010-12-07T22:01:42.367" Score="2" ViewCount="206" Body="&lt;p&gt;Are there solutions in $\mathbb{Q}-\mathbb{Z}$ to $X^2+2y^2=1$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm pretty sure that there isnt but Im not sure how to show this.  I don't have much experience (yet) with $p$-adics.&lt;/p&gt;&#xA;" OwnerUserId="2212" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-08T03:33:20.517" LastActivityDate="2010-12-08T03:33:20.517" Title="Does there exist $X$, $Y \in\mathbb{Q}-\mathbb{Z}$ such that $X^2+2Y^2=1$?" Tags="&lt;number-theory&gt;&lt;diophantine-equations&gt;" AnswerCount="2" CommentCount="6" />
  <row Id="13431" PostTypeId="1" AcceptedAnswerId="13435" CreationDate="2010-12-07T22:17:46.847" Score="1" ViewCount="114" Body="&lt;p&gt;I came across an interesting problem in Jacod's probability book. But have no idea how to approach it. Should I approach it using induction? Any ideas?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $X_1, X_2, \cdots$ be an infinite sequence of iid sequence of integrable random variables&#xA;and let $N$ be a positive, integer-valued integrable random variable which is independent&#xA;from the sequence. Define $S_n = \sum_{k=1}^{n} X_k$ and assume that $S_0 = 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(a) Show that $E[S_N] = E[N]E[X_1]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(b) Show that the characteristic function of $S_N$ is given by $E[\phi_{X_{1}}(t)^N]$, where $\phi_{X_{1}}$ is the characteristic function of $X_1$.&lt;/p&gt;&#xA;" OwnerUserId="957" LastActivityDate="2010-12-07T22:32:15.167" Title="distribution of iid sequence of integrable random variables" Tags="&lt;probability-theory&gt;&lt;measure-theory&gt;&lt;probability&gt;&lt;distribution-theory&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="13432" PostTypeId="2" ParentId="13430" CreationDate="2010-12-07T22:18:16.143" Score="9" ViewCount="" Body="&lt;p&gt;Yes, there are solutions. For example $({7 \over 11}, {6 \over 11})$ solves it. In general one can find such solutions to $x^2 + ry^2 = 1$ whenever $r$ is a rational number. One can do this by intersecting a line $ax + by = c$ containing $(1,0)$ ($a,b,$ and $c$ rational) with your ellipse $x^2 + ry^2 = 1$; the second intersection point will also have rational coordinates. Since only finitely many of these will have an integer coordinate the rest of such points will have both their entries in $Q - Z$. &lt;/p&gt;&#xA;" OwnerUserId="3035" LastActivityDate="2010-12-07T22:18:16.143" />
  <row Id="13433" PostTypeId="2" ParentId="13421" CreationDate="2010-12-07T22:19:22.923" Score="2" ViewCount="" Body="&lt;p&gt;From section 7.4 of Grimmett and Stirzaker's &lt;em&gt;Probability and Random Processes (3rd edition)&lt;/em&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The independent and identically distributed sequence $(X_n)$, with common distribution function $F$, satisfies $${1\over n} \sum_{i=1}^n X_i\to \mu$$ in probability for some constant $\mu$ if and only if the characteristic function $\phi$ of $X_n$ is differentiable at $t=0$ and $\phi^\prime(0)=i \mu$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, the weak law holds but the strong law fails for $\mu=0$ and symmetric random variables with $1-F(x)\sim 1/(x\log(x))$ as $x\to\infty$.&lt;/p&gt;&#xA;" OwnerUserId="940" LastActivityDate="2010-12-07T22:19:22.923" />
  <row Id="13434" PostTypeId="2" ParentId="13430" CreationDate="2010-12-07T22:19:44.350" Score="8" ViewCount="" Body="&lt;p&gt;Yes.  E.g. $X = 1/3$, $Y = 2/3$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;[Added: Carrying out the procedure suggested in Zaricuses's answer, i.e. intersecting the line with slope $-p/q$ through $(1,0)$&#xD;&#xA;with the conic $X^2 + 2 Y^2 =1$, we find that the general solution&#xA;is $X = (2p^2 - q^2)/(2 p^2 + q^2), Y = 2pq/(2 p^2 + q^2).$ &#xA;The solution above comes form taking $p = 1, q = 1$.  The solution in Zaricuse's answer comes from taking $p = 3, q = 2$.]&lt;/p&gt;&#xA;" OwnerUserId="221" LastEditorUserId="221" LastEditorDisplayName="" LastEditDate="2010-12-08T03:32:48.393" LastActivityDate="2010-12-08T03:32:48.393" />
  <row Id="13435" PostTypeId="2" ParentId="13431" CreationDate="2010-12-07T22:32:15.167" Score="2" ViewCount="" Body="&lt;p&gt;For part (a) use $E[S_N]= \sum\nolimits_n {E[S_N |N = n]P(N = n)}$. This leads straightforwardly to the result.&#xA;For part (b) use $E[e^{tS_N } ] = \sum\nolimits_n {E[e^{tS_N } |N = n]P(N = n)} $. Again, this leads straightforwardly to the result.&lt;/p&gt;&#xA;" OwnerUserId="2810" LastActivityDate="2010-12-07T22:32:15.167" CommentCount="2" />
  <row Id="13436" PostTypeId="1" AcceptedAnswerId="13445" CreationDate="2010-12-07T22:35:50.357" Score="3" ViewCount="220" Body="&lt;p&gt;I need to find the two roots (x-intercepts) from the equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(x + 7)(x - 3) = (x + 7)(5 - x)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How would I go about this? &lt;strong&gt;&lt;em&gt;I would greatly appreciate a step-by-step explanation.&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="4526" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2010-12-07T23:24:42.860" LastActivityDate="2011-04-20T11:25:17.347" Title="How would I solve $(x + 7)(x - 3) = (x + 7)(5 - x)$?" Tags="&lt;algebra-precalculus&gt;&lt;roots&gt;" AnswerCount="4" CommentCount="5" />
  <row Id="13437" PostTypeId="2" ParentId="13436" CreationDate="2010-12-07T22:42:16.557" Score="10" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\rm\ \ A\ B\ =\ A\ C\ \iff\ A\ (B-C)\ =\ 0\ \iff\ A=0\ \ or\ \ B-C = 0 $&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-12-07T22:42:16.557" />
  <row Id="13438" PostTypeId="1" AcceptedAnswerId="13440" CreationDate="2010-12-07T22:49:22.547" Score="1" ViewCount="120" Body="&lt;p&gt;$$\int_0^2 \dfrac{\mathrm dx}{\sqrt{x}(x-1)}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to determine whether this integral converges or diverges. Now usually problems like these are easy, but this one is kind of tricky since it is discontinuous at both 0 and 1. Whereas with one situations where the integral is only discontinuous at 1 value, I could just set up 2 integrals and use the &quot;lim of t&quot; method, but here I cant set up two integrals. The only way that two integrals could be set up is if the first one is from 0 to 1 and the second one is from 1 to 2, but the first one is discontinuous at both values, so not sure what the work around is.  &lt;/p&gt;&#xA;" OwnerUserId="2985" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-08T02:09:11.140" LastActivityDate="2010-12-08T02:09:11.140" Title="Question on improper integrals" Tags="&lt;calculus&gt;&lt;integral&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="13439" PostTypeId="2" ParentId="13438" CreationDate="2010-12-07T22:53:16.013" Score="3" ViewCount="" Body="&lt;p&gt;Consider the change of variables $t=\sqrt x$. It transforms the interval $[0,2]$ into $[0,\sqrt 2]$ and removes one of the problem zeroes. It is easy to see from the resulting expression that the integral diverges.&lt;/p&gt;&#xA;" OwnerUserId="462" LastActivityDate="2010-12-07T22:53:16.013" />
  <row Id="13440" PostTypeId="2" ParentId="13438" CreationDate="2010-12-07T22:53:52.863" Score="6" ViewCount="" Body="&lt;p&gt;You can break up the integral at &lt;em&gt;any&lt;/em&gt; point or points you like. In this case, you could break it up into &lt;em&gt;three&lt;/em&gt; integrals: pick a point $c$ strictly between $0$ and $1$, and consider:&#xA;\begin{align*}&#xA;\int_0^2 \frac{dx}{\sqrt{x}(x-1)} &amp;amp;= \int_0^c\frac{dx}{\sqrt{x}(x-1)}+\int_c^1\frac{dx}{\sqrt{x}(x-1)} + \int_1^2\frac{dx}{\sqrt{x}(x-1)}\\&#xA;&amp;amp;= \lim_{a\to 0^+}\int_a^c \frac{dx}{\sqrt{x}(x-1)} + \lim_{b\to 1^-}\int_c^b\frac{dx}{\sqrt{x}(x-1)} + \lim_{d\to 1^+}\int_d^2\frac{dx}{\sqrt{x}(x-1)}.&#xA;\end{align*}&#xA;The original improper integral exists if and only if each of the three improper integrals exist.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-12-07T22:53:52.863" CommentCount="3" />
  <row Id="13441" PostTypeId="1" AcceptedAnswerId="13443" CreationDate="2010-12-07T23:01:02.253" Score="2" ViewCount="138" Body="&lt;p&gt;The book I am studying uses the notation $w_k$ to denote the number of roots of unity contained in $K$, an imaginary quadratic number field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Supposedly, this is the same as the size of the unit group in $O$, the ring of integers in $K$.  I am well familiar with the standard arguments for finding the units in $O$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Im not really sure exactly what the phrase &quot;the number of roots of unity in $K$&quot; means precisely.  Does anyone know?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, I was thinking that it is probibly the case that the only elements of $K$ with finite multiplicative order are the elements of the unit group.  Is this correct?  If so, then is this the correct line of reasoning to show that $w_k=|U(O)|$?&lt;/p&gt;&#xA;" OwnerUserId="2212" LastActivityDate="2010-12-07T23:17:50.303" Title="The number of roots of unity in an imaginary quadratic number field." Tags="&lt;number-theory&gt;" AnswerCount="2" />
  <row Id="13442" PostTypeId="2" ParentId="13441" CreationDate="2010-12-07T23:05:40.100" Score="2" ViewCount="" Body="&lt;p&gt;You need Dirichlet's unit theorem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.math.uconn.edu/~kconrad/blurbs/gradnumthy/unittheorem.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.math.uconn.edu/~kconrad/blurbs/gradnumthy/unittheorem.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3431" LastActivityDate="2010-12-07T23:05:40.100" CommentCount="5" />
  <row Id="13443" PostTypeId="2" ParentId="13441" CreationDate="2010-12-07T23:08:58.147" Score="2" ViewCount="" Body="&lt;p&gt;A root of unity is a complex number $a$ such that $a^n = 1$ for some positive integer $n$. The &quot;number of roots of unity&quot; is the cardinality of the set of roots of unity. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, if $K=\mathbb{Q}(i)$, then the roots of unity in $K$ are $1$, $-1$, $i$, and $-i$, so the number of roots of unity contained in $K$ is $4$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yes: a complex number is a root of unity if and only if it has finite multiplicative order in $\mathbb{C}^*$. In fact, the group of roots of unity is the torsion subgroup of $\mathbb{C}^*$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To prove that in the imaginary quadratic case the unit group $\mathcal{O}_K$ coincides with the group of roots of unity in $K$ you could use Dirichlet's Unit Theorem (as Timothy Wagner suggests), but you don't have to. You can simply show that every unit in $\mathcal{O}_K$ must in fact be a root of unity. This is not hard, since you (probably) know exactly what $\mathcal{O}_K$ for $K=\mathbb{Q}(\sqrt{d})$ ($d$ square free, $d\neq 1$), and you can use the norm map. For example, to show it in the case of $K=\mathbb{Q}(i)$, note that an element in $\mathcal{O}_K = \mathbb{Z}[i]$ is a unit if and only if $N(a+bi) = a^2+b^2 = 1$ (since $N((a+bi)(c+di)) = N(a+bi)N(c+di)$). But since $a$ and $b$ are integers, that requires $a=\pm 1$ and $b=0$ or $a=0$ and $b=\pm 1$, giving you roots of unity. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(In fact, imaginary quadratic fields are the only number fields in which the group of units in the number field is finite and torsion; this follows from the aforementioned Dirichlet Unit Theorem)&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-07T23:17:50.303" LastActivityDate="2010-12-07T23:17:50.303" CommentCount="2" />
  <row Id="13444" PostTypeId="2" ParentId="13387" CreationDate="2010-12-07T23:13:52.243" Score="6" ViewCount="" Body="&lt;p&gt;You're asking about a very wide range of numbers!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For all numbers, you should do ECM first.  I recommend GMP-ECM (&lt;a href=&quot;http://gilchrist.ca/jeff/factoring/index.html&quot; rel=&quot;nofollow&quot;&gt;Windows&lt;/a&gt;, &lt;a href=&quot;http://gforge.inria.fr/projects/ecm&quot; rel=&quot;nofollow&quot;&gt;source&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For numbers of about 100 digits, your best bet (after doing an appropriate amount of trial factorization and ECM) is to use a SIQS. &lt;a href=&quot;http://www.boo.net/~jasonp/qs.html&quot; rel=&quot;nofollow&quot;&gt;msieve&lt;/a&gt; would be a good choice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For numbers of between 100 and 200 digits, you should do a large amount of ECM, at the very least a CPU-day's worth (and possibly 100+ CPU-days).  Once those complete, if no factors were found, you should move to GNFS.  GGNFS (&lt;a href=&quot;http://gilchrist.ca/jeff/factoring/index.html&quot; rel=&quot;nofollow&quot;&gt;Windows&lt;/a&gt;, &lt;a href=&quot;http://ggnfs.svn.sourceforge.net/viewvc/ggnfs/trunk/&quot; rel=&quot;nofollow&quot;&gt;source&lt;/a&gt;) is good and does not require a cluster, though for the upper end of the range you'll need at least a year if you're running on a single (multicore, one hopes) machine.  Here are some &lt;a href=&quot;http://gilchrist.ca/jeff/factoring/nfs_beginners_guide_perl.html&quot; rel=&quot;nofollow&quot;&gt;guidelines on using NFS&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have no hope of factoring hard numbers of 200 to 300 digits without many machines.  Thus, for large general-form numbers, you should focus entirely on ECM in hopes of finding a small factor (up to about 60 digits, with patience, or about 70 digits, with patience and good fortune).  If you find a factor, the cofactor may well be susceptible to GNFS (or it might be prime).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;No one has a reasonable chance to factor hard numbers of 300 to 1000 digits, unless possibly the NSA.  You should use ECM on such numbers in hopes of finding a factor, but you're unlikely to finish the factorization even if you do.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now if you have access to a CUDA-enabled GPU (or several), you can speed the ECM with &lt;a href=&quot;http://eecm.cr.yp.to/&quot; rel=&quot;nofollow&quot;&gt;EECM&lt;/a&gt; .  I don't know of any tools that do NFS on GPUs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;In short&lt;/strong&gt;: ECM a lot, then finish off with SIQS or NFS if the number is small enough (below 180-200 digits depending on your resources and patience).  If you want to throw money at the problem, buy more hardware, possibly GPUs -- don't buy commercial software, that has no bang for your buck.&lt;/p&gt;&#xA;" OwnerUserId="1778" LastEditorUserId="1778" LastEditorDisplayName="" LastEditDate="2010-12-10T18:36:05.197" LastActivityDate="2010-12-10T18:36:05.197" CommentCount="10" />
  <row Id="13445" PostTypeId="2" ParentId="13436" CreationDate="2010-12-07T23:26:07.253" Score="4" ViewCount="" Body="&lt;p&gt;Generally speaking, in an equation like yours with polynomial-type expressions (even in factored form, as yours are) on both sides, you probably want to collect everything on one side of the equals sign so that you have some polynomial-type expression equal to zero.  You might do this by multiplying out both sides, then collecting all the terms on one side; or you might rewrite your equation $A=B$ as $A-B=0$ and factor the $A-B$ part, since your $A$ and $B$ are already factored.&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-12-07T23:26:07.253" />
  <row Id="13446" PostTypeId="2" ParentId="4894" CreationDate="2010-12-07T23:29:51.383" Score="5" ViewCount="" Body="&lt;p&gt;I suppose a question to ask here is : what kinds of applications? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of the areas where symplectic geometry is used is Hamiltonian dynamics, where the geometric framework allows one to generalize away from cotangent bundles, but this direction is directly related to the origins of the subject.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another direction comes from algebraic geometry, where the fact that complex and Riemannian structure determine a symplectic one  (see Aaron's answer) makes all complex projective (and affine) algebraic varieties symplectic. Viewing them as such allows one to use differential-geometric methods, while the symplectic form provides some control (in the theory of holomorphic curves, symplectic form gives the crucial energy bounds without which things fall apart; this is why there is no Gromow-Witten theory of almost-complex manifolds). Add a bit of mirror symmetry to the mix and you have interaction with many other areas of math.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is also the fact that the wedge product is anti-symmetric, which leads the space of connections on a principal bundle over 2-d Riemannian manifold to be an (infinite dimensional) symplectic manifold (the gauge group is Hamiltonian, and curvature is the moment map, leading to flat connections being the reduction!). This leads, for G=SO(3) and the surface coming from a Heegard splitting, to Atiyah-Floer conjecture and some related stuff leads to Heegard-Floer theory. Thus connections to low-dimensional topology. The &quot;naive&quot; connection here is the anti-symmetry underlying symplectic form and wedge product, but probably there is a deeper reason known to wise people...&lt;/p&gt;&#xA;" OwnerUserId="2633" LastActivityDate="2010-12-07T23:29:51.383" />
  <row Id="13447" PostTypeId="1" CreationDate="2010-12-07T23:34:38.683" Score="1" ViewCount="94" Body="&lt;p&gt;Let $K$ be an imaginary quadratic field and $U$ denote the unit group in the ring of integers in $K$.  Are there $\alpha \in K-U$ with finite multiplicative order?  That is, is there $n \in N$ such that $\alpha^n=1$?  &lt;/p&gt;&#xA;" OwnerUserId="2212" LastActivityDate="2010-12-08T04:33:38.640" Title="Multiplicative order of elements in an imaginary quadratic field." Tags="&lt;number-theory&gt;" AnswerCount="1" CommentCount="6" />
  <row Id="13448" PostTypeId="2" ParentId="13402" CreationDate="2010-12-07T23:35:43.293" Score="1" ViewCount="" Body="&lt;p&gt;I'm not entirely sure that this answers what you're asking--if not, please comment so that I can revise it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have two points $A$ and $B$, the set of points $P=(1-t)\cdot A+t\cdot B$, where $t$ is a real number, is the line through $A$ and $B$, parameterized with $A=P|_{t=0}$ and $B=P|_{t=1}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you know for sure that $C$ is on the line through $A$ and $B$, then set $C=(1-t)\cdot A+t\cdot B$ and solve for $t$ in any one of the components (if $C$ is in fact on the line, then the value of $t$ will be the same, regardless of which component you use; if $C$ is not on the line, then the value of $t$ will not be the same for each component).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;note&lt;/em&gt;: edited to fix equation $P=$ in the second paragraph, and correspondingly in the third paragraph.&lt;/p&gt;&#xA;" OwnerUserId="72" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-12-08T01:37:37.167" LastActivityDate="2010-12-08T01:37:37.167" CommentCount="15" />
  <row Id="13449" PostTypeId="2" ParentId="13414" CreationDate="2010-12-07T23:36:34.650" Score="19" ViewCount="" Body="&lt;p&gt;Yes it is true! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle I = \int_{0}^{\infty} \dfrac{x^2}{x^2 + (x^2-a^2)^2} \ \text{dx}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Make the substitution $\displaystyle x = \dfrac{a^2}{t}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle I = \int_{0}^{\infty} \dfrac{a^6}{t^4\left(\dfrac{a^4}{t^2} + \left(\dfrac{a^4}{t^2} - a^2\right)^2\right)} \ \text{dt} = \int_{0}^{\infty} \dfrac{a^2}{t^2 + (a^2 - t^2)^2} \ \text{dt}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i.e.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle I = \int_{0}^{\infty} \dfrac{a^2}{x^2 + (a^2 - x^2)^2} \ \text{dx}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore&#xA;$$\displaystyle 2I = \int_{0}^{\infty} \dfrac{x^2}{x^2 + (x^2-a^2)^2} \ \text{dx} + \int_{0}^{\infty} \dfrac{a^2}{x^2 + (a^2 - x^2)^2}\ \text{dx}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ = \int_{0}^{\infty} \dfrac{x^2 + a^2}{x^2 + (x^2-a^2)^2} \ \text{dx}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle = \int_{0}^{\infty} \dfrac{1 + \dfrac{a^2}{x^2}}{1 + \left(x-\dfrac{a^2}{x}\right)^2} \ \text{dx}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Making the substitution $\displaystyle t = x - \dfrac{a^2}{x}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gives us&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle 2I = \int_{-\infty}^{\infty} \dfrac{\text{dt}}{1 + t^2} = \pi$$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-12-08T07:04:14.350" LastActivityDate="2010-12-08T07:04:14.350" CommentCount="0" />
  <row Id="13450" PostTypeId="2" ParentId="13412" CreationDate="2010-12-07T23:41:11.320" Score="0" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\ $ If $\rm\ a &amp;lt; b\ $ then $\rm\  (x-a)\ (x-b) &amp;lt; 0\ \iff\ a &amp;lt; x &amp;lt; b $&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-12-07T23:41:11.320" />
  <row Id="13451" PostTypeId="2" ParentId="13436" CreationDate="2010-12-07T23:50:49.997" Score="4" ViewCount="" Body="&lt;p&gt;Hint: If $x+7\neq 0$, your equation is equivalent to $x-3=5-x$, because you can divide both sides by a quantity different from zero. And what does happen to your equation when $x+7=0$?&lt;/p&gt;&#xA;" OwnerUserId="752" LastActivityDate="2010-12-07T23:50:49.997" />
  <row Id="13452" PostTypeId="1" AcceptedAnswerId="13457" CreationDate="2010-12-07T23:53:35.173" Score="9" ViewCount="613" Body="&lt;p&gt;Hello,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to solve this equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$y&amp;#39;&amp;#39;+2y&amp;#39;-3y=0$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I did this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$y&amp;#39; = z$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$y&amp;#39;&amp;#39; = z\dfrac{dz}{dy}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$z\dfrac{dz}{dy}+2z-3y=0$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$zdz+2zdy-3ydy=0$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$zdz=(3y-2z)dy$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$z=3y-2z$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$z=y$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$y=y&amp;#39;=y&amp;#39;&amp;#39;$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;???&lt;/p&gt;&#xA;&#xA;&lt;p&gt;now, I'm pretty sure I did something wrong. could you please correct.&lt;/p&gt;&#xA;" OwnerUserId="4528" LastEditorUserId="4528" LastEditorDisplayName="" LastEditDate="2010-12-08T00:40:01.777" LastActivityDate="2011-05-06T06:30:31.123" Title="how do you solve $y''+2y'-3y=0$?" Tags="&lt;differential-equations&gt;" AnswerCount="7" CommentCount="5" FavoriteCount="1" />
  <row Id="13453" PostTypeId="1" AcceptedAnswerId="13456" CreationDate="2010-12-07T23:54:05.357" Score="2" ViewCount="62" Body="&lt;p&gt;The product of two univariate generating functions is simply given by the Cauchy product. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ A(x) = \sum_{n=0} a_n x^n $$&#xD;&#xA;$$ B(x) = \sum_{n=0} b_n x^n $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ A(x)B(x) = C(x) = \sum_{n=0} x^n c_n $$&#xD;&#xA;with $c_n = \sum_{k=0}^n a_k b_{n-k}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the resulting generating function for the bivariate case?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ A(x,y) = \sum_{n=0}\sum_{m=0} a_{nm} x^n y^m $$&#xD;&#xA;$$ B(x,y) = \sum_{n=0}\sum_{m=0} b_{nm} x^n y^m $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ A(x,y)B(x,y) = C(x,y) = \sum_{n=0}\sum_{m=0} x^n y^m c_{nm} $$&#xD;&#xA;What is $c_{nm}$, and how does this generalize to multivariate generating functions?&lt;/p&gt;&#xA;" OwnerUserId="196" LastActivityDate="2010-12-08T00:00:50.597" Title="Product of bivariate generating functions" Tags="&lt;generating-functions&gt;&lt;power-series&gt;" AnswerCount="1" />
  <row Id="13455" PostTypeId="2" ParentId="13452" CreationDate="2010-12-08T00:00:44.103" Score="2" ViewCount="" Body="&lt;p&gt;$zdz=(3y−2z)dy$ does not imply $z=(3y−2z)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See this&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Linear_differential_equation#Homogeneous_equations_with_constant_coefficients&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Linear_differential_equation#Homogeneous_equations_with_constant_coefficients&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3431" LastActivityDate="2010-12-08T00:00:44.103" />
  <row Id="13456" PostTypeId="2" ParentId="13453" CreationDate="2010-12-08T00:00:50.597" Score="2" ViewCount="" Body="&lt;p&gt;$$c_{nm} = \sum_{i+j=n, k+l=m} a_{ik} b_{jl}.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This follows directly from basic properties of addition and multiplication and you should make sure to thoroughly understand this, for example by writing out the first few coefficients by hand.  The generalization to any finite number of variables should be clear.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-12-08T00:00:50.597" CommentCount="1" />
  <row Id="13457" PostTypeId="2" ParentId="13452" CreationDate="2010-12-08T00:01:12.683" Score="9" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\rm\ \ \ 0\  =\  y&amp;#39;&amp;#39; + 2\ y&amp;#39; - 3\ y\ =\  (D^2 + 2\ D -3)\ y\ =\ (D+3)\ (D-1)\ y, \quad D = d/dx $  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;i.e. factor the differential operator as you would any polynomial.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-12-08T01:20:49.247" LastActivityDate="2010-12-08T01:20:49.247" CommentCount="8" />
  <row Id="13458" PostTypeId="1" AcceptedAnswerId="13462" CreationDate="2010-12-08T00:23:35.657" Score="3" ViewCount="96" Body="&lt;p&gt;$\displaystyle\sum_{n=0}^{\infty}\sqrt{n}(\sqrt{n^4+1}-n^2)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ok I was really hesitant to post this because it's such an elementary question, and I really should be able to do this. However, I've been looking at this for over half an hour and I've tried many methods, but I'm not able to get the series into a nice form where I can take the limit. Any ideas on how to tackle this problem?&lt;/p&gt;&#xA;" OwnerUserId="2985" LastActivityDate="2010-12-08T00:38:35.883" Title="Determining convergence of a series" Tags="&lt;calculus&gt;&lt;sequences-and-series&gt;&lt;convergence&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="13459" PostTypeId="1" AcceptedAnswerId="13461" CreationDate="2010-12-08T00:25:39.050" Score="1" ViewCount="79" Body="&lt;p&gt;Let $\rho$ be a representation of $G$ on $V$.  Why are its eigenvalues roots of unity?&lt;/p&gt;&#xA;" OwnerUserId="314" LastActivityDate="2010-12-08T00:27:40.677" Title="Eigenvalues of representations" Tags="&lt;representation-theory&gt;" AnswerCount="1" />
  <row Id="13460" PostTypeId="1" AcceptedAnswerId="13467" CreationDate="2010-12-08T00:27:29.953" Score="9" ViewCount="284" Body="&lt;p&gt;I hope that this question is on-topic, though it is not quite technical.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am curious to hear from people how they approach reading a mathematical paper.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not asking specific questions on purpose, though at first I had a few. But I want to keep it rather open-ended.&lt;/p&gt;&#xA;" OwnerUserId="7076" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-08T00:36:40.580" LastActivityDate="2010-12-08T06:34:07.320" CommunityOwnedDate="2010-12-08T06:31:13.470" Title="how to read a mathematical paper?" Tags="&lt;soft-question&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="7" />
  <row Id="13461" PostTypeId="2" ParentId="13459" CreationDate="2010-12-08T00:27:40.677" Score="4" ViewCount="" Body="&lt;p&gt;I assume $G$ is finite.  In that case any $g \in G$ has some finite order $n$, hence $\rho(g)^n = 1$.  It follows that the characteristic polynomial of $\rho(g)$ divides $x^n - 1$.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-12-08T00:27:40.677" />
  <row Id="13462" PostTypeId="2" ParentId="13458" CreationDate="2010-12-08T00:28:14.810" Score="9" ViewCount="" Body="&lt;p&gt;Just note that $$\sqrt{n^4 + 1} - n^2 = \frac{( \sqrt{n^4 + 1} - n^2 )( \sqrt{n^4 + 1} + n^2 )}{\sqrt{n^4 + 1} + n^2} = \frac{n^4 + 1 - n^4}{\sqrt{n^4 + 1} + n^2} =  \frac{1}{\sqrt{n^4 + 1} + n^2}$$ then you can use the comparison test.&lt;/p&gt;&#xA;" OwnerUserId="900" LastEditorUserId="900" LastEditorDisplayName="" LastEditDate="2010-12-08T00:38:35.883" LastActivityDate="2010-12-08T00:38:35.883" CommentCount="4" />
  <row Id="13463" PostTypeId="2" ParentId="13460" CreationDate="2010-12-08T00:46:06.157" Score="4" ViewCount="" Body="&lt;p&gt;How I read a paper really depends on why I'm reading the paper.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A lot of papers I go to because I have a specific goal.  Maybe they have been cited elsewhere as containing a proof of something I want to understand.  Or a different proof of something I already know how to prove.  Or maybe someone refers to the paper as having a particularly lucid explanation of something.  If that's my goal, I go straight to whatever is in the paper that I want, assuming the organization of the paper makes this possible (not always the case, but very common, at least with articles written after say 1950).  In doing my close reading of the particular thing I want, I often take notes.  Depending on the results of my reading the thing I set out to read, I &lt;em&gt;might&lt;/em&gt; go on to read other parts of the paper.  But I generally don't.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I'm reading a paper without a specific goal, I generally start at the beginning and read the introduction.  Ideally this contains the statements of the main results, but if it doesn't, I skim the paper for those, and also the statements of any lemmas, corollaries, etc.  19 times out of 20 my reading stops there, because I find that, beyond the statements of the theorems adding to my general awareness of the world, I'm not all that interested in the paper.  That 1 time out of 20, I will probably try to come up with the proofs on my own, using the paper's introduction as a really big hint.  Whether I succeed or not, I usually end up giving the paper a pretty thorough reading.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So most of the time, for me, &quot;reading&quot; a paper does not mean reading the entire paper, or even reading a large contiguous chunk of the paper.  I could probably fit all of the papers I have ever read beginning to end in a regular sized binder with tons of room to spare.  I think I am far from alone in this.  In fact I think it is probably counterproductive to insist on reading things beginning to end, and I would advise that people who are new to the mathematical literature (e.g. students) suppress the desire to do this.  Generally speaking, if you are new to the literature in a given subject, you will waste a lot of time if you read this way.&lt;/p&gt;&#xA;" OwnerUserId="3840" LastEditorUserId="3840" LastEditorDisplayName="" LastEditDate="2010-12-08T00:52:18.337" LastActivityDate="2010-12-08T00:52:18.337" CommunityOwnedDate="2010-12-08T06:31:13.470" />
  <row Id="13464" PostTypeId="2" ParentId="13414" CreationDate="2010-12-08T00:48:54.693" Score="10" ViewCount="" Body="&lt;p&gt;With apologies to Robin Chapman.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The integrand is an even function of $x$, so we can integrate from $-\infty$ to $\infty$ and take half. The integrand tends to $1/z^2$ for large $z$ and as the length of a large arc is $\pi z$ the contribution of the arc tends to zero. So we just need to integrate over the upper half plane. The residues are solutions to $0=(x^2-a^2)^2+x^2= (x^2-i x -a^2)(x^2+i x-a^2)$. If $x_1$ and $x_2$ are solutions to the first quadratic, then at $x_1$ the second polynomial is equal to $2i x_1$, and the residue at $x_1$ is $x_1^2/2ix_1(x_2-x_1)=x_1/2i(x_2-x_1)$. The sum of residues at $x_1$ and $x_2$ is therefore $1/2i$. Now we just note that the two poles in the upper half plane are indeed the solutions $x_1$ and $x_2$ (which are  $i (\pm\sqrt{-a^2+1/4} + 1/2)$)). Hence the contour integral is $\pi$, and the original integral is $\pi/2$.&lt;/p&gt;&#xA;" OwnerUserId="2633" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2010-12-08T05:45:42.947" LastActivityDate="2010-12-08T05:45:42.947" CommentCount="1" />
  <row Id="13465" PostTypeId="2" ParentId="13396" CreationDate="2010-12-08T01:14:45.323" Score="2" ViewCount="" Body="&lt;p&gt;In a somewhat more general setting, let $A(R)$ denote the area of region $R$. If the number of flaws found on region $R$ follows a Poisson distribution, then the mean is proportional to $A(R)$. That is, if $X(R)$ denotes the number of flaws found on region $R$, then $X(R)$ is Poisson distributed with mean $\lambda A(R)$, for some fixed $\lambda &amp;gt; 0$.&#xA;Now, if $X_1,\ldots,X_n$ are i.i.d. Poisson$(\lambda)$ rv's (corresponding to the number of flaws found on a unit-area region, say square yard), then $\frac{1}{n}\sum\nolimits_{i = 1}^n {X_i }$ is an unbiased estimator for $\lambda$, and in turn, $\frac{1}{n}\sum\nolimits_{i = 1}^n {A(R) X_i }$ is an unbiased estimator for&#xA;$\lambda A(R) = {\rm E}[X(R)]$. So, in order to estimate ${\rm E}[X(R)]$, it suffices to estimate $\lambda$ (and then just multiply by $A(R)$).&lt;/p&gt;&#xA;" OwnerUserId="2810" LastActivityDate="2010-12-08T01:14:45.323" />
  <row Id="13466" PostTypeId="2" ParentId="13447" CreationDate="2010-12-08T01:23:24.547" Score="3" ViewCount="" Body="&lt;p&gt;If $\alpha^n=1$, then $\alpha$ satisfies the polynomial $x^n-1$, hence is an algebraic integer. Thus, $\alpha\in \mathcal{O}_K$; once you have that, you can make the easy observation Qiaochu did in the comments that $\alpha^{-1}=\alpha^{n-1}$ is also in $\mathcal{O}_K$ to show $\alpha\in U$ (or you can use the fact that if an algebraic integer satisfies a monic polynomial with integer coefficients and constant term $1$, then it must be a unit). So the answer is &quot;no.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, every root of unity is integral (satisfying $x^n-1$ for some $n$, or the appropriate cyclotomic polynomial if you insist on getting the minimal polynomial), so roots of unity in a number field are always in the ring of integers.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-08T04:33:38.640" LastActivityDate="2010-12-08T04:33:38.640" CommentCount="5" />
  <row Id="13467" PostTypeId="2" ParentId="13460" CreationDate="2010-12-08T01:32:04.577" Score="8" ViewCount="" Body="&lt;p&gt;I have a couple of pieces of advice:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Do not be intimidated by the introduction. This is where the whole paper is laid out,so it typically won't be easy to follow. The author will expand sentences into whole pages later on.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Take notes. I usually get a legal pad out and start copying down the paper, and fill in things, and make notes to myself or work out small examples.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Talk to people about the material. If you have an asdvisor or someone you are going through the paper with it is often helpful to present portions of the paper if not all of it.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Skip confusing bits. Feel free to black box certain parts. If you get stuck on something, don't worry, one sentence should not keep you from progressing through the paper. Box it off, pretend you understand the conclusion and make a note of it. Take that piece as a fact and move on to the next part of the paper. The goal is to get something out of a paper not understand it line for line the first time through.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Come back to the paper. You may want to go over the paper multiple times, take several passes at it. At least this is what I have to do when I am reading something. The amount of time it takes me to assimilate different levels/types of arguments gradually decreases, but it does take time.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I don't spend a lot of time reading papers right now, but this is what I try to do if I really want to understand something.&lt;/p&gt;&#xA;" OwnerUserId="627" LastEditorUserId="856" LastEditorDisplayName="" LastEditDate="2010-12-08T02:55:38.287" LastActivityDate="2010-12-08T02:55:38.287" CommunityOwnedDate="2010-12-08T06:31:13.470" />
  <row Id="13468" PostTypeId="1" AcceptedAnswerId="13472" CreationDate="2010-12-08T01:33:53.527" Score="2" ViewCount="65" Body="&lt;p&gt;Given two random samples a and b of a normal distribution, how can I calculate the probability that their difference |a-b| is bigger (or lesser) than c?&lt;/p&gt;&#xA;" OwnerUserId="4530" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-08T02:01:42.270" LastActivityDate="2010-12-08T02:01:42.270" Title="what is the formula for P(|a-b| &gt; c) in a normal distribution?" Tags="&lt;statistics&gt;&lt;normal-distribution&gt;" AnswerCount="1" />
  <row Id="13469" PostTypeId="1" AcceptedAnswerId="13474" CreationDate="2010-12-08T01:35:57.480" Score="4" ViewCount="252" Body="&lt;p&gt;I seem to remember reading somewhere an &quot;abstract nonsense&quot; proof of the well-known fact that the fundamental group of a connected topological group is abelian.  By &quot;abstract nonsense&quot; I mean that the proof used little more than the fact that topological groups are the group objects in the category of topological spaces and the fact that $\pi_1$ is a homotopy functor.  Does anybody remember how this works?  References are fine.&lt;/p&gt;&#xA;" OwnerUserId="1509" LastActivityDate="2010-12-08T02:18:11.170" Title="&quot;Abstract nonsense&quot; proof that the fundamental group of a topological group is abelian" Tags="&lt;topology&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" />
  <row Id="13470" PostTypeId="1" CreationDate="2010-12-08T01:46:54.010" Score="1" ViewCount="113" Body="&lt;p&gt;$$\sum_{i=0}^{n-1} (i+1)(n-1)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is that $O(n^2)$ or $O(n^3)$? Can you explain me how you found it? Thanks.&lt;/p&gt;&#xA;" OwnerUserId="4531" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-08T01:59:35.783" LastActivityDate="2010-12-08T03:56:15.823" Title="is this expression O(n^2) or O(n^3)?" Tags="&lt;sequences-and-series&gt;&lt;discrete-mathematics&gt;" AnswerCount="2" CommentCount="5" />
  <row Id="13471" PostTypeId="2" ParentId="13470" CreationDate="2010-12-08T01:52:00.793" Score="6" ViewCount="" Body="&lt;p&gt;Hint: Factor out the (n-1) and then use that the average term in the sum has size n/2.&lt;/p&gt;&#xA;" OwnerUserId="8" LastActivityDate="2010-12-08T01:52:00.793" CommentCount="4" />
  <row Id="13472" PostTypeId="2" ParentId="13468" CreationDate="2010-12-08T01:58:36.190" Score="2" ViewCount="" Body="&lt;p&gt;Suppose that $X_1 \sim {\rm N}(\mu_1,\sigma_1^2)$ and $X_2 \sim {\rm N}(\mu_2,\sigma_2^2)$ are independent. Then, $X_1 - X_2 \sim {\rm N}(\mu_1 - \mu_2 , \sigma_1^2 + \sigma_2^2)$. Set $Y = X_1 - X_2$, $\mu = \mu_1 - \mu_2$, and $\sigma^2 = \sigma_1^2 + \sigma_2^2$. Then, for any $c &amp;gt; 0$,&#xA;$$&#xD;&#xA;P(|Y| \le c) = P( - c \le Y \le c) = P\bigg(\frac{{ - c - \mu }}{{\sigma }} \le \frac{{Y - \mu }}{{\sigma }} \le \frac{{c - \mu }}{{\sigma }}\bigg).&#xD;&#xA;$$&#xD;&#xA;Noting that $(Y - \mu)/\sigma \sim {\rm N}(0,1)$, we are actually done.&lt;/p&gt;&#xA;" OwnerUserId="2810" LastActivityDate="2010-12-08T01:58:36.190" />
  <row Id="13473" PostTypeId="1" AcceptedAnswerId="13499" CreationDate="2010-12-08T02:07:52.857" Score="4" ViewCount="130" Body="&lt;p&gt;A version of the Riesz Representation Theorem says that a continuous linear functional on the space of continuous real-valued mappings on a compact metric space, $C(X)$, can be identified with a signed Borel measure on the set $X$. Are there any similar results when we replace $C(X)$ by the space of continuous functions of $X$ (compact metric) into $Y$ when (1) $Y=R^N$ or in general (2) $Y$ is a Banach space? I suspect the answer is yes, but I would like to find the right reference to start looking at. Thanks.&lt;/p&gt;&#xA;" OwnerUserId="4532" LastActivityDate="2010-12-08T07:30:37.400" Title="Riesz representation and vector-valued functions" Tags="&lt;real-analysis&gt;&lt;reference-request&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" />
  <row Id="13474" PostTypeId="2" ParentId="13469" CreationDate="2010-12-08T02:18:11.170" Score="4" ViewCount="" Body="&lt;p&gt;There is a mostly nonsensical proof in Robert M. Switzer's &lt;em&gt;Algebraic topology--homotopy and homology&lt;/em&gt;, at the end of the first chapter. (You can see it in googlebooks)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;He proves that if $X$ is an $H$-cogroup, then $[X,Y]$ is a group, and that if $Y$ is an $H$-group, then $[X,Y]$ is also a group, so when $X$ and $Y$ are an $H$-cogroups and an $H$-group, respectively, then $[X,Y]$ is a group in two ways. And so on.&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2010-12-08T02:18:11.170" CommentCount="2" />
  <row Id="13475" PostTypeId="2" ParentId="13452" CreationDate="2010-12-08T02:57:51.870" Score="4" ViewCount="" Body="&lt;p&gt;For this kind of differential equations Laplace transform is handy:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;{\cal L}[y&amp;#39;&amp;#39;] + 2 {\cal L}[y&amp;#39;] -3{\cal L}[y] = 0 \ \Longleftrightarrow \ (s^2{\cal L}[y] -sy(0) - y&amp;#39;(0)) +2(s{\cal L}[y] -y(0)) -3{\cal L}[y] = 0  \ .&#xD;&#xA;$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now put $F(s) = {\cal L}[y]$ and $a= y(0), b= y&amp;#39;(0)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;(s^2F(s) -as - b) +2(sF(s) -a) -3F(s) = 0 \ \Longleftrightarrow \ (s^2 +2s -3)F(s) -as -2a -b = 0 \ .&#xD;&#xA;$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We solve this equation for $F(s)$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;F(s) = \dfrac{as + 2a +b}{s^2 +2s -3} = \dfrac{as + 2a +b}{(s-1)(s+3)} = \dfrac{A}{s-1} + \dfrac{B}{s+3} \ ,&#xD;&#xA;$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $A$ and $B$ are constants depending on $a$ and $b$, which I let you the pleasure to compute  :-)  and we get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;y(t) = A{\cal L}^{-1}\left[ \dfrac{1}{s-1}\right] + B{\cal L}^{-1}\left[ \dfrac{1}{s+3} \right] = A e^t + B e^{-3t} \ .&#xD;&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="664" LastEditorUserId="664" LastEditorDisplayName="" LastEditDate="2010-12-09T02:34:00.027" LastActivityDate="2010-12-09T02:34:00.027" />
  <row Id="13476" PostTypeId="1" AcceptedAnswerId="13488" CreationDate="2010-12-08T03:01:13.473" Score="3" ViewCount="260" Body="&lt;p&gt;Is there a good explanation of a manifold on the web somewhere?  The wikipedia article isn't really working for me.  I was actually hoping for a whiteboard lecture on youtube, but can't find one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My math experience is calculus through differential equations, twenty five years ago.  I also have some computer science related math; discrete structures and numerical methods.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;My&lt;/em&gt; problem with the Wikipedia article is that I just can't visualize what they're saying.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any sources (not just web based) are appreciated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit 2&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I stumbled upon the Wikipedia article while trying to brush up on my math.  I had been talking to my daughter about primes, we wandered there from her 4th grade homework.  We ventured as far as Mersene primes and perfect numbers.  It was at this point that I knew I had to brush up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Somehow I chased a link to Riemann manifolds and this was completely new to me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Having looked at some of the suggested material from the answers, the Wikipedia article makes a lot more sense.  In fact I'm trying to figure out why I found it unclear, maybe it can be improved.&lt;/p&gt;&#xA;" OwnerUserId="4181" LastEditorUserId="4181" LastEditorDisplayName="" LastEditDate="2010-12-23T14:57:57.913" LastActivityDate="2010-12-23T14:57:57.913" Title="Manifold Explained" Tags="&lt;manifold&gt;" AnswerCount="2" CommentCount="5" />
  <row Id="13477" PostTypeId="1" CreationDate="2010-12-08T03:03:22.490" Score="5" ViewCount="341" Body="&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/G9pG2.png&quot; alt=&quot;diagram&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given a circle, a point $H$ outside the circle, segments $\overline{HE}$ and $\overline{HT}$ tangent to the circle at $E$ and $T$, respectively, and points $I$ and $G$ on the circle such that $I$, $G$, and $H$ are collinear (all as shown above), knowing the measures of $\angle EHG$ and $\angle GHT$ (call them $\alpha$ and $\beta$, respectively, for convenience) determines the measures of each of the four arcs on the circle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it possible to compute the measures of the minor arcs $EI$, $IT$, $TG$, and $GE$ in terms of $\alpha$ and $\beta$ without using trigonometry?&lt;/p&gt;&#xA;" OwnerUserId="72" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-12-09T21:02:24.340" LastActivityDate="2010-12-09T21:42:59.857" Title="How to determine arc measures from angles between secant and tangents (without trigonometry)" Tags="&lt;geometry&gt;&lt;trigonometry&gt;&lt;euclidean-geometry&gt;&lt;circle&gt;" AnswerCount="3" CommentCount="9" />
  <row Id="13478" PostTypeId="2" ParentId="13476" CreationDate="2010-12-08T03:05:08.180" Score="3" ViewCount="" Body="&lt;p&gt;I liked Lee's book on Manifolds. You can find the first chapter here on his website linked below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.math.washington.edu/~lee/Books/Manifolds/c1.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.math.washington.edu/~lee/Books/Manifolds/c1.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3431" LastActivityDate="2010-12-08T03:05:08.180" CommentCount="1" />
  <row Id="13479" PostTypeId="1" CreationDate="2010-12-08T03:19:14.593" Score="0" ViewCount="82" Body="&lt;p&gt;I am having issues solving the following equation:&#xA;$$\frac{x}{2x-3} - \frac{1}{2x} = \frac{3}{4x-6}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The resolution of this is 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is in the section of parabolas and it should be pretty easy to solve. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My steps are as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{x}{2x-3} - \frac{1}{2x} = \frac{3}{4x-6}$$&#xD;&#xA;Multiply by $2x(4x-6)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$4x^2 -1(4x-6) = 6x$$&#xD;&#xA;gets me here, remove $6x$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$4x^2 - 10x + 6 = 0$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solving this with the parabola formula I ended up getting $\{4,6\}$.&lt;/p&gt;&#xA;" OwnerUserId="3034" LastEditorUserId="3034" LastEditorDisplayName="" LastEditDate="2010-12-08T03:44:24.753" LastActivityDate="2010-12-08T03:57:54.377" Title="Solving an equation of terms containing a variable in their denominator" Tags="&lt;homework&gt;&lt;algebra-precalculus&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="13480" PostTypeId="2" ParentId="13479" CreationDate="2010-12-08T03:44:51.600" Score="1" ViewCount="" Body="&lt;p&gt;@J. M. :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After checking out the formula I found out that I dividing by $2$ instead of $4a$/$8$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{10 \pm \sqrt{100-96}}{2}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With the fixed form I get the results $1.5$ and $1$ of which $1$ is matching result.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for the kind support!&lt;/p&gt;&#xA;" OwnerUserId="3034" LastActivityDate="2010-12-08T03:44:51.600" CommentCount="2" />
  <row Id="13481" PostTypeId="2" ParentId="13357" CreationDate="2010-12-08T03:46:48.963" Score="2" ViewCount="" Body="&lt;p&gt;Let $G$ be a filtered commutative group and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;\dots \subset F^{s+1} \subset F^s \subset \dots&#xD;&#xA;$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;its decreasing filtration. Quoting &lt;a href=&quot;http://hopf.math.purdue.edu/Boardman/ccspseq.pdf&quot; rel=&quot;nofollow&quot;&gt;Boardman&lt;/a&gt; &quot;(...) we topologize $G$ by taking the cosets $x + F^s$ of $F^s$ for all $x \in G$ and all $s$ as basic open sets (...) A Cauchy sequence $n \mapsto x_n$ is one in which $x_m − x_n \mapsto 0$ as $m, n \mapsto \infty$.&quot; And $G$ is said to be &lt;em&gt;complete&lt;/em&gt; if every Cauchy sequence has a limit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This condition can be written in more algebraic terms as follows: for each $s$, consider the exact sequence &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;0 \longrightarrow F^s \longrightarrow G \longrightarrow G/F^s \longrightarrow 0 &#xD;&#xA;$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and apply $\varprojlim_s$ to it to obtain the long exact sequence&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;0 \longrightarrow \varprojlim_s F^s \longrightarrow G \longrightarrow \varprojlim_s G/F^s \longrightarrow \varprojlim_s{}^1F^s \longrightarrow 0 \ .&#xD;&#xA;$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then the filtration is complete if and only if $\varprojlim_s^1F^s = 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moreover: a filtration is called &lt;em&gt;Hausdorff&lt;/em&gt; if so is the topological space $G$. In Hausdorff spaces convergent sequences have unique limits. This is easily seen to be equivalent to the fact that $\varprojlim_s F^s = \bigcap_s F^s = 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in $G$, with the topology induced by the filtration, every Cauchy sequence has a unique limit if and only if $G \cong \varprojlim_s G/F^s$. This last group is called the &lt;em&gt;completion&lt;/em&gt; of $G$. (Beware: Weibel in his introduction to Homological Algebra uses the term with a different meaning; for him, &quot;complete&quot; filtration means complete and Hausdorff.)&lt;/p&gt;&#xA;" OwnerUserId="664" LastEditorUserId="664" LastEditorDisplayName="" LastEditDate="2010-12-09T03:38:19.087" LastActivityDate="2010-12-09T03:38:19.087" CommentCount="1" />
  <row Id="13482" PostTypeId="1" AcceptedAnswerId="13675" CreationDate="2010-12-08T03:51:30.113" Score="3" ViewCount="119" Body="&lt;p&gt;Let $X = (X_t: t \in [0,T])$ be a stochastic process satisfying a &lt;a href=&quot;http://en.wikipedia.org/wiki/Cox%E2%80%93Ingersoll%E2%80%93Ross_model&quot; rel=&quot;nofollow&quot;&gt;CIR model&lt;/a&gt;&#xA;$$&#xD;&#xA;dX_t = \beta (X_t - \gamma) dt + \sigma\sqrt{X_t} dB_t,&#xD;&#xA;$$&#xD;&#xA;where $B_t$ is a standard Brownian motion, $\beta$ is a negative constant, $\gamma, \sigma$ are positive constants. In order for the SDE to make sense, assume that $X_t &amp;gt; 0$ for all $t \in [0,T]$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider following two ways to simulate the model based on discretization of $t$ with Ito-Taylor expansion:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;the Euler scheme:&#xA;$$                                                                &#xD;&#xA;X_{t + \Delta} \approx X_t + \beta(X_t - \gamma)\Delta + \sigma \sqrt{X_t} Z \Delta, &#xD;&#xA;$$&#xD;&#xA;where $Z$ is $N(0, 1)$ Gaussian variable. &lt;/li&gt;&#xA;&lt;li&gt;the Milstein scheme&#xA;$$&#xD;&#xA;X_{t + \Delta} \approx X_t + \beta(X_t - \gamma)\Delta + \sigma \sqrt{X_t}Z\sqrt{\Delta} + \frac{1}{4}\sigma^2 \Delta (Z^2-1)&#xD;&#xA;$$&#xD;&#xA;where $Z$ is $N(0, 1)$ Gaussian variable. &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I was wondering why these two schemes have a positive probability of generating negative values of $X_t$ and therefore cannot be used without suitable modifications?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;References (book, tutorial and/or paper) will be helpful too!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks and regards!&lt;/p&gt;&#xA;" OwnerUserId="3348" LastEditorUserId="3348" LastEditorDisplayName="" LastEditDate="2010-12-09T15:03:28.287" LastActivityDate="2010-12-09T16:13:18.470" Title="Negativity in a CIR model discretized by Ito-Taylor expansion" Tags="&lt;probability&gt;&lt;stochastic-processes&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="13483" PostTypeId="2" ParentId="13470" CreationDate="2010-12-08T03:56:15.823" Score="2" ViewCount="" Body="&lt;p&gt;ok .. so from what you guys say: that will be: n-1*Σ from 0 to n-1 of (n/2). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;the inner bracket is O(n^2), and the outer is N, so this is the O(n^3) complexity&#xA;right?&lt;/p&gt;&#xA;" OwnerUserId="4531" LastActivityDate="2010-12-08T03:56:15.823" CommentCount="1" />
  <row Id="13484" PostTypeId="2" ParentId="13479" CreationDate="2010-12-08T03:57:54.377" Score="2" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\rm\displaystyle\quad \frac{1}{2\ x}\ =\ \frac{2\ x}{4\ x-6} - \frac{3}{4\ x-6}\ =\ \frac{1}2\ \ \Rightarrow\ \ x\ =\ \ldots\quad$ Note: no quadratic formula needed.&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-12-08T03:57:54.377" />
  <row Id="13485" PostTypeId="2" ParentId="13452" CreationDate="2010-12-08T03:58:11.970" Score="1" ViewCount="" Body="&lt;p&gt;Start by assuming a solution of the form&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$y(t)=e^{ct}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence $y&amp;#39;(t)=ce^{ct}$ and $y&amp;#39;&amp;#39;(t)=c^2e^{ct}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Substitute into the original equation. You will find a quadratic for $c$. Solve it, and note that both solutions work, and as the ODE is linear and homogeneous, the sum of solutions also works. &lt;/p&gt;&#xA;" OwnerUserId="2219" LastActivityDate="2010-12-08T03:58:11.970" />
  <row Id="13486" PostTypeId="2" ParentId="13477" CreationDate="2010-12-08T04:26:40.067" Score="1" ViewCount="" Body="&lt;p&gt;I'm within one, so I hope it prompts your thinking.  Basic geometry was a lot of years ago.  The theorems I remember are $\stackrel{\frown}{EGT} = 2\angle EIT$, $\beta=(\stackrel{\frown}{IT} - \stackrel{\frown}{TG})/2$,  and $ET=\pi-\alpha-\beta$, which you can see by drawing radii from E and T and the line from H to the center of the circle.  Then if $\angle EIG=\gamma$ we have $EG=2\gamma, GT=\pi-\alpha-\beta-2\gamma, TI=\pi-\alpha+\beta-2\gamma, IE=2\gamma+2\alpha$&lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-08T04:36:44.243" LastActivityDate="2010-12-08T04:36:44.243" CommentCount="3" />
  <row Id="13487" PostTypeId="1" CreationDate="2010-12-08T04:55:01.303" Score="7" ViewCount="86" Body="&lt;p&gt;Consider the set $A$ of natural numbers which are of the form $k^3-p$, for $k$ a positive integer and $p$ a positive prime. Does $A$ have a density (of any of the usual kinds for sets of natural numbers) and if so, what is it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$A$ contains a lot of numbers, and it seems somewhat difficult to me to prove that a particular number is not in $A$, except that most cubes are not in $A$.&lt;/p&gt;&#xA;" OwnerUserId="4535" LastActivityDate="2010-12-08T05:34:23.930" Title="Density of set of differences of cubes and primes" Tags="&lt;number-theory&gt;" AnswerCount="1" />
  <row Id="13488" PostTypeId="2" ParentId="13476" CreationDate="2010-12-08T05:31:09.570" Score="5" ViewCount="" Body="&lt;p&gt;This isn't in any way a thorough or rigorous introduction to manifolds, but as far as visualization it's quite nice: Weeks' &lt;a href=&quot;http://books.google.com/books?id=Lurp6nB4LtQC&amp;amp;printsec=frontcover&amp;amp;dq=shape+of+space&amp;amp;source=bl&quot; rel=&quot;nofollow&quot;&gt;The Shape of Space&lt;/a&gt;.  &lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-12-08T05:31:09.570" />
  <row Id="13489" PostTypeId="2" ParentId="13487" CreationDate="2010-12-08T05:34:23.930" Score="4" ViewCount="" Body="&lt;p&gt;A naive approach would be to say that the chance $n$ is equal to $k^3-p$ for a given $k$ is about $\frac{1}{\ln(k^3-n)}$.  Then the chance $n$ is not equal to $k^3-p$ for any $k$ is $$\prod_{k=\sqrt[3]{n}}^\infty {1-\frac{1}{\ln(k^3-n)}}$$  as this goes to zero, &quot;all&quot; numbers should be in A.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-12-08T05:34:23.930" />
  <row Id="13490" PostTypeId="1" AcceptedAnswerId="13494" CreationDate="2010-12-08T06:14:37.940" Score="8" ViewCount="226" Body="&lt;p&gt;Here is an exercise, on analysis which i am stuck. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;How do I prove that if $F_{n}(x)=\sum\limits_{k=1}^{n} \frac{\sin{kx}}{k}$, then the sequence $\{F_{n}(x)\}$ is boundedly convergent on $\mathbb{R}$?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-08T06:33:03.857" LastActivityDate="2010-12-08T06:54:02.943" Title="Proving that the sequence $F_{n}(x)=\sum\limits_{k=1}^{n} \frac{\sin{kx}}{k}$ is boundedly convergent on $\mathbb{R}$" Tags="&lt;real-analysis&gt;&lt;analysis&gt;&lt;sequences-and-series&gt;" AnswerCount="1" FavoriteCount="3" />
  <row Id="13491" PostTypeId="2" ParentId="5944" CreationDate="2010-12-08T06:31:59.413" Score="1" ViewCount="" Body="&lt;p&gt;With $x=\frac{1}{2}\tan\theta$ and $dx=\frac{1}{2}\sec^2\theta d\theta$, $2\pi\int x^2\sqrt{1+4x^2}dx=\frac{\pi}{4}\int\tan^2\theta\sec^3\theta d\theta$.  Now, let&#xA;$$\begin{align}&#xD;&#xA;I_1&amp;amp;=\int\tan^2\theta\sec^3\theta d\theta&#xD;&#xA;\\&#xD;&#xA;&amp;amp;=\int(\sec^3\theta-\sec\theta)\sec^2\theta d\theta&#xD;&#xA;\\&#xD;&#xA;&amp;amp;=\tan\theta(\sec^3\theta-\sec\theta)-\int3\sec^3\theta\tan^2\theta-\sec\theta\tan^2\theta d\theta&#xD;&#xA;\\&#xD;&#xA;&amp;amp;=\tan\theta(\sec^3\theta-\sec\theta)-3I_1+\int\sec\theta\tan^2\theta d\theta&#xD;&#xA;\end{align}$$&#xD;&#xA;(using the integration by parts as you suggested) so that $I_1=\frac{1}{4}\left(\tan\theta(\sec^3\theta-\sec\theta)+\int\sec\theta\tan^2\theta d\theta\right)$.  Let&#xA;$$\begin{align}&#xD;&#xA;I_2&amp;amp;=\int\sec\theta\tan^2\theta d\theta&#xD;&#xA;\\&#xD;&#xA;&amp;amp;=\sec\theta\tan\theta-\int\sec^3\theta d\theta&#xD;&#xA;\\&#xD;&#xA;&amp;amp;=\sec\theta\tan\theta-\int\sec^3\theta-\sec\theta d\theta-\int\sec\theta d\theta&#xD;&#xA;\\&#xD;&#xA;&amp;amp;=\sec\theta\tan\theta-I_2+\log|\sec\theta-\tan\theta|&#xD;&#xA;\end{align}$$&#xD;&#xA;(using integration by parts with $u=\tan\theta$ and $dv=\sec\theta\tan\theta d\theta$) so that $I_2=\frac{1}{2}\left(\sec\theta\tan\theta+\log|\sec\theta-\tan\theta|\right)$.  Now,&#xA;$$\begin{align}&#xD;&#xA;\frac{\pi}{4}\int\tan^2\theta\sec^3\theta d\theta&#xD;&#xA;&amp;amp;=\frac{\pi}{4}I_1=\frac{\pi}{16}\left(\tan\theta(\sec^3\theta-\sec\theta)+\int\sec\theta\tan^2\theta d\theta\right)&#xD;&#xA;\\&#xD;&#xA;&amp;amp;=\frac{\pi}{16}\left(\tan\theta(\sec^3\theta-\sec\theta)+\frac{1}{2}\left(\sec\theta\tan\theta+\log|\sec\theta-\tan\theta|\right)\right)&#xD;&#xA;\\&#xD;&#xA;&amp;amp;=\frac{\pi}{16}\left(\tan\theta\sec^3\theta-\frac{1}{2}\sec\theta\tan\theta+\frac{1}{2}\log|\sec\theta-\tan\theta|\right)&#xD;&#xA;\\&#xD;&#xA;&amp;amp;=\frac{\pi}{16}\left(2x(1+4x^2)^{3/2}-x\sqrt{1+4x^2}+\frac{1}{2}\log\left|\sqrt{1+4x^2}-2x\right|\right)&#xD;&#xA;\end{align}$$&#xD;&#xA;which is similar to what you have, but not the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Of course, doing this by hand and covering a sheet of paper with scribbles to do the two instances of over-and-back integration by parts, I may well have made a mistake along the way, but my result does appear to check by the method described in &lt;a href=&quot;http://math.stackexchange.com/questions/5944/lateral-surface-area-by-hand/5978#5978&quot;&gt;aschepler's answer&lt;/a&gt;.)&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-12-08T06:31:59.413" />
  <row Id="13492" PostTypeId="2" ParentId="13460" CreationDate="2010-12-08T06:34:07.320" Score="2" ViewCount="" Body="&lt;p&gt;I generally have a &quot;top down approach&quot; to reading papers. More often than not, I am driven to a particular paper looking for a particular result. So I would start by looking at the statement of the result and try to see what background is needed in order to parse the statements which appear there. Once I have understood the philosophy of the result, I think about/look at the proof. Again, if I come across something unfamiliar, I try to acquaint myself with the relevant material on an as needed basis. I find this approach works better for me if I am interested in a paper for isolated results. Reading such papers from introduction to conclusion might not be of immediate (or even future) value.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other kinds of papers I read would include survey papers (exposition of a particular new idea) and papers which essentially develop a new idea/concept/structure for the first time. A lot of these papers, I find, can (and eventually should) be read from front matter to bibliography. So I would essentially read these papers as I would read a textbook.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One thing which I find especially important while reading papers is to look at the cross references. This can be useful for motivation, alternate proofs and related ideas. It also gets me acquainted with who else is working in the area and the exposure to ideas from different mathematicians achieved thus is a nice side effect. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, I feel it is important to ask a lot of questions while I read papers. Why is hypothesis X necessary. Can condition Y be weakened? Can this result be generalized to Z structure? This is eventually useful when I write my own results, to conceive the most general form that I can.&lt;/p&gt;&#xA;" OwnerUserId="3431" LastActivityDate="2010-12-08T06:34:07.320" CommunityOwnedDate="2010-12-08T06:34:07.320" />
  <row Id="13493" PostTypeId="2" ParentId="6268" CreationDate="2010-12-08T06:43:12.357" Score="1" ViewCount="" Body="&lt;p&gt;Note, you also need the fact that $f(\mathbb{R})$ is borel. This fortunately is so, as $\mathbb{R}$ is $\sigma$-compact, and the continuous image of a compact set is compact, thus closed (since we are working in hausdorff spaces), thus borel.&lt;/p&gt;&#xA;" OwnerUserId="4537" LastActivityDate="2010-12-08T06:43:12.357" />
  <row Id="13494" PostTypeId="2" ParentId="13490" CreationDate="2010-12-08T06:47:10.750" Score="10" ViewCount="" Body="&lt;p&gt;First, let's note that for $x\in(0,2\pi)$&#xD;&#xA;$$\sum\limits_{k=1}^{n}\frac{\sin kx}{k}=\int_{0}^{x}\sum\limits_{k=1}^{n}\cos kt\ dt=&#xD;&#xA;-\frac{x}{2}+\int_{0}^{x}\frac{\sin \frac{(2n+1)t}{2}}{2\sin{\frac{t}{2}}}\ dt$$&#xD;&#xA;$$=-\frac{x}{2}+\int_{0}^{x}\left(\frac{1}{2\sin{\frac{t}{2}}}-\frac{1}{t}\right)\sin \frac{(2n+1)t}{2}\ dt +\int_{0}^{x}\frac{\sin \frac{(2n+1)t}{2}}{t}dt.$$&#xD;&#xA;Now, the first integral at the right-hand side tends to zero by the Riemann-Lebesgue lemma.&#xA;The second one is equal to (via a substitution $s=(2n+1)t/2$) the integral&#xA;$$\int_{0}^{\frac{(2n+1)x}{2}}\frac{\sin s}{s}\ ds\to\int_{0}^{\infty}\frac{\sin s}{s}\ ds=\frac{\pi}{2}.$$&#xD;&#xA;Therefore&#xA;$$\lim\limits_{n\to\infty}\ \sum\limits_{k=1}^{n}\frac{\sin kx}{k}=\frac{\pi-x}{2}=f(x),\qquad x\in(0,2\pi).$$&#xD;&#xA;The series converges on $\mathbb R$ to the periodic extension of $f(x)$. &lt;/p&gt;&#xA;" OwnerUserId="723" LastEditorUserId="723" LastEditorDisplayName="" LastEditDate="2010-12-08T06:54:02.943" LastActivityDate="2010-12-08T06:54:02.943" CommentCount="3" />
  <row Id="13495" PostTypeId="1" CreationDate="2010-12-08T06:49:05.960" Score="1" ViewCount="133" Body="&lt;p&gt;Here is a problem in Griffiths Introduction to Electrodynamics as follows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Check the divergence theorem for the function $\mathbf{v} = r^2\mathbf{\hat{r}}$, using as your volume, the sphere of radius R, centered at the origin?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is my calculation for the surface intergral part, please help me finding the error in it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathbf{v} = r^2(\sin\theta \cos\phi \ \mathbf{\hat{x}} + \sin\theta \sin\phi \ \mathbf{\hat{y}} + \cos \theta \ \mathbf{\hat{z}})$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;\oint_s \mathbf{v} \cdot d\mathbf{a} &amp;amp;= 2 \iint r^2 \cos\theta\ dxdy  \\&lt;br&gt;&#xA;&amp;amp;= 2 \iint r^2 \cdot \frac{z}{r} dxdy \\&lt;br&gt;&#xA;&amp;amp;= 2R \iint \sqrt{R^2 - x^2 - y^2} dxdy \\&lt;br&gt;&#xA;&amp;amp;= 2R \int_{r=0}^{R}\sqrt{R^2 - r^2} rdr \int_{\theta=0}^{2\pi} d\theta \\&lt;br&gt;&#xA;&amp;amp;= 4\pi R\int_{\theta=0}^{\frac{\pi}{2}}\sqrt{R^2 - R^2{\sin^2\theta}} R\sin\theta\ R\cos\theta\ d\theta  \\&lt;br&gt;&#xA;&amp;amp;= \frac{4\pi R^4}{3}&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the right answer should be $4\pi R^4$ considering the left part of the divergence theorem $\int_V \nabla\cdot\mathbf{v}\ d\tau$. I have check my answer serveral times but could not find error, could you help me?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;thanks.&lt;/p&gt;&#xA;" OwnerUserId="1382" LastEditorUserId="1382" LastEditorDisplayName="" LastEditDate="2010-12-08T07:39:59.953" LastActivityDate="2010-12-08T07:39:59.953" Title="What's wrong with my calculation for checking the divergence law?" Tags="&lt;vector-calculus&gt;" CommentCount="4" />
  <row Id="13496" PostTypeId="2" ParentId="8808" CreationDate="2010-12-08T06:53:06.703" Score="2" ViewCount="" Body="&lt;p&gt;I'm interested in differential topology, so that baises my answer.  $\mathbb{R}$, and Euclidean spaces generally, are interesting because you can do calculus on them.  Manifolds are interesting because, not only as OrbiculaR says, you can do standard analysis in a chart, but you can sensibly extend the concepts of analysis to the entire manifold.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can also do analysis in the p-adics, so you might wonder why aren't studying p-adic manifolds.  (I don't actually know.)&lt;/p&gt;&#xA;" OwnerUserId="4074" LastActivityDate="2010-12-08T06:53:06.703" CommentCount="1" />
  <row Id="13497" PostTypeId="2" ParentId="10593" CreationDate="2010-12-08T07:07:36.730" Score="1" ViewCount="" Body="&lt;p&gt;For the 2-sphere, it isn't hard to show it directly.  However, in general, you'll need more powerful tools.  A manifold $M$ is simply-connected if it is path-connected and if $\pi_1(M) = 1$.  The standard tool to compute $\pi_1$ is the van Kampen theorem.&lt;/p&gt;&#xA;" OwnerUserId="4074" LastActivityDate="2010-12-08T07:07:36.730" />
  <row Id="13498" PostTypeId="2" ParentId="13412" CreationDate="2010-12-08T07:17:14.550" Score="2" ViewCount="" Body="&lt;p&gt;The Zero-Product Property ($ab=0\implies a=0\text{ or }b=0$) is the mechanics behind going from $(7y-1)(y-7)=0$ to $y=\frac{1}{7}$ or $y=7$.  There is no directly analogous property for inequalities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The method that I'd suggest for examining inequalities that compare an expression to 0 is the boundary algorithm—find the values that make the expression equal to 0, which are the boundary points of a set of intervals on the number line, then test each interval to see if it satisfies the original equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your specific example, the boundary points are $y=\frac{1}{7}$ and $y=7$, so test some value of $y$ below $y=\frac{1}{7}$ (for example, $y=0$), some value of $y$ between $y=\frac{1}{7}$ and $y=7$ (for example, $y=\frac{1}{2}$), and some value of $y$ above $y=7$ (for example, $y=10$).  The original inequality is false below $y=\frac{1}{7}$ and above $y=7$ and true between $y=\frac{1}{7}$ and $y=7$.  Since the original inequality included = (it was ≤), the solution includes the boundary points that solved the corresponding equation, so the solution is $\frac{1}{7}\le y\le 7$.&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-12-08T07:17:14.550" />
  <row Id="13499" PostTypeId="2" ParentId="13473" CreationDate="2010-12-08T07:23:42.367" Score="2" ViewCount="" Body="&lt;p&gt;Yes, there are similar results in the vector-valued case.  Dunford and Schwartz is a standard reference for this kind of thing.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some notation: $X$ is a fixed compact Hausdorff space.  For a Banach space $Y$, the space of continuous functions from $X$ to $Y$, endowed with the supremum norm coming from the norm of $Y$, I denote by $C(X,Y)$.  For a Banach space $Z$, I denote its dual by $Z&amp;#39;$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is one way to think about the dual of $C(X,Y)$ for $Y$ a Banach space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An element $\phi$ of $C(X,Y)&amp;#39;$ gives rise to a family of measures on $X$ parametrized by $Y$ in the following way.  Fixing $\xi \in Y$, one can define a linear functional $L_{\phi,\xi}$ on $C(X)$ by sending the function $f$ on $X$, to the value of $\phi$ on the function $X \to Y$ given by $x \mapsto f(x) \xi$.  In symbols:&#xA;$$&#xD;&#xA;L_{\phi,\xi}(f) = \phi(x \mapsto f(x) \xi).&#xD;&#xA;$$&#xD;&#xA;From the usual Riesz theorem, there is then a measure $m_{\phi, \xi}$ defined on the Borel subsets of $X$ satisfying&#xA;$$&#xD;&#xA;L_{\xi,\phi}(f) = \int_X f \, dm_{\phi, \xi}.&#xD;&#xA;$$&#xD;&#xA;So from $\phi$ we have produced a family of measures on $X$, one for each $\xi$ in $Y$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now define a map $m_{\phi}$ from the Borel subsets of $X$ to $Y&amp;#39;$ as follows: for any Borel subset $E$ of $X$, define $m_{\phi}(E)$ to be the linear functional on $Y$ given by&#xA;$$&#xD;&#xA;m_{\phi}(E)(\xi) = \int_E 1 \, dm_{\phi, \xi}.&#xD;&#xA;$$&#xD;&#xA;The map $m_{\phi}$ has various nice properties (it is a $Y&amp;#39;$-valued analogue of a regular signed Borel measure on $X$).  Since the functions of the form $x \mapsto f(x) \xi$, with $f \in C(X)$ and $\xi \in Y$, are dense in $C(X,Y)$, it is easy to show that $\phi$ is uniquely determined by $m_{\phi}$.  (The intuition is to think of $\phi$ as coming from $m_{\phi}$ as follows: for each $f \in C(X,Y)$, the number $\phi(f)$ is obtained &quot;by integrating, over $X$, the values of $f$ with respect to the $Y&amp;#39;$-valued measure $m_{\phi}$, so that $\phi(f) = \int_X f \, dm_{\phi}$.&quot;  You can think of this just as a formal thing, or, think enough about the integration of vector-valued functions with respect to vector-valued set mappings like $m_{\phi}$ to formalize this and remove the quotation marks.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, you can reverse this whole chain of reasoning: starting with a map from the Borel sets of $X$ to $Y&amp;#39;$ with nice enough properties, you can show that it must be $m_{\phi}$ for some $\phi$ in $C(X,Y)&amp;#39;$.  There is a natural notion of norm for these things (a &quot;variation&quot; norm) and it turns out to coincide with the norm you'd get from $C(X,Y)&amp;#39;$.  So the dual of $C(X,Y)$, in this picture, is a space of nicely behaved $Y&amp;#39;$-valued mappings on the Borel subsets of $X$, with a certain variation norm.   When $Y$ is the scalars this is turns into the original Riesz theorem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More generally you can think of any bounded linear map from $C(X,Y)$ into a Banach space $Z$ in similar terms, but things get more complicated (the &quot;measure-like&quot; things you integrate over $X$ to represent maps $C(X,Y) \to Z$ take values in the linear operators from $Y$ to $Z''$).  You can also weaken various hypotheses here (e.g. you can drop the compactness hypothesis on $X$, or replace $Y$ with a more general topological linear space, provided that you are willing to make additional complicated hypotheses in order to state a decent theorem).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is another direction you can go.  If $X$ is compact Hausdorff and $Y$ is a Banach space, the space $C(X,Y)$ is isometrically isomorphic to a certain tensor product, namely the &lt;em&gt;injective Banach space tensor product&lt;/em&gt;, of $C(X)$ with $Y$.  So identifying the dual of $C(X,Y)$ is a special case of identifying the dual of an injective tensor product $A \otimes_i B$ of Banach spaces $A$ and $B$.  The dual of this tensor product has various characterizations.  One is in terms of Borel measures on the Cartesian product of the compact topological spaces $(A&amp;#39;)_1$ and $(B&amp;#39;)_1$ (the unit balls of the duals of $A$ and $B$, given the weak-$*$ topology).  Any book on Banach spaces that discusses the tensor product theory will have theorems about the injective Banach space tensor product and how duality interacts with it.&lt;/p&gt;&#xA;" OwnerUserId="3840" LastEditorUserId="3840" LastEditorDisplayName="" LastEditDate="2010-12-08T07:30:37.400" LastActivityDate="2010-12-08T07:30:37.400" CommentCount="4" />
  <row Id="13500" PostTypeId="1" AcceptedAnswerId="13527" CreationDate="2010-12-08T07:31:34.580" Score="8" ViewCount="271" Body="&lt;p&gt;I was Looking at another questions title, and given the tag of DG, I thought it would read a little more like this one. Or at least that answers to this question would be answers to that question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are many different techniques one might use to show a manifold is simply conencted, I am only interested in a specific brand of them: those involving Differential Geometry. I borrowed a book from my friends physics professor, I think it was called Comparison theorems in Riemannian Geometry by Cheeger and Ebin, and the content was foreign to me. The goal of the book, if I recall properly, seemed to be proving theorems about $\pi_1(M)$ by examining geometric quantities like curvature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is amazing to me, and I am curious what the main idea could be. I currently only know of one relationship between geometric quantities and homotopy invariants, that is the fabled Chern-Weil theory. (fabled because I don't understand it... yet)&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;What are the precise statements of such types  of results?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Which parts of the hypothesis do what? (for example what does compactness help you do here since it doesn't do a whole lot on the homotopy side(maybe that is wrong))&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Other than other evidence/theorems, like Gauss-Bonnet, why would someone expect such results?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Can we get interesting results going the other way? How does $\pi_1$ affect things like curvature?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Can we say anything about higher homotopy groups?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;thanks for your patience!&lt;/p&gt;&#xA;" OwnerUserId="627" LastActivityDate="2011-04-19T08:28:34.420" Title="How to prove a manifold is simply connected?... using geometry" Tags="&lt;differential-geometry&gt;&lt;algebraic-topology&gt;&lt;homotopy-theory&gt;&lt;curvature&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="1" />
  <row Id="13501" PostTypeId="2" ParentId="13500" CreationDate="2010-12-08T08:07:43.473" Score="12" ViewCount="" Body="&lt;p&gt;One example of how geometry affects $\pi_1$: a compact negatively curved manifold will have infinite fundamental group.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proof: Negative curvature implies that the universal cover of such a manifold $M$ is diffeomorphic to $\mathbb R^n$.  If $\Gamma = \pi_1(M)$,&#xA;then $M = \mathbb R^n/\Gamma$, and so since $M$ is compact, the group $\Gamma$ must be infinite.  QED&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Speaking more vaguely, negative curvature makes geodesics diverge, and so tends to create a lot of $\pi_1$ (when the manifold is also compact, so that geodesics are forced to wrap around, and so have a chance to close up).  On the other hand, positive curvature focuses geodesics, and so will tend to make $\pi_1$ smaller.&#xA;A result somewhat related to this philosophy (that positive curvature implies simpler topology) is Gromov's theorem that the Betti numbers of  a non-negatively curved manifold are uniformly bounded (the bound depends on the dimension, but nothing else).  (&lt;a href=&quot;http://www.aimath.org/WWN/nnsectcurvature/WilkingSurvey.pdf&quot; rel=&quot;nofollow&quot;&gt;This survey paper by Wilking&lt;/a&gt; gives&#xA;a nice discussion of this result, and many related results about non-negative curvature and its implications for topology.)&lt;/p&gt;&#xA;" OwnerUserId="221" LastEditorUserId="221" LastEditorDisplayName="" LastEditDate="2011-04-19T08:28:34.420" LastActivityDate="2011-04-19T08:28:34.420" CommentCount="1" />
  <row Id="13502" PostTypeId="2" ParentId="13345" CreationDate="2010-12-08T08:16:16.730" Score="0" ViewCount="" Body="&lt;p&gt;Hello Pete,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;thanks for the (counter-)example. Yes, there's a few basic results (i) if $X$ is compact dann es sei metrisable $\Leftrightarrow$ es sei hausdorff and second countable; (ii) compact metrisable implies polish (obviously!); (iii) any uncountable polish space has cardinality $2^{\aleph_{0}}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So your claim runs through, provided $Ult(\omega)$ the space of ultrafilters over $\omega$, does indeed have cardinality $&amp;gt;2^{\aleph_{0}}$. However I'll have to run a few calculations to check this is the case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nevertheless, it is certain that it is consistent to assume that compact hausdorff separable does not imply polish ($\Leftrightarrow$ second countable in this context).&lt;/p&gt;&#xA;" OwnerUserId="4537" LastActivityDate="2010-12-08T08:16:16.730" CommentCount="1" />
  <row Id="13503" PostTypeId="1" CreationDate="2010-12-08T08:36:31.227" Score="2" ViewCount="185" Body="&lt;p&gt;I have a circle of radius $r$, and I have a bounding box with 4 sides, $a$, $b$, $c$ and $d$ as illustrated in the below diagram:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://docs.google.com/drawings/pub?id=1_TelXrp1fyoK1XExygptKEl3ogy_jKPpFCpAhNKYoaA&amp;amp;w=960&amp;amp;h=720&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are a few constraints&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$r \geq a$$&#xD;&#xA;$$r \geq b$$&#xD;&#xA;$$c &amp;gt;&amp;gt; a$$&#xD;&#xA;$$d &amp;gt;&amp;gt; a$$&#xD;&#xA;$$c &amp;gt;&amp;gt; b$$&#xD;&#xA;$$d &amp;gt;&amp;gt; b$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a$ and $b$ must be perpendicular to $d$ segment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another condition is that the the circular sectors must touch one of the long line segment ( the top line segment in this case), base on another long segment ( the bottom line segment in this case) and must not go out from it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the algorithm/equation that allows me to generate all the circular sectors that &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Touch one of the line segments&lt;/li&gt;&#xA;&lt;li&gt;Won't go out from it&lt;/li&gt;&#xA;&lt;li&gt;Fill the bounding box ( as defined by $a$, $b$, $c$ and $d$) as compactly as possible?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="550" LastEditorUserId="550" LastEditorDisplayName="" LastEditDate="2010-12-09T02:25:03.897" LastActivityDate="2011-05-08T04:50:31.747" Title="Algorithm to Pack as Many Circular Sectors into a Rectangular Box" Tags="&lt;geometry&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="13504" PostTypeId="1" AcceptedAnswerId="13512" CreationDate="2010-12-08T08:39:05.407" Score="30" ViewCount="718" Body="&lt;p&gt;This is a very simple question but I believe it's nontrivial.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to know if the following is true: If $R$ and $S$ are rings and $R[x]$ and $S[x]$ are isomorphic as rings, then $R$ and $S$ are isomorphic. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there isn't a proof (or disproof) of the general result, I would be interested to know if there are particular cases when this claim is true. &lt;/p&gt;&#xA;" OwnerUserId="4457" LastActivityDate="2010-12-11T01:52:34.893" Title="Does $R[x] \cong S[x]$ imply $R \cong S$?" Tags="&lt;abstract-algebra&gt;" AnswerCount="4" FavoriteCount="7" />
  <row Id="13506" PostTypeId="2" ParentId="13504" CreationDate="2010-12-08T08:49:46.230" Score="24" ViewCount="" Body="&lt;p&gt;I found &lt;a href=&quot;http://www.springerlink.com/content/n41677w1xj417462/&quot; rel=&quot;nofollow&quot;&gt;this paper&lt;/a&gt; by Brewer and Rutter that discusses&#xA;related matters. They cite a forthcoming paper by Hochster&#xA;which proves there are non-isomorphic&#xA;commutative integral domains $R$ and $S$ with $R[x]\cong S[x]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hochster's paper is&#xA;M. Hochster, Nonuniqueness of coefficient rings in a polynomial ring&#xA;Proc. Amer. Math. Soc. 34 (1972), 81-82, and is &lt;a href=&quot;http://www.ams.org/journals/proc/1972-034-01/S0002-9939-1972-0294325-3/home.html&quot; rel=&quot;nofollow&quot;&gt;freely available&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="226" LastEditorUserId="226" LastEditorDisplayName="" LastEditDate="2010-12-08T09:34:27.707" LastActivityDate="2010-12-08T09:34:27.707" />
  <row Id="13507" PostTypeId="1" AcceptedAnswerId="13517" CreationDate="2010-12-08T09:28:21.233" Score="5" ViewCount="230" Body="&lt;p&gt;I saw an exercise which was meant to find integers all integers $m,n$ satisfying $2m^2 + 5n^2 = 11(mn-11)$. I found them using the factorization $(m-5n)(2m-n)=-11\cdot 11$. However, what kind of methods there are to solve the original problem, find all $(m,n)\in\mathbb{Z}\times\mathbb{Z}$ satisfying $2m^2+5n^3=11(mn-11)$? I haven't solved many cubic Diophantine equations so I was just wondering if there is some birational transformation to convert the equation to a Weierstrass form of an elliptic curve.&lt;/p&gt;&#xA;" OwnerUserId="2856" LastEditorUserId="2856" LastEditorDisplayName="" LastEditDate="2010-12-08T13:25:25.780" LastActivityDate="2010-12-13T01:58:47.853" Title="Third degree Diophantine equation" Tags="&lt;number-theory&gt;&lt;diophantine-equations&gt;" AnswerCount="1" />
  <row Id="13508" PostTypeId="2" ParentId="13503" CreationDate="2010-12-08T12:18:14.137" Score="0" ViewCount="" Body="&lt;p&gt;There is quite a bit of information on circle packing on &lt;a href=&quot;http://www.packomania.com/&quot; rel=&quot;nofollow&quot;&gt;packomania&lt;/a&gt;, including packings in rectangles. So some of it may be of use to you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's &lt;a href=&quot;http://www.critticall.com/SQU_cir.html&quot; rel=&quot;nofollow&quot;&gt;another page&lt;/a&gt; on the same theme.&lt;/p&gt;&#xA;" OwnerUserId="1301" LastActivityDate="2010-12-08T12:18:14.137" CommentCount="2" />
  <row Id="13509" PostTypeId="1" CreationDate="2010-12-08T13:37:02.357" Score="0" ViewCount="32" Body="&lt;p&gt;Let $X$ be a compact orientable surface, and let $x \in X$. Let $P_{x,x}X$ be the space of loops in $X$ based at $x$.  Then we have the pairing of the Chen complex with the group algebra for the fundamental group:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$  \int: H^0(Ch^{\leq n}(P_{x,x}X), \mathbb Q) \otimes \mathbb{Q}[\pi_1(X,x)]/J_x^{n+1} \to  \mathbb{C} .$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a nice theorem; but here the choice of a base point is involved. Can a more general statement can be made without preference to any particular basepoint? For instance, considering the fundamental groupoid instead of the fundamental group?&lt;/p&gt;&#xA;" OwnerUserId="4543" LastActivityDate="2010-12-08T13:37:02.357" Title="Base-point free version of Chen's theorem" Tags="&lt;algebraic-topology&gt;" />
  <row Id="13510" PostTypeId="2" ParentId="13504" CreationDate="2010-12-08T13:46:14.147" Score="1" ViewCount="" Body="&lt;p&gt;Along the lines of &quot;particular cases where the claim is true,&quot; if $R$ and $S$ are fields, then they must be isomorphic since they are distinguishable as the units (excepting zero of course) of $R[x]$ and $S[x]$, respectively, and an isomorphism conserves units.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More generally if we only know that one of R or S is a field, the claim is still true.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose $R$ is a field and $R[X] \cong S[Y]$, where $X,Y$ are distinct indeterminates for clarity, and $f:R[X] \rightarrow S[Y]$ is an isomorphism.  Since the nonzero elements of $f(R)$ are units in $S[Y]$, they must be degree zero, i.e. elements of $S$.  Also the inverse image of $S$ is a subring of Euclidean domain $R[X]$, so $S$ is an integral domain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Knowing that $f(R) \subseteq S$, we need only show $f$ maps $R$ onto $S$.  Suppose $s \in S$.  Then there exists polynomial $p(X) \in R[X]$ s.t. $f(p(X)) = s$.  Let $p(X) = \sum_{i=0}^n \enspace r_i X^i$ where coefficients $r_i \in R$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now $f(X) = q(Y)$ for some polynomial $q(Y) \in S[Y]$, and degree of $q(Y)$ must be positive for $f(R[X]) = S[Y]$.  Appy $f$ to $p(X)$ and compare degrees:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$s = \sum_{i=0}^n \enspace f(r_i) q(Y)^i$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $S$ is an integral domain, the degrees of $q(Y)^i$ are positive for $i \gt 0$.  Thus the only nonzero coefficient of $p(X)$ is $r_0$, which shows $f(r_0) = s$.  Therefore $f$ maps $R$ onto $S$ and $R \cong S$.&lt;/p&gt;&#xA;" OwnerUserId="3111" LastEditorUserId="3111" LastEditorDisplayName="" LastEditDate="2010-12-11T01:52:34.893" LastActivityDate="2010-12-11T01:52:34.893" />
  <row Id="13511" PostTypeId="1" AcceptedAnswerId="13561" CreationDate="2010-12-08T13:52:27.997" Score="1" ViewCount="71" Body="&lt;p&gt;Hello All, I have just discussed with student and he told me him task, and I think it is interesting.&#xA;The task.&#xA;There are file with points like:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Point0: x=1; y=4;&#xA;Point1: x=199; y=45;&#xA;Point2: x=42; y=333;&#xA;Point3: x=444; y=444;&#xA;...&#xA;PointN: x=nnn; y=mmm;&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;You should find polygons and draw them. Each polygon present as internal&#xA;I mean something like this:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;---------&#xA;| -----  |&#xA;| |    | |&#xA;| |----| | &#xA;|        |&#xA;|--------|&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;And question what algorithm can you advice to use in this case?&#xA;I understand this is from graph theory , but want opinion of other.&#xA;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="4544" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-08T13:57:41.727" LastActivityDate="2010-12-08T20:59:42.807" Title="Algorithm to find figures using input points." Tags="&lt;geometry&gt;&lt;algorithms&gt;" AnswerCount="5" CommentCount="2" />
  <row Id="13512" PostTypeId="2" ParentId="13504" CreationDate="2010-12-08T13:54:20.233" Score="16" ViewCount="" Body="&lt;p&gt;Here is a counterexample.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $R=\frac{\mathbb{C}[x,y,z]}{(xy(1-z^2))}$, $S=\frac{\mathbb{C}[x,y,z]}{(x^2y(1-z^2))}$. Then, $R$ is not isomorphic to $S$ but, $R[T]\cong S[T]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In many variables, this is called the Zariski problem or cancellation of indeterminates and is largely open. Here is a discussion by Hochster (problem 3)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.math.lsa.umich.edu/~hochster/Lip.text.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.math.lsa.umich.edu/~hochster/Lip.text.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3431" LastEditorUserId="3431" LastEditorDisplayName="" LastEditDate="2010-12-09T00:18:05.843" LastActivityDate="2010-12-09T00:18:05.843" />
  <row Id="13513" PostTypeId="2" ParentId="13500" CreationDate="2010-12-08T14:00:17.387" Score="3" ViewCount="" Body="&lt;p&gt;Brendle and Schoen use Ricci flow to prove a smooth version of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Sphere_theorem&quot; rel=&quot;nofollow&quot;&gt;Sphere Theorem&lt;/a&gt;: An n-manifold with pinched positive curvature is &lt;em&gt;diffeomorphic&lt;/em&gt; to a quotient of the $n$-sphere. (In particular, the fundamental group is finite.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Very much in the other direction (ie regarding part four of your question) &lt;a href=&quot;http://www.math.uchicago.edu/~shmuel/&quot; rel=&quot;nofollow&quot;&gt;Shmuel Weinberger&lt;/a&gt; and co-authors have results of the form: Suppose that $M$ is a compact Riemannian manifold so that $\pi_1(M)$ has unsolvable word problem.  Then the set of geodesics in $M$ is &quot;wildly complicated&quot;.  Some of this material is discussed in his book &quot;Computers, Rigidity, and Moduli: The Large Scale Geometry of Riemannian Moduli Space&quot;.&lt;/p&gt;&#xA;" OwnerUserId="1307" LastEditorUserId="1307" LastEditorDisplayName="" LastEditDate="2010-12-08T15:35:31.370" LastActivityDate="2010-12-08T15:35:31.370" CommentCount="5" />
  <row Id="13514" PostTypeId="1" AcceptedAnswerId="13516" CreationDate="2010-12-08T14:02:13.813" Score="4" ViewCount="138" Body="&lt;p&gt;In &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/1568810938&quot; rel=&quot;nofollow&quot;&gt;Goldstern and Judah's The Incompleteness Phenomenon&lt;/a&gt; we are asked to prove that any model of the first two Peano Axioms: &#xA;$$\forall x [Sx\neq0]$$  $$\forall x\forall y[Sx=Sy\implies x=y]$$&#xD;&#xA;must be infinite.  S is a unary function intended as successor.  If the model is finite there must be $m$ and $n$ with $S^m0=S^n0$ where the superscript represents iteration.  Then $|n-m|$ applications of the second axiom lead to a contradiction with the first.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We are clearly arguing in the metatheory, as our axioms have no ideas of infinite sets.  Is there a clear definition of what axioms and logic are allowed here?  Can you suggest a book on the subject?&lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2010-12-08T14:44:14.430" LastActivityDate="2010-12-12T07:14:34.147" Title="Logic in the metatheory" Tags="&lt;logic&gt;&lt;reference-request&gt;&lt;model-theory&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="1" />
  <row Id="13515" PostTypeId="2" ParentId="13511" CreationDate="2010-12-08T14:34:34.553" Score="0" ViewCount="" Body="&lt;p&gt;This is not an answer, since the problem statement is kind of vague, here is what I assume it is.Assuming a polygon immediately inside a polygon to be the child of the polygon (and similarly for parent)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;maximize the number of points utilized to draw polygons such that&lt;br&gt;&lt;br&gt;&#xA;1. each polygon contains only one child polygon (no siblings)&lt;br&gt;&#xA;2. each polygon has only one immediate parent (doesn't lie in the intersection of 2 or more parents)&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, outermost and innermost polygons do not need to satify conditions 2 and 1 respectively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again J.M.'s question is important - do you mean polygons or quadrilateral?&lt;/p&gt;&#xA;" OwnerUserId="4545" LastEditorUserId="4545" LastEditorDisplayName="" LastEditDate="2010-12-08T14:42:57.177" LastActivityDate="2010-12-08T14:42:57.177" CommentCount="2" />
  <row Id="13516" PostTypeId="2" ParentId="13514" CreationDate="2010-12-08T14:38:09.193" Score="4" ViewCount="" Body="&lt;p&gt;There are two ways to handle this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most common way is simply to assume some form of set theory in the metatheory, e.g. ZFC. Then you can use all the normal semantic and model-theoretic methods that you are used to. The advantages of this method are that it matches the way that mathematicians actually think about things, and that it allows us to use all the theorems we're used to. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another option is to reinterpret claims made in the metatheory as syntactic claims. For example, the claim &quot;the model must be infinite&quot; can be reinterpreted to mean that the theory itself (!) proves each of the sentences&#xA;$$ \lnot (\exists c_0) (\forall x) (x = c_0) $$&#xD;&#xA;$$ \lnot (\exists c_0, c_1) (\forall x) (x = c_0 \lor x = c_1) $$&#xD;&#xA;$$ \lnot (\exists c_0, c_1, c_2) (\forall x) (x = c_0 \lor x = c_1 \lor c = c_2) $$&#xD;&#xA;and more generally, for each $k &amp;gt; 0$, &#xA;$$ &#xD;&#xA;\lnot (\exists c_0, \ldots, c_k ) (\forall x) (x = c_0 \lor \cdots \lor c = c_k) $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fact that each of these sentences is provable in the theory can be verified using a very weak metatheory. There's a simple and uniform way, for any fixed $k$, to generate a proof from your two axioms for that particular $k$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This reinterpretation process is well known to logicians, and it can be applied very generally. The main limitation is whether the property in question can be stated in the language of the theory itself (possibly as an infinite set of sentences, as with the example above). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gödel's completeness theorem says that if a sentence in the language of a theory is true (semantically) in every model of the theory, then that sentence is provable (syntactically) within the theory itself. So we could, if we wanted, ignore the &quot;true&quot; part in those cases and focus just on the &quot;provable&quot; part. The cost of doing that is that it arguably misses the point, since we usually think semantically about models, and that it does not apply to properties of models that cannot be expressed within the language of the theory itself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because this reinterpretation process is usually very routine, it's rare for logicians to bother to mention it, even though we're well aware it could be done. We just write in the normal model-theoretic (set theoretic) way unless there is a particular reason that we need to focus on the syntactic interpretation. &lt;/p&gt;&#xA;" OwnerUserId="630" LastActivityDate="2010-12-08T14:38:09.193" CommentCount="1" />
  <row Id="13517" PostTypeId="2" ParentId="13507" CreationDate="2010-12-08T14:45:08.237" Score="3" ViewCount="" Body="&lt;p&gt;The equation $2y^2+5x^3=11(xy−11)$ describes an elliptic curve. Given any elliptic curve, you can perform a change of variables to put it into Weierstrass form, and if the field of definition has characteristic different from 2 or 3, you can even put it into the form $y^2 = x^3 + ax + b$. In this particular case, to find the transformation is easy: first scale $x$ and $y$ to get rid of make the coefficients of $x^3$ and of $y^2$ equal to 1. Then write $y&amp;#39; = y +  \alpha x$ to get rid of the $xy$-term. This will introduce an $x^2$ term. So now, translate $x$ to get rid of the $x^2$ term (I haven't actually done the computation).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An important thing to note is that while the notion of rational points does not depend on the model over $\mathbb{Q}$ that you are working with (as long as you only change variables over $\mathbb{Q}$), the notion of integral points does depend on the integral model. A theorem of Siegel says that on any Weierstrass model of an elliptic curve, there are only finitely many integral points, and there are &lt;a href=&quot;http://en.wikipedia.org/wiki/Elliptic_curve#Integral_points&quot; rel=&quot;nofollow&quot;&gt;bounds on their height&lt;/a&gt; in terms of the coefficients of the model. But as far as I know, to actually find the points comes down to brute force methods: e.g. if you manage to compute generators of the Mordell-Weil group, i.e. of the group of the rational points, and the torsion subgroup, then you just check all possible linear combinations up to the given bound. There are better methods using elliptic logarithms, but they are theoretically more involved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: some more info on your concrete curve:&#xA;To get a Weierstrass model you can replace $y$ by $y&amp;#39;/20$ and $x$ by $-x&amp;#39;/10$ in the original equation, resulting in $y&amp;#39;^2+11x&amp;#39;y&amp;#39; = x&amp;#39;^3 - 2^35^211^2$, confirming your hint. The elliptic curve has rank 1, as you say, and no torsion. The point $P=(22,-88)$ generates all the rational points on the curve under the group law. Integral solutions of the original equation correspond to points on the Weierstrass model with integral $x$-coordinate divisible by 10 and integral $y$-coordinate divisible by 20 (see our transformation). So the naive way of ruling out integral solutions to the original equation is to check all multiples of the generator $P=(22,-88)$ under the group law up to the given bound and convince yourself that no point satisfies the divisibility criteria. However, the bound of Baker, referred to in the wikipedia article, is huge and the computation might not actually be feasible. A possibly more promising approach is to write down the polynomial that computes $Q\mapsto Q\oplus P$ and see whether this operation always introduces higher and higher denominators that never cancel the numerators.&lt;/p&gt;&#xA;" OwnerUserId="3212" LastEditorUserId="3212" LastEditorDisplayName="" LastEditDate="2010-12-13T01:58:47.853" LastActivityDate="2010-12-13T01:58:47.853" CommentCount="3" />
  <row Id="13518" PostTypeId="1" AcceptedAnswerId="13540" CreationDate="2010-12-08T14:45:19.220" Score="3" ViewCount="235" Body="&lt;p&gt;Can someone suggest an idea for a school project in knot-theory for a 13 year old?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="4545" LastActivityDate="2010-12-08T18:44:02.897" Title="School project in knot theory" Tags="&lt;middle-school&gt;&lt;knot-theory&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="13519" PostTypeId="1" AcceptedAnswerId="14028" CreationDate="2010-12-08T14:46:15.833" Score="7" ViewCount="215" Body="&lt;p&gt;My basic question is: &quot;Do longer tautologies take longer to prove?&quot; But obviously this is underdetermined. If you are allowed an inference rule &quot;Tautological Implication&quot; then any tautology has a 1 line proof. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But let's say we're working in natural deduction, is it true that longer tautologies (tautologies with more letters and connectives in them) take more proof steps?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But this is still not quite a good enough question, since we can add superfluous steps to a proof and still prove the result. So the question is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In Natural Deduction, is the shortest proof of a tautology related to its length? Or perhaps: does a shorter tautology always admit a shorter proof?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And how robust is this result across different sets of connectives/inference rules?&lt;/p&gt;&#xA;" OwnerUserId="172" LastActivityDate="2010-12-12T08:15:30.170" Title="Is there a connection between length of sentence and length of proof?" Tags="&lt;logic&gt;&lt;proof-theory&gt;" AnswerCount="4" FavoriteCount="4" />
  <row Id="13520" PostTypeId="2" ParentId="13511" CreationDate="2010-12-08T14:55:20.593" Score="0" ViewCount="" Body="&lt;p&gt;If what I posted below is a correct assumption, then you could think of a solution on these lines.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;find centroid of all points in set S (expensive)&lt;br&gt;&lt;/li&gt;&#xA;&lt;li&gt;obtain farthest points from the centroid. &lt;br&gt;&lt;/li&gt;&#xA;&lt;li&gt;use these to draw the outermost polygon. &lt;br&gt;&lt;/li&gt;&#xA;&lt;li&gt;remove from this points from set S&lt;br&gt;&lt;/li&gt;&#xA;&lt;li&gt;repeat step 1 until (no more inner polygons can be drawn)&lt;br&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="4545" LastActivityDate="2010-12-08T14:55:20.593" />
  <row Id="13521" PostTypeId="2" ParentId="13519" CreationDate="2010-12-08T15:00:13.520" Score="3" ViewCount="" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;does a shorter tautology always admit a shorter proof?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;it seems like there should be a tautology whose statement is longer than fermats last theorem but whose shortest proof is shorter&lt;/p&gt;&#xA;" OwnerUserId="4488" LastActivityDate="2010-12-08T15:00:13.520" />
  <row Id="13522" PostTypeId="2" ParentId="13504" CreationDate="2010-12-08T15:07:15.930" Score="6" ViewCount="" Body="&lt;p&gt;There's been much work on this problem since the mentioned seminal work in the early seventies. Searching on the buzzword &quot;stably equivalent&quot; should help locate most of it. Below is a helpful introduction from Jon L Johnson: &lt;a href=&quot;http://books.google.com/books?id=m7MKj5tLAJwC&amp;amp;lpg=PA277&amp;amp;ots=Wc0BuJo04-&amp;amp;dq=brewer%20rutter%20hochster&amp;amp;pg=PA277#v=onepage&amp;amp;q=picard&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;Cancellation and Prime Spectra&lt;/a&gt;&#xA;&lt;img src=&quot;http://i.imgur.com/AiI7j.jpg&quot; alt=&quot;alt text&quot;&gt;&#xA;&lt;img src=&quot;http://i.imgur.com/IQGPZ.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-12-08T15:07:15.930" />
  <row Id="13523" PostTypeId="2" ParentId="13518" CreationDate="2010-12-08T15:11:16.050" Score="2" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Borromean_rings&quot; rel=&quot;nofollow&quot;&gt;Borromean rings&lt;/a&gt;? Can the same thing be done with more than three loops?&lt;/p&gt;&#xA;" OwnerUserId="1149" LastActivityDate="2010-12-08T15:11:16.050" />
  <row Id="13524" PostTypeId="2" ParentId="13519" CreationDate="2010-12-08T15:15:17.363" Score="1" ViewCount="" Body="&lt;p&gt;Suppose that $\phi$ is any sentence of the form $\lnot (\psi \land \lnot \psi)$. Then there is a pattern for a very short proof of $\phi$ that works for any formula $\psi$. I can't format it here, but the idea is to obtain $\bot$ from $\psi \land \lnot \psi$ in a fixed number of steps, and then apply the rule for proof by contradiction to get $\phi$. The original contradiction will take three steps in natural deduction: two are conjunction elimination and the third derives $\bot$.  So the entire proof pattern has four steps in natural deduction regardless of $\psi$.  If you tweak the setup for natural deduction the number of steps may go up to five or six, but it will be constant in any case. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure whether there is any system where longer tautologies always take longer proofs and where length is measured in proof steps. Certainly it would have to have modified versions of the rules relating to $\bot$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One way to change the question is to change the way you measure the length of a proof. If the size of the proof is the total symbol count then there is clearly a sense in which longer facts require longer proofs.&lt;/p&gt;&#xA;" OwnerUserId="630" LastEditorUserId="630" LastEditorDisplayName="" LastEditDate="2010-12-09T17:39:42.983" LastActivityDate="2010-12-09T17:39:42.983" CommentCount="1" />
  <row Id="13525" PostTypeId="1" AcceptedAnswerId="13758" CreationDate="2010-12-08T15:21:17.627" Score="3" ViewCount="187" Body="&lt;p&gt;I just attended a seminar on the chromatic filtration of the Burnside ring.  I understood it relatively well, but at no point did anyone give an explicit definition of what a chromatic filtration actually is, and I have had some trouble finding one.  Assuming I have the necessary categorical background, i.e. knowledge of filtrations, etc. can anyone give me an explanation of what makes a filtration &quot;chromatic&quot;?&lt;/p&gt;&#xA;" OwnerUserId="3595" LastEditorUserId="627" LastEditorDisplayName="" LastEditDate="2010-12-13T05:43:07.610" LastActivityDate="2010-12-16T21:53:18.190" Title="Chromatic Filtration of Burnside Ring" Tags="&lt;group-theory&gt;&lt;algebraic-topology&gt;&lt;category-theory&gt;&lt;homotopy-theory&gt;&lt;algebraic-k-theory&gt;" AnswerCount="4" CommentCount="2" FavoriteCount="1" />
  <row Id="13526" PostTypeId="2" ParentId="13424" CreationDate="2010-12-08T15:26:44.537" Score="0" ViewCount="" Body="&lt;p&gt;Im trying to work out a similair question, but I am unsure as to how to expand it as a power series?&#xA;I have taken the log of my function and differentiated it, but what is my final answer for the power series expansion? &lt;/p&gt;&#xA;" OwnerUserId="4547" LastActivityDate="2010-12-08T15:26:44.537" CommentCount="3" />
  <row Id="13527" PostTypeId="2" ParentId="13500" CreationDate="2010-12-08T15:32:46.943" Score="8" ViewCount="" Body="&lt;p&gt;One also has the classical Bonnet-Myers and Synge theorems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Bonnet-Myers theorem states that if $M$ is complete with Ricci curvature bounded below $\delta &amp;gt; 0$, then $M$ is compact with finite fundamental group.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The idea of the proof is the opposite of the one in Matt E's answer.  In positive (sectional) curvature, geodesics tend to get stuck together.  One uses this to show that geodesics past a certain length (something like $\pi/\delta^2$) cannot continue to minimize.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It follows that if $B_r(0)\subseteq T_p M$ is any closed ball with radius $r &amp;gt; \pi/\delta^2$, then $exp(B_r(0)) = M$.  In particular, you've written $M$ as the compact image of a continuous function so it's compact.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The theorem about the fundamental group is an immediate corollary:  look at the universal cover $\tilde{M}$.  One can &quot;pull back&quot; the metric so that the covering map is a local isometry.  Hence, the same curvature estimates apply to $\tilde{M}$, so it, too, must be compact.  But a compact manifold can only finitely cover something, so the fundamental group of $M$ is finite.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To push this to Ricci, one uses a nice trick:  if $\text{Ric} &amp;gt; 0$, then the sectional curvature in some directions must be $&amp;gt; 0$, and one uses these directions only to show geodesics eventually stop minimizing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Synge theorems have a different style of proof.  Here's one version of the theorem:  suppose $M$ is compact with positive curvature, and $f:M\rightarrow M$ is an isometry.  Suppose $\text{dim}(M)$ is even and $f$ is orientation preserving OR $\text{dim}(M)$ is odd and $f$ is orientation reversing.  Then $f$ has a fixed point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The proof is straightforward:  suppose not.  Choose $p\in M$ with  $d(p,f(p))$ as small as possible.  Choose a minimal geodesic from $p$ to $f(p)$.  One then computes variations of the geodesic (this uses the parity of the dimension, if I recall correctly), and shows that some nearby geodesic is smaller, contradicting your choice of $\gamma$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a corollary, one learns that the fundamental group of an even dimension positively curved compact manifold is either trivial or $\mathbb{Z}/2\mathbb{Z}$.  For, suppose $M$ satisfies all the hypothesis.  Look at the deck group action on $\tilde{M}$.  If it's orientation preserving, it has a fixed point, but the only deck transformation with a fixed point is the identity.  If it's orientation reversing, then it's square is orientation preserving, and hence is the identity, so all elements are of order 2 in $\pi_1$.  If there are 2 orientation reversing maps, then their product is orientation preserving and hence the identity, so there is at most one element of order 2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Likewise, in odd dimensions, the corollary is that $M$ must be orientable, since the deck group must acts on $\tilde{M}$ by orientation preserving maps.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, to just randomly answer one of your other questions, curvature can affect higher homotopy groups.  As Matt E pointed out, in negative (really, nonpositive) curvature, the universal cover is $\mathbb{R}^n$, implying all higher homotopy groups vanish.  In positive curvature, this is a very important and currently unsolved problem.&lt;/p&gt;&#xA;" OwnerUserId="331" LastEditorUserId="4322" LastEditorDisplayName="" LastEditDate="2011-04-19T06:03:23.457" LastActivityDate="2011-04-19T06:03:23.457" CommentCount="1" />
  <row Id="13528" PostTypeId="1" CreationDate="2010-12-08T15:38:54.920" Score="0" ViewCount="109" Body="&lt;p&gt;I have been asked to expand (1-2t)^(-n/2). So far I have found the log of this function and then differentiated twice in order to find my mean and variance as n and 2n respectively. &#xA;How do I write my final answer as a power series in t as far as the term in t^2?&lt;/p&gt;&#xA;" OwnerUserId="4547" LastActivityDate="2010-12-08T17:56:55.463" Title="Expansion of an MGF as a power series" Tags="&lt;statistics&gt;" AnswerCount="2" CommentCount="8" />
  <row Id="13529" PostTypeId="2" ParentId="13511" CreationDate="2010-12-08T15:56:41.540" Score="0" ViewCount="" Body="&lt;p&gt;There are several algorithms for curve reconstruction from point samples. See for instance &lt;a href=&quot;http://www.cse.ohio-state.edu/~tamaldey/curverecon.htm&quot; rel=&quot;nofollow&quot;&gt;http://www.cse.ohio-state.edu/~tamaldey/curverecon.htm&lt;/a&gt; and &lt;a href=&quot;http://valis.cs.uiuc.edu/~sariel/research/CG/applets/Crust/Crust.html&quot; rel=&quot;nofollow&quot;&gt;http://valis.cs.uiuc.edu/~sariel/research/CG/applets/Crust/Crust.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="589" LastActivityDate="2010-12-08T15:56:41.540" />
  <row Id="13530" PostTypeId="2" ParentId="13528" CreationDate="2010-12-08T16:08:40.663" Score="0" ViewCount="" Body="&lt;p&gt;If you want to expand your function in a power series on the argument $t$ you can write its &lt;a href=&quot;http://en.wikipedia.org/wiki/Taylor%27s_theorem&quot; rel=&quot;nofollow&quot;&gt;Taylor Polynomials&lt;/a&gt; .&lt;/p&gt;&#xA;" OwnerUserId="1178" LastActivityDate="2010-12-08T16:08:40.663" />
  <row Id="13531" PostTypeId="2" ParentId="13525" CreationDate="2010-12-08T16:14:48.470" Score="3" ViewCount="" Body="&lt;p&gt;It appears to be non-technical, and roughly equivalent to &quot;periodic&quot;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;An explanation is given &lt;a href=&quot;http://books.google.com/books?id=RA18_pxdPK4C&amp;amp;pg=PA23&quot; rel=&quot;nofollow&quot;&gt;by Ravenel&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The &quot;original paper&quot; appears to be:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Miller, Haynes R.; Ravenel, Douglas C.; Wilson, W. Stephen. &quot;Periodic phenomena in the Adams-Novikov spectral sequence.&quot;&#xA;  Ann. Math. (2) 106 (1977), no. 3, 469–516.&#xA;  MR&lt;a href=&quot;http://www.ams.org/mathscinet-getitem?mr=458423&quot; rel=&quot;nofollow&quot;&gt;458423&lt;/a&gt;&#xA;  &lt;a href=&quot;http://www.jstor.org/stable/1971064&quot; rel=&quot;nofollow&quot;&gt;JSTOR&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;but the term is not used there.  These sorts of things were for homotopy of spheres, a major topic in algebraic topology.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Presumably the filtration discussed for the Burnside ring had a similar ambience.&lt;/p&gt;&#xA;" OwnerUserId="583" LastActivityDate="2010-12-08T16:14:48.470" />
  <row Id="13533" PostTypeId="2" ParentId="13518" CreationDate="2010-12-08T16:59:27.600" Score="3" ViewCount="" Body="&lt;p&gt;There are some elementary ideas in Colin Adam's book: The Knot Book that could be adapted for investigation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another problem to think about would be the number of sides for regions that occur when the same knot (different drawings) is considered embedded in the plane as a 4-valent (degree 4) graph. The number of faces with k sides (knot drawings can have 1-gons, 2-gons, etc.) for such a drawing obeys a relationship that can be derived from Euler's polyhedral formula using the fact that the graph of a projection of a knot is 4-valent.&lt;/p&gt;&#xA;" OwnerUserId="1369" LastActivityDate="2010-12-08T16:59:27.600" />
  <row Id="13534" PostTypeId="1" AcceptedAnswerId="13537" CreationDate="2010-12-08T17:47:04.250" Score="2" ViewCount="221" Body="&lt;p&gt;We have a group $G$ of order $2p$, $p$ an odd prime, which has a nontrivial center. We wish to show the group is cyclic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a simple exercise which can be instantly killed using heavy cannons. However, since I was asked about it by students which barely reached quotients, I wonder what simple ways are to prove this; I couldn't find any.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="1818" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-09T10:17:29.643" LastActivityDate="2010-12-09T10:17:29.643" Title="Simple proof that group of order $2p$ with nontrivial center is cyclic?" Tags="&lt;homework&gt;&lt;group-theory&gt;" AnswerCount="3" />
  <row Id="13535" PostTypeId="2" ParentId="13528" CreationDate="2010-12-08T17:56:55.463" Score="1" ViewCount="" Body="&lt;p&gt;You have already done most of the work. If the mean $\mu&amp;#39;_1$ and variance $\sigma^2$ are equal to $n$ and $2n$, respectively, then the second moment $\mu&amp;#39;_2$ is given by $\mu&amp;#39;_2 = \sigma^2 + (\mu&amp;#39;_1)^2 = n(n+2)$. Now, the moment-generating function can be expanded as follows:&#xA;$$&#xD;&#xA;M(t) = 1 + \mu&amp;#39; _1 t + \frac{1}{{2!}}\mu&amp;#39; _2 t^2  + \cdots = 1 + nt + \frac{{n(n + 2)}}{2}t^2  \cdots.&#xD;&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="2810" LastActivityDate="2010-12-08T17:56:55.463" />
  <row Id="13536" PostTypeId="1" CreationDate="2010-12-08T17:57:49.010" Score="0" ViewCount="316" Body="&lt;p&gt;I have a 3d object which in its simplest form consists of an origin in 3d space and a set of vertices that are all local to this origin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I then transform this 3d origin into 2d camera coordinates using a perspective transform, but I need some way to also transform the local vertices of this object which has the effect of moving this 3d object from 3d world coordinates into 3d camera coordinates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;How can I determine the scale of an object based on it's distance from the viewer/camera?&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope this makes sense, the title is probably the most concise explanation I can give.  Any help would be very appreciated.&lt;/p&gt;&#xA;" OwnerUserId="4549" LastActivityDate="2011-05-10T02:56:12.527" Title="Determine scale factor of an object given its distance from a viewer/camera" Tags="&lt;geometry&gt;&lt;transformation&gt;" AnswerCount="1" CommentCount="6" />
  <row Id="13537" PostTypeId="2" ParentId="13534" CreationDate="2010-12-08T18:19:40.770" Score="3" ViewCount="" Body="&lt;p&gt;Let $g$ be a nontrivial element of the center.  If $g$ generates the whole group, we are done; otherwise $g$ generates a cyclic subgroup of order $2$ (resp. $p$).  Hence $G/\langle g \rangle$ is a cyclic subgroup of order $p$ (resp. $2$).  It follows that $G$ is generated by $g$ and one other element $h$ such that $h^p = g^k$ for some $k$ (resp. $h^2 = g^k$ for some $k$).  Since $g$ is in the center, $gh = hg$.  But this implies $h$ is also in the center.  Hence $G$ is abelian.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $h$ generates the whole group we are done.  Otherwise $h$ has order $2$ (resp. $p$), since it has order dividing $2p$ and cannot have order $p$ (resp. $2$), and $G \simeq C_2 \times C_p$ as desired.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-12-08T18:19:40.770" CommentCount="3" />
  <row Id="13539" PostTypeId="2" ParentId="13511" CreationDate="2010-12-08T18:24:04.203" Score="0" ViewCount="" Body="&lt;p&gt;I will use &lt;a href=&quot;http://en.wikipedia.org/wiki/Convex_hull&quot; rel=&quot;nofollow&quot;&gt;Convex_hull&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks to all for answers.&lt;/p&gt;&#xA;" OwnerUserId="4544" LastActivityDate="2010-12-08T18:24:04.203" CommentCount="1" />
  <row Id="13540" PostTypeId="2" ParentId="13518" CreationDate="2010-12-08T18:31:44.743" Score="3" ViewCount="" Body="&lt;p&gt;You may wish to take a look at the readable and well-illustrated &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/1931914222&quot; rel=&quot;nofollow&quot;&gt;Why Knot?&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Something to consider about knots is how 'basic' they are -- children will often learn to construct simple knots before learning arithmetic. This demonstrates an ability to manipulate (and distinguish between classes of) closed curves in three dimensions, and amounts to an intuitive theory of topology. We can use this as a staging ground to ask more interesting and difficult questions about knots and their classification.&lt;/p&gt;&#xA;" OwnerUserId="4300" LastEditorUserId="4300" LastEditorDisplayName="" LastEditDate="2010-12-08T18:44:02.897" LastActivityDate="2010-12-08T18:44:02.897" CommentCount="1" />
  <row Id="13541" PostTypeId="2" ParentId="13534" CreationDate="2010-12-08T18:38:03.697" Score="2" ViewCount="" Body="&lt;p&gt;Qiaochu's answer is simpler, but the following almost classifies all groups of order $2p$, so I'll leave it here as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lemma 1: The order of an element divides order of the group. &#xA;No new proof, but easy enough. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lemma 2: There exists an element of order $p$. &#xA;Can circumvent usual proof as follows: If there is an element of order $2p$ G is cyclic. If no element of order $p$ or $2p$ then $a^2=e$ for all $a$, so $G$ is abelian ($abab=e$, so $ab=b^{-1}a^{-1}=ba$). Then either use the fact that you know all abelian groups, or the fact that then $G$ is a vector space over the field $Z_2$, so has basis and so has order $2^k$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Call this element $a$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lemma 3: Any subgroup of index 2 is normal. &#xA;There are two cosets, and one of them is the subgroup itself. So right and left cosets coincide. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Applying this to the subgroup generated by $a$, for $b$ not a power of $a$, we have $bab^{-1}=a^m$.  The group contains only $a^k$ and $ba^l$'s, which are all distinct. If $m=1$ they all commute and $ba$ generates the cyclic group. If $m\neq1$ there is no center. (The only step in classification left is to show that different $m \neq 1$ lead to the same group, which is not hard either.)&lt;/p&gt;&#xA;" OwnerUserId="2633" LastActivityDate="2010-12-08T18:38:03.697" CommentCount="2" />
  <row Id="13542" PostTypeId="1" AcceptedAnswerId="13544" CreationDate="2010-12-08T19:02:09.660" Score="3" ViewCount="135" Body="&lt;p&gt;Suppose that $f(z)$ is complex analytic on $|z| \leq 1$ and satisfies&#xA;$|f(z)| &amp;lt; 1$ for $|z|=1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(a) Prove that the equation $f(z)=z$ has exactly one root&#xA;(counting multiplicities) in $|z|&amp;lt;1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(b) Prove that if $|z_0| \leq 1$, then the sequence $z_n$ defined recursively&#xA;by $z_n= f(z_{n-1}) , n=1,2,...$, converges to the fixed point of $f$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was able to prove (a) using Rouche's theorem, but (b) stumps me.&#xA;I know that (b) is true for analytic fuctions such that $f(0)=0$ or&#xA;$|f&amp;#39;(z)|&amp;lt;1$ on the disc, neither of which are necessarily true in general.&#xA;The farthest I was able to get was $|f(z)-z^*|&amp;lt;\frac{1}{1-|z*|}|z-z^*|$,&#xA;where $z^*$ is the fixed point of $f$, but&#xA;$\frac{1}{1-|z^*|}&amp;gt;1$, so I don't think this helps me.&#xA;Can someone please point me in the right direction?&lt;/p&gt;&#xA;" OwnerUserId="2029" LastActivityDate="2010-12-08T19:12:14.610" Title="Fixed point iteration for analytic functions on the unit disc" Tags="&lt;homework&gt;&lt;complex-analysis&gt;&lt;numerical-methods&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" />
  <row Id="13543" PostTypeId="1" AcceptedAnswerId="13943" CreationDate="2010-12-08T19:11:47.310" Score="2" ViewCount="277" Body="&lt;p&gt;This question is motivated by my previous &lt;a href=&quot;http://math.stackexchange.com/questions/13344/proof-for-an-integral-involving-sinc-function&quot;&gt;post&lt;/a&gt; about sinc function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Prove or disprove that $\frac{\sin x}{x}$ is the only nonzero entire (i.e. analytic everywhere)&#xA;function $f(x)$ on $\mathbb{R}$ such that $$\int_0^\infty f(x) dx=\int_0^\infty f(x)^2 dx$$&#xD;&#xA;or $$\int_{-\infty}^\infty f(x) dx=\int_{-\infty}^\infty f(x)^2 dx.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $f$ is required only to be continuous, then other examples are possible, e.g. the even extension of the following function:&#xA;$$ f(x)=\left\{\begin{array}{ll} -2(5+\sqrt{65})x^2+(7+\sqrt{65})x-1 &amp;amp; 0\le x\le \frac{1}{2}\\&#xD;&#xA;2(5+\sqrt{65})x^2-(13+3\sqrt{65})x+4+\sqrt{65} &amp;amp; \frac{1}{2}\le x\le 1\\&#xD;&#xA;\frac{1}{x^2} &amp;amp; x\ge 1 \end{array}\right. $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As commented below, it turns out that there are easy answers to the above question. AD also showed a function below that also satisfies $$ \sum_{n=1}^\infty f(n)=\sum_{n=1}^\infty f(n)^2=0.  $$&#xD;&#xA;In view of these answers, my question is now revised to:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Prove or disprove that $\frac{\sin x}{x}$ is the only nonzero entire&#xA;function, $f(x)$ on $\mathbb{R}$ such that $$\int_{-\infty}^\infty f(x) dx=\int_{-\infty}^\infty f(x)^2 dx=\sum_{-\infty}^\infty f(n) =\sum_{-\infty}^\infty f(n)^2 $$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="3249" LastEditorUserId="3249" LastEditorDisplayName="" LastEditDate="2010-12-09T14:55:03.370" LastActivityDate="2010-12-14T02:24:55.977" Title="Are there other analytic functions with this property of sinc function?" Tags="&lt;analysis&gt;&lt;integral&gt;" AnswerCount="3" CommentCount="4" />
  <row Id="13544" PostTypeId="2" ParentId="13542" CreationDate="2010-12-08T19:12:14.610" Score="3" ViewCount="" Body="&lt;p&gt;One can reduce it to the case of $f(0)=0$ (i.e. $0$ is the fixed point) by making a suitable linear fractional transformation of the disk. Namely, if $z^*$ is the fixed point, apply the above argument to $L \circ f \circ L^{-1}$ where $L$ sends $z^* \to 0$ and is a LFT.&lt;/p&gt;&#xA;" OwnerUserId="536" LastActivityDate="2010-12-08T19:12:14.610" CommentCount="2" />
  <row Id="13547" PostTypeId="1" AcceptedAnswerId="13550" CreationDate="2010-12-08T19:40:06.827" Score="6" ViewCount="145" Body="&lt;p&gt;When one writes $\zeta_n$ which of the n roots of unity is meant here?  Does it matter? &lt;/p&gt;&#xA;" OwnerUserId="2212" LastActivityDate="2010-12-09T00:52:19.070" Title="When one writes $\zeta_n$ which of the n roots of unity is meant here?" Tags="&lt;number-theory&gt;" AnswerCount="3" />
  <row Id="13548" PostTypeId="2" ParentId="13547" CreationDate="2010-12-08T19:41:45.333" Score="1" ViewCount="" Body="&lt;p&gt;Probably a primitive one.&lt;/p&gt;&#xA;" OwnerUserId="589" LastActivityDate="2010-12-08T19:41:45.333" />
  <row Id="13549" PostTypeId="2" ParentId="13547" CreationDate="2010-12-08T19:42:00.590" Score="5" ViewCount="" Body="&lt;p&gt;Usually this means $e^{ \frac{2\pi i}{n} }$.  Depending on the context, it can just mean any primitive $n^{th}$ root of unity (and sometimes it doesn't matter which one, since they're all taken to each other under the action of the Galois group).&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-12-08T19:42:00.590" CommentCount="1" />
  <row Id="13550" PostTypeId="2" ParentId="13547" CreationDate="2010-12-08T19:49:59.853" Score="10" ViewCount="" Body="&lt;p&gt;The context is important here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is rather standard for $\zeta_n$ to at least denote a &lt;a href=&quot;http://en.wikipedia.org/wiki/Primitive_nth_root_of_unity&quot; rel=&quot;nofollow&quot;&gt;primitive&lt;/a&gt; $n$th root of unity in the algebraic closure of the field $k$ which is currently being considered (this is possible iff the characteristic of $k$ does not divide $n$; in particular it is true for all fields of characteristic zero).  When the characteristic of $k$ does not divide $n$ (i.e., when any primitive $n$th roots of unity exist) there are precisely $\varphi(n)$ primitive $n$th roots of unity, where $\varphi$ is Euler's phi function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a context in which $k$ is a subfield of the complex numbers, it is also rather standard for $\zeta_n$ to denote the specific primitive $n$th root of unity $e^{\frac{2 \pi i}{n}}$, i.e., the one of minimal argument in the complex plane.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does it matter?  For algebraic purposes, probably not: the primitive $n$th roots of unity in $\mathbb{C}$ are algebraically conjugate over $\mathbb{Q}$: i.e., different roots of a common irreducible polynomial over $\mathbb{Q}$, the cyclotomic polynomial $\Phi_n(t)$.  Sometimes in number theory one considers systems of $n$th roots of unity for varying $n$, and in this case it is necessary to make a consistent (in a certain sense) choice of $\zeta_n$'s.  Taking $\zeta_n = e^{\frac{2 \pi i}{n}}$ for all positive integers $n$ is such a consistent choice, but there are (many!) others.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course there are always situations when confusing one complex number for another would lead to trouble, so yes, in principle it might matter, especially in analytic or metric arguments.  &lt;/p&gt;&#xA;" OwnerUserId="299" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2010-12-09T00:52:19.070" LastActivityDate="2010-12-09T00:52:19.070" CommentCount="3" />
  <row Id="13551" PostTypeId="1" AcceptedAnswerId="13553" CreationDate="2010-12-08T19:57:48.280" Score="3" ViewCount="223" Body="&lt;p&gt;I have been trying to do this for quite a while, but generally speaking the partially relevant information I could find on the internet only dealt with the question: &quot;How does on convert a recurrence relation into... well, a non-recurrence relation?&quot; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's start of with a simple example: $f(n) = 34n^3+51n^2+27n+5$. How do we find $f_{n}$? &#xA;I'd really like to see this solved in analogy with the following: Consider $g(n)=n^6$ We can then find the recursion formula: $g_n=((g_{n-1})^{1/6}+1)^6$. What about $f_{n}$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We could generalize this question in a number of ways. For instance, is it (also) possible too turn an infinite polynomial, like the Taylor Series expansion of a trigonometric formula, into a recursion formula? Furthermore, what happens when we allow the coefficients of the polynomial to be real and even complex?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Max&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bonus side-question: How, if at all, are &quot;generating functions&quot; useful in this context?&lt;/p&gt;&#xA;" OwnerUserId="1055" LastEditorUserId="1055" LastEditorDisplayName="" LastEditDate="2010-12-08T20:45:40.517" LastActivityDate="2010-12-09T17:48:56.933" Title="Is it possible to convert a polynomial into a recurrence relation? If so, how?" Tags="&lt;functions&gt;&lt;polynomials&gt;&lt;generating-functions&gt;" AnswerCount="4" CommentCount="6" />
  <row Id="13552" PostTypeId="2" ParentId="8808" CreationDate="2010-12-08T20:00:31.727" Score="0" ViewCount="" Body="&lt;p&gt;In topology people study intensely some naturally occurring abut quite abstract and not &quot;locally Euclidean&quot; spaces. Examples are path spaces(space of paths in a given topological space or manifold) and some natural $H$-spaces in the homotopy category.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway there a whole lot of new kind of &quot;geometry&quot; in the field of algebraic geometry, which is not at all Euclidean.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But your point is good, that the Euclidean space is quite specially useful in physics. With this in mind, some people are actually doing adelic physics these days.&lt;/p&gt;&#xA;" OwnerUserId="1119" LastActivityDate="2010-12-08T20:00:31.727" />
  <row Id="13553" PostTypeId="2" ParentId="13551" CreationDate="2010-12-08T20:03:33.653" Score="11" ViewCount="" Body="&lt;p&gt;Any polynomial $f(n)$ of degree $d$, over an arbitrary commutative ring, satisfies the recursion&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{k=0}^{d+1} (-1)^{d+1-k} {d+1 \choose k} f(x+k) = 0$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for any $x$.  This is an application of the method of &lt;a href=&quot;http://en.wikipedia.org/wiki/Finite_difference&quot; rel=&quot;nofollow&quot;&gt;finite differences&lt;/a&gt;.  In terms of generating functions, it says that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{n \ge 0} f(n) x^n = \frac{P(x)}{(1 - x)^{d+1}}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for some polynomial $P(x)$.  This should be thoroughly covered in a book like Wilf's &lt;a href=&quot;http://www.math.upenn.edu/~wilf/DownldGF.html&quot; rel=&quot;nofollow&quot;&gt;generatingfunctionology&lt;/a&gt;; otherwise, I discuss it in my &lt;a href=&quot;http://web.mit.edu/~qchu/Public/TopicsInGF.pdf&quot; rel=&quot;nofollow&quot;&gt;notes on generating functions&lt;/a&gt;.  More generally, if $f(n)$ has a generating function of the form $\frac{P(x)}{Q(x)}$ where $P$ is a polynomial, then the coefficients of $Q$ describe a recurrence that $f(n)$ eventually satisfies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure I understand your more general question.  Do you have a specific application in mind?  &lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-12-08T20:11:03.527" LastActivityDate="2010-12-08T20:11:03.527" CommentCount="9" />
  <row Id="13554" PostTypeId="1" AcceptedAnswerId="13565" CreationDate="2010-12-08T20:06:22.827" Score="3" ViewCount="104" Body="&lt;p&gt;If $(X, \lt)$ is a well-ordering I can show by transfinite recursion over the ordinals that the function $f(x) = \text{ran} f |_{\hat{x}}$ exists (where $\hat{x} = \{ y : y \lt x\}$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have obtained $f$ this way, Let $V$ be the class of all sets and $F:V \to V$ be a class-function, then there is a unique $G:ON \to V$ where $ON$ is the class of all ordinals such that $F(\alpha) = F(G|_\alpha)$. So I can apply this to get a function $f$ such that $f(x) = F(f|_\hat{x})$. Now I let $F = \{(x, \text{ran} x) : x \in V\}$ and then I get the function as above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, this should be an isomorphism (order preserving bijection) between $X$ and the set of true initial segments of $X$, $I_X$ ordered by inclusion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, when I have $x &amp;lt; y$, then I see that $\text{ran} f|_\hat{x} \subset \text{ran} f|_\hat{y}$. So $f(x) \leq f(y)$. Why do I have $f(x) \neq f(y)$?&lt;/p&gt;&#xA;" OwnerUserId="1120" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-08T21:04:53.223" LastActivityDate="2010-12-08T21:44:57.627" Title="Recursive definition isomorphism." Tags="&lt;set-theory&gt;" AnswerCount="2" CommentCount="17" />
  <row Id="13555" PostTypeId="1" AcceptedAnswerId="13588" CreationDate="2010-12-08T20:26:56.343" Score="2" ViewCount="63" Body="&lt;p&gt;The n-dimensional cone, with vertex at the origin, central angle, $\alpha$ and central axis in the direction of the unit vector $\xi$ is defined to be all those points, $x\in {R^n}$ whose dot product with $\xi$ is |$x$|$cos(\alpha)$.  How would I find parametric equations for this surface? &lt;/p&gt;&#xA;" OwnerUserId="3971" LastActivityDate="2010-12-09T00:52:48.167" Title="Parametric Equations for a Hypercone" Tags="&lt;differential-geometry&gt;&lt;parametric&gt;" AnswerCount="1" />
  <row Id="13556" PostTypeId="1" CreationDate="2010-12-08T20:34:38.147" Score="8" ViewCount="141" Body="&lt;p&gt;Given a long exact sequence of the form&#xA;$$&#xD;&#xA;\dots\to A&amp;#39;_n \to B&amp;#39;_n \to C&amp;#39;_n&#xD;&#xA;\,\xrightarrow{\omega_n}\,&#xD;&#xA;A&amp;#39;_{n-1} \to B&amp;#39;_{n-1} \to C&amp;#39;_{n-1}\to \dots\qquad (*)&#xD;&#xA;$$&#xD;&#xA;is there a way to recover a short exact sequence of complexes $\mathcal A=\{A_n,\partial_n^A\}$, $\mathcal B=\{B_n,\partial_n^B\}$, $\mathcal C=\{C_n,\partial_n^C\}$ such that the sequence (*) &quot;is&quot; the long exact sequence in homology induced by&#xA;$$&#xD;&#xA;0\to \mathcal A\to  \mathcal B\to  \mathcal C \to 0&#xD;&#xA;$$&#xD;&#xA;and the morphisms $\omega_n$ are in fact the connection morphisms of that homology? I mean $A&amp;#39;_n\cong H_n(\mathcal A)$ for all $n\ge 0$ and similarly for $B&amp;#39;_n$, $C&amp;#39;_n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I expect the answer will be &quot;obviously no&quot;, but then is there a case in which it is possible?&lt;/p&gt;&#xA;" OwnerUserId="685" LastActivityDate="2010-12-09T01:49:07.383" Title="Does every les &quot;in homology&quot; come from a ses of complexes?" Tags="&lt;homological-algebra&gt;" AnswerCount="1" FavoriteCount="2" />
  <row Id="13557" PostTypeId="2" ParentId="13551" CreationDate="2010-12-08T20:34:50.170" Score="1" ViewCount="" Body="&lt;p&gt;Expand $f(n+1)$ and you'll get $f(n+1)=f(n)+g(n)$ for some $g$. For $f(n)=34n^3+51n^2+27n+5$, you get $f(n+1)=f(n)+102n^2+204n+112$. &lt;/p&gt;&#xA;" OwnerUserId="589" LastActivityDate="2010-12-08T20:34:50.170" CommentCount="2" />
  <row Id="13558" PostTypeId="2" ParentId="13551" CreationDate="2010-12-08T20:45:48.547" Score="3" ViewCount="" Body="&lt;p&gt;Max, I'm just detailing Qiaochu's answer, using Pari/GP.&#xA;First define the function&lt;br&gt;&#xA;$ f(x) = 34*x^3+51*x^2+27*x+5$    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then the first idea to decompose f into a 2-term-recursion is&lt;br&gt;&#xA;$ f(x)-1*f(x-1) $&lt;br&gt;&#xA;Pari/GP: $ &amp;lt;out&amp;gt; = 102*x^2 + 10 $    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the result is already a reduced polynomial. Now subtract from this $f(x-1)-f(x-2) $ because this will be a polynomial of the same degree, only &quot;shifted&quot; by some coefficients.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This gives&lt;br&gt;&#xA;$ f(x)-2*f(x-1)+f(x-2) $&lt;br&gt;&#xA;Pari/GP: $ &amp;lt;out&amp;gt;  = 204*x - 102 $    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again the resulting polynomial is of reduced order. Now we finish by subtracting $ f(x-1)-2*f(x-2)+f(x-3) $     &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We get:&lt;br&gt;&#xA;$ f(x)-3*f(x-1)+3*f(x-2)-f(x-3) $&lt;br&gt;&#xA;Pari/GP: $ &amp;lt;out&amp;gt; = 204 $     &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we have a constant expression only. We are finished and write, reordering the terms:   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ f(x) = 3*f(x-1)-3*f(x-2)+f(x-3)+204 $    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;[update] For completeness one can increase the degree. According to Qiaochu's answer/comment we can subtract again the x-1-shifted polynomial $ f(x-1)-3*f(x-2)+3*f(x-3)-f(x-4) $ and of course getting    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ f(x)-4*f(x-1)+6*f(x-2)-4*f(x-3)+f(x-4) $&lt;br&gt;&#xA;Pari/GP: $ &amp;lt;out&amp;gt; = 0 $     &lt;/p&gt;&#xA;&#xA;&lt;p&gt;completing the answer &#xA;$ f(x) = 4*f(x-1)-6*f(x-2)+4*f(x-3)-f(x-4) $    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;[end update]&lt;/p&gt;&#xA;" OwnerUserId="1714" LastEditorUserId="1714" LastEditorDisplayName="" LastEditDate="2010-12-08T21:56:48.620" LastActivityDate="2010-12-08T21:56:48.620" />
  <row Id="13559" PostTypeId="2" ParentId="13519" CreationDate="2010-12-08T20:57:58.697" Score="2" ViewCount="" Body="&lt;p&gt;The size of proofs is almost totally unrelated to the size of the statements they prove (in the worst case).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Lower Bound: constant size&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The size of the smallest proof for statements of size N is constant. Just make your statements of the form &quot;True or S&quot;. (Unless you count the input statement as part of the size, in which case it is linear.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Upper Bound: incomputable size&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The size of the largest proof proving a statement of size less than N grows so fast it is not computable from N.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Proof&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the size of a proof that a Turing Machine does not halt within S steps. Some turing machines run forever but can't be proven to run forever. Suppose you have such a machine M. Proving M doesn't halt before S steps must take more steps as S grows. Otherwise we have a proof for all S and have contradicted ourselves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lets call that number-of-steps to proof-size function G. I am going to make one assumption, which is that there is a computable function G' such that G(G'(X)) &gt;= X.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose you have some computable function F you claim is an asymptotic upper bound on proof sizes given the statement size. Then I just generate a sequence of statements where the Nth statement is &quot;M does not halt within G'(N*F(N)) steps&quot;. The statements are all provable (worst case: simulate M in the proof; that's why I assumed G' is computable) and grow in size logarithmically. But the proof sizes grow asymptotically faster than F, which is exponentially faster than F with respect to the statement sizes. So F is not actually an asymptotic upper bound.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore no computable function is an asymptotic upper bound on proof size given the statement size, which means the upper bound is incomputable.&lt;/p&gt;&#xA;" OwnerUserId="3820" LastEditorUserId="3820" LastEditorDisplayName="" LastEditDate="2010-12-09T17:37:30.633" LastActivityDate="2010-12-09T17:37:30.633" />
  <row Id="13560" PostTypeId="2" ParentId="13554" CreationDate="2010-12-08T20:58:44.263" Score="2" ViewCount="" Body="&lt;p&gt;You need to use the fact that a well-ordering can't be isomorphic to any of its initial segments.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-12-08T20:58:44.263" />
  <row Id="13561" PostTypeId="2" ParentId="13511" CreationDate="2010-12-08T20:59:42.807" Score="2" ViewCount="" Body="&lt;p&gt;Use the Convex Layers algorithm of Bernard Chazelle, which is optimal: $\mathcal{O}(n \log n)$ time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See here: &lt;a href=&quot;http://www.cs.princeton.edu/~chazelle/pubs/ConvexLayers.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.cs.princeton.edu/~chazelle/pubs/ConvexLayers.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a snapshot of part of the first page of the paper:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/a6nnh.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-12-08T20:59:42.807" CommentCount="1" />
  <row Id="13562" PostTypeId="1" CreationDate="2010-12-08T21:10:50.160" Score="2" ViewCount="110" Body="&lt;p&gt;let $S(k), k\geq 0$ a discrete random process. Suppose $S(N)$ is with probability one either 100 or 0 and that $S(0)=50$. Suppose further there is at least a sixty percent probability that the price will at some point dip below 40 and then subsequently rise above 60 before time $N$. &#xA;How do you prove that $S(k)$ cannot be a martingale?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By advance, thank you very much for your help.&lt;/p&gt;&#xA;" OwnerUserId="4552" LastActivityDate="2010-12-08T21:10:50.160" Title="An application of the Optional Sampling Theorem" Tags="&lt;probability&gt;&lt;stochastic-processes&gt;" CommentCount="1" FavoriteCount="1" />
  <row Id="13563" PostTypeId="1" AcceptedAnswerId="13564" CreationDate="2010-12-08T21:19:26.333" Score="1" ViewCount="134" Body="&lt;p&gt;Hello.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am finding with some formulas for my web application. I need a formula that counts the number of values between two number, e.g. there are four numbers (3,4,5,6) between 2 and 7&lt;/p&gt;&#xA;" OwnerUserId="4553" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-09T19:45:51.647" LastActivityDate="2010-12-09T19:45:51.647" Title="Count number of values between two numbers" Tags="&lt;arithmetic&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="13564" PostTypeId="2" ParentId="13563" CreationDate="2010-12-08T21:22:03.390" Score="5" ViewCount="" Body="&lt;p&gt;larger - smaller - 1. If you don't count the larger and the smaller (like your example).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the numbers are $n, n+1, \dots, n+k , n+k+1$, the numbers you want are $n+1, n+2, \dots, n+k$, which are $k$ in number = $n+k+1 - n - 1$ = larger - smaller - 1.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-12-08T21:22:03.390" CommentCount="4" />
  <row Id="13565" PostTypeId="2" ParentId="13554" CreationDate="2010-12-08T21:35:16.583" Score="2" ViewCount="" Body="&lt;p&gt;Okay, it looks like you are thinking of your $(X,\lt)$ as an ordinal, rather than an arbitrary set. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I claim that for all $y\in X$, if $x\lt y$, then  $f(x)\neq f(y)$ and $f(x)\subseteq f(y)$. You have already shown $f(x)\subseteq f(y)$, so we just need to show the inequality.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $y=\emptyset$, the least element of $X$, then there is nothing to do and the claim holds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume the claim holds for all $z\lt y$. Let $x\lt y$. If $x^+$, the successor of $x$, is also less than $y$, then $f(x)\subseteq f(x^+)\subseteq f(y)$, and $f(x)\neq f(x^+)$ by the induction hypothesis, so $f(x)\neq f(y)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $y=x^+$, then $\hat{y} = \hat{x}\cup\{x\}$. So $f(y) = f(x)\cup\{f(x)\}$. If $f(x)\cup\{f(x)\} = f(x)$, then $f(x)\in f(x)$, which is impossible since ordinals are well-founded relative to $\in$. Therefore, $f(y)=f(x)\cup\{f(x)\}\neq f(x)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;By transfinite induction, the claim holds for all $y\in X$.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-08T21:44:57.627" LastActivityDate="2010-12-08T21:44:57.627" />
  <row Id="13566" PostTypeId="2" ParentId="13543" CreationDate="2010-12-08T21:53:11.060" Score="2" ViewCount="" Body="&lt;p&gt;In the spirit of Zaricuse's solution without the sum requirement, take any sufficiently  well behaved functions f and g.  Then you should be able to find a linear combination af+bg that satisfies both equations.  If you let &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$if=\int_{-\infty}^\infty f(x) dx$$&#xD;&#xA;  $$if2=\int_{-\infty}^\infty f(x)^2 dx$$&#xD;&#xA;  $$sf=\sum_{n=1}^\infty f(n)$$&#xD;&#xA;  $$sf2=\sum_{n=1}^\infty f(n)^2$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and similarly for fg and g, we have &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$a*if + b*ig=a^2*if2+2ab*ifg+b^2ig2$$ and&lt;br&gt;&#xA;$$a*sf + b*sg=a^2*sf2+2ab*sfg+b^2sg2$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which can be solved for a and b in most cases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added&lt;/strong&gt; In response to the new request that the values of the two integrals and two sums all match, I just need enough knobs to turn.  Define $g(k,x)=\exp(-kx^2)$ and take $f(x)=g(1,x)+ag(2,x)+bg(3,x)+cg(4,x)$  The nice thing about this $f$ is that $f^2$ is written in terms of $g(k,x)$, though k goes up to 8.  The integral of $g(k,x)$ is just $\sqrt{\frac{\pi}{k}}$ and the sum is calculated by Wolfram Alpha as $\vartheta_3(0,\exp(-k))$. We can make a table:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{array}{ccc}k&amp;amp;\int g(k,x)&amp;amp;\sum g(k,x)\\1&amp;amp;1.772453851&amp;amp;1.77264\\2&amp;amp;1.253314137&amp;amp;1.27134\\3&amp;amp;1.023326708&amp;amp;1.09959\\4&amp;amp;0.886226925&amp;amp;1.03663\\5&amp;amp;0.79266546&amp;amp;1.01348\\6&amp;amp;0.723601255&amp;amp;1.00496\\7&amp;amp;0.669924586&amp;amp;1.00182\\8&amp;amp;0.626657069&amp;amp;1.00067\end{array}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the integral of f is $\sqrt{\pi}(1+a/\sqrt{2}+b/\sqrt{3}+c/\sqrt{4})$  The integral of f^2 is $\sqrt{\pi}(1/\sqrt{2}+2a/\sqrt{3}+(a^2+2b)/\sqrt{4}+(2c+2ab)/\sqrt{5}+(b^2+2ac)/\sqrt{6}+2bc/\sqrt{7}+c^2/\sqrt{8})$ with similar expressions for the sum in terms of theta3.  We want to find a,b,c so that the integrals and sums all match.  Unless my matrix of coefficients has a very unlikely dependence it will be available.  Excel claims $a=-3.782590725, b=4.503400057, c=-1.83137936$ is very close to a solution, and there should be more.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-10T03:41:30.973" LastActivityDate="2010-12-10T03:41:30.973" CommentCount="5" />
  <row Id="13567" PostTypeId="2" ParentId="13543" CreationDate="2010-12-08T21:58:19.353" Score="3" ViewCount="" Body="&lt;p&gt;Similar to the suggestion of Zaricuse, that is take an entire function $f$ such that $\int_{-\infty}^\infty f(x) dx \ne0$ then solve for &#xA;$$\int_{-\infty}^\infty af(x)dx = \int_{-\infty}^\infty (af(x))^2dx$$&#xD;&#xA;Then $g(z)=af(z)$ solves half of the problem. To reach &#xA;$$\sum g(n)=\sum g(n)^2$$&#xD;&#xA;we may for example start with $f(z)=\sin (\pi z) \cdot h(z)$  where $h$ is an other integrable entire function.&lt;/p&gt;&#xA;" OwnerUserId="1154" LastActivityDate="2010-12-08T21:58:19.353" CommentCount="2" />
  <row Id="13568" PostTypeId="1" AcceptedAnswerId="13570" CreationDate="2010-12-08T21:58:30.607" Score="1" ViewCount="91" Body="&lt;p&gt;If i know the means and standard deviations of 2 sets of data, and i know the slope of the regression line, how can I find the correlation?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;edit&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sample 1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;SD: 12.37&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sample 2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;SD: 7.00&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Slope of regression line: 0.789&lt;/p&gt;&#xA;" OwnerUserId="1630" LastEditorUserId="1630" LastEditorDisplayName="" LastEditDate="2010-12-08T22:13:25.183" LastActivityDate="2010-12-08T22:20:53.990" Title="Statistics - Correlation between 2 sets of data" Tags="&lt;statistics&gt;&lt;correlation&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="13569" PostTypeId="1" AcceptedAnswerId="13571" CreationDate="2010-12-08T22:02:48.490" Score="1" ViewCount="161" Body="&lt;p&gt;Find an equation of the parabola with focus at point $(0,5)$ whose directrix is the line $y=0$. (Derive this equation using the definition of the parabola as a set of points that are equidistant from the directrix and the focus)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ok this one is killing me. My textbook has this&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;An equation of the parabola with focus $(0,p)$ and directrix $y=-p$ is $x^2=4py$&#xD;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;However, my problem has $(0,p)$ with p being 5, yet the directrix is not $p=-y=-5$, it's $0$. And so I tried deriving it manually as they said by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sqrt{x^2+(y-p)^2}=|y+p|$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But that just takes me back to the original equation I mentioned, $x^2=4py$. Not sure what I'm supposed to do. &lt;/p&gt;&#xA;" OwnerUserId="2985" LastActivityDate="2010-12-09T02:58:54.970" Title="Finding equation of parabola" Tags="&lt;conic-sections&gt;" AnswerCount="3" />
  <row Id="13570" PostTypeId="2" ParentId="13568" CreationDate="2010-12-08T22:06:03.793" Score="3" ViewCount="" Body="&lt;p&gt;The formula is $$r = b_1 \frac{s_x}{s_y},$$ where $r$ is the correlation, $b_1$ is the slope, and $s_x$ and $s_y$ are the standard deviations of the independent $(x)$ and dependent $(y)$ variables, respectively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A reference is Wikipedia's page on &lt;a href=&quot;http://en.wikipedia.org/wiki/Simple_linear_regression#Fitting_the_regression_line&quot; rel=&quot;nofollow&quot;&gt;simple linear regression&lt;/a&gt;.  See the formula for $\hat{\beta}$. &lt;/p&gt;&#xA;" OwnerUserId="2370" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-12-08T22:20:53.990" LastActivityDate="2010-12-08T22:20:53.990" CommentCount="7" />
  <row Id="13571" PostTypeId="2" ParentId="13569" CreationDate="2010-12-08T22:06:33.837" Score="8" ViewCount="" Body="&lt;p&gt;&lt;em&gt;Edited to try to clarify OP's confusion:&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The distance between $(x,y)$ and $(0,5)$ is indeed $\sqrt{x^2+(y-5)^2}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The distance between $(x,y)$ and the line $y=0$ is $|y|$ (if $y&amp;gt;0$, then it is above the $X$-axis, and the distance is just the $y$ coordinate; if $y&amp;lt;0$, then the distance is $-y = |y|$). So the points on the parabola are exactly the points for which the two distances are equal, that is, all $(x,y)$ for which:&#xA;$$\sqrt{x^2+(y-5)^2} = |y|.$$&#xD;&#xA;Now square both sides, and you'll get the equation of the parabola you want.&lt;/p&gt;&#xA;&#xA;&lt;hr/&gt;&#xA;&#xA;&lt;p&gt;How is this related to the formula you quote? It's really all just the same process. If the directrix is either horizontal (a line of the form $y=k$) or vertical (a line of the form $x=\ell$), then it is very easy to compute the distance from a point to the directrix: the point $(x,y)$ is $|y-k|$ away from the line $y=k$, and is $|x-\ell|$ away from the line $x=\ell$. If the focus of the parabola is at $(a,b)$, then you want the distance from $(x,y)$ to $(a,b)$ to equal the distance to the directrix, so you get:&#xA;\begin{align*}&#xA;\sqrt{(x-a)^2 + (y-b)^2} &amp;amp;= |y-k| &amp;amp;\qquad&amp;amp;\mbox{if the directrix is $y=k$,}\\&#xA;\sqrt{(x-a)^2 + (y-b)^2} &amp;amp;= |x-\ell| &amp;amp;&amp;amp;\mbox{if the directrix is $x=\ell$.}&#xA;\end{align*}&#xA;Then squaring both sides yields the equation of the parabola. Doing it correctly will cancel out the $y^2$ term when the directrix is horizontal, and the $x^2$ term when the directrix is vertical. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For more general parabolas, when the directrix is neither horizontal nor vertical but an arbitrary line $y=mx+b$, you need to work a bit harder, because the distance from $y=mx+b$ to $(x,y)$ is not so simple to compute. Not &lt;em&gt;too&lt;/em&gt; hard, but not as simple. Once you have a formula for the distance, you set it equal to the distance to the focus, square both sides, and get the equation of the parabola.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-09T02:58:54.970" LastActivityDate="2010-12-09T02:58:54.970" CommentCount="4" />
  <row Id="13572" PostTypeId="6" CreationDate="2010-12-08T22:10:57.063" Score="0" ViewCount="0" Body="&lt;p&gt;I discovered math.stackexchange.com less than 4 weeks (at the time of writing this) ago and I have been a regular follower/contributer since then. I am relatively new to online mathematics and have not been a particularly active member anywhere else, partly because I am relatively new to mathematics itself (I used to be an engineer and switched to math a couple of years ago). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have been especially interested in this community and also math overflow (which I also joined only recently) because of the absence of discussion format which is probably the reason for the high signal to noise ratio. If elected, maintaining and improving this efficiency would be my main goal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe in a largely &lt;em&gt;laissez faire&lt;/em&gt; operation of such sites and as such would trust the invisible hand of the majority to take care of most issues. I generally prefer to keep my discussions technical (as can be seen from my posts) and as such would avoid imposing my opinions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wish to see this site prosper and be of permanent value to active and amateur mathematicians at every level. I will try my best to play my part in this.&lt;/p&gt;&#xA;" OwnerUserId="3431" LastEditorUserId="3431" LastEditorDisplayName="" LastEditDate="2010-12-08T22:45:23.930" LastActivityDate="2010-12-08T22:45:23.930" CommunityOwnedDate="2010-12-08T22:10:57.063" />
  <row Id="13573" PostTypeId="2" ParentId="13569" CreationDate="2010-12-08T22:19:44.913" Score="1" ViewCount="" Body="&lt;p&gt;From &lt;a href=&quot;http://en.wikipedia.org/wiki/Parabola&quot; rel=&quot;nofollow&quot;&gt;wikipedia&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Equation for a general parabola with a focus point $F =(u, v)$, and a directrix in the form&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$n_1x+n_2y+c=0 \,$$&#xD;&#xA;is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\left|n_1x+n_2y+c\right|}{\sqrt{{n_1}^{2}+{n_2}^{2}}}=\sqrt{\left(x-u\right)^2+\left(y-v\right)^2} $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So make the substitutions and verify that the what you get agrees with &lt;a href=&quot;http://math.stackexchange.com/questions/13569/finding-equation-of-parabola/13571#13571&quot;&gt;Arturo Magidin's derivation&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1119" LastActivityDate="2010-12-08T22:19:44.913" />
  <row Id="13574" PostTypeId="2" ParentId="13569" CreationDate="2010-12-08T22:38:26.060" Score="3" ViewCount="" Body="&lt;p&gt;So, you have a method for writing an equation for a parabola where the focus is at $(0,p)$ and the directrix is $y=-p$, but now you want to write an equation for a parabola with focus $(0,5)$ and directrix $y=0$, which doesn't fit the formula.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you take the conditions you are given and translate them down 2.5 units, the focus would be at $(0,2.5)$ and the directrix would be at $y=-2.5$, so you should be able to write an equation for that parabola.  Now, translate that equation up 2.5 units (replace $y$ with $(y-2.5)$).&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-12-08T22:38:26.060" />
  <row Id="13575" PostTypeId="1" CreationDate="2010-12-08T22:42:57.350" Score="17" ViewCount="1006" Body="&lt;p&gt;I have a hazy notion of some stuff in differential geometry and a better, but still not quite rigorous understanding of basics of differential topology.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have decided to fix this lacuna once for all. Unfortunately I cannot attend a course right now. I must teach myself all the stuff by reading books.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Towards this purpose I want to know what are the most important basic theorems in differential geometry and differential topology. For a start, for differential topology, I think I must read Stokes' theorem and de Rham theorem with complete proofs. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Differential geometry is a bit more difficult. What is a connection? Which notion should I use? I want to know about parallel transport and holonomy. What are the most important and basic theorems here? Are there concise books which can teach me the stuff faster than the voluminous Spivak books?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also finally I want to read into some algebraic geometry and Hodge/Kähler stuff.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suggestions about important theorems and concepts to learn, and book references, will be most helpful.&lt;/p&gt;&#xA;" OwnerUserId="4555" LastActivityDate="2011-03-02T02:10:12.243" Title="Teaching myself differential topology and differential geometry." Tags="&lt;reference-request&gt;&lt;differential-geometry&gt;&lt;differential-topology&gt;" AnswerCount="7" CommentCount="4" FavoriteCount="17" />
  <row Id="13576" PostTypeId="2" ParentId="13575" CreationDate="2010-12-08T22:53:23.670" Score="5" ViewCount="" Body="&lt;p&gt;Guillemin and Pollack's &quot;Differential Topology&quot; is about the friendliest introduction to the subject you could hope for.  It's an excellent non-course book.  Good supplementary books would be Milnor's &quot;Topology from a differentiable viewpoint&quot; (much more terse), and Hirsch's &quot;Differential Topology&quot; (much more elaborate, focusing on the key analytical theorems). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For differential geometry it's much more of a mixed bag as it really depends on where you want to go.  I've always viewed Ehresmann connections as the fundamental notion of connection.  But it suits my tastes.   But I don't know much in the way of great self-learning differential geometry texts, they're all rather quirky special-interest textbooks or undergraduate-level grab-bags of light topics.  I haven't spent any serious amount of time with the Spivak books so I don't feel comfortable giving any advice on them. &lt;/p&gt;&#xA;" OwnerUserId="642" LastActivityDate="2010-12-08T22:53:23.670" />
  <row Id="13577" PostTypeId="2" ParentId="13477" CreationDate="2010-12-08T23:00:31.150" Score="2" ViewCount="" Body="&lt;p&gt;&lt;em&gt;With&lt;/em&gt; trig, I get that the arcs have measure $\theta \pm \alpha$ and $\pi - \theta \pm \beta$, where &#xA;$$\cos\theta = \frac{\sin\frac{\beta-\alpha}{2}}{\sin\frac{\alpha+\beta}{2}} = \frac{\cos\beta-\cos\alpha}{1-\cos\left(\alpha+\beta\right)}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the &lt;a href=&quot;http://www.mathwarehouse.com/geometry/circle/tangents-secants-arcs-angles.php&quot; rel=&quot;nofollow&quot;&gt;&quot;far arc / near arc&quot; principle&lt;/a&gt;, $\stackrel{\frown}{EI}-\stackrel{\frown}{EG} = 2\alpha$, so we can write $\stackrel{\frown}{EI} = \theta+\alpha$ and $\stackrel{\frown}{EG}=\theta-\alpha$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's go analytical ... &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let the circle to have radius $1$, with its center, $O$, at the origin; and let $H$ lie on the positive $x$-axis. In quadrilateral $OEHT$, angles at $E$ and $T$ are right angles, and the angle at $H$ has measure $\alpha+\beta$, so that the angle at $O$ has measure $2 \gamma := \pi-(\alpha+\beta)$. Thus, $\angle HOE = \gamma$ and $|OH| = \sec\gamma$. Moving around the circle by arcs of $\theta \pm \alpha$ from $E$ gets us to $I$ and $G$, and we determine coordinates for these points, writing &quot;$\rm{cis}\cdot$&quot; for &quot;$(\cos\cdot,\sin\cdot)$&quot;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$I = \rm{cis}\left(\gamma + ( \theta + \alpha )\right) = \rm{cis}\left(\gamma+\alpha+\theta\right)$$ &#xA;$$G = \rm{cis}\left(\gamma-(\theta-\alpha)\right) = \rm{cis}\left(\gamma+\alpha-\theta\right)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $H$, $G$, and $I$ are collinear, we must have equal slopes for segments $HI$ and $HG$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{ \sin(\gamma+\alpha+\theta) }{ \cos(\gamma+\alpha+\theta)-\sec\gamma } = \frac{ \sin(\gamma+\alpha-\theta) }{ \cos(\gamma+\alpha-\theta)-\sec\gamma }$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sin(\gamma+\alpha+\theta)\left(\cos(\gamma+\alpha-\theta)-\sec\gamma\right) = \sin(\gamma+\alpha-\theta) \left( \cos(\gamma+\alpha+\theta)-\sec\gamma\right) $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{eqnarray*}&#xD;&#xA;&amp;amp;&amp;amp;\sin(\gamma+\alpha+\theta)\cos(\gamma+\alpha-\theta)- \sin(\gamma+\alpha-\theta) \cos(\gamma+\alpha+\theta)  \\&#xD;&#xA;&amp;amp;=&amp;amp; \sec{\gamma} \; \left( \sin(\gamma+\alpha+\theta) -\sin(\gamma+\alpha-\theta) \right) \\&#xD;&#xA;\sin 2\theta &amp;amp;=&amp;amp; 2 \sec{\gamma} \; \sin \theta \cos\left(\gamma+\alpha\right) \\&#xD;&#xA;2 \sin\theta \cos\theta &amp;amp;=&amp;amp; 2 \sec{\gamma} \; \sin \theta \cos\left(\gamma+\alpha\right)&#xD;&#xA;\end{eqnarray*}$$&#xD;&#xA;If $\theta\ne 0$ and $\theta \ne \pi$ [*], then we can cancel $\sin\theta$ and finish-up:&#xA;$$\begin{eqnarray*}&#xD;&#xA;\cos\theta &amp;amp;=&amp;amp; \frac{\cos\left(\gamma+\alpha\right)}{\cos\gamma} = \frac{\cos\frac{\pi+\alpha-\beta}{2}}{\cos\frac{\pi-(\alpha+\beta)}{2}}= \frac{\sin\frac{\beta-\alpha}{2}}{\sin\frac{\alpha+\beta}{2}} \\&#xD;&#xA;\end{eqnarray*}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's certainly a way to arrive at this result without using coordinates and slopes (I got lazy), but if there's a way to get there &quot;without using trigonometry&quot;, I'm not seeing it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit.&lt;/strong&gt;&#xA;I'll just point out that, if $F$ is the foot of the perpendicular from the circle's center to segment $HI$, then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{|OF|}{|OE|}=\frac{|OH| \sin\angle FHO}{|OH|\sin\angle EHO}=\frac{ \sin\frac{\alpha-\beta}{2}}{\sin\frac{\alpha+\beta}{2}}=\cos\left(\pi-\theta\right)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, there are points $P$ and $Q$ on the circle, with $P$ on $\stackrel{\frown}{IE}$ and $G$ on $\stackrel{\frown}{QE}$, such that $\stackrel{\frown}{IP} = \stackrel{\frown}{GQ} = \alpha$, whence $\stackrel{\frown}{PE}=\stackrel{\frown}{QE}=\theta$. So, all the components of the final relation appear in the figure somewhere. I haven't looked hard enough to determine if it's possible to &quot;see&quot; the relation geometrically.&#xA;&lt;strong&gt;End edit.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[*] If $\theta = 0$, then $\alpha$ itself must be $0$ (so that $\theta-\alpha$ is a non-negative arc measure), with $I$ and $G$ coinciding with $E$. Likewise, $\theta = \pi$ implies $\beta = 0$, with $I$ and $G$ coinciding with $T$. Analysis of these cases is straightforward, with the same formula resulting as in the general case.&lt;/p&gt;&#xA;" OwnerUserId="409" LastEditorUserId="409" LastEditorDisplayName="" LastEditDate="2010-12-09T13:18:12.850" LastActivityDate="2010-12-09T13:18:12.850" CommentCount="1" />
  <row Id="13578" PostTypeId="1" AcceptedAnswerId="13579" CreationDate="2010-12-08T23:03:39.250" Score="-5" ViewCount="148" Body="&lt;p&gt;What is a primitive nth root of unity?&lt;/p&gt;&#xA;" OwnerUserId="2212" LastActivityDate="2010-12-08T23:11:24.393" Title="What is a primitive nth root of unity?" Tags="&lt;number-theory&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="13579" PostTypeId="2" ParentId="13578" CreationDate="2010-12-08T23:11:24.393" Score="3" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Primitive_nth_root_of_unity&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt; says:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An nth root of unity, where n = 1,2,3,··· is a positive integer, is a complex number z satisfying the equation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$z^n = 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An nth root of unity is primitive if it is not a kth root of unity for some smaller k:&#xA;$z^k \neq 1  (k = 1, 2, 3, \ldots, n-1 )$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-12-08T23:11:24.393" />
  <row Id="13580" PostTypeId="2" ParentId="13556" CreationDate="2010-12-08T23:14:04.753" Score="3" ViewCount="" Body="&lt;p&gt;This is not a direct answer, but related. This paper by Jan Stovicek addresses the question &quot;which long exact sequences can arise from the snake lemma&quot; so it might be of help here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.mathjournals.org/mrl/2010-017-001/2010-017-001-011.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.mathjournals.org/mrl/2010-017-001/2010-017-001-011.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3431" LastEditorUserId="3431" LastEditorDisplayName="" LastEditDate="2010-12-09T01:49:07.383" LastActivityDate="2010-12-09T01:49:07.383" CommentCount="2" />
  <row Id="13581" PostTypeId="2" ParentId="13575" CreationDate="2010-12-08T23:14:11.237" Score="8" ViewCount="" Body="&lt;p&gt;For differential topology, I would add Poincare duality to something you may want to know. A good textbook is Madsen and Tornehave's &lt;em&gt;From Calculus to Cohomology&lt;/em&gt;. Another nice book is John Lee's &lt;em&gt;Introduction to Smooth Manifolds&lt;/em&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For differential geometry, I don't really know any good texts. Besides the standard Spivak, the other canonical choice would be Kobayashi-Nomizu's &lt;em&gt;Foundations of Differential Geometry&lt;/em&gt;, which is by no means easy going. There is a new book by Jeffrey Lee called &lt;em&gt;Manifolds and Differential Geometry&lt;/em&gt; in the AMS Graduate Studies series. I have not looked at it personally in depth, but it has some decent reviews. It covers a large swath of the differential topology, and also the basic theory of connections. (As a side remark, if you like doing computations, Kobayashi's original paper &quot;Theory of connections&quot; is not very hard to read, and may be a good starting place before you jump into some of the more special-topic/advanced texts like Kolar, Slovak, and Michor's &lt;em&gt;Natural operations in differential geometry&lt;/em&gt;.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A book I've enjoyed and found useful (though not so much as a textbook) is Morita's &lt;em&gt;Geometry of differential forms&lt;/em&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can't help you with algebraic geometry. &lt;/p&gt;&#xA;" OwnerUserId="1543" LastActivityDate="2010-12-08T23:14:11.237" CommunityOwnedDate="2010-12-08T23:14:11.237" CommentCount="8" />
  <row Id="13582" PostTypeId="1" AcceptedAnswerId="13584" CreationDate="2010-12-08T23:23:02.103" Score="0" ViewCount="292" Body="&lt;p&gt;Question: Let $a, b \in \mathbb{R}$ with $a &amp;lt; b$ and let $f: [a,b] \rightarrow [a,b]$ continuous. Show: $f$ has a fixed point, that is, there is an $x \in [a,b]$ with $f(x)=x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suppose this has to do with the basic definition of continuity. The definition I am using is that $f$ is continuous at $a$ if $\displaystyle \lim_{x \to a} f(x)$ exists and if $\displaystyle \lim_{x \to a} f(x) = f(a)$. I must not be understanding it, since I am not sure how to begin showing this... Should I be trying to show that $x$ is both greater than or equal to and less than or equal to $\displaystyle \lim_{x \to a} f(x)$ ?&lt;/p&gt;&#xA;" OwnerUserId="3711" LastEditorUserId="3431" LastEditorDisplayName="" LastEditDate="2010-12-08T23:28:13.173" LastActivityDate="2010-12-08T23:43:25.073" Title="Show that a continuous function has a fixed point" Tags="&lt;analysis&gt;" AnswerCount="4" CommentCount="1" />
  <row Id="13583" PostTypeId="2" ParentId="13582" CreationDate="2010-12-08T23:27:17.793" Score="6" ViewCount="" Body="&lt;p&gt;Consider $x-f(x)$ and use Intermediate Value Theorem.&lt;/p&gt;&#xA;" OwnerUserId="3249" LastActivityDate="2010-12-08T23:27:17.793" />
  <row Id="13584" PostTypeId="2" ParentId="13582" CreationDate="2010-12-08T23:28:30.680" Score="2" ViewCount="" Body="&lt;p&gt;Let $g(x) = f(x) - x$.  Then $g$ is continuous and $g(a) \geq 0$ while $g(b) \leq 0$.  By the Intermediate Value Theorem, $g$ has at least one zero on $[a, b]$.&lt;/p&gt;&#xA;" OwnerUserId="506" LastActivityDate="2010-12-08T23:28:30.680" CommentCount="3" />
  <row Id="13585" PostTypeId="2" ParentId="13582" CreationDate="2010-12-08T23:31:03.420" Score="2" ViewCount="" Body="&lt;p&gt;Consider $g(x)=f(x)-x$.  $f(a)\ge a$ so $g(a)=f(a)-a\ge 0$.  $f(b)\le b$ so $g(b)=f(b)-b\le 0$.  By the Intermediate Value Theorem, since $g$ is continuous and $0\in[g(b),g(a)]$ there exists $c\in[a,b]$ such that $g(c)=f(c)-c=0$, so $f(c)=c$ for some $c\in[a,b]$.&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-12-08T23:31:03.420" CommentCount="9" />
  <row Id="13586" PostTypeId="2" ParentId="13582" CreationDate="2010-12-08T23:33:12.140" Score="4" ViewCount="" Body="&lt;p&gt;For a different approach then the ones above, let us take $a = 0$ and $b = 1$. So assume $f:[0,1] \to [0,1]$ has no fixed point. Then $[0,1] = \{x \in [0,1] : f(x) &amp;lt; x\} \cup \{x \in [0,1] : f(x) &amp;gt; x \}$. Now argue that this is not possible.&lt;/p&gt;&#xA;" OwnerUserId="1120" LastEditorUserId="1120" LastEditorDisplayName="" LastEditDate="2010-12-08T23:43:25.073" LastActivityDate="2010-12-08T23:43:25.073" CommentCount="3" />
  <row Id="13587" PostTypeId="2" ParentId="12644" CreationDate="2010-12-09T00:35:28.517" Score="9" ViewCount="" Body="&lt;p&gt;For the sake of completeness: $\mathbb{C}^{\ast}$ is isomorphic to $\mathbb{R}^{+} \oplus \mathbb{R}/\mathbb{Z}$ in the obvious way, and $\mathbb{C}^{\ast}$ modulo the roots of unity is then isomorphic to $\mathbb{R}^{+} \oplus \mathbb{R}/\mathbb{Q}$.  As a $\mathbb{Q}$-vector space this is abstractly isomorphic to $\mathbb{R}^{+}$, but the construction of such an isomorphism is likely to require the axiom of choice.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-12-09T00:35:28.517" CommunityOwnedDate="2010-12-09T00:35:28.517" CommentCount="1" />
  <row Id="13588" PostTypeId="2" ParentId="13555" CreationDate="2010-12-09T00:52:48.167" Score="4" ViewCount="" Body="&lt;p&gt;The easiest answer is if $\xi$ is along one of the axes, say the first.  Then the cone can be parameterized as  $(t\cos(\alpha ),t\sin(\alpha){\bf\hat{x}})$, where ${\bf\hat{x}}$ is a unit vector in $\mathbb{R^{n-1}}$.  In coordinates, this is $(t\cos(\alpha ),t\sin(\alpha )\cos(\phi_1),t\sin(\alpha )\sin(\phi_1)\cos(\phi_2),t\sin(\alpha )\sin(\phi_1)\sin(\phi_2)\cos(\phi_3),\ldots)$ where you have n-1 angular terms and the last has no cosine.  If n&gt;3, $\phi_1$ runs from 0 to $\pi$ while the others run from 0 to $2\pi$.  Then you can apply a rotation matrix to take $(1,0,0,\ldots)$ to $\xi$.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-12-09T00:52:48.167" />
  <row Id="13589" PostTypeId="1" AcceptedAnswerId="13590" CreationDate="2010-12-09T01:16:45.033" Score="1" ViewCount="204" Body="&lt;p&gt;Let Y1...Yn be independent and identically distributed random vairables such that for 0 &amp;lt; p &amp;lt; 1, P(Yi = 1) = p and P(Yi = 0) = q = 1-p.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A. Find the moment-generating functions for the random variable Y1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;B. Find the moment-generating functions for W = Y1 + ... + Yn.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;C. What is the distribution of W?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have started to try A.&#xA;My book stays that m(t) = E(e^(tY)). But i'm sure sure what that is. I think that expected value of Y1 is p. But i'm not sure where to go from here. I'm completely clueless, statistics is not my area of expertise(I'm a computer science guy).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="3049" LastActivityDate="2010-12-09T02:50:49.793" Title="statistics - moment-generating function" Tags="&lt;probability&gt;&lt;homework&gt;&lt;statistics&gt;" AnswerCount="1" FavoriteCount="0" />
  <row Id="13590" PostTypeId="2" ParentId="13589" CreationDate="2010-12-09T01:30:10.380" Score="4" ViewCount="" Body="&lt;p&gt;If those are the only two values that $Y_i$ takes on then you are correct that $E[Y_i]=p$. The definition of the moment generating function is what you have described as $M_{Y_i}(t)=E[e^{tY_i}]$. So you compute this by multiplying $e^{ty_i}$ by your density function and summing over all of the appropriate values. So in this case $M_{Y_i}(t)=(e^{t(0)})(P(Y_i=0))+(e^{t(1)})(P(Y_i=1))$ which gives you $M_{Y_i}(t)=1-p+pe^t$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For part B you should use the fact that the moment generating function of a sum of independent random variables is the product of the moment generating functions. That gives you $M_W(t)=(1-p+pe^t)^n$ which i believe is the moment generating function for a Binomial random variable with parameters $p$ and $n$. This makes sense if we do a quick mental check and note that $Y_i$ can be thought of as the success or failure of the $i$th trial, the indicator functions. So the total number of successes would be $W$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me know if there are parts where I need to be more specific.&lt;/p&gt;&#xA;" OwnerUserId="627" LastEditorUserId="627" LastEditorDisplayName="" LastEditDate="2010-12-09T02:50:49.793" LastActivityDate="2010-12-09T02:50:49.793" CommentCount="2" />
  <row Id="13591" PostTypeId="1" AcceptedAnswerId="13594" CreationDate="2010-12-09T01:36:37.943" Score="13" ViewCount="460" Body="&lt;p&gt;It can be shown by simple induction that $\dfrac{\mathrm d^n}{\mathrm dx^n}\left(\dfrac1{x}\right) =  \frac{(-1)^n n!}{x^{n+1}}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But what about the nth integral of $\displaystyle \frac1{x}$? Finding the first few primitives, I can't discern a pattern. &lt;/p&gt;&#xA;" OwnerUserId="4558" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-09T02:31:08.277" LastActivityDate="2010-12-09T16:38:49.513" Title="What's the nth integral of $\displaystyle \frac1{x}$?" Tags="&lt;calculus&gt;&lt;integral&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="1" />
  <row Id="13592" PostTypeId="2" ParentId="13591" CreationDate="2010-12-09T01:43:09.317" Score="1" ViewCount="" Body="&lt;p&gt;Remembre that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int \frac{1}{x} dx= \ln (x) + C $$  Now integrate again&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint :&lt;/strong&gt;&lt;br&gt;&#xA;$\displaystyle\int \ln(x) dx = x \ln(x)-x +C$&lt;/p&gt;&#xA;" OwnerUserId="3421" LastActivityDate="2010-12-09T01:43:09.317" CommentCount="4" />
  <row Id="13593" PostTypeId="2" ParentId="13591" CreationDate="2010-12-09T01:52:55.703" Score="9" ViewCount="" Body="&lt;p&gt;If we use the &lt;a href=&quot;http://mathworld.wolfram.com/RepeatedIntegral.html&quot; rel=&quot;nofollow&quot;&gt;repeated integral&lt;/a&gt; formula for the reciprocal function, with 1 as the lower limit, we get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\underbrace{\int_1^x\int_1^{t_{n-1}}\cdots\int_1^{t_1}}_{n} \frac1{t}\;\mathrm dt\cdots\mathrm dt_{n-2}\mathrm dt_{n-1}=\frac1{(n-1)!}\int_1^x\frac{(x-t)^{n-1}}{t}\mathrm dt=(-1)^n \frac{x^{n-1}}{(n-1)!}B_{1-x}(n,1-n)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $B_x(a,b)$ is the &lt;a href=&quot;http://functions.wolfram.com/GammaBetaErf/Beta3/&quot; rel=&quot;nofollow&quot;&gt;incomplete beta function&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Letting&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$g_n(x)=(-1)^n \frac{x^{n-1}}{(n-1)!}B_{1-x}(n,1-n)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the following more &quot;elementary&quot; representation can be derived:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$g_n(x)=\frac{x^{n-1}}{(n-1)!}(\ln\;x-H_{n-1})-\sum_{j=1}^{n-1}\frac{(-1)^j}{j\cdot j!}\frac{x^{n-j-1}}{(n-j-1)!}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $H_n=\displaystyle \sum_{j=1}^n\frac1{j}$ is a harmonic number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As Moron mentions in his answer,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\mathrm d^n}{\mathrm dx^n}\left(\frac{x^{n-1}}{(n-1)!}\ln\;x+p_{n-1}(x)\right)=\frac1{x}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $p_{n-1}(x)$ is any polynomial of degree $n-1$; $g_n(x)$, however, has the special property (by virtue of how it was constructed) that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left.\frac{\mathrm d^k}{\mathrm dx^k}g_n(x)\right|_{x=1}=0\qquad \mathrm{if}\qquad k &amp;lt; n$$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="498" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-09T16:38:49.513" LastActivityDate="2010-12-09T16:38:49.513" CommentCount="4" />
  <row Id="13594" PostTypeId="2" ParentId="13591" CreationDate="2010-12-09T02:03:49.673" Score="15" ViewCount="" Body="&lt;p&gt;As Isaac noted, repeated integration seems to give the following pattern&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \frac{x^{n-1} \log x}{(n-1)!} - C_n x^{n-1}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the value of $\displaystyle C_n$ does not really matter, as differentiating $n$ times nukes it. Also, note that, we can add any arbitrary $\displaystyle (n-1)^{th}$ degree polynomial to this, without changing the $\displaystyle n^{th}$ derivative.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to prove that the $\displaystyle n^{th}$ derivative of $\displaystyle  \frac{x^{n-1} \log x}{(n-1)!}$ is $\displaystyle \frac{1}{x}$, we can use induction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \frac{1}{(n-1)!} \frac{d (x^{n-1} \log x)}{dx} = \frac{x^{n-2}}{(n-1)!} + \frac{x^{n-2} \log x}{(n-2)!}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since adding an arbitrary $\displaystyle (n-2)^{th}$ degree polynomial does not change the $\displaystyle (n-1)^{th}$ derivative of $\displaystyle \frac{x^{n-2} \log x}{(n-2)!}$ we are done using induction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that if $\displaystyle f(x)$ is another function such that $\displaystyle \frac{d^n f}{dx} = \frac{1}{x}$, then we have that $\displaystyle \phi(x) = f(x) - \frac{x^{n-1} \log x}{(n-1)!}$ has it's $\displaystyle n^{th}$ derivative to be zero, and hence it is a polynomial of degree $\displaystyle n-1$ or lower (can be proved using induction, again).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus all the functions you are looking for are of the form &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{x^{n-1} \log x}{(n-1)!} + \sum_{j=0}^{n-1} c_j x^j$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\displaystyle c_j$ are arbitrary constants.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-12-09T03:49:22.833" LastActivityDate="2010-12-09T03:49:22.833" CommentCount="4" />
  <row Id="13595" PostTypeId="2" ParentId="13575" CreationDate="2010-12-09T02:14:04.540" Score="1" ViewCount="" Body="&lt;p&gt;I would like to recommend &lt;em&gt;Modern Differential Geometry of curves and surfaces with Mathematica&lt;/em&gt;, by Alfred Gray, Elsa Abbena, and Simon Salamon.  You can look at &lt;a href=&quot;http://books.google.com/books?id=owEj9TMYo7IC&amp;amp;printsec=frontcover&quot; rel=&quot;nofollow&quot;&gt;it on Google books&lt;/a&gt; to decide if it fits your style.  If you are a Mathematica user, I think this is a wonderful avenue for self-study, for you can see and manipulate all the central constructions yourself.  I use Gray's code frequently; I was a fan.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS. Here is how he died: &quot;of a heart attack which occurred while working with students in a computer lab at 4 a.m.&quot;!&lt;/p&gt;&#xA;" OwnerUserId="237" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-09T03:23:11.880" LastActivityDate="2010-12-09T03:23:11.880" CommentCount="1" />
  <row Id="13596" PostTypeId="1" AcceptedAnswerId="13598" CreationDate="2010-12-09T02:25:53.507" Score="0" ViewCount="107" Body="&lt;p&gt;Let Y be a binomial random variable with n trials and probability of success given by p. Show that n-Y is a binomial random variable with n trials and probability of success given by 1-p.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand why this is true, and can explain in logically, but I just don't know how to prove it in mathematical terms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any help is greatly appreciated!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="3049" LastActivityDate="2010-12-09T06:09:33.073" Title="binomial random variable" Tags="&lt;probability&gt;&lt;homework&gt;&lt;statistics&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="13597" PostTypeId="2" ParentId="13596" CreationDate="2010-12-09T02:47:22.247" Score="1" ViewCount="" Body="&lt;p&gt;I am not really sure how I qould go about giving a full prove right now, but here is a suggestion. Rewrite $Y=\Sigma_{i=1}^n Y_i$ where the $Y_i$ are from your other question, they are the indicator functions for each trial. &lt;/p&gt;&#xA;" OwnerUserId="627" LastActivityDate="2010-12-09T02:47:22.247" CommentCount="1" />
  <row Id="13598" PostTypeId="2" ParentId="13596" CreationDate="2010-12-09T03:17:30.077" Score="2" ViewCount="" Body="&lt;p&gt;A discrete random variable is determined by the probabilities of all possible outcomes. If $Y$ is $n$ trials of a binomial random variable with probability $p$ of success, then the probability that $Y = i$ is given by ${\displaystyle{n \choose i} p^i(1 - p)^{n-i}}$. Thus the probability that $n - Y = i$ is given by the probability that $Y = n - i$, or ${\displaystyle {n \choose  n - i} p^{n-i}(1 - p)^i}$. Since ${\displaystyle{n \choose n - i} = {n \choose i}}$, this is the same as ${\displaystyle{n \choose  i} (1 - p)^ip^{n-i}}$. But if $Z$ denotes $n$ trials of a binomial random variable with probability $1 - p$ of success, this is exactly the probability that $Z = i$. Thus $n - Y$ and $Z$ have the same distribution (i.e. the same probability of each outcome), which is equivalent to saying that $n - Y$ is also $n$ trials of a binomial random variable with probability $1 - p$ of success.&lt;/p&gt;&#xA;" OwnerUserId="3035" LastActivityDate="2010-12-09T03:17:30.077" />
  <row Id="13599" PostTypeId="1" CreationDate="2010-12-09T03:38:52.673" Score="1" ViewCount="117" Body="&lt;p&gt;I feel like this should have a rather simple solution using logarithms, but I don't quite know where to start.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically, every day I am going to multiply a given quantity by $r$, where $r &amp;lt; 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously, if $r$ is 0.5, then my half-life $n$ is 1 day. (This is also stated to remove any ambiguities on what counts as a day/range)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the relationship between $r$ and $n$?&lt;/p&gt;&#xA;" OwnerUserId="51" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-09T04:50:55.783" LastActivityDate="2010-12-09T04:50:55.783" Title="If I wanted something to decay with a half-life of $n$ days, what should I multiply it by every day?" Tags="&lt;algebra-precalculus&gt;" AnswerCount="2" CommentCount="8" />
  <row Id="13600" PostTypeId="2" ParentId="13599" CreationDate="2010-12-09T03:45:40.123" Score="5" ViewCount="" Body="&lt;p&gt;Suppose you start with a quantity of 1 (whatever that means).  You know that after $n$ days, there should be $\frac{1}{2}$ left.  This should give you an equation that you can solve for $r$ (in terms of $n$ or with a specific value of $n$).&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-12-09T03:45:40.123" CommentCount="4" />
  <row Id="13602" PostTypeId="2" ParentId="13599" CreationDate="2010-12-09T03:48:36.093" Score="6" ViewCount="" Body="&lt;p&gt;Suppose $x$ is the quantity, and you want it to decay to $\frac{1}{2}$ after $n$ discrete steps (whether they be minutes, hours, days, years, eons, or nanoseconds), and you want to model that by multiplication by $r$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You start with $x$. After one time interval, you want to have $rx$. After two time intervals, you want $r(rx) = r^2x$. After three time intervals, $r(r^2x)=r^3x$. After $n$ times intervals, $r^nx$. Since you want $r^nx = \frac{1}{2}x$, you can now solve for $r$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is &lt;em&gt;discrete&lt;/em&gt; decay, though, not continuous decay. It is exactly analogous to the difference between continuous interest compounding, and interest compounding in discrete time intervals. &quot;Half-life&quot; is a term usually reserved for continuous decay, so you'll want to be careful with that.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-09T03:54:07.990" LastActivityDate="2010-12-09T03:54:07.990" />
  <row Id="13603" PostTypeId="1" AcceptedAnswerId="13604" CreationDate="2010-12-09T03:57:53.103" Score="2" ViewCount="59" Body="&lt;p&gt;Suppose, $k,k^&amp;#39;$ are fields such that $k\subset k&amp;#39;$ and $x_1,x_2,...,x_n,t$ are indeterminates over $k^&amp;#39;$. Suppose we have a homomorphism of rings given by $f:k[x_1,x_2,...,x_n]\to k[t]$, where $$x_1\to f_1(t)$$ $$x_2\to f_2(t)$$ $$.$$ $$.$$ $$x_n\to f_n(t)$$ Let the kernel of this map be $I$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What can be said about the kernel $I^&amp;#39;$ of the map $f^&amp;#39;:k^&amp;#39;[x_1,x_2,...,x_n]\to k^&amp;#39;[t]$, where restriction of $f^&amp;#39;$ to $k[x_1,...,x_n]$ is $f$ (note this completely determines $f^&amp;#39;$ since it determines its action on the indeterminates). The ideal $I$ represents the &quot;ideal of relations&quot; for the polynomials $f_1,...,f_n$. So it seems there may be a compact description for $I^&amp;#39;$ in terms of the generators of $I$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular what can we say when $k=\mathbb{Q}$ and $k^&amp;#39;=\mathbb{C}$. This case is important because Macaulay2 can compute kernels of such maps when the field of coefficients is $\mathbb{Q}$. So we can use those results to compute the kernels of such maps when the field is $\mathbb{C}$. (For the same reasons what can be said when $k$ is a finite field?)&lt;/p&gt;&#xA;" OwnerUserId="3431" LastEditorUserId="3431" LastEditorDisplayName="" LastEditDate="2010-12-09T04:19:28.770" LastActivityDate="2010-12-09T04:19:28.770" Title="Behavior of kernels of homomorphisms of polynomial rings under a change of the field of coefficients" Tags="&lt;commutative-algebra&gt;&lt;ring-theory&gt;" AnswerCount="1" />
  <row Id="13604" PostTypeId="2" ParentId="13603" CreationDate="2010-12-09T04:04:46.047" Score="3" ViewCount="" Body="&lt;p&gt;If $I$ is the original kernel, you have a short exact sequence $$0\to I\to k[x_1,\dots,x_n]\to k[t]\to0$$ Apply the $(\mathord-)\otimes_k{k&amp;#39;}$ functor, which is exact.&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2010-12-09T04:04:46.047" CommentCount="1" />
  <row Id="13605" PostTypeId="1" AcceptedAnswerId="19533" CreationDate="2010-12-09T04:18:44.373" Score="1" ViewCount="237" Body="&lt;p&gt;As I understand, martingale is a stochastic process (i.e., a sequence of random variables) such that the conditional expected value of an observation at some time $t$, given all the observations up to some earlier time $s$, is equal to the observation at that earlier time $s$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A sequence $Y_1, Y_2, Y_3 ...$ is said to be a martingale with respect to another sequence $X_1, X_2, X_3 ...$ if for all $n$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$E(Y_{n+1}|X_1,...,X_n) = Y_n$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I don't understand how it is defined in terms of filtration. Does filtration discretize the time space of a stochastic process so that we can analyze the process as a martingale? A simple explanation or an example on what is filtration and how it relates to martingale theory would be very helpful. I can then read more detailed content.&lt;/p&gt;&#xA;" OwnerUserId="957" LastActivityDate="2011-03-31T07:38:50.130" Title="martingale and filtration" Tags="&lt;probability-theory&gt;&lt;measure-theory&gt;&lt;probability&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="1" />
  <row Id="13606" PostTypeId="1" AcceptedAnswerId="13607" CreationDate="2010-12-09T05:28:05.033" Score="1" ViewCount="72" Body="&lt;p&gt;In a set of lecture notes, there is an example of calculating the group $E_\text{tors}$ of an elliptic curve. This is the example:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Let $E$ be the elliptic curve &#xA;  $$y^2=x^3-5x+4.$$&#xD;&#xA;  The curve has discriminant&#xA;  $$\Delta=16(4\cdot 125 - 27\cdot 16)=2^6\cdot 17.$$&#xD;&#xA;  Therefore, if $[a:b:1]$ is a torsion point, we must have&#xA;  $$b=0,1,2,4,\text{or}\ 8.$$&#xD;&#xA;  Now the case $b=0$ clearly occurs with the point $[1:0:1]$. Let $f(x)$ denote the function $x^3-5x+4$. This function has derivative $3x^2-5$, which is positive for $|x|\geq 2$. Also, we have &#xA;  $$f(-3)=-8,\quad f(5)=104.$$&#xD;&#xA;  Combining these two statements we see the only possible integer values for $x$ for which $f(x)$ is between $0$ and $64$ are&#xA;  $$x=-2,-1,0,1,2,3,4.$$&#xD;&#xA;  Computing each of these two statements we see that the only possible torsion points are &#xA;  $$[0:\pm 2: 1],[3:\pm 4:1],[0:1:0],[1:0:1].$$&#xD;&#xA;  Also, one computes that &#xA;  $$[0:2:1]+[1:0:1]=[3:4:1].$$&#xD;&#xA;  However, if you compute &#xA;  $$[0:2:1]+[0:2:1]$$&#xD;&#xA;  then you find that the $x$-coordinate is $25/16$. As this is not an integer, by the Lutz-Nagell thereom $[0:2:1]$ is not a torsion point, and therefore $[3:4:1]$ is not a torsion point either. &#xA;  This implies that $E_\text{tors}=\{[0:1:0],[1:0:1]\}\simeq\mathbb{Z}/(2)$.&#xA;  This also show that the rank of $E$ is $\geq 1$. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So I understand why $[0:2:1]+[1:0:1]=[3:4:1]$, but it seems to me that there's the implicit assumption that $[0:1:0]$ and $[1:0:1]$ are already in the torsion group. Thus we know that since $[0:2:1]$ is not in the group, neither is $[3:4:1]$, for when composed with the inverse of $[1:0:1]$, it would give an element not in $E_\text{tors}$, which would mean $E_\text{tors}$ is not closed under composition. My question is, is it obvious that $[0:1:0]$ and $[1:0:1]$ are in the torsion group? Otherwise, I don't see why it would follow from $[0:2:1]+[1:0:1]=[3:4:1]$ that $[3:4:1]$ is not in torsion group, if we don't necessarily already know that $[1:0:1]$ is, and hence it's inverse, is in $E_\text{tors}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, is it generally true that if $[a:b:1]$ is not a torsion point, then neither is $[a,-b:1]$? This seems to be assumed as well. Thanks!&lt;/p&gt;&#xA;" OwnerUserId="1583" LastActivityDate="2010-12-09T06:42:22.617" Title="Understanding how to calculate $E_\text{tors}$ of an elliptic curve." Tags="&lt;number-theory&gt;&lt;elliptic-curves&gt;" AnswerCount="2" />
  <row Id="13607" PostTypeId="2" ParentId="13606" CreationDate="2010-12-09T05:37:23.513" Score="4" ViewCount="" Body="&lt;p&gt;I think that pretty much everything you're asking will become clear if you notice what the neutral element is here (hint: it's the point at infinity). &lt;/p&gt;&#xA;" OwnerUserId="308" LastActivityDate="2010-12-09T05:37:23.513" CommentCount="14" />
  <row Id="13608" PostTypeId="1" AcceptedAnswerId="13609" CreationDate="2010-12-09T05:48:24.953" Score="4" ViewCount="204" Body="&lt;p&gt;How can I prove that $\gcd(7^{79}+5,7^{78}+3) = 4$ ? This was a question on a past exam, so the naive euclidean algorithm doesn't seem to suffice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not really sure where to start with this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: This is exam prep, not homework.&lt;/p&gt;&#xA;" OwnerUserId="181" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-09T06:09:07.073" LastActivityDate="2010-12-09T16:21:36.150" Title="Show that $\gcd(7^{79}+5,7^{78}+3) = 4$" Tags="&lt;number-theory&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="13609" PostTypeId="2" ParentId="13608" CreationDate="2010-12-09T05:52:01.057" Score="7" ViewCount="" Body="&lt;p&gt;For a first step, $7^{79}+5 - 7*(7^{78}+3) = -16$, which gets you a long way.  Then you only need to study the factors of 2 in the numbers.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2010-12-09T06:02:35.100" LastActivityDate="2010-12-09T06:02:35.100" CommentCount="3" />
  <row Id="13611" PostTypeId="2" ParentId="13608" CreationDate="2010-12-09T06:02:25.320" Score="4" ViewCount="" Body="&lt;p&gt;Since $\gcd(a,b)=\gcd(a-b,b)$, $$\begin{align}&#xD;&#xA;\gcd(7^{79}+5,7^{78}+3)&#xD;&#xA;&amp;amp;=\gcd(7\cdot 7^{78}+5-7^{78}-3,7^{78}+3)&#xD;&#xA;\\&#xD;&#xA;&amp;amp;=\gcd(6\cdot 7^{78}+2,7^{78}+3)&#xD;&#xA;\\&#xD;&#xA;&amp;amp;=\gcd(6\cdot 7^{78}+2-7^{78}-3,7^{78}+3)&#xD;&#xA;\\&#xD;&#xA;&amp;amp;=\gcd(5\cdot 7^{78}-1,7^{78}+3)&#xD;&#xA;\\&#xD;&#xA;&amp;amp;\vdots&#xD;&#xA;\\&#xD;&#xA;&amp;amp;=\gcd(7^{78}-13,7^{78}+3)&#xD;&#xA;\\&#xD;&#xA;&amp;amp;=\gcd(7^{78}-13,7^{78}+3-7^{78}+13)&#xD;&#xA;\\&#xD;&#xA;&amp;amp;=\gcd(7^{78}-13,16)&#xD;&#xA;\\&#xD;&#xA;\end{align}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From there, I'd determine the remainder when $7^{78}$ is divided by 16 and use that to see how $7^{78}-13$ compares to 16.&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-12-09T06:02:25.320" />
  <row Id="13612" PostTypeId="2" ParentId="13575" CreationDate="2010-12-09T06:03:10.137" Score="1" ViewCount="" Body="&lt;p&gt;I'm doing exactly the same thing as you right now.  I'm self-learning differential topology and differential geometry.  To those ends, I really cannot recommend John Lee's &quot;Introduction to Smooth Manifolds&quot; and &quot;Riemannian Manifolds: An Introduction to Curvature&quot; highly enough.  &quot;Smooth Manifolds&quot; covers Stokes Theorem, the de Rham theorem and more, while &quot;Riemnannian Manifolds&quot; covers connections, metrics, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The attention to detail that Lee writes with is so fantastic.  When reading his texts that you know you're learning things the standard way with no omissions.  And of course, the same goes for his proofs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Plus, the two books are the second and third in a triology (the first being his &quot;Introduction to Topological Manifolds&quot;), so they were really meant to be read in this order.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, I also agree that Guilleman and Pollack, Hirsch, and Milnor are great supplements, and will probably emphasize some of the topological aspects that Lee doesn't go into.&lt;/p&gt;&#xA;" OwnerUserId="640" LastActivityDate="2010-12-09T06:03:10.137" />
  <row Id="13613" PostTypeId="2" ParentId="13596" CreationDate="2010-12-09T06:09:33.073" Score="1" ViewCount="" Body="&lt;p&gt;$Y$ counts the number of successes in $n$ trials. $n - Y$ counts the number of failures in $n$ trials. Notion of success and failure is arbitrary. $P[Y = i] = P[n - Y = n - i]$, so $n-Y$ is binomially distributed with &quot;success&quot; probability $1 - p$.&lt;/p&gt;&#xA;" OwnerUserId="1527" LastActivityDate="2010-12-09T06:09:33.073" />
  <row Id="13614" PostTypeId="2" ParentId="13575" CreationDate="2010-12-09T06:11:35.527" Score="1" ViewCount="" Body="&lt;p&gt;Like the other posters, I think Lee's books are fantastic.  I'd start with his &lt;em&gt;Introduction to Smooth Manifolds&lt;/em&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For differential geometry, I'd go on to his &lt;em&gt;Riemannian Manifolds&lt;/em&gt; and then follow up with do Carmo's &lt;em&gt;Riemannian Geometry&lt;/em&gt;.  (That's what I did.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For differential topology, after Lee's &lt;em&gt;Smooth Manifolds&lt;/em&gt;, I'd suggest &lt;em&gt;Differential Forms in Algebraic Topology&lt;/em&gt; by Bott, Tu and anything (and everything) by Milnor.&lt;/p&gt;&#xA;" OwnerUserId="4074" LastActivityDate="2010-12-09T06:11:35.527" />
  <row Id="13615" PostTypeId="2" ParentId="13606" CreationDate="2010-12-09T06:28:39.203" Score="1" ViewCount="" Body="&lt;p&gt;The point here is that the elliptic curve is not just the affine line whose equation you wrote; actually it is the projective line whose equation is given by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ y^z = x^3 + axz^2 + bz^3$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;whose solutions are triples of rational numbers $[a,b,c]$ not all of which are zero, with the additional condition that the two triples $[ta,tb,tc]$ is treated as the same as $[a,b,c]$ for rational $t \neq 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The line you wrote is an affine segment of the projective line. The projective point $[a,b,c]$ on the projective line with $c\neq 0$ is equivalent to $[a/c,b/c,1]$ on the affine line. But this transformation works only for $c \neq 0$, and to treat $c = 0$, you must include the equivalence class of one additional point, $[0,1,0]$, which is called the point at infinity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That point happens to be the identity for the addition in the group law.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, you can write an explicit formula for addition and verify everything.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, the inverse of the element $[a,b,1]$ is $[a,-b,1]$. You can check that in any abelian group $G$, if $x \in G$ is torsion, then $-x$ is also torsion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More generally, to compute the torsion points on an elliptic curve, you can start with the &lt;a href=&quot;http://en.wikipedia.org/wiki/Nagell%E2%80%93Lutz_theorem&quot; rel=&quot;nofollow&quot;&gt;Nagell-Lutz theorem&lt;/a&gt; for a set of possible points and then do computation on each of those points to make an explicit check. The &lt;a href=&quot;http://pari.math.u-bordeaux.fr/&quot; rel=&quot;nofollow&quot;&gt;PARI&lt;/a&gt; routine &lt;em&gt;elltors&lt;/em&gt; can do it automatically for you; for the syntax just download and check some reference for PARI.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Here in your particular example, you have to check manually that $[1,0,1]$ is in the torsion group. Write down the formula and check, or use the appropriate PARI routine. I have explained that $[0,1,0]$ is the identity. The rest, you seem to have got right.&lt;/p&gt;&#xA;" OwnerUserId="1119" LastEditorUserId="1119" LastEditorDisplayName="" LastEditDate="2010-12-09T06:42:22.617" LastActivityDate="2010-12-09T06:42:22.617" CommentCount="1" />
  <row Id="13616" PostTypeId="1" CreationDate="2010-12-09T06:42:57.503" Score="5" ViewCount="94" Body="&lt;p&gt;Consider large (100,000+ vertices, say) graphs, which we think of as representing some population with edges representing some form of symmetric relation.  They might be the Friend graph of Facebook, mathematicians with the collaboration relation, or a large computer network.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;These networks have the property that they are neither highly structured, nor totally random.  No information about other edges on the graph can tell me for certain whether a given pair of vertices is connected.  That said, if a given pair of vertices have many common neighbors, then it is considerably more likely that they are connected by an edge (so it is not entirely random).  I've seen some lectures on graphs like this, and I understand they are a productive area of research (see, for instance, Kleinberg or Lovasz).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am curious about the following phenomenon (my description is vague, but part of my question is asking for a good definition).  These networks tend to have subsets (which I will call '&lt;strong&gt;clumps&lt;/strong&gt;') which are significantly more connected to each other than to the average vertex in the graph.  Consider a college in Facebook or a research group in mathematics, for example.  If the graphs were small enough to draw in a reasonable way, such clumps would be obvious to the naked eye.  For very large graphs, this is impractical; so instead, I ask,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;1) What is a graph-theoretic way to characterize these clumps?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Clearly, there won't be a yes-no criterion, but I am hoping for some quantity that measures how much a given subset is a clump.  This should also factor in the statistical significance of the clump.  Very small subsets which are highly connected will happen even in totally random graphs, whereas a large subset which is even moderately well-connected is unlikely in a random graph, and would be interesting to find.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;2) Given a graph (and a definition of a clump), how does one find the clumps?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Is there a definition and an algorithm so robust that it can take networks like Facebook or the collaboration graph, and return the clumps that we know are there, like colleges or research discplines?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Oh, and I am not looking for the Szemeredi partition of a graph, which has some similarities to the kinds of partitions I am looking for, but is explicitly a partition of the graph into similar sized chunks.  The clumps in a graph don't have to be the same size, disjoint, or contain every element.&lt;/p&gt;&#xA;" OwnerUserId="1062" LastActivityDate="2010-12-17T03:08:44.013" Title="How to Characterize Clumps in a Large, Semi-Random Graph" Tags="&lt;graph-theory&gt;&lt;pattern-recognition&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="1" />
  <row Id="13617" PostTypeId="1" AcceptedAnswerId="13622" CreationDate="2010-12-09T06:46:57.933" Score="15" ViewCount="333" Body="&lt;p&gt;Background: Let $n$ be an integer and let $p$ be a prime. If $p^{e} || n$, we write $v_{p}(n) = e$. A natural number $n$ is a sum of two integer squares if and only if for each prime $p \equiv 3 \pmod 4$, $v_{p}(n)$ is even. Every natural number is a sum of four squares. A natural number $n$ is a sum of three squares if and only if it is not of the form $4^{k}u$ where $u \equiv 7 \pmod 8$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to know why it is harder to prove the above result for sums of three squares as opposed to sums of two squares or four squares.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've heard somewhere that one way to see this involves modular forms... but I don't remember any details. I would also like to know if there is a formula for the number of ways of representing a natural number n as a sum of three squares (or more generally, $m$ squares) that is similar in spirit to the formulas for the number of ways of representing a natural number as the sum of two squares and four squares. &lt;/p&gt;&#xA;" OwnerUserId="4457" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-09T06:52:19.593" LastActivityDate="2010-12-09T16:39:48.927" Title="Why is it harder to prove which integers are sums of three squares rather than sums of two squares or four squares?" Tags="&lt;number-theory&gt;&lt;algebraic-number-theory&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="5" />
  <row Id="13618" PostTypeId="2" ParentId="13617" CreationDate="2010-12-09T06:56:30.477" Score="14" ViewCount="" Body="&lt;p&gt;An explanation from a different direction than you may be expecting: one reason why the sum-of-three-squares problem is so much harder is that there's no normed division algebra of dimension three.  A key element of both the two-squares and four-squares proof is expressing the product of two numbers in the relevant form as another number in that form; these product formulas 'come from' the formulas for multiplying complex and quaternionic numbers respectively.  That is, taking norms of $(a+bi)(c+di) = (ac-bd) + (ad+bc)i$ gives rise to the formula $(a^2+b^2)(c^2+d^2) = (ac-bd)^2+(ad+bc)^2$, and the equivalent formula for quaternions gives Euler's four-square identity.  John Baez wrote a particularly interesting article on why you don't keep getting division algebras as you keep doubling the dimension; you should be able to find it on his homepage if you're curious.&lt;/p&gt;&#xA;" OwnerUserId="785" LastActivityDate="2010-12-09T06:56:30.477" CommentCount="6" />
  <row Id="13619" PostTypeId="2" ParentId="13608" CreationDate="2010-12-09T06:58:34.580" Score="7" ViewCount="" Body="&lt;p&gt;Modulo the gcd: $\rm\ 7^{78}\ \equiv -3\ \Rightarrow\ 0\ \equiv\ 7^{79}+5\ \equiv\ 7(-3)+5\ \equiv -16\:,\ $ i.e. the gcd dvides $16\:$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now $\rm\ mod\ 8:\ 7^{78}+3\ \equiv\ (-1)^{78}+3\ \equiv\ 4\ \equiv\ (-1)^{79}+5\ \equiv\ 7^{79}+5\:,\ $  i.e. $\ $ gcd $\rm\equiv 4\ (mod\ 8)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence:$\ $ gcd $\rm = 4 + 8\ k\ $ divides $16\:$ implies $\rm\ k = 0\:,\ $ so $\ $ gcd $ = 4\:$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice how the calculations become more intuitive by working in various rings $\rm\ \mathbb Z/m\:.$ Doing such allows us to reuse our well-honed intuition of arithmetic &lt;em&gt;operations&lt;/em&gt; (ring laws), versus the much more cumbersome $\ $ and $\ $ much less intuitive divisibility &lt;em&gt;relation&lt;/em&gt;, i.e. calculating in equational algebras is simpler than calculating in relational algebras, so whenever a problem can be converted from relational to equational it usually yields a simplification.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-12-09T16:21:36.150" LastActivityDate="2010-12-09T16:21:36.150" CommentCount="1" />
  <row Id="13620" PostTypeId="2" ParentId="13477" CreationDate="2010-12-09T07:50:36.327" Score="1" ViewCount="" Body="&lt;p&gt;There is no solution without inverse trigonometric functions, because any elementary geometric argument would, for a given value of the angle between the tangent lines, calculate all 4 arc lengths as linear functions of $\alpha$.  This would imply that, as the (secant) line through $H$ is rotated at uniform angular speed, it sweeps out arclength on the circle at a uniform rate.  But this is impossible because the distance from $H$ to the intersection point, $|HG|$, is not constant.  It is not hard to make this argument rigorous using calculus.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the arcs, in the order listed in the question, are denoted $A, B, b, a$, then  elementary geometry does express $(A-a)$, $(B-b)$, and $(A+B+b+a)$ as linear functions of $\alpha, \beta$ and $1$, as in the other solutions.   I think this set of 3 equations in 4 unknowns is what the OP meant by &quot;not enough equations&quot; to solve everything purely geometrically.&lt;/p&gt;&#xA;" OwnerUserId="467" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-12-09T21:42:59.857" LastActivityDate="2010-12-09T21:42:59.857" />
  <row Id="13621" PostTypeId="2" ParentId="13575" CreationDate="2010-12-09T08:09:39.660" Score="0" ViewCount="" Body="&lt;p&gt;I would recommend Jost's book &quot;Riemannian geometry and geometric analysis&quot; as well as Sharpe's &quot;Differential geometry&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first book is pragmatically written and guides the reader to a lot of interesting stuff, like Hodge's theorem, Morse homology and harmonic maps.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second book is mainly concerned with Cartan connection, but before that it has an excellent chapter on differential topology. Furthermore it treats Ehresmann connections in appendix A.&lt;/p&gt;&#xA;" OwnerUserId="4520" LastActivityDate="2010-12-09T08:09:39.660" />
  <row Id="13622" PostTypeId="2" ParentId="13617" CreationDate="2010-12-09T08:14:19.563" Score="15" ViewCount="" Body="&lt;p&gt;The modular forms explanation is basically due to the fact&#xA;that $3$ is odd and so the generating function for representations&#xA;of sums of three squares is a modular form of half-integer weight.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general if $r_k(n)$ is the number of representations of $n$&#xD;&#xA;as a sum of $k$ squares then&#xA;$$\sum_{n=0}^\infty r_k(n)q^n=\theta(z)^k$$&#xD;&#xA;where $q=\exp(\pi i z)$ and &#xA;$$\theta(z)=1+2\sum_{n=1}^\infty q^{n^2}.$$&#xD;&#xA;Then $f_k(z)=\theta(z)^k$ is a modular form of weight $k/2$ for the group&#xA;$\Gamma_0(4)$. This means that&#xA;$$f_k((az+b)/(cz+d))=(cz+d)^{k/2}f_k(z)$$&#xD;&#xA;whenever the matrix $\begin{pmatrix}a&amp;amp;b\\\\c&amp;amp;d\end{pmatrix}$&#xD;&#xA;lies in $\Gamma_0(4)$, that is $a$, $b$, $c$ and $d$ are integers, $4\mid c$&#xD;&#xA;and $ad-bc=1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This definition is easy to understand when $k$ is even, but for odd&#xA;$k$ one needs to take the correct branch of $(cz+d)^{k/2}$, and&#xA;this is awkward. The space of modular forms of weight $k/2$ is&#xA;finite-dimensional for all $k$, and is one-dimensional for small enough $k$.&#xA;For these small $k$ the space is spanned by an &quot;Eisenstein series&quot;.&#xA;Computing the Eisenstein series isn't too hard for even $k$, but is&#xA;much nastier for odd $k$ where again square roots need to be&#xA;dealt with. See Koblitz's book on modular forms and elliptic functions&#xA;for the calculation for $k\ge5$ odd. The calculation for $k=3$&#xD;&#xA;is even nastier as the Eisenstein series does not converge absolutely.&#xA;In fact the cases where $k$ is divisible by $4$ are even easier, as&#xA;even weight modular forms behave nicer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For large $k$, Eisenstein series are no longer enough, one needs&#xA;also &quot;cusp forms&quot;. While fascinating, cusp forms have coefficients&#xA;which aren't given by nice formulae unlike Eisenstein series.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course there is a formula for $r_3(n)$, due to Gauss in his&#xA;&lt;em&gt;Disquisitiones Arithmeticae&lt;/em&gt;. It involves class numbers of quadratic&#xA;fields (or to Gauss numbers of classes of integral quadratic forms).&lt;/p&gt;&#xA;" OwnerUserId="226" LastEditorUserId="226" LastEditorDisplayName="" LastEditDate="2010-12-09T16:39:48.927" LastActivityDate="2010-12-09T16:39:48.927" CommentCount="4" />
  <row Id="13624" PostTypeId="1" CreationDate="2010-12-09T08:45:34.767" Score="2" ViewCount="83" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Let $X$ be a topological space such that every open cover has a finite refinement. Then is $X$ compact, or is there a counterexample?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Let $X$ be a topological space such that every open cover has a locally finite subcover. Then is $X$ compact, or is there a counterexample?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="4561" LastActivityDate="2010-12-25T07:01:29.440" Title="Weakening paracompactness condition." Tags="&lt;general-topology&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="13625" PostTypeId="1" CreationDate="2010-12-09T08:45:55.577" Score="0" ViewCount="238" Body="&lt;p&gt;I was reading &lt;a href=&quot;http://math.stackexchange.com/questions/8808/why-the-interest-in-locally-euclidean-spaces&quot;&gt;this question&lt;/a&gt;, and I idly wondered if there could be spaces that were globally Euclidean, but locally not, and couldn't think of any circumstances in which that might be the case.&lt;/p&gt;&#xA;" OwnerUserId="890" LastActivityDate="2010-12-11T01:47:54.680" Title="Are there spaces that are globally Euclidean, but locally not?" Tags="&lt;general-topology&gt;" AnswerCount="3" CommentCount="3" />
  <row Id="13627" PostTypeId="1" AcceptedAnswerId="13668" CreationDate="2010-12-09T09:13:31.177" Score="6" ViewCount="294" Body="&lt;p&gt;&lt;strong&gt;The Question:&lt;/strong&gt; Are there any ways that &quot;applied&quot; mathematicians can use Homology theory?  Have you seen any good applications of it to the &quot;real world&quot; either directly or indirectly?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Why do I care?&lt;/strong&gt;  Topology has appealed to me since beginning it in undergrad where my university was more into pure math.  I'm currently in a program where the mathematics program is geared towards more applied mathematics and I am constantly asked, &quot;Yeah, that's cool, but what can you use it for in the real world?&quot;  I'd like to have some kind of a stock answer for this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Full Disclosure.&lt;/strong&gt; I am a first year graduate student and have worked through most of Hatcher, though I am not by any means an expert at any topic in the book.  This is also my first post on here, so if I've done something wrong just tell me and I'll try to fix it.&lt;/p&gt;&#xA;" OwnerUserId="2959" LastEditorUserId="1543" LastEditorDisplayName="" LastEditDate="2010-12-09T13:04:18.547" LastActivityDate="2011-04-24T10:18:02.010" Title="Applications for Homology." Tags="&lt;topology&gt;&lt;algebraic-topology&gt;&lt;applications&gt;" AnswerCount="2" CommentCount="6" FavoriteCount="2" />
  <row Id="13628" PostTypeId="2" ParentId="13111" CreationDate="2010-12-09T09:15:49.477" Score="0" ViewCount="" Body="&lt;p&gt;Ok..I got something here..John.this is betty from class..let me know what you guys think of this.&#xA;Proof:&#xA;Let G be a simple group of order 300. i.e, |G|=300&#xA;assume ----&gt;&amp;lt;----- that G is simple. &#xA;Since |G|=300=2^2*3*5^2, then there exists a sylow 5-subgroup, say P that has order 25. Now s_5 is congruent to 1(mod 5) and s_5 divides 300/25=12. Hence s_5=1 or 6. But if s_5=1, then P&amp;lt;|G, which is impossible. Thus s_5=6 and |G:N_G(P)|=6. Now recall a thm, proved from class. &quot;Let G be a group. If there exists a subgroup of G s.t [G:H]=n, then there exists a N&amp;lt;|G, s.t N C H, and [G:N] | n!.&quot; We apply this thm to |G:N_G(P)|=6 now. If |G:N_G(P)|=6, then there exists a normal subgroup of G that is contained in N_G(P), say this subgroup is P (again), where P&amp;lt;|G and P C N_G(P), then [G:P] | 6!. But this is impossible, since |G|=300 does not divide 6!. Also, if G were simple, then P would be 1 or G, and we already showed above that P cannot be equal to 1, and P is not equal to G. Thus we have reached a contradiction. Therefore G cannot be simple, and there is no simple group of order 300.&lt;/p&gt;&#xA;" OwnerUserId="4562" LastActivityDate="2010-12-09T09:15:49.477" />
  <row Id="13630" PostTypeId="1" AcceptedAnswerId="13645" CreationDate="2010-12-09T09:41:23.547" Score="0" ViewCount="45" Body="&lt;p&gt;Let $X1,X2$ be independent random variables with $P[Xi = 0] = P[Xi = 1] = \frac{1}{2}$ for $i = 1, 2$. What is the random variable set $A1 = \{X1 = X2\}$ mean? I came across this representation in the book but not sure what this means.&lt;/p&gt;&#xA;" OwnerUserId="957" LastActivityDate="2010-12-09T13:14:25.287" Title="what does this set definition mean defined on independent random variables?" Tags="&lt;probability-theory&gt;&lt;measure-theory&gt;" AnswerCount="3" />
  <row Id="13631" PostTypeId="1" AcceptedAnswerId="13696" CreationDate="2010-12-09T09:41:46.680" Score="1" ViewCount="212" Body="&lt;p&gt;One of false beliefs in &lt;a href=&quot;http://mathoverflow.net/questions/23478/23640#23640&quot; rel=&quot;nofollow&quot;&gt;this question on Math Overflow&lt;/a&gt; is &quot;If f is a smooth function with df=0, then f is constant&quot;. What is a counterexample to this statement? Can it be made correct by adding some restriction, e.g. that f is a function from reals to reals?&lt;/p&gt;&#xA;" OwnerUserId="947" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-09T09:46:47.723" LastActivityDate="2010-12-09T18:58:07.143" Title="Example of a smooth function with zero derivative that is not constant." Tags="&lt;calculus&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="13633" PostTypeId="2" ParentId="13630" CreationDate="2010-12-09T09:48:49.330" Score="0" ViewCount="" Body="&lt;p&gt;It's the event that $X_1 = X_2$.  So $X_1 = 0, X_2 = 0$ or $X_1 = 1, X_2 = 1$. &lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-12-09T09:48:49.330" />
  <row Id="13634" PostTypeId="2" ParentId="13534" CreationDate="2010-12-09T09:56:26.953" Score="1" ViewCount="" Body="&lt;p&gt;Actually, this can be done very simply using the following two facts, both of which follow from things the students should know at this point (though they may not have seen a centralizer before, it is not too hard to understand).&#xA;First, any subgroup of prime index is maximal. This follows directly from Lagrange.&#xA;Second, the center is never maximal. This follows by observing that if the center is not the entire group, let x be an element not in the center. Then the centralizer of x is not the entire group, since then x would be in the center. But clearly the center is contained in the centralizer of x and does not contain x, so the centralizer of x is a subgroup properly between the center and the entire group.&#xA;Now it is easy to see that the group must be abelian, and then it is easy to show that it must in fact be cyclic (it has an element of order 2, which is simpler to show than Cauchy, and then it just needs to be shown that it cannot have 2 elemens of order 2, since then it would have a subgroup of order 4 contradicting Lagrange).&#xA;(As a side note, Cauchy's theorem for abelian groups is fairly simple to show just from knowledge of the order of a product of subgroups, and in fact the general theorem then follows from the class equation)&lt;/p&gt;&#xA;" OwnerUserId="2538" LastActivityDate="2010-12-09T09:56:26.953" />
  <row Id="13635" PostTypeId="2" ParentId="13624" CreationDate="2010-12-09T09:59:52.223" Score="2" ViewCount="" Body="&lt;p&gt;I quote from Wikipedia's article on &lt;a href=&quot;http://en.wikipedia.org/wiki/Paracompact_space&quot; rel=&quot;nofollow&quot;&gt;paracompact spaces&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;Note the similarity between the definitions of compact and paracompact: for paracompact, we replace &quot;subcover&quot; by &quot;open refinement&quot; and &quot;finite&quot; by &quot;locally finite&quot;. Both of these changes are significant: if we take the above definition of paracompact and change &quot;open refinement&quot; back to &quot;subcover&quot;, or &quot;locally finite&quot; back to &quot;finite&quot;, we end up with the compact spaces in both cases.&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The page lists Willard's &lt;em&gt;General Topology&lt;/em&gt; in its references, so that seems like a good place to go hunting for a proof - or an exercise.&lt;/p&gt;&#xA;" OwnerUserId="1269" LastActivityDate="2010-12-09T09:59:52.223" CommentCount="2" />
  <row Id="13636" PostTypeId="1" CreationDate="2010-12-09T10:04:31.250" Score="0" ViewCount="40" Body="&lt;p&gt;The category of simplicial presheaves on a small Grothendieck site $\mathcal{C}$ can be given a model structure by defining weak equivalences and cofibrations sectionwise. It's called the (global) injective model structure and has a mapping space functor $Hom$, given by $Hom(X, Y)_n = hom(X\times \Delta^n, Y)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the given topology, one can actually define 'local' weak equivalences and Jardine showed that the left Bousfield localization at the class of local weak equivalence exists. This model category structure is called local injective model structure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, it's often mentioned that this is also the same as localizing the injective model structure at the class $S := \{X\rightarrow L^2X\}$ (where $L^2$ is the sheafification functor).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if $W$ denotes the class of local weak equivalences, this amounts to say that $S$-local is the same as $W$-local, that is for an injective fibrant simplicial presheav $A$ the following are equivalent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) $Hom(L^2X, A) \cong Hom(X, A)$ for all $X$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) $Hom(Y, A) \cong Hom(X, Y)$ for all local weak equivalences $X\rightarrow Y$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here, $\cong$ stands for weak equivalence of simplicial sets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) implies 1), of course. But how does 1) imply 2)? Or am I mistaken?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; In general 1) does not imply 2), and this is not stated anywhere. I mixed things up.&lt;/p&gt;&#xA;" OwnerUserId="4563" LastEditorUserId="4563" LastEditorDisplayName="" LastEditDate="2010-12-09T19:35:44.637" LastActivityDate="2010-12-09T19:35:44.637" Title="Local injective model structure for simplicial presheaves" Tags="&lt;homotopy-theory&gt;" CommentCount="3" />
  <row Id="13637" PostTypeId="2" ParentId="7949" CreationDate="2010-12-09T10:22:23.303" Score="2" ViewCount="" Body="&lt;p&gt;Considering the category $\Sigma_R(C) $ of the exat seqences like: $0\to A\to B\to C$ this category is complete and the inclusion $\Sigma(C)  \subset  \Sigma _R(C)$ is coreflexive (the coreflector come from the coker of the last arrow), then limits in $\Sigma(C) $ exixst and are gived by the limit in $\Sigma_R(C) $ followed by the coreflection.&#xA;WHen the inclusion $\Sigma(C)  \subset  \Sigma _R(C)$ preserve (directed) limits is argument of Mittag-Leffler teorem's.&lt;/p&gt;&#xA;" OwnerUserId="4565" LastActivityDate="2010-12-09T10:22:23.303" CommentCount="1" />
  <row Id="13638" PostTypeId="1" CreationDate="2010-12-09T10:25:06.450" Score="4" ViewCount="66" Body="&lt;p&gt;The definition of &lt;a href=&quot;http://en.wikipedia.org/wiki/Gauss-Manin_connection&quot; rel=&quot;nofollow&quot;&gt;Gauss-Manin&lt;/a&gt; connection involves de Rham cohomology. Surely, Gauss didn't work with de Rham cohomology as we know it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, what was the context in which Gauss came up with this idea?&lt;/p&gt;&#xA;" OwnerUserId="4566" LastActivityDate="2010-12-09T10:25:06.450" Title="What is the Gauss part of Gauss-Manin connection?" Tags="&lt;algebraic-geometry&gt;&lt;differential-geometry&gt;" CommentCount="1" />
  <row Id="13639" PostTypeId="1" CreationDate="2010-12-09T11:12:35.300" Score="1" ViewCount="61" Body="&lt;p&gt;Let Let $E \rightarrow X$ be a vector bundle on a manifold $X$. Let $\cal E$ be the sheaf of sections of $E$. Let $\cal F$ be a subsheaf of $\cal E$, and let $F$ be the etale space of $\cal F$. What is an example that the map $F \rightarrow E$ might not be an injection on all fibers?&lt;/p&gt;&#xA;" OwnerUserId="4561" LastEditorUserId="4561" LastEditorDisplayName="" LastEditDate="2010-12-09T11:16:15.170" LastActivityDate="2010-12-09T16:22:38.233" Title="Subbundles and subsheaves." Tags="&lt;differential-topology&gt;&lt;sheaf-theory&gt;" AnswerCount="1" CommentCount="0" />
  <row Id="13640" PostTypeId="1" AcceptedAnswerId="13650" CreationDate="2010-12-09T11:20:43.357" Score="1" ViewCount="148" Body="&lt;p&gt;Classify up to similarity all 3 x 3 complex matrices $A$ such that $A^n$ = $I$.&lt;/p&gt;&#xA;" OwnerUserId="4433" LastActivityDate="2011-04-17T20:18:33.883" Title="classification up to similarity of complex n-by-n matrices" Tags="&lt;linear-algebra&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="13641" PostTypeId="1" AcceptedAnswerId="13643" CreationDate="2010-12-09T11:48:55.787" Score="8" ViewCount="330" Body="&lt;p&gt;I'm reading the Algebra book by Knapp and he mentions in passing that an orthonormal set in an infinite dimension vector space is &quot;never large enough&quot; to be a vector-space basis (i.e. that every vector can be written as a &lt;strong&gt;finite&lt;/strong&gt; sum of vectors from the basis; such bases do exist for infinite dimension vector spaces by virtue of the axiom of choice, but usually one works with orthonormal sets for which &lt;strong&gt;infinite&lt;/strong&gt; sums yield all the elements of the vector space).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So my question is - how can we prove this claim? (That an orthonormal set in an infinite dimension vector space is not a vector space basis).&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Edit (by Jonas Meyer): Knapp says in a footnote on &lt;a href=&quot;http://books.google.com/books?id=NSXCaGSVaX4C&amp;amp;lpg=PR1&amp;amp;pg=PA92&quot; rel=&quot;nofollow&quot;&gt;page 92&lt;/a&gt; of &lt;em&gt;Basic algebra&lt;/em&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In the infinite-dimensional theory the term &quot;orthonormal basis&quot; is used for an orthonormal set that spans $V$ when &lt;em&gt;limits&lt;/em&gt; of finite sums are allowed, in addition to finite sums themselves; when $V$ is infinite-dimensional, an orthonormal basis is never large enough to be a vector-space basis.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Thus, without explicitly using the word, Knapp is referring only to &lt;em&gt;complete&lt;/em&gt; infinite-dimensional inner product spaces.&lt;/p&gt;&#xA;" OwnerUserId="1818" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2010-12-10T03:44:17.727" LastActivityDate="2010-12-10T03:44:17.727" Title="An orthonormal set cannot be a basis in an infinite dimension vector space?" Tags="&lt;linear-algebra&gt;&lt;orthonormal&gt;" AnswerCount="3" CommentCount="6" FavoriteCount="2" />
  <row Id="13642" PostTypeId="1" AcceptedAnswerId="13670" CreationDate="2010-12-09T11:53:01.720" Score="1" ViewCount="55" Body="&lt;p&gt;I'm modeling a process with a Markov chain like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;Pr(i \rightarrow i) = \alpha,&#xD;&#xA;Pr(i \rightarrow i+1) = \beta,&#xD;&#xA;Pr(i \rightarrow i+2) = \gamma&#xD;&#xA;$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All other probabilities are zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any way to calculate the expected number of steps it takes to reach state $n$ from state $0$? So far, all I've come up is binomial distribution in the case of $\alpha = 0$, but I'm unsure of even that.&lt;/p&gt;&#xA;" OwnerUserId="4567" LastEditorUserId="4567" LastEditorDisplayName="" LastEditDate="2010-12-09T11:58:36.133" LastActivityDate="2010-12-09T15:42:22.697" Title="Persistent stationary Markov chain properties" Tags="&lt;statistics&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="13643" PostTypeId="2" ParentId="13641" CreationDate="2010-12-09T12:01:05.393" Score="10" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Edit.&lt;/strong&gt; The answer assumes that one works in a Hilbert space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $\mathcal A$ denote the set of orthonormal vectors. Then for any orthonormal sequence $(e_n)\subset\mathcal A $,  the element &#xA;$$u=\sum_{n=1}^{\infty}\frac{1}{n}e_n$$&#xD;&#xA;cannot be represented as a finite sum of vectors from  $\mathcal A$.&lt;/p&gt;&#xA;" OwnerUserId="723" LastEditorUserId="723" LastEditorDisplayName="" LastEditDate="2010-12-09T12:41:21.537" LastActivityDate="2010-12-09T12:41:21.537" CommentCount="4" />
  <row Id="13644" PostTypeId="2" ParentId="13641" CreationDate="2010-12-09T12:19:55.093" Score="12" ViewCount="" Body="&lt;p&gt;Take any infinite dimensional inner product space $V$ and any orthonormal sequence $(w_n : n \in \mathbb{N})$. Let $W$ be the subspace generated by this sequence. Then $W$ is certainly an infinite dimensional vector space (because it has an infinite independent subset). Also $W$ has an orthonormal basis, because the inner product on $W$ is inherited from $V$ and thus $(w_n)$ is still an orthonormal sequence in $W$.  This means that the theorem you have suggested, &quot;an orthonormal set in an infinite dimension vector space is not a vector space basis&quot;, is not true.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I believe might be true is that no infinite dimensional &lt;em&gt;complete&lt;/em&gt; inner product space has a orthonormal basis. This is the question that Andrey Rekalo addressed in another answer.&lt;/p&gt;&#xA;" OwnerUserId="630" LastActivityDate="2010-12-09T12:19:55.093" CommentCount="1" />
  <row Id="13645" PostTypeId="2" ParentId="13630" CreationDate="2010-12-09T12:19:57.587" Score="3" ViewCount="" Body="&lt;p&gt;Although this question may look very simple, a full understanding of it is not. For general purposes, I give the following elaborated answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, if $X_1$ and $X_2$ are any two random variables on a common probability space $(\Omega,\mathcal{F},{\rm P})$, then $\lbrace X_1 = X_2 \rbrace$ stands as a shorthand for the event $\lbrace \omega \in \Omega : X_1 {(\omega)} = X_2 {(\omega)} \rbrace$. For your specific example, we can define $(\Omega,\mathcal{F},{\rm P})$ as follows:&#xA;$\Omega = \lbrace (0,0),(0,1),(1,0),(1,1) \rbrace$, $\mathcal{F}=2^\Omega$ (the power set of $\Omega$), and, for any $\omega = (i,j) \in \Omega$, ${\rm P}(\lbrace \omega \rbrace) = 1/4$. Note that, by additivity, this determines (the probability measure) ${\rm P}$ on $\mathcal{F}$; for example,&#xA;$$&#xD;&#xA;{\rm P}(\lbrace (0,0),(1,0),(1,1)\rbrace) = {\rm P}(\lbrace (0,0)\rbrace) + {\rm P}(\lbrace (1,0)\rbrace) + {\rm P}(\lbrace (1,1)\rbrace) = 3/4.&#xD;&#xA;$$&#xD;&#xA;Now, we can define the random variables $X_1$ and $X_2$ as follows. For any $\omega = (i,j) \in \Omega$, $X_1 {(\omega)} = i$ and $X_2 {(\omega)} = j$. Thus, &#xA;$$&#xD;&#xA;{\rm P}(X_1 = 0) := {\rm P}(\lbrace \omega \in \Omega : X_1 {(\omega)} = 0 \rbrace) = {\rm P}(\lbrace (0,0),(0,1) \rbrace) = 1/2&#xD;&#xA;$$&#xD;&#xA;and&#xA;$$&#xD;&#xA;{\rm P}(X_1 = 1) := {\rm P}(\lbrace \omega \in \Omega : X_1 {(\omega)} = 1 \rbrace) = {\rm P}(\lbrace (1,0),(1,1) \rbrace) = 1/2,&#xD;&#xA;$$&#xD;&#xA;and analogously for $X_2$. As for the event in question,&#xA;$$&#xD;&#xA;\lbrace X_1 = X_2 \rbrace : = \lbrace \omega \in \Omega : X_1 {(\omega)} = X_2 {(\omega)} \rbrace = \lbrace (0,0),(1,1) \rbrace,&#xD;&#xA;$$&#xD;&#xA;and so&#xA;$$&#xD;&#xA;{\rm P}(X_1 = X_2) := {\rm P}(\lbrace \omega \in \Omega : X_1 {(\omega)} = X_2 {(\omega)} \rbrace) = {\rm P}(\lbrace (0,0),(1,1) \rbrace) = 1/2.&#xD;&#xA;$$&#xD;&#xA;Finally, let's see that $X_1$ and $X_2$ are indeed independent (this should be clear from our construction). This amounts to showing that, for any $i,j \in \lbrace 0,1 \rbrace$, &#xA;$$&#xD;&#xA;{\rm P}(X_1 = i, X_2 = j) = {\rm P}(X_1 = i) {\rm P}(X_2 = j), &#xD;&#xA;$$&#xD;&#xA;that is,&#xA;$$&#xD;&#xA;{\rm P}(\lbrace \omega \in \Omega : X_1 {(\omega)} = i, X_2 {(\omega)} = j \rbrace) = {\rm P}(\lbrace \omega \in \Omega : X_1 {(\omega)} =i  \rbrace ){\rm P}(\lbrace \omega \in \Omega : X_2 {(\omega)} =j  \rbrace ).&#xD;&#xA;$$&#xD;&#xA;Indeed, both sides are equal to $1/4$.&lt;/p&gt;&#xA;" OwnerUserId="2810" LastActivityDate="2010-12-09T12:19:57.587" CommentCount="1" />
  <row Id="13646" PostTypeId="1" AcceptedAnswerId="13647" CreationDate="2010-12-09T12:42:09.437" Score="1" ViewCount="179" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The book &lt;a href=&quot;http://books.google.com/books?id=g9tH8M3NvDIC&amp;amp;printsec=frontcover&amp;amp;dq=irresistible+integrals&amp;amp;hl=pt-PT&amp;amp;ei=W8sATYbBLoSXOu2craYB&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=1&amp;amp;ved=0CCUQ6AEwAA#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;Irresistible Integrals by George Boros and Victor Moll&lt;/a&gt; on page 204 has the following identity &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle \frac{1}{1+x}=\prod_{k=1}^{\infty}\left(\frac{k+x+1}{k+x} \times \frac{k}{k+1}\right)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How does one derive this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="1747" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2010-12-09T16:11:11.700" LastActivityDate="2010-12-09T16:11:11.700" Title="How to derive the following identity?" Tags="&lt;real-analysis&gt;&lt;complex-analysis&gt;" AnswerCount="2" />
  <row Id="13647" PostTypeId="2" ParentId="13646" CreationDate="2010-12-09T12:54:57.213" Score="9" ViewCount="" Body="&lt;p&gt;Multiply out&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\prod_{k=1}^N\left(\frac{k+x+1}{k+x} \times \frac{k}{k+1}\right)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and cancel like terms. You will be left with&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{1}{1+x}\left(\frac{N+x+1}{N+1}\right) .$$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="1301" LastActivityDate="2010-12-09T12:54:57.213" CommentCount="4" />
  <row Id="13648" PostTypeId="1" AcceptedAnswerId="13649" CreationDate="2010-12-09T12:56:06.743" Score="15" ViewCount="587" Body="&lt;p&gt;How do I show that the product $$\biggl(4 - \frac21\biggr) \cdot \biggl(4 - \frac22\biggr) \cdot \biggl(4 - \frac23\biggr) \cdots \biggl(4 - \frac2{n}\biggr)$$ is an integer for any $n \in \mathbb{N}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Source:&lt;/strong&gt; &lt;a href=&quot;http://www.math.muni.cz/~bulik/vyuka/pen-20070711.pdf&quot; rel=&quot;nofollow&quot;&gt;www.math.muni.cz/~bulik/vyuka/pen-20070711.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="9413" LastEditorUserId="9413" LastEditorDisplayName="" LastEditDate="2011-05-22T11:32:27.807" LastActivityDate="2011-05-22T11:32:27.807" Title="Proving $\prod_{j=1}^n \left(4-\frac2{j}\right)$ is an integer" Tags="&lt;number-theory&gt;&lt;elementary-number-theory&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="4" />
  <row Id="13649" PostTypeId="2" ParentId="13648" CreationDate="2010-12-09T13:05:15.317" Score="21" ViewCount="" Body="&lt;p&gt;Show that it's equal to $ { 2n \choose n } . $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: Okay, here's how it goes: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left( 4 - \frac{2}{1} \right)&#xD;&#xA;\left( 4 - \frac{2}{2} \right)&#xD;&#xA;\left( 4 - \frac{2}{3} \right) \cdots&#xD;&#xA;\left( 4 - \frac{2}{n} \right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$= \frac{(4 \cdot 1 - 2)}{1}&#xD;&#xA;\frac{(4 \cdot 2 - 2)}{2}&#xD;&#xA;\frac{(4 \cdot 3 - 2)}{3} \cdots \frac{(4n - 2)}{n}$$&#xA;$$= \frac{2^n (2 \cdot 1 - 1)(2 \cdot 2 - 1)( 2 \cdot 3 - 1) \cdots (2n-1)}{n!}$$&#xA;$$= \frac{2^n \cdot 1 \cdot 3 \cdot 5 \cdots (2n-1)}{n!}$$&#xA;$$= \frac{2^n \cdot 1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdots (2n-1)(2n)}{n! \cdot 2 \cdot 4 \cdots (2n)}$$&#xA;$$ = \frac{2^n (2n)!}{ 2^n (n!)^2} = { 2n \choose n}.$$&lt;/p&gt;&#xA;" OwnerUserId="1301" LastEditorUserId="1301" LastEditorDisplayName="" LastEditDate="2011-01-11T19:30:04.360" LastActivityDate="2011-01-11T19:30:04.360" CommentCount="4" />
  <row Id="13650" PostTypeId="2" ParentId="13640" CreationDate="2010-12-09T13:08:03.540" Score="7" ViewCount="" Body="&lt;p&gt;If you define $p=X^n-1\in\mathbb C[X]$, then $p(A)=0$. This tells you that the minimal polinomial $m_A$ of $A$ divides $p$ and, in particular, that $m_A$ has all its roots simple, because the same is true of $p$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It follows that $A$ is diagonalizable, so, up to similarity, you can suppose that it is diagonal. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you see which are the diagonal matrices $A$ which satisfy the condition $A^n=I$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;NB:&lt;/strong&gt; This argument does not depend on your knowing about Jordan canonical forms.&lt;/p&gt;&#xA;" OwnerUserId="274" LastEditorUserId="274" LastEditorDisplayName="" LastEditDate="2010-12-09T13:19:50.820" LastActivityDate="2010-12-09T13:19:50.820" CommentCount="2" />
  <row Id="13651" PostTypeId="2" ParentId="13630" CreationDate="2010-12-09T13:14:25.287" Score="3" ViewCount="" Body="&lt;p&gt;To add to Shai Covo's answer, it's worth noting that the convention of writing something like $\{X_1 = X_2\}$ for $\{\omega \in \Omega : X_1(\omega) = X_2(\omega)\}$, essentially &quot;suppressing the $\omega$s&quot;, is very common in probability, and you will see it a lot as you continue your studies.  Essentially, any &quot;statement&quot; written inside braces should be interpreted as the set of all $\omega$ for which the statement (which is generally something in terms of random variables, which are really functions of $\omega$) is true.  Other common examples are things like $\{X_n \to X\}$, which is the set of all $\omega$ for which $X_n(\omega)$ converges to $X(\omega)$.  At first you may find it helpful to &quot;fill in the $\omega$s&quot; when reading.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, for intuition, it is often easy to understand such expressions.  If you are doing an experiment where you flip two coins, with heads and tables labeled 0 and 1, $\{X_1 = X_2\}$ is just the event that the two coins come up the same.&lt;/p&gt;&#xA;" OwnerUserId="822" LastActivityDate="2010-12-09T13:14:25.287" />
  <row Id="13652" PostTypeId="1" AcceptedAnswerId="13653" CreationDate="2010-12-09T13:19:04.220" Score="3" ViewCount="62" Body="&lt;p&gt;&lt;strong&gt;Problem:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&lt;strong&gt;From the deck of the 52 cards,cards are drawn randomly without replacement.What is the probability of drawing a king of hearts at the third attempt? If it was drawn at the 15th attempt, what is the probability ?&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I have solved the first by finding P(getting the card at third attempt) = P(not getting at first) $\times$ P(not getting at second) $\times$ P(getting at third attempt) = $\frac{51}{52} \times \frac{50}{51} \times \frac{1}{50} = \frac{1}{52}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This method seems fine when the times to be drawn is less but for 15 or say for $x$ ... I have solved for $15$ by just writing just two lines in &lt;a href=&quot;http://en.wikipedia.org/wiki/C_%28programming_language%29&quot; rel=&quot;nofollow&quot;&gt;C&lt;/a&gt; and got  $\frac{1}{52}$, i do agree that 15 is not much big and it may be possible to use this during exam but does there any tricky method exist?  &lt;/p&gt;&#xA;" OwnerUserId="2109" LastActivityDate="2010-12-09T13:47:02.423" Title="Doubts on a probability problem" Tags="&lt;probability&gt;" AnswerCount="2" />
  <row Id="13653" PostTypeId="2" ParentId="13652" CreationDate="2010-12-09T13:29:44.653" Score="6" ViewCount="" Body="&lt;p&gt;No calculation is required and certainly no C. As you should observe when you start writing out the same product, the denominator of each next term will cancel the numerator of the previous, so the only terms left will be the denominator of the first and the numerator of the last terms. So the answer will always be $\frac{1}{52}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An easy way to see this is to stand the problem on its head: at any attempt, you are equally likely to draw any given card from the deck (why should the kind of hearts be any different from the 9 of clubs?). Since the are 52 cards, you get your probability.&lt;/p&gt;&#xA;" OwnerUserId="3212" LastActivityDate="2010-12-09T13:29:44.653" CommentCount="3" />
  <row Id="13654" PostTypeId="1" AcceptedAnswerId="13678" CreationDate="2010-12-09T13:31:37.007" Score="2" ViewCount="212" Body="&lt;p&gt;It seems like I am facing some confusion while handling with inequalities,I was doing some tasks where it is asked to find the interval of the variable,after some steps I deduced the the necessary expression but I am facing some problem while trying to deduce the final expression.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider for example:&#xA;    $$a^2 \ge 8;\qquad 0 &amp;lt; a &amp;lt; 6; \qquad a &amp;lt; \frac{11}{3}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the start I write $a^2 \ge 8$ as $ a \lt -2\sqrt{2} \quad\text{or}\quad a &amp;gt; 2\sqrt{2} $ but the final answer is $ 2\sqrt{2} &amp;lt; a &amp;lt; \frac{11}{3}$. How can this be deduced?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The original problem is as follows:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If both the roots of the equation $x^2-ax+2 = 0$ lie in the interval $(0,3)$, then find the value of $a$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Since the two roots lie in the same interval, those three conditions follow.&lt;/p&gt;&#xA;" OwnerUserId="3744" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-10T01:22:08.240" LastActivityDate="2010-12-10T02:59:54.770" Title="Problems with Inequalities" Tags="&lt;algebra-precalculus&gt;" AnswerCount="4" CommentCount="2" FavoriteCount="2" />
  <row Id="13655" PostTypeId="2" ParentId="13625" CreationDate="2010-12-09T13:35:07.473" Score="7" ViewCount="" Body="&lt;p&gt;Yes, $\mathbb{Z}$ is one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(&lt;strong&gt;Note&lt;/strong&gt;: As pointed out in the comments, $\mathbb{Z}$ is &lt;em&gt;technically&lt;/em&gt; locally Euclidean as it is discrete.  If you are bothered by such things, replace $\mathbb{Z}$ by something like $\mathbb{Q}$, or the union of copies of the Cantor set.  I think that, despite this flaw, $\mathbb{Z}$ is a better example of the point I wish to make so I'm going to stick with it.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, as Qiaochu and Alex effectively say in the comments, this all depends on what you mean by &quot;globally&quot;.  But before I say what that means, let me point out that &quot;locally X&quot; is not so straightforward.  In topology, we say that a space is &quot;locally &lt;strong&gt;X&lt;/strong&gt;&quot; if whenever we have $p \in U$ with $U$ open (or just a neighbourhood of $p$) then there is some neighbourhood $V$ of $p$ with $p \in V \subseteq U$ such that $V$ has property &lt;strong&gt;X&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This works fine with lots of properties such as connected, path connected, compact, and similar.  When we put the word &quot;Euclidean&quot; in it, something is a bit weird.  What does it mean to say that a space is Euclidean?  We don't tend to define that.  Rather we define &quot;locally Euclidean&quot; all in one go.  It's like the definition of the limit of a sequence.  We write $\lim_{n \to \infty} s_n = s$ but we never decompose that funny symbol &quot;$\lim_{n \to \infty}$&quot; and talk about $n$ &lt;em&gt;actually&lt;/em&gt; going to $\infty$ in the same way that we talk about $x$ going to $0$ in the expression $\lim_{x \to 0} \sin(x) = 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, except that we do.  There &lt;em&gt;is&lt;/em&gt; a way to make sense of $n$ going to $\infty$, but it comes second, not first.  We could happily deal with limits and the like without ever knowing how to really make sense of $n$ going to $\infty$.  But to know what $n$ going to $\infty$ means, we have to know about limits.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So it is with &quot;locally Euclidean&quot;.  We &lt;em&gt;can&lt;/em&gt; decompose this, but it comes second.  And the decomposition, into &quot;locally&quot; + &quot;Euclidean&quot;, is almost determined by what &quot;locally Euclidean&quot; means.  I say almost, and this is the point of the ramble, because there are actually two things that it could mean which both give the same answer for &quot;locally Euclidean&quot;.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;A space is &quot;Euclidean&quot; if it is homeomorphic to $\mathbb{R}^n$ for some $n$.&lt;/li&gt;&#xA;&lt;li&gt;A space is &quot;Euclidean&quot; if it is homeomorphic to an open subset of $\mathbb{R}^n$ for some $n$.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;For &quot;locally Euclidean&quot;, both of these give the same answer because any open subset of $\mathbb{R}^n$ contains lots of small open subsets that are homeomorphic to $\mathbb{R}^n$.  But if we replace &quot;Euclidean&quot; by some other space, this equivalence will break.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And so back to the original question.  Why is $\mathbb{Z}$ globally Euclidean but not locally Euclidean?  Again, we need to know what everything means.  But now on a slightly higher level.  What are we doing when we talk about &quot;locally &lt;strong&gt;X&lt;/strong&gt;&quot;?  We're saying that we can get out our microscopes and examine our space near $p$ and find that property &lt;strong&gt;X&lt;/strong&gt; holds.  By doing this, we effectively ignore a huge part of our space.  In fact, we can ignore almost as much as we like (within reason) and focus down only on a tiny, tiny bit near $p$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what happens if we reverse this process?  Instead of taking a microscope and examining our space really close up, let's put it in a rocket, shoot it off to the Moon and look at it through a telescope.  Then, not only do we not see any of the small detail, we can't really tell if all that detail is there in the first place.  The resolution of our telescope means that we cannot examine our space very closely at all.  Indeed, we can't even tell if it is all there or not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So when put on the Moon, $\mathbb{R}$ and $\mathbb{Z}$ are indistinguishable.  So $\mathbb{Z}$ is Euclidean in the opposite sense to &quot;locally Euclidean&quot; but is (clearly) not locally Euclidean.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you look for this concept in the literature, you won't find it under the heading of &quot;globally &lt;strong&gt;X&lt;/strong&gt;&quot;.  That's because that would be too confusing a term.  Rather, what I'm talking about is &lt;strong&gt;coarse geometry&lt;/strong&gt;.  And it's a lot of fun, and I wish I knew much, much more about it than I do.&lt;/p&gt;&#xA;" OwnerUserId="2907" LastEditorUserId="2907" LastEditorDisplayName="" LastEditDate="2010-12-10T09:24:43.970" LastActivityDate="2010-12-10T09:24:43.970" CommentCount="8" />
  <row Id="13656" PostTypeId="2" ParentId="13654" CreationDate="2010-12-09T13:35:46.867" Score="5" ViewCount="" Body="&lt;p&gt;Everything you have done so far is correct. Once you get down to linear constraints, you essentially have to intersect all the permitted ranges. E.g. since $0&amp;lt;a&amp;lt;6$, the negative solutions $a&amp;lt;-2\sqrt{2}$ are out. Thus the only possibility is $a&amp;gt;2\sqrt{2}$. Now the remaining inequalities also give you an upper bound.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A good strategy is, as I said at the beginnig, to sketch all the ranges and then intersect them to find the range in which &lt;em&gt;all&lt;/em&gt; inequalities are satisfied. If you had some &quot;or&quot;-conditions, you would take unions of ranges, rather than intersections.&lt;/p&gt;&#xA;" OwnerUserId="3212" LastActivityDate="2010-12-09T13:35:46.867" CommentCount="3" />
  <row Id="13657" PostTypeId="1" CreationDate="2010-12-09T13:36:29.940" Score="2" ViewCount="109" Body="&lt;p&gt;To state the context, I am familiar with the Hopf-Rinow theorem. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My request is three fold,&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;I would like to know of general classes of geodesically incomplete spaces. I basically want to see lots of examples for this. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I want to know of techniques of proving and testing for geodesic incompleteness or completeness. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I want to know of methods of proving and testing for existence of maximal extension of curves. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;{Confusingly in quite a bit of literature I have seen the adjective &quot;inextensible&quot; being used when actually they seem to want to mean &quot;maximally extended&quot;. Is there some subtle point here that I am missing?} &lt;/p&gt;&#xA;" OwnerUserId="965" LastActivityDate="2011-02-07T18:11:26.017" Title="Intuition about geodesic incompleteness" Tags="&lt;differential-geometry&gt;" AnswerCount="2" CommentCount="6" />
  <row Id="13658" PostTypeId="1" CreationDate="2010-12-09T13:37:25.883" Score="0" ViewCount="44" Body="&lt;p&gt;I would like to gain intuition how &quot;singular&quot; a distribution can be.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $X \in \mathbb R^n$ be open, $\mathcal D(X)$ be the smooth test functions on $X$ with canonical LF topology, and let $\mathcal D&amp;#39;(X)$ be the topological dual of $D(X)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $\Psi \in D&amp;#39;(X)$. Can the singular support of $ \Psi$, i.e. the complement in $X$ of the largest open subset of $X$, on which $\Psi$ is a locally integrable function, be open?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, the Dirac delta has singular support in $0$. Similarly, integrals with respect to Hausdorff measures on finite measure submanifolds of $X$  may be distributions, whose singular support is not open.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All examples of distributions I know constitute of a function &quot;disturbed&quot; by a &quot;small&quot; singularity.&lt;/p&gt;&#xA;" OwnerUserId="3557" LastActivityDate="2010-12-09T13:37:25.883" Title="Distribution whose singular support is open" Tags="&lt;measure-theory&gt;&lt;functional-analysis&gt;" CommentCount="4" />
  <row Id="13659" PostTypeId="2" ParentId="13652" CreationDate="2010-12-09T13:47:02.423" Score="1" ViewCount="" Body="&lt;p&gt;A combinatorial approach is useful here: 51!/52! = 1/52.&lt;/p&gt;&#xA;" OwnerUserId="2810" LastActivityDate="2010-12-09T13:47:02.423" CommentCount="2" />
  <row Id="13660" PostTypeId="1" CreationDate="2010-12-09T13:49:03.693" Score="1" ViewCount="82" Body="&lt;p&gt;Suppose that $R$ is a ring, $I$ and $J$ are ideals, and $R=I+J$. Can we conclude that $R/J$ is free?&lt;/p&gt;&#xA;" OwnerUserId="4570" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-09T16:22:20.427" LastActivityDate="2010-12-09T16:22:20.427" Title="If $R=I+J$, can we conclude that $R/J$ is free?" Tags="&lt;abstract-algebra&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="13661" PostTypeId="1" CreationDate="2010-12-09T13:49:11.097" Score="1" ViewCount="65" Body="&lt;p&gt;It's well known that under some &quot;weak&quot; hypothesis, such as finitely generation, the support of an A-module is closed in Spec(A). It is true also in the most general case? &lt;/p&gt;&#xA;" OwnerUserId="4571" LastActivityDate="2010-12-09T19:07:33.443" Title="The support of a module is closed?" Tags="&lt;commutative-algebra&gt;" AnswerCount="1" />
  <row Id="13662" PostTypeId="1" AcceptedAnswerId="13810" CreationDate="2010-12-09T13:49:35.550" Score="2" ViewCount="235" Body="&lt;p&gt;&lt;strong&gt;Problem:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&lt;strong&gt;The probability that a man who is 85 years. old will die before attaining the age of 90 is $\frac13$. A,B,C,D are four person who are 85 years old. what is the probability that A will die before attaining the age of 90 and will be the first to die ?&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="2109" LastEditorUserId="2109" LastEditorDisplayName="" LastEditDate="2010-12-09T18:08:06.373" LastActivityDate="2011-03-15T13:49:26.843" Title="Probability Problems" Tags="&lt;probability&gt;" AnswerCount="4" CommentCount="10" />
  <row Id="13663" PostTypeId="2" ParentId="13661" CreationDate="2010-12-09T13:52:22.613" Score="4" ViewCount="" Body="&lt;p&gt;Pick a non-closed subset $S$ of $\operatorname{Spec}\mathbb Z$, and let $\displaystyle M=\bigoplus_{\mathfrak p\in S}\:\mathbb Z/{\mathfrak p}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;NB:&lt;/strong&gt; Editing this answer required solving a captcha.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Thanks go to Akhil for catching a stupid mistake)&lt;/p&gt;&#xA;" OwnerUserId="274" LastEditorUserId="274" LastEditorDisplayName="" LastEditDate="2010-12-09T19:07:33.443" LastActivityDate="2010-12-09T19:07:33.443" CommentCount="2" />
  <row Id="13664" PostTypeId="2" ParentId="13660" CreationDate="2010-12-09T13:53:20.540" Score="4" ViewCount="" Body="&lt;p&gt;$\mathbb{Z} = 2\mathbb{Z} + 3\mathbb{Z}$. I am sure you can now answer your question yourself.&lt;/p&gt;&#xA;" OwnerUserId="3212" LastActivityDate="2010-12-09T13:53:20.540" CommentCount="2" />
  <row Id="13665" PostTypeId="1" CreationDate="2010-12-09T14:00:03.397" Score="3" ViewCount="60" Body="&lt;p&gt;Hi &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a procedure/algorithm for calculating sums of the form&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \sum_{n_1,\ldots,n_r &amp;gt;0} \frac1{L_1(n_1,\ldots,n_r)^{m_1} \ldots L_r(n_1,\ldots,n_r)^{m_r}} $$&#xD;&#xA;where &#xA;$$   L_i(n_1,\ldots, n_r) := a_{1,i} n_1 + \ldots + a_{r,i} n_r $$&#xD;&#xA;for some $a_{i,j} \in \mathbb C$. How would one go about evaluating this sum to say 20 or 30 digits?&lt;/p&gt;&#xA;" OwnerUserId="4572" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-09T14:03:35.843" LastActivityDate="2010-12-09T14:03:35.843" Title="Shintani cone zeta function" Tags="&lt;number-theory&gt;&lt;sequences-and-series&gt;&lt;numerical-methods&gt;&lt;zeta-function&gt;" CommentCount="1" />
  <row Id="13667" PostTypeId="2" ParentId="13657" CreationDate="2010-12-09T14:23:29.867" Score="0" ViewCount="" Body="&lt;p&gt;The equation for a geodesic is just an ODE with an initial value. The classical Picard-Lindeloef theorem gives the existence of a maximal solution (defined from 0 to t_+). Thus, for instance, if you can show that some &quot;energy&quot; (for instance the L^2 integral of the derivative from 0 to t_+) is finite, then t_+=\infty.&lt;/p&gt;&#xA;" OwnerUserId="4520" LastActivityDate="2010-12-09T14:23:29.867" CommentCount="2" />
  <row Id="13668" PostTypeId="2" ParentId="13627" CreationDate="2010-12-09T14:57:42.770" Score="3" ViewCount="" Body="&lt;p&gt;There are definite real world applications. I would look at the website/work of Gunnar Carlsson (http://comptop.stanford.edu/) and Robert Ghrist (http://www.math.uiuc.edu/~ghrist/). Both are excellent mathematicians. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The following could be completely wrong: Carlsson is one of the main proponents of Persistent Homology which is about looking at what homology can tell you about large data sets, clouds, as well as applications of category theory to computer science. Ghrist works on stuff like sensor networks. I don't understand any of the math behind these things.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also there are some preprints by Phillipe Gaucher you might want to check out. Peter Bubenik at cleveland state might also have some fun stuff on his website.&lt;/p&gt;&#xA;" OwnerUserId="627" LastActivityDate="2010-12-09T14:57:42.770" />
  <row Id="13669" PostTypeId="2" ParentId="13660" CreationDate="2010-12-09T15:21:12.400" Score="2" ViewCount="" Body="&lt;p&gt;You can almost never conclude that $R/J$ is free.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $R$ is a nonzero commutative ring and $J$ is an ideal of $R$, then $R/J$ is a free $R$-module iff $J = 0$.  Indeed, if $J \neq 0$, let $x$ be a nonzero element of $J$.  Then for all $y \in R/J$, $xy = 0$.  On the other hand, consider $\bigoplus_{i \in I} R$ for some nonempty index set $I$: taking $y$ to be an element of the form $(1,0,\ldots,0)$ -- i.e., which has one coordinate equal to $1$ and all the other coordinates equal to $0$ -- we find that $xy = 0$ implies $x = 0$, contradiction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, if $J \neq 0$, then $R/J$ has a nonzero annihilator, which is not the case for any nontrivial free module. &lt;/p&gt;&#xA;" OwnerUserId="299" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2010-12-09T15:30:35.587" LastActivityDate="2010-12-09T15:30:35.587" />
  <row Id="13670" PostTypeId="2" ParentId="13642" CreationDate="2010-12-09T15:42:22.697" Score="4" ViewCount="" Body="&lt;p&gt;If $\beta,\gamma &amp;gt; 0$ then the expected number is infinity (unless $n = 0$), because either $n &amp;lt; 0$ and then you never reach $n$, or $n &amp;gt; 0$ and then with some positive probability you &quot;hop&quot; over $n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose that state $n+1$ is also good, and $n &amp;gt; 0$. Let $E_i$ be the expected number of steps to get to states $n$ or $n+1$ starting at state $i$. Then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ E_i = \alpha E_i + \beta E_{i+1} + \gamma E_{i+2} + 1. $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a non-homogeneous recurrence relation that you can solve by first finding some non-homogeneous solution (try $E_i = a i + b$) and then solving the corresponding homogeneous recurrence (without the $+1$) in the usual way (this amounts to solving the quadratic $(1-\alpha)x^2-\beta x-\gamma$; note the singular solution if this is a square). You will get a two-dimensional solution space which can you collapse by using $$E_n = E_{n+1} = 0.$$ Then the answer is $E_0$.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-12-09T15:42:22.697" />
  <row Id="13671" PostTypeId="2" ParentId="13625" CreationDate="2010-12-09T15:57:23.043" Score="7" ViewCount="" Body="&lt;p&gt;As the comments have suggested, the question is a bit under-determined, but let me supply some definitions in a plausible way and answer it based on those definitions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $P$ be a property of topological spaces.  (Formally speaking this could be any &quot;predicate&quot;, but it would only be asking for trouble if $P$ were not a &lt;em&gt;topological property&lt;/em&gt; of topological spaces -- i.e., such that if $X$ and $Y$ are homeomorphic topological spaces, then $X$ has property $P$ iff $Y$ has property $P$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then we say that $X$ is &lt;strong&gt;locally P&lt;/strong&gt; if for every point of $x$ there is a neighborhood base $\{U_x\}$ each of whose elements $U_x$ has property $P$ in the subspace topology.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We say that $X$ is &lt;strong&gt;globally P&lt;/strong&gt; if it has property $P$.  Why would we say this?  Well, it's a matter of emphasis, presumably by way of comparison to something which locally has property $P$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now define a space to be &lt;strong&gt;Euclidean&lt;/strong&gt; if it is homeomorphic to $\mathbb{R}^n$ for some $n \in \mathbb{N}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I claim that any space which is globally Euclidean is also locally Euclidean.  In other words, I claim that for all $n \in \mathbb{N}$, every $x \in \mathbb{R}^n$ admits a neighborhood base, each of whose members is homeomorphic to Euclidean $m$-space for some $m$.  But indeed, for any $\epsilon &amp;gt; 0$, the open $\epsilon$-ball around $x$ is isomorphic to $\mathbb{R}^n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that in general &quot;globally P&quot; &lt;em&gt;need not&lt;/em&gt; imply &quot;locally P&quot;.  For instance a connected space need not be locally connected, and a quasi-compact (non-Hausdorff!) space need not be locally quasi-compact.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fairness I should say that some mathematicians don't like the idea that globally $P$ need not imply locally $P$, so they weaken the definition of locally $P$ to &quot;each point of $X$ admits &lt;em&gt;some&lt;/em&gt; neighborhood which has property $P$&quot;.  I would call this &lt;strong&gt;weakly locally P&lt;/strong&gt; and then it is trivial that globally $P$ implies weakly globally $P$: for &#xA;each $x$, take $X$ itself as a neighborhood which has property $P$.  &lt;/p&gt;&#xA;" OwnerUserId="299" LastActivityDate="2010-12-09T15:57:23.043" />
  <row Id="13672" PostTypeId="2" ParentId="13639" CreationDate="2010-12-09T15:59:04.790" Score="2" ViewCount="" Body="&lt;p&gt;Take $E$ to be the trivial bundle $X \times \mathbb{R}$. Let $f: X \to \mathbb{R}$ to be a function vanishing only at $x_0$. Then multiplication by the global function $f$ defines a morphism $E \to E$ which is injective as a morphism of sheaves (because it is injective on $E(U)$ for any open set $U$), but the map on fibers at $x_0$ is not injective.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(To avoid uninteresting cases, assume $x_0$ is not an isolated point of $X$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is that for an injection of sheaves $E \to F$, the map on &lt;em&gt;stalks&lt;/em&gt; $E_x \to F_x$ is always injective. The map on &lt;em&gt;fibers&lt;/em&gt; $E(x) \to F(x)$ obtained by tensoring the map on stalks with the residue field is generally not, because tensoring is not an exact functor. If the map on stalks is a &lt;em&gt;split&lt;/em&gt; injection, then the map on fibers will be an injection still.&lt;/p&gt;&#xA;" OwnerUserId="536" LastEditorUserId="536" LastEditorDisplayName="" LastEditDate="2010-12-09T16:22:38.233" LastActivityDate="2010-12-09T16:22:38.233" />
  <row Id="13673" PostTypeId="2" ParentId="13646" CreationDate="2010-12-09T16:03:36.413" Score="5" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\ $ It &lt;strong&gt;telescopes&lt;/strong&gt;: $\rm\displaystyle\ \ \prod_{k=a}^b\ \frac{f(k+1)}{f(k)}\ \ = \ \ \frac{\rlap{----}f(a+1)}{f(a)}\frac{\rlap{----}f(a+2)}{\rlap{----}f(a+1)}\ \ \cdots\ \ \frac{f(b+1)}{\rlap{--}f(b)}\ =\ \frac{f(b+1)}{f(a)}\:.\ $&lt;br&gt;&#xA;Apply that to both factors in the product then take the limit as $\rm\ b\to\infty\:.$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For some other examples of additive/multiplicative telescopy see &lt;a href=&quot;http://math.stackexchange.com/questions/6581/proof-of-inequalities-by-induction/6586#6586&quot;&gt;here&lt;/a&gt; or &lt;a href=&quot;http://math.stackexchange.com/questions/7441/summation-of-sum-limits-n1-infty-fracxx1-cdots-xn-1yy1-cdo/7453#7453&quot;&gt;here&lt;/a&gt; or &lt;a href=&quot;http://math.stackexchange.com/questions/6805/simple-proof-by-induction-9-divides-n3-n13-n23/6814#6814&quot;&gt;here.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-12-09T16:03:36.413" />
  <row Id="13674" PostTypeId="1" AcceptedAnswerId="13684" CreationDate="2010-12-09T16:11:12.667" Score="0" ViewCount="97" Body="&lt;p&gt;We have a random variable Y that has a uniform distribution on the interval[1,5]. The cost of delay is given by U = 2Y^2 + 3. Use the method of transformations to derive the density of function of U.&lt;/p&gt;&#xA;" OwnerUserId="3049" LastActivityDate="2010-12-09T16:50:31.370" Title="density function by the method of transformations" Tags="&lt;probability&gt;&lt;homework&gt;&lt;statistics&gt;&lt;distribution-theory&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="13675" PostTypeId="2" ParentId="13482" CreationDate="2010-12-09T16:13:18.470" Score="1" ViewCount="" Body="&lt;p&gt;Take the Euler scheme. Let's assume that you start with $X_0=\epsilon$. Let's take $\beta=-1$, $\gamma =1$ and $\sigma=1$ for the sake of definiteness. Then in the next step you'll have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$X_{\Delta} = \epsilon + (1-\epsilon) \Delta + \sqrt{\epsilon} Z \Delta$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What's the probability that this is smaller than zero?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathbb{P}\left[X_{\Delta}&amp;lt;0\right]= \mathbb{P}\left[\epsilon + (1-\epsilon) \Delta + \sqrt{\epsilon} Z \Delta &amp;lt; 0\right]$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rearranging you'll get:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathbb{P}\left[X_{\Delta}&amp;lt;0\right]= \mathbb{P}\left[ Z &amp;lt; -\frac{\sqrt{\epsilon}}{\Delta}(1+(1-\frac{1}{\epsilon})\Delta)\right]$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using Chebyshev for an estimate this is smaller than&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathbb{P}\left[X_{\Delta}&amp;lt;0\right] \leq \frac{\Delta^2}{2\epsilon}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the closer you are to zero, the higher the probability of crossing that line. But taking time steps sufficiently small lowers the chance of crossing quadratically. That should give you an idea of how to show it for the other scheme.&lt;/p&gt;&#xA;" OwnerUserId="3567" LastActivityDate="2010-12-09T16:13:18.470" CommentCount="1" />
  <row Id="13676" PostTypeId="1" AcceptedAnswerId="13677" CreationDate="2010-12-09T16:14:04.820" Score="5" ViewCount="134" Body="&lt;p&gt;This might be an elementary question, but I am just beginning to learn logic theory. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From wikipedia article on Godel's incompleteness theorems&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Any effectively generated theory capable of expressing elementary arithmetic cannot be both consistent and complete. In particular, for any consistent, effectively generated formal theory that proves certain basic arithmetic truths, there is an arithmetical statement that is true, but not provable in the theory (Kleene 1967, p. 250).&#xA;  The true but unprovable statement referred to by the theorem is often referred to as “the Gödel sentence” for the theory. The true but unprovable statement referred to by the theorem is often referred to as “the Gödel sentence” for the theory.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My question: Is a Godel statement &lt;a href=&quot;http://en.wikipedia.org/wiki/Validity#Validity_of_statements&quot; rel=&quot;nofollow&quot;&gt;logically valid&lt;/a&gt;?.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: As Carl answers below, if the Godel statement is valid, then by completeness theorem, it is provable, which leads to a contradiction. So there exists a model in which the statement is false. Can we construct such a model?&lt;/p&gt;&#xA;" OwnerUserId="3431" LastEditorUserId="3431" LastEditorDisplayName="" LastEditDate="2010-12-09T16:30:04.613" LastActivityDate="2010-12-09T16:30:04.613" Title="Is a Godel sentence logically valid?" Tags="&lt;logic&gt;" AnswerCount="1" />
  <row Id="13677" PostTypeId="2" ParentId="13676" CreationDate="2010-12-09T16:18:40.247" Score="7" ViewCount="" Body="&lt;p&gt;No, a Gödel sentence is not logically valid. Because the Gödel sentence for a theory $T$ is unprovable from $T$, it follows from the completeness theorem for first-order logic that there is a model of $T$ in which the Gödel sentence is false. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the text you quoted says &quot;true&quot; you should read that as &quot;true in the standard model of arithmetic&quot;. Logical validity would correspond to truth in &lt;em&gt;all&lt;/em&gt; models. An example of a logically valid sentence is $(\forall x) (x=x)$. &lt;/p&gt;&#xA;" OwnerUserId="630" LastActivityDate="2010-12-09T16:18:40.247" CommentCount="3" />
  <row Id="13678" PostTypeId="2" ParentId="13654" CreationDate="2010-12-09T16:20:42.610" Score="9" ViewCount="" Body="&lt;p&gt;What you are forgetting is that you have more than &lt;em&gt;one&lt;/em&gt; inequality.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The inequality $a^2\geq 8$, by &lt;em&gt;itself&lt;/em&gt;, only gives you $|a|\geq \sqrt{8}=2\sqrt{2}$, or equivalently as you (almost) write:&#xA;$$a\geq 2\sqrt{2}\qquad\text{or}\qquad a\leq -2\sqrt{2}.$$&#xD;&#xA;&lt;strong&gt;Added:&lt;/strong&gt; Note that you took $a^2\geq 8$, a &lt;em&gt;weak&lt;/em&gt; inequality, and &quot;translated&quot; it into &quot;$a\gt 2\sqrt{2}$ or $a\lt -2\sqrt{2}$.&quot; Technically, this is incorrect because you have now excluded the possitibility of equality. Since you were starting with with $\geq$, you should have continued with $\geq$ and $\leq$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, you have &lt;em&gt;other&lt;/em&gt; inequalities that need to be satisfied as well.The second inequality tells you that $0\lt a \lt 6$ must be satisfied. How does this mesh with the inequalities you already have?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $0\lt a \lt 6$  &lt;em&gt;and&lt;/em&gt; $a\leq -2\sqrt{2}\lt 0$, then that tells you that $a$ must be both greater than &lt;em&gt;and&lt;/em&gt; less than zero. This is a &lt;em&gt;very&lt;/em&gt; big problem for $a$, which decides to solve this problem by ceasing to exist. There is no solution to the inequality $0\lt a\lt 6$ that &lt;em&gt;also&lt;/em&gt; satisfy $a\lt -2\sqrt{2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So putting these two together, you know that you have $a\geq 2\sqrt{2}$, and $0\lt a \lt 6$. Since satisfying $a\geq 2\sqrt{2}$ automatically guarantees you that $0\lt a$, you can combine them all by saying that the first two inequalities you have are equivalent to&#xA;$$ 2\sqrt{2} \leq a \lt 6.$$&#xD;&#xA;Finally, we need to add in the condition that $a\lt \frac{11}{3}$. Since $\frac{11}{3}\lt 6$, if $a$ is less than $\frac{11}{3}$ it will also automatically be less than $6$. It &lt;em&gt;still&lt;/em&gt; needs to be greater than or equal to $2\sqrt{2}$, though. So combining them all you get that you must have&#xA;$2\sqrt{2} \leq a \lt 6$ in order to satisfy all three (four, if you count &quot;$0\lt a\lt 6$&quot; as two inequalities) inequalities you have.&lt;/p&gt;&#xA;&#xA;&lt;hr/&gt;&#xA;&#xA;&lt;p&gt;Okay, the &lt;strong&gt;original&lt;/strong&gt; problem was: &quot;if the two roots of $x^2-ax+2=0$ lie in the interval $(0,3)$, then find the values of $a$.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The roots are $\frac{a\pm\sqrt{a^2-8}}{2}$. So you assumed that you only needed to ask for $a^2-8\geq 0$ in order for the roots to be real. However, the use of the word &quot;&lt;strong&gt;both&lt;/strong&gt;&quot; suggests that in fact they want there to be two &lt;em&gt;distinct&lt;/em&gt; roots, which would require $a^2-8\gt 0$. If you take $a=2\sqrt{2}$, the lower limit I suggested earlier, then you have a single (repeated) root at $\sqrt{2}$ (which satisfies the condition). If they want &lt;strong&gt;two&lt;/strong&gt; roots, then you need the discriminant &lt;em&gt;positive&lt;/em&gt;, not merely nonnegative.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-10T02:59:54.770" LastActivityDate="2010-12-10T02:59:54.770" CommentCount="12" />
  <row Id="13679" PostTypeId="1" AcceptedAnswerId="13680" CreationDate="2010-12-09T16:27:55.593" Score="5" ViewCount="240" Body="&lt;p&gt;Is there a general way to show whether the limit $$\lim_{n \to \infty}\frac{\mathrm d^n}{\mathrm dx^n} f(x)$$ converges to some expression?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What about repeatedly integrating an expression $n$ times as $n \to \infty$?&lt;/p&gt;&#xA;" OwnerUserId="4576" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-10T16:27:19.583" LastActivityDate="2010-12-10T16:27:19.583" Title="Does $\lim_{n \to \infty}\frac{\mathrm d^n}{\mathrm dx^n} f(x)$ converge?" Tags="&lt;calculus&gt;&lt;limit&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="13680" PostTypeId="2" ParentId="13679" CreationDate="2010-12-09T16:38:54.693" Score="10" ViewCount="" Body="&lt;p&gt;For the notion of limit, you need a notion of convergence or of &quot;distance.&quot; Here, you are talking about &lt;em&gt;functions&lt;/em&gt;, not numbers, so you need to specify how you measure the &quot;distance&quot; between two functions. Turns out, there are many different ways of doing it, and the answer for a particular $f(x)$ &lt;em&gt;may&lt;/em&gt; depend on which way you do it. So you would need to specify that in general. Among the different kinds of &quot;convergence&quot; you can read about there is &lt;a href=&quot;http://en.wikipedia.org/wiki/Uniform_convergence&quot; rel=&quot;nofollow&quot;&gt;uniform convergence&lt;/a&gt;, &lt;a href=&quot;http://en.wikipedia.org/wiki/Pointwise_convergence&quot; rel=&quot;nofollow&quot;&gt;pointwise convergence&lt;/a&gt;, &lt;a href=&quot;http://en.wikipedia.org/wiki/Convergence_in_measure&quot; rel=&quot;nofollow&quot;&gt;convergence in measure&lt;/a&gt;, and &lt;a href=&quot;http://en.wikipedia.org/wiki/Lp_space&quot; rel=&quot;nofollow&quot;&gt;convergence in $\mathcal{L}_p$-norm&lt;/a&gt;. But this hardly exhausts the possibilities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you have the notion of convergence of functions, then you are dealing with a sequence of functions, and one checks convergence the &quot;usual&quot; way (that is, in essentially the same ways as you do with sequences of real numbers, substituting the absolute value of the difference with whatever way you have decided to describe distance of functions). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for &quot;repeated integrals&quot;, that's not going to help much. The $n$th iterated integral of $\frac{d^n}{dx^n}f(x)$ is $f(x)+p(x)$, where $p(x)$ is an arbitrary polynomial of degree $n-1$ (which you get through the constants of integration). Taking &quot;the limit&quot; as $n$ goes to infinity just drops you back into the problem of &quot;what does it mean for a sequence of functions to converge&quot;? Plus, an &quot;infinite polynomial&quot; (a power series) is probably not what you want anyway.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, for &lt;em&gt;specific&lt;/em&gt; functions one may be able to discern either a pattern, or an answer. For polynomials, it is plain that $f^{(n)}(x)$ will be zero for all sufficiently large $n$, and in that case no matter how you define your distance between functions, the answer will be the zero function. If you have a function with periodic derivatives, such as $\sin(x)$, you know there will be no limit. And so on. &lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-12-09T16:38:54.693" CommentCount="2" />
  <row Id="13681" PostTypeId="2" ParentId="13654" CreationDate="2010-12-09T16:39:30.387" Score="5" ViewCount="" Body="&lt;p&gt;Any polynomial inequality over the reals can be reduced to testing it on the intervals between its roots - where the polynomial has constant sign. This will yield a &lt;em&gt;finite&lt;/em&gt; number of intervals where the inequality holds true. When you have multiple such inequalities you simply intersect the intervals for each polynomial to obtain the &lt;em&gt;finite&lt;/em&gt; number of intervals where all of the inequalities are true. In fact the same idea generalizes to arbitrary first order multivariate statements about the reals - as Tarski has shown by his celebrated method of &lt;em&gt;quantifier elimination&lt;/em&gt; for real-closed fields. It has an algorithmic form that is essentially a higher-dimensional &lt;em&gt;cylindrical&lt;/em&gt; version of the finite interval decomposition for the above one dimensional case - google &quot;cylindrical algebraic decomposition&quot; or CAD algorithm.&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-12-09T16:39:30.387" />
  <row Id="13682" PostTypeId="1" CreationDate="2010-12-09T16:45:29.860" Score="0" ViewCount="30" Body="&lt;p&gt;Hi my apologies first, I'm a biologist and not so good in statistics. In my study I'm studying the effect of concentration on growth of a certain specimen. I have with me the different concentrations and the growth data for each concentration. What do you suggest is the best statistical test to find out the best concentration (i.e the one having maximum growth). Thanks in advance.&lt;/p&gt;&#xA;" OwnerUserId="4578" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-09T17:01:08.247" LastActivityDate="2010-12-09T17:01:08.247" Title="Statistics Question -which test to find out best concentration (the one having maximum effect)?" Tags="&lt;statistics&gt;" CommentCount="2" />
  <row Id="13683" PostTypeId="2" ParentId="13648" CreationDate="2010-12-09T16:48:34.687" Score="3" ViewCount="" Body="&lt;p&gt;Using Pascal identity&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle\binom{2n}{n}=\left(4-\frac{2}{n}\right)\cdot\binom{2n-2}{n-1}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;one gets for your product&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle\frac{\binom{2}{1}}{\binom{0}{0}}\cdot\frac{\binom{4}{2}}{\binom{2}{1}}\cdot\cdot\cdot\frac{\binom{2n}{n}}{\binom{2n-2}{n-1}}=\binom{2n}{n}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and this is an integer as required.&lt;/p&gt;&#xA;" OwnerUserId="1747" LastEditorUserId="1747" LastEditorDisplayName="" LastEditDate="2010-12-09T18:45:45.460" LastActivityDate="2010-12-09T18:45:45.460" />
  <row Id="13684" PostTypeId="2" ParentId="13674" CreationDate="2010-12-09T16:50:31.370" Score="2" ViewCount="" Body="&lt;p&gt;First, if $Y$ takes values in $[1,5]$, then $U$ takes values in the range...? So, for $x$ in that range, you should consider the distribution function $F_U {(x)} := {\rm P}(U \le x)$. Substituting $U=2Y^2+3$ leads straightforwardly to $F_U {(x)} = F_Y {(\varphi(x))}$ for some function $\varphi(x)$, where $F_Y$ is the distribution function of $Y$. What is $\varphi(x)$? Note that $\varphi(x)$ takes values in $[1,5]$. Now, the density $f_U$ of $U$ is given by&#xA;$$&#xD;&#xA;f_U (x) = \frac{{\rm d}}{{{\rm d}x}}F_U (x) = \frac{{\rm d}}{{{\rm d}x}}F_Y (\varphi (x)) =...?&#xD;&#xA;$$ &#xA;Final hint: the density $f_Y$ of $Y$ is constant on $[1,5]$; what is that constant?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suggestion: verify that the $f_U {(x)}$ you find integrates to $1$.&lt;/p&gt;&#xA;" OwnerUserId="2810" LastActivityDate="2010-12-09T16:50:31.370" CommentCount="1" />
  <row Id="13685" PostTypeId="1" CreationDate="2010-12-09T16:51:12.697" Score="2" ViewCount="362" Body="&lt;p&gt;Is the function $\frac{e^z-1}{e^z+1}$ analytic?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried using $\mathop{\lim}\limits_{z\to a} \frac{f(z)-f(a)}{z-a}$, but I didnt really know what to use for $a$.&lt;/p&gt;&#xA;" OwnerUserId="4577" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-09T18:29:54.260" LastActivityDate="2011-05-15T07:14:49.593" Title="Is $\frac{e^z-1}{e^z+1}$ analytic?" Tags="&lt;calculus&gt;&lt;complex-analysis&gt;" AnswerCount="2" CommentCount="9" />
  <row Id="13686" PostTypeId="2" ParentId="13563" CreationDate="2010-12-09T17:28:46.113" Score="0" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\ $ The problem admits an obvious shift &lt;strong&gt;symmetry&lt;/strong&gt;, e.g. considering the example that you gave, a shift of $-2\:$ to $[3, 6]$ yields $[1,4]$, which clearly has $4$ integers. Innate symmetries should always be investigated since they may play a key role in simplifying the search for a solution. Moreover, it's best to learn these symmetry-based problem solving techniques for simple problems like this, since you may not be able to see the forest for the trees in much more complex problems.&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-12-09T17:28:46.113" />
  <row Id="13687" PostTypeId="1" CreationDate="2010-12-09T17:32:18.797" Score="2" ViewCount="224" Body="&lt;p&gt;Hello,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a non-linear ODE that I need to linearise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I could go about linearising the trig terms using Taylor series...but not sure about a square root of the form:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sqrt{(A+C)^2 + B -2 * \cos (\alpha - \theta)}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="4581" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-09T19:22:18.267" LastActivityDate="2010-12-09T20:32:20.033" Title="How to linearise a square root expression?" Tags="&lt;differential-equations&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="13688" PostTypeId="2" ParentId="13551" CreationDate="2010-12-09T17:48:56.933" Score="0" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\rm\ \ \Delta\ f \ :=\ f(n+1) - f(n)\ $ reduces $\rm\ d = deg\ f\:,\ $ so $\rm\ \Delta^{d+1}\ f\ =\ 0\ $ yields a recurrence for $\rm\: f\:$. Functions admitting recurrences are known as &lt;strong&gt;P-recursive&lt;/strong&gt;. Google &lt;strong&gt;holonomic function&lt;/strong&gt; to learn about their rich theory - which has widespread applications.&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-12-09T17:48:56.933" />
  <row Id="13689" PostTypeId="1" CreationDate="2010-12-09T17:57:32.630" Score="4" ViewCount="110" Body="&lt;p&gt;Suppose we have a countable set of objects $\{x_i|i \in [1..m]\}$ in a metric space $(\mathbb R^n,d_n)$ and a map ($F$) mapping the objects to objects in a metric space $(\mathbb R^1,d_1)$. For each pair of objects we can define an error:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;  e_{ij} = \frac{d_m(x_i,x_j)}{d_1(F(x_i),F(x_j))}&#xD;&#xA;$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $e_{ii}$ be 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is obvious that we can define the countable set in a way that any projection has an error ($e_{ij}\ne 1$) at least for one pair of objects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But for given set of objects we can define the map trying to minimize the errors. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I had met this problem developing a web site's rating system based on several criteria. At the moment I use the following algorithm for calculating F (images of objects):&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;select random $\{y_i|i \in [1..m]\}$ set in space $(\mathbb R^1,d_1)$ and think that $y_i=F(x_i)$ where F is defining map&lt;/li&gt;&#xA;&lt;li&gt;for each dot calculate $\delta_i=\sum\limits_{j}k\frac{1}{2}e_{ij}(y_i-y_j)$ where $k$ is a parameter ($0 \le k \le 1$) and them change items in the set $y_i \to y_i + \delta_i$&#xD;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I repeat the step two over and over several times, but at each step I make the parameter $k$ smaller.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I infer this algorithm from physics intuition: if two objects are too close they push away, if they are too far away they attract and $k$ is a resistance of an environment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to get deeper knowledge about this problem, so could you provide me some links to articles and literature about it or the official name of the problem.&lt;/p&gt;&#xA;" OwnerUserId="2016" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-10T02:53:52.597" LastActivityDate="2011-01-09T17:44:40.127" Title="Projection to space with smaller dimension that saves a distance" Tags="&lt;reference-request&gt;&lt;metric-spaces&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="13690" PostTypeId="2" ParentId="13687" CreationDate="2010-12-09T18:07:48.713" Score="0" ViewCount="" Body="&lt;p&gt;If $(A+C)^2+B$ is large compared with 1 you can divide it out to get $$\sqrt{(A+C)^2+B}\sqrt{1-\frac{2cos(\alpha-\theta)}{(A+C)^2+B}}$$ and use $\sqrt{1-x}\approx 1-\frac{x}{2}$ for $x&amp;lt;&amp;lt;1$.  If you are specifically interested in $\theta$ you can expand the cosine term  and still use the approximation to the square root considering $\alpha$ to be fixed.  Does that help?&lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2010-12-09T20:32:20.033" LastActivityDate="2010-12-09T20:32:20.033" />
  <row Id="13691" PostTypeId="2" ParentId="13687" CreationDate="2010-12-09T18:22:59.923" Score="1" ViewCount="" Body="&lt;p&gt;Since you have an ODE, you probably want to be linearizing around an equilbrium point $\theta_0$ (which might not be zero).  Note that I'm assuming that $\theta$ is the variable. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then think about the general form of a Taylor series expansion -- the linearization is given by&#xA;$$&#xD;&#xA;f(\theta) \approx f(\theta_0) + (\theta-\theta_0) f&amp;#39;(\theta_0)&#xD;&#xA;$$&#xD;&#xA;where&#xA;$$&#xD;&#xA;f(\theta) = \sqrt{(A+C)^2 + B - 2\cos(\alpha - \theta)}&#xD;&#xA;$$ &lt;/p&gt;&#xA;" OwnerUserId="4229" LastEditorUserId="4229" LastEditorDisplayName="" LastEditDate="2010-12-09T18:57:03.353" LastActivityDate="2010-12-09T18:57:03.353" />
  <row Id="13692" PostTypeId="2" ParentId="13641" CreationDate="2010-12-09T18:24:11.203" Score="4" ViewCount="" Body="&lt;p&gt;As already mentioned in the other answers and comments, this is true if the space is assumed to be complete (see Andrey Rekalo's answer), but not necessarily otherwise (see Carl Mummert's answer).  In the complete case, more can be said.  If the Hilbert space dimension (cardinality of a maximal orthonormal set) is infinite, then the linear dimension (cardinality of a maximal linearly independent set) is at least $\mathfrak{c}=2^{\aleph_0}=|\mathbb{R}|$.  Of course, this doesn't directly tell you anything if the Hilbert space dimension is $\mathfrak{c}$ or greater, but in the case of separable Hilbert spaces like $\ell^2$ (as defined &lt;a href=&quot;http://en.wikipedia.org/wiki/Hilbert_space#Second_example%3a_sequence_spaces&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;), it tells you that not only does an orthonormal set not span, but &lt;em&gt;no&lt;/em&gt; subset of cardinality less than $\mathfrak{c}$ can span.  One way to see this is to consider the linearly independent set $\{(1,t,t^2,t^3,\ldots):-1\lt t\lt 1\}\subset\ell^2$.  Since $\ell^2$ imbeds into every infinite dimensional Hilbert space as the closed linear span of any countably infinite orthonormal set, this also demonstrates the general fact.  This and other proofs can be found in the solutions to &lt;a href=&quot;http://books.google.com/books?id=S57XLkgbf0oC&amp;amp;lpg=PP1&amp;amp;dq=hilbert%20space%20problem%20book%20halmos&amp;amp;pg=PA6#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;Problem 7&lt;/a&gt; of Halmos's &lt;em&gt;Hilbert Space Problem Book&lt;/em&gt;, which I highly recommend. This statement also &lt;a href=&quot;http://www.jstor.org/stable/2318458&quot; rel=&quot;nofollow&quot;&gt;extends&lt;/a&gt; to Banach spaces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Strictly speaking, what I have said so far hasn't answered your question in the case of Hilbert space dimension greater than or equal to $\mathfrak{c}$.  However, you can get an overkill solution to your question by taking a countably infinite subset $A$ of an orthonormal subset of an arbitrary Hilbert space $H$.   If $M_0$ is the linear span of $A$ and $M=\overline{M_0}$ is the closed linear span of $A$, then $H=M\oplus M^\perp$, while the linear span of your orthonormal set is contained in $M_0\oplus M^\perp$, which is properly contained in $H$ because $M$ has higher dimension than $M_0$.  ($M^\perp$ denotes the set of vectors orthogonal to every element of $M$, and $\oplus$ is being used here to denote internal direct sums.)  All that was really needed here is that the linear dimension of $M$ is not countable, and this also follows from &lt;a href=&quot;http://en.wikipedia.org/wiki/Baire_category_theorem&quot; rel=&quot;nofollow&quot;&gt;Baire's theorem&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Andrey Rekalo gives a better, nonoverkill answer for the complete case.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-12-09T18:24:11.203" />
  <row Id="13693" PostTypeId="1" AcceptedAnswerId="13714" CreationDate="2010-12-09T18:45:46.253" Score="1" ViewCount="62" Body="&lt;p&gt;Hi all&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose that we have two varieties of algebras $A$ and $B$, whose operators all have arities less than some regular cardinal, and such that every $B$-algebra (please correct me if this is not the usual notation) is also an $A$-algebra. It is known that the forgetful functor from the category of $B$-algebras to that of $A$-algebras has a left adjoint $F$, which takes every $A$-algebra $a$ to the relatively free $B$-algebra $Fa$ generated by $a$. My question is, if $a$ and $b$ are $A$-algebras such that $Fa$ is isomorphic (as a $B$-algebra) to $Fb$, are $a$ and $b$ necessarily isomorphic? If not in general, how about the special case where $A$-algebras are Boolean algebras and $B$-algebras are countably complete Boolean algebras? &lt;/p&gt;&#xA;" OwnerUserId="2491" LastActivityDate="2010-12-09T21:37:58.207" Title="Can nonisomorphic algebras generate isomorphic relatively free algebras?" Tags="&lt;boolean-algebra&gt;&lt;universal-algebra&gt;" AnswerCount="1" CommentCount="7" FavoriteCount="2" />
  <row Id="13694" PostTypeId="2" ParentId="13662" CreationDate="2010-12-09T18:53:43.683" Score="2" ViewCount="" Body="&lt;p&gt;I'm not sure from the comments above if you've solved this or not already, but I had this idea when I first read your post. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think you can partition the outcome space, based on when $B,C,D$ die. There are four disjoint cases:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$B_0:$ none of them dies before 90,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$B_1:$ exactly one of them dies before 90,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$B_2:$ exactly two...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$B_3:$ exactly three...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So $$P(B_n)=\binom{3}{n}(1/3)^n(2/3)^{3-n}.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then let $F$ be the event that $A$ dies first, and $E$ be the event that it happens before $90$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So $$P(F\cap E)=\sum_{n=0}^3 P((F\cap E)|B_n)P(B_n)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So for example, $P((F\cap E)|B_0)$ is the probability that $A$ dies first and before $90$, given that the other three die after $90$. This occurs with probability $(1/3)$, since all the others are given to die after $90$. Then $P((F\cap E)|B_1)$ is the probability that $A$ dies first and before $90$, when exactly one other person dies before $90$. This probability is $(1/3)(1/2)$. The other cases are similar, and then one can just add them up.&lt;/p&gt;&#xA;" OwnerUserId="1583" LastEditorUserId="1583" LastEditorDisplayName="" LastEditDate="2010-12-10T14:34:32.173" LastActivityDate="2010-12-10T14:34:32.173" />
  <row Id="13695" PostTypeId="1" AcceptedAnswerId="13699" CreationDate="2010-12-09T18:56:58.457" Score="0" ViewCount="294" Body="&lt;p&gt;What is the rationale behind saying that $|x|=1$ implies $x=\pm 1$?&lt;/p&gt;&#xA;" OwnerUserId="3744" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-09T19:20:10.653" LastActivityDate="2010-12-14T22:46:22.897" Title="If $|x| = 1 $ why this implies that $ x = \pm 1 $" Tags="&lt;algebra-precalculus&gt;" AnswerCount="5" CommentCount="4" />
  <row Id="13696" PostTypeId="2" ParentId="13631" CreationDate="2010-12-09T18:58:07.143" Score="4" ViewCount="" Body="&lt;p&gt;This is described in the comments; $f$ need only be locally constant, so as a counterexample take $f : (0, 1) \cup (2, 3) \to \mathbb{R}$ which is equal to $0$ on the first component and $1$ on the second.  It is certainly true for functions from $\mathbb{R}$ to $\mathbb{R}$, for example by the mean value theorem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This seems trite, but it's still an important point to keep in mind.  If you ever feel like you're missing something, then of course you should attempt to carefully prove the result that you thought was true and see where it fails (if it does).&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-12-09T18:58:07.143" />
  <row Id="13697" PostTypeId="2" ParentId="13695" CreationDate="2010-12-09T19:02:23.400" Score="1" ViewCount="" Body="&lt;p&gt;$|1| = 1$ and $|-1| = 1$, and there are no other real numbers $x$ such that $|x|=1$.&lt;/p&gt;&#xA;" OwnerUserId="3638" LastActivityDate="2010-12-09T19:02:23.400" />
  <row Id="13699" PostTypeId="2" ParentId="13695" CreationDate="2010-12-09T19:05:17.157" Score="6" ViewCount="" Body="&lt;p&gt;Absolute value of a number is its distance from zero on the number line. The absolute value of a number n is denoted by, |n|. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So here if $|x| = 1$ implies, in the number line the distance from zero on the number is 1.$\text{ Hence, } x = \pm 1$.&lt;/p&gt;&#xA;" OwnerUserId="2109" LastEditorUserId="2109" LastEditorDisplayName="" LastEditDate="2010-12-09T19:33:38.920" LastActivityDate="2010-12-09T19:33:38.920" CommentCount="1" />
  <row Id="13700" PostTypeId="2" ParentId="13695" CreationDate="2010-12-09T19:12:05.107" Score="3" ViewCount="" Body="&lt;p&gt;You should think of the absolute value as being the distance a point is away from 0. So with $|x| = 1$, we want all of  the points that are one unit away from 0. We see that 1 and -1 are the only values for x that satisfies this. Therefore, with $|x| = 1$ we have that $x \in\{-1,1\}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope this helps.&lt;/p&gt;&#xA;" OwnerUserId="3791" LastEditorUserId="3791" LastEditorDisplayName="" LastEditDate="2010-12-14T22:46:22.897" LastActivityDate="2010-12-14T22:46:22.897" CommentCount="2" />
  <row Id="13701" PostTypeId="1" AcceptedAnswerId="13704" CreationDate="2010-12-09T19:15:05.733" Score="4" ViewCount="209" Body="&lt;p&gt;Bill Dubuque has answered several questions by indicating that some form of &quot;telescoping&quot; is taking place.  See &lt;a href=&quot;http://math.stackexchange.com/questions/13646/how-to-derive-the-following-identity&quot;&gt;this&lt;/a&gt; post and the links provided by Bill for more information.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have never heard of &quot;telescoping&quot; until I read a few answers on here by Bill which refer to this notion.  It seems fairly straightforward, basically you expand some expression using basic arithmetic, there is a minor miracle and lots of terms cancel out in a systematic way, and we are then able to solve the problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suppose &quot;telescoping&quot; in this sense was something I always took for granted, and considered a low level &quot;trick&quot; to keep in my back pocket.  However, considering the importance Bill seems to attach to this notion of telescoping, and considering that I have a great deal of respect for Bill based on the post's by him I have read, I was wondering if I'm not missing something about telescoping.  There is no wiki article on the subject, and a Google search directs me to Bill's answers on SE!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore I would like to ask: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) What unintuitive results can I achieve with telescoping? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Is there a good reference which only discusses telescoping and applications, or is this concept too narrow for anyone to write a paper/book like this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) More trivially, am I missing something about what telescoping actually means?  If not, then why is this called telescoping, because I don't see what this has to do with a telescope?&lt;/p&gt;&#xA;" OwnerUserId="1417" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-09T19:18:44.900" LastActivityDate="2010-12-09T19:37:25.627" Title="Mathematical Telescoping" Tags="&lt;soft-question&gt;&lt;proof-strategy&gt;" AnswerCount="2" CommentCount="5" FavoriteCount="1" />
  <row Id="13702" PostTypeId="2" ParentId="13695" CreationDate="2010-12-09T19:16:47.120" Score="5" ViewCount="" Body="&lt;p&gt;As in other's answers, the absolute value of $x$ can be defined as the distance between $x$ and 0.  When $x$ is a real number, we can think of $x$ as being on a number line, and there are two locations on a number line that are 1 unit away from 0: 1 unit in the positive direction (called 1) and 1 unit in the negative direction (called &quot;-1&quot;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $x$ can be a complex number, we can think of $x$ as being a point in the plane, and there are a whole lot more locations that are a distance of 1 from 0, such as $\frac{3}{5}+\frac{4}{5}i$ (1 and -1 are also still 1 unit away from 0, so they are still possible values of $x$).&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-12-09T19:16:47.120" CommentCount="2" />
  <row Id="13703" PostTypeId="2" ParentId="13701" CreationDate="2010-12-09T19:19:40.567" Score="2" ViewCount="" Body="&lt;p&gt;The book &lt;a href=&quot;http://www.math.upenn.edu/~wilf/AeqB.html&quot; rel=&quot;nofollow&quot;&gt;A=B&lt;/a&gt; by Petkovsek, Wilf and Zeilberger&#xA;is an extended exposition of the Wilf-Zeilberger method&#xA;for summing series often called &quot;creative telescoping&quot;.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-12-09T19:19:40.567" />
  <row Id="13704" PostTypeId="2" ParentId="13701" CreationDate="2010-12-09T19:19:48.193" Score="2" ViewCount="" Body="&lt;p&gt;1, 2)  Telescoping is one of the ideas behind modern algorithms to automatically prove hypergeometric identities.  These algorithms allow you, for example, to automatically prove binomial coefficient identities.  The standard reference here is Petkovsek, Wilf, and Zeilberger's &lt;a href=&quot;http://www.math.upenn.edu/~wilf/AeqB.html&quot; rel=&quot;nofollow&quot;&gt;A=B&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3)  The name comes from the process of collapsing a telescope, which is analogous to the collapsing of a telescoping sum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Philosophically telescoping is the same as &quot;discrete integration&quot;: telescoping a sum $\sum f(n)$ is the same as finding $g(n)$ such that $f(n) = g(n+1) - g(n)$.  In that sense it is part of the theory of &lt;a href=&quot;http://en.wikipedia.org/wiki/Finite_difference&quot; rel=&quot;nofollow&quot;&gt;finite differences&lt;/a&gt;, although people probably don't call it &quot;telescoping&quot; in this context.  The context in which I hear the term &quot;telescoping&quot; being used is high school math competitions.  It's one of those basic ideas that everyone has in the back of their head, I suppose.  It's elementary and effective when it applies, but usually there are more sophisticated methods available.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt;  Some specific examples.  The ur-example of a telescoping sum is probably&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{k=1}^n \frac{1}{k(k+1)} = \sum_{k=1}^{n} \left( \frac{1}{k} - \frac{1}{k+1} \right) = 1 - \frac{1}{n+1}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and many people have seen this application, but probably far fewer have seen its generalization:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{k=1}^n \frac{1}{k(k+1)...(k+r)} = \frac{1}{r} \sum_{k=1}^n \left( \frac{1}{k(k+1)...(k+r-1)} - \frac{1}{(k+1)...(k+r)} \right) = \frac{1}{r} \left( \frac{1}{r!} - \frac{n!}{(n+r)!} \right).$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other classic example I remember from my competition days is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{k=1}^n \frac{k}{k^4 - k^2 + 1} = \frac{1}{2} \sum_{k=1}^n \left( \frac{1}{k^2 - k + 1} - \frac{1}{(k+1)^2 - (k+1) + 1} \right) = \frac{1}{2} \left( 1 - \frac{1}{n^2 + n + 1} \right)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;although I have to admit I always found it a little contrived.  Finally, telescoping was put to good use to solve &lt;a href=&quot;http://math.stackexchange.com/questions/1431/prove-that-this-function-is-bounded&quot;&gt;this math.SE question&lt;/a&gt; I posed.&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-12-09T19:37:25.627" LastActivityDate="2010-12-09T19:37:25.627" CommentCount="1" />
  <row Id="13705" PostTypeId="1" CreationDate="2010-12-09T19:25:29.900" Score="2" ViewCount="84" Body="&lt;p&gt;Let $m$ be a positive integer. Define $N_m:=\{x\in \mathbb{Z}: x&amp;gt;m\}$. I was wondering when does $N_m$ have a &quot;basis&quot; of two elements. I shall clarify what I mean by a basis of two elements: We shall say the positive integers $a,b$ generate $N_m$ and denote $N_m=&amp;lt;a,b&amp;gt;$ if every element $x\in N_m$ can be written as $x=\alpha a+\beta b$ where $\alpha,\beta$ are nonnegative integers (not both zero) &lt;STRIKE&gt;and no nonnegative linear combination (with at least one coefficient nonzero) of $a,b$ gives an element not in $N_m$.&lt;/STRIKE&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More specifically, I want to understand the set $\{(a,b,m):N_m=&amp;lt;a,b&amp;gt;\}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An example would be the triple $(2,3,1)$. Every positive integer greater than 1 can be written as $2\alpha +3\beta$ for nonnegative integers $\alpha,\beta$, for if it is even we can write it as $2\alpha$ for some positive integer $\alpha$ and if it is odd and greater than $3$ then we can write it as $3+2\alpha$ for some positive integer $\alpha$ and of course $3=1\cdot 3$. FInally the smallest integer a positive linear combination of $2,3$ can generate is $2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS: Not quite sure what to tag this. Feel free to retag.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: After Jason's answer, it seems the first version of this question is more interesting, where the last condition in paragraph 1 is relaxed.&lt;/p&gt;&#xA;" OwnerUserId="3431" LastEditorUserId="3431" LastEditorDisplayName="" LastEditDate="2010-12-09T20:11:19.547" LastActivityDate="2010-12-09T20:24:38.330" Title="Subsets of natural numbers generated by two generators." Tags="&lt;arithmetic&gt;" AnswerCount="3" CommentCount="9" />
  <row Id="13706" PostTypeId="2" ParentId="13705" CreationDate="2010-12-09T19:32:10.467" Score="2" ViewCount="" Body="&lt;p&gt;If $m$ and $n$ are coprime postive integers, then every integer&#xA;with $&amp;gt;mn-m-n$ is expressible in the form $am+bn$ where $a$&#xD;&#xA;and $b$ are nonnegative integers, but $mn-m-n$ isn't.&#xA;This is due to &lt;a href=&quot;http://en.wikipedia.org/wiki/Coin_problem&quot; rel=&quot;nofollow&quot;&gt;Sylvester&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added&lt;/strong&gt; The key to the proof is the lemma&#xA;that if $0\le k&amp;lt;(m-1)(n-1)$ then $k$ is representable iff $mn-n-m-k$&#xD;&#xA;isn't. ISTR this involves a cunning use of the Chinese remainder theorem.&lt;/p&gt;&#xA;" OwnerUserId="226" LastEditorUserId="226" LastEditorDisplayName="" LastEditDate="2010-12-09T19:51:00.597" LastActivityDate="2010-12-09T19:51:00.597" CommentCount="1" />
  <row Id="13708" PostTypeId="2" ParentId="13705" CreationDate="2010-12-09T19:39:09.323" Score="1" ViewCount="" Body="&lt;p&gt;(Now that you have relaxed the last condition of paragraph 1)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suppose you mean non negative $\alpha$, $\beta$? (You seem to be allowing them to be zero).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In which case, this looks like the &lt;a href=&quot;http://mathworld.wolfram.com/FrobeniusNumber.html&quot; rel=&quot;nofollow&quot;&gt;Frobenius Problem&lt;/a&gt; which says that for $a,b$ such that $(a,b)=1$, every number greater than $(a-1)(b-1) - 1 $ is representable in the for $a\alpha + b \beta$ for $\alpha \ge 0, \beta \ge 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus given an $m$, if a factorization of $m+1$ as $(a-1)(b-1)$ with $(a,b) = 1$ exists, then $N_m  \subset \lt a, b \gt$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are also looking for $m$ such that $m \notin \lt a, b \gt$ (the least such $m$) then there are $m$ for which there are no such $a,b$, as $m$ must necessarily be of the form $(a-1)(b-1) - 1$ with $(a,b) = 1$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact we can characterize all such $m$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $m$ is even, there is no such representation (as both $a-1$ and $b-1$ must be odd).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $m$ is odd, then $m+1 = (2-1)(m+2 - 1)$ and thus $a = 2, b = m+2$ will work.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-12-09T20:24:38.330" LastActivityDate="2010-12-09T20:24:38.330" CommentCount="13" />
  <row Id="13709" PostTypeId="2" ParentId="13705" CreationDate="2010-12-09T19:41:55.497" Score="3" ViewCount="" Body="&lt;p&gt;This can only happen for the triple you found.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For, if $m=1$, then we must be able to generate 2, so we must take 2 in the basis.  Likewise, we must be able to generate 3, so we must allow 3 in the basis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $m&amp;gt;1$, then we see by the same reasoning that $N_m$ must be generated by $m+1$ and $m+2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But then we can't express $m+3$: since $m&amp;gt;1$, we see if $\alpha &amp;gt;1$, then that $\alpha(m+1)\geq 2m + 2 &amp;gt; m+3$ .  So, we must take $\alpha = 0$ or $1$.  Then same reasoning applies to $\beta$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we take either of $\alpha$ and $\beta$ equal to $0$, then we just generate $m+1$ or $m+2$.  Hence, we must take them both one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then $m+1 + m+2 = 2m+3 &amp;gt; m+3$.&lt;/p&gt;&#xA;" OwnerUserId="331" LastActivityDate="2010-12-09T19:41:55.497" CommentCount="1" />
  <row Id="13710" PostTypeId="2" ParentId="13563" CreationDate="2010-12-09T19:43:32.273" Score="0" ViewCount="" Body="&lt;p&gt;The bigger number minus the smaller number minus 1.&#xA;Or if $b$ is the bigger number, $s$ the smaller. We have:&#xA;$b - s - 1$&lt;/p&gt;&#xA;" OwnerUserId="2533" LastActivityDate="2010-12-09T19:43:32.273" />
  <row Id="13711" PostTypeId="1" AcceptedAnswerId="13985" CreationDate="2010-12-09T19:49:50.043" Score="2" ViewCount="125" Body="&lt;p&gt;I'm looking for a good book on distribution theory (in the Schwartz sense), I have the basic knowledge as given in Grafakos' Classical Fourier Analysis, but I want to know more about it. Is the reference still Laurent Schwartz' Théorie des distributions? That book is hard to obtain, it seems to be only available in France.&lt;/p&gt;&#xA;" OwnerUserId="1120" LastActivityDate="2010-12-12T00:38:16.777" Title="Distribution theory book." Tags="&lt;reference-request&gt;&lt;distribution-theory&gt;" AnswerCount="1" CommentCount="8" />
  <row Id="13712" PostTypeId="1" AcceptedAnswerId="13720" CreationDate="2010-12-09T19:59:12.280" Score="1" ViewCount="158" Body="&lt;p&gt;Hello,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've read that the formula A if and only if B is a formulation of the equivalence&#xA;that can be decomposed in A if B (the implication B -&gt; A) and in A only if B (the implication A -&gt; B). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't agree with the second part : for me it looks like that A only if B is the translation of ¬B -&gt; ¬A (I know that it's the contraposition of A -&gt; B and consequently it means the same but it's not the same formulation).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So can anyone tell me the correct interpretation please.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;thanks,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bruno&lt;/p&gt;&#xA;" OwnerUserId="4475" LastEditorUserId="4475" LastEditorDisplayName="" LastEditDate="2010-12-09T20:18:10.257" LastActivityDate="2010-12-12T20:18:57.750" Title="How to decompose A if and only if B ?" Tags="&lt;logic&gt;" AnswerCount="4" CommentCount="5" />
  <row Id="13713" PostTypeId="2" ParentId="13712" CreationDate="2010-12-09T20:09:10.877" Score="0" ViewCount="" Body="&lt;p&gt;I am not sure I am understanding what you are asking, so my answer my not address what you are asking.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For an if an only if (iff) theorem, you have $A \Leftrightarrow B$. This means that A only occurs when B occurs and B only occurs when A occurs. If you are trying to prove an iff theorem, you must prove it in both directions. That is, you must show that $A \Rightarrow B$ and then show that $B \Rightarrow B$.&lt;/p&gt;&#xA;" OwnerUserId="3791" LastActivityDate="2010-12-09T20:09:10.877" />
  <row Id="13714" PostTypeId="2" ParentId="13693" CreationDate="2010-12-09T20:16:54.610" Score="1" ViewCount="" Body="&lt;p&gt;Your left adjoint is not quite correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, take the case where $A$ is the variety of all groups, and $B$ is the variety of &lt;em&gt;abelian&lt;/em&gt; groups. The forgetful functor from $\mathcal{A}b$ to $\mathcal{G}roup$  indeed has a left adjoint, but the left adjoint $\mathcal{F}$ takes a group $G$ to $G^{\rm ab}$, the abelianization of $G$. This easily yields examples of objects $a$ and $b$ such that $\mathcal{F}(a)\cong \mathcal{F}(b)$, even though $a\not\cong b$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strike&gt;For the specific case of Boolean algebras I think the answer is still negative as stated: take a boolean algebra $a$ that is not [countably] complete and such that its [countable] completion has strictly larger cardinality than $a$ (I assume such things exist, but if this is not the case, then of course I'm falling flat on my case here). Let $b$ be the image of the [countable] completion under the forgetful functor. Since $a$ and $b$ have distinct cardinality, $a\not\cong b$, but $\mathcal{F}(b) = \mathcal{F}(\mathcal{U}(\mathcal{F}(a)))\cong \mathcal{F}(a)$, with the last equality using the unit and counits of the adjunction. &lt;/strike&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One way to construct $\mathcal{F}(a)$ if your varieties have a nice underlying set functor with an adjoint is to first consider $U(a)$, the underlying set of $a$, then take $\mathfrak{F}_B(U(a))$, the free $B$-algebra on the underlying set of $a$, and then let $\mathcal{F}(a) = \mathfrak{F}_B(U(a))/\Phi$, where $\Phi$ is the $B$-congruence generated by the relations corresponding to the $A$-operations on $a$. For instance, if $B$ is $\mathcal{G}roups$ and $A$ is $\mathcal{M}onoids$, given a monoid $M$, you first take $U(M)=\{[m]\mid m\in M\}$, the underlying set of $M$ (I'm using $[m]$ to denote the element $m$ in the &lt;em&gt;set&lt;/em&gt; $U(M)$). Then you take the free group on $U(M)$, and you mod out by the the normal subgroup generated by the relations $[m][n]=[mn]$ (and since $[mn]=[nm]$, you end up with an abelian group). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Added:&lt;/em&gt; I need to think a bit more about the Boolean algebra case; you are correct that my assertion about $\mathcal{F}(\mathcal{U}(\mathcal{F}(a)))$ does not hold in general, and likely not in this case either. My gut feeling is that you can have $\mathcal{F}(a)\cong\mathcal{F}(b)$ even with $a\not\cong b$, though.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added 2:&lt;/strong&gt; Heh; reading the MO question you link to in the comments suggests that I am facing exactly what the long discussion between Joel David Hamkins and Todd Trimble was about, and I was making the same kind of assumptions Joel made to reach my conclusion that my construction would necessarily yield an example. &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-09T21:37:58.207" LastActivityDate="2010-12-09T21:37:58.207" CommentCount="7" />
  <row Id="13715" PostTypeId="2" ParentId="13712" CreationDate="2010-12-09T20:23:40.520" Score="0" ViewCount="" Body="&lt;p&gt;I would expand &quot;A only if B&quot; to &quot;A can only be true if B is also true&quot;, then go to &quot;if A then B&quot; and represent that as $A \implies B$.  As you say, that is English processing, not logic processing.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-12-09T20:23:40.520" CommentCount="3" />
  <row Id="13716" PostTypeId="1" AcceptedAnswerId="13724" CreationDate="2010-12-09T20:30:37.130" Score="1" ViewCount="86" Body="&lt;p&gt;I came across this problem. If $X_1,X_2, \ldots, X_n$ is an iid sequence with uniform distribution on [-1,1] and we define a new random variable like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$Y_n = \frac{\sum X_k}{\sum(X_k^2 + X_k^3)}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do I use the strong law of large numbers to show that $Y_n$ is normal and find its mean and variance? The uniform distribution $X_k$ will converge to a normal distribution with mean 0 and variance 3. But I am having trouble finding expectation and variance for $Y_n$.&lt;/p&gt;&#xA;" OwnerUserId="957" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-12-09T22:12:58.353" LastActivityDate="2010-12-09T22:12:58.353" Title="problem on strong law of large numbers" Tags="&lt;probability-theory&gt;&lt;measure-theory&gt;&lt;convergence&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="13717" PostTypeId="2" ParentId="13695" CreationDate="2010-12-09T20:44:20.973" Score="1" ViewCount="" Body="&lt;p&gt;If thinks in this as $f(x)= |x|$, you need to see which are the preimages of $1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Graphicaly you could to get this:&#xA;&lt;img src=&quot;http://i.imgur.com/9xJ4J.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Blue&lt;/em&gt; : $f(x)=|x|$&lt;br&gt;&#xA;&lt;em&gt;Red&lt;/em&gt; : $g(x)=1$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, $f(x)=1$ only on the intersection points of these two functions, is say on $x=\pm 1$&lt;/p&gt;&#xA;" OwnerUserId="3421" LastEditorUserId="3421" LastEditorDisplayName="" LastEditDate="2010-12-09T21:08:11.733" LastActivityDate="2010-12-09T21:08:11.733" />
  <row Id="13718" PostTypeId="1" CreationDate="2010-12-09T20:50:50.990" Score="0" ViewCount="107" Body="&lt;p&gt;This is part of a chapter from Spivak's Calculus on Manifolds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The initial conditions are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $a\in A \subset \mathbb{R}^n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $g:\mathbb{R}^n \to \mathbb{R}^n$.  Assume $g&amp;#39;(a)=I$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Define $h:A\to\mathbb{R}^n$ by $h(x)=(g^1(x),\dots,g^{n-1}(x),x^n)$.  Then $h&amp;#39;(a)=I$.  Then there is some open $U&amp;#39;$ with $a\in U&amp;#39; \subset A$ where $h$ is 1-1 and $\det h&amp;#39;(x) \ne 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Define $k:h(U&amp;#39;)\to\mathbb{R}^n$ by $k(x)=(x^1,\dots,x^{n-1},g^n(h^{-1}(x)))$.  Then $g = k \circ h$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Spivak shows that $k&amp;#39;(h(a))=I$.  Then there is some open set $V$ where $h(a)\in V\subset h(U&amp;#39;)$ where $k$ is 1-1 and $\det k&amp;#39;(x)\ne 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $U=k^{-1}(V)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is where my problem is: supposedly, this is sufficient to show that $g=k \circ h$ where $h:U \to \mathbb{R}^n$ and $k:V \to\mathbb{R}^n$ and $h(U) \subset V$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is $h(U) \subset V$?  Obviously $h(a)\in V$ and $k(U) \subset V$, but I don't see the connection between the two.&lt;/p&gt;&#xA;" OwnerUserId="326" LastActivityDate="2010-12-09T20:50:50.990" Title="Problem understanding proof of change of variable in integration" Tags="&lt;calculus&gt;&lt;real-analysis&gt;" CommentCount="1" />
  <row Id="13719" PostTypeId="1" CreationDate="2010-12-09T21:03:13.887" Score="1" ViewCount="111" Body="&lt;p&gt;The pullback is a subset of the cartesian product in the category of commutative rings with unit. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the pullback in the category of commutative $k$-algebras? Is it the same set as in rings?&lt;/p&gt;&#xA;" OwnerUserId="3789" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-09T21:11:58.597" LastActivityDate="2010-12-09T21:17:27.843" Title="What is the pullback in the category of commutative algebras?" Tags="&lt;abstract-algebra&gt;&lt;algebraic-geometry&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="13720" PostTypeId="2" ParentId="13712" CreationDate="2010-12-09T21:04:36.877" Score="1" ViewCount="" Body="&lt;p&gt;If I understand your question correctly, you are bothered by the semantics of the statement &quot;$A$ only if $B$&quot;.  You can take this statement as idiomatic, that is, if taken literally it has one meaning, but social convention (in this case the social convention of the mathematics community) leads us to interpret it as always meaning $A \rightarrow B$.  There is certainly room in English to take certain phrases as being somewhat illogical.  My point is that &quot;$A$ only if $B$&quot; doesn't actually have to stand up to logical dissection of the semantics.  Extracting meaning from English sentences is not linear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I always look at that the statement &quot;$A$ only if $B$&quot; as meaning A is true only if B is true.  If we look at a truth table for a statement where A is true only if B is true, then we see that this is equivalent to the statement $A\rightarrow B$. &lt;/p&gt;&#xA;" OwnerUserId="1004" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2010-12-09T21:07:55.053" LastActivityDate="2010-12-09T21:07:55.053" CommentCount="1" />
  <row Id="13721" PostTypeId="2" ParentId="13719" CreationDate="2010-12-09T21:17:27.843" Score="2" ViewCount="" Body="&lt;p&gt;Yes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More generally, fix a category $\mathcal{C}$ that contains finite limits. Given an object $A \in \mathcal{C}$, we can construct the category $\mathcal{C}_A$ of objects &quot;under $A$.&quot; That is, an object of this category $\mathcal{C}_A$ is a morphism $A \to B$ and a  morphism is a commutative triangle. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The claim is that $\mathcal{C}_A$ admits finite limits, and the limits are the same as in $\mathcal{C}$. This is basically formal. Given a functor $F:I \to \mathcal{C}_A$ from a finite category $I$, we get a functor $G: I \to \mathcal{C}$ which has a limit by assumption. Moreover, for each $i \in I$, we have a morphism $A \to Gi$. By the universal property, this becomes a morphism $A \to \lim G$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I claim that the object $A \to \lim G$ in $\mathcal{C}_A$ is a limit of $F$. This can be checked directly from the definitions.&lt;/p&gt;&#xA;" OwnerUserId="536" LastActivityDate="2010-12-09T21:17:27.843" />
  <row Id="13722" PostTypeId="1" CreationDate="2010-12-09T21:33:37.573" Score="0" ViewCount="104" Body="&lt;p&gt;I need to find the function c(k), knowing that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{k=0}^{\infty} \frac{c(k)}{k!}=1$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{k=0}^{\infty} \frac{c(2k)}{(2k)!}=0$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{k=0}^{\infty} \frac{c(2k+1)}{(2k+1)!}=1$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{k=0}^{\infty} \frac{(-1)^k c(2k+1)}{(2k+1)!}=-1$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{k=0}^{\infty} \frac{(-1)^k c(2k)}{(2k)!}=0$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it possible?&lt;/p&gt;&#xA;" OwnerUserId="2513" LastEditorUserId="2513" LastEditorDisplayName="" LastEditDate="2010-12-09T21:45:45.250" LastActivityDate="2010-12-09T22:25:53.923" Title="How to solve the following system?" Tags="&lt;analysis&gt;&lt;sequences-and-series&gt;&lt;functional-equations&gt;&lt;generating-functions&gt;&lt;power-series&gt;" AnswerCount="2" CommentCount="12" />
  <row Id="13724" PostTypeId="2" ParentId="13716" CreationDate="2010-12-09T22:10:25.063" Score="4" ViewCount="" Body="&lt;p&gt;In order to have the desired convergence to the ${\rm N}(0,3)$ distribution, we must multiply $Y_n$ by $\sqrt{n}$ (otherwise, $Y_n \to 0$ almost surely).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose that $X_1,X_2,\ldots$ is a sequence of i.i.d. uniform$[-1,1]$ rv's.&#xA;Define a sequence $(Y_n)$ by&#xA;$$&#xD;&#xA;Y_n = &#xD;&#xA;\sqrt{n} \frac{{\sum\nolimits_{k = 1}^n {X_k } }}{{\sum\nolimits_{k = 1}^n {(X_k^2  + X_k^3 )} }}.&#xD;&#xA;$$ &#xA;Then $Y_n$ can be written as&#xA;$$&#xD;&#xA;Y_n = \frac{{n^{ - 1/2} \sum\nolimits_{k = 1}^n {X_k } }}{{n^{ - 1} \sum\nolimits_{k = 1}^n {X_k^2 }  + n^{ - 1} \sum\nolimits_{k = 1}^n {X_k^3 } }}.&#xD;&#xA;$$&#xD;&#xA;Now, $X_k^3$ has mean zero, and so by the strong law of large numbers (SLLN), $n^{ - 1} \sum\nolimits_{k = 1}^n {X_k^3 }  \to 0$ almost surely. $X_k^2$, on the other hand, has mean equal to $\int_{ - 1}^1 {x^2 (1/2){\rm d}x}  = 1/3$; hence, by SLLN, $n^{ - 1} \sum\nolimits_{k = 1}^n {X_k^2 }  \to 1/3$ almost surely. Next, $X_k$ has mean zero and (hence) variance equal to ${\rm E}(X_k^2) = 1/3$. So, by the central limit theorem, &#xA;$$&#xD;&#xA;\frac{{\sum\nolimits_{k = 1}^n {X_k } }}{{\sqrt {1/3} \sqrt n }} \to {\rm N}(0,1)&#xD;&#xA;$$&#xD;&#xA;(in distribution). Combining it all, we see that $Y_n$ converges in distribution to $(3/\sqrt{3})Z$, where $Z \sim {\rm N}(0,1)$. Put it another way, $Y_n \to {\rm N}(0,3)$.&lt;/p&gt;&#xA;" OwnerUserId="2810" LastActivityDate="2010-12-09T22:10:25.063" CommentCount="1" />
  <row Id="13725" PostTypeId="2" ParentId="13722" CreationDate="2010-12-09T22:13:51.950" Score="3" ViewCount="" Body="&lt;p&gt;Following Rotwang, define $b(k)=c(k)/k!$, then note that both the second and fifth can be satisfied as long as $\sum{b(2k)}=0$.  Then the first and third are redundant and define $d(k)=b(2k+1)$.  Now we have $\sum{d(k)}=1, \sum{(-1)^kd(k)}=-1$.  Adding and subtracting, $\sum{d(2k)}=0, \sum{d(2k+1)}=1.$  So the final is that we must have $$\begin{align} \sum\frac{c(2k)}{k!}&amp;amp;=0\\&#xD;&#xA;   \sum\frac{c(4k+1)}{(4k+1)!}&amp;amp;=0\\&#xD;&#xA;   \sum\frac{c(4k+3)}{(4k+3)!}&amp;amp;=1  \end{align}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and any $c(k)$ that satisfies this will work.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2010-12-09T22:19:41.027" LastActivityDate="2010-12-09T22:19:41.027" CommentCount="4" />
  <row Id="13726" PostTypeId="2" ParentId="13722" CreationDate="2010-12-09T22:25:53.923" Score="3" ViewCount="" Body="&lt;p&gt;You are looking for a function $\displaystyle f(z) = \sum_{k \ge 0} \frac{c(k)}{k!} z^k$ satisfying&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f(1) = 1$$&#xD;&#xA;$$f(-1) = -1$$&#xD;&#xA;$$f(i) = -i.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Infinitely many functions have this property.  There is a unique quadratic polynomial $p(z)$ with this property (for example by &lt;a href=&quot;http://en.wikipedia.org/wiki/Lagrange_polynomial&quot; rel=&quot;nofollow&quot;&gt;Lagrange interpolation&lt;/a&gt;), and for any entire function $q(z)$ the function $p(z) + (x - 1)(x + 1)(x - i) q(z)$ has this property.  In fact these are all entire functions with this property.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More generally I think the theory of interpolation by entire functions is fairly well-understood, but I don't know of a good reference.  If the set of $z$ at which you fix the value of $f$ has a limit point, then $f$ is unique by the identity theorem.  If the set of $z$ at which you fix the value of $f$ is countable and does not have a limit point, then $f$ is non-unique by &lt;a href=&quot;http://en.wikipedia.org/wiki/Weierstrass_factorization_theorem&quot; rel=&quot;nofollow&quot;&gt;Weierstrass factorization&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-12-09T22:25:53.923" />
  <row Id="13727" PostTypeId="1" AcceptedAnswerId="13729" CreationDate="2010-12-09T23:09:50.567" Score="2" ViewCount="137" Body="&lt;p&gt;WolframAlpha is confusing me. I'm working on two different integral problems, and with each one, for some reason, Wolfram switches the signs of the answer.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;http://www.wolframalpha.com/input/?i=integral+dx/(x^.5(x-1))&lt;/li&gt;&#xA;&lt;li&gt;http://www.wolframalpha.com/input/?i=integral+(x-1)/(x^2-4)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;For the first one, they get $\log[1 - \sqrt{x}] - \log[(1 + \sqrt{x})]$ (see under alternate forms, thats the one I'm using).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I get:&#xA;$\log[\sqrt{x}-1] - \log[(\sqrt{x}+1)]$ And so it differs by a sign, but its really bothering me, since I'm sure I'm doing it right. And the same applies to the second one:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They get:&#xA;$3/4 \log(-x-2)+1/4 \log(x-2)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I get:&#xA;$3/4 \log(x+2)+1/4 \log(x-2)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And so it differs by a sign again. (If this is a common mistake, then maybe you can point it out, but if this is just me and you can't see where I could have gone wrong, I can list my steps)&lt;/p&gt;&#xA;" OwnerUserId="2985" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-10T00:49:34.383" LastActivityDate="2010-12-10T00:49:34.383" Title="Integration and natural log question" Tags="&lt;calculus&gt;&lt;integral&gt;" AnswerCount="1" />
  <row Id="13728" PostTypeId="1" CreationDate="2010-12-09T23:11:39.613" Score="3" ViewCount="121" Body="&lt;p&gt;Is it correct to say that the number of circles in an Apollonian gasket is countable becuase we can form a correspondence with a Cantor set, as their methods of construction are similar?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What about if we apply the Apollonian gasket construction inside of a fractal like the Koch snowflake? (I think that will still be countable.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What if you did the Apollonian gasket construction between f(x) = sin(1/x) and  g(x) = 2 - sin(1/x) between -1 and 1?(I think that will still be countable too, but it's not matching my intuition... which says &quot;no way is that countable!&quot;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any closed curve that would result in the number of circles being uncountable? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What if we consider the Apollonian gasket made of spheres in $\mathbb{R}^3$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Please keep in mind I have only had two courses in Analysis. My apologies if any of this is too naive.) &lt;/p&gt;&#xA;" OwnerUserId="2563" LastEditorUserId="2563" LastEditorDisplayName="" LastEditDate="2010-12-09T23:50:27.130" LastActivityDate="2010-12-09T23:56:46.580" Title="Is the number of circles in the Apollonian gasket countable?" Tags="&lt;geometry&gt;&lt;real-analysis&gt;&lt;infinity&gt;&lt;fractals&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="13729" PostTypeId="2" ParentId="13727" CreationDate="2010-12-09T23:15:26.477" Score="3" ViewCount="" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The (real) indefinite integral is actually different in $(0, 1)$ and in $(1, \infty)$.  WolframAlpha gave you the answer in the first interval and your answer works for the second.  (Remember that in the real numbers, the logarithm of a non-positive number is undefined.)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;This time WolframAlpha's answer is not defined for any real $x$.  Your answer works in $(2, \infty)$ but there is a different antiderivative in $(-2, 2)$ and yet another antiderivative in $(-\infty, -2)$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;This is a fairly subtle property of the logarithm.  Remember that any two antiderivatives of a function differ by a constant.  The problem in this case is that the constant is complex, so if you aren't working in the complex numbers you get genuinely different answers.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-12-09T23:15:26.477" CommentCount="5" />
  <row Id="13730" PostTypeId="2" ParentId="13728" CreationDate="2010-12-09T23:30:12.897" Score="6" ViewCount="" Body="&lt;p&gt;The reason a collection of disjoint open sets in $\mathbb{R}^n$ is at most countable is because $\mathbb{R}^n$ is because $\mathbb{R}^n$ is &lt;a href=&quot;http://en.wikipedia.org/wiki/Separable_space&quot; rel=&quot;nofollow&quot;&gt;separable&lt;/a&gt;: it has a countable dense subset given by the points with rational coordinates, and any open set must contain such a point that the others don't contain, so there can be at most countably many of them.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is also a measure-theoretic argument relying on the fact that $\mathbb{R}^n$ with the Lebesgue measure is &lt;a href=&quot;http://en.wikipedia.org/wiki/%CE%A3-finite_measure&quot; rel=&quot;nofollow&quot;&gt;$\sigma$-finite&lt;/a&gt;, which might come closer to agreeing with geometric intuition: if you could fit uncountably many open balls into $\mathbb{R}^n$, some closed $n$-cube (WLOG $[0, 1]^n$) would contain uncountably many of them, each of which would have positive measure.  But a standard argument shows that this is impossible:  if $S_k$ is the set of balls in $[0, 1]^n$ of radius at least $\frac{1}{k}$, then since $\bigcup S_k$ is uncountable it follows that some $S_k$ is uncountable, hence countable unions of its elements have unbounded measure; contradiction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The argument in your first sentence works to show that the number of circles in the standard construction of an Apollonian gasket (start with some circles and add circles) is countable, since at each step only finitely many circles are constructed.  But maybe you are thinking of a more general construction. &lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-12-09T23:56:46.580" LastActivityDate="2010-12-09T23:56:46.580" />
  <row Id="13732" PostTypeId="1" CreationDate="2010-12-09T23:47:15.400" Score="1" ViewCount="91" Body="&lt;p&gt;Suppose I have acquired 103,304 customers for a transaction-based business at 12 months into the business and their number has increased at a monthly rate of 40% (from month 1, in which 2,551 customers were acquired). Assume 103,304 represents the total number of new customers (active or dead) acquired over the 12 month period. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there is one fixed-price product/service worth $10 that each customer makes 4 purchases of across a 6 month period (1 every 1.5 months) and then ceases to be a customer, how can I calculate my total revenue by month 12? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that I'm something of an interloper in this forum, but I'll absolutely accept an answer and up-vote all good ones.&lt;/p&gt;&#xA;" OwnerUserId="4588" LastEditorUserId="4588" LastEditorDisplayName="" LastEditDate="2010-12-10T02:00:38.300" LastActivityDate="2010-12-10T02:00:38.300" Title="Tricky Finance-related Question" Tags="&lt;homework&gt;&lt;algebra-precalculus&gt;&lt;finance&gt;" AnswerCount="1" />
  <row Id="13733" PostTypeId="2" ParentId="13732" CreationDate="2010-12-09T23:58:03.770" Score="1" ViewCount="" Body="&lt;p&gt;You don't give enough data to find an answer.  Each customer is worth 40 in revenue, so if they all signed up by month 6 (and therefore already spend their 40) you have 4M.  If they all signed up in month 11.5 and haven't bought yet you have zero.  You need something about the rate of increase of customers and about when in the six months they buy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It would simplify things to assume they spend evenly over time, so every customer is worth 40/6 per month.  Then if n(i) is the number of customers who join in month i, the number of customers active in month i is the sum over the last six months of n(i) and your revenue will be 4M less what you will get from the current customer base in the next months.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-12-09T23:58:03.770" CommentCount="2" />
  <row Id="13734" PostTypeId="1" AcceptedAnswerId="13738" CreationDate="2010-12-10T00:18:22.717" Score="0" ViewCount="337" Body="&lt;p&gt;If given 2 lines $\alpha$ and $\beta$, &lt;s&gt;that are created by&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;2 points: A and B &lt;/li&gt;&#xA;&lt;li&gt;2 plane intersection&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I want to find shortest distance between them. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left\{\begin{array}{c} P_1=x_1X+y_1Y+z_1Z+C=0 \\ P_2=x_2X+y_2Y+z_2Z+C=0\end{array}\right.$$&lt;br&gt;&#xA;$$A=\left(x_3;y_3;z_3\right)$$&lt;br&gt;&#xA;$$B=\left(x_4;y_4;z_4\right)$$&lt;br&gt;&#xA;$$\alpha =n_1\times n_2=\left(\left|\begin{array}{cc} y_1 &amp;amp; z_1 \\ y_2 &amp;amp; z_2\end{array}\right|;-\left|\begin{array}{cc} x_1 &amp;amp; z_1 \\ x_2 &amp;amp; z_2\end{array}\right|;\left|\begin{array}{cc} x_1 &amp;amp; y_1 \\ x_2 &amp;amp; y_2\end{array}\right|\right)$$&lt;br&gt;&#xA;$$\beta =$$ &lt;/s&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From here I tried:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The question of &quot;shortest distance&quot; is&#xA;  only interesting in the skew case.&#xA;  Let's say p0 and p1 are points on the&#xA;  lines L0 and L1, respectively. Also d0&#xA;  and d1 are the direction vectors of L0&#xA;  and L1, respectively. The shortest&#xA;  distance is (p0 - p1) * , in which *&#xA;  is dot product, and is the normalized&#xA;  cross product. The point on L0 that is&#xA;  nearest to L1 is p0 + d0(((p1 - p0) *&#xA;  k) / (d0 * k)), in which k is d1 x d0&#xA;  x d1.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Read more: &lt;a href=&quot;http://wiki.answers.com/Q/If_the_shortest_distance_between_two_points_is_a_straight_line_what_is_the_shortest_distance_between_two_straight_lines#ixzz17fAWKFst&quot; rel=&quot;nofollow&quot;&gt;http://wiki.answers.com/Q/If_the_shortest_distance_between_two_points_is_a_straight_line_what_is_the_shortest_distance_between_two_straight_lines#ixzz17fAWKFst&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried, but failed. &lt;/p&gt;&#xA;" OwnerUserId="2317" LastEditorUserId="2317" LastEditorDisplayName="" LastEditDate="2010-12-10T00:54:28.867" LastActivityDate="2010-12-10T01:00:02.923" Title="How to find shortest distance between two skew lines in 3D?" Tags="&lt;geometry&gt;" AnswerCount="1" CommentCount="6" />
  <row Id="13735" PostTypeId="1" AcceptedAnswerId="13759" CreationDate="2010-12-10T00:20:41.170" Score="15" ViewCount="512" Body="&lt;p&gt;Hi,&#xA;I'm curious to whether you guys have any tips on book concerning classical euclidean geometry. I'd like somewhat of an advanced treatment, around the same level as Coxeter's &quot;Geometry revisited&quot;. I'd however want something more comprehensive, therefore I'm asking you for tips!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hopeful for answers ^^&lt;/p&gt;&#xA;" OwnerUserId="2394" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-10T00:39:14.967" LastActivityDate="2011-05-24T15:44:32.500" Title="Books on classical geometry" Tags="&lt;geometry&gt;&lt;books&gt;" AnswerCount="7" CommentCount="5" FavoriteCount="6" />
  <row Id="13736" PostTypeId="1" AcceptedAnswerId="13741" CreationDate="2010-12-10T00:21:12.140" Score="4" ViewCount="256" Body="&lt;p&gt;So, I am a graduate student who is certain that he does not want to do analysis (I think...).  What are the most exciting fields in mathematics right now?  It seems to me that very generally, they are algebraic geometry and algebraic topology, with the latter being specifically concerned with homotopy theory (something I find very interesting).  Perhaps this is not the correct place to post such a question and I'm welcome to redirection on that front.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ultimately, I would like to do something highly abstract and categorical.  I love the ideas of sheaves, homological algebra, and various categorical concepts.  Does anyone have any good knowledge of where to begin?  Also, do algebraic geometry and homotopy theory have a great deal of common ground?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks, and sorry if this question belongs somewhere else.  &lt;/p&gt;&#xA;" OwnerUserId="3595" LastActivityDate="2010-12-10T01:49:12.157" Title="Geometry or Topology" Tags="&lt;soft-question&gt;&lt;education&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="1" />
  <row Id="13737" PostTypeId="2" ParentId="13735" CreationDate="2010-12-10T00:38:24.203" Score="7" ViewCount="" Body="&lt;p&gt;Well I only took one course on elementary geometry so I don't know much about this, but we used Edwin Moise's book &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0201508672&quot; rel=&quot;nofollow&quot;&gt;Elementary Geometry From an Advanced Standpoint&lt;/a&gt;. It is very well written and probably has a level that will suit you. We used the first edition so I'm not sure what may have changed in the other editions and I can't seem to find a google preview of the third edition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In any case I highly recommend you take a look at it to see if you like it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;&#xA;I found the table of contents for the third edition &lt;a href=&quot;http://www.pearsonhighered.com/educator/product/Elementary-Geometry-from-an-Advanced-Standpoint/9780201508673.page&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="900" LastEditorUserId="900" LastEditorDisplayName="" LastEditDate="2010-12-10T04:37:07.693" LastActivityDate="2010-12-10T04:37:07.693" />
  <row Id="13738" PostTypeId="2" ParentId="13734" CreationDate="2010-12-10T01:00:02.923" Score="3" ViewCount="" Body="&lt;p&gt;Per &lt;a href=&quot;http://en.wikipedia.org/wiki/Skew_lines#Distance_between_two_skew_lines&quot; rel=&quot;nofollow&quot;&gt;this wikipedia article&lt;/a&gt;, if your lines are $\vec{X}=\vec{X_1}+t\vec{D_1}$ and $\vec{X}=\vec{X_2}+t\vec{D_2}$, the distance between them is $$\left|\frac{\vec{D_1}\times\vec{D_2}}{|\vec{D_1}\times\vec{D_2}|}\cdot(\vec{X_1}-\vec{X_2})\right|.$$&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-12-10T01:00:02.923" />
  <row Id="13739" PostTypeId="1" AcceptedAnswerId="15000" CreationDate="2010-12-10T01:26:21.463" Score="1" ViewCount="130" Body="&lt;p&gt;I'm trying to solve exercise 2614.1 from Demidovich's famous book of exercises on Analysis. Having solved the bit about convergence, I'm now stuck in trying to prove that a series of terms $a_n\geq 0$ is divergent if there exists $N$ such that $(1-\sqrt[n]{a_n})\frac{n}{\log{n}} \leq 1$ for all $n&amp;gt;N$. I've been trying to find another divergent series with which to compare it, or to prove that $a_n$ does not tend to zero. Any help will be appreciated.&lt;/p&gt;&#xA;" OwnerUserId="1313" LastActivityDate="2011-01-19T21:28:53.600" Title="Test for divergence of series." Tags="&lt;sequences-and-series&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="1" />
  <row Id="13740" PostTypeId="1" AcceptedAnswerId="13746" CreationDate="2010-12-10T01:29:02.280" Score="0" ViewCount="107" Body="&lt;p&gt;I've been spinning my wheels on this for an embarrassing amount of time. This is not for an assignment, I'm just studying.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Three balls are placed at random in&#xA;  three boxes. There is no restriction&#xA;  on the number of balls per box; list&#xA;  the 27 possible outcomes of this&#xA;  experiment.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I could only come up with 10 where order of placing the balls into the boxes doesn't matter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;{(0,0,3), (0,1,2), (0,2,1), (0,3,0), (1,0,2), (1,1,1), (1,2,0), (2,0,1), (2,1,0), (3,0,0)}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What am I doing wrong?&lt;/p&gt;&#xA;" OwnerUserId="71" LastActivityDate="2010-12-10T01:50:22.613" Title="How to Count - 3 balls in 3 boxes" Tags="&lt;counting&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="13741" PostTypeId="2" ParentId="13736" CreationDate="2010-12-10T01:40:08.230" Score="6" ViewCount="" Body="&lt;p&gt;Wow, this is great! I would love to expound a bit on some of the things you have mentioned. While I agree with Ryan that what is exciting and hot is entirely subjective, I too find homotopy theory and algebraic geometry enthralling. I really only know things about homotopy theory though, so I can only talk on that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are loads of places to go abstract crazy under the general heading of homotopy theory. A lot of the model category theory is being put to use in many many places, like Algebraic Geometry! (This is the work of Dugger-Isaksen that Ryan mentioned above). There is a lot of beautiful abstract framework that people work with that falls under the umbrella of homotopy theory. Personally I am a bit more on the computational side, or rather old fashioned &quot;how do we compute $\pi_n^S(\mathbb{S})$?&quot; So I am interested in different computational aspects of the stable homotopy category of spectra. It really is a big field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I just realized I did not address your question about where to begin. There are some amazing deep results that are really cool that you can get to in a finite amount of time. For homotopy theory I would work on trying to get to come classical results of Adams, like vector fields on spheres or Hopf invariant one. Both of these are addressed in Mosher and Tangora (now a dover book). It is a good book, but you should skip certain bits, like you don't need their construction of the steenrod operations. There is also the theory of formal group laws and how those relate to stable homotopy theory, that stuff is awesome. Being at JHU, I would begin by asking Jack Morava or Andrew Salch. They are both super nice guys that know a whole lot, but they are really smart and might be hard to keep up with. So maybe ask them what got them started in being interested in such things, or what they think is something that is really cool to work towards. I also think that Boardman and Wilson would be excellent people to ask, but I have had no interaction with them. They are also deep vats of information. I learned so much going through the first third of Boardmans CCSS paper, it was great!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as the interaction, it is large! As Mathew pointed out there is a whole &quot;new&quot; field called motivic homotopy theory that asks &quot;what can homotopy theory tell us about schemes?&quot; There are other interactions though, a lot of number theory and stackiness is interesting via the chromatic picture. A lot of people study the moduli stack of formal group (laws) in order to get at the stable homotopy groups of spheres. That moduli stack necessarily has a lot of AG in it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the best thing to do would be to talk to people around about things that you are learning/want to learn. In fact, feel free to drop me a line at first intial last name at wayne.edu. Seriously drop me a line if you want to get into a bit more detail, it is unclear what will be beneficial for me to say without a bit more background.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS: there are great people at JHU that know loads about homotopy theory, as well as its interactions with algebraic geometry. Morava pioneered the relationship with class field theory, and now it seems like Salch is taking it all the way to Langlands Land. It would be hard to be at a better place to study homotopy theory!&lt;/p&gt;&#xA;" OwnerUserId="627" LastEditorUserId="627" LastEditorDisplayName="" LastEditDate="2010-12-10T01:49:12.157" LastActivityDate="2010-12-10T01:49:12.157" CommentCount="2" />
  <row Id="13744" PostTypeId="2" ParentId="13654" CreationDate="2010-12-10T01:45:30.840" Score="2" ViewCount="" Body="&lt;p&gt;For your original problem, here's the approach I'd take:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know what the set of graphs of $y=x^2-ax+2$ looks like as $a$ varies:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/U3UQ8.gif&quot; alt=&quot;animation of graphs as a varies&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Okay, the picture in my head isn't that clear, but I know that they all have y-intercept 2 and move in more or less that fashion as $a$ varies (the locus of the vertices of those parabolas is itself a parabola).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that I want the interval from when the equation starts to have real roots up to when the greater of its two roots is 3.  Starting with the latter condition, $3=\frac{a+\sqrt{a^2-8}}{2}\implies(6-a)^2=a^2-8\implies 36-12a=-8\implies a=\frac{11}{3}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The earlier condition happens when $x^2-ax+2$ is a perfect square, so when $\frac{a}{2}=\pm\sqrt{2}$.  The negative value of $a$ leads to the parabola that touches the x-axis on the negative side, so the relevant value is $a=2\sqrt{2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From these two bounding values and the animated image in my head, the desired values of $a$ are $2\sqrt{2}&amp;lt;a&amp;lt;3$.&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-12-10T01:45:30.840" CommentCount="2" />
  <row Id="13745" PostTypeId="2" ParentId="13739" CreationDate="2010-12-10T01:48:28.933" Score="1" ViewCount="" Body="&lt;p&gt;Try to show that $(1 - \frac{{\log n}}{n})^n $ is asymptotically equal to $b_n$, as $n \to \infty$, where &#xA;$\sum\nolimits_{n = 1}^\infty  {b_n }$ is some very well known divergent series...&lt;/p&gt;&#xA;" OwnerUserId="2810" LastActivityDate="2010-12-10T01:48:28.933" />
  <row Id="13746" PostTypeId="2" ParentId="13740" CreationDate="2010-12-10T01:50:22.613" Score="1" ViewCount="" Body="&lt;p&gt;Using &lt;a href=&quot;http://math.stackexchange.com/questions/686/combinations-of-selecting-n-objects-with-k-different-types/691#691&quot;&gt;the stars-and-bars technique&lt;/a&gt;, for your interpretation of the problem, there are in fact ${3+3-1\choose 3}=10$ possible outcomes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As Qiaochu Yuan said in his comment, 27 is the result when the three balls are distinguishable (and I agree with him that it is not clear in the problem statement)—for each of the 3 balls, there are 3 possible locations, so $3^3=27$ possible outcomes.&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-12-10T01:50:22.613" />
  <row Id="13748" PostTypeId="2" ParentId="13525" CreationDate="2010-12-10T02:24:18.543" Score="3" ViewCount="" Body="&lt;p&gt;Expanding on Jack's answer, there is a whole exciting field known as &quot;chromatic homotopy theory&quot;.  I don't know much about it, I'll say a few slogans since I noticed you also asked the question &quot;geometry or topology&quot;.  Hopefully none of them are wrong.  (If so, somebody please correct me!)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;s&gt;Chromatic homotopy theory somehow slices the &lt;a href=&quot;http://en.wikipedia.org/wiki/Adams_spectral_sequence&quot; rel=&quot;nofollow&quot;&gt;Adams-Novikov spectral sequence&lt;/a&gt; based on levels of periodicity (hence the name &quot;chromatic&quot; -- it's supposed to conjure up the image of light refracting through a prism or something).&lt;/s&gt; [cf. Sean's answer for a much better informed account.]&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Level 0 is ordinary cohomology with $\mathbb{Q}$ coefficients.  There, the interaction of geometry and homotopy theory is given by integration.&lt;/li&gt;&#xA;&lt;li&gt;Level 1 is K-theory.  There, the interaction of geometry and homotopy theory is given by the Atiyah-Singer index theorem.&lt;/li&gt;&#xA;&lt;li&gt;Level 2 is elliptic cohomology (and something called topological modular forms, which somehow loosely is the &quot;universal elliptic cohomology theory&quot;).  It is unknown how to bridge the gap with geometry, but this is something people are working on.&lt;/li&gt;&#xA;&lt;li&gt;All the other finite levels are unknown.  A few people are trying to figure out what level 3 should be.&lt;/li&gt;&#xA;&lt;li&gt;Level $\infty$ is complex cobordism.&lt;/li&gt;&#xA;&lt;li&gt;All the levels (so far?) are &lt;a href=&quot;http://en.wikipedia.org/wiki/Complex_orientation&quot; rel=&quot;nofollow&quot;&gt;complex-oriented&lt;/a&gt;.  The complex cobordism spectrum $MU$ carries the universal formal group law, and complex orientations of a spectrum $E$ are obtained as multiplicative maps of spectra $MU\rightarrow E$.  (In fact, $(MU_*,MU_*MU)$ corepresents the moduli stack of formal &lt;em&gt;groups&lt;/em&gt;!)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;If you want to actually learn some things about this, try:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Hopkins' notes, &lt;a href=&quot;http://www.math.rochester.edu/u/faculty/doug/otherpapers/coctalos.pdf&quot; rel=&quot;nofollow&quot;&gt;&quot;Complex Oriented Cohomology Theories and the Language of Stacks&quot;&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.math.harvard.edu/~lurie/252x.html&quot; rel=&quot;nofollow&quot;&gt;class on chromatic homotopy theory taught by Lurie&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;Lurie: &lt;a href=&quot;http://www.math.harvard.edu/~lurie/papers/survey.pdf&quot; rel=&quot;nofollow&quot;&gt;&quot;A Survey of Elliptic Cohomology&quot;&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="2159" LastEditorUserId="2159" LastEditorDisplayName="" LastEditDate="2010-12-10T18:27:27.723" LastActivityDate="2010-12-10T18:27:27.723" />
  <row Id="13749" PostTypeId="2" ParentId="13679" CreationDate="2010-12-10T02:29:56.603" Score="2" ViewCount="" Body="&lt;p&gt;In terms of pointwise convergence, you cannot ensure the limit to exist at a given point, even if all the derivatives of $f$ exist (everywhere).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, by a theorem of Borel, &lt;em&gt;any&lt;/em&gt; sequence $a_0,a_1,\dots$ is $f(0),f&amp;#39;(0),f&amp;#39;&amp;#39;(0),\dots$ for some function $f$, see this &lt;a href=&quot;http://math.stackexchange.com/questions/10643/every-real-sequence-is-the-derivative-sequence-of-some-function/&quot;&gt;question&lt;/a&gt;. &lt;/p&gt;&#xA;" OwnerUserId="462" LastActivityDate="2010-12-10T02:29:56.603" CommentCount="5" />
  <row Id="13750" PostTypeId="2" ParentId="13735" CreationDate="2010-12-10T02:34:09.313" Score="12" ViewCount="" Body="&lt;p&gt;For a careful and very comprehensive treatment:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Robin Hartshorne: Geometry: Euclid and Beyond; Springer, 2000.&lt;/p&gt;&#xA;" OwnerUserId="1369" LastActivityDate="2010-12-10T02:34:09.313" CommentCount="1" />
  <row Id="13751" PostTypeId="1" AcceptedAnswerId="13753" CreationDate="2010-12-10T02:38:53.960" Score="0" ViewCount="62" Body="&lt;p&gt;$x=t^2-3$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$y=9t$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume that a point moves in (x,y) plane according to the above parametric equations. Find for which $t$ the velocity is equal to 9. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have an idea about this problem, but I'm not sure how to work with it. If we are given the position functions, then the derivative is the velocity. The derivative of both of those with respect to t is $x=2t, y=9$, but I'm not sure what to do after that. &lt;/p&gt;&#xA;" OwnerUserId="2985" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-10T04:02:49.940" LastActivityDate="2010-12-10T04:02:49.940" Title="Parametric Equation Question" Tags="&lt;calculus&gt;" AnswerCount="2" />
  <row Id="13752" PostTypeId="2" ParentId="13751" CreationDate="2010-12-10T02:44:27.553" Score="4" ViewCount="" Body="&lt;p&gt;Hint: position vector $r=&amp;lt;x(t),y(t)&amp;gt;$. Velocity vector is $dr/dt$ and speed is the magnitude of this vector.&lt;/p&gt;&#xA;" OwnerUserId="3431" LastActivityDate="2010-12-10T02:44:27.553" />
  <row Id="13753" PostTypeId="2" ParentId="13751" CreationDate="2010-12-10T02:46:12.317" Score="4" ViewCount="" Body="&lt;p&gt;$v(t)=(2t,9)$ ( Vector ), then in magnitude get $\|v(t)\|  = \sqrt{(2t)^2+9^2}$ ( speed ), thus &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\|v(t)\| = 9 \Rightarrow \sqrt{(2t)^2+9^2}= 9$$ Now solve for $t$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Magnitude &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/IiDdP.gif&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$A_{y}$ and $A_{x}$ are the components of vector $A$, the magnitude of vector $A$ is the distance from $ A $ to the origin. By pytagorean theorem the length of $ A $ is given by: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$A^2=A_{x}^2+A_{y}^2 \Rightarrow A=\sqrt{A_{x}^2+A_{y}^2}$$  ( In magnitude)&lt;/p&gt;&#xA;" OwnerUserId="3421" LastEditorUserId="3421" LastEditorDisplayName="" LastEditDate="2010-12-10T03:47:58.980" LastActivityDate="2010-12-10T03:47:58.980" CommentCount="3" />
  <row Id="13754" PostTypeId="1" AcceptedAnswerId="13756" CreationDate="2010-12-10T03:32:41.433" Score="4" ViewCount="148" Body="&lt;p&gt;The permanent can be interpreted as the number of perfect matchings in bipartite graphs.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Is there a similar graph-theoretic interpretation of the determinant? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2459" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-10T03:34:20.950" LastActivityDate="2010-12-10T07:49:50.377" Title="Graph-theoretic interpretation of determinant?" Tags="&lt;graph-theory&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" />
  <row Id="13755" PostTypeId="2" ParentId="12054" CreationDate="2010-12-10T03:42:22.207" Score="2" ViewCount="" Body="&lt;p&gt;I think there is a problem with the above solution.  In the estimate, $$|f^{(k+1)}(x)| \leq \left(\frac{1}{1-r}\right)^{k+1},$$ there is a dropped $k!$.  Indeed, it should read,&#xA;$$|f^{(k+1)}(x)| \leq \left(\frac{k!}{1-r}\right)^{k+1},$$ &#xA;and thus&#xA;$$&#xD;&#xA;|r_k(x)| \leq \left( \frac{r}{1-r} \right)^{k+1} \cdot \frac{k!}{(k+1)!} = \left( \frac{r}{1-r} \right)^{k+1} \cdot \frac{1}{k+1}.&#xD;&#xA;$$&#xD;&#xA;Unfortunately now this expression won't go to $0$ if $r&amp;gt;0$ (now the exponential term will dominate $\frac{1}{k+1}$).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The above solution does work for x in (-1/2,1).  Here's a way to handle the remaining cases.  In fact, let's just take x in (-1,0).  Now Lagrange's form of the remainder gives:&#xA;$$&#xD;&#xA;r_k(x) = \int_0^x \frac{f^{(k+1)}(t)}{k!} (x-t)^{k} dt &#xD;&#xA;= \int_0^x \frac{(-1)^k}{(1+t)^{k+1}} (x-t)^{k} dt &#xD;&#xA;$$&#xD;&#xA;Note that for $x&amp;lt;0$, the above integrand has the same sign for every $t$.  In particular,&#xA;$$&#xD;&#xA;|r_k(x)| &#xD;&#xA;= \int_x^0 \frac{1}{(1+t)^{k+1}} (t-x)^{k} dt. &#xD;&#xA;$$&#xD;&#xA;Consider $\frac{t-x}{1+t}$ as a function in $t$ with $x$ fixed.  It then is an increasing function on [x,0] with maximal value of $-x$ when $t=0$.  Thus,&#xA;$$&#xD;&#xA;|r_k(x)| \leq \int_x^0 (-x)^k \frac{1}{1+t} dt \leq \int_x^0 (-x)^k \frac{1}{1+x} dt = \frac{(-x)^{k+1}}{1+x}.&#xD;&#xA;$$&#xD;&#xA;As desired, this last expression does go to $0$ as $k \to \infty$ as $-1&amp;lt;x&amp;lt;0$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that in Spivak's Calculus, the stronger bound of &#xA;$$&#xD;&#xA;|r_k(x)| \leq \frac{(-x)^{k+1}}{(1+x)(1+k)}&#xD;&#xA;$$&#xD;&#xA;is left as Exercise 16 in Chapter 20.  I don't see how to get this, but I'm anxious that I'm just making some mistake in the computations above.&lt;/p&gt;&#xA;" OwnerUserId="4593" LastEditorUserId="4593" LastEditorDisplayName="" LastEditDate="2010-12-10T17:00:14.863" LastActivityDate="2010-12-10T17:00:14.863" CommunityOwnedDate="2010-12-10T15:48:51.773" CommentCount="5" />
  <row Id="13756" PostTypeId="2" ParentId="13754" CreationDate="2010-12-10T03:45:59.277" Score="10" ViewCount="" Body="&lt;p&gt;I'm aware of a few.  There is the &lt;a href=&quot;http://qchu.wordpress.com/2009/11/17/the-lindstrom-gessel-viennot-lemma/&quot; rel=&quot;nofollow&quot;&gt;Lindström-Gessel-Viennot lemma&lt;/a&gt;, and there is also the &lt;a href=&quot;http://en.wikipedia.org/wiki/Kirchhoff%27s_theorem&quot; rel=&quot;nofollow&quot;&gt;matrix-tree theorem&lt;/a&gt;.  If $A$ is the adjacency matrix of a finite graph $G$ then $\frac{1}{\det(I - At)}$ describes a kind of &quot;zeta function&quot; of $G$.  I describe some of how this works in &lt;a href=&quot;http://qchu.wordpress.com/2009/11/04/newtons-sums-necklace-congruences-and-zeta-functions-ii/&quot; rel=&quot;nofollow&quot;&gt;this blog post&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You may also be interested in Kuperberg's &lt;a href=&quot;http://www.combinatorics.org/Volume_5/PDF/v5i1r46.pdf&quot; rel=&quot;nofollow&quot;&gt;An exploration of the permanent-determinant method&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-12-10T07:49:50.377" LastActivityDate="2010-12-10T07:49:50.377" CommentCount="3" />
  <row Id="13757" PostTypeId="2" ParentId="13735" CreationDate="2010-12-10T03:54:51.427" Score="4" ViewCount="" Body="&lt;p&gt;Try &lt;em&gt;Elementary Geometry&lt;/em&gt; by Ilka Agricola and Thomas Friedrich. It has nice expositions of two-dimensional spherical and hyperbolic geometries as well.&lt;/p&gt;&#xA;" OwnerUserId="4594" LastEditorUserId="4594" LastEditorDisplayName="" LastEditDate="2011-05-24T09:13:00.103" LastActivityDate="2011-05-24T09:13:00.103" />
  <row Id="13758" PostTypeId="2" ParentId="13525" CreationDate="2010-12-10T03:56:39.647" Score="5" ViewCount="" Body="&lt;p&gt;There is no, as far as i know, general chromatic filtration for object of type X. There is a chromatic filtration of the stable homotopy groups of spheres. You could try to make sense of some directed system that is &quot;chromatic&quot; associated to a space/spectrum by looking at the maps between its localizations with respect to Morava K-theory (or Morava E-theory or Johnson-Wilson, or what have you) and how they fit together, but i think this is essentially the chromatic convergence thm, (there are finite type and p-local hypotheses).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An important ingredient to note is Carlsson's Thm/the Segal conjecture, which states, i believe, that stable cohomotopy is the completed Burnside ring. Maybe you can feed some of the chromatic ideas through all the hard work that went into the Segal Conjecture and get something that way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But maybe that is not it at all. A note about Aarons answer, the chromatic filtration does not come from the ANSS, although it does break up the way you say though. Ravenel spends some time in Nilpotence and Periodicity talking about this stuff. I think it has to do with self maps inducing multiplication by $v_n$ in $K(n)_*(X)$, and you filter the self maps of X that way. The $v_n$ are tied to the theory of formal group laws intimately. The way I think of the chromatic filtration, and this could be off, is by what level of (complex oriented) cohomology theory i need to detect that a self map is not trivial, where by level i mean height of the formal group law corresponding to the particular complex oriented theory. The height of the formal group laws govern the filtration and how it behaves. (This is how I think of it, but in light of the below, it is sort of junk, I will leave it because it is less technical.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe the paper Jack linked to is where they develop the chromatic spectral sequence which converges to the $E_2$ term of ANSS which is crazy hard to compute (i dont think anyone knows all the 2 line, and if they do then no one knows all of the 3 line, and by all i mean out to $t-s=50$) So there is a filtration on the $E_2$ term of ANSS that gives rise to this spectral sequence, and I think this is the chromatic filtration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Also, it is a technical term, and it does not just mean periodic.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt;&#xA;I just found the following sentence while reading the review of &quot;$v_n$ periodic elements in ring spectra and applications to bordism theory&quot; by Hovey (the review is by stong):&#xA;&quot;Let $R$ be a ring spectrum. If $v \in \pi_k R$, $v$ is called a $v_n$ element if $K(n)_*(v) \in K(n)_*(R)$ is a unit and $K(i)_*(v) \in K(i)_*(R)$ is nilpotent for all $i \neq n$.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I will be back in a bit to check it against a reading of Ravenel's orange book.&lt;/p&gt;&#xA;" OwnerUserId="627" LastEditorUserId="627" LastEditorDisplayName="" LastEditDate="2010-12-16T21:53:18.190" LastActivityDate="2010-12-16T21:53:18.190" />
  <row Id="13759" PostTypeId="2" ParentId="13735" CreationDate="2010-12-10T04:03:11.923" Score="7" ViewCount="" Body="&lt;p&gt;A book I like very much is the classic &quot;College Geometry: An Introduction to the Modern Geometry of the Triangle and the Circle&quot; by Nathan Altshiller-Court. The (revised in 1952) book has been reissued recently (2007) by Dover in a very affordable paperback. There is also a companion book on spatial geometry, but I do not know how easy it would be to find a copy. In Colombia, training for the Math Olympiads used for a while this book and Coxeter's, so I have first hand experience of learning from it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For completely different reasons, another book I like a lot is &quot;Higher Geometry&quot; by N.V. Efimov, originally from Mir eds. The book begins with an interesting discussion of axiomatic systems, talks about the attempts to prove the 5th postulate, Hilbert's axiomatization of Euclidean geometry, and then develops it synthetically. The book then does the same with hyperbolic geometry, and this is the only place where I have seen this done (it is well worth reading this part). There is also a treatment of projective geometry which is very nice, and (I think) of differential geometry. (The description in this book of the foundational approach was one of the reasons I decided I wanted to specialize in logic.)&lt;/p&gt;&#xA;" OwnerUserId="462" LastEditorUserId="462" LastEditorDisplayName="" LastEditDate="2010-12-10T04:17:39.540" LastActivityDate="2010-12-10T04:17:39.540" />
  <row Id="13760" PostTypeId="2" ParentId="13404" CreationDate="2010-12-10T04:05:09.033" Score="0" ViewCount="" Body="&lt;p&gt;I've been wrestling with a very similar problem in order to determine gradients in an irregular quad grid and needing to map points within arbitrary quadrilaterals to a unit square.  In addition, I require the inverse mapping of the x and y axis at the mapped normalized coordinate location back into the quad so I can determine the orientation of the quad grid at that point.  i.e. if &lt;code&gt;[x',y']&lt;/code&gt; are the transformed coordinates, I need to be able to do an inverse transform on &lt;code&gt;[0,y'],[1,y']&lt;/code&gt; and &lt;code&gt;[x',0],[x',1]&lt;/code&gt;. Here is what I've come up with:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can divide the quad into two tris, and use affine maps on these individually.  This is not difficult.  This will create a noticable effect at the division between the two tris, however.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want a smooth mapping from a quad to a square (or rectangle), you need to use a non-affine transform such as a projective transform.  There are other transforms other than projective that will also work, and also be colinear (preserve straight lines).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If &lt;code&gt;[x1,y1],[x2,y2],[x3,y3],[x4,y4]&lt;/code&gt; are the four points in the quad, then the 4x4 matrix &lt;strong&gt;B&lt;/strong&gt; in the the following will yield a mapping into the square (on the RHS) that seems to work and may be easier to compute than the proper 3x3 projective matrix.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;%     [x1 y1 x1*y1 1]              [0 0 0 1]&#xA;%     [x2 y2 x2*y2 1]   X  B  =    [1 0 0 1]&#xA;%     [x3 y3 x3*y3 1]              [0 1 0 1]&#xA;%     [x4 y4 x4*y4 1]              [1 1 1 1]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The question I have is that if one does this, and then wants to use the inverse of &lt;strong&gt;B&lt;/strong&gt; to do the inverse transform, how do you calculate the third elements of the location vectors for the orthogonal coordinates.  (They are no longer x*y.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;NOTE: If you want to map into any other (arbitrary) quadrilateral (such as a rectangle), then just replace the RHS of what I have above with the new coordinates.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;%     [x1 y1 x1*y1 1]              [x1' y1' x1'*y1' 1]&#xA;%     [x2 y2 x2*y2 1]   X  B  =    [x2' y2' x2'*y2' 1]&#xA;%     [x3 y3 x3*y3 1]              [x3' y3' x3'*y3' 1]&#xA;%     [x4 y4 x4*y4 1]              [x4' y4' x4'*y4' 1]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="4592" LastEditorUserId="4592" LastEditorDisplayName="" LastEditDate="2010-12-10T04:15:35.923" LastActivityDate="2010-12-10T04:15:35.923" />
  <row Id="13761" PostTypeId="1" AcceptedAnswerId="13765" CreationDate="2010-12-10T04:06:58.627" Score="4" ViewCount="261" Body="&lt;p&gt;Using only four 8's and only addition, subtraction, multiplication, and division.. how can you come up with the number 24? You may use fractions and decimals! But remember, you can ONLY USE four 8's&lt;/p&gt;&#xA;" OwnerUserId="4595" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-10T04:17:45.497" LastActivityDate="2010-12-10T08:09:02.427" Title="Using 4 '8's, get the number 24" Tags="&lt;puzzle&gt;" AnswerCount="3" CommentCount="9" FavoriteCount="2" />
  <row Id="13762" PostTypeId="2" ParentId="13761" CreationDate="2010-12-10T04:32:17.073" Score="3" ViewCount="" Body="&lt;p&gt;I usually recommend &lt;a href=&quot;http://mrob.com/pub/ries/index.html&quot; rel=&quot;nofollow&quot;&gt;RIES&lt;/a&gt; for this sort of problem.&lt;/p&gt;&#xA;" OwnerUserId="1778" LastActivityDate="2010-12-10T04:32:17.073" CommentCount="3" />
  <row Id="13763" PostTypeId="1" AcceptedAnswerId="13767" CreationDate="2010-12-10T05:22:06.750" Score="11" ViewCount="298" Body="&lt;p&gt;My undergraduate number theory class decided to dip into a bit of algebraic geometry to finish up the semester. I'm having trouble understanding this bit of information that the instructor presented in his notes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here it is in paraphrase (assume we are over an abstract field k)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We take a polynomial in k, $f  =Y^2 - X^3 -aX -b$ and homogenize the polynomial to $F = Y^2Z - X^3 -aXZ^2 - bZ^3$. &lt;em&gt;Note that the points at infinity of V(F) consist of triples&lt;/em&gt; $[\alpha : \beta : 0]$ s.t $ -\alpha^3 = 0$, hence the only point at infinity is $[0 : 1 :0]$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The part I'm confused about is in italics. He introduces the terms &quot;points at infinity&quot; without defining it. After some google time, I understand what a point at infinity means in the context of a projective space/projective line but am having trouble understanding how the professor came to his conclusion about the point at infinity in this particular example &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is my question. In general, are all points in the locus of vanishing points for a homogeneous polynomials considered points at infinity? If not, is there a general procedure for calculating these point if we are given an arbitrary polynomial?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More abstractly, How do I understand that a finite point in the projective space is a &quot;point at infinity&quot; for this polynomial.&lt;/p&gt;&#xA;" OwnerUserId="1906" LastEditorUserId="8575" LastEditorDisplayName="" LastEditDate="2011-05-20T19:15:45.113" LastActivityDate="2011-05-20T19:15:45.113" Title="Elliptic Curves and Points at Infinity" Tags="&lt;algebraic-geometry&gt;&lt;elliptic-curves&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="4" />
  <row Id="13765" PostTypeId="2" ParentId="13761" CreationDate="2010-12-10T05:32:13.377" Score="6" ViewCount="" Body="&lt;p&gt;$ 88 - (8 \cdot 8 ) $&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Much easier with five 8's:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ 8(8+8+8)/8 $&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;:)&lt;/p&gt;&#xA;" OwnerUserId="3384" LastActivityDate="2010-12-10T05:32:13.377" CommentCount="2" />
  <row Id="13766" PostTypeId="2" ParentId="13763" CreationDate="2010-12-10T05:36:53.313" Score="4" ViewCount="" Body="&lt;p&gt;The term &quot;point at infinity&quot; is not actually a well-defined term from the point of view of your projective model. You should think of it this way: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose you are given a homogeneous polynomial and you are interested in its zero set as a subset of projective space.&#xA;The projective space is made up of several so-called affine pieces, and accordingly your zero set is made up of several affine pieces. Here is how it works: let's take the concrete polynomial $F=Y^2Z - X^3-aXZ^2 - bZ^3$. Since you can take any point $(x:y:z)$ satisfying this polynomial and scale the coordinates by any non-zero scalar, $\alpha$ say, and the result still represents the same point, you may scale the point by $\alpha=z^{-1}$ and write it as $(x/z,y/z,1)$. But now, you have to be careful: this only works for points, at which $z\neq 0$. For those points, you are setting $z=1$, so the resulting affine model is given by your polynomial $f$ (just set $Z=1$ in $F$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If on the other hand $z$ happened to be 0, then you have attempted to divide by 0, so in this sense, you obtained a &quot;point at infinity&quot;. Now, what is this point at infinity? Set $Z=0$ in $F$ and see what points satisfy the resulting polynomial. For the polynomial to be 0, you also need $-X^3=0$, so $X=0$. Now, since we are in projective space, $Y$ mustn't be 0 (recall that (0:0:0) is not a point in projective space), so you can scale the coordinates so that $Y=1$. That's your &quot;point at infinity&quot; for this particular affine piece: $(X:Y:Z)=(0:1:0)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the procedure was not canonical. Instead, you might have chosen to look at the affine piece $Y\neq 0$, say. Then, you would have written all points in this affine model as $(x/y:1:z/y)$ and the dehomogenised polynomial would have been different. Also, the &quot;points at infinity&quot; would have been different ones, namely all those projective points, for which $Y=0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope this makes sense. Otherwise, feel free to ask for clarifications.&lt;/p&gt;&#xA;" OwnerUserId="3212" LastEditorUserId="3212" LastEditorDisplayName="" LastEditDate="2010-12-10T05:41:55.147" LastActivityDate="2010-12-10T05:41:55.147" CommentCount="4" />
  <row Id="13767" PostTypeId="2" ParentId="13763" CreationDate="2010-12-10T06:40:16.760" Score="11" ViewCount="" Body="&lt;p&gt;Here's another way to think about the &quot;line at infinity&quot; and the &quot;points at infinity&quot;...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Think of the usual $XY$-plane as sitting inside of $3$-space, but instead of it sitting in its usual place, $\{(x,y,0) : x,y\in\mathbb{R}\}$, shift it up by $1$ so that it sits as the $z=1$ plane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, you are sitting at the origin with a very powerful laser pointer. Whenever you want a point on the $XY$-plane, you shine your laser pointer at that point. So, if you want the point $(x,y)$, you are actually pointing your laser pointer at the point $(x,y,1)$; since you are sitting at the origin, the laser beam describes a (half)-line, joining $(0,0,0)$ to $(x,y,1)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, for example, look at the point $(x,0,1)$, and imagine $x$ getting larger. The angle your laser pointer makes with the $z=0$ plane gets smaller and smaller, until &quot;as $x$ goes to infinity&quot;, your laser pointer is just pointing along the line $x$ axis (at the point $(1,0,0)$), and the same thing happens if you let $x$ go to $-\infty$. More generally, if you start pointing to points that are further and further away from the &quot;origin&quot; in your plane (away from $(0,0,1)$), the laser beam's angle with $Z=0$ gets smaller and smaller, until, &quot;at the limit&quot; as $||(x,y)||\to\infty$, you end up with the laser beam pointing along the $z=0$ plane in some direction. We can represent the direction with the slope of the line, so that we are pointing at $(1,m,0)$ for some $m$ (or perhaps to $(-1,-m,0)$, but that's the same direction), or perhaps to the point $(0,1,0)$. So we &quot;add&quot; these &quot;points at infinity&quot; (so called because we get them by letting the point we are shining the laser beam on &quot;go to infinity&quot;), one for each direction away from the &quot;origin&quot;: $(1,m,0)$ for arbitrary $m$ for nonvertical lines, and $(0,1,0)$ corresponding to the direction of $x=0$, $y\to\pm\infty$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So: the &quot;usual&quot;, affine points, are the ones in the $z=1$ plane, and they correspond to laser beams coming from the origin; they are each of the form $(x,y,1)$ for some $x,y$ in $\mathbb{R}$. In addition, for each &quot;direction&quot; we want to include that limiting laser beam which does &lt;em&gt;not&lt;/em&gt; intersect the plane $z=1$; those correspond to points $(1,m,0)$, or the point $(0,1,0)$ when you do it with the line $x=0$. So we get one point for every real $m$, $(1,m,0)$, and another for $(0,1,0)$. You are adding one point for every direction of lines through the origin; these points are the &quot;points at infinity&quot;, and together they make the &quot;line at infinity&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, put your elliptic curve/polynomial $F=Y^2 - X^3 - aX-b$, and draw the points that correspond to it on the $z=1$ plane; that's the &quot;affine piece&quot; of the curve. But do you also get any of those &quot;points at infinity&quot;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, even though we are thinking of the points as being on the $XY$-plane, they &quot;really&quot; are in the $Z=1$ plane; so our equation actually has a &quot;hidden&quot; $Z$ that we lost sight of when we evaluated at $Z=1$. We use the homogenization $f = Y^2Z - X^3 - aXZ^2 - bZ^3$ to find it. Why that? Well, for any fixed point $(x,y,1)$ in our &quot;$XY$-plane&quot;, the laser pointer points to all points of the form $(\alpha x,\alpha y,\alpha)$. If we were to shift up our copy of the plane from $Z=1$ to $Z=\alpha$, we'll want to scale everything so that it still corresponds to what I'm tracing from the origin; this requires that every monomial have the same total degree, which is why we put in factors of $Z$ to complete them to degree $3$, the smallest we can (making it bigger would give you the point $(0,0,0)$ as a solution, and we &lt;em&gt;do&lt;/em&gt; need to stay away from that because we cannot point the laser pointer in our eye). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once we do that, we find the &quot;directions&quot; that also correspond to our curve by setting $Z=0$ and solving, to find those points $(1,m,0)$ and $(0,1,0)$ that may also lie in our curve. But the only one that works is $(0,1,0)$, which is why the elliptic curve $F$ only has one &quot;point at infinity&quot;. &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-11T07:33:00.487" LastActivityDate="2010-12-11T07:33:00.487" CommentCount="2" />
  <row Id="13769" PostTypeId="2" ParentId="13735" CreationDate="2010-12-10T06:41:28.787" Score="4" ViewCount="" Body="&lt;p&gt;Christopher Gibson's &lt;a href=&quot;http://books.google.com/books?id=P7WPwxKB1q8C&amp;amp;printsec=frontcover&amp;amp;dq=elementary+euclidean+geometry+gibson&amp;amp;source=bl&amp;amp;ots=8mCsRnNAFP&amp;amp;sig=OUPFdlGdjOrzAIu-8az4wdqf-AU&amp;amp;hl=en&amp;amp;ei=tEiBTZq_DJCWOtjcpLsI&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=1&amp;amp;ved=0CBgQ6AEwAA#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;Elementary Euclidean Geometry&lt;/em&gt;&lt;/a&gt; is a new (published in 2004) and fairly advanced introduction to the geometry of lines and conics in Euclidean space.&lt;/p&gt;&#xA;" OwnerUserId="1702" LastEditorUserId="1702" LastEditorDisplayName="" LastEditDate="2011-03-16T23:33:48.303" LastActivityDate="2011-03-16T23:33:48.303" />
  <row Id="13770" PostTypeId="2" ParentId="13761" CreationDate="2010-12-10T06:47:48.550" Score="2" ViewCount="" Body="&lt;p&gt;The 24game solver found &lt;a href=&quot;http://code.google.com/p/solve24game/&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; couldn't find any solutions with 4 8s. I don't think the software uses decimals to create 8.8 etc but its hard to see how that path leads to a solution. The other solution is as friedo suggests (88-8*8).&lt;/p&gt;&#xA;" OwnerUserId="618" LastEditorUserId="618" LastEditorDisplayName="" LastEditDate="2010-12-10T08:09:02.427" LastActivityDate="2010-12-10T08:09:02.427" />
  <row Id="13771" PostTypeId="1" AcceptedAnswerId="13775" CreationDate="2010-12-10T06:51:20.343" Score="2" ViewCount="152" Body="&lt;p&gt;According to wikipedia the &lt;a href=&quot;http://en.wikipedia.org/wiki/Right_quotient&quot; rel=&quot;nofollow&quot;&gt;right quotient of a regular language&lt;/a&gt; with ANY other language is regular. I have not been able to find a proof of this fact. All the sources talk about quotient with another regular language. Can somebody point me to a proof?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S. Definition from Wikipedia: The right quotient (or simply quotient) of a formal language $L_1$ with a formal language $L_2$ is the language consisting of strings $w$ such that $wx$ is in $L_1$ for some string $x$ in $L_2$. In symbols, we write:&#xA;$$L_1 / L_2 = \{w \ | \ \exists x ((x \in L_2) \land (wx \in L_1)) \}$$&#xD;&#xA;In other words, each string in L1 / L2 is the prefix of a string wx in L1, with the remainder of the word being a string in L2.&lt;/p&gt;&#xA;" OwnerUserId="58" LastActivityDate="2010-12-10T07:39:40.990" Title="Quotient of a regular language." Tags="&lt;computer-science&gt;&lt;computability&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="13772" PostTypeId="1" CreationDate="2010-12-10T07:01:29.713" Score="0" ViewCount="133" Body="&lt;p&gt;I have two faces ($S_1$ and $S_2$), each bounded by a series of Vertex $V$. These two faces may or may not intersect. In addition to that there are two vertical faces $S_3$ and $S_4$, that connect between $S_1$ and $S_2$ vertically, as shown in the below diagram. The question is how to find the polyhedrons that are formed by the bounding of the four faces?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An example of this is shown below:&#xA;&lt;img src=&quot;http://i.imgur.com/HXLRD.gif&quot; alt=&quot;alt text&quot;&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the above example, there should be two polyhedrons returned, one above $S_2$, another below $S_1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both $S_1$ and $S_2$ are defined by 3D coordinates. Let's say the coordinates for $S_1$ is ($P_1$,$P_2$, $P_3$, $P_4$) and $S_2$ is $P_5$,$P_6$, $P_7$, $P_8$, and the intersected points are $P_9$ and $P_{10}$. I want a program that returns me two polyhedrons:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$P_1,P_2,P_9,P_{10},P_5,P_6$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$P_3,P_4,P_{10},P_9,P_7,P_8$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One way to do this is to find the intersecting line of the two faces, and then construct polyhedron from there. But is there a more elegant method?&lt;/p&gt;&#xA;" OwnerUserId="550" LastEditorUserId="550" LastEditorDisplayName="" LastEditDate="2010-12-11T15:27:27.613" LastActivityDate="2011-05-10T16:56:20.143" Title="Find the Polyhedron Enclosed by Multiple Faces" Tags="&lt;geometry&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="13773" PostTypeId="1" CreationDate="2010-12-10T07:23:49.810" Score="1" ViewCount="77" Body="&lt;p&gt;I am wondering how to find the Zariski Cover of the Grassmannian over $\mathbb{C}$. I was hoping some reference would go through this calculation. I was recently told that it was not too hard, so I wondered if it was written somewhere. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any reference that does this, or at least gives an introduction for how to calculate Zariski Covers of specific schemes, would be much appreciated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you!&lt;/p&gt;&#xA;" OwnerUserId="16" LastActivityDate="2010-12-10T12:44:03.900" Title="How to find a Zariski Cover of the Grassmannian" Tags="&lt;algebraic-geometry&gt;&lt;reference-request&gt;&lt;grassmannian&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="13774" PostTypeId="1" AcceptedAnswerId="13776" CreationDate="2010-12-10T07:38:55.033" Score="3" ViewCount="162" Body="&lt;p&gt;such a RSA algorithm.&lt;/p&gt;&#xA;" OwnerUserId="4599" LastActivityDate="2010-12-10T07:50:59.963" Title="What is the importance of prime numbers in cryptographic algorithms ?" Tags="&lt;prime-numbers&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="13775" PostTypeId="2" ParentId="13771" CreationDate="2010-12-10T07:39:40.990" Score="2" ViewCount="" Body="&lt;p&gt;You can find a proof in this book: Introduction to Automata Theory, Languages and Computation by Hopcroft and Ullman, 1979 edition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The short and non-constructive proof appears on page 63.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a snapshot (which I got from google books by searching for quotient in that book):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/gswKZ.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Only the last line of the proof is incomplete in the above snapshot, which reads: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Thus $M&amp;#39;$ accepts $R/L$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here $R$ is a regular language and $L$ is an arbitrary language.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-12-10T07:39:40.990" />
  <row Id="13776" PostTypeId="2" ParentId="13774" CreationDate="2010-12-10T07:50:59.963" Score="2" ViewCount="" Body="&lt;p&gt;The short answer is that what makes primes useful is that it is easy to multiply two primes, but difficult to algorithmically factorise a given number into prime factors (i.e. takes a long time, if the number is big). So multiplying primes is an operation that is easy to perform but difficult to reverse. That makes them a perfect candidate for implementing &lt;a href=&quot;http://en.wikipedia.org/wiki/Public-key_cryptography&quot; rel=&quot;nofollow&quot;&gt;public-private key encryption&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A long answer would involve explaining RSA, which has been done in lots of places and you will have no difficulty finding them. You need to know Fermat's little theorem to understand how RSA works.&lt;/p&gt;&#xA;" OwnerUserId="3212" LastActivityDate="2010-12-10T07:50:59.963" CommentCount="4" />
  <row Id="13777" PostTypeId="1" CreationDate="2010-12-10T08:05:42.583" Score="5" ViewCount="319" Body="&lt;p&gt;I am trying to understand why $1^{x}=1$ for any $x\in\mathbb{R}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it OK to write $1^{x}$? As the base 1 should not equal 1 for $1^{x}$ to be an exponential function? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is $1^{x}=1$ just because it is defined to be so?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If possible please refer me to a book or article that discusses this topic.&lt;/p&gt;&#xA;" OwnerUserId="4601" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-11T02:39:48.217" LastActivityDate="2010-12-11T02:39:48.217" Title="What is the value of $1^x$ ?" Tags="&lt;calculus&gt;&lt;functions&gt;&lt;notation&gt;" AnswerCount="3" />
  <row Id="13778" PostTypeId="2" ParentId="13777" CreationDate="2010-12-10T08:23:29.310" Score="6" ViewCount="" Body="&lt;p&gt;I think you will at least agree that $1^x$=$1$ if $x$ is any natural number (since this is just 1 times itself x times).  We can extend this to all integer values of $x$ by using the facts $a^0$=$1$ for all non-zero $a$ and $x^{-c}$=$1\over{x^c}$.  Then we have $1^x$=1 for all rational numbers $x$ by the fact $1^{b\over{c}}$ is the cth root of $1^b$ which is 1 by previous results.  Finally we exdend this to all real values of $x$ by using the following definition of the exponential function from calculus.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;If $t$ is irational, then $a^t$ is defined to be the limit as n approaches infinity of $a^{t_n}$ where {$t_n$} is any sequence of rational numbers converging to $t$ (assuming this limit exists).&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of $1^t$ we already have that $1^{t_n}=1$ for all rational $t_n$ so any such sequence as described above is the constant sequence, 1,1,1,..... which converges to 1.&lt;/p&gt;&#xA;" OwnerUserId="3971" LastEditorUserId="3971" LastEditorDisplayName="" LastEditDate="2010-12-10T08:39:20.890" LastActivityDate="2010-12-10T08:39:20.890" CommentCount="1" />
  <row Id="13779" PostTypeId="2" ParentId="13777" CreationDate="2010-12-10T08:34:19.207" Score="15" ViewCount="" Body="&lt;p&gt;You don't have to specifically define $1^x$ to be something. It follows from the definition of exponentiation in general.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The story is roughly as follows: first there was exponentiation with exponents being natural numbers. That is just repeated multiplication of the base with itself. Then people noticed that this exponentiation obeys the rule&#xA;\begin{eqnarray}&#xA;a^{n+m} = a^na^m.&#xA;\end{eqnarray} This suggests that if we want to define $a^0$, then we want it to satisfy $a^m = a^{m+0} = a^ma^0$, so that it is natural to define $a^0 = 1$. Now, we can use the same rule to extend this operation to all integral exponents, i.e. positive and negative. Since we want $a^na^{-n} = a^{n-n} = a^0 = 1$, we are forced to define $a^{-n} = \frac{1}{a^n}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, this way you have defined exponentiation with arbitrary integral exponents. Now, going back to the original exponentiation with natural numbers, you also notice that&#xA;\[(a^n)^m = a^{nm}&#xA;\] and $a^1 = a$. Applying the same considerations as above, you are forced to define $a^{\frac{1}{n}} = \sqrt[n]{a}$. This way, you have now extended exponentiation to all rational exponents. Note that all these rules &lt;em&gt;imply&lt;/em&gt; that $1^x = 1$ for $x\in \mathbb{Q}$. Now, you extend all this to all real exponents using continuity, so you still have $1^x = 1\;\forall x\in \mathbb{R}$.&lt;/p&gt;&#xA;" OwnerUserId="3212" LastEditorUserId="3212" LastEditorDisplayName="" LastEditDate="2010-12-10T13:48:40.400" LastActivityDate="2010-12-10T13:48:40.400" CommentCount="3" />
  <row Id="13781" PostTypeId="1" CreationDate="2010-12-10T09:53:05.893" Score="3" ViewCount="135" Body="&lt;p&gt;If you sum an expression over an uncountable set&#xA;$\sum_{x\in \mathbb{R}}f(x)$, then do we need $f(x)=0$ on all but a countable subset in order for the sum to have a finite value?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If not can you give an example of a function everywhere nonzero that has a transfinite sum with a finite value?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Possible keywords: Transseries, Écalle–Borel Summation, analyzable function&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://arxiv.org/pdf/0801.4877&quot; rel=&quot;nofollow&quot;&gt;Transseries for beginners&lt;/a&gt;, GA Edgar, 2009&lt;/p&gt;&#xA;" OwnerUserId="276" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-10T17:40:16.763" LastActivityDate="2010-12-10T17:40:16.763" Title="Transfinite series: Uncountable sums" Tags="&lt;analysis&gt;&lt;real-analysis&gt;" AnswerCount="1" />
  <row Id="13783" PostTypeId="2" ParentId="13781" CreationDate="2010-12-10T10:03:05.770" Score="3" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given $\varepsilon&amp;gt;0$, can you measure the set $\{x:f(x)&amp;gt;\varepsilon\}$?&lt;/p&gt;&#xA;" OwnerUserId="1154" LastEditorUserId="1154" LastEditorDisplayName="" LastEditDate="2010-12-10T10:21:07.097" LastActivityDate="2010-12-10T10:21:07.097" CommentCount="2" />
  <row Id="13784" PostTypeId="2" ParentId="154" CreationDate="2010-12-10T10:07:37.933" Score="1" ViewCount="" Body="&lt;p&gt;Re: &quot;What's the best way to explain to a non-mathematician that complex numbers are necessary and meaningful, in the same way that real numbers are?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am a non-mathematician...so fully qualified to answer this question...&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;first explain imaginary numbers&lt;/li&gt;&#xA;&lt;li&gt;then explain that complex numbers are simply numbers that use imaginary numbers&lt;/li&gt;&#xA;&lt;li&gt;then tell 'em about how they are applied in weather forecasting&lt;/li&gt;&#xA;&lt;li&gt;bang!....I geddit&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I got all this from an amazing radio show :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&quot;&lt;a href=&quot;http://www.bbc.co.uk/iplayer/episode/b00tt6b2/In_Our_Time_Imaginary_Numbers/&quot; rel=&quot;nofollow&quot;&gt;In Our Time - Imaginary Numbers&lt;/a&gt;&quot;&lt;/em&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Melvyn Bragg and his guests discuss imaginary numbers - important mathematical phenomena which provide us with useful tools for understanding the world.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="4603" LastActivityDate="2010-12-10T10:07:37.933" CommunityOwnedDate="2011-04-15T20:45:29.080" />
  <row Id="13785" PostTypeId="1" AcceptedAnswerId="13786" CreationDate="2010-12-10T10:08:25.777" Score="2" ViewCount="211" Body="&lt;p&gt;If I have term -5 and I want to change its sign to opposite can I say that I'm negating its sign? &lt;/p&gt;&#xA;" OwnerUserId="4366" LastEditorUserId="4366" LastEditorDisplayName="" LastEditDate="2010-12-10T15:10:52.987" LastActivityDate="2011-05-22T18:49:56.517" Title="Does saying &quot;to negate a sign of a const term&quot; has a meaning" Tags="&lt;terminology&gt;" AnswerCount="3" CommentCount="3" />
  <row Id="13786" PostTypeId="2" ParentId="13785" CreationDate="2010-12-10T10:30:59.497" Score="2" ViewCount="" Body="&lt;p&gt;You could say that, you could also say &quot;to negate the constant term&quot; -- both seem grammatically correct. In your expression, 'negate' changes the object of negation, while in the &quot;to negate the constant term&quot; 'negate' changes the sign in front of the object of negation, but it makes sense both ways.&lt;/p&gt;&#xA;" OwnerUserId="3731" LastActivityDate="2010-12-10T10:30:59.497" />
  <row Id="13787" PostTypeId="2" ParentId="13785" CreationDate="2010-12-10T10:52:39.237" Score="3" ViewCount="" Body="&lt;p&gt;&lt;em&gt;Negating its sign&lt;/em&gt; sounds alright.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A more fanciful way of putting it is &lt;em&gt;changing its polarity&lt;/em&gt;. Note that &lt;em&gt;polarity&lt;/em&gt; is not to be confused with &lt;em&gt;parity&lt;/em&gt;, whether a number is odd or even.&lt;/p&gt;&#xA;" OwnerUserId="4594" LastEditorUserId="4594" LastEditorDisplayName="" LastEditDate="2011-05-22T18:49:56.517" LastActivityDate="2011-05-22T18:49:56.517" CommentCount="1" />
  <row Id="13788" PostTypeId="2" ParentId="13785" CreationDate="2010-12-10T10:52:42.153" Score="2" ViewCount="" Body="&lt;p&gt;I think it is much more standard to say 'Multiplying equation (..) by -1 gives' than 'negating equation (...)'.  As for 'negating the constant term', I'm not even sure of what that would mean, and without some serious context seems a bad idea.&lt;/p&gt;&#xA;" OwnerUserId="4322" LastActivityDate="2010-12-10T10:52:42.153" />
  <row Id="13789" PostTypeId="2" ParentId="13773" CreationDate="2010-12-10T11:28:13.837" Score="2" ViewCount="" Body="&lt;p&gt;Consider Grassmannian of $d$-dimensional subspaces of $n$-dimensional vector space&#xA;$V$. Then you have Pluker inbeding of grassmanian $G=Gr(d,V)$ of d-planes of $V$ in $\mathbb P(\Lambda ^d V)=\mathbb  P^N, N=\binom{n}{d}-1$,  as follows. For every $d-plane$ D chose base $v_1,...,v_d$ of $D$ and consider exterior product $\omega = v_1\wedge...\wedge v_d \in \wedge ^d V$. Pluker inbeding send $D$ to line:  Span $\omega \in \mathbb P(\Lambda ^d V)$  .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For find  Zariski covering, fix base $e_1,...,e_n$ of $V$ and receive base $e_I=e_{i_1}\wedge...\wedge e_{i_d}$ of $\wedge ^d V $ ($I$ is increasing sequence $I=(1\leq i_1 \lt ...\lt i_d \leq n$)). Zariski covering consist of $U_I$  formed by these $d-$ planes $D$ whose Pluker image in $\mathbb P(\wedge ^d V)$ has coordinate $z_I\neq 0$. This mean that when you chose base $v_1,...,v_d$ of $D$ and developp $\omega = v_1\wedge...\wedge v_d \in \wedge ^d V$ in base $e_I$ of $\Lambda ^d V$, then coefficient of $e_I$ is non-null. This Zariski cover $U_I$ of grasmanian $Gr(d,V)$ possesses $N+1=\binom{n}{d}$ open sets (Attention: $\mathbb P^N$&#xD;&#xA;has $N+1$ coordinates, not $N$)&lt;/p&gt;&#xA;" OwnerUserId="2872" LastEditorUserId="2872" LastEditorDisplayName="" LastEditDate="2010-12-10T12:44:03.900" LastActivityDate="2010-12-10T12:44:03.900" CommentCount="1" />
  <row Id="13790" PostTypeId="1" CreationDate="2010-12-10T11:49:02.907" Score="2" ViewCount="181" Body="&lt;p&gt;The problem:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;&lt;pre&gt;&lt;code&gt;               Given a Fibonacci number,find its index.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am aware of the standard solution '&lt;strong&gt;generate-hash-find&lt;/strong&gt;'. I am just curious if there is some inverse system of matrix exponentiation or some other mathematical method that gives the solution.&lt;/p&gt;&#xA;" OwnerUserId="2109" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-10T13:25:32.490" LastActivityDate="2010-12-20T17:57:11.563" Title="Finding index of a Fibonacci number: any mathematical solution possible?" Tags="&lt;algorithms&gt;&lt;fibonacci-number&gt;" AnswerCount="2" CommentCount="16" />
  <row Id="13791" PostTypeId="1" AcceptedAnswerId="13797" CreationDate="2010-12-10T12:14:51.897" Score="3" ViewCount="67" Body="&lt;p&gt;If a Diophantine equation is: exists v, p(x,v) = 0 (where v is a vector of finitely many integers) for some polynomial p, is there a Turing machine which prints out all values of x?&lt;/p&gt;&#xA;" OwnerUserId="4290" LastActivityDate="2010-12-10T12:32:47.050" Title="Diophantine equation and Turing Machine" Tags="&lt;diophantine-equations&gt;&lt;computability&gt;" AnswerCount="1" />
  <row Id="13792" PostTypeId="2" ParentId="13790" CreationDate="2010-12-10T12:18:00.210" Score="6" ViewCount="" Body="&lt;p&gt;From wikipedia:&#xA;&quot;Similarly, if we already know that the number F is a Fibonacci number, we can determine its index within the sequence by&#xA;$n = \bigg\lfloor \log_\varphi \left(F\cdot\sqrt{5}\right) + \frac{1}{2} \bigg\rfloor$&quot;.&lt;/p&gt;&#xA;" OwnerUserId="3212" LastActivityDate="2010-12-10T12:18:00.210" />
  <row Id="13793" PostTypeId="1" AcceptedAnswerId="13796" CreationDate="2010-12-10T12:18:17.553" Score="14" ViewCount="397" Body="&lt;p&gt;A while ago I was dreaming up point-set topology exam questions, and this one came to mind:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is $\mathbb Q\setminus \{0\}$ homeomorphic to $\mathbb Q$? (Where both sets have the subspace topology induced from the standard topology on $\mathbb R$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I couldn't figure this out at the time, and I'm curious to see whether anyone has a nice argument. I'm not even willing to take a guess as to whether they are or aren't homeomorphic.&lt;/p&gt;&#xA;" OwnerUserId="2941" LastActivityDate="2011-05-05T21:00:40.137" Title="Are the rationals minus a point homeomorphic to the rationals?" Tags="&lt;topology&gt;" AnswerCount="3" FavoriteCount="2" />
  <row Id="13794" PostTypeId="1" AcceptedAnswerId="13828" CreationDate="2010-12-10T12:18:56.747" Score="2" ViewCount="136" Body="&lt;p&gt;There is a direct relationship between equivalence relations and partitions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a way to simply use an equivalence relation's definition to get the matching partition? And what about the other way around?&lt;/p&gt;&#xA;" OwnerUserId="3016" LastEditorUserId="622" LastEditorDisplayName="" LastEditDate="2010-12-10T20:02:58.710" LastActivityDate="2010-12-12T19:48:53.387" Title="&quot;Converting&quot; equivalence relations to partitions" Tags="&lt;elementary-set-theory&gt;&lt;relations&gt;&lt;partitions&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="13795" PostTypeId="2" ParentId="13777" CreationDate="2010-12-10T12:20:31.673" Score="7" ViewCount="" Body="&lt;p&gt;Once you move past elementary topics, definitions become much more fundamental in mathematics. So, in a formal sense, you're right that the reason $1^x = 1$ for all $x \in \mathbb{R}$ is that the definition of $1^x$ makes this so.  The emphasis on definitions comes from the use of mathematical proofs; the only way to make a rigorous proof about exponentiation is to start with a rigorous definition.  So, in a sense, all formal mathematical propositions are true because the definitions have been chosen to make them true. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, we have a clear motivation behind exponentiation, and if the definition of real number exponentiation did not make $1^x = 1$ for all real $x$, then the definition would have been changed.  We don't make up mathematical definitions at random - they are motivated by our informal ideas about the objects we are studying. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The role of this motivation can be seen more clearly by considering complex exponentiation.  Unlike natural number exponentiation, complex number exponentiation is not based on repeated multiplication; it's based on logarithms and the function $\exp(z)$. So the definition of complex exponentiation does not imply that $1^i = 1$, and mathematicians are OK with that. There are several possible values for $1^i$, only one of which is $1$. I explained this in &lt;a href=&quot;http://math.stackexchange.com/questions/3668/what-is-the-value-of-1i/3674#3674&quot;&gt;this answer&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One thing that is often confusing at first is that there are really several different exponentiation functions, with different domains, all of which are denoted with the notation $x^y$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, you asked whether $f(x) = 1^x$, as a function from $\mathbb{R}$ to itself, is an exponential function. Many calculus books seem to include a special clause in their definitions that makes this not be an exponential function. However, things would work just as well if you did call it an exponential function. It's just a question of terminology. The only downside to calling $1^x$ an exponential function is that, when stating some results, you might have to add an exception to get rid of $1^x$. Instead of saying &quot;Every exponential function&quot; you would say &quot;Every exponential function except $1^x$&quot;.  Of course, students in a class need to adopt the conventions of the class so that everyone can understand them. But if you were writing a math book alone on a desert island you could adopt whatever terminology you wanted. &lt;/p&gt;&#xA;" OwnerUserId="630" LastActivityDate="2010-12-10T12:20:31.673" />
  <row Id="13796" PostTypeId="2" ParentId="13793" CreationDate="2010-12-10T12:24:26.370" Score="18" ViewCount="" Body="&lt;p&gt;A well-known theorem of Cantor says that any two countable dense linear orderings without endpoints are isomorphic as linear orders. So in particular there is an order-preserving bijection between $\mathbb{Q}$ and $\mathbb{Q} - \{0\}$. This bijection will be a homeomorphism if you give each space the order topology, which is the standard topology inherited from $\mathbb{R}$. &lt;/p&gt;&#xA;" OwnerUserId="630" LastActivityDate="2010-12-10T12:24:26.370" CommentCount="3" />
  <row Id="13797" PostTypeId="2" ParentId="13791" CreationDate="2010-12-10T12:32:47.050" Score="4" ViewCount="" Body="&lt;p&gt;Given a polynomial $p(\bar{x},\bar{v})$ with integer coefficients, there is indeed a Turing machine that enumerates all tuples of integers $\bar{x}$ for which there exists a tuple of integers $\bar{v}$ with $p(\bar{x},\bar{v}) = 0$. In other words, the set of such tuples $\bar{x}$ is recursively enumerable. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One way to do the enumeration is to start by making a machine that enumerates &lt;em&gt;all&lt;/em&gt; pairs of integer tuples $(\bar{x}, \bar{v})$ (where each tuple has the appropriate length). Then feed the output of this machine into a machine that checks whether $p(\bar{x},\bar{v}) = 0$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, there is no Turing machine that takes a polynomial $p$ as input, and a tuple $\bar{x}$ as a second input, and decides whether there is a tuple $\bar{v}$ of integers with $p(\bar{x},\bar{v}) = 0$. This is the MRDP theorem, which solved &lt;a href=&quot;http://en.wikipedia.org/wiki/Hilbert%27s_tenth_problem&quot; rel=&quot;nofollow&quot;&gt;Hilbert's 10th problem&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="630" LastActivityDate="2010-12-10T12:32:47.050" CommentCount="1" />
  <row Id="13798" PostTypeId="2" ParentId="13794" CreationDate="2010-12-10T12:59:14.883" Score="3" ViewCount="" Body="&lt;p&gt;Suppose you have an equivalence relation $\sim$ on $X$. For each $x\in X$, the set&#xA;$S_x=\lbrace y\in X: y\sim x \rbrace$ is one of the sets in the partition, and every set in the &#xA;partition arises this way. But you should know that $S_x$ and $S_y$ are identical (the same)&#xA;if and only if $x\sim y$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, if you have a partition $\lbrace C_i: i\in I \rbrace$ of $X$. Then&#xA;the relation $\sim$ defined by $$ x\sim y \mbox{ iff } x,y \in C_i \mbox { for some } i $$&#xD;&#xA;is an equivalence relation on $X$.&lt;/p&gt;&#xA;" OwnerUserId="3249" LastActivityDate="2010-12-10T12:59:14.883" />
  <row Id="13799" PostTypeId="2" ParentId="13794" CreationDate="2010-12-10T13:02:26.770" Score="1" ViewCount="" Body="&lt;p&gt;In addition to my comment above, see also &lt;a href=&quot;http://www.proofwiki.org/wiki/Fundamental_Theorem_on_Equivalence_Relations&quot; rel=&quot;nofollow&quot;&gt;this post at the ProofWiki&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1543" LastActivityDate="2010-12-10T13:02:26.770" />
  <row Id="13800" PostTypeId="1" AcceptedAnswerId="13837" CreationDate="2010-12-10T13:13:28.150" Score="6" ViewCount="63" Body="&lt;p&gt;I have been looking for a proof of the statement:&#xA;&quot;Assume the Generalized Riemann Hypothesis. Let $d$ be a fundamental discriminant and $\chi_d$ the associated primitive quadratic character. Then, $$L(\tfrac{1}{2},\chi_d)\geq 0.&amp;quot;$$&#xD;&#xA;Can anyone point me in the right direction or give a reference? Thanks a lot!&lt;/p&gt;&#xA;" OwnerUserId="3244" LastActivityDate="2010-12-10T18:55:45.937" Title="Nonnegativity of the quadratic Dirichlet L-function $L(\tfrac{1}{2},\chi)$ under GRH" Tags="&lt;number-theory&gt;&lt;analytic-number-theory&gt;&lt;riemann-hypothesis&gt;" AnswerCount="1" />
  <row Id="13801" PostTypeId="1" AcceptedAnswerId="13808" CreationDate="2010-12-10T13:25:33.600" Score="4" ViewCount="863" Body="&lt;p&gt;What's the thing with $\sqrt{-1} = i$? Do they really teach this in the US? It makes very little sense, because $-i$ is also a square root of $-1$, and the choice of which root to label as $i$ is arbitrary. So saying $\sqrt{-1} = i$ is plainly false!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So why do people say $\sqrt{-1} = i$? Is this how it's taught in the US?&lt;/p&gt;&#xA;" OwnerUserId="3793" LastActivityDate="2011-04-19T21:07:40.630" Title="What's the thing with $\sqrt{-1} = i$" Tags="&lt;complex-numbers&gt;" AnswerCount="7" CommentCount="15" FavoriteCount="1" />
  <row Id="13802" PostTypeId="1" AcceptedAnswerId="13813" CreationDate="2010-12-10T13:32:58.397" Score="4" ViewCount="125" Body="&lt;p&gt;When asked to evaluate $\int_{a}^{\infty}f(x)dx$, you split the interval based on the improper points.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there is another improper point other than $\infty$, at $b$, we will write: $\int_{a}^{\infty}f(x)dx=\int_{a}^{b}f(x)dx+\int_{b}^{c}f(x)dx+\int_{c}^{\infty}f(x)dx$ and ask whether each of the integrals on the right hand side converge. If they all converge, so does the original one.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if at least one of them diverges, the original one doesn't. What is the justification for this conclusion? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can see that if $\int_{c}^{b}f(x)dx$ diverges, we can assume that the original integral converges and move the integrals around like this: $\int_{a}^{\infty}f(x)dx-\int_{a}^{b}f(x)dx-\int_{c}^{\infty}f(x)dx=\int_{b}^{c}f(x)dx$ then we'll get a contradiction. But what if both $\int_{c}^{b}f(x)dx$ and $\int_{a}^{b}f(x)dx$ diverge? What is the argument then?&lt;/p&gt;&#xA;" OwnerUserId="3237" LastEditorUserId="3237" LastEditorDisplayName="" LastEditDate="2010-12-10T14:00:15.800" LastActivityDate="2010-12-10T14:43:01.957" Title="Diverging improper integral" Tags="&lt;calculus&gt;&lt;improper-integrals&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="13803" PostTypeId="2" ParentId="13801" CreationDate="2010-12-10T13:41:08.610" Score="3" ViewCount="" Body="&lt;p&gt;Alex it is just a notion. We use the value of $i = \sqrt{-1}$ so that we can work more freely. For example consider the Quadratic Equation $$x^{2}+x+1=0$$ The discriminant for this is $D=b^{2}-4ac=-3$. So the roots have the value $$x = \frac{-1 \pm{\sqrt{3}i}}{2}$$ which looks better when written with an $i$ notation. That's all. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know how this is taught in the US but to me, i encountered this when i was at high school, learning how to solve for Quadratic Equations when the &lt;em&gt;Discriminant&lt;/em&gt; is less than $0$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next, note that $\mathbb{C}$ doesn't have the same ordering as $\mathbb{R}$. That is for any 2 real numbers $a,b$ we have either:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;$a&amp;gt;b$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$a&amp;lt; b$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$a=b$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;But for complex number's this is not true. Since if you take $i$ and $0$, we must have either $i &amp;gt; 0$ or $i &amp;lt; 0$, but this isn't true.&lt;/p&gt;&#xA;" LastActivityDate="2011-01-19T22:15:33.883" CommentCount="5" />
  <row Id="13804" PostTypeId="1" CreationDate="2010-12-10T13:54:17.917" Score="3" ViewCount="74" Body="&lt;p&gt;I was thinking about some problem involving quantifiers (the existencial and universal quantifiers) and I noticed how it might resemble probability in a sense. They both assume a variable and its domain, and put restrictions on some predicate of that variable over that domain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For an example if you didn't get me yet, we can put the following statement involving quantified formula in a probability format:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$&#xD;&#xA;( \exists x \in X : g(x) ) \equiv ( Pr_{x \in X} [g(x)=1] &amp;gt; 0 ),&#xD;&#xA;$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $g(x) \in \{0,1\}$ is a predicate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly defined as well for the universal quantifier as $Pr=1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However it gets trickier if we nest the quantifiers, for instance I am not sure if the following is correct:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Update: &#xA;This equation is wrong: $ \exists x \in X : ( \forall y \in Y : g(x,y) ) \equiv Pr[g(X,Y)=1 | Y=y] &amp;gt; 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I propose this one instead: &#xA;$ \exists x \in X : ( \forall y \in Y : g(x,y) ) \equiv Pr_{x\in X} \left[ Pr_{y\in Y}[g(x,y)=1 | X=x] = 1 \right] &amp;gt; 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-end update&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If that line of intuition was correct as I hope, I was wondering if such relationship has been explored before and if not I was hoping to know if it would be possible or not to put a general framework for converting a quantified formula of any alternation depth into a probability equivalent.&lt;/p&gt;&#xA;" OwnerUserId="558" LastEditorUserId="558" LastEditorDisplayName="" LastEditDate="2010-12-10T15:44:10.907" LastActivityDate="2010-12-11T01:05:04.170" Title="Reducing quantification to probability" Tags="&lt;probability&gt;&lt;quantifiers&gt;&lt;quantifier-elimination&gt;" AnswerCount="1" CommentCount="6" />
  <row Id="13805" PostTypeId="2" ParentId="13801" CreationDate="2010-12-10T14:11:12.733" Score="4" ViewCount="" Body="&lt;p&gt;This is answered quite succinctly in the MathWorld article on &lt;a href=&quot;http://mathworld.wolfram.com/PrincipalSquareRoot.html&quot; rel=&quot;nofollow&quot;&gt;Principal Square Root&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="4604" LastActivityDate="2010-12-10T14:11:12.733" CommentCount="2" />
  <row Id="13806" PostTypeId="2" ParentId="13801" CreationDate="2010-12-10T14:12:10.813" Score="2" ViewCount="" Body="&lt;p&gt;The proper way to define $i$ is to define the laws for complex addition and multiplication the usual way using ordered pairs of real numbers, and then define $i$ to be $(0,1)$. This distinguishes it from $-i$ which is $(0,-1)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One can then set the root of $-1$ to be $i$, but of course this last step is pretty arbitrary. You can see it either as &lt;em&gt;wrong&lt;/em&gt; or as &lt;em&gt;not wrong&lt;/em&gt; if you want to work with the last step as an arbitrary definition. The important thing is that we have an unambiguous definition of $i$ using the initial steps specified above.&lt;/p&gt;&#xA;" OwnerUserId="4594" LastEditorUserId="4594" LastEditorDisplayName="" LastEditDate="2011-04-19T20:29:31.063" LastActivityDate="2011-04-19T20:29:31.063" CommentCount="2" />
  <row Id="13807" PostTypeId="1" AcceptedAnswerId="13812" CreationDate="2010-12-10T14:13:02.623" Score="0" ViewCount="137" Body="&lt;p&gt;Let's say I have two numbers that I know to have been normalized- 39, 12.  How do I calculate the percentage difference between them?  Is it the same as if they hadn't been normalized?  ie: (12/39) * 100&lt;/p&gt;&#xA;" OwnerUserId="4608" LastActivityDate="2010-12-10T14:33:24.717" Title="Percentage difference in normalized number set" Tags="&lt;normal-distribution&gt;" AnswerCount="1" />
  <row Id="13808" PostTypeId="2" ParentId="13801" CreationDate="2010-12-10T14:15:46.613" Score="14" ViewCount="" Body="&lt;p&gt;I believe this is a common misconception. Here in Sweden we (or at least I) was taught that $i^2 = -1$, &lt;em&gt;not&lt;/em&gt; that $\sqrt{-1} = i$. They are two fundamentally different statements. Of course most of the time one chooses $\sqrt{-1} = i$ as the principal branch of the square root.&lt;/p&gt;&#xA;" OwnerUserId="708" LastActivityDate="2010-12-10T14:15:46.613" CommentCount="12" />
  <row Id="13809" PostTypeId="2" ParentId="13772" CreationDate="2010-12-10T14:28:33.227" Score="0" ViewCount="" Body="&lt;p&gt;There is a large literature about how to take a drawing in the plane which tries to give a 3-dimensional view of a polyhedron and extract a polyhedron from this drawing, in contrast to applying Steinitz's Theorem which gives necessary and sufficient conditions that a graph be the vertex-edge graph of a convex 3-dimensional polyhedron:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Steinitz%27s_theorem&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Steinitz%27s_theorem&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a sample of the other kind of work:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://linkinghub.elsevier.com/retrieve/pii/S009784930900020X&quot; rel=&quot;nofollow&quot;&gt;http://linkinghub.elsevier.com/retrieve/pii/S009784930900020X&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://linkinghub.elsevier.com/retrieve/pii/S009784930900020X&quot; rel=&quot;nofollow&quot;&gt;http://linkinghub.elsevier.com/retrieve/pii/S009784930900020X&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://portal.acm.org/citation.cfm?id=1052355&quot; rel=&quot;nofollow&quot;&gt;http://portal.acm.org/citation.cfm?id=1052355&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1369" LastActivityDate="2010-12-10T14:28:33.227" CommentCount="1" />
  <row Id="13810" PostTypeId="2" ParentId="13662" CreationDate="2010-12-10T14:31:04.390" Score="7" ViewCount="" Body="&lt;p&gt;The probability that A dies first is 1/4 (since it's the same probability for everyone).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The probability that they all die after the age of 90 is $(2/3)^4 = 16/81$. The two events are independent, so the probability that A dies first &lt;em&gt;and&lt;/em&gt; they all die after the age of 90, or equivalently that A dies first and that that happens after he is 90, is $\frac{16}{81}\cdot\frac{1}{4}=\frac{4}{81}$. But if A dies first, then either he dies first and is above 90 or he dies first and is younger than 90, and the two are mutually exclusive. Therefore (or equivalently using the rule that $P(X) = P(X\cap Y) + P(X\cap Y^C)$), we deduce that the probability that A dies first and &lt;em&gt;before&lt;/em&gt; the age of 90 is 1/4 - 4/81 = 65/324.&lt;/p&gt;&#xA;" OwnerUserId="3212" LastActivityDate="2010-12-10T14:31:04.390" CommentCount="4" />
  <row Id="13811" PostTypeId="1" CreationDate="2010-12-10T14:32:14.627" Score="1" ViewCount="69" Body="&lt;p&gt;I just read through a proof of the Compact Uniformization Theorem, and I follow it up to the very last line. The proof is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Compact Uniformization Threorem. If $X$ is a compact regular space, then the neighborhood filter $\mathcal{N}_\Delta$ of the diagonal $\Delta\subset X\times X$ is a uniform structure on $X$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose that $\mathcal{N}_\Delta$ does not have the property that for any entourage $E$, there exists an entourage $D$ such that $D\circ D\subseteq E$. Then there exists an open neighborhood $E\in\mathcal{N}_\Delta$ such that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(D\circ D)\setminus E\neq\emptyset\qquad (D\in\mathcal{N}_\Delta)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, the family&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathcal{C}=\{(D\circ D)\setminus E\ | \ D\in\mathcal{N}_\Delta\}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;consists of nonempty subsets and is a filter-base on $X\times X\setminus E$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The latter begin a closed subset of the product of two compact spaces is compact by Tychonoff's Theorem. It follows that $\mathcal{C}$ has a cluster point $(p,q)\in X\times X\setminus E$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note the neighborhood $E(q)$ of $q$ does not contain $p$, and thus the neighborhood $E^{-1}(p)$ of $p$ does not contain $q$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So by a previous Lemma, there exists an open neighborhood $D$ of $\Delta$ such that $D\circ D$ is disjoint with some neighborhood of $(p,q)$. It follows that $(p,q)$ cannot be a cluster point of $\mathcal{C}$, a contradiction. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I just don't see why it immediately follows that $(p,q)$ is not a cluster point of $\mathcal{C}$, simply because there is some open neighborhood of the diagonal that is disjoint from one of the neighborhoods of $(p,q)$. Can someone please explain?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The previous lemma is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $p$ and $q$ be a pair of points in a regular topological space $X$ such that $\mathcal{N}_p\neq\mathcal{N}_q$. Then there exists an open cover $\mathcal{U}$ of $X$ and a neighborhood $W$ of $(p,q)$ such that &#xA;$$W\cap(\Delta_\mathcal{U}\circ\Delta_\mathcal{U})=\emptyset.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it's not clear, $$\Delta_\mathcal{U}=\bigcup_{U\in\mathcal{U}}U\times U.$$&lt;/p&gt;&#xA;" OwnerUserId="1583" LastEditorUserId="1583" LastEditorDisplayName="" LastEditDate="2010-12-10T14:57:57.200" LastActivityDate="2010-12-10T14:57:57.200" Title="On the Compact Uniformization Theorem" Tags="&lt;topology&gt;&lt;general-topology&gt;" CommentCount="6" FavoriteCount="1" />
  <row Id="13812" PostTypeId="2" ParentId="13807" CreationDate="2010-12-10T14:33:24.717" Score="0" ViewCount="" Body="&lt;p&gt;Shouldn't the percentage difference be [(12-39)/39]*100 or [(39-12)/12]*100?  Note that you get different answers, not just the negative-when the two numbers are unequal the percentage difference depends upon which you regard as the base.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whether normalization changes the percentage depends upon what you mean by normalization.  Sometimes it is multiplication by a constant to bring the range to something you like.  This will not impact the percentage difference-try multiplying 39 and 12 by a and note that it divides out of the expression.  Sometimes it involves adding or subtracting a constant-that will make a difference.  Compare [(39-12)/39]*100 to [(30-3)/30]*100.  The answer is different.  Sometimes it is a more complicated expression, which will almost always change the percentages.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-12-10T14:33:24.717" />
  <row Id="13813" PostTypeId="2" ParentId="13802" CreationDate="2010-12-10T14:36:04.100" Score="5" ViewCount="" Body="&lt;p&gt;This is a good question. One way to think about the issue is that the &quot;convergence&quot; of your integral is not just about whether the value is finite, but also about whether the finite value is well defined. This is somewhat related to &lt;a href=&quot;http://en.wikipedia.org/wiki/Indeterminate_form&quot; rel=&quot;nofollow&quot;&gt;indeterminant forms&lt;/a&gt; where $+\infty + (-\infty)$ is not a well-defined quantity. (You can't say they cancel out, since, heuristically, $\infty - \infty = (1+\infty)-\infty = 1+\infty -\infty = 1 + (\infty - \infty)$...)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So as long as one of the integrals on the right hand side diverges, the entire algebraic expression becomes indeterminate, and hence we say the integral diverges. (Diverges doesn't necessarily mean that the value must run-off to infinity; it can just mean that the value does not converge to a definite number/expression.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(A similar issue also crops up when summing infinite series that doesn't converge absolutely. &lt;a href=&quot;http://en.wikipedia.org/wiki/Riemann_series_theorem&quot; rel=&quot;nofollow&quot;&gt;Riemann rearrangement theorem&lt;/a&gt; tells you that, depending on &quot;how&quot; you sum the series you can get the final number to be anything you want.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sometimes, however, it is advantageous to try to make sense of an integral which can be split into two divergent improper integrals, but also where one can argue that there should be some natural cancellation. For example, one may want to argue that $\int_{-a}^a \frac{1}{x^3} dx$ evaluates to 0 since it is the integral of an odd function. For this kind of situations, the notion of &lt;a href=&quot;http://en.wikipedia.org/wiki/Cauchy_principal_value&quot; rel=&quot;nofollow&quot;&gt;Cauchy principal value&lt;/a&gt; is useful. But notice that the definition is taken in the sense of a limit that rely on some cancellation, and so much in the same way of Riemann rearrangement theorem, &quot;how&quot; you take the limit can affect what value you get as the end result. (This is compatible with the notion that the integral diverges; as I said above, divergence should be taken to mean the lack of a well-defined, unique convergence.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;. Let me add another example of an integral that remains finite but does not converge. What is the value of $\int_{0}^\infty \sin(x) dx$? For every fixed $a &amp;gt; 0$, $\int_0^a\sin(x) dx = 1 - \cos(a) $ is a number between 0 and 2. But the limit as $a\to \infty$ doesn't exist! If you pick a certain way to approach $\infty$, say choose a sequence $a_n = 2\pi n$, then you'll come to the conclusion that the &quot;limit&quot; is 0; but if you choose $a_n = (2n + 1)\pi$, then you get the conclusion that the limit is $2$. The idea here is roughly similar: you take a left limit and a right limit approaching the improper point, and depending on how you choose your representative points (by an algebraic relation between the speed at which the left and right limits approach the improper point, say), you can get different answers. &lt;/p&gt;&#xA;" OwnerUserId="1543" LastEditorUserId="1543" LastEditorDisplayName="" LastEditDate="2010-12-10T14:43:01.957" LastActivityDate="2010-12-10T14:43:01.957" CommentCount="4" />
  <row Id="13814" PostTypeId="1" AcceptedAnswerId="13816" CreationDate="2010-12-10T14:55:43.147" Score="2" ViewCount="264" Body="&lt;p&gt;I have to test for convergence of the series:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\displaystyle \sum\limits_{n=1}^{\infty} \sin\Bigl(\frac{\pi}{n}\Bigr)$&#xD;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;What i did was &lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align*}&#xA;\sin\Bigl(\frac{\pi}{n}\Bigr)+ \sin\Bigl(\frac{\pi}{n+1}\Bigr) + \cdots  &amp;amp; &amp;lt; \pi \biggl( \frac{1}{n+1} + \frac{1}{n+2} + \cdots \biggr) \\ &amp;amp;= \pi \biggl( \sum\limits_{r=1}^{\infty} \frac{1}{n+r}\biggr) =\int\limits_{0}^{1} \frac{1}{1+x} \ dx \\ &amp;amp;= \pi\log{2}&#xA;\end{align*}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think this proves the &lt;em&gt;convergence&lt;/em&gt; of the series.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;I am interesting in knowing some more methods which can be used to prove the convergence so that i can apply them. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;ADDED:&lt;/strong&gt; Note that $$\lim_{n \to \infty} \sum\limits_{r=1}^{n} \frac{1}{n} \cdot f\Bigl(\frac{r}{n}\Bigr) = \int\limits_{0}^{1} f(x) \ \textrm{dx}$$&lt;/p&gt;&#xA;" LastActivityDate="2010-12-10T15:56:16.393" Title="Is $\sum \sin{\frac{\pi}{n}}$ convergent?" Tags="&lt;real-analysis&gt;&lt;analysis&gt;&lt;sequences-and-series&gt;" AnswerCount="1" CommentCount="8" />
  <row Id="13815" PostTypeId="1" AcceptedAnswerId="13819" CreationDate="2010-12-10T14:58:44.203" Score="2" ViewCount="219" Body="&lt;p&gt;How can one find if a function $f$ is infinitely differentiable?&lt;/p&gt;&#xA;" OwnerUserId="1069" LastActivityDate="2010-12-10T15:47:05.983" Title="Infinitely differentiable" Tags="&lt;calculus&gt;" AnswerCount="4" FavoriteCount="1" />
  <row Id="13816" PostTypeId="2" ParentId="13814" CreationDate="2010-12-10T15:02:46.310" Score="10" ViewCount="" Body="&lt;p&gt;You observed correctly that for small angles $\frac{\pi}{n}$, sin$\left(\frac{\pi}{n}\right)$ is very close to $\frac{\pi}{n}$. As for convergence of $\pi\sum_{n=1}^\infty\frac{1}{n}$...&lt;/p&gt;&#xA;" OwnerUserId="3212" LastActivityDate="2010-12-10T15:02:46.310" CommentCount="2" />
  <row Id="13817" PostTypeId="2" ParentId="13815" CreationDate="2010-12-10T15:05:46.850" Score="2" ViewCount="" Body="&lt;p&gt;There is no magic recipe to tell whether a function is infinitely differentiable, or even once differentiable, or even continuous indeed. It really depends on the function itself. Of course there are really general and trivial statements like all polynomials are infinitely differentiable and so on.&lt;/p&gt;&#xA;" OwnerUserId="4594" LastActivityDate="2010-12-10T15:05:46.850" />
  <row Id="13818" PostTypeId="2" ParentId="13815" CreationDate="2010-12-10T15:11:53.123" Score="4" ViewCount="" Body="&lt;p&gt;Depending on how your function is given, that can be very easy or very hard. For functions given by expressions in closed form, you can try deriving a general formula for the $n$-th derivative. E.g. this way you can prove that all polynomials are smooth (=infinitely differentiable), so are sin, cos and exponential functions and any concatenation, product and sum of smooth functions is smooth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Often, you will have to work with the actual derivative as a limit and differentiate until you find either a non-smooth point or some closed form that you recognise, you maybe some periodicity (i.e. so that some $n$-derivative is equal to some $m$-the derivative, like with trigonometric functions) or you have to get really clever.&lt;/p&gt;&#xA;" OwnerUserId="3212" LastActivityDate="2010-12-10T15:11:53.123" />
  <row Id="13819" PostTypeId="2" ParentId="13815" CreationDate="2010-12-10T15:13:53.443" Score="9" ViewCount="" Body="&lt;p&gt;By differentiating it an infinite number of times? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But seriously, that's what you do. Just that usually you can infer what the higher order derivatives will be, so you don't have to compute it one by one. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Example&lt;/strong&gt; To see that $\sin(x)$ is infinitely differentiable, you realize the following: $\frac{d}{dx}\sin(x) = \cos(x)$, and $\frac{d}{dx} \cos(x) = -\sin(x)$. So you see that $(\frac{d}{dx})^4\sin(x) = \sin(x)$, and so the derivatives are periodic. Therefore by continuity of $\sin(x)$ and its first three derivatives, $\sin(x)$ must be infinitely differentiable. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Example&lt;/strong&gt; To see that $(1 + x^2)^{-1}$ is infinitely differentiable, you realize that $\frac{d}{dx}(1+x^2)^{-n} = -2n x (1+x^2)^{-n-1}$. So therefore by induction you have the following statement: all derivatives of $(1+x^2)^{-1}$ can be written as a polynomial in $x$ multplied by $(1 +x^2)^{-1}$ to some power. Then you can use the fact that (a) polynomial functions are continuous and (b) quotients of polynomial functions are continuous away from where the denominator vanishes to conclude that all derivatives are continuous. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The general philosophy at work is that in order to show all derivatives are bounded and continuous, you can take advantage of some sort of recursive relationship between the various derivatives to inductively give a general form of the derivatives. Then you reduce the problem to showing that all functions of that general form are continuous. &lt;/p&gt;&#xA;" OwnerUserId="1543" LastActivityDate="2010-12-10T15:13:53.443" CommentCount="2" />
  <row Id="13820" PostTypeId="1" CreationDate="2010-12-10T15:29:06.957" Score="2" ViewCount="58" Body="&lt;p&gt;Define a &lt;strong&gt;quadtope&lt;/strong&gt; P as a polytope which is the convex hull of a set Q of convex quadrilaterals, any pair of which intersect in at most one vertex, and are in maximally general position, or, a k-fold pyramid over such a convex hull. (So, a pair of quadrilaterals from Q with a common vertex subtends a 4-dimensional subpolytope of P, and a pair of disjoint quadrilaterals from Q subtends a five-dimensional subpolytope of P.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A &lt;strong&gt;pure quadtope&lt;/strong&gt; is a quadtope each of whose vertices is a vertex of an element of Q.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I assert that the number of combinatorial types of pure quadtopes for dimensions 1 through 4 are 0, 1, 0, 1, respectively. How many combinatorial types of pure quadtopes of dimension 5 are there?&lt;/p&gt;&#xA;" OwnerUserId="1782" LastEditorUserId="1782" LastEditorDisplayName="" LastEditDate="2010-12-10T18:38:37.547" LastActivityDate="2011-01-18T16:47:48.867" Title="How many combinatorial types of pure quadtopes of dimension 5 are there?" Tags="&lt;geometry&gt;&lt;combinatorics&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="13821" PostTypeId="2" ParentId="13815" CreationDate="2010-12-10T15:47:05.983" Score="1" ViewCount="" Body="&lt;p&gt;As Willie Wong points out, that's the only way to do it. By the way $\sin{x},\cos{x}$ are examples of infinitely differentiable functions, and most famous example is the function $f(x)=e^{x}$. This function is infinitely differentiable and it's derivative, is $e^{x}$. To see this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align*}&#xA;e^{x}  = 1 + \frac{x}{1!} + \frac{x^{2}}{2!} + \cdots \\ \frac{\textrm{d}} {\textrm{dx}}e^{x} = 1 + \frac{x}{1!} + \frac{x^{2}}{2!} + \cdots &#xA;\end{align*}&lt;/p&gt;&#xA;" LastActivityDate="2010-12-10T15:47:05.983" />
  <row Id="13822" PostTypeId="1" CreationDate="2010-12-10T16:24:57.843" Score="7" ViewCount="180" Body="&lt;p&gt;I am trying to figure out the fundamental group (actually simply connected or not will suffice) of the following quotient space of $SO(3)$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $X = SO(3)/E$, where $E$ is the equivalence relation defined as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$E \equiv M \sim {S_{A}}^{i} * M * {S_{B}}^{j}$ where ${S_{A}}^{i} \in $ Crystallographic point group $A$ and ${S_{B}}^{j} \in $ Crystallographic point group $B$, $M \in SO(3)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$*$ represents the multiplication operation (matrix multiplication if rotations are represented as $3 \times 3$ special orthogonal matrices)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Crystallographic point group is defined here: &lt;a href=&quot;http://en.wikipedia.org/wiki/Crystallographic_point_group&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Crystallographic_point_group&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;In crystallography, a crystallographic point group is a set of symmetry operations, like rotations or reflections, that leave the crystal invariant (hence a symmetry).&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Only consider the point groups with rotational symmetries. There exist 11 crystallographic point groups for three-dimensional crystals. Let us start with just considering cyclic point groups. They are the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(a) $C_1 = \{I \}$   where I is the identity rotation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $Z_{\omega}$ is a rotation of angle $\omega$ about the $Z-$axis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(b) $C_2 = \{I, Z_{\pi} \}$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(c) $C_3 = \{I, Z_{\frac{2\pi}{3}}, Z_{\frac{4\pi}{3}} \}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(d) $C_4 = \{I, Z_{\frac{\pi}{2}}, Z_{\pi}, Z_{\frac{3\pi}{2}} \}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(e) $C_6 = \{I, Z_{\frac{\pi}{3}}, Z_{\frac{2\pi}{3}}, Z_{\pi}, Z_{\frac{4\pi}{3}},  Z_{\frac{5\pi}{3}} \}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example if Crystallographic point group of A is $C_2$ and B is $C_3$, the equivalence relations are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ M \sim M * Z_{\frac{2\pi}{3}} \sim M * Z_{\frac{4\pi}{3}} \sim Z_{\pi}* M \sim Z_{\pi} * M * Z_{\frac{2\pi}{3}}  \sim Z_{\pi} * M * Z_{\frac{4\pi}{3}} $&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There exists literature for cases when one of the point groups is $C_1 = \{ I \}$. In this case it is a group action on $SO(3)$ and the space $X = SO(3) / G$ where $G$ is one of the 11 crystallographic point groups. These spaces fall under the so-called spherical 3-manifolds ( &lt;a href=&quot;http://en.wikipedia.org/wiki/Spherical_3-manifold&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Spherical_3-manifold&lt;/a&gt; ). When $G$ is one of the cyclic groups above, $X$ is a lens space. $L(2n,1) \cong SO(3)/ C_{n}$. I am not able to figure out how to think of these spaces when there are two point groups involved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Progress so far: Even when there are two point groups acting on $SO(3)$, which if we refer to as $ G_{1} \backslash SO(3)/ G_{2}$ where $G_{1}$ is crystallographic point group of $A$ and $G_2$ refers to system $B$, there exists a finite subgroup of $\Gamma$ of $SO(4)$ such that  $ G_{1} \backslash SO(3)/ G_{2} \cong S^{3}/ \Gamma$. In the case $G_{1} = C_{1}$, it turns out that $\Gamma$  acts properly discontinuously and hence the fundamental group of that space is $\Gamma$ itself (from literature). But I am not sure how to check if $\Gamma$ acts &quot;properly discontinuously&quot; or not. And I am not sure how to check if the space is simply connected or not if $\Gamma$  does not act properly discontinuously. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any help appreciated. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To answer Aaron's question: do you know how to obtain this group $\Gamma$ explicitly? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I use the quaternion representation for rotations. Let $M = (q_0, q_1, q_2, q_3)$. I also use the following fact: Each 4D rotation $R$ is in two ways the product of left- and right-isoclinic rotations $R_L$ and $R_R$. $R_L$ and $R_R$ are together determined up to the central inversion, i.e. when both $R_L$ and $R_R$ are multiplied by the central inversion their product is $R$ again. (from wiki: &lt;a href=&quot;http://en.wikipedia.org/wiki/SO(4&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/SO(4&lt;/a&gt;) ). So for any operation ${S_{A}}^{i} * M * {S_{B}}^{j}$, I can find $\Gamma_{ij}$ such that &#xA;${S_{A}}^{i} * M * {S_{B}}^{j} = \Gamma_{ij}*[q_0 \ q_1 \ q_2 \ q_3]&amp;#39;$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The collection of all such $\Gamma_{ij}$ (and $\Gamma_{ij} * -I_4$, where $I_4$ is the $4 \times 4$ identity matrix) forms a finite subgroup $\Gamma$ of $SO(4)$. I used the definition provided by Aaron below for free action and found that at least for the cases where crystallographic point groups of $A$ and $B$ are the same (e.g. $C_2 \backslash SO(3) / C_2$), the action of finite subgroup $\Gamma_{C_2,C_2}$ is not properly discontinuous. Even though I tried to generalize the problem I am more interested in the cases when the crystallographic point groups of A and B are the same. So, how can I say whether this space $C_2 \backslash SO(3) / C_2$ is simply connected or not, especially now that i know that the action of $\Gamma$ is not properly discontinuous ??&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any ideas ?&lt;/p&gt;&#xA;" OwnerUserId="3050" LastEditorUserId="3050" LastEditorDisplayName="" LastEditDate="2010-12-10T19:33:25.893" LastActivityDate="2010-12-10T19:33:25.893" Title="Fundamental group of quotient spaces of $SO(3)$" Tags="&lt;algebraic-topology&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="13823" PostTypeId="1" AcceptedAnswerId="13824" CreationDate="2010-12-10T17:02:09.667" Score="2" ViewCount="379" Body="&lt;p&gt;I know this is a silly question, but I've tried to find an answer using my TI-89 calculator, Maple and wolframalpha but none of those could tell me whether&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_0^\infty |\cos(x^2)| \mathrm dx$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;converges or diverges.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, I'd be very happy if someone could help me out and tell me, whether the given integral converges or not (and why?). Thanks a lot. &lt;/p&gt;&#xA;" OwnerUserId="3787" LastActivityDate="2010-12-10T22:46:29.037" Title="Does $\int_0^\infty |\cos(x^2)| \mathrm dx$ converge?" Tags="&lt;integral&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="13824" PostTypeId="2" ParentId="13823" CreationDate="2010-12-10T17:13:04.687" Score="10" ViewCount="" Body="&lt;p&gt;It diverges.  You should be able to prove that $|\cos(x^2)|&amp;gt;0.1$ for most x.  If you let x=$\sqrt{\pi}u$ it is easier to assess the range of u where the cosine is close to zero.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-12-10T17:13:04.687" />
  <row Id="13826" PostTypeId="2" ParentId="13823" CreationDate="2010-12-10T17:19:55.223" Score="10" ViewCount="" Body="&lt;p&gt;Another method: You could write it as the sum of the integrals on the intervals $\left[\sqrt{\frac{\pi}{2}+k\pi},\sqrt{\frac{\pi}{2}+(k+1)\pi}\right]$, and make a substitution $u=x^2$ to bound the integral on such an interval below by $\frac{1}{\sqrt{\frac{\pi}{2}+(k+1)\pi}}$.  (I'm ignoring the interval $\left[0,\sqrt{\frac{\pi}{2}}\right]$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the inequality $\frac{\pi}{2}+(k+1)\pi\leq(k+2)\pi$ along with an integral comparison then leads to the estimate &#xA;$$\int_0^\sqrt{\frac{\pi}{2}+N\pi}|\cos(x^2)|dx\geq\frac{2}{\sqrt{\pi}}(\sqrt{N}-\sqrt{2}).$$&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2010-12-10T21:17:42.263" LastActivityDate="2010-12-10T21:17:42.263" CommentCount="1" />
  <row Id="13827" PostTypeId="2" ParentId="13689" CreationDate="2010-12-10T17:21:04.940" Score="1" ViewCount="" Body="&lt;p&gt;Your problem is almost exactly &lt;a href=&quot;http://en.wikipedia.org/wiki/Multidimensional_scaling&quot; rel=&quot;nofollow&quot;&gt;multidimensional scaling&lt;/a&gt;: given a set of $n$ objects with given dissimilarities $\delta_{ij}$, find $n$ corresponding points $y_i$ in a low-dimensional space such that $\lVert y_i - y_j\rVert \approx \delta_{ij}$. I'm not intimately familiar with this area, but the Wikipedia article mentions a number of algorithms you could look at. You should be aware that there will often be many locally optimal configurations for a given problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, multidimensional scaling is typically used for &lt;em&gt;visualization&lt;/em&gt; applications, where you want to map your data to 2 or 3 dimensions and see what objects are similar to each other. I'm not sure how you plan to use it for computing &lt;em&gt;ratings&lt;/em&gt;, because replacing all $y_i$ with $-y_i$ is exactly as good a solution but will turn low ratings into high ratings and vice versa.&lt;/p&gt;&#xA;" OwnerUserId="856" LastActivityDate="2010-12-10T17:21:04.940" />
  <row Id="13828" PostTypeId="2" ParentId="13794" CreationDate="2010-12-10T17:39:18.997" Score="5" ViewCount="" Body="&lt;p&gt;Most of the time, an equivalence relation is hiding an &quot;equality&quot; somewhere; that is, $x\sim y$ if and only if $x$ and $y$ have &lt;em&gt;something&lt;/em&gt; that you are trying to isolate which is &quot;equal&quot;. With integers, the equivalence $a\equiv b\pmod{m}$ means that $a$ and $b$ have &lt;strong&gt;equal&lt;/strong&gt; &lt;em&gt;remainder when divided by $m$&lt;/em&gt;. We usually start from a notion of the thing that we want to be &quot;the same&quot;, and define the equivalence relation accordingly, which makes it easier to think about just what are the equivalence classes: they correspond to all objects with the same &quot;&lt;em&gt;thingie&lt;/em&gt;&quot; that we are focusing on in the first place.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But suppose you were walking down the street and you found an equivalence relation lying on the ground. Just how easy is it to figure out what this &quot;equality&quot; that is hinding behind it? How easy is it to describe all elements of the equivalence class of a given $x$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer is that it depends greatly on the equivalence relation you find, and sometimes on the $x$. For some equivalence reations, it is fairly easy to figure it out, and then to partition the objects into equivalence classes. But for others, it may be more mysterious.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, suppose you only know about positive integers, their order, and their addition and multiplication. You don't know about negative numbers (as most of humanity did not for most of its history). I can define an equivalence relation on pairs of positive integers by&#xA;$$(a,b)\sim (r,s) \Longleftrightarrow a+s = b+r.$$&#xD;&#xA;It is easy to verify that this is an equivalence relation. But what is the partition it induces? What is that &quot;equality&quot; that is hiding behind that equivalence relation?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's perhaps not so obvious. In fact, it comes from thinking of an ordered pair $(a,b)$ as corresponding to the equation $a+x=b$, so that the pair $(a,b)$ represents what we want to be a solution to this equation, which may or may not exist in the positive integers. Then $(a,b)\sim(r,s)$ means that the solution of $a+x=b$ should be &quot;the same&quot; as the solution of $r+x=s$. It seems obvious, then, that since $x=b-a=s-r$, this gives the condition I give, but the point is that one makes this definition among positive integers, where $b-a$ may be undefined, e.g., if $a\geq b$. It corresponds to a way of trying to think about negative numbers without having to use subtraction. But if you've never even thought about negative numbers, it would be rather difficult to figure it out, thought it would be pretty easy to check, given two pairs, whether they are equivalent or not; and you may even be able to describe all the pairs that are equivalent to a given one, at least sometimes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But you really have little hope of having some general method that will work easily and always. The reason is precisely because equivalence relations correspond to partitions. Pick &lt;strong&gt;any&lt;/strong&gt; partition, and define an equivalence relation by &quot;$a\sim b$ if and only if they are in the same piece of the partition.&quot; That will be an equivalence relation, but if you find it lying on the quad you would be hard pressed to figure out where it could have come from, precisely because it came from an arbitrary partition that happened to catch your fancy at the moment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the moral is that most of the time equivalence relations come from a very specific idea we are trying to isolate, or a specific issue that is somewhat troublesome and we want to avoid (for instance, two functions being identical except for some negligible set of points where they differ, which may lead to some 'obvious' conclusions being technically false but true in 'spirit', so we define an equivalence relation that puts two such functions into the same class so that our &quot;true in spirit&quot; becomes &quot;technically true&quot; by now refering to &lt;em&gt;equivalence classes of functions&lt;/em&gt; instead of functions themselves). In such cases, it is very often easy to figure out the equivalence classes, or at least the bulk of each equivalence class (with some outliers being a bit troublesome from time to time). Practice and experience will let you spot them as they show up.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-12T19:48:53.387" LastActivityDate="2010-12-12T19:48:53.387" CommentCount="1" />
  <row Id="13829" PostTypeId="1" CreationDate="2010-12-10T17:42:27.263" Score="2" ViewCount="109" Body="&lt;p&gt;I was trying to understand the notion of a connection. I have heard in seminars that a connection is more or less a differential equation. I read the definition of Kozsul connection and I am trying to assimilate it. So far I cannot see why a connection is a differential equation. Please help me with some clarification.&lt;/p&gt;&#xA;" OwnerUserId="4611" LastActivityDate="2010-12-10T18:55:24.790" Title="Connections and differential equations" Tags="&lt;differential-geometry&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="2" />
  <row Id="13830" PostTypeId="1" CreationDate="2010-12-10T18:05:57.787" Score="1" ViewCount="190" Body="&lt;p&gt;The Mathematica code is &lt;a href=&quot;http://demonstrations.wolfram.com/TransientHeatConductionWithNoSteadyState/&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the heat conduction problem with Neumann (constant flux) at both boundaries of a solid slab. A constant radiant heat flux is imposed on one surface (derivative = -1) and the other surface is thermally insulated (derivative = 0).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you increase the flux of the heat at one boundary, the temperature inside the system should raise faster over time. However, with the current code, if you increase the flux from 1 to 100, the system behavior doesn't change at all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the code, &lt;code&gt;Quiet[NDSolve[...]]&lt;/code&gt; is used to suppress the error message: &lt;code&gt;&quot;NDSolve::ibcinc: Warning: Boundary and initial conditions are inconsistent.&quot;&lt;/code&gt; The simulation might be incorrect due to this error message.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to have the ability to change the system behavior by changing the value of the flux at the boundary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a very important question to me and I couldn't figure it out. Any help will be highly appreciated.&lt;/p&gt;&#xA;" OwnerUserId="4612" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-10T18:09:24.463" LastActivityDate="2010-12-11T06:23:20.343" Title="Help with the Neumann boundary conditions" Tags="&lt;differential-equations&gt;&lt;mathematica&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="13831" PostTypeId="2" ParentId="13829" CreationDate="2010-12-10T18:06:37.047" Score="3" ViewCount="" Body="&lt;p&gt;A connection is not a differential equation-- it is a differential operator.  Where the differential aspect comes from is that it is a derivation.  Remember in calculus the product rule says that $\frac{d}{dx} fg = \frac{df}{dx} g + f\frac{dg}{dx}$.  Well for a vector field $\sigma$, a function $f$, and Koszul connection $\nabla$, the property&#xA;$$&#xD;&#xA;\nabla_V (f\sigma) = df(V) \sigma + f\nabla_V \sigma&#xD;&#xA;$$&#xD;&#xA;is satisfied.  This is what makes it differential.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What makes it a &quot;connection&quot; is that with $\nabla$ you can define parallel translation.  If you have a curve $c$, you can say that a vector field $\sigma(t)$ defined on the curve is parallel if $\nabla_{c&amp;#39;(t)} \sigma = 0$.  Think of this as $\frac{d}{dt} \sigma = 0$.  Now take a path $c$ starting at a point $x$.  Then by basic theorems of differential equations, for any vector at $x$ there is a unique extension to a parallel vector field along $c$.  This allows you to connect the tangent space at $x$ with the tangent space at any point on $c$.&lt;/p&gt;&#xA;" OwnerUserId="9" LastActivityDate="2010-12-10T18:06:37.047" />
  <row Id="13832" PostTypeId="1" AcceptedAnswerId="13842" CreationDate="2010-12-10T18:20:06.407" Score="2" ViewCount="66" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given an elliptic curve $y^2 = x^3 + ax + b$ over a finite field $\mathbf{F}_p$, how can I retrieve the value of $y$ given the value of $x$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My knowledge in this area is quite limited, so I understand this question might seems silly. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm currently working on an application that uses an elliptic curves library and I saw at some point that this value was retrieved via: &#xA;$$(x^3 + ax + b)^{(p+1) &amp;gt;&amp;gt; 2} \mod p$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the $&amp;gt;&amp;gt;$ means shift right though I am not sure what this means mathematically.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does this make sense?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="4613" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-10T20:07:27.007" LastActivityDate="2010-12-10T20:15:28.917" Title="Extracting the value of $y$ from $x$ in an elliptic curve over a finite field" Tags="&lt;group-theory&gt;&lt;elliptic-curves&gt;&lt;finite-fields&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="1" />
  <row Id="13833" PostTypeId="1" AcceptedAnswerId="13835" CreationDate="2010-12-10T18:35:51.550" Score="0" ViewCount="73" Body="&lt;p&gt;I am attempting to write a function (programming, not mathematics) which calculates the shortest distance between two points on a toroid.  I got the following formula from &lt;a href=&quot;http://stackoverflow.com/questions/2123947/calculate-distance-between-two-x-y-coordinates&quot;&gt;StackOverflow&lt;/a&gt;, but I don't recognize the &lt;code&gt;min&lt;/code&gt; portion.  Can someone explain that to me, please?&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;sqrt(min(|x1 - x2|, w - |x1 - x2|)^2 + min(|y1 - y2|, h - |y1-y2|)^2)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="4614" LastActivityDate="2010-12-10T18:42:05.147" Title="Can someone tell me what &quot;min&quot; means in this context?" Tags="&lt;geometry&gt;&lt;algorithms&gt;" AnswerCount="1" CommentCount="0" />
  <row Id="13834" PostTypeId="2" ParentId="13822" CreationDate="2010-12-10T18:40:52.980" Score="2" ViewCount="" Body="&lt;p&gt;This is cool.  Are you doing chemistry?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, a slightly more standard notation might be $G_1\backslash SO(3)/G_2$ -- this indicates that you're left-multiplying by elements of $G_1$ and right-multiplying by elements of $G_2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A &quot;properly discontinuous&quot; action of a group $G$ on a space $X$ is one where every point $x\in X$ has a neighborhood $U$ such that $g(U)\cap U=\emptyset$ unless $g=1$; that is, $G$ not only acts &lt;em&gt;freely&lt;/em&gt; (no nonidentity element has fixed points), but the nonidentity elements of $G$ take every point &quot;sufficiently far away from itself&quot;.  This may seem vacuous, but in fact when you're working with Kleinian groups and things this can be an important condition.  However, in your case what is immediately true (since your space $SO(3)$ is Hausdorff and your groups are finite) is that any free action is automatically properly discontinuous.  So the only thing you need to check is that your actions have no fixed points.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I guess the question becomes, do you know how to obtain this group $\Gamma$ explicitly?  If it's cyclic, it probably will be taken to be a subgroup of the circle, which acts freely on $S^3$.  (This is $S^1\subseteq \mathbb{C}$ acting on $S^3\subseteq \mathbb{C}^2$ by multiplication.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $\Gamma$ doesn't act properly discontinuously, things are harder.&lt;/p&gt;&#xA;" OwnerUserId="2159" LastActivityDate="2010-12-10T18:40:52.980" CommentCount="3" />
  <row Id="13835" PostTypeId="2" ParentId="13833" CreationDate="2010-12-10T18:42:05.147" Score="3" ViewCount="" Body="&lt;p&gt;As explained in the question you reference, this toroid is created by identifying the sides of a rectangle w units wide in the x direction and h units high in the y direction.  So assume you have two points (x1,y) and (x2,y), at the same height and let x2&gt;x1.  To get from (x2,y) to (x1,y) you can either go left a distance x2-x1, or you can go right (off the edge) and come around a distance of (w-x2)+x1.  The min just picks the smaller of these.  Similarly in y.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-12-10T18:42:05.147" CommentCount="1" />
  <row Id="13836" PostTypeId="2" ParentId="13829" CreationDate="2010-12-10T18:55:24.790" Score="2" ViewCount="" Body="&lt;p&gt;I'm going to be bold and go against the wisdom of the rest of the posters here. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A connection &lt;em&gt;can&lt;/em&gt; be regarded as a differential equation, through the notion of parallel transport and geodesics. In the parallel transport formulation, a connection can be associated with the following map: give a point $p$ in your manifold $M$, a tangent vector $v \in T_pM$ in the tangent space at $p$, and a curve $\gamma:[0,1] \to M$ such that $\gamma(0) = p$, you have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ (p,v,\gamma) \mapsto \tilde{v}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\tilde{v}$ is a vector field on $M$ defined along $\gamma$. That is, $\tilde{v}(\gamma(s)) \in T_{\gamma(s)}M$ for every $s\in [0,1]$. The specification of $\tilde{v}$ is given by the ordinary differential equation &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \nabla_{\dot{\gamma}}\tilde{v} = 0, \qquad \tilde{v}(0) = v$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or, more explicitly&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \frac{d}{ds} \tilde{v}(\gamma(s)) = 0, \qquad \tilde{v}(\gamma(0)) = v$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In terms of the geodesic formulation, in local coordinates, a specification of a parallel transport is equivalent to the specification of the geodesics for that connection. More precisely, fixing a coordinate system $\{x_1, \ldots, x_n\}$ for your manifold $M$, the connection can be represented by its Christoffel symbols (relative to the coordinate system) $\Gamma_{ij}^k$, such that the geodesics satisfy &lt;strong&gt;Newton's Equations&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \frac{d^2 x_i}{dt^2} = -\sum_{k,j} \Gamma^i_{jk} \frac{d x_j}{dt} \frac{d x_k}{dt} $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and in this way the connection precisely is associated to a particular set of second order differential equations, such that for each prescribed initial data $x_i(0) = x_{i,0}$ and $dx_i/dt (0) = y_{i,0}$ you have a corresponding geodesic. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;In a more abstract setting, a connection on the tangent space $TM$ of a manifold $M$ can be associated to a vector field $v$ define &lt;em&gt;over&lt;/em&gt; the total space $TM$ (so that $v$ is a section of $TTM$) so that the natural projection $\pi_*v(x) = x$ where $\pi$ is the projection map from $TM$ to $M$. The vector $v$ defines what is known as the &lt;a href=&quot;http://en.wikipedia.org/wiki/Geodesic#Geodesic_spray&quot; rel=&quot;nofollow&quot;&gt;geodesic spray&lt;/a&gt; of the connection, which can be written as a first order ordinary differential equation on $TM$ (or a second order ordinary differential equation on $M$). So yes, it is possible to associate a connection to a differential equation. (The reverse association, however, is more delicate.)&lt;/p&gt;&#xA;" OwnerUserId="1543" LastActivityDate="2010-12-10T18:55:24.790" CommentCount="5" />
  <row Id="13837" PostTypeId="2" ParentId="13800" CreationDate="2010-12-10T18:55:45.937" Score="3" ViewCount="" Body="&lt;p&gt;For real $s$, $L(s,\chi_d)$ is real, and it is certainly positive for&#xA;large $s$. It is nonzero for $s&amp;gt;1$ by the Euler product and $L(1,\chi_d)\ne0$.&#xA;If $L(1/2,\chi_s)&amp;lt;0$ what would happen for some $1/2&amp;lt;s&amp;lt;1$?&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-12-10T18:55:45.937" CommentCount="1" />
  <row Id="13838" PostTypeId="1" AcceptedAnswerId="13871" CreationDate="2010-12-10T18:57:10.077" Score="1" ViewCount="117" Body="&lt;p&gt;I'm doing a multiple choice exercise and I'm stuck at a question, where there are two possible answers (may be both or none correct).&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Consider the differential equation&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$$y&amp;#39;&amp;#39;+y&amp;#39;+y=0$$&#xD;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;for $y(x)$. Which of the following statements are true?&lt;/p&gt;&#xA;  &#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;&lt;p&gt;The set of complex solutions forms a two-dimensional complex vector space.&lt;/p&gt;&lt;/li&gt;&#xA;  &lt;li&gt;&lt;p&gt;The set of real solutions forms a two-dimensional real vector space.&lt;/p&gt;&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm really not comfortable with vector spaces and the differentiation of real and complex vector space makes me even more unconfident. I solved the differential equation with the ansatz $y(x) = e^{\lambda x}$ and got to $\lambda^2 + \lambda + 1 = 0$, an equation which does not possess any real solutions. However, when I proceeded, I figured that I could rewrite the exponential function (with complex exponent) with the help of the sine and the cosine, so in the end I got a general solution with no complex coefficients.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What kind of vector space is this?&lt;/p&gt;&#xA;" OwnerUserId="3787" LastActivityDate="2010-12-11T04:21:36.920" Title="What kind of vector space does the set of (complex/real) solutions of the following differential equation form?" Tags="&lt;differential-equations&gt;" AnswerCount="2" CommentCount="14" />
  <row Id="13839" PostTypeId="2" ParentId="13373" CreationDate="2010-12-10T19:26:51.563" Score="6" ViewCount="" Body="&lt;p&gt;Sergei Ivanov has given a positive answer for the existence of such functions on MO:&#xA;&lt;a href=&quot;http://mathoverflow.net/questions/48910/smooth-functions-for-which-fx-is-rational-if-and-only-if-x-is-rational&quot; rel=&quot;nofollow&quot;&gt;http://mathoverflow.net/questions/48910/smooth-functions-for-which-fx-is-rational-if-and-only-if-x-is-rational&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="3776" LastActivityDate="2010-12-10T19:26:51.563" />
  <row Id="13840" PostTypeId="2" ParentId="886" CreationDate="2010-12-10T19:52:34.887" Score="1" ViewCount="" Body="&lt;p&gt;22 boxes. 2 at the final stage.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-Assume you will definately swap.&#xA;If you've picked 250k, which is 1/22 chance, you lose.&#xA;If you've not picked 250k, which is 21/22 chance, you win.&#xA;Every 22 times you play, you would expect to pick 250k once and expect to have 250k in the last 2 boxes twice (22 x 2 = 44).&#xA;Of the two times that 250k is in the last 2 boxes, 1 time you will swap (and lose), 1 time you will swap (and win). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;-Assume you will definately not swap.&#xA;If you've picked 250k, which is 1/22 chance, you win.&#xA;If you've not picked 250k, which is 21/22 chance, you lose.&#xA;Every 22 times you play, you would expect to pick 250k once and expect to have 250k in the last 2 boxes twice (22 x 2 = 44).&#xA;Of the two times that 250k is in the last 2 boxes, 1 time you will not swap (and win), 1 time you will not swap (and lose).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you swap (or not), half the time you will win, half the time you will lose.&#xA;With the monty hall problem, 2/3 or the time if you swap you will win, 1/3 of the time you will lose. &#xA;The difference is that in EVERY game of the MH problem, a losing box is revealed and the player reaches the final stage, even though there are more than 2 boxes to start with.&#xA;In deal or no deal, 250k would only be at the final stage 2/22 or 1/13 times.&lt;/p&gt;&#xA;" OwnerUserId="4617" LastActivityDate="2010-12-10T19:52:34.887" />
  <row Id="13841" PostTypeId="1" AcceptedAnswerId="13843" CreationDate="2010-12-10T20:12:06.117" Score="7" ViewCount="299" Body="&lt;p&gt;Anyone know of an online tool available for making graphs (as in graph theory - consisting of edges and vertices)? I have about 36 vertices and even more edges that I wish to draw. &#xA;(why do I have so many? It's for pathing in a game)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Only tool available to me right now is MS Paint and it would be very messy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not actually looking for a 'standard' - just some way to neatly draw a set of vertices and edges, preferably without having to write code. I was planning to screenshot the pic and use it in a question on gamedev but I managed to solve the problem for now. Thanks for all the answers. I still prefer GeoGebra because it accessible online, neat, accurate, aligns elements to a grid, has consistent shapes, and has a high degree of customization available.&lt;/p&gt;&#xA;" OwnerUserId="4618" LastEditorUserId="4618" LastEditorDisplayName="" LastEditDate="2010-12-12T05:20:10.357" LastActivityDate="2010-12-12T05:20:10.357" Title="Online tool for making graphs (vertices and edges)?" Tags="&lt;graph-theory&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="3" />
  <row Id="13842" PostTypeId="2" ParentId="13832" CreationDate="2010-12-10T20:15:28.917" Score="4" ViewCount="" Body="&lt;p&gt;Call C = x&lt;sup&gt;3&lt;/sup&gt;+ax+b.  You want to solve y&lt;sup&gt;2&lt;/sup&gt; = C, which is a quadratic equation, specifically extracting a square root.  There are lots of ways to extract square roots.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, the way you mention is the first method described in &lt;a href=&quot;http://en.wikipedia.org/wiki/Quadratic_residue#Prime_or_prime_power_modulus&quot; rel=&quot;nofollow&quot;&gt;this section of the wikipedia article&lt;/a&gt; on numbers that are squares (quadratic residues).  This only works if p+1 is divisible by 4, but is a consequence of &lt;a href=&quot;http://en.wikipedia.org/wiki/Fermat%27s_little_theorem&quot; rel=&quot;nofollow&quot;&gt;Fermat's little theorem&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It also links to a few efficient algorithms, including a &lt;a href=&quot;http://en.wikipedia.org/wiki/Shanks-Tonelli_algorithm&quot; rel=&quot;nofollow&quot;&gt;full algorithm&lt;/a&gt; that uses the one you mentioned if p+1 is divisible by 4.&lt;/p&gt;&#xA;" OwnerUserId="583" LastActivityDate="2010-12-10T20:15:28.917" />
  <row Id="13843" PostTypeId="2" ParentId="13841" CreationDate="2010-12-10T20:20:50.077" Score="9" ViewCount="" Body="&lt;p&gt;For drawing graph,the best is &lt;a href=&quot;http://www.texample.net/tikz/examples/prims-algorithm/&quot; rel=&quot;nofollow&quot;&gt;TIKZ&lt;/a&gt; but it does take some time to study,an alternate is &lt;a href=&quot;http://www.graphviz.org/&quot; rel=&quot;nofollow&quot;&gt;graphviz&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Rollbacked:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For drawing 2D geometry, try &lt;a href=&quot;http://www.geogebra.org/cms/&quot; rel=&quot;nofollow&quot;&gt;GeoGebra&lt;/a&gt; or &lt;a href=&quot;http://www.dynamicgeometry.com/&quot; rel=&quot;nofollow&quot;&gt;Sketchpad&lt;/a&gt; (licensed).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For drawing 3D geomtry, try &lt;a href=&quot;http://www.cabri.com/download-cabri-3d.html&quot; rel=&quot;nofollow&quot;&gt;Cabri 3d&lt;/a&gt; (licensed).&lt;/p&gt;&#xA;" OwnerUserId="2109" LastEditorUserId="2109" LastEditorDisplayName="" LastEditDate="2010-12-10T20:49:28.253" LastActivityDate="2010-12-10T20:49:28.253" CommentCount="4" />
  <row Id="13844" PostTypeId="1" AcceptedAnswerId="13851" CreationDate="2010-12-10T20:21:34.880" Score="1" ViewCount="77" Body="&lt;p&gt;&lt;strong&gt;Motivation&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a prior on a random variable $X\sim \beta(\alpha,\beta)$ but I need to transform the variable to $Y=\frac{X}{1-X}$, for use in an analysis and I would like to know the distribution of $Y$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Beta_prime_distribution#Related_distributions&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt; states:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;if $X\sim\beta(\alpha,\beta)$ then $\frac{X}{1-X} \sim\beta^\prime(\alpha,\beta)$&#xD;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Thus, the distribution is $Y\sim\beta^\prime(\alpha,\beta)$. The software that I am using, JAGS, does not support the $\beta^\prime$ distribution. So I would like to find an equivalent of a distribution that is supported by JAGS, such as the $F$ or $\beta$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In addition to the above relationship between the $\beta$ and $\beta^\prime$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Beta_prime_distribution#Related_distributions&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt; states:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;if $X\sim\beta^\prime(\alpha,\beta)$ then $\frac{X\beta}{\alpha}\sim F(2\alpha, 2\beta)$&#xD;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, neither of these statements are referenced. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Questions&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;1) Can I find $c$, $d$ for $Y\sim\beta^\prime(\alpha,\beta)$ where $Y\sim\beta(c,d)$&#xD;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;2) Are these transformations correct? If so, are there limitations to using them, or a reason to use one versus the other (I presume $\beta$ is a more direct transformation, but why)? &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;3) Where can I find such a proof or how would one demonstrate the validity of these relatively simple transformations?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="3733" LastEditorUserId="3733" LastEditorDisplayName="" LastEditDate="2010-12-13T19:34:07.110" LastActivityDate="2010-12-13T19:34:07.110" Title="Are these transformations of the $\beta^\prime$ distribution from $\beta$ and to $F$ correct?" Tags="&lt;probability&gt;&lt;distribution-theory&gt;" AnswerCount="2" />
  <row Id="13845" PostTypeId="2" ParentId="13841" CreationDate="2010-12-10T20:23:09.470" Score="3" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://creately.com/&quot; rel=&quot;nofollow&quot;&gt;http://creately.com/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://dabbleboard.com/draw&quot; rel=&quot;nofollow&quot;&gt;http://dabbleboard.com/draw&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3330" LastActivityDate="2010-12-10T20:23:09.470" CommentCount="2" />
  <row Id="13846" PostTypeId="2" ParentId="13823" CreationDate="2010-12-10T20:29:03.693" Score="4" ViewCount="" Body="&lt;p&gt;There is a standard technique we learned for dealing with integrals of the form $(*) \int_{a}^{\infty}f(x)\cos (\alpha x)dx$ or $\int_{a}^{\infty}f(x)\sin (\alpha x)dx$ when $f(x)$ is continuous and positive in $[a, \infty)$.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;If $\int_{a}^{\infty}f(x)dx$ converges, then (*) converges absolutely by the comparison test because $|f(x)\sin (\alpha x)|\leq f(x)$ (or $|f(x)\cos (\alpha x)|\leq f(x)$).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If $\int_{a}^{\infty}f(x)dx$ diverges while $f(x)$ is decreasing and $\lim_{x\to\infty}f(x)=0$ then the integrals (*) converge conditionally by &lt;a href=&quot;http://math.nyu.edu/student_resources/wwiki/index.php/Category%3aDirichlet_Test#Dirichlet_Test_for_Integrals#Warning_About_Improper_Integrals&quot; rel=&quot;nofollow&quot;&gt;Dirichlets test&lt;/a&gt;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I'll illustrate the technique on your integral:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, lets substitute $t=x^2$, then $(**) \int_{0}^{\infty} \cos(x^2) \mathrm dx=\int_{0}^{\infty} \frac{\cos(t)}{2 \sqrt t} \mathrm dt$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $f(x)=\frac{1}{2\sqrt x}$ then $f(x)$ is decreasing and $\lim_{x\to\infty}f(x)=0$. Also $cos(x)$ has a bounded anti-derivative. Therefore (**) converges by Dirichlets test.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we observe that $|\frac{\cos(x)}{2 \sqrt x}|\geq \frac{\cos^2(x)}{2 \sqrt x}=\frac{1}{4\sqrt x}+\frac{\cos(2x)}{4\sqrt x}$. By the same arguments above $\int_{0}^{\infty}\frac{\cos(2x)}{4\sqrt x}dx$ converges. But then if we assume that $\int_{0}^{\infty}\frac{\cos^2(x)}{2 \sqrt x}$ converges, we get that $\int_{0}^{\infty}\frac{1}{4\sqrt x}$ converges and that's obviously not true. &lt;/p&gt;&#xA;" OwnerUserId="3237" LastEditorUserId="3237" LastEditorDisplayName="" LastEditDate="2010-12-10T22:46:29.037" LastActivityDate="2010-12-10T22:46:29.037" />
  <row Id="13847" PostTypeId="1" AcceptedAnswerId="13848" CreationDate="2010-12-10T20:31:19.603" Score="6" ViewCount="150" Body="&lt;p&gt;It seems so. $\mathbb{Q}_p(\zeta_p)$ is a $p-1^{th}$ extension of $\mathbb{Q}_p$ which doesn't extend the residue field; and so is $\mathbb{Q}_p(p^{\frac{1}{p-1}})$. However I can't see how to express $\zeta_p$ in $\mathbb{Q}_p(p^{\frac{1}{p-1}})$ or how to express $p^{\frac{1}{p-1}}$ in $\mathbb{Q}_p(\zeta_p)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you help?&lt;/p&gt;&#xA;" OwnerUserId="4619" LastActivityDate="2010-12-11T02:36:12.933" Title="Is $\mathbb{Q}_p(\zeta_p)$ the same as $\mathbb{Q}_p(p^{\frac{1}{p-1}})$?" Tags="&lt;number-theory&gt;&lt;field-theory&gt;&lt;finite-fields&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="3" />
  <row Id="13848" PostTypeId="2" ParentId="13847" CreationDate="2010-12-10T20:45:33.260" Score="10" ViewCount="" Body="&lt;p&gt;The right result is that $\mathbb{Q}_p(\zeta_p) = \mathbb{Q}_p((-p)^{1/(p-1)})$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I will show that $\zeta_p$ is in $\mathbb{Q}_p((-p)^{1/(p-1)})$. Since both fields are of dimension $p-1$ over $\mathbb{Q}_p$ (compute ramification degrees), this shows the fields are equal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $K=\mathbb{Q}_p((-p)^{1/(p-1)})$ and let $\pi$ be a $(p-1)$st root of $-p$ in $K$. We want to show that the equation $x^p=1$ has $p$ roots in $K$. Make the change of variable $x=1+\pi y$. So $\pi^{p} y + p \pi^{p-1} y^{p-1} + \cdots + p \pi y + 1 =1$. &#xA;Dividing out $\pi^{p} = -p \pi$, we get&#xA;$$y^{p} - \pi^{p-1} y^{p-1} - \frac{1}{p} \binom{p}{2} \pi^{p-2} y^{p-2} - \frac{1}{p} \binom{p}{3} \pi^{p-3} y^{p-3} - \cdots - y=0 \quad (*).$$&#xD;&#xA;Notice that $(1/p) \binom{p}{k}$ is an integer for $1 \leq k \leq p-1$. So, when we reduce $(*)$ modulo $\pi$, we get&#xA;$$y^p - y =0.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The equation $y^p-y=0$ has $p$ distinct roots in $\mathbb{F}_p$. So, by &lt;a href=&quot;http://en.wikipedia.org/wiki/Hensel%27s_lemma&quot; rel=&quot;nofollow&quot;&gt;Hensel's lemma&lt;/a&gt;, equation $(*)$ has $p$ roots in $K$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have shown that $K$ contains a nontrivial $p$-th root of unity, as desired.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Here is a direct argument for the other direction, when $p &amp;gt;2$. Let $1-\zeta = \lambda$ (where $\zeta$ is a primitive $p$th root of unity.) We know that&#xA;$$\prod_{j=1}^{p-1} (1-\zeta^j) = p.$$&#xD;&#xA;The left hand side is&#xA;$$\prod_{j=1}^{p-1} (j \lambda - \binom{j}{2} \lambda^2 + \cdots) = (p-1)! \lambda^{p-1} \cdot \prod_{j=1}^{p-1} \cdot \left( 1 - \frac{j}{2} \lambda + \cdots \right) =$$ &#xA;$$(p-1)! \cdot \lambda^{p-1} \cdot  \left( 1 - \frac{p(p-1)}{4} \lambda + \cdots \right)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By &lt;a href=&quot;http://en.wikipedia.org/wiki/Wilson%27s_theorem&quot; rel=&quot;nofollow&quot;&gt;Wilson's theorem&lt;/a&gt;, $(p-1)! \equiv -1 \mod p$. And, since $p$ is odd, $p(p-1)/4 \equiv 0 \mod p$. So we deduce that&#xA;$$p = (-1) \lambda^{p-1} \left( 1+ \mbox{something divisible by} \lambda^2 \right).$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anything of the form $ \left( 1+ \mbox{something divisible by} \lambda^2 \right)$ has a $(p-1)$st root in $\mathbb{Q}_p(\zeta)$, as the Taylor series for $(1+u)^{1/(p-1)}$ will converge for $u$ that small. So we see that &#xA;$$(-p)^{1/(p-1)} = \lambda  \left( 1+ \mbox{something divisible by} \lambda^2 \right)^{1/(p-1)}$$&#xD;&#xA;is in $\mathbb{Q}_p(\zeta)$ as desired.&lt;/p&gt;&#xA;" OwnerUserId="448" LastEditorUserId="448" LastEditorDisplayName="" LastEditDate="2010-12-11T02:36:12.933" LastActivityDate="2010-12-11T02:36:12.933" />
  <row Id="13850" PostTypeId="2" ParentId="13662" CreationDate="2010-12-10T21:07:20.553" Score="5" ViewCount="" Body="&lt;p&gt;This is essentially the same approach as Alex Bartel's, expressed perhaps more simply.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The probability that &lt;em&gt;at least one&lt;/em&gt; dies by age 90 is $1 - (2/3)^4 = 65/81$, and when that happens the probability that it's A who dies first is (by symmetry) a quarter of that, or $65/324$.&lt;/p&gt;&#xA;" OwnerUserId="2682" LastActivityDate="2010-12-10T21:07:20.553" CommentCount="1" />
  <row Id="13851" PostTypeId="2" ParentId="13844" CreationDate="2010-12-10T21:08:56.700" Score="2" ViewCount="" Body="&lt;p&gt;2 and 3) Both of these transformations are correct; you can prove them with the cdf (cumulative distribution function) technique.  I don't see any limitations on using them. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is the derivation for the first transformation using the cdf technique.  The derivation for the other will be similar.  Let $X \sim \beta(\alpha, \beta)$.  Let $Y = \frac{X}{1-X}.$  Then&#xA;$$P\left(Y \leq y\right) = P\left(\frac{X}{1-X} \leq y\right) = P\left(X \leq y(1-X)\right) = P\left(X(1+y) \leq y\right) = P\left(X \leq \frac{y}{1+y}\right)$$&#xD;&#xA;$$= \int_0^{\frac{y}{1+y}} \frac{x^{\alpha - 1} (1-x)^{\beta-1}}{B(\alpha,\beta)} dx.$$&#xD;&#xA;Differentiating both sides of this equation then yields the pdf (probability density function) of $Y$.  We have &#xA;$$f_Y(y)=\frac{\left(\frac{y}{1+y}\right)^{\alpha - 1} \left(1-\frac{y}{1+y}\right)^{\beta-1}}{B(\alpha,\beta)} \frac{d}{dy} \left(\frac{y}{1+y}\right)$$&#xD;&#xA;$$= \frac{\left(\frac{y}{1+y}\right)^{\alpha - 1} \left(\frac{1}{1+y}\right)^{\beta-1}}{B(\alpha,\beta)} \left(\frac{1}{1+y}\right)^2$$&#xD;&#xA;$$= \frac{y^{\alpha - 1} \left(1+y\right)^{-\alpha-\beta}}{B(\alpha,\beta)},$$&#xD;&#xA;which is the pdf of a $\beta&amp;#39;(\alpha,\beta)$ random variable.&lt;/p&gt;&#xA;" OwnerUserId="2370" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-12-10T21:15:13.983" LastActivityDate="2010-12-10T21:15:13.983" CommentCount="2" />
  <row Id="13852" PostTypeId="1" AcceptedAnswerId="13853" CreationDate="2010-12-10T21:19:13.323" Score="1" ViewCount="151" Body="&lt;p&gt;I am Computer science student so pardon me if I am asking this in a wrong place!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My inspiration of learning maths is mainly due to algorithms and its feels to me that without mathematics I am not justifying my computer science degree, I am  trying to work on my rusty math since last two years but until now I am not much confident enough the main problem that I am facing is that I no matter whatever I do I am making silly mistakes, I will be presenting a college level maths entrance exams next year, and am preparing for this for two years; but now there is a feeling that I might lack ingenuity, which is required to be able to crack any competitive exam. So, is there any way to develop mathematical thinking? Does a mathematician usually/(at some part of their life) follow any exercise to concentrate?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would also mention that I am almost alone in my home and I am rather compelled to learn maths all my own. But is this even  possible? I have taken some correspondence classes and I am trying with all my efforts but is it really possible to learn on my own? Please help me!&lt;/p&gt;&#xA;" OwnerUserId="4622" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-10T21:49:28.440" LastActivityDate="2010-12-10T21:58:25.323" Title="Advice: How can one prepare for a maths entrance exam? How can one develop mathematical thinking?" Tags="&lt;soft-question&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="13853" PostTypeId="2" ParentId="13852" CreationDate="2010-12-10T21:30:54.360" Score="6" ViewCount="" Body="&lt;p&gt;Whether or not &quot;ingenuity&quot; is &quot;required to be able to crack&quot; (do well?) the exam you will present depends enormously on the kind of exam it is. Ingenuity is not particularly required, say, in the United States' GRE examination (at least, when I took it, even the Mathematics Subject test). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The ability to concentrate has little to do with mathematics per se; it is a requirement of &lt;strong&gt;any&lt;/strong&gt; serious field of study or endeavor. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The best way to develop &quot;ingenuity&quot; (really, ability to solve certain kinds of problems) is practice. Practice, practice, practice. And then, more practice. The vast majority of problems you will encounter are problems that have something in common with other problems, that can be analyzed with the same kind of ideas, the same kind of methods. So what you &lt;em&gt;really&lt;/em&gt; want is not ways to concentrate, or ways to become &quot;ingenious&quot;, but a big &quot;portfolio&quot; of problems that you know how to solve and that you can refer to when you encounter a new problem. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is few advice better than that given by Polya decades ago in &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/4871878309&quot; rel=&quot;nofollow&quot;&gt;his book &lt;strong&gt;How to Solve It&lt;/strong&gt;&lt;/a&gt;. See some of the reviews there for some of the questions to ask. And always ask: &quot;Is this like something I've seen before? If so, can I try doing things similar to what I did then?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Added:&lt;/em&gt; Yes, it is possible to learn on your own, though it is always much easier to learn with immediate feedback and help from others. From your description of the exam (&quot;only 30 people selected in all of India&quot;) things may be very hard &lt;em&gt;even if&lt;/em&gt; you had immediate feedback and the best tutor money could rent available 24 hours a day at your beck and call. If what you want is reassurance that you are doing things correctly when you are in doubt, there are sites such as this one that can help; try not to get carried away and flood the site with all the questions that have accummulated over the past two years (you'll turn off people), but reasoned questions showing both interest and that you've tried to work them through on your own and are genuinely confused are likely to be well received. None of this, however, has to do with developing &quot;ingenuity&quot;, or the ability to concentrate, or &quot;mathematical thinking&quot;. &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-10T21:58:25.323" LastActivityDate="2010-12-10T21:58:25.323" CommentCount="7" />
  <row Id="13854" PostTypeId="1" CreationDate="2010-12-10T21:32:00.027" Score="2" ViewCount="118" Body="&lt;p&gt;Suppose I know the values of a continuous function on $[a,b]$ in some finite number of points $x_0,x_1 \ldots x_n$. I can form the Lagrange interpolating polynomial, $p$. I am curious if there is any interesting estimate of the expression $|f(x)-p(x)|$ for an arbitrary point $x$ that does not assume $f$ is smooth? (Perhaps it would involve moduli of continuity of higher order.) The textbooks on approximation theory I have seen only treat differentiable case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To clarify what I mean by interesting: I can rather trivially bound $|f(x)-p(x)|$ from above by $\sum |\omega(x-x_i)||L_i(x)|$, where $\omega$ is the modulus of continuity and $L_i$ are the normal Lagrange basis polynomials. I do not view this as interesting, among other things because it does not reduce to the normal Lagrange error term when $f$ is smooth. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose now further that I know that the function is in fact nowhere differentiable and I want to interpolate between the given values of the function. Could one give lower bounds for the error now when using Lagrange interpolation of various orders?   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The background to the question is that I been thinking about if one could interpolate in a table of the Weierstrass function, such as the one given here: &lt;a href=&quot;http://www.ams.org/journals/mcom/1961-15-074/S0025-5718-1961-0122011-X/home.html&quot; rel=&quot;nofollow&quot;&gt;http://www.ams.org/journals/mcom/1961-15-074/S0025-5718-1961-0122011-X/home.html&lt;/a&gt;. (Do not ask me why I have been thinking about that.) &lt;/p&gt;&#xA;" OwnerUserId="4623" LastActivityDate="2010-12-10T21:32:00.027" Title="Error term when Lagrange interpolating continuous non-differentiable functions" Tags="&lt;approximation-theory&gt;" CommentCount="1" />
  <row Id="13855" PostTypeId="2" ParentId="13844" CreationDate="2010-12-10T21:50:36.057" Score="1" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Answer to 1:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is no $c$, $d$ for $Y~\beta^\prime(\alpha,\beta)$ where $Y~\beta(c,d)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is because $Y&amp;gt;0$ but $X\in(0; 1)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Use the $F$ distribution which has support for $Y$ which is defined for positive real numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Answer to 2 (simulation approach):&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;although this is not a mathematical proof, it provides an example of how to sufficiently support the postulate that $X\frac{\beta}{\alpha}\sim F(2*\alpha,2*\beta)$. Feedback appreciated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Choose arbitrary parameter values&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;alpha &amp;lt;- 2&#xA;beta  &amp;lt;- 4&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Create a vector of a $\beta^\prime(\alpha, \beta)$ variate &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;X &amp;lt;- rbeta(1000000,alpha, beta)&#xA;Y1 &amp;lt;- X/(1-X)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Create a vector of an $F(2*\alpha, 2*\beta)$ variate and divide by $\frac{\beta}{\alpha}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Y2 &amp;lt;- rf(1000000, 2*alpha, 2*beta) / (beta / alpha)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Compare the variables at decile intervals across their range.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;testY1 &amp;lt;- quantile(Y1, seq(0.1,0.9,0.1))&#xA;testY2 &amp;lt;- quantile(Y2, seq(0.1,0.9,0.1))&#xA;&#xA;signif(testY1,2) == signif(testY2,2)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Answer to 3:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.math.wm.edu/~leemis/2008amstat.pdf&quot; rel=&quot;nofollow&quot;&gt;Leemis and McQuestion (2008)&lt;/a&gt; is an excellent peer-reviewed reference detailing the inter-relationships among distributions &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Thanks Mike for providing the proper answer to 2 and 3.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Leemis, L.M. and J.T. McQuestion. 2008 Univariate Distribution Relationships. The American Statistician 62(1) 45:53&lt;/p&gt;&#xA;" OwnerUserId="3733" LastEditorUserId="3733" LastEditorDisplayName="" LastEditDate="2010-12-10T23:04:31.480" LastActivityDate="2010-12-10T23:04:31.480" CommentCount="1" />
  <row Id="13856" PostTypeId="2" ParentId="13801" CreationDate="2010-12-10T21:50:36.603" Score="12" ViewCount="" Body="&lt;p&gt;As even the capitalistic &lt;a href=&quot;http://en.wikipedia.org/wiki/Dialectical_materialism%22Wikipedia%22&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt; acknowledges, there are three laws of dialectics:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The law of the unity and conflict of opposites;&lt;/li&gt;&#xA;&lt;li&gt;The law of the passage of quantitative changes into qualitative changes;&lt;/li&gt;&#xA;&lt;li&gt;The law of the negation of the negation.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Thus, from law 1, we see that i and -i are both equal and opposite. From law 2, we learn that $\mathbb C$ is qualitatively different from $\mathbb R$. From law 3, we attain the dynamic equilibrium between i and -i that was so powerfully expounded by the revolutionary hero Évariste Galois.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This at least is how we taught such matters in the USSR.&lt;/p&gt;&#xA;" OwnerUserId="1508" LastActivityDate="2010-12-10T21:50:36.603" CommentCount="2" />
  <row Id="13857" PostTypeId="2" ParentId="13801" CreationDate="2010-12-10T22:12:45.890" Score="15" ViewCount="" Body="&lt;p&gt;$\sqrt{-1}$ is an imprecise notation.  There are several ways of making it precise, some of which involve what comes down to an arbitrary choice of square root of $-1$ and some of which don't.  This is because there are several related ways to construct $\mathbb{C}$ from $\mathbb{R}$:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;As the ring $F = \mathbb{R}[x]/(x^2 + 1)$ (edit: following in the discussion in the comments, perhaps it would be better to say &quot;an algebraic closure of $\mathbb{R}$.&quot;).  The Galois group $\text{Gal}(F/\mathbb{R})$ has order $2$, and complex conjugation $x \mapsto -x$ is its only nontrivial element.  The Galois symmetry here forbids us from distinguishing between $x$ and $-x$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;As the ring $\mathbb{C} = \mathbb{R}[\imath]/(\imath^2 + 1)$, where we fix the choice $\imath$ of generator.  This is what is typically meant by $\mathbb{C}$.  This field has a distinguished element $\imath$ satisfying $\imath^2 = -1$, and this is what we usually mean by $i$.  $\mathbb{C}$ is isomorphic to $F$, but &lt;strong&gt;not canonically so&lt;/strong&gt;, since $\imath$ can be sent to either $x$ or $-x$ and there is no way to choose between these.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;As the ring $R$ of linear endomorphisms $\mathbb{R}^2 \to \mathbb{R}^2$ of non-negative determinant preserving an inner product $\langle \cdot, \cdot \rangle$.  There are two elements of $R$ squaring to $-1$, corresponding to a rotation and its inverse, and there is no way to choose between them unless $\mathbb{R}^2$ is also equipped with an &lt;strong&gt;orientation&lt;/strong&gt;, which is an identification of its exterior square with $\mathbb{R}$.  This choice of orientation corresponds to the difference between the first and the second constructions above.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;But most people will not bother to stop and talk about such subtleties as $F, \mathbb{C}, R$ being non-canonically isomorphic, so they say $\sqrt{-1} = i$ because the truth is (for practical purposes) unnecessarily complicated.&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-12-11T06:42:09.430" LastActivityDate="2010-12-11T06:42:09.430" CommentCount="5" />
  <row Id="13858" PostTypeId="2" ParentId="13838" CreationDate="2010-12-10T22:38:02.397" Score="2" ViewCount="" Body="&lt;p&gt;Both statements are true.&lt;/p&gt;&#xA;" OwnerUserId="3787" LastActivityDate="2010-12-10T22:38:02.397" CommentCount="1" />
  <row Id="13859" PostTypeId="2" ParentId="13793" CreationDate="2010-12-10T23:35:11.630" Score="12" ViewCount="" Body="&lt;p&gt;Since the technique proving the homeomorphism is useful in many other situations, it may be worth adding some details to Carl's answer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose $A,B$ are two countable, dense linear orders without end points. We show that they are isomorphic by building an isomorphism $f:A\to B$. This is done by what we call a &lt;em&gt;back-and-forth&lt;/em&gt; argument. Say that $A=\{a_n\mid n\in{\mathbb N}\}$ and $B=\{b_n\mid n\in{\mathbb N}\}$. We build $f$ by stages. At the end of stage $2n$ we have ensured that $a_n\in{\rm dom}(f)$, and at the end of stage $2n+1$, we have ensured that $b_n\in{\rm ran}(f)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The construction is simple. Begin by picking any $b\in B$ and letting $f(a_0)=b$. This completes stage 0. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then we do stage 1: If $b=b_0$ we are done and go to stage 2. If $b&amp;lt;b_0$, we pick an $a\in A$ and set $f(a)=b_0$. Of course, since $f$ is to be an isomorphism, we better ensure that $a_0&amp;lt;a$. But this is trivial to accomplish, since $A$ has no endpoints. Similarly, if $b_0&amp;lt;b$, then we pick $a$ so that $a&amp;lt;a_0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, at stage $2n$ do the following: If $a_n$ is already in the domain we have built, we are done with this stage. Otherwise, if $a_n$ is larger than all the elements in the domain of $f$ so far, pick an element $c$ of $B$ larger than all the elements in the range of $f$ so far and set $f(a_n)=c$; this is possible since $B$ has no largest element. If $a_n$ is smaller than all elements in the current domain of $f$, pick $c$ in $B$ smaller than all elements in the current range of $f$, and set $f(a_n)=c$. Again, this is possible, since $B$ has no smallest element. Finally, if $a_n$ is between elements of the current domain of $f$, pick $d,e$ in the current domain of $f$ so $d&amp;lt;a_n&amp;lt;e$, $d$ is largest below $a_n$, and $e$ is smallest above $a_n$. Then pick in $B$ some $c$ between $f(d)$ and $f(e)$ and set $f(a_n)=c$. This is possible, since $B$ is dense in itself. This completes this stage.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At stage $2n+1$ we do the same, but now ensuring that $b_n$ is put in the range of $f$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This construction gives us an isomorphism $f$ at the end: Even stages ensure the domain of $f$ is all of $A$, odd stages that the range is all of $B$. The construction is designed so for $\alpha,\beta$ in the domain of $f$, $\alpha&amp;lt;\beta$ iff $f(\alpha)&amp;lt;f(\beta)$. But this is precisely what it means to be an isomorphism.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The method of back-and-forth is very flexible. For example, it shows that any two countable random graphs are isomorphic. There are plenty of applications of this technique.  &lt;/p&gt;&#xA;" OwnerUserId="462" LastEditorUserId="1583" LastEditorDisplayName="" LastEditDate="2011-05-05T21:00:40.137" LastActivityDate="2011-05-05T21:00:40.137" CommentCount="3" />
  <row Id="13860" PostTypeId="2" ParentId="13830" CreationDate="2010-12-10T23:43:06.470" Score="1" ViewCount="" Body="&lt;p&gt;This is really a question about Mathematica, but the short answer is that the code that you're looking at is not correctly handling the boundary and initial conditions.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The PDE in question is the 1-d heat equation on the unit interval,&#xA;$$\theta_t = \theta_{xx}$$&#xD;&#xA;with Neumann boundary conditions $\theta_x(0,t) = -1$ and $\theta_x(1,t) = 0$ and with initial condition $\theta(x,0) = e^{-1000 x}/1000$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is that the default method that Mathematica is using to numerically solve this problem does not like the fact that the boundary conditions are inconsistent with the initial condition in the sense that the initial temperature distribution does not satisfy the Neumann condition at $x=1$.  The problem is compounded if one changes the boundary condition at $x=0$, as then the initial temperature satisfies neither boundary condition.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This issue is addressed &lt;a href=&quot;http://reference.wolfram.com/mathematica/tutorial/NDSolvePDE.html#51169332&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="4229" LastEditorUserId="4229" LastEditorDisplayName="" LastEditDate="2010-12-11T06:23:20.343" LastActivityDate="2010-12-11T06:23:20.343" />
  <row Id="13861" PostTypeId="1" AcceptedAnswerId="16977" CreationDate="2010-12-11T00:04:31.593" Score="7" ViewCount="297" Body="&lt;p&gt;Given a collection of points $P \subset \mathbb R^3$, a crude characterization of the &quot;shape&quot; of $P$ is sometimes given by the principal components. We construct a covariance matrix, e.g., if $P$ is discrete, $C = \displaystyle\sum_{p\in P} (p - \mu)(p - \mu)^\intercal$, where $\mu = \displaystyle\frac1{|P|}\sum_{p\in P}\, p$ is the centre of mass. This defines an ellipsoid whose semi-axis are defined by the unit eigenvectors of $C$, scaled by the associated eigenvalues.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is concerned with the following statement:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The ellipsoid described by the principal components is the best fit&#xA;  ellipsoid for $P$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, I don't know of any author or resource that I can accuse of explicitly making such a claim$^*$.  Anyway, my question is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Is there a natural geometric definition of &quot;best fit ellipsoid&quot; for&#xA;  which the above statement becomes true? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For example, some kind of least squares or other variational characterization of this same ellipsoid is what I am looking for. I would also accept an answer that convinces me that this is the wrong way to be looking at the principal components, but that will be a tough sell.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we do a coordinate translation, so that $v_p = (p - \mu)$, and let&#xA;$\hat{v}_p = \frac{v_p}{\left\|v_p\right\|}$, and look at $C$ as a&#xA;linear transformation which is the sum of the rank one operators in&#xA;this coordinate system, $C = \displaystyle\sum_{p \in P} \left\|v_p\right\|^2&#xD;&#xA;\hat{v}_p\hat{v}_p^\intercal$, then the ellipsoid in question is the&#xA;image of the unit ball. From this characterization I gain some&#xA;intuition as to why this particular ellipsoid is a good one. I am&#xA;looking for a better understanding, preferably from a geometric perspective.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;* Wikipedia comes close to such a claim in the description of the&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Moment_%28mathematics%29&quot; rel=&quot;nofollow&quot;&gt;moments&lt;/a&gt;: &quot;The 'second moment', ... in higher dimensions measures the shape of a cloud of points as it could be fit by an ellipsoid.&quot;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt;&#xA;Although I feel that the observation that the ellipsoid reflects the variance of the  Gaussian distribution that has maximal likelihood to produce $P$ (I haven't rolled up my sleeves and checked), this is not the kind of answer I am looking for. Perhaps I should remove all tags that refer to probability or regression.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I will make the question very specific then. From stuff I've seen elsewhere on the web, I get the feeling that this ellipsoid is different from the one that minimizes the sum of squared distances to the points, but I don't know for sure. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How about this then: the &lt;em&gt;radial distance&lt;/em&gt; from a point $p$ to the ellipsoid is the distance as measured along the line that contains $p$ and $\mu$ (the latter being the origin in our new coordinate system). So let this be my question:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Does the ellipsoid defined above minimize the sum of the squared radial distances?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1728" LastEditorUserId="1728" LastEditorDisplayName="" LastEditDate="2011-01-10T06:28:15.970" LastActivityDate="2011-01-10T13:09:54.847" Title="Best fit ellipsoid" Tags="&lt;geometry&gt;&lt;probability-theory&gt;&lt;optimization&gt;&lt;regression&gt;" AnswerCount="2" CommentCount="5" FavoriteCount="1" />
  <row Id="13862" PostTypeId="2" ParentId="13804" CreationDate="2010-12-11T01:05:04.170" Score="2" ViewCount="" Body="&lt;p&gt;There's a recent book on the subject by Jan Krajicek, &lt;a href=&quot;http://www.karlin.mff.cuni.cz/~krajicek/fdraft.pdf&quot; rel=&quot;nofollow&quot;&gt;draft here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instead of defining the probabilities directly, the truth value should be the corresponding event itself, i.e. the points in which the statement is true. So for a proposition $P(x)$, the truth value is the points in which $x$ is true. The Boolean connectives correspond to set operations: negation to complementation, and to intersection, or to union. Quantifiers correspond to supremum (exists) and infimum (forall). Having defined the truth values, you can calculate the probability at the end. You then get that a statement is tautological iff it holds with probability $1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The preceding has assumed that the probability space is finite; for infinite spaces the picture is more complicated, and you can consult the draft to see how to do it with non-standard models of the naturals (endowed with the Loeb measure, which is an extension of the counting measure).&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-12-11T01:05:04.170" />
  <row Id="13864" PostTypeId="2" ParentId="13841" CreationDate="2010-12-11T01:31:01.330" Score="3" ViewCount="" Body="&lt;p&gt;You can also use &lt;a href=&quot;http://www.lsv.ens-cachan.fr/~gastin/gastex/gastex.html&quot; rel=&quot;nofollow&quot;&gt;GasTeX&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-12-11T01:31:01.330" />
  <row Id="13865" PostTypeId="2" ParentId="13536" CreationDate="2010-12-11T01:31:49.953" Score="0" ViewCount="" Body="&lt;p&gt;This is an interesting optics/physics/math problem.  I believe that your question could be asking for a number of distinct answers:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Scale in image space is determined according to the magnification of the camera: &lt;a href=&quot;http://en.wikipedia.org/wiki/Magnification#Calculating_the_magnification_of_optical_systems&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Magnification#Calculating_the_magnification_of_optical_systems&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Calculate the distance from the 'lens' to the image using the distance from the object to the image, information about the camera, and lens equations - note that complicated optical systems can be represented by a single lens with a defined focal length.  (Also, note that the focal length may vary as the camera is focused.)  This is most easily accomplished for the pinhole camera, for which the distance from the 'lens' (the pinhole) to the image is defined entirely by the placement of a surface behind the pinhole.  More complicated cameras can be translated into a pinhole camera for the purposes of performing a raytrace.  To calculate the 'scale' of features, project each of the vertices through the pinhole and calculate their intersections with the image surface - when using a simplified representation of a camera, this will be a plane a defined distance beyond the pinhole, but a rotated image can be calculated by placing the plane between the object and the pinhole.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Finally, it is possible to image each of the vertices through an optical system individually, taking into account the depth of the scene to account for depth of the image, but typically image sensors are planar and this type of analysis is only useful for depth-of-field type calculations.  I recommend #1 or #2 unless there is a compelling reason to do this.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="4628" LastActivityDate="2010-12-11T01:31:49.953" />
  <row Id="13866" PostTypeId="2" ParentId="9594" CreationDate="2010-12-11T01:37:40.340" Score="2" ViewCount="" Body="&lt;p&gt;There are 7,618,438 diferent positions after six movements according to &lt;a href=&quot;http://cube20.org/&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; site, but they use face movements. By the way they show that the Rubik's cube can be solved in 20 face movements or less. &lt;/p&gt;&#xA;" OwnerUserId="3274" LastEditorUserId="3274" LastEditorDisplayName="" LastEditDate="2010-12-11T02:01:14.983" LastActivityDate="2010-12-11T02:01:14.983" />
  <row Id="13867" PostTypeId="2" ParentId="13625" CreationDate="2010-12-11T01:47:54.680" Score="0" ViewCount="" Body="&lt;p&gt;(disclaimer: I'm not writing anything deep in this answer, but it might be useful for people starting to learn manifolds/topology)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I guess it would be safe to say no, globally euclidean always implies locally euclidean, although of course it all depends on how you define these concepts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A (not very useful) definition could be: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;X is globally euclidean if it is homeomorphic to $\mathbb{R}^n$ for some n.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;X is euclidean at x $\in X$ if there exists an open neighbourhood of x homeomorphic to an open subset of a globally euclidean space.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;X is locally euclidean if it is euclidean at every point $x \in X$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;From this (tautological) definition then it is obvious that&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If X is globally euclidean then X is locally euclidean.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The converse is of course false: take the real line $\mathbb{R}$ and the circle $S^1$, the first is globally euclidean, the second is locally euclidean but not globally. If $S^1$ were globally euclidean then (by a general result of algebraic topology &quot;invariance of dimension&quot;) it would be the one-dimensional euclidean space $\mathbb{R}$. But $\mathbb{R}$ is contractible and $S^1$ is not (or $S^1$ is compact and $\mathbb{R}$ is not) (or if you remove a point from $S^1$ it stays connected but if you remove a point from $\mathbb{R}$ it splits in two).&lt;/p&gt;&#xA;" OwnerUserId="1007" LastActivityDate="2010-12-11T01:47:54.680" />
  <row Id="13868" PostTypeId="2" ParentId="460" CreationDate="2010-12-11T01:58:42.147" Score="0" ViewCount="" Body="&lt;p&gt;Here is a cheap answer which takes the stance that semigroups are inherently interesting :)&#xA;Consider the following definition of a group:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\textbf{Definition}$ A semigroup $S$ is said to be a &lt;em&gt;group&lt;/em&gt; if the following hold:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;There is an $e \in S$ such that $ea=a$ for all $a\in S$&#xD;&lt;/li&gt;&#xA;&lt;li&gt;For each $a \in S$ there is an element $a^{-1} \in S$ with $a^{-1}a=e$&#xD;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;At one point in my life, it seemed natural to ask what happens if we replace axiom 2 with the very similar axiom&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2$^\prime$. For each $a \in S$ there is an element $a^{-1} \in S$ with $aa^{-1} = e$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is a fun exercise to work out some of the consequences that result from this. Here are a few facts about a semigroup $S$ which satisfies $1$ and $2^\prime$:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If $e$ is the unique element of $S$ satisfying axiom 1, then $S$ is a group&lt;/li&gt;&#xA;&lt;li&gt;If $S$ has an identity (in the usual sense) then $S$ is a group&lt;/li&gt;&#xA;&lt;li&gt;The principal left ideal $Sa = \{sa \mid s \in S\}$ is a group for all $a \in S$, and in fact all principal left ideals of $S$ are isomorphic as groups.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;It is not difficult to find examples of such semigroups that are not groups. For example,&#xA;consider the following set of $2\times 2$ matrices (with matrix multiplication as the operation):&#xA;$$\left\{\begin{pmatrix} a &amp;amp; b \\ 0 &amp;amp; 0\end{pmatrix} \mid a,b \in \mathbb{R}, a \neq 0\right\}$$&#xD;&#xA;Or, an example that appears as exercise 30 in section 4 of Fraleigh's abstract algebra text:&#xA;the nonzero real numbers under the operation $\ast$ defined by $a\ast b = |a|b$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Certainly it is debatable whether or not semigroups satisfying axioms 1 and 2$^\prime$ are &quot;interesting&quot; or &quot;natural&quot;. But I guess I think they are. And, I am not the only one (or the first one, by a long shot!) to think this. See&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.ams.org/journals/bull/1944-50-12/S0002-9904-1944-08256-6/S0002-9904-1944-08256-6.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.ams.org/journals/bull/1944-50-12/S0002-9904-1944-08256-6/S0002-9904-1944-08256-6.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(A bit of googling will turn up more results if you are interested.)&lt;/p&gt;&#xA;" OwnerUserId="4630" LastActivityDate="2010-12-11T01:58:42.147" CommunityOwnedDate="2010-12-11T01:58:42.147" />
  <row Id="13869" PostTypeId="2" ParentId="13861" CreationDate="2010-12-11T03:13:59.843" Score="0" ViewCount="" Body="&lt;p&gt;Just fleshing out Rahul's comment: probably this choice of $\mu$ and $\Sigma$ maximizes the likelihood of the points given a multivariate Gaussian distribution $N(\mu,\Sigma)$.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-12-11T03:13:59.843" CommentCount="1" />
  <row Id="13870" PostTypeId="1" AcceptedAnswerId="13872" CreationDate="2010-12-11T04:19:17.997" Score="2" ViewCount="102" Body="&lt;p&gt;A question I recall from Carl Linderholm's &quot;Mathematics Made Difficult&quot;, Chapter 3 Exercise 8.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;A farmer acquires and algebraically closed field by extending his field finitely. &#xA;What can be said about the original field?&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;What can be said? I immediately see two fields with this property, ${\Bbb R}$, and $\overline{\Bbb Q}\cap {\Bbb R}$. Is this it? or is there a better characterization? Do these fields have to contain ${\Bbb Q}$?&lt;/p&gt;&#xA;" OwnerUserId="2809" LastActivityDate="2010-12-11T04:31:48.447" Title="Field extension question" Tags="&lt;group-theory&gt;&lt;field-theory&gt;&lt;galois-theory&gt;" AnswerCount="1" />
  <row Id="13871" PostTypeId="2" ParentId="13838" CreationDate="2010-12-11T04:21:36.920" Score="2" ViewCount="" Body="&lt;p&gt;As Huy says, both statements are true.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, if you are a hammer, all the problems seem nails to you, but since this is a linear differential equation, Laplace transform &lt;a href=&quot;http://math.stackexchange.com/questions/13452/how-do-you-solve-y2y-3y0/13475#13475&quot;&gt;again&lt;/a&gt; gives us the general solution in both, real and complex, cases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;{\cal L}[y&amp;#39;&amp;#39;] + {\cal L}[y&amp;#39;] + {\cal L}[y] = 0 \ \Longleftrightarrow \ (s^2 F(s) -sy(0) -y&amp;#39;(0) ) + (sF(s) -y(0)) + F(s) = 0 \ ,&#xD;&#xA;$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $F(s) = {\cal L}[y]$. Now, put $a= y(0)$ and $b= y&amp;#39;(0)$ and solve this equation for $F(s)$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;F(s) = \dfrac{as + (a+b)}{s^2+s+1} \ .&#xD;&#xA;$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the polynomial $s^2+s+1$ has no real roots, the distinction between the real and complex cases arrives at this point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;In the complex case&lt;/strong&gt;, you write &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;s^2 + s + 1 = (s-\alpha)(s-\beta)  \ ,&#xD;&#xA;$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\alpha = \dfrac{-1+i\sqrt{3}}{2}$ and $\beta = \dfrac{-1-i\sqrt{3}}{2}$ are the roots of your characteristic polynomial. Hence the solution of your differential equation is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;y(t) = A{\cal L}^{-1} \left[\dfrac{1}{s-\alpha} \right] + B {\cal L}^{-1} \left[\dfrac{1}{s-\beta} \right] = A e^{\alpha t} + B e^{\beta t} \ ,&#xD;&#xA;$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $A$ and $B$ are complex numbers, depending on $a , b$, which again I let you the pleasure to compute  :-) .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, the fact is that the solution of your differential equation is the $\mathbb{C}$-linear span of two linearly independent complex functions &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;e^{\alpha t} \qquad \text{and} \qquad e^{\beta t} \ .&#xD;&#xA;$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus a $\mathbb{C}$-vector space of dimension two.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;In the real case&lt;/strong&gt;, you write&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;s^2 + s + 1 = \left(s + \frac{1}{2} \right)^2 + \frac{3}{4}&#xD;&#xA;$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and this time, the solution set of your differential equation is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;y(t) = &#xD;&#xA;A {\cal L}^{-1} \left[ \frac{s + \frac{1}{2}}{ \left( s+ \frac{1}{2} \right)^2+ \frac{3}{4} }  \right] +&#xD;&#xA;B {\cal L}^{-1} \left[ \frac{ \frac{\sqrt{3}}{2}}{ \left( s+ \frac{1}{2} \right)^2+ \frac{3}{4} }  \right] &#xD;&#xA;= A e^{-\frac{t}{2}}\cos \left( \frac{\sqrt{3}}{2}t\right) +&#xD;&#xA;B e^{-\frac{t}{2}}\sin \left( \frac{\sqrt{3}}{2}t\right)  \ ,&#xD;&#xA;$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $A$ and $B$ are real numbers, depending on $a,b$, which... Anyway again: the solution set is the $\mathbb{R}$-linear span of two linearly independent real functions &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;e^{-\frac{t}{2}}\cos \left( \frac{\sqrt{3}}{2} t \right) \qquad \text{and} \qquad e^{-\frac{t}{2}}\sin \left( \frac{\sqrt{3}}{2} t \right)  \ .&#xD;&#xA;$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, a real vector space of dimension two.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Alternatively&lt;/strong&gt;, you could read what &lt;a href=&quot;http://en.wikipedia.org/wiki/Linear_differential_equation#Summary_of_General_Solutions&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt; says about linear differential equations.&lt;/p&gt;&#xA;" OwnerUserId="664" LastActivityDate="2010-12-11T04:21:36.920" />
  <row Id="13872" PostTypeId="2" ParentId="13870" CreationDate="2010-12-11T04:31:48.447" Score="5" ViewCount="" Body="&lt;p&gt;Your question was asked and answered &lt;a href=&quot;http://mathoverflow.net/questions/8756/examples-of-algebraic-closures-of-finite-index&quot; rel=&quot;nofollow&quot;&gt;on MO&lt;/a&gt; before. The examples you give are essentially it, in the sense that the field has to be &lt;a href=&quot;http://en.wikipedia.org/wiki/Real_closed_field&quot; rel=&quot;nofollow&quot;&gt;real closed&lt;/a&gt; and the algebraic closure is obtained by adding a square root of -1. This is the Artin-Schreier theorem.&lt;/p&gt;&#xA;" OwnerUserId="3212" LastActivityDate="2010-12-11T04:31:48.447" CommentCount="1" />
  <row Id="13873" PostTypeId="1" AcceptedAnswerId="13875" CreationDate="2010-12-11T04:53:43.507" Score="1" ViewCount="147" Body="&lt;p&gt;I need help with this recursion question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/njBJi.jpg&quot; alt=&quot;help&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4351" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2011-06-03T20:02:45.290" LastActivityDate="2011-06-03T20:02:45.290" Title="Homework question dealing with recursion" Tags="&lt;homework&gt;&lt;recursive-algorithms&gt;" AnswerCount="2" CommentCount="9" />
  <row Id="13874" PostTypeId="2" ParentId="12362" CreationDate="2010-12-11T04:54:58.200" Score="1" ViewCount="" Body="&lt;p&gt;Recall that $\sqrt{4} = \pm 2$. Since $-4 = (4)(-1)$, we have $\sqrt{-4} = \sqrt{4} \sqrt{-1} = \pm 2 i$, where we define $i = \sqrt{-1}$. Note that these are the two roots of the quadratic $x^{2} + 4 = 0$, which has no real roots (positive or negative) by the Descartes Rule of Signs. This last statement rules out the case that the root is $-2$, which is a common mistake and misconception.&lt;/p&gt;&#xA;" OwnerUserId="2720" LastActivityDate="2010-12-11T04:54:58.200" CommentCount="1" />
  <row Id="13875" PostTypeId="2" ParentId="13873" CreationDate="2010-12-11T04:59:58.960" Score="6" ViewCount="" Body="&lt;p&gt;Don't code it! Instead compute the first 4 or so values by hand and see if you understand what is going on. You don't even have to look at the numbers, you just need to see what you are doing when you evaluate the code. E.g. try to say in words what you do when you follow the code. Pretend that you are explaining the program to a friend: &quot;so this thing first checks whether the second parameter is less than 100...&quot; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you don't see what you are doing, then do some more iterations (by hand) and look at the numbers, you will easily recognise them. If you don't recognise them, please leave a comment here, telling us what the numbers are.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But let me repeat: if you want to practice reading code, try to &lt;em&gt;understand&lt;/em&gt; what you are doing, rather than guess a pattern from the numbers. Therefore, &lt;em&gt;do not code it&lt;/em&gt;!&lt;/p&gt;&#xA;" OwnerUserId="3212" LastActivityDate="2010-12-11T04:59:58.960" CommentCount="26" />
  <row Id="13876" PostTypeId="1" CreationDate="2010-12-11T05:50:04.670" Score="2" ViewCount="270" Body="&lt;p&gt;If $E$ is Lebesgue measurable in $\mathbb{R}^n$ and $I=[a,b]$ how do I show that $E\times I$ is measurable in $\mathbb{R}^{n+1}$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Jonas:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm using $\mu^*(E)=\inf \{ \sum \mathrm{Vol}(I_k) \mid E\subseteq \cup I_k\}$ and for every $\epsilon \gt 0$ there exists an open set $G$ containing $E$ such that $\mu^*(G-E)\lt\epsilon$ ($\mu^*$ is the outer measure).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried using the first definition since I think it would be easier, but I don't know how to make it fit together.&lt;/p&gt;&#xA;" OwnerUserId="2143" LastEditorUserId="1543" LastEditorDisplayName="" LastEditDate="2010-12-17T00:00:11.400" LastActivityDate="2010-12-17T00:00:11.400" Title="Lebesgue Measure of the Cartesian Product" Tags="&lt;measure-theory&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" />
  <row Id="13877" PostTypeId="1" AcceptedAnswerId="13879" CreationDate="2010-12-11T05:58:36.097" Score="2" ViewCount="57" Body="&lt;p&gt;&lt;a href=&quot;http://my.unix-center.net/~martian/math%20album/%E6%95%B0%E5%AD%A6%E4%B8%93%E8%BE%91%202%20-%20%E5%86%B0%E5%B2%9B%E7%81%AB%E5%B1%B1/%E7%9B%B4%E5%B8%83%E7%BD%97%E9%99%80%E7%9A%84%E5%B7%A8%E5%B2%A9%20%28On%20twin%20prime%29.html&quot; rel=&quot;nofollow&quot;&gt;a sequence on twin primes&lt;/a&gt; In the diagram, How do the stripes come from? Can the prime numbers also interfere  like light and wave? When zoom in or zoom out the diagram in Mathematica, the stripes are changing.&lt;/p&gt;&#xA;" OwnerUserId="1498" LastActivityDate="2010-12-11T06:37:18.643" Title="numbers interference" Tags="&lt;sequences-and-series&gt;&lt;prime-numbers&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="13878" PostTypeId="2" ParentId="13877" CreationDate="2010-12-11T06:32:04.913" Score="3" ViewCount="" Body="&lt;p&gt;The stripes come mostly because there are more primes among the small numbers than large.  The density goes as about n/log(n).  Also because people are good at seeing patterns, whether they are present or not (that is, distinguishing true patterns from clumps in randomness.)  &lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-12-11T06:32:04.913" />
  <row Id="13879" PostTypeId="2" ParentId="13877" CreationDate="2010-12-11T06:37:18.643" Score="3" ViewCount="" Body="&lt;p&gt;The fact that the stripes change when you change the zoom level supports the idea that this is a graphical problem with your implementation of Mathematica.  Notice that, for instance, every value between 1 and 20 is attained numerous times for n between 1 and 10000, so there should be no &quot;blank stripe&quot; for $0 \le y \le 20$ in your picture.  It is just a graphical anomaly.  This explains the white stripes.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The stripes that occur as a result of higher concentrations occur where $k$ has a significant number of small prime factors besides 2 and 3, as such a $k$ makes $6pk+1$ and $6pk-1$ more likely to be prime, as they will be relatively prime to a higher density of numbers.  So for instance, you see clumping at $y=35=(5)(7)$, and multiples of 5 generally, to varying degrees.&lt;/p&gt;&#xA;" OwnerUserId="2937" LastActivityDate="2010-12-11T06:37:18.643" />
  <row Id="13880" PostTypeId="1" AcceptedAnswerId="13882" CreationDate="2010-12-11T07:20:21.453" Score="3" ViewCount="259" Body="&lt;p&gt;What does &lt;em&gt;general&lt;/em&gt; in &lt;em&gt;general topology&lt;/em&gt; really refer to? We may use the phrase all the time without thinking about its origin.&lt;/p&gt;&#xA;" OwnerUserId="4594" LastEditorUserId="4594" LastEditorDisplayName="" LastEditDate="2011-05-22T15:52:18.030" LastActivityDate="2011-05-22T15:52:18.030" Title="What does 'general' in 'general topology' really refer to?" Tags="&lt;general-topology&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="13881" PostTypeId="2" ParentId="13876" CreationDate="2010-12-11T07:24:21.627" Score="5" ViewCount="" Body="&lt;p&gt;An equivalent criterion for measurability of a set $E$ is the existence of a $G_\delta$ set $G$ containing $E$ such that $\mu^*(G\setminus E)=0$.  (If you haven't already seen this, you can prove it.)  You can use this along with the fact that $(G_1\times G_2)\setminus(E_1\times E_2)=((G_1\setminus E_1)\times G_2)\cup(G_1\times(G_2\setminus E_2))$ to show that if $E_1$ is measurable in $\mathbb{R^n}$ and $E_2$ is measurable in $\mathbb{R^m}$, then $E_1\times E_2$ is measurable in $\mathbb{R^{n+m}}$.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2010-12-11T07:42:02.083" LastActivityDate="2010-12-11T07:42:02.083" />
  <row Id="13882" PostTypeId="2" ParentId="13880" CreationDate="2010-12-11T07:33:40.323" Score="7" ViewCount="" Body="&lt;p&gt;When I use it I mean to distinguish the subject matter from &quot;specific&quot; applications of topology where there is more structure than just that specified by the axioms for a topological space.  (For example, a metric space, or a manifold or Riemannian manifold, or an algebraic variety, or you name it...)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would add that in general, investigating terms like these for what they &quot;really refer to&quot; is a little misleading, because it sort of presumes that these terms were consciously designed by somebody (or a group of people) with a specific purpose in mind, instead of arrived at over the years sort of by accident, with no precise meaning--- only a general, vague, descriptive intent--- behind them.&lt;/p&gt;&#xA;" OwnerUserId="3840" LastActivityDate="2010-12-11T07:33:40.323" />
  <row Id="13883" PostTypeId="2" ParentId="13880" CreationDate="2010-12-11T07:42:43.327" Score="7" ViewCount="" Body="&lt;p&gt;To distinguish &lt;a href=&quot;http://en.wikipedia.org/wiki/General_topology&quot; rel=&quot;nofollow&quot;&gt;(general) topology&lt;/a&gt; from more specific branches of topology? -&lt;a href=&quot;http://en.wikipedia.org/wiki/Differential_topology&quot; rel=&quot;nofollow&quot;&gt;Differential topology&lt;/a&gt;, &lt;a href=&quot;http://en.wikipedia.org/wiki/Algebraic_topology&quot; rel=&quot;nofollow&quot;&gt;algebraic topology&lt;/a&gt;, &lt;a href=&quot;http://en.wikipedia.org/wiki/Geometric_topology&quot; rel=&quot;nofollow&quot;&gt;geometric topology&lt;/a&gt;, &lt;a href=&quot;http://en.wikipedia.org/wiki/Low-dimensional_topology&quot; rel=&quot;nofollow&quot;&gt;low-dimensional topology&lt;/a&gt;...&lt;/p&gt;&#xA;" OwnerUserId="664" LastEditorUserId="664" LastEditorDisplayName="" LastEditDate="2010-12-11T09:39:12.747" LastActivityDate="2010-12-11T09:39:12.747" />
  <row Id="13884" PostTypeId="1" CreationDate="2010-12-11T07:42:51.283" Score="9" ViewCount="598" Body="&lt;p&gt;I've noticed when sweeping the floor that dust gathers particularly in the corners. I assume there is a fluid mechanics reason for this. Does anyone know what it is?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; No, really, this is a mathematical question. Air blows around the room, which constitutes a vector field. Let's say it blows through a square room with doors near bottom-left and top-right. Air blows from bottom-left to top-right.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are dust particles in the room, let's say uniformly distributed at first. But after application of this flow they are not uniformly distributed. They pile in the corners. Maybe that's because vortices form more in the corners, maybe some other reason.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's &lt;em&gt;not&lt;/em&gt; because of where I start sweeping.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is a physics question, but obviously knowledge of Newtonian mechanics won't solve it. It comes down to fluid flow and vector fields, which is math.&lt;/p&gt;&#xA;" OwnerUserId="1457" LastEditorUserId="1457" LastEditorDisplayName="" LastEditDate="2010-12-11T08:05:01.997" LastActivityDate="2010-12-14T02:30:34.100" Title="Why does dust gather in corners?" Tags="&lt;physics&gt;&lt;applications&gt;&lt;vector-calculus&gt;" AnswerCount="4" CommentCount="12" FavoriteCount="6" />
  <row Id="13885" PostTypeId="1" CreationDate="2010-12-11T08:22:07.070" Score="6" ViewCount="135" Body="&lt;p&gt;I was thinking about Feynman integrals the other day and in particular about discretizing the paths.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anyone know the lay of the land about what happens when you do path integrals over, say, a lattice or graph?&lt;/p&gt;&#xA;" OwnerUserId="1457" LastEditorUserId="1457" LastEditorDisplayName="" LastEditDate="2011-03-19T08:26:00.033" LastActivityDate="2011-03-19T08:49:57.033" Title="Path (Feynman) Integrals over Graphs" Tags="&lt;integral&gt;&lt;graph-theory&gt;&lt;lattices&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="2" />
  <row Id="13886" PostTypeId="1" AcceptedAnswerId="13973" CreationDate="2010-12-11T08:24:40.833" Score="4" ViewCount="69" Body="&lt;p&gt;&lt;a href=&quot;http://my.unix-center.net/~martian/math%20album/%E6%95%B0%E5%AD%A6%E4%B8%93%E8%BE%91%202%20-%20%E5%86%B0%E5%B2%9B%E7%81%AB%E5%B1%B1/%E7%9B%B4%E5%B8%83%E7%BD%97%E9%99%80%E7%9A%84%E5%B7%A8%E5%B2%A9%20%28On%20twin%20prime%29.html&quot; rel=&quot;nofollow&quot;&gt;another sequence on twin primes&lt;/a&gt; The maximal prime factor of average of twin prime pair:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;n = 100000;&#xA;averageList = Select[Prime[Range[n]], PrimeQ[# + 2] &amp;amp;] + 1;&#xA;mpfList = FactorInteger[#][[-1, 1]] &amp;amp; /@ averageList;&#xA;ListPlot[%]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why does this diagram look like a series of trajectories? &#xA;&lt;img src=&quot;http://i.imgur.com/SouNY.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1498" LastActivityDate="2010-12-12T07:39:14.873" Title="trajectories in ListPlot[the maximal prime factor of average of twin prime pair]" Tags="&lt;sequences-and-series&gt;&lt;prime-numbers&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="13887" PostTypeId="2" ParentId="2448" CreationDate="2010-12-11T08:50:59.150" Score="9" ViewCount="" Body="&lt;p&gt;Here is the full question from Halmos's &lt;em&gt;Problems for mathematicians, young and old&lt;/em&gt;:&#xA;&lt;img src=&quot;http://i.imgur.com/6JWk6.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's the hint:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/RhFoy.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And here's the solution:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/Jt39W.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-12-11T08:50:59.150" CommentCount="2" />
  <row Id="13888" PostTypeId="1" AcceptedAnswerId="13894" CreationDate="2010-12-11T09:39:01.593" Score="3" ViewCount="294" Body="&lt;p&gt;&lt;strong&gt;The Problem:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac1{1\cdot 2\cdot 3\cdot 4} + \frac4{3\cdot 4\cdot 5\cdot 6} + \frac9{5\cdot 6\cdot 7\cdot 8} + \frac{16}{7 \cdot 8 \cdot 9 \cdot 10} + \cdots$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any smarter way to solve this may be within a minute or two?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I am adding my solution,If you are a student please don't read-on I find this particular problem to some sort of interesting so just try it once :-)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I started off with trying to find the $T_n$ th term&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$T_n = \frac{n^2}{(2n-1)(2n)(2n+1)(2n+2)}$$&#xD;&#xA;$$ = \frac1{24} \times \biggl( \frac1{2n-1} + \frac3{2n+1} - \frac4{2n+2} \biggr)$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ = \frac1{24} \times \biggl[\biggl( \frac1{2n-1} + \frac1{2n+1}\biggr) + 4 \times \biggl(\frac1{2n+1} - \frac1{2n+2} \biggr) \biggr]$$   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;After this this becomes really easy,&#xA;  $$ S_\infty =   \frac1{24} \times \biggl[( 1 - \frac13 + \frac13 -  \frac15 +  \frac15 + \cdots ) + 4 \times (\ln 2 - \frac12)\biggr]$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ = \frac16 \log_e 2 - \frac1{24}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But as you can see this approach is a bit tedious and it took me some 30 minutes to reduce the $T_n$ to that form. Any other smart approaches?&lt;/p&gt;&#xA;" OwnerUserId="2109" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-11T11:04:41.360" LastActivityDate="2010-12-17T01:23:33.943" Title="How to sum $\frac1{1\cdot 2\cdot 3\cdot 4} + \frac4{3\cdot 4\cdot 5\cdot 6} + \frac9{5\cdot 6\cdot 7\cdot 8} + \cdots$ quickly?" Tags="&lt;sequences-and-series&gt;&lt;algebra-precalculus&gt;" AnswerCount="2" CommentCount="5" FavoriteCount="0" />
  <row Id="13889" PostTypeId="1" AcceptedAnswerId="13906" CreationDate="2010-12-11T09:46:14.097" Score="12" ViewCount="402" Body="&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; Are there intuitive ways to introduce cohomology?  Pretend you're talking to a high school student; how could we use pictures and easy (even trivial!) examples to illustrate cohomology?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Why do I care:&lt;/strong&gt; For a number of math kids I know, doing algebraic topology is fine until we get to homology, and then it begins to get a bit hazy: why does all this quotienting out work, why do we make spaces up from other spaces, how do we define attaching maps, etc, etc.  I try to help my peers do basic homological calculations through a sequence of easy examples (much like the ones Hatcher begins with: taking a circle and showing how &quot;filling it in&quot; with a disk will make the &quot;hole&quot; disappear --- ) and then begin talking about what kinds of axioms would be nice to have in a theory like this.  I have attempted to begin studying co-homology through &quot;From Calculus to Cohomology&quot; and Hatcher's text, but I cannot see the &quot;picture&quot; or imagine easy examples of cohomology to start with.  &lt;/p&gt;&#xA;" OwnerUserId="2959" LastActivityDate="2011-01-21T02:31:22.080" Title="Motivating Cohomology." Tags="&lt;topology&gt;&lt;algebraic-topology&gt;&lt;cohomology&gt;" AnswerCount="3" CommentCount="6" FavoriteCount="3" />
  <row Id="13890" PostTypeId="1" CreationDate="2010-12-11T09:57:00.060" Score="4" ViewCount="176" Body="&lt;p&gt;How to prove that in every power of 3, with natural exponent, the tens digit&#xA;is an even number?&#xA;(For example, 3 ^ 5 = 243, 4 is even.)&lt;/p&gt;&#xA;" OwnerUserId="2938" LastEditorUserId="2938" LastEditorDisplayName="" LastEditDate="2010-12-11T10:02:30.727" LastActivityDate="2010-12-12T23:12:10.840" Title="In every power of 3 the tens digit is an even number" Tags="&lt;number-theory&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="13891" PostTypeId="1" AcceptedAnswerId="13927" CreationDate="2010-12-11T10:02:52.437" Score="6" ViewCount="180" Body="&lt;p&gt;$A$ and $B$ are finite groups.&#xA;$H \leq A \times B$.&#xA;Can we find some $C \leq A$, $D \leq B$ such that $H \cong C \times D$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In case the statement is not true: is it true under further assumptions about A and B, such as solvability, nilpotency, etc? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Special cases I can prove:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;$A$ and $B$ are abelian (following ideas from another discussion: &lt;a href=&quot;http://math.stackexchange.com/questions/10687/g-finite-abelian-a-times-b-embedded-in-g-is-g-c-times-d-such-that-a&quot;&gt;$G$ finite abelian. $A \times B$ embedded in $G$. Is $G=C \times D$ such that $A$ embedded in $C$, $B$ embedded in $D$?&lt;/a&gt;)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$(|A|,|B|)=1$. In this case we even have $H = C \times D$. By using the Chinese remainder theorem for instance.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="3533" LastEditorUserId="5465" LastEditorDisplayName="" LastEditDate="2011-05-17T12:48:57.350" LastActivityDate="2011-05-17T12:48:57.350" Title="Finite groups: $H \leq A \times B$. Is $H \cong C \times D$ for some $C \leq A$, $D \leq B$?" Tags="&lt;group-theory&gt;&lt;finite-groups&gt;" AnswerCount="3" />
  <row Id="13892" PostTypeId="2" ParentId="13890" CreationDate="2010-12-11T10:05:41.607" Score="11" ViewCount="" Body="&lt;p&gt;Use induction on the exponent:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The last digit is either 1, 3, 7 or 9.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If 1 or 3, multiplying a power of 3 will not change the evenness of the second last digit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If 7 or 9, multiplying will result in a carry over of 2, which again does not change the evenness of the second last digit.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-12-11T10:05:41.607" />
  <row Id="13894" PostTypeId="2" ParentId="13888" CreationDate="2010-12-11T10:34:24.020" Score="8" ViewCount="" Body="&lt;p&gt;Problems like these are usually amenable to partial fraction decompositions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From my experience with exam problems like these, once we do the partial fractions (which can be done quite quickly as described below), either we can do some sort of a telescoping sum, or write the resulting as a combination of well known series. In some cases, you can quickly and mechanically apply known estimates for well known series and find the sum. Of course, each problem is different and you might have to get creative.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let us try that approach to your problem.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Step1: Splitting into Partial Fractions&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can try the following which should give you the partial fraction decomposition pretty quickly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Set&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \dfrac{n}{2(2n-1)(2n+1)(2n+2)} = \dfrac{A}{2n-1} + \dfrac{B}{2n+1} + \dfrac{C}{2n+2}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Multiplying by $\displaystyle 2n-1$ and setting $\displaystyle n=1/2$ gives&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle A = \dfrac{1/2}{2 \times 2 \times 3} = \dfrac{1}{24}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Multiply the original by $\displaystyle 2n+1$ and set $\displaystyle n=-1/2$. This gives us $B$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle B = \dfrac{-1/2}{2 \times -2 \times 1} = \dfrac{1}{8}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Multiply the original by $\displaystyle 2n+2$ and set $\displaystyle n = -1$. This gives us&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle C = \dfrac{-1}{ 2 \times -3 \times -1 } = \dfrac{-1}{6}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \dfrac{n}{2(2n-1)(2n+1)(2n+2)} = \dfrac{1}{24}(\dfrac{1}{2n-1} + \dfrac{3}{2n+1} - \dfrac{4}{2n+2})$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Apparently this is due to Heaviside: &lt;a href=&quot;http://en.wikipedia.org/wiki/Heaviside_cover-up_method&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Heaviside_cover-up_method&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Step2: Summing the series&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now to find the sum, you can use the following, which does not require any clever algebraic manipulations and so can be done quite quickly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$H_n = \sum_{j=1}^{n} \dfrac{1}{j} = \log n + \gamma + \mathcal{O}(\dfrac{1}{n})$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This gives us&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{j=1}^{n} \dfrac{1}{2j - 1} = H_{2n} - \dfrac{1}{2} H_n = \log 2 + \frac{\log n}{2} + \gamma/2 + \mathcal{O}(\dfrac{1}{n})$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus your sum to $n$ terms is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\dfrac{1}{24}(\log 2 + \dfrac{\log n}{2} + \gamma/2 + 3\log 2 + 3 \dfrac{\log (n+1)}{2} - 3 + 3 \gamma/2  - 2\log (n+1)  - 2\gamma +2) + \mathcal{O}(\dfrac{1}{n})$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $\log(n+1)  = \log n + \mathcal{O}(\dfrac{1}{n})$ we get that the sum to n terms is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\dfrac{1}{24}(4 \log 2  -1) + \mathcal{O}(\dfrac{1}{n})$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As $\displaystyle n \to \infty$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the limit is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \dfrac{1}{24}(4 \log 2 - 1)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: As &lt;a href=&quot;http://math.stackexchange.com/users/498/j-m&quot;&gt;J.M&lt;/a&gt; points out in the comments below, a very general method similar to the above can be found in Abramowitz and Stegun's book:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The technique you used for summing the&#xA;  series is described in &lt;a href=&quot;http://people.math.sfu.ca/~cbm/aands/page_264.htm&quot; rel=&quot;nofollow&quot;&gt;Abramowitz and&#xA;  Stegun&lt;/a&gt;. (Harmonic numbers and&#xA;  digamma functions are trivially&#xA;  related).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Note1: As &lt;a href=&quot;http://math.stackexchange.com/users/2370/mike-spivey&quot;&gt;Mike&lt;/a&gt; points, out, there is an updated version of the book available here: &lt;a href=&quot;http://dlmf.nist.gov/&quot; rel=&quot;nofollow&quot;&gt;http://dlmf.nist.gov/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-12-17T01:23:33.943" LastActivityDate="2010-12-17T01:23:33.943" CommentCount="13" />
  <row Id="13895" PostTypeId="2" ParentId="13884" CreationDate="2010-12-11T10:37:03.693" Score="0" ViewCount="" Body="&lt;p&gt;If there is a net airflow in any particular direction at floor-level for any reason (doors/windows/radiators etc.), then the dust will tend to accumulate in the corner that is furthest 'down-stream'.&lt;/p&gt;&#xA;" OwnerUserId="1508" LastEditorUserId="1508" LastEditorDisplayName="" LastEditDate="2010-12-11T12:00:33.450" LastActivityDate="2010-12-11T12:00:33.450" CommentCount="1" />
  <row Id="13896" PostTypeId="2" ParentId="13880" CreationDate="2010-12-11T10:54:06.617" Score="2" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/General_topology&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt; says:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In mathematics, general topology or point-set topology is the branch of topology which studies properties of topological spaces and structures defined on them. It is distinct from other branches of topology in that the topological spaces may be very general, and do not have to be at all similar to manifolds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other main branches of topology are algebraic topology, geometric topology, and differential topology. As the name implies, general topology provides the common foundation for these areas.&lt;/p&gt;&#xA;" OwnerUserId="367" LastActivityDate="2010-12-11T10:54:06.617" />
  <row Id="13897" PostTypeId="2" ParentId="13884" CreationDate="2010-12-11T12:13:59.460" Score="9" ViewCount="" Body="&lt;p&gt;Consider the life of a single dust particle that randomly gets kicked up into the air, at which point it floats around for a while before being deposited on the floor again. One (very) simple model of this process might be as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) the probability of the particle getting kicked up in 1 second increases with the airspeed at it's location, and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) once it's in the air, where it falls is random.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then one would expect the particle to spend more time on the ground where the airspeed is low, and less where the airspeed is high, therefore accumulating in low-speed locations. If this model is legit, the question becomes &quot;why does the air move slower in corners?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Empirically, (at least from walking around my apartment) it seems that the airflow is indeed slower in the corners, so this is a good sanity-check.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To really do this justice, you'd have to consider the full Navier-stokes equations, but those are hard, so lets consider a simplified model. Suppose the flow is an incompressible, slow, pressure-driven flow with a few sources and sinks (air conditioner, cracks under the door, etc), and no-flow boundary conditions at the walls. In this case, the pressure field should solve the Laplace equation, which we have some intuition about it as some sort of smooth rubber surface trying to be &lt;a href=&quot;http://www.math.hmc.edu/~jacobsen/sirev-flat-as-possible.pdf&quot; rel=&quot;nofollow&quot;&gt;&quot;as flat as possible&quot;&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The no-flow condition means that the normal derivative of the pressure at a wall is zero. Now consider a corner between one wall along x=0, and another wall along y=0. If we are near enough to the corner, we expect to be within the realm of influence of both walls, so we expect both derivatives dP/dx and dP/dy to be small, since they are zero at the y=0 and x=0 walls respectively. Therefore the pressure gradient is small, so the velocity is small in the corner, as expected.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Summary:&lt;/strong&gt; a simple particle model predicts dust will accumulate where the air is moving slowly, and a simple pressure-driven flow model predicts flow to be slow near corners. Whether these models are legit or not is worthy of debate, but  they do predict the right behavior in this case.&lt;/p&gt;&#xA;" OwnerUserId="3060" LastEditorUserId="3060" LastEditorDisplayName="" LastEditDate="2010-12-11T14:51:39.327" LastActivityDate="2010-12-11T14:51:39.327" />
  <row Id="13898" PostTypeId="1" AcceptedAnswerId="13903" CreationDate="2010-12-11T12:37:56.767" Score="3" ViewCount="332" Body="&lt;p&gt;Here is a problem from Griffith's book Introduction to E&amp;amp;M.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $\theta(x)$ be the step function&#xA;$$\theta = \begin{cases}&#xD;&#xA;0 &amp;amp; x \leq 0 \\&#xD;&#xA;1 &amp;amp; x &amp;gt; 0 &#xD;&#xA;\end{cases}&#xD;&#xA;$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is how to prove $\frac{d\theta}{dx} = \delta(x)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think since the function is discontinous at x = 0, there is no definition of $\frac{d\theta}{dx}$ at the point x = 0 at all. Thus, how could we show the equivalence of a equation if the left part of equation is not defined at the point x = 0?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;thanks.&lt;/p&gt;&#xA;" OwnerUserId="1382" LastEditorUserId="1382" LastEditorDisplayName="" LastEditDate="2010-12-11T14:49:12.323" LastActivityDate="2011-01-06T16:40:44.250" Title="How to prove $\frac{d\theta}{dx} = \delta(x)$?" Tags="&lt;calculus&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="13899" PostTypeId="1" AcceptedAnswerId="13932" CreationDate="2010-12-11T12:45:51.953" Score="2" ViewCount="146" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Hypercube&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt; says, A hypercube is an n-dimensional analog of a square/cube. What is a hypercylinder then? An n-dimensional analog of a cylinder?&#xA;Constant Approximation Algorithm for MST in Resource Constrained Wireless&#xA;Sensor Networks gives a definition,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;The Isolation Property&lt;/strong&gt;. Let c&gt; 0 be a constant. Let E be a set of edges in k−dimensional space, and let e ∈ E be an edge of length l. If it is possible to place a hypercylinder B of radius and height c .l each, such that the axis of B is a subedge of e and B ∩ (E − e)= φ, then e is said to be isolated. If all the edges in E are isolated, then E is said to satisfy the isolation property.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Does it mean, if I have edge sets E and then take a subset e out E, then if I'm able to place, a some sorts of martian object, B in the so called k space along with an element of e, then B cannot contain any elements from E-e ?! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance&lt;/p&gt;&#xA;" OwnerUserId="1869" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-11T12:49:16.453" LastActivityDate="2010-12-12T02:58:53.483" Title="What is a hypercylinder ?" Tags="&lt;geometry&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="13900" PostTypeId="2" ParentId="13889" CreationDate="2010-12-11T12:55:26.090" Score="3" ViewCount="" Body="&lt;p&gt;For simplicial/cellular cohomology, one way to think of it is in terms of dual cell structures: if $a_1,a_2,\dotsc$ are your $k$-simplices or $k$-cells, then they generate the $k$th chain group, and the $k$th cochain group is generated by their duals $\alpha_1,\alpha_2,\dotsc$, where $\alpha_i(a_j)=\delta_{ij}$.  You can then draw a dual cell structure which has an $n-k$-cell for every $k$-cell in your original space, with each cell representing a cochain, and with that cochain sending the chains it intersects to $1$ and the rest to $0$ (and extending linearly).  So if your space is a surface, you put a vertex inside every face, draw an edge between two of those vertices if there's an edge between the corresponding faces, and add a face between a set of edges if the dual edges all intersect in a vertex.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then the homology of the new cell structure is the cohomology of the original structure.  With field coefficients on a manifold, at least.  But it does allow you to visualize, for example, the coboundary map: in the case of our surface, it sends $C^1$ to $C^2$, but $C^2$ &quot;looks 0-dimensional&quot;, and so the coboundary map &quot;looks like a boundary map,&quot; which is something your students are hopefully familiar with.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, if you do it with Platonic polyhedra, you get other Platonic polyhedra.  A cube becomes an octahedron, etc.  Of course, they all have the same (co)homology but they have different (co)chain groups so they're nice trivial examples, and more interesting than just an arbitrary sphere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hatcher goes over this very briefly in the beginning of his chapter on cohomology.  He also gives a thing you can do with $\mathbb{Z}$ coefficients that's similar, though in this case you can run into torsion, so I don't know if it's as good an example.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other thing you can do is just take the no-nonsense algebraic tack.  We like cohomology because sometimes we want maps going the wrong direction.  For example, its ring structure is easier to work with than homology's coalgebra structure (if your kids are familiar with homology, at this point you show them the coalgebra structure induced by the diagonal map and how it's a bitch to work with).  And you get so much information for free just by knowing that certain maps are ring homomorphisms rather than graded abelian group homomorphisms.  I can't think of a good example off the top of my head, but I know there is one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Oh, I was taught de Rham cohomology before I even knew what homology ways.  I think it's pretty easy to understand.  That's another option.&lt;/p&gt;&#xA;" OwnerUserId="2215" LastActivityDate="2010-12-11T12:55:26.090" CommentCount="3" />
  <row Id="13901" PostTypeId="2" ParentId="13891" CreationDate="2010-12-11T13:51:26.913" Score="2" ViewCount="" Body="&lt;p&gt;It's not possible if you want the isomorphism to be compatible with the product structure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Namely, choose $A=B$ and $H = \lbrace (a,a) : a\in A\rbrace$ the &quot;diagonal subgroup&quot;. Clearly, the subgroup $H$ does &lt;em&gt;not&lt;/em&gt; have the form $H = C\times D \subseteq A\times A$.&lt;/p&gt;&#xA;" OwnerUserId="434" LastActivityDate="2010-12-11T13:51:26.913" CommentCount="4" />
  <row Id="13902" PostTypeId="1" AcceptedAnswerId="13952" CreationDate="2010-12-11T14:00:30.140" Score="10" ViewCount="213" Body="&lt;p&gt;Hello community,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the Fourier transformation of a scalar function with respect to one variable might be defined as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\mathcal{F}\left[w\right](\omega )\equiv \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}w(t)e^{-\mathrm{i}\omega t}dt$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In physics, this transformation along with its generalization, the Laplace transformation, has a tremendous importance because of its feature to turn linear partial differential equations into algebraic ones.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, suppose that we have a pseudo-Riemannian manifold $\mathcal{M}$ where $\det{g_{\mu\nu}} = -1$ holds like in special and general relativity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am wondering, what would be the generalization of the Fourier transformation of scalar functions or forms? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The difficulty I have with this question is that a three-dimensional slicing of $\mathcal{M}$ is not unique, so how to take the integral in an invariant form? What will happen to the differntials $dx^{\mu}$, e.g. $dt,dx^i\rightarrow d\omega dx^i$ in some sense?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any insight would be well appreciated.&lt;br&gt;&#xA;Thank you in advance&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Robert&lt;/p&gt;&#xA;" OwnerUserId="2671" LastActivityDate="2010-12-11T20:10:48.430" Title="Does a Fourier transformation on a (pseudo-)Riemannian manifold make sense?" Tags="&lt;differential-geometry&gt;&lt;functional-analysis&gt;&lt;integral-transforms&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="5" />
  <row Id="13903" PostTypeId="2" ParentId="13898" CreationDate="2010-12-11T14:10:18.957" Score="8" ViewCount="" Body="&lt;p&gt;This is a place where physicists and mathematicians would phrase the question differently. A mathematician would say that $d \theta/dx$ is undefined at $0$, and that $\delta$ is not a function. However, as Mariano says, the statement is true &quot;in the sense of distributions&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What does that mean? A distribution is a gadget which takes as input a smooth function $g(x)$, which is zero for $|x|$ sufficiently large, and returns some real number. When we think of an ordinary function as a distribution, that makes we think of $f$ as corresponding to the gadget $F: g \mapsto \int f(x) g(x) dx$. Notice that changing $f$ at a finite number of points leaves $F$ unaltered. From a physical point of view, if $f$ is something like the value of an electric field at a point, we would never know if it had a finite discontinuity at some point, so the gadget $F$ captures everything that is physically measurable about $f$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, how can we see differentiation in terms of distributions? By integration by parts, we have $\int f&amp;#39;(x) g(x) dx = - \int f(x) g&amp;#39;(x)$. So, for any distribution $F$, we define the derivative of $F$ to be the gadget $g \mapsto -F(g&amp;#39;)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, let $F$ correspond to $\theta$, so $F(g) = \int_{- \infty}^0 g(x) dx$. The Dirac delta distribution is $\delta(g) = g(0)$. I leave it to you to show that $F&amp;#39;(g) = \delta(g)$, with the definitions above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd be curious to see how a physicist would answer this question. I suspect that I have acted &lt;a href=&quot;http://thinkexist.com/quotation/mathematicians_are_like_frenchmen-whatever_you/179777.html&quot; rel=&quot;nofollow&quot;&gt;like a Frenchman&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="448" LastEditorUserId="448" LastEditorDisplayName="" LastEditDate="2011-01-06T16:40:44.250" LastActivityDate="2011-01-06T16:40:44.250" CommentCount="4" />
  <row Id="13904" PostTypeId="1" AcceptedAnswerId="13908" CreationDate="2010-12-11T14:11:41.577" Score="1" ViewCount="111" Body="&lt;p&gt;Is it possible to characterize the distributional derivative as some sort of &quot;best linear approximation&quot; of a distribution (&lt;em&gt;a la&lt;/em&gt; the Fréchet/Gâteaux derivatives), if viewed in the appropriate spaces?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, let $f \in D&amp;#39;$, $\phi \in D$, and let $\psi_x \in D$ be a mollified ramp function so that $\psi_x(x,y,z,...)=x$ on the support of $\phi$. What can we say about the quantity:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{f(\phi)-f(\phi+s\psi_x)}{s}-\frac{\partial f}{\partial x}(\phi)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\frac{\partial f}{\partial x}$ is the usual distributional derivative?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Going further, if we relax possibilities for &quot;derivative direction&quot;, so that simply $\psi \in D$, does it even make sense to talk about the quantity &quot;$Df:D'\rightarrow D'$&quot;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$Df(\phi)(\psi):=\lim_{s\rightarrow 0} \frac{f(\phi)-f(\phi+s\psi)}{s}$$&#xD;&#xA;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the space of distributions we don't have a norm, but we still have a vector space structure and a notion of convergence, so it seems that something like this is at least plausible.&lt;/p&gt;&#xA;" OwnerUserId="3060" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-11T14:49:35.773" LastActivityDate="2010-12-11T14:59:49.913" Title="Distributional derivative as linear approximation" Tags="&lt;distribution-theory&gt;" AnswerCount="1" />
  <row Id="13905" PostTypeId="1" AcceptedAnswerId="13934" CreationDate="2010-12-11T14:23:03.347" Score="2" ViewCount="132" Body="&lt;p&gt;Does anybody know how to derive &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\det(\mathbf A)\cdot \det(\mathbf D + \mathbf E \cdot \mathbf A^{-1} \cdot \mathbf B) = \det(\mathbf D)\cdot \det(\mathbf A + \mathbf B \cdot \mathbf D^{-1} \cdot \mathbf E)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(most likely using &lt;a href=&quot;http://en.wikipedia.org/wiki/Sylvester%27s_determinant_theorem&quot; rel=&quot;nofollow&quot;&gt;Sylvester determinant&lt;/a&gt; theorem), where $\mathbf A$,$\mathbf B$,$\mathbf C$,$\mathbf D$,$\mathbf E$ are non-singular matrices?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;UPDATE: corrected &lt;/p&gt;&#xA;&#xA;&lt;p&gt;UPDATE2:&#xA;Thank you all for your answers and suggestions&lt;/p&gt;&#xA;" OwnerUserId="1996" LastEditorUserId="1996" LastEditorDisplayName="" LastEditDate="2010-12-11T18:12:23.710" LastActivityDate="2010-12-19T22:18:11.457" Title="How to derive a certain determinantal identity" Tags="&lt;linear-algebra&gt;&lt;matrices&gt;" AnswerCount="2" CommentCount="5" />
  <row Id="13906" PostTypeId="2" ParentId="13889" CreationDate="2010-12-11T14:31:36.520" Score="7" ViewCount="" Body="&lt;p&gt;If one operates on an open subset $U$ of Euclidean space $\mathbb{R}^n$, then de Rham cohomology falls out of trying to solve some differential equations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The starting observation is that a locally constant smooth function will be constant on connected components, so the dimension of the vector space of locally constant smooth functions is the number of connected components. This space is the 0th de Rham cohomology group.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The next step, as is the key in a lot of topology, is to start throwing paths all over the place: replace the space $U$ with the paths inside it (not required to be loops in this case). So we view a 1-form $\omega$ as a function $P_{\omega}$ on paths via the integration map which sends a path to the integral of the 1-form $\omega$ along it. And again we look for those functions $P_{\omega}$ which are locally constant, meaning they remain unchanged under small deformations of the path leaving the endpoints fixed. We know that if $\omega = df$ for $f$ a smooth function, then $P_{\omega}$ satisfies this property trivially because it sends a path to the difference of the values of $f$ evaluated at the path's endpoints. We consider two locally constant path functions $P_{\omega_1}, P_{\omega_2}$ the same if their difference is $df$ for some smooth function $f$. This space is the first de Rham cohomology group.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And so on. For the $k$th group, we consider locally constant $k$-dimensional integrals modulo trivially constant ones. The precise definition is that $H^k(U)$ is the space of closed $k$-forms modulo exact $k$-forms, where a form is closed if its exterior derivative is zero and it's exact if it's the exterior derivative of a $(k-1)$-form. For instance, the statement that on a space $U$ every closed form is exact now becomes the statement that all the higher cohomology groups are trivial. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To finally phrase this in the standard language of homological algebra, you can define de Rham cohomology as the cohomology of the chain complex $(\Omega^k(U), d)$ where $\Omega^k(U)$ is the set of $k$-forms on $U$ and $d$ is the exterior derivative.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems that this could be explained to anybody who has studied multivariable calculus. It has the advantage that computing some basic examples of cohomology groups requires one to think only about some calculus, nothing hard to visualize.&lt;/p&gt;&#xA;" OwnerUserId="251" LastEditorUserId="251" LastEditorDisplayName="" LastEditDate="2010-12-11T18:57:55.747" LastActivityDate="2010-12-11T18:57:55.747" />
  <row Id="13907" PostTypeId="2" ParentId="13884" CreationDate="2010-12-11T14:41:46.960" Score="3" ViewCount="" Body="&lt;p&gt;Air flow in a room will change direction from time to time.  (Just the simple act of walking through the room in random directions will shift the air flow.)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The dust particle will be carried in the direction of the air current each time.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Eventually the dust particle will hit a wall and can obviously be carried no further in that direction. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;New air currents, in slightly different directions, will then carry the dust particle in the direction of other walls.  Because the dust particle is next to the wall, it will be difficult to create an air current that will move the dust particle in a direction 180 degrees in the opposite direction; therefore, the air current will have a tendency to move in the direction of one of the two perpendicular walls (assuming a square room for example's sake).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The interplay of air currents will continue to push the dust particle back and forth between the two perpendicular walls into it eventually arrives in the corner.  This will happen because the closer the dust particle gets to the corner (as when it is pushed directly against a wall), the surface area available to capture air currents from the general direction of the corner DECREASES, while the sufrace area available to capture air currents from the perpendicular corner INCREASES.  To say another way, the ability to capture an air current that carries the dust particle 180 degrees away from the corner exponentially decreases the closer it gets closer to the corner, while the ability to capture an air current that carries the dust particle 180 degrees towards the corner exponentially increases as it gets closer to the corner.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EVERYTHING in life is a matter of mathematics.  There is not an organic or inorganic process anywhere in the known universe that can not be illustrated by mathematics (of course, there are many processes we do not yet know HOW to quantify mathematically, but that is a limitation on our part; not within the inherent reach of mathematics to explain it).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Okay, that makes for an interesting Saturday morning ... cheers.&lt;/p&gt;&#xA;" OwnerUserId="4639" LastActivityDate="2010-12-11T14:41:46.960" CommentCount="3" />
  <row Id="13908" PostTypeId="2" ParentId="13904" CreationDate="2010-12-11T14:43:17.737" Score="2" ViewCount="" Body="&lt;p&gt;&lt;em&gt;(I use below $f$ for the distribution and $\phi$ for the test function...)&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $\phi\in D&amp;#39;$ and $f\in D$. Let $\tau_x:D\to D$ be translation, so that $\tau_x(f)(y)=f(y+x)$, and let $\tau^t_x:D&amp;#39;\to D&amp;#39;$ be the transpose of $\tau_x$. Then $\frac{\tau_{h}f-f}h\to f&amp;#39;$ in $D$ when $h\to0$, so $$\langle\frac{\tau^t_h\phi-\phi}h,f\rangle=\langle\phi,\frac{\tau_{h}f-f}h\rangle\to\langle\phi,f&amp;#39;\rangle=\langle \phi&amp;#39;,f\rangle$$ as $h\to 0$. Since the topology in $D&amp;#39;$ is that of pointwise convergence on $D$, it follows that $$\lim_{h\to0}\frac{\tau^t_h\phi-\phi}h=\phi&amp;#39;,$$ which is something along the lines of what you wanted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;NB:&lt;/strong&gt; Differentialting the actual map $\phi:D\to\mathbb R$ is not going to produce anything interesting, because the map is linear itself.&lt;/p&gt;&#xA;" OwnerUserId="274" LastEditorUserId="274" LastEditorDisplayName="" LastEditDate="2010-12-11T14:59:49.913" LastActivityDate="2010-12-11T14:59:49.913" CommentCount="1" />
  <row Id="13909" PostTypeId="2" ParentId="13898" CreationDate="2010-12-11T14:54:06.890" Score="3" ViewCount="" Body="&lt;p&gt;By definition, $\delta(x)$ satisfies $$\int_{-\infty}^\infty f(x)\delta(x) dx=f(0)$$ for any continuous function $f$ on $\mathbb{R}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;By definition of derivative in distribution theory,  $$ \int_{-\infty}^\infty f(x)\theta&amp;#39;(x) dx=-\int_{-\infty}^\infty f&amp;#39;(x) \theta(x) dx$$ for any $C^1$ function $f(x)$ which vanishes outside of a bounded interval.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now $$-\int_{-\infty}^\infty f&amp;#39;(x) \theta(x) dx=-\int_0^\infty f&amp;#39;(x) dx= -f(\infty)+f(0)=f(0)=\int_{-\infty}^\infty f(x)\delta(x) dx.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So $\theta&amp;#39;(x)=\delta(x)$. &lt;/p&gt;&#xA;" OwnerUserId="3249" LastActivityDate="2010-12-11T14:54:06.890" CommentCount="2" />
  <row Id="13910" PostTypeId="1" AcceptedAnswerId="13913" CreationDate="2010-12-11T15:09:24.527" Score="2" ViewCount="72" Body="&lt;p&gt;These two appear in my module (without any proof):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{r = 1}^{n} C(n-2,r-2) = 2^{n-1}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{r = 0}^{n-1} C(n-2,r) = 2^{n-2}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the first one when when $r=1 \Rightarrow  C(n-2,1-2) =  C(n-2,-1)$ ?!The same thing for $r=n-1$ in the second one,are they valid if yes,please explain how?&lt;/p&gt;&#xA;" OwnerUserId="2109" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-12-11T17:23:56.513" LastActivityDate="2010-12-11T20:06:58.583" Title="Are these two expressions for sums of binomial coefficients valid?" Tags="&lt;algebra-precalculus&gt;&lt;binomial-coefficients&gt;" AnswerCount="2" CommentCount="8" />
  <row Id="13911" PostTypeId="2" ParentId="13376" CreationDate="2010-12-11T15:11:24.263" Score="2" ViewCount="" Body="&lt;p&gt;Is the equator of the Earth a line or a circle? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, it's a circle of course. But as you know by now, from Willie Wong's answer and from others, it's also a geodesic, which is very much like a straight line: unless you are ready for some serious digging, the equator is the shortest path between any two places on it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your Portal picture is nothing like the Earth, mind you. But in my kitchen I have a cutting board that's a flat rectangle with a thin rectangular hole cut out near one side, as a handle for carrying. The corners are rounded somewhat. Now an ant walking along the top side, if it passed through the handle, would wind up on the bottom side with its momentum reflected. The same if it walked over an edge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine an orthographic projection of my cutting board. (By the way I'm also an engineer. I learned drafing the old-school way, with a T-square.) I draw a top view which is a rectangle with a thin vertical rectangular hole near the right side. To the right I place a side view which is just a vertical rectangle with a couple of hidden lines, the thickness of the cutting board. It's not normal &quot;third-angle&quot; to also draw the bottom view, but I will, so it is a rectange with a thin vertical rectangular hole near the left side.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An ant walking over the right edge would, on the drawing, pass from the top view to the side view to the bottom view with its momentum unchanged. (From a 2-D perspective, the edges are not actually special. The corners are another matter.) If it walked through the handle, it would pass from one rectangle on the top view through an &quot;interspace&quot; - hidden lines on the side view - to the other rectangle on the bottom view, again with its momentum again unchanged. Imagine if we tied a string through around the handle and the right edge of the cutting board. On the drawing, it would be a straight line between the two rectangles passing through the three views. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the area around the handle and right edge of my cutting board is very much like a 2-D version of Portal. 3-D Portal can then be thought of as being played on the hypersurface of a higher dimensional cutting board. (To visualise a higher dimensional cutting board, ponder an ordinary cutting board while enjoying a &quot;herbal cigarette&quot;.)&lt;/p&gt;&#xA;" OwnerUserId="4640" LastEditorUserId="4640" LastEditorDisplayName="" LastEditDate="2010-12-11T19:21:08.693" LastActivityDate="2010-12-11T19:21:08.693" CommentCount="2" />
  <row Id="13912" PostTypeId="1" AcceptedAnswerId="13977" CreationDate="2010-12-11T15:17:16.833" Score="22" ViewCount="685" Body="&lt;p&gt;My algebra background: I've had 2 undergrad semesters of algebra, a reading course in Galois Theory, a graduate course in commutative algebra and one in algebraic geometry, and I've done (most of) MacLane and Birkhoff's Algebra on my own.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is that I feel like I still don't have any idea how to do algebra. I do well in my classes and don't have any problem with most exercises in M&amp;amp;B I do. But my process consists pretty much entirely of fiddling around with symbols until I figure out how to apply theorems I know in a completely straightforward way. I'm able to do exercises from a book, but rarely able to prove theorems in the text on my own. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a complete contrast to how I feel in topology and analysis (not that I really know any topology or analysis), where I have a halfway decent intuition and really think I know why things are true (and moreover, why anyone should care). I'm able to prove theorems. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In topology and analysis, I am able to visualize things pretty directly in a way that I can get insight into how things work. For algebra, I have no picture. I've tried learning about Cayley graphs to visualize groups. I think these are neat, but I have yet to successfully apply any insight from them. I hoped learning about algebraic geometry would help me visualize rings. But the geometry in algebraic geometry is dictated entirely by the algebra. So how can you use geometry to help you with the algebra, when you have to do the algebra first to figure out what the geometry looks like? I don't get it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the question I'm trying to get at is: How do I develop some insight or intuition about algebra? I don't really know what form answers might take; maybe a reading suggestion, or just a general way to look at things. Maybe this isn't a good question, but I'm kind of at the end of my rope with this stuff. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A particular user on MathOverflow said he fell in love with algebra the first time he saw the axioms for a group. When I first saw the axioms for a group, I spent the next year trying to figure out why the heck anybody cared about groups (and frankly still only know this in a detached and academic way). So it's possible the only answer is: I'm barking up the wrong tree; algebra isn't for me and I should move on to something that comes more naturally. &lt;/p&gt;&#xA;" OwnerUserId="566" LastEditorUserId="3431" LastEditorDisplayName="" LastEditDate="2010-12-11T15:19:29.743" LastActivityDate="2010-12-14T07:59:37.877" Title="Intuition in algebra?" Tags="&lt;abstract-algebra&gt;&lt;soft-question&gt;&lt;intuition&gt;" AnswerCount="7" CommentCount="2" FavoriteCount="18" />
  <row Id="13913" PostTypeId="2" ParentId="13910" CreationDate="2010-12-11T15:17:47.063" Score="5" ViewCount="" Body="&lt;p&gt;Hint: Use the binomial expansion for $(1+x)^{n-2}$ and choose an appropriate value for $x$. Binomial coefficients with a negative value of $r$ are zero. You should see what the LHS ought to be. &lt;/p&gt;&#xA;" OwnerUserId="3431" LastActivityDate="2010-12-11T15:17:47.063" CommentCount="3" />
  <row Id="13914" PostTypeId="2" ParentId="13912" CreationDate="2010-12-11T15:21:30.693" Score="2" ViewCount="" Body="&lt;p&gt;It seems you have a kind of geometrical intuition. If that's the case, why don't you construct geometrical representations of the algebraic entities you encounter? Have you had any representation theory till now? If not, that might help you figure out why algebra is &quot;important&quot; or why anyone, even outside the field of algebra, cares.&lt;/p&gt;&#xA;" OwnerUserId="3567" LastActivityDate="2010-12-11T15:21:30.693" />
  <row Id="13915" PostTypeId="1" AcceptedAnswerId="13921" CreationDate="2010-12-11T15:24:59.783" Score="3" ViewCount="131" Body="&lt;p&gt;Let $$a_n=\left(\frac{1+\sqrt{5}}{2}\right)^n.$$&#xD;&#xA;For a real number $r$, denote by $\langle r\rangle$ the fractional part of $r$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is the sequence $$\langle a_n\rangle$$ not equidistributed in $[0,1]$?&lt;/p&gt;&#xA;" OwnerUserId="3249" LastActivityDate="2010-12-11T17:59:33.460" Title="A sequence not equidistributed in [0,1]" Tags="&lt;analysis&gt;&lt;fourier-analysis&gt;&lt;dynamical-systems&gt;&lt;classical-analysis&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="0" />
  <row Id="13916" PostTypeId="2" ParentId="13912" CreationDate="2010-12-11T15:26:24.353" Score="7" ViewCount="" Body="&lt;p&gt;A lot of intuition about abstract algebra and commutative algebra can be developed by visualizing by means of &lt;a href=&quot;http://en.wikipedia.org/wiki/Commutative_diagram&quot; rel=&quot;nofollow&quot;&gt;commutative diagrams&lt;/a&gt;. &lt;/p&gt;&#xA;" OwnerUserId="3431" LastActivityDate="2010-12-11T15:26:24.353" />
  <row Id="13917" PostTypeId="2" ParentId="309" CreationDate="2010-12-11T15:32:59.133" Score="1" ViewCount="" Body="&lt;p&gt;In Top (topological speces) take the subspaces of the real line:  $B=[0,1]$ and $A=[0,1]\cup \{3/4\}$ considering the inclusion $A \subset B$ and imbedding map $f: B\to A: x \mapsto x/2$. Then $A$ isnt isomorphic  (i.e. homeomorphic) to $B$&lt;/p&gt;&#xA;" OwnerUserId="4565" LastActivityDate="2010-12-11T15:32:59.133" CommentCount="1" />
  <row Id="13918" PostTypeId="2" ParentId="13873" CreationDate="2010-12-11T15:50:31.047" Score="1" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\ \ $ Show the sequence of Current values has form $\rm\ 1,\ 1,\ 2,\ \cdots\ a,\ b,\ a+b$&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-12-11T15:50:31.047" />
  <row Id="13919" PostTypeId="2" ParentId="13915" CreationDate="2010-12-11T15:56:03.630" Score="3" ViewCount="" Body="&lt;p&gt;If you just try it, the fractional part becomes very close to 0 or 1 quickly.  This is because $\phi ^2=\phi +1$.  &lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-12-11T15:56:03.630" />
  <row Id="13920" PostTypeId="2" ParentId="13884" CreationDate="2010-12-11T15:56:51.087" Score="3" ViewCount="" Body="&lt;p&gt;First, we must assume a more or less constant airflow in the room, for as everyone knows,in a room that is closed up for a long time with no airflow, dust accumulates evenly throughout the room. The simplest air flow type is random. Like water, or anything tbat flows, air flows through the path of least resistance. For air, the path of least resistance is a straight line. Resistance to flow is proportional to the angular change of direction produced by an obstacle; in this case a corner, which does not neccessarily imply a 90 degree corner. There are two types of corners: inside corners (less than 180 degrees and outside corners (greater than 180 degrees). Outside corners do not gather dust, simply because they do not force the air flow to change its direction. Inside corners, however present a physical barrier to straight-line air flow. The smaller the angle of deflection, the greater the resistance. The air simply avoids the highly flow resistant corner by curving around it. In doing so, it loses some of its energy, slowing its flowrate or speed.  Of course, some of the air particles pass closer to the corner than others. The closer particles have to turn more sharply to avoid the black hole of the corner, thus slowing down more, thus losing more of their ability to hold a dust particle aloft, thus losing more dust particles before returning to the (assumed) random air flow of the room. In a simple rectangular or square room, this action results in dusty corners. This is the simplest and least technical answer I can think of. Such things as  Navier-stokes equations, Laplace equations and pressure derivatives have their place, but are superflous, irrelevant overkill for answering such a simple question. &quot;KISS&quot; always applies. If you want a solution to the dusty corner problem (and seeking the cause of a problem is, after all, the first step in seeking a solution), simply place a contantly running vaccumn inlet in each corner with the (filtered) outlet in the center of the room, directed in all directions, and you will have constantly self-dusting corners. Then, (not so simply) eliminate all dust generators and dust attractors, and replace all objects that have inside corners (or modify them so that they have inside curves instead), and you will have a constantly self-dusting room. However, you should not complain if, upon enterimg this room, you friends declare, &quot;This room really sucks!&quot;. I hope this clears the air (and the corners, of course).&lt;/p&gt;&#xA;" OwnerUserId="4642" LastActivityDate="2010-12-11T15:56:51.087" CommentCount="3" />
  <row Id="13921" PostTypeId="2" ParentId="13915" CreationDate="2010-12-11T16:07:32.803" Score="10" ViewCount="" Body="&lt;p&gt;Let $\phi&amp;#39;=(1-\sqrt 5)/2$ denote the Galois conjugate of the golden mean $\phi$. &#xA;Then $\phi^n+\phi&amp;#39;^{n}$ is an integer for every $n\in\mathbb N$, i.e.&#xA;$$\phi^n+\phi&amp;#39;^{n}\equiv 0\ (\mbox{mod}\ 1).$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But $|\phi&amp;#39;|&amp;lt;1$, so $\phi&amp;#39;^{n}\to 0$. This implies that $\phi^n\to 0\ (\mbox{mod}\ 1)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The property that the sequence  $\{ \mbox{frac} (x^n)\}$ is not equidistributed is shared by other &lt;a href=&quot;http://mathworld.wolfram.com/PisotNumber.html&quot; rel=&quot;nofollow&quot;&gt;Pisot numbers&lt;/a&gt;. There is quite a lot of research publications devoted to them. &lt;/p&gt;&#xA;" OwnerUserId="723" LastEditorUserId="723" LastEditorDisplayName="" LastEditDate="2010-12-11T17:59:33.460" LastActivityDate="2010-12-11T17:59:33.460" />
  <row Id="13922" PostTypeId="1" AcceptedAnswerId="13923" CreationDate="2010-12-11T16:10:58.210" Score="2" ViewCount="120" Body="&lt;p&gt;How to sum up this series :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$2C_o + \frac{2^2}{2}C_1 +  \frac{2^3}{3}C_2 + \cdots +  \frac{2^{n+1}}{n+1}C_n$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any hint that will lead me to the correct solution will be highly appreciated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Here $C_i = ^nC_i $&lt;/p&gt;&#xA;" OwnerUserId="2109" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-12-11T17:24:23.423" LastActivityDate="2010-12-11T19:07:33.640" Title="How to sum up this series?" Tags="&lt;algebra-precalculus&gt;&lt;binomial-coefficients&gt;" AnswerCount="4" CommentCount="3" />
  <row Id="13923" PostTypeId="2" ParentId="13922" CreationDate="2010-12-11T16:19:39.607" Score="5" ViewCount="" Body="&lt;p&gt;Hint: Use binomial expansion for $(1+x)^n$ and integrate once. Then choose an appropriate value for $x$.&lt;/p&gt;&#xA;" OwnerUserId="3431" LastActivityDate="2010-12-11T16:19:39.607" CommentCount="3" />
  <row Id="13924" PostTypeId="2" ParentId="13922" CreationDate="2010-12-11T16:23:32.557" Score="3" ViewCount="" Body="&lt;p&gt;Let's assume $C_i=\binom ni$. I'll give a solution that is not precalculus level. Consider first the equality&#xA;$$ (1+x)^n=C_0+xC_1+x^2C_2+\dots+x^nC_n. $$&#xD;&#xA;This is the &lt;em&gt;binomial theorem&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Integrate from 0 to t. On the left hand side we get $\frac{(1+t)^{n+1}-1}{n+1}$ and on the right hand side $\sum \frac1{i+1}t^{i+1}C_i$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now set $t=2$, and a bit of algebra gives you the answer you want.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pretty sure there is an elementary approach as well. &lt;/p&gt;&#xA;" OwnerUserId="462" LastActivityDate="2010-12-11T16:23:32.557" CommentCount="3" />
  <row Id="13925" PostTypeId="1" CreationDate="2010-12-11T16:55:39.097" Score="3" ViewCount="281" Body="&lt;p&gt;In other words, is their brain state such that its similar to taking a shower? Or is it always in a state of trying new things which mentally exhausts the brain?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, when one drives a car he doesn't have to think about all the mechanics of it. Is this true for mathematicians with regards to mathematics?&lt;/p&gt;&#xA;" OwnerUserId="4644" LastActivityDate="2010-12-11T17:51:03.980" ClosedDate="2010-12-11T20:06:17.490" Title="Do mathematicians think hard?" Tags="&lt;soft-question&gt;" AnswerCount="1" CommentCount="6" />
  <row Id="13926" PostTypeId="2" ParentId="2687" CreationDate="2010-12-11T17:25:19.837" Score="0" ViewCount="" Body="&lt;p&gt;show that a subset of a metric space X is  closed iff its intersection with every compact subset of X is closed&lt;/p&gt;&#xA;" OwnerUserId="4646" LastActivityDate="2010-12-11T17:25:19.837" />
  <row Id="13927" PostTypeId="2" ParentId="13891" CreationDate="2010-12-11T17:28:21.923" Score="10" ViewCount="" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;No, a subgroup of a direct product need not be a direct product of subgroups, even up to isomorphism, even if the factors are almost abelian.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Examples are plentiful.  Here is a small nearly-abelian example (nilpotent and solvable and hamiltonian) and a small easy to write down example:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Take A=B to be quaternion of order 8. A×B has 15 maximal subgroups of order 32, 9 of which are directly indecomposable, and so no such C or D exist.  Hence nilpotent of class at most 2 with all subgroups normal (just barely not-abelian) is not sufficient.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Take A=B to be non-abelian of order 6. A×B has a maximal subgroup H of order 18 that is directly indecomposable, so again no such C or D exist. H can be generated by (1,2,3), (4,5,6) and (1,2)(4,5).  Its elements of order 2 are all of the form (a,b)(c,d) where a≠b in {1,2,3} and c≠d in {4,5,6}, and so all of them are self-centralizing, and so none of them can be part of a proper direct factor (which is centralized by the entire other non-identity factor).  One of the direct factors has to have even order, and so that one has to be all of H.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="583" LastEditorUserId="583" LastEditorDisplayName="" LastEditDate="2010-12-11T18:42:09.083" LastActivityDate="2010-12-11T18:42:09.083" CommentCount="2" />
  <row Id="13928" PostTypeId="2" ParentId="13925" CreationDate="2010-12-11T17:35:11.777" Score="1" ViewCount="" Body="&lt;p&gt;Well, imagine you are driving on a road with a lot of forks (that you only see after you solve a puzzle), and you have a general idea where you're driving at, but only the vaguest idea of how do you get there. That's your brain state, I think. Sometimes the road itself is rough, too, with an occasional pit.&lt;/p&gt;&#xA;" OwnerUserId="3793" LastEditorUserId="3793" LastEditorDisplayName="" LastEditDate="2010-12-11T17:51:03.980" LastActivityDate="2010-12-11T17:51:03.980" CommunityOwnedDate="2010-12-11T17:35:32.150" />
  <row Id="13929" PostTypeId="2" ParentId="13922" CreationDate="2010-12-11T17:41:51.173" Score="6" ViewCount="" Body="&lt;p&gt;For an elementary method which does not use calculus:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice that $\displaystyle \dfrac{{n \choose k}}{k+1} = \dfrac{1}{n+1} {n+1 \choose k+1}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus your sum is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{k=0}^{n} \dfrac{1}{n+1} {n+1 \choose k+1} 2^{k+1} = \dfrac{\sum_{k=0}^{n+1} {n+1 \choose k} 2^k -1}{n+1} = \dfrac{3^{n+1} - 1}{n+1}$$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-12-11T17:41:51.173" CommentCount="4" />
  <row Id="13930" PostTypeId="2" ParentId="1132" CreationDate="2010-12-11T17:47:37.827" Score="1" ViewCount="" Body="&lt;p&gt;In the following, $W_t$ is a Wiener process with increments having mean zero and standard deviation 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically, the Stratonovich formula does the following in terms of the Itô formula:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f(W_t) \circ dW_t = \frac{f(W_t+dW_t)+f(W_t)}{2}dW_t \; ,$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which can be rewritten as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f(W_t) \circ dW_t = \frac{f(W_t+dW_t)-f(W_t)}{2}dW_t + f(W_t)dW_t \; .$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using Taylor expansion on the first term and Itô calculus rules, this can be simplified to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f(W_t) \circ dW_t = \frac{f&amp;#39;(W_t)}{2}dt + f(W_t)dW_t \; .$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, we can replace $f$ with $f&amp;#39;$ in the formula to give&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f&amp;#39;(W_t) \circ dW_t = \frac{f&amp;#39;&amp;#39;(W_t)}{2}dt + f&amp;#39;(W_t)dW_t \; .$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But you can easily check that the right hand side is nothing but $df(W_t)$ according to Itô calculus rules, therefore&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$df(W_t) = f&amp;#39;(W_t) \circ dW_t \; .$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which is just a special case of the chain rule. I guess from here on out you can generalize the argument.&lt;/p&gt;&#xA;" OwnerUserId="3567" LastActivityDate="2010-12-11T17:47:37.827" />
  <row Id="13931" PostTypeId="2" ParentId="13905" CreationDate="2010-12-11T17:48:34.737" Score="5" ViewCount="" Body="&lt;p&gt;A neat way to prove this is by observing that these are two ways to write the determinant of the block matrix&#xA;$$ \pmatrix{ A &amp;amp; B \\ -E &amp;amp; D } $$&#xD;&#xA;(see &lt;a href=&quot;http://en.wikipedia.org/wiki/Determinant#Block_matrices&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Determinant#Block_matrices&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The proof using Sylvester's determinant theorem is done by simply plugging in the right &quot;pairs&quot; of matrices into the right spots.&#xA;That is, given the theorem as&#xA;$$ \det(I_p + XY ) = \det(I_n + YX) $$&#xD;&#xA;Just let $X = EA^{-1}$ and $Y = BD^{-1}$. From there I think you can figure it out on your own.&lt;/p&gt;&#xA;" OwnerUserId="3638" LastEditorUserId="3638" LastEditorDisplayName="" LastEditDate="2010-12-19T22:18:11.457" LastActivityDate="2010-12-19T22:18:11.457" CommentCount="2" />
  <row Id="13932" PostTypeId="2" ParentId="13899" CreationDate="2010-12-11T17:51:15.633" Score="2" ViewCount="" Body="&lt;p&gt;First, in n dimensional geometry, it is common to define an object using it's features in 2 or 3 dimensions (i.e., by writing an n dimensional definition, which,when restricted to the case of $n=2$ or $n=3$ is constant with the 2 and 3 dimensional objects we know).  In the case of a cylinder, this object, in 3D is a circular disk or radius $r$, lying on a plane, that has been extended by a distance, $h$, along a vertical, perpendicular axis.  In n dimensions, we proceed as follows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take an n dimensional plane $a_1x_1 +a_2x_2+...+a_nx_n=c$ (where the x's are variables and all the other letters are constants).  Choose a point,$t$ on this plane to be the centre of the base.  The base will be all those points on the plane whose distance from $t$ is less than or equal to $r$ (where distance is measured using the Euclidean metric).  From linear algebra, we can construct a vector, $v$ which is perpendicular to this plane.  I will refer to the &quot;top&quot; side of the plane as the side in the direction of $v$.  The cylinder is all those $x\in{R^n}$ that are &quot;above&quot; the plane, whose projection onto the plane, along $v$ is a point on the base, and the minimum distance from $x$ to a point on the base is less than or equal to $h$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For you second question, the &quot;Isolation Property&quot; seems like a topological definition.  It sounds very similar to the following:&#xA; &quot;For every $e\in E$ there exists an open neighborhood of $e$ not intersecting any other element of $E$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is an n dimensional way of saying that each edge has a little bit of space around it on all sides (while this definition is clear in 3 dimensions, you need a rigorous description in higher dimensions).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does that help?&lt;/p&gt;&#xA;" OwnerUserId="3971" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-12T02:58:53.483" LastActivityDate="2010-12-12T02:58:53.483" CommentCount="3" />
  <row Id="13933" PostTypeId="2" ParentId="13912" CreationDate="2010-12-11T17:57:45.880" Score="1" ViewCount="" Body="&lt;p&gt;You could try looking at axiomatic planar geometry for a new perspective on rings and fields. Given enough axioms of abstract planar geometry, one can construct a model of that geometry as $R^2$ for some ring $R$ (it's the same thing as happens when we construct vectors out of the affine plane). It turns out that various geometric theorems are then equivalent to algebraic statements about the ring $R$, like that it be commutative, or finite, or a field, etc. For example, if I remember correctly, then Desargue's theorem holds iff the ring $R$ is a division ring. I don't have references for this at hand, but I can look some up at work next week.&lt;/p&gt;&#xA;" OwnerUserId="3225" LastActivityDate="2010-12-11T17:57:45.880" />
  <row Id="13934" PostTypeId="2" ParentId="13905" CreationDate="2010-12-11T18:02:33.177" Score="7" ViewCount="" Body="&lt;p&gt;$$\begin{align}&#xD;&#xA;\det(A)\cdot\det(D+E\cdot A^{-1}\cdot B) &amp;amp; = \det(A)\cdot\det(D)\cdot\det(1+D^{-1}\cdot E\cdot A^{-1}\cdot B) \\&#xD;&#xA;&amp;amp; = \det(A)\cdot\det(D)\cdot\det(B)\cdot\det(B^{-1}+D^{-1}\cdot E\cdot A^{-1}) \\&#xD;&#xA;&amp;amp; = \det(A)\cdot\det(D)\cdot\det(1+B\cdot D^{-1}\cdot E\cdot A^{-1}) \\&#xD;&#xA;&amp;amp; = \det(D)\cdot\det(A+B\cdot D^{-1}\cdot E)&#xD;&#xA;\end{align}$$&lt;/p&gt;&#xA;" OwnerUserId="3567" LastActivityDate="2010-12-11T18:02:33.177" />
  <row Id="13935" PostTypeId="1" AcceptedAnswerId="13966" CreationDate="2010-12-11T18:03:24.703" Score="5" ViewCount="107" Body="&lt;p&gt;For example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What conclusions can be drawn about the relations between two objects with the same group of endomorphism?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can we tell from End(A) if A is Abelian or not?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does End(A) contain information about the sub-objects of A?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any information or references to information about this is highly appreciated.&lt;/p&gt;&#xA;" OwnerUserId="4643" LastActivityDate="2010-12-11T22:24:20.320" Title="What does the endomorphism group of an object tell us about the object in question?" Tags="&lt;abstract-algebra&gt;&lt;group-theory&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="0" />
  <row Id="13936" PostTypeId="2" ParentId="13935" CreationDate="2010-12-11T18:13:24.643" Score="4" ViewCount="" Body="&lt;p&gt;Here is a partial answer (although it uses the ring structure on the set of endomorphisms). A torsion abelian group is cyclic if and only if any two elements in $End(A)$ commute with respect to composition (in other words, $End(A)$ is a commutative ring).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.springerlink.com/content/m72222448q6j7327/&quot; rel=&quot;nofollow&quot;&gt;http://www.springerlink.com/content/m72222448q6j7327/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a whole book on this subject: &lt;a href=&quot;http://books.google.com/books?id=iy4sVSgzrvYC&amp;amp;printsec=frontcover&amp;amp;dq=endomorphism+rings+of+abelian+groups&quot; rel=&quot;nofollow&quot;&gt;Endomorphism rings of abelian groups&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="3431" LastEditorUserId="3431" LastEditorDisplayName="" LastEditDate="2010-12-11T18:31:55.493" LastActivityDate="2010-12-11T18:31:55.493" CommentCount="3" />
  <row Id="13937" PostTypeId="1" AcceptedAnswerId="14319" CreationDate="2010-12-11T18:21:07.583" Score="2" ViewCount="232" Body="&lt;p&gt;Prove that $$\sum_{\substack{x,y \in \mathbb N \\ ax-by \ne 0 }}\frac1{|ax-by|xy}$$ converges, where $a, b \in \mathbb N$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This 2-D problem can be proved by an integral test. I'm looking for some other proofs that can be easily generalized to higher dimensional cases like below,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Prove that $$\sum_{\substack{x_1,x_2, \cdots, x_d \in \mathbb N \\ \bf a \cdot \bf x \ne 0 }}\frac1{|{\bf a} \cdot {\bf x}|x_1 x_2 \cdots x_d}$$ converges, with ${\bf a} \in \mathbb Z^d$, and $\forall \; 1 \le i \le d, a_i \ne 0$.&lt;/p&gt;&#xA;" OwnerUserId="4649" LastEditorUserId="4649" LastEditorDisplayName="" LastEditDate="2010-12-13T14:03:34.303" LastActivityDate="2010-12-15T15:09:53.383" Title="Prove that $\sum_{\substack{x,y \in N \\ ax-by \ne 0 }}\frac{1}{|ax-by|xy}$ (and its generalization to higher dimensions) converges." Tags="&lt;sequences-and-series&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="1" />
  <row Id="13938" PostTypeId="2" ParentId="13888" CreationDate="2010-12-11T18:29:56.433" Score="2" ViewCount="" Body="&lt;p&gt;Again this is trivial employing &lt;strong&gt;telescopy&lt;/strong&gt;. After partial fraction decomposition the &#xA;summand has the form $\rm\ a_1(n-1) - a_1(n) + c\ (a_1(n)+a_0(n))\ $ where $\rm\ a_0(n),\ a_1(n)\ $ are the coefficients of the even, odd part of the  $\rm log\ x\ $ power series, for $\rm\ x = 2\:.\:$ The sum of the first part $\rm\ a_1(n-1) - a_1(n)\ $ telescopes to $\rm\:a(0)\:$ and the sum of second part combines all the odd and even terms into the complete power series for $\rm\: log\ 2\:.\ $  Notice how very simple this approach is. In particular it does not require any knowledge of asymptotics of harmonic series  - as in the approach in Moron's answer.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Generally no ingenuity is required to find such telescopy. It can be done  mechanically as follows.  Let $\rm\:S\:$ be the shift $\rm\: S\ a(n)\: \to\: a(n+1)\:,\: $ and let $\rm\ P(S)\ $ be a polynomial in $\rm\:S\:$ whose coefficients $\rm\:c\:$ are constant w.r.t. $\rm\:n\:,\ $ i.e. $\rm\ S\ c = c\:.\: $ Then $\rm\ \sum\ P(S)\ a(n)\ =\ \sum\ (P(S)-P(1))\ a(n)\ +\ P(1)\: \sum\: a(n)\:.\ $ The first sum telescopes since the Factor Theorem $\rm\: \Rightarrow\ S-1 \:|\: P(S)-P(1)\:.\ $ Therefore the sum computation reduces to that of the simpler $\rm\ \sum\: a(n)\:.$&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-12-11T18:29:56.433" CommentCount="0" />
  <row Id="13940" PostTypeId="2" ParentId="13891" CreationDate="2010-12-11T18:40:09.697" Score="5" ViewCount="" Body="&lt;p&gt;You might be interested in the expository article &quot;&lt;a href=&quot;http://books.google.com/books?id=J-tyGidCI-UC&amp;amp;lpg=PP1&amp;amp;pg=PA1#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;Subgroups of direct products of groups, ideals and subrings of direct products of rings, and Goursat's lemma&lt;/a&gt;&quot; by Anderson and Camillo.  A couple of excerpts:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/vE50x.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/GIVMC.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-12-11T18:40:09.697" CommentCount="1" />
  <row Id="13941" PostTypeId="1" AcceptedAnswerId="13962" CreationDate="2010-12-11T18:58:39.313" Score="3" ViewCount="49" Body="&lt;p&gt;I am reading Andrew Baker's notes on p-adic number theory. I have a question regarding a result he is using. In Lemma 2.19 he says in the third line the &#xA;limit&#xA;$$\lim_{n\to\infty}N(a_n-b)$$&#xD;&#xA;is positive, where $\{a_n\}$ is Cauchy, $N$ is a non-archimedean norm, and $b$ is not the limit of $a_n$ in the $N$-norm. I am not quite sure why this is true. The notes are obtainable &lt;a href=&quot;http://www.maths.gla.ac.uk/~ajb/dvi-ps/padicnotes.pdf&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="3947" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-11T21:36:24.827" LastActivityDate="2010-12-11T21:38:37.213" Title="If ${a_n}$ is Cauchy and $N$ is a nonarchimedean norm, why is the limit of $N(a_n-b)$ positive." Tags="&lt;number-theory&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="13942" PostTypeId="2" ParentId="13922" CreationDate="2010-12-11T19:07:33.640" Score="1" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;REMARK&lt;/strong&gt; $\ $ The various approaches are all equivalent. Namely, suppose that we desire to prove without calculus the identity arising from integrating the binomial formula, viz.  $$\rm (1 + x)^{n+1}\ =\ 1 + \sum_{k=1}^{n+1}\: \frac{n+1}{k+1} {n\choose k}\ x^{k+1}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Comparing coefficients reduces it to the identity&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\rm \quad\quad\  {n+1 \choose k+1}\ =\ \frac{n+1}{k+1} {n\choose k} $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is precisely the identity employed in Moron's &quot;calculus free&quot; approach.&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-12-11T19:07:33.640" CommentCount="6" />
  <row Id="13943" PostTypeId="2" ParentId="13543" CreationDate="2010-12-11T19:16:25.657" Score="0" ViewCount="" Body="&lt;p&gt;The revised question has been answered in this &lt;a href=&quot;http://mathoverflow.net/questions/48761/continuous-or-analytic-functions-with-this-property-of-sinc-function&quot; rel=&quot;nofollow&quot;&gt;post&lt;/a&gt; at MO. In particular, it was shown that $\frac{\sin ax}{ax}$ satisfies this equality for each $0&amp;lt;a\le \pi$.&lt;/p&gt;&#xA;" OwnerUserId="3249" LastEditorUserId="3249" LastEditorDisplayName="" LastEditDate="2010-12-14T02:24:55.977" LastActivityDate="2010-12-14T02:24:55.977" />
  <row Id="13944" PostTypeId="2" ParentId="9999" CreationDate="2010-12-11T19:19:22.690" Score="2" ViewCount="" Body="&lt;p&gt;I can only answer a basic part of it. Namely that Fibonacci numbers have indeed this property. This can be shown by using Lucas numbers. Lucas numbers have the same &#xA;recurrence relation like Fibonacci numbers:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$L_{n+1} = L_n + L_{n-1}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But their initial conditions are $L_0 = 2$ and $L_1 = 1$. As result we get a closed form for Lucas numbers similar to Binet's formula:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$L_n = \alpha^n + \beta^n$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$F_n = {\alpha^n - \beta^n \over \alpha - \beta}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $\alpha$, $\beta$ are the roots of $x^2 - x - 1 = 0$. Now oberve by some basic polynom arithmetic:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(\alpha - \beta)^2 F_n^2 + 4(\alpha^n \beta^n) = L_n^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because $(\alpha - \beta) = \sqrt{5}$ and $\alpha \beta = -1$ we are done. But &#xA;I am stuck with the question whether only Fibonacci numbers have this property.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Credits: J. H. E. Cohn. &quot;On Square Fibonacci Numbers.&quot; J. London Math Soc. 39 (1964):537-40.&lt;/p&gt;&#xA;" OwnerUserId="4414" LastEditorUserId="4414" LastEditorDisplayName="" LastEditDate="2010-12-26T23:51:24.940" LastActivityDate="2010-12-26T23:51:24.940" CommentCount="4" />
  <row Id="13945" PostTypeId="1" AcceptedAnswerId="13950" CreationDate="2010-12-11T19:19:26.350" Score="2" ViewCount="122" Body="&lt;p&gt;I saw to prove that a^2+b^2=c^2 you can produce the algebraic solutions a = x^2-y^2, b = 2xy, c = x^2+y^2. To check that works just multiply it out but it produces infinitely many solutions when you set x and y to be integers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you please show me more examples of this, especially if they show how to find the polynomials!&lt;/p&gt;&#xA;" OwnerUserId="4290" LastActivityDate="2010-12-11T20:08:10.757" Title="Examples of finding algebraic solutions to diophantine equations." Tags="&lt;algebra&gt;&lt;diophantine-equations&gt;" AnswerCount="2" />
  <row Id="13946" PostTypeId="2" ParentId="13890" CreationDate="2010-12-11T19:38:22.357" Score="5" ViewCount="" Body="&lt;p&gt;Moron's answer is very elegant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is an easy but tedious way of dealing with many similar questions: For any $n$ and $k$, the last $k$ digits of the numbers in the sequence $n^1,n^2,n^3,\dots$ eventually repeat, so we only need to check a few cases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the question at hand, note that $3^2=9=10-1$, so $3^{20}=(10-1)^{10}$ has the form $100k+1$ for some $k$. This means that the last 2 digits of powers of 3 repeat each 20 numbers. So we just need to check the last two digits of $3^0,3^1,\dots,3^{19}$, and this can be done very quickly even by hand: $$01, 03, 09, 27, 81, 43, 29, 87,\dots,89,67.$$&lt;/p&gt;&#xA;" OwnerUserId="462" LastActivityDate="2010-12-11T19:38:22.357" />
  <row Id="13947" PostTypeId="1" AcceptedAnswerId="13960" CreationDate="2010-12-11T19:53:28.010" Score="2" ViewCount="80" Body="&lt;p&gt;Let $L \subseteq M$ be A-modules and let $\mathfrak{p}$ be an associated prime of $M/L$ and assume that the annihilator of $L$ is not contained in $\mathfrak{p}$. How would one go about proving that $\mathfrak{p}$ is an associated prime of $M$?&lt;/p&gt;&#xA;" OwnerUserId="3620" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-11T21:07:08.420" LastActivityDate="2010-12-11T21:19:02.183" Title="Proving that $\mathfrak{p} \in Ass(M/L)$ and $Ann(L) \nsubseteq\mathfrak{p}$ implies $\mathfrak{p} \in Ass(M)$" Tags="&lt;commutative-algebra&gt;" AnswerCount="1" />
  <row Id="13948" PostTypeId="2" ParentId="13945" CreationDate="2010-12-11T20:00:46.320" Score="2" ViewCount="" Body="&lt;p&gt;This site: &lt;a href=&quot;http://sites.google.com/site/tpiezas/Home&quot; rel=&quot;nofollow&quot;&gt;A Collection of Algebraic Identities&lt;/a&gt;, seems to have a very good list of such solutions. Unfortunately, it does not seem to have any derivations, but it does give the name of the author in most cases (and sometimes, pdf files), so you should be able to track down the derivation if needed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are looking for a general way to find such solutions, you are out of luck, I think. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Hilbert%27s_tenth_problem&quot; rel=&quot;nofollow&quot;&gt;Hilbert's tenth problem&lt;/a&gt; has been answered in the negative: there is no algorithm which determines if a solution to a general Diophantine equation exists.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope that helps.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-12-11T20:00:46.320" />
  <row Id="13949" PostTypeId="1" AcceptedAnswerId="13969" CreationDate="2010-12-11T20:02:42.387" Score="0" ViewCount="217" Body="&lt;p&gt;If one case is true but the other false, does that mean that the statement is true or false?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There exists a positive integer $n$ such that $n^2-79n+1601$ is composite&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Case 1 - $n$ is even - false.&lt;/li&gt;&#xA;&lt;li&gt;Case 2 - $n$ is odd  - true. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="3691" LastEditorUserId="1543" LastEditorDisplayName="" LastEditDate="2011-05-03T10:10:43.380" LastActivityDate="2011-05-03T10:10:43.380" Title="Proof by cases..." Tags="&lt;elementary-number-theory&gt;&lt;proof-strategy&gt;" AnswerCount="6" CommentCount="3" />
  <row Id="13950" PostTypeId="2" ParentId="13945" CreationDate="2010-12-11T20:03:08.173" Score="4" ViewCount="" Body="&lt;p&gt;Read the section &quot;Rational points on conics&quot; in the motivational chapter in the very beautiful book &lt;a href=&quot;http://books.google.com/books?id=mAJei2-JcE4C&amp;amp;pg=PA9&quot; rel=&quot;nofollow&quot;&gt;Rational points on elliptic curves&lt;/a&gt; by Silverman and Tate and/or google &quot;rational parametrization of the circle&quot;. Also, once you have some knowledge of elliptic curves don't miss Franz Lemmermeyer's &lt;a href=&quot;http://www.fen.bilkent.edu.tr/~franz/publ/conics.pdf&quot; rel=&quot;nofollow&quot;&gt;Conics - a poor man's ellitpic curves&lt;/a&gt; which gives a beautiful treatment of Pell conics unified with theory of elliptic curves. Previously fragments of this theory were folklore scattered in pieces throughout the literature. &lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-12-11T20:08:10.757" LastActivityDate="2010-12-11T20:08:10.757" />
  <row Id="13951" PostTypeId="2" ParentId="13910" CreationDate="2010-12-11T20:06:58.583" Score="3" ViewCount="" Body="&lt;p&gt;In addition to the calculation implied by Timothy's answer, let me describe a way to see it directly.  Consider $\sum_{r=0}^{n-1}C(n-2,r)$.  This is the sum of&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(the number of ways of choosing a subset of $0$ elements out of a set of $n-2$ elements)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;+ (the number of ways of choosing a subset of $1$ element out of a set of $n-2$ elements)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;+ (the number of ways of choosing a subset of $2$ elements out of a set of $n-2$ elements)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;. . . &#xA;+ (the number of ways of choosing a subset of $n-1$ elements out of a set of $n-2$ elements)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since any subset of $n-2$ elements has either 0, 1, 2, . . . , or $n-2$ elements, the total is just the number of ways of choosing &lt;em&gt;any&lt;/em&gt; subset of $n-2$ elements.  And this is just $2^{n-2}$.&lt;/p&gt;&#xA;" OwnerUserId="2682" LastActivityDate="2010-12-11T20:06:58.583" />
  <row Id="13952" PostTypeId="2" ParentId="13902" CreationDate="2010-12-11T20:10:48.430" Score="11" ViewCount="" Body="&lt;p&gt;There are three main ways of interpreting the Fourier transform. &lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Decomposition relative to eigenfunctions of the Laplacian&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;On $\mathbb{R}^n$, the plane waves $E_\xi(x) = \exp( i \xi\cdot x)$ can be interpreted as generalized eigenfunctions of the Laplacian. That is, let &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \triangle = \sum_{i=1}^n \left(\frac{\partial}{\partial x_i}\right)^2 $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;then we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \triangle E_\xi(x) = -|\xi|^2 E_\xi(x) $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice that I say &quot;generalized eigenfunctions&quot;. This is because usually it is preferred to do linear algebra/functional analysis on a complete inner product space (a Hilbert space), and in the case of functions on Euclidean space, it is usually preferred to work over the Hilbert space of square integrable functions. The functions $E_\xi(x)$, however, do not belong in the Hilbert space. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But roughly speaking, the Fourier representation of a function&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ f(x) = \int_{\mathbb{R}^n} \hat{f}(\xi) E_\xi(x) d\xi $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;can be thought of as a decomposition of $f(x)$ in terms of a linear combination of eigenfunctions of the Laplacian. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this form, as Raskolnikov mentioned in the comment, there is a natural generalization to &lt;strong&gt;compact Riemannian manifolds&lt;/strong&gt;. Let $(M,g)$ be a compact Riemannian manifold, and let $\triangle_g$ denote the associated Laplace-Beltrami operator. One has that $\triangle_g$ is a densely defined, unbounded, self-adjoint operator on $L^2(M)$, the space of square integrable functions on the manifold $M$. As it turns out (using a bit of linear partial differential equations theory and a bit of functional analysis), $\triangle_g$ is invertible if we restrict the co-domain to functions of mean 0 on $M$. The operator $\triangle_g^{-1}$ furthermore turns out to be a compact operator, and thus has discrete eigenvalues with finite dimensional eigen-spaces in $L^2$ by the spectral theorem for compact operators. It is clear by definition that the eigenfunctions of $\triangle_g^{-1}$ will be also eigenfunctions for $\triangle_g$ (just change the eigenvalue $\lambda \to \lambda^{-1}$. Using the spectral theorem we can write an arbitrary square integrable function on $M$ as a unique linear combination of the set of eigenfunctions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is one interpretation of the Fourier transform. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can further refine this theory in two directions. For functions (and to a small extent tensors), instead of explicitly decomposing using an eigenbasis of $L^2$, you can study &quot;smooth frequency cut-offs&quot; by considering the heat-flow on the manifold. That is, given a function $f$, consider the solution to the heat equation $u(t,x)$ defined on $\mathbb{R}_+ \times M$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \partial_t u = \triangle_g u, \quad u(0,x) = f(x) $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and study, the cut-off of $f$ by a bump-function $\zeta$ in frequencies in the following sense: take a non-negative function $\zeta\in \C^\infty_0(\mathbb{R}_+)$ and consider the one-parameter family of functions &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ f_\lambda(x) = \int_0^\infty \zeta(\lambda t) u(t,x) dt $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;roughly speaking $f_\lambda(x)$ here corresponds to, in the Euclidean space case, the restriction&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \int_{\frac12 \lambda &amp;lt;|\xi|&amp;lt; 2\lambda} \hat{f}(\xi) E_\xi(x) d\xi $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;of the function to the spatial frequencies that are close, in absolute value, to $\lambda$. This theory is developed more in Eli Stein's book &quot;Topics in Harmonic Analysis related to Littlewood-Paley Theory&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another generalization of this decomposition is applicable to not just functions, but also differential forms. In this generalization, instead of looking at the Laplace-Beltrami operator, we consider the Hodge-Laplace operator defined for differential forms. As it turns out, essentially the same argument as for the scalar case give rise to a decomposition of square-integrable differential forms into three categories, the harmonic forms, the exact forms, and the co-exact forms. This goes under the name of &lt;a href=&quot;http://en.wikipedia.org/wiki/Hodge_theory&quot; rel=&quot;nofollow&quot;&gt;Hodge theory&lt;/a&gt;, and I refer you to Wikipedia, and also S. Agmon's &quot;Lectures on Elliptic Boundary Value Problems&quot;, among many other very good books in this subject. &lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Fourier transform and representation theory&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Another way of treating the Fourier transform on Euclidean space is from the point of view of Euclidean space as an abelian Lie group (in particular a locally compact topological group) which is &quot;self-dual&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Very roughly speaking, for an abelian locally compact topological group $G$, you can consider the set of group homomorphisms from $G\to T$, where $T$ is the one-dimensional circle group. In the case $G = \mathbb{R}^n$, we see that each of the functions $E_\xi(x)$ is one such group homomorphism. This set of group homomorphisms can be made into another topological group, known as the dual group of $G$. In the case of Euclidean space, its dual group has the same structure as another copy of the Euclidean space, hence we say it is self-dual. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, given an invariant measure $\mu$ on $G$, in the sense that group translations in $G$ leaves the measure the same (I am being very rough here and sweeping a lot of stuff under the rug), it turns out that we can define the analogue of the Fourier transform by the same formula&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \int_G f(x) E_{-\xi}(x) d\mu(x) $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\xi$ is a natural parameter of the dual group of $G$. So in particular, this theory is applicable to the case when we start with a Riemannian manifold $(M,g)$ that is either a Lie group or a symmetric space. This notion is widely applicable in representation theory and analytic number theory, among other things. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;See &lt;a href=&quot;http://en.wikipedia.org/wiki/Pontryagin_duality&quot; rel=&quot;nofollow&quot;&gt;this Wikipedia article&lt;/a&gt;, Walter Rudin's &quot;Fourier analysis on groups&quot;, S. Helgason's &quot;Differential geometry, Lie groups, and symmetric spaces&quot;, and the article by Atle Selberg called &quot;Harmonic analysis and discontinuous groups in weakly symmetric riemannian spaces, with applications to Dirichlet series&quot; for some standard references. &lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Symmetries on the manifold and partial Fourier transforms&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;A much more pedestrian way of getting the Fourier transform for functions on manifolds is to just integrate using the usual formula. But as you noted in your question, depending on how you &quot;slice&quot; the manifold, you can get different representations, and whether those representations are meaningful can be questionable. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fortunately, in one special case we can have a meaningful formula. Let $(M,g)$ be a (pseudo)Riemannian manifold (note that this is the only case I am familiar with that has a natural extension to the case of pseudo-Riemannian manifolds), and assume that it admits a continuous symmetry. That is, there exists a nowhere vanishing vector field $T$ such that the Lie derivative $\mathcal{L}_Tg = 0$; in other words $T$ is a Killing vector. Assume further that the integral curves of $T$ are diffeomorphic to $\mathbb{R}$ (so in particular $T$ does not have closed orbits; you can also similarly assume all integral curves are closed, so $T$ does not have open orbits, in this latter case you will deal with Fourier series instead of Fourier transform, I'll leave the obvious changes necessary as an exercise). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then the action of $T$ defines an equivalence relation on $M$, and we can consider the quotient manifold $Q = M / T$. (Here's where we used the condition on the topology of the orbits of $T$. If $T$ has fixed points of orbits of different topologies, $Q$ will in general be a manifold with boundary or an orbifold.) The quotient manifold also inherits a Riemannian structure, but that is not too important right now. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The point is the following, we can write a point $x\in M$ as $(t,q)\in \mathbb{R}\times Q$, where $t$ is a natural coordinate on $T$ so that $\partial_t$ corresponds to $\mathcal{L}_T$ in $M$. Using this splitting, you can take the Fourier transform in $t$ the usual way, writing $f(x) = f(t,q)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \hat{f}(\tau,q) = \int f(t,q) \exp( - it\tau) dt $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That $T$ is a Killing vector is used here so that if you pick a coordinate system adapted to the coordinate $t$, the actual volume form &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \sqrt{|g|} dt dq $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;will be independent of $t$, so when you try to do analysis with this notion of the Fourier transform, the formula above will not look like it is missing a factor coming from the volume form (since anything independent of $t$ can be factored out, and taken outside of the integral, so the worst-case scenario is the entire expression is off be a fixed factor depending on $q$, which can always be inserted back later). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This formula can also be easily generalized to the case where $(M,g)$ admits several mutually commuting Killing vector fields $T_1, T_2, \ldots, T_k$. In the case of $\mathbb{R}^n$, the standard coordinate vector fields form a complete set of $n$ mutually commuting Killing vector fields that span the tangent space at every point, and this allows us to take the Fourier transform &quot;in all directions&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This last formulation is useful especially in studying linear partial differential equations on a space-time that is stationary (for example, a stationary black-hole background). See the papers of Dafermos and Rodnianski for example (&lt;a href=&quot;http://arxiv.org/abs/1010.5132&quot; rel=&quot;nofollow&quot;&gt;paper 1&lt;/a&gt;, &lt;a href=&quot;http://arxiv.org/abs/1010.5137&quot; rel=&quot;nofollow&quot;&gt;paper 2&lt;/a&gt; ).&lt;/p&gt;&#xA;" OwnerUserId="1543" LastActivityDate="2010-12-11T20:10:48.430" CommentCount="5" />
  <row Id="13953" PostTypeId="2" ParentId="13949" CreationDate="2010-12-11T20:15:12.527" Score="4" ViewCount="" Body="&lt;p&gt;The statement you give in your example doesn't really have cases.  It is true if some such integer exists, and false otherwise.  I don't know what your cases are supposed to mean, because there are both odd and even examples showing that the statement is true.  However, you only need to give one example.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for actually proving this, a hint is that you can look for a value of $n$ such that the 3 terms in your expression have a common factor.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2010-12-11T20:20:43.470" LastActivityDate="2010-12-11T20:20:43.470" />
  <row Id="13954" PostTypeId="2" ParentId="13949" CreationDate="2010-12-11T20:15:54.210" Score="4" ViewCount="" Body="&lt;p&gt;The statement is that there exists &lt;strong&gt;some&lt;/strong&gt; positive integer $n$ such that $n^2 - 79n + 1601$ is composite.  You need to find only one $n$ for this to be a true statement.  To prove the statement is false, you would need to show that $n^2 - 79n + 1601$ is prime for all $n$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It turns out the statement is true, but not for all odd $n$.  Take $n = 1$, for example.  $1 - 79 + 1601 = 1523$, which is prime.  I'll leave it to you to find $n$ such that the expression is composite.&lt;/p&gt;&#xA;" OwnerUserId="3537" LastActivityDate="2010-12-11T20:15:54.210" CommentCount="6" />
  <row Id="13955" PostTypeId="2" ParentId="13949" CreationDate="2010-12-11T20:34:07.060" Score="5" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\ $ The polynomial takes many successive prime values because it is &lt;a href=&quot;http://mathworld.wolfram.com/Prime-GeneratingPolynomial.html&quot; rel=&quot;nofollow&quot;&gt;Euler's famous prime generating polynomial&lt;/a&gt; $\rm\ x^2 + x + 41\ $ shifted by $\rm\ x\to x - c\ $ for $\rm\ c = \ldots$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-12-11T20:39:27.270" LastActivityDate="2010-12-11T20:39:27.270" />
  <row Id="13956" PostTypeId="1" AcceptedAnswerId="13957" CreationDate="2010-12-11T20:40:39.543" Score="4" ViewCount="103" Body="&lt;p&gt;Hey there,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need to find the domain of the Gamma function, that is to say all $z \in \mathbb{C}$, for which the integral:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Gamma(z) = \int_0^\infty t^{z-1} e^{-t} \mathrm dt$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;converges. I started by splitting up the integral into an integral running from $0$ to $1$ and another one from $1$ to $\infty$. I first tried to figure out for what $z \in \mathbb{C}$ the integral from $0$ to $1$ converges and I came to the conclusion, that $Re(z) &amp;gt; 0$ is the condition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other integral, I believe, converges for every $z$, as the exponential function dominates the monomial eventually. So I concluded:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\exists \Gamma(z) \iff Re(z) &amp;gt; 0$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I just learnt that this is wrong. I found out that the integral only diverges for non-positive integers. What did I do wrong or what is a better way to find the domain of the Gamma function?&lt;/p&gt;&#xA;" OwnerUserId="3787" LastActivityDate="2010-12-11T21:47:03.427" Title="Domain of the Gamma function" Tags="&lt;analysis&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="13957" PostTypeId="2" ParentId="13956" CreationDate="2010-12-11T20:48:41.113" Score="4" ViewCount="" Body="&lt;p&gt;The integral does converge iff $\Re z &amp;gt; 0$. However, it defines a function that can be analytically extended to the whole complex plane except the non-positive integers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are analogous integral representations for $\Gamma(z)$ which hold true for $\Re z &amp;lt; 0$. For instance, it is not difficult to show that for any $k\in\mathbb N$&#xD;&#xA;$$\Gamma(z)=\int_{0}^{\infty} t^{z-1}\left(e^{-t}-1+t-\frac{t^2}{2!}+\dots+&#xD;&#xA;(-1)^{k+1}\frac{t^k}{k!}\right)dt,$$&#xD;&#xA;where $-k &amp;lt; \Re z &amp;lt; - k+1$ (the Cauchy–Saalschütz integral). &lt;/p&gt;&#xA;" OwnerUserId="723" LastEditorUserId="723" LastEditorDisplayName="" LastEditDate="2010-12-11T21:47:03.427" LastActivityDate="2010-12-11T21:47:03.427" CommentCount="3" />
  <row Id="13958" PostTypeId="2" ParentId="13949" CreationDate="2010-12-11T20:56:27.453" Score="5" ViewCount="" Body="&lt;p&gt;Trevor, you are confusing what is generally thought of as &quot;Proof by cases&quot;, and a search for an example that proceeds by narrowing the possibilities/examining different cases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A &lt;strong&gt;Proof by Cases&lt;/strong&gt; is one that proceeds along the following lines: you want to show that a statement is true for all $x$, but you do so by first breaking up all the possible $x$'s into different &quot;camps&quot;, and then showing the property holds in each &quot;camp&quot;. For instance, take&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;For all real numbers $x$, $|x|\geq 0$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now, since the absolute value is defined as&#xA;$$|x|=\left\{\begin{array}{ll}&#xA;x &amp;amp; \mbox{if $x\gt 0$,}\\&#xA;-x &amp;amp; \mbox{if $x\leq 0$.}\\&#xA;\end{array}\right.$$&#xA;then we can proceed in the proof &lt;em&gt;by cases&lt;/em&gt;: every real number $x$ is &lt;em&gt;either&lt;/em&gt; positive, or less than or equal to zero. We examine each possibility separately, and if the conclusion holds in each of them, we'll know that the conclusion always holds (since every real number is in one of the two camps, and everyone in each of the camps has the desired property). If $x\gt 0$. then $|x|=x\gt 0$, and if $|x|\gt 0$, then $|x|\geq 0$ is also true. So the conclusion holds for the gang of positive numbers. If $x\leq 0$, then $|x|=-x$, and since $x\leq 0$, then $-x\geq 0$, so $|x|\geq 0$. Thus, the conclusion &lt;em&gt;also&lt;/em&gt; holds for elements of the &quot;other&quot; gang, the nonpositives.  Since every real number belongs to either the Positives or the Nonpositives, and every positive has the desired property and every nonpositive has the desired property, then every real number has the desired property.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's a &lt;strong&gt;Proof by cases.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can also do a &quot;proof by cases&quot; to show that something is &lt;em&gt;false&lt;/em&gt;; but in that case, you are usually trying to negate an &lt;em&gt;existential&lt;/em&gt; statement, such as &quot;There exists an integer $n$ such that $2n=1$.&quot; You could say: &quot;every integer is either even or odd; if $n$ is even, then $2n$ is a multiple of an even number $n$, so it is even, hence not equal to $1$. If $n$ is odd, then $2n$ is a multiple of $2$, which is even, so $2n$ is even, hence not equal to $1$.  Since no even number has the property, no odd number has the property, and every integer is either even or odd, then no integer has the property; the statement is false&quot;  (It's a bit convoluted, and you don't &lt;em&gt;need&lt;/em&gt; to do it by cases, but hopefully you see the idea). In fact, you are really proving the negation, &quot;For all integers $n$, $2n\neq 1$,&quot; so it is the same as the previous example.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By contrast, what you have here is an &lt;strong&gt;existential&lt;/strong&gt; statement: it says &quot;there is an integer $n$ with a property.&quot; In order to prove that the statement is &lt;em&gt;false&lt;/em&gt;, you could proceed by cases as above; but in order to prove that the statement is &lt;em&gt;true&lt;/em&gt;, what you need to do is produce a single integer $n$ for which the condition holds. Your answer, if the statement is true, should be of the form: &quot;Take $n=xx$; this integer has the property&quot;, followed by the computations to justify the assertion that it has the property (in this case, you would plug it into $n^2 -79n + 1601$, compute the answer, and then factor the answer to show it is composite.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you search for an $n$ for which the property will hold, you don't want to examine every possible integer until you hit one that works; because you may run the risk that it takes a long time to hit the first example (or maybe there isn't  one? Just because you haven't found one after searching for a long time, you don't know if that's because there isn't an example, or because there &lt;em&gt;is&lt;/em&gt; an example but is much bigger than what you've checked so far). So you are certainly smart to consider different possibilities for $n$ to see how they behave; e.g.,  if $n$ is even, then the expression does not have any obvious factors; if $n$ is a multiple of $3$, does it have any obvious factors? And so on. But that's not a &quot;proof by cases&quot;, because in the end your proof will be of the form: &quot;Here is an $n$ with the property&quot;, and not a case-by-case analysis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Did that make sense and help? Feel free to ask questions, I can add to the answer to clarify or give other examples.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-12-11T20:56:27.453" />
  <row Id="13959" PostTypeId="1" AcceptedAnswerId="13972" CreationDate="2010-12-11T20:59:07.650" Score="19" ViewCount="530" Body="&lt;p&gt;I got this question as a mathematical puzzle from a friend. It looks similar to this one &lt;a href=&quot;http://mathoverflow.net/questions/2014/if-you-break-a-stick-at-two-points-chosen-uniformly-the-probability-the-three-re&quot; rel=&quot;nofollow&quot;&gt;http://mathoverflow.net/questions/2014/if-you-break-a-stick-at-two-points-chosen-uniformly-the-probability-the-three-re&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, in this case, I have to find the expected length of the smallest segment. The two points where the rope is cut are selected uniformly at random. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried simulating it and I got an average value of 0.1114. I suspect the answer is 1/9 but I don't have any rigorous math to back it up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do I solve this problem?&lt;/p&gt;&#xA;" OwnerUserId="2641" LastActivityDate="2011-01-19T21:23:07.460" Title="If a 1 meter rope is cut at two uniformly randomly chosen points (to give three pieces), what is the average length of the smallest piece?" Tags="&lt;probability&gt;" AnswerCount="6" CommentCount="3" FavoriteCount="4" />
  <row Id="13960" PostTypeId="2" ParentId="13947" CreationDate="2010-12-11T21:10:30.253" Score="4" ViewCount="" Body="&lt;p&gt;Recall that $\mathfrak p$ being an associated prime of $M/L$ precisely means that&#xA;$\mathfrak p$ is a prime ideal which is the annihilator of some element of $M/L$.&#xA;If we think what this means in terms of the module $M$, it means that we can&#xA;find an element $m \in M$ such that $\mathfrak p = \{a \in A \,| \, a m \in L\}.$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now you want to replace $m$ by some other element $m&amp;#39; \in M$ with the property&#xA;that $\mathfrak p = \{ a\in A \, | \, a m&amp;#39; = 0\},$ i.e. such that $\mathfrak p$ is the annihilator of $m&amp;#39;$.  How do we find $m&amp;#39;$?  Well, what we have to work with&#xA;is the fact that $\mathfrak p$ does not contain the annihilator of $L$, so&#xA;we may find $a \in A \setminus \mathfrak p$ such that $a L = 0.$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do you see how to use $a$ and $m$ together to find the required $m&amp;#39;$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S.  From the wording of the question I wasn't sure if you wanted a hint or if you needed the complete solution, so to be on the safe side I'm giving a hint.&lt;/p&gt;&#xA;" OwnerUserId="221" LastEditorUserId="221" LastEditorDisplayName="" LastEditDate="2010-12-11T21:19:02.183" LastActivityDate="2010-12-11T21:19:02.183" CommentCount="3" />
  <row Id="13961" PostTypeId="2" ParentId="13959" CreationDate="2010-12-11T21:30:21.653" Score="2" ViewCount="" Body="&lt;p&gt;If you think of this as uniformly randomly choosing points from $[0,1]\times[0,1]$ and break the square into the triangles sharing the diagonal $y=x$, the expected value can be expressed as twice the integral of the function $f(x,y)=\min(x,y-x,1-y)$ over the region $0\leq x\leq y\leq 1$.  You can use &lt;a href=&quot;http://www.missingbrace.com/2009/03/07/mathematical-formulas-for-max-and-min/&quot; rel=&quot;nofollow&quot;&gt;explicit formulas&lt;/a&gt; for the minimum value of 2 numbers to compute the integral.  There is probably a lot more simplifying that can be done.  If this seems too tedious, it is at least something you could put into Mathematica to find the answer.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-12-11T21:30:21.653" />
  <row Id="13962" PostTypeId="2" ParentId="13941" CreationDate="2010-12-11T21:38:37.213" Score="4" ViewCount="" Body="&lt;p&gt;I think you are missing the fact that $b$ is not the $N$-limit of the $a_n$, which you did not include in your original post (I just added it).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By definition, $N(a_n-b)$ is greater than or equal to $0$ (it is a norm). Since you are also assuming that $b\neq{\lim\limits_{n\to\infty}}^{(N)}a_n$, that means that $N(a_n-b)\not\to 0$ as $n\to\infty$; that is, the distance from $a_n$ to $b$ is bounded away from $0$; and since the terms are positive and bounded away from zero, the limit (which exists since we are dealing with a Cauchy sequence) must be strictly positive.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-12-11T21:38:37.213" CommentCount="1" />
  <row Id="13963" PostTypeId="2" ParentId="13959" CreationDate="2010-12-11T21:39:17.707" Score="9" ViewCount="" Body="&lt;p&gt;If we divide the rope into $n$ segments then the $k$th largest segment will have expected length $$\frac{1}{n} \left( \frac{1}{n} + \cdots + \frac{1}{k} \right),$$ but I don't remember why. In particular, the smallest segment has expected length $1/n^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that equivalently we could throw $n$ random points on a circle of unit circumference and consider lengths of segments on the circle.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Derivation of the general formula from the formula for the smallest segment&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Len $n$ points be thrown uniformly on the circle. Suppose the smallest segment has length $x$. We can remove an initial prefix of length $x$ from each segment to get a uniformly generated ensemble of $n-1$ segments on a circle of length $1-nx$. In this new arrangement, the smallest segment has expected size $(1-nx)/(n-1)^2$. Therefore the expected size of the second smallest segment is $$Ex + \frac{1-nEx}{(n-1)^2} = \frac{1}{n^2} + \frac{1-1/n}{(n-1)^2} = \frac{1}{n^2} + \frac{1}{n(n-1)}.$$&#xD;&#xA;We can get the general formula in the same way.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Derivation of the formula for the smallest segment&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Use the same method and come up with one more equation that will enable you to find this unknown.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2010-12-11T22:27:55.627" LastActivityDate="2010-12-11T22:27:55.627" CommentCount="3" />
  <row Id="13965" PostTypeId="1" AcceptedAnswerId="13984" CreationDate="2010-12-11T22:14:59.453" Score="1" ViewCount="46" Body="&lt;p&gt;A finite abstract simplicial complex is a pair $D=(S,D)$ where $S$ is a finite set and $D$ is a non-empty subset of the power set of $S$ closed under the subset operation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What's the name for the following:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$D=(S,D)$ defined as above except that $D$ is closed under the superset operation?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Crossposted from &lt;a href=&quot;http://mathoverflow.net/questions/49084/dual-of-an-abstract-simplicial-complex&quot; rel=&quot;nofollow&quot;&gt;MO&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1856" LastEditorUserId="1856" LastEditorDisplayName="" LastEditDate="2010-12-11T23:41:32.813" LastActivityDate="2010-12-12T00:41:02.067" Title="What's the name for a dual of an abstract simplicial complex" Tags="&lt;combinatorics&gt;&lt;reference-request&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="13966" PostTypeId="2" ParentId="13935" CreationDate="2010-12-11T22:16:38.487" Score="4" ViewCount="" Body="&lt;p&gt;For abelian groups, the ring End(A) is very important.  As far as non-abelian groups A go, End(A) is not even (usually considered) a group.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Adding&quot; homomorphisms doesn't work in the non-abelian case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you define (f+g)(x) = f(x) + g(x), then (f+g)(x+y) = f(x+y) + g(x+y) = f(x) + f(y) + g(x) + g(y), but (f+g)(x) + (f+g)(y) = f(x) + g(x) + f(y) + g(y).  To conclude that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;    f(y) + g(x) = g(x) + f(y)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;are equal, you use that + is commutative, that A is abelian.  More precisely, if you take f=g to be the identity endomorphism, then f+g is an endomorphism iff A is abelian.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Composing&quot; homomorphisms doesn't work to form a group, since they are not invertible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Aut(A), the group of invertible endomorphisms, does form a group.  Aut(A) does not determine if a group is abelian or not: 4×2 and the dihedral group of order 8 have isomorphic automorphism groups.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instead of a ring, End(A) sits inside the &quot;near-ring&quot; of self-maps.  See &lt;a href=&quot;http://en.wikipedia.org/wiki/Nearring#Mappings_from_a_group_to_itself&quot; rel=&quot;nofollow&quot;&gt;the wikipedia article on nearring&lt;/a&gt; for an explanation.&lt;/p&gt;&#xA;" OwnerUserId="583" LastEditorUserId="583" LastEditorDisplayName="" LastEditDate="2010-12-11T22:24:20.320" LastActivityDate="2010-12-11T22:24:20.320" CommentCount="1" />
  <row Id="13967" PostTypeId="2" ParentId="13959" CreationDate="2010-12-11T22:30:01.707" Score="6" ViewCount="" Body="&lt;p&gt;Here you have two independent random variables $X,Y$, both uniform on $[0,1]$. Let &#xA;$A=\min (X,Y), B=\max (X,Y)$ and $C=\min (A, 1-B, B-A)$. First you want to find the probability&#xA;density function $f_C(a)$ of $C$. Let $F_C(a)$ be the cumulative distribution function. Then&#xA;$$ 1- F_C(a) = P(C\ge a)=P(A\ge a, 1-B\ge a, B-A\ge a)$$ $$=P(a\le X\le 1-a, a\le Y\le 1-a, |X-Y|\ge a)$$&#xD;&#xA;This is equal to the total area of the two sets&#xA; $$\lbrace (x,y): 0\le x,y\le 1, y\le x-a, x\le 1-a, y\ge a\rbrace$$ and &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\lbrace (x,y): 0\le x,y\le 1, y\ge x+a, x\ge a, y\le 1-a\rbrace$$&#xD;&#xA;which is $(1-3a)^2$. Note that for these sets to be nonempty, one must have $0\le a\le 1/3$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence $F_C(a)=1-(1-3a)^2$, from which it follows $f_C(a)=F_C&amp;#39;(a)=6(1-3a)$. Therefore&#xA;the expected value of $C$ is&#xA;$$\int_0^{1/3} 6a(1-3a) da= \frac{1}{9}.$$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="3249" LastActivityDate="2010-12-11T22:30:01.707" CommentCount="1" />
  <row Id="13968" PostTypeId="1" CreationDate="2010-12-11T22:35:29.530" Score="5" ViewCount="144" Body="&lt;p&gt;The proof that I know of the theorem goes like this:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Any module $M$ is a quotient of a free module $F$ (over any ring).&lt;/li&gt;&#xA;&lt;li&gt;Any submodule $K$ of a free module $F$ over a PID $R$ is a free module, so in particular the kernel of the above quotient map is free.&lt;/li&gt;&#xA;&lt;li&gt;For any free submodule $K$ of a free module $F$ over a PID $R$ we can find a $y\in K, x\in F$ and $a\in R$ such that $F=\left&amp;lt;x\right&amp;gt;\oplus F&amp;#39;$ and $K=\left&amp;lt;y\right&amp;gt;\oplus K&amp;#39;$ with $K&amp;#39;=K\cap F&amp;#39;$. Furthermore, $\left&amp;lt;a\right&amp;gt;$ is an ideal maximal among images of $K$ under homomorphisms $F\to R$.&lt;/li&gt;&#xA;&lt;li&gt;If $K$ is finitely generated (which would follow from $F$ being finitely generated), we iterate this construction, which gives us sequences $x_i\in F$, $y_i\in K$ and $a_i\in R$ such that $F=F&amp;#39;\oplus_{i} \left&amp;lt;x_i\right&amp;gt;$ and $K=\oplus_{i}\left&amp;lt;y_i\right&amp;gt;$ where $a_ix_i=y_i$ and $a_i$ divides $a_j$ for $i&amp;lt;j$ (which follows from choosing a module homomorphism $f$ from $F$ to $R$ that is $1$ on all the $x_i$ and on the basis elements of the $F&amp;#39;$, then exploiting the fact that $f(y_j)=f(a_jx_j)=a_jf(x_j)=a_j\in f(\oplus_{i\leq j}y_j)\subset\left&amp;lt;a_i\right&amp;gt;$ by $\left&amp;lt;a_i\right&amp;gt;$ being maximal among images of $\oplus_{i\leq j}a_j$ under homomorphisms from $\oplus_{i\leq j}\left&amp;lt;x_j\right&amp;gt;\oplus F&amp;#39;$ to $R$).&lt;/li&gt;&#xA;&lt;li&gt;Taking quotients, we obtain that $M=F&amp;#39;\oplus_i R/(a_i)$ where $a_i$ divides $a_j$ if $i&amp;lt;j$.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;My question is why does this break down if we drop the assumption in 4 that $K$ is finitely generated? Does iterating the process transfinitely no longer give us a basis for $K$? If so, why not?&lt;/p&gt;&#xA;" OwnerUserId="400" LastEditorUserId="400" LastEditorDisplayName="" LastEditDate="2010-12-11T22:46:41.330" LastActivityDate="2010-12-12T04:17:00.763" Title="Why does the structure theorem for finitely generated modules over PIDs fail for arbitrary modules over a PID?" Tags="&lt;commutative-algebra&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="13969" PostTypeId="2" ParentId="13949" CreationDate="2010-12-11T22:38:59.193" Score="4" ViewCount="" Body="&lt;p&gt;It is not appropriate to use proof by cases here.  The statement, &quot;There exists a positive integer n such that $n^2−79n+1601$ is composite&quot; means exactly what it says.  It states that there is at least one positive integer that makes this statement true.  It is no different from any other similar statement such as &quot;I have at least one dollar in my pocket&quot;.  All I have to do to prove this is reach into my pocket and show you the dollar.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;All you have to do to show your original proposition is to find one number, that makes it true.  Being too abstract is counterproductive here.  You can find one number such as 1601. Then, you just have to do the computations that show that the number produced is composite. Literally, the same computations you might do in grade school.  In the case of 1601,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$1601^2-79\times1601+1601 $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;gives a number that is clearly composite, which is why I picked it.&lt;/p&gt;&#xA;" OwnerUserId="1004" LastActivityDate="2010-12-11T22:38:59.193" CommentCount="2" />
  <row Id="13970" PostTypeId="1" AcceptedAnswerId="13982" CreationDate="2010-12-11T22:44:42.893" Score="1" ViewCount="60" Body="&lt;p&gt;Having: $P = N^{CN + 1}$;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I simplify this equation to $N = \cdots$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried using logarithms but I'm stucked...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any ideas?&lt;/p&gt;&#xA;" OwnerUserId="4652" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-12T04:00:15.340" LastActivityDate="2010-12-12T04:00:15.340" Title="Is possible to simplify $P = N^{ CN + 1}$ in terms of $N$?" Tags="&lt;analysis&gt;&lt;logarithms&gt;" AnswerCount="1" />
  <row Id="13971" PostTypeId="1" CreationDate="2010-12-11T22:45:43.280" Score="0" ViewCount="154" Body="&lt;p&gt;How do I apply the Doleans-Dade exponential formula for the following levy stochastic differential equation:&#xA;$$dZ_t=Z_t\left(\theta_1(t)dW_t^{(1)} +\theta_2(t)dW_t^{(2)}+\int_\mathbb R \theta(s,x)\mu(ds,st)-v(ds,dt)\right),$$ where $W_t^{(1)}$ and $W_t^{(2)}$ are two independent Brownian motions, $\mu(ds,st)$ is the jump measure and $v(ds,dt)$ is the levy measure&lt;/p&gt;&#xA;" OwnerUserId="626" LastActivityDate="2010-12-11T22:45:43.280" Title="Doleans-Dade exponential formula" Tags="&lt;stochastic-integrals&gt;" />
  <row Id="13972" PostTypeId="2" ParentId="13959" CreationDate="2010-12-11T22:49:43.200" Score="12" ViewCount="" Body="&lt;p&gt;Update: The current version of this answer is more intuitive (IMHO) than the previous one.  See also &lt;a href=&quot;http://math.stackexchange.com/questions/14190/average-length-of-the-longest-segment/14194#14194&quot;&gt;this similar answer&lt;/a&gt; to a similar question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The generalization of this problem to $n$ pieces is discussed extensively in David and Nagaraja's &lt;em&gt;&lt;a href=&quot;http://books.google.com/books?id=bdhzFXg6xFkC&amp;amp;printsec=frontcover&amp;amp;dq=david+and+nagaraja+order+statistics&amp;amp;source=bl&amp;amp;ots=OaK_k-BbTf&amp;amp;sig=vAtEAURNfZTe3NDV9WodQZ-3ip4&amp;amp;hl=en&amp;amp;ei=DfsDTdiXPML88AafsKnqAg&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=1&amp;amp;ved=0CBgQ6AEwAA#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;Order Statistics&lt;/a&gt;&lt;/em&gt; (pp. 133-135, and p. 153).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $X_1, X_2, \ldots, X_{n-1}$ denote the positions on the rope where the cuts are made, let $V_i = X_i - X_{i-1}$, where $X_0 = 0$ and $X_n = 1$.  So the $V_i$'s are the lengths of the pieces of rope.&#xA;&lt;BR&gt;&lt;BR&gt;&#xA;The key idea is that the probability that any particular $k$ of the $V_i$'s simultaneously have lengths longer than $c_1, c_2, \ldots, c_k$, respectively (where $\sum_{i=1}^k c_i \leq 1$), is $$(1-c_1-c_2-\ldots-c_k)^{n-1}.$$  This is proved formally in David and Nagaraja's &lt;em&gt;&lt;a href=&quot;http://books.google.com/books?id=bdhzFXg6xFkC&amp;amp;pg=PA133&amp;amp;lpg=PA133&amp;amp;dq=david+and+nagaraja+order+statistics+division+of+random+interval&amp;amp;source=bl&amp;amp;ots=OaK_l7B9-j&amp;amp;sig=por_GsntbxBDII72xTWISQ9Keas&amp;amp;hl=en&amp;amp;ei=urkGTevVIZTmsQOM_7HsBw&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=1&amp;amp;ved=0CBMQ6AEwAA#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;Order Statistics&lt;/a&gt;&lt;/em&gt;, p. 135.  Intuitively, the idea is that in order to have pieces of size at least $c_1, c_2, \ldots, c_k$, all $n-1$ of the cuts have to occur in intervals of the rope of total length $1 - c_1 - c_2 - \ldots - c_k$.  For example, $P(V_1 &amp;gt; c_1)$ is the probability that all $n-1$ cuts occur in the interval $(c_1, 1]$, which, since the cuts are randomly distributed in $[0,1]$, is $(1-c_1)^{n-1}$.&#xA;&lt;BR&gt;&lt;BR&gt;&#xA;If $V_{(1)}$ denotes the shortest piece of rope, then for $x \leq \frac{1}{n}$, (following Raskolnikov's comment) $$P(V_{(1)} &amp;gt; x) = P(V_1 &amp;gt; x, V_2 &amp;gt; x, \ldots, V_n &amp;gt; x) = (1 - nx)^{n-1}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, &#xA;$$E[V_{(1)}] = \int_0^{\infty} P(V_{(1)} &amp;gt; x) dx = \int_0^{1/n} (1-nx)^{n-1} = \frac{1}{n^2}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;David and Nagaraja also give the formula Yuval Filmus mentions (as Problem 6.4.2):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$E[V_{(r)}] = \frac{1}{n} \sum_{j=1}^r \frac{1}{n-j+1}.$$&lt;/p&gt;&#xA;" OwnerUserId="2370" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2011-01-19T21:23:07.460" LastActivityDate="2011-01-19T21:23:07.460" CommentCount="1" />
  <row Id="13973" PostTypeId="2" ParentId="13886" CreationDate="2010-12-11T22:50:28.987" Score="0" ViewCount="" Body="&lt;p&gt;The nth twin prime pair appears to be around $c n \log^2 n$ for a constant $c$, large $n$.  The largest prime factor of any non-prime $m$ is equal to $n/m&amp;#39;$ where $m&amp;#39;\ge2$.  What you see in the plot are approximations of the curves $y=\frac{c}{k} x \log^2 x$ for $k=6,12,24,...$, since the mean of a twin pair is a multiple of 6.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Added: Notice the same effect when plotting simply the maximum prime factor of 6n versus n:&#xA;&lt;img src=&quot;http://i.imgur.com/b5Ldm.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="2937" LastEditorUserId="2937" LastEditorDisplayName="" LastEditDate="2010-12-12T07:39:14.873" LastActivityDate="2010-12-12T07:39:14.873" CommentCount="5" />
  <row Id="13974" PostTypeId="2" ParentId="13949" CreationDate="2010-12-11T22:53:10.060" Score="1" ViewCount="" Body="&lt;p&gt;It may actually be worth pointing out that if $p(x)$ is a polynomial with integer coefficients, the only way $p(n)$ is never composite as $n$ varies over the integers is if $p$ is constant (and $p(x)=c$ for some $c$ that is either prime, the negative of a prime, $1$, or $-1$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For suppose that $p(x)=ax^n+bx^{n-1}+\dots+c$. Then clearly $c$ divides $p(c)$. Also, $c$ divides $p(2c)$, $p(3c)$, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose first $|c|&amp;gt;1$. The only way we do not find a composite number in the list $p(c),p(2c),p(3c),\dots$ is if all these numbers are $c$ or $-c$ and $c$ or $-c$ itself is prime. But then infinitely many of the $p(nc)$ are equal to each other. But polynomials that take the same value too often are constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $c=0$, we are done because $p(0)=0$ is composite.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, finally, suppose that $|c|=1$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Write $p(x)=\alpha(x-1)^n+\beta(x-1)^{n-1}+\dots$, $p(x)=A(x-2)^n+B(x-2)^{n-1}+\dots$, etc. If at least one of these expansions has constant term of absolute value larger than 1, and we are done by the previous analysis, or else all of them have constant term $1$, $-1$, or $0$. Note that the constant term in the first case is $p(1)$, in the second is $p(2)$, etc. So again we have that either $p(n)=0$ for some $n$ (and so it is composite), or else $p$ takes the same value ($1$ or $-1$) infinitely often, and we conclude that $p$ is constant.&lt;/p&gt;&#xA;" OwnerUserId="462" LastActivityDate="2010-12-11T22:53:10.060" />
  <row Id="13975" PostTypeId="1" AcceptedAnswerId="13978" CreationDate="2010-12-11T22:56:55.000" Score="3" ViewCount="212" Body="&lt;p&gt;Prove that $\left \{ 0,1 \right \}^{\mathbb{N}}\sim \left \{ 0,1,2,3 \right \}^{\mathbb{N}}$ and find a direct bijection function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I got the first part by showing that $\left \{ 0,1 \right \}^{\mathbb{N}} \subseteq \left \{ 0,1,2,3 \right \}^{\mathbb{N}} \subseteq {\mathbb{N}}^{\mathbb{N}}$, which implies that $|\left \{ 0,1 \right \}^{\mathbb{N}}| \leq |\left \{ 0,1,2,3 \right \}^{\mathbb{N}}| \leq |{\mathbb{N}}^{\mathbb{N}}|$ and since $|{\mathbb{N}}^{\mathbb{N}}| = |\left \{ 0,1 \right \}^{\mathbb{N}} | = 2^{\aleph_0} $ and Cantor-Bernstein you get that $\left \{ 0,1 \right \}^{\mathbb{N}}\sim \left \{ 0,1,2,3 \right \}^{\mathbb{N}}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I'm stuck with formulating a bijection function. More generally, what approach do you use when you need a formulate an exact function?&lt;/p&gt;&#xA;" OwnerUserId="3637" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-11T23:04:17.270" LastActivityDate="2011-01-16T15:15:26.417" Title="$\left \{ 0,1 \right \}^{\mathbb{N}}\sim \left \{ 0,1,2,3 \right \}^{\mathbb{N}}$ bijection function" Tags="&lt;elementary-set-theory&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="1" />
  <row Id="13976" PostTypeId="2" ParentId="13975" CreationDate="2010-12-11T23:02:10.110" Score="9" ViewCount="" Body="&lt;p&gt;Think in terms of how you convert between binary and base 4 representations of numbers.  Break up a sequence of 0s and 1s into pairs, and to each pair (for which there are the 4 possibilities) assign one of 0, 1, 2, or 3.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-12-11T23:02:10.110" CommentCount="2" />
  <row Id="13977" PostTypeId="2" ParentId="13912" CreationDate="2010-12-11T23:05:52.207" Score="13" ViewCount="" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;But the geometry in algebraic geometry is dictated entirely by the algebra. So how can you use geometry to help you with the algebra, when you have to do the algebra first to figure out what the geometry looks like? I don't get it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Roughly speaking, geometric intuition suggests to you what should be true, and suggests a strategy for proving it; and algebra is how you carry out the proofs.  For example, it's geometry that should inform your intuitions about things like intersections of varieties, but algebra that you use to actually rigorously define intersections and prove things about them.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the &quot;principle of continuity.&quot;  If you intersect a circle and an ellipse, you'll generically get four points, although sometimes you might get two or one.  If you intersect a circle and a parabola the same thing happens.  In a perfect world, you might suspect that the intersection of two conic sections is always &quot;four points,&quot; but this clearly isn't true for the usual definition of &quot;four points.&quot;  But if you broaden your definitions (count the points with multiplicity; count the complex points; count the points at infinity), you will eventually be led to complex projective varieties, where something like the &quot;principle of continuity&quot; holds: if you have two varieties that intersect in $m$ points and you nudge one of them or the other continuously, they will still intersect in $m$ points, if you count the points properly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's important to keep in mind that historically a major incentive for developing commutative algebra was to rigorize parts of algebraic geometry; some algebraic geometers (the Italian school) had begun relying too heavily on geometric intuition and had been ignoring special cases, etc. and commutative algebra was one way to fix their proofs.  But the point is that they were doing algebraic geometry &lt;em&gt;first&lt;/em&gt;!  There is a long and interesting history here which I think it is very instructive to learn; you should try to find Dieudonne's &lt;em&gt;History of Algebraic Geometry&lt;/em&gt;, as well as (for a personal perspective) Parikh's &lt;em&gt;The Unreal Life of Oscar Zariski.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;When I first saw the axioms for a group, I spent the next year trying to figure out why the heck anybody cared about groups (and frankly still only know this in a detached and academic way).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The group axioms are an abstraction of the notion of symmetry, and symmetry is a natural and beautiful idea, and symmetries are everywhere in mathematics.  Perhaps you aren't acquainted with enough examples; it's hard for me to give more specific advice here without knowing what you find unsatisfying about groups.  But you might be interested in Conway, Burgiel, and Goodman-Strauss's &lt;em&gt;The Symmetries of Things&lt;/em&gt;, as well as in Mumford, Series, and Wright's &lt;em&gt;Indra's Pearls: the Vision of Felix Klein.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-12-12T02:52:52.363" LastActivityDate="2010-12-12T02:52:52.363" CommentCount="6" />
  <row Id="13978" PostTypeId="2" ParentId="13975" CreationDate="2010-12-11T23:08:37.287" Score="5" ViewCount="" Body="&lt;p&gt;There are at least two ways to proceed: Either you start as you did, and then you follow the argument of Cantor-Bernstein, which explicitly gives you how to build a bijection from the two given injections.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other way is to directly argue in the case at hand. For example, identify the sequence $(a_0,a_1,a_2,a_3,...)$ in $\{0,1\}^{\mathbb N}$ with the sequence $(b_0,b_1,b_2,\dots)$ in $\{0,1,2,3\}^{\mathbb N}$ as follows: Replace $a_{2n},a_{2n+1}$ with $b_n$, where $0,0$ is replaced with $0$; $0,1$ is replaced with $1$; $1,0$ with $2$; and $1,1$ with $3$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[&lt;strong&gt;Edit&lt;/strong&gt;: I see Jonas wrote the same explicit bijection as I was typing this.]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a slightly more challenging exercise, pick any two positive integers $n&amp;lt;m$, and build a &quot;combinatorial&quot; bijection between $\{0,1,\dots,n\}^{\mathbb N}$ and $\{0,1,\dots,m\}^{\mathbb N}$. &lt;em&gt;Combinatorial&lt;/em&gt; meaning here something in the same spirit of the explicit bijection above.&lt;/p&gt;&#xA;" OwnerUserId="462" LastActivityDate="2010-12-11T23:08:37.287" CommentCount="1" />
  <row Id="13979" PostTypeId="2" ParentId="13959" CreationDate="2010-12-11T23:38:23.940" Score="8" ViewCount="" Body="&lt;p&gt;My approach is maybe more naive than the others posted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Break the unit interval at $x$ and $y$ where $x &amp;lt; y$.  Our lengths are then $x$, $y - x$, and $1 - y$.  It's not hard to show that they all have probability $1/3$ of being the shortest.  In any case, our joint PDF is given by $f(x,y) = 6$ (since $x$ and $y$ remain uniform random variables on $1/6$th of the square $[0,1] \times [0,1]$).  Each triangle in the diagram below corresponds to the domain of the PDF for one of the three cases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/obVJs.png&quot; alt=&quot;Each triangle corresponds to which length is shortest&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll take care of the case when $x$ is shortest, that is, $x \leq y - x$ and $x \leq 1 - y$.  This is the leftmost triangle.  Since we're assuming $x$ is least, we are looking for&#xA;$$E[x] = \int_0^{1/3} \int_{2x}^{1 - x} 6x \;dy \;dx = 1/9$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The cases when $y - x$ and $1 - y$ are shortest are similar.&lt;/p&gt;&#xA;" OwnerUserId="3537" LastEditorUserId="3537" LastEditorDisplayName="" LastEditDate="2010-12-11T23:53:34.367" LastActivityDate="2010-12-11T23:53:34.367" CommentCount="1" />
  <row Id="13980" PostTypeId="1" AcceptedAnswerId="13994" CreationDate="2010-12-11T23:39:37.010" Score="1" ViewCount="52" Body="&lt;p&gt;Currently I am implementing a DTW (Dynamic Time Warping) algorithm for my project. As some will know it has the complexity of O(n^2). Considering a sound file of length 1-h with 44100 sampling. How long will it take to compute it? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I know we need some other knowledge such as cpu speed, length of the word we are trying to match, our library etc. but you can assume them with your own variables.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, I am aware of the fact that Fast DTW exists and it is way lighter in terms of O complexity but using it is not a case. &lt;/p&gt;&#xA;" OwnerUserId="4653" LastEditorUserId="1016" LastEditorDisplayName="" LastEditDate="2010-12-12T00:18:17.663" LastActivityDate="2010-12-12T02:06:32.813" Title="Question regarding Dynamic Programming for Audio Recognition" Tags="&lt;algorithms&gt;&lt;pattern-recognition&gt;" AnswerCount="1" CommentCount="6" />
  <row Id="13981" PostTypeId="1" AcceptedAnswerId="13983" CreationDate="2010-12-12T00:12:07.433" Score="4" ViewCount="109" Body="&lt;p&gt;Let $\omega$ denote the countably infinite cardinality and $A$ a cardinality that is strictly larger (i.e., an uncountable one). Is it true that $A^\omega$ has strictly larger cardinality than $A$?&lt;/p&gt;&#xA;" OwnerUserId="4654" LastEditorUserId="4654" LastEditorDisplayName="" LastEditDate="2010-12-12T00:55:31.047" LastActivityDate="2010-12-12T02:47:03.183" Title="$\aleph_0$ to $\aleph_0$" Tags="&lt;set-theory&gt;" AnswerCount="1" />
  <row Id="13982" PostTypeId="2" ParentId="13970" CreationDate="2010-12-12T00:13:41.487" Score="3" ViewCount="" Body="&lt;p&gt;The equation $P=N^N$ doesn't have an elementary solution, but taking logarithms you can find approximate solutions (or even solutions in the form of transseries). In your case, take logarithms to find $(cn+1)\log n = p$. Assuming $p$ and so $n$ are large, $n\log n \approx p/c$. Therefore $n \approx p/c$, and so $$n \approx \frac{p/c}{\log (p/c)}.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: the following is wrong, as J.M. pointed out.&#xA;&lt;i&gt;Again ignoring the $1$ in the exponent, we have $P^{1/c} = N^N$ and so $N = W(P^{1/c})$, where $W$ is the Lambert function (see Wikipedia).&lt;/i&gt;&lt;/p&gt;&#xA;" OwnerUserId="1277" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2010-12-12T02:46:12.803" LastActivityDate="2010-12-12T02:46:12.803" CommentCount="1" />
  <row Id="13983" PostTypeId="2" ParentId="13981" CreationDate="2010-12-12T00:29:00.183" Score="8" ViewCount="" Body="&lt;p&gt;No, it is not necessarily true. For example, let $A$ be the cardinality of the Cantor set, $2^{\aleph_0}$. Then $A^\omega=(2^{\aleph_0})^{\aleph_0}=2^{(\aleph_0\times\aleph_0)}=2^{\aleph_0}=A$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, it is not necessarily false. For example, if $A$ is the supremum of the sequence $\aleph_0,2^{\aleph_0},2^{2^{\aleph_0}},\dots$, then $A^\omega&amp;gt;A$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is because of a general fact due to König: The &lt;em&gt;cofinality&lt;/em&gt; ${\rm cf}(A)$ of a cardinal $A$ is the smallest $\kappa$ such that $A$ can be written as a union of $\kappa$ sets, each of size smaller than $A$. König showed that $A^{{\rm cf}(A)}&amp;gt;A$ for any $A$ infinite--this is a generalization of Cantor's easier result that $2^\kappa&amp;gt;\kappa$. The $A$ in the previous paragraph explicitly has cofinality $\omega$.&lt;/p&gt;&#xA;" OwnerUserId="462" LastActivityDate="2010-12-12T00:29:00.183" CommentCount="5" />
  <row Id="13984" PostTypeId="2" ParentId="13965" CreationDate="2010-12-12T00:33:01.930" Score="0" ViewCount="" Body="&lt;p&gt;I've found &lt;a href=&quot;http://www.albany.edu/~lenart/articles/ucpol.ps&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; for simplicial compleces - down closed set system and, respectively, for the objects I asked about - up closed set system.&lt;/p&gt;&#xA;" OwnerUserId="1856" LastEditorUserId="1856" LastEditorDisplayName="" LastEditDate="2010-12-12T00:41:02.067" LastActivityDate="2010-12-12T00:41:02.067" />
  <row Id="13985" PostTypeId="2" ParentId="13711" CreationDate="2010-12-12T00:38:16.777" Score="0" ViewCount="" Body="&lt;p&gt;Since I want to close this question, I will post an answer myself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A Guide to Distribution Theory and Fourier Transforms - Strichartz is a nice introduction but it contains almost nothing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Théorie des distributions by Laurent Schwartz is written by the master and father on the subject and therefore I say it is a good book.&lt;/p&gt;&#xA;" OwnerUserId="1120" LastActivityDate="2010-12-12T00:38:16.777" CommentCount="2" />
  <row Id="13986" PostTypeId="1" AcceptedAnswerId="13989" CreationDate="2010-12-12T00:44:15.030" Score="4" ViewCount="105" Body="&lt;p&gt;Greetings!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On a test recently I ended up having to solve this for y:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ x = \frac{2y}{y + 1} $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I kept getting stuck in cirlces...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ &#xD;&#xA;\begin{aligned}&#xD;&#xA;x(y + 1 ) = 2y  \\&#xD;&#xA;xy + x = 2y \\&#xD;&#xA;\frac{xy + x}{2} = y \\&#xD;&#xA;\end{aligned} &#xD;&#xA;$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That didn't get me anywhere, so then I started over and tried multiplying both sides by the reciprocal:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;\begin{aligned}&#xD;&#xA;(\frac{y+1}{2y})(x) = 1 \\ &#xD;&#xA;\frac{xy+x}{2y} = 1 \\&#xD;&#xA;\end{aligned}&#xD;&#xA;$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And still I can't see a way to isolate y. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The worst part of this is that I remember being specifically taught a trick for this particular conundrum, but I can't remember the trick!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wolfram gives &lt;a href=&quot;http://www.wolframalpha.com/input/?i=solve+for+y%3A+x+%3D+2y%2F%28y%2B1%29&quot; rel=&quot;nofollow&quot;&gt;the answer&lt;/a&gt; as &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;y = - \frac{x}{x-2} &#xD;&#xA;$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but it doesn't show the steps.&lt;/p&gt;&#xA;" OwnerUserId="3384" LastActivityDate="2010-12-12T02:09:25.317" Title="How to solve this rational equation for y" Tags="&lt;algebra-precalculus&gt;&lt;fractions&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="1" />
  <row Id="13987" PostTypeId="1" AcceptedAnswerId="13988" CreationDate="2010-12-12T01:03:08.233" Score="2" ViewCount="66" Body="&lt;p&gt;I have:&#xA;$$ln(ax + b)^4$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After calculating (u'/u) I get:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$4/(ax+b)$$ but the solution is $$4a/(ax+b)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What am I doing wrong? Thanks!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS: a and b are IR+.&lt;/p&gt;&#xA;" OwnerUserId="2950" LastActivityDate="2010-12-12T01:06:21.750" Title="Derivative log problem" Tags="&lt;calculus&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="13988" PostTypeId="2" ParentId="13987" CreationDate="2010-12-12T01:06:21.750" Score="1" ViewCount="" Body="&lt;p&gt;You need to apply the chain rule, using $\frac{d}{dx}(ax+b)=a$.  (That is your missing factor of $a$.) I can't tell whether you did so, but you can also simplify the first expression to $4\log(ax+b)$.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-12-12T01:06:21.750" CommentCount="1" />
  <row Id="13989" PostTypeId="2" ParentId="13986" CreationDate="2010-12-12T01:08:03.880" Score="5" ViewCount="" Body="&lt;p&gt;Thanks to Timothy and Yuan for pointing out how to factor it. Despite my best efforts, I have never, ever, ever, ever been able to magically see factors appear the way everybody else seems to do so immediately. Sigh.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So that gives&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;\begin{aligned}&#xD;&#xA;xy + x = 2y \\&#xD;&#xA;x = 2y - xy \\&#xD;&#xA;x = y( 2 - x ) \\&#xD;&#xA;y = \frac{x}{2-x} \\&#xD;&#xA;\end{aligned}&#xD;&#xA;$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is the same as the answer given by Wolfram. All is right with the world again.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="3384" LastActivityDate="2010-12-12T01:08:03.880" CommentCount="2" />
  <row Id="13990" PostTypeId="2" ParentId="13514" CreationDate="2010-12-12T01:08:26.520" Score="0" ViewCount="" Body="&lt;p&gt;I guess an alternative standard way to proof it, is to proof sentences that express that there at least n elements. Let denote these entences by $\phi_{\geq n}$ where $\phi_{\geq n} :\Leftrightarrow \exists x_1 ... \exists x_n \rho_n$. The matrix $\rho_n$ can be inductively defined as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\rho_2 :\Leftrightarrow x_1 \neq x_2$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\rho_{n+1} :\Leftrightarrow \rho_n \land x_1 \neq x_{n+1} \land ... \land x_n \neq x_{n+1}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each existence sentence $\phi_{\geq n}$ for $n \geq 2$ can be proved without induction via some trivial witness numerals. Therefore the models are infinite. Induction happens on the meta level, but not on the object level.&lt;/p&gt;&#xA;" OwnerUserId="4414" LastEditorUserId="4414" LastEditorDisplayName="" LastEditDate="2010-12-12T01:32:26.430" LastActivityDate="2010-12-12T01:32:26.430" />
  <row Id="13991" PostTypeId="1" AcceptedAnswerId="14349" CreationDate="2010-12-12T01:21:30.017" Score="7" ViewCount="384" Body="&lt;p&gt;A lot of &quot;big&quot; names in analysis (and other fields) seem to be doing some form of combinatorics (without any order, some examples are Tim Gowers, Terence Tao and Jean Bourgain).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, looking a bit around makes me conclude that combinatorics is a huge field. There must be one &quot;kind&quot; which is the most fruitful in analysis. What kind is this? What is a good introduction to this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: I forgot, analysis is also a big field. I mean more in the direction of harmonic analysis and PDE.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="1120" LastEditorUserId="1120" LastEditorDisplayName="" LastEditDate="2010-12-12T01:28:03.063" LastActivityDate="2010-12-14T23:50:58.017" Title="Combinatorics and analysis." Tags="&lt;combinatorics&gt;&lt;reference-request&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="3" />
  <row Id="13992" PostTypeId="2" ParentId="13991" CreationDate="2010-12-12T01:34:16.597" Score="6" ViewCount="" Body="&lt;p&gt;Your question seems to ask what combinatorics is fruitful in analysis.  I think the trend (at least with those you have mentioned) is the application of analysis to combinatorics&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might be most interested in arithmetic combinatorics.  I'd recommend the page from Tao's course &lt;a href=&quot;http://www.math.ucla.edu/~tao/254a.1.03w/&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.  Alan Frieze has a page with many good links &lt;a href=&quot;http://www.math.cmu.edu/~af1p/Teaching/AdditiveCombinatorics/AC.html&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have also heard good things (and keep intending to read more closely) Richard Stanley's text &lt;a href=&quot;http://www-math.mit.edu/~rstan/ec/&quot; rel=&quot;nofollow&quot;&gt;Enumerative Combinatorics&lt;/a&gt;, which is different &quot;flavor&quot; of combinatorics than the links above.&lt;/p&gt;&#xA;" OwnerUserId="3537" LastActivityDate="2010-12-12T01:34:16.597" CommentCount="1" />
  <row Id="13993" PostTypeId="1" AcceptedAnswerId="14053" CreationDate="2010-12-12T01:46:17.637" Score="1" ViewCount="177" Body="&lt;p&gt;Hi! I'm studying physics, so I'm sorry if I'll write some inexact things in this post. I wish you can understand me. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we have 1D wave equation: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\partial^2 \psi}{\partial x^2}=\frac{1}{c^2}\frac{\partial^2 \psi}{\partial t^2}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;we say that it's always possible (for physics situations...) to decompose the generic solution $f(x+ct)+g(x-ct)$ using Fourier Transforms in $e^{i\omega(x/c\mp t)}.$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the FT is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{1}{2\pi} \int_R \psi(\omega)e^{i\omega t}d\omega$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and we have two variables, $x$ and $t$, how can we use the FT with &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \int_R \psi(\omega)e^{i(k(\omega)x\mp\omega t)}d\omega \; ?$$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have two different types of temporal evolution, $+ct$ and $-ct$... &#xA;I have seen situations with no dispersion law $\omega=kc$, where we consider only a wave $\psi=f(x+ct)$  and use the point $x=0$ to find the spectrum of waves $f(\omega)$, : &#xA;$$&#xD;&#xA;f(\omega)=\frac{1}{2\pi}\int_R f(0,t)e^{-i\omega t}&#xD;&#xA;$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;f(x,t)=\int_R f(\omega) e^{i\omega(x/c+t)}&#xD;&#xA;$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but can someone explain me the general method? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In 2D or 3D I haven't any idea about how I can do all this. Someone can explain me this too?&#xA;Can you link me a .pdf online where this is explained?&#xA;Thank you very much!&lt;/p&gt;&#xA;" OwnerUserId="3871" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-12-12T04:11:47.807" LastActivityDate="2010-12-12T16:10:11.210" Title="Fourier analysis for waves" Tags="&lt;differential-equations&gt;&lt;fourier-analysis&gt;&lt;physics&gt;" AnswerCount="3" />
  <row Id="13994" PostTypeId="2" ParentId="13980" CreationDate="2010-12-12T02:06:32.813" Score="0" ViewCount="" Body="&lt;p&gt;The asymptotic complexity of the algorithm can not provide any information about the time it will take absolutely, but instead shows how running time scales with input size. What I mean is that if you know that it takes $x$ seconds to process $y$ samples of audio, then it will take $4x$ seconds to process $2y$ samples of audio given that the algorithm runs in $O(n^2)$ with $n$ being number of samples.&lt;/p&gt;&#xA;" OwnerUserId="2256" LastActivityDate="2010-12-12T02:06:32.813" CommentCount="1" />
  <row Id="13995" PostTypeId="2" ParentId="13986" CreationDate="2010-12-12T02:09:25.317" Score="3" ViewCount="" Body="&lt;p&gt;Another way to arrive at the conclusion, if you don't &quot;see&quot; factors jumping out at you is this: When you see a fraction like $\frac{2y}{y+1}$, I am sure you feel the urge to split into summands. Now, of course, you cannot do it with the sum in the denominator, so that suggests inverting the whole thing (assuming that everything that you want to be non-zero is non-zero):&#xA;$$&#xD;&#xA;\frac{1}{x} = \frac{y+1}{2y} = \frac{y}{2y} + \frac{1}{2y} = \frac{1}{2} + \frac{1}{2y}.&#xD;&#xA;$$&#xD;&#xA;Now subtract 1/2 and invert back again to get the result.&lt;/p&gt;&#xA;" OwnerUserId="3212" LastActivityDate="2010-12-12T02:09:25.317" />
  <row Id="13996" PostTypeId="1" AcceptedAnswerId="14003" CreationDate="2010-12-12T02:41:21.833" Score="3" ViewCount="269" Body="&lt;p&gt;Hello, which ways to compute this integral:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;\int_{-\infty}^{\infty} e^{tx} \frac{1}{\pi(1+x^2)} dx&#xD;&#xA;$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for different cases of t. t can be any value in $\mathbb{R}$.  I was stuck at how to get the antiderivative of the integrand. Thank you!&lt;/p&gt;&#xA;" OwnerUserId="4658" LastEditorUserId="4658" LastEditorDisplayName="" LastEditDate="2010-12-12T19:25:36.333" LastActivityDate="2010-12-12T19:25:36.333" Title="How to compute this integral?" Tags="&lt;integral&gt;" AnswerCount="3" CommentCount="4" FavoriteCount="1" />
  <row Id="13997" PostTypeId="2" ParentId="13912" CreationDate="2010-12-12T02:44:31.120" Score="5" ViewCount="" Body="&lt;p&gt;Dear Mike,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One suggestion is that you learn some algebraic geometry from a different perspective, and, having done so, return to commutative algebra and see if&#xA;you have more insight by virtue of the intuition in algebraic geometry that you (hopefully) will have developed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You wrote in your question that &quot;the geometry in algebraic geometry is dictated entirely by the algebra&quot;, but this need not be true (and certainly is not true historically).  If you find it is this way because of the particular books you've been reading, I suggest that you try some other books.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming that you are a graduate student (based on your posts), you might try Mumford's book on complex projective varieties.  (I forget the precise title, but it won't be hard to find.)  It is quite geometric.  You could also try looking at Mumford's Red Book, which tries to unify the geometry and algebra.&#xA;There are other books, too: Griffith's and Harris is one, although it requires&#xA;a reasonably strong background in differential and algebraic topology.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As I've tried to emphasize in some other posts of mine here and on MO, algebraic geometry needn't be done via commutative algebra at all; some of the most important recent work in algebraic geometry has relied almost entirely on analytic methods.  There are also topological approaches, via Morse theory and its generalizations and variations.  (Lefschetz, who was one of the founders of algebraic topology, was motivated by the applications to algebraic geometry.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, you needn't focus your study on algebraic geometry at all: I am mentioning it mainly because it came up in your post, and it is an area of mathematics that is very central, and very multifaceted, so that there are lots of ways to approach it, which in turn makes it a good way to build up intuition in lots of different areas of mathematics.  (In your case I am suggesting it as a way to build up intuition in algebra; for myself, algebra was always one of my stronger suits, but through studying algebraic geometry I was able to strengthen&#xA;my geometric and topological intuition.)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Depending on what other answers appear here, I might write some more later which connects with some of other aspects of your post.&lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2010-12-12T02:44:31.120" />
  <row Id="13999" PostTypeId="2" ParentId="13996" CreationDate="2010-12-12T02:52:18.840" Score="1" ViewCount="" Body="&lt;p&gt;First, check whether they converge.  You will find that the first one doesn't unless $t=0$, and the second one doesn't unless $t\leq0$.  For the first integral, when $t=0$, you can use the arctangent function.  For the second, when $t\lt0$, you can try to use &lt;a href=&quot;http://en.wikipedia.org/wiki/Methods_of_contour_integration#Example_.28I.29&quot; rel=&quot;nofollow&quot;&gt;complex contour integration&lt;/a&gt; methods as seen in many texts on complex analysis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: I was hasty in posting this, and hadn't actually thought through how the contour integration would go.  Based on the other answers, it looks like it wouldn't be so straightforward.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2010-12-12T08:45:47.537" LastActivityDate="2010-12-12T08:45:47.537" CommentCount="8" />
  <row Id="14000" PostTypeId="1" AcceptedAnswerId="14024" CreationDate="2010-12-12T02:56:23.223" Score="2" ViewCount="230" Body="&lt;p&gt;Is there any intuition behind surface integrals and their applications? I get that there's some stuff with fluid flow but it doesn't really stick that well, especially when it comes to surface orientation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any surefire way to figure out the orientation of a surface? And is there any sort of intuition behind that or surface integrals?&lt;/p&gt;&#xA;" OwnerUserId="1562" LastActivityDate="2010-12-17T09:21:24.920" Title="Intuition for Surface Integrals" Tags="&lt;calculus&gt;&lt;soft-question&gt;&lt;intuition&gt;" AnswerCount="3" CommentCount="4" FavoriteCount="1" />
  <row Id="14001" PostTypeId="1" CreationDate="2010-12-12T03:07:11.037" Score="0" ViewCount="62" Body="&lt;p&gt;If E={E1, E2, E3} is an orthonormal frame defined on some open set in R^3&lt;br&gt;&#xA;Show that if the dual frame O={O1, O2, O3} satisfies:&lt;br&gt;&#xA;dOi = 0 for i = 1, 2, 3&lt;br&gt;&#xA;Then O is a constant rotation or reflection of E on R^3.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that since the frame is orthonormal, that means the connection matrix w is skew symmetric. I really have no idea how to start this problem.&lt;/p&gt;&#xA;" OwnerUserId="4660" LastActivityDate="2010-12-12T03:07:11.037" Title="Working with orthonormal frames and dual frames" Tags="&lt;homework&gt;&lt;differential-geometry&gt;" CommentCount="3" FavoriteCount="1" />
  <row Id="14003" PostTypeId="2" ParentId="13996" CreationDate="2010-12-12T03:36:23.157" Score="4" ViewCount="" Body="&lt;p&gt;When $t \neq 0$, the first integral diverges and the second one converges only if $t &amp;lt; 0$. In the latter case, if you call your integral $f(t)$, then differentiating under the integral sign gives that &#xA;$$f&amp;#39;&amp;#39;(t) + f(t) = \int_0^{\infty} {e^{tx} + x^2e^{tx} \over \pi(1 + x^2)}$$&#xD;&#xA;$$ = {1\over \pi} \int_0^{\infty}e^{tx}$$&#xD;&#xA;The latter integral integrates to ${\displaystyle-{1 \over t}}$, so that you have ${\displaystyle f&amp;#39;&amp;#39;(t) + f(t) = -{1 \over \pi t}}$. This can be solved via variation of parameters, and (according to wolframalpha at least) it does not have an elementary expression. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand the nonelementary functions in the solution just involve indefinite integrals of ${\displaystyle {\sin(t) \over t}}$ and ${\displaystyle{\cos(t) \over t}}$, so if you want to do something this advanced you might be able to get something satisfactory using initial conditions for this differential equation. &lt;/p&gt;&#xA;" OwnerUserId="3035" LastEditorUserId="3035" LastEditorDisplayName="" LastEditDate="2010-12-12T03:41:43.667" LastActivityDate="2010-12-12T03:41:43.667" />
  <row Id="14004" PostTypeId="1" AcceptedAnswerId="14008" CreationDate="2010-12-12T03:44:00.213" Score="13" ViewCount="355" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Possible Duplicate:&lt;/strong&gt;&lt;br&gt;&#xA;  &lt;a href=&quot;http://math.stackexchange.com/questions/3215/convergence-of-sqrtnx-n-where-x-n1-sinx-n&quot;&gt;Convergence of $\\sqrt{n}x_{n}$ where $x_{n+1} = \sin(x_{n})$&lt;/a&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;p&gt;Hi all,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've been trying to show the convergence or divergence of&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \sum_{n=1}^\infty \frac{\sin^n 1}{n} = \frac{\sin 1}{1} + \frac{\sin \sin 1}{2} + \frac{\sin \sin \sin 1}{3} + \ \cdots $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where the superscript means iteration (&lt;em&gt;not&lt;/em&gt; multiplication, so it's not simply less than a geometric series -- I couldn't find the standard notation for this).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Problem is,&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;$ \sin^n 1 \to 0 $ as $ n \to \infty $ (which I eventually proved by assuming a positive limit and having $ \sin^n 1 $ fall below it, after &lt;a href=&quot;http://en.wikipedia.org/wiki/Monotone_convergence_theorem&quot; rel=&quot;nofollow&quot;&gt;getting its existence&lt;/a&gt;) helps the series to converge,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but at the same time&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$ \sin^{n+1} 1 = \sin \sin^n 1 \approx \sin^n 1 $ for large $ n $ makes it resemble the divergent harmonic series.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I would appreciate it if someone knows a helpful convergence test or a proof (or any kind of advice, for that matter).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In case it's useful, here are some things I've tried:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Show $ \sin^n 1 = O(n^{-\epsilon}) $ and use the p-series.  I'm not sure that's even true.&lt;/li&gt;&#xA;&lt;li&gt;Computer tests and looking at partial sums.  Unfortunately, $ \sum 1/n $ diverges very slowly, which is hard to distinguish from convergence.&lt;/li&gt;&#xA;&lt;li&gt;Somehow work in the related series&#xA;$$ \sum_{n=1}^\infty \frac{\cos^n 1}{n} = \frac{\cos 1}{1} + \frac{\cos \cos 1}{2} + \frac{\cos \cos \cos 1}{3} + \ \cdots $$&#xD;&#xA;which I know diverges since the numerators approach a fixed point.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="336" LastActivityDate="2010-12-12T04:48:25.280" Title="Convergence of series involving iterated $ \sin $" Tags="&lt;analysis&gt;&lt;sequences-and-series&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="1" />
  <row Id="14005" PostTypeId="2" ParentId="13968" CreationDate="2010-12-12T03:58:56.323" Score="3" ViewCount="" Body="&lt;p&gt;Consider the case of $R=\mathbb{Z}$ and $M=\mathbb{Z}_{p^{\infty}}$, the Prufer group (the subgroup of $\mathbb{Q}/\mathbb{Z}$ generated by $\frac{1}{p^n}$ for all $n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take $F$ be freely generated on countably many generators $x_n$, with $x_n$ mapping to $\frac{1}{p^n}$. Then $K$ is freely generated by $px_1$, $x_1-px_2$, $x_2-px_3,\ldots$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is that as you try to perform the procedure in step $4$, at every step you get $a_i = 1$. To see this, simply notice $M$ is divisible, so you can never have a finite direct summand. The subgroup generated by the $y_i$ does not generate $K$, but only a proper submodule of $K$. The process does not capture all of $K$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the real question is what is &quot;morally&quot; going wrong? I'm not sure I can give a good answer to that right now, other than to point out that it &lt;em&gt;does&lt;/em&gt; go wrong. My feeling is that the process described is trying to arrange things so that we can describe $K$ as a &quot;nice&quot; submodule by a good change of basis; we arrange things so that the first coordinate is chosen nicely, and then &quot;push&quot; the rest of the problem to the remaining coordinates. Because at each step we can retain the things we already fixed, and we can only take so many steps before we run out of room, we eventually have fixed the problem. But in the infinitely generated case, not only do we not &quot;run out of room&quot;, we may be unable to even fix that first coordinate in the first place! I think it comes down to the fact that even though $\mathbb{Z}$ is a PID, you can still have an infinite &lt;em&gt;descending&lt;/em&gt; chain of ideals, like $(p)\supset (p^2)\supset (p^3)\supset\cdots$, which in a sense is what we have with $M$ above. Trying to go &quot;transfinite&quot; doesn't solve the problem because $\cap (p^n) = (0)$ loses all the information about the torsion of $M$. In the finitely generated case this cannot occur, because we have a definite &quot;floor&quot; below which we cannot go; the infinitely generated case does not have that reassurance, which is what is going wrong here and why you keep getting $a_i = 1$, so you are never actually finding a way to describe $K$ effectively this way.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-12T04:17:00.763" LastActivityDate="2010-12-12T04:17:00.763" />
  <row Id="14006" PostTypeId="1" AcceptedAnswerId="14011" CreationDate="2010-12-12T04:02:26.333" Score="2" ViewCount="98" Body="&lt;p&gt;I don't have enough memory to simply create diagonal matrix D x D, getting an 'out of memory' error.&#xA;Instead of performing M x D x D operations in the first multiplication, I do M x D operations, but still my code takes ages to run.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anybody find a more effective way to multiply these matrices in matlab?&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;D=20000&#xA;M=25&#xA;&#xA;A = floor(rand(D,M)*10);&#xA;B = floor(rand(1,M)*10);&#xA;&#xA;&#xA;for i=1:D&#xA; for j=1:M&#xA;  result(i,j) = A(i,j) * B(1,j);&#xA; end&#xA;end&#xA;&#xA;&#xA;manual = result * A';&#xA;auto = A*diag(B)*A';&#xA;isequal(manual,auto)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="1996" LastEditorUserId="1996" LastEditorDisplayName="" LastEditDate="2010-12-12T18:01:39.040" LastActivityDate="2010-12-12T18:01:39.040" Title="Efficient multiplication of super-size matrices in matlab" Tags="&lt;linear-algebra&gt;&lt;matrices&gt;&lt;matlab&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="14007" PostTypeId="2" ParentId="13993" CreationDate="2010-12-12T04:11:54.713" Score="1" ViewCount="" Body="&lt;p&gt;The general idea behind transform methods for solving PDEs is that they turn the differential equation into an ODE.  In the case of the wave equation, the Fourier transform acts on the space variable $x$ only, not $t$.  Since derivatives with respect to $x$ get transformed into multipliation by factors of $2\pi i \xi$ under the action of the transform, the result is a second order ODE in the variable $t$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that there are multiple conventions about the definition of the Fourier transform, having to do with where the factor of $2\pi$ goes, but for instance, one can define it by:&#xA;$$&#xD;&#xA;F(\xi) = \int_{-\infty}^\infty f(x) e^{-2\pi i \xi x}\, dx&#xD;&#xA;$$&#xD;&#xA;In particular, notice that the transform is a function of $\xi$ (usually thought of as frequency).  Compare this with the definition you stated, in which the independent variable in the transformed function is $t$.  I think this may be the source of your confusion.&#xA;The method is basically the same in higher dimensions, letting $\xi$ be an $n$-vector.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A good reference (although not online) is Strauss's text on Partial Differential Equations.  I also feel compelled to point out that one does not need Fourier transforms to derive the general solution of the wave equation.&lt;/p&gt;&#xA;" OwnerUserId="4229" LastActivityDate="2010-12-12T04:11:54.713" />
  <row Id="14008" PostTypeId="2" ParentId="14004" CreationDate="2010-12-12T04:12:48.337" Score="10" ViewCount="" Body="&lt;p&gt;A Google search has turned up an analysis of the asymptotic behavior of the iterates of $\sin$ on &lt;a href=&quot;http://books.google.com/books?id=_tnwmvHmVwMC&amp;amp;lpg=PA157&amp;amp;pg=PA157#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;page 157&lt;/a&gt; of de Bruijn's &lt;em&gt;Asymptotic methods in analysis&lt;/em&gt;.  Namely, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sin^n(1)=\frac{\sqrt{3}}{\sqrt{n}}\left(1+O\left(\frac{\log(n)}{n}\right)\right),$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which implies that your series converges.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit:  Moron has pointed out in a comment that the problem of showing that $\sqrt{n}\sin^n(1)$ converges to $\sqrt{3}$ already appeared in the &lt;a href=&quot;http://math.stackexchange.com/questions/3215/convergence-of-sqrtnx-n-where-x-n1-sinx-n&quot;&gt;question&lt;/a&gt; &quot;Convergence of $\sqrt{n}x_n$ where $x_{n+1}=\sin(x_n)$&quot; (asked by Moron in August).  I had missed or forgot about it.  David Speyer gave a great self contained answer, and he also referenced de Bruijn's book.  De Bruijn gives a reference to a 1945 work of Pólya and Szegő for this result.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2010-12-12T04:48:25.280" LastActivityDate="2010-12-12T04:48:25.280" CommentCount="2" />
  <row Id="14009" PostTypeId="1" CreationDate="2010-12-12T04:37:19.343" Score="2" ViewCount="117" Body="&lt;p&gt;Let $\mu$ be a finite positive measure on the borel sets in $\mathbb{R}$ and suppose $\mathcal{L}^{1}(\mathbb{R},\mu) \subset \mathcal{L}^{\infty}(\mathbb{R},\mu)$. &#xA;Show that there exists $c&amp;gt;0$ such that if A is a Borel set with $\mu(A) &amp;gt; 0$ then $\mu(A) \geq c$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;No idea at all. Can you please help?&lt;/p&gt;&#xA;" OwnerUserId="4663" LastActivityDate="2010-12-16T06:11:51.387" Title="Integration problem, existence of a constant in a inclusion between integrable functions and essentially bounded functions" Tags="&lt;measure-theory&gt;" AnswerCount="3" />
  <row Id="14010" PostTypeId="1" AcceptedAnswerId="14015" CreationDate="2010-12-12T04:41:37.297" Score="1" ViewCount="97" Body="&lt;p&gt;What interesting properties of convex sets are retained by star-convex sets?&lt;/p&gt;&#xA;" OwnerUserId="1457" LastActivityDate="2010-12-12T05:19:17.253" Title="How close are star-convex sets to convex sets?" Tags="&lt;euclidean-geometry&gt;&lt;analytic-geometry&gt;&lt;discrete-geometry&gt;&lt;convex-sets&gt;" AnswerCount="1" />
  <row Id="14011" PostTypeId="2" ParentId="14006" CreationDate="2010-12-12T04:51:26.253" Score="5" ViewCount="" Body="&lt;p&gt;Your problem is discussed &lt;a href=&quot;http://www.mathworks.com/matlabcentral/newsreader/view_thread/248346&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;. They suggest one of the following solutions:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;A*bsxfun(@times,diag(D),A')&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;OR&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;A*sparse(D)*A'&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Matlab is an interpreted language and so explicit loops are very slow (even though in the latest version there's supposedly in-place compilation). So everything must be vectorized.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an aside, did you search the internet at all? It took Google 0.16s to come up with that web page as the first result (for &lt;i&gt;matlab multiplying by a diagonal matrix&lt;/i&gt;).&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-12-12T04:51:26.253" CommentCount="3" />
  <row Id="14012" PostTypeId="1" AcceptedAnswerId="14819" CreationDate="2010-12-12T04:58:08.653" Score="1" ViewCount="176" Body="&lt;p&gt;If you dropped &lt;strong&gt;two rocks in a pond&lt;/strong&gt;, the concentric circles emanating from the two spots would osculate $\infty$ times. The locus of osculating points would form a line.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now imagine that instead of concentric circles, &lt;strong&gt;concentric ellipses emanate&lt;/strong&gt; from the rocks' entry points. Again the concentric ellipses will osculate and again let's consider the locus of points where they do.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What is the shape of the locus?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="1457" LastEditorUserId="1457" LastEditorDisplayName="" LastEditDate="2010-12-13T00:59:14.747" LastActivityDate="2010-12-18T23:52:59.840" Title="Locus of osculation of concentric ellipses (elliptic pond ripples)" Tags="&lt;calculus&gt;&lt;differential-geometry&gt;&lt;euclidean-geometry&gt;&lt;analytic-geometry&gt;&lt;vector-calculus&gt;" AnswerCount="1" CommentCount="12" FavoriteCount="1" />
  <row Id="14013" PostTypeId="2" ParentId="14009" CreationDate="2010-12-12T04:58:11.897" Score="4" ViewCount="" Body="&lt;p&gt;The rough idea is that if there are sets of arbitrarily small measure, then you can define unbounded integrable functions by making them get larger and larger but on subsets that are more quickly getting smaller and smaller.  This is why $L^1$ is not contained in $L^\infty$ in the case of $(0,1)$ with Lebesgue measure, for example.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll try to give you a little more of the idea without giving everything away.  You can prove this with contraposition, so your assumption will be that for all $c\gt0$ there exists a Borel set $A$ with $0\lt\mu(A)\lt c$.  Use this to construct an infinite sequence of disjoint Borel sets with positive measure converging to zero, say at a rate faster than $2^{-n}$.  Then define a function whose value on the $n^\text{th}$ set is $n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All the work is in constructing the sequence of sets.  I'd rather not spoil your fun by saying more, unless you get stuck along the way and have a specific question.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;December 15 update:  To enjoy the new &lt;a href=&quot;http://meta.stackoverflow.com/questions/1191/add-markdown-support-for-hidden-until-you-click-text-aka-spoilers/71396#71396&quot;&gt;spoiler feature&lt;/a&gt;, I decided to post a sketch of one way to construct the sequence in the box below.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Suppose that for all $c\gt0$ there exists a Borel set $A$ with $0\lt\mu(A)\lt c$.  Start with any Borel set $A_0$ such that $0\lt\mu(A_0)\lt \infty$.  (Technically the finiteness is redundant here, because $\mu$ is assumed to be finite.)  For each positive integer $n$, the hypothesis guarantees that given $A_{n-1}$ with positive measure, there is a Borel set $A_n$ such that $0\lt\mu(A_n)\lt \frac{1}{3}\mu(A_{n-1})$.  Do this for every positive integer (countable AC).  Then define a sequence of Borel sets $B_0,B_1,\ldots$ by $B_n=A_n\setminus\cup_{k\gt n}A_k$.   The inequality $\mu(A_k)\lt 3^{n-k}\mu(A_n)$ for all $k\gt n$ implies that each $B_n$ has positive measure (at least half of $\mu(A_n)$) and that $\sum_{n\geq 0}n\mu(B_n)$ converges.  If $m\gt n$, then $B_m$ is contained in $A_m$, while $B_n$ is contained in $A_n\setminus A_m$, so $B_m$ and $B_n$ are disjoint.  The function $\sum_{n\geq1}n\chi_{B_n}$ is thus in $L^1\setminus L^\infty$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2010-12-16T06:11:51.387" LastActivityDate="2010-12-16T06:11:51.387" />
  <row Id="14014" PostTypeId="2" ParentId="13959" CreationDate="2010-12-12T05:05:43.503" Score="0" ViewCount="" Body="&lt;p&gt;Hans and Jonas pretty much tell you how to do it. You are basically trying to calculate the volume on the square with side 0:1, and coordinates of each of the random variables, and height equal to the minimum segment. So you have several regions, as shown by Jonas, each of which you could obtain an analytic integral for. There will be symmetry above/below the diagonal, so you can compute just one and double it. The problem is it is quite a bit of algebra, and it is just too easy to make a mistake. A wrote a simple monte-carlo program, pich two random values, r1,r2, compute the minimum, and do this N times. My results for the differing values of N:&#xA; n=1K .111988&#xA; n=10k .111647&#xA; n=100k .111101&#xA; n=1m .111298&#xA; n=10m .111121&#xA; n=100m .111129&#xA; n=1G .111111&#xA;   So it sure seems to be converging on 1/9.&#xA;  Monte Carlo, is a good way to gain confidence in your result, i.e. if you did&#xA;an analytic solution and it agrees with the Monte Carlo result, it is probably correct. If it doesn't agree you have probably made a mistake. Programming the Monte Carlo for this problem only took a few minutes.&lt;/p&gt;&#xA;" OwnerUserId="4664" LastActivityDate="2010-12-12T05:05:43.503" />
  <row Id="14015" PostTypeId="2" ParentId="14010" CreationDate="2010-12-12T05:05:47.537" Score="5" ViewCount="" Body="&lt;p&gt;&quot;Star-convex set&quot; is a bit of a misnomer. The prevailing term probably should have been &quot;&lt;a href=&quot;http://en.wikipedia.org/wiki/Star_domain&quot; rel=&quot;nofollow&quot;&gt;star domain&lt;/a&gt;&quot; instead (but star-convex is so common that the ship has sailed on that one) since that is what you use it for: An open star domain is a simply connected domain - a handy fact for proving simple cases of theorems in e.g. complex analysis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In short, they have very few of the properties of convex sets. For example: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The intersection of two star-convex sets need not be connected (and thus in particular not star-convex). You can take two L-shaped sets and their intersection will be two disjoint squares.&lt;/li&gt;&#xA;&lt;li&gt;The interior of a star-convex set need not be connected. An example is $\mathbb{C} \setminus \lbrace x+iy \;\vert\; x = 0 \text{ and } y \neq 0 \rbrace$, i.e. the right half-plane plus the left half-plane plus the origin.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1269" LastEditorUserId="1269" LastEditorDisplayName="" LastEditDate="2010-12-12T05:19:17.253" LastActivityDate="2010-12-12T05:19:17.253" CommentCount="2" />
  <row Id="14016" PostTypeId="2" ParentId="14009" CreationDate="2010-12-12T05:26:23.250" Score="3" ViewCount="" Body="&lt;p&gt;What Jonas Meyer did is definitely the most natural approach, but there's also a functional analysis approach to such problems I find amusing. In this case, for each $A$ with $\mu(A) &amp;gt; 0$ define&#xA;$$T_A(f) = {1 \over \mu(A)} \int_A f$$&#xD;&#xA; Then $T_A$ is a bounded linear map from $L^1({\mathbb R},\mu)$ to ${\mathbb C}$, and by your condition that $L^1({\mathbb R},\mu) \subset L^{\infty}({\mathbb R},\mu)$, for each $f$ there is a constant $C_f = ||f||_{\infty}$ such that  $||T_A f|| \leq C_f$ for all $A$. Hence by the uniform boundedness principle there is a single constant $C$ such that $||T_A f|| \leq C||f||_1$ for all $A$. Taking $f$ to be the characteristic function of $A$ this leads to $1 \leq C\mu(A)$, which is equivalent to what you are trying to prove.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And those of you who are familiar with the proof of the uniform boundedness principle may see the connection between this proof and Jonas's.&lt;/p&gt;&#xA;" OwnerUserId="3035" LastActivityDate="2010-12-12T05:26:23.250" CommentCount="1" />
  <row Id="14017" PostTypeId="2" ParentId="13993" CreationDate="2010-12-12T05:34:01.760" Score="2" ViewCount="" Body="&lt;p&gt;The things that you wanted to know is on pages 184--189 of Stein-Shakarchi 's Fourier Analysis. In particular, you will see how the solution is written as $$f(x+ct)+g(x-ct)$$ on p. 188. (Substitute $\frac{1}{2}(f(x+t)+G(x+t))$ and $\frac{1}{2}(f(x-t)-G(x-t))$ in the book for your $f(x+ct)$ and $g(x-ct)$ respectively), where $G$ denotes the integral (primitive) of $g$. &lt;/p&gt;&#xA;" OwnerUserId="3249" LastEditorUserId="3249" LastEditorDisplayName="" LastEditDate="2010-12-12T16:10:11.210" LastActivityDate="2010-12-12T16:10:11.210" CommentCount="1" />
  <row Id="14018" PostTypeId="2" ParentId="3399" CreationDate="2010-12-12T06:33:58.330" Score="1" ViewCount="" Body="&lt;p&gt;Instead of thinking about the graph of $f: \mathbb{R} \to \mathbb{R}$ as a 2-D object in the plane, think about $f$ mapping one number line onto another. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/hFH2J.jpg&quot; alt=&quot;drawing of a convex function from ℝ to ℝ&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll return to that picture, just notice that there is plenty of &quot;room&quot; so to speak and that the arrows mapping point to image will never &quot;fall back&quot; on each other, overlap, or &quot;crowd in&quot; so long as the function is convex.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine a closed loop in the plane whose interior is non-convex. It's like a deflated balloon. You need to &quot;blow it up&quot; until it's at least modestly ($\leq$) full of air for the interior to be convex. Similarly if you had a non-convex polygon and &quot;blew air&quot; inside of it, you would get a convex shape. So it's like convex shapes have to be sufficiently &quot;inflated&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, the arrows in an $\mathbb{R} \to \mathbb{R}$-type picture like the above would be &quot;flopping&quot; inward onto each other, deflated if you will. In the convex mapping the arrows don't overlap at all -- they've got &quot;pressure&quot; or &quot;energy&quot; pushing them outward enough so that they don't overlap. So the image is properly inflated, if you will.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So @NateEldridge, the epigraph being a convex set is a red herring. Think about just the right-most point of a graph as it's being generated by a s-l-o-w graphing calculator. The image has to &quot;outrun&quot; the domain it comes from by $\geq$ each $dt$. And there you have your $f(\mathrm{interior\ of\ domain}) \leq \mathrm{image}_1 + \mathrm{image}_2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is meant as an elaboration on @whuber's answer.&lt;/p&gt;&#xA;" OwnerUserId="1457" LastEditorUserId="1457" LastEditorDisplayName="" LastEditDate="2010-12-12T06:39:39.963" LastActivityDate="2010-12-12T06:39:39.963" CommunityOwnedDate="2010-12-12T06:33:58.330" CommentCount="4" />
  <row Id="14019" PostTypeId="1" CreationDate="2010-12-12T06:42:25.387" Score="1" ViewCount="79" Body="&lt;p&gt;Is the singular locus of the singular locus of a variety somehow &quot;more singular&quot; in the original variety than the rest of the singular locus?  I know there are many complicated tools to measure singularity, but is there any elementary way to see it in this case?&lt;/p&gt;&#xA;" OwnerUserId="4665" LastActivityDate="2010-12-12T06:42:25.387" Title="Singular locus of the singular locus of a variety" Tags="&lt;algebraic-geometry&gt;" CommentCount="2" FavoriteCount="0" />
  <row Id="14020" PostTypeId="1" AcceptedAnswerId="14021" CreationDate="2010-12-12T06:58:44.907" Score="4" ViewCount="72" Body="&lt;p&gt;Is there a good online reference that lists known bounds on Ramsey numbers (and is relatively up to date)? The &lt;a href=&quot;http://en.wikipedia.org/wiki/Ramsey%27s_theorem&quot; rel=&quot;nofollow&quot;&gt;wikipedia page&lt;/a&gt; only has numbers for $R_2(n,m)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am specifically interested in known bounds and values for hypergraph Ramsey Numbers, i.e. 2 colorings of k-subsets (or 2 colorings of the edges of a complete k-uniform hypergraph).  These are commonly denoted $R_k(m,n)$.  A shallow internet search has yielded only a couple sets of papers and notes on bounds.&lt;/p&gt;&#xA;" OwnerUserId="565" LastActivityDate="2010-12-12T07:08:49.740" Title="Known bounds and values for Ramsey Numbers." Tags="&lt;graph-theory&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="14021" PostTypeId="2" ParentId="14020" CreationDate="2010-12-12T07:08:49.740" Score="6" ViewCount="" Body="&lt;p&gt;The best resource online is the (frequently updated) survey &quot;Small Ramsey Numbers&quot; by Stanisław Radziszowski, in The Electronic Journal of Combinatorics. Go to &lt;a href=&quot;http://www.combinatorics.org/&quot; rel=&quot;nofollow&quot;&gt;http://www.combinatorics.org/&lt;/a&gt; and click on the link for &quot;dynamic surveys&quot;. The most recent update on the Ramsey numbers paper is from August 2009. The paper gives extensive references where one can find complete proofs or details of the computations involved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is some recent work on asymptotic bounds that is of significance, but the survey has not addressed yet. For that, you may want to look at the papers in the pages of Jacob Fox, David Conlon, and their collaborators.&lt;/p&gt;&#xA;" OwnerUserId="462" LastActivityDate="2010-12-12T07:08:49.740" CommentCount="1" />
  <row Id="14022" PostTypeId="2" ParentId="13514" CreationDate="2010-12-12T07:14:34.147" Score="0" ViewCount="" Body="&lt;p&gt;You are doing it indirectly, the direct way to do this is by defining a one to one function from $N$ to the members of the model.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So let $M$ be a model of the first two axioms. (To able to to talk about the model you need a theory that can talk about sets to some extent or a theory that can interpret that amount of set theory.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The next thing is defining the function inductively in the obvious way and then use induction and the fact that $M$ satisfies the first and second axiom to prove that the function is one to one.&lt;/p&gt;&#xA;" OwnerUserId="468" LastActivityDate="2010-12-12T07:14:34.147" />
  <row Id="14023" PostTypeId="2" ParentId="13898" CreationDate="2010-12-12T07:17:19.820" Score="1" ViewCount="" Body="&lt;p&gt;I'd like to add an answer that will try to justify the statement with an example. Consider the function sequence&#xA;$$&#xD;&#xA;f_n = \left\{\begin{align} 0 \ \ &amp;amp;\text{ if } |x|&amp;gt;1/n \\&#xD;&#xA;n^2x^2/2 + nx + 1/2 &amp;amp;\text{ if } -1/n\leq x \leq 0\\&#xD;&#xA;-n^2x^2/2 + nx + 1/2 &amp;amp;\text{ if } 0&amp;lt;x\leq 1/n&#xD;&#xA;\end{align}\right..&#xD;&#xA;$$&lt;br&gt;&#xA;These functions are differentiable everywhere, and they approximate the unit step function better and better as $n$ gets larger. If we look at the derivatives of these functions, we'll see that they are a sequence of &lt;em&gt;tent functions&lt;/em&gt; that are getting narrower and taller, as Griffith uses to introduce the delta function.&lt;br&gt;&#xA;&lt;img src=&quot;http://i.imgur.com/I79Ct.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Actually, with an appropriate mode of convergence, when a sequence of differentiable functions converge to the unit step, it can be shown that, their derivatives converge to the delta function. That's why, one may take the derivative of the unit step function to be &lt;em&gt;defined&lt;/em&gt; as the limit of the derivatives, which is the delta function. &lt;/p&gt;&#xA;" OwnerUserId="145" LastEditorUserId="145" LastEditorDisplayName="" LastEditDate="2010-12-12T07:50:04.937" LastActivityDate="2010-12-12T07:50:04.937" />
  <row Id="14024" PostTypeId="2" ParentId="14000" CreationDate="2010-12-12T07:36:34.253" Score="5" ViewCount="" Body="&lt;p&gt;The surface integral of a vector field, $F$ over a surface $S$ (or flux as it is called) is, as stated above, the net flow of fluid through the surface.  The purpose of orientation is to simply indicate which side of the surface is the &quot;positive&quot; side.  For instance, if I calculate flux to be +10, this means there is a net movement of 10 units of fluid towards whichever side of the surface is indicated as the positive side.  The answer would be -10 if the orientation was reversed (since the &quot;positive&quot; side is now losing 10 units of fluid).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are asked to find the flux of $F$ over $S$, they have to tell you the orientation of $S$ in the question.  For instance&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Find the flux of $F(x,y,z) = &amp;lt;2xz,z-y,y^3&amp;gt;$ DOWNWARD through the section of the plane $x+y+z=6$ in the first octant.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this question, the region under the plane is the positive side, and a positive flux would indicate a net movement of fluid downward through the plane.  A negative answer indicates a net upward movement of fluid.  So, how do you figure out orientation for a surface?  You don't.  It's usual stated in the question. In some cases they don't state orientation because they will accept either the positive or negative answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some theorem's such as Stokes' and Gauss' Divergence Theorem require that the surface in question has a particular orientation.  Stokes' theorem for instance, allows us to calculate a line integral along a closed curve by first choosing any surface enclosed by this curve.  If the curve is traversed counterclockwise as viewed from above, the orientation of the surface must be &quot;Upward&quot; to use this theorem.  So, whatever surface you choose, it's normal vector must point upward. In this case, you are the one choosing the surface so you so have to figure out the orientation yourself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I should finally point out that I am assuming you are only working with surfaces in $R^3$.  Orientation of a surface in $R^n$ (or hypersurfaces in $R^n$) is a very difficult concept since these are not real world objects that you can look at and decide the orientation of.  Also, some 2d surfaces such as a mobius strip, have no orientation since they only have one side.&lt;/p&gt;&#xA;" OwnerUserId="3971" LastEditorUserId="3971" LastEditorDisplayName="" LastEditDate="2010-12-17T07:49:42.317" LastActivityDate="2010-12-17T07:49:42.317" />
  <row Id="14026" PostTypeId="1" AcceptedAnswerId="14029" CreationDate="2010-12-12T07:57:52.330" Score="3" ViewCount="336" Body="&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; Which computer programs are useful for a pure mathematician to familiarize themselves with?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Less Briefly:&lt;/strong&gt; I was once told that, now-a-days, any new mathematician worth his beans knows how to TeX up work; when I began my graduate work one of the fourth year students told me he couldn't TeX, I was horrified!  Similarly, a number of my peers were horrified when I told them I'd never used Matlab or Mathematica before.  Currently, I can &quot;get by&quot; in all three of these programs, but it made me think: what programs are popular with pure mathematicians?  I don't mean to limit this to computing things: programs which help to draw pictures and things can also be included.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Lest I Start A Flame War:&lt;/strong&gt; This is not meant to be a &quot;what is your favorite computer language&quot; question or a poll on who thinks Mathematica is worse than Sage.  This is meant to be a survey on what programs and resources are currently being used by pure mathematicians and may be good to look at and read up on.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm also unsure of how to tag this.&lt;/p&gt;&#xA;" OwnerUserId="2959" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-12-12T08:02:21.490" LastActivityDate="2010-12-12T22:46:48.310" Title="Computer Programs for Pure Mathematicians." Tags="&lt;math-software&gt;&lt;computer-algebra-systems&gt;" AnswerCount="4" CommentCount="8" FavoriteCount="3" />
  <row Id="14027" PostTypeId="1" CreationDate="2010-12-12T08:04:12.560" Score="0" ViewCount="98" Body="&lt;h2&gt;Some background&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;I have the following homogeneous transformation matrix (transpose by 2 on Y axis) in raytracing calculations to transform vectors from object space to world space after a ray-sphere intersection:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$M_{OW} = &#xD;&#xA;\left[ {\begin{array}{*{40}c}&#xD;&#xA;   1 &amp;amp; 0 &amp;amp; 0 &amp;amp; 0  \\&#xD;&#xA;   0 &amp;amp; 1 &amp;amp; 0 &amp;amp; 2  \\&#xD;&#xA;   0 &amp;amp; 0 &amp;amp; 1 &amp;amp; 0  \\&#xD;&#xA;   0 &amp;amp; 0 &amp;amp; 0 &amp;amp; 1  \\&#xD;&#xA; \end{array} } \right]&#xD;&#xA;$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Corresponding matrix to transform from world space to object space is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$M_{WO}=M_{OW}^{-1} = &#xD;&#xA;\left[ {\begin{array}{*{40}c}&#xD;&#xA;   1 &amp;amp; 0 &amp;amp; 0 &amp;amp; 0  \\&#xD;&#xA;   0 &amp;amp; 1 &amp;amp; 0 &amp;amp; -2  \\&#xD;&#xA;   0 &amp;amp; 0 &amp;amp; 1 &amp;amp; 0  \\&#xD;&#xA;   0 &amp;amp; 0 &amp;amp; 0 &amp;amp; 1  \\&#xD;&#xA; \end{array} } \right]&#xD;&#xA;$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whereas regular vectors are transformed from object space to world space like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$v_{WS} = M_{WO} v_{OS}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Normals to the surface are transformed different way using a plane perpendicular to the normal instead of the normal itself, see &lt;a href=&quot;http://ocw.mit.edu/courses/electrical-engineering-and-computer-science/6-837-computer-graphics-fall-2003/lecture-notes/5_1_trans_hier.pdf&quot; rel=&quot;nofollow&quot;&gt;MIT's lecture slides on computer graphics&lt;/a&gt; [PDF warning].&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$n_{WS} = (M_{OW}^{-1})^T  n_{OS} = M_{WO}^T  n_{OS}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This gives us a transformation matrix for normals from object space to world space:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$M_{WO}^T = &#xD;&#xA;\left[ {\begin{array}{*{40}c}&#xD;&#xA;   1 &amp;amp; 0 &amp;amp; 0 &amp;amp; 0  \\&#xD;&#xA;   0 &amp;amp; 1 &amp;amp; 0 &amp;amp; 0  \\&#xD;&#xA;   0 &amp;amp; 0 &amp;amp; 1 &amp;amp; 0  \\&#xD;&#xA;   0 &amp;amp; -2 &amp;amp; 0 &amp;amp; 1  \\&#xD;&#xA; \end{array} } \right]&#xD;&#xA;$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Question&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Because this particular transformation is translation-only, I expect all vectors represented as points to infinity (where $w = 0$) to be unaffected, including normals (since planes perpendicular to normals are unaffected). But when I transform the following normal, it gets a non-zero $w$ component:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$n_{OS} = &#xD;&#xA;\left[ {\begin{array}{*{10}c}&#xD;&#xA;   -0.9 \\&#xD;&#xA;   0.99 \\&#xD;&#xA;   -0.17 \\&#xD;&#xA;   0 \\&#xD;&#xA; \end{array} } \right]$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$n_{WS} = &#xD;&#xA;\left[ {\begin{array}{*{40}c}&#xD;&#xA;   1 &amp;amp; 0 &amp;amp; 0 &amp;amp; 0  \\&#xD;&#xA;   0 &amp;amp; 1 &amp;amp; 0 &amp;amp; 0  \\&#xD;&#xA;   0 &amp;amp; 0 &amp;amp; 1 &amp;amp; 0  \\&#xD;&#xA;   0 &amp;amp; -2 &amp;amp; 0 &amp;amp; 1  \\&#xD;&#xA;\end{array} } \right]&#xD;&#xA;\left[ {\begin{array}{*{10}c}&#xD;&#xA;   -0.9 \\&#xD;&#xA;   0.99 \\&#xD;&#xA;   -0.17 \\&#xD;&#xA;   0 \\&#xD;&#xA; \end{array} } \right] = &#xD;&#xA;\left[ {\begin{array}{*{10}c}&#xD;&#xA;   -0.9 \\&#xD;&#xA;   0.99 \\&#xD;&#xA;   -0.17 \\&#xD;&#xA;   -1.98 \\&#xD;&#xA; \end{array} } \right]&#xD;&#xA;$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I fail to visualize how a normal in a ray-sphere intersection would get a non-zero $w$. Intuitively it should remain the same if I only translate along the Y axis, but obviously this is not the case here.&lt;/p&gt;&#xA;" OwnerUserId="2360" LastActivityDate="2010-12-12T08:04:12.560" Title="How to interpret normals transformation in homogeneous coordinates?" Tags="&lt;linear-algebra&gt;&lt;matrices&gt;" />
  <row Id="14028" PostTypeId="2" ParentId="13519" CreationDate="2010-12-12T08:06:50.750" Score="4" ViewCount="" Body="&lt;p&gt;This question is one of the main topics studied in &lt;a href=&quot;http://en.wikipedia.org/wiki/Proof_complexity&quot; rel=&quot;nofollow&quot;&gt;proof complexity&lt;/a&gt;. The measure you are talking about is called the number of &lt;em&gt;steps&lt;/em&gt; in the proof or proof-length.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The size of a tautology is related to the number of steps of its shortest proof, but this does not mean that the number of steps is a monotone function of size of tautology. As mentioned by others, the length of a long tautology can be very short. Coming up with arbitrary long tautologies having proofs with constant (i.e. independent from the size of tautology) number of steps is not difficult. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, the task of coming up with tautologies with large number of steps is not obvious and is open for many propositional proof systems. Note that it follows from the (constructive) proof of completeness theorem that every tautology of size $s$ has a proof of height $O(s)$ with $O(2^s)$ steps and size $O(s2^s)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have some very weak lowerbounds for maximum number of steps needed for a tautology of size $s$ and the question that this upperbound is tight is wide open for sequent calculus and also for natural deduction which are essentially equivalent to Hilbert style proof systems. The bound is tight if we remove cut rule from sequent calculus, i.e. there are tautologies which require exponential number of steps.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can find out more about the relation between different measure studied in proof complexity and the relation between different propositional proof systems by checking surveys/books on the topic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For first order logic, note that the first order logic is undecidable, i.e. given a formula there is no algorithm to check if it is a valid formula. Any computable upperbound on the proof size would give an algorithm for deciding first order logic, check all possible proofs up to that size. (Note that an upper bound on proof length in natural deduction/sequent calculus will give an upper bound on the size because of normalization/cut elimination). Since there is no such algorithm, there cannot be any computable upperbound.&lt;/p&gt;&#xA;" OwnerUserId="468" LastEditorUserId="468" LastEditorDisplayName="" LastEditDate="2010-12-12T08:15:30.170" LastActivityDate="2010-12-12T08:15:30.170" />
  <row Id="14029" PostTypeId="2" ParentId="14026" CreationDate="2010-12-12T08:46:43.760" Score="7" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Mathematica&lt;/strong&gt; is the de-facto standard for symbolic computation. I would argue that there is no other CAS that matches the sheer amount of stuff that Mathematica can do (this is not to say that there aren't other systems out there that do certain things better - like graphing/plotting). Part of the power of Mathematica's strength is the ability to write and evaluate expressions just like you would in your notebook (hence the &quot;notebook&quot; UI paradigm) , however, its reliance on symbolic computation also is its greatest weekness (in terms of strength as a programming language, portability, and speed). Just because of its ubiquity, if there is one thing you should learn, its this. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Matlab&lt;/strong&gt; is the de-facto standard for numerical computations, under the broad category of numerical analysis. It is less designed for symbolic analysis, but its numerical analysis tools are industry standard. With countless additional packages it can seamlessly (well almost seamlessly, and for several thousand dollars) incorporate data capture from everything from webcams to particle physics experiments. Its usefulness as a &lt;em&gt;pure mathematics&lt;/em&gt; tool might not be as as obvious as Mathematica, but if you ever plan on collaborating with applied mathematicians, physicists, or engineers, working knowledge of Matlab is a must.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Python + Scipy + numpy&lt;/strong&gt; . By combining the free programming/scripting language with powerful (C-based) packages for scientific computation, this suite can rival the numerical analysis tools of Matlab. If you extend this combination with Matlplotlib (numpy based package for plotting) you can get publication quality plots and graphs with a familiar syntax (matlab) for $0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;GnuPlot&lt;/strong&gt;. With the ability to graph/plot pretty much anything you want and a very powerful text based input, this software allows you to create publication quality plots with ease (after learning the syntax of course). Plus its, free and open source.&lt;/p&gt;&#xA;" OwnerUserId="1906" LastActivityDate="2010-12-12T08:46:43.760" CommentCount="6" />
  <row Id="14030" PostTypeId="2" ParentId="13996" CreationDate="2010-12-12T08:48:51.507" Score="1" ViewCount="" Body="&lt;p&gt;Correspondingly to Zaricuse's second paragraph, according to &lt;a href=&quot;http://www.efunda.com/math/laplace_transform/forward.cfm?FuncName=Basic&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; (see the formula before the last one), for any $s &amp;gt; 0$,&#xA;$$&#xD;&#xA;\int_0^\infty  {e^{ - sx} \frac{1}{{\pi (1 + x^2 )}}\,{\rm d}x}  = \frac{1}{\pi } \Big \lbrace \cos (s)\Big[\frac{\pi }{2} - {\rm Si}(s)\Big] - \sin (s){\rm Ci}(s)\Big \rbrace,&#xD;&#xA;$$&#xD;&#xA;where ${\rm Si}(s) = \int_0^s {\frac{{\sin u}}{u}} \,{\rm d}u$ is the Sine Integral, and ${\rm Ci}(s) = \int_s^\infty  {\frac{{\cos u}}{u}\,{\rm d}u}$ the Cosine Integral.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;: Probabilistic interpretation of the integrals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The function $f(x) = \frac{1}{{\pi (1 + x^2 )}}$, $x \in \mathbb{R}$, is the density function of a standard Cauchy random variable $X$. The divergent integral&#xA;$$&#xD;&#xA;{\rm E}[e^{tX}] = \int_{ - \infty }^\infty  {e^{tx} f(x)\,{\rm d}x}  = \infty, \;\; t \neq 0, &#xD;&#xA;$$&#xD;&#xA;corresponds exactly to the (elementary) fact that the moment-generating function of the Cauchy distribution does not exist.&#xA;On the other hand, the characteristic function is well known to be&#xA;$$&#xD;&#xA;{\rm E}[e^{{\rm i}{\rm t}X}] = \int_{ - \infty }^\infty  {e^{{\rm i}tx} f(x)\,{\rm d}x}  = e^{ - |t|}, \;\; t \in \mathbb{R}.&#xD;&#xA;$$&#xD;&#xA;Finally, the function $\tilde f(x) = \frac{2}{{\pi (1 + x^2 )}}$, $x &amp;gt; 0$, is the density function of $|X|$. Then, the Laplace transform of $|X|$ is &#xA;$$&#xD;&#xA;{\rm E}[e^{-t|X|}] = \int_0^\infty  {e^{ - tx} \tilde f(x)\,{\rm d}x},\;\; t &amp;gt; 0,&#xD;&#xA;$$&#xD;&#xA;and, according to the first paragraph, can be expressed in terms of the Sine Integral and Cosine Integral functions.&lt;/p&gt;&#xA;" OwnerUserId="2810" LastEditorUserId="2810" LastEditorDisplayName="" LastEditDate="2010-12-12T10:11:45.370" LastActivityDate="2010-12-12T10:11:45.370" CommentCount="2" />
  <row Id="14031" PostTypeId="2" ParentId="14000" CreationDate="2010-12-12T09:55:08.667" Score="0" ViewCount="" Body="&lt;p&gt;Maybe some trivial examples could show you the importance of orientation. Try to compute the fluid flow through the unit square&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;S = \left\{(x,y,z) \ | \ 0\leq x,y \leq 1 , z=0 \right\}&#xD;&#xA;$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;of some fluid moving upwards at constant speed of $1$m/s.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;F(x,y,z) = (0,0,1) \ .&#xD;&#xA;$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then you parametrize your surface, for instance with&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;\varphi (x,y) = (x,y,0) , \quad (x,y) \in D = \left\{ (x,y) \ | \ 0\leq x,y \leq 1  \right\}&#xD;&#xA;$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The normal field to this parametrization is $\varphi_x \times \varphi_y = (0,0,1)$ and we compute the fluid flow:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;\int_\varphi \langle F, dS\rangle = \int_0^1\int_0^1 \langle (0,0,1) , (0,0,1) \rangle dxdy = 1 \ .&#xD;&#xA;$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But why using that particular parametrization? Why not trying with this one?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;\psi (u,v) = (v,u,0) , \quad (u,v) \in D = \left\{ (u,v) \ | \ 0\leq x,y \leq 1  \right\} \ .&#xD;&#xA;$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We compute the normal field again: $\psi_u \times \psi_v = (0,1,0)\times (1,0,0) = (0,0,-1)$. And the fluid flow:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;\int_\psi \langle F, dS \rangle = \int_0^1\int_0^1 \langle (0,0,1) , (0,0,-1) \rangle dudv = -1 \ .&#xD;&#xA;$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Astonishing, isn't it? -The fluid flow seems to depend on something secondary: the coordinates you use on the surface $S$. This could be even worse, since some surface integrals are used to compute the area of the surface: if the area of a surface depended on the coordinates you put in it, this would be a nonsense.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fortunately, this isn't true: you can prove that surface integrals do NOT depend on the coordinates you put on the surface EXCEPT (in the case of fluid flows) on the &lt;em&gt;orientation&lt;/em&gt; of the coordinates -parametrizations. There are two kinds of orientations and before computing a fluid flow you must choose one of them. For instance, in our examples, if we have said that the fluid is going &lt;em&gt;upwards&lt;/em&gt; we have to choose $(0,0,1)$ as the orientation of our surface (the $\varphi$ parametrization). Otherwise, the result we get -with the $\psi$ parametrization- makes no sense: in one second $-1$ litre has gone trhough the surface!?&lt;/p&gt;&#xA;" OwnerUserId="664" LastActivityDate="2010-12-12T09:55:08.667" CommentCount="1" />
  <row Id="14032" PostTypeId="1" CreationDate="2010-12-12T10:13:04.810" Score="2" ViewCount="283" Body="&lt;p&gt;I'm having trouble with this integral&lt;br&gt;&#xA;$$\int\arcsin(\sin x)\,\mathrm dx$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is with the intervals of definition for each function :/ if someone could dumb it down for me.   &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\arcsin\colon [-1, 1] \to [-\pi/2, \pi/2]$.  &lt;/li&gt;&#xA;&lt;li&gt;$\sin:\mathbb{R}\to [-1, 1]$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;right?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But what about $\arcsin(\sin x)$ ?&lt;/p&gt;&#xA;" OwnerUserId="2227" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-15T05:52:05.243" LastActivityDate="2010-12-15T05:52:05.243" Title="integral of $\arcsin(\sin(x))$" Tags="&lt;calculus&gt;&lt;trigonometry&gt;&lt;integral&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="14033" PostTypeId="2" ParentId="14032" CreationDate="2010-12-12T10:20:32.083" Score="1" ViewCount="" Body="&lt;p&gt;&lt;em&gt;My previous hint was a bit misleading - hence here is a (the comments are still appropriate)&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;CORRECTED HINT&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On $I= [-\pi/2,\pi/2]$ the sine function grows from $\sin(-\pi/2)=-1$ to $\sin(\pi/2)=1$, hence it is invertible on $I$. The inverse, $\arcsin$, is defined on $[-1,1]$ and has the property $\arcsin(\sin x) = x$ if $-\pi/2\le x\le \pi/2$.&#xA;Since $\sin$ is $2\pi$-peiodic we must have $\arcsin(\sin (x+2n\pi))=\arcsin(\sin(x))$ for all $x$ and integers $n$. &#xA;Hence it is sufficient to understand $\arcsin(\sin x)$ for $\pi/2\le x\le 3\pi/2$, to achieve that use $\sin(x+\pi)=\sin (-x)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you see the function there will be not problem to perform any integration.&lt;/p&gt;&#xA;" OwnerUserId="1154" LastEditorUserId="1154" LastEditorDisplayName="" LastEditDate="2010-12-14T21:16:34.130" LastActivityDate="2010-12-14T21:16:34.130" CommentCount="3" />
  <row Id="14034" PostTypeId="2" ParentId="14032" CreationDate="2010-12-12T10:23:29.983" Score="1" ViewCount="" Body="&lt;p&gt;Since $\sin(x)$ can take $ \mathbb R$ and outputs $[-1, 1]$ and $\arcsin(x)$ can take $[-1, 1]$ and output $[-\pi/2, \pi/2]$ , then it stands to reason that the constraints for $\arcsin(\sin(x))$ would be $\mathbb R \rightarrow [-\pi/2, \pi/2]$.&lt;/p&gt;&#xA;" OwnerUserId="3384" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-12T10:28:42.903" LastActivityDate="2010-12-12T10:28:42.903" />
  <row Id="14035" PostTypeId="1" CreationDate="2010-12-12T10:44:50.737" Score="4" ViewCount="129" Body="&lt;p&gt;I am able to measure my distance to a set of (about 6 or 7) fixed but unknown points from many positions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The difference in position between measurements is also unknown.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe that I should be able to work out the relative position of the fixed points, and therefore where I measured from and the path I took.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have looked at the wiki page for trilateration, but it only gives examples working from known points.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any help?&lt;/p&gt;&#xA;" OwnerUserId="4668" LastActivityDate="2011-01-11T21:44:43.497" Title="Trilateration with unknown fixed points" Tags="&lt;euclidean-geometry&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="14036" PostTypeId="1" CreationDate="2010-12-12T11:01:39.693" Score="0" ViewCount="115" Body="&lt;p&gt;&lt;strong&gt;Short version:&lt;/strong&gt; if two spherical caps of the same sphere intersect, how can I determined the coordinates of the two &quot;singular points&quot; of this intersection&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Long version:&lt;/strong&gt; On a unit sphere, centered at the original, we consider two spherical cap. Each cap is defined by:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;the (unit) vector indicating the top of the spherical cap, $\mathbf{n}$&#xD;&lt;/li&gt;&#xA;&lt;li&gt;the angle $\theta$, defined as is &lt;a href=&quot;http://wapedia.mobi/en/File%3aSteradian_cone_and_cap.svg&quot; rel=&quot;nofollow&quot;&gt;done there&lt;/a&gt;; any point $\mathbf{r}$ on the sphere belongs to the cap iff $\mathrm{angle}(\mathbf{n},\mathbf{r}) \le \theta$&#xD;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The question is: given two spherical caps $(\mathbf{n}_1,\theta_1)$ and $(\mathbf{n}_2,\theta_2)$, their intersection can be:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;empty&lt;/li&gt;&#xA;&lt;li&gt;a single point&lt;/li&gt;&#xA;&lt;li&gt;a surface delimited by two circular arcs, intersecting in two points&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The question is: in the third case, how can I determine the coordinates of the two points of interest? I'm sure it's not so hard if you find the right way to go at it.&lt;/p&gt;&#xA;" OwnerUserId="3406" LastActivityDate="2010-12-12T13:12:07.270" Title="Intersection of spherical caps" Tags="&lt;geometry&gt;&lt;intersection&gt;&lt;3d&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="14037" PostTypeId="1" CreationDate="2010-12-12T11:20:19.660" Score="3" ViewCount="203" Body="&lt;p&gt;In a And\Or-graph induced by the transition function, each node of G corresponds to a state q belonging to set Q of the state of the Automaton, for q with $\delta(q,a)=q1*q2$, the node is a $*-node$ with two successor q1 and q2. For q={true,false}, the node q is a sink-node. Hence, if * is $\vee$ then q is $\vee-node$, else q is $\wedge-node$. My problem is this : since that the result of transition function of a alternating automata includes more nodes in $\wedge$ and $\vee$ (example: $\delta(q_0,a)=(q1 \vee q2) \wedge q3 \wedge q4 )$ ), how to build the graph And\Or on transition function of an alternating automaton?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Graph And/Or graph is defined as following : A form of graph or tree used in problem solving and problem decomposition. The nodes of the graph represent states or goals and their successors are labeled as either AND or OR branches. The AND successors are subgoals that must all be achieved to satisfy the parent goal, while OR branches indicate alternative subgoals, any one of which could satisfy the parent goal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instead the alternating automata is a automata with transition function defined  as following : $\delta: S \times \Sigma \longrightarrow B^+(S)$ where S are the states of the automata ,$\Sigma$ the alphabet and $B^+(S)$ is the set of positive Boolean formulas over S.&lt;/p&gt;&#xA;" OwnerUserId="4670" LastEditorUserId="4670" LastEditorDisplayName="" LastEditDate="2010-12-19T15:02:56.680" LastActivityDate="2010-12-19T15:02:56.680" Title="Constructor And\Or-graph on function transition of the alternating automata" Tags="&lt;graph-theory&gt;&lt;computer-science&gt;&lt;propositional-calculus&gt;" CommentCount="6" FavoriteCount="1" />
  <row Id="14039" PostTypeId="2" ParentId="12091" CreationDate="2010-12-12T11:43:56.160" Score="0" ViewCount="" Body="&lt;p&gt;You can try to treat your fuzzy logic algebraically, so as to have Tarskian non-boolean truth definition based on some fuzzy connectives. Some interesting choice is to use a t-norm for conjunction, and then view implication as a residuum:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x * y : [0,1] \times [0,1] \rightarrow [0,1]$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x \rightarrow y := max\{z| x*z \leq y\}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The above allows an easy comparison of fuzzy algebras with boolean algebras by the usual methods. But I am not sure whether fuzzy logic provides more or less insight or does use more or less parameters. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The original motivation for fuzzy logic was a linguistic observation, namely that in every day language we typically use attributions that express a degree of agreement, and that we even can reason with such degrees. So fuzzy logic is more a convenience for this machinery.&lt;/p&gt;&#xA;" OwnerUserId="4414" LastActivityDate="2010-12-12T11:43:56.160" />
  <row Id="14040" PostTypeId="1" AcceptedAnswerId="14050" CreationDate="2010-12-12T11:44:57.033" Score="2" ViewCount="102" Body="&lt;p&gt;Hey there,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need to find out for what $\alpha, \beta$ the following sum converges:&#xA;$$\sum_{n=2}^\infty n^\alpha (\log n)^\beta$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I thought I'd do that with the help of the integral criterion, that is to say I considered (after I had substituted $x = e^u$):&#xA;$$\int_2^\infty e^{u(\alpha+1)} \cdot u^\beta \mathrm du$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, I realised that for $\alpha + 1 &amp;gt; 0$, the exponential function will always dominate the monomial, so I only considered $\alpha + 1 \leq 0$. First, I let $\alpha + 1=0$, then I would be left with:&#xA;$$\int_2^\infty u^\beta \mathrm du$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I solved the integral and substituted back and ended up with:&#xA;$$\lim_{b \to \infty} \frac{(\log b)^{\beta+1}}{\beta+1}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I figured this would only converge for $\beta+1 \leq 0$ and thought: &quot;Ok, now I have some cases, where the sum converges: $\alpha = -1 \land \beta \leq -1$&quot;. To be sure, I checked the integral with the help of wolframalpha, and it told me the integral would not converge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What did I do wrong? Is there an easier way to find $\alpha$ and $\beta$ so that the sum converges?&lt;/p&gt;&#xA;" OwnerUserId="3787" LastActivityDate="2010-12-12T12:29:24.793" Title="When does $\sum_{n=2}^\infty n^\alpha (\log n)^\beta$ converge?" Tags="&lt;analysis&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="14041" PostTypeId="1" CreationDate="2010-12-12T11:50:34.047" Score="8" ViewCount="161" Body="&lt;p&gt;Given a bijection $f:\mathbb Z \to \mathbb Z$ where $\mathbb Z$ is the set of all integers, does there always exist two bijections $g:\mathbb Z \to \mathbb Z$ and $h:\mathbb Z \to \mathbb Z$ which satisfy $f(n) = g(n) + h(n), ~ \forall n \in \mathbb Z$ ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Evidently it does while $\mathbb Z$ is replaced with the set of all rational numbers $\mathbb Q$, since $f(q) = 2f(q) - f(q), ~ \forall q \in \mathbb Q$. Nevertheless, $2f$ doesn't maps $\mathbb Z$ onto $\mathbb Z$, thus it's not a bijection on $\mathbb Z$; therefor we need a new construction&#xA;for the $\mathbb Z$ case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Firstly we note that &amp;nbsp; $0 = 0 + 0$ &amp;nbsp; and&#xA;\begin{align*}&#xA;1 &amp;amp;= (-2) + 3, &amp;amp;\qquad -1 &amp;amp;= 2 + (-3),\\&#xA;2 &amp;amp;= 3 + (- 1), &amp;amp;-2 &amp;amp;= (-3) + 1,\\&#xA;3 &amp;amp;= 1 + 2, &amp;amp;-3 &amp;amp;= (-1) + (-2).&#xA;\end{align*}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's easy to show that any bijections on the symmetric integer interval $[-3,3]$ can be represented as a sum of two bijections; or equivalently, all integers $a \in [-3,3]$ can be represented as a sum of two others, i.e., $a = b + c$ where $b, c \in [-3, 3]$, while $a,b,c$ exhausted the integer interval $[-3,3]$ respectively, and keeps $0 = 0 + 0$. For convenience, let's denote this fact as $[-3,3] = [-3,3] + [-3,3]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose now we have $[-m, m] = [-m, m] + [-m, m]$, then we may obtain&#xA;$$[-m + aL, ~ m + aL] = [-m + bL, ~ m + bL] + [-m + cL, ~ m + cL],\qquad L = 2m+1,$$&#xD;&#xA;where $a,b,c \in [-n,n]$ for some positive integers $n$, provided $a = b + c$. Well, if $a,b,c$ form a representation, say, $[-n,n] = [-n,n] + [-n,n]$, then we may get a &quot;larger&quot; representation&#xA;$$[-m - nL, ~ m + nL] = [-m - nL, ~ m + nL] + [-m - nL, ~ m + nL], \qquad L = 2m+1,$$&#xD;&#xA;through the repeatedly use of &quot;small&quot; representations; and obviously, the larger representation &quot;contains&quot; the small representation as a &quot;subrepresentation&quot;, since $0 = 0 + 0$. From now on, we may construct a representation $\mathbb Z = \mathbb Z + \mathbb Z$ by induction, start from $[-3,3] = [-3,3] + [-3,3]$. Thus we've proved that any bijections on $\mathbb Z$ can be represented as a sum of two bijections on $\mathbb Z$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now a problem rises:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;for any integers&lt;/strong&gt; $n &amp;gt; 2$, &lt;strong&gt;is there always a representation&lt;/strong&gt; $[-n, n] = [-n, n] + [-n, n]$ ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that by our means of &quot;representation&quot;, $0 = 0 + 0$ is required; otherwise it's trivial; note also the above approach we've used for resolving the &quot;$\mathbb{Z}$-representation problem&quot; can't reach all cases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyone can answer this question? Please help me.&lt;/p&gt;&#xA;" OwnerUserId="4666" LastEditorUserId="4666" LastEditorDisplayName="" LastEditDate="2010-12-13T07:47:39.280" LastActivityDate="2010-12-13T11:39:31.333" Title="bijection = bijection + bijection on symmetric integer intervals" Tags="&lt;number-theory&gt;&lt;combinatorics&gt;&lt;algebra&gt;&lt;discrete-mathematics&gt;" AnswerCount="3" FavoriteCount="2" />
  <row Id="14042" PostTypeId="1" AcceptedAnswerId="14047" CreationDate="2010-12-12T11:52:32.550" Score="5" ViewCount="131" Body="&lt;p&gt;Is $\mathbf{Grp}$ a concrete category? I thought it is, but then the group of symmetries of a square and the quaternion group are both of the order 8, and they are not isomorphic as groups. But sets of the same cardinality are isomorphic as sets, so a standard forgetful functor $(G, \cdot) \mapsto G$ is not faithful! Am I missing something here?&lt;/p&gt;&#xA;" OwnerUserId="3793" LastActivityDate="2010-12-12T12:03:02.947" Title="Is $\mathbf{Grp}$ a concrete category?" Tags="&lt;category-theory&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="14043" PostTypeId="1" AcceptedAnswerId="14046" CreationDate="2010-12-12T11:56:46.117" Score="1" ViewCount="106" Body="&lt;p&gt;&lt;em&gt;&lt;strong&gt;Problem:&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&lt;strong&gt;Alex flip a fair coin three times. what is the probability that she gets two heads given that the first is a head?&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;My solution is based on the argument that from the given information the sample space can be limited to four elements {$HHH,HHT,HTH,HTT$} Of these {$HHH,HHT,HTH$} satisfies the required event.Thus the probability is equal to $\frac{3}{4} = 0.75$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the given answer is $0.5$,where exactly is the problem in my approach ?&lt;/p&gt;&#xA;" OwnerUserId="2109" LastEditorUserId="2109" LastEditorDisplayName="" LastEditDate="2010-12-12T12:08:30.603" LastActivityDate="2010-12-12T12:14:16.823" Title="Doubt in a probability problem" Tags="&lt;probability&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="14044" PostTypeId="1" AcceptedAnswerId="14057" CreationDate="2010-12-12T11:58:18.590" Score="5" ViewCount="261" Body="&lt;p&gt;I'm going through an &lt;a href=&quot;http://ocw.mit.edu/courses/mathematics/18-781-theory-of-numbers-spring-2003/exams/final.pdf&quot; rel=&quot;nofollow&quot;&gt;old final&lt;/a&gt; from 2003 on MIT's Opencourseware, and problem 6b is giving me a little trouble.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It asks for which primes $p$ is $34$ a square modulo $p$. I approached it like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left(\frac{34}{p}\right)=\left(\frac{2}{p}\right)\left(\frac{17}{p}\right)=(-1)^{(p^2-1)/8}\left(\frac{p}{17}\right).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I figure I can break it down into cases where $p\equiv 1,3,5,7\pmod{8}$. So if $p\equiv 1\pmod{8}$, then $(2|p)=1$, and thus I want $(p|17)=1$ as well. I calculated all the squares modulo $17$, and found them to be $1,2,4,8,9,13,15,16$. I suppose I could then go through all cases where $p\equiv 1\pmod{8}$, and $p\equiv 1,2,4,8,\dots\pmod{17}$, and then use the Chinese remainder theorem to find what $p$ is congruent to modulo $8\cdot 17$, but this seems very tedious to do for each case. First of all, is my method correct, and also, is there a better way to solve this question? Thank you.&lt;/p&gt;&#xA;" OwnerUserId="1583" LastEditorUserId="1583" LastEditorDisplayName="" LastEditDate="2011-01-08T01:37:39.447" LastActivityDate="2011-01-08T01:37:39.447" Title="Trick to showing for which primes $p$ is $34$ a square?" Tags="&lt;number-theory&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="1" />
  <row Id="14045" PostTypeId="2" ParentId="14040" CreationDate="2010-12-12T12:01:40.817" Score="1" ViewCount="" Body="&lt;p&gt;For $\alpha = \beta = -1$, you have something like $\int \frac{1}{x \log x}$, which is $\log \log x$, which diverges. For $\beta &amp;lt; -1$, though, it's safe.&lt;/p&gt;&#xA;" OwnerUserId="3406" LastActivityDate="2010-12-12T12:01:40.817" CommentCount="1" />
  <row Id="14046" PostTypeId="2" ParentId="14043" CreationDate="2010-12-12T12:02:41.690" Score="2" ViewCount="" Body="&lt;p&gt;Given that the first flip is a head, the problem is equivalent to asking, what is the probability of getting exactly $1$ head in $2$ flips. There are two possibilities, that the first flip is a head, and the second is tails, and vice versa. This gives the probability the given answer says.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;To address your question more directly, you should use conditional probability. Let $A$ be the event that Alex flips exactly $2$ heads, and $B$ be the event that the first flip is heads.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Based on the sample space you wrote out, note $P(A\cap B)=2/8$, as the two possibilities are $\{HHT,HTH\}$. Also, $P(B)=1/2$, as you listed the four possibilities. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now use the formula&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$P(A|B)=\frac{P(A\cap B)}{P(B)}.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think your main error is that $\{HHH\}$ does not satisfy the condition that the first flip is a head, and that there are exactly two heads, and thus should not be included when calculating the probability $P(A\cap B)$.&lt;/p&gt;&#xA;" OwnerUserId="1583" LastEditorUserId="1583" LastEditorDisplayName="" LastEditDate="2010-12-12T12:10:00.997" LastActivityDate="2010-12-12T12:10:00.997" CommentCount="6" />
  <row Id="14047" PostTypeId="2" ParentId="14042" CreationDate="2010-12-12T12:03:02.947" Score="9" ViewCount="" Body="&lt;p&gt;Yes, you have the wrong definition of a faithful functor. A functor &lt;em&gt;F&lt;/em&gt; is faithful precisely when &lt;em&gt;F&lt;/em&gt; sends &lt;strong&gt;morphisms&lt;/strong&gt; injectively, but nowhere is there required any kind of injectivity on &lt;strong&gt;objects&lt;/strong&gt;.&lt;/p&gt;&#xA;" OwnerUserId="677" LastActivityDate="2010-12-12T12:03:02.947" CommentCount="1" />
  <row Id="14048" PostTypeId="2" ParentId="14043" CreationDate="2010-12-12T12:14:16.823" Score="2" ViewCount="" Body="&lt;p&gt;The problem here is that if the answer is half, then what the question means is exactly two heads, so three heads are not allowed. In your solution you included the case with three heads.&lt;/p&gt;&#xA;" OwnerUserId="4594" LastActivityDate="2010-12-12T12:14:16.823" CommentCount="5" />
  <row Id="14049" PostTypeId="1" CreationDate="2010-12-12T12:22:56.877" Score="4" ViewCount="191" Body="&lt;p&gt;Many years ago I wrote this fractal generator:&#xA;&lt;a href=&quot;http://uberto.fractovia.org/&quot; rel=&quot;nofollow&quot;&gt;http://uberto.fractovia.org/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It was shareware but then I put it as open source.&#xA;It's written in Delphi, a language that I don't use anymore. So I'm playing lately with the idea of rewriting on some other lang (scala maybe?).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the mathematical part I'm not expert at all, so I don't know if there is something a mathematician would find interesting. Maybe none because of software like Mathematica. I really don't know.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the graphic oriented people I implemented some algos for continuous coloring, including the ones listed here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://math.stackexchange.com/questions/4035/continuous-coloring-of-a-mandelbrot-fractal&quot;&gt;Continuous coloring of a Mandelbrot fractal&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but I'd like to add some new ones.&lt;/p&gt;&#xA;" OwnerUserId="4615" LastActivityDate="2011-05-10T23:56:24.503" Title="Which features would be interesting for a mathematician in a fractal program?" Tags="&lt;algorithms&gt;&lt;applications&gt;&lt;fractals&gt;" AnswerCount="2" CommentCount="9" />
  <row Id="14050" PostTypeId="2" ParentId="14040" CreationDate="2010-12-12T12:29:24.793" Score="4" ViewCount="" Body="&lt;p&gt;Since, for any $\varepsilon &amp;gt; 0$ fixed, $(\log n)^\beta / n^\varepsilon \to 0$ as $n \to \infty$, it is obvious that the sum converges for $\alpha &amp;lt; -1$ and diverges for $\alpha &amp;gt; -1$ (regardless of $\beta$). When $\alpha = -1$ and $\beta = - 1$ the sum diverges, as indicated by FX (hence also for all $\beta &gt; -1$). We are left with the case $\alpha =-1$ and $\beta &amp;lt; -1$. Then you should compare with the integral&#xA;$$&#xD;&#xA;\int_2^\infty  {\frac{1}{{x(\log x)^{1 + \delta } }}\,{\rm d}x} ,&#xD;&#xA;$$&#xD;&#xA;where $\delta &amp;gt; 0$ fixed. Make a change of variable $x&amp;#39; = \log x$ to find out that the integral (and hence the sum) converges.&lt;/p&gt;&#xA;" OwnerUserId="2810" LastActivityDate="2010-12-12T12:29:24.793" />
  <row Id="14051" PostTypeId="1" AcceptedAnswerId="14061" CreationDate="2010-12-12T13:11:01.017" Score="3" ViewCount="230" Body="&lt;p&gt;I want to write&#xA;$P(x,y,z)=yx^{3}+zx^{3}+xy^{3}+zy^{3}+xz^{3}+yz^{3}$&#xD;&#xA;in terms of elementary symmetric polynomials, but I'm getting stuck at the first step. I know I should follow the proof of the fundamental theorem of symmetric polynomials using the Newton identities. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;First I pick out the 'biggest' monomial according to the lexicographical ordering: $yz^{3}$. Now I want to rewrite this as a polynomial in the elementary symmetric polynomials. I don't quite understand how to do this.&lt;/p&gt;&#xA;" OwnerUserId="2298" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-12-12T22:00:58.757" LastActivityDate="2010-12-12T22:00:58.757" Title="symmetric polynomials and the Newton identities" Tags="&lt;algebra&gt;&lt;polynomials&gt;" AnswerCount="2" />
  <row Id="14052" PostTypeId="2" ParentId="14036" CreationDate="2010-12-12T13:12:07.270" Score="1" ViewCount="" Body="&lt;p&gt;Put ${\bf n}:={\bf n}_1\times{\bf n}_2/|{\bf n}_1\times{\bf n}_2|$ and write ${\bf x}:=\xi_1{\bf n}_1+ \xi_2 {\bf n_2} + \xi_3 {\bf n}$ with unknown $\xi_i$. Then&#xA;$$\cos\theta_1={\bf x}\bullet {\bf n}_1=\xi_1 +\xi_2({\bf n}_1\bullet{\bf n}_2)$$ and similarly $\cos\theta_2=\ldots\ $. After $\xi_1$ and $\xi_2$ have been computed the value of $\xi_3$ follows from the condition $|{\bf x}|=1$.&lt;/p&gt;&#xA;" OwnerUserId="1303" LastActivityDate="2010-12-12T13:12:07.270" />
  <row Id="14053" PostTypeId="2" ParentId="13993" CreationDate="2010-12-12T13:51:05.770" Score="5" ViewCount="" Body="&lt;p&gt;There are actually two separate questions in your post.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Fourier transform of the travelling waves&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The first question is about Fourier decomposition of the travelling wave solutions $\psi(x,t) = f(x + ct) + g(x-ct)$. It suffices to notice the following change of variables: let $u = x + ct$ and $v = x - ct$. Then inverting the change of variables you get $x = \frac12 (u+v)$ and $t = \frac1{2c} (u - v)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then we can re-write $\psi$ in this new coordinate system as $f(u) + g(v)$. Notice that $f$ is constant along $v$ direction, and $g$ along the $u$ direction. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since now $f$ is a function of one variable only, you can take the Fourier transform in the $u$ variable:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ f(u) = \int \hat{f}(\omega_u) \exp ( i u \omega_u ) d\omega_u $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So with a further change of variables, you have that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ f(x,t) = f(x + ct) = \int \hat{f}(\omega) \exp (i\omega(x + ct)) d\omega $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;as desired (notice that I normalized the exponential differently. You can rescale coordinates once more to get it to be the form with $x/c + t$ if you so wish). By similarly playing with the $v$ variable you get a Fourier representation of $g$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, to find $\hat{f}(\omega_u)$, we take the Fourier transform in the $u$ variable&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \hat{f}(\omega_u) = \frac{1}{2\pi} \int f(u) \exp (i u \omega_u) du $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now remember that $f$ is independent of $v$! So writing $f = f(u,v) = f(u)$, for any curve $v = v(u)$ the integral is the same, that is, we can re-write&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \hat{f}(\omega_u) = \frac{1}{2\pi} \int f(u,v(u)) \exp(i u \omega_u) du $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, choose the curve to be $v = -u$, which translates to $x = 0$ by our change of variable formula before. And do a change of variable back to $t,x$, we have that along the curve $v = -u$, $u = ct$. So the integral becomes&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \hat{f}(\omega_u) = \frac{1}{2\pi} \int f(0,t) \exp (i ct \omega_u) c dt$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;again, the extra factor of $c$ can be removed by a further change of variables on $\omega_u$. &lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Two and higher dimensions&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;In higher dimensions, the situation is analogous, but we do not have the method of characteristics for us to start the first step. So instead, we start with the general high dimensional Fourier transform. (What I am about to write is mathematically non-rigorous, but morally correct.) Consider space-time coordinates $(x_1,\ldots,x_n,t)$ and space-time frequencies $(\xi_1,\ldots,\xi_n, \omega)$. We take the full Fourier transform of a solution to the wave equation in the following way. Suppose&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \partial_t^2 \psi = c\triangle \psi $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\triangle$ is the Laplacian in $n$ dimensions. Consider the Fourier transform&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \hat{\psi}(\xi_1,\ldots,\xi_n,\omega) = \int_{\mathbb{R}^{n+1}} f(x_1,\ldots,x_n,t) \exp [ -2\pi i (x_1\xi_1 + \cdots + x_n\xi_n + ct\omega) ] c dx_1\cdots dx_n dt $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(this is the slightly questionable step. In general this integral will not converge; but it captures the idea of the thing.) Using the properties of the Fourier transform to replace derivatives by multiplication of the frequencies, you get that $\hat{\psi}$ solves the algebraic equation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ - \omega^2 \hat{\psi} = -(\xi_1^2 + \cdots + \xi_n^2) \hat{\psi} $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ (\xi_1^2 + \cdots + \xi_n^2 - \omega^2) \hat{\psi} = 0 $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an algebraic equation, this means that if the number inside the first parenthesis is not zero, we must have $\hat{\psi} = 0$. And that $\hat{\psi}$ is only non-zero where the term in the parenthesis vanishes. (In mathematical jargon, this indicates that $\hat{\psi}$ is a distribution/measure, and not strictly speaking a function.) And in particular, we have the &quot;dispersion law&quot; that the only space-time frequencies contributing to $\psi$ are those where&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \omega = \pm \sqrt{ \xi_1^2 + \cdots + \xi_n^2 } $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we can write $\hat{\psi} = \hat{\psi}_+ + \hat{\psi}_-$, the forward and backward waves, with $\hat{\psi}_\pm$ only functions of $\xi_1,\ldots, \xi_n$. This gives us the &lt;em&gt;plane wave decomposition&lt;/em&gt; of a solution to the wave equation by taking the inverse Fourier transform (to save space, I write $x$ for the vector $(x_1,\ldots,x_n)$ and ditto $\xi$, using $\cdot$ to denote the scalar product)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \psi(x,t) = \int_{\mathbb{R}^n} \hat{\psi}_+(\xi) e^{2\pi i(\xi\cdot x + c|\xi|t)} + \hat{\psi}_-(\xi) e^{2\pi i(\xi\cdot x - c|\xi| t)} d\xi $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Remark&lt;/strong&gt; this is only one of the very many ways of solving the linear wave equation. (In a recent PDE course where I work, the lecturer allegedly gave 6 different methods...) Maths textbook I know tend not to want to go into the above Fourier transform derivation, because it is not very rigorous (it can be made rigorous with &lt;em&gt;a lot&lt;/em&gt; of additional preparations). The above idea should be treated more as a morally correct guiding principle.&lt;/p&gt;&#xA;" OwnerUserId="1543" LastActivityDate="2010-12-12T13:51:05.770" CommentCount="2" />
  <row Id="14054" PostTypeId="2" ParentId="14051" CreationDate="2010-12-12T14:13:14.493" Score="6" ViewCount="" Body="&lt;p&gt;Edit: As per Bill's comment I would like to clarify that this is not related to Gauss' proof.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$P(x,y,z)=yx^{3}+zx^{3}+xy^{3}+zy^{3}+xz^{3}+yz^{3}$$&#xD;&#xA;$$=x^3(y+z+x-x)+y^3(x+z+y-y)+z^3(x+y+z-z)$$&#xD;&#xA;$$=x^3(x+y+z)+y^3(x+y+z)+z^3(x+y+z)-x^4-y^4-z^4$$&#xD;&#xA;$$=(x+y+z)(x^3+y^3+z^3)-(x^4+y^4+z^4)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now you can use identities for &lt;a href=&quot;http://en.wikipedia.org/wiki/Newton%27s_identities#Formulation_in_terms_of_symmetric_polynomials&quot; rel=&quot;nofollow&quot;&gt;power sums&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="3431" LastEditorUserId="3431" LastEditorDisplayName="" LastEditDate="2010-12-12T19:25:29.377" LastActivityDate="2010-12-12T19:25:29.377" CommentCount="11" />
  <row Id="14055" PostTypeId="2" ParentId="14026" CreationDate="2010-12-12T14:39:14.803" Score="3" ViewCount="" Body="&lt;p&gt;If you do any commutative algebra or algebraic geometry, &lt;a href=&quot;http://www.math.uiuc.edu/Macaulay2/&quot; rel=&quot;nofollow&quot;&gt;Macaulay2&lt;/a&gt; is amazing. It requires minimal system resources and is free. There is a springer book about computations in Macaulay2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.math.uiuc.edu/Macaulay2/Book/&quot; rel=&quot;nofollow&quot;&gt;http://www.math.uiuc.edu/Macaulay2/Book/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, here is a comparison of computer algebra systems on wikipedia.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Comparison_of_computer_algebra_systems&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Comparison_of_computer_algebra_systems&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3431" LastActivityDate="2010-12-12T14:39:14.803" />
  <row Id="14056" PostTypeId="2" ParentId="14026" CreationDate="2010-12-12T15:23:49.193" Score="6" ViewCount="" Body="&lt;p&gt;I found &lt;a href=&quot;http://www.sagemath.org/&quot; rel=&quot;nofollow&quot;&gt;Sage&lt;/a&gt; pretty useful for number theory and algebraic geometry. But it can do lots of other stuff, too.&lt;/p&gt;&#xA;" OwnerUserId="1766" LastActivityDate="2010-12-12T15:23:49.193" />
  <row Id="14057" PostTypeId="2" ParentId="14044" CreationDate="2010-12-12T15:33:12.500" Score="7" ViewCount="" Body="&lt;p&gt;You have all that you need right there:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left(\frac{34}{p}\right)=\left(\frac{2}{p}\right)\left(\frac{17}{p}\right)=(-1)^{(p^2-1)/8}\left(\frac{p}{17}\right).$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if $p \equiv \pm 1 \pmod{8}$ then $p$ must be a quadratic residue mod 17, otherwise it must be a quadratic non-residue mod 17. (But delete 14 from your list first!)&lt;/p&gt;&#xA;" OwnerUserId="1508" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-12T23:11:34.400" LastActivityDate="2010-12-12T23:11:34.400" CommentCount="1" />
  <row Id="14058" PostTypeId="1" CreationDate="2010-12-12T15:52:50.423" Score="0" ViewCount="204" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Possible Duplicate:&lt;/strong&gt;&lt;br&gt;&#xA;  &lt;a href=&quot;http://math.stackexchange.com/questions/283/is-0-a-natural-number&quot;&gt;Is 0 a natural number?&lt;/a&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;p&gt;There seems to be no consensus, although perhaps one is gathering over the centuries to say yes to the first question and identify $\mathbb{N}$ with $\omega$ : the set of finite Von Neumann ordinal numbers. That leaves $\textbf{N}$ = {$1$, $2$, ...} for the counting numbers (who counts with $0$?). You can dodge the issue by writing $\mathbb{Z}_{\geq0}$, $\mathbb{Z}_+$, etc., but it seems awkward to use derived and complicated notation for something so basic.&lt;/p&gt;&#xA;" OwnerUserId="875" LastActivityDate="2011-05-23T16:23:53.803" ClosedDate="2010-12-12T17:45:56.347" Title="Do the natural numbers include zero? And what should we call them, with or without zero?" Tags="&lt;notation&gt;" AnswerCount="0" CommentCount="10" />
  <row Id="14060" PostTypeId="1" CreationDate="2010-12-12T16:25:23.250" Score="2" ViewCount="111" Body="&lt;hr&gt;&#xA;&#xA;&lt;p&gt;$ \qquad \qquad $ &lt;strong&gt;&lt;em&gt;The greatest term in $\left(1+x\right)^{2n}$ has the greatest cofficient if $\frac{n}{n+1} \lt x \lt \frac{n+1}{n}$&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Can we derive something like this for $n$ in general? Please give me some hints.&lt;/p&gt;&#xA;" OwnerUserId="2109" LastEditorUserId="2109" LastEditorDisplayName="" LastEditDate="2010-12-12T16:33:13.587" LastActivityDate="2010-12-12T16:52:54.347" Title="How can we prove this property?" Tags="&lt;algebra-precalculus&gt;&lt;binomial-coefficients&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="14061" PostTypeId="2" ParentId="14051" CreationDate="2010-12-12T16:46:28.017" Score="5" ViewCount="" Body="&lt;p&gt;By Gauss's algorithm, if  $\rm\ z^a\ y^b\ x^c\ $ is the highest w.r.t. lex order $\rm\ z &amp;gt; y &amp;gt; x\ $ then you subtract $\rm\ s_1^{a-b}\ s_2^{b-c}\ s_3^c\:.\:$ Thus since $\rm\ z^3\ y $ is highest you subtract $\rm s_1^{3-1}\ s_2^{1-0}\ s_3^0\ = (x+y+z)^2\ (xy+yz+zx)$ from $\rm\:P\:$. The result is smaller in lex-order, so iterating this reduction yields a representation of $\rm\:P\:$ in terms of elementary symmetric polynomials $\rm\:s_i\:.\:$ Here the algorithm terminates in two more steps.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As I mentioned in a &lt;a href=&quot;http://math.stackexchange.com/questions/3550/using-grobner-bases-for-solving-polynomial-equations/3562#3562&quot;&gt;prior post&lt;/a&gt;, Gauss's algorithm is the earliest known example of using lex-order reduction as in the Grobner basis algorithm. For a nice exposition see Chapter 7 of Cox, Little, O'Shea: Ideals, Varieties and Algorithms. They also give generalizations to  the ring of invariants of a finite matrix group $\rm G \subset  GL(n,k)$. Here's an excerpt which, coincidentally, presents this example. You might find it helpful to first read the example at the end before reading the proof.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/tx23R.jpg&quot; alt=&quot;alt text&quot;&gt;&#xA;&lt;img src=&quot;http://i.imgur.com/cGSvO.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-12-12T17:32:13.807" LastActivityDate="2010-12-12T17:32:13.807" CommentCount="4" />
  <row Id="14062" PostTypeId="2" ParentId="14060" CreationDate="2010-12-12T16:52:54.347" Score="4" ViewCount="" Body="&lt;p&gt;Hint: I think you can solve this by looking at the adjacent terms of the middle term. If the middle term is the biggest term, it has to be bigger than both adjacent terms. This will give you the bounds.&lt;/p&gt;&#xA;" OwnerUserId="3567" LastActivityDate="2010-12-12T16:52:54.347" />
  <row Id="14063" PostTypeId="1" CreationDate="2010-12-12T17:28:26.870" Score="4" ViewCount="221" Body="&lt;p&gt;How can I figure out the classical construction (direct sum, product, pullbacks, and in general direct and inverse limits) in the category made by chain complexes and chain maps (of abelian groups or any abelian stuff)? Because of this category is abelian it &lt;em&gt;must&lt;/em&gt; have (co)limits, isn't it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular take&#xA;$$&#xD;&#xA;\begin{gather*}&#xD;&#xA;\dots\to A_n\to A_{n-1}\to\dots\\&#xD;&#xA;\dots\to B_n\to B_{n-1}\to\dots&#xD;&#xA;\end{gather*}&#xD;&#xA;$$&#xD;&#xA;I would like to say that $\mathcal A\oplus\mathcal B$ is &quot;what I want it to be&quot;, $A_n\oplus B_n$ with obvious maps. But a standard argument doesn't allow me to conclude it: maybe it is false? A single word with the right reference will be enough to close the topic; I am now reading Hilton &amp;amp; Stammbach.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;: I would like to add what I've tried to do, but seems difficult not to invoke some diagrams, which I'm not able to draw without a suitable package here. However, first of all both complexes inject in the sum by maps which are part of a chian complex, $\iota_n^A,\iota_n^B$. Then, consider another complex $\{C_n,\partial_n^C\}$ and a couple of chain maps $\{a_n\colon A_n\to C_n\}$, $\{b_n\colon B_n\to C_n\}$. For each $n$ there exists a map $\alpha_n\colon A_n\oplus B_n\to C_n$ factoring the $a_n$ and the $b_n$s. Then I would like to show that the maps $\alpha_n$ are part of a chain map between the sum and the complex $\mathcal C$, but trying to prove it I can only conclude that in the diagram (vertical rows are the $\alpha_n$s)&#xA;$$&#xD;&#xA;\begin{array}{ccc}&#xD;&#xA;A_n\oplus B_n &amp;amp;\xrightarrow{\partial_n^\oplus}&amp;amp; A_{n-1}\oplus B_{n-1} \\&#xD;&#xA;\downarrow &amp;amp;&amp;amp; \downarrow\\&#xD;&#xA;C_n &amp;amp;\xrightarrow[\partial_n^C]{}&amp;amp; C_{n-1}&#xD;&#xA;\end{array}&#xD;&#xA;$$&#xD;&#xA;which I want to be commutative, aka $\alpha_{n-1}\partial_n^\oplus=\partial_n^C\alpha_n$, I have $\alpha_{n-1}\partial_n^\oplus\iota_n^A=\partial_n^C\alpha_n\iota_n^A$. How can I remove the &lt;em&gt;iotas&lt;/em&gt;?&lt;/p&gt;&#xA;" OwnerUserId="685" LastEditorUserId="685" LastEditorDisplayName="" LastEditDate="2010-12-12T19:16:43.190" LastActivityDate="2010-12-13T06:00:35.267" Title="Direct sum of complexes" Tags="&lt;homological-algebra&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="14064" PostTypeId="1" CreationDate="2010-12-12T18:11:10.360" Score="1" ViewCount="38" Body="&lt;p&gt;I'm computing an LDA (linear discriminant analysis) transform, for an application I'm working on, and I've been following these &lt;a href=&quot;http://www.ed.uiuc.edu/courses/EdPsy494/lectures/DiscriminantAnalysis-online.pdf&quot; rel=&quot;nofollow&quot;&gt;notes&lt;/a&gt; (starting at page 36, especially slide 47 in green).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm doing this in Python (with numpy and scipy), and this is what I have come up with:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;import numpy as np&#xA;from scipy.linalg import sqrtm&#xA;...&#xA;sw_inv_sqrt = np.linalg.inv(sqrtm(self.sigma_within))&#xA;self.d, self.v = np.linalg.eig(&#xA;    np.dot(&#xA;        np.dot(sw_inv_sqrt, self.sigma_between),&#xA;        sw_inv_sqrt&#xA;        ))&#xA;self.v = np.dot(sw_inv_sqrt, self.v)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I know that this implementation is correct as I have compared it to others.  My concern is whether this is &lt;strong&gt;good&lt;/strong&gt; solution in the &lt;strong&gt;numerical sense&lt;/strong&gt;.  In comparing my solution to others, they match only to about 6 decimal places.  Is there a better way to do this numerically?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;NB. I originally posted this to Stack Overflow and it was recommended to me to post it here.&lt;/p&gt;&#xA;" OwnerUserId="4676" LastEditorUserId="4676" LastEditorDisplayName="" LastEditDate="2010-12-12T18:19:30.617" LastActivityDate="2010-12-12T18:19:30.617" Title="good numerical solution for LDA transformation" Tags="&lt;numerical-methods&gt;&lt;numerical-optimization&gt;" />
  <row Id="14065" PostTypeId="1" AcceptedAnswerId="14067" CreationDate="2010-12-12T18:11:23.713" Score="3" ViewCount="159" Body="&lt;p&gt;I have a question about whether the following two sets on the two sides are the same:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;\lim_{a \rightarrow \infty} \, \limsup_{n \rightarrow \infty} \, \{ x \in S \, | \, f_n(x)  &amp;gt; a \}&#xD;&#xA;=&#xD;&#xA;\left\{ x \in S \, \Bigm| \, \lim_{n \rightarrow \infty} f_n(x) =\infty \right\}\quad ?&#xD;&#xA;$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\{f_n\}$ is a sequence of real-valued functions defined on a set $S$. How can you explain it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance!&lt;/p&gt;&#xA;" OwnerUserId="4674" LastEditorUserId="4674" LastEditorDisplayName="" LastEditDate="2010-12-12T23:28:34.800" LastActivityDate="2010-12-13T02:09:21.893" Title="Are these two sets same" Tags="&lt;functions&gt;&lt;elementary-set-theory&gt;&lt;limit&gt;" AnswerCount="2" CommentCount="7" FavoriteCount="1" />
  <row Id="14066" PostTypeId="1" AcceptedAnswerId="14117" CreationDate="2010-12-12T19:06:36.160" Score="15" ViewCount="315" Body="&lt;p&gt;Does there exist a good reference on the algorithms used by calculators, especially on the trig and transcendental funtions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would still like to know how Casio generates its random numbers. I still wonder if they are any good.&lt;/p&gt;&#xA;" OwnerUserId="2809" LastActivityDate="2010-12-13T14:51:16.730" Title="Calculator algorithms" Tags="&lt;algorithms&gt;&lt;calculator&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="6" />
  <row Id="14067" PostTypeId="2" ParentId="14065" CreationDate="2010-12-12T19:12:04.777" Score="4" ViewCount="" Body="&lt;p&gt;Consider a sequence of functions $f_n:\mathbb{R} \to \mathbb{R}$ defined by&#xA;$f_n {(x)} = n$ if $x=0$ and $n$ is even, and $f_n {(x)} = 0$ otherwise. The right-hand side is obviously $\emptyset$, while the left-hand should  be $\lbrace 0 \rbrace$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Answering the additional questions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose that  $f_n{(x)}$ is an increasing sequence for each $x$. If $x$ belongs to the left set, then it must belong to the set $\lim \sup _{n \to \infty } \{ x \in S|f_n (x) &amp;gt; a\}$ for any $a &amp;gt; 0$ fixed. This means that for any $a&amp;gt;0$, there are infinitely many $n$ such that $f_n (x) &amp;gt; a$. Since $f_n (x)$ in increasing, it must converge to a positive number, or diverge to $\infty$. But $a&amp;gt;0$ is arbitrary, hence $f_n (x)$ must diverge to infinity. That is, $x$ belongs to the right set. Since the right set is contained in the left one, we conclude that both sets are equal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: The fact that the right set is contained in the left one, is proved as follows. Suppose that $x$ belongs to the right set, and let $a&amp;gt;0$ be arbitrary but fixed. Then, for all sufficiently large $n$, $f_n {(x)} &amp;gt; a$. In particular, $x \in \lim \sup _{n \to \infty } \{ x \in S|f_n (x) &amp;gt; a\}$. Since this is true for any $a &amp;gt; 0$, $x$ belongs to the left set. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: The following point should be stressed. Denote by $E_a$ the set $\lim \sup _{n \to \infty } \{ x \in S|f_n (x) &amp;gt; a\}$.&#xA;If $x \in E_a$, then $x \in E_{a&amp;#39;}$ for any $a&amp;#39; &amp;lt; a$. It follows that $\lim \sup _{a \to \infty} E_a = \lim \inf _{a \to \infty} E_a$; hence, by definition, the limit $\lim  _{a \to \infty} E_a$ exists, and is equal to $\lim \sup _{a \to \infty} E_a = \lim \inf _{a \to \infty} E_a$. So, the left set in the question is indeed properly defined, and $x \in \lim  _{a \to \infty} E_a$ means, in our case, that $x \in E_a$ for every $a$. Finally, note that always the $\lim \inf$ is a subset of the $\lim \sup$ (in analogy with the case of sequences of real numbers, where $\leq$ plays the role of $\subseteq$). &lt;/p&gt;&#xA;" OwnerUserId="2810" LastEditorUserId="2810" LastEditorDisplayName="" LastEditDate="2010-12-13T02:09:21.893" LastActivityDate="2010-12-13T02:09:21.893" CommentCount="25" />
  <row Id="14068" PostTypeId="2" ParentId="14063" CreationDate="2010-12-12T19:26:21.450" Score="2" ViewCount="" Body="&lt;p&gt;In a preadditive category (it's clear that chain complexes are preadditive), the idea of a  direct sum can be described directly. Given two objects $A, B$, a direct sum $C$ has morphisms $i_1: A \to C, i_2: B \to C$ and projections $p_1: C \to A,  p_2: C \to B$ that satisfy $i_1 p_1 + i_2 p_2 = 1_C$ and $p_1 i_1 = 1_A, p_2 i_2 = 1_B$. Indeed, it is easy to get such maps in a direct sum (inclusion and projection); conversely if we get such maps one can check that $C$ is a coproduct and a direct product by explicitly using these maps. Namely, one checks that such maps guarantee a direct sum in the case of abelian groups, and then applies hom-sets and Yoneda nonsense (if we have $\hom(X, C) = \hom(X, A) \oplus \hom(X, B)$ for all $X$ naturally, then we're done). I'm being a little vague here and only sketching it; if it's not clear, I can clarify.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is clearly the case for the construction of chain complexes that you described above. The maps $p_1, p_2, i_1, i_2$ are the natural projections and  inclusions piecewise. They are maps of complexes, as one can check, and they satisfy the required identities because they do for abelian groups.&lt;/p&gt;&#xA;" OwnerUserId="536" LastEditorUserId="536" LastEditorDisplayName="" LastEditDate="2010-12-13T06:00:35.267" LastActivityDate="2010-12-13T06:00:35.267" />
  <row Id="14069" PostTypeId="1" CreationDate="2010-12-12T19:46:55.357" Score="5" ViewCount="204" Body="&lt;p&gt;If a Skoda sets off at an average speed of 20 miles per hour at 7 am and a Porsche sets off at an average speed of 40 miles per hour at 4 pm in the same direction, at what time will the Porsche catch the Skoda?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could you please explain how it works?&lt;/p&gt;&#xA;" OwnerUserId="4678" LastEditorUserId="2109" LastEditorDisplayName="" LastEditDate="2010-12-13T07:36:16.417" LastActivityDate="2011-05-23T16:20:16.100" Title="If a second car leaves later at a higher speed, how long will it take to catch up?" Tags="&lt;algebra-precalculus&gt;&lt;quantitative-aptitude&gt;" AnswerCount="5" CommentCount="4" />
  <row Id="14070" PostTypeId="2" ParentId="14066" CreationDate="2010-12-12T19:47:06.243" Score="6" ViewCount="" Body="&lt;p&gt;See &lt;a href=&quot;http://en.wikipedia.org/wiki/CORDIC&quot; rel=&quot;nofollow&quot;&gt;CORDIC&lt;/a&gt;. See also this question: &lt;a href=&quot;http://stackoverflow.com/questions/2169641/where-to-find-algorithms-for-standard-math-functions/2169666&quot;&gt;http://stackoverflow.com/questions/2169641/where-to-find-algorithms-for-standard-math-functions/2169666&lt;/a&gt; and this review: &lt;a href=&quot;http://mathdl.maa.org/mathDL/19/?pa=reviews&amp;amp;sa=viewBook&amp;amp;bookId=65790&quot; rel=&quot;nofollow&quot;&gt;http://mathdl.maa.org/mathDL/19/?pa=reviews&amp;amp;sa=viewBook&amp;amp;bookId=65790&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="589" LastEditorUserId="589" LastEditorDisplayName="" LastEditDate="2010-12-12T19:52:18.860" LastActivityDate="2010-12-12T19:52:18.860" CommentCount="1" />
  <row Id="14071" PostTypeId="2" ParentId="14063" CreationDate="2010-12-12T19:54:35.643" Score="4" ViewCount="" Body="&lt;p&gt;The category of complexes in an abelian category $\mathcal{A}$ is a full subcategory of $\text{Fun}({\mathbb{Z}},\mathcal{A})$, where $\mathbb{Z}$ is partially ordered under reverse inequality.  So if we know how these constructions are performed in the category of functors from $\mathbb{Z}$ to $\mathcal{A}$, we'll have the natural candidates to the category of complexes. The standard result is that (co)limits in $\text{Fun}({\mathcal{D}},\mathcal{C})$, where $\mathcal{D}$ is a small category and $\mathcal{C}$ is a category, are computed pointwise. Take a look at Borceux's 'Handbook of Categorical Algebra, Vol.$1$' section $2.15$. There he explains the precise meaning of being computed pointwise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since any abelian category is finitely (co)complete, we can compute any finite (co)limit in $\text{Fun}({\mathbb{Z}},\mathcal{A})$ pointwise. If we consider a (co)complete category, e.g., the category of modules over a ring, we can compute any (co)limit pointwise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you think Borceux's book is too terse, there is a similar discussion in Rotman's 'An Introduction to Homological Algebra' on page $317$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added:&lt;/strong&gt; In order to remove the iotas you will need to prove that $\alpha_{n-1}\partial_n^\oplus\iota_n^B=\partial_n^C\alpha_n\iota_n^B$. Now use the fact that there is only one morphism $\varphi: A_n \oplus B_n \rightarrow C_{n-1}$ such that $\varphi \iota_n^A = \alpha_{n-1}\partial_n^\oplus\iota_n^A$ and $\varphi \iota_n^B = \alpha_{n-1}\partial_n^\oplus\iota_n^B $.&lt;/p&gt;&#xA;" OwnerUserId="329" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2010-12-12T22:26:45.837" LastActivityDate="2010-12-12T22:26:45.837" CommentCount="4" />
  <row Id="14072" PostTypeId="2" ParentId="14069" CreationDate="2010-12-12T19:55:28.513" Score="6" ViewCount="" Body="&lt;p&gt;Hint: Distance covered is speed$\times$ time traveled. If you assume the time at which the Skoda leaves to be $t=0$, distance covered by Skoda after $t$ hours is $20t$ and distance covered by Porsche after 4pm is $40(t-9)$. &lt;/p&gt;&#xA;" OwnerUserId="3431" LastActivityDate="2010-12-12T19:55:28.513" CommentCount="5" />
  <row Id="14073" PostTypeId="2" ParentId="14069" CreationDate="2010-12-12T20:06:38.077" Score="2" ViewCount="" Body="&lt;p&gt;We could look at this problem slightly differently without resorting to algebraic variables. From 7 am to 4 pm the Skoda would have travelled 180 miles. Thus the Porsche has to cover 180 miles extra to overtake the Skoda. In each hour the Porsche can only cover 20 miles extra, so the Porsche would need another extra 9 hours to do that, thus finishing at 1 am.&lt;/p&gt;&#xA;" OwnerUserId="4594" LastEditorUserId="4594" LastEditorDisplayName="" LastEditDate="2011-05-23T16:20:16.100" LastActivityDate="2011-05-23T16:20:16.100" />
  <row Id="14074" PostTypeId="2" ParentId="14065" CreationDate="2010-12-12T20:07:33.520" Score="1" ViewCount="" Body="&lt;p&gt;For question number two, construct the following sequence of $f_n(x)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;f_n(x)  = \left\{\begin{array}{ll} &#xA;k+\epsilon_n, &amp;amp; \mbox{for $x=1/k$ with $k=1,\ldots,n$,} \\&#xA;\epsilon_n, &amp;amp; \mbox{otherwise.}&lt;br&gt;&#xA;\end{array}\right.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With the sequence of $\epsilon_n=1-\frac{1}{2^n}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I did not make any mistake, the right-hand side should be empty, while the left-hand side contains $0$. (This example was constructed with the &lt;a href=&quot;http://en.wikipedia.org/wiki/Limit_superior_and_limit_inferior#General_set_convergence&quot; rel=&quot;nofollow&quot;&gt;general set convergence&lt;/a&gt; in mind.)&lt;/p&gt;&#xA;" OwnerUserId="3567" LastEditorUserId="3567" LastEditorDisplayName="" LastEditDate="2010-12-12T21:27:15.213" LastActivityDate="2010-12-12T21:27:15.213" CommentCount="1" />
  <row Id="14075" PostTypeId="1" AcceptedAnswerId="14079" CreationDate="2010-12-12T20:08:55.623" Score="7" ViewCount="342" Body="&lt;p&gt;I have two $2\times 2$ matrices, $A$ and $B$, with the same determinant. I want to know if they are similar or not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I solved this by using a matrix called $S$:&#xA;$$\left(\begin{array}{cc}&#xD;&#xA;a&amp;amp; b\\&#xD;&#xA;c&amp;amp; d&#xD;&#xA;\end{array}\right)$$&#xD;&#xA;and its inverse in terms of $a$, $b$, $c$, and $d$, then showing that there was no solution to $A = SBS^{-1}$. That worked fine, but what will I do if I have $3\times 3$ or $9\times 9$ matrices? I can't possibly make system that complex and solve it.  How can I know if any two matrices represent the &quot;same&quot; linear transformation with different bases?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is, how can I find $S$ that change of basis matrix?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried making $A$ and $B$ into linear transformations... but without the bases for the linear transformations I had no way of comparing them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I have read that similar matrices will have the same eigenvalues... and the same &quot;trace&quot; --but my class has not studied these yet. Also, it may be the case that some matrices with the same trace and eigenvalues are not similar so this will not solve my problem.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have one idea. Maybe if I look at the reduced col. and row echelon forms that will tell me something about the basis for the linear transformation? I'm not really certain how this would work though? Please help.&lt;/p&gt;&#xA;" OwnerUserId="4681" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-13T06:04:12.430" LastActivityDate="2010-12-13T06:04:12.430" Title="How do I tell if matrices are similar?" Tags="&lt;linear-algebra&gt;&lt;matrices&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="14076" PostTypeId="2" ParentId="14069" CreationDate="2010-12-12T20:10:20.537" Score="1" ViewCount="" Body="&lt;p&gt;Let's say Ds = distance traveled by the Skoda and Dp = distance traveled by the Porsche.&#xA;If we set t to be the time after 4pm that the cars are traveling, then we set up the formulas as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dp = 40t&#xA;Ds = 20t + 20(9) *the Skoda has been travelling from 7am to 4pm, or 9 hours.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the cars meet, the distance traveled should be the same.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;40t = 20t + 180&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Subtract 20t from both sides:  20t = 180&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Divide both sides by 20: t = 9&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, after 4pm, the Porsche should catch up with the Skoda 9 hours later, or at 1pm.&lt;/p&gt;&#xA;" OwnerUserId="4680" LastActivityDate="2010-12-12T20:10:20.537" CommentCount="0" />
  <row Id="14077" PostTypeId="2" ParentId="13912" CreationDate="2010-12-12T20:14:27.150" Score="1" ViewCount="" Body="&lt;p&gt;I have one small insight just for finite groups which works for my own personal intuition.  Since you seem to feel like topology and analysis make sense and are well motivated, then why not think about the elements of a finite group as points on a continuous manifold, in other words, represent the elements of a finite group using $N\times N$ matricies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lie groups are continuous manifolds with a group structure, and they are manifestly topological.  Furthermore, a finite subgroup of $GL(\mathbb{R}^N)$ is, by definition, a &lt;em&gt;representation&lt;/em&gt; of some finite group.  So the elements of a finite group are represented by points on a manifold associated with some subgroup of $GL(\mathbb{R}^N)$, and these elements are connected to each other by continuous paths.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the way, if anyone is looking for a good introduction to linear groups and representation theory, I highly recommend &lt;a href=&quot;http://books.google.com/books?id=m1VQi8HmEwcC&amp;amp;printsec=frontcover&amp;amp;dq=brian+hall+lie+groups&amp;amp;hl=en&amp;amp;ei=lzIFTYbnN4OclgeS5t2DCA&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=1&amp;amp;ved=0CCsQ6AEwAA#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;B. Hall&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1417" LastEditorUserId="1417" LastEditorDisplayName="" LastEditDate="2010-12-12T21:40:39.370" LastActivityDate="2010-12-12T21:40:39.370" />
  <row Id="14078" PostTypeId="1" CreationDate="2010-12-12T20:18:36.517" Score="5" ViewCount="166" Body="&lt;p&gt;If your salary is increased by 10% and then after 6 months by a further 50%, what is the overall % increase?&lt;/p&gt;&#xA;" OwnerUserId="4678" LastEditorUserId="2109" LastEditorDisplayName="" LastEditDate="2010-12-13T07:36:58.180" LastActivityDate="2011-05-22T18:45:32.087" Title="Word based problem with percentages" Tags="&lt;algebra-precalculus&gt;&lt;quantitative-aptitude&gt;" AnswerCount="4" CommentCount="1" />
  <row Id="14079" PostTypeId="2" ParentId="14075" CreationDate="2010-12-12T20:18:57.080" Score="5" ViewCount="" Body="&lt;p&gt;There is something called &quot;canonical forms&quot; for a matrix; they are special forms for a matrix that can be obtained intrinsically from the matrix, and which will allow you to easily compare two matrices of the same size to see if they are similar or not. They are indeed based on eigenvalues and eigenvectors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At this point, without the necessary machinery having been covered, the answer is that it is difficult to know if the two matrices are the same or not. The simplest test you can make is to see whether their &lt;a href=&quot;http://en.wikipedia.org/wiki/Characteristic_polynomial&quot; rel=&quot;nofollow&quot;&gt;characteristic polynomials&lt;/a&gt; are the same. This is &lt;em&gt;necessary&lt;/em&gt;, but &lt;strong&gt;not sufficient&lt;/strong&gt; for similarity (it is related to having the same eigenvalues).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you have learned about canonical forms, one can use either the &lt;a href=&quot;http://en.wikipedia.org/wiki/Jordan_canonical_form&quot; rel=&quot;nofollow&quot;&gt;Jordan canonical form&lt;/a&gt; (if the characteristic polynomial splits) or the &lt;a href=&quot;http://en.wikipedia.org/wiki/Frobenius_normal_form&quot; rel=&quot;nofollow&quot;&gt;rational canonical form&lt;/a&gt; (if the characteristic polynomial does not split) to compare the two matrices. They will be similar if and only if their rational forms are equal (up to some easily spotted differences; exactly analogous to the fact that two diagonal matrices are the same if they have the same diagonal entries, though the entries don't have to appear in the same order in both matrices).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reduced row echelon form and the reduced column echelon form will not help you, because any two invertible matrices have the same forms (the identity), but need not have the same determinant (so they will not be similar). &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-12T20:28:10.920" LastActivityDate="2010-12-12T20:28:10.920" CommentCount="3" />
  <row Id="14080" PostTypeId="2" ParentId="13712" CreationDate="2010-12-12T20:18:57.750" Score="0" ViewCount="" Body="&lt;p&gt;In classical logic, $B\rightarrow A$ is equivalent to $\lnot A \rightarrow \lnot B$, so you will simply use the simplest definition of $A\leftrightarrow B$, which is to use $B\rightarrow A$. In intuitionistic logic, the contraposition is not generally derivable. You have $(B\rightarrow A)\rightarrow(\lnot A\rightarrow \lnot B)$, but not $(\lnot A\rightarrow \lnot B)\rightarrow(B\rightarrow A)$. Hence, your definition of equivalence is weaker - and doesnt really make sense, it only makes sense if you assume $A\vee \lnot A$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And if you look at it computationally, $A\leftrightarrow B$ means you have terms $u^{A\rightarrow B}$, $v^{B\rightarrow A}$, such that if you have an object $a^A$ you always get an object $b^B$ and vice versa. With your definition, you would have $u^{A\rightarrow B}$, but $v^{(A\rightarrow\bot)\rightarrow(B\rightarrow\bot)}$ where $\lnot A = A\rightarrow\bot$ and $\bot$ can be defined as $0=1$ (or any other type of which no term is derivable), hence, $v$ maps terms $t^{A\rightarrow\bot}$ that map terms of type $A$ on something not derivable onto terms $u^{B\rightarrow\bot}$ mapping terms of type $B$ on something not derivable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This isnt what we usually want from equivalence.&lt;/p&gt;&#xA;" OwnerUserId="1858" LastActivityDate="2010-12-12T20:18:57.750" CommentCount="1" />
  <row Id="14081" PostTypeId="1" CreationDate="2010-12-12T20:25:45.437" Score="1" ViewCount="50" Body="&lt;p&gt;The current price of a stock can be modeled by $P_0 = \frac{D_{1}}{r-g}$ where $D_1$ is the expected dividend, $r$ is the rate of return, and $g$ is the expected growth rate in the perpetuity. If $r&amp;lt;g$, then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle P_0 = \sum_{t=1}^{N} \frac{D_{0}(1+g)^{t}}{(1+r)^{t}} + \frac{P_N}{(1+r)^{N}}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there anyway to simplify this?  &lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2010-12-12T20:25:45.437" Title="Dividend Discount Model" Tags="&lt;finance&gt;" CommentCount="3" />
  <row Id="14082" PostTypeId="2" ParentId="14078" CreationDate="2010-12-12T20:28:35.987" Score="5" ViewCount="" Body="&lt;p&gt;When it is first increased you are multiplying it by $\frac{110}{100}$. When it is further increased you are then multiplying the new salary by $\frac{150}{100}$. Find out how much you are multiplying the original by in all and from there deduce the increase.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here we have assumed that in the question the further increase is based on the new salary and not the original one.&lt;/p&gt;&#xA;" OwnerUserId="4594" LastEditorUserId="4594" LastEditorDisplayName="" LastEditDate="2011-05-22T18:45:32.087" LastActivityDate="2011-05-22T18:45:32.087" CommentCount="4" />
  <row Id="14083" PostTypeId="2" ParentId="14066" CreationDate="2010-12-12T20:57:49.140" Score="3" ViewCount="" Body="&lt;p&gt;The GNU Multiple Precision Arithmetic Library (GMP) has a very good documentation describing how they implemented all the arithmetic functions for their multiple precision library:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://gmplib.org/manual/&quot; rel=&quot;nofollow&quot;&gt;http://gmplib.org/manual/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Give it a try!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;UPDATE: I know the GMP is used for big number computation, but arithmetic is arithmetic, and to make big number computations you must understand small number computations, so I am sure (or at least &quot;almost&quot; sure) the documentation mention what is enough to for calculator arithmetic.&lt;/p&gt;&#xA;" OwnerUserId="4671" LastEditorUserId="4671" LastEditorDisplayName="" LastEditDate="2010-12-13T14:51:16.730" LastActivityDate="2010-12-13T14:51:16.730" CommentCount="5" />
  <row Id="14084" PostTypeId="1" AcceptedAnswerId="14086" CreationDate="2010-12-12T21:00:57.780" Score="3" ViewCount="204" Body="&lt;p&gt;Hi all,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fermat's Little Theorem, in Fermat-Euler form, states that: if $\gcd(a,m)=1$, then $a^{\varphi(m)} \equiv 1 \pmod{m}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, I've been asked to prove it via modular arithmetic. In order to do this I understand that I'm to use two lemmas:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$\varphi(p^n) = p^{n} - p^{n-1}$. This I can prove by working out that there are $p^n$ numbers less than $p^n$ of which $p^{n-1}$ of them are divisible by $p$.&lt;/li&gt;&#xA;&lt;li&gt;Given $gcd(r,s) = 1$, $\varphi(rs) = \varphi(r)\varphi(s)$. This I'm having problems with.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;My lecturer suggested the proposition that  $\varphi(n) = n\prod_{p|n}\left(1-\frac{1}{p}\right)$. I can re-arrange this to equal lemma 1, but what I can't understand how that proposition proves 2, which my lecturer claims it does, and why these two points together prove the theorem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suspect I might follow the argument correctly if I fully understood lemma 2. &lt;/p&gt;&#xA;" OwnerUserId="892" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-12T21:17:47.100" LastActivityDate="2010-12-13T23:54:55.527" Title="Proof of the Euler Generalisation of Fermat's Little Theorem using modular arithmetic" Tags="&lt;elementary-number-theory&gt;" AnswerCount="2" CommentCount="5" />
  <row Id="14085" PostTypeId="2" ParentId="14078" CreationDate="2010-12-12T21:03:10.087" Score="2" ViewCount="" Body="&lt;p&gt;If x is your original salary,&#xA;10 % increase in salary is obtained by multiplying 1.1 (since 0.1 is 10% of 1)&#xA;Similarly 50 % increase in salary is obtained by multiplying 1.5 ( since 0.5 is 50% of 1)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so with two consecutive increments of 10% and 50%. your new salary becomes 1.5 * (1.1*(x)) = 1.65*x.&lt;BR&gt;&#xA;Work from here on...&lt;/p&gt;&#xA;" OwnerUserId="3964" LastActivityDate="2010-12-12T21:03:10.087" />
  <row Id="14086" PostTypeId="2" ParentId="14084" CreationDate="2010-12-12T21:10:12.927" Score="3" ViewCount="" Body="&lt;p&gt;Part 2 can be proven in many different ways; what your lecturer suggests, if you already know the identity, will do it.  Just notice that if $p|rs$, then &lt;em&gt;either&lt;/em&gt; $p|r$ &lt;em&gt;or&lt;/em&gt; $p|s$, &lt;em&gt;and not both&lt;/em&gt;, since $\gcd(r,s)=1$. So we have:&#xA;$$\varphi(rs) = rs\prod_{p|rs}\left(1 - \frac{1}{p}\right) = rs\left(\prod_{p|r}\left(1 - \frac{1}{p}\right)\right)\left(\prod_{p|s}\left(1-\frac{1}{p}\right)\right),$$&#xD;&#xA;and go from there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One you have this, the proof of the Fermat-Euler Theorem reduces to proving that if $\gcd(a,p)=1$, then $a^{\varphi(p^n)} \equiv 1 \pmod{p^n}$ for every positive integer $n$; you can use 2 and the Chinese Remainder Theorem to prove that this suffices, and then you can use 1 to try to establish the condition modulo a prime power.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-12T21:30:10.610" LastActivityDate="2010-12-12T21:30:10.610" CommentCount="1" />
  <row Id="14087" PostTypeId="2" ParentId="14035" CreationDate="2010-12-12T21:19:39.513" Score="3" ViewCount="" Body="&lt;p&gt;For each distance measured, you can write an equation for the distance between that point on your path and that fixed point, both of which are unknown.  In two dimensions (on a flat plane), each unknown point has two unknown coordinates.  If you have $n$ fixed points and you measure from $k$ points on your path, you will have $nk$ distances and equations and $2n+2k$ unknowns.  As long as $nk\ge 2n+2k$, this should yield a system that can be solved (if the distances are exact), but it probably helps to have $nk$ be more than $2n+2k$ because the system when $nk=2n+2k$ might have multiple solutions (because the equations are nonlinear).  (If you are working in three dimensions, then $2n+2k$ changes to $3n+3k$.)&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-12-12T21:19:39.513" />
  <row Id="14088" PostTypeId="2" ParentId="3399" CreationDate="2010-12-12T21:27:05.663" Score="4" ViewCount="" Body="&lt;p&gt;The primary concept is convexity, not concavity. It applies to geometric figures, originally lenses, and this usage was adapted to functions.  There is no comparable concept of concavity for, say, 2-dimensional regions, except as the absence of the property of convexity.  There is also no property for figures in general corresponding to the anti-convexity inequality, because most non-convex figures will be locally convex.  It is a matter of historical convention that a function is called &quot;convex&quot; if the region &lt;em&gt;above&lt;/em&gt; the graph of the function is convex, and it would have caused no mathematical problem to use the opposite convention based on the region &lt;em&gt;below&lt;/em&gt;, but concavity is a more limited concept that is defined in terms of convexity (or only defined for functions) and not the other way around. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The terms &quot;concave up&quot; and &quot;concave down&quot; appear mainly in non-specialist US college textbooks on calculus.  They are nonstandard terminology and, I think, bad practice that should be discouraged (with luck and sufficient ruthlessness maybe they can be squelched in a generation...).  As far as I know the etymology went as follows:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Like &quot;convex&quot;, the word &quot;concave&quot; has a prior use in optics. Concave (inward-curved) lenses are the opposite of convex lenses, so there is a pre-existing word for &quot;not convex&quot; or &quot;convex in the opposite direction&quot;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Convex has an absolutely entrenched mathematical use to denote convex figures as well as functions (and sequences) with increasing derivative.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Functions whose negative is convex occur frequently and &quot;concave [function]&quot; came into use as a convenient description of this situation.  The linguistic logic was clear enough to make this immediately understandable. It's not clear whether it was more or less favored compared to statements involving the negative, such as saying that $-f(u)$ is convex, or $f$ is anti-convex, or that is it the negative of a convex function.  I don't have data at hand from web searches or anything like that, but I think concavity is less common as a description of negatively convex sequences.  For functions the ability to draw a graph makes the resemblance to lenses clearer so that both words seem sensible.&#xA;(added: concavity as a counterpart to convexity for functions and sequences also gained momentum as its own term once log-convexity and log-convex became standard usage. Because the relationship between log-convex and log-concave functions is not simply change of sign but a multiplicative inverse, using only the words based on convexity might lead to confusion or circumlocution.) &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Authors of US college calculus textbooks, writing for an audience not familiar with or necessarily interested in convex figures and optics, and aware of potential for confusion (e.g., the graph of a concave function still bounds a convex-shaped region, or the subsequent use of convex to describe functions of several variables and the regions on which those are defined) cooked up a terminology based on &quot;concavity&quot; as a stand-alone concept, limited to the one-variable context where $f(x)$ is graphed with the $y$-axis direction being upward.  It's not clear how consistent this concave-up and concave-down terminology is between books and whether it agrees with the earlier, non-confusing use of concave to denote negative convexity.   &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="467" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-12-13T05:43:05.053" LastActivityDate="2010-12-13T05:43:05.053" CommunityOwnedDate="2010-12-12T21:27:05.663" CommentCount="6" />
  <row Id="14089" PostTypeId="2" ParentId="13793" CreationDate="2010-12-12T21:54:47.093" Score="6" ViewCount="" Body="&lt;p&gt;The ordered approach is fine. A classical theorem by Sierpinski says that all countable metric spaces without isolated points are homeomorphic. Q is such a space.&#xA;It also implies $\mathbf{Q} \setminus \{0\}$ is homeomorphic to $\mathbf{Q}$ and $\mathbf{Q} \times \mathbf{Q}$ e.g., or any finite product for that matter. &#xA;A proof is at &lt;a href=&quot;http://at.yorku.ca/p/a/c/a/25.htm&quot; rel=&quot;nofollow&quot;&gt;the topology atlas, topology explained&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4280" LastActivityDate="2010-12-12T21:54:47.093" CommentCount="2" />
  <row Id="14090" PostTypeId="2" ParentId="1064" CreationDate="2010-12-12T21:57:02.770" Score="4" ViewCount="" Body="&lt;p&gt;It is well-known that $C$ is homeomorphic to $C \times C$, where $C$ is the Cantor set, as both are zero-dimensional compact metric spaces without isolated points. So $C$ contains uncountably many disjoint homeomorphic copies of $C$ and all but countably many of them can contain rationals...&lt;/p&gt;&#xA;" OwnerUserId="4280" LastActivityDate="2010-12-12T21:57:02.770" />
  <row Id="14091" PostTypeId="1" CreationDate="2010-12-12T22:12:46.547" Score="1" ViewCount="106" Body="&lt;p&gt;$Z(t) = \int_0^t  g(s)\,dW(s)$, where $g$ is an adapted stochastic process.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Find $dZ$ ?&lt;/p&gt;&#xA;" OwnerUserId="3964" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-12T22:38:17.607" LastActivityDate="2010-12-12T22:38:17.607" Title="Ito's Lemma application" Tags="&lt;stochastic-integrals&gt;" CommentCount="3" />
  <row Id="14092" PostTypeId="2" ParentId="9523" CreationDate="2010-12-12T22:14:13.440" Score="0" ViewCount="" Body="&lt;p&gt;According to my professor the approach I've laid out here is fine.  It does appear, though, that the more complicated technique that involves splitting (fg)'/fg into f'/f + g'/g has some uses, so I can see why they'd bring that up.&lt;/p&gt;&#xA;" OwnerUserId="3296" LastActivityDate="2010-12-12T22:14:13.440" />
  <row Id="14093" PostTypeId="2" ParentId="14026" CreationDate="2010-12-12T22:46:48.310" Score="0" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://code.google.com/p/mpmath/&quot; rel=&quot;nofollow&quot;&gt;mpmath&lt;/a&gt;, which is a part of sage has great special functions support.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.nickle.org&quot; rel=&quot;nofollow&quot;&gt;nickle&lt;/a&gt; is good for quick things, and has C like syntax.&lt;/p&gt;&#xA;" OwnerUserId="890" LastActivityDate="2010-12-12T22:46:48.310" />
  <row Id="14094" PostTypeId="2" ParentId="14078" CreationDate="2010-12-12T22:51:07.327" Score="4" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;NOTE:&lt;/strong&gt; $\text{orig-salary}$ represents the salary you had before it was changed.&#xA;$\text{next-salary}$ represents the salary you had after it was increased by 10%.&#xA;$\text{final-salary}$ represents the salary you had after it was increased by 50%.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whenever we say &quot;x was increased by y%&quot; we mean this:&#xA;$$x + (x * y\%) = (x * 100\%) + (x * y\%) = x * (100\% + y\%)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, using the information in your problem, we have the following two equations:&#xA;\begin{align*}&#xA;\text{next-salary} &amp;amp;= \text{orig-salary} * (100\% + 10\%)\\&#xA;\text{final-salary} &amp;amp;= \text{next-salary} * (100\% + 50\%)&#xA;\end{align*}&#xA;Replace $\text{next-salary}$ with $\text{orig-salary} * (100\% + 10\%)$&#xD;&#xA;\begin{align*}&#xA;\text{final-salary} &amp;amp;= \text{next-salary} * (100\% + 50\%)\\&#xA;\text{final-salary} &amp;amp;= (\text{orig-salary} * (100\% + 10\%)) * (100\% + 50\%)\\&#xA;\text{final-salary} &amp;amp;= \text{orig-salary} * (110\%) * (150\%)\\&#xA;\text{final-salary} &amp;amp;= \text{orig-salary} * 1.10 * 1.50\\&#xA;\text{final-salary} &amp;amp;= \text{orig-salary} * 1.650&#xA;\end{align*}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The overall increase would be $1.650 - 100\%$&#xD;&#xA;OR&#xA;$165.0\% - 100\% = 65\%$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The overall increase is $65\%$.&lt;/p&gt;&#xA;" OwnerUserId="2736" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-13T06:04:17.850" LastActivityDate="2010-12-13T06:04:17.850" />
  <row Id="14095" PostTypeId="2" ParentId="13890" CreationDate="2010-12-12T23:12:10.840" Score="3" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\rm\quad 3^4 = 1\ (mod\ 20)\ \Rightarrow\ 3^n\ =\ 20\ k + \{1,3,9,27\}$&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-12-12T23:12:10.840" />
  <row Id="14096" PostTypeId="1" CreationDate="2010-12-12T23:14:23.163" Score="3" ViewCount="105" Body="&lt;p&gt;Hello.&#xA;We know that $H^p(S^2 \vee S^4) = H^p(S^2)\oplus H^p(S^4)$ for $p\neq 0$. I want to show that this space has different ring structure than $CP^2$. So, given a generator in $H^2(S^2 \vee S^4)$ I want to cup it with itself and get 0. My idea is to use the generator from $H^2(S^2)$(which obviously is zero when squared). How should I go from here? Is this even the correct idea? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;($\vee$ here is one-point intersection).&lt;/p&gt;&#xA;" OwnerUserId="2900" LastEditorUserId="2900" LastEditorDisplayName="" LastEditDate="2010-12-13T00:34:35.993" LastActivityDate="2010-12-13T00:34:35.993" Title="Ring structure of $H^2(S^2 \vee S^4)$" Tags="&lt;algebraic-topology&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="14097" PostTypeId="1" AcceptedAnswerId="14099" CreationDate="2010-12-12T23:16:23.130" Score="3" ViewCount="90" Body="&lt;p&gt;Define $(a,b) &amp;lt; (a&amp;#39;,b&amp;#39;)$ if $\max(a,b) &amp;lt; \max(a&amp;#39;,b&amp;#39;)$ or $\max(a,b) = \max(a&amp;#39;,b&amp;#39;)$ and $b &amp;lt; b&amp;#39;$ or $\max(a,b) = \max(a&amp;#39;,b&amp;#39;)$ and $b = b&amp;#39;$ and $a &amp;lt; a&amp;#39;$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I want to prove that the order-type of $(\{(b,c) : \max(b,c) = a\}, &amp;lt;)$ is equal to $a + a + 1$, does someone have a hint how to do this? I can't find the bijection. All elements are ordinals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="1120" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-13T04:26:14.790" LastActivityDate="2010-12-13T04:26:14.790" Title="Ordering on pairs." Tags="&lt;set-theory&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="14098" PostTypeId="2" ParentId="14096" CreationDate="2010-12-12T23:19:38.023" Score="2" ViewCount="" Body="&lt;p&gt;Yes, that is the right idea. Use that the cup product is &quot;natural&quot; with respect to pull backs.&lt;/p&gt;&#xA;" OwnerUserId="2633" LastActivityDate="2010-12-12T23:19:38.023" CommentCount="6" />
  <row Id="14099" PostTypeId="2" ParentId="14097" CreationDate="2010-12-12T23:30:22.830" Score="5" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Hint:&lt;/strong&gt; The set naturally breaks up into three disjoint parts: $\{(a,a)\}$, the set $\{(b,a)\mid b\lt a\}$; and the set $\{(a,b)\mid b\lt a\}$. Every element of the latter is less than each element of the middle one, and each element of the middle one is stricly smaller than $(a,a)$. &lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-12-12T23:30:22.830" CommentCount="1" />
  <row Id="14100" PostTypeId="1" AcceptedAnswerId="14110" CreationDate="2010-12-12T23:30:32.390" Score="7" ViewCount="230" Body="&lt;p&gt;Here is what I need to prove: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Let $d_1,d_2,...,d_n$ be a sequence of natural numbers (&gt;0). Show that $d_i$ is a degree sequence of some tree if and only if $\sum d_i = 2(n-1)$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I know that:&lt;br&gt;&#xA;1. for any graph $\sum_{v \in V}\ deg(v) = 2e$;&lt;br&gt;&#xA;2. for any tree $e=v-1$.&lt;br&gt;&#xA;From 1 and 2 it follows that for any tree $\sum_{v \in V}\ deg(v) = 2(v-1)$.&lt;br&gt;&#xA;If I understand it correctly, this is only a half of the proof ($\rightarrow$), isn't it? &#xA;Any hints on how to prove it the other way?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit (induction attempt):&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;$n=1$, we have $d_1 = 2(1-1) = 0$ and $d_1$ is a degree sequence of a tree. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Let's assume the theorem holds for all $k&amp;lt;n$. So we know that $d_1 + d_2 + ... + d_{n-1} = 2(n-2) $ and that it's a degree sequence of a tree. In order to add one vertex to this tree, so that it remains a tree we only can add a vertex of degree one (we connect it to any existing vertex with a single edge). By doing this our new sequence is $1,d_1,...,d_j+1,...,d_{n-1}$, where $j$ is the vertex to which we attached the new vertex (it's degree increments). Clearly this sums to $2(n-1)$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Is this proof correct or am I missing the point?&lt;/p&gt;&#xA;" OwnerUserId="4687" LastEditorUserId="4687" LastEditorDisplayName="" LastEditDate="2010-12-13T16:52:17.140" LastActivityDate="2010-12-13T16:52:17.140" Title="Tree related proof problem" Tags="&lt;graph-theory&gt;&lt;trees&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="14101" PostTypeId="1" CreationDate="2010-12-12T23:30:47.807" Score="0" ViewCount="248" Body="&lt;p&gt;Can you tell me why the following is true? &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A ring $R$ is local if and only if every principal left $R$-module is indecomposable.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;(Edit by KennyTM: The above is OP's original question. The latest, completely changed question follows:)&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;what is the relation between regular ring and this property:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ab=1$ then $ba=1$ when $a,b\in R$&lt;/p&gt;&#xA;" OwnerUserId="4688" LastEditorUserId="171" LastEditorDisplayName="" LastEditDate="2010-12-16T09:24:42.660" LastActivityDate="2010-12-16T09:24:42.660" Title="Ring Theory (Local Rings)" Tags="&lt;ring-theory&gt;" AnswerCount="1" CommentCount="7" />
  <row Id="14102" PostTypeId="2" ParentId="14041" CreationDate="2010-12-12T23:32:44.373" Score="0" ViewCount="" Body="&lt;p&gt;I think you're over-thinking it slightly. With regards to your original question,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given a bijection $f:\mathbb{Z}\to\mathbb{Z}$ where $\mathbb{Z}$ is the set of all integers, does there always exist two bijections $g:\mathbb{Z} \to \mathbb{Z}$ and $h:\mathbb{Z} \to \mathbb{Z}$ which satisfy $f(n)=g(n)+h(n), \forall n∈\mathbb{Z}$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer is yes. Note that the function $g:\mathbb{Z}\to\mathbb{Z}$ defined by $g(n) = n$ is a bijection; moreover, since $f$ and $g$ are both bijections, it's pretty easy to show that the function $h:\mathbb{Z}\to\mathbb{Z}$ defined by $h(n) = f(n) - g(n)$ is also a bijection; just go through the usual steps to show it's both injective and surjective. Finally, $g(n) + h(n) = g(n) + f(n) - g(n) = f(n)$, as desired.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;EDIT: It turns out that my argument doesn't hold in general. Apologies!&lt;/p&gt;&#xA;" OwnerUserId="4689" LastEditorUserId="4689" LastEditorDisplayName="" LastEditDate="2010-12-12T23:45:17.897" LastActivityDate="2010-12-12T23:45:17.897" CommentCount="2" />
  <row Id="14103" PostTypeId="1" CreationDate="2010-12-12T23:33:53.090" Score="4" ViewCount="128" Body="&lt;p&gt;Let $K$ be a field and $f(x)$ be an irreducible polynomial over $K$. Suppose, $f(x)$ has degree at least $2$. Is it possible that if $a,b$ are two roots of $f(x)$ with $a\neq b$, then $K(a)=K(b)$. Note I need equality, not isomorphism.&lt;/p&gt;&#xA;" OwnerUserId="3947" LastActivityDate="2010-12-13T04:06:57.683" Title="Can two different roots of an irreducible polynomial generate the same extension?" Tags="&lt;field-theory&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" />
  <row Id="14104" PostTypeId="2" ParentId="14103" CreationDate="2010-12-12T23:40:07.140" Score="6" ViewCount="" Body="&lt;p&gt;For the question to &lt;em&gt;really&lt;/em&gt; make sense we need to fix an algebraic closure of $K$ and work always inside of it; otherwise, we can even make two constructions of $K(a)$ that would not be &quot;equal&quot; (e.g., constructing $\mathbb{Q}(i)$ as the set of pairs $(a,b)\in\mathbb{Q}^2$ with suitable addition/multiplication; or as $\mathbb{Q}[x]/(x^2+1)$; or as the set of formal symbols $a+bi$ with $a,b\in\mathbb{Q}$, etc.) &lt;strong&gt;Added:&lt;/strong&gt; See Qiaochu's answer for a much better explanation of this issue.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given that, yes, it is possible: for a trivial example, take $K$ to be a finite field with $q$ elements. Then both $K(a)$ and $K(b)$ are extensions of degree $\deg(f)$ over $K$; if $\deg(f)=n$, then this is a field with $q^n$ elements, and since there is a unique field of $q^n$ elements sitting inside the algebraic closure of $K$, $K(a)=K(b)$. In fact, in this case you &lt;em&gt;always&lt;/em&gt; have that situation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a nonfinite example in a field of characteristic $0$, take either $\mathbb{Q}(\sqrt{d})$: the irreducible polynomial is $x^2-d$, which has roots $\sqrt{d}$ and $-\sqrt{d}$; naturally, $\mathbb{Q}(\sqrt{d})=\mathbb{Q}(-\sqrt{d})$ (always inside the same fixed algebraic closure)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For another example, take $f(x)$ to the cyclotomic polynomial $\Phi_n(x)$ with $K=\mathbb{Q}$; the roots are precisely the primitive $n$-roots of unity, and all the extensions are equal to $\mathbb{Q}(e^{2\pi i/n})$ (again, always sitting inside the same fixed algebraic closure of $\mathbb{Q}$). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added:&lt;/strong&gt; More generally, any simple extension $K(a)$ which is Galois over $K$ will have the property, since the irreducible polynomial of $a$ necessarily splits in $K(a)$. That means that if $f(x)$ is the irreducible polynomial of $a$, then for any root $r$ of $f(x)$ in the fixed algebraic closure of $K$ containing $K(a)$ we have $r\in K(a)$, so $K(r)\subseteq K(a)$; but applying the usual isomorphism we get that $f(x)$ must also split in $K(r)$, so $a\in K(r)$ hence $K(a)\subseteq K(r)$, giving equality. In particular, this holds for any finite Galois extension of a perfect field. This includes all the examples I gave above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conversely, I claim that if $f(x)$ is an irreducible polynomial over $K$ with degree at least $2$, and for any two roots $a$ and $b$ of $f(x)$ in a fixed algebraic closure of $K$ we have that $K(a) = K(b)$, then $K(a)$ is the splitting field of $f(x)$ (and so if $f(x)$ is separable, $K(a)$ is a finite Galois extension of $K$). Indeed, this is trivial, since for any root $b$ of $f(x)$ we have $b\in K(b)=K(a)$, so $K(a)$ contains all the roots of $f(x)$; since $K(a)$ is the least field containing $a$, $K(a)$ is contained in the splitting field of $f(x)$ over $K$, thus showing $K(a)$ &lt;em&gt;is&lt;/em&gt; the splitting field.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-13T04:06:57.683" LastActivityDate="2010-12-13T04:06:57.683" CommentCount="2" />
  <row Id="14105" PostTypeId="1" CreationDate="2010-12-12T23:47:03.433" Score="5" ViewCount="117" Body="&lt;p&gt;It is possible to find the shortest route thanks to algorithms like A*, bread first search, depth first search, etc..&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any known algorithm to find how many routes are available if there are more than 1 optimum solution (for example routes with same costs) or any solution at all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am working on some graph algorithms but I don't think I'm on the right direction. Any advices?&lt;/p&gt;&#xA;" OwnerUserId="4653" LastActivityDate="2011-05-08T18:50:45.730" Title="Question about path-finding" Tags="&lt;graph-theory&gt;&lt;optimization&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="14106" PostTypeId="2" ParentId="14100" CreationDate="2010-12-12T23:57:08.307" Score="3" ViewCount="" Body="&lt;p&gt;Hints for the other way round: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Do you know about mathematical induction?&lt;/li&gt;&#xA;&lt;li&gt;Can every $d_i$ be $\gt 1$?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-12-12T23:57:08.307" CommentCount="2" />
  <row Id="14107" PostTypeId="1" AcceptedAnswerId="14114" CreationDate="2010-12-13T00:06:34.620" Score="2" ViewCount="79" Body="&lt;p&gt;Say I have two algorithms that I don't know how they work but I know what they are meant to achieve. I tried algorithm A once and it succeeded. And tried algorithm B 100 times and succeeded 99 times. So A succeeded 100% times and B succeeded 99% times, but my common sense tells me B is still more &quot;reliable&quot; than A. How can I express this mathematically, or, how can I calculate which algorithm should I choose if I have to choose one when all I know about each of them is how many times it has been tested and how many times it succeeded (if it didn't that just means it failed for what I need)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please be as simple as possible I'm a newbie at math.&lt;/p&gt;&#xA;" OwnerUserId="1542" LastEditorUserId="1542" LastEditorDisplayName="" LastEditDate="2010-12-13T00:31:36.557" LastActivityDate="2010-12-13T01:26:32.397" Title="How to calculate likelihood to succeed knowing attempts and successful attempts amounts?" Tags="&lt;probability&gt;&lt;statistics&gt;&lt;probability-theory&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="14108" PostTypeId="1" CreationDate="2010-12-13T00:12:43.527" Score="3" ViewCount="145" Body="&lt;p&gt;Let $R$ be a commutative ring with unity and let $B(R)$ be the set of all idempotent elements in $R$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Show for $b\in B(R)$, the $R$-modules $R$ and $Rb \times R(1-b)$ are isomorphic to one another. &lt;/p&gt;&#xA;" OwnerUserId="4697" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-13T00:16:56.580" LastActivityDate="2011-05-10T21:56:22.937" Title="Show the $R$-module $R$ is isomorphic to $Rb \times R(1-b)$ where $b$ is an idempotent of a commutative ring with unity" Tags="&lt;abstract-algebra&gt;&lt;modules&gt;" AnswerCount="1" CommentCount="7" />
  <row Id="14109" PostTypeId="1" AcceptedAnswerId="14122" CreationDate="2010-12-13T00:16:06.113" Score="8" ViewCount="104" Body="&lt;p&gt;Let $R$ be a commutative ring with identity and $I,J,K$ be ideals of $R$. If $I\supseteq J$ or $I\supseteq K$, we have the following modular law &#xA;$$ I\cap (J+K)=I\cap J + I\cap K$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was wondering if there are situations in which the modular law holds in which the hypothesis that $I$ contains at least one of $J,K$ is relaxed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One example is when $R$ is a polynomial ring or power series ring and $I,J,K$ are monomial ideals. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course one containment always holds $I\cap (J+K)\supseteq I\cap J +I\cap K$. In what other situations does the other containment hold?&lt;/p&gt;&#xA;" OwnerUserId="3431" LastEditorUserId="3431" LastEditorDisplayName="" LastEditDate="2010-12-13T01:43:10.753" LastActivityDate="2010-12-13T04:52:15.147" Title="When does the modular law apply to ideals in a commutative ring" Tags="&lt;abstract-algebra&gt;&lt;commutative-algebra&gt;" AnswerCount="1" FavoriteCount="2" />
  <row Id="14110" PostTypeId="2" ParentId="14100" CreationDate="2010-12-13T00:37:56.883" Score="2" ViewCount="" Body="&lt;p&gt;From your fact (1) and the hypothesis $\sum d_i = 2(n-1)$, you know $e = n - 1$.  You need now only prove that any connected graph with $n - 1$ edges is a tree.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt; - Since you are going down an alternate path, I have a new suggestion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Order the $d_i$ so that $d_1 \leq d_2 \leq \ldots \leq d_n$.  Construct a tree with degree sequence $d_2, \ldots, d_n$. You know $\sum_{i \neq 1} d_i = 2(n - 2)$ (why?).  Now try to construct your tree with degree sequence $d_1, \ldots, d_n$.  You have to be careful about how you add in your vertex with degree $d_1$ so that it remains a tree, but this can be done.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt; - The above is incorrect.  The idea remains to add in a vertex of degree $1$ to a tree with a specific degree sequence related to $d_1, \ldots, d_n$. But $\sum_{i \neq 1} d_i = 2n - 3$, so we cannot claim $d_2, \ldots, d_n$ is a degree sequence of a tree by the inductive hypothesis.  Of course, when we add in the vertex of degree $1$, we alter the degree of another vertex, right?&lt;/p&gt;&#xA;" OwnerUserId="3537" LastEditorUserId="3537" LastEditorDisplayName="" LastEditDate="2010-12-13T16:30:02.830" LastActivityDate="2010-12-13T16:30:02.830" CommentCount="8" />
  <row Id="14112" PostTypeId="2" ParentId="14041" CreationDate="2010-12-13T00:57:38.260" Score="7" ViewCount="" Body="&lt;p&gt;&lt;em&gt;Observation&lt;/em&gt;:  It is sufficient to resolve the question for $f(n)=n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Proof idea&lt;/em&gt;:  If $n=g(n)+h(n)$ for all $n$, then $f(n)=g(f(n))+h(f(n))$.  Conversely, if $f(n)=g(n)+h(n)$ then $n=f(f^{-1}(n))=g(f^{-1}(n))+h(f^{-1}(n))$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, if &lt;em&gt;some&lt;/em&gt; $f$ can be decomposed into two bijections $h$ and $g$, then &lt;em&gt;every&lt;/em&gt; $f$ can be decomposed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So resolving this question amounts to finding a permutation $g$ of $\mathbb{Z}$ for which $n \mapsto g(n)-n$ is also a permutation.  Such permutations are called &lt;em&gt;orthomorphisms&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;UPDATE&lt;/em&gt;: I asked both Ian Wanless and Tony Evans about this, and they simultaneously pointed out that the existence of such an orthomorphism was proved in:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;MR0040293 (12,670f) &#xA;Bateman, P. T.&#xA;&lt;a href=&quot;http://www.jstor.org/stable/2308116&quot; rel=&quot;nofollow&quot;&gt;A remark on infinite groups&lt;/a&gt;. &#xA;Amer. Math. Monthly 57, (1950). 623–624. &#xA;20.0X &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we can conclude that, yes, it's always possible.&lt;/p&gt;&#xA;" OwnerUserId="139" LastEditorUserId="139" LastEditorDisplayName="" LastEditDate="2010-12-13T02:43:20.050" LastActivityDate="2010-12-13T02:43:20.050" CommentCount="7" />
  <row Id="14113" PostTypeId="1" AcceptedAnswerId="14179" CreationDate="2010-12-13T00:58:16.170" Score="4" ViewCount="506" Body="&lt;p&gt;I know the series, $4-{4\over3}+{4\over5}-{4\over7}...$ converges to $\pi$ but I have heard many people say that while this is a classic example, there are series that converge much faster.  Does anyone know of any?&lt;/p&gt;&#xA;" OwnerUserId="3971" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-13T01:32:31.180" LastActivityDate="2010-12-15T09:07:35.300" Title="Series that converge to $\pi$ quickly" Tags="&lt;calculus&gt;&lt;sequences-and-series&gt;&lt;convergence&gt;&lt;pi&gt;" AnswerCount="7" CommentCount="7" FavoriteCount="5" />
  <row Id="14114" PostTypeId="2" ParentId="14107" CreationDate="2010-12-13T01:11:40.313" Score="2" ViewCount="" Body="&lt;p&gt;Perhaps what you are looking for is &quot;Confidence Interval for Proportion&quot;. There are calculators for that online, see e.g. &lt;a href=&quot;http://www.dimensionresearch.com/resources/calculators/conf_prop.html&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; or &lt;a href=&quot;http://www.causascientia.org/math_stat/ProportionCI.html&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt;. &lt;a href=&quot;http://www.dimensionresearch.com/resources/calculators/sample_size.html&quot; rel=&quot;nofollow&quot;&gt;Here&lt;/a&gt; is also a calculator that determines the minimum sample size you would need in order for your sample to meet the desired precision requirements for a study .&lt;/p&gt;&#xA;" OwnerUserId="2810" LastEditorUserId="2810" LastEditorDisplayName="" LastEditDate="2010-12-13T01:26:32.397" LastActivityDate="2010-12-13T01:26:32.397" />
  <row Id="14115" PostTypeId="1" AcceptedAnswerId="14422" CreationDate="2010-12-13T01:28:27.437" Score="17" ViewCount="643" Body="&lt;p&gt;The following formula for $\pi$ was discovered by Ramanujan:&#xA;$$\frac1{\pi} = \frac{2\sqrt{2}}{9801} \sum_{k=0}^\infty \frac{(4k)!(1103+26390k)}{(k!)^4 396^{4k}}\!$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anyone know how it works, or what the motivation for it is?&lt;/p&gt;&#xA;" OwnerUserId="3060" LastEditorUserId="9413" LastEditorDisplayName="" LastEditDate="2011-05-22T11:27:53.890" LastActivityDate="2011-05-26T14:10:25.410" Title="Motivation for Ramanujan's mysterious $\pi$ formula" Tags="&lt;calculus&gt;&lt;sequences-and-series&gt;&lt;approximation&gt;&lt;pi&gt;&lt;ramanujan&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="4" />
  <row Id="14116" PostTypeId="2" ParentId="14113" CreationDate="2010-12-13T01:36:46.327" Score="9" ViewCount="" Body="&lt;p&gt;The series $$ \sum_{n=0}^{\infty} \frac{(2n)!!}{(2n+1)!!} \left(\frac{1}{2}\right)^n = \frac{\pi}{2}$$ converges quickly. Here $!!$ is the double factorial defined by $0!! = 1!! = 1$ and $n!! = n (n-2)!!$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is series is not too hard to derive. Start by defining $$f(t) = \sum _{n=0}^{\infty } \frac{(-1)^n}{(2n+1)}t^n.$$ Note that $f(1) = \pi/4$ is the series you referenced. Now we take what is called the Euler Transform of the series which gives us $$ \left(\frac{1}{1-t}\right)f\left(\frac{t}{1-t}\right) = \sum _{n=0}^{\infty } \left(\sum _{k=0}^n {n \choose k}\frac{(-1)^k}{(2k+1)}\right)t^n.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now $$\sum _{k=0}^n {n \choose k}\frac{(-1)^k}{(2k+1)} = \frac{(2n)!!}{(2n+1)!!}$$ for hints on how to prove this identity see &lt;a href=&quot;http://math.stackexchange.com/questions/4261/proving-a-binomial-sum-identity&quot;&gt;Proving a binomial sum identity&lt;/a&gt;. Now put $t = 1/2$ and the identity follows. Showing the error term for the nth partial sum is less than $(1/2)^n$ is not too difficult.&lt;/p&gt;&#xA;" OwnerUserId="899" LastEditorUserId="899" LastEditorDisplayName="" LastEditDate="2010-12-13T01:57:49.257" LastActivityDate="2010-12-13T01:57:49.257" CommentCount="2" />
  <row Id="14117" PostTypeId="2" ParentId="14066" CreationDate="2010-12-13T01:40:15.823" Score="4" ViewCount="" Body="&lt;p&gt;I would recommend reading Gerald Rising's &lt;a href=&quot;http://books.google.com/books?id=ok_PE_WHCREC&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;Inside your Calculator&lt;/em&gt;&lt;/a&gt; (which has a &lt;a href=&quot;http://www.acsu.buffalo.edu/~insrisg/InsideYourCalculator.htm&quot; rel=&quot;nofollow&quot;&gt;supplementary website&lt;/a&gt;); there is a nice discussion of the methods used by some calculators that is suitable at the undergraduate level.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Otherwise, to really figure out what methods they are using, it might help to search the technical notes of the manufacturer's websites. For instance, Texas Instruments has notes like &lt;a href=&quot;http://epsstore.ti.com/OA_HTML/csksxvm.jsp?nSetId=74414&quot; rel=&quot;nofollow&quot;&gt;this one&lt;/a&gt; on their &lt;a href=&quot;http://epsstore.ti.com/OA_HTML/cskmbasicsrch.jsp&quot; rel=&quot;nofollow&quot;&gt;&quot;knowledge base&quot;&lt;/a&gt; that discuss &quot;what's under the hood&quot;, though not in detail of course. (Sometimes, hobbyist sites like &lt;a href=&quot;http://tibasicdev.wikidot.com/68k%3arandseed&quot; rel=&quot;nofollow&quot;&gt;this one&lt;/a&gt; also discuss calculator algorithms.)&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-12-13T01:40:15.823" CommentCount="2" />
  <row Id="14118" PostTypeId="1" AcceptedAnswerId="14121" CreationDate="2010-12-13T01:40:48.730" Score="4" ViewCount="113" Body="&lt;p&gt;This sounds like a simple problem, but I can't get it done. Given the general equation&#xA;$ax^2 + by^2 + cz^2 + dxy + exz + fyz + gx + hy + iz + j = 0,$ &#xA;what are the requirements on the coefficients so that this equation represents an ellipsoid? All coefficients are real.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some more context. To make the representation unique, we establish another constraint: $a + c = 1$. I'm not sure how important that might be. The question is in homework where we have to fit this equation to a set of points, using QR-decomposition. That isn't a problem, the small followup question &quot;how can you see the coefficients determine an ellipsoid&quot; is the problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried working on the equation to get something usable, from basic term regrouping to putting stuff in matrices and decomposing these, which didn't really help. One theoretic way to do it, I think, is to find the center of mass of the surface, translate to the origin, then find the principal axes, transform again, and show the radii are all larger than 0. Sounds like a lot of work for a simple question, and I'm kind of stuck on the &quot;center of mass&quot; part, I don't feel the idea here is to integrate the formula over $\mathbb{R}^3$. I've also been thinking about calculating curvature or something with the Frenet-Serret formulas, but my geometry courses have been too long ago to know what I want here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a simple test to determine this? Something along the lines of &quot;the eigenvalues of matrix foo are all positive&quot; or &quot;the value of bar is real&quot; or I don't know what...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks a lot for any insights you can give me!&lt;/p&gt;&#xA;" OwnerUserId="4692" LastActivityDate="2010-12-13T03:22:39.760" Title="Is a general (non-homogeneous) quadratic equation in $\mathbb{R}^3$ an ellipsoid?" Tags="&lt;geometry&gt;&lt;analytic-geometry&gt;&lt;quadratic-forms&gt;" AnswerCount="1" />
  <row Id="14119" PostTypeId="2" ParentId="14084" CreationDate="2010-12-13T01:45:14.123" Score="3" ViewCount="" Body="&lt;p&gt;Here is another way to prove Euler's generalization. You do not need to know the formula of $\varphi(n)$ for this method which I think makes this method elegant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the set of all numbers less than $n$ and relatively prime to it. Let $\{a_1,a_2,...,a_{\varphi(n)}\}$ be this set.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider a number $c &amp;lt; n$ and relatively prime to it i.e. $c \in \{a_1,a_2,\ldots,a_{\varphi(n)}\}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First observe that for any $a_i$, $c a_{i} \equiv a_{j} (\bmod n)$ for some $j$.&#xA;(True since $c$ and $a_i$ are themselves relatively prime to $n$, their product has to be relatively prime to $n$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And if $c a_{i} \equiv c a_{j} (\bmod n)$ then $a_i = a_j$.&#xA;(True as cancellation can be done since $c$ is relatively prime to $n$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, if we now consider the set $\{ca_1,ca_2,...,ca_{\varphi(n)}\}$ this is just a permutation of the set $\{a_1,a_2,...,a_{\varphi(n)}\}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thereby, we have $\displaystyle \prod_{k=1}^{\varphi(n)} ca_k \equiv \prod_{k=1}^{\varphi(n)} a_k (\bmod n)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, we get $\displaystyle c^{\varphi(n)} \prod_{k=1}^{\varphi(n)} a_k \equiv \prod_{k=1}^{\varphi(n)} a_k (\bmod n)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, note that $\displaystyle \prod_{k=1}^{\varphi(n)} a_k$ is relatively prime to $n$ and hence you can cancel them on both sides to get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$c^{\varphi(n)} \equiv 1 (\bmod n)$$ whenever $(c,n) = 1$&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2010-12-13T23:54:55.527" LastActivityDate="2010-12-13T23:54:55.527" CommentCount="0" />
  <row Id="14121" PostTypeId="2" ParentId="14118" CreationDate="2010-12-13T01:59:32.950" Score="6" ViewCount="" Body="&lt;p&gt;There are two steps involved. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Step 1: translate to centre of mass&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The goal of translation to the centre of mass is to remove the &lt;em&gt;linear terms&lt;/em&gt; in your coefficients. You want to turn your expression into the form&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ AX^2 + BY^2 + CZ^2 + D XY + E YZ + F XZ + G = 0$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $X = (x - x_0)$, $Y = (y - y_0)$ and $Z = (z-z_0)$, where $x_0, y_0, z_0$ are the coordinates for the centre of mass. To do so you can solve forward by completing the square, or solve backwards by plugging in the expressions for $X,Y,Z$ and matching coefficients. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Step 2: check for positivity of eigenvalues&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Relative to the centre of mass, you can write your equation as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ Q(v) = -G $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $Q$ is the quadratic form &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ Q(v) = AX^2 + BY^2 + \cdots + F XZ, \qquad v = (X,Y,Z) $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then recall that the level surfaces of a quadratic form define ellipsoids if and only if it is positive definite. So computing the eigenvalues (among the many ways to check that a quadratic form is positive definite) will get you the answer. (Of course, this will also require $G$ to be a negative number for there to be a solution.)&lt;/p&gt;&#xA;" OwnerUserId="1543" LastEditorUserId="1543" LastEditorDisplayName="" LastEditDate="2010-12-13T03:22:39.760" LastActivityDate="2010-12-13T03:22:39.760" CommentCount="8" />
  <row Id="14122" PostTypeId="2" ParentId="14109" CreationDate="2010-12-13T02:02:25.497" Score="12" ViewCount="" Body="&lt;p&gt;Such domains are known as Prüfer domains. They are non-Noetherian generalizations of Dedekind domains. Their ubiquity stems from a remarkable confluence of interesting characterizations. For example, they are those domains satisfy either the Chinese Remainder Theorem for ideals, or Gauss's Lemma for polynomial content ideals, or for ideals: $\rm\ A\cap (B + C) = A\cap B + A\cap C\:,\  $ or $\rm\ (A + B)\ (A \cap B) = A\ B\:,\ $ or $\rm\ A\supset B\ \Rightarrow\ A\:|\:B\ $ for fin. gen. $\rm\:A\:$ etc. It's been estimated that there are close to 100 such characterizations known, e.g. see my &lt;a href=&quot;http://groups.google.com/group/sci.math/msg/ba5ac69dddfc4a67&quot; rel=&quot;nofollow&quot;&gt;sci.math post&lt;/a&gt; for 30 odd characterizations. Below is an excerpt:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;THEOREM&lt;/strong&gt;  $\ \ $ Let $\rm\:D\:$ be a domain. The following are equivalent:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) $\rm\:D\:$ is a Prufer domain, i.e. every nonzero f.g. (finitely generated) ideal is invertible.&lt;br&gt;&#xA;(2) Every nonzero two-generated ideal of $\rm\:D\:$ is invertible.&lt;br&gt;&#xA;(3) $\rm\:D_P\:$ is a  Prufer   domain for every  prime  ideal $\rm\:P\:$ of $\rm\:D.\:$&lt;br&gt;&#xA;(4) $\rm\:D_P\:$ is a valuation domain for every  prime  ideal $\rm\:P\:$ of $\rm\:D.\:$&lt;br&gt;&#xA;(5) $\rm\:D_P\:$ is a valuation domain for every maximal ideal $\rm\:P\:$ of $\rm\:D.\:$&lt;br&gt;&#xA;(6) Every nonzero f.g. ideal $\rm\:I\:$ of $\rm\:D\:$ is cancellable, i.e. $\rm\:I\:J = I\:K\ \Rightarrow\ J = K\:$&lt;br&gt;&#xA;(7) $\: $ (6) restricted to f.g. $\rm\:J,K.$&lt;br&gt;&#xA;(8) $\rm\:D\:$ is integrally closed and there is an $\rm\:n &amp;gt; 1\:$ such that for all $\rm\: a,b \in D,\ (a,b)^n = (a^n,b^n).$&lt;br&gt;&#xA;(9) $\rm\:D\:$ is integrally closed and there is an $\rm\: n &amp;gt; 1\:$ such that for all $\rm\:a,b \in D,\ a^{n-1} b \ \in\ (a^n, b^n).$&lt;br&gt;&#xA;(10) Each ideal $\rm\:I\:$ of $\rm\:D\:$ is complete, i.e. $\rm\:I = \cap\ I\: V_j\:$ as $\rm\:V_j\:$ run over all the valuation overrings of $\rm\:D.\:$&lt;br&gt;&#xA;(11) Each f.g. ideal of $\rm\:D\:$ is an intersection of valuation ideals.&lt;br&gt;&#xA;(12) If $\rm\:I,J,K\:$ are nonzero ideals of $\rm\:D,\:$ then  $\rm\:I \cap (J + K) = I\cap J + I\cap K.$&lt;br&gt;&#xA;(13) If $\rm\:I,J,K\:$ are nonzero ideals of $\rm\:D,\:$ then  $\rm\:I\ (J \cap K) = I\:J\cap I\:K.$&lt;br&gt;&#xA;(14) If $\rm\:I,J\:$ are nonzero ideals of $\rm\:D,\:$ then $\rm\:(I + J)\ (I \cap J) = I\:J.\ $  ($\rm LCM\times GCD$ law)&lt;br&gt;&#xA;(15) If $\rm\:I,J,K\:$ are nonzero ideals of $\rm\:D,\:$ with $\rm\:K\:$ f.g. then $\rm\:(I + J):K = I:K + J:K.$&lt;br&gt;&#xA;(16) For any two elements $\rm\:a,b \in D,\ (a:b) + (b:a) = D.$&lt;br&gt;&#xA;(17) If $\rm\:I,J,K\:$ are nonzero ideals of $\rm\:D\:$ with $\rm\:I,J\:$ f.g. then $\rm\:K:(I \cap J) = K:I + K:J.$&lt;br&gt;&#xA;(18) $\rm\:D\:$ is integrally closed and each overring of $\rm\:D\:$ is the intersection of localizations of $\rm\:D.\:$&lt;br&gt;&#xA;(19) $\rm\:D\:$ is integrally closed and each overring of $\rm\:D\:$ is the intersection  of quotient rings of $\rm\:D.\:$&lt;br&gt;&#xA;(20) Each overring of $\rm\:D\:$ is integrally closed.&lt;br&gt;&#xA;(21) Each overring of $\rm\:D\:$ is flat over $\rm\:D.\:$&lt;br&gt;&#xA;(22) $\rm\:D\:$ is integrally closed and prime ideals of overrings of are extensions of  prime ideals of $\rm\:D.$&lt;br&gt;&#xA;(23) $\rm\:D\:$ is integrally closed and for each prime ideal $\rm\:P\:$ of $\rm\:D,\:$ and each overring $\rm\:S\:$ of $\rm\:D,\:$ there is at most one prime ideal of $\rm\:S\:$ lying over $\rm\:P.\:$&lt;br&gt;&#xA;(24) For polynomials $\rm\:f,g \in D[x],\ c(fg) = c(f)\: c(g)\:$ where for a polynomial $\rm\:h \in D[x],\ c(h)\:$ denotes the &quot;content&quot; ideal of $\rm\:D\:$ generated by the coefficients of $\rm\:h.\:$ (Gauss' Lemma)&lt;br&gt;&#xA;(25) Ideals in $\rm\:D\:$ are integrally closed.&lt;br&gt;&#xA;(26) If $\rm\:I,J\:$ are ideals with $\rm\:I\:$ f.g. then $\rm\: I\supset J\ \Rightarrow\ I|J.$ (contains $\:\Rightarrow\:$ divides)&lt;br&gt;&#xA;(27) the Chinese Remainder Theorem $\rm(CRT)$ holds true in $\rm\:D\:,\:$ i.e. a system of congruences $\rm\:x\equiv x_j\ (mod\ I_j)\:$ is solvable iff $\rm\:x_j\equiv x_k\ (mod\ I_j + I_k).$  &lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-12-13T04:52:15.147" LastActivityDate="2010-12-13T04:52:15.147" CommentCount="9" />
  <row Id="14123" PostTypeId="1" CreationDate="2010-12-13T02:14:51.970" Score="2" ViewCount="180" Body="&lt;p&gt;Whilst answering my homework, I have noticed that a question has said that I need to consider the subgroups of $A=\mathbb{Z}_{135}$, including $W=\langle 18,81\rangle$. I was under the impression that for a subset to be a subgroup $a+b$ in the subgroup had to be in the subgroup as well. $18+81 = 99\mod{135}$ which isn't in the subgroup.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it my reasoning or the problem sheet that is wrong? Thank you in advance. &lt;/p&gt;&#xA;" OwnerUserId="4697" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-13T02:26:21.880" LastActivityDate="2011-03-13T03:32:38.467" Title="Is $\langle 18,81\rangle$ a subgroup of the abelian group $\mathbb{Z}_{135}$?" Tags="&lt;group-theory&gt;" AnswerCount="0" CommentCount="3" />
  <row Id="14124" PostTypeId="1" AcceptedAnswerId="14129" CreationDate="2010-12-13T02:20:00.453" Score="2" ViewCount="261" Body="&lt;p&gt;I am working through a review problem asking to find the inverse of &#xA;$4\bmod 9 $.&#xA;Through examples I know that I first need to verify that the gcd is equal to 1 and write it as a linear combination of 4 and 9 to find the inverse.    I can do this in just one step:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;gcd(4,9)&#xA;9 = 2 * 4 + 1&#xA;1 = 9 - 2 * 4&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This would suggest that the inverse is 1 if I am understanding this correctly.  However, the solution manual  doesn't show the work but says the LC should actually be&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;1 = 7 * 4 - 3 * 9&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;making the answer to the question 7.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone explain to me what is going on here and how to properly find the inverse?&#xA;Thanks!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S.  wish I could add tags for congruency, gcd, and inverse.  I can't believe their isn't an inverse tag already :(&lt;/p&gt;&#xA;" OwnerUserId="1631" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-13T19:29:44.990" LastActivityDate="2010-12-13T19:29:44.990" Title="finding inverse of $x\bmod y$" Tags="&lt;modular-arithmetic&gt;&lt;elementary-number-theory&gt;" AnswerCount="3" CommentCount="22" FavoriteCount="1" />
  <row Id="14126" PostTypeId="2" ParentId="14124" CreationDate="2010-12-13T02:22:23.010" Score="5" ViewCount="" Body="&lt;p&gt;$-2$ is congruent to $7$ mod $9$.  Your arithmetic steps are correct, but the conclusion should be that the congruence class of $-2$ is the inverse of $4$ mod $9$, and you probably want to use representatives in $\{1,2,3,4,5,6,7,8\}$ for your inverses, for simplicity, by adding appropriate multiples of $9$.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2010-12-13T02:29:17.203" LastActivityDate="2010-12-13T02:29:17.203" CommentCount="8" />
  <row Id="14127" PostTypeId="2" ParentId="14101" CreationDate="2010-12-13T02:33:23.823" Score="2" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT for $\Leftarrow$&lt;/strong&gt;: Do the contrapositive; if $\mathfrak{m}_1$ and $\mathfrak{m}_2$ are two distinct maximal ideals of $R$, then look at the left $R$-module $R/(\mathfrak{m}_1\cap\mathfrak{m}_2)$; it is principal (generated by $1+(\mathfrak{m}_1\cap\mathfrak{m}_2)$. Is it indecomposable?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;HINT for $\Rightarrow$&lt;/strong&gt;: Suppose $M\neq\{0\}$ is principal, and $M=N_1\times N_2$. Let $m=(n_1,n_2)$ generate $M$. Then there are $a,b\in R$ such that $am = (n_1,0)$ and $bm=(0,n_2)$; so $(a+b)m = m$. Then $a+b-1\in\mathrm{Ann}(m)$. Can $a$ and $b$ both lie in the maximal ideal of $R$? What does that mean?&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-12-13T02:33:23.823" />
  <row Id="14128" PostTypeId="1" AcceptedAnswerId="30559" CreationDate="2010-12-13T02:47:00.883" Score="1" ViewCount="112" Body="&lt;p&gt;I want to create a function $f : [0,255]^3 \rightarrow [0,255]^3$ over the integers such that if you pass in three RGB color values, a triple with a darker or lighter color will be returned, and I want the values to increase in some sort of proportion to how dark or light they are already; lightness and darkness being determined by, say, taking the average of the values.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a relatively simple function that will do this? If not, is there a way to emulate it?&lt;/p&gt;&#xA;" OwnerUserId="1562" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-12-13T03:16:27.213" LastActivityDate="2011-04-02T21:00:53.743" Title="Creating a function to brighten or darken colors" Tags="&lt;functions&gt;" AnswerCount="1" CommentCount="7" FavoriteCount="1" />
  <row Id="14129" PostTypeId="2" ParentId="14124" CreationDate="2010-12-13T03:02:25.813" Score="3" ViewCount="" Body="&lt;p&gt;I know you've already gotten lots of responses, but instead of trying to fit responses into the comments under an answer I thought I'd just reiterate exactly &lt;em&gt;why&lt;/em&gt; you need to do what you're doing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An inverse to $4 \mod 9$ is an integer $a$ such that $4a \equiv 1 \mod 9$. If we rewrite this, it means precisely that $9| (4a-1)$, or that there is another integer $b$ so that $9b = 4a-1$. What this says is that you need to find integer solutions to $4x-9y=1$. If you find that $x=a$ and $y=b$ works then $a$ is an inverse.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what you do to solve something like this is run the Euclidean algorithm for $9$ and $4$, and then &quot;reverse&quot; the steps.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$9=4\cdot 2 + 1$, so it is actually over in 1 step in this case. Just subtract over to get $9-4\cdot 2 = 1$, or in the form we wrote it above $4\cdot (-2)-9\cdot (-1)=1$. So a solution is $x=-2$ and $y=-1$ and we said the $x$-value was the inverse, so the inverse is $-2\mod 9$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hopefully if you understand the process of why you're doing these things, then if you get confused about which one is which on the final you can always just rederive it from these steps.&lt;/p&gt;&#xA;" OwnerUserId="2000" LastActivityDate="2010-12-13T03:02:25.813" CommentCount="1" />
  <row Id="14130" PostTypeId="2" ParentId="14103" CreationDate="2010-12-13T03:08:56.810" Score="4" ViewCount="" Body="&lt;p&gt;I think it's worth elaborating on this distinction between equality and isomorphism.  The problem occurs if the extension you're considering isn't &lt;a href=&quot;http://en.wikipedia.org/wiki/Normal_extension&quot; rel=&quot;nofollow&quot;&gt;normal&lt;/a&gt; and can therefore embed into an algebraic closure in more than one way.  For example, the abstract field $\mathbb{Q}[x]/(x^3 - 2)$ embeds in three ways into $\mathbb{C}$, corresponding to the three roots of $x^3 - 2$.  So it doesn't make sense to ask whether $\mathbb{Q}[x]/(x^3 - 2)$ is &quot;equal to&quot; $\mathbb{Q}[y]/(y^3 - 2)$ without specifying an embedding into a larger field.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a fairly subtle point which I don't think is addressed particularly well in introductions to Galois theory (at least the ones I've seen).  There are three categories one might work in when studying fields (where $K$ is a fixed field):&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;The category of fields.  Here it is &quot;evil&quot; (really, impossible) to speak of the &quot;equality&quot; of two fields and one can only speak of isomorphism.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The category of (say, algebraic) field extensions $K \to L$.  (The morphisms are morphisms $\phi : L \to L&amp;#39;$ making the obvious triangle commute.)  Here it is still &quot;evil&quot; to speak of the &quot;equality&quot; of two extensions, and one can only speak of isomorphism, but the isomorphism type of an extension $K \to L$ is not determined by the isomorphism type of $L$ (that is, $L$ can be an extension of $K$ in more than one way).  &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The category of subfields of $\bar{K}$ containing $K$ (for a fixed embedding $K \to \bar{K}$).  This is the category in which it makes sense to take intersections and composita of fields; you can't do either of these constructions in the above category.  Here, at last, one can speak of equality (as subfields of $\bar{K}$), and it is not the same as isomorphism of $K$-extensions, which is in turn not the same as isomorphism of abstract fields.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;People sometimes don't specify which of the above categories they're working in, and until you do this you can't be precise: there are three different notions of equality or isomorphism in the last category.&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-12-13T03:32:14.100" LastActivityDate="2010-12-13T03:32:14.100" />
  <row Id="14132" PostTypeId="2" ParentId="14124" CreationDate="2010-12-13T03:16:16.000" Score="1" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\ $ Congruences preserve inverses, i.e. $\rm\ A\equiv a\ \Rightarrow\ A^{-1}\equiv a^{-1}\:.\ $ This follows from the fact that congruences preserve products, i.e. it's the special case $\rm\: AB \equiv 1\: $ in this &lt;strong&gt;congruence product rule&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;LEMMA&lt;/strong&gt; $\rm\ \ A\equiv a,\ B\equiv b\ \Rightarrow\ AB\equiv ab\ \ (mod\ m)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof&lt;/strong&gt; $\rm\ \ m\: |\: A-a,\:\:\ B-b\ \Rightarrow\ m\ |\ (A-a)\ B + a\ (B-b)\ =\ AB - ab $&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This congruence product rule is at the heart of many other similar product rules, for example Leibnitz's product rule for derivatives in calculus, e.g. see my &lt;a href=&quot;http://math.stackexchange.com/questions/3101/derivative-of-a-product-and-derivative-of-quotient-of-functions-theorem-i-dont/3178#3178&quot;&gt;post here.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-12-13T03:16:16.000" CommentCount="3" />
  <row Id="14133" PostTypeId="1" AcceptedAnswerId="14136" CreationDate="2010-12-13T03:22:39.953" Score="4" ViewCount="231" Body="&lt;p&gt;One of the rules of logarithms that I've been taught is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\log_b(x)$ is equivalent to $\frac{\log_k(x)}{\log_k(b)}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Recently I've also seen another rule that says:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\log_b(x)$ is equivalent to $\log_b(k)\log_k(x)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are these equivalent (perhaps via some refactoring) ?&lt;/p&gt;&#xA;" OwnerUserId="4247" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-13T03:29:50.413" LastActivityDate="2010-12-13T15:37:34.063" Title="Clarify why all logarithms differ by a constant" Tags="&lt;logarithms&gt;" AnswerCount="4" CommentCount="1" FavoriteCount="1" />
  <row Id="14134" PostTypeId="2" ParentId="14133" CreationDate="2010-12-13T03:27:55.283" Score="5" ViewCount="" Body="&lt;p&gt;Yes. Your second equation is $\log_b k\cdot \log_k x = \log_b x$. Dividing by $\log_b k$ we get $\log_k x=\frac{\log_b x}{\log_b k}$. This is the same as your first equation with different symbols (an interchange of $k$ and $b$).&lt;/p&gt;&#xA;" OwnerUserId="3431" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-13T03:36:49.220" LastActivityDate="2010-12-13T03:36:49.220" CommentCount="1" />
  <row Id="14136" PostTypeId="2" ParentId="14133" CreationDate="2010-12-13T03:30:00.700" Score="4" ViewCount="" Body="&lt;p&gt;Divide the second equation by $\log_b(k)$ and shuffle the variables to see that these are the same thing.  The reason they're true can be expressed in terms of properties of exponents. Start with the observation that $\log_b(x)=y$ means that $b^y=x$.  Then use $b=a^{\log_a(b)}$ and properties of exponents to get $b^y=a^{y\log_a(b)}=x$.  The last equation says that $\log_a(x)=y\log_a(b)$.  Now substitute $\log_b(x)$ back in for $y$ to get the desired identity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This fits into the general principle that each of the fundamental properties of logarithms can be seen as a translation of a property of their inverses, the exponential functions.  For example, logarithms convert products to sums ($\log_b(xy)=\log_b(x)+\log_b(y)$) because exponential functions convert sums to products ($b^{x+y}=b^xb^y$).  Here, the fact that $b^x=a^{x\log_a(b)}$ says that you can change bases of your exponential function by multiplying the input variable by the constant $\log_a(b)$.  In other words, exponential functions with different bases are related by horizontal stretches or shrinks of the graphs.  When expressed in terms of the inverse functions, input and output switch, and the horizontal stretch (or shrink) changes to a vertical stretch (or shrink).  That is, instead of multiplying the input by $\log_a(b)$, you divide the output by $\log_a(b)$ to get $\log_b(x)=\frac{\log_a(x)}{\log_a(b)}$.  More succinctly, $g(x)=f(cx)$ implies $g^{-1}(x)=\frac{1}{c}f^{-1}(x)$.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2010-12-13T04:00:11.730" LastActivityDate="2010-12-13T04:00:11.730" CommentCount="1" />
  <row Id="14137" PostTypeId="1" CreationDate="2010-12-13T03:33:56.597" Score="0" ViewCount="153" Body="&lt;p&gt;Can anyone suggest at least verbally how it is that a torus is constructed from two cells?  (Total beginner to topology, been away from math as a whole for awhile...)  Thanks for any insights!&lt;/p&gt;&#xA;" OwnerUserId="4696" LastActivityDate="2010-12-13T23:26:55.440" Title="Please demonstrate decomposition of a torus into two cells" Tags="&lt;topology&gt;" AnswerCount="2" CommentCount="7" />
  <row Id="14138" PostTypeId="2" ParentId="14133" CreationDate="2010-12-13T03:36:25.383" Score="9" ViewCount="" Body="&lt;p&gt;First: They are not &quot;equivalent&quot;, they are &lt;em&gt;equal&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That said: remember the meaning of &quot;$\log_b(a) = r$&quot;. It means that $r$ is the exponent to which you must raise $b$ in order to get $a$; that is, $\log_b(a)=r$ is equivalent to $b^r=a$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So... why is $\log_b(x)=\frac{\log_k(x)}{\log_k(b)}$? Because if you let $r=\log_b(x)$, $s=\log_k(x)$ and $t=\log_k(b)$, then that means that $b^r = x$, $k^s = x$, and $k^t=b$. Therefore, $$k^s = x = b^r = (k^t)^r = k^{tr}.$$ &#xA;So $\log_k(x) = s = rt = \log_b(x)\log_k(b)$, from which you get the equality you want by dividing through by $\log_k(b)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the second equality, $\log_b(x) = \log_b(k)\log_k(x)$, let $r=\log_b(x)$, $s=\log_b(k)$, and $t=\log_k(x)$. Then $b^r = x$, $b^s = k$, and $k^t=s$. So we have:&#xA;$$b^r = x = k^t = (b^s)^t = b^{st},$$&#xD;&#xA;which means we must have $\log_b(x) = r=st=\log_b(k)\log_k(x)$, as desired.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-13T15:37:34.063" LastActivityDate="2010-12-13T15:37:34.063" CommentCount="1" />
  <row Id="14139" PostTypeId="2" ParentId="14069" CreationDate="2010-12-13T03:53:32.863" Score="1" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\: $ The Porsche travels $\rm 20\ t$ more miles than the Skoda during the time $\rm\:t\:$ they're both driving. This must be equal to the $\ 20\cdot 9$ miles of the Skoda's head start. $\ $ Hence $\rm\ t\ =\ \ldots$&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-12-13T03:53:32.863" />
  <row Id="14140" PostTypeId="1" CreationDate="2010-12-13T04:11:12.493" Score="2" ViewCount="108" Body="&lt;p&gt;Hello &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could you please help me in this problem?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have 3 independent variables, $(T,H,t)$, as inputs and&#xA;one output $P$  ( I have all data for these inputs and the output, done&#xA;experimentally measured every hour during one year).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to find a formula of this form:&#xA;$$P = f ( T , H , t )$$&#xD;&#xA;where $t$ is the time in hours and it is always in the $x$-axis (index),&#xA;$T$ is temperature, $H$ is humidity, and $P$ is power.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have all the data, and when I draw them in the same graph during one year, meaning that $P$, $T$,  and $H$ vs. hours. I found that the behavior of $P$ is oscillating, making a sinusoidal shape over the entire, year as you can see from the following ($P$ and time only):&lt;img src=&quot;http://i.imgur.com/3FdYF.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, if I make a zoom view to this figure, for example from&#xA;the $2000$th hour of the year to the $3000$th hour, it is clear that it almost has the same shape but it is oscillating.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, it keeps oscillating and increasing up to the peak point and then&#xA;it starts decreasing till the end of the year.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But this is only for only one independent variable which is time in hours.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, how if we include the effect of temperature and humidity and draw &#xA;these vectors along with the power vs. time, to see how power is changing&#xA;with respect to $T$, $H$, and $t$  rather than $t$ only:&#xA;&lt;img src=&quot;http://i.imgur.com/GwBxw.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, how can I predict the structure of the formula that relates $P$ with $T$, $H$, and $t$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any approach that you advise me to follow?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sorry for this long question and any help from you is highly appreciated. I read many papers but I could not know how to solve the problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;thanks&lt;/p&gt;&#xA;" OwnerUserId="4700" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-13T04:16:53.430" LastActivityDate="2011-04-12T19:40:14.117" Title="Find formula to changing cyclic behavior" Tags="&lt;statistics&gt;&lt;functions&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="14141" PostTypeId="1" AcceptedAnswerId="14143" CreationDate="2010-12-13T04:30:04.207" Score="3" ViewCount="112" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Euler%27s_rotation_theorem&quot; rel=&quot;nofollow&quot;&gt;Euler's rotation theorem&lt;/a&gt; (very informally stated) says that an rotation of a sphere in three dimensional space always has two fixed points. I believe this claim generalizes to (&lt;em&gt;n&lt;/em&gt;-1)-hyperspheres in &lt;em&gt;n&lt;/em&gt;-dimensional space for odd &lt;em&gt;n&lt;/em&gt;, but I am unable to patch the final gap in the proof. My strategy so far is as follows; any ideas on how to proceed would be greatly appreciated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, note that an arbitrary rotation of our hypersphere can be characterized by rotations about n-1 different axes. Since a rotation by theta in $\mathbb{R}^2$ is characterized by $\left[\begin{array}{cc}&#xD;&#xA;\cos\theta &amp;amp; -\sin\theta\\&#xD;&#xA;\sin\theta &amp;amp; \cos\theta\end{array}\right]$, it is not hard to see that a rotation about &lt;em&gt;n&lt;/em&gt;-1 different axes can be characterized by the matrix product&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$&#xD;&#xA;\left[\begin{array}{ccccccc}&#xD;&#xA;\cos\theta_{1} &amp;amp; -\sin\theta_{1} &amp;amp; 0 &amp;amp; \cdots &amp;amp;  &amp;amp;  &amp;amp; 0\\&#xD;&#xA;\sin\theta_{1} &amp;amp; \cos\theta_{1} &amp;amp; 0 &amp;amp; \cdots &amp;amp;  &amp;amp;  &amp;amp; 0\\&#xD;&#xA;0 &amp;amp; 0 &amp;amp; 1\\&#xD;&#xA;\vdots &amp;amp; \vdots &amp;amp;  &amp;amp;  &amp;amp;  &amp;amp;  &amp;amp; \vdots\\&#xD;&#xA; &amp;amp;  &amp;amp;  &amp;amp;  &amp;amp; \ddots\\&#xD;&#xA; &amp;amp;  &amp;amp;  &amp;amp;  &amp;amp;  &amp;amp;  &amp;amp; 0\\&#xD;&#xA;0 &amp;amp; 0 &amp;amp;  &amp;amp; \cdots &amp;amp;  &amp;amp; 0 &amp;amp; 1\end{array}\right]\left[\begin{array}{ccccccc}&#xD;&#xA;1 &amp;amp; 0 &amp;amp; 0 &amp;amp; \cdots &amp;amp;  &amp;amp;  &amp;amp; 0\\&#xD;&#xA;0 &amp;amp; \cos\theta_{2} &amp;amp; -\sin\theta_{2} &amp;amp; \cdots &amp;amp;  &amp;amp;  &amp;amp; 0\\&#xD;&#xA;0 &amp;amp; \sin\theta_{2} &amp;amp; \cos\theta_{2} &amp;amp; \cdots &amp;amp;  &amp;amp;  &amp;amp; 0\\&#xD;&#xA;\vdots &amp;amp; \vdots &amp;amp; \vdots &amp;amp; 1 &amp;amp;  &amp;amp;  &amp;amp; \vdots\\&#xD;&#xA; &amp;amp;  &amp;amp;  &amp;amp;  &amp;amp; \ddots\\&#xD;&#xA; &amp;amp;  &amp;amp;  &amp;amp;  &amp;amp;  &amp;amp;  &amp;amp; 0\\&#xD;&#xA;0 &amp;amp; 0 &amp;amp; 0 &amp;amp; \cdots &amp;amp;  &amp;amp; 0 &amp;amp; 1\end{array}\right]&#xD;&#xA;$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$&#xD;&#xA;\cdots\left[\begin{array}{ccccccc}&#xD;&#xA;\cos\theta_{n-1} &amp;amp; 0 &amp;amp;  &amp;amp; \cdots &amp;amp;  &amp;amp; 0 &amp;amp; -\sin\theta_{n-1}\\&#xD;&#xA;0 &amp;amp; 1 &amp;amp;  &amp;amp; \cdots &amp;amp;  &amp;amp;  &amp;amp; 0\\&#xD;&#xA;\\\vdots &amp;amp; \vdots &amp;amp;  &amp;amp; \ddots &amp;amp;  &amp;amp;  &amp;amp; \vdots\\&#xD;&#xA;\\0 &amp;amp;  &amp;amp;  &amp;amp;  &amp;amp;  &amp;amp; 1 &amp;amp; 0\\&#xD;&#xA;\sin\theta_{n-1} &amp;amp; 0 &amp;amp;  &amp;amp; \cdots &amp;amp;  &amp;amp; 0 &amp;amp; \cos\theta_{n-1}\end{array}\right]&#xD;&#xA;$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then notice that we have two fixed points if the matrix product just described has an eigenvalue of 1. The eigenvalues of a matrix are given by the roots of the characteristic polynomial; the characteristic polynomial here has real coefficents; therefore any complex roots of the characteristic polynomial must come in conjugate pairs; therefore if &lt;em&gt;n&lt;/em&gt; is odd, we must have a real eigenvalue. But since rotations preserve length, any real eigenvalues must have an absolute value of one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we've shown that for odd &lt;em&gt;n&lt;/em&gt;, there must be an eigenvalue of 1 or -1, but &lt;strong&gt;to complete the proof, we need to show that there can't be a lone eigenvalue of -1&lt;/strong&gt;, and I haven't been able to figure out how to do that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I thought I might be able to glean some intuition from the three-dimensional case, where a half-rotation about one axis gives us two eigenvalues of -1. Perhaps eigenvalues of -1 must also come in pairs? That seems plausible---dare I say even obvious?---on intuitive and physical grounds, but how might one prove it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This excursion was inspired by an exercise in &lt;a href=&quot;http://wps.prenhall.com/esm_bretscher_linear_3/&quot; rel=&quot;nofollow&quot;&gt;Bretscher &lt;em&gt;Linear Algebra&lt;/em&gt;&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2773" LastEditorUserId="2773" LastEditorDisplayName="" LastEditDate="2010-12-13T04:37:43.870" LastActivityDate="2010-12-14T05:22:12.550" Title="Ruling out a lone eigenvalue of -1 to complete this proof?" Tags="&lt;linear-algebra&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="14142" PostTypeId="2" ParentId="14009" CreationDate="2010-12-13T04:34:41.023" Score="0" ViewCount="" Body="&lt;p&gt;Beautiful, thank you both!&lt;/p&gt;&#xA;" OwnerUserId="4701" LastActivityDate="2010-12-13T04:34:41.023" CommentCount="1" />
  <row Id="14143" PostTypeId="2" ParentId="14141" CreationDate="2010-12-13T04:39:47.363" Score="5" ViewCount="" Body="&lt;p&gt;You need to show that the determinant of your transformation is equal to $1$ (or at least is positive; but then it will necessarily equal $1$).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given this:&#xA;Recall that this determinant is the product of the eigenvalues.&#xA;Now the (possibly complex) eigenvalues are all of absolute value 1, and there are an odd number of them.  The non-real eigenvalues come in conjugate pairs, which have product 1, so there are also an odd number of real eigenvalues, which are $\pm 1$, and whose product equals $1$.  It follows that at least one of them equals $1$, as desired.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll let you work out the determinant fact.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[Added: Several people wrote down the same argument in comments, reflecting (I think) the fact that this is the standard approach to this quesion.]&lt;/p&gt;&#xA;" OwnerUserId="221" LastEditorUserId="221" LastEditorDisplayName="" LastEditDate="2010-12-13T04:46:21.567" LastActivityDate="2010-12-13T04:46:21.567" CommentCount="2" />
  <row Id="14144" PostTypeId="2" ParentId="14141" CreationDate="2010-12-13T05:01:36.477" Score="9" ViewCount="" Body="&lt;p&gt;Here is an approach which gives a stronger result.  Let $g$ be a rotation of $\mathbb{R}^n$ and suppose that $g$ has a complex eigenvalue $\lambda$.  By the spectral theorem, there is a complex eigenvector $v$ associated to $\lambda$, hence a conjugate eigenvector $\bar{v}$ associated to $\bar{\lambda}$.  It follows that on the real subspace $V$ of $\mathbb{R}^n$ spanned by $v + \bar{v}$ and $\frac{v - \bar{v}}{i}$, $g$ acts as a 2-dimensional rotation, and we can break $\mathbb{R}^n$ up into a direct sum $V \oplus V^{\perp}$, and this respects the action of $g$ (so $g$ acts as a rotation on $V^{\perp} \simeq \mathbb{R}^{n-2}$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By induction, we obtain a structure theorem for rotations: any rotation is a direct sum of 2-dimensional rotations and 1-dimensional rotations (multiplication by $\pm 1$).  The determinant condition then tells us that the multiplications by $-1$ must occur in pairs, hence must pair up as 2-dimensional rotations by $\pi$.  So we can say more: any rotation has an invariant subspace $V$ of even codimension, and on $V^{\perp}$ it acts as a direct sum of nontrivial 2-dimensional rotations.  &lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-12-14T05:22:12.550" LastActivityDate="2010-12-14T05:22:12.550" />
  <row Id="14145" PostTypeId="2" ParentId="14133" CreationDate="2010-12-13T05:10:44.343" Score="3" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\: $ The first equation is $\rm\ log_K\ $ of $\rm\ B^{log_B\ X}\ =\ X$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;and the $\:$ second $\:$ equation is $\rm\  log_B\ $ of $\rm\ K^{log_K\ X}\ =\ X$&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-12-13T05:10:44.343" />
  <row Id="14146" PostTypeId="1" AcceptedAnswerId="14192" CreationDate="2010-12-13T06:23:07.577" Score="3" ViewCount="139" Body="&lt;p&gt;Hi everyone,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It would be great if someone could explain me the following. If I understand the &lt;a href=&quot;http://ocw.mit.edu/courses/mathematics/18-06-linear-algebra-spring-2010/video-lectures/lecture-1-the-geometry-of-linear-equations/&quot; rel=&quot;nofollow&quot;&gt;lecture&lt;/a&gt; correctly, given a system of, say, following equations:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x+2y-2z=1$ &lt;br&gt;&#xA;$2x+4y+z=3$ &lt;br&gt;&#xA;$4x+8y+4z=10$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can say that the they do not describe the whole 3D space because the vectors formed by the first and second columns ( $\begin{pmatrix}&#xD;&#xA; 1 \\ &#xD;&#xA; 2 \\ &#xD;&#xA; 4 &#xD;&#xA; \end{pmatrix}&#xD;&#xA;$ and $\begin{pmatrix}&#xD;&#xA; 2 \\ &#xD;&#xA; 4 \\ &#xD;&#xA; 8 &#xD;&#xA; \end{pmatrix}&#xD;&#xA;$ respectively) are not linearly independent (meaning we can't solve the given equations for &lt;em&gt;all possible&lt;/em&gt; RHS). From what I understand, it makes sense geometrically -- if we start at an arbitrary point in 3D, we can cover only a plane as we have only two directions we could move in. However, could someone please explain the meaning of the statement above in algebraic terms? What is it about the coefficients of a given variable that when they can be expressed via multiplication of some $c$ (where $c=2$ for the two aforementioned columns) by coefficients of some &lt;em&gt;other&lt;/em&gt; variable, not all RHS values generate solutions? Thanks a lot!&lt;/p&gt;&#xA;" OwnerUserId="3731" LastEditorUserId="3731" LastEditorDisplayName="" LastEditDate="2010-12-13T17:50:26.970" LastActivityDate="2010-12-13T17:58:53.777" Title="Linear independence of columns and covering the whole space (algebraic rationale)" Tags="&lt;linear-algebra&gt;" AnswerCount="4" CommentCount="4" FavoriteCount="2" />
  <row Id="14147" PostTypeId="2" ParentId="14146" CreationDate="2010-12-13T06:34:44.887" Score="3" ViewCount="" Body="&lt;p&gt;If you understand what it means in geometric terms, then you are most of the way there.  I'll try to elaborate a little on the algebra involved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The equation can be rewritten as $xv_1+yv_2+zv_3=b$, where $v_1$, $v_2$, and $v_3$ are the column vectors.  The question is, based on the relationship among the columns, which vectors $b$ can you plug in to make the equation have a solution.  Since there is a linear relationship among the columns, the answer, as you said, is that not all choices of $b$ will work.  You have $v_2=2v_1$, so the equation can also be written as $(x+2y)v_1+zv_3=b$.  Thus, $b$ must lie in the $2$-dimensional subspace spanned by $v_1$ and $v_3$, your first and third columns.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More generally, if there is some nontrivial linear relationship among the columns of the form $a_1v_1+a_2v_2+a_3v_3=0$, where $a_1$, $a_2$, and $a_3$ are scalars and at least one is nonzero, then you can write one of your vectors as a linear combination of the other $2$, and hence any vector in the span (possible &quot;right-hand sides&quot; in your question) will have to be in the span of these $2$.  The same idea generalizes to all (finite) dimensions.  If the set of column vectors is not linearly independent, then the span of the columns has lower dimension.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to see explicitly what choices of $b$ will not work in your example, you can write out what the span of $v_1$ and $v_3$ looks like, and choose any vectors not in this span.  For a quick example in this case, you could choose the cross product $v_1\times v_3$, since it is orthogonal to both $v_1$ and $v_3$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another way to look at this is that if the columns are not linearly independent, then neither are the rows, and a linear relationship among the row vectors translates to a linear relationship among the entries of the right hand side.  That is, if $A$ has row vectors $r_1$, $r_2$, and $r_3$, then &#xA;$A\mathbf{x}=&#xD;&#xA;\begin{bmatrix}&#xD;&#xA;r_1\cdot\mathbf{x}\\&#xD;&#xA;r_2\cdot\mathbf{x}\\&#xD;&#xA;r_3\cdot\mathbf{x}&#xD;&#xA;\end{bmatrix}&#xD;&#xA;$, and if there are scalars $a_1$, $a_2$, and $a_3$, not all zero, such that $a_1r_1+a_2r_2+a_3r_3=0$, then it follows from the linearity of the dot product in each variable that $a_1(r_1\cdot\mathbf{x})+a_2(r_2\cdot\mathbf{x})+a_3(r_3\cdot\mathbf{x})=0$.  That is, the right-hand side must lie in the plane described by the equation $a_1x+a_2y+a_3z=0$.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2010-12-13T06:55:05.547" LastActivityDate="2010-12-13T06:55:05.547" />
  <row Id="14148" PostTypeId="1" AcceptedAnswerId="14150" CreationDate="2010-12-13T06:49:08.700" Score="0" ViewCount="83" Body="&lt;p&gt;I am looking at the first two examples in Paul Garretts&lt;a href=&quot;http://www.math.umn.edu/~garrett/m/number_theory/newton_polygon.pdf&quot; rel=&quot;nofollow&quot;&gt; notes&lt;/a&gt; on p-adic number theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first example is computing the Newton polygon of $x^5+2x^2+5$ over $\mathbb{Q}_2$. I think this is the lower convex hull of $(0,1),(1,0),(2,0.5),(3,0),(4,0),(5,1)$. So I thought the Newton polygon should just be a horizontal line segment going from $(0,1)$ to $(5,1)$. Of course, I am completely wrong, but I am not sure what is the source of error. Are the points whose convex hull I am supposed to consider correctly identified?&lt;/p&gt;&#xA;" OwnerUserId="3947" LastActivityDate="2010-12-13T07:09:14.213" Title="Computing Newton polygons" Tags="&lt;p-adic-number-theory&gt;" AnswerCount="2" />
  <row Id="14149" PostTypeId="2" ParentId="14146" CreationDate="2010-12-13T06:53:36.493" Score="1" ViewCount="" Body="&lt;p&gt;With $\rm\ c_i = i$'th column, the system is $\rm\ c_1\ x + c_2\ y + c_3\ z\ =\ d\:.\:$ This has a solution iff $\rm\:d\:$ lies in the space spanned by the columns. Since two of the three columns are linearly dependent, the columns can span at most a two-dimensional space. Therefore the  system is not solvable for all $\rm\:d\:,\:$ i.e. the associated map is not onto.&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-12-13T06:53:36.493" />
  <row Id="14150" PostTypeId="2" ParentId="14148" CreationDate="2010-12-13T07:03:19.400" Score="2" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\ $ Below is the Newton polygon graphed by Filaseta's &lt;a href=&quot;http://www.math.sc.edu/~filaseta/newton/newton.html&quot; rel=&quot;nofollow&quot;&gt;Java applet&lt;/a&gt;. His web pages have much of interest on Newton polygons.&#xA;&lt;img src=&quot;http://i.imgur.com/N4HPL.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-12-13T07:03:19.400" CommentCount="1" />
  <row Id="14151" PostTypeId="2" ParentId="14148" CreationDate="2010-12-13T07:09:14.213" Score="2" ViewCount="" Body="&lt;p&gt;You have miscalculated the $2$-adic orders of the coefficients, and probably confused the concept of $p$-adic order with something else. In particular, the $p$-adic order of a $p$-adic number is an integer, but your third point has $y$-coordinate 0.5. As a reminder: the $p$-adic order of an integer $x$ in $\mathbb{Z}_p$ is the highest integer $v$ such that $x\in {\mathfrak m}^v$, where ${\mathfrak{m}}=p\mathbb{Z}_p$ is the maximal ideal of $\mathbb{Z}_p$. E.g. the $2$-adic order of 5 is 0, since 5 is a 2-adic unit, so the first point of the Newton polygon has coords (0,0). The $p$-adic order of 0 is $\infty$ by convention. The $2$-adic order of 2 is 1, since 2 is a uniformiser. And so on.&lt;/p&gt;&#xA;" OwnerUserId="3212" LastActivityDate="2010-12-13T07:09:14.213" />
  <row Id="14152" PostTypeId="2" ParentId="14078" CreationDate="2010-12-13T07:16:13.323" Score="2" ViewCount="" Body="&lt;p&gt;I would like to suggest three quick formulas for tackling similar kind of problems:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If there is successive &lt;strong&gt;increase&lt;/strong&gt; of $x\%$ and $y\%$, then the net change will be $$x + y + \frac{x \cdot y}{100}\%$$&#xD;&lt;/li&gt;&#xA;&lt;li&gt;If there is successive &lt;strong&gt;decrease&lt;/strong&gt; of  $x\%$ and  $y\%$, then the net change will be $$x + y - \frac{x \cdot y}{100}\%$$&#xD;&lt;/li&gt;&#xA;&lt;li&gt;If there is  &lt;strong&gt;$x\%$ increase and then  $y\%$ decrease&lt;/strong&gt;, then the net change is $$ x - y -  \frac{x \cdot y}{100}\%$$&#xD;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Exercise:  How does the last result change if there is &lt;strong&gt;$x\%$ decrease followed by $y\%$ increase&lt;/strong&gt;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These results can  be  easily verified, but it's always better to memorize them before  any quantitative aptitude test.&lt;/p&gt;&#xA;" OwnerUserId="2109" LastEditorUserId="2109" LastEditorDisplayName="" LastEditDate="2010-12-13T09:17:58.227" LastActivityDate="2010-12-13T09:17:58.227" CommentCount="1" />
  <row Id="14153" PostTypeId="2" ParentId="14146" CreationDate="2010-12-13T07:18:20.837" Score="2" ViewCount="" Body="&lt;p&gt;It seems that whatever algebraic way you try to look at it is, upon some thought, really just a paraphrase of the geometric one. One way of answering your question is that to generate all RHS values, we must have sufficient variation of the variables. However since one of the variables in this case has coefficients a multiple of another, as it varies to try to change the RHS it really is just doing the work of the other variable and so there is just not enough variation to go around to give what you want on the RHS.&lt;/p&gt;&#xA;" OwnerUserId="4594" LastActivityDate="2010-12-13T07:18:20.837" />
  <row Id="14154" PostTypeId="1" CreationDate="2010-12-13T07:53:29.960" Score="3" ViewCount="131" Body="&lt;p&gt;I have a list of several entities, all of each have a numerical relationship to each other that defines an abstract distance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a mathematical way to plot all of these on a 2D space, turning the abstract distance into a 2D euclidian distance, and preserving all of their distances?&lt;/p&gt;&#xA;" OwnerUserId="51" LastActivityDate="2011-05-22T22:39:52.330" Title="Given a metric function between a set of abstract points, what is the best way to plot them on a 2D space?" Tags="&lt;metric-spaces&gt;" AnswerCount="4" CommentCount="1" />
  <row Id="14155" PostTypeId="1" AcceptedAnswerId="14203" CreationDate="2010-12-13T07:54:27.030" Score="3" ViewCount="188" Body="&lt;p&gt;Can it be done?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For an arbitrary quadratic field $Q[\sqrt{d}]$, it's easy to show the equations are simply $ f_x = -\sqrt{d} f_y $, where $ f : Q[\sqrt{d}] \to Q[\sqrt{d}]$. I'm working on the case of $Q[\theta]$, when $\theta$ is a root of $\theta^3 - a\theta - b$, but I'm not sure if it's even possible. Has there been any mathematical research done on this topic? What do you think about it?&lt;/p&gt;&#xA;" OwnerUserId="3122" LastActivityDate="2010-12-13T22:58:55.377" Title="Generalizing Cauchy-Riemann Equations to Arbitrary Algebraic Fields" Tags="&lt;algebraic-number-theory&gt;" AnswerCount="3" CommentCount="8" />
  <row Id="14156" PostTypeId="2" ParentId="14154" CreationDate="2010-12-13T08:01:10.377" Score="0" ViewCount="" Body="&lt;p&gt;If you want to keep the distances exactly, then no, not really. If you fix 2 points in the plane, and want a third point with prescribed distances from the first two, you only have two options for that third point (symmetric across the line connecting the first two points). This severely constrains even three points; more will almost surely be impossible to graph exactly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to alter the distances in a way to preserve an order on them (in which I mean: if the distance between a pair of points is greater than that of a second pair of points, this will remain the case after altercation), or see them in more than three dimensions but from various angles or perspectives, then there might be mathematics to help you out there.&lt;/p&gt;&#xA;" OwnerUserId="3122" LastActivityDate="2010-12-13T08:01:10.377" />
  <row Id="14157" PostTypeId="2" ParentId="14154" CreationDate="2010-12-13T08:19:03.233" Score="3" ViewCount="" Body="&lt;p&gt;The question may be paraphrased as whether there is an isometry of the metric space into the plane. The answer is: in general, no.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the counter-example of a metric space with 4 points in which the distance between any 2 distinct points is 1. After placing 2 points in the plane, the third must lie in the intersection of the circles of radius 1 centred at these 2 points. The fourth must lie in the intersection of the circles of radius 1 centred at the other 3 points, and there is no way to place it.&lt;/p&gt;&#xA;" OwnerUserId="4594" LastEditorUserId="4594" LastEditorDisplayName="" LastEditDate="2011-05-22T18:47:13.507" LastActivityDate="2011-05-22T18:47:13.507" />
  <row Id="14159" PostTypeId="2" ParentId="14069" CreationDate="2010-12-13T09:02:42.463" Score="3" ViewCount="" Body="&lt;p&gt;If you are geometrically inclined, like I am, you can visualize the path of the Skoda as a line on a graph where the X axis is the time of day (in military time) and the Y axis is distance traveled in miles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are some points for the Skoda -- &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;code&gt;(7,0)&lt;/code&gt;, because at 7 am, the Skoda has traveled 0 miles.&lt;/li&gt;&#xA;&lt;li&gt;&lt;code&gt;(8,20)&lt;/code&gt;, because at 8 am, the Skoda has traveled 20 miles.&lt;/li&gt;&#xA;&lt;li&gt;&lt;code&gt;(12,100)&lt;/code&gt;, because at 12 pm, the Skoda has traveled 100 miles.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;If you plot these points on your graph, you'll see that it makes a straight line with slope 20 and X-intercept of 7.  From that you could figure out that the Y intercept is -140.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, the Porsche would have a slope of 40 and an X-intercept of 16 (we're using military time).  From there you could find out that the Y intercept is -640.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you have&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Skoda's traveled distance = $20*t - 140$&#xD;&lt;/li&gt;&#xA;&lt;li&gt;Porsche traveled distance = $40*t - 640$&#xD;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;We can graph these on a chart -- &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://img718.imageshack.us/img718/6264/msp136919dbih50965e08ac.gif&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you want to know when their traveled distance is equal.  So --&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$Skoda \, distance = Porsche \, distance$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$20*t - 140 = 40*t - 640$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$20*t = 40*t - 500$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$20*t = 500$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$t = 25$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, at 25 o' clock, or 1 am the next day, their traveled distances will be the same.&lt;/p&gt;&#xA;" OwnerUserId="51" LastActivityDate="2010-12-13T09:02:42.463" />
  <row Id="14160" PostTypeId="1" AcceptedAnswerId="14161" CreationDate="2010-12-13T09:11:09.973" Score="0" ViewCount="99" Body="&lt;p&gt;I need a function (a curve -- preferably a simple one) that, given $n$ points of a 2D space ($R^2$) passes (interpolates) through all points in a smooth/continuous way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Found out that what I need is a &lt;a href=&quot;http://en.wikipedia.org/wiki/Spline_curve&quot; rel=&quot;nofollow&quot;&gt;spline&lt;/a&gt;, however cannot find one that behaves how I need. $n$-degree Bezier curves and B-splines don't pass through the $n$ points, just move from the first to the last one using the others as control points.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A Bezier spline of $n-1$ Bezier curves (defined between each pair of adjacent points) should do the trick, but to have a smooth result I would need some control points which I don't have and don't know how to compute.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any hints?&lt;/p&gt;&#xA;" OwnerUserId="3846" LastActivityDate="2010-12-13T09:20:43.710" Title="Continuous curve interpolating a list of points" Tags="&lt;functions&gt;&lt;plane-curves&gt;&lt;bezier-curve&gt;" AnswerCount="1" />
  <row Id="14161" PostTypeId="2" ParentId="14160" CreationDate="2010-12-13T09:20:43.710" Score="0" ViewCount="" Body="&lt;p&gt;A parametric cubic spline is what you want; I've talked about them at some length in &lt;a href=&quot;http://math.stackexchange.com/questions/5166/5192#5192&quot;&gt;this answer&lt;/a&gt; (and in probably a few other places on this site, since it seems to be a common question), so I won't repeat myself now. Suffice it to say that yes, Bézier isn't what you need here.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-12-13T09:20:43.710" CommentCount="2" />
  <row Id="14162" PostTypeId="1" CreationDate="2010-12-13T10:14:25.650" Score="0" ViewCount="157" Body="&lt;p&gt;This image show a histogram (200 bins) of accumulated distances from a radar distance meter (very noisy).&#xA;The peak around 7 meters is an object. At thought this looked kind of like a normal distribution, at least if you ignore values &amp;lt;4m (which for this application is reasonable).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I am trying to do is to filter out true distances based on the probability distribution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/tYgBf.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="2367" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2011-01-12T13:54:42.820" LastActivityDate="2011-02-11T15:22:37.960" Title="What probability distribution is this?" Tags="&lt;probability&gt;&lt;signal-processing&gt;&lt;normal-distribution&gt;" AnswerCount="3" CommentCount="4" FavoriteCount="0" />
  <row Id="14163" PostTypeId="2" ParentId="14137" CreationDate="2010-12-13T10:27:57.897" Score="2" ViewCount="" Body="&lt;p&gt;The torus can be made from a single 0-cell, two 1-cells and one 2-cell. The 1-cells are attached along their boundary to the 0-cell, making two circles identified at a point. Then the 2-cell is attached to this 1-skeleton, as illustrated in the comments above. &lt;/p&gt;&#xA;" OwnerUserId="2941" LastActivityDate="2010-12-13T10:27:57.897" />
  <row Id="14164" PostTypeId="2" ParentId="14115" CreationDate="2010-12-13T10:30:08.723" Score="7" ViewCount="" Body="&lt;p&gt;The explanation for the existence of this series is given &lt;a href=&quot;http://mathworld.wolfram.com/PiFormulas.html&quot; rel=&quot;nofollow&quot;&gt;here.&lt;/a&gt; Search for the phrase&#xA;&quot;The general form of the series is&quot; to locate it. The series cited in the question appears immediately before the explanation.&lt;/p&gt;&#xA;" OwnerUserId="1301" LastActivityDate="2010-12-13T10:30:08.723" CommentCount="1" />
  <row Id="14165" PostTypeId="2" ParentId="14041" CreationDate="2010-12-13T10:30:27.317" Score="1" ViewCount="" Body="&lt;p&gt;Firstly, consider the case when we are not obliged to have the equation 0=0+0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question becomes: given $n \geq 1$ does there exist two permutations $g,h$ of $A:=\{-n,-n+1,\ldots,n\}$ such that $n=g(i)+h(i)$ where addition is addition over the integers (and therefore is not closed on $A$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can interpret an instance of this question as a set of $n$ non-attacking semiqueens on an $(2n+1) \times (2n+1)$ chessboard (with rows and columns indexed by $A$) such that no semiqueen is in cell $(i,j)$ whenever $i+j \not\in A$.  [semiqueens move up-and-down, left-and-right and along one diagonal]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the instance you cite above is illustrated by this figure:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/XEf0N.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just as with orthomorphisms of $\mathbb{Z}_n$ we can find linear constructions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/u9G5X.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we can take the equations&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$0=n+(-n)$, $1=(n-1)+(-n+2)$, up to $n=0+n$, along with&lt;/li&gt;&#xA;&lt;li&gt;$-1=-n+(n-1)$, $-2=(-n+1)+(n-3)$, up to $-n=-1+(-n+1)$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So, g(n) takes the values $\{n,n-1,\ldots,0\}$ in the first step and $\{-n,-n+1,\ldots,-1\}$ in the second and h(n) takes the values $\{-n,-n+2,\ldots,n\}$ in the first step and $\{n-1,n-3,\ldots,-n+1\}$ in the second step.  Hence both g and h are bijections of $A$, and we can clearly see that $i \mapsto g(i)-h(i)$ is also a bijection.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, yes, I guess it is easy in this case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Warning&lt;/em&gt;: Although the motivation for this question was to prove the existence of an orthomorphism of $\mathbb{Z}$, this is insufficient to prove it.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Now lets consider the case when the equation 0=0+0 is required.  We can generate some small random examples, for example, in GAP.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;RandomPermutationList:=function(n)&#xA;  local A,i,j,temp;&#xA;  A:=List([1..n],k-&amp;gt;k);&#xA;  i:=n;&#xA;  while(i&amp;gt;=2) do&#xA;    j:=Random([1..i]);&#xA;    if(j&amp;lt;i) then&#xA;      temp:=A[i];&#xA;      A[i]:=A[j];&#xA;      A[j]:=temp;&#xA;    fi;&#xA;    i:=i-1;&#xA;  od;&#xA;  return A;&#xA;end;;&#xA;&#xA;TripleJQuestion2:=function(n)&#xA;  local A,B,p,q;&#xA;  A:=Concatenation([-n..-1],[1..n]);&#xA;  B:=List([1..2*n+1],i-&amp;gt;-n-1+i);&#xA;  while(true) do&#xA;    q:=RandomPermutationList(2*n);&#xA;    p:=Concatenation(List([1..n],i-&amp;gt;A[q[i]]),[0],List([1..n],i-&amp;gt;A[q[n+i]]));&#xA;    if(Set(List([1..2*n+1],i-&amp;gt;p[i]-B[i]))=B and p[n+1]=0) then return p; fi;&#xA;  od;&#xA;  return fail;&#xA;end;;&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This was successful for $4 \leq n \leq 8$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/mwRrJ.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/1YYqR.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/FjR7k.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/z3EZB.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/qxNAa.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, combined with the fact that there are many orthomorphisms of $\mathbb{Z}_n$, it seems very likely that an instance of the desired construction exists for all $n \geq 3$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, continuing further (i.e. finding a constructions for all $n \geq 3$) could require considerable effort.  If you're really keen on resolving this problem, you could generate lots of random examples, and hope you could generalise enough of them to cover all $n \geq 3$.  But why would you want to do this?  It's not likely to give a surprising result, and it seems to be a fairly contrived problem (yes, the problem is harder if we assume the equation 0=0+0 is required, but why would anyone be interested in this problem?).  And also, unless you somehow build these constructions recursively, you're probably not going to solve the orthomorphisms of $\mathbb{Z}$ existence question.&lt;/p&gt;&#xA;" OwnerUserId="139" LastEditorUserId="139" LastEditorDisplayName="" LastEditDate="2010-12-13T11:39:31.333" LastActivityDate="2010-12-13T11:39:31.333" CommentCount="1" />
  <row Id="14167" PostTypeId="1" AcceptedAnswerId="14169" CreationDate="2010-12-13T10:55:24.610" Score="2" ViewCount="343" Body="&lt;p&gt;I randomly pick a natural number &lt;em&gt;n&lt;/em&gt;. Assuming that I would have picked each number with the same probability, what was the probability for me to pick &lt;em&gt;n&lt;/em&gt; before I did it?&lt;/p&gt;&#xA;" OwnerUserId="4707" LastEditorUserId="123" LastEditorDisplayName="" LastEditDate="2010-12-18T21:50:47.030" LastActivityDate="2011-04-03T07:49:52.183" Title="Probability of picking a random natural number" Tags="&lt;probability&gt;" AnswerCount="4" CommentCount="1" />
  <row Id="14168" PostTypeId="2" ParentId="14113" CreationDate="2010-12-13T11:00:10.657" Score="5" ViewCount="" Body="&lt;p&gt;Here is a really nice one due to &lt;a href=&quot;http://www.lacim.uqam.ca/~plouffe/inspired2.pdf&quot; rel=&quot;nofollow&quot;&gt;Simon Plouffe.&lt;/a&gt; There are many similar examples in his linked paper.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\pi = 72\sum_{n=1}^\infty \frac{1}{n(e^{n\pi} - 1)} - &#xD;&#xA;96\sum_{n=1}^\infty \frac{1}{n(e^{2n\pi} - 1)} +&#xD;&#xA;24\sum_{n=1}^\infty \frac{1}{n(e^{4n\pi} - 1)} .$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I like about it is that I can see at a glance that the series converge rapidly without having to make some mental estimate of the size of factorials.&lt;/p&gt;&#xA;" OwnerUserId="1301" LastActivityDate="2010-12-13T11:00:10.657" CommentCount="2" />
  <row Id="14169" PostTypeId="2" ParentId="14167" CreationDate="2010-12-13T11:07:47.227" Score="6" ViewCount="" Body="&lt;p&gt;There is no uniform distribution on the natural numbers. So the question itself is self-contradictory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some explanation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You say that each number is chosen with the same probability. Say probability c.&#xA;Let's denote the probability to choose n by p(n). So you say p(n)=c for any natural n.&#xA;For any distribution, the sum of probabilities always equals 1: p(1)+p(2)+p(3)+... = 1.&#xA;So c+c+c+... = 1.&#xA;But if c &gt; 0 you get c+c+c+... = $\infty$.&#xA;And if c=0 you get c+c+c+... = 0.&#xA;There's no value of c for which this sum equals 1.&lt;/p&gt;&#xA;" OwnerUserId="3533" LastEditorUserId="3533" LastEditorDisplayName="" LastEditDate="2010-12-13T11:17:45.897" LastActivityDate="2010-12-13T11:17:45.897" CommentCount="4" />
  <row Id="14171" PostTypeId="2" ParentId="14167" CreationDate="2010-12-13T11:24:12.210" Score="4" ViewCount="" Body="&lt;p&gt;If you want to choose a natural number from the set $\lbrace 1,2,3,\ldots \rbrace$ of all natural numbers, then user3533 is right. But if you pick a natural number from a finite set, say $A = \lbrace 1,2,\ldots,N \rbrace$, each number with the same probability, then this probability must be the reciprocal of the number of elements in that set. So, if $A = \lbrace 1,2,\ldots,N \rbrace$, then $p=1/N$. This follows immediately from &#xA;$\sum\nolimits_{i = 1}^N {p_i }  = 1$ with $p_1 = p_2 = \cdots = p_N = p$.&lt;/p&gt;&#xA;" OwnerUserId="2810" LastEditorUserId="2810" LastEditorDisplayName="" LastEditDate="2010-12-13T11:44:13.823" LastActivityDate="2010-12-13T11:44:13.823" />
  <row Id="14172" PostTypeId="1" CreationDate="2010-12-13T11:44:32.107" Score="0" ViewCount="180" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Possible Duplicate:&lt;/strong&gt;&lt;br&gt;&#xA;  &lt;a href=&quot;http://math.stackexchange.com/questions/12328/rsa-fast-factorization-of-n-if-d-and-e-are-know&quot;&gt;RSA: Fast factorization of N if d and e are know&lt;/a&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;p&gt;Hello!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I factor n that is a product of p and q which are large primes? The setting is an RSA encryption/decryption where you know the multiple of (p-1)(q-1), because the n has been shared through a group of users.&#xA;Also if someone can provide more info about this kind of attack and it's usage would be great.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;best regards,&#xA;Rok&lt;/p&gt;&#xA;" OwnerUserId="4708" LastActivityDate="2010-12-13T11:44:32.107" ClosedDate="2010-12-17T02:49:50.683" Title="RSA: Factorization of n=pq when we know multiple of (p-1)(q-1)" Tags="&lt;number-theory&gt;&lt;prime-numbers&gt;" CommentCount="4" />
  <row Id="14173" PostTypeId="1" AcceptedAnswerId="14184" CreationDate="2010-12-13T11:56:51.370" Score="3" ViewCount="249" Body="&lt;p&gt;In high school, when we talked about &quot;combinatorics,&quot; we solely meant &quot;mathematics of choice.&quot; For instance:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;There are 10 people who want to sit around a table. In how many ways is this possible?&lt;/li&gt;&#xA;&lt;li&gt;We have 50 balls and 20 boxes. In how many ways can we distribute balls into the boxes?&lt;/li&gt;&#xA;&lt;li&gt;There are 10 apples and 5 oranges. In how many ways can we select 7 fruits?&lt;/li&gt;&#xA;&lt;li&gt;...&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I did have this certain wrong view of combinatorics unless I read about combinatorial objects (lists, sets, graphs, etc.) and combinatorial proofs. I tried reading several sources (Wikipedia, books, papers), but still don't have a clear understanding of when something is called &lt;strong&gt;combinatorial&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Could you please elaborate it?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2420" LastActivityDate="2011-05-22T18:42:18.657" Title="What's Combinatorial Proof/Object/etc.?" Tags="&lt;combinatorics&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="2" />
  <row Id="14174" PostTypeId="2" ParentId="14162" CreationDate="2010-12-13T12:24:03.593" Score="0" ViewCount="" Body="&lt;p&gt;The density function of distance-from-mean of a bivariate normal distribution takes the form $Ax e^{-Bx^2}$, which look pretty much like your noisy graph.&lt;/p&gt;&#xA;" OwnerUserId="1508" LastActivityDate="2010-12-13T12:24:03.593" CommentCount="3" />
  <row Id="14175" PostTypeId="1" AcceptedAnswerId="14176" CreationDate="2010-12-13T13:11:47.453" Score="1" ViewCount="69" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am reading a text on maths applied to naturally occurring networks, eg. social networks. The section I am on is &quot;Walks&quot; networks. The text says:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Walks:  which allow both nodes and&#xA;  edges to be used more than once. For&#xA;  example, in Figure 1,&#xA;  Mary-Bob-Sue-Ramona-Bob-Mary-Joe is a&#xA;  walk, but not a trail or a path. A&#xA;  particular dollar bill flows through&#xA;  the person-person network along walks.&#xA;  Mary might give the bill to Bob in one&#xA;  transaction, and Bob might return it&#xA;  to Mary in another.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A trail is when a node i may be visited multiple times, but a particular edge connecting it with a node j, can be visited no more than 1 time.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;An identity from linear algebra is used which I cannot find is used &#xA;Suppose an undirected network has an adjacency matrix A, where $a_{i,j}=1$ if nodes i and j are connected, and $a_{i,j}=0$ otherwise. Then the basic identity&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(A^{n})_{ij} = \sum^{n}_{k_{1} = 1}\sum^{n}_{k_{2}=1}\ldots\sum^{n}_{k_{n-1}=1}a_{i,k_{1}}a_{k_{1},k_{2}}\ldots a_{k_{n-1},j}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;shows that $(A^{n})_{ij}$ counts the total number of walks of length n from node i to node j.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The indice k is not explicitly defined in the text or given more description. What makes it hard to understand is that the index k is not being summed over the possible nodes the matrix since n is the walk length as it is written, unless there is a mistake.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to see if I can be bold and say that the equation should have been better written as (making N be the node set or square matrix dimension):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(A^{n})_{ij} = \sum^{N}_{k_{1}=1}(\sum^{N}_{k_{2}=1}\ldots(\sum^{N}_{k_{n-1}=1}a_{i,k_{1}}a_{k_{1},k_{2}}\ldots a_{k_{n-1},j}))$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because I read the equation as a brute force summation over the total set of combinations of length n.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this interpretation correct?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and is $(A^{n})_{ij}$ the result of the entry for ij in doing the matrix exponentiation of the adjacency matrix n times because in simple examples I tried with small graphs I did not get the correct answer (tried simple fully connected graph with n=2,3,4)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Best,&lt;/p&gt;&#xA;" OwnerUserId="1587" LastActivityDate="2010-12-13T13:29:51.353" Title="calculating walks in an undirected graph using linear algebra identity" Tags="&lt;linear-algebra&gt;&lt;sequences-and-series&gt;&lt;graph-theory&gt;&lt;searching&gt;&lt;network-flow&gt;" AnswerCount="1" />
  <row Id="14176" PostTypeId="2" ParentId="14175" CreationDate="2010-12-13T13:29:51.353" Score="0" ViewCount="" Body="&lt;p&gt;Yes, you're right (the $k$'s should run from 1 to $N$, the size of the matrix), and yes, $(A^n)_{ij}$ is the $(i,j)$ entry in $A^n=AA\dots A$ ($n$ factors).&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2010-12-13T13:29:51.353" />
  <row Id="14177" PostTypeId="1" AcceptedAnswerId="14180" CreationDate="2010-12-13T13:31:17.203" Score="1" ViewCount="79" Body="&lt;p&gt;I know the following:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;my own position&lt;/li&gt;&#xA;&lt;li&gt;my own facing (the angle im turned)&lt;/li&gt;&#xA;&lt;li&gt;my targets position&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;What i would like some help with is how i calculate the shortes way to turn and the angle to turn. If i my position was (4,4) and i faced 90 degrees and my target was to face (1,1) where my target is, i know by drawing it on some paper that the shortest way to rotate is counter-clockwise 135 degrees.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But how do i calulate this? Ive tried using tangent, but using sine seems just as easy. Any ideas?&lt;/p&gt;&#xA;" OwnerUserId="4710" LastActivityDate="2010-12-13T19:08:28.223" Title="Help with calculating the angle to turn towards a target in a coordinatesystem" Tags="&lt;trigonometry&gt;" AnswerCount="1" />
  <row Id="14178" PostTypeId="2" ParentId="14167" CreationDate="2010-12-13T13:40:11.490" Score="0" ViewCount="" Body="&lt;p&gt;In case you  pick numbers every-time and thus when the number of trials are indefinitely large we use &lt;strong&gt;statistical or empirical&lt;/strong&gt; definition of probability&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;If trials are to be repeated a great number of times under essentially the same condition then the limit of the ratio of the number of times the an event happens to the total number of trials number of trials, as the number of trials increases indefinitely is called the probability of the happening of the even. In is assumed that the limit exists and finite uniquely. Symbolically&lt;/strong&gt;&#xA;$$\lim_{n-&amp;gt; \infty} \frac{m}{n} \text{ provided it is finite and unique}$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However it is worth of mentioning that this definition might appear to be different from the classical definition but both of them can be reconciled to the same sense.&lt;/p&gt;&#xA;" OwnerUserId="2109" LastEditorUserId="2109" LastEditorDisplayName="" LastEditDate="2010-12-14T07:48:40.170" LastActivityDate="2010-12-14T07:48:40.170" CommentCount="2" />
  <row Id="14179" PostTypeId="2" ParentId="14113" CreationDate="2010-12-13T13:55:58.900" Score="6" ViewCount="" Body="&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Bailey%E2%80%93Borwein%E2%80%93Plouffe_formula&quot; rel=&quot;nofollow&quot;&gt;BBP formula&lt;/a&gt; is another nice one:&#xA;$$&#xD;&#xA;  \pi = \sum_{k=0}^\infty \left[&#xD;&#xA;  \frac{1}{16^k} \! \left(&#xD;&#xA;  \frac{4}{8k+1} - \frac{2}{8k+4} - \frac{1}{8k+5} - \frac{1}{8k+6}&#xD;&#xA;  \right) \right]&#xD;&#xA;$$&#xD;&#xA;It can be used to compute the $n$th hexadecimal digit of $\pi$ without computing the preceding $n{-}1$ digits.&lt;/p&gt;&#xA;" OwnerUserId="2575" LastActivityDate="2010-12-13T13:55:58.900" />
  <row Id="14180" PostTypeId="2" ParentId="14177" CreationDate="2010-12-13T14:03:50.187" Score="3" ViewCount="" Body="&lt;p&gt;Most computer systems have a function Atan2(x,y) which returns the polar angle of a point in the range $(-\pi,\pi]$ (check the endpoints-I think you get $+\pi$ for (-1,0).  If you subtract your position from the target position and take Atan2 of the difference it will be the angle from you to target.  Now subtract your heading and you have the angle to turn.  If the result is outside  $(-\pi,\pi]$, add or subtract $2\pi$ to get into the range.  The nice thing of Atan2, as opposed to the regular atan, is it sorts out for you the signs and the branch of tangent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can use arcsin as well, but that requires measuring the distance to the target, which has an extra square root in it.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2010-12-13T19:08:28.223" LastActivityDate="2010-12-13T19:08:28.223" CommentCount="2" />
  <row Id="14181" PostTypeId="2" ParentId="14155" CreationDate="2010-12-13T14:05:12.757" Score="2" ViewCount="" Body="&lt;p&gt;I don't know if this will help, but I thought about something like this when I was an undergrad. I was thinking about the &lt;a href=&quot;http://en.wikipedia.org/wiki/Kronecker_Jugendtraum&quot; rel=&quot;nofollow&quot;&gt;Jugendtraum&lt;/a&gt;: the fact that abelian extensions of imaginary quadratic fields can be described by values of analytic functions on $\mathbb{C}$. My thought was the following: Let $K=\mathbb{Q}(\sqrt{-D})$ be a quadratic imaginary field. Then $\mathbb{C} \cong K \otimes \mathbb{R}$ and we can write the Cauchy-Riemmann equations as $(D \partial_x^2 + \partial_y^2) f=0$, which seems to be built from the norm form $K \to \mathbb{Q}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore (I thought), if we want to generalize the Jugendtraum to a real quadratic field $\mathbb{Q}(\sqrt{D})$, we should consider functions on $K \otimes \mathbb{R}$ which obey $(D \partial_x^2 - \partial_y^2) f=0$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, this didn't get anywhere. But I did show that a function $f: \mathbb{R}^2 \to \mathbb{R}$ obeying $(D \partial_x^2 - \partial_y^2) f=0$ is of the form $g(x+\sqrt{D}y) + h(x-\sqrt{D}y)$. So, if that's the road you're gong down, I can tell you where it ends.&lt;/p&gt;&#xA;" OwnerUserId="448" LastActivityDate="2010-12-13T14:05:12.757" CommunityOwnedDate="2010-12-13T14:05:12.757" CommentCount="5" />
  <row Id="14182" PostTypeId="2" ParentId="13937" CreationDate="2010-12-13T14:15:22.797" Score="1" ViewCount="" Body="&lt;p&gt;In the two dimensional case there is a simple proof. consider the set $U = \{ ax-by = 0 : x,y\in\mathbb{N}\}$. Let $A = (a,-b)$ and $v \in \mathbb{N}^2$, then for any $u\in U$, $A\cdot (v + u) = A\cdot v$. Furthermore, we have that, since $A\in\mathbb{Z}^2$, the inner product of $A$ against any vector in $\mathbb{N}^2$ is integral. So we can foliate $\mathbb{N}^2$ by lines $U_k$ such that $A\cdot u_k = k$ for $u_k \in U_k$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \sum_{U_k} \frac{1}{|A\cdot u| u_1 u_2} = \frac{1}{|k|} \sum_{U} \frac{1}{(x_k + u_1)(y_k + u_2)} $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where to get to the right hand side we used the fact that on each $U_k$, as long as $a,b &amp;gt; 0$, we have a linear ordering of the elements by distance to the origin, and we can pick $x_k, y_k$ to be the infimum. Now, as long as $a,b \neq 0$, we have that $u_1,u_2$ are both increasing, so the inner sum is summable. Furthermore, a little bit of geometry shows you that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \sum_U \frac{1}{(x_k + u_1)(y_k + u_2)} &amp;lt; \frac{2}{\sqrt{x_ky_k}} \sum_U \frac{1}{u_1u_2} $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Edit for clarification: Using the fact that $(u_1,u_2)$ are integer multiples of a unique vector $(\hat{u}_1,\hat{u}_2)$, you have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \sum_U \frac{1}{(x_k + u_1)(y_k + u_2)} \leq \frac{1}{x_ky_k}\sum_j \frac{1}{1 + j^2\frac{\hat{u}_1\hat{u}_2}{x_ky_k}} $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[here we also need to use that $\hat{u}_1\hat{u}_2 \neq 0$] then you use the fact that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \sum_j \frac{1}{1 + \alpha j^2} \leq 1 + (\alpha)^{-1/2} \sum_k \frac{1}{1 + k^2}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which I leave as an exercise [if you run out of ideas, you can always prove this using integral comparison].)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The last thing to note is that $\frac{1}{x_ky_k} = O(|k|^{-1})$ using that the points $(1,k)$ and $(k,1)$ are the infimum points for the slices $U_{-bk}$ and $U_{ak}$ respectively. So &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \sum_{k\neq 0} \sum_{U_k} \frac{1}{|A\cdot u| u_1u_2} \leq \sum_{k\neq 0} \frac{C}{k^{3/2}} \sum_U \frac{1}{u_1 u_2} &amp;lt; \infty$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;With some addition condition on what $A$ you allow, I think the claim may also be true in higher dimensions. I'll add a proof to that effect if I can think of it. &lt;/p&gt;&#xA;" OwnerUserId="1543" LastEditorUserId="1543" LastEditorDisplayName="" LastEditDate="2010-12-14T16:57:55.243" LastActivityDate="2010-12-14T16:57:55.243" CommentCount="3" />
  <row Id="14183" PostTypeId="1" AcceptedAnswerId="14185" CreationDate="2010-12-13T14:33:18.303" Score="7" ViewCount="409" Body="&lt;p&gt;I'm a lazy type of person. I love experimenting with things before learning the theory. I like to practice, and learn from my mistakes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In 3-4 months I have a basic University exam on Calculus, and I'd like to learn it in a fun way. I have found these videos to be quite entertaining:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.calculus-help.com/tutorials&quot; rel=&quot;nofollow&quot;&gt;http://www.calculus-help.com/tutorials&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However they are not exhaustive enough for my needs. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any references you would like to share? I'm not sure if it's a valid type of question for this group, so I'm sorry in advanced if it's off-topic.&lt;/p&gt;&#xA;" OwnerUserId="4713" LastActivityDate="2011-04-21T07:46:41.813" Title="Can I learn Calculus on the web, for free, using simple tutorials?" Tags="&lt;calculus&gt;&lt;learning&gt;" AnswerCount="4" CommentCount="5" FavoriteCount="3" />
  <row Id="14184" PostTypeId="2" ParentId="14173" CreationDate="2010-12-13T14:44:41.630" Score="7" ViewCount="" Body="&lt;p&gt;There are several different branches of combinatorics but in general they deal with discrete structures.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Enumerative combinatorics&lt;/em&gt;, as the name suggests, deals with counting, so the combinatorics you learn in school mostly falls into this category, asking you for the number of permutations or combinations in a particular situation. &lt;em&gt;Extremal combinatorics&lt;/em&gt;, for another example, asks for the largest or smallest structure satisfying certain properties. These terms are deliberately vague to allow for generality.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A &lt;em&gt;combinatorial proof&lt;/em&gt; is simply a proof using a combinatorial argument. For example, one can prove the binomial theorem using mathematical induction or using a combinatorial argument, in which case what is to be justified is the coefficient of the various terms which is to be obtained by counting in some way.&lt;/p&gt;&#xA;" OwnerUserId="4594" LastEditorUserId="4594" LastEditorDisplayName="" LastEditDate="2011-05-22T18:42:18.657" LastActivityDate="2011-05-22T18:42:18.657" CommentCount="1" />
  <row Id="14185" PostTypeId="2" ParentId="14183" CreationDate="2010-12-13T15:05:05.647" Score="15" ViewCount="" Body="&lt;p&gt;The MIT courseware is one of the first places to look for such materials. Here is a video series on a first course in Calculus.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://ocw.mit.edu/courses/mathematics/18-01-single-variable-calculus-fall-2006/video-lectures/&quot; rel=&quot;nofollow&quot;&gt;http://ocw.mit.edu/courses/mathematics/18-01-single-variable-calculus-fall-2006/video-lectures/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a video series on multivariable calculus&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://ocw.mit.edu/courses/mathematics/18-02-multivariable-calculus-fall-2007/video-lectures/&quot; rel=&quot;nofollow&quot;&gt;http://ocw.mit.edu/courses/mathematics/18-02-multivariable-calculus-fall-2007/video-lectures/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: I found a few more such video series. I haven't personally looked at these though.&#xA;&lt;a href=&quot;http://press.princeton.edu/video/banner/&quot; rel=&quot;nofollow&quot;&gt;http://press.princeton.edu/video/banner/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.online.math.uh.edu/HoustonACT/videocalculus/index.html&quot; rel=&quot;nofollow&quot;&gt;http://www.online.math.uh.edu/HoustonACT/videocalculus/index.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.tutor-homework.com/Math_Help/Calculus.html&quot; rel=&quot;nofollow&quot;&gt;http://www.tutor-homework.com/Math_Help/Calculus.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3431" LastEditorUserId="8476" LastEditorDisplayName="" LastEditDate="2011-04-21T07:39:06.523" LastActivityDate="2011-04-21T07:39:06.523" CommentCount="3" />
  <row Id="14186" PostTypeId="2" ParentId="14113" CreationDate="2010-12-13T15:16:48.213" Score="2" ViewCount="" Body="&lt;p&gt;The convergence can be arbitrary fast unless you don't specify what kind of series you are looking. Let $k$ be a positive integer, $a_n=\pi/k$ if $n\leq k$ and zero elsewhere. Then $\sum_{n=1}^\infty a_n$ converges to $\pi$ after $k$ summands.&lt;/p&gt;&#xA;" OwnerUserId="2856" LastActivityDate="2010-12-13T15:16:48.213" CommentCount="13" />
  <row Id="14187" PostTypeId="1" AcceptedAnswerId="14188" CreationDate="2010-12-13T15:35:43.477" Score="1" ViewCount="79" Body="&lt;p&gt;Can anybody give me a hint how to approach the following problem, please?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I actually am having a hard time stating the problem. I think an example would help you understand what the problem is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose I have 5 numbers each can be max 3 digits. I sum them together. What's the probability of the sum having the last 3 digits 500?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example of 5 numbers having the last 3 digits 500 are (100, 100, 100, 100, 100) and (0, 0, 0, 0, 1500).&lt;/p&gt;&#xA;" OwnerUserId="4102" LastActivityDate="2010-12-13T15:45:37.813" Title="Probability of the sum of n numbers giving the same last d digits" Tags="&lt;probability&gt;" AnswerCount="1" />
  <row Id="14188" PostTypeId="2" ParentId="14187" CreationDate="2010-12-13T15:45:37.813" Score="4" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\ $ For $\rm\ x + y + z\equiv 500\ \ (mod\ 1000) $ you have $1000$ choices for $\rm\:x\:$, $1000$ choices for $\rm\:y\:$, and then $\rm\:z\:$ is uniquely determined by the linear equation.&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-12-13T15:45:37.813" CommentCount="9" />
  <row Id="14189" PostTypeId="1" AcceptedAnswerId="14196" CreationDate="2010-12-13T16:22:04.127" Score="2" ViewCount="161" Body="&lt;p&gt;I am a scientist writing my first manuscript with a substantial amount of mathematical methodological documentation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am using LaTeX, but this is not my question. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to find a list of usage rules similar to &lt;a href=&quot;http://en.wikipedia.org/wiki/The_Elements_of_Style&quot; rel=&quot;nofollow&quot;&gt;Strunk and White&lt;/a&gt;. I can follow the practices that I see in my field, but it would be helpful to know the underlying rules and recommendations that will help me explain mathematical concepts in a reproducible and understandable way. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="3733" LastActivityDate="2010-12-13T19:25:01.033" Title="What are or where can I find style guidelines for writing math?" Tags="&lt;reference-request&gt;&lt;teaching&gt;" AnswerCount="5" CommentCount="2" FavoriteCount="1" />
  <row Id="14190" PostTypeId="1" AcceptedAnswerId="14194" CreationDate="2010-12-13T16:41:43.873" Score="2" ViewCount="139" Body="&lt;p&gt;This post is related to a previous SE post &lt;a href=&quot;http://math.stackexchange.com/questions/13959/if-a-1-meter-rope-is-cut-at-two-uniformly-randomly-chosen-points-to-give-three-p/13967#13967&quot;&gt; If a 1 meter rope ….   &lt;/a&gt; concerning average length of a smallest segment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A rope of 1m is divided into three pieces by two random points. Find the average length of the largest segment.&#xA;My answer is 11/18.  Here is how I do it:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here we have two independent random variables $X,Y$, both uniform on $[0,1]$. Let &#xA;$A=\min (X,Y), B=\max (X,Y)$ and $C=\max (A, 1-B, B-A)$. First we want to find the probability&#xA;density function $f_C(a)$ of $C$. Let $F_C(a)$ be the cumulative distribution function. Then&#xA;$$ F_C(a) = P(C\le a)=P(A\le a, 1-B\le a, B-A\le a).$$  By rewriting this probability as area in the unit square, I get&#xA;$$F_C(a)=\left\{\begin{array}{ll} (3a-1)^2 &amp;amp; \frac{1}{3}\le a\le \frac{1}{2}\\ 1-3(1-a)^2 &amp;amp; \frac{1}{2}\le a\le 1\end{array}\right.$$ from which it follows that &#xA;$$f_C(a)=\left\{\begin{array}{ll} 6(3a-1) &amp;amp; \frac{1}{3}\le a\le \frac{1}{2}\\ 6(1-a) &amp;amp; \frac{1}{2}\le a\le 1\end{array}\right.$$ Therefore&#xA;the expected value of $C$ is&#xA;$$\int_{1/3} ^{1/2}6a(3a-1) da+\int_{1/2} ^{1}6a(1-a) da= \frac{11}{18}.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My questions are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(A) Is there a &quot;clever&quot; way to figure out this number 11/18?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(B) What is the answer if the rope is divided into $n&amp;gt;3$ segments?&lt;/p&gt;&#xA;" OwnerUserId="3249" LastActivityDate="2011-01-19T21:23:47.727" Title="Average length of the longest segment" Tags="&lt;probability&gt;" AnswerCount="1" />
  <row Id="14191" PostTypeId="2" ParentId="14189" CreationDate="2010-12-13T16:43:51.480" Score="1" ViewCount="" Body="&lt;p&gt;This may not be exactly what you're looking for, but &lt;a href=&quot;http://terrytao.wordpress.com/advice-on-writing-papers/&quot; rel=&quot;nofollow&quot;&gt;Tao&lt;/a&gt; has good advice and includes on his page any other source I would consider referencing.&lt;/p&gt;&#xA;" OwnerUserId="3537" LastActivityDate="2010-12-13T16:43:51.480" />
  <row Id="14192" PostTypeId="2" ParentId="14146" CreationDate="2010-12-13T16:46:16.147" Score="4" ViewCount="" Body="&lt;p&gt;You seem to want some explanation that does not involve vectors, subspaces spanned by columns, etc. I have to say that, in my personal opinion, vectors, subspaces, spans, etc. are the best way to think about it, but let me give it a try.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose you have a $3\times 3$ linear system in $x$, $y$, and $c$,&#xA;$$\begin{array}{rcccccl}&#xD;&#xA;a_1x &amp;amp; + &amp;amp; b_1y &amp;amp; + &amp;amp; c_1z &amp;amp; = &amp;amp; r_1\\&#xD;&#xA;a_2x &amp;amp; + &amp;amp; b_2y &amp;amp; + &amp;amp; c_2z &amp;amp; = &amp;amp; r_2\\&#xD;&#xA;a_3x &amp;amp; + &amp;amp; b_3y &amp;amp; + &amp;amp; c_3z &amp;amp; = &amp;amp; r_3&#xD;&#xA;\end{array}$$&#xD;&#xA;and that there is a constant $k$ such that $b_1=ka_1$, $b_2=ka_2$, and $b_3=kc_3$. Why does this mean that the system will not necessarily have solutions for any given $r_1$, $r_2$, and $r_3$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Set $w=x+ky$. Then we can rewrite $a_ix + b_iy = a_ix + ka_iy = a_i(x+ky) = a_iw$. So we can rewrite the system as a system of three equations in &lt;em&gt;two&lt;/em&gt; unknowns:&#xA;$$\begin{array}{rcccl}&#xD;&#xA;a_1w &amp;amp; + &amp;amp; c_1z &amp;amp; = &amp;amp; r_1\\&#xD;&#xA;a_2w &amp;amp; + &amp;amp; c_2z &amp;amp; = &amp;amp; r_2\\&#xD;&#xA;a_3w &amp;amp; + &amp;amp; c_3z &amp;amp; = &amp;amp; r_3&#xD;&#xA;\end{array}$$&#xD;&#xA;This is an overdetermined system (more equations than unknowns), so we know that there are ways to make the right hand side inconsistent and for the system to have no solutions. So it is not the case that the system has solutions for all possible choices of $r_1$, $r_2$, and $r_3$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that a solution to the original system gives a solution to this system by simply letting $w=x+ky$. Does a solution to this system correspond to a solution of the original system? Yes. Suppose that $w=a$, $z=c$ is a solution. Set $x=a$ and $y=0$ to get a solution to the original system (in fact, you can let $x=u$ and $y = (a-u)/k$ if $k\neq 0$ and get a solution, for any real $u$; if $k=0$, then you take $x=a$ and $y$ arbitrary). So solutions to this system are equivalent to solutions of the original system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same thing happens if what you have is that there exist nonzero $k$ and $\ell$ such that $a_i = kb_i+\ell c_i$ (another way for the columns to be linearly independent). Then we can rewrite &#xA;$$a_ix + b_iy + c_iz = b_i(kx+y) + c_i(\ell x + z)$$&#xD;&#xA;so letting $w_1 = kx+y$ and $w_2=\ell x+z$ you get an overdetermined system again. And again, solutions of this system correspond to solutions of the original, in the sense that the original has a solution if and only if this one has a solution.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-13T17:58:53.777" LastActivityDate="2010-12-13T17:58:53.777" CommentCount="1" />
  <row Id="14193" PostTypeId="2" ParentId="14189" CreationDate="2010-12-13T16:50:58.560" Score="4" ViewCount="" Body="&lt;p&gt;An excellent reference is &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Krantz, Steven. A primer of mathematical writing. AMS (1997).    &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;It covers much more than just papers, and I recommend it even to people with some experience. (There is a more technical book by Krantz in the same vein that treats books more extensively, &quot;Mathematical publishing, a guidebook&quot;.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another excellent source, particularly because of Halmos's essay, is the short pamphlet &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Steenrod, Norman; Halmos, Paul; Schiffer, Menahem; Dieudonné, Jean. How to write mathematics. AMS (1973).&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Finally, some colleagues have recommended the following as well: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Higham, Nicholas. Handbook of writing for the mathematical sciences. SIAM (1993).&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Hope this helps!&lt;/p&gt;&#xA;" OwnerUserId="462" LastActivityDate="2010-12-13T16:50:58.560" />
  <row Id="14194" PostTypeId="2" ParentId="14190" CreationDate="2010-12-13T16:51:17.200" Score="4" ViewCount="" Body="&lt;p&gt;The answer to (B) is actually given in both Yuval Filmus' and my answers to the question about the &lt;a href=&quot;http://math.stackexchange.com/questions/13959&quot;&gt;average length of the shortest segment&lt;/a&gt;.  It's $$\frac{1}{n} H_n,$$ where $H_n = \sum_{k=1}^n \frac{1}{k},$ i.e., the $n$th &lt;a href=&quot;http://en.wikipedia.org/wiki/Harmonic_number&quot; rel=&quot;nofollow&quot;&gt;harmonic number&lt;/a&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;HR&gt;&#xA;&quot;Clever&quot; is of course subjective, but here's an argument for (A) in the $n$-piece case.  At least there's only one (single-variable) integration in it. :)  &lt;BR&gt; &lt;BR&gt;&#xA;If $X_1, X_2, \ldots, X_{n-1}$ denote the positions on the rope where the cuts are made, let $V_i = X_i - X_{i-1}$, where $X_0 = 0$ and $X_n = 1$.  So the $V_i$'s are the lengths of the pieces of rope.&#xA;&lt;BR&gt;&lt;BR&gt;&#xA;The key idea is that the probability that any particular $k$ of the $V_i$'s simultaneously have lengths longer than $c_1, c_2, \ldots, c_k$, respectively (where $\sum_{i=1}^k c_i \leq 1$), is $$(1-c_1-c_2-\ldots-c_k)^{n-1}.$$  This is proved formally in David and Nagaraja's &lt;em&gt;&lt;a href=&quot;http://books.google.com/books?id=bdhzFXg6xFkC&amp;amp;pg=PA133&amp;amp;lpg=PA133&amp;amp;dq=david+and+nagaraja+order+statistics+division+of+random+interval&amp;amp;source=bl&amp;amp;ots=OaK_l7ycZk&amp;amp;sig=BADI1rALE7hGm3-tvR6du1XKKzk&amp;amp;hl=en&amp;amp;ei=G68GTcqPEIeusAOZmdC4Bw&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=1&amp;amp;ved=0CBMQ6AEwAA#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;Order Statistics&lt;/a&gt;&lt;/em&gt;, p. 135.  Intuitively, the idea is that in order to have pieces of size at least $c_1, c_2, \ldots, c_k$, all $n-1$ of the cuts have to occur in intervals of the rope of total length $1 - c_1 - c_2 - \ldots - c_k$.  For example, $P(V_1 &amp;gt; c_1)$ is the probability that all $n-1$ cuts occur in the interval $(c_1, 1]$, which, since the cuts are randomly distributed in $[0,1]$, is $(1-c_1)^{n-1}$.&#xA;&lt;BR&gt;&lt;BR&gt;&#xA;If $V_{(n)}$ denotes the largest piece of rope, then &#xA;$$P(V_{(n)} &amp;gt; x) = P(V_1 &amp;gt; x \text{ or } V_2 &amp;gt; x \text{ or } \cdots \text{ or } V_n &amp;gt; x).$$  This calls for the &lt;a href=&quot;http://en.wikipedia.org/wiki/Inclusion%E2%80%93exclusion_principle#In_probability&quot; rel=&quot;nofollow&quot;&gt;principle of inclusion/exclusion&lt;/a&gt;.  Thus we have, using the &quot;key idea&quot; above,&#xA;$$P(V_{(n)} &amp;gt; x) = n(1-x)^{n-1} - \binom{n}{2} (1 - 2x)^{n-1} + \cdots + (-1)^{k-1} \binom{n}{k} (1 - kx)^{n-1} + \cdots,$$ &#xA;where the sum continues until $kx &amp;gt; 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, &#xA;$$E[V_{(n)}] = \int_0^{\infty} P(V_{(n)} &amp;gt; x) dx = \sum_{k=1}^n \binom{n}{k} (-1)^{k-1} \int_0^{1/k} (1 - kx)^{n-1} dx = \sum_{k=1}^n \binom{n}{k} (-1)^{k-1} \frac{1}{nk} $$&#xA;$$= \frac{1}{n} \sum_{k=1}^n \frac{\binom{n}{k}}{k} (-1)^{k-1} = \frac{H_n}{n},$$&#xA;where the last step applies a known &lt;a href=&quot;http://en.wikipedia.org/wiki/Harmonic_number#Calculation&quot; rel=&quot;nofollow&quot;&gt;binomial sum identity&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2370" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2011-01-19T21:23:47.727" LastActivityDate="2011-01-19T21:23:47.727" CommentCount="1" />
  <row Id="14195" PostTypeId="2" ParentId="14189" CreationDate="2010-12-13T16:53:47.197" Score="1" ViewCount="" Body="&lt;p&gt;See &lt;a href=&quot;http://math.stackexchange.com/questions/7011/book-about-technical-and-academical-writting/7032#7032&quot;&gt;this&lt;/a&gt; previous post.&lt;/p&gt;&#xA;" OwnerUserId="1747" LastActivityDate="2010-12-13T16:53:47.197" />
  <row Id="14196" PostTypeId="2" ParentId="14189" CreationDate="2010-12-13T17:19:23.793" Score="8" ViewCount="" Body="&lt;p&gt;Here's what is in my shelf, in the order in which I picked them up:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;Mathematics into Type&lt;/em&gt; (Updated Edition). American Mathematical Society, Providence, RI, 1999, ISBN 0-8218-1961-5. It includes what the usual submission/refereeing process is, how to mark manuscripts, how to space symbols, in-line equations, and display equations (and how to break long equations across lines). It does not discuss matters of &lt;em&gt;style&lt;/em&gt; (how a mathematician usually &lt;em&gt;says&lt;/em&gt; certain things), however.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;Handbook of Writing for the Mathematical Sciences&lt;/em&gt; by Nicholas J. Higham. Society for Industrial and Applied Mathematics, Philadelphia PA, 1993. ISBN 0-89871-314-5. It includes a section on Mathematical Writing (Chapter 3) and one on English Usage (Chapter 4), as well as tips on organizing a paper and the like. I'll add that Chapter 2 (&quot;Writer's Tools and Recommended Reading&quot;) contains a wealth of references.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;A Primer of Mathematical Writing. Being a disquisition on having your ideas recorded, typeset, published, read, and appreciated&lt;/em&gt; by Steven G. Krantz. American Mathematical Society, Providence RI, 1997, ISBN 0-8218-0635-1. One full chapter devoted specifically to mathematical writing (&quot;How to State a Theorem&quot;, &quot;How to Prove a Theorem&quot;, &quot;How to State a Definition&quot;, etc).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Paul R. Halmos. &lt;em&gt;How to write mathematics&lt;/em&gt;. Enseign. Math. &lt;strong&gt;16&lt;/strong&gt; (1970), pp. 123-152. A very good read, cited by every other work mentioned above. Also available on the web, as a quick google search will reveal; e.g., &lt;a href=&quot;http://www.math.uga.edu/~azoff/courses/halmos.pdf&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;How to write mathematics&lt;/em&gt;, corrected edition, by Norman E. Steenrod, Paul R. Halmos, Menahem M. Schiffer, and Jean A. Dieudonn&amp;eacute; American Mathematical Society, Providence RI 1981, ISBN 0-8218-0055-8. A reprint of four papers on writing mathematics, including Halmos's paper mentioned above.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;Writing Mathematics Well: A Manual for Authors&lt;/em&gt; by Leonard Gillman. Mathematical Association of America, 1987, ISBN 0-88385-443-0 (page ix reads: &quot;This manuscript was prepared by the author on an Apple Macintosh with the help of Mac$\Sigma$qn, a program for symbols and equations&quot;; how times have changed...)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;Mathematical Writing&lt;/em&gt;, by Donald E. Knuth, Tracy Larrabee, and Paul M. Roberts. MAA Notes no. 14, The Mathematical Association of America, 1989, ISBN 0-88385-063-X; this is a bit of an odd duck, in my opinion. They are the class notes for a course on mathematical writing taught by Knuth.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;There are other, more general guides; I keep a copy of Fowler's &quot;Modern English Usage&quot; and the &quot;Oxford Guide to English Usage&quot; always available, as well as a copy of the Chicago Manual of Style (14th edition) and of Strunk and White at home (recently joined by &lt;em&gt;Eats, Shoots and Leaves: the zero tolerance approach to punctuation&lt;/em&gt; by Lynne Truss; now I can tell my students that the semi-colon was invented by the same guy who invented italics). Mary-Claire van Leunen's &lt;em&gt;A Handbook for Scholars&lt;/em&gt; (revised edition) is a classic as well. But these are not specific to mathematics, and a lot of the advice has to be actively disobeyed to follow standard mathematical phraseology or uses (many a copywriter unaccostumed to mathematics has choked on &quot;a Green's function&quot;). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Added:&lt;/em&gt; I don't think you will necessarily pick up the traditional jargon/cadence of mathematics from any of the above sources, however. That's something that is picked up by osmosis, through hearing and reading a lot of mathematics. But this even happens across fields, not just across disciplines; one can often spot that someone is new to a particular field within mathematics by how he or she phrases certain statements, which is at odds with the usual practices of the sub-field in question. But the advice you get in any of the above (my recommendations, in no particular order, are Highman, Krantz, and Halmos for writing content, and &lt;em&gt;Mathematics into Type&lt;/em&gt; for final preparation of the manuscript for submission, especially if you are submitting to a mathematics journal) should carry you through.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-13T19:25:01.033" LastActivityDate="2010-12-13T19:25:01.033" CommentCount="4" />
  <row Id="14197" PostTypeId="2" ParentId="14189" CreationDate="2010-12-13T17:21:37.317" Score="3" ViewCount="" Body="&lt;p&gt;And if you have time, you should watch J.P. Serre's lecture &lt;a href=&quot;http://wstein.org/edu/basic/serre/&quot; rel=&quot;nofollow&quot;&gt;&quot;How to Write Mathematics Badly&quot;&lt;/a&gt; for what &lt;em&gt;not&lt;/em&gt; to do. &lt;/p&gt;&#xA;" OwnerUserId="1543" LastActivityDate="2010-12-13T17:21:37.317" CommentCount="1" />
  <row Id="14199" PostTypeId="2" ParentId="14155" CreationDate="2010-12-13T17:51:14.490" Score="3" ViewCount="" Body="&lt;p&gt;To answer the three literal questions asked:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A wise man said once, &quot;define PDE over a number field and you will be a rich man&quot;.  This was meant not as research advice but to indicate that everyone would love to have such a thing, many have tried to create it, and obvious approaches don't work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(p.s.  if you search this site for &quot;arithmetic Kodaira-Spencer class&quot; you will find an earlier discussion with a pointer from Matt E to a paper by Faltings on the nonexistence of (any linear version of) such an object over number fields.  In other words, the elusiveness of arithmetic differentiation is not just a sociological observation.)&lt;/p&gt;&#xA;" OwnerUserId="467" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-12-13T18:02:21.600" LastActivityDate="2010-12-13T18:02:21.600" CommentCount="1" />
  <row Id="14200" PostTypeId="1" AcceptedAnswerId="16198" CreationDate="2010-12-13T18:19:24.790" Score="1" ViewCount="116" Body="&lt;p&gt;Is there a way to (efficiently) enumerate all simple graphs with a given number of vertices up to isomorphism? That is, I don't care about the degeneracy of a graph, only the set of isomorphicly isomorphically unique ones of a given size.&lt;/p&gt;&#xA;" OwnerUserId="196" LastActivityDate="2011-01-03T03:13:41.813" Title="Enumerating all simple graphs" Tags="&lt;graph-theory&gt;&lt;combinatorics&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="14201" PostTypeId="2" ParentId="14140" CreationDate="2010-12-13T18:41:02.080" Score="1" ViewCount="" Body="&lt;p&gt;It seems you have one independent variable (t) and three dependent variables (P, H, and T) in that you do not have independent control of H and T.  This does not prevent finding a relation that would allow you to predict P based on H and T.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Looking at your data from year 2000, P and T overlay very closely.  A naive point of view would be to predict that P is a linear function of T and H doesn't matter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, you need to decide if you think the oscillations are data or noise.  If they are noise, you should filter them out.  The easiest, but far from the best, approach is just to average a number of data points.  There is discussion of filtering in &lt;a href=&quot;http://nr.com/&quot; rel=&quot;nofollow&quot;&gt;Numerical Recipes&lt;/a&gt; as well as any numeric analysis book.  As suggested by J.M., doing an FFT makes this much easier, and there are routines to do that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After any filtering, you can write P in some functional form involving T, H, and t and use a multidimensional minimizer to find the best fit.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-12-13T18:41:02.080" />
  <row Id="14202" PostTypeId="2" ParentId="14173" CreationDate="2010-12-13T19:11:13.803" Score="2" ViewCount="" Body="&lt;p&gt;A combinatorial argument often consists in giving a bijection between two sets or at least has such an observation as its key step. More precisely, one often starts with one set $A$ whose elements one wishes to count. You then find a bijection from $A$ to some other set $B$ whose elements are easily counted, leading to some expression for the number of elements of $A$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A very simple example would be the problem of counting the number of subsets of some finite set of size $n$; call this set of subsets $A$. We first notice there's a bijection from $A$ to the set $B$ of length $n$ binary strings. We can easily count the number of elements of $B$; it's just $2^n$. Hence $A$ also has $2^n$ elements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More generally, a combinatorial argument proceeds by noticing that some statement to be proved is equivalent to the assertion that two sets are in bijection and then by constructing an explicit bijection.&lt;/p&gt;&#xA;" OwnerUserId="251" LastActivityDate="2010-12-13T19:11:13.803" />
  <row Id="14203" PostTypeId="2" ParentId="14155" CreationDate="2010-12-13T19:40:21.540" Score="7" ViewCount="" Body="&lt;p&gt;If you want to take derivatives in a rather general context, you can.  For instance, let &#xA;$K$ be any topological field.  Then for any function $f: K \rightarrow K$ and any &#xA;$x \in K$, we say the derivative of $f$ exists at $x$ if the usual limit &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f&amp;#39;(x) = \lim_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;exists.  So if your topology on $\mathbb{Q}$ is the usual Archimedean one coming from the restricting the Euclidean metric on $\mathbb{R}$, you can speak of continuous and differentiable functions $f: \mathbb{Q} \rightarrow \mathbb{Q}$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;However&lt;/strong&gt; these functions lack most of the nice properties of the corresponding functions on $\mathbb{R}$ or $\mathbb{C}$, due to the lack of completeness of $\mathbb{Q}$.  That is, a continuous (and even differentiable) function on a closed &#xA;interval in $\mathbb{Q}$ need not satisfy the intermediate value property, need not be bounded, if it is bounded need not assume a maximum or minimum value, and need not be uniformly continuous, need not satisfy the Mean Value Theorem or Taylor's Theorem, and so forth.  So it is fair to ask why one would want to study differentiable functions on $\mathbb{Q}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I should say that it's not completely clear that there is no good answer to this.  For instance, in the case of the $p$-adic field $\mathbb{Q}_p$, it is not so common to speak of or study differentiable functions.  However, there &lt;em&gt;is&lt;/em&gt; a nontrivial theory here, as I learned from Alain Robert's book on $p$-adic analysis.  While it is not as essential as in the real or complex case, it has definitely been studied and written about.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The issue of defining partial derivatives over number fields is a bit more subtle, and here I think there are problems that the OP has yet to appreciate.  Think about the Cauchy-Riemann equations on $\mathbb{C}$: Step 0 here is identifying $\mathbb{C}$ with $\mathbb{R}^2$ and a function $f: \mathbb{C} \rightarrow \mathbb{C}$ as a function &#xA;$f: \mathbb{R}^2 \rightarrow \mathbb{R}^2$, i.e., a real function in several variables.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More generally, let $(K,| \ |)$ be a normed field, i.e., $| \ |: K \rightarrow \mathbb{R}^{\geq 0}$ is such that $\rho(x,y) := |x-y|$ is gives a metric on $K$ with the additional property that $|xy| = |x||y|$ for all $x,y \in K$.  And let $V$ be a finite-dimensional normed $K$-vector space, say of dimension $n$  Then a basic but perhaps underappreciated result is that the &lt;strong&gt;completeness&lt;/strong&gt; of $K$ is essential to make an identification of normed $K$-linear spaces $V \cong K^n$.  So, for example, the Euclidean norm $| \ |$ on $\mathbb{C}$ is &lt;em&gt;equivalent&lt;/em&gt; to any product metric on $\mathbb{R}^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This property does not hold in general when we extend the Archimedean norm $| \ |$ on $\mathbb{Q}$ to an arbitrary number field $K$.  In fact, things work out okay exactly if $K = \mathbb{Q}$ or $K$ is an imaginary quadratic field.  So let's look at the next &#xA;simplest case, that of a real quadratic field $K = \mathbb{Q}(\sqrt{D})$.  In this case &#xA;there are two norms on $K$ extending the Euclidean norm, say $| \ |_1$ and &#xA;$| \ |_2$ corresponding to the two different embeddings of $K$ into $\mathbb{R}$.  (So, for instance, if $|\sqrt{D}|_1$ is the positive square root of $D$, $|\sqrt{D}|_2$ is &#xA;the negative square root of $D$.)  Neither $(K,| \ |_1)$ nor $(K,| \ |_2)$ is equivalent, as a normed $\mathbb{Q}$-vector space, to $\mathbb{Q}^2$ with the product norm.  Indeed, here is an even stronger statement: consider the (unique) embedding $\iota$ of $\mathbb{Q}$ into $K$.  Then, with respect to the topology induced by either $| \ |_1$ or $|\ |_2$ $\iota(\mathbb{Q})$ is dense, since indeed both are dense in their completions, which are isomorphic to $\mathbb{R}$.  On the other hand, the embedding of $\mathbb{Q}$ into $\mathbb{Q}^2$ via the diagonal, $x \mapsto (x,x)$, has closed image.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the very idea of partial derivatives here makes me nervous.  An upshot of the above discussion is that choosing the basis $\{1,\sqrt{D} \}$ for $\mathbb{Q}(\sqrt{D})$ over $\mathbb{Q}$, the ``directions'' $1$ and $\sqrt{D}$ are not metrically/topologically independent, even though they are independent in the sense of linear algebra.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A final remark to make is that, to a number theorist like myself, it is very unnatural to choose a particular Archimedean norm $| \ |$ on a number field $K$.  Rather, there is a finite set of equivalence classes of such norms (&quot;Archimedean places&quot;) which can be determined by looking at the factorization of any polynomial $P(t) \in \mathbb{Q}[t]$ &#xA;such that $K \cong \mathbb{Q}[t]/(P)$ over $\mathbb{R}$: if $P$ has $r$ real roots and &#xA;$s$ complex-conjugate pairs of complex roots, then there are $r + s$ Archimedean places of $K$, and one needs to work with all of them at once in order to do topologically useful things.  In particular, the natural embedding here is really from $K$ into $K \otimes_{\mathbb{Q}} \mathbb{R} \cong \mathbb{R}^r \oplus \mathbb{C}^s$.  Note that this latter object is a field in exactly two cases: when $(r,s) = (1,0)$ (i.e., $K = \mathbb{Q})$ or when $(r,s) = (0,1)$ (i.e., $K$ is an imaginary quadratic field).&lt;/p&gt;&#xA;" OwnerUserId="299" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2010-12-13T22:58:55.377" LastActivityDate="2010-12-13T22:58:55.377" />
  <row Id="14204" PostTypeId="1" AcceptedAnswerId="14205" CreationDate="2010-12-13T19:45:25.653" Score="2" ViewCount="79" Body="&lt;p&gt;What is the rate of growth of the partial sums of the reciprocals of the odd numbers?&lt;/p&gt;&#xA;" OwnerUserId="1856" LastEditorUserId="1016" LastEditorDisplayName="" LastEditDate="2010-12-13T19:50:21.273" LastActivityDate="2010-12-13T19:54:17.653" Title="The rate of growth of the partial sums of the reciprocals of the odd numbers" Tags="&lt;sequences-and-series&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="14205" PostTypeId="2" ParentId="14204" CreationDate="2010-12-13T19:54:17.653" Score="6" ViewCount="" Body="&lt;p&gt;$\sum_{1}^{n} \frac{1}{2i-1} = \sum_{1}^{2n} \frac{1}{i} - \frac{1}{2}\sum_{1}^{n} \frac{1}{i}$, and this is approximately $\ln(2n) - \frac{1}{2}\ln(n)+\frac{1}{2} \gamma = \frac{1}{2} \ln(n) + \ln(2) + \frac{1}{2} \gamma$ for large $n$.&lt;/p&gt;&#xA;" OwnerUserId="4253" LastActivityDate="2010-12-13T19:54:17.653" CommentCount="2" />
  <row Id="14206" PostTypeId="1" CreationDate="2010-12-13T20:14:46.387" Score="2" ViewCount="104" Body="&lt;p&gt;&lt;strong&gt;How does a technical and theoretical study of a project of implementing an algorithm on a specific hardware architecture?&lt;/strong&gt;&#xA;Function example:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;%Expresion 1:&#xA;y1 = exp (- (const1 + x) ^ 2 / ((conts2 ^ 2)),&#xA;y2 = y1 * const3&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Where x is the input variable,&#xA;y2 is the output and const1, const3 const2 and are constant.&#xA;I need to determine the error you get in terms of architecture that decides to develop, for example it suppose that is not the same architecture with 11 bits for the exponent and 52 bits for mantissa.&#xA;This is the concept of error I handle:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Relative Error = (Real Data - Architecture Data) / (Real Data) * 100&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I consider as 'Real Data' the ouput of my algorithm I get from Matlab  (Matlab use double precission float point, IEEE 754, 52 bits to mantisa, 11 bits for exponent, one bit fo sign ) with the expression 1 and I consider as 'Architecture Data' the ouput of my algorithm running in an particular architecture (For instance a architecture that uses 12 bits for mantisa, 5 bits for exponent and 1 bit for sign)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt;&#xA;NOTE: The kind of algorithm I am refering to are all those which use matemathical functions that can be descompose in terms of addition, multiplication, subtrations and division.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you!&lt;/p&gt;&#xA;" OwnerUserId="4255" LastEditorUserId="4255" LastEditorDisplayName="" LastEditDate="2010-12-16T17:39:44.323" LastActivityDate="2010-12-23T11:02:30.000" Title="Development of a specific hardware architecture for a particular algorithm" Tags="&lt;analysis&gt;&lt;algorithms&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="14207" PostTypeId="1" CreationDate="2010-12-13T20:35:20.547" Score="1" ViewCount="220" Body="&lt;p&gt;Suppose $g:\mathbb{R} \rightarrow [0,\infty)$ is a strictly increasing function such that $\lim g(x) = \infty$ as $x \rightarrow \infty$.&#xA;Suppose $h:\mathbb{R} \rightarrow [0,\infty)$ is a strictly decreasing function such that $\lim h(x) = 0$ as $x \rightarrow \infty$.&#xA;Consider the product function $f(x)=g(x)h(x)$. Is it possible to construct such $g,h$ such that $f$ does not have a limit as $x \rightarrow \infty$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even better, can such an $f$ be constructed if we insist that $g(x)=x$?&lt;/p&gt;&#xA;" OwnerUserId="4719" LastActivityDate="2010-12-14T05:48:53.420" Title="Product of a monotone increasing function and a monotone decreasing function" Tags="&lt;calculus&gt;&lt;functions&gt;" AnswerCount="3" CommentCount="4" />
  <row Id="14208" PostTypeId="1" AcceptedAnswerId="14223" CreationDate="2010-12-13T20:35:23.107" Score="1" ViewCount="96" Body="&lt;p&gt;I will be using relativistic terminology for pseudo-Riemannian manifolds as in the book by Barret O' Neil. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;If one can show that the chronal future and the chronal past of some set are disjoint then why is this enough for showing that the set is achronal? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I mean one can have a set such that there is precisely one time-like curve intersecting it at just two distinct points. Then clearly this set is not achronal (because this one curve exists) but chronal future and past of this set will include one each of these points and will be disjoint. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does something rule out this possibility? &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Let $p$ be a point on the boundary of some future set. Why is showing that chronal future and chronal past of $p$ are disjoint enough to prove that this future set is edgless? &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;How generic is the situation that Cauchy Horizon of a set separates that part of the causal future of a set which is not a part of its domain of dependence? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this something on which intuition can be built or is this a very special scenario?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Consider the interval $(0,1)$ on the x-axis on the $2-$dimensional Minkowski space where time is the y-axis. For this interval is the Cauchy Horizon the union of the set of points on the lines, $y=-x$ and $y=x-1$ ?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;If yes, then I guess this is an example where the Cauchy Horizon is not a part of the domain of dependence of the interval which is the set of points above the given interval and contained in between the above two straight lines. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this right? &lt;/p&gt;&#xA;" OwnerUserId="965" LastEditorUserId="965" LastEditorDisplayName="" LastEditDate="2010-12-13T21:15:40.937" LastActivityDate="2010-12-13T22:55:51.647" Title="Some questions about geometry on pseudo-Riemannian manifolds" Tags="&lt;differential-geometry&gt;" AnswerCount="1" />
  <row Id="14209" PostTypeId="1" AcceptedAnswerId="14323" CreationDate="2010-12-13T21:03:27.033" Score="5" ViewCount="115" Body="&lt;p&gt;Hey there,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm trying to get familiar with the Newton-iteration over here but I got stuck at the proof of the error estimate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $f: [a,b] \rightarrow \mathbb{R}$ be continuously differentiable twice, concave or convex and $f&amp;#39; \neq 0 \;\; \forall x \in [a,b]$. Let $\xi$ be the root of $f$. We define the Newton-iteration for $k \in \mathbb{Z}_{\geq 0}$:&#xA;$$x_{k+1} = x_k - \frac{f(x_k)}{f&amp;#39;(x_k)}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, we assume $x_1 \in [a,b]$ for $x_0 = a$ and $x_0 = b$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I already showed that the sequence $(x_k)_{k \in \mathbb{N}}$ converges to $\xi$. Now, I want to show the following error estimate:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$|\xi - x_{k+1}| \leq \frac{\max_{a \leq x \leq b} |f&amp;#39;&amp;#39;(x)|}{2 \min_{a \leq x \leq b} |f&amp;#39;(x)|} (x_{k+1}-x_k)^2$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am quite sure I will have to combine the mean value theorem and Taylor's theorem (and Lagrange's remainder), but I have no idea, how to. I don't quite know at what point I should use Taylor's theorem, also, I don't know between which two points I should apply the mean value theorem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd be very happy if somebody could give me a little hint so that I can proceed.&lt;/p&gt;&#xA;" OwnerUserId="3787" LastEditorUserId="2549" LastEditorDisplayName="" LastEditDate="2010-12-14T16:34:17.730" LastActivityDate="2010-12-14T19:52:28.307" Title="How to prove the error estimate of the Newton-iteration?" Tags="&lt;analysis&gt;&lt;numerical-methods&gt;" AnswerCount="2" />
  <row Id="14210" PostTypeId="1" AcceptedAnswerId="14214" CreationDate="2010-12-13T21:07:54.803" Score="5" ViewCount="238" Body="&lt;p&gt;Every finitely generated free group is a subgroup of $F_2$, the free group on two generators.  This is an elementary fact, as is the fact that $G$, finitely presented, is the quotient of $F(|S|)$ the free group on some set of generators $S$ for $G$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is whether $F_3$, and hence any finitely presented group, is a quotient of $F_2$.&lt;/p&gt;&#xA;" OwnerUserId="4721" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-13T21:34:52.970" LastActivityDate="2010-12-15T23:27:39.507" Title="The free group $F_3$ being a quotient of $F_2$" Tags="&lt;group-theory&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="14212" PostTypeId="1" CreationDate="2010-12-13T21:16:26.733" Score="1" ViewCount="190" Body="&lt;p&gt;I am interested in learning (and implementing) an alternative to polynomial interpolation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I am having trouble finding a good description of how these methods work, how they relate, and how they compare.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would appreciate your input on the pros/cons/conditions under which these methods or alternatives would be useful, but some good references to texts, slides, or podcasts would be sufficient.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;note:&lt;/em&gt;&#xA;I asked this &lt;a href=&quot;http://stats.stackexchange.com/q/5015/1381&quot;&gt;question&lt;/a&gt; on stats.SE a few weeks ago and received the recommendation to ask here since I have not received an answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="3733" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-14T00:49:09.053" LastActivityDate="2010-12-14T00:49:09.053" Title="What are the advantages / disadvantages of using splines, smoothed splines, and gaussian process emulators?" Tags="&lt;statistics&gt;&lt;approximation&gt;" CommentCount="3" />
  <row Id="14213" PostTypeId="1" CreationDate="2010-12-13T21:29:56.237" Score="2" ViewCount="172" Body="&lt;p&gt;I found a contradiction I couldn't resolve by my self.&#xA;It's about a &quot;Uniform White Noise&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let ${x}_{t}$ be a &quot;White Noise&quot; i.i.d. Random Process:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ \forall t \in  \mathbb{R}, \ {x}_{t} \sim U[-1, \  1] $&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we chose to go by the PSD definition of &quot;White Noise&quot; (Constant all over the Frequencies) we get:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ {R}_{xx}( \tau ) = var({x}_{t}) \delta ( \tau ) $&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yet, Clearly:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ E[{x}_{t} {x}_{t +  \tau}] \underset{ \tau = 0}{=}E[{x}_{t} {x}_{t}]= \frac{1}{3} $&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Intuitively, a Process with bounded variance and values can't be &quot;White Noise&quot;.&lt;br&gt;&#xA;Please mind this is a Continuous Random Process. We don't have such problem in the Discrete case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What am I missing here?&#xA;Either there's no such &quot;White Noise&quot; (Why?) or There's a good explanation (Could someone derive it Mathematically) how to get the Delta in The Variance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="33" LastActivityDate="2010-12-14T00:02:49.307" Title="Uniform White Noise" Tags="&lt;probability&gt;" AnswerCount="1" />
  <row Id="14214" PostTypeId="2" ParentId="14210" CreationDate="2010-12-13T21:30:50.160" Score="11" ViewCount="" Body="&lt;p&gt;If $F_3$ were a quotient of $F_2$, then $\mathbb{Z}^3$ would be, but $\mathbb{Z}^3$ cannot be generated by fewer than $3$ elements. To me it seems easier to see directly that $\mathbb{Z}^3$ needs at least $3$ generators than the corresponding statement for $F_3$, perhaps because it's easy to visualize.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rank of a group is the smallest cardinality of a generating set.  Here's a list of some facts about ranks of groups (including that the rank of $F_3$ is $3$) on &lt;a href=&quot;http://en.wikipedia.org/wiki/Rank_of_a_group#Known_facts_and_examples&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2010-12-13T21:37:59.263" LastActivityDate="2010-12-13T21:37:59.263" CommentCount="1" />
  <row Id="14215" PostTypeId="2" ParentId="14113" CreationDate="2010-12-13T21:51:02.590" Score="3" ViewCount="" Body="&lt;p&gt;I think you may find interesting to browse the webpage of &lt;a href=&quot;http://www.carma.newcastle.edu.au/~jb616/&quot; rel=&quot;nofollow&quot;&gt;Jon Borwein&lt;/a&gt;, which I would call the standard reference for your question. In particular, take a look at the latest version of his talk on &quot;&lt;a href=&quot;http://www.carma.newcastle.edu.au/~jb616/pi-2010.pdf&quot; rel=&quot;nofollow&quot;&gt;The life of pi&lt;/a&gt;&quot; (and its references!), which includes many of the fast converging algorithms and series used in practice for high precision computations of $\pi$, such as the one from this &lt;a href=&quot;http://www.google.com/hostednews/afp/article/ALeqM5i8R8sLPPRJvax-BjAtH0-lsV3kGw&quot; rel=&quot;nofollow&quot;&gt;Summer&lt;/a&gt;. &lt;/p&gt;&#xA;" OwnerUserId="462" LastActivityDate="2010-12-13T21:51:02.590" />
  <row Id="14216" PostTypeId="2" ParentId="14209" CreationDate="2010-12-13T22:01:07.430" Score="5" ViewCount="" Body="&lt;p&gt;Check out the proof &lt;a href=&quot;http://en.wikipedia.org/wiki/Newton%27s_method#Proof_of_quadratic_convergence_for_Newton.27s_iterative_method&quot; rel=&quot;nofollow&quot;&gt;on wikipedia&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you just want a hint: call $g(x) = x - \frac{f(x)}{f&amp;#39;(x)}$. Consider $g(x) - g(\xi)$ with the Taylor expansion. You will get something related to $g&amp;#39;&amp;#39;(c)$, which is your desired result, upon expanding $g&amp;#39;&amp;#39;$. &lt;/p&gt;&#xA;" OwnerUserId="2549" LastEditorUserId="2549" LastEditorDisplayName="" LastEditDate="2010-12-13T22:44:34.117" LastActivityDate="2010-12-13T22:44:34.117" CommentCount="8" />
  <row Id="14217" PostTypeId="1" AcceptedAnswerId="14225" CreationDate="2010-12-13T22:08:06.657" Score="0" ViewCount="208" Body="&lt;p&gt;I'm confused about the concept of &lt;em&gt;pivoting&lt;/em&gt; a Matrix element.  This operation seems to be a fundamental operation which is used as part of more complex operations such as inverting a matrix.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the definition of pivoting, and what it entails, seems to vary depending on the context.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the context of inverting a matrix, for example, pivoting entails changing the pivot element to &lt;code&gt;1&lt;/code&gt;, and then all other elements in the same column to &lt;code&gt;0&lt;/code&gt; (and appropriately adjusting the other elements in the same row/column.)  However, other times the definition of pivoting only involves changing other elements in the same column to &lt;code&gt;0&lt;/code&gt;, while leaving the pivot element as it is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a universal definition of &lt;em&gt;pivoting&lt;/em&gt;, or is it simply a casual term which changes depending on context?  Or, to put it another way: must the pivot element always be a &lt;code&gt;1&lt;/code&gt;?&lt;/p&gt;&#xA;" OwnerUserId="4724" LastActivityDate="2010-12-13T23:06:21.920" Title="Pivoting on a matrix element" Tags="&lt;linear-algebra&gt;&lt;matrices&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="14218" PostTypeId="1" CreationDate="2010-12-13T22:17:40.857" Score="5" ViewCount="165" Body="&lt;p&gt;Let's say, we have an array/matrix $n \times m$ and we need to find, how many ways can we fill this array with numbers from $\{ 1, \ldots , m\cdot n \}$, but:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) every number can be used only 1 time&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) every column and every row should be sorted in increasing order&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, if we have $n = m = 3$, there are 42 different ways to do this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was thinking about this problem, but I can't find out any &lt;em&gt;simple&lt;/em&gt; formula to compute this for every $n, m$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can solve this problem by writnig a program that uses backtracking method/algorithm [1] but it has huge complexity and problem is solved by computer, and I want to know, how to do this &lt;em&gt;by hand&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sombody told me, that in this problem I can use some kind of Catalan numbers [2] [3], but I'm not sure about this method.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, what do You think about this problem and its solution?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[1] Backtracking (Wikipedia): &lt;a href=&quot;http://en.wikipedia.org/wiki/Backtracking&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Backtracking&lt;/a&gt; &lt;br/&gt;&#xA;[2] Catalan numbers (Wikipedia): &lt;a href=&quot;http://en.wikipedia.org/wiki/Catalan_number&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Catalan_number&lt;/a&gt; &lt;br/&gt;&#xA;[3] Catalan numbers (OEIS): &lt;a href=&quot;http://oeis.org/A000108&quot; rel=&quot;nofollow&quot;&gt;http://oeis.org/A000108&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3856" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-12-14T01:21:18.443" LastActivityDate="2010-12-14T01:21:18.443" Title="Array of numbers, how many solutions/ways?" Tags="&lt;combinatorics&gt;&lt;matrices&gt;&lt;permutations&gt;" AnswerCount="2" />
  <row Id="14219" PostTypeId="2" ParentId="14207" CreationDate="2010-12-13T22:28:18.940" Score="5" ViewCount="" Body="&lt;p&gt;Take&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ g(x) = x 2^{\lfloor x/2\rfloor}, h(x) = \frac{1}{x} 2^{-\lfloor (x+1)/2\rfloor} $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The product oscillates between 1 and 1/2: let $f = g\cdot h$, $f(2n) = 1$, $f(2n+1) = 1/2$ for $n$ an integer. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Edit&lt;/em&gt;: If you want $g(x) = x$. Take $h(x)$ to be the following (continuous) function: let $h(e^{2n}) = e^{-2n}$, and let $h(e^{2n+1}) = e^{-2n - \frac12}$, and linearly interpolate. Then $f(e^{2n}) = 1$ and $f(e^{2n+1}) =\sqrt{e} \neq 1$ for all natural numbers $n$. So $f$ does not have a limit. &lt;/p&gt;&#xA;" OwnerUserId="1543" LastEditorUserId="1543" LastEditorDisplayName="" LastEditDate="2010-12-13T22:35:09.043" LastActivityDate="2010-12-13T22:35:09.043" CommentCount="3" />
  <row Id="14220" PostTypeId="2" ParentId="14218" CreationDate="2010-12-13T22:33:54.450" Score="6" ViewCount="" Body="&lt;p&gt;These are Young tableaux.  OEIS has the numbers.  For the square case, see &lt;a href=&quot;http://oeis.org/A039622&quot; rel=&quot;nofollow&quot;&gt;A039622&lt;/a&gt;.   For the rectangular case, see &lt;a href=&quot;http://oeis.org/A060854&quot; rel=&quot;nofollow&quot;&gt;A060854&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-12-13T22:33:54.450" CommentCount="1" />
  <row Id="14221" PostTypeId="2" ParentId="14218" CreationDate="2010-12-13T22:36:35.277" Score="5" ViewCount="" Body="&lt;p&gt;You are counting the number of &lt;a href=&quot;http://en.wikipedia.org/wiki/Young_tableau&quot; rel=&quot;nofollow&quot;&gt;standard Young tableaux&lt;/a&gt; of shape $(m, m, m, ...)$ (where $m$ occurs $n$ times).  The formula that counts these is called the &lt;a href=&quot;http://en.wikipedia.org/wiki/Young_tableau#Dimension_of_a_representation&quot; rel=&quot;nofollow&quot;&gt;hook length formula&lt;/a&gt;.  It correctly reproduces the answer&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{9!}{5 \cdot 4 \cdot 4 \cdot 3 \cdot 3 \cdot 3 \cdot 2 \cdot 2} = 42$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for the example you gave.  The general formula seems annoying to write down.  In the special case $n = 2$ you get Catalan numbers; this is a nice exercise and it is worth trying to prove this with and then without the hook length formula.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-12-13T22:36:35.277" CommentCount="6" />
  <row Id="14222" PostTypeId="2" ParentId="14207" CreationDate="2010-12-13T22:38:42.797" Score="2" ViewCount="" Body="&lt;p&gt;Define $h(x) = 3e^{-x}$ for $x &amp;lt; 1$, and $h(x) = 1/x$ for $x \geq 1$. Hence, $h$ satisfies the conditions in the question. Define $g(x) = (1/3) e^x$ for $x&amp;lt;1$, and for $x \geq 1$ define it piecewise as follows:&#xA;$g(x) = x + a_n$ for $n \leq x &amp;lt; n+1$, $n=1,2,3,\ldots$, where $a_n$ is a strictly increasing sequence with $a_1 = 0$. Hence, $g$ satisfies the conditions in the question. Now define $f(x)=g(x)h(x)$. Thus, for any $n \in \mathbb{N}$, $f(n)=(n+a_n)/n = 1 + a_n / n$. We can choose $a_n$ so that $a_n / n$ does not have a limit. For example, consider arbitrarily large $n$, and set $a_n = 2n$. Then, let $a_{n+k} = (2n+k)$, for $k = 1,\ldots,n^2$. Then, $a_n / n = 2$, but $a_{n+n^2}/(n+n^2) \approx 1$. Continue the same way to construct a sequence with $\lim \sup _{n \to \infty}(a_n/n) = 2$ but $\lim \inf _{n \to \infty}(a_n/n) = 1$. We conclude that $f(x)$ does not have a limit as $x \to \infty$.&lt;/p&gt;&#xA;" OwnerUserId="2810" LastActivityDate="2010-12-13T22:38:42.797" />
  <row Id="14223" PostTypeId="2" ParentId="14208" CreationDate="2010-12-13T22:55:51.647" Score="2" ViewCount="" Body="&lt;p&gt;For the first question, I fear that you may have not understood the definition completely. Let $p,q\in U$ be the two points. Let $\gamma$ be the future time-like curve joining $p = \gamma(0)$ and $q= \gamma(1)$. I claim that for any $s\in (0,1)$, $\gamma(s)$ is in both the future and past of $U$. Indeed, $\gamma(s)$ is in the future of $p\in U$, hence is in the chronal future of $U$. And $\gamma(s)$ is in the past of $q\in U$, hence is in the chronal past of $U$. Hence the chronological future and the chronological past of $U$ must intersect non-emptily, i.e. they are not disjoint. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Indeed, this shows that if a set $U$ is &lt;em&gt;not&lt;/em&gt; achronal, then $I^+(U)$ and $I^-(U)$ must intersect. Hence if the chronological future and past are disjoint, the set $U$ must be achronal. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For question two, I don't think your statement is right. Can you give me a reference (page number in O'Neill will do)? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For question three, isn't that just the definition? The Cauchy Horizon is generally regarded as the boundary of the domain of dependence, no? If that's not the case, can you provide the definition you are using for the Cauchy horizon?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For question four: not quite. The future Cauchy horizon of the set is the union $ \{ y = x: x \in [0,1/2] \} \cup \{ y + x = 1: x \in [1/2,1]\} $. The past Cauchy horizon is $\{ y = -x: x \in[0,1/2] \} \cup \{ x-y = 1: x\in[1/2,1]\} $. &lt;/p&gt;&#xA;" OwnerUserId="1543" LastActivityDate="2010-12-13T22:55:51.647" CommentCount="7" />
  <row Id="14224" PostTypeId="2" ParentId="14154" CreationDate="2010-12-13T22:57:53.267" Score="2" ViewCount="" Body="&lt;p&gt;If your purpose is visualization, then often a good approximate solution is useful.  This is called &lt;a href=&quot;http://en.wikipedia.org/wiki/Multidimensional_scaling&quot; rel=&quot;nofollow&quot;&gt;Multidimensional Scaling&lt;/a&gt; (MDS).&lt;/p&gt;&#xA;" OwnerUserId="1489" LastEditorUserId="1489" LastEditorDisplayName="" LastEditDate="2010-12-13T23:10:38.717" LastActivityDate="2010-12-13T23:10:38.717" CommentCount="1" />
  <row Id="14225" PostTypeId="2" ParentId="14217" CreationDate="2010-12-13T23:06:21.920" Score="2" ViewCount="" Body="&lt;p&gt;One can better understand the role that pivoting plays in Gaussian elimination by viewing it from more general perspectives. For example one can compare analogous elimination algorithms over rings (vs. fields), e.g. Hermite / Smith normal forms. Additionally one can compare the the more general choice of &quot;critical pair&quot; pivots in non-linear elimination algorithms such as &lt;a href=&quot;http://chiragwiki.wikispaces.com/file/view/publication_190.pdf&quot; rel=&quot;nofollow&quot;&gt;Grobner basis algorithms&lt;/a&gt;, or the more general &lt;a href=&quot;http://en.wikipedia.org/wiki/Knuth%E2%80%93Bendix_completion_algorithm&quot; rel=&quot;nofollow&quot;&gt;Knuth-Bendix equational completion&lt;/a&gt;, etc. Here an optimal choice of &quot;pivoting&quot; / &lt;a href=&quot;http://dx.doi.org/10.1016/S0747-7171%2888%2980018-X&quot; rel=&quot;nofollow&quot;&gt;critical pair strategy,&lt;/a&gt; can prove crucial to tractable computation (e.g.  to avoid combinatorial explosion). The Knuth-Bendix algorithm provides a fairly universal point of view that encompasses all these elimination algorithms.&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-12-13T23:06:21.920" />
  <row Id="14226" PostTypeId="2" ParentId="14206" CreationDate="2010-12-13T23:09:22.030" Score="2" ViewCount="" Body="&lt;p&gt;You have several issues to consider.  First, your output format allows very many fewer output values than the double standard.  If your exponent is 2-based, your output will be of the form $\pm m*2^e$ where m has 2048 values available and e has 32 values (maybe they range from -15 to +16).  The restricted range of e means that you will underflow/overflow quite often.  If you are within range, you will still have an error as large as one part in 4096 assuming your calculation is perfect and the only error is representing the output value.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A second issue is precision of the input and error propagation.  If your input values are represented in the same format, they will be in error by up to 1 part in 4096 for the same reason.  Then when you calculate $f(x)=c_3 \exp {\frac{-(c_1+x)^2}{c_2}}$ (where I absorbed the squaring of $c_2$ into its definition) the error will propagate.  You can assess this by evaluating $\frac{\text{d}\ln f(x)}{\text{d}x}=\frac{\text{d}\ln (c_3 \exp {\frac{-(c_1+x)^2}{c_2}})}{\text{d}x}$ to get the fractional error in $f(x)$ depending on the error in $x$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All this assumes your calculation is perfect and the only problem is the accuracy of representing the data.  If not, you also need to assess the error that is inherent in the computation.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-12-13T23:09:22.030" CommentCount="2" />
  <row Id="14227" PostTypeId="2" ParentId="14137" CreationDate="2010-12-13T23:26:55.440" Score="0" ViewCount="" Body="&lt;p&gt;The point of view offered by Jim is good because it can be generalized: if you take any surface with $n$ holes (say a connected sum of $n$ tori), it has a CW-complex structure made by a single 0-cell, $2n$ 1-cells 'wedged' to the 0-cell and a single 2-cell applied on the 1-cells. It is page 5 of Hatcher's &lt;em&gt;Algebraic Topology&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="685" LastActivityDate="2010-12-13T23:26:55.440" />
  <row Id="14228" PostTypeId="1" CreationDate="2010-12-13T23:48:53.223" Score="1" ViewCount="210" Body="&lt;p&gt;A module $M_{R}$ is called semi-artinian if every nonzero image of $M$ contains a simple submodule. Given $m\in M$ and $a_1,a_2,...$ in J(R). Why $ma_1a_2...a_{n-1}a_n=0$ for some $n\geq 1$.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;(Edit by KennyTM: The above is OP's original question. The latest, completely changed question follows:)&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;if $R$ is regular what is the relation between:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) J(R)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) a left $R$-module has a projective cover?&lt;/p&gt;&#xA;" OwnerUserId="4688" LastEditorUserId="171" LastEditorDisplayName="" LastEditDate="2010-12-16T09:23:08.807" LastActivityDate="2010-12-16T09:23:08.807" Title="Ring Theory (semi-artinian)" Tags="&lt;ring-theory&gt;" AnswerCount="1" CommentCount="7" />
  <row Id="14229" PostTypeId="1" AcceptedAnswerId="14242" CreationDate="2010-12-13T23:53:53.063" Score="3" ViewCount="264" Body="&lt;p&gt;I was rather confused about this proof I came across in Rudin Chapter 7.  The premise is: If K is a compact metric space, if $f_n \in C(K)$ for $n =  1, 2, 3, ...,$ ($C(K)$ being a set of complex-valued, continuous and bounded functions), and if $\{f_n\}$ converges uniformly on $K$, then $\{f_n\}$ is equicontinuous on $K$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proof:&#xA;Let $\epsilon &amp;gt; 0$ be given.  Since $\{f_n\}$ converges uniformly, there is an integer $N$ such that $$||f_n - f_N|| &amp;lt; \epsilon (n &amp;gt; N).$$ (See Definition 7.14).  Since continuous functions are uniformly continuous on compact sets, there is a $\delta &amp;gt; 0$ such that $$|f_i(x) - f_i(y)| &amp;lt; \epsilon$$ if $1 \leq i \leq N$ and $d(x,y) &amp;lt; \delta$.*&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $n &amp;lt; N$ and $d(x,y) &amp;lt; \delta$, it follows that $$|f_n(x) - f_n(y)| \leq |f_n(x) - f_n(y)| + |f_n(x) - f_N(y)| + |f_N(x) - f_N(y)| &amp;lt; 3\epsilon.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$QED$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;**&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I don't understand is the line ending with *.  I understand that when a family of functions are uniformly continuous, for every $\epsilon &amp;gt; 0$, there exists an integer $N$ such that $n \geq N$ implies $|f_n(x) - f_n(y)| \leq \epsilon$, but it doesn't say anything about the case when $n \leq N$.  Perhaps it has something to do with the finiteness of the number of functions?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It might be something really obvious I'm overlooking.&lt;/p&gt;&#xA;" OwnerUserId="2919" LastEditorUserId="2919" LastEditorDisplayName="" LastEditDate="2010-12-14T01:36:27.093" LastActivityDate="2010-12-14T02:06:09.250" Title="Don't Understand Proof in Rudin Principles of Mathematical Analysis Book Thm 7.24?" Tags="&lt;real-analysis&gt;&lt;analysis&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="1" />
  <row Id="14230" PostTypeId="2" ParentId="14213" CreationDate="2010-12-14T00:02:49.307" Score="1" ViewCount="" Body="&lt;p&gt;Apparently, saying that $(x_t)$ is a continuous white noise process simply refers to the fact that $(x_t)$ is a continuous time process, that is $(x_t)$ is indexed by a continuous parameter set. That is, the sample paths of $(x_t)$ are not assumed to be continuous, and in fact may be expected to be discontinuous at every fixed point (almost surely). Indeed, consider a continuous Gaussian white noise process. Then, $E[x_s x_t]=0$, $s \neq t$, implies that $x_s$ and $x_t$ are independent, and hence the sample paths must be discontinuous at every point (almost surely). The case of &quot;Uniform White Noise&quot; is essentially the same.&lt;/p&gt;&#xA;" OwnerUserId="2810" LastActivityDate="2010-12-14T00:02:49.307" CommentCount="9" />
  <row Id="14231" PostTypeId="1" CreationDate="2010-12-14T00:07:11.100" Score="1" ViewCount="183" Body="&lt;p&gt;The question asks me to determine if $4\mathbb{Z}$ and $5\mathbb{Z}$ (with standard addition) are isomorphic and if so to give the isomorphism.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My attempts:&#xA;What I am having difficulty with is showing a mapping that preserves the operation.  i.e., $\phi(a+b) = \phi(a) + \phi(b)$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I have so far is: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\phi(a+b) = \phi(5a/4) + \phi(5b/4)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\phi(a+b) = 5/4(a+b) = 5/4a + 5/4b = \phi(a) + \phi(b)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, my conclusion is $4\mathbb{Z}$ and $5\mathbb{Z}$ are isomorphic and the isomorphism is $\phi(n) = \frac{5}{4}n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can someone either confirm this or point me in the right direction?  I really appreciate everyone willingness to help one another!  Thank you! &lt;/p&gt;&#xA;" OwnerUserId="4179" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-14T03:11:26.030" LastActivityDate="2010-12-14T03:14:28.303" Title="$(4\mathbb{Z}, +)$ isomorphic to $(5\mathbb{Z}, +)$" Tags="&lt;homework&gt;&lt;abstract-algebra&gt;&lt;group-theory&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="14233" PostTypeId="2" ParentId="14183" CreationDate="2010-12-14T00:36:55.203" Score="3" ViewCount="" Body="&lt;p&gt;I don't have a full calculus 1 course, but I have a lot  on youtube at  Professor Elvis Zap . You can learn to rap the laws of differentiation from the calculus rap. &lt;/p&gt;&#xA;" OwnerUserId="722" LastActivityDate="2010-12-14T00:36:55.203" />
  <row Id="14234" PostTypeId="2" ParentId="14228" CreationDate="2010-12-14T00:49:26.227" Score="2" ViewCount="" Body="&lt;p&gt;This is proved in Askar A. Tuganbaev' s &lt;em&gt;Semidistributive modules and rings&lt;/em&gt;, page 153, which you can read on googlebooks.&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2010-12-14T00:49:26.227" CommentCount="2" />
  <row Id="14235" PostTypeId="1" AcceptedAnswerId="14247" CreationDate="2010-12-14T00:59:01.793" Score="6" ViewCount="407" Body="&lt;p&gt;I was reading over this in my notes, but it doesnt make any sense to me:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;However, suppose we ﬁx an alphabet for&#xA;  writing our programs (e.g. 8- bit&#xA;  ASCII). Since each individual program&#xA;  is ﬁnite in length, we can put all&#xA;  possible programs into a (very long)&#xA;  ordered list. For any ﬁxed character&#xA;  length k, there are only a ﬁnite set&#xA;  of possible programs. So, we can write&#xA;  down all programs by ﬁrst ﬁrst writing&#xA;  down all the 1-character programs,&#xA;  then all the 2-character programs, and&#xA;  so forth. In other words, there’s a&#xA;  bijection between the integers and the&#xA;  total set of programs. But this means&#xA;  that the number of functions is&#xA;  uncountable, whereas the number of&#xA;  programs is only countably inﬁnite. So&#xA;  there must be mathematical functions&#xA;  that we can’t compute with any&#xA;  (ﬁnite-length) program.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What exactly is this saying? Why does it hold? It seems like it shouldn't. In some cases a program of only 30-40 characters can compute billions of numbers! Maybe with current hardware it may not be possible to computer some numbers, but theoretically, this shouldn't be a problem, right? &lt;/p&gt;&#xA;" OwnerUserId="2985" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-14T03:31:16.113" LastActivityDate="2010-12-14T17:04:00.980" Title="Mathematical functions that can't be computed" Tags="&lt;counting&gt;" AnswerCount="4" CommentCount="2" FavoriteCount="1" />
  <row Id="14236" PostTypeId="1" CreationDate="2010-12-14T01:10:17.427" Score="0" ViewCount="62" Body="&lt;p&gt;Suppose I have a list of keywords, and number of times that keywords were mentioned this week and last week.  How can I rank them by &quot;percent increase&quot; if some of the keywords were &quot;new&quot; this week (and so were mentioned zero times last week)?  Is there a &quot;proper&quot; way to normalize the numbers so that I can rank them by some metric that is similar to &quot;percent increase&quot;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="4725" LastActivityDate="2010-12-14T01:10:17.427" Title="Ranking by percent increase" Tags="&lt;statistics&gt;" CommentCount="1" />
  <row Id="14237" PostTypeId="2" ParentId="14231" CreationDate="2010-12-14T01:11:18.910" Score="4" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\rm\ \ \ \mathbb Z\ \cong n\ \mathbb Z\ $ via $\rm\ x\to\ n\ x\ $ for $\rm\: n\ne 0\:.\ $ Thus $\rm\ n\ \mathbb Z\ \cong\ \mathbb Z\ \cong\: n&amp;#39;\ \mathbb Z\ $ by transitivity of $\rm\: \cong\:$&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-12-14T02:24:03.843" LastActivityDate="2010-12-14T02:24:03.843" />
  <row Id="14238" PostTypeId="1" CreationDate="2010-12-14T01:19:18.400" Score="0" ViewCount="124" Body="&lt;p&gt;Alright, so this is how I am doing it:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;float xrot = 0;&#xA;           float yrot = 0;&#xA;           float zrot = 0;&#xA;&#xA;           Quaternion q = new Quaternion().fromRotationMatrix(player.model.getRotation());&#xA;           if(q.getW() &amp;gt; 1){&#xA;            q.normalizeLocal();&#xA;           }&#xA;&#xA;           float angle = (float) (2 * Math.acos(q.getW()));&#xA;           double s = Math.sqrt(1-q.getW()*q.getW());&#xA;           if (s &amp;lt; 0.001) { // test to avoid divide by zero, s is always positive due to sqrt&#xA;            // if s close to zero then direction of axis not important&#xA;            xrot = q.getXf(); // if it is important that axis is normalised then replace with x=1; y=z=0;&#xA;            yrot = q.getYf();&#xA;            zrot = q.getZf();&#xA;           // z = q.getZ();&#xA;            } else {&#xA;             xrot = (float) (q.getXf() / s); // normalise axis&#xA;             yrot = (float) (q.getYf() / s);&#xA;             zrot = (float) (q.getZf() / s);&#xA;            }&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;But it doesn't seem to work when I try to put it into use:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;player.model.addTranslation(zrot * player.speed, 0, xrot * player.speed);&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;AddTranslation takes 3 numbers to move my model by than many spaces (x, y, z), but hen I give it the numbers above it doesn't move the model in the direction it has been rotated (on the XZ plane)    &lt;/p&gt;&#xA;" OwnerUserId="4464" LastActivityDate="2011-01-13T04:44:49.097" Title="quaternion to angle" Tags="&lt;geometry&gt;&lt;trigonometry&gt;&lt;vector-space&gt;" AnswerCount="1" CommentCount="6" />
  <row Id="14239" PostTypeId="2" ParentId="14235" CreationDate="2010-12-14T01:19:53.390" Score="13" ViewCount="" Body="&lt;p&gt;Cantor's &lt;a href=&quot;http://en.wikipedia.org/wiki/Cantor%27s_diagonal_argument&quot; rel=&quot;nofollow&quot;&gt;diagonal argument&lt;/a&gt; can be used to show that the number of functions from $\mathbb{N}$ to $\mathbb{N}$ is uncountable.  However, of those functions, the number of functions which can be computed by any computer program is countable because there are only countably many computer programs.  So there exist functions which cannot be computed by a computer program - in fact, &quot;most&quot; functions have this property.  An explicit example is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Busy_beaver&quot; rel=&quot;nofollow&quot;&gt;busy beaver function&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is not a matter of computing numbers but of computing &lt;em&gt;sequences&lt;/em&gt; of numbers, and is a fundamental limitation of computation.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-12-14T01:19:53.390" />
  <row Id="14240" PostTypeId="1" AcceptedAnswerId="14243" CreationDate="2010-12-14T01:26:02.057" Score="1" ViewCount="112" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Justin needs to pick 17 toy animals to&#xA;  give to children at a party. The&#xA;  animals come in 5 kinds: dogs,&#xA;  dinosaurs, cows, lizards, and ﬁsh. How&#xA;  many diﬀerent ways can he choose his&#xA;  set of toys&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Ok the answer for this is 21 choose 4. However, I dont understand how they arrived at this answer. If there are 17 slots, and for each slot, you can choose between 5 animals, then you should have $5^{17}$. Where has my reasoning gone wrong?&lt;/p&gt;&#xA;" OwnerUserId="2985" LastActivityDate="2010-12-14T16:12:06.500" Title="Simple combinations question" Tags="&lt;combinatorics&gt;" AnswerCount="2" CommentCount="5" />
  <row Id="14241" PostTypeId="2" ParentId="14235" CreationDate="2010-12-14T01:37:32.357" Score="8" ViewCount="" Body="&lt;p&gt;Each string (that parses) represents a program that computes a function on $\rm\:\mathbb N$. The argument shows that the number of functions representable by such a language is countable because the number of program strings in the language is countable (note that some programs may compute the same function). But there are uncountably many functions on $\rm\:\mathbb N\:$ by  diagonalization (du-Bois-Reymond, Cantor). Thus there are functions not computable by any program in the language.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might find it helpful to consider a more explicit language. For example, consider a language that represents polynomials with natural coefficients. An analogous argument shows that such polynomials are countable, so we can enumerate them, i.e. we can index the polynomials by naturals: $\rm\ f_i(x),\ \:i\in \mathbb N\:.\ $ By diagonalization we may construct a non-polynomial function on $\rm\:\mathbb N\:,\:$ e.g. $\rm\;\ g(n)\ :=\ f_n(n)+1\:.\ $ Notice that if $\rm\ f_i = g\ $ then $\rm\ f_i(i) = g(i) = f_i(i) + 1\:,\ $ a contradiction. Therefore  $\rm\:g\:$ is not equal to any polynomial function $\rm\:f_i\:.$&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-12-14T02:18:41.650" LastActivityDate="2010-12-14T02:18:41.650" />
  <row Id="14242" PostTypeId="2" ParentId="14229" CreationDate="2010-12-14T01:42:24.153" Score="6" ViewCount="" Body="&lt;p&gt;Now that you have updated with the filled in lines, it seems the confusion is mixing up the conditions for continuity and convergence. Uniform continuity has nothing to do with taking $n≥N$; that would be convergence. And the line ending with an asterisk has nothing to do with the convergence. What you can do is take $N$ different $\delta$s corresponding to the uniform continuity of each of the first $N$ functions, then take their minimum. As you said, it is possible because there are only finitely many.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice that for convergence, you are talking about functions getting closer together &lt;em&gt;at each point&lt;/em&gt; (uniformly in this case) as $n$ gets larger, whereas for continuity you are talking about &lt;em&gt;each function&lt;/em&gt; taking close values at nearby points.  Equicontinuity says that the you can choose your definition of &quot;nearby&quot; the same for all functions in the family simultaneously.  The step you are asking about is basically saying that a finite family of uniformly continuous functions is equicontinuous.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2010-12-14T02:06:09.250" LastActivityDate="2010-12-14T02:06:09.250" />
  <row Id="14243" PostTypeId="2" ParentId="14240" CreationDate="2010-12-14T01:43:37.693" Score="3" ViewCount="" Body="&lt;p&gt;As Jonas Meyer indicated, the way you counted will include separately different orderings of the same numbers of each animal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instead, you can have 21 slots. Four slots are to be filled with &quot;dividers&quot;. You put dogs to the slots (if any) to the left of divider 1, dinosaurs in slots (if any) between divider 1 and divider 2, and so on until you place fish to the right of the 4th divider. There are ${\displaystyle {21 \choose 4}}$ ways to place the four dividers, and each placement of the four dividers corresponds to exactly one way of choosing the number of each animal brought to the party.&lt;/p&gt;&#xA;" OwnerUserId="3035" LastActivityDate="2010-12-14T01:43:37.693" />
  <row Id="14244" PostTypeId="2" ParentId="14240" CreationDate="2010-12-14T02:06:12.887" Score="1" ViewCount="" Body="&lt;p&gt;Following the advice from Jonas, there are $A_{r,n}=\displaystyle \binom{n+r-1}{r}$ ways to obtain indistinguishable distributions where each distribution is described in terms of n-tuples $(r_{1},...,r_{n})$  satisfying $$r_{1}+r_{2}+...+r_{n}=r$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your particular problem, you want to enumerate the number of solutions for the equation $$r_{1}+r_{2}+...+r_{5}=17$$ where $r_{i}$ represents the number of elements contained in a box/cell labeled as $i$ where $i=1=\text{dogs}$, $i=2=\text{dinosaurs}$,..., $i=5=\text{fishes}$. For example, one such a way is to pick $(17,0,0,0,0)$ (say, 17 dogs and nothing else), other choice would be $(13,1,1,1,1)$ (13 dogs, 1 dinosaur, 1 cow, 1 lizard and 1 ﬁsh), etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, you have $\displaystyle \binom{5+17-1}{17}$ which is equivalent to $\displaystyle \binom{21}{4}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reference: Feller. &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0471257087&quot; rel=&quot;nofollow&quot;&gt;An Introduction to Probability Theory and Its Applications&lt;/a&gt;. Chapter II.5&lt;/p&gt;&#xA;" OwnerUserId="2569" LastEditorUserId="2569" LastEditorDisplayName="" LastEditDate="2010-12-14T16:12:06.500" LastActivityDate="2010-12-14T16:12:06.500" CommentCount="2" />
  <row Id="14245" PostTypeId="1" AcceptedAnswerId="14251" CreationDate="2010-12-14T02:24:43.343" Score="3" ViewCount="340" Body="&lt;p&gt;I tried to prove this, but was unsuccessful for a long time.. &#xA;Any ideas?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Prove that $(\exp(x))^y=\exp(xy)$ using the identities, &#xA;$$\exp(x)=\sum_{n\geq0} \frac{x^n}{n!}, \quad (1+x)^\alpha=\sum_{n\geq0} {\alpha \choose n} x^n$$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="4728" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-14T03:08:18.887" LastActivityDate="2010-12-14T05:10:48.647" Title="Formal Proof of Exponential rule" Tags="&lt;sequences-and-series&gt;&lt;power-series&gt;" AnswerCount="2" CommentCount="8" />
  <row Id="14247" PostTypeId="2" ParentId="14235" CreationDate="2010-12-14T02:32:16.373" Score="3" ViewCount="" Body="&lt;p&gt;Here is one analogy which may prove helpful: there are rational numbers, which can be represented as the ratio of two integers: 1/2, 3, 5/6 etc. This can be seen as all the numbers which can be printed out by &quot;programs&quot; which only have access to the &quot;division&quot; operator. Clearly some numbers cannot be printed out by a program with only this operation - $\pi$, for example cannot.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now suppose we add in the full power of arithmetic: you can specify a number like &quot;the root of $x^2-2x+3$&quot;. These are the algebraic numbers. We can print out more numbers this way, but not all of them. $\pi$ again cannot be printed out by this type of program, as it is transcendental.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a last hurrah, we allow any operation which can print out an arbitrary number of digits in a finite amount of time. We can now find values like $\pi$ and $e$, since there are formulae for these. But can we print &lt;em&gt;all&lt;/em&gt; values?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer is no. As you have indicated, there are &quot;too many&quot; numbers and too few programs. It is not merely that we have not discovered (or created) some operation - even if we double the number of symbols in our language that only increases the amount of numbers we can write by a finite amount, and there is an infinite amount of space to make up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rather than talking about programs, another way to phrase it is: &quot;for every real number, is there some &lt;em&gt;finitary&lt;/em&gt; description of it?&quot; E.g. $e=\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^n$ - even though $e$ has an infinite number of digits, we can describe it using only a finite number of symbols. The proof you are discussing shows that there must be some numbers which admit no finitary description.&lt;/p&gt;&#xA;" OwnerUserId="2549" LastActivityDate="2010-12-14T02:32:16.373" CommentCount="4" />
  <row Id="14249" PostTypeId="2" ParentId="14231" CreationDate="2010-12-14T03:14:28.303" Score="4" ViewCount="" Body="&lt;p&gt;Bill's argument is of course the simplest, but to discuss your effort...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Two comments: your argument only shows that $\phi$ is a homomorphism; you haven't shown it is an isomorphism (one-to-one and onto). This is a major thing: you cannot conclude it is an isomorphism just form knowing it is a homomorphism. You &lt;em&gt;need&lt;/em&gt; to show it is invertible, or show it is one-to-one and show it is onto.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second comment is more of a nit-pick: you should not say that $\phi$ is &lt;strong&gt;the&lt;/strong&gt; isomorphism; there may be more than one isomorphism (in fact, there is; you can also take $\psi(4n) = -5n$). You should say (once you establish it, of course), that $\phi$ is &lt;em&gt;an&lt;/em&gt; isomorphism.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-12-14T03:14:28.303" CommentCount="1" />
  <row Id="14250" PostTypeId="1" CreationDate="2010-12-14T03:22:08.340" Score="5" ViewCount="120" Body="&lt;p&gt;I've been studying the finite-dimensional representations of the lie algebra sl(2,C). I've read that these representations are related to the harmonic oscillator and the associated raising and lowering operators, but I'm not really sure how they are related. I've read you can generate a new energy state from an old one, but I'm not really sure how it works, and the energy states do not differ by 2 like the eigenspaces do for sl(2,C). I'm having trouble finding references about this, especially ones that are accessible. If anyone knows where I can find more info, or has some knowledge of the material, I would greatly appreciate their help.&lt;/p&gt;&#xA;" OwnerUserId="4238" LastActivityDate="2010-12-14T04:30:52.000" Title="sl(2,C) and the harmonic oscillator" Tags="&lt;reference-request&gt;&lt;physics&gt;&lt;representation-theory&gt;&lt;lie-algebra&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="2" />
  <row Id="14251" PostTypeId="2" ParentId="14245" CreationDate="2010-12-14T03:44:53.303" Score="9" ViewCount="" Body="&lt;p&gt;I'll assume you want an equality of formal power series&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{n \ge 0} \frac{x^n y^n}{n!} = \sum_{n \ge 0} {y \choose n} (\exp(x) - 1)^n.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The proof of this is easier to explain combinatorially than via series manipulation, if you allow me to first assume that $y$ is a non-negative integer.  In that case I claim that the coefficient of $\frac{x^n}{n!}$ on the RHS is the number of functions from $[n]$ to $[y]$ (where $[k] = { 1, 2, ... k }$), which is clearly $y^n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To see this, note that $\exp(x) - 1$ is the exponential generating function for non-empty sets, hence $(\exp(x) - 1)^k$ is the exponential generating function for partitions of a set into $k$ non-empty blocks, and ${y \choose k}$ is the number of ways to choose $k$ elements out of $y$ elements.  But a function from $[n]$ to $[y]$ is nothing more than a partition of $[n]$ into $k$ non-empty blocks (the preimages of each element of the range), together with a choice of $k$ elements of $[y]$ (the image of each of the blocks).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To see how this implies the identity for arbitrary $y$, note that the coefficient of $\frac{x^n}{n!}$ on the RHS (regarded as a formal power series in $x$ and $y$) must be a polynomial of degree at most $n$, and this polynomial agrees with $y^n$ for all non-negative integers, so it must be identically equal to $y^n$.  (What I've proven essentially amounts to the &lt;a href=&quot;http://en.wikipedia.org/wiki/Stirling_number&quot; rel=&quot;nofollow&quot;&gt;inversion formulas&lt;/a&gt; relating the Stirling numbers of the first and second kinds.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want a proof without this extra simplification then you need to do some messy inclusion-exclusion or some messy series manipulation and I don't really see the point either way.&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-12-14T03:50:03.880" LastActivityDate="2010-12-14T03:50:03.880" CommentCount="1" />
  <row Id="14252" PostTypeId="1" AcceptedAnswerId="14256" CreationDate="2010-12-14T03:56:50.647" Score="3" ViewCount="88" Body="&lt;p&gt;A two dimensional space (eg. $\mathbb{R}^2$ ) could be a flat or it could be the surface of sphere (or of any shape) in a 3-dimensional space. How do we distinguish between such spaces without invoking higher dimensions.(here I am using the term 'space' in a general term and not strictly in a manner used in mathematics notations). What is the subject of math which deals with such properties ? Also please give the mathematical term closer to the term 'space'.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS: please feel free to edit this post if it could make it more meaningful and clear.&lt;/p&gt;&#xA;" OwnerUserId="2987" LastActivityDate="2010-12-14T04:17:03.600" Title="property of a space." Tags="&lt;soft-question&gt;" AnswerCount="1" CommentCount="6" />
  <row Id="14253" PostTypeId="1" CreationDate="2010-12-14T04:04:54.497" Score="2" ViewCount="284" Body="&lt;p&gt;Let $X$ be a subspace of $\ell_1^4$ (i.e. $\mathbb{R}^4$ equipped with the $\ell^1$ norm). Can one always extend a linear operator $l:X\rightarrow \ell_1^4$&#xA;to $L:\ell_1^4\rightarrow \ell_1^4$ such that $L$ has the same operator norm as $l$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="4730" LastEditorUserId="822" LastEditorDisplayName="" LastEditDate="2011-05-13T19:12:36.017" LastActivityDate="2011-05-13T19:12:36.017" ClosedDate="2011-05-14T02:57:12.240" Title="Can a linear operator from a normed space to itself be extended with operator norm preserved?" Tags="&lt;linear-algebra&gt;&lt;real-analysis&gt;&lt;norm&gt;" AnswerCount="2" CommentCount="9" />
  <row Id="14255" PostTypeId="2" ParentId="14238" CreationDate="2010-12-14T04:11:28.753" Score="1" ViewCount="" Body="&lt;p&gt;You first need to determine which local (i.e. body-fixed) axis points out of the front of the front of your model.  It is likely the positive X or positive Y axis (depending on your convention).  You then take the first column if it is the X or the second column if it is the Y.  These will give you a unit vector pointing in the direction that you want to travel.  You can then update your position in time with &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;P_{k+1} = P_k + U*speed*dt&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;where P is your position and U is the unit vector.&lt;/p&gt;&#xA;" OwnerUserId="4726" LastActivityDate="2010-12-14T04:11:28.753" />
  <row Id="14256" PostTypeId="2" ParentId="14252" CreationDate="2010-12-14T04:17:03.600" Score="2" ViewCount="" Body="&lt;p&gt;I recommend reading up on differential geometry.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think what you are asking is how to determine if a space is curved without access to the embedding space.  That depends on curvature of the space.  There are some spaces that are equivalent intrinsically but different extrinsically (i.e. a cylinder or cone to a flat plane).  In these spaces you cannot tell the difference from within the space. &lt;/p&gt;&#xA;" OwnerUserId="4726" LastActivityDate="2010-12-14T04:17:03.600" />
  <row Id="14257" PostTypeId="2" ParentId="14250" CreationDate="2010-12-14T04:30:52.000" Score="4" ViewCount="" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;The Quantum Mechanical Harmonic Oscillator&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you look at the &lt;a href=&quot;http://en.wikipedia.org/wiki/Quantum_harmonic_oscillator&quot; rel=&quot;nofollow&quot;&gt;Wikipedia page&lt;/a&gt; on the quantum mechanical harmonic oscillator, you will see that it is not the Lie algebra $\mathfrak{sl}_2$ that enters the picture, but rather an algebra generated by&#xA;two operators $a$ and $a^{\dagger}$ satisfying $[a,a^{\dagger}] = 1$ (see also Eric's comment above).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a familiar algebra in disguise, the so called Weyl algebra of differential operators with polynomial coefficients $\mathbb C[x,\partial_x]$.  (Note that $[\partial_x,x] = 1$, by the Leibniz rule.)  The product $x\partial x$ is the so-called Euler operator (it acts on $x^n$ as multiplication by $n$); you can see that in the harmonic oscillator interpretation, it acts as the Hamiltonian (up to a shift of $1/2$, to add non-zero groundstate energy, and a rescaling by $\hbar \omega$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This algebra has a natural representation, namely on the space of polynomials $\mathbb C[x]$, and we see that this is isomorphic to the representation on the span of the eigenstates that occurs in the harmonic oscillator picture.  (The polynomial $x^n$ corresponds to the eigenstate with energy $\hbar \omega(n + 1/2).$)&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The Lie algebra $\mathfrak sl_2$ in quantum mechanics&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If my memory serves, the first time $\mathfrak sl_2$ entered the picture (in my undergrad QM class, at least) is in the analysis of the hydrogen atom.  The point is that&#xA;the Schrodinger equation in this case has a rotational symmetry, and differentiating this action, one gets an action of the Lie algebra of $SO(3)$ on the space of states.  Complexifying, this gives an action of $\mathfrak sl_2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you separate variables in the equation (using spherical coordinates, which are natural in view of the rotational symmetry), the radial part and the spherical part separate, the radial part is easily dealt with, and one is left&#xA;studying a certain differential equation on the space of functions of the&#xA;sphere $S^2$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mathematically, one is looking at $L^2(S^2)$, with its natural action of $\mathfrak sl_2$ (in the guise of the complexified Lie algebra of $SO(3)$).&#xA;The Shrodinger equation involves the Laplacian, which is also the Casimir in the&#xA;enveloping algebra of $\mathfrak sl_2$, and so if you fix the energy (eigenvalue of the Laplacian), one is trying to understand the corresponding eigenspace of the Casimir on $L^2(S^2)$.  This is a special case of the Peter--Weyl theorem,&#xA;which (in this case) also goes under the name of the the theory of spherical harmonics.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The idea, roughly, is:  $S^2 = SO(3)/SO(2)$.  Now by Peter--Weyl,&#xA;$L^2(SO(3))$ is the Hilbert space direct sum of $V\otimes V^*$, where&#xA;$V$ runs over the irreps. of $SO(3)$.&#xA;Thus $L^2(S^2)$ is the Hilbert space direct sum of $V\otimes (V^*)^{SO(2)}$.&#xA;The space of $SO(2)$ invariants in each irrep. is one-dimensional, so in&#xA;fact $L^2(S^2)$ is the Hilbert space direct sum of $V$, as $V$ runs over&#xA;the irreps. of $SO(3)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $SO(3) = SU(2)/\{\pm 1\}$, these correspond on the Lie algebra level&#xA;to the &lt;em&gt;odd-dimensional&lt;/em&gt; irreps. of $\mathfrak sl_2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each $V$ is an eigenspace for the Casimir (if $V$ has dimension $n$ then&#xA;the Casimiar eigenvalue is some quadratic expression in $n$ that you can&#xA;easily figure out, or look up), and distinct $V$ give distinct eigenvalues.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, going back to the hydrogen atom, if you fix the energy of the electron,&#xA;then the collection of states has dimension equal to the dimension of the&#xA;corresponding $V$. (The fact that this is typically greater than one-dimensional is referred to in physics as &lt;em&gt;degeneracy&lt;/em&gt;; the energy does not uniquely determine the state.) This has a natural basis corresponding to the $SO(2)$&#xD;&#xA;eigenvectors.  Physically, the derivative of this $SO(2)$ (the traditional&#xA;element $H$ in the Lie algebra $\mathfrak sl_2$, perhaps up some scaling,&#xA;and often denoted $L_z$ in physics literature) is the operator corresponding to angular momentum around the axis in $S^2$ which is fixed by $SO(2)$ (often taken to be the $z$-axis, by convention, in the physics literature), and so the different eigenvectors for $SO(2)$ in the given $V$ are states with the same energy, but with different angular momenta around the $z$-axis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Physically, one can apply a magnetic field along this axis which will then affect these different $L_z$-eigenstates differently (because its effect on&#xA;the electron depends on the angular momentum of the electron around the field&#xA;lines), and so one can &lt;em&gt;split the degeneracy&lt;/em&gt;.  This is the Zeeman effect.  One sees the different energy lines in the spectrum of hydrogen split into a number of lines ($n$ lines when the energy is proportional to $n^2$).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I haven't looked, but this must be discussed in many, many places online, and is also in any standard intro. to quantum text.  (But probably with less appearance of the symbols $L^2(S^2)$ and $\mathfrak sl_2$ then a mathematician would use!)&lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2010-12-14T04:30:52.000" CommentCount="1" />
  <row Id="14258" PostTypeId="2" ParentId="14245" CreationDate="2010-12-14T05:00:42.380" Score="4" ViewCount="" Body="&lt;p&gt;One can prove the sought power series equality via the uniqueness theorem for first-order ODE's. It's easy to show that both power series satisfy $\rm\ f&amp;#39;(x) = y\ f(x),\ \ f(0) = 1\:.\ $ Put $\rm\ f(x) = \sum_{n \ge 0} \frac{x^n y^n}{n!},\:$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\displaystyle\ \ g(x)\ =\ \sum_{n \ge 0}\ \ {y \choose n}\ (\exp(x) - 1)^n\:.\ \ $ Clearly $\rm\:f\:$ satisfies the ODE. $\ $ For $\rm\:g\:$ we have  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\displaystyle\ g&amp;#39;(x)\ =\ \sum_{n \ge 1}\ n\: {y \choose n}\ exp(x)\ (\exp(x) - 1)^{n-1}$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\displaystyle\quad\quad\quad\quad\quad =\ y\ \exp(x)\ \sum_{n \ge 1}\ {y-1\choose n-1}\ (\exp(x)-1)^{n-1}$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\displaystyle\quad\quad\quad\quad\quad =\ y\ \exp(x)\ \sum_{n \ge 0}\ {y-1\choose n}\   (\exp(x)-1)^n $  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\displaystyle\quad\quad\quad\quad\quad =\ y\ \exp(x)\ {\exp(x)}^{y-1}$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\displaystyle\quad\quad\quad\quad\quad =\ y\ g(x)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;NOTE&lt;/strong&gt; $\ $ The above is simply $\rm\ (exp(x)^y)&amp;#39; =\ y\ exp(x)\ exp(x)^{y-1}\ $ expressed in power series form. Note that the coefficient identity at the core is the same binomial identity as a &lt;a href=&quot;http://math.stackexchange.com/questions/13922/how-to-sum-up-this-series/13942#13942&quot;&gt;couple days ago&lt;/a&gt;, viz.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\displaystyle\quad\quad\quad\quad\quad\  {y \choose n}\ =\ \frac{y}{n}\ {y-1\choose n-1}$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-12-14T05:10:48.647" LastActivityDate="2010-12-14T05:10:48.647" CommentCount="3" />
  <row Id="14259" PostTypeId="1" AcceptedAnswerId="14260" CreationDate="2010-12-14T05:00:53.180" Score="5" ViewCount="164" Body="&lt;p&gt;There are Artin $L$-series and Dirichlet $L$-series, and zeta functions for varieties and for number fields; there are a slew of objects named after Hecke... There are also various kinds of characters in these areas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've always had a hard time keeping all of them straight. I would be very grateful if someone could put them in some sort of perspective that would make it clear what the role of each of those objects is with respect to the rest.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you need to use the language of Langlands, that would be acceptable. (perhaps it is best to see it in those terms?)&lt;/p&gt;&#xA;" OwnerUserId="4732" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-14T17:33:02.280" LastActivityDate="2010-12-14T17:33:02.280" Title="Getting my number theoretic series straight" Tags="&lt;number-theory&gt;&lt;zeta-function&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="2" />
  <row Id="14260" PostTypeId="2" ParentId="14259" CreationDate="2010-12-14T05:33:35.840" Score="12" ViewCount="" Body="&lt;p&gt;First of all, one shouldn't distinguish too much between the terms $L$-series and $\zeta$-function; it is more or less a matter of history which term you use in a given context.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The true distinction is between objects on the &quot;motivic&quot; (Diophantine equation or algebraic number theory) side, and objects on the &quot;automorphic&quot; side.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, Artin $L$-functions, and $\zeta$-functions and $L$-functions attached to varieties, are on the motivic side.   They don't have obvious analytic continuations or functional equations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dirichlet $L$-functions (i.e. the ones built out of characters of the multiplicative group mod $N$ for some $N$), Hecke $L$-functions (built out &#xA;of ideal class group characters), $L$-functions of modular forms, or more generally automorphic forms, are automorphic $L$-functions.  Except for the&#xA;last (general automorphic $L$-functions) these &lt;em&gt;can&lt;/em&gt; be proved to have analytic continuation with functional equations.  The natural limit of these analytic continuation/functional equation arguments (which begin with Riemann, with Tate's thesis being another major point en route) is the fact that &lt;em&gt;standard $L$-functions&lt;/em&gt; for automorphic forms on $GL_n$ have analytic continuation/functional equations.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Langlands's &lt;em&gt;functoriality conjecture&lt;/em&gt; in particular conjectures that any (&lt;em&gt;a priori&lt;/em&gt; more general) automorphic $L$-function is in fact a standard $L$-function.  Many special cases are known (and Ngo got the Fields medal for proving the fundamental lemma, which is one tool in proving certain cases, namely those arising from &lt;em&gt;endoscopy&lt;/em&gt;), but it is wide open in general.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;em&gt;reciprocity conjecture&lt;/em&gt; then states that any of the motivic $L$-functions are also actually standard $L$-functions.  (E.g. for $L$-functions of elliptic curves over $\mathbb Q$, this becomes the statement that any such elliptic curve is associated to a weight 2 modular form, which was proved by Wiles et. al.)  This is also wide open in general.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The special case of Artin $L$-functions, and some other special cases, are actually covered by both conjectures.  (This is because, if you broaden your perspective sufficiently, as Langlands did, by allowing arbitrary reductive algebraic groups into the picture, then Artin $L$-functions can be thought of as&#xA;a particular kind of automorphic $L$-functions.  This is not so easy to see when you are just entering the subject, but one can think of just the Riemann $\zeta$-function: this is certainly motivic, being the $\zeta$-function of the&#xA;variety Spec $\mathbb Q$, but is also automorphic, being the Dirichlet $L$-function for the trivial character.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But in general, the conjectures have a slightly different nature: functoriality can be thought of (if you want) purely in terms of harmonic analysis on adelic groups, whereas by its very nature reciprocity involves arithmetic geometry.  In practice, at least at the moment, the two seem to be fairly intertwined. Functoriality certainly is a tool that can be very helpful in proving reciprocity; also, of those cases of the Artin conjecture which are known, some are proved using the functoriality view-point (e.g. Langlands--Tunnell) and some using the reciprocity view-point (e.g. Buzzard--Dickinson--Shepherd-Barron--Taylor).&lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2010-12-14T05:33:35.840" />
  <row Id="14261" PostTypeId="2" ParentId="14235" CreationDate="2010-12-14T05:42:16.173" Score="0" ViewCount="" Body="&lt;p&gt;You are looking for the &lt;a href=&quot;http://en.wikipedia.org/wiki/Halting_problem&quot; rel=&quot;nofollow&quot;&gt;Halting Problem&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/Higher-order_logic&quot; rel=&quot;nofollow&quot;&gt;Higher-order logic&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can't really count the functions via their representations as programs, because you can neither compare all program behaviours nor find out all those programs that won't halt at all. You could define a function which is not only undefined on some or many of uncountably many points, but also can't be proven to be undefined at those points. If you could, then you could count the functions of First-Order Logic, but still not those of Higher Order Logic (Second Order Logic?) and above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Aditionally, there are functions, that cannot be expressed as programs at all, like the &quot;busy beaver function&quot;: It describes for every n the largest finite number of steps after which any ever halting program of length n halts. The program of this function itself would need to have an infinite description length to be correct. The busy beaver function could solve the Halting Problem via timeout, iff it would be expressible as a (finite) program.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you can count, that are the source codes of all programs. In a similar way can all functions be represented as a finite amount of chalk – in infinite different ways. But even with this, they can only be described but not always in all their properties proven, because there exist theorems that are true even though they can't be proven, and because the possibility of a prove does not always imply the existance of a proven example. To compare it with the busy beaver function: There exist functions that would need an infinite amount of chalk to be proven (in other words: they can't be proven) but not to be described; to map those functions to programs would either require to prove them first (→ impossible or program won't halt) or imply to prove them (→ infinite program description). As an example: Think about a program H that receives the description of another program P as input and needs to calculate an existing but unprovable property of that program P (like halting or not) and therefore would not halt itself; the function could simply be defined on that existing property, which makes it to something else than this non-halting program H.&lt;/p&gt;&#xA;" OwnerUserId="3226" LastEditorUserId="3226" LastEditorDisplayName="" LastEditDate="2010-12-14T17:04:00.980" LastActivityDate="2010-12-14T17:04:00.980" CommentCount="4" />
  <row Id="14262" PostTypeId="2" ParentId="14207" CreationDate="2010-12-14T05:48:53.420" Score="0" ViewCount="" Body="&lt;p&gt;You are greatly helped by the fact that $f$ and $g$ are not required to be analytic so you can easily analize $f$ and $g$ in some neighbourhood of infinity and don't really care about their behaviour outside that neighbourhood :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, you need a function $f(x)$ such that $\lim_{x\to\infty} f(x)g(x) &amp;lt; \infty$, which is to say that $f(x) = \frac{\alpha(x)}{g(x)}$, where $\alpha$ is bounded ($\alpha = h/g^2$). It is not very hard to prove that such $f$ exists for any continuous $g$. Just construct $f$ as $\frac{\alpha(x)}{g(x)}$ in the neighbourhood of infinity where $g(x) \neq 0$ and then 'glue' this $f$ to some other function on the rest of the domain (it is possible to do this in $C^\infty$ way).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This covers all monotonic functions in $C^0, C^1, \ldots, C^\infty$. The most interesting stuff is about the rest of them :)&lt;/p&gt;&#xA;" OwnerUserId="3793" LastActivityDate="2010-12-14T05:48:53.420" />
  <row Id="14263" PostTypeId="1" AcceptedAnswerId="14295" CreationDate="2010-12-14T06:28:51.217" Score="0" ViewCount="146" Body="&lt;p&gt;Are all groups of order $156$ solvable?&lt;/p&gt;&#xA;" OwnerUserId="4728" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-12-14T06:39:03.667" LastActivityDate="2010-12-14T14:35:12.017" Title="Groups of order 156" Tags="&lt;homework&gt;&lt;group-theory&gt;" AnswerCount="1" CommentCount="16" />
  <row Id="14264" PostTypeId="1" AcceptedAnswerId="14460" CreationDate="2010-12-14T06:44:38.780" Score="5" ViewCount="104" Body="&lt;p&gt;I have an epimorphism $f:B_4\longrightarrow S_4$, from the braid group on 4 strands &lt;em&gt;onto&lt;/em&gt; the symmetric group on 4 elements.  Is it possible the kernel is not isomorphic to $P_4$, the pure braid group on 4 strands? &lt;/p&gt;&#xA;" OwnerUserId="4721" LastActivityDate="2010-12-15T23:16:17.980" Title="Braid Group, B_4-&gt;&gt; S_4 onto, do I know kernel is P_4, pure braid group?" Tags="&lt;group-theory&gt;" AnswerCount="1" CommentCount="8" />
  <row Id="14265" PostTypeId="1" CreationDate="2010-12-14T07:39:53.203" Score="1" ViewCount="130" Body="&lt;p&gt;I want to construct a nowhere dense set of positive measure by using the Smith-Volterra-Cantor method but subtracting closed sets intead of open sets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose we start with the open interval (0,1).  Let $\alpha=1/2$.  Now remove $U_1=[1/2-\alpha/4,1/2+\alpha/4]$.  The measure $\mu(U_1)=\alpha/2$.  For each of the two remaining open sets, subtract a closed from the center of width $\alpha/8$, thus a total measure of $\alpha/4$.  Repeating this process results in removing a total measure of $\alpha/2+\alpha/4+\dots=\alpha$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does this give me a set of positive measure just as with subtracting closed intervals?  I suspect not, since I have never heard of this, but what's the exact function of removing a closed set instead of an empty set?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or, if I can do this, can I now consider this to be an open set that I can choose an admissible partition of unity for?&lt;/p&gt;&#xA;" OwnerUserId="326" LastActivityDate="2010-12-14T14:01:15.163" Title="Constructing a Cantor-like set by subtracting closed intervals." Tags="&lt;calculus&gt;&lt;analysis&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="14266" PostTypeId="2" ParentId="14265" CreationDate="2010-12-14T07:52:43.717" Score="6" ViewCount="" Body="&lt;p&gt;&quot;Does this give me a set of positive measure just as with subtracting closed intervals?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yes, as follows from the basic property of &lt;a href=&quot;http://en.wikipedia.org/wiki/Measure_%28mathematics%29#Measures_of_infinite_intersections_of_measurable_sets&quot; rel=&quot;nofollow&quot;&gt;continuity of measure&lt;/a&gt;: if $E_1 \supset E_2 \supset \ldots \supset E_n \supset \ldots$ is a nested sequence of subsets with $\mu(E_1) &amp;lt; \infty$, then if $E = \bigcap_{n=1}^{\infty} E_n$, we have $\mu(E) = \lim_{n \rightarrow \infty} \mu(E_n)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;[C]an I now consider this to be an open set...?&quot; No.  Here the variation you've made on the usual construction affects the answer: any intersection of closed sets is necessarily closed, whereas (even) a countably infinite intersection of open sets need not be open.  I think it would be good for you to figure out for yourself exactly why the intersection $E$ is not open here.  Think for instance about the fact that if you do the usual construction by removing open intervals you get a set which is nowhere dense and how your modified set $E$ compares to that closed, nowhere dense set.  &lt;/p&gt;&#xA;" OwnerUserId="299" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2010-12-14T08:01:47.763" LastActivityDate="2010-12-14T08:01:47.763" />
  <row Id="14267" PostTypeId="2" ParentId="13912" CreationDate="2010-12-14T07:59:37.877" Score="0" ViewCount="" Body="&lt;p&gt;This may be a bit fundamental, but i found that Exploring Abtract Alegebra Package for Mathematica had some nice visualizations that helped me develop an intuitive understanding.&lt;/p&gt;&#xA;" OwnerUserId="4735" LastActivityDate="2010-12-14T07:59:37.877" />
  <row Id="14268" PostTypeId="1" CreationDate="2010-12-14T08:47:09.250" Score="4" ViewCount="345" Body="&lt;p&gt;How to check if two multiplications are equal to each other or greater or lesser without actually multiplying them?&lt;br&gt;&#xA;For example, compare (254)(847) and (383)(536)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT:&lt;br&gt;&#xA;While trying to find a rule i got one&lt;br&gt;&#xA;(5)(11) &amp;lt; (6)(10)&lt;br&gt;&#xA;or&lt;br&gt;&#xA;(x)(y) &amp;lt; (x+1)(y-1) when y &gt; x &gt; 0&lt;br&gt;&#xA;and another rule is  that if adding and subtracting 1 equates them the difference is one&lt;br&gt;&#xA;(3)(5) + 1 = (4)(4)&lt;br&gt;&#xA;(x)(y) + 1 = (x+1)(y-1) when y + 2 = x , y &gt; x &gt;= 0&lt;/p&gt;&#xA;" OwnerUserId="1515" LastEditorUserId="1515" LastEditorDisplayName="" LastEditDate="2010-12-15T06:36:07.780" LastActivityDate="2010-12-31T00:52:33.823" Title="How to Compare two multiplications without multiplying?" Tags="&lt;arithmetic&gt;" AnswerCount="4" CommentCount="8" FavoriteCount="2" />
  <row Id="14269" PostTypeId="2" ParentId="13525" CreationDate="2010-12-14T08:53:05.007" Score="3" ViewCount="" Body="&lt;p&gt;There are joint papers by Benedict Gross (a number theorist) and Mike Hopkins (a topologist) on formal groups and the chromatic tower, that may be more accessible as a quick introduction to the concepts. They try to explain some of this material to non-topologists, and their exposition from almost 20 years ago does not rely on the machinery of stacks, &quot;brave new algebra&quot; or infinity-categories.  The summary paper in the AMS Bulletin has a dictionary relating stable homotopy and quasicoherent sheaves, in particular they explain that the chromatic filtration is analogous to the Cousin complex.  See:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.ams.org/journals/bull/1994-30-01/S0273-0979-1994-00438-0/&quot; rel=&quot;nofollow&quot;&gt;http://www.ams.org/journals/bull/1994-30-01/S0273-0979-1994-00438-0/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="467" LastActivityDate="2010-12-14T08:53:05.007" CommentCount="1" />
  <row Id="14270" PostTypeId="1" AcceptedAnswerId="14274" CreationDate="2010-12-14T08:57:10.370" Score="4" ViewCount="191" Body="&lt;p&gt;Thank you very much!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My problem is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If R is a commutative ring with identity, and a, b are its elements that are divisible by each other, is it true that they must be associates?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here, a being divisible by b means there exists an r \in R, such that a=rb;&#xA;and a and b being associates means there exists a invertible element u \in R such that a=ub.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was told that this not always true. But I encountered some difficulties in finding a counterexample.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many thanks!&lt;/p&gt;&#xA;" OwnerUserId="4736" LastActivityDate="2010-12-14T16:43:09.417" Title="If R is a commutative ring with identity, and a, b are its elements that are divisible by each other, is it true that they must be associates?" Tags="&lt;divisibility&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="1" />
  <row Id="14272" PostTypeId="1" CreationDate="2010-12-14T09:07:27.543" Score="0" ViewCount="81" Body="&lt;p&gt;Let $f, g$ be two paths in a space such that their concatenation $f * g$ is nullhomotopic. Prove that $f$ is homotopic to $g$ rel $\{0,1\}$.&lt;/p&gt;&#xA;" OwnerUserId="4737" LastActivityDate="2010-12-14T11:20:16.050" Title="Homotopy of two paths whose composition is nullhomotopic" Tags="&lt;algebraic-topology&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="14273" PostTypeId="1" AcceptedAnswerId="14277" CreationDate="2010-12-14T10:03:12.010" Score="4" ViewCount="98" Body="&lt;p&gt;I'm trying to prove an identity involving the &lt;a href=&quot;http://en.wikipedia.org/wiki/Digamma_function&quot; rel=&quot;nofollow&quot;&gt;digamma function&lt;/a&gt; $\psi(z)$, but I can't seem to figure out a way to do it. Can anyone help me out? The identity is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\psi\left(\frac{m}{2} + iy\right) + \psi\left(\frac{m}{2} - iy\right) = \psi\left(\frac{n}{2} + iy\right) + \psi\left(\frac{n}{2} - iy\right)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $m$ and $n$ are integers satisfying $m + n = 2$, and $y$ is any nonzero real number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've tried looking at a couple of the integral representations of $\psi(z)$ listed on the Wikipedia page, but I haven't been able to figure it out from those. I think there's probably some simple integral identity I'm forgetting that would make the whole thing work out - for instance, if I could show that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_0^\infty \frac{e^{nt/2}\cos(yt)}{\sinh(t/2)}\mathrm{d}t = \int_0^\infty \frac{e^{-nt/2}\cos(yt)}{\sinh(t/2)}\mathrm{d}t$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for $n \in \mathbb{Z}, y \neq 0$, I think I'd be set. So what I'm hoping to get is a pointer to some relation like that which I could use. Of course I'd be happy with a full proof if you prefer ;-)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S. I know this sounds kind of &quot;homework-y,&quot; but it's not for a homework assignment; I'm trying to verify a calculation in a physics paper.&lt;/p&gt;&#xA;" OwnerUserId="1190" LastEditorUserId="1190" LastEditorDisplayName="" LastEditDate="2010-12-14T19:09:53.947" LastActivityDate="2010-12-14T19:09:53.947" Title="How can I prove this identity involving the digamma function?" Tags="&lt;special-functions&gt;" AnswerCount="1" CommentCount="0" />
  <row Id="14274" PostTypeId="2" ParentId="14270" CreationDate="2010-12-14T10:18:18.883" Score="9" ViewCount="" Body="&lt;p&gt;See the following paper,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When are Associates Unit Multiples?&#xA;D.D. Anderson, M. Axtell, S.J. Forman, and Joe Stickles&#xA;Rocky Mountain J. Math. Volume 34, Number 3 (2004), 811-828.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is &lt;a href=&quot;http://projecteuclid.org/DPubS?verb=Display&amp;amp;version=1.0&amp;amp;service=UI&amp;amp;handle=euclid.rmjm/1181069828&amp;amp;page=record&quot; rel=&quot;nofollow&quot;&gt;freely available&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is mostly concerned with finding sufficent conditions on&#xA;commutative ringsthat ensure that $Ra=Rb$ imply $a$ and $b$ associates,&#xA;but they do give some examples of $R$ where this fails. In particular&#xA;this simple example of Kaplansky. Let $R=C[0,3]$, the set of continuous&#xA;function from the interval $[0,3]$ to the reals. Let $f(t)$ and $g(t)$&#xD;&#xA;equal $1-t$ on $[0,1]$, zero on $[1,2]$ but let $f(t)=t-2$&#xD;&#xA;on $[2,3]$ and $g(t)=2-t$ on $[2,3]$. Then $f$ and $g$ are non-associates&#xA;in $R$ but each divides the other.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-12-14T10:18:18.883" CommentCount="2" />
  <row Id="14275" PostTypeId="1" AcceptedAnswerId="14330" CreationDate="2010-12-14T10:23:42.220" Score="3" ViewCount="172" Body="&lt;p&gt;Let $a$ and $b$ be elements of the polynomial ring $\mathbb{Z}/n\mathbb{Z}[x]$. If $a$ and $b$ generate the same ideal, must they be associates? If $n$ is prime, then it is easy to see that the answer is &quot;yes&quot;. But what if $n$ is not prime?&lt;/p&gt;&#xA;" OwnerUserId="746" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-14T20:23:16.053" LastActivityDate="2010-12-15T06:16:19.003" Title="Associates in the Ring of Polynomials mod n" Tags="&lt;ring-theory&gt;&lt;elementary-number-theory&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="14276" PostTypeId="1" CreationDate="2010-12-14T10:32:07.717" Score="3" ViewCount="149" Body="&lt;p&gt;Is the functor from the category of projective varieties over a field $k$ to the category of pure motives over $k$, faithful? (Perhaps it is not full).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ditto:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is the functor from the category of &lt;em&gt;affine&lt;/em&gt; varieties over a field $k$ to the category of mixed motives over $k$, faithful? (Perhaps it is not full).&lt;/p&gt;&#xA;" OwnerUserId="4739" LastActivityDate="2010-12-15T00:05:45.690" Title="Does a motive capture everything about an algebraic variety?" Tags="&lt;algebraic-geometry&gt;" AnswerCount="4" />
  <row Id="14277" PostTypeId="2" ParentId="14273" CreationDate="2010-12-14T10:39:01.750" Score="6" ViewCount="" Body="&lt;p&gt;The first relation can be rewritten as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\psi\left(\frac{m}{2} + iy\right) + \psi\left(\frac{m}{2} - iy\right) = \psi\left(\frac{2-m}{2} + iy\right) + \psi\left(\frac{2-m}{2} - iy\right)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\psi\left(\frac{m}{2} + iy\right) + \psi\left(\frac{m}{2} - iy\right) = \psi\left(1-\left(\frac{m}{2} + iy\right)\right) + \psi\left(1-\left(\frac{m}{2} - iy\right)\right)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we use the reflection formula for the digamma function (which can be derived from the reflection formula for the gamma function)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\psi(1-z)=\psi(z)+\pi\cot\pi z$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;your equation simplifies as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\cot\left(\frac{m\pi}{2}+i\pi y\right)+\cot\left(\frac{m\pi}{2}-i\pi y\right)=0$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or since the cotangent is an odd function,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\cot\left(\frac{m\pi}{2}+i\pi y\right)=\cot\left(-\frac{m\pi}{2}+i\pi y\right)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which can be rewritten as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\cot\left(\frac{m\pi}{2}+i\pi y\right)=\cot\left(\frac{m\pi}{2}+i\pi y-m\pi\right)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and it is clear that both sides of the equation will agree only if $m\in\mathbb Z$ due to the periodicity of the cotangent.&lt;/p&gt;&#xA;" OwnerUserId="498" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-14T12:37:08.963" LastActivityDate="2010-12-14T12:37:08.963" CommentCount="1" />
  <row Id="14278" PostTypeId="2" ParentId="14272" CreationDate="2010-12-14T11:11:19.413" Score="1" ViewCount="" Body="&lt;p&gt;As Rasmus points out, once you state the result correctly, it's easy -assuming you've already proved some elementary properties of homotopy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Namely, let $f,g$ be two loops with $f(0) = g(0) = f(1)= g(1) = x_0$ and such that $g^{-1}*f \sim 1$, where $1$ is the constant path at $x_0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then $f\sim 1* f \sim (g*g^{-1})*f \sim g* (g^{-1}*f) \sim g*1  \sim g$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Exercise.&lt;/strong&gt; Which &quot;elementary properties of homotopy&quot; have we used and where?&lt;/p&gt;&#xA;" OwnerUserId="664" LastEditorUserId="664" LastEditorDisplayName="" LastEditDate="2010-12-14T11:20:16.050" LastActivityDate="2010-12-14T11:20:16.050" />
  <row Id="14279" PostTypeId="1" CreationDate="2010-12-14T11:13:54.933" Score="0" ViewCount="64" Body="&lt;p&gt;Let $\psi_{j,k}(t)=\{ \begin{array}{ll} 2^{j/2} &amp;amp;, 2^{-j}k &amp;lt; t &amp;lt; 2^{-j}(k+1/2) \\-2^{j/2} &amp;amp;,  2^{-j}(k+1/2) &amp;lt; t &amp;lt; 2^{-j}(k+1) \\ 0, \textrm{otherwise} \end{array} $. How to prove that it is orthogonal? In other words how to prove that $\langle \psi_{j_1,k_1}(t), \psi_{j_2, k_2}(t) \rangle = 0$, $j_1&amp;lt;j_2$ and $k_1&amp;lt;k_2$. I have already proved that it's norm is one $|| \psi_{j,k} ||_2 = 1$, but now I'm stuck in this orthogonality issue, because it involves many nontrivial computations. &lt;/p&gt;&#xA;" OwnerUserId="2723" LastEditorUserId="2723" LastEditorDisplayName="" LastEditDate="2010-12-14T11:32:48.233" LastActivityDate="2010-12-14T13:43:31.877" Title="Let $\{ \psi_{j,k}(t)\}$ haar system. How to prove that it is orthogonal?" Tags="&lt;linear-algebra&gt;" AnswerCount="2" />
  <row Id="14280" PostTypeId="2" ParentId="14113" CreationDate="2010-12-14T11:22:39.840" Score="3" ViewCount="" Body="&lt;p&gt;Just to give people an idea on convergence rates, here is a plot of $-\log_{10}\left|\frac{S_n-\pi}{\pi}\right|$ versus $n$ , where $S_n$ is the nth partial sum of the series in question, for three of the series featured in the answers to this question (note the vertical scale):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/PdR9f.png&quot; alt=&quot;partial sum plots&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The three series are, from top to bottom, $\arctan(1)$ (the series mentioned by the OP), $2\arcsin\left(\sqrt{\frac12}\right)$ (the series mentioned by yjj in his answer), and the series by Ramanujan I mentioned in the comments (I didn't include the series by the Chudnovsky brothers, since that converges even faster than the Ramanujan series, and that makes for boring plots).&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-12-14T11:22:39.840" CommunityOwnedDate="2010-12-14T11:22:39.840" />
  <row Id="14281" PostTypeId="1" AcceptedAnswerId="14454" CreationDate="2010-12-14T11:26:04.487" Score="3" ViewCount="99" Body="&lt;p&gt;Say $M := \begin{pmatrix} A &amp;amp; B\\ C &amp;amp; D \end{pmatrix}$ is a block matrix with $A, D$ being square matrices and this $B$ and $C^T$ having the same shape. Is there any norm characterizing the collection of $B$ and $C$ that is invariant under all block diagonal similarity transformations $M\to S M S^{-1}$ with $S=\begin{pmatrix}E &amp;amp; 0\\ 0 &amp;amp; F\end{pmatrix}$ of the whole block matrix? Were $B$ and $C$ square matrices I'd use e.g. the sum of squared eigenvalues, but is there something similar for non-square matrices?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In brief, the requirements for the block norms I seek are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;the norm of the square blocks $A$ and $D$ must be invariant (under these similarity transformations of $M$) respectively&lt;/li&gt;&#xA;&lt;li&gt;either there are individual norms for $B$ and $C$ which are invariant, or the is one combined norm depending on $B$ and $C$ which is invariant&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Is the latter requirement possible?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Due to the block diagonal structure of $S$, all blocks transform independently, so there should be independent norms for $B$ and $C$. But since $B\to F B E^{-1}$ and $C\to E C F^{-1}$ are no similarity transformations, I couldn't even use eigenvalues if $B$ and $C$ &lt;em&gt;were&lt;/em&gt; square matrices. However, $BC \to F BC F^{-1}$ and $CB \to E CB E^{-1}$ are similarity transformations, the remaining question is what kind of norm to use and maybe also how to decide whether $BC$ or $CB$ or both are &quot;significant&quot;...&lt;/p&gt;&#xA;" OwnerUserId="163" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-14T17:29:03.517" LastActivityDate="2010-12-15T21:20:12.937" Title="What are norms of sub-matrices invariant under a block diagonal similarity transformation of a block matrix?" Tags="&lt;linear-algebra&gt;&lt;matrices&gt;&lt;norm&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="14282" PostTypeId="1" AcceptedAnswerId="14289" CreationDate="2010-12-14T11:54:20.500" Score="6" ViewCount="464" Body="&lt;p&gt;Let $G \in \mathbf{Grp}$, $H \leq G$, $G/H := \lbrace gH: g \in G \rbrace$. We can then introduce group operation on $G/H$ as $(xH)*(yH) := (xy)H$, so that $G/H$ becomes a quotient group when $H$ is a normal subgroup.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But why do we only work with quotient groups by normal subgroups? If we introduce the notion of &lt;em&gt;left quotient group&lt;/em&gt; in the above manner, how much good properties of a quotient group do we lose?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;UPD: fixed confusing notation: from $(xH)(yH)$ to $(xH)*(yH)$.&lt;/p&gt;&#xA;" OwnerUserId="3793" LastEditorUserId="3793" LastEditorDisplayName="" LastEditDate="2010-12-14T13:07:53.323" LastActivityDate="2010-12-16T18:18:38.393" Title="Why do we define quotient groups for normal subgroups only?" Tags="&lt;group-theory&gt;" AnswerCount="3" CommentCount="6" FavoriteCount="5" />
  <row Id="14283" PostTypeId="1" CreationDate="2010-12-14T12:08:11.267" Score="0" ViewCount="79" Body="&lt;p&gt;I'm working on data structure for graph cut algorithm. Problem is to make different cuts on shortest paths. I made data structure for which I'm not sure about properties.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Input is directed graph of shortest paths, which is bounded lattice, partially ordered set with minimum and maximum element.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Define &lt;em&gt;next node N(n)&lt;/em&gt; of node n as a set of nodes b for which a &amp;lt; b and there is no c with a &amp;lt; c &amp;lt; b. Similar define &lt;em&gt;previous node P(n)&lt;/em&gt;. Extend definitions on sets, N(S) union of N(n) for $n \in S$, similar for P(S).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is easy to make different cuts on list of set of nodes $L, N(L), N(N(L)), ...$ where for each neighboring pair of sets A, N(A)=B holds that there is no partition of $A=A_1 \cup A_2$ and $B=B_1 \cup B_2$ with $B_i = N(A_i)$, $A_i = P(B_i)$ for $i=1,2$.&#xA;With this property create new lattice with mapping:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;sub-lattice to one node&lt;/li&gt;&#xA;&lt;li&gt;if upper partition is found create partition cardinality number of edges.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In simple, algorithm for lattice -&gt; lattice mapping is:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;A = {minimum node}&#xA;new_node = [A]&#xA;1:&#xA;while A, N(A) don't have partitions&#xA;  append N(A) to new_node&#xA;  A = N(A)&#xA;for each partition $B_i$&#xD;&#xA;  last_new_node = new_node&#xA;  create new_node = [B_i]&#xA;  create edge last_new_node to new_node&#xA;  go to 1&#xA;At the end fix maximum node in new lattice if needed&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This algorithm can be called repeatedly on new lattices. I'm concern about:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Is there guaranty to reach one-node lattice?&lt;/li&gt;&#xA;&lt;li&gt;Is there any measure of number of iterations to reach one-node lattice? It seams to me that bound is diameter of input graph. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I appreciate link to any similar data structure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Tnx&lt;/p&gt;&#xA;" OwnerUserId="3037" LastActivityDate="2010-12-14T12:08:11.267" Title="Lattice simplification" Tags="&lt;algorithms&gt;&lt;graph-theory&gt;&lt;lattices&gt;" />
  <row Id="14284" PostTypeId="1" CreationDate="2010-12-14T12:15:13.913" Score="1" ViewCount="69" Body="&lt;p&gt;We are talking about &lt;a href=&quot;http://en.wikipedia.org/wiki/Non-uniform_rational_B-spline&quot; rel=&quot;nofollow&quot;&gt;Non-uniform rational B-spline&lt;/a&gt;. We have some simple 3 dimensional array like &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;{1,1,1}&#xA;{1,2,3}&#xA;{1,3,3}&#xA;{2,4,5}&#xA;{2,5,6}&#xA;{4,4,4}&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Which are points from a plane created by some B-spline&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How to find control points of spline that created that plane? (I know its a hard task because of weights that need to be calculated but I really hope it is solvable)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/lOuL8.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="2065" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-14T17:45:38.743" LastActivityDate="2010-12-14T17:45:38.743" Title="How to get NURBS control points from an array of points that should be part of its solution from controll points we are searching for?" Tags="&lt;geometry&gt;&lt;numerical-methods&gt;&lt;approximation&gt;" CommentCount="1" />
  <row Id="14285" PostTypeId="2" ParentId="14282" CreationDate="2010-12-14T12:34:45.183" Score="3" ViewCount="" Body="&lt;p&gt;(Sorry, new here, can't comment.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To expand on Alex's answer, if we demand that $(xh)(yh) \in (xy)H$ for any coset representative $h$, then this is equivalent to asking that $y^{-1}hyh \in H$ for all $h \in H$. But the conjugate $y^{-1}hy$ need not lie in $H$ if $H$ isn't normal, hence the claim that the operation in your &quot;left quotient group&quot; isn't well defined.&lt;/p&gt;&#xA;" OwnerUserId="4743" LastActivityDate="2010-12-14T12:34:45.183" CommentCount="8" />
  <row Id="14286" PostTypeId="2" ParentId="14279" CreationDate="2010-12-14T12:48:42.340" Score="1" ViewCount="" Body="&lt;p&gt;$(j_1,k_1)\ne(j_2,k_2)$ means (a) $j_1\ne j_2$ or (b) $j_1=j_2$ and $k_1\ne k_2$. In each of these cases the statement follows immediately by looking at the graphs of two such $\psi$'s.&lt;/p&gt;&#xA;" OwnerUserId="1303" LastActivityDate="2010-12-14T12:48:42.340" CommentCount="2" />
  <row Id="14287" PostTypeId="2" ParentId="14279" CreationDate="2010-12-14T13:13:17.203" Score="0" ViewCount="" Body="&lt;p&gt;If you are working in $L_2(0,1)$ space it is then quite easy. The scalar product then is the integral&#xA;\begin{align*}&#xA;\int_0^1\psi_{j_1,k_1}(t)\psi_{j_2,k_2}(t)dt&#xA;\end{align*}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rewrite $\psi_{j,k}(t)$ as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align*}&#xA;\psi_{j,k}(t)=2^j1\left\{t\in\left(\frac{k}{2^j},\frac{k+1/2}{2^j}\right)\right\}-&#xA;2^{-j}1\left\{t\in\left(\frac{k+1/2}{2^j},\frac{k+1}{2^j}\right)\right\}&#xA;\end{align*}&#xA;then&#xA;\begin{align*}&#xA;\psi_{j,k}(t)\psi_{l,m}(t)&amp;amp;=2^{j+l}1\left\{t\in\left(\frac{k}{2^j},\frac{k+1/2}{2^j}\right)\cap\left(\frac{m}{2^l},\frac{m+1/2}{2^l}\right)\right\}\\&#xA;&amp;amp;-2^{j+l}1\left\{t\in\left(\frac{k}{2^j},\frac{k+1/2}{2^j}\right)\cap\left(\frac{m+1/2}{2^l},\frac{m+1}{2^l}\right)\right\}\\&#xA;&amp;amp;-2^{j+l}1\left\{t\in\left(\frac{k+1/2}{2^j},\frac{k+1}{2^j}\right)\cap\left(\frac{m}{2^l},\frac{m+1/2}{2^l}\right)\right\}\\&#xA;&amp;amp;+2^{j+l}1\left\{t\in\left(\frac{k+1/2}{2^j},\frac{k+1}{2^j}\right)\cap\left(\frac{m+1/2}{2^l},\frac{m+1}{2^l}\right)\right\}&#xA;\end{align*}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now if $j&amp;lt;l$, we have that $2^{l-j-1}\ge 1$ and we need to investigate the following cases:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$m&amp;lt;2^{l-j}k$. The product is immediately zero, since the intervals do not overlap&lt;/li&gt;&#xA;&lt;li&gt;$2^{l-j}k\le m &amp;lt;2^{l-j}k+2^{l-j-1}$. Then the 3rd and 4th terms in the sum are zero and the product is equal to $2^j\psi_{l,m}$, whose integral is zero.&lt;/li&gt;&#xA;&lt;li&gt;$2^{l-j}k+2^{l-j-1}\le m&amp;lt;2^{l-j}(k+1)$. Then the 1st and 2nd terms in the sum are zero and the product is equal to $-2^{j}\psi_{l,m}$, whose integral is again zero.&lt;/li&gt;&#xA;&lt;li&gt;$m\ge 2^{l-j}(k+1)$. The product is zero, since the intervals do not overlap.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="4742" LastEditorUserId="4742" LastEditorDisplayName="" LastEditDate="2010-12-14T13:43:31.877" LastActivityDate="2010-12-14T13:43:31.877" CommentCount="2" />
  <row Id="14288" PostTypeId="1" AcceptedAnswerId="14291" CreationDate="2010-12-14T13:17:36.510" Score="0" ViewCount="68" Body="&lt;p&gt;I need to understend how  black box optimizer work. I need a real life example of how it hlps. So what I need: What was the task, what were parameters, what was the function to minimize and how it all worked toogether. I need deteiled info on this topic, please - &lt;a href=&quot;http://en.wikipedia.org/wiki/Optimization_%28mathematics%29&quot; rel=&quot;nofollow&quot;&gt;wiki&lt;/a&gt; does not give lot of help...&lt;/p&gt;&#xA;" OwnerUserId="2065" LastActivityDate="2010-12-14T13:53:42.287" Title="Wanted: Examples of how black box optimizer works - step by step?" Tags="&lt;optimization&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="14289" PostTypeId="2" ParentId="14282" CreationDate="2010-12-14T13:33:39.783" Score="11" ViewCount="" Body="&lt;p&gt;I cannot comment yet, but I thought I would elaborate on what is meant by being well-defined.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your definition of the multiplication in $G/H$, you take cosets $xH$ and $yH$ and multiply by setting thir product equal to $(xy)H$. But this means that you have chosen $x$ and $y$, and there will in general be other choices of $x$ and $y$ that give the same cosets. So if you have $xH$ = $x&amp;#39;H$ and $yH$ = $y&amp;#39;H$, you want to be sure that you also have $(xy)H = (x&amp;#39;y&amp;#39;)H$, and it can be shown that this holds for all elements of $G$ iff $H$ is normal in $G$, so this is the only case where this multiplication is well-defined.&lt;/p&gt;&#xA;" OwnerUserId="2538" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-15T17:49:31.490" LastActivityDate="2010-12-15T17:49:31.490" />
  <row Id="14290" PostTypeId="2" ParentId="14268" CreationDate="2010-12-14T13:43:25.170" Score="6" ViewCount="" Body="&lt;p&gt;So, you want to compare (254)(847) and (383)(536) without actually computing the products, look at&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{847}{383} \; ? \; \frac{536}{254}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Multiplying both denominators by 2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{847}{766} \; ? \; \frac{536}{508}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$1+\frac{81}{766} \; ? \; 1+\frac{28}{508}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are now left with comparing&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{81}{766} \; ? \; \frac{28}{508}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Applying the same trick as before, consider&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{81}{28} \; ? \; \frac{766}{508}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The left hand side is obviously bigger than $2$ while the right hand side is smaller, therefore, you can conclude that the original left hand side product was the bigger one, since none of the operations performed inverted the order of the inequalities.&lt;/p&gt;&#xA;" OwnerUserId="3567" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2010-12-14T13:46:11.407" LastActivityDate="2010-12-14T13:46:11.407" CommentCount="4" />
  <row Id="14291" PostTypeId="2" ParentId="14288" CreationDate="2010-12-14T13:53:42.287" Score="2" ViewCount="" Body="&lt;p&gt;Chapter 10 of &lt;a href=&quot;http://nr.com/&quot; rel=&quot;nofollow&quot;&gt;Numerical Recipes&lt;/a&gt; has a good description of how the golden section search works in 1D, as well as description of multidimensional cases.  You could see figure 10.1.1 in the obsolete C version (available free online)&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-12-14T13:53:42.287" />
  <row Id="14293" PostTypeId="1" AcceptedAnswerId="14307" CreationDate="2010-12-14T14:11:16.690" Score="1" ViewCount="82" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm developing a computer program, and need an algorithm to solve the following problem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How to select the optimum combination of numbers from a random list that add to up to a certain total (or as close to as possible).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Example:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;List of random numbers:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1.0,&#xA;2.1,&#xA;5.3&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Required number = 6.5&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here the answer would be 6.3 (from 5.3 + 1.0) I believe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The program needs to be able to operate on much bigger number sets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have some maths qualifications (up to a-level), but not expert, so if you could temper your respose to my level if possible, that would be appreciated!&lt;/p&gt;&#xA;" OwnerUserId="4745" LastActivityDate="2010-12-14T16:10:17.577" Title="How to select the optimum combination of numbers from a random list that add to up to a certain total (or as close to)." Tags="&lt;computer-science&gt;&lt;probability&gt;" AnswerCount="2" CommentCount="5" />
  <row Id="14294" PostTypeId="2" ParentId="14293" CreationDate="2010-12-14T14:18:50.857" Score="3" ViewCount="" Body="&lt;p&gt;This is close to the &lt;a href=&quot;http://en.wikipedia.org/wiki/Subset_sum_problem&quot; rel=&quot;nofollow&quot;&gt;subset sum problem&lt;/a&gt;, which is not easy.  But some approaches are in the Wikipedia article.  The section on the approximate algorithm seems quite applicable.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-12-14T14:18:50.857" CommentCount="3" />
  <row Id="14295" PostTypeId="2" ParentId="14263" CreationDate="2010-12-14T14:35:12.017" Score="2" ViewCount="" Body="&lt;p&gt;Ask GAP:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;gap&amp;gt; AllSmallGroups(156, IsSolvable, false);&#xA;[  ]&#xA;gap&amp;gt; &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This gives the list of all groups of order 156 which are not solvable.&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2010-12-14T14:35:12.017" />
  <row Id="14296" PostTypeId="1" AcceptedAnswerId="14321" CreationDate="2010-12-14T15:14:31.947" Score="1" ViewCount="51" Body="&lt;p&gt;I came across this problem in the context of spectroscopy today.  Because of it's simplicity, I'm sure it's a question that's been posed by some mathematician ages ago, I just can't figure out how to formulate it mathematically:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose you have a mirror that forms a perfect ring and has an aperture in it with width $\delta w$.  A ray of light enters the aperture at angle $\theta$ relative to the tangent at the aperture.  There are several questions that I have regarding this system, some of which would be of practical use and some of which are mere curiosity:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Is this a chaotic system?  I know a little about chaotic systems from my undergrad math major, and this looks like one to me, but I don't know where to start working on it if it is.&lt;/li&gt;&#xA;&lt;li&gt;How can this problem be formulated mathematically?  I would assume that there is some sort of iterated function approach, but I can't figure out what the iterated function would be (probably because as a chemist I'm more comfortable with cartesian coordinates than polar coordinates).&lt;/li&gt;&#xA;&lt;li&gt;What is the optimal angle for achieving the most passes through the area within some $\epsilon$ of the center of the ring before exiting through the aperture, assuming (a) $\epsilon$ is small compared to $r$, the radius of the ring, or (b) $\epsilon$ is not small compared to the radius of the ring?  (This is one of the questions of practical importance)&lt;/li&gt;&#xA;&lt;li&gt;Bonus:  Suppose there is a second aperture placed at an angle $\phi$ away from the first (with $\phi$ relative to the center of the circle and to the first aperture).  Can this angle be optimized in conjunction with the variables in $\phi$?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;To summarize, the variables are $r$, $\theta$, $\delta w$, $\epsilon$, and (if we're lucky) $\phi$.  Assume that $\epsilon$ and $\delta w$ are given (obviously the optimal size for $\delta w$ is infinitesimal, but this isn't practical of course).  Also (for the physicists out there), assume $\delta w$ is wide enough that the beam can be treated classically, i.e. no diffraction.  Also, there needs to be a reasonable way to calculate the exit angle based on the entry angle (if this is a truly chaotic system, I realize this will be practically impossible).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance for any help, and let me know if I've left out any important details.  I'm comfortable with any math that a typical undergraduate math major should know by the time they graduate; anything more and I'll need a reference to read up on.&lt;/p&gt;&#xA;" OwnerUserId="1590" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2010-12-14T15:23:27.007" LastActivityDate="2010-12-14T19:10:04.507" Title="Ring-shaped mirror chaos points" Tags="&lt;geometry&gt;&lt;chaotic-systems&gt;&lt;dynamical-systems&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="14297" PostTypeId="1" AcceptedAnswerId="14316" CreationDate="2010-12-14T15:34:15.323" Score="7" ViewCount="240" Body="&lt;p&gt;You are allowed to roll a die up to six times. Anytime you stop, you get the dollar amount of the face value of your last roll.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Question: What is the best strategy?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to my calculation, for the strategy 6,5,5,4,4, the expected value is $142/27\approx 5.26$, which I consider quite high. So this might be the best strategy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here,  6,5,5,4,4 means in the first roll you stop only when you get a 6; if you did not get a 6 in the first roll, then in the second roll you stop only when you roll a number 5 or higher (i.e. 5 or 6), etc.&lt;/p&gt;&#xA;" OwnerUserId="3249" LastActivityDate="2010-12-14T17:39:57.733" Title="Best Strategy for a die game" Tags="&lt;probability&gt;&lt;game-theory&gt;" AnswerCount="3" FavoriteCount="1" />
  <row Id="14298" PostTypeId="1" AcceptedAnswerId="14308" CreationDate="2010-12-14T15:38:22.707" Score="2" ViewCount="114" Body="&lt;p&gt;We solve the pell equation using the continued fraction for square root of 2. What equations can we solve using the continued fraction of cube roots (and other numbers too)?&lt;/p&gt;&#xA;" OwnerUserId="4290" LastActivityDate="2010-12-14T16:27:43.370" Title="Solving equations using continued fractions?" Tags="&lt;diophantine-equations&gt;&lt;continued-fractions&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="14299" PostTypeId="2" ParentId="14297" CreationDate="2010-12-14T15:45:59.113" Score="1" ViewCount="" Body="&lt;p&gt;On the next to last roll, the expectation on the last roll is 3.5, so accepting 4 and rejecting 3 is clearly right.  On the third to last roll, you should calculate the expectation if you decline.  You have 1/2 chance of rolling 4 or better and accepting it, with an expectation of 5.  You have 1/2 chance of rolling 3 or worse and rejecting it, with an expectation of 3.5.  So if you decline the third to last, you expect 4.25.  So you should reject a 4 and accept a 5.  If you keep working backwards like this, you may find you should bump at least one of your 5's to a 6.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-12-14T15:45:59.113" />
  <row Id="14300" PostTypeId="2" ParentId="11703" CreationDate="2010-12-14T15:47:57.007" Score="0" ViewCount="" Body="&lt;p&gt;Since $$\frac1{1-x}=1+x+x^2+x^3+\cdots,$$ we have&#xA;$$&#xD;&#xA;\frac{1-x^{n+1}}{1-x}=(1+x+x^2+x^3+\cdots)-(x^{n+1}+x^{n+2}+x^{n+3}+x^{n+4}+\cdots)&#xD;&#xA;$$&#xD;&#xA;And on the right hand side every thing cancels except $1+x+x^2+\cdots+x^n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(This argument is probably circular! :) )&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2010-12-14T15:47:57.007" CommentCount="1" />
  <row Id="14301" PostTypeId="2" ParentId="14296" CreationDate="2010-12-14T15:54:28.700" Score="1" ViewCount="" Body="&lt;p&gt;I'm not an expert, but circular billiards is a special case of elliptic billiards, and what you want to know can probably be found in &lt;a href=&quot;http://www.math.psu.edu/tabachni/Books/billiardsgeometry.pdf&quot; rel=&quot;nofollow&quot;&gt;Billiards and Geometry (pdf file)&lt;/a&gt; by Serge Tabachnikov.&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2010-12-14T15:54:28.700" />
  <row Id="14302" PostTypeId="2" ParentId="14297" CreationDate="2010-12-14T15:54:48.510" Score="9" ViewCount="" Body="&lt;p&gt;Just work backwards. At each stage, you accept a roll that is &gt;= the expected gain from the later stages:&lt;br&gt;&#xA;Expected gain from 6th roll: 7/2&lt;br&gt;&#xA;Therefore strategy for 5th roll is: accept if &gt;= 4&lt;br&gt;&#xA;Expected gain from 5th roll: (6 + 5 + 4)/6 + (7/2)(3/6) = 17/4&lt;br&gt;&#xA;Therefore strategy for 4th roll is: accept if &gt;= 5&lt;br&gt;&#xA;Expected gain from 4th roll: (6 + 5)/6 + (17/4)(4/6) = 14/3&lt;br&gt;&#xA;Therefore strategy for 3rd roll is: accept if &gt;= 5&lt;br&gt;&#xA;Expected gain from 3rd roll: (6 + 5)/6 + (14/3)(4/6) = 89/18&lt;br&gt;&#xA;Therefore strategy for 2nd roll is: accept if &gt;= 5&lt;br&gt;&#xA;Expected gain from 2nd roll: (6 + 5)/6 + (89/18)(4/6) = 277/54&lt;br&gt;&#xA;Therefore strategy for 1st roll is: accept only if 6&lt;br&gt;&#xA;Expected gain from 1st roll: 6/6 + (277/54)(5/6) = 1709/324  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So your strategy is 6,5,5,5,4 for an expectation of $5.27469...&lt;/p&gt;&#xA;" OwnerUserId="1508" LastActivityDate="2010-12-14T15:54:48.510" CommentCount="4" />
  <row Id="14303" PostTypeId="2" ParentId="11703" CreationDate="2010-12-14T15:58:54.113" Score="5" ViewCount="" Body="&lt;p&gt;Since $1-x^{n+1}$ has $1$ as a root, the quotient $\frac{1-x^{n+1}}{1-x}$ is a polynomial.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $\mathbb F_q$ is a finite field with $q$ elements and $V$ is a $\mathbb F_q$-vector space of dimension $n+1$, then $\frac{1-q^{n+1}}{1-q}=|P(V)|$ is the cardinal of the projective space attached to $V$. Now $P(V)$ can be described as a disjoint union $$P(V)=\mathbb A^0\sqcup\mathbb A^1\sqcup \mathbb A^2\sqcup\cdots\sqcup\mathbb A^n$$ where $\mathbb A^k$ is, for each $k$, an affine space of dimension $k$ over $\mathbb F_q$ (which is a complicated way of saying, as far as our purposes go, a vector space over $\mathbb F_q$ of dimension $k$) Since $|\mathbb A^k|=q^k$, we find that&#xA;$$\frac{1-q^{n+1}}{1-q}=1+q+q^2+q^3+\cdots+q^n$$&#xD;&#xA;for all numbers $q$ which are powers of prime numbers. It follows that&#xA;$$\frac{1-x^{n+1}}{1-x}=1+x+x^2+x^3+\cdots+x^n$$&#xD;&#xA;as &lt;em&gt;polynomials&lt;/em&gt;, because the equality holds for infinitely many values of $x$ (and we are working over $\mathbb Z$...)&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2010-12-14T15:58:54.113" CommentCount="6" />
  <row Id="14304" PostTypeId="2" ParentId="14268" CreationDate="2010-12-14T16:01:50.173" Score="1" ViewCount="" Body="&lt;p&gt;In the worst case, where the comparison is close, you will always end up having to do as much work as multiplication. The performance gains from any non-worst-case improvements would have to be pretty good to make the programming effort worthwhile. Why do you need this?&lt;/p&gt;&#xA;" OwnerUserId="1508" LastActivityDate="2010-12-14T16:01:50.173" />
  <row Id="14305" PostTypeId="2" ParentId="14270" CreationDate="2010-12-14T16:04:20.793" Score="3" ViewCount="" Body="&lt;p&gt;This is true if $R$ is a domain. More generally this is true if $a$ or $b$ is not a zero divisor in $R$. Suppose, $a$ is not a zero divisor and there exist $r,s\in R$ such that $a=rb$ and $b=sa$, then $a=rsa$, so $a(1-rs)=0$. Since $a$ is not a zero divisor, $1-rs=0$, so $rs=1$. So, $r,s$ are units. So $a$ and $b$ are associates.&lt;/p&gt;&#xA;" OwnerUserId="3431" LastActivityDate="2010-12-14T16:04:20.793" CommentCount="1" />
  <row Id="14306" PostTypeId="1" AcceptedAnswerId="14336" CreationDate="2010-12-14T16:07:00.133" Score="0" ViewCount="266" Body="&lt;p&gt;Hey there,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have to prove Simpson's rule including the error with the help of the integral remainder. However, I have practically no idea how to start.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $f: [a,b] \rightarrow \mathbb{R}$ be continuously differentiable four times. Let $n \in 2\mathbb{N}$ and $h = (b-a)/n$. Let $y_k = f(a+kh)$ for $0 \leq k \leq n$. Show:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left|\int_a^b f(x) \mathrm dx - \frac{h}{3} [(y_0 + y_n) + 4(y_1+y_3+...+y_{n-1})+2(y_2+y_4+...+y_{n-2})]\right|$$&#xD;&#xA;$$\leq \frac{1}{180} \max_{a \leq x \leq b} |f^{(4)}(x)|(b-a)h^4$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I thought I'd do a Taylor expansion for $\int_a^b f(x) \mathrm dx$ and then apply the integral remainder, but then all terms vanish and I get $0=0$ which is not very helpful...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I only need a good advice how to start! Thanks in advance!&lt;/p&gt;&#xA;" OwnerUserId="3787" LastEditorUserId="3787" LastEditorDisplayName="" LastEditDate="2011-03-19T20:15:58.087" LastActivityDate="2011-03-19T20:15:58.087" Title="Prove Simpson's rule (including error) using the integral remainder" Tags="&lt;homework&gt;&lt;analysis&gt;&lt;numerical-methods&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" />
  <row Id="14307" PostTypeId="2" ParentId="14293" CreationDate="2010-12-14T16:10:17.577" Score="0" ViewCount="" Body="&lt;p&gt;As Ross Millikan notes, this is a standard &lt;a href=&quot;http://en.wikipedia.org/wiki/Subset_sum_problem&quot; rel=&quot;nofollow&quot;&gt;Subset Sum&lt;/a&gt; or &lt;a href=&quot;http://en.wikipedia.org/wiki/Partition_problem&quot; rel=&quot;nofollow&quot;&gt;Number Partition Problem&lt;/a&gt; which are both &lt;a href=&quot;http://en.wikipedia.org/wiki/NP-complete&quot; rel=&quot;nofollow&quot;&gt;NP-Complete&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If your Subset Sum instance is $(a_0, a_1, \dots, a_{n-1})$, where each $a_k \in \mathbb{Z}$, and the numbers in the list are small compared with the list size (i.e. $\log_2 ( \max(a_k) ) &amp;lt;&amp;lt; n$), then you can either use some brute force search, as in the Complete Karmarkar Karp algorithm (a review of which can be found &lt;a href=&quot;http://arxiv.org/pdf/0802.4040v2&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;) or some dynamic programming method (&lt;a href=&quot;http://www.cse.ohio-state.edu/~tamaldey/course/794/subset-sum.pdf&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; for example).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can always create an instance of Subset Sum/Number Partition out of a list that has it's numbers drawn from the reals by multiplying each element in the list by some large number to make it integer (for example, multiplying each number in your above list by 10 would turn this into a standard Subset Sum/Number Partition Problem).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Being NP-Complete means, in general, this problem is probably hard, in that there is nothing better than an exponential/brute force approach.  Not all brute force algorithms are created equal, though, and depending on what your needs are, you might find a better algorithm suited to the particular type of Subset Sum/Number Partition Problem instances you're creating.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some researchers have had some luck &lt;a href=&quot;http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.57.3331&amp;amp;rep=rep1&amp;amp;type=ps&quot; rel=&quot;nofollow&quot;&gt;with lattice reduction techniques&lt;/a&gt;, so that's another avenue of attack, though lattice reduction techniques (&lt;a href=&quot;http://en.wikipedia.org/wiki/Lenstra%E2%80%93Lenstra%E2%80%93Lov%C3%A1sz_lattice_basis_reduction_algorithm&quot; rel=&quot;nofollow&quot;&gt;LLL&lt;/a&gt; and &lt;a href=&quot;http://arminstraub.com/files/pslq.pdfhttp://arminstraub.com/files/pslq.pdf&quot; rel=&quot;nofollow&quot;&gt;PSLQ&lt;/a&gt; for example) are a bit advanced and, from my understanding, don't fare all that well for Subset Sum/Number Partition Problem instances past a certain size.&lt;/p&gt;&#xA;" OwnerUserId="4143" LastActivityDate="2010-12-14T16:10:17.577" CommentCount="1" />
  <row Id="14308" PostTypeId="2" ParentId="14298" CreationDate="2010-12-14T16:27:43.370" Score="4" ViewCount="" Body="&lt;p&gt;Voronoi developed a generalized continued fraction algorithm for computing  units in cubic fields. You can find a beautiful geometric interpretation of Voronoi's algorithm in Delone and Faddeev's: &lt;em&gt;Theory of Irrationalities of Third Degree&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-12-14T16:27:43.370" CommentCount="2" />
  <row Id="14309" PostTypeId="2" ParentId="14253" CreationDate="2010-12-14T16:30:32.727" Score="0" ViewCount="" Body="&lt;p&gt;If $X$ is a closed subspace of $\ell_1^4$, then it is true if the latter has the metric extension property.  If you are working over the reals, then Google :&quot;binary intersection property Nachbin&quot;. I think Grothendieck proved that if a space has the metric extension property, then it is either finite dimensional or non-separable. I hope this helps!.&lt;/p&gt;&#xA;" OwnerUserId="4749" LastActivityDate="2010-12-14T16:30:32.727" CommentCount="1" />
  <row Id="14310" PostTypeId="2" ParentId="14270" CreationDate="2010-12-14T16:43:09.417" Score="2" ViewCount="" Body="&lt;p&gt;Related questions arise frequently on sci.math, where I &lt;a href=&quot;http://groups.google.com/group/sci.math/msg/d9512c74d9ce4acf&quot; rel=&quot;nofollow&quot;&gt;often&lt;/a&gt; point out the following standard reference on the topic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Beware that this equivalence, i.e. $\rm\ aR = bR \iff a/b\ $ is a unit in $\rm\:R\:$, &#xA;generally fails when $\rm\:R\:$ has zero-divisors, so that there are at least a few different notions of &quot;associate&quot; that are of interest.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When are Associates Unit Multiples?&lt;br&gt;&#xA;D.D. Anderson, M. Axtell, S.J. Forman, and Joe Stickles&lt;br&gt;&#xA;Rocky Mountain J. Math. Volume 34, Number 3 (2004), 811-828. &#xA;&lt;a href=&quot;http://math.la.asu.edu/~rmmc/rmj/vol34-3/andepage1.pdf&quot; rel=&quot;nofollow&quot;&gt;http://math.la.asu.edu/~rmmc/rmj/vol34-3/andepage1.pdf&lt;/a&gt; &#xA;&lt;a href=&quot;http://projecteuclid.org/handle/euclid.rmjm/1181069828&quot; rel=&quot;nofollow&quot;&gt;http://projecteuclid.org/handle/euclid.rmjm/1181069828&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-12-14T16:43:09.417" CommentCount="1" />
  <row Id="14311" PostTypeId="1" CreationDate="2010-12-14T16:59:42.010" Score="5" ViewCount="474" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Can a CFG over large alphabet describe a language, where each terminal appears even number of times?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If yes, would the Chomsky Normal Form be polynomial in |Σ| ?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;EDIT: What about a language where each terminal appears 0 or 2 times?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Solving this may give interesting extension to &lt;a href=&quot;http://math.stackexchange.com/questions/14506/restricted-read-twice-bdds-and-context-free-grammars&quot;&gt;&quot;Restricted read twice BDDs and context free grammars&quot;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Find a context-free grammar (CFG) for the language L of all&#xA;words such that each terminal in a word occurs even number of times over a possibly large alphabet Σ&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My long aproach is (the only nonterminal is S):&#xA;&lt;strike&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;S ⟶ ε | SS&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;x ∈ Σ : S ⟶ xSx&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;x,y ∈ Σ : S ⟶ xxSyy | yySxx | xySxy | xySyx | yxSyx | yxSxy&#xA;  &lt;/strike&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strike&gt;&#xA;Second try, incremental approach.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;S ⟶ ε | SS&lt;br&gt;&#xA;  Terminal productions (suboptimal): x,y ∈ Σ : S ⟶ xx | yy |xxyy | yyxx | xyxy | xyyx | yxyx | yxxy&lt;br&gt;&#xA;  Incremental: x ∈ Σ : S ⟶ SxSxS &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;By the way, the homework is over now.&#xA;&lt;/strike&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Technically&lt;/em&gt; it wasn't a homework assigned to me. I don't know the answer and thought the question was easy. &#xA;Best.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: I didn't award the bounty :-)&lt;/p&gt;&#xA;" OwnerUserId="4750" LastEditorUserId="4750" LastEditorDisplayName="" LastEditDate="2010-12-28T10:23:38.807" LastActivityDate="2010-12-31T21:58:47.790" Title="Can a polynomial size CFG over large alphabet describe a language, where each terminal appears even number of times?" Tags="&lt;homework&gt;&lt;computer-science&gt;&lt;computability&gt;&lt;formal-systems&gt;" AnswerCount="5" CommentCount="21" FavoriteCount="2" />
  <row Id="14312" PostTypeId="2" ParentId="14276" CreationDate="2010-12-14T17:06:21.730" Score="3" ViewCount="" Body="&lt;p&gt;What equivalence relation are you using to define motives?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The finest that people normally consider is rational equivalence, in which case&#xA;two morphisms whose graphs are rationally equivalent (which is to say, roughly speaking, two morphisms which can be deformed one into the other via a family parameterized by $\mathbb P^1$) will give the same morphism.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A trivial, but concrete, example: all constant morphisms from any variety to&#xA;$\mathbb P^1$ will be identified in the category of motives.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you use a coarser relation (homological equivalence or numerical equivalence, which are conjecturally the same, and lead to what are called &lt;em&gt;Grothendieck motives&lt;/em&gt;) than even more morphisms of varieties will be identified in the category of motives.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Incidentally, these functors are certainly not full, since one takes a Karoubian closure as part of the construction of the category of motives, and then furthermore inverts the Lefschetz motive.&lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2010-12-14T17:06:21.730" />
  <row Id="14313" PostTypeId="1" CreationDate="2010-12-14T17:16:32.840" Score="0" ViewCount="184" Body="&lt;p&gt;Let $R$ be a ring, $A$ is an ideal of $R$, and assume that $A\subseteq J(R)$.&#xA;Let $r\longmapsto \overline{r}$  denote the coset map $R\rightarrow \overline{R}=R/A$. The following are equivalent:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) $A$ is lifting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) If $\overline{R}=\overline{K_1}\oplus...\oplus \overline{K_n}$ where the $\overline{K_i}$ are left ideals of $\overline{R}$, there exists a decomposable $R=L_1\oplus...\oplus L_n$ where the $L_i$ are left ideals of $R$ such that $\overline{L_i}=\overline{K_i}$ for each $i=1,2,...,n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(3) Same as $(2)$ except $n=2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do you have any idea or any source for finding a prove?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;hr/&gt;&#xA;&lt;em&gt;Edit by Arturo Magidin:&lt;/em&gt; Above is the original question. The new phrasing is below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $R$ be a ring, $A$ is an ideal of $R$, and assume that $A\subseteq J(R)$. Let $r\longmapsto \overline{r}$  denote the coset map $R\rightarrow \overline{R}=R/A$&#xD;&#xA;If $A$ is lifting, for $\overline{R}=\overline{K_1}\oplus \overline{K_2}$ where the $\overline{K_i}$ are left ideals of $\overline{R}$, there exists a decomposable $R=L_1\oplus L_2$ where the $L_i$ are left ideals of $R$ such that $\overline{L_i}=\overline{K_i}$ for each $i=1,2$.&lt;/p&gt;&#xA;" OwnerUserId="4688" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-16T17:24:42.533" LastActivityDate="2010-12-16T17:24:42.533" Title="Ring Theory (Lift)" Tags="&lt;ring-theory&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="14314" PostTypeId="2" ParentId="14268" CreationDate="2010-12-14T17:16:59.610" Score="12" ViewCount="" Body="&lt;p&gt;Generally such comparisons can be done efficiently via continued fractions, e.g.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\rm\displaystyle a = \frac{847}{383}\ =\ 2 + \cfrac{1}{4 + \cfrac{1}{1 + \cdots}}\ \ \Rightarrow\ \ 2+\frac{1}{4+1} &amp;lt; a &amp;lt; 2+\frac{1}4$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\rm\displaystyle b = \frac{536}{254}\ =\ 2 + \cfrac{1}{9 + \cdots}\ \ \Rightarrow\ \ 2 &amp;lt; b &amp;lt; 2 + \frac{1}9 &amp;lt; 2+\frac{1}5 &amp;lt; a$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The comparison of the continued fraction coefficients can be done in parallel with the computation of the continued fraction. Namely, compare the integer parts. If they are unequal then that will determine the inequality. Otherwise, recurse on the (inverses of) the fractional parts (and note that inversion reverses the inequality). For the above example this yields:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\rm\displaystyle\ \frac{847}{383} &amp;gt; \frac{536}{254}\ \iff\  \frac{81}{383}&amp;gt;\frac{28}{254}\ \iff\ \frac{254}{28}&amp;gt;\frac{383}{81}\ \Leftarrow\ \ 9 &amp;gt; 4$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In words: to test if $\rm\:847/383 &amp;gt; 536/254\: $ we first compare their integer parts (floor). They both have integer part $\:2\:$ so we subtract $\:2\:$ from both and reduce to comparing their fractional parts $\rm\  81/383,\ \ 28/254\:.$ To do this we invert them and recurse. But since  inversion reverses inequalities $\rm\ x &amp;lt; y\ \iff\ 1/y &amp;lt; 1/x\ $ (for $\rm\:xy&gt;0\:$),  the equivalent inequality to test is if $\rm\ 254/28 &amp;gt; 383/81\:.\ $ Comparing their integer parts $\rm\:m,\:n\:$ we see since $\rm\ m &amp;gt; 5 &amp;gt; n\:$ so the inequality is true. This leads to the following simple algorithm that essentially compares any two real numbers via the lex order of their continued fraction coefficients (note: it will not terminate if given two equal reals with infinite continued fraction).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm compare\_reals\:(A,\: B)\ := \quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\color{blue}{\ // \ computes\ \  sgn(A - B) }$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\ let\ [\ n_1\leftarrow \lfloor A\rfloor\:;\ \ \ n_2\leftarrow \lfloor B\rfloor\ ] \ \quad\quad\quad\quad\quad\quad\ \color{blue}{\ //\ compare\  integer\ parts}$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\quad if\ \ n_1 \ne n_2\ \ then\ \ return\ \ sgn(n_1 - n_2)\:;$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\quad let\ [\ a \leftarrow A - n_1\:;\ \ \ b \leftarrow B - n_2\ ] \quad\quad\quad\quad \color{blue}{//\ compute\ fractional\ parts\ \ a,\: b }$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\quad\quad if\ \ a\:b=0\ \ then\ \ return\ \ sgn(a-b)\:;$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\quad\quad compare\_reals(b^{-1}\:, a^{-1})\:;\ \quad\quad\quad\quad\quad\quad\color{blue}{\ //\ \text{recurse on inverses of fractional parts}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Equivalently one can employ Farey fractions and mediants. Generally such approaches will be quite efficient due to the best approximations properties of the algorithm. For a nice example see my &lt;a href=&quot;http://groups.google.com/group/sci.math/msg/51d8687167920ebd?hl=en&quot; rel=&quot;nofollow&quot;&gt;post here&lt;/a&gt; using continued fractions to solve the old chestnut of comparing $\ 7^\sqrt 8\ $ to $\ 8^\sqrt 7\ $ and see also &lt;a href=&quot;http://groups.google.com/group/sci.math/browse_frm/thread/2182a45f22ed5d48&quot; rel=&quot;nofollow&quot;&gt;this thread&lt;/a&gt; where some folks struggled to prove this by calculus.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-12-30T20:21:31.627" LastActivityDate="2010-12-30T20:21:31.627" CommentCount="3" />
  <row Id="14315" PostTypeId="2" ParentId="14282" CreationDate="2010-12-14T17:26:47.587" Score="21" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Added.&lt;/strong&gt; So, what is the problem? Let's look at the simplest example of a non-normal subgroup. Take $G=S_3$, and $H=\{e,(1,2)\}$. If we compose permutations right to left, the left cosets of $H$ in $G$ are:&#xA;\begin{align*}&#xA;eH = (1,2)H &amp;amp;= \{ e, (1,2)\};\\&#xA;(1,2,3)H = (1,3)H &amp;amp;= \{(1,2,3), (1,3)\};\\&#xA;(1,3,2)H = (2,3)H &amp;amp;= \{(1,3,2), (2,3)\}.&#xA;\end{align*}&#xA;If we try multiplying cosets term-by-term, we run into problems. Multiplying by $eH$ is not a problem, but take $(1,2,3)H$ multiplied by itself. The products are:&#xA;$$\{ (1,2,3)(1,2,3), (1,2,3)(1,3), (1,3)(1,2,3), (1,3)(1,3)\} = \{(1,3,2), (2,3), (1,2), e\}$$&#xD;&#xA;which is not a coset. If we multiply using representatives, as in the original question,we also run into problems: if we multiply $(1,2,3)H\cdot(1,3,2)H$ as $(1,2,3)(1,3,2)H$, we get $eH$. But $(1,2,3)H = (1,3)H$, and $(1,3,2)H = (2,3)H$, and if we multiply them by looking at these alternative representatives/names, we get $(1,3)H\cdot (2,3)H = (1,3)(2,3)H = (1,3,2)H\neq eH$. That is, the multiplication rule depends on the &lt;em&gt;name&lt;/em&gt; we give the coset, rather than on what the coset &lt;em&gt;is&lt;/em&gt;. This means that the rule is not well-defined.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can let $G$ act on the set of left cosets of $H$, by $g(xH) = gxH$; this is not a group, but a group action. We can also, separtely, put group structures on the set of cosets of $H$ in $G$, since it will be a set, and every nonempty set can be given a group structure (assuming the Axiom of Choice; in fact, equivalent to the Axiom of Choice). But that structure will generally have nothing to do with the operation in $G$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we want an operation on the set of cosets, and have the operation in the cosets to be induced by the operation in $G$, then we have:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;THEOREM.&lt;/strong&gt; &lt;em&gt;Let $G$ be a group and $H$ a subgroup. Let $G/H = \{ gH\mid g\in G\}$ be the set of left cosets of $H$ in $G$. The operation&#xA;$$\cdot \colon G/H \times G/H \to G/H\qquad\text{defined by}\qquad (xH)\cdot(yH) = (xy)H$$&#xD;&#xA;is &lt;strong&gt;well defined&lt;/strong&gt; (that is, if $xH = x&amp;#39;H$ and $yH=y&amp;#39;H$ then $xyH = x'y'H$) if and only if $H$ is normal in $G$.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Proof.&lt;/em&gt; Suppose $H$ is normal. Then $gH=Hg$ for every $g\in G$. If $xH=x&amp;#39;H$ and $yH=y&amp;#39;H$, then &#xA;$$xyH = x(yH) = x(Hy) = x(Hy&amp;#39;) = (xH)y&amp;#39; = (x&amp;#39;H)y&amp;#39; = x&amp;#39;(Hy&amp;#39;) = x&amp;#39;(y&amp;#39;H) = x&amp;#39;y&amp;#39;H$$&#xD;&#xA;so the operation is well defined.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conversely, suppose the operation is well defined so that whenever $xH=x&amp;#39;H$ and $yH=y&amp;#39;H$, then $xyH = x&amp;#39;y&amp;#39;H$. We want to show that $g^{-1}Hg\subseteq H$ for all $g\in G$. For each $h\in H$, since $hH = eH$, we have $eHgH = egH=gH$ is equal to $hHgH = hgH$. So $gH = hgH$, hence $H=g^{-1}hgH$, so $g^{-1}hg\in H$ for each $h\in H$. Thus, $g^{-1}Hg\subseteq H$. This holds for any $g\in G$, so $g^{-1}Hg\subseteq H$ for all $g\in G$, hence $H\triangleleft G$, as claimed. &lt;strong&gt;QED&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And now I'm going to hijack your question a bit and go on a long excursion to a closely related subject which is not necessary to answer your question, but which I think is interesting. (Sorry!)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is really connected to the notion of a &lt;em&gt;congruence&lt;/em&gt; on an algebra (in the sense of universal algebra). A subgroup $H$ defines an equivalence relation on $G$ by $x\sim y$ if and only if $xH = yH$ (it also defines the equivalence relation of right congruence, but ignore that for now). Being an equivalence relation, it corresponds to a subset $\Phi$ of $G\times G$, with $(x,y)\in\Phi$ if and only if $x\sim y$, if and only if $xH=yH$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But $G\times G$ also has a natural group structure inherited from $G$. Since $\Phi$ is a subset of the group $G\times G$, it makes sense to ask whether $\Phi$ is a subgroup of $G\times G$ or not. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;THEOREM.&lt;/strong&gt; &lt;em&gt;$\Phi$ is a subgroup of $G\times G$ if and only if $H$ is normal in $G$.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Proof.&lt;/em&gt; Note that $\Phi$ always contains $(e,e)$ (in fact, the entire diagonal). Assume first that $H$ is normal. If $(x,y)\in \Phi$, then $xH=yH$. Hence, $Hx = Hy$, and taking inverses of each element we get $x^{-1}H = y^{-1}H$, so $(x^{-1},y^{-1})\in\Phi$; thus, $\Phi$ is closed under inverses. If $(x,y),(z,w)\in\Phi$, then $xH=yH$, $zH=wH$, so by what we proved above we have $xzH = ywH$, so $(xz,yw)=(x,y)(z,w)\in\Phi$. Thus, $\Phi$ is closed under products, so $\Phi$ is a subgroup of $G\times G$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conversely, suppose that $\Phi$ is a subgroup. If $xH=yH$ and $zH=wH$, then $(x,y),(z,w)\in \Phi$, hence $(x,y)(z,w)=(xz,yw)\in\Phi$; thus, $xzH = ywH$. This proves that the operation $xH\cdot zH \mapsto xzH$ is well-defined, so by the theorem above $H$ is a normal subgroup of $G$. &lt;strong&gt;QED&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;hr/&gt;&#xA;&lt;strong&gt;Added.&lt;/strong&gt; (Stuff between the two horizontal lines was added in this edit). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Okay, how about this? Can we define &lt;em&gt;some&lt;/em&gt; equivalence relation $\sim$ on a group $G$ so that we can multiply equivalence classes through representatives, have it be well-defined, but not have $\sim$ &quot;come from&quot; a normal subgroup?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is: suppose $\sim$ is an equivalence relation on the group $G$ (&lt;em&gt;any&lt;/em&gt; equivalence relation; all we require is that it be reflexive, symmetric, and transitive on the underlying set of $G$), and let $[x]$ denote the equivalence class of $x$ for each $x$ in $G$. We want to define a multiplication on the quotient set $G/\sim$ by $[x][y] = [xy]$. When is this well defined?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;THEOREM.&lt;/strong&gt; &lt;em&gt;Let $G$ be a group, and let $\sim$ be an equivalence relation on $G$. Then the multiplication on $G/\sim$ defined by $[x][y]=[xy]$ is well defined if and only if $\sim$ is a subgroup of $G\times G$.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Proof.&lt;/em&gt; If $\sim$ is a subgroup of $G\times G$ and $[x]=[z]$, $[y]=[w]$, then $(x,z),(y,w)\in \sim$, so $(x,z)(y,w) = (xy,zw)\in \sim$, hence $[xy]=[zw]$, so multiplication is well defined.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conversely, suppose the multiplication is well defined. Since $\sim$ is an equivalence relation, $(e,e)\in\sim$. Also, if $(x,z), (y,w)\in\sim$, then $[x]=[z]$, $[y]=[w]$, so $[x][y]=[xy]$ equals $[z][w]=[zw]$, so $(xy,zw)=(x,y)(z,w)\in\sim$. Thus, $\sim$ is a submonoid of $G\times G$. It only remains to show that $\sim$ is closed under inverses. Suppose that $(x,y)\in\sim$. Multiplying by $(x^{-1},x^{-1})$ on the left and $(y^{-1},y^{-1})$ on the right (both elements of $\sim$) we have $(y^{-1},x^{-1})\in\sim$. Since $\sim$ is an equivalence relation, it is reflexive, so $(x^{-1},y^{-1})\in\sim$. Thus, $\sim$ is a subgroup of $G\times G$, as claimed. &lt;strong&gt;QED&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Okay, so we can define the multiplication on the quotient set if and only if the equivalence relation is a subgroup of $G\times G$. Do all such equivalence relations &quot;come&quot; from subgroups? Yes:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;THEOREM.&lt;/strong&gt; &lt;em&gt;Let $\sim$ be an equivalence relation on $G$ that is a subgroup of $G\times G$, and let $H=\{g\in G\mid g\sim e\}$. Then $H$ is a normal subgroup of $G$, and $x\sim y$ if and only if $xH=yH$.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Proof.&lt;/em&gt; We can identify $G$ with the subgroup $G\times\{e\}$ of $G\times G$; then $H$ corresponds to the intersection of the subgroup $\sim$ and the subgroup $G\times\{e\}$, hence is a subgroup of $G$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To see that $\sim$ is in fact left congruence modulo $H$, simply note that $xH=yH$ if and only if $y^{-1}x\in H$, if and only if $(y^{-1}x,e)\in \sim$, if and only if $(y,y)(y^{-1}x,e)\in\sim$, if and only if $(x,y)\in \sim$, if and only if $x\sim y$. That $H$ is normal now follows because multiplication of cosets (equivalently, of equivalence classes modulo $\sim$) is well defined. &lt;strong&gt;QED&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;hr/&gt;&#xA;When we generalize this to an arbitrary &lt;em&gt;algebra&lt;/em&gt; (in the sense of universal algebra; examples include magmas, monoids, semigroups, groups, rings, $K$-algebras, lattices, etc), we get the notion of &lt;em&gt;congruence&lt;/em&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Definition.&lt;/strong&gt; Let $A$ be an &lt;em&gt;algebra&lt;/em&gt; (in the sense of universal algebra). A &lt;em&gt;congruence on $A$&lt;/em&gt; is an equivalence relation $\Phi$ on $A$ such that $\Phi$ is a subalgebra of $A\times A$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;THEOREM.&lt;/strong&gt; &lt;em&gt;Let $A$ and $B$ be algebras of the same type, and let $f\colon A\to B$ be a homomorphism; define $\Phi_f = \{(x,y)\in A\times A\mid f(x)=f(y)\}$. Then $\Phi_f$ is a congruence on $A$. Conversely, if $\Psi$ is a congruence on $A$, then there exists an algebra $\mathcal{B}$ and a homomorphism $\pi\colon A\to \mathcal{B}$ such that $\Psi=\Phi_{\pi}$.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Special case.&lt;/em&gt; Let me do the proof with semigroups. First, let $f\colon A\to B$ be a semigroup homomorphism; then $\Phi_f$ is an equivalence relation on $A$. To show it is a subsemigroup of $A\times A$, suppose $(x,y),(z,w)\in \Phi_f$. Then $f(xz) = f(x)f(z) = f(y)f(w) = f(yw)$, so $(xz,yw)\in\Phi_f$. Thus, $\Phi_f$ is a subsemigroup of $A\times A$, hence a congruence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conversely, suppose that $\Psi$ is a congruence. Let $\mathcal{B}=A/\Psi$, the set of equivalence classes of $\Psi$ in $A$. Let $[x]$ denote the class of $x$. We define a multiplication on $\mathcal{B}$ by $[a][b]=[ab]$. We must prove that this is well defined: suppose $[x]=[y]$ and $[z]=[w]$. Then $(x,y),(z,w)\in\Psi$, so $(xz,yw)\in\Psi$ (since $\Psi$ is a congruence). Therefore, $[xz]=[yw]$, so multiplication is well-defined. It is now easy to verify that multiplication is associative:&#xA;$$[x]([y][z]) = [x][yz] = [x(yz)]=[(xy)z] = [xy][z] = ([x][y])[z].$$&#xD;&#xA;Thus, $\mathcal{B}$ is a semigroup. The map $\pi\colon A\to \mathcal{B}$ given by $\pi(a)=[a]$ is easily verified to be a homomorphism, and $\Phi_{\pi}=\Psi$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The proof for an arbitrary algebra with operations $f_1,\ldots,f_k$ of arities $a_1,\ldots,a_k$ (or even an infinite number of operations, each with finite arities) is essentially the same.&#xA;&lt;strong&gt;QED&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One can then prove the Isomorphism Theorems for &quot;congruences&quot; in exactly the analogous manner as for groups and rings, for any class of algebras.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In Groups and Rings, we can &quot;get away&quot; with not studying congruences because if $\Phi$ is a congruence on $G$, then we can let $N = \{ g\in G\mid (g,e)\in\Phi\}$, and it turns out that $\Phi$ is the equivalence relation induced by $N$. Essentially, a congruence tells you that certain things are equal, but in groups (and rings), instead of saying &quot;$x$ is equal to $y$&quot;, we can say &quot;$xy^{-1}$ is the identity&quot;. The normal subgroup (resp. the ideal) is the set of elements equivalent to the identity. In more general classes, such as semigroups, we don't have the luxury of expressing any identity in the form $w=e$, so we must actually keep track of both sides of the equality.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is all part of &lt;em&gt;Universal Algebra&lt;/em&gt;. I cannot recommend &lt;a href=&quot;http://math.berkeley.edu/~gbergman/245/&quot; rel=&quot;nofollow&quot;&gt;George Bergman's book&lt;/a&gt; highly enough if you want a great introduction to the subject (with a Category Theory flavor).&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-16T18:18:38.393" LastActivityDate="2010-12-16T18:18:38.393" CommunityOwnedDate="2010-12-16T18:18:38.393" CommentCount="4" />
  <row Id="14316" PostTypeId="2" ParentId="14297" CreationDate="2010-12-14T17:31:08.380" Score="3" ViewCount="" Body="&lt;p&gt;Let $X_n$ be your winnings in a game of length $n$ (in your case $n = 6$), if you are playing optimally.  Here, &quot;optimally&quot; means that at roll $m$, you will accept if the value is greater than $\mathbb{E} X_{n-m}$, which is your expected winnings if you continued to play with this strategy.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $X \sim \mathrm{Unif}(1,2,3,4,5,6) $ (you can also insert any distribution you like here).  Then $X_n$ can be defined as $X_1 = X$ and for $n \geq 2$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ X_n = \begin{cases}  X_{n-1}, \quad \mathrm{if} \quad X &amp;lt; \mathbb{E}X_{n-1} \\&#xD;&#xA;X \quad \mathrm{if} \quad X \geq \mathbb{E}X_{n-1} \end{cases} $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So your decisions can be determined by computing $\mathbb{E} X_n$ for each $n$ recursively.  For the dice case,  $\mathbb{E} X_1 = \mathbb{E}X = 7/2$ (meaning on the fifth roll, accept if you get &gt;7/2, or 4,5 or 6), and so,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathbb{E} X_2 = \mathbb{E} X_1 \mathrm{P}[X = 1,2,3] + \mathbb{E} [X | X \geq 4] \mathrm{P}[X = 4,5,6]$$&#xD;&#xA;$$ = (7/2)(3/6) + (4 + 5  + 6)/3 (1/2) = 17/4 $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So on the fourth roll, accept if you get &gt; 17/4, or 5 or 6,  and so on (you need to round the answer up at each step, which makes it hard to give a closed form for $\mathbb{E} X_n$ unfortunately).  &lt;/p&gt;&#xA;" OwnerUserId="2515" LastEditorUserId="2515" LastEditorDisplayName="" LastEditDate="2010-12-14T17:39:57.733" LastActivityDate="2010-12-14T17:39:57.733" />
  <row Id="14317" PostTypeId="2" ParentId="14313" CreationDate="2010-12-14T18:03:02.987" Score="3" ViewCount="" Body="&lt;p&gt;Suppose you can lift idempotents modulo $A$. If $\bar R=I_1\oplus I_2$ as a left $\bar R$-module, there is an idempotent $e\in\operatorname{End}_{\bar R}(\bar R)=\bar R$, given by projecting onto $I_1$, and we have $I_1=\bar R\cdot e$ and $I_2=\bar R\cdot(1-e)$. Lift $e$ to an idempotent $\tilde e\in R$, and let $J_1=R\cdot \tilde e$ and $J_2=R\cdot(1-\tilde e)$. Etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This will give $(1)\implies(3)$. The rest of the implications are of the same spirit.&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2010-12-14T18:03:02.987" />
  <row Id="14318" PostTypeId="2" ParentId="14276" CreationDate="2010-12-14T18:09:00.090" Score="1" ViewCount="" Body="&lt;p&gt;Different varieties with the same zeta function would likely (and maybe provably in the case of varieties over a finite field) provide examples of non-faithfulness on objects.  The zetas are defined without direct reference to motives or &quot;motivic structures&quot; on cohomology.     &lt;/p&gt;&#xA;" OwnerUserId="467" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-12-14T18:14:03.597" LastActivityDate="2010-12-14T18:14:03.597" />
  <row Id="14319" PostTypeId="2" ParentId="13937" CreationDate="2010-12-14T18:19:08.903" Score="1" ViewCount="" Body="&lt;p&gt;Let $x_{d+1}$ denote $|a \cdot x|$, so that ${\displaystyle {1 \over |a \cdot x| x_1....x_d} = {1 \over x_1....x_{d+1}}}$. &#xA;Then by the arithmetic-geometric inequality, for any $x_1,...,x_{d+1}$ one has&#xA;$${1 \over x_1....x_{d+1}} \leq {1 \over d+1} \sum_{i = 1}^{d+1} \prod_{j \neq i} x_j^{-{d+1 \over d}}$$&#xD;&#xA;So summing one has&#xA;$$\sum_{x_1,...,x_d}{1 \over x_1....x_{d+1}} \leq {1 \over d+1} \sum_{i = 1}^{d+1} \sum_{x_1,...,x_d}&#xD;&#xA;\prod_{j \neq i} x_j^{-{d+1 \over d}}$$&#xD;&#xA;We examine a given term ${\displaystyle \sum_{x_1,...,x_d}\prod_{j \neq i} x_j^{-{d+1 \over d}}}$. If $i = d+1$, then one is &#xA;adding ${\displaystyle \prod_{j = 1}^d x_j^{-{d+1 \over d}}}$ over positive integers $x_1,..x_d$ and one gets a finite result since &#xA;${d + 1 \over d} &amp;gt; 1$. If $i \neq d+1$, then we sum ${\displaystyle \sum_{x_1,...,x_d}\prod_{j \neq i} x_j^{-{d+1 \over d}}}$&#xD;&#xA;in the $x_i$ variable first, fixing the other variables. Only the $x_{d+1}$ &#xA;factor depends on $x_i$, and this $x_i$ variable sum is&#xA;$$\sum_{x_i} {1 \over |a_1x_1 + ... + a_nx_n|^{d+1 \over d}} \prod_{j \neq i} x_j^{-{d+1 \over d}}$$&#xD;&#xA;The sum is a portion of the sum of $|x|^{-{d+1 \over d}}$ over nonzero integers $x$, so the result is bounded by some fixed &#xA;constant $C$. So the overall sum is at most&#xA;$$C \sum_{x_1,...,x_d}  \prod_{j \neq i} x_j^{-{d+1 \over d}}$$&#xD;&#xA;In the sum above the $x_i$ variable is excluded.&#xA;Like above in the $i = d+1$ case, since the exponent ${d+1 \over d}$ is greater than one the sum is finite and we are done.&lt;/p&gt;&#xA;" OwnerUserId="3035" LastEditorUserId="3035" LastEditorDisplayName="" LastEditDate="2010-12-15T15:09:53.383" LastActivityDate="2010-12-15T15:09:53.383" CommentCount="6" />
  <row Id="14320" PostTypeId="1" CreationDate="2010-12-14T18:25:58.163" Score="19" ViewCount="312" Body="&lt;p&gt;It is known that, for $n \geq 5$, it is possible to partition the integers $\{1, 2, \ldots, n\}$ into two disjoint subsets such that the product of the elements in one set equals the sum of the elements in the other.  One solution is the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $N = \{1, 2, \ldots, n\}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $n$ is even, take $P = \{1, \frac{n-2}{2}, n\}$ and $N-P$ as the two sets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $n$ is odd, take $P = \{1, \frac{n-1}{2}, n-1\}$ and $N-P$ as the two sets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is this: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Is this partition unique for infinitely many $n$?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;One might be able to prove an even stronger result, as I don't know of any other solutions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Background: The problem of proving that the partition is possible was posed several years ago as Problem 2826 in the journal &lt;em&gt;Crux Mathematicorum&lt;/em&gt;, with solutions in the &lt;a href=&quot;http://math.ca/crux/v30/n3/page177-192.pdf&quot; rel=&quot;nofollow&quot;&gt;April 2004 issue&lt;/a&gt;.  Every one of the 20 or so solvers (including me, which is why I'm interested in the question) came up with the partition given here.  The person who originally posed the problem also asked if the partition is unique for infinitely many $n$.  I don't think anyone ever submitted an answer to that latter question to &lt;em&gt;Crux&lt;/em&gt; (although I cannot verify that, as I no longer have a subscription).  I thought someone here might be able to give an answer.&lt;/p&gt;&#xA;" OwnerUserId="2370" LastActivityDate="2010-12-15T22:04:01.737" Title="Partitioning the integers $1$ through $n$ so that the product of the elements in one set is equal to the sum of the elements in the other" Tags="&lt;number-theory&gt;&lt;discrete-mathematics&gt;" AnswerCount="1" CommentCount="8" FavoriteCount="4" />
  <row Id="14321" PostTypeId="2" ParentId="14296" CreationDate="2010-12-14T19:10:04.507" Score="3" ViewCount="" Body="&lt;p&gt;The dynamics are those of a repeated constant rotation, $\theta \to \theta + c$ where $c$ is the angle on the circle (completed to fill in the aperture) cut out by the line on which the light entered.  Exit will occur after $n$ reflections when $\theta_1 + nc$ is in the aperture interval, where $\theta_1$ is the point of first impact between light and mirror.  This type of dynamics is integrable, not chaotic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the ring-shaped mirror is not a perfect circle but a more general convex curve this goes under the name &quot;convex billiards&quot;, or elliptical billiards in the case of an ellipse-shaped mirror.  The dynamics can be more interesting (e.g., for polygonal mirrors) in those cases and the book by Tabachnikov cited in the other answer discusses all of these more general problems. &lt;/p&gt;&#xA;" OwnerUserId="467" LastActivityDate="2010-12-14T19:10:04.507" CommentCount="1" />
  <row Id="14322" PostTypeId="2" ParentId="14276" CreationDate="2010-12-14T19:49:21.290" Score="5" ViewCount="" Body="&lt;p&gt;First of all, I wouldn't equate &quot;captures everything&quot; with &quot;faithful functor&quot;.  The functor from groups to sets which forgets the multiplication structure is faithful, but I wouldn't say that knowing a group has 512 elements tells you everything about it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;An example of nonisomorphic varieties with isomorphic motives: Consider $\mathbb{P}^2$ blown up at $5$ points. Blowing up four points (in general position) rigidifies the $\mathbb{P}^2$. There is then a two dimensional space of moduli for where you choose the fifth point. All of these nonisomorphic surfaces have isomorphic motives. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;An example of the functor to motives not being faithful: The automorphism group of $\mathbb{P}^1$ is $PGL_2$. All of these nontrivial automorphisms act trivially on the motive.&lt;/p&gt;&#xA;" OwnerUserId="448" LastActivityDate="2010-12-14T19:49:21.290" />
  <row Id="14323" PostTypeId="2" ParentId="14209" CreationDate="2010-12-14T19:52:28.307" Score="1" ViewCount="" Body="&lt;p&gt;We first apply the Taylor-expansion for $f(x_{k+1})$ around $x_k$:&#xA;$$f(x_{k+1})=f(x_k) + f&amp;#39;(x_k)(x_{k+1}-x_k)+R_2$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $R_2$ is the remainder. We'll take Lagrange's remainder and we get to:&#xA;$$f(x_{k+1}) = f&amp;#39;(x_k) \cdot (\frac{f(x_k)}{f&amp;#39;(x_k)} - x_k + x_{k+1}) + \frac{1}{2} f&amp;#39;&amp;#39;(x_0)(x_{k+1} - x_k)^2$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where per definition $\frac{f(x)}{f&amp;#39;(x)}-x_k = -x_{k+1}$ and $x_0$ is some point between $x_{k+1}$ and $x_k$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the first term vanishes and we can write:&#xA;$$|f(x_{k+1})| \leq \frac{1}{2} \max_{a \leq x \leq b} |f&amp;#39;&amp;#39;(x)| (x_{k+1}-x_k)^2$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, from the mean value theorem we know that:&#xA;$$\min_{a \leq x \leq b} |f&amp;#39;(x)| \leq \frac{|f(x_{k+1} - f(\xi)|}{|x_{k+1} - \xi|} = \frac{|f(x_{k+1}|}{|x_{k+1} - \xi|}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It follows:&#xA;$$|\xi - x_{k+1}| \leq \frac{1}{2} \frac{\max_{a \leq x \leq b} |f&amp;#39;&amp;#39;(x)|}{\min_{a \leq x \leq b} |f&amp;#39;(x)|} (x_{k+1}-x_k)^2$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;q.e.d.&lt;/p&gt;&#xA;" OwnerUserId="3787" LastActivityDate="2010-12-14T19:52:28.307" CommentCount="1" />
  <row Id="14324" PostTypeId="1" AcceptedAnswerId="14331" CreationDate="2010-12-14T20:09:09.143" Score="0" ViewCount="83" Body="&lt;p&gt;If i have a graph in this form (adjacency matrix): &lt;/p&gt;&#xA;&#xA;&lt;p&gt;0 0 1 0 0 0 1&lt;br /&gt;&#xA;0 0 0 1 0 1 0&lt;br /&gt;&#xA;1 0 0 1 0 0 0&lt;br /&gt;&#xA;0 1 1 0 0 0 0&lt;br /&gt;&#xA;0 0 0 0 0 1 1&lt;br /&gt;&#xA;0 1 0 0 1 0 0&lt;br /&gt;&#xA;1 0 0 0 1 0 0&lt;br /&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a 2-regular graph with 7 vertices. What are examples of vertex invariants? Usually i would look at the degrees of each vertex, but they are all the same in this case.&lt;/p&gt;&#xA;" OwnerUserId="43" LastActivityDate="2011-01-13T22:44:52.010" Title="Vertex Invariants for a 2-regular Graph" Tags="&lt;graph-theory&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="14325" PostTypeId="2" ParentId="13053" CreationDate="2010-12-14T20:09:55.413" Score="1" ViewCount="" Body="&lt;p&gt;This seems to work:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;dzeta[x_] := Derivative[1, 0][Zeta][y, x]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Plot[{Exp[dzeta[0, x] + 1/2 Log[2 Pi]] , Gamma[x]}, {x, -3, 3}]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The derivative of a function returns a function. I think that in your version, maybe Mathematica binds x to specific numerical values during the plot and then tries to take the derivative. The version above computes the derivative once and for all and then does the plot.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Oops, Annix is right. This is not the first time I have been undone by mistaking the graph color for two graphs instead of one. Anyway, I'm sorry for the false starts, but the following code properly displays both functions and the graphs coincide.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clear[x, y, dzeta]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;dzeta[y_, x_] := Derivative[1, 0][HurwitzZeta][y, x]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Plot[{Exp[dzeta[0, x] + 1/2 Log[2 Pi]], Gamma[x]}, {x, -3, 3}]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Changing &quot;Zeta&quot; to &quot;HurwitzZeta&quot; in your original expression also works:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Plot[{Exp[&#xA;   Derivative[1, 0][HurwitzZeta][0, x] + 1/2 Log[2 Pi]]}, {x, -3, 3}]&lt;/p&gt;&#xA;" OwnerUserId="1123" LastEditorUserId="1123" LastEditorDisplayName="" LastEditDate="2010-12-15T07:10:54.083" LastActivityDate="2010-12-15T07:10:54.083" CommentCount="4" />
  <row Id="14326" PostTypeId="2" ParentId="14311" CreationDate="2010-12-14T20:14:23.573" Score="4" ViewCount="" Body="&lt;p&gt;There is a deterministic finite state machine that accepts a word if and only if each terminal appears an even number of times. Assume that there are $k$ symbols in the alphabet. The machine has $2^k$ states, each of which is associated with a subset of the alphabet, and we identify the states with the subsets. The machine is defined so that: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The initial state corresponds to the empty set, and this is the only accepting state.&lt;/li&gt;&#xA;&lt;li&gt;The transition rules are defined as follows. From state $A$, on input $l$, if $l \in A$ then let $B = A - \{l\}$ and move to state $B$. Otherwise let $B = A \cup \{l\}$ and move to state $B$. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The idea is that a state $B$ corresponds to the set of terminals that have appeared an odd number of times so far. Being in state $\emptyset$ corresponds to every terminal appearing an even number of times. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the language is accepted by a DFSM, the language is regular.  This means that there is, trivially, a CFG that accepts the language. The grammar has one nonterminal for each state of the machine, and the rules of the grammar match up with transitions of the machine. I don't know yet whether there is a context free grammar for this language that is polynomial size compared to the size of the alphabet, though. &lt;/p&gt;&#xA;" OwnerUserId="630" LastEditorUserId="630" LastEditorDisplayName="" LastEditDate="2010-12-14T21:20:47.177" LastActivityDate="2010-12-14T21:20:47.177" CommentCount="10" />
  <row Id="14327" PostTypeId="1" AcceptedAnswerId="14468" CreationDate="2010-12-14T20:31:12.427" Score="7" ViewCount="136" Body="&lt;p&gt;Consider the game described &lt;a href=&quot;http://math.stackexchange.com/questions/14297/best-strategy-for-a-die-game&quot;&gt;here&lt;/a&gt;, but for a sequence $X_1,\ldots,X_n$ of i.i.d. uniform rv's on $\lbrace 1,\ldots,n \rbrace$ (in the original game $n=6$). Using the original notation, let $a_n$ denote the first element in the best strategy $a_n,\ldots,a_2$. We saw that for $n=6$, $a_n = n$. Can you provide a heuristic explanation as to why $a_n &amp;lt; n$ for all sufficiently large $n$ (this is indicated by numerical results), or even much better, can you determine the behavior of $n - a_n$ as $n \to \infty$? No rigorous proof is required, only heuristic ideas.&lt;/p&gt;&#xA;" OwnerUserId="2810" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-12-15T05:18:46.987" LastActivityDate="2010-12-16T03:54:30.380" Title="Asymptotic behavior of the first step in a best strategy" Tags="&lt;probability&gt;&lt;sequences-and-series&gt;&lt;asymptotics&gt;" AnswerCount="3" FavoriteCount="3" />
  <row Id="14328" PostTypeId="1" AcceptedAnswerId="14344" CreationDate="2010-12-14T20:47:46.293" Score="11" ViewCount="135" Body="&lt;p&gt;A Lasker ring is a ring in which every ideal has a primary decomposition. The Lasker-Noether theorem states that every commutative Noetherian ring is a Lasker ring (as an easy consequence of the ascending chain condition). And I've found the statement that there are non-Noetherian Lasker rings, but I can't find an example.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any ideas?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;edit: As the tag already suggested, I'm particularly interested in a commutative non-Noetherian Lasker ring, but noncommutative examples are also welcome. It never hurts to know more counterexamples.&lt;/p&gt;&#xA;" OwnerUserId="77" LastEditorUserId="77" LastEditorDisplayName="" LastEditDate="2010-12-14T21:59:17.573" LastActivityDate="2010-12-16T16:21:43.283" Title="A non-Noetherian Lasker ring" Tags="&lt;commutative-algebra&gt;&lt;ring-theory&gt;" AnswerCount="2" FavoriteCount="2" />
  <row Id="14329" PostTypeId="2" ParentId="13991" CreationDate="2010-12-14T21:01:42.240" Score="2" ViewCount="" Body="&lt;p&gt;Phillipe Flajolet's &lt;a href=&quot;http://algo.inria.fr/flajolet/Publications/books.html&quot; rel=&quot;nofollow&quot;&gt;Analytic Combinatorics&lt;/a&gt; is a good resource for this. (first book on the page, pdf)&lt;/p&gt;&#xA;" OwnerUserId="890" LastActivityDate="2010-12-14T21:01:42.240" CommentCount="1" />
  <row Id="14330" PostTypeId="2" ParentId="14275" CreationDate="2010-12-14T21:31:17.577" Score="3" ViewCount="" Body="&lt;p&gt;Looking at &lt;a href=&quot;http://projecteuclid.org/DPubS/Repository/1.0/Disseminate?view=body&amp;amp;id=pdf_1&amp;amp;handle=euclid.rmjm/1181069828&quot; rel=&quot;nofollow&quot;&gt;&#xA;the paper&lt;/a&gt; referenced by Robin Chapman and Bill Dubuque in response to &lt;a href=&quot;http://math.stackexchange.com/questions/14270/if-r-is-a-commutative-ring-with-identity-and-a-b-are-its-elements-that-are-divi&quot;&gt;the recent closely related question&lt;/a&gt;, it seems to me that the answer is &quot;yes&quot;, though it may very well fall into mosquito-nuking territory. I hope what follows is not in error!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A commutative ring with identity $R$ is called &lt;em&gt;pr&amp;eacute;simplifiable&lt;/em&gt; if and only if $xy=x$ implies $x=0$ or $y\in U(R)$ ($y$ is a unit). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $R$ is pr&amp;eacute;simplifiable, then $(a)=(b)$ implies $a$ and $b$ are associates: we know that $a=bx$ for some $x$, and $b=ay$ for some $y$, so $a=bx=a(xy)$. Since $R$ is pr&amp;eacute;simplifiable, either $a=0$ or $xy$ is a unit. If $a=0$, then $(a)=(0)=(b)$, so $b=0$, hence $b=a1$. If $xy$ is a unit, then $y$ is a unit, so $b=ay$ is an associate of $y$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The paper cited above cites another paper for the following result:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Theorem.&lt;/strong&gt; $R[X]$ is pr&amp;eacute;simplifiable if and only if $R$ is pr&amp;eacute;simplifiable and $(0)$ is a primary ideal of $R$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Getting back to your question. First, consider the case in which $n$ is a prime power, $n=p^n$. Note that $R=\mathbb{Z}/p^n\mathbb{Z}$ is pr&amp;eacute;simplifiable: if $ab\equiv a \pmod{p^n}$, then $p^n|a(b-1)$. If $p^n|a$, then $\overline{a}=0$ (where $\overline{a}$ is the image of $a$ in $R$). If $p^n$ does not divide $a$, then $b-1$ must be a multiple of $p$. But if $b-1$ is a multiple. of $p$, then $b$ is relatively prime to $p$, hence $\overline{b}$ is a unit in $R$. Thus, if $\overline{a}\overline{b}=\overline{a}$ in $R$, then either $\overline{a}=0$ or $\overline{b}$ is a unit. Also, $(0)$ is a primary ideal of $\mathbb{Z}/p^n\mathbb{Z}$. If $\overline{a}\overline{b}\in(0)$ and $\overline{a}\notin (0)$, then $p|b$, hence some power of $b$ is a multiple of $p^n$, so there exists $j\gt 0$ with $\overline{b}^j\in(0)$. By the theorem quoted above, it follows that $\mathbb{Z}/p^n\mathbb{Z}[X]$ is pr&amp;eacute;simplifiable. In particular, if $a$ and $b$ in $\mathbb{Z}/p^n\mathbb{Z}[x]$ generate the same ideal, then they are associates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the property holds for $n$ a prime power, because then the polynomial ring is pr&amp;eacute;simplifiable, which is a stronger condition. However, when trying to go from prime powers to arbitrary composites, we are not so lucky. For $n$ composite but not a prime power, the ideal $(n)$ of $\mathbb{Z}$ is not primary, so the ideal $(0)$ of $\mathbb{Z}/n\mathbb{Z}$ is not primary, hence $\mathbb{Z}/n\mathbb{Z}[X]$ is not pr&amp;eacute;simplifiable. For instance, $2=2(4)$ in $\mathbb{Z}/6\mathbb{Z}[X]$, but we have neither $2=0$ nor $4$ a unit. But nonetheless I believe the result holds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To try to get the result for arbitrary $n$, the Chinese Remainder Theorem suggests itself instead. Factor $n$ as a product of prime powers, $n=p_1^{a_1}\cdots p_r^{a_r}$. Suppose that $(a)=(b)$ in $\mathbb{Z}/n\mathbb{Z}[X]$. Moding out by the ideal $(p_i^{a_i})$, we get $(\overline{a})=(\overline{b})$ in $\mathbb{Z}/p_i^{a_i}\mathbb{Z}[X]$, so by the result for prime powers we know that there is a unit $\overline{u}_i$ of $\mathbb{Z}/p_i^{a_i}\mathbb{Z}[X]$ such that $\overline{a}\overline{u}_i=\overline{b}$. Using the Chinese Remainder Theorem we obtain a unit $u\in\mathbb{Z}/n\mathbb{Z}[X]$ such that $u\equiv u_i \pmod{(p_i^{a_i})}$ for each $i$, hence $au=b$ (to see it is a unit, just lift the inverses to get an inverse for $u$; to see that we have equality $au=b$, note that the Chinese Remainder Theorem guarantees it holds modulo the gcd of the ideals, but $(p_1^{a_1})\cdots(p_r^{a_r}) = (0)$ in $\mathbb{Z}/n\mathbb{Z}[X]$, so we actually have equality). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So it would seem, if I did not make some mistake, that for these rings you do have that $(a)=(b)$ if and only if $b=au$ for some unit $u$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;hr/&gt;&#xA;I think we can avoid using the sledgehammer theorem quoted above to show that if $p$ is a prime, then $\mathbb{Z}/p^n\mathbb{Z}[X]$ is pr&amp;eacute;simplifiable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose that $a,b\in\mathbb{Z}/p^s\mathbb{Z}[X]$ are such that $ab=a$ and $a\neq 0$. We have $a(b-1)=0$; lift them to $\mathbb{Z}[X]$ and write:&#xA;\begin{align*}&#xA;a &amp;amp;= a_0 + a_1x + \cdots +a_mx^m\\&#xA;b-1 &amp;amp;= b_0 + b_1x + \cdots b_mx^m.&#xA;\end{align*}&#xA;I claim that $p|b_i$ for each $i$. Indeed, since the content of $a(b-1)$ is a multiple of $p^n$, but the content of $a$ is not a multiple of $p^n$, then the content of $b-1$ must be a multiple of $p$ (since the content of the product is the product of the contents). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, the content of $(b-1)^{n}$ is divisible by $p^n$, hence projecting down to $\mathbb{Z}/p^n\mathbb{Z}[X]$ we have that $(b-1)^n = 0$. Therefore,&#xA;$$b\left(b^{n-1} - \binom{n}{1}b^{n-2} + \cdots + (-1)^{n-1}\binom{n}{n-1}\right) = (-1)^{n}$$&#xD;&#xA;hence $b$ is a unit in $\mathbb{Z}/p^n\mathbb{Z}[X]$. Thus, if $ab=a$, then either $a=0$ or $b$ is a unit, and we conclude that $\mathbb{Z}/p^n\mathbb{Z}[X]$ is pr&amp;eacute;simplifiable.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-15T06:16:19.003" LastActivityDate="2010-12-15T06:16:19.003" CommentCount="4" />
  <row Id="14331" PostTypeId="2" ParentId="14324" CreationDate="2010-12-14T21:56:28.920" Score="3" ViewCount="" Body="&lt;p&gt;2-regular (simple) graphs on n vertices are determined (up to isomorphism) by the partition of n induced by the size of its connected components.  This is because connected components of 2-regular graphs must be cycles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conversely, given a partition $d_1,d_2,\ldots,d_k$ of n into parts of size at least 3 (you can't have cycles of length 1 or 2, which would induce loops and multiedges) then you can construct a 2-regular graph consisting of a $d_i$-cycle for each $d_i$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the above case, the partition is 7 (since the graph is a 7-cycle).&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;0 0 1 0 0 0 1&#xA;0 0 0 1 0 1 0&#xA;1 0 0 1 0 0 0&#xA;0 1 1 0 0 0 0&#xA;0 0 0 0 0 1 1&#xA;0 1 0 0 1 0 0&#xA;1 0 0 0 1 0 0&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;There exist an automorphism of the graph that maps vertex u to vertex v provided u and v both belong to cycles of the same length.  To illustrate, you can fix everything outside of the components containing u and v, then&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;if u and v are in the same cycle, just rotate the cycle, or otherwise,&lt;/li&gt;&#xA;&lt;li&gt;swap the cycles containing u and v, then rotate each appropriately.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="139" LastEditorUserId="139" LastEditorDisplayName="" LastEditDate="2010-12-14T22:10:56.043" LastActivityDate="2010-12-14T22:10:56.043" />
  <row Id="14333" PostTypeId="1" CreationDate="2010-12-14T22:09:06.060" Score="3" ViewCount="129" Body="&lt;p&gt;I am truly lost as to what this problem is asking.  I did post this on another forum and received what my have been wonderful advice.  However, even after multiple hours and many &quot;Google&quot; searches I am still confused.  Please anyone....&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem states:&#xA;If $\sigma$ is a permutation of a set $A$, we say that &lt;em&gt;$\sigma$ moves $a\in A$&lt;/em&gt; iff $\sigma(a)\neq a$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the symmetric group $S_{36}$ of all permutations of 36 elements, let $H$ be a subset of $S_{36}$ containing all permutations that move no more than four elements.  Is $H$ a subgroup of $S_{36}$?  Prove.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not looking for an answer but more an explanation of what the problem is saying.  I do not know what &quot;move no more than four elements&quot; means. I am feeling really stupid.  Any help appreciated.   &lt;/p&gt;&#xA;" OwnerUserId="4755" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2010-12-15T04:32:26.823" LastActivityDate="2010-12-15T04:32:26.823" Title="Is the set of permutations in $S_{36}$ that move at most 4 elements a subgroup of $S_{36}$?" Tags="&lt;homework&gt;&lt;group-theory&gt;&lt;permutations&gt;" AnswerCount="2" />
  <row Id="14334" PostTypeId="2" ParentId="14333" CreationDate="2010-12-14T22:14:47.227" Score="2" ViewCount="" Body="&lt;p&gt;Imagine you have 36 volumes of an encyclopedia in front of you. A permutation $\sigma\in S_{36}$ is a set of instructions of how to swap positions of the volumes of that encyclopedia. For instance, the permutation $\sigma=(1,2,3,4)$  (this is &lt;em&gt;cycle notation&lt;/em&gt;: it says element 1 is sent to element 2, element 2 is sent to element 3, 3 to 4, and 4 to 1; if you are used to two-line or abbreviated two-line notation, then this is the permutation that has&#xA;$$\left(\begin{array}{cccccccc}&#xD;&#xA;1 &amp;amp; 2 &amp;amp; 3 &amp;amp; 4 &amp;amp; 5 &amp;amp; 6 &amp;amp; \cdots &amp;amp; 36\\&#xD;&#xA;2 &amp;amp; 3 &amp;amp; 4 &amp;amp; 1 &amp;amp; 5 &amp;amp; 6 &amp;amp; \cdots &amp;amp; 36&#xD;&#xA;\end{array}\right).$$&#xD;&#xA;as its two-line description) &#xA;tells you to take volumes $1$ through $4$, and move volume $1$ to the second position, volume $2$ to the third position, volume $3$ to the fourth position, and volume $4$ to the first position. Even though $\sigma$ is a permutation of all 36 volumes, in fact you only need to move four of the volumes to perform the permutation $\sigma$. So we say that $\sigma$ &lt;em&gt;fixes&lt;/em&gt; 32 volumes (volumes 5 through 36 are left untouched) and &lt;em&gt;moves&lt;/em&gt; 4 volumes (volumes 1 through 4). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For each permutation $\sigma\in S_n$, which acts on the set $[n]=\{1,2,\ldots,n\}$, you can ask for the set $\mathrm{Fix}(\sigma) = \{ a\in[n]\mid \sigma(a)=a\}$, the set of elements fixed by $\sigma$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are the elements &quot;moved&quot; by $\sigma$? They are the elements of $[n]-\mathrm{Fix}(\sigma)$, the complement of $\mathrm{Fix}(\sigma)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A $\sigma\in S_{36}$ will lie in $H$ if and only if $[36]-\mathrm{Fix}(\sigma)$ has at most $4$ elements; that is, if and only if $\sigma$ fixes &lt;em&gt;at least&lt;/em&gt; 32 of the elements of the numbers in $[36]=\{1,2,3,\ldots,36\}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is whether $H$ is a subgroup. It is easy to check that it is nonempty and closed under inverses (though you should do that). The trick is going to be figuring out if it is closed under products: if you have $\sigma$ and $\tau$, and each of them fix at least $32$ elements (though possibly not the &lt;em&gt;same&lt;/em&gt; 32 elements), can $\sigma\tau$ fix &lt;em&gt;fewer&lt;/em&gt; than 32 elements?&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-14T22:20:49.527" LastActivityDate="2010-12-14T22:20:49.527" CommentCount="4" />
  <row Id="14335" PostTypeId="2" ParentId="14333" CreationDate="2010-12-14T22:18:12.420" Score="3" ViewCount="" Body="&lt;p&gt;Let's look at $S_3$ instead.  There are six permutations:  $\left(\begin{array}{ccc}&#xD;&#xA;1 &amp;amp; 2 &amp;amp; 3  \\&#xD;&#xA;1 &amp;amp; 2 &amp;amp; 3&#xD;&#xA;\end{array}\right)$, $\left(\begin{array}{ccc}&#xD;&#xA;1 &amp;amp; 2 &amp;amp; 3  \\&#xD;&#xA;1 &amp;amp; 3 &amp;amp; 2&#xD;&#xA;\end{array}\right)$, $\left(\begin{array}{ccc}&#xD;&#xA;1 &amp;amp; 2 &amp;amp; 3  \\&#xD;&#xA;2 &amp;amp; 1 &amp;amp; 3&#xD;&#xA;\end{array}\right)$, $\left(\begin{array}{ccc}&#xD;&#xA;1 &amp;amp; 2 &amp;amp; 3  \\&#xD;&#xA;2 &amp;amp; 3 &amp;amp; 1&#xD;&#xA;\end{array}\right)$, $\left(\begin{array}{ccc}&#xD;&#xA;1 &amp;amp; 2 &amp;amp; 3  \\&#xD;&#xA;3 &amp;amp; 1 &amp;amp; 2&#xD;&#xA;\end{array}\right)$, and $\left(\begin{array}{ccc}&#xD;&#xA;1 &amp;amp; 2 &amp;amp; 3  \\&#xD;&#xA;3 &amp;amp; 2 &amp;amp; 1&#xD;&#xA;\end{array}\right)$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\left(\begin{array}{ccc}&#xD;&#xA;1 &amp;amp; 2 &amp;amp; 3  \\&#xD;&#xA;1 &amp;amp; 2 &amp;amp; 3&#xD;&#xA;\end{array}\right)$ moves no elements.  $\left(\begin{array}{ccc}&#xD;&#xA;1 &amp;amp; 2 &amp;amp; 3  \\&#xD;&#xA;1 &amp;amp; 3 &amp;amp; 2&#xD;&#xA;\end{array}\right)$, $\left(\begin{array}{ccc}&#xD;&#xA;1 &amp;amp; 2 &amp;amp; 3  \\&#xD;&#xA;3 &amp;amp; 2 &amp;amp; 1&#xD;&#xA;\end{array}\right)$, and $\left(\begin{array}{ccc}&#xD;&#xA;1 &amp;amp; 2 &amp;amp; 3  \\&#xD;&#xA;2 &amp;amp; 1 &amp;amp; 3&#xD;&#xA;\end{array}\right)$ move two elements.  $\left(\begin{array}{ccc}&#xD;&#xA;1 &amp;amp; 2 &amp;amp; 3  \\&#xD;&#xA;3 &amp;amp; 1 &amp;amp; 2&#xD;&#xA;\end{array}\right)$ and $\left(\begin{array}{ccc}&#xD;&#xA;1 &amp;amp; 2 &amp;amp; 3  \\&#xD;&#xA;2 &amp;amp; 1 &amp;amp; 3&#xD;&#xA;\end{array}\right)$ move three elements.  Just look at how many are left where they started.  With Arturo's answer, you should be on your way.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2010-12-15T00:21:02.670" LastActivityDate="2010-12-15T00:21:02.670" CommentCount="6" />
  <row Id="14336" PostTypeId="2" ParentId="14306" CreationDate="2010-12-14T23:01:21.270" Score="5" ViewCount="" Body="&lt;p&gt;A nice, very readable reference for this question is a recent paper in the Monthly, &quot;Simpson's rule is exact for quintics&quot;. It is available for download at the author's (&lt;a href=&quot;http://rowdy.mscd.edu/~talmanl/&quot; rel=&quot;nofollow&quot;&gt;Louis A. Talman&lt;/a&gt;) webpage. It appeared American Mathematical Monthly, 113(2006), 144–155. I recommend it; I have used it a couple of times when covering this material in class. Here is the abstract: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In this article, we use tools accessible to freshman calculus students to develop exact—though usually uncomputable—expressions for the error that results in replacing a definite integral with its midpoint rule, trapezoidal rule, or Simpson's rule approximation. Among the tools we use is an extended version of the first mean value theorem for integrals. We obtain not only the classical estimates that appear in calculus books, but estimates for functions less smooth than the classical results require. We show, in particular, how to compute the exact error for a Simpson's rule approximation to an integral of a quintic polynomial. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you visit Talman's page, you may also enjoy taking a look at the nice &quot;The Mother of All Calculus Quizzes.&quot;&lt;/p&gt;&#xA;" OwnerUserId="462" LastActivityDate="2010-12-14T23:01:21.270" />
  <row Id="14343" PostTypeId="1" AcceptedAnswerId="14439" CreationDate="2010-12-14T23:32:09.080" Score="0" ViewCount="187" Body="&lt;p&gt;I have a triangle ABC in the x-y coordinate plane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Corners A and B lie on x-axis. Length of AB is known. Angles A and B are also known.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question&lt;/strong&gt;: Find (x,y) coordinates of corner C.&lt;/p&gt;&#xA;" OwnerUserId="4760" LastActivityDate="2010-12-15T19:34:56.980" Title="Finding X and Y coordinates from two angles" Tags="&lt;trigonometry&gt;" AnswerCount="4" CommentCount="1" />
  <row Id="14344" PostTypeId="2" ParentId="14328" CreationDate="2010-12-14T23:35:34.900" Score="6" ViewCount="" Body="&lt;p&gt;I don't know of any easier examples, but here is one. Let $k$ be a field and $R$ be the set equivalence classes of elements of the form $\frac{f(x,y)}{g(x,y)}$ where $x,y$ are indeterminates over $k$, $f,g\in k[x,y]$, $x$ does not divide $g$ (in $k[x,y]$) and $\frac{f(0,y)}{g(0,y)}\in k$. Make $R$ into a ring by the usual addition and multiplication of rational functions. Then, it can be shown that $R$ is a commutative ring that is not noetherian and every ideal of $R$ has a primary decomposition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is an example from Gilmer's paper linked here &lt;a href=&quot;http://www.jstor.org/pss/2034992&quot; rel=&quot;nofollow&quot;&gt;http://www.jstor.org/pss/2034992&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;He also gives a characterization of rings in which every ideal has a unique primary decomposition.&lt;/p&gt;&#xA;" OwnerUserId="3431" LastEditorUserId="3431" LastEditorDisplayName="" LastEditDate="2010-12-16T16:21:43.283" LastActivityDate="2010-12-16T16:21:43.283" />
  <row Id="14345" PostTypeId="1" AcceptedAnswerId="14351" CreationDate="2010-12-14T23:39:04.457" Score="2" ViewCount="161" Body="&lt;p&gt;Question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(a) Determine sign$(\tau)$ for $$\left( \begin{array}{ccccccc} 1&amp;amp;2&amp;amp;3&amp;amp;4&amp;amp;5&amp;amp;6&amp;amp;7 \\ 2&amp;amp;3&amp;amp;5&amp;amp;7&amp;amp;1&amp;amp;6&amp;amp;4 \end{array} \right )$$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(b) Let $A_{n} = \{\tau \in S_{n} | \mbox{sign}(\tau) = 1\}$. Show that $A_{n}$ is a sub group of $S_{n}$ and calculate the number of elements (Tip: Show that for any transposition $\tau$, $S_{n}$ is the disjoint union of $A_{n}$ and $A_{n}\circ\tau$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My attempt so far:&#xA;(a) So I have this formula for sign$(\sigma) = \displaystyle \prod_{i &amp;lt; j} \frac{\sigma(j)-\sigma(i)}{j-i}$. This wouldn't be so difficult, however it's apparently ok to count the number of elements for which $i&amp;lt;j$ but $\sigma(i)&amp;gt;\sigma(j)$ (these are called inversions, I believe). If this number is even, sign$(\sigma)=1$ if odd, $-1$. In this case I named $(1,5),(2,5),(3,5),(3,7),(4,5),(4,6),(4,7),(6,7) \Rightarrow \mbox{sign}(\tau) =1$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(b) To begin with, if I want to show that something is a subgroup, I have to show that: (i)$ a,b \in U \Rightarrow a \circ b \in U$ and that (ii)$a \in U \Rightarrow a^{-1} \in U$ correct? (I'm actually a little confused at why we can just assume that the identity element is still part of a subgroup, for instance if the exercise were a little different and I was checking if $A_{n} = \{\tau \in S_{n} | \mbox{sign}(\tau) = -1\}$ was a sub group, I would say that the identity element for permutations $123...n$ has no inversions $\Rightarrow$ sign$(\tau) = 1 \Rightarrow \tau \notin A_{n}$... I mean, subgroups are groups themselves, right?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well anyway, for (i) I tried to show it by writing: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $\tau, \sigma \in A_{n}$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Rightarrow$ sign$(\tau) = 1$ and sign$(\sigma)=1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a polynomial (I honestly don't totally understand this or know if I need to put this stuff with 'g' in...) $g=g(x_{1},...,x_{n})$, sign$(\tau \circ \sigma)g= (\tau \circ \sigma)(g)=\tau(\sigma(g))=\tau((\mbox{sign}\sigma)g)=(\mbox{sign}\tau)(\mbox{sign}\sigma)g$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Rightarrow$ sign$(\tau \circ \sigma) = (1)(1) = 1 \Rightarrow (\tau \circ \sigma) \in A_{n}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with (ii) my attempt was:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $\sigma \in A_{n}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $\sigma \circ \sigma^{-1} = \epsilon$ (this should represent the identity element)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Rightarrow$ sign$(\sigma \circ \sigma^{-1}) = \mbox{sign}\epsilon$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(\mbox{sign}\sigma)(\mbox{sign}\sigma^{-1})=1$ (Because sign$(\epsilon)=1$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(1)(\mbox{sign}(\sigma^{-1}))=1$ (Because $\sigma \in A_{n}\Rightarrow \mbox{sign}(\sigma) = 1$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;sign$(\sigma^{-1})=1 \Rightarrow \sigma^{-1} \in A_{n}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Out of (i) and (ii) follows the conclusion that $A_{n}$ is a subgroup of $S_{n}$...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So hopefully that would all be on the right track so far... With the calculate the number of elements part, I don't know what to do or why. Following the tip I would try to show that $S_{n}$ is the disjoint union (not the same as showing that the intersection is empty, right?) of $A_{n}$ and $A_{n} \circ \tau$ (where $\tau$ is a transposition). I don't understand this, but directly out of the book is the statement &quot;We call $\tau$ a transposition. If $i&amp;lt;j$, then there are $2(j-i=1)+1$ inversions in $\tau$, and hence the transposition $\tau$ is odd.&quot; Accordingly I think I could use sign$(\tau \circ \sigma)=(\mbox{sign}\tau)(\mbox{sign}\sigma)$ to show that the product of an odd and even permutation is odd and therefore $A_{n} \circ \tau$ is odd and $A_{n}$ has no elements in common with $A_n \circ \tau$. But even then how does one calculate a number of elements?&lt;/p&gt;&#xA;" OwnerUserId="3711" LastEditorUserId="3711" LastEditorDisplayName="" LastEditDate="2010-12-15T00:19:49.117" LastActivityDate="2010-12-15T00:19:49.117" Title="Determine sign of a permutation, calculate number of elements in the subgroup of permutations with sign = 1" Tags="&lt;homework&gt;&lt;group-theory&gt;&lt;permutations&gt;&lt;symmetric-groups&gt;" AnswerCount="1" CommentCount="0" />
  <row Id="14346" PostTypeId="2" ParentId="14327" CreationDate="2010-12-14T23:45:31.347" Score="2" ViewCount="" Body="&lt;p&gt;If we let $E(n,m)$ be the expectation with m throws of an n sided die, we will accept the $m^{th}$ to last throw if it is $\geq E(n,m-1)$, so $E(n,1)=\frac{n+1}{2}$ and $$E(n,m)=\frac{(n-\lceil E(n,m-1)\rceil +1)}{n}\frac{(n+\lceil E(n,m-1)\rceil )}{2}+\frac{\lceil E(n,m-1)\rceil -1}{n}E(n,m-1)$$  where the first term comes from accepting the throw and the second is from rejecting it.  Letting $n-\lceil E(n,m-1)\rceil=d$ (the whole part of the difference) and $\lceil E(n,m-1)\rceil-E(n,m-1)=e$ (the fractional part) we find $$E(n,m)-E(n,m-1)=\frac{d^2+(1+2e)d+2e}{2n}$$  If we figure that e averages about 1/2, this means we increase E by $\frac{(d+1)^2}{2n}$ per step.  The number of steps to get the cut off up to n is about $$\sum_{d=1}^{n}{\frac{2n}{(d+1)^2}}$$  For n=100, I find by explicit calculation we hit a cutoff of 99 (E&gt;98) at m=75 and a cutoff of 100 (E&gt;99) at m=129.  The sum comes out just about 128.  We can find lower cutoffs by raising the lower limit of the sum.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2010-12-15T04:16:41.570" LastActivityDate="2010-12-15T04:16:41.570" CommentCount="6" />
  <row Id="14347" PostTypeId="1" CreationDate="2010-12-14T23:46:29.380" Score="2" ViewCount="122" Body="&lt;p&gt;Can anyone give a nonlinear regular function from C^2 to C^2 with a constant nonzero Jacobian? It seems to me that the only such functions are linear. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to the Jacobian conjecture, a function from C^2 to C^2 with a constant nonzero Jacobian must have an inverse. &lt;/p&gt;&#xA;" OwnerUserId="3163" LastActivityDate="2010-12-15T03:02:51.720" Title="Example of nonlinear regular function with constant nonzero Jacobian" Tags="&lt;analysis&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="14348" PostTypeId="2" ParentId="14343" CreationDate="2010-12-14T23:46:53.607" Score="2" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Hint:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With law of sines you can find the other sides.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $A= (x_{1},y_{1})$ and $B=(x_{2},y_{2})$, then $|AB|=\sqrt{(x_{1}-x_{2})^2+(y_{1}-y_{2})^2}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if $C=(x,y)$ get :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$|AC| = \sqrt{(x-x_{1})^2+(y-y_{1})^2}$$  $$|BC| = \sqrt{(x-x_{2})^2+(y-y_{2})^2}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remember that you know $|AB|,|AC|,|AB|$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now you can find $x,y$&lt;/p&gt;&#xA;" OwnerUserId="3421" LastEditorUserId="3421" LastEditorDisplayName="" LastEditDate="2010-12-15T18:52:16.270" LastActivityDate="2010-12-15T18:52:16.270" />
  <row Id="14349" PostTypeId="2" ParentId="13991" CreationDate="2010-12-14T23:50:58.017" Score="2" ViewCount="" Body="&lt;p&gt;With regard to combinatorics &amp;amp; harmonic analysis, you might find &lt;a href=&quot;http://books.google.com/books?id=qjVaD1OQbxEC&amp;amp;lpg=PP1&amp;amp;pg=PA13#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; interesting. &lt;a href=&quot;http://arxiv.org/PS_cache/math/pdf/0008/0008098v1.pdf&quot; rel=&quot;nofollow&quot;&gt;This&lt;/a&gt; work by Terence Tao (whom you mentioned) sheds light on all three of combinatorics, analysis, and PDEs. Hope this helps!&lt;/p&gt;&#xA;" OwnerUserId="3731" LastActivityDate="2010-12-14T23:50:58.017" />
  <row Id="14351" PostTypeId="2" ParentId="14345" CreationDate="2010-12-15T00:00:19.687" Score="1" ViewCount="" Body="&lt;p&gt;Your work on (a) got cut off somehow.  For your comment on (b), if you have $a,b \in U \Rightarrow a \circ b \in U$ and $a \in U \Rightarrow a^{-1} \in U$ then you have $a \circ a^{-1} \in U$ so you don't have to check the identity (as long as $U$ is nonempty).  Showing the identity is not in U is another way to prove U is not a subgroup.  You are right that subgroups are groups themselves, but I think you would have trouble showing the product to two odd permutations is odd.  To say $S_n$ is a disjoint union of $A_n$ and $A_{n} \circ \tau$, you need the intersection to be empty and that every element of $S_n$ is in one of them.  In calculating the number of elements, if every element is either odd or even and there are the same number of odd ones and even ones...&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-12-15T00:00:19.687" CommentCount="7" />
  <row Id="14352" PostTypeId="2" ParentId="14276" CreationDate="2010-12-15T00:05:45.690" Score="0" ViewCount="" Body="&lt;p&gt;If you work with Grothendieck motives modulo rational equivalence, you even have simpler examples : the motive of any split (projective) quadric of odd dimension $n$ is the same as the motive of $\mathbb{P}^n$.&lt;/p&gt;&#xA;" OwnerUserId="4763" LastActivityDate="2010-12-15T00:05:45.690" />
  <row Id="14353" PostTypeId="1" CreationDate="2010-12-15T00:29:42.653" Score="7" ViewCount="243" Body="&lt;p&gt;Twin, cousin, and sexy primes are of the forms $(p,p+2)$, $(p,p+4)$, $(p,p+6)$ respectively, for $p$ a prime.&#xA;The Wikipedia &lt;a href=&quot;http://en.wikipedia.org/wiki/Cousin_prime&quot; rel=&quot;nofollow&quot;&gt;article on cousin primes&lt;/a&gt; says that,&#xA;&quot;It follows from the first Hardy–Littlewood conjecture that cousin primes have the same asymptotic density as twin primes,&quot; but the analogous &lt;a href=&quot;http://en.wikipedia.org/wiki/Sexy_prime&quot; rel=&quot;nofollow&quot;&gt;article on sexy primes&lt;/a&gt; does not make a similar claim. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;b&gt;Q1&lt;/b&gt;. Are the sexy primes expected to have the same density as twin primes?&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;b&gt;Q2&lt;/b&gt;. Is it conjectured that there are an infinite number of cousin and sexy prime&#xA;  pairs?&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;b&gt;Q3&lt;/b&gt;. Have prime pairs of the form $(p,p+2k)$ been studied for $k&amp;gt;3$?&#xA;  If so, what are the conjectures?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Thanks for information or pointers!&lt;/p&gt;&#xA;" OwnerUserId="237" LastActivityDate="2011-04-20T07:55:36.053" Title="Twin, cousin, sexy, ... primes" Tags="&lt;prime-numbers&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" />
  <row Id="14354" PostTypeId="1" CreationDate="2010-12-15T00:46:02.637" Score="6" ViewCount="103" Body="&lt;p&gt;I've just been thinking about for what values of n we can place n points in the plane so that any three of those points define an isosceles triangle. A triangle, square and pentagon work for 3,4 and 5, and to get 6 just place a point in the centre of the pentagon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is can we do this for 7 points in the plane?&lt;/p&gt;&#xA;" OwnerUserId="4764" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-15T01:03:15.850" LastActivityDate="2011-01-14T02:44:52.080" Title="Collections of points containing only isosceles triangles" Tags="&lt;geometry&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="2" />
  <row Id="14355" PostTypeId="2" ParentId="14347" CreationDate="2010-12-15T00:47:09.967" Score="3" ViewCount="" Body="&lt;p&gt;Some trial and error should get it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I figured $\left[\begin{array}{cc} x &amp;amp; 1\\ x-1 &amp;amp; 1\end{array}\right]$ was a good target Jacobian matrix, and so got $f(x,y)=(\frac 1 2 x^2 + y,\frac 1 2 x^2-x + y)$.  Is that what you're looking for?  Evidently it has an inverse, $g(a,b)=(a-b,a-\frac 1 2(a-b)^2)$.&lt;/p&gt;&#xA;" OwnerUserId="4721" LastActivityDate="2010-12-15T00:47:09.967" CommentCount="1" />
  <row Id="14356" PostTypeId="1" CreationDate="2010-12-15T00:57:04.653" Score="2" ViewCount="227" Body="&lt;p&gt;I am in graduate school for chemical engineering and the method of characteristics appears in many journal articles.  Does anyone have a good reference: either online or a book that provides a good explanation with examples of the method of characteristics.&lt;/p&gt;&#xA;" OwnerUserId="4766" LastEditorUserId="1543" LastEditorDisplayName="" LastEditDate="2011-04-21T22:10:31.003" LastActivityDate="2011-04-21T22:10:31.003" Title="Good reference for method of characteristics" Tags="&lt;differential-equations&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="14357" PostTypeId="2" ParentId="14343" CreationDate="2010-12-15T01:10:50.570" Score="2" ViewCount="" Body="&lt;p&gt;Another possibility is as follows: this solution to your problem involves using the fact that the slope of a line making an angle $\theta$ with the positive x-axis is $\tan\;\theta$; now, what you have are two lines with slopes $\tan\;A$ and $\tan(\pi-B)$ in one case, and slopes $\tan\;B$ and $\tan(\pi-A)$ in the other. Construct the point-slope form for the two lines through $A$ and $B$, and find their intersection through the usual methods.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-12-15T01:10:50.570" />
  <row Id="14358" PostTypeId="1" AcceptedAnswerId="14860" CreationDate="2010-12-15T01:19:40.527" Score="3" ViewCount="259" Body="&lt;p&gt;Let $G$ be a $n \times n$ matrix with real entries and let $\Lambda = \{x^n \colon \exists i^n \in \mathbb{Z}^n \text{ such that } x^n = G \cdot i^n\}$ define a lattice. I am interested in projecting the lattice points onto a $k$-dimensional subspace $U$ with $k &amp;lt; n$. Let $A$ be a $n \times k$ matrix of vectors that span $U$. Then, the projection of $\Lambda$ onto $U$ has the generator matrix $P_A \cdot G$ where $P_A = A (A^TA)^{-1}A^T$ is the projection operator. Lets also assume that the entries of $G$ have no structure between them - i.e., they are randomly chosen from the reals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am interested in finding out when the projection of $\Lambda$ onto $U$ (call it $\Lambda_U$) is also a $k$-dimensional lattice. It seems to me that in most cases, $\Lambda_U$ will fill out the $k$-dimensional space $U$ completely, i.e., the shortest distance between neighboring points in $\Lambda_U$ will be arbitrarily small (obs. 1). For certain subspaces $U$ that seem to be &quot;aligned&quot; correctly with $\Lambda$, we get a well-defined $\Lambda_U$. By well-defined, I mean that the spacing between the nearest neighbors of $\Lambda_U$ is bounded away from $0$. As far as I can tell, this happens when $A$ is chosen such that the columns of $P_A \cdot G$ are rational multiples of one another (obs. 2).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Questions:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Are the observations (obs. 1) and (obs. 2) correct?&lt;/li&gt;&#xA;&lt;li&gt;If so, is this a well-studied concept? I didn't have much success googling with the obvious keywords such as lattice &quot;projection&quot; or &quot;matrix column rational multiples&quot; etc.&lt;/li&gt;&#xA;&lt;li&gt;Assuming the observations are correct, given $G$, is there a way to choose $A$ such that the columns of $P_A \cdot G$ are rational multiples of one another?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="618" LastActivityDate="2010-12-19T18:23:31.823" Title="Projection of a lattice onto a subspace" Tags="&lt;linear-algebra&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="14359" PostTypeId="1" CreationDate="2010-12-15T01:23:29.147" Score="2" ViewCount="91" Body="&lt;p&gt;The problem is stated-&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do the following two sets of points have the same cardinality and if so, establish a bijection:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A line segment of length four and half of the circumference of radius one (including both endpoints).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My reasoning is they do have the same cardinality and my bijection is a picture in which I drew a horizontal line and a semicircle under it  (separated by approx. 2cm with the semicircle's open side facing down).  At what would be the center of the circle made by the semi-circle I drew a point P.  I then drew lines vertical from the original horizontal line to point P.  This shows, goes my reasoning, that for every point on the line there is a corresponding point on the semi-circle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is is this drawing enough to show a bijection or do I need to do more?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you for your thoughts.&lt;/p&gt;&#xA;" OwnerUserId="4179" LastActivityDate="2010-12-15T01:38:42.587" Title="Cardinality of Two Sets of Points" Tags="&lt;homework&gt;&lt;abstract-algebra&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="14360" PostTypeId="2" ParentId="14328" CreationDate="2010-12-15T01:25:08.383" Score="3" ViewCount="" Body="&lt;p&gt;Many interesting examples are in: $\ $ Barucci and Montana: &lt;a href=&quot;http://www.springerlink.com/content/7752947723p76557/fulltext.pdf&quot; rel=&quot;nofollow&quot;&gt;When are $\rm\ D +  \frak M\ $ rings Laskerian?&lt;/a&gt; They construct (non-Noetherian) Laskerian or strongly Laskerian rings and domains either integrally closed or not, of any dimension. They employ a frequently used powerful tool for constructing counterexamples - the  $\rm\ D +  \frak M\ $ construction. For more on this construction see Anderson: &lt;a href=&quot;http://rmmc.asu.edu/TO%20DOUGLAS/JCA/anderson/anderson.pdf&quot; rel=&quot;nofollow&quot;&gt;Some remarks on the  $\rm\ D +  \frak M\ $ construction&lt;/a&gt; and see this very informative survey on more general constructions: $\ $ Zafrullah: &lt;a href=&quot;http://www.lohar.com/researchpdf/axbx.pdf&quot; rel=&quot;nofollow&quot;&gt;Various facets of rings between $\rm\:D[X]\:$ and $\rm\:K[X]\:$&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-12-15T01:25:08.383" />
  <row Id="14361" PostTypeId="2" ParentId="14347" CreationDate="2010-12-15T01:30:06.773" Score="1" ViewCount="" Body="&lt;p&gt;If $u(x,y)$ is of the form $af(x-y)+bx$, where $a$ and $b$ are constants and $f$ is a one-variable differentiable function, and if $v(x,y)=y-x$, then&#xA;$$&#xD;&#xA;\frac{{\partial u(x,y)}}{{\partial x}} \frac{{\partial v(x,y)}}{{\partial y}} - \frac{{\partial u(x,y)}}{{\partial y}}\frac{{\partial v(x,y)}}{{\partial x}} = [af&amp;#39;(x-y)+b]+[-af&amp;#39;(x-y)] = b.&#xD;&#xA;$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: More simply, letting $u(x,y)=af(x+y)+bx$ and $v(x,y)=x+y$ gives the same result as above.&lt;/p&gt;&#xA;" OwnerUserId="2810" LastEditorUserId="2810" LastEditorDisplayName="" LastEditDate="2010-12-15T03:02:51.720" LastActivityDate="2010-12-15T03:02:51.720" />
  <row Id="14362" PostTypeId="2" ParentId="14359" CreationDate="2010-12-15T01:33:05.273" Score="2" ViewCount="" Body="&lt;p&gt;I think your lines are not vertical, right?  With your construction, there are points on the semi-circle that do not have a corresponding point on the line.  But you are quite close.  You can 1) modify the construction so that every line from P through the semicircle hits the line (and still have every line from the line segment hit the semi-circle.  What does that tell you about the endpoints of each?) Or 2)find another construction that injects the semi-circle into the line, then argue from the Cantor–Bernstein–Schroeder theorem. &lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2010-12-15T01:38:42.587" LastActivityDate="2010-12-15T01:38:42.587" />
  <row Id="14363" PostTypeId="1" AcceptedAnswerId="14385" CreationDate="2010-12-15T01:36:08.573" Score="1" ViewCount="86" Body="&lt;p&gt;Given a multi-undirected graph, how can I find the number of different paths from Node A to B using every node in the graph (means shortest path algorithms are useless)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In some cases there may be no way to achieve this. I am looking for the number of available different paths.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: You have to use every node and use them once.&lt;/p&gt;&#xA;" OwnerUserId="4653" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-15T01:54:22.407" LastActivityDate="2010-12-15T06:16:42.303" Title="Find the number of different paths across two nodes in a multi-undirected graph" Tags="&lt;graph-theory&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="14364" PostTypeId="1" CreationDate="2010-12-15T01:40:11.670" Score="1" ViewCount="171" Body="&lt;p&gt;Given $R$ is a ring, and $b$ is some positive integer such that $x + x^{2b + 1} = x^{2b} + x^{10b + 1}$ for all $x \in R$, prove that $R$ is Boolean, i.e. $x =x^2$ for all $x$ in $R$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not sure where to begin with the problem, and what avenue of approach I should take such that my methodology is inclusive to all $x$ in R. &#xA;Would it be feasible to convert this problem into matrix form, and then prove $A = A^2$? &lt;/p&gt;&#xA;" OwnerUserId="4767" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-15T03:11:28.173" LastActivityDate="2010-12-15T18:01:59.230" Title="Proving a ring is Boolean" Tags="&lt;abstract-algebra&gt;&lt;ring-theory&gt;" AnswerCount="1" CommentCount="12" />
  <row Id="14365" PostTypeId="2" ParentId="14347" CreationDate="2010-12-15T01:53:13.373" Score="0" ViewCount="" Body="&lt;p&gt;Any linear function has constant Jacobian determinant, as does any map of the form $(z_1,z_2) \rightarrow (z_1, z_2 - f(z_1))$ or $(z_1,z_2) \rightarrow (z_1 - f(z_2), z_2)$. As a result, any finite composition of maps of these forms will have constant Jacobian. These include the examples that Shai Covo and Andrew Marshall listed. I forget if there are known examples outside this category.&lt;/p&gt;&#xA;" OwnerUserId="3035" LastActivityDate="2010-12-15T01:53:13.373" />
  <row Id="14367" PostTypeId="1" AcceptedAnswerId="14370" CreationDate="2010-12-15T02:39:37.620" Score="2" ViewCount="132" Body="&lt;p&gt;So, we have a function&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f(x) = \frac{3\sin(x)}{2 + \cos(x)} $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and we fix a point $$ x_0 \in \left(0, \frac{2\pi}{3}\right] $$ and define a sequence by setting $$ x_{n+1} = f(x_n) $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I need to show that $$ \lim_{n\to\infty} n^{1/4}x_n = 45^{1/4}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am pretty sure this has something to do with Cesàro means. In a previous part of the problem we proved that the sequence is strictly decreasing and has a limit of zero, and we also found the 5th order Taylor approximation of the function (centered at 0), which happens to be $$ p_5(f, x) = x - \frac{x^5}{180} $$ but we have not proved very many theorems involving Taylor polynomials (and none that seem to apply), and the section in Rudin (our textbook) is quite brief. Any help or suggestions would be welcome.&lt;/p&gt;&#xA;" OwnerUserId="2465" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-15T02:48:26.430" LastActivityDate="2010-12-15T03:33:27.910" Title="recursively defined trigonometric sequence and Cesàro means." Tags="&lt;real-analysis&gt;&lt;homework&gt;&lt;sequences-and-series&gt;&lt;trigonometry&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="14368" PostTypeId="2" ParentId="14354" CreationDate="2010-12-15T02:42:12.390" Score="2" ViewCount="" Body="&lt;p&gt;Here is a recent paper related to this topic:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.cims.nyu.edu/~pach/publications/isosceles.ps&quot; rel=&quot;nofollow&quot;&gt;http://www.cims.nyu.edu/~pach/publications/isosceles.ps&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1369" LastActivityDate="2010-12-15T02:42:12.390" />
  <row Id="14370" PostTypeId="2" ParentId="14367" CreationDate="2010-12-15T02:47:05.987" Score="2" ViewCount="" Body="&lt;p&gt;Check David Speyer's excellent answer here: &lt;a href=&quot;http://math.stackexchange.com/questions/3215/convergence-of-sqrtnx-n-where-x-n1-sinx-n/3220#3220&quot;&gt;Convergence of $\sqrt{n}x_{n}$ where $x_{n+1} = \sin(x_{n})$&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Quoted from that answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Let $f$ be a function $0 \leq f(u)&#xD;&#xA;&amp;gt; \leq u$ on an interval $[0,c]$ then&#xA;  the sequence $x_n:=f(f(f(\cdots&#xD;&#xA;&amp;gt; f(c)\cdots)$ approaches $0$. If&#xA;  $f(u)=u-a u^{k+1} + O(u^{k+2})$ (with&#xA;  $a&gt;0$) then $x_n \approx \alpha n^{-1/k}$ and you can prove that by&#xA;  the same methods here.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-12-15T03:33:27.910" LastActivityDate="2010-12-15T03:33:27.910" />
  <row Id="14372" PostTypeId="2" ParentId="8933" CreationDate="2010-12-15T03:17:33.547" Score="1" ViewCount="" Body="&lt;p&gt;By the way, your equation can be written in form &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$H^{(x)}_3=11$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where H is the generalized harmonic number: &lt;a href=&quot;http://www.wolframalpha.com/input/?i=HarmonicNumber%5b3%2C+x%5d%3D%3D11&quot; rel=&quot;nofollow&quot;&gt;http://www.wolframalpha.com/input/?i=HarmonicNumber[3%2C+x]%3D%3D11&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So to find x you should investigate the inverse function of generalized harmonic number.&lt;/p&gt;&#xA;" OwnerUserId="2513" LastActivityDate="2010-12-15T03:17:33.547" />
  <row Id="14373" PostTypeId="2" ParentId="14343" CreationDate="2010-12-15T04:08:42.227" Score="0" ViewCount="" Body="&lt;p&gt;A system of linear equations:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume A is at the origin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can form basis vectors for AC and BC using the angles A and B&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;U_{AC} = [cosA  sinA]^T&#xA;U_{BC} = [cosB  sinB]^T&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Then we know geometrically&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;w*U_{AC} - x*U_{BC} = [ AB  0 ]^T&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This is a 2x2 linear system.  Solve for w and x.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;x = w*cosA&#xA;Y = w*sinA&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;best regards.&lt;/p&gt;&#xA;" OwnerUserId="4726" LastActivityDate="2010-12-15T04:08:42.227" />
  <row Id="14374" PostTypeId="2" ParentId="14356" CreationDate="2010-12-15T04:14:47.350" Score="2" ViewCount="" Body="&lt;p&gt;Off the top of my head, I've found both of these to be well-written and easy to follow.  Cavet:  I don't have them in front of me, so my recommendation is based on general impressions rather than their particular treatment of characteristics.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;em&gt;Partial Differential Equations&lt;/em&gt;, by L. C. Evans. Aimed at the intro grad level.&lt;/li&gt;&#xA;&lt;li&gt;&lt;em&gt;Partial Differential Equations&lt;/em&gt;, by Walter Strauss.  Aimed at an advanced undergrad audience, but good for someone new to the field.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The External Links listed at the end of the wikipedia article on characteristics also look like they might be worth your time.&lt;/p&gt;&#xA;" OwnerUserId="4229" LastActivityDate="2010-12-15T04:14:47.350" CommentCount="1" />
  <row Id="14375" PostTypeId="1" AcceptedAnswerId="14377" CreationDate="2010-12-15T04:30:50.603" Score="2" ViewCount="96" Body="&lt;p&gt;I'm trying to solve the following problem:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The probability $p_{i}$ that a given cell contains exactly i balls is given by the binomial distribution $p_{i}=\displaystyle \frac{\binom{r}{i}(n-1)^{r-i}}{n^{r}}$. Prove that the most probable number is the integer $v$ such that $\displaystyle \frac{(r-n+1)}{n}&amp;lt; v \leq \displaystyle \frac{(r+1)}{n}$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I've tried to get the inequality using first $\displaystyle \frac{p_{i}}{p_{i-1}}$ (which didn't work quite well) and later $\displaystyle \frac{p_{i}}{p_{i+1}}$. For example, for the first attempt the result I obtained was: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{p_{i}}{p_{i-1}}=\frac{r-i+1}{i(n-1)}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this kind of problem, I'm expecting to obtain a $1$ in the resulting expression in order to figure out for which values of $i$ there is a maximum in the function, however, I can't get the correct result.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any help will be appreciated.&lt;/p&gt;&#xA;" OwnerUserId="2569" LastEditorUserId="2569" LastEditorDisplayName="" LastEditDate="2010-12-15T22:46:09.993" LastActivityDate="2010-12-15T22:46:09.993" Title="Binomial distribution $\binom{r}{i}(n-1)^{r-i}/n^{r}$" Tags="&lt;combinatorics&gt;" AnswerCount="1" />
  <row Id="14376" PostTypeId="1" CreationDate="2010-12-15T04:39:15.133" Score="6" ViewCount="162" Body="&lt;p&gt;Suppose that every open interval in $\mathbb{R}$ is a connected set. Does this implies the least upper bound axiom? (i.e every non-empty subset of $\mathbb{R}$ which is bounded above has a least upper bound) Is this true? in such case, how would you prove this? &lt;/p&gt;&#xA;" OwnerUserId="4771" LastEditorUserId="4771" LastEditorDisplayName="" LastEditDate="2010-12-15T05:11:03.687" LastActivityDate="2010-12-15T05:59:53.867" Title="Does the fact that every interval in $\mathbb{R}$ is connected implies that $\mathbb{R}$ is order-complete?" Tags="&lt;topology&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="14377" PostTypeId="2" ParentId="14375" CreationDate="2010-12-15T05:07:56.887" Score="2" ViewCount="" Body="&lt;p&gt;See &lt;a href=&quot;http://mathforum.org/library/drmath/view/52222.html&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; (also for a better formulation of the problem, using common notation).&lt;/p&gt;&#xA;" OwnerUserId="2810" LastEditorUserId="2810" LastEditorDisplayName="" LastEditDate="2010-12-15T05:25:34.197" LastActivityDate="2010-12-15T05:25:34.197" CommentCount="7" />
  <row Id="14378" PostTypeId="2" ParentId="14376" CreationDate="2010-12-15T05:14:11.297" Score="7" ViewCount="" Body="&lt;p&gt;The answer is yes, in the following sense:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Theorem: For a linear order on a set $X$, the following are equivalent:&lt;br&gt;&#xA;(i) $X$ is connected in the order topology.&lt;br&gt;&#xA;(ii) The ordering is dense and $X$ is Dedekind complete -- every subset which is bounded above admits a least upper bound.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See Theorem 13 of &lt;a href=&quot;http://www.math.uga.edu/~pete/realinduction.pdf&quot; rel=&quot;nofollow&quot;&gt;these notes&lt;/a&gt; for a proof which also explores the connection to induction in totally ordered sets.&lt;/p&gt;&#xA;" OwnerUserId="299" LastActivityDate="2010-12-15T05:14:11.297" />
  <row Id="14379" PostTypeId="1" AcceptedAnswerId="14381" CreationDate="2010-12-15T05:24:11.643" Score="5" ViewCount="128" Body="&lt;p&gt;I am a bit confused on how to go about this problem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let G be a finite group and $p$ the smallest prime dividing its order.  Assume $P$ is a $p$-Sylow subgroup which is cyclic.  Show $P$ is contained in the center of its normalizer.&lt;/p&gt;&#xA;" OwnerUserId="4772" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-15T05:30:24.130" LastActivityDate="2010-12-19T20:44:46.580" Title="Containment of small P-sylow subgroups in their Normalizers." Tags="&lt;group-theory&gt;" AnswerCount="3" />
  <row Id="14380" PostTypeId="2" ParentId="14379" CreationDate="2010-12-15T05:39:12.253" Score="4" ViewCount="" Body="&lt;p&gt;You have to show that if an element $g$ of $G$ normalises $P$, then in fact it commutes with all the elements in $P$, i.e. if the conjugation by $g$ preserves the set, then it actually preserves all the elements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if $g$ normalises $P$, then conjugation by $g$ gives you an automorphism of $P$. What is the automorphism group of a cyclic $p$-group? (Hint: an automorphism is determined by what it does to a generator.) Don't forget the restriction, that $p$ is the smallest prime divisor of $|G|$.&lt;/p&gt;&#xA;" OwnerUserId="3212" LastActivityDate="2010-12-15T05:39:12.253" />
  <row Id="14381" PostTypeId="2" ParentId="14379" CreationDate="2010-12-15T05:40:07.050" Score="3" ViewCount="" Body="&lt;p&gt;Let $N$ be the normalizer of $P$, and let $N$ act on $P$ by conjugation; this induces a homomorphism $N\to\mathrm{Aut}(P)$; since $P$ is abelian, it is contained in the kernel of that map, so we have that $N/P$ maps into $\mathrm{Aut}(P)$. But what is the order of the automorphism group of a cyclic group of order $p^n$, and what is the order of $N/P$? What does that tell you about the action of $N$ on $P$?&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-12-15T05:40:07.050" CommentCount="0" />
  <row Id="14382" PostTypeId="2" ParentId="14376" CreationDate="2010-12-15T05:59:53.867" Score="2" ViewCount="" Body="&lt;p&gt;Suppose not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take two intervals which witness that fact - namely $(a,b)$ and $(c,d)$ such that $b&amp;lt;c$ however $(b,c)=\emptyset$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Choose $\varepsilon$ small enough and look at the interval $(b-\varepsilon , c+\varepsilon)$ which is nonempty. By our assumption this interval is also connected, there is someone between $b$ and $c$ otherwise you'd have a non-empty interval which is not connected. A contradiction.&lt;/p&gt;&#xA;" OwnerUserId="622" LastActivityDate="2010-12-15T05:59:53.867" />
  <row Id="14383" PostTypeId="1" AcceptedAnswerId="14394" CreationDate="2010-12-15T06:13:49.100" Score="5" ViewCount="368" Body="&lt;p&gt;Here's what I recall of the question from CNML Grade 11, 2010/2011 Contest #3, Question 7:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;There are 2010 points on a circle,&#xA;  evenly spaced.&lt;/strong&gt; Ford Prefect will*&#xA;  randomly choose three points on the&#xA;  circle. He will* connect these points&#xA;  to form a shape. What is the&#xA;  probability that the resulting shape&#xA;  will* form a right angled triangle?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/9isGJ.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I answered $\frac{1}{4} = 25\%$, but that's probably incorrect. (Right?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I got home, I thought it out in my head, and I got this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{2010 * (2010/1005)}{2010 \choose 3}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{2020050}{1351414120} = \frac{3015}{2017036} = 0.149476756984010201\%$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm probably wrong ...again. Can anyone tell how to get the right answer (if I'm not wrong :) )?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;*in the past of the future of the perfect present present time double into ripple fluctuater byer doininger of the past future continuum...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Realized my mistake in copying the question.&lt;/p&gt;&#xA;" OwnerUserId="2736" LastEditorUserId="2736" LastEditorDisplayName="" LastEditDate="2010-12-15T07:32:32.370" LastActivityDate="2010-12-15T12:23:19.443" Title="How many right angled triangles can a circle have?" Tags="&lt;geometry&gt;&lt;probability&gt;&lt;circle&gt;" AnswerCount="3" CommentCount="19" FavoriteCount="2" />
  <row Id="14384" PostTypeId="1" AcceptedAnswerId="14397" CreationDate="2010-12-15T06:13:51.210" Score="1" ViewCount="150" Body="&lt;p&gt;I was talking with a friend and we were discussing a math problem disguised as a social situation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the chance that someone to accept your request to go out with them was 1% and you asked 1 person only, then the probability they would say yes is .01&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you were to ask 100 people, then your chances are still .01 according to my friend.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now my question is, why does it seem intuitive that the chances would increase when asking 100 people rather than 1?  Is it the expected value that's causing this intuition?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can this situation be modeled using a balls &amp;amp; bins problem?&lt;/p&gt;&#xA;" OwnerUserId="1630" LastActivityDate="2010-12-15T07:55:13.637" Title="Probability increases as sample size increases?" Tags="&lt;probability&gt;&lt;statistics&gt;" AnswerCount="1" />
  <row Id="14385" PostTypeId="2" ParentId="14363" CreationDate="2010-12-15T06:16:42.303" Score="1" ViewCount="" Body="&lt;p&gt;This problem is #P-complete. Informally, that essentially means that the best you can do is try every possible path.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See &lt;a href=&quot;http://cstheory.stackexchange.com/questions/2396/counting-the-number-of-hamiltonian-cycles-in-cubic-hamiltonian-graphs&quot;&gt;http://cstheory.stackexchange.com/questions/2396/counting-the-number-of-hamiltonian-cycles-in-cubic-hamiltonian-graphs&lt;/a&gt; for a more complete discussion of this problem.&lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2010-12-15T06:16:42.303" />
  <row Id="14386" PostTypeId="1" AcceptedAnswerId="14398" CreationDate="2010-12-15T06:17:50.893" Score="0" ViewCount="216" Body="&lt;p&gt;I have the following overdetermined linear equation system:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$Ax=b$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $A$ is a matrix of $n \times k$, $x$ is of $k \times 1$,$b$ is of $n \times 1$, where $n&amp;gt;k$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We all know this is an overdetermined linear equation system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is how to check whether the solve for $x$, and check that the vector is consistent in this case? Consistent as in the sense that when we plug in the $x$ vector value into the above linear equation systems, then the above matrix will be satisfied.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can separate out the $k$ linear equations and find $x$ from $x_1$ to $x_k$, and then substitute in the remaining equations to check for consistency. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I afraid that  this method can be numerically unstable; I would like to implement this on a computer, so I would prefer a solution that fully works here. Let us consider one pitfall of my above solution:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$A=\begin{bmatrix} 10 &amp;amp; 10 \\ 0 &amp;amp; 0 \\0 &amp;amp; 10 \end{bmatrix}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that if you separate the $1$ and $2$ rows out, and compute the solution, you may not be able to even solve it ( equation $2$ is an equation here with no unknown terms, after you times in the $0$ factor)!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there other method?&lt;/p&gt;&#xA;" OwnerUserId="550" LastEditorUserId="550" LastEditorDisplayName="" LastEditDate="2010-12-16T07:07:08.177" LastActivityDate="2010-12-16T10:50:17.730" Title="Check Whether an Overdetermined Linear Equation System is Consistent: General Approach?" Tags="&lt;linear-algebra&gt;" AnswerCount="3" CommentCount="7" />
  <row Id="14387" PostTypeId="1" AcceptedAnswerId="14391" CreationDate="2010-12-15T06:18:14.847" Score="1" ViewCount="103" Body="&lt;p&gt;Hi there, just a quick question dealing with probability.  The annual returns on stocks and treasury bonds over the next 12 months are uncertain. Suppose that these returns can be described by normal distributions with stocks having a mean of 15% and a standard deviation of 20%, and bonds having a mean of 6% and a standard deviation of 9%. Which asset is more likely to have a negative return?  Thanks for any help!&lt;/p&gt;&#xA;" OwnerUserId="4773" LastActivityDate="2010-12-15T06:54:29.893" Title="Normal distribution probability" Tags="&lt;probability&gt;&lt;homework&gt;&lt;economics&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="14389" PostTypeId="1" CreationDate="2010-12-15T06:32:46.160" Score="6" ViewCount="439" Body="&lt;p&gt;I was asked to compute the Fourier series for $\sin^2(x)$ on $[0,\pi]$. Now this is what I did and I'd like to know if I'm right.  $\sin^2(x)=\frac12-\frac12\cos(2x)$ . I got the right hand side using trig identities. I'm wondering If I can do this without using the formulas. Thanks.&lt;/p&gt;&#xA;" OwnerUserId="4774" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-15T09:31:49.103" LastActivityDate="2010-12-16T07:09:59.407" Title="Fourier series for $\sin^2(x)$" Tags="&lt;homework&gt;&lt;fourier-series&gt;" AnswerCount="4" CommentCount="2" />
  <row Id="14390" PostTypeId="2" ParentId="14389" CreationDate="2010-12-15T06:50:03.237" Score="2" ViewCount="" Body="&lt;p&gt;Sure. That is the Fourier series for the function.&lt;/p&gt;&#xA;" OwnerUserId="3249" LastActivityDate="2010-12-15T06:50:03.237" CommentCount="8" />
  <row Id="14391" PostTypeId="2" ParentId="14387" CreationDate="2010-12-15T06:54:29.893" Score="5" ViewCount="" Body="&lt;p&gt;I don't know if that's what you are looking for, but if $X$ is normally distributed with mean $\mu$ and standard deviation $\sigma &amp;gt; 0$, then &#xA;$$&#xD;&#xA;{\rm P}(X&amp;lt;0) = {\rm P}\bigg(\frac{{X - \mu }}{\sigma } &amp;lt; \frac{{0 - \mu }}{\sigma }\bigg) = {\rm P}\bigg(Z &amp;lt; \frac{{ - \mu }}{\sigma }\bigg),&#xD;&#xA;$$&#xD;&#xA;where $Z$ is a standard normal random variable, i.e. having mean zero and unit variance (or standard deviation). The probability ${\rm P}(Z&amp;lt;z)$ can be found using a Normal Distribution Calculator, and many are available online (check, e.g., &lt;a href=&quot;http://davidmlane.com/hyperstat/z_table.html&quot; rel=&quot;nofollow&quot;&gt;this one&lt;/a&gt;). &lt;/p&gt;&#xA;" OwnerUserId="2810" LastActivityDate="2010-12-15T06:54:29.893" CommentCount="3" />
  <row Id="14392" PostTypeId="2" ParentId="14386" CreationDate="2010-12-15T07:35:31.143" Score="0" ViewCount="" Body="&lt;p&gt;I am not sure why I cannot comment, but anyway, you can always do an RREF and then you should be fine.&lt;/p&gt;&#xA;" OwnerUserId="4728" LastActivityDate="2010-12-15T07:35:31.143" CommentCount="4" />
  <row Id="14393" PostTypeId="2" ParentId="14383" CreationDate="2010-12-15T07:45:32.823" Score="3" ViewCount="" Body="&lt;p&gt;Hint: You can choose $3$ points in ${2010}\choose {3}$ ways. To form a right angled triangle, we need to choose a pair of diametrically opposite points (how many ways to do this?) and then any third point will form a right angled triangle with the chosen pair. &lt;/p&gt;&#xA;" OwnerUserId="3431" LastActivityDate="2010-12-15T07:45:32.823" CommentCount="8" />
  <row Id="14394" PostTypeId="2" ParentId="14383" CreationDate="2010-12-15T07:45:44.583" Score="11" ViewCount="" Body="&lt;p&gt;Let the points be $P_1$, $P_2$, and $P_3$.  Letting $P_1$ be arbitrary, we win if $P_2$ is the point diametrically opposite to $P_1$, probability $1/2009$.  Otherwise, with probability $2008/2009$, we still win if $P_3$ is diametrically opposite either $P_1$ or $P_2$, which it is with probability $2/2008$.  So the probability of a win is $1/2009 + (2008/2009)(2/2008) = 3/2009$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can also get this as follows:  The number of possible wins is the number of diameters times the number of remaining points, or 1005*2008. The number of possible triples is $2010\choose 3$. Dividing the first by the second gives $3/2009$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even easier:  The probability that a given pair of points lie on a diameter is $1/2009$.  With three points, you have three pairs, hence $3/2009$.  Here it's OK to add probabilities because it's not possible to have overlap; if one of the pairs lies on a diameter, no other pair can lie on a diameter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;BTW, I do not agree with the comment thread that says we must consider the possibility that the points are not distinct.  Sometimes there can be an ambiguity, but in common language &quot;choose three points from 2010 points&quot; means &quot;without replacement&quot;.  For example, I am not incorrect in saying that $n\choose r$ is the number of ways of choosing $r$ objects out of a set of $n$.  Although I admit that if I were grading the test and this issue were pointed out to me, I might accept the &quot;with replacement&quot; answer as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only other alternative answer I might accept is 1, i.e., that the points are certain to lie on a right triangle.  After all, it is Ford Prefect, so I have no way of knowing whether the Infinite Improbability Drive has been activated.&lt;/p&gt;&#xA;" OwnerUserId="2682" LastEditorUserId="2682" LastEditorDisplayName="" LastEditDate="2010-12-15T12:23:19.443" LastActivityDate="2010-12-15T12:23:19.443" CommentCount="3" />
  <row Id="14395" PostTypeId="2" ParentId="14383" CreationDate="2010-12-15T07:51:18.967" Score="1" ViewCount="" Body="&lt;p&gt;As you know, the triangle will be right-angled if and only if two of the three points lie on opposite ends of a diameter. There are $2010^3$ ways of choosing three points (your problem didn't specify that the points are distinct). Let's count, how many of the choices give a right-angled triangle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are lots of ways of doing this counting, here is one of them: I will first count the right-angled triangles whose right angle lies on vertex number 1. All the others will be obtained from these by rotations, so the number will have to by multiplied by 2010. Next, I can choose the vertices of any triangle in any order, so I will have to multiply by 3!.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, once I have fixed the vertex 1, there are 2008 ways of choosing the second vertex (I am not allowed to take the same vertex, nor the opposite one, since I want my right angle to be at vertex 1). But that determines the third vertex uniquely, since it has to lie opposite the second, so I get 2008 possibilities. But wait a minute, I have counted each triangle twice, since putting the second vertex at 2 and the third at the opposite end gives the same triangle as swapping the second and the third vertex. So I get 1004 right-angled triangles with the right angle at vertex 1, so by rotating I get 1004*2010 right angled triangles altogether.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any such triangle can be obtained by 6 different choices of vertices, since I can choose the vertices in any order, so the probability is $\frac{6*1004*2010}{2010^3}=\frac{6024}{2010^2}$.&lt;/p&gt;&#xA;" OwnerUserId="3212" LastActivityDate="2010-12-15T07:51:18.967" CommentCount="1" />
  <row Id="14397" PostTypeId="2" ParentId="14384" CreationDate="2010-12-15T07:55:13.637" Score="5" ViewCount="" Body="&lt;p&gt;The problem stems from ambiguous use of term chance and probability. Probability can be defined as a ratio of successful outcomes and number of trials. So if we increase the number of trials and let the probability remain the same, the number of successful outcomes must increase in order that the ratio remains the same. So if we say that number of successful outcomes are chances, then yes, chances are increasing, yet the probability remains the same. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another way to look at that is to look at the probability of getting a no. With one person the probability is $0.99$. If you ask 100 persons, then the probability that they all say no is $0.99^{100}\approx 0.36$. So the probability that you will get at least one positive answer after asking 100 persons is actualy $1-0.99^{100}\approx0.64$. If you ask 1000 persons, then the probability of getting a no from all of them is $0.99^{1000}\approx 4.31\times 10^{-5}$, so probability of one yes is $1-0.99^{1000}\approx0.99999$. But that is not surprising, since saying that probability of a yes is 0.01 you are effectively saying that 1% of population will say yes. So among 1000 persons there should be 10 who say yes, so getting only one yes from 1000 persons is less likely (the event has smaller probability) than getting 10.&lt;/p&gt;&#xA;" OwnerUserId="4742" LastActivityDate="2010-12-15T07:55:13.637" CommentCount="1" />
  <row Id="14398" PostTypeId="2" ParentId="14386" CreationDate="2010-12-15T08:09:12.293" Score="2" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Updated according to comments.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are worried about the numerical stability do &lt;a href=&quot;http://en.wikipedia.org/wiki/QR_decomposition&quot; rel=&quot;nofollow&quot;&gt;QR decomposition&lt;/a&gt; of matrix $A$. Then $A=QR$, where $Q$ is orthogonal and $R$ is triangular. Then you need to check whether $x$ satisfies the equation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$Rx=Q^Tb$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now since $R$ is triangular and $n&amp;gt;k$ we will have that the last $n-k$ rows of $R$ are zero. Since $Ax=b$ it follows that the last $n-k$ elements of $Q^Tb$ should be zero also. If they are not, then $x$ is not a solution. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore since we have an overdetermined matrix the solution exists only if $b$ lies in the linear space spanned by columns of $A$. So the real question is, how do we reliably check whether $b$ is in linear space spanned by columns of $A$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update 2&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $n&amp;gt;k$, the $R$ matrix will look like:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align*}&#xA;R=\begin{bmatrix}&#xA;R_1\\&#xA;0&#xA;\end{bmatrix}&#xA;\end{align*}&#xA;where $R_1$ is $k\times k$ upper triangular matrix. If the solution exists, then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align*}&#xA;Q^Tb=\begin{bmatrix}&#xA;b_1\\&#xA;0&#xA;\end{bmatrix}&#xA;\end{align*}&#xA;where $b_1$ is $k\times 1$ vector. The solution for our system is then&#xA;$$x=R_1^{-1}b_1$$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="4742" LastEditorUserId="4742" LastEditorDisplayName="" LastEditDate="2010-12-15T11:41:08.443" LastActivityDate="2010-12-15T11:41:08.443" CommentCount="11" />
  <row Id="14399" PostTypeId="1" AcceptedAnswerId="14404" CreationDate="2010-12-15T08:10:26.397" Score="0" ViewCount="226" Body="&lt;p&gt;I am using math as a tool and for entertainment, I am not a pro, I do not understand most of the posts here but I am glad such site exists.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, a friend asked me a question the other day, and I was able to redefine it in terms of a directed graph where all nodes in and out degree equal to 1. I know that If if each node's indegree equals to outdegree that is called an Eulerian graph, what I am asking is a special case of these graphs. Do they have a name or are their properties known?&lt;/p&gt;&#xA;" OwnerUserId="4776" LastActivityDate="2010-12-15T09:28:19.107" Title="Directed graph with all indegree = outdegree = 1" Tags="&lt;graph-theory&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="14401" PostTypeId="2" ParentId="14389" CreationDate="2010-12-15T08:26:52.053" Score="1" ViewCount="" Body="&lt;p&gt;@AD&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even though the question has been answered... the Fourier expansion is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f(x) = \frac{a_0}{2} + \sum_{n=1}^{\infty}\left[a_n\cos(nx) + b_n\sin(nx)\right]$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is a decomposition over the set of orthogonal trigonometric functions $\{1, \sin(kx), \cos(kx)\}, \forall k \in \mathbb{Z}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, that means that really the Fourier expansion of a function is unique. And so, simply by identifying $a_0 = \frac{1}{2}$, $a_2 = -\frac{1}{2}$ and all of the other coefficients 0, we find a pairing that has the desired form of a Fourier Series, and which therefore &lt;em&gt;is&lt;/em&gt; the Fourier expansion of $f(x)$.&lt;/p&gt;&#xA;" OwnerUserId="4139" LastEditorUserId="4139" LastEditorDisplayName="" LastEditDate="2010-12-15T09:08:05.217" LastActivityDate="2010-12-15T09:08:05.217" CommentCount="3" />
  <row Id="14402" PostTypeId="2" ParentId="14113" CreationDate="2010-12-15T08:36:00.337" Score="1" ViewCount="" Body="&lt;p&gt;You should take a look at the paper: &lt;a href=&quot;http://projecteuclid.org/euclid.em/1087568020&quot; rel=&quot;nofollow&quot;&gt;Some New Formulas for π&lt;/a&gt; by Gert Almkvist, Christian Krattenthaler, and Joakim Petersson, &lt;em&gt;Experiment. Math.&lt;/em&gt; Volume 12, Number 4 (2003), 441-456.&lt;/p&gt;&#xA;" OwnerUserId="1154" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-15T09:07:35.300" LastActivityDate="2010-12-15T09:07:35.300" />
  <row Id="14403" PostTypeId="1" AcceptedAnswerId="14641" CreationDate="2010-12-15T09:00:15.290" Score="0" ViewCount="121" Body="&lt;p&gt;It is referred to in &lt;a href=&quot;http://ifile.it/niok3wy/jeffreys-an_invariant_form-1946.pdf&quot; rel=&quot;nofollow&quot;&gt;An invariant form for the prior probability in estimation problems (Jeffreys)&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="774" LastActivityDate="2010-12-17T07:17:42.670" Title="What is the $d\sigma/\sigma$ rule?" Tags="&lt;probability-theory&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" />
  <row Id="14404" PostTypeId="2" ParentId="14399" CreationDate="2010-12-15T09:28:19.107" Score="3" ViewCount="" Body="&lt;p&gt;These types of graphs are equivalent to &lt;a href=&quot;http://en.wikipedia.org/wiki/Permutation&quot; rel=&quot;nofollow&quot;&gt;permutations&lt;/a&gt; (or permutations without fixed points, if you're not allowed a vertex to have an edge to itself).  If there is an edge from a to b, then a is mapped to b.  These are extremely well-studied in mathematics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[As Prometheus remarks, permutations are equivalent to a collection of cycles]&lt;/p&gt;&#xA;" OwnerUserId="139" LastActivityDate="2010-12-15T09:28:19.107" CommentCount="2" />
  <row Id="14406" PostTypeId="1" AcceptedAnswerId="14414" CreationDate="2010-12-15T11:27:25.777" Score="13" ViewCount="349" Body="&lt;p&gt;I was watching this &lt;a href=&quot;http://www.ted.com/talks/nicholas_christakis_how_social_networks_predict_epidemics.html&quot; rel=&quot;nofollow&quot;&gt;TED talk&lt;/a&gt;, which suggested that on average your friends tend to individually have more friends than you do. To define this more formally, we are comparing the average number of friends with:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;average over each person p of:&#xA;    friend popularity, defined as:&#xA;    average over each friend f of p:&#xA;        number of friends f has&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Intuitively, this seems to make sense. After all, if someone has a high number of friends, they will tend to increase friend popularity and affect a high number of people, while those people who decrease friend popularity only affect a low number of people. Does this result hold for all graphs?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given a person &lt;code&gt;p&lt;/code&gt;, let &lt;code&gt;t&lt;/code&gt; stand for:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;sum over each friend f of p:&#xA;    number of friends f has&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;It is pretty clear that &lt;code&gt;sum(t)=sum(f^2)&lt;/code&gt; as a person with &lt;code&gt;f&lt;/code&gt; friends has value of &lt;code&gt;f&lt;/code&gt; towards their &lt;code&gt;f&lt;/code&gt; friends value of &lt;code&gt;t&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We are then trying to determine whether: &lt;code&gt;sum(t/f)&amp;gt;sum(f)&lt;/code&gt; holds for all graphs.&lt;/p&gt;&#xA;" OwnerUserId="123" LastEditorUserId="123" LastEditorDisplayName="" LastEditDate="2010-12-15T11:47:45.143" LastActivityDate="2010-12-15T14:12:12.140" Title="Do your friends on average have more friends than you do?" Tags="&lt;graph-theory&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="3" />
  <row Id="14407" PostTypeId="2" ParentId="13053" CreationDate="2010-12-15T11:34:42.273" Score="3" ViewCount="" Body="&lt;p&gt;To expand on the comments I made (and Anixx has confirmed that it works):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem, as noted in the &lt;em&gt;Mathematica&lt;/em&gt; documentation for &lt;a href=&quot;http://reference.wolfram.com/mathematica/ref/Zeta.html&quot; rel=&quot;nofollow&quot;&gt;&lt;code&gt;Zeta[]&lt;/code&gt;&lt;/a&gt; and &lt;a href=&quot;http://reference.wolfram.com/mathematica/ref/HurwitzZeta.html&quot; rel=&quot;nofollow&quot;&gt;&lt;code&gt;HurwitzZeta[]&lt;/code&gt;&lt;/a&gt; is that those two functions are defined slightly differently:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathrm{Zeta}(s,a):=\sum_{j=0}^\infty\frac1{((j+a)^2)^{\frac{s}{2}}}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\zeta(s,a):=\sum_{j=0}^\infty\frac1{(j+a)^s}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Though the two expressions are equivalent for $\Re a &amp;gt; 0$, they behave rather differently in the left half-plane:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;N[Zeta[1/3, -1/5], 20]&#xA;0.99113858338730120895&#xA;&#xA;N[HurwitzZeta[1/3, -1/5], 20]&#xA;0.1361506100489527143 - 1.4808826096823642385 I&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;and this manifests itself when you use &lt;code&gt;Zeta[]&lt;/code&gt; instead of &lt;code&gt;HurwitzZeta[]&lt;/code&gt; for your gamma function expression in the left half-plane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Not knowing how they implemented &lt;code&gt;HurwitzZeta[]&lt;/code&gt; internally, however, (it seems to have only been implemented starting with version 7 of &lt;em&gt;Mathematica&lt;/em&gt;), I cannot say what, exactly, are the differences in the behavior of those two functions in the left half of the complex plane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an aside, a similar issue arises with the Lerch transcendent $\Phi(z,s,a)$, the trivariate generalization of the zeta function and the related polylogarithms; &lt;em&gt;Mathematica&lt;/em&gt; from version 7 onwards provides for two different functions, &lt;a href=&quot;http://reference.wolfram.com/mathematica/ref/LerchPhi.html&quot; rel=&quot;nofollow&quot;&gt;&lt;code&gt;LerchPhi[]&lt;/code&gt;&lt;/a&gt; and &lt;a href=&quot;http://reference.wolfram.com/mathematica/ref/HurwitzLerchPhi.html&quot; rel=&quot;nofollow&quot;&gt;&lt;code&gt;HurwitzLerchPhi[]&lt;/code&gt;&lt;/a&gt; which also agree for $\Re a &amp;gt; 0$ but behave differently otherwise.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-12-15T11:34:42.273" CommentCount="5" />
  <row Id="14408" PostTypeId="1" AcceptedAnswerId="14419" CreationDate="2010-12-15T12:02:59.000" Score="1" ViewCount="59" Body="&lt;p&gt;It's possible, that count of unique algebras (up to isomorphism) with one unary operation on set with $n$ elements is $2^n-1$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $n=1,2,3,4$ is this hypothesis true (I still have not verified it on the computer). Is there any counterexample, or idea for proof?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="4195" LastEditorUserId="4195" LastEditorDisplayName="" LastEditDate="2010-12-15T12:10:53.023" LastActivityDate="2010-12-15T15:32:07.513" Title="Count of unique algebras with one unary operation (on finite set)" Tags="&lt;algebra&gt;&lt;universal-algebra&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="14409" PostTypeId="1" CreationDate="2010-12-15T12:41:19.163" Score="0" ViewCount="166" Body="&lt;p&gt;I was trying to expand $\displaystyle \left(1+x\right)^{y}$ as a power series in terms of $y$ using the Generalized Binomial Theorem, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle \left(1+x\right)^{y}=\sum_{k=0}^{\infty}\binom{y}{k}x^{k}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume $x,y \in \mathbb{R}$ and $\left | x \right | &amp;lt; 1$. With this in mind I would get something like &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle \left(1+x\right)^{y}=\sum_{k=0}^{\infty}a_{k}\cdot y^{k}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is, what is the general form of the $a_{k}$? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;thanks.&lt;/p&gt;&#xA;" OwnerUserId="1747" LastActivityDate="2010-12-15T12:41:19.163" Title="Generalized Binomial Expansion of $\left(1+x \right )^{y}$" Tags="&lt;real-analysis&gt;&lt;sequences-and-series&gt;&lt;power-series&gt;" CommentCount="10" />
  <row Id="14410" PostTypeId="2" ParentId="14406" CreationDate="2010-12-15T13:13:49.847" Score="4" ViewCount="" Body="&lt;p&gt;Trivial comment, but I'm new so I have to post it as an answer: The strict inequality definitely doesn't hold in general, since the sums are equal for regular graphs.&lt;/p&gt;&#xA;" OwnerUserId="4743" LastActivityDate="2010-12-15T13:13:49.847" />
  <row Id="14411" PostTypeId="1" AcceptedAnswerId="14413" CreationDate="2010-12-15T13:37:48.860" Score="1" ViewCount="185" Body="&lt;p&gt;Hi,&#xA;It is well known that any conformal mapping of the unit disk onto itself extends to the unit cirle. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, is there an homeomorphism of the unit disk onto itself which does not extend to a continuous function on the closed unit disk?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If yes, can you give an explicit one?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you,&#xA;Malik&lt;/p&gt;&#xA;" OwnerUserId="1197" LastActivityDate="2011-05-02T23:56:09.003" Title="Homeomorphism of the unit disk onto itself which does not extend to the boundary" Tags="&lt;analysis&gt;" AnswerCount="2" />
  <row Id="14412" PostTypeId="1" CreationDate="2010-12-15T13:54:55.610" Score="4" ViewCount="142" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $\Omega$ be a domain of the complex plane. The Hardy space $H^p(\Omega)$ is defined, for $1 \leq p&amp;lt;\infty$, as the class of functions $f$ that are holomorphic on $\Omega$ such that $|f|^p$ has a harmonic majorant on $\Omega$, i.e. there is a function $u$ harmonic on $\Omega$ such that&#xA;$$|f(z)|^p \leq u(z) $$&#xD;&#xA;for all $z \in \Omega$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $p=\infty$, $H^\infty(\Omega)$ is the class of bounded holomorphic functions on $\Omega$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm interested in cases when $H^p(\Omega)$ consist only of constant functions. For example, this is the case when $\Omega$ is the whole plane, because positive harmonic functions on $\mathbb{C}$ are constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I came upon the following question :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $E$ be a compact subset of the real line, and suppose that $E$ has zero length. Let $\Omega$ be the complement of $E$. Does $H^p(\Omega)$ consist only of the constant functions?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $p=\infty$, the answer is yes : one can use Cauchy's formula to extend any bounded holomorphic function on $\Omega$ to a bounded holomorphic function on $\mathbb{C}$, and that function is now constant by Liouville's theorem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $1 \leq p&amp;lt;\infty$, I am pretty sure the answer is also yes. However, I can't seem to find a way to extend $f$ or the harmonic majorant of $|f|^p$ to the whole plane. Is there any way to do so?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you,&#xA;Malik&lt;/p&gt;&#xA;" OwnerUserId="1197" LastActivityDate="2010-12-26T15:43:06.340" Title="Removable sets for harmonic functions and hardy spaces of general domains" Tags="&lt;complex-analysis&gt;&lt;harmonic-analysis&gt;" FavoriteCount="1" />
  <row Id="14413" PostTypeId="2" ParentId="14411" CreationDate="2010-12-15T14:06:29.490" Score="11" ViewCount="" Body="&lt;p&gt;Yes, suppose we have in polar coordinates $f(r,\theta) = (r,\theta + r/(1-r))$.  In effect the disk is rotated by increasing amounts as $r$ tends to 1.  It is still a homeomorphism on the interior of the unit disk, but it's impossible to extend continuously to the boundary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added:&lt;/strong&gt; This map cannot be extended continuously to any point of the boundary.  If it could be, then for any &quot;target&quot; open neighborhood of the image of that point, we should be able to find an open neighborhood of the point whose image is contained in the &quot;target&quot;.  But the open neighborhood of any boundary point will be mapped to points arbitrarily close to every point on the boundary (and so cannot be contained in any sufficiently small &quot;target&quot;).&lt;/p&gt;&#xA;" OwnerUserId="3111" LastEditorUserId="3111" LastEditorDisplayName="" LastEditDate="2010-12-15T18:23:27.797" LastActivityDate="2010-12-15T18:23:27.797" CommentCount="1" />
  <row Id="14414" PostTypeId="2" ParentId="14406" CreationDate="2010-12-15T14:12:12.140" Score="13" ViewCount="" Body="&lt;p&gt;The answer is yes, this holds for any graph (with weak inequality, as Jon points out).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's set up some notation. The graph of friendships is $G$. The set of vertices of $G$ (the people) is $V$; the set of edges (the friendships) is $E$. For a person $v$, the number of friends that person has is $\deg v$. The total number of people is $n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We want to show that&#xA;$$\frac{1}{n} \sum_{v \in V} \deg v \leq \frac{1}{n} \sum_{v \in V} \frac{1}{\deg v} \sum_{(u,v) \in E} \deg(u).$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cancel the $1/n$'s from both sides. After a little rewriting, we want to show that&#xA;$$\sum_{v \in V} \sum_{(u,v) \in E} 1 \leq \sum_{v \in V} \sum_{(u,v) \in E} \frac{\deg u}{\deg v}. \quad (*)$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's consider what a given edge $(u,v)$ contributes to each side of $(*)$. On the left, it contributes $1+1=2$. On the right, it contributes $(\deg u)/(\deg v) + (\deg v)/(\deg u)$. For any two positive numbers $x$ and $y$, we have $2 \leq x/y+y/x$. So every edge contributes more to the right hand side of $(*)$ than to the left, and we have the claimed result.&lt;/p&gt;&#xA;" OwnerUserId="448" LastActivityDate="2010-12-15T14:12:12.140" CommentCount="2" />
  <row Id="14415" PostTypeId="2" ParentId="14411" CreationDate="2010-12-15T14:20:34.210" Score="0" ViewCount="" Body="&lt;p&gt;This following gives an example, &lt;em&gt;probably&lt;/em&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $H=\{(x,y)\in\mathbb R^2:y&amp;gt;0\}$ be the upper half plane in $\mathbb R^2$ and let $\bar H$ be its closure. Let me give you an homeo $f:H\to H$ which does not extend to an homeo $\bar f:\bar H\to\bar H$, and which fixes the point of infinity on the $x$-axis. Then you can conjugate with an homeo $D\to H$ from the open unit disk which extends to the boundary (apart from one point...)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $\phi:\mathbb R\times(0,\infty)\to\mathbb R$ be given by $\phi(x,t)=\frac1{\sqrt{4\pi t}}\exp\left(-\frac{x^2}{4t}\right)$, and let $h:\mathbb R\to\mathbb R$ be a positive smooth function with support on $[-2,2]$ and such that it is constantly $1$ on $[-1,1]$. Define now $g:H\to H$ so that when $x\in\mathbb R$ and $y&amp;gt;0$ we have $$&#xD;&#xA;g(x,y)=\int_{-\infty}^\infty\phi(x-\xi,y)(1-h(\xi))\,\mathrm d\xi.&#xD;&#xA;$$ This is a solution of the one-dimensional head equation. Properties of that equation (which one can easily show in this case!) imply that $f$ extends continuously to a map $\bar g:\bar H\to\mathbb H$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now consider the map $f:(x,t)\in H\mapsto (xg(x,t),t)\in H$. This is an homeomorphism, and it extends to the map $\bar f:(x,t)\in\bar H\mapsto (x\bar g(x,t),t)\in\bar H$, which is not an homeo.&lt;/p&gt;&#xA;" OwnerUserId="274" LastEditorUserId="274" LastEditorDisplayName="" LastEditDate="2010-12-15T14:27:04.660" LastActivityDate="2010-12-15T14:27:04.660" />
  <row Id="14416" PostTypeId="2" ParentId="14327" CreationDate="2010-12-15T14:33:17.497" Score="2" ViewCount="" Body="&lt;p&gt;Thinking more, as n gets large we can pass to the continuous case and get rid of the ceilings and e.  So imagine we are pulling numbers from [0,1] n times and each time you can accept it and take the value or decline it and take your future chances.  Let E(n) be the expectation value of n pulls.  Again, when you have n pulls left you should accept anything greater than E(n-1).  So the recurrence is $E(n)=(1-E(n-1))\frac{1+E(n-1)}{2}+E(n-1)^2$ or $E(n)=\frac{1+E(n-1)^2}{2}$  I think we were told &lt;a href=&quot;http://math.stackexchange.com/questions/12485/whats-known-about-recurrences-involving-a-n2&quot;&gt;here&lt;/a&gt;  that sequences involving E(n-1)^2 are very hard.  In the current case the E(n-1)^2 is divided by 2 but we can take that out by rescaling to D=2E and we have $D(n)=D(n-1)^2+\frac{1}{2}$ with D(1)=1&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-12-15T14:33:17.497" CommentCount="2" />
  <row Id="14417" PostTypeId="1" AcceptedAnswerId="14539" CreationDate="2010-12-15T15:16:28.970" Score="5" ViewCount="417" Body="&lt;p&gt;Can this sum be simplified: $\sum_{k=0}^{n} \binom{n}{k} 2^{\sqrt{k}}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or at least is there a simple fairly tight upperbound?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;&#xA;So I think this sum is more easily bounded than I previously thought:&#xA;Clearly, $\sum_{k=0}^{n} \binom{n}{k} 2^{\sqrt{k}} \leq 2^{\sqrt{n}} \sum_{k=0}^{n} \binom{n}{k} = 2^{n+\sqrt{n}}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, along the same lines as Shai Covo's answer, $\sum_{k=0}^{n} \binom{n}{k} 2^{\sqrt{k}} \geq \binom{n}{n/2} 2^{\sqrt{n/2}}$. The central binomial coefficient: $\binom{n}{n/2}$ is at least $\frac{2^n}{\sqrt{2n}}$ hence $\sum_{k=0}^{n} \binom{n}{k} 2^{\sqrt{k}} \geq 2^{\sqrt{n/2}} \cdot \frac{2^n}{\sqrt{2n}} = \frac{2^{n + \sqrt{n/2}}}{2n}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So $\sum_{k=0}^{n} \binom{n}{k} 2^{\sqrt{k}}$ is $O(2^{n + \sqrt{n}})$ and $\Omega(\frac{2^{n+\sqrt{n/2}}}{n})$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What about expressions of the form:&#xA;$\sum_{k=0}^{n} \binom{n}{k} a^{\sqrt{k}} b^{\sqrt{n-k}}$&lt;/p&gt;&#xA;" OwnerUserId="1487" LastEditorUserId="1487" LastEditorDisplayName="" LastEditDate="2010-12-15T19:28:07.063" LastActivityDate="2010-12-17T06:45:13.783" Title="Simplify $\sum_{k=0}^{n} \binom{n}{k} 2^{\sqrt{k}}$" Tags="&lt;combinatorics&gt;&lt;binomial-coefficients&gt;" AnswerCount="3" CommentCount="5" FavoriteCount="0" />
  <row Id="14418" PostTypeId="2" ParentId="14356" CreationDate="2010-12-15T15:17:11.550" Score="1" ViewCount="" Body="&lt;p&gt;Since you have more of an engineering background, I would suggest the textbook of Robert McOwen, &lt;em&gt;Partial Differential Equations: Methods and Applications&lt;/em&gt;. Chapter 1 is about solving first order partial differential equations, where the method of characteristics is most applicable. Chapters 2 and 3 also contains some application of the notion of characteristics to higher order equations and to the wave equation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Compare to the treatments in other standard, more theory-oriented textbooks, I find McOwen's treatment of the method of characteristics to be more detailed and complete, and easier to teach from. As a remark: section 1.3, the application to fully non-linear first order equations, is very difficult to wrap your head around. So don't worry if you feel like you suddenly hit a wall when you get to that section. &lt;/p&gt;&#xA;" OwnerUserId="1543" LastActivityDate="2010-12-15T15:17:11.550" />
  <row Id="14419" PostTypeId="2" ParentId="14408" CreationDate="2010-12-15T15:32:07.513" Score="3" ViewCount="" Body="&lt;p&gt;In brief: no. There are 19 algebras on 4 elements, and the pattern breaks down after that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instead of algebras with one unary operation, we might as well speak of nonisomorphic functions on a set of $n$ elements. The answer is then given by Theorem 6 of &quot;The Number of Structures of Finite Relations&quot; by Davis, R.L. A pdf is available &lt;a href=&quot;http://www.ams.org/journals/proc/1953-004-03/S0002-9939-1953-0055294-2/S0002-9939-1953-0055294-2.pdf&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="4743" LastActivityDate="2010-12-15T15:32:07.513" />
  <row Id="14420" PostTypeId="2" ParentId="639" CreationDate="2010-12-15T16:00:57.083" Score="3" ViewCount="" Body="&lt;p&gt;Pour all the wine down the drain and nip down to the nearest liquor store. Result, 1000 cups of untainted wine and 0 servant related deaths.&lt;/p&gt;&#xA;" OwnerUserId="4780" LastActivityDate="2010-12-15T16:00:57.083" CommentCount="2" />
  <row Id="14421" PostTypeId="1" AcceptedAnswerId="14425" CreationDate="2010-12-15T16:16:13.607" Score="3" ViewCount="209" Body="&lt;p&gt;Hello.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know there are some encrypting algorithm that once the text (the message) is encrypted there is no way to decrypt it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can think one way of doing this: using as encryption key the current date at the encryption moment. But there is one problem wiht that: Each time you encrypt the same message the result will be different.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think there must be some &quot;math-algorithm&quot; to encrypt without out the posibility to decrypt and that always the same message turns out with the same encrypted message, but I realy can't imagine how to do it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Note: I'm posting this question in this forum and not in &lt;a href=&quot;http://stackoverflow.com&quot;&gt;stackoverflow.com&lt;/a&gt; intentionally, it is not referred to &quot;how to programm the algorithm&quot;, it is refered to &quot;how is the math algorithm&quot;.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="3314" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-15T16:18:35.803" LastActivityDate="2011-01-17T22:20:43.047" Title="Encrypting algorithm" Tags="&lt;algorithms&gt;&lt;cryptography&gt;" AnswerCount="3" CommentCount="6" FavoriteCount="1" />
  <row Id="14422" PostTypeId="2" ParentId="14115" CreationDate="2010-12-15T16:22:08.093" Score="7" ViewCount="" Body="&lt;p&gt;Here's an easy introduction to the basics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Pi Formulas and the Monster Group&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://sites.google.com/site/tpiezas/0013&quot; rel=&quot;nofollow&quot;&gt;http://sites.google.com/site/tpiezas/0013&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4781" LastActivityDate="2010-12-15T16:22:08.093" />
  <row Id="14423" PostTypeId="1" CreationDate="2010-12-15T16:38:40.067" Score="4" ViewCount="196" Body="&lt;p&gt;I am studing for exams and am stuck on this problem.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt; &#xA;Suppose $f$ is an entire function s.t. $f(z) =f(z+1)$ and $|f(z)| &amp;lt; e^{|z|}$.&#xA;Show $f$ is constant.&#xA;&lt;/blockquote&gt; &#xA;&#xA;&lt;p&gt;I've deduced so far that:&#xA;a) $f$ is bounded on every horizontal strip &#xA;b) for every bounded horizontal strip of length greater than 1 a maximum modulus must occur on a horizontal boundary.&lt;/p&gt;&#xA;" OwnerUserId="832" LastActivityDate="2010-12-15T18:01:33.990" Title="Entire &quot;periodic&quot; function" Tags="&lt;complex-analysis&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="1" />
  <row Id="14424" PostTypeId="2" ParentId="14421" CreationDate="2010-12-15T16:43:49.993" Score="5" ViewCount="" Body="&lt;p&gt;As J.M. commented above, the only totally secure cryptographic algorithm is the one-time pad. Because this relies on both the sender and receiver of a message to be able to have access to the &lt;em&gt;same&lt;/em&gt; arbitrarily long random sequence, it is often not feasible. Cryptography originally relied on shared keys, from which simple ciphers, such as &lt;a href=&quot;http://en.wikipedia.org/wiki/Vigen%C3%A8re_cipher&quot; rel=&quot;nofollow&quot;&gt;Vigenere&lt;/a&gt;, &lt;a href=&quot;http://en.wikipedia.org/wiki/Caesar_cipher&quot; rel=&quot;nofollow&quot;&gt;Caesar&lt;/a&gt;, and &lt;a href=&quot;http://en.wikipedia.org/wiki/Substitution_cipher&quot; rel=&quot;nofollow&quot;&gt;substitution&lt;/a&gt;, come from. Unfortunately, all of these algorithms are so simple that using frequency analysis can often be used to break them very quickly. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Modern cryptography is generally split into two core areas: symmetric and asymmetric algorithms. Symmetric seems to be what you are describing, so I will describe that first. In symmetric algorithms, both users must share a key, which they often pass to one another via asymmetric cryptography or using &lt;a href=&quot;http://en.wikipedia.org/wiki/Diffie%E2%80%93Hellman_key_exchange&quot; rel=&quot;nofollow&quot;&gt;Diffie-Hellman key exchange&lt;/a&gt;. Once the key is known between the two parties, eavesdroppers are generally going to have a very difficult time breaking the encryption. As with all things, certain algorithms are stronger than others. Older encryption schemes like &lt;a href=&quot;http://en.wikipedia.org/wiki/Data_Encryption_Standard&quot; rel=&quot;nofollow&quot;&gt;DES&lt;/a&gt; are not as secure as some of the newer ones like &lt;a href=&quot;http://en.wikipedia.org/wiki/Advanced_Encryption_Standard&quot; rel=&quot;nofollow&quot;&gt;AES&lt;/a&gt;, and it's not infeasible to try to break the weaker ones with sufficient computing power. Asymmetric cryptography, or public key cryptography, eliminates the need to share keys between users by admitting a public key and private key for each user. To send something to someone else, you encrypt using their public key (which they publish) and only they are able to decrypt using their private key. &lt;a href=&quot;http://en.wikipedia.org/wiki/RSA&quot; rel=&quot;nofollow&quot;&gt;RSA&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/Elliptic_curve_cryptography&quot; rel=&quot;nofollow&quot;&gt;ECC&lt;/a&gt; are two of the more common public key protocols.&lt;/p&gt;&#xA;" OwnerUserId="1016" LastActivityDate="2010-12-15T16:43:49.993" CommentCount="3" />
  <row Id="14425" PostTypeId="2" ParentId="14421" CreationDate="2010-12-15T16:50:01.953" Score="7" ViewCount="" Body="&lt;p&gt;Such algorithms do exist. Which one you mean depends on what you mean by &quot;no way to decrypt it&quot;. You refer to either:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;A function which takes an input and generates a unique output, with there being ideally no known way to derive the input from the output. These are called hashing algorithms; examples of which include &lt;a href=&quot;http://en.wikipedia.org/wiki/MD5&quot; rel=&quot;nofollow&quot;&gt;MD5&lt;/a&gt;, &lt;a href=&quot;http://en.wikipedia.org/wiki/SHA256&quot; rel=&quot;nofollow&quot;&gt;SHA256&lt;/a&gt;. Those links provide more detail than I possibly can.&lt;/li&gt;&#xA;&lt;li&gt;A cryptosystem that is impossible for any other person but the keyholder to decrypt. This is the &lt;a href=&quot;http://en.wikipedia.org/wiki/One-time_pad&quot; rel=&quot;nofollow&quot;&gt;one time pad&lt;/a&gt; &lt;a href=&quot;http://math.stackexchange.com/users/498/j-m&quot;&gt;J.M.&lt;/a&gt; mentions. This is the only currently &lt;a href=&quot;http://netlab.cs.ucla.edu/wiki/files/shannon1949.pdf&quot; rel=&quot;nofollow&quot;&gt;proven-secure algorithm&lt;/a&gt;.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Edited in because this is probably too long for comments:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Diego, no problem. They're not, as they're two different things and you're comparing apples and oranges, so to speak. A cryptosystem is designed to be hard to undo without the key, but easy if you have the key. By contrast, a hashing function should only require 1 input, be one way and give a unique output for each input. The idea is to be able to generate a unique representation of the output without being able to deduce what the output is. Why?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well aside from verifying the unique nature of your input, a hashing function is used as part of a digital signature. In a public key cryptosystem, you may well be aware that Alice can send Bob an encrypted message using Bob's public key. However, anyone can send Bob a message using Bob's public key. If Eve has broken the cryptosystem, she could intercept Alice's message, decrypt it, change it, re-encrypt it and send it on to Bob.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do we prevent this? Well, let $H: x\rightarrow y$ be a map from $x$ some input to $y$ some output. It is trivial to compute $H$ but impossible (ideally, the truth is just very very unlikely) to compute $H^{-1}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then let $C:x,k$ be some assymetric cryptosystem Alice and Bob know, with Alice's keys $a_p$ and $a_s$ (public and secret) and likewise Bob's keys $b_p$ and $b_s$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we know that as this cryptosystem is assymetric, that $C:x,b_p \rightarrow z$ where $z$ is some ciphertext (encrypted secret) and that $C:z,b_s \rightarrow x$ i.e. if you use Bob's public key to encrypt, Bob's private key using exactly the same algorithm will decrypt the process.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This we know already. However, as discussed, a man in the middle attack is possible if Eve knows Bob's secret. However, Alice can do two things:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Compute $C(H(x), a_s)$&lt;/li&gt;&#xA;&lt;li&gt;Compute $C(x+H(x), b_p)$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;And send the second one to Bob. What's the point in this? Well, Bob can reverse both &quot;encryptions&quot;, because he holds the opposite keys (his own private and Alice's public. He is left with two things: the ciphertext and a unique hash of the ciphertext. He can also compute H(x) and compare results. If they differ, he knows the message has been tampered with.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why? Well, consider Eve sat gleefully with Bob's keys. She intercept's Alices message and sees it contains the two parts (she can decrypt the outer layer, Bob's message). However, now she's left with an impossible situation. She can compute the hash of her newly inserted message, but she does not know Alice's private key (if she does, the cryptosystem is truly broken). Therefore she cannot encrypt the new hash. So, if she tries to send the new message onwards, Bob will immediately be aware the system has been tampered with.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is called a digital signature or digitally signing and is used to identify the message author as who they truly say they are. After all, only they can encrypt with their private key (allegedly). The importance of hashing in this scenario is that it generates a unique output for each given output. However, not being reversible is also desired so that it is impossible to determine what a secret is given the output. This also makes it a highly used password-storage mechanism (because in theory, you're not storing the passwords, just a unique representation of them).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As always in cryptography, someone is always trying to break hash functions. &lt;a href=&quot;http://en.wikipedia.org/wiki/Rainbow_table&quot; rel=&quot;nofollow&quot;&gt;Rainbow tables&lt;/a&gt; are the technique used to pre-compute parts of hashes to make hash reversing easier. Note however that the longer the input, the larger the tables need to be. Hence why passwords should always be greater than 6 characters...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, you may be interested to know in Hash weaknesses. If two hashes with unique input produce the same output, this is called a collision and is a serious weakness, especially if someone can work out any pattern to said collisions. MD5 is thought to suffer from some, as is SHA1. SHA2 may, also. In any case, NIST has launched SHA-3, a competition to find future hash functions offering security now and in many years to come. &lt;a href=&quot;http://csrc.nist.gov/groups/ST/hash/sha-3/Round3/submissions_rnd3.html&quot; rel=&quot;nofollow&quot;&gt;The finalists&lt;/a&gt; are available to view here.&lt;/p&gt;&#xA;" OwnerUserId="892" LastEditorUserId="892" LastEditorDisplayName="" LastEditDate="2011-01-17T22:20:43.047" LastActivityDate="2011-01-17T22:20:43.047" CommentCount="5" />
  <row Id="14429" PostTypeId="1" AcceptedAnswerId="14434" CreationDate="2010-12-15T17:15:43.347" Score="4" ViewCount="164" Body="&lt;p&gt;My motivation for asking this question is that a classmate of mine asked me some kind of question that made me think of this one. I can't recall his exact question because he is kind of messy (both when talking about math and when thinking about math). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm kind of stuck though. I feel like the set $A^{\mathbb{N}} = \{f: \mathbb{N} \rightarrow A, f \text{ is a function} \}$ should have the same cardinality as the power set of A, if A is infinite. On the other hand, in this &lt;a href=&quot;http://math.stackexchange.com/questions/477/cardinality-of-set-of-real-continuous-functions&quot;&gt;post&lt;/a&gt;, it is stated that the sequences with real coefficients have the same cardinality as the reals. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's easy to see that $A^{\mathbb{N}} \subseteq P(A)$, but (obviously) I got stuck on the other inclusion. Is there any general result that says anything else? References would be appreciated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT To clarify the intetion of this question: I want to know if there are any general results on the cardinality of $A^{\mathbb{N}}$ other that it is strictly less than that of the power set of A. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, I was aware that the other inclusion isn't true in general (as the post on here I linked to gave a counterexample), but thanks for pointing out why too. :)&lt;/p&gt;&#xA;" OwnerUserId="1415" LastEditorUserId="630" LastEditorDisplayName="" LastEditDate="2010-12-15T18:19:07.377" LastActivityDate="2010-12-15T18:31:56.200" Title="What's the cardinality of all sequences with coefficients in an infinite set?" Tags="&lt;set-theory&gt;&lt;cardinals&gt;" AnswerCount="4" CommentCount="5" />
  <row Id="14430" PostTypeId="2" ParentId="14423" CreationDate="2010-12-15T17:23:16.173" Score="0" ViewCount="" Body="&lt;p&gt;The following lemma may help.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;If $f$ is an entire function and and $Re(f)$ is bounded, then $f$ is constant.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Write $f(z)=g(z)+ih(z)$ where $g$ and $h$ are real valued functions.  Now use the information in the question to prove that $g$ is bounded which completes the answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To prove the lemma consider $e^{f(z)}$ and use Liouville's theorem.&lt;/p&gt;&#xA;" OwnerUserId="3971" LastActivityDate="2010-12-15T17:23:16.173" CommentCount="1" />
  <row Id="14431" PostTypeId="2" ParentId="14429" CreationDate="2010-12-15T17:28:05.530" Score="3" ViewCount="" Body="&lt;p&gt;Take $A = \mathbb{R}$ then you have:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$|A| = 2^{\aleph_0}$ and so $|A^\mathbb{N}| = (2^{\aleph_0})^{\aleph_0} = 2^{\aleph_0\cdot\aleph_0} = 2^{\aleph_0} = |A| &amp;lt; |\mathcal{P}(A)|$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Addendum&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To the comment, as well as a small &quot;obvious&quot; remark that should probably be mentioned - if $\kappa&amp;gt;1$ and $\lambda\ge\aleph_0$ are two cardinal numbers, to &quot;generate&quot; a relatively small (although this is a bad term, as it is consistent that this size is pretty much unbounded) cardinal which is invariant under powers of $\lambda$, one can always take $\kappa^\lambda$. The above computation is as good for this case to show that the result has this property.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is also obvious that if $\kappa$ is already invariant under powers of $\lambda$ then $\kappa^\lambda = \kappa$.&lt;/p&gt;&#xA;" OwnerUserId="622" LastEditorUserId="622" LastEditorDisplayName="" LastEditDate="2010-12-15T18:31:56.200" LastActivityDate="2010-12-15T18:31:56.200" CommentCount="2" />
  <row Id="14432" PostTypeId="2" ParentId="14429" CreationDate="2010-12-15T17:39:36.960" Score="2" ViewCount="" Body="&lt;p&gt;If $A$ is countably infinite, then the cardinality of $A^{\Bbb N}$ equals the cardinality of the reals: $|A^{\Bbb N}|=|A|^{|\Bbb N|}=\aleph_0^{\aleph_0}\leq (2^{\aleph_0})^{\aleph_0}=2^{\aleph_0\aleph_0}=2^{\aleph_0}=|\Bbb R|$ and $\aleph_0^{\aleph_0}\geq 2^{\aleph_0}$ together gives $|A^{\Bbb N}|=|\Bbb R|$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See &lt;a href=&quot;http://en.wikipedia.org/wiki/Cardinal_number&quot; rel=&quot;nofollow&quot;&gt;the Wikipedia page on cardinal numbers&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As shown by Asaf in his answer, $|A^{\Bbb N}|=2^{\aleph_0}$ if $|A|=2^{\aleph_0}$. It follows easily that if $\aleph_0\leq|A|\leq2^{\aleph_0}$, then still $|A^{\Bbb N}|=2^{\aleph_0}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When $A&amp;gt;2^{\aleph_0}$, things start to become complicated, and depends on whether you assume the Generalised Continuum Hypothesis or not. See for example &lt;a href=&quot;http://www.ucl.ac.uk/~ucahcjm/ast/ast_notes_4.pdf&quot; rel=&quot;nofollow&quot;&gt;these notes by Charles Morgan&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1021" LastEditorUserId="1021" LastEditorDisplayName="" LastEditDate="2010-12-15T17:49:55.753" LastActivityDate="2010-12-15T17:49:55.753" />
  <row Id="14433" PostTypeId="2" ParentId="14417" CreationDate="2010-12-15T17:47:00.183" Score="8" ViewCount="" Body="&lt;p&gt;I will only give some simple and useful estimate (confirmed by numerical results, based on some mathematical idea). It is essentially a lower bound, but can probably be easily modified to be an upper bound. Moreover, it might give you further ideas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Denote your expression by $\varphi_n$, and consider a random variable $X$ with binomial$(n,1/2)$ distribution. Then&#xA;$$&#xD;&#xA;{\rm E}\big[2^{\sqrt X } \big] = \sum\limits_{k = 0}^n {2^{\sqrt k } {n \choose k}\frac{1}{{2^n }}}  = \frac{1}{{2^n }}\varphi _n. &#xD;&#xA;$$&#xD;&#xA;The left-hand side is actually the moment-generating function $M(t)$ of $\sqrt{X}$ at $t=\ln2$, but we will not use this fact.&#xA;I have plotted the function $2^{\sqrt{x}}$, $x &amp;gt; 0$, and it seems that this function is typically convex (it is not convex near $0$). So, heuristically, Jensen's inequality suggests the following result:&#xA;$$&#xD;&#xA;{\rm E}\big[2^{\sqrt X } \big] &amp;gt; 2^{\sqrt {{\rm E}[X]} }, &#xD;&#xA;$$&#xD;&#xA;for all sufficiently large $n$. (If $n$ is large, then $X$ typically takes values in a set on which $2^{\sqrt{x}}$ is convex.) Finally, ${\rm E}[X]=n/2$ leads to $2^{ - n} \varphi _n  &amp;gt; 2^{\sqrt {n/2} }$, that is&#xA;$$&#xD;&#xA;\sum\limits_{k = 0}^n {{n \choose k} 2^{\sqrt k } }  &amp;gt; 2^{n + \sqrt {n/2} }, &#xD;&#xA;$$&#xD;&#xA;for all sufficiently large $n$. Numerical results indicate that this inequality holds for all $n \geq 6$. Moreover, the results suggest that the (lower-)bound is quite tight (I checked for $n \leq 150$). So, a tight upper-bound is likely to be something not far from $2^{n + \sqrt {n/2}}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;: We can also employ the strong law of large numbers (SLLN) to show that $\varphi_n$ may be expected to behave something like $2^{n + \sqrt {n/2}}$ for large $n$. Indeed, writing $X$ as $Y_1 + \cdots + Y_n$, where $Y_i$ are i.i.d. rv's with ${\rm P}(Y_i = 0) = {\rm P}(Y_i = 1) = 1/2$, we have&#xA;$$&#xD;&#xA;\varphi _n  = 2^n {\rm E}\bigg[2^{\sqrt n \sqrt {\frac{{Y_1  +  \cdots  + Y_n }}{n}} } \bigg].&#xD;&#xA;$$&#xD;&#xA;By SLLN, $\frac{{\sum\nolimits_{i = 1}^n {Y_i } }}{n}$ converges to $1/2 \,(= {\rm E}[Y_1])$ almost surely, and the conclusion follows.&lt;/p&gt;&#xA;" OwnerUserId="2810" LastEditorUserId="2810" LastEditorDisplayName="" LastEditDate="2010-12-15T19:52:21.513" LastActivityDate="2010-12-15T19:52:21.513" />
  <row Id="14434" PostTypeId="2" ParentId="14429" CreationDate="2010-12-15T17:48:30.820" Score="7" ViewCount="" Body="&lt;p&gt;From Jech's &lt;strong&gt;Set Theory&lt;/strong&gt;, we have the following theorems on cardinal exponentiation (a Corollary on page 49):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Theorem.&lt;/strong&gt; For all $\alpha,\beta$, the value of $\aleph_{\alpha}^{\aleph_{\beta}}$  is always either:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$2^{\aleph_{\beta}}$; or&lt;/li&gt;&#xA;&lt;li&gt;$\aleph_{\alpha}$; or&lt;/li&gt;&#xA;&lt;li&gt;$\aleph_{\gamma}^{\mathrm{cf}\;\aleph_{\gamma}}$ for some $\gamma\leq\alpha$ where $\aleph_{\gamma}$ is such that $\mathrm{cf}\;\aleph_{\gamma}\leq\aleph_{\beta}\lt\aleph_{\gamma}$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Here, $\mathrm{cf}\;\aleph_{\gamma}$ is the cofinality of $\aleph_{\gamma}$: the cofinality of a cardinal $\kappa$ (or of any limit ordinal) is the least limit ordinal $\delta$ such that there is an increasing $\delta$-sequence $\langle \alpha_{\zeta}\mid \zeta\lt\delta\rangle$ with $\lim\limits_{\zeta\to\delta} = \kappa$. The cofinality is always a cardinal, so it makes sense to understand the operations above as cardinal operations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Corollary.&lt;/strong&gt; If the Generalized Continuum Hypothesis holds, then&#xA;$$\aleph_{\alpha}^{\aleph_{\beta}} = \left\{\begin{array}{lcl}&#xA;\aleph_{\alpha} &amp;amp;\quad &amp;amp; \mbox{if $\aleph_{\beta}\lt\mathrm{cf}\;\aleph_{\alpha}$;}\\&#xA;\aleph_{\alpha+1} &amp;amp;&amp;amp;\mbox{if $\mathrm{cf}\;\aleph_{\alpha}\leq\aleph_{\beta}\leq\aleph_{\alpha}$;}\\&#xA;\aleph_{\beta+1} &amp;amp;&amp;amp;\mbox{if $\aleph_{\alpha}\leq\aleph_{\beta}$.}&#xA;\end{array}\right.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, under GCH, for all cardinals $\kappa$ with cofinality greater than $\aleph_0$ have $\kappa^{\aleph_0} = \kappa$, and for cardinals $\kappa$ with cofinality $\aleph_0$ (e.g., $\aleph_0$, $\aleph_{\omega}$), we have $\kappa^{\aleph_0} = 2^{\kappa}$. (In particular, it is &lt;em&gt;not&lt;/em&gt; the case the cardinality of $A^{\mathbb{N}}$ is necessarily less than the cardinality of $\mathcal{P}(A)$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then again, GCH is usually considered &quot;boring&quot; by set theorists, from what I understand. &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-15T18:28:33.157" LastActivityDate="2010-12-15T18:28:33.157" CommentCount="2" />
  <row Id="14435" PostTypeId="2" ParentId="14423" CreationDate="2010-12-15T18:01:33.990" Score="4" ViewCount="" Body="&lt;p&gt;I'm a little wary of Liouville Theorem approaches... if you choose $f(z) = {1 \over 2}\sin(2\pi z)$ then it satisfies the conditions of the problem except $|f(z)| &amp;lt; e^{2\pi |z|}$ instead of $|f(z)| &amp;lt; e^{|z|}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A suggestion: try showing $f(z) = g(e^{2\pi iz})$ where $g(z)$ is analytic except at $z = 0$. Then translate the condition $|f(z)| &amp;lt; e^{|z|}$ into growth conditions of $|g(z)|$ as $z \rightarrow \infty$ and $z \rightarrow 0$ and show that if they occur $g(z)$ must be constant.  &lt;/p&gt;&#xA;" OwnerUserId="3035" LastActivityDate="2010-12-15T18:01:33.990" />
  <row Id="14436" PostTypeId="2" ParentId="14364" CreationDate="2010-12-15T18:01:59.230" Score="1" ViewCount="" Body="&lt;p&gt;This kind of problem usually succumbs to some sort of &quot;clever choice&quot; of $x$'s to plug in and manipulate. I don't have a full answer for you, but here's Bill's suggestion for showing that such a ring has characteristic $2$ (a necessary condition for being Boolean). Let $a\in R$. Evaluating the identity at $a$ and at $-a$, we have:&#xA;\begin{align*}&#xA;a + a^{2b+1} &amp;amp;= a^{2b} + a^{10b+1}\\&#xA;-a -a^{2b+1} &amp;amp;= a^{2b} - a^{10b+1}&#xA;\end{align*}&#xA;Adding both equalities we obtain $2a^{2b}=0$. Now, using the fact that $a=a^{2b}+a^{10b-1}-a^{2b+1}$, we have:&#xA;$$2a = 2a^{2b} + 2a^{10b-1}-2a^{2b+1} = 0 + 2a^{2b}a^{8b-1} - 2a^{2b}a = 0,$$&#xD;&#xA;so for all $a\in R$ we have $a+a=0$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't have any clever manipulations to suggest for proving $a=a^2$ for all $a$, alas. I tried a few last night, and didn't get anywhere.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-12-15T18:01:59.230" CommunityOwnedDate="2010-12-15T18:18:18.583" CommentCount="1" />
  <row Id="14437" PostTypeId="2" ParentId="14429" CreationDate="2010-12-15T18:08:26.803" Score="3" ViewCount="" Body="&lt;p&gt;Arturo Magidin's answer has the general theorem. Here are two more facts that can be useful:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;If $\aleph_0 \leq \lambda$ and $2 \leq \kappa \leq \lambda$ then $\kappa^\lambda = 2^\lambda = |P(\lambda)|$&#xD;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If $\aleph_0 \leq \lambda \leq \kappa$ then $\kappa^\lambda = |\{ X \subseteq \kappa : |X| = \lambda \}|$&#xD;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="630" LastActivityDate="2010-12-15T18:08:26.803" />
  <row Id="14438" PostTypeId="2" ParentId="14389" CreationDate="2010-12-15T19:03:25.327" Score="3" ViewCount="" Body="&lt;p&gt;One small point...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The way the question is stated, there may be a slight ambiguity.  One way (and almost certainly the intended way) to read the question is:  given the (periodic) function $\sin^2(x)$, find its Fourier series on the interval $[0, \pi]$.  In this case, $(1 - \cos(2x))/2$ is correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, we could also read it as follows:  given the function $\sin^2(x)$ defined on the interval $[0, \pi]$, find its Fourier series.  In this case, we must first decide how to extend the function to be periodic.  Of course the natural choice is to extend it to equal $\sin^2(x)$ for all $x$.  Again, the cosine series is correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But... we do have the freedom to extend $\sin^2(x)$ to an &lt;em&gt;odd&lt;/em&gt; function on $[-\pi, \pi]$ instead, in which case the Fourier series will contain only sine functions (with the coefficients computed in the usual way). The point being that there is in fact another series, featuring only sines, that converges to $\sin^2(x)$ on the interval $[0, \pi]$.  Of course, the convergence isn't as fast ;-).&lt;/p&gt;&#xA;" OwnerUserId="4229" LastEditorUserId="4229" LastEditorDisplayName="" LastEditDate="2010-12-16T07:09:59.407" LastActivityDate="2010-12-16T07:09:59.407" CommentCount="1" />
  <row Id="14439" PostTypeId="2" ParentId="14343" CreationDate="2010-12-15T19:34:56.980" Score="0" ViewCount="" Body="&lt;p&gt;$c = || AB ||;$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\theta_c = 360 - \theta_a - \theta_b;$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$b = {\sin(\theta_b) \over \sin(\theta_c)} c;$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$C.x = b \cos(\theta_a)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$C.y = b \sin(\theta_a)$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="4760" LastActivityDate="2010-12-15T19:34:56.980" CommentCount="1" />
  <row Id="14440" PostTypeId="2" ParentId="14320" CreationDate="2010-12-15T19:42:39.503" Score="5" ViewCount="" Body="&lt;p&gt;This does not answer the question but it is progress towards.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $n=4m$ we can obtain a partition with the required property by taking $P=\lbrace 8,m-1,m+1 \rbrace $ for $m&amp;gt;1 \textrm{ and } m \ne 7 \textrm{ or } 9.$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This partition is distinct from the one given in the question for all $m&amp;gt;2$ and hence the partition in the question is not unique for $n$ a multiple of $4$ and greater than $8.$&#xD;&#xA;(Other solutions exist for $n=28$ and $n=36.$)&lt;/p&gt;&#xA;" OwnerUserId="1301" LastEditorUserId="1301" LastEditorDisplayName="" LastEditDate="2010-12-15T22:04:01.737" LastActivityDate="2010-12-15T22:04:01.737" CommentCount="3" />
  <row Id="14441" PostTypeId="1" CreationDate="2010-12-15T19:42:50.960" Score="0" ViewCount="71" Body="&lt;p&gt;I've been assigned to do a computer model of a simple pendulum. The model uses Runge-Kutta to do the integration to find the position and velocity of the pendulum. I am currently having problems with using the correct intial conditions that would allow me to achieve chaos, was wondering if any one has any pointers?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The pendulum equation is :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a = F cos(omega*t) - c*v - (x-x^3)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where: &#xA;a is acceleration&#xA;F is the driving force&#xA;c*v is the damping force&#xA;(x-X^3) is the duffing oscillator eq&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="4789" LastActivityDate="2010-12-15T19:42:50.960" Title="What are the intial conditions for Chaotic Pendulum behavior?" Tags="&lt;calculus&gt;&lt;chaotic-systems&gt;" CommentCount="5" />
  <row Id="14442" PostTypeId="1" AcceptedAnswerId="14603" CreationDate="2010-12-15T19:49:28.513" Score="7" ViewCount="282" Body="&lt;p&gt;It's pretty easy to construct an (R-S) bi-module M and a left S-module N such that neither M_S nor N is a projective S-module, but the tensor product is a non-zero projective R-module.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, taking R=S=Z defies my zoo of examples.  Here projective=free is closed under direct sums and summands, and so it seems like M and N can be chosen indecomposable.  The indecomposable abelian groups I know don't seem to fit.  I don't know how to &quot;uninvert&quot; using a tensor product.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Can the tensor product of two non-free abelian groups be non-zero free?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Inspired by &lt;a href=&quot;http://www.artofproblemsolving.com/Forum/viewtopic.php?f=61&amp;amp;t=382311&quot; rel=&quot;nofollow&quot;&gt;an AoPS question&lt;/a&gt; and finals week.&lt;/p&gt;&#xA;" OwnerUserId="583" LastActivityDate="2010-12-17T02:02:36.880" Title="Can the tensor product of two non-free abelian groups be non-zero free?" Tags="&lt;abstract-algebra&gt;&lt;modules&gt;&lt;abelian-groups&gt;" AnswerCount="2" CommentCount="0" />
  <row Id="14443" PostTypeId="2" ParentId="14389" CreationDate="2010-12-15T19:57:57.493" Score="1" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Regarding the uniqueness theorem for Fourier series.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I felt the obligation to say something due to all comments and the directed answer. What I meant is that it is in general one can not just say that a trigonometric expansion is a Fourier expansion, you must relay on something (for example that you have a &lt;em&gt;trigonometric polynomial&lt;/em&gt; of basis elements, hence it is the Fourier series of the function). &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;If we look at $I=(0,2\pi)$, then $\Sigma=\{1,\sin(nx),\cos(nx)\}$ makes up an &lt;em&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Orthonormal_basis&quot; rel=&quot;nofollow&quot;&gt;orthonormal basis&lt;/a&gt;&lt;/em&gt; for $L^2(I)$ &#xA;(or piecewise continuous complex valued functions on $I$ below denoted by $C_p(I)$) in the sense&#xA;$$(1)\qquad\langle e,f\rangle=0$$&#xD;&#xA;for all $e_1,e_2\in\Sigma$ where $e_1\ne e_2$, and if $f\in L^2(I)$ (or $f\in C_p(I)$ satisfy) &#xA;$$(2)\qquad \langle e,f\rangle=0$$&#xD;&#xA;for all $e\in\Sigma$ then $f=0$. &#xA;Here &#xA;$$ \langle e,f\rangle =\frac{1}{\pi}\int_0^{2\pi}e(x)\bar{f(x)}dx.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The property (2) is called the uniquness of Fourier expansion for $L^2(I)$ (or $C_p(I)$) and is a cosecuence of the more general statement &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Theorem.&lt;/em&gt; If $f\in L^1(I)$ then $\langle f,e\rangle=0$ for all $e\in\Sigma$ if and only if $f=0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The theorem disurves some comment even though it seams easy. Let us write &#xA;$$ a_n=\frac{1}{2\pi}\int_0^{\pi}f(x)\cos(nx)dx\qquad\text{and}\qquad b_n=\frac{1}{\pi}\int_0^{2\pi}f(x)\sin(nx)dx.$$&#xD;&#xA;The difficult part of the theorem is that in general we do not know if the &lt;a href=&quot;http://en.wikipedia.org/wiki/Fourier_series&quot; rel=&quot;nofollow&quot;&gt;Fourier series&lt;/a&gt; $\frac{a_0}{2}+\sum_k=1^\infty a_n\cos(nx)+b_n\sin(nx)$ &#xA;converges to $f$, and also if it is not obvious that it is it possible to interchange integration and summation like this&#xA;$$\int_0^{2\pi}f(x)\sin(kx)dx=\int_0^{2\pi}(\frac{a_0}{2}+\sum_{k=1}^\infty a_n\cos(nx)+b_n\sin(nx))\sin(kx)dx =$$&#xD;&#xA;$$\qquad\sum_{k=1}^\infty \int_0^{2\pi}(a_n\cos(nx)+b_n\sin(nx))\sin(kx).$$&#xD;&#xA;Also, there are convergent trigonometric series that are &lt;em&gt;not&lt;/em&gt; Fourier series an example is (see Katznelson, Yitzhak (1976) &lt;em&gt;An introduction to harmonic analysis&lt;/em&gt;.) $$\sum_{k=2}^\infty \frac{\sin(nx)}{\log n}$$&#xD;&#xA;A way to prove the theorem is to use that $K_n*f\to f$ in $L^1$-norm where $K_n$ is a trigonometric summation kernel such as the &lt;a href=&quot;http://en.wikipedia.org/wiki/Fej%C3%A9r_kernel&quot; rel=&quot;nofollow&quot;&gt;Fejér kernel&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1154" LastEditorUserId="1154" LastEditorDisplayName="" LastEditDate="2010-12-16T05:13:55.617" LastActivityDate="2010-12-16T05:13:55.617" />
  <row Id="14444" PostTypeId="2" ParentId="14421" CreationDate="2010-12-15T20:25:05.040" Score="1" ViewCount="" Body="&lt;p&gt;I can give an answer, which may seem like a joke. But you can map all text (the set of all codes) to the same result. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Say for example I map the codes &#xA;ABC to 1&#xA;BDCD to 1 &#xA;ABDGR to 1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;there is no opportunity for decoding here. :)&lt;/p&gt;&#xA;" OwnerUserId="4728" LastActivityDate="2010-12-15T20:25:05.040" CommentCount="3" />
  <row Id="14445" PostTypeId="1" CreationDate="2010-12-15T20:30:43.287" Score="5" ViewCount="143" Body="&lt;p&gt;Hello,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;this question may be shameful, but nevertheless I can't help myself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $U \subset \mathbb R^n$ be arbitrary, in particular not the whole of the space itself. I wonder about the dual of the space $W^{1,p}(U)$, for $p &amp;lt; \infty$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $U = \mathbb R^n$, we have $(W^{1,p})&amp;#39; = (W^{1,p&amp;#39;})$ with $p&amp;#39; = \frac{p}{p-1}$. How about different $U$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, in case $U = B_1(0)$ being the closed $1$-Ball, it seems the dual is not a function space. Just recall that the trace is well-defined, linear and continuous on $W^{1,p}(U)$ and, with $S_1$ the boundary of $B_1(0)$ and $w \in L^p(S_1)$, we are given are continuous linear functional by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ W^{1,p}(B_1(0)) \longrightarrow \mathbb C \, , f \mapsto \int_{S_1} w \cdot tr f dx $.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, I wouldn't be surprised if the above example were somehow prototypical, but I have no clue how to proceed from this point. I regard this relevant, as these spaces are ubiquitous in analysis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you!&lt;/p&gt;&#xA;" OwnerUserId="3557" LastEditorUserId="3557" LastEditorDisplayName="" LastEditDate="2010-12-15T20:40:35.800" LastActivityDate="2010-12-16T18:25:22.333" Title="Dual of Sobolev space $W^{1,p}(U)$ for $U$ an arbitrary subset of $\mathbb R^n$" Tags="&lt;functional-analysis&gt;&lt;distribution-theory&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="14446" PostTypeId="1" CreationDate="2010-12-15T20:38:16.443" Score="5" ViewCount="150" Body="&lt;p&gt;Prove that if $\kappa$ is an inaccessible cardinal, then $V_{\kappa}$ satisfies all the axioms of ZFC.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How is this done for the axiom of choice and for regularity?&lt;/p&gt;&#xA;" OwnerUserId="4792" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2010-12-15T23:05:51.700" LastActivityDate="2010-12-16T13:37:04.983" Title="Proving $V_{\kappa}$ is a model of ZFC for inaccessible $\kappa$" Tags="&lt;set-theory&gt;&lt;cardinals&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="14447" PostTypeId="2" ParentId="14445" CreationDate="2010-12-15T20:48:30.817" Score="2" ViewCount="" Body="&lt;p&gt;I think you can find the answer yourself - I will just give you an hint: You are considering the wrong pairing between W^{1,p}(U) and W^{1,p'}(U). (The pairing you consider doesn't give you the desired isomorphism even in the R^n case.)&lt;/p&gt;&#xA;" OwnerUserId="4520" LastActivityDate="2010-12-15T20:48:30.817" CommentCount="1" />
  <row Id="14448" PostTypeId="1" CreationDate="2010-12-15T20:59:24.857" Score="0" ViewCount="153" Body="&lt;p&gt;Is there an explicit expression for &#xA;$$\int_a^b \sqrt{(k-x)^2 + (f(x))^2} dx$$ &#xA;in terms of $\int_a^b f(x)dx$ for constant k ?&lt;/p&gt;&#xA;" OwnerUserId="4793" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-15T21:02:04.637" LastActivityDate="2010-12-15T21:12:29.207" Title="Integration question" Tags="&lt;calculus&gt;" AnswerCount="0" CommentCount="7" />
  <row Id="14449" PostTypeId="2" ParentId="14446" CreationDate="2010-12-15T21:03:35.287" Score="2" ViewCount="" Body="&lt;p&gt;The idea is that you need to remember that whenever ZFC requires you something to be a set, you want it to be an element of $V_\kappa$, as this is the model.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So to expand Arturo's comment, the axiom of choice says that when you have a set of non-empty sets the product is non-empty. Since the sets of $V_\kappa$ are elements of it in the world, you can take a product of them, from the inaccessibility of $\kappa$ you'd have that the rank of the product (in the big universe) is still less than $\kappa$ and thus the product remains within $V_\kappa$ as you'd like it to.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for regularity, well... basically the relation $\in$ is the same as in the world, so it's well-founded and everything holds.&lt;/p&gt;&#xA;" OwnerUserId="622" LastActivityDate="2010-12-15T21:03:35.287" />
  <row Id="14450" PostTypeId="1" AcceptedAnswerId="14453" CreationDate="2010-12-15T21:10:27.277" Score="2" ViewCount="154" Body="&lt;p&gt;I'm trying to understand this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Working in $\mathbb{Z}_p[x]$, $(x-c)^p \equiv x-c \pmod{x^p-x}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From Fermat Little theorem we know that $(x-c)^p \equiv (x-c) \pmod{p}$ but I don't see a connection. Actually, I'm not sure if I'm mixing things up because $\mathbb{Z}_p[x]$ is a polynomial ring.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also I tried to show it from: $x^p \equiv x \pmod{x^p-x}$ holds because $x^p-p \mid x^p-x$. But I don't know how substitute $x-c$ instead of $x$.&lt;/p&gt;&#xA;" OwnerUserId="2468" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-15T21:23:46.763" LastActivityDate="2010-12-16T03:40:22.387" Title="Why is $(x-c)^p \equiv x-c \pmod{x^p-x}$?" Tags="&lt;elementary-number-theory&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="14451" PostTypeId="1" CreationDate="2010-12-15T21:11:02.407" Score="4" ViewCount="117" Body="&lt;p&gt;Given four coordinates that define the corners of an irregular quadrilateral and a point defined by its coordinates, what is the simplest way to determine if the point is within or outside of the quadrilateral?&lt;/p&gt;&#xA;" OwnerUserId="4794" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-16T00:38:28.717" LastActivityDate="2010-12-20T15:24:48.403" Title="Determine if the coordinates of a point are within an irregular quadrilateral whose corners are defined by coordinates" Tags="&lt;geometry&gt;&lt;computational-geometry&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="14453" PostTypeId="2" ParentId="14450" CreationDate="2010-12-15T21:19:34.880" Score="10" ViewCount="" Body="&lt;p&gt;Working in $\mathbb{Z}_p[x]$, you have the &quot;Freshman's Dream&quot; binomial theorem:&#xA;$$(a+b)^p = a^p + b^p.$$&#xD;&#xA;(In fact, this holds in any commutative ring of characteristic $p$, simply by the Binomial Theorem, as mentioned by Qiaochu: all the mixed coefficients are multiples of $p$). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So $(x-c)^p = x^p - c^p$. By Fermat's Little Theorem, you also know that $c^p \equiv c \pmod{p}$, so $(x-c)^p = x^p - c$ in $\mathbb{Z}_p[x]$. That's pretty much all you need.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;hr/&gt;&#xA;&lt;strong&gt;Addendum&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me add that you were very perceptive to notice that even though the &lt;em&gt;function&lt;/em&gt; $x\mapsto (x-c)^p$ is the same as the function $x\mapsto x-c$ in $\mathbb{Z}_p$, this does not mean that the polynomials $(x-c)^p$ and $x-c$ are equal. We are used to (in the integers, rationals, reals, etc) the fact that two polynomials give the same function if and only if they are identical, but this does not occur over finite rings; for example, Fermat's Little Theorem tells you that the polynomial $x^p$ and the polynomial $x$ both define the same function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, what you &lt;em&gt;can&lt;/em&gt; prove is that if $p(x)$ and $q(x)$ are two polynomials with coefficients in $\mathbb{Z}_p$, and they define the same function (that is, $p(a) = q(a)$ for all $a\in\mathbb{Z}_p$), then they differ by a multiple of the polynomial $x^p-x$; i.e., $p(x)\equiv q(x)\pmod{x^p-x}$ in $\mathbb{Z}_p[x]$. What you have here is a special case. See if you can figure out how to prove the more general statement.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-16T02:36:00.890" LastActivityDate="2010-12-16T02:36:00.890" CommentCount="1" />
  <row Id="14454" PostTypeId="2" ParentId="14281" CreationDate="2010-12-15T21:20:12.937" Score="2" ViewCount="" Body="&lt;p&gt;Let $X$ be an $m \times n$ matrix. A norm $\|X\|$ is called &lt;em&gt;unitarily invariant&lt;/em&gt; if $\|UXV\| = \|X\|$ for appropriately shaped unitary matrices $U$ and $V$ (i.e., $U^*U=I_m$ and $V^*V=I_n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the singular value decomposition of $X$ tells us that if $\|\cdot\|$ is a unitarily invariant norm, it will be a function &lt;em&gt;only&lt;/em&gt; of the singular values of $X$ (a so-called spectral function). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A simple observation is that if $X$ is square and diagonalizable, then we can select $S=V$, the matrix of eigenvectors of $X$, and consequently the norms that you are looking for again must be unitarily (here orthogonally) invariant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please have a look in the book &lt;em&gt;Matrix Analysis&lt;/em&gt; by Horn and Johnson, who devote an entire chapter (or more) to matrix norms.&lt;/p&gt;&#xA;" OwnerUserId="1709" LastActivityDate="2010-12-15T21:20:12.937" CommentCount="1" />
  <row Id="14455" PostTypeId="2" ParentId="14451" CreationDate="2010-12-15T21:30:43.137" Score="5" ViewCount="" Body="&lt;p&gt;Take a look at &lt;a href=&quot;http://en.wikipedia.org/wiki/Point_in_polygon&quot; rel=&quot;nofollow&quot;&gt;this wikipedia entry&lt;/a&gt;, or an introductory book on computational geometry.&lt;/p&gt;&#xA;" OwnerUserId="2937" LastActivityDate="2010-12-15T21:30:43.137" CommentCount="1" />
  <row Id="14456" PostTypeId="2" ParentId="14442" CreationDate="2010-12-15T21:38:39.833" Score="5" ViewCount="" Body="&lt;p&gt;Any example must arise from infinitely generated groups.  Suppose $A$ and $B$ are finitely generated, non-zero, non-free, abelian groups.  Being non-free means there is torsion (since we're in the finitely generated case).  If every element of $B$ has finite order, then every element of the tensor product has finite order.  So, assume $b \in B$ has infinite order, and let $a \neq 0$ be an element of order $n$.  Then $a \otimes b$ is non-zero and has order dividing $n$.  Hence, $A \otimes B$ has torsion and, consequently, it is non-free.&lt;/p&gt;&#xA;" OwnerUserId="4009" LastActivityDate="2010-12-15T21:38:39.833" CommentCount="1" />
  <row Id="14457" PostTypeId="1" AcceptedAnswerId="14459" CreationDate="2010-12-15T22:56:09.143" Score="2" ViewCount="109" Body="&lt;p&gt;While I tagged this question homework, it's not really homework. I'm really studying for a test and don't understand how to do something. I have the answer to the question, so the important part is for my to understand how to get there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;X has distribution &lt;code&gt;Gamma(2,4)&lt;/code&gt;&#xA;&lt;code&gt;X = Y-3&lt;/code&gt; Find: &lt;code&gt;P(Y&amp;gt;11|Y&amp;gt;7)&lt;/code&gt; Which I know is equals to:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;(1-P(X&amp;lt;8))/(1-P(X&amp;lt;4)) = (1-Fx(8))/(1-Fx(4))&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have been told that &lt;code&gt;Fx(a) = 1-e^(-a/4)-(a/4)*e^(-a/4)&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I can clearly solve the problem now it's just algebra. &lt;strong&gt;But how do we know what Fx(a) equals???&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="3049" LastActivityDate="2010-12-15T23:16:04.047" Title="Finding conditional probability with Gamma distribution" Tags="&lt;homework&gt;&lt;statistics&gt;" AnswerCount="1" />
  <row Id="14458" PostTypeId="1" AcceptedAnswerId="14607" CreationDate="2010-12-15T23:06:28.583" Score="3" ViewCount="235" Body="&lt;p&gt;Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be strictly monotonically increasing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(i) Is $f$ not continuous at $p \in \mathbb{R}$, there exists a non-empty, open interval $(a_p, b_p) \subset \mathbb{R}$ such that $f(x)\leq a_p$ for all $x &amp;lt; p$ and $f(x) \geq b_p$ for all $x &amp;gt; p$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(ii) The set of discontinuity points $$ \{ p \in \mathbb{R} | f \; \mbox{is not continuous at} \; p \}$$ is countable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For (i) I could be way off, but I am picturing a graph with $p$ on the $x$-axis for which the value $f(p)$ on the $y$-axis is undefined. Am I correct to interpret the open interval $(a_p, b_p)$ as an interval on the $y$-axis which should be contained within the distance between two $f(x)$'s (one for $x&amp;lt;p$ and the other for $x&gt;p$)? If that is so far correct, there could be 2 types of discontinuous points $p$, a jump or removable type. For the jump it would be easier to show that somehow the interval $(a_p, b_p)$ is smaller than the vertical jump... For a removable discontinuity $c$, I would think that the interval could contain just the $y$-axis value $\displaystyle \lim_{x\to c}f(x)$ but I don't really know how to express that the upper and lower bounds would be just above and below that...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With (ii) I am currently trying to understand a proof, what exactly does the notation $f(p-)$ or $f(p+)$ mean in this context? Is it simply the value when approached from the left or the right(respectively)?&lt;/p&gt;&#xA;" OwnerUserId="3711" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-04T06:17:10.633" LastActivityDate="2011-01-04T06:17:10.633" Title="Showing properties of discontinuous points of a strictly increasing function" Tags="&lt;analysis&gt;" AnswerCount="1" CommentCount="6" FavoriteCount="1" />
  <row Id="14459" PostTypeId="2" ParentId="14457" CreationDate="2010-12-15T23:16:04.047" Score="3" ViewCount="" Body="&lt;p&gt;$F_x(a)$ is the cumulative distribution function of a &lt;a href=&quot;http://en.wikipedia.org/wiki/Gamma_distribution#Probability_density_function&quot; rel=&quot;nofollow&quot;&gt;gamma$(2,4)$ distribution&lt;/a&gt;.  This is given by $$F_x(a) = \int_0^a \frac{t e^{-t/4}}{(4^2)\Gamma(2)} dt.$$ &#xA;Integrating (by parts) yields the given expression for $F_x(a)$, since $\Gamma(2) = (2-1)! = 1$.  &lt;/p&gt;&#xA;" OwnerUserId="2370" LastActivityDate="2010-12-15T23:16:04.047" CommentCount="5" />
  <row Id="14460" PostTypeId="2" ParentId="14264" CreationDate="2010-12-15T23:16:17.980" Score="6" ViewCount="" Body="&lt;p&gt;By Ryan Budney's suggestion, I went ahead and proved the general case.  When $B_n$ onto $S_n$ the kernel is isomorphic to $P_n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A proof sketch is this: relations for the Artin generators in $B_n$ must be satisfied in the image.  Relations $b_ib_{i+1}b_i=b_{i+1}b_ib_{i+1}$ can be rewritten as in terms of conjugation so that every $b_i$ has image of a fixed cycle structure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The relations which impose commutativity of non-adjacent generators imply that non-adjacent generators get sent to permutations with cycles either coincidental or disjoint.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It can be shown that for $n&amp;gt;4$ the images of non-adjacent generators must actually be disjoint: mildly technical, but not difficult. (The $n=4$ case being easily solved by hand or GAP).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then counting every other odd generator $b_1,b_3,\ldots$, of which there are $\lceil \frac n 2\rceil$ we have that they must be transpositions, since $3\lceil \frac n 2\rceil&amp;gt;n$.  Essentially that's it: up to isomorphism of $S_n$ the images of the generators for $B_n$ are the usual transpositions they induce, so kernel is $P_n$.&lt;/p&gt;&#xA;" OwnerUserId="4721" LastActivityDate="2010-12-15T23:16:17.980" CommentCount="1" />
  <row Id="14461" PostTypeId="1" AcceptedAnswerId="14464" CreationDate="2010-12-15T23:18:58.030" Score="4" ViewCount="95" Body="&lt;p&gt;Is there a closed form expression for the number of terms in a &lt;a href=&quot;http://en.wikipedia.org/wiki/Symmetric_polynomial#Monomial_symmetric_polynomials&quot; rel=&quot;nofollow&quot;&gt;monomial symmetric polynomial&lt;/a&gt; in a given number of variables for a particular partition of exponents, in terms of which/how many exponents are distinct?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I feel the answer should be straightforward, and it's probably just a contrived statement of a more elementary number theoretic question, but I'm just drawing a blank at the moment. I at first thought that the answer might be&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{N!}{(N-k+1)!}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for $N$ variables and $k$ distinct exponents, which works for all $N=2$ and $N=3$ cases, but predicts 4 terms for a partition $\alpha=(1,1,0,0)$, while there are in fact 6:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$m_\alpha(a,b,c,d) = ab+ac+ad+bc+bd+cd$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Am I being an idiot?&lt;/p&gt;&#xA;" OwnerUserId="67" LastEditorUserId="67" LastEditorDisplayName="" LastEditDate="2010-12-15T23:29:28.790" LastActivityDate="2010-12-15T23:45:01.463" Title="Number of terms in a monomial symmetric polynomial" Tags="&lt;combinatorics&gt;&lt;polynomials&gt;&lt;permutations&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="14462" PostTypeId="2" ParentId="14210" CreationDate="2010-12-15T23:27:39.507" Score="5" ViewCount="" Body="&lt;p&gt;Here's a slightly different way, perhaps a little more sophisticated, to see this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Free groups are &lt;em&gt;Hopfian&lt;/em&gt;, meaning that every surjective endomorphism is an isomorphism.  There are a variety of ways to prove this.  It's proved in Lyndon &amp;amp; Schupp, using Nielsen transformations. Alternatively, you can appeal to an (easy) result of Malcev, which states that every finitely generated, residually finite group is Hopfian.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, there is an obvious epimorphism $F_3\to F_2$ with non-trivial kernel, given by killing a generator.  If $F_3$ were a quotient of $F_2$, the composition of these two maps would give an epimorphism $F_2\to F_2$ with non-trivial kernel.&lt;/p&gt;&#xA;" OwnerUserId="2287" LastActivityDate="2010-12-15T23:27:39.507" />
  <row Id="14463" PostTypeId="2" ParentId="14461" CreationDate="2010-12-15T23:42:01.457" Score="4" ViewCount="" Body="&lt;p&gt;Suppose the number of indeterminates is $n$. Suppose the monomial symmetric polynomial is of type $(a_1,a_2,...,a_n)$ where the first $b_1$ exponents are identical, the next $b_2$ are identical and so on until $b_r$, where $1\leq b_j \leq n$ and $\sum_{1}^{r} b_j = n$. Then, number of terms in a monomial symmetric polynomial is the same as number of permutations of the set of exponents. The number of permutations when they are all distinct is $n!$. If not, the every permutations of any collection of identical exponents amongst themselves should be treated the same. So factoring out such permutations, we get the total number to be $\frac{n!}{b_1!\cdot b_2!\cdot...\cdot b_r!}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, consider the examples on the wikipedia page.&#xA;We have, $n=3$. In the first case, we have 2 identical exponents, so number of terms is $\frac{3!}{2!}=3$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the second case, all exponents are distinct, so the number of terms is $3!=6$&lt;/p&gt;&#xA;" OwnerUserId="3431" LastActivityDate="2010-12-15T23:42:01.457" CommentCount="3" />
  <row Id="14464" PostTypeId="2" ParentId="14461" CreationDate="2010-12-15T23:45:01.463" Score="5" ViewCount="" Body="&lt;p&gt;If the smallest exponent appears $k_1$ times, the second smallest appears $k_2$ times, etc. then the answer is the multinomial coefficient: ${n \choose k_1,k_2,...,k_n}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The multinomial coefficient $n \choose k_1,k_2,...,k_n$ counts how many ways you have to partition n objects into disjoint subsets of sizes $k_1,...,k_n$ which is just what you need.&lt;/p&gt;&#xA;" OwnerUserId="3533" LastActivityDate="2010-12-15T23:45:01.463" />
  <row Id="14465" PostTypeId="1" CreationDate="2010-12-16T00:11:59.877" Score="-2" ViewCount="147" Body="&lt;p&gt;This is a question on my homework could you help me figure it out?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Write a non-recursive procedure that takes an array and its length and prints its elements in the reverse order. For example, if the array has elements 77, 12, 90, 34, 45, the procedure should print the values 45, 34, 90, 12, 77. Here is the header for the procedure:&#xA;Procedure printReverse(array[], length)&lt;/p&gt;&#xA;" OwnerUserId="4351" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2011-06-03T13:53:26.650" LastActivityDate="2011-06-03T13:53:26.650" ClosedDate="2010-12-16T03:57:25.890" Title="help writing a non-recursive procedure" Tags="&lt;homework&gt;&lt;algorithms&gt;&lt;recursive-algorithms&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="14466" PostTypeId="2" ParentId="14465" CreationDate="2010-12-16T00:23:30.340" Score="0" ViewCount="" Body="&lt;p&gt;Can you just print array[i] for i counting down from length to 1?  Python can do it in one line.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-12-16T00:23:30.340" CommentCount="2" />
  <row Id="14467" PostTypeId="1" AcceptedAnswerId="14469" CreationDate="2010-12-16T00:26:52.773" Score="2" ViewCount="169" Body="&lt;p&gt;In Hatcher's &quot;Algebraic Topology&quot; in the Poincare Duality section he introduces the subject by doing orientable surfaces.  He shows that there is a dual cell structure to each cell structure and it's easy to see that the first structure gives the cellular chain complex, while the other gives the cellular cochain complex.  He goes on to say that this generalizes for manifolds of higher dimension, but that &quot;requires a certain amount of manifold theory&quot;.  Is there a good book or paper where I can read about this formulation of Poincare Duality?&lt;/p&gt;&#xA;" OwnerUserId="2838" LastActivityDate="2010-12-16T01:16:07.813" Title="Poincare Duality Reference" Tags="&lt;reference-request&gt;&lt;algebraic-topology&gt;&lt;cohomology&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="2" />
  <row Id="14468" PostTypeId="2" ParentId="14327" CreationDate="2010-12-16T00:44:28.743" Score="3" ViewCount="" Body="&lt;p&gt;Update: See the last section for a possible proof (**) that $a_n &amp;lt; n$ for all $n \geq 18$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;HR&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are some bounds on $E(n,m)$ that turn out to be useful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using $E_{n,m}$ for $E(n,m)$ in Ross Millikan's notation, let $R_{n,m} = \lceil E_{n,m} \rceil - E_{n,m}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Simplifying the expression for $E_{n,m}$ in Ross's first answer yields (thanks to some nice cancellation)&#xA;$$E_{n,m} = \frac{n+1}{2} + \frac{E_{n,m-1}(E_{n,m-1}-1)}{2n} - \frac{R_{n,m-1}(R_{n,m-1}-1)}{2n}.$$&#xD;&#xA;Since $0 \leq R_{n,m-1} &amp;lt; 1$, we have $$0 \leq - \frac{R_{n,m-1}(R_{n,m-1}-1)}{2n} \leq \frac{1}{8n}.$$  This can be seen easily by the fact that the expression being bounded is quadratic in $R_{n,m-1}$ with vertex at $R_{n,m-1} = \frac{1}{2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, $F_{n,m} \leq E_{n,m} \leq G_{n,m}$, where $F_{n,1} = G_{n,1} = \frac{n+1}{2}$, and&#xA;$$F_{n,m} =  \frac{n+1}{2} + \frac{F_{n,m-1}(F_{n,m-1}-1)}{2n},$$&#xD;&#xA;$$G_{n,m} =  \frac{n+1}{2} + \frac{G_{n,m-1}(G_{n,m-1}-1)}{2n} + \frac{1}{8n}.$$&#xD;&#xA;Thus we have recurrences that give upper and lower bounds on $E_{n,m}$ without having to deal with the problem of taking ceilings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Numerical experiments indicate that &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$F_{n,n}$ and $G_{n,n}$ are very close to each other,&lt;/li&gt;&#xA;&lt;li&gt;$G_{n,n} - F_{n,n}$ is decreasing,&lt;/li&gt;&#xA;&lt;li&gt;$n - G_{n,n}$ is increasing, &lt;/li&gt;&#xA;&lt;li&gt;$n - G_{n,n} &amp;gt; 1$ for $n \geq 15$.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;A proof of 3 or 4 would imply $a_n &amp;lt; n$ for $n \geq 15$.  A close analysis of $G_{n,n} - F_{n,n}$, together with an asymptotic estimate of $F_{n,n}$ or $G_{n,n}$, would help with the requested behavior of $n - a_n$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, it is easy to see that $F_{n,m} = n$ is an equilibrium solution for the $F_{n,m}$ recurrence.  That should be helpful as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;HR&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It turns out that $G_{n,m} = \frac{1}{2} + n a_m$, where $a_1 = \frac{1}{2}$ and $a_m$ satisfies the recurrence $$a_m = \frac{a^2_{m-1}+1}{2}.$$  This is easy to verify once one has the conjectured expression.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It also turns out that &lt;a href=&quot;http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1710384#p1710384&quot; rel=&quot;nofollow&quot;&gt;the $a_m$ recurrence has been studied&lt;/a&gt; (**), with the following bounds:&#xA;$$ 1 - \frac{2}{m} + \frac{2}{m^2} \ln \frac{m}{3} + \frac{417}{128m^2} \leq a_m \leq 1 - \frac{2}{m} + \frac{5 \ln m + 3}{2m^2}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The upper bound implies &#xA;$$G_{n,n} \leq \frac{1}{2} + n \left(1 - \frac{2}{n} + \frac{5 \ln n + 3}{2n^2}\right) = n - \frac{3}{2} + \frac{5 \ln n + 3}{2n} (*) $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, since $E_{n,n} \leq G_{n,n}$, $G_{n,n} &amp;lt; n-1$ implies $a_n &amp;lt; n$.  The expression on the right in $(*)$ is less than $n-1$ when $$\frac{5 \ln n + 3}{2n} &amp;lt; \frac{1}{2},$$&#xD;&#xA;which is true for all $n \geq 18$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(**) The bounds required for my argument are given in a post in the &quot;Real Analysis Unsolved and Proposed Problems&quot; forum at the Art of Problem Solving.  I cannot tell whether the bounds are conjectured and the poster is asking for a proof, or whether the poster has a proof and is merely posing the problem for others to solve.  So I cannot claim that this is a complete proof.&lt;/p&gt;&#xA;" OwnerUserId="2370" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-12-16T03:54:30.380" LastActivityDate="2010-12-16T03:54:30.380" CommentCount="3" />
  <row Id="14469" PostTypeId="2" ParentId="14467" CreationDate="2010-12-16T00:47:42.670" Score="7" ViewCount="" Body="&lt;p&gt;As far as I know, the only textbook reference for this approach, which is Poincare's original approach, is Seifert and Threlfall's text &quot;A textbook of topology&quot;.  It's available in English translation but the original was in German.  Moreover, Seifert and Threlfall's proof isn't as efficient as it could be, since they're working &lt;em&gt;entirely&lt;/em&gt; with simplicial homology.  It's much more efficient to work with both simplicial homology and CW-cohomology (together with the knowledge that simplicial and CW homology / cohomology are canonically isomorphic via the relation to singular homology / cohomology).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This version of the proof only works in the context where your manifold is triangulable, and here it goes:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The idea of the dual cell decomposition in general goes like this. Let $\Delta_n$ be an $n$-simplex, and let $F$ be a facet of $\Delta_n$, meaning the convex hull of some collection of $\Delta_n$'s vertices. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;em&gt;dual polyhedral bit&lt;/em&gt; corresponding to $F$ is the convex hull of the barycentres of &lt;em&gt;all&lt;/em&gt; facets $F&amp;#39;$ of $\Delta_n$ which contain $F$ (including $\Delta_n$ itself).  So in a tetrahedron $\Delta_3$, if $F$ was an edge, the dual polyhedral bit would be a quadrilateral that intersects $F$ in a single point. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given a triangulated manifold $M$, if $F$ is a simplex of the triangulation, the &lt;em&gt;dual cell&lt;/em&gt; corresponding to $F$ is the union of all the dual polyhedral bits to $F$ in all the top-dimensional simplices containing $F$.  If $M$ is $m$-dimensional and $F$ is $k$-dimensional, a little geometry later and you'll see the dual cell is an $(m-k)$-dimensional cell in a genuine CW-decomposition of $M$.  Again, in the 3-manifold case, if $F$ were an edge, the dual cell would be a $2$-cell with a single vertex at its centre, decomposed into squares. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's the basic idea.  From there the proof of Poincare duality is very much &quot;follow your nose&quot;.  It's a fun chase and I encourage you to try to work it out on your own, rather than looking it up. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moreover, spend as much time as you can thinking about evaluating a homology class $X$ on the dual of a homology class $Y$ (provided $X$ and $Y$ have complementary dimensions).  You'll have to be careful about thinking of the simplicial vs. CW-homology when thinking this through, of course. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/2G38W.png&quot; alt=&quot;dual polyhedral bits in a tetrahedron&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="642" LastEditorUserId="642" LastEditorDisplayName="" LastEditDate="2010-12-16T01:16:07.813" LastActivityDate="2010-12-16T01:16:07.813" />
  <row Id="14470" PostTypeId="2" ParentId="14465" CreationDate="2010-12-16T00:54:24.990" Score="1" ViewCount="" Body="&lt;p&gt;Ross is correct.  Remember that a recursive function is one that calls itself.  Writing a for loop as follows involves no such recursion:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for(int i = length - 1; i &gt;= 0; i--) cout &amp;lt;&amp;lt; array[i] &amp;lt;&amp;lt; &quot; &quot;;&lt;/p&gt;&#xA;" OwnerUserId="4693" LastActivityDate="2010-12-16T00:54:24.990" CommentCount="1" />
  <row Id="14473" PostTypeId="1" CreationDate="2010-12-16T03:17:14.927" Score="0" ViewCount="59" Body="&lt;p&gt;Let $R$ be a ring and assume that for every left $R$-module ${}_RM$, if $M$ has a projective cover, then $M$ is projective.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can someone help me prove that in that case, $J(R)=0$?&lt;/p&gt;&#xA;" OwnerUserId="4797" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-16T03:56:02.163" LastActivityDate="2010-12-16T03:56:02.163" Title="projective-projective cover" Tags="&lt;ring-theory&gt;" CommentCount="2" />
  <row Id="14474" PostTypeId="2" ParentId="14450" CreationDate="2010-12-16T03:40:22.387" Score="3" ViewCount="" Body="&lt;p&gt;Alternatively, by Fermat's little Theorem every $\rm\ a\in \mathbb F_p\ $ is a root of $\rm\  f(x)^p - f(x)\:,\ $ therefore $\rm\ x\ (x-1)\ (x-2)\cdots (x-p+1)\ |\ f(x)^p-f(x)\:.\ $ Making the specialization $\rm\ f(x) = x\ $ this yields $\rm\ x \ (x-1)\ (x-2)\cdots (x-p+1)\ |\ x^p-x\:.\ $ This divisibility must be an equality since both are monic polynomials of equal degree. Therefore we deduce that $\rm\: x^p - x\ |\ f(x)^p - f(x),\ \forall\: f\in \mathbb F_p\:[x]\:.\ $ The sought result is the special case $\rm\ f(x) = x-c\:. $ &lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-12-16T03:40:22.387" />
  <row Id="14475" PostTypeId="1" AcceptedAnswerId="14487" CreationDate="2010-12-16T03:50:59.870" Score="2" ViewCount="459" Body="&lt;p&gt;I have already taken up to Multivariable Calculus, Linear Algebra and Diff Eq.&#xA;I want to learn Calculus on Manifolds by myself, could you recommend a good introductory book on this subject?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Should I look at other things first, like topology, to get a better background?&lt;/p&gt;&#xA;" OwnerUserId="4800" LastActivityDate="2011-02-02T00:09:45.290" Title="Good introductory book on Calculus on Manifolds" Tags="&lt;manifold&gt;" AnswerCount="6" CommentCount="2" />
  <row Id="14476" PostTypeId="2" ParentId="14475" CreationDate="2010-12-16T03:58:29.927" Score="4" ViewCount="" Body="&lt;p&gt;May be this :  &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0805390219&quot; rel=&quot;nofollow&quot;&gt;Spivak,Calculus-Manifolds-Approach&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is an reprint from year 1995, although maybe an more new could also you serve.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/TQkRk.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="3421" LastEditorUserId="3421" LastEditorDisplayName="" LastEditDate="2010-12-16T18:29:46.027" LastActivityDate="2010-12-16T18:29:46.027" CommentCount="1" />
  <row Id="14477" PostTypeId="1" AcceptedAnswerId="14479" CreationDate="2010-12-16T04:13:31.533" Score="6" ViewCount="282" Body="&lt;p&gt;Any hints about how to prove &#xA;$$!n = n!- \sum_{i=1}^{n} {{n} \choose {i}} \quad!(n-i)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;from &lt;a href=&quot;http://en.wikipedia.org/wiki/Subfactorial&quot; rel=&quot;nofollow&quot;&gt;&#xA;Wikipedia's article on derangements&lt;/a&gt;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here, $!n$ is the number of derangements of a set with $n$ elements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not looking for proofs, just nudges in the right direction.&lt;/p&gt;&#xA;" OwnerUserId="4728" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-16T04:29:26.553" LastActivityDate="2010-12-16T05:35:46.160" Title="Proof of subfactorial formula $!n = n!- \sum_{i=1}^{n} {{n} \choose {i}} \quad!(n-i)$" Tags="&lt;combinatorics&gt;" AnswerCount="4" CommentCount="1" />
  <row Id="14478" PostTypeId="2" ParentId="14475" CreationDate="2010-12-16T04:22:41.170" Score="0" ViewCount="" Body="&lt;p&gt;If you are looking for a book with tons of proofs, I would strongly recommend J.H. Hubbard and B.B. Hubbard. Vector Calculus, Linear Algebra, and Differential Forms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Super good as an introduction book if you can find someone to go to when you are stuck. &#xA;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0130414085&quot; rel=&quot;nofollow&quot;&gt;http://www.amazon.com/Vector-Calculus-Linear-Algebra-Differential/dp/0130414085&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4728" LastEditorUserId="4728" LastEditorDisplayName="" LastEditDate="2010-12-16T04:37:12.523" LastActivityDate="2010-12-16T04:37:12.523" />
  <row Id="14479" PostTypeId="2" ParentId="14477" CreationDate="2010-12-16T04:29:34.890" Score="8" ViewCount="" Body="&lt;p&gt;Hint: ${{n} \choose {i}} \cdot!(n-i)$ counts the number of permutations that fix exactly $i$ elements.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-12-16T04:29:34.890" CommentCount="2" />
  <row Id="14480" PostTypeId="2" ParentId="14477" CreationDate="2010-12-16T04:29:35.467" Score="4" ViewCount="" Body="&lt;p&gt;I actually prove a generalization of this in my paper &quot;&lt;a href=&quot;http://math.pugetsound.edu/~mspivey/DT.pdf&quot; rel=&quot;nofollow&quot;&gt;Deranged Exams&lt;/a&gt;&quot; (&lt;em&gt;College Mathematics Journal&lt;/em&gt;, 41 (3): 197-202, 2010).  See Theorem 7.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The generalization is the following.  Let $S_{n,k}$ be the number of permutations on $n$ elements in which none of the first $k$ elements remains in its original position.  Thus $S_{n,0} = n!$, and the number of derangements on $n$ elements, $D_n$, is $S_{n,n}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$S_{n+k,k} = \sum_{j=0}^n \binom{n}{j} D_{k+j}.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The OP's question is the case $k = 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strike&gt;I'll extract the essence of the proof and post it in the next few minutes.&lt;/strike&gt;&#xA;Since you want hints rather than a full proof, I'll just leave this as a reference in case you (or anyone else reading this) is interested.  Jonas Meyer's answer gives a good hint.&lt;/p&gt;&#xA;" OwnerUserId="2370" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-12-16T04:46:46.327" LastActivityDate="2010-12-16T04:46:46.327" CommentCount="2" />
  <row Id="14481" PostTypeId="2" ParentId="14477" CreationDate="2010-12-16T04:30:45.367" Score="1" ViewCount="" Body="&lt;p&gt;The number of derangements is the number of permutations less the number that leave some numbers fixed.  So for example the term i=5 in the sum is all the ways to pick 5 out of n to leave fixed and derange all the rest.  Then sum over all the numbers of ones that can be fixed.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-12-16T04:30:45.367" />
  <row Id="14482" PostTypeId="2" ParentId="14477" CreationDate="2010-12-16T04:32:43.367" Score="3" ViewCount="" Body="&lt;p&gt;Here's a proof, obscured using spoiler space.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If $d_n$ is the number of derangements on $n$ elements, then the number of permutations on $n$ elements with &lt;em&gt;exactly&lt;/em&gt; $i$ fixed points is ${n \choose i} d_{n-i}$ (choose i points to fix, then any permutation that fixes exactly those $i$ points (and nothing else) determines a derangement on the non-fixed points, and there are $d_{n-i}$ such derangements).  Hence, $n!=\sum_{i=0}^n {n \choose i} d_{n-i}$, which can be rearranged to give the above formula.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;PS. I'm not a fan of the $!n$ notation, I'm pretty sure it's not standard in combinatorics.&lt;/p&gt;&#xA;" OwnerUserId="139" LastEditorUserId="139" LastEditorDisplayName="" LastEditDate="2010-12-16T05:35:46.160" LastActivityDate="2010-12-16T05:35:46.160" CommentCount="2" />
  <row Id="14483" PostTypeId="1" CreationDate="2010-12-16T04:44:25.077" Score="11" ViewCount="253" Body="&lt;p&gt;I have a question about Hatcher's proof that the fundamental group of a circle is Z. Specifically, halfway through, ( &lt;a href=&quot;http://www.math.cornell.edu/~hatcher/AT/ATch1.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.math.cornell.edu/~hatcher/AT/ATch1.pdf&lt;/a&gt; , page 30), he proves an important a lemma stated rather generally:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given a map $Y\times I\rightarrow S^1$ (Y is any space, and I is an interval), there is a unique lifting to  $Y\times I\rightarrow \mathbb{R}$ once we've specified an initial condition  $Y\times \{0\}\rightarrow \mathbb{R}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the proof of this, he breaks $Y\times I$ into smaller pieces $N\times \{y_0\}$, and keeps adjusting the size of the $N$. In the corresponding proofs in Munkres or Fulton, $Y$ is fixed as just another copy of $I$, and there is no need to modify the pieces we're looking at as the proof progresses. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question: what is different in Hatcher's more general proof? Specifically, why does he keep modifying the $N$? My feeling is that this is because $Y$ could be disconnected or bad in some other sense. My question seems a little vague, but (a) if you look at Hatcher's proof, you'll see what I mean (near the bottom of p. 30), and (b) I'm hoping that maybe this situation, and the difficulties with it, are well known enough for someone to answer anyway.&lt;/p&gt;&#xA;" OwnerUserId="3656" LastEditorUserId="3656" LastEditorDisplayName="" LastEditDate="2010-12-19T21:15:11.053" LastActivityDate="2010-12-19T21:15:11.053" Title="Question about the first proof in Hatcher's Algebraic Topology" Tags="&lt;algebraic-topology&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="3" />
  <row Id="14484" PostTypeId="2" ParentId="14475" CreationDate="2010-12-16T04:50:32.173" Score="7" ViewCount="" Body="&lt;p&gt;Munkres (the author of the very clear text on Topology) wrote a book called &lt;em&gt;Analysis on Manifolds&lt;/em&gt;. This is basically like Spivak, but twice as long and with more pictures. Everyone has fond memories of Spivak &lt;em&gt;after&lt;/em&gt; they've used it, because of how small and &quot;cool&quot; it is. But at the end of the day (especially for self-learning) I think Munkres is probably more straightforward and less frustrating.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0201315963&quot; rel=&quot;nofollow&quot;&gt;http://www.amazon.com/Analysis-Manifolds-Advanced-Books-Classics/dp/0201315963&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3656" LastEditorUserId="3656" LastEditorDisplayName="" LastEditDate="2010-12-16T05:01:35.647" LastActivityDate="2010-12-16T05:01:35.647" />
  <row Id="14485" PostTypeId="1" CreationDate="2010-12-16T05:18:40.600" Score="1" ViewCount="135" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Possible Duplicate:&lt;/strong&gt;&lt;br&gt;&#xA;  &lt;a href=&quot;http://math.stackexchange.com/questions/8337/different-methods-to-compute-sum-n-1-infty-frac1n2&quot;&gt;Different methods to compute $\sum_{n=1}^\infty \frac{1}{n^2}$.&lt;/a&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;p&gt;I just got the &quot;New and Revised&quot; edition of &quot;Mathematics: The New Golden Age&quot;, by Keith Devlin. On p. 64 it says the sum is $\pi^2/6$, but that's way off. $\pi^2/6 \approx 1.64493406685$ whereas the sum in question is $\approx 1.29128599706$. I'm expecting the sum to be something interesting, but I've forgotten how to do these things. &lt;/p&gt;&#xA;" OwnerUserId="1964" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-16T06:12:26.580" LastActivityDate="2010-12-16T06:12:26.580" ClosedDate="2010-12-16T05:32:57.037" Title="Sum of 1/n*n for n = 1 ,2 ,3, ... ?" Tags="&lt;sequences-and-series&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="14486" PostTypeId="2" ParentId="14485" CreationDate="2010-12-16T05:23:10.987" Score="2" ViewCount="" Body="&lt;p&gt;The answer is indeed pretty interesting!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ \sum_{n=1}^\infty \frac{1}{n^2} = \frac{\pi^2}{6} $&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This can be proven using complex analysis or calculus, or probably in many hundreds of other ways. One example of how to prove this is given here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.math.uu.se/~bjorklund/euler.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.math.uu.se/~bjorklund/euler.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1089" LastActivityDate="2010-12-16T05:23:10.987" />
  <row Id="14487" PostTypeId="2" ParentId="14475" CreationDate="2010-12-16T05:37:27.780" Score="2" ViewCount="" Body="&lt;p&gt;A good place to start learning about manifolds is to read a book by Spivak called &quot;Calculus on Manifolds.&quot; This book is very small, and the first three chapters are a short review on multivariate calculus, but I should say that the focus is much more on mathematical rigor than a normal calculus course so it would be worth your time to look through that I imagine. The last chapter introduces manifolds, how to integrate on them, and eventually culminates in the modern version of Stoke's Theorem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want however to get a much more in depth view on manifolds, you will have to learn some topology. A good free online book to learn from, that I myself originally used, is called &quot;Topology Without Tears.&quot; It can be downloaded for free at: &lt;a href=&quot;http://uob-community.ballarat.edu.au/~smorris/topology.htm&quot; rel=&quot;nofollow&quot;&gt;http://uob-community.ballarat.edu.au/~smorris/topology.htm&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It covers about everything a first course in topology covers, while being fairly straightforward and easy to understand. It is especially good for people who have not dealt with mathematical proof very much yet, because the author is very clear and detailed in his explanations and often explains about common proof techniques.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you prefer a different book, or a more commonly used book than &quot;Topology Without Tears,&quot; Munkre's Topology book is fine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can recommend two books on introductory manifolds which will be a lot more in depth than Spivak, but are still at the introductory level:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;John Lee's &quot;Introduction to Smooth Manifolds&quot; and&#xA;Boothby's &quot;Introduction to Differentiable manifolds&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These books are about Smooth Manifolds, which are a type of manifold in which calculus can be done.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lastly, if you learn topology and think you want to learn about manifolds from a topological point of view&lt;/p&gt;&#xA;&#xA;&lt;p&gt;William Fulton's &quot;Algebraic Topology&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is a good place to look.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope this helps! I've read a lot of books, so if you are looking for anything in particular that is slightly different from what I've said please feel free to email me.&lt;/p&gt;&#xA;" OwnerUserId="1089" LastActivityDate="2010-12-16T05:37:27.780" />
  <row Id="14488" PostTypeId="1" AcceptedAnswerId="14490" CreationDate="2010-12-16T05:44:15.843" Score="1" ViewCount="85" Body="&lt;p&gt;Hi, we are a group of three, and we've got this question in an assignment (the question is originally in French, so bear with me) :&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Let $X_1 \dots X_4$ be a random sample drawn from a population of average $\mu$ and variance $\sigma^2$. We define the following estimators:&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$$\hat{\mu}_1=\frac{X_1+2X_2-2X_3+5X_4}{6} \qquad \text{and} \qquad \hat{\mu}_2=\frac{X_1-2X_2+X_3+5X_4}{5}$$&#xD;&lt;/p&gt;&#xA;  &#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;&lt;p&gt;Show that $\hat{\mu}_1$ and $\hat{\mu}_2$ are both unbiased estimators&lt;/p&gt;&lt;/li&gt;&#xA;  &lt;li&gt;&lt;p&gt;Which one is the best? Justify.&lt;/p&gt;&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is the only number (out of 9) that we still have to complete, and we have no clue on how to start this up. Please, none of us are actually studying in a math degree (we're all programmers), so please give examples in your answers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you!&lt;/p&gt;&#xA;" OwnerUserId="4803" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-16T06:32:13.317" LastActivityDate="2010-12-16T06:32:13.317" Title="Unbiased estimators ($\hat{\mu}_1$ and $\hat{\mu}_2$); proof and which is best" Tags="&lt;homework&gt;&lt;statistics&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="14489" PostTypeId="1" AcceptedAnswerId="14492" CreationDate="2010-12-16T05:49:31.273" Score="4" ViewCount="161" Body="&lt;p&gt;This is from one of the exercises in &quot;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0201558025&quot; rel=&quot;nofollow&quot;&gt;Concrete Mathematics&lt;/a&gt;&quot;, and is something I'm doing privately, not homework.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a variant on the classic tower of hanoi, where all moves must be made clockwise only. The recursive solution is to consider Qn to be the number of moves to move a tower size n one step across(e.g. from start-&gt;end post), and Rn to be the number of moves to move it two(e.g start-&gt;end-&gt;spare)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The recursive solution is(sorry about the formatting, just getting to grips with tex)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$Q_n = 0 \quad \text{if n = 0}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$Q_n = 2R_{n-1} + 1\quad \text{if n &amp;gt; 0}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$R_n = 0 \quad \text{if n = 0}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$R_n = Q_n + Q_{n-1} + 1 \quad \text{if n &amp;gt; 0}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I prove these? I presume an inductive proof is required but everything I've tried to do just gets garbled into a mess of further decreasing n values.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Explaining why these work is simple enough: $Q_n$ can be explained by the fact that you need to move a stack of size n-1 two steps forward followed by the n piece, followed by moving the n-1 stack two steps again onto the n piece=&gt; $Q_n = R_{n-1} + 1 + R_{n-1} = 2R_{n-1}+1$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The wordy proof for $R_n$ is similar: It consists of moving $R_{n-1}$ then 1 followed by $Q_n$, 1 and $R_{n-1}$ which will end us up with &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$R_n = 2R_{n-1}+1+Q_{n-1}+1 = Q_n+Q_{n-1}+1$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But these are all just words that don't form any real proof... I've gone over the leading chapter and looked up some resources on proof by induction, but doing it with a recursion with two components has put me for a spin(Having forgotten a lot of the math studied in the past certainly isn't helping either). If someone could run me through a rigorous proof for n &gt; 0 it would be much appreciated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an aside, any recommended learning resources on proof by induction would be appreciated as I'm still barely &quot;getting it&quot;&lt;/p&gt;&#xA;" OwnerUserId="4801" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-16T06:00:23.877" LastActivityDate="2010-12-17T21:53:23.007" Title="Proof of clockwise tower of hanoi variant recursive solution" Tags="&lt;discrete-mathematics&gt;" AnswerCount="1" />
  <row Id="14490" PostTypeId="2" ParentId="14488" CreationDate="2010-12-16T05:52:04.063" Score="4" ViewCount="" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Show that $E[\hat{\mu_{1}}] = \mu$ and $E[\hat{\mu_2}] = \mu$. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Find $\min \left[\text{Var}(\hat{\mu_{1}}), \text{Var}(\hat{\mu_{2}}) \right]$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2010-12-16T05:52:04.063" CommentCount="5" />
  <row Id="14492" PostTypeId="2" ParentId="14489" CreationDate="2010-12-16T06:09:36.083" Score="5" ViewCount="" Body="&lt;p&gt;The &quot;word&quot; description is in fact the guide to a proof. The only problem is showing that the process you describe is in fact optimal. This is done by induction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;we want to prove the formulas you give are the optimal solution. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The case $n=0$ is immediate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now: to prove the $k+1$ case assuming the $k$th case (so our induction hypothesis is that the formula you give is optimal for $R_k$ and for $Q_k$), you take a stack of $k+1$ disks. To move them all forward (to perform $Q_{k+1}$) you must get the top $k$ out of the way; that is done, by the induction assumption, with $R_k$ as optimal. Then you move the $k+1$ disk to the target, taking one move; then you move the disks from where they are to the target. Again, by the induction assumption, the best possible way of doing this is $R_k$. So you get that $Q_{k+1}=R_k+1+R_k=2R_k+1$, on the assumption that $Q_k$ and $R_k$ are the optimal moves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same argument works for $R_{k+1}$; the only thing that your &quot;just words&quot; is missing is to be clear about what the induction hypothesis is, and to apply it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added:&lt;/strong&gt; Why is the given strategy actually the best? In order to move all the disks from the original rod to the target rod, we must move the largest disk at least once; note that the position of the largest disk does not affect the validity or lack thereof of any move, so the position of the largest disk is immaterial for any other moves (as opposed to any other disk). So the optimal solution will only require us to move the largest disk exactly once, from the original rod to the target rod. In order to do this, we must first remove all disks from above the largest disk, and have no disks on the target rod so we can perform the move; that is $R_k$, because the only configuration that allows us to move the largest disk from the original rod to the target rod is to have all other disks in the non-target rod. Then we move the largest disk, which is the $+1$. And now, we just want to move the remaining disks to their final position in the quickest possible way, which again is $R_k$. A similar argument focusing on placing the largest disk in its final position, works for the computation of $R_{k+1}$. &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-17T21:53:23.007" LastActivityDate="2010-12-17T21:53:23.007" CommentCount="2" />
  <row Id="14493" PostTypeId="2" ParentId="14268" CreationDate="2010-12-16T06:26:08.117" Score="0" ViewCount="" Body="&lt;p&gt;Here is a simple example that does the job without any multiplication:&#xA;$5 \times 12$ vs $6 \times 11$ , rewrite both sides as $5\times(11+1)$ and $(5+1)\times11$, &#xA;then we get $5\times11+5\times1$ vs $5\times11 + 1 \times 11$ subtracting $5\times11$ from both sides we get : $1\times5 vs 1\times11$ which does not require any multiplication at all. The same technique can be used on (254)(847) and (383)(536).&#xA;$ 254 \times 847 vs 383 \times 536 $&lt;br&gt;&#xA;$ 254 \times ( 536 + 310 ) vs (254+129)\times536$&lt;br&gt;&#xA;$ 254 \times 310 vs 129\times536$ continuing in similar fashion we end up with&lt;br&gt;&#xA;$125\times 80 vs 4 \times 101$ although obvious but just for fun :&lt;br&gt;&#xA;$ (101+24)\times(76+4) vs 4 \times 101$ where in the next line we get :&lt;br&gt;&#xA;$ 4 \times 101 + 101\times 76 + 24\times76 + 24\times 4 vs 4\times 101 $  &lt;/p&gt;&#xA;" OwnerUserId="3936" LastEditorUserId="3936" LastEditorDisplayName="" LastEditDate="2010-12-16T07:37:02.000" LastActivityDate="2010-12-16T07:37:02.000" />
  <row Id="14494" PostTypeId="1" AcceptedAnswerId="14535" CreationDate="2010-12-16T08:02:15.070" Score="4" ViewCount="125" Body="&lt;p&gt;I am trying to work through Boothby's An Introduction to Differentiable Manifolds on my own and, embarassingly, have got stuck at the very first chapter. At the end of section 4, chapter 1 (called: Further examples of manifolds: cutting and pasting), there's this question: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Prove that adding a handle to a 2-manifolds in the fashion described above for $S^2$ and $T^2$ actually does give a 2 manifold&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, Boothby gives no formal definitions for cutting and pasting so I have no idea about how to show that a sphere with a handle will be locally euclidean. For example, what if the handle is attached to the surface such that at the joint (a circle) the two surfaces meet sharply, so that for points on the joint we cannot find a normal. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, my questions are:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Where can I find the operations of cutting formalised (I realise that pasting is connected to quotient topology, so I am studying that now). And, how does one answer this question rigorously?  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Edit&lt;/b&gt;&#xA;Upon further thought, 'cutting' could be this: since a manifold $M$ is metrizable, we can remove an open disk (open so that I get a boundary in the resulting manifold) around a point, i.e., we consider $M - B(p,\epsilon)$. So, given two dissimilar points, we remove two non-intersecting open disks (we can do this because a manifold is hausdorff). Now, how do I 'paste' the ends of the cylinder 'smoothly'? &lt;/p&gt;&#xA;" OwnerUserId="4806" LastEditorUserId="4806" LastEditorDisplayName="" LastEditDate="2010-12-16T08:54:38.937" LastActivityDate="2010-12-16T17:25:55.777" Title="Adding handles to a sphere" Tags="&lt;differential-geometry&gt;" AnswerCount="2" />
  <row Id="14495" PostTypeId="1" AcceptedAnswerId="16689" CreationDate="2010-12-16T08:11:18.893" Score="0" ViewCount="157" Body="&lt;p&gt;Let's say I have a list of points( in 2D coordinates!) with the vertex $V$, how to find all the triangular faces $F$ ( which has three edges $E$) that covers &lt;strong&gt;all&lt;/strong&gt; of the points? In other words, I'm looking for a mesh that connects all the points together with well behaved triangles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The conditions are that the faces must not overlap or cut through one another. And it would help if all the $F$ has not-too-narrow angles in it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also note that the meshing algorithm above &lt;em&gt;must not introduced additional points&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suspect that the &lt;a href=&quot;http://en.wikipedia.org/wiki/Delaunay_triangulation&quot; rel=&quot;nofollow&quot;&gt;delaunay algorithm&lt;/a&gt; is meant for something like this. But I am not too clear about this. Am I right? Or is it that there are other algorithms that work for this problem?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit: A point cloud is just a list of points, scattered in 2D. The condition is all the points must have distinct ${x,y}$ coordinates.&lt;/strong&gt; &lt;/p&gt;&#xA;" OwnerUserId="550" LastEditorUserId="550" LastEditorDisplayName="" LastEditDate="2010-12-17T01:16:45.627" LastActivityDate="2011-02-27T18:21:34.220" Title="Given a Set of Points, Find All the Triangular Faces that Connect Them Together" Tags="&lt;geometry&gt;&lt;computational-geometry&gt;" AnswerCount="2" CommentCount="12" />
  <row Id="14497" PostTypeId="1" CreationDate="2010-12-16T08:53:01.637" Score="1" ViewCount="89" Body="&lt;p&gt;How can I construct a CW complex A with H_0(A) = Z, H_2(A) = Z/4Z, H_4(A) =&#xA;Z ⊕ Z and all other homology groups trivial? &#xA;Any idea?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="4808" LastActivityDate="2010-12-16T13:13:36.577" Title="construct CW complex" Tags="&lt;algebraic-topology&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="14499" PostTypeId="1" AcceptedAnswerId="14545" CreationDate="2010-12-16T10:10:01.340" Score="6" ViewCount="206" Body="&lt;p&gt;&lt;strong&gt;Definition&lt;/strong&gt;: n is said to be a highly composite number if and only if $d(n)&amp;gt;d(m)$ for all $m&amp;lt;n$, where $d(n)$ denotes number of divisors of n.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Questions&lt;/strong&gt;:&lt;br&gt;&#xA;1) Are there any theorems about highest prime that divides n, assuming that n is highly composite?. It seems that the highest prime is of the order of ln(n).&lt;br&gt;&#xA;2) Are there any theorems about the highest power of 2 which divides n, assuming that n is highly composite?&lt;/p&gt;&#xA;" OwnerUserId="4811" LastEditorUserId="1301" LastEditorDisplayName="" LastEditDate="2010-12-16T14:44:09.167" LastActivityDate="2010-12-17T06:07:06.710" Title="Highly composite number" Tags="&lt;number-theory&gt;&lt;prime-numbers&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="2" />
  <row Id="14500" PostTypeId="2" ParentId="14386" CreationDate="2010-12-16T10:50:17.730" Score="0" ViewCount="" Body="&lt;p&gt;Consider the related least squares question: find x minimizing $||Ax-b||_2$. In the unlikely scenario that the overdetermined system has a solution, then we in fact have $||Ax-b||_2=0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The x solving the least squares problem can be found via the &lt;a href=&quot;http://en.wikipedia.org/wiki/Moore-Penrose_pseudoinverse&quot; rel=&quot;nofollow&quot;&gt;Moore-Penrose pseudoinverse&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x=(A^tA)^{-1}A^tb$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, one algorithm would be as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Solve $(A^tA)x=A^tb$ via your favorite linear solver.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Check if $Ax-b=0$.&lt;/p&gt;&#xA;" OwnerUserId="3060" LastActivityDate="2010-12-16T10:50:17.730" CommentCount="3" />
  <row Id="14501" PostTypeId="1" CreationDate="2010-12-16T11:44:08.030" Score="2" ViewCount="190" Body="&lt;p&gt;You are given three non-negative integers $A$, $B$ and $C$, find a number $X$ (say) satisfy&#xA;$X^A \equiv B\pmod{2C + 1}$ and $0 \le X \le 2C$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am inquisitive about how to approach this one?&lt;/p&gt;&#xA;" OwnerUserId="2109" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-16T18:07:37.533" LastActivityDate="2010-12-16T18:07:37.533" Title="A problem on Number theory" Tags="&lt;number-theory&gt;&lt;algorithms&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="1" />
  <row Id="14502" PostTypeId="1" AcceptedAnswerId="14504" CreationDate="2010-12-16T12:48:37.807" Score="0" ViewCount="47" Body="&lt;p&gt;Common form of system of linear equations is A*X = B, X is unknown. But how to find A, if X and B are known?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A is MxN matrix, X is column vector(N), B is column vector(M)   &lt;/p&gt;&#xA;" OwnerUserId="3532" LastEditorUserId="3532" LastEditorDisplayName="" LastEditDate="2010-12-16T13:04:20.533" LastActivityDate="2010-12-16T13:08:40.593" Title="System of linear equations" Tags="&lt;linear-algebra&gt;" AnswerCount="2" CommentCount="6" />
  <row Id="14503" PostTypeId="2" ParentId="14497" CreationDate="2010-12-16T12:55:13.287" Score="6" ViewCount="" Body="&lt;p&gt;Let $f:S^2\to S^2$ be a map of degree $4$. Attach a $3$-cell $B^3$ to $Y=S^2$ using $f$ to glue the boundary $\partial B^3=S^2$ to $Y$. Then $Z=Y\cup_f B^3$ has homology given by $$H_q(Z)=\begin{cases}&#xD;&#xA;\mathbb Z, &amp;amp; \mbox{if }q=0; \\&#xD;&#xA;\mathbb Z/\mathbb 4Z, &amp;amp; \mbox{if }q=2; \\&#xD;&#xA;0, &amp;amp; \mbox{otherwise.}&#xD;&#xA;\end{cases}&#xD;&#xA;$$&#xD;&#xA;Now consider the connected sum $Z\vee S^4\vee S^4$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;NB:&lt;/strong&gt; The interesting part is of course the construction of $Z$. If I recall correctly, you can find in Hatcher's book information regarding &lt;em&gt;Moore spaces&lt;/em&gt;, of which $Z$ is a simple example.&lt;/p&gt;&#xA;" OwnerUserId="274" LastEditorUserId="274" LastEditorDisplayName="" LastEditDate="2010-12-16T13:13:36.577" LastActivityDate="2010-12-16T13:13:36.577" />
  <row Id="14504" PostTypeId="2" ParentId="14502" CreationDate="2010-12-16T12:57:57.390" Score="2" ViewCount="" Body="&lt;p&gt;If all matrices are square, then $A=BX^{-1}$.&lt;/p&gt;&#xA;" OwnerUserId="4742" LastActivityDate="2010-12-16T12:57:57.390" />
  <row Id="14505" PostTypeId="2" ParentId="14502" CreationDate="2010-12-16T13:08:40.593" Score="2" ViewCount="" Body="&lt;p&gt;Do it row by row. Row $k$ in $A$ multiplied by the column vector $X$ equals the $k$th entry in the vector $B$. This is a single equation for the $N$ entries in that row of $A$ (so unless $X$ is zero, you get an $(N-1)$-parameter set of solutions for each row).&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2010-12-16T13:08:40.593" />
  <row Id="14506" PostTypeId="1" CreationDate="2010-12-16T13:18:59.937" Score="2" ViewCount="157" Body="&lt;p&gt;Several papers give poly-time algorithms for constrained paths on labelled graphs, e.g. [1]&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Quote:&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Given an alphabet Σ, a (directed) graph G whose edges are weighted and Σ-labeled,&#xA;and a formal language L ⊆ Σ∗ , the formal-language-constrained shortest/simple path problem con-&#xA;sists of finding a shortest (simple) path p in G complying with the additional constraint that l(p) ∈ L.&#xA;Here l(p) denotes the unique word obtained by concatenating the Σ-labels of the edges along the&#xA;path p.&#xA;(1) We show that the formal-language-constrained shortest path problem is solvable efficiently&#xA;in polynomial time when L is restricted to be a context-free language (CFL).&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Can this be used for restricted read twice BDDs (Binary Decision Diagram)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider BDD where every variables occurs twice and the order of all paths root-&gt;T is restricted to:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1 .. n n .. 1 (this is palindrome)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Treat the BDD as edge labelled digraph where the true edge of variable is labelled as +variable and the false edge as -variable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some paths in the digraph are &quot;BDD inconsistent&quot; - if a variable takes both values.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Looks like CFG constraint can help to find &quot;consistent&quot; paths.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is possible to construct CFG that accepts all palindromes and only palindromes (over the alphabet U {+variable,-variable}.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finding a path from the root to &quot;true&quot; constrained by the above CFG will give a consistent satisfying assignment - the first half of the path is correct because a variable occurs only once and the second part is the reverse.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this correct (computer experiments suggest it is)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A different grammar may give different restriction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this published somewhere?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To my humble knowledge this is one of the very few examples of read twice BDD supporting polytime SAT query.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: The problem I am trying to solve is to find types of read twice BDDs which allow polytime SAT query with the help of CFGs. The above construction gives example for the case when the order of the variables is restricted to a palindrome, hence the (possibly incorrect) description &quot;restricted&quot; (because of the restriction of the order). I am not familiar with CFGs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[1]&#xA;&lt;a href=&quot;http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.105.446&amp;amp;rep=rep1&amp;amp;type=pdf&quot; rel=&quot;nofollow&quot;&gt;FORMAL-LANGUAGE-CONSTRAINED PATH PROBLEMS∗&#xA;CHRIS BARRETT† , RIKO JACOB‡ , AND MADHAV MARATHE†&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Simple read twice BDD on x,y. The concatenation of edge labels give the palindrome word: &quot;+x -y -y +x&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://ll0x.wdfiles.com/local--files/admin%3Amanage/aa.png&quot; alt=&quot;BDD&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The graph is at: &lt;a href=&quot;http://ll0x.wdfiles.com/local--files/admin%3Amanage/aa.png&quot; rel=&quot;nofollow&quot;&gt;http://ll0x.wdfiles.com/local--files/admin%3Amanage/aa.png&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(the graph may need copy/paste of the link)&lt;/p&gt;&#xA;" OwnerUserId="4813" LastEditorUserId="4813" LastEditorDisplayName="" LastEditDate="2011-01-03T11:33:21.777" LastActivityDate="2011-01-03T11:33:21.777" Title="Restricted read twice BDDs and context free grammars" Tags="&lt;graph-theory&gt;&lt;computer-science&gt;&lt;computability&gt;&lt;formal-systems&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="1" />
  <row Id="14507" PostTypeId="1" AcceptedAnswerId="14508" CreationDate="2010-12-16T13:27:13.893" Score="3" ViewCount="241" Body="&lt;p&gt;I am helping my teenager son. So far I have been able to answer all his questions, and to solve all the problems he has had problems with. Except this one. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\int{{3^x}{e^x}dx}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please, help me keep my success track and my son's confidence.&lt;/p&gt;&#xA;" OwnerUserId="4257" LastActivityDate="2010-12-17T00:52:02.463" Title="solving integral $\int{{3^x}{e^x}dx}$" Tags="&lt;calculus&gt;&lt;homework&gt;&lt;integral&gt;" AnswerCount="2" />
  <row Id="14508" PostTypeId="2" ParentId="14507" CreationDate="2010-12-16T13:28:35.190" Score="9" ViewCount="" Body="&lt;p&gt;Hint: try rewriting $3^x$ as $e^{x \ln 3}$.&lt;/p&gt;&#xA;" OwnerUserId="4743" LastActivityDate="2010-12-16T13:28:35.190" CommentCount="8" />
  <row Id="14509" PostTypeId="2" ParentId="14501" CreationDate="2010-12-16T13:36:00.197" Score="1" ViewCount="" Body="&lt;p&gt;Assuming X coprime to 2C+1. Then Euler totient theorem gives $X^{\varphi(2C+1)} \equiv 1 \mod 2C+1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming further that A coprime to $\varphi(2C+1)$. Then (By Bezout's identity) there's a D such that $AD \equiv 1 \mod \varphi(2C+1)$. In which case $(X^A)^D \equiv X^{(AD)} \equiv X \equiv B^D \mod 2C+1$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In steps:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1- Compute $\varphi(2C+1)$ (Which should be very time consuming if it's a product of large primes). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2- Obtain D (through extended euclidean algorithm!)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3- Compute $B^D \mod 2C+1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note, if a fast solution to your problem exists, then RSA cryptography would be insecure.&lt;/p&gt;&#xA;" OwnerUserId="1746" LastEditorUserId="1746" LastEditorDisplayName="" LastEditDate="2010-12-16T14:27:22.737" LastActivityDate="2010-12-16T14:27:22.737" CommentCount="0" />
  <row Id="14510" PostTypeId="2" ParentId="14446" CreationDate="2010-12-16T13:37:04.983" Score="6" ViewCount="" Body="&lt;p&gt;Let's solve the general problem surrounding the question,&#xA;with a few observations, each of them easy to see.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Every $V_\alpha$ for any ordinal $\alpha$ satisfies&#xA;Extensionality and Foundation, since all transitive sets satisfy&#xA;Extensionality and Foundation.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Every $V_\alpha$ satisfies Separation, for the simple reason that&#xA;$A\subset B\in V_\alpha\implies A\in V_\alpha$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Every $V_\alpha$ satisfies Union, since $A\in&#xD;&#xA; V_\alpha\implies \bigcup A\in V_\alpha$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Every $V_\lambda$ for a limit ordinal $\lambda$&#xD;&#xA;satisfies Pairing and Powerset, since the required set is&#xA;added at the next stage below $\lambda$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Every $V_\alpha$ satisfies the Axiom of Choice (assuming&#xA;this holds in $V$) in the choice-set version, since if&#xA;${\cal A}\in V_\alpha$ is a family of disjoint sets , then all choice&#xA;sets $B\subset \bigcup {\cal A}$ selecting one element from each $A\in{\cal A}$&#xD;&#xA;have the same or lower rank than ${\cal A}$, and so $B\in&#xD;&#xA; V_\alpha$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Every $V_\alpha$ with $\omega\lt\alpha$ satisfies&#xA;Infinity, since $\omega\in V_\alpha$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The only remaining axiom is Replacement, and this is the&#xA;only one that makes use of inaccessibility. But if&#xA;$\kappa$ is inaccessible, then $V_\kappa$ satisfies&#xA;Replacement, since if $A\in V_\kappa$ and $F:A\to&#xD;&#xA; V_\kappa$ is definable over $V_\kappa$, then $F&amp;#39;&amp;#39;A$ has&#xA;bounded rank below $\kappa$, since $|A|\lt\kappa$ and&#xA;$\kappa$ is regular. Thus, $F&amp;#39;&amp;#39;A\in V_\kappa$ as desired.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Finally, one can also consider the question about&#xA;$H_\delta$, the sets of hereditary size less than $\delta$,&#xA;and things are a bit nicer here in several ways.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;For any regular uncountable cardinal $\delta$, the set $H_\delta$ of&#xA;sets having hereditary size less than $\delta$, satisfies&#xA;$ZFC^-$, that is, all of ZFC except the Powerset axiom.&#xA;One gets the easy axioms easily; Separation is easy since the subset also has small hereditary size;&#xA;and Replacment follows from the fact that the union of fewer&#xA;than $\delta$ many sets in $H_\delta$ still has size less&#xA;than $\delta$ by the regularity of $\delta$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In particular, this shows that ZFC proves that there are&#xA;numerous transitive models of $ZFC^-$.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If $\kappa$ is inaccessible, then $V_\kappa=H_\kappa$,&#xA;and this satisfies full ZFC, since we get $ZFC^-$ in&#xA;$H_\kappa$, and we get power set since $\kappa$ is a&#xA;strong limit.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="413" LastActivityDate="2010-12-16T13:37:04.983" />
  <row Id="14511" PostTypeId="1" AcceptedAnswerId="14529" CreationDate="2010-12-16T13:37:16.167" Score="1" ViewCount="259" Body="&lt;p&gt;Let A be a 4 × 4 matrix with eigenvalues -5, -2, 1, 4. Which of the following is an&#xA;eigenvalue of&lt;br&gt;&#xA;A  I&lt;br&gt;&#xA;I  A  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;where I is the 4 × 4 identity matrix?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(A) -5 (B) -7 (C) 2 (D) 1  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not able to find out relation between eigenvalues of matrix A and of the given block matrix.&lt;/p&gt;&#xA;" OwnerUserId="4814" LastActivityDate="2010-12-16T15:34:25.197" Title="Eigenvalue of Block matrix" Tags="&lt;matrices&gt;&lt;eigenvectors&gt;" AnswerCount="2" CommentCount="0" />
  <row Id="14512" PostTypeId="1" AcceptedAnswerId="14513" CreationDate="2010-12-16T13:39:42.907" Score="0" ViewCount="62" Body="&lt;p&gt;Let x,y,z be three random variables. How can you show that cov(x+y,z) = cov(x,z) + cov(y,z) by using the definition of covariance.&lt;/p&gt;&#xA;" OwnerUserId="4815" LastActivityDate="2010-12-16T13:54:02.280" Title="simple question about covariance" Tags="&lt;probability&gt;" AnswerCount="1" />
  <row Id="14513" PostTypeId="2" ParentId="14512" CreationDate="2010-12-16T13:46:44.303" Score="2" ViewCount="" Body="&lt;p&gt;Use the fact that $cov(X,Y)=E(X-EX)(Y-EY)$ and rearrange the terms.&#xA;I've included the full solution below, just move the mouse on the grey area.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  \begin{align*}&#xA;   cov(x+y,z)&amp;amp;=E[(x+y-E(x+y))(z-Ez)]\\&#xA;   &amp;amp;=E[((x-Ex)+(y-Ey))(z-Ez)]\\&#xA;   &amp;amp;=E(x-Ex)(z-Ez)+E(y-Ey)(z-EZ)=cov(x,z)+cov(y,z)&#xA;   \end{align*}&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4742" LastEditorUserId="4742" LastEditorDisplayName="" LastEditDate="2010-12-16T13:54:02.280" LastActivityDate="2010-12-16T13:54:02.280" />
  <row Id="14514" PostTypeId="2" ParentId="14475" CreationDate="2010-12-16T14:06:41.207" Score="0" ViewCount="" Body="&lt;p&gt;Differential forms and connections by Darling is not too bad. I can't remember how basic it is though. He does spend a lot of time going through what differential forms are. My approach to learning is to grab a bunch of books and read each one in turn until i get confused or stuck and then to go to the other to see if it makes more sense. There is also topology from the differentiable viewpoint by milnor. but this is more on smooth manifolds than calculus.&lt;/p&gt;&#xA;" OwnerUserId="627" LastActivityDate="2010-12-16T14:06:41.207" />
  <row Id="14515" PostTypeId="1" AcceptedAnswerId="14518" CreationDate="2010-12-16T14:07:41.617" Score="3" ViewCount="187" Body="&lt;p&gt;My problem is:&#xA;What is the expression in $n$ that equals to $\sum_{i=1}^n \frac{1}{i^2}$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you very much~&lt;/p&gt;&#xA;" OwnerUserId="4736" LastEditorUserId="1543" LastEditorDisplayName="" LastEditDate="2010-12-16T14:11:44.813" LastActivityDate="2010-12-16T15:52:52.927" Title="What is the expression of $n$ that equals to $\sum_{i=1}^n \frac{1}{i^2}$?" Tags="&lt;sequences-and-series&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="14516" PostTypeId="2" ParentId="14511" CreationDate="2010-12-16T14:24:43.320" Score="1" ViewCount="" Body="&lt;p&gt;Show that if $\alpha$ is an eigenvalue of $A$ and $\beta$ is&#xA;an eigenvalue of&#xA;$$B=\begin{pmatrix}&#xD;&#xA;\alpha&amp;amp;1\\&#xD;&#xA;1&amp;amp;\alpha&#xD;&#xA;\end{pmatrix}$$&#xD;&#xA;then $\beta$ is also an eigenvalue of&#xA;$$C=\begin{pmatrix}&#xD;&#xA;A&amp;amp;I\\&#xD;&#xA;I&amp;amp;A&#xD;&#xA;\end{pmatrix}.$$&#xD;&#xA;(you should be able to write down an eigenvector for $C$in terms&#xA;of ones for $A$ and for $B$).&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-12-16T14:24:43.320" CommentCount="1" />
  <row Id="14518" PostTypeId="2" ParentId="14515" CreationDate="2010-12-16T14:31:08.927" Score="6" ViewCount="" Body="&lt;p&gt;I don't think there is a &quot;closed&quot; form. You can give a good approximation using the &lt;a href=&quot;http://en.wikipedia.org/wiki/Euler%E2%80%93Maclaurin_formula&quot; rel=&quot;nofollow&quot;&gt;Euler-McLaurin Summation&lt;/a&gt; formula though:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{j=1}^{n} \dfrac{1}{j^2}  = \dfrac{\pi^2}{6} - \dfrac{1}{n} - \dfrac{1}{2n^2} + \mathcal{O}(\dfrac{1}{n^3})$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(If you need more accuracy you can include more terms from the summation formula to give the coefficients of the lower order terms)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: The Euler McLaurin Summation formula only tells us that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{j=1}^{n} \dfrac{1}{j^2}  = C - \dfrac{1}{n} - \dfrac{1}{2n^2} + \mathcal{O}(\dfrac{1}{n^3})$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for some constant $\displaystyle C$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know by other means that $\displaystyle C = \dfrac{\pi^2}{6}$, for instance, see this for a multitude of ways: &lt;a href=&quot;http://math.stackexchange.com/questions/8337/different-methods-to-compute-sum-n-1-infty-frac1n2&quot;&gt;Different methods to compute $\sum_{n=1}^\infty \frac{1}{n^2}$.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-12-16T14:48:17.657" LastActivityDate="2010-12-16T14:48:17.657" CommentCount="1" />
  <row Id="14520" PostTypeId="2" ParentId="14483" CreationDate="2010-12-16T14:39:14.627" Score="2" ViewCount="" Body="&lt;p&gt;This steps seems to be unnecessary.  In what follows, I'll assume you have p. 30 in front of you (since it's available freely online).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Start with the part where he has constructed neighborhood $N$ of $y_0 \in Y$ and a partition $0 = t_0 &amp;lt; \dots t_n = 1$ such that $F(N \times [t_i,t_{i+1}]) \subset U_i$ for each $i = 0, \dots, n-1$.  The lift is constructed by defining&#xA;$\widetilde{F}:N \times I \to \mathbb{R}$ to be $h_{i} \circ F$, where $h_{i}$ is the homeomorphism from the evenly covered neighborhood $U_i$ to the sheet $\widetilde{U_i}$ which contains $h_{i-1} \circ F(y_0,t_i)$.  (The homeomorphism $h_0$ is determined by the given lift $\widetilde{F}$ at time $t=0$.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only question is whether this is well-defined.  One might worry that $N \times \{t_i\}$ is mapped by the $(i-1)$-st lift into the sheet $\widetilde{U}_{i-1}$ but that the neighborhood $N$ needs to be modified so that the $i$-th lift glues to this.  In other words, could it be that we need to replace $N$ with a smaller neighborhood $N&amp;#39;$ so that $h_{(i-1)} \circ F(N&amp;#39; \times \{t_i\}) \subset \widetilde{U}_{i-1} \cap \widetilde{U}_{i}$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is not the case because, by design, $F(N \times \{t_i\}) \subset U_{i-1} \cap U_i$. Therefore, the map above is well-defined.  The lift is continuous by the glueing lemma for open sets.&lt;/p&gt;&#xA;" OwnerUserId="4009" LastActivityDate="2010-12-16T14:39:14.627" />
  <row Id="14521" PostTypeId="2" ParentId="14515" CreationDate="2010-12-16T14:39:40.013" Score="4" ViewCount="" Body="&lt;p&gt;I not sure of the thrust of your question but maybe the generalised harmonic numbers are what you want&#xA;$$ H_{n,r} = \sum_{k=1}^n \frac{1}{k^r} , $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and in particular $H_{n,2}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can find more information &lt;a href=&quot;http://mathworld.wolfram.com/HarmonicNumber.html&quot; rel=&quot;nofollow&quot;&gt;here,&lt;/a&gt; including a very nice identity for $H_{n,2}$ by B. Cloitre.&lt;/p&gt;&#xA;" OwnerUserId="1301" LastActivityDate="2010-12-16T14:39:40.013" CommentCount="12" />
  <row Id="14522" PostTypeId="1" AcceptedAnswerId="14532" CreationDate="2010-12-16T14:49:01.743" Score="10" ViewCount="131" Body="&lt;p&gt;In our analysis course, the following question came up and could, up to now, not be solved:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Let $a: \mathbb{N} \to \mathbb{C}$ be a sequence of complex numbers. What are necessary and sufficient conditions for the existence of a function $\mu: 2^\mathbb{N} \to \mathbb{C}$ satisfying the properties&lt;/p&gt;&#xA;  &#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;$\mu(\{i\}) = a_i$&#xD;&lt;/li&gt;&#xA;  &lt;li&gt;$\mu$ is finitely additive, i.e. $A\cap B = \emptyset\implies \mu(A\cup B) = \mu(A)+\mu(B)$?&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h2&gt;Partial Results&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;If we are given such a function defined on a subset of $2^\mathbb{N}$, we can of course extend it to all sets of the form $A\cup B, A\cap B = \emptyset$ and $A\setminus B, B\subset A$. This invites the use of Zorn's Lemma; but it seems impossible to prove that a maximal set closed under these operations must be $2^\mathbb{N}$. However, this approach strongly suggests that $\mu$ exists for all $(a_i)$, as the problem only depends on $2^\mathbb{N}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, if $a_i$ converges absolutely, one can set $\mu(I) = \sum_{i\in I} a_i$ which fulfills the required properties, but this approach does not generalise at all.&lt;/p&gt;&#xA;" OwnerUserId="4817" LastActivityDate="2010-12-16T16:30:05.913" Title="Finitely additive measures on $2^\mathbb{N}$" Tags="&lt;measure-theory&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="2" />
  <row Id="14523" PostTypeId="1" CreationDate="2010-12-16T14:57:20.950" Score="0" ViewCount="176" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is ; How can I solve the following integral question?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int\tan\;x\;\cos\;2x\;\mathrm dx$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance,&lt;/p&gt;&#xA;" OwnerUserId="3669" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-12-17T01:46:08.627" LastActivityDate="2010-12-17T01:46:08.627" Title="How can I find $\int\tan\;x\;\cos\;2x\;\mathrm dx$?" Tags="&lt;calculus&gt;&lt;trigonometry&gt;&lt;integral&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="14524" PostTypeId="2" ParentId="14523" CreationDate="2010-12-16T15:03:58.120" Score="6" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) $\cos 2x = \cos^2x-\sin^2x=2\cos^2x-1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) $2\sin x\cos x = \sin 2x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(3) $\frac{d}{dx}\log(f(x))=?$&lt;/p&gt;&#xA;" OwnerUserId="1154" LastActivityDate="2010-12-16T15:03:58.120" CommentCount="3" />
  <row Id="14525" PostTypeId="2" ParentId="14523" CreationDate="2010-12-16T15:08:27.337" Score="1" ViewCount="" Body="&lt;p&gt;Suppose I gave you an integral of the form&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle \int \cot x \ \ f(\sin x) \ \text{dx}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you think of a substitution to get rid of the $\cot x$ term?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a concrete example, can you try evaluating&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle \int \cot x  \ \ (1 + \sin^5 x) \ \ \text{dx}$ ?&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-12-16T15:20:49.363" LastActivityDate="2010-12-16T15:20:49.363" />
  <row Id="14527" PostTypeId="1" CreationDate="2010-12-16T15:13:48.093" Score="0" ViewCount="28" Body="&lt;p&gt;If $N$ is even, how to prove that $(D(z))\hat{}(n)=1/2(\hat{z}(n)+\hat{z}(n+N/2))$, for all $n$. Hint: Show that the discrete Fourier transformation(DFT) on $N/2$ points of $D(z)$ agrees with the DFT on $N$ points of $(z+z\ast)/2$,  for $z\ast(n)=(-1)^nz(n)$, DFT: $\hat{z}(m)=\sum_{n=0}^{N-1} z(n)e^{-2\pi imn/N}$, $D(z)(n) = z(2n)$(i.e. if $z=(z(0),z(1),\dots,z(N-1))$ then $D(z)=(z(0),z(2),\dots,z(N-2))$) also remember $(z+z\ast)(n)=z(n)(1+(-1)^n))= { \begin{array}{ll} 2 z(n) &amp;amp;, n \textrm{is even} \0 &amp;amp;, \textrm{is odd}\end{array} &lt;/p&gt;&#xA;" OwnerUserId="2723" LastActivityDate="2010-12-16T15:13:48.093" Title="If $N$ is even, how to prove that $(D(z))\hat{}(n)=1/2(\hat{z}(n)+\hat{z}(n+N/2))$" Tags="&lt;linear-algebra&gt;" />
  <row Id="14528" PostTypeId="2" ParentId="14523" CreationDate="2010-12-16T15:32:40.513" Score="1" ViewCount="" Body="&lt;p&gt;I'm going to tell you that by parts done directly isn't the way to approach this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int \tan(x)\cos(2x)dx = -\ln(\cos(x))\cos(2x) - 2\int \ln(\cos(x))\sin(2x)dx$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you can see, this expression is not likely to become any more manageable by solving the next integral.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In short, your problem comes down to simplifying the expression $\tan(x)cos(2x)$. Big hint. The other answers have shown you how to do this. Once you simplify it, you will have a much easier job of integrating said expression and you most certainly won't need integration by parts.&lt;/p&gt;&#xA;" OwnerUserId="892" LastEditorUserId="892" LastEditorDisplayName="" LastEditDate="2010-12-16T15:50:42.777" LastActivityDate="2010-12-16T15:50:42.777" />
  <row Id="14529" PostTypeId="2" ParentId="14511" CreationDate="2010-12-16T15:34:25.197" Score="2" ViewCount="" Body="&lt;p&gt;The block matrix can be written as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$C=A \otimes I_2 + I_4 \otimes J_2$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$J_2=\left (\begin{array}{cc} 0 &amp;amp;\ 1\\ 1 &amp;amp;\ 0 \end{array} \right )$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Denote the eigenvectors of $A$ by:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$Av_{\lambda} = \lambda v_{\lambda}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and $J_2$ by:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$J_2w_{\pm} = \pm w_{\pm}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(The eigenvalues of $J_2$ are $1$ and $-1$). Clearly, the eight combinations:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$v_{\lambda} \otimes w_{\pm}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;are all eigenvectors of $C$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus the eigenvalues of C are $\lambda \pm 1$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus the right answer is C corresponding to $\lambda = 1$&lt;/p&gt;&#xA;" OwnerUserId="1430" LastActivityDate="2010-12-16T15:34:25.197" CommentCount="1" />
  <row Id="14530" PostTypeId="1" AcceptedAnswerId="30939" CreationDate="2010-12-16T15:39:16.890" Score="2" ViewCount="248" Body="&lt;p&gt;Looking on Internet, I mostly found this definition:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given quantities on a cyclic domain D, first rescale the domain to [0;2$\pi$[, then, find $z_n$ the point on the unit circle corresponding to the $n$th value, and compute the average by:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$z_m = \sum_{n=1}^N z_n$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The average angle is then $\theta_m = \arg z_m$ and to obtain the average value you scale back to your original domain D.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I must say, I have problems with this definition. For simplicity, I will use oriented angles in degree for my examples (i.e. D = [0;360[). With this formula, having angles -90, 90 and 40 will give 40 as mean angle, when I would expect 13.33 as an answer (i.e. (90-90+40)/3).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For my own problems, I usually use:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$v_m = \mathop{\rm arg\,min}_{v\in D} \sum_{n=1}^{N} d(v_n,v)^2$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $d(x,y)$ is the distance in the cyclic domain, and $\{v_1, v_2, \ldots, v_n\}$ is the set of cyclic data I want to average of.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It has the advantage to work the same way whatever the domain (replace D by a non-cyclic domain and $d$ with the usual euclidean distance, and you find the usual definition of an average). However, it is expensive to compute and I don't know any exact method to do it in general.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So my question is: what is the appropriate way to deal with average of cyclic data? And do you have good pointers that explain the problem and its solutions?&lt;/p&gt;&#xA;" OwnerUserId="4709" LastEditorUserId="4709" LastEditorDisplayName="" LastEditDate="2010-12-16T17:53:09.050" LastActivityDate="2011-04-04T18:40:20.290" Title="How to average cyclic quantities?" Tags="&lt;algebra&gt;" AnswerCount="4" CommentCount="3" />
  <row Id="14531" PostTypeId="2" ParentId="14515" CreationDate="2010-12-16T15:52:52.927" Score="0" ViewCount="" Body="&lt;p&gt;I am not sure if this will work or not, but maybe you could try writing the expression in terms of falling factorials[1]. Then maybe use summation by parts[2]. I am not sure how nicely this will work, but you could try it. Let me know what you find out!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[1] &lt;a href=&quot;http://mathworld.wolfram.com/FallingFactorial.html&quot; rel=&quot;nofollow&quot;&gt;http://mathworld.wolfram.com/FallingFactorial.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[2] &lt;a href=&quot;http://mathworld.wolfram.com/SummationbyParts.html&quot; rel=&quot;nofollow&quot;&gt;http://mathworld.wolfram.com/SummationbyParts.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3791" LastActivityDate="2010-12-16T15:52:52.927" CommentCount="2" />
  <row Id="14532" PostTypeId="2" ParentId="14522" CreationDate="2010-12-16T15:58:20.213" Score="6" ViewCount="" Body="&lt;p&gt;Given any such sequence $a_i$, you can use it to define a finitely additive measure on the collection of &lt;em&gt;finite&lt;/em&gt; subsets of $\mathbb N$.  Now consider the vector&#xA;space $V:= \mathbb C^{\oplus \mathbb N}$, i.e. the direct sum of countably many copies of $\mathbb C$, with the copies of $\mathbb C$ being indexed by elements $i \in \mathbb N$.  Alternatively, this is the space of sequences $(z_i)_{i \in \mathbb N}$ with $z_i =0$ for all but finitely many $i$.&#xA;Or, if you want to think a little more analytically, you can think&#xA;of this as the space of $\mathbb C$-valued functions on $\mathbb N$ with &lt;em&gt;finite support&lt;/em&gt;, i.e. which vanish outside a finite set.   (The function attached to a sequence is just $i \mapsto z_i$,  of course.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Our choice of finitely addivite measure defines a functional on $V$, given by&#xA;$(z_i) \mapsto \sum_{i \in \mathbb N} a_i z_i.$ (More analytically, this is &lt;em&gt;integration&lt;/em&gt; of the finitely supported function corresponding to $(z_i)$ against our finitely addivite measure.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let $W$ be the vector space of &lt;em&gt;all&lt;/em&gt; $\mathbb C$-valued functions on $V$.&#xA;Certainly $V \subset W$, and we may always extend linear functionals from a subspace to the whole space; thus we may extend our given functional to a&#xA;functional $I: W \to \mathbb C$.  (The label $I$ is chosen to suggest &lt;em&gt;integration&lt;/em&gt;.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now if $S$ is any subset of $\mathbb N$, let $\chi_S$ be the characteristic function of $S$.  Define $\mu(S) = I(\chi_S)$.  Then $\mu$ is a finitely additive measure on $\mathbb N$ satisfying the required properties.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(This is essentially the Zorn's lemma argument suggested in the original posting, but reformulated in terms of extending functionals on vector spaces,&#xA;which makes it more transparent.)&lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2010-12-16T15:58:20.213" CommentCount="3" />
  <row Id="14533" PostTypeId="2" ParentId="14507" CreationDate="2010-12-16T16:06:34.150" Score="7" ViewCount="" Body="&lt;p&gt;Why not simply $\int 3^x e^x \mathrm dx = \int (3e)^x \mathrm dx = \frac{(3e)^x}{\ln(3e)} + Constant$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To see that $\int (3e)^x \mathrm dx = \frac{(3e)^x}{\ln(3e)}+C$,  You can differentiate $(3e)^x$ i.e. Let&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$y = (3e)^x$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ln y = x \ln(3e)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \mathrm dy/\mathrm dx = y \ln(3e) = (3e)^x \ln(3e)$$&lt;/p&gt;&#xA;" OwnerUserId="4728" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-17T00:52:02.463" LastActivityDate="2010-12-17T00:52:02.463" CommentCount="3" />
  <row Id="14534" PostTypeId="1" CreationDate="2010-12-16T16:06:37.453" Score="3" ViewCount="90" Body="&lt;p&gt;Is it true that $O(M^3 + NM^2) \, = \, O(M^3 + N)$, where $M$ and $N$ are variables of the function?&lt;/p&gt;&#xA;" OwnerUserId="4820" LastActivityDate="2010-12-16T18:00:31.240" Title="Big-oh for function of two variables" Tags="&lt;computer-science&gt;&lt;asymptotics&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="14535" PostTypeId="2" ParentId="14494" CreationDate="2010-12-16T16:17:12.193" Score="4" ViewCount="" Body="&lt;p&gt;This is a well-known issue, which one resolves as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, it is better not to work with some given metric on the manifold, because you don't know precisely how the metric structure (e.g. the balls in the given metric) interact with the locally Euclidean structure.  So it is better to &lt;em&gt;first&lt;/em&gt; choose a n.h. of a point $p$ which is diffeomorphic to an open disk in $\mathbb R^2$, and then to work with that n.h., which one can now think of very&#xA;concretely as an open disk in the plane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For concreteness, suppose it is the open disk of radius $r$ around $0$.  Now &#xA;delete the open disk of radius $r- \epsilon$, for some positive $\epsilon$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we are in a situation very similar to you one you described (i.e. $M\setminus B(p,\epsilon)$) except that we know exactly what it is that we deleted (i.e. we are certain that it is a small disk), &lt;em&gt;and&lt;/em&gt; we know that there is an open annulus of inner radius $r-\epsilon$ and outer radius $r$ left over.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can do the same thing at the other point (using Hausdorffness, as you point out).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now when we glue in the cylinder, think of the cylinder as a circle cross an open interval, say $I$.  Now we can choose  small open subintervals $I_1$ and $I_2$ at either end of the interval $I$ (so that altogether $I = I_1 \cup \text{ a closed interval } \cup I_2$, with the unions being disjoint).&#xA;Notice that a circle cross $I_1$ is diffeomorphic to our open annulus from above,&#xA;and the same for a circle cross $I_2$.  (This is just using the fact that the $I_i$ are open intervals, and I leave it as an exercise for you to find the&#xA;diffeomorphism.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So now we can glue our cylinder to $M$ minus the two disks by identifying&#xA;the circle cross $I_1$ with the annulus around one of our deleted points,&#xA;and the circle cross $I_2$ with the other annulus, using the diffeos of the preceding paragraph.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In summary: we have carefully glued in our cylinder to that there are no sharp corners, by making sure that each end of the cylinder is glued smoothly to an annulus around each of the points that is being deleted.&lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2010-12-16T16:17:12.193" />
  <row Id="14538" PostTypeId="2" ParentId="14530" CreationDate="2010-12-16T16:33:15.537" Score="0" ViewCount="" Body="&lt;p&gt;The angle is supposed to be the independent variable, not the dependent variable.  If your function is cyclic (using degrees), z(-90)=z(270) so it doesn't matter which you use.  Then the average value of the function is $$z_m=\sum_{n=1}^Nz(\theta_n)$$  You are right that if you average the angles, it matters which lap of the circle you use (-90 vs 270) because the difference of 360 gets divided by N.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-12-16T16:33:15.537" CommentCount="3" />
  <row Id="14539" PostTypeId="2" ParentId="14417" CreationDate="2010-12-16T16:47:49.827" Score="4" ViewCount="" Body="&lt;p&gt;For the more general question at the end, let's first consider the weighted inner product and weighted norm on $\mathbb{R}^n$, which are defined by &#xA;$$&#xD;&#xA;\langle u,v\rangle  = \sum\limits_{i = 1}^n {c_i u_i v_i }, \;\; ||u|| = \sqrt {\langle u,u\rangle }  = \sqrt {\sum\limits_{i = 1}^n {c_i u_i^2 } },  &#xD;&#xA;$$&#xD;&#xA;where $c_1,\ldots,c_n$ are positive constants (called weights). By Cauchy-Schwartz inequality, $|\langle u,v\rangle | \le ||u||||v||$. Hence, letting ${n \choose k}$ play the role of weights (and using the symmetry of binomial coefficients), we have&#xA;$$&#xD;&#xA;\sum\limits_{k = 0}^n {{n \choose k}a^{\sqrt k } b^{\sqrt {n - k} } }  \le \sqrt {\sum\limits_{k = 0}^n {{n \choose k}(a^2 )^{\sqrt k } } } \sqrt {\sum\limits_{k = 0}^n {{n \choose k}(b^2 )^{\sqrt k } } } .&#xD;&#xA;$$&#xD;&#xA;So, if we assume that $a,b \geq 1$, the problem essentially reduces to the original one (as far as an upper-bound is concerned). In particular, when $a=b$, we get&#xA;$$&#xD;&#xA;\sum\limits_{k = 0}^n {{n \choose k} a^{\sqrt k } a^{\sqrt {n - k} } }  \le \sum\limits_{k = 0}^n {{n \choose k} a^{2\sqrt k } }.&#xD;&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="2810" LastActivityDate="2010-12-16T16:47:49.827" />
  <row Id="14540" PostTypeId="1" CreationDate="2010-12-16T16:51:19.060" Score="3" ViewCount="198" Body="&lt;p&gt;Find Fourier cosine transform of $e^{-a^2 x^2}$ and hense evaluate Fourier sine transform of $x\cdot e^{-a^2x^2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can solve this question only if there is $x$ instead of $x^2$ in the exponential function $e^{-a^2x^2}$. Because in this situation i can use integral formula :- &#xA;$$\int e^{-ax} \cos sx\,dx = \frac{e^{-ax}}{a^2+x^2} \left( s\sin sx-a\cos sx \right)$$&#xD;&#xA;but what should i do if there is $x^2$ in exponential function $e$??&lt;/p&gt;&#xA;" OwnerUserId="4822" LastEditorUserId="171" LastEditorDisplayName="" LastEditDate="2010-12-16T18:52:44.780" LastActivityDate="2010-12-16T20:37:59.313" Title="Fourier cosine transform" Tags="&lt;fourier-analysis&gt;&lt;fourier-series&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="14541" PostTypeId="1" CreationDate="2010-12-16T17:05:07.160" Score="2" ViewCount="104" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know this is a quite general question , but I've always heard about creating mathematical models  in economics, social sciences, engineering,... And I would want to know what are the starting points and roadmap to understand what a mathematical model is and how to create them. Thanks&lt;/p&gt;&#xA;" OwnerUserId="4824" LastActivityDate="2010-12-16T18:09:21.253" Title="How to start creating Mathematical models?" Tags="&lt;mathematical-modeling&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="14542" PostTypeId="2" ParentId="14495" CreationDate="2010-12-16T17:10:35.577" Score="3" ViewCount="" Body="&lt;p&gt;Typically, in modeling problems of the kind you are suggesting there are a variety of ways that the triangles could be selected, and one's goal would govern how the triangles might be chosen. Sometimes these are referred to as a triangular mesh:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Triangle_mesh&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Triangle_mesh&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the analogous problem of selecting triangles based on a point set in the plane you can find some ideas here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Point_set_triangulation&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Point_set_triangulation&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1369" LastActivityDate="2010-12-16T17:10:35.577" CommentCount="1" />
  <row Id="14543" PostTypeId="2" ParentId="14541" CreationDate="2010-12-16T17:13:49.380" Score="1" ViewCount="" Body="&lt;p&gt;COMAP, the Consortium for Mathematics and Its Applications has developed lots of materials at different levels that address your question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.comap.com/&quot; rel=&quot;nofollow&quot;&gt;http://www.comap.com/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1369" LastActivityDate="2010-12-16T17:13:49.380" CommentCount="2" />
  <row Id="14544" PostTypeId="2" ParentId="14494" CreationDate="2010-12-16T17:25:55.777" Score="0" ViewCount="" Body="&lt;p&gt;I cannot add a comment yet, so I am posting this as a separate answer.&#xA;Firstly, thank you Matt E for the response. &#xA;Now, supposing I want to find a diffeomorphism from $S^1\times I_1$ to the annulus, shouldn't the annulus itself be open, i.e., I have to remove a closure of $B(p,r-\epsilon)$. In which case, I can think of the annulus as $S^1 \times (r-\epsilon,r)$ and then the diffeomorphism is just the lift of the usual diffeomorphism between open intervals.&lt;/p&gt;&#xA;" OwnerUserId="4828" LastActivityDate="2010-12-16T17:25:55.777" CommentCount="1" />
  <row Id="14545" PostTypeId="2" ParentId="14499" CreationDate="2010-12-16T17:41:16.957" Score="7" ViewCount="" Body="&lt;p&gt;This &lt;a href=&quot;http://www.ams.org/journals/tran/1944-056-00/S0002-9947-1944-0011087-2/S0002-9947-1944-0011087-2.pdf&quot; rel=&quot;nofollow&quot;&gt;paper&lt;/a&gt; by Alaoglu and Erdős proves that if $p$ is the largest prime divisor of a highly composite number (HCN) $n$ then the exponent $k_q$ on another prime $q$ dividing&#xA;$n$ satisfies &#xA;$$\log\left(1 + \frac{1}{k_q}\right) &amp;gt; \frac{\log q \log 2}{\log p} + O\left(\frac{(\log \log p)^3}{(\log p)^3}\right),$$&#xD;&#xA;$$\log\left(1 + \frac{1}{k_q+1}\right) &amp;lt; \frac{\log q \log 2}{\log p} + O\left(\frac{(\log \log p)^3}{(\log p)^3}\right).$$&#xD;&#xA;(See Theorem 12.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For sufficiently large $q$ they claim that $k_q$ is either $\Lambda$, $\Lambda + 1$, or $\Lambda - 1$, where $$\Lambda = \left\lfloor \left(2^{\log q/\log p} - 1\right)^{-1} \right\rfloor.$$  (See Theorem 13.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When $q = 2$ this latter formula appears to be close, although it is not exactly correct, as there are more than three possible values for the power of 2 that divides a highly composite number with largest prime factor $p$.  See, for example, &lt;a href=&quot;http://wwwhomes.uni-bielefeld.de/achim/highly.txt&quot; rel=&quot;nofollow&quot;&gt;this list of the first 1200 HCN's&lt;/a&gt;, which includes their prime factorizations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This, together with Derek Jennings' answer on the order of $p$, should give you the estimate of the highest power of 2 that divides $n$.&lt;/p&gt;&#xA;" OwnerUserId="2370" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-12-17T06:07:06.710" LastActivityDate="2010-12-17T06:07:06.710" CommentCount="1" />
  <row Id="14546" PostTypeId="2" ParentId="14540" CreationDate="2010-12-16T17:42:28.810" Score="2" ViewCount="" Body="&lt;p&gt;HINT 1:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;\int_{ - \infty }^\infty  {\cos (sx)e^{ - a^2 x^2 } \,{\rm d}x}  = \int_{ - \infty }^\infty  {[\cos (sx) + {\rm i}\sin (sx)]e^{ - a^2 x^2 } \,{\rm d}x}.&#xD;&#xA;$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;HINT 2:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Characteristic function of the (centered) normal distribution: see &lt;a href=&quot;http://en.wikipedia.org/wiki/Normal_distribution#Characteristic_function_and_moment_generating_function&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;: The first hints didn't help. So,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;\int_{ - \infty }^\infty  {\cos (sx)e^{ - a^2 x^2 } \,{\rm d}x}  = \int_{ - \infty }^\infty  {[\cos (sx) + {\rm i}\sin (sx)]e^{ - a^2 x^2 } \,{\rm d}x} = \int_{ - \infty }^\infty  {e^{{\rm i}sx} e^{ - a^2 x^2 } {\rm d}x} = ? &#xD;&#xA;$$&#xD;&#xA;In order to calculate the right-most integral, bring it to the form(s) &#xA;$$&#xD;&#xA;b \int_{ - \infty }^\infty  {e^{ - a^2(x - {\rm i}c )^2  } \, {\rm d}x} = &#xD;&#xA;b \int_{ - \infty }^\infty  {e^{ - a^2 x^2 } \,{\rm d}x} = b&amp;#39; \int_{ - \infty }^\infty  {e^{ - x^2 } \,{\rm d}x} =...&#xD;&#xA;$$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="2810" LastEditorUserId="2810" LastEditorDisplayName="" LastEditDate="2010-12-16T20:37:59.313" LastActivityDate="2010-12-16T20:37:59.313" CommentCount="5" />
  <row Id="14547" PostTypeId="2" ParentId="14540" CreationDate="2010-12-16T17:43:15.990" Score="2" ViewCount="" Body="&lt;p&gt;Find the Fourier transform of $e^{-a^2x^2}$. The real part of that is relevant to your cosine transform. &lt;/p&gt;&#xA;" OwnerUserId="3249" LastActivityDate="2010-12-16T17:43:15.990" CommentCount="2" />
  <row Id="14548" PostTypeId="2" ParentId="14501" CreationDate="2010-12-16T17:49:46.497" Score="0" ViewCount="" Body="&lt;p&gt;Generally it's difficult to compute such $\rm\:A$'th roots in $\rm \mathbb Z/m\ $ except for simple special cases, e.g. when $\rm\ gcd(A,\phi(m)) = 1\:.\ $ Otherwise one generally has to resort to factoring $\rm\:m\:$ to get $\rm\: \phi(m)\: $ and hence the factorization of $\rm\ Z/m^*\ $ into a product of cyclic groups. Then one may apply generalizations of Shanks's  square-root algorithm. See for example Section 3.2: Extensions of Shanks's algorithm to r'th roots in cyclic groups in Lindhurst: &lt;a href=&quot;http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.144.2770&amp;amp;rep=rep1&amp;amp;type=pdf&quot; rel=&quot;nofollow&quot;&gt;Computing roots in finite fields and groups, with a jaunt through sums of digits&lt;/a&gt;. One can also generalize the Euler criterion for square-roots - see my &lt;a href=&quot;http://math.stackexchange.com/questions/5424/how-can-i-find-out-whether-a-number-is-a-quadratic-residue-in-a-large-modulo/5433#5433&quot;&gt;post here&lt;/a&gt; and Chapter 4 of Ireland and Rosen, A Classical Introduction to Modern Number Theory.&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-12-16T17:49:46.497" CommentCount="3" />
  <row Id="14549" PostTypeId="1" AcceptedAnswerId="14636" CreationDate="2010-12-16T17:56:43.983" Score="1" ViewCount="78" Body="&lt;p&gt;Consider the &lt;a href=&quot;http://en.wikipedia.org/wiki/Chernoff_bound&quot; rel=&quot;nofollow&quot;&gt;Chernoff bound&lt;/a&gt;: It assumes $X_1, \ldots, X_n$ are independent Bernoulli random variables, each having probability $p &amp;gt; 1/2$, and seeks the probability of simultaneous occurrence of more than $n/2$ of the events.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I once heard that, in the proof of the Chernoff bound, parameters $p$ and $n$ must be constant. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this true? for instance, can we let $p = 1/2 + n^{-3}$?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h3&gt;Edit (As per moron's request)&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;I'll elaborate on how I wan to use it. The &lt;a href=&quot;http://en.wikipedia.org/wiki/BPP&quot; rel=&quot;nofollow&quot;&gt;complexity class BPP&lt;/a&gt; is defined as below:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A language L is in BPP if and only if there exists a probabilistic Turing machine M, such that&lt;/p&gt;&#xA;  &#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;M runs for polynomial time on all inputs;&lt;/li&gt;&#xA;  &lt;li&gt;&lt;strong&gt;Completeness:&lt;/strong&gt; For all $x \in L$, M outputs 0 with probability less than or equal to 1/3;&lt;/li&gt;&#xA;  &lt;li&gt;&lt;strong&gt;Soundness:&lt;/strong&gt; For all $x \not \in L$, M outputs 1 with probability less than or equal to 1/3.&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Again, from Wikipedia:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In practice, an error probability of 1/3 might not be acceptable, ... but if the algorithm is run many times, the chance that the majority of the runs are wrong drops off exponentially as a consequence of the &lt;strong&gt;Chernoff bound&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This seems to be true as long as the error probability is a constant. However:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;It does not even have to be constant: the same class of problems is defined by allowing error as high as $1/2 − n^{−c}$ on the one hand, or requiring error as small as $2^{−n^c}$ on the other hand, where $c$ is any positive constant, and $n$ is the length of input.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;How is Chernoff bound applied in such cases?&lt;/p&gt;&#xA;" OwnerUserId="2420" LastEditorUserId="2420" LastEditorDisplayName="" LastEditDate="2010-12-16T21:19:49.450" LastActivityDate="2010-12-17T04:52:40.090" Title="Non-constant parameters in Chernoff bound" Tags="&lt;inequality&gt;&lt;probability&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="14550" PostTypeId="1" AcceptedAnswerId="14587" CreationDate="2010-12-16T17:59:22.947" Score="1" ViewCount="74" Body="&lt;p&gt;Let $G$ be a group of permutations of  $n$ objects. Let subsets $A$,$B$ of the objects be equivalent if $A=g(B)$ for some $g \in G$. &#xA;Is there a way to show that a bijection exists between the equivalence classes and the $2$-colorings of the objects that are invariant under those permutations in $G$? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So Far, I have verified that it works(their cardinalities are the same) for a few cases I tried, but I do not see how it can be generalized. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I will post an example, the definition should be clear from the example. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;let $G=\{e,(2 3)\}$, $n=3$&#xD;&#xA;then, we have 8 subsets and 6 equivalent subsets. namely $\{\}, \{1\},\{\{2\},\{3\}\}, \{\{1,2\},\{1,3\}\},\{2,3\},\{1,2,3\}$ &#xA;and the invarient transformations are. &#xA;$\{1,1,1\},\{1,1,0\} \sim \{1,0,1\},\{0,1,1\},\{1,0,0\},\{0,0,1\} \sim \{0,1,0\},\{0,0,0\}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this example applying (2 3) to {0,1,0} changes it to {0,0,1} and applying e, leaves it as it is. &lt;/p&gt;&#xA;" OwnerUserId="4728" LastEditorUserId="4728" LastEditorDisplayName="" LastEditDate="2010-12-16T19:38:54.607" LastActivityDate="2010-12-16T21:31:07.990" Title="Bijection of 2-colorings and permutation equivalence classes" Tags="&lt;combinatorics&gt;" AnswerCount="2" CommentCount="10" />
  <row Id="14551" PostTypeId="2" ParentId="14534" CreationDate="2010-12-16T18:00:31.240" Score="5" ViewCount="" Body="&lt;p&gt;[Throughout this answer, I'm going to use set notation in order to be pedantic.]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is true that $O(M^3+N)\subset O(M^3+NM^2)$, since for $M$ sufficiently large, $c(M^3+N)&amp;lt;c(M^3+NM^2)$. It's worth noting that for $M$ sufficiently close to $0$, $M^3+N&amp;gt;M^3+NM^2$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the reverse inclusion, consider $f(M,N)=M^3+NM^2$. Clearly $f\in O(M^3+NM^2)$. We now need to show that for all $c$ and arbitrarily large $M,N$ that $f(M,N)&amp;gt;c(M^3+N)$. We can do this by setting $M=\log N$. We then have $f(\log N,N)=\log^3 N + N\log^2 N$, which is clearly larger than $c(\log^3 N + N)$ for all sufficiently large $N$. So $f(M,N)\not\in O(M^3+N)$.&lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2010-12-16T18:00:31.240" />
  <row Id="14552" PostTypeId="2" ParentId="14499" CreationDate="2010-12-16T18:06:27.230" Score="4" ViewCount="" Body="&lt;p&gt;In this paper &lt;a href=&quot;http://www.renyi.hu/~p_erdos/1944-04.pdf&quot; rel=&quot;nofollow&quot;&gt;On Highly Composite Numbers&lt;/a&gt; P. Erdös uses Bertrand's postulate and the Prime Number Theorem to show that if $p$ is the highest prime that divides $n,$ a highly composite number, then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ c_1 \log n &amp;lt; p &amp;lt; c_2 \log n $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for some constants $c_1$ and $c_2 .$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="1301" LastActivityDate="2010-12-16T18:06:27.230" CommentCount="1" />
  <row Id="14553" PostTypeId="2" ParentId="14541" CreationDate="2010-12-16T18:09:21.253" Score="1" ViewCount="" Body="&lt;p&gt;The last time I taught mathematical modeling we used &lt;em&gt;&lt;a href=&quot;http://books.google.com/books?id=XPxrItVsXpQC&amp;amp;printsec=frontcover&amp;amp;dq=a+first+course+in+mathematical+modeling&amp;amp;source=bl&amp;amp;ots=4EcHhIII_M&amp;amp;sig=K1KZ1lq9OmZWj-aMKMdJiv5wg64&amp;amp;hl=en&amp;amp;ei=pVQKTeTAI4nUtQPioODDCg&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=6&amp;amp;ved=0CEMQ6AEwBQ#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;Introduction to Mathematical Modeling&lt;/a&gt;,&lt;/em&gt; by Giordano, Fox, Horton, and Weir.  The first part of Chapter 2 contains a nice introduction to the modeling process.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A &lt;a href=&quot;http://www.google.com/search?tbs=bks%3A1&amp;amp;tbo=1&amp;amp;q=mathematical+modeling&amp;amp;btnG=Search+Books&quot; rel=&quot;nofollow&quot;&gt;search on Google books for mathematical modeling texts&lt;/a&gt; should yield some more good results.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My &lt;a href=&quot;http://math.stackexchange.com/questions/11231/probability-in-single-lane-traffic-flow-what-are-the-odds-of-choke-points-bein/11316#11316&quot;&gt;answer to a recent math.SE question&lt;/a&gt; also contains some discussion of the modeling process.&lt;/p&gt;&#xA;" OwnerUserId="2370" LastActivityDate="2010-12-16T18:09:21.253" />
  <row Id="14554" PostTypeId="1" AcceptedAnswerId="14555" CreationDate="2010-12-16T18:17:18.190" Score="5" ViewCount="238" Body="&lt;p&gt;I am a calculus TA a bit stumped at a question a student has posed me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the method for finding the sum of the series $$\sum_{n=0}^\infty\frac{1+n}{3^n}$$ If you split the numerator up, one becomes a geometric series summing to 3/2, and the other sums to 3/4 by Wolfram Alpha, but I can't explain why.&lt;/p&gt;&#xA;" OwnerUserId="541" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-12-16T19:32:24.853" LastActivityDate="2010-12-16T19:32:24.853" Title="Sum of a Series (Calculus)" Tags="&lt;calculus&gt;&lt;sequences-and-series&gt;" AnswerCount="4" CommentCount="3" />
  <row Id="14555" PostTypeId="2" ParentId="14554" CreationDate="2010-12-16T18:21:24.257" Score="8" ViewCount="" Body="&lt;p&gt;Using the usual geometric series, you have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{x}{1-x} = \sum_{n=1}^{\infty} x^n $$ thus differentiating you get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{n=1}^{\infty} n x^{n-1} = \frac{1}{(1-x)^2}$$ thus &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{n=1}^{\infty} n x^{n} = \frac{x}{(1-x)^2}$$ Then &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{n=1}^{\infty} n x^{n} = \frac{x}{(1-x)^2}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So your sum is just&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{n=1}^{\infty} \frac{n}{3^n} = \frac{1/3}{(1-1/3)^2} = \frac{3}{4}$$&lt;/p&gt;&#xA;" OwnerUserId="900" LastEditorUserId="900" LastEditorDisplayName="" LastEditDate="2010-12-16T18:30:26.570" LastActivityDate="2010-12-16T18:30:26.570" CommentCount="4" />
  <row Id="14556" PostTypeId="2" ParentId="14554" CreationDate="2010-12-16T18:24:37.510" Score="4" ViewCount="" Body="&lt;p&gt;Consider for $|x| &amp;lt; 1$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{n=1}^\infty x^n = \frac{x}{1-x} .$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Differentiate and set $x=1/3.$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="1301" LastActivityDate="2010-12-16T18:24:37.510" />
  <row Id="14557" PostTypeId="2" ParentId="14445" CreationDate="2010-12-16T18:25:22.333" Score="1" ViewCount="" Body="&lt;p&gt;How about embedding $W^{1,p}$ in $L^p\times (L^p)^n$ and using Hahn-Banach and Riesz' representation theorem to get a nice characterization of elements in the dual &lt;/p&gt;&#xA;" OwnerUserId="4829" LastActivityDate="2010-12-16T18:25:22.333" CommentCount="3" />
  <row Id="14559" PostTypeId="2" ParentId="14530" CreationDate="2010-12-16T19:06:07.320" Score="0" ViewCount="" Body="&lt;p&gt;For angles, one can adapt the iterative way of computing means to angles, that is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;given angles v[1] .. v[n]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;m[1] = v[1]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;m[i] = remainder( m[i-1] + remainder( v[i]-m[i-1], C)/i, C)   (i=2..n)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where remainder(x,y) gives the signed remainder on dividing x by y and C is the measure of a circle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suspect this gives your vm, but haven't been able to prove it.&lt;/p&gt;&#xA;" OwnerUserId="4830" LastActivityDate="2010-12-16T19:06:07.320" CommentCount="1" />
  <row Id="14560" PostTypeId="1" AcceptedAnswerId="14565" CreationDate="2010-12-16T19:11:32.010" Score="8" ViewCount="308" Body="&lt;p&gt;I understand the basic idea of Conditionally Convergent (some infinitely long series can be made to converge to any value by reordering the series). I just do not understand how this could possibly be true. I think it defies common sense and seems like a clear violation of the Commutative Law.&lt;/p&gt;&#xA;" OwnerUserId="4831" LastActivityDate="2010-12-17T03:01:11.483" Title="Please explain how Conditionally Convergent can be valid?" Tags="&lt;sequences-and-series&gt;&lt;convergence&gt;" AnswerCount="4" CommentCount="2" FavoriteCount="1" />
  <row Id="14561" PostTypeId="2" ParentId="14554" CreationDate="2010-12-16T19:12:37.740" Score="6" ViewCount="" Body="&lt;p&gt;Michael, there is a general trick here that can be very useful. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Think of $xD$ as an operator on differentiable functions $p(x)$ that &quot;differentiates and then multiplies by $x$&quot;. Then polynomials in $xD$ are also operators on differentiable functions, if one interprets powers of $xD$ by composition. For example if $f(x)=2x^{2}+3$ then $f(xD)$ applied to a differentiable function p(x) gives&#xA;$2xD(xD(p(x)))+3p(x)$. (Think of 3 as the operator &quot;multiplication by 3&quot;. )&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now you should verify the following very useful fact: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The operator $p(xD)$ applied term by term to the power series $\sum_{n}a_{n}x^{n}$ gives the power series $\sum_{n}p(n)a_{n}x^{n}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, your series has the form $\sum f(n)x^{n}$ evaluated at $x=1/3$, where $f(n)=n+1$. So to compute the sum, apply $1+xD$ to the function $1/(1-x)=1+x+x^{2}+\ldots$ and then evaluate at $x=1/3$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think I learned this from Polya and Szego's &quot;Problems and Theorems in Analysis&quot;, which is absolutely stuffed with mathematical wisdom.&lt;/p&gt;&#xA;" OwnerUserId="746" LastEditorUserId="746" LastEditorDisplayName="" LastEditDate="2010-12-16T19:25:55.877" LastActivityDate="2010-12-16T19:25:55.877" CommentCount="1" />
  <row Id="14562" PostTypeId="2" ParentId="14554" CreationDate="2010-12-16T19:25:06.383" Score="6" ViewCount="" Body="&lt;p&gt;I believe the other answers about differentiating and such are much cleaner, but I thought I would add that there is an even more elementary way of doing the sum:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ S = \sum_{k=1}^{\infty} k b^{-k} $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;by using the standard trick of multiplying by $b^{-1}$ then subtracting:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ b^{-1} S = \sum_{k=2}^{\infty} (k-1) b^{-k} $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ S - b^{-1} S = S (1 - b^{-1}) = \sum_{k=1}^{\infty} b^{-k} $$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="4143" LastActivityDate="2010-12-16T19:25:06.383" />
  <row Id="14563" PostTypeId="2" ParentId="14560" CreationDate="2010-12-16T19:26:26.810" Score="15" ViewCount="" Body="&lt;p&gt;When it comes to infinity, most common sense goes out the window. Cantor had a hard time convincing some of his peers...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For me what was surprising is that absolutely convergent series can be reordered in any fashion and still give the same result!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When begining to learn, one naturally considers the series $\displaystyle 1 - 1 + 1 - 1 \dots$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This can be written as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle  1 - (1-1) - (1-1) \dots = 1$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle (1-1) + (1-1) + \dots = 0$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now it should not be a (big) surprise if we could reorder some series differently to give multiple possible values. From there it should not be a big surprise if there were some series which could give &lt;em&gt;any&lt;/em&gt; possible value. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The proof of &lt;a href=&quot;http://en.wikipedia.org/wiki/Riemann_series_theorem&quot; rel=&quot;nofollow&quot;&gt;Riemann Rearrangement Theorem&lt;/a&gt; might help reduce the surprise :-)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope that helps.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-12-16T19:26:26.810" />
  <row Id="14564" PostTypeId="2" ParentId="14560" CreationDate="2010-12-16T19:31:57.077" Score="10" ViewCount="" Body="&lt;p&gt;The commutative law holds for finitely many terms.  There's no reason to expect it to hold for infinitely many terms unless the series converges absolutely.  That's what absolute convergence is for.  In mathematics, you can't assume that the infinite analogues of finite statements are always true; you have to prove it, because often they aren't.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-12-16T19:31:57.077" CommentCount="1" />
  <row Id="14565" PostTypeId="2" ParentId="14560" CreationDate="2010-12-16T19:38:49.387" Score="19" ViewCount="" Body="&lt;p&gt;What Riemann's Rearrangement Theorem on convergent series tells you is &lt;em&gt;precisely&lt;/em&gt; that these &quot;infinite sums&quot; don't behave in the same way as the &quot;usual&quot; sums do. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The key is to understand that just because we can add two numbers together, and from that (through the use of the associative law) we can add any &lt;em&gt;finite&lt;/em&gt; quantity of numbers together, we do &lt;em&gt;not&lt;/em&gt; have a way to &quot;add an infinite quantity of numbers&quot; together. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So series are &lt;em&gt;not&lt;/em&gt; really sums in the usual sense, they are &lt;em&gt;limits&lt;/em&gt;, and they are limits of &lt;em&gt;sequences&lt;/em&gt;. So commutativity and associativity of sums are not really at play, because series are not really sums in the usual sense. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remember, $\sum_{i=1}^{\infty}a_n = L$ does not mean that you add infinitely many numbers and you get $L$, what it means is that the sequence&#xA;$s_1,s_2,s_3,\ldots,s_n,\ldots$ converges to $L$, where $s_k = a_1+a_2+\cdots+a_k$ is the $k$th partial sum. So, for every $\epsilon\gt 0$ there exists $N\gt 0$ such that for all $n\geq N$, $|s_n-L|\lt \epsilon$. We are not adding up the infinitely  many terms, we are saying that we can get arbitrarily close to $L$ if we add enough (but still only finitely many) of the terms. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, we can certainly play the usual games of commutativity and associativity so long as we only modify finitely many of the terms of the sequence $s_1,s_2,s_3,\ldots$, without modifying the limit: remember that the limit of the sequence only depends on what happens &quot;eventually&quot;, not any changes you make to the beginning of the sequence. But that means that you need to keep the sequence intact from some point on. So you can only move around your original terms $a_n$ up to some point (up to some $n_0$) without having to worry about potentially changing the limit. But if you allow yourself to move the numbers around so much that you change the values of arbitrarily large partial sums, then you no longer have any reason to expect the new sequence to have the same limit as the old one (it &lt;em&gt;could&lt;/em&gt;, but it doesn't &lt;em&gt;have to&lt;/em&gt;, because you are now looking at an entirely different sequence). Which means that the usual ideas of &quot;associativity&quot; and &quot;commutativity&quot; of &lt;strong&gt;finite&lt;/strong&gt; sums (you can add them up in any order and still get the same answer) no longer necessarily work for &quot;&lt;strong&gt;infinite&lt;/strong&gt; sums&quot; (because they are &lt;em&gt;not really&lt;/em&gt; sums, they are limits). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is how Riemann's Rearrangement Theorem works: it modifies &lt;em&gt;all&lt;/em&gt; the terms of the sequence so that you get a new limit. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Moron has given you the standard example of why you cannot play associative and commutative games with infinite sums the way you do with finite sums, so I don't have to)&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-16T20:35:20.050" LastActivityDate="2010-12-16T20:35:20.050" />
  <row Id="14566" PostTypeId="1" CreationDate="2010-12-16T19:46:05.943" Score="1" ViewCount="71" Body="&lt;p&gt;I am running an image classification with 5 classes. There is a large set of training data from which I randomly select a training subset, yielding slightly different classifications. I run the classification three times, and obtain the following user's accuracies for each class (the user's accuracy is the probability that a certain pixel of a given class is correctly classified):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1 | .794 | .802 | .796 &lt;br&gt;&#xA;2 | .511 | .506 | .503 &lt;br&gt;&#xA;3 | .938 | .941 | .940 &lt;br&gt;&#xA;4 | .909 | .907 | .912 &lt;br&gt; &#xA;5 | .768 | .765 | .765 &lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For each pixel, there are three outcomes:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) All runs have the same class (either none are wrong or all are wrongly classified)&lt;br&gt;&#xA;2) 2 runs have the same class (either 1, 2 or all were wrongly classified)&lt;br&gt;&#xA;3) All runs are different (either 2 or all were wrongly classified)&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to calculate the probability for each combination of results, of each of these outcomes:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1, 1, 1:  P none wrong, P all wrong &lt;br&gt;&#xA;1, 1, 2:  P 1 one wrong, P 2 wrong, P all wrong&lt;br&gt;&#xA;1, 2, 3:  P 2 wrong, P all wrong&lt;br&gt;&#xA;etc...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure how to approach the problem? Could any one lead me in the right direction? Is it simpler to assume that the accuracies are the same for each run?&lt;/p&gt;&#xA;" OwnerUserId="4833" LastEditorUserId="4833" LastEditorDisplayName="" LastEditDate="2010-12-17T18:41:10.827" LastActivityDate="2010-12-17T18:41:10.827" Title="Probability of multiple classification results being incorrect" Tags="&lt;probability&gt;" CommentCount="2" />
  <row Id="14567" PostTypeId="1" AcceptedAnswerId="14571" CreationDate="2010-12-16T19:51:19.853" Score="4" ViewCount="91" Body="&lt;p&gt;I'm working my way through &lt;em&gt;Linear Algebra Done Right&lt;/em&gt;. To help with one proof, I want to prove the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given $\mathbf{V}$, a vector space and $T$, a linear operator on it, then:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $\mathbf{W}_1$ and $\mathbf{W}_2$ are subspaces of $\mathbf{V}$ such that:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;$\mathbf{V}$ is a direct sum of $\mathbf{W}_1$ and $\mathbf{W}_2$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$\mathbf{W}_1$ and $\mathbf{W}_2$ are invariant under $T$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The restriction of $T$ to $\mathbf{W}_1$ has at most $k$ eigenvalues.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The restriction of $T$ to $\mathbf{W}_2$ has at most $p$ eigenvalues.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Then $T$ has at most $k+p$ eigenvalues.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've done a sketch of a proof using determinants, but it was based on old knowledge about the properties of determinants with regards to eigenvalues, so it may not be correct. The book doesn't emphasize using them though, and maybe there's a proof of this without using determinants.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've tried a proof by contradiction, trying to find something weird by assuming that T can have more than $k+p$ eigenvalues, but I haven't been able to find anything.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any help would be appreciated.&lt;/p&gt;&#xA;" OwnerUserId="4832" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-16T20:03:48.433" LastActivityDate="2010-12-16T20:06:43.757" Title="Help with proof about maximum number of eigenvalues." Tags="&lt;linear-algebra&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="14568" PostTypeId="2" ParentId="14550" CreationDate="2010-12-16T19:59:44.290" Score="0" ViewCount="" Body="&lt;p&gt;So, what you are actually doing is this: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $X$ be a set, and let $G$ be a subgroup of $S_X$, the group of permutations on $X$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We define two equivalence relations, one, let's call it $\sim$, on $\mathcal{P}(X)$, the power set of $X$, and one, let's call it $\equiv$, on the set of functions $2^X = \{f\colon X\to \{0,1\}\}$, which we think of as the $2$-colorings of $X$. We define:&#xA;\begin{align*}&#xA;A\sim B &amp;amp;\Longleftrightarrow \text{there exists $\sigma\in G$ such that $\sigma(A)=B$}\\&#xA;f\equiv g &amp;amp;\Longleftrightarrow \text{there exists $\sigma\in G$ such that $f\circ\sigma = g$.}&#xA;\end{align*}&#xA;The question is whether there is a bijection between $\mathcal{P}(X)/\sim$ and $2^X/\equiv$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The key is that there is a natural identification of $2^X$ with $\mathcal{P}(X)$, by mapping $f\colon X\to\{0,1\}$ to the subset $f^{-1}(1) = \{x\in X\mid f(x)=1\}$, and mapping $A\subseteq X$ to $\chi_A$, the characteristic function of $A$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Define $\mathcal{F}\colon \mathcal{P}(X)/\sim \to 2^X/\equiv$ as follows: given a class $[A] = \{ B\in\mathcal{P}(X)\mid A\sim B\}$, we let&#xA;$$\mathcal{F}([A]) = \{ \chi_B\mid B\in [A]\}.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, I claim this is well defined: that is, I claim that $\mathcal{F}([A])$ is an equivalence class modulo $\equiv$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose $A\sim B$. Then there exists $\sigma\in G$ such that $\sigma(A)=B$. I claim that $\chi_B\circ\sigma = \chi_A$. Indeed, let $x\in X$. If $x\in A$, then $\sigma(x)\in B$, so $\chi_B\circ\sigma(x) = 1 = \chi_A(x)$. And if $x\notin A$, then $\sigma(x)\notin B$ (since $A=\sigma^{-1}(B)$), so $\chi_B\sigma(x) = 0 = \chi_A(x)$. Thus, $\chi_B\equiv \chi_A$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conversely, suppose that $f\in 2^X$ is equivalent to $\chi_A$. Then there exists $\tau\in G$ such that $f\circ\tau = \chi_A$. Let $B=\tau(A)$. I claim that $\chi_B=f$. Indeed, let $x\in X$. Then $f(\tau(x))=1$ if and only if $x\in A$, if and only if $\tau(x)\in B$. So $f(y)=1$ if and only if $y\in B$. Note that $B\sim A$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, every element of $[A]$ gets mapped to an element of&#xA;$$\langle \chi_A \rangle = \{ f\in 2^X\mid f\equiv \chi_A\}$$&#xD;&#xA;and every element of $\langle\chi_A\rangle$ comes from an element of $[A]$. Thus, $\mathcal{F}$ is well defined.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, $\mathcal{F}$ is one-to-one: if $\mathcal{F}([A]) = \mathcal{F}([B])$, then &#xA;$$\chi_B\in \langle \chi_B\rangle = \mathcal{F}([B]) = \mathcal{F}([A])=\langle \chi_A\rangle$$&#xD;&#xA;so there exists $\sigma\in G$ such that $\chi_B\sigma = \chi_A$. that means that $x\in A$ if and only if $\sigma(x)\in B$, so $B=\sigma(A)$, hence $B\sim A$, so $[A]=[B]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And $\mathcal{F}$ is onto: let $\langle f\rangle$ be an equivalence class of $2^X/\equiv$. Let $A = f^{-1}(1)$. I claim that $\mathcal{F}([A])=\langle \chi_A\rangle = \langle f\rangle$. Indeed, $\chi_A = f$ by construction, so $f\in\langle\chi_A\rangle$, hence $\langle f\rangle = \langle \chi_A\rangle = \mathcal{F}([A])$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, $\mathcal{F}$ is a bijection between $\mathcal{P}(X)/\sim$ and $2^X/\equiv$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;hr/&gt;&#xA;However, I want to add that I think it is fair to say that your nomenclature is misleading/nonstandard. I would take that a $2$-coloring of $X$ is &lt;em&gt;invariant under $G$&lt;/em&gt; to mean that if $f\colon X\to\{0,1\}$ is the $2$-coloring, then $f\circ \sigma = f$ for all $\sigma\in G$. &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-16T21:05:22.547" LastActivityDate="2010-12-16T21:05:22.547" CommentCount="2" />
  <row Id="14569" PostTypeId="1" AcceptedAnswerId="14590" CreationDate="2010-12-16T20:00:26.193" Score="8" ViewCount="333" Body="&lt;p&gt;The Collatz Conjecture is a famous conjecture in mathematics that has lasted for over 70 years. It goes as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Define $f(n)$ to be as a function on the natural numbers by:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f(n) = n/2$ if $n$ is even and&#xA;$f(n) = 3n+1$ if $n$ is odd&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The conjecture is that for all $n \in \mathbb{N}$, $n$ eventually converges under iteration by $f$ to $1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was wondering if the &quot;5n+1&quot; problem has been solved. This problem is the same as the Collatz problem except that in the above one replaces $3n+1$ with $5n+1$.&lt;/p&gt;&#xA;" OwnerUserId="1089" LastActivityDate="2010-12-16T21:44:46.947" Title="The $5n+1$ Problem" Tags="&lt;analysis&gt;&lt;combinatorics&gt;&lt;algebra&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="14570" PostTypeId="2" ParentId="14567" CreationDate="2010-12-16T20:01:30.143" Score="3" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Hint&lt;/strong&gt;: $ker(T) = ker(T|W_1)\ \oplus\ ker(T|W_2)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here $ker$ stands for kernel, $T|W_i$ stands for $T$ restricted to $W_i$, etc.. Actually you have to do this with the transformation $T -\lambda I$ instead of $T$.&lt;/p&gt;&#xA;" OwnerUserId="1119" LastEditorUserId="1119" LastEditorDisplayName="" LastEditDate="2010-12-16T20:06:43.757" LastActivityDate="2010-12-16T20:06:43.757" CommentCount="4" />
  <row Id="14571" PostTypeId="2" ParentId="14567" CreationDate="2010-12-16T20:05:37.293" Score="3" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Hint:&lt;/strong&gt; Suppose $\lambda$ is an eigenvalue. Then there exists an eigenvector $\mathbf{v}$ corresponding to $\lambda$. Write $\mathbf{v}=\mathbf{w}_1+\mathbf{w}_2$; since $\mathbf{v}$ is nonzero, at least one of $\mathbf{w}_1$ and $\mathbf{w}_2$ is nonzero. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, evaluate $T(\mathbf{v})$; since $\mathbf{V}$ is a &lt;em&gt;direct sum&lt;/em&gt;, and each $\mathbf{W}_i$ is invariant, what can you say about $T(\mathbf{w}_1)$ and $T(\mathbf{w}_2)$?&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-12-16T20:05:37.293" CommentCount="4" />
  <row Id="14573" PostTypeId="1" AcceptedAnswerId="14578" CreationDate="2010-12-16T20:16:25.723" Score="3" ViewCount="66" Body="&lt;p&gt;Let's say that I have a tax of 17% over my services and I want to receive $20,000.00 regardless (after paying) the tax.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I want something like: &#xA;$$X - 17\% = \$20,000.00$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $X$ is the final value that I need to charge to receive $\$20,000$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do I calculate this?&lt;/p&gt;&#xA;" OwnerUserId="1425" LastEditorUserId="1425" LastEditorDisplayName="" LastEditDate="2010-12-16T22:53:50.477" LastActivityDate="2010-12-16T22:53:50.477" Title="Calculating percentages for taxes." Tags="&lt;algebra-precalculus&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="14574" PostTypeId="2" ParentId="14569" CreationDate="2010-12-16T20:21:10.077" Score="5" ViewCount="" Body="&lt;p&gt;you have the wrong problem, it is solved by 13. The generalization next is:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;if n = 3k then k&#xA;if n = 2k then k&#xA;else 5n+1&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="4290" LastActivityDate="2010-12-16T20:21:10.077" />
  <row Id="14575" PostTypeId="2" ParentId="14573" CreationDate="2010-12-16T20:25:38.580" Score="4" ViewCount="" Body="&lt;p&gt;Removing 17% leaves (100-17)%=83%, so you want 83%*x=20.000.  Divide both sides by 83% to get your answer.&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-12-16T20:25:38.580" CommentCount="1" />
  <row Id="14578" PostTypeId="2" ParentId="14573" CreationDate="2010-12-16T20:31:28.087" Score="6" ViewCount="" Body="&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Rule_of_three_%28mathematics%29#Rule_of_Three&quot; rel=&quot;nofollow&quot;&gt;Rule of Three&lt;/a&gt; is your friend.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You want the 83% that is left to be $\$20,000$. If $X$ is 100% of what you need to charge, then you have:&#xA;$$\begin{array}{rcl}&#xD;&#xA;X &amp;amp;\longrightarrow &amp;amp;100\%\\\&#xD;&#xA;20,000 &amp;amp;\longrightarrow&amp;amp;83\%&#xD;&#xA;\end{array}$$&#xD;&#xA;By the Rule of Three, to get the value of $X$ you multiply across the known diagonal and divide by the remaining quantity; so $X = \frac{(20,000)(100)}{83}\approx \$24,096.39$.  &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-16T20:57:47.867" LastActivityDate="2010-12-16T20:57:47.867" CommentCount="1" />
  <row Id="14579" PostTypeId="1" CreationDate="2010-12-16T20:46:49.297" Score="1" ViewCount="176" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Possible Duplicate:&lt;/strong&gt;&lt;br&gt;&#xA;  &lt;a href=&quot;http://math.stackexchange.com/questions/14124/finding-inverse-of-x-bmod-y&quot;&gt;finding inverse of $x\bmod y$&lt;/a&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;p&gt;Hello all  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Me and some friends are studying for a discrete exam and we are having some troubles finding the inverse modulus of things. The question we are suck on is &quot;what is the inverse of 8 mod 29&quot;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks a lot!&lt;/p&gt;&#xA;" OwnerUserId="4835" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-16T20:52:22.123" LastActivityDate="2010-12-16T23:37:46.680" ClosedDate="2010-12-17T02:49:35.270" Title="Troubles finding inverse modulus" Tags="&lt;discrete-mathematics&gt;&lt;elementary-number-theory&gt;" AnswerCount="2" CommentCount="0" />
  <row Id="14580" PostTypeId="2" ParentId="14579" CreationDate="2010-12-16T20:51:52.103" Score="4" ViewCount="" Body="&lt;p&gt;The inverse of $8$ modulo $29$ is an integer $k$ such that $8k \equiv 1\pmod{29}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are many ways of finding such an integer. The standard way is to use the &lt;a href=&quot;http://en.wikipedia.org/wiki/Extended_euclidean_algorithm&quot; rel=&quot;nofollow&quot;&gt;Extended Euclidean Algorithm&lt;/a&gt;, but sometimes you can do it without using this algorithm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, since $8\times 3 = 24 = -5\bmod 29$, and $-5\times -6 =30=1\bmod 29$, then $8\times(3\times-6) = 1\bmod 29$. Now just adjust by adding multiples of $29$ to get an answer between $0$ and $28$.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-12-16T20:51:52.103" CommentCount="1" />
  <row Id="14581" PostTypeId="1" AcceptedAnswerId="14609" CreationDate="2010-12-16T20:52:49.327" Score="3" ViewCount="226" Body="&lt;p&gt;Draw two balls centered at the origin in R^3, where $r_1 &amp;lt; r_2$. What is the topological name for the boundary created? Does it even make sense to talk about a figure with an inside boundary and an outside one?&lt;/p&gt;&#xA;" OwnerUserId="196" LastEditorUserId="196" LastEditorDisplayName="" LastEditDate="2010-12-16T21:48:23.277" LastActivityDate="2011-04-01T02:51:10.700" Title="What is topological name of a sphere with a hole in it?" Tags="&lt;topology&gt;" AnswerCount="2" CommentCount="5" />
  <row Id="14582" PostTypeId="1" AcceptedAnswerId="14596" CreationDate="2010-12-16T21:11:09.873" Score="4" ViewCount="272" Body="&lt;p&gt;I have a problem with solution of this limit:&#xA;$$\lim_{x\to 0}{\frac{\tan{x}-x}{x^2}}$$&#xD;&#xA;Of course, it's a very easy to solve, using (twice) L'Hôpital's rule, but I need to find out, how to do this without this rule.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I stuck in this point:&#xA;$$\lim_{x\to 0}{\frac{\frac{\sin^2{x}}{x^2}\cdot\frac{1}{\cos^2{x}} - 1}{2x}}$$&#xD;&#xA;Everything I need to know is how to eliminate $\frac{\sin^2{x}}{x^2}$, because - as my tutor said - I can't simply substitute $1$ for this expression.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for help.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS: It's not a homework. My tutor showed this problem as a puzzle and said, that it would be a good exercise to solve this without L'Hôpital's rule.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Here is a way I got to the point $\lim_{x\to 0}{\frac{\frac{\sin^2{x}}{x^2}\cdot\frac{1}{\cos^2{x}} - 1}{2x}}$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align*}&#xA;\lim_{x-&gt;0}{\frac{\tan{x}-x}{x^2}} &amp;amp;= \lim_{x-&gt;0}{\frac{(\frac{\sin{x}}{\sqrt{1-\sin^2{x}}}-x)\cdot (\frac{\sin{x}}{\sqrt{1-\sin^2{x}}}+x)}{x^2(\frac{\sin{x}}{\sqrt{1-\sin^2{x}}}+x)}}\\&#xA; &amp;amp;= \lim_{x-&gt;0}{\frac{\frac{\sin^2{x}}{x^2\cos^2{x}}-1}{x(\frac{\sin{x}}{x}\frac{1}{\cos{x}}+1)}}= \lim_{x\to 0}{\frac{\frac{\sin^2{x}}{x^2}\cdot\frac{1}{\cos^2{x}} - 1}{2x}}&#xA;\end{align*}&lt;/p&gt;&#xA;" OwnerUserId="3856" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-17T04:34:27.280" LastActivityDate="2010-12-17T04:39:41.483" Title="Limit, solution in unusual way" Tags="&lt;calculus&gt;&lt;trigonometry&gt;&lt;limit&gt;" AnswerCount="5" CommentCount="5" />
  <row Id="14583" PostTypeId="2" ParentId="14569" CreationDate="2010-12-16T21:14:15.133" Score="7" ViewCount="" Body="&lt;p&gt;In Part I of Lagarias' &lt;a href=&quot;http://arxiv.org/abs/math/0309224&quot; rel=&quot;nofollow&quot;&gt;extensive, annotated bibliography&lt;/a&gt; of the 3x+1 problem, he notes a 1999 paper by Metzger (reference 112) regarding the 5x+1 problem:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;For the 5x + 1 problem he shows that on the positive integers there is no cycle of size 1, a unique cycle of size 2, having smallest element n = 1, and exactly two cycles of size 3, having smallest elements n = 13 and n = 17, respectively.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It is unclear from the notes whether the paper shows that these are the &lt;em&gt;only&lt;/em&gt; cycles of the 5x+1 problem or whether there may exist longer cycles.&lt;/p&gt;&#xA;" OwnerUserId="2643" LastActivityDate="2010-12-16T21:14:15.133" CommentCount="2" />
  <row Id="14585" PostTypeId="1" AcceptedAnswerId="14599" CreationDate="2010-12-16T21:15:53.010" Score="7" ViewCount="125" Body="&lt;p&gt;Ok, so I just finished a Discrete Math class where we learned about all types of proofs and counting and graphs, etc. So I know the basics, but never thought I'd need to use it in the real world (not anytime soon at least). However, a friend of mine showed me a puzzle game, and he said that it's &lt;em&gt;always&lt;/em&gt; possible to win this game, even though it seems unlikely to me. I will explain the rules, which are pretty simple, and I want to know how I could prove this, or if the answer is evident using combinatorics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's how the game works:&#xA;The game is a slight variation of a game called 'Poker Squares'. Anyway, you lay down (all at once) 25 &lt;strong&gt;random&lt;/strong&gt; cards arranged in a 5x5 grid (no jokers). The goal of the game is to arrange all the cards so that each row is a poker hand of the following:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;A straight: 5 cards arranged in order of their number (i.e. 9,10,J,Q,K, irregardless of their suite.)&lt;/li&gt;&#xA;&lt;li&gt;A flush: any 5 cards in any order of the same suite (i.e. five hearts)&lt;/li&gt;&#xA;&lt;li&gt;A full house: A triple and a double (i.e. three 10's and two J's.)&lt;/li&gt;&#xA;&lt;li&gt;A straight flush: 5 cards of the same suite arranged in increasing or decreasing order (i.e. 4,5,6,7,8 all diamonds)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;And thats it. The claim is that given any random 25 cards, these cards can ALWAYS be arranged so that each row is a winning poker hand (of the above mentioned hands only). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any ideas?&lt;/p&gt;&#xA;" OwnerUserId="2985" LastEditorUserId="2985" LastEditorDisplayName="" LastEditDate="2010-12-16T21:47:32.450" LastActivityDate="2010-12-16T22:24:11.657" Title="Proving a certain type of poker game can always be won" Tags="&lt;combinatorics&gt;&lt;card-games&gt;" AnswerCount="1" CommentCount="6" FavoriteCount="2" />
  <row Id="14586" PostTypeId="2" ParentId="14549" CreationDate="2010-12-16T21:23:08.530" Score="3" ViewCount="" Body="&lt;p&gt;The Chernoff bound states that a certain function $f(p, n)$ of $p$ and $n$ is greater than a certain other function $g(p, n)$ of $p$ and $n$, subject to the condition that $p &amp;gt; \frac{1}{2}$.  I see no reason why you can't pick $p$ and $n$ to be whatever you want; in particular, I see no reason why you can't let $p$ depend on $n$.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-12-16T21:23:08.530" />
  <row Id="14587" PostTypeId="2" ParentId="14550" CreationDate="2010-12-16T21:31:07.990" Score="0" ViewCount="" Body="&lt;p&gt;I think this will work. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the bijection where coloring $\{...,x_i=1,...\}$ means that object $i$ is included in our set, then the rest follows. &lt;/p&gt;&#xA;" OwnerUserId="4728" LastActivityDate="2010-12-16T21:31:07.990" CommentCount="3" />
  <row Id="14588" PostTypeId="1" AcceptedAnswerId="14732" CreationDate="2010-12-16T21:33:15.500" Score="3" ViewCount="125" Body="&lt;p&gt;If we have a two lattices (partially ordered) - one for subgroups, one for factor groups, and we know order of the group we want to have these subgroup and factor group lattices, is such a group unique up to isomorphism (if exists)? Or is there a counterexample?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If that's true, are sufficient conditions on the order and subgroup lattices to guarantee uniqueness? Another way - what if we now lattice for subgroup and group of automorphism of group; is that group uniquely determined by that information?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for help. (sorry for english)&lt;/p&gt;&#xA;" OwnerUserId="4195" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-17T02:58:22.733" LastActivityDate="2010-12-18T06:23:09.310" Title="Does the order, lattice of subgroups, and lattice of factor groups, uniquely determine a group up to isomorphism?" Tags="&lt;group-theory&gt;&lt;algebra&gt;" AnswerCount="1" CommentCount="6" FavoriteCount="1" />
  <row Id="14590" PostTypeId="2" ParentId="14569" CreationDate="2010-12-16T21:44:46.947" Score="15" ViewCount="" Body="&lt;p&gt;You shouldn't expect this to be true. Here is a nonrigorous argument. Let $n_k$ be the sequence of odd numbers you obtain. So (heuristically), with probability $1/2$, we have $n_{k+1} = (5n_k+1)/2$, with probability $1/4$, we have $n_{k+1} = (5 n_k+1)/4$, with probability $1/8$, we have $n_{k+1} = (5 n_k+1)/8$ and so forth. Setting $x_k = \log n_k$, we approximately have $x_{k+1} \approx x_k + \log 5 - \log 2$ with probability $1/2$, $x_{k+1} \approx x_k + \log 5 - 2 \log 2$ with probability $1/4$, $x_{k+1} \approx x_k + \log 5 - 3 \log 2$ with probability $1/8$ and so forth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the expected change from $x_{k}$ to $x_{k+1}$ is &#xA;$$\sum_{j=1}^{\infty} \frac{ \log 5 - j \log 2}{2^j} = \log 5 - 2 \log 2.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is positive! So, heurisitically, I expect this sequence to run off to $\infty$. This is different from the $3n+1$ problem, where $\log 3 - 2 \log 2 &amp;lt;0$, and so you heurisitically expect the sequence to decrease over time. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a numerical example. I started with $n=25$ and generated $25$ odd numbers. Here is a plot of $(k, \log n_k)$, versus the linear growth predicted by my heuristic. Notice that we are up to 4 digit numbers and show no signs of dropping down.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/8nRFy.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="448" LastActivityDate="2010-12-16T21:44:46.947" />
  <row Id="14591" PostTypeId="1" AcceptedAnswerId="14623" CreationDate="2010-12-16T21:50:55.660" Score="6" ViewCount="197" Body="&lt;p&gt;Is it possible to construct a non-measurable set in $[0,1]$ of a given outer measure $x \in [0,1]$? This will probably require the axiom of choice. Does anyone have a suggestion?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; I forgot to mention that the set has to be in $[0,1]$ which seemed &quot;clear&quot; in my head, but which obviously isn't.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit 2:&lt;/strong&gt; Now it is clear that I just need a non-measurable set with outer measure 1, from that I can make any other one. I have once seen somewhere that they do this with transfinite recursion (if I recall correctly), but I don't remember where and how because I didn't know that transfinite recursion was back then.&lt;/p&gt;&#xA;" OwnerUserId="1120" LastEditorUserId="1120" LastEditorDisplayName="" LastEditDate="2010-12-16T23:32:07.563" LastActivityDate="2010-12-17T03:37:34.670" Title="Vitali-type set with given outer measure." Tags="&lt;measure-theory&gt;" AnswerCount="2" />
  <row Id="14592" PostTypeId="2" ParentId="14582" CreationDate="2010-12-16T21:51:52.417" Score="5" ViewCount="" Body="&lt;p&gt;The Maclaurin series for $\tan x$ begins with $x$, and there's no $x^2$ term since it's an odd function. Thus $\tan x = x + O(x^3)$, and therefore $(\tan x - x)/x^2 = O(x) \to 0$ as $x\to 0$.&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2010-12-16T21:51:52.417" CommentCount="13" />
  <row Id="14593" PostTypeId="1" CreationDate="2010-12-16T21:55:16.813" Score="1" ViewCount="87" Body="&lt;p&gt;Anyone know how to convert from lat/lng to EPSG:27700?&lt;/p&gt;&#xA;" OwnerUserId="4840" LastActivityDate="2011-02-15T12:22:54.077" Title="Formula to convert from lat/lng to EPSG:27700?" Tags="&lt;geometry&gt;" AnswerCount="2" />
  <row Id="14594" PostTypeId="2" ParentId="14582" CreationDate="2010-12-16T21:59:09.557" Score="1" ViewCount="" Body="&lt;p&gt;I am not sure what you can use; if you know of Maclaurin series, the argument is very easy. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe that what your tutor meant is that you cannot rewrite $\displaystyle \frac{\tan x-x}{x^2}=\frac{\frac{\sin x}x-\cos x}{x\cos x}$ as $\displaystyle \frac{1-\cos x}{x\cos x}$. The argument below uses that $\sin x/x\to 1$, but not by substituting it this way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that $\displaystyle \frac{1-\cos x}{x^2}=\frac{1-\cos^2x}{x^2(1+\cos x)}=\left(\frac{\sin x}x\right)^2\frac1{1+\cos x}\to\frac12$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is enough to find the limit of $\displaystyle \frac{\frac{\sin x}x-\cos x}x$. Using the limit above, all you need is to find the limit of $\displaystyle\frac{\sin x -x}{x^2}$, because $$\frac{\frac{\sin x}x-\cos x}x=\frac{\frac{\sin x}x+\left(\frac{1-\cos x}{x^2}\right)x^2-1}x.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The limit of $\displaystyle\frac{\sin x -x}{x^2}$ is 0. This is fairly easy to evaluate using the Maclaurin expansion of $\sin x$. All you really need is that $\sin x=x+O(x^3)$, but I am not sure you are familiar with this notation. Also, you could define $f(x)=\sin x/x$ if $x\ne 0$ and $f(0)=1$, and verify that this function is differentiable at 0. But I am not certain you have the tools to do that. &lt;/p&gt;&#xA;" OwnerUserId="462" LastActivityDate="2010-12-16T21:59:09.557" />
  <row Id="14595" PostTypeId="2" ParentId="14417" CreationDate="2010-12-16T22:11:06.073" Score="1" ViewCount="" Body="&lt;p&gt;The mass in the sum is approximately proportional to a (shifted) Gaussian centered at $k = n/2 + C\sqrt{n}$ and with standard deviation $A \sqrt{n}$ for explicitly calculable constants $C$ and $A$.  This implies asymptotics of the form $(M + o(1))2^{n + \sqrt{n/2}}$ where $M$ is another computable constant. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same is true for the $a,b$ version of the problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[edit: my calculations give $M = 2^{(\ln 2)/8} = 1.0618966...$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[edit-2: similar calculations give $M=(a/b)^{\ln (a/b)}/8$ and asymptotics $(M+o(1))2^n (ab)^\sqrt{n/2}$ for the $a,b$ sum.]  &lt;/p&gt;&#xA;" OwnerUserId="467" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-12-17T06:45:13.783" LastActivityDate="2010-12-17T06:45:13.783" CommentCount="6" />
  <row Id="14596" PostTypeId="2" ParentId="14582" CreationDate="2010-12-16T22:16:48.873" Score="13" ViewCount="" Body="&lt;p&gt;Here is an attempt at a geometric proof.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/8AjCc.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Figure thanks to J.M)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider $\triangle ABC$ such that $\angle{BCA} = x$. Let $BC=1$ and so $AB = \tan x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $BE$ be the arc of radius 1 and angle $x$ drawn with $C$ as the center (note that $E$ is on AC, between $A$ and $D$ and is kind of hidden in the brown region). Note that $CE = 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the area of the gray region is $\dfrac{\tan x}{2} - \dfrac{x}{2}$ (area of $\triangle ABC$ - area of the sector $CBE$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $D$ be the perpendicular on the hypotenuse $AC$ from $B$. It can be seen that $CD = \cos x$ and thus distance from $D$ to $C$ is less than distance from $E$ to $C$ (which is $1$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus the area of the gray region is less than the area of $\triangle BAD$ (gray + brown).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now $AD = \dfrac{\sin^2 x}{\cos x}$ and thus we have that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ 0 &amp;lt; \dfrac{\tan x - x}{2} \lt \dfrac{\sin^3 x}{2\cos x}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And so &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ 0 &amp;lt; \dfrac{\tan x - x}{2x^2} \lt \dfrac{\sin^3 x}{2x^2 \cos x}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since we know that $\lim_{x \to 0+} \dfrac{\sin x}{x} = 1$, and that $\dfrac{\tan x - x}{x^2}$ is an odd function, that the limit is $0$, follows.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Previous answer, which was a feeble attempt at being pedantic:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a way to find the limit without using more advanced concepts like McLaurin series etc...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\dfrac{\tan(2x) - 2x}{(2x)^2} = \dfrac{ \frac{2\tan x}{1-\tan^2 x} - 2x}{4x^2}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ = \dfrac{(2\tan x - 2x) + 2x \tan^2 x}{4x^2(1 - \tan^2 x)} = (\dfrac{\tan x - x}{2x^2} +  \dfrac{x\tan^2 x}{2x^2}) \dfrac{1}{1-\tan^2 x}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, taking limits as $\displaystyle x \to 0$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ L = (\dfrac{L}{2} + 0)\dfrac{1}{1-0}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$L = 0$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is one problem with the above, though. Can you tell what that is?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Or rather more simply, replace $\displaystyle x$ with $\displaystyle -x$)&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-12-17T04:39:41.483" LastActivityDate="2010-12-17T04:39:41.483" CommentCount="8" />
  <row Id="14597" PostTypeId="2" ParentId="14593" CreationDate="2010-12-16T22:19:26.777" Score="0" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://osgeo-org.1803224.n2.nabble.com/Mouse-position-in-lat-long-td5567114.html&quot; rel=&quot;nofollow&quot;&gt;http://osgeo-org.1803224.n2.nabble.com/Mouse-position-in-lat-long-td5567114.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;May be useful&lt;/p&gt;&#xA;" OwnerUserId="4728" LastActivityDate="2010-12-16T22:19:26.777" />
  <row Id="14598" PostTypeId="2" ParentId="14591" CreationDate="2010-12-16T22:23:42.357" Score="4" ViewCount="" Body="&lt;p&gt;This was my answer to the question as originally phrased: Once you have one nonmeasurable set with finite outer measure, you can get all positive values by rescaling.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;With the additional requirement that the set is contained in $[0,1]$, this argument will work as long as you have a nonmeasurable subset of outer measure $1$.  Sierpiński and Lusin showed that $[0,1]$ can be decomposed into uncountably many nonmeasurable sets of outer measure $1$ in an &lt;a href=&quot;http://www.emis.de/cgi-bin/JFM-item?46.0294.01&quot; rel=&quot;nofollow&quot;&gt;article&lt;/a&gt; published in 1917, but I do not know the details.  Robert Israel sketches a construction of a Vitali subset of [0,1]  with outer measure 1  in a &lt;a href=&quot;http://groups.google.com/group/sci.math/browse_frm/thread/a1f91aa3b8ae80d8?pli=1&quot; rel=&quot;nofollow&quot;&gt;1997 sci.math post&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Oh, and strictly speaking the answer is no, you can't have $x=0$; I suppose you meant $x\in(0,1]$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the axiom of choice, I'm no expert but have been told that it is consistent with ZF that all subsets of the real line are Lebesgue measurable.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Peter LeFanu Lumsdaine points out in a comment that the result is due to Robert Solovay.  The &lt;a href=&quot;http://www.jstor.org/stable/1970696&quot; rel=&quot;nofollow&quot;&gt;1970 article&lt;/a&gt; is on JSTOR, and begins as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/b5H9D.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2010-12-16T23:53:23.420" LastActivityDate="2010-12-16T23:53:23.420" CommentCount="14" />
  <row Id="14599" PostTypeId="2" ParentId="14585" CreationDate="2010-12-16T22:24:11.657" Score="11" ViewCount="" Body="&lt;p&gt;Suppose you have one two of clubs, and the rest of the cards are all four and up, and all diamonds, hearts and spades (there are 33 such cards, which is more than enough). You can't use the two of clubs in any winning hand. So the claim is not true. In fact, this counterexample works even if you allow four of a kind.&lt;/p&gt;&#xA;" OwnerUserId="2501" LastActivityDate="2010-12-16T22:24:11.657" CommentCount="2" />
  <row Id="14600" PostTypeId="1" AcceptedAnswerId="14619" CreationDate="2010-12-16T22:27:28.097" Score="7" ViewCount="306" Body="&lt;p&gt;In multivariable calculus, I was taught to compute the chain rule by drawing a &quot;tree diagram&quot; (a directed acyclic graph) representing the dependence of one variable on the others. I now want to understand the theory behind it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Examples:&#xA;Let $y$ and $x$ both be functions of $t$.&#xA;Let $z$ be a function of both $x$ and $y$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The derivative of z with respect to t is:&#xA;$$\frac{dz}{dt} = \frac{\partial z}{\partial x} \frac{dx}{dt} + \frac{\partial z}{\partial y} \frac{dy}{dt}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To compute this derivative, I was taught to draw a graph with the following edges:&#xA;$x \to z$, $y \to z$, $t \to x$, and $t \to y$.&#xA;Source: &lt;a href=&quot;http://www.math.hmc.edu/calculus/tutorials/multichainrule/&quot; rel=&quot;nofollow&quot;&gt;http://www.math.hmc.edu/calculus/tutorials/multichainrule/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/lwPwl.gif&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These tree diagrams can be constructed for arbitrarily complex functions with many variables. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, to find a derivative of a dependent variable with respect to an independent variable, you need to take the sum of all of the different paths to reach the dependent variable from the independent variable. Traveling down a path, you multiply the functions (e.g. $\frac{\partial z}{\partial x} \cdot \frac{dx}{dt}$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why does this work? What is the connection to the sum-product algorithm for semi-rings? If there's a connection, what is the semi-ring? What group operation does function composition play?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If possible, I'm looking for an algebra (group, semi-ring, etc.) based explanation. Does one exist?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm hoping for something along the lines of a semiring $(R, \cdot, +)$, where $R$ is the set of functions (invariant under some transformation (possibly scaling?)), $\cdot$ is function composition, $+$ does something like combine the arguments of functions: $f(x) + f(y) = f(x, y)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The $0$ element would be the constant function $f_0(x) = c$, and the $1$ element would be the identity function: $f_1(x) = x$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note, $f_0(x)$ is typically unique, so we need to treat $R$ such that all constant functions are in the same equivalence class (hence the invariance mentioned earlier).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, we can see differentiation as an operation that maps between $(R, \cdot, +)$ and $(S, \cdot&amp;#39;, +&amp;#39;)$, where $\cdot&amp;#39;$ and $+&amp;#39;$ are the standard times and plus over functions that we learn in grammar school.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this a pipe-dream, or is there an explanation along these lines?&lt;/p&gt;&#xA;" OwnerUserId="4841" LastEditorUserId="4841" LastEditorDisplayName="" LastEditDate="2010-12-17T04:32:19.187" LastActivityDate="2010-12-17T04:32:19.187" Title="chain rule using tree diagram, why does it work?" Tags="&lt;calculus&gt;&lt;graph-theory&gt;&lt;functions&gt;" AnswerCount="2" CommentCount="13" FavoriteCount="1" />
  <row Id="14601" PostTypeId="2" ParentId="14582" CreationDate="2010-12-16T22:31:14.377" Score="0" ViewCount="" Body="&lt;p&gt;My inclination is to rewrite the original limit as:$$\lim_{x\to 0}{\frac{\tan{x}-x}{x^2}}=\lim_{x\to 0}{\frac{\frac{\tan{x}}{x}-1}{x-0}},$$ with the thinking that if $f(x)=\frac{\tan x}{x}$, $\lim_{x\to 0}f(x)=1$ and $\lim_{x\to 0}f&amp;#39;(x)=0$ and $$f&amp;#39;(a)=\lim_{x\to a}{\frac{f(x)-f(a)}{x-a}}=\lim_{x\to a}{\frac{\frac{\tan x}{x}-\frac{\tan a}{a}}{x-a}}.$$&#xD;&#xA;Unfortunately, to proceed from there, I'm getting stuck in some ugliness with double limits.&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-12-16T22:31:14.377" />
  <row Id="14603" PostTypeId="2" ParentId="14442" CreationDate="2010-12-16T22:48:24.677" Score="7" ViewCount="" Body="&lt;p&gt;The tensor product $M\otimes N$ of abelian groups cannot be non-zero and free unless both $M$ and $N$ are free. This follows from the following facts.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Any subgroup of a free abelian group is free. This holds even in the infinitely generated case. Wikipedia has a proof in its article on &lt;a href=&quot;http://en.wikipedia.org/w/index.php?title=Free_abelian_group&amp;amp;oldid=400178458#Subgroup_closure&quot; rel=&quot;nofollow&quot;&gt;free abelian groups&lt;/a&gt;.&lt;/li&gt;&#xA;&lt;li&gt;Under certain conditions on $M$ and $N$, for non-zero $n\in N$ the homomorphism&#xA;$$&#xD;&#xA;\begin{align}&#xD;&#xA;&amp;amp;M\to M\otimes N,\\&#xD;&#xA;&amp;amp;m\mapsto m\otimes n,&#xD;&#xA;\end{align}&#xD;&#xA;$$&#xD;&#xA;is injective. It is easy to see that this holds when $N$ is free and $n$ is a basis element. It also holds when $M$ and $N$ are both torsion free. You could prove this by &lt;a href=&quot;http://en.wikipedia.org/w/index.php?title=Extension_of_scalars&amp;amp;oldid=303425028&quot; rel=&quot;nofollow&quot;&gt;extension of scalars&lt;/a&gt; to reduce it to the case of vector spaces over the rationals (and vector spaces are always free). This &lt;em&gt;does not&lt;/em&gt; hold if you merely assume that $N$ is torsion free, as pointed out by Jack Schmidt in a comment (and I must apologize for making the mistake of assuming this in my original answer).&lt;/li&gt;&#xA;&lt;li&gt;Tensor products are &lt;a href=&quot;http://en.wikipedia.org/w/index.php?title=Exact_functor&amp;amp;oldid=377038315&quot; rel=&quot;nofollow&quot;&gt;right exact&lt;/a&gt;. That is, if $0\to A\to B\to C\to 0$ is an exact sequence of abelian groups, then&#xA;$$&#xD;&#xA;M\otimes A\to M\otimes B\to M\otimes C\to 0&#xD;&#xA;$$&#xD;&#xA;is exact. In particular, if $M\otimes A$ maps to zero, this shows that $M\otimes B$ and $M\otimes C$ are isomorphic, which I will use to quotient out the torsion subgroups before applying 2.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Now, suppose that $M\otimes N$ is non-zero and free.&#xA;Let $T$ be the torsion subgroup of $N$, so that $N/T$ is torsion free and $0\to T\to N\to N/T\to 0$ is an exact sequence. By 3,&#xA;$$&#xD;&#xA;M\otimes T\to M\otimes N\to M\otimes(N/T)\to 0&#xD;&#xA;$$&#xD;&#xA;is exact. However, as $T$ is torsion, $M\otimes T$ will also be torsion, so its image in the torsion free group $M\otimes N$ is zero. This gives an isomorphism $M\otimes N\cong M\otimes(N/T)$. Applying the same argument with the torsion subgroup $S$ of $M$ shows that $(M/S)\otimes(N/T)$ is isomorphic to the free group $M\otimes N$. Then, as $M/S$, $N/T$ are torsion free, picking any non-zero $n\in N/T$ and using 2 gives an injection&#xA;$$&#xD;&#xA;\begin{align}&#xD;&#xA;&amp;amp;M/S\to (M/S)\otimes(N/T),\\&#xD;&#xA;&amp;amp; m\mapsto m\otimes n.&#xD;&#xA;\end{align}&#xD;&#xA;$$&#xD;&#xA;So $M/S$ is isomorphic to a subgroup of the free group $(M/S)\otimes(N/T)$ which, as stated in 1, means that $M/S$ is free. Similarly, $N/T$ is free. Applying 2 again, the homomorphism&#xA;$$&#xD;&#xA;\begin{align}&#xD;&#xA;&amp;amp;M\to M\otimes(N/T),\\&#xD;&#xA;&amp;amp; m\mapsto m\otimes n&#xD;&#xA;\end{align}&#xD;&#xA;$$&#xD;&#xA;is injective for any basis element $n\in N/T$. So $M$ is isomorphic to a subgroup of the free abelian group $M\otimes(N/T)\cong M\otimes N$ and, using 1 once more, must be free. Similarly, $N$ is free.&lt;/p&gt;&#xA;" OwnerUserId="1321" LastEditorUserId="1321" LastEditorDisplayName="" LastEditDate="2010-12-17T02:02:36.880" LastActivityDate="2010-12-17T02:02:36.880" CommentCount="7" />
  <row Id="14604" PostTypeId="1" AcceptedAnswerId="14616" CreationDate="2010-12-16T23:04:53.837" Score="3" ViewCount="131" Body="&lt;p&gt;Is the projective closure of an infinite affine variety (over an algebraically closed field, I only care about the classical case right now) always strictly larger than the affine variety? I know it is an open dense subset of its projective closure, but I don't think it can actually be its own projective closure unless it is finite. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I guess my intuition has me worried about cases like the plane curve $X^2 + Y^2 - 1$, since the real part is compact, but such a curve must still &quot;escape to infinity&quot; over an algebraically closed field, right?&lt;/p&gt;&#xA;" OwnerUserId="4842" LastActivityDate="2010-12-18T05:22:05.953" Title="Projective closure" Tags="&lt;algebraic-geometry&gt;" AnswerCount="2" />
  <row Id="14605" PostTypeId="2" ParentId="14604" CreationDate="2010-12-16T23:10:43.033" Score="1" ViewCount="" Body="&lt;p&gt;For your own example: the projectivization of your curve is given by $X^2 + Y^2 - Z^2= 0$. There is the point corresponding to the projective equivalence class of $(1,i,0)$, which does not belong to the affine curve.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;For more general varieties given by a single equation, you can always similarly find a projective point that is part of the affine curve. For more general varieties, I should imagine that it is still the case; but for a proper argument a better person in algebraic geometry is needed.&lt;/p&gt;&#xA;" OwnerUserId="1119" LastEditorUserId="1119" LastEditorDisplayName="" LastEditDate="2010-12-16T23:22:20.393" LastActivityDate="2010-12-16T23:22:20.393" CommentCount="1" />
  <row Id="14606" PostTypeId="2" ParentId="14582" CreationDate="2010-12-16T23:11:37.463" Score="2" ViewCount="" Body="&lt;p&gt;Note that ${\displaystyle {\tan(x) - x \over x^2}}$ is an odd function, so it suffices to show the limit from either side is zero. So we focus on the right limit, and changing $x$ to $\sqrt{x}$ it suffices to show that&#xA;$$\lim_{x \rightarrow 0^+} {\tan(\sqrt{x}) - \sqrt{x} \over x} = 0$$&#xD;&#xA;We apply the mean-value theorem to $f(y) = \tan(\sqrt{y}) - \sqrt{y}$ on $[0,x]$, which we can do since the mean-value theorem only requires that $f(x)$ is differentiable on the interior of the interval. We obtain that there is a $y \in (0,x)$ such that&#xA;$$f&amp;#39;(y) = {\tan(\sqrt{x}) - \sqrt{x} \over x}$$&#xD;&#xA;But using the chain rule we have&#xA;$$f&amp;#39;(y) = {\sec^2(\sqrt{y}) - 1 \over 2 \sqrt{y}}$$&#xD;&#xA;$$= {\tan^2(\sqrt{y}) \over \sqrt{y}}$$&#xD;&#xA;Note that we have&#xA;$$\lim_{y \rightarrow 0^+} {\tan^2(\sqrt{y}) \over \sqrt{y}} = \lim_{y \rightarrow 0^+}\tan(\sqrt{y})\,\,\,\times \,\,\lim_{y \rightarrow 0^+}{\tan(\sqrt{y}) \over \sqrt{y}}$$&#xD;&#xA;$$ = 0*1 = 0$$&#xD;&#xA;Thus we conclude that ${\displaystyle \lim_{x \rightarrow 0^+} {\tan(\sqrt{x}) - \sqrt{x} \over x} = 0}$ as needed.&lt;/p&gt;&#xA;" OwnerUserId="3035" LastActivityDate="2010-12-16T23:11:37.463" />
  <row Id="14607" PostTypeId="2" ParentId="14458" CreationDate="2010-12-16T23:27:20.947" Score="0" ViewCount="" Body="&lt;p&gt;Revised answer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(i)$x &amp;lt; p \Rightarrow f(x) &amp;lt; f(p)$ (because $f$ is strictly monotonically increasing).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because this set of values is bounded from above there exists a supremum, namely $f(p-)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Likewise, $x &amp;gt; p \Rightarrow f(x) &amp;gt; f(p)$ (because $f$ is strictly monotonically increasing).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because this set of values is bounded from below there exists an infinimum, namely $f(p+)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $a_p = f(p-)$ and $b_p = f(p+)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Rightarrow f(x) &amp;lt; a_p \forall x &amp;lt; p$ and $f(x) &amp;gt; b_p \forall x &amp;gt;p$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore, $p$ discontinuous $\Rightarrow f(p-) \neq f(p+)$, $f$ strictly monotonically increasing $\Rightarrow f(p+) - f(p-) &amp;gt; 0 \Rightarrow (a_p, b_p) \neq \emptyset$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(ii)For every point of discontinuity $p_i$, there exists (after (i)) an interval $(a_{p_i}, b_{p_i})$. These intervals are by construction disjoint. Because $\mathbb{R}$ is dense, in every interval there exists an element from $\mathbb{Q} \Rightarrow$ the set of discontinuity points has cardinality at most equal to that of $\mathbb{Q} \Rightarrow$ the set of discontinuity points is countable.&lt;/p&gt;&#xA;" OwnerUserId="3711" LastEditorUserId="3711" LastEditorDisplayName="" LastEditDate="2011-01-04T06:12:50.687" LastActivityDate="2011-01-04T06:12:50.687" CommentCount="12" />
  <row Id="14608" PostTypeId="2" ParentId="14579" CreationDate="2010-12-16T23:31:22.633" Score="1" ViewCount="" Body="&lt;p&gt;Using &lt;a href=&quot;http://math.stackexchange.com/questions/2991/not-understanding-simple-modulus-congruency/3230#3230&quot;&gt;Gauss's algorithm&lt;/a&gt; we have $\rm\displaystyle\ \frac{1}8\ \to\ \frac{4}{32}\ \equiv\ \frac{4}{3}\ \to\ \frac{40}{30}\ \equiv\ \frac{11}{1}\ \ (mod\ 29)$   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I.e. scale $\rm A/B\ \to AN/BN\ $ by the least $\rm\:N\:$ so that $\rm\ BN &amp;gt; 29\:.\ $ Then reduce the numerator and denominator $\rm\ mod\ 29,\:$ and iterate. You will eventually obtain a denominator of $1$ since each step reduces the denominator. Isn't that sweet? That's they key idea that led Gauss to the first proof of the Fundamental Theorem of Arithmetic, i.e. unique factorization of integers.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-12-16T23:37:46.680" LastActivityDate="2010-12-16T23:37:46.680" />
  <row Id="14609" PostTypeId="2" ParentId="14581" CreationDate="2010-12-16T23:32:47.307" Score="2" ViewCount="" Body="&lt;p&gt;Please let me know if the following answer is at the correct level. Thanks!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have a &quot;nice&quot; subset $X \subset R^3$ then it makes sense to talk about the outer boundary of $X$ and the inner boundaries.  Here &quot;nice&quot; means a connected, compact submanifold of dimension three.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's break this down into easy pieces. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $F \subset R^3$ is a compact, connected surface without boundary then $F$ cuts $R^3$ into exactly two pieces $A$ and $B$ so that the closure of $A$ is compact and the closure of $B$ is not compact. Let's refer to $A$ as the piece &quot;inside&quot; of $F$ and to $B$ as the piece &quot;outside&quot; of $F$.  (For help with the exercise please see &lt;a href=&quot;http://www.math.cornell.edu/~hatcher/3M/3Mfds.pdf&quot; rel=&quot;nofollow&quot;&gt;Hatcher's notes&lt;/a&gt; on three-manifolds.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Useful Lemma: If $F$ and $F&amp;#39;$ are surfaces (as above), and $F$ meets the outside of $F&amp;#39;$, then $F&amp;#39;$ meets the inside of $F$.  Proof: Exercise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now suppose that $X$ is as above.  Let $F_0, F_1, \ldots, F_k$ be the connected components of $\partial X$, the boundary of $X$.  Let $A_i$ and $B_i$ be the inner and outer pieces of $R^3 - F_i$.  Say that $F_i$ is &quot;inside&quot; $X$ if $X$ meets $B_i$ (and so is disjoint from $A_i$).  Otherwise say that $F_i$ is &quot;outside&quot; of $X$.  (Exercise: check everything is well-defined using connectedness of $X$!) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Theorem: exactly one of the $F_i$ is outside $X$ and the rest are inside.  Proof: Exercise. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ok, this is a long row to hoe.  But this gives precise meanings to inside and outside and then verifies your intuition about those words.  It is also comforting that it all comes down to understanding just a few facts -- the Schoneflies theorem (Alexander's theorem in dimension three) and properties of connected sets. &lt;/p&gt;&#xA;" OwnerUserId="1307" LastActivityDate="2010-12-16T23:32:47.307" CommentCount="1" />
  <row Id="14610" PostTypeId="2" ParentId="14600" CreationDate="2010-12-17T01:26:50.893" Score="1" ViewCount="" Body="&lt;p&gt;Think about the differentiation as the derivatives along different axes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what you have is in essense &#xA;$\frac{dz}{dt} = \frac{\partial z}{\partial  x}|_y \frac{dx}{dt}+ \frac{\partial  z}{\partial  y}|_x \frac{dy}{dt}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sum exists when you are not travelling in either of those axes, then you are travelling along a path that is 'shared' by the two axes, and their sum tells you the gradient of that path. &lt;/p&gt;&#xA;" OwnerUserId="4728" LastActivityDate="2010-12-17T01:26:50.893" CommentCount="1" />
  <row Id="14611" PostTypeId="2" ParentId="13137" CreationDate="2010-12-17T01:31:58.707" Score="4" ViewCount="" Body="&lt;p&gt;For the sake of completeness, the answer is no; a counterexample is given by the &lt;a href=&quot;http://en.wikipedia.org/wiki/Smith%E2%80%93Volterra%E2%80%93Cantor_set&quot; rel=&quot;nofollow&quot;&gt;Smith-Volterra-Cantor set&lt;/a&gt;, or fat Cantor set.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-12-17T01:31:58.707" CommunityOwnedDate="2010-12-17T01:31:58.707" />
  <row Id="14612" PostTypeId="1" AcceptedAnswerId="14615" CreationDate="2010-12-17T01:37:52.070" Score="3" ViewCount="104" Body="&lt;p&gt;Hello,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;in standard text books on (smooth) manifolds, for example the known series by John M. Lee or Jeffrey Lee, you either deal with continuous manifolds, or with smooth manifolds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, neither in these books nor in lectures I have encountered real examples when a manifold may be $C^k$, but not $C^{k+1}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Intuitively, I would suppose the $|\cdot|_\infty$-Ball with radius $1$ to be a merely continuous, non-smooth manifold, because smoothness fails at the edges of the cube. In contrast to this, polar coordinates show the $|\cdot|_{2}$ with radius $1$ is in fact a smooth manifold.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd be thankful for some examples with clues to basic techniques, how the different degrees of smoothnesses manifest 'in real life'.&lt;/p&gt;&#xA;" OwnerUserId="3557" LastActivityDate="2010-12-17T01:56:58.863" Title="Intuition for not-so-smooth manifolds" Tags="&lt;intuition&gt;&lt;differential-topology&gt;&lt;manifold&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" />
  <row Id="14613" PostTypeId="1" CreationDate="2010-12-17T01:42:34.683" Score="0" ViewCount="142" Body="&lt;p&gt;Hello,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What would be the best way to calculate some sort of correlation or similarity factor between two lists of time intervals :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;List1 :&lt;br&gt;&#xA;1. 2010-06-06 to 2010-12-12&lt;br&gt;&#xA;2. 2010-05-04 to 2010-11-02&lt;br&gt;&#xA;3. 2010-02-04 to 2010-10-08&lt;br&gt;&#xA;4. 2010-04-01 to 2010-08-02&lt;br&gt;&#xA;5. 2010-01-03 to 2010-02-02&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and List2 :&lt;br&gt;&#xA;1. 2010-06-08 to 2010-12-14&lt;br&gt;&#xA;2. 2010-04-04 to 2010-10-10&lt;br&gt;&#xA;3. 2010-02-02 to 2010-12-16&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This question was previously asked at &lt;a href=&quot;http://stackoverflow.com/questions/4466255/how-to-calculate-correlation-between-time-periods&quot;&gt;StackExchange&lt;/a&gt; and the answer provided was a link to &lt;a href=&quot;http://downloads.hindawi.com/journals/asp/2007/089150.pdf&quot; rel=&quot;nofollow&quot;&gt;this article about Feature-List Cross-Correlation&lt;/a&gt; in the field of medical imaging. While it is plenty of useful information, it is a bit daunting for me so it was suggested to try and find a more simple intro to &lt;strong&gt;Vector Cross Correlation&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyone could provide some help on this please?&lt;/p&gt;&#xA;" OwnerUserId="4845" LastEditorUserId="4845" LastEditorDisplayName="" LastEditDate="2010-12-17T02:13:00.553" LastActivityDate="2011-02-15T18:22:53.927" Title="How to calculate correlation between time periods" Tags="&lt;sequences-and-series&gt;&lt;statistics&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="14614" PostTypeId="2" ParentId="14612" CreationDate="2010-12-17T01:55:32.360" Score="6" ViewCount="" Body="&lt;p&gt;One can show that any $C^k$-manifold, for $k \geq 1$, has a unique enrichment to a $C^{\infty}$-manifold.  (I.e. given $M$ with its $C^k$-atlas, we can find a $C^{\infty}$-atlas on $M$, compatible with the given $C^k$-atlas, and this $C^{\infty}$-atlas is unique up to equivalence; see &lt;a href=&quot;http://en.wikipedia.org/wiki/Differentiable_manifold&quot; rel=&quot;nofollow&quot;&gt;wikipedia&lt;/a&gt; for more details.)&#xA;So there is not much point in considering $C^k$-manifolds other than for $k = 0$ or $\infty$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;With regard to your unit ball examples, note that the $| \cdot |_{\infty}$-unit ball, although it has&#xA;corners, is homeomorphic to the $| \cdot |_2$-unit ball; one says that it can be &lt;em&gt;smoothed&lt;/em&gt;.   There &lt;em&gt;are&lt;/em&gt; topological manifolds that cannot be smoothed (in dimension 4 and higher), in the sense that they are not homeomorphic to a smooth manifold.  There are also smooth manifolds that are homeomorphic, but not diffeomorphic.  (E.g. when $n \geq 7,$ one can find smooth manifolds that are homeomorphic to $S^n$, but not diffeomorphic to it; these are so-called &lt;em&gt;exotic spheres&lt;/em&gt;.)  Again, the wikipedia entry has more details.&lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2010-12-17T01:55:32.360" CommentCount="1" />
  <row Id="14615" PostTypeId="2" ParentId="14612" CreationDate="2010-12-17T01:56:58.863" Score="6" ViewCount="" Body="&lt;p&gt;Take a look at Morris Hirsch's text &quot;Differential Topology&quot;, in particular Theorem 2.9 on page 51.  The upshot is that if $M$ is a $C^k$-manifold for $k \geq 1$, then it admits the structure of a $C^j$-manifold for any $j \geq k$, and that structure is unique up to diffeomorphism.   So in that sense the different degrees of smoothness is largely just an artefact of how your manifold is constructed -- you can always &quot;do better&quot; provided the manifold is at least $C^1$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I finished this as Emerton's answer appeared, anyhow, take this as a reference for the result. &lt;/p&gt;&#xA;" OwnerUserId="642" LastActivityDate="2010-12-17T01:56:58.863" />
  <row Id="14616" PostTypeId="2" ParentId="14604" CreationDate="2010-12-17T02:00:04.150" Score="6" ViewCount="" Body="&lt;p&gt;One can show that if the dimension of the affine variety is positive (equivalently, if it is infinite in the sense of our question) then the projective closure is strictly larger.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are probably lots of ways to prove this, but one is by a version of Noether normalization.   If you would like a detailed proof, let me know by a comment and I can give one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Another way to phrase this result is to say that a variety that is simultaneously affine and projective is necessarily finite.  Scheme-theory mavens will recognize this as a special case of the more general statement that a morphism which is simultaneously affine and proper is finite.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Added: Here is a sketch of a proof, as promised:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me begin with Noether normalization in a geometric form.  Fix an afffine variety $V$ contained in $\mathbb A^n$, with projective closue $\overline{V}$.  (Here and below I am always working over an algebraically closed field $k$.)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming that $V$ is not all of $\mathbb A^n$, we see that $\overline{V}$ does not contain the hyperplane at infinity, and so we may choose a point $P$ lying in the hyperplane at infinity, but not lying in $\overline{V}$.  We may also choose a different hyperplane $H$ (i.e. not the hyperplane at infinity) which doesn't contain $P$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With $P$ and $H$ in hand, we may define the projection map  $\pi: \mathbb P^n \setminus P \to H$, which maps any $Q \neq P$ to the intersection of the line $\ell$ joining $P$ and $Q$ with the hyperplane $H$.  Restricting $\pi$ to $\overline{V}$, we obtain a map $V \to H$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now since $P$ is not contained in $\overline{V}$, none of the lines $\ell$ appearing in the projection map are contained in $V$, and so each of them meets $\overline{V}$ in only finitely many points.  In particular, if $\overline{V}$ is infinite, so is its image under $\pi$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now by elimination theory, i.e. the fact that projective varieties are proper, we know that $\pi(\overline{V})$ is closed in $H$, i.e. is a projective variety in $H$,&#xA;which is a projective space of dimension $n-1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, our choice of $P$ ensures that for any $Q \in \overline{V}$, the image $\pi(Q)$ lies at infinity if and only if $Q$ itself does.  So $\pi(V)$ is an affine variety (in the affine space $H \cap \mathbb A^n$ of dimension $n-1$), and $\pi(\overline{V})$ is its projective closure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I have just done is prove Noether normalization, in a geometric form.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The result we want now follows immeidately, by induction on $n$.  (Basically, the case when&#xA;$V = \mathbb A^n$ is clear, and if $V$ is not all of $\mathbb A^n$, the preceding argument allows us to reduce the dimension of the ambient affine space&#xA;by one.)&lt;/p&gt;&#xA;" OwnerUserId="221" LastEditorUserId="221" LastEditorDisplayName="" LastEditDate="2010-12-18T05:22:05.953" LastActivityDate="2010-12-18T05:22:05.953" CommentCount="7" />
  <row Id="14617" PostTypeId="1" AcceptedAnswerId="14620" CreationDate="2010-12-17T02:19:26.773" Score="6" ViewCount="226" Body="&lt;p&gt;There's a question in Spivak's &lt;em&gt;Calculus&lt;/em&gt; (&lt;strike&gt;I don't happen to have the question number in front of me&lt;/strike&gt; in the 2nd Edition, it's Chapter 21, Problem 7) that develops the concept of continued fraction, specifically the continued fraction representation of $\sqrt{2}$.  Having only recently become aware of continued fractions, I'm trying to work my way through this problem, but I'm stuck at the very last stage.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's what I've got so far:  Let $\{a_n\}$ be the sequence defined recursively as $$a_1 = 1, a_{n + 1} = 1 + \frac{1}{1 + a_n}$$  Consider the two subsequences $\{a_{2n}\}$ and $\{a_{2n - 1}\}$.  I've already shown that $\{a_{2n}\}$ is monotonic, strictly decreasing, and bounded below by $\sqrt{2}$, and similarly I've shown that $\{a_{2n - 1}\}$ is monotonic, strictly increasing, and bounded above by $\sqrt{2}$.  Obviously, both of these subsequences converge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Although of course in general, if two subsequences of a sequence happen to converge to the same value, that doesn't guarantee that the sequence itself converges at all (much less to that same value), in the case where the subsequences are $\{a_{2n}\}$ and $\{a_{2n - 1}\}$, it's easy to show that if they both converge to the same value, then so will $\{a_n\}$ (since every term of $\{a_n\}$ is a term of one of the two subsequences).  So no problem there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, it remains only to show that not only do the subsequences converge, but they converge to $\sqrt{2}$ in particular.  Take, for starters, $\{a_{2n - 1}\}$ (if I can get $\{a_{2n - 1}\}$ to converge to $\sqrt{2}$, I'm sure getting $\{a_{2n}\}$ to converge to $\sqrt{2}$ won't be very different).  Because it's strictly increasing and bounded above by $\sqrt{2}$, it converges to some number $x \leq \sqrt{2}$.  Suppose that $x &amp;lt; \sqrt{2}$.  We want to show this doesn't happen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But this is where I'm getting stuck.  I feel like I want to take $\epsilon = \sqrt{2} - x$ and show that there exists some $N$ such that $a_{2N - 1} &amp;gt; x$, which would finish the problem due to the monotonicity of $\{a_{2n - 1}\}$.  But this isn't working.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any hints?  Thanks a ton.&lt;/p&gt;&#xA;" OwnerUserId="506" LastEditorUserId="506" LastEditorDisplayName="" LastEditDate="2010-12-17T15:12:53.237" LastActivityDate="2010-12-17T15:12:53.237" Title="Proving the continued fraction representation of $\sqrt{2}$" Tags="&lt;calculus&gt;&lt;sequences-and-series&gt;&lt;continued-fractions&gt;" AnswerCount="3" />
  <row Id="14618" PostTypeId="2" ParentId="13137" CreationDate="2010-12-17T02:22:19.370" Score="3" ViewCount="" Body="&lt;p&gt;What is actually true is this: for every set $C$ of positive measure and every $\epsilon &amp;lt; 1$ there is some open interval $(a,b)$ such that $\mu(C \cap (a,b)) \geq \epsilon |b-a|$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have always viewed this as an instance of one of &lt;a href=&quot;http://en.wikipedia.org/wiki/Littlewood%27s_three_principles_of_real_analysis&quot; rel=&quot;nofollow&quot;&gt;Littlewood's three principles&lt;/a&gt; for analysis: a measurable set is almost an open set. &lt;/p&gt;&#xA;" OwnerUserId="630" LastEditorUserId="630" LastEditorDisplayName="" LastEditDate="2010-12-17T02:27:25.063" LastActivityDate="2010-12-17T02:27:25.063" />
  <row Id="14619" PostTypeId="2" ParentId="14600" CreationDate="2010-12-17T02:23:01.320" Score="7" ViewCount="" Body="&lt;p&gt;The point of derivatives in one variable is to provide linear approximations $f(x) = f(p) + f&amp;#39;(p) (x - p) + o(|x - p|)$ to nice functions.  Multivariate derivatives work the same way, except &quot;linear approximation&quot; here means approximation by a general linear transformation (a matrix) instead of a scalar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is made precise by the following definition: we say that a function $f : \mathbb{R}^n \to \mathbb{R}^m$ has &lt;a href=&quot;http://en.wikipedia.org/wiki/Total_derivative&quot; rel=&quot;nofollow&quot;&gt;total derivative&lt;/a&gt; a linear transformation $df_p : \mathbb{R}^n \to \mathbb{R}^m$ at a point $p$ if there exists $\epsilon &amp;gt; 0$ and a function $E_p(h)$ defined for $|h| &amp;lt; \epsilon$ such that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f(p + h) = f(p) + df_p(h) + |h| E_p(h)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\lim_{h \to 0} E_p(h) = 0$.  The matrix $df_p$ is sometimes called the &lt;a href=&quot;http://en.wikipedia.org/wiki/Jacobian_matrix_and_determinant&quot; rel=&quot;nofollow&quot;&gt;Jacobian&lt;/a&gt;.  In &lt;a href=&quot;http://en.wikipedia.org/wiki/Big_O_notation#Little-o_notation&quot; rel=&quot;nofollow&quot;&gt;little-o notation&lt;/a&gt;, we write this&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f(p + h) = f(p) + df_p(h) + o(|h|).$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This might seem unnecessarily complicated, but it is the key to understanding the multivariate chain rule.  Suppose that in addition to $f$ we have another function $g : \mathbb{R}^m \to \mathbb{R}^k$ with a total derivative $dg_q$ at some point $q$, and suppose that $f(p) = q$.  Then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$gf(p + h) = g \left( f(p) + df_p(h) + o(|h|) \right) = gf(p) + dg_q df_p(h) + o(|h|)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or, in other words, &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The total derivative $d(gf)_p$ of $gf$ at $p$ is the (matrix) product of the total derivatives $dg_q$ and $df_p$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is the most general statement of the multivariate chain rule.  The relationship to tree diagrams is that one can model matrix multiplication using composition of &lt;a href=&quot;http://en.wikipedia.org/wiki/Incidence_matrix&quot; rel=&quot;nofollow&quot;&gt;incidence matrices&lt;/a&gt;, which come from graphs depicting incidence relationships between sets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your particular example, you have a function $t \mapsto (x, y) : \mathbb{R}^1 \to \mathbb{R}^2$ and another function $(x, y) \mapsto z : \mathbb{R}^2 \to \mathbb{R}^1$.  The total derivative of the first function is $\left[ \begin{array}{c} \frac{dx}{dt} \\\ \frac{dy}{dt} \end{array} \right]$ and the total derivative of the second function is $\left[ \frac{dz}{dx}, \frac{dz}{dy} \right]$, so the total derivative of their composition is the product&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{dz}{dt} = \left[ \frac{dz}{dx}, \frac{dz}{dy} \right] \left[ \begin{array}{c} \frac{dx}{dt} \\\ \frac{dy}{dt} \end{array} \right]$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and this is precisely the formula you give.  The connection to diagrams is that one can represent a composition of linear transformations $\mathbb{R}^1 \to \mathbb{R}^2$ and $\mathbb{R}^2 \to \mathbb{R}^1$ using a pair of incidence matrices, one to represent incidences between a $1$-element set and a $2$-element set, and the other to represent incidences between that $2$-element set and another $1$-element set.&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-12-17T02:29:46.433" LastActivityDate="2010-12-17T02:29:46.433" CommentCount="4" />
  <row Id="14620" PostTypeId="2" ParentId="14617" CreationDate="2010-12-17T02:31:57.760" Score="6" ViewCount="" Body="&lt;p&gt;You have shown that the sequence of $a_{2n}$ converges. Call $e$ (for &lt;em&gt;even&lt;/em&gt;) its limit, and call $o$ the limit of $a_{2n+1}$. Then the recursive relation giving $a_{n+1}$ in terms of $a_n$ implies that $$e=1+1/(1+o)$$ and $$o=1+1/(1+e),$$ from which it follows that $e=o=\sqrt2$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;[For example, take the limit as $n\to\infty$ of $a_{2n}=1+1/(1+a_{2n-1})$ to get the first relation, and similarly you get the second exchanging odd and even. From the first equation, you get $2+o=e+eo$ and from the second $2+e=o+oe$. Subtracting, you get $o-e=e-o$, so $o=e$. From the first equation, you now get $2=o^2$.]&lt;/p&gt;&#xA;" OwnerUserId="462" LastEditorUserId="462" LastEditorDisplayName="" LastEditDate="2010-12-17T02:48:12.880" LastActivityDate="2010-12-17T02:48:12.880" CommentCount="0" />
  <row Id="14621" PostTypeId="2" ParentId="14617" CreationDate="2010-12-17T02:34:59.863" Score="3" ViewCount="" Body="&lt;p&gt;As I explained in a &lt;a href=&quot;http://math.stackexchange.com/questions/7266/why-does-this-process-when-iterated-tend-towards-a-certain-number-the-golden/7275#7275&quot;&gt;previous similar question&lt;/a&gt;, one can use the Banach fixed point theorem to solve these kinds of problems.  For the special case of iterating a fractional linear transformation you can also (as I think is also explained in answers to that question) write down an explicit closed form for the $n^{th}$ term by diagonalizing a $2 \times 2$ matrix.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-12-17T02:34:59.863" CommentCount="1" />
  <row Id="14622" PostTypeId="1" AcceptedAnswerId="14628" CreationDate="2010-12-17T02:37:39.030" Score="5" ViewCount="92" Body="&lt;p&gt;For a two-sphere embedded in $\mathbb{R}^4$,how can you check whether or not there is an ambient isotopy to the &quot;standard&quot; 2-sphere (the set of points $(x,y,z,0)$ in $\mathbb{R}^4$ distance 1 from the origin)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Knot theory was discussed in an intro topology course I took and I'm wondering about further generalizations of the concept.  It seems to me that such &quot;knotted&quot; 2-spheres could be created by rotating a knotted arc about a plane in $\mathbb{R}^4$; and knotted tori by rotating a standard knot about a nonintersecting plane.  However I cannot think of a way to show which of these constructions, if any, are indeed nonisotopic to their standard counterparts.&lt;/p&gt;&#xA;" OwnerUserId="4772" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-17T02:43:39.507" LastActivityDate="2010-12-17T12:36:44.767" Title="The construction of knotted surfaces in $\mathbb{R}^4$" Tags="&lt;topology&gt;&lt;knot-theory&gt;" AnswerCount="2" />
  <row Id="14623" PostTypeId="2" ParentId="14591" CreationDate="2010-12-17T02:42:18.473" Score="7" ViewCount="" Body="&lt;p&gt;One can proceed directly with the Vitali construction, without need for any scaling. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Namely, just carry out the Vitali construction, but ensure that the resulting Vitali set is contained in the interval $[0,a]$. That is, declare that two reals are equivalent if their difference is rational, and observe that every real is equivalent to a real in the interval $[0,a]$. Let $V\subset [0,a]$ select exactly one element from each equivalence class. Observe that the rational translations $V+q$, working modulo $1$ so as to regard $V+q\subset [0,1]$, are disjoint and union up to the whole interval $[0,1]$. It follows easily that $V$ is not measurable and has inner measure $0$, since otherwise the translates modulo $1$ would have infinite measure inside $[0,1]$, which is impossible. Thus, the complement $[0,a]-V$ is non-measurable and has outer measure $a$, as desired. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to note that this argument shows that the outer measure of the classical Vitali set is not determined by the usual features of that set. For example, if all you know about a Vitali set $V$ is that it is contained in $[0,1]$ and contains exactly one element of each equivalence class, then it follows that $V$ is non-measurable and has inner measure $0$, but for all you know, $V$ is actually contained in a very tiny interval $[0,\epsilon]$, and could have tiny outer measure. &lt;/p&gt;&#xA;" OwnerUserId="413" LastEditorUserId="413" LastEditorDisplayName="" LastEditDate="2010-12-17T03:37:34.670" LastActivityDate="2010-12-17T03:37:34.670" CommentCount="0" />
  <row Id="14624" PostTypeId="1" CreationDate="2010-12-17T02:56:34.453" Score="4" ViewCount="294" Body="&lt;p&gt;The adelics seem counter-intuitive. I wonder how they came up originally, and what was the immediate reward for introducing them. What was the politics of introducing the adelics into mathematical mainstream?&lt;/p&gt;&#xA;" OwnerUserId="4732" LastActivityDate="2011-04-08T01:29:48.193" Title="Politics of the Adelics" Tags="&lt;number-theory&gt;&lt;math-history&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="0" />
  <row Id="14625" PostTypeId="1" CreationDate="2010-12-17T02:59:16.927" Score="22" ViewCount="667" Body="&lt;p&gt;Consider the curve $\frac{1}{x}$  where $x \geq 1$. Rotate this curve around the x-axis. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One Dimension - Clearly this structure is infinitely long.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Two Dimensions - Surface Area = $2\pi\int_∞^1\frac{1}{x}dx = 2\pi(\ln ∞ - \ln 1) = ∞$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Three Dimensions -  Volume = $\pi\int_∞^1{x}^{-2}dx = \pi(-\frac{1}{∞} + \frac{1}{1}) = \pi$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So this structure has infinite length and infinite surface area. However it has finite volume, which just does not make sense.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even more interesting, the &quot;walls&quot; of this structure are infinitely thin. Since the volume is finite, we could fill this structure with a finite amount of paint. To fill the structure the paint would need to cover the complete surface area of the inside of this structure. Since the &quot;walls&quot; are infinitely thin, why would a finite amount of paint not be able to cover the outside of the &quot;walls&quot; too?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please help me make sense of this whole thing.&lt;/p&gt;&#xA;" OwnerUserId="4831" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-12-17T16:41:06.983" LastActivityDate="2011-03-30T17:35:30.687" Title="How can a structure have infinite length and infinite surface area, but have finite volume?" Tags="&lt;calculus&gt;&lt;infinity&gt;&lt;paradox&gt;" AnswerCount="5" CommentCount="7" FavoriteCount="1" />
  <row Id="14626" PostTypeId="2" ParentId="14560" CreationDate="2010-12-17T03:01:11.483" Score="3" ViewCount="" Body="&lt;p&gt;It deserves to be better known that there are simple cases where one can give closed forms for some rearrangements of alternating series. Here are a couple of interesting examples based on results of Schlömilch in 1873. Many further results can be found in classical textbooks on infinite series, e.g. those by Bromwich and Knopp.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $\rm\ H^m_n\ $ be the rearrangement of the alternating harmonic series $\rm\ 1 - 1/2 + 1/3 - 1/4 +\: \cdots\ $ obtained by taking consecutive groups of $\rm\:m\:$ positive terms and $\rm\:n\:$ negative terms. Then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\rm H^m_n\ =\ log\ 2 + \frac{1}2\ \lim_{k\to\infty}\ \int^{\:mk}_{nk}\frac{1}x\ dx\ =\ \log 2 + \frac{1}2 \log\frac{m}n $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly rearranging Lebniz's sum $\rm\ L\ =\ \pi/4\ =\ 1 - 1/3 + 1/5 - 1/7 +\: \cdots\ $ yields&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\rm L^m_n\ =\ \frac{\pi}4 + \frac{1}2\ \lim_{k\to\infty}\ \int^{\:mk}_{nk}\frac{1}{2x-1}\ dx\ =\ \frac{\pi}4 + \frac{1}4 \log\frac{m}n $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus as $\rm\:m\:$ varies we obtain infinitely many rearrangements with distinct sums. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The proof of the general theorem underlying these results is quite simple - using nothing deeper than the integral test. See Beigel: &lt;a href=&quot;http://www.jstor.org/stable/2689984&quot; rel=&quot;nofollow&quot;&gt;Rearranging Terms in Alternating Series&lt;/a&gt;, Math. Mag. 1981.&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-12-17T03:01:11.483" />
  <row Id="14627" PostTypeId="2" ParentId="13616" CreationDate="2010-12-17T03:08:44.013" Score="2" ViewCount="" Body="&lt;p&gt;Here is one example I stumbled upon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Random walks have been used to identify clustering in networks in Rosvall and Bergstrom, &lt;a href=&quot;http://octavia.zoology.washington.edu/publications/RosvallAndBergstrom08.pdf&quot; rel=&quot;nofollow&quot;&gt;Maps of random walks on complex networks reveal community structure&lt;/a&gt;.  In particular, this technique was used by &lt;a href=&quot;http://www.eigenfactor.org/map/methods.htm&quot; rel=&quot;nofollow&quot;&gt;Eigenfactor&lt;/a&gt; (who publish journal rankings) to deduce clusterings in research communities.&lt;/p&gt;&#xA;" OwnerUserId="139" LastActivityDate="2010-12-17T03:08:44.013" />
  <row Id="14628" PostTypeId="2" ParentId="14622" CreationDate="2010-12-17T03:09:10.223" Score="5" ViewCount="" Body="&lt;p&gt;The Alexander polynomial is one of the more interesting isotopy-invariants for co-dimension two embedded things in spheres. So it applies to both of the things you're interested in -- knotted $2$-spheres and tori in $S^4$.  Take a look at Akio Kawauchi's &lt;em&gt;Survey of Knot Theory&lt;/em&gt;, or Jonathan Hillman's &lt;em&gt;Algebraic Invariants of Knots and Links&lt;/em&gt; for some details.  The Alexander polynomials have more subtle, associated invariants like Tristram-Levine invariants and such, these are induced from Poincare duality. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &quot;rotating about a plane&quot; construction you mention is called Artin spinning.  It produces non-trivial knots for fairly simple reasons.  Given a knot $K$ in $\mathbb S^3$, let $G$ be the fundamental group of its complement.  Then if you do Artin spinning on $K$ to produce a knotted $S^2$ in $S^4$, call it $K&amp;#39;$, then the fundamental group of the complement of $K&amp;#39;$ is also isomorphic to $G$. &lt;/p&gt;&#xA;" OwnerUserId="642" LastEditorUserId="642" LastEditorDisplayName="" LastEditDate="2010-12-17T07:51:26.263" LastActivityDate="2010-12-17T07:51:26.263" />
  <row Id="14629" PostTypeId="1" AcceptedAnswerId="14893" CreationDate="2010-12-17T03:11:51.767" Score="7" ViewCount="160" Body="&lt;p&gt;I'm having some trouble proving the following statement: let $g:S^{2n-1} \to S^{2n-1}, f: S^{2n-1} \to S^n$.  Then $H(f\circ g) = \deg g H(f)$ where $H(f)$ is the Hopf invariant.  The definition I am using for Hopf invariant is as follows:  let $C_f = D^{2n} \sqcup_f S^{n}$ where $D^{2n}$ is the $2n$ disk which we attach to $S^{n}$ via $f$.  Let $\alpha, \beta$ be generators for $H^n(C_f)$ and $H^{2n}(C_f)$ respectively.  Then $H(f)$ is defined by $\alpha \cup \alpha = H(f) \beta$.  I can prove the statement using the integral formula for $H(f)$ but would like to prove it cohomologically.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think I should consider the map $G: C_{f\circ g} \to C_f$ that takes $x\in \partial D^{2n}$ to $g(x)$ and acts as the identity on everything else.  But then I don't know how to compute what $G^* \alpha$ is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="9" LastEditorUserId="9" LastEditorDisplayName="" LastEditDate="2010-12-19T14:45:06.080" LastActivityDate="2010-12-20T06:17:41.787" Title="Hopf invariant formula" Tags="&lt;algebraic-topology&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" />
  <row Id="14630" PostTypeId="1" CreationDate="2010-12-17T03:18:07.800" Score="4" ViewCount="217" Body="&lt;p&gt;Hello,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a random variable X that is a mixture of a binomial and two normals (see what the &lt;a href=&quot;http://stackoverflow.com/questions/4454513/combining-two-normal-random-variables/4454941#4454941&quot;&gt;probability density function would look like&lt;/a&gt; (first chart))&lt;br&gt;&#xA;and I have another random variable Y of similar shape but with different values for each normally distributed side.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;X and Y are also correlated, here's an example of data that could be plausible :&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;    X     Y&#xA;1.  0    -20&#xA;2. -5     2&#xA;3. -30    6&#xA;4.  7    -2&#xA;5.  7     2&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;As you can see, that was simply to represent that my random variables are either a small positive (often) or a large negative (rare) and have a certain covariance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My problem is : I would like to be able to sample correlated and random values from these two distributions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I could use Cholesky decomposition for generating correlated &lt;em&gt;normally distributed&lt;/em&gt; random variables, but the random variables we are talking here are not normal but rather a mixture of a binomial and two normals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many thanks!&lt;/p&gt;&#xA;" OwnerUserId="4845" LastActivityDate="2011-05-16T10:58:04.880" Title="Generating random values from non-normal and correlated distributions" Tags="&lt;probability&gt;&lt;statistics&gt;&lt;probability-theory&gt;&lt;distribution-theory&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="14631" PostTypeId="1" CreationDate="2010-12-17T03:43:38.053" Score="4" ViewCount="62" Body="&lt;p&gt;Let $S$ be a ruled surface over a curve of genus $g$. Is it possible to compute the second Chern class of $S$ in terms of $g$? &lt;/p&gt;&#xA;" OwnerUserId="4654" LastActivityDate="2010-12-17T03:43:38.053" Title="Second Chern class of a ruled surface" Tags="&lt;algebraic-geometry&gt;" FavoriteCount="1" />
  <row Id="14632" PostTypeId="2" ParentId="14625" CreationDate="2010-12-17T03:45:41.110" Score="39" ViewCount="" Body="&lt;p&gt;It seems you already have made sense of this whole thing from a mathematical point of view (it has infinite surface area and finite volume -- there's no contradiction here).  The paradox is that infinity, at times, does not match well with our day-to-day experiences.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a related example (and hopefully a little bit easier to picture):  Take a unit square.  Cut it in half vertically down the middle -- forming two pieces.  The area clearly remains the same, but the total perimeter (counting both pieces) has increased by 2.  Keep cutting the right-most piece vertically down the middle forever.  The area remains unchanged throughout, whereas the perimeter is $4+\lim_{n \rightarrow \infty} \sum_{i=1}^n 2=\infty$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/s22uq.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if you wanted to make it look more like a horn, you could move the pieces around, for example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/2CgVv.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which still has area 1 and perimeter infinity.  If you wanted to make it have volume 1 and surface area infinity, replace &quot;square&quot; with &quot;cube&quot;.&lt;/p&gt;&#xA;" OwnerUserId="139" LastActivityDate="2010-12-17T03:45:41.110" CommentCount="3" />
  <row Id="14633" PostTypeId="2" ParentId="14625" CreationDate="2010-12-17T03:50:24.493" Score="5" ViewCount="" Body="&lt;p&gt;The answer to the paint question is as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To define terms, let's say you are rotating about the $x$ axis and the volume is between the surface and the $y-z$ plane.  If we take a small (but finite) patch of the $y-z$ plane of $\Delta A$ area then the area above it is roughly $\Delta A$ as $\sqrt{y^2 + z^2} \rightarrow \infty$ and the volume is roughly $\frac{\Delta A}{\sqrt{y^2 + z^2}}$.  So as we move away from the $x$ axis, the trapped volume goes to zero but the surface area does not.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now back to the paint analogy.  It takes infinite paint because a fixed volume of paint covers a fixed surface area (i.e. one gallon / 400 sq.ft.).  This is true even as $\sqrt{y^2 + z^2} \rightarrow \infty$, so we find ourselves having to paint a surface area equivalent to the $y-z$ plane.  Now if we tried to fill the volume with the paint, we are no longer required to cover at a given rate of 1 gal. / 400 sq.ft.  In other words our film thickness is $\frac{1}{\sqrt{y^2 + z^2}}$ which tends to zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in summary, your assumption regarding the walls being infinitely thin is true, but you relate volume and surface area using the metaphor of paint.  Paint always has a finite spread rate no matter how thin the film.  This film does not get thinner as we move away from the $x$ axis.  The volume between the curve and the $y-z$ plane however does.&lt;/p&gt;&#xA;" OwnerUserId="4726" LastActivityDate="2010-12-17T03:50:24.493" />
  <row Id="14634" PostTypeId="2" ParentId="14625" CreationDate="2010-12-17T04:05:57.900" Score="15" ViewCount="" Body="&lt;p&gt;The mathematics has already been explained; when the mathematics contradicts your intuition the answer is to ignore your old intuition and find new intuition.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the physics, Gabriel's horn is in no meaningful sense a physically realistic object.  One way to summarize the lesson here is that the assumption that very thin objects have zero thickness is only reasonable if this causes a reasonable amount of error in one's computations, but in the construction of Gabriel's horn the thickness of the membrane comes to dominate the computation and it is no sense reasonable to ignore this error.  Any time a mathematical model fails to accurately reflect reality one should question the assumptions that go into building that model, and to the extent that this is a mathematical model of anything, the assumption that thickness is negligible is the one that should be thrown out.  If you try to repeat the construction of Gabriel's horn with a fixed finite thickness you will, of course, find that the resulting object has infinite volume.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-12-17T04:05:57.900" />
  <row Id="14635" PostTypeId="1" CreationDate="2010-12-17T04:40:10.847" Score="0" ViewCount="88" Body="&lt;p&gt;Just wondering:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;How to solve ultrahyperbolic PDEs? Is there any analytical solution for linear ultrahyperbolic PDEs?&lt;/li&gt;&#xA;&lt;li&gt;If there are only numerical solutions, are the solutions' behavior similar to those of nonlinear eqns? I mean, like an anharmonic oscillator, chaotic but deterministic?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Information of reference books is also welcomed - but better not too specialized in math. I am an engineering student with very very limited math skills.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks :)&lt;/p&gt;&#xA;" OwnerUserId="4846" LastEditorUserId="1543" LastEditorDisplayName="" LastEditDate="2011-04-18T12:14:18.633" LastActivityDate="2011-04-18T12:14:18.633" Title="ultrahyperbolic PDE" Tags="&lt;differential-equations&gt;&lt;pde&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="14636" PostTypeId="2" ParentId="14549" CreationDate="2010-12-17T04:52:40.090" Score="2" ViewCount="" Body="&lt;p&gt;the wikipedia link you give has an inequality that allows the success probabilities for the different bernoulli trials to be different - and the bound does not depend explicitly on the number of trials n. rather, n is replaced by $\mu = \sum_1^n p_i$, where $p_i$ is the success probability for the i$^{th}$ trial. look at the section &quot;Theorem for multiplicative form of Chernoff bound (relative error)&quot;. in this form, it is clear, as qiaochu states, that the $\{p_i\}$ can depend on n - and can all be the same - or not.&lt;/p&gt;&#xA;" OwnerUserId="1547" LastActivityDate="2010-12-17T04:52:40.090" />
  <row Id="14637" PostTypeId="1" AcceptedAnswerId="14640" CreationDate="2010-12-17T05:31:20.400" Score="0" ViewCount="100" Body="&lt;p&gt;I've always had the following silly(?) doubt about convergent sequences. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given a finite sequence $A_{1:n}=(a_1,a_2\ldots ,a_n)$, define its &lt;i&gt; reverse &lt;/i&gt; as $rev(A) = (a_n,a_{n-1},..a_1)$.  Further, if $A$ is infinite, define its reverse as the limit of the reverses of its initial subsequences, i.e. as the limit of the sequence of sequences $ \{ (a_1), (a_2,a_1), (a_3,a_2,a_1), ..\}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Q:&lt;i&gt; Is this notion well defined? Why or why not? &lt;/i&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suspect not, because the reverse of $(1,\frac{1}{2},\frac{1}{3},...)$ would be $(0,0,... , \frac{1}{3},\frac{1}{2},1)$? This seems absurd!&lt;/p&gt;&#xA;" OwnerUserId="1585" LastActivityDate="2010-12-17T07:12:14.193" Title="Reversing a convergent sequence" Tags="&lt;real-analysis&gt;" AnswerCount="1" CommentCount="6" FavoriteCount="1" />
  <row Id="14638" PostTypeId="2" ParentId="14624" CreationDate="2010-12-17T06:01:40.247" Score="13" ViewCount="" Body="&lt;p&gt;The ideles came first, and were introduced by Chevalley as part of his program of giving a (then) new proof of class field theory which avoided the analysis of $L$-functions.  The adeles came a little later; they were originally called &lt;em&gt;valuation vectors&lt;/em&gt;, and were introduced by Artin and Whaples.  It took a little time for the&#xA;details to be sorted out (e.g. Chevalley orginally used a slightly different topology on the ideles to the one used currently; see &lt;a href=&quot;http://mathoverflow.net/questions/41253/who-fixed-the-topology-on-ideles&quot; rel=&quot;nofollow&quot;&gt;this MO question&lt;/a&gt; for more details); when Tate wrote his thesis in 1950, he used the current definitions (although I think he still used the term valuation vectors,&#xA;rather than adeles).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as I know, the name ideles was invented by Chevalley, as a pun on the word ideals, since from the ideles one could recover all the generalized ideal class groups of traditional algebraic number theory.  The term adeles (when it was finally settled on, maybe by Weil?) was then a pun on ideles, together with the fact that the adeles have an additive structure.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The analogue of the adeles in which the number field is replaced by the function field of a smooth projective irreducible algebraic curve (so that the places of the number field are replaced by the points of the curve) appeared in Weil's approach to proving the Riemann--Roch theorem.  (In this context the adeles were also called &lt;em&gt;repartitions&lt;/em&gt;, although I doubt anyone uses that terminology anymore.)   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If one looks in Tate's thesis, there is a result there called the Riemann--Roch theorem, which really does reduce to the usual result with that name in the function field case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The consideration of adelic points of algebraic groups was first undertaken by Weil, I think, in his reformuation of the work of Tamagawa in the study of what are now called Tamagawa numbers.   The use of this point of view in the study of automorphic forms was brought to the fore in the work of Langlands in the late 1960s and early 1970s, although I don't know whether he was the first one who realized that it was the natural way to encode the notion of Hecke eigenform in representation theoretic terms.  Chevalley's original concept of idele is subsumed in this general formalism; they are precisely the adelic points of the group $GL_1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The overarching reason for introducing the adelic viewpoint, beginning with Chevalley's original work and continuing through to the modern treatment of automorphic forms, has always been the same: it provides the most convenient and flexible framework for studying the interaction of local and global phenomena.  All the traditional concepts (e.g. finiteness of the class number, Dirichlet's unit theorem, finite volume of $\Gamma\backslash G$ for semi-simple groups $G$ and congruence subgroups $\Gamma$) are easily expressed in adelic terms in a natural and uniform way, and for studying certain concepts (local reciprocity laws and their interaction with global reciprocity) it is hard to imagine how one could sensibly formulate things in anything other than the adelic manner. &lt;/p&gt;&#xA;" OwnerUserId="221" LastEditorUserId="221" LastEditorDisplayName="" LastEditDate="2011-04-08T01:29:48.193" LastActivityDate="2011-04-08T01:29:48.193" CommentCount="2" />
  <row Id="14640" PostTypeId="2" ParentId="14637" CreationDate="2010-12-17T07:12:14.193" Score="4" ViewCount="" Body="&lt;p&gt;This &quot;reverse sequence&quot; notion is either going to be meaningless, or be identical to sequences (but just written backwards).  For it to be well-defined, you need somehow to index the entries.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As it is defined, the sequence of sequences $(a_1)$, $(a_2,a_1)$, etc., would converge pointwise to the sequence $(\ell,\ell,\ldots)$, where $\ell = \lim_{n \rightarrow \infty} a_n$ (if it exists).  So this is not particularly exciting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem with the $(\ell,\ell,\ldots,a_N,a_{N-1},\ldots,a_1)$ idea, is that you need to take $\infty-\infty=N$ (which will give rise to nonsense).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you defined the sequence $\mathrm{rev}(A)$ such that the $(-i)$-th entry of $\mathrm{rev}(A)$ is the $i$-th entry of $A$, then it will also be well-defined, but identical to $A$ itself (just written backwards).  So this is also not particularly exciting.&lt;/p&gt;&#xA;" OwnerUserId="139" LastActivityDate="2010-12-17T07:12:14.193" />
  <row Id="14641" PostTypeId="2" ParentId="14403" CreationDate="2010-12-17T07:17:42.670" Score="2" ViewCount="" Body="&lt;p&gt;don't know exactly what you want to know about $d\sigma/\sigma$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;mathematically, it is like lebesgue measure $dx$ on $\mathbb{R}$ - in the following sense:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$dx$ does not change under a simple additive change [or shift] of a variable: $y= x+a$, where $a$ is a constant. that is, $dy=dx$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;similarly, $d\sigma/\sigma$ does not change under a multiplicative change [or rescaling] of a variable: $y=c\sigma$, where $c&amp;gt;0$. that is, $dy/y=d\sigma/\sigma$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;both $dx$ on $\mathbb{R}$ and $d\sigma/\sigma$ on $\mathbb{R}^+$ are examples of haar measures [&lt;em&gt;i.e.&lt;/em&gt;, invariant measures] on $\mathbb{R}$ and $\mathbb{R}^+$, respecctively, considered as groups acting on themselves, as indicated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;as for jeffries: many [non-orthodox] bayesians consider $dx$ to be like a uniform [prior] distribution on $\mathbb{R}$. [it is the limit - in some sense - of the uniform distribution on [-A, A], as A $\to\infty$.]  it is considered to be an 'objective prior',&#xA;in that everyone [supposedly] agrees that it is THE uniform distribution on $\mathbb{R}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;that $dx$ is improper - it is not actually a probability distribution on $\mathbb{R}$ - doesn't seem to faze these folks; as long as one comes up with a proper posterior distribution when using it [as is often the case], it can be taken as a 'prior'. an example is inference about the unknown mean $\mu$ of a normal distribution with $\sigma = 1$. if $X$ is one observation from this distribution, the conditional pdf of $X|\mu$ is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f_{X|\mu}(x|\mu) = \frac{1}{\sqrt{2\pi}}e^{\frac{1}{2}(x-\mu)^2},\quad -\infty &amp;lt; x &amp;lt; \infty.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;multiplying by the 'uniform' prior 'pdf' of $\mu$, the [improper] joint pdf of $(X,\mu)$ is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f_{X,\mu}(x,\mu) = f_{X|\mu}(x|\mu){\mathbb{1}}_{(-\infty.\infty)}(\mu) = \frac{1}{\sqrt{2\pi}}e^{\frac{1}{2}(x-\mu)^2},\quad -\infty &amp;lt; x, \mu &amp;lt;\infty.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;note that the [improper] prior 'pdf' for $\mu$ is $f_\mu(\mu)={\mathbb{1}}_{(-\infty,\infty)}(\mu)$, in that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathrm{&amp;#39;prob&amp;#39;}(a&amp;lt; \mu&amp;lt; b) = \int_a^b f_\mu(\mu) d\mu = \int_a^b d\mu = b-a.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;then, integrating out the $\mu$ in the joint pdf, one gets that the marginal 'pdf' of $X$ is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f_X(x) = \int_{-\infty}^{\infty}  f_{X|\mu}(x|\mu)d\mu = \int_{-\infty}^{\infty} \frac{1}{\sqrt{2\pi}}e^{\frac{1}{2}(x-\mu)^2}d\mu = {\mathbb{1}}_{(-\infty.\infty)}(x),$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so that $X$ marginally has an improper distribution as well. finally, the posterior pdf for $\mu$ is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f_{\mu|X}(\mu|x)=  f_{X,\mu}(x,\mu)/f_{X}(x) = \frac{1}{\sqrt{2\pi}}e^{\frac{1}{2}(x-\mu)^2},\quad -\infty &amp;lt; \mu &amp;lt; \infty,$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so that $\mu|X$ is normal with mean $X$ and SD = 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with this example in mind, perhaps you can persuade yourself that $d\sigma/\sigma$ is some sort of analog of a uniform distribution on $\mathbb{{R}^+}$. [if one considers the parameter $\nu=\log\sigma $, this turns into the [lebesgue] uniform distribution for $\nu$, since [as pointed out by @hardmath], $d\nu=d(\log\sigma)=d\sigma/\sigma$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;whether or not you buy into this depends, i guess, on your propensity for being a 'true believer'.&lt;/p&gt;&#xA;" OwnerUserId="1547" LastActivityDate="2010-12-17T07:17:42.670" CommentCount="1" />
  <row Id="14642" PostTypeId="2" ParentId="14630" CreationDate="2010-12-17T07:21:20.997" Score="1" ViewCount="" Body="&lt;p&gt;If you want to sample $(X,Y)$ you need to find out what is their distribution function. If you know the the conditional densities $p(X|Y)$ and $p(Y|X)$ you can use &lt;a href=&quot;http://en.wikipedia.org/wiki/Gibbs_sampling&quot; rel=&quot;nofollow&quot;&gt;Gibbs sampling&lt;/a&gt;. &lt;/p&gt;&#xA;" OwnerUserId="4742" LastActivityDate="2010-12-17T07:21:20.997" CommentCount="3" />
  <row Id="14645" PostTypeId="1" CreationDate="2010-12-17T08:12:22.367" Score="6" ViewCount="110" Body="&lt;p&gt;Is there an algorithm to find out all possible resistance values of series, parallel, and series-parallel arrangements given $n$ identical resistors, $R$? All of them must be used.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This might even extend to differently-valued resistors, but I'll just focus on identical resistors.&lt;/p&gt;&#xA;" OwnerUserId="1323" LastEditorUserId="1323" LastEditorDisplayName="" LastEditDate="2011-02-26T01:16:53.270" LastActivityDate="2011-02-26T01:16:53.270" Title="Given $n$ identical resistors $R$, find combinations of series, parallel, and series-parallel arrangements" Tags="&lt;combinatorics&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="14646" PostTypeId="1" CreationDate="2010-12-17T08:17:13.533" Score="2" ViewCount="163" Body="&lt;p&gt;I am studying QR decomposition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could you explain the geometric intuition for what the Householder transformation does in that context, and why it's sometimes referred to as the Householder &lt;em&gt;reflection&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="4373" LastActivityDate="2010-12-17T14:39:25.303" Title="Geometric intuition for the Householder transformation" Tags="&lt;linear-algebra&gt;&lt;matrices&gt;&lt;reflection&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="14648" PostTypeId="2" ParentId="14593" CreationDate="2010-12-17T08:28:00.570" Score="1" ViewCount="" Body="&lt;p&gt;hi, you can refer this link &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.ehow.com/how_5169102_convert-latitude-longitude-gps-coordinates.html&quot; rel=&quot;nofollow&quot;&gt;http://www.ehow.com/how_5169102_convert-latitude-longitude-gps-coordinates.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="4851" LastActivityDate="2010-12-17T08:28:00.570" />
  <row Id="14649" PostTypeId="1" CreationDate="2010-12-17T08:30:54.483" Score="2" ViewCount="88" Body="&lt;p&gt;In my lecture today, the lecturer mentioned in passing that there are unusual results when you look at combination of elements of a ring, and whether they form a subring or not.&lt;br&gt;&#xA;More specifically he said that if a and b are subrings of a ring, this implies that a-2b, 2a-b and 3a-b are subrings, while a-3b is not.  He didn't go into any detail why this is the case (perhaps we haven't learnt enough to explain it yet) but I was wondering whether anyone could shed any light on why this is the case?&lt;br&gt;&#xA;Thanks! &lt;/p&gt;&#xA;" OwnerUserId="4697" LastActivityDate="2010-12-17T08:30:54.483" Title="When is a combinations of subrings a subring?" Tags="&lt;ring-theory&gt;" CommentCount="3" />
  <row Id="14650" PostTypeId="2" ParentId="14624" CreationDate="2010-12-17T08:58:11.977" Score="2" ViewCount="" Body="&lt;p&gt;Recall the process of completing $\mathbb Q$ to obtain $\mathbb R$. Here the metric is coming from the absolute value norm. It turns out that in addition to the usual absolute value, one can put the $p$-adic valuations on $\mathbb Q$. One can do a similar completion and complete $\mathbb Q$ to $\mathbb Q_p$. You can also prove that these are the only valuations on $\mathbb Q$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More generally, on number fields and function fields(together called global fields), one can similarly classify all the valuations,aka &lt;em&gt;places&lt;/em&gt;, and some of them are Archimedean and others non-Archmidean. Now according to Weil in his book &lt;em&gt;Basic Number Theory&lt;/em&gt;, the theory of adèles come up with the point of view that from the arithmetician's perspective, no valuation is preferred over the other; and the global field must be seen at once as completing to infinitely many places all at once, with $\mathbb R$ being no better or worse than any other. He uses the theory of adèles to achieve this end. The nomenclature was perhaps influenced by Chevalley's &lt;em&gt;ideles&lt;/em&gt;, which can be realized as a subset of adèles. The name is supposed to be a pun on a French girls's name.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you consult the preface of Weil's &lt;em&gt;Basic Number Theory&lt;/em&gt;, you can read exactly what he says. My paraphrase here might be a bit jaded by memory.&lt;/p&gt;&#xA;" OwnerUserId="1119" LastActivityDate="2010-12-17T08:58:11.977" />
  <row Id="14651" PostTypeId="2" ParentId="14000" CreationDate="2010-12-17T09:12:41.620" Score="2" ViewCount="" Body="&lt;p&gt;A surface integral measures the &quot;average outflow&quot; of the vector field from the surface. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One illuminating application is &lt;a href=&quot;http://en.wikipedia.org/wiki/Gauss%27s_law&quot; rel=&quot;nofollow&quot;&gt;Gauss's Law&lt;/a&gt; for electric fields, i.e.,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The electric flux through any closed surface is proportional to the enclosed electric charge.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;More precisely, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\oint_S \mathbf{E} \cdot \mathrm{d}\mathbf{A}  = \frac{Q}{\varepsilon_0},$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where the left-hand side of the equation is a surface integral denoting the electric flux through a closed surface $S$, and the right-hand side of the equation is the total charge $Q$ enclosed by $S$, divided by the electric constant $\varepsilon_0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gauss's Law itself has very interesting physical consequences. See the wikipedia page for examples. For instance, the renowned &lt;a href=&quot;http://en.wikipedia.org/wiki/Coulomb%27s_law&quot; rel=&quot;nofollow&quot;&gt;Coulomb's law&lt;/a&gt; is equivalent to it. For another: all the charges on a conductor migrate to its surface.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is also a &lt;a href=&quot;http://en.wikipedia.org/wiki/Gauss%27s_law_for_magnetism&quot; rel=&quot;nofollow&quot;&gt;Gauss's Law for magnetism&lt;/a&gt;. It states that the net flux of magnetic fields through any surface is always zero. The difference with electric fields is that there are no magnetic &quot;monopoles&quot;; you always find dipoles. The equation is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \oint_S \mathbf{B} \cdot \mathrm{d}\mathbf{A} = 0 $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$S$ is any closed surface and $B$ is the magnetic field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both these Gauss's laws are part of the important &lt;a href=&quot;http://en.wikipedia.org/wiki/Maxwell%27s_equations&quot; rel=&quot;nofollow&quot;&gt;Maxwell's equations&lt;/a&gt;. Many applications are through integration.&lt;/p&gt;&#xA;" OwnerUserId="1119" LastEditorUserId="1119" LastEditorDisplayName="" LastEditDate="2010-12-17T09:21:24.920" LastActivityDate="2010-12-17T09:21:24.920" />
  <row Id="14652" PostTypeId="2" ParentId="14646" CreationDate="2010-12-17T09:57:17.230" Score="1" ViewCount="" Body="&lt;p&gt;Assume that $V$ is an $n$-dimensional Euclidean space. Then, given an orthonormal basis $(e_i)_{i=1}^n$ and an arbitrary n-tuple of vectors $(u_i)_{i=1}^n$, there exists a sequence of $n$ isometries $H_1,...,H_n$ such that &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$H_i$ is either &lt;em&gt;a hyperplane reflection&lt;/em&gt; or the identity;&lt;/li&gt;&#xA;&lt;li&gt;every vector of the form &#xA;$$r_j=H_nH_{n-1}\dots H_1(u_j),\qquad j=1,\dots, n,$$&#xD;&#xA;is a linear combination of the vectors $e_1,\dots, e_j$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In other words, the matrix $R$ whose columns are the components of the vectors $r_j$ written over the basis $(e_i)_{i=1}^n$ is upper triangular. The isometries $H_1,\dots, H_n$ which are not identities (i.e. the hyperplane reflections) correspond to Householder matrices. The composition &#xA;$$R=H_n\dots H_1A$$ directly yields the $QR$-decomposition of a given matrix $A$ (whose columns are the components of $u_j$ over the given basis).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The geometric interpretation of the $QR$-decomposition can be probably found in several places. I like the presentation in &lt;a href=&quot;http://books.google.com.ua/books?id=B4JtblR1lkMC&amp;amp;printsec=frontcover&amp;amp;dq=books%09+++Geometric+Methods+and+Applications%3a+For+Computer+Science+and+Engineering&amp;amp;source=bl&amp;amp;ots=clsb3DcEQF&amp;amp;sig=-LHtZ9JiiwlgZKWtcgILchZuhHY&amp;amp;hl=en&amp;amp;ei=1zcLTaaLG4Wq8QO6k7H-DQ&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=2&amp;amp;ved=0CCAQ6AEwAQ#v=onepage&amp;amp;q=books%09%20%20%20Geometric%20Methods%20and%20Applications%3A%20For%20Computer%20Science%20and%20Engineering&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;Geometric Methods and Applications&lt;/em&gt;&lt;/a&gt; by Jean Gallier (have a look at Section 7.3).&lt;/p&gt;&#xA;" OwnerUserId="723" LastEditorUserId="723" LastEditorDisplayName="" LastEditDate="2010-12-17T10:17:37.937" LastActivityDate="2010-12-17T10:17:37.937" />
  <row Id="14653" PostTypeId="1" AcceptedAnswerId="14665" CreationDate="2010-12-17T10:10:50.603" Score="4" ViewCount="139" Body="&lt;p&gt;I've been looking for less trivial examples of computing Ext than finitely generated abelian groups, which tends to be the standard example (and often the only example). Here's an interesting exercise I found in some notes:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $M = \mathbb{C}[x,y] / (x,y), N = \mathbb{C}[x,y] / (x-1)$. My question is how to compute $\text{Ext}_v(M,N)$ in the category of $\mathbb{C}[x,y]$-modules.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, first of all $\text{Ext}_0(M,N) = \text{Hom}(M,N)$. However, I'm not sure how to identify what this $\text{Hom}$ is! More generally, we have the short exact sequence&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$0\rightarrow K \rightarrow \mathbb{C}[x,y] \rightarrow M \rightarrow 0$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where the second map is the inclusion, the third map is the quotient projection, and $K$ is the kernel of the projection. This sequence gives the exact sequence (a piece of the long exact sequence)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$0\rightarrow \text{Ext}_1(M,N) \rightarrow \text{Hom}(K,N) \rightarrow \text{Hom}(\mathbb{C}[x,y],N) \rightarrow \text{Hom}(M,N) \rightarrow 0$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which means that $\text{Ext}_1(M,N)$ is the kernel of the map $\text{Hom}(K,N) \rightarrow \text{Hom}(\mathbb{C}[x,y], N)$. But again I'm having trouble determining this kernel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally I think the projective resolution $0 \rightarrow \mathbb{C}[x,y] \rightarrow \mathbb{C}[x,y] \rightarrow \mathbb{C}[x,y]/(x-1)$ shows that the higher Ext's are zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any help would be greatly appreciated.&lt;/p&gt;&#xA;" OwnerUserId="251" LastActivityDate="2010-12-17T12:58:53.483" Title="An example of computing Ext" Tags="&lt;homological-algebra&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="1" />
  <row Id="14654" PostTypeId="1" AcceptedAnswerId="14677" CreationDate="2010-12-17T10:21:50.490" Score="0" ViewCount="198" Body="&lt;p&gt;Concerning &lt;a href=&quot;http://math.stackexchange.com/questions/13646&quot;&gt;this&lt;/a&gt; question of mine which involves telescoping in the solution, I was wondering if it is possible to express&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ln(x)=\lim_{n\to\infty}n\left(x^{\frac1{n}}-1\right)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$e^x=\lim_{n\to\infty}\left(1+\frac{x}{n}\right)^n$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;as infinite telescoping products.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it possible?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="1747" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-17T14:19:32.337" LastActivityDate="2010-12-17T16:29:56.530" Title="Can $\ln(x)=\lim_{n\to\infty} n\left(x^{\frac1{n}}-1 \right)$ be expressed as an infinite telescoping product?" Tags="&lt;real-analysis&gt;&lt;analysis&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="14655" PostTypeId="1" CreationDate="2010-12-17T10:33:49.400" Score="3" ViewCount="124" Body="&lt;p&gt;Can a polynomial size CFG over large alphabet describe any of these languages:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Each terminal appears 0 or 2 times&lt;/li&gt;&#xA;&lt;li&gt;Word repetition {www* | w \in \Sigma *} (word repetition of an arbitrary word w)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&quot;Polynomial size&quot; Context Free Grammar is defined as Chomsky Normal Form polynomial in |Σ|.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;References to poly-size CFGs over large alphabets will be appreciated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: This is not a homwork. Positive answer can give extension to &lt;a href=&quot;http://math.stackexchange.com/questions/14506/restricted-read-twice-bdds-and-context-free-grammars&quot;&gt;Restricted read twice BDDs and context free grammars&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4750" LastEditorUserId="4750" LastEditorDisplayName="" LastEditDate="2010-12-17T15:15:07.847" LastActivityDate="2011-02-19T08:23:10.287" Title="Can a polynomial size CFG over large alphabet describe any of these languages:" Tags="&lt;computer-science&gt;&lt;computability&gt;&lt;formal-systems&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="14656" PostTypeId="2" ParentId="14654" CreationDate="2010-12-17T11:24:05.797" Score="2" ViewCount="" Body="&lt;p&gt;Well, we could produce a very contrived example for $e^x$ by defining&#xA;$S_m(x) = \sum_{k=0}^m x^k/k! $ and let&#xA;$$ P_N(x) = \prod_{k=0}^N \frac{S_{k+1}(x)}{S_k(x)} = S_{N+1}(x).$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And so $\lim_{N \rightarrow \infty} P_N(x) = e^x .$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More concretely, $e^x$ cannot be represented as an infinite product, such as we can do for $\sin x$ for example, because any zeros of the factors would also be zeros of $e^x$ but $e^x$ is positive for all $x \in \mathbb{R}.$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Here's a slightly less contrived product formula for $e^x .$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ P_n(x) = \prod_{k=1}^n \left( 1+ \frac{x}{n} \right)^{n/k(k+1)} $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;then $e^x = \lim_{n \rightarrow \infty} P_n(x),$ although there's no telescoping going on here.&lt;/p&gt;&#xA;" OwnerUserId="1301" LastEditorUserId="1301" LastEditorDisplayName="" LastEditDate="2010-12-17T14:21:04.350" LastActivityDate="2010-12-17T14:21:04.350" />
  <row Id="14657" PostTypeId="2" ParentId="14645" CreationDate="2010-12-17T11:27:01.927" Score="2" ViewCount="" Body="&lt;p&gt;Following the Wikipedia link given by Douglas, and the references from there to OEIS, leads to this paper which seems to answer your question:&#xA;&lt;a href=&quot;http://dx.doi.org/10.1119/1.19396&quot; rel=&quot;nofollow&quot;&gt;Antoni Amengual, The intriguing properties of the equivalent resistances of n equal resistors combined in series and in parallel, American Journal of Physics, 68(2), 175-179 (February 2000)&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The author points out that asking for the possible values of the resistance is not the same as asking for the total number of networks. The answer to the former question is claimed to be approximately $2.55^n$ (for large $n$, I presume).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[EDIT: A look in the paper reveals that the formula $2.55^n$ was simply obtained by straight-line fitting of the logarithm of the exact values for $6 \le n \le 16$, so (being a mathematician) I'm not quite convinced about its validity for large $n$...]&lt;/p&gt;&#xA;" OwnerUserId="1242" LastEditorUserId="1242" LastEditorDisplayName="" LastEditDate="2010-12-17T13:38:26.023" LastActivityDate="2010-12-17T13:38:26.023" CommentCount="2" />
  <row Id="14658" PostTypeId="1" AcceptedAnswerId="14661" CreationDate="2010-12-17T11:46:28.397" Score="3" ViewCount="140" Body="&lt;p&gt;The infinite-length sequence $x_1[n]$ defined by &#xA;\begin{multline}&#xA;x_1[n]=&#xA;\begin{cases}&#xA;\dfrac{1}{n}&amp;amp; \text{if $n \geq $1},\&#xA;0&amp;amp; \text{if $n \leq $0}.&#xA;\end{cases}&#xA;\end{multline}&#xA;has an energy equal to &lt;br/&gt;&#xA;$\mathcal{E _x {_1}} = \sum^\infty_{n=1}(\dfrac{1}{n})^{2}$&#xD;&#xA;&lt;br/&gt;which converges to $\pi^2/6$ indicating that $x_1[n]$ has finite energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't get where we find $\pi^2/6$. It would be great if anyone can help me out.&lt;/p&gt;&#xA;" OwnerUserId="4653" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-12-18T02:59:26.087" LastActivityDate="2010-12-18T02:59:26.087" Title="Easy question about finite energy due to convergence" Tags="&lt;sequences-and-series&gt;&lt;signal-processing&gt;" AnswerCount="2" CommentCount="5" />
  <row Id="14659" PostTypeId="2" ParentId="14617" CreationDate="2010-12-17T11:56:58.907" Score="1" ViewCount="" Body="&lt;p&gt;I don't think you need to break it up into two cases. In stead we can do this:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Establish that the sequence $\langle a_i \rangle$ is cauchy and therefore convergent to some $L$. For this establish the sequence $\langle f^i(0)\rangle$ converges to zero, where $f^i$ is the 'i'th iterate of $\frac{1}{2+x}$.&lt;/li&gt;&#xA;&lt;li&gt;Notice that the relation: $a_{n+1}=1+\frac{1}{1+a_n}$ is the same as $a_{n+1}a_n + a_{n+1} - a_n -2 = 0$.&lt;/li&gt;&#xA;&lt;li&gt;Take limits at both ends to get: $L^2+L-L-2=0$. &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="4806" LastActivityDate="2010-12-17T11:56:58.907" />
  <row Id="14660" PostTypeId="1" AcceptedAnswerId="14664" CreationDate="2010-12-17T12:01:09.650" Score="6" ViewCount="147" Body="&lt;p&gt;One partition of 8 is 5 + 3, but if we then partition each of the 5 and 3 we could get (3+2) + (2+1), and then partition again to get ((2+1)+(1+1)) + ((1+1)+1) and finally (((1+1)+1)+(1+1)) + ((1+1)+1). 5+3 could also be expanded as (4+1)+(2+1), then ((2+2)+1)+((1+1)+1), then (((1+1)+(1+1))+1)+((1+1)+1).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This question is about viewing &quot;+&quot; as a binary operation , so 1+1+1+1 would have to written as either (1+1)+(1+1) or ((1+1)+1)+1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Every partition can be written as a such nested partition of 1s. It is still order independent, but associative dependent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a given number $n$, how many associative dependent binary-operation nested partitions of 1s are there of $n$ ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is known about this function ?&lt;/p&gt;&#xA;" OwnerUserId="276" LastActivityDate="2011-05-20T03:49:36.757" Title="Counting nested integer partitions" Tags="&lt;combinatorics&gt;" AnswerCount="2" />
  <row Id="14661" PostTypeId="2" ParentId="14658" CreationDate="2010-12-17T12:02:42.330" Score="5" ViewCount="" Body="&lt;p&gt;The sum of the series $\displaystyle\sum^\infty_{n=1}(\dfrac{1}{n})^{2}=\dfrac{\pi^2}{6}$ is a &lt;a href=&quot;http://en.wikipedia.org/wiki/Basel_problem&quot; rel=&quot;nofollow&quot;&gt;classical result due to Euler&lt;/a&gt;. Several proofs are given in the answers to &lt;a href=&quot;http://math.stackexchange.com/questions/8337/different-methods-to-compute-sum-n-1-infty-frac1n2&quot;&gt;this question&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;PS. &lt;a href=&quot;http://empslocal.ex.ac.uk/people/staff/rjchapma/etc/zeta2.pdf&quot; rel=&quot;nofollow&quot;&gt;Here&lt;/a&gt; Robin Chapman collects 14 proofs.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;PPS. The improper double integral $$\int_{0}^{1}\int_{0}^{1}\left(\dfrac{1}{1-xy}\right) \mathrm{d}x\mathrm{d}y=\int_{0}^{1}\int_{0}^{1}\left(\sum_{n=1}^{\infty }\left( xy\right)^{n-1}\right) \mathrm{d}x\mathrm{d}y=\sum^\infty_{n=1}\dfrac{1}{n^2} =\dfrac{\pi^2}{6}=\zeta(2)$$ is finite, as pointed out in &lt;em&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/3642008550&quot; rel=&quot;nofollow&quot;&gt;Proofs from THE BOOK&lt;/a&gt;&lt;/em&gt; by Martin Aigner and Günter Ziegler. The original article by Tom Apostol is &lt;a href=&quot;http://www.springerlink.com/content/045l250v885v033u/fulltext.pdf&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;. &lt;/p&gt;&#xA;" OwnerUserId="752" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2010-12-18T01:46:30.120" LastActivityDate="2010-12-18T01:46:30.120" CommentCount="8" />
  <row Id="14662" PostTypeId="2" ParentId="14622" CreationDate="2010-12-17T12:36:44.767" Score="2" ViewCount="" Body="&lt;p&gt;Kawauchi's and Hillman's books are indeed good references. Many known knotted surfaces have non-trivial and easily computable fundamental groups. You should also look at R.H. Fox's &quot;Quick Trip through Knot Theory,&quot; in Fort's Georgia Topology Conference Proceedings, republished by Dover. In addition, there are three books dedicated to studying knotted surfaces from a diagrammatic point of view. Kamada's book on Surface Braids published by the AMS, my book with Saito, &quot;Knotted Surfaces and Their Diagrams,&quot; and my book with Kamada and Saito, &quot;Surfaces in 4-space.&quot; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The quandle cocycle invariants discovered by Jelsovsky, Langford, Kamada, and me are known to be powerful invariants of knotted surfaces. &lt;/p&gt;&#xA;" OwnerUserId="722" LastActivityDate="2010-12-17T12:36:44.767" />
  <row Id="14663" PostTypeId="2" ParentId="14635" CreationDate="2010-12-17T12:47:17.690" Score="3" ViewCount="" Body="&lt;p&gt;The answer depends on what you mean by &quot;solving&quot; the PDE. The initial value problem for the ultrahyperbolic PDEs are ill-posed. In particular, there is a theorem (the version I know is due to Hormander, you can find it in his &lt;em&gt;Analysis of Linear Partial Differential Operators&lt;/em&gt;; but presumably some versions go back earlier) which states that:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Theorem&lt;/strong&gt; Let $L$ be a linear partial differential operator with smooth coefficients of order $m$ on $\mathbb{R}^{1+n}$. Consider the Cauchy problem for $Lu = F$ on the upper-half-space $\{ x_0 \geq 0 \}$ with initial data $u_0, u_1, \ldots, u_{m-1}$ (such that $(\partial_0)^k u|_{x_0 = 0} = u_k$ ). Then the following are equivalent. (a) The Cauchy problem has a unique smooth solution $u$ for every prescribed smooth data $u_0,\ldots,u_{m-1}$ and source $F$ and (b) $L$ is a hyperbolic operator.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So in particular, in general there cannot be well-defined solutions to the initial value problem for ultrahyperbolic PDEs. (They either don't exist or aren't unique. And in the case you do have a solution, the solution is unstable.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, you cannot even construct approximate solutions to the initial value problem reliably using numerical methods, since ultrahyperbolic equations do not have finite-speed of propagation, so you cannot &quot;localise&quot; the problem, and small changes around a point $x$ may almost instantaneously affect the solution at a far-away point $y$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In certain special cases you can produce some semblance of a solution. In the constant coefficient case in second order, where the equation can be written as $(\triangle_X - \triangle_Y)u = 0$, you have what is known as Asgeirsson's Mean Value Theorem which is sort of a generalisation of the mean value theorem for harmonic functions, and also a generalisation for the Green's function formula for the linear, constant coefficient wave equation. In this particular case you can also consider solutions using Fourier analytic methods. From there one sees that if one were to assume certain restrictions on the allowed wave-numbers (which leads to a non-local constraint on the initial data), one can &lt;a href=&quot;http://dx.doi.org/10.1098/rspa.2009.0097&quot; rel=&quot;nofollow&quot;&gt;recover well-posedness of the initial value problem&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As to references, perhaps an easy way to do a reverse search on the classical paper of Fritz John &lt;a href=&quot;http://projecteuclid.org/euclid.dmj/1077490637&quot; rel=&quot;nofollow&quot;&gt;on the subject&lt;/a&gt;. John's various textbooks also contain some information about it; in particular you may want to consult his &lt;em&gt;Partial Differential Equations&lt;/em&gt; book, and his book on &lt;em&gt;Plane Waves and Spherical Means Applied to Partial Differential Equations&lt;/em&gt;. &lt;/p&gt;&#xA;" OwnerUserId="1543" LastActivityDate="2010-12-17T12:47:17.690" />
  <row Id="14664" PostTypeId="2" ParentId="14660" CreationDate="2010-12-17T12:57:47.660" Score="9" ViewCount="" Body="&lt;p&gt;Every nested partition defines a full binary tree (all nodes have degree 0 or 2) with $n$ leaves (the 1s) and $n-1$ internal nodes (the binary operators). The number of different full binary trees with $n+1$ leaves is related to the $n$-th &lt;a href=&quot;http://en.wikipedia.org/wiki/Catalan_number&quot; rel=&quot;nofollow&quot; title=&quot;Catalan Number&quot;&gt;Catalan Number&lt;/a&gt;. This is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$C_n = \frac{1}{n+1} {2n \choose n}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Catalan Numbers occur in many counting problems (see the wikipedia article for more details).&lt;/p&gt;&#xA;" OwnerUserId="4857" LastActivityDate="2010-12-17T12:57:47.660" />
  <row Id="14665" PostTypeId="2" ParentId="14653" CreationDate="2010-12-17T12:58:53.483" Score="3" ViewCount="" Body="&lt;p&gt;First, as Aaron mentioned, a homomorphism from $M$ to $N$ is uniquely defined by its image on the generator 1 of $M$, and it must commute with the action $x\cdot 1=y\cdot 1=0$. In particular, if $f$ is such a homomorphism and $\bar{f(1)}$ is a coset representative of $f(1)$, then you must have $x\cdot \bar{f(1)} \in (x-1)$. Since $\mathbb{C}[x,y]$ is a UFD and $x$ and $x-1$ are both irreducible, this implies that $\bar{f(1)}\in (x-1)$, i.e. that $f(1) = 0\in N$. So, that hom-space is 0, and that's good news for the next computation (see below).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for your computation of $\text{Ext}^1$, you have forgotten that $N\mapsto \text{Hom}(,N)$ is a&#xA;contraviariant functor and you have to turn your long exact sequence around accordingly (indeed,&#xA;there is no obvious way of defining the map $\text{Hom}(K,N)\rightarrow \text{Hom}(\mathbb{C}[x,y],N)$,&#xA;while it's perfectly clear how to define the map the other way round - namely by restriction; similarly $\text{Hom}(M,N)\rightarrow \text{Hom}(\mathbb{C}[x,y],N)$ should be defined by composing homs with projection). See if that gets you anywhere&#xA;and feel free to report back.&lt;/p&gt;&#xA;" OwnerUserId="3212" LastActivityDate="2010-12-17T12:58:53.483" />
  <row Id="14666" PostTypeId="1" CreationDate="2010-12-17T13:05:03.393" Score="6" ViewCount="137" Body="&lt;p&gt;I am trying to figure out how many permutations exist in a set where none of the numbers equal their own position in the set; for example, 3,1,5,2,4 is an acceptable permutation where 3,1,2,4,5 is not because 5 is in position 5.  I know that the number of total permutations is n!. Is there a formula for how many are acceptable given the case that no position holds its own number?&lt;/p&gt;&#xA;" OwnerUserId="4858" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-17T14:31:46.657" LastActivityDate="2010-12-17T14:31:46.657" Title="Number of permutations where n ≠ position n" Tags="&lt;combinatorics&gt;&lt;permutations&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="1" />
  <row Id="14667" PostTypeId="1" AcceptedAnswerId="14680" CreationDate="2010-12-17T13:09:08.610" Score="2" ViewCount="161" Body="&lt;p&gt;Suggest an algorithm for factorizing an integer such that each of its factor is &lt;a href=&quot;http://en.wikipedia.org/wiki/Square-free_integer&quot; rel=&quot;nofollow&quot;&gt;square-free&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks,&lt;/p&gt;&#xA;" OwnerUserId="2109" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-17T17:08:01.613" LastActivityDate="2010-12-17T17:08:01.613" Title="Square-free Integers Factorization" Tags="&lt;number-theory&gt;&lt;algorithms&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="14668" PostTypeId="2" ParentId="14666" CreationDate="2010-12-17T13:16:54.020" Score="7" ViewCount="" Body="&lt;p&gt;Check out &lt;a href=&quot;http://en.wikipedia.org/wiki/Derangement&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Derangement&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1609" LastActivityDate="2010-12-17T13:16:54.020" />
  <row Id="14669" PostTypeId="2" ParentId="14666" CreationDate="2010-12-17T13:19:41.723" Score="8" ViewCount="" Body="&lt;p&gt;What you are looking for is known as &lt;a href=&quot;http://en.wikipedia.org/wiki/Derangement&quot; rel=&quot;nofollow&quot;&gt;derangement&lt;/a&gt;.However,for counting the number of derangement for say $n$-elements you could possible use a trick,compute$\frac{n!}{e}$ and then round off to an integer and this will give you the desired result. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is actually another application of $e$,which was discovered by Jacob Bernoulli in the problem of derangement,also known as the &lt;a href=&quot;http://www.jstor.org/pss/2315337&quot; rel=&quot;nofollow&quot;&gt;hat check problem&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2109" LastEditorUserId="2109" LastEditorDisplayName="" LastEditDate="2010-12-17T13:41:13.947" LastActivityDate="2010-12-17T13:41:13.947" />
  <row Id="14670" PostTypeId="1" CreationDate="2010-12-17T13:23:46.963" Score="0" ViewCount="73" Body="&lt;p&gt;Hello! I would like to know a way to check if my version of Matlab support object oriented programming.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By, the way this question is related with other I asked before in a stackexchange site.&#xA;&lt;a href=&quot;http://stackoverflow.com/questions/4462441/matlab-creating-customized-numeric-data-types&quot;&gt;Matlab: Creating customized numeric data types&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you!&lt;/p&gt;&#xA;" OwnerUserId="4255" LastActivityDate="2010-12-17T13:23:46.963" ClosedDate="2010-12-17T15:01:40.127" Title="How can I check if my version of matlab support object oriented programming?" Tags="&lt;matlab&gt;" CommentCount="6" />
  <row Id="14671" PostTypeId="1" AcceptedAnswerId="14684" CreationDate="2010-12-17T13:34:18.817" Score="3" ViewCount="125" Body="&lt;p&gt;Recall that complex topological $K$-theory is representable on reasonable spaces by the space $BU \times \mathbb{Z}$ (where $BU$ is a colimit of various infinite Grassmannians), and that the total Chern class provides  a natural map $\mathrm{Vect}(B) \to H^*(B)$ for every such space $B$. By the multiplicativity property, this map factors through the K-group and leads to a natural transformation $K(B) \to H^*(B)$. $H^*(B)$ is also representable by a product of Eilenberg-MacLane spaces $K(\mathbb{Z}, n)$ over all $n$. There is thus a map, unique up to homotopy &#xA;$$BU \times\mathbb{Z} \to \prod_n K(\mathbb{Z},n).$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is this map?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As Mariano observes, one can simply define the individual Chern classes on the $K$-group as well, albeit not immediately through the universal property, so the question reduces to the determination of the (homotopy class of the) map $BU \times \mathbb{Z} \to K(\mathbb{Z},n)$ induced by each Chern class.&lt;/p&gt;&#xA;" OwnerUserId="536" LastEditorUserId="536" LastEditorDisplayName="" LastEditDate="2010-12-17T15:29:45.497" LastActivityDate="2010-12-17T19:58:39.387" Title="What's the map $BU \times \mathbb{Z} \to \prod K(\mathbb{Z},n)$ representing the total Chern class?" Tags="&lt;algebraic-topology&gt;&lt;k-theory&gt;&lt;characteristic-classes&gt;" AnswerCount="1" CommentCount="12" FavoriteCount="1" />
  <row Id="14672" PostTypeId="1" CreationDate="2010-12-17T13:37:50.330" Score="4" ViewCount="162" Body="&lt;p&gt;Hi,&lt;br /&gt;&#xA;Given two positive rational number $\frac{a_1}{b_1}$ and $\frac{a_3}{b_3}$ (written in lowest terms) such that $$\frac{a_1}{b_1} &amp;lt; \frac{a_3}{b_3},$$&#xD;&#xA;I want to find a rational number&#xA;$\frac{a_2}{b_2}$ such that $$\frac{a_1}{b_1} &amp;lt; \frac{a_2}{b_2} &amp;lt; \frac{a_3}{b_3}$$&#xD;&#xA;and $a_2$ and $b_2$ are the smallest possible numbers&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've come with the solution $a_2 = a_1 + a_3$ and $b_2 = b_1 + b_3$ then dividing $a_2$ and $b_2$ by their greatest common divisor but there must be better solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks by advance for your suggestions.&lt;/p&gt;&#xA;" OwnerUserId="4861" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-17T19:56:39.953" LastActivityDate="2010-12-18T07:36:31.947" Title="Computing a rational number between two others, minimizing numerator and denominator" Tags="&lt;number-theory&gt;&lt;fractions&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="14673" PostTypeId="2" ParentId="14672" CreationDate="2010-12-17T13:57:24.657" Score="3" ViewCount="" Body="&lt;p&gt;See Wikipedia's article on the &lt;a href=&quot;http://en.wikipedia.org/wiki/Mediant_%28mathematics%29&quot; rel=&quot;nofollow&quot;&gt;mediant&lt;/a&gt;, which explains when your solution is the best.&lt;/p&gt;&#xA;" OwnerUserId="589" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-12-17T16:03:10.700" LastActivityDate="2010-12-17T16:03:10.700" CommentCount="2" />
  <row Id="14674" PostTypeId="2" ParentId="14646" CreationDate="2010-12-17T14:39:25.303" Score="1" ViewCount="" Body="&lt;p&gt;We start with a square matrix $M$ of dimension $n$. We can think of its $n$ columns as vectors in $\mathbb{R}^n$. We consider the hyperplane generated by the first column (for example the orthogonal complement of that vector). Next, we reflect each of the columns about this hyperplane. In symbols: $H_1M= [ H_1(v_1) \ldots H_1(v_n)]$, where on the RHS we use functional notation for $H_1$. Now, because $v_1$ is normal to the hyperplane, $H_1(v_1)$ looks simple. The rest of the vectors transform like: &lt;img src=&quot;http://i.imgur.com/Arm68.png&quot; alt=&quot;alt text&quot;&gt;     &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is we subtract twice their projections onto $v_1$ (this gives me the formula for householder reflections).&#xA;Then we consider the $n-1$ dimensional submatrix of $H_1M:=M_2$, and repeat. The submatrix takes me into the hyperplane, since the first reflection leaves that plane invariant. What we are doing is changing the basis (since Reflections have $det \neq 0$) of the underlying space progressively so that the vectors have a nice representation (Thats what QR decomposition is, The Q contains the orthonormal vectors, while the R tracks all the changes we have made).  &lt;/p&gt;&#xA;" OwnerUserId="4806" LastActivityDate="2010-12-17T14:39:25.303" />
  <row Id="14675" PostTypeId="1" AcceptedAnswerId="14676" CreationDate="2010-12-17T14:44:52.280" Score="1" ViewCount="39" Body="&lt;p&gt;I'm trying to compute the variational distance between two negative exponential distributions. If $a$ denotes the value of $x$ where both curves are equal, then:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_0^{+\infty}\left|\lambda_1e^{-\lambda_1x}-\lambda_2e^{-\lambda_2x}\right|dx$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$=-\int_{0}^a(\lambda_1e^{-\lambda_1x}-\lambda_2e^{-\lambda_2x})dx+\int^{+\infty}_a(\lambda_1e^{-\lambda_1x}-\lambda_2e^{-\lambda_2x})dx$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$=\left[e^{-\lambda_2x} - e^{-\lambda_1x}\right]_{0}^a +\left[e^{-\lambda_1x} - e^{-\lambda_2x}\right]^{+\infty}_a$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$=e^{-\lambda_2a} - e^{-\lambda_1a} -1 +1  +e^{-\infty} - e^{-\infty}-e^{-\lambda_1a}+e^{-\lambda_2a}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$=2\left(e^{-\lambda_2a}-e^{-\lambda_1a}\right).$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Computing $a$ is not difficult:&#xA;\begin{eqnarray*}&#xA; \lambda_1e^{-\lambda_1x}=\lambda_2e^{-\lambda_2x} \Leftrightarrow\ln(\lambda_1)-\lambda_1x&amp;amp;=&amp;amp;\ln(\lambda_2)-\lambda_2x\&#xA; \Leftrightarrow x&amp;amp;=&amp;amp;\frac{\ln(\lambda_2)-\ln(\lambda_1)}{\lambda_2-\lambda_1}.&#xA;\end{eqnarray*}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, what puzzles me is that I am integrating a positive function, yet depending on how parameters are related to each other, the value of the above result may be negative ... what am I missing?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(sorry for the messy layout, it seems that my &amp;amp;'s are ignored ...)&lt;/p&gt;&#xA;" OwnerUserId="4863" LastActivityDate="2010-12-17T15:17:59.317" Title="Variational distance between negative exponential distributions" Tags="&lt;integral&gt;" AnswerCount="1" />
  <row Id="14676" PostTypeId="2" ParentId="14675" CreationDate="2010-12-17T15:17:59.317" Score="3" ViewCount="" Body="&lt;p&gt;In the change from the first line of your equation to the second line, you implicitly used the fact that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \lambda_1 e^{-\lambda_1x} &amp;gt; \lambda_2 e^{-\lambda_2x} $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for $x &amp;gt; a$. (Which is equivalent to assuming $0 &amp;lt; \lambda_1 &amp;lt; \lambda_2$.) You'll see that the parameter ranges that gives you a negative answer from your computation precisely corresponds to the case where the above assumption is false&lt;/p&gt;&#xA;" OwnerUserId="1543" LastActivityDate="2010-12-17T15:17:59.317" CommentCount="1" />
  <row Id="14677" PostTypeId="2" ParentId="14654" CreationDate="2010-12-17T15:34:27.120" Score="4" ViewCount="" Body="&lt;p&gt;Let $A_n(x) = n(x^{1 \over n} - 1)$. Then if $B_n$ denotes $A_{2^{n}}$, then $B_n(x) = 2^n(x^{1 \over 2^n} - 1)$, and $\lim_{n \rightarrow \infty} B_n(x)$ equals $\ln(x)$ because $B_n(x)$ is a subsequence of $A_n(x)$. Note that &#xA;$${B_n(x) \over B_{n-1}(x)} = {2^n(x^{1 \over 2^n} - 1)   \over 2^{n-1}(x^{1 \over 2^{n-1}} - 1)}$$&#xD;&#xA;$$= {2 (x^{1 \over 2^n} - 1) \over (x^{1 \over 2^n} - 1)(x^{1 \over 2^n} + 1)}$$&#xD;&#xA;$$= {2 \over x^{1 \over 2^n} + 1}$$&#xD;&#xA;So $B_n$ can be written as a finite telescoping product&#xA;$$B_n = (x- 1)\prod_{i = 1}^{n} {2 \over x^{1 \over 2^i} + 1}$$&#xD;&#xA;So taking the limits as $n$ goes to $\infty$ one has&#xA;$$\ln(x) = (x-1)\prod_{i = 1}^{\infty} {2 \over x^{1 \over 2^i} + 1}$$&#xD;&#xA;So at least a subsequence of the partial products gives a telescoping product.&lt;/p&gt;&#xA;" OwnerUserId="3035" LastActivityDate="2010-12-17T15:34:27.120" />
  <row Id="14678" PostTypeId="2" ParentId="14672" CreationDate="2010-12-17T15:46:11.297" Score="1" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\rm\displaystyle\quad \frac{A}B &amp;lt; \frac{X}Y &amp;lt; \frac{C}D\ \iff\ X = \frac{MA+NC}{BC-AD},\ Y = \frac{MB+ND}{BC-AD}\quad$ for some integers $\rm\:M,N&amp;gt;0$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This has a nice geometric interpretation in terms of plane lattices - which provides a vivid geometric interpretation of good approximations by Farey mediant searching - see my &lt;a href=&quot;http://groups.google.com/groups?selm=y8zpsy7ixpf.fsf@nestle.csail.mit.edu&quot; rel=&quot;nofollow&quot;&gt;post here&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-12-17T15:46:11.297" />
  <row Id="14679" PostTypeId="1" AcceptedAnswerId="14681" CreationDate="2010-12-17T15:57:18.877" Score="2" ViewCount="221" Body="&lt;p&gt;Let $f(x)$ be two times differentiable in $[a, b]$. The Mean Value Theorem tells us that there's such a point $a\lt c\lt b$, such that ${f}&amp;#39;(c)=\frac{f(b)-f(a)}{b-a}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $\forall x \in [a, b], {f}&amp;#39;&amp;#39;(x)\gt 0$ and $f&amp;#39;&amp;#39;(x)$ is strictly increasing then $c\gt \frac{a+b}{2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm having a hard time figuring this one out. More so, I can't seem to understand why this is true, intuitively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried the following:&#xA;Write $f(x)$ Taylor's expansion of degree 1, around $a$. Then, for $x=b$ we get: $f(b)-f(a)=f&amp;#39;(a)(b-a)+\frac{1}{2}f&amp;#39;&amp;#39;(\xi)(b-a)^2 \implies f&amp;#39;(c)=f&amp;#39;(a)+\frac{1}{2}f&amp;#39;&amp;#39;(\xi)(b-a)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I still haven't used the fact that $f&amp;#39;&amp;#39;(x)$ is positive and strictly increasing in $[a, b]$ (which btw also implies that $f&amp;#39;(x)$ is strictly increasing). I tried playing with the expression above but I just don't see which direction to go.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please don't give a full solution. Hints and intuition are appreciated.&lt;/p&gt;&#xA;" OwnerUserId="3237" LastEditorUserId="3237" LastEditorDisplayName="" LastEditDate="2010-12-17T16:06:56.747" LastActivityDate="2010-12-18T08:33:46.127" Title="Mean value theorem and a strictly increasing, positive second derivative" Tags="&lt;calculus&gt;&lt;homework&gt;&lt;taylor-expansion&gt;" AnswerCount="3" />
  <row Id="14680" PostTypeId="2" ParentId="14667" CreationDate="2010-12-17T16:03:10.540" Score="5" ViewCount="" Body="&lt;p&gt;It's trivial given the complete factorization into primes. Currently we do not know any other way to compute squarefree parts and it is widely suspected that it cannot be computed in any simpler way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This problem is important because one of the main tasks of computational algebraic number theory reduces to it (in deterministic polynomial time). Namely the problem of computing the ring of integers of an algebraic number field depends upon the square-free decomposition of the polynomial discriminant when computing an integral basis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Contrast this difficulty with the trivial squarefree decomposition of polynomials by way of gcd with its derivative. The availability of derivatives for polynomials opens up a powerful toolbox that is not available for integers. For example once derivatives are available so are Wronskians - which provide powerful measures of dependence in transcendence theory and diophantine approximation. A simple yet stunning example is the elementary proof of the polynomial case of Mason's ABC theorem, which yields as a very special case a high-school-level proof of FLT for polynomials.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For references see my &lt;a href=&quot;http://math.stackexchange.com/questions/2239/importance-of-determining-whether-a-number-is-squarefree-using-geometry/2249#2249&quot;&gt;post here.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-12-17T16:03:10.540" />
  <row Id="14681" PostTypeId="2" ParentId="14679" CreationDate="2010-12-17T16:07:06.463" Score="1" ViewCount="" Body="&lt;p&gt;Try sketching a graph, since you are looking for intuition only. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A sketch of the f' will be easier as then you can use the Intermediate Value Theorem too. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Say we want to see that the MVT holds for the case of $f(x)$ in [a,b] &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We need to prove that the slope $\frac{f(b)-f(a)}{b-a}$ is inside $[f&amp;#39;(a),f&amp;#39;(b)]$&#xD;&#xA;This is not too difficult to prove.&lt;/p&gt;&#xA;" OwnerUserId="4728" LastEditorUserId="4728" LastEditorDisplayName="" LastEditDate="2010-12-17T19:00:45.303" LastActivityDate="2010-12-17T19:00:45.303" CommentCount="7" />
  <row Id="14682" PostTypeId="2" ParentId="14654" CreationDate="2010-12-17T16:29:56.530" Score="2" ViewCount="" Body="&lt;p&gt;This is a comment on generalities(therefore just community wiki).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For infinite products of expressions of the form $(1 + \mathrm{something})$, it is often more productive to take the logarithm and convert it into an infinite sum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a little bit of an issue of convergence here. If one of the expressions is $0$, then there would be trouble, etc.. A precise handling of such stuff can be found in the book of Ahlfors on Complex Analysis.&lt;/p&gt;&#xA;" OwnerUserId="1119" LastActivityDate="2010-12-17T16:29:56.530" CommunityOwnedDate="2010-12-17T16:29:56.530" />
  <row Id="14684" PostTypeId="2" ParentId="14671" CreationDate="2010-12-17T16:44:42.327" Score="4" ViewCount="" Body="&lt;p&gt;One way, my favorite, of defining the chern classes goes as follows: first compute $H^*(BU; \mathbb{Z})=\mathbb{Z}[c_1, c_2, ...]$ by using your favorite method. I know two methods, one is using some cellular description of the grassmanians (see Milnor and Stasheff), the other is to first compute the cohomology of the lie groups $U(n)$ and then use a path-loop space fibration and the Serre SS to get at the cohomology of $H^*(BU(n);\mathbb{Z})$ (see &lt;a href=&quot;http://math.stackexchange.com/questions/13260/homology-and-euler-characteristics-of-the-classical-lie-groups/13271#13271&quot;&gt;Homology and Euler characteristics of the classical Lie groups&lt;/a&gt;). Next you notice that $BU$ and $BU(n)$ have the same cell structure through a range and so you essentially take the limit (colimit, and there are some subtleties here, the place I recall seeing these is in Jacob Lurie's survey on elliptic cohomology, at the beginning). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next suppose you have a complex vector bundle $\xi : E \to B$ then it is classified by a map $f: B \to BU(n)$ where n is the dimension of $\xi$. You can get a map on all of $BU$ by just adding on trivial bundles to get a map out of $BU(k)$ for all $k$ larger than $n$. This gives a map out of BU (although technically you don't really need this, $BU(n)$ works fine for defining the chern classes of an $n$ dimensional complex vector bundle). Now define $c_n(\xi):=f^*(c_n)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From this perspective we started with the universal case, so maybe it is a bit of a cheat. For me this even clarifies how I should think about characteristic classes in general. Suppose you want to look at $G$-bundles and see what you can tell about them from $E$-theory ($E$ some ring spectrum and $G$ some compact lie group or whatever you need for $BG$ to be nice, I am not sure if there are other restrictions for this to work). Now compute $E^*BG$ and use the fact that $G$-bundles over $X$ are classified by homotopy classes $X \to BG$. You should check out some of the threads on MO about characteristic classes, I think I can learn something from each of Rezk's answers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;please let me know if I can make some of the above clearer or if there are any mistakes.&lt;/p&gt;&#xA;" OwnerUserId="627" LastEditorUserId="627" LastEditorDisplayName="" LastEditDate="2010-12-17T19:58:39.387" LastActivityDate="2010-12-17T19:58:39.387" CommentCount="11" />
  <row Id="14685" PostTypeId="1" CreationDate="2010-12-17T17:15:40.463" Score="1" ViewCount="116" Body="&lt;p&gt;This is a quick (not to scale) drawing of what I mean: the floor gradually slopes from the wall to the drain at about 7/16( 0.4375) inch per foot from the short sides of the wall (the 7 ft from wall to the drain) and exactly 3/16(.1875&quot;) of a inch per foot on the long sides ( the 16ft from the wall to the drain).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the base is a rectangle 14'x32' but the shape is most related to a cone ,.... but all cone formulas assume the base is circular. I'm at a loss of how to pull out only the rectangular portion out from the middle of a cone shape. I hope this made it more visual &lt;img src=&quot;http://i119.photobucket.com/albums/o141/Elder_Paul/area.jpg&quot; alt=&quot;&quot;&gt; it has 3&quot; of Drop in the Center, length is 32' and width is 14'. just looking for a formula to calculate it. Any help in the right direction would be appreciated.&lt;/p&gt;&#xA;" OwnerUserId="4870" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-17T18:15:12.607" LastActivityDate="2010-12-17T19:14:59.990" Title="How do I find the area or the volume (or a formula) for a region that is rectangular but has a cone slope to a drain ?" Tags="&lt;calculus&gt;&lt;trigonometry&gt;" AnswerCount="3" CommentCount="0" />
  <row Id="14686" PostTypeId="1" CreationDate="2010-12-17T17:18:23.787" Score="1" ViewCount="202" Body="&lt;p&gt;Why is integral of $x$ from $-1$ to $1$ $x^2/2$ rather than $x^2$?&lt;/p&gt;&#xA;" OwnerUserId="4871" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-17T20:09:46.770" LastActivityDate="2010-12-18T03:27:09.293" Title="Why is the integral of $x$ from $-1$ to $1$ $x^2/2$ rather than $x^2$?" Tags="&lt;calculus&gt;&lt;integral&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="14687" PostTypeId="1" AcceptedAnswerId="14690" CreationDate="2010-12-17T17:19:34.970" Score="1" ViewCount="126" Body="&lt;p&gt;Suppose $f$ is analytic on the domain $D = \{z \big| |z|&amp;gt;1\}$ in $\mathbb{C}$.   I know that if the line integral of $f$ along any closed path is 0 then $f$ has an anti-derivative.  Is this also a necessary condition.&lt;/p&gt;&#xA;" OwnerUserId="832" LastEditorUserId="832" LastEditorDisplayName="" LastEditDate="2010-12-17T17:26:56.103" LastActivityDate="2010-12-17T17:29:04.120" Title="Necessary condition for existence of anti-derivative" Tags="&lt;complex-analysis&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="14688" PostTypeId="2" ParentId="14686" CreationDate="2010-12-17T17:23:49.733" Score="2" ViewCount="" Body="&lt;p&gt;Integration is the opposite of differentiation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you differentiate  x^2/2 you get x (you bring down the power of 2, which cancels with the 1/2 out front) which means that when you integrate x you must get x^2/2.&lt;/p&gt;&#xA;" OwnerUserId="4873" LastActivityDate="2010-12-17T17:23:49.733" />
  <row Id="14689" PostTypeId="1" CreationDate="2010-12-17T17:27:12.103" Score="7" ViewCount="268" Body="&lt;p&gt;Two players in a dice game to see who can roll a total of 60 first, taking turns each rolling 2 dice. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For one player, one die is 4 sided and one is 6 sided. Therefore the average roll for this player will be a 6. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other player has two 6 sided dice. So this players average roll will be a 7. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If they take turns, each rolling both dice, what % of the time will the first player (a 4-6 dice combo) get to 60 first and what % of the time will the second player (6-6 dice combo) get to 60 first. If they get to 60 on the same roll, the player who goes the furthest wins.&lt;/p&gt;&#xA;" OwnerUserId="4872" LastActivityDate="2010-12-20T00:34:01.337" Title="Thought experiment for dice game.." Tags="&lt;probability&gt;&lt;probability-theory&gt;" AnswerCount="3" CommentCount="5" FavoriteCount="1" />
  <row Id="14690" PostTypeId="2" ParentId="14687" CreationDate="2010-12-17T17:29:04.120" Score="3" ViewCount="" Body="&lt;p&gt;Yes.  If $F&amp;#39;=f$ and $\gamma:[a,b]\to D$ is a curve, then $\int_\gamma f = F(\gamma(b))-F(\gamma(a))$.  For example, see Theorem 1.18 on &lt;a href=&quot;http://books.google.com/books?id=9LtfZr1snG0C&amp;amp;lpg=PP1&amp;amp;ots=OePFw5q3Ex&amp;amp;dq=conway%20complex&amp;amp;pg=PA65#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;page 65&lt;/a&gt; of John B. Conway's &lt;em&gt;Functions of one complex variable&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-12-17T17:29:04.120" />
  <row Id="14691" PostTypeId="2" ParentId="14679" CreationDate="2010-12-17T17:31:44.477" Score="0" ViewCount="" Body="&lt;p&gt;I prefer words, really. What the mean value theorem tells us is that at some point on the interval $[a,b]$ the function $f(x)$ (continuous and differentiable on this interval) has a gradient equal to the slope of the line connecting $a$ and $b$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why should this be? Well, the trivial case we can handle, namely that if $f\prime(x) = k$ some constant and passes through $a,b$ then k can only be that slope.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So let's take your case. $f\prime\prime(x) &amp;gt; 0$ and strictly increasing, so the rate of change of gradient is positive and going up. So this means the gradient is increasing in increasingly big chunks. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sketch what that means, then take the line from $a$ to $b$ and find out where it hits your sketched curve. Why can it not appear before $\frac{a+b}{2}$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To ask a different question - the rate of change of gradient is strictly increasing. Given then that the gradient between $a,b$ gets bigger in bigger chunks, where do you expect the gradient to change most between? If I said compare this to something like a probability distribution, would it make sense why it's called the mean value theorem?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does that help at all?&lt;/p&gt;&#xA;" OwnerUserId="892" LastActivityDate="2010-12-17T17:31:44.477" />
  <row Id="14692" PostTypeId="2" ParentId="14613" CreationDate="2010-12-17T17:35:00.883" Score="1" ViewCount="" Body="&lt;p&gt;Here's something off the top of my head.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Pick a statistic for your data which might measure how correlated they are (traditional notions of correlation don't apply here, because you don't have any observations -- you only have trade times). For example, you might calculate the average difference between each point in series A and its closest point in series B.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Generate some new data from a known distribution. In your example, you might generate series A' as five random draws from a uniform distribution and series B' as three random draws from a uniform distribution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) Compute your chosen statistic on your artificial data. The idea is to work out how the statistic behaves on 'random' data with no correlation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4) Repeat steps 2 and 3 until you have a good idea what the distribution of your statistic is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;5) Compare your result to the generated distribution. If it's lying way out in the tails, you can be relatively certain that there is significant correlation between your two strategies. If it lies near the center of the distribution, then you can't reject the hypothesis that there is no correlation.&lt;/p&gt;&#xA;" OwnerUserId="4873" LastActivityDate="2010-12-17T17:35:00.883" />
  <row Id="14693" PostTypeId="1" CreationDate="2010-12-17T17:44:24.470" Score="3" ViewCount="186" Body="&lt;p&gt;I have this equation $3-x-x^2$ which I want to convert into the form $a^2 - y^2$, so that then I can suppose $y$ = $a\sin\theta$, differentiate and then integrate it. But I can't convert that equation into this $a^2 - y^2$ form.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S. this is a homework question. My question here is not to solve this for me, but tell me how to do such (if there exists a formula) so that I may learn and prosper :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt; $a^2 - y^2$, in this equation $y$ may be anything which involves $x$ e.g. $(x-2)^2$&lt;/p&gt;&#xA;" OwnerUserId="4875" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-17T19:40:03.017" LastActivityDate="2010-12-17T20:19:39.597" Title="Rewrite a general quadratic as $a^2-x^2$" Tags="&lt;homework&gt;&lt;algebra&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="14694" PostTypeId="2" ParentId="14693" CreationDate="2010-12-17T17:55:17.950" Score="11" ViewCount="" Body="&lt;p&gt;What you are trying to do is called &lt;a href=&quot;http://en.wikipedia.org/wiki/Completing_the_square&quot; rel=&quot;nofollow&quot;&gt;completing the square&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-12-17T17:55:17.950" CommunityOwnedDate="2010-12-17T17:55:17.950" />
  <row Id="14696" PostTypeId="2" ParentId="14686" CreationDate="2010-12-17T17:59:02.580" Score="8" ViewCount="" Body="&lt;p&gt;Consider $f(x)=x$ for x equals 0 to a, it is a right-angled triangle (the blue region in figure below). &#xA;&lt;img src=&quot;http://i.imgur.com/M8vb6.jpg&quot; alt=&quot;alt text&quot;&gt;&#xA;The area of the square (having sides equal a) is $a^2$. Since there are two identical triangles each filling half the area, the area under the curve of f(x)=x for x=0 to a is $\frac{a^2}{2}$, which is precisely what you get from computing the definite integral, since a definite integral calculates area under the curve.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another example, consider f(x)=x from a to b, with a&gt;=0. The definite integral from a to b calculated the area of the blue region illustrated below. It can be reduced to the previous case as the area of the bigger triangle from 0 to b minus the area of the smaller triangle from 0 to a. Hence $\frac{b^2}{2} -\frac{a^2}{2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/KnyNM.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps I should also mention, when the area is below the x-axis, such as f(x) for x=a to 0 (a&amp;lt;0), a change of sign is needed, i.e $\frac{-a^2}{2}$.&lt;/p&gt;&#xA;" OwnerUserId="1746" LastEditorUserId="1746" LastEditorDisplayName="" LastEditDate="2010-12-17T18:40:33.527" LastActivityDate="2010-12-17T18:40:33.527" />
  <row Id="14697" PostTypeId="1" AcceptedAnswerId="14700" CreationDate="2010-12-17T18:02:21.377" Score="4" ViewCount="199" Body="&lt;p&gt;How to prove $\{t\} \notin t$ using the axiom of foundation (aka axiom of regularity):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$A = \emptyset \vee \exists x \in A \forall y \in x : y \not\in A$&lt;/p&gt;&#xA;" OwnerUserId="4876" LastActivityDate="2010-12-17T18:12:45.700" Title="How to prove $\{t\} \notin t$" Tags="&lt;logic&gt;&lt;set-theory&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" />
  <row Id="14698" PostTypeId="2" ParentId="14693" CreationDate="2010-12-17T18:06:02.667" Score="6" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\rm\quad A^2 - (X-B)^2\ =\ 3 - X - X^2\ \Rightarrow\ B = \ldots\ \Rightarrow\ A = \ldots\ $ (expand and compare coeff's)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Generally if you want to check that some specific polynomial is an instance of some general polynomial form  you may treat the coefficients of the general form as variables or indeterminates (a.k.a. &quot;undetermined coefficients&quot;). With luck, the system of equations obtained by comparing coefficients might be easily solvable (e.g. it might be triangular). This is sometimes referred to as the &lt;em&gt;method of undetermined coefficients&lt;/em&gt; and it works more generally for objects that are uniquely determined by their &quot;coefficients&quot;, e.g. vectors, matrices, (formal) power series, radix notation, partial fraction expansions, etc. &lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-12-17T20:19:39.597" LastActivityDate="2010-12-17T20:19:39.597" CommentCount="2" />
  <row Id="14699" PostTypeId="2" ParentId="14672" CreationDate="2010-12-17T18:08:37.467" Score="2" ViewCount="" Body="&lt;p&gt;Your solution is not the best. (Take, for instance, 1/2 and 9/10.) One way to solve your problem is to calculate the &lt;a href=&quot;http://en.wikipedia.org/wiki/Continued_fraction&quot; rel=&quot;nofollow&quot;&gt;continued fractions&lt;/a&gt; of the two numbers until they differ. Say the continued fractions start $[x_0;x_1,x_2,x_3]$ and $[x_0;x_1,x_2,y_3]$, with $x_3 \ne y_3$. Then you can just try all values $z_3$ between $x_3$ and $y_3$ inclusive to get your answer. There might be a cleverer way to find the best $z_3$, but this way should work: it relies on the fact that continued-fraction convergents are alternately greater and less than the value that they are approximating, so one of $[x_0;x_1,x_2,x_3]$ and $[x_0;x_1,x_2,y_3]$ will lie between $a1/b1$ and $a3/b3$.&lt;br&gt;&#xA;There is a slight complication: suppose $x_3$ and $y_3$ differ by one, and one of the convergents is exact. Then you might not be able to find $z_3$ such that $[x_0;x_1,x_2,z_3]$ is &lt;em&gt;strictly&lt;/em&gt; between $a1/b1$ and $a3/b3$. But in this case you can use your formula on the two convergents (instead of on $a1/b1$ and $a3/b3$) to get the answer, because the condition that $x_3$ and $y_3$ differ by one is sufficient to guarantee that your mediant is the best approximation.&lt;br&gt;&#xA;For example, 1/2=[0;2] and 9/10=[0;1,9]. So they differ at $x_1$, which means we should try [0;1] = 1 and [0;2] = 1/2. Neither is satisfactory, so we take their mediant, which is 2/3.&lt;/p&gt;&#xA;" OwnerUserId="1508" LastEditorUserId="1508" LastEditorDisplayName="" LastEditDate="2010-12-18T07:36:31.947" LastActivityDate="2010-12-18T07:36:31.947" />
  <row Id="14700" PostTypeId="2" ParentId="14697" CreationDate="2010-12-17T18:12:45.700" Score="14" ViewCount="" Body="&lt;p&gt;If you are okay with Choice, the Axiom of Foundation implies that there can be no infinite descending chain of sets ordered by $\in$. You would have $t\in\{t\}\in t$, leading to such an infinite chain, and thus contradicting AF.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But you can do it directly. Let $t$ be a set, and suppose that $\{t\}\in t$. Let&#xA;$$X = \{t,\{t\}\}.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, by Foundation there exists $x\in X$ such that $x\cap X=\emptyset$. So either $t\cap X=\emptyset$ or $\{t\}\cap X=\emptyset$. Now, $t\in \{t\}\cap X$, so the latter cannot hold. And since $\{t\}\in t$ and $\{t\}\in X$, then $\{t\}\in t\cap X\neq\emptyset$. So neither holds, hence $X$ contradicts Foundation.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-12-17T18:12:45.700" CommentCount="1" />
  <row Id="14701" PostTypeId="2" ParentId="14689" CreationDate="2010-12-17T18:19:49.413" Score="6" ViewCount="" Body="&lt;p&gt;A rough answer, from Monte Carlo simulation, is:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;&quot;Player 1 wins  83.85 %, Player 2 wins  14.28 %, and  1.87 % are draws.&quot;&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The following R code generates this output:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;N &amp;lt;- 60&#xA;&#xA;p1Wins  &amp;lt;- 0&#xA;p2Wins  &amp;lt;- 0&#xA;draws   &amp;lt;- 0&#xA;&#xA;n.sim   &amp;lt;- 10000&#xA;&#xA;p1Roll &amp;lt;- function() {&#xA;    d1 &amp;lt;- sample(1:6,1)&#xA;    d2 &amp;lt;- sample(1:6,1)&#xA;    return(d1+d2)&#xA;}&#xA;&#xA;p2Roll &amp;lt;- function() {&#xA;    d1 &amp;lt;- sample(1:6,1)&#xA;    d2 &amp;lt;- sample(1:4,1)&#xA;    return(d1+d2)&#xA;}&#xA;&#xA;&#xA;for (i in 1:n.sim) {&#xA;    p1Total &amp;lt;- 0&#xA;    p2Total &amp;lt;- 0&#xA;    while (p1Total &amp;lt; 60 &amp;amp; p2Total &amp;lt; 60) {&#xA;        p1Total &amp;lt;- p1Total + p1Roll()&#xA;        p2Total &amp;lt;- p2Total + p2Roll()&#xA;    }&#xA;    if (p1Total &amp;gt; p2Total) {&#xA;        p1Wins &amp;lt;- p1Wins + 1&#xA;    } else&#xA;    if (p2Total &amp;gt; p1Total) {&#xA;        p2Wins &amp;lt;- p2Wins + 1&#xA;    } else {&#xA;        draws &amp;lt;- draws + 1&#xA;    }&#xA;}&#xA;&#xA;print(paste(&quot;Player 1 wins &quot;,p1Wins/n.sim*100,&quot;%, Player 2 wins &quot;,p2Wins/n.sim*100,&quot;%, and &quot;,draws/n.sim*100,&quot;% are draws.&quot;))&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="4873" LastActivityDate="2010-12-17T18:19:49.413" CommentCount="3" />
  <row Id="14702" PostTypeId="2" ParentId="14685" CreationDate="2010-12-17T18:21:06.087" Score="-1" ViewCount="" Body="&lt;p&gt;I am not completely sure about the actual geometry of the problem, so I'll make a few assumptions. Correct me if the assumptions are wrong.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;You have a rectangle of 168&quot; by 384&quot;. &lt;/li&gt;&#xA;&lt;li&gt;The center of the rectangle is depressed by 3&quot;&lt;/li&gt;&#xA;&lt;li&gt;(most importantly) The shape is such that if you draw a line segment from the &quot;drain&quot; to the &quot;wall&quot;, it will be straight (not curved in the vertical direction). &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The third assumption may be different from what you want, since this assumption will require there be something like a &quot;fold/corner&quot; along the diagonals. To get more precise answers, you need to specify exactly what the shape of the floor is (how it curves towards the drain). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then in this case you don't need calculus at all. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each &quot;trianglular piece&quot; in your drawing can be computed directly. For the triangles with base on the short sides, the base is 168&quot;. The &quot;height&quot; can be found using Pythagorean theorem applied to the vertical dip:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$h^2 = 192^2 + 3^2 \implies h \sim 192.0234 $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so the area of the triangle piece is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ A = h \times 168 \sim 16130.0 \textrm{sq in}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly the area of the triangle piece based on the long side of the wall is roughly 16130.6 sq in. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the total area is roughly 64521 sq in, which is 9 square inches more than if you don't factor in the vertical drop to the drain hole. &lt;/p&gt;&#xA;" OwnerUserId="1543" LastActivityDate="2010-12-17T18:21:06.087" CommentCount="2" />
  <row Id="14703" PostTypeId="2" ParentId="14685" CreationDate="2010-12-17T18:30:58.267" Score="1" ViewCount="" Body="&lt;p&gt;It looks like this is a truncated rectangular pyramid or &lt;em&gt;pyramidal frustum&lt;/em&gt;. This is probably what frustrated your searches.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The volume for a square base is given in &lt;a href=&quot;http://mathworld.wolfram.com/TruncatedSquarePyramid.html&quot; rel=&quot;nofollow&quot;&gt;mathworld&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the rectangular case, you can compute the area of the full pyramid and then of the top that you cut off; the formulas are also available at &lt;a href=&quot;http://mathworld.wolfram.com/RegularPyramid.html&quot; rel=&quot;nofollow&quot;&gt;mathworld&lt;/a&gt;. A quick Google search also provided &lt;a href=&quot;http://wiki.answers.com/Q/Volume_of_a_truncated_rectangular_based_pyramid&quot; rel=&quot;nofollow&quot;&gt;this page.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It can be done with Calculus fairly straightforward (a standard Calculus II exercise/problem); but you don't need it. The egyptians already knew the formula. &lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-12-17T18:30:58.267" CommentCount="1" />
  <row Id="14704" PostTypeId="2" ParentId="9524" CreationDate="2010-12-17T18:33:04.570" Score="2" ViewCount="" Body="&lt;p&gt;Good question, and one that is timely for me as I'm teaching calc II next semester for the first time in several years.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;An application of Taylor series that sometimes comes up in an intro to ODEs course is that of series solutions to differential equations.  I usually don't spend more than a day or two on the topic, but I think the students appreciate seeing how a series representation of a solution can be found in cases that can't be solved by any of the other methods that they've learned.  I especially like the example of Airy's equation&#xA;$$&#xD;&#xA;y&amp;#39;&amp;#39; - t \, y = 0&#xD;&#xA;$$&#xD;&#xA;because it's simple enough that students feel like there ought to be an explicit solution plus the series coefficients are fairly easy to express.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure if this kind of application would be helpful or meaningful for calc II students, but maybe if they've seen some simple differential equations (separation of variables, for example) already?&lt;/p&gt;&#xA;" OwnerUserId="4229" LastActivityDate="2010-12-17T18:33:04.570" CommentCount="1" />
  <row Id="14705" PostTypeId="2" ParentId="14686" CreationDate="2010-12-17T18:36:41.083" Score="12" ViewCount="" Body="&lt;p&gt;First, the integral of $x$ from $-1$ to $1$ is &lt;strong&gt;a number&lt;/strong&gt;, not a function. Definite integrals give you the net signed area between the $X$-axis and the function. So the integral of $x$ from $-1$ to $1$ is &lt;em&gt;neither&lt;/em&gt; $x^2$  &lt;em&gt;nor&lt;/em&gt; $x^2/2$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second, the &lt;em&gt;indefinite&lt;/em&gt; integral of $x$ is a &lt;em&gt;family&lt;/em&gt; of functions, namely, the family of all functions $F(x)$ such that $F&amp;#39;(x)=x$; that is, the family of antiderivatives of $x$. Since the indefinite integral is a &lt;em&gt;family&lt;/em&gt; of functions and not a single function, the indefinite integral of $x$ is &lt;em&gt;neither&lt;/em&gt; $x^2/2$  &lt;em&gt;nor&lt;/em&gt; $x^2$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, by the Fundamental Theorem of Calculus, if you are trying to compute the definite integral of a function $f(x)$ that is continuous on $[a,b]$, and you are lucky enough to have &lt;em&gt;any&lt;/em&gt; antiderivative $F(x)$ of $f(x)$, then you can compute the definite integral as the Total Change of $F(x)$ on $[a,b]$. That is,&#xA;$$\int_a^b f(x)\,dx = F(b)-F(a).$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, in order to compute &#xA;$$\int_{-1}^1 x\,dx$$&#xD;&#xA;you have several choices:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Find the net signed area geometrically. This is very easy in this case, because all you have is two triangles that cancel each other, so the net signed area is $0$. That is, $\int_{-1}^1 x\,dx = 0$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Use the definition of the integral as a limit of Riemann sums. Again, not hard to do: the function $y=x$ is increasing, the left hand and right hand sums are easy to compute, and the limit turns out to be $0$. For example, the left hand Riemann sum with $n$ intervals, each of equal length $\frac{2}{n}$, is:&#xA;\begin{align*}&#xA;\mathrm{LHS}(n) &amp;amp;= \frac{-n}{n}\left(\frac{2}{n}\right) + \frac{-n+2}{n}\left(\frac{2}{n}\right) + \cdots + \frac{n-2}{n}\left(\frac{2}{n}\right)\\&#xA;&amp;amp;= \frac{2}{n^2}\Bigl(-n + (-n+2) + (-n+4)+\cdots+\bigl(-n+(2n-2)\bigr)\Bigr)\\&#xA;&amp;amp;= \frac{2}{n^2}\Bigl( -n(n) + 2\bigl(0 + 1 + 2 + \cdots + (n-1)\bigr)\Bigr)\\&#xA;&amp;amp;= \frac{2}{n^2}(-n^2 + n^2 - n) = \frac{-2n}{n^2} = -\frac{2}{n}.&#xA;\end{align*}&#xA;Thus, &#xA;$$\lim\limits_{n\to\infty}\mathrm{LHS}(n) = \lim_{n\to\infty}\left(-\frac{2}{n}\right) = 0.$$&#xD;&#xA;The right hand sums also converge to $0$, so the integral is equal to $0$. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Use the Fundamental Theorem of Calculus and find some function $F(x)$ whose derivative is $x$ on $[-1,1]$. Now, $F(x)=x^2$  &lt;strong&gt;does not work&lt;/strong&gt; because $F&amp;#39;(x) = 2x\neq x$. But $\mathcal{F}(x) = \frac{1}{2}x^2$  &lt;em&gt;does&lt;/em&gt; work because $\mathcal{F}&amp;#39;(x) = \frac{1}{2}(2x) = x$. That means (by the Constant Function Theorem) that&#xA;$$\int x\,dx = \frac{1}{2}x^2 + C,$$&#xD;&#xA;(the family of functions that are the antiderivatives of $x$ are the vertical translates of $\frac{1}{2}x^2$) so you can use any one of them (say, the one with $C=0$) to compute $\int_{-1}^1x\,dx$. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-18T03:27:09.293" LastActivityDate="2010-12-18T03:27:09.293" />
  <row Id="14706" PostTypeId="1" CreationDate="2010-12-17T19:14:30.613" Score="2" ViewCount="59" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm trying to understand a bit about chern classes/chern characters (in the algebraic setting, for varieties over C say) and was hoping to find some notes describing some well-known properties of ch (e.g. the relation between the support of a coherent sheaf and the component of ch of the same dimension).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm really hoping of something in the middle between Hartshorne's appendix and say Fulton's intersection theory (or Fulton-Lang RR algebra), which concentrates more on coherent sheaves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I know, I'm asking too much)&lt;/p&gt;&#xA;" OwnerUserId="1007" LastActivityDate="2010-12-17T19:14:30.613" CommunityOwnedDate="2010-12-17T19:21:42.383" Title="notes on properties of chern character of coherent sheaves" Tags="&lt;soft-question&gt;&lt;algebraic-geometry&gt;&lt;reference-request&gt;" CommentCount="2" FavoriteCount="1" />
  <row Id="14707" PostTypeId="2" ParentId="14685" CreationDate="2010-12-17T19:14:59.990" Score="1" ViewCount="" Body="&lt;p&gt;The volume of a pyramid is one-third the height times the area of the base. In your case you have sliced off a tiny (approximate) pyramid from the top, to create the drain hole, so subtract that volume from the total.&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-12-17T19:14:59.990" />
  <row Id="14708" PostTypeId="2" ParentId="14679" CreationDate="2010-12-17T19:28:05.547" Score="0" ViewCount="" Body="&lt;p&gt;After changing $f$ by a suitable linear function you may assume $f(a)=f(b)=f&amp;#39;(c)=0$. Expanding $f$ at $c$ by Taylor's theorem you get $f(x)=f(c)+{1\over2}(x-c)^2 f&amp;#39;&amp;#39;(\xi)$ where $\xi$ lies between $c$ and $x$. Now put $x:=a$ and $x:=b$ and draw conclusions.&lt;/p&gt;&#xA;" OwnerUserId="1303" LastEditorUserId="1303" LastEditorDisplayName="" LastEditDate="2010-12-18T08:33:46.127" LastActivityDate="2010-12-18T08:33:46.127" />
  <row Id="14709" PostTypeId="1" AcceptedAnswerId="14710" CreationDate="2010-12-17T19:31:27.710" Score="3" ViewCount="175" Body="&lt;p&gt;Same as title. &#xA;What is the difference between Godel's Completeness and Incompleteness Theorems?&lt;/p&gt;&#xA;" OwnerUserId="4878" LastActivityDate="2010-12-17T19:47:33.163" Title="What is the difference between Godel's Completeness and Incompleteness Theorems?" Tags="&lt;logic&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="14710" PostTypeId="2" ParentId="14709" CreationDate="2010-12-17T19:47:33.163" Score="8" ViewCount="" Body="&lt;p&gt;First, note that, in spite of their names, one is not a negation of the other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The completeness theorem applies to any first order theory: If $T$ is such a theory, and $\phi$ is a sentence (in the same language) and any model of $T$ is a model of $\phi$, then there is a (first-order) &lt;em&gt;proof&lt;/em&gt; of $\phi$ using the statements of $T$ as axioms. One sometimes says this as &quot;anything true is provable.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The incompleteness theorem is more technical. It says that if $T$ is a first-order theory that is:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Recursively enumerable (i.e., there is a computer program that can list the axioms of $T$),&lt;/li&gt;&#xA;&lt;li&gt;Consistent, and &lt;/li&gt;&#xA;&lt;li&gt;Capable of interpreting some amount of Peano arithmetic (typically, one requires the fragment known as Robinson's Q), &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;then $T$ is not complete, i.e., there is at least one sentence $\phi$ in the same language as $T$ such that there is a model of $T$ and $\phi$, and there is also a model of $T$ and $\lnot\phi$. Equivalently (by the completeness theorem), $T$ cannot prove $\phi$ and also $T$ cannot prove $\lnot\phi$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One usually says this as follows: If a theory is reasonable and at least modestly strong, then it is not complete.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second incompleteness theorem is more striking. If in fact we require that $T$ interprets Peano Arithmetic, then in fact $T$ cannot prove its own consistency. So: There is no way of proving the consistency of a reasonably strong mathematical theory, unless we are willing to assume an even stronger setting to carry out the proof. Or: If a reasonably strong theory can prove its own consistency, then it is in fact inconsistent. (Note that any inconsistent theory proves anything, in particular, if its language allows us to formulate this statement, then it can prove that it is consistent). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The requirement that $T$ is recursively enumerable is reasonable, I think. Formally, a theory is just a set of sentences, but we are mostly interested in theories that we can write down or, at least, for which we can recognize whether something is an axiom or not. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The interpretability requirement is usually presented in a more restrictive form, for example, asking that $T$ is a theory about numbers, and it contains Peano Arithmetic. But the version I mentioned applies in more situations; for example, to set theory, which is not strictly speaking a theory about numbers, but can easily interpret number theory. The requirement of interpreting Peano Arithmetic is two-fold. First, we look at theories that allows us (by coding) to carry out at least some amount of common mathematical practice, and number theory is singled out as the usual way of doing that. More significantly, we want some amount of &quot;coding&quot; within the theory to be possible, so we can talk about sentences, and proofs. Number theory allows us to do this easily, and this is way we can talk about &quot;the theory is consistent&quot;, a statement about proofs, although our theory may really be about numbers and not about first order formulas.&lt;/p&gt;&#xA;" OwnerUserId="462" LastActivityDate="2010-12-17T19:47:33.163" CommentCount="1" />
  <row Id="14711" PostTypeId="1" CreationDate="2010-12-17T19:57:28.873" Score="-4" ViewCount="199" Body="&lt;p&gt;In &lt;a href=&quot;http://www.cs.dal.ca/~eem/fea-knuth.pdf&quot; rel=&quot;nofollow&quot;&gt;this article&lt;/a&gt; Knuth talks about the bottom 10 problems rather than top 10. The problems which are not solved but they are most ready to be solved of all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are they?&lt;/p&gt;&#xA;" OwnerUserId="4290" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-17T20:02:23.957" LastActivityDate="2010-12-17T23:10:27.420" ClosedDate="2010-12-18T03:06:13.463" Title="What are the bottom 10 theorems of number theory?" Tags="&lt;number-theory&gt;" AnswerCount="1" CommentCount="10" />
  <row Id="14712" PostTypeId="2" ParentId="14693" CreationDate="2010-12-17T20:13:32.510" Score="0" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Hint&lt;/strong&gt;.  step one is to change the 3 into 3.25  minus   1/4&lt;/p&gt;&#xA;" OwnerUserId="4879" LastActivityDate="2010-12-17T20:13:32.510" />
  <row Id="14713" PostTypeId="2" ParentId="14658" CreationDate="2010-12-17T20:16:16.860" Score="3" ViewCount="" Body="&lt;p&gt;Since you are into signal processing, you might like the proof using Parseval's theorem. I have paraphrased this into the signal processing language from the original proof #4 of &lt;a href=&quot;http://empslocal.ex.ac.uk/people/staff/rjchapma/etc/zeta2.pdf&quot; rel=&quot;nofollow&quot;&gt;Robin Chapman's collection of proofs&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $e_n = e^{2\pi inx}$ where $n \in \mathbb Z$. Let $f(x) = x$ in the interval $[0,1]$ and we compute its Fourier series&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ f (x) = \sum_n a_n e^{2\pi inx}.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, in your terms, Parseval's theorem would mean that the energy computation in the time domain is identical to the energy computation in the frequence domain. To compute energy in the time domain, we integrate the square of the abs. value of the function, and to compute energy in the Frequency domain, we sum the squares of the abs. values of the Fourier coefficients. So,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \int_0^1 x^2 dx\ =\ \sum |a_n|^2$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As R. Chapman remarks, the left side is $1/3$, and we have $a_0 = 1/2$ and $a_n = 1/2\pi in$ for $n \neq 0$. So the above simplifies to &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \frac{1}{3} = \frac{1}{4} + \underset{n\in \mathbb Z , n \neq 0}{\sum} \frac{1}{4\pi n^2}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;from which the result follows.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Incidentally, note here that the explicit calculation was not necessary to prove finite energy. The much simpler way is to note that your signal can be bounded by the Fourier series of some other signal, and then observe that energy of that signal as seen in the time domain is finite.&lt;/p&gt;&#xA;" OwnerUserId="1119" LastEditorUserId="1119" LastEditorDisplayName="" LastEditDate="2010-12-17T20:24:22.643" LastActivityDate="2010-12-17T20:24:22.643" CommentCount="3" />
  <row Id="14714" PostTypeId="1" AcceptedAnswerId="14720" CreationDate="2010-12-17T20:47:52.980" Score="2" ViewCount="176" Body="&lt;p&gt;I've recently studied characteristic functions in my probability course and I can't get &lt;i&gt;why&lt;/i&gt; we define it to be the Fourier transform of the distribution (if the random variabile is continuous). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I mean that if $X$ is a random variable, $\varphi_X (t) = \mathbb{E}(e^{i t X}) = \int_{-\infty}^{+\infty} e^{i t x}f_X(x) dx$ where $f_X(x)$ is the distrubution function of $X$, and I can't see any motivation for doing this. I asked my professor but he wasn't clear at all; he said something like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Since we proved the theorem that if $\varphi_X (t) = \varphi_Y (t)$ then $X \sim Y$ (or $P_X \equiv P_Y)$, it is natural to define it this way&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But of course, to prove that we need the definition! So I couldn't really make up my mind about it, if you could provide some help in this sense (motivation for defining the characteristic function of a random variable as the Fourier transform of its distribution) it would be much appreciated.&lt;/p&gt;&#xA;" OwnerUserId="1415" LastEditorUserId="1415" LastEditorDisplayName="" LastEditDate="2010-12-17T20:53:33.617" LastActivityDate="2010-12-18T00:18:27.023" Title="Characteristic Functions and motivations" Tags="&lt;probability-theory&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="14715" PostTypeId="1" AcceptedAnswerId="14716" CreationDate="2010-12-17T20:56:19.790" Score="3" ViewCount="195" Body="&lt;p&gt;First,&#xA;to understand my question, checkout this one:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://math.stackexchange.com/questions/14573/calculating-percentages-for-taxes&quot;&gt;Calculating percentages for taxes.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second, &#xA;consider that I'm a layman in math.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, after trying to understand the logic used to get the final result. I was wondering:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why multiply $20,000 by 100 and then divide by 83? I know this is the rule of three, but, I can't understand the &quot;internals&quot;  of this approach. It isn't intuitive as think in this way:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Say 100% of one value, is the same of divide this value by 100. In other words: I have 100 separeted parts of this integer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's intuitive think about the taxes like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$X - 17\% = \$20.000$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\$20.000 = 83\%$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For me, the most easy and compreensive way to solve this is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\$20.000 / 83 = 240.96$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's the same as think, if 100% is 100 parts of one integer, 83% of that integer is the same of divide this integer by 83.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And finally to get the result:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\$20.000 + 17 * 240.96$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My final question is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I think intuitively like this using the Rule of Three?&#xA;In other words, why multiply 20.000 by 100 and then divide by 83 is a shorcut to get the result? &lt;/p&gt;&#xA;" OwnerUserId="1425" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-17T21:36:21.000" LastActivityDate="2010-12-20T01:10:44.737" Title="The logic behind the rule of three on this calculation." Tags="&lt;algebra-precalculus&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="2" />
  <row Id="14716" PostTypeId="2" ParentId="14715" CreationDate="2010-12-17T21:07:33.343" Score="6" ViewCount="" Body="&lt;p&gt;The Rule of Three is just doing the same thing you are doing: if $\$20,000$ is $83$%, then $\frac{20000}{83}$ is $1$% of the original. Since $X$ is the original, then $\frac{X}{100}$ is &lt;strong&gt;also&lt;/strong&gt; $1$% of the original. That means that the two quantities are equal, so we have:&#xA;$$\frac{20,000}{83} = \frac{X}{100}.$$&#xD;&#xA;From here it is easy to solve for $X$:  &lt;em&gt;Edited:&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since both sides are equal, if you do the same thing to both sides you will get equal quantities. Or, intuitively, since each of these quantities is 1% of what we want, multiplying them by 100 will give us 100% (which is what we want). So we have&#xA;$$100\left(\frac{20,000}{83}\right) = 100\left(\frac{X}{100}\right).$$&#xD;&#xA;Now, on the right hand side, the $100$ cancels with the $100$ on the denominator (which again, makes sense: remember that $X$ was the total, $\frac{X}{100}$ was 1% of $X$, so a hundred times 1% of $X$ gives us back $X$), so we are left with&#xA;$$\frac{(100)(20,000)}{83} = X.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, the Rule of Three works because you have proportionality: if $a$ is to $b$ like $x$ is to $c$, (if $\$20,000$ is to $83$% like $X$ is to $100$%)&#xA;$$\begin{array}{ccc}&#xD;&#xA;a &amp;amp; \longrightarrow &amp;amp;b\\\&#xD;&#xA;x &amp;amp; \longrightarrow &amp;amp;c&#xD;&#xA;\end{array}$$&#xD;&#xA;then that means that&#xA;$$\frac{a}{b} = \frac{x}{c}.$$&#xD;&#xA;which is solved by multiplying through by $c$, which cancels the $c$ in the denominator on the right hand side; it comes out to exactly the same as the Rule of Three says: multiply across the diagonal of known quantities (in this case, $a$ and $c$) and divide by the remaining one (in this case, $b$). The point of the Rule of Three is to do it by rote and not having to think about setting everything up carefully to get the answer. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It also works if your unknown were the percentage: if $\$20,000$ is $87$%, then what percentage of the original is $\$15,575$? (I just made that number up) Well, using the logic from above, since $\frac{20,000}{87}$ is $1$%, and $\frac{15,575}{p}$ is $1$% (where $p$ is the percentage we are looking for) then we have&#xA;$$\frac{20,000}{87} = \frac{15,575}{p}$$&#xD;&#xA;which solving for $p$ by first multiplying across by $p$, then multiplying across by $87$, and finally  dividing through by $20,000$ (aka &quot;cross-multiplying&quot;):&#xA;\begin{align}&#xA;\frac{20,000}{87} &amp;amp;= \frac{15,575}{p}\\&#xA;\frac{20,000p}{87} &amp;amp;= \frac{15,575p}{p}\\&#xA;\frac{20,000p}{87} &amp;amp;= 15,575\\&#xA;\frac{20,000p(87)}{87} &amp;amp;= (15,575)(87)\\&#xA;20,000p &amp;amp;= (15,575)(87)\\&#xA;\frac{20,000p}{20,000} &amp;amp;= \frac{(15,575)(87)}{20,000}\\&#xA;p &amp;amp;= \frac{(15,575)(87)}{20,000} \approx 67.75&#xA;\end{align}&#xA;so the percentage here is approximately $67.75$%. Using the Rule of Three, you have that $20,000$ is to $87$ like $15,575$ is to $p$, so&#xA;$$\begin{array}{ccc}&#xD;&#xA;20,000 &amp;amp;\longrightarrow &amp;amp; 87\\\&#xD;&#xA;15,575 &amp;amp; \longrightarrow &amp;amp; p&#xD;&#xA;\end{array}$$&#xD;&#xA;so to get $p$, you multiply across the known diagonal ($15,575$ times $87$) and divide by the remaining quantity ($20,000$), same operation as before.&#xA;&lt;em&gt;Added:&lt;/em&gt; Again, intuitively, what we are doing is noting that if $20,000$ is 87%, then $\frac{20,000}{87}$ is 1% of the total; we want to know the $p$ such that $15,575$ is $p$%; whatever it is, $\frac{15,575}{p}$ will also be $1$%, so the two quantities are equal. Multiplying both sides by $p$ makes both sides equal to $p$%. of the total. Multiplying both sides by 87 makes both sides $87p$&amp;amp;%. Since $20,000$ is $87$%, dividing both sides by $20,000$ will give $p$, the percentage we are looking for (since we have $87p$% divided by $87$%). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(But it's important that the Rule of Three only works when you &lt;em&gt;do&lt;/em&gt; have proportionality. It works for percentages, it works for linear aggregates, but it does not work for things like exponential growth or decay, because then you don't have proportionality between growth and time elapsed.)&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-20T01:10:44.737" LastActivityDate="2010-12-20T01:10:44.737" CommentCount="6" />
  <row Id="14717" PostTypeId="2" ParentId="14715" CreationDate="2010-12-17T21:10:33.163" Score="3" ViewCount="" Body="&lt;p&gt;I think that the form to see this more easy,is to see it as proportionally; is say:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{X}{100}=\frac{20000}{83}$$ (Are equals as they have the same proportion)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus only, you must solve the equation for $X$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then $$X= \frac{(20000)(100)}{83}$$ Passing to multiply to 100, remember this is solved in cross multiplying .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Arturo had been clear in the other post.&lt;/p&gt;&#xA;" OwnerUserId="3421" LastEditorUserId="3421" LastEditorDisplayName="" LastEditDate="2010-12-17T21:18:58.150" LastActivityDate="2010-12-17T21:18:58.150" />
  <row Id="14718" PostTypeId="1" CreationDate="2010-12-17T21:12:37.750" Score="3" ViewCount="118" Body="&lt;p&gt;I am looking for the the method of constructing the midpoint of two points in spherical geometry. The only tools allowed for the construction are a pair of spherical compasses and a spherical ruler.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In Eclidean geometry constructing the midpoint is relatively easy. We are looking for the midpoint of points A and B. We construct two circles on A and B with the radiuses of AB. Then we construct two straight lines. One is through the two intersections of the two circles and one is through A and B. The intersection of these two lines will give the midpoint of A and B.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is clear that the Euclidean method of construction does not work in Spherical geometry. The circles do not intersect when the distance of our two points exceeds 120°. There is also no solution when their distance is exactly 90°.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How would you construct the midpoint of two points in spherical geomerty?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you&lt;/p&gt;&#xA;" OwnerUserId="4881" LastActivityDate="2010-12-17T22:40:25.707" Title="How to construct the midpoint in spherical geometry?" Tags="&lt;geometry&gt;&lt;spherical-geometry&gt;" AnswerCount="1" CommentCount="0" />
  <row Id="14719" PostTypeId="2" ParentId="5510" CreationDate="2010-12-17T21:19:40.990" Score="1" ViewCount="" Body="&lt;p&gt;I have used the KKT conditions numerous times, and am working on a project now that involves using the KKT conditions along with convexity in non-negative matrix factorization, but I'm having trouble finding a good, two-dimensional, image of why the KKT conditions work.  Although I understand what is going on, I have never been one who is able to draw what makes sense.  Does anyone know where I can find a good image to show why these conditions work with a convex function?&lt;/p&gt;&#xA;" OwnerUserId="4882" LastActivityDate="2010-12-17T21:19:40.990" CommentCount="2" />
  <row Id="14720" PostTypeId="2" ParentId="14714" CreationDate="2010-12-17T21:20:48.227" Score="4" ViewCount="" Body="&lt;p&gt;The reason we care about the Fourier transform of a distribution is that it has some useful properties.  The term &quot;characteristic function&quot; is, of course, just a label.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe a comparison with some of the other kinds of generating functions for random variables would be helpful.  These include:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Probability-generating_function&quot; rel=&quot;nofollow&quot;&gt;Probability generating function&lt;/a&gt;, $E[t^X]$ (also known as the factorial moment generating function),&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Moment-generating_function&quot; rel=&quot;nofollow&quot;&gt;Moment generating function&lt;/a&gt;, $E[e^{tX}]$ (which is the Laplace transform if $X$ is nonnegative),&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Characteristic_function_%28probability_theory%29&quot; rel=&quot;nofollow&quot;&gt;Characteristic function&lt;/a&gt;, $E[e^{itX}]$ (which, as you have stated, is the Fourier transform).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;To quote from Casella and Berger's &lt;em&gt;Statistical Inference&lt;/em&gt; (first edition, p. 84), &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Perhaps the most useful of all these types of functions is the characteristic function...  The characteristic function does much more than the mgf [moment generating function] does.  When the moments of $F_X$ exist, $\phi_X$ can be used to generate them, much like an mgf.  The characteristic function always exists [unlike the mgf] and it completely determines the distribution.  That is, every cdf has a unique characteristic function.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They then give an example of a theorem that, for mgf's, has some qualifications to it, but that doesn't require any for characteristic functions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To elaborate on the examples in the quote, the characteristic function has the two following useful properties (although it has more):&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;It can be used to find the moments $E[X^n]$,&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;It can be used to show that a transformation of a random variable from a particular distribution has some other known distribution.  This method is often much easier than using the cdf technique for transformations.  (See, for example, &lt;a href=&quot;http://math.stackexchange.com/questions/10318/sum-of-two-independent-geometric-random-variables/10371#10371&quot;&gt;this answer&lt;/a&gt; to a recent math.SE question.  The answer uses mgf's rather than characteristic functions, but the idea is similar.)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;See also the Wikipedia page for the &lt;a href=&quot;http://en.wikipedia.org/wiki/Characteristic_function_%28probability_theory%29&quot; rel=&quot;nofollow&quot;&gt;characteristic function&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2370" LastActivityDate="2010-12-17T21:20:48.227" CommentCount="0" />
  <row Id="14721" PostTypeId="1" AcceptedAnswerId="14727" CreationDate="2010-12-17T22:21:55.943" Score="7" ViewCount="278" Body="&lt;p&gt;I haven't found any useful method to solve the following problem: Prove that if $x,y,z\in\mathbb{Z}$ and $x^3+y^3=3z^3$ then $xyz=0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Source: &lt;a href=&quot;http://www.artofproblemsolving.com/Forum/viewtopic.php?f=56&amp;amp;t=382377&quot; rel=&quot;nofollow&quot;&gt;http://www.artofproblemsolving.com/Forum/viewtopic.php?f=56&amp;amp;t=382377&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2856" LastEditorUserId="2856" LastEditorDisplayName="" LastEditDate="2010-12-17T22:27:55.223" LastActivityDate="2010-12-17T23:34:26.267" Title="Three variable, third degree Diophantine equation" Tags="&lt;number-theory&gt;&lt;diophantine-equations&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="14722" PostTypeId="2" ParentId="14718" CreationDate="2010-12-17T22:40:25.707" Score="2" ViewCount="" Body="&lt;p&gt;Draw a great circle $L$ between $A$ and $B$. Set your compass to a small distance, say $r$.  Mark off distance $r, 2r, 3r, \ldots$ along $L$, moving from $A$ to $B$.  Now do the same from $B$ to $A$.  Count to find the middle interval and subdivide that. &lt;/p&gt;&#xA;" OwnerUserId="1307" LastActivityDate="2010-12-17T22:40:25.707" CommentCount="3" />
  <row Id="14723" PostTypeId="1" AcceptedAnswerId="14726" CreationDate="2010-12-17T22:44:06.563" Score="4" ViewCount="323" Body="&lt;p&gt;There's a puzzle where you have 3 houses and 3 utilities. You must draw lines so that each house is connected to all three utilities, but the lines cannot overlap. However, I'm fairly sure that the puzzle is impossible. How is this proved?&lt;/p&gt;&#xA;" OwnerUserId="4885" LastEditorUserId="4885" LastEditorDisplayName="" LastEditDate="2010-12-17T23:12:03.403" LastActivityDate="2010-12-18T02:28:38.790" Title="3 Utilities | 3 Houses puzzle?" Tags="&lt;graph-theory&gt;&lt;puzzle&gt;" AnswerCount="4" CommentCount="7" FavoriteCount="2" />
  <row Id="14724" PostTypeId="2" ParentId="14711" CreationDate="2010-12-17T23:10:27.420" Score="2" ViewCount="" Body="&lt;p&gt;Well... this is somewhat subjective but...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think Knuth is hinting that most mathematicians (and I'm guessing most computer scientists) work on problems that are not particularly glamorous and not immediately important, nor do they have particularly elegant proofs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The use of the phrase &quot;bottom ten&quot; appears to be a metaphor, referring to problems that could be solved (using well-established techniques), but haven't yet been solved.  There's no actual list.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &quot;stones that make up the wall&quot; refers to that important mathematical discoveries are typically not made by one person working hard on a problem, but by a host of contributors over many years, each making incremental progress on the problem.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If I have seen a little further it is by standing on the shoulders of Giants. -- Isaac Newton&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;PS.  I can understand that Knuth's response might be surprising to people without research experience.  Why would he want to work on the least interesting problems?&lt;/p&gt;&#xA;" OwnerUserId="139" LastActivityDate="2010-12-17T23:10:27.420" />
  <row Id="14725" PostTypeId="2" ParentId="9524" CreationDate="2010-12-17T23:22:30.023" Score="1" ViewCount="" Body="&lt;p&gt;Some CS / computer engineering things: &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Division_%28digital%29#Newton.E2.80.93Raphson_division&quot; rel=&quot;nofollow&quot;&gt;Newton-Raphson&lt;/a&gt; division is a clever way that you can do division with only using multiplication / addition. Telling people that that's how calculators do division might be interesting, and it is obviously relevant to ECE majors. (See &lt;a href=&quot;http://en.wikipedia.org/wiki/Newton%27s_method&quot; rel=&quot;nofollow&quot;&gt;the analysis of its error&lt;/a&gt; for how it relates to Taylor series.)&lt;/li&gt;&#xA;&lt;li&gt;You can use infinite sums to calculate pi, e, etc.&lt;/li&gt;&#xA;&lt;li&gt;More relevant to calc is the fact that numerical integration methods use e.g. Simpson's method, the error analysis of which is related to Taylor series.&lt;/li&gt;&#xA;&lt;li&gt;(Also, here Calc II is the same course people get introduced to linear algebra, and I wish someone had told me that PageRank was an eigenvector problem. So some unasked-for advice if it's the same where you are is to introduce eigenvectors this way)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="2549" LastEditorUserId="2549" LastEditorDisplayName="" LastEditDate="2010-12-19T02:34:08.720" LastActivityDate="2010-12-19T02:34:08.720" CommentCount="4" />
  <row Id="14726" PostTypeId="2" ParentId="14723" CreationDate="2010-12-17T23:25:31.300" Score="4" ViewCount="" Body="&lt;p&gt;It's not hard to see this is impossible. Connect two houses to the three utilities, and you will essentially have a square with one diagonal drawn. The two corners joined are two of the houses, the other two corners and the midpoint of the diagonal are the utilities. (The actual shape may look distorted from this, but it is &lt;em&gt;essentially&lt;/em&gt; this).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/kF0IT.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where is the third house? If the house is &quot;outside&quot; the square, you cannot connect it to the utility in the middle. If the house is inside the square, then it is on one of the two sides of the diagonal, and therefore you cannot connect it to the vertex (utility) that is &quot;across&quot; the diagonal. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Postscript.&lt;/em&gt; As Moron points out, I am implicitly using &lt;a href=&quot;http://en.wikipedia.org/wiki/Jordan_curve_theorem&quot; rel=&quot;nofollow&quot;&gt;the Jordan curve theorem&lt;/a&gt; which says that a simple closed curve divides the plane into two disjoint regions, so that any path joining a point from the &quot;outside&quot; to a point &quot;inside&quot; has to cross the boundary. I use it when I argue that the house &quot;outside&quot; cannot be connected to the utility &quot;inside&quot; (without crossing the lines), or that you cannot go from one side of the diagonal to the other without crossing the diagonal.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-18T02:28:38.790" LastActivityDate="2010-12-18T02:28:38.790" CommentCount="7" />
  <row Id="14727" PostTypeId="2" ParentId="14721" CreationDate="2010-12-17T23:25:52.437" Score="9" ViewCount="" Body="&lt;p&gt;The technique used to prove that $x^3 + y^3 + z^3 = 0$ has no non-trivial solutions in $\mathbb{Z}(\sqrt{-3})$ is also applicable to showing that $x^3 + y^3 = 3z^3$ has no non-trivial solutions (in $\mathbb{Z}(\sqrt{-3})$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, this appears (with proof) in section 13.5 as Theorem 232 in the excellent book, &quot;An Introduction to the Theory of Numbers&quot;, by Hardy &amp;amp; Wright, 5th Edition (I have the Indian Edition, so might be a bit different from yours).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a snapshot I managed to scrape (though the notation is quite old, and you would need parts of the rest of the book to make sense of it).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/Xxz8L.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-12-17T23:34:26.267" LastActivityDate="2010-12-17T23:34:26.267" CommentCount="2" />
  <row Id="14728" PostTypeId="2" ParentId="14723" CreationDate="2010-12-17T23:41:42.137" Score="2" ViewCount="" Body="&lt;p&gt;So, the question seems to be asking for an embedding of $K_{3,3}$ in the plane.  This is well-known to be impossible.  But, we live on a sphere, rather than a plane.  However, it's also impossible in this case, which we can visualise using &lt;a href=&quot;http://en.wikipedia.org/wiki/Stereographic_projection&quot; rel=&quot;nofollow&quot;&gt;stereographic projection&lt;/a&gt;.  You can get pretty close, $K_{3,3} \setminus e$ is planar (i.e. delete an edge).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But... thinking outside the box, there are situations where it is possible:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;We live in three dimensions, where an embedding of $K_{3,3}$ is easily possible.&lt;/li&gt;&#xA;&lt;li&gt;There doesn't seem to be anything to prevent one of the utilities also being a house.  So in the following graph, red vertices are utilities, blue vertices are houses and the green vertex is both a house and a utility.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/vQQul.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="139" LastActivityDate="2010-12-17T23:41:42.137" />
  <row Id="14729" PostTypeId="2" ParentId="14723" CreationDate="2010-12-17T23:42:40.623" Score="1" ViewCount="" Body="&lt;p&gt;Being able to draw in the plane without edges crossing is the property of planar graphs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That you cannot solve the puzzle follows readily from the following theorem on planar graphs&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If $G(V,E)$ is a planar graph and has&#xA;  no cycles of length $3$, then $|E| \le 2|V| - 4$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In our case $|V| = 6$ (number of nodes) and we need $|E| = 9$ (number of edges). The graph we have cannot have any triangles, as it is bipartite, and any bipartite graph can only have even length cycles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The above theorem can be proven using &lt;a href=&quot;http://www.ics.uci.edu/~eppstein/junkyard/euler/&quot; rel=&quot;nofollow&quot;&gt;Euler's Formula&lt;/a&gt; very easily.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The graph you have is known as $K_{3,3}$ and appears in the statement of &lt;a href=&quot;http://mathworld.wolfram.com/KuratowskiReductionTheorem.html&quot; rel=&quot;nofollow&quot;&gt;Kuratowski's Theorem&lt;/a&gt; on planar graphs. This theorem characterizes all planar graphs in terms on forbidding $K_5$ (complete graph on 5 vertices) and $K_{3,3}$ as minors.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-12-17T23:49:53.640" LastActivityDate="2010-12-17T23:49:53.640" />
  <row Id="14730" PostTypeId="2" ParentId="14723" CreationDate="2010-12-18T00:00:46.380" Score="2" ViewCount="" Body="&lt;p&gt;Here's a short proof that $K_{3,3}$ is nonplanar.  (It's borrowed from Bondy and Murty, &lt;em&gt;Graph Theory with Applications&lt;/em&gt;, pp. 144-145.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some notation: Let $F$ and $\phi$ denote the set and number, respectively, of faces of a planar embedding of a graph.  Let $V$ and $\nu$, denote the set and number, respectively, of vertices of a graph.  Let $\epsilon$ denote the number of edges in the graph.  Let $d(f)$ denote the degree of the face $f$; i.e., the number of edges incident on $f$.  Let $G^*$ denote the dual graph.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Claim: $\sum_{f \in F} d(f) = 2\epsilon$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proof: $\sum_{f \in F} d(f) = \sum_{f^* \in V(G^*)} d(f^*) = 2\epsilon^* = 2\epsilon$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(The first step follows from the fact that faces in $G$ correspond to vertices in $G^*$, the second from the well-known fact that the sum of the vertex degrees of a graph is twice the number of edges, and the last from the fact that edges in $G$ correspond to edges in $G^*$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, suppose $K_{3,3}$ is planar, and let $G$ be a planar embedding of $K_{3,3}$.  Since $K_{3,3}$ has no cut edges and no cycles of length less than four, every face of $G$ must have degree at least four.  Thus &#xA;$$4\phi \leq \sum_{f \in F} d(f) = 2\epsilon  = 18.$$&#xD;&#xA;Thus $\phi \leq 4$.  But, by &lt;a href=&quot;http://en.wikipedia.org/wiki/Euler_characteristic&quot; rel=&quot;nofollow&quot;&gt;Euler's theorem&lt;/a&gt;, this means $2 = \nu - \epsilon + \phi \leq 6 - 9 + 4 = 1$, a contradiction.&lt;/p&gt;&#xA;" OwnerUserId="2370" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-12-18T00:06:24.820" LastActivityDate="2010-12-18T00:06:24.820" />
  <row Id="14731" PostTypeId="2" ParentId="14714" CreationDate="2010-12-18T00:18:27.023" Score="2" ViewCount="" Body="&lt;p&gt;Practically speaking, the short answer is that it's convenient.  The characteristic function has better analytic properties than the moment generating function, lets you study all of the moments of a random variable at once, and has the extremely convenient property that $\phi_{X+Y}(t) = \phi_X(t) \phi_Y(t)$ if $X, Y$ are independent.  This makes the characteristic function an amazing tool for understanding sums of independent random variables, and indeed a standard proof of the central limit theorem proceeds via a computation of characteristic functions.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many constructions in mathematics translate problems in one domain (understanding distribution functions) to problems in another domain (understanding characteristic functions), and these constructions are useful because different tools apply in the second domain.  That is exactly what happens in the Fourier-theoretic proof of the central limit theorem.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-12-18T00:18:27.023" />
  <row Id="14732" PostTypeId="2" ParentId="14588" CreationDate="2010-12-18T00:24:45.910" Score="5" ViewCount="" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;No&lt;/strong&gt;, the lattice of subgroups, the lattice of normal subgroups, the order of the group, and the automorphism group do not (even taken together) determine the isomorphism type of a finite group.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Take G = SmallGroup(243,19) and H = SmallGroup(243,20).  There is a bijection f:L(G)→L(H) between their lattices of subgroups such that:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;|X| = |f(X)|&lt;/li&gt;&#xA;&lt;li&gt;X ≅ f(X) unless X = G&lt;/li&gt;&#xA;&lt;li&gt;X ≤ Y iff f(X) ≤ f(Y)&lt;/li&gt;&#xA;&lt;li&gt;X ⊴ G iff f(X) ⊴ f(G) = H&lt;/li&gt;&#xA;&lt;li&gt;G/X ≅ H/f(X) whenever X≠1 is normal&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Additionally Aut(G) ≅ Aut(H).  The fourth bullet shows in particular, that f induces an isomorphism between the lattice of quotient groups of G and the lattice of quotient groups of H.  The second and fifth bullets show the isomorphism respects everything about the subgroups' properties as abstract groups.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The groups have presentations:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align*}&#xA; G &amp;amp;= \bigl\langle a,b,c \mid a^{27} = b^{3} = c^{3} = 1,\ ba = abc,\ ca = acz,\ cb = bcz \bigr\rangle\text{ where }z = a^9\\&#xA;H &amp;amp;= \bigl\langle a,b,c \mid a^{27} = b^3 = c^{3} = 1,\ ba = abc,\ ca = acz,\ cb = bcz \bigr\rangle\text{ where }z = a^{-9}&#xA;\end{align*}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The function is induced by a bijection of the underlying sets:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;f(a&lt;sup&gt;i&lt;/sup&gt; b&lt;sup&gt;j&lt;/sup&gt; c&lt;sup&gt;k&lt;/sup&gt;) = a&lt;sup&gt;i&lt;/sup&gt; b&lt;sup&gt;j&lt;/sup&gt; c&lt;sup&gt;k&lt;/sup&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;There are no such groups of order dividing 64 (even just having an isomorphism of subgroup lattices respecting normal subgroups).&lt;/p&gt;&#xA;" OwnerUserId="583" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-18T06:23:09.310" LastActivityDate="2010-12-18T06:23:09.310" CommentCount="5" />
  <row Id="14733" PostTypeId="2" ParentId="14689" CreationDate="2010-12-18T00:25:23.700" Score="1" ViewCount="" Body="&lt;p&gt;Here's how I &lt;strong&gt;approximated&lt;/strong&gt; an answer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know player-2 usually wins. The expected value of player-2 after 9 games is 63, just over 60. So the expected number of turns the game will last is about 9.&#xA;Let's consider another version of this game which is played for 9 turns exactly and the winner is the one who got the higher score.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How is the score of each player distributed after 9 games? We need to get the distribution of a sum of 18 random variables (because 9 turns correspond to 18 dice throws). The distribution of a sum of a random variable is exactly the &lt;strong&gt;convolution&lt;/strong&gt; of the distributions. This can be efficiently computed by the &lt;strong&gt;convolution theorem&lt;/strong&gt; which says: compute the &lt;strong&gt;Discrete Fourier Transform&lt;/strong&gt; of each of the distributions, multiply the resulting distributions element by element, then compute the Inverse Discrete Fourier Transform of the resulting distribution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I computed that in python. Now that we got the distribution of the scores for each player after 9 turns. To compute the probability player-2 wins, I computed the cumulative distribution of the score for each player (i.e: the probability of getting score &gt;= i for each i). Now I just had to sum of each score i, the probability of player-1 getting score &amp;lt; i, multiplied by the probability of player-2 getting score &gt;= i.&#xA;The probability for a draw is computed similarly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Repeating the calculation I get about: 81%,16%,3% (but it's just an approximation anyway).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To calculate the accurate numbers you should do this calculation for any number of turns, and then compute the average of these numbers, weighted by the probability of the game to end in any specific number of turns.&#xA;But I expect the calculation for 9 turns to give a very good approximation, because the distribution of the number of turns is centered around 9 and I expect to get similar results of 8 or 10 turns.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Some notes in case you're not familiar with convolutions and Fourier Transform&lt;/strong&gt;:&#xA;Look at a random variable X over some finite probability space of n elements. Think of X as a vector of n real numbers (representing its distribution). Then the operator T, defined as T(Y)=X+Y for any Y (this is &lt;strong&gt;not&lt;/strong&gt; element-wise addition. this is addition of random variables, also called &lt;strong&gt;convolution&lt;/strong&gt;). Then T is a linear operator in $\mathbb{R}^n$. It happens to be diagonizable. The Discrete Fourier Transform is just a change of basis, after which T is a diagonal operator. It may be a little surprising, but this basis does not depend of X (although T does). The i,i element in the diagonal of the matrix representing T in this new basis happens to be Pr(X=i). Computing the compositions of diagonal operators in very easy - just multiply the diagonals element-wise. This is the justification for the convolution theorem. After you're done adding random variables in this new basis, just return to the original basis (i.e: do the inverse Fourier transform) and get the desired probability vector of the addition of all the random variables you added.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's how the probabilities are calculated in python (I'm supplying all the routines you need):&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;def dft(x):&#xA;     N = len(x)&#xA;     ret = [0]*N&#xA;     for k in range(N):&#xA;         for n in range(N):&#xA;             ret[k] += x[n]*math.e**((-2*math.pi*(1j)/N)*k*n)&#xA;     return ret&#xA;def idft(x):&#xA;     N = len(x)&#xA;     ret = [0]*N&#xA;     for k in range(N):&#xA;         for n in range(N):&#xA;             ret[k] += x[n]*math.e**((2*math.pi*(1j)/N)*k*n)&#xA;         ret[k] /= N&#xA;     return ret&#xA;def mul_vec(x1,x2):&#xA;     return [a1*a2 for (a1,a2) in zip(x1,x2)]&#xA;def pow_vec(x,n):&#xA;     ret = [1]*len(x)&#xA;     for i in range(n):&#xA;         ret = mul_vec(ret, x)&#xA;     return ret&#xA;def pad(x,n):&#xA;     return x+[0]*(n-len(x))&#xA;p1_probs=idft(pow_vec(dft(pad([0]+[1./6]*6,200)),18))&#xA;p2_probs=idft( mul_vec(pow_vec(dft(pad([0]+[1./6]*6,200)),9) ,  pow_vec(dft(pad([0]+[1./4]*4,200)),9)    )  )&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="3533" LastEditorUserId="3533" LastEditorDisplayName="" LastEditDate="2010-12-18T12:53:04.927" LastActivityDate="2010-12-18T12:53:04.927" />
  <row Id="14734" PostTypeId="1" CreationDate="2010-12-18T00:35:17.777" Score="0" ViewCount="38" Body="&lt;p&gt;Alright, so I'm trying to check if Quaternion a's y axis is &gt; quaternion b's y axis. How would I do this?&lt;/p&gt;&#xA;" OwnerUserId="4464" LastActivityDate="2010-12-18T00:35:17.777" Title="Check if Quaternion a's y axis angle is greater than Quaternion b?" Tags="&lt;geometry&gt;&lt;trigonometry&gt;&lt;quaternions&gt;" CommentCount="1" />
  <row Id="14735" PostTypeId="1" AcceptedAnswerId="14739" CreationDate="2010-12-18T00:38:05.610" Score="2" ViewCount="245" Body="&lt;p&gt;Let $g: \mathbb{R} \rightarrow \mathbb{R}$ be uniformly continous with $g(0)=0,c\geq 0, c \in \mathbb{R}$. Show: $\exists a\geq 0 \in \mathbb{R}: \forall x \in \mathbb{R}: |g(x)| \leq a \cdot |x|+c$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I could also say $g(x) \in \mathcal{O}(x)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notes: I could not make up any counterexample so I guess it could be true, all uniformly continous functions I know grow to slow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My approach:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given an $\epsilon &amp;gt; 0$, we have because of the continuity of $g$, that: $\exists \delta(\epsilon): |x-y|&amp;lt;\delta=&amp;gt; |g(x)-g(y)|&amp;lt;\epsilon$. Now choose $n=\text{max}\{n \in \mathbb{N}: (n-1)\delta/2\leq|x|\}$. Obviously such a $n$ exists, and $n &amp;gt; 0$. We also easily see that an upper bound for $n$ is $n \leq \frac{2}{\delta}|x|+1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we use this to seperate $|x|$ into $n-1$ distinct parts of size $s&amp;lt;\delta/2$, and the last part thats smaller than $\delta$ :&#xA;$|x|=|x_1-x_0|+|x_2-x_1|+|x_3-x_2|+...+|x_n-x_{n-1}| &amp;lt; (n-1)\delta/2 + \delta = (n+1)\delta/2$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Rightarrow |g(x)|=|g(x_1)-g(x_0)+g(x_2)-g(x_1)+g(x_3)-g(x_2)+...+g(x_n)-g(x_{n-1})| $&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\leq |g(x_1)-g(x_0)|+|g(x_2)-g(x_1)|+|g(x_3)-g(x_2)|+...+|g(x_n)-g(x_{n-1})|$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$&amp;lt;n \cdot \epsilon \leq (\frac{2}{\delta}|x|+1) \cdot \epsilon = \frac{2\epsilon}{\delta} \cdot |x|+\epsilon$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we can see that the constant $c$ we were given can be set as the $\epsilon := c$, and that was also the reason why general speaking $c&amp;gt;0$. Then we can choose $a := \frac{2\epsilon}{\delta}$, as our $\delta$ only depends on the $\epsilon$ and we have that $|g(x)| \leq a \cdot |x| + c$ for $c &amp;gt; 0$. $\square$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="3123" LastEditorUserId="3123" LastEditorDisplayName="" LastEditDate="2010-12-18T22:25:35.347" LastActivityDate="2010-12-18T22:25:35.347" Title="Is a uniformly continuous function vanishing at $0$ bounded by $a|x|+c$?" Tags="&lt;real-analysis&gt;&lt;homework&gt;&lt;analysis&gt;" AnswerCount="1" CommentCount="12" FavoriteCount="1" />
  <row Id="14736" PostTypeId="1" AcceptedAnswerId="14737" CreationDate="2010-12-18T00:39:09.733" Score="2" ViewCount="75" Body="&lt;p&gt;The solution for the number of distributions leaving none of the $n$ cells empty (with unlike cells and $r$ unlike objects) is given by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$A(r,n)=\sum_{\nu=0}^{n-1}(-1)^{\nu}\binom{n}{\nu}(n-\nu)^{r}$$ (although I have seen the same expression summing from $\nu=0$ to $n$). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;By the way, if anyone knows which one is correct, please let me know. However, that's not the question.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have read that this expression provides a solution to a famous problem. I'd like to know where can I find more information about this solution and which famous problem solves. I already tried &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0691023654&quot; rel=&quot;nofollow&quot;&gt;An Introduction to Combinatorial Analysis&lt;/a&gt; by John Riordan and &lt;a href=&quot;http://www.jstor.org/pss/2688836&quot; rel=&quot;nofollow&quot;&gt;Certain distributions of unlike objects into cells&lt;/a&gt; by Morton Abramson but their treatment of this particular expression is very brief.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance.&lt;/p&gt;&#xA;" OwnerUserId="2569" LastEditorUserId="2569" LastEditorDisplayName="" LastEditDate="2010-12-18T00:44:35.440" LastActivityDate="2010-12-18T00:58:19.017" Title="Number of distributions leaving none of $n$ cells empty" Tags="&lt;combinatorics&gt;&lt;partitions&gt;" AnswerCount="1" CommentCount="6" />
  <row Id="14737" PostTypeId="2" ParentId="14736" CreationDate="2010-12-18T00:58:19.017" Score="3" ViewCount="" Body="&lt;p&gt;These are the &lt;a href=&quot;http://en.wikipedia.org/wiki/Stirling_numbers_of_the_second_kind&quot; rel=&quot;nofollow&quot;&gt;Stirling numbers of the second kind&lt;/a&gt;.  I am not sure what you mean by a famous problem that this sequence solves; you seem to already have given the standard combinatorial interpretation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Stirling numbers of the second kind occur in a certain formula for the sum $\sum_{k=1}^n k^r$; perhaps that's the problem you're thinking of?&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-12-18T00:58:19.017" CommentCount="10" />
  <row Id="14739" PostTypeId="2" ParentId="14735" CreationDate="2010-12-18T01:18:16.817" Score="4" ViewCount="" Body="&lt;p&gt;It is false if $c=0$.  To see this, try to think of a continuous function that grows very rapidly near $0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is true if $c\gt 0$.  One way to show it is by taking a number of very small steps from $0$ to $x$, small enough to guarantee (using uniform continuity) that the function changes no more than a certain fixed amount at each step.  Trying to write out the details should lead you to what this fixed amount is, and to what value of $a$ will work.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-12-18T01:18:16.817" CommentCount="12" />
  <row Id="14740" PostTypeId="1" CreationDate="2010-12-18T02:36:11.697" Score="0" ViewCount="188" Body="&lt;p&gt;Linear Algebra - True or False if $Ay = B$ and $Az=B$ then y = z? Why or why not? you may give a counter example to prove it false.&lt;/p&gt;&#xA;" OwnerUserId="4887" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-18T02:38:43.253" LastActivityDate="2010-12-18T09:34:23.940" Title="Linear Algebra - True or False if Ay = B and Az=B then y = z? Why or why not?" Tags="&lt;linear-algebra&gt;" AnswerCount="5" CommentCount="3" />
  <row Id="14741" PostTypeId="2" ParentId="14740" CreationDate="2010-12-18T02:40:21.847" Score="1" ViewCount="" Body="&lt;p&gt;It is not true if the null space of $A$ is nontrivial: If $v$ is any nonzero vector such that $Av=0$, then $Av=A0$ and, more generally, $Ay$ and $A(y+v)$ are equal, no matter what $y$ is. On the other hand, it should be easy to check now that the question is true if the null space of $A$ is trivial (i.e., if $Av=0$ only for $v=0$).&lt;/p&gt;&#xA;" OwnerUserId="462" LastActivityDate="2010-12-18T02:40:21.847" />
  <row Id="14742" PostTypeId="1" AcceptedAnswerId="14744" CreationDate="2010-12-18T02:41:36.230" Score="1" ViewCount="199" Body="&lt;p&gt;I am looking at the following problem:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Find all values of $m$ for which the equation $(3m-2)x^2 + 2mx + 3m = 0$ shas only one root between $-1$ and $0$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I don't know what only one root between -1 and 0 actually means. I count these possibilities:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$-1&amp;lt;x_1=x_2&amp;lt;0$ &lt;/li&gt;&#xA;&lt;li&gt;$x_1&amp;lt;-1&amp;lt;x_2&amp;lt;0$&#xD;&lt;/li&gt;&#xA;&lt;li&gt;$-1&amp;lt;x_1&amp;lt;0&amp;lt;x_2$&#xD;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Let $a=3m-2$, $\Delta$ be the discriminant, and $S/2$ the $x$-coordinate of the vertex ($S/2 = -m/(3m-2)$) and also the arithmetic mean value of the roots $(x_1+x_2)/2$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conditions to solve 1 alone would be:&#xA;$af(-1)&amp;lt;0$,&#xA;$af(0)&amp;lt;0$ and&#xA;$\Delta \geq 0$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conditions to solve 2 alone would be:&#xA;$af(-1)&amp;lt;0$,&#xA;$af(0)&amp;gt;0$,&#xA;$\Delta&amp;gt;0$ and&#xA;$S/2&amp;lt;0$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conditions to solve 3 alone would be:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$af(-1)&amp;gt;0$,&#xA;$af(0)&amp;lt;0$,&#xA;$\Delta&amp;gt;0$, and&#xA;$S/2&amp;gt;-1$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then it could also mean that it is all conditions intersected $ 1 \cap 2 \cap 3$&#xD;&#xA;or just $2 \cap 3$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What am I doing wrong? the correct answer for this problem according to the book is $0&amp;lt;m&amp;lt;1/2$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS.: Although this question was already answered correctly I would like to see how to use the theorems that I brought in, and the book as well, to solve this. Solving questions any way or another is always welcomed and helpful, what should certainly be done. Yet the purpose in the book is not a general math challenge, it's intended to teach the theorems and aplly them. So if I can, let me try it also:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I will put the theorems here because no one seems to know what theorems I am refering to:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $f(x)=ax^2+bx+c$ presents two real roots $x_1 \leq x_2$ and $ \alpha $ is the real number that will be compared to $x_1$ and $x_2$, we have:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$af(\alpha)&amp;lt;0 \implies x_1 &amp;lt; \alpha &amp;lt;x_2$.&lt;/li&gt;&#xA;&lt;li&gt;$af(\alpha)=0 \implies \alpha$ is one of the roots.&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If $af(\alpha)&amp;gt;0$ and $\Delta \geq 0$, then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(a)  $\alpha &amp;lt;x_1 \leq x_2$ if $\alpha &amp;lt; S/2$;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(b) $x_1 \leq x_2 &amp;lt; \alpha$ if $\alpha &amp;gt; S/2$&#xD;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;There is two situations in this problem: $-1&amp;lt;x_1&amp;lt;0&amp;lt;x_2$ and $x_1&amp;lt;-1&amp;lt;x_2&amp;lt;0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;I.&lt;/strong&gt;&#xA;$$-1&amp;lt;x_1&amp;lt;0&amp;lt;x_2 \implies af(-1)&amp;gt;0 \wedge af(0)&amp;lt;0 \wedge \Delta &amp;gt; 0 \wedge S/2 &amp;gt; -1.$$&#xD;&#xA;\begin{align*}&#xA;af(-1)&gt;0 &amp;amp;\implies (3m-2)[(3m-2)(-1)^2+2m(-1)+3m]&gt;0\\&#xA; &amp;amp;\implies (3m-2)(4m-2)&gt;0 \implies 12m^2-14m+4&gt;0\\&#xA; &amp;amp;\implies m&amp;lt;1/2\text{ or }m&gt;2/3.\\&#xA;af(0)&amp;lt;0 &amp;amp;\implies (3m-2)[(3m-2)0^2+2m0+3m]&amp;lt;0 \\&#xA;&amp;amp;\implies 9m^2-6m\lt 0 \implies 0\lt m\lt 2/3.\\&#xA;\Delta \gt 0 &amp;amp;\implies (2m)^2-4(3m-2)3m\gt 0 \implies 4m^2-12m(3m-2)\gt 0\\&#xA; &amp;amp;\implies 4m^2-36m^2+24\gt 0 \implies -32m^2+24\gt 0 \implies 0 \lt m \lt 3/4.\\&#xA;S/2\gt -1 &amp;amp;\implies -b/2a\gt -1 \implies -m/(3m-2) \gt -1\\&#xA; &amp;amp;\implies (-m+3m-2)/(3m-2) \gt 0 \implies (2m-2)/(3m-2) \gt 0\\&#xA; &amp;amp;\implies m \lt 2/3 \text{ or } m \gt 1.&#xA;\end{align*}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(af(-1) \gt 0) \wedge (af(0) \lt 0) \wedge (\Delta \gt 0) \wedge (S/2 \gt -1) \implies 0 \lt m \lt 1/2.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;II.&lt;/strong&gt;&#xA;$$x_1&amp;lt;-1&amp;lt;x_2&amp;lt;0 \implies af(-1)&amp;lt;0 \wedge af(0)&amp;gt;0 \wedge \Delta &amp;gt; 0 \wedge S/2&amp;lt;0.$$&#xD;&#xA;\begin{align*}&#xA;af(-1)&amp;lt;0 &amp;amp;\implies 1/2 \lt m \lt 2/3.\\&#xA;af(0)&gt;0 &amp;amp;\implies m \lt 0\text{ or }m \gt 2/3.\\&#xA;\Delta \gt 0 &amp;amp;\implies 0 \lt m \lt 3/4.\\&#xA;S/2 \lt 0 &amp;amp;\implies -b/2a&amp;lt;0 \implies -m/(3m-2)&amp;lt;0 \\&#xA;&amp;amp;\implies m \lt 0 \text{ or }m \gt 2/3.&#xA;\end{align*}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(af(-1)&amp;lt;0 \wedge af(0)&amp;gt;0 \wedge \Delta \gt 0 \wedge S/2 \lt 0) = \emptyset$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As both set of solutions are possible answers, non-exclusory, I think the answer is to unite the two sets: $I \cup II = (0 \lt m \lt 1/2) \cup \emptyset = 0 \lt m \lt 1/2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Final answer $0 \lt m \lt 1/2$.&lt;/p&gt;&#xA;" OwnerUserId="3509" LastEditorUserId="3509" LastEditorDisplayName="" LastEditDate="2010-12-20T15:02:28.303" LastActivityDate="2010-12-20T15:02:28.303" Title="Find the values of $m$ in the 2nd degree equation $(3m-2)x^2+2mx+3m=0$ so that it has only one root between -1 and 0." Tags="&lt;homework&gt;&lt;algebra-precalculus&gt;" AnswerCount="2" CommentCount="14" />
  <row Id="14743" PostTypeId="2" ParentId="14740" CreationDate="2010-12-18T02:50:05.127" Score="1" ViewCount="" Body="&lt;p&gt;If $A=B=0$, then we can find many $y\ne z$ such that the equations are satisfied!&lt;/p&gt;&#xA;" OwnerUserId="2764" LastActivityDate="2010-12-18T02:50:05.127" />
  <row Id="14744" PostTypeId="2" ParentId="14742" CreationDate="2010-12-18T03:13:11.230" Score="3" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\rm\ \ 0\ &amp;gt;\ f(-1)\ f(0)\ =\ 3\:m\:(4m-2)\ \iff\ a &amp;lt; m &amp;lt; b\ $ for $\rm\ a = \ldots,\ b = \ldots$&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-12-18T03:13:11.230" CommentCount="5" />
  <row Id="14745" PostTypeId="2" ParentId="14742" CreationDate="2010-12-18T03:13:33.220" Score="1" ViewCount="" Body="&lt;p&gt;Some suggestions... If $ax^2 + bx + c$ is any quadratic with $a \neq 0$, it's a parabola (possibly upside down). The center of the parabola is at $x = -{b \over 2a}$, corresponding to $x = {m \over 2 - 3m}$ in your case. So if ${m \over  2 - 3m}$ lies outside the range $(-1,0)$ then the function is monotone on $[-1,0]$. As a result it has exactly one root in $(-1,0)$ if and only if one of $f(0)$ and $f(-1)$ is positive and one is negative. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $-1 &amp;lt; {m \over  2 - 3m} &amp;lt; 0$ on the other hand then your function is monotone from $-1$ to ${m \over  2 - 3m}$ and also monotone from ${m \over  2 - 3m}$ to $0$. So you just have to look at the signs of $f(-1)$, $f({m \over  2 - 3m})$, and $f(0)$ and consider the various possibilities.&lt;/p&gt;&#xA;" OwnerUserId="3035" LastEditorUserId="3035" LastEditorDisplayName="" LastEditDate="2010-12-18T03:28:01.710" LastActivityDate="2010-12-18T03:28:01.710" CommentCount="3" />
  <row Id="14746" PostTypeId="2" ParentId="14740" CreationDate="2010-12-18T03:33:12.267" Score="5" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\rm\quad A\ y\ =\ A\ z\ \Rightarrow\ y = z\ \ $ is equivalent to $\rm\ A\ x = 0\ \Rightarrow\ x = 0,\ \ for\ \ \ x = y-z\:$&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-12-18T03:33:12.267" />
  <row Id="14747" PostTypeId="1" CreationDate="2010-12-18T03:42:25.863" Score="1" ViewCount="135" Body="&lt;p&gt;Is the intersection of a decreasing sequence of countably many stationary subsets of $\omega_1$ always stationary?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems elementary, but I fail to find the answer in textbooks.&lt;/p&gt;&#xA;" OwnerUserId="4892" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-18T04:30:22.303" LastActivityDate="2010-12-18T06:18:33.140" Title="Is the intersection of a decreasing sequence of countably many stationary subsets of $\omega_1$ always stationary?" Tags="&lt;set-theory&gt;" AnswerCount="1" />
  <row Id="14748" PostTypeId="2" ParentId="14740" CreationDate="2010-12-18T03:43:03.883" Score="3" ViewCount="" Body="&lt;p&gt;If we have $Ay = Az$, all we could conclude is $A(y-z) = 0 \Rightarrow y-z \in $Nullspace$(A)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To be more explicit in terms of matrices, if $A$ is a square or a skinny matrix, and if $A$ is full rank, then we can conclude that $y=z$. If not, $y$ need not be equal to $z$. All we can say is $y-z \in $Nullspace$(A)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A simplest example is $2 + 3 = 1 + 4$ i.e. $[1,1][2,3]^T = [1,1][1,4]^T$&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2010-12-18T03:48:49.857" LastActivityDate="2010-12-18T03:48:49.857" />
  <row Id="14749" PostTypeId="1" CreationDate="2010-12-18T03:48:20.150" Score="2" ViewCount="62" Body="&lt;p&gt;I'm preparing a talk on lattice point enumeration in polytopes (Ehrhart-Macdonald Theory), and I'd like to have an introduction with a few motivational problems/results which arise from the application of lattice point methods. (The audience consists of junior and senior faculty across the mathematical disciplines.) A few that immediately come to mind are the following: Pick's theorem, Ehrhart reciprocity, Minkowski's bound in algebraic number theory and Hilbert series in Stanley-Reisner rings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question&lt;/strong&gt;: What kinds of interesting problems have been tackled (or generated some widespread interest) by counting points in lattices?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="2720" LastActivityDate="2010-12-18T04:39:52.497" Title="Mathematical Results from Counting Points in Lattices" Tags="&lt;number-theory&gt;&lt;combinatorics&gt;&lt;soft-question&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="14750" PostTypeId="2" ParentId="14747" CreationDate="2010-12-18T03:57:42.387" Score="2" ViewCount="" Body="&lt;p&gt;The answer is no. Here is an example: Take $T_n$ for $n&amp;lt;\omega$ disjoint stationary sets. Let $S_m$, for $m&amp;lt;\omega$, be the union of $T_n$ with $n\ge m$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That there are $\omega$ (in fact, $\omega_1$) disjoint stationary subsets of $\omega_1$ is an immediate consequence of the existence of Ulam matrices. (See for example this &lt;a href=&quot;http://caicedoteaching.wordpress.com/2009/02/13/580-cardinal-arithmetic-5/&quot; rel=&quot;nofollow&quot;&gt;blog post&lt;/a&gt; of mine. You can also find this in Kunen's book or in Jech's book.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the thing is that the construction suggests itself: Suppose $S_0\supseteq S_1\supseteq\dots$ are stationary sets. If each $S_i\setminus S_{i+1}$ is non-stationary, then $\bigcap_n S_n$ is stationary; in fact, $S_0\setminus \bigcap_n S_n$ is a non-stationary set. So we would need that infinitely often we have $S_i\setminus S_{i+1}$ stationary. By passing to a subsequence if necessary, we may as well assume that it happens for all $i$. But now the construction above is immediate.&lt;/p&gt;&#xA;" OwnerUserId="462" LastEditorUserId="462" LastEditorDisplayName="" LastEditDate="2010-12-18T06:18:33.140" LastActivityDate="2010-12-18T06:18:33.140" CommentCount="1" />
  <row Id="14752" PostTypeId="2" ParentId="14749" CreationDate="2010-12-18T04:39:52.497" Score="1" ViewCount="" Body="&lt;p&gt;You may well know this paper already but I think it is very nice:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.87.2267&amp;amp;rep=rep1&amp;amp;type=pdf&quot; rel=&quot;nofollow&quot;&gt;http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.87.2267&amp;amp;rep=rep1&amp;amp;type=pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1369" LastActivityDate="2010-12-18T04:39:52.497" />
  <row Id="14753" PostTypeId="2" ParentId="14715" CreationDate="2010-12-18T05:16:50.170" Score="7" ViewCount="" Body="&lt;p&gt;The &quot;rule of three&quot; is an ancient ad-hoc mindless rote rule of inference that is best ignored. Instead, you should strive to learn the general principles behind it - namely, the laws of fraction arithmetic. Let's consider the example at hand. You  seek the number of dollars $\rm\:X\:$ such that when decremented by $\:17\%\:$ yields $\rm\:N\:$ dollars. First, recall that $\rm\  17\%\:$ of $\rm\:X\:$ means $\rm\displaystyle \frac{17}{100}\ X\:.\ $ Thus your equation is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\rm N\ =\ X - \frac{17}{100}\ X\ =\ \bigg(1 - \frac{17}{100}\bigg)\ X\ =\ \bigg(\frac{100}{100}-\frac{17}{100}\bigg)\ X\ =\ \frac{83}{100}\ X $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus $\rm\displaystyle\ \frac{83}{100}\ X\ =\ N\ \ \Rightarrow\ \ X\ =\ \frac{100}{83}\ N\ $ follows by multiplying both sides by $\rm\displaystyle\ \frac{100}{83}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that we applied no ad-hoc rules above - just the basic laws of the arithmetic of fractions. &lt;em&gt;These&lt;/em&gt; are the laws that are worthy of mastering.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's interesting to look at the decline of the use of the &quot;rule of three&quot; over the last two centuries as the knowledge of general (abstract) algebra evolved. This is very easy using the recently-released &lt;a href=&quot;http://booksearch.blogspot.com/2010/12/find-out-whats-in-word-or-five-with.html&quot; rel=&quot;nofollow&quot;&gt;Google Books Ngram viewer&lt;/a&gt; - which searches for phrases over 5 million books back to 1500. Browsing one of the &lt;a href=&quot;http://books.google.com/books?id=1P0nX_ld-nMC&amp;amp;dq=%22rule%20of%20three%22&amp;amp;pg=PA60#v=onepage&amp;amp;q=%22rule%20of%20three%22&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;earliest textbooks&lt;/a&gt; in the Google corpus containing the rule of three I noticed that it is immediately followed by a section titled &quot;method of making taxes&quot;. So it seems this was a big application in the old days. Also notice how &quot;fraction arithmetic&quot; really ramped up circa 1960 (perhaps due to &quot;new math&quot; programs?).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/FMJK8.jpg&quot; alt=&quot;alt text&quot;&gt;&#xA;&lt;img src=&quot;http://i.imgur.com/bHyZw.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-12-18T07:22:28.113" LastActivityDate="2010-12-18T07:22:28.113" CommentCount="1" />
  <row Id="14754" PostTypeId="1" CreationDate="2010-12-18T05:59:09.357" Score="2" ViewCount="302" Body="&lt;p&gt;Is it true or false that a group of order 12 always has a normal 2-sylow subgroup? &#xA;I have a hunch it is false.. &lt;/p&gt;&#xA;" OwnerUserId="4896" LastActivityDate="2011-05-26T06:39:20.173" Title="Group of order 12" Tags="&lt;group-theory&gt;" AnswerCount="3" CommentCount="8" />
  <row Id="14755" PostTypeId="1" AcceptedAnswerId="14758" CreationDate="2010-12-18T06:30:41.093" Score="0" ViewCount="42" Body="&lt;p&gt;Consider an sequence of unique values. Each value is composed of a surname and a first name in that order of significance. Based on that, I can define a complete ordering of values in the sequence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I can also usefully define less fine-grained equivalence classes of items that have a &quot;consistent&quot; ordering, but where some values compare as equivalent (and have no particular ordering relative to each other) even though in the &quot;full&quot; ordering they compare as greater and lesser.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, sometimes, I may want to search an index for a particular surname, not caring about the first name.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the most general case, there are finer-grain and coarser-grain but otherwise consistent ways to divide a set of values into equivalence classes and order those classes. There are generally many (mutually incompatible) finer grained orderings for any one coarser-grained ordering. It is possible to have a partitioning that is neither finer-grain nor coarser-grain than another different partitioning.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The trouble is that when I use the terminology of being &quot;consistent&quot;, people get confused. A lot of people won't accept that an ordering in which $a=b$ is consistent with (but coarser grained) than one in which $a&amp;lt;b$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I imagine that there are likely to be official terms to describe these relationships in abstract algebra, but my abstract algebra is pretty limited.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Am I right? If so, which is the relevant field of abstract algebra, and what are the relevant terms?&lt;/p&gt;&#xA;" OwnerUserId="510" LastEditorUserId="510" LastEditorDisplayName="" LastEditDate="2010-12-18T07:07:00.830" LastActivityDate="2010-12-18T07:07:00.830" Title="Are there standard terms for more fine/coarse grained (but otherwise consistent) ways of ordering values?" Tags="&lt;terminology&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="14756" PostTypeId="2" ParentId="14755" CreationDate="2010-12-18T06:41:11.020" Score="3" ViewCount="" Body="&lt;p&gt;I would call the relevant maps here &lt;a href=&quot;http://en.wikipedia.org/wiki/Order_theory#Functions_between_orders&quot; rel=&quot;nofollow&quot;&gt;order-preserving surjections&lt;/a&gt;.  Alternately, if instead of using the language of partial orders you want to use the language of &lt;a href=&quot;http://en.wikipedia.org/wiki/Pre-order&quot; rel=&quot;nofollow&quot;&gt;preorders&lt;/a&gt;, you could say that a preorder $\le$ &lt;em&gt;refines&lt;/em&gt; a preorder $\le&amp;#39;$ on the same set if whenever $x \le y$ it is also true that $x \le&amp;#39; y$.  &lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-12-18T06:41:11.020" />
  <row Id="14757" PostTypeId="2" ParentId="14754" CreationDate="2010-12-18T06:46:48.983" Score="3" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Hint&lt;/strong&gt; : Use the second of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Sylow_theorems&quot; rel=&quot;nofollow&quot;&gt;Sylow theorems&lt;/a&gt;. All the $p$-Sylow subgroups are conjugate to each other and a group is normal iff it is equal to each of its conjugates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some semidirect product would give you an explicit example.&lt;/p&gt;&#xA;" OwnerUserId="1119" LastActivityDate="2010-12-18T06:46:48.983" />
  <row Id="14758" PostTypeId="2" ParentId="14755" CreationDate="2010-12-18T06:53:54.980" Score="3" ViewCount="" Body="&lt;p&gt;The concepts you want are those associated with the notion of partial orders, part of &lt;a href=&quot;http://en.wikipedia.org/wiki/Order_theory&quot; rel=&quot;nofollow&quot;&gt;Order Theory&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When everything is ordered, you have a &lt;a href=&quot;http://en.wikipedia.org/wiki/Total_order&quot; rel=&quot;nofollow&quot;&gt;totally ordered set&lt;/a&gt; (also called &lt;em&gt;linearly ordered set&lt;/em&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you only worry about, say, last names, if you do not specify any order relations between the people with same last name (neither is greater than or smaller than any other), then you have a &lt;a href=&quot;http://en.wikipedia.org/wiki/Partial_order&quot; rel=&quot;nofollow&quot;&gt;partially ordered set&lt;/a&gt;. The fact that your total order is &quot;consistent&quot; with the partial order is expressed saying that the total order is a &quot;linear extension&quot; of the partial order. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are considering people with the same last name but different first name &quot;equal&quot;, then you are dealing with either a &lt;a href=&quot;http://www.wikipedia.org/wiki/Preorder&quot; rel=&quot;nofollow&quot;&gt;preorder&lt;/a&gt;, or a (partial or total) order on the set of &lt;a href=&quot;http://en.wikipedia.org/wiki/Equivalence_class&quot; rel=&quot;nofollow&quot;&gt;equivalence classes&lt;/a&gt; induced by the preorder. &lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-12-18T06:53:54.980" />
  <row Id="14759" PostTypeId="1" AcceptedAnswerId="14760" CreationDate="2010-12-18T06:58:07.130" Score="2" ViewCount="184" Body="&lt;p&gt;This may seem like an overly trivial question, but I've just recently become confused about Langrange's 'prime' notation for derivatives (for example $f'(x)$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know for sure that $f&amp;#39;(x) = \frac{\delta f(x)}{\delta x}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But suppose we replace x with an expression, like 2x+1. Do we write $f&amp;#39;(x^2+1) = \frac{\delta f(x^2+1)}{\delta x}$ or $f&amp;#39;(x^2+1) = \frac{\delta f(x^2+1)}{\delta (x^2+1)}$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does putting the prime around the function instead of between its letter and parentheses make a difference? For example what does $(f(x^2+1))&amp;#39;$ mean?&lt;/p&gt;&#xA;" OwnerUserId="181" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-18T07:12:54.427" LastActivityDate="2010-12-18T09:14:38.813" Title="Prime notation for derivatives" Tags="&lt;calculus&gt;&lt;notation&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="14760" PostTypeId="2" ParentId="14759" CreationDate="2010-12-18T07:01:34.240" Score="8" ViewCount="" Body="&lt;p&gt;$f&amp;#39;$ is a function, so $f&amp;#39;(2x + 1)$ denotes $f&amp;#39;$ applied to $2x + 1$, or $\frac{df}{dx}(2x + 1)$.  For example if $f = x^2$ then $f&amp;#39; = 2x$ and $f&amp;#39;(2x + 1) = 4x + 2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(f(x^2 + 1))&amp;#39;$ is the derivative of the function $f(x^2 + 1)$, which is $2x f&amp;#39;(x^2 + 1)$ by the chain rule.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This question highlights a weakness of the $&amp;#39;$ notation, which is that it always comes with an implied variable with respect to which you're differentiating.  If this variable is clear from context there's no problem, but sometimes it isn't.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-12-18T07:01:34.240" CommentCount="2" />
  <row Id="14761" PostTypeId="2" ParentId="2293" CreationDate="2010-12-18T07:52:59.350" Score="2" ViewCount="" Body="&lt;p&gt;Kakutani's fixed point theorem allowed John Nash to prove his crucial result that started the whole field of game theory in like 9 pages or something -- a very sexy maths thesis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Brouwer's theorem appears in economics as well in &quot;satisficing&quot; solutions (behavior based on choosing &quot;better&quot; options rather than &quot;the best&quot;), for example &lt;a href=&quot;http://www.springerlink.com/index/39Y2VUHUDC749Q8D.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.springerlink.com/index/39Y2VUHUDC749Q8D.pdf&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And, Brouwer's fixed-point theorem was used to Arrow &amp;amp; Debreu to establish general equilibrium theory. That theory is widely despised but mathematically beautiful, however nowadays it's considered more believable if your paper says &quot;I rely on no fixed-point theorems&quot;.&lt;/p&gt;&#xA;" OwnerUserId="1457" LastActivityDate="2010-12-18T07:52:59.350" CommunityOwnedDate="2010-12-18T07:52:59.350" />
  <row Id="14762" PostTypeId="2" ParentId="14759" CreationDate="2010-12-18T09:14:38.813" Score="4" ViewCount="" Body="&lt;p&gt;Often people write something like $(x^2)&amp;#39;=2x$ or $(e^x)&amp;#39;=e^x$, but as you noticed yourself, this is ambiguous, and I never use this notation. You can write $\frac{d}{dx}x^2=2x$ instead, i.e., you don't have to put the $x^2$ into the numerator. This way it looks much clearer, I think (though not shorter, unfortunately). So for me it would be&#xA;$$&#xD;&#xA;  \tfrac{d}{dx}f(x^2+1) = 2xf&amp;#39;(x^2+1).&#xD;&#xA;$$&#xD;&#xA;(However, $\tfrac{df}{dx}(x^2+1) = f&amp;#39;(x^2+1)$.)&lt;/p&gt;&#xA;" OwnerUserId="2575" LastActivityDate="2010-12-18T09:14:38.813" />
  <row Id="14764" PostTypeId="2" ParentId="14740" CreationDate="2010-12-18T09:34:23.940" Score="0" ViewCount="" Body="&lt;p&gt;You'd have y=z only if A is an injective linear map. In fact, that's the definition of being injective: different values in the domain map through A to different values in the codomain or if two values in the domain have identical images through A then they are equal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the general case, this does not hold. As an example, consider the operator T on $\mathbb{R}^3$ &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;T(x,y,z) = (0,y,z)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;T maps, for example, every triple of the form (a,2,2) to (0,2,2). Therefore, you not only have two, but the whole infinite set {a * (1,2,2) | a $\in$ $\mathbb{R}$} mapping to (0,2,2) through T.&lt;/p&gt;&#xA;" OwnerUserId="4832" LastActivityDate="2010-12-18T09:34:23.940" />
  <row Id="14765" PostTypeId="2" ParentId="3977" CreationDate="2010-12-18T09:53:15.143" Score="0" ViewCount="" Body="&lt;p&gt;Try kd3-surf if you haven't yet found it at &lt;a href=&quot;http://k3dsurf.sourceforge.net/&quot; rel=&quot;nofollow&quot;&gt;http://k3dsurf.sourceforge.net/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Free software and small file size, easy to manipulate.&lt;/p&gt;&#xA;" OwnerUserId="4899" LastActivityDate="2010-12-18T09:53:15.143" CommunityOwnedDate="2010-12-18T09:53:15.143" />
  <row Id="14766" PostTypeId="1" CreationDate="2010-12-18T10:24:50.720" Score="2" ViewCount="69" Body="&lt;p&gt;I am mostly asking about terminology here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Background&lt;/strong&gt; A generalized linear eigenvalue problem is of the form $Ax=\lambda B x$, where $A$ and $B$ are square matrices, $\lambda$ is the eigenvalue and $x$ is the eigenvector. We can write this in the form that makes it evident that it is linear in $\lambda$: $(A-\lambda B) x = 0$. A polynomial eigenvalue problem is of the form&#xA;$$ \left[\sum_{k=0}^N \lambda^k A_k\right] x = 0$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question&lt;/strong&gt; Now, I am wondering what you would call an eigenvalue problem which is &quot;bilinear&quot; in its eigenparameters:&#xA;$$ \left[ A\lambda\mu + B\lambda + C\mu + D \right] x = 0 $$&#xD;&#xA;where $\lambda$ and $\mu$ are the eigenvalues, $x$ is the eigenvector and $A,B,C,D$ are square matrices.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Has this kind of problem been studied? I know this is generally a nonlinear eigenvalue problem of the form $M(\lambda,\mu)x = 0$, but it seems like there is important additional structure here.&lt;/p&gt;&#xA;" OwnerUserId="398" LastActivityDate="2010-12-18T10:24:50.720" Title="Multiparameter eigenvalue problem" Tags="&lt;linear-algebra&gt;&lt;matrices&gt;" CommentCount="2" />
  <row Id="14767" PostTypeId="1" AcceptedAnswerId="14799" CreationDate="2010-12-18T10:55:01.493" Score="4" ViewCount="187" Body="&lt;p&gt;&lt;strong&gt;Let $F(n)$ be $n$-th Fibonacci number.$$F(0) = 0, F(1) = 1, F(2) = 1, F(3) = 2, F(4) = 3 \text{ and so on. }$$Given a positive integer $n \gt 2$,Find the smallest prime number $P$ such that $P$ divides $F(n)$ but it does not divide any $F(k)$ smaller than $F(n)$ ?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now,my question: Is it possible to find the answer  without actually computing the value of $F(n)$? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am looking for an interesting algorithm for this purpose.&lt;/p&gt;&#xA;" OwnerUserId="2109" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-12-18T21:17:26.417" LastActivityDate="2010-12-19T16:09:18.567" Title="Finding the smallest prime number that divides $Fib(n)$ but not any other $Fib(k)$ smaller than $Fib(n)$" Tags="&lt;number-theory&gt;&lt;algorithms&gt;&lt;fibonacci-number&gt;" AnswerCount="1" CommentCount="7" />
  <row Id="14768" PostTypeId="1" CreationDate="2010-12-18T11:51:55.097" Score="2" ViewCount="41" Body="&lt;p&gt;Suppose we have a Hadamard matrix $H$ with rows $H_1,H_2,\ldots$ and look at the following system&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$H_j \cdot \mathbf{x} \le 1 \ \forall j$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_i x_i=1$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a name for polytopes defined in this way?&lt;/p&gt;&#xA;" OwnerUserId="998" LastActivityDate="2011-04-19T09:10:54.173" Title="Polytopes defined by Hadamard matrices?" Tags="&lt;geometry&gt;" AnswerCount="1" />
  <row Id="14771" PostTypeId="1" AcceptedAnswerId="14788" CreationDate="2010-12-18T12:59:01.753" Score="2" ViewCount="102" Body="&lt;p&gt;The invariant subspace problem states something along the following lines: Let $H$ be a Banach space of dimension greater than one. Does every bounded linear operator $T:H \to H$ have a non-trivial closed $T$-invariant subspace?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now my question is: Why is this a big problem? What does the solution give us? It has been solved in most of the special cases which occur in practice.&lt;/p&gt;&#xA;" OwnerUserId="1120" LastEditorUserId="1120" LastEditorDisplayName="" LastEditDate="2010-12-18T13:26:47.257" LastActivityDate="2010-12-18T17:46:17.463" Title="Invariant Subspace Problem." Tags="&lt;reference-request&gt;&lt;functional-analysis&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="14772" PostTypeId="2" ParentId="14771" CreationDate="2010-12-18T13:17:46.513" Score="1" ViewCount="" Body="&lt;p&gt;The invariant subspace problem asks whether &lt;strong&gt;every&lt;/strong&gt;  bounded linear operator has a non-trivial invariant subspace. The existence of some operator with invariant subspaces is trivial. (Consider finite-rank operators.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And it is known that the theorem is not true for all Banach spaces so the current formulation asks whether it is true for Hilbert spaces. I do not have any good references however. &lt;/p&gt;&#xA;" OwnerUserId="4623" LastActivityDate="2010-12-18T13:17:46.513" CommentCount="1" />
  <row Id="14774" PostTypeId="2" ParentId="12526" CreationDate="2010-12-18T13:35:44.517" Score="1" ViewCount="" Body="&lt;p&gt;It's wrong: Take $x=0$ and $y=e_1=(1,0,0,\dots)$. Then $f(x)=e_1$ and $f(y)=e_2=(0,1,0,\dots)$, so $r(x) = 1/2$ and $r(y) = \sqrt2/2$. Now it is not hard to see that $p = (-\sqrt7/4,3/4,0,\dots)$ is the point in $Ty = B(f(y),r(y))\cap B$ farthest away from $f(x)$. From this one deduces that&#xA;$$&#xD;&#xA;  D(Tx,Ty) = \sqrt{(1+\sqrt7/4)^2+(3/4)^2} - 1/2 = \sqrt{2+\sqrt7/2} - 1/2  &amp;gt; 1 = \|x-y\|.&#xD;&#xA;$$&#xD;&#xA;(As $Ty$ is the larger chunk, geometric considerations show that $Ty \subseteq N_r(Tx)$ implies $Tx \subseteq N_r(Ty)$, for all $r&amp;gt;0$.)&lt;/p&gt;&#xA;" OwnerUserId="2575" LastEditorUserId="2575" LastEditorDisplayName="" LastEditDate="2010-12-18T17:38:16.397" LastActivityDate="2010-12-18T17:38:16.397" CommentCount="4" />
  <row Id="14775" PostTypeId="1" AcceptedAnswerId="14784" CreationDate="2010-12-18T13:50:48.087" Score="5" ViewCount="315" Body="&lt;p&gt;Fiction &quot;Division by Zero&quot; By Ted Chiang&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I read the fiction story &lt;a href=&quot;http://www.fantasticmetropolis.com/i/division/full/&quot; rel=&quot;nofollow&quot;&gt;&quot;Division by Zero&quot; By Ted Chiang&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My interpretation is the character finds a proof that arithmetic is inconsistent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a formal proof the fiction can't come true? (I don't suggest the fiction can come true).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: I see someone &lt;a href=&quot;http://mathoverflow.net/questions/40920&quot; rel=&quot;nofollow&quot;&gt;tried&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4750" LastEditorUserId="4750" LastEditorDisplayName="" LastEditDate="2010-12-19T08:07:29.193" LastActivityDate="2010-12-20T02:43:19.067" Title="Fiction &quot;Division by Zero&quot; By Ted Chiang" Tags="&lt;meta-math&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="1" />
  <row Id="14776" PostTypeId="2" ParentId="14771" CreationDate="2010-12-18T13:54:33.493" Score="4" ViewCount="" Body="&lt;p&gt;A similar question was asked on MathOverflow a week ago. Take a look at the enlightening answer by Bill Johnson and the discussion in the comments &lt;a href=&quot;http://mathoverflow.net/questions/48908/is-the-invariant-subspace-problem-interesting&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="723" LastActivityDate="2010-12-18T13:54:33.493" CommentCount="1" />
  <row Id="14777" PostTypeId="1" AcceptedAnswerId="14779" CreationDate="2010-12-18T14:08:29.123" Score="1" ViewCount="238" Body="&lt;p&gt;Apparently, the solution to the &lt;a href=&quot;http://math.stackexchange.com/questions/234/card-doubling-paradox&quot;&gt;Card Doubling Paradox&lt;/a&gt; is that a uniform probability distribution over the positive real numbers doesn't exist. Can anyone explain why this is the case and what probability distributions can exist over the positive real numbers (it seems that this would be quite limited, given that such a simple distribution is impossible)?&lt;/p&gt;&#xA;" OwnerUserId="123" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-18T14:19:27.770" LastActivityDate="2010-12-18T21:04:13.963" Title="Why isn't there a uniform probability distribution over the positive real numbers?" Tags="&lt;probability-theory&gt;" AnswerCount="4" CommentCount="1" />
  <row Id="14779" PostTypeId="2" ParentId="14777" CreationDate="2010-12-18T15:40:34.400" Score="6" ViewCount="" Body="&lt;p&gt;A uniform probability over the positive real numbers would not satisfy all three axioms of probability. Specifically, there is a conflict between the second axiom ($P(\Omega) = 1$) and the third axiom (countable additivity).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Check out the wikipedia entry &lt;a href=&quot;http://en.wikipedia.org/wiki/Axioms_of_probability&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Axioms_of_probability&lt;/a&gt;. I think it should be clear how a uniform probability over the positive reals would cause problems.&lt;/p&gt;&#xA;" OwnerUserId="2641" LastActivityDate="2010-12-18T15:40:34.400" />
  <row Id="14780" PostTypeId="1" AcceptedAnswerId="14781" CreationDate="2010-12-18T16:19:37.773" Score="3" ViewCount="143" Body="&lt;p&gt;I recently took my Honors Pre-Calc. exam. Wanting to get as many points as I possibly could, I circled and skipped several problems. However, I forgot to do this one, and wanted to know if I'm on the right track, because the answer I get for x doesn't work correctly. I've come up with a separate solution that comes close, but is off by .4 as well. &#xA;\begin{align*}&#xA;e^{2x} - 7e^x + 12 &amp;amp;= 0\\&lt;br&gt;&#xA;e^{2x} - 7e^x &amp;amp;= -12 \\ &#xA;e^x(1^2 - 7) &amp;amp;= -12\\&lt;br&gt;&#xA;e^x &amp;amp;= \frac{-12}{1^2 - 7}\\&lt;br&gt;&#xA;e^x &amp;amp;= 2\\&lt;br&gt;&#xA;x &amp;amp;= \ln 2&#xA;\end{align*}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Putting this back into the equation doesn't work. I think I might have a problem with how I'm factoring, but I can't seem to fix it. The only other way I've tried comes closer, but still gives me a wrong answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your help and suggestions are appreciated :).&lt;/p&gt;&#xA;" OwnerUserId="4888" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-18T17:20:32.097" LastActivityDate="2010-12-18T17:21:49.083" Title="Solving an equation with $e^x$" Tags="&lt;algebra-precalculus&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="14781" PostTypeId="2" ParentId="14780" CreationDate="2010-12-18T16:24:00.077" Score="14" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\rm\quad  e^{\:2\:x} = (e^x)^2\:.\ $ Thus for $\rm\ y = e^x\ $ the equation is $\rm\ y^2 - 7\ y + 12 = 0\:$&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-12-18T16:24:00.077" CommentCount="1" />
  <row Id="14782" PostTypeId="1" CreationDate="2010-12-18T17:03:47.753" Score="1" ViewCount="89" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Is there a finite group which is contained in infinitely many core-free finite groups as the centralizer of a central involution?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The Brauer–Fowler results show that if a finite group has no non-identity normal subgroups of odd order (so is core-free), then its order is bounded by a function of the size of a few centralizers (of involutions, of strongly real elements, etc.).  In particular, the size of a core-free group is bounded by a function of the size of its largest and smallest involution centralizers, and a simple group by a function of the size of any of its involution centralizers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sometimes one can apparently do better: there are only finitely many core-free groups with a central involution centralizer of size 4.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there an explicit example where one cannot do better?  In other words:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Is the some finite group X such that there are infinitely many finite groups G with no non-identity normal subgroups of odd order, and with a Sylow 2-subgroup P and involution t in Z(P) such that C&lt;sub&gt;G&lt;/sub&gt;(t) = X?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="583" LastEditorUserId="583" LastEditorDisplayName="" LastEditDate="2011-02-10T02:54:45.750" LastActivityDate="2011-02-10T02:54:45.750" Title="Involution centralizer does not determine the group" Tags="&lt;group-theory&gt;" CommentCount="1" FavoriteCount="1" />
  <row Id="14783" PostTypeId="2" ParentId="14777" CreationDate="2010-12-18T17:03:59.580" Score="3" ViewCount="" Body="&lt;p&gt;For every probability density $f$, $\lim_{x\to \pm \infty} f(x) = 0$ must hold.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is clear that for a uniform distribution, the density has to be constant on the considered interval.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Combining both requirements, only $g(x) = 0$ remains as a choice for a uniform density on $(-\infty, \infty)$. But $\int_{-\infty}^\infty g(x)dx = 0 \neq 1$, that is $g$ is no proper probability density.&lt;/p&gt;&#xA;" OwnerUserId="3330" LastActivityDate="2010-12-18T17:03:59.580" />
  <row Id="14784" PostTypeId="2" ParentId="14775" CreationDate="2010-12-18T17:16:51.837" Score="4" ViewCount="" Body="&lt;blockquote&gt;Is there a formal proof the fiction can't come true?&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No, by &lt;a href=&quot;http://en.wikipedia.org/wiki/G%C3%B6del%27s_incompleteness_theorems#Second_incompleteness_theorem&quot; rel=&quot;nofollow&quot;&gt;Gödel's second incompleteness theorem&lt;/a&gt;, formal systems can prove their own consistency if and only if they are inconsistent. So given that arithmetic is consistent, we'll never be able to &lt;i&gt;prove&lt;/i&gt; that it is. (&lt;b&gt;EDIT&lt;/b&gt;: Actually not quite true; see Alon's clarification below.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an aside, if you liked &quot;Division by Zero,&quot; you might also like Greg Egan's pair of stories in which arithmetic isn't consistent: &quot;Luminous&quot; and &quot;&lt;a href=&quot;http://www.asimovs.com/_issue_0805/DarkINtegers.shtml&quot; rel=&quot;nofollow&quot;&gt;Dark Integers&lt;/a&gt;&quot;.&lt;/p&gt;&#xA;" OwnerUserId="2773" LastEditorUserId="2773" LastEditorDisplayName="" LastEditDate="2010-12-20T02:43:19.067" LastActivityDate="2010-12-20T02:43:19.067" CommentCount="2" />
  <row Id="14785" PostTypeId="2" ParentId="14780" CreationDate="2010-12-18T17:21:49.083" Score="4" ViewCount="" Body="&lt;p&gt;Your error is in the third line. $e^{2x}$ is not equal to $e^x$ times $1^2$ (that would make $e^{2x}$ equal to $e^x\times 1 = e^x$). That line would read&#xA;$$e^x(e^x - 7) = -12,$$&#xD;&#xA;and that won't really help you. Bill's solution is the correct approach.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-12-18T17:21:49.083" />
  <row Id="14786" PostTypeId="1" CreationDate="2010-12-18T17:31:06.293" Score="8" ViewCount="498" Body="&lt;p&gt;I was looking at movie times today and was struck by the oddly-spaced showing times. For example, at the local Loew's Theater &quot;Tron: Legacy 3D&quot; (127 min.) is playing on two screens at the following interlaced times: 1:00 pm, 1:45 pm, 4:00 pm, 4:45 pm, 7:00 pm, 7:45 pm, 10:00 pm and 10:45 pm. Why not space the times equally? Is there an algorithm at work here? Other than optimizing food sales by cleverly keeping a pool of waiters, the strange times might have to do with overbooking and accommodating johnnys-come-lately. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the following idealized scenario. Suppose only $1$ movie is a playing at a theater with $n$ screens and free popcorn and refreshments is given upon sitting in the theater, so no other factors are relevant for spacing movie times but ticket sales. Suppose each showing can accommodate at most $N$ people. Suppose $N \pm M$ arrive at the kiosk reasonably before any particular showing time, where $0 &amp;lt; M &amp;lt; N$, and $0 &amp;lt; L &amp;lt; M$ people show up just a little too late for any particular show -- the same number of latecomers come by each time. If any person has to wait for more than some fraction $0 &amp;lt; R &amp;lt; 1$ of the time $t$ of the movie in question to watch the next movie in the cue, then he/she returns the ticket and goes home. Suppose the $\pm$ sign above is governed by tossing a fair coin, $+$ for heads, $-$ for tails.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; Given the above data, what is the optimal spacing of $X$ movie times, each movie of the same length $t$, on $n$ different screens that maximizes the total number of ticket purchases &lt;em&gt;and&lt;/em&gt; (happy) moviegoers?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If this question is too easy, then generalize the above scenario to multiple movies showing at the same theater. If this question is too hard, then simplify it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Of course, feel free to edit and improve.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(&lt;em&gt;Added Thoughts&lt;/em&gt;) The constraints above are in place to try to model the scenario as closely as possible while keeping the mathematics simple. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd like to account for a little randomness, and the simplest truly non-trivial random event is the tossing of a fair coin. If $N - M$ or $N + M$ people come every time, then the problem is trivial or cumulatively impossible, respectively. What makes this problem tractable is that there are &lt;em&gt;some&lt;/em&gt; occasions when some people are left out of a showing. These people are either at the end of a long cue or literally late; either way they must wait but few, if any, will wait longer than the length of the movie. &lt;em&gt;I believe the answer of spacing depends heavily on the amount of wait time.&lt;/em&gt; That is, if $R = 0$, $L + M &amp;gt; 0$ people go home every time (not optimal). If $R = 1$, then any reasonable spacing should suffice to accommodate the extremely patient moviegoers. I think this possibility oversimplifies the problem, unless I'm missing something crucial or obvious. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suppose also that the condition $L &amp;lt; M$ could be relaxed to $L &amp;lt; N$, but my reasoning is that latecomers seem to be rarer than overbookers. Are these constraints reasonable?&lt;/p&gt;&#xA;" OwnerUserId="2720" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2011-04-28T09:44:30.320" LastActivityDate="2011-04-28T09:44:30.320" Title="Going to the Movies!" Tags="&lt;probability&gt;&lt;stochastic-processes&gt;&lt;recreational-mathematics&gt;" CommentCount="12" FavoriteCount="2" />
  <row Id="14787" PostTypeId="1" AcceptedAnswerId="14790" CreationDate="2010-12-18T17:34:06.200" Score="4" ViewCount="603" Body="&lt;p&gt;I'm investigating an idea in cryptography that requires irreducible polynomials with coefficients of either 0 or 1 (e.g. over GF[2]). Essentially I am mapping bytes to polynomials. For this reason, the degree of the polynomial will always be an integer multiple of 8.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to build a table of such irreducible polynomials for various degrees (e.g. degrees from 8, 16, 24, ..., 1024). Because there are multiple irreducible polynomials for a given degree, I'd like the one with the fewest number of terms since I will hard code the non-zero terms. For example, for degree 16, both of these polynomials are irreducible:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x^{16} + x^{14} + x^{12} + x^7 + x^6 + x^4 + x^2 + x + 1$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x^{16} + x^5 + x^3 + x + 1$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously, the latter one is preferred because it requires less space in code and is more likely to be right (e.g. that I wouldn't have made a copy/paste error).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore, I've noticed that to at least degree 1024 where the degree is a multiple of 8, there are irreducible polynomials of the form:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x^n + x^i + x^j + x^k + 1$ where $n = 8*m$ and $0 &amp;lt; i,j,k &amp;lt; 25$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there an good algorithmic way of finding these polynomials (or ones that have even fewer terms)? Again, the purpose is to keep the non-zero terms in a look-up table in code.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance for any help!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;UPDATE&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This Mathematica code generates all pentanomials for degrees that are multiples of 8 up to degree 1024:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;IrreducibleInGF2[x_] := IrreduciblePolynomialQ[x, Modulus -&amp;gt; 2]&#xA;&#xA;ParallelTable[&#xA; Select[&#xA;   Sort[&#xA;    Flatten[&#xA;        Table[&#xA;           x^n + x^a + x^b + x^c + 1, &#xA;           {a, 1, Min[25, n - 1]}, {b, 1, a - 1}, {c, 1, b - 1}&#xA;             ]&#xA;           ]&#xA;       ], &#xA;       IrreducibleInGF2, 1], &#xA;       {n, 8, 1024, 8}]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;(I sorted the list of polynomials to make sure I always got the one with the overall smallest degrees first). However, it takes quite a bit of time to run. For example, it took over 26 minutes for the case of $x^{984} + x^{24} + x^9 + x^3 + 1$ .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;UPDATE #2&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The HP paper &quot;&lt;a href=&quot;http://www.hpl.hp.com/techreports/98/HPL-98-135.pdf&quot; rel=&quot;nofollow&quot;&gt;Table of Low-Weight Binary Irreducible Polynomials&lt;/a&gt;&quot; has been incredibly helpful. It lists up to $x^{10000}$ and reiterates a proof by Swan that there are no trinomials when the degree is a multiple of 8 (which matches my findings). I've spot checked that their results match mine up to $x^{1024}$ so I'll just need to double check their results up to 10000 which should be much easier than finding them myself.&lt;/p&gt;&#xA;" OwnerUserId="872" LastEditorUserId="872" LastEditorDisplayName="" LastEditDate="2010-12-20T04:01:42.080" LastActivityDate="2010-12-20T17:05:20.637" Title="Finding irreducible polynomials over GF(2) with the fewest terms" Tags="&lt;algorithms&gt;&lt;polynomials&gt;&lt;finite-fields&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="2" />
  <row Id="14788" PostTypeId="2" ParentId="14771" CreationDate="2010-12-18T17:46:17.463" Score="3" ViewCount="" Body="&lt;p&gt;One of the primary reasons for studying the invariant subspace problem is the hope that it will lead to useful structural theories. For example, in their interesting expository paper $\ $ &lt;a href=&quot;http://www.springerlink.com/content/9393t358126466x4/fulltext.pdf&quot; rel=&quot;nofollow&quot;&gt;The invariant subspace problem,&lt;/a&gt; $\ $ Radjavi and Rosenthal wrote:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;It is hoped  that knowledge of the  invariant subspaces of operators will shed light on their structure.  In  the case of operators on finite-dimensional spaces,  for example, the Jordan canonical form theorem shows that operators &#xA;  are direct sums of particularly well-behaved operators on certain invariant subspaces.  Also, normal operators on infinite-dimensional spaces can be represented as integrals with respect to some invariant subspaces, (their spectral subspaces -  this is the  spectral theorem). To prove that every operator has a non-trivial invariant subspace might be the beginning of a general structure theory. $\ \:$ On the other hand, $\ $ a concrete counterexample would also be very interesting. To say that $A$ has no non-trivial invariant subspace  is equivalent to saying that linear combinations of $\ {A^n\:f\:: n \in \mathbb N}\ $ are dense in the space for each nonzero vector $f$. Representing such an $A$  on various spaces would give a number of approximation theorems.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This paper, combined with Yadav's $\ $ &lt;a href=&quot;http://www.math.leidenuniv.nl/~naw/serie5/deel06/jun2005/pdf/yadav.pdf&quot; rel=&quot;nofollow&quot;&gt;The Invariant Subspace Problem&lt;/a&gt; $\ $provide nice expositions on the problem and its rich history. Neither of these introductions are mentioned on said MO page. IMO they provide a much nicer overview of the problem than the little that is said there.&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-12-18T17:46:17.463" />
  <row Id="14789" PostTypeId="1" CreationDate="2010-12-18T17:58:17.487" Score="3" ViewCount="299" Body="&lt;p&gt;In numerical analysis, I was asked whether Bairstow's algorithm convergence rate is quadratic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My initial feeling was that it does, since it is essentially Newton's method for a system of non-linear equations, and Newton's method converges quadratically in one dimension (When f is from R to R).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A quick search on the internet seemed to indicate that indeed, the rate is quadratic as long as the roots are of multiplicity 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I have real trouble proving this. I was trying to imitate the proof of Newton's method in one dimesnsion found in Kincaid and Cheney's book &quot;Numerical Analysis&quot;, but ran into some trouble since there isn't a simple &quot;second derivative&quot; for a vector valued function (unless I want to use tensors, and that's way beyond our course material).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I'm stuck. How do I show that the rate of convergence is quadratic (or that it's not if the roots aren't simple)?&lt;/p&gt;&#xA;" OwnerUserId="4209" LastActivityDate="2011-04-07T04:39:38.810" Title="Bairstow's method: Rate of convergence" Tags="&lt;homework&gt;&lt;numerical-methods&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="14790" PostTypeId="2" ParentId="14787" CreationDate="2010-12-18T18:00:50.213" Score="4" ViewCount="" Body="&lt;p&gt;According to a paper &quot;Optimal Irreducible Polynomials for GF(2m) Arithmetic&quot; by M. Scott,&#xA;&quot;it is in practise always possible to chooose as an irreducible polynomial either a trinomial... or a pentanomial.&quot; &lt;a href=&quot;http://www.hyperelliptic.org/SPEED/slides/Scott_optimal_polynomials.pdf&quot; rel=&quot;nofollow&quot;&gt; [talk slides]&lt;/a&gt; &lt;a href=&quot;http://eprint.iacr.org/2007/192&quot; rel=&quot;nofollow&quot;&gt; [PDF link]&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In random number generators irreducible trinomials of various degrees with three nonzero binary coefficients are associated with the names &lt;a href=&quot;http://www.ciphersbyritter.com/RES/RNGENS.HTM&quot; rel=&quot;nofollow&quot;&gt; Tausworthe-Lewis-Payne&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added:&lt;/strong&gt; It has been known &lt;a href=&quot;http://arxiv.org/abs/1001.0409&quot; rel=&quot;nofollow&quot;&gt; since Gauss&lt;/a&gt; that there are lots of irreducible polynomials over a finite field, basically &lt;a href=&quot;http://en.wikipedia.org/wiki/Prime_number_theorem#Analogue_for_irreducible_polynomials_over_a_finite_field&quot; rel=&quot;nofollow&quot;&gt; the analog of the Prime Number Theorem&lt;/a&gt; for such polynomials.  Among the $2^m$ (monic) polynomials over Z/2Z of degree $m$, approximately $1/m$ of them are irreducible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can eliminate the possibility of first degree factors by inspection, for divisibility by $x$ or $x+1$ would imply respectively a zero constant term or an even number of nonzero terms in the polynomial.  So the first case to test for irreducibility is the trinomials of degree $m$.  With leading and constant coefficients accounting for two of the three nonzero terms, there are but $m-1$ possibilities to test, and by symmetry of $x$ &amp;rarr; $1/x$ substitution, we can restrict the middle terms to degree &amp;le; $m/2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If none of those pan out, we have the richer supply of pentanomials to test.  Indeed you seem to have hit upon a seam of cases where &lt;strong&gt;trinomials will never work out&lt;/strong&gt;, namely degree $m$ &lt;a href=&quot;http://u.math.biu.ac.il/~vishne/publications/trinom.ps&quot; rel=&quot;nofollow&quot;&gt; a multiple of 8 [PS]&lt;/a&gt; (Swan, 1962).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The work then comes down to testing all the $\binom{m-1}{3}$ binary pentanomials $p(x)$ until we find one that's irreducible.  Your application might make other conditions, perhaps similar to those considered in Scott's paper above, attractive.  Given the modest degrees you are working with, trial division (taking $p(x) \; mod \; q(x)$ for all $q(x)$ of degree &amp;le; $m/2$) should be fast enough. [Remember, we shouldn't have to test more than O(m) possibilities before we find success.]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is &lt;a href=&quot;http://wwwmaths.anu.edu.au/~brent/pd/ANZMC08t4.pdf&quot; rel=&quot;nofollow&quot;&gt; a fancier way [PDF]&lt;/a&gt; to test polynomials for irreducibility over GF(2).  A &lt;strong&gt;necessary&lt;/strong&gt; condition for binary polynomial $p(x)$ to be irreducible over $GF(2)$ is that:&lt;br&gt;&#xA;$$x^{2^m} = x \mod p(x)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact Gauss showed for prime q that $x^{q^m} - x$ is precisely the product&#xA;of all monic irreducible polynomials over $GF(q)$ whose degrees divide&#xA;$m$.  [From this he deduced the count of monic irreducible polynomials of&#xA;degree &lt;strong&gt;exactly&lt;/strong&gt; $m$ is asymptotically $q^m/m$ as $m \rightarrow \infty$.]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $q = 2$ it follows that if $p(x)$ is irreducible of degree $m$,&#xA;it divides $x^{2^m} - x$ over $GF(2)$, i.e. the congruence above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rubin (1980) proposed a necessary and sufficient test for irreducibility,&#xA;combining the above with some additional steps to rule out the possibility&#xA;that $p(x)$ might be the product of some irreducible factors whose degrees&#xA;properly divide $m$.  [While the degrees of the irreducible factors would&#xA;naturally &lt;em&gt;sum&lt;/em&gt; to $m$, having all the irreducible factors' degrees &lt;em&gt;divide&lt;/em&gt;&#xA;$m$ would be somewhat special, unless of course there is only one factor.]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The additional &quot;sufficiency&quot; steps are to check for each &lt;em&gt;prime&lt;/em&gt; factor&#xA;$d_i$ of $m$ that:&#xA;$$GCD(x^{2^{m/d}} - x, p(x)) = 1$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is, if $p(x)$ were to have an irreducible factor of degree $k$ properly&#xA;dividing $m$, it would crop up when taking the gcd of $x^{2^{m/d}} - x$ and&#xA;$p(x)$ if $k$ divides $m/d$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since then a lot of ingenuity has been applied to efficiently doing these&#xA;steps.  Of course the necessary condition lends itself to &lt;strong&gt;repeated squaring&lt;/strong&gt;,&#xA;computing $x^{2^{k+1}} = (x^{2^k})^2$ mod $p(x)$, for $k$ up to $m-1$.  We&#xA;can take advantage here of the fact that the multiplication we're doing&#xA;is a squaring and advantage of the sparsity of our $p(x)$ as a pentanomial&#xA;when doing reduction mod $p(x)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the report by Brent of work with Zimmerman (linked above) points out,&#xA;this repeated squaring gives with each (fairly inexpensive step) linear&#xA;progress toward the &quot;exponent of the exponent&quot; $m$.  There is also a way&#xA;to progress farther with greater computational effort by &lt;strong&gt;modular&#xA;composition&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is, suppose we've already arrived at:&#xA;$$f(x) = x^{2^k} \mod p(x)$$&#xD;&#xA;and&#xA;$$g(x) = x^{2^j} \mod p(x)$$&#xD;&#xA;Then:&#xA;$$f(g(x)) = x^{2^{k+j}} \mod p(x)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus composition of two polynomials $f(x)$ and $g(x)$ mod $p(x)$ can&#xA;replace a number of repeated squaring steps.  But composition mod $p(x)$&#xD;&#xA;is more expensive than squaring or even than multiplication generally&#xA;mod $p(x)$.  So as Brent points out, practical advantage for using&#xA;modular composition lies at the final stage(s) of evaluating the&#xA;necessary condition.  E.g. at the end one modular composition might&#xA;replace $m/2$ repeated squarings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as the &quot;sufficiency&quot; conditions go, Gao and Panario outlined&#xA;an improvement over a naive implementation of Rubin's tests in this&#xA;&lt;a href=&quot;http://www.math.clemson.edu/faculty/Gao/papers/GP97a.pdf&quot; rel=&quot;nofollow&quot;&gt;&#xA;1997 paper [PDF]&lt;/a&gt;, basically sequencing the gcd computations in&#xA;an expedient order.&lt;/p&gt;&#xA;" OwnerUserId="3111" LastEditorUserId="3111" LastEditorDisplayName="" LastEditDate="2010-12-20T17:05:20.637" LastActivityDate="2010-12-20T17:05:20.637" CommentCount="3" />
  <row Id="14791" PostTypeId="2" ParentId="14787" CreationDate="2010-12-18T18:10:42.423" Score="0" ViewCount="" Body="&lt;p&gt;The abstract of the following paper indicates that it may be of interest:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;An Improved Method for the Table Lookup Multiplication Algorithm over $\rm\: GF(2^m\:)$.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Author: Hadama Yoichi (Kobe Univ., Graduate School of Sci. and Technol., JPN)&lt;br&gt;&#xA;Nakamura Daisuke (Kobe Univ., Graduate School of Sci. and Technol., JPN)&lt;br&gt;&#xA;Hirotomo Masanori (Hyogojohokyoikukiko)&lt;br&gt;&#xA;Morii Masakatsu (Kobe Univ., Fac. Engineering, JPN)    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Journal Title: IEIC Technical Report (Institute of Electronics, Information and Communication Engineers)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Journal Code: S0532B; ISSN:0913-5685 VOL.106;NO.175(ISEC2006 9-39);PAGE.95-102(2006)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Abstract:&lt;/em&gt; As effective arithmetic methods in the finite field $\rm\: GF(2^m\:)$, M.A. Hasan has presented a look-up table-based algorithm for $\rm\: GF(2^m\:)$ multiplication. In Hasan's method, the number of memory access and the table size are depend on the form of the irreducible polynomial which is used in the $\rm\: GF(2^m\:)$ multiplication. In this paper, we propose a lookup-up table-based algorithm which can be performed as an efficient $\rm\: GF(2^m\:)$ multiplication using the irreducible polynomial which has $m$ degree and $m$ terms. In this method, the algorithm is performed as the multiplication modulo quadnomial which is obtained by the irreducible polynomial which has m degree and m terms. Using this method, the number of memory access and the table size in the algorithm is smaller than that in Hasan's method, and $\rm\: GF(2^m\:)$ multiplication is faster than Hasan's method.&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-12-18T18:10:42.423" CommentCount="4" />
  <row Id="14792" PostTypeId="1" AcceptedAnswerId="14941" CreationDate="2010-12-18T18:14:19.313" Score="2" ViewCount="76" Body="&lt;p&gt;This is part of Exercise III.1.16(c) of Bourbaki's set theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $E$ be a lattice. The condition &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(x\vee y)\wedge(z\vee(x\wedge y))=(x\wedge y)\vee(y\wedge z)\vee(z\wedge x)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for all $x,y,z\in E$ implies that $E$ is distributive. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I already know that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(x\vee y)\wedge(y\vee z)\wedge(z\vee x)=(x\wedge y)\vee(y\wedge z)\vee(z\wedge x)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;implies distributivity, so it suffices to prove&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(x\vee y)\wedge(z\vee(x\wedge y))=(x\vee y)\wedge(y\vee z)\wedge(z\vee x)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can't find a proof of this last statement. Either it is true in any lattice, then the proof should be so simple I must be blind not to see it, or I have to use the condition above again, but I can't imagine in what way. &lt;/p&gt;&#xA;" OwnerUserId="1878" LastActivityDate="2010-12-20T12:37:45.443" Title="Condition for a lattice to be distributive" Tags="&lt;lattices&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="14793" PostTypeId="1" CreationDate="2010-12-18T18:34:42.893" Score="1" ViewCount="172" Body="&lt;p&gt;Let me try to split the question in a few parts,&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;I would like to understand the claim that all non-degenerate bilinear symmetric forms are equivalent over the complex while for the reals they can be distinguished by the signature. Hence there is just one Clifford algebra over $\mathbb{C}^n$ but over $\mathbb{R}^n$ there are as many Clifford algebras as integral solutions to $p+q = n$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Is the above thing that I am puzzled about also related to another fact that I want to understand that any complex square matrix can be similarity transformed to a symmetric complex matrix?)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If I have a bilinear form on a real vector space with a non-trivial signature and then I complexify the vector space then why does the bilinear form always lift to something with a positive definite signature? &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;How does one see that the complexified Clifford algebra over a real vector space with a non-trivial signature is isomorphic to the Clifford algebra over the complexified vector space with a positive definite norm? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is the above something special about the Clifford algebra or is there something more general?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I want to understand how on this complexified algebra there exists a natural automorphism which leaves a subalgebra invariant and hence gives a definition for ``real subalgebra&quot; &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;In this context I would want to know about the notion of &quot;even subalgebras&quot;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;How to understand the constraints between dimensions and signatures about taking real subalgebras? &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="965" LastEditorUserId="965" LastEditorDisplayName="" LastEditDate="2010-12-18T19:20:38.907" LastActivityDate="2011-02-16T21:22:57.237" Title="Complexifying representations" Tags="&lt;abstract-algebra&gt;&lt;physics&gt;&lt;representation-theory&gt;&lt;lie-algebra&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="14794" PostTypeId="1" AcceptedAnswerId="14796" CreationDate="2010-12-18T18:42:25.937" Score="11" ViewCount="310" Body="&lt;p&gt;This question came up when trying to trying to find a $3\times3$ real matrix $A$ such that &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$Ax$ is nonzero for nonzero $x$&#xD;&lt;/li&gt;&#xA;&lt;li&gt;$Ax$ is orthogonal to $x$ for any $x$ in $\mathbb{R}^3$&#xD;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;We know such a matrix cannot exist because $A$ must have an eigenvalue (thus there is some $x$ such that either $Ax = 0$ or $Ax$ is parallel to $x$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Is there a nice, &lt;strong&gt;purely geometric&lt;/strong&gt; way to justify that every operator on $\mathbb{R}^3 $ has a (real) eigenvalue?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To clarify: I'm looking an intuitive way to &lt;em&gt;visualize&lt;/em&gt; why an eigenvalue must exist in this case.  In particular, no polynomials and no determinants are allowed!&lt;/p&gt;&#xA;" OwnerUserId="4904" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2010-12-19T04:12:52.063" LastActivityDate="2010-12-20T18:57:36.887" Title="Geometric argument that operators on $\mathbb{R}^3$ have an eigenvalue?" Tags="&lt;linear-algebra&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="2" />
  <row Id="14795" PostTypeId="2" ParentId="14789" CreationDate="2010-12-18T18:43:39.587" Score="0" ViewCount="" Body="&lt;p&gt;You can find a nice concise exposition of Newton's method for algebraic systems of equations in the paper below. There it is applied to the algebraic equations that arise from (multivariate) polynomial factorization - but the ideas are quite general (Newton-Hensel methods are fundamental for effective algebraic computation). For a more general and more detailed presentation see the chapter on &quot;Hensel algorithms&quot; in Zippel's textbook &lt;em&gt;Effective Polynomial computation,&lt;/em&gt; and see also my &lt;a href=&quot;http://groups.google.com/groups?selm=y8zohi1mgg1.fsf_-_@martigny.ai.mit.edu&quot; rel=&quot;nofollow&quot;&gt;sci.math post&lt;/a&gt; for references to expositions of Mora, Ribenboim, von zur Gathen et. al.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Zippel, R. $\ $ &lt;a href=&quot;http://portal.acm.org/citation.cfm?id=806372&quot; rel=&quot;nofollow&quot;&gt;Newton's Iteration and the Sparse Hensel Algorithm&lt;/a&gt;,&lt;br&gt;&#xA;Proc. 1981 ACM Sympos. Symbolic and Algebraic Computation, Utah, 1981, pp. 68-72&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-12-18T18:43:39.587" />
  <row Id="14796" PostTypeId="2" ParentId="14794" CreationDate="2010-12-18T19:10:31.177" Score="20" ViewCount="" Body="&lt;p&gt;If $A\colon\mathbb{R}^3\to\mathbb{R}^3$ was a linear map with no eigenvectors, then $x\mapsto Ax/\Vert Ax\Vert$ would give a map on the unit sphere with no fixed points and without taking any point to its antipode, contradicting the &lt;a href=&quot;http://en.wikipedia.org/w/index.php?title=Hairy_ball_theorem&amp;amp;oldid=402767345#Corollary&quot; rel=&quot;nofollow&quot;&gt;hairy ball theorem&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1321" LastActivityDate="2010-12-18T19:10:31.177" />
  <row Id="14797" PostTypeId="2" ParentId="14794" CreationDate="2010-12-18T19:55:12.590" Score="3" ViewCount="" Body="&lt;p&gt;Just using linear algebra but no determinants, the closer to what you ask that I know uses that polynomials of degree 3 with real coefficients have roots: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's argue by contradiction, and assume that $A$ has no eigenvalues. First, I claim that any $v$ is contained in a proper $A$-invariant subspace of ${\mathbb R}^3$. This is clear if $v=0$. Otherwise, consider $v,Av,A^2v,A^3v$. They are linearly dependent, so there are real coefficients $a,b,c,d$ not all zero with $(aA^3+bA^2+cA+dI)v=0$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we examine the polynomial $p(x)=ax^3+bx^2+cx+d$. If $a\ne0$, $p$ has a real root $r$, and we can write $p(x)=(x-r)(ax^2+ex+f)$ for some real numbers $e,f$. This gives us $(A-rI)(aA^2+eA+fI)v=0$, but $A-rI$ is invertible (since $r$ is not an eigenvalue of $A$), so in fact $(aA^2+eA+fI)v=0$. If $a=0$, then directly we have $(bA^2+cA+dI)v=0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This shows that $v$ is contained in a proper $A$-invariant subspace (consider the span $S$ of $v,AV$, and note that the argument above shows that either $A^2v$ is in this span, from which it follows that $S$ is $A$-invariant, or else, the coefficient of $A^2$ above is 0, and in fact $v$ is an eigenvalue, a contradiction).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we may assume that any non-zero $v$ is in an $A$-invariant &lt;em&gt;plane&lt;/em&gt; $P_v$. Now, if $w$ is a vector not in $P_v$, then any vector in $P_v\cap P_w$ is mapped by $A$ to another vector in the same line, so $P_v\cap P_w$ is an $A$-invariant line, i.e., $A$ has a real eigenvalue after all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Note that I presented the argument as a contradiction for brevity, but it can be rearranged as a direct proof.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;David Milovich found a nice way of extending this argument so we have a short nice &quot;determinant free&quot; proof that $n\times n$ matrices with real entries and $n$ odd admit a real eigenvalue, see this &lt;a href=&quot;http://caicedoteaching.wordpress.com/2010/02/28/403503-eigenvectors-for-operators-on-real-odd-dimensional-spaces/&quot; rel=&quot;nofollow&quot;&gt;blog post&lt;/a&gt; of mine. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was interested in this because this is the base case of a nice inductive argument (that actually can be traced back to one of Gauss' first proofs of the fundamental theorem of algebra) that allows us to show that &lt;em&gt;any&lt;/em&gt; square matrix with real coefficients admits a (perhaps complex) eigenvalue, from which we can deduce the fundamental theorem of algebra. I refer to this in the post above, but it comes from &quot;The fundamental theorem of algebra and linear algebra&quot; by Harm Derksen, American Mathematical Monthly, 110 (7) (2003), 620-623. (The issue is that in that paper, the odd-dimensional case is done by appealing to determinants.) &lt;/p&gt;&#xA;" OwnerUserId="462" LastActivityDate="2010-12-18T19:55:12.590" CommentCount="2" />
  <row Id="14798" PostTypeId="1" AcceptedAnswerId="14805" CreationDate="2010-12-18T20:14:29.047" Score="2" ViewCount="222" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not so well up on my math, but I was wondering if the set of &quot;[EDIT: &lt;em&gt;finitely&lt;/em&gt;] decimal expressible&quot; real numbers $R$ between $[0,1]$ is countable?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wikipedia gives the definition:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A set $S$ is called countable if there exists an injective function $f$ from $S$ to the natural numbers $\mathbb{N}$&#xD;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It would seem to my mathematically na&amp;iuml;ve brain that such an injective function exists&amp;mdash;for any element $r \in R$ you can simply &quot;reverse the number&quot; and remove the decimal point to end up with, e.g.,:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f(0.0001) = 10000$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This leads me to three questions:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Is this intuition correct?&lt;/li&gt;&#xA;&lt;li&gt;Would this constitute a proof? &lt;/li&gt;&#xA;&lt;li&gt;Can the result be extended to general real numbers between $[0,1]$?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Any help with terminology would also be great... E.g., I don't know a mathematical function for &quot;reversing&quot; a number...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(P.S., I've just noted a specific question &lt;a href=&quot;http://math.stackexchange.com/questions/12088/real-numbers-and-countability&quot;&gt;here&lt;/a&gt; which is related and gives the general principle, but still doesn't answer my specific examples.)&lt;/p&gt;&#xA;" OwnerUserId="2331" LastEditorUserId="2331" LastEditorDisplayName="" LastEditDate="2010-12-18T21:19:36.463" LastActivityDate="2011-01-10T05:16:00.333" Title="(Finitely) decimal expressible real numbers between [0,1] countable?" Tags="&lt;elementary-set-theory&gt;&lt;discrete-mathematics&gt;" AnswerCount="6" CommentCount="7" />
  <row Id="14799" PostTypeId="2" ParentId="14767" CreationDate="2010-12-18T20:14:56.643" Score="5" ViewCount="" Body="&lt;p&gt;According to this article &lt;a href=&quot;http://www.fq.math.ca/Scanned/8-1/daykin-a.pdf&quot; rel=&quot;nofollow&quot;&gt;Factorisation of Fibonacci Numbers,&lt;/a&gt; if $F_n$ is the entry point in the Fibonacci sequence of the prime $p$ (i.e. $F_n$ is the smallest Fibonacci number divisible by the prime $p$) then for $p&amp;gt;5$ we have: for odd $n$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$p=4kn+1 \quad \textrm{ or } \quad p=(4k+2)n-1,$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for $n \equiv 2 \pmod{4}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$p=kn+1,$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and for $n \equiv 0 \pmod{4}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$p=2kn+1 \quad \textrm{ or } \quad p=(2k+1)n-1 .$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This should significantly reduce the search of possible candidates.&lt;/p&gt;&#xA;" OwnerUserId="1301" LastActivityDate="2010-12-18T20:14:56.643" />
  <row Id="14800" PostTypeId="2" ParentId="14798" CreationDate="2010-12-18T20:24:33.560" Score="1" ViewCount="" Body="&lt;p&gt;No they are not countable. Any real number can be expressed in decimal notation, and this representation may be infinite (non-terminating). And that's where your problem lies: how can you reverse an infinite decimal?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In my interpretation of your question there is no difference between what you are calling &quot;decimal expressible&quot; and real numbers, since any real number can be expressed in decimal notation. So the answer to part 3 is no, since the argument doesn't work as you can't reverse an infinite representation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: In the light of your recent comment, if you are taking your expression &quot;decimal expressible&quot; to mean a terminating decimal, then yes these are countable. But unfortunately your argument cannot be extend to the reals in $[0,1]$ because of the problem of reversing a non-terminating decimal.&lt;/p&gt;&#xA;" OwnerUserId="1301" LastEditorUserId="1301" LastEditorDisplayName="" LastEditDate="2010-12-18T20:29:43.260" LastActivityDate="2010-12-18T20:29:43.260" CommentCount="3" />
  <row Id="14801" PostTypeId="1" AcceptedAnswerId="14818" CreationDate="2010-12-18T20:29:40.780" Score="3" ViewCount="78" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Is there a commutative integral domain R in which:&lt;/p&gt;&#xA;  &#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;every nonzero prime ideal Q is maximal, and&lt;/li&gt;&#xA;  &lt;li&gt;there are maximal ideals Q with R/Q of sizes 3, 11, and 27?&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This doesn't happen with number rings of Galois extensions of &lt;b&gt;Q&lt;/b&gt;, as far as I know, since what happens to 3 seriously happens to 3.  You cannot have residue fields of sizes both 3 and 27.  Now &lt;b&gt;Z&lt;/b&gt;/3&lt;b&gt;Z&lt;/b&gt;[X] has residue fields of sizes 3 and 27, but not 11.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; I'll separate the next into its own question.&lt;/p&gt;&#xA;" OwnerUserId="583" LastEditorUserId="583" LastEditorDisplayName="" LastEditDate="2010-12-20T14:42:58.057" LastActivityDate="2010-12-20T14:42:58.057" Title="A variety of residue fields of size 3 ≡ q mod 8" Tags="&lt;abstract-algebra&gt;&lt;algebraic-number-theory&gt;" AnswerCount="1" />
  <row Id="14802" PostTypeId="2" ParentId="14798" CreationDate="2010-12-18T20:32:38.050" Score="1" ViewCount="" Body="&lt;p&gt;When I was taking my first course in set theory I asked my prof. (now my advisor) the same thing really.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First of all, the function is not well-defined, since there are numbers with several different decimal expansions (for example $\frac{1}{2} = 0.49999\ldots = 0.5$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let us assume that we only take the shortest finite expansion, it is possible but only for a very small portion of the numbers, for example $\frac{1}{3}$ has no finite decimal expansion, despite being a very simple rational number. So in fact, you don't really use even all the rational numbers in the interval.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now assume that a number $x$ has a finite representation at all, that means that for some $n\in\mathbb{N}$ we have that $x$ has $n$ digits after the decimal point, namely $x\cdot 10^n\in\mathbb{N}$, so you only define your function on a small portion of the rational numbers in the interval $[0,1]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As pointed by Derek Jennings, some (in fact almost all) numbers doesn't have a finite decimal representation (and actually most numbers has no finite representation in any base you can choose), so even if you do define this function it will only cover a countable subset of the interval, whereas the interval itself is uncountable and much much bigger than $|\mathbb{N}|$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So to answer your question, yes it is countable and your function (after correcting the minor problem of the uniqueness in the expression) is a proof for that.&lt;/p&gt;&#xA;" OwnerUserId="622" LastActivityDate="2010-12-18T20:32:38.050" CommentCount="4" />
  <row Id="14803" PostTypeId="2" ParentId="14798" CreationDate="2010-12-18T20:33:04.130" Score="1" ViewCount="" Body="&lt;p&gt;No, they are not countable. The proof uses an argument by contradiction. If you at all manage to list the real numbers into a countable list, you can always construct one which is not in the list. You can find a very neat exposition of the argument at the &lt;a href=&quot;http://planetmath.org/encyclopedia/CantorsDiagonalArgument.html&quot; rel=&quot;nofollow&quot;&gt;planetmath page&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1119" LastActivityDate="2010-12-18T20:33:04.130" CommentCount="1" />
  <row Id="14804" PostTypeId="2" ParentId="14777" CreationDate="2010-12-18T20:36:01.383" Score="1" ViewCount="" Body="&lt;p&gt;&quot;Uniform probability distribution&quot; always refers to a pre-existent natural measure on a space with a transitive group of &quot;translations&quot;, so that any two points resp. neighbourhoods of two points are comparable which each other. The simplest space of this sort would be the set $\mathbb Z$ of integers, and already there the problem you address becomes manifest. Of course it is easy to prove that there doesn't exist a uniform probability distribution on $\mathbb Z$; but you want an intuitive reasoning. While it is easy to come up with a random mechanism that selects any number between $-10^6$ and $10^6$ with equal probability, it is just unimaginable to fathom a mechanism that selects 5 with the same probability as any power of the prime with number $67212345$.&lt;/p&gt;&#xA;" OwnerUserId="1303" LastActivityDate="2010-12-18T20:36:01.383" />
  <row Id="14805" PostTypeId="2" ParentId="14798" CreationDate="2010-12-18T20:38:55.660" Score="3" ViewCount="" Body="&lt;p&gt;Your intuition is correct: the set of &quot;decimal expressible&quot; real numbers in [0, 1] such that the decimal expression has only finitely many nonzero digits is &lt;strong&gt;countable&lt;/strong&gt;, and the &quot;injective function&quot; you gave works.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if you allow decimal expressions with an infinite number of decimals (e.g. 0.333...), then this no longer works.  First of all, which natural number are you going to associate with 0.333...?  In fact, if you allow decimal expressions with an infinite number of decimals, then the set you described becomes uncountable; for the justification of that, I'll point you to &lt;a href=&quot;http://en.wikipedia.org/wiki/Cantor%27s_diagonal_argument&quot; rel=&quot;nofollow&quot;&gt;Cantor's Diagonal Argument&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To answer your three questions:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Yes.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Yes, all you needed to do is associate every element of your set with an element of the natural numbers (and vice versa), which you've done.  If I give you a decimal expression with finitely many terms, you can give me the unique natural number that you've associated with that decimal expression, and if I give you a natural number, you can give me the unique (finite) decimal expression it corresponds to.  This is exactly what it means for a set to be countable.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;No; again, see &lt;a href=&quot;http://en.wikipedia.org/wiki/Cantor%27s_diagonal_argument&quot; rel=&quot;nofollow&quot;&gt;Cantor's Diagonal Argument&lt;/a&gt;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="4904" LastEditorUserId="4904" LastEditorDisplayName="" LastEditDate="2010-12-18T20:51:31.583" LastActivityDate="2010-12-18T20:51:31.583" CommentCount="2" />
  <row Id="14806" PostTypeId="2" ParentId="14793" CreationDate="2010-12-18T20:40:14.803" Score="1" ViewCount="" Body="&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;The signature of a real non-degenerate bilinear form on an $n$-dimensional vector space can be defined as $(p,n-p)$ where $p$ is the largest dimension of a subspace on which the form restricts to be positive definite.  But over the complex numbers you can never have a positive definite symmetric form:  if $\langle v ,v \rangle &amp;gt; 0$ then $\langle iv ,iv\rangle &amp;lt; 0$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;To address your third question, let $Cl(p,q)$ be the Clifford algebra associated to a real inner product of signature $(p,q)$.  To show that $Cl(p,q) \otimes C \simeq Cl(C^{p+q})$, we just need to show that $Cl(p,q) \otimes C$ has elements $\{E_1, \ldots E_{p+q}\}$ that anti-commute and satisfy $E_i^2 = -1$.  But $\{e_1 \otimes 1, \ldots, e_p \otimes 1, e_{p+1} \otimes i, \ldots, e_{p+q}\otimes i\}$ is such a set, where $\{e_i\}$ is an orthonormal basis for $R^{p,q}$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I think the automorphism is just complex conjugation: $u \otimes z \mapsto u \otimes \bar z$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I'm not sure what is meant by even algebra in this context.  Clifford algebras are $Z_2$ graded so there is an even subalgebra.  This is the subalgebra generated by even products of vectors.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="9" LastActivityDate="2010-12-18T20:40:14.803" />
  <row Id="14807" PostTypeId="2" ParentId="2487" CreationDate="2010-12-18T20:45:40.807" Score="1" ViewCount="" Body="&lt;p&gt;The answer by Akhil Mathew is incomplete. It remains to check that $\sum_{n&amp;gt;k} x_n/n!\neq 0$ for infinitely many $k$. This is not obviously true here, it might even be false for some sequence $(x_n)$. Akhil Mathew's argument could be used for the series $\sum_{n\ge 0} n/(n+1)!$ which is similar to $\sum_{n\ge 0} 1/n!$. But its sum is equal to 1.&lt;/p&gt;&#xA;" OwnerUserId="4906" LastActivityDate="2010-12-18T20:45:40.807" CommentCount="1" />
  <row Id="14808" PostTypeId="2" ParentId="14798" CreationDate="2010-12-18T20:46:41.243" Score="2" ViewCount="" Body="&lt;p&gt;Yes, that works for terminating decimals (which does not include all rationals). More generally, it's simpler to note that the rationals are countable by mapping the reduced rational $\rm\ m/n\ \to\ 2^m\ 3^n\:.$ The same idea easily extends from pairs of naturals to all eventually zero sequences of naturals. Such arguments cannot be extended to all reals since the well-know diagonalization argument due to du-Bois-Reymond (and, later, Cantor) easily proves that that they are uncountable.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-12-18T20:51:46.167" LastActivityDate="2010-12-18T20:51:46.167" CommentCount="1" />
  <row Id="14809" PostTypeId="2" ParentId="14777" CreationDate="2010-12-18T21:04:13.963" Score="4" ViewCount="" Body="&lt;p&gt;If a uniform probability distribution did exist, then for any integer $n$ the probability a real number $x$ satisfies $n \leq x &amp;lt; n+1$ would have to be the same for all $n$. Call this probability $p$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of the rules that a probability distribution has to satisfy is that if $\{E_n\}$ are a countably infinite collection of disjoint events then $P(\cup_n E_n) = \sum_n P(E_n)$. Let $E_n$ be the event &quot;the real number $x$ satisfies $n \leq x &amp;lt; n+1$&quot;. Since every real number is between some $n$ and $n + 1$, $P(\cup_n E_n) = 1$. On the other hand, $\sum_n P(E_n) = p + p + p + ....$. If $p &amp;gt; 0$, this gives infinity. If $p = 0$ it gives zero. In either case, you'll never add up to $1$. Hence you cannot have a uniform probability distribution over the reals. &lt;/p&gt;&#xA;" OwnerUserId="3035" LastActivityDate="2010-12-18T21:04:13.963" />
  <row Id="14811" PostTypeId="1" AcceptedAnswerId="14812" CreationDate="2010-12-18T21:37:56.483" Score="10" ViewCount="318" Body="&lt;p&gt;Suppose that $g$ is a bijection on the real line, and $g^{-1} \circ f \circ g$ is a $C^\infty$ function whenever $f$ is $C^\infty$.  It seems howlingly obvious that this can only happen if $g$ is itself $C^\infty$.  But I can't figure out how to prove it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you help me, Internets?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Context: I want to show that in a manifold of dimension at least 2, the facts about which unparameterized curves are smooth suffice to determine the differential structure.  I think I can sort of see how to prove this provided I can rely on the claim above, but that's where I'm stuck.  However, if anyone here knows of a proof of the geometric claim, pointers would be most welcome.)&lt;/p&gt;&#xA;" OwnerUserId="4909" LastEditorUserId="4909" LastEditorDisplayName="" LastEditDate="2010-12-18T22:09:15.430" LastActivityDate="2010-12-20T08:35:34.280" Title="If $g^{-1} \circ f \circ g$ is $C^\infty$ whenever $f$ is $C^\infty$, must $g$ be $C^\infty$?" Tags="&lt;real-analysis&gt;&lt;differential-geometry&gt;" AnswerCount="1" FavoriteCount="6" />
  <row Id="14812" PostTypeId="2" ParentId="14811" CreationDate="2010-12-18T21:53:40.380" Score="14" ViewCount="" Body="&lt;p&gt;This is a result due to Floris Takens:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Let $\Phi \colon M_1 \to M_2$ be a bijection between two smooth $n$-manifolds such that $\lambda \colon M_2 \to M_2$ is a diffeomorphism iff $\Phi^{-1} \circ \lambda \circ \Phi$ is a diffeomorphism.  Then $\Phi$ is a diffeomorphism.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It can be found in &lt;em&gt;Characterization of a differentiable structure by its group of diffeomorphisms&lt;/em&gt; (Math Reviews number: MR552032).  I remember that I found an online version of it (but don't remember where off the top of my head, probably via MathSciNet).  I don't have it in front of me, but I do recall that the argument was fairly involved.  (I found it for a fairly similar reason: I wanted to show that the only Fr&amp;ouml;licher space with endomorphism monoid $C^\infty(\mathbb{R},\mathbb{R})$ was $\mathbb{R}$ itself, this is Proposition 8.4 of &lt;a href=&quot;http://www.math.ntnu.no/~stacey/Research/Preprints/smthcat.html&quot; rel=&quot;nofollow&quot;&gt;Comparative Smootheology&lt;/a&gt;.)&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update&lt;/strong&gt;: After reading the comments, I realised that the quoted theorem is not quite what is wanted.  Actually, what you need is Proposition 8.4 of &lt;a href=&quot;http://www.math.ntnu.no/~stacey/Research/Preprints/smthcat.html&quot; rel=&quot;nofollow&quot;&gt;Comparative Smootheology&lt;/a&gt;, which uses Takens' result at the key stage but nonetheless is about a page long in the proving.  The result in the paper is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The only Fr&amp;ouml;licher structures on $\mathbb{R}$ whose endomorphism monoid contains $C^\infty(\mathbb{R},\mathbb{R})$ are the standard, the discrete, and the indiscrete structures.  In particular, the only Fr&amp;ouml;licher structure on $\mathbb{R}$ whose endomorphism monoid is precisely $C^\infty(\mathbb{R},\mathbb{R})$ is the standard structure.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You can almost replace &quot;Fr&amp;ouml;licher structure&quot; by &quot;smooth structure&quot; in the above, except that there's no such thing as the &quot;indiscrete smooth structure&quot;.  The &lt;em&gt;discrete&lt;/em&gt; smooth/Fr&amp;ouml;licher structure on $\mathbb{R}$ views $\mathbb{R}$ as a load of disjoint points: a zero-dimensional manifold.  This is characterised by the fact that the only smooth functions $\mathbb{R} \to \mathbb{R}_{disc}$ are the constant functions (here the unadorned $\mathbb{R}$ is the usual manifold $\mathbb{R}$).  The indiscrete structure is the opposite: the only smooth functions $\mathbb{R}_{indisc} \to \mathbb{R}$ are the constant functions.  So we could rephrase the above result as:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The only smooth structures on $\mathbb{R}$ whose endomorphism monoid contains $C^\infty(\mathbb{R},\mathbb{R})$ are the standard and the discrete structures.  In particular, the only smooth structure on $\mathbb{R}$ whose endomorphism monoid is precisely $C^\infty(\mathbb{R},\mathbb{R})$ is the standard structure.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Let us connect that to your question.  You ask:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If $g \colon \mathbb{R} \to \mathbb{R}$ is a bijection such that $g^{-1} \circ f \circ g$ is $C^\infty$ whenever $f$ is $C^\infty$, must $g$ be $C^\infty$?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;First, we can use $g$ to put a smooth structure on $\mathbb{R}$ by post-composing $g$ with charts (or pre-composing with $g^{-1}$ if your charts go &lt;em&gt;from&lt;/em&gt; your manifold).  Let us write this as $\mathbb{R}_g$.  For clarity, let us write $\mathbb{R}_s$ for the standard smooth structure on $\mathbb{R}$.  Then $g \colon \mathbb{R}_s \to \mathbb{R}_g$ is a diffeomorphism, by construction.  Given a smooth function $f \colon \mathbb{R}_s \to \mathbb{R}_s$ we get a smooth function $f_g \colon \mathbb{R}_g \to \mathbb{R}_g$ by $g \circ f \circ g^{-1}$.  Conversely, given a smooth function $h \colon \mathbb{R}_g \to \mathbb{R}_g$, $g^{-1} \circ h \circ g$ is smooth on $\mathbb{R}_s$ and thus is $C^\infty$.  Hence the endomorphism monoid of $\mathbb{R}_g$, $C^\infty(\mathbb{R}_g,\mathbb{R}_g)$, is $g C^\infty(\mathbb{R}_s,\mathbb{R}_s) g^{-1}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The condition imposed is that $g^{-1} \circ f \circ g \in C^{\infty}(\mathbb{R},\mathbb{R})$ for all $f \in C^\infty(\mathbb{R},\mathbb{R})$.  The $\mathbb{R}$s here are $\mathbb{R}_s$ in our notation.  Translated into a single statement (rather than one for each $f \in C^\infty(\mathbb{R},\mathbb{R})$, this is: $g^{-1} C^\infty(\mathbb{R}_s,\mathbb{R}_s) g \subseteq C^\infty(\mathbb{R}_s,\mathbb{R}_s)$.  Now here's the sneaky bit; the inclusion here is going the wrong way.  So we simply apply $g \circ  - \circ g^{-1}$ to both sides to get $C^\infty(\mathbb{R}_s,\mathbb{R}_s) \subseteq g C^\infty(\mathbb{R}_s,\mathbb{R}_s) g^{-1}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we apply my result to deduce that $\mathbb{R}_g$ must be either the discrete smooth structure (which it isn't) or the standard one: $\mathbb{R}_g = \mathbb{R}_s$.  But then $g$ is a diffeomorphism $\mathbb{R}_s \to \mathbb{R}_g = \mathbb{R}_s$ and hence is $C^\infty$.&lt;/p&gt;&#xA;" OwnerUserId="2907" LastEditorUserId="2907" LastEditorDisplayName="" LastEditDate="2010-12-20T08:35:34.280" LastActivityDate="2010-12-20T08:35:34.280" CommentCount="11" />
  <row Id="14814" PostTypeId="2" ParentId="1037" CreationDate="2010-12-18T22:36:49.877" Score="1" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://www.friesian.com/calculus.htm&quot; rel=&quot;nofollow&quot;&gt;This&lt;/a&gt; page, (which quotes from &quot;Yearning for the Impossible, The Surprising Truths of Mathematics&quot;) has:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;...It has to be explained that dy/dx is not the ratio of infinitesimal differences dy and dx -- since infinitesimals do not exist -- but is rather a symbol for the limit of the ratio y/x as x tends to zero...&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1069" LastActivityDate="2010-12-18T22:36:49.877" />
  <row Id="14815" PostTypeId="1" AcceptedAnswerId="14817" CreationDate="2010-12-18T22:58:09.783" Score="6" ViewCount="502" Body="&lt;p&gt;The infinite series...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\pi/4 = 1 - 1/3 + 1/5 - 1/7 ...$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;...is very intriguing to me and seems like a crazy coincidence (its relationship to $\pi$).  Is it actually crazy or does it have an easy to explain, logical reasoning behind it that would make it seem not so magical?&lt;/p&gt;&#xA;" OwnerUserId="4914" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-12-18T23:06:22.330" LastActivityDate="2010-12-19T14:26:45.343" Title="Why does this converge to $\pi/4$?" Tags="&lt;sequences-and-series&gt;&lt;pi&gt;" AnswerCount="6" CommentCount="3" FavoriteCount="5" />
  <row Id="14816" PostTypeId="2" ParentId="14815" CreationDate="2010-12-18T23:09:26.727" Score="3" ViewCount="" Body="&lt;p&gt;It comes from evaluating &lt;a href=&quot;http://en.wikipedia.org/wiki/Leibniz_formula_for_pi&quot; rel=&quot;nofollow&quot;&gt;the arctan of 1&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1543" LastActivityDate="2010-12-18T23:09:26.727" CommentCount="1" />
  <row Id="14817" PostTypeId="2" ParentId="14815" CreationDate="2010-12-18T23:14:02.217" Score="17" ViewCount="" Body="&lt;p&gt;The standard proof is that this follows from the &lt;a href=&quot;http://en.wikipedia.org/wiki/Taylor_series&quot; rel=&quot;nofollow&quot;&gt;Taylor series&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\arctan x = x - \frac{x^3}{3} + \frac{x^5}{5} \mp ...$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for the arctangent.  This Taylor series is closely related to the Taylor series of the logarithm &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\log (1 + x) = x - \frac{x^2}{2} + \frac{x^3}{3} \mp ...$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and this is because the tangent function can be written in terms of complex exponentials, so the arctangent function can be written in terms of &lt;a href=&quot;http://en.wikipedia.org/wiki/Complex_logarithm&quot; rel=&quot;nofollow&quot;&gt;complex logarithms&lt;/a&gt;.  So the appearance of $\pi$ in this formula is morally due to &lt;a href=&quot;http://en.wikipedia.org/wiki/Euler%27s_formula&quot; rel=&quot;nofollow&quot;&gt;Euler's formula&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But there is also the following beautiful proof, which I learned from Gabor Toth's &lt;a href=&quot;http://books.google.com/books?id=lJxOgVcHBlkC&amp;amp;printsec=frontcover&amp;amp;dq=toth+algebra+and+geometry&quot; rel=&quot;nofollow&quot;&gt;Glimpses of Algebra and Geometry&lt;/a&gt;.  Consider the number $N(r)$ of integer lattice points inside the circle of radius $r$ centered at the origin, or in other words the number of pairs of integers $x, y$ satisfying $x^2 + y^2 \le r^2$.  It is not hard to see that $N(r) \sim \pi r^2$ for large $r$; in fact, it is not hard to see that $N(r) = \pi r^2 + O(r)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $r_2(n)$ denote the number of pairs of integers $(x, y)$ such that $x^2 + y^2 = n$.  Then $N(r) = 1 + r_2(1) + ... + r_2(r^2)$ (if $r$ is an integer).  On the other hand, a &lt;a href=&quot;http://en.wikipedia.org/wiki/Fermat%27s_theorem_on_sums_of_two_squares&quot; rel=&quot;nofollow&quot;&gt;classic result of number theory&lt;/a&gt; implies that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$r_2(n) = 4(d_1(n) - d_3(n))$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $d_k(n)$ is the number of divisors of $n$ congruent to $k \bmod 4$.  It follows that we can evaluate $N(r)$ by counting how many numbers between $1$ and $r^2$ are divisible by each number congruent to $1, 3 \bmod 4$ with the appropriate sign.  This gives&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{N(r) - 1}{4} = r^2 - \left\lfloor \frac{r^2}{3} \right\rfloor + \left\lfloor \frac{r^2}{5} \right\rfloor \mp ...$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and the result follows by taking the limit as $r \to \infty$.&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-12-19T07:31:45.060" LastActivityDate="2010-12-19T07:31:45.060" CommentCount="1" />
  <row Id="14818" PostTypeId="2" ParentId="14801" CreationDate="2010-12-18T23:35:04.387" Score="4" ViewCount="" Body="&lt;p&gt;The answer to your first question is yes.  Indeed, for any $d \geq 4$, you can construct infinitely many degree $d$ algebraic number fields $K/\mathbb{Q}$ such that the ring of &#xA;integers $\mathbb{Z}_K$ has this property. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a basic philosophy here that you can build a global object that has prescribed local behavior at any finite set of primes.  In this particular case, Theorem 23 of &lt;a href=&quot;http://math.uga.edu/~pete/8410Chapter3.pdf&quot; rel=&quot;nofollow&quot;&gt;these notes&lt;/a&gt; does what you want: you want to take as a $\mathbb{Q}_3$ algebra &#xA;$L \oplus \mathbb{Q}_3 \oplus A&amp;#39;$ where $L/\mathbb{Q}_3$ is an unramified cubic extension (and $A&amp;#39;$ is any separable $\mathbb{Q}_p$-algebra of dimension $d-4$), and similarly you &#xA;want to choose your $\mathbb{Q}_{11}$-algebra  to have a $\mathbb{Q}_{11}$ factor.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are right that these conditions cannot be realized by a global Galois extension, since then the local invariants of ramification index and inertial degree would have to agree for each of the factors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am still parsing the rest of your question...&lt;/p&gt;&#xA;" OwnerUserId="299" LastActivityDate="2010-12-18T23:35:04.387" CommentCount="3" />
  <row Id="14819" PostTypeId="2" ParentId="14012" CreationDate="2010-12-18T23:52:59.840" Score="1" ViewCount="" Body="&lt;p&gt;Just a remark on how to find the general answer to problems of this type. The &quot;ripples&quot; you are describing are level sets of some function $f_1, f_2$. Corresponding to them are their gradient vector fields $\nabla f_1, \nabla f_2$. The set of &quot;osculating points&quot; that you are looking for in your question is the set of points where $\nabla f_1$ and $\nabla f_2$ are parallel. So we can express it using the two-dimensional cross product $v \times w = v_x w_y - v_y w_x$ as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ C = \{ p\in\mathbb{R}^2 | \nabla f_1 \times \nabla f_2 = 0 \} $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;For families of confocal ellipses, the level function $f$ is of the form&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ f(x,y) = \sqrt{(x-x_1)^2 + (y-y_1)^2} + \sqrt{(x-x_2)^2 + (y-y_2)^2} $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;since the family of ellipses are the loci of points where sum of distances to the two foci are constant. So &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \nabla f =  \frac{(x-x_1, y-y_1)}{\sqrt{(x-x_1)^2 + (y-y_1)^2}} + \frac{(x-x_2, y-y_2)}{\sqrt{(x-x_2)^2 + (y-y_2)^2}} $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you see that for generic focal points, the expression defining the algebraic curve $C$ is often very complicated. You can try plotting some examples in Mathematica to see what they will look like.  &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;In your final comment, however, it looks like you are not considering confocal ellipses. Instead, it appears you are considering functions $f$ of the form &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ f(z) = Q(z-z_0) $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $z = (x,y)$ is a point in the plane, and $Q$ is a positive definite quadratic form. $z_0$ is the centre of the ellipse. (In other words, you are considering a family of ellipses that all share a centre and are dilations of each other.) In this case the answer is easier. Remember that a quadratic form can be represented by a matrix: $Q(z-z_0) = (z-z_0)^T A (z-z_0)$. Let $\epsilon$ denote the totally antisymmetric matrix&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \epsilon = \begin{pmatrix} 0 &amp;amp; 1 \\ -1 &amp;amp; 0 \end{pmatrix} $$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;and take $f_1(z) = (z-z_1)^T A_1 (z-z_1)$, $f_2(z) = (z-z_2)^T A_2(z-z_2)$, we have easily that $\nabla f_1 = 2 A_1(z-z1)$ and $\nabla f_2 = 2 A_2(z-z_2)$. So the set $C$ is given by &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \nabla f_1 \times \nabla f_2 = 0 \iff (z-z_2)^T A_2^T \epsilon A_1 (z-z_1) = 0 $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which implies that $C$ is described by a quadratic expression in $z$. That is, $C$ is in general given by a conic section. &lt;/p&gt;&#xA;" OwnerUserId="1543" LastActivityDate="2010-12-18T23:52:59.840" CommentCount="8" />
  <row Id="14820" PostTypeId="2" ParentId="14689" CreationDate="2010-12-18T23:58:40.130" Score="6" ViewCount="" Body="&lt;p&gt;Here is the exact answer, which I used Mathematica to compute. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$P(player\ 1\ wins) = \frac{118598889714523902216022358617928917633636253614645787377890526452587094871641057161197}{778560366535929033488842048429259732340012411410736514701931793474234814991339289051136}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$P(player\ 2\ wins) = \frac{858777828556435297558093986193070674512712644399583464104126713012616584348576395258605}{1038080488714572044651789397905679643120016548547648686269242391298979753321785718734848}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$P(players\ tie) = \frac{63512421616314632416996800666111235287366697985612516983784929048741127433063741783941}{3114241466143716133955368193717038929360049645642946058807727173896939259965357156204544}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Numerically, that's &lt;br/&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;.15233101351178373415...&#xA;.82727479987591082853...&#xA;.02039418661230543732...&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Here are the Mathematica rules I used to compute the probability of player 1 winnng. The other two cases are similar.&#xA;&lt;br /&gt;&#xA;(sorry for the image) &lt;br /&gt;&lt;br /&gt;&#xA;&lt;img src=&quot;http://i.imgur.com/C2eot.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With these rules in place I then computed &lt;code&gt;p[60,60]&lt;/code&gt;. I thought it was simpler to start the players at 60 and count down to 0, rather than the other way round.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first rule just says if player 1 has reached 0 and he player 2 hasn't, then the probability of player 1 winning is 1. The second rule is similar for player 2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The third rule handles the case where both players have crossed the finish line. Note that despite the game description, there aren't really turns involved—there are rounds. It doesn't matter which player goes first, so me might as well consider all four dice being rolled simultaneously.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fourth and final rule is where all the action is. The recursive sum is over all possible outcomes of the four dice (two for each player). The constant $864$ is merely $4\cdot6\cdot 6\cdot 6$. The &lt;code&gt;p[x,y] =&lt;/code&gt; thing in the center is a kind of memoization (caching). Without it, the simplistic recursion would take forever.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT:&#xA;Here is a plot of $P(n)$, the probability that the weaker player wins, when playing to a total of $n$ (instead of $60$), for $n = 1\dots80$. To my surprise, it's not quite monotonic! But a little thought will explain why.&#xA;&lt;img src=&quot;http://i.imgur.com/9Rk55.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="130" LastEditorUserId="130" LastEditorDisplayName="" LastEditDate="2010-12-20T00:34:01.337" LastActivityDate="2010-12-20T00:34:01.337" CommentCount="3" />
  <row Id="14821" PostTypeId="1" AcceptedAnswerId="14822" CreationDate="2010-12-19T00:41:01.390" Score="4" ViewCount="186" Body="&lt;p&gt;Hello, I was given the question:what is 9+99+999+9999+...+999..99(30 digits) After noticing a trend, I came with the conclusion that the answer would be 28 1's 080. Can anyone confirm my answer and give a reason as to why?&lt;/p&gt;&#xA;" OwnerUserId="4918" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-19T01:33:26.313" LastActivityDate="2010-12-19T17:17:22.760" Title="help confirm my answer for the following question" Tags="&lt;homework&gt;&lt;sequences-and-series&gt;" AnswerCount="2" CommentCount="9" />
  <row Id="14822" PostTypeId="2" ParentId="14821" CreationDate="2010-12-19T00:45:08.607" Score="17" ViewCount="" Body="&lt;p&gt;Note that&#xA;$$\underbrace{99\cdots 9}_{k\text{ digits}} = 10^k - 1.$$&#xD;&#xA;So your sum is the same as&#xA;$$(10-1) + (10^2-1) + (10^3-1) + \cdots + (10^{30}-1),$$&#xD;&#xA;which is equal to &#xA;$$(10 + 10^2 + 10^3 + \cdots + 10^{30}) - 30.$$&#xD;&#xA;The first sum is easy to do, the difference is easy to do, and it gives your answer.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-12-19T00:45:08.607" CommentCount="6" />
  <row Id="14823" PostTypeId="1" CreationDate="2010-12-19T01:22:07.820" Score="0" ViewCount="130" Body="&lt;p&gt;Imagine the varying frequency generator, which feeds this frequency to the oscilloscope.&#xA;Oscilloscope is turned to show signal spectra with FFT.&lt;br&gt;&#xA;The image on the screen of the oscilloscope is as expected: single peak, traveling from 1 kHz to 50 kHz. Of course, my generator is not ideal, it gives a number of harmonics with less magnitude. But the strange thing happens: if my peak goes from 1 kHz to 50 kHz, the other peak appears at 50-60 kHz, and goes to lower frequencies towards mine peak. At time of appear, it's hardly noticeable, but in the end it's almost half of main frequency. Whenever varying frequency comes to it's maximum (50 kHz), ghost frequency almost approaches it. At this moment my generator drops back frequency to 1 kHz, ghost frequency disappears and process starts again.&lt;br&gt;&#xA;What causes such a behaviour?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS: I think it's more mathematical question, rather then physical or electronic.&lt;br&gt;&#xA;PPS: I have some decisions, but don't want to spoil the question.&lt;/p&gt;&#xA;" OwnerUserId="2715" LastEditorUserId="2715" LastEditorDisplayName="" LastEditDate="2010-12-19T04:19:13.937" LastActivityDate="2010-12-23T05:39:07.410" Title="FFT gives ghost frequency." Tags="&lt;fourier-analysis&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="14824" PostTypeId="1" CreationDate="2010-12-19T01:47:14.767" Score="1" ViewCount="105" Body="&lt;p&gt;I have an old Spanish CG book that calls Perlin Noise a &quot;fractal structure&quot;. After reading &lt;a href=&quot;http://webstaff.itn.liu.se/~stegu/TNM022-2005/perlinnoiselinks/perlin-noise-math-faq.html&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; I couldn't deny it or confirm it. Is it a fractal structure? What would it Hausdorff dimension be?&lt;/p&gt;&#xA;" OwnerUserId="4474" LastActivityDate="2011-04-10T10:44:03.737" Title="Is Perlin Noise a &quot;fractal&quot;?" Tags="&lt;fractals&gt;&lt;noise&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="14825" PostTypeId="1" AcceptedAnswerId="14829" CreationDate="2010-12-19T02:05:00.693" Score="8" ViewCount="79" Body="&lt;blockquote&gt;&#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;G = { x,y : x&lt;sup&gt;2&lt;sup&gt;n+1&lt;/sup&gt;&lt;/sup&gt; = 1, y&lt;sup&gt;4&lt;/sup&gt; = 1, xy = yx&lt;sup&gt;−1&lt;/sup&gt; }&lt;/li&gt;&#xA;  &lt;li&gt;H = { x,y : x&lt;sup&gt;2&lt;sup&gt;n+1&lt;/sup&gt;&lt;/sup&gt; = 1, y&lt;sup&gt;4&lt;/sup&gt; = 1, xy = yx&lt;sup&gt;2&lt;sup&gt;n&lt;/sup&gt;−1&lt;/sup&gt; }&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;  &#xA;  &lt;p&gt;Can these groups of order 2&lt;sup&gt;n+3&lt;/sup&gt; and nilpotency class n+1 be distinguished by any reasonable (computable) set of invariants?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;While for any set of reasonable invariants, there is surely a pair of groups that share the invariants, it is also surely true that for any pair of groups there is some invariant that distinguishes them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I personally have not found any significant differences in the conjugacy classes, centralizers, proper subgroups, normalizers, normal subgroups, etc., but perhaps I overlooked something.  I've been looking at simple groups lately, and have forgotten how muddy p-groups can be.&lt;/p&gt;&#xA;" OwnerUserId="583" LastActivityDate="2010-12-19T05:29:38.717" Title="Can invariants distinguish these two 2-groups?" Tags="&lt;group-theory&gt;" AnswerCount="2" />
  <row Id="14826" PostTypeId="1" AcceptedAnswerId="14838" CreationDate="2010-12-19T02:16:48.600" Score="9" ViewCount="258" Body="&lt;p&gt;Let $X$ be a CW complex. The (isomorphism classes of) complex line bundles on $X$ are classified by the homotopy classes of maps $X \to \mathbb{CP}^\infty$, that is by the elements of $H^2(X, \mathbb{Z})$. &#xA;It is also true that the tensor product of line bundles corresponds to adding cohomology classes. It follows that if $n \in \mathbb{N}$, then the line bundles of $\mathbb{CP}^n$ are generated by the tensor powers of the tautological line bundle, or, equivalently, by the tensor powers of the sheaf typically denoted  $\mathcal{O}(1)$ in algebraic geometry (because the dual of $\mathcal{O}(1)$ is the tautological bundle).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is also true that the isomorphism classes of algebraic line bundles on $\mathbb{CP}^n$ forms  a group isomorphic to $\mathbb{Z}$, given by the powers of $\mathcal{O}(1)$, as one can see by studying the Weil class group. It follows that the isomorphism classes of line bundles are the same in both the algebraic and the topological category.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The above observation is also true for affine $n$-space, because the topological line bundles are trivial ($\mathbb{C}^n$ being contractible), and the algebraic ones are as well (the polynomial ring being a UFD).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To what extent is this true in general?&lt;/p&gt;&#xA;" OwnerUserId="536" LastActivityDate="2011-03-14T19:12:32.047" Title="Algebraic versus topological line bundles" Tags="&lt;algebraic-geometry&gt;&lt;algebraic-topology&gt;" AnswerCount="2" CommentCount="6" FavoriteCount="3" />
  <row Id="14827" PostTypeId="2" ParentId="14794" CreationDate="2010-12-19T03:02:30.210" Score="-1" ViewCount="" Body="&lt;p&gt;Your first condition &quot;(1) $Ax$ is nonzero for nonzero $x$&quot; implies that the matrix is full rank.  We hence conclude that $A$ has non-vanishing eigenvalues.  We then have two possibilities:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(a) There exist three non-vanishing real eigenvalues (possibly degenerate).&#xA;(b) There exist a complex conjugate pair of eigenvalues and one real non-vanishing eigenvalue.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In either case you have you have at least one real eigenvalue with an associated eigenvector.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can also reference &#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Euler%27s_rotation_theorem&quot; rel=&quot;nofollow&quot;&gt;Euler's Rotation Theorem&lt;/a&gt; which states that in ${\mathbb R}^3$ there will always be one fixed axis (i.e. eigen value of 1) under any rotation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the way, your objective is possible in ${\mathbb R}^4$.&lt;/p&gt;&#xA;" OwnerUserId="4726" LastActivityDate="2010-12-19T03:02:30.210" CommentCount="2" />
  <row Id="14828" PostTypeId="1" AcceptedAnswerId="14842" CreationDate="2010-12-19T03:31:32.513" Score="11" ViewCount="289" Body="&lt;p&gt;I am reading the book by Goldrei on Classic Set Theory. My question is more of a clarification. It is on if we are overloading symbols in some cases. For instance, when we define $2$ as a natural number, we define $$2_{\mathbb{N}} = \{\emptyset,\{\emptyset\} \}$$ When we define $2$ as an integer, $2_{\mathbb{Z}}$ is an equivalence class of ordered pair $$2_{\mathbb{Z}} = \{(n+_{\mathbb{N}}2_{\mathbb{N}},n):n \in \mathbb{N}\}$$ Similarly, when we define $2$ as a rational number, $2_{\mathbb{Q}}$ is an equivalence class of ordered pair $$2_{\mathbb{Q}} = \{(a \times_{\mathbb{Z}} 2_{\mathbb{Z}},a):a \in \mathbb{Z}\backslash\{0\}\}$$ and as a real number we define it as the left Dedekind cut of rationals less than $2_{\mathbb{Q}}$, i.e. $$2_{\mathbb{R}} = \{q \in \mathbb{Q}: q &amp;lt;_{\mathbb{Q}} 2_{\mathbb{Q}}\}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The clarification is each of the above are different objects right? So when we say $2$, it depends on the context? Also, if the above is true, is it correct or incorrect to say that &quot;The set of natural numbers is a subset of reals&quot;? Should we take the statement with a pinch of salt and understand accordingly?&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2010-12-19T04:14:00.830" LastActivityDate="2010-12-24T23:17:39.900" Title="Set Theoretic Definition of Numbers" Tags="&lt;set-theory&gt;" AnswerCount="3" CommentCount="4" FavoriteCount="2" />
  <row Id="14829" PostTypeId="2" ParentId="14825" CreationDate="2010-12-19T03:47:00.457" Score="5" ViewCount="" Body="&lt;p&gt;I have played around with these two groups for $n=3$ and $n=4$, and for those, one distinction is that the first group has no 2-dimensional irreducible representations with non-real characters (i.e. the Frobenius-Schur indicators of all the 2-dim irreducibles are $\pm1$), while the second has plenty 2-dim irreducible non-real characters. That should be easily provable for all $n$, although I haven't tried yet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not sure whether this can be turned into a purely group theoretic criterion. I briefly tried something in the spirit of &lt;a href=&quot;http://mathoverflow.net/questions/41784/roots-of-permutations/41788#41788&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; and the answer to &lt;a href=&quot;http://mathoverflow.net/questions/42646/square-roots-of-elements-in-a-finite-group-and-representation-theory&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt;, but to no avail: both groups have the same number of involutions and for both of them, the square roots counting function assumes its maximum at a non-trivial central element.&lt;/p&gt;&#xA;" OwnerUserId="3212" LastEditorUserId="3212" LastEditorDisplayName="" LastEditDate="2010-12-19T05:25:41.597" LastActivityDate="2010-12-19T05:25:41.597" CommentCount="1" />
  <row Id="14830" PostTypeId="2" ParentId="14828" CreationDate="2010-12-19T04:54:22.657" Score="3" ViewCount="" Body="&lt;p&gt;Yes, in the sense that in ZF, each of the above definitions defines a different collection of sets.  Also yes, in the sense that one should always be aware of context when discussing a mathematical object.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The mathematical fact that allows us to overload the symbol $2$ here is the existence of a sequence of natural injections $\mathbb{N} \to \mathbb{Z} \to \mathbb{Q} \to \mathbb{R}$, and elements in any term in this sequence can be pushed forward under the next injection or, if they lie in the range of the previous injection, pulled back.  Most people do this automatically and so don't usually point it out explicitly.  More generally, $1$ refers to the multiplicative identity in any ring, $0$ to the additive identity in any ring, and $2 = 1 + 1, 3 = 1 + 1 + 1$, and so forth.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-12-19T04:54:22.657" CommentCount="1" />
  <row Id="14831" PostTypeId="1" AcceptedAnswerId="14927" CreationDate="2010-12-19T04:56:18.920" Score="3" ViewCount="235" Body="&lt;p&gt;If &lt;em&gt;G&lt;/em&gt; is a free group generated by &lt;em&gt;n&lt;/em&gt; elements, is it possible to find an isomorphism of &lt;em&gt;G&lt;/em&gt; with a free group generated by &lt;em&gt;n-1&lt;/em&gt; (or any fewer number) of elements?&lt;/p&gt;&#xA;" OwnerUserId="4772" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-19T06:02:53.410" LastActivityDate="2010-12-20T08:03:16.980" Title="Generators of a free group." Tags="&lt;abstract-algebra&gt;&lt;group-theory&gt;" AnswerCount="4" CommentCount="1" />
  <row Id="14832" PostTypeId="2" ParentId="9416" CreationDate="2010-12-19T04:59:38.107" Score="4" ViewCount="" Body="&lt;p&gt;Remember that the FFT is circular.  Inputs which contain an integer number of cycles will come out clean as a single point, in the corresponding bin.  Those which do not, act as if they are multiplied by a rectangular pulse in the time domain, which creates convolution by a sinc function in the frequency domain.  Since sinc has unlimited support, your supposition that all bins except the closest two would be zero is incorrect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finding a closed-form analytic solution may be impossible, in which case your best bet would be to start with the center frequency for the two strongest bins and use binary search to find the frequency in-between that most closely corresponds to your actual spectrum.&lt;/p&gt;&#xA;" OwnerUserId="4923" LastActivityDate="2010-12-19T04:59:38.107" />
  <row Id="14833" PostTypeId="2" ParentId="14831" CreationDate="2010-12-19T05:08:03.523" Score="8" ViewCount="" Body="&lt;p&gt;No, if $F_n$ and $F_m$ are isomorphic, then their abelianizations are also isomorphic, which only happens when n=m (for example, reduce mod p and count points). &lt;/p&gt;&#xA;" OwnerUserId="8" LastEditorUserId="8" LastEditorDisplayName="" LastEditDate="2010-12-19T05:15:46.687" LastActivityDate="2010-12-19T05:15:46.687" />
  <row Id="14834" PostTypeId="1" AcceptedAnswerId="14837" CreationDate="2010-12-19T05:13:02.287" Score="1" ViewCount="166" Body="&lt;p&gt;For example, saying that $T$ is a k-tensor one might see $T\in 3^k(V)$, of course it's not actually a  3.  It looks somewhat like Fraktur font Z: $\frak{Z}$.  I couldn't detexify it, and it doesn't appear to be a standard notation for the set of tensors.  Other resources seem to just use $T$.&lt;/p&gt;&#xA;" OwnerUserId="326" LastActivityDate="2010-12-19T06:09:28.027" Title="In tensor notation in Spivak's Calculus on Manifolds, what is that character that looks like a 3?" Tags="&lt;soft-question&gt;&lt;tensors&gt;" AnswerCount="2" />
  <row Id="14835" PostTypeId="2" ParentId="14831" CreationDate="2010-12-19T05:17:44.923" Score="2" ViewCount="" Body="&lt;p&gt;To add to Noah Snyder's answer: abelian groups are $\mathbb{Z}$-modules, and the ring $\mathbb{Z}$ has the &lt;a href=&quot;http://en.wikipedia.org/wiki/Invariant_basis_number&quot; rel=&quot;nofollow&quot;&gt;Invariant Basis Number&lt;/a&gt; property. That means that for free $\mathbb{Z}$-modules, the rank is well-defined. More generally, all commutative rings have this property, as well as all left-Noetherian rings ($\mathbb{Z}$ is both). The wikipedia article gives an example of a ring that doesn't have this property, so modules over that ring can exhibit the strange behaviour that you are looking for.&lt;/p&gt;&#xA;" OwnerUserId="3212" LastEditorUserId="3212" LastEditorDisplayName="" LastEditDate="2010-12-19T05:23:51.177" LastActivityDate="2010-12-19T05:23:51.177" CommentCount="2" />
  <row Id="14836" PostTypeId="2" ParentId="14825" CreationDate="2010-12-19T05:29:38.717" Score="1" ViewCount="" Body="&lt;p&gt;A completely different method of recognition is just a random constructive recognition:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Choose a random element z&lt;/li&gt;&#xA;&lt;li&gt;If z has order |G|/4, call it x [ probability 25% ]&lt;/li&gt;&#xA;&lt;li&gt;If z has order 4 and if it has more than 2 conjugates, call it y [ probability 50% ]&lt;/li&gt;&#xA;&lt;li&gt;Otherwise, go to 1 [ probability 25% ]&lt;/li&gt;&#xA;&lt;li&gt;If we are missing x or y, go to 1.&lt;/li&gt;&#xA;&lt;li&gt;If x^y = x^-1, return &quot;G&quot;, else x^y = x^(|G|/8-1) and so return &quot;H&quot;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;This is easy to do since the automorphism groups of G and H act quite transitively on the possible generators.  This works well even for large n.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is using the normally &quot;cheating&quot; purely-group-theoretic invariant of &quot;how many isomorphisms to G?&quot;  In this case, there are just so many, it is easy to find one by random chance.&lt;/p&gt;&#xA;" OwnerUserId="583" LastActivityDate="2010-12-19T05:29:38.717" />
  <row Id="14837" PostTypeId="2" ParentId="14834" CreationDate="2010-12-19T05:34:37.737" Score="1" ViewCount="" Body="&lt;p&gt;As far as I can tell, it's fraktur I: $\mathfrak{I}$.  I don't think this is standard.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-12-19T05:34:37.737" />
  <row Id="14838" PostTypeId="2" ParentId="14826" CreationDate="2010-12-19T05:54:31.677" Score="5" ViewCount="" Body="&lt;p&gt;Let $X$ be a complex analytic space.  There is an exact sequence, the &lt;em&gt;exponential exact sequence&lt;/em&gt;, which is of fundamental importance for analyzing this (and related) questions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ 0 \to 2 \pi i \mathbb Z \to \mathcal O_X \buildrel \exp \over \longrightarrow \mathcal O^{\times}_X \to 1 .$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume now that $X$ is proper and connected. When we pass to cohomology, the sequence of $H^0$s is then short exact, but we obtain the following crucial long exact sequence:&#xA;$$H^1(X, 2 \pi i \mathbb Z) \to H^1(X,\mathcal O_X) \to Pic(X) \to&#xD;&#xA;H^2(X,2 \pi i \mathbb Z) \to H^2(X,\mathcal O_X).$$&#xD;&#xA;Here I am writing (as is usual) $Pic(X)$ to denote $H^1(X,\mathcal O_X^{\times})$, the group of isomorphism classes of analytic line bundles on $X$.  If $X$ is algebraic, then by GAGA this is the same as the group of algebraic line bundles on $X$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The boundary map $Pic(X) \to H^2(X,2 \pi i \mathbb Z)$ is the Chern class map&#xA;(with a $2\pi i$ twist, or &lt;em&gt;Tate twist&lt;/em&gt;; this is natural in the algebraic context, and to get the topological Chern class you just divide through by&#xA;$2 \pi i$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So that we see that the kernel of the Chern class map can be identified with&#xA;$H^1(X,\mathcal O_X)/H^1(X,2 \pi i\mathbb Z)$, and vanishes when $H^1(X,\mathcal O_X) = 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The image of $Pic(X)$ under the Chern class map is called the Neron--Severi group; its kernel is denoted $Pic^0(X)$ or $Pic^{\tau}(X)$.   When $X$ is algebraic, $Pic(X)$ is naturally an algebraic group,  $Pic^0(X)$ is the connected component of the identity, and $H^1(X,\mathcal O_X)$ is the tangent space to the identity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $X$ is a smooth projective curve, then $Pic^0(X)$ is usually called the &lt;em&gt;Jacobian&lt;/em&gt; of $X$.  You can look at the section of Hartshorne in Chapter IV to get some sense of it, although you may not realize from reading that how fundamental the role of the Jacobian is in the theory of algebraic curves.  If you google Torrelli theorem, Abel--Jacobi theorem, and theta divisor (just to give some sample search terms) you will get some sense of it.  Griffiths and Harris also has a detailed discussion, which gives a better sense of its significance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $X$ is algebraic but not proper, then you can compacitify it by adding a divisor at infinity.  Let me write $\overline{X}$ for the compacification, and let me assume that $\overline{X}$ is in fact smooth, so then the divisor $D := \overline{X}\setminus X$ is a Cartier divisor, and gives rise to an associated line bundle $\mathcal O(D) \in  Pic(\overline{X}).$  (This is denoted&#xA;$\mathcal L(D)$ in Hartshorne, I think, and in some other texts, especially older ones, but $\mathcal O(D)$ is more common notation these days, and is better notation too.)  If $D$ is reducible (as can happen; in general it can be taken to be a normal crossings divisor, but no better --- e.g. to compactify a curve to a smooth projective curve, we have to add in a finite number of points, but one point will not be enough, typically), write it as &#xA;$D_1 \cup \cdots \cup D_n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then we also have associated line bundles $\mathcal O(D_i)$ for each $i$, whose product is $\mathcal O(D)$, and note that each of these is trivial when restricted to $X$ (because $X = \overline{X} \setminus D$).  One now sees&#xA;that $Pic(X) = Pic(\overline{X})/\langle \mathcal O(D_1),\ldots,\mathcal O(D_n)&#xD;&#xA;\rangle,$ and so if $Pic^0(\overline{X})$ is non-trivial, then $Pic(X)$ will also have a non-discrete part (because we can't kill a connected algebraic group by quotienting out a finitely generated subgroup).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the answer to your question is, at least for smooth $X$, is: compactify $X$ to $\overline{X}$, and then compute $H^1(\overline{X}, \mathcal O)$; if this is non-trivial, then the Chern class map has a (huge!) kernel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[Added later: As an example, if $X$ is a hypersurface in $\mathbb P^n$ for $n &amp;gt; 2$ (so $X$ has dimension $&gt; 1$), then $H^1(X,\mathcal O_X) = 0$ (exercise!), and so hypersurfaces give interesting examples.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a surfaces, the dimension $H^1(X,\mathcal O_X)$ was classically (i.e. by the Italians) known as the irregularity of the surface $X$.  (The reason being that they knew formulas, like Riemann--Roch, for surfaces in space, which when they tried to extend to more general surfaces became false unless the extra quantity $\dim H^1(X,\mathcal O_X)$ was introduced --- although of course they didn't describe it this way.)   See &lt;a href=&quot;http://quomodocumque.wordpress.com/2009/08/05/professor-severi-is-a-jerk/&quot; rel=&quot;nofollow&quot;&gt;my comment here&lt;/a&gt;, as well as the notes of Kleiman linked to by Jason Starr, which will tell you a lot about Picard varieties and much more.]&lt;/p&gt;&#xA;" OwnerUserId="221" LastEditorUserId="221" LastEditorDisplayName="" LastEditDate="2010-12-19T06:48:43.193" LastActivityDate="2010-12-19T06:48:43.193" CommentCount="10" />
  <row Id="14839" PostTypeId="2" ParentId="14834" CreationDate="2010-12-19T06:09:28.027" Score="2" ViewCount="" Body="&lt;p&gt;I always thought it was a script T, as is used in John Lee's &quot;Introduction to Smooth Manifolds&quot; (though admittedly for tensor &lt;em&gt;fields&lt;/em&gt;).&lt;/p&gt;&#xA;" OwnerUserId="640" LastActivityDate="2010-12-19T06:09:28.027" CommentCount="2" />
  <row Id="14840" PostTypeId="2" ParentId="14815" CreationDate="2010-12-19T06:16:59.887" Score="4" ViewCount="" Body="&lt;p&gt;The function $\sec(x)=\frac{1}{\cos(x)}$ also has a series expansion that will give you this result, but it is not its Taylor series.  Namely, its &lt;a href=&quot;http://en.wikipedia.org/wiki/Partial_fractions_in_complex_analysis&quot; rel=&quot;nofollow&quot;&gt;partial fraction expansion&lt;/a&gt; is $$\sec(x)=\sum_{n=1}^\infty \frac{(-1)^n(2n-1)\pi }{x^2-\left (n-\frac{1}{2}\right )^2\pi^2},$$ &#xA;as derived in Example 1 at &lt;a href=&quot;http://books.google.com/books?id=DYCOCBCBwoIC&amp;amp;lpg=PA64&amp;amp;dq=partial%20fraction%20series%20csc%20cot%20sec%20complex&amp;amp;pg=PA59#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;this link&lt;/a&gt; to Markushevich and Silverman, for example.  The series converges to $\sec(x)$ everywhere the latter is defined (and uniformly on compact subsets of the domain).  Plugging in $x=0$ and multiplying both sides by $\frac{\pi}{4}$ yields $$\frac{\pi}{4}=1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\cdots.$$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-12-19T06:16:59.887" />
  <row Id="14841" PostTypeId="1" CreationDate="2010-12-19T06:33:42.950" Score="10" ViewCount="1270" Body="&lt;p&gt;What is the geometric, physical or other meaning of the third derivative of a function at a point?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Originally &lt;a href=&quot;http://mathoverflow.net/questions/49820/whats-meaning-of-geometrical-meaning-of-third-derivative-of-a-function-at-a-poin&quot; rel=&quot;nofollow&quot;&gt;asked on MO&lt;/a&gt; by AJAY)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have interesting things to say about the meaning of the first, and second derivatives, please do.&lt;/p&gt;&#xA;" OwnerUserId="1594" LastEditorUserId="1594" LastEditorDisplayName="" LastEditDate="2010-12-19T07:20:19.633" LastActivityDate="2011-05-14T17:32:48.170" Title="What is the meaning of the third derivative of a function at a point" Tags="&lt;calculus&gt;&lt;geometry&gt;&lt;physics&gt;&lt;education&gt;" AnswerCount="5" CommentCount="2" FavoriteCount="4" />
  <row Id="14842" PostTypeId="2" ParentId="14828" CreationDate="2010-12-19T06:34:11.683" Score="9" ViewCount="" Body="&lt;p&gt;Yes. And no. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You start with $\mathbb{N}$, and define $+$ and $\times$ (and $\lt$ and so on) appropriately.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then you define an equivalence relation on $\mathbb{N}\times\mathbb{N}$ given by &#xA;$$(a,b)\sim(c,d) \Longleftrightarrow a+d=b+c,$$&#xD;&#xA;and call the quotient set $(\mathbb{N}\times\mathbb{N})/\sim$ by the name $\mathbb{Z}$. (Behind the scenes, we are thinking of $(a,b)$ as meaning &quot;the solution to $a=x+b$&quot;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can then define an addition $+_{\mathbb{Z}}$ and a product $\times_{\mathbb{Z}}$ on $\mathbb{Z}$, as well as an order $\leq_{\mathbb{Z}}$ by&#xA;$$\begin{array}{rcl}&#xD;&#xA;[(a,b)]+_{\mathbb{Z}}[(c,d)] &amp;amp;=&amp;amp; [(a+c,b+d)]\\\&#xD;&#xA;[(c,d)]\times_{\mathbb{Z}}[(c,d)] &amp;amp;=&amp;amp; [(ac+bd,ad+bc)]\\\&#xD;&#xA;[(a,b)] \leq [(c,d)] &amp;amp;\Leftrightarrow&amp;amp; a+d\leq b+c,&#xD;&#xA;\end{array}$$&#xD;&#xA;and show that this is well defined. (I am using $[(a,b)]$ to denote the equivalence class of the pair $(a,b)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Certainly, $\mathbb{N}$ and $\mathbb{Z}$ are entirely different animals; set-theoretically, you can even show that they are disjoint.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But we can define a map $f\colon \mathbb{N}\to\mathbb{Z}$ by $f(n) = [(n,0)]$. This map is one-to-one, and for all $n,m\in\mathbb{N}$,&#xA;$$\begin{array}{rcl}&#xD;&#xA;f(n+m) &amp;amp;=&amp;amp; f(n)+_{\mathbb{Z}}f(m),\\\&#xD;&#xA;f(n\times m) &amp;amp;=&amp;amp; f(n)\times_{\mathbb{Z}}f(m)\\\&#xD;&#xA;n\leq m &amp;amp;\Leftrightarrow&amp;amp; f(n)\leq_{\mathbb{Z}} f(m)&#xD;&#xA;\end{array}$$&#xD;&#xA;That means that even though $\mathbb{N}$ and $\mathbb{Z}$ are disjoint, there is a &quot;perfect copy&quot; of $\mathbb{N}$ (in so far as its operations $+$ and $\times$ are concerned, and as far as the order $\leq$ is concerned) sitting inside of $\mathbb{Z}$. (&lt;em&gt;Added:&lt;/em&gt; In fact, this $f$ not only gives us perfect copy, it is the &lt;em&gt;only&lt;/em&gt; map from $\mathbb{N}$ to $\mathbb{Z}$ that is one-to-one and respects all the operations; we say it is a &quot;canonical embedding&quot;). Since we have this perfect copy, and a very specific map identifying this copy with the original, we can think of $\mathbb{N}$ as being a subset of $\mathbb{Z}$ by &quot;identifying it with its copy&quot;. So we do that. We can then introduce notation by showing that for every $[(a,b)]\in\mathbb{Z}$, either $a=b$, or there exists $n\in\mathbb{N}$, $n\neq 0$, such that $[(a,b)]=f(n)=[(n,0)]$, or there exists $n\in\mathbb{N}$, $n\neq 0$, such that $[(b,a)]=f(n)=[(n,0)]$; and then using $0$ to denote the class with $a=b$, $n$ to denote the class with $[(a,b)]=[(n,0)]$, and $-n$ to denote the class $[(c,d)]$ with $[(d,c)]=[(n,0)]$. This notation makes the identification clearer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, once we have $\mathbb{Z}$, we define $\mathbb{Q}$ as the quotient of $\mathbb{Z}\times(\mathbb{Z}-\{0\}$ modulo $\cong$, where &#xA;$$(a,b)\cong (c,d) \Longleftrightarrow ad=bc$$&#xD;&#xA;(behind the scenes, we think of $(a,b)$ as meaning &quot;the solution to $a=xb$&quot;). &#xA;We can then proceed as before, defining&#xA;$$\begin{array}{rcl}&#xD;&#xA;[(a,b)]+_{\mathbb{Q}}[(c,d)] &amp;amp;=&amp;amp; [(ad+bc,bd)]\\\&#xD;&#xA;[(a,b)]\times_{\mathbb{Q}}[(c,d)] &amp;amp;=&amp;amp; [(ac,bd)]&#xD;&#xA;\end{array}$$&#xD;&#xA;and showing this is well defined; defining an order, etc. Again, $\mathbb{Q}$ and $\mathbb{Z}$ (and the original $\mathbb{N}$) are completely different &lt;em&gt;sets&lt;/em&gt;. But we have a function $g\colon\mathbb{Z}\to\mathbb{Q}$ defined by $g(a) = [(a,1)]$. This is one-to-one, $g(a+_{\mathbb{Z}}b) = g(a)+_{\mathbb{Q}}g(b)$, and $g(a\times_{\mathbb{Z}}b) = g(a)\times_{\mathbb{Q}}g(b)$. (&lt;em&gt;Added:&lt;/em&gt; And again, this is the &lt;em&gt;only&lt;/em&gt; map from $\mathbb{Z}$ to $\mathbb{Q}$ that satisfies these conditions.) So again, we have a &quot;perfect copy&quot; of $\mathbb{Z}$ sitting inside of $\mathbb{Q}$ (and so also a perfect copy of the perfect copy of $\mathbb{N}$ that is sitting inside of $\mathbb{Z}$). So once again we &quot;identify&quot; $\mathbb{Z}$ with its image inside $\mathbb{Q}$ (and so we identify $\mathbb{N}$ with its image inside the image of $\mathbb{Z}$). Because, via $f$ and $g$, we have perfect copies of them anyway.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We do the same thing with $\mathbb{Q}$ as being &quot;inside of $\mathbb{R}$&quot;, by identifying elements of $\mathbb{Q}$ with specific Dedekind cuts or with specific equivalence classes of Cauchy sequences, showing the identification is one-to-one and respects all the operations (and is essentially unique), and so obtaining a &quot;perfect copy&quot; of $\mathbb{Q}$ inside of $\mathbb{R}$ (and by extension, perfect copies of $\mathbb{N}$ and of $\mathbb{Z}$ also sitting inside of $\mathbb{R}$). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can keep going, of course: define $\mathbb{C}$ as the set of all pairs $\mathbb{R}\times\mathbb{R}$; then identify $\mathbb{R}$ with the pairs $\mathbb{R}\times\{0\}$, and you have a copy of $\mathbb{N}$ sitting inside a copy of $\mathbb{Z}$ sitting inside a copy of $\mathbb{Q}$ sitting inside a copy of $\mathbb{R}$ sitting inside $\mathbb{C}$. (And then you can stick $\mathbb{C}$ inside the quaternions, the quaternions inside the octonions).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So even though they are actually very different sets, we have copies of each sitting inside the &quot;next one&quot;, copies that respect all the structures we are interested in, so we can still think of them as being &quot;subsets&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You used to do that all the time without the formalism: we think of &quot;fractions&quot; as being made up of an integer, a solidus, and a nonzero integer, so that &quot;$3$&quot; is not a fraction; but when needed, we are perfectly happy writing &quot;$3 = \frac{3}{1}$&quot; and working with either version of $3$ (the integer, or the fraction) depending on context.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-19T19:02:56.857" LastActivityDate="2010-12-19T19:02:56.857" CommentCount="1" />
  <row Id="14843" PostTypeId="2" ParentId="14841" CreationDate="2010-12-19T07:01:47.557" Score="1" ViewCount="" Body="&lt;p&gt;By asking the question, you can interpret the 1st and 2nd derivatives in some meaningful ways. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the 3rd as the combination of them. In other words, consider the third derivative as the 'acceration' if the velocity was infact the displacement. &lt;/p&gt;&#xA;" OwnerUserId="4728" LastActivityDate="2010-12-19T07:01:47.557" CommentCount="2" />
  <row Id="14844" PostTypeId="1" AcceptedAnswerId="14874" CreationDate="2010-12-19T07:18:04.157" Score="4" ViewCount="167" Body="&lt;p&gt;cross-post from &lt;a href=&quot;http://stackoverflow.com/questions/4481973/permutations-distinct-under-given-symmetry-mathematica-8-group-theory&quot;&gt;stack overflow&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a list of numbers like {2,1,1,0} and I'd like to list all permutations of that list that are not equivalent under given symmetry group. So for instance if it was a dihedral group of order 4, result would be {{2, 1, 1, 0}, {2, 1, 0, 1}}. Is there a practical way to do this when the list is too large to generate all permutations?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mathematica 8 seems to have a few group theory functions, but I don't have any group theory background, so any pointers are appreciated.&lt;/p&gt;&#xA;" OwnerUserId="998" LastActivityDate="2010-12-20T23:15:49.550" Title="Listing permutations distinct under given symmetry in Mathematica" Tags="&lt;group-theory&gt;&lt;mathematica&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" />
  <row Id="14845" PostTypeId="2" ParentId="14841" CreationDate="2010-12-19T07:25:48.320" Score="9" ViewCount="" Body="&lt;p&gt;It is a common theme in applied math that you can easily interpret first and second derivative or moment (in case of probability theory), but after that trouble begin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That being said, the third derivative is used in calculating torsion of a curve, see &lt;a href=&quot;http://en.wikipedia.org/wiki/Torsion_of_a_curve&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Torsion_of_a_curve&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's review an example (rather poor one, I admit). Let's work in $\mathbb{R}^3$ with a Cartesian coordinate system $x$, $y$, $z$ and the associated basis $\mathbf{e}_1$, $\mathbf{e}_2$, $\mathbf{e}_3$. Then let $\gamma$ be a unit circle given by $x = \cos \varphi$, $y = \sin \varphi$, $z = z_0$, parametrized naturally (by $\varphi$). Then $\dot{\gamma} \times \ddot{\gamma} = \mathbf{e}_3$, and then $\tau = {\dddot{\gamma}}^3 = 0$. I guess this particular curve is not very instructive, but I can't think of a better one off the top of my head :)&lt;/p&gt;&#xA;" OwnerUserId="3793" LastActivityDate="2010-12-19T07:25:48.320" CommentCount="2" />
  <row Id="14846" PostTypeId="2" ParentId="6009" CreationDate="2010-12-19T07:45:59.363" Score="0" ViewCount="" Body="&lt;p&gt;the second problem is the one i too hv got my idea wud be arranging the men first in order and thn finding derangement of the 8 wommen put the answer dosnt match&lt;/p&gt;&#xA;" OwnerUserId="4924" LastActivityDate="2010-12-19T07:45:59.363" />
  <row Id="14847" PostTypeId="2" ParentId="14841" CreationDate="2010-12-19T07:52:34.817" Score="7" ViewCount="" Body="&lt;p&gt;For the position function $p=f(t)$, you probably know that the first derivative $f&amp;#39;(t)$ gives the instantaneous velocity, and that the second derivative $f&amp;#39;&amp;#39;(t)$ gives the instantaneous acceleration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rate of change of the acceleration is called the &lt;a href=&quot;http://en.wikipedia.org/wiki/Jerk_%28physics%29&quot; rel=&quot;nofollow&quot;&gt;&quot;jerk&quot;&lt;/a&gt; (also known as the &quot;surge&quot;, the &quot;jolt&quot;, or the &quot;lurch&quot;). So the third derivative $f&amp;#39;&amp;#39;&amp;#39;(t)$ would give the instantaneous jerk.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-12-19T07:52:34.817" CommentCount="3" />
  <row Id="14848" PostTypeId="2" ParentId="14815" CreationDate="2010-12-19T08:01:57.027" Score="0" ViewCount="" Body="&lt;p&gt;I think this is relevant, and an explanation of QY's proof. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$1+i\,\tan\,\theta=\sec\,\theta(\cos\,\theta+i\,\sin\,\theta)$$&lt;br&gt;&#xA;$$1+i\,\tan\,\theta=\sec\,\theta\cdot e^{i\theta}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Taking logs, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \ln(1+i\,\tan\,\theta) = \ln(\sec\,\theta\cdot e^{i\theta})$$&lt;br&gt;&#xA;$$ \ln(1+i\,\tan\,\theta) = i\theta + \ln(\sec\,\theta)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rest follows easily. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are other interesting identities which stem from this, and are easily proved.&#xA;$$\arctan(1/2)+\arctan(1/3) = \pi/4$$&#xD;&#xA;$$4\arctan(1/5)-\arctan(1/239) = \pi/4$$&#xD;&#xA;$$4\arctan(1/5)-\arctan(1/70)+\arctan(1/99) = \pi/4$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These final equations make it progressively easier to evaluate $\pi$, and that is what they are used for.&lt;/p&gt;&#xA;" OwnerUserId="4728" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-19T09:04:51.323" LastActivityDate="2010-12-19T09:04:51.323" CommentCount="1" />
  <row Id="14850" PostTypeId="2" ParentId="14841" CreationDate="2010-12-19T09:04:00.387" Score="11" ViewCount="" Body="&lt;p&gt;The rate of change of acceleration is studied in various situations in physics, mechanics and engineering design.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From &lt;a href=&quot;http://en.wikipedia.org/wiki/Jerk_%28physics%29&quot; rel=&quot;nofollow&quot;&gt;wikipedia&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In physics, &lt;strong&gt;jerk&lt;/strong&gt;, also known as &lt;strong&gt;jolt&lt;/strong&gt; (especially in British English), &lt;strong&gt;surge&lt;/strong&gt; and &lt;strong&gt;lurch&lt;/strong&gt;, is the rate of change of acceleration; that is, the derivative of acceleration with respect to time, the second derivative of velocity, or the third derivative of position. Jerk is defined by any of the following equivalent expressions:&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$$ \vec j=\frac {\mathrm{d} \vec a} {\mathrm{d}t}=\frac {\mathrm{d}^2 \vec v} {\mathrm{d}t^2}=\frac {\mathrm{d}^3 \vec s} {\mathrm{d}t^3}$$&#xA;  where&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$\vec a$ is acceleration, $\vec v$ is velocity, $\vec s$ is position and $\mathit{t}$ is time.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Jerk is a vector, and there is no generally used term to describe its scalar magnitude (e.g. &quot;speed&quot; as the scalar magnitude for velocity). &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Think of roller coaster designs. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Indeed, in mechanics, the fourth derivative is also studied. It is called Jounce or Snap. From &lt;a href=&quot;http://en.wikipedia.org/wiki/Jounce&quot; rel=&quot;nofollow&quot;&gt;wikipedia&lt;/a&gt;: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In physics, &lt;strong&gt;jounce&lt;/strong&gt; or &lt;strong&gt;snap&lt;/strong&gt; is the fourth derivative of the position vector with respect to time, with the first, second, and third derivatives being velocity, acceleration, and jerk, respectively; in other words, the jounce is the rate of change of the jerk with respect to time.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$$\vec s =\frac {d \vec j} {dt}=\frac {d^2 \vec a} {dt^2}=\frac {d^3 \vec v} {dt^3}=\frac {d^4 \vec r} {dt^4}$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sometimes even the higher derivatives are nontrivial and come into play. Think of a sudden impact, an earthquake, a &lt;a href=&quot;http://en.wikipedia.org/wiki/Shock_%28mechanics%29&quot; rel=&quot;nofollow&quot;&gt;shock&lt;/a&gt;, or a lightning effect in electrical systems. Constructs like &lt;a href=&quot;http://en.wikipedia.org/wiki/Dirac_delta_function&quot; rel=&quot;nofollow&quot;&gt;Dirac Delta function&lt;/a&gt; are very convenient for dealing with such situations. For instance, for a lightning, there is approximately a very high surge of current for a very brief instant and for any smooth approximation all the higher derivatives are nonzero. So you take the limit and manipulate it as if everything were concentrated at a single point. &lt;/p&gt;&#xA;" OwnerUserId="1119" LastActivityDate="2010-12-19T09:04:00.387" CommentCount="1" />
  <row Id="14851" PostTypeId="1" AcceptedAnswerId="14884" CreationDate="2010-12-19T09:42:27.567" Score="3" ViewCount="130" Body="&lt;p&gt;The recent post &lt;a href=&quot;http://math.stackexchange.com/questions/14540/fourier-cosine-transform/14546#14546&quot;&gt;here&lt;/a&gt; has led to the following question (consider $\cos(sx) = \sum\nolimits_{k = 1}^\infty  {\frac{{( - 1)^k (sx)^{2k} }}{{(2k)!}}}$). I find it instructive, whether it will turn out to be easy/well-known or not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, suppose that $(f_k)_{k \geq 1}$ is a sequence of continuous functions on a non-compact interval $I$. Suppose that $&#xD;&#xA;\sum\nolimits_{k = 1}^\infty  {f_k (x)}$ converges uniformly on every compact subinterval of $I$. Further, suppose that both $\sum\nolimits_{k = 1}^\infty  {\int_I {f_k (x)\,{\rm d}x} }$ and $\int_I {[\sum\nolimits_{k = 1}^\infty  {f_k (x)]} \,{\rm d}x}$ converge (i.e., are finite).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Under these assumptions, can you find an example where &#xA;$$&#xD;&#xA;\int_I {\bigg[\sum\limits_{k = 1}^\infty  {f_k (x)\bigg]} {\rm d}x} \neq \sum\limits_{k = 1}^\infty  {\int_I {f_k (x)\,{\rm d}x} }?&#xD;&#xA;$$&#xD;&#xA;Consider also the case where we are only given that the right-hand side converges.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If relevant: What about the case where $\sum\nolimits_{k = 1}^\infty  {f_k (x)}$ and its partial sums are analytic functions? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any relevant discussion is welcome here.&lt;/p&gt;&#xA;" OwnerUserId="2810" LastActivityDate="2010-12-19T19:10:51.173" Title="Term-by-term integration under a certain condition" Tags="&lt;real-analysis&gt;&lt;sequences-and-series&gt;&lt;integral&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="14852" PostTypeId="1" AcceptedAnswerId="14856" CreationDate="2010-12-19T10:21:41.080" Score="3" ViewCount="103" Body="&lt;p&gt;&lt;em&gt;&lt;strong&gt;How many ways of giving out 6 pieces of candy to 3 children if each child must receive at-least one piece ?&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think, $ 5 \choose 2$ is the answer,consistent to $ n-1 \choose r $ as this is similar to  finding the  number of non-zero positive integer solution of the equation $X_1 + X_2 + X_3 = 6$ ,but the correct answer (as given in my module) is $540$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where exactly am I wrong?&lt;/p&gt;&#xA;" OwnerUserId="3744" LastActivityDate="2010-12-19T11:07:51.667" Title="About a combinatoric problem" Tags="&lt;combinatorics&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="14853" PostTypeId="1" AcceptedAnswerId="14859" CreationDate="2010-12-19T10:34:05.240" Score="0" ViewCount="201" Body="&lt;p&gt;How to show that:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ (-i\nabla-eA)\times(-i\nabla-eA) = (ie\nabla \times A) $  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;i and e are constants&lt;br&gt;&#xA;A is a vector field&lt;br&gt;&#xA;$\nabla$ = vector differential operator&lt;/p&gt;&#xA;" OwnerUserId="4929" LastEditorUserId="4929" LastEditorDisplayName="" LastEditDate="2010-12-19T12:02:54.270" LastActivityDate="2010-12-19T12:02:54.270" Title="Cross product of operators" Tags="&lt;differential-geometry&gt;&lt;vector-calculus&gt;&lt;operator-theory&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="1" />
  <row Id="14854" PostTypeId="1" CreationDate="2010-12-19T10:39:10.817" Score="1" ViewCount="166" Body="&lt;p&gt;I have a question about abelian group (or rather X-group):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose A is abelian group (or rather X-group) and it's artinian and noetherian. f is an endomorphism (or rather X-endomorphism) on A, i.e. $f\in \mathrm{End}(A)$. Also know that f(f(A))=f(A).&#xA;Prove $A=ker(f)\dot{+}f(A)$ i.e. A is direct sum of those two subgroup.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My idea is&#xA;firstly show $A=ker(f)+f(A)$ which I have proved. So next is to prove that $ker(f)\bigcap f(A)$ is trivial, i.e. =0. This part I got astray.&#xA;I thought I should use the noetherian condition here. Since, A is noetherian, then there exists an subgp C of A and $C\subset f(A)$, s.t. C is maximal in the sense of $ker(f)+C=0$, then find an subgp T, s.t. $C&amp;lt;T&amp;lt;f(A)$ satisfy $T\bigcap f(A)=0$ then contradict with maximality of C to reach a contradiction. But I get astray here to find this T. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe I use the wrong method. Can anyone help me?&lt;/p&gt;&#xA;" OwnerUserId="4900" LastEditorUserId="4900" LastEditorDisplayName="" LastEditDate="2010-12-19T12:36:15.513" LastActivityDate="2010-12-19T12:36:15.513" Title="a problem about abelian group" Tags="&lt;homework&gt;&lt;abstract-algebra&gt;&lt;group-theory&gt;" CommentCount="12" />
  <row Id="14855" PostTypeId="1" AcceptedAnswerId="14903" CreationDate="2010-12-19T10:40:20.527" Score="3" ViewCount="297" Body="&lt;p&gt;Please give me a link to a reference for an example of a continuous function whose Fourier series diverges at a dense set of points. (given by Du Bois-Reymond). I couldn't find this in Wikipedia.&lt;/p&gt;&#xA;" OwnerUserId="2987" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-20T02:05:15.237" LastActivityDate="2010-12-20T02:05:15.237" Title="an example of a continuous function whose Fourier series diverges at a dense set of points." Tags="&lt;reference-request&gt;&lt;fourier-series&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="14856" PostTypeId="2" ParentId="14852" CreationDate="2010-12-19T10:45:50.787" Score="4" ViewCount="" Body="&lt;p&gt;You have assumed that $6$ candies are identical but that is not the case here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This problem can be modeled be letting A represent the set of candies and B represent the set of children.Then a function $f \colon A \to B $ can be interpreted as giving candy $a_i$ to $c_J$. Since each child must receive at-least one candy, we are considering only onto functions.So your problem boils down to counting number of onto functions in $f \colon A \to B $.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To calculate  this value we could use a formula which is actually derived by using the principle of inclusion-exclusion:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If A,B are non-empty sets of cardinality $m$,$n$ with $m \ge n$.Then there are &#xA;$$\sum_{i=0}^{(n-1)} (-1)^i {n \choose i} (n-i)^m \text{ onto functions in } f \colon A \to B $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case we have $m = 6$ and $n = 3$ then we get : $3^6 - {3 \choose 1} (3 - 1)^6 + {3 \choose 2} (3 - 2)^6  = 540$ &lt;/p&gt;&#xA;" OwnerUserId="2109" LastEditorUserId="2109" LastEditorDisplayName="" LastEditDate="2010-12-19T11:07:51.667" LastActivityDate="2010-12-19T11:07:51.667" CommentCount="1" />
  <row Id="14857" PostTypeId="2" ParentId="14852" CreationDate="2010-12-19T11:02:53.410" Score="2" ViewCount="" Body="&lt;p&gt;If all you care about is how many pieces of candy each chuld gets, then you are right. We would say in this case that the pieces of candy are &quot;indistinguishable&quot;. But if you care WHICH piece of candy each child gets, e.g if the individual candies are numbered from 1 to 6, then there are many solutions corresponding to each solution of the equation $X_1+X_2+X_3=6$, In this case the problem is to count the number of surjective (or &quot;onto&quot;) functions from a six-element set to a three-element set.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general there is no &quot;nice&quot; formula (comparable to the one for the binomial coefficients) for the number of surjective functions from an $n$ element set to an $m$ element set, but this number is $m!S(n,m)$, where the expression $S(n,m)$ refers to a Stirling number of the second kind, and is defined recursively for $n,m&amp;gt;1 $ by  $S(n,m)=mS(n-1,m)+S(n-1,m-1)$. The basis cases are a bit messy: $S(n,1)=1$, $S(n,n)=1$, and  $S(n,m)=0$ if $n&amp;lt;m$ or $n=m=0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Actually $S(n,m)$ counts the number of partitions of $n$ distinguishable elements into $m$ non-empty sets. All onto functions from an $n$ element set to an $m$ element set are obtained choosing such a partition $p$ arbitrarily, and then assigning each of the $m$ elements of $p$  to one of the $m$ elements of the $m$-element set. There are $m!$ ways to do this, hence the formula $m!S(n,m)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A nice discussion of all this can be found in Section 1.4, &quot;The Twelvefold Way&quot;, in Stanley's &quot;Enumerative Combinatorics&quot; volume 1.&lt;/p&gt;&#xA;" OwnerUserId="746" LastActivityDate="2010-12-19T11:02:53.410" CommentCount="1" />
  <row Id="14858" PostTypeId="2" ParentId="14855" CreationDate="2010-12-19T11:05:04.760" Score="4" ViewCount="" Body="&lt;p&gt;Actually, such an almost-everywhere divergent Fourier series was constructed by Kolmogorov.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For an explicit example, you can consider a &lt;a href=&quot;http://books.google.com/books?id=W9AxAjSiIaUC&amp;amp;pg=PA302&quot; rel=&quot;nofollow&quot;&gt;Riesz product&lt;/a&gt; of the form:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \prod_{k=1}^\infty \left( 1+ i \frac{\cos 10^k x}{k}\right)$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is divergent. For more examples, see &lt;a href=&quot;http://projecteuclid.org/euclid.pja/1195523374&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; and &lt;a href=&quot;http://www.ams.org/journals/proc/1958-009-02/S0002-9939-1958-0096068-0/S0002-9939-1958-0096068-0.pdf&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Edit: (response to comment). Yes, you are right, du Bois Reymond did indeed construct the examples of Fourier series diverging at a dense set of points. However the result of Kolmogorov is stronger in that it gives almost everywhere divergence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The papers of du Bois Reymond are:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Ueber die Fourierschen Reihen&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;available for free download &lt;a href=&quot;http://gdz.sub.uni-goettingen.de/dms/load/img/?PPN=GDZPPN002515881&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; also another one &lt;a href=&quot;http://gdz.sub.uni-goettingen.de/dms/load/img/?PPN=GDZPPN00224361X&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;. &lt;/p&gt;&#xA;" OwnerUserId="1119" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-19T12:08:26.827" LastActivityDate="2010-12-19T12:08:26.827" CommentCount="10" />
  <row Id="14859" PostTypeId="2" ParentId="14853" CreationDate="2010-12-19T11:11:30.107" Score="4" ViewCount="" Body="&lt;p&gt;As I suggested in my hint, apply the operator on a test function $f$. Then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(-i\nabla-eA)\times(-i\nabla-eA)f = (-i\nabla-eA)\times(-i\nabla f-e A f)$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Working this out we get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(-i\nabla-eA)\times(-i\nabla f-e A f)= -\nabla \times \nabla f + ie A\times \nabla f +i e \nabla \times (A f) + e^2 A \times A f$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first and last term are zero. The third term can be expanded using &lt;a href=&quot;http://en.wikipedia.org/wiki/Curl_%28mathematics%29#Identities&quot; rel=&quot;nofollow&quot;&gt;the last identity&lt;/a&gt; (or simply working it out in terms of partial derivatives). This will give the result.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(-i\nabla-eA)\times(-i\nabla-eA) = (ie\nabla \times A)$$ &lt;/p&gt;&#xA;" OwnerUserId="3567" LastActivityDate="2010-12-19T11:11:30.107" CommentCount="6" />
  <row Id="14860" PostTypeId="2" ParentId="14358" CreationDate="2010-12-19T11:40:26.837" Score="1" ViewCount="" Body="&lt;p&gt;I disagree with observation 2. It gives a sufficient condition that is not necessary unless you only consider projections onto a 1-dimensional subspace.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A weaker sufficient condition, where I am not sure whether its also necessary&#xA;is the following: if there is a decomposition $P_A G = BC$, where $B$ is&#xA;any regular matrix and $C$ is a matrix with only integer entries then $\Lambda_U$&#xD;&#xA;also must be a lattice. This must be because $C$ represents a map $\mathbb{Z}^n \rightarrow \mathbb{Z}^n$, and $B$ represents a vector space isomorphism. (note&#xA;that it does not matter if $C$ can be integer or rational, for any common denominator can be moved into $B$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems clear to me for geometric reasons that for any regular $G$ a suitable&#xA;1-dimensional $U$ can be found. Consider the basis vectors of the lattice, i.e.&#xA;the columns of $G$. There must be a hyperplane through them and a line through&#xA;the origin perpendicular to that hyperplane. An orthogonal projection onto that&#xA;line will move points parallel to the hyperplane. In particular, the basis&#xA;vectors will all be projected to the intersection of the line with the hyperplane. It follows that integer combinations of the basis&#xA;vectors will be mapped to integer multiples of that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not sure whether there is a simple way to find a base for this&#xA;$U$, but it should always possible to take differences of basis vectors as a&#xA;basis for the hyperplane and use Gramm-Schmidt orthonogonalization to find a&#xA;normal vector.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are interested, it can be proved easily, at least in $\mathbb{R}^2$, that there is in fact an (countable) infinity of suitable $U$s.&#xA;This can be done by setting up a sine/cosine parametrization of the unit&#xA;vectors and showing that the ratio of the projection of the basis vectors of the&#xA;lattice is a continuous non-constant function of the angle.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Addendum:&lt;/strong&gt; Actually, it is algebraically not as complicated as I imagined.&#xA;Supposing that $G$ is invertible, let $H = G^{-1}$ and take for $D$ an&#xA;$n \times 1$ matrix of all ones.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now if we let $A = H^TD$ then $A^TA$ becomes scalar and we can simplify&#xA;$$&#xD;&#xA;P_A = \frac{H^TDD^TH}{D^THH^TD}&#xD;&#xA;$$&#xD;&#xA;This means that&#xA;$$&#xD;&#xA;P_AG = \frac{H^TD}{D^THH^TD} D^T&#xD;&#xA;$$&#xD;&#xA;which means that the columns are all equal.&#xA;Alternatively, you could of course replace the ones in D with any rationals.&lt;/p&gt;&#xA;" OwnerUserId="3457" LastEditorUserId="3457" LastEditorDisplayName="" LastEditDate="2010-12-19T18:23:31.823" LastActivityDate="2010-12-19T18:23:31.823" CommentCount="1" />
  <row Id="14861" PostTypeId="2" ParentId="14815" CreationDate="2010-12-19T12:01:20.497" Score="1" ViewCount="" Body="&lt;p&gt;For a historical perspective (i.e. if you want to see how geniuses struggled through things that would be natural today, mostly because of things they later discovered), the nice article &lt;a href=&quot;http://mathdl.maa.org/mathDL/46/?pa=content&amp;amp;sa=viewDocument&amp;amp;nodeId=1408&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;The Discovery of the Series Formula for π by Leibniz, Gregory and Nilakantha&lt;/em&gt;&lt;/a&gt;&lt;sup&gt;1&lt;/sup&gt; by Ranjan Roy (1990) describes how the formula was discovered independently thrice:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The series (2) was obtained independently by Gottfried Wilhelm Leibniz (1646-1716), James Gregory (1638-1675) and an Indian mathematician of the fourteenth century or probably the fifteenth cen­tury whose identity is not definitely known. Usually ascribed to Nilakantha, the Indian proof of (2) ap­pears to date from the mid-fifteenth century and was a consequence of an effort to rectify the circle. […] Leibniz's work, in fact, was primarily concerned with quadrature; the π/4 series resulted (in 1673) when he applied his method to the circle. Gregory, by comparison, was interested in finding an infinite series representation of any given func­tion and discovered the relationship between this and the successive derivatives of the given function. Gre­gory's discovery, made in 1671, is none other than the Taylor series; note that Taylor was not born un­til 1685. […]&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Finally, although the proofs of (2) by Leibniz, Gregory and Nilakantha are very different in approach and motivation, they all bear a relation to the modem proof given above.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Perhaps I'll come back and edit this post for a summary of their methods if I actually read the article. :-)&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;1&lt;/sup&gt;: Ranjan Roy, &lt;em&gt;Mathematics Magazine&lt;/em&gt;, Vol. 63 (1990), pp. 291-306. I found this while flipping through the book &lt;a href=&quot;http://books.google.com/books?id=BKRE5AjRM3AC&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;Sherlock Holmes in Babylon: and other tales of mathematical history&lt;/em&gt;&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="205" LastActivityDate="2010-12-19T12:01:20.497" />
  <row Id="14862" PostTypeId="1" AcceptedAnswerId="14871" CreationDate="2010-12-19T12:46:29.773" Score="5" ViewCount="180" Body="&lt;p&gt;I had this problem on my last exam in calculus, I still don't know, how to solve this, so I would appreciate your help.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $a_{ij} \in \mathbb{R}$ for $i,j \in \mathbb{N}$. Suppose, that for every $j$ series $S_j = \sum_{i=1}^{\infty}{a_{ij}}$ is absolutely convergent and, for every $i$ exists finite limit (limit $\not= \pm\infty$) $c_i=\lim_{j\to\infty}{a_{ij}}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(a) Prove, that if exists such absolutely convergent series $\sum_{i=1}^{\infty}{b_i}$ that $|a_{ij}| \leq |b_i|$ for every $i,j$ then $S_j \to S = \sum_{i=1}^{\infty}{c_i}$, $j \to \infty$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(b) Is argument in (a) true without assumption that series $\sum{}b_i$  with properties given in (a) exists?&lt;/p&gt;&#xA;" OwnerUserId="3856" LastActivityDate="2010-12-19T14:47:22.977" Title="Series absolute convergence, proof" Tags="&lt;calculus&gt;&lt;sequences-and-series&gt;&lt;limit&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="14864" PostTypeId="2" ParentId="14358" CreationDate="2010-12-19T13:23:01.057" Score="1" ViewCount="" Body="&lt;p&gt;The technique you suggest is sometimes used to construct &lt;a href=&quot;http://en.wikipedia.org/wiki/Aperiodic_tiling#Cut-and-project_method&quot; rel=&quot;nofollow&quot;&gt;quasi-crystaline structures&lt;/a&gt; or in mathematical parlance, aperiodic tilings. If you take a high dimensional lattice and you project on an appropriate subspace U, the result can be the lattice structure of a quasi-crystal. &lt;/p&gt;&#xA;" OwnerUserId="3567" LastActivityDate="2010-12-19T13:23:01.057" CommentCount="1" />
  <row Id="14865" PostTypeId="1" AcceptedAnswerId="14866" CreationDate="2010-12-19T13:25:38.320" Score="0" ViewCount="187" Body="&lt;p&gt;I've looked but cannot find the mathematical way to represent the following:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;r = Max(x1, x2, x3)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I want to mathematically show that r = max value of the set (x1, x2, x3). To make sure I'm being clear, I want to write the math formula of the code equivalent.&lt;/p&gt;&#xA;" OwnerUserId="4930" LastActivityDate="2010-12-19T13:53:35.760" Title="Mathematical representation of the largest element in a set" Tags="&lt;set-theory&gt;&lt;notation&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="14866" PostTypeId="2" ParentId="14865" CreationDate="2010-12-19T13:29:01.220" Score="2" ViewCount="" Body="&lt;p&gt;As I responded on stackoverflow...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can take the &lt;a href=&quot;http://en.wikipedia.org/wiki/Infinity_norm&quot; rel=&quot;nofollow&quot;&gt;infinity/uniform norm&lt;/a&gt; of the corresponding tuple, which is defined as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\lim_{n \rightarrow \infty} (|x_1|^n + |x_2|^n + |x_3|^n + ...)^{\frac{1}{n}}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or you can just have a maximum/infimum function... what exactly is your problem with using that?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively, you can define $f\mbox{ = max}$ recursively as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f(\{a_1,a_2,a_3,\cdots\}) = \begin{cases} a_1 &amp;amp; \mbox{if the sequence is singleton} \\ f(\{a_2,a_3,a_4,\cdots\}) &amp;amp; \mbox{if } a_1 \leq a_2 \\ f(\{a_1,a_3,a_4,\cdots\}) &amp;amp; \mbox{otherwise} \end{cases}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This can be made a lot nicer if you allow $f$ to be a two variable function with an accumulator.&lt;/p&gt;&#xA;" OwnerUserId="1507" LastEditorUserId="1507" LastEditorDisplayName="" LastEditDate="2010-12-19T13:52:37.987" LastActivityDate="2010-12-19T13:52:37.987" CommentCount="3" />
  <row Id="14868" PostTypeId="2" ParentId="14865" CreationDate="2010-12-19T13:53:35.760" Score="1" ViewCount="" Body="&lt;p&gt;I wouldn't write $\max(a,b)$ or $\max(a,b,c)$. For me, max assigns to a &lt;em&gt;set&lt;/em&gt; its maximal element (if it exists), so I use $\max\{a,b\}$ and $\max\{a,b,c\}$. In general, if $A$ is a set of real numbers, I write $\max A := \sup A$ (the supremum of $A$) if $\sup A$ is an element of $A$.&lt;/p&gt;&#xA;" OwnerUserId="2575" LastActivityDate="2010-12-19T13:53:35.760" CommentCount="1" />
  <row Id="14869" PostTypeId="2" ParentId="14826" CreationDate="2010-12-19T14:10:11.230" Score="3" ViewCount="" Body="&lt;p&gt;Deat Akhil, I have nothing to add to Matt E's masterful survey on the algebraic/analytic comparison. However since you ask about &lt;em&gt;topological&lt;/em&gt; line bundles, you have to modify his answer in the following way. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You must replace everywhere $\mathcal O$ by $\mathcal C$, the sheaf of continuous functions. Now things are very easy: since $\mathcal C$ is soft (fine if you prefer), it is acyclic and so the map &#xA;$Pic^{top}(X) \to H^2(X,&#xD;&#xA; \mathbb Z)$ is an isomorphism: a topological line bundle is classified by its Chern class, which lives in the second cohomology group of the space. To put it dramatically: the continuous jacobian is trivial ! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example topological line bundles on a compact Riemann surface are classified by  $\mathbb Z$ in stark contrast to the huge Picard variety classifying its algebraic= analytic line bundles. So to answer your question &quot;To what extent is this true in general?&quot; [natural bijection between  algebraic and topological line bundles] I would answer, just for the pleasure of using the anglicism : &quot;once in a blue moon&quot;.&lt;/p&gt;&#xA;" OwnerUserId="3217" LastEditorUserId="3217" LastEditorDisplayName="" LastEditDate="2010-12-19T18:37:05.063" LastActivityDate="2010-12-19T18:37:05.063" CommentCount="4" />
  <row Id="14870" PostTypeId="2" ParentId="14815" CreationDate="2010-12-19T14:26:45.343" Score="1" ViewCount="" Body="&lt;p&gt;This is a follow-up to the comment about Abel's theorem below Willie Wong's post.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You don't need Abel's theorem if you proceed as follows (taken from &quot;What is Mathematics?&quot; by Courant and Robbins):  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sum_{k=0}^{n} (-1)^n x^2 = \frac{1+(-1)^nx^{2n+2}}{1+x^2} = \frac{1}{1+x^2} + \frac{(-1)^nx^{2n+2}}{1+x^2}$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;and you integrate each side from $x=0$ to $x=1$ and then take a limit, bounding the rightmost term in the right integral by discarding the $x^2$ term in the denominator.  This is nice as it allows a rigorous presentation of this to (a suitably motivated) audience of first year calculus students.&lt;/p&gt;&#xA;" OwnerUserId="4009" LastActivityDate="2010-12-19T14:26:45.343" />
  <row Id="14871" PostTypeId="2" ParentId="14862" CreationDate="2010-12-19T14:47:22.977" Score="5" ViewCount="" Body="&lt;p&gt;First let us look at part (b) to extract some ideas about proving part (a). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the double sequence $a_{ij} = \delta_{ij}$, that is, $a_{ij} = 1$ if $i = j$ and $a_{ij} = 0$ otherwise. Then $S_j = 1$ for any $j$. On the other hand, $c_i = 0$ for any $i$. So $S = \sum c_i = 0 \neq \lim S_j$. So without the dominating sequence $b_i$ the statement is false.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Like Pete says, this is another reflection of something in the continuous case, namely the principle of weak convergence and also Fatou's lemma.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we prove part (a). Fix $\epsilon &amp;gt; 0$. Let $N$ be chosen large enough such that $\sum_{i = N}^\infty |b_i| &amp;lt; \epsilon/4 $, this $N$ exists since $(b_i)$ is absolutely convergent. Consider the partial sums $S_{j,N} = \sum_{i = 1}^{N-1} a_{ij}$. We can then pick $M$ sufficiently large such that for all $j &amp;gt; M$ and  $i &amp;lt; N$, $|a_{ij} - c_i| &amp;lt; \epsilon / (4N)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in particular, $|S_{j,N} - S_N| &amp;lt; \epsilon / 4$ for $j &amp;gt; M$ where $S_N = \sum_{i = 1}^{N-1} c_i$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, by assumption of convergence, and the domination of $(b_i)$, you have $|c_i| &amp;lt; |b_i|$. So in particular &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ |S - S_N| = | \sum_{N}^{\infty} c_i | \leq \sum_{N}^\infty |b_i| \leq \epsilon/4 $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We also have &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ |S_N - S_{j,N} | &amp;lt; \epsilon / 4, \quad j &amp;gt; M $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(where $M$ implicitly depends on $N$, and hence on $\epsilon$)  and &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ |S_j - S_{j,N} | &amp;lt; \epsilon / 4 $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we have that, applying the triangle inequality, that &lt;em&gt;for every fixed $\epsilon &amp;gt; 0$, we can pick $M$ sufficiently large such that&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ |S_j - S| &amp;lt; \epsilon \qquad \forall j &amp;gt; M$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Q.E.D.&lt;/p&gt;&#xA;" OwnerUserId="1543" LastActivityDate="2010-12-19T14:47:22.977" CommentCount="4" />
  <row Id="14872" PostTypeId="2" ParentId="9999" CreationDate="2010-12-19T15:10:23.837" Score="1" ViewCount="" Body="&lt;p&gt;Using integers only, I would use &lt;a href=&quot;http://en.wikipedia.org/wiki/Binary_search&quot; rel=&quot;nofollow&quot;&gt;Binary Search&lt;/a&gt;. Certainly you can compute $F_n$ only with integers, the simplest way is matrix exponentiation. Using Binary Search you can find numbers ``near'' your number $x$ and you will find either $x = F_n$ or $F_{n+1} &amp;gt; x &amp;gt; F_n$. I suppose this method is generic for anything monotone you can compute fast. To initiliaze the binary search, just keep doubling $ F_{2n} $&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Binary search allows you to search for a number x in a sorted &quot;array&quot; F[] (in the programming sense). Use this method to search for your number. When you need F[n] just compute $F_n$. This will work because the sequence is strictly increasing except for the initial 1,1.&lt;/p&gt;&#xA;" OwnerUserId="4750" LastEditorUserId="4750" LastEditorDisplayName="" LastEditDate="2010-12-19T15:26:56.887" LastActivityDate="2010-12-19T15:26:56.887" />
  <row Id="14873" PostTypeId="2" ParentId="154" CreationDate="2010-12-19T15:35:30.110" Score="2" ViewCount="" Body="&lt;p&gt;Ordered pairs exist.&#xA;If you define an operation on something that exists, the thing 'with the operation' exists.&#xA;Certain operations can be defined on ordered pairs&#xA;Complex numbers are ordered pairs with those operations defined on them.&#xA;∴ Complex numbers exist.&lt;/p&gt;&#xA;" OwnerUserId="4934" LastActivityDate="2010-12-19T15:35:30.110" CommunityOwnedDate="2011-04-15T20:45:29.080" />
  <row Id="14874" PostTypeId="2" ParentId="14844" CreationDate="2010-12-19T15:55:19.737" Score="3" ViewCount="" Body="&lt;p&gt;To do this in (the free, open-source, group-theory software) &lt;a href=&quot;http://www.gap-system.org/&quot; rel=&quot;nofollow&quot;&gt;GAP&lt;/a&gt;, you could use:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;g := Group( (1,2)(3,4), (1,3)(2,4), (1,2,3,4) );;&#xA;d := PermutationsList( [2,1,1,0] );;&#xA;o := OrbitsDomain( g, d, Permuted );;&#xA;r := List( o, orbit -&amp;gt; Reversed( AsSet( orbit ) )[1] );&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The first command defines a dihedral group (acting on the vertices of a square).  The second command defines the domain on which the group will act: the set of all permutations of the list [2,1,1,0].  This list is constructed in memory, so this limits the size of the list.  You can calculate the size of the list with:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;NrPermutationsList([2,1,1,0]);&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;and look for a different method if the list is too long (up to ten thousand will be very quick, up to a million will be fine, after ten or twenty million you'll need to try something else).  The third command sorts the domain into &quot;orbits&quot;, the sets of equivalent permutations, with the group acting on the domain via the function &quot;Permuted&quot; which permutes the indices of a list.  The fourth command asks for orbit representatives.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If g is small compared to d, then you are stuck.  This is simply how hard the problem is.  If g is reasonably large compared to d you can avoid computing d explicitly, using a transversal of an overgroup of g in the symmetric group.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, try this version first.  If you get to an example that doesn't work, post a specific group and list and I'll describe how the transversal works (it is often going to be slower, just more efficient in memory usage).&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;To work more specifically with your problem here are some GAP commands.  For many partitions, GAP takes too long (so memory efficiency is not the problem).  To get a random partition you could use:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;x:=Random(Partitions(16));;&#xA;Append(x,ListWithIdenticalEntries(16-Size(x),0));;&#xA;x; NrPermutationsList(x);&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I'll take:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;x := [ 2, 2, 2, 2, 2, 2, 2, 1, 1, 0, 0, 0, 0, 0, 0, 0 ];&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Now construct the symmetry group of the hypercube, the set of permutations of the partition, the orbits, and the orbit representatives as before:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;g := WreathProductProductAction(SymmetricGroup(2),SymmetricGroup(4));;&#xA;d := PermutationsList( x );;&#xA;o := OrbitsDomain( g, d, Permuted );;&#xA;r := List( o, orbit -&amp;gt; Reversed( AsSet( orbit ) )[1] );;&#xA;Size( r );&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;After about a minute, it should print 1292.  For 16 = 1+1+2+3+4+5 it has run for a day or so, and still isn't finished.  No trouble with memory usage, it is just sifting through 3 million orderings.  So, it is not practical for all the partitions, but for many it will work just fine.&lt;/p&gt;&#xA;" OwnerUserId="583" LastEditorUserId="583" LastEditorDisplayName="" LastEditDate="2010-12-20T23:15:49.550" LastActivityDate="2010-12-20T23:15:49.550" CommentCount="5" />
  <row Id="14876" PostTypeId="1" CreationDate="2010-12-19T16:58:56.543" Score="4" ViewCount="318" Body="&lt;p&gt;Hi, I am a Software Engineer. With interest on computers I have chosen this path. But I don't know anything about great maths. I want to learn from basic to some level(which would help me in my field..like identifying solution to a problem, designing algorithms, designing shapes and some animation on them). So, please suggest me where do I need to start. Please suggest me/list out the concepts that I need to go through.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you very much&lt;/p&gt;&#xA;" OwnerUserId="4937" LastEditorUserId="1543" LastEditorDisplayName="" LastEditDate="2010-12-19T19:15:38.627" LastActivityDate="2010-12-20T14:03:20.613" Title="Suggest me the path to learn Maths" Tags="&lt;soft-question&gt;&lt;self-learning&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="14877" PostTypeId="2" ParentId="14821" CreationDate="2010-12-19T17:17:22.760" Score="0" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\ $ Exploit the linearity of $\Sigma\:$: $\rm\ \Sigma\ (f(k)+c)\ = \ \Sigma\ f(k) + \Sigma\ c\ $ to reduce to a geometric sum.&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-12-19T17:17:22.760" />
  <row Id="14878" PostTypeId="1" AcceptedAnswerId="14916" CreationDate="2010-12-19T17:24:01.417" Score="12" ViewCount="395" Body="&lt;p&gt;My algebra professor gave me this puzzle a while back.  I'm pretty sure I've found the right solution; nonetheless, I wanted to share it and see if you come up with anything really nice or unexpected.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Prove that if you take a unit square and cut it into a finite number of smaller squares (in any way you can think of), the side lengths of the smaller squares are all rational.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;P.S. The first tag was my professor's hint.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[Edit] Just to be clear, every piece must actually be a square (e.g. no gluing two triangles into a square).&lt;/p&gt;&#xA;" OwnerUserId="4904" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-12-20T03:21:53.763" LastActivityDate="2011-01-03T04:53:28.177" Title="Cutting a unit square into smaller squares" Tags="&lt;linear-algebra&gt;&lt;combinatorics&gt;&lt;puzzle&gt;" AnswerCount="4" CommentCount="8" FavoriteCount="3" />
  <row Id="14879" PostTypeId="1" AcceptedAnswerId="14901" CreationDate="2010-12-19T18:13:56.870" Score="-16" ViewCount="1215" Body="&lt;p&gt;I know 0.9999999999... =1   , cause of its infinite nines&lt;br&gt;&#xA;But what about 0.9999999..(infinite nines)...999998&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it also 1?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Notice the 8 in the end)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Update:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So there is a difference, as we see clearly, but they are also the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As T.. answer says, it seems I have invented new math field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We naturally call these super-numbers for kalle-numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I want to know what are the generall consequences for this extension of math, or I want a proof that they are useless (Which I highly doubt).&lt;/p&gt;&#xA;" OwnerUserId="4929" LastEditorUserId="1543" LastEditorDisplayName="" LastEditDate="2010-12-21T17:47:46.460" LastActivityDate="2010-12-21T17:47:46.460" ClosedDate="2010-12-20T19:45:13.227" Title="What are the implications of the discovery of the kalle-numbers ?" Tags="&lt;algebra-precalculus&gt;" AnswerCount="3" CommentCount="33" FavoriteCount="2" />
  <row Id="14880" PostTypeId="2" ParentId="14879" CreationDate="2010-12-19T18:20:46.523" Score="29" ViewCount="" Body="&lt;p&gt;What does it mean to have &quot;infinitely many 9's and then an 8&quot;?  Well, you're putting the 8 after the last 9, but there is no last 9!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One possible way to get around this is to consider the sequence 0.98, 0.998, 0.9998, etc., where the n-th term of the sequence has n 9's and then an 8.  Then your question becomes, &quot;is the limit of this sequence, as n goes to infinity, equal to 1?&quot; to which the answer is, &quot;yes&quot;.&lt;/p&gt;&#xA;" OwnerUserId="4904" LastActivityDate="2010-12-19T18:20:46.523" CommentCount="10" />
  <row Id="14882" PostTypeId="2" ParentId="14876" CreationDate="2010-12-19T18:30:24.320" Score="3" ViewCount="" Body="&lt;p&gt;Perhaps what you mean is that you want a book like Polya's &quot;How to solve it&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0691023565&quot; rel=&quot;nofollow&quot;&gt;http://www.amazon.com/How-Solve-Aspect-Mathematical-Method/dp/0691023565&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4728" LastActivityDate="2010-12-19T18:30:24.320" />
  <row Id="14883" PostTypeId="1" CreationDate="2010-12-19T18:56:18.737" Score="-4" ViewCount="156" Body="&lt;p&gt;Does this equation define a function with independent variable x?&lt;br&gt;&#xA;xlyl=x+9&lt;/p&gt;&#xA;" OwnerUserId="4942" LastActivityDate="2010-12-19T19:29:57.253" Title="Defining a function" Tags="&lt;functions&gt;" AnswerCount="1" CommentCount="6" />
  <row Id="14884" PostTypeId="2" ParentId="14851" CreationDate="2010-12-19T19:04:22.523" Score="4" ViewCount="" Body="&lt;p&gt;Here I use notation $\chi_N$ to be the characteristic function of $[N,N+1]$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $f_1 = \chi_1$. Let $f_k = \chi_k - \chi_{k-1}$ for $k\geq 2$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On any compact set of the reals, $\sum f_i$ converges uniformly to $f = 0$. So $\int \sum f_i = 0$. Yet $\int f_1 = 1$ and $\int f_k = 0$ for $k \geq 2$. So $\sum \int f_i = 1\neq \int\sum f_i$. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Analyticity doesn't matter. Let $g(x)$ be any arbitrary, positive, integrable function on the real line which decays faster then, say, $|x|^{-1}$. Then if we take $f_1(x) = g(x-1) |_{\mathbb{R}_+}$ and $f_k(x) = g(x-k) - g(x-k+1) |_{\mathbb{R}_+}$ for $k\geq 2$. You get the same result that on any compact subset of the reals $\sum f_i$ converges uniformly to 0. Yet $\sum\int f_i \geq \int_0^\infty g(x) dx$. You can no insert in an arbitrary analytic $g(x)$ (say the Gaussian for example) and get the same conclusion. &lt;/p&gt;&#xA;" OwnerUserId="1543" LastEditorUserId="1543" LastEditorDisplayName="" LastEditDate="2010-12-19T19:10:51.173" LastActivityDate="2010-12-19T19:10:51.173" CommentCount="3" />
  <row Id="14885" PostTypeId="2" ParentId="13054" CreationDate="2010-12-19T19:08:35.580" Score="1" ViewCount="" Body="&lt;p&gt;Perhaps you could do something with the taylor series? Would be pretty nasty but youd only have to worry about the first few terms.&lt;/p&gt;&#xA;" OwnerUserId="2484" LastActivityDate="2010-12-19T19:08:35.580" CommentCount="12" />
  <row Id="14886" PostTypeId="2" ParentId="14883" CreationDate="2010-12-19T19:19:48.600" Score="3" ViewCount="" Body="&lt;p&gt;assuming that you mean $x|y|=x+9$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;then $|y|=1+9/x$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if $-9 &amp;lt; x &amp;lt; 0 $ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;then $y$ is not defined. &lt;/p&gt;&#xA;" OwnerUserId="4728" LastEditorUserId="4728" LastEditorDisplayName="" LastEditDate="2010-12-19T19:29:57.253" LastActivityDate="2010-12-19T19:29:57.253" CommentCount="10" />
  <row Id="14887" PostTypeId="1" AcceptedAnswerId="14889" CreationDate="2010-12-19T19:42:51.367" Score="6" ViewCount="199" Body="&lt;p&gt;I seem to have a misunderstanding of how to work with a Taylor series. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose I want to write $f(x)=x e^x$'s Taylor expansion of $n$ degree around $0$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I see two ways:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Find the $n$th derivative of $f(x)$, this is quite easy: $f^{(n)} (x)=n e^x+x e^x$.&lt;br&gt;&#xA;And from the formula I get: &#xA;$$f(x)=\sum_{k=1}^{n} \frac{x^k}{(k-1)!} + R_{n}(x).$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Since $e^x$ Taylors expansion is already known, it's also possible to do this:&#xA;$$f(x)=xe^x=x\left(\sum_{k=0}^{n-1} \frac{x^k}{k!} + R_{n-1}(x)\right)=\sum_{k=0}^{n-1} \frac{x^{k+1}}{k!} + xR_{n-1}(x)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But how do I interpret $xR_{n-1}(x)$?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I feel like I'm missing something fundamental about the meaning of $R_{n}(x)$.&lt;/p&gt;&#xA;" OwnerUserId="3237" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-19T20:00:18.440" LastActivityDate="2010-12-20T04:59:07.480" Title="Basic Taylor expansion question" Tags="&lt;calculus&gt;&lt;taylor-expansion&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" />
  <row Id="14888" PostTypeId="2" ParentId="11076" CreationDate="2010-12-19T19:49:05.200" Score="1" ViewCount="" Body="&lt;p&gt;1) Sean Carroll claims that the essentials of the vector concept emerged with &lt;strong&gt;Galileo Galilei&lt;/strong&gt;. Galileo observed that if you kick a ball forwards off a building, or drop a ball off a building, that the balls land at the same time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore horizontal motion (&lt;strong&gt;kick&lt;/strong&gt;) is &lt;strong&gt;separable&lt;/strong&gt; from vertical motion (&lt;strong&gt;fall&lt;/strong&gt;). From there you get the separation of the force into two parts: the 2 components of the vector.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) Regarding the dot product, it is the most natural way to numerically capture the concept of &quot;angle between two things&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) The pattern continues like this: $\sqrt{a^2 + b^2 + c^2 + d^2 + e^2 + \ldots}$ for a multi-dimensional right triangle with sides $a, b, c, d, e, \ldots$. Surprising, isn't it?&lt;/p&gt;&#xA;" OwnerUserId="1457" LastEditorUserId="1457" LastEditorDisplayName="" LastEditDate="2010-12-20T01:15:40.627" LastActivityDate="2010-12-20T01:15:40.627" />
  <row Id="14889" PostTypeId="2" ParentId="14887" CreationDate="2010-12-19T19:54:45.217" Score="16" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Edited and rewritten, hopefully to make it clearer.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is something that the notation does not help and which you need to keep straight (this may be the source of your confusions). $R_n(x)$ is the &quot;remainder&quot; at $x$: the error between the actual value of the function and the value you get from evaluating the Taylor polynomial of degree $n$ instead. That is, the $n$th remainder is equal to&#xA;$$f(x) - \sum_{k=0}^n \frac{f^{(k)}(0)}{k!}x^k$$&#xD;&#xA;(I'm assuming the polynomial is being expanded around $a=0$). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the remainder really needs three pieces of information to make it precise: the value of $n$, the value of $x$, and the function $f$ in question. It would be best to write it as $R(n,x,f)$, but since the function $f$ is usually clear from context, we ignore it. But I will use $R(n,x,f)$ in what follows. So notice that in your question, in (1) what you have is $R(n,x,xe^x)$, while in (2) what you have is $xR(n-1,x,e^x)$.  And so the question is what is the relation between the two expressions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Before addressing this in more detail, let me take a slight detour to Taylor series.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Say we have the Taylor &lt;em&gt;series&lt;/em&gt; for $f$,&#xA;$$\sum_{k=0}^{\infty}\frac{f^{(k)}(0)}{k!}x^k.$$&#xD;&#xA;We know this power series converges inside the radius of convergence (which can be found using, say, the Ratio Test); for $f(x)=e^x$, the radius of convergence is infinite, so the series converges for &lt;em&gt;all&lt;/em&gt; values of $x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, there is still the question of whether the value of the series at a given $x$ &lt;em&gt;equals&lt;/em&gt; the value of the function at that point. This is where the remainders come in. The Taylor series for $f$ converges to $f(x)$ at the point $x_0$ if and only if $R(n,x_0,f)\to 0$ as $n\to\infty$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of $f(x)=e^x$, one can show that the Taylor series not only converges everywhere, but also that it converges &lt;strong&gt;to&lt;/strong&gt; $e^x$. For example, this can be done using the Cauchy estimate for the remainder: given $r\gt 0$, and $x$ in $(-r,r)$, if $M_n$ is a number such that $|f^{(n+1)}(x)|\leq M_n$ for all $x$ in $(-r,r)$, then&#xA;$$|R(n,x,f)|\leq \frac{M_nr^{n+1}}{(n+1)!}.$$&#xD;&#xA;For $f(x) = e^x$, you can take $M=e^r$ (or even $3^r$), so you get an exponential divided by a factorial, and that goes to $0$ as $n\to\infty$. This holds for any $x$ (by changing the $r$), so that $R(n,x,e^x)\to 0$ as $n\to\infty$ for all $x$. So the Taylor series for $e^x$ converges to $e^x$ at every $x$. That is, for each $x$,&#xA;$$e^x = \sum_{k=0}^{\infty}\frac{x^k}{k!} = 1 + x + \frac{x^2}{2!} + \cdots + \frac{x^n}{n!}+\cdots$$&#xD;&#xA;in the sense that the value of the limit on the right hand side &lt;em&gt;is&lt;/em&gt; exactly equal to $e^x$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &quot;Taylor series&quot; for $x$ is also very easy: it's just $x$ itself (not even a real series). It also has infinite radius of convergence, and $R(n,x,x) = 0$ for any $n\gt 1$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is a theorem that if the Taylor series for $f$ converges to $f$ in $(-r,r)$, and the Taylor series for $g$ converges to $g$ in $(-R,R)$, then the product of the series will converge to $fg$ in $(-\min(r,R),\min(r,R))$ (that is, on the interval where they both converge). For $e^x$ and $x$, this tells you that indeed you have that &#xA;$$xe^x = x\left(\sum_{k=0}^{\infty}\frac{x^k}{k!}\right) = \sum_{k=0}^{\infty}\frac{x^{k+1}}{k!} = \sum_{k=1}^{\infty}\frac{x^k}{(k-1)!}.$$&#xD;&#xA;Because the series converges, and converges to $xe^x$, that means that if you write&#xA;$$xe^x = \sum_{k=1}^n \frac{x^k}{(k-1)!} + R(n,x,xe^x)$$&#xD;&#xA;then you have that &#xA;$$R(n,x,xe^x) = \sum_{k=n+1}^{\infty} \frac{x^k}{(k-1)!}$$&#xD;&#xA;(in the sense that the value of $R(n,x,xe^x)$ is the limit of the partial sums of that series) and moreover that&#xA;$$\sum_{k=n+1}^{\infty}\frac{x^k}{(k-1)!} \to 0\text{ as }n\to\infty.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you are doing in (2) is to essentially deal with the Taylor polynomials instead of the series as I did above. We have that the Taylor series for $e^x$ converges to $e^x$ at every $x$, so that if we write&#xA;$$e^x = \sum_{k=0}^{n}\frac{x^k}{k!} + R(n,x,e^x)$$&#xD;&#xA;then&#xA;$$R(n,x,e^x) = \sum_{k=n+1}^{\infty}\frac{x^k}{k!}$$&#xD;&#xA;and $R(n,x,e^x)\to 0$ as $n\to \infty$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Multiplying the expression for $e^x$ by $x$, we get&#xA;$$xe^x = x\left(\sum_{k=0}^n\frac{x^k}{k!} + R(n,x,e^x)\right) = \sum_{k=0}^n\frac{x^{k+1}}{k!}+xR(n,x,e^x).$$&#xD;&#xA;Except that note that this time $xR(n,x,e^x)$ is giving the $(n+1)$st remainder, since the polynomial we have is of degree $n+1$. So, equating this with the expression we had before, this suggests that we should have&#xA;$$R(n+1,x,xe^x) = xR(n,x,e^x).$$&#xD;&#xA;And indeed, this is what we find when we express them as series and limits:&#xA;\begin{align}&#xA;xR(n,x,e^x) &amp;amp;= x\left(\sum_{k=n+1}^{\infty}\frac{x^k}{k!}\right)\\&#xA;&amp;amp;=x\left(\lim_{m\to\infty}\sum_{k=n+1}^m\frac{x^k}{k!}\right)\\&#xA;&amp;amp;= \lim_{m\to\infty}\left(x\sum_{k=n+1}^m \frac{x^k}{k!}\right)\\&#xA;&amp;amp;= \lim_{m\to\infty}\sum_{k=n+1}^m \frac{x^{k+1}}{k!}\\&#xA;&amp;amp;=\lim_{m\to\infty}\sum_{k=n+2}^{m-1}\frac{x^k}{(k-1)!}\\&#xA;&amp;amp;= \lim_{m\to\infty}\sum_{k=n+2}^{m}\frac{x^k}{(k-1)!}\\&#xA;&amp;amp;= R(n+1,x,xe^x).&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What this tells you is that the $n+1$st remainder at $x$ for the function $xe^x$ equals $x$ times the $n$th remainder at $x$ for the function $e^x$.  So your equations both make sense and they are telling you correct things, once you put back the necessary information into the remainder.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-20T04:59:07.480" LastActivityDate="2010-12-20T04:59:07.480" CommentCount="8" />
  <row Id="14891" PostTypeId="2" ParentId="14379" CreationDate="2010-12-19T20:44:46.580" Score="1" ViewCount="" Body="&lt;p&gt;Since P is abelian, when acting upon P by conjugation from P, say you have p,q in P, then pqp^{-1} = q.  Thus P is indeed contained in the kernel of N -&gt; Aut(P), and hence induces a map N/P -&gt; Aut(P).  Examine the order of Aut(P) and this allows you to conclude that the image of N/P in Aut(P) must be trivial.  From this you can conclude the desired result.&lt;/p&gt;&#xA;" OwnerUserId="4943" LastActivityDate="2010-12-19T20:44:46.580" />
  <row Id="14892" PostTypeId="2" ParentId="14876" CreationDate="2010-12-19T20:52:36.003" Score="1" ViewCount="" Body="&lt;p&gt;The &lt;b&gt;program ``sage''&lt;/b&gt; helped me a lot. You can experiment with math objects in a programming environment. It is free software - GPL2. You can try it online in a browser. The site is &lt;a href=&quot;http://sagemath.org/&quot; rel=&quot;nofollow&quot;&gt;http://sagemath.org/&lt;/a&gt;. More information: &lt;a href=&quot;http://en.wikipedia.org/wiki/Sage_%28mathematics_software%29&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Sage_%28mathematics_software%29&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: This will not teach you math, but may help you.&lt;/p&gt;&#xA;" OwnerUserId="4750" LastEditorUserId="4750" LastEditorDisplayName="" LastEditDate="2010-12-20T08:37:15.153" LastActivityDate="2010-12-20T08:37:15.153" />
  <row Id="14893" PostTypeId="2" ParentId="14629" CreationDate="2010-12-19T20:54:49.803" Score="2" ViewCount="" Body="&lt;p&gt;So, if for $x\in \partial D^{2n}$ you take $G(x)=g(x)$, then probably you want to do &quot;the same&quot; map over the whole cone.  It follows from the definition of the union topology (or whatever it's called) that this extends to a continuous map.  Then, look at the induced map on the pairs $(K,K-D^{2n})$ (for either complex $K$), and here the map is just (up to confusing maps on pairs with maps on their quotients) the suspension of $g$.  Note that for both complexes, the inclusion $(K,\emptyset)\rightarrow (K,K-D^{2n})$ is a cohomology isomorphism in degree $2n$, and use the fact that suspension induces an isomorphism $\pi_n(S^n)\rightarrow \pi_{n+1}(S^{n+1})$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A diagram might help:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;    (Cfg,o)    ---&amp;gt;     (Cf,o)&#xA;       |                  |&#xA;       |                  |&#xA;       V                  V&#xA;(Cfg,Cfg-D^2n) ---&amp;gt; (Cf, Cf-D^2n)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Here, the o's stand for $\emptyset$.  The vertical maps are inclusions of pairs, and the horizontal lines are really all just $G$ (as I defined it).  The top arrow induces $\alpha_{Cf} \mapsto \alpha_{Cfg}$, and the vertical maps are cohomology isomorphisms in degree $2n$.  On the bottom row, since $H^*(X,A)=\tilde{H}^*(X/A)$, this is basically a map $S^{2n}\rightarrow S^{2n}$, which by construction is exactly the suspension $Sg$ of $g$.  Then, $\deg(Sg)=\deg(g)$; that is, $S:\pi_{2n-1}(S^{2n-1})\rightarrow \pi_{2n}(S^{2n})$ is an isomorphism.  So this induces $\beta_{Cf}\mapsto \deg(g)\cdot \beta_{Cfg}$.  By naturality,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ H(fg)\cdot \beta_{Cfg} = \alpha_{Cfg}^2=G^*(\alpha_{Cf}^2)=G^*(H(f)\cdot \beta_{Cf})=H(f)\cdot (\deg(g)\cdot \beta_{Cfg}).$$&lt;/p&gt;&#xA;" OwnerUserId="2159" LastEditorUserId="2159" LastEditorDisplayName="" LastEditDate="2010-12-20T06:17:41.787" LastActivityDate="2010-12-20T06:17:41.787" CommentCount="4" />
  <row Id="14894" PostTypeId="1" CreationDate="2010-12-19T21:14:15.917" Score="5" ViewCount="127" Body="&lt;p&gt;Does anyone have a good explanation for why Knot invariants tend to be well organized  as polynomials?  What exactly is going on and why don't we often see polynomial invariants for classifying other geometric objects?  For instance, why don't we just say that the Alexander invariants are a finite set of numbers?  Presumably, the organizing of them into polynomials shows off some inherent geometric properties of knot description and combination which is easily encapsulated in the multiplication and addition of polynomials.  I would be happy (and not surprised) if this also helps me understand why Lie algebra representations also often arise, especially in more recently discovered invariants.&lt;/p&gt;&#xA;" OwnerUserId="948" LastActivityDate="2010-12-19T21:57:14.447" Title="Why are knot invariants best organized as polynomials?" Tags="&lt;algebraic-topology&gt;&lt;lie-algebra&gt;&lt;knot-theory&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="14895" PostTypeId="1" CreationDate="2010-12-19T21:21:17.013" Score="0" ViewCount="138" Body="&lt;p&gt;Here is my set up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am running a simulation of a gambler betting over time&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are the givens I am using.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Probability of win = 35%   which is in B1 table&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Odds = 2  which is in B3 table&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Kelly # (amount of current bankroll to bet each time) = 2.5%  which is in D1 table&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bankroll = $10,000  which is in A6 table&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am going to put in formulas I have for other cells in my simulation of bets, but will provide the data that those tables hold as well&lt;/p&gt;&#xA;&#xA;&lt;p&gt;SOME OTHER COLUMNS&#xA;B6:&#xA;=$D$1*A6&lt;/p&gt;&#xA;&#xA;&lt;p&gt;C6:&#xA;=randbetween(1,000) column&lt;/p&gt;&#xA;&#xA;&lt;p&gt;D6:&#xA;=if(C6&amp;lt;36,&quot;Win&quot;,&quot;Loss&quot;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;E6:&#xA;=if(D6=&quot;Win&quot;,B6*$B$3,(-B6))&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A7:&#xA;=A6+E6&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then I drag these down to run 2000 simulation of random bets &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Near the bottom I have a table that has the capital gain&#xA;the final bankroll - the beginning $10,000 bankroll.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(is there a way to graph these wins on a line graph just the start of $10,000 to my final bankroll?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;also what i need most&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How to make a what if analysis... comparing different outcomes if the Kelly # (the percentage of my current bankroll I am betting) was different.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;0.01     0.015   0.02       0.025      0.03     0.035    0.04&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and under each column of the different percentages I want the final capital gain in them... and I want to have like 200 examples of this new capital gain...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then I will average each column above for the 200 capital gains of the different percentages.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My confusion is this...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First off what is the value on the left of the percentages that I need to put in that block that will be concrete and unchanging? &#xA;Is it $10,000 my beginning bankroll?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then what are the values under that number?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know I need to highlight the grid beginning with that concrete block (set as $10,000?) and then highlight down the 200 simulations I want.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then I go to data &gt; what if analysis &gt; data table&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now what do i put in the table for row input cell?&#xA;what about for the column input cell? &amp;lt;--- I believe I am supposed to leave this as a blank box - so I will just chose $G$1  an unused cell&lt;/p&gt;&#xA;&#xA;&lt;p&gt;HELP!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also here is a screen cap of my excel sheet (I am unable to link to my second screen cap... here it is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i6.photobucket.com/albums/y249/spittballz/Capture2.jpg&quot; alt=&quot;Excel screenshot&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4945" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-20T00:56:50.070" LastActivityDate="2010-12-20T00:56:50.070" Title="What if analysis of an Excel spreadsheet." Tags="&lt;statistics&gt;&lt;probability-theory&gt;&lt;simulation&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="14896" PostTypeId="1" CreationDate="2010-12-19T21:24:41.517" Score="2" ViewCount="170" Body="&lt;p&gt;In Wikipedia and MathPlanet an equivalent definition of a symplectic matrix is given:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left( \begin{array}{ccc}&#xD;&#xA;A &amp;amp; B \\&#xD;&#xA;C &amp;amp; D \end{array} \right)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is symplectic if and only if:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$A^TD-C^TB=I, A^TC=C^TA, D^TB=B^TD$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but it seems wrong, since, for example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left( \begin{array}{ccc}&#xD;&#xA;0 &amp;amp; 0 &amp;amp; -1 &amp;amp; 1 \\&#xD;&#xA;0 &amp;amp; 0 &amp;amp; 0 &amp;amp; -1 \\&#xD;&#xA;1 &amp;amp; 1 &amp;amp; 0 &amp;amp; 0 \\&#xD;&#xA;0 &amp;amp; 1 &amp;amp; 0 &amp;amp; 0 \end{array} \right)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is symplectic but doesn't satisfy the conditions. Or have I mixed everything up?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: this is crazy talk. That matrix isn't symplectic! (not for the form defined in Wikipedia or MathPlanet.&lt;/p&gt;&#xA;" OwnerUserId="4946" LastEditorUserId="4946" LastEditorDisplayName="" LastEditDate="2010-12-19T21:39:25.147" LastActivityDate="2011-05-07T16:46:18.910" Title="Definition of Symplectic Matrix" Tags="&lt;matrices&gt;&lt;definition&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="14897" PostTypeId="2" ParentId="14896" CreationDate="2010-12-19T21:28:52.420" Score="7" ViewCount="" Body="&lt;p&gt;This is another question which highlights the problems with not thinking about things in a coordinate-free manner.  Symplectic transformations are defined relative to a &lt;a href=&quot;http://en.wikipedia.org/wiki/Symplectic_form&quot; rel=&quot;nofollow&quot;&gt;symplectic form&lt;/a&gt;, and symplectic matrices in turn are defined relative to some &quot;canonical&quot; symplectic form with respect to the standard basis.  The problem is that there are at least two reasonable choices for such a &quot;canonical&quot; form (both of which are described at the Wikipedia article), and the resulting symplectic matrices you get from each form are different.  So you are probably just using a different one.  &lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-12-19T21:28:52.420" CommentCount="5" />
  <row Id="14898" PostTypeId="2" ParentId="14894" CreationDate="2010-12-19T21:57:14.447" Score="6" ViewCount="" Body="&lt;p&gt;I disagree with the hypothesis of the question -- knot invariants do not tend to be organized as polynomials.  The Alexander module is not a polynomial.  The fundamental group of the complement is not a polynomial.  The complement of a knot is not a polynomial.  These are some of the most fundamental invariants of a knot, and they're not polynomials. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That said, polynomial invariants do occur in knot theory and there are a few good reasons.  One is that knots up to isotopy have the structure of a commutative monoid under the connect-sum operation. Invariably, when you start talking about functions on an object $O$, you like to talk about the vector space/module structure of such a function space, so you like to talk about formal linear combinations of objects from $O$.  Formal addition and connect-sum turns knots into a commutative ring.  Polynomial rings are free commutative rings, so they're fundamental in discussions about the structure of commutative rings.  In the special case of knots in $\mathbb R^3$, it's a theorem of Schubert's, that knots under these two operations &lt;em&gt;are&lt;/em&gt; a free commutative ring (the free generators being the &quot;prime&quot; knots).   So in some sense knots can be thought of as a polynomial ring. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the upshot of this is the study of &quot;knot invariants&quot; together with the connect-sum operation on knots prejudices the discussion towards one where you are concerned with commutative rings, and so commutative-ring valued invariants.  So that's why polynomial invariants occur so often. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you were to study different internal structures on knots, you perhaps prejudice the types of invariants you use to study knots.  Tangle categories and categorical invariants would be another example.  The knot complement and geometrization would be another. &lt;/p&gt;&#xA;" OwnerUserId="642" LastActivityDate="2010-12-19T21:57:14.447" />
  <row Id="14900" PostTypeId="2" ParentId="14879" CreationDate="2010-12-19T22:14:10.477" Score="8" ViewCount="" Body="&lt;p&gt;I'll try and give an answer that needs a bit of calculus. Whenever a mathematician writes something down, he has to define what he means by what he writes. For example, when I write $1.3$, then by definition this is the number $1+\frac{3}{10}$. When I write $5.402$, then this is $5 + \frac{1}{10} + \frac{0}{100} + \frac{2}{1000}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now what is this with infinitely many digits? Let's assume for simpliciticy that we have $0$ before the decimal point. If I write&#xA;$$&#xD;&#xA;  0.a_1a_2a_3a_4\dots\,,&#xD;&#xA;$$&#xD;&#xA;with $a_n\in\{0,1,2,3,4,5,6,7,8,9\}$ for all $n\in\mathbb N$, then this mean the &lt;em&gt;infinite series&lt;/em&gt;&#xA;$$&#xD;&#xA;  \sum_{n=1}^\infty a_n(\tfrac{1}{10})^n = \lim_{N\to\infty} \sum_{n=1}^N a_n(\tfrac{1}{10})^n\,.&#xD;&#xA;$$&#xD;&#xA;If, for example, $a_n=9$ for all $n\in\mathbb N$, then by the formula for the geometric series you get&#xA;$$&#xD;&#xA;  0.9999\dots = \sum_{n=0}^\infty 9\cdot(\tfrac{1}{10})^n = 9\cdot\frac{1}{1-1/10} = 1.&#xD;&#xA;$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;OK, now you want an additional $8$ after all the nines. Well, you'll have to say what you &lt;em&gt;mean&lt;/em&gt; when you write that. I don't know of any definition of such an object, so my answer is &lt;em&gt;it's not defined&lt;/em&gt;. (By the way, I also think that it will be very difficult to give it a sensible definition, but strictly speaking this is only my belief.)&lt;/p&gt;&#xA;" OwnerUserId="2575" LastActivityDate="2010-12-19T22:14:10.477" />
  <row Id="14901" PostTypeId="2" ParentId="14879" CreationDate="2010-12-19T22:37:49.423" Score="10" ViewCount="" Body="&lt;p&gt;You could certainly have a number system that includes decimal expansions with a string of digits that is &quot;transfinitely long&quot;, indexed by an ordinal higher than $\omega$, such as &#xA;$$0.\overline{999}3216\overline{12345}623$$ (where the bar denotes infinite repetition of a block) which has an ordinal length of $\omega + \omega + 3$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this system there are objects between $0.9999....$ and $1.000...$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is that other than ordering these &quot;numbers&quot; (i.e., creating them for the sake of having things between 0.999... and 1) it is unclear whether there is anything useful that can be done with them, such as measuring some other type of object, or extending the arithmetic operations on ordinary decimals to the realm of trans-decimals.  It's not that these superobjects can't exist, it's that there is nothing to be done with them except point to their existence.&lt;/p&gt;&#xA;" OwnerUserId="467" LastActivityDate="2010-12-19T22:37:49.423" CommentCount="10" />
  <row Id="14902" PostTypeId="2" ParentId="9524" CreationDate="2010-12-19T23:32:32.717" Score="7" ViewCount="" Body="&lt;p&gt;Thanks for all the responses so far.  I thought I would summarize them for anyone else who might be interested in this question.  Rather than continuing to update the summary in the original question, though, it seems better from an organizational standpoint for the summary to appear in a community wiki answer than I then accept so that it appears at the top of the list of answers.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Others seem to agree that focusing on Taylor series is the right approach.  See the answers and comments for justifications.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Examples of the use of infinite series &lt;/p&gt;&#xA;&#xA;&lt;p&gt;a. General: Zeno's paradoxes&lt;/p&gt;&#xA;&#xA;&lt;p&gt;b. Physics: Using the first order Taylor approximation $\sin \theta \approx \theta$ in solving the pendulum differential equation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;c. Chemistry: Extending the ideal gas law to apply to high pressure and low temperature situations&lt;/p&gt;&#xA;&#xA;&lt;p&gt;d. Economics: Calculating fiscal multipliers involves geometric series&lt;/p&gt;&#xA;&#xA;&lt;p&gt;e. Computer science, 1: Several uses for generating functions (see examples by robinhoode and Raphael)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;f. Computer science, 2: Taylor series are involved in the error analysis of some numerical methods, such as Newton-Raphson and Simpson's rule.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;g. Mathematics, 1: Taylor series show that calculations involving functions like $e^x$ and $\sin x$ can all be computed using just addition, subtraction, multiplication, and division.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;h. Mathematics, 2: Power series, and Euler products in number theory in particular, as most people find number theory intrinsically interesting whether they have the background or not&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i. Mathematics, 3: Taylor series can be used to solve differential equations.  (Often students will have seen a brief introduction to differential equations earlier in the course.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;j. Mathematics, 4: There are infinite series expressions for interesting constants such as $\pi$ and $e$.  Also, any nonterminating decimal representation of a real number is an infinite series.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;k. Mathematics, 5: Using Taylor polynomials to approximate integrands in definite integrals.  (This fits well in a course like Calculus II that spends a lot of time on the integral.)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Links to other resources &lt;/p&gt;&#xA;&#xA;&lt;p&gt;a. Graphs of Taylor polynomials converging to a function (Brandon Carter's answer)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;b. Graphs of Taylor polynomials in the complex plane (Hans Lundmark's comment on Brandon's answer)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Other&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a. Emphasize similarity of series and improper integrals&lt;/p&gt;&#xA;&#xA;&lt;p&gt;b. Emphasize that convergence must be dealt with carefully.  Use geometric series as a way to introduce convergence issues.  Mention that even Euler was not always good about handling convergence.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="2370" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2011-01-06T23:22:44.117" LastActivityDate="2011-01-06T23:22:44.117" CommunityOwnedDate="2010-12-19T23:32:32.717" CommentCount="3" />
  <row Id="14903" PostTypeId="2" ParentId="14855" CreationDate="2010-12-19T23:51:28.237" Score="3" ViewCount="" Body="&lt;p&gt;As I mentioned in comments below, Kolmogorov's example is for a discontinuous function in $L^1$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a continuous function whose Fourier series diverges at all rational multiples of $2\pi$ (and hence on a dense set) see &lt;a href=&quot;http://www.amazon.com/s/ref=nb_sb_ss_i_2_28?url=search-alias%3Daps&amp;amp;field-keywords=katznelson+harmonic+analysis&amp;amp;sprefix=katznelson+harmonic+analysis&quot; rel=&quot;nofollow&quot;&gt;Katznelson's book: An Introduction to Harmonic Analysis&lt;/a&gt; Chapter 2, Remark after proof of Theorem 2.1. Note that the Fourier series of such a continuous function still converges almost everywhere by Carleson's theorem.&lt;/p&gt;&#xA;" OwnerUserId="3249" LastEditorUserId="3249" LastEditorDisplayName="" LastEditDate="2010-12-19T23:56:34.673" LastActivityDate="2010-12-19T23:56:34.673" />
  <row Id="14904" PostTypeId="1" AcceptedAnswerId="14913" CreationDate="2010-12-20T00:35:42.783" Score="3" ViewCount="117" Body="&lt;p&gt;The vector force field F=(yi,-xj) has a curl of -2. The acceleration of a particle in space is given by:&#xA;ax=y/m&#xA;ay=-x/m&#xA;This vector field has a divergence of 0. Will particles in this vector FORCE field move in circles or will they spiral outwards in a crazy fashion? My conundrum is this, the acceleration of the particles is never towards the center, therefor there is no centripetal acceleration, so how can these particles move in discreet circles?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does this reasoning make sense or am I doing a nested substitution:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;r=Sqrt[x^2+y^2]&#xA;r=Sqrt[((1/2)ax*t^2)^2 + ((1/2)ay*t^2)^2]&#xA;r=(1/2)Sqrt[t^4 (ax^2 + ay^2)]&#xA;r=(1/2)Sqrt[t^4 a^2]&#xA;r=(1/2)at^2&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="4951" LastEditorUserId="4951" LastEditorDisplayName="" LastEditDate="2010-12-20T01:57:44.680" LastActivityDate="2010-12-20T02:01:57.297" Title="Vector Force Fields and Their Physical Interpretations" Tags="&lt;calculus&gt;&lt;physics&gt;&lt;vector-space&gt;&lt;simulation&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="14905" PostTypeId="2" ParentId="14895" CreationDate="2010-12-20T00:44:59.543" Score="1" ViewCount="" Body="&lt;p&gt;I am a big fan of programming in Excel, as I find it easier to see what is going on during the debug phase.  But you are stretching what is easy.  Each run of your simulation will use 2000 rows by 6 columns.  You could cut that down to 2 if you just have the current bankroll in one and the random number in the next.  If you want to run 200 cases of each of seven different Kelly numbers, you will fill 6*200*7=8400 columns.  I think it will do that, but it will be unwieldy.  This would be enough to drive me to Python, though I might keep the spreadsheet as a check.  You might be able to set up one tab per Kelly number, and just copy from one tab to the next.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It basically sounds like you are doing the right thing.  I have an older Excel that doesn't have randbetween.  I would use rand() which gives 0-1 and compare with .35.  Column A will then have the bankroll after each play and you can look at the bottom for the result.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can get an excellent prediction of what you should see.  For a bet of 1, your expected return is .35*3=1.05 (it looks like you are paying 2:1 plus returning the bet), so you want to have as much money on the table as possible.  For a run of 2000 plays, you should win about 700 and lose 1300.  The standard deviation in wins is sqrt(2000*.35*.65)=21.33.  And if we write the Kelly number, K,  as a decimal and the number of wins as W, the bankroll at the end should be $10000(1+2K)^W(1-K)^{(2000-W)}$&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-12-20T00:44:59.543" CommentCount="2" />
  <row Id="14906" PostTypeId="1" AcceptedAnswerId="14907" CreationDate="2010-12-20T01:05:31.857" Score="0" ViewCount="70" Body="&lt;p&gt;Hi i am trying to understand a proof, I dont understand one of the steps.&#xA;could someone shed some light why the following is true.&#xA;if $f(x)$ is a non-increasing and continuous function and $a&amp;gt;b$ then $\{x\colon f(x) &amp;gt; f(b) \} \subset \{x\colon f(x) &amp;gt; a \} $ &#xA;i dont see why this is true. any insights ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;thanks&lt;/p&gt;&#xA;" OwnerUserId="4952" LastActivityDate="2010-12-20T01:09:41.103" Title="non increasing function" Tags="&lt;real-analysis&gt;" AnswerCount="1" CommentCount="8" />
  <row Id="14907" PostTypeId="2" ParentId="14906" CreationDate="2010-12-20T01:09:41.103" Score="6" ViewCount="" Body="&lt;p&gt;To say that $f$ is nonincreasing means that for all $x,y$ in the domain, if $x \leq y$ then $f(x) \geq f(y)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are given that $b &amp;lt; a$, so $f(b) \geq f(a)$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally if $A \leq B$ are two real numbers, then for any function $f$, the set of $x$ in the domain such that $f(x) &amp;gt; B$ is a subset of the set of $x$ in the domain such that &#xA;$f(x) &amp;gt; A$, just because any number which is greater than $B$ is also greater than $A$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that continuity of $f$ was not used here.&lt;/p&gt;&#xA;" OwnerUserId="299" LastActivityDate="2010-12-20T01:09:41.103" />
  <row Id="14908" PostTypeId="1" AcceptedAnswerId="14910" CreationDate="2010-12-20T01:14:20.400" Score="2" ViewCount="234" Body="&lt;p&gt;Hello, I was given the following question:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Calculate &#xA;  $$1-\frac12+\frac14-\frac18+\frac1{16}- \frac1{32}+\frac1{64}-\cdots -\frac1{512}+\cdots;$$&#xD;&#xA;  and express the answer as a simple fraction.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My approach was to use the following formula: $\frac1{1-r}$ where $r$ is the common ratio.  In the end I got $2/3$.  Am I correct?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;...Edit...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, how would i be able to explain this by words?&lt;/p&gt;&#xA;" OwnerUserId="4918" LastEditorUserId="4918" LastEditorDisplayName="" LastEditDate="2010-12-20T02:04:47.330" LastActivityDate="2010-12-21T00:21:35.053" Title="Does the following series converge to 2/3?" Tags="&lt;homework&gt;&lt;sequences-and-series&gt;" AnswerCount="3" CommentCount="10" />
  <row Id="14909" PostTypeId="1" CreationDate="2010-12-20T01:15:05.857" Score="1" ViewCount="142" Body="&lt;p&gt;I need to find out $\log_g {-1}$ in $\mathbb{Z}_n$ where $n$ is an odd prime and $g$ is a primitive root mod $n$. How do I do that? Thanks.&lt;/p&gt;&#xA;" OwnerUserId="2868" LastActivityDate="2010-12-20T14:39:16.150" Title="Discrete logarithm to a primitive root base" Tags="&lt;number-theory&gt;&lt;logarithms&gt;" AnswerCount="1" CommentCount="6" />
  <row Id="14910" PostTypeId="2" ParentId="14908" CreationDate="2010-12-20T01:17:03.483" Score="4" ViewCount="" Body="&lt;p&gt;Yes, and your reason is correct.  If you want a simple way to check your work on such things, you could use a computer algebra system like Mathematica, or simply Wolfram Alpha:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.wolframalpha.com/input/?i=sum+%28-1%2F2%29%5En&quot; rel=&quot;nofollow&quot;&gt;http://www.wolframalpha.com/input/?i=sum+%28-1%2F2%29%5En&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2010-12-20T01:22:10.440" LastActivityDate="2010-12-20T01:22:10.440" CommentCount="5" />
  <row Id="14911" PostTypeId="1" CreationDate="2010-12-20T01:32:50.853" Score="3" ViewCount="47" Body="&lt;p&gt;&lt;strong&gt;Question&lt;/strong&gt;:&#xA;We say that a convex lattice polytope $P\subset \mathbb{R}^d$ is a &lt;em&gt;smooth Fano polytope&lt;/em&gt; if:  &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The origin is contained in the interior of $P$  &lt;/li&gt;&#xA;&lt;li&gt;The vertices of every facet of $P$ are a $\mathbb{Z}$-basis of $\mathbb{Z}^d$&#xD;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Now, suppose we have a set $V=\{v_1,\ldots,v_n\}\subset \mathbb{Z}^d$. What is an (preferentially efficient) algorithm for deciding if the convex hull $\text{conv}(V)$ is a smooth Fano polytope?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Motivation&lt;/strong&gt;:&lt;br&gt;&#xA;I've read a paper that gives an algorithm to classify all smooth Fano polytopes given the dimension $d$ as input (&lt;a href=&quot;http://arxiv.org/abs/0704.0049v1&quot; rel=&quot;nofollow&quot;&gt;An algorithm for the classification of smooth Fano polytopes&lt;/a&gt; by Mikkel Øbro), and while trying to implement said algorithm I discovered I don't know how to solve this question.&lt;br&gt;&#xA;For those who are curious, I want to use the algorithm to help me gain some intuition about toric varieties, and use  it to compute known invariants and test conjectures.&lt;/p&gt;&#xA;" OwnerUserId="446" LastActivityDate="2010-12-20T01:32:50.853" Title="How to check if a polytope is a smooth Fano polytope?" Tags="&lt;geometry&gt;&lt;combinatorics&gt;&lt;algorithms&gt;&lt;computational-geometry&gt;" CommentCount="1" FavoriteCount="1" />
  <row Id="14912" PostTypeId="2" ParentId="14904" CreationDate="2010-12-20T01:57:17.767" Score="2" ViewCount="" Body="&lt;p&gt;From some basic facts of systems of linear differential equations, it follows that the particles will spiral like your simulation showed.  Indeed, solutions of the differential equation have terms involving things like $e^{t/\sqrt{2}} \sin(t/\sqrt{2})$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The system of differential equations you have can be made a first order system by introducing new variables $w$ and $z$ with $x&amp;#39; = w$ and $y&amp;#39; = z$.  Then $w&amp;#39; = y$ and $z&amp;#39; = -x$ (here I'm taking $m=1$).  We can then write the system in matrix form as&#xA;$$&#xD;&#xA;\left(\begin{array}{c} w \\ x \\ y \\ z \end{array}\right)&amp;#39; = \left(\begin{array}{cccc} 0 &amp;amp; 0 &amp;amp; 1 &amp;amp; 0 \\ 1 &amp;amp; 0 &amp;amp; 0 &amp;amp; 0 \\ 0 &amp;amp; 0 &amp;amp; 0 &amp;amp; 1 \\ 0 &amp;amp; -1 &amp;amp; 0 &amp;amp; 0 \end{array}\right)\left(\begin{array}{c} w \\ x \\ y \\ z \end{array}\right).&#xD;&#xA;$$&lt;br&gt;&#xA;The type of solutions are then determined by the eigenvalues of the $4 \times 4$ matrix.  If an eigenvalue if $a + ib$, then a solution will be given by $v e^{a+ib} = v e^a + v(\cos b + i \sin b)$ where $v$ is the corresponding eigenvector.  In your system, the eigenvalues are $\frac{1}{\sqrt 2}(\pm 1 \pm i)$, so we get spiral solutions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Its pretty cool that you made a flash simulation and I wish I could give you an answer without invoking ODE theory, but I can't really think of one.&lt;/p&gt;&#xA;" OwnerUserId="9" LastActivityDate="2010-12-20T01:57:17.767" CommentCount="1" />
  <row Id="14913" PostTypeId="2" ParentId="14904" CreationDate="2010-12-20T02:01:57.297" Score="1" ViewCount="" Body="&lt;p&gt;Err, right after I said I coulnd't think of a more elementary way I did.  Suppose a solution is a circle, then we'd have $x^2 + y^2 = const$.  Differentiating both sides gives $2x x&amp;#39; + 2y y&amp;#39; = 0$.  Dividing by 2 and differentiating again gives $x&amp;#39;^2 + xy + y&amp;#39;^2 - xy = 0$.  So $x&amp;#39;^2 + y&amp;#39;^2 = 0$.  This means $x&amp;#39; = y&amp;#39; = 0$.  So the solutions are constant, but only $(x,y) = (0,0)$ is a constant solution to the differential equation.&lt;/p&gt;&#xA;" OwnerUserId="9" LastActivityDate="2010-12-20T02:01:57.297" />
  <row Id="14914" PostTypeId="2" ParentId="14908" CreationDate="2010-12-20T02:06:25.407" Score="0" ViewCount="" Body="&lt;p&gt;You could also have split it up into two sums. &#xA;$$(1+\frac{1}{4}+\frac{1}{16}+...)-(\frac{1}{2}+\frac{1}{8}+...)$$&#xD;&#xA;$$(1+\frac{1}{4}+\frac{1}{16}+...)-\frac{1}{2}(1+\frac{1}{4}+\frac{1}{16}+...)$$&#xD;&#xA;$$\frac{1}{2}(1+\frac{1}{4}+\frac{1}{16}+...)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;After reading Moron's comment, I decided to post the alternative way to get to my final equation. &#xA;$$(1-\frac{1}{2})+(\frac{1}{4}-\frac{1}{8})+...$$&#xD;&#xA;$$\frac{1}{2}(1+\frac{1}{4}+\frac{1}{16}+...)$$&lt;/p&gt;&#xA;" OwnerUserId="4728" LastEditorUserId="4728" LastEditorDisplayName="" LastEditDate="2010-12-20T02:44:06.683" LastActivityDate="2010-12-20T02:44:06.683" CommentCount="2" />
  <row Id="14915" PostTypeId="1" AcceptedAnswerId="14923" CreationDate="2010-12-20T02:12:52.180" Score="10" ViewCount="228" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was reading up on class field theory, and I have a question. On wiki (http://en.wikipedia.org/wiki/Artin_reciprocity), one formulation is that there's &lt;em&gt;some&lt;/em&gt; modulus for which $I^c_K/i(K_{c,1})Nm_{L/K}(I^c_L)$ is isomorphic to $Gal(L/K)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another formulation on the same page is:&#xA;$C_K/N_{L/K}(C_L)$ is isomorphic to $Gal(L/K)$.&#xA;(where $C_{number\,field}$ is the idele class group of that number field).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How does one relate these two formulations? Is it true that for some modulus, $c$, $I^c_K$ is $C_K$? I don't really see how this fits into one picture.&lt;/p&gt;&#xA;" OwnerUserId="4953" LastEditorUserId="8575" LastEditorDisplayName="" LastEditDate="2011-05-25T12:37:08.093" LastActivityDate="2011-05-25T12:37:08.093" Title="Different formulations of Class Field Theory" Tags="&lt;algebraic-number-theory&gt;" AnswerCount="2" FavoriteCount="3" />
  <row Id="14916" PostTypeId="2" ParentId="14878" CreationDate="2010-12-20T02:36:01.483" Score="9" ViewCount="" Body="&lt;p&gt;This theorem was proved by Max Dehn in early 20th century.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose there is some square whose side $s$ is not rational. Pick some $\mathbb{Q}$-homomorphism $h$ such that $h(1) = 0$ and $h(s) = 1$. Define the &lt;i&gt;area&lt;/i&gt; $\sigma(R)$ of a rectangle $R$ with sides $a,b$ by $h(a)h(b)$. We show below that if a rectangle $R$ is partitioned into rectangles $R_i$ then $$\sigma(R) = \sum \sigma(R_i).$$ Take $R$ to be the unit square, and $R_i$ the squares into which it is partitioned. If $R_i$ has side $s_i$ then we get $$0 = h(1)^2 = \sigma(R) = \sum \sigma(R_i) = \sum h(s_i)^2 \geq h(s)^2 = 1.$$ This contradiction shows that all sides must be rational.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It remains to show that $\sigma$ is additive. Extend all sides in the partition of $R$ to get a grid $G_j$. From additivity of $h$ we easily get $$\sigma(R) = \sum\sigma(G_j) = \sum\sigma(R_i).$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using very similar methods, you can show the following slightly stronger theorem: Suppose that a rectangle is partitioned into squares. Then all edge lengths are rationally related.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Indeed, if the outside rectangle had sides $a,b$ not rationally dependent then we could pick a homomorphism $h$ such that $h(a)=1,h(b)=-1$, and then get that $-1$ is a sum of squares. Wlog, $a,b$ are both rational, and the previous proof completes the argument.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-12-20T02:36:01.483" CommentCount="3" />
  <row Id="14917" PostTypeId="1" CreationDate="2010-12-20T02:38:07.373" Score="4" ViewCount="90" Body="&lt;p&gt;One of the perks of my research topic (Latin squares) is that it's somewhat possible to explain what I do to those with a fairly minimal mathematics background.  I'll pose a typical question that comes up in one of my main research topics as a problem here  -- I'm hoping that this will whet the appetite of someone considering doing postgrad mathematics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A &lt;a href=&quot;http://en.wikipedia.org/wiki/Latin_square&quot; rel=&quot;nofollow&quot;&gt;Latin square&lt;/a&gt; is an $n \times n$ matrix with symbols from $[n]:=\{1,2,\ldots,n\}$ such that each row and each column contains every symbol.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $L=(l_{ij})$ be a Latin square.  For any row $i$ and column $j$, We say $(i,j,l_{ij})$ is a &lt;em&gt;entry&lt;/em&gt; of $L$.  The set of entries $O=\{(i,j,l_{ij}):i \in [n], j \in [n]\}$ is called the &lt;em&gt;orthogonal array&lt;/em&gt; of $L$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An &lt;em&gt;automorphism&lt;/em&gt; of $L$ is a permutation $\alpha$ of $\{1,2,\ldots,n\}$ such that $O$ is preserved under $(i,j,l_{ij}) \mapsto (\alpha(i),\alpha(j),\alpha(l_{ij}))$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This definition might seem a little complicated, but we can view this property reasonably easily.  Here's an example for order 15:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\left(\begin{array}{cccccccccccc|ccc}&#xD;&#xA;13 &amp;amp; 8 &amp;amp; 14 &amp;amp; 9 &amp;amp; 15 &amp;amp; \mathbf{10} &amp;amp; 4 &amp;amp; 11 &amp;amp; 5 &amp;amp; 12 &amp;amp; 6 &amp;amp; 1 &amp;amp; 7 &amp;amp; 3 &amp;amp; 2 \\&#xD;&#xA;2 &amp;amp; 14 &amp;amp; 9 &amp;amp; 15 &amp;amp; 10 &amp;amp; 13 &amp;amp; \mathbf{11} &amp;amp; 5 &amp;amp; 12 &amp;amp; 6 &amp;amp; 1 &amp;amp; 7 &amp;amp; 3 &amp;amp; 8 &amp;amp; 4 \\&#xD;&#xA;8 &amp;amp; 3 &amp;amp; 15 &amp;amp; 10 &amp;amp; 13 &amp;amp; 11 &amp;amp; 14 &amp;amp; \mathbf{12} &amp;amp; 6 &amp;amp; 1 &amp;amp; 7 &amp;amp; 2 &amp;amp; 5 &amp;amp; 4 &amp;amp; 9 \\&#xD;&#xA;3 &amp;amp; 9 &amp;amp; 4 &amp;amp; 13 &amp;amp; 11 &amp;amp; 14 &amp;amp; 12 &amp;amp; 15 &amp;amp; \mathbf{1} &amp;amp; 7 &amp;amp; 2 &amp;amp; 8 &amp;amp; 10 &amp;amp; 6 &amp;amp; 5 \\&#xD;&#xA;9 &amp;amp; 4 &amp;amp; 10 &amp;amp; 5 &amp;amp; 14 &amp;amp; 12 &amp;amp; 15 &amp;amp; 1 &amp;amp; 13 &amp;amp; \mathbf{2} &amp;amp; 8 &amp;amp; 3 &amp;amp; 6 &amp;amp; 11 &amp;amp; 7 \\&#xD;&#xA;4 &amp;amp; 10 &amp;amp; 5 &amp;amp; 11 &amp;amp; 6 &amp;amp; 15 &amp;amp; 1 &amp;amp; 13 &amp;amp; 2 &amp;amp; 14 &amp;amp; \mathbf{3} &amp;amp; 9 &amp;amp; 8 &amp;amp; 7 &amp;amp; 12 \\&#xD;&#xA;10 &amp;amp; 5 &amp;amp; 11 &amp;amp; 6 &amp;amp; 12 &amp;amp; 7 &amp;amp; 13 &amp;amp; 2 &amp;amp; 14 &amp;amp; 3 &amp;amp; 15 &amp;amp; \mathbf{4} &amp;amp; 1 &amp;amp; 9 &amp;amp; 8 \\&#xD;&#xA;\mathbf{5} &amp;amp; 11 &amp;amp; 6 &amp;amp; 12 &amp;amp; 7 &amp;amp; 1 &amp;amp; 8 &amp;amp; 14 &amp;amp; 3 &amp;amp; 15 &amp;amp; 4 &amp;amp; 13 &amp;amp; 9 &amp;amp; 2 &amp;amp; 10 \\&#xD;&#xA;14 &amp;amp; \mathbf{6} &amp;amp; 12 &amp;amp; 7 &amp;amp; 1 &amp;amp; 8 &amp;amp; 2 &amp;amp; 9 &amp;amp; 15 &amp;amp; 4 &amp;amp; 13 &amp;amp; 5 &amp;amp; 11 &amp;amp; 10 &amp;amp; 3 \\&#xD;&#xA;6 &amp;amp; 15 &amp;amp; \mathbf{7} &amp;amp; 1 &amp;amp; 8 &amp;amp; 2 &amp;amp; 9 &amp;amp; 3 &amp;amp; 10 &amp;amp; 13 &amp;amp; 5 &amp;amp; 14 &amp;amp; 4 &amp;amp; 12 &amp;amp; 11 \\&#xD;&#xA;15 &amp;amp; 7 &amp;amp; 13 &amp;amp; \mathbf{8} &amp;amp; 2 &amp;amp; 9 &amp;amp; 3 &amp;amp; 10 &amp;amp; 4 &amp;amp; 11 &amp;amp; 14 &amp;amp; 6 &amp;amp; 12 &amp;amp; 5 &amp;amp; 1 \\&#xD;&#xA;7 &amp;amp; 13 &amp;amp; 8 &amp;amp; 14 &amp;amp; \mathbf{9} &amp;amp; 3 &amp;amp; 10 &amp;amp; 4 &amp;amp; 11 &amp;amp; 5 &amp;amp; 12 &amp;amp; 15 &amp;amp; 2 &amp;amp; 1 &amp;amp; 6 \\&#xD;&#xA;\hline&#xD;&#xA;12 &amp;amp; 2 &amp;amp; 1 &amp;amp; 3 &amp;amp; 5 &amp;amp; 4 &amp;amp; 6 &amp;amp; 8 &amp;amp; 7 &amp;amp; 9 &amp;amp; 11 &amp;amp; 10 &amp;amp; 15 &amp;amp; 14 &amp;amp; 13 \\&#xD;&#xA;11 &amp;amp; 1 &amp;amp; 3 &amp;amp; 2 &amp;amp; 4 &amp;amp; 6 &amp;amp; 5 &amp;amp; 7 &amp;amp; 9 &amp;amp; 8 &amp;amp; 10 &amp;amp; 12 &amp;amp; 14 &amp;amp; 13 &amp;amp; 15 \\&#xD;&#xA;1 &amp;amp; 12 &amp;amp; 2 &amp;amp; 4 &amp;amp; 3 &amp;amp; 5 &amp;amp; 7 &amp;amp; 6 &amp;amp; 8 &amp;amp; 10 &amp;amp; 9 &amp;amp; 11 &amp;amp; 13 &amp;amp; 15 &amp;amp; 14 \\&#xD;&#xA;\end{array}\right)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which admits the automorphism $\alpha=(1,2,3,4,5,6,7,8,9,10,11,12)(13,14,15)$.  The horizontal and vertical lines indicate where the cycles &quot;break&quot; in $\alpha$.  I highlight an orbit of an entry [i.e. starting with the entry (1,6,10), we apply $\alpha$ repeatedly] -- it traces out a broken diagonal of the upper-left block, and the symbol &quot;increases by one, wrapping around after 12&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, here's a sample question from my research:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Question&lt;/strong&gt;: Let $\alpha=(1,2,3,4,5,6)(7,8,9,10,11,12)(13,14)(15,16)$.  Is $\alpha$ an automorphism of some Latin square of order $16$?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is a question that has come up in my research, and the more general question &quot;Given $\alpha$, is $\alpha$ an automorphism of some Latin square of order $n$?&quot; is one of my main research topics.&lt;/p&gt;&#xA;" OwnerUserId="139" LastActivityDate="2010-12-20T02:38:07.373" Title="Does there exist a Latin square of order 16 that admits a specific automorphism?" Tags="&lt;combinatorics&gt;&lt;latin-square&gt;" FavoriteCount="1" />
  <row Id="14918" PostTypeId="2" ParentId="14915" CreationDate="2010-12-20T02:39:19.743" Score="4" ViewCount="" Body="&lt;p&gt;The generalized ideal and (normal) idele class groups (by the generalized ideal class group I really mean ideals prime to $\mathfrak{c}$ for $\mathfrak{c}$ an &quot;admissible cycle,&quot; which more or less means divisible by enough primes such that anything congruent to 1 mod it is a local norm) are isomorphic. The short answer for why this is is that given an element in the idele class group, you can get an element of the (generalized) ideal class group since an idele always gives you a fractional ideal. By choosing a suitable representative, you can assume that the ideal you get is actually prime to $\mathfrak{c}$. It is then necessary (and nontrivial) to prove that the corresponding norm groups are mapped to each other, but this is done in Lang's &quot;Algebraic number theory.&quot; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are (at least) two approaches to class field theory:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Define the Artin map globally, on the generalized &lt;em&gt;ideal&lt;/em&gt; class group. This has the benefit that you only have to define it on ideals, so just on primes which are unramified (in which case you just use the Frobenius). In this case, it is a lot of work to show that it actually factors not just through the norm group, but also through a group of principal ideals.&lt;/li&gt;&#xA;&lt;li&gt;Define the Artin map on ideles by piecing together the local Artin maps.  This then requires some construction of the local Artin maps (which can be done using more-or-less pure group cohomology theory and a little bit of analysis of local fields). One has to check again that it is trivial on principal ideles, which is very nontrivial -- since, again the Artin map was defined by piecing together local things.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;So, in a sense, 1 is the one which gives you the ideal form of Artin reciprocity, and 2 gives you the idele one.&lt;/p&gt;&#xA;" OwnerUserId="536" LastEditorUserId="536" LastEditorDisplayName="" LastEditDate="2010-12-20T03:00:15.693" LastActivityDate="2010-12-20T03:00:15.693" CommentCount="5" />
  <row Id="14919" PostTypeId="1" AcceptedAnswerId="14944" CreationDate="2010-12-20T02:44:52.707" Score="5" ViewCount="141" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;How many stripes can you paint on a 2-group of fixed size?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A group of order 2&lt;sup&gt;a&lt;/sup&gt; p&lt;sup&gt;b&lt;/sup&gt; is solvable, by Burnside's theorem, so its chief factors are either abelian 2-groups or abelian p-groups.  Such a group is a &lt;strong&gt;zebra group&lt;/strong&gt; if its chief factors in any chief series alternate between the 2 and the p.  In other words, if you take a chain of normal subgroups 1 = N&lt;sub&gt;1&lt;/sub&gt; ⊲ N&lt;sub&gt;2&lt;/sub&gt; ⊲ &amp;hellip; ⊲ N&lt;sub&gt;n&lt;/sub&gt; = G of maximal length, then the quotient groups N&lt;sub&gt;i+1&lt;/sub&gt;/N&lt;sub&gt;i&lt;/sub&gt; alternate between being abelian 2-groups (the stripes) and abelian p-groups (the background).&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If we fix a, say a=8, then how many stripes can a zebra group have?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Obviously no more than 8 2-stripes, but with a little work one can see that it can have no more than 4 2-stripes.  Unfortunately, I'm having trouble getting even 3 stripes.&lt;/p&gt;&#xA;" OwnerUserId="583" LastActivityDate="2010-12-20T16:02:52.870" Title="Zebra groups and counting stripes" Tags="&lt;group-theory&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" />
  <row Id="14920" PostTypeId="2" ParentId="14878" CreationDate="2010-12-20T03:14:52.960" Score="5" ViewCount="" Body="&lt;p&gt;I believe there is also a proof by associating the tiling with an electric network!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a section in the book &quot;Modern Graph Theory&quot; by B&amp;eacute;la Bollob&amp;aacute;s.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is titled &quot;Squaring the Square&quot; and is numbered section II.2. See this: &lt;a href=&quot;http://books.google.com/books?id=SbZKSZ-1qrwC&amp;amp;pg=PA46&quot; rel=&quot;nofollow&quot;&gt;http://books.google.com/books?id=SbZKSZ-1qrwC&amp;amp;pg=PA46&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A snapshot of the page which talks about Dehn's theorem and it's proof (also found here: &lt;a href=&quot;http://books.google.com/books?id=SbZKSZ-1qrwC&amp;amp;pg=PA48&quot; rel=&quot;nofollow&quot;&gt;http://books.google.com/books?id=SbZKSZ-1qrwC&amp;amp;pg=PA48&lt;/a&gt;):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;br/&gt;&#xA;&lt;br/&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/vz7kV.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note, even though the section starts off talking about the tiling squares being of distinct lengths, the proof of Dehn's theorem (and the problem you seek) is essentially there.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-12-20T03:24:09.157" LastActivityDate="2010-12-20T03:24:09.157" CommentCount="0" />
  <row Id="14921" PostTypeId="1" AcceptedAnswerId="14922" CreationDate="2010-12-20T03:20:36.047" Score="19" ViewCount="424" Body="&lt;p&gt;It is well-known that there is only one &quot;kind&quot; of line, and that there are three &quot;kinds&quot; of quadratic curves (the nature of which depends on the sign of a so-called &quot;discriminant&quot;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is noteworthy that many of the named cubic curves look rather similar: the &lt;a href=&quot;http://mathworld.wolfram.com/FoliumofDescartes.html&quot; rel=&quot;nofollow&quot;&gt;folium of Descartes&lt;/a&gt;, the &lt;a href=&quot;http://mathworld.wolfram.com/MaclaurinTrisectrix.html&quot; rel=&quot;nofollow&quot;&gt;trisectrix of Maclaurin&lt;/a&gt;, the &lt;a href=&quot;http://mathworld.wolfram.com/RightStrophoid.html&quot; rel=&quot;nofollow&quot;&gt;(right) strophoid&lt;/a&gt;, and the &lt;a href=&quot;http://mathworld.wolfram.com/TschirnhausenCubic.html&quot; rel=&quot;nofollow&quot;&gt;Tschirnhausen cubic&lt;/a&gt; look very similar in form; the &lt;a href=&quot;http://mathworld.wolfram.com/SemicubicalParabola.html&quot; rel=&quot;nofollow&quot;&gt;semicubical parabola&lt;/a&gt; and the &lt;a href=&quot;http://mathworld.wolfram.com/CissoidofDiocles.html&quot; rel=&quot;nofollow&quot;&gt;cissoid of Diocles&lt;/a&gt; resemble each other as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have deliberately placed the word &quot;kind&quot; in quotes since there does not seem to me an intuitive way of defining the term, so an answer to my question might have to define &quot;kind&quot; rigorously in the context of cubic curves. (An algebraic invariant, for instance... it is a pity that there does not seem to be an analogue of &quot;eccentricity&quot; for cubics!)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In &lt;a href=&quot;http://mathworld.wolfram.com/CubicCurve.html&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;, it is noted that Newton classified cubics into 72 &quot;kinds&quot;, and Plücker after him described 219 &quot;kinds&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, how does one algebraically distinguish one cubic curve from another, and with a rigorous definition of &quot;kind&quot;, how many cubics are there?&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2011-01-01T00:52:56.590" Title="How many cubic curves are there?" Tags="&lt;algebraic-geometry&gt;&lt;plane-curves&gt;" AnswerCount="2" CommentCount="7" FavoriteCount="2" />
  <row Id="14922" PostTypeId="2" ParentId="14921" CreationDate="2010-12-20T03:38:29.743" Score="16" ViewCount="" Body="&lt;p&gt;The most standard notion of &quot;kind&quot; is that of isomorphism in algebraic geometry, that is, there are polynomial maps from one curve to the other which, when composed, are the identity on the curves (they don't have to be the identity in other places).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For real, affine plane curves, this notion reproduces the fact that there is a unique line and exactly three smooth conics (over the complex numbers, ellipses and hyperbolas turn out to be the same, and in the projective plane, all are the same).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, regardless of ground field (well, regardless of working over the real or complex numbers), there are infinitely many different cubics up to isomorphism, determined by the &lt;a href=&quot;http://en.wikipedia.org/wiki/J-invariant#Algebraic_definition&quot; rel=&quot;nofollow&quot;&gt;j-invariant&lt;/a&gt;, one for every element of the field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A conclusion you can draw from this is that there really isn't a good algebraic way to say that there are only finitely many types of plane cubic, unless you want to lump all smooth cubics together (which you clearly don't, as you consider parabolas and hyperbolas and ellipses to be distinct).  I don't know what method Newton and Plücker used to classify them, but isomorphism is the standard method in modern algebraic geometry.&lt;/p&gt;&#xA;" OwnerUserId="176" LastActivityDate="2010-12-20T03:38:29.743" CommentCount="2" />
  <row Id="14923" PostTypeId="2" ParentId="14915" CreationDate="2010-12-20T04:26:18.570" Score="11" ViewCount="" Body="&lt;p&gt;Dear Nicole, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you read the initial section or two of the chapter by Tate in Cassels and Frolich, he gives a nice explanation of how to pass from the classical formulation in terms of generalized ideal class groups w.r.t a modulus, and the more modern formulation in terms of idele class groups.  As Tate explains,&#xA;the two formulations are indeed equivalent, but it is not quite as simple as saying that $I^c_K = C_K$.   Here is a sketch of the equivalence (of course it is the same is in Akhil Mathew's answer, just slightly more detailed):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $I_K^c$ has already been taken to denote the ideals prime to $c$ (at least, this is how I interpret your notation), let me use $J_K$ to denote the ideles for&#xA;$K$.  Then we can consider the subgroup $J_K^c$ of the ideles whose entries are all $1$ at any finite place dividing $c$, and at any infinite place.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then there is a natural surjection $J_K^c \to I_K^c$ given by sending any element&#xA;$(a_{\wp})$ of the former to the ideal $\prod_{\wp} \wp^{v_{\wp}(a_{\wp})}$&#xD;&#xA;of the  latter (where the product is over finite places, i.e. prime ideals,&#xA;$\wp$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now one can show that $K^{\times} J_K^c$ is dense in $J_K$, so&#xA;the image of $J_K^c$ is dense in $C_K$.  Since $N_{L/K}(C_L)$ is open in $C_K$,&#xA;we see that $J_K^c$ surjects onto $C_K/N_{L/K}(C_L)$. &#xA;Now one checks that this map factors through the surjection $J_K^c \to I_K^c$&#xD;&#xA;described in the preceding paragraph, and in fact induces an isomorphism&#xA;$I_K^c/i(K_{c,1}) N_{L/K}(I_L^c) \buildrel \sim \over \longrightarrow&#xD;&#xA;C_K/N_{L/K}(C_L)$, as required.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In practice, suppose you want to compute the Artin map on an element of $J_K$:&#xA;the algorithm is you first multiply by a principal idele so that the resulting element is in $J_K^c$ times $N_{L/K}(C_L)$.  (You may not know exactly what this&#xA;group is, but its not hard to at least identify an open subgroup of it: for example, at any complex infinite place $v$ the norm map is surjective, at any real place $v$ the image of the norm map at least contains the positive reals, and at any finite place $\wp$ the image of the norm map will contain elements which are congruent to $1$ modulo the power of $\wp$ dividing the relevant modulus $c$.)&#xA;Now the Artin map on $J_K^c$ factors through the surjection $J_K^c \to I_K^c$,&#xA;and is computed on the target using Frobenius elements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Indeed, this was the argument via which local class field theory was originally proved; one took a local extension, embedded it into a global context (so that the original local situation was realized as $L_{\wp}/K_{\wp}$ for some abelian&#xA;extension of number fields $L/K$), and then defined the Artin map via the above computation (which means concretely that one passes from the possibly ramified&#xA;situation at $\wp$ to a consideration just at the unramified primes, where everything is easily understood just in terms of ideals and Frobenius elements).&#xA;Of course, one then had to check that the resulting local Artin map was well-defined independent of the choice of &quot;global context&quot;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nowadays, one can define the local Artin maps at &lt;em&gt;all&lt;/em&gt; places (unramified or ramified) first.  However, in generalizations to the non-abelian situation (i.e. local and global Langlands) one generally uses the old-fashioned technique of proving certain global results first, and then establishing the precise local results by passing to a well-chosen global context and reducing to a calculation at unramified primes. (This is a bit of an oversimplification, but I think it is correct in spirit.)   So (if one has an eventual aim of understanding modern algebraic number theory and the Langlands program) it is well worth understanding the passage between the idelic and ideal-theoretic view-points on class field theory, and practicing how to use the algorithm described above.&lt;/p&gt;&#xA;" OwnerUserId="221" LastEditorUserId="221" LastEditorDisplayName="" LastEditDate="2010-12-20T07:57:07.620" LastActivityDate="2010-12-20T07:57:07.620" />
  <row Id="14924" PostTypeId="2" ParentId="14878" CreationDate="2010-12-20T04:48:09.460" Score="5" ViewCount="" Body="&lt;p&gt;Since your professor's hint was to use linear algebra, the intended proof is probably that of Hadwiger - employing a Hamel basis of $\:\mathbb R\:$ over $\:\mathbb Q\:$ to construct additive &quot;area&quot; functions in order to deduce a contradiction (as in Yuval's answer). Related results were proved by Dehn circa 1900 by different complicated methods. Although Hamel's work on Hamel bases occurred shortly later in 1905, it was not until much later in the 1950's that Hadwiger and his students noticed that Hamel bases could be employed to greatly simplify Dehn's proofs. For a nice exposition see Freiling; Rinne: &lt;a href=&quot;http://www.mathjournals.org/mrl/1994-001-005/1994-001-005-003.pdf&quot; rel=&quot;nofollow&quot;&gt;Tiling a square with similar rectangles&lt;/a&gt;. It's worth remarking that use of the axiom of choice in the construction of a Hamel basis can easily be eliminated for applications of this type, e.g. see the discussion in section 2 of Feshchenko et. al.: &lt;a href=&quot;http://arxiv.org/PS_cache/arxiv/pdf/0809/0809.1883v1.pdf&quot; rel=&quot;nofollow&quot;&gt;Dissecting a brick into bars.&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-12-20T14:37:56.760" LastActivityDate="2010-12-20T14:37:56.760" CommentCount="1" />
  <row Id="14925" PostTypeId="1" AcceptedAnswerId="14935" CreationDate="2010-12-20T06:11:00.490" Score="10" ViewCount="351" Body="&lt;p&gt;Is it true that $\frac1{n!} \int_0^\infty x^{n-k} (x-1)^k e^{-x}\,dx \approx e^{-k/n}$ when $k$ and $n$ are large integers with $k \le n$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This quantity is the probability that a random permutation of $n$ elements does not fix any of the first $k$ elements.&lt;/p&gt;&#xA;" OwnerUserId="857" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-12-20T21:30:24.557" LastActivityDate="2011-03-07T03:50:34.430" Title="Probability that a random permutation has no fixed point among the first $k$ elements." Tags="&lt;calculus&gt;&lt;probability&gt;&lt;analysis&gt;&lt;permutations&gt;" AnswerCount="5" CommentCount="2" FavoriteCount="3" />
  <row Id="14926" PostTypeId="1" CreationDate="2010-12-20T06:48:39.667" Score="11" ViewCount="238" Body="&lt;p&gt;What are the formal series $f$ with non-negative integer coefficients and constant term equal to $1$ whose multiplicative inverse $1/f$ has all coefficients, apart from a finite subset, all non-positive?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, assume the series converges in some disk if you want...&lt;/p&gt;&#xA;" OwnerUserId="274" LastEditorUserId="274" LastEditorDisplayName="" LastEditDate="2010-12-20T16:02:48.517" LastActivityDate="2011-05-05T02:56:50.620" Title="Multiplicative inverses of formal series with non-negative coefficients" Tags="&lt;sequences-and-series&gt;&lt;power-series&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="2" />
  <row Id="14927" PostTypeId="2" ParentId="14831" CreationDate="2010-12-20T07:18:57.113" Score="5" ViewCount="" Body="&lt;p&gt;Here is another approach. Let $G$ be a free group on $m$ generators, and let $H$ be a free group on $n$ generators. There are exactly $2^m$ homomorphisms from $G$ to a group of order two, since each generator can be mapped in two ways. Likewise, there are $2^n$ homomorphisms from $H$ to a group of order two.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $G$ and $H$ are isomorphic, then they have the same number of homomorphisms to a group of order two. Therefore $2^m = 2^n$, which implies $m=n$.&lt;/p&gt;&#xA;" OwnerUserId="857" LastActivityDate="2010-12-20T07:18:57.113" />
  <row Id="14928" PostTypeId="2" ParentId="14831" CreationDate="2010-12-20T07:55:01.663" Score="2" ViewCount="" Body="&lt;p&gt;Dave R's answer highlights a more general principle: by the &lt;a href=&quot;http://en.wikipedia.org/wiki/Yoneda_lemma&quot; rel=&quot;nofollow&quot;&gt;Yoneda lemma&lt;/a&gt;, an object $X$ in a category is determined up to isomorphism by the behavior of the functor $F_X = \text{Hom}(X, -)$, so to show that two objects $X, Y$ are &lt;em&gt;not&lt;/em&gt; isomorphic it suffices to show that the corresponding functors $F_X, F_Y$ are not isomorphic.  (In particular it suffices to show the existence of an object $Z$ such that $\text{Hom}(X, Z)$ has a different size from $\text{Hom}(Y, Z)$.)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The free groups $F_n$ represent some special functors $\text{Grp} \to \text{Set}$: namely, $\text{Hom}(F_n, G)$ is precisely the set $G^n$ of $n$-tuples of elements of $G$.  This is a manifestation of the adjunction between the free group functor $\text{Set} \to \text{Grp}$ and the underlying set functor $\text{Grp} \to \text{Set}$, and by setting $G$ to any nontrivial finite group (Dave R's answer uses $\mathbb{Z}/2\mathbb{Z}$) it is not hard to see that these functors are all nonisomorphic. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;More generally, I believe one can say the following.  Let $F : D \to C$ and $G : C \to D$ be an adjunction, hence&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\text{Hom}_C(FX, Y) \simeq \text{Hom}_D(X, GY)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and suppose that the functor $G$ is essentially surjective.  (In this case $G$ is the forgetful functor $\text{Grp} \to \text{Set}$ and $F$ is the free group functor $\text{Set} \to \text{Grp}$; then it is a classical result that $G$ is essentially surjective.)  Then by another application of the Yoneda lemma, I believe the functors $\text{Hom}_C(FX_1, -)$ and $\text{Hom}_C(FX_2, -)$ are isomorphic if and only if the objects $X_1, X_2$ are isomorphic in $D$.  Can anyone confirm this?&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-12-20T08:03:16.980" LastActivityDate="2010-12-20T08:03:16.980" />
  <row Id="14929" PostTypeId="1" CreationDate="2010-12-20T07:57:17.450" Score="0" ViewCount="42" Body="&lt;p&gt;If $A_n$ is minuend,$B_n$ is subtrahend and $C_{n-1}$ are the three inputs and $D_n$ (difference), $C_n$ (borrow) are the two outputs of a full substractor,we know the $C_n$ is given by:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$C_n = \overline{A_n}B_n + \left(\overline{A_n \bigoplus B_n}\right)C_{n-1}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can we reduce it  to $C_n = \overline{A_n} B_n + \overline{A_n}C_{n-1} + B_nC_{n-1}$ ?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I have reduced upto $C_n = \overline{A_n}B_n + \left(A_n B_n + \overline{A_n} \overline{B_n}\right)C_{n-1}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also tried using Karnaugh map for further reduction but not quite there.Any pointer in the correct direction would be appreciated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt; Just now I solved it! Earlier it was a mistake in my Karnaugh Map,I also got the same answer using Boolean algebra substitution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks,&lt;/p&gt;&#xA;" OwnerUserId="2109" LastEditorUserId="2109" LastEditorDisplayName="" LastEditDate="2010-12-20T08:29:08.260" LastActivityDate="2010-12-20T08:29:08.260" Title="Boolean Algebra :: Full substactor" Tags="&lt;boolean-algebra&gt;" CommentCount="1" />
  <row Id="14930" PostTypeId="1" AcceptedAnswerId="14959" CreationDate="2010-12-20T08:48:43.307" Score="0" ViewCount="175" Body="&lt;p&gt;And how to express $\{ x \} = x - \lfloor x \rfloor$ as function of $sin(x)$ and $sign(x)$?&lt;/p&gt;&#xA;" OwnerUserId="1041" LastEditorUserId="367" LastEditorDisplayName="" LastEditDate="2010-12-20T15:06:17.143" LastActivityDate="2010-12-20T18:58:39.300" Title="How to express the whole part $\lfloor x \rfloor$ as analytical function or Taylor/Fourier series?" Tags="&lt;fourier-series&gt;&lt;taylor-expansion&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="14931" PostTypeId="2" ParentId="14876" CreationDate="2010-12-20T09:10:45.233" Score="2" ViewCount="" Body="&lt;p&gt;The most useful mathematical areas for a programmer are&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;trigonometry and linear algebra, especially to draw and recognise shapes, ray tracing, animations and so on,&lt;/li&gt;&#xA;&lt;li&gt;combinatorics, particularly to cleverly enumerate sets, estimate running times of simple loops and recursions etc.,&lt;/li&gt;&#xA;&lt;li&gt;some elementary group theory and particularly permutation groups, for algorithms that involve tracing out graphs (particularly trees) and also for clever enumeration, as above.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;There are lots of introductory books on each of the above subjects, although maybe other people can comment on which ones are best suited for non-mathematicians. Note that the three areas are fairly independent, at least at the beginning, and you can learn them in any order or in parallel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: I forgot the possibly best way to get started: once you know the definition and some basic facts about groups, you should read the book &lt;a href=&quot;http://books.google.co.uk/books?id=kC5kdUirHHoC&amp;amp;lpg=PP1&amp;amp;ots=2m5l2_shyb&amp;amp;dq=indra%27s%20pearls&amp;amp;pg=PP1#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;Indra's Pearls&lt;/a&gt;. It is a very beautiful book that explains how to draw certain types of fractals with the computer. It introduces all the relevant mathematics pretty much from zero and it provides the actual programs in pseudo-code, ready for you to implement in your favourite language. The mathematics and the algorithms it introduces are very relevant to other situations, but since you are particularly interested in graphics, you will hardly find a better way to get started! There is even a &lt;a href=&quot;http://www.josleys.com/galleries.php?catid=7&quot; rel=&quot;nofollow&quot;&gt;computer graphics artist&lt;/a&gt; who bases some of his work on that book.&lt;/p&gt;&#xA;" OwnerUserId="3212" LastEditorUserId="3212" LastEditorDisplayName="" LastEditDate="2010-12-20T14:03:20.613" LastActivityDate="2010-12-20T14:03:20.613" CommentCount="1" />
  <row Id="14932" PostTypeId="1" CreationDate="2010-12-20T09:29:04.987" Score="7" ViewCount="385" Body="&lt;p&gt;Can the number of solutions $xy(x-y-1)=n$ for $x,y,n \in Z$ be unbounded as n varies?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;x,y are integral points on an Elliptic Curve and are easy to find using enumeration of divisors of n (assuming n can be factored).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If yes, will large number of solutions give moderate rank EC?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If one drops $-1$ i.e. $xy(x-y)=n$ the answer is &quot;yes&quot; via multiples of rational point(s) and then multiplying by a cube.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Suppose it is an open question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: I would be interested in this computational experiment: find $n$ that gives a lot of solutions, say $100$ (I can't do it), check which points are linearly independent and this is a lower bound on the rank.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: What I find intriguing is that &lt;strong&gt;all integral points&lt;/strong&gt; in this model come from factorization/divisors only.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strike&gt;&#xA;EDIT: Current record is n=&lt;strong&gt;179071200&lt;/strong&gt; with 22 solutions with positive x,y. Due to Matthew Conroy.&#xA;Current record is n=&lt;strong&gt;391287046550400&lt;/strong&gt; with 26 solutions with positive x,y. Due to  Aaron Meyerowitz&#xA;Current record is n=8659883232000 with 28 solutions with positive x,y. Found by Tapio Rajala.&#xA;&lt;/strike&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Current record is n=2597882099904000 with 36 solutions with positive x,y. Found by Tapio Rajala.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Is it possible some relation in the primes or primes of certain form to produce records? Read an article I didn't quite understand about maximizing the Selmer rank by chosing the primes carefully.&lt;/p&gt;&#xA;" OwnerUserId="4750" LastEditorUserId="4750" LastEditorDisplayName="" LastEditDate="2011-01-01T19:36:20.143" LastActivityDate="2011-01-01T19:36:20.143" CommunityOwnedDate="2010-12-28T06:42:53.157" Title="Can the number of solutions $xy(x-y-1)=n$ for $x,y,n \in Z$ be unbounded as n varies?" Tags="&lt;number-theory&gt;&lt;algebraic-geometry&gt;" CommentCount="18" />
  <row Id="14933" PostTypeId="1" CreationDate="2010-12-20T09:53:38.237" Score="2" ViewCount="110" Body="&lt;p&gt;I need to solve following system of nonlinear equations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are some characteristics of this system:&lt;br&gt;&#xA;It consists of n equation and n variables.&lt;br&gt;&#xA;Every equation is in similar form -&gt; sum of products = constant.&lt;br&gt;&#xA;The lenght of every product is the same (it will be denoted as k).&lt;br&gt;&#xA;The number of elements in sum might be different in each equation.&lt;br&gt;&#xA;In every product in equation $i$ one of elements is $x_{i}$ - this is very important.&lt;br&gt;&#xA;This system is &quot;symmetrical&quot;, it means that if $x_{i} \cdot x_{j} \cdot ...$ is one of elements of equation i, then it is also in equaton j.&lt;br&gt;&#xA;$b_{i} &amp;gt; 0$ - where $b_{i}$ is intercept in equation i.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll write an example of such equation for k=3 and n=6:&lt;br&gt;&#xA;$x_{1} \cdot x_{3} \cdot x_{6} + x_{1} \cdot x_{2} \cdot x_{4} = b_{1}$&lt;br&gt;&#xA;$x_{2} \cdot x_{1} \cdot x_{4} + x_{2} \cdot x_{5} \cdot x_{6} = b_{2}$&lt;br&gt;&#xA;$x_{3} \cdot x_{1} \cdot x_{6} + x_{3} \cdot x_{4} \cdot x_{6} = b_{3}$&lt;br&gt;&#xA;$x_{4} \cdot x_{1} \cdot x_{2} + x_{4} \cdot x_{3} \cdot x_{6} = b_{4}$&lt;br&gt;&#xA;$x_{5} \cdot x_{2} \cdot x_{6} = b_{5}$&lt;br&gt;&#xA;$x_{6} \cdot x_{1} \cdot x_{3} + x_{6} \cdot x_{2} \cdot x_{5} + x_{6} \cdot x_{3} \cdot x_{4} = b_{6}$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is very easy to transform this equation to following form:&lt;br&gt;&#xA;$x_{i} = b_{i} / something$ , $x_{i}$ is only on left-hand side of ith equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we have all equation in such form then the fixed point is solution of it.&lt;br&gt;&#xA;I've experimentally checked that algorithm analogic to Gauss-Seidel is covergent (i've checked ~100 random examples, and in every case it was convergent).&lt;br&gt;&#xA;By analogic to Gauss-Seidel algorithm I mean:&lt;br&gt;&#xA;1) Choose any initial solution $[x_{1}^{0} , ... , x_{n}^{0}]$&lt;br&gt;&#xA;2.1) Calculate value of $x_{1}^{i+1}$ using $[x_{2}^{i} , ... , x_{n}^{i}]$&lt;br&gt;&#xA;2.2) Calculate value of $x_{2}^{i+1}$ using $[x_{1}^{i+1} , ... , x_{n}^{i}]$&lt;br&gt;&#xA;...&lt;br&gt;&#xA;2.n) Calculate value of $x_{n}^{i+1}$ using $[x_{1}^{i+1} , ... , x_{n-1}^{i+1}]$&lt;br&gt;&#xA;3) If solution is good enough stop, otherwise go to 2.1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've tried Banach fixed point theorem, but is hard to say anything about spectral radius.&#xA;Does anyone have a clue how to prove convergence of this algorithm?&lt;/p&gt;&#xA;" OwnerUserId="4811" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-20T10:25:45.673" LastActivityDate="2010-12-20T10:25:45.673" Title="System of nonlinear equations" Tags="&lt;numerical-methods&gt;" CommentCount="6" FavoriteCount="1" />
  <row Id="14934" PostTypeId="2" ParentId="14925" CreationDate="2010-12-20T10:27:22.940" Score="5" ViewCount="" Body="&lt;p&gt;This is not an answer but an observation that the quantity is equal to $\sum_{i=0}^{k} \frac{{(-1)}^i}{n!}{k \choose i}(n-i)!$.&lt;/p&gt;&#xA;" OwnerUserId="4959" LastActivityDate="2010-12-20T10:27:22.940" />
  <row Id="14935" PostTypeId="2" ParentId="14925" CreationDate="2010-12-20T10:27:52.900" Score="6" ViewCount="" Body="&lt;p&gt;Here's a heuristic idea, based on the strong law of large numbers (SLLN). A rigorous proof can be made using the central limit theorem (CLT), as in my new answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First write &#xA;$$&#xD;&#xA;I = \frac{1}{{n!}}\int_0^\infty  {x^{n - k} (x - 1)^k e^{ - x} \,{\rm d}x}  = \int_0^\infty  {\bigg(1 - \frac{1}{x}} \bigg)^k \frac{{x^n e^{ - x} }}{{\Gamma (n + 1)}}\,{\rm d}x.&#xD;&#xA;$$&#xD;&#xA;Now, if $X_1,\ldots,X_{n+1}$ are i.i.d. exponential(1) random variables, then their sum $X_1 + \cdots + X_{n+1}$ has gamma density $x^n e^{-x} / \Gamma(n+1)$, $x &amp;gt; 0$. Hence,&#xA;$$&#xD;&#xA;I = {\rm E} \bigg[ 1 - \frac{1}{{X_1  +  \cdots  + X_{n + 1} }}\bigg]^k = {\rm E}\bigg[1 - \frac{{n + 1}}{{X_1  +  \cdots  + X_{n + 1} }}\frac{1}{{n + 1}}\bigg]^k .&#xD;&#xA;$$&#xD;&#xA;By SLLN, $(n + 1)^{ - 1} \sum\nolimits_{i = 1}^{n + 1} {X_i }$ converges almost surely to  ${\rm E}[X_1 ] = 1$ as $n \to \infty$. So this suggests the approximation&#xA;$$&#xD;&#xA;I \approx \bigg(1 - \frac{1}{{n + 1}}\bigg)^k  = \bigg(1 - \frac{1}{{n + 1}}\bigg)^{n(k/n)}  \approx e^{ - k/n}.&#xD;&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="2810" LastEditorUserId="2810" LastEditorDisplayName="" LastEditDate="2010-12-20T15:44:22.207" LastActivityDate="2010-12-20T15:44:22.207" />
  <row Id="14936" PostTypeId="1" AcceptedAnswerId="14937" CreationDate="2010-12-20T10:33:53.070" Score="0" ViewCount="78" Body="&lt;p&gt;This may have more to do with computing than Mathematics in its application, however this has been giving me a headache for some time and I see no other recourse than to ask...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given a natural integer T &gt; 9 (and probably between 1,000 and 1,000,000,000), and a choice N in {0,1,2,3,4}, provide two algorithms (encoding and decoding):&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;One which take T and N as inputs and outputs R, a number which has no more digits than T. [encoding]&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The other which takes an (encoded) number R and outputs the original (unencoded) T and N. [decoding]&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The complexity of the algorithms should at best allow to be implemented by modern computing. Apart from that, &lt;em&gt;everything and anything&lt;/em&gt; is allowed :).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a side note, do advise me if this is impossible.&lt;/p&gt;&#xA;" OwnerUserId="4958" LastActivityDate="2010-12-20T11:11:19.580" Title="&quot;Encode&quot; a single digit in a multi-digit number in the smallest way possible" Tags="&lt;algorithms&gt;&lt;natural-number&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="14937" PostTypeId="2" ParentId="14936" CreationDate="2010-12-20T10:47:23.080" Score="6" ViewCount="" Body="&lt;p&gt;This is impossible for the same reason a compression algorithm that reduces the size of every file is impossible. Suppose $T$ has $k$ decimal digits. There are $9\times10^{k-1}$ possible values of $T$, and $5$ possible values of $N$, making $45\times10^{k-1}$ possible $(T,N)$ pairs. But since you're only allowed up to $k$ digits in $R$, there are only $10\times10^{k-1}$ possible values of $R$ to encode them in. So no matter what you do, you won't be able to decode $R$ back into $T$ and $N$.&lt;/p&gt;&#xA;" OwnerUserId="856" LastEditorUserId="856" LastEditorDisplayName="" LastEditDate="2010-12-20T11:11:19.580" LastActivityDate="2010-12-20T11:11:19.580" CommentCount="2" />
  <row Id="14938" PostTypeId="2" ParentId="11076" CreationDate="2010-12-20T10:52:40.393" Score="1" ViewCount="" Body="&lt;p&gt;If vectors are thought of as triplets of numbers (a,b,c) then why should these triplets be added by the parallelogram law? It all depends what these triplets are being used to represent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you start with addition of positive numbers on the number line, say 5+3, this can be seen as laying down two line segments of length 5 and 3 along a line starting at the origin. This can be generalized to line segments in a plane by allowing them to point in different directions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you start with two adjoined line segments then you can expand the drawing to a parallelogram and it is due to Euclidean geometry that opposite sides of a parallelogram have the same length and point in the same direction. So the parallelogram gives two different paths to move from the origin to the resulting point which is saying that the paralleogram law is equivalent to the commutative law A+B=B+A.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even in one dimension, the line segments can be given a direction, so that 5+(-3) means putting down a line of length 5 in the +direction, and then turning around and drawing a line of length 3 in the opposite direction to arrive at the number 2. Being commutative you could also start with a line of length 3 in the negative direction from the origin along the real number line and then turn around and draw a line of length 5 in the positive direction again arriving at 2. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the parallelogram law of vector addition is a straightforward extension of addition of ordinary numbers as directed line segments and is compatible with addition of negative numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Long before cartesian coordinates were developed, people would have used geometry including the pythagorean theorem to calculate the addition of distances and velocities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then special relativity was discovered and it was found that the parallelogram law no longer holds for velocity addition, so to answer your question: other laws are indeed sometimes used. The question is: how do we know which laws apply to which aspects of nature? We don't know. We make observations and then make assumptions about how things behave based on those observations. We try to use mathematical structures that accurately describe the observed behaviour, and assume that if our models are accurate enough then any calculations derived from those models will also describe nature accurately.&lt;/p&gt;&#xA;" OwnerUserId="276" LastActivityDate="2010-12-20T10:52:40.393" />
  <row Id="14939" PostTypeId="2" ParentId="14925" CreationDate="2010-12-20T11:05:36.323" Score="3" ViewCount="" Body="&lt;p&gt;Continuing Aravind's observation, write it like&#xA;$$&#xD;&#xA;1 - \frac{k}{n} + \frac{1}{2!} \left( \frac{k}{n} \right)^2 \frac{1-1/k}{1-1/n}&#xD;&#xA;- \frac{1}{3!} \left( \frac{k}{n} \right)^3 \frac{(1-1/k)(1-2/k)}{(1-1/n)(1-2/n)}&#xD;&#xA;$$&#xD;&#xA;$$+ \dots&#xD;&#xA;+ (-1)^k \frac{1}{k!} \left( \frac{k}{n} \right)^k \frac{(1-1/k)(1-2/k)\dots (1-(k-1)/k)}{(1-1/n)(1-2/n) \dots (1-(k-1)/n)}.&#xD;&#xA;$$&#xD;&#xA;If $k$ and $n$ are large, this seems to be close to the $k$th degree Maclaurin polynomial&#xA;for $e^z$ evaluated at $z=-k/n$ (although I don't have an estimate of the error).&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2010-12-20T11:05:36.323" />
  <row Id="14940" PostTypeId="1" AcceptedAnswerId="14943" CreationDate="2010-12-20T12:22:19.020" Score="3" ViewCount="161" Body="&lt;p&gt;I am unable to understand the motivation behind the set theoretic definition of a natural number. The definition given in the book by Goldrei is as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First he defines an inductive set:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A set $y$ is &lt;em&gt;inductive&lt;/em&gt; if $\emptyset \in y$ and $x^{+} \in y$ whenever $x \in y$, where $x^{+} = x \cup \{x\}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So far fine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But then the set of natural number is defined as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The set of natural numbers $\mathbb{N}$ is the intersection of all inductive subsets of any inductive set $y$, i.e.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\mathbb{N} = \cap \{z:z \text{ is an inductive subset of }y\} = \{x: x \in z, \forall \text{ inductive }z \subseteq y\}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am actually lost with this definition. I feel any inductive set is isomorphic to $\mathbb{N}$. What is the reason/motivation for defining $\mathbb{N}$ as the intersection of all the &lt;em&gt;inductive&lt;/em&gt; subsets of an &lt;em&gt;inductive&lt;/em&gt; set $y$? What are we missing if we were to define $\mathbb{N}$ as just an inductive set?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If my statements/questions don't make sense, it is because I am confused. I would appreciate if someone could throw light on this.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastActivityDate="2010-12-20T12:42:09.263" Title="Set theoretic definition of a Natural Number" Tags="&lt;set-theory&gt;&lt;natural-number&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="14941" PostTypeId="2" ParentId="14792" CreationDate="2010-12-20T12:31:36.803" Score="1" ViewCount="" Body="&lt;p&gt;$(x\vee y)\wedge(z\vee(x\wedge y))=(x\wedge y)\vee(y\wedge z)\vee(z\wedge x)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;implies&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x\vee((x\vee y)\wedge(z\vee(x\wedge y)))=x\vee((x\wedge y)\vee(y\wedge z)\vee(z\wedge x)).$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The right side equals $x\vee(y\wedge z)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the left side, we use the condition above again, substituting $x\vee y$ for $y$ and $z\vee(x\wedge y)$ for $z$, and get &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x\vee((x\vee y)\wedge(z\vee(x\wedge y)))=(x\vee y)\wedge((z\vee(x\wedge y))\vee x)=(x\wedge y)\vee(x\wedge z).$&lt;/p&gt;&#xA;" OwnerUserId="1878" LastEditorUserId="1878" LastEditorDisplayName="" LastEditDate="2010-12-20T12:37:45.443" LastActivityDate="2010-12-20T12:37:45.443" />
  <row Id="14942" PostTypeId="1" AcceptedAnswerId="14954" CreationDate="2010-12-20T12:32:17.803" Score="0" ViewCount="184" Body="&lt;p&gt;Hello,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm trying to make one body fly around another, using Coulomb law $F=\frac{q_1 q_2}{r^2}$ and second Newton law $ma=F$. Now I'm doing it this intuitive way:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Move body1 according to current speed and $dt$;&lt;/li&gt;&#xA;&lt;li&gt;Calc new Coulomb force then find acceleration $a=\frac{F}{m}$;&lt;/li&gt;&#xA;&lt;li&gt;Add acceleration to current speed;&lt;/li&gt;&#xA;&lt;li&gt;Repeat.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;This works fine. I know this is very simple Euler method and it's very inaccurate. I'd like to use Runge-Kutta method, but I can't figure out how I should implement it. &lt;a href=&quot;http://en.wikipedia.org/wiki/Runge%96Kutta_methods&quot; rel=&quot;nofollow&quot;&gt;Here&lt;/a&gt; it's described, but&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;What is my $f(x, y)$?&lt;/li&gt;&#xA;&lt;li&gt;What is my $y&amp;#39;$ and $y&amp;#39;&amp;#39;$?&lt;/li&gt;&#xA;&lt;li&gt;Where $dt$ goes?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Thank you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;UPD. Thanks for help! Also I've found this nice little article that was quite helpful. Can't insert direct link, but it's first result if google &quot;Many-Body Gravity Simulation Using Multivariable Numerical Integration&quot;.&lt;/p&gt;&#xA;" OwnerUserId="4961" LastEditorUserId="4961" LastEditorDisplayName="" LastEditDate="2011-02-02T03:26:24.557" LastActivityDate="2011-02-02T03:26:24.557" Title="Runge-Kutta method for Newton law" Tags="&lt;differential-equations&gt;&lt;numerical-methods&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="14943" PostTypeId="2" ParentId="14940" CreationDate="2010-12-20T12:42:09.263" Score="9" ViewCount="" Body="&lt;p&gt;Usually in set theory we write $\omega$ instead of $\mathbb{N}$. A concrete example of an inductive set other than $\omega$ is the ordinal $\omega + \omega$, which is the closure of $\omega \cup \{\omega\}$ under ${}^+$. This set has the structure of two copies of $\omega$, one after the other. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given any set $x$, if you take the closure of $x \cup \{\emptyset\}$ under ${}^+$ then you get an inductive set containing every member of $x$. So it is not true that any inductive set is even isomorphic to $\omega$. For example there are uncountable inductive sets. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you really want is a $\subseteq$-*&lt;em&gt;minimal&lt;/em&gt;* inductive set. There is only one of these, which is $\omega$ (which, as before, you are identifying with $\mathbb{N}$). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most likely reason that Goldrei says &quot;the intersection of all inductive subsets of a fixed inductive set&quot; rather than &quot;the intersection of all inductive sets&quot; is that the latter of those is an intersection of a proper class worth of sets, which he may not have defined formally. But the result is the same: if you take the intersection of all inductive sets, or the intersection of all inductive subsets of an inductive set, either way you get the unique $\subseteq$-minimal inductive set. &lt;/p&gt;&#xA;" OwnerUserId="630" LastActivityDate="2010-12-20T12:42:09.263" CommentCount="1" />
  <row Id="14944" PostTypeId="2" ParentId="14919" CreationDate="2010-12-20T12:45:14.443" Score="2" ViewCount="" Body="&lt;p&gt;I doubt whether 3 stripes is possible with $a=8$, although I have not tried to prove it. The problem is that, if you construct the group from the top downwards, then each new layer needs to be a faithful module for the group you have already, which means that the dimensions grow rapidly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The obvious way to start is a 2 at the top, followed by a 3 and then $2^2$, so we now have $S_4$. The smallest faithful module for $S_4$ over any field has dimension 3, so we can put a $3^3$ under the $S_4$ to get a group $G$ of order $2^33^4$. Now the smallest faithful irreducible module for $G$ over the field of order 2 has dimension 6 (I did that calculation in Magma, but I expect you can do it in GAP), so we get a group of order $2^93^4$ with three stripes, which is a subgroup of ${\rm AGL}(6,2)$, and I would be surprised if you can do better.&lt;/p&gt;&#xA;" OwnerUserId="2820" LastActivityDate="2010-12-20T12:45:14.443" CommentCount="3" />
  <row Id="14945" PostTypeId="1" CreationDate="2010-12-20T12:47:58.700" Score="2" ViewCount="56" Body="&lt;p&gt;Thanks for your attention,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm an undergraduate. I'm reading the book of Dieter Happel, &lt;em&gt;Triangulated categories in the representation theory of finite dimensional algebras&lt;/em&gt;, &#xA;I cannot prove this sentence:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;The maps factoring through an injective object are precisely the null-homotopic maps.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here, the object is in the complex category.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please give a completely proof , or some references which take it!&#xA;Thanks a lot!&lt;/p&gt;&#xA;" OwnerUserId="4962" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-20T18:48:38.533" LastActivityDate="2010-12-20T18:48:38.533" Title="How to prove &quot;The maps factoring through an injective object are precisely the null-homotopic maps&quot;" Tags="&lt;category-theory&gt;&lt;representation-theory&gt;" CommentCount="2" />
  <row Id="14946" PostTypeId="1" AcceptedAnswerId="14950" CreationDate="2010-12-20T12:57:13.097" Score="2" ViewCount="125" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $0 \leq \alpha &amp;lt; \beta \leq 1$. I'm looking for an example of a Lebesgue measurable subset $E$ of $\mathbb{R}$ such that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\liminf_{\delta \rightarrow 0} \frac{m(E \cap (-\delta,\delta))}{2\delta} = \alpha$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\limsup_{\delta \rightarrow 0} \frac{m(E \cap (-\delta,\delta))}{2\delta} = \beta$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $m$ is the Lebesgue measure on $\mathbb{R}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can someone give an example?&#xA;Thank you,&#xA;Malik&lt;/p&gt;&#xA;" OwnerUserId="1197" LastActivityDate="2010-12-20T13:20:44.350" Title="Lebesgue measurable subset of $\mathbb{R}$ with given metric density at zero" Tags="&lt;measure-theory&gt;" AnswerCount="3" CommentCount="4" FavoriteCount="0" />
  <row Id="14947" PostTypeId="1" CreationDate="2010-12-20T13:02:07.923" Score="1" ViewCount="200" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider an ellipsoid of semi-axes a, b, c (possibly prolate, b=c). I am interested in the &quot;shadow&quot; of this solid onto a distant plane, in a given direction d=(k,l,m) orthogonal to that plane. By shadow I mean the projected area onto the plane: each point on the surface of the ellipsoid is translated in the same direction until it intersects with a plane normal to it; the shadow is defined by the envelope of the intersection points.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First question: Is the projected curve an ellipse? (and what is its equation in terms of a,b,c and the direction vector d)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second question: What is the mean area of this shadow when averaged over all orientations of the ellipsoid (or, equivalently, the plane of projection)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm guessing this problem has been solved in the past; any references would be very welcome. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;baptiste&lt;/p&gt;&#xA;" OwnerUserId="4964" LastActivityDate="2010-12-21T09:05:15.497" Title="Average projected area of an ellipsoid" Tags="&lt;analytic-geometry&gt;" AnswerCount="1" CommentCount="6" />
  <row Id="14948" PostTypeId="1" AcceptedAnswerId="14953" CreationDate="2010-12-20T13:09:23.297" Score="2" ViewCount="105" Body="&lt;p&gt;I was asked to prove:&#xA; no group can have a minimal normal subgroup isomorphic to a $\mathrm{Syl}_2(A_7)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think I should find some property that $\mathrm{Syl}_2(A_7)$ has but not a minimal normal subgroup. So then I thought that it a $\mathrm{Syl}_2(A_7)$ always has characteristic subgroup, i.e. exist $G\ char\ \mathrm{Syl}_2(A_7)$, then the image of G under isomorphism will contradict minimality of &quot;group can have a minimal normal subgroup isomorphic to a $\mathrm{Syl}_2(A_7)$&quot;. But I get stuck then about whether it is true that $\mathrm{Syl}_2(A_7)$ always has characteristic subgroup. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe this is not the correct way to prove it. Can anyone help me?&lt;/p&gt;&#xA;" OwnerUserId="4900" LastActivityDate="2010-12-20T13:28:38.417" Title="Sylow Subgroup problem" Tags="&lt;abstract-algebra&gt;&lt;group-theory&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="14949" PostTypeId="2" ParentId="14942" CreationDate="2010-12-20T13:15:13.523" Score="4" ViewCount="" Body="&lt;p&gt;Write $u(t)=(x(t),v(t)$, where $x$ is the position and $v$ is the velocity. Then Newton's law implies $u&amp;#39; = (v,F/m)$. That's your differential equation, and you can apply RK to it.&lt;/p&gt;&#xA;" OwnerUserId="589" LastActivityDate="2010-12-20T13:15:13.523" />
  <row Id="14950" PostTypeId="2" ParentId="14946" CreationDate="2010-12-20T13:16:03.880" Score="2" ViewCount="" Body="&lt;p&gt;Try the duplication across zero of $$\bigcup_{n \geq 1} \left[\frac{1}{(2n)!}-\alpha\left(\frac{1}{(2n)!} - \frac{1}{(2n+1)!}\right),\frac{1}{(2n)!}\right] \cup \left[\frac{1}{(2n+1)!}-\beta\left(\frac{1}{(2n+1)!} - \frac{1}{(2n+2)!}\right),\frac{1}{(2n+1)!}\right].$$&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-12-20T13:16:03.880" CommentCount="1" />
  <row Id="14952" PostTypeId="1" AcceptedAnswerId="14955" CreationDate="2010-12-20T13:25:56.140" Score="4" ViewCount="1036" Body="&lt;p&gt;What is Jacobian matrix?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are its applications? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is its physical and geometrical meaning?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can someone explain with examples?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks! :)&lt;/p&gt;&#xA;" OwnerUserId="705" LastActivityDate="2011-05-03T11:22:20.920" Title="What is Jacobian Matrix?" Tags="&lt;calculus&gt;" AnswerCount="4" CommentCount="3" FavoriteCount="1" />
  <row Id="14953" PostTypeId="2" ParentId="14948" CreationDate="2010-12-20T13:28:38.417" Score="2" ViewCount="" Body="&lt;p&gt;Minimal normal subgroups are direct products of isomorphic simple groups.  If they have order a power of a prime, like 8, then they are abelian of exponent p.  The dihedral group of order 8 is not abelian.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your idea works.  Just take the derived subgroup of the dihedral group; it has order 2 and is normal in any group containing the dihedral subgroup as a normal subgroup.&lt;/p&gt;&#xA;" OwnerUserId="583" LastActivityDate="2010-12-20T13:28:38.417" />
  <row Id="14954" PostTypeId="2" ParentId="14942" CreationDate="2010-12-20T13:39:30.980" Score="4" ViewCount="" Body="&lt;p&gt;I wanted to note that the method you are using is &lt;strong&gt;not&lt;/strong&gt; foward Euler. What you are doing is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$v(t+\delta t) = v(t) + \delta t \cdot a(t) $$&#xD;&#xA;$$x(t+\delta t) = x(t) + \delta t \cdot v(t+\delta t)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is called symplectic Euler method, and has much better convergence properties than the standard Forward Euler method. Forward Euler would be:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$v(t+\delta t) = v(t) + \delta t \cdot a(t) $$&#xD;&#xA;$$x(t+\delta t) = x(t) + \delta t \cdot v(t)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For your answer, the trick is indeed to consider the velocity as a final variable of your system and write what @lhf said:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$u(t) = (x(t), v(t))$$&#xD;&#xA;$$u&amp;#39; = (v, F/m)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have to solve a single system with six equations per points now. But each equation is only a first order ODE.&lt;/p&gt;&#xA;" OwnerUserId="4709" LastActivityDate="2010-12-20T13:39:30.980" />
  <row Id="14955" PostTypeId="2" ParentId="14952" CreationDate="2010-12-20T14:32:26.163" Score="13" ViewCount="" Body="&lt;p&gt;Here is an example. Suppose you have two implicit differentiable functions&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$F(x,y,z,u,v)=0,\qquad G(x,y,z,u,v)=0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and the functions, also differentiable, $u=f(x,y,z)$ and $v=g(x,y,z)$ such&#xA;that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$F(x,y,z,f(x,y,z),g(x,y,z))=0,\qquad G(x,y,z,f(x,y,z),g(x,y,z))=0.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you differentiate $F$ and $G$, you get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{eqnarray*}&#xA;\frac{\partial F}{\partial x}+\frac{\partial F}{\partial u}\frac{\partial u}{&#xA;\partial x}+\frac{\partial F}{\partial v}\frac{\partial v}{\partial x} &amp;amp;=&amp;amp;0\qquad&#xA;\&#xA;\frac{\partial G}{\partial x}+\frac{\partial G}{\partial u}\frac{\partial u}{&#xA;\partial x}+\frac{\partial G}{\partial v}\frac{\partial v}{\partial x} &amp;amp;=&amp;amp;0.&#xA;\end{eqnarray*}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solving this system you obtain&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\partial u}{\partial x}=-\frac{\det \begin{pmatrix}&#xD;&#xA;\frac{\partial F}{\partial x} &amp;amp;  &amp;amp; \frac{\partial F}{\partial v} \\ &#xD;&#xA;&amp;amp;  &amp;amp;  \\ \frac{\partial G}{\partial x} &amp;amp;  &amp;amp; \frac{\partial G}{\partial v}%&#xD;&#xA;\end{pmatrix}}{\det \begin{pmatrix}&#xD;&#xA;\frac{\partial F}{\partial u} &amp;amp;  &amp;amp; \frac{\partial F}{\partial v} \\ &#xD;&#xA;&amp;amp;  &amp;amp;  \\ &#xD;&#xA;\frac{\partial G}{\partial u} &amp;amp;  &amp;amp; \frac{\partial G}{\partial v}%&#xD;&#xA;\end{pmatrix}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and similar for $\dfrac{\partial u}{\partial y}$, $\dfrac{\partial u}{\partial&#xD;&#xA;z}$, $\dfrac{\partial v}{\partial x}$, $\dfrac{\partial v}{\partial y}$, $%&#xD;&#xA;\dfrac{\partial v}{\partial z}$. The compact notation for the denominator is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\partial (F,G)}{\partial (u,v)}=\det &#xD;&#xA;\begin{pmatrix}&#xD;&#xA;\frac{\partial F}{\partial u} &amp;amp;  &amp;amp; \frac{\partial F}{\partial v} \\ &#xD;&#xA;&amp;amp;  &amp;amp;  \\ &#xD;&#xA;\frac{\partial G}{\partial u} &amp;amp;  &amp;amp; \frac{\partial G}{\partial v}&#xD;&#xA;\end{pmatrix}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and similar for the numerator. Then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\dfrac{\partial u}{\partial x}=-\dfrac{\dfrac{\partial (F,G)}{\partial (x,v)}}{%&#xD;&#xA;\dfrac{\partial (F,G)}{\partial (u,v)}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\dfrac{\partial (F,G)}{\partial (x,y)},\dfrac{\partial (F,G)}{\partial&#xD;&#xA;(u,v)}$ are &lt;em&gt;Jacobians&lt;/em&gt; (after the 19th century German mathematician Carl Jacobi).&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The absolute value of the Jacobian of a coordinate system transformation is also used to convert a multiple integral from one  system into another. In $\mathbb{R}^2$ it measures how much the unit area is distorted by the given transformation, and in $\mathbb{R}^3$ this factor measures the unit volume distortion, etc. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Another example: the following coordinate transformation (due to Beukers, Calabi and Kolk)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x=\frac{\sin u}{\cos v}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$y=\frac{\sin v}{\cos u}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;transforms (see &lt;a href=&quot;http://math.stackexchange.com/questions/7807/a-limit-and-a-coordinate-trigonometric-transformation-of-the-interior-points-of-a&quot;&gt;this&lt;/a&gt; question of mine) the square domain $0\lt x\lt 1$ and $0\lt y\lt 1$ into the triangle domain $u,v&amp;gt;0,u+v&amp;lt;\pi /2$ (in &lt;em&gt;&lt;a href=&quot;http://books.google.pt/books?id=CuTpgZaLahkC&amp;amp;printsec=frontcover&amp;amp;dq=proofs+from+the+book&amp;amp;source=bl&amp;amp;ots=swepaSdSxh&amp;amp;sig=j1gB1zxabJE8qxEgbFh1ig8RA3Q&amp;amp;hl=pt-PT&amp;amp;ei=Pnj2TKKcCZOKhQfb67yABg&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=3&amp;amp;ved=0CDIQ6AEwAg#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;Proofs from the BOOK&lt;/a&gt;&lt;/em&gt; by M. Aigner and G. Ziegler).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For this transformation you get (see Proof 2 in &lt;a href=&quot;http://empslocal.ex.ac.uk/people/staff/rjchapma/etc/zeta2.pdf&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; collection of proofs by Robin Chapman) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\dfrac{\partial (x,y)}{\partial (u,v)}=1-x^2y^{2}.$$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Jacobian sign and orientation of closed curves. Assume you have two small closed curves, one around $(x_0,y_0)$ and another around $u_0,v_0$, this one being the image of the first under the mapping $u=f(x,y),v=g(x,y)$. If the sign of $\dfrac{\partial (x,y)}{\partial (u,v)}$ is positive, then both curves will be travelled in the same sense. If the sign is negative, they will have opposite senses. (See &lt;a href=&quot;http://kr.cs.ait.ac.th/~radok/math/mat9/05a.htm&quot; rel=&quot;nofollow&quot;&gt;Oriented Regions and their Orientation&lt;/a&gt;.) &lt;/p&gt;&#xA;" OwnerUserId="752" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2010-12-21T17:59:10.257" LastActivityDate="2010-12-21T17:59:10.257" />
  <row Id="14956" PostTypeId="2" ParentId="14909" CreationDate="2010-12-20T14:39:16.150" Score="1" ViewCount="" Body="&lt;p&gt;Seems you are looking for $x$ such that $g^x = -1$.&lt;br&gt;&#xA;$x = \frac{n-1}{2}$ seems to work because $\phi(n)=n-1$ and g is primitive root and $g^{\phi(n)}=1$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="4750" LastActivityDate="2010-12-20T14:39:16.150" />
  <row Id="14957" PostTypeId="1" CreationDate="2010-12-20T14:46:26.273" Score="4" ViewCount="74" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Is there a commutative integral domain R in which:&lt;/p&gt;&#xA;  &#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;every nonzero prime ideal Q is maximal, and&lt;/li&gt;&#xA;  &lt;li&gt;for every prime power q congruent to 3 mod 8, there is a maximal ideal Q of R with R/Q a field of size q?&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am looking for a ring where &quot;q = 3 mod 8&quot; describes finite fields, rather than just finite local rings like in the ring &lt;strong&gt;Z&lt;/strong&gt;.  The only examples I can think of that satisfy the first condition have a finite number of residue fields of each characteristic.  The only examples I can think of that satisfy the second condition have a ton of non-maximal non-proper prime ideals with all sorts of bizarre fields associated to them.&lt;/p&gt;&#xA;" OwnerUserId="583" LastActivityDate="2010-12-20T14:46:26.273" Title="Is there an integral domain with a lot of residue fields of the same characteristic?" Tags="&lt;abstract-algebra&gt;&lt;algebraic-number-theory&gt;" CommentCount="4" />
  <row Id="14958" PostTypeId="1" AcceptedAnswerId="14964" CreationDate="2010-12-20T14:47:09.260" Score="1" ViewCount="170" Body="&lt;p&gt;I am looking for a symbol to represent the operation of taking unique values from a vector. So, say the symbol was $\theta$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$v = [0, 0, 0, 1, 1, 1, 3, 1, 2, 0]$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\theta(v) = [0, 1, 3, 2]$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or is this something that just isn't defined?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks all for your help here. Your answers and comments have led me to realise that doing this operation on a vector is not really what I want to do; using a set makes much more sense logically.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Background: I have many coordinates $\boldsymbol{p}_i = [x_i, y_i, z_i]$ in 3D space, which I then perform a rounding function on so they are equally sampled. However, there will be many duplicates points at each new coordinate, so I remove the duplicated $\boldsymbol{p}$ points to get the final set of points. So it makes sense to define a set $\boldsymbol{P} = \{\boldsymbol{p}_1, ... \boldsymbol{p}_i, \cdot \boldsymbol{p}_n\}$).&lt;/p&gt;&#xA;" OwnerUserId="4965" LastEditorUserId="4965" LastEditorDisplayName="" LastEditDate="2010-12-20T17:36:16.423" LastActivityDate="2010-12-20T17:36:16.423" Title="What is the mathematical symbol for the unique values in a vector?" Tags="&lt;notation&gt;" AnswerCount="2" CommentCount="6" />
  <row Id="14959" PostTypeId="2" ParentId="14930" CreationDate="2010-12-20T14:53:52.777" Score="8" ViewCount="" Body="&lt;p&gt;To express ⌊x⌋ in terms of Fourier series for $x \in \mathbb{R}/\mathbb{Z}$, let f(x) = x - ⌊x⌋, then the Fourier series for f(x) is quite straightforward, and so we start by showing that f(x) is indeed periodic:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For any integer $n \in \mathbb{Z}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;f(x) = x - ⌊x⌋ = x + n - n - ⌊x⌋ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;= x + n - ⌊x + n⌋ = f(x+n) = f(x + n*1)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence f(x) is periodic with fundamental period $T_0 = 1$, so $\omega_0 = \frac{2\pi}{T_0} = 2\pi$, and f(x)=x at $0 \leq x &amp;lt; 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/dWIIN.png&quot; alt=&quot;A plot of f(x)&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The periodic function f(x) can be expressed by the Trigonometric Fourier series,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f(x) = a_0 + \sum_{m=1}^{\infty} (a_{m}\cos(m\omega_0 x) + b_{m}\sin(m\omega_0 x))$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ = a_0 + \sum_{m=1}^{\infty} (a_{m}\cos(2\pi mx) + b_{m}\sin(2\pi m x))$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where the coefficients are calculated as follows&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a_0 = \frac{1}{T_0}\int_{&amp;lt;T_0&amp;gt;}f(x)dx = \int_{0}^{1}x dx = \frac{1}{2}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a_m = \frac{2}{T_0}\int_{&amp;lt;T_0&amp;gt;}f(x)\cos(m 2\pi  x)dx = 0$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$b_m = \frac{2}{T_0}\int_{&amp;lt;T_0&amp;gt;}f(x)\sin(m 2\pi  x)dx = -\frac{1}{\pi m}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint. Use integration by parts...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Plot of $f(x) = a_0 + \sum_{m=1}^{10} b_m\sin(2\pi m x)$, (summation from m=1 to 10).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/XshIE.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now that we obtained the Fourier series representation of f(x) = x - ⌊x⌋ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then ⌊x⌋ $ = x - f(x) = x - 0.5 + \frac{1}{\pi}\sum_{m=1}^{\infty} \frac{\sin(2\pi m x)}{m}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A plot of  ⌊x⌋, where summation from m=1 to m=10&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/e4JYv.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note this is only valid for $x \in \mathbb{R}/\mathbb{Z}$. Because although ⌊x⌋ is continuous on $\mathbb{R}/\mathbb{Z}$. It has a jump of one unit at each integer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;⌊x⌋ consists of an infinite number of steps, and also can be represented as a summation of shifted unit step functions u(x).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;⌊x⌋ $ = \sum_{m=1}^{\infty}(u(x-m)-u(-x-m+1))$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that u(x) = Sign(x)/2 + 0.5&lt;/p&gt;&#xA;" OwnerUserId="1746" LastEditorUserId="1746" LastEditorDisplayName="" LastEditDate="2010-12-20T18:58:39.300" LastActivityDate="2010-12-20T18:58:39.300" CommentCount="4" />
  <row Id="14960" PostTypeId="2" ParentId="14925" CreationDate="2010-12-20T15:09:47.973" Score="7" ViewCount="" Body="&lt;p&gt;A rigorous proof can be made using the central limit theorem (CLT), as follows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since &#xA;$$&#xD;&#xA;\frac{1}{{n!}}\int_0^1  {x^{n - k} (x - 1)^k e^{ - x} \,{\rm d}x} \to 0  &#xD;&#xA;$$&#xA;as $n \to \infty$, it suffices to consider the integral from $1$ to $\infty$. Let $c&amp;gt;0$ be arbitrary but fixed, and let $X_i$ be as in my previous answer. Define $f_n {(x)} = x^n e^{-x} / \Gamma(n+1)$, $x&amp;gt;0$, and put $n&amp;#39;=n+1$. Then, since $f_n$ is the density of $\sum\nolimits_{i = 1}^{n&amp;#39;} {X_i }$,&#xA;$$&#xD;&#xA;\frac{1}{{n!}}\int_1^{n&amp;#39; - c\sqrt {n&amp;#39;} } {x^{n - k} (x - 1)^k e^{ - x} \,{\rm d}x}  \leq \int_0^{n&amp;#39; - c\sqrt {n&amp;#39;} } {f_n (x)\,{\rm d}x} = {\rm P}\bigg(\frac{{\sum\nolimits_{i = 1}^{n&amp;#39;} {X_i }  - n&amp;#39;}}{{\sqrt {n&amp;#39;} }} \le  - c \bigg).&#xD;&#xA;$$&#xA;By the CLT, since the $X_i$ have unit mean and unit variance, the probability on the right-hand side tends to $\Phi(-c)$ as $n \to \infty$, where $\Phi$ denotes the distribution function of the standard normal distribution. &#xA;Similarly,&#xA;$$&#xD;&#xA;\frac{1}{{n!}}\int_{n&amp;#39; + c\sqrt {n&amp;#39;} }^\infty  {x^{n - k} (x - 1)^k e^{ - x} \,{\rm d}x}  \leq \int_{{n&amp;#39;} + c\sqrt {n&amp;#39;} }^\infty  {f_n (x)\,{\rm d}x} = {\rm P}\bigg(\frac{{\sum\nolimits_{i = 1}^{n&amp;#39;} {X_i }  - n&amp;#39;}}{{\sqrt {n&amp;#39;} }} &amp;gt; c \bigg),&#xD;&#xA;$$&#xA;and, by the CLT, the probability on the right-hand side tends to $1 - \Phi(c)$ as $n \to \infty$.&#xA;Next, note that&#xA;$$&#xD;&#xA;\bigg(1 - \frac{1}{{n&amp;#39; - c\sqrt {n&amp;#39;} }}\bigg)^k \int_{n&amp;#39; - c\sqrt {n&amp;#39;} }^{n&amp;#39; + c\sqrt {n&amp;#39;} } {f_n (x)\,{\rm d}x} \leq&#xD;&#xA;\frac{1}{{n!}}\int_{n&amp;#39; - c\sqrt {n&amp;#39;} }^{n&amp;#39; + c\sqrt {n&amp;#39;} } {x^{n - k} (x - 1)^k e^{ - x} \,{\rm d}x}&#xD;&#xA;$$&#xA;and &#xA;$$&#xD;&#xA;\frac{1}{{n!}}\int_{n&amp;#39; - c\sqrt {n&amp;#39;} }^{n&amp;#39; + c\sqrt {n&amp;#39;} } {x^{n - k} (x - 1)^k e^{ - x} \,{\rm d}x} \leq \bigg(1 - \frac{1}{{n&amp;#39; + c\sqrt {n&amp;#39;} }}\bigg)^k \int_{n&amp;#39; - c\sqrt {n&amp;#39;} }^{n&amp;#39; + c\sqrt {n&amp;#39;} } {f_n (x)\,{\rm d}x}.&#xD;&#xA;$$&#xA;By the CLT,&#xA;$$&#xD;&#xA; \int_{n&amp;#39; - c\sqrt {n&amp;#39;} }^{n&amp;#39; + c\sqrt {n&amp;#39;} } {f_n (x)\,{\rm d}x} = {\rm P}\bigg(-c \leq \frac{{\sum\nolimits_{i = 1}^{n&amp;#39;} {X_i }  - n&amp;#39;}}{{\sqrt {n&amp;#39;} }} \leq c \bigg) \to 2 \Phi(c) -1&#xD;&#xA;$$&#xA;as $n \to \infty$. Combining it all, and noting that &#xA;$$&#xD;&#xA;\bigg(1 - \frac{1}{{n&amp;#39; \pm c\sqrt {n&amp;#39;} }}\bigg)^k  = \bigg(1 - \frac{1}{{n&amp;#39; \pm c\sqrt {n&amp;#39;} }}\bigg)^{n(k/n)}  \approx e^{ - k/n}, &#xD;&#xA;$$&#xA;the result follows by choosing $c$ sufficiently large. [Note that first we choose $c$ sufficiently large, then we let $n \to \infty$.]&lt;/p&gt;&#xA;" OwnerUserId="2810" LastEditorUserId="2810" LastEditorDisplayName="" LastEditDate="2011-03-06T16:30:26.737" LastActivityDate="2011-03-06T16:30:26.737" CommentCount="4" />
  <row Id="14962" PostTypeId="2" ParentId="14451" CreationDate="2010-12-20T15:24:48.403" Score="4" ViewCount="" Body="&lt;p&gt;Although the links provided in some sense answer the question, the specific question can be answered without the full force of a point-in-polygon computation.  I would recommend this.&#xA;Compute whether each angle of your quad $(a,b,c,d)$ is convex or reflex.  If one is reflex&#xA;(say $a$), connect it to the opposite vertex $c$.  If all are convex, choose any diagaonal; e.g., $(a,c)$.  Now you have partitioned your quad into two triangles.  Check if your point is in either triangle, by checking if it is left-of-or-on each of its three edges. &lt;/p&gt;&#xA;" OwnerUserId="237" LastActivityDate="2010-12-20T15:24:48.403" />
  <row Id="14963" PostTypeId="1" AcceptedAnswerId="14966" CreationDate="2010-12-20T15:43:42.733" Score="3" ViewCount="82" Body="&lt;p&gt;Does there exist a bi-Invariant metric on $SL_n(\mathbb{R})$. I tried to google a bit but I didn't find anything helpful. &lt;/p&gt;&#xA;" OwnerUserId="707" LastActivityDate="2010-12-20T16:20:52.107" Title="Bi invariant metrics on $SL_n(\mathbb{R})$" Tags="&lt;lie-groups&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" />
  <row Id="14964" PostTypeId="2" ParentId="14958" CreationDate="2010-12-20T15:49:05.710" Score="3" ViewCount="" Body="&lt;p&gt;A common way to express that you are throwing away duplicates is to translate a sequence into a set. A set is commonly understood as an unordered collection of &lt;em&gt;distinct&lt;/em&gt; objects. It is usually denoted using curly braces. So you could write $\theta(v) = \{v_i\}_{i\in\{1,\ldots,n\}}$, where $v=[v_1,\ldots,v_n]$.&lt;/p&gt;&#xA;" OwnerUserId="3212" LastActivityDate="2010-12-20T15:49:05.710" CommentCount="3" />
  <row Id="14965" PostTypeId="2" ParentId="14919" CreationDate="2010-12-20T15:52:15.517" Score="1" ViewCount="" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;2&lt;/strong&gt; is the maximum number of stripes possible on a zebra group with a=8.  S stripes require &lt;b&gt; a ≥ C⋅9&lt;sup&gt;S&lt;/sup&gt; &lt;/b&gt; for large S.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The number of stripes S of a zebra group gives bounds on the derived length D of a zebra group: 2S - 1 ≤ D ≤ 2S + 1.  The action of G/H on a chief factor H/K must be irreducible and faithful, and so G/H embeds in GL(H/K) = GL(n,q) for q in {2,p} the prime dividing the order q^n of H/K.  However, for large n, the maximal derived length of a solvable subgroup of GL(n,q) is about log(n-2)/log(9), or in other words, n ≥ 9^D.  In particular, as S increases, the &quot;a&quot; from 2&lt;sup&gt;a&lt;/sup&gt; p&lt;sup&gt;b&lt;/sup&gt; increases exponentially.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For small n, the maximal derived length of soluble subgroups of GL(n,2) are: 2, 3, 4, 4.  The maximal derived length of irreducible soluble subgroups of GL(n,2) appear to be 2,2,3,2,6,2,5,4,4.  Hence the minimum dimensions for derived length D=2 is n=2, D=3 is n=4, D=4 is n=6, D=5 is n=6, D=6 is n=6, and D≥7 is n≥11.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The top stripe has order at least 2^1.  By the time H/K is the second stripe, G/H has derived length 2, and so H/K has order at least 2^2.  By the time H/K is the third stripe, G/H has derived length 4, and so H/K has order at least 2^6.&#xA;  Since 1+2+6 ≥ 8, this shows there is no three-striped zebra group with a Sylow 2-subgroup of order 2^8.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Three stripes is attainable at a=9.  Four stripes is not attainable for a≤14 and is attainable for a=27.  Roughly speaking, S stripes require a ≥ C⋅9&lt;sup&gt;S&lt;/sup&gt; (though my proof needs S really large).&lt;/p&gt;&#xA;" OwnerUserId="583" LastEditorUserId="583" LastEditorDisplayName="" LastEditDate="2010-12-20T16:02:52.870" LastActivityDate="2010-12-20T16:02:52.870" />
  <row Id="14966" PostTypeId="2" ParentId="14963" CreationDate="2010-12-20T16:11:52.173" Score="5" ViewCount="" Body="&lt;p&gt;A connected Lie group has a biinvariant metric iff it is isomorphic to the dirtect product of a compact one and a vector space---see Milnor's &lt;em&gt;Curvatures of left invariant metrics&lt;/em&gt;. Such a factorization would carry over to the Lie algebras, etc.&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2010-12-20T16:11:52.173" />
  <row Id="14967" PostTypeId="2" ParentId="14963" CreationDate="2010-12-20T16:20:52.107" Score="6" ViewCount="" Body="&lt;p&gt;It's enough to show the answer is &quot;no&quot; for $SL_2(\mathbb{R})$, because $SL_2(\mathbb{R})$ is naturally embedded in $SL_n(\mathbb{R})$, and if $SL_n(\mathbb{R})$ has a biinvariant metric, then the induced metric on $SL_2(\mathbb{R})$ would be biinvariant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So let's focus on $SL_2(\mathbb{R})$.  Finding a biinvariant metric on $SL_2(\mathbb{R})$ is equivalent to finding an $Ad(SL_2(\mathbb{R})$ invariant inner product on $\mathfrak{sl_2(\mathbb{R})}$, so I'll show there is no such $Ad$ invariant metric.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, $\mathfrak{sl_2(\mathbb{R})}$ is 3 dimensional with basis $E_{12}, E_{21}, E_{11} - E_{22}$ where $E_{ij}$ denotes the matrix with a 1 in the $ij$ slot and a 0 elsewhere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider $A = diag(2,1/2)$ ,the diagonal matrix with diagonal entries 2 and 1/2.  This is clearly in $SL_2(\mathbb{R})$.  Note then that $AE_{12}A^{-1} = 4 E_{12}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if an inner product were $Ad$ invariant, it would have to preserve the length of $E_{12}$, giving a contradiction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, there is no $Ad$ invariant inner product on $\mathfrak{sl_2(\mathbb{R})}$, and hence there is no biinvariant metric on $SL_n(\mathbb{R})$.&lt;/p&gt;&#xA;" OwnerUserId="331" LastActivityDate="2010-12-20T16:20:52.107" />
  <row Id="14968" PostTypeId="2" ParentId="14947" CreationDate="2010-12-20T16:23:54.147" Score="1" ViewCount="" Body="&lt;p&gt;When a surface element $d\omega$ at a point $P$ in three-space is projected orthogonally into all possible directions then the average area of the image is $\rho\thinspace d\omega$ where $\rho$ is a universal constant. To determine the value of $\rho$ we argue as follows: The unit sphere $S^2$ has a surface area of $4\pi$, and the area of its projection is $2\pi$ in all directions (note that the &quot;shadow&quot; is doubly covered); whence also the average area of the projection is $2\pi$. So $\rho$ must be ${1\over2}$. What we have found out is valid for any surface element $d\omega$ of your ellipsoid $E$. It follows that the average area of the shadow of $E$ is equal to $\omega(E)/4$, where $\omega(E)$ is the total surface area of $E$ and we have taken into account that the shadow is covered twice. The value so obtained can be expressed as an elliptic integral in terms of $a$, $b$, $c$.&lt;/p&gt;&#xA;" OwnerUserId="1303" LastEditorUserId="1303" LastEditorDisplayName="" LastEditDate="2010-12-21T09:05:15.497" LastActivityDate="2010-12-21T09:05:15.497" CommentCount="7" />
  <row Id="14969" PostTypeId="2" ParentId="14958" CreationDate="2010-12-20T16:41:04.230" Score="1" ViewCount="" Body="&lt;p&gt;There is no standard notation for the &lt;em&gt;set&lt;/em&gt; of coefficients of a vector. Indeed, there's not even any standard notation for the set of coefficients of a polynomial (which would be useful when taking the &lt;em&gt;content&lt;/em&gt;, i.e. the gcd / ideal generated by the coefficients).&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-12-20T16:41:04.230" />
  <row Id="14970" PostTypeId="1" AcceptedAnswerId="14974" CreationDate="2010-12-20T17:14:57.207" Score="7" ViewCount="423" Body="&lt;p&gt;Suppose $A$ is the set of all rational numbers $p$ such that $p^2 &amp;lt;2$ and $B$ is the set of all rational numbers $p$ such that $p^2 &amp;gt; 2$. We want to show that $A$ contains no largest element and $B$ contains no smallest element.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In Rudin's &lt;em&gt;Principles of Mathematical Analysis&lt;/em&gt;, he associates $q = p- \frac{p^2-2}{p+2} = \frac{2p+2}{p+2}$. Did he just come up with this using trial and error?&lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2010-12-21T01:16:09.827" Title="Rational Numbers" Tags="&lt;real-analysis&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="14971" PostTypeId="1" AcceptedAnswerId="14993" CreationDate="2010-12-20T17:22:51.870" Score="5" ViewCount="74" Body="&lt;p&gt;I know this seems like a juvenile question, but for some reason I can't recall how to do this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lets say I want to find 87 reduced by 99.75%. What is the equation to do that?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="3621" LastEditorUserId="2109" LastEditorDisplayName="" LastEditDate="2010-12-20T19:15:03.070" LastActivityDate="2010-12-20T19:15:03.070" Title="What is the equation for finding a number that is X% less than another number?" Tags="&lt;algebra-precalculus&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="14972" PostTypeId="2" ParentId="14926" CreationDate="2010-12-20T17:28:29.210" Score="0" ViewCount="" Body="&lt;p&gt;Here is a non-rigorous attempt at an answer, but perhaps someone can make this into a formal argument.  I will assume your series is centered at $0$.  I will also denote your function by $g$ for reasons you will see as you read on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose $g(x) = \sum_{n=0}^{\infty} a_n x^n$, where $a_n \geq0$ for every $n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We look for a series for $h(x) = \frac{1}{g(x)} = \sum_{n=0}^{\infty} b_n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By taking formal derivatives $b_n = D^n(h(0))/n!$ (as in the Taylor expansion).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And then do a Google search... Behold the Faá di Bruno's Formula for derivatives of composite functions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://mathworld.wolfram.com/FaadiBrunosFormula.html&quot; rel=&quot;nofollow&quot;&gt;http://mathworld.wolfram.com/FaadiBrunosFormula.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $f(x) = 1/x$ so that $h(x) = f(g(x))$.  Comparing with the terms in this formula, we see that the signs of the $b_n$ must alternate (or be zero).  This is because the derivatives of $1/x$ alternate in sign and the derivatives of $g(x)$ are non-negative by your hypothesis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, I don't trust any of this, however, since I've completely neglected matters of convergence; but when dealing with formal power series, maybe this point is moot?  &lt;/p&gt;&#xA;" OwnerUserId="4009" LastActivityDate="2010-12-20T17:28:29.210" CommentCount="1" />
  <row Id="14973" PostTypeId="2" ParentId="14971" CreationDate="2010-12-20T17:33:19.017" Score="6" ViewCount="" Body="&lt;p&gt;If you want $X$% of 87, multiply the latter by $\frac{X}{100}$. (So, for example, to get 25 percent, you multiply by $0.25$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If by &quot;reduced by 99.75%&quot; you mean 99.75% of the total, then, multiply by $.9975$. If &quot;reduced by 99.75%&quot; means taking 99.75% &lt;em&gt;off&lt;/em&gt; (so you are left with only 0.25%), then multiply by $0.0025$. &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-20T18:18:29.320" LastActivityDate="2010-12-20T18:18:29.320" CommentCount="2" />
  <row Id="14974" PostTypeId="2" ParentId="14970" CreationDate="2010-12-20T17:42:15.953" Score="19" ViewCount="" Body="&lt;p&gt;The point here is that iterations of the &lt;a href=&quot;http://en.wikipedia.org/wiki/M%C3%B6bius_transformation&quot; rel=&quot;nofollow&quot;&gt;Mobius transformation&lt;/a&gt; $p \mapsto \frac{2p+2}{p+2}$ converge to $\sqrt{2}$, so every time you apply the transformation you get closer to $\sqrt{2}$.  This can be thought of as describing a &lt;a href=&quot;http://en.wikipedia.org/wiki/Generalized_continued_fraction&quot; rel=&quot;nofollow&quot;&gt;generalized continued fraction&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$2 - \frac{2}{4 - \frac{2}{4 - \frac{2}{...}}}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for $\sqrt{2}$.  The dynamics of Mobius transformations in general are fairly well-understood; for the transformation $p \mapsto \frac{ap+b}{cp+d}$ the possible fixed points are the roots of the quadratic polynomial $cp^2 + dp = ap + b$, and using the &lt;a href=&quot;http://en.wikipedia.org/wiki/Banach_fixed_point_theorem&quot; rel=&quot;nofollow&quot;&gt;Banach fixed point theorem&lt;/a&gt; (or a more specific closed form of iterations of Mobius transformations using linear algebra) one can determine which of the fixed points are attractive or repellent.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if one wanted to design a Mobius transformation converging to $\sqrt{n}$ for some non-square $n$, this would require that $d = a, c = 1, b = n$, giving&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$p \mapsto \frac{ap + n}{p + a}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for some $a$, and it should not be hard to find a value of $a$ such that the fixed point $\sqrt{n}$ is attractive.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;While this technique is a nice trick, because the polynomial $cp^2 + dp = ap + b$ is quadratic, it does not generalize to prove the corresponding result for roots of cubic or higher degree polynomials.  &lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-12-20T17:42:15.953" CommentCount="5" />
  <row Id="14975" PostTypeId="1" CreationDate="2010-12-20T17:51:01.357" Score="3" ViewCount="124" Body="&lt;p&gt;Hi Folks,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm trying to track down the source of some wonky data. The data are response times (RTs) collected from humans using a computer keyboard. Here's a histogram of the RTs, binned to 1ms:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/54QBH.png&quot; alt=&quot;1hist&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obvious is a skew in the data, which is expected/known for such data. However, also noticeable is a slight &quot;ghosting&quot; of the image. Zooming in we see:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/mLVww.png&quot; alt=&quot;2hist&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There appears to be some phenomenon such that the probability of observing an RT at a given time point fluctuates across time in a cyclical manner. This is also evident in the empirical cumulative density function (suggesting that the phenomenon is about data, not the binning algorithms of the histograms):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/n9RLH.png&quot; alt=&quot;3ecdf&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To determine the nature of this cyclicality, I computed the 2nd derivative of the ecdf (actually, I've never taken calculus, so I simply computed diff(diff(y)) in R), yielding the following waveform (again, zoomed in to show details):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/q9P1h.png&quot; alt=&quot;4ecdf_deriv&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I actually encountered a similar phenomenon in the past and at that time my googling seemed to suggest that a periodogram would help me determine the Hz values at which something was happening to cause this phenomenon. I found some R code for a periodogram:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;periodogram &amp;lt;- function(x) {&#xA;    n &amp;lt;- length(x)&#xA;    Mod(fft(x))[1:(n%/%2 + 1)]^2 / (2*pi*n)&#xA;}&#xA;&#xA;plot(&#xA;    x = 1:((length(rt)/2)+1)&#xA;    , y = periodogram(ddy)&#xA;    , type = 'l'&#xA;)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;which yields:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/CAwJJ.png&quot; alt=&quot;5periodogram&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I'm not sure what the x-axis is supposed to be. Hz seems reasonable, but doesn't seem to fit with my impression from the histograms that the phenomenon has a cycle of about 100Hz. Any help providing clarification on the x-axis of this plot would be greatly appreciated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(For those with interest in this over-and-above the math, if I do find that the phenomenon has a 100Hz or thereabouts frequency, then I'm pretty sure it can be attributable to the polling rate of the USB keyboard, plus a little random error here and there in other aspects of timing)&lt;/p&gt;&#xA;" OwnerUserId="1361" LastActivityDate="2011-02-18T20:23:03.960" Title="Can you provide help with interpreting this periodogram?" Tags="&lt;fourier-analysis&gt;" AnswerCount="1" />
  <row Id="14976" PostTypeId="2" ParentId="13790" CreationDate="2010-12-20T17:57:11.563" Score="2" ViewCount="" Body="&lt;p&gt;Using this &lt;a href=&quot;http://math.stackexchange.com/questions/9999/checking-if-a-number-is-a-fibonacci-or-not/14872#14872&quot;&gt;other my answer&lt;/a&gt;&lt;br&gt;&#xA;Using integers only, I would use &lt;a href=&quot;http://en.wikipedia.org/wiki/Binary_search&quot; rel=&quot;nofollow&quot;&gt;Binary Search&lt;/a&gt;. Certainly you can compute $F_n$ only with integers, the simplest way is matrix exponentiation. Using Binary Search you can find numbers ``near'' your number $x$ and you will find $x = F_n$ (and $n$). I suppose this method is generic for anything monotone you can compute fast. To initiliaze the binary search, just keep doubling $ F_{2n} $&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Binary search allows you to search for a number x in a sorted &quot;array&quot; F[] (in the programming sense). Use this method to search for your number. When you need F[n] just compute $F_n$. This will work because the sequence is strictly increasing except for the initial 1,1.&lt;/p&gt;&#xA;" OwnerUserId="4750" LastActivityDate="2010-12-20T17:57:11.563" />
  <row Id="14977" PostTypeId="1" AcceptedAnswerId="14978" CreationDate="2010-12-20T18:01:57.073" Score="3" ViewCount="228" Body="&lt;p&gt;Let $A$ be a nonempty set of real numbers which is bounded below. Let $-A$ be the set of all numbers $-x$, where $x \in A$. Prove that $\inf A = -\sup(-A)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know that $-A$ is bounded above. Hence $\sup(-A)$ exists. So to proceed, one would show that $\inf A \leq -\sup(-A)$ and $\inf A \geq -\sup(-A)$? Or $-\inf A &amp;gt; \sup(-A)$ and $-\inf A &amp;lt; \sup(-A)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I get the sense that one can use the following fact as well: (i) Let $\alpha = \sup S$. If $\beta &amp;lt; \alpha$ then $\beta$ is not an upper bound for $S$. &lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2010-12-20T18:23:53.573" Title="Supremum and Infimum" Tags="&lt;real-analysis&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="14978" PostTypeId="2" ParentId="14977" CreationDate="2010-12-20T18:23:12.263" Score="5" ViewCount="" Body="&lt;p&gt;This is just a matter of using the definitions of supremum and infimum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just let $x = \inf{A}$ and $y = \sup{(-A)}$. So since $x = \inf{A}$ then for every $a \in A$ you know that $x \leq a$. But then this implies that $-x \geq -a$ for all $a \in A$, so $-x = - \inf{A}$ is an upper bound for the set $-A$. But since $y = \sup{(-A)}$ is the least upper bound for $-A$, then this means that $y \leq -x$ or that $\sup{(-A)} \leq - \inf{A}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can prove the other inequality.&lt;/p&gt;&#xA;" OwnerUserId="900" LastActivityDate="2010-12-20T18:23:12.263" />
  <row Id="14979" PostTypeId="2" ParentId="14977" CreationDate="2010-12-20T18:23:53.573" Score="1" ViewCount="" Body="&lt;p&gt;You can do it by inequalities; or you can use the definitions of supremum and infimum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remember that a real number $S$ is the supremum of $A$ if and only if:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$a\leq S$ for all $a\in A$ ($S$ is an upper bound for $A$); and&lt;/li&gt;&#xA;&lt;li&gt;For all $\epsilon\gt 0$ there exists $a_{\epsilon}\in A$ such that $S-\epsilon \lt a_{\epsilon}$ (no number strictly smaller than $S$ is an upper bound for $A$).&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;(2 above can be written slightly differently so that it applies to arbitrary partially ordered sets: you require that for all $s$, if $s\lt S$, then there exists $a\in A$ such that $s\lt a$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dually, a real number $T$ is the infimum of $A$ if and only if:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$a\geq T$ for all $a\in A$ ($T$ is a lower bound for $A$); and&lt;/li&gt;&#xA;&lt;li&gt;For all $\epsilon \gt 0$ there exists $a_{\epsilon}\in A$ such that $T+\epsilon \gt a_{\epsilon}$ (no number strictly larger than $T$ is a lower bound for $A$).&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;So, you can use the fact that $T=\inf(A)$ is the infimum to show that $-T$ is the supremum of $-A$ by verifying the properties of the supremum. It should be straightforward.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-12-20T18:23:53.573" />
  <row Id="14980" PostTypeId="1" AcceptedAnswerId="14983" CreationDate="2010-12-20T18:30:54.757" Score="2" ViewCount="132" Body="&lt;p&gt;Hello!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does someone know if the following is true:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $\mathbb{X}$ be some arbitrary Banach space.&#xA;$\{x_k \}_{k=1}^{\infty} \in \mathbb{X}$ is a sequence chosen from $\mathbb{X}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if the series &#xA;$$\sum_{k=1}^\infty \|x_k\|_X$$&#xD;&#xA;converges, would the &quot;more generic&quot; series&lt;br&gt;&#xA;$$\sum_{k=1}^\infty x_k$$ &#xA;also converge?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If yes, could you please give the proof (or just mark the proof steps) ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you.&lt;/p&gt;&#xA;" OwnerUserId="4971" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-20T18:36:40.593" LastActivityDate="2010-12-20T18:59:58.737" Title="Converging series in Banach space" Tags="&lt;convergence&gt;&lt;banach-spaces&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="14982" PostTypeId="2" ParentId="14952" CreationDate="2010-12-20T18:34:25.600" Score="12" ViewCount="" Body="&lt;p&gt;The Jacobian $df_p$ of a differentiable function $f : \mathbb{R}^n \to \mathbb{R}^m$ at a point $p$ is its best linear approximation at $p$, in the sense that $f(p + h) = f(p) + df_p(h) + o(|h|)$ for small $h$.  This is the &quot;correct&quot; generalization of the derivative of a function $f : \mathbb{R} \to \mathbb{R}$, and everything we can do with derivatives we can also do with Jacobians.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, when $n = m$, the determinant of the Jacobian at a point $p$ is the factor by which $f$ locally dilates volumes around $p$ (since $f$ acts locally like the linear transformation $df_p$, which dilates volumes by $\det df_p$).  This is the reason that the Jacobian appears in the change of variables formula for multivariate integrals, which is perhaps the basic reason to care about the Jacobian.  For example this is how one changes an integral in rectangular coordinates to cylindrical or spherical coordinates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Jacobian specializes to the most important constructions in multivariable calculus.  It immediately specializes to the gradient, for example.  When $n = m$ its trace is the divergence.  And a more &lt;a href=&quot;http://en.wikipedia.org/wiki/Hodge_dual&quot; rel=&quot;nofollow&quot;&gt;complicated construction&lt;/a&gt; gives the curl.  The rank of the Jacobian is also an important local invariant of $f$; it roughly measures how &quot;degenerate&quot; or &quot;singular&quot; $f$ is at $p$.  This is the reason the Jacobian appears in the statement of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Implicit_function_theorem&quot; rel=&quot;nofollow&quot;&gt;implicit function theorem&lt;/a&gt;, which is a fundamental result with applications everywhere.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-12-20T18:34:25.600" CommentCount="1" />
  <row Id="14983" PostTypeId="2" ParentId="14980" CreationDate="2010-12-20T18:35:18.713" Score="4" ViewCount="" Body="&lt;p&gt;Yes.  You can use the hypothesis that $\sum\|x_k\|$ converges to show that the sequence of partial sums of the series $\sum x_k$ is a Cauchy sequence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conversely, if this property holds in a normed space, then the space is complete; that direction isn't as straightforward.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-12-20T18:35:18.713" />
  <row Id="14984" PostTypeId="1" AcceptedAnswerId="14987" CreationDate="2010-12-20T18:47:22.900" Score="5" ViewCount="319" Body="&lt;p&gt;I understand the principle of finite induction, but my book then mentions that there is a variant of the first where requirement b is changed to &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If k is a positive integer such that 1,2...,k belong to S, then k + 1 must also be in S.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The sample problem is proving that the inequality about the Lucas numbers &#xA;l&lt;sub&gt;n&lt;/sub&gt; &amp;lt; (7/4)&lt;sup&gt;n&lt;/sup&gt;&#xA;I understand the algebra of the proof, but I do no understand how it uses the second principle or how it is different from the first principle?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The proof for the lucas numbers&#xA;$$&#xD;&#xA;a_1 = 1 &amp;lt; \left(\frac{7}{4}\right)^1 = \frac{7}{4}\text{ and }a_2 = 3 \lt \left(\frac{7}{4}\right)^2 = \frac{49}{16}.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Choose an integer $k\geq 3$ and assume that the inequality is valid for $n = 1,2,\ldots,k$.  Then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$a_{k-1} \lt \left(\frac{7}{4}\right)^{k-1}\text{ and }a_{k-2} \lt \left(\frac{7}{4}\right)^{k-2},$$&#xD;&#xA;so &#xA;\begin{align*}&#xA;a_k &amp;amp;= a_{k-1} + a_{k-2}\\&#xA;&amp;amp;\lt \left(\frac{7}{4}\right)^{k-1} + \left(\frac{7}{4}\right)^{k-2}\\&#xA;&amp;amp;= \left(\frac{7}{4}\right)^{k-2}\left(\frac{7}{4} + 1\right)\\&#xA;&amp;amp;= \left(\frac{7}{4}\right)^{k-2}\left(\frac{11}{4}\right)\\&#xA;&amp;amp;\lt \left(\frac{7}{4}\right)^{k-2}\left(\frac{7}{4}\right)^2\\&#xA;&amp;amp;= \left(\frac{7}{4}\right)^k&#xA;\end{align*}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The responses are all helpful, but if someone could point out what the difference is when you go about actually using strong induction instead of weak induction. (You could use the lucas number proof or something else)&lt;/p&gt;&#xA;" OwnerUserId="735" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-20T20:45:28.123" LastActivityDate="2010-12-20T21:04:52.687" Title="What is the second principle of finite induction?" Tags="&lt;induction&gt;" AnswerCount="4" CommentCount="7" FavoriteCount="3" />
  <row Id="14985" PostTypeId="1" AcceptedAnswerId="14992" CreationDate="2010-12-20T18:48:58.887" Score="9" ViewCount="215" Body="&lt;p&gt;This problem came from a Putnam problem solving seminar.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If each person in a group of n people is a friend of at least half the people in the group, then show that it is possible to seat the n people in a circle so that everyone sits next to friends only.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My idea was to use induction on $n$; if $n$ is odd we can remove a person, note that the remaining $n-1$ people all have at least $n/2$ friends left, use our inductive hypothesis to seat them, and then use the pigeonhole principle to seat the last person.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, this doesn't work when $n$ is even because after removing a person, some of the remaining people might have less than $n/2$ friends left.  In fact, my friend who actually participated in the seminar said they had the same issue, but didn't address it because they ran out of time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this sort of induction a reasonable approach?  If so, how would we deal with the case when $n$ is even?  If not, what's a better way to think about the problem?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S. I'm not sure of the best tags for this question, so please feel free to re-tag if necessary.&lt;/p&gt;&#xA;" OwnerUserId="4904" LastActivityDate="2010-12-20T19:57:11.473" Title="Seating friends around a dinner table" Tags="&lt;combinatorics&gt;&lt;graph-theory&gt;&lt;contest-math&gt;" AnswerCount="2" />
  <row Id="14987" PostTypeId="2" ParentId="14984" CreationDate="2010-12-20T18:58:13.483" Score="14" ViewCount="" Body="&lt;p&gt;There are two basic differences:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;In ordinary induction, we need a &lt;strong&gt;base case&lt;/strong&gt; (proving it for $k=1$; that is, proving that $1\in S$); in the second principle of induction (also called &quot;strong induction&quot;) you do not need a base case (but see the &lt;em&gt;caveat&lt;/em&gt; below).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;In ordinary induction, the &quot;Induction Hypothesis&quot; is that the $k\in S$; in the second principle, the &quot;Induction Hypothesis&quot; is that &lt;em&gt;all&lt;/em&gt; the natural numbers strictly less than $k+1$ are in $S$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;For example, if you want to prove that every positive integer can be factored into primes, if you use ordinary induction you would have to first show that $1$ can be factored into primes, and then you would show that if $k$ can be factored into primes, then $k+1$ can be factored into primes. This is actually pretty hard to do, because the factorization of $k+1$ has little or nothing to do with the factorization of $k$. This is a very difficult proposition to prove directly using ordinary induction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you wanted to prove the same proposition using the second principle of induction, you would instead assume that &lt;strong&gt;all&lt;/strong&gt; numbers strictly less than $k$ can be factored into primes, and try to show from this assumption that $k$ can be factored into primes (which is much simpler: if $k=1$ or $k$ is prime, there is nothing to do; otherwise, you can write $k=ab$, each $a$ or $b$ strictly smaller than $k$ and greater than $1$, so you can apply the induction hypothesis to both to get a factorization).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, in the second principle, you get to assume &lt;em&gt;more&lt;/em&gt; in the inductive step.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Caveat:&lt;/em&gt; Formally, the second principle does not require a &lt;em&gt;base&lt;/em&gt;; this is because if you can actually prove that &quot;if all positive integers strictly smaller than $k$ are in $S$ then $k$ is in $S$&quot;, then for $k=1$ the premise is vacuously true (&lt;em&gt;all&lt;/em&gt; positive integer smaller than $1$ &lt;em&gt;are&lt;/em&gt; in $S$, because there is no such number at all), so the implication would yield that $k=1$ must also be in $S$. However, in most actual applications, the inductive argument does not work for $k=1$ (or for other special cases), so that these have to be proven separately. They amount to &quot;special cases&quot; instead of &quot;bases&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, both principles are &lt;em&gt;equivalent&lt;/em&gt; for the positive integers. That is, if you can do a proof using ordinary induction, then you can translate that into a proof using the second induction principle in a fairly mechanical manner. And if you can do a proof using the second induction principle, you can transform that into a proof using the first induction principle (though in an indirect manner), again in a fairly mechanical way (provided you take for granted some properties of the positive integers; for example, that every positive integer is either equal to $1$, or else is $k+1$ for some positive integer $k$). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Note:&lt;/strong&gt; The reason the second principle is called &quot;strong induction&quot; is that you can use it in other contexts to prove more than ordinary induction would. For example, imagine a situation in which you have a set $T$ that consists of &lt;em&gt;two&lt;/em&gt; copies of the integers: a blue copy and a green copy, with the blue copy going &quot;first&quot;. If you prove that $S$ is a subset of $T$ such that blue $1$ is in $S$, and whenever $k$ ($k$ either a blue positive integer or a green positive integer) is in $S$ then $k+1$ is in $S$ (ordinary induction), then you can conclude that all blue integers are in $S$, but the argument by itself does not tell you whether there are any green integers in $S$ (you would have to separately prove that green-$1$ is in $S$). But if you use strong induction, and prove that whenever all integers, blue or green, that are strictly smaller than $k$ are in $S$ then $k$ is in $S$, then you &lt;em&gt;can&lt;/em&gt; conclude that $S$ will contain all blue integers &lt;em&gt;and&lt;/em&gt; all green integers too. So you can actually get a &lt;em&gt;stronger&lt;/em&gt; conclusion from using strong induction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Caveat the Second&lt;/em&gt;: Special cases show up far more often in proofs using strong induction than in proofs using ordinary induction, but they sometimes occur in the latter as well. As a famous example of a &quot;proof&quot; using regular induction that would require a special case (and that in fact is false because that special case fails) is the fake proof of the proposition that &quot;If, in a group of people, at least one of them has blue eyes, then everyone in the group has blue eyes.&quot; The proof is as follows: Let $k$ denote the number of people in the group. We prove it by induction on $k$. If $k=1$ then the result obviously holds. Assume the result holds for any group with $k$ people, and take a group with $k+1$ people, $(p_1,\ldots,p_{k},p_{k+1})$, in which at least one has blue eyes, say $p_1$. Looking at the first $k$, $(p_1,\ldots,p_k)$, we can use the induction hypothesis to conclude that $p_1,p_2,\ldots,p_k$ each have blue eyes. Then looking at the last $k$, $(p_2,\ldots,p_k,p_{k+1})$, since at least one has blue eyes (say $p_k$), then they all have blue eyes by the induction hypothesis. In particular, $p_{k+1}$ has blue eyes. Therefore, they al have blue eyes, QED.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason this is incomplete is that the inductive step only works if $k\geq 3$, so that a proof would require the &lt;em&gt;special case&lt;/em&gt; of showing that $1\in S$ implies $2\in S$ (the general argument we have does not cover that particular implication). That's where it all breaks down, of course. But the point is that sometimes ordinary induction also requires &quot;special cases&quot;, though this tends to happen less often than with strong induction.&lt;/p&gt;&#xA;&#xA;&lt;hr/&gt;&#xA;&#xA;&lt;p&gt;The step where you use the inductive hypothesis in the proof above is in the second line of the long display, where you are using that the inequality you want holds for both $a_{k-1}$  &lt;strong&gt;and&lt;/strong&gt; for $a_{k-2}$. This requires you to assume that it holds for two numbers strictly smaller than $k$ (you don't need the fact that it holds for &lt;em&gt;all&lt;/em&gt; numbers less than $k$ here, but you do end up needing it for more than just the immediately previous one).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In regular induction, your only assumption is that the result holds for the immediately previous number (for $k-1$, when you are trying to prove it for $k$; or for $k$, when you are trying to prove it for $k+1$). You make no assumptions about what happens for $\ell$ with $\ell\lt k-1$. In strong induction, you assume that it holds for &lt;em&gt;all&lt;/em&gt; $\ell\lt k$, not just for $k-1$. Here, because you need the assumption for &lt;em&gt;two&lt;/em&gt; smaller values of $k$, this argument would not work if you were only assuming that $a_{k-1}\lt\left(\frac{7}{4}\right)^{k-1}$ but were making no assumptions about $a_{k-2}$. There &lt;em&gt;are&lt;/em&gt; ways to prove the result using ordinary induction, but this particular argument would not work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice that since your proof requires you to have at least two smaller values than $k$, the argument will not work for either $k=1$ or $k=2$ (the first has no smaller values, the second has only one); that is why you end up having to prove the &lt;em&gt;two&lt;/em&gt; special cases of $a_1$ and $a_2$. Again, this is not a &quot;basis for the induction&quot;, but rather special cases of the inductive argument.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-20T21:04:52.687" LastActivityDate="2010-12-20T21:04:52.687" />
  <row Id="14989" PostTypeId="2" ParentId="14984" CreationDate="2010-12-20T18:59:27.570" Score="4" ViewCount="" Body="&lt;p&gt;The second principle of induction to which you refer is sometimes called the strong principle of induction, as opposed to the usual one which is also sometimes called the weak principle of induction. This distinction is only apparent at first since they are both equivalent, so there's no harm in using the strong version of it and thinking you're using something much more powerful. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you can see in this &lt;a href=&quot;http://mathcircle.berkeley.edu/BMC4/Handouts/induct/node6.html&quot; rel=&quot;nofollow&quot;&gt;post&lt;/a&gt;, the difference is basically that in the weak principle of induction, your inductive hypothesis is just that the property you want to prove is true for some integer $n$, and then you show that it also holds for $n+1$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But in the strong version you assume that the property holds for all integers up to $n$, that is, you assume it holds for $0, 1, 2, \dots , n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An example of a theorem that is sometimes proved using the strong version is the so called &lt;a href=&quot;http://en.wikipedia.org/wiki/Fundamental_theorem_of_arithmetic&quot; rel=&quot;nofollow&quot;&gt;Fundamental Theorem of Arithmetic&lt;/a&gt;. The theorem basically consists of two parts, one is showing the existence of a factorization into primes and the other is showing that &quot;the&quot; factorization is unique. So for instance if you're showing that a factorization exists by induction, then in the inductive step you'll have to prove that if $n$ can be factored as a product of primes, the so can $n+1$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem now is that you have basically two options, that $n+1$ is a prime in which case you're done, or that $n+1$ is composite, thus it can be factored as $n+1 = ab$ where $1 &amp;lt; a \leq n$ and $1 &amp;lt; b \leq n$. Now you can see where the strong principle comes in handy because if you assume not only that $n$ can be factored as a product of primes, but also that all the integers greater than $1$ can (up to $n$ of course), then you immediately know that both $a$ and $b$ are products of primes, say $a = p_1 \cdots p_k$ and $b = q_1 \cdots q_s$ with $p_i$ and $q_j$ primes, then $n+1 = ab = p_1 \cdots p_k \cdot q_1 \cdots q_s$ is also a product of primes, so the induction is complete.&lt;/p&gt;&#xA;" OwnerUserId="900" LastEditorUserId="900" LastEditorDisplayName="" LastEditDate="2010-12-20T19:28:53.040" LastActivityDate="2010-12-20T19:28:53.040" CommentCount="2" />
  <row Id="14990" PostTypeId="2" ParentId="14980" CreationDate="2010-12-20T18:59:58.737" Score="3" ViewCount="" Body="&lt;p&gt;It is indeed true. It is in fact almost the same as the statement of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Weierstrass_M-test&quot; rel=&quot;nofollow&quot;&gt;Weierstrass M-test&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1269" LastActivityDate="2010-12-20T18:59:58.737" />
  <row Id="14991" PostTypeId="2" ParentId="14984" CreationDate="2010-12-20T19:01:41.140" Score="7" ViewCount="" Body="&lt;p&gt;That's not a very good example.  This principle usually goes by the name of &lt;strong&gt;strong induction&lt;/strong&gt;, and the point is that while it's equivalent to ordinary induction, it's often more convenient to use in proofs.  For proving a sequence $P(k)$ of identities, it's often straightforward to write $P(k+1)$ in terms of $P(k)$, but for a more complicated sequence of statements one might need the stronger assumption that $P(1), ... P(k)$ all hold in order to readily deduce $P(k+1)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, if one is proving a sequence of statements $P(k)$ about finite &lt;a href=&quot;http://en.wikipedia.org/wiki/Graph_theory&quot; rel=&quot;nofollow&quot;&gt;graphs&lt;/a&gt; with $k$ vertices, one general proof strategy is to break a graph up into two smaller graphs, and if the statement for each of the smaller graphs implies the statement for the larger graph, we are done by strong induction, but not necessarily by weak induction, since the smaller graphs do not necessarily have sizes $1$ and $k-1$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As another example, here's a nice exercise.  Consider the sequence $a_n$ with $a_1 = 1$ and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$a_{2n} = a_n$$&#xD;&#xA;$$a_{2n+1} = a_n + 1.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Try proving by strong induction that $a_n$ is the sum of the digits of $n$ in base $2$.  Can you see why this proof is difficult to phrase using weak induction?&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-12-20T19:01:41.140" CommentCount="2" />
  <row Id="14992" PostTypeId="2" ParentId="14985" CreationDate="2010-12-20T19:06:14.883" Score="11" ViewCount="" Body="&lt;p&gt;This is known as &lt;a href=&quot;http://en.wikipedia.org/wiki/Hamiltonian_path#Bondy.E2.80.93Chv.C3.A1tal_theorem&quot; rel=&quot;nofollow&quot;&gt;Dirac's theorem&lt;/a&gt;.  I don't know if an inductive proof is possible; the standard proof is extremal.  You consider the longest sequence of friends and show by contradiction that it must be a Hamiltonian circuit.  I can edit with more details if you want the full argument; it's quite nice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: the argument is as follows.  It can be found, for example, in Bollobas' &lt;em&gt;Modern Graph Theory&lt;/em&gt; (p. 106).  First let's establish the graph-theoretic language.  Let $G$ be a graph on $n$ vertices, $n \ge 3$.  For each $v \in G$ let $d_v$ denote the number of neighbors of $v$.  Then the claim is that if $d_v \ge \frac{n}{2}$ for each $v \in G$, then $G$ is Hamiltonian.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We claim that $G$ is connected.  Indeed, if $v, w \in G$ then $v$ and $w$ each have $\frac{n}{2}$ neighbors out of a possible $n-1$, so they must share some neighbor.  (This step works under the weaker assumption that $d_v + d_w \ge n$ for each $v \neq w$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now suppose that $G$ is not Hamiltonian.  Let $v_1, ... v_k$ be a path all of whose vertices are distinct with $k$ maximal.  Then $v_1$ and $v_k$ have no neighbors outside the path (by maximality), so all of their neighbors must be one of $v_1, ... v_k$.  In addition, $v_1$ cannot be connected to $v_k$; otherwise we would have a cycle of length $k$, and by connectivity this cycle would be connected to some $w$ not in the cycle, giving a path of length $k+1$, which contradicts maximality.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More generally, $v_1$ cannot be connected to a vertex $v_{i+1}$ with the property that $v_i$ is connected to $v_k$, since otherwise $v_{i+1} v_1 v_2 ... v_i v_k v_{k-1} ... v_{i+2}$ is a cycle of length $k$.  It follows that the sets $\{ v_i : v_1 \sim v_i \}$ and $\{ v_{i+1} : v_i \sim v_k \}$ are disjoint subsets of $\{ v_2, ... v_k \}$, but by assumption their union must have size at least $n$ (this step also works under the weaker assumption that $d_v + d_w \ge n$ for each $v \neq w$); contradiction.&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-12-20T19:57:11.473" LastActivityDate="2010-12-20T19:57:11.473" CommentCount="3" />
  <row Id="14993" PostTypeId="2" ParentId="14971" CreationDate="2010-12-20T19:10:11.153" Score="5" ViewCount="" Body="&lt;p&gt;A simple formula for this is: $$ \text{ New Number } = \text{ Old Number } \times \left(\frac{100 - x}{100}\right)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general if a number is increased or decreased by $x\%$ then the new number is given by: $$ \text{ Old Number } \times \left(\frac{100 \pm x}{100}\right)$$&lt;/p&gt;&#xA;" OwnerUserId="2109" LastActivityDate="2010-12-20T19:10:11.153" CommentCount="1" />
  <row Id="14994" PostTypeId="2" ParentId="14975" CreationDate="2010-12-20T19:21:33.753" Score="1" ViewCount="" Body="&lt;p&gt;What is your sampling interval?  Looking at your second graph, it appears there is something in your acquisition that makes it much more probable to measure times that are at about 7.5 or 8 millisecond intervals than times in between.  One approach is to figure out the spacing and bin at this interval.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The periodogram shows frequencies where there is lots of energy in the data.  It looks like the peak at 1000 would correspond to 1/(8 msec) or about 120 Hz.  Does that correspond to your sampling?&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-12-20T19:21:33.753" CommentCount="1" />
  <row Id="14995" PostTypeId="2" ParentId="14925" CreationDate="2010-12-20T19:40:48.687" Score="6" ViewCount="" Body="&lt;p&gt;Update: This argument only holds for some cases.  See italicized additions below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $S_{n,k}$ denote the number of permutations in which the first $k$ elements are not fixed.  I published an expository paper on these numbers earlier this year.  See &quot;&lt;a href=&quot;http://math.pugetsound.edu/~mspivey/DT.pdf&quot; rel=&quot;nofollow&quot;&gt;Deranged Exams&lt;/a&gt;,&quot; (College Mathematics Journal, 41 (3): 197-202, 2010).  Aravind's formula is in the paper, as are several others involving $S_{n,k}$ and related numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Theorem 7 (which I also mention in &lt;a href=&quot;http://math.stackexchange.com/questions/14477/&quot;&gt;this recent math.SE question&lt;/a&gt;) is relevant to this question.  It's&#xA;$$S_{n+k,k} = \sum_{j=0}^n \binom{n}{j} D_{k+j},$$&#xA;where $D_n$ is the number of derangements on $n$ elements.  See the paper for a simple combinatorial proof of this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $D_n$ grows as $n!$ via $D_n = \frac{n!}{e} + O(1)$ (see Wikipedia's page on the &lt;a href=&quot;http://en.wikipedia.org/wiki/Derangement#Counting_derangements&quot; rel=&quot;nofollow&quot;&gt;derangement numbers&lt;/a&gt;), and &lt;em&gt;if $k$ is much larger than $n$&lt;/em&gt;,&#xA;the dominant terms in the probability&#xA;$\frac{S_{n+k,k}}{(n+k)!}$ are the $j = n$ and $j = n-1$ terms from the Theorem 7 expression.  Thus we have &#xA;$$\frac{S_{n+k,k}}{(n+k)!} \approx \frac{D_{n+k} + n D_{n+k-1}}{(n+k)!} \approx \frac{1}{e}\left(1  + \frac{n}{n+k}\right) \approx e^{-1} e^{\frac{n}{n+k}} = e^\frac{-k}{n+k},$$&#xA;where the second-to-last step uses the first two terms in the Maclaurin series expansion for $e^x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Again, this argument holds only for (in my notation) $k$ much larger than $n$.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="2370" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2011-03-07T03:50:34.430" LastActivityDate="2011-03-07T03:50:34.430" CommentCount="6" />
  <row Id="14996" PostTypeId="1" AcceptedAnswerId="15032" CreationDate="2010-12-20T19:41:41.883" Score="3" ViewCount="166" Body="&lt;p&gt;Let $C: x=x(t), y=y(t), a\le t\le b$ be a $C^1$ closed curve (not necessarily simple).The isoperimetric inequality says that&#xA; $$ A\le \frac{\ell^2}{4\pi},$$ where $$A=\left|\int_C y(t)x&amp;#39;(t) dt\right|$$ is the area enclosed by $C$, and &#xA;$\ell=\int_a^b \sqrt{(x&amp;#39;(t))^2+(y&amp;#39;(t))^2} dt$ is the arc length of $C$. &#xA;My question is how to use this theorem to prove  Wirtinger theorem: If $f(t)$ is a $T$-periodic $C^1$ real-valued function such that&#xA;$\int_0^T f(t) dt=0,$ then $$\int_0^T |f(t)|^2 dt\le \frac{T^2}{4\pi^2}\int_0^T |f&amp;#39;(t)|^2 dt.$$&lt;/p&gt;&#xA;" OwnerUserId="3249" LastEditorUserId="3249" LastEditorDisplayName="" LastEditDate="2010-12-20T22:05:33.280" LastActivityDate="2010-12-21T15:33:35.920" Title="Isoperimetric inequality implies Wirtinger's inequality" Tags="&lt;analysis&gt;&lt;fourier-analysis&gt;&lt;inequality&gt;&lt;fourier-series&gt;" AnswerCount="1" />
  <row Id="14997" PostTypeId="1" AcceptedAnswerId="15003" CreationDate="2010-12-20T19:42:04.367" Score="1" ViewCount="80" Body="&lt;p&gt;Fix $b&amp;gt;1$. Show the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(a) If $m,n,p,q$ are integers, $n&amp;gt;0, q&amp;gt;0$, and $r = m/n = p/q$, prove that $(b^{m})^{1/n} = (b^p)^{1/q}$. So $(b^r) = (b^m)^{1/n}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(b) Prove that $b^{r+s} = b^{r}b^{s}$ for $r,s \in \mathbf{Q}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(c) If $x$ is real, define $B(x)$ to be the set of all numbers $b^t$, there $t$ is rational and $t \leq x$. Prove that $b^r = \sup B(r)$ where $r$ is rational. So $b^x = \sup B(x)$ for every real $x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(d) Show that $b^{x+y} = b^{x}b^{y}$ for all real $x$ and $y$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For (a), we know that $mq = np$. Thus $b^{mq} = b^{np}$. Then $(b^{m})^{q} = (b^{n})^{p}$. Or $(b^{m})^{q} = (b^{p})^{n}$. So the result follows if we note that $m = np/q$ and $n = mq/p$? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For (b) just start with $b^{\frac{mq+pn}{qn}}$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For (c) and (d), we just need to show that $B(x)$ is non-empty. Then use a proof by contradiction? &lt;/p&gt;&#xA;" OwnerUserId="2548" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-12-20T20:11:34.663" LastActivityDate="2010-12-20T20:23:44.030" Title="Exponents and Suprema" Tags="&lt;real-analysis&gt;&lt;homework&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="14998" PostTypeId="1" AcceptedAnswerId="14999" CreationDate="2010-12-20T19:47:50.087" Score="1" ViewCount="103" Body="&lt;p&gt;Hello!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was particularly interested in the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I read this proof, everything seemed fine and logical except one detail (the proof is located &lt;a href=&quot;http://planetmath.org/?method=l2h&amp;amp;from=objects&amp;amp;name=ProofOfNecessaryAndSufficientConditionsForANormedVectorSpaceToBeABanachSpace&amp;amp;op=getobj&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Right after we prove, that the series $\sum_{k=1}^\infty x_{N_m} - x_{N_m+1}$ converges, there is a statement which tells us that the &lt;em&gt;limit&lt;/em&gt; of that series (let's name it $s$) definitely belongs to the initial space $\mathbb{X}$: $s \in \mathbb{X}$.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Why is that? That could probably be very obvious, but unfortunately I can't get it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why should a $lim \hspace{2 mm} \sum_{k=1}^\infty (x_{i} - x_{j})$, where $x_i \in \mathbb{X}$ belong to $\mathbb{X}$ itself?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;What am I missing? :)&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="4971" LastActivityDate="2010-12-20T20:06:22.113" Title="Series sum question" Tags="&lt;functional-analysis&gt;" AnswerCount="1" />
  <row Id="14999" PostTypeId="2" ParentId="14998" CreationDate="2010-12-20T19:52:08.117" Score="5" ViewCount="" Body="&lt;p&gt;Because that is what it means for a series to converge in $X$.  It means that the sequence of partial sums converges to an element of $X$.  In general, convergence of a sequence means that there is some element of your space to which the sequence converges.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To elaborate a bit, to say that a sequence $(y_n)$ in $X$ converges without adding any qualification is another way to say that the sequence converges &lt;em&gt;in&lt;/em&gt; $X$, which means that it has a limit in $X$.  That is, there is an element $L$ of $X$ such that for all $\varepsilon\gt0$ there exists an $N$ such that $n\gt N$ implies $\|y_n-L\|\lt\varepsilon$.  That is precisely the notion used in the hypothesis at your link, and it is the convergence referred to when we say that Cauchy sequences converge in complete spaces.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2010-12-20T20:06:22.113" LastActivityDate="2010-12-20T20:06:22.113" CommentCount="4" />
  <row Id="15000" PostTypeId="2" ParentId="13739" CreationDate="2010-12-20T19:54:05.347" Score="3" ViewCount="" Body="&lt;p&gt;I'll prove both parts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose that for some $N$ and some $p$,&#xA;$(1-\sqrt[n]{a_n})\frac{n}{\log{n}}\geq p&amp;gt;1$ for $n&amp;gt;N$. From here it&#xA;follows that it's enough to show the convergence of $\sum&#xD;&#xA;(1-p\frac{\log{n}}{n})^n$. Since&#xA;$(1-p\frac{\log{n}}{n})^n=\exp(n\log(1-p\frac{\log{n}}{n}))$, I'll use&#xA;an order argument using the series for $\log(1-x)$ and $\exp{x}$.&#xA;These are&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\log(1-x)=-x-\frac{x^2}{2}-\frac{x^3}{3}-...-\frac{x^n}{n} + o(x^n), x\to 0$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\exp{x}=1+x+\frac{x^2}{2}+\frac{x^3}{3!}+...+\frac{x^n}{n!} + o(x^n), x\to 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, $\log(1-p\frac{\log{n}}{n})=-p\frac{\log{n}}{n}-p^2\frac{(\log{n})^2}{n^2}+o(\frac{(\log{n})^2}{n^2})$,&#xA;for $\frac{\log{n}}{n}\to 0$, so&#xA;$n\log(1-p\frac{\log{n}}{n})=-p\log{n}-p^2\frac{(\log{n})^2}{n}+o(\frac{(\log{n})^2}{n})$.&#xA;Therefore&#xA;$\exp(n\log(1-p\frac{\log{n}}{n}))=\exp(-p\log{n}-p^2\frac{(\log{n})^2}{n}+o(\frac{(\log{n})^2}{n}))$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$=\exp(-p\log{n})\exp(-p^2\frac{(\log{n})^2}{n}+o(\frac{(\log{n})^2}{n}))=\frac{1}{n^p}\exp(-p^2\frac{(\log{n})^2}{n}+o(\frac{(\log{n})^2}{n}))$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$=\frac{1}{n^p}\left(&#xD;&#xA;1-p^2\frac{(\log{n})^2}{n}+o(\frac{(\log{n})^2}{n})+o\left(-p^2\frac{(\log{n})^2}{n}+o(\frac{(\log{n})^2}{n})&#xD;&#xA;\right) \right)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$=\frac{1}{n^p}\left(&#xD;&#xA;1-p^2\frac{(\log{n})^2}{n}+o(\frac{(\log{n})^2}{n}) \right)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we have $(1-p\frac{\log{n}}{n})^n=\frac{1}{n^p}-p^2\frac{(\log{n})^2}{n^{p+1}}+o(\frac{(\log{n})^2}{n^{p+1}})$.&#xA;Dividing the LHS of the last equation by&#xA;$\frac{1}{n^p}-p^2\frac{(\log{n})^2}{n^{p+1}}$ and evaluating the&#xA;limit, we find that&#xA;$\lim_{n\to\infty}\frac{(1-p\frac{\log{n}}{n})^n}{\frac{1}{n^p}-p^2\frac{(\log{n})^2}{n^{p+1}}}=1$.&#xA;So the series of the terms in the denominator converges iff if the&#xA;series of the terms in the numerator does. But since $p&amp;gt;1$, we know&#xA;that $\sum\frac{1}{n^p}$ converges, and&#xA;$\sum\frac{(\log{n})^2}{n^{p+1}}$ converges too by comparing with the&#xA;former. Hence $\sum (1-p\frac{\log{n}}{n})^n$ converges and so does&#xA;$\sum a_n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On to the divergence part. It's enough to prove that $\sum&#xD;&#xA;(1-\frac{\log{n}}{n})^n$ diverges. All steps above are still valid for&#xA;$p=1$, so $\lim_{n\to\infty}\frac{(1-\log{n}/n))^n}{1/n-(\log{n})^2/n^2}=1$.&#xA;But $\sum(\log{n})^2/n^2$ can be shown to converge, by comparison&#xA;with the series $\sum\frac{1}{n^\frac{3}{2}}$. This implies that the series in the&#xA;denominator diverges and so does that in the numerator, which is what&#xA;we wanted to prove.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the key here involves manipulation of $o-$little.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;this test is known in&#xA;Russian circles as &quot;Zhame test&quot; &lt;/p&gt;&#xA;" OwnerUserId="4786" LastEditorUserId="4786" LastEditorDisplayName="" LastEditDate="2011-01-19T21:28:53.600" LastActivityDate="2011-01-19T21:28:53.600" CommentCount="2" />
  <row Id="15001" PostTypeId="2" ParentId="14985" CreationDate="2010-12-20T19:54:39.143" Score="2" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://www.math.uri.edu/~eaton/GT4F03.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.math.uri.edu/~eaton/GT4F03.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Has a simple proof: Theorem 6&lt;/p&gt;&#xA;" OwnerUserId="4728" LastActivityDate="2010-12-20T19:54:39.143" />
  <row Id="15002" PostTypeId="2" ParentId="14984" CreationDate="2010-12-20T19:59:05.193" Score="4" ViewCount="" Body="&lt;p&gt;There is a common misconception that strong induction is somehow &quot;stronger&quot; than normal induction, but it isn't really. Strong induction is just the special case of normal induction in which the induction hypothesis $P(n)$ takes the form &quot;for all $m &amp;lt; n$, $Q(m)$&quot;. Thus strong induction is actually a &lt;em&gt;limited&lt;/em&gt; form of the principle of induction, because not all induction hypotheses are of that special form. On the other hand, any proof by strong induction can be trivially rephrased as a proof by &quot;weak&quot; induction. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One reason for the terminological difficulty is that the only place that people talk about &quot;strong induction&quot; is in introductory courses. There, &quot;use strong induction&quot; can be a hint about what sort of induction hypothesis to choose. In these classes, the principle of induction is stated somewhat informally, so it may not be obvious what counts as a legal induction hypothesis in the first place. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In contexts where the principle of induction is stated formally, e.g. Peano arithmetic, nobody talks about strong induction at all. The axiom scheme for induction in Peano arithmetic simply includes all axioms of the form&#xA;$$&#xD;&#xA;(P(0) \land (\forall n)(P(n) \to P(n+1)) \to (\forall n)(P(n))&#xD;&#xA;$$&#xD;&#xA;where $P$ is any formula of Peano arithmetic. This includes formulas of the form $(m &amp;lt; n)\to Q(m)$ as well as all other formulas. &lt;/p&gt;&#xA;" OwnerUserId="630" LastActivityDate="2010-12-20T19:59:05.193" />
  <row Id="15003" PostTypeId="2" ParentId="14997" CreationDate="2010-12-20T20:23:44.030" Score="0" ViewCount="" Body="&lt;p&gt;I've been working on this exercise today as well. I managed to prove (I think) (a), but it's a bit complicated. Let's see:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Lemma 1:&lt;/strong&gt; $(a^x)^{1/x} = a$,  for all $a&amp;gt;0,x\in\mathbb{Q}$. This follows from the uniqueness of positive roots.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Lemma 2:&lt;/strong&gt; $(a^x)^{1/y} = (a^{1/y})^x$, for all $a&amp;gt;0\in\mathbb{Q}$, $x,y\in\mathbb{Z}$. This follows from Theorem 1.2.1 from Rudin's.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Lemma 3:&lt;/strong&gt; $(a^x)^y = a^{xy}$, for all $a\in\mathbb{Q}$, $x,y\in\mathbb{Z}$, since $\prod_1^y\prod_1^x a = \prod_1^{xy} a$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To prove (a), we start from $b = (b^{1/q})^q$ because of (1).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then we raise both sides to the $m$th power to get $b^m = ((b^{1/q})^q)^m= (b^{1/q})^{qm}$, because of (3).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $m/n = p/q$, then $qm = pn$. Then $b^m = (b^{1/q})^{pn}=((b^{1/q})^p)^n$, once again because of (3).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then $b^m = ((b^p)^{1/q})^n$ because of (2). Taking the $n$th root to both sides we get $(b^m)^{1/n} = (b^p)^{1/q}$, which is what we wanted to prove.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To prove (b), consider $b^{mq+np}=b^{mq}b^{np}$ (which holds because $m,n,p,q\in\mathbb{Z}$). This is equivalent to $(b^{m/n})^{nq}(b^{p/q})^{qn}$ because of the previously mentioned lemmas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Taking the $nq$th root to both sides, we end up with $b^{\frac{mq+np}{nq}} = b^{m/n}b^{p/q}$, which is what we wanted to prove.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(c) is proved easily since $b^r \geq x\in B(r)$ and $b^r\in B(r)$, therefore $b^r=\sup(B(r))$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I haven't proved (d) yet but will edit this if I do.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, I'm not sure about this so feedback is really appreciated.&lt;/p&gt;&#xA;" OwnerUserId="1523" LastActivityDate="2010-12-20T20:23:44.030" />
  <row Id="15004" PostTypeId="1" CreationDate="2010-12-20T20:43:42.963" Score="1" ViewCount="79" Body="&lt;p&gt;I completely understand how to find the likelihood functions of simple pdfs. However, how would you attempt to find the likelihood function of a pdf with a negative exponential function with two different means, where the overall mean is just the two means added together??? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The method to do this would be appreciated. thanks.&lt;/p&gt;&#xA;" OwnerUserId="4522" LastActivityDate="2010-12-20T21:41:56.650" Title="Likelihood function estimating two different means." Tags="&lt;statistics&gt;&lt;estimation-theory&gt;" AnswerCount="1" CommentCount="0" />
  <row Id="15005" PostTypeId="2" ParentId="15004" CreationDate="2010-12-20T21:02:32.067" Score="1" ViewCount="" Body="&lt;p&gt;I suspect that you are describing &lt;a href=&quot;http://en.wikipedia.org/wiki/Mixture_model&quot; rel=&quot;nofollow&quot;&gt;mixture model&lt;/a&gt;. There are a lot of ways of estimating these types of models.&lt;/p&gt;&#xA;" OwnerUserId="4742" LastEditorUserId="4742" LastEditorDisplayName="" LastEditDate="2010-12-20T21:41:56.650" LastActivityDate="2010-12-20T21:41:56.650" />
  <row Id="15006" PostTypeId="2" ParentId="14970" CreationDate="2010-12-20T21:03:25.887" Score="9" ViewCount="" Body="&lt;p&gt;Rudin's result is very simply explained: $\ $ given lower and upper approximations to a square-root, we may obtain a better lower approximation $\rm\ p&amp;#39;\ $ by $\:$ &quot;composing&quot; $\:$ them,$\ $ namely:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;THEOREM&lt;/strong&gt; $\rm\displaystyle\quad\ \  q\  &amp;gt;\ \sqrt d\  &amp;gt; \ p\ \ \:\Rightarrow\:\ \ \sqrt d\ &amp;gt; \ p&amp;#39;\ &amp;gt;\ p\quad\ \ for\quad\ p&amp;#39; \:=\ \frac{p\:q+d}{p+q} $&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof:&lt;/strong&gt; $\rm\quad\displaystyle  0\ \: &amp;gt;\ (q-\sqrt d)\ \ (p-\sqrt d)\ =\  p\:q+d - (p+q)\:\sqrt d\ \ \Rightarrow\ \  \sqrt d\ &amp;gt;\ p&amp;#39;$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally $\rm\quad\quad\displaystyle p&amp;#39;-p\ =\ \frac{p\:q+d}{p+q} - p\ =\ \frac{\ d - p^2}{p+q}\: &amp;gt;\ 0\ \ \Rightarrow\ \ p&amp;#39;\ &amp;gt;\ p\quad\quad\ \ $ &lt;strong&gt;QED&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally specializing $\rm\: q = 2 = d\: $ yields Rudin's approximation $\rm\displaystyle\  p&amp;#39;\ =\ \frac{2\:p+2}{\ \:p+2}$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-12-21T01:16:09.827" LastActivityDate="2010-12-21T01:16:09.827" />
  <row Id="15007" PostTypeId="1" CreationDate="2010-12-20T21:14:29.660" Score="3" ViewCount="361" Body="&lt;p&gt;I'm having a little trouble grasping the definition of a topological space in terms of the open-sets. Suppose that we're defining a topology on the real line, in theory could we arbitrarily select the elements of the open-sets, so long as the set of all open sets satisfies the properties of a topological space?&lt;/p&gt;&#xA;" OwnerUserId="4820" LastActivityDate="2010-12-21T09:18:59.600" Title="Topological space definition in terms of open-sets" Tags="&lt;topology&gt;&lt;general-topology&gt;" AnswerCount="3" CommentCount="8" />
  <row Id="15008" PostTypeId="1" AcceptedAnswerId="15026" CreationDate="2010-12-20T22:14:19.137" Score="4" ViewCount="267" Body="&lt;p&gt;Is it possible to find closed form of following sum: $\sum\limits_{k=1}^{n-1}\frac{x^k}{n-k}$ ?&lt;br&gt;&#xA;If not then maybe there is at least some good aproximation?&lt;/p&gt;&#xA;" OwnerUserId="4811" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-12-21T00:45:56.973" LastActivityDate="2010-12-21T12:00:30.550" Title="Sum of series $\frac{x^k}{n-k}$" Tags="&lt;sequences-and-series&gt;" AnswerCount="2" CommentCount="7" />
  <row Id="15009" PostTypeId="2" ParentId="15007" CreationDate="2010-12-20T22:21:43.153" Score="6" ViewCount="" Body="&lt;p&gt;The main drive in mathematics is usually to look at some known object and to generalize the notion. Once you defined &quot;open intervals&quot; and saw that the union of them remains open (in the sense that each point has some interval around it) and that a finite intersection is also open, and so on...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you have, say, a list of properties that some space you know has, and you want things to look more or less like your known space (in this case $\mathbb{R}$). So you start looking for a &quot;right&quot; generalization. In topology it was found out that it is enough to request that the open sets have these three properties, namely - any union, and a finite intersection of open sets is still open, the empty set and the entire space are open as well. Once you have these properties you can talk about convergence and continuity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How did we define convergence of sequences on the real line? $a_n \to a$ if and only if, for an interval around $a$ which is arbitrarily small we have only finitely many elements outside the interval. So we do the same with open sets. Eventually it was apparent that sequences are not always enough, and you need something stronger (namely &lt;a href=&quot;http://en.wikipedia.org/wiki/Net_%28mathematics%29&quot; rel=&quot;nofollow&quot;&gt;nets&lt;/a&gt;). The main idea here, that you took something that well-defined to begin with and sought a way to generalize it so it would be both consistent and useful. And when groups, rings and things like function spaces (that is all the functions from some set $X$ to $\mathbb{R}$ or $\mathbb{C}$) are useful to you, in one way or another - you might want to define some general notion that would capture the essence in all the spaces that you encounter, granted that they have certain properties.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;About continuity, it turns out that one can simply require that the preimage of an open set is an open set, and that is enough for a function to be continuous. Combine this with convergence and you have that by defining the open sets you already know a lot about the space at hand (with the given topology), because you know to say when something converges and when functions are continuous, and that is quite a lot.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To your question, you can define several different topologies on the real line. The standard one is generated by open intervals, while others might have different properties and would be generated by other kind of sets (e.g. sets of the form $[a,b)$) or maybe stated explicitly (e.g. &quot;Every subset is open&quot;), but different topologies might have different characteristics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have two topologies, and you have a basis for each one, if you can show that each open set in one topology contains an open set from the other - you can say that the topologies are essentially the same, and would have the same properties. A good example is when you take $\mathbb{R}^2$ and you take the topology generated by open balls (that is given a point and a radius under the familiar Euclidean distance) and by squares, for example. These two families would generate the same topology. That is, both the topologies will have the same open sets and the same properties.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The main issue, when you approach a new space and try to define topology on it, is just one: You want the topology not to be &quot;too loose&quot; (i.e. not enough open sets, so nothing really converges, and nothing is really continuous) and not &quot;too tight&quot; (i.e. too many open sets, so not enough functions are continuous and not enough sequences/nets would converge.)&lt;br&gt;&#xA;This is not always an easy task, although many times there is a natural way to define things, so not all is lost.&lt;/p&gt;&#xA;" OwnerUserId="622" LastActivityDate="2010-12-20T22:21:43.153" />
  <row Id="15010" PostTypeId="2" ParentId="14311" CreationDate="2010-12-20T22:31:03.100" Score="1" ViewCount="" Body="&lt;p&gt;Edit: This solution is wrong, as explained in the comments below. The proof breaks in the italicized phrase.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $\Sigma = \{ \sigma_i \}$. Take the following grammar of size $O(|\Sigma|)$: $$S \longrightarrow \sigma_i S \sigma_i S | \varepsilon$$&#xA;Clearly in every word produced, every terminal appears an even number of times (proof by induction). In order to generate a word $w$ matching this description, suppose that $w = \sigma_i w&amp;#39;$. Then $\sigma_i$ must appear again in $w&amp;#39;$, say $w = \sigma_i z_1 \sigma_i z_2$. &lt;i&gt;Note that $z_1,z_2$ also match the description.&lt;/i&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now in CNF: $$\begin{align*} S &amp;amp;\longrightarrow \varepsilon \\ S &amp;amp;\longrightarrow S_i S_i \\ S_i &amp;amp;\longrightarrow \sigma_i | A_i T \\ A_i &amp;amp;\longrightarrow \sigma_i \\ T &amp;amp;\longrightarrow S_i S_i \end{align*}$$&lt;/p&gt;&#xA;" OwnerUserId="1277" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2010-12-21T18:36:41.447" LastActivityDate="2010-12-21T18:36:41.447" CommentCount="8" />
  <row Id="15011" PostTypeId="1" CreationDate="2010-12-20T23:11:02.387" Score="7" ViewCount="137" Body="&lt;p&gt;I was sorting out the coins in my loose-change jar the other day, and the following thought crossed my head:  Is it possible to deduce the number of each type of coin in this pile by simply weighing them?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The coins I were counting were Australian 5c, 10c, 20c and 50c.  According to &lt;a href=&quot;http://en.wikipedia.org/wiki/Coins_of_the_Australian_dollar&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;, they weights are:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;5c     2.83 grams&#xA;10c    5.65 grams&#xA;20c   11.30 grams&#xA;50c   15.55 grams&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;But this might be a particularly bad choice of weights for the coins.  Since 565 divides 1130, we can't tell the difference (through weighing) between two 10c coins and one 20c coin.  It seems that two 5c coins and one 10c coin would be hard to distinguish also.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So my question is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Question&lt;/strong&gt;: What would be a better way to designate the weights of these coins so that we could (in most cases) uniquely determine the number of each type of coin in a single weighing?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This would be subject to some practical constraints:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Each coin is a reasonably light, but not too light (e.g. between 2 and 20 grams).&lt;/li&gt;&#xA;&lt;li&gt;If A and B are two multisets of coins, and A and B have equal weights, then |A| and |B| should be very large (more than is likely to be in a typical jar).&lt;/li&gt;&#xA;&lt;li&gt;The scales do not measure with infinite precision.  Coins are not minted with infinitely accurate weights.&lt;/li&gt;&#xA;&lt;li&gt;The weights of the coins must differ by a reasonable amount (e.g. by at least 3 grams).&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="139" LastActivityDate="2010-12-21T00:35:52.787" Title="How can I assign weights to coins so that they can be counted in a single weighing?" Tags="&lt;number-theory&gt;&lt;combinatorics&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="1" />
  <row Id="15012" PostTypeId="2" ParentId="15008" CreationDate="2010-12-20T23:20:40.730" Score="3" ViewCount="" Body="&lt;p&gt;Tomek: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a suggestion, perhaps it is not giving you as explicit a formula as you'd like. Let $f_n(x)$ the sum that you are interested in. Then $$g_n(x):=x^{-n}f_n(x)=\sum_{k=1}^{n-1}\frac{x^{k-n}}{n-k}=\sum_{k=1}^{n-1}\frac{x^{-k}}{k}.$$&#xD;&#xA;Let $y=1/x$, and $h_n(y):=g_n(1/y)$. Differentiate to get $$\sum_{k=1}^{n-1}y^{k-1}=\frac{1-y^{n-1}}{1-y}.$$ This gives you a formula for $h_n(y)$ after integrating (if you try to find it explicitly rather than leaving it in terms of the integral, this formula is a bit messy because it depends on $n$). From this, you easily get the formula for $f_n(x)$.&lt;/p&gt;&#xA;" OwnerUserId="462" LastActivityDate="2010-12-20T23:20:40.730" CommentCount="5" />
  <row Id="15013" PostTypeId="1" AcceptedAnswerId="15014" CreationDate="2010-12-20T23:25:42.787" Score="0" ViewCount="198" Body="&lt;p&gt;I am supposed to prove that $(1/1) + (1/3) + (1/6) + \dots + (1/t_n) &amp;lt; 2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The hint is that &#xA;$$&#xD;&#xA;\frac2{n(n+1)} = 2\left(\frac1{n} - \frac1{n+1}\right)&#xD;&#xA;$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I was thinking that if you take $ 2\left(\frac{1}{t_n} + \frac{1}{t_{n+1}}\right) = \frac1{t_{\frac{n}2}}$ where $n$ is even.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So my logic is you can subtract 1 from both sides multiply by 2 and you are back where you started.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;eg&#xA;$$ 1-1 + \frac{1}{3} + \frac{1}{6} + \frac{1}{10} + \frac{1}{15} + ... &amp;lt; 2 -1$$&#xD;&#xA;$$ 2\left(\frac{1}{2} + \frac{1}{6} + ... \right) &amp;lt; 1 \times 2 $$&#xD;&#xA;$$ 1 - 1 + \frac{1}{3} + ... &amp;lt; 2 -1$$&#xD;&#xA;etc&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this correct logic or am I missing something?  Obviously by the hint this was not the proof they had in mind.  So how would I prove this using the hint?&lt;/p&gt;&#xA;" OwnerUserId="735" LastEditorUserId="735" LastEditorDisplayName="" LastEditDate="2010-12-21T05:12:03.763" LastActivityDate="2010-12-21T15:10:47.817" Title="The sum of the reciprocal of the triangular numbers up to $\frac1{t_n}$ is &lt; 2" Tags="&lt;algebra-precalculus&gt;" AnswerCount="3" />
  <row Id="15014" PostTypeId="2" ParentId="15013" CreationDate="2010-12-20T23:38:54.973" Score="6" ViewCount="" Body="&lt;p&gt;You're supposed to add the reciprocals of the triangular numbers, which are given by $$t_n = \frac{n(n+1)}{2} $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then the hint is just telling you to write in a more convenient way the reciprocal of $t_n$, so that when you add them something nice will happen. For example, try writing the first few sums using the suggested identity and try to discern a pattern from that.&lt;/p&gt;&#xA;" OwnerUserId="900" LastActivityDate="2010-12-20T23:38:54.973" CommentCount="3" />
  <row Id="15015" PostTypeId="2" ParentId="15013" CreationDate="2010-12-20T23:50:50.787" Score="4" ViewCount="" Body="&lt;p&gt;Your statement that $\frac{1}{t_n} + \frac{1}{t_{n+1}} = \left(\frac{1}{t_{n-1}}\right)\left(\frac{1}{2}\right)$ where n&gt;1 is not correct. $\frac{1}{6}+\frac{1}{10}=\frac{4}{15}$ which is not equal to $\frac{1}{3}\cdot \frac {1}{2}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is not clear what you mean &quot;you can always get back to the first term 1 as long as you multiply the inequality by 2.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you say &quot;$(1/t_4)+(1/t_5)\dots(1/t_n) &amp;lt; 2$. Because $(1/t_1) = 1$ and $(1/t_2) + (1/t_3) = 1$&quot;, in fact $(1/t_2) + (1/t_3) = 1/2$.  If you add the three inequalities together, the right side is $3 \frac{1}{2}$ with this correction, but you were asked to prove it with $2$ on the right side.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As Adrian Barquero said, you need to follow the direction of the hint.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-12-20T23:50:50.787" CommentCount="4" />
  <row Id="15016" PostTypeId="1" AcceptedAnswerId="15020" CreationDate="2010-12-20T23:51:07.680" Score="2" ViewCount="85" Body="&lt;p&gt;Another problem on combinatorics. This time I'm asking for a hint and if it is possible a general strategy when dealing with this kind of problems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Show without using the preceding results * that the probability $p_{m}(r,n)=n^{-1}E_{m}(r,n)$ of finding exactly $m$ cells empty satisfies&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$p_{m}(r+1,n)=p_{m}(r,n)\frac{n-m}{n}+p_{m+1}(r,n)\frac{m+1}{n} \qquad \qquad (1)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;* The results which I can't use must be &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$E_{m}(r,n)=\binom{n}{m}A(r,n-m)=\binom{n}{m}\sum_{\nu=0}^{n-m}(-1)^{\nu}\binom{n-m}{\nu}(n-m-\nu)^{r}$$ and by association $$A(r,n+1)=\sum_{k=1}^{r}\binom{r}{k}A(r-k,n)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the way, $A(r,n)$ is equal to $n!S(r,n)$ where $S(r,n)$ are the Stirling numbers of the second kind.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure how to proceed since the 'preceding result' I can't use is basically the definition for $E_{m}(r,n)$. Even if I use such definition in the left hand side of equation $(1)$, I end up with the first term $p_{m}(r,n)\frac{n-m}{m}$ and a nasty second term $\binom{n}{m}\sum_{s=0}^{n-m}(-1)^{s}\binom{n-m}{s}(1-\frac{m+s}{n})^{r}(-\frac{s}{n})$which doesn't resemble the equation $(1)$.&lt;/p&gt;&#xA;" OwnerUserId="2569" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2011-06-01T11:46:22.463" LastActivityDate="2011-06-01T11:46:22.463" Title="Probability of finding exactly $m$ cells empty" Tags="&lt;probability&gt;&lt;combinatorics&gt;&lt;stirling-numbers&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="15017" PostTypeId="1" CreationDate="2010-12-20T23:56:25.030" Score="5" ViewCount="105" Body="&lt;p&gt;$X$ is a random variable, which is not constant. $E[X]=0$. $E[X^4] \leq 2(E[X^2])^2$.&#xA;Let $Y$ be given by: $P(Y=E[X|X \geq 0]) = P(X \geq 0)$ and $P(Y=E[X|X \lt 0]) = P(X \lt 0)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do we necessarily have $E[Y^4] \leq 2(E[Y^2])^2$?&lt;/p&gt;&#xA;" OwnerUserId="4979" LastActivityDate="2010-12-23T15:17:36.993" Title="Effects of condensing a random variable to only 2 possible values" Tags="&lt;probability&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="15018" PostTypeId="2" ParentId="14908" CreationDate="2010-12-21T00:21:35.053" Score="1" ViewCount="" Body="&lt;p&gt;Couldn't resist that the 'solution' that instantly came to mind was to observe that if x is the supplied series, then 2x = 2 - x, from which x = 2/3 follows easily. Also, x = 1 - 2 + 4 - 8 + 16 ... implies 2x + x = 1 implies x = 1/3. BZZZT!!!&lt;/p&gt;&#xA;" OwnerUserId="4980" LastActivityDate="2010-12-21T00:21:35.053" />
  <row Id="15020" PostTypeId="2" ParentId="15016" CreationDate="2010-12-21T00:23:23.743" Score="4" ViewCount="" Body="&lt;p&gt;So (looking at your last problem) $p_m(r,n)$ is the probability of finding m cells out of n empty when you scatter r objects randomly among the bins.  Now you are trying to find an expression for $p_m(r+1,n)$.  Think about how you can get there with $r+1$ objects.  You can either have $m$ cells empty with $r$ objects and put the new one in an occupied bin, or you can have $m+1$ cells empty with $r$ objects and put the new one in an unoccupied bin.  This should lead you to the equation you want.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-12-21T00:23:23.743" CommentCount="3" />
  <row Id="15021" PostTypeId="2" ParentId="15011" CreationDate="2010-12-21T00:35:52.787" Score="1" ViewCount="" Body="&lt;p&gt;A small observation.  For two coins let's suppose we arbitrarily fix the precision at somewhat better than one gram, so let's make the weights $w_1, w_2$ positive integers.  Then the way to minimize redundancy is to require that the weights $w_1, w_2$ are relatively prime.  This is to maximize the size of the smallest nontrivial solution to $w_1 x + w_2 y = 0$.  Within the constraints you describe this means we should take $w_1 = 17, w_2 = 20$; then we can decide the number of coins unambiguously up to the point where we have either $20$ of the first coin or $17$ of the second.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we up the precision to somewhat better than a tenth of a gram then we should instead take something like $w_1 = 16.9, w_2 = 20$; then I think we can decide the number of coins unambiguously up to the point where we have either $200$ of the first coin or $169$ of the second, which is surely good enough for all practical purposes.  So it seems to me that the answer to the problem depends strongly on what precision we can measure weights to.  &lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-12-21T00:35:52.787" CommentCount="2" />
  <row Id="15022" PostTypeId="1" AcceptedAnswerId="15027" CreationDate="2010-12-21T00:44:24.173" Score="0" ViewCount="152" Body="&lt;p&gt;How do I find generators of &lt;a href=&quot;http://en.wikipedia.org/wiki/Octahedral_symmetry&quot; rel=&quot;nofollow&quot;&gt;Octahedral&lt;/a&gt; group when viewing it as a subgroup of S8?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Update&lt;/b&gt;: Alex Bartel's suggestion gives the following permutations and Cayley graph&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://yaroslavvb.com/upload/save/math.se-cube.png&quot;&gt;&#xA;&lt;img src=&quot;http://yaroslavvb.com/upload/save/math.se-cayley.png&quot;&gt;&#xA;&lt;img src=&quot;http://yaroslavvb.com/upload/save/math.se-cayley3d.png&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Update 2&lt;/b&gt;: second suggestion by Alex gives following Cayley graph, which &lt;a href=&quot;http://11011110.livejournal.com/116782.html&quot; rel=&quot;nofollow&quot;&gt;seems to be isomorphic&lt;/a&gt; to the Cayley graph of symmetric group given elsewhere, check!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://yaroslavvb.com/upload/save/math.se-cayley2.png&quot;&gt;&#xA;&lt;img src=&quot;http://yaroslavvb.com/upload/save/math.se-cayley23d.png&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="998" LastEditorUserId="998" LastEditorDisplayName="" LastEditDate="2010-12-21T05:13:00.297" LastActivityDate="2010-12-21T05:13:37.410" Title="How to get generators for a finite point group?" Tags="&lt;group-theory&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="15023" PostTypeId="1" AcceptedAnswerId="15036" CreationDate="2010-12-21T00:45:45.700" Score="3" ViewCount="114" Body="&lt;p&gt;There are many examples in which making more &quot;precise&quot; predictions gives worse performance (e.g. Runge's phenomenon). My professor implied that there was a sound basis for choosing &quot;simple&quot; functions over complex ones in the general case, and that it had to do with information theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anyone know what he was referring to?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an example: consider least square's. Obviously we could find a polynomial of very high degree which has zero error, but we prefer a linear equation with higher error. Why should this be?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I am familiar with some basic notions like entropy, but not much more than that, so simpler explanations would be much preferred. Although I understand that if it's complex, it's complex.)&lt;/p&gt;&#xA;" OwnerUserId="2549" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2010-12-21T06:18:22.123" LastActivityDate="2010-12-21T06:18:22.123" Title="Theoretical basis for overfitting" Tags="&lt;information-theory&gt;&lt;machine-learning&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="15024" PostTypeId="1" CreationDate="2010-12-21T01:41:04.207" Score="1" ViewCount="204" Body="&lt;p&gt;I have a series of tasks where when one task finishes the next task runs, until all of the tasks are done. I need to find the probability that everything will be finished at different points in time. How should I approach this? Is there a way to find this in polynomial time?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The pdfs for how long individual tasks will run have been found experimentally and are not guaranteed to follow any particular type of distribution.&lt;/p&gt;&#xA;" OwnerUserId="4981" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2011-02-19T18:42:53.980" LastActivityDate="2011-05-20T19:59:37.713" Title="Convolution of multiple probability density funtions" Tags="&lt;probability&gt;&lt;probability-theory&gt;" AnswerCount="1" />
  <row Id="15026" PostTypeId="2" ParentId="15008" CreationDate="2010-12-21T01:54:56.020" Score="4" ViewCount="" Body="&lt;p&gt;Here's a very simple yet quite good approximation to $\sum\nolimits_{k = 1}^{n - 1} {\frac{{x^k }}{{n - k}}}$, suitable for $|x|&amp;gt;1$ and $n$ sufficiently large (which is the interesting case, in my opinion).&#xA;First write &#xA;$$&#xD;&#xA;S(n,x) := \sum\limits_{k = 1}^{n - 1} {\frac{{x^k }}{{n - k}}} = x^n \sum\limits_{k = 1}^{n - 1} {\frac{{u^k }}{k}},&#xD;&#xA;$$&#xA;where $u=1/x$. If $|x|&amp;gt;1$, then $|u|&amp;lt;1$; hence $\sum\nolimits_{k = 1}^\infty  {\frac{{u^k }}{k}}  =  - \log (1 - u)$. &#xA;So, noting that $ - \log (1 - \frac{1}{x}) = \log (\frac{x}{{x - 1}})$, this gives rise to the approximation&#xA;$$&#xD;&#xA;S(n,x) \approx x^n \log \bigg(\frac{x}{{x - 1}}\bigg),&#xD;&#xA;$$&#xA;provided that $n$ is sufficiently large. It turns out that this simple approximation is quite good. (You can check by yourself.) See the table for an example.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Rounded values of $S(n,x)$ and its approximation, for $n=10$ and several values of $x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{array}{ccc}x&amp;amp;S(n,x)&amp;amp;x^n \log\left(\frac{x}{x-1}\right)\\-5&amp;amp;-1780484.04&amp;amp;-1780483.95\\-3&amp;amp;-16987.42&amp;amp;-16987.34\\2&amp;amp;709.598&amp;amp;709.783\\&#xD;&#xA;4&amp;amp;301656.39&amp;amp;301656.52\\6&amp;amp;1.102428722 \times 10^7&amp;amp;1.102428734 \times 10^7\end{array}$$&lt;/p&gt;&#xA;" OwnerUserId="2810" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-21T12:00:30.550" LastActivityDate="2010-12-21T12:00:30.550" CommentCount="4" />
  <row Id="15027" PostTypeId="2" ParentId="15022" CreationDate="2010-12-21T02:22:13.360" Score="5" ViewCount="" Body="&lt;p&gt;Observation number one: the group we are talking about is $S_4$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Observation number two: $S_4$ is generated by any 4-cycle and any 3-cycle. Indeed, the group generated by them must be of size at least 12, and it cannot be $A_4$, since the 4-cycle is an odd permutation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So now, just write down what a rotation of order 4 and a rotation of order 3 do to your vertices. E.g. if the vertices around one face are labelled by 1,...,4, and the vertices around the opposite face are 5,...,8, then a rotation of order 4 sits as (1 2 3 4)(5 6 7 8) inside $S_8$. I am sure you can write down yourself what a rotation about a long diagonal looks like.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, you can do it for any other set of generators, but others might be harder to spot. As I wrote in a comment, the group of rotations is realised as $S_4$ by its action on the 4 long diagonals of the cube. This should enable you to write down any given element of $S_4$ in terms of its action on the vertices.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: Since there still seems to be some confusion, I will add this for clarification: when I say that the rotations permute the diagonals, I regard a diagonal as a line segment, which is unchanged if you swap the two end points. In particular, a rotation about an axis that goes through the mid points of two edges (and of course through the centroid of the cube) doesn't fix any vertices, but it fixes two diagonals by swapping the end points, so it induces a 2-cycle in $S_4$. A rotation about a long diagonal of the cube fixes only that diagonal and induces a 3-cycle.&lt;/p&gt;&#xA;" OwnerUserId="3212" LastEditorUserId="3212" LastEditorDisplayName="" LastEditDate="2010-12-21T05:13:37.410" LastActivityDate="2010-12-21T05:13:37.410" CommentCount="8" />
  <row Id="15028" PostTypeId="2" ParentId="15017" CreationDate="2010-12-21T02:52:00.410" Score="3" ViewCount="" Body="&lt;p&gt;Here is a start.  Let $f(x)$ be the probability density function of $X$.  Let $p$ be the probability that $X&amp;lt;0,  p=\int_{-\infty }^0f(x)dx$  $Y$ has only two values, $y_1=\frac{1}{p}\int_{-\infty }^0xf(x)dx$ with probability $p$ and $y_2=\frac{1}{1-p}\int_0^{\infty }xf(x)dx$ with probability $1-p$.  The condition that $E[X]=0$ tells us that the integrals in $y_1$ and $y_2$ are the negative of each other, call them $-a$ and $a$.  So $Y$ is $\frac{-a}{p}$ with probability $p$ and $\frac{a}{1-p}$ with probability $1-p$.  $E[Y^2]=\frac{a^2}{p}+\frac{a^2}{1-p}=\frac{a^2}{p(1-p)}$.  $E[Y^4]=\frac{a^4}{p^3}+\frac{a^4}{(1-p)^3}=\frac{a^4(p^3+(1-p)^3)}{p^3(1-p)^3}=\frac{a^4(1-3p+3p^2)}{p^3(1-p)^3}$&lt;br&gt;&#xA;Wolfram Alpha says $E[Y^4]\leq 2(E[Y^2])^2$ if $\frac{5-\sqrt{5}}{10}\leq p \leq \frac{5+\sqrt{5}}{10}$ or about $0.276393 \leq p \leq 0.723607$ .  Presumably the condition $E[X^4] \leq 2(E[X^2])^2$ can show us that, but I need to think more on it.  &lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-12-21T02:52:00.410" />
  <row Id="15029" PostTypeId="2" ParentId="11468" CreationDate="2010-12-21T03:00:26.217" Score="1" ViewCount="" Body="&lt;p&gt;I'm too new here to comment on @Sivaram's answer, but I believe it to be correct and well explained. For further reference see Stanley's &quot;Twelvefold Way&quot; in Combinatorics at either [1] or the Wikipedia page.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[1] &lt;a href=&quot;http://mathsci.kaist.ac.kr/~drake/pdf/twelvefold-way.pdf&quot; rel=&quot;nofollow&quot;&gt;http://mathsci.kaist.ac.kr/~drake/pdf/twelvefold-way.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4689" LastActivityDate="2010-12-21T03:00:26.217" />
  <row Id="15030" PostTypeId="2" ParentId="15024" CreationDate="2010-12-21T03:01:03.877" Score="0" ViewCount="" Body="&lt;p&gt;I don't know if that is what you are looking for, but if $X_1,\ldots,X_n$ are i.i.d. random variables with mean $\mu$ and (finite) variance $\sigma^2$, then&#xA;$$&#xD;&#xA;{\rm P}\bigg(\sum\limits_{i = 1}^n {X_i }  \le t \bigg) = {\rm P}\bigg(\frac{{\sum\nolimits_{i = 1}^n {X_i }  - n\mu }}{{\sigma \sqrt n }} \le \frac{{t - n\mu }}{{\sigma \sqrt n }}\bigg),&#xD;&#xA;$$&#xD;&#xA;and the random variable $\frac{{\sum\nolimits_{i = 1}^n {X_i }  - n\mu }}{{\sigma \sqrt n }}$ converges to the standard normal distribution as $n \to \infty$. Now, you can estimate the unknown parameters $\mu$ and $\sigma^2$ (assuming the variance is finite), so if $n$ is sufficiently large, you are actually done.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For further details see &lt;a href=&quot;http://en.wikipedia.org/wiki/Central_limit_theorem&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; (note the subsection &lt;strong&gt;Density functions&lt;/strong&gt;). &lt;/p&gt;&#xA;" OwnerUserId="2810" LastActivityDate="2010-12-21T03:01:03.877" CommentCount="4" />
  <row Id="15031" PostTypeId="1" AcceptedAnswerId="15038" CreationDate="2010-12-21T03:06:30.267" Score="2" ViewCount="171" Body="&lt;p&gt;Fix $b&amp;gt;1, y&amp;gt;0$ and prove that there is a unique real $x$ such that $b^x = y$. So for uniqueness, $x_1 &amp;lt; x_2 \Rightarrow b^{x_1} &amp;lt; b^{x_2}$. Then consider the set $S = \{b^t: t \leq x \}$ and work from there? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Source: Chapter 1, Problem 7, &lt;em&gt;Principles of Mathematical Analysis&lt;/em&gt; by Rudin&lt;/p&gt;&#xA;" OwnerUserId="2548" LastEditorUserId="2548" LastEditorDisplayName="" LastEditDate="2010-12-21T03:56:43.860" LastActivityDate="2010-12-21T16:54:12.400" Title="Logarithm and Uniqueness" Tags="&lt;real-analysis&gt;" AnswerCount="2" CommentCount="5" />
  <row Id="15032" PostTypeId="2" ParentId="14996" CreationDate="2010-12-21T03:24:22.390" Score="1" ViewCount="" Body="&lt;p&gt;First thing to note that if we re-parametrize $t = ks$, we have, by the chain rule,  $\frac{d}{ds}f = k (\frac{d}{dt}f)\circ t$. Choose $k = T/(2\pi)$, then the change of variable shows that it is sufficient to prove the claim for the period being $2\pi$. That is, if we can show for any $2\pi$ periodic function with mean 0&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \int_0^{2\pi} f^2 ds \leq \int_0^{2\pi} |f&amp;#39;|^2 ds $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;we'll be done by a re-scaling argument.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $f$ has mean 0, you can write $f = F&amp;#39;$ for $F$ another $2\pi$ periodic function. So the isoperimetric inequality implies&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \left|\int_0^{2\pi} f F&amp;#39; ds \right| \leq \frac{1}{4\pi} \left( \int_0^{2\pi} \sqrt{ (F&amp;#39;)^2 + (f&amp;#39;)^2 } ds \right)^2 $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \int_0^{2\pi} f^2 ds \leq \frac{1}{4\pi} \left( \int_0^{2\pi} \sqrt{f^2 + (f&amp;#39;)^2} ds \right)^2 $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now use Holder's inequality on the finite interval $[0,2\pi]$, we get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \int_0^{2\pi} \sqrt{f^2 + (f&amp;#39;)^2} ds \leq \sqrt{2\pi} \left( \int f^2 + (f&amp;#39;)^2 ds \right) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we get &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \int_0^{2\pi} f^2 ds \leq \frac{1}{2}\int_0^{2\pi} f^2 + (f&amp;#39;)^2 ds $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which, subtracting $\frac12 \int f^2$ from both sides yields the desired inequality. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Now, a short remark on why it is necessary to first use the scaling argument. The principle is the following: Wirtinger's inequality, as discussed in the first paragraph above, is scale invariant: changing the scale of parameter $t$ changes the terms $f^2$ and $T^2 (f&amp;#39;)^2$ equally. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The isoperimetric inequality, however, is not scale invariant in the same way: using the ansatz where $x$ corresponds to $F$ and $y$ corresponds to $F&amp;#39; = f$, you see that a change of parametrization $t \to ks$ will leave $x$ the same while changing $y$ by a multiple factor. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, this means that the depending on scale, the inequality may not be sharp. In other words, the changing of scale corresponds, morally speaking, to changing the $x$ and $y$ directions in different proportions. So if we start out with a circle, which is a maximizer of the isoperimetric inequality, after this funny reparametrisation which scales $x$ and $y$ differently, we end up with an ellipse, which no longer is a maximizer. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The change of scale in the first paragraph allows us to &quot;use a version of the isoperimetric inequality as close to the circle version as possible&quot;. In other words, by isolating the scale you can most efficiently use the isoperimetric inequality, which then allows for a simple proof of Wirtinger's inequality.   &lt;/p&gt;&#xA;" OwnerUserId="1543" LastEditorUserId="1543" LastEditorDisplayName="" LastEditDate="2010-12-21T15:33:35.920" LastActivityDate="2010-12-21T15:33:35.920" CommentCount="5" />
  <row Id="15033" PostTypeId="1" AcceptedAnswerId="15035" CreationDate="2010-12-21T03:29:58.430" Score="1" ViewCount="73" Body="&lt;p&gt;I have one incident edges and multiple outgoing Edges, for which I want to pick an outgoing edge such that the angles between the outgoing edge and the incoming edge is the smallest of all. We know the coordinates for the vertex $V$ .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The angle must start from the incoming edge ($e_1$) and ends at another edge ($e_2$, $e_3$, $e_4$). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, the angle must be form in such a way that the face constructed from $e_1$ to $e_2$ must be in counter-close-wise direction. In other words, the face that you construct when you link all the vertexes $V$ in $e_1$ and $e_i$ together, must be a counter clockwise cycle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in the case below, edge $e_2$ must be picked because the $\theta$ is the smallest.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://docs.google.com/drawings/pub?id=1ducKqeABywKY4A1WS17exNytbvNnDxSAovK0cZnUs3Y&amp;amp;w=960&amp;amp;h=720&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there an algorithm that does this elegantly? The few algorithms I devise are just plain ugly and requires a lot of hacks. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Update: One method that I think of ( and &lt;a href=&quot;http://math.stackexchange.com/questions/15033/find-the-outgoing-edge-with-the-smallest-angles-given-one-incident-edges-and-mul/15035#15035&quot;&gt;proposed below&lt;/a&gt;) is to use the cross product for each edge against the incoming edge $e_1$, and get the minimum $\theta_{i}$. i.e.,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$e_{1} \times e_{i} = |e_{1}||e_{i}| \sin\, \theta_{i}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the problem of this is it doesn't really work. Consider the following test cases:&#xA;$$e_1=(1,0)$$&#xA;$$e_2=(-1/\sqrt(2),1/\sqrt(2) )$$&#xA;$$e_3=(1/\sqrt(2),-1/\sqrt(2))$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One would find that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\theta_2 = -\angle45$$&#xA;$$\theta_3=-\angle45$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both are equal-- even though we know that $e_2$ should be selected. &lt;/p&gt;&#xA;" OwnerUserId="550" LastEditorUserId="550" LastEditorDisplayName="" LastEditDate="2010-12-21T07:33:19.250" LastActivityDate="2011-01-02T09:20:04.803" Title="Find the Outgoing Edge with the Smallest Angles, Given one Incident Edges and Multiple Outgoing Edges" Tags="&lt;geometry&gt;&lt;trigonometry&gt;&lt;euclidean-geometry&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="15034" PostTypeId="2" ParentId="15031" CreationDate="2010-12-21T04:51:04.693" Score="1" ViewCount="" Body="&lt;p&gt;Yes, if you prove that $x\mapsto b^x$ is strictly increasing, that will show that there is at most one solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, your proposal with $S$ does not really work because what you are looking for is $x$; since you do not know ahead of time what $x$ is, you cannot define $S$ in the first place!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instead, consider the set $S = \{ r\in\mathbb{Q} \mid b^r\leq y\}$ (or $\{r\in\mathbb{R}\mid b^r\leq y\}$). &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-21T05:31:24.877" LastActivityDate="2010-12-21T05:31:24.877" />
  <row Id="15035" PostTypeId="2" ParentId="15033" CreationDate="2010-12-21T05:08:32.937" Score="2" ViewCount="" Body="&lt;p&gt;Assuming that finding the coordinates(of your vectors) is not too difficult. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can use the definition of dot and or cross products, and then solve for the angles. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you can for example use $| a \times b | = |a||b| \sin\, \theta$&lt;/p&gt;&#xA;" OwnerUserId="4728" LastEditorUserId="4728" LastEditorDisplayName="" LastEditDate="2010-12-21T07:29:48.597" LastActivityDate="2010-12-21T07:29:48.597" CommentCount="6" />
  <row Id="15036" PostTypeId="2" ParentId="15023" CreationDate="2010-12-21T06:17:28.007" Score="1" ViewCount="" Body="&lt;p&gt;Look up VC theory (named after Vapnik and Chervonenkis), and especially Structural Risk Minimization (SRM).&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-12-21T06:17:28.007" CommentCount="1" />
  <row Id="15037" PostTypeId="2" ParentId="14655" CreationDate="2010-12-21T07:06:05.003" Score="1" ViewCount="" Body="&lt;p&gt;The language described in (2) is not context free. Take the word $a^nba^nba^nb$ for $n$ large enough, and use the pumping lemma to get a word not in the language.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for (1), consider a grammar in CNF (Chomsky Normal Form) generating that language. Let $n = |\Sigma|$. Consider all $S=(2n)!/2^n$ saturated words containing each character twice. Looking at the tree generating each of these words $w$, there must be a symbol $A_w$ which generates a subword $s_w$ of length between $1/3$ of $2/3$ of the entire word (of length $2n$). Suppose $A_u = A_v$. Then $|s_u| = |s_v|$ and $s_u = s_v$ up to permutation, since otherwise we can replace one occurrence by the other to get a word not in the language.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How many times can a given $A_u$ appear? Equivalently, in how many saturated words does $s_u$ appear, up to permutation, somewhere in the word? Any such word can be obtained from $u$ by permuting $s_u$, permuting the rest of the word, any rotating the result, to a total of $2n|s_u|!(2n-|s_u|)! \leq 2n(2n/3)!(4n/3)!$. Therefore the number of different symbols in the language is at least&#xA;$$ \frac{(2n)!/2^n}{2n(2n/3)!(4n/3)!} = O\left(\frac{c^n}{n^{1.5}}\right),$$&#xD;&#xA;where $c = 9/4\sqrt[3]{2} \approx 1.78583$ (asymptotics courtesy of Wolfram Alpha).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to Wikipedia, one can transform a grammar $G$ into a CNF grammar of size $O(|G|^2)$. Therefore this language requires exponentially many symbols for any grammar.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-12-21T07:06:05.003" CommentCount="2" />
  <row Id="15038" PostTypeId="2" ParentId="15031" CreationDate="2010-12-21T07:13:10.753" Score="2" ViewCount="" Body="&lt;p&gt;Actually, the proof is quite easy following his outline:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(a) &lt;strong&gt;Lemma:&lt;/strong&gt; For any positive integer $n$, $b^n -1 \geq n(b-1)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proof: Suppose this holds for $n$. Then $b^n-1\geq n(b-1)\geq n/b(b-1)$, since $b&amp;gt;1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then $b^{n+1} \geq n(b-1)+b$, and so $b^{n+1}-1 \geq n(b-1)+(b-1)$, therefore $b^{n+1}-1\geq (n+1)(b-1)$. Since $b^n -1 \geq n(b-1)$ holds for $n=1$, by induction it holds for any $n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(b) Hence, $b-1 \geq n(b^{1/n}-1)$ if we apply the lemma setting $b&amp;#39; =b^n$ ($b'&gt;1$ since $b&gt;1$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(c) If $t&amp;gt;1$ and $n&amp;gt;\frac{b-1}{t-1}$, then $b^{1/n} &amp;lt; t$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proof: Since $n&amp;gt;\frac{b-1}{t-1}$, then $n(t-1)&amp;gt;(b-1)\geq n(b^{1/n}-1)$. It follows that $b^{1/n} &amp;lt;t$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(d) If $b^w&amp;lt;y$, then $b^{w+(1/n)}&amp;lt;y$ for sufficiently large $n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we set $t=y/b^w$, we see that $t&amp;gt;1$ and we can choose a sufficiently large $n$ such that $n &amp;gt;\frac{b-1}{t-1}$. Because of (c) we know that $b^{1/n}&amp;lt;t$. Multiply both sides by $b^w$ to get $b^{w+1/n}&amp;lt;y$, which is what we wanted to prove.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(e) If $b^w&amp;gt;y$, then $b^{w-(1/n)}&amp;lt;y$ for sufficiently large $n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we set $t=b^w/y$, we see that $t&amp;gt;1$ and we can choose a sufficiently large $n$ such that $n &amp;gt;\frac{b-1}{t-1}$. Because of (c) we know that $b^{1/n}&amp;lt;t$. Then $1&amp;lt;\frac{t}{b^{1/n}}$, so $1&amp;lt;\frac{b^w}{yb^{1/n}}$ and therefore $y&amp;lt;b^{w-(1/n)}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(f) Let $A = \{w : b^w&amp;lt;y\}$, then if $x=\sup A$, $b^x = y$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $b&amp;gt;1, y&amp;gt;0$, choosing a sufficiently small $w$ will get us arbitrarily close to $0$, so $A$ is non-empty. Since $b^w&amp;gt;y$ for a sufficiently large $w$, $A$ is bounded above, and because of $A$ being a subset of $\mathbb{R}$ then $\exists\sup A$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fix $x = \sup A$. Suppose $b^x &amp;lt; y$, then because of (d), $b^{x+(1/n)}&amp;lt;y$ for some $n$. But then $x+1/n &amp;gt; x$ and $x+1/n \in A$, which is a contradiction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose $b^x &amp;gt; y$, then because of (e), $b^{x-(1/n)}&amp;gt;n$ for some $n$. But then $x-(1/n)$ is an upper bound of $A$ smaller than $x$, which is a contradiction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore if $x=\sup A$, then $b^x=y$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(g) The uniqueness of $x$ follows from the fact that there can't be two different least upper bounds for the same set $A$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Feedback is appreciated, thanks.&lt;/p&gt;&#xA;" OwnerUserId="1523" LastEditorUserId="1523" LastEditorDisplayName="" LastEditDate="2010-12-21T16:54:12.400" LastActivityDate="2010-12-21T16:54:12.400" />
  <row Id="15039" PostTypeId="1" CreationDate="2010-12-21T07:41:15.703" Score="-3" ViewCount="650" Body="&lt;p&gt;Dear Professors, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am looking for any &lt;em&gt;&lt;strong&gt;reference articles on the uses of the preposition &quot;of&quot;&lt;/strong&gt;&lt;/em&gt; that are applied most often in mathematics. I once read in the teacher's edition of Pre-algebra: An Accelerated Course by Dolciani that students should be taught, as early as possible, the many uses of the preposition &quot;of&quot; in mathematics, but I have been unable to find anything on this particular question/topic. &lt;em&gt;&lt;strong&gt;Could you please help me find something? (books, articles, etc.)&lt;/strong&gt;&lt;/em&gt;  &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Please note:&#xA; I am &lt;strong&gt;&lt;em&gt;NOT&lt;/em&gt;&lt;/strong&gt; saying that the preposition &quot;of&quot; has a &quot;unique-mathematical-meaning&quot; when used by mathematicians. What I am saying is that of all the general usages that can be found at dictionary.com for &quot;of&quot;, mathematicians &lt;em&gt;CHOOSE&lt;/em&gt; to use a limited subset of them  and then asking &lt;em&gt;WHY&lt;/em&gt; did they choose this specific subset of meanings? This is, in my opinion, a question worth trying to answer because it will help us to more clearly explain things in mathematics.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The teacher's edition of Pre-algebra: An Accelerated Course by Dolciani that I was referencing is the 2006 edition. I am unsure of the exact page number since I had to borrow it through interlibrary loan.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Examples of the uses of this preposition as applied most often in Algebra under the topic heading of &quot;Variables and Equations&quot; include: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) values &lt;em&gt;&lt;strong&gt;of&lt;/strong&gt;&lt;/em&gt; a variable&lt;br&gt;&#xA;2) value &lt;em&gt;&lt;strong&gt;of&lt;/strong&gt;&lt;/em&gt; a numerical expression&lt;br&gt;&#xA;3) sides &lt;em&gt;&lt;strong&gt;of&lt;/strong&gt;&lt;/em&gt; an equation&lt;br&gt;&#xA;4) domain &lt;em&gt;&lt;strong&gt;of&lt;/strong&gt;&lt;/em&gt; a variable&lt;br&gt;&#xA;5) solution &lt;em&gt;&lt;strong&gt;of&lt;/strong&gt;&lt;/em&gt; a sentence&lt;br&gt;&#xA;and my personal favourite,&lt;br&gt;&#xA;6) solution set &lt;em&gt;&lt;strong&gt;of&lt;/strong&gt;&lt;/em&gt; an open sentence: The set &lt;em&gt;&lt;strong&gt;of&lt;/strong&gt;&lt;/em&gt; all solutions &lt;em&gt;&lt;strong&gt;of&lt;/strong&gt;&lt;/em&gt; the sentence. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please note these examples have been taken from Chapter 1 sections 1-1 and 1-3 in &quot;Algebra Structure and Method Book 1&quot; by Brown, Dolciani et al. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If on the other hand, you find it useless that I am trying to explain to others the beauty of mathematics through its choice of vocabulary and why that choice was made as demonstrated by comments about counting occurrences of &quot;of&quot; in mathematical writing, being unduly obsessed, and categorizing what I am trying to do as the subject of a &quot;crank.&quot;  May I respond by a description of mathematics as the art of creating images and establishing relationships, is it not so that the &quot;material&quot;  of this activity are thoughts? And since words convey those thoughts we must &quot;measure&quot; our choice of words carefully for the sake of lucidity. The sheer number of occurrences of &quot;of&quot; in academic writing and in mathematical writing in particular justifies, in my opinion, at least an honest attempt at &quot;looking for any &lt;em&gt;&lt;strong&gt;reference articles on the uses of the preposition &quot;of&quot;&lt;/strong&gt;&lt;/em&gt;...I have been unable to find anything on this particular question/topic. &lt;em&gt;&lt;strong&gt;Could you please help me find something? (books, articles, etc.)&lt;/strong&gt;&lt;/em&gt;&quot; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please note: The primary tag for this thread is a &lt;strong&gt;REFERENCE-REQUEST&lt;/strong&gt;. Reference-requests are for questions that the poster wants to seek for references &lt;strong&gt;(books, articles, etc.)&lt;/strong&gt; in a subject.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yours respectfully.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;" OwnerUserId="5030" LastEditorUserId="5030" LastEditorDisplayName="" LastEditDate="2010-12-25T09:38:53.360" LastActivityDate="2010-12-25T09:38:53.360" CommunityOwnedDate="2010-12-23T12:10:27.240" ClosedDate="2010-12-24T00:14:28.970" Title="Uses of &quot;of&quot; applied most often in math and why this specific choice was made?" Tags="&lt;reference-request&gt;" AnswerCount="3" CommentCount="12" />
  <row Id="15040" PostTypeId="2" ParentId="15007" CreationDate="2010-12-21T07:45:46.190" Score="0" ViewCount="" Body="&lt;p&gt;As others have commented and answered, the answer is, roughly, &quot;yes&quot;. I might add only that to see examples of 'other' topologies, ones in which open sets are not what you may be used to, check out (a) the problems in a point-set topology book or (b) &lt;a href=&quot;http://en.wikipedia.org/wiki/Counterexamples_in_Topology&quot; rel=&quot;nofollow&quot;&gt;the book &lt;em&gt;Counterexamples in Topology&lt;/em&gt;&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="4992" LastActivityDate="2010-12-21T07:45:46.190" />
  <row Id="15041" PostTypeId="1" CreationDate="2010-12-21T07:59:02.607" Score="-2" ViewCount="226" Body="&lt;p&gt;i need to know which number divides 566?&lt;/p&gt;&#xA;" OwnerUserId="4993" LastEditorUserId="2109" LastEditorDisplayName="" LastEditDate="2010-12-21T11:53:53.227" LastActivityDate="2010-12-21T12:34:10.350" ClosedDate="2010-12-21T14:39:18.760" Title="which number divides 566?" Tags="&lt;arithmetic&gt;" AnswerCount="2" CommentCount="6" />
  <row Id="15042" PostTypeId="1" CreationDate="2010-12-21T08:17:26.967" Score="3" ViewCount="211" Body="&lt;p&gt;Hi all!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is the problem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have a square with some numbers from 1..N in some cells. It's needed to determine if it can be completed to a magic square.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Examples:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;2 _ 6       2 7 6&#xA;_ 5 1  &amp;gt;&amp;gt;&amp;gt;  9 5 1&#xA;4 3 _       4 3 8&#xA;&#xA;7 _ _ &#xA;9 _ _  &amp;gt;&amp;gt;&amp;gt;  NO SOLUTION &#xA;8 _ _&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Is this problem NP-complete? If yes, how can I prove it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;P.S.:&lt;/strong&gt;&lt;br&gt;&#xA;I know that I should reduce one of the known NP-complete problems to this one to prove its NP-completeness. The main question is which NP-complete probleme to choose and how to buld polynomial reduction algorithm. Any ideas?&lt;/p&gt;&#xA;" OwnerUserId="4003" LastEditorUserId="4003" LastEditorDisplayName="" LastEditDate="2010-12-21T09:28:02.700" LastActivityDate="2011-04-07T12:34:50.120" Title="Is half-filled magic square problem NP-complete?" Tags="&lt;homework&gt;&lt;np-complete&gt;" AnswerCount="1" CommentCount="11" FavoriteCount="1" />
  <row Id="15044" PostTypeId="2" ParentId="15007" CreationDate="2010-12-21T08:52:11.473" Score="6" ViewCount="" Body="&lt;p&gt;The answer to your specific question is &lt;em&gt;yes&lt;/em&gt;: if you are given a set $X$, then (even in practice!) you may choose any collection $\{U_i\}_{i \in I}$ of subsets of $X$ which (i) contains the empty set and $X$ itself, (ii) is closed under the operation of finite intersections and (iii) is closed under the operation of arbitrary unions, and then this family will be the open sets for a unique topology on $X$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, all of the above is nothing else than the definition of a topology on a set $X$.  But let's be honest: this is quite a profound and general definition, especially at first. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can absolutely define &quot;strange&quot; topologies on the real line.  For instance, a famous one is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Lower_limit_topology&quot; rel=&quot;nofollow&quot;&gt;lower limit topology&lt;/a&gt; in which the open sets are decreed to be unions of half-open intervals of the form $[a,b)$.  It is easy to see that this gives a strictly larger family of open sets than the usual &quot;Euclidean&quot; topology on $\mathbb{R}$, i.e., every open interval $(c,d)$ is an infinite union of half-open intervals, but no $[a,b)$ is an infinite union of open intervals!  This lower limit topology has all sorts of bizarre properties, many of which are detailed in the wikipedia article linked to above.  In fact it is primarily of interest as a &quot;counterexample&quot;, i.e., an example of a topological space that does not have many of the properties of more familiar spaces -- like metrizable spaces -- do.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A brief discussion of the family of all topologies on a fixed set $X$ can be found &lt;a href=&quot;http://math.uga.edu/~pete/TopSection4.pdf&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; (although I do not necessarily recommend these notes for a true beginner; they are somewhat more sophisticated than is necessary for a first pass).  In particular, right at the beginning I have left as a (not that easy) exercise that on any infinite set $X$, there are $2^{2^{\# X}}$ different topologies on $X$.  When $X = \mathbb{R}$, this is a really, really large infinite number, larger than one usually encounters outside of set-theoretic considerations.  One is certainly not going to meet most of these topologies personally.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note though that if you change the topology on &quot;the real numbers&quot;, it loses much of the structure that made it the real numbers in the first place.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On a first pass through topological spaces, you should definitely keep your eye on the &lt;strong&gt;metric topologies&lt;/strong&gt;: i.e., the topology associated to a metric space $(X,d)$ by decreeing a set to be open if it contains an $\epsilon$-ball around each of its points.  For these types of topologies one can use the spatial and analytic intuition developed in one's previous studies in a useful way.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One may well ask: why do we speak of topological spaces rather than just metric spaces?  There are (at least) two good reasons for this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Even if the spaces that we want to work with are &lt;em&gt;metrizable spaces&lt;/em&gt; -- i.e., the topology is derived from a metric -- often it turns out to be the case that the properties of the space that one wants to study depend only on the topology induced by the metric.  For instance, continuous functions and convergent sequences -- two basic concepts in elementary analysis -- are purely topological in this way.  (Arguably the only important concept of basic analysis which is not purely topological is that of Cauchy sequence and the related notion of completeness.  Even these depend only a structure intermediate between the metric and the induced topology, namely the &lt;strong&gt;uniform structure&lt;/strong&gt; of the space.)  If one is really working with topological notions like convergence, after a certain point it becomes simpler not to carry around the metric structure.  The standard example I like to give of this is that of product spaces: if $(X_1,d_1),\ldots,(X_n,d_n)$ are finitely many metric spaces, one would like to discuss concepts like convergence on the product set $X = X_1 \times \ldots \times X_n$.  Experience shows that we want e.g. a sequence in the product to converge iff each of its components converges.  There is one natural topology on the product which gives this convergence, the &lt;strong&gt;product topology&lt;/strong&gt;.  The product topology on a finite product of metric spaces &lt;em&gt;is&lt;/em&gt; a metric topology, but not in a canonical way: even to put a metric on $\mathbb{R}^N$ there are plenty of choices, e.g. the $L^p$ norm $||(x_1,\ldots,x_n)||_p := (|x_1|^p + \ldots + |x_n|^p)^{\frac{1}{p}}$ for every $1 \leq p &amp;lt; \infty$ and also the infinity norm $||(x_1,\ldots,x_n)||_{\infty} = \max_i |x_i|$.  (Kaplansky in his book &lt;em&gt;Set Theory and Metric Spaces&lt;/em&gt; describes this situation as an &quot;embarrassment of riches&quot;.)  When one wants the product of a countably infinite family of metric spaces, there is again an obvious topology which turns out to be induced by a metric, but now it becomes slightly tricky to write down a single metric.  But the product topology exists -- and is very natural and well-behaved -- for &lt;em&gt;any&lt;/em&gt; family of topological spaces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Remarkably -- even spookily -- in many parts of mathematics topological spaces show up which are far from being metrizable.  For instance, non-Hausdorff topologies are useful in functional analysis, algebraic geometry and order theory.  A few examples:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a) As above, any family $\{X_i\}_{i \in I}$ of (say nonempty) topological spaces admits a product space $X = \prod_i X_i$.  Even if all of the spaces $X_i$ are metrizable, then as long as uncountably many of them consist of more than a single point, the product topology will not even be first countable, let alone metrizable.  But still any product of Hausdorff spaces is Hausdorff and any product of compact spaces is compact: these are immensely useful facts.  Taking &quot;big&quot; products of spaces -- and in particular of the interval $[0,1]$ -- comes up much more often in topology and related branches of mathematics than one might at first suspect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;b) A space is &lt;strong&gt;Boolean&lt;/strong&gt; if it is compact, Hausdorff and totally disconnected.  From the perspective of classical geometric topology, such spaces are of rather limited interest.  Indeed if such a space has no isolated points and is metrizable, then it must be homeomorphic to the classical Cantor set.  But M. Stone proved that to every Boolean space one can associate a Boolean ring and conversely: there is a categorical (anti-)equivalence between Boolean spaces and Boolean rings.  (More generally, there is an equivalence between locally compact Hausdorff totally disconnected spaces and Boolean rings-without-multiplicative-identity.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;c) A space is &lt;strong&gt;Alexandroff&lt;/strong&gt; if the intersection of any collection of open sets is again open.  The obvious examples are discrete spaces and finite spaces.  Indeed it is not hard to show that a Hausdorff space is Alexandroff iff it is discrete, so in studying interesting Alexandroff spaces we are leaving Hausdorff spaces far behind. Every $T_0$-Alexandroff space determines a partial ordering on the underlying set, and conversely given a partially ordered set one can define an Alexandroff topology.  In this way one gets an equivalence between $T_0$-Alexandroff spaces and partially ordered sets.  (More generally, there is an equivalence between Alexandroff spaces and quasi-ordered sets, i.e., with the axiom of anti-symmetry removed.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;d) There is up to homeomorphism a unique 2-point space in which exactly one of the points is closed.  In general topology this is sometimes called the &lt;strong&gt;Sierpinski space&lt;/strong&gt;.  It is arguably the single most important topological space in arithmetic geometry, since it is the spectrum (in the sense of Zariski) of any discrete valuation ring.  The germ of the idea of &lt;em&gt;specialization&lt;/em&gt; in algebraic geometry is already contained in this two-point non-Hausdorff space!&lt;/p&gt;&#xA;" OwnerUserId="299" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2010-12-21T09:18:59.600" LastActivityDate="2010-12-21T09:18:59.600" />
  <row Id="15045" PostTypeId="1" AcceptedAnswerId="15056" CreationDate="2010-12-21T09:36:36.690" Score="1" ViewCount="105" Body="&lt;p&gt;I am looking for a generalized normal distribution with mean $\mu$ and variance $\sigma^2$ but with an additional parameter for the kurtosis to add fat-tails. I can't find it - can anybody help?&lt;/p&gt;&#xA;" OwnerUserId="346" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-12-22T04:42:54.460" LastActivityDate="2010-12-22T04:42:54.460" Title="generalized normal distribution with additional kurtosis parameter" Tags="&lt;probability-distributions&gt;&lt;normal-distribution&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="1" />
  <row Id="15047" PostTypeId="2" ParentId="11650" CreationDate="2010-12-21T10:16:22.013" Score="0" ViewCount="" Body="&lt;p&gt;About the connection with  Schrodinger equations of quantum mechanics and certain models in QFT: you can have look to the papers: arXiv:hep-th/9812211,  arXiv:hep-th/9906219&lt;/p&gt;&#xA;" OwnerUserId="4995" LastActivityDate="2010-12-21T10:16:22.013" />
  <row Id="15048" PostTypeId="1" CreationDate="2010-12-21T10:44:56.330" Score="4" ViewCount="113" Body="&lt;p&gt;Suppose I have an inscribed simplex which has n+1 vertices, and the diameter of the hypersphere is d. I have a point x inside this simplex, is it true that the distance between x and x's nearest vertex is not larger than d/2 ?&lt;/p&gt;&#xA;" OwnerUserId="4998" LastEditorUserId="4998" LastEditorDisplayName="" LastEditDate="2010-12-21T11:53:35.173" LastActivityDate="2010-12-27T08:55:33.877" Title="inscribed simplex." Tags="&lt;geometry&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="15050" PostTypeId="2" ParentId="15041" CreationDate="2010-12-21T11:34:39.163" Score="4" ViewCount="" Body="&lt;p&gt;To find out which number divides a certain number,you need to compute the &lt;a href=&quot;http://en.wikipedia.org/wiki/Prime_factor&quot; rel=&quot;nofollow&quot;&gt;prime factorization&lt;/a&gt; of that number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have access to the internet you could try &lt;a href=&quot;http://www.wolframalpha.com/&quot; rel=&quot;nofollow&quot;&gt;wolfram alpha&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For &lt;a href=&quot;http://www.wolframalpha.com/input/?i=factor+566&quot; rel=&quot;nofollow&quot;&gt;$566$&lt;/a&gt;,only $2$ and $283$ divides $566$.Also any number is divisible by $1$ and itself.&lt;/p&gt;&#xA;" OwnerUserId="2109" LastActivityDate="2010-12-21T11:34:39.163" CommentCount="3" />
  <row Id="15051" PostTypeId="1" AcceptedAnswerId="15054" CreationDate="2010-12-21T11:43:57.410" Score="4" ViewCount="80" Body="&lt;p&gt;Please help me in understanding the difference between this two question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;How many ways can $4$ men and $4$ women stand in a line if the women and men alternate ?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;How many ways can $4$ men and $4$ women stand in a line if no two women are together ?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We employ a bit of different strategies for solving them and the answer is also different,but I am not understanding why is this differences.Kindly explain.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I am in bit of a doubt in another similar problem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Find the number of ways of arranging $21$ white balls and $3$ black balls in a row so that no two black balls are together.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the answer should be $21! \times ^{24}P_3$ but this is incorrect.Could somebody explain?&lt;/p&gt;&#xA;" OwnerUserId="3744" LastEditorUserId="3744" LastEditorDisplayName="" LastEditDate="2010-12-21T11:59:10.803" LastActivityDate="2010-12-21T12:07:31.663" Title="Understanding differences" Tags="&lt;combinatorics&gt;" AnswerCount="1" />
  <row Id="15052" PostTypeId="2" ParentId="15041" CreationDate="2010-12-21T11:53:45.737" Score="2" ViewCount="" Body="&lt;p&gt;If you are interested what divides other numbers you can use the free software &quot;sage&quot; or &quot;gp/pari&quot;. The &lt;em&gt;function&lt;/em&gt; you are looking for is &lt;strong&gt;divisors&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;? divisors(566)&lt;br&gt;&#xA;  %1 = [1, 2, 283, 566]&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Links:&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Sage_%28mathematics_software%29&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Sage_%28mathematics_software%29&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/PARI/GP&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/PARI/GP&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: If the computer fail to answer for other numbers, I would pick &lt;strong&gt;1&lt;/strong&gt; or the number itself: &lt;strong&gt;566&lt;/strong&gt; in your case.&lt;/p&gt;&#xA;" OwnerUserId="4750" LastEditorUserId="4750" LastEditorDisplayName="" LastEditDate="2010-12-21T12:34:10.350" LastActivityDate="2010-12-21T12:34:10.350" />
  <row Id="15053" PostTypeId="1" AcceptedAnswerId="15062" CreationDate="2010-12-21T11:53:48.483" Score="2" ViewCount="189" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These are questions given for an assignment and I REALLY need to get a good grade so please could you guys check over them and give some pointers where I've gone wrong? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;thanks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ed: I hope it's ok to overwrite the original post: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_0^\infty {\frac{(2x-8)^{n}}{4^{n}(n^{2}+2)}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;a_n &amp;amp;= \frac{(2x-8)^{n}}{4^{n}(n^{2}+2)}\\&#xA;a_{n+1} &amp;amp;=&#xA;\frac{(2x-8)^{n+1}}{4^{n+1}((n+1)^{2}+2)}.\\&#xA;\left|\frac{a_n}{a_{n+1}}\right| &amp;amp;= \left|\frac{(2x-8)^{n+1}}{4^{n+1}((n+1)^{2}+2)} \times&#xA;\frac{4^{n}(n^{2}+2)}{(2x-8)^{n}}\right|.\\&#xA;\lim_{n \to \infty}\left|\frac{a_n}{a_{n+1}}\right|&amp;amp;= |2x-8| \lim_{n \to&#xA;\infty}\frac{n^{2}+2}{4(n^{2}+2n+3)}\\&#xA;&amp;amp;= |2x-8| \times\frac{1}{4}.&#xA;\end{align}&#xA;Therefore $L = \frac{1}{4}$ and series converges when&#xA;$|2x-8| \times\frac{1}{4} &amp;lt; 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Radius of convergence is $2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Endpoints; $|2x-8| = 4 \rightarrow x = 6$ and $x=2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Testing endpoints. &#xA;When $x = 6$:&#xA;$$\sum_0^\infty {\frac{(2x-8)^{n}}{4^{n}(n^{2}+2)}} = \sum_0^\infty&#xD;&#xA;\frac{1}{n^{2}+2}.$$&#xA;convergent by comparison test with $\sum\frac{1}{n^{2}}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When x = 2:&#xA;$$\sum_0^\infty {\frac{(2x-8)^{n}}{4^{n}(n^{2}+2)}} =&#xD;&#xA;\sum_0^\infty\frac{(-1)^n}{n^{2}+2},$$&#xA;convergent by alternating sequence test and is absolutely convergent. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore the series $$\sum_0^\infty {\frac{(2x-8)^{n}}{4^{n}(n^{2}+2)}}$$&#xA;is convergent for $x\in [2,6]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2)b&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_1^\infty\frac{(-1)^n}{n4^{2n}}(x+3)^n$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;a_n &amp;amp;= \frac{(-1)^n}{n4^{2n}},\\&#xA;a_{n+1} &amp;amp;= \frac{(-1)^{n+1}}{(n+1)4^{2(n+1)}}.\\&#xA;\left|\frac{a_{n+1}}{a_n}\right| &amp;amp;= \left|\frac{(-1)^{n+1}}{(n+1)4^{2(n+1)}} \times&#xA;\frac{n4^{2n}}{(-1)^n}\right|\\&#xA;\lim_{n \to \infty}\left|\frac{a_{n+1}}{a_n}\right| &amp;amp;= |x+3|\lim_{n \to&#xA;\infty}\frac{n}{4^2(n+1)}\\&#xA; &amp;amp;= |x+3|\times\frac{1}{16}.&#xA;\end{align} &#xA;Series converges when $|x+3|\times\frac{1}{16} \lt 1$.&#xA;Radius of convergence \is $16$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Endpoints: $|x+3| = 16 \rightarrow x = 13$ or $-19$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Testing endpoints:&#xA;When $x = 13$, &#xA;$$\sum_1^\infty\frac{(-1)^n(16)^n}{n4^{2n}} =&#xD;&#xA;\sum_1^\infty\frac{(-1)^n)}{n}.$$&#xA;This is an alternating harmonic series and therefore convergent. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When x = -19, &#xA;$$\sum_1^\infty\frac{(-1)^n(-16)^n}{n4^{2n}}{} =&#xD;&#xA;\sum_1^\infty\frac{(-1)^{2n}}{n}.$$&#xA;Harmonic series and therefore divergent. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore the series $$\sum_1^\infty\frac{(-1)^n}{n4^{2n}}(x+3)^n$$&#xA;is convergent for $x \in(-19, 13]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) c)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_0^\infty\frac{(1+3^{n-1})x^n}{(n+1)!}$$&#xA;\begin{align}&#xA;a_n &amp;amp;= \frac{(1+3^{n-1})x^n}{(n+1)!}\\&#xA;a_n+1 &amp;amp;=&#xA;\frac{(1+3^{(n+1)-1})x^{n+1}}{((n+1)+1)!}.\\&#xA;\left|\frac{a_{n+1}}{a_n}\right| &amp;amp;= \left|\frac{(1+3^n)x^{n+1}}{(n+2)!} \times&#xA;\frac{(n+1)!}{(1+3^{n-1})x^n}\right|\\&#xA;\lim_{n \to \infty}\left|\frac{a_{n+1}}{a_n}\right| &amp;amp;= &#xA;|x|\lim_{n \to \infty}&#xA;\frac{1+3^n}{1+3^{n-1}} \times \lim_{n \to \infty} \frac{1}{n+2}\\&#xA;&amp;amp;= |x|\lim_{n \to \infty} \frac{3^{-n}+ 1}{3^{-n}+3^{-1}} \times 0\\&#xA;&amp;amp;=&#xA;|x|\lim_{n \to \infty} \frac{0 + 1}{0 +3^{-1}} \times 0\\&#xA;&amp;amp;= |x|\times3\times0 = 0.&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, $L = 0$ and the series converges absolutely for all $x$. &#xA;The radius of convergence is $\infty$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle\sum_0^\infty\frac{(1+3^{n-1})x^n}{(n+1)!}$ converges absolutely for $x\in(-\infty,\infty)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cheers, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gregg&lt;/p&gt;&#xA;" OwnerUserId="4362" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-22T21:45:22.900" LastActivityDate="2010-12-22T21:45:22.900" Title="radius of convergence of a power series" Tags="&lt;calculus&gt;&lt;homework&gt;&lt;sequences-and-series&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="15054" PostTypeId="2" ParentId="15051" CreationDate="2010-12-21T11:56:15.890" Score="8" ViewCount="" Body="&lt;p&gt;In the first point, the problems are different because the second problem would allow e.g. WMW*MM*WMW while the first would not. (The first is essentially saying &quot;no two women are together &lt;em&gt;and no two men are together&lt;/em&gt;&quot;, while the second only says &quot;no two women are together&quot;.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the second point, I think you're making two mistakes:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;By multiplying by $21!$, and by using $P$ instead of $C$, you're assuming you can tell the difference beween the white balls, and can tell the difference between the black balls. (i.e. B1 B2 W1 W2... is different from B2 B1 W1 W2...) Usually when dealing with balls, they're indistinguishable.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The answer $^{24}C_3$ (i.e. choosing any 3 balls out of the 24 to be black) would be incorrect because it doesn't exclude answers such as BWBBW... where two black balls appear next to each other.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="5001" LastEditorUserId="5001" LastEditorDisplayName="" LastEditDate="2010-12-21T12:07:31.663" LastActivityDate="2010-12-21T12:07:31.663" CommentCount="4" />
  <row Id="15055" PostTypeId="1" AcceptedAnswerId="15269" CreationDate="2010-12-21T12:17:16.163" Score="18" ViewCount="1206" Body="&lt;p&gt;In a family with two children, what are the chances, if one of the children is a girl, that both children are girls?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I just dipped into a book, &lt;em&gt;The Drunkard's Walk - How Randomness Rules Our Lives&lt;/em&gt;, by Leonard Mlodinow, Vintage Books, 2008. On p.107 Mlodinow says the chances are 1 in 3. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems obvious to me that the chances are 1 in 2. Am I correct? Is this not exactly analogous to having a bowl with an infinite number of marbles, half black and half red? Without looking I draw out a black marble. The probability of the second marble I draw being black is 1/2.&lt;/p&gt;&#xA;" OwnerUserId="1964" LastEditorUserId="5363" LastEditorDisplayName="" LastEditDate="2011-04-18T12:52:47.667" LastActivityDate="2011-04-18T12:52:47.667" Title="In a family with two children, what are the chances, if one of the children is a girl, that both children are girls?" Tags="&lt;probability&gt;&lt;faq&gt;" AnswerCount="8" CommentCount="4" FavoriteCount="4" />
  <row Id="15056" PostTypeId="2" ParentId="15045" CreationDate="2010-12-21T12:17:59.453" Score="5" ViewCount="" Body="&lt;p&gt;I think &lt;a href=&quot;http://en.wikipedia.org/wiki/Pearson_distribution&quot; rel=&quot;nofollow&quot;&gt;Pearson type VII distribution&lt;/a&gt; might be what your are looking for. You will need to reparametrize it, but wikipedia page has the necessary formulas.&lt;/p&gt;&#xA;" OwnerUserId="4742" LastActivityDate="2010-12-21T12:17:59.453" CommentCount="2" />
  <row Id="15057" PostTypeId="2" ParentId="15048" CreationDate="2010-12-21T12:24:33.703" Score="6" ViewCount="" Body="&lt;p&gt;Center the hypersphere at the origin, and let $h$ be the (non-negative!) distance between the point $P$ and the center. We can position our hypersphere so that $P$ lies on the $x$ axis with $x$-coordinate $-h$. I'll write $r$ for $d/2$, the radius of the hypersphere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the case $n=3$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose that the distance from a vertex, $V(x,y,z)$, to $P$ is strictly greater than $r$. Then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\text{dist}(V,P)^2=(x+h)^2+y^2+z^2&amp;gt;r^2$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But $V$ is on the (hyper)sphere, so &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x^2 + y^2 + z^2 = r^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xD;&#xA;2hx+h^2 &amp;amp;&amp;gt; 0 \\&#xD;&#xA;\Rightarrow h(x+\frac{h}{2}) &amp;amp;&amp;gt; 0&#xD;&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We see, then, that our non-negative $h$ must be strictly positive, and that $x &amp;gt; -\frac{h}{2} &amp;gt; -h$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if ALL vertices are further than $r$ away from $P$, then the above shows that $P$ is separated from those vertices --and the simplex they determine-- by the (hyper)plane $x=-\frac{h}{2}$. The point cannot lie inside the simplex.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consequently, for a point within the simplex, the distance to &lt;em&gt;at least one&lt;/em&gt; vertex (in particular, the closest) must be no greater than $r$. In the case where $P$ lies at the center of the hypersphere, the distance to any vertex is exactly $r$, so this upper bound on the minimum distance cannot be improved.&lt;/p&gt;&#xA;" OwnerUserId="409" LastEditorUserId="409" LastEditorDisplayName="" LastEditDate="2010-12-22T00:38:37.460" LastActivityDate="2010-12-22T00:38:37.460" CommentCount="1" />
  <row Id="15058" PostTypeId="2" ParentId="15055" CreationDate="2010-12-21T12:42:27.283" Score="18" ViewCount="" Body="&lt;p&gt;In a family with 2 children there are four possibilities:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) the first child is a boy and the second child is a boy (bb)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) the first child is a boy and the second child is a girl (bg)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) the first child is a girl and the second child is a boy (gb)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4) the first child is a girl and the second child is a girl (gg)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since we are given that at least one child is a girl there are three possibilities: bg, gb, or gg. Out of those three possibilities the only one with two girls is gg. Hence the probability is $\frac{1}{3}$.&lt;/p&gt;&#xA;" OwnerUserId="1487" LastActivityDate="2010-12-21T12:42:27.283" CommentCount="5" />
  <row Id="15059" PostTypeId="2" ParentId="15055" CreationDate="2010-12-21T12:47:59.120" Score="7" ViewCount="" Body="&lt;p&gt;Here's some idea.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose you sample $4n$ families with two children, where $n$ is a very large integer. Then, with a very high probability, about $n$ of them have two boys, about $n$ have two girls, and about $2n$ have one girl and one boy.&#xA;Since you ignore the families with two boys, the desired probability is given by $n/(3n) = 1/3$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: The above solution corresponds to the following situation. You visit a large number of families. In each family, you check if there are two kids. If no, you ignore this family. If yes, you check if one of the kids is a girl. If no, you ignore this family. If yes, you check if both kids are girls.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Among all the OP's examples, only one is relevant here. Namely, &quot;I visit this family. I know they have 2 kids. One of them, a girl, comes into the room. The probability that the 2nd kid is also a girl is 1/2, no?&quot;. Well, in this situation the probability is indeed $1/2$, assuming the following interpretation. You visit a large number of families. In each family, you check if there are two kids. If no, you ignore this family. If yes, you wait until one of the kids comes into the room. If it is a boy, you ignore this family; otherwise you check if both kids are girls. It is readily understood that this leads to a probability of $1/2$. &lt;/p&gt;&#xA;" OwnerUserId="2810" LastEditorUserId="2810" LastEditorDisplayName="" LastEditDate="2010-12-21T13:52:48.960" LastActivityDate="2010-12-21T13:52:48.960" CommentCount="16" />
  <row Id="15060" PostTypeId="1" CreationDate="2010-12-21T12:59:42.207" Score="3" ViewCount="102" Body="&lt;p&gt;Hello,&#xA;we all know that 31,331,3331,33331,333331,3333331,33333331 all are primes. Here we prepend the digit 3 to 31, to get a list of 7 primes.This gives me the following thought:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $D = \{\text{all possible nonnull finite digit strings}\}$,  $D&amp;#39; = \{\text{all things in D that do not start with &amp;quot;0&amp;quot;}\}$.  Define a function $m: D&amp;#39; \times D -&amp;gt; N \cup {\infty}$  by: $m(A,B)= |${all prime members of the list AB, AAB, AAAB, &#xA;...up until but not including the first composite member}| (the size of the set).Then:&#xA;Does m ever take the value $\infty$ ?  If not, is it an unbounded function?&lt;/p&gt;&#xA;" OwnerUserId="2394" LastActivityDate="2010-12-21T12:59:42.207" Title="Prepending strings and primes?" Tags="&lt;elementary-number-theory&gt;&lt;number&gt;" CommentCount="7" />
  <row Id="15061" PostTypeId="1" AcceptedAnswerId="15121" CreationDate="2010-12-21T13:09:11.413" Score="8" ViewCount="205" Body="&lt;p&gt;I could use some help with proving the following irreducibility criterion. (It came up in class and got me interested.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let &lt;em&gt;p&lt;/em&gt; be a prime. For an integer $n = p^k n_0$, where &lt;em&gt;p&lt;/em&gt; doesn't divide $n_0$, set: $e_p(n) = k$. Let $f(x) = a_n x^n + \cdots + a_1 x + a_0$ be a polynomial with integer coefficients. If:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$e_p(a_n) = 0$, &lt;/li&gt;&#xA;&lt;li&gt;$e_p(a_i) \geq n - i$, where $i = 1, 2, \ldots, n-1$,&lt;/li&gt;&#xA;&lt;li&gt;$e_p(a_0) = n - 1$,&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;then &lt;em&gt;f&lt;/em&gt; is irreducible over the rationals. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reducing mod &lt;em&gt;p&lt;/em&gt; and mimicking the proof of Eisenstein's criterion doesn't cut it (I think). I also tried playing with reduction mod $p^k$, but got stuck since $Z_{p^k}[X]$ is not a UFD.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, does this criterion has a name?&lt;/p&gt;&#xA;" OwnerUserId="1875" LastActivityDate="2010-12-21T22:56:40.207" Title="An Eisenstein-like irreducibility criterion" Tags="&lt;polynomials&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="15062" PostTypeId="2" ParentId="15053" CreationDate="2010-12-21T13:09:19.167" Score="3" ViewCount="" Body="&lt;p&gt;It seems mostly OK (at a quick glance), but there are few minor issues.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Don't write $|a_n/a_{n+1}|=(\dots)=\lim (\dots)$, because the elements of the sequence are not equal to the limit of the sequence. Either you write &quot;lim&quot; in &lt;em&gt;each&lt;/em&gt; step,&#xA;like $\lim (\dots) =\lim (\dots) = A$, or else use the arrow notation $(\dots) = (\dots) \to A$. (But don't mix the two notations as many people do; that is, don't write &quot;$\lim(\dots) \to A$&quot;!).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Write $x\in [2,6]$, not $x=(2,6)$. Also: $x \in (-19,13]$ (the smallest number first).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the last problem, you've made a mistake: $(n+1)!/(n+2)! = 1/(n+2)$, not $(n+1)/(n+2)$.&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2010-12-21T13:09:19.167" CommentCount="1" />
  <row Id="15063" PostTypeId="2" ParentId="15053" CreationDate="2010-12-21T13:11:50.100" Score="2" ViewCount="" Body="&lt;p&gt;In the first problem, you have the wrong radius of convergence. You forgot that you have a factor of $2$ in the term $|2x-8|$. Second problem is done correctly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the last problem, I would suggest doing the arithmetic correctly. You have a mistake in the computation of the limits (as well as, though it is moot, a mistake in the computation of the series when $x=\frac{1}{3}$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was going to mention the issues Hans Lundmark raised, about confusing the terms of the sequence with the limits and the value of the variable with the interval in which it may lie for convergence.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-12-21T13:11:50.100" CommentCount="3" />
  <row Id="15064" PostTypeId="1" CreationDate="2010-12-21T14:00:47.500" Score="1" ViewCount="162" Body="&lt;p&gt;Hello! How to prove that the group of order 28 which have normal subgroup of order 4 is Abelian group?&lt;/p&gt;&#xA;" OwnerUserId="5004" LastEditorUserId="5004" LastEditorDisplayName="" LastEditDate="2010-12-21T15:05:25.610" LastActivityDate="2010-12-21T15:37:55.723" Title="Proof - Abelian group?" Tags="&lt;homework&gt;&lt;group-theory&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="15065" PostTypeId="2" ParentId="15055" CreationDate="2010-12-21T14:03:29.243" Score="3" ViewCount="" Body="&lt;p&gt;It really depends on what knowledge you have to restrict the probability space.  (The original question is a little ambiguous, but I agree that 1/3 is the best answer.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The marble example and the example of the firstborn daughter are the same situation-- the second trial is independent, and the odds are 1/2.  In these cases you know the first trial is independent, and you can ignore it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The original question intends you to imagine a randomly chosen 2 child family from among all families with at least 1 daughter.  For example, suppose there is a social science study on 2 child families with at least 1 daughter-- in this situation, about 1/3 of the families will be daughter-daughter, 1/3 will be daughter-son, and 1/3 will be son-daughter.  You have to consider the full probability space of two trials (d-d,d-s,s-d,s-s) and eliminate the s-s possibility.  This leaves a 1/3 chance of 2 daughters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For other variations on this question, you have to decide the appropriate probability space before answering.  (One common, confusing variation is to suppose the family has two children, one of which is a son born on a Tuesday-- what are the odds of two sons?)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your example where the first child in the room is a daughter, if we suppose that the order the children enter has nothing to do with sex, then the probability the other child is a daughter is 1/2.  However, maybe the family encourages their daughters to play inside and their sons outside-- then you could assume the other child is a son, since he hasn't come in yet.  Maybe the culture has a tradition that the oldest daughter in the household (if there is a daughter) offers guests drinks-- then you know the family has at least one daughter, and the odds that they have another daughter is 1/3.  Once you've chosen your assumptions, you can define the probability space and compute the answer.&lt;/p&gt;&#xA;" OwnerUserId="1461" LastActivityDate="2010-12-21T14:03:29.243" CommentCount="6" />
  <row Id="15066" PostTypeId="2" ParentId="15013" CreationDate="2010-12-21T14:07:26.210" Score="2" ViewCount="" Body="&lt;p&gt;Using your approach, we can write the proof as follows:&#xA;Choose $k$ such that $2^k+1 \geq n$.  Then $$\sum_{i=1}^n \frac{2}{i(i+1)} \leq \sum_{i=1}^{2^k+1} \frac{2}{i(i+1)}$$  Using the summation formula lets us talk about how many terms are involved and gives a name for the dummy variable.  $$\sum_{i=1}^{2^k+1} \frac{2}{i(i+1)}=1+\frac{1}{2}\sum_{i=1}^{2^{k-1}+1} \frac{2}{i(i+1)}$$ Here is where we used your relation cutting the number of terms in half.  I chose the upper limit of the sum so we can continue the process.  Define $$f(2^k+1) = \sum_{i=1}^{2^k+1} \frac{2}{i(i+1)}$$  We have shown $f(2^k+1)=1+\frac{1}{2}f(2^{k-1}+1)$.  We can continue the recursion to find $$f(2^k+1)=\sum_{i=0}^{k-2} 2^{-i} + \frac{1}{2^{k-1}}f(3)$$  As $f(3)=\frac{3}{2}$, we have $$f(2^k+1)=\sum_{i=0}^{k} 2^{-i} =2-\frac{1}{2^k}\lt 2$$ &lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2010-12-21T15:10:47.817" LastActivityDate="2010-12-21T15:10:47.817" CommentCount="2" />
  <row Id="15067" PostTypeId="2" ParentId="15064" CreationDate="2010-12-21T14:23:05.150" Score="4" ViewCount="" Body="&lt;p&gt;Let $G$ be the group of order $28$, $N$ the normal subgroup of order $4$, and let $K = G/N$ be the quotient group, which has order $7$ and is therefore cyclic.  By the &lt;a href=&quot;http://en.wikipedia.org/wiki/Schur%E2%80%93Zassenhaus_theorem&quot; rel=&quot;nofollow&quot;&gt;Schur-Zassenhaus Theorem&lt;/a&gt;, $G$ is a semidirect product of $N$ and $K$.  It will be abelian iff the semidirect product is actually direct.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to give a nontrivial semidirect product here, it is necessary and sufficient to find a nontrivial homomorphism from &#xA;$K$ into the automorphism group $\operatorname{Aut}(N)$ of $N$, or equivalently an element of order $7$ in $\operatorname{Aut}(N)$.  But $N$ is either cyclic of order $4$ -- in which case $\operatorname{Aut}(N)$ has order $2$ -- or is the Klein group $C_2 \times C_2$, in which case $\operatorname{Aut}(N)$ has order $6$.  Either way there is no element of order $7$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternately, consider what Sylow theory has to say about the Sylow $7$-subgroups. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Note: of course the second method, which is only hinted at, is probably what is intended.  But the two methods interact in an interesting way with regard to the following more general problem.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Suppose $G$ is a finite group of order $np$ with $p$ &#xA;  a prime number which does not divide $n$, and that there is an abelian normal subgroup $N$ of order $n$.  Under what conditions on $N$ and $p$ is $G$ necessarily abelian?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you use Sylow theory, you get that it is enough for $p$ to be sufficiently large compared to $n$.  (One can be explicit; I don't want to be yet to give the OP some time to think about his particular problem.)  Now comparing to the approach of showing that any semidirect product of $N$ with the cyclic group of order $p$ is necessarily direct one gets information about the largest prime $p$ which divides the order of &#xA;the automorphism group $\operatorname{Aut}(N)$.  It could be fun to explore this further.  (More fun than proving something about groups of order $28$, anyway.  This type of overly specific question tends to bore me.)  &lt;/p&gt;&#xA;" OwnerUserId="299" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2010-12-21T15:37:55.723" LastActivityDate="2010-12-21T15:37:55.723" CommentCount="1" />
  <row Id="15068" PostTypeId="2" ParentId="15055" CreationDate="2010-12-21T14:36:32.757" Score="1" ViewCount="" Body="&lt;p&gt;One possible solution for this problem might be though using conditional probability.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $B_i$ and $G_i$ represent $i$th boy and $i$th girl respectively. Here the sample space could be expressed as $S =$ {$(B_1,B_2),(B_1,G_2),(G_1,B_2),(G_1,G_2)$}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $A$ = An event of having both girl chil.&#xA;$B$ =   An event of having at-least one girl child.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$A = ${$(G_1,G_2)$} and $B =${$(B_1,G_2),(G_1,B_2),(G_1,G_2)$}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what you are asking is $$P(A / B) = \frac{P(A \cap B)}{P(B)} = \frac{\frac{1}{4}}{\frac{3}{4}} = \frac{1}{3} $$&lt;/p&gt;&#xA;" OwnerUserId="2109" LastActivityDate="2010-12-21T14:36:32.757" />
  <row Id="15069" PostTypeId="2" ParentId="8140" CreationDate="2010-12-21T14:51:12.283" Score="4" ViewCount="" Body="&lt;p&gt;In &lt;a href=&quot;http://math.stackexchange.com/questions/12488/for-a-planar-graph-find-the-algorithm-that-constructs-a-cycle-basis-with-each-e&quot;&gt; your previous question&lt;/a&gt; better use was made of graph theory terminology:  you refer to the graph as planar and to a cycle basis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming you understand those terms, perhaps I can point out how Mariano's answer builds on the earlier problem.  In fact here you asked, &quot;Why is it that the algorithm to find the face in the case without coordinate is different from the one with coordinates?&quot;  Although this is probably meant rhetorically to explain your assumption &quot;All the coordinates for the [vertices] are known,&quot; it's important to point out that &quot;faces&quot; you've asked to find will depend on where the edges are as well as where the vertices are.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's illustrate with the complete graph on four vertices $K_4$.  Define planar coordinates for the vertices to be corners of the unit square {(0,0),(1,0),(1,1),(0,1)}.  There will be four edges, but we cannot take all of them to be straight line segments in the plane, since doing so produces an intersection (something a planar embedding of graph does not allow).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact depending on how you connect the vertices, you will get different faces.  Draw two squares (with four edges each), and connect one pair of &quot;diagonal&quot; vertices inside one square and the other pair inside the other square.  Now complete both graphs with an outside edge that connects the remaining diagonal pairs, looping over the square in one case and under the square in the other.  Two (or more) cases with different arrangments of &quot;faces&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The point is that knowing a planar embedding well enough to make the &quot;faces&quot; cycle basis you're looking for unique requires you to &quot;know the coordinates&quot; of the edges as well as the vertices.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mariano's answer focuses on the edges.  This has the nice effect of allowing us to count each edge as belonging to exactly two cycles or faces, once for each direction that the edge is travelled.  [No such simple statement will hold for the number of faces that a vertex appears in.]  However to make that &quot;two cycles per edge&quot; work, we must include the &quot;unbounded face&quot; or the cycle that runs all around the outside of the planar graph.  In the simplest case, a triangle say, we have a cycle that goes clockwise around it (following the direction you've used in your diagrams at top).  To get each edge appearing twice, once in each direction, we need to get a counterclockwise cycle.  You can think of one being the &quot;finite&quot; face bounded by the triangle, and the other being the face corresponding to the unbounded region outside the triangle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you ask Mariano to tell you how to &quot;sort the edge[s] cyclically,&quot; I guess there are two aspects of this.  One is to understand what he means by that, and the other is to connect this task with how you &quot;know the coordinates&quot; of the graph including its edges.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What Mariano means is just to know, when we arrive at a vertex along one &quot;inbound edge&quot;, what will be the right &quot;outbound edge&quot; to take next?  Let's suppose you want to construct all the &quot;finite&quot; faces going in the &lt;strong&gt;clockwise&lt;/strong&gt; direction, as in your diagrams.  Then you would need to sort the &lt;strong&gt;edges&lt;/strong&gt; around each vertex into a cyclically linked &quot;list&quot; that goes around the vertex in the &lt;strong&gt;counterclockwise&lt;/strong&gt; direction.  This &quot;list&quot; wraps around; it has no specific beginning or ending edge.  It also does not distinguish the directions of travel along the edge.  It just tells us, for all the edges at a given vertex, which follows which going counterclockwise around the vertex.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we can elaborate Mariano's solution to your question.  For each edge and &lt;em&gt;each direction of travel&lt;/em&gt; along that edge, build a cycle by this algorithm:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Algorithm FollowYourNose&lt;/strong&gt;&lt;br&gt;&#xA;(1) Follow the given edge in the given direction to its end vertex.&lt;br&gt;&#xA;(2) Consult the cyclically linked list of edges at that vertex to determine the next edge.&lt;br&gt;&#xA;(3) Repeat steps (1) and (2) until you return to the original edge.&lt;br&gt;&#xA;(4) Stop.  You have built a cycle that uses the given edge and given direction.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you've done this for each edge and each direction, you will have a list of cycles that contains duplicates.  If a cycle contains $k$ edges, you will have built that same cycle $k$ times.  Discard the duplicates, leaving just one copy of each cycle.  [If it's important for performance reasons, we could keep track of which edges have been travelled in which directions and preemptively avoid creating the duplicates.]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You will then have one copy left of the cycle that corresponds to the unbounded face, the outside of the planar graph.  You want to &lt;strong&gt;remove&lt;/strong&gt; that also.  Where all the cycles that correspond to the bounded faces are going in the clockwise direction, the cycle that corresponds to the unbounded face goes in the counterclockwise direction.  In fact it is algebraically &lt;em&gt;minus&lt;/em&gt; the sum of all the clockwise/bounded face cycles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added:&lt;/strong&gt; A cyclically linked list (or some would write, cyclic linked list) is a special case of &lt;a href=&quot;http://en.wikipedia.org/wiki/Linked_list&quot; rel=&quot;nofollow&quot;&gt; a linked list&lt;/a&gt;.  It means that the linked list comes around in cycle or loop.  No confusion with the &quot;cycles&quot; of the graph is intended here.  We are simply asking for some concrete way to systematically determine how to continue a path in the graph with each vertex we come to.  An informal description would be to say &quot;bear right at each intersection&quot;.  Doing so will of course take us around the block in terms of ordinary street traffic, and it has a similar effect in navigating a planar graph.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What Mariano proposed was to answer the question of &quot;which way to go&quot; by sorting the edges that are incident to each vertex in an order corresponding to the angles at which they reach the vertex.  Think of standing in a intersection (vertex) of two or more streets (edges).  List them (the edges or streets) in the order they appear as you turn counterclockwise through one revolution, and use that list/ordering in the algorithm outlined by Mariano, so that when you enter an intersection (vertex) on one street (edge), you leave along the next one on the list (bear right!).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We should also be a bit careful about the kind of planar graph that the algorithm applies to.  If you are willing to accept certain degeneracies, then planar graph is good enough.  But consider the case of a single edge connecting two vertices.  What is the face?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you wish to avoid such degeneracies, then assume a simple planar graph that is 2-connected.  What 2-connected means is that the graph is connected and remains connected even after any one edge is removed.  &quot;Simple&quot; means that there are no &quot;self-edges&quot; (edge that connects a vertex to itself) nor multiple edges connecting a single pair of vertices.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regarding Update 3:  It is not suggested that you sort edges &quot;according to the angles made between the incoming edge and the outgoing edge.&quot;  You are asked to make for each vertex a list of the edges ordered going counterclockwise around that vertex (a circular list if you will), without any regard to incoming or outgoing.  In your butterfly graph there is only one vertex with more than two edges, so only the list for that vertex requires care in sorting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Worked Examples Involving Butterfly Graph&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's consider first a planar embedding of the butterfly graph similar to what was once part of the Question:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;&lt;pre&gt;&lt;code&gt;   v2    v3    +e1: (x,v1)  -e1: (v1,x)  &#xA;    |\   /|    +e2: (x,v2)  -e2: (v2,x)  &#xA;    | \ / |    +e3: (x,v3)  -e3: (v3,x)  &#xA;    |  X  |    +e4: (x,v4)  -e4: (v4,x)  &#xA;    | / \ |    +e5: (v1,v2) -e5: (v2,v1)  &#xA;    |/   \|    +e6: (v3,v4) -e6: (v4,v3)  &#xA;   v1    v4  &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I've used plus and minus signs to (arbitrarily) assign directions for each of the six edges.  Only the center vertex &lt;strong&gt;x&lt;/strong&gt; has more than two incident edges, so I'll be explicit about ordering the edges (without regard to incoming or outgoing direction) in counterclockwise order around &lt;strong&gt;x&lt;/strong&gt;:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;...e4,e3,e2,e1,e4,...&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It should be clear that this is a &lt;em&gt;circular&lt;/em&gt; list, that it repeats, looping through the same order of edges over and over again, which in data structures we would represent as a cyclically linked list.  Nothing more than the standard apparatus for linked lists is required to do this.  One simply makes the final node of the linked list point back to the first node of the linked list, and voila, a cyclically linked list.  But let's not confuse the details of the implementation with the details of the algorithm.  All that the algorithm requires is how to find the next edge at a vertex which is counterclockwise from a given edge at that vertex.  A cyclically linked list does that function, in that we would search the list linearly for the given edge at the given vertex, then take the next edge on the list.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since there are six edges, there are twelve directed edges, and thus we will apply the algorithm &lt;strong&gt;FollowYourNose&lt;/strong&gt; twelve times (once for each directed edge).  Each time we will get a cycle (a loop on edges).  Cycles may visit a vertex more than once but will never use an edge more than once (assuming the planar graph is simple and 2-connected).  However the twelve results will contain duplicates.  Once the duplicates are removed, there are just three unique cycles left:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;[+e1,+e5,-e2]&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;[+e3,+e6,-e4]&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;[-e1,+e4,-e6,-e3,+e2,-e5]&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It appears from the diagrams provided in the Question, that &quot;faces&quot; are meant to be the bounded components of the plane after the edges of the embedded graph are removed (thereby disconnecting the plane into one unbounded and finitely many bounded components).  So we should remove from the list of three cycles above that one which corresponds to the boundary of the unbounded component.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This distinction can be made algorithmically on the clockwise or counterclockwise orientation of the cycles.  The cycles that give &quot;faces&quot; are going clockwise around their bounded components.  The cycles which is a boundary of the unbounded component goes counterclockwise.  In this case it is the third cycle that should be discarded.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To illustrate how the distinction is made on the basis of clockwise orientation, and not on some other criterion such as which cycle has greatest length, let's look at another embedding of the butterfly graph:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;&lt;pre&gt;&lt;code&gt;           _ x _              +e1: (x,v1)  -e1: (v1,x)  &#xA;         _/ / \ \_            +e2: (x,v2)  -e2: (v2,x)  &#xA;       _/  /   \  \_          +e3: (x,v3)  -e3: (v3,x)  &#xA;     _/   v2___v1   \_        +e4: (x,v4)  -e4: (v4,x)  &#xA;    /                 \       +e5: (v1,v2) -e5: (v2,v1)  &#xA;   v4_________________v3      +e6: (v3,v4) -e6: (v4,v3)  &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have taken care to make the labelling of the edges with respect to the vertices the same here, as well as the orientations of the two 3-cycles in the plane the same.  The algorithm will now produce a different result based on the change in how the edges incident to vertex &lt;strong&gt;x&lt;/strong&gt; are ordered going counterclockwise around that vertex:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;...e4,e2,e1,e3,e4,...&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again we will apply algorithm &lt;strong&gt;FollowYourNose&lt;/strong&gt; to each of the twelve directed edges and obtain twelve cycles, including duplicates.  After eliminating duplicates the following three cycles remain:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;[+e1,+e5,-e2]&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;[+e3,+e6,-e4,+e2,-e5,-e1]&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;[+e4,-e6,-e3]&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is left to the reader to verify which of these has counterclockwise orientation and that it is not the longest of the three cycles.  The one with the counterclockwise orientation is the &quot;outermost&quot; and thus the boundary of the unbounded component of the plane\graph.&lt;/p&gt;&#xA;" OwnerUserId="3111" LastEditorUserId="3111" LastEditorDisplayName="" LastEditDate="2010-12-24T18:49:30.593" LastActivityDate="2010-12-24T18:49:30.593" CommentCount="11" />
  <row Id="15070" PostTypeId="1" AcceptedAnswerId="15071" CreationDate="2010-12-21T15:18:51.957" Score="2" ViewCount="108" Body="&lt;p&gt;Given an alphabet of $k$ symbols, $\lbrace s_1,s_2, \ldots,s_k \rbrace$, how many words of length $n$, $w(n)$, can be generated knowing that, &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Only two of the $k$ available symbols participate;&lt;/li&gt;&#xA;&lt;li&gt;The symbol $s_i$ must appear exactly $n_i$ times and the symbol $s_j$ must appear exactly $n_j$ times and $n_i+n_j=n$ (the length of the word);&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;To ilustrate the problem suppose that the alphabet is $\lbrace A, B, C\rbrace$ and that I want words of length $4$ with two $A$ and two $B$. With the stated restrictions all of the possible words are&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\lbrace AABB, ABAB,ABBA, BBAA, BAAB, BABA \rbrace$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, I need to know what this $w(n)=w(n_i,n_j)$ is. Some references on aproaches on how to solve this kind of problems and related algorithms to generate such words would be apreciated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="1747" LastEditorUserId="1747" LastEditorDisplayName="" LastEditDate="2010-12-22T08:42:47.563" LastActivityDate="2010-12-22T08:42:47.563" Title="How many words of length $n$ can be generated from an alphabet of $k$ symbols with restrictions on the number of symbols." Tags="&lt;combinatorics&gt;" AnswerCount="1" />
  <row Id="15071" PostTypeId="2" ParentId="15070" CreationDate="2010-12-21T15:24:47.177" Score="3" ViewCount="" Body="&lt;p&gt;Hint:  For this problem, you first have to select two of the symbols from your list.  How many ways can that be done?  Should the order matter? Second, you have to select which $n_1$ positions in the word can take the first of the two symbols.  How many ways can that be done?  As the two selections are independent, you then multiply the two numbers.  You also have to worry about factors of 2-is there some way that interchanging the selected letters in the first choice and interchanging the selected positions in the second can get you to the same word in different ways?  This is particularly a problem when $n_1$=$n_2$.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2010-12-21T22:55:51.527" LastActivityDate="2010-12-21T22:55:51.527" CommentCount="0" />
  <row Id="15072" PostTypeId="2" ParentId="942" CreationDate="2010-12-21T15:38:30.933" Score="1" ViewCount="" Body="&lt;p&gt;Sorry for posting again on this old thread but I'd like to know what's wrong, if something, with the following, because I think it should works.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have an open affine covering $X = \bigcup_{i} U_{i}$ with $U_{i} = \mbox{Spec}(A_{i})$. Let $\xi_{1}$ be a closed point in $U_{1}$ (just take a maximal ideal). If $\xi_{1}$ is still closed in $X$ we are done. Otherwise, take $\xi_{2} \neq \xi_{1}$ with $\xi_{2} \in \overline{\left\{\xi_{1}\right\}}$. Hence $\xi_{2} \in U_{i}$ for some $i \neq 1$, say $i = 2$. Then repeat the reasoning above again. The process must stop since there are only finitely many $U_{i}$ and so $X$ has a closed point.&lt;/p&gt;&#xA;" OwnerUserId="5006" LastActivityDate="2010-12-21T15:38:30.933" CommentCount="2" />
  <row Id="15073" PostTypeId="1" AcceptedAnswerId="15099" CreationDate="2010-12-21T16:52:48.843" Score="4" ViewCount="93" Body="&lt;p&gt;Consider the case where we have a bag of 'N' unique marbles, and we randomly select all 'S' possible sets of size 'm' from the bag (with replacement) under the following constraints:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) - Any set of 'm' marbles must have all unique elements, i.e. there must be no duplicate copies of a marble in any given set. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) - The maximum intersection between any two sets, i.e. the number of unique marbles they have in common, can be of &lt;em&gt;at most&lt;/em&gt; size 'k'. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Provided (1) &amp;amp; (2), what is 'S'?  Equivalently, what is the number of possible sets of size 'm' obeying the aforementioned constraints?&lt;/p&gt;&#xA;" OwnerUserId="5009" LastActivityDate="2010-12-21T20:01:10.813" Title="Counting all possible size 'm' sets of 'N' marbles under intersection and membership constraints" Tags="&lt;probability&gt;&lt;counting&gt;" AnswerCount="1" />
  <row Id="15074" PostTypeId="1" CreationDate="2010-12-21T17:16:04.853" Score="0" ViewCount="86" Body="&lt;p&gt;Let $f:(a+bi) \to (a^2+b)+ai$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there an expression for f such that does not use functions Re and Im?&lt;/p&gt;&#xA;" OwnerUserId="2513" LastActivityDate="2010-12-21T17:41:17.410" Title="Expression for function f:(a+bi) -&gt; (a^2+b)+ai" Tags="&lt;complex-analysis&gt;&lt;complex-numbers&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="15075" PostTypeId="1" CreationDate="2010-12-21T17:29:04.793" Score="7" ViewCount="603" Body="&lt;p&gt;&lt;a href=&quot;http://mathoverflow.net/questions/49647/finite-or-infinite-closed&quot; rel=&quot;nofollow&quot;&gt;Crossposted on Mathoverflow.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given a natural number $a$, are there infinitely many natural numbers &lt;i&gt;not&lt;/i&gt; of the form $anm \pm m \pm n$, $n, m$ positive natural?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I give a proof that for $a=6$ the question is equivalent to the twin prime conjecture so it is known that we don't have any proof. But what about other values of $a$?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;There are infinitely many twin primes if and only if there are infinitely many natural numbers that are not of the form $6nm \pm n \pm m$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Proof:&lt;/b&gt; Every number that is not a multiple of $2$ or $3$ is of the form $6N\pm 1$. So the only pairs that are not divisible by $2$ or $3$ are $(6N-1,6N+1)$ for any $N$. Now are there infinitely many such prime pairs (twin primes)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the number $6N-1$ is prime it should not be written as a product of some numbers $6n+1,6m-1$ for any $n,m &amp;gt; 0$. So $(6n+1)(6m-1)=6(6nm-n+m)-1$, which means that $N$ should not be of the form $6nm-n+m$ for any $n,m&amp;gt;0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, if $6N+1$ is a prime it should not be a product of some numbers $(6n-1)(6m-1) =6(6nm-n-m)+1$, or $(6n+1)(6m+1) =6(6nm+n+m)+1$. Which means that we have a prime couple of the form $(6N-1,6N+1)$ if and only if $N$ is not of the form $6nm \pm n \pm m$ for any $n,m$.&lt;/p&gt;&#xA;" OwnerUserId="5010" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-12-22T16:27:58.483" LastActivityDate="2010-12-22T19:06:08.187" Title="Do we have a proof of the infiniteness?" Tags="&lt;number-theory&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="0" />
  <row Id="15076" PostTypeId="2" ParentId="15074" CreationDate="2010-12-21T17:41:17.410" Score="7" ViewCount="" Body="&lt;p&gt;No expression as an analytic function because it does not satisfy the Cauchy-Riemann equations.&lt;/p&gt;&#xA;" OwnerUserId="589" LastActivityDate="2010-12-21T17:41:17.410" CommentCount="0" />
  <row Id="15077" PostTypeId="1" AcceptedAnswerId="15104" CreationDate="2010-12-21T17:49:35.180" Score="1" ViewCount="137" Body="&lt;p&gt;Suppose $z = a+bi$ and $w = u+iv$. Let $\displaystyle a = \left(\frac{|w|+u}{2} \right)^{1/2}$ and $\displaystyle b = \left(\frac{|w|-u}{2} \right)^{1/2}$. Show that $z^2 = w$ if $v \geq 0$ and $(\bar{z})^{2} = w$ if $v \leq 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So $|w| = \sqrt{u^2+v^2}$. So this is just a matter of computing $(a+bi)^2$ and $(a-bi)^2$ and substituting in the values?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Source: Chapter 1, Problem 10 from &lt;em&gt;Principles of Mathematical Analysis&lt;/em&gt; by Rudin&lt;/p&gt;&#xA;" OwnerUserId="2548" LastEditorUserId="2548" LastEditorDisplayName="" LastEditDate="2010-12-21T18:01:19.990" LastActivityDate="2010-12-21T20:46:52.317" Title="Complex Numbers and Square Roots" Tags="&lt;real-analysis&gt;" AnswerCount="2" CommentCount="5" />
  <row Id="15078" PostTypeId="2" ParentId="15039" CreationDate="2010-12-21T18:00:22.707" Score="1" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://www.scribd.com/doc/3573726/ENGLISH-for-Mathematics&quot; rel=&quot;nofollow&quot;&gt;http://www.scribd.com/doc/3573726/ENGLISH-for-Mathematics&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the risk of getting voted down,I am posting a reference used to teach English to Mathematics majors. It may be the resource you are looking for, though I highly doubt that &quot;of&quot; is such a big issue that people have sleepless nights over. However, you certainly seem to, so that is the reason for this reply. &lt;/p&gt;&#xA;" OwnerUserId="4728" LastActivityDate="2010-12-21T18:00:22.707" CommentCount="2" />
  <row Id="15079" PostTypeId="1" CreationDate="2010-12-21T18:20:09.023" Score="11" ViewCount="325" Body="&lt;p&gt;I think it's an interesting coincidence that the &lt;a href=&quot;http://gato-docs.its.txstate.edu/mathworks/Locker_Puzzle.pdf&quot; rel=&quot;nofollow&quot;&gt;locker puzzle&lt;/a&gt; and this &lt;a href=&quot;http://www.msri.org/people/staff/levy/files/Emissary/0308/puzzles.pdf&quot; rel=&quot;nofollow&quot;&gt;puzzle about duplicate array entries (see problem 6b)&lt;/a&gt; have such similar solutions.  &lt;em&gt;Spoiler alert!&lt;/em&gt; Don't read on if you want to solve these puzzles yourself first (they're two of the best puzzles I've ever seen).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In both solutions we consider a collection of labeled boxes, each with a number inside, and then &quot;traverse through boxes&quot; by starting at a given box and using the number inside to decide where to go next.  For example, we might start at box $1$, find the number $5$ inside, proceed to box $5$, find the number $2$ inside, proceed to box $2$, and so forth.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore, in both solutions we &quot;traverse through boxes&quot; for the same reason: we're interested in finding cycles.  More specifically, for the locker puzzle, we're interested in the question, &lt;em&gt;&quot;If we start at box $n$, how many steps does it take to get back to box $n$?&quot;&lt;/em&gt;, and for the duplicate array puzzle, we're interested in the question, &lt;em&gt;&quot;Does there exist $n$ such that (1) if we start at box $1$, we will eventually get to box $n$, and (2) if we start at box $n$, we will eventually get back to box $n$?&quot;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the two puzzles seem quite unrelated at first glance, I'll pose the following question:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Is there a deep reason why &quot;traversing through boxes&quot; (described in the 2nd paragraph above) shows up in the solution for both of these puzzles?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In addition,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Are there other interesting problems for which &quot;traversing through boxes&quot;, whether to find cycles or for any other reason, shows up in the solution?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;[Edit] I mistakenly said that for the locker puzzle, we're interested in the question, &lt;em&gt;&quot;If we start at box $n$, will we eventually get back to box $n$?&quot;&lt;/em&gt;  Instead, the question should be, &lt;em&gt;&quot;If we start at box $n$, how many steps does it take to get back to box $n$?&quot;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[Edit] Thanks for all the great answers so far!  However, I'm still not completely convinced there's nothing more going on here...&lt;/p&gt;&#xA;" OwnerUserId="4904" LastEditorUserId="4904" LastEditorDisplayName="" LastEditDate="2010-12-23T17:11:41.603" LastActivityDate="2010-12-23T17:11:41.603" Title="Two seemingly unrelated puzzles have very similar solutions; what's the connection?" Tags="&lt;group-theory&gt;&lt;graph-theory&gt;&lt;puzzle&gt;" AnswerCount="3" FavoriteCount="1" />
  <row Id="15080" PostTypeId="1" AcceptedAnswerId="15084" CreationDate="2010-12-21T18:24:39.137" Score="1" ViewCount="71" Body="&lt;p&gt;What's the reasoning behind assigning a proposition to a set, like $A=p(x)$ and $B=q(x)$? Is it because any set of elements has some property that defines all elements in it?&#xA;$$&#xD;&#xA;\begin{align*}&#xD;&#xA;A \cup (B \cap C) &amp;amp;= \{x|p(x) \vee (q(x) \wedge r(x))\}= \\\&#xD;&#xA;&amp;amp;= \{x| (p(x)\vee q(x)) \wedge (p(x) \vee r(x)) \} \\\&#xD;&#xA;&amp;amp;= \{x| p(x) \vee q(x)\} \cap  \{x|p(x) \vee r(x) \} \\\&#xD;&#xA;&amp;amp;= (A \cup  B) \cap (A \cup C)&#xD;&#xA;&#xD;&#xA;\end{align*}&#xD;&#xA;$$&#xA;How exactly it works to prove things like this?&lt;/p&gt;&#xA;" OwnerUserId="3509" LastActivityDate="2010-12-21T18:58:42.857" Title="What's the reasoning behind assigning a proposition to a set, like $A=p(x)$ and $B=q(x)$?" Tags="&lt;homework&gt;&lt;elementary-set-theory&gt;" AnswerCount="2" />
  <row Id="15081" PostTypeId="1" AcceptedAnswerId="15083" CreationDate="2010-12-21T18:27:53.713" Score="0" ViewCount="89" Body="&lt;p&gt;Suppose $z$ is a complex number. Prove that there exists an $r \geq 0$ and a complex number $w$ (with $|w| = 1$) such that $z = rw$. Does $z$ uniquely determine $r$ and $w$?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $z = a+bi$. Then $|z| = \sqrt{a^2+b^2}$. So take $r = |z|$. It seems like one can take $w$ to be any complex number such that $|w| = 1$. So I think $z$ uniquely determines $r$ but not $w$. &lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2010-12-22T07:21:04.980" Title="Existence of factors of a complex number" Tags="&lt;real-analysis&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" />
  <row Id="15082" PostTypeId="2" ParentId="15081" CreationDate="2010-12-21T18:33:29.047" Score="1" ViewCount="" Body="&lt;p&gt;No, if you take $w=1+0i$ for example, $wr$ will always be real.  So $w$ must have the same phase as $z$.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-12-21T18:33:29.047" CommentCount="2" />
  <row Id="15083" PostTypeId="2" ParentId="15081" CreationDate="2010-12-21T18:34:25.003" Score="2" ViewCount="" Body="&lt;p&gt;Let $k\in\mathbb{R}, z\in\mathbb{C}, z=a+bi$. Then $kz = ka+kbi$. Therefore $w$ is uniquely determined by $z$, and $w = \frac{a}{r} + \frac{b}{r}i$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can also think of this using vectors in $\mathbb{R}^2$; any vector can be represented as the unit vector with its same angle multiplied by its length.&lt;/p&gt;&#xA;" OwnerUserId="1523" LastEditorUserId="1523" LastEditorDisplayName="" LastEditDate="2010-12-22T07:21:04.980" LastActivityDate="2010-12-22T07:21:04.980" CommentCount="2" />
  <row Id="15084" PostTypeId="2" ParentId="15080" CreationDate="2010-12-21T18:34:58.327" Score="2" ViewCount="" Body="&lt;p&gt;Firstly I want to make clear that the only way to ensure that every set has a property defining it is some trivial property like $A = \{x|x\in A\}$, because there are sets which you cannot define by a simple formula just like that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, to prove equality of sets, one usually shows that both the left hand side is included in the right hand side, and vice versa.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, you take some arbitrary $x\in A\cup (B\cap C)$ and then that means either $x\in A$ or $x\in B\cap C$ that is either $p(x)$ or both $q(x)$ and $r(x)$, that means that $p(x) \vee q(x) \wedge r(x)$ is true, from here you need to either prove it directly or by DeMorgan's laws show that this is equivalent that $(p(x) \vee q(x)) \wedge (p(x) \vee r(x))$ and from here follows that $x\in A\cup B$ and $x\in A\cup C$, that is to say $x\in (A\cup B)\cap (A\cup C)$.&lt;/p&gt;&#xA;" OwnerUserId="622" LastActivityDate="2010-12-21T18:34:58.327" />
  <row Id="15085" PostTypeId="2" ParentId="15055" CreationDate="2010-12-21T18:36:39.293" Score="7" ViewCount="" Body="&lt;p&gt;Just like the better-known &lt;a href=&quot;http://en.wikipedia.org/wiki/Monty_Hall_problem&quot; rel=&quot;nofollow&quot;&gt;Monty Hall problem&lt;/a&gt;, this is a question about knowledge, and knowledge-of-knowledge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the Monty Hall problem, &lt;a href=&quot;http://math.stackexchange.com/questions/886/deal-or-no-deal-does-one-switch-to-avoid-a-goat-should-deal-or-no-deal-be-10/888#888&quot;&gt;the probability depends on whether or not Monty &lt;em&gt;knows&lt;/em&gt; where the car is&lt;/a&gt;.  If Monty knows where the car is, he will &lt;em&gt;always&lt;/em&gt; open a door with a goat, and switching will give you better odds.  If Monty doesn't know where the car is, then he will only open a goat-door 2/3rds of the time, and switching will not give you any better odds.  If you, as the contestant, don't know whether or not &lt;em&gt;Monty&lt;/em&gt; knows, you cannot assign a probability - you must assume one way or the other first.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Likewise, let's change this problem slightly:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Your friend tells you that Mrs. Smith has two children, and at least one is a girl.  What is the probability that Mrs. Smith has two girls?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The important question here is:  &lt;strong&gt;Does your friend know whether or not Mrs. Smith has two girls?&lt;/strong&gt;  There are two possibilities:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Yes, your friend knows the gender of both of Mrs. Smith's children.&lt;/strong&gt;  In this case, there are three equally likely possibilities: {BG, GB, GG} &lt;em&gt;(where BG means she has an older boy and a younger girl)&lt;/em&gt;.  Thus, the probability of two girls is 1/3.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;No, your friend does not know the gender of both of Mrs. Smith's children.&lt;/strong&gt;  In this case, let's say he met only one of the children, and she was a girl.  Whether she was the older or younger sibling, the chances of the other sibling also being a girl are 1/2, so the overall probability is 1/2.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Thus, we have another question about knowledge-of-knowledge.  Like the Monty Hall problem, without knowledge of the asker's knowledge, there is no way to assign a probability.  Most other posts have assumed the person asking has knowledge of the sex of both of the children; however, NotSuper's counter-examples (in the comments above) assume the asker &lt;em&gt;does not&lt;/em&gt; have knowledge of both children, which is where the apparant contradiction lies.&lt;/p&gt;&#xA;" OwnerUserId="136" LastEditorUserId="136" LastEditorDisplayName="" LastEditDate="2010-12-21T23:19:23.767" LastActivityDate="2010-12-21T23:19:23.767" CommentCount="5" />
  <row Id="15086" PostTypeId="1" AcceptedAnswerId="15089" CreationDate="2010-12-21T18:41:58.840" Score="4" ViewCount="240" Body="&lt;p&gt;Are there any advantages or disadvantages in defining a real number in the following ways:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Definition 1&lt;/strong&gt;&#xA;A real number is an object of the form $\lim\limits_{n \to \infty} a_n$ where $(a_n)_{n}^{\infty}$ is a cauchy sequence of rational numbers. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Definition 2&lt;/strong&gt; A real number is a cut in $\mathbb{Q}$. &lt;/p&gt;&#xA;" OwnerUserId="2548" LastEditorUserId="2548" LastEditorDisplayName="" LastEditDate="2010-12-21T18:57:17.607" LastActivityDate="2010-12-21T20:25:09.897" Title="Dedekind Cuts versus Cauchy Sequences" Tags="&lt;real-analysis&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="15087" PostTypeId="2" ParentId="15077" CreationDate="2010-12-21T18:45:31.647" Score="1" ViewCount="" Body="&lt;p&gt;HINT:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The key is:  if $v \geq 0 \Longrightarrow  (v^{2})^\frac{1}{2}=v$ and if $v \leq 0 \Longrightarrow  (v^{2})^\frac{1}{2}=-v$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the first part Note that :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$z^{2}=a^{2}-b^{2}+2ab \imath = \frac{|w|+u}{2}-\frac{|w|-u}{2}+2\imath \left(\frac{|w|^{2}-u^{2}}{4} \right)^\frac{1}{2} = $$ &#xA;$$ u + 2 \imath \left(\frac{\left(\sqrt{u^{2}+v^{2}}\right)^{2}-u^{2}}{4}\right)^\frac{1}{2}= u+2 \imath \left(\frac{v^{2}}{4}\right)^\frac{1}{2} = u + \imath (v^{2})^\frac{1}{2} = u+v\imath = w$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other is same, using $-v$ (in the key) and expanding $\bar{z}$.&lt;/p&gt;&#xA;" OwnerUserId="3421" LastEditorUserId="3421" LastEditorDisplayName="" LastEditDate="2010-12-21T20:36:53.433" LastActivityDate="2010-12-21T20:36:53.433" CommentCount="6" />
  <row Id="15088" PostTypeId="1" AcceptedAnswerId="15093" CreationDate="2010-12-21T18:49:19.413" Score="6" ViewCount="347" Body="&lt;p&gt;I know that if $f$ is a Lebesgue measurable function on $[a,b]$ then there exists a continuous function $g$ such that $|f(x)-g(x)|&amp;lt; \epsilon$ for all $x\in [a,b]\setminus P$ where the measure of $P$ is less than $\epsilon$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This seems to imply that every Lebesgue measurable function on $\mathbb{R}$ is the pointwise limit of continuous functions.  Is this correct? &lt;/p&gt;&#xA;" OwnerUserId="832" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2010-12-21T19:01:12.357" LastActivityDate="2010-12-21T19:02:42.330" Title="Is every Lebesgue measurable function on $\mathbb{R}$ the pointwise limit of continuous functions?" Tags="&lt;real-analysis&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" />
  <row Id="15089" PostTypeId="2" ParentId="15086" CreationDate="2010-12-21T18:52:12.330" Score="6" ViewCount="" Body="&lt;p&gt;They are the same (up to order-preserving isomorphism). I defer to Theorem 15 of a &lt;a href=&quot;http://math.stackexchange.com/questions/11923/completion-of-rational-numbers-via-cauchy-sequences/12009#12009&quot;&gt;previous answer of mine&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moreover, your first definition is wrong. It's supposed to be &lt;em&gt;equivalence classes&lt;/em&gt; of rational Cauchy sequences - otherwise you have a ring but not a field.&lt;/p&gt;&#xA;" OwnerUserId="1269" LastActivityDate="2010-12-21T18:52:12.330" CommentCount="3" />
  <row Id="15091" PostTypeId="2" ParentId="15088" CreationDate="2010-12-21T18:55:42.100" Score="6" ViewCount="" Body="&lt;p&gt;No. Pointwise limits of continuous functions are &lt;a href=&quot;http://en.wikipedia.org/wiki/Baire_function&quot; rel=&quot;nofollow&quot;&gt;Baire functions of class 1&lt;/a&gt; (or 0 it they are simply continuous). The characteristic function of the rationals is measurable but not of Baire class 1. &lt;/p&gt;&#xA;" OwnerUserId="308" LastActivityDate="2010-12-21T18:55:42.100" />
  <row Id="15092" PostTypeId="2" ParentId="15080" CreationDate="2010-12-21T18:58:42.857" Score="2" ViewCount="" Body="&lt;p&gt;As Asaf said, not all sets have a good definition.  But many of the ones we care about do.  In your example, you did not make use of the definition of A as $\{x|p(x)\}$, so could bypass that step and not even define it.  But it is often useful, as when we want $A\cap B$ we can look at p(x) and q(x) and ask what would x have to be to satisfy both.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-12-21T18:58:42.857" />
  <row Id="15093" PostTypeId="2" ParentId="15088" CreationDate="2010-12-21T19:02:42.330" Score="7" ViewCount="" Body="&lt;p&gt;I thought of a worse example.  A pointwise limit of a sequence of continuous functions is Borel measurable, and there are Lebesgue measurable functions that are not Borel measurable.  The characteristic function of any non-Borel set of measure $0$ will do, for example.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem with &quot;this seems to imply&quot; is that &quot;almost everywhere&quot; and &quot;everywhere&quot; are different.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-12-21T19:02:42.330" CommentCount="5" />
  <row Id="15094" PostTypeId="2" ParentId="14952" CreationDate="2010-12-21T19:31:37.013" Score="1" ViewCount="" Body="&lt;p&gt;I found the most beautiful usage of jacobian matrices in studying differential geometry, when one abandons the idea that analysis can be done &quot;only on balls of $\mathbb{R}^n$&quot;. The definition of tangent space in a point $p$ of a manifold $M$ can be given via the kernel of the jacobian of a suitable submersion, or via the image of the differential of a suitable immersion from an open set $U\subseteq\mathbb{R}^{\dim M}$. Quite a simple example, but when I was an undergrad four years ago it gave me the &quot;right&quot; idea of what a linear transformation &lt;em&gt;does&lt;/em&gt; in a differential (analytical) framework.&lt;/p&gt;&#xA;" OwnerUserId="685" LastActivityDate="2010-12-21T19:31:37.013" CommentCount="1" />
  <row Id="15095" PostTypeId="2" ParentId="14311" CreationDate="2010-12-21T19:35:21.327" Score="2" ViewCount="" Body="&lt;p&gt;Edit: Here is a more verbose proof. The original proof is retained below.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Introduction&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Let $\Sigma$ be some alphabet with $n$ symbols. Let $L$ be the language of words in which each symbol appears an even number of times. Suppose $G$ is a grammar for $L$. Our goal is to given an exponential lower bound on the size of $G$ in terms of $n$.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Transition to Chomsky Normal Form&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;According to Wikipedia (and not difficult to prove), there is a grammar $G&amp;#39;$ for $L$ in Chomsky Normal Form such that $|G&amp;#39;| = O(|G|^2)$. We will show an exponential lower bound on $|G&amp;#39;|$, and this will imply an exponential lower bound on $|G|$ (if $|G&amp;#39;| = \Omega(c^n)$ for $c &amp;gt; 1$ then $|G| = \Omega(\sqrt{c}^n)$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For our purposes, a grammar is in Chomsky Normal Form (CNF) if the productions are of the following two types: $A \rightarrow BC$ and $A \rightarrow a$, where $A,B,C$ are non-terminals and $a$ is terminal. Some people also place the restriction that $B,C$ cannot equal the starting symbol $S$, but we do not need this condition.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Word derivation in CNF grammars&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Let $w \in L$. Choose some derivation of $w$ in $G&amp;#39;$ (there might be several sharing the same &lt;i&gt;derivation tree&lt;/i&gt;). The derivation can be thought of as a binary tree (&lt;i&gt;derivation tree&lt;/i&gt;) where each non-leaf node is labeled by a terminal, and each leaf is labeled by a non-terminal; a non-leaf node either has two non-leaf children or one leaf child.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an example, consider the derivation $$S \rightarrow AB \rightarrow ACD \rightarrow^* xyz.$$ The corresponding derivation tree can be described as follows (sorry for the missing graphics):&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The root is labeled $S$.&lt;/li&gt;&#xA;&lt;li&gt;The root has two children labeled $A,B$.&lt;/li&gt;&#xA;&lt;li&gt;The node labeled $B$ has two children labeled $C,D$.&lt;/li&gt;&#xA;&lt;li&gt;The node labeled $A$ has one child labeled $x$, which is a leaf.&lt;/li&gt;&#xA;&lt;li&gt;The node labeled $C$ has one child labeled $y$, which is a leaf.&lt;/li&gt;&#xA;&lt;li&gt;The node labeled $D$ has one child labeled $z$, which is a leaf.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Note that in general, several nodes can share a label (indeed, this is how one proves the pumping lemma).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each node in the tree generates a subword of the original word, which is formed from its descendant leaf nodes in order. Continuing our example:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The root generates the entire word $xyz$.&lt;/li&gt;&#xA;&lt;li&gt;The node labeled $A$ generates the subword $x$.&lt;/li&gt;&#xA;&lt;li&gt;The node labeled $B$ generates the subword $yz$.&lt;/li&gt;&#xA;&lt;li&gt;The node labeled $C$ generates the subword $y$.&lt;/li&gt;&#xA;&lt;li&gt;The node labeled $D$ generates the subword $z$.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;h3&gt;A subword lemma&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Lemma.&lt;/b&gt; Suppose $w \in L$ is of length $m$, and choose some positive $\ell \leq m$. Then there is a node in the derivation tree of $w$ that generates a subword of length $\alpha$ satisfying $\ell \leq \alpha &amp;lt; 2\ell$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Proof.&lt;/b&gt; Run the following algorithm:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Set $v$ to the root of the derivation tree.&lt;/li&gt;&#xA;&lt;li&gt;While the subword generated by $v$ has size at least $2\ell$, replace $v$ by its child whose generated subword is longest.&lt;/li&gt;&#xA;&lt;li&gt;Output the subword generated by $v$.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Note that the node $v$ always has two children, since otherwise it generated a subword of length $1$. If the loop in step (2) never runs then certainly $v$ generates a subword of the requisite length, by our assumption that $\ell \leq m$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Otherwise, consider the last iteration of the loop. Denote by $w(u)$ the subword generated by a node $u$. In the last iteration of the loop, there is a node $v$ with $|w(v)| \geq 2\ell$, with children $v_1,v_2$ such that $|w(v_1)| \geq |w(v_2)|$; we move to $v_1$. Note that $$2\ell \leq |w(v)| = |w(v_1)| + |w(v_2)| \leq 2|w(v_1)|$$ so that $|w(v_1)| \geq \ell$. Since the loop stopped, $|w(v_1)| &amp;lt; 2\ell$ and $w(v_1)$ is of the requisite size.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;The proof&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;We consider the set $P$ of words in $L$ of the form $\pi(\Sigma) \pi(\Sigma)$, where $\pi$ runs over all permutations of $n$ symbols. For example, if $\Sigma = \{0,1\}$ then $$P = \{0101,1010\}.$$ Note that indeed $P \subseteq L$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since all words in $P$ are of length $2n$ (recall $n$ is the size of the alphabet $\Sigma$), using the lemma with $\ell = n/3$ (assume for simplicity that $n$ is divisible by $3$) we get that some node in the derivation tree of any $w \in P$ generates a subword $x(w)$ whose size satisfies $$n/3 \leq |x(w)| &amp;lt; 2n/3.$$ Denote by $s(w)$ the non-terminal generating $x(w)$ (it is the label of the node).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $|x(w)| \leq n$, all symbols in $x(w)$ are distinct: indeed, any subword of $w$ of length $n$ is a cyclic rotation of the original permutation (think of all subwords of length $3$ in $abcabc$: $abc,bca,cab,abc$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Our strategy now will be to show that any single non-terminal can generate only a relatively small number of the $x(w)$; since $P$ is big, it will follow that $G&amp;#39;$ must have lots of non-terminals. More explicitly, let $$S = \{ s(w) : w \in P\}.$$ We can find a list of representative, distinct words $w_1,\ldots,w_{|S|}$ such that $$S = \{ s(w_i) : 1 \leq i \leq |S| \}.$$ We will show that for each $w_i$, the set of words $w \in P$ with $s(w) = s(w_i)$ consists of at most $B$ words ($B$ will be determined below). It will follow that the number of distinct non-terminals is at least $$\frac{|P|}{B} = \frac{n!}{B}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So let $w \in P$, and suppose that $s(w) = s(w_i)$ for some fixed $i$. What this means is that we can replace $x(w)$ in $w$ with $x(w_i)$ and still get a word in $L$. The word $w$ has a decomposition $w = lx(w)r$. Every symbol appearing in $x(w)$ appears exactly once in $lr$, and every other symbol appears exactly twice in $lr$. Note that this defines the word $x(w)$ up to permutation, i.e. since we assumed that $lx(w_i)r \in L$, then $x(w_i)$ must be a permutation of $x(w)$ (more accurately, the set of (unique) symbols occurring in $x(w),x(w_i)$ must be the same).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We will now bound the number of words in $P$ that have a subword of length $x(w_i)$ which is a permutation of $x(w_i)$. Any such word $w$ can be broken as $lyzr$ where $yz$ is a permutation of $x(w_i)$ and $|ly|=|zr|=n$; in fact, since $w \in P$ then $ly = zr$. There are $|x(w_i)|!$ possible choices for $yz$. Together, these define $|x(w_i)|$ points of the permutation $\pi(\Sigma) = ly$. For the rest of the points we have $(n-|x(w_i)|)!$ possible choices. Finally, $yz$ can start in at most $n$ &quot;cyclic&quot; locations inside $\pi$, and so $$B \leq n|x(w_i)|!(n-|x(w_i)|)!.$$ When is this expression maximized, as a function of $\alpha = |x(w_i)|$? Taking the logarithm of the bound, $$\log B \leq \log 2n + \log \alpha! + \log (n-\alpha)!.$$ The factorial function (in fact, the Gamma function) is log-convex, and so subject to the condition $n/3 \leq \alpha \leq 2n/3$, this expression is maximized for $\alpha=n/3$ or $\alpha =2n/3$ (the value is the same). That is $$B \leq n(n/3)!(2n/3)!.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Plugging the value of $B$ that we have laboriously estimated, we conclude that the number of different non-terminals in $G&amp;#39;$ must be at least $$\frac{|P|}{B} = \frac{n!}{n(n/3)!(2n/3)!} = \frac{1}{n} \binom{n}{n/3}.$$ Now we will use Stirling's approximation $$m! \sim \sqrt{2\pi m} (m/e)^m.$$ Plugging the approximation, we get&#xA;$$\frac{n!}{2n(n/3)!(2n/3)!} \sim \frac{\sqrt{2\pi n}(n/e)^n}{2n\sqrt{2\pi(n/3)}(n/3e)^{n/3} \sqrt{2\pi(2n/3)}(2n/3e)^{2n/3}}.$$ All powers of $n$ and $e$ cancel, and we're left with the following $$\frac{n!}{n(n/3)!(2n/3)!} \sim \frac{1}{n\sqrt{4\pi n/9}} 3^{n/3} (3/2)^{2n/3} = \Omega\left(n^{-3/2} c^n \right),$$ where the constant $c$ is $$c = 3^{1/3} (3/2)^{2/3} = \frac{3}{2^{2/3}} &amp;gt; 1.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, the original grammar $G$ must be of size at least $\Omega(n^{-3/4} \sqrt{c}^n)$.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Exactly 0 or 2 occurrences&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Another questioned concerned the language $L&amp;#39;$ of all words which contain exactly $0$ or $2$ occurrences of each symbol in $\Sigma$. The proof given in the preceding section applies equally well to $L&amp;#39;$ since the set of words $P$ we used belongs to $L&amp;#39;$; the rest of the proof uses the fact that certain words are not in $L$, and since $L \supset L&amp;#39;$, they are a fortiori not in $L&amp;#39;$.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Generalization&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;We have seen that the lower bound holds for both $L$ and $L&amp;#39;$. When does the proof hold for a language $\Lambda$? We used only two properties of $\Lambda$:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$\Lambda$ contains $P$.&lt;/li&gt;&#xA;&lt;li&gt;If $w_1,w_2 \in P$, $x_1,x_2$ are the subwords of $w_1,w_2$ chosen using the subword lemma, and $w_1(x_1:=x_2),w_2(x_2:=x_1) \in \Lambda$ (i.e. the word resulting by replacing $x_1$ with $x_2$ in $w_1$, and the word resulting by replacing $x_2$ with $x_1$ in $w_2$) then $x_2$ is a permutation of $x_1$.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;When will the second condition fail? The words $x_1,x_2$ both have no repeated symbols. If they are not permutations of each other then $w_1(x_1:=x_2)$ will have a symbol repeated $1$ or $3$ times. So words containing such symbols must lie out of $\Lambda$. Considering $w_2(x_2:=x_1)$, we see that it is enough to exclude one of the possibilities. We get the following theorem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Theorem.&lt;/b&gt; Suppose $\Lambda$ is a language containing $P$ in which no word contains any symbol exactly once, or in which no word contains any symbol exactly $3$ times. Then any grammar for $\Lambda$ must have size $\Omega(n^{-3/4} C^n)$, where $C = \sqrt{3}/\sqrt[3]{2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an example, the theorem applies for the language in which the number of occurrences of each symbol lies in $S$ as long as $2 \in S$ and $1,3 \notin S$. For example, $L$ corresponds to $S = \{0,2,4,6,\ldots\}$ and $L&amp;#39;$ corresponds to $S = \{0,2\}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We leave the reader the following easy generalizations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Theorem.&lt;/b&gt; Suppose $\Lambda$ is a language containing $\{ \pi(\Sigma)^\ell \}$ for some $\ell &amp;gt; 0$, in which no word contains any symbol exactly $\ell + \epsilon$ times, where $\epsilon$ is either $1$ or $-1$. Then any grammar for $\Lambda$ must have size $\Omega(n^{-3/4} C^n)$, where $C = \sqrt{3}/\sqrt[3]{2}$.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h3&gt;Original, condensed proof&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Let $L$ be the language of words in which each symbol appears an even number of times. We give an exponential lower bound (in $n = |\Sigma|$) on the size of a Chomsky Normal Form (CNF) grammar for $L$. This implies an exponential lower bound on the size of any context-free grammar for $L$, since (according to Wikipedia) the blowup to CNF is at most quadratic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given any derivation of a word $w$ of length $m$ and any $\ell &amp;lt; m/2$ we can always find a non-terminal symbol in the derivation which derives a subword $x$ of $w$ of length $\ell \leq |x| &amp;lt; 2\ell$: start from the root of the derivation, and always pick the larger branch. The first time we hit a branch smaller than $2\ell$, it must be at least $\ell$ in size.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the set of all $S = n!$ words in $L$ of the form $\pi\pi$ (all of size $2n$), where $\pi$ is a permutation of $\Sigma$. For each such word $w$ we can find a symbol $s_w$ generating a subword $x_w$ of size $n/3 \leq |x_w| &amp;lt; 2n/3$. Note that all symbols in $x_w$ are distinct. Therefore if $s_u = s_v$ then $x_u,x_v$ must be permutations of each other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given $w$, for how many words $u$ can it be that $s_w = s_u$? To get from $w$ to $u$ we need to permute $x_w$, to permute the rest of the permutation, and to choose the starting location of $x_u$ within the permutation generating $w_u$. All in all, the number of possibilities is at most $$M = n|x_w|!(n-|x_w|)! \leq n(n/3)!(2n/3)!.$$ Therefore the number of distinct symbols in the grammar is at least $$\frac{S}{M} = \frac{n!}{n(n/3)!(2n/3)!} = O\left(\frac{c^n}{n^{1.5}}\right),$$&#xA;where $c = 3/2^{2/3} \approx 1.88988$ (thanks to Wolfram Alpha).&lt;/p&gt;&#xA;" OwnerUserId="1277" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2010-12-31T21:58:47.790" LastActivityDate="2010-12-31T21:58:47.790" CommentCount="4" />
  <row Id="15096" PostTypeId="2" ParentId="15039" CreationDate="2010-12-21T19:42:06.960" Score="2" ViewCount="" Body="&lt;p&gt;When students are learning to solve word problems, they are usually taught to look for keywords that identify the operation. The word &quot;of&quot; sometimes signals a multiplication, especially multiplication by a fraction. Examples: half of the books are missing; 70% of Earth's surface is covered in water. But it is used in many other ways: a function of x; a set of natural numbers; 4 out of 5 dentists agree.&lt;/p&gt;&#xA;" OwnerUserId="857" LastActivityDate="2010-12-21T19:42:06.960" CommentCount="1" />
  <row Id="15097" PostTypeId="1" AcceptedAnswerId="15103" CreationDate="2010-12-21T19:43:35.007" Score="6" ViewCount="91" Body="&lt;p&gt;When solving questions like these:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Let $f(x)$ be a real function. Find&#xA;  $f(0.1)$ using its Taylor expansion&#xA;  such that the error is less than&#xA;  $10^{-3}$. Find the lowest degree of&#xA;  Taylor polynomial needed.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(if someone can rephrase the question so its more clear that would be great. I didn't quite nail the translation to English)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We were explained that usually the process of solving this is finding some degree that works. Then, depending on how 'far' you are from the error term, you start trying lower degrees. When some degree doesn't work anymore, you say the one before it is the minimal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was wondering though, why the implicit assumption that the remainder function is decreasing? i.e. if degree $k$ doesn't suffice, $1, 2, ..., k-2, k-1$ won't work either. Our professor said that we can rely on this because most functions we're dealing with comply with this property. Why is that?&lt;/p&gt;&#xA;" OwnerUserId="3237" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-21T23:43:42.960" LastActivityDate="2010-12-22T00:17:53.063" Title="Why is the remainder function $R_{n}(x)$ decreasing?" Tags="&lt;calculus&gt;&lt;sequences-and-series&gt;&lt;taylor-expansion&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="15098" PostTypeId="1" AcceptedAnswerId="15106" CreationDate="2010-12-21T19:50:19.323" Score="5" ViewCount="255" Body="&lt;p&gt;I have an HP 50g graphing calculator and I am using it to calculate the standard deviation of some data. In the statistics calculation there is a type which can have two values:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sample&#xA;Population&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I didn't change it, but I kept getting the wrong results for the standard deviation. When I changed it to &quot;Population&quot; type, I started getting correct results!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is that? As far as I know, there is only one type of standard deviation which is to calculate the root-mean-square of the values!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Did I miss something?&lt;/p&gt;&#xA;" OwnerUserId="4671" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-12-21T21:05:23.643" LastActivityDate="2010-12-21T21:06:06.467" Title="Sample Standard Deviation vs. Population Standard Deviation" Tags="&lt;statistics&gt;&lt;standard-deviation&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" />
  <row Id="15099" PostTypeId="2" ParentId="15073" CreationDate="2010-12-21T20:01:10.813" Score="2" ViewCount="" Body="&lt;p&gt;You're looking for a family of subsets of $[n]$ of size $m$, the intersection of any two of which is at most $k$. If $k \geq m$ then we can take all subsets of size $m$. If $k &amp;lt; m$ then a well-known theorem of Ray-Chaudhuri and Wilson shows an upper bound of $\binom{n}{k}$. A trivial construction gives a lower bound of $\binom{n-(m-k)}{k}$. So for $k \leq m$ constant, the answer is $\Theta(n^k)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the theorem mentioned above, see for example Babai and Frankl's book &quot;Linear Algebra Methods in Combinatorics with Applications to Geometry and Computer Science&quot;.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-12-21T20:01:10.813" CommentCount="1" />
  <row Id="15100" PostTypeId="1" AcceptedAnswerId="15116" CreationDate="2010-12-21T20:05:30.453" Score="3" ViewCount="163" Body="&lt;p&gt;IF $A$,$B$,$C$ are finite sets then,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Number of elements in exactly one of sets $A$,$B$,$C$:$$n(A)+n(B)+n(C)-2 \times n(A \cap B)-2 \times n(A \cap C)-2 \times n(C \cap B) + 3 \times n(A \cap B \cap C)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Number of elements in exactly two of the sets $A$,$B$,$C$: $$n(A \cap B)+(A \cap C)+n(C \cap B) - 3 \times n(A \cap B \cap C)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could somebody explain how can we prove them? I think it's something with inclusion and exclusion but am not getting how to get those results.Also I think we can have a generalized results for these kinds ?!&lt;/p&gt;&#xA;" OwnerUserId="2109" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-22T05:57:58.843" LastActivityDate="2010-12-22T05:57:58.843" Title="Finding the cardinality of exactly one or two elements of two or three finite sets" Tags="&lt;elementary-set-theory&gt;&lt;counting&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="15101" PostTypeId="1" CreationDate="2010-12-21T20:09:25.470" Score="0" ViewCount="161" Body="&lt;p&gt;I have two images. One is the reference image and the other is the rotated version of the reference image. I would like to find what is the angle by which the second image is rotated with respect to the first one. &#xA;  i want this to work automatic.&lt;/p&gt;&#xA;" OwnerUserId="5018" LastEditorUserId="5018" LastEditorDisplayName="" LastEditDate="2010-12-22T18:00:03.537" LastActivityDate="2011-03-29T11:04:43.873" Title="find the rotation of one image with respect to other in MATLAB" Tags="&lt;matlab&gt;&lt;image-processing&gt;" AnswerCount="3" />
  <row Id="15102" PostTypeId="2" ParentId="15086" CreationDate="2010-12-21T20:25:09.897" Score="4" ViewCount="" Body="&lt;p&gt;In the real numbers case, the construction is isomorphic, as kahen said in his answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you look at this fact closely enough, you realize that this is because the backbone of the real line is actually the integers, which has no accumulation points within itself, that is if a sequence converges it has to be constant from some point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This fact, while seems unrelated, helps us avoid the following case:&#xA;Consider some dense linear order (like the rationals) only bigger. For example, put two copies of the rationals one on top of the other, of course this can be embedded back into the rationals themselves, but consider the Dedekind closure of this order. It would have a point which is the exact point in which you switch from the first copy of the rationals to the second one. (This might not be such a good example, if you have a better example - this post is CW for this very reason)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So now let us try and prove something (together, I got stuck halfway through and although the intuition is clear to me, I'm uncertain how to put it into words - and this might as well be complete rubbish too.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Theorem:&lt;/strong&gt; Suppse $F$ is an ordered field (so it has characteristics 0, and we can assume $\mathbb{Q}\subseteq F$ without the loss of generality) such that its Cauchy completion (denoted by $C(F)$) and Dedekind completion (denoted by $D(F)$), both form a field - and it is the same field (namely there is a ring isomorphism between them).&lt;br&gt;&#xA;Then $F\subseteq\mathbb{R}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Proof:&lt;/em&gt; Suppose not. Since $\mathbb{Q}\subseteq F$ then $\mathbb{R}\subseteq C(F)$ (and from our assumption, $\mathbb{R}\subseteq D(F)$ as well - they are isomorphic). Now let $a\in C(F)\setminus\mathbb{R}$ (without loss of generality take $a&gt;0$), that is it is not a Cauchy limit of rational sequences. Meaning every $a_n\to a$ has only finitely many rational elements, so again without the loss of generality $a_n$ is not rational for all $n$. Since $D(F)$ is an ordered field so is $C(F)$, if there was $r\in\mathbb{R}$ such that $a&amp;lt;r$ then we have that $a\in (0,r)$ and so we can find a rational Cauchy sequence that converges to $a$. That to say that $a&amp;gt;r$ for any real number $r$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once more, we use the isomorphism, $a$ defines a cut which means the set $A = \{x\in F| x&amp;gt;q,\ \forall q\in\mathbb{Q}\}$ is non-empty, take $z$ to be the cut defined as the number between $\mathbb{Q}$ and $A$ (usually denoted by $z=\langle \mathbb{Q}\big| A\rangle$). Take some $\epsilon&amp;gt;0$ then $z-\epsilon$ is already smaller than some rational number, and therefore can be expressed as a limit of rational numbers, ergo $z-\epsilon\in\mathbb{R}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take $z_n \to z$ some Cauchy sequence, since $z$ cannot be expressed as a real number, we can assume $z_n$ is not rational for all $z$, and therefore even if we take real numbers we cannot express $z$ as a Cauchy limit of such real numbers. However, for all $\epsilon&amp;gt;0$ we can express $z-\epsilon$ as a Cauchy limit of real numbers, so we can choose some sequence that would converge to $z$ Cauchy wise, which is a contradiction.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I am quite certain that I made a big mess in this proof, and would like it very much if someone will clean it up a bit. It should probably be easier to prove it by some model theoretic method, or maybe by using equivalent topologies of order and metric. I'm not sure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That being said, I completely resent the whole idea that we based metric spaces on the real line and then we say there are no metric-complete ordered fields except $\mathbb{R}$.&lt;/p&gt;&#xA;" OwnerUserId="622" LastActivityDate="2010-12-21T20:25:09.897" CommunityOwnedDate="2010-12-21T20:25:09.897" />
  <row Id="15103" PostTypeId="2" ParentId="15097" CreationDate="2010-12-21T20:38:56.323" Score="5" ViewCount="" Body="&lt;p&gt;I presume you have a theorem like this:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Suppose $f \in C^k(I)$ for some open interval $I$. Let $a,b \in I$ and $M$ be numbers such that $|f^{(n+1)}(t)| \leq M$ for all $t$ between $a$ and $b$. Then we have that&#xA;  $$ |R_n f(b)| \leq \frac{M}{(n+1)!} |b-a|^{n+1}.$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Obviously you'll want $|b-a| &amp;lt; 1$ which you almost always can arrange for. Now if $f$ is analytic (and thus equal to its own Taylor series), we can use the third of these &lt;a href=&quot;http://en.wikipedia.org/wiki/Analytic_function#Alternate_characterizations&quot; rel=&quot;nofollow&quot;&gt;alternate characterizations&lt;/a&gt; to see that the approximations get better. This should give you a good idea of how the approximations might get worse.&lt;/p&gt;&#xA;" OwnerUserId="1269" LastActivityDate="2010-12-21T20:38:56.323" />
  <row Id="15104" PostTypeId="2" ParentId="15077" CreationDate="2010-12-21T20:46:52.317" Score="0" ViewCount="" Body="&lt;p&gt;This is related to what Gunnar Magnusson said above. First suppose $|w| = 1$. Then $u = \cos(\theta)$ and $v = \sin(\theta)$ for some $\theta \in (-\pi,\pi]$, so that $w = e^{i\theta}$. Then $a = ({1 + \cos(\theta) \over 2})^{1 \over 2}$ and $b = ({1 - \cos(\theta) \over 2})^{1 \over 2}$. By the half-angle formulas, $a = \cos{\theta \over 2}$. Also, $ b= \sin{\theta \over 2}$ when $v \geq 0$, while $b =&#xD;&#xA;-\sin{\theta \over 2}$ when $v &amp;lt; 0$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if $v \geq 0$, $z = a + ib = \cos{\theta \over 2}+ i\sin{\theta \over 2} = e^{i{\theta \over 2}}$, while if $v &amp;lt; 0$, $\bar{z} = a - ib = \cos{\theta \over 2}+ i\sin{\theta \over 2} = e^{i{\theta \over 2}}$. Thus in the first case, $z^2 = w$, while in the second case $(\bar{z})^2 = w$. This is what you want.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $|w| \neq 1$, the above proves it for ${w \over |w|}$ in place of $w$. Multiplying everything through by $|w|$ then gives your result.&lt;/p&gt;&#xA;" OwnerUserId="3035" LastActivityDate="2010-12-21T20:46:52.317" />
  <row Id="15105" PostTypeId="2" ParentId="15061" CreationDate="2010-12-21T20:58:33.197" Score="5" ViewCount="" Body="&lt;p&gt;One way to prove the irreducibility seems to be to use the &lt;a href=&quot;http://en.wikipedia.org/wiki/Newton_polygon&quot; rel=&quot;nofollow&quot;&gt;Newton Polygon&lt;/a&gt;.&#xA;The condition on the coefficients of $f$ means that the Newton Polygon has a side of slope $\dfrac{1}{n}-1$ and hence that $f$ has a root $\alpha$ in some algebraic closure $F$ of $\mathbf{Q}_p$ having valuation $1 - \dfrac{1}{n}$ (there is a unique way to prolong $e_p$ to a valuation of $F$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But then the extension $\mathbf{Q}_p \subset \mathbf{Q}_p(\alpha)$ is totally ramified&#xA;of degree $n$ and $f$ must be irreducible over $\mathbf{Q}_p$ hence &lt;em&gt;a fortiori&lt;/em&gt; irreducible over $\mathbf{Q}$.&lt;/p&gt;&#xA;" OwnerUserId="5020" LastActivityDate="2010-12-21T20:58:33.197" CommentCount="3" />
  <row Id="15106" PostTypeId="2" ParentId="15098" CreationDate="2010-12-21T21:00:41.323" Score="11" ViewCount="" Body="&lt;p&gt;There are, in fact, two different formulas for standard deviation here: The population standard deviation $\sigma$ and the sample standard deviation $s$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $x_1, x_2, \ldots, x_N$ denote all $N$ values from a population, then the (population) standard deviation is &#xA;$$\sigma = \sqrt{\frac{1}{N} \sum_{i=1}^N (x_i - \mu)^2},$$&#xA;where $\mu$ is the mean of the population.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $x_1, x_2, \ldots, x_N$ denote $N$ values from a sample, however, then the (sample) standard deviation is &#xA;$$s = \sqrt{\frac{1}{N-1} \sum_{i=1}^N (x_i - \bar{x})^2},$$&#xA;where $\bar{x}$ is the mean of the sample.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason for the change in formula with the sample is this: When you're calculating $s$ you are normally using $s^2$ (the sample variance) to estimate $\sigma^2$ (the population variance).  The problem, though, is that if you don't know $\sigma$ you generally don't know the population mean $\mu$, either, and so you have to use $\bar{x}$ in the place in the formula where you normally would use $\mu$.  Doing so introduces a slight bias into the calculation: Since $\bar{x}$ is calculated from the sample, the values of $x_i$ are on average closer to $\bar{x}$ than they would be to $\mu$, and so the sum of squares $\sum_{i=1}^N (x_i - \bar{x})^2$ turns out to be smaller on average than $\sum_{i=1}^N (x_i - \mu)^2$.  It just so happens that that bias can be corrected by dividing by $N-1$ instead of $N$.  (Proving this is a standard exercise in an advanced undergraduate or beginning graduate course in statistical theory.)  The technical term here is that $s^2$ (because of the division by $N-1$) is an &lt;em&gt;unbiased estimator&lt;/em&gt; of $\sigma^2$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another way to think about it is that with a sample you have $N$ independent pieces of information.  However, since $\bar{x}$ is the average of those $N$ pieces, if you know $x_1 - \bar{x}, x_2 - \bar{x}, \ldots, x_{N-1} - \bar{x}$, you can figure out what $x_N - \bar{x}$ is.  So when you're squaring and adding up the residuals $x_i - \bar{x}$, there are really only $N-1$ independent pieces of information there.  So in that sense perhaps dividing by $N-1$ rather than $N$ makes sense.  The technical term here is that there are $N-1$ &lt;em&gt;degrees of freedom&lt;/em&gt; in the residuals $x_i - \bar{x}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For more information, see Wikipedia's article on the &lt;a href=&quot;http://en.wikipedia.org/wiki/Standard_deviation#With_sample_standard_deviation&quot; rel=&quot;nofollow&quot;&gt;sample standard deviation&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2370" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-12-21T21:06:06.467" LastActivityDate="2010-12-21T21:06:06.467" />
  <row Id="15107" PostTypeId="1" AcceptedAnswerId="15171" CreationDate="2010-12-21T21:03:40.790" Score="1" ViewCount="150" Body="&lt;p&gt;&lt;strong&gt;What is the integral of a $k$-dimensional (multivariate) Gaussian over a convex $k$-polytope?&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here I am specifically interested in $k\in\{2,3\}$, but insight on the general problem would also be appreciated.  I am guessing that there is no closed form in the general case, but what if we restrict the number of vertices of the polytope to at most $k+1$?  If no closed form exists even under this constraint, are there efficient ways to approximate the integral?&lt;/p&gt;&#xA;" OwnerUserId="5022" LastActivityDate="2010-12-22T08:19:40.100" Title="Integrating a $k$-dimensional (multivariate) Gaussian over a convex $k$-polytope" Tags="&lt;probability-theory&gt;&lt;differential-geometry&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="15108" PostTypeId="1" CreationDate="2010-12-21T21:08:13.777" Score="4" ViewCount="250" Body="&lt;p&gt;Hello,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How many symmetric relations are there for an $n$-element set?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you.&lt;/p&gt;&#xA;" OwnerUserId="5023" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-12-21T21:47:08.890" LastActivityDate="2010-12-21T22:10:57.113" Title="How Many Symmetric Relations on a Finite Set?" Tags="&lt;combinatorics&gt;" AnswerCount="1" CommentCount="10" />
  <row Id="15109" PostTypeId="1" AcceptedAnswerId="15117" CreationDate="2010-12-21T21:15:45.270" Score="3" ViewCount="71" Body="&lt;p&gt;If $K_1$ and $K_2$ are subfields of a pre-chosen $\overline{\mathbb{Q}_p}$, and if they're both unramified at $p$, and $[K_1:\mathbb{Q}_p]=[K_2:\mathbb{Q}_p]$, does that imply that $K_1=K_2$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My intuition says that this is true because all that's happening is that we're extending the residue field, and there's only one way to do that if we fix the degree of the extension. But that's not a proof...&lt;/p&gt;&#xA;" OwnerUserId="4953" LastActivityDate="2010-12-21T22:05:13.403" Title="Is an unramified cover of the p-adics determined by its degree?" Tags="&lt;number-theory&gt;&lt;field-theory&gt;&lt;galois-theory&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="15110" PostTypeId="1" AcceptedAnswerId="15111" CreationDate="2010-12-21T21:40:15.733" Score="3" ViewCount="211" Body="&lt;p&gt;Hello, during a test i was given the following question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is 10+15+20+25+...+1490+1495+1500=?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After process of elimination and guessing, i came up with an answer.  Can anyone tell me a simple way to calculate this problem without having to try to actually add the numbers?&lt;/p&gt;&#xA;" OwnerUserId="4918" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-12-22T00:45:46.150" LastActivityDate="2010-12-22T00:45:46.150" Title="How to find $10 + 15 + 20 + 25 + \dots + 1500$." Tags="&lt;homework&gt;&lt;algebra-precalculus&gt;" AnswerCount="4" CommentCount="5" />
  <row Id="15111" PostTypeId="2" ParentId="15110" CreationDate="2010-12-21T21:45:33.230" Score="15" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Hint 1&lt;/strong&gt; -  Is there anything you can factor out of all the numbers?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint 2&lt;/strong&gt; -  Do you know how to sum $1 + 2 + \cdots + n$ in a simple way? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt; - Your process in the comments is not wrong, but I don't feel like it's very intuitive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I do think it's immediately clear that all the numbers in your sum above are divisible by 5.  That is, we can rewrite&#xA;$$10 + 15 + 20 + \cdots + 1500 = 5(2 + 3 + 4 + \cdots + 300)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe not as obvious, but incredibly useful to know, is that&#xA;$$1 + 2 + 3 + \cdots + n = \frac{n(n+1)}{2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally notice $2 + 3 + \cdots + 300$ is &lt;em&gt;almost&lt;/em&gt; $1 + 2 + \cdots + 300$, and you can use the above formula to finish the problem.&lt;/p&gt;&#xA;" OwnerUserId="3537" LastEditorUserId="3537" LastEditorDisplayName="" LastEditDate="2010-12-22T00:11:56.420" LastActivityDate="2010-12-22T00:11:56.420" CommentCount="7" />
  <row Id="15113" PostTypeId="2" ParentId="15110" CreationDate="2010-12-21T21:47:23.917" Score="5" ViewCount="" Body="&lt;p&gt;HINT: &#xA;$(10 + 1500) + (15 + 1495) + (20 + 1490) +  \cdots + (750 + 760) + 755$.&lt;/p&gt;&#xA;" OwnerUserId="2810" LastEditorUserId="2810" LastEditorDisplayName="" LastEditDate="2010-12-21T21:54:18.843" LastActivityDate="2010-12-21T21:54:18.843" />
  <row Id="15115" PostTypeId="2" ParentId="15110" CreationDate="2010-12-21T21:47:45.443" Score="0" ViewCount="" Body="&lt;p&gt;&lt;b&gt;Hint 3&lt;/b&gt; (Gauss): $10 + 1500 = 15 + 1495 = 20 + 1490 = \cdots = 1500 + 10$.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-12-21T21:47:45.443" />
  <row Id="15116" PostTypeId="2" ParentId="15100" CreationDate="2010-12-21T21:55:25.377" Score="2" ViewCount="" Body="&lt;p&gt;You're right about inclusion and exclusion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the number of elements in exactly one of the sets, you first count all $n(A) + n(B) + n(C)$ elements.  You disregard the $n(A \cap B)$ elements twice, as you need to disregard them from your count for both $A$ and $B$.  The logic is similar for $A \cap C$ and $B \cap C$.  Finally, you've disregarded the $n(A \cap B \cap C)$ elements in all three sets twice, so you have to correct the count by adding this quantity back in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I bet you can figure out the second one once you've carefully understood the first.&lt;/p&gt;&#xA;" OwnerUserId="3537" LastActivityDate="2010-12-21T21:55:25.377" CommentCount="1" />
  <row Id="15117" PostTypeId="2" ParentId="15109" CreationDate="2010-12-21T22:05:13.403" Score="3" ViewCount="" Body="&lt;p&gt;This is slightly outside my domain.. but Proposition 5.4.11 of Gouvea's (excellent) book on p-adics says &quot;For each $f$ there is exactly one unramified extension of degree $f$. It can be obtained by adjoining to ${\mathbb Q}_p$ a primitive $p^f - 1$-st root of unity&quot;. So in your case $K_1$ and $K_2$ would both have to be extensions of ${\mathbb Q}_p$ by a primitive $p^f - 1$-st root of unity. Since any two such extensions contained within the same $\bar{{\mathbb Q}_p}$ must be the same, $K_1$ must be $K_2$.&lt;/p&gt;&#xA;" OwnerUserId="3035" LastActivityDate="2010-12-21T22:05:13.403" CommentCount="5" />
  <row Id="15118" PostTypeId="2" ParentId="15108" CreationDate="2010-12-21T22:10:57.113" Score="2" ViewCount="" Body="&lt;p&gt;This problem is very similar to &lt;a href=&quot;http://math.stackexchange.com/questions/12139/number-of-relations-that-are-both-symmetric-and-reflexive/&quot;&gt;Number of relations that are both symmetric and reflexive&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A symmetric relation $R$ on a set $A$ is a subset $A\times A$. We can write $R$ as $B\cup C$, where $B$ is a subset of $\{(a,a)\mid a\in A\}$ and $C$ is a subset of $\{(b,c)\in A\times A\mid b\ne c\}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note there are as many choices for $B$ as subsets of $A$, namely $2^n$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To count the number of possible sets $C$, we use that $C$ is symmetric, meaning, if $(b,c)\in C$ then also $(c,b)\in C$. Now, let ${}[A]^2$ be the collection of subsets of $A$ of size 2. Note ${}[A]^2$ consists of subsets rather than ordered pairs. Given any $D$ subset of ${}[A]^2$, we can associate to it a set $C$ by setting $C=\{(b,c)\mid \{b,c\}\in D\}$. Note that $C$ is symmetric, since $\{b,c\}=\{c,b\}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conversely, any $C$ symmetric corresponds to a unique $D\subseteq[A]^2$, namely $\{\{b,c\}\mid (b,c)\in C\}$. This is why in $C$ we only allow pairs $(b,c)$ with $b\ne c$, so the resulting $D$ is a subset of ${}[A]^2$ and we have a correspondence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have shown that there are as many $C$ as there are subsets $D$ of ${}[A]^2$. But $\displaystyle |[A]^2|={n\choose 2}=\frac{n(n-1)}2$, so the number of subsets is $2^{n\choose 2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, since we can pair any $B$ with any $C$, we have that the number of binary symmetric relations on a set $A$ of size $n$ is precisely $$ 2^{n+{n\choose 2}}=2^{{n+1}\choose 2}.$$ &lt;/p&gt;&#xA;" OwnerUserId="462" LastActivityDate="2010-12-21T22:10:57.113" CommentCount="9" />
  <row Id="15119" PostTypeId="1" CreationDate="2010-12-21T22:15:13.457" Score="1" ViewCount="110" Body="&lt;p&gt;Taken from Rudin's &lt;em&gt;Principles of Mathematical Analysis&lt;/em&gt;, 3rd edition (Chapter 1, Exercise 16):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose $k\geq 3, x,y\in\mathbb{R}^k, |x-y|=d&amp;gt;0$, and $r&amp;gt;0$. Prove:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(a) If $2r &amp;gt; d$, there are infinitely many $z\in\mathbb{R}^k$ such that $|z-x|=|z-y|=r$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(b) If $2r=d$ there is exactly one such $z$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(c) If $2r &amp;lt; d$, there is no such $z$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Geometrically (at least in $\mathbb{R}^3$), if $2r&amp;gt;d$ then there exists a circumference of points $z$ such that $|z-x|=|z-y|=r$. I guess that for $\mathbb{R}^k$ the set of such points forms a $(k-2)$-sphere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More formally, let $m$ be the midpoint between $x$ and $y$: $m=\frac{x+y}{2}$. Then we define the plane $\Pi$ as the set of $w\in\mathbb{R}^k$ that satisfy $(\lambda(x-y))w = (\lambda&amp;#39;(x-y))m$ for some $\lambda, \lambda&amp;#39;\in\mathbb{R}$. Let $S = \{p\in\Pi : |p-x|=r\}$. Because of its geometric interpretation it's obvious that $|p-x|=|p-y|$ for every $p\in S$, but I can't figure out how to prove it formally. $S$ can also be defined as the intersection of $\Pi$ and a $(k-1)$-sphere centered at $m$ with radius $\sqrt{r^2-(d/2)^2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(b) is obvious since $m$ is the only possible point that fulfills what's asked.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(c) follows from the fact that, if we set $S$ as the $(k-1)$-sphere centered at $m$ with radius $\sqrt{r^2-(d/2)^2}$, its radius is non-existent, therefore no such sphere exists.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this approach correct? How can I prove (a)? Thanks in advance.&#xA;(c) &lt;/p&gt;&#xA;" OwnerUserId="1523" LastEditorUserId="1523" LastEditorDisplayName="" LastEditDate="2010-12-21T23:12:14.557" LastActivityDate="2010-12-21T23:12:14.557" Title="Proving a statement related to norms in euclidean spaces." Tags="&lt;real-analysis&gt;" CommentCount="5" />
  <row Id="15120" PostTypeId="2" ParentId="15079" CreationDate="2010-12-21T22:42:55.617" Score="3" ViewCount="" Body="&lt;p&gt;I'm not sure of a deep connection; from what I gather, it is merely the case that the locker puzzle and problem 6 have a common subproblem, which is: prove that &quot;traversing through boxes&quot; produces a cycle, i.e., leads us to visit a node we have already visited. In the locker puzzle, there is an extra restriction that this cycle include the starting node. This restriction holds because the locker puzzle involves a permutation, while problem 6 does not necessarily involve a permutation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unless I'm missing something, problems 6a and 6b can be solved simultaneously using the &quot;traverse through boxes&quot; technique starting at array index 1, despite the comment that &quot;Part (b) seems to be a lot harder than part (a), and it apparently requires entirely different techniques&quot;. Since the hedge words &quot;seems&quot; and &quot;apparently&quot; were used, we can't say that the comment is false, but nonetheless, I believe it would be false without those hedge words. The key idea is that, as explained in the PDF on the locker puzzle, &quot;player $ B_i $ never opens a locker (other than locker i) without first finding its number, so each time he opens a new locker he must find either his own number or the number of another unopened locker&quot;. The same basic reasoning applies to problem 6, with just slight modifications.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Problem 6 can be made to look a little more like the locker puzzle by taking a[1] out of the picture. We can consider b = {a[2], ... , a[n]} as the &quot;real&quot; array with m = n - 1 elements, and the value a[1] as the index of the start node. We can further relabel {2, ... , n} as {1, ... , m} if desired. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;All we care about is finding a cycle, that is, coming across a value that is an index we have already visited. This is equivalent to finding a duplicate in array a, since the indices we visited are precisely array values in a. We can fulfill the time and memory requirements using a boolean array of size m. The boolean array allows us to see if we have visited an index yet, and &lt;strike&gt;of course has constant memory and&lt;/strike&gt; is fast. (&lt;strong&gt;Correction:&lt;/strong&gt; the boolean array would have linear memory, not constant. Evidently there is a better way to solve this.) We are guaranteed not to visit any array index twice, so the algorithm has linear time. If we didn't think of a boolean array, we might be tempted to use some sort of a list structure with fast insert and find, but a boolean array is faster.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To tie the problems together further, we can interpret the data as a directed graph where each vertex has outdegree 1. In the locker puzzle, the indegree for each vertex is also 1, but problem 6 has no such restriction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The outdegree being 1 guarantees that we can traverse from one vertex to another without any ambiguity. Beyond that, we need to guarantee a cycle will occur regardless of where we start.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A permutation is partitioned into cycles, so this allows us to conclude for the locker puzzle that we will find our cycle precisely when we get back to the starting node. For problem 6, we are guaranteed to find some cycle, because otherwise we would have to go on endlessly, but the array is finite, so that is absurd.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe my answer should be made more rigorous, but I hope it is at least satisfactory and accurate. If not, let me know, and I'll try to fix it. I admit that after looking at this problem for an extended period of time, my mind is getting a bit foggy..&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;: Oh I think I figured out how to solve 6b with constant memory. You can follow n - 1 links, thus covering n (not necessarily distinct) array indices of a, and by this time you are guaranteed to be in the middle of a cycle. Note your current array index. Then go around until you see that index again, which will give you the period of the cycle. Then using this you can find the &quot;entry node&quot; of the cycle, which will be your duplicate. (Use modular arithmetic.) I haven't worked it out in detail but I believe that's an accurate sketch.&lt;/p&gt;&#xA;" OwnerUserId="4604" LastEditorUserId="4604" LastEditorDisplayName="" LastEditDate="2010-12-22T00:26:12.907" LastActivityDate="2010-12-22T00:26:12.907" CommentCount="8" />
  <row Id="15121" PostTypeId="2" ParentId="15061" CreationDate="2010-12-21T22:56:40.207" Score="4" ViewCount="" Body="&lt;p&gt;Apply Eisenstein's criterion to ${1 \over p^{n-1}}x^nf({p \over x})$.&lt;/p&gt;&#xA;" OwnerUserId="3035" LastActivityDate="2010-12-21T22:56:40.207" CommentCount="3" />
  <row Id="15123" PostTypeId="2" ParentId="15110" CreationDate="2010-12-22T00:28:52.340" Score="1" ViewCount="" Body="&lt;p&gt;This is probably the method C. F. Gauss would have used ;)&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Factor out 5:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$5(2+3+4+5+...+297+298+299+300)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Notice: $302/2 = 151$&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$5((2+300)+(3+299)+(4+298)+(5+297)+...(149+153)+(150+152)+(151))$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$5(302+302+302+302+...+302+302+151)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;There are $(150-2+1)$ 302s, so:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$5((150-2+1)(302) + 151)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$5((149)(302) + 151)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$5(44998 + 151)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$5(45149)$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The answer is $225745$.&lt;/p&gt;&#xA;" OwnerUserId="2736" LastActivityDate="2010-12-22T00:28:52.340" CommentCount="2" />
  <row Id="15124" PostTypeId="1" CreationDate="2010-12-22T00:52:10.653" Score="4" ViewCount="112" Body="&lt;p&gt;Let $X \geq 1$ be an integer r.v. with $E[X]=\mu$. Let $X_i$ be a sequence of iid rvs with the distribution of $X$. On the integer line, we start at $0$, and want to know the expected position after we first cross $K$, which is some fixed integer. Each next position is determined by adding $X_i$ to the previous position. So the question is, if we stop this process after the first time $\tau$ for which $Y_{\tau}=\sum_{i=1}^{\tau}X_i &amp;gt; K$, that is, after the first time it crosses $K$, then what is $E[Y_{\tau}-K]$?.   Can we get a bound of $O(\mu)$?&lt;/p&gt;&#xA;" OwnerUserId="5031" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2010-12-22T02:18:57.000" LastActivityDate="2010-12-22T15:45:38.377" Title="A type of stochastic jump process" Tags="&lt;stochastic-processes&gt;" AnswerCount="2" />
  <row Id="15125" PostTypeId="1" CreationDate="2010-12-22T01:24:46.087" Score="-1" ViewCount="206" Body="&lt;p&gt;If $f(x) = \sqrt{ 4\sin x + 2 }$, then $f&amp;#39;(0) =$?&lt;/p&gt;&#xA;" OwnerUserId="4918" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-12-22T01:34:32.333" LastActivityDate="2010-12-22T03:20:34.763" Title="Derivative of $\sqrt{4 \sin x + 2}$ at $0$" Tags="&lt;calculus&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="15126" PostTypeId="2" ParentId="15125" CreationDate="2010-12-22T01:28:55.307" Score="6" ViewCount="" Body="&lt;p&gt;Do you know the &lt;a href=&quot;http://en.wikipedia.org/wiki/Chain_rule&quot; rel=&quot;nofollow&quot;&gt;chain rule?&lt;/a&gt;  What have you tried?&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-12-22T01:28:55.307" />
  <row Id="15127" PostTypeId="2" ParentId="15125" CreationDate="2010-12-22T01:42:41.110" Score="4" ViewCount="" Body="&lt;p&gt;By definition, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f&amp;#39;(0)=\lim_{h\to0}\frac{f(h)-f(0)}{h}=\lim_{h\to0}\frac{\sqrt{4\sin(h)+2}-\sqrt2}{h}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can &quot;rationalize&quot; the numerator, use the fact that a limit of a product is the product of the limits, use continuity of the square root and sine functions, and use the fact that $\lim_{h\to0}\frac{\sin(h)}{h}=1$ to finish.  Of course, you would get the same answer by deriving a general formula for $f&amp;#39;(x)$ using the chain rule and plugging in $x=0$, as Ross hints.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-12-22T01:42:41.110" CommentCount="4" />
  <row Id="15128" PostTypeId="2" ParentId="15079" CreationDate="2010-12-22T01:44:59.563" Score="1" ViewCount="" Body="&lt;p&gt;Since you requested for more puzzles, here is one which is an algorithmic puzzle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are given an array of $\displaystyle 2n$ elements: $$a_1, a_2, \dots, a_n , b_1, b_2, \dots, b_n$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You need to rearrange the elements of the array to look like: $$b_1, a_1, b_2, a_2, \dots, b_n , a_n$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is trivial by using an extra array, but the hard part of the puzzle is that you need to do it in $\displaystyle \mathcal{O}(1)$ (or to be pedantic, $\displaystyle \mathcal{O}(\log n))$ space and $\displaystyle \mathcal{O}(n)$ time. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Assume each element of the array takes up constant space)&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-12-22T02:10:36.407" LastActivityDate="2010-12-22T02:10:36.407" CommentCount="2" />
  <row Id="15129" PostTypeId="1" AcceptedAnswerId="15130" CreationDate="2010-12-22T01:45:21.120" Score="1" ViewCount="381" Body="&lt;p&gt;how can be prove that $\max(f(n),g(n)) = \theta(f(n)+g(n))$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;though the big O case is simple since $\max(f(n),g(n)) \leq f(n)+g(n)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;edit : where $f(n)$ and $g(n)$ are asymptotically nonnegative functions.&lt;/p&gt;&#xA;" OwnerUserId="3545" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-22T02:09:42.757" LastActivityDate="2010-12-22T02:09:42.757" Title="how can be prove that $\max(f(n),g(n)) = \theta(f(n)+g(n))$" Tags="&lt;algorithms&gt;&lt;asymptotics&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="15130" PostTypeId="2" ParentId="15129" CreationDate="2010-12-22T01:58:58.040" Score="2" ViewCount="" Body="&lt;p&gt;Hint...: $2 + 100 \le 100 + 100$&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-12-22T01:58:58.040" CommentCount="6" />
  <row Id="15131" PostTypeId="2" ParentId="15124" CreationDate="2010-12-22T02:18:47.747" Score="0" ViewCount="" Body="&lt;p&gt;Edit: This answer shows what happens when $X$ could be zero (contrary to the question).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $K = 1$, and $X = M$ ($M &gt; 1$) with probability $p$ and $0$ otherwise. Thus $E[Y_\tau - K] = M-1$ whereas $E[X] = pM$. Choosing $M = n+1$ and $p = 1/n(n+1)$ we get $E[X] = 1/n$ and $E[Y_\tau - K] = n$, so a bound of $O(\mu)$ isn't possible.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2010-12-22T15:45:38.377" LastActivityDate="2010-12-22T15:45:38.377" CommentCount="1" />
  <row Id="15132" PostTypeId="2" ParentId="15101" CreationDate="2010-12-22T02:21:54.313" Score="1" ViewCount="" Body="&lt;p&gt;Enumerate over all possible angles (under some reasonable discretization), rotate one of the images, and calculate the $L_2$ distance (sum of squared differences); the angle at which the distance is minimal is the one you seek.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An even better idea is to calculate the two-dimensional Fourier transform, and now do the same with shifts instead of rotations (rotation in the time domain corresponds to shift in the frequency domain). Maybe you can reduce the amount of shifts tested by looking at some prominent peeks of the spectrum and trying to match them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have no idea how to implement these suggestions in Matlab, but I'm sure you can find out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both solutions assume that the image is &quot;continuous&quot; with reasonable &quot;modulus of continuity&quot;. If the image is random that the methods wouldn't work.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-12-22T02:21:54.313" CommentCount="2" />
  <row Id="15133" PostTypeId="1" AcceptedAnswerId="15140" CreationDate="2010-12-22T02:21:59.323" Score="1" ViewCount="92" Body="&lt;p&gt;Let $G$ be a set of elements and $*$ a binary operation defined on the elements of $G$. Then $G$ is a group with respect to $*$ if the following holds:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$a*(b*c) = (a*b)*c$ for all $a,b,c \in G$.&lt;/li&gt;&#xA;&lt;li&gt;There exists a left identity in $G$.&lt;/li&gt;&#xA;&lt;li&gt;For each $a \in G$, there exists a left inverse.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;We could replace &quot;left&quot; with &quot;right.&quot; Is there any relationship between this formulation and &lt;a href=&quot;http://en.wikipedia.org/wiki/Burnside%27s_lemma&quot; rel=&quot;nofollow&quot;&gt;Burnside's Lemma&lt;/a&gt;? In other words, does the &quot;left&quot; or &quot;right&quot; criteria for groups come in handy when dealing with Burnside's Lemma (e.g could the number of orbits be the &quot;left&quot; part and the number of fixed points be the &quot;right&quot; part)?&lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2010-12-22T03:28:47.417" Title="Criteria for Groups and Burnside's Lemma" Tags="&lt;group-theory&gt;" AnswerCount="1" CommentCount="6" />
  <row Id="15134" PostTypeId="1" AcceptedAnswerId="15175" CreationDate="2010-12-22T02:31:27.850" Score="2" ViewCount="193" Body="&lt;p&gt;Are there any good algorithms which can be used to construct a prime greater than $n$, for arbitrary $n$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are some brute force approaches: for example, factoring $n!+1$. However, I'm looking for something significantly more efficient; preferably, polynomial in $n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, if anyone knows of a proof showing a bound on the speed of such an algorithm (say, $n^4$), I would like to hear about it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: It appears that polynomial in $ln(n)$ makes more sense. &lt;/p&gt;&#xA;" OwnerUserId="796" LastEditorUserId="796" LastEditorDisplayName="" LastEditDate="2010-12-22T17:04:14.773" LastActivityDate="2010-12-22T17:04:14.773" Title="Algorithm for constructing primes" Tags="&lt;prime-numbers&gt;&lt;elementary-number-theory&gt;" AnswerCount="3" CommentCount="6" />
  <row Id="15135" PostTypeId="1" AcceptedAnswerId="15137" CreationDate="2010-12-22T02:39:05.087" Score="2" ViewCount="94" Body="&lt;p&gt;I keep trying to solve it but I seem to be making an error somewhere: &#xA; $$4-\left|\frac{5y}{3}+4\right| &amp;gt; \frac25$$&lt;/p&gt;&#xA;" OwnerUserId="5032" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-22T03:01:58.733" LastActivityDate="2010-12-22T03:54:47.373" Title="what is the process to solving this absolute value inequality: $4-\left|\frac{5y}{3}+4\right| &gt; \frac25$?" Tags="&lt;algebra-precalculus&gt;&lt;inequality&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="15136" PostTypeId="2" ParentId="15134" CreationDate="2010-12-22T02:40:46.210" Score="5" ViewCount="" Body="&lt;p&gt;Well theoretically speaking, there is a &lt;em&gt;deterministic&lt;/em&gt; $\mathcal{O}(n \log ^{12} n)$ time algorithm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By Bertrand's postulate, there is a prime $p$ guaranteed to satisfy $n \le p \lt2n$. (In practice, you should hit one much earlier than $2n$, owing to the prime number theorem).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Use the AKS primality testing program to test each of the numbers in the range $[n, 2n)$ and you will find one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can also try some sieving etc as suggested in the answers to this other question here: &lt;a href=&quot;http://math.stackexchange.com/questions/8865/the-least-prime-greater-than-2000&quot;&gt;The least prime greater than 2000&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-12-22T02:40:46.210" CommunityOwnedDate="2010-12-22T02:44:43.677" CommentCount="2" />
  <row Id="15137" PostTypeId="2" ParentId="15135" CreationDate="2010-12-22T03:14:39.747" Score="2" ViewCount="" Body="&lt;p&gt;$4-\left|\frac{5y}{3}+4\right| &amp;gt; \frac25$ can be rewritten as $-\left|\frac{5y}{3}+4\right| &amp;gt; \frac{-18}{5}$ or $\left|\frac{5y}{3}+4\right| &amp;lt; \frac{18}{5}$. From here, represent the expression as $\frac{-18}{5} &amp;lt; \frac{5y}{3}+4\ &amp;lt; \frac{18}{5}$ and proceed to solve it using arithmetic.&lt;/p&gt;&#xA;" OwnerUserId="3731" LastEditorUserId="3731" LastEditorDisplayName="" LastEditDate="2010-12-22T03:54:47.373" LastActivityDate="2010-12-22T03:54:47.373" CommentCount="2" />
  <row Id="15138" PostTypeId="2" ParentId="15125" CreationDate="2010-12-22T03:20:34.763" Score="1" ViewCount="" Body="&lt;p&gt;$$f&amp;#39;(x)=\frac{4\cos x}{2\sqrt{4\sin x+2}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f&amp;#39;(0)=\frac{4\cdot1}{2\sqrt{4\cdot0+2}}=\frac{4}{2\sqrt{2}}=\sqrt{2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;what is the difficulty? It does not even involve a $\frac{0}{0}$ like $\sin{x}/x$ does.&lt;/p&gt;&#xA;" OwnerUserId="3301" LastActivityDate="2010-12-22T03:20:34.763" />
  <row Id="15139" PostTypeId="1" AcceptedAnswerId="15168" CreationDate="2010-12-22T03:27:15.800" Score="5" ViewCount="164" Body="&lt;p&gt;What values of $Z_1$ and $Z_2$ make the five regions of the unit circle, shown below, equal in area? $\overline{Z_1}$ and $\overline{Z_2}$ are conjugates of $Z_1$ and $Z_2$; in other words they lie directly across the real axis from their counterparts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/8cmP9.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="130" LastActivityDate="2010-12-22T07:56:20.363" Title="Two points on circle resulting in 5 equal regions" Tags="&lt;geometry&gt;&lt;euclidean-geometry&gt;&lt;complex-numbers&gt;&lt;circle&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="15140" PostTypeId="2" ParentId="15133" CreationDate="2010-12-22T03:28:47.417" Score="0" ViewCount="" Body="&lt;p&gt;First, I assume you know that the two definitions yield objects which are not just isomorphic, but identical: from the 'left' definition it follows that the 'left' identity is also a 'right' identity and ergo it is 'the' identity, and the 'left' inverse is also a 'right' inverse and ergo 'the' inverse; and vice versa follows for the 'right' definition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I would say that the formulation of the axioms has little to do with Burnside's lemma per se; and instead is intended to provide a minimum set of requirements: only one side, left or right, is required to deduce that the resulting entities (identity, inverse) are actually /both-sided/. This is in contrast to other types of structures; e.g., modules, where a left module is not necessarily a right module.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, historically, the notation for the application of a group operation has varied between right to left and left to right over time. For example in W. R. Scott's &quot;Group Theory&quot; (1964), assuming x in S and g,h in G where G acts on S, the author writes &quot;xhg&quot; for what would now more typically would be written as &quot;ghx&quot;. This has other terminological effects such as the precise meaning of &quot;left coset&quot; or &quot;right coset&quot;, but of course the essential theory is the same.&lt;/p&gt;&#xA;" OwnerUserId="4980" LastActivityDate="2010-12-22T03:28:47.417" />
  <row Id="15141" PostTypeId="1" CreationDate="2010-12-22T03:55:20.507" Score="4" ViewCount="97" Body="&lt;p&gt;I'm asking in the spirit of these &lt;a href=&quot;http://math.stackexchange.com/questions/13237&quot;&gt;two&lt;/a&gt; &lt;a href=&quot;http://math.stackexchange.com/questions/13253&quot;&gt;questions&lt;/a&gt;: can bitwise operations (AND, OR, XOR) be expressed in terms of other (more familiar?) functions?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I had been playing around with the bitwise operations in &lt;em&gt;Mathematica&lt;/em&gt; a while back, and was at first struck by this identity for bitwise NOT: &lt;code&gt;~ n == -1 - n&lt;/code&gt; (I'll use C-ish notation/syntax for this question, but I'm using &lt;a href=&quot;http://reference.wolfram.com/mathematica/tutorial/IntegerAndNumberTheoreticalFunctions.html#28221&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;Mathematica&lt;/em&gt;'s definitions&lt;/a&gt;, which assume a two's complement representation for negative integers.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Try as I might, I have not managed to figure out alternative ways of expressing the other bitwise operations. I have also tried Using The Fabulous Search Engine to see if other people have looked into this, but I probably am not using the right keywords.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have noticed that the implementation of the bitwise operators in &lt;em&gt;Mathematica&lt;/em&gt; satisfies &lt;a href=&quot;http://mathworld.wolfram.com/deMorgansLaws.html&quot; rel=&quot;nofollow&quot;&gt;de Morgan's theorem&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;i | j == -1 - ((-1 - i) &amp;amp; (-1 - j))&#xA;&#xA;i &amp;amp; j == -1 - ((-1 - i) | (-1 - j))&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;and the following identity for bitwise XOR is satisfied as well:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;i ^ j == (i | j) &amp;amp; (-1 - (i &amp;amp; j))&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I suppose then that any alternate expression for the bitwise operators would have to satisfy these identities as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I am not intending to replace the bitwise expressions in actual programming, of course; they are already optimized, and there is really no practical reason to replace them with &quot;more mathematical&quot; expressions. I am asking merely for curiosity's sake.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;References and sundry information will be appreciated.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-12-22T04:55:32.470" Title="Expressing bitwise operations in terms of other functions" Tags="&lt;functions&gt;" AnswerCount="1" CommentCount="10" />
  <row Id="15142" PostTypeId="2" ParentId="15139" CreationDate="2010-12-22T03:57:15.527" Score="2" ViewCount="" Body="&lt;p&gt;We have segment $Z_2\overline{Z_2}$ is $\frac15$ of the area.  The area of a segment is $\frac{\theta-\sin(\theta)}{2}$.  Setting this to $\frac{\pi}{5}$ gives the angle at the center to be $\theta$=2.113139.  Segment $Z_1\overline{Z_1}\overline{Z_2}$ then has area $\frac{2\pi}{5}$, so the angle is 2.824797.  So $Z_2=\exp{(\pi-1.056569)i}$ and $Z_1=\exp{(.74890i)}$&lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-22T06:38:33.223" LastActivityDate="2010-12-22T06:38:33.223" CommentCount="3" />
  <row Id="15143" PostTypeId="1" AcceptedAnswerId="27822" CreationDate="2010-12-22T04:00:48.330" Score="8" ViewCount="99" Body="&lt;p&gt;Differential equations of the form $M\,dx+N\,dy=0$ such that $\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}$ are said to be &lt;i&gt;exact&lt;/i&gt;, because the left hand side of our equation is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Exact_differential&quot; rel=&quot;nofollow&quot;&gt;exact differential&lt;/a&gt; of some function &lt;i&gt;f&lt;/i&gt;, and the DE has a solution of the form $f(x,y)=c$, where &lt;i&gt;c&lt;/i&gt; is a constant. Standard textbooks give us a procedure for solving such equations that essentially amounts to using the identity  $f=\int\frac{\partial f}{\partial x}\: dx+\int(\frac{\partial f}{\partial y}-\frac{\partial}{\partial y}\int\frac{\partial f}{\partial x}\: dx)\: dy$. Specifically, we integrate &lt;i&gt;M&lt;/i&gt; (respectively &lt;i&gt;N, &amp;amp;c.&lt;/i&gt;) with respect to &lt;i&gt;x&lt;/i&gt; to get $f=\int\frac{\partial f}{\partial x}\: dx+C(y)$, and then differentiate this with respect to &lt;i&gt;y&lt;/i&gt; and do algebra to find &lt;i&gt;C'(y)&lt;/i&gt;, which we can then integrate to find &lt;i&gt;C(y)&lt;/i&gt;, which we can substitute back into our expression for &lt;i&gt;f&lt;/i&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It occurred to me that this technique should generalize---that is, when instead of $M\,dx+N\,dy=0$  such that $\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}$, we have $\sum_{i}M_{i}\,dx_{i}=0$ where the $M_{i}$ satisfy the criteria for being an exact differential, then we can iteratively apply the same procedure to solve the equation. And so I came up with the following---&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Theorem (?).&lt;/b&gt; A differential equation of the form $\sum_{i=1}^{n}\frac{\partial f}{\partial x_{i}}=0$ for some function &lt;i&gt;f&lt;/i&gt; has an implicit solution of the form $f=\sum_{i=1}^{n}a_{i}=c$ where $a_{0}=0$, $a_{i}=\int\left(\frac{\partial f}{\partial x_{i}}-\frac{\partial}{\partial x_{i}}\left(\sum_{j=1}^{i-1}a_{j}\right)\right)\: dx_{i}$ for $i\in\mathbb{N}_{+}$, and &lt;i&gt;c&lt;/i&gt; is an arbitrary constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;&lt;i&gt;Proof (?).&lt;/i&gt;&lt;/b&gt; By induction on the number of variables &lt;i&gt;n&lt;/i&gt;. The theorem is true for &lt;i&gt;n&lt;/i&gt;=1 because $f=\int\,(\frac{\partial f}{\partial x}-0)\,dx$ by the fundamental theorem of calculus.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To complete the induction, we need to show that if $f(x_{1},...,x_{n})=\sum_{i=1}^{n}a_{i}=c$&#xA;is a solution to $\sum_{i=1}^{n}\frac{\partial f}{\partial x_{i}}=0$, then $f(x_{1},...,x_{n+1})=\sum_{i=1}^{n+1}a_{i}=c$ is a solution to $\sum_{i=1}^{n+1}\frac{\partial f}{\partial x_{i}}=0$. Think of $f(x_{1},...,x_{n})$ as being the &quot;special case&quot; of $f(x_{1},...,x_{n+1})$ where $x_{n+1}$ is being &quot;treated like a constant.&quot; (My thinking here is not nearly as precise as it should be and I'm overloading the function name &lt;i&gt;f&lt;/i&gt;, but hopefully you understand the intuition to which I am appealing.) So we can say that $f(x_{1},...,x_{n+1})=\sum_{i=1}^{n}a_{i}+C(x_{n+1})$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then we proceed analogously as in the two-variable case. Differentiating by $x_{n+1}$ and applying algebra yields $C&amp;#39;(x_{n+1})=\frac{\partial f}{\partial x_{n+1}}-\frac{\partial}{\partial x_{n+1}}\left(\sum_{i=1}^{n}a_{i}\right)$, and then integrating with respect to $x_{n+1}$ yields $C(x_{n+1})=\int\left(\frac{\partial f}{\partial x_{n+1}}-\frac{\partial f}{\partial x_{n+1}}\left(\sum_{i=1}^{n}a_{i}\right)\right)\: dx_{n+1}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then substituting into our earlier expression for $f(x_{1},...,x_{n+1})$ we get $f(x_{1},...,x_{n+1})=\sum_{i=1}^{n}a_{i}+\int\left(\frac{\partial f}{\partial x_{n+1}}-\frac{\partial f}{\partial x_{n+1}}\left(\sum_{i=1}^{n}a_{i}\right)\right)\: dx_{n+1}$, which by virtue of the definition of $a_{i}$ is equivalent to $f(x_{1},...,x_{n+1})=\sum_{i=1}^{n+1}a_{i}$. But by the principle of induction, this is &lt;i&gt;quod erat demonstrandum&lt;/i&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Example&lt;/b&gt;. In the &lt;i&gt;n&lt;/i&gt;=3 case (and naming our variables x, y, and z), we get $f(x,y,z)=\int\frac{\partial f}{\partial x}\: dx+\int(\frac{\partial f}{\partial y}-\frac{\partial}{\partial y}\int\frac{\partial f}{\partial x}\: dx)\: dy$&#xA;$+\int(\frac{\partial f}{\partial z}-\frac{\partial}{\partial z}(\int\frac{\partial f}{\partial x}\: dx+\int(\frac{\partial f}{\partial y}-\frac{\partial}{\partial y}\int\frac{\partial f}{\partial x}\: dx)\: dy))\, dz$, which is seen to be an identity by performing the integrations. What's going on is that $a_{1}=\int\frac{\partial f}{\partial x}\: dx=f$, and all following terms are zero by design, &lt;i&gt;e.g.&lt;/i&gt;, $a_{2}=f-f$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So my question is (and I offer my sincerest apologies if this is not an appropriate question, or if my poor exposition has rendered my intentions virtually unreadable) &lt;b&gt;does this seem basically correct&lt;/b&gt;, or am I doing something wrong? And &lt;b&gt;how do I fix my sloppy reasoning in the inductive step&lt;/b&gt; (&quot;special case&quot; ... &quot;treated like a constant&quot;)? I would be much obliged for any input.&lt;/p&gt;&#xA;" OwnerUserId="2773" LastActivityDate="2011-03-18T16:28:40.447" Title="Does the method for solving exact DEs generalize like this?" Tags="&lt;differential-equations&gt;" AnswerCount="1" />
  <row Id="15144" PostTypeId="2" ParentId="15141" CreationDate="2010-12-22T04:55:32.470" Score="1" ViewCount="" Body="&lt;p&gt;Exclusive OR is addition in $\mathbb{Z}_2^n$. The AND operation corresponds to set intersection, OR to set union, NOT to complementation (and XOR to symmetric difference).&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-12-22T04:55:32.470" CommentCount="1" />
  <row Id="15145" PostTypeId="1" AcceptedAnswerId="15146" CreationDate="2010-12-22T05:14:48.110" Score="5" ViewCount="364" Body="&lt;p&gt;What is a &quot;simpler&quot; formula for&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{3}^{n} \frac{(k-1)(k-2)(k-3)}{6}$$&lt;/p&gt;&#xA;" OwnerUserId="5035" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-12-22T06:03:05.513" LastActivityDate="2010-12-22T16:58:59.850" Title="A general formula for $\sum (k-1)(k-2)(k-3)$?" Tags="&lt;algebra-precalculus&gt;&lt;arithmetic&gt;" AnswerCount="6" CommentCount="10" FavoriteCount="1" />
  <row Id="15146" PostTypeId="2" ParentId="15145" CreationDate="2010-12-22T05:23:04.393" Score="12" ViewCount="" Body="&lt;p&gt;Hint:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$k(k-1)(k-2)(k-3) - (k-1)(k-2)(k-3)(k-4) = 4(k-1)(k-2)(k-3)$$&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-12-22T05:23:04.393" CommentCount="5" />
  <row Id="15147" PostTypeId="2" ParentId="15145" CreationDate="2010-12-22T05:23:51.053" Score="4" ViewCount="" Body="&lt;p&gt;There is no need for a general formula for this special case, all one needs is to use the formulas for $\sum {k} , \sum{k^2} \text{ and } \sum{k^3}$. Just expand the expression and use the simpler known sums.&lt;/p&gt;&#xA;" OwnerUserId="3936" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2010-12-22T06:14:24.790" LastActivityDate="2010-12-22T06:14:24.790" CommentCount="3" />
  <row Id="15148" PostTypeId="2" ParentId="15145" CreationDate="2010-12-22T05:28:19.383" Score="6" ViewCount="" Body="&lt;p&gt;I really like Moron's hint, but another way to simplify the sum is to reindex with $j=k-2$, and use $(j+1)j(j-1)=j^3-j$.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-12-22T05:28:19.383" CommentCount="7" />
  <row Id="15149" PostTypeId="2" ParentId="15079" CreationDate="2010-12-22T05:29:40.707" Score="2" ViewCount="" Body="&lt;p&gt;You might also enjoy this one, which is, I think, easier than the two you have linked to: you are starting with a pack of $n$ cards, numbered from 1 to $n$, but in a random order. You now play the following game: you look at the top card, numbered $i$, say, then you put the card into the $i$-th position from top. Repeat the whole procedure. Clearly, if at some point the top card it numbered 1, nothing interesting happens any more and the position of the cards never changes after that point. Prove that, independently of the starting configuration, after finitely many steps, the 1 will turn up on the top.&lt;/p&gt;&#xA;" OwnerUserId="3212" LastActivityDate="2010-12-22T05:29:40.707" CommentCount="2" />
  <row Id="15150" PostTypeId="1" CreationDate="2010-12-22T05:49:28.787" Score="2" ViewCount="96" Body="&lt;p&gt;I want to show that $G$ is open in $X$ iff $\overline{G \cap \overline{A}} = \overline{G \cap A}$ for every $A \subset X$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This direction $\rightarrow$ is easy. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm stuck in showing that $\overline{G \cap \overline{A}} = \overline{G \cap A}$ for every $A \subset X$ implies $G$ is open in $X$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I guess the game is to find suitable A and apply the hypothesis, I tried taking A as the interior of G and also as the interior of the complement of G but didn't work (at least I didn't see it). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any ideas?&lt;/p&gt;&#xA;" OwnerUserId="5036" LastActivityDate="2010-12-22T05:54:39.813" Title="Prove G is an open set" Tags="&lt;topology&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="15151" PostTypeId="2" ParentId="15150" CreationDate="2010-12-22T05:54:33.683" Score="3" ViewCount="" Body="&lt;p&gt;Here are some hints:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$G$ is open if and only if the complement $G^C$ is closed.&lt;/li&gt;&#xA;&lt;li&gt;A set is closed if and only if it contains all its limit points.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Now, see whether the assumption that $\overline{G \cap \overline{A}} = \overline{G \cap A}$ for all $A$ and that $G^C$ is not closed (i.e. that there exists a limit point of $G^C$ that is not in $G^C$) leads you to a contradiction.&lt;/p&gt;&#xA;" OwnerUserId="3212" LastActivityDate="2010-12-22T05:54:33.683" CommentCount="1" />
  <row Id="15152" PostTypeId="2" ParentId="15150" CreationDate="2010-12-22T05:54:39.813" Score="3" ViewCount="" Body="&lt;p&gt;Trying $A=X\setminus G$ should do the trick.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-12-22T05:54:39.813" />
  <row Id="15153" PostTypeId="2" ParentId="15145" CreationDate="2010-12-22T06:11:03.543" Score="8" ViewCount="" Body="&lt;p&gt;Show that ${k+3 \choose 3}$ is the number of solutions to $x_1 + ... + x_4 = k$ in non-negative integers.  Then $\sum_{k=0}^{n-4} {k+3 \choose 3}$ is the number of solutions to $x_1 + ... + x_5 = n-4$ in non-negative integers, which is...?&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-12-22T06:11:03.543" CommentCount="1" />
  <row Id="15154" PostTypeId="1" CreationDate="2010-12-22T06:11:50.697" Score="-1" ViewCount="95" Body="&lt;p&gt;Ronald's federal income tax for the year 2000 was $34597. Find his exact taxable income m for 2000 by solving the equation 2371.7 + 0.31(m - 10570)=34597. (Please round your answer to two decimal places.) i got M=114522.58  but my online test is saying its incorrect ! HELP !&lt;/p&gt;&#xA;" OwnerUserId="5037" LastActivityDate="2011-01-21T15:46:41.083" Title="algebra 2 solving linear equations" Tags="&lt;algebra&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="15155" PostTypeId="2" ParentId="15145" CreationDate="2010-12-22T06:12:53.650" Score="3" ViewCount="" Body="&lt;p&gt;Consider the following sum $$\sum_{m=0}^n \binom{m}{k}.$$ It counts the number of possibilities to select $k$ elements from at most $n$ elements. Some choices are counted multiple times, for example $\{0,\ldots,k-1\}$ is counted once for each $m \in [k-1,n]$. So it's natural to distinguish among those by tagging them with $m$ somehow. The best way to do that is to &lt;i&gt;add&lt;/i&gt; the element $m+1$. The result is a choice of $k+1$ elements from $n+1$, and so $$\sum_{m=0}^n \binom{m}{k} = \binom{n+1}{k+1}.$$ From this formula one can extract (using linear algebra) the usual formulas for $\sum_{m=0}^n m^k$.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-12-22T06:12:53.650" />
  <row Id="15156" PostTypeId="1" CreationDate="2010-12-22T06:26:09.490" Score="1" ViewCount="77" Body="&lt;p&gt;A question from Dugundji's book which I don't even understand the statement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Describe the relative topology of $\{z: |z|=1\}$ as a subspace of ${\mathbb{R}}^{2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What do they mean by &quot;describe&quot;? I don't understand what they are asking for. Can you please clarify?&lt;/p&gt;&#xA;" OwnerUserId="5036" LastActivityDate="2010-12-22T06:33:43.523" Title="Describe the relative topology of the unit circle as a subspace of the plane." Tags="&lt;topology&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="15157" PostTypeId="2" ParentId="15156" CreationDate="2010-12-22T06:33:43.523" Score="2" ViewCount="" Body="&lt;p&gt;I agree that the exercise is not very precisely stated, and I don't myself know exactly what Dugundji was intending here (and not, I think, due to any lack of understanding of basic topology).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The best I can do is to suggest a reasonable problem which &lt;em&gt;could&lt;/em&gt; be what Dugundji had in mind: show that &quot;open intervals&quot; (&lt;b&gt;edit&lt;/b&gt;: Jonas Meyer's terminology &lt;em&gt;open arcs&lt;/em&gt; is better) on the unit circle form a base for the subspace topology.&lt;/p&gt;&#xA;" OwnerUserId="299" LastActivityDate="2010-12-22T06:33:43.523" CommentCount="2" />
  <row Id="15158" PostTypeId="1" AcceptedAnswerId="15163" CreationDate="2010-12-22T06:37:48.200" Score="0" ViewCount="79" Body="&lt;p&gt;(1/8)(3m - 5) = (6/5)(6m - 2) + 5...Type your answer as a fraction. Do not use decimals!  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;**i got -129/273 is this correct ?&lt;/p&gt;&#xA;" OwnerUserId="5037" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-22T07:45:40.420" LastActivityDate="2010-12-22T07:45:40.420" Title="algebra 2 fraction and linear equations" Tags="&lt;algebra-precalculus&gt;" AnswerCount="1" CommentCount="11" />
  <row Id="15159" PostTypeId="2" ParentId="8140" CreationDate="2010-12-22T06:41:52.297" Score="0" ViewCount="" Body="&lt;p&gt;This was my project not so long ago. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I solved it in more than one way. But I will present you the easiest of the methods I used. Use quick fill inside of a region. Then find the lines that border the region. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It fails in certain situations, however, in most cases, those situations are not that serious anyway. If this method fails for you, post back and I will share the other methods that I used. &lt;/p&gt;&#xA;" OwnerUserId="4728" LastActivityDate="2010-12-22T06:41:52.297" CommentCount="3" />
  <row Id="15160" PostTypeId="1" CreationDate="2010-12-22T06:49:30.653" Score="2" ViewCount="111" Body="&lt;p&gt;Two questions from Dugundji's book (not hw, just practice).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Let $Y_{1}, Y_{2}$ be subspaces of $X$ and $A \subset Y_{1} \cap Y_{2}$. Assume that $A$ is open in $Y_{1}$ and open in $Y_{2}$. Prove A is open in $Y_{1} \cup Y_{2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you please give a hint for this one?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) a. Let $D$ be dense in $X$. Give an example to show that $D \cap A$ need not to be dense in $A$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can't we take $X = \mathbb{R}$, $D=\mathbb{Q}$ then $D$ in dense in $X$. Now take $A =$ irrationals, since the empty set is closed then it cannot be dense in $A$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;b. If $A$ is dense in $B \subset X$ then $A$ is dense in $\overline{B}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Attempt:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $V \subset \overline{B}$ be an open set, then by definition of subspace topology we have $V = C \cap \overline{B}$ where $C$ is an open subset of $X$. Now consider $C \cap B$ ,this is an open set in $B$ so $A$ intersects this set and hence $A$ is dense in $\overline{B}$. What bothers me, is how do we know that $C \cap B$ is non-empty?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="5036" LastEditorUserId="5036" LastEditorDisplayName="" LastEditDate="2010-12-22T07:34:27.417" LastActivityDate="2010-12-22T07:34:27.417" Title="Closed, open subspaces, dense subsets" Tags="&lt;topology&gt;" AnswerCount="1" CommentCount="7" />
  <row Id="15161" PostTypeId="2" ParentId="15160" CreationDate="2010-12-22T07:04:39.743" Score="1" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; For the first one Recall the definition of $A$ being open in the relative topology: if there exists some $A&amp;#39;$ open in $X$ such that $A&amp;#39;\cap Y_1 = A$.&lt;/p&gt;&#xA;" OwnerUserId="622" LastActivityDate="2010-12-22T07:04:39.743" CommentCount="5" />
  <row Id="15162" PostTypeId="1" AcceptedAnswerId="15167" CreationDate="2010-12-22T07:17:27.120" Score="3" ViewCount="112" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First off: I barely have any set theoretic knowledge, but I read a bit about cardinal arithmetic today and the following idea came to me, and since I found it kind of funny, I wanted to know a bit more about it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $A$ is the set of all real positive sequences that either converge to $0$ or diverge to $\infty$, we put an equivalence relation &quot;$\sim$&quot; on $A$ defined as $a \sim b$ iff $\lim \frac a b \in \mathbb R ^+$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $B$ is the set of all infinite cardinals, can we associate to every equivalence class $[a]$ in $A/\sim$ a cardinal $p([a])$ or to every cardinal $\lambda$ an equivalence class $[q(\lambda)]$ in such a way that the map $p: A/\sim \to B$, or $q: B \to A/\sim$ is a &quot;homomorphism&quot;? That is, so that we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ p([a] + [b]) = p([a]) + p([b]) $$ or $$ [q(\lambda + \mu)] = [q(\lambda)] + [q(\mu)]$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If yes, could this map even be surjective, injective or an &quot;isomorphism&quot;? (I don0t know how many cardinals there are of course...)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It at least superficially seems to make some sense, since for cardinals $\lambda, \mu$ we have $\lambda + \mu = \max\{\lambda, \mu\}$ and the same is true for the classes of sequences, if we order them by $a &amp;lt; b \Leftrightarrow \lim \frac a b = 0$&lt;/p&gt;&#xA;" OwnerUserId="3208" LastEditorUserId="622" LastEditorDisplayName="" LastEditDate="2010-12-22T07:54:37.783" LastActivityDate="2010-12-25T21:29:36.747" Title="&quot;Homomorphism&quot; from set of sequences to cardinals?" Tags="&lt;set-theory&gt;&lt;cardinals&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="15163" PostTypeId="2" ParentId="15158" CreationDate="2010-12-22T07:19:53.417" Score="0" ViewCount="" Body="&lt;p&gt;Assuming you mean: $\frac{1}{8}(3m - 5) = \frac{6}{5}(6m - 2) + 5$&lt;br&gt;&#xA;You would solve as:&lt;br&gt;&#xA;$$40 * \frac{1}{8}(3m - 5) = 40 * (\frac{6}{5}(6m - 2) + 5)$$&#xA;$$5 * (3m - 5) = (48 * (6m - 2)) + 200$$&#xA;$$15m - 25 = 288m - 96 + 200$$&#xA;$$15m - 25 = 288m + 104$$&#xA;$$-129 = 273m$$&#xA;$$m = -\frac{129}{273} = -\frac{43}{91}$$    &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Assuming you mean $\frac{1}{8(3m - 5)} = \frac{6}{5(6m - 2)} + 5$&lt;br&gt;&#xA;You would solve as:&lt;br&gt;&#xA;$$\frac{1}{24m - 40} = \frac{6}{30m - 10} + 5$$&#xA;$$(30m - 10)(24m - 40) * \frac{1}{24m - 40} = (30m - 10)(24m - 40) * (\frac{6}{30m - 10} + 5)$$&#xA;$$30m - 10 = 24m - 40 + (5*(30m - 10)(24m - 40))$$&#xA;$$30m - 10 = 24m - 40 + (5 * (720m^2 - 1140m + 400))$$&#xA;$$30m - 10 = 24m - 40 + 3600m^2 - 7200m + 2000$$&#xA;$$0 = 3600m^2 - 7206m + 1970$$&#xA;$$m = \frac{1201 \pm \sqrt{654401}}{1200}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I would double check the work on both of these though, I'm tired =/) &lt;/p&gt;&#xA;" OwnerUserId="5038" LastEditorUserId="5038" LastEditorDisplayName="" LastEditDate="2010-12-22T07:33:32.613" LastActivityDate="2010-12-22T07:33:32.613" CommentCount="5" />
  <row Id="15164" PostTypeId="1" AcceptedAnswerId="15816" CreationDate="2010-12-22T07:38:13.647" Score="1" ViewCount="124" Body="&lt;p&gt;This &lt;a href=&quot;http://math.stackexchange.com/questions/15033/find-the-outgoing-edge-with-the-smallest-angles-given-one-incident-edges-and-mul&quot;&gt;is a similar question&lt;/a&gt; to this one, but slightly different. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is given two edges ($e_1$ and $e_2$, with the vertex coordinates known), how to find the angles from $e_1$ to $e_2$, with the angles measured in anti clock wise direction?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A diagram is shown below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://docs.google.com/drawings/pub?id=1qQx7aHPuQv3wX9UULf-6a082MTUlODFhv2iIBogAWyA&amp;amp;w=960&amp;amp;h=720&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One way I can think of is to compute the cross and dot product of the two edge's unit vector&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sin\,\theta=\frac{|e_1\times e_2|}{|e_1||e_2|}$$&#xA;$$\cos\,\theta=\frac{e_1\cdot e_2}{||e_1|| ||e_2||}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And try to find the $\theta$, taken into account of whether $sin\theta$ and $cos\theta$ is $&amp;gt;0$ or $&amp;lt;0$. But this is very, very tedious and error prone. Not to mention I'm not too sure whether the angle I get is &lt;em&gt;always&lt;/em&gt; measured in counter clock wise direction or not. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a single, clean formula that allows me to do what I want to do?&lt;/p&gt;&#xA;" OwnerUserId="550" LastActivityDate="2010-12-29T13:09:11.507" Title="Find the Angle ( as Measured in Counter Clock Wise Direction) Between Two Edges" Tags="&lt;geometry&gt;&lt;trigonometry&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="1" />
  <row Id="15165" PostTypeId="2" ParentId="15101" CreationDate="2010-12-22T07:42:53.620" Score="0" ViewCount="" Body="&lt;p&gt;i want this to work automatic.&lt;/p&gt;&#xA;" OwnerUserId="5039" LastActivityDate="2010-12-22T07:42:53.620" CommentCount="1" />
  <row Id="15166" PostTypeId="1" AcceptedAnswerId="15187" CreationDate="2010-12-22T07:45:40.550" Score="4" ViewCount="298" Body="&lt;p&gt;Could you a give a intuitive interpretation of curl, geometrical interpretation, real-world example or physical interpretation would be ok.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Consider a specific vector field &#xA;$$\mathbf{v} = -y\mathbf{\hat{x}} + x\mathbf{\hat{y}}$$&#xA;$$\nabla\times\mathbf{v} = 2\mathbf{\hat{z}}$$&#xA;Could you give a intuitive interpretation about why all points in this field has this curl w? What if &#xA;$\mathbf{v} = y\mathbf{\hat{x}} - x\mathbf{\hat{y}}$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="1382" LastEditorUserId="346" LastEditorDisplayName="" LastEditDate="2010-12-22T10:29:04.137" LastActivityDate="2010-12-23T01:02:11.163" Title="Could you a give a intutive interpretation of curl?" Tags="&lt;intuition&gt;&lt;vector-calculus&gt;" AnswerCount="3" FavoriteCount="1" />
  <row Id="15167" PostTypeId="2" ParentId="15162" CreationDate="2010-12-22T07:52:51.467" Score="5" ViewCount="" Body="&lt;p&gt;I think that there are a handful of points that might need clarification here.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;There is no &quot;set of all infinite cardinals&quot;, the family of all infinite cardinals is a proper class, and if it were a set it would imply the &lt;a href=&quot;http://en.wikipedia.org/wiki/Burali-Forti_paradox&quot; rel=&quot;nofollow&quot;&gt;Burali-Forti paradox&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;Secondly, consider the meaning of this equivalence relation: $a\sim b$ if and only if they behave the same asymptotically, up to some finite constant. You want, now to take this and say &quot;cardinals and sequences behave similarly&quot; in the manner that the &quot;stronger&quot; wins. This intuition is not too bad, although I'd have taken multiplication and not addition. However, when you want some structure to be preserved, it might take a bit more than just &quot;behaving the same&quot;, because certain interpretation of the same idea could have very different properties, as in my next point&lt;/li&gt;&#xA;&lt;li&gt;Cardinals are well ordered. There is a minimal cardinal, and there is a next one after that, and a next one after that. In your equivalence classes take $n^{-x}$ and $n^{-y}$, for choice of different $x,y\in\mathbb{R}^+$ those will give you different equivalence classes, it means you have continuum many equivalence classes and in terms of ordering - you're a lot like the order of $\mathbb{R}^{\ge 0}$ which is dense, which makes the idea of taking a cardinal for each equivalence class (even if you do only limit yourself to the first continuum-many cardinals) a tad problematic. &lt;/li&gt;&#xA;&lt;li&gt;And another point would be infinite summations. Cardinal arithmetic is slightly different when it comes to infinite sums, for example take the first $\aleph$ number which has an uncountable index (denoted by $\aleph_{\omega_1}$) then every addition of the form $\sum_{i\in\mathbb{N}} \lambda_i$ where all the $\lambda_i&amp;lt;\aleph_{\omega_1}$ is still a lot smaller than $\aleph_{\omega_1}$. So unlike the real numbers, adding uncountably many cardinals would have meaning if the &lt;a href=&quot;http://en.wikipedia.org/wiki/Continuum_hypothesis&quot; rel=&quot;nofollow&quot;&gt;Continuum Hypothesis&lt;/a&gt; fails, &lt;/li&gt;&#xA;&lt;li&gt;If however CH holds, and you only speak of the first $\omega_1$ cardinals then summation has a different trick up its sleeve: $\sum_{i\in\mathbb{N}}\lambda_i$ is just the biggest $\lambda_i$ or the first one which is bigger than all of them. I'm not quite certain how that would be handled by the equivalent classes.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;That been said, &quot;homomorphism&quot; is not dealing with infinite summations so the two points about that might be redundant, but these are things to consider when you go between two structures, especially infinite ones that have a very strong meaning when it comes to sequences and convergence (because what is an infinite sum? It is the limit of its finite partial sums).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is probably a &quot;nice&quot; way disproving the existence of such function, but I will leave this for Future-Asaf as well the other members of the site.&lt;/p&gt;&#xA;" OwnerUserId="622" LastActivityDate="2010-12-22T07:52:51.467" CommentCount="2" />
  <row Id="15168" PostTypeId="2" ParentId="15139" CreationDate="2010-12-22T07:56:20.363" Score="7" ViewCount="" Body="&lt;p&gt;The problem is overdetermined/impossible.  The values of $Z_1$ and $Z_2$ are forced, &lt;a href=&quot;http://math.stackexchange.com/questions/15139/two-points-on-circle-resulting-in-5-equal-regions/15142#15142&quot;&gt;as in Ross's answer&lt;/a&gt;, but the segments with area $\frac{2\pi}{5}$ are not divided into regions of equal area.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $0&amp;lt;\alpha&amp;lt;\pi$ be the solution to $\frac{\alpha-\sin\alpha}{2}=\frac{\pi}{5}$ (so $\alpha\approx 2.11314$).  $\arg(Z_2)=\pi-\frac{\alpha}{2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $0&amp;lt;\beta&amp;lt;\pi$ be the solution to $\frac{\beta-\sin\beta}{2}=\frac{2\pi}{5}$ (so $\beta\approx 2.8248$).  $\arg(\overline{Z_1})=\arg(Z_2)-\beta=\pi-\frac{\alpha}{2}-\beta$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any point on the line segment joining $Z_2$ and $\overline{Z_1}$ is a linear combination of these two points, $t\cdot Z_2+(1-t)\overline{Z_1}$ with $0&amp;lt;t&amp;lt;1$.  The intersection of that line segment and the line segment joining $\overline{Z_2}$ and $Z_1$ has imaginary part $0$.  Solving for $t$ in terms of $\alpha$ and $\beta$ yields $t=\frac{1}{2}\csc\frac{\beta}{2}\sec\frac{\alpha+\beta}{2}\sin(\frac{\alpha}{2}+\beta)$ ($t\approx 0.436382$).  The location of the point of intersection of the two lines is at $-\cos\frac{\beta}{2}\sec\frac{\alpha+\beta}{2}+0i$ ($\approx 0.20166$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the area of the region determined by this point of intersection, $Z_1$, $\overline{Z_1}$, and the circle is the area of the segment of the circle plus the area of the triangle determined by those three points.  The area of the segment of the circle is $\frac{2\arg(Z_1)-\sin(2\arg(Z_1))}{2}=\frac{1}{2}(\alpha+2\beta-2\pi-\sin(\alpha+2\beta))$ $\approx 0.241854$; the area of the triangle is $|\sin^2(\frac{\alpha}{2}+\beta)\tan\frac{\alpha+\beta}{2}|\approx 0.361977$ (found using shoelace method with the three vertices).  The total area of the region is thus $\approx 0.603831&amp;lt;0.628319\approx\frac{\pi}{5}$.&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-12-22T07:56:20.363" />
  <row Id="15169" PostTypeId="2" ParentId="15166" CreationDate="2010-12-22T08:01:29.577" Score="5" ViewCount="" Body="&lt;p&gt;From Wikipedia:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If the vector field represents the flow velocity of a moving fluid, then the curl is the circulation density of the fluid.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The circulation is defined as the integral of the velocity around a closed curve. An intuitive way of thinking about this is that if the fluid flows in the same direction everywhere, then integrating the velocity around any closed curve will give you 0. Indeed, you will integrate in the direction of the flow on some interval of the curve, but you will also have to &quot;come back&quot; and integrate against the flow, so the contributions will cancel. If on the other hand the fluid rotates in a vortex, then as you integrate the velocity around a circle centred on the axis of rotation, you pick up a positive contribution to the integral at each point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Further from wikipedia:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Suppose the vector field describes the velocity field of a fluid flow (maybe a large tank of water or gas) and a small ball is located within the fluid or gas (the centre of the ball being fixed at a certain point). If the ball has a rough surface it will be made to rotate by the fluid flowing past it. The rotation axis (oriented according to the right hand rule) points in the direction of the curl of the field at the centre of the ball, and the angular speed of the rotation is half the value of the curl at this point.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="3212" LastActivityDate="2010-12-22T08:01:29.577" CommentCount="1" />
  <row Id="15170" PostTypeId="1" CreationDate="2010-12-22T08:12:57.253" Score="0" ViewCount="111" Body="&lt;p&gt;$-2(y-2)= 7(5-y)-9$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;does y= 1/8 in the equation above&lt;/p&gt;&#xA;" OwnerUserId="5037" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-12-22T08:18:39.557" LastActivityDate="2010-12-22T15:17:54.037" ClosedDate="2010-12-22T15:27:33.657" Title="Is my answer of $-2(y-2)= 7(5-y)-9$ correct?" Tags="&lt;algebra-precalculus&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="15171" PostTypeId="2" ParentId="15107" CreationDate="2010-12-22T08:19:40.100" Score="1" ViewCount="" Body="&lt;p&gt;So you want to compute the following integral:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align*}&#xA;\frac{1}{(2\pi|\Sigma|)^{n/2}}\int_{V_k}\exp\left(-\frac{1}{2}(x-a)^T\Sigma^{-1}(x-a)\right)dx&#xA;\end{align*}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $V_k\subset \mathbb{R}^k$ is a $k$-polytope, $a\in \mathbb{R}^k$ is the mean and $\Sigma$ is the $k\times k$ covariance matrix of $k$-variate normal vector. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My approach would be doing a change of variables $y=\Sigma^{1/2}(x-a)$ and then integrate by one variable at a time, since the integrand function will be a product of one-variable functions. If $\Sigma^{1/2}V_k$ is a $k$-cube then the integration is straightforward. For the general case it will not be easy. Take $k=2$, normal with zero mean and unit covariance matrix, and $V_2$ a triangle with vertices in $(0,0)$, $(1,0)$, $(0,1)$. Then your integral will be&#xA;\begin{align}&#xA;\frac{1}{2\pi}\int_{V_2}e^{-\frac{x_1^2}{2}}e^{-\frac{x_2^2}{2}}dx_1dx_2&amp;amp;=\frac{1}{2\pi}\int_{0}^{1}e^{-\frac{x_1^2}{2}}dx_1\int_{0}^{1-x_1}e^{-\frac{x_2^2}{2}}dx_2\\&#xA;&amp;amp;=\frac{1}{\sqrt{2\pi}}\int_0^1e^{-\frac{x_1^2}{2}}(\Phi(1-x_1)-\Phi(0))dx_1&#xA;\end{align}&#xA;which I do not think has closed formula. (I might be wrong).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for approximation, I do not see why standard numerical integral solving methods would not work.&lt;/p&gt;&#xA;" OwnerUserId="4742" LastActivityDate="2010-12-22T08:19:40.100" CommentCount="1" />
  <row Id="15172" PostTypeId="1" CreationDate="2010-12-22T08:47:08.150" Score="0" ViewCount="63" Body="&lt;p&gt;Describe the open sets if all straight lines in the plane parallel to the x-axis are&#xA;used for subbasis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, wouldn't all open sets just be horizontal lines? Or whatelse should I specify?&lt;/p&gt;&#xA;" OwnerUserId="5036" LastActivityDate="2010-12-22T11:36:26.440" Title="Straight lines in the plane parallel to the x-axis as a subbasis for a topology" Tags="&lt;topology&gt;" AnswerCount="0" CommentCount="1" />
  <row Id="15173" PostTypeId="2" ParentId="9851" CreationDate="2010-12-22T09:17:55.817" Score="0" ViewCount="" Body="&lt;p&gt;I'm not sure if this applies, but for minimum-phase functions, the phase and magnitude of the Fourier Transform are related.  See &lt;a href=&quot;http://en.wikipedia.org/wiki/Minimum_phase#Relationship_of_magnitude_response_to_phase_response&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; for a brief overview.  I've never actually used this relationship in practice, so can't really give you much more information.&lt;/p&gt;&#xA;" OwnerUserId="5042" LastActivityDate="2010-12-22T09:17:55.817" />
  <row Id="15174" PostTypeId="1" AcceptedAnswerId="15457" CreationDate="2010-12-22T09:21:50.943" Score="2" ViewCount="112" Body="&lt;p&gt;How do I find generators of group of rotations of d-dimensional cube into itself when viewed as a subgroup of symmetric group of order $2^d$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is related to my previous &lt;a href=&quot;http://math.stackexchange.com/questions/15022/how-to-get-generators-for-a-finite-point-group&quot;&gt;question&lt;/a&gt; but not specializing to $d=3$&lt;/p&gt;&#xA;" OwnerUserId="998" LastEditorUserId="998" LastEditorDisplayName="" LastEditDate="2010-12-22T09:35:57.387" LastActivityDate="2010-12-24T18:29:15.017" Title="How to get group of hypercube rotations as a subgroup of symmetric group?" Tags="&lt;group-theory&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="0" />
  <row Id="15175" PostTypeId="2" ParentId="15134" CreationDate="2010-12-22T09:23:35.793" Score="1" ViewCount="" Body="&lt;p&gt;Terence Tao has set up a &quot;polymath project&quot; which addresses exactly your question. Here is the relevant page:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://michaelnielsen.org/polymath1/index.php?title=Finding_primes&quot; rel=&quot;nofollow&quot;&gt;http://michaelnielsen.org/polymath1/index.php?title=Finding_primes&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It contains a list of conjectures, partial results and references. To sum it all up: At the moment the answer to your question is No.&lt;/p&gt;&#xA;" OwnerUserId="1303" LastActivityDate="2010-12-22T09:23:35.793" />
  <row Id="15177" PostTypeId="1" CreationDate="2010-12-22T10:01:48.507" Score="1" ViewCount="111" Body="&lt;p&gt;I have two numbers $x$ and $y$ which are $\log_2$ of two other numbers $P$ and $V$.  I'm trying to calculate $\log_2 (P-V)$ without transforming $x$ and $y$ back to the linear world.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $P = AV$, then $P-V = V(A-1)$, and $x-y = \log_2 A$.  So, if I could calculate $\log_2 (A-1)$ I could just add it to $y$ and be done.  Is there a relationship I can use to do this?&lt;/p&gt;&#xA;" OwnerUserId="5042" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-22T10:36:03.610" LastActivityDate="2010-12-22T10:36:03.610" Title="Given $\log_2 A$, is there an easy way to calculate $\log_2 (A-1)$?" Tags="&lt;logarithms&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="15178" PostTypeId="1" AcceptedAnswerId="15182" CreationDate="2010-12-22T10:09:53.333" Score="0" ViewCount="200" Body="&lt;p&gt;Hola, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm currently creating a program to plot simple graphs in 2D. In order to draw those properly I have to know if a function graph is continuous between 2 points on the graph.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming the graph for f(x)=tan(x) and two points on the graph P1(1,tan(1)) and P2(2,tan(2)). Is there an algorithmic way to tell if the graph is continuous for 1 &amp;lt;= x &amp;lt;= 2?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;thanks&lt;/p&gt;&#xA;" OwnerUserId="5043" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-22T10:40:26.697" LastActivityDate="2010-12-22T11:04:33.983" Title="How to tell if a function is continuous in an interval" Tags="&lt;analysis&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="15179" PostTypeId="2" ParentId="15177" CreationDate="2010-12-22T10:23:34.247" Score="3" ViewCount="" Body="&lt;p&gt;One way is to write (am using natural logs for now): $\log(A-1) = \log A + \log (1-1/A)$, and to then use the power series for $\log(1-x)$ to approximate the latter term. But this is not really a good solution.&lt;/p&gt;&#xA;" OwnerUserId="1709" LastActivityDate="2010-12-22T10:23:34.247" CommentCount="1" />
  <row Id="15180" PostTypeId="2" ParentId="15124" CreationDate="2010-12-22T10:24:29.883" Score="4" ViewCount="" Body="&lt;p&gt;Let $\tau  = \min \{ n \geq 1:X_1  +  \cdots  + X_n  &amp;gt; K \}$. Then $\tau$ is an integer-valued random variable, bounded from above by $K+1$ (since $X_i \geq 1$). Note that $\tau = n$ if and only if $\sum\nolimits_{i = 1}^{n - 1} {X_i }  \le K$ and $\sum\nolimits_{i = 1}^{n} {X_i } &amp;gt; K$. Thus, the event $\lbrace \tau = n \rbrace$ depends only on the values $X_1,\ldots,X_n$. So, by definition, $\tau$ is a stopping time with respect to the sequence $X_1,X_2,\ldots$.  Now, $X_1,X_2,\ldots$ are i.i.d. with finite expectation $\mu$, and $\tau$ is a stopping time for them. Moreover, ${\rm E}(\tau) &amp;lt; \infty$ since $\tau \leq K+1$. Hence, by Wald's identity,&#xA;$$&#xD;&#xA;{\rm E}\bigg(\sum\limits_{i = 1}^\tau  {X_i } \bigg) = {\rm E}(\tau )\mu \leq (K+1)\mu.&#xD;&#xA;$$&#xA;So if we put $Y_\tau = \sum\nolimits_{i = 1}^\tau  {X_i }$, we get&#xA;$$&#xD;&#xA;{\rm E}(Y_\tau - K) = {\rm E}(Y_\tau) - K \leq (K+1)\mu - K.&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $\tau \geq 1$, we have&#xA;$$&#xD;&#xA;\mu - K \leq {\rm E}(Y_\tau - K) \leq (K+1)\mu - K.&#xD;&#xA;$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As we have seen above, the problem reduces to calculating ${\rm E}(\tau)$. Put $S_n = \sum\nolimits_{i = 1}^n {X_i }$ ($S_0 = 0$).&#xA;Note that &#xA;$$&#xD;&#xA;{\rm P}(\tau  = n) = {\rm P}(S_{n - 1}  \le K,S_n  &amp;gt; K) = {\rm P}(S_{n - 1}  \le K) - {\rm P}(S_n  \le K).&#xD;&#xA;$$&#xA;Hence,&#xA;$$&#xD;&#xA;{\rm E}(\tau) = \sum\limits_{n = 1}^{K + 1} {n{\rm P}(\tau  = n)}  = \sum\limits_{n = 1}^{K + 1} n[{\rm P}(S_{n - 1}  \le K) - {\rm P}(S_n  \le K)] = \sum\limits_{n = 0}^K {{\rm P}(S_n  \le K)}. &#xD;&#xA;$$&#xA;So, we can write &#xA;$$&#xD;&#xA;{\rm E}(\tau) = 1 + \sum\limits_{n = 1}^\infty  {{\rm P}(S_n  \le K)} = 1 + \sum\limits_{n = 1}^\infty  {F^{(n)}(K)},&#xD;&#xA;$$&#xA;where $F^{(n)}$ is the distribution function of $S_n$. For $t&amp;gt;0$ real, define $m(t) = \sum\nolimits_{n = 1}^\infty  {F^{(n)}(t)}$.&#xA;From the theory of renewal processes, we know that $m(t) = {\rm E}(N_t)$, where $\lbrace N_t:t \geq 0 \rbrace$ is a renewal process with inter-arrival times distributed according to the distribution of $X$. $m(t)$ is called the renewal function. It may be worth noting that by the Elementary Renewal Theorem, &#xA;$$&#xD;&#xA;\mathop {\lim }\limits_{t \to \infty } \frac{{m(t)}}{t} = \frac{1}{\mu }.&#xD;&#xA;$$&#xA;Returning to our original setting, we have&#xA;$$&#xD;&#xA;{\rm E}(Y_\tau  ) = {\rm E}(\tau )\mu  = (1 + m(K))\mu.&#xD;&#xA;$$&#xA;So, the problem reduces to calculating $m(K)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, &lt;a href=&quot;http://www.postech.ac.kr/class/ie272/ie666_temp/Renewal.pdf&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; is some useful link concerning renewal theory, which is very relevant to this answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.postech.ac.kr/class/ie272/ie666_temp/Renewal.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.postech.ac.kr/class/ie272/ie666_temp/Renewal.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2810" LastEditorUserId="2810" LastEditorDisplayName="" LastEditDate="2010-12-22T14:16:22.937" LastActivityDate="2010-12-22T14:16:22.937" CommentCount="5" />
  <row Id="15181" PostTypeId="1" AcceptedAnswerId="15185" CreationDate="2010-12-22T11:03:30.817" Score="2" ViewCount="105" Body="&lt;p&gt;For example, I have set of points in 3D. Points lie on straight line. Transformed set of points lies on straight line too. How to check if transformation is affine? &lt;/p&gt;&#xA;" OwnerUserId="3532" LastActivityDate="2010-12-22T12:17:17.097" Title="How to check if transformation is affine?" Tags="&lt;linear-algebra&gt;" AnswerCount="2" />
  <row Id="15182" PostTypeId="2" ParentId="15178" CreationDate="2010-12-22T11:04:33.983" Score="5" ViewCount="" Body="&lt;p&gt;You can use interval arithmetic to compute reliable plots. See for instance this paper: Jeff Tupper, &lt;a href=&quot;http://www.dgp.toronto.edu/~mooncake/papers/SIGGRAPH2001_Tupper.pdf&quot; rel=&quot;nofollow&quot;&gt;Reliable Two-Dimensional Graphing Methods for Mathematical Formulae with Two Free Variables&lt;/a&gt;, SIGGRAPH 2001. The excellent &lt;a href=&quot;http://www.peda.com/grafeq/&quot; rel=&quot;nofollow&quot;&gt;GrafEq&lt;/a&gt; software uses this technique.&lt;/p&gt;&#xA;" OwnerUserId="589" LastActivityDate="2010-12-22T11:04:33.983" CommentCount="1" />
  <row Id="15184" PostTypeId="2" ParentId="15181" CreationDate="2010-12-22T11:44:40.500" Score="0" ViewCount="" Body="&lt;p&gt;Dear qutron,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;an &lt;a href=&quot;http://en.wikipedia.org/wiki/Affine_transformation&quot; rel=&quot;nofollow&quot;&gt;affine transformation&lt;/a&gt; between two vector spaces $$F:X\rightarrow Y$$ (one might define it more general) is defined as&#xA;$$y = F(x) = Ax + y_0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $A$ is a constant map (might be represented as matrix) and $y_0\in Y$ is a constant element.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, to check if a transformation is affine you might try to proof that such $A$ and $y_0$ exist.&lt;br&gt;&#xA;Sincerely&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Robert&lt;/p&gt;&#xA;" OwnerUserId="2671" LastActivityDate="2010-12-22T11:44:40.500" CommentCount="1" />
  <row Id="15185" PostTypeId="2" ParentId="15181" CreationDate="2010-12-22T11:56:16.743" Score="3" ViewCount="" Body="&lt;p&gt;According to the fundamental theorem of affine geometry, it suffices to check that the transformation is a bijective &lt;em&gt;collineation&lt;/em&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Theorem.&lt;/strong&gt; Let $\mathbb A^{n}$ be an affine space over $\mathbb R$ with $n &amp;gt; 2$ and fix&#xA;  $a \in A$. Let $\phi :\mathbb A^{n}\to \mathbb A^{n}$ be a bijection which takes each three collinear points into collinear points. Then there exists a point $b\in \mathbb A^{n}$ and an&#xA;  invertible linear map $F$ such that $\phi(x) = F(x-a) + b$ for all $x \in\mathbb  A^n$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The proof can be found in Berger's &lt;a href=&quot;http://books.google.com/books?id=iER421ZjkqcC&amp;amp;pg=PA320&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;Geometry 1&lt;/em&gt;&lt;/a&gt;  (Springer, 1987, pp. 52-56).&lt;/p&gt;&#xA;" OwnerUserId="723" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-22T12:17:17.097" LastActivityDate="2010-12-22T12:17:17.097" CommentCount="1" />
  <row Id="15186" PostTypeId="2" ParentId="15134" CreationDate="2010-12-22T13:07:39.887" Score="0" ViewCount="" Body="&lt;p&gt;Polynomial in $n$ is not efficient because the input length (measured in bits) is $\log_2 n$ and so $n$ is exponentially large in relation to your input length. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So it is easy to find a prime number greater than $n$ in time polynomial in $n$ (e.g. with the answer from Moron). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are interested in a deterministic algorithm than the polymath project gives you the current known results.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But there is a randomized algorithm with time polynomial in $\log_2 n$ that gives you a prime greater than $n$. You pick a random number from the set $n, \ldots, 2n$ and check this number with AKS or &lt;a href=&quot;http://en.wikipedia.org/wiki/Miller%E2%80%93Rabin_primality_test&quot; rel=&quot;nofollow&quot;&gt;Miller Rabin primality test&lt;/a&gt;. With the &lt;a href=&quot;http://en.wikipedia.org/wiki/Prime_number_theorem&quot; rel=&quot;nofollow&quot;&gt;prime number theorem&lt;/a&gt; you can proof that there are enough primes in the set (for large $n$) and if you repeat this algorithm $O(\log_2 n)$ times you find a prime with high probability.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the moment I cannot find a reference for a detailed proof for the randomized algorithm. But if you are interested I can look for it when I have more time.&lt;/p&gt;&#xA;" OwnerUserId="4857" LastActivityDate="2010-12-22T13:07:39.887" />
  <row Id="15187" PostTypeId="2" ParentId="15166" CreationDate="2010-12-22T13:36:53.063" Score="8" ViewCount="" Body="&lt;p&gt;Here's a &lt;a href=&quot;http://www.sciencelive.org/component/option,com_mediadb/task,view/idstr,Open-podcast-feeds_mst209_grad_div_curl_rss2_xml/Itemid,97&quot; rel=&quot;nofollow&quot;&gt;video&lt;/a&gt; explaining curl.&lt;/p&gt;&#xA;" OwnerUserId="247" LastActivityDate="2010-12-22T13:36:53.063" CommentCount="1" />
  <row Id="15188" PostTypeId="1" CreationDate="2010-12-22T13:50:14.890" Score="1" ViewCount="165" Body="&lt;p&gt;Let $A,B \text{ and } C$ are three sets then if $ A \subset B, B \subset C, C \subset A \Rightarrow B = C $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How could we prove this ?&lt;/p&gt;&#xA;" OwnerUserId="5045" LastEditorUserId="5045" LastEditorDisplayName="" LastEditDate="2010-12-22T13:57:10.877" LastActivityDate="2010-12-22T14:01:14.503" Title="How to prove :: $ A \subset B, B \subset C, C \subset A \Rightarrow B = C $" Tags="&lt;elementary-set-theory&gt;" AnswerCount="1" CommentCount="13" />
  <row Id="15189" PostTypeId="2" ParentId="15188" CreationDate="2010-12-22T14:01:14.503" Score="3" ViewCount="" Body="&lt;p&gt;This is effectively asking to prove $B \subseteq C \land C \subseteq B \implies B = C$. The usual way to prove this is to use the &lt;a href=&quot;http://en.wikipedia.org/wiki/Axiom_of_extensionality&quot; rel=&quot;nofollow&quot;&gt;Axiom of Extensionality&lt;/a&gt; - i.e. take an element $b \in B$ and show that it is in $C$. Then show that $c \in C \implies c \in B$. Extensionality now tells you that the two sets are identical. &lt;/p&gt;&#xA;" OwnerUserId="1269" LastActivityDate="2010-12-22T14:01:14.503" CommentCount="5" />
  <row Id="15190" PostTypeId="1" CreationDate="2010-12-22T14:21:56.950" Score="1" ViewCount="138" Body="&lt;p&gt;Hi,&#xA;i have $P_1(x_1, y_1, z_1)$ and $P_2(x_2, y_2, z_2)$ that lies on a line, $L$,&#xA;and another point $P_0(x_0, y_0, z_0)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i want to fine the distance between $P_0$ and $L$.&#xA;some helps?&lt;/p&gt;&#xA;" OwnerUserId="5046" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2010-12-22T15:38:20.760" LastActivityDate="2010-12-22T15:38:20.760" Title="Distance between a point and a line in space" Tags="&lt;euclidean-geometry&gt;" CommentCount="9" />
  <row Id="15191" PostTypeId="2" ParentId="8263" CreationDate="2010-12-22T14:44:13.733" Score="1" ViewCount="" Body="&lt;p&gt;For the second question, for single digits $a+b\geq ab$ when a=0 or 1, b=0 or 1, or a=b=2.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-12-22T14:44:13.733" />
  <row Id="15192" PostTypeId="1" AcceptedAnswerId="15230" CreationDate="2010-12-22T14:59:09.063" Score="2" ViewCount="79" Body="&lt;p&gt;Hello,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;let $X$ be some Banach space. Let $L(X)$ be the set of continuous operators on $X$ into $X$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $(\tau_i)_i$ be a set of topologies on $L(X)$ s.t. $L(X)$ is topological vector space (i.e. addition and scalar multiplication continuous, not necessarily Hausdorff).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $\tau$ be the initial topology on $L(X)$ induced by the linear identities $f_i : L(X) \longrightarrow (L(X),\tau_i)$. Then $(L(X),\tau)$ is a topological vector space, too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the sake of illustration: if we restrict $(\tau_i)_i$ to the topologies that are coarser than the operator norm topology on $X$, $\tau_{op}$, then obviously $\tau = \tau_{op}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; Does $\tau = \tau_{op}$ hold even in general? Alternatively, is there at least a &quot;nice&quot; set of restrictions on $(\tau_i)_i$, or a characterization of &quot;reasonable&quot; topologies, for which we have $\tau = \tau_{op}$?&lt;/p&gt;&#xA;" OwnerUserId="3557" LastEditorUserId="3557" LastEditorDisplayName="" LastEditDate="2010-12-22T19:13:52.007" LastActivityDate="2010-12-22T21:16:54.757" Title="Finest topology on a space of banach space operators" Tags="&lt;functional-analysis&gt;&lt;topological-vector-spaces&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="15193" PostTypeId="1" CreationDate="2010-12-22T15:07:45.773" Score="0" ViewCount="54" Body="&lt;p&gt;Apologies if the title is unclear, I couldn't think of a good way to express what I want.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given the usual 2D representation of the complete graph $K_n$ as a regular $n$-gon with each pair of vertices joined, let $U_n$ be the set of the disjoint regions of space delimited by the edges of the graph (so $|U_3|=1, |U_4|=4, |U_5|=11...$). Further, label the nodes of $K_n$ as $p_1, ..., p_n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the size of the largest subset $S_n \subset U_n$ such that $\forall s \in S_n \exists i, j, k \in K_n$ with $s \subset \triangle ijk$ and $t \not\subset \triangle ijk \forall t \in S_n, t \ne s$? That is, How many points can I place inside drawing of $K_n$ such that each triangle formed by three points on the exterior of $K_n$ contains at most one point?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In case it helps clarify what I'm trying to do, I'm working on a puzzle where letters are encoded by giving three points from a discrete set defining a triangle in which the letter is written. For instance, for $K_4$:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;1-------2&#xA;|\  A  /|&#xA;| \   / |&#xA;|  \ /  | &#xA;|   X   |&#xA;|  / \  |&#xA;| /   \ |&#xA;|/  B  \|&#xA;3-------4&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&quot;A&quot; can be encoded as 123 or 124; B as 234 or 134. No further letters can be unambiguously encoded.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Playing around with pencil and paper, it seems as though $K_n$ can define $n-2$ letters unambiguously, but I haven't investigated far, and I can't immediately come up with a proof.&lt;/p&gt;&#xA;" OwnerUserId="2432" LastActivityDate="2011-01-02T01:57:48.957" Title="Largest uniquely-identifiable subset of regions of space within complete graph $K_n$" Tags="&lt;geometry&gt;&lt;graph-theory&gt;" AnswerCount="1" />
  <row Id="15194" PostTypeId="1" CreationDate="2010-12-22T15:11:47.863" Score="12" ViewCount="333" Body="&lt;p&gt;A while back, I wanted to see if the notion of the &lt;a href=&quot;http://mathworld.wolfram.com/Arithmetic-GeometricMean.html&quot; rel=&quot;nofollow&quot;&gt;arithmetic-geometric mean&lt;/a&gt; could be extended to a pair of symmetric positive definite matrices. (I considered positive definite matrices only since the notion of the matrix square root is a bit intricate for other kinds of matrices.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I expected that some complications would arise since, unlike scalar multiplication, matrix multiplication is noncommutative. Another complication would be that the product of two symmetric matrices need not be symmetric (though the positive definiteness is retained, so one can still speak of a matrix square root).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By analogy with the scalar AGM, I considered the iteration&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathbf A_0=\mathbf A \; ; \; \mathbf B_0=\mathbf B$$ $$\mathbf A_{i+1}=\frac12(\mathbf A_i+\mathbf B_i) \; ; \; \mathbf B_{i+1}=\sqrt{\mathbf A_i \mathbf B_i}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I cranked up a short &lt;em&gt;Mathematica&lt;/em&gt; routine:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;matAGM[u_, v_] := First[FixedPoint[({Apply[Plus, #]/2, MatrixPower[Apply[Dot, #], 1/2]}) &amp;amp;, {u, v}]] /; MatrixQ[u, InexactNumberQ] &amp;amp;&amp;amp; MatrixQ[v, InexactNumberQ]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;and decided to try it out on randomly generated SPD matrices.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(A numerical note: &lt;em&gt;Mathematica&lt;/em&gt; uses the numerically stable Schur decomposition in computing matrix functions like the matrix square root.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found that for all of the randomly generated pairs of SPD matrices I tried, the process was convergent (though the rate of convergence is apparently not as fast as the scalar AGM). As expected, the order matters: &lt;code&gt;matAGM[A, B]&lt;/code&gt; and &lt;code&gt;matAGM[B, A]&lt;/code&gt; are usually not equal (and both results are &lt;em&gt;unnsymmetric&lt;/em&gt;) unless &lt;code&gt;A&lt;/code&gt; and &lt;code&gt;B&lt;/code&gt; commute (for the special case of diagonal &lt;code&gt;A&lt;/code&gt; and &lt;code&gt;B&lt;/code&gt;, the result is the diagonal matrix whose entries are the arithmetic-geometric means of the corresponding entries of the pair.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I now have three questions:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;How do I prove or disprove that this process converges for any pair of SPD matrices? If it is convergent, what is the rate of convergence?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Is there any relationship between &lt;code&gt;matAGM[A, B]&lt;/code&gt; and &lt;code&gt;matAGM[B, A]&lt;/code&gt; if the two matrices &lt;code&gt;A&lt;/code&gt; and &lt;code&gt;B&lt;/code&gt; do not commute?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Is there any relationship between this matrix arithmetic-geometric mean and the usual scalar arithmetic-geometric mean? Would, say, arithmetic-geometric means of the eigenvalues of the two matrices have anything to do with this?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="498" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-23T00:19:48.887" LastActivityDate="2011-01-21T10:47:08.060" Title="The arithmetic-geometric mean for symmetric positive definite matrices" Tags="&lt;linear-algebra&gt;&lt;matrices&gt;" AnswerCount="4" FavoriteCount="1" />
  <row Id="15195" PostTypeId="2" ParentId="15170" CreationDate="2010-12-22T15:17:54.037" Score="2" ViewCount="" Body="&lt;p&gt;No. You can check this by performing the following calculation (Plug in the answer and check for equality):&#xA;$$-2(y - 2) = 7(5 - y) - 9, y = \frac{1}{8}$$&#xA;$$-2(\frac{1}{8} - 2) = 7(5 - \frac{1}{8}) - 9$$&#xA;$$4 - \frac{2}{8} = 35 - \frac{7}{8} - 9$$&#xA;$$\frac{7}{8} - \frac{2}{8} = 35 - 9 - 4$$&#xA;$$\frac{5}{8} = 22$$&#xA;$$5 = 176$$&#xA;$$5 \ne 176 \therefore y \ne \frac{1}{8}$$    &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;To solve for the correct answer: &#xA;$$−2(y−2)=7(5−y)−9$$&#xA;$$4 - 2y = 35 - 7y - 9$$&#xA;$$5y = 35 - 9 - 4$$&#xA;$$5y = 22$$&#xA;$$y = \frac{22}{5}$$&lt;/p&gt;&#xA;" OwnerUserId="5038" LastActivityDate="2010-12-22T15:17:54.037" />
  <row Id="15196" PostTypeId="1" CreationDate="2010-12-22T15:42:19.283" Score="0" ViewCount="66" Body="&lt;p&gt;In the definition of the complexity class NP, we allow the certificate to be polynomial in the input size. But do we ever need a certificate that is longer than the input? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In all the NP problems I could think of, we don't. The &quot;worst&quot; problem was factorization&lt;s&gt;, where the obvious incoding is $n+O(\lg(n))$ (where n is length of the number we wish factor; I just realized that the input also contain a maximal number, as the decision problem is: Do there exists a factor less than some number), because we want to tell where the first factor end and the second begins. However we don't need to tell this, the verifier could just try all possibilities. Furthermore, we could decide not to write the last bit of the factors, and this way we get a certificate that is shorter than the number we wish to factor. &lt;/s&gt; Do'h division is fast, so you could just give one of the factors as certificate.&lt;/p&gt;&#xA;" OwnerUserId="3078" LastEditorUserId="3078" LastEditorDisplayName="" LastEditDate="2010-12-23T15:11:36.573" LastActivityDate="2010-12-23T15:11:36.573" Title="NP-problem that needs superlinear certificates" Tags="&lt;computer-science&gt;" AnswerCount="1" />
  <row Id="15197" PostTypeId="2" ParentId="15154" CreationDate="2010-12-22T15:44:50.910" Score="1" ViewCount="" Body="&lt;p&gt;Sorry, I'd comment but I'm new here. I'm also getting (once we round to two decimal places) $114522.58$. Perhaps the machine is expecting rounding up (to $114522.59$) instead of rounding to the nearest hundredth?&lt;/p&gt;&#xA;" OwnerUserId="4689" LastActivityDate="2010-12-22T15:44:50.910" />
  <row Id="15198" PostTypeId="2" ParentId="15075" CreationDate="2010-12-22T15:53:50.477" Score="6" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Added&lt;/strong&gt;: Unfortunately, the answer below as originally written was wrong.  It answered a related, but easier, question.  I have added some bracketed remarks to point out where the blunder happens, and appended some remarks about the expected answer to the actual question.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;It might help to explicitly reformulate the question.  (This is implicit in the proof that the question is equivalent to the twin prime conjecture for $a = 6$, but it seems worthwhile to make it explicit.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $k = a m n \pm m \pm n,$ &#xA;then &#xA;$a k \pm 1 = (a m \pm 1) (a n \pm 1).$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $a = 2$ then we can write $1 = (2\cdot 1 - 1)$, and so for any $k$,&#xA;we have $2 k - 1 = (2 \cdot 1 - 1)(2 k  -1)$,&#xA;and (as the OP noted) every integer $k$ can be written in the desired form.&#xA;Thus we suppose from now on that $a &amp;gt; 2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is then equivalent to asking if we can find infinitely many&#xA;numbers congruent to $\pm 1 \bmod a$ which have no proper factors congruent&#xA;to $\pm 1 \bmod a$.  [&lt;strong&gt;Added&lt;/strong&gt;: This is wrong.  The problem is rather to&#xA;find infintely many multiples $a k$ of $a$ so that &lt;em&gt;both&lt;/em&gt; $a k +1 $ and&#xA;$a k - 1$ have no proper factors congruent to $\pm 1 \bmod a$.]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This shows that the answer to the question depends a lot  on whether or not&#xA;&lt;em&gt;every&lt;/em&gt; residue class mod $a$ which is coprime to $a$ is congruent to $\pm 1 \bmod a$, i.e. on whether or not $\varphi(a) = 2$.  [&lt;strong&gt;Added&lt;/strong&gt;: It is not so clear to me now what difference the condition on $\varphi(a)$ makes, if any.]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $\varphi(a) = 2$, then we are simply asking whether we can find infinitely many numbers coprime to $a$, which differ by $2$, and which have no proper factors, i.e. we asking for infinitely many twin primes.  This is a well-known open problem!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand,&#xA;suppose that $\varphi(a) &amp;gt; 2$ (which in particular is the case if $a &gt; 6$).  We may then choose &#xA;$b$ and $c$ so that $b, c \not\equiv \pm 1 \bmod a$, and $b c \equiv 1 \bmod a$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By Dirichlet's theorem, we may choose infinitely primes $p$ and $q$ such that&#xA;$p \equiv b \bmod a$ and $q \equiv c \bmod a$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For any such $p$ and $q$, the product $pq \equiv 1 \bmod a$, but by construction has no proper factors congruent to $\pm 1 \bmod a$.&#xA;So if we set $k = (p q - 1)/a$, then $k$ cannot be written in the given form.&#xA;This gives infinitely many $k$, as desired.&#xA;[&lt;strong&gt;Added&lt;/strong&gt;: Rather, this gives infinitely many $k$ that are not of the form&#xA;$a m n + m + n$ or $a m n - m - n$, but doesn't address whether these can be written in the form $a m n - m + n$, since we don't have any control over $a k - 1 = p q - 2.] &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added&lt;/strong&gt;:&#xA;It is not clear to me how to gain control over the factors of $p q - 2$, other than to assume it is prime.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So one way to solve the problem would be to find infinitely many primes $r \equiv -1 \bmod a,$ such that $r + 2$ is either prime, or is of the form&#xA;$r + 2 = p q,$ where $p,q \not\equiv \pm 1 \bmod a$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is then related to Chen's theorem, indeed is a strengthening of it.  Thus&#xA;we are led to &lt;a href=&quot;http://mathoverflow.net/questions/49751/chens-theorem-with-congruence-conditions&quot; rel=&quot;nofollow&quot;&gt;MO Scribe's question&lt;/a&gt; on MO which was inspired by the initial posting of the present question on MO.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is no doubt that this result (on the existence of such primes $r$) is true. (Indeed, standard conjectures predict that there should be infinitely many twin primes $r, r+2$ with $r\equiv -1 \bmod a$.)  But it is not clear (to me) whether or not it is in reach of current methods; that is the thrust of MO Scribe's question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what is the conclusion:  Well, there is no doubt that one should be able to find infintely many such $k$, since it follows from standard conjectures on twin primes satisfying congruence conditions.  On the other hand, proving this may be tricky, since it seems to require results that are at the edge of what is currently possible via seiving techniques.&lt;/p&gt;&#xA;" OwnerUserId="221" LastEditorUserId="221" LastEditorDisplayName="" LastEditDate="2010-12-22T19:06:08.187" LastActivityDate="2010-12-22T19:06:08.187" CommentCount="14" />
  <row Id="15199" PostTypeId="1" AcceptedAnswerId="15206" CreationDate="2010-12-22T15:56:02.397" Score="1" ViewCount="97" Body="&lt;p&gt;Three fellows accused of stealing CDs make the following statements:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) Ed: “Fred did it, and Ted is innocent.”&lt;br&gt;&#xA;(2) Fred: “If Ed is guilty, then so is Ted.”&lt;br&gt;&#xA;(3) Ted: “I’m innocent, but at least one of the others is guilty.”&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the innocent told the truth and the guilty lied, who is guilty?&#xA;(Remember that false statements imply anything).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think Ed and Ted are innocent and Fred is guilty.&#xA;Is it in contradiction with statement 2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What do you say?&lt;/p&gt;&#xA;" OwnerUserId="5049" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-22T16:44:14.213" LastActivityDate="2010-12-22T16:44:14.213" Title="Implication of three statements" Tags="&lt;logic&gt;&lt;puzzle&gt;" AnswerCount="2" CommentCount="0" />
  <row Id="15200" PostTypeId="2" ParentId="15196" CreationDate="2010-12-22T15:57:20.663" Score="1" ViewCount="" Body="&lt;p&gt;In the definition of NP, no attempt is made to shorten the size of the certificate. Since the running time is polynomial and the machine might &quot;guess&quot; at each step, the certificate is polynomial.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can artificially create an NP problem which seems to require longer certificates: given an instance $x$, answer some predicate on $G(x)$, where $G$ is some PRNG that takes an input of size $n$ and outputs a string of size $n^2$. Now the certificate would be quadratic (though probably no-one knows whether it has to be that large).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an aside, &quot;factorization&quot; is not an NP predicate. To make it a predicate, you need to ask some &quot;hard-core&quot; question on one of the factors.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-12-22T15:57:20.663" CommentCount="3" />
  <row Id="15201" PostTypeId="1" CreationDate="2010-12-22T16:03:44.500" Score="7" ViewCount="537" Body="&lt;p&gt;Hello,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is your recommendation for an in-depth introductory combinatoric book? A book that doesn't just tell you about the multiplication principle, but rather shows the whole logic behind the questions with full proofs. The book should be for a first-year-student in college. Do you know a good book on the subject?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="5023" LastEditorUserId="622" LastEditorDisplayName="" LastEditDate="2010-12-22T16:07:53.297" LastActivityDate="2011-05-24T09:11:40.367" Title="Good Book On Combinatorics" Tags="&lt;combinatorics&gt;&lt;reference-request&gt;" AnswerCount="14" CommentCount="6" FavoriteCount="4" />
  <row Id="15202" PostTypeId="2" ParentId="15201" CreationDate="2010-12-22T16:20:35.127" Score="7" ViewCount="" Body="&lt;p&gt;If the student's leaning towards computer science at all, I'd recommend Knuth et. al.'s &lt;a href=&quot;http://en.wikipedia.org/wiki/Concrete_Mathematics&quot; rel=&quot;nofollow&quot;&gt;Concrete Mathematics&lt;/a&gt;. It's full of solid math and has the aim of building mathematical tools for CS. Other than that, a newer addition that looks promising is Russell Merris's &quot;Combinatorics, 2nd ed.&quot; It gives a pretty broad introduction while also giving in-depth work and examples.&lt;/p&gt;&#xA;" OwnerUserId="4689" LastActivityDate="2010-12-22T16:20:35.127" CommentCount="1" />
  <row Id="15203" PostTypeId="2" ParentId="15199" CreationDate="2010-12-22T16:21:28.173" Score="0" ViewCount="" Body="&lt;p&gt;You are right.  According to the problem, the guilty lie, so Fred's statement should be false.  If the problem is well constructed (and I think this one is) there should be only one consistent assignment of truth values to the atomic sentences.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a large body of problems of this nature.  One of my favorite authors if Raymond Smullyan.  A couple of his books lead you through the logic of Goedel's incompleteness proof.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-12-22T16:21:28.173" CommentCount="2" />
  <row Id="15204" PostTypeId="1" AcceptedAnswerId="15211" CreationDate="2010-12-22T16:24:33.420" Score="1" ViewCount="75" Body="&lt;p&gt;Please forgive me if this is not the right Stack Exchange (I also posted it at &lt;a href=&quot;http://stats.stackexchange.com/questions/5698/can-the-mutual-information-of-a-cell-be-negative&quot;&gt;Cross Validated&lt;/a&gt; and &lt;a href=&quot;http://cstheory.stackexchange.com/questions/3939/can-the-mutual-information-of-a-cell-be-negative&quot;&gt;Theoretical Computer Science&lt;/a&gt;).  Please also forgive me for inventing terms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For discrete random variables X and Y, the mutual information of X and Y can be defined as follows:&#xA;$I(X;Y) = \sum_{y \in Y} \sum_{x \in X} &#xD;&#xA;                 p(x,y) \log{ \left( \frac{p(x,y)}{p_1(x)\,p_2(y)}&#xD;&#xA;                              \right) }, \,\!$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I will define the mutual information of a &quot;cell&quot; $x_0$ to be:&#xA;$CI(x_0,Y) = \sum_{y \in Y} p(x_0,y) \log{ \left( \frac{p(x_0,y)}{p_1(x_0)\,p_2(y)}&#xD;&#xA;                              \right) }, \,\!$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure if this quantity goes by another name.  Essentially I'm restricting focus to a single state of variable X (and then the full MI can be calculated by summing all the cell MIs).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question: is it guaranteed that $CI(x_0,Y) \ge 0$?  We know $I(X;Y)\ge0$ and we know that the pointwise mutual information can be negative.  I feel like CI should be nonnegative and that I might be missing some obvious proof.&lt;/p&gt;&#xA;" OwnerUserId="5048" LastEditorUserId="5048" LastEditorDisplayName="" LastEditDate="2010-12-22T16:49:28.757" LastActivityDate="2010-12-22T18:16:43.827" Title="Can the mutual information of a &quot;cell&quot; be negative?" Tags="&lt;probability-theory&gt;&lt;information-theory&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="15205" PostTypeId="2" ParentId="15201" CreationDate="2010-12-22T16:33:04.080" Score="6" ViewCount="" Body="&lt;p&gt;I had my first intro graph theory and combinatorics class last semester. The book we were using was pretty terrible so I looked around and found a copy of &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0387797106&quot; rel=&quot;nofollow&quot;&gt;Combinatorics and Graph Theory&lt;/a&gt; by Harris et. al. and I really enjoyed it.  The book contains a lot of topics and the explanations are very to the point.  I especially liked the sections on Ramsey numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The questions are all to the point and illustrate some important concept which is also nice.  For example in the section on the happy ending problem the exercises reconstruct several historical proofs and introduce you to other problems like the empty polygon problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a very positive review I read recently: &lt;a href=&quot;http://www.maa.org/reviews/combinatoricsgraphs.html&quot; rel=&quot;nofollow&quot;&gt;http://www.maa.org/reviews/combinatoricsgraphs.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: I'm not sure if this book is appropriate for your situation specifically, but I highly recommend it none the less.&lt;/p&gt;&#xA;" OwnerUserId="565" LastEditorUserId="565" LastEditorDisplayName="" LastEditDate="2010-12-22T16:44:33.697" LastActivityDate="2010-12-22T16:44:33.697" CommentCount="2" />
  <row Id="15206" PostTypeId="2" ParentId="15199" CreationDate="2010-12-22T16:38:59.770" Score="2" ViewCount="" Body="&lt;p&gt;Let us write the following propositions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$F_g$ means Fred is guilty, and $F_i$ means Fred is innocent, $T_g$ and $T_i$ for Ted and $E_g$ and $E_i$ for Ed.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Ed says: $F_g \wedge T_i$&lt;/li&gt;&#xA;&lt;li&gt;Fred says: $E_g \rightarrow T_g$&lt;/li&gt;&#xA;&lt;li&gt;Ted says: $T_i \wedge (F_g \vee E_g)$&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;We know that the guilty is lying and the innocent tells the truth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume Fred did it, he is guilty and lying. Meaning $E_g \rightarrow T_g$ is false, meaning $E_g$ is true and $T_g$ is false, that is Ed is guilty and Ted is innocent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In turn it means that Ed is also lying, that is either Fred is innocent or Ted is guilty. If Fred is innocent then he is telling the truth - contradiction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore we know that Ted is guilty as well, another contradiction because it follows from Fred's lie that Ted is innocent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alright, then we have that Fred is definitely not guilty. So either Ed is innocent or Ted is guilty.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume that Ed is innocent, this is a right out contradiction, as this implies Fred is guilty. So we have that Ed has to be guilty, which means Ted is also guilty - therefore he is lying. Which means he either guilty or both the other guys are innocent. Since he is guilty we have a good solution:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fred is innocent, Ted and Ed are guilty.&lt;/p&gt;&#xA;" OwnerUserId="622" LastActivityDate="2010-12-22T16:38:59.770" CommentCount="2" />
  <row Id="15207" PostTypeId="1" AcceptedAnswerId="15210" CreationDate="2010-12-22T16:44:08.003" Score="8" ViewCount="268" Body="&lt;p&gt;I would like to prove $$\frac{\Gamma(n+\frac{1}{2})}{\Gamma(n+1)} \le \frac{1}{\sqrt{n}}$$ for all natural $n \ge 1$. The inequality does seem to be true numerically, but the proof eludes me.&lt;/p&gt;&#xA;" OwnerUserId="5051" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-23T14:04:40.393" LastActivityDate="2010-12-23T16:16:14.837" Title="A gamma function inequality" Tags="&lt;inequality&gt;&lt;special-functions&gt;&lt;gamma-function&gt;" AnswerCount="5" CommentCount="7" FavoriteCount="1" />
  <row Id="15208" PostTypeId="2" ParentId="15145" CreationDate="2010-12-22T16:51:39.033" Score="1" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\ $ The sum &lt;strong&gt;telescopes&lt;/strong&gt; since the falling factorial summand is a perfect difference: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad (k+1)^{[n]} - k^{[n]}\ =\ (k+1)\ k\ \cdots\ (k-n+2)\ -\ k\ (k-1)\ \cdots\ (k-n+1)$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\phantom{(k+1)^{[n]} - k^{[n]}\ }  =\ (k+1 - (k-n+1))\ \ \ k\ (k-1)\ \cdots\ (k-n+2)$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\phantom{(k+1)^{[n]} - k^{[n]}\ }\ =\ n\ k^{[n-1]}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For other examples of additive/multiplicative telescopy see &lt;a href=&quot;http://math.stackexchange.com/q/13701/242&quot;&gt;here&lt;/a&gt; and &lt;a href=&quot;http://math.stackexchange.com/questions/13646/how-to-derive-the-following-identity/13673#13673&quot;&gt;here&lt;/a&gt; or &lt;a href=&quot;http://math.stackexchange.com/questions/6581/proof-of-inequalities-by-induction/6586#6586&quot;&gt;here&lt;/a&gt; or &lt;a href=&quot;http://math.stackexchange.com/questions/7441/summation-of-sum-limits-n1-infty-fracxx1-cdots-xn-1yy1-cdo/7453#7453&quot;&gt;here&lt;/a&gt; or &lt;a href=&quot;http://math.stackexchange.com/questions/6805/simple-proof-by-induction-9-divides-n3-n13-n23/6814#6814&quot;&gt;here.&lt;/a&gt; For much more on the falling factorials see Steven Roman's textbook &lt;em&gt;The Umbral Calculus&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-12-22T16:58:59.850" LastActivityDate="2010-12-22T16:58:59.850" CommentCount="1" />
  <row Id="15209" PostTypeId="2" ParentId="15201" CreationDate="2010-12-22T17:07:49.700" Score="7" ViewCount="" Body="&lt;p&gt;A good suggestion is Combinatorial Problems and Exercises by Laszlo Lovasz.&lt;/p&gt;&#xA;" OwnerUserId="5010" LastActivityDate="2010-12-22T17:07:49.700" CommentCount="2" />
  <row Id="15210" PostTypeId="2" ParentId="15207" CreationDate="2010-12-22T17:08:02.647" Score="7" ViewCount="" Body="&lt;p&gt;This follows directly from &lt;a href=&quot;http://dlmf.nist.gov/5.6.E4&quot; rel=&quot;nofollow&quot;&gt;Gautschi's inequality&lt;/a&gt;, valid for $0 &amp;lt; s &amp;lt; 1$ and $x &amp;gt; 0$: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x^{1-s} &amp;lt; \frac{\Gamma(x+1)}{\Gamma(x+s)} &amp;lt; (x+1)^{1-s}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a side note, the link is to NIST's new (as of this year) &lt;a href=&quot;http://dlmf.nist.gov/&quot; rel=&quot;nofollow&quot;&gt;Digital Library of Mathematical Functions&lt;/a&gt;.  The DLMF is an update of the classic Abramowitz and Stegun text, &lt;em&gt;&lt;a href=&quot;http://people.math.sfu.ca/~cbm/aands/&quot; rel=&quot;nofollow&quot;&gt;Handbook of Mathematical Functions&lt;/a&gt;&lt;/em&gt;.  &lt;/p&gt;&#xA;" OwnerUserId="2370" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-23T08:33:47.420" LastActivityDate="2010-12-23T08:33:47.420" CommentCount="2" />
  <row Id="15211" PostTypeId="2" ParentId="15204" CreationDate="2010-12-22T17:17:46.973" Score="1" ViewCount="" Body="&lt;p&gt;Answered on &lt;a href=&quot;http://cstheory.stackexchange.com/questions/3939/can-the-mutual-information-of-a-cell-be-negative/3943#3943&quot;&gt;Theoretical Computer Science&lt;/a&gt;.  CI cannot be negative.&lt;/p&gt;&#xA;" OwnerUserId="5048" LastEditorUserId="5048" LastEditorDisplayName="" LastEditDate="2010-12-22T18:16:43.827" LastActivityDate="2010-12-22T18:16:43.827" CommentCount="2" />
  <row Id="15212" PostTypeId="1" AcceptedAnswerId="15215" CreationDate="2010-12-22T17:25:47.267" Score="8" ViewCount="428" Body="&lt;p&gt;First some context, I'm not a mathematician, not even close (as you will soon see) I do grasp some things about it but in a need to know basis, so plain english answers are appreciated (too).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can't figure out this one even though some have tried very hard to explain/convince me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So let's say I'm on a roulette (let's leave out the zero, or make it a coin for that matter).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The roulette hits 5 (or any amount of) times red &lt;strong&gt;on a row&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is why should't I bet to black (or better said why it is the same the one I pick).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I DO UNDERSTAND THE BASIC PRINCIPLE, that there is a material reality which makes the ball randomly fall on any of the two colors (which gives the options a 50-50 chance).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, I also understand that if you &quot;say&quot;: I bet red will come out six times on a row, you do have a very low probability of that happening so:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How is it that if you have seen the ball fall 5 times on a row on red, and you bet on red, you are NOT betting on 6 times on a row on red (that does have lower probability).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance!&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I hope I was clear enough with my question, if not, please ask for any clarifications needed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sorry for the VERY plain English, feel free to modify or suggest a change to anything that may be misleading.&lt;/p&gt;&#xA;" OwnerUserId="4256" LastActivityDate="2010-12-23T03:15:15.713" Title="When the roulette has hit 5 reds why shouldn't I bet to black?" Tags="&lt;probability&gt;" AnswerCount="5" CommentCount="2" />
  <row Id="15213" PostTypeId="2" ParentId="15194" CreationDate="2010-12-22T17:26:24.283" Score="2" ViewCount="" Body="&lt;p&gt;Nice question!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What follows below is not really an answer, but too long to fit as a comment. It would be helpful if you could write your Mathematica algorithm is traditional mathematical notation so it is easier to see what is going on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, I am sure you are aware of the standard definitions of matrix means. I recall them below for the benefit of others who consider your question.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The arithmetic mean is: $(A+B)/2$ as expected. It satisfies several of the standard desirable properties of a mean of spd matrices (nonnegative, if $A \le B$, then $A \le AM(A,B) \le B$, etc)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The standard geometric mean is: $$GM(A,B) = A^{1/2}(A^{-1/2}BA^{-1/2})^{1/2}A^{1/2}$$&#xA;This particular choice of GM satisfies: $GM(A,B)=GM(B,A)$. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Now if you iterate the AGM, using the above two definitions of the means, then I think convergence, etc. can be proven---though you might have to resort to some kind of Fixed-point theorem. If I get time, I will think more about this.&lt;/p&gt;&#xA;" OwnerUserId="1709" LastActivityDate="2010-12-22T17:26:24.283" CommentCount="3" />
  <row Id="15214" PostTypeId="1" CreationDate="2010-12-22T17:30:46.397" Score="0" ViewCount="181" Body="&lt;p&gt;Greetings,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm performing research that will help determine the size of observed space and the time elapsed since the big bang. Hopefully you can help!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have data conforming to a piecewise linear function on which I want to perform two linear regressions. There is a point at which the slope and intercept change, and I need to (write a program to) find this point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thoughts?&lt;/p&gt;&#xA;" OwnerUserId="5053" LastEditorUserId="2548" LastEditorDisplayName="" LastEditDate="2010-12-23T06:01:21.140" LastActivityDate="2010-12-23T06:01:21.140" Title="Finding the change point in data from a piecewise linear function" Tags="&lt;statistics&gt;&lt;regression&gt;&lt;linear-algebra&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="15215" PostTypeId="2" ParentId="15212" CreationDate="2010-12-22T17:33:58.793" Score="13" ViewCount="" Body="&lt;p&gt;Lots of people have trouble with this.  If you bet before any spins on six in a row, the chance of winning is 1/2^6=1/64.  But having seen the five reds, as you say, the chances are 1/2 each on red and black for the next spin.  It is also true for &lt;strong&gt;any&lt;/strong&gt; specific series of six spins, at the start the chance of that series is 1/64.  If you were to bet on RBBRRB, that is also 1/64.  Now that you have seen five reds, there are only two series of six that are possible:  RRRRRR and RRRRRB.  They each started out 1/64 and are now each 1/2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope this helps&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-12-22T17:33:58.793" CommentCount="1" />
  <row Id="15216" PostTypeId="2" ParentId="15212" CreationDate="2010-12-22T17:36:10.977" Score="3" ViewCount="" Body="&lt;p&gt;The events &quot;given that the roulette hits $5$ reds, the next hit is red&quot; and &quot;the roulette hits $6$ reds&quot; are different.  The first has probability $\frac{1}{2}$ because different spins of the roulette are independent and the second has probability $\frac{1}{64}$ by the multiplication principle.  (This is the same as the probability of &quot;the roulette hits $5$ reds and a black,&quot; which is exactly the point of the first computation.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For questions like these the general principle for getting to grips with how your intuition is failing is &quot;when in doubt, list out all the possibilities.&quot;&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-12-22T17:36:10.977" CommentCount="1" />
  <row Id="15217" PostTypeId="2" ParentId="15212" CreationDate="2010-12-22T17:36:57.823" Score="3" ViewCount="" Body="&lt;p&gt;There's an analogy that I like. Suppose I'm afraid to take planes as it's possible that some crazy terrorist brings a bomb onto the plane. Now it's really improbable that there are &lt;em&gt;two&lt;/em&gt; bombs on the plane that I'm going to take. So I'll bring a bomb myself (which I of course won't detonate); then it's virtually impossible that anything happens. Well ...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know, but maybe this helps.&lt;/p&gt;&#xA;" OwnerUserId="2575" LastActivityDate="2010-12-22T17:36:57.823" CommentCount="1" />
  <row Id="15218" PostTypeId="2" ParentId="15201" CreationDate="2010-12-22T17:40:04.210" Score="8" ViewCount="" Body="&lt;p&gt;My personal favorites are the following:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Introduction to Combinatorial analysis --- Riordan&lt;/li&gt;&#xA;&lt;li&gt;Concrete Mathematics --- Graham, Knuth, Patashnik&lt;/li&gt;&#xA;&lt;li&gt;Enumerative Combinatorics vol. 1 --- Richard Stanley (is not always that introductory, but for those who like counting, it is a must have)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;If you want really easy, but still interesting books, you might like Brualdi's book (though apparently, that book has many mistakes). Also interesting might be some chapters from Feller's book on Probability (volume 1).&lt;/p&gt;&#xA;" OwnerUserId="1709" LastActivityDate="2010-12-22T17:40:04.210" CommentCount="1" />
  <row Id="15219" PostTypeId="2" ParentId="15214" CreationDate="2010-12-22T17:48:02.203" Score="1" ViewCount="" Body="&lt;p&gt;A simple answer would be to consider the break point a variable, call it x.  Collect the points below x and perform a linear fit, returning the error (say sum of squared errors over the points).  Collect the points above x and do the same.  Then you can define f(x) as the total error over the two fits.  Consider this a function in one variable and minimize it.  &lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-12-22T17:48:02.203" />
  <row Id="15221" PostTypeId="2" ParentId="15212" CreationDate="2010-12-22T18:06:36.810" Score="2" ViewCount="" Body="&lt;p&gt;Why is it so unlikely to get 6 reds in a row?  Well, you have to cross your fingers and say &quot;Hope the first one's red!&quot;  And if you get lucky and that one comes up red, then you have to cross your fingers and say &quot;Hope the second one's red!&quot;  And if you get really lucky and that one also comes up red, then you have to cross your fingers and say &quot;Hope the third one's red!&quot;  And so on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, if five reds already came up, all you have to do is cross your fingers and say &quot;Hope this one's red!&quot; a single time.  In one case, you're hoping for &quot;red red red red red red&quot;; in the other, you don't care about the first five (they're already done with); you're just hoping to get &quot;red&quot; once.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To put it another way: would you rather go for six reds &lt;em&gt;from scratch&lt;/em&gt;, or would you rather go for six reds &lt;em&gt;after five have already come up red&lt;/em&gt;?&lt;/p&gt;&#xA;" OwnerUserId="4904" LastEditorUserId="4904" LastEditorDisplayName="" LastEditDate="2010-12-22T18:13:00.177" LastActivityDate="2010-12-22T18:13:00.177" />
  <row Id="15223" PostTypeId="2" ParentId="15207" CreationDate="2010-12-22T18:10:07.150" Score="3" ViewCount="" Body="&lt;p&gt;Since&#xA;$$&#xD;&#xA;\Gamma \bigg(n + \frac{1}{2}\bigg) = \frac{{(2n)!}}{{4^n n!}}\sqrt \pi  &#xD;&#xA;$$&#xA;(see e.g. &lt;a href=&quot;http://en.wikipedia.org/wiki/Gamma_function&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;),&#xA;we have&#xA;$$&#xD;&#xA;\frac{{\Gamma (n + \frac{1}{2})}}{{\Gamma (n+1)}} = \frac{{{2n \choose n}}}{{4^n }}\sqrt \pi  .&#xD;&#xA;$$&#xA;This might be very useful, but I have to check.&lt;/p&gt;&#xA;" OwnerUserId="2810" LastEditorUserId="2810" LastEditorDisplayName="" LastEditDate="2010-12-22T18:44:14.623" LastActivityDate="2010-12-22T18:44:14.623" CommentCount="3" />
  <row Id="15224" PostTypeId="2" ParentId="15194" CreationDate="2010-12-22T18:26:24.563" Score="0" ViewCount="" Body="&lt;p&gt;I'm sorry that I'm putting this in an answer rather than a comment, but I'm too new to comment on your question. My thoughts for proving this process converges (question 1.) were along the lines of:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$A\,\&amp;amp;\,B$ both SPD $\implies AB$ and $A+B$ are also SPD&lt;/li&gt;&#xA;&lt;li&gt;If $A$ and $B$ are symmetric $n\times n$ matrices, $A$'s eigenvalues $\{\lambda_n^A\} \subset [0, a]$, and $B$'s $\{\lambda_n^B\} \subset [0, b]$, then the eigenvalues $\{\lambda_N^{AB}\}$ of $AB$ are $\subset [0, ab]$. See, for instance, &lt;a href=&quot;http://mathforum.org/kb/thread.jspa?threadID=1347224&amp;amp;messageID=4543523&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; discussion. Moreover, $\{\lambda_n^{A+B}\} \subset [0, a + b]$; see Terry Tao's &lt;a href=&quot;http://terrytao.wordpress.com/2010/01/12/254a-notes-3a-eigenvalues-and-sums-of-hermitian-matrices/&quot; rel=&quot;nofollow&quot;&gt;notes&lt;/a&gt; on this.&lt;/li&gt;&#xA;&lt;li&gt;I think convergence might follow by examining the AGM of the eigenvalues; that is, look at $AGM(\lambda_k^A, \lambda_k^B)$ for $k = 1,\ldots,n$. The difference between the arithmetic and geometric means gets arbitrarily small, and the eigenvalues are all positive; this iteration leads to a sequence of SPD matrices whose largest eigenvalue $\lambda_1 \to 0$, so the AGM in question does converge (perhaps).&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Not sure, but thought I'd give my $\$0.02$. As @Slowsolver said, nice question! I also second the request of writing your algorithm in more traditional notation, since I am not as well-versed in Mathematica as others may be. Finally, I might add the tag &quot;numerical-linear-algebra&quot; to your question.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;EDIT: turns out I was wrong in part 1, as others have pointed out to me. My apologies! In light of their corrections, it may make sense to use the geometric mean given by SlowSolver.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;" OwnerUserId="4689" LastEditorUserId="4689" LastEditorDisplayName="" LastEditDate="2010-12-23T15:20:10.693" LastActivityDate="2010-12-23T15:20:10.693" CommentCount="3" />
  <row Id="15225" PostTypeId="1" AcceptedAnswerId="15227" CreationDate="2010-12-22T18:34:07.953" Score="5" ViewCount="159" Body="&lt;p&gt;I was trying to write a poker program yesterday, and trying to figure out a way to tell the computer how to detect a straight. I realized that if you add 5 consecutive numbers and divide by 5, you'll get no remainder. I tried for 6 consecutive numbers, but it didn't work. It turns out it only works for odd numbers. Really silly question, but does this 'method' have a name? Has it been proven that for all odd $n$, the sum of any $n$ consecutive numbers divided by $n$ yields a remainder of zero?&lt;/p&gt;&#xA;" OwnerUserId="2985" LastActivityDate="2010-12-22T20:20:50.037" Title="Sum of $n$ consecutive numbers divided by $n$" Tags="&lt;number-theory&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="15226" PostTypeId="2" ParentId="15225" CreationDate="2010-12-22T18:41:06.103" Score="3" ViewCount="" Body="&lt;p&gt;Yes, If $n=2k+1$, write the numbers as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a - k, a-(k-1), \dots,a-1, a, a + 1, \dots, a+(k-1), a+k$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Adding gives you $na$ which is divisible by $n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Try something similar for even $n =2m$ and you see that the sum is $m \mod n$, which is not divisible by $n$.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-12-22T18:41:06.103" />
  <row Id="15227" PostTypeId="2" ParentId="15225" CreationDate="2010-12-22T18:45:07.433" Score="4" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://www.jimloy.com/number/consec0.htm&quot; rel=&quot;nofollow&quot;&gt;This page&lt;/a&gt; have a proof for the same.It also includes the the prove that when $n$ is even then this is not true.&lt;/p&gt;&#xA;" OwnerUserId="2109" LastActivityDate="2010-12-22T18:45:07.433" />
  <row Id="15229" PostTypeId="2" ParentId="15214" CreationDate="2010-12-22T18:59:19.030" Score="1" ViewCount="" Body="&lt;p&gt;I am assuming that you want a &lt;strong&gt;continuous&lt;/strong&gt; piecewise linear fit of the form&#xA;$$&#xD;&#xA;a(x-\xi)+c\hbox{ if }x\le\xi,\quad b(x-\xi)+c\hbox{ if }x&amp;gt;\xi.&#xD;&#xA;$$&#xA;There are 4 variables to determine: the slopes $a$, $b$, the point $\xi$ and $c$, the ordinate at $\xi$. Let $\{(x_i,y_i)\}$, $1\le i\le N$, be your data. For simplicity I will assume that $x_i&amp;lt;x_{i+1}$. For each $i=1,\dots,N-1$ minimize the expression&#xA;$$&#xD;&#xA;O_i=\sum_{j=1}^i(y_j-a(x_j-\xi)-c)^2+\sum_{j=i+1}^N(y_j-b(x_j-\xi)-c)^2.&#xD;&#xA;$$&#xA;This can be done by solving the system of 4 equations with 4 unknowns&#xA;$$&#xD;&#xA;\frac{\partial O_i}{\partial a}=\frac{\partial O_i}{\partial b}=\frac{\partial O_i}{\partial c}=\frac{\partial O_i}{\partial \xi}=0.&#xD;&#xA;$$&#xA;(This will be nonlinear, and will have more than one solution)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, choose the index $i$ that gives the minimum $O_i$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried this in a mock example, and it came out quite right.&lt;/p&gt;&#xA;" OwnerUserId="4791" LastActivityDate="2010-12-22T18:59:19.030" />
  <row Id="15230" PostTypeId="2" ParentId="15192" CreationDate="2010-12-22T19:02:28.960" Score="2" ViewCount="" Body="&lt;p&gt;Assuming that your $f_i$ is just the identity map with $\tau_i$ on the codomain, $\tau$ is just the coarsest topology finer than all $\tau_i$.  This explains the assertion that restricting to those $\tau_i$ coarser than $\tau_{op}$ yields $\tau=\tau_{op}$.  But then to ask whether $\tau=\tau_{op}$ in general is to ask whether every TVS topology on $L(X)$ is coarser than $\tau_{op}$.  To contradict it you would just need an example of a TVS topology that is not comparable to $\tau_{op}$ or one that is strictly finer than $\tau_{op}$.  If $X$ is infinite dimensional, then there are Banach space norms on $L(X)$ that are not equivalent to the operator norm, which by the open mapping theorem means the topologies are not comparable. So the answer is no.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2010-12-22T21:16:54.757" LastActivityDate="2010-12-22T21:16:54.757" />
  <row Id="15231" PostTypeId="1" AcceptedAnswerId="15265" CreationDate="2010-12-22T19:05:31.237" Score="0" ViewCount="177" Body="&lt;p&gt;I have asked this question on &lt;a href=&quot;http://cstheory.stackexchange.com/questions/3592/can-we-partition-np-complete-problem-into-finite-number-of-polynomially-solvable&quot;&gt;cstheory&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $\Pi$ be NP-complete problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can we partition the set of instances of $\Pi$ into finite number of subsets (subproblems) each of which is polynomially solvable (and not necessarily polynomially recognizable)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On cstheory I have obtained two answers on my question: &quot;trivially yes&quot; and no, unless $P=NP$. Trivially yes, since we could just take the sets of instances with equal value of objective function for each of subset. And no, since if we had such subproblems then we could solve the general problem, but that's not possible unless $P=NP$. So the question is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;how to formulate two questions such that for each of them corresponding answers would hold.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1856" LastEditorUserId="1856" LastEditorDisplayName="" LastEditDate="2010-12-22T19:15:12.973" LastActivityDate="2010-12-23T00:51:45.250" Title="Can we partition NP-complete problem into finite number of polynomially solvable problems?" Tags="&lt;computer-science&gt;" AnswerCount="2" CommentCount="17" />
  <row Id="15232" PostTypeId="2" ParentId="15201" CreationDate="2010-12-22T19:09:39.843" Score="3" ViewCount="" Body="&lt;p&gt;To add two more to the list already here, try &lt;em&gt;Principles and Techniques in Combinatorics&lt;/em&gt; by Chen Chuan Chong and Koh Khee Meng or &lt;em&gt;Combinatorics&lt;/em&gt; by Peter Cameron. The latter is more advanced and has more topics.&lt;/p&gt;&#xA;" OwnerUserId="4594" LastEditorUserId="4594" LastEditorDisplayName="" LastEditDate="2011-05-24T09:11:40.367" LastActivityDate="2011-05-24T09:11:40.367" CommentCount="1" />
  <row Id="15233" PostTypeId="2" ParentId="15201" CreationDate="2010-12-22T19:22:44.180" Score="1" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://books.google.com/books?id=60rb-NF-uvYC&amp;amp;printsec=frontcover&amp;amp;dq=A+Path+to+Combinatorics+for+Undergraduates,+by+Titu+Andreescu&amp;amp;source=bl&amp;amp;ots=wutA-jSIvk&amp;amp;sig=9g6WslFnhgcBzhSAD8c7d8H2dWg&amp;amp;hl=en&amp;amp;ei=EkmBTaDZCYqBOt_LibsI&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=1&amp;amp;ved=0CBcQ6AEwAA#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;A Path to Combinatorics for Undergraduates&lt;/em&gt;,&lt;/a&gt; by Titu Adreescu and Zuming Feng introduces the subject by presenting a large number of problems (many from Olympiads and other competitions), and covers a broad range of methods and results.&lt;/p&gt;&#xA;" OwnerUserId="1702" LastEditorUserId="1702" LastEditorDisplayName="" LastEditDate="2011-03-16T23:35:20.130" LastActivityDate="2011-03-16T23:35:20.130" CommentCount="1" />
  <row Id="15234" PostTypeId="2" ParentId="15207" CreationDate="2010-12-22T19:41:26.827" Score="5" ViewCount="" Body="&lt;p&gt;To elaborate on Shai Covo's answer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$2n\left[{2n\choose n} 4^{-n}\right]^2&#xD;&#xA;= {1\over2}\,{3\over2}\,{3\over4}\,{5\over4}\cdots {{2n-1}\over{2n-2}}\,{{2n-1}\over{2n}}&#xD;&#xA;={1\over 2}\prod_{j=2}^n\left(1+{1\over4j(j-1)}\right).&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By Wallis's formula, the middle expression converges to $2\over\pi$.&#xA;The right hand expression shows that it is strictly increasing. Therefore, multiplying by $\pi/2$ and taking square roots shows that&lt;br&gt;&#xA;$$\sqrt{n\pi}{2n\choose n} 4^{-n}\uparrow 1.$$&lt;/p&gt;&#xA;" OwnerUserId="940" LastEditorUserId="940" LastEditorDisplayName="" LastEditDate="2010-12-23T16:16:14.837" LastActivityDate="2010-12-23T16:16:14.837" />
  <row Id="15235" PostTypeId="1" AcceptedAnswerId="17489" CreationDate="2010-12-22T19:50:01.713" Score="7" ViewCount="170" Body="&lt;p&gt;Suppose we have a square real $n\times n$ matrix $X=[x_1,...,x_n]$, where $x_i$ is $i$-th column of the matrix. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now define  $X_k=[x_1,..,x_k]$, i.e. matrix $X_k$ columns are the first $k$ columns of the matrix $X$. Define $\lambda_k$ as the maximal eigenvalue of $(X_k^TX_k)^{-1}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it possible to prove that $\lambda_1\le \lambda_2\le ... \le \lambda_n$? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $X$ is orthogonal, then the answer is yes. Maybe this holds only for certain matrices? Any pointers would be greatly appreciated.&lt;/p&gt;&#xA;" OwnerUserId="4742" LastActivityDate="2011-01-14T22:44:34.860" Title="Do the maximal eigenvalues of $(X^TX)^{-1}$ matrix increase when number of columns of $X$ increases?" Tags="&lt;linear-algebra&gt;&lt;matrices&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="3" />
  <row Id="15236" PostTypeId="2" ParentId="15207" CreationDate="2010-12-22T19:58:16.993" Score="9" ViewCount="" Body="&lt;p&gt;Here's a direct proof:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The gamma function satisfies $\Gamma(n+1) = n \Gamma(n)$,&#xA;so the function $g(x) = \ln \Gamma(x)$ satisfies&#xA;$g(n+1) = \ln n + g(n)$.&#xA;In other words, the line segment between the points $(n,g(n))$&#xA;and $(n+1,g(n+1))$ has slope $\ln n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moreover, $g$ is known to be a convex function for $x&amp;gt;0$&#xA;(cf. the &lt;a href=&quot;http://en.wikipedia.org/wiki/Bohr%E2%80%93Mollerup_theorem&quot; rel=&quot;nofollow&quot;&gt;Bohr–Mollerup theorem&lt;/a&gt;),&#xA;so the point $(n+1/2, g(n+1/2))$ lies &lt;em&gt;below&lt;/em&gt; that line segment:&#xA;$g(n+1/2) &amp;lt; g(n+1) - \frac12 \ln n$.&#xA;Exponentiating both sides gives $\Gamma(n+1/2) &amp;lt; \Gamma(n+1) / \sqrt{n}$,&#xA;as desired.&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2010-12-22T19:58:16.993" CommentCount="2" />
  <row Id="15237" PostTypeId="2" ParentId="15225" CreationDate="2010-12-22T20:11:59.860" Score="2" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\ $ The sum is $\rm\ 0 + 1 +\:\cdots\:+\: n-1\ \ (mod\ n)\:.\: $ Pairing each summand $\ne 0\:$ with its negation shows that the sum is $\rm\: \equiv 0\ \ (mod\ n)\:.\ \ $ Note: $\rm\ - x\ \equiv\ x\ \Rightarrow\ 2\ x \equiv 0\ \Rightarrow\ x\equiv 0\ \:$ since $\rm\:n\:$ is odd.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $\rm\:n\:$ is even then the above argument demonstrates simply that the sum reduces to the sum over the &lt;em&gt;fixed points&lt;/em&gt; of the negation map, viz. $\rm\ 0,\ n/2\:.\ $ Therefore the sum is  $\rm\:\equiv n/2\ \ (mod\ n)\:.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a special case of &lt;strong&gt;Wilson's theorem&lt;/strong&gt; for groups, e.g. see &lt;a href=&quot;http://math.stackexchange.com/questions/2260/proof-for-formula-for-sum-of-sequence-123-n/2267#2267&quot;&gt;here&lt;/a&gt; and &lt;a href=&quot;http://math.stackexchange.com/questions/9311/product-of-all-elements-in-an-odd-finite-abelian-group-is-1/9345#9345&quot;&gt;here&lt;/a&gt; -&#xA;which highlight the key role played by the reflection / involution symmetry (here negation $\rm\ x\to -x\:$).&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-12-22T20:20:50.037" LastActivityDate="2010-12-22T20:20:50.037" />
  <row Id="15238" PostTypeId="1" CreationDate="2010-12-22T20:13:27.520" Score="-6" ViewCount="654" Body="&lt;p&gt;The Green-Tao Theorem states that for any natural number k, there exist k-term arithmetic progressions of primes. Suppose this were not true. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then there must exist a minimum k such that there are no k-term arithmetic progressions of primes. So if a-(k-2)b,a-(k-3)b,...,a-2b,a-b,a is a (k-1)-term arithmetic progression of primes and a&gt;(k-1)b, then c=a-(k-1)b cannot be prime. Notice that we have determined that c is not prime without knowing anything about the factors of c. This is impossible since the factors of c determine whether c is prime or not. Hence, the Green-Tao theorem must be true.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where is the flaw in this proof?&lt;/p&gt;&#xA;" OwnerUserId="3163" LastEditorUserId="3163" LastEditorDisplayName="" LastEditDate="2010-12-23T01:04:04.153" LastActivityDate="2010-12-23T03:28:15.893" ClosedDate="2010-12-23T04:16:36.123" Title="Puzzle - A strange proof of the Green-Tao Theorem" Tags="&lt;number-theory&gt;" AnswerCount="2" CommentCount="27" />
  <row Id="15239" PostTypeId="2" ParentId="15207" CreationDate="2010-12-22T20:17:45.963" Score="7" ViewCount="" Body="&lt;p&gt;For a completely elementary proof let&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$I_n= \int_0^{ \pi/2} \sin^n x \textrm{ d}x $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;then integration by parts gives &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$I_n = \frac{n-1}{n} I_{n-2} . \quad (1) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From which we get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$I_{2n}= \frac{2n-1}{2n} \frac{2n-3}{2n-2} \cdots \frac{1}{2} \frac{\pi}{2}  \quad (2) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$I_{2n+1}= \frac{2n}{2n+1} \frac{2n-2}{2n-1} \cdots \frac{2}{3} .  \quad (3) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $ \sin^{2n+1} x &amp;lt; \sin^{2n} x $ for $ x \in (0,\pi/2) $ we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$I_{2n+1}&amp;lt; I_{2n} . \quad (4) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, from $(1)$ we have $(2n+1)I_{2n+1} = 2nI_{2n-1} &amp;gt; 2nI_{2n}$ since $I_{2n-1} &amp;gt;I_{2n}.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore using this and $(4)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \frac{2n+1}{2n} &amp;gt; \frac{ I_{2n} }{ I_{2n+1} } &amp;gt; 1 .  \quad (5) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But from $(2)$ and $(3)$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$  \frac{ I_{2n} }{ I_{2n+1} }  = \frac{2n+1}{2} \frac{ (2n!)^2 }{ 4^{2n} (n!)^4 } \pi .$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Putting this in $(5)$ gives&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \frac{1}{ \sqrt{n} } &amp;gt; \frac{1}{4^n} { 2n \choose n } \sqrt{ \pi } &amp;gt; \frac{1}{ \sqrt{ n + 1/2} }.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i.e.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \frac{1}{ \sqrt{n} } &amp;gt; \frac{ \Gamma(n + 1/2) }{ \Gamma(n+1) } &amp;gt; \frac{1}{ \sqrt{ n + 1/2} }.$$&lt;/p&gt;&#xA;" OwnerUserId="1301" LastEditorUserId="1301" LastEditorDisplayName="" LastEditDate="2010-12-22T21:34:55.240" LastActivityDate="2010-12-22T21:34:55.240" CommentCount="3" />
  <row Id="15240" PostTypeId="1" AcceptedAnswerId="15296" CreationDate="2010-12-22T20:26:12.727" Score="9" ViewCount="555" Body="&lt;p&gt;Suppose you have a double sequence $\displaystyle a_{nm}$.  What are sufficient conditions for you to be able to say that $\displaystyle \lim_{n\to \infty}\,\lim_{m\to \infty}{a_{nm}} = \lim_{m\to \infty}\,\lim_{n\to \infty}{a_{nm}}$?  Bonus points for necessary and sufficient conditions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For an example of a sequence where this is not the case, consider $\displaystyle a_{nm}=\left(\frac{1}{n}\right)^{\frac{1}{m}}$.  $\displaystyle \lim_{n\to \infty}\,\lim_{m\to \infty}{a_{nm}}=\lim_{n\to \infty}{\left(\frac{1}{n}\right)^0}=\lim_{n\to \infty}{1}=1$, but $\displaystyle \lim_{m\to \infty}\,\lim_{n\to \infty}{a_{nm}}=\lim_{m\to \infty}{0^{\frac{1}{m}}}=\lim_{m\to \infty}{0}=0$.&lt;/p&gt;&#xA;" OwnerUserId="781" LastEditorUserId="781" LastEditorDisplayName="" LastEditDate="2010-12-23T07:08:01.947" LastActivityDate="2010-12-27T08:30:21.430" Title="When can you switch the order of limits?" Tags="&lt;real-analysis&gt;&lt;sequences-and-series&gt;&lt;limit&gt;" AnswerCount="3" CommentCount="8" FavoriteCount="7" />
  <row Id="15241" PostTypeId="2" ParentId="9659" CreationDate="2010-12-22T20:41:47.783" Score="0" ViewCount="" Body="&lt;p&gt;The basic idea here is that of &lt;a href=&quot;http://en.wikipedia.org/wiki/Vacuous_truth&quot; rel=&quot;nofollow&quot; title=&quot;Vacuous Truth&quot;&gt;Vacuous Truth&lt;/a&gt;.  The universal quantifier applied to the empty set is, by definition, true.  In symbols, $\forall{x}{\in}\emptyset \, P(x)$ is true, regardless of the statement $P(x)$ (see also &lt;a href=&quot;http://en.wikipedia.org/wiki/Universal_quantifier#The_empty_set&quot; rel=&quot;nofollow&quot;&gt;this Wikipedia page&lt;/a&gt;).  This is mainly defined this way for convenience, because otherwise you would always have to consider the empty set as a special case every time you used the universal quantifier.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This apples to subsets because a set $A\subseteq B$ by definition means $\forall x\in A \, x\in B$.  If $A=\emptyset$, then here $P(x)=x\in B$, and we see from the above that this must be true, i.e., $\emptyset \subseteq B$ for any set $B$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the way, &quot;&lt;a href=&quot;http://en.wikipedia.org/wiki/Proper_subset#proper_subset&quot; rel=&quot;nofollow&quot;&gt;proper subset&lt;/a&gt;&quot; almost always means $A\subset B;\,A\neq B$, so of course $\emptyset$ is a proper subset of any set, except for itself.  &lt;/p&gt;&#xA;" OwnerUserId="781" LastActivityDate="2010-12-22T20:41:47.783" CommentCount="2" />
  <row Id="15242" PostTypeId="1" AcceptedAnswerId="15244" CreationDate="2010-12-22T20:51:54.867" Score="3" ViewCount="183" Body="&lt;p&gt;Wikipedia &lt;a href=&quot;http://en.wikipedia.org/wiki/Midpoint_method&quot; rel=&quot;nofollow&quot;&gt;says&lt;/a&gt; the midpoint formula for numerical integration has error of order $h^3 f&amp;#39;&amp;#39;(\xi)$. I am trying to replicate this result.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm guessing that I want to use Lagrange's formulation of the remainder for Taylor series. Let $x_0=\frac{a+b}{2}$ (i.e. the midpoint). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The midpoint method says $\int_a^b f(x)dx \approx (b-a)f(\frac{a+b}{2})$, so to get the error I find $(b-a) f(\frac{a+b}{2}) - \int_a^bf(x)dx$. If I expand this using Taylor's theorem I get:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ \begin{aligned}&#xD;&#xA;error &amp;amp; =(b-a) f(x_0) - \int_a^bf(x_0)+\frac{f&amp;#39;(\xi)(x-x_0)}{2}dx \\&#xD;&#xA;&amp;amp; =\frac{f&amp;#39;(\xi)}{2}\int_a^b(x-x_0)dx \\&#xD;&#xA;&amp;amp; =\frac{f&amp;#39;(\xi)}{2}\int_a^b(x-\frac{a+b}{2})dx \\&#xD;&#xA;&amp;amp; = 0&#xD;&#xA;\end{aligned}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So apparently I have just proven that it has zero error? Any hints as to what I did wrong? (I realize that since wikipedia gives it in terms of $f&amp;#39;&amp;#39;$ I probably want to take the expansion one level further to match them, but I don't understand why this doesn't work.)&lt;/p&gt;&#xA;" OwnerUserId="2549" LastEditorUserId="2549" LastEditorDisplayName="" LastEditDate="2010-12-22T21:24:28.677" LastActivityDate="2010-12-22T21:24:28.677" Title="The error of the midpoint rule for quadrature" Tags="&lt;calculus&gt;&lt;numerical-methods&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="15244" PostTypeId="2" ParentId="15242" CreationDate="2010-12-22T21:12:36.783" Score="3" ViewCount="" Body="&lt;p&gt;You are right that you need one more term in your Taylor series.  If you write $f(x)=f(x_0)+(x-x_0)f&amp;#39;(x_0)+(x-x_0)^2f&amp;#39;&amp;#39;(\xi)/2$ and plug that into the midpoint method, the term in $f&amp;#39;&amp;#39;$ will survive.  What you have done essentially shows that if $f$ is linear, the method is exact.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-12-22T21:12:36.783" CommentCount="2" />
  <row Id="15245" PostTypeId="1" AcceptedAnswerId="15247" CreationDate="2010-12-22T21:19:30.107" Score="0" ViewCount="153" Body="&lt;p&gt;Suppose $w(z)$ is an analytic function in a domain $G$ which is symmetric with respect to the real axis. Show that $f(z) = \overline{w(\bar{z})}$ is then an analytic function of $z$ in $G$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So this is interesting because we are looking at the conjugate of a function of a conjugate. Let $w(z) = u(x,y)+iv(x,y)$. Then $f(z) = \overline{u(x,y)-iv(x,y)}$. So then just see that the Cauchy-Riemann equations are satisfied? But this is a necessary condition. So we need a sufficient condition to show analyticity right?&lt;/p&gt;&#xA;" OwnerUserId="2548" LastEditorUserId="2548" LastEditorDisplayName="" LastEditDate="2010-12-22T21:38:48.233" LastActivityDate="2010-12-22T21:38:48.233" Title="Analytic Functions" Tags="&lt;complex-analysis&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="15246" PostTypeId="2" ParentId="12167" CreationDate="2010-12-22T21:21:46.343" Score="0" ViewCount="" Body="&lt;p&gt;Just because the integers are a proper subset of the rationals doesn't mean that the rationals have a higher cardinality than the integers.  Actually, there is a theorem that says that a set is infinite if and only if it has the same cardinality to a proper subset of itself (so your logic would only apply to a finite set).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A set is countable, or has the same cardinality as the integers, if you can count the elements.  In other words, you can label each element by a unique positive integer.  We can see from the diagonals argument (see &lt;a href=&quot;http://en.wikipedia.org/wiki/File%3aDiagonal_argument.svg&quot; rel=&quot;nofollow&quot;&gt;this image&lt;/a&gt; on Wikipedia for a good illustration) that this holds for that rational numbers.  Once you get the hang of it, you can see that a lot of sets that seem to be a lot bigger than the set of integers are in fact the same &quot;size&quot; (have the same cardinality) as the integers.  See &lt;a href=&quot;http://en.wikipedia.org/wiki/Hilbert%27s_paradox_of_the_Grand_Hotel&quot; rel=&quot;nofollow&quot;&gt;Hilbert's Paradax of the Grand Hotel&lt;/a&gt; for a good example of this.&lt;/p&gt;&#xA;" OwnerUserId="781" LastActivityDate="2010-12-22T21:21:46.343" />
  <row Id="15247" PostTypeId="2" ParentId="15245" CreationDate="2010-12-22T21:25:24.983" Score="3" ViewCount="" Body="&lt;p&gt;See &lt;a href=&quot;http://mathworld.wolfram.com/SchwarzReflectionPrinciple.html&quot; rel=&quot;nofollow&quot;&gt;Reflection Principle&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-12-22T21:25:24.983" />
  <row Id="15248" PostTypeId="1" CreationDate="2010-12-22T21:33:26.163" Score="8" ViewCount="220" Body="&lt;p&gt;Suppose $\Omega$ is a bounded domain in the plane whose boundary consist of $m+1$ disjoint analytic simple closed curves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $f$ be holomorphic and nonconstant on a neighborhood of the closure of $\Omega$ such that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$|f(z)|=1$$ for all $z$ in the boundary of $\Omega$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $m=0$, then the maximum principle applied to $f$ and $1/f$ implies that $f$ has at least one zero in $\Omega$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What about the general case? I believe that $f$ must have at least $m+1$ zeros in $\Omega$, but I'm not able to prove it...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you&lt;/p&gt;&#xA;" OwnerUserId="3550" LastActivityDate="2010-12-23T02:42:13.123" Title="Zeros of a holomorphic function" Tags="&lt;complex-analysis&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="5" />
  <row Id="15249" PostTypeId="1" AcceptedAnswerId="15653" CreationDate="2010-12-22T21:35:11.933" Score="2" ViewCount="429" Body="&lt;p&gt;I read somewhere that a set is infinite if and only if it has a proper infinite subset.  I also remember seeing someones name attached to this theorem on Wikipedia once, but I can't even find that now.  I haven't been able to find a proof of this theorem, nor been able to generate one myself.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can prove that if a set has a proper infinite subset then it is itself infinite by proving the contrapositive that if a set is finite then it does not have any proper infinite subsets (this is a simple contradiction proof).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I can't figure out how to, given an arbitrary infinite set, construct a proper infinite subset.  Does this require the Axiom of Choice?  I can't really figure out how to do it with that either.  A proof or reference to a proof would be much appreciated.  &lt;/p&gt;&#xA;" OwnerUserId="781" LastActivityDate="2010-12-27T18:40:52.170" Title="Proof that a set is infinite if and only if it has an infinite proper subset" Tags="&lt;set-theory&gt;" AnswerCount="3" CommentCount="14" />
  <row Id="15250" PostTypeId="2" ParentId="15249" CreationDate="2010-12-22T21:41:09.400" Score="5" ViewCount="" Body="&lt;p&gt;It doesn't require the axiom of choice.  Remove one point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want a &lt;em&gt;countably&lt;/em&gt; infinite subset of every infinite set, I think you need to use (or at least the usual proof uses) the axiom of countable choice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, maybe you are thinking of the condition of being Dedekind infinite, as mentioned in Adrián Barquero's comment.  There you want not just an &lt;em&gt;infinite&lt;/em&gt; proper subset, but a proper subset that is in bijection with the whole set.  This can be proved using the existence of a countably infinite subset, so again uses countable choice.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-12-22T21:41:09.400" CommentCount="6" />
  <row Id="15251" PostTypeId="1" CreationDate="2010-12-22T21:52:35.710" Score="18" ViewCount="453" Body="&lt;p&gt;Doing a little reading over the break (The Probabilistic Method by Alon and Spencer); can't come up with the solution for this seemingly simple (and perhaps even a little surprising?) result:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(A-S 1.6.3) Prove that for every two independent, identically distributed real random variables X and Y,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$Pr[|X-Y| \leq 2] \leq 3 Pr[|X-Y| \leq 1]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for your time.&lt;/p&gt;&#xA;" OwnerUserId="5056" LastActivityDate="2010-12-28T08:04:38.377" Title="Random Variable Inequality" Tags="&lt;probability-theory&gt;&lt;measure-theory&gt;&lt;inequality&gt;&lt;probability&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="6" />
  <row Id="15252" PostTypeId="1" AcceptedAnswerId="15273" CreationDate="2010-12-22T21:57:05.210" Score="0" ViewCount="169" Body="&lt;p&gt;Are there any analogues of Galois Theory for complex numbers (with non-zero imaginary part)? This is motivated by &lt;a href=&quot;http://mathworld.wolfram.com/SchwarzReflectionPrinciple.html&quot; rel=&quot;nofollow&quot;&gt;Schwartz Principle&lt;/a&gt;. It says that if $f$ is analytic, $f$ is defined in the upper-half disk, and $f$ extends to a continuous function on the real axis, then $f$ can be extended to an analytic function on the whole disk by the formula $f(\bar{z}) = \overline{f(z)}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This reminds me of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Isomorphism_extension_theorem&quot; rel=&quot;nofollow&quot;&gt;Isomorphism Extension Theorem&lt;/a&gt;.  &lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2010-12-23T02:13:04.323" Title="Analogues of Galois Theory for Complex Numbers" Tags="&lt;complex-analysis&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="15253" PostTypeId="2" ParentId="14311" CreationDate="2010-12-22T22:09:33.910" Score="1" ViewCount="" Body="&lt;p&gt;I have no idea how to comment. But I do know how to answer. Sorry about that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I considered the pumping lemma. And I considered how many words are necessary to pump the language.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the language over {0,1}.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We need the following words: 00, 11, 0101, and 1010. To pump we can combine any words by concatenation or insertion. For this example a grammar would be&lt;/p&gt;&#xA;&#xA;&lt;p&gt;S =&gt;  | S0S0S | S1S1S | S0S1S0S1S | S1S0S1S0S&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A quick count shows the number of words necessary for the grammar with n terminals is exponential in n. I guess that shows there is no polynomial sized grammar.&lt;/p&gt;&#xA;" OwnerUserId="5055" LastActivityDate="2010-12-22T22:09:33.910" CommentCount="3" />
  <row Id="15254" PostTypeId="1" AcceptedAnswerId="15255" CreationDate="2010-12-22T22:34:56.003" Score="5" ViewCount="169" Body="&lt;p&gt;I am under the impression that the standard convention for the homology (singular) of the empty set is 0 in all non negative degrees and $\mathbb{Z}$ in degree $-1$. I have no problem with this convention, I am just curious what role it plays. Most conventions helps something sensible remain true in a particular case, what is this convention doing?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Doe this convention depend on which version of cohomology we use? Is there a different convention for Cech or what have cohomology?&lt;/p&gt;&#xA;" OwnerUserId="627" LastActivityDate="2010-12-22T22:46:49.257" Title="Homology of the Empty set" Tags="&lt;algebraic-topology&gt;&lt;cohomology&gt;" AnswerCount="1" CommentCount="0" />
  <row Id="15255" PostTypeId="2" ParentId="15254" CreationDate="2010-12-22T22:46:49.257" Score="2" ViewCount="" Body="&lt;p&gt;Here's an idea. The &lt;em&gt;reduced homology&lt;/em&gt;  of the empty set is what you describe: one $\mathbb Z$ in degree $-1$ and $0$ otherwise. This is because the chain complex computing the reduced homology groups has an extra $\mathbb Z$ in degree $-1$, whose role is to kill one $\mathbb Z$ in degree $0$. However, if we are looking at the empty set, all chain groups are $0$ so there is only the extra $\mathbb Z$ in degree $-1$. I'm just guessing here.&lt;/p&gt;&#xA;" OwnerUserId="2941" LastActivityDate="2010-12-22T22:46:49.257" />
  <row Id="15256" PostTypeId="2" ParentId="15055" CreationDate="2010-12-22T22:47:35.800" Score="1" ViewCount="" Body="&lt;p&gt;This is really a continuation of BlueRaja's answer, where I'm changing the question slightly to give another illustration of the underlying point.  Consider the following two procedures:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Procedure A: I flip two coins, a nickel and a penny.  After the flip, I&#xA;pick one at random, and report its value to you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Procedure B: I flip two coins, a nickel and a penny.  If exactly one lands&#xA;heads, I report that one's value to you, i.e. I say either &quot;the nickel&#xA;landed heads&quot; or &quot;the penny landed heads&quot; as appropriate.  If both land&#xA;heads, I report on one of them chosen at random.  If neither lands&#xA;heads, I say &quot;both landed tails.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if I follow Procedure A and say &quot;the nickel landed&#xA;heads&quot;, then the probability that the penny landed heads is 50/50.  But&#xA;if I say &lt;em&gt;the very same thing&lt;/em&gt; after following Procedure B, the&#xA;probability that the penny landed heads is 1 in 3.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now take the following problem:  I flip two coins and say to you &quot;the&#xA;nickel landed heads&quot;.  What is the probability that the penny landed&#xA;heads? Clearly there's no way to answer without knowing or assuming something about the procedure I'm following.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(All this having been said, I agree that the best answer to the original question is 1/3.)&lt;/p&gt;&#xA;" OwnerUserId="2682" LastActivityDate="2010-12-22T22:47:35.800" />
  <row Id="15257" PostTypeId="2" ParentId="15240" CreationDate="2010-12-22T23:00:24.460" Score="7" ViewCount="" Body="&lt;p&gt;Here's some results.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let us say that a sequence $a_{nm}$ of real numbers indexed by pairs of positive integers $(n,m)$ &lt;em&gt;converges to $L$&lt;/em&gt; if and only if for every $\epsilon\gt 0$ there exists $N\gt 0$ such that for all $n,m\geq N$, $|a_{nm}-L|\lt \epsilon$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a fairly strong condition. We have:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Theorem.&lt;/strong&gt; &lt;em&gt;Suppose that $\lim\limits_{(n,m)\to\infty}a_{nm}$ exists and equals $L$. Then the following are equivalent:&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;em&gt;For each (sufficiently large) $n_0$, $\lim\limits_{m\to\infty}a_{n_0m}$ exists;&lt;/em&gt;&lt;/li&gt;&#xA;&lt;li&gt;$\lim\limits_{n\to\infty}\lim\limits_{m\to\infty}a_{nm} = L$.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Proof.&lt;/em&gt; If 2 holds, then we must have 1 (otherwise the expression in 2 does not even make sense). Now assume that 1 holds, and let $\lim\limits_{m\to\infty}a_{nm} = L_{n}$. We want to prove that $\lim\limits_{n\to\infty}L_n=L$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $\epsilon\gt 0$. Then there exists $N\gt 0$ such that for all $n,m\geq N$, $|a_{nm}-L| \lt \epsilon$. Let $M_N\gt N$ be such that for all $m\geq M_N$, $|a_{M_Nm}-L_{M_N}|\lt \epsilon$. Since $M_n\gt N$, we have&#xA;$$|L_{M_N} - L| \leq |L_{M_N}-a_{M_Nm}| + |a_{M_Nm}-L| \leq 2\epsilon,$$&#xA;so this proves that $L_n\to L$ as $n\to\infty$. In particular, we have for the iterated limit &#xA;$$\lim\limits_{n\to\infty}\lim\limits_{m\to\infty} a_{nm} = \lim_{n\to\infty}L_n = L.\ \Box$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A symmetric argument shows that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Theorem.&lt;/strong&gt; &lt;em&gt;Suppose that $\lim\limits_{(n,m)\to\infty}a_{nm}$ exists and equals $L$. Then the following are equivalent:&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;em&gt;For each (sufficiently large) $m_0$, $\lim\limits_{n\to\infty}a_{nm_0}$ exists;&lt;/em&gt;&lt;/li&gt;&#xA;&lt;li&gt;$\lim\limits_{m\to\infty}\lim\limits_{n\to\infty}a_{nm} = L$.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;However, $\lim\limits_{n,m\to\infty}a_{nm}=L$ does not imply the existence of the iterated limits; in fact, existence of the double limit and the existence of &lt;em&gt;one&lt;/em&gt; iterated limit does not suffice to imply that the other iterated limits exists. Take $a_{nm}=\frac{(-1)^n}{m}$. Then $\lim\limits_{n,m\to\infty}a_{nm}=0$ (given $\epsilon\gt 0$, pick $N$ such that $\frac{1}{N}\lt\epsilon$), and the limit as $m\to\infty$ of $\frac{(-1)^n}{m}$ exists for each fixed $n$, but $\lim\limits_{n\to\infty}\frac{(-1)^n}{m}$ does not exist for any $m$. (By taking $a_{nm} = \frac{(-1)^n}{m} + \frac{(-1)^m}{n}$ you get one in which neither iterated limit exists.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, a sufficient conditions for the iterated limits to exist is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Corollary.&lt;/strong&gt; &lt;em&gt;Suppose that $\lim\limits_{(n,m)\to\infty}a_{nm}=L$. Then the iterated limits&#xA;$$\lim_{n\to\infty}\lim_{m\to\infty}a_{nm}\text{ and }\lim_{m\to\infty}\lim_{n\to\infty}a_{nm}$$&#xA;both exist and are equal to $L$ if and only if $\lim\limits_{n\to\infty}a_{nm}$ exists for almost all $m$ and $\lim\limits_{m\to\infty}a_{nm}$ exists for almost all $n$.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Here, &quot;almost all&quot; means &quot;all except perhaps for a finite number&quot;). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As I said, the condition above is pretty strong. You can have both iterated limits exist &lt;em&gt;and&lt;/em&gt; be equal and yet for the double limit not to exist. To adapt the standard two-variable example, take $a_{nm}=\frac{nm}{n^2+m^2}$. The iterated limits both exist and are equal to $0$, but the double limit does not exist (for any $N\gt 0$ there exist $n,m\geq N$ such that $a_{nm}=\frac{1}{2}$ and there exist $n,m\geq N$ such that $a_{nm}=\frac{2}{5}$; just take $n=m=N$ for the first, and $n=2m=2N$ for the second). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can get that the double limit exists and is equal to (one of) the iterated limits if you have some uniformness conditions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know of any necessary and sufficient conditions, and I suspect there won't generally be without some other overarching conditions. This is essentially the same problem as the problem of iterated limits in functions of two variables (just as the problem of finding the limit of a real function of real variable is closely connected to the problem of finding limits of sequences of real numbers). They are connected to the double limits, and you often have conditions based on uniform convergence that guarantee good things happen, but to state some general, simple condition for the iterated limits to exist and be equal seems difficult in general.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-22T23:37:36.317" LastActivityDate="2010-12-22T23:37:36.317" CommentCount="3" />
  <row Id="15258" PostTypeId="2" ParentId="15231" CreationDate="2010-12-22T23:02:18.390" Score="4" ViewCount="" Body="&lt;p&gt;If you allow exponentially many machines and each can know its serial number, try this.  For travelling salesman, you have n! machines.  Each one attempts to verify the cost of a trip for a given permutation, generated in a reasonable way from its serial number.  This can be done in polynomial time.  Each machine raises a bit if it succeeds.  Can't you then combine these in log(n!) time, which is polynomial.  You are still doing exponentially much computation, but spreading it over many machines.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-12-22T23:02:18.390" CommentCount="3" />
  <row Id="15259" PostTypeId="2" ParentId="15238" CreationDate="2010-12-22T23:30:50.077" Score="5" ViewCount="" Body="&lt;p&gt;The problem is in the claim &quot;Notice that we have determined that c is not prime without proving that there are other factors of c besides c and 1. This is impossible.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The absence of knowledge is not the knowledge of absence.  If you actually show that c has no other factors than 1 and c, you would have a contradiction to c being composite.  But I hope you will agree that you haven't shown that; you just note that you don't know anything specific about other factors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By definition, &quot;determining that c is not prime&quot; = &quot;determining that c has other factors than 1 and c.&quot;  The definition of prime numbers is simple, but the study of prime numbers is a deep and active part of mathematics.  It involves many long and (initially) mysterious arguments.  One nice thing about theorems is that you don't have to think about all of the details of the proof to use the result-- if you cite any theorem about prime numbers, you know something about the factors or non-factors of the numbers involved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your example argument, you assume that no k-term arithmetic progression of primes exists.  Then you automatically know that the terms preceding and following a k-1 term progression of primes are composite, because of your assumption.  In other words, your attempted proof doesn't actually do anything-- you assume something and never get anything more than the assumption you started with.&lt;/p&gt;&#xA;" OwnerUserId="1461" LastActivityDate="2010-12-22T23:30:50.077" />
  <row Id="15260" PostTypeId="1" CreationDate="2010-12-22T23:52:40.007" Score="6" ViewCount="299" Body="&lt;p&gt;Hi all&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For an university assignment I have to visualize some curves in 3 dimensional space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Until now I've implemented Bézier, helix and conical spiral.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could you give me some advice about some famous curves in geometry history?&lt;/p&gt;&#xA;" OwnerUserId="5046" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-23T06:47:42.077" LastActivityDate="2010-12-23T12:48:38.587" Title="famous space curves in geometry history?" Tags="&lt;homework&gt;&lt;differential-geometry&gt;" AnswerCount="6" CommentCount="3" />
  <row Id="15261" PostTypeId="1" AcceptedAnswerId="15462" CreationDate="2010-12-22T23:58:39.480" Score="6" ViewCount="244" Body="&lt;p&gt;For improper single integrals with positive integrands of the first, second or mixed type there are the comparison and the limit tests to determine their convergence or divergence. There is also the absolute convergence theorem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For multiple integrals I know that the comparison test can be used as well. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question&lt;/strong&gt;: But can the limit or other tests be generalized to multiple integrals? Could you provide some references?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added&lt;/strong&gt;: I have thought of Riemann integration only. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Example&lt;/em&gt;: When we evaluate (Proof 1 of &lt;a href=&quot;http://empslocal.ex.ac.uk/people/staff/rjchapma/etc/zeta2.pdf&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; and &lt;a href=&quot;http://www.springerlink.com/content/045l250v885v033u/fulltext.pdf&quot; rel=&quot;nofollow&quot;&gt;Apostol's article&lt;/a&gt;)  this improper double integral $$\int_{0}^{1}\int_{0}^{1}\left(\dfrac{1}{1-xy}\right) \mathrm{d}x\mathrm{d}y,$$ we conclude that it is finite. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question&lt;/strong&gt; 2: Which test should we apply to know in advance that it converges? &lt;/p&gt;&#xA;" OwnerUserId="752" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2010-12-23T23:17:13.627" LastActivityDate="2010-12-24T19:30:24.557" Title="Convergence tests for improper multiple integrals" Tags="&lt;calculus&gt;&lt;real-analysis&gt;&lt;analysis&gt;&lt;reference-request&gt;&lt;multivariable-calculus&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="15262" PostTypeId="2" ParentId="15260" CreationDate="2010-12-23T00:01:44.900" Score="8" ViewCount="" Body="&lt;p&gt;How about &lt;a href=&quot;http://en.wikipedia.org/wiki/Viviani%27s_curve&quot; rel=&quot;nofollow&quot;&gt;Viviani's curve&lt;/a&gt;?  It is the intersection of a sphere with a cylinder. Viviani was a student (or disciple) of Gallileo.&#xA;Wikipedia has a beautiful image:&#xA;&lt;img src=&quot;http://upload.wikimedia.org/wikipedia/commons/thumb/0/05/Viviani_curve.png/200px-Viviani_curve.png&quot; width=&quot;300&quot; /&gt;&lt;/p&gt;&#xA;" OwnerUserId="237" LastActivityDate="2010-12-23T00:01:44.900" />
  <row Id="15263" PostTypeId="2" ParentId="15260" CreationDate="2010-12-23T00:09:38.313" Score="9" ViewCount="" Body="&lt;p&gt;Space-filling curves, in particular the Hilbert curve (in 3D):&#xA;&lt;img src=&quot;http://i.imgur.com/OoCvm.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="723" LastActivityDate="2010-12-23T00:09:38.313" />
  <row Id="15264" PostTypeId="2" ParentId="15238" CreationDate="2010-12-23T00:12:29.653" Score="3" ViewCount="" Body="&lt;p&gt;Use [false-proof] or whatever the tag is to ask about proofs whose flaws are the subject of discussion.  This would help prevent the impression of claiming an easy 3-sentence proof of a result that took 40 pages and decades of prior work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The logic is identical, but leads to a false conclusion, if you replace &quot;prime&quot; by &quot;not divisible by $k$&quot;.   Given a $(k-1)$-term arithmetic progression [with difference having no factor in common with $k$] of not-divisible-by-k integers, its extension to a $k$ term progression has to have the last term divisible by $k$, even though we were given no divisibility information in advance about that last term.   The difference between this and the Green-Tao theorem is that &lt;em&gt;sometimes&lt;/em&gt; a string of $(k-1)$ primes can be extended to a string of $k$, and sometimes not. There is a relationship between the first $(k-1)$ items and the $k$th one, but it is more statistical in nature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Green-Tao proof, loosely speaking, does use the logic you outlined, but in a statistical, on-average sense.  A deterministic guarantee that any $k$ term progression can't have all terms primes (so e.g., if the first (k-1) are prime the last cannot be) would create too much correlation, too much &quot;shared information&quot; about the prime factorization of the different terms.  There are theorems in analytic number theory saying that there is at least a certain amount of pseudorandomness in the primes and in the prime factorizations of integers, and knowing in advance that every string of $(k-1)$ primes in progression cannot be extended to $k$ would enforce too high a degree of structure.  The reason 40 pages of cutting-edge analytic number theory and Fourier analysis are needed, is to convert the simple heuristics about &quot;information&quot; and &quot;correlation&quot; into provable and well-defined mathematical assertions, which is complicated because the primes themselves are deterministically defined, not a sequence of random numbers to which the probabilistic concepts most naturally apply.&lt;/p&gt;&#xA;" OwnerUserId="467" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-12-23T03:28:15.893" LastActivityDate="2010-12-23T03:28:15.893" CommentCount="17" />
  <row Id="15265" PostTypeId="2" ParentId="15231" CreationDate="2010-12-23T00:45:44.587" Score="5" ViewCount="" Body="&lt;p&gt;There are two different possible questions here. When you ask for the solution of an NP-complete problem, you can either (a) require the computer to give you a witness in the &quot;yes&quot; cases or (b) just require the computer to give you the answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose you are just asking for the answer, as in (b). Then you can partition the set of instances of $\Pi$ into &quot;yes&quot; instances and &quot;no&quot; instances. For each set of instances, there is a one-line program which gives you the answer for that set of instances. So in this case, the answer to your question is &quot;yes.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, suppose, as in (a), you require a witness in the &quot;yes&quot; cases as well as the answer. Let the algorithms for the different subclasses be $A_1$, $A_2$, $A_3$, $\ldots$, $A_k$. Now we give a master algorithm $A$ that, given an instance $x$, runs all $k$ algorithms on it. $A$ checks the witnesses that the algorithms $A_i$ output to see whether one of them verifies that $x$ is a &quot;yes&quot; instances. If one of them does, then $A$ outputs &quot;yes&quot;. Otherwise, $A$ outputs &quot;no&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $x$ is a &quot;yes&quot; instance, one of the algorithms $A_i$ must output a valid witness, and so $A$ will accept.  If $x$ is a &quot;no&quot; instance, there are no valid witnesses, and so $A$ will reject. Thus, for case (a), if the answer to your question is &quot;yes&quot;, $A$ solves an NP-complete problem in polynomial time, and we have P$=$NP.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;None of this argument is original; I'm just consolidating and summarizing the answers that this question got on &lt;a href=&quot;http://cstheory.stackexchange.com/questions/3592/can-we-partition-np-complete-problem-into-finite-number-of-polynomially-solvable&quot;&gt;cstheory&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2501" LastEditorUserId="2501" LastEditorDisplayName="" LastEditDate="2010-12-23T00:51:45.250" LastActivityDate="2010-12-23T00:51:45.250" CommentCount="7" />
  <row Id="15266" PostTypeId="2" ParentId="15039" CreationDate="2010-12-23T00:49:55.500" Score="0" ViewCount="" Body="&lt;p&gt;Falling in to the trap of thinking that mathematical usage should correspond directly with usages in English is very dangerous.  I would advise focusing on understanding the mathematics in whatever way you can and accepting that it might occasionally offend your sense of proper English grammar or semantics.  English is just another language, albeit a very successful global language.  There is no reason it should be perfectly conducive to conveying mathematical meaning.  Note that most English speakers are not mathematicians and yet they are the population that has the largest effect on the evolution of the English language.  Mathematical language is functional, meant to help us get greater insight into mathematical objects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Examining 'of' as defined by dictionary.com, I was able to clean the following relevant possible usages: used to indicate distance or direction from, separation, deprivation; used to indicate derivation, origin, or source; used to indicate cause, motive, occasion, or reason; used to indicate material, component parts, substance, or contents; used to indicate apposition or identity; used to indicate specific identity or a particular item within a category; used to indicate possession, connection, or association; used to indicate inclusion in a number, class, or whole; used to indicate the objective relation, the object of the action noted by the preceding noun or the application of a verb or adjective; used to indicate reference or respect; used to indicate qualities or attributes; used to indicate a specified time; on the part of; in respect to; and set aside for or devoted to.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Having looked at the list, I was not convinced that the preposition 'of' had unique meaning in mathematics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The usage 'f of x' seems like the only usage that has even the slightest hint of uniqueness to mathematics. However, we hear sentences like 'this is the average of', 'this is the maximum of' or &quot;this is the total of&quot;.  This seems to be the sense in which 'f of x' is meant. &lt;/p&gt;&#xA;" OwnerUserId="1004" LastActivityDate="2010-12-23T00:49:55.500" CommentCount="2" />
  <row Id="15267" PostTypeId="2" ParentId="15260" CreationDate="2010-12-23T00:54:54.223" Score="7" ViewCount="" Body="&lt;p&gt;I'd propose the curve of Archytas, the space curve formed by the intersection of a horn torus and a cylinder:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/sxAR5.png&quot; alt=&quot;curve of Archytas&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The parametric equations for the curve of Archytas are&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left(\cos^2t\qquad\frac12\sin\;2t\qquad\pm\sqrt{2\cos\;t}\sin\frac{t}{2}\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.mlahanas.de/Greeks/Bios/Archytas.html&quot; rel=&quot;nofollow&quot;&gt;Archytas of Tarentum&lt;/a&gt; used this curve to help him solve the &lt;a href=&quot;http://mathworld.wolfram.com/CubeDuplication.html&quot; rel=&quot;nofollow&quot;&gt;Delian problem&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-12-23T00:54:54.223" />
  <row Id="15268" PostTypeId="2" ParentId="15166" CreationDate="2010-12-23T01:02:11.163" Score="2" ViewCount="" Body="&lt;p&gt;The whole &quot;paddlewheel&quot; business was explained to me the first time I took multivariable calculus (I was 17), and I confess I found it entirely mysterious.  When I taught multivariable calculus myself for the first time (I was 27), I was delighted to find that with a little thought one could make sense of these geometric and physical explanations in a fairly satisfying way.  (Note: I knew less about physics at age 27 than age 17, so that's not the difference.  In between I had gotten a PhD in mathematics, which probably helped, although perhaps mostly in a psychological way.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wrote these observations up in notes for my class.  See Section 2 &lt;a href=&quot;http://math.uga.edu/~pete/handoutfive.pdf&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; and Section 3 &lt;a href=&quot;http://math.uga.edu/~pete/handouteight.pdf&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="299" LastActivityDate="2010-12-23T01:02:11.163" CommentCount="1" />
  <row Id="15269" PostTypeId="2" ParentId="15055" CreationDate="2010-12-23T01:05:05.577" Score="7" ViewCount="" Body="&lt;p&gt;I think this question confuses a lot of people because there's a lack of intuitive context -- I'll try to supply that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose there is a birthday party to which all of the girls (and none of the boys) in a small town are invited. If you run into a mother who has dropped off a kid at this birthday party and who has two children, the chance that she has two girls is $1/3$. Why? $3/4$ of the mothers with two children will have a daughter at the birthday party, the ones with two girls ($1/4$ of the total mothers with two children) and the ones with one girl and one boy ($1/2$ of the total mothers with two children). Out of these $3/4$ of the mothers, $1/3$ have two girls.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, if the birthday party is only for fifth-grade girls, you get a different answer. Assuming there are no siblings who are both in the fifth grade, the answer in this case is $1/2$. The child in fifth grade is a girl, but the other child has probability $1/2$ of being a girl. Situations of this kind arise in real life much more commonly than situations of the other kind, so the answer of $1/3$ is quite nonintuitive.&lt;/p&gt;&#xA;" OwnerUserId="2501" LastEditorUserId="2501" LastEditorDisplayName="" LastEditDate="2010-12-30T01:54:48.190" LastActivityDate="2010-12-30T01:54:48.190" CommentCount="2" />
  <row Id="15270" PostTypeId="2" ParentId="15260" CreationDate="2010-12-23T01:18:37.547" Score="2" ViewCount="" Body="&lt;p&gt;Let $t \in [0,2\pi]$. And let $r(t)=\frac{1}{2-\sin (2t)}[\cos(3t),\sin(3t),\cos(2t)]$. Then $r(t)$ parametrizes a 3-dimensional curve with no three-tangent plane (I must admit I'm not sure of the English terminology here) - that is, every tangent plane of the curve meets the curve in no more than two points.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I have no clue how to prove such things, however)&lt;/p&gt;&#xA;" OwnerUserId="4284" LastActivityDate="2010-12-23T01:18:37.547" CommentCount="2" />
  <row Id="15271" PostTypeId="2" ParentId="15194" CreationDate="2010-12-23T01:49:13.207" Score="3" ViewCount="" Body="&lt;p&gt;I'll use the characterisation of the geometric mean as given by Slowsolver. As I mentioned in a comment to his answer, the geometric mean $GM(A,B)$ is the geodesic midpoint corresponding to the Riemannian metric on the space of symmetric positive definite matrices with&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ g(X,Y) = \mathop{tr}(A^{-1}XA^{-1}Y)~,\quad X,Y\in T_A(SPD) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where the tangent space $T_A(SPD)$ is the space of symmetric matrices. This interpretation has the following advantage: both the AM and GM are &quot;translation invariant&quot;. Let $P$ be a non-singular matrix, then $AM(PAP^T,PBP^T) = P(AM(A,B))P^T$ trivially, and similarly for $GM$ using that $$g_A(X,Y) = g_{PAP^T}(PXP^T, PYP^T)$$ so that a geodesic $\gamma$ joining $A,B$ becomes a geodesic $P\gamma P^T$ joining $PAP^T$ and $PBP^T$. By first conjugating with $P = A^{-1/2}$, we can WLOG assume $A = I$. By then conjugating with a suitable orthogonal matrix $O$ to diagonalize $B$, we can WLOG assume $A = I$ and $B$ is diagonal. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The result on convergence now follows from the convergence of the AGM for the single, real variable case.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For your specific questions:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The rate of convergence is the same as the scalar case, up to a constant factor depending on the initial values of $A,B$ (which determines $P$). &lt;/li&gt;&#xA;&lt;li&gt;In this definition $GM(A,B) = GM(B,A)$, so the question is moot. &lt;/li&gt;&#xA;&lt;li&gt;This is in some sense the natural generalisation of the scalar case. Notice that the scalar geometric mean corresponds to the geodesic midpoint for the multiplicative Lie group $(\mathbb{R}_+,\times)$ with the invariant metric. A first generalisation is to the abelian Lie group $\mathbb{R}^k_+$ with componentwise multiplication. (This corresponds to the diagonal case.) What we showed above is that the entire AGM sequence for arbitrary starting SPD matrices $A$ and $B$ is contained, up to conjugation by $P\cdot P^T$, in one such abelian case. Note that this is slightly less related to eigenvalues per se, since the transformation we use above does not preserve eigenvalues. &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;EDIT: It just occurred to me to do a literature search. In &lt;a href=&quot;http://www.jstor.org/stable/2695553&quot; rel=&quot;nofollow&quot;&gt;Lawson, J. D. &amp;amp; Lim, Y. &quot;The geometric mean, matrices, metrics, and more&quot;. Amer. Math. Monthly, 2001, 108, 797-812&lt;/a&gt;, many properties of this geometric mean on SPD matrices are derived in Sections 2 and 3. In particular, the crucial property (as slowsolver remarks below) of being able to set $A = I$ and $B$ diagonal, is the content of Lemma 3.1 of that paper. That is followed by the harmonic-geometric-arithmetic-mean inequality for SPD matrices. By starting with that and playing with monotonicity properties, you presumably also get another proof of the convergence result. &lt;/p&gt;&#xA;" OwnerUserId="1543" LastEditorUserId="1543" LastEditorDisplayName="" LastEditDate="2010-12-23T16:10:41.600" LastActivityDate="2010-12-23T16:10:41.600" CommentCount="8" />
  <row Id="15272" PostTypeId="1" CreationDate="2010-12-23T01:51:07.700" Score="2" ViewCount="122" Body="&lt;p&gt;How to prove that the Sorgenfrey line is hereditarily separable?&lt;/p&gt;&#xA;" OwnerUserId="5059" LastActivityDate="2011-02-27T08:20:55.453" Title="Sorgenfrey line is hereditarily separable" Tags="&lt;topology&gt;" AnswerCount="2" />
  <row Id="15273" PostTypeId="2" ParentId="15252" CreationDate="2010-12-23T02:13:04.323" Score="0" ViewCount="" Body="&lt;p&gt;Galois theory is useful when you have some algebraic object, and a list of tools you are allowed to use within that object. The purpose of Galois theory is to explain how far one can go only using those tools.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, it is impossible to create, using only the tools of +, -, *, / and nth roots, a formula for the zeroes of a general 5th degree polynomial. This is the most basic application of Galois theory. This is the Abel-Ruffini theorem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Abel%E2%80%93Ruffini_theorem&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Abel%E2%80%93Ruffini_theorem&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another application is Differential Galois theory, which tells you which integrals can be expressed in terms of elementary functions. But once again, it comes down to having only a few basic tools with which to construct an anti-derivative for a function using only elementary functions, and that is why Galois theory is useful in this application.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, its hard to understand what your question is asking about, because you haven't really phrased your question as a Galois Theory question.&lt;/p&gt;&#xA;" OwnerUserId="1089" LastActivityDate="2010-12-23T02:13:04.323" />
  <row Id="15274" PostTypeId="1" AcceptedAnswerId="15288" CreationDate="2010-12-23T02:18:34.833" Score="2" ViewCount="118" Body="&lt;p&gt;I'm going through Enderton's Elements of Set Theory, as I heard it is a gentle introduction to set theory. I'm a little confused on how the subset axioms are used. In the text, the axiom is given as:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Subset Axioms&lt;/strong&gt; For each formula $\underline{\hspace{1cm}}$ not containing $B$, the following is an axiom:&#xA;  $$&#xD;&#xA;\forall t_1\cdots\forall t_k\forall c\exists B\forall x(x\in B\Leftrightarrow x\in c \wedge \underline{\hspace{1cm}}).&#xD;&#xA;$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Later, the text asserts the existence of the set $\{A\cup X\ | \ X\in\mathscr{B}\}$, call it $\mathscr{D}$, saying that since $A\cup X\subseteq A\cup\bigcup\mathscr{B}$, $\mathscr{D}$ is a subset of $\mathscr{P}(A\cup\bigcup\mathscr{B})$. A subset axiom produces &#xA;$$&#xD;&#xA;\{t\in\mathscr{P}(A\cup\bigcup\mathscr{B})\ |\ t=A\cup X\ \text{for some}\ X\in\mathscr{B}\}.&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is, what exactly is that subset axiom? Would it be something of the form&#xA;$$&#xD;&#xA;\forall t_1\cdots\forall t_k\forall c\exists B\forall x(x\in B\Leftrightarrow x\in c \wedge [(x=t_1\cup t_2)\vee(x=t_1\cup t_3)\vee\cdots\vee(x=t_1\cup t_k)])&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where we take $A$ to be a particular instance of $t_1$, $\mathscr{P}(A\cup\bigcup\mathscr{B})$ to be a particular instance of $c$, and $t_2,\dots,t_k$ to be particular instances of members of $\mathscr{B}$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or perhaps something like&#xA;$$&#xD;&#xA;\forall t_1\forall t_2\forall t_3\forall c\exists B\forall x(x\in B\Leftrightarrow x\in c \wedge [\exists t_3\in t_2(x=t_1\cup t_3)])&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where in this case $\mathscr{B}$ is a particular instance of $t_2$?, and then $\mathscr{D}$ would be the set $B$ which has been proven to exist. Thanks, right now I'm just not used to what should go in that blank spot for the formula.&lt;/p&gt;&#xA;" OwnerUserId="1583" LastActivityDate="2010-12-23T03:58:44.367" Title="Using the Subset Axioms to prove the existence of a set" Tags="&lt;elementary-set-theory&gt;&lt;axioms&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="15275" PostTypeId="1" AcceptedAnswerId="15277" CreationDate="2010-12-23T02:32:01.130" Score="4" ViewCount="174" Body="&lt;p&gt;Almost everyone is familiar with the famous Taylor Series:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!} (x-a)^n $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which, if it converges at more than one point, will converge in some interval about $a$. Has anyone considered the &quot;Reverse&quot; Taylor Series:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ g(x) = \sum_{n=0}^\infty \frac{f^{(n)}(x-a)}{n!} a^n $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I call it reverse, because its what you get for symbolically taking $a \rightarrow (x-a)$. You might be saying that there is absolutely no reason to believe that this series should converge to $f(x)$, but there are two big examples where it does.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $e^x$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$g(x) = \sum_{n=0}^\infty \frac{e^{(x-a)}}{n!} a^n = \frac{e^x}{e^a} \sum_{n=0}^\infty \frac{a^n}{n!} = e^x $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $x^k$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$g(x) = \sum_{n=0}^k \frac{k(k-1)...(k-n+1)(x-a)^{k-n} a^n}{n!} = \sum_{n=0}^k {k \choose n} (x-a)^{k-n} a^n = ((x-a) + a)^k $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have yet to find a counter-example for when the Reverse Taylor Series does not give back the original function for an analytic function, and I also have yet to think of a way to prove that the Reverse Taylor Series should converge for a given function. Does anyone have any ideas?&lt;/p&gt;&#xA;" OwnerUserId="1089" LastActivityDate="2010-12-23T02:57:52.947" Title="Reverse Taylor Series" Tags="&lt;calculus&gt;&lt;real-analysis&gt;&lt;analysis&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="15276" PostTypeId="2" ParentId="15248" CreationDate="2010-12-23T02:42:13.123" Score="-3" ViewCount="" Body="&lt;p&gt;Since each of the $m+1$ simple closed curves are disjoint, it is clear that $\Omega$ has $m+1$ connected components. Let $U$ be any one of these.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then $f(z)$ restricted to $U$ is a holomorphic function. Furthermore, it is non-constant and holomorphic on some neighborhood of the closure of $U$, because this closure is contained in $\Omega$. Furthermore, the boundary of $U$ is contained in the boundary of $\Omega$, so $|f(z)| = 1$ on the boundary of $U$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Like you outlined in your explanation of the problem, if we assume that $f$ is non-zero in $U$ then both $f(z)$ and $1/f(z)$ are holomorphic in $U$, and thus must achieve their maximums on the boundary of $U$. But its given that $|f(z)| = 1$ on the boundary of $U$, and so we get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$|f(z)| \leq 1$ and $|1/f(z)| \leq 1$ for $z \in U$. Thus we must have that $|f(z)| \equiv 1$ on $U$, which is a contradiction as we assumed that $f(z)$ is non-constant. Thus $f$ has at least one zero in $U$. We assumed nothing about which connected component we used, so this is true in each component of $\Omega$, and therefore $f(z)$ has at least $m+1$ zeroes.&lt;/p&gt;&#xA;" OwnerUserId="1089" LastActivityDate="2010-12-23T02:42:13.123" CommentCount="2" />
  <row Id="15277" PostTypeId="2" ParentId="15275" CreationDate="2010-12-23T02:48:50.303" Score="6" ViewCount="" Body="&lt;p&gt;Your two series are the same thing pointwise when they converge. $g(x)$ is the Taylor series for $f$ expanded around the point $x-a$. So it will converge as long as the Taylor series for $f$ at $x-a$ has a radius of convergence at least $a$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, in $f(x)$, moving $x$ around may make the Taylor series fail to converge due to your $x$ moving out of the radius of convergence. Whereas in $g(x)$, moving $x$ around will make the series fail to converge due to &lt;em&gt;moving the point&lt;/em&gt;  $x - a$  &lt;em&gt;somewhere with a radius of convergence smaller than&lt;/em&gt;  $a$. That is, once you fix $a$, in the Taylor series for $f$ around $a$, the convergence radius is fixed, independent of $x$. But in the series $g(x)$, the convergence of the series depends on how the radius of convergence at $x-a$ (which depends on $x$) compares with $a$. So that the function $g(x)$ converges at more than one point is insufficient to guarantee that $g(x)$ converges in a small interval about $a$. &lt;/p&gt;&#xA;" OwnerUserId="1543" LastActivityDate="2010-12-23T02:48:50.303" />
  <row Id="15278" PostTypeId="1" AcceptedAnswerId="15402" CreationDate="2010-12-23T02:49:24.030" Score="9" ViewCount="231" Body="&lt;p&gt;&lt;i&gt;I've revised my question a bit in response to the (very helpful) advice so far--&lt;/i&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have an engineering background but am interested in learning abstract harmonic analysis. My interest is rather unstructured; the Fourier transform sparked my interest in mathematics, and I've been following since I first learned of it as a sort of guide through mathematics. I've decided it's time I delved into the abstract, and from my readings so far it's become clear to me that this will require a sojourn into a swathe of unfamiliar mathematics. My background in classical analysis, linear algebra, and probability is rather good (probably the rough equivalent of a 1st year american grad student), and so I'm familiar as well with point-set topology. I'm also familiar with measure theory (via probability, and some real analysis). I have essentially no abstract algebra. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would be interested in three bits of advice:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) What algebra is essential for the canon of harmonic analysis? Perhaps this is a vacuous question (i.e., that an answer is as simple &quot;just google&quot; or &quot;open a book&quot;)? It strikes me that techniques or background might work their way in unnamed. I'm hoping to prevent the sinking feeling that I don't know something I should. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Specific examples (theorems, objects, counterexamples) which illustrate important applications of harmonic analysis, especially in relation to algebra. I'm not looking for explanations, just statements of what's thought to be important. I'm hoping that this information will help me sift through the different branches of harmonic analysis, especially where I am most unfamiliar with the language. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) Canonical/important texts which I might find useful. I'm especially interested in those good for self study (well-edited with many examples and end-of-chapter exercises, and ideally a conversational tone with lengthy remarks and some historical content). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks very much!&lt;/p&gt;&#xA;" OwnerUserId="5060" LastEditorUserId="5060" LastEditorDisplayName="" LastEditDate="2010-12-24T08:47:25.130" LastActivityDate="2010-12-24T14:02:12.033" Title="learning algebra and harmonic analysis" Tags="&lt;soft-question&gt;&lt;fourier-analysis&gt;&lt;harmonic-analysis&gt;" AnswerCount="3" FavoriteCount="1" />
  <row Id="15279" PostTypeId="2" ParentId="15275" CreationDate="2010-12-23T02:49:27.993" Score="4" ViewCount="" Body="&lt;p&gt;The general idea:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Put $y=x-a$.&#xA;Then, &#xA;$$&#xD;&#xA;g(x) = \sum\limits_{n = 0}^\infty  {\frac{{f^{(n)} (y)}}{{n!}}(x - y)^n }  = f(x).&#xD;&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="2810" LastEditorUserId="2810" LastEditorDisplayName="" LastEditDate="2010-12-23T02:57:52.947" LastActivityDate="2010-12-23T02:57:52.947" />
  <row Id="15280" PostTypeId="1" CreationDate="2010-12-23T02:53:10.370" Score="2" ViewCount="165" Body="&lt;p&gt;I became interested in how numbers actually work recently. I want a source that covers number theory. Explaining how and why numbers react the way they do. A friend recommended Mathematical Proofs: A Transition to Advanced Mathematics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to learn about the concept of numbers, how they interact etc. I am a high school student, but most college level material will be appropriate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-- Also, in my search I have found Mathematics in 10 lessons: The Grand Tour, and Foundations and Fundamentals of Mathematics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If anyone could confirm that those two resources are adequate, I would appreciate it, and other resources are still welcome.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S. I do not like the Khan Academy. His presentations do not... serve my needs.&lt;/p&gt;&#xA;" OwnerUserId="5061" LastEditorUserId="5061" LastEditorDisplayName="" LastEditDate="2010-12-23T17:34:53.543" LastActivityDate="2010-12-23T17:57:02.690" Title="Fundamental Math Theory Resources?" Tags="&lt;soft-question&gt;&lt;reference-request&gt;&lt;learning&gt;" AnswerCount="3" CommentCount="4" FavoriteCount="3" />
  <row Id="15281" PostTypeId="2" ParentId="15280" CreationDate="2010-12-23T03:03:38.547" Score="2" ViewCount="" Body="&lt;p&gt;For an overview of the many parts of mathematics and the goals and accomplishments of various subareas within mathematics I recommend Timothy Gowers (ed.) The Princeton Companion to Mathematics. The book has pointers to more technical issues if you want to go into more depth after having a broad view. &lt;/p&gt;&#xA;" OwnerUserId="1369" LastActivityDate="2010-12-23T03:03:38.547" />
  <row Id="15282" PostTypeId="2" ParentId="15280" CreationDate="2010-12-23T03:11:16.383" Score="2" ViewCount="" Body="&lt;p&gt;It seems like you're interested in seeing how mathematics is used beyond calculation in a general sort of way. Here are some recommendations - &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dr. Euler's Fabulous Formula&lt;br/&gt;&#xA;Paul Nahin &#xA;(and check out his other work as well)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Number: The Language of Science &lt;br/&gt;&#xA;Tobias Dantzig&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Pleasures of Counting &lt;br/&gt;&#xA;T.W. Korner&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gödel, Escher, Bach&lt;br/&gt;&#xA;Douglas R. Hofstadter&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These texts try to tease out interesting ideas in an accessible way. They're not exactly going to pave the road to learning proof (for that, I think a teacher is best), but they might help you see where you're going. &lt;/p&gt;&#xA;" OwnerUserId="5060" LastActivityDate="2010-12-23T03:11:16.383" />
  <row Id="15283" PostTypeId="2" ParentId="15212" CreationDate="2010-12-23T03:15:15.713" Score="3" ViewCount="" Body="&lt;p&gt;The best explanation I have heard for this is: The roulette wheel has no memory. When you spin the wheel it does not know that there have been 5 reds in a row and it is due for a black. The odds are 1/2 each time.&lt;/p&gt;&#xA;" OwnerUserId="5062" LastActivityDate="2010-12-23T03:15:15.713" />
  <row Id="15284" PostTypeId="2" ParentId="15272" CreationDate="2010-12-23T03:22:44.007" Score="2" ViewCount="" Body="&lt;p&gt;Hint: Let $S$ be a subset of the Sorgenfrey line. From each interval of the form $[q,r)$ in the original line, where $q,r \in \mathbb{Q}$, pick one point from $S$ if possible. Then characterize the points in $S$ that are not limits of the points you just chose. &lt;/p&gt;&#xA;" OwnerUserId="630" LastActivityDate="2010-12-23T03:22:44.007" />
  <row Id="15285" PostTypeId="1" AcceptedAnswerId="15287" CreationDate="2010-12-23T03:24:42.167" Score="6" ViewCount="247" Body="&lt;p&gt;It's famously unknown whether the natural log of 2 is rational or not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How about the natural log of other numbers. Is it known/unknown whether these are rational?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously ln(1) is 0, and ln(2^n) is n*ln(2) (and is thus rational iff ln(2) is rational), but how about other cases?&lt;/p&gt;&#xA;" OwnerUserId="2469" LastEditorUserId="2548" LastEditorDisplayName="" LastEditDate="2010-12-23T05:29:06.960" LastActivityDate="2011-05-07T17:15:35.143" Title="Is the natural log of n rational?" Tags="&lt;real-analysis&gt;&lt;logarithms&gt;" AnswerCount="2" CommentCount="8" FavoriteCount="1" />
  <row Id="15287" PostTypeId="2" ParentId="15285" CreationDate="2010-12-23T03:29:21.780" Score="8" ViewCount="" Body="&lt;p&gt;To shoot a bird with a cannon...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the &lt;a href=&quot;http://en.wikipedia.org/wiki/Lindemann%E2%80%93Weierstrass_theorem&quot; rel=&quot;nofollow&quot;&gt;Lindemann-Weierstrass theorem&lt;/a&gt;, $e^a$ is transcendental for all $a$ algebraic and non-zero. In particular if $a$ is rational, $e^a$ cannot be rational. Hence $\ln(n)$ is always irrational. &lt;/p&gt;&#xA;" OwnerUserId="1543" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2011-05-07T17:15:35.143" LastActivityDate="2011-05-07T17:15:35.143" CommentCount="4" />
  <row Id="15288" PostTypeId="2" ParentId="15274" CreationDate="2010-12-23T03:32:47.297" Score="4" ViewCount="" Body="&lt;p&gt;If you want to form the set &#xA;$$&#xD;&#xA;\{t\in\mathscr{P}(A\cup\bigcup\mathscr{B})\ |\ t=A\cup X\ \text{for some}\ X\in\mathscr{B}\}.&#xD;&#xA;$$&#xA;then a natural instance of the axiom of separation (the &quot;subset axiom&quot;) is&#xA;$$&#xD;&#xA;(\exists B)(\forall t) ( t \in B \Leftrightarrow t\in\mathscr{P}(A\cup\bigcup\mathscr{B}) \land (\exists X)[X \in \mathscr{B} \land (\forall s) (s \in t \Leftrightarrow s \in A \lor s \in X)]).&#xD;&#xA;$$&#xA;This formula is obtained by simply unpacking the original abbreviated formula, for example by expanding the &quot;=&quot; using its set-theoretic definition. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a general principle that if you have assumed a formula of the form $(\forall z)\phi(z)$ then, for any set $Y$, the formula $\phi(Y)$ will be true as well. This is called (universal) instantiation. The way you get to the separation axiom above is to start with &#xA;$$&#xD;&#xA;(\forall C_1)(\forall C_2)(\forall C)(\exists B)(\forall t) ( t \in B \Leftrightarrow t\in C \land (\exists X)[X \in C_1 \land (\forall s) (s \in t \Leftrightarrow s \in C_2 \lor s \in X)]).&#xD;&#xA;$$&#xA;and then instantiate it with $C_1 = \mathscr{B}$, $C_2 = A$, $C = \mathscr{P}(A\cup\bigcup \mathscr{B}) $. &lt;/p&gt;&#xA;" OwnerUserId="630" LastEditorUserId="630" LastEditorDisplayName="" LastEditDate="2010-12-23T03:58:44.367" LastActivityDate="2010-12-23T03:58:44.367" CommentCount="1" />
  <row Id="15289" PostTypeId="1" CreationDate="2010-12-23T03:50:32.903" Score="3" ViewCount="113" Body="&lt;p&gt;Let $T: E \rightarrow E$ be an endomorphism of a finite-dimensional vector space, and let $S$ be a circle in the complex plane that does not intersect any eigenvalues of $T$. Now let $Q = \frac{1}{2\pi i} \int_S (z-T)^{-1} \, dz$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is $Q$ a projection operator?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The motivation behind this question is that the above situation occurs in a proof of Bott's periodicity theorem, but it's not clear to me that $Q$ is a projection...&lt;/p&gt;&#xA;" OwnerUserId="423" LastEditorUserId="423" LastEditorDisplayName="" LastEditDate="2010-12-23T04:27:50.780" LastActivityDate="2010-12-23T04:27:50.780" Title="why is the following thing a projection operator?" Tags="&lt;linear-algebra&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="15290" PostTypeId="2" ParentId="15289" CreationDate="2010-12-23T03:53:53.727" Score="3" ViewCount="" Body="&lt;p&gt;Show that the integral depends continuously on $T$, and show that $Q^2=Q$ when $T$ is diagonalizable, by finding how $Q$ changes change you change $T$ by a similar matrix, and then reducing to the one dimensional case. Then use the fact that diagonalizable matrices are dense in the space of all matrices, and that $Q^2$ and $Q$ are continuous functions of $Q$.&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2010-12-23T03:53:53.727" CommentCount="5" />
  <row Id="15291" PostTypeId="2" ParentId="15289" CreationDate="2010-12-23T04:05:14.137" Score="2" ViewCount="" Body="&lt;p&gt;If $A$ is any Banach algebra (such as the algebra of endomorphisms of a finite dimensional  complex vector space), then for each subset $\Omega$ of the complex plane and each element $T$ of $A$ whose spectrum is contained in $\Omega$, &lt;a href=&quot;http://en.wikipedia.org/wiki/Holomorphic_functional_calculus&quot; rel=&quot;nofollow&quot;&gt;holomorphic functional calculus&lt;/a&gt; yields a homomorphism $f\mapsto f(T)$ from the algebra of functions holomorphic in an open set containing $\Omega$ (identified if they agree on some neighborhood of $\Omega$) into $A$.  Since the function $f:(\mathbb{C}\setminus S)\to\mathbb{C}$ defined by $f(w)=\frac{1}{2\pi i}\int_S(z-w)^{-1}dz$ takes on only the values $0$ and $1$ (it gives the winding number of $S$ about $w$), $f$ is idempotent (i.e., $f(w)^2=f(w)$ for all $w\in\mathbb{C}\setminus S$), and thus $f(T)$ is an idempotent element of $A$ for each $T$ whose spectrum is disjoint from $S$.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2010-12-23T04:18:29.757" LastActivityDate="2010-12-23T04:18:29.757" CommentCount="5" />
  <row Id="15293" PostTypeId="2" ParentId="15240" CreationDate="2010-12-23T04:28:59.047" Score="1" ViewCount="" Body="&lt;p&gt;The &lt;a href=&quot;http://www.mathresources.com/products/mathresource/maa/mooreosgood_theorem.html&quot; rel=&quot;nofollow&quot;&gt;Moore-Osgood Theorem&lt;/a&gt; might also be relevant here.&lt;/p&gt;&#xA;" OwnerUserId="640" LastActivityDate="2010-12-23T04:28:59.047" />
  <row Id="15294" PostTypeId="1" CreationDate="2010-12-23T04:47:57.227" Score="6" ViewCount="434" Body="&lt;p&gt;I understand how derivatives work based on the definition, and the fact that my professor explained it step by step until the point where I can derive it myself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However when it comes to the area under a curve for some reason when you break it up into an infinite amount of rectangles, magically it turns into the anti-derivative. Can someone explain why that is the definition of the integral and how Newton figured this out?&lt;/p&gt;&#xA;" OwnerUserId="5063" LastEditorUserId="2548" LastEditorDisplayName="" LastEditDate="2010-12-23T05:30:45.383" LastActivityDate="2010-12-23T06:40:29.690" Title="Why is the area under a curve the integral?" Tags="&lt;calculus&gt;&lt;integral&gt;&lt;intuition&gt;" AnswerCount="2" CommentCount="6" FavoriteCount="6" />
  <row Id="15295" PostTypeId="1" AcceptedAnswerId="15305" CreationDate="2010-12-23T04:50:35.530" Score="2" ViewCount="82" Body="&lt;p&gt;Suppose $\kappa$ is an uncountable cardinal. Then $S\subseteq\kappa$ is club (CLosed UnBounded) if $S$ is unbounded in $\kappa$ and is a closed subset of $\kappa$ under the order topology. My question is the following: suppose $S\subseteq\kappa$ is club, and $S=S_1\cup S_2$. Is there necessarily some club set $C$ such that either $C\subseteq S_1$ or $C\subseteq S_2$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Note that, since the intersection of two clubs is again club, this question amounts to asking whether, for $\kappa$ an infinite cardinal, the set $\{S\subseteq\kappa: S$ is a superset of some club set $\}$ is an ultrafilter.) &lt;/p&gt;&#xA;" OwnerUserId="5064" LastEditorUserId="622" LastEditorDisplayName="" LastEditDate="2010-12-23T06:54:43.563" LastActivityDate="2010-12-23T06:54:43.563" Title="Splitting a club set" Tags="&lt;set-theory&gt;" AnswerCount="2" />
  <row Id="15296" PostTypeId="2" ParentId="15240" CreationDate="2010-12-23T04:57:30.503" Score="4" ViewCount="" Body="&lt;p&gt;If you want to avoid hypotheses that involve uniform convergence, you can always cheat and use the counting measure on {0, 1, 2,...} and then use either the Monotone or Dominated Convergence Theorem from integration theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, using the &lt;a href=&quot;http://en.wikipedia.org/wiki/Monotone_convergence_theorem&quot; rel=&quot;nofollow&quot;&gt;Monotone Convergence Theorem&lt;/a&gt;, we get the following (perhaps silly) sufficient criterion:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proposition:&lt;/strong&gt; If $a_{mn}$ is monotonically increasing in $n$, and is such that $c_{mn} = a_{m,n+1} - a_{mn}$ is also monotonically increasing in $n$, then $\lim\limits_{m \to \infty} \lim\limits_{n \to \infty} a_{mn} = \lim\limits_{n\to\infty}\lim\limits_{m\to\infty} a_{mn}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Proof:&lt;/em&gt; Our two hypotheses really just amount to saying that each $c_{mn} \geq 0$ and that the $c_{mn}$ are monotonically increasing in $n$.  So we can use the Monotone Convergence Theorem with respect to the counting measure:&#xA;$$\lim_{m\to \infty} \int c_{mn} = \int \lim_{m\to \infty} c_{mn}$$&#xA;But really, these integrals are sums, so:&#xA;$$\lim_{m\to \infty} \sum_{n=0}^\infty c_{mn} = \sum_{n=0}^\infty \lim_{m\to \infty} c_{mn}$$&#xA;Since infinite sums are just limits of partial sums, we have:&#xA;$$\lim_{m\to \infty} \lim_{N \to \infty} \sum_{n=0}^N c_{mn} = \lim_{N\to\infty}\lim_{m\to \infty} \sum_{n=0}^N c_{mn}$$&#xA;By our construction of $c_{mn}$, the left side is $\lim\limits_{m\to\infty} \lim\limits_{N\to\infty} a_{mN}$, and the right side is $\lim\limits_{N\to\infty} \lim\limits_{m\to\infty}a_{mN}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the &lt;a href=&quot;http://en.wikipedia.org/wiki/Dominated_convergence_theorem&quot; rel=&quot;nofollow&quot;&gt;Dominated Convergence Theorem&lt;/a&gt;, we could probably get something slightly more useful.&lt;/p&gt;&#xA;" OwnerUserId="640" LastEditorUserId="640" LastEditorDisplayName="" LastEditDate="2010-12-27T08:30:21.430" LastActivityDate="2010-12-27T08:30:21.430" CommentCount="4" />
  <row Id="15297" PostTypeId="1" CreationDate="2010-12-23T05:09:02.693" Score="0" ViewCount="157" Body="&lt;p&gt;Hi Guys, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is part of an assignment that I need to get a good mark for - I'd appreciate it if you guys could look over it and give some pointers where I've gone wrong. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(apologies for the italics)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\prod_{n=1}^\infty\left(1+\frac{2}{n}\right)\; converges \;when&#xD;&#xA;\sum_1^\infty ln\left(1+\frac{2}{n}\right)\;converges$$&#xA; $$\sum_1^\infty ln\left(1+\frac{2}{n}\right)\;=\;\sum_1^\infty ln(n+2)-ln(n)$$&#xA; $$ let\, f(x)=ln(x+2)-ln(x) \rightarrow f&amp;#39;(x)=\frac{1}{x+2} - \frac{1}{x}$$&#xA; $$ = \frac{x-x-2}{x(x+2)} = \frac{-2}{x(x+2)}&amp;lt;0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f(x)\; is\; a\; decreasing \;function.$$&#xA; $$f(x) \;is \;a \;positive \;function \;for \;x&amp;gt;=1$$ &#xA; $$f(x)\;is \;a \;continuous \;function \;for \;x&amp;gt;=1$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;using integration test. \newline&#xA; $$\int_1^\infty ln(x+2) - ln(x) = \lim_{t \to \infty}\int_1^t ln(x+2)dx -&#xD;&#xA; \lim_{t \to \infty}\int_1^t lnx dx$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int ln(x)dx = xlnx - x + c \;\;it\;follows\;that\; \int ln(x+2) =&#xD;&#xA;(x+2)ln(x+2) - (x+2) + c$$&#xA;Therefore&#xA;$$\int ln(x+2) - ln(x)dx  = (x+2)ln(x+2)-x - 2 - xln(x) + x + c$$&#xA;$$ = xln(\frac{x+2}{x})+ 2ln(x+2)-2 + c $$&#xA;Therefore, &#xA;$$\int_1^\infty ln(x+2) - ln(x)dx = \lim_{t \to&#xD;&#xA;\infty}\left[xln(\frac{x+2}{x}) + 2ln(x+2) - 2\right]_1^t$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ = \lim_{t \to&#xD;&#xA;\infty}\left[t ln(\frac{t+2}{t}) + 2ln(t+2) - 2\right] - \lim_{t \to&#xD;&#xA;\infty}\left[ln(\frac{3}{1}) + 2ln(3) - 2\right] $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ =\lim_{t \to&#xD;&#xA;\infty}\left[t ln(\frac{t+2}{t}) + 2ln(t+2) - 3ln(3)\right]$$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ As\; t\rightarrow\infty, \; \lim_{t \to \infty}t ln\left(\frac{t+2}{t}\right)&#xD;&#xA;+ 2ln(t+2) = \infty. $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$Therefore\; the \; series\; \sum_1^\infty ln\left(1+\frac{2}{n}\right)\; is\;&#xD;&#xA;divergent$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$Similarly\; the\; infinite\; product\;&#xD;&#xA;\prod_{n=1}^\infty\left(1+\frac{2}{n}\right)\; is\; also\; divergent$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gregg&lt;/p&gt;&#xA;" OwnerUserId="4362" LastEditorUserId="1301" LastEditorDisplayName="" LastEditDate="2010-12-23T08:36:26.417" LastActivityDate="2010-12-23T09:01:22.193" Title="integral test to show that infinite product $ \prod_{n=1}^\infty\left(1+\frac{2}{n}\right)$ diverges" Tags="&lt;calculus&gt;&lt;homework&gt;&lt;sequences-and-series&gt;" AnswerCount="2" CommentCount="7" />
  <row Id="15298" PostTypeId="2" ParentId="7296" CreationDate="2010-12-23T05:14:58.913" Score="0" ViewCount="" Body="&lt;p&gt;i would like to know the solution of the abel equation from either the integral form or the differintegral one&lt;/p&gt;&#xA;" OwnerUserId="5065" LastActivityDate="2010-12-23T05:14:58.913" />
  <row Id="15299" PostTypeId="2" ParentId="15285" CreationDate="2010-12-23T05:19:53.260" Score="5" ViewCount="" Body="&lt;p&gt;We can also use a non-simple continued fraction expansion of $\displaystyle e^{2x/y}$ to prove the irrationality of $\displaystyle e^{2x/y}$ when $\displaystyle x,y$ are positive integers. Thus if $\displaystyle \log n = x/y$, then $\displaystyle e^{2x/y} = n^2 $ is rational, contradicting irrationality of $\displaystyle  e^{2x/y}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Incidentally, the first proof of irrationality of $\pi$ by Lambert used a continued fraction expansion (of $\tan x$, I believe).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The expansion we use:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/RA92t.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and the theorem we use to prove irrationality is quoted in the wiki page for Generalized Continued Fractions here: &lt;a href=&quot;http://en.wikipedia.org/wiki/Generalized_continued_fraction#Conditions_for_irrationality&quot; rel=&quot;nofollow&quot;&gt;Conditions of Irrationality&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By this theorem, it is enough that for all sufficiently large positive integers $\displaystyle m$ we have that $\displaystyle (2m+1)y \gt x^2$, which is true for fixed $\displaystyle x,y$.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-12-23T05:19:53.260" />
  <row Id="15300" PostTypeId="1" CreationDate="2010-12-23T05:20:56.893" Score="1" ViewCount="103" Body="&lt;p&gt;one of the early applications of fractional calculus is the tautochronus problem set up by Abel in the integral form or its fractional derivative one. i wish to know its solution&lt;/p&gt;&#xA;" OwnerUserId="5065" LastEditorUserId="2548" LastEditorDisplayName="" LastEditDate="2010-12-23T05:26:58.100" LastActivityDate="2011-03-08T21:24:54.600" Title="a problem in fractional calculus" Tags="&lt;calculus&gt;&lt;analysis&gt;" AnswerCount="0" CommentCount="1" />
  <row Id="15301" PostTypeId="2" ParentId="15294" CreationDate="2010-12-23T05:29:16.647" Score="20" ViewCount="" Body="&lt;p&gt;First: the integral is &lt;em&gt;defined&lt;/em&gt; to be the (net signed) area under the curve. The definition in terms of Riemann sums is precisely designed to accomplish this. The integral is a limit, a number. There is, a priori, no connection whatsoever with derivatives. (That is one of the things that makes the Fundamental Theorems of Calculus such a potentially surprising things). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why does the limit of the Riemann sums actually give the area under the graph? The idea of approximating a shape whose area we don't know both from &quot;above&quot; and from &quot;below&quot; with areas we &lt;em&gt;do&lt;/em&gt; know goes all the way back to the Greeks. Archimedes gave bounds for the value of $\pi$ by figuring out areas of inscribed and circumscribed polygons in a circle, knowing that the area of the circle would be somewhere between the two; the more sides to the polygons, the closer the inner and outer polygons are to the circle, the closer the areas are to the area of the circle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The way Riemann tried to formalize this was with the &quot;upper&quot; and &quot;lower&quot; Riemann sums: assuming the function is relatively &quot;nice&quot;, so that on each subinterval it has a maximum and a minimum, the &quot;lower Riemann sum&quot; is done by taking the largest &quot;rectangle&quot; that will lie completely under the graph by looking at the minimum value of the function on the interval and using that as height; and the &quot;upper Riemann sum&quot; is done by taking the smallest rectangle for which the graph will lie completely &lt;em&gt;under&lt;/em&gt; it (by taking the maximum value of the function as the height). Certainly, the exact area under the graph on that interval will be somewhere between the two. If we let $\underline{S}(f,P)$ be the lower sum corresponding to some fixed partition $P$ of the interval, and $\overline{S}(f,P)$ be the upper sum, we will have that &#xA;$$\underline{S}(f,P) \leq \int_a^b f(x)\,dx \leq \overline{S}(f,P).$$&#xA;(Remember that $\int_a^bf(x)\,dx$ is just the symbol we use to denote the exact (net signed) area under the graph of $f(x)$ between $a$ and $b$, whatever that quantity may be.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, intuitively, the &lt;em&gt;more&lt;/em&gt; intervals we take, the closer these two approximations (one from below and one from above) will be. This does not always work out if &lt;em&gt;all&lt;/em&gt; we do is take &quot;more&quot; intervals. But one thing we &lt;em&gt;can&lt;/em&gt; show is that if $P&amp;#39;$ is a &lt;em&gt;refinement&lt;/em&gt; of $P$ (it includes all the dividing points that $P$ had, &lt;em&gt;and&lt;/em&gt; possibly more points) then&#xA;$$\underline{S}(f,P)\leq \underline{S}(f,P&amp;#39;)\text{ and } \overline{S}(f,P&amp;#39;)\leq \overline{S}(f,P)$$&#xA;so at least the approximations are heading in the right direction. To see why this happens, suppose you split one of the subintervals $[t_i,t_{i+1}]$ in two, $[t_i,t&amp;#39;]$ and $[t&amp;#39;,t_{i+1}]$. The minimum of $f$ on $[t_i,t&amp;#39;]$ and on $[t&amp;#39;,t_{i+1}]$ are each greater than or equal to the minimum over the whole of $[t_i,t_{i+1}]$, but it may be that the minimum in one of the two bits is actually strictly larger than the minimum over $[t_i,t_{i+1}]$. The areas we get after the split can be no smaller, but they can be larger than the ones we had before the split. Similarly for the upper sums.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, let's consider one particular sequence of partitions: divide the interval into 2 equal parts; then into 4; then into 8; then into 16; then into 32; and so on; then into $2^n$, etc. If $P_n$ is the partition that divides $[a,b]$ into $2^n$ equal parts, then $P_{n+1}$ is a refinement of $P_n$, and so we have:&#xA;$$\underline{S}(f,P_1) \leq\cdots \leq \underline{S}(f,P_n)\leq\cdots \leq\int_a^b f(x)\,dx \leq\cdots \leq\overline{S}(f,P_n)\leq\cdots \leq \overline{S}(f,P_2)\leq\overline{S}(f,P_1).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the sequence of numbers $\underline{S}(f,P_1) \leq \underline{S}(f,P_2)\leq\cdots \leq \underline{S}(f,P_n)\leq\cdots$ is increasing and bounded above (by the area). So the numbers have a supremum; call it $\underline{S}$. This number is no more than $\int_a^b f(x)\,dx$. And the numbers $\overline{S}(f,P_1) \geq \overline{S}(f,P_2)\geq\cdots \geq \overline{S}(f,P_n)\geq\cdots$ are decreasing and bounded below, so they have a minimum; call this $\overline{S}$; again, it is no less than $\int_a^bf(x)\,dx$. So we have:&#xA;$$\lim_{n\to\infty}\underline{S}(f,P_n) = \underline{S} \leq \int_a^b f(x)\,dx \leq \overline{S} = \lim_{n\to\infty}\overline{S}(f,P_n).$$&#xA;What if we are lucky? What if actually we have $\underline{S}=\overline{S}$? Then it &lt;em&gt;must&lt;/em&gt; be the case that this common value is the value of $\int_a^b f(x)\,dx$. It just doesn't have a choice! It's definitely trapped between the two, and if there is no space between them, then it's equal to them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What Riemann proved was several things:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;If $f$ is &quot;nice enough&quot;, then you will &lt;em&gt;necessarily&lt;/em&gt; get that $\underline{S}=\overline{S}$. In particular, continuous functions happen to be &quot;nice enough&quot;, so it will definitely work for them (in fact, continuous functions turn out to be &quot;very nice&quot;, not just &quot;nice enough&quot;). &lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If $f$ is &quot;nice enough&quot;, then you don't &lt;em&gt;have&lt;/em&gt; to use the partitions we used above. You can use &lt;em&gt;any&lt;/em&gt; sequence of partitions, so lonthe &quot;mesh size&quot; (the size of the largest subinterval in the partition) gets smaller and smaller, and has limit of $0$ as $n\to\infty$; if it works for the partitions &quot;divide-into-$2^n$-equal-intervals&quot;, then it works for &lt;em&gt;any&lt;/em&gt; sequence of partitions whose mesh size goes to zero. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, for example, we can take $P_n$ to be the partition that divides $[a,b]$ into $n$ equal parts, even though $P_{n+1}$ is not a refinement of $P_n$ in this case.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;In fact, you don't have to do $\underline{S}(f,P)$ and $\overline{S}(f,P)$. For the partition $P$, just pick any rectangle that has as its height &lt;em&gt;any&lt;/em&gt; value of the function in the subinterval (that is, pick an arbitrary $x_i^*$ in the subinterval $[t_i,t_{i+1}]$, and use $f(x_i^*)$ as the height). Call the resulting sum $S(f,P,x_1^*,\ldots,x_n^*)$. Then you have&#xA;$$\underline{S}(f,P) \leq S(f,P,x_1^*,\ldots,x_n^*)\leq \overline{S}(f,P)$$&#xA;because $\underline{S}(f,P)$ is computed using the &lt;em&gt;smallest&lt;/em&gt; possible values of $f$ throughout, and $\overline{S}(f,P)$ is computedusing the &lt;em&gt;largest&lt;/em&gt; possible values of $f$ throughout. But since we already know, from 1 and 2 above, that $\underline{S}(f,P)$ and $\overline{S}(f,P)$ have the same limit, then the sums $S(f,P,x_1^*,\ldots,x_n^*)$  &lt;em&gt;also&lt;/em&gt; get squeezed and must have &lt;em&gt;that&lt;/em&gt; same limit, which equals the integral. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, we can always take the left endpoint (and get a &quot;Left Hand Sum&quot;) or we can always take the right endpoint (and get a &quot;Right Hand Sum&quot;), and you will nevertheless get the same limit.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;So in summary, you can pick &lt;em&gt;any&lt;/em&gt; sequence of partitions, whichever happens to be convenient, so long as the mesh size goes to $0$, and you can pick &lt;em&gt;any&lt;/em&gt; points on the subintervals (say, ones which make the calculations simpler) at each stage, and so long as the function is &quot;nice enough&quot; (for example, if it is continuous), everything will work out and the limit will be the number which &lt;em&gt;must&lt;/em&gt; be the value of the area (because it was trapped between the lower and upper sums, and they both got squeezed together trapping the limit and the integral both between them). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, (1) and (2) above are the hardest part of what Riemann did. Don't be surprised if it sounds a bit magical at this point. But I hope that you agree that &lt;em&gt;if&lt;/em&gt; the lower and upper sums for the special partitions have the same limits then that limit &lt;em&gt;must be&lt;/em&gt; the area that lies under the graph. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks to that work of Riemann, then (at least for continuous functions) we can &lt;em&gt;define&lt;/em&gt; $\int_a^b f(x)\,dx$ to be the limit of, say, the left hand sums of the partitions we get by dividing $[a,b]$ into $n$ equal parts, because these partitions have mesh size going to $0$, we can pick any points we like (say, the left end points), and we know the limit is going to be that common value of $\underline{S}$ and $\overline{S}$, which &lt;em&gt;has&lt;/em&gt; to be the area. So that, under this definition, $\int_a^b f(x)\,dx$ really &lt;em&gt;is&lt;/em&gt; the net signed area under the graph of $f(x)$. It just doesn't have a choice but to be that, when $f$ is &quot;nice enough&quot;. &lt;/p&gt;&#xA;&#xA;&lt;hr/&gt;&#xA;&#xA;&lt;p&gt;Second, the area does &lt;em&gt;not&lt;/em&gt; turn into &quot;the&quot; antiderivative. What happens is that it turns out (perhaps somewhat magically) that the area can be &lt;em&gt;computed&lt;/em&gt; using &lt;em&gt;an&lt;/em&gt; antiderivative. I'll go into some more details below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As to how Newton figured this out, &lt;em&gt;his&lt;/em&gt; teacher, Isaac Barrow, was the one who discovered there was a connection between derivatives and tangents; some of the basic ideas were his. They came from studying some simple functions and some simple formulas for tangents he had discovered.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, the tangents to the parabola $y=x^2$ were interesting (there was generally geometric interest in tangents and in &quot;squaring&quot; regions, also known as finding the &quot;quadrature&quot; of a region, that is, finding a way to construct a square or rectangle that had the same area as the region you were considering), and let to associate the parabola $y=x^2$ to lines of the form $y=2x$. It does not take too much experimentation to realize that if you look at the area under $y=2x$ from 0 to a, you end up with $a^2$, establishing a connection. Barrow did this with arguments with infinitesimals (which were a bit fuzzy and not set on entirely correct  and solid logical foundation until well into the 20th century), which were generally laborious, and only for &lt;em&gt;some&lt;/em&gt; curves. When Newton extended Barrow's methods to more general curves and tangents, he also extended the discovery of the connection with areas, and was able to prove what is essentially the Fundamental Theorem of Calculus.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, here is one way to approach the connection. We want to figure out the value of, say, &#xA;$$\int_0^a f(x)\,dx$$&#xA;for some $a$. This can be done using limits and Riemann sums (Newton and Leibnitz had similar methods, though not set up quite as precisely as Riemann sums are). But here is an absolutely crazy suggestion: suppose you can find a &quot;master function&quot; $\mathcal{M}$, which, when given &lt;em&gt;any&lt;/em&gt; point $b$ between $0$ and $a$, will give you the value of $\int_0^b f(x)\,dx$. If you have such a master function, then you can use it to find the value of the integral you want just by taking $\mathcal{M}(a)$!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, this is the approach Barrow had taken: his great insight was that instead of trying to find the quadratire a particular area, he was trying to solve the problem of squaring several different (but related) areas at the same time. So he was looking for, for instance, a &quot;master function&quot; for the region was like a triangle except that the top was a parabola instead of a line (like the area from $0$ to $a$ under $y=x^2$), and so on. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On its face, this is a completely ludicrous suggestion. It's like telling someone who is trying to know how to get from building A to building B that if he only memorizes the map for the entire city &lt;em&gt;first&lt;/em&gt;, then he can use that knowledge to figure out how to get form A to B. If we are having trouble finding the integral $\int_0^a f(x)\,dx$, then the &quot;master function&quot; seems to require us to find not just &lt;em&gt;that&lt;/em&gt; area, but also &lt;em&gt;all&lt;/em&gt; areas in between! It's like telling someone who is having trouble walking that he should just run very slowly when he wants to walk.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, again, the interesting thing is that even though we may not be able to say what the &quot;master function&quot; &lt;em&gt;is&lt;/em&gt;, we &lt;em&gt;can&lt;/em&gt; say how it changes as b changes (remember, $\mathcal{M}(b) = \int_0^b f(x)\,dx$ is a number that depends on $b$, so $\mathcal{M}$ is a function of $b$). Because figuring out how functions change is easier than computing their values (just thing about derivatives, and how we can easily figure out the rate of change of $\sin(x)$, but we have a hard time actually &lt;em&gt;computing&lt;/em&gt; specific values of $\sin(x)$ that are not among some very simple ones). (This is also something Barrow already knew, as did Newton).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For &quot;nice functions&quot; (if $f$ is &lt;em&gt;continuous&lt;/em&gt; on an interval that contains $0$ and $a$), we can do it using limits and some theorems about &quot;nice&quot; functions:&#xA;Using limits, we have:&#xA;\begin{align}&#xA;\lim_{h\to 0}\frac{\mathcal{M}(b+h)-\mathcal{M}}{h} &amp;amp;= \frac{1}{h}\left(\int_0^{b+h}f(x)\,dx - \int_0^bf(x)\,dx\right)\\&#xA;&amp;amp;= \frac{1}{h}\int_b^{b+h}f(x)\,dx.&#xA;\end{align}&#xA;Since we are assuming that $f$ is continuous on $[0,a]$, it is continuous on the interval with endpoints $b$ and $b+h$ (I say it this way because $h$ could be negative). So it has a maximum and a minimum (continuous function on a finite closed interval). Same the maximum is $M(h)$ and the minimum is $m(h)$. Then $m(h) \leq f(x) \leq M(h)$ for all $x$ in the interval, so we know, since the integral &lt;em&gt;is&lt;/em&gt; the area, that&#xA;$$hm(h) \leq \int_b^{b+h}f(x)\,dx \leq hM(h).$$&#xA;That means that&#xA;$$m(h) \leq \frac{1}{h}\int_b^{b+h}f(x)\,dx \leq M(h)\text{ if $h\gt 0$}$$&#xA;and&#xA;$$ M(h) \leq \frac{1}{h}\int_b^{b+h}f(x)\,dx \leq m(h)\text{ if $h\lt 0$.}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As $h\to 0$, the interval gets smaller, the difference between the minimum and maximum value gets smaller. One can prove that both $M$ and $m$ are continuous functions, and that $m(h)\to f(b)$ as $h\to 0$, and likewise that $M(h)\to f(b)$ as $h\to 0$. So we can use the Squeeze Theorem to conclude that since the limit of $\frac{1}{h}\int_b^{b+h}f(x)\,dx$ is squeezed between two functions that both have the same limit as $h\to 0$, then $\frac{1}{h}\int_b^{b+h}f(x)\,dx$ also has a limit as $h\to 0$ and is in fact that same quantity, namely $f(b)$. That is&#xA;$$\frac{d}{db}\mathcal{M}(b) = \lim_{h\to 0}\frac{\mathcal{M}(b+h)-\mathcal{M}(b)}{h} = \lim_{h\to 0}\frac{1}{h}\int_{b}^{b+h}f(x)\,dx = f(b).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is: when $f$ is continuous, the &quot;Master function&quot; for areas turns out to have a rate of change equal to $f$. This is not that crazy, if you think about it: how is the area under $y=f(x)$ from $x=0$ to $x=b$ changing? Well, it's changing by whatever $f$ is. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means that, whatever the &quot;Master function&quot; turns out to be, it will be an antiderivative of $f(x)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We also know, because we are very good with derivatives, that if $\mathcal{F}(x)$ and $\mathcal{G}(x)$ are two functions, and $\mathcal{F}&amp;#39;(x) = \mathcal{G}&amp;#39;(x)$ for all $x$, then $\mathcal{F}$ and $\mathcal{G}$ differ by a constant: there exists a constant $k$ such that $\mathcal{F}(x) = \mathcal{G}(x)+k$ for all $x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, we know that the &quot;Master function&quot; is &lt;em&gt;an&lt;/em&gt; antiderivative. If, by some sheer stroke of luck, we happen to find &lt;em&gt;any&lt;/em&gt; antiderivative $F(x)$ for $f(x)$, then we know that the only possible difference between $\mathcal{M}(b)$ and $F(b)$ is a constant. What constant? Well, luckily we know &lt;em&gt;one&lt;/em&gt; value of $\mathcal{M}(b)$: we know that $\mathcal{M}(0) = \int_0^0f(x)\,dx$ &lt;em&gt;should&lt;/em&gt; be $0$.  So, $M(0) = 0 = F(0)-F(0)$, which means the constant has to be $-F(0)$. That is, we must have $M(b) = F(b)-F(0)$ for all $b$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, if we find &lt;em&gt;any&lt;/em&gt; antiderivative $F$ of $f$, then $\mathcal{M}(b) = F(b)-F(0)$ is in fact the &quot;Master function&quot; we were looking for, the one that gives &lt;em&gt;all&lt;/em&gt; the integrals between $0$ and $a$, including $0$ and including $a$. So that we have that two very different processes (computing areas using limits of Riemann sums, and derivatives) are connected: if $f(x)$ is continuous, and $F(x)$ is &lt;em&gt;any&lt;/em&gt; antiderivative for $f(x)$ on $[0,a]$, then&#xA;$$\int_0^a f(x)\,dx = \mathcal{M}(a) = F(a)-F(0).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the integral did not &quot;magically turn&quot; into an antiderivative. It's that the &quot;Master function&quot; which can be used to keep track of &lt;em&gt;all&lt;/em&gt; integrals of $f(x)$ has rate of change equal to $f$, which gives us a &quot;back door&quot; to computing integrals. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Newton was able to prove this because he had the guide of Barrow's insight that this was happening for the functions he worked with. Barrow's insight was achieved because he had the brilliant idea of trying to come up with a &quot;Master function&quot; instead of trying to rectify lots of different areas one at a time, and he noticed the connection because he had already worked with tangents/derivatives for those functions. Leibnitz likewise had access to Barrow's ideas, so the connection between the two was also known to him.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-23T06:40:29.690" LastActivityDate="2010-12-23T06:40:29.690" CommentCount="4" />
  <row Id="15302" PostTypeId="2" ParentId="15294" CreationDate="2010-12-23T05:31:52.793" Score="13" ViewCount="" Body="&lt;p&gt;One way you can perhaps &quot;justify&quot;/give an intuitive reason is to consider the following figure:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/ZhB8N.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$A(x)$ is the area under the curve from $0$ to $x$, the brown region.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$A(x+dx)$ is the area under the curve from $0$ to $x + dx$, the brown + gray.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now for for really small $dx$, we can consider the gray region to be a rectangle of side width $dx$ and height $f(x)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus $\dfrac{A(x+dx) - A(x)}{dx} = f(x)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus as $dx \to 0$, we see that $A&amp;#39;(x) = f(x)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is kind of intuitive to define area by approximating by very thin rectangles. The above gives an intuition as to why the derivative of the area gives the curve.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-12-23T05:31:52.793" CommentCount="1" />
  <row Id="15303" PostTypeId="2" ParentId="14823" CreationDate="2010-12-23T05:39:07.410" Score="3" ViewCount="" Body="&lt;p&gt;@PF4Public, As stated in the comment and also mentioned by @mtrw this is likely aliasing phenomenon.  Your intuition that it is a mathematical issue is correct.  There are many websites that describe the mechanics of aliasing such as this &lt;a href=&quot;http://en.wikipedia.org/wiki/Aliasing&quot; rel=&quot;nofollow&quot;&gt;Wikipedia article&lt;/a&gt;.  Basically, the phenomenon occurs when an analog signal is sampled at a rate that is too slow as compared to the frequency content.  You will get this phenomenon whenever the sampling frequency is less than 2x the highest frequency component (known as the Nyquist rate).  When aliasing occurs, higher frequencies &quot;appear&quot; to be lower.  Furthermore, when a signal frequency exceeds 1/2 the Nyquist rate its apparent frequency starts to go down, exactly the phenomenon that you are seeing.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To solve this problem you need to make sure that the sampling frequency of you oscilloscope is at least 2x your largest harmonic.  I am not sure how the FFT function works on you oscilloscope, but sometimes the sampling rate is tied to the sec/div setting.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope that this helps.&lt;/p&gt;&#xA;" OwnerUserId="4726" LastActivityDate="2010-12-23T05:39:07.410" />
  <row Id="15304" PostTypeId="2" ParentId="15214" CreationDate="2010-12-23T05:43:52.773" Score="0" ViewCount="" Body="&lt;p&gt;(This was supposed to be a comment, but it got too long.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem of piecewise linear regression has been looked into many times before; I do not currently have access to these papers (and thus cannot say more about them), but you might want to look into these:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.ncbi.nlm.nih.gov/pubmed/2759968&quot; rel=&quot;nofollow&quot;&gt;This paper&lt;/a&gt; (published in a &lt;em&gt;physiology&lt;/em&gt; journal, of all places) discusses how to fit a piecewise linear function to certain data sets encountered in neurological research. A FORTRAN routine is included.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.jstor.org/pss/2285594&quot; rel=&quot;nofollow&quot;&gt;This paper&lt;/a&gt; also includes a FORTRAN routine for fitting piecewise linear functions to data.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.jstor.org/pss/2287599&quot; rel=&quot;nofollow&quot;&gt;This paper&lt;/a&gt; relies on maximum likelihood to find the best fit piecewise linear function.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-12-23T05:43:52.773" />
  <row Id="15305" PostTypeId="2" ParentId="15295" CreationDate="2010-12-23T06:10:03.140" Score="3" ViewCount="" Body="&lt;p&gt;Hi Noah,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer is no, this is never an ultrafilter. This uses choice. We assume that the cofinality of $\kappa$ is bigger than $\omega$, else, $C$ has a subset of type $\omega$ cofinal (which is automatically club), and this can be split into two disjoint cofinal (therefore club) sets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the cofinality of $\kappa$ is larger than $\omega_1$, simply note that for any regular $\alpha&amp;lt;\kappa$, $\{\beta&amp;lt;\kappa:\beta$ has cofinality $\alpha\}$ is &lt;em&gt;stationary&lt;/em&gt;, i.e., it meets every club (consider the $\alpha$-th member of the increasing enumeration of the club. These sets are disjoint for different $\alpha$, and if $\kappa&amp;gt;\omega_1$, there are at least two regular $\alpha$ below $\kappa$, namely $\omega$ and $\omega_1$. This shows that we can split $C$ into disjoint stationary sets (and therefore, neither is club, since a stat. set must meet every club).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All I've used so far is that $\omega_1$ is regular.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To show that clubs on ordinals of cofinality $\omega_1$ can be split into disjoint stationary sets  requires more choice. In fact, it is consistent that the club filter on $\omega_1$ is an ultrafilter. This is the case, for example, in models where the axiom of determinacy holds. (But it is strictly weaker than determinacy. A measurable suffices in consistency strength.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The typical argument from choice uses the existence of Ulam matrices. See definition 12 and the following paragraph in this &lt;a href=&quot;http://caicedoteaching.wordpress.com/2009/02/13/580-cardinal-arithmetic-5/&quot; rel=&quot;nofollow&quot;&gt;blog entry&lt;/a&gt;. An $\omega\times\omega_1$ Ulam matrix gives us that every stat. subset of $\omega_1$, in particular every club, can be split into $\omega_1$ disjoint stat. sets. So the club filter is not an ultrafilter, and we are done.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Ulam's argument also shows that any stat. subset of $\kappa^+$ can be split into $\kappa^+$ stat. sets, and Solovay proved the same for any reglar $\lambda$, not just a successor. This, of course, gives the result that no club filter is an ultrafilter, but I wanted to point out that only in cofinality $\omega_1$ we need that much choice to carry out the argument.&lt;/p&gt;&#xA;" OwnerUserId="462" LastEditorUserId="462" LastEditorDisplayName="" LastEditDate="2010-12-23T06:36:01.103" LastActivityDate="2010-12-23T06:36:01.103" CommentCount="1" />
  <row Id="15306" PostTypeId="2" ParentId="15295" CreationDate="2010-12-23T06:54:20.553" Score="1" ViewCount="" Body="&lt;p&gt;If you are familiar with the idea of clubs and stationary, then there is a good measure theoretic intuition around it:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clubs are of measure $1$. They are almost everywhere.&lt;br&gt;&#xA;Non-stationary sets are of measure $0$ and are almost nowhere.&lt;br&gt;&#xA;Stationary sets are of positive measure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now to your question, look at the unit interval, and take some subset of measure $1$, you can split it into two positive measured sets, none of which is of measure $1$, namely intersect it with $[0,\frac{1}{2})$ and with $[\frac{1}{2},1]$, and you have two disjoint subsets of positive measure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now it is important to say that this is just the intuition at heart, and since Andres gave a good answer with a formal backbone I decided to add this one as well.&lt;/p&gt;&#xA;" OwnerUserId="622" LastActivityDate="2010-12-23T06:54:20.553" />
  <row Id="15307" PostTypeId="1" AcceptedAnswerId="15308" CreationDate="2010-12-23T07:16:04.083" Score="3" ViewCount="109" Body="&lt;p&gt;OK, I know that Matiyasevich's solution to Hilbert's 10th problem shows that there is no algorithm to decide whether or not a polynomial $p(x_1,\ldots,p_n)$ with integer coefficients has a solution over the integers.  I have no doubts about the truth of that.  However, I have been unable to figure out why the following algorithm doesn't work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Program one computer to plug every $(x_1,\ldots,x_n) \in \mathbb{Z}^n$ into $p(x_1,\ldots,p_n)$.  If there is an integer solution, then this will halt eventually.  Program a second computer to enumerate every possible proof.  There are only countably many, so if there is a proof that $p(x_1,\ldots,x_n)$ has no integer solutions, then this computer will eventually find one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given Matiyasevich's theorem, the only conclusion I can draw is that there are polynomials that don't have integral zeros, but for which there is no proof that they don't (so it's independent of ZFC).  This seems very odd to me.  What am I missing?&lt;/p&gt;&#xA;" OwnerUserId="5067" LastActivityDate="2010-12-23T07:24:36.993" Title="Algorithms to prove that polynomials don't have integer solutions" Tags="&lt;algorithms&gt;&lt;algebraic-geometry&gt;&lt;computability&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="15308" PostTypeId="2" ParentId="15307" CreationDate="2010-12-23T07:24:36.993" Score="6" ViewCount="" Body="&lt;p&gt;(Usual disclaimers about ZFC being consistent, etc.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are not missing anything, this is precisely the case. For example, there is a polynomial $p(x_1,\dots,x_n)\in{\mathbb Z}[x_1,\dots,x_n]$ for some $n$ so that any integer solution codes a proof that ZFC is inconsistent. There are no integer solutions, and there are no proofs in ZFC of this fact. The point is that the 10th problem shows that the r.e. sets are precisely the Diophantine sets.&lt;/p&gt;&#xA;" OwnerUserId="462" LastActivityDate="2010-12-23T07:24:36.993" CommentCount="2" />
  <row Id="15309" PostTypeId="1" AcceptedAnswerId="15311" CreationDate="2010-12-23T07:35:04.897" Score="4" ViewCount="137" Body="&lt;p&gt;I got an assignment to prove certain things about right triangles in Lobachevsky geometry, but so far I don't know where to start. What model is the best for studying these objects? What is the general approach to this kind of task?&lt;/p&gt;&#xA;" OwnerUserId="3793" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-23T08:49:34.237" LastActivityDate="2010-12-23T09:15:38.290" Title="How to analyze triangles in Lobachevsky geometry?" Tags="&lt;geometry&gt;&lt;soft-question&gt;&lt;hyperbolic-geometry&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" />
  <row Id="15310" PostTypeId="2" ParentId="6090" CreationDate="2010-12-23T07:38:57.383" Score="1" ViewCount="" Body="&lt;p&gt;De La Vallee-Pousin's theorem has been improved by Korobov and Vinogradov in the 50's and I believe their result is the strongest known asymptotic zero free region, cf. &lt;em&gt;The Riemann zeta-function: Theory and applications&lt;/em&gt; by Alexandar Ivic. One can find more recent papers but my impression is that they don't touch the main exponents and so relatively tractable problems seem to be improving the constants, or giving explicit bounds on the constants etc.&lt;/p&gt;&#xA;" OwnerUserId="2473" LastEditorUserId="2473" LastEditorDisplayName="" LastEditDate="2010-12-24T23:30:36.527" LastActivityDate="2010-12-24T23:30:36.527" CommentCount="2" />
  <row Id="15311" PostTypeId="2" ParentId="15309" CreationDate="2010-12-23T07:48:40.943" Score="2" ViewCount="" Body="&lt;p&gt;Lobachevsky's geometry also goes by the more common name &lt;a href=&quot;http://en.wikipedia.org/wiki/Hyperbolic_geometry&quot; rel=&quot;nofollow&quot;&gt;hyperbolic geometry&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I will assume that we are talking about the 2-dimensional case. There are two particularly convenient models: the &lt;a href=&quot;http://en.wikipedia.org/wiki/Poincar%C3%A9_half-plane_model&quot; rel=&quot;nofollow&quot;&gt;Poincaré upper half plane&lt;/a&gt; and the &lt;a href=&quot;http://en.wikipedia.org/wiki/Poincar%C3%A9_disc_model&quot; rel=&quot;nofollow&quot;&gt;Poincaré disc&lt;/a&gt;. The linked pages have plenty of identities between angles and sides in hyperbolic triangles. Many of the usual trigonometric identities hold (I imagine that you will have to prove that if you want to use it), but it is no longer true that angles of a triangle sum to $\pi$. Indeed, the sum is always strictly smaller (see formula for the area of a triangle).&lt;/p&gt;&#xA;" OwnerUserId="3212" LastEditorUserId="3212" LastEditorDisplayName="" LastEditDate="2010-12-23T08:03:12.220" LastActivityDate="2010-12-23T08:03:12.220" />
  <row Id="15312" PostTypeId="1" AcceptedAnswerId="15336" CreationDate="2010-12-23T07:54:30.557" Score="0" ViewCount="105" Body="&lt;p&gt;Everything is in $\mathbb{Z}$. Let $v_1 &amp;lt; v_2 &amp;lt; ... &amp;lt; v_n = k$, and $v_1 = 1$ for $k &amp;gt;&amp;gt; n$. Let $ P = \Pi_{i &amp;lt; j} (v_j - v_i)$. How can I show that $P \le k^{n^2}$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are $n + (n-1) + ... + 1 = \frac{n(n+1)}{2}$ terms in the product. Starting from $v_n - v_{n-1} = 1$, etc. Clearly, $P = 1(1*2)(1*2*3) ...(k-1)! = \Pi_{i=1}^{i=k-1} i!$. But I'm not sure about this superfactorial(?).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, I noticed that the product $P$ is very similar to the determinant of a Vandermonde matrix.&lt;/p&gt;&#xA;" OwnerUserId="2468" LastActivityDate="2010-12-23T15:09:23.030" Title="Show that $\Pi_{i &lt; j} (v_i - v_j) \le k^{n^2}$ for $1 \le v_1 &lt; v_2 &lt; ... &lt; v_n = k$" Tags="&lt;sequences-and-series&gt;&lt;factorial&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="15313" PostTypeId="2" ParentId="15297" CreationDate="2010-12-23T08:23:11.687" Score="1" ViewCount="" Body="&lt;p&gt;For $a_n \ge 0 $ the infinite product $\prod_{n=1}^\infty (1+ a_n)$ converges precisely when the infinite sum $\sum_{n=1}^\infty a_n $ converges, since&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$1+ \sum_{n=1}^N a_n \le \prod_{n=1}^N (1+ a_n) \le \exp \left( \sum_{n=1}^N a_n \right) . $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you only need consider $ \sum_{n=1}^\infty 2/n $ and you can use your integral test for that, or just quote the standard result.&lt;/p&gt;&#xA;" OwnerUserId="1301" LastActivityDate="2010-12-23T08:23:11.687" />
  <row Id="15314" PostTypeId="1" CreationDate="2010-12-23T08:29:24.130" Score="0" ViewCount="63" Body="&lt;p&gt;Hello! I have n points that determine a polygon. I need to color the interior of this polygon with a color . How can I do this?&lt;/p&gt;&#xA;" OwnerUserId="2777" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-23T08:46:59.793" LastActivityDate="2010-12-23T08:51:04.703" Title="color polygon interior in different color" Tags="&lt;geometry&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="15315" PostTypeId="1" AcceptedAnswerId="15322" CreationDate="2010-12-23T08:37:25.017" Score="4" ViewCount="255" Body="&lt;p&gt;The eigenvalues of a rotation matrix are complex numbers. I understand that they cannot be real numbers because when you rotate something no direction stays the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My question&lt;/strong&gt;&lt;br&gt;&#xA;What is the intuition that the eigenvalues are complex? Why do they exist at all for rotation matrices? I mean it is not so that every time a calculation is not possible the result is complex (dividing by 0 is not possible at all - the result is not real, but it is not complex either!). The complex numbers seem to cover some middle ground here, but I don't understand how and why they come into play - there don't seem to be any square roots taken from negative numbers...)&lt;/p&gt;&#xA;" OwnerUserId="346" LastEditorUserId="346" LastEditorDisplayName="" LastEditDate="2010-12-23T08:48:18.520" LastActivityDate="2010-12-23T09:50:30.730" Title="intuition for complex eigenvalues" Tags="&lt;linear-algebra&gt;&lt;matrices&gt;&lt;intuition&gt;&lt;complex-numbers&gt;&lt;eigenvalues&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="1" />
  <row Id="15316" PostTypeId="2" ParentId="15314" CreationDate="2010-12-23T08:41:14.013" Score="2" ViewCount="" Body="&lt;p&gt;Here is one naive approach: triangulate the polygon (that's pretty easy to do and I will leave that to you to figure out) and then colour the triangles. That reduces the problem to determining the interior of a triangle that is given by three points. To do that, you could express the triangle as an intersection of three half planes. Then, each half plane is determined by a linear inequality, so given any point, you can determine whether it's in the interior by checking all three inequalities. The rest is accomplished by obvious loops, with judiciously chosen start and end values.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can optimise some of that by putting more thought into how to set up the loops. E.g. if the outer loop goes through rows and the inner through the columns, then the starting point of the inner loop can be set to the left most point of the interior.&lt;/p&gt;&#xA;" OwnerUserId="3212" LastEditorUserId="3212" LastEditorDisplayName="" LastEditDate="2010-12-23T08:51:04.703" LastActivityDate="2010-12-23T08:51:04.703" />
  <row Id="15318" PostTypeId="2" ParentId="15297" CreationDate="2010-12-23T09:01:22.193" Score="1" ViewCount="" Body="&lt;p&gt;If You dont have to use integration test then You can do something like this, it is much easier.&#xA;$\prod\limits_{n=1}^{k}(1+\frac{2}{n})=\frac{(k+1)(k+2)}{2}$&lt;br&gt;&#xA;It is very easy to show since it is equal to: $\frac{1}{3}\cdot\frac{2}{4}\cdot\frac{3}{5}\cdot\frac{4}{6}\cdot\frac{5}{7}\cdot\cdot\cdot\cdot$&lt;br&gt;&#xA;You can reduce almost all this fractions. Denominator of nth fraction can be reduced with numerator of n+2th fraction.&lt;/p&gt;&#xA;" OwnerUserId="4811" LastActivityDate="2010-12-23T09:01:22.193" />
  <row Id="15319" PostTypeId="2" ParentId="15309" CreationDate="2010-12-23T09:15:38.290" Score="2" ViewCount="" Body="&lt;p&gt;There are a few models, and the great thing is that (once you've shown that they're equivalent) different things are easy to show in different models.  If you're clever about it, you can juggle back and forth and do as little work as possible!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As Alex mentions, there are the Poincare disc and UHP models.  Both of these &quot;respect angles&quot; (e.g. two hyperbolic geodesics are perpendicular iff they look perpendicular in either of these models).  There's another disc model which does &lt;em&gt;not&lt;/em&gt; respect angles: hyperbolic geodesics are now just chords of the disc.  In this model, for instance, it's just a tad easier to prove that the postulate &quot;two lines intersect in at most one point&quot; holds than it is in either of the others (not that it's really so hard in any of them).  I'm sure there are less trivial things that are easiest to show in this model, but none are springing to mind.  But then, for instance, if you want to find a regular octagon whose angles are each 45 degrees (which comes up in topology as the fundamental domain of a 2-holed torus), you're going to be by far best off with the Poincare disc model.  As I said, each model has its own perks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In my opinion, if you're dealing with right triangles it might be easiest to use the Poincare disc model, because then you can take two of the three edges of your triangle to lie in the x- and y-axes.  Or similarly in the UHP model you could let one edge lie in the y-axis and one edge lie in the unit circle.&lt;/p&gt;&#xA;" OwnerUserId="2159" LastActivityDate="2010-12-23T09:15:38.290" />
  <row Id="15321" PostTypeId="1" CreationDate="2010-12-23T09:47:22.487" Score="0" ViewCount="69" Body="&lt;p&gt;Cartan for beginner book page 6, 7, 8 not have enough examples for understanding.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;how to rewrite ODE or PDE as exterior differential equation?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If further, please give example to solve these exterior differential form in maple.&lt;/p&gt;&#xA;" OwnerUserId="5070" LastActivityDate="2010-12-23T09:47:22.487" Title="Please give exmaples of rewriting ODE as ideal of differential forms and PDE case as well?" Tags="&lt;differential-geometry&gt;&lt;differential-forms&gt;" CommentCount="1" />
  <row Id="15322" PostTypeId="2" ParentId="15315" CreationDate="2010-12-23T09:50:30.730" Score="2" ViewCount="" Body="&lt;p&gt;Heuristically, I suppose we could see this being because the standard action of the complex numbers on $V = \mathbb R^{2n}$ is by rotation. That is, $(e_1, \ldots, e_{2n})$ is a basis for $V$, then we define multiplication by $i$ as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ i e_{2k-1} = e_{2k}, \quad i e_{2k} = - e_{2k-1},$ for $k = 1, \ldots, n$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so multiplying a vector $v$ by a complex number $\lambda$ will correspond to a scaling by a real and a rotation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if we have a rotation $A$ on the space $V$ and we want to find a line $l$ &quot;invariant&quot; under $A$, then we can try to look for a complex number $\lambda$ such that rotation of $l$ under $A$ is equivalent to the action of $\lambda$ on $l$. Thus, we can try and look for complex eigenvalues $\lambda$ of $A$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This line of &quot;reasoning&quot; completely breaks down in the odd-dimensional case, because we can't define a complex structure on odd-dimensional spaces, but it might give a hint into why we'd look for complex eigenvalues at all. Then it becomes a matter of algebra to figure out that it actually works, and that it does so in a vector space of any finite dimension.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, for the existence, as Alex already pointed out, we look for eigenvalues by finding roots of polynomials. All polynomials admit a root over the complex numbers, which translates into the existence of a complex eigenvalue.&lt;/p&gt;&#xA;" OwnerUserId="3225" LastActivityDate="2010-12-23T09:50:30.730" />
  <row Id="15323" PostTypeId="1" CreationDate="2010-12-23T10:04:39.483" Score="0" ViewCount="144" Body="&lt;p&gt;Given the Right ascension in h m s, Declination in deg ' &quot; and the Trigonometric parallax How can I get the cartesian (x,y,z) coordinates of a star? I'm guessing I need 3 separate formulas to get each x, y and z values.&lt;/p&gt;&#xA;" OwnerUserId="5071" LastActivityDate="2010-12-23T10:04:39.483" Title="How do I calculate the cartesian coordinates of stars" Tags="&lt;trigonometry&gt;" CommentCount="3" />
  <row Id="15324" PostTypeId="2" ParentId="15260" CreationDate="2010-12-23T10:35:02.307" Score="0" ViewCount="" Body="&lt;p&gt;wow! thanks all, everyone got very interesting answer&#xA;i've found also &lt;a href=&quot;http://mathworld.wolfram.com/Slinky.html&quot; rel=&quot;nofollow&quot;&gt;slinky&lt;/a&gt; that is quite funny cause it is also a toy&lt;/p&gt;&#xA;" OwnerUserId="5046" LastActivityDate="2010-12-23T10:35:02.307" />
  <row Id="15326" PostTypeId="1" AcceptedAnswerId="15329" CreationDate="2010-12-23T10:51:38.213" Score="2" ViewCount="426" Body="&lt;p&gt;there is a prove that there is no $2^n$ that is divisible by 5?&lt;/p&gt;&#xA;" OwnerUserId="5046" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-12-23T20:09:33.843" LastActivityDate="2010-12-23T20:09:33.843" Title="prove: $2^n$ is not divisible by 5 for any $n$" Tags="&lt;number-theory&gt;" AnswerCount="4" CommentCount="1" />
  <row Id="15327" PostTypeId="2" ParentId="15326" CreationDate="2010-12-23T10:54:20.240" Score="7" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT:&lt;/strong&gt; Use the fact that the powers of $2$ end with $2,4,8,6$ and for divisibility for 5 last digit has to end with either 5 or $0$.&lt;/p&gt;&#xA;" LastActivityDate="2010-12-23T11:01:08.457" />
  <row Id="15328" PostTypeId="2" ParentId="14206" CreationDate="2010-12-23T11:02:30.000" Score="1" ViewCount="" Body="&lt;p&gt;I specially recomend you reading &lt;a href=&quot;http://docs.sun.com/source/806-3568/ncg_goldberg.html&quot; rel=&quot;nofollow&quot;&gt;What Every Computer Scientist Should Know About Floating-Point Arithmetic&lt;/a&gt; by David Goldberg. It should illuminate a lot of issues regarding floating-point computations Errors. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;This paper is a tutorial on those aspects of floating-point arithmetic (floating-point hereafter) that have a direct connection to systems building. It consists of three loosely connected parts. The first section, Rounding Error, discusses the implications of using different rounding strategies for the basic operations of addition, subtraction, multiplication and division. It also contains background information on the two methods of measuring rounding error, ulps and relative error. The second part discusses the IEEE floating-point standard, which is becoming rapidly accepted by commercial hardware manufacturers. Included in the IEEE standard is the rounding method for basic operations. The discussion of the standard draws on the material in the section Rounding Error. The third part discusses the connections between floating-point and the design of various aspects of computer systems. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I specially recomend you to design your algorithm according The IEEE 754 rounding methods.&#xA;I also recomend you to read the third part because it discussed the implications of floating point arithmetic on the design of computer systems.&lt;/p&gt;&#xA;" OwnerUserId="4255" LastActivityDate="2010-12-23T11:02:30.000" />
  <row Id="15329" PostTypeId="2" ParentId="15326" CreationDate="2010-12-23T11:10:05.613" Score="12" ViewCount="" Body="&lt;p&gt;This is a special case of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Fundamental_theorem_of_arithmetic&quot; rel=&quot;nofollow&quot;&gt;fundamental theorem of arithmetic&lt;/a&gt;: every integer can be &lt;em&gt;uniquely&lt;/em&gt; written as a product of primes. Since both 2 and 5 are primes, an identity like $2^n = 5k$ for $k\in \mathbb{N}$ would contradict the uniqueness of prime factorisation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Exercise: more generally, find a necessary and sufficient criterion on integers $m$ and primes $p$ for $m^n$ to be divisible by $p$ for some $n$. Further prove that if divisibility holds for some $n$, then it holds for all natural $n&amp;gt;0$.&lt;/p&gt;&#xA;" OwnerUserId="3212" LastActivityDate="2010-12-23T11:10:05.613" CommentCount="2" />
  <row Id="15330" PostTypeId="1" CreationDate="2010-12-23T11:25:36.790" Score="0" ViewCount="76" Body="&lt;p&gt;Hello!I´m trying to develop a architecture hardware to make a implementation of an algorithm that can be descompose in terms of sums, multiplications, substrations and exponential functions. I,m trying to modelling, exp(-x) through Taylor series. The domain of my function is bounded between 0 and 1500, but I want to use a particular Taylor approximation whose domain is bounded between 0 and 0.5. &lt;strong&gt;Is there any way to get an approximation using the my tailor series whose domain is bounded between 0 and 0.5 to modelling the function whose domain is bounded between 0 and 1500?&lt;/strong&gt; &#xA;The function I want to modelling for bounded domain is exp(-x).&#xA;Thank you for your help. &lt;/p&gt;&#xA;" OwnerUserId="4255" LastActivityDate="2011-02-03T13:55:08.510" Title="Development of a specific hardware architecture for a particular algorithm. Modelling fuctions by Taylor Series." Tags="&lt;algorithms&gt;&lt;functions&gt;&lt;approximation&gt;&lt;mathematical-modeling&gt;" AnswerCount="1" />
  <row Id="15331" PostTypeId="2" ParentId="15260" CreationDate="2010-12-23T12:48:38.587" Score="2" ViewCount="" Body="&lt;p&gt;If I may be allowed to give another example, there is the spherical epitrochoid, the locus of a circle rolling on another circle. It was studied by Johann Bernoulli and Franz Reuleaux in the context of the kinematics of gear systems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Letting $p$ be the radius of the fixed circle (&quot;pitch circle&quot;), $r$ the radius of the rolling circle, $hr$ the distance of the tracing point from the center of the rolling circle, and $\omega$ the inclination of the rolling circle from the x-y coordinate plane, the parametric equations for the spherical epitrochoid are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align*}x&amp;amp;=(p+r\cos\;\omega)\cos\;t-h r\left(\cos\;\omega\cos\left(\frac{p}{r}t\right)\cos\;t-\sin\left(\frac{p}{r}t\right)\sin\;t\right)\\y&amp;amp;=(p+r \cos\;\omega)\sin\;t-h r\left(\cos\;\omega\cos\left(\frac{p}{r}t\right)\sin\;t+\sin\left(\frac{p}{r}t\right)\cos\;t\right)\\z&amp;amp;=r\sin\;\omega\left(1-h\cos\left(\frac{p}{r}t\right)\right)\end{align*}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is the case $p=1$, $r=\frac15$, $h=1$, $\omega=\frac{2\pi}{3}$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/4m4Tf.gif&quot; alt=&quot;three-dimensional epitrochoid&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is the case $p=1$, $r=\frac15$, $h=1$, $\omega=\frac{\pi}{6}$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/BkZdu.gif&quot; alt=&quot;three-dimensional epitrochoid&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-12-23T12:48:38.587" />
  <row Id="15332" PostTypeId="1" AcceptedAnswerId="15370" CreationDate="2010-12-23T13:27:16.247" Score="10" ViewCount="171" Body="&lt;p&gt;It is well known how to use algebraic geometry (differentials, divisors, and Riemann-Roch) in order to prove the finite dimensionality of the vector space of modular forms of some fixed weight and level.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it possible to prove the finite dimensionality of modular forms using representation theory? If so, any references?&lt;/p&gt;&#xA;" OwnerUserId="4946" LastActivityDate="2011-02-12T20:39:49.600" Title="Proving finite dimensionality of modular forms using representation theory?" Tags="&lt;reference-request&gt;&lt;representation-theory&gt;&lt;modular-forms&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="15333" PostTypeId="2" ParentId="15326" CreationDate="2010-12-23T13:43:27.523" Score="5" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\ $ By &lt;a href=&quot;http://math.stackexchange.com/questions/14579/troubles-finding-inverse-modulus/14608#14608&quot;&gt;Gauss's algorithm&lt;/a&gt; for computing inverses modulo a prime $\rm\: 2\:n+1\:$ (here $5$) we have:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\ \ \displaystyle \frac{1}2\ \equiv\ \frac{n+1}{2\:(n+1)}\ \equiv\ n+1\ \ (mod\:\ 2\:n+1)\:.\quad $ Hence $\rm\ 2^n \equiv 0\ $ times $\ (1/2)^n\ $ yields $\rm\ 1\equiv 0\ \Rightarrow\Leftarrow$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;More generally any integer $\rm\:a\:$ coprime to $\rm\:d\:$ can be constructively inverted $\rm\:(mod\ d)\:$ by employing the &lt;a href=&quot;http://en.wikipedia.org/wiki/Extended_Euclidean_algorithm&quot; rel=&quot;nofollow&quot;&gt;extended Euclidean algorithm&lt;/a&gt; to find a Bezout relation $\rm\ a\ b + c\ d = 1\ \Rightarrow\ a\ b \equiv 1\ \:(mod\ d)\:.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is very closely related to &quot;rationalizing denominators&quot;, e.g. see my &lt;a href=&quot;http://groups.google.com/groups?selm=y8zwtid4b0u.fsf@nestle.csail.mit.edu&quot; rel=&quot;nofollow&quot;&gt;sci.math post&lt;/a&gt; for further remarks on rationalizing denominators &#xA;and computing inverses via (minimal) polynomials, the extended Euclidean &#xA;algorithm, Grobner bases, resultants, norms, etc. &lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-12-23T18:42:58.197" LastActivityDate="2010-12-23T18:42:58.197" CommentCount="6" />
  <row Id="15334" PostTypeId="1" AcceptedAnswerId="15379" CreationDate="2010-12-23T14:29:07.750" Score="2" ViewCount="160" Body="&lt;p&gt;If we start with $\int\frac{v}{g-2v} dv$ how would we go about integrating it? (g is a constant)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answers to the past exam paper I have tell me to rearrange it to $\frac{1}{2}\int ( -1 + \frac{g}{g-2v} ) dv$ which integrates to $\frac{1}{2}v-\frac{1}{4}g\ln(g-2v) + c$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can see that the rearrangement works 'in reverse', but I'm not sure how I would go about rearranging it the right way round.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Note: Part of question 3(iii) in January 2006 MEI Differential Equations paper.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="1699" LastEditorUserId="1699" LastEditorDisplayName="" LastEditDate="2010-12-23T16:25:17.570" LastActivityDate="2010-12-23T21:11:07.567" Title="Rearranging an Integral" Tags="&lt;algebra&gt;&lt;integral&gt;" AnswerCount="2" CommentCount="5" />
  <row Id="15335" PostTypeId="1" AcceptedAnswerId="15348" CreationDate="2010-12-23T15:06:41.863" Score="3" ViewCount="303" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is ; How can I solve the following integral question?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int{(1 + \cos x)^3\mathrm dx}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance,&lt;/p&gt;&#xA;" OwnerUserId="3669" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-23T15:16:43.170" LastActivityDate="2010-12-24T03:46:38.150" Title="How can I find $ \int{(1 + \cos x)^3\mathrm dx} $ ?" Tags="&lt;calculus&gt;&lt;trigonometry&gt;&lt;integral&gt;" AnswerCount="4" CommentCount="2" FavoriteCount="1" />
  <row Id="15336" PostTypeId="2" ParentId="15312" CreationDate="2010-12-23T15:09:23.030" Score="2" ViewCount="" Body="&lt;p&gt;First note that there are a total of $1 + 2 + \cdots + (n-1)$ terms i.e. a total of $\frac{n(n-1)}{2}$ terms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And $\forall n \in \mathbb{N}$, $\frac{n(n-1)}{2} &amp;lt; n^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And $|v_i - v_j| &amp;lt; k$, $\forall i,j$ since $v_{i} \in [1,k]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, we get $\displaystyle \prod_{i&amp;lt;j} |v_i - v_j| &amp;lt; k^{\frac{n(n-1)}{2}} &amp;lt; k^{n^2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The last inequality follows from the fact that $\frac{n(n-1)}{2} &amp;lt; n^2$ and $k&amp;gt;1$&lt;/p&gt;&#xA;" OwnerUserId="3302" LastActivityDate="2010-12-23T15:09:23.030" CommentCount="3" />
  <row Id="15337" PostTypeId="1" AcceptedAnswerId="15351" CreationDate="2010-12-23T15:15:53.787" Score="2" ViewCount="150" Body="&lt;p&gt;Is  $$\left |\int_1^N e^{ix^\alpha} \ \text{dx} \right| = \mathcal{O}(N^\alpha)?$$ &#xA;where $\displaystyle 0\lt \alpha \lt 1$ and $\displaystyle i=\sqrt{-1}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;${\bf Edit.}$ As Willie Wong shows below, the correct equation is &#xA;$$\left |\int_1^N e^{ix^\alpha} \ \text{dx} \right| = \mathcal{O}(N^{1-\alpha}).$$&lt;/p&gt;&#xA;" OwnerUserId="3249" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-12-23T20:32:09.167" LastActivityDate="2010-12-24T06:12:31.660" Title="Is this integral of order $N^{\alpha}$?" Tags="&lt;analysis&gt;&lt;integral&gt;" AnswerCount="3" CommentCount="5" />
  <row Id="15338" PostTypeId="2" ParentId="15017" CreationDate="2010-12-23T15:17:36.993" Score="2" ViewCount="" Body="&lt;p&gt;No. Here is a counterexample. Define $X$ as follows. ${\rm P}(X=a) = p_1$, ${\rm P}(X=0) = p_2$, and ${\rm P}(X=\frac{{ - ap_1 }}{{1 - p_1  - p_2 }}) = 1-p_1-p_2$,&#xA;where $a$ is a positive constant. Then, ${\rm E}(X)=0$. Denote ${\rm E}(X^4)-2 [{\rm E}(X^2)]^2$ by $\xi$. Then,&#xA;$$&#xD;&#xA;\xi = a^4 p_1  + \Big(\frac{{ap_1 }}{{1 - p_1  - p_2 }}\Big)^4 (1 - p_1  - p_2 ) - 2\Big[a^2 p_1  + \Big(\frac{{ap_1 }}{{1 - p_1  - p_2 }}\Big)^2 (1 - p_1  - p_2 )\Big]^2.&#xD;&#xA;$$&#xA;To find ${\rm E}(Y^4)-2 [{\rm E}(Y^2)]^2$, which we denote by $\eta$, we first find &#xA;$$&#xD;&#xA;{\rm E}[X|X \ge 0] = a{\rm P}(X = a|X \ge 0) = a\frac{{{\rm P}(X = a)}}{{{\rm P}(X \ge 0)}} = \frac{{ap_1 }}{{p_1  + p_2 }}&#xD;&#xA;$$&#xA;and&#xA;$$&#xD;&#xA;{\rm E}[X|X &amp;lt; 0] = \frac{{ - ap_1 }}{{1 - p_1  - p_2 }}.&#xD;&#xA;$$&#xA;Hence, by definition, ${\rm P}(Y = \frac{{ap_1 }}{{p_1  + p_2 }}) = p_1  + p_2 $ and ${\rm P}(Y = \frac{{ - ap_1 }}{{1 - p_1  - p_2 }}) = 1 - p{}_1 - p_2$. &#xA;Thus, &#xA;$$&#xD;&#xA;{\rm E}(Y^i) = \Big(\frac{{ap_1 }}{{p_1  + p_2 }}\Big)^i (p_1  + p_2) + \Big(\frac{{ap_1 }}{{1 - p_1  - p_2 }}\Big)^i (1 - p_1  - p_2),&#xD;&#xA;$$&#xA;from which we get an explicit expression for $\eta$. Now, to furnish a counterexample, it suffices to find a triple $(a,p_1,p_2)$ for which $\xi \leq 0$ and $\eta &amp;gt; 0$. This is most easily done using a computer. Here is a concrete example.&#xA;Letting $a=4$, $p_1=0.36$, and $p_2=0.4$, we have $-ap_1/(1-p_1-p_2) = -6$. Here,&#xA;$$&#xD;&#xA;a^4 p_1  + \Big(\frac{{ap_1 }}{{1 - p_1  - p_2 }}\Big)^4 (1 - p_1  - p_2 ) = 403.2&#xD;&#xA;$$&#xA;and&#xA;$$&#xD;&#xA;2 \Big[a^2 p_1  + \Big(\frac{{ap_1 }}{{1 - p_1  - p_2 }}\Big)^2 (1 - p_1  - p_2 )\Big]^2 = 414.72,&#xD;&#xA;$$&#xA;so ${\rm E}(X^4) \leq 2 [{\rm E}(X^2)]^2$; on the other hand,&#xA;$$&#xD;&#xA;\Big(\frac{{ap_1 }}{{p_1  + p_2 }}\Big)^4 (p_1  + p_2) + \Big(\frac{{ap_1 }}{{1 - p_1  - p_2 }}\Big)^4 (1 - p_1  - p_2) \approx 320.835&#xD;&#xA;$$&#xA;and&#xA;$$&#xD;&#xA;2\Big[\Big(\frac{{ap_1 }}{{p_1  + p_2 }}\Big)^2 (p_1  + p_2) + \Big(\frac{{ap_1 }}{{1 - p_1  - p_2 }}\Big)^2 (1 - p_1  - p_2)\Big]^2 \approx 258.482,&#xD;&#xA;$$&#xA;so ${\rm E}(Y^4) &amp;gt; 2 [{\rm E}(Y^2)]^2$.&lt;/p&gt;&#xA;" OwnerUserId="2810" LastActivityDate="2010-12-23T15:17:36.993" />
  <row Id="15339" PostTypeId="2" ParentId="15335" CreationDate="2010-12-23T15:19:55.500" Score="3" ViewCount="" Body="&lt;p&gt;Expand $(1+\cos(x))^3$. And rewrite $\cos^n(x)$ in terms of $\cos(nx)$ and other similar terms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(1+\cos(x))^3 = 1 + 3 \cos(x) + 3 \cos^2(x) + \cos^3(x) = \frac{\cos(3x)+6 \cos(2x)+15 \cos(x)+10}{4}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, integrate term by term to get the desired answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\int \cos(nx) = \frac{\sin(nx)}{n}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, $\int (1+\cos(x))^3 = \frac{\frac{\sin(3x)}{3}+6\frac{\sin(2x)}{2}+15 \sin(x) + 10x}{4} + c = \frac{\sin(3x)}{12} + \frac{3\sin(2x)}{4} + \frac{15 \sin(x)}{4} + \frac{5x}{2} + c$.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastActivityDate="2010-12-23T15:19:55.500" />
  <row Id="15340" PostTypeId="2" ParentId="15337" CreationDate="2010-12-23T15:27:34.460" Score="1" ViewCount="" Body="&lt;p&gt;$\alpha = 0$ gives the integral as $(N-1)e^{i} = \mathcal{O}(N)$ and $\alpha = 1$ gives the integral to be $\frac{e^{iN}-e^{i}}{i} = \mathcal{O}(1)$.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2010-12-23T15:50:35.117" LastActivityDate="2010-12-23T15:50:35.117" CommentCount="3" />
  <row Id="15341" PostTypeId="2" ParentId="5377" CreationDate="2010-12-23T15:35:42.033" Score="-1" ViewCount="" Body="&lt;p&gt;Why not replace &quot;ZFC is inconsistent&quot; by false? Namely you can prove anything this way.&#xA;&lt;hr&gt;&#xA;But this argument is not a proof because it has an extra assumption that A halts iff there is a proof that A does not halt. You mean A is constructible.&#xA;&lt;hr&gt;&#xA;What does consistent mean? Someone defines consistent as ~provable(0 = 1), but this only implies the axioms are not against our intuition. But we could have 0 = 1 as an axiom.&lt;/p&gt;&#xA;" OwnerUserId="5079" LastEditorUserId="5079" LastEditorDisplayName="" LastEditDate="2010-12-24T03:53:16.367" LastActivityDate="2010-12-24T03:53:16.367" CommentCount="2" />
  <row Id="15342" PostTypeId="1" CreationDate="2010-12-23T15:42:33.363" Score="0" ViewCount="192" Body="&lt;p&gt;Let $A \subset X$ be closed and $U \subset A$ open in $A$. Let $V$ be any open&#xA;set in $X$ with $U \subset V$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Prove $U \cup (V \setminus A)$ is open in $X$.&lt;/p&gt;&#xA;" OwnerUserId="5080" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-23T15:44:36.077" LastActivityDate="2010-12-23T16:01:44.070" Title="Prove this set is open" Tags="&lt;topology&gt;" AnswerCount="1" CommentCount="10" />
  <row Id="15343" PostTypeId="2" ParentId="15326" CreationDate="2010-12-23T15:44:00.433" Score="0" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT:&lt;/strong&gt; &lt;a href=&quot;http://en.wikipedia.org/wiki/Multiplicative_group_of_integers_modulo_n&quot; rel=&quot;nofollow&quot;&gt;For every modulus, all relatively prime integers are a multiplicative group.&lt;/a&gt;&lt;br&gt;&#xA;(Without this, number theory would be quite pointless)&lt;/p&gt;&#xA;" OwnerUserId="5078" LastActivityDate="2010-12-23T15:44:00.433" />
  <row Id="15344" PostTypeId="2" ParentId="15055" CreationDate="2010-12-23T15:47:04.217" Score="5" ViewCount="" Body="&lt;p&gt;I think that the reason that these puzzles are so often confusing is that they rely on the limitations of the English language rather than on any mathematical difficulties.  Of course this is not unique to English, and I think you should be able to find similar similar puzzles in pretty much any natural language.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is an example that brings out the difficulty more clearly.  First, consider the following similar puzzle: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;A family has two children, Robin and Lindsay. Lindsay is a girl. What is the probability that both children are girls? &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In elementary probability class, they teach you to answer this by making a table of the four options&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&#xA;  Robin    Lindsay&#xA;  B        B&#xA;* B        G&#xA;  G        B&#xA;* G        G&#xA;&lt;/pre&gt; &#xA;&#xA;&lt;p&gt;The starred rows are the ones where Lindsay is a girl, and we compute from them that the chance that both children are girls is 1/2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now consider this puzzle&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;A family has two children, Robin and Lindsay. At least one of them is a girl. What is the probability that both children are girls? &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The elementary probability method gives the following table, and a probability of 1/3.  The difference is that we gain one more row, compared to the previous puzzle. &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&#xA;  Robin    Lindsay&#xA;  B        B&#xA;* B        G&#xA;* G        B&#xA;* G        G&#xA;&lt;/pre&gt; &#xA;&#xA;&lt;p&gt;After looking at these, you can see that the difficulty of the original puzzle comes because, in English, &quot;one of the children&quot; can mean several different things:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&quot;one&quot; can mean &quot;a particular one&quot;. If you read the original puzzle like this, it becomes analogous to the first puzzle I wrote, and the answer will be 1/2.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&quot;one&quot; can mean &quot;at least one&quot;. If you read the original puzzle like this, it become analogous to the second puzzle I wrote, and the answer is 1/3.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&quot;one&quot; can mean &quot;exactly one&quot;. If you read the original puzzle like this, the answer is 0.  &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;There is a common convention in mathematics that &quot;one&quot; usually means &quot;at least one&quot;. For example, this is the sense intended in the following sentence, which is a typical example of mathematical English: &quot;if a natural number $n$ is a multiple of a prime number $p$, and $n = ab$, then one of $a$ and $b$ is divisible by $p$.&quot;  We would not read this as saying that exactly one of $a$ and $b$ is divisible by $p$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't believe this convention is very common in non-mathematical English. If I say, &quot;one of my children is a girl&quot;, in normal English this means that the other is a boy. Similar discrepancies between mathematical and non-mathematical English come up with our use of the word &quot;or&quot; and our use of the phrase &quot;if/then&quot;.  When we teach mathematics, we have to spend time explaining this mathematical argot to students, so they can use the same English conventions that we do. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Probability puzzles like the one you're asking about rely on these differences of English meaning, rather than on any logical or mathematical problem.  In that sense, they aren't really puzzles, they're just tricks. &lt;/p&gt;&#xA;" OwnerUserId="630" LastActivityDate="2010-12-23T15:47:04.217" CommentCount="4" />
  <row Id="15346" PostTypeId="2" ParentId="15334" CreationDate="2010-12-23T15:53:02.533" Score="4" ViewCount="" Body="&lt;p&gt;Here's how you can find your numerator. Set&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ v \equiv B + A(g-2v).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This immediately gives $A = -1/2 $ and hence $ B = g/2 . $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Try $\int \frac{5g - 3v}{g+v} \textrm{ d}v$ to check that you have the idea.&lt;/p&gt;&#xA;" OwnerUserId="1301" LastEditorUserId="1301" LastEditorDisplayName="" LastEditDate="2010-12-23T16:01:09.263" LastActivityDate="2010-12-23T16:01:09.263" CommentCount="3" />
  <row Id="15347" PostTypeId="2" ParentId="15342" CreationDate="2010-12-23T16:01:44.070" Score="3" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT:&lt;/strong&gt; Since $U$ is open in $A$, there is an open set $\mathscr{U}$ of $X$ such that $U = A\cap\mathscr{U}$. Consider the open sets $\mathscr{U}\cap V$ and $V\setminus A$.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-12-23T16:01:44.070" />
  <row Id="15348" PostTypeId="2" ParentId="15335" CreationDate="2010-12-23T16:06:08.163" Score="7" ViewCount="" Body="&lt;p&gt;If you don't know how to rewrite $\cos^n(x)$ in terms of $\cos(nx)$, here is an alternate solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Expand $(1+\cos(x))^3 = 1 + 3\cos(x) + 3\cos^2(x) + \cos^3(x)$.&#xA;Now, you know how to integrate the first two terms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The third term can be integrated by using the half-angle formula&#xA;$$\cos^2(x) = \frac{1+\cos(2x)}{2},$$&#xA;so that&#xA;$$\int 3\cos^2(x)\,dx = 3\left(\frac{x}{2} + \frac{\sin(2x)}{4}\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The last term can be integrated using the pythagorean identity $\cos^2(x) = 1 - \sin^2(x),$ so&#xA;$$\cos^3(x) = \cos(x) - \cos(x)\sin^2(x).$$&#xA;And now $\int \cos(x)\,dx = \sin x$ and $\int \cos(x)\sin^2(x)\,dx = \frac{\sin^3(x)}{3}$, so&#xA;$$\int \cos^3(x)\,dx = \sin x - \frac{\sin^3(x)}{3}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now just put the pieces together.&lt;/p&gt;&#xA;" OwnerUserId="640" LastEditorUserId="640" LastEditorDisplayName="" LastEditDate="2010-12-23T16:16:46.413" LastActivityDate="2010-12-23T16:16:46.413" CommentCount="1" />
  <row Id="15349" PostTypeId="1" AcceptedAnswerId="15467" CreationDate="2010-12-23T16:10:14.270" Score="0" ViewCount="74" Body="&lt;p&gt;Given a directed graph $G\langle V,E\rangle$, and weight $w[u][v]$ of each $\langle u,v \rangle$ representing an edge of $E$ the question is, try to divide the whole graph into several &quot;rings&quot;, making each vertex $u$ belong to a certain &quot;ring&quot; and a &quot;ring&quot; contains at least two vertices, it's guaranteed that the given graph must be able to be divided, satisfying the restrictions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, please make the total weight of all the &quot;ring&quot;s minimum .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a task assigned by my tutor, I don't know if it's an old problem because I'm new on algorithm, my tutor told me that it could be solved within polynomial time complexity, I wanna know how it can be done, I just have no idea about that...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS: I'm terribly sorry for my forgetting to post the correct question at the first time... &lt;/p&gt;&#xA;" OwnerUserId="5066" LastEditorUserId="5066" LastEditorDisplayName="" LastEditDate="2010-12-24T14:15:47.630" LastActivityDate="2010-12-24T21:36:32.173" Title="algorithm problem: Prove minimum ring division on directed graph can be solved in polynomial time" Tags="&lt;graph-theory&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="15350" PostTypeId="1" AcceptedAnswerId="15424" CreationDate="2010-12-23T16:41:02.683" Score="6" ViewCount="248" Body="&lt;p&gt;(I'll keep this one short)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given a Dirichlet series&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$g(s)=\sum_{k=1}^\infty\frac{c_k}{k^s}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $c_k\in\mathbb R$ and $c_k \neq 0$ (i.e., the coefficients are a sequence of arbitrary nonzero real numbers), and assuming that $g(s)$ can be analytically continued, does it follow that $g(s)$ possesses a critical strip containing its nontrivial zeroes?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If this does not generally hold, what restrictions should there be on the $c_k$ for $g(s)$ to possess a critical strip?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(My attempts at searching bring too much stuff on Riemann $\zeta$, with only a quick mention of general Dirichlet series; pointers to the literature would be appreciated.)&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-12-24T09:13:10.720" Title="On Dirichlet series and critical strips" Tags="&lt;analytic-number-theory&gt;" AnswerCount="3" CommentCount="4" />
  <row Id="15351" PostTypeId="2" ParentId="15337" CreationDate="2010-12-23T16:45:10.073" Score="5" ViewCount="" Body="&lt;p&gt;I think what you should get is, for each fixed $\alpha$, asymptotically $O(N^{1-\alpha})$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's why: do the change of variables $y = x^\alpha$. Your integral becomes&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \alpha \int_1^{N^\alpha} y^{\frac{1-\alpha}\alpha} e^{iy} dy $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now integrate by parts to lower the exponent on $y$. The first step would be&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ = \frac{\alpha}{i}\left( N^{1-\alpha} e^{i N^\alpha} - e^i \right) - \frac{1-\alpha}{i} \int_1^{N^\alpha} y^{\frac{1-2\alpha}\alpha} e^{iy} dy$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Repeating the process $M = \lceil \frac{1}{\alpha} \rceil$ number of times, you get a string of terms that estimates to &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ = O(N^{1-\alpha} + N^{1-2\alpha} + \ldots + N^{1-(M-1)\alpha}) + O(\int_1^{N^\alpha} y^{\frac{1-M\alpha}{\alpha}} e^{iy} dy) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $M\alpha \geq 1$, the final integral is dominated by a constant. Noting that for large $N$, the first term inside the first $O(\cdot)$ dominates, you get the result. &lt;/p&gt;&#xA;" OwnerUserId="1543" LastActivityDate="2010-12-23T16:45:10.073" />
  <row Id="15352" PostTypeId="2" ParentId="15350" CreationDate="2010-12-23T16:52:35.590" Score="1" ViewCount="" Body="&lt;p&gt;Possible keywords include &lt;a href=&quot;http://en.wikipedia.org/wiki/Dirichlet_L-function&quot; rel=&quot;nofollow&quot;&gt;Dirichlet L-functions&lt;/a&gt; or the generalized Riemann hypothesis. For the case where $c_k$ are the values of a &lt;a href=&quot;http://en.wikipedia.org/wiki/Dirichlet_character&quot; rel=&quot;nofollow&quot;&gt;Dirichlet character&lt;/a&gt;, in certain situations the existence of a critical strip is known. See the Wikipedia article for references. &lt;/p&gt;&#xA;" OwnerUserId="1543" LastActivityDate="2010-12-23T16:52:35.590" CommentCount="13" />
  <row Id="15353" PostTypeId="2" ParentId="15350" CreationDate="2010-12-23T17:14:14.833" Score="4" ViewCount="" Body="&lt;p&gt;$g$ will always have a half-plane free from zeroes (cf. e.g. Titchmarsh, &lt;em&gt;Theory of functions&lt;/em&gt;, Section 9.6). This means a functional equation would be sufficient to guarantee a critical strip (essentially calling the other zeroes trivial by definition).&lt;/p&gt;&#xA;" OwnerUserId="2473" LastActivityDate="2010-12-23T17:14:14.833" />
  <row Id="15354" PostTypeId="1" AcceptedAnswerId="15374" CreationDate="2010-12-23T17:45:57.250" Score="7" ViewCount="235" Body="&lt;p&gt;How may one go about proving&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle\frac{\Gamma&amp;#39;(s)}{\Gamma(s)}=O(\log|s|)$, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(away from the poles) directly? By a direct proof, I mean not to go through the usual Stirling formula with the exact error term. The use of a rough form of Stirling's formula is welcome.&lt;/p&gt;&#xA;" OwnerUserId="2473" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-24T00:57:28.590" LastActivityDate="2010-12-24T00:57:28.590" Title="Stirling-type formula for the logarithmic derivative of the Gamma function" Tags="&lt;special-functions&gt;&lt;gamma-function&gt;" AnswerCount="1" />
  <row Id="15355" PostTypeId="2" ParentId="15337" CreationDate="2010-12-23T17:53:03.800" Score="4" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Van_der_Corput_lemma_%28harmonic_analysis%29&quot; rel=&quot;nofollow&quot;&gt;Van der Corput's lemma&lt;/a&gt; gives&#xA;$$&#xD;&#xA;\left|\int_1^N e^{ix^\alpha}dx\right|\le \frac{C}{\alpha}\,N^{1-\alpha},&#xD;&#xA;$$&#xA;where the constant $C$ is independent of $\alpha$ (and of $N$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&lt;strong&gt;Edited to answer TCL's comment&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Van der Corput's lemma&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $\phi\colon(a,b)\to\mathbb{R}$ is smooth and $|\phi^{(k)}(x)|\ge1$ for $a&amp;lt;x&amp;lt;b$, then&#xA;$$&#xD;&#xA;\left|\int_1^N e^{i\lambda\phi(x)}dx\right|\le C_k\lambda^{-1/k}.&#xD;&#xA;$$&#xA;This valid if $k\ge2$ or if $k\ge1$ and $\phi&amp;#39;$ is monotone.&#xA;It is clearly equivalent to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $\phi\colon(a,b)\to\mathbb{R}$ is smooth and $|\phi^{(k)}(x)|\ge\lambda$ for $a&amp;lt;x&amp;lt;b$, then&#xA;$$&#xD;&#xA;\left|\int_a^b e^{i\phi(x)}dx\right|\le C_k\lambda^{-1/k}.&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To aply it to $\int_1^N e^{ix^\alpha}dx$, choose $a=1$, $b=N$, $\phi(x)=x^\alpha$, and $k=1$. Then $\phi&amp;#39;(x)=\alpha\,x^{\alpha-1}$ is decrasing, so that $|\phi&amp;#39;(x)|\ge\alpha\,N^{\alpha-1}(=\lambda)$.&lt;/p&gt;&#xA;" OwnerUserId="4791" LastEditorUserId="3035" LastEditorDisplayName="" LastEditDate="2010-12-24T06:12:31.660" LastActivityDate="2010-12-24T06:12:31.660" CommentCount="3" />
  <row Id="15356" PostTypeId="2" ParentId="15201" CreationDate="2010-12-23T17:54:27.570" Score="2" ViewCount="" Body="&lt;p&gt;There are some excellent combinatorics books which also look at the applicability of combinatorics both within mathematics and outside of mathematics:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a. Applied Combinatorics by Fred Roberts&lt;/p&gt;&#xA;&#xA;&lt;p&gt;b. Applied Combinatorics by Alan Tucker&lt;/p&gt;&#xA;" OwnerUserId="1369" LastActivityDate="2010-12-23T17:54:27.570" CommentCount="1" />
  <row Id="15357" PostTypeId="2" ParentId="15280" CreationDate="2010-12-23T17:57:02.690" Score="0" ViewCount="" Body="&lt;p&gt;You should try reading &quot;Mathematics - form and function&quot; by Saunders MacLane. The author tries to convey what Mathematics is &quot;all about&quot; and at the same time giving a overview over several different areas. (including manifolds, number theory, groups/fiels, algebra, geometry, etc)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Mathematics,_Form_and_Function&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Mathematics,_Form_and_Function&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0387962174&quot; rel=&quot;nofollow&quot;&gt;http://www.amazon.com/Mathematics-Function-Saunders-Mac-Lane/dp/0387962174&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The presentation may at times be too dense for someone who has not encountered the material before, but if you're prepared to read slow and careful at times, the book is excellent.&lt;/p&gt;&#xA;" OwnerUserId="4284" LastActivityDate="2010-12-23T17:57:02.690" />
  <row Id="15358" PostTypeId="2" ParentId="15278" CreationDate="2010-12-23T18:01:43.627" Score="4" ViewCount="" Body="&lt;p&gt;Harmonic analysis is a very broad field. A nice introduction to harmonic analysis at the level of a beginning graduate student would be Grafakos' Classical and Modern Fourier analysis (two books).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Elias Stein also has many nice books, but they are quite hard as he leaves out a lot of details.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The books require at least knowledge in measure and integration theory (and so real analysis), some basic functional analysis and for Grafakos you might want to know what a group is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, maybe you should also state what kind of harmonic analysis you want to do. The contents of the first book of Grafakos (Classical Fourier Analysis) is at least something you should know if you want to do harmonic analysis (as I have been told).&lt;/p&gt;&#xA;" OwnerUserId="1120" LastActivityDate="2010-12-23T18:01:43.627" CommentCount="1" />
  <row Id="15359" PostTypeId="1" AcceptedAnswerId="15361" CreationDate="2010-12-23T18:26:48.247" Score="0" ViewCount="37" Body="&lt;p&gt;I'm working with percentage splits and I'm trying to add extra players into the split but I'm coming unstuck...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Scenario:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have agreements with a number of parties that set out what type of split we get from a pot of cash.  With some it's 50/50, with others it may be 60/40 in favour of either party or perhaps even 70/30.  In a scenario where there are only 2 players, myself and someone else, this is simple.  What happens however when I add a third party?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So lets say I am already working with someone that gets an equal cut (50/50) with me, and we have to bring someone else on board whom I have a 70/30 agreement.  So now that adjusts the equation, two parties must make equal amounts and the third party must make some factor less to balance the pot to 100%.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can figure out that this is about ratios - 30%/70% = 0.428571.  By trial and error, I can figure out that each of the parties getting an equal cut now get 41.2% of the pot and the remaining party (30%) gets 17.6% of the pot.  I'd like to figure out a formula that gives me the values but I can only seem to get so far...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming that the splits are as follows:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Me     Other    Ratio&#xA;--     -----    -----&#xA;70%    30%      0.428571&#xA;50%    50%      1&#xA;&#xA;X  = First player   (Me/Player 1)&#xA;Y  = Second player  (Player 2)&#xA;Z  = Third player   (Player 3)&#xA;R1 = Ratio split between X and Y&#xA;R2 = Ratio split between X and Z&#xA;&#xA;Y + Z = X / 0.428571&#xA;&#xA;Given that Y = Z in this scenario, we could say that:&#xA;&#xA;  2Y = X / 0.428571 or &#xA;  Y (or Z) = X / 0.428571 / 2&#xA;&#xA;Or&#xA;&#xA;  Y (or Z) = X / (2 x 0.428571)&#xA;           = X / 0.857143&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;It seems like X / R1 = Y, which divided by R2 = Z, the sum of which adds up to 1.  But I come unstuck when I calculate different splits using this kind of algorithm:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Me    Other    Ratio&#xA;--    -----    -----&#xA;60%   40%      0.666667&#xA;30%   70%      2.333333&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;When I calculate using the second ratio, I should be multiplying(?)  In the first scenario, R1 is 1, so it makes no difference whether I divide or multiply, so I guess I should be multiplying in this case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps I've take completely the wrong path.  Can anyone show me either where I'm going wrong or show me the next leap I need to make to get to a result?  Given the original splits, is there a formula to quickly tell me what the adjusted split would be when the extra party is added to the equation?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any help would be greatly appreciated, I've been out of school so long, this kind of theoretical math is way outside my normal range of expertise.&lt;/p&gt;&#xA;" OwnerUserId="5084" LastActivityDate="2010-12-23T18:35:21.753" Title="How can I factor in an extra player on a percentage split?" Tags="&lt;linear-algebra&gt;&lt;algebra&gt;" AnswerCount="1" />
  <row Id="15360" PostTypeId="1" AcceptedAnswerId="15405" CreationDate="2010-12-23T18:30:52.157" Score="7" ViewCount="197" Body="&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Integral_test_for_convergence&quot; rel=&quot;nofollow&quot;&gt;integral test of convergence&lt;/a&gt; states that, if $f:[1,+\infty)\to[0,+\infty)$ is a monotonically decreasing nonnegative function, then the series $\sum_1^\infty f(n)$ converges iff $\int_1^\infty f(n) dn$ is finite.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is the high-dimensional generalization also true? That is, given $f:[1,+\infty)^N \to[0,+\infty)$, and $f(\dotsc,n_i,\dotsc) \ge f(\dotsc,n_i+\epsilon,\dotsc)$ for all $1\le i\le N$ and $n_i\in[1,+\infty)$ and $\epsilon&amp;gt;0$, then the sum&#xA;$$ \sum_{n_1=1}^\infty  \cdots \sum_{n_N=1}^\infty f(n_1,\dotsc,n_N) $$&#xA;converges iff the multiple integral&#xA;$$ \int_1^\infty \cdots \int_1^\infty f(n_1,\dotsc,n_N) dn_1 \dotsm dn_N $$&#xA;is finite.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;(This is just for checking if &lt;a href=&quot;http://physics.stackexchange.com/questions/2196/anti-gravity-in-an-infinite-lattice-of-point-masses/2202#2202&quot;&gt;my answer over physics.SE&lt;/a&gt; is reasonable.)&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="171" LastEditorUserId="171" LastEditorDisplayName="" LastEditDate="2010-12-23T19:12:45.193" LastActivityDate="2010-12-26T23:51:13.707" Title="Does the integral test work on higher dimensions?" Tags="&lt;real-analysis&gt;&lt;sequences-and-series&gt;&lt;convergence&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="15361" PostTypeId="2" ParentId="15359" CreationDate="2010-12-23T18:35:21.753" Score="2" ViewCount="" Body="&lt;p&gt;If you have a two way deal with all the other players, you can just normalize them all.  Say you have a 70/30 split with A and a 40/60 split with B.  Then if you get 1, A gets 7/3 and B gets 4/6.  That would call for a total of 7/3+1+2/3=4, so divide everybody by 4.  A gets 7/12, you get 1/4=3/12, B gets 1/6=2/12.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-12-23T18:35:21.753" CommentCount="1" />
  <row Id="15362" PostTypeId="2" ParentId="15201" CreationDate="2010-12-23T18:51:20.357" Score="3" ViewCount="" Body="&lt;p&gt;I really like &quot;A Course in Enumeration&quot; by Martin Aigner.&lt;/p&gt;&#xA;" OwnerUserId="2710" LastActivityDate="2010-12-23T18:51:20.357" />
  <row Id="15363" PostTypeId="2" ParentId="15278" CreationDate="2010-12-23T18:57:02.347" Score="1" ViewCount="" Body="&lt;p&gt;Worth having a look at is &quot;Harmonic analysis on finite groups&quot; for applications to probability theory (one of your stated strengths). That way you'll be going from what you already know to the new stuff. &lt;a href=&quot;http://books.google.com/books?id=3trud4weQ3AC&quot; rel=&quot;nofollow&quot;&gt;Here&lt;/a&gt; is the Google Books link.&lt;/p&gt;&#xA;" OwnerUserId="2710" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-24T10:53:49.933" LastActivityDate="2010-12-24T10:53:49.933" />
  <row Id="15364" PostTypeId="1" CreationDate="2010-12-23T18:57:16.000" Score="0" ViewCount="69" Body="&lt;p&gt;Frequently a company wants to match its assets and liabilities. However, perfect matching is not practical due to fluctuations in interest rates. So they hedge their risk using &lt;a href=&quot;http://en.wikipedia.org/wiki/Immunization_%28finance%29&quot; rel=&quot;nofollow&quot;&gt;immunization&lt;/a&gt;. This can be achieved using present value matching, duration matching, and greater convexity for assets. In particular:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$PV_{A}(i_0) = PV_{L}(i_0)$&lt;/li&gt;&#xA;&lt;li&gt;$\frac{d}{di} PV_{A}(i)|_{i_0} = \frac{d}{di} PV_{L}(i)|_{i_0}$&lt;/li&gt;&#xA;&lt;li&gt;$\frac{d^2}{di^2} PV_{A}(i)|_{i_0} &amp;gt; \frac{d^2}{di^2} PV_{L}|_{i_0}$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Could we add any extra conditions so that the risk is reduced even more(e.g. could be look at $\frac{d^3}{di^3} PV_{A}(i)$ etc..)?&lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2010-12-23T18:57:16.000" Title="Immunization and Sensitivity Analysis" Tags="&lt;finance&gt;&lt;mathematical-modeling&gt;" CommentCount="1" />
  <row Id="15365" PostTypeId="1" AcceptedAnswerId="15369" CreationDate="2010-12-23T19:22:12.090" Score="0" ViewCount="104" Body="&lt;p&gt;I'm doing a physics course in my spare time and the problem set is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Ball A is dropped from rest from a building of height H exactly as ball B is thrown up vertically from the ground. When they collide A has twice the speed of B. If the collision occurs at height h, what is h/H?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It also gives a hint: &lt;em&gt;Write equations for heights $y_A, y_B$ and velocities $v_A, v_B$. What can you say about them at the time of the collision?&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, I'm stuck at the beginning. I'm trying to find a displacement function for ball A, and I realized that I need to find a velocity function first, and the only thing I know is that the ball is moving under gravity so then $a_A(t) = -g$ and then integrated this means $v_A(t) = -1/2gt^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So far so good, except that now I got confused. Should I integrate the velocity function? It doesn't feel like I should, and even so I'm not really sure how - so I looked at the problem solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem solution says the displacement function is $y_A(t) = H - 1/2gt^2$ - which looks a lot like my velocity function! I am confused. The gravitational force &lt;em&gt;accelerates&lt;/em&gt; mass ($F = ma$ amirite), so how can this be?&lt;/p&gt;&#xA;" OwnerUserId="5085" LastEditorUserId="5085" LastEditorDisplayName="" LastEditDate="2010-12-23T19:38:40.583" LastActivityDate="2010-12-23T19:51:03.027" Title="Newtonian Physics: ball A is dropped, ball B is thrown - where do they coincide?" Tags="&lt;calculus&gt;&lt;physics&gt;" AnswerCount="2" />
  <row Id="15366" PostTypeId="1" AcceptedAnswerId="15377" CreationDate="2010-12-23T19:24:21.090" Score="16" ViewCount="195" Body="&lt;p&gt;This is prompted by &lt;a href=&quot;http://math.stackexchange.com/questions/15312/show-that-pi-i-j-v-i-v-j-le-kn2-for-1-le-v-1-v-2-v-n&quot;&gt;question 15312&lt;/a&gt;, but moved to the reals.  It must be solved already.  Pick n points $x_i \in [0,1]$ to maximize $\prod_{i &amp;lt; j} (x_i - x_j)$.  A little playing shows you don't want them evenly distributed-they need to push out to the ends.  With four points, Alpha says to use $\{0,\frac{1}{2}\pm\frac{1}{2\sqrt{5}},1\}$ and with five, $\{0,\frac{1}{2}-\frac{\sqrt{\frac{3}{7}}}{2},\frac{1}{2},\frac{1}{2}+\frac{\sqrt{\frac{3}{7}}}{2},1\}$&lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2010-12-23T20:21:44.430" LastActivityDate="2010-12-23T21:01:08.773" Title="Spreading points in the unit interval to maximize the product of pairwise distances" Tags="&lt;real-analysis&gt;&lt;analysis&gt;" AnswerCount="1" CommentCount="7" FavoriteCount="1" />
  <row Id="15367" PostTypeId="2" ParentId="639" CreationDate="2010-12-23T19:26:16.583" Score="-1" ViewCount="" Body="&lt;p&gt;I would do it by changing my base. Instead of using binary for all the problems, I'd use base 3 for two poisoned bottles, base 4 for 3 poisoned bottles, and so forth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the general problem, with &lt;em&gt;k&lt;/em&gt; poisoned bottles and &lt;em&gt;N&lt;/em&gt; total bottles, optimal number is:&#xA;roundup[log_base{&lt;em&gt;k&lt;/em&gt;+1}(&lt;em&gt;N&lt;/em&gt;) &lt;em&gt;k&lt;/em&gt;]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the two bottle case, &lt;em&gt;N&lt;/em&gt;=1000, the solution is as follows:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Label the 1000 bottles in base 3 from 0 to 999.&lt;/li&gt;&#xA;&lt;li&gt;Calculate log_base{3}(1000): 6.29 =&gt; we will have 7 places.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;For i going from 0 to 6 (0th place being right-most)&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Allocate a servant for bottles whose base 3 number has 1 in ith place.&lt;/li&gt;&#xA;&lt;li&gt;Allocate a servant for bottles whose base 3 number has 2 in ith place.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Repeat. You may not need a servant for step 2 if the highest number is 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Following that method, the final tally of servants required is 13, which is much better than the proposed 65.&lt;/p&gt;&#xA;" OwnerUserId="5086" LastActivityDate="2010-12-23T19:26:16.583" CommentCount="1" />
  <row Id="15368" PostTypeId="2" ParentId="15365" CreationDate="2010-12-23T19:49:18.970" Score="1" ViewCount="" Body="&lt;p&gt;I see my fault: the integral of $a_A(t) = -g$ is $v_A(t) = -gt$, and therefore the displacement function is $y_A(t) = H - 1/2gt^2$.&lt;/p&gt;&#xA;" OwnerUserId="5085" LastActivityDate="2010-12-23T19:49:18.970" />
  <row Id="15369" PostTypeId="2" ParentId="15365" CreationDate="2010-12-23T19:51:03.027" Score="2" ViewCount="" Body="&lt;p&gt;You double integrated to get the -(1/2)gt^2 !&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Integrate(-g) = -gt + C  (C=0 for starting at rest)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Integrate(-gt) = -(1/2)gt^2 + C  (C=H for starting at top of building)&lt;/p&gt;&#xA;" OwnerUserId="5087" LastActivityDate="2010-12-23T19:51:03.027" />
  <row Id="15370" PostTypeId="2" ParentId="15332" CreationDate="2010-12-23T19:55:01.450" Score="7" ViewCount="" Body="&lt;p&gt;It is possible to &lt;em&gt;express&lt;/em&gt; the finite-dimensionality in representation-theoretic terms: it is amounts to the fact that automorphic forms on $\mathrm{GL}_2(\mathbb A)$ form an &lt;em&gt;admissible&lt;/em&gt; representation of $\mathrm{GL}_2(\mathbb A)$.  (Actually, this statement is a little stronger, because it includes the Maass form case.)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I'm not sure that one can prove this purely representation-theoretically.  At least, it depends on analytic facts (or algebro-geometric equivalents, if one thinks in terms of Riemann--Roch), which have to be brought into play in one manner or another.&lt;/p&gt;&#xA;" OwnerUserId="221" LastEditorUserId="221" LastEditorDisplayName="" LastEditDate="2010-12-23T21:29:30.217" LastActivityDate="2010-12-23T21:29:30.217" />
  <row Id="15371" PostTypeId="1" CreationDate="2010-12-23T19:56:16.960" Score="4" ViewCount="173" Body="&lt;p&gt;The hint for this problem is $a^{n+1} - 1 = (a + 1)(a^n - 1) - a(a^{n-1} - 1)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I see that the problem is true because if you distribute the $a$ and the $-1$ the terms cancel out to equal the left side.  However, since it is telling me to use strong induction I am guessing there is more I am supposed to be doing.  On the hint I can see that it is a true statement, but I am not sure how to use that to prove the equation or how the right side of the hint relates to the right side of the problem.  Also, I do realize that in the case of the hint $n = 1$ would be the special case. &lt;/p&gt;&#xA;" OwnerUserId="735" LastEditorUserId="735" LastEditorDisplayName="" LastEditDate="2010-12-23T20:36:08.143" LastActivityDate="2010-12-23T22:51:54.640" Title="Using the second principle of finite induction to prove $a^n -1 = (a-1)(a^{n-1} + a^{n-2} + ... + a + 1)$ for all $n \geq 1$" Tags="&lt;algebra-precalculus&gt;&lt;induction&gt;" AnswerCount="4" CommentCount="5" FavoriteCount="1" />
  <row Id="15373" PostTypeId="1" CreationDate="2010-12-23T20:04:38.140" Score="1" ViewCount="148" Body="&lt;p&gt;I was reading a comment on MathOverflow and did not understand what was being suggested. I also don't understand Anton's response:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://mathoverflow.net/questions/17732/difference-between-measures-and-distributions/18062#18062&quot; rel=&quot;nofollow&quot;&gt;http://mathoverflow.net/questions/17732/difference-between-measures-and-distributions/18062#18062&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I initially thought Regenbogen's comment was being applied to the derivative of the dirac distribution, i.e. he wanted to extend this continuous linear functional from $D$ to $C$. But I don't understand what dominating semi-norm (sublinear functional) he had in mind... And I don't understand Anton's response - How are the derivatives of the delta distribution not bounded linear functionals on $D$? They seem to have operator norm equal to 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then I thought perhaps he literally was referring to the measure mentioned earlier, i.e. a Radon measure acting as a distribution on $C$ and trivially extending to $D$... but this also seems to be a bounded linear functional on both $C$ and $D$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can someone explain what was being suggested, and how Anton's response makes sense?&lt;/p&gt;&#xA;" OwnerUserId="1260" LastEditorUserId="1260" LastEditorDisplayName="" LastEditDate="2010-12-24T16:29:38.740" LastActivityDate="2010-12-24T16:29:38.740" Title="Misapplication of the Hahn-Banach theorem" Tags="&lt;functional-analysis&gt;&lt;distribution-theory&gt;" CommentCount="3" />
  <row Id="15374" PostTypeId="2" ParentId="15354" CreationDate="2010-12-23T20:18:42.897" Score="7" ViewCount="" Body="&lt;p&gt;You can use the product representation of $\Gamma(z)$, take logs and differentiate the resulting series.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Gamma(z) = \dfrac{e^{-\gamma z}}{z} \prod_{n=1}^{\infty} \left(1 + \dfrac{z}{n}\right)^{-1} \ e^{\frac{z}{n}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can find more information here: &lt;a href=&quot;http://en.wikipedia.org/wiki/Polygamma_function&quot; rel=&quot;nofollow&quot;&gt;Polygamma Function&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/Digamma_function&quot; rel=&quot;nofollow&quot;&gt;Digamma Function&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-12-23T20:18:42.897" CommentCount="6" />
  <row Id="15375" PostTypeId="2" ParentId="15371" CreationDate="2010-12-23T20:29:07.940" Score="2" ViewCount="" Body="&lt;p&gt;$\displaystyle a^{n+1} - 1 = (a + 1)(a^n + 1) - a(a^{n-1} - 1)$ is not right.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Try putting $a=1$, the right hand side is $\displaystyle 4$, while the left is $\displaystyle 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps the hint was $\displaystyle a^{n+1} - 1 = (a-1)(a^n + 1) + a(a^{n-1} - 1)$&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-12-23T20:29:07.940" CommentCount="1" />
  <row Id="15376" PostTypeId="2" ParentId="15335" CreationDate="2010-12-23T20:40:02.450" Score="7" ViewCount="" Body="&lt;p&gt;You can also use complex numbers to make it mechanical.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Write $\displaystyle \cos x = \dfrac{e^{ix} + e^{-ix}}{2}$ and just expand out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This method makes finding integrals of the form&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int P(\sin x, \cos x) \ \text{dx}$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(where $P(x,y)$ is a polynomial in the variables $x,y$) purely mechanical, without having to do any clever algebra etc.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-12-23T21:00:35.077" LastActivityDate="2010-12-23T21:00:35.077" CommentCount="1" />
  <row Id="15377" PostTypeId="2" ParentId="15366" CreationDate="2010-12-23T20:46:50.477" Score="13" ViewCount="" Body="&lt;p&gt;These points are known as Fekete points. A general Fekete problem is to maximize the product &#xA;$$\max_{z_1,...,z_n\in E}\prod\limits_{\quad 1\leq i &amp;lt; j \leq n}|z_i-z_j|$$&#xA;where $E\subset \mathbb C$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In case $E=[-1,1]$, there is a unique solution and the corresponding points  coincide with the zeros of $(1-x^2)P&amp;#39;_{n-1}(x)$, where $P_{n-1}$ is the Legendre polynomial of degree $n-1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I cannot give a precise reference at the moment, but this can be probably found in Szegő's book on orthogonal polynomials.  &lt;/p&gt;&#xA;" OwnerUserId="723" LastEditorUserId="723" LastEditorDisplayName="" LastEditDate="2010-12-23T21:01:08.773" LastActivityDate="2010-12-23T21:01:08.773" CommentCount="10" />
  <row Id="15379" PostTypeId="2" ParentId="15334" CreationDate="2010-12-23T21:11:07.567" Score="2" ViewCount="" Body="&lt;p&gt;Couldn't one just do a u-substitution with u=g-2v?&lt;/p&gt;&#xA;" OwnerUserId="1260" LastActivityDate="2010-12-23T21:11:07.567" CommentCount="1" />
  <row Id="15380" PostTypeId="2" ParentId="15371" CreationDate="2010-12-23T21:18:24.977" Score="3" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\ $ Put $\rm\ f(n) = a^n+a^{n-1}+\:\cdots\:+1\:.\ \ $  Then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\ a^{n+1}-1\ = \ (a+1)\ (a^n-1) - a\ (a^{n-1}-1)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\phantom{\ a^{n+1}-1\ }  =\ (a+1)\ ((a-1)\ f(n-1) - a\ (a-1)\ f(n-2))\quad $ by strong induction&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\phantom{\ a^{n+1}-1\ } =\ (a-1)\ ((a+1)\ f(n-1)- a\ f(n-2))\quad$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\phantom{\ a^{n+1}-1\ } =\ (a-1)\ (\:f(n-1) + a\ (f(n-1)-f(n-2)) $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\phantom{\ a^{n+1}-1\ } =\ \ldots $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\phantom{\ a^{n+1}-1\ } =\ (a-1)\ f(n)$&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-12-23T21:37:44.203" LastActivityDate="2010-12-23T21:37:44.203" />
  <row Id="15381" PostTypeId="2" ParentId="11492" CreationDate="2010-12-23T21:23:13.623" Score="1" ViewCount="" Body="&lt;p&gt;Since the parameter $\sigma$ is positive, the quantity &#xA;$$&#xD;&#xA;-2\sigma \left ( rx^2 + y^2 + b(z-r)^2 - br^2 \right )&#xD;&#xA;$$&#xA;is negative if &#xA;$$&#xD;&#xA;rx^2 + y^2 + b(z-r)^2 &amp;gt; br^{2}.&#xD;&#xA;$$ &#xA;This inequality defines the exterior of an ellipsoid (call it $E_1$); note that the size of this ellipsoid is fixed by the parameters (that's a hint).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the equation&#xA;$$&#xD;&#xA;rx^2 + \sigma y^2 + \sigma \left ( z - 2r \right )^2 = C&#xD;&#xA;$$&#xA;defines a different ellipsoid, $E_2$, the size of which is determined by your choice of $C$ (another hint).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At this point, remind yourself what it is that you want to show.  Typically, the goal is to show that there exists a $C$ such that $E_2$ defines a trapping region for the Lorenz equations, in which case it suffices to show that  $E_2$ can be made large enough so that it contains $E_1$.  There's really no additional calculation necessary to do this -- you just need to understand what you've done so far.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A related (but different) question is to find an explicit lower bound on $C$ in terms of the parameters.  In this case, you can find bounds on each of $x$, $y$, and $z$ separately for points inside of $E_1$.  This will then give you a bound on the quantity&#xA;$$&#xD;&#xA;rx^2 + \sigma y^2 + \sigma \left ( z - 2r \right )^2&#xD;&#xA;$$&#xA;which then defines $C$.&lt;/p&gt;&#xA;" OwnerUserId="4229" LastActivityDate="2010-12-23T21:23:13.623" />
  <row Id="15382" PostTypeId="2" ParentId="15371" CreationDate="2010-12-23T21:24:41.750" Score="2" ViewCount="" Body="&lt;p&gt;Sometimes the easiest way to figure out an induction argument like this is to prove a particular case.  I'll take care of proving $n = 3$, assuming you've already proven $n = 1$ and $n = 2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the hint, we have&#xA;$$a^{3} - 1 = (a + 1)(a^2 - 1) - a(a - 1)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the cases $n = 1$ and $n = 2$ hold, so we rewrite this as &#xA;$$a^{3} - 1 = (a + 1)(a - 1)(a + 1) - a(a - 1)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now by factoring $(a - 1)$ on the right hand side, we have&#xA;$$a^3 - 1 = (a - 1)(a^2 + 2a + 1 - a) = (a - 1)(a^2 + a + 1)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is less interesting than the case $n = 4$; try to work that out on your own! After that, you should be able to work out a general $n + 1$ case.&lt;/p&gt;&#xA;" OwnerUserId="3537" LastActivityDate="2010-12-23T21:24:41.750" />
  <row Id="15383" PostTypeId="1" AcceptedAnswerId="15386" CreationDate="2010-12-23T21:27:21.000" Score="5" ViewCount="189" Body="&lt;p&gt;I have found the following ellipse representation $(x,y)=(x_0\cos(\theta+d/2),y_0\cos(\theta-d/2))$, $\theta \in [0,2\pi]$. This is a contour of bivariate normal distribution with uneven variances and correlation $\rho=\cos(d)$. I know that this is a rotated ellipse with centre $(0,0)$. How to find the lengths of major and minor axes and the angle between x-axis and major axis? &lt;/p&gt;&#xA;" OwnerUserId="4742" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-25T11:53:52.650" LastActivityDate="2010-12-25T11:53:52.650" Title="Finding standard ellipse characteristics from specific ellipse parametrisation" Tags="&lt;geometry&gt;&lt;plane-curves&gt;&lt;conic-sections&gt;" AnswerCount="2" />
  <row Id="15385" PostTypeId="2" ParentId="15371" CreationDate="2010-12-23T21:52:08.817" Score="1" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\ \ $ Show that $\rm\ (a^{n+1}-1)/(a-1)\ $ and $\rm\ a^n+a^{n-1}+\:\cdots\:+1\ $ are both solutions of the recurrence $\rm\ f(n+2)\ = (a+1)\ f(n+1) - a\ f(n)\ $ with identical initial conditions $\rm\ f(1),\ f(0)\:.\ \ \ $&lt;br&gt;&#xA;Now use strong induction to prove the &lt;strong&gt;uniqueness theorem&lt;/strong&gt; for solutions of such difference equations (which is very easy). This is the essence of the calculation in my other answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As I've emphasized in &lt;a href=&quot;http://math.stackexchange.com/search?tab=newest&amp;amp;q=user:242%20uniqueness&quot;&gt;many posts here&lt;/a&gt; uniqueness theorems provide powerful tools for proving equalities. Luckily, here, the uniqueness theorem for difference (vs. differential) equations has an absolutely trivial proof by way of (strong) induction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the corresponding first-order case is essentially &lt;a href=&quot;http://math.stackexchange.com/search?tab=newest&amp;amp;q=user:242%20telescopy&quot;&gt;&lt;strong&gt;telescopy&lt;/strong&gt;&lt;/a&gt;, also known as the fundamental theorem of difference calculus.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-12-23T22:51:54.640" LastActivityDate="2010-12-23T22:51:54.640" />
  <row Id="15386" PostTypeId="2" ParentId="15383" CreationDate="2010-12-23T22:10:40.057" Score="6" ViewCount="" Body="&lt;p&gt;The semi-major axis will be the distance from the origin to the farthest point on the ellipse.  So $r^2=x_0^2\cos^2(\theta+d/2)+y_0^2\cos^2(\theta-d/2)$.  As $r^2$ is a monotonic function of $r$, you can maximize it instead, which simplifies things a bit.  The angle between the $X$ axis and the major axis will then be the arctangent of $y/x$ at this maximum point.  The semi-minor axis will be the distance from the origin to the closest point on the ellipse, and will be $\pi/2$ away from the semi-major axis.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-12-23T22:10:40.057" />
  <row Id="15387" PostTypeId="2" ParentId="15383" CreationDate="2010-12-23T22:36:39.350" Score="7" ViewCount="" Body="&lt;p&gt;This formula rescales a standard ellipse $(\cos(\theta + d/2), \cos(\theta - d/2))$ (inscribed within the unit square) by the diagonal matrix $(x_0, y_0)$.  By symmetry, the values $\theta = 0$ and $\theta = \pi/2$ correspond to vertices of this standard ellipse, allowing us to find their coordinates (and thus the lengths of the semi-axes), whence we easily deduce its equation is $x^2 + y^2 - 2 \rho x y = 1 - \rho^2$.  Applying the diagonal matrix gives the conventional implicit form&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left(\frac{x}{x_0}\right)^2 + \left(\frac{y}{y_0}\right)^2 - 2 \rho \frac{x}{x_0} \frac{y}{y_0} = 1 - \rho^2 \text{.}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From here you can &lt;a href=&quot;http://en.wikipedia.org/wiki/Ellipse&quot; rel=&quot;nofollow&quot;&gt;look up&lt;/a&gt; anything you want.&lt;/p&gt;&#xA;" OwnerUserId="1489" LastActivityDate="2010-12-23T22:36:39.350" CommentCount="2" />
  <row Id="15388" PostTypeId="1" AcceptedAnswerId="15390" CreationDate="2010-12-23T22:54:45.450" Score="3" ViewCount="183" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I just started real analysis.  I don't have a background in proofs or logic, simply calculus.  So I'm trying to learn more about proofs--so forgive the basic question, please.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do you go about proving this theorem: If a and b are any numbers, then there is one and only one number x such that a + x=b.  This number is given by x=b+(-a).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First part of the proof I understand--it's simple.  We simply use some axioms and do the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a+b-a=b, which of course is true.  This is just true from plugging in the b+(-a) for x.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But how about the uniqueness issue?  In my text, it says:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(a+x)+(-a)=b+(-a)&#xA;x=b+(-a).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How does this prove uniqueness?  Is this equation essentially saying that the ONLY possible value of x is expressed by b+(-a)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How would I have known to set this up as part of the proof if I was asked to prove something is unique?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sorry for the basic question--have to start somewhere.&lt;/p&gt;&#xA;" OwnerUserId="5088" LastActivityDate="2010-12-24T19:56:17.983" Title="Basic question about proving uniqueness" Tags="&lt;real-analysis&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="15390" PostTypeId="2" ParentId="15388" CreationDate="2010-12-23T23:10:02.263" Score="5" ViewCount="" Body="&lt;p&gt;Usually when you want to prove something is unique you start by assuming there are two things that satisfy the given property, and then you show that they are the same. In this case, if you want to prove that there is exactly one $x$ such that $a + x = b$, you could start by assuming there are two numbers, say $x$ and $y$ that satisfy the identity, that is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$a + x = b \quad \text{and} \quad a + y = b$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but then this implies that $a + x = a + y$ since both are equal to $b$. Now you can add $-a$ to both sides of the equation to get&#xA;$$-a + (a + x) = -a + (a + y) \implies (-a + a) + x = (-a + a) + y $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\implies 0 + x = 0 + y \implies x = y$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This takes care of the uniqueness of the number and since you already know one such $x$ then you're done. I suppose you're doing this in the real number field and that you know which axioms take place in the argument. Hope this helps a little.&lt;/p&gt;&#xA;" OwnerUserId="900" LastActivityDate="2010-12-23T23:10:02.263" CommentCount="5" />
  <row Id="15391" PostTypeId="1" AcceptedAnswerId="15393" CreationDate="2010-12-24T00:07:40.190" Score="3" ViewCount="97" Body="&lt;p&gt;Let $\kappa$ be a measurable cardinal, we say that $\mathcal{D}$ is a normal ultrafilter iff whenever $g\in\kappa^\kappa$ such that $g&amp;lt;_\mathcal{D} Id$, we have some $\alpha&amp;lt;\kappa$ such that $\{\gamma&amp;lt;\kappa | g(\gamma) = \alpha\}\in\mathcal{D}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given $\kappa$ a measurable cardinal and some $M$ which is a transitive collapse of an ultrapower by some $\kappa$-complete ultrafilter, $\mathcal{U}$, on $\kappa$, and $j_\mathcal{U}$ the canonical embedding of $V$ into $M$, we can reconstruct a normal ultrafilter by: $\mathcal{D} = \{X\subseteq\kappa | \kappa\in j_\mathcal{U}(X)\}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, given a set $A\in\mathcal{U}$ then $\kappa\in j_\mathcal{U}(A)$ iff $[Id]_\mathcal{U} E [C_A]_\mathcal{U}$ iff $\{\alpha&amp;lt;\kappa | \alpha\in A\}\in\mathcal{U}$. (where $E$ is the relation on the ultraproduct, and $C_x$ is the constant function giving $x$ for every $\alpha$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, clearly if $A\in\mathcal{U}$ then $\{\alpha&amp;lt;\kappa | \alpha\in A\}\in\mathcal{U}$ and therefore $\kappa\in j_\mathcal{U}(A)$ and therefore $A\in\mathcal{D}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But that means the ultraproduct was given by a normal ultrafilter, even though we took some general ultrafilter, and I was told by my prof. that not every ultrafilter can be normalized through a permutation on $\kappa$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where and what am I doing wrong here?&lt;/p&gt;&#xA;" OwnerUserId="622" LastActivityDate="2010-12-24T01:19:21.650" Title="Normal ultrafilters on measurable cardinals" Tags="&lt;set-theory&gt;&lt;large-cardinals&gt;" AnswerCount="1" />
  <row Id="15393" PostTypeId="2" ParentId="15391" CreationDate="2010-12-24T01:19:21.650" Score="2" ViewCount="" Body="&lt;p&gt;Hi Asaf, the problem with your argument is when you say that $\kappa\in j_{\mathcal U}(A)$ iff the class of the identity is a &quot;member&quot; of the class of the constantly-equal-to-$A$ function. This assumes that the identity represents $\kappa$, which is equivalent to the normality of ${\mathcal U}$.&lt;/p&gt;&#xA;" OwnerUserId="462" LastActivityDate="2010-12-24T01:19:21.650" CommentCount="2" />
  <row Id="15395" PostTypeId="2" ParentId="15251" CreationDate="2010-12-24T01:50:26.160" Score="3" ViewCount="" Body="&lt;p&gt;I can prove it for the case when $Z = |X-Y|$ takes only integer values. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $q_i = P(Z=i)$ for $i=0,1,\dots$. Then, we need to show that $\frac{q_0+q_1}{q_0+q_1+q_2} \geq \frac{1}{3}$. This follows from the observation that $2q_0 \geq q_i$ for all $i$. This follows from Cauchy Schwarz inequality. Then,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\begin{aligned}&#xD;&#xA;3(q_0+q_1) &amp;amp;\geq (q_0+q_1+q_2) \\&#xD;&#xA;2(q_0+q_1) &amp;amp;\geq q_2 \\&#xD;&#xA;\end{aligned}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is true since $2q_0 \geq q_2$. Thanks to iMath for this last observation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of $Z$ being real, I tried mimicking the proof above but the details don't quite work out. In this case, Cauchy-Schwarz still implies that $f_Z(z) \leq 2f_Z(0)$ for all $z$. However, the proof seems to need one more estimation along the lines of $\int_0^1 f_Z(z) dz \geq  f_Z(0)$.&lt;/p&gt;&#xA;" OwnerUserId="618" LastEditorUserId="618" LastEditorDisplayName="" LastEditDate="2010-12-24T18:42:59.483" LastActivityDate="2010-12-24T18:42:59.483" CommentCount="3" />
  <row Id="15396" PostTypeId="1" CreationDate="2010-12-24T01:53:02.490" Score="1" ViewCount="151" Body="&lt;p&gt;I'm looking for an algorithm for the transitive reduction of a digraph (a DAG in fact).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Transitive_reduction#Graph_algorithms_for_transitive_reduction&quot; rel=&quot;nofollow&quot;&gt;wikipedia article&lt;/a&gt; displays a formula R&lt;sup&gt;-&lt;/sup&gt; = R - (R &amp;#215; R&lt;sup&gt;+&lt;/sup&gt;) &amp;mdash; where R&lt;sup&gt;-&lt;/sup&gt; is the transitive reduction of R and R&lt;sup&gt;+&lt;/sup&gt; is the transitive closure of R &amp;mdash; but that Cartesian product doesn't form an edge (unless I'm missing something), so how can I take away from R?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, say 3 depends on 2 and 1, and 2 depends on 1.  The edges of this graph G are (2, 1), (3, 1), and (3, 2).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm expecting the reduction to be G' = {(2, 1), (3, 2)}.&#xA;The transitive closure of G is G, right?&lt;/p&gt;&#xA;" OwnerUserId="623" LastActivityDate="2010-12-24T01:53:02.490" Title="Algorithm for the transitive reduction of a DAG" Tags="&lt;algorithms&gt;&lt;graph-theory&gt;" CommentCount="2" />
  <row Id="15399" PostTypeId="2" ParentId="15335" CreationDate="2010-12-24T03:46:38.150" Score="0" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can use the &lt;em&gt;Half-angle&lt;/em&gt; formula. $$\cos{2x} = 2\cos^{2}{x}-1$$ and so you have $$1+\cos{x}= 1 + 2\cos^{2}\frac{x}{2} -1 =2\cos^{2}\frac{x}{2}$$ So the required integral is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align*}&#xA;\int (1+\cos{x})^{3} \rm{dx} &amp;amp;= \int 8 \cdot \cos^{6}\Bigl(\frac{x}{2}\Bigr) \ \rm{dx} \\ &amp;amp;= 8 \int \cos^{6}\Bigl(\frac{x}{2}\Bigr) \ \rm{dx}&#xA;\end{align*}&lt;/p&gt;&#xA;" LastActivityDate="2010-12-24T03:46:38.150" />
  <row Id="15400" PostTypeId="1" AcceptedAnswerId="15497" CreationDate="2010-12-24T03:53:27.317" Score="3" ViewCount="222" Body="&lt;p&gt;While combing around my notes looking for other possible examples for &lt;a href=&quot;http://math.stackexchange.com/questions/15260&quot;&gt;this question&lt;/a&gt;, I chanced upon another one of my unsolved problems:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cycloidal curves are curves generated by a circle rolling upon a plane or space curve. It's not too hard to derive the required parametric equations if the fixed curve is a plane curve, but I've had some trouble deriving the corresponding expression for space curves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More specifically, here is the particular problem I was concerned with: consider a (cylindrical) helix:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align*}x&amp;amp;=a\cos\;t\\y&amp;amp;=a\sin\;t\\z&amp;amp;=ct\end{align*}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and imagine a circle of radius $r$ whose plane is always perpendicular to the x-y plane rolling on the helix, starting at the point $(a,0,0)$ ($t=0$). Imagine a point in the plane of the circle at a distance $hr$ from the center. What are the parametric equations for the locus of the point?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The two obvious pieces of information I have are that &lt;strike&gt;the center of the circle also traces a helix, whose parametric equation differs from the original by a vertical shift of $r$&lt;/strike&gt; (per Tony, that was an erroneous assumption), and that the expression for the arclength of the helix, $s=\sqrt{a^2+c^2}t$, should figure into the final parametric equations. Otherwise, I'm not sure how to start.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How does one derive parametric equations for the &quot;helical cycloid&quot;?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The physical model I had in mind was a screw ramp winding around a cylinder. Supposing that there was a car that needed to go to the top of the cylinder by driving on the ramp, and supposing that a spot is placed on one of the car's wheels, what are the equations for the locus of the spot?&lt;/p&gt;&#xA;" OwnerUserId="498" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-25T09:33:22.773" LastActivityDate="2011-01-16T02:26:14.143" Title="A helical cycloid?" Tags="&lt;geometry&gt;" AnswerCount="3" CommentCount="5" />
  <row Id="15401" PostTypeId="1" AcceptedAnswerId="15403" CreationDate="2010-12-24T03:56:38.483" Score="0" ViewCount="58" Body="&lt;p&gt;I'm developing an open-ended strategy game.  I am using the following formula to calculate damage (In PHP 'cause I don't know MathJaX!):&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;$rand = rand($a, $b) + $c;&#xA;$damage = $rand * sqrt(($d / 20) * $c));&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;a&lt;/code&gt;, &lt;code&gt;b&lt;/code&gt;, &lt;code&gt;c&lt;/code&gt;, and &lt;code&gt;d&lt;/code&gt; are all values that can be modified by the user over the course of play, either by buying a better item (&lt;code&gt;a&lt;/code&gt; and &lt;code&gt;b&lt;/code&gt;), investing in the item (&lt;code&gt;c&lt;/code&gt;), or investing in their character &lt;code&gt;d&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I want to do now is add a bit of randomness to the outcome of the equation.  Because the game is open ended:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;a static value would become unnoticeable/negligible over time.&lt;/li&gt;&#xA;&lt;li&gt;a percentage based value would allow for too much noise over time.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;So, I want to add a random value that is small at first, and grows with increased input, but has diminishing returns.  I'm sure I need some kind of logarithmic formula, but I'm not sure how to go about it!&lt;/p&gt;&#xA;" OwnerUserId="5092" LastActivityDate="2010-12-24T04:26:21.013" Title="How can I add random &quot;fuzz&quot; to this formula, so that the randomness range diminishes as values increase?" Tags="&lt;probability&gt;&lt;logarithms&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="15402" PostTypeId="2" ParentId="15278" CreationDate="2010-12-24T04:25:08.033" Score="7" ViewCount="" Body="&lt;p&gt;Dear Albert,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me suggest that you look at the book &quot;An introduction to topology and modern analysis&quot;, by Simmons.  It covers concepts of point set topology that you presumably already know, but also gives a fairly concise, but quite readable, introduction to abstract algebra.  It then brings these together in its final sections into a discussion of Banach algebras and related topics, culminating in a proof of the Gelfand--Naimark theorem.  [See below for a brief remark about this theorem, and the other theorems mentioned in the subsequent paragraphs.]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since you say that you are interested in &lt;em&gt;abstract&lt;/em&gt; harmonic analysis, the Gelfand--Naimark theorem is a good place to start.  (For example, it is not so far to go from there to the abstact form of Wiener's Tauberian theorem.)&#xA;Note: I am interpreting &lt;em&gt;abstract harmonic analysis&lt;/em&gt; to mean something like harmonic analysis on locally compact abelian groups (and related topics).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Simmons also has exercises, I think.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I was studying this stuff, the next place I went to after Simmons was Naimark's tome &lt;em&gt;Normed rings&lt;/em&gt;.  (There are various editions, and some of the later ones might be called &lt;em&gt;Normed algebras&lt;/em&gt; instead, if I'm not misremembering.  They are translated from Russian, so the slighly unusual, and changing, name may be an artefact of this; I'm not sure.  In any case, they are basically about the theory of Banach algebras and its applications to abstract harmonic analysis.)&#xA;This is a place where one can read about various group rings of topological groups, Haar measure, the general form of Wiener's Tauberian theorem, and other concepts of abstract harmonic analysis.  It is essentially too long to read from&#xA;start to finish, but in my experience one can dip into it in bits and pieces, and having a firm understanding of the material from Simmons helps a lot.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Naimark's book is a monograph, not a textbook as such, and although it has many historical comments and illustrative examples (although the examples are often at a theoretically fairly high level), I don't remember it as having exercises.  But in any case, I am not suggesting it as a first point of call, but as somewhere to go after you have some basics under your belt.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is also a book by Loomis, &lt;em&gt;An introduction to abstract harmonic analysis&lt;/em&gt;,&#xA;which also treats Haar measure, various group rings, and so on.  If I remember correctly it is  less condensed than Naimark and also less comprehensive.  My memory is that I preferred Naimark, but probably for idiosyncratic reasons.  I don't remember whether Loomis's book has exercises.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All the books I'm mentioning are probably out of print, so I'm also assuming that you have access to a university library or something similar.  (Any decent such library should have them.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, some fundamental results that I would recommend aiming for, which combine algebra&#xA;and analysis nicely:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Gelfand's generalization of Wiener's theorem (that if $f$ is a nowhere zero continuous periodic function whose Fourier series is absolutely convergent, than&#xA;the Fourier series of $1/f$ is also absolutely convergent).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;the Gelfand--Naimark theorem (identifying certain commutative Banach algebras with extra&#xA;structure as being algebras of continuous functions on compact topological space; it is a beautiful generalization of the classical spectral theorems for matrices).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The generalization of Wiener's Tauberian theorem to arbitrary commutative locally compact groups.  (Wiener's original theorem says that if $f$ is an $L^1$-function on the real line whose Fourier transform is nowhere zero, then&#xA;the translates of $f$ span a dense subspace of $L^1$.)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;More abstract, but basic to the previous example and lots of other things, is&#xA;the existence of Haar measure for any locally compact group.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;As already mentioned,&#xA;the first two results are in Simmons, and are easier, but already involve a very nice interplay between analysis and algebra.  (The general framework is that of Banach algebras, which combine the analysis of Banach spaces with the&#xA;algebra of rings, ideals, and so on.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second two results are in Naimark, and Haar measure is also in Loomis (and many other places) (and the general form of Wiener may be in Loomis too; I forget now).&lt;/p&gt;&#xA;" OwnerUserId="221" LastEditorUserId="221" LastEditorDisplayName="" LastEditDate="2010-12-24T14:02:12.033" LastActivityDate="2010-12-24T14:02:12.033" CommentCount="1" />
  <row Id="15403" PostTypeId="2" ParentId="15401" CreationDate="2010-12-24T04:26:21.013" Score="2" ViewCount="" Body="&lt;p&gt;Let me expand upon my comment:&#xA;you can pick any positive increasing concave function $f$(damage) and produce a random disturbance in the range [-$f$(damage),+$f$(damage)]. That will give you a disturbance that increases with the damage, but it increases more and more slowly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are lots of concave functions: log (to any base), n-th root for any integer n, arctan,... and of course you can scale them in any way you like. Moreover, any composition of concave functions is concave, so you can do e.g. s.th. like $log(\sqrt{x})$. Note that these functions behave very differently, as $x$ gets large: log falls below any $n$-root eventually, but is unbounded. arctan on the other hand asymptotically approaches $2\pi$, so for very large damage it will behave almost like a constant disturbance. You can influence when that happens by dilating, scaling and translating along the $x$-axis.&lt;/p&gt;&#xA;" OwnerUserId="3212" LastActivityDate="2010-12-24T04:26:21.013" />
  <row Id="15404" PostTypeId="2" ParentId="2949" CreationDate="2010-12-24T04:51:11.767" Score="3" ViewCount="" Body="&lt;p&gt;The use of compactness to show existence of solutions to differential equations. It was not surprising as in unexpected, but in the sense that it opened so much possibility almost unreal. It felt like I was given an amazing super toy that can never be destructed.&lt;/p&gt;&#xA;" OwnerUserId="2473" LastActivityDate="2010-12-24T04:51:11.767" CommunityOwnedDate="2010-12-24T04:51:11.767" />
  <row Id="15405" PostTypeId="2" ParentId="15360" CreationDate="2010-12-24T04:55:42.630" Score="5" ViewCount="" Body="&lt;p&gt;It isn't true in general, but the direction you used in your physics.SE answer is.  That is, if the sum converges, then the integral does too.  The decreasing hypothesis implies that the maximum value of $f$ on the cube $[n_1,n_1+1]\times\cdots\times[n_N,n_N+1]$ is $f(n_1,\ldots,n_N)$, so that the integral over that cube is less than or equal to $f(n_1,\ldots,n_N)$.  Adding up the integrals over all such cubes yields the result.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem with the other direction is that the function may drop off to zero in some directions rapidly enough to make the integral converge, while staying too big in another direction for the sum to converge.  For example, $N=2$, $f(x,y)=1$ if $x=1$, $f(x,y)=0$ otherwise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With an additional &quot;shift&quot; you can go from integral convergence to sum convergence.  The decreasing hypothesis also implies that the minimum value of $f$ on the cube $[n_1,n_1+1]\times\cdots\times[n_N,n_N+1]$ is $f(n_1+1,\ldots,n_N+1)$, so that the integral over that cube is greater than or equal to $f(n_1+1,\ldots,n_N+1)$.  By adding up the integrals over all such cubes, this implies that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;\sum_{n_1=2}^\infty  \cdots \sum_{n_N=2}^\infty f(n_1,\ldots,n_N)&#xD;&#xA;\leq&#xD;&#xA;\int_1^\infty \cdots \int_1^\infty f(x_1,\ldots,x_N) dx_1 \dotsm dx_N.&#xD;&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2010-12-26T23:51:13.707" LastActivityDate="2010-12-26T23:51:13.707" />
  <row Id="15406" PostTypeId="1" CreationDate="2010-12-24T05:10:48.560" Score="1" ViewCount="240" Body="&lt;p&gt;How do we numerically integrate a rapidly decaying exponential function? A simple Gauss quadrature which is based on approximating the function by polynomial, I think will not work, since rapidly decaying exponential function will require more terms in the series. Is there a series approximation for the function?&lt;/p&gt;&#xA;" OwnerUserId="3892" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-25T00:55:32.450" LastActivityDate="2010-12-25T00:55:32.450" Title="Numerical integration - Gauss quadrature rule" Tags="&lt;integral&gt;&lt;numerical-methods&gt;&lt;approximation&gt;" AnswerCount="1" CommentCount="7" />
  <row Id="15407" PostTypeId="2" ParentId="3273" CreationDate="2010-12-24T05:29:11.530" Score="2" ViewCount="" Body="&lt;p&gt;Perhaps one should mention &lt;a href=&quot;http://en.wikipedia.org/wiki/Runge%27s_theorem&quot; rel=&quot;nofollow&quot;&gt;Runge's theorem&lt;/a&gt;, and &lt;a href=&quot;http://en.wikipedia.org/wiki/Mergelyan%27s_theorem&quot; rel=&quot;nofollow&quot;&gt;Mergelyan's theorem&lt;/a&gt; which deal with approximation by rational functions and polynomials, respectively.&lt;/p&gt;&#xA;" OwnerUserId="2473" LastActivityDate="2010-12-24T05:29:11.530" />
  <row Id="15408" PostTypeId="2" ParentId="15400" CreationDate="2010-12-24T05:43:04.727" Score="2" ViewCount="" Body="&lt;p&gt;I'm adjusting my answer, which as TonyK points out was based on an incorrect assumption.  Hopefully it's correct now.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The center of the circle is not directly above the point of contact with the helix as my incorrect answer had asserted.  To see where it is, first picture the starting point at $(a,0,0)$, where the circle is parallel to the $y$-$z$ plane.  The tangent vector to the helix is in the direction of $(0,a,c)$, so the radius is perpendicular to this in the direction of $(0,-c,a)$.  So you have to add $\frac{r}{\sqrt{a^2+c^2}}(0,-c,a)$ to get to the center.  As the circle rolls along, the vector you add to get to the center will rotate around the $z$ axis to $\frac{r}{\sqrt{a^2+c^2}}(c\sin(t),-c\cos(t),a)$.  So the center of the circle is parametrized by $(a\cos(t),a\sin(t),ct)+\frac{r}{\sqrt{a^2+c^2}}(c\sin(t),-c\cos(t),a)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The arclength formula allows you to write down the rotation angle of the point about the center of the circle in terms of $t$ as $\frac{\sqrt{a^2+c^2}}{r}t$.  Let $b=\frac{\sqrt{a^2+c^2}}{r}$ from now on. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we pretend for a minute that the point was just going to rotate about the center of the circle at this rate while the circle remains fixed, and simplify the problem for now by assuming that the point starts above the center of the circle, then the vector we would add to the center of the circle to parametrize the point would be $(0,hr\sin(bt),hr\cos(bt))$.  When looking toward the $y$-$z$ plane from the positive $x$ axis, this motion would appear clockwise, as it should because the circle will be starting off to the right from this perspective.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the vector we actually add to the center will also be rotated by $t$ about the $z$-axis, so putting it all together (with minimal simplification) we get &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\begin{align*}&#xD;&#xA;x&amp;amp;=a\cos(t)+\frac{c}{b}\sin(t)-hr\sin(t)\sin(bt)\\&#xD;&#xA;y&amp;amp;=a\sin(t)-\frac{c}{b}\cos(t)+hr\cos(t)\sin(bt)\\&#xD;&#xA;z&amp;amp;=ct+\frac{a}{b}+hr\cos(bt).&#xD;&#xA;\end{align*}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now to adjust the starting point, you can change $bt$ to $b(t-t_0)$ for suitable $t_0$.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2011-01-16T02:26:14.143" LastActivityDate="2011-01-16T02:26:14.143" CommentCount="2" />
  <row Id="15409" PostTypeId="2" ParentId="9085" CreationDate="2010-12-24T05:45:48.790" Score="1" ViewCount="" Body="&lt;p&gt;Maybe &lt;a href=&quot;http://en.wikipedia.org/wiki/Lagrangian_and_Eulerian_specification_of_the_flow_field&quot; rel=&quot;nofollow&quot;&gt;Lagrangian coordinates&lt;/a&gt; is what you are after.&lt;/p&gt;&#xA;" OwnerUserId="2473" LastActivityDate="2010-12-24T05:45:48.790" />
  <row Id="15411" PostTypeId="2" ParentId="14789" CreationDate="2010-12-24T06:41:12.523" Score="0" ViewCount="" Body="&lt;p&gt;Actually, a straightforward componentwise analysis should do the job. First prove the quadratic rate for the fixed point iteration $x\leftarrow\phi(x)$, where $\phi:\mathbb{R}^n\to\mathbb{R}^n$ and the jacobian of $\phi$ is zero at the solution $\alpha=\phi(\alpha)$. Simply do what you do in 1 dimension, e.g., use Taylor's theorem with second order remainder. Then for the Newton method one has to substitute $\phi(x)=x-J(x)f(x)$ where $J$ is the jacobian of $f$, and show that the jacobian of such $\phi$ is zero at a solution $\alpha$ of $f(\alpha)=0$ if $J(\alpha)\neq0$.&lt;/p&gt;&#xA;" OwnerUserId="2473" LastEditorUserId="2473" LastEditorDisplayName="" LastEditDate="2010-12-24T22:36:26.660" LastActivityDate="2010-12-24T22:36:26.660" />
  <row Id="15412" PostTypeId="2" ParentId="7554" CreationDate="2010-12-24T07:10:56.393" Score="0" ViewCount="" Body="&lt;p&gt;In many cases you have a good idea of the limiting behavior of the function. For instance, you may have a conjecture and you want to test it. Then you can do some transformation on your function to a convenient form. For example, I have seen people plot $f(1/x)$ with $x\to0$ and guess the shape of $g(x)=f(1/x)$ from there. You can repeat this process to get more and more accurate pictures.&lt;/p&gt;&#xA;" OwnerUserId="2473" LastActivityDate="2010-12-24T07:10:56.393" />
  <row Id="15413" PostTypeId="2" ParentId="15406" CreationDate="2010-12-24T07:21:47.483" Score="1" ViewCount="" Body="&lt;p&gt;Variable substitution to make $k=1$, and use &lt;a href=&quot;http://en.wikipedia.org/wiki/Gauss%E2%80%93Laguerre_quadrature&quot; rel=&quot;nofollow&quot;&gt;Gauss-Laguerre&lt;/a&gt;. For more advanced method, the &lt;a href=&quot;http://math.stackexchange.com/questions/2949/which-one-result-in-maths-has-surprised-you-the-most/2977#2977&quot;&gt;double exponential transformation&lt;/a&gt; maybe relevant.&lt;/p&gt;&#xA;" OwnerUserId="2473" LastActivityDate="2010-12-24T07:21:47.483" CommentCount="2" />
  <row Id="15414" PostTypeId="1" AcceptedAnswerId="15416" CreationDate="2010-12-24T07:28:41.770" Score="3" ViewCount="142" Body="&lt;p&gt;How do I show that $\text{Hom}(\mathbb{Z},G)$ and $G$ are isomorphic?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The method suggested by the book is to define a map $k$: $\text(\mathbb{Z},G) \to G$ by $f \to f(1)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am stuck on how exactly to show that it is an isomorphism. Could anyone shed some light into this? Is $f(1)$ an element of G?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(For those who wanted to know, this is not homework)&lt;/p&gt;&#xA;" OwnerUserId="1683" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2010-12-24T07:44:01.500" LastActivityDate="2010-12-24T07:44:01.500" Title="How to show that for any abelian group $G$, $\text{Hom}(\mathbb{Z},G)$ is isomorphic to $G$." Tags="&lt;group-theory&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="15415" PostTypeId="1" AcceptedAnswerId="15442" CreationDate="2010-12-24T07:36:24.667" Score="0" ViewCount="210" Body="&lt;p&gt;Prove that there is a largest integer $n$ such that $n \le x$ for any fixed rational $x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What about any fixed real number $x$?&lt;/p&gt;&#xA;" OwnerUserId="5035" LastEditorUserId="5035" LastEditorDisplayName="" LastEditDate="2010-12-24T13:22:07.747" LastActivityDate="2010-12-24T17:56:49.383" Title="Prove the existence of a largest integer less than or equal to a rational number" Tags="&lt;real-analysis&gt;" AnswerCount="4" CommentCount="3" FavoriteCount="1" />
  <row Id="15416" PostTypeId="2" ParentId="15414" CreationDate="2010-12-24T07:37:49.923" Score="6" ViewCount="" Body="&lt;p&gt;Any homomorphism $f$ from $\mathbb{Z}$ to $G$ is determined by its value at 1: indeed $f(n) = f(1)^n$ for any $n\in \mathbb{Z}$, since $f$ is a group homomorphism (I am using multiplicative notation for the group operation on $G$). So it is natural to consider the map suggested by the book: given $f\in \text{Hom}(\mathbb{Z},G)$, evaluate it at 1: $k:f\mapsto f(1)$. It sends $f$ to an element of $G$, namely $f(1)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To check that this is a group homomorphism, you need to understand how $\text{Hom}(\mathbb{Z},G)$ is a group in the first place. That's very simple: the group operation is pointwise multiplication: the product of two homs $f$ and $g$ is given by $fg(n) = f(n)g(n)$ (to explain what $fg$ is you need to evaluate it on an arbitrary input, since you are describing a homomorphism). So now, you should have no difficulties verifying that $k$ respects the group structure.&lt;/p&gt;&#xA;" OwnerUserId="3212" LastActivityDate="2010-12-24T07:37:49.923" CommentCount="7" />
  <row Id="15417" PostTypeId="2" ParentId="15415" CreationDate="2010-12-24T07:49:12.760" Score="2" ViewCount="" Body="&lt;p&gt;(Sorry to post this as an answer, but I can't comment yet). Do you mean that given a fixed rational number $x$, the set $\lbrace n: n\leq x, n \in \mathbb Z\rbrace$ has a maximum? (Hint: Use the well-ordering principle)&lt;/p&gt;&#xA;" OwnerUserId="4806" LastActivityDate="2010-12-24T07:49:12.760" CommentCount="3" />
  <row Id="15418" PostTypeId="1" AcceptedAnswerId="15419" CreationDate="2010-12-24T08:04:50.660" Score="2" ViewCount="273" Body="&lt;p&gt;If we define $x$ = displacement, $v$ = velocity and $a$ = acceleration then I am used to the ideas that $a= \frac{dv}{dt} = \frac{d^2x}{dt^2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However I also understand $a=v \frac{dv}{dx}$. Can someone explain to me why this is?&lt;/p&gt;&#xA;" OwnerUserId="1699" LastEditorUserId="723" LastEditorDisplayName="" LastEditDate="2010-12-28T15:11:15.143" LastActivityDate="2010-12-28T15:11:15.143" Title="Why does acceleration = $v\frac{dv}{dx}$" Tags="&lt;differential-equations&gt;&lt;physics&gt;&lt;classical-mechanics&gt;" AnswerCount="1" CommentCount="0" />
  <row Id="15419" PostTypeId="2" ParentId="15418" CreationDate="2010-12-24T08:13:17.967" Score="9" ViewCount="" Body="&lt;p&gt;Notice: $\frac{dv}{dt}=\frac{dx}{dt}\frac{dv}{dx}$ by Chain rule. &lt;/p&gt;&#xA;" OwnerUserId="4806" LastActivityDate="2010-12-24T08:13:17.967" CommentCount="1" />
  <row Id="15420" PostTypeId="1" CreationDate="2010-12-24T08:35:21.303" Score="0" ViewCount="83" Body="&lt;p&gt;Given set of points in 3D ( X = (x1, x2, x3), Y = (y1, y2, y3) ), how can I fit transformation from X to Y?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as I know this is called projective transformation.&#xA;Here is example of X and Y.&#xA;Blue and red lines in X are parallel, but they are not parallel in Y.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/YGEfS.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/LzjsM.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="3532" LastActivityDate="2010-12-24T08:35:21.303" Title="Fitting projective transformation" Tags="&lt;linear-algebra&gt;&lt;transformation&gt;" CommentCount="3" />
  <row Id="15421" PostTypeId="1" CreationDate="2010-12-24T08:52:57.303" Score="0" ViewCount="87" Body="&lt;p&gt;Hello! I have this function, $e^{-x}$ bounded between 0 and 1500 and I have an approximation by Taylor Series of the same function bounded between 0 and 0.5. I would like to express my function $e^{-x}$ bounded between 0 and 15oo in terms of the last approximation I described, what is bounded between 0 and 0.5.&#xA;Does anyone know how I can do it?&#xA;Thank you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;NOTE: This question is related with &lt;a href=&quot;http://math.stackexchange.com/questions/14206/development-of-a-specific-hardware-architecture-for-a-particular-algorithm&quot;&gt;Development of a hardware arquitecture for a particular algorithm.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4255" LastActivityDate="2010-12-24T16:14:42.817" Title="Taylor Series. Reusing an approximation of a function" Tags="&lt;functions&gt;&lt;approximation&gt;&lt;taylor-expansion&gt;&lt;mathematical-modeling&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="15422" PostTypeId="2" ParentId="15400" CreationDate="2010-12-24T08:56:30.290" Score="3" ViewCount="" Body="&lt;p&gt;The tangent to the helix at $P = (a \cos t, a \sin t, ct)$ is  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$T = (-a \sin t, a \cos t, c)$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The vector  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$H = (\cos t, \sin t, 0)$   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;is horizontal, and perpendicular to $T$. The radius vector $R$ from $P$ to the centre $C$ of the circle is perpendicular to both of these, so it is proportional to  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$H \wedge T = (c \sin t, -c \cos t, a)$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which has length $\sqrt{a^2 + c^2}$. So the centre of the circle is at $P + h H \wedge T$, with $h = r/\sqrt{a^2 + c^2}$.&lt;/p&gt;&#xA;" OwnerUserId="1508" LastEditorUserId="1508" LastEditorDisplayName="" LastEditDate="2010-12-24T09:23:04.850" LastActivityDate="2010-12-24T09:23:04.850" CommentCount="2" />
  <row Id="15423" PostTypeId="1" AcceptedAnswerId="15448" CreationDate="2010-12-24T08:57:57.547" Score="5" ViewCount="143" Body="&lt;p&gt;&lt;em&gt;&lt;strong&gt;Say we have $N$ [$3 \le N \le 100,000$] spheres indexed as $1,2,3,\cdots N$,all of them have identical weight apart from one.We have to determine which sphere it is (index) by using only the pair of scales.&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We could solve this problem by weighing repeatedly,but I am obviously interested weighing as minimum as possible so my question is what could be the optimal algorithm for this task? &lt;/p&gt;&#xA;" OwnerUserId="2109" LastEditorUserId="2109" LastEditorDisplayName="" LastEditDate="2010-12-24T09:12:51.533" LastActivityDate="2010-12-27T07:40:32.587" Title="Optimal algorithm for finding the odd spheres" Tags="&lt;algorithms&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="15424" PostTypeId="2" ParentId="15350" CreationDate="2010-12-24T09:07:14.513" Score="8" ViewCount="" Body="&lt;p&gt;In the comments section to Willie Wong's answer, the following Dirichlet series came up: the Riemann $\zeta$-function, Dirichlet $L$-functions, and Ramanujans series $\sum_{n \geq 1}\tau(n) n^{-s}$, where $\tau(n)$ is the coefficient of $q^n$ in $\Delta(q) = q\prod_{n=1}^{\infty} (1-q^n)^{24}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First note that the $\zeta$-function is a special case of a Dirichlet $L$-function (it is the $L$-function of the trivial character).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now what is it that Dirichlet $L$-functions and Ramanujan's series have in common? Well, they are all automorphic $L$-functions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An automorphic form (for the group $\mathrm{GL}_n$ over $\mathbb Q$; there are generalizations where $\mathbb Q$ is replaced by an arbitrary number field $F$ and  $\mathrm{GL}_n$ is replaced by an arbitrary reductive group, but to simplify the explanations, I will focus just on the simplest level of generality here) is a function on the product $\mathrm{GL}_n(\mathbb R)\times \mathrm{GL}_n(\mathbb Z/N\mathbb Z)$ for some integer $N \geq 1$ which&#xA;is &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;invariant under the natural (diagonal) action of $\mathrm{GL}_n(\mathbb Z)$;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;grows moderately at infinity with respect to the $\mathrm{GL}_n(\mathbb R)$-coordinates;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;satisfies a suitable differential equation in the $\mathrm{GL}_n(\mathbb R)$-coordinates.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Rather than explaining the generalities in more detail (they can be found in many places), I think it's better to illustrate them:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;E.g. Dirichlet characters arise in the case $n = 1$: they are defined as functions&#xA;on $(\mathbb Z/N\mathbb Z)^{\times} =: \mathrm{GL}_1(\mathbb Z/N\mathbb Z)$,&#xA;and so we can make them into functions on $\mathrm{GL}_1(\mathbb R)\times&#xD;&#xA;\mathrm{GL}_1(\mathbb Z/N\mathbb Z)$ by defining them to be trivial on the&#xA;$\mathbb R^{\times}$-coordinate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;E.g. If $f(\tau)$ is a holomorphic modular form of weight $k$ and level one (where $\tau$ is an upper half-plane variable as usual), we can make $f$ into a function on&#xA;$\mathrm{GL}_2(\mathbb R)$ by first identifying this matrix group with the collection of bases of $\mathbb R^2$, then identifying $\mathbb R^2$ with $\mathbb C$,&#xA;and then defining, for any $\mathbb R$-basis $\omega_1,\omega_2$ of $\mathbb C$,&#xA;$f(\omega_1,\omega_2) := \omega_2^{-k} f(\omega_1/\omega_2)$.  (This presumes&#xA;that $\omega_1/\omega_2$ is in the upper half-plane rather than the lower,&#xA;for simplicity.)  Thus we get a function of the required kind (with $N = 1$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The usual modularity condition becomes invariance under $\mathbb GL_2(\mathbb Z)$.  The moderate growth condition becomes the condition that the Fourier expansion of $f$ involves only non-negative powers of $e^{2 \pi i \tau}$.  The differential equation is the Cauchy--Riemann equation expressing holomorphy of $f$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Higher level modular forms will involve values of $N$ that are $&amp;gt; 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;E.g.  Maass forms are similar to the preceding example, except that now the&#xA;differential equation expresses that a Maass form is an eigenvector of the Laplacian.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For any fixed $n$ and fixed $N$, we have Hecke operators acting on the space&#xA;of automorphic forms, labelled by primes $p$ not dividing $N$, and so we can&#xA;consider Hecke eigenforms.  In the case of Dirichlet characters, the fact that they are &lt;em&gt;characters&lt;/em&gt; of $(\mathbb Z/N\mathbb Z)^{\times}$ (rather than just arbitrary functions) can be reinterpreted as saying that they are Hecke eigenforms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, Ramanujan's $\Delta$ is well-known to be a Hecke eigenform of weight $12$ and level $1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given an automorphic Hecke eigenform we can use the Hecke eigenvalues to make&#xA;an Euler product Dirichlet series, which will give Dirichlet $L$-functions&#xA;for Dirichlet characters, and Ramanujan's Dirichlet series for $\Delta$.&#xA;(In the Dirichlet character case, if a prime $p$ divides the conductor $N$, we just have a trivial factor in the Euler product for that prime; when $n &amp;gt; 1$,&#xA;and $N &amp;gt; 1$, it is a bit more of a battle to figure out what Euler factors&#xA;to put in at the primes dividing $p$, but it can be done.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Actually, it is better to restrict to &lt;em&gt;cuspidal&lt;/em&gt; automorphic Hecke eigenforms.&#xA;Cuspidal is a vacuous condition when $n = 1$ (i.e. in that case we agree to call&#xA;everything cuspidal), and when $n &amp;gt; 1$ we replace &quot;moderate growth at infinity&quot; by &quot;rapid decay at infinity&quot;, suitably interpreted.  I'll assume that all my eigenforms are cuspidal&#xA;form now on.  (E.g. $\Delta$ is cuspidal.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this way we get a natural class of $L$-functions which have:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;meromorphic continuation to the whole complex plane, which is in fact&#xA;holomorphic with the sole exception of Riemann's $\zeta$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Functional equation with completely understood $\Gamma$-factors.  E.g. for a&#xA;weight $k$ modular form of level one, if the $p$the Hecke eigenvalue is $a_p$,&#xA;then the $L$-function is $\prod_p (1 - a_p p^{-s} + p^{k - 1 - 2s})^{-1},$ and the functional equation relates $s$ and $k - s$.  For $\Delta,$ I've already noted that $k = 12$.  (In general the functional equation relates the $L$-series of an automorphic eigenform with $L$-series of its &quot;complex conjugate&quot; suitably understood, just as in the case of Dirichlet characters that are not necessarily real values.)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;Conjecturally&lt;/em&gt;, they should all satisfy the analogue of RH, i.e. all non-trivial zeroes should line down the critical line, in the centre of the critical strip.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Note incidentally that it is easy to change the apparent form of the functional equation.  E.g. if we make a change of variable $s \mapsto s + 11/2$ in Ramanujan's series, then the functional equation will become $s \mapsto 1 -s$&#xA;rather than $s \mapsto 12 -s $, and the critical line will be $\Re s = 1/2$, just as in the $\zeta$-function case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All cuspidal automorphic $L$-functions can be renormalized in a similar way, so that the symmetry of the functional equation is $s \mapsto 1 -s$.  This is called &lt;em&gt;unitary&lt;/em&gt; normalization, and is common in the automorphic forms literature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Up to rescaling, there are only countably many automorphic eigenforms altogether (just because if we fix the level $N$ and (appropriately generalized version of) the weight the space of automorphic forms is finite dimensional) and so altogether we are talking about a very special class of just countably many Dirichlet series, but these seem to be the ones that naturally generalize $\zeta(s)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the way, this general point of view is due to Langlands, and forms a part of the general Langlands program.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another point of view was given by Selberg, which focuses more on capturing the analytic properties necessary for getting good properties of a Dirichlet series, rather than beginning from a conceptual construction (as in the automorphic point of view).  Namely, he introduced the &lt;a href=&quot;http://en.wikipedia.org/wiki/Selberg_class&quot; rel=&quot;nofollow&quot;&gt;Selberg class&lt;/a&gt; of Dirichlet series.  Note that part of his axioms are an Euler product, analytic continuation, and functional equation.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My sense is, though, that people expect the Selberg class of Dirichlet series to more-or-less coincide with the class of automorphic $L$-functions, so I think it is just two points of view on the same question: Langlands is showing how to &lt;em&gt;construct&lt;/em&gt; &quot;good&quot; Dirichlet series, and Selberg is writing down the &lt;em&gt;properties&lt;/em&gt; a &quot;good&quot; Dirichlet series should satisfy.  It turns out that &quot;good&quot; Dirichlet series are so special, though, that however you try to pick them out, you seem to end up with the same very special collection, namely the automorphic ones.&lt;/p&gt;&#xA;" OwnerUserId="221" LastEditorUserId="221" LastEditorDisplayName="" LastEditDate="2010-12-24T09:13:10.720" LastActivityDate="2010-12-24T09:13:10.720" CommentCount="2" />
  <row Id="15425" PostTypeId="1" AcceptedAnswerId="15909" CreationDate="2010-12-24T09:08:16.957" Score="6" ViewCount="212" Body="&lt;p&gt;For an abelian variety $A_{/\mathbb{Q}}$, its volume $vol(A(\mathbb{R}))$ appears in the conjectured Birch Swinnerton-Dyer formula for the L-series at 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am having trouble in understanding the size of this volume, as a function of the (minimal) equations defining the variety. Hence the question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Say we have a hyperelliptic curve given by $$y^2 = x^{2g+1}+a_{2g}x^{2g}+\ldots +a_0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for simplicity, assume that $a_i\approx M &amp;gt; 0$ for all $i$. What is a good crude estimate for the volume of the Jacobian?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: for an elliptic curve this is easy, since there is only one differential and hence only one simple integral.&lt;/p&gt;&#xA;" OwnerUserId="4946" LastActivityDate="2010-12-30T21:02:23.683" Title="Approximating the volume of the Jacobian of a hyperelliptic curve" Tags="&lt;number-theory&gt;&lt;algebraic-geometry&gt;&lt;differential-geometry&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="15426" PostTypeId="1" AcceptedAnswerId="15428" CreationDate="2010-12-24T09:29:25.003" Score="4" ViewCount="181" Body="&lt;p&gt;Situation A: Once only, I toss 2 identical fair coins and don't look at the outcomes. A truthful observer looks at one of the coins and tells me that at least one of the coins is a head. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Situation B: Once only, I toss 2 identical fair coins and don't look at the outcomes. A truthful observer looks at both of the coins and tells me that at least one of the coins is a head. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In A, what is the probability that there are 2 heads?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In B, what is the probability that there are 2 heads?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Aren't both probabilities 1/2? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me refine the question. The agreement I have with the observer is this:&#xA;1) I will toss 2 identical coins.&lt;br&gt;&#xA;2) In situation A a third party will cover the coins with a cloth. The observer will look only at the outcome of the coin that comes to rest nearest him and report it to me. What is the probability that the second outcome will be the same as the first?&lt;br&gt;&#xA;3) In situation B the observer will look at both outcomes and report only the state (heads or tails) of the coin that came to rest nearest him. What is the probability that the second outcome will be the same as the first?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2nd EDIT&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Changing 3) above to:&#xA;3) In situation B the observer will look at both outcomes, choose one of them, and truthfully report, &quot;There is at least one heads&quot;, or &quot;There is at least one tails&quot;, whichever is the case. What is the probability that the second outcome (that of the other coin) will be the same as the first?  &lt;/p&gt;&#xA;" OwnerUserId="1964" LastEditorUserId="1964" LastEditorDisplayName="" LastEditDate="2010-12-24T19:05:03.980" LastActivityDate="2010-12-25T08:08:41.083" Title="2 slightly different situations in which 2 coins are tossed. Does the knowledge of an observer effect the probabilities of the outcomes?" Tags="&lt;probability&gt;" AnswerCount="3" CommentCount="3" />
  <row Id="15427" PostTypeId="2" ParentId="15426" CreationDate="2010-12-24T09:45:40.627" Score="2" ViewCount="" Body="&lt;p&gt;A is easy.&lt;br&gt;&#xA;In B, if you know in advance that the observer is just going to let you know whether the number of heads is zero or non-zero, then the probablity is 1/3.&lt;br&gt;&#xA;If you don't know in advance what the observer is going to reveal, then the probability calculation involves the observer's strategy (what information the observer intends to reveal depending on the number of heads). So you can't calculate the probability. This is at the heart of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Monty_Hall_problem&quot; rel=&quot;nofollow&quot;&gt;Monty Hall problem&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1508" LastActivityDate="2010-12-24T09:45:40.627" CommentCount="1" />
  <row Id="15428" PostTypeId="2" ParentId="15426" CreationDate="2010-12-24T09:46:52.427" Score="7" ViewCount="" Body="&lt;p&gt;No. There are four possibilities with two coins. For the time being identify one coin as X and one coin as Y. You have either &lt;/p&gt;&#xA;&#xA;&lt;p&gt;X = 1, Y = 1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;X = 0, Y = 1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;X = 1, Y = 0&lt;/p&gt;&#xA;&#xA;&lt;p&gt;X = 0, Y = 0&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(where 1 is heads and 0 is tails). A truthful observer who only looks at coin X and ignores coin Y will say: &quot;there is at least one head&quot; in cases 1 and 3. A truthful observer who only looks at coin Y and ignores coin X will say: &quot;there is at least one head&quot; in cases 1 and 2. In either of those situations, only one of the two cases have the other coin as a head. So the probability of two heads is 1/2. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, a truthful observer who looks at both coins will say: &quot;there is at least one head&quot; in cases 1, 2, and 3. Out of those only 1 case has two heads. So the probability of that is 1/3. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;By choosing only one of the coins to look at, you are looking at the probability &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \mathbb{P}\{ X = 1 | \mbox{ Observer chooses }Y\mbox{ and }Y = 1\} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;plus the same with $X$ and $Y$ swapped. Since $X$ and $Y$ are independent, it is equal to &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \mathbb{P}\{X = 1\} \cdot \mathbb{P}\{\mbox{ Observer chooses } Y\} + \mathbb{P}\{Y = 1\} \cdot \mathbb{P}\{\mbox{ Observer chooses } X\} = 1/2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case where the observer sees both coins, you are looking at the conditional probability&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \mathbb{P}\{ Y + X = 2 | Y + X \geq 1 \} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;that the sum of the variables $Y$ and $X$ is 2 when we know their sum is at least 1. Clearly the conditioning is not independent of what you are testing! You can establish that the above is 1/3 by counting the total number of ways $A+B$ can be at least 1 as was done above.&lt;/p&gt;&#xA;" OwnerUserId="1543" LastActivityDate="2010-12-24T09:46:52.427" CommentCount="11" />
  <row Id="15429" PostTypeId="1" CreationDate="2010-12-24T10:28:28.640" Score="1" ViewCount="235" Body="&lt;p&gt;I am wondering whether there are any high resolution photos of Alexander Grothendieck available online. I did a search on Google Images, but most of the results are low quality and/or low resolution. &lt;a href=&quot;http://seven.ihes.fr/IMAGE/grothendieck.jpg&quot; rel=&quot;nofollow&quot;&gt;Here is a nice one&lt;/a&gt;, but I am wondering if there are any other good ones out there. Thanks!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(P.S. Feel free to close this question if you think it's inappropriate... Obviously this is not really a mathematical question... I'm not so familiar with what the standards of appropriateness are here. Also, I guess this question should be community wiki but I don't see any option anywhere to make it such.)&lt;/p&gt;&#xA;" OwnerUserId="263" LastEditorUserId="263" LastEditorDisplayName="" LastEditDate="2010-12-24T10:34:11.277" LastActivityDate="2010-12-24T10:34:11.277" ClosedDate="2010-12-25T20:22:40.897" Title="High resolution photos of Grothendieck" Tags="&lt;algebraic-geometry&gt;" CommentCount="10" />
  <row Id="15430" PostTypeId="2" ParentId="15423" CreationDate="2010-12-24T10:40:35.967" Score="2" ViewCount="" Body="&lt;p&gt;If $N = 3^n$ is a power of three, then we can do it $n+1$ weighings:&#xA;First, assume that the odd ball is heavier than the others (this is just to simplify the exposition; we will come back to this). Now label all the balls from $0$ to $N-1$, in ternary notation. For each ternary digit position, weigh all the balls with $0$ in that position against all the balls with $2$ in that position, leaving out the remaining balls (with 1 in that position). The result tells you (assuming that the odd ball is heavier) the ternary digit of the odd ball in that position: $0$ if the $0$'s outweigh the $2$'s; $2$ if the $2$'s outweigh the $0$'s; and $1$ if they are equal.&lt;br&gt;&#xA;So now you have the ternary expansion of the odd ball, assuming that it's heavy. If that assumption is wrong, and the odd ball is light, then the number of the odd ball is obtained by flipping all the $0$'s to $2$ and vice versa. To find out which of these two candidates is the odd ball, just weigh one of them against a ball that is known not to be odd.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This doesn't work if $N$ is not a power of three, because in general the number of balls with $0$ in a given position is not the same as the number of balls with $2$ in that position. So we can't weigh them against each other. But we can fix this by the simple device of splitting the balls into two equal heaps, numbering one heap upwards from $0$ and the other heap downwards from $3^n-1$, where $3^n$ is the smallest power of three that is $\ge N$. (If the number of balls is odd, label the left-over ball with $11...11$.)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My guess is that this procedure is optimal when $N$ is a power of three, but otherwise we can sometimes do better; for instance, if $N=4$, we only need two weighings to identify the odd ball (but we won't always know whether it is heavier or lighter).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt; This solution shows how to solve the problem in $n$ weighings for any $N \le 3^n$ when we know whether the odd ball is heavier or lighter. Now we can put this together with Moron's proposed solution (linked to in Moron's answer) to get a complete solution in $n$ weighings when $N \le (3^n-3)/2$. (This solution also tells us whether the odd ball is heavier or lighter.)&lt;/p&gt;&#xA;" OwnerUserId="1508" LastEditorUserId="1508" LastEditorDisplayName="" LastEditDate="2010-12-27T07:40:32.587" LastActivityDate="2010-12-27T07:40:32.587" CommentCount="3" />
  <row Id="15431" PostTypeId="1" CreationDate="2010-12-24T10:45:54.977" Score="4" ViewCount="198" Body="&lt;p&gt;Hi all, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is a good introductory book  book for Markov chains and Markov processes?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you. &lt;/p&gt;&#xA;" OwnerUserId="1268" LastEditorUserId="3731" LastEditorDisplayName="" LastEditDate="2010-12-24T10:57:55.760" LastActivityDate="2010-12-24T13:51:48.720" Title="Good introductory book for Markov processes" Tags="&lt;reference-request&gt;&lt;books&gt;" AnswerCount="3" FavoriteCount="4" />
  <row Id="15432" PostTypeId="2" ParentId="15415" CreationDate="2010-12-24T10:53:48.733" Score="1" ViewCount="" Body="&lt;p&gt;First, as a clarification pointed out by Dactyl, you probably want a proof of:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For any rational (or real) $x$, there is a largest integer $n$ such that $n \leq x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reversal of words makes the claim false because it says that there is a fixed largest integer $n$ less than or equal to arbitrarily large $x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also for Dactyl's hint, you have to be careful when $x$ is negative because the well-ordering principle is for Natural numbers.  Specifically, not every nonempty subset of the integers has a smallest element (e.g., the integers themselves have no least element).  However, every nonempty subset of the integers that has a &lt;em&gt;lower bound&lt;/em&gt; has a least element, and you can prove this by induction on the absolute value of an integral lower bound.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, $n$ is typically used to denote a Natural number so maybe you're restricting yourself to nonnegative $x$ anyway.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Edit (in response to request for different proof):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider $\{n \in \mathbb{N}| |x| &amp;lt; n\}$.  The &lt;B&gt;Archimedean property&lt;/B&gt; tells us that for any Real number, we can find a Natural number that's larger so this set is non-empty.  Then by the well-ordering of the Natural numbers, there exists a least element in this set, call it $n_0$.  If $x \geq 0$, verify that $n_0 - 1$ is the desired integer, and if $x &amp;lt; 0$, verify that that $1 - n_0$ is the one you want if $x$ is an integer and $-n_0$ otherwise.&lt;/p&gt;&#xA;" OwnerUserId="4693" LastEditorUserId="4693" LastEditorDisplayName="" LastEditDate="2010-12-24T17:56:49.383" LastActivityDate="2010-12-24T17:56:49.383" />
  <row Id="15433" PostTypeId="2" ParentId="15431" CreationDate="2010-12-24T10:57:35.010" Score="1" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://www.goodreads.com/book/show/6110201-an-introduction-to-markov-chain-analysis&quot; rel=&quot;nofollow&quot;&gt;An Introduction to Markov Chain Analysis&lt;/a&gt; and &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/3540234519&quot; rel=&quot;nofollow&quot;&gt;An Introduction to Markov Processes&lt;/a&gt; would be a good start.&lt;/p&gt;&#xA;" OwnerUserId="3731" LastActivityDate="2010-12-24T10:57:35.010" />
  <row Id="15434" PostTypeId="1" AcceptedAnswerId="15481" CreationDate="2010-12-24T11:22:55.823" Score="3" ViewCount="172" Body="&lt;p&gt;This question is about Schauder bases in Banach spaces.&#xA;Suppose $(x_n)$ is a basic sequence in a Banach space $E$. Prove that $0$ is the only possible weak limit point of $\{x_n \in E \;\vert\; n \in \mathbb{N}\}$.&lt;/p&gt;&#xA;" OwnerUserId="4706" LastEditorUserId="1269" LastEditorDisplayName="" LastEditDate="2010-12-24T12:18:31.177" LastActivityDate="2010-12-25T05:12:45.337" Title="Limit point of basic sequence in Banach space" Tags="&lt;functional-analysis&gt;&lt;banach-spaces&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" />
  <row Id="15435" PostTypeId="1" AcceptedAnswerId="15436" CreationDate="2010-12-24T11:25:07.757" Score="0" ViewCount="116" Body="&lt;p&gt;&lt;img src=&quot;http://i51.tinypic.com/2rfx91v.png&quot; alt=&quot;sketch&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(sorry for the draw)&#xA;I try to get the blue angle , i thought it was easy but...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the curve is a Bézier curve.&#xA;the black dot shows the tangent of D&#xA;I know position of A B C D.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried to get the angle by addition of angle a and angle b &lt;/p&gt;&#xA;&#xA;&lt;p&gt;the point F is found by multiplying AB per the time factor (between 0 and 1)&#xA;the point E is found using the y coordinate of B and x coordinate of A&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i get a with Al-Kashi method with the triangle FBD.&#xA;i get b with Al-Kashi method too with the triangle EBA.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;this is for rotation of an object who follow the curve&#xA;and i always get weird result.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any better way to do it and if not what I'm doing wrong ?&#xA;thanks&lt;/p&gt;&#xA;" OwnerUserId="4279" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-24T11:31:55.720" LastActivityDate="2010-12-24T11:34:58.120" Title="getting angle between tangent of Bézier curve and an y axis" Tags="&lt;bezier-curve&gt;" AnswerCount="1" />
  <row Id="15436" PostTypeId="2" ParentId="15435" CreationDate="2010-12-24T11:34:58.120" Score="0" ViewCount="" Body="&lt;p&gt;It's not too hard: if you know the slope $m$ of the tangent (assuming it's not vertical), then the angle it makes with the positive x-axis is $\arctan(m)$; thus, the angle it makes with the positive y-axis should be the complement of that, or $\frac{\pi}{2}-\arctan(m)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Better yet, since your Bézier curve is parametrically defined anyway, it would be a good idea to use the two-argument arctangent here to reckon the angle; I'll leave it to you to figure out the details.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-12-24T11:34:58.120" CommentCount="1" />
  <row Id="15437" PostTypeId="2" ParentId="15431" CreationDate="2010-12-24T11:35:15.347" Score="2" ViewCount="" Body="&lt;p&gt;&quot;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0126848874&quot; rel=&quot;nofollow&quot;&gt;An Introduction to Stochastic Modeling&lt;/a&gt;&quot; by Karlin and Taylor is a very good introduction to Stochastic processes in general. Bulk of the book is dedicated to Markov Chain. This book is more of applied Markov Chains than Theoretical development of Markov Chains. This book is one of my favorites especially when it comes to applied Stochastics.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastActivityDate="2010-12-24T11:35:15.347" />
  <row Id="15438" PostTypeId="1" AcceptedAnswerId="18345" CreationDate="2010-12-24T12:55:58.103" Score="4" ViewCount="499" Body="&lt;p&gt;I want to make an origami of a sphere, so I planned to print some net of a pentakis icosahedron, but I have a image of another sphere with more polygons:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://www.webkinesia.com/online/graphics/images/icos.gif&quot; alt=&quot;icos&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to find the net of such model (I know it will be very fun to cut).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do you know if it has a name ?&lt;/p&gt;&#xA;" OwnerUserId="3709" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-24T13:04:27.980" LastActivityDate="2011-01-20T19:06:13.173" Title="How to make a sphere-ish shape with triangle faces ?" Tags="&lt;geometry&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="1" />
  <row Id="15441" PostTypeId="2" ParentId="15431" CreationDate="2010-12-24T13:44:57.200" Score="2" ViewCount="" Body="&lt;p&gt;&lt;em&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0486453057&quot; rel=&quot;nofollow&quot;&gt;Theory of Markov Processes&lt;/a&gt;&lt;/em&gt; by Eugene Dynkin is a paperback published by Dover, so it has the advantage of being inexpensive. The author has made many contributions to the subject. Dynkin's lemma, the Dynkin diagram and the Dynkin system are named after him.&lt;/p&gt;&#xA;" OwnerUserId="1702" LastEditorUserId="1702" LastEditorDisplayName="" LastEditDate="2010-12-24T13:51:48.720" LastActivityDate="2010-12-24T13:51:48.720" />
  <row Id="15442" PostTypeId="2" ParentId="15415" CreationDate="2010-12-24T15:19:12.577" Score="4" ViewCount="" Body="&lt;p&gt;The answer to your question depends on the set of facts you take for granted. In terms of elementary number theory I would argue as follows: Let $x=p/q$ with $q&amp;gt;0$. Then the $q$ consecutive numbers $p-q+1$, $p-q+2$, $\ldots$, $p$ form a complete residue system mod$\thinspace q$, so there is exactly one among them, say $r$, that is divisible by $q$. It follows that $n:=r/q$ is the largest integer $≤x$.&lt;/p&gt;&#xA;" OwnerUserId="1303" LastActivityDate="2010-12-24T15:19:12.577" CommentCount="2" />
  <row Id="15443" PostTypeId="1" CreationDate="2010-12-24T15:24:33.537" Score="10" ViewCount="266" Body="&lt;p&gt;This is a very basic question that I feel that I should know the answer to but haven't been able to think through clearly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In linear algebra, we learn that the basis of a finite dimensional vector space can be thought of as 'co-ordinates' of that space. And, we model what we intuitively understand as the euclidean plane, using the vector space $\mathbb{R^2}$ equipped with the standard inner product and metric etc. The underlying space is taken to be independent of the choice of basis, that is we understand that properties that are inherent to the space are those that will be invariant under change of basis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, $\mathbb{R^2}$ comes with a canonical basis: this can be understood as saying that given any arbitrary two dimensional vector space and any basis, the vectors $v_i$ of the basis under the co-ordinate map maps to $e_i$. Since, we have also introduced inner products and thus a notion of parallel, the intuitive picture we now have of the co-odrinate grid is a criss-cross of lines. and the 'co-ordinates' are called, imaginatively, 'rectangular co-ordinates'.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In school, we also learn about polar 'co-ordinates' of the plane. The associated picture is of concentric circles and rays fanning out of the origin.&#xA;However, these 'co-ordinates' do not fit within the 'Basic Linear Algebra' framework (since among other things, $0$ has no unique representation and the functions that change the variables are not linear). The one way of seeing the transformation of the 'rulings' of the plane is to consider $\mathbb{R^2}$ as $\mathbb{C}$ and the change as the map $\exp: \mathbb C \to \mathbb C$.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What is the framework in which the notion of 'co-ordinates' subsumes both these pictures (likewise for cylindrical, spherical, etc in dimension(?) three). My second and connected question: is there a linear algebraic connection for using two numbers to represent points in polar co-ordinate, i.e. is it because the vector space dimension of $\mathbb{E^2}$ is two. My third question is: am I confusing different concepts of 'dimension' here?      &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added&lt;/strong&gt;: Thanks for all the replies. They are all great but given my continuing dissatisfaction, either I have not really understood the answers or haven't been able to communicate the question properly (or perhaps and quite likely, I don't know what I want to ask).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, when we think of a (topological) manifold, we think of some object that is locally euclidean, i.e., we already have a 'handle' on $\mathbb{E^n}$ and want to use our knowledge of being able to do things, such as calculus, onto the new object . In my question, I am looking at ways we can 'handle' $\mathbb{E^2}$ itself, so invoking manifolds, seems a bit like putting the cart before the horse. I want to say something like this: The point of polar co-ordinates is to represent the plane by looking at it as $S^1\times \mathbb{R}$, where $S^1$ is a basic object like $\mathbb R$.&lt;/p&gt;&#xA;" OwnerUserId="4806" LastEditorUserId="4806" LastEditorDisplayName="" LastEditDate="2010-12-25T23:16:33.320" LastActivityDate="2010-12-25T23:16:33.320" Title="What are polar co-ordinates?" Tags="&lt;calculus&gt;&lt;linear-algebra&gt;" AnswerCount="4" CommentCount="2" FavoriteCount="2" />
  <row Id="15444" PostTypeId="2" ParentId="15443" CreationDate="2010-12-24T15:37:51.147" Score="10" ViewCount="" Body="&lt;p&gt;Great observation; it is quite annoying that nobody bothers to explicitly tell you this.  The relevant concept here is the dimension of a &lt;a href=&quot;http://en.wikipedia.org/wiki/Manifold&quot; rel=&quot;nofollow&quot;&gt;manifold&lt;/a&gt;, not the dimension of a vector space (although one is defined in terms of the other).  What &quot;coordinate&quot; means here is that you are specifying a certain isomorphism (say, differentiable) between $\mathbb{R}^n$ (or some nice submanifold thereof, for example restricting the values of some of the coordinates) and your manifold $M$, and you are using the points of $\mathbb{R}^n$ to parameterize points of $M$.  $M$ itself may be of the form $\mathbb{R}^m$, but the isomorphism need not be linear; it is an isomorphism of (say, smooth) manifolds.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The value of requiring the isomorphism to be differentiable is that a differentiable isomorphism $\phi : M \to N$ between smooth manifolds induces an isomorphism in the ordinary vector space sense between their &lt;a href=&quot;http://en.wikipedia.org/wiki/Tangent_space&quot; rel=&quot;nofollow&quot;&gt;tangent spaces&lt;/a&gt; at any point.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-12-24T15:37:51.147" CommentCount="4" />
  <row Id="15445" PostTypeId="2" ParentId="15443" CreationDate="2010-12-24T15:54:57.753" Score="4" ViewCount="" Body="&lt;p&gt;Another way of looking at coordinate systems is they are a way of naming points.  There are many different systems which have been created because they make it easy to work on certain kinds of problem.  Polar coordinates make it easy to express things that happen around a center (compare $r=c$ to $\sqrt{x^2+y^2}=c$), but hard to express things in straight lines unless they pass through the origin and addition of vectors.  &lt;a href=&quot;http://en.wikipedia.org/wiki/Dimension&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt; has a discussion of various definitions of dimension.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-12-24T15:54:57.753" />
  <row Id="15446" PostTypeId="2" ParentId="15272" CreationDate="2010-12-24T15:56:42.387" Score="0" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://at.yorku.ca/cgi-bin/bbqa?forum=ask_a_topologist_2005;task=show_msg;msg=2905.0001.0002.0001&quot; rel=&quot;nofollow&quot;&gt;This&lt;/a&gt; has a proof that for any ordered space separable implies hereditarily separable.&#xA;And the Sorgenfrey line is a subspace of a separable ordered space (e.g. the double arrow).&lt;/p&gt;&#xA;" OwnerUserId="4280" LastEditorUserId="4280" LastEditorDisplayName="" LastEditDate="2011-02-27T08:20:55.453" LastActivityDate="2011-02-27T08:20:55.453" />
  <row Id="15447" PostTypeId="2" ParentId="15421" CreationDate="2010-12-24T16:14:42.817" Score="3" ViewCount="" Body="&lt;p&gt;Where is the Taylor series for (0,1500) centered?  If at zero, it is the same Taylor series as for (0,0.5).  But you really don't want a Taylor series for (0,1500)-it would take an enormous number of terms for any reasonable accuracy.  &lt;a href=&quot;http://www.nrbook.com/c/&quot; rel=&quot;nofollow&quot;&gt;Numerical Recipes&lt;/a&gt; (chapters 5 and 6) and all other numerical analysis books have suggestions for better models.  &lt;a href=&quot;http://people.math.sfu.ca/~cbm/aands/page_69.htm&quot; rel=&quot;nofollow&quot;&gt;Abramowitz and Stegun&lt;/a&gt; has a polynomial fit over (0,1) that is much more accurate than you need.  Given what you had in the previous question, you can't represent $\exp(-x)$ for $x$ much over 11 because you will underflow.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-12-24T16:14:42.817" CommentCount="3" />
  <row Id="15448" PostTypeId="2" ParentId="15423" CreationDate="2010-12-24T16:46:28.643" Score="5" ViewCount="" Body="&lt;p&gt;This looks like a generalization of the classic $12$ ball problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You should be able to modify &lt;a href=&quot;http://www.cut-the-knot.org/blue/OddCoinProblems.shtml&quot; rel=&quot;nofollow&quot;&gt;Jack Wert's&lt;/a&gt; wonderful algorithm, (which was designed for the case when $N= \dfrac{3^m - 3}{2}$) to work for any $N$. I believe I had made an (incomplete) &lt;a href=&quot;http://stackoverflow.com/questions/3234790/algorithm-to-find-minimum-number-of-weightings-required-to-find-defective-ball-fr/3234846#3234846&quot;&gt;attempt&lt;/a&gt; when someone asked this on stackoverflow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the numbers $\dfrac{3^m - 3}{2}$ are special, in the sense that they are the turning points.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the variant of the problem where you are also required to tell if the odd sphere is heavier or lighter, for $\dfrac{3^m -3}{2} \lt N \le \dfrac{3^{m+1}-1}{2}$, the optimal number of weighings can be shown to be $m+1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are only required to find the odd sphere and not necessarily figure out if it is heavier or lighter, the turning points are $\dfrac{3^m -3}{2} + 1$.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-12-24T17:21:09.910" LastActivityDate="2010-12-24T17:21:09.910" />
  <row Id="15449" PostTypeId="2" ParentId="15251" CreationDate="2010-12-24T16:53:35.967" Score="2" ViewCount="" Body="&lt;p&gt;Dinesh, you seem to have the answer (for integer distributions) - need to show&#xA;$3(q_0+q_1)\geq q_0+q_1+q_2$, ie, $2(q_0+q_1)\geq q_2$, which is true since $2q_0\geq q_i$ for any $i$.&lt;/p&gt;&#xA;" OwnerUserId="5031" LastEditorUserId="5031" LastEditorDisplayName="" LastEditDate="2010-12-24T17:08:28.507" LastActivityDate="2010-12-24T17:08:28.507" CommentCount="1" />
  <row Id="15450" PostTypeId="2" ParentId="15443" CreationDate="2010-12-24T17:08:38.200" Score="2" ViewCount="" Body="&lt;p&gt;I think the key point is understanding the notion of &quot;howto parametrize points&quot; in a manifold, and in particular the &lt;em&gt;big question&lt;/em&gt; is &quot;How can I pass from delightful, amenable coordinates in $\mathbb{R}^n$ to those fancy 'things' with a lot of cosines, logarithms and so on?&quot;. &#xA;Well, the answer (IMO) is &quot;Because &lt;em&gt;I know&lt;/em&gt; how to do analysis nowhere but in open balls of $\mathbb {R}^n$&quot;. Studying the beginning of Mathematical Physics teached me that anytime I'm considering the speed of a motion over a (hopefully smooth) surface I'm considering a vector in the tangent space (plane) of that surface in a point: in this fashion &quot;local&quot; description of a motion is the description of a &quot;straight&quot; motion of a point $p$, with rules established, instant per instant, by the velocity vector $\vec v$ of the point $p$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to be more precise. Fix a surface $\mathcal M$ in the euclidean space (if you don't know the precise definition of a 2-dimensional manifold don't worry and go on thinking to $\mathcal M$ as a sheet on which you can smoothly skate) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You want to mathematically capture the idea of $p$ to be &lt;em&gt;bonded to move on $\mathcal M$&lt;/em&gt;. So you define its motion to be not the entire three dimensional space $\mathbb R^3$ but only a precise subset of it, which is &lt;em&gt;parametrized&lt;/em&gt; by some diffeomorphism (if you don't know what a diffeomorphism is, then learn it on wikipedia, or visualize it as a &quot;highly differentiable homeomorphism&quot; ;) ) $f\colon U\to W\subseteq \mathcal M$, $U$ being an oper set of $\mathbb{R}^2$ and $W$ being an open set (with the subspace topology induced by $\mathbb R^3$) of $\mathcal M$, containing the point $p$. It means in a few words that you're taking a point $q=(q_1,q_2)$ in the affine space of $\mathbb{R}^2$ and you're smoothly sending it in a point $f(q_1,q_2)=(f_1(q),f_2(q),f_3(q))$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this neighborhood of $p$, every happens &quot;as if you were on $\mathbb R^2$, upto go back with the inverse of $f$ on your starting $U$. A motion on $\mathcal M$ is totally captured by a motion on $U$, because you can think about a motion on $\mathcal M$ as a &quot;suitably deformed&quot; motion on $U$, where you know how to define the &lt;em&gt;position&lt;/em&gt; of a point, using coordinates, and other useful things like (piecewise) smoothness of a curve, tangent lines, et cetera. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The idea to be 'bonden on $\mathcal M$' is finally captured by asking that, given the curve $\gamma\colon ]-\epsilon,\epsilon[\to \mathcal M\subseteq\mathbb{R}^3$ describing the motion of $p$ (i.e., for example $\gamma(0)=p$), the velocity vector $\gamma&amp;#39;(t)$ lies in $\mathbb{R}^3$ &lt;em&gt;but also&lt;/em&gt; in the tangent plane $T_p(\mathcal M)$, which can in fact be defined as the collection of &lt;em&gt;all&lt;/em&gt; possible tangent vectors to curves lying on $\mathcal M$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now notice that the number '2' of independent coordinate you are using to uniquely describe the motion of $p$ on $\mathcal M$ is not at all a random number. It is the &lt;em&gt;dimension&lt;/em&gt; of the tangent plane in $p$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, let's find a basis of $T_p(\mathcal M)$. Nothing easier: do you remember the function $f\colon U\to W$? Brutally derive it, taking its differential $\text{d}f_q$, where I'm writing $q=f^{-1}(p)$, and you will get a very nice invertible matrix whose columns form a basis of $T_p(\mathcal M)$. Don't you like the canonical basis of $\mathbb{R}^2$, do you? Well, change it and see what's happen (the answer is: upto a linear automorphism of $T_p(\mathcal M)$ I'm doing nothing).&lt;/p&gt;&#xA;" OwnerUserId="685" LastActivityDate="2010-12-24T17:08:38.200" />
  <row Id="15451" PostTypeId="2" ParentId="7570" CreationDate="2010-12-24T17:23:18.287" Score="0" ViewCount="" Body="&lt;p&gt;Is the limit of the $f_j$'s continuous? If so I think your statement is true. Let $g_N(x) = 1/N \ \sum f_j(x)$. Now if $f_j \to f$, then $g_N \to f$ as well (if each $f_j$ is within an epsilon of $f$, then $g_N$ is within an epsilon of $f$ as well; go far enough out in the sequence to make this happen). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now say $x_n \to x$. Then the RHS is&#xA;$$ \lim_{N \to \infty} g_N (x) = f(x)$$, and the LHS is&#xA;$$ \lim_{n \to \infty} f(x_n) = f(x)$$ if $f$ is continuous.&lt;/p&gt;&#xA;" OwnerUserId="5100" LastActivityDate="2010-12-24T17:23:18.287" />
  <row Id="15452" PostTypeId="2" ParentId="15415" CreationDate="2010-12-24T17:28:56.560" Score="0" ViewCount="" Body="&lt;p&gt;Since the problem is tagged &quot;real analysis&quot;, I'd just use the supremum of the set $\{ n \in \mathbf{Z} : n \le x \}$. Of course, you need to prove that this set is not empty and that the supremum is actually a maximum.&lt;/p&gt;&#xA;" OwnerUserId="589" LastActivityDate="2010-12-24T17:28:56.560" />
  <row Id="15453" PostTypeId="1" AcceptedAnswerId="15454" CreationDate="2010-12-24T17:37:17.363" Score="5" ViewCount="215" Body="&lt;p&gt;I ran into this expression in a paper I was reading, and I'm confused about part of the meaning.  Here $u$ and $v$ are two polynomials.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$u, v \in \mathbb{R}[x]$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not really familiar with usage of $[x]$ here, but if it means &quot;nearest integer&quot;, then isn't this expression equivalent to simply:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$u, v \in \mathbb{Z}$$&lt;/p&gt;&#xA;" OwnerUserId="4724" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-25T00:49:26.447" LastActivityDate="2010-12-25T00:49:26.447" Title="Meaning of $\mathbb{R}[x]$" Tags="&lt;polynomials&gt;&lt;notation&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="15454" PostTypeId="2" ParentId="15453" CreationDate="2010-12-24T17:40:08.187" Score="6" ViewCount="" Body="&lt;p&gt;$\mathbb{R}[x]$ is the set of polynomials (with variable $x$) whose coefficients are taken from $\mathbb{R}$, the set of real numbers. It has got nothing to do with the greatest integer function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, people talk about polynomials in $\mathbb{Q}[x]$, where coefficients are rational, or more generally, when the coefficients are taken from an arbitrary &lt;a href=&quot;http://en.wikipedia.org/wiki/Ring_%28mathematics%29&quot; rel=&quot;nofollow&quot;&gt;Ring&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance of a usage:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Definition&lt;/strong&gt;: A real number $r$ is transcendental if and only if, for every $P \in \mathbb{Q}[x]$, we have that $P(r) \neq 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can find more information here: &lt;a href=&quot;http://en.wikipedia.org/wiki/Polynomial_ring&quot; rel=&quot;nofollow&quot;&gt;Polynomial Ring&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-12-24T17:54:20.633" LastActivityDate="2010-12-24T17:54:20.633" />
  <row Id="15455" PostTypeId="1" AcceptedAnswerId="15456" CreationDate="2010-12-24T18:12:33.777" Score="5" ViewCount="275" Body="&lt;p&gt;What does the $\ni$ (backwards element of) symbol mean?  It doesn't appear in the &lt;a href=&quot;http://en.wikipedia.org/wiki/Mathematical_symbols&quot; rel=&quot;nofollow&quot;&gt;Wikipedia list of mathematical symbols&lt;/a&gt;, and a Google search for &quot;backwards element of&quot; or &quot;backwards epsilon&quot; turns up contradictory (or unreliable) information.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems it can mean both &quot;such that&quot;, or &quot;contains as an element&quot;.  Is this correct, and if so, which is the more common usage?&lt;/p&gt;&#xA;" OwnerUserId="4724" LastActivityDate="2010-12-24T21:44:04.473" Title="Backwards epsilon" Tags="&lt;notation&gt;" AnswerCount="4" CommentCount="3" />
  <row Id="15456" PostTypeId="2" ParentId="15455" CreationDate="2010-12-24T18:21:13.547" Score="4" ViewCount="" Body="&lt;p&gt;I believe the most common usage is &quot;such that&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This seems to agree: &lt;a href=&quot;http://www.math.ucdavis.edu/~anne/WQ2007/mat67-Common_Math_Symbols.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.math.ucdavis.edu/~anne/WQ2007/mat67-Common_Math_Symbols.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While this pdf here: &lt;a href=&quot;http://faculty.ssfs.org/~doschel/2008-2009%20School%20Year/Calculus/Math%20Symbols.pdf&quot; rel=&quot;nofollow&quot;&gt;http://faculty.ssfs.org/~doschel/2008-2009%20School%20Year/Calculus/Math%20Symbols.pdf&lt;/a&gt; seems to indicate a growing trend in using it as &quot;contains as an element&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Historically, it was first used to mean &quot;such that&quot; (see the second pdf link above).&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-12-24T18:43:10.837" LastActivityDate="2010-12-24T18:43:10.837" CommentCount="3" />
  <row Id="15457" PostTypeId="2" ParentId="15174" CreationDate="2010-12-24T18:29:15.017" Score="4" ViewCount="" Body="&lt;p&gt;I'll describe the cube, the symmetries of the cube, and the generators of the symmetry of the cube.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The plus–minus–stars description of the faces of a cube:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The d-dimensional cube can be thought of as the set of points in &lt;b&gt;R&lt;/b&gt;&lt;sup&gt;d&lt;/sup&gt; with coordinates between −1 and +1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Its vertices are all sequences of plusses and minuses of length d.  For d=2 you get the four vertices { ++, +−, −+, −− }.  For d=3 you get the eight vertices { +++, ++−, …, −−+, −−−}.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Its edges are obtained from vertices by allowing one of the coordinates to vary.  For d=2 you get the four edges { ☆+, ☆−, +☆, −☆ } where for instance +☆ = { (1,y) : −1 ≤ y ≤ 1 } and ☆− = { (x,−1) : −1 ≤ x ≤ 1 }.  For d=3 you get the twelve edges { ☆++, ☆+−, …, −+☆, −−☆ }.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Its 2-faces are obtained from vertices by allowing two of the coordinates to vary.  For d=2 you get the single face { ☆☆ } where ☆☆ = { (x,y) : −1 ≤ x,y ≤ 1 }.  For d=3 you get the six faces { ☆☆+, ☆☆−, ☆+☆, ☆−☆, +☆☆, −☆☆ }.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Its k-faces for k ≤ d are obtained from vertices by allowing k of the coordinates to vary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For some people the cube is defined by its k-faces for 0 ≤ k ≤ d.  A symmetry of &lt;b&gt;R&lt;/b&gt;&lt;sup&gt;d&lt;/sup&gt; is defined by what it does to an affine basis such as the all −1s vector combined with the vertices of the cube with exactly 1 coordinate with a +1 and d−1 coordinates with a −1.  So a symmetry is uniquely determined by where it sends the 0-faces (the vertices).&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The symmetries of the cube as signed permutations:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A symmetry of the cube must take k-faces to k-faces.  In particular, it must take edges to edges.  Since it is determined by what it does to the vertices, we get two distinct operations:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;for each coordinate, we must send it to another coordinate&lt;/li&gt;&#xA;&lt;li&gt;for each coordinate, we can either reverse it or not&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;For instance ☆++ can be sent to any of the other edges, like −☆+.  For the first point, we see that the first coordinate has been sent to the second coordinate (permuting the stars).  For the second point we have two possibilities based on what happens to the vertices inside ☆++: we must send { +++, −++ } to { −++, −−+ }, but we can do this in two distinct ways (negating the sign or not).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the first point, we get the action of the symmetry group on certain &quot;blocks&quot; gives rise to the symmetric group on d objects.  For the second point, we get that the action of the symmetry group on each block gives rise to the symmetric group on 2 objects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, for every coordinate we have to decide where to send it and whether to negate it.  The collection of all such decisions is called the group of &lt;a href=&quot;http://en.wikipedia.org/wiki/Signed_permutations&quot; rel=&quot;nofollow&quot;&gt;signed permutations&lt;/a&gt;, also known as the &lt;a href=&quot;http://en.wikipedia.org/wiki/Hyperoctahedral_group&quot; rel=&quot;nofollow&quot;&gt;hyperoctahedral group&lt;/a&gt;.  Using the matrix or coordinate representation it is easy to check that all such signed permutations are in fact symmetries of the cube.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, for d=2 we could say &quot;switch the coordinates, then negate the first coordinate,&quot; which can be written simply as (x,y)↦(−y,x) in function notation, or in matrix notation: \begin{bmatrix}0&amp;amp;1\\-1&amp;amp;0\end{bmatrix}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More compactly, we could say [1,2]↦[-2,1] where we permute the coordinates and mark which coordinates get negated.  The action on the vertices is easy to describe: it sends [ ++, +−, −+, -- ] to [ −+, ++, −−, +− ].&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, to describe such a symmetry we need to choose a permutation of the vertices (an element of the symmetric group on d objects), and then d choices of ± to decide whether to negate a coordinate or not.  The size of the group is thus d!⋅2&lt;sup&gt;d&lt;/sup&gt; and the structure of the group is the wreath product of Sym(d) acting on Sym(2), and the permutation structure of the group on the vertices is more specifically the product action.  In GAP this is written as:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;WreathProductProductAction( SymmetricGroup(2), SymmetricGroup(d) );&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The generators of the symmetry group on the vertices:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The part of the symmetry group that just permutes the coordinates without doing any negating is called the &lt;a href=&quot;http://en.wikipedia.org/wiki/Symmetric_group&quot; rel=&quot;nofollow&quot;&gt;symmetric group&lt;/a&gt; and has two common generating sets:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;{ (1,2), (2,3), …, (d−1,d) }, the &lt;strong&gt;coxeter generators&lt;/strong&gt;, and&lt;/li&gt;&#xA;&lt;li&gt;{ (1,2,3,…,d), (1,2) }, the d-cycle generating set.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The Coxeter generators are better for many theoretical reasons, and the d-cycle generating set is very short and good for computational reasons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This part of the symmetry group can switch the first coordinate with any other coordinate, and so to finish generating the group it in fact suffices to merely switch the sign of one coordinate.  Thus we have the two generating sets:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;{ (1,2), (2,3), …, (d−1,d), −d }, the coxeter generators, and&lt;/li&gt;&#xA;&lt;li&gt;{ (1,2,3,…,d), (1,2), −d }, a d-cycle generating set&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Presentation_of_a_group&quot; rel=&quot;nofollow&quot;&gt;presentation of the group&lt;/a&gt; on the Coxeter generators is very nice.  Call them g(1), g(2), … g(d−1), g(d).  Then we always have g(k)&lt;sup&gt;2&lt;/sup&gt; = 1 is the identity, g(i)⋅g(j) = g(j)⋅g(i) as long as 1 ≤ i ≤ d−2, and i+2 ≤ j ≤ d, that is, as long as i and j are not adjacent.  Furthermore, g(k)⋅g(k+1)⋅g(k) = g(k+1)⋅g(k)⋅g(k+1) for 1 ≤ k ≤ d−2, but for k = d−1 we have the longer: g(d−1)⋅g(d)⋅g(d−1)⋅g(d) = g(d)⋅g(d−1)⋅g(d)⋅g(d−1).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you label the vertices from 1 to 2&lt;sup&gt;d&lt;/sup&gt;, then it is a little bit of a hassle to write down the generators, just because there are so many points.  I would probably just use:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;GeneratorsOfGroup( symmetry_group );&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;in GAP myself, but you could write them out if you wanted.&lt;/p&gt;&#xA;" OwnerUserId="583" LastActivityDate="2010-12-24T18:29:15.017" />
  <row Id="15458" PostTypeId="2" ParentId="15455" CreationDate="2010-12-24T18:33:11.233" Score="2" ViewCount="" Body="&lt;p&gt;It is for example used for functions that are inconvenient to refer to with a symbol. E.g. if $G$ is a group, then the composition function can be referred to like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ G \times G \owns (x,y) \mapsto xy \in G $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fact that the LaTeX command for it is &lt;code&gt;\owns&lt;/code&gt; should also give a clue as to what it's used for...&lt;/p&gt;&#xA;" OwnerUserId="1269" LastActivityDate="2010-12-24T18:33:11.233" CommentCount="1" />
  <row Id="15459" PostTypeId="2" ParentId="15443" CreationDate="2010-12-24T18:33:17.497" Score="2" ViewCount="" Body="&lt;p&gt;Another question implicit in what you ask is why use other co-ordinate systems. One reason is that the form of many equations is dramatically simplified in other co-ordinate systems.  For example, the rectangular representation of the curve defined by the polar coordinate equation  r = 1 + cos(2(theata)) would make it much harder to understand than in its polar form. r= 3 is a circle of radius 3 in polar coordinates. For modeling of certain phenomena polar coordinates makes things much easier, including work involving orbit calculations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Orbit_equation&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Orbit_equation&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1369" LastActivityDate="2010-12-24T18:33:17.497" />
  <row Id="15460" PostTypeId="2" ParentId="15455" CreationDate="2010-12-24T18:41:10.950" Score="7" ViewCount="" Body="&lt;p&gt;The backwards epsilon notation for &quot;such that&quot; was introduced by Peano in 1898, e.g. from Jeff Miller's &lt;a href=&quot;http://jeff560.tripod.com/set.html&quot; rel=&quot;nofollow&quot;&gt;Earliest Uses of Various Mathematical Symbols&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Such that&lt;/strong&gt;.  According to Julio González Cabillón, Peano introduced the backwards lower-case epsilon for &quot;such that&quot; in &quot;Formulaire de Mathematiques vol. II, #2&quot; (p. iv, 1898).&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Peano introduced the backwards lower-case epsilon for &quot;such that&quot; in his 1889 &quot;Principles of arithmetic, presented by a new method,&quot; according van Heijenoort's From Frege to Gödel: A Source Book in Mathematical Logic, 1879--1931 [Judy Green].&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-12-24T18:41:10.950" CommentCount="1" />
  <row Id="15461" PostTypeId="1" CreationDate="2010-12-24T19:09:40.130" Score="3" ViewCount="93" Body="&lt;p&gt;Hi,&#xA;I'm stuck on this problem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a &quot;truth-table&quot; (well, I don't know if it can be called truth table, if there aren't true/false values only):&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; string |  a |  b&#xA;---------------------&#xA;      x |  1 |  0&#xA;      z |  1 |  0&#xA;     xx |  1 |  1&#xA;     xz |  1 |  1&#xA;     zx |  1 | -1&#xA;     zz |  1 | -1&#xA;    xxx |  0 |  1&#xA;    xxz |  0 |  1&#xA;    xzx |  2 |  1&#xA;    xzz |  2 |  1&#xA;    zxx |  2 | -1&#xA;    zxz |  2 | -1&#xA;    zzx |  0 | -1&#xA;    zzz |  0 | -1&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;... the list goes on for string of any length, e.g. xxzxxzxxzzxzzxx -&gt; a = -4, b = -1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can calculate the &lt;strong&gt;a&lt;/strong&gt; and &lt;strong&gt;b&lt;/strong&gt; values from a given string, but the number of steps grows with the length of the string. I am hoping for some kind of better algorithm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;hr&gt;&#xA;&lt;em&gt;Edit: Here is my original algorithm:&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;rotation = 0 // 0 -&amp;gt; up, 1 -&amp;gt; right, 2 -&amp;gt; down, 3 -&amp;gt; left&#xA;pos_x = 0    // this is the &quot;b&quot;&#xA;pos_y = 0    // this is the &quot;a&quot;&#xA;&#xA;function rotate (n):&#xA;    rotation += n&#xA;    rotation %= 4 // result is positive integer, e.g. -1 % 4 = 3&#xA;&#xA;function forward (n):&#xA;    if n == 0: pos_y++&#xA;    if n == 1: pos_x++&#xA;    if n == 2: pos_y--&#xA;    if n == 3: pos_x--&#xA;&#xA;for char in string:&#xA;    forward()&#xA;    if char == &quot;x&quot;: rotate(1)  // to the right&#xA;    else:           rotate(-1) // to the left&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I can easily get the final rotation from a string:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;function final_rotation (n):&#xA;    rot = (number of occurences of &quot;x&quot;) - (number of occurences of &quot;z&quot;)&#xA;    rot %= 4&#xA;    return rot&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;But the pos_x and pos_y (or a and b)? No idea :/&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The Karnaugh-maps came to my mind - but I'm unsure how to deal with this, since the values of &lt;strong&gt;a&lt;/strong&gt; and &lt;strong&gt;b&lt;/strong&gt; aren't true/false, but integers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also it looks like it's ternary logic, because of the three possible states of string[position] - &lt;strong&gt;x, z, none&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as I understand this, the last (rightmost) letter of a string doesn't have a say in what the &lt;strong&gt;a&lt;/strong&gt; and &lt;strong&gt;b&lt;/strong&gt; values will be. But what does?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, my question is - how to find the algorithm for &lt;strong&gt;a&lt;/strong&gt; and &lt;strong&gt;b&lt;/strong&gt;?&lt;/p&gt;&#xA;" OwnerUserId="5102" LastEditorUserId="5102" LastEditorDisplayName="" LastEditDate="2010-12-24T19:57:01.010" LastActivityDate="2010-12-24T19:57:01.010" Title="Definition of functions based on &quot;fuzzy&quot; truth table" Tags="&lt;algorithms&gt;&lt;functions&gt;&lt;fuzzy-logic&gt;" CommentCount="7" />
  <row Id="15462" PostTypeId="2" ParentId="15261" CreationDate="2010-12-24T19:30:24.557" Score="4" ViewCount="" Body="&lt;h3&gt;Introduction&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;The following response attempts to address two aspects I perceive to be of interest in this question: one concerns whether there are some relatively rote or standard procedures for evaluating the convergence of certain kinds of improper multivariate integrals; the other is whether there is some simple intuition behind the subject.  To keep the discussion from becoming too abstract, I will use the double integral (introduced in the question) as a running example.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Synopsis&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Many integrals that are improper due to singular behavior of their integrand can be analyzed, both rigorously and intuitively, with a simple comparison.  The idea is that any singularity of the integrand that doesn't blow up too fast compared to the codimension of the manifold of singular points can be &quot;smoothed over&quot; by the integral, provided the domain of integration doesn't contact the singular points too &quot;tightly.&quot;  It remains to make these ideas precise.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Analysis&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;When the domain of integration is relatively compact, as the one in the example is, the problems with convergence occur only at the possible singularities of the integrand within the closure of the domain, which in this case is the isolated point $(x,y) = (1,1)$.  However, it is evident that if the domain of integration were to be expanded, any zero of the curve $1 - x y = 0$ could be a singularity.  In general, many singularities occur this way: an integrand $f(\bf{x})$ is locally of the form $g(||h(\bf{x})||)$ where $h: \mathbb{R}^n \to \mathbb{R}^k$ and $g(r) = r^{-s} u(r)$ for some function $u$ that is bounded in some nonnegative neighborhood of $0$.  In this case $h(\bf{x}) = 1 - x y$ and $g(r) = r^{-1}$, so $s=1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When $h$ is differentiable at $\bf{0}$ with nonsingular derivative there, &lt;strong&gt;principles of Calculus&lt;/strong&gt; invite us to linearize the situation and &lt;strong&gt;geometric ideas&lt;/strong&gt; suggest a simple form for the linearization.  Specifically, the &lt;a href=&quot;http://en.wikipedia.org/wiki/Implicit_function_theorem&quot; rel=&quot;nofollow&quot;&gt;Implicit Function Theorem&lt;/a&gt; guarantees that local coordinates can be found near such a singularity in which the derivative of $h$ is in the form $\left( \bf{1}_{k \times k} ; \bf{0}_{k \times n-k} \right)$.  The singularity itself can be translated to the origin where, to first order, the zeros of $h$ locally correspond with the vector subspace generated by the last $n-k$ coordinate axes.  The effect on the integral is to introduce a factor given by the determinant of the Jacobian, which is locally bounded and so does not affect the convergence.  In this example we can explicitly make these changes of coordinates by translating the singularity to $\bf{0}$, computing the gradient of $h$ there, and rotating it to point along the first axis.  This amounts to the change of variables $(u, v)$ = $(2 - x - y, y - x)$, in which $h(u, v) = u - \frac{1}{4}(u^2 - v^2)$, equal to $u$ to first order.  Within a small neighborhood of the origin, the domain of integration becomes a 90-degree wedge containing positive $(u,v)$ for which $u \le |v|$ and the zeros of $h$ coincide with the $v$ axis to first order.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's consider the case where the domain of integration locally contains an isolated point of the singular set $H$ (the zeros of $h$), which we translate to $\bf{0}$.  &lt;strong&gt;Estimate the integral near this singularity by adopting spherical coordinates there&lt;/strong&gt;.  To do this, we need a closer look at the domain of integration in spherical coordinates.  Any $\epsilon \gt 0$ determines the set of all possible unit direction vectors from $\bf{0}$ toward elements of the domain of integration that lie within a distance $\epsilon$ of $\bf{0}$.  In the example, this corresponds to the set of points on the unit circle with angles between $-\pi/4$ and $\pi/4$.  Suppose that for sufficiently small $\epsilon$ the closure (in the unit sphere) of this set of direction vectors does not include any of the tangent vectors of $H$ at $\bf{0}$.  Then an easy estimate shows that the angular part of the integral in spherical coordinates is bounded.  This reduces the task of integration to estimating the radial part.  Because the volume element in spherical coordinates $(\rho, \Omega)$ is $\rho^{n-1} d\rho d\Omega$, the radial integral is proportional to $\rho^{n-1} g(\rho) d \rho$.  &lt;strong&gt;This is the key calculation:&lt;/strong&gt; it shows how integration in $n$ dimensions can &quot;cancel&quot; a factor of $\rho^{1-n}$ in the integrand.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We're reduced to evaluating a 1D integral, improper at $0$, whose integrand behaves like  $\rho^{n-1} g(\rho)$.  By virtue of the assumptions about $g$, this is bounded above by a multiple of $\rho^{n-1-s}$.  &lt;em&gt;Provided&lt;/em&gt; $n-1-s \gt -1$, this converges (at a rate proportional to $\rho^{n-s}$).  In the example, $n=2$, $s=1$, so convergence is assured.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Generalizations&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;When $n-1-s \le -1$, the behavior of the original integral depends on exactly how the domain &quot;pinches out&quot; as it approaches the singularity, so more detailed analysis is needed.  The spirit of the exercise doesn't change, though: with Calculus we linearize the situation (relying on simple estimates to take care of the second order remainder term), we use spherical coordinates centered at an isolated singularity in the domain (or more generally, cylindrical coordinates near a more complicated singularity) to perform the integration, and we estimate the rate of growth of the integral near the singularity in terms of a radial component and a contribution from a spherical angle.  The behavior of the integral over the spherical angle as we approach the singularity is determined by the shape of the domain near the singularity, so often some careful (and interesting) geometric analysis is needed.  This approach is characteristic of the proofs of many theorems in Several Complex Variables, such as the &lt;a href=&quot;http://en.wikipedia.org/wiki/Edge-of-the-wedge_theorem&quot; rel=&quot;nofollow&quot;&gt;Edge of the Wedge Theorem&lt;/a&gt; (if I recall correctly--I'm reaching back to memories now a quarter century old).  One good reference is &lt;a href=&quot;http://books.google.com/books?id=uZZsh1w_RJcC&amp;amp;pg=PR7&amp;amp;source=gbs_selected_pages&amp;amp;cad=3#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;Steven Krantz's book on Function Theory of SCV&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Summary&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;The behavior of the integral of $(1-x y)^{-1}$ near $(1,1)$ can be analyzed by adopting polar coordinates $(\rho, \theta)$ centered at $(1,1)$, observing that the domain here forms a wedge whose point contacts the singular set $1 - x y = 0$ &quot;transversely,&quot; noticing that the radial behavior of the integrand is $\rho^{-1}$, remembering that the area element in polar coordinates has a $\rho^1 d \rho$ term, and noting that the integral of $\rho^{-1} \rho^1 d \rho$ converges at $0$.  Indeed, what originally looked like a singularity has entirely disappeared.  This procedure generalizes to multidimensional integrals subject to fairly mild restrictions on the nature of the singularities of their integrands and on the geometry of the domain of integration near those singularities.&lt;/p&gt;&#xA;" OwnerUserId="1489" LastActivityDate="2010-12-24T19:30:24.557" CommentCount="1" />
  <row Id="15463" PostTypeId="2" ParentId="15453" CreationDate="2010-12-24T19:32:31.650" Score="3" ViewCount="" Body="&lt;p&gt;More generally, if $\rm\:R \subset S\:$ are rings and $\rm\:s\in S\:$ then $\rm\:R[s]\:$ denotes the &lt;strong&gt;ring-adjunction&lt;/strong&gt; of $\rm\:s\:$ to $\rm\:R\:,\:$ i.e. the smallest subring of $\rm\:S\:$ containing both $\rm\:R\:$ and $\rm\:s\:.\:$ The notation for the polynomial ring $\rm\:R[x]\:$ is  the special case where $\rm\:x\:$ is transcendental over $\rm\:R\ $ (an &quot;indeterminate&quot; in old-fashioned language),$\ $ i.e. $\rm\: x\:$ isn't a root of any polynomial with coefficients in $\rm\:R\:$. One may view $\rm\:R[x]\:$ as the adjunction of a &lt;em&gt;universal&lt;/em&gt; (or generic) element $\rm\:x\:$, in the sense that any other adjunction $\rm\:R[s]\:$ is a ring-image of $\rm\:R[x]\:$ under the evaluation homomomorphism $\rm\: x\to s\:.\ $ For example, if $\rm\:R \subset S\:$  are fields then $\rm\:R[s]\cong R[x]/(f(x))\:$ where $\rm\:f(x)\:$ is the minimal polynomial of $\rm\:s\:$ over $\rm\:R\:.\:$ Essentially this serves to faithfully ring-theoretically model $\rm\:s\:$ as a &quot;generic&quot; root $\rm\:x\:$ of the minimal polynomial $\rm\:f(x)\:$ for $\rm\:s\:.\:$ Polynomial rings may be characterized by the existence and uniqueness of such evaluation maps (&quot;universal mapping property&quot;), e.g. see any textbook on Universal Algebra, e.g. &lt;a href=&quot;http://math.berkeley.edu/~gbergman/245/index.html&quot; rel=&quot;nofollow&quot;&gt;Bergman.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-12-24T21:23:47.603" LastActivityDate="2010-12-24T21:23:47.603" />
  <row Id="15464" PostTypeId="2" ParentId="15438" CreationDate="2010-12-24T19:44:42.820" Score="4" ViewCount="" Body="&lt;p&gt;Here is a net of a buckyball, from &lt;a href=&quot;http://www.goldennumber.net/buckyball.htm&quot; rel=&quot;nofollow&quot;&gt;GoldenNumber.net&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/8Bwto.gif&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It should be possible to turn this into the kind of net you're looking for by replacing the pentagons and hexagons with 5 and 6 isosceles triangles (the heights of the triangles determine the &quot;elevation&quot; of the center vertex from the original pentagonal/hexagonal faces and thus affect the sphericality of the result). &lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-12-24T19:44:42.820" CommentCount="1" />
  <row Id="15465" PostTypeId="2" ParentId="15388" CreationDate="2010-12-24T19:56:17.983" Score="3" ViewCount="" Body="&lt;p&gt;Simply use the (additive group) axioms to prove $\rm\ a + x = b\ \Rightarrow\ x = b + (-a)\:.\ $ That yields both existence &lt;em&gt;and&lt;/em&gt; uniqueness. That the uniqueness doesn't require further proof is a subtlety that sometimes confuses students. This is discussed at length in my posts in &lt;a href=&quot;http://groups.google.com/group/sci.math/browse_frm/thread/1bbf3ac81e6df7f9&quot; rel=&quot;nofollow&quot;&gt;this thread&lt;/a&gt; on such a proof in Max Rosenlicht's &lt;em&gt;Introduction to Analysis&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-12-24T19:56:17.983" />
  <row Id="15467" PostTypeId="2" ParentId="15349" CreationDate="2010-12-24T21:36:32.173" Score="0" ViewCount="" Body="&lt;p&gt;Use any algorithm for &lt;a href=&quot;http://en.wikipedia.org/wiki/Matching_%28graph_theory%29#Maximum_matchings_in_bipartite_graphs&quot; rel=&quot;nofollow&quot;&gt;Maximum Weighted Bipartite Matching&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-12-24T21:36:32.173" CommentCount="1" />
  <row Id="15468" PostTypeId="2" ParentId="15455" CreationDate="2010-12-24T21:44:04.473" Score="1" ViewCount="" Body="&lt;p&gt;I have always used $\ni$ to represent &quot;such that.&quot; Many times people do not use this notation though; rather, they use s.t. or just say &quot;such that.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, you can see that Wikipedia uses the &quot;:&quot; notation, which is also correct; however, I usually only use that notation when in curly brackets (ie { }).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can visit [1] to see an actual discussion of it in a list of mathematical symbols.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This was a very good question, by the way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[1] &lt;a href=&quot;http://www.math.ucdavis.edu/~anne/WQ2007/mat67-Common_Math_Symbols.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.math.ucdavis.edu/~anne/WQ2007/mat67-Common_Math_Symbols.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3791" LastActivityDate="2010-12-24T21:44:04.473" />
  <row Id="15469" PostTypeId="1" AcceptedAnswerId="15490" CreationDate="2010-12-24T22:00:01.693" Score="9" ViewCount="296" Body="&lt;p&gt;Does anyone know a combinatorial proof of the following identity, where $F_n$ is the $n$th Fibonacci number?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$n F_1 + (n-1)F_2 + \cdots + F_n = F_{n+4} - n - 3$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's not in the place I thought it most likely to appear: Benjamin and Quinn's &lt;em&gt;&lt;a href=&quot;http://books.google.com/books?id=FXLGzXwbwIAC&amp;amp;printsec=frontcover&amp;amp;dq=proofs+that+really+count&amp;amp;source=bl&amp;amp;ots=cLasNhVboM&amp;amp;sig=cgpcvX12VJy4hzDjlM8buMY1O1w&amp;amp;hl=en&amp;amp;ei=1RUVTY2vAY7EsAPklv3GAg&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=3&amp;amp;ved=0CCkQ6AEwAg#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;Proofs That Really Count&lt;/a&gt;&lt;/em&gt;.  In fact, this may be a hard problem, as they say the similar identity &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ F_1 + 2F_2 + \cdots + nF_n = (n+1)F_{n+2} - F_{n+4} +2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is &quot;in need of a combinatorial proof.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For reference, &lt;a href=&quot;http://books.google.com/books?id=n8Xx2oSlz4MC&amp;amp;pg=PA14&amp;amp;lpg=PA14&amp;amp;dq=benjamin+and+quinn+uncounted+fibonacci&amp;amp;source=bl&amp;amp;ots=UDfzehytLs&amp;amp;sig=mcWgV3BP2_Jyl5zi1fhL2EHqqfU&amp;amp;hl=en&amp;amp;ei=ThIVTcmsLo7ksQPcvMXIAg&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=3&amp;amp;ved=0CCIQ6AEwAg#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; (from Benjamin and Quinn's text) are several combinatorial interpretations of the Fibonacci numbers.&lt;/p&gt;&#xA;" OwnerUserId="2370" LastActivityDate="2010-12-25T06:16:12.707" Title="Combinatorial proof of a Fibonacci identity: $n F_1 + (n-1)F_2 + \cdots + F_n = F_{n+4} - n - 3.$" Tags="&lt;combinatorics&gt;&lt;fibonacci-number&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="3" />
  <row Id="15470" PostTypeId="1" AcceptedAnswerId="15488" CreationDate="2010-12-24T22:39:09.373" Score="3" ViewCount="87" Body="&lt;p&gt;Let $f: X \to Y$ be a proper morphism of noetherian schemes. If the natural map $\mathcal{O}_Y \to f_*(\mathcal{O}_X)$ is an isomorphism, then a version of Zariski's main theorem states that the fibers $X_y, y \in Y$ are all connected. (The case listed as &quot;Zariski's main theorem&quot;  in Hartshorne is the case of $f$ birational and $Y$ normal.) This may be proved via the formal function theorem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fancier version of ZMT (EGA IV-8) is that a quasifinite separated morphism of finite presentation between quasicompact, quasiseparated schemes factors as the composite of an open immersion and a finite morphism. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a direct way to deduce the connectedness theorem from the more general ZMT?&lt;/p&gt;&#xA;" OwnerUserId="536" LastActivityDate="2010-12-25T05:58:47.473" Title="How to get the connectedness theorem from the quasi-finite version of ZMT?" Tags="&lt;algebraic-geometry&gt;" AnswerCount="1" />
  <row Id="15471" PostTypeId="1" AcceptedAnswerId="15474" CreationDate="2010-12-24T22:46:51.917" Score="3" ViewCount="251" Body="&lt;p&gt;The sum of a finite number of rational numbers is of course a rational number, but the sum of an infinite number of rational numbers might be an irrational number.  Can someone give me some intuition why this sum might be irrational?  I just &quot;don't feel it.&quot;&lt;/p&gt;&#xA;" OwnerUserId="4811" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2011-01-05T17:36:00.240" LastActivityDate="2011-01-05T17:36:00.240" Title="Sum of rational numbers" Tags="&lt;number-theory&gt;&lt;irrational-numbers&gt;" AnswerCount="4" CommentCount="4" />
  <row Id="15472" PostTypeId="2" ParentId="15471" CreationDate="2010-12-24T22:55:13.000" Score="2" ViewCount="" Body="&lt;p&gt;Any irrational number $x$ is the limit of a sequence of rational numbers $a_n$ (take for example the decimal expansion truncated after $n$ decimals, for $n=1,2,3,\dots$). Then&#xA;$$x = a_1 + (a_2-a_1) + (a_3-a_2) + \dots$$&#xA;is a sum of rational numbers, but it is irrational by construction.&lt;/p&gt;&#xA;" OwnerUserId="1242" LastEditorUserId="1242" LastEditorDisplayName="" LastEditDate="2010-12-24T23:00:18.757" LastActivityDate="2010-12-24T23:00:18.757" />
  <row Id="15473" PostTypeId="2" ParentId="15471" CreationDate="2010-12-24T22:55:43.593" Score="1" ViewCount="" Body="&lt;p&gt;An irrational number is a &quot;gap&quot; inside the rationals. The decimal expansion of any irrational number is an infinite sum converging to it. It gives better and better approximations to the irrational number, which is unfortunately just not &quot;there&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are many more examples of such limiting constructions. Sometimes you get new objects and sometimes you don't. An example is the delta function which can be approximated using bona fide functions. So there are functions arbitrarily &quot;close&quot; (in some sense) to the delta function, but the delta function itself is just &quot;too good&quot; to be an actual function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an aside, if these approximations are good enough, you can prove that the number is transcendental (Legendre's or Roth's theorem).&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-12-24T22:55:43.593" />
  <row Id="15474" PostTypeId="2" ParentId="15471" CreationDate="2010-12-24T22:56:36.573" Score="13" ViewCount="" Body="&lt;p&gt;Look at &lt;a href=&quot;http://mathforum.org/library/drmath/view/52829.html&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So: $3.14159 \dots = 3 + \frac{1}{10} + \frac{4}{10^2} +\frac{1}{10^3} + \frac{5}{10^4}+ \frac{9}{10^5} + \dots$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The above expression is $\pi$.&lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2010-12-24T22:56:36.573" />
  <row Id="15475" PostTypeId="2" ParentId="14828" CreationDate="2010-12-24T23:17:39.900" Score="2" ViewCount="" Body="&lt;p&gt;In addition to the other answers, it's also noteworthy to know that there is a number system (more-or-less) integrating N, Z, Q and R (and &lt;em&gt;a lot&lt;/em&gt; more, but not C!). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Surreal_number&quot; rel=&quot;nofollow&quot;&gt;surreal numbers&lt;/a&gt; take the basic idea from Dedekind cuts, assigning to each number a so-called &lt;em&gt;left set&lt;/em&gt; ('smaller') and &lt;em&gt;right set&lt;/em&gt; ('larger') of numbers (constraint to certain rules), bootstrapping the whole process from the empty set and ending up with the reals and weird infinitesimals like $\frac{1}{\sqrt{\omega - \pi}}$!&lt;/p&gt;&#xA;" OwnerUserId="360" LastActivityDate="2010-12-24T23:17:39.900" />
  <row Id="15476" PostTypeId="1" AcceptedAnswerId="15477" CreationDate="2010-12-25T00:06:36.170" Score="4" ViewCount="188" Body="&lt;p&gt;Let $0&amp;lt;a\le 1$ be fixed. Is the sequence $$a_N=\int_1^N x^{-a}e^{ix} dx$$ bounded?&lt;/p&gt;&#xA;" OwnerUserId="3249" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-27T22:09:51.967" LastActivityDate="2010-12-27T22:09:51.967" Title="Is this a bounded sequence?" Tags="&lt;analysis&gt;&lt;sequences-and-series&gt;&lt;fourier-analysis&gt;" AnswerCount="2" />
  <row Id="15477" PostTypeId="2" ParentId="15476" CreationDate="2010-12-25T00:17:40.390" Score="5" ViewCount="" Body="&lt;p&gt;Yes, and it converges to $\int_1^\infty x^{-a}e^{ix}dx$.  To see that this improper Riemann integral exists, you could break up the integrand into real and imaginary parts, break up the resulting real integrals over intervals of length $\pi$, and apply the alternating series test.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-12-25T00:17:40.390" />
  <row Id="15478" PostTypeId="2" ParentId="15476" CreationDate="2010-12-25T00:57:06.340" Score="7" ViewCount="" Body="&lt;p&gt;I just thought of another answer. Using integration by parts, we get&#xA;$$\frac{1}{i}(N^{-a}e^{iN}-e^i)+\frac{a}{i}\int_1^N y^{-(1+a)}e^{iy} dy$$&#xA;The first term is bounded and the second term is dominated by the integral $\int_1^\infty y^{-(1+a)} dy$.&lt;/p&gt;&#xA;" OwnerUserId="3249" LastActivityDate="2010-12-25T00:57:06.340" />
  <row Id="15479" PostTypeId="1" AcceptedAnswerId="15482" CreationDate="2010-12-25T01:57:02.287" Score="8" ViewCount="213" Body="&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; What are some good resources for a student who has taken algebraic and point-set topology and who wishes to learn how filters and ultrafilters are applied in topology?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Motivation:&lt;/strong&gt; I've seen in a number of papers where ultrafilters are briefly mentioned as tools which can be used to generalize an argument or something, but the few papers I've found which introduce filters mainly use them as as tools in Logic and mention topology only in passing.  The wikipedia page gives some good starting points, but I felt that it was more of a reference list than an introduction to the topic.&lt;/p&gt;&#xA;" OwnerUserId="2959" LastActivityDate="2011-03-04T06:27:12.207" Title="Introduction to Filters in Topology." Tags="&lt;topology&gt;&lt;online-resources&gt;" AnswerCount="3" FavoriteCount="1" />
  <row Id="15480" PostTypeId="1" AcceptedAnswerId="15483" CreationDate="2010-12-25T04:11:02.357" Score="5" ViewCount="100" Body="&lt;p&gt;This question is vaguely related to: &lt;a href=&quot;http://math.stackexchange.com/questions/14915/different-formulations-of-class-field-theory&quot;&gt;Different formulations of Class Field Theory&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As I said there, I'm currently learning class field theory. For some motivation, I've also read a little about Langlands. Everything I've read about that program seems to be a generalization of the global formulation of CFT. Is the local language (and theorems) preserved as well? Are they obviously/with difficulty or/conjecturally equivalent?&lt;/p&gt;&#xA;" OwnerUserId="4953" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2011-01-30T01:56:26.327" LastActivityDate="2011-01-30T01:56:26.327" Title="Does the Langlands program preserve CFT's distinction between local and global theories?" Tags="&lt;number-theory&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="15481" PostTypeId="2" ParentId="15434" CreationDate="2010-12-25T04:18:11.867" Score="2" ViewCount="" Body="&lt;p&gt;Edit: I was initially assuming that $E$ is the closed linear span of the basic sequence, but then I realized that from your question there is no justification for me having done so.  Since my initial argument applied only to that case and I didn't know how to extend it, I did a Google search to find that one way to reduce to this case is to apply Mazur's theorem, and that a nice and succinct answer to your question is given as &lt;a href=&quot;http://books.google.com/books?id=FkjCulO5xg0C&amp;amp;lpg=PP1&amp;amp;pg=PA23#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;Lemma 1.6.1&lt;/a&gt; in &lt;em&gt;Topics in Banach space theory&lt;/em&gt; by Albiac and Kalton.  (This is something I should have known, but I had apparently not thought enough about weak topologies.)&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;For what it's worth, here's my original argument that applies after reducing to the case where $E$ is the closed linear span of $\{x_n\}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus every element $x$ of $E$ has a unique representation as an infinite sum $x=\sum_n a_n(x)x_n$, and the coefficient functionals $x\mapsto a_n(x)$ are bounded.  Let $B=\{x_n\}$ denote the basis, and suppose that $x$ is in $\overline{B}\setminus B$.  You want to show that $x=0$, which is equivalent to showing that $a_n(x)=0$ for all $n$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For each $n$ and each $\varepsilon\gt0$, consider the subbasic neigborhood $U_{n,\varepsilon}$ of $x$ defined by  $U_{n,\varepsilon}= \{y:|a_n(y)-a_n(x)|\lt \epsilon \}$.  Suppose to get a contradiction that $x\neq 0$, so there exists a $k$ such that $a_k(x)\neq 0$.  Then $U_{k,\varepsilon}$ is disjoint from $B\setminus\{x_k\}$ for sufficiently small $\epsilon$, and since $x$ is in $\overline B$ you can deduce that $a_k(x)=1$.  Since $x\neq x_k$, there exists $m\neq k$ such that $a_m(x)\neq0$, and the same argument indicated above shows that $a_m(x)$ must be $1$.  Then $U_{k,1}\cap U_{m,1}$ is a neighborhood of $x$ that doesn't intersect $B$, a contradiction.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2010-12-25T05:12:45.337" LastActivityDate="2010-12-25T05:12:45.337" />
  <row Id="15482" PostTypeId="2" ParentId="15479" CreationDate="2010-12-25T04:26:33.347" Score="4" ViewCount="" Body="&lt;p&gt;Bourbaki's &lt;em&gt;General Topology, Volume 1&lt;/em&gt; devotes a chapter to filters. The notion of net and equivalence between filters and nets is developed (partly in a set of guided exercises) in Kelley's &lt;em&gt;General Topology&lt;/em&gt;. &lt;em&gt;The use of ultraproducts in Commutative algebra&lt;/em&gt; by Schoutens might interest you. &lt;em&gt;The theory of ultrafilters&lt;/em&gt; by Comfort and Negrepontis is encyclopedic. Jech's &lt;em&gt;Set theory&lt;/em&gt; has a chapter on ultrafilters. Sections 4 and 6 of Chang and Keisler's &lt;em&gt;Model Theory&lt;/em&gt; is also worth a look. These are all available on the website library.nu.&lt;/p&gt;&#xA;" OwnerUserId="746" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2011-03-04T05:57:31.033" LastActivityDate="2011-03-04T05:57:31.033" />
  <row Id="15483" PostTypeId="2" ParentId="15480" CreationDate="2010-12-25T04:57:54.570" Score="6" ViewCount="" Body="&lt;p&gt;Dear Nicole, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a local aspect to the Langlands program, known as the local Langlands correspondence.  In fact, Langlands conjectured the existence of such a correspondence for each local field $F$ and each reductive group $G$ over $F$.&#xA;He proved his local conjecture when $F$ is archimedean and $G$ is arbitrary.  The case when $G = \mathrm{GL}_n$ and $F$ is non-archimedean is now solved (by Laumon, Rapoport, and Stuhler in the function field case, and by Harris and Taylor in the case of $p$-adic fields).   There are many results known for other groups $G$ as well, but the full local conjectures for arbitrary $G$ are not yet settled, as far as I know.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the $\mathrm{GL}_n$ case, the correspondence, roughly speaking, gives a bijeciton between (typically infinite dimensional!) irreducible representations of the group $\mathrm{GL}_n(F)$ and $n$-dimensional representations of the Galois group of $F$.  (So, unlike the abelian case, there is no isomorphism of groups, but rather a certain bijection between certain kinds of representations of rather different groups.)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The case of general $G$ is more involved to state, and indeed, it is not so easy to find the precise conjecture in the literature.  In any case, it involves many complications, the most significant of which is probably so-called &lt;em&gt;endoscopy&lt;/em&gt;.  Note that Ngo won the Fields medal this summer for his work on this topic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just as with local CFT, the motivations for the local conjecture come from the global theory.  On the one hand, representations of matrix groups over local fields arise naturally in the theory of automorphic forms (this is a generalization/reformulation of the classical theory of Hecke operators in the theory of modular forms), and, on the other hand, (at least certain) automorphic Hecke eigenforms are suppose to be related to representations of Galois groups of global fields by global reciprocity laws.  Restricting these global Galois representations to decomposition groups, one should recover the conjectural local correspondence.  (And as I noted in my answer to your earlier question, this is in fact the mechanism by which those local correspondences are normally constructed, in the cases when they can be constructed.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, I'm not sure if I understand your last question about equivalence correctly, but if you are asking whether the existence of the global Langlands correspondence (whatever exactly that means) should follow from the local correspondence, the answer is no.  Consider the case of CFT: knowing all the local Artin maps lets you write down the global Artin map, but you still have to prove the global Artin reciprocity law.  Similarly, knowing the local Langlands correspondence allows one to write down a candidate for the global correspondence, but to prove that this candidate actually does the job is another, even more difficult, matter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(As one example: before the modularity theorem for elliptic curves was proved, people knew how to write down the &#xA;candidate $q$-expansion that should be the weight $2$ modular form attached to an elliptic curve over $\mathbb Q$; this was because the relevant local issues were all completely understood.  The problem was then to prove that this actually &lt;em&gt;was&lt;/em&gt; a weight $2$ modular form; this was a global issue, which was completely open until Wiles and Taylor, Breuil, Conrad and Diamond solved it.)&lt;/p&gt;&#xA;" OwnerUserId="221" LastEditorUserId="221" LastEditorDisplayName="" LastEditDate="2010-12-25T05:06:38.233" LastActivityDate="2010-12-25T05:06:38.233" />
  <row Id="15484" PostTypeId="1" CreationDate="2010-12-25T05:31:44.120" Score="1" ViewCount="172" Body="&lt;p&gt;Can you help me to prove this inequality&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{aligned}&#xA;|\sqrt{3} - m/n| \geq 1/(5n^2)&#xA;\end{aligned}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where m and n are integers.&#xA;Hint:$sqrt(3)$ is irrational.&lt;/p&gt;&#xA;" OwnerUserId="5049" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-12-25T06:56:39.637" LastActivityDate="2011-01-03T02:25:17.317" Title="Showing that $|\sqrt{3} - m/n| \geq 1/(5n^2)$" Tags="&lt;number-theory&gt;&lt;inequality&gt;" AnswerCount="1" />
  <row Id="15485" PostTypeId="2" ParentId="15479" CreationDate="2010-12-25T05:45:42.473" Score="2" ViewCount="" Body="&lt;p&gt;As far as online resources:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://math.uga.edu/~pete/convergence.pdf&quot; rel=&quot;nofollow&quot;&gt;Notes on convergence&lt;/a&gt; by Pete Clark,&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://topologicalmusings.wordpress.com/2008/07/18/ultrafilter-convergence-compactness-and-the-spectrum-of-a-boolean-algebra/&quot; rel=&quot;nofollow&quot;&gt;Blog post&lt;/a&gt; by Todd Trimble,&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://terrytao.wordpress.com/tag/ultrafilters/&quot; rel=&quot;nofollow&quot;&gt;Blog posts&lt;/a&gt; by Terence Tao (see also the posts tagged ultralimit-analysis and nonstandard-analysis)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I am also writing a &lt;a href=&quot;http://qchu.wordpress.com/tag/ultrafilters/&quot; rel=&quot;nofollow&quot;&gt;series of blog posts&lt;/a&gt; (unfinished) on the subject.  &lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-12-25T05:45:42.473" CommentCount="1" />
  <row Id="15486" PostTypeId="1" AcceptedAnswerId="15492" CreationDate="2010-12-25T05:51:50.333" Score="6" ViewCount="455" Body="&lt;p&gt;How to understand the compactness in topology space in intuitive way? &lt;/p&gt;&#xA;" OwnerUserId="1080" LastEditorUserId="6296" LastEditorDisplayName="" LastEditDate="2011-03-19T00:00:52.357" LastActivityDate="2011-03-19T00:00:52.357" Title="How to understand compactness?" Tags="&lt;intuition&gt;&lt;general-topology&gt;" AnswerCount="6" CommentCount="1" FavoriteCount="3" />
  <row Id="15487" PostTypeId="2" ParentId="15471" CreationDate="2010-12-25T05:53:44.933" Score="6" ViewCount="" Body="&lt;p&gt;The problem is psychological: you think of the &quot;infinite sum&quot; of rational numbers as an obvious, intuitive concept, but it's not.  It has a precise mathematical meaning, and that precise mathematical meaning only works if you allow the sums to be real numbers (which themselves have a precise mathematical meaning).  The definitions which allow these &quot;infinite sums&quot; to make sense are much less trivial than someone who's never worked through them would think.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-12-25T05:53:44.933" />
  <row Id="15488" PostTypeId="2" ParentId="15470" CreationDate="2010-12-25T05:58:47.473" Score="1" ViewCount="" Body="&lt;p&gt;Dear Akhil,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure that this is directly possible.  If memory serves, Zariski's original version of his main theorem showed something like the following: if $f: X \to Y$ is birational with $Y$ normal, and if $y \in Y$ is a point where $f^{-1}$ is not defined (I think this is what Zariski calls a fundamental point), then each component of $f^{-1}(y)$ is positive dimensional.  &lt;em&gt;This&lt;/em&gt; result can be proved using Grothendieck's form of ZMT, I think: if $f^{-1}(y)$ contains an isolated point, we can choose a n.h. of this point such that the restriction of $f$ to this n.h. has finite fibres, hence is an isomorphism (here we are using Grothendieck's ZMT together with normality of $Y$), and so in fact&#xA;$f^{-1}$ can be defined at $y$ after all.  (This is quite possibly not quite correct, but hopefully is not totally bogus either, both in the argument and in the claim to some historical accuracy.  Also, I think that if you look e.g. in the commutative algebra part of the stacks project a version of ZMT along these lines is discussed; at least, there are results in which the hypothesis of an isolated point in the fibre plays a prominent role.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Zariski's connectedness theorem came quite a bit later, in his monograph on formal functions, and the techniques were quite a bit more involved.  (They were a precursor to formal scheme techniques, in which one can complete in some directions but not in others, as opposed to earlier techniques with complete local rings, in which one completes in every direction around a point at once.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A quick glance over Mumford's discussion of ZMT in the Red Book (which is always a good place to go to for learning ZTM intuition) suggests that I'm not blundering here.  (Grothendieck's formulation of ZMT is what he calls version IV of ZMT, while the connectedness theorem is his version V, and he singles out version V as being &quot;more global&quot; than the other versions, and doesn't discuss any implications from the other versions to this one.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[On the other hand, this contradicts the wikipedia entry, which suggests that Grothendieck's formulation does imply the connectedness theorem.  (But doesn't quite say how.)  There are plenty of people at Harvard you could ask for clarification on this point, of course ... .]&lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2010-12-25T05:58:47.473" CommentCount="1" />
  <row Id="15489" PostTypeId="2" ParentId="12057" CreationDate="2010-12-25T06:01:03.390" Score="2" ViewCount="" Body="&lt;p&gt;Simply repeatedly apply Euclid's lemma:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\quad\quad\quad\quad 1 = (pb+qa,qb)$&lt;br&gt;&#xA;$\rm\quad \iff\ 1 = (pb+qa,q) = (pb,q) = (b,q)\quad\ via\quad\ (p,q) = 1$&lt;br&gt;&#xA;$\quad\ $ and $\rm\ \ 1 = (pb+qa,b) = (qa,b) = (q,b)\quad\ via\quad\ (a,b) = 1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence $\rm\ \ 1 = (pb+qa,qb)\ \iff\ 1 = (b,q)$&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-12-28T02:15:51.310" LastActivityDate="2010-12-28T02:15:51.310" />
  <row Id="15490" PostTypeId="2" ParentId="15469" CreationDate="2010-12-25T06:10:32.523" Score="9" ViewCount="" Body="&lt;p&gt;Recall that $F_{n+1}$ is the number of ways to tile a board of length $n$ with tiles of length $1$ and $2$.  So $F_{n+4}$ is the number of ways to tile a board of length $n+3$ with tiles of length $1$ and $2$.  Note that $n+3$ such tilings use at most one tile of length $2$, so $F_{n+4} - (n+3)$ such tilings use at least two tiles of length $2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given such a tiling, look at where the second-to-last tile of length $2$ is used.  The part after this tile is a tiling of some section of length $k+1$ where exactly one tile of length $2$ is used (which can be done in $k$ ways), and the part before this tile is a tiling of the remaining portion of length $n-k$ (which can be done in $F_{n-k+1}$ ways).  Sum over $k$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(The bigger lesson to take away here is that convolution is much easier to deal with than Hadamard product.  Also, since the bijection I described above preserves the number of tiles of each type, the identity can be upgraded to an identity of $q$-Fibonacci numbers.)&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-12-25T06:16:12.707" LastActivityDate="2010-12-25T06:16:12.707" CommentCount="2" />
  <row Id="15491" PostTypeId="2" ParentId="15486" CreationDate="2010-12-25T06:40:58.547" Score="0" ViewCount="" Body="&lt;p&gt;Look at as many examples of what sets are compact and which ones are not, then you get a feeling of what compactness tries to encapsulate. Then try to see why the examples of compact sets you looked at are compact and why the non compact examples are not. Human mind learns a lot by comparing and mixing stuff, so just compare mix, add repeat.  &lt;/p&gt;&#xA;" OwnerUserId="3936" LastActivityDate="2010-12-25T06:40:58.547" />
  <row Id="15492" PostTypeId="2" ParentId="15486" CreationDate="2010-12-25T06:52:11.477" Score="10" ViewCount="" Body="&lt;p&gt;Maybe you should think about compactness, as something that takes local properties to global properties. For example, if $f:K\rightarrow \mathbb{R}$ is continuous, $K$ is compact, and $f(x)&amp;gt;t_x&amp;gt;0$ for all x, then you can find $t&amp;gt;0$ such that $f(x)&amp;gt;t&amp;gt;0$ for all x - so from $f(x)&amp;gt;t_x&amp;gt;0$ point wise, you know that $f&amp;gt;t&amp;gt;0$ as a function. (This is a simple consequence of Weierstrass theorem in $[a,b]\rightarrow \mathbb{R}$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Usually we find some property that is true for every &quot;small&quot; enough open sets, then use compactness to reduce the case to finitely many open sets and use induction to show that the property is true for all of the space. This is at least how I understand compactness.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;As the commenters below this message wrote (and I didn't emphasize enough), we usually use compactness to reduce infinite problems\conditions\restraints to a finite subset that cover the entire space, and then use some argument that works only for finite cases (like induction, taking max\min, take finite sums etc). In my example above, we wanted to find a minimum over all the lower bounds, but in the infinite case this is usually just an infimum (and can be zero), but when we reduce to a finite case there is a minimum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this way we can think of compactness as something that let us use some finite argument on infinite covers (and many times to transfer some property from the cover to the entire space). &lt;/p&gt;&#xA;" OwnerUserId="2125" LastEditorUserId="2125" LastEditorDisplayName="" LastEditDate="2010-12-26T06:38:06.327" LastActivityDate="2010-12-26T06:38:06.327" CommentCount="6" />
  <row Id="15493" PostTypeId="2" ParentId="15484" CreationDate="2010-12-25T06:53:58.300" Score="3" ViewCount="" Body="&lt;p&gt;Hint:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Show that if $\displaystyle f(x) = x^2 -3$, then $$\left|f\left(\dfrac{m}{n}\right)\right| \ge \dfrac{1}{n^2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ok, I guess Vinod isn't interested in following up on the hints, here is almost a full proof.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For any rational number $\displaystyle m/n$, we have that $\displaystyle |3 - \frac{m^2}{n^2}|  = \frac{|m^2 - 3n^2|}{n^2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now since $\displaystyle \sqrt{3}$ is irrational, $\displaystyle |m^2 - 3n^2| \ge 1$ (it is an integer).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus $\displaystyle |3 - \frac{m^2}{n^2} | \ge \frac{1}{n^2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now if $\displaystyle |\frac{m}{n} + \sqrt{3}| \le 5$, then we have that $\displaystyle |\frac{m}{n} - \sqrt{3}| \ge \frac{1}{5n^2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The case $\displaystyle |\frac{m}{n} + \sqrt{3}| \ge 5$, is easier, as then $\displaystyle \frac{m}{n}$ will be not be very close to $\displaystyle \sqrt{3}$.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2011-01-03T02:25:17.317" LastActivityDate="2011-01-03T02:25:17.317" CommentCount="4" />
  <row Id="15494" PostTypeId="2" ParentId="13624" CreationDate="2010-12-25T07:01:29.440" Score="2" ViewCount="" Body="&lt;p&gt;For 1., clearly yes: if a cover has a finite refinement, then the sets it is a refinement of also cover $X$, and form a finite subcover.&#xA;For 2., see Nuno's comment on the other answer.&lt;/p&gt;&#xA;" OwnerUserId="4280" LastActivityDate="2010-12-25T07:01:29.440" />
  <row Id="15495" PostTypeId="1" CreationDate="2010-12-25T07:32:59.570" Score="0" ViewCount="50" Body="&lt;p&gt;From example, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Chevalley basis for L consist of E, H, F in Book Lie Groups : A problem-oriented ... Harriet&lt;/p&gt;&#xA;&#xA;&lt;p&gt;E := matrix([[0,0,1],[0,0,-i],[-1,i,0]]);&#xA;X := matrix([[0,1,0],[-1,0,0],[0,0,0]]);&#xA;ad := E&amp;amp;*X - X&amp;amp;*E;&#xA;evalm(ad);&lt;/p&gt;&#xA;&#xA;&lt;p&gt;above can not get the answer below in maple&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Answer is&#xA;ad(E) := matrix([[0,-1,0],[0,0,-2],[0,0,0]]);   // How to get this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moreover for H&lt;/p&gt;&#xA;&#xA;&lt;p&gt;H := matrix([[1,0,0],[0,0,0],[0,0,-1]]);&#xA;Y := matrix([[0,0,1],[0,0,0],[-1,0,0]]);&#xA;ad := H&amp;amp;*Y - Y&amp;amp;*H;&#xA;evalm(ad);&lt;/p&gt;&#xA;" OwnerUserId="5070" LastActivityDate="2010-12-25T07:32:59.570" Title="How to calculate ad(X) for exp(t*ad(X))?" Tags="&lt;lie-groups&gt;" CommentCount="1" />
  <row Id="15496" PostTypeId="2" ParentId="15426" CreationDate="2010-12-25T08:08:41.083" Score="0" ViewCount="" Body="&lt;p&gt;I'm posting this as a separate answer so as to stop bothering Willie with incessant comment notifications. Of course, the other answers that have been posted are completely correct, and I'm only trying to help you reconcile your intuitive reasoning with their results.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In situation A, the information you obtain about the state of the coins is, &quot;The first coin is heads.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In situation B, the information you obtain is, &quot;At least one of the coins is heads.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These are not equivalent; the former is not implied by the latter, because in B you cannot conclude that the first coin is heads. In other words, the observer's statement &lt;em&gt;in conjunction with your knowledge of the observer's strategy&lt;/em&gt; enables you to make a stronger conclusion about the coins in situation A.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you wish, I can clarify (a) how the observer's strategy affects things, as TonyK has also mentioned, and/or (b) why in situation B the probability is 1/3 and not 1/2 as you expect.&lt;/p&gt;&#xA;" OwnerUserId="856" LastActivityDate="2010-12-25T08:08:41.083" CommentCount="16" />
  <row Id="15497" PostTypeId="2" ParentId="15400" CreationDate="2010-12-25T08:12:52.537" Score="2" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://math.stackexchange.com/questions/392/intuitive-understanding-of-the-derivatives-of-sinx-and-cosx/1093#1093&quot;&gt;I like to think of helices on cylinders as images of lines under planar curling&lt;/a&gt;, so here's an approach incorporating that idea.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine the circle rolling up a line drawn in the plane, with $P$ the point of tangency along the line and $Q$ a distinguished point on the circumference of the circle. (So, $Q$ traces out a cycloid in the plane.) Now, imagine the plane is made of thin paper, but the circle is made of stiff cardboard. If we curl the plane into a cylinder but the circle remains flat, the plane of the circle will be tangent to the cylinder along the (vertical) line passing through the point $P$. The point $P$ will follow a helix, and the point $Q$ --which lies in the plane of the circle-- traces out the helical cycloid in space. While the path that $P$ (and $Q$) takes is decidedly different after curling than before, the &lt;em&gt;displacement vector&lt;/em&gt; between $P$ and $Q$ in the plane of the circle is the same throughout. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Before curling, if the circle of radius $r$ rolls along a horizontal track in the $uv$-plane, then we have $P=(r t, 0)$, and the point $Q$ traces out the standard cycloid:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xD;&#xA;u &amp;amp;= r ( t - \sin t ) \\&#xD;&#xA;v &amp;amp;= r ( 1 - \cos t )&#xD;&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The displacement vector from $P$ to $Q$ at time $t$ is given by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$d := PQ = [m, n] = r[-\sin t, 1 - \cos t ] = - 2 r \sin\frac{t}{2} \; [\cos\frac{t}{2},-\sin\frac{t}{2}]$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rotating the plane through an angle, say, $\theta := \rm{atan\frac{c}{a}}$, and writing $b$ for $\sqrt{a^2+c^2}$, the circle's point of tangency along the tilted track is given by ...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$P_2 = (r t \cos\theta,r t \sin\theta) = \left( \frac{r t a}{b}, \frac{r t c}{b}\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;... and the displacement vector, $d_2$, to the point $Q_2$ tracing out the tilted cycloid is given by ...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xD;&#xA;d_2 :&amp;amp;= [m \cos\theta - n \sin \theta, m \sin\theta + n \cos \theta ] \\&#xD;&#xA;&amp;amp;= \frac{1}{b} [a m - c n, c m + a n] \\&#xD;&#xA;&amp;amp;= -\frac{2r}{b}\sin\frac{t}{2}\left[\cos\left(\theta-\frac{t}{2}\right), \sin\left(\theta - \frac{t}{2} \right) \right] \\&#xD;&#xA;\end{align}$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the fun part: Curl the $(u,v)$-plane around a cylinder of radius $a$, such that the $uv$-origin aligns with the $xyz$-point $(a,0,0)$ and the $v$-axis runs parallel to the $z$ axis. The curled, tilted track will coincide with a helix. The horizontal distance travelled by $P_2$ in the plane becomes the length of horizontal circular &lt;em&gt;arc&lt;/em&gt; travelled by $P_3$ around the cylinder; upon dividing by the radius, $a$, of the cylinder, this length becomes the &lt;em&gt;angular&lt;/em&gt; &quot;distance&quot; --$s := \frac{rt}{b}$-- travelled by $P_3$; the vertical distances match. Therefore: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$P_3 = \left(a\cos s, a\sin s, \frac{rtc}{b} \right)=(a\cos s,a\sin s,c s)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the displacement vector: The $u$ direction of the tangent plane coincides with the horizontal vector tangent to the cylinder at $P_3$; the $v$ direction coincides with the $z$ direction. Thus, the transformation from $uv$-coordinates to $xyz$-coordinates is given by &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$[1,0]\to[-\sin s, \cos s, 0]\hspace{0.5in}[0,1]\to[0,0,1]$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The image, $d_3$, of the displacement vector $d_2$, then, is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$d_3 = -\frac{2r}{b}\sin\frac{t}{2}\left[-\sin s \cos\left(\theta-\frac{t}{2}\right), \cos s \cos\left(\theta-\frac{t}{2}\right), \sin\left(\theta - \frac{t}{2} \right) \right] $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and the path of $Q_3$, which traces the helical cycloid, is given by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;Q_3 = P_3 + d_3 = \left\{&#xD;&#xA;\begin{align}&#xD;&#xA;x &amp;amp;= a \cos s &amp;amp;+ \frac{2 r}{b} \sin s \sin\frac{bs}{2r} \cos\left(\theta-\frac{bs}{2r}\right)\\&#xD;&#xA;y &amp;amp;= a \sin s &amp;amp;- \frac{2 r}{b} \cos s \sin\frac{bs}{2r} \cos\left(\theta-\frac{bs}{2r}\right)\\&#xD;&#xA;z &amp;amp;= c s &amp;amp;- \frac{2 r}{b} \sin\frac{b s}{2r}\sin\left(\theta-\frac{bs}{2r}\right)&#xD;&#xA;\end{align}&#xD;&#xA;\right.&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a picture with $a=c=1$ and $r=1/2$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/vIRtM.jpg&quot; alt=&quot;Helical Cycloid&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: The above does not simply curl the planar cycloid around the cylinder. Since&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x^2 + y^2 = a^2 + \frac{4r^2}{b^2} \sin^2\frac{bs}{2r} \cos^2\left(\theta-\frac{bs}{2r}\right) \ge a^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;we see that most of the helical cycloid lies outside the surface of the cylinder.&lt;/p&gt;&#xA;" OwnerUserId="409" LastActivityDate="2010-12-25T08:12:52.537" />
  <row Id="15498" PostTypeId="2" ParentId="15486" CreationDate="2010-12-25T09:17:41.053" Score="3" ViewCount="" Body="&lt;p&gt;As I mentioned in &lt;a href=&quot;http://math.stackexchange.com/questions/9978/real-analysis-continuity&quot;&gt;another question&lt;/a&gt;, I tend to intuit compactness as a sort of &quot;super closedness.&quot;  There are a few reasons for this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, in &lt;a href=&quot;http://en.wikipedia.org/wiki/Hausdorff_space&quot; rel=&quot;nofollow&quot;&gt;Hausdorff spaces&lt;/a&gt;, every compact subset will automatically be closed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second, in &lt;a href=&quot;http://en.wikipedia.org/wiki/First-countable_space&quot; rel=&quot;nofollow&quot;&gt;first-countable&lt;/a&gt; spaces, one can think about &quot;closedness&quot; as &quot;closed under limit operations.&quot;  That is, a set $E$ is closed if and only if it contains the limits of all sequences in $E$.  By analogy, every &lt;em&gt;compact&lt;/em&gt; subset of a first-countable space is &lt;a href=&quot;http://en.wikipedia.org/wiki/Sequential_compactness&quot; rel=&quot;nofollow&quot;&gt;sequentially compact&lt;/a&gt; -- that is, every sequence has a convergent subsequence.  (Note, however, that the converse of this is not generally true in first-countable spaces.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Third, in metric spaces, compactness is equivalent to being complete and totally bounded.  This can be viewed as a generalization of the Heine-Borel Theorem which classifies the compact sets of $\mathbb{R}^n$.  In such a scheme, completeness corresponds to closedness, and total boundedness to boundedness.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that metric spaces are automatically Hausdorff and first-countable, so all of the above certainly applies to metric spaces.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;For general topological spaces (without any assumption of &quot;Hausdorffness&quot; or first countability), one can also intuit compact topological spaces as those with relatively few open sets.  Indeed, if a set $X$ is compact under a given topology, then it will also be compact under any weaker (coarser) topology.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;EDIT: To add to Prometheus' answer, I would also say that compactness is a property that plays especially well with continuity (&lt;a href=&quot;http://en.wikipedia.org/wiki/Extreme_value_theorem&quot; rel=&quot;nofollow&quot;&gt;Extreme-Value Theorem&lt;/a&gt;, &lt;a href=&quot;http://en.wikipedia.org/wiki/Heine-Cantor_theorem&quot; rel=&quot;nofollow&quot;&gt;Heine-Cantor Theorem&lt;/a&gt;) and convergence of functions (&lt;a href=&quot;http://en.wikipedia.org/wiki/Dini%27s_theorem&quot; rel=&quot;nofollow&quot;&gt;Dini's Theorem&lt;/a&gt;).  As Prometheus said, this is in part because of its role in reducing global problems to local ones -- or, perhaps, infinite ones to finite ones.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I'm thinking particularly of the proof of Heine-Cantor which relies on the fact that the infimum of a finite number of positive distances is positive -- whereas an infinite number of positive distances may have an infimum of zero.)&lt;/p&gt;&#xA;" OwnerUserId="640" LastEditorUserId="640" LastEditorDisplayName="" LastEditDate="2010-12-25T09:27:51.997" LastActivityDate="2010-12-25T09:27:51.997" CommentCount="2" />
  <row Id="15499" PostTypeId="1" CreationDate="2010-12-25T09:53:40.260" Score="6" ViewCount="151" Body="&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since I did not quite get the responses I would have liked when I asked this question four months ago, let me reformulate it slightly:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What are some examples of $\text{Isom}(M)$ and $\text{Conf}(M)$?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For example, Aaron mentions in his (very helpful) answer that $\text{Isom}(\mathbb{S}^n) \cong O(n+1)$.  What about hyperbolic n-space?  Or the n-torus?  The more examples the better.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For precision, I am hoping that we can express $\text{Isom}(M)$ or $\text{Conf}(M)$ as some sort of &quot;recognizable&quot; Lie group, by which I mean a product, quotient, and/or connected sum of linear groups, euclidean spaces, or spheres.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Original Question:&lt;/strong&gt; (What are some examples of automorphism groups of manifolds which turn out to be Lie groups?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I recently read that the group of diffeomorphisms of a smooth manifold which preserve some sort of geometric structure (e.g. Riemannian structure, conformal structure, etc.) frequently turn out to be a Lie group.  What are some examples of this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've read about the euclidean group $E(n)$ which are the isometries of $\mathbb{R}^n$, and also about the conformal automorphisms of the complex plane, upper half plane, and unit disc.  What others are there?  Do the groups Diff(M), Iso(M), etc. frequently turn out to be something recognizable?  What are the isometry groups or conformal groups of n-spheres, say, or some common 2-manifolds?&lt;/p&gt;&#xA;" OwnerUserId="640" LastEditorUserId="640" LastEditorDisplayName="" LastEditDate="2011-04-10T21:21:01.147" LastActivityDate="2011-04-14T08:11:43.337" Title="What are some examples of $\text{Isom}(M)$ and $\text{Conf}(M)$?" Tags="&lt;reference-request&gt;&lt;differential-geometry&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="1" />
  <row Id="15501" PostTypeId="2" ParentId="15486" CreationDate="2010-12-25T10:05:29.183" Score="0" ViewCount="" Body="&lt;p&gt;For me, the compactness of a topological space means that it has &lt;em&gt;enough points&lt;/em&gt; to provide exact solutions to continuous equations. More precisely,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;compactness = Any equation that can be approximated by a consistent system of $\leq$ inequalities of continuous functions has a solution.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For instance, being a solution to the equation $x^2 = 2$ is equivalent to being a simultaneous solution to the infinite system of inequalities $\lbrace |x^2-2| \leq \frac 1 n \rbrace_{n\in\mathbb N}$. This system is consistent in the sense that every &lt;em&gt;finite&lt;/em&gt; subsystem has solutions; after all, we can approximate $\sqrt 2$ by rational numbers arbitrarily well. Then, it is the &lt;em&gt;compactness&lt;/em&gt; of bounded intervals of &lt;em&gt;real numbers&lt;/em&gt; that tells us that there exists a simultaneous solution of all of these inequalities, and hence of the equation itself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In contrast, trying to encircle $\sqrt 2$ as a &lt;em&gt;rational number&lt;/em&gt; fails, you can enclose it by smaller and smaller intervals, but when you try to catch it with an infinite number of intervals, it &quot;falls through the cracks&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The principle discussed above is also known as the &lt;a href=&quot;http://en.wikipedia.org/wiki/Finite_intersection_property&quot; rel=&quot;nofollow&quot;&gt;finite intersection property&lt;/a&gt;. The inequalities, or &quot;conditions&quot;, define a collection of closed sets and the exact solution we seek is contained in the intersection of infinitely many closed sets. If every finite intersection of these closed sets is non-empty, i.e. if every finite collection of conditions can be fulfilled, then one could expect that all of the conditions can be satisfied simultaneously. This is precisely what compactness ensures.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other examples:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Peano's theorem about the existence of solutions to ordinary differential equations relies on compactness to construct a good candidate.&lt;/li&gt;&#xA;&lt;li&gt;A commonly used technique for showing the existence of solution for partial differential equations is the calculus variations. In this context, the weak compactness of the unit ball in a Hilbert space is often used to get a weak solution to the PDE which is then upgraded to a strong solution.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="434" LastEditorUserId="434" LastEditorDisplayName="" LastEditDate="2010-12-26T19:36:00.803" LastActivityDate="2010-12-26T19:36:00.803" CommentCount="5" />
  <row Id="15502" PostTypeId="2" ParentId="9999" CreationDate="2010-12-25T11:30:59.483" Score="0" ViewCount="" Body="&lt;p&gt;To answer the other direction of the question, i.e. to show that $F_n, L_n$ are fundamental for the following equations:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$L_{2n}^2 - 5 F_{2n}^2 = 4$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$L_{2n+1}^2 - 5 F_{2n+1}^2 = -4$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;we do not need to invoke the full theory of Pell equations, algebraic extensions, continued fraction, etc.. since we have some concrete parameters at hand. Lets turn to the Pell equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$y^2 - a x^2 = 1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $(x_1, y_1)$ and $(x_2,y_2)$ be solutions of this equation. We see that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(y_1^2 - a x_1^2)*(y_2^2 - a x_2^2) = y_1^2 * y_2^2 - a y_1^2 x_2^2 - a x_1^2 y_2^2 + a^2 x_1^2 * x_1^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$= (y_1 y_2)^2 + 2 a y_1 y_2 x_1 x_2 + (a x_1 x_2)^2 - a ((y_1 x_2)^2 + 2 y_1 x_2 x_1 y_2 +(x_1 y_2)^2)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$= (y_1 y_2 + a x_1 x_2)^2 - a (y_1 x_2 + x_1 y_2)^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so we can go to a third solution $(x_3,y_3)$ from the given two solutions as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x_3 := y_1 x_2 + x_1 y_2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$y_3 := y_1 y_2 + a x_1 x_2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lets call this operations $\times$. It spans an abelian group with the following unit and inverse:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(0,1) \times (x,y) = (x,y)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(-x,y) \times (x,y) = (0,1)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The group has a t-norm $|(x,y)|$ in $Q$. We define for $x&amp;gt;0$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$|(x,y)| := 1/x$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can show that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$|p \times q| \leq |p| |q|$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The inverse of the t-norm can be used to show that if $p_0$ is the smallest non-unit solution than any non-unit solution $q$ has the form $p_0^n = p_0 \times ... \times p_0$. Just observe that for $q\neq (0,1)$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$q=p_0 \lor |p_0^{-1} \times q|^{-1} \leq {|q|^{-1} \over |p_0|^{-1}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When $|p_0|^{-1} \neq 1$ this gives a decreasing terminating chain. Now exhibiting $p_0$ we should be done more or less.&lt;/p&gt;&#xA;" OwnerUserId="4414" LastEditorUserId="4414" LastEditorDisplayName="" LastEditDate="2010-12-25T11:40:52.717" LastActivityDate="2010-12-25T11:40:52.717" CommentCount="3" />
  <row Id="15503" PostTypeId="1" AcceptedAnswerId="15544" CreationDate="2010-12-25T12:47:42.630" Score="1" ViewCount="109" Body="&lt;p&gt;I have a functor $F\colon \mathbf{Rng}\to\mathbf{Grp}$, and a correspondence on objects which assigns to every group $F(R)$ a suitable subgroup $G_o(R)\subseteq F(R)$. Is there a way to turn $G$ into a functor, defining $G_o(R)\to G_o(S)$ via the maps I have between $F(R)$ and $F(S)$?&#xA;$$&#xD;&#xA;\begin{array}{ccc}&#xD;&#xA;F(R) &amp;amp;\to^{F(f)}&amp;amp; F(S) \\\&#xD;&#xA;\uparrow_{\iota_R} &amp;amp;&amp;amp; \uparrow_{\iota_S}\\\&#xD;&#xA;G_o(R) &amp;amp; &amp;amp;G_o(S)&#xD;&#xA;\end{array}&#xD;&#xA;$$&#xA;In this diagram vertical arrows are simply the existing injections. I thought to define $G_o(R)\to G_o(S)$ taking the obvious left inverse going from the copy of $G_o(-)$ into $F(-)$ to $G_o(-)$, (call $\pi_S$ this map, thn $G(R)\to G(S)$ is defined by $\pi_s\circ F(f)\circ \iota_R$) but I'm not sure this is going to work...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you think you'll find it useful, $F$ is the functor which assigns to every ring its group of unities.&lt;/p&gt;&#xA;" OwnerUserId="685" LastActivityDate="2010-12-26T08:59:12.163" Title="Defining &quot;subfunctors&quot;" Tags="&lt;category-theory&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="15504" PostTypeId="1" AcceptedAnswerId="15506" CreationDate="2010-12-25T16:27:36.727" Score="3" ViewCount="167" Body="&lt;p&gt;In my course there's a paragraph: Taylor polynomial with Lagrange remainder,&lt;br&gt;&#xA;The paragraph starts with a theorem (I left out the constraints):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ ( \exists \theta \in ]0,1[)(f(a +h) = T_{f,a,n}(a + h) + \frac{h^{n+1}}{(n+1)!}D^{n+1}f(a + \theta h)) $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With $f$ an  $R-R$ function and $a$ and $a + h$ defined over an interval $I$.&lt;br&gt;&#xA;And the only other thing in the paragraph is a proof of the above theorem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Before I'm starting with the proof, could someone please explain the above theorem in &lt;em&gt;human language&lt;/em&gt;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand what a Taylor polynomial is, and what it's good for, but I can't turn this into anything that makes sense...&lt;/p&gt;&#xA;" OwnerUserId="5109" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-12-25T17:08:51.090" LastActivityDate="2010-12-26T10:04:38.440" Title="Taylor polynomial with Lagrange remainder" Tags="&lt;calculus&gt;&lt;analysis&gt;&lt;sequences-and-series&gt;&lt;taylor-expansion&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="15505" PostTypeId="1" AcceptedAnswerId="15599" CreationDate="2010-12-25T16:30:34.860" Score="4" ViewCount="229" Body="&lt;p&gt;Show that $(n-r) \binom{n+r-1}{r} \binom{n}{r} = n \binom{n+r-1}{2r} \binom{2r}{r}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the LHS $\binom{n+r-1}{r}$ counts the number of ways of selecting $r$ objects from a set of size $n$ where order is not significant and repetitions are allowed. So you have $n$ people you form $r$ teams and select $r$ captains and select $(n-r)$ players.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The RHS divides up a team into 2 sets? &lt;/p&gt;&#xA;" OwnerUserId="2548" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-12-25T20:16:39.867" LastActivityDate="2010-12-27T00:35:22.657" Title="Combinatorial Identity" Tags="&lt;combinatorics&gt;&lt;binomial-coefficients&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="15506" PostTypeId="2" ParentId="15504" CreationDate="2010-12-25T17:02:22.470" Score="2" ViewCount="" Body="&lt;p&gt;I think what you've got there is the $n$th-degree Taylor polynomial for a function $f$ centered at $a$ and being evaluated at $x=a+h$.  Basically, the error term $\frac{h^{n+1}}{(n+1)!}D^{n+1}f(a + \theta h))$ looks like the last term in the $(n+1)$st degree Taylor polynomial would, except that instead of evaluating the $(n+1)$st derivative at $a$, it's evaluated somewhere between $a$ and $a+h$ ($\theta$ measures how much of the way from $a$ to $a+h$).&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-12-25T17:02:22.470" />
  <row Id="15507" PostTypeId="1" CreationDate="2010-12-25T17:22:17.720" Score="19" ViewCount="544" Body="&lt;p&gt;We did in class $x^2+y^2$, which was easy, and we had for homework $2x^2+2xy+3y^2$, which I did (its values (if not square) must be divisible by form primes, or of the form $x^2+5y^2$, and clearly they can't all be the latter!).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just to test my method I took the middle coefficient down by one - and ran into trouble...&lt;/p&gt;&#xA;" OwnerUserId="4946" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-29T19:07:17.030" LastActivityDate="2011-03-30T15:02:20.443" Title="Elementary proof that $2x^2+xy+3y^2$ represents infinitely many primes?" Tags="&lt;number-theory&gt;" AnswerCount="2" CommentCount="8" FavoriteCount="4" />
  <row Id="15508" PostTypeId="1" CreationDate="2010-12-25T17:57:51.353" Score="3" ViewCount="177" Body="&lt;p&gt;Hey.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I got a quick question on riemannian geometry. I'm not quite sure whether this is the right place to ask this question, since it might be a rather elementary one from a research point of view. Nontheless I think it might be too advanced for a typical homework question (which by the way it isn't) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So let $M^m$ be a manifold embedded into euclidean space $R^n$. Let L($\gamma$) be the &#xA;the lenght of a smooth curve $\gamma$: [0,1] --&gt; $R^n$ which is the number &#xA;$\int_0^1 \ |d/dt \gamma(t)| \ \mathrm{d}t$ . Define the distance function&#xA;d: M x M --&gt; [0,$\infty$) by d(p,q):=inf L ($\gamma$). The infimum is taken over all smooth paths connecting p and q. This formula defines a metric on M. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wan't to show that &#xA;$\forall$ $p_0 \in M^m \exists U \subset M^m $ open neighborhood of $p_0$ such that (1-$\epsilon$) |p-q| &amp;lt; d(p,q) &amp;lt; (1+$\epsilon$) |p-q| .&#xA;Note that |.| denotes the usual euclidean distance on $M^m$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we fix an arbitrary $p_0 \in M^m$. Without loss of generality (Translations and rotations preserves the length of a curve) we assume that &#xA;$p_0$ = 0 and $T_{p_{0}}M$ = $R^m$ x {0} .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now comes the step that I don't understand. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now there's a smooth function f: $\Omega$ --&gt; $R^{n-m}$ defined on an open neighborhood $\Omega \subset R^m$ of the origin such that &#xA;{ (x,y) $\in R^m x R^{n-m}$ | x $\in \Omega$ , y=f(x) } $\subset M^m$  , f(0)=0, df(0)=0 &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I don't really see where this is coming from. &#xA;It seems to me that it might be some combination of a characterisation of manifolds&#xA;and tangent spaces we introduced, which is the following: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $M^m \subset R^n$, p $\in M^m$ &quot;smooth manifold&quot; , then&#xA;$\exists U \subset R^n$ and f: U --&gt; $R^{n-m}$ smooth such that p $\in$ U , df(q) surjective for all q $\in$ U $\cap$ M and U $\cap$ M = $f^{-1}$(0). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone help me?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance.&lt;/p&gt;&#xA;" OwnerUserId="5110" LastEditorUserId="1543" LastEditorDisplayName="" LastEditDate="2011-04-21T22:10:16.993" LastActivityDate="2011-05-21T23:00:03.917" Title="Quick question on Riemannian geometry" Tags="&lt;differential-geometry&gt;&lt;riemannian-geometry&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="15509" PostTypeId="1" AcceptedAnswerId="15511" CreationDate="2010-12-25T18:31:45.457" Score="3" ViewCount="124" Body="&lt;p&gt;For $n \geq 4$, show that $\binom{\binom{n}{2}}{2} = 3 \binom{n}{3}+ 3 \binom{n}{4}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;LHS: So we have a set of $\binom{n}{2}$ elements, and we are choosing a $2$ element subset.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;RHS: We are choosing a $3$ element subset and a $4$ element subset (each from a set of $n$ elements). But we multiply by $3$ by the multiplication principle for some reason. &lt;/p&gt;&#xA;" OwnerUserId="2548" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-12-25T20:13:13.113" LastActivityDate="2010-12-25T20:13:13.113" Title="Combinatorial Proof" Tags="&lt;combinatorics&gt;&lt;binomial-coefficients&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="15510" PostTypeId="1" CreationDate="2010-12-25T18:39:37.920" Score="14" ViewCount="389" Body="&lt;p&gt;What is the shortest string $S$ over an alphabet of size $n$, such that every permutation of the alphabet is a substring of $S$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I thought of this problem while reading a open problem on &lt;a href=&quot;http://garden.irmacs.sfu.ca/?q=op/smallest_universal_supersequence&quot; rel=&quot;nofollow&quot;&gt;shortest supersequence of all permutations&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="96" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-12-25T21:39:50.200" LastActivityDate="2010-12-26T06:59:19.260" Title="What is the shortest string that contains all permutations of an alphabet?" Tags="&lt;combinatorics&gt;&lt;permutations&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="2" />
  <row Id="15511" PostTypeId="2" ParentId="15509" CreationDate="2010-12-25T18:50:31.297" Score="7" ViewCount="" Body="&lt;p&gt;LHS: The $\binom{n}{2}$ is the number of pairs you can form of n distinct elements, so the LHS counts the number of ways to choose two distinct pairs. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;RHS: Notice that you can choose two pairs that have a common element (but only one). If the two pairs are disjoint, then you need to choose four elements and then ask how you pair them. If the pairs have a common element, then you need to choose only three elements and then choose which is the common element.&lt;/p&gt;&#xA;" OwnerUserId="2125" LastEditorUserId="2125" LastEditorDisplayName="" LastEditDate="2010-12-25T19:58:56.420" LastActivityDate="2010-12-25T19:58:56.420" CommentCount="2" />
  <row Id="15512" PostTypeId="2" ParentId="15510" CreationDate="2010-12-25T19:10:20.697" Score="3" ViewCount="" Body="&lt;p&gt;Here are some thoughts: given a permutation $abcde$, you can generate all its cyclic rotations with no additional cost: $abcdeabcd$. Then you'd want to capitalize of the largest number symbols, so we repeat the deepest symbol, in this case $a$, complete the permutation with $e$, and generate all cyclic rotations: $abcdeabcdaebcda$. So far we've generated all cyclic rotations of $abcde$ and $bcdae$. Notice that the first $n-1$ characters are rotated. We continue to generate all cyclic rotations of $cdabe,dabce$, and then we have to bring back one symbol shallower: $\ldots dabcedabcadebcad$. So we moved from $dabce$ to $bcade$. This corresponds to rotation of the middle $n-2$ characters, followed by the usual rotation of $n-1$ characters. Now we get back to the earlier operation (rotation of $n-1$), applying occasionally the new rotation, until we get stuck; we would then need to do a rotation of the middle $n-3$ characters, and so on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would conjecture that the preceding scheme is optimal; the diligent reader can recursively calculate its length.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-12-25T19:10:20.697" />
  <row Id="15513" PostTypeId="2" ParentId="15510" CreationDate="2010-12-25T20:29:45.660" Score="4" ViewCount="" Body="&lt;p&gt;[NB] This argument as stated is incorrect, and the error is pointed out by Yuval in the comments below. One can indeed use the De Bruijn graph to get all permutations, exactly as I described below, but as Yuval's comment implies, one will also get plenty of non-permutations. The length of the string $S$ is not what my argument says  it is, and the problem of the MINIMAL LENGTH sequence seems to require a different argument. &lt;/p&gt;&#xA;&#xA;&lt;h2&gt;This is very dangerous because all those undeserved points might cause someone to dismiss what might be an interesting and difficult question.&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Since there are $n!$ permutations of $n$ symbols, the string $S$ must have length&#xA;at least $n!+n-1$. That is, there must be $n!$ starting positions for a  permutation, and the last starting position must be succeeded by $n-1$ symbols. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is in fact such a string, which is obtained from an Eulerian path in the De Bruijn graph $B(n,n)$. This is a directed graph whose vertices are all strings of $n-1$ distinct symbols, with  a directed edge from $u$ to $v$ if there is a symbol $s$ such that $v$ has the form $u_1s$, where $u_1$ is $u$ with the left-most symbol deleted. We label the edge with $s$. Every vertex has the same in-degree and out degree so the graph has an eulerian path. The labels of the edges on any such path, together with the string where the path ends, gives a minimal sequence of the kind you want. See this Wikipedia article &lt;a href=&quot;http://en.wikipedia.org/wiki/De_Bruijn_sequence&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/De_Bruijn_sequence&lt;/a&gt; for terminology, references and a nice explanation. &lt;/p&gt;&#xA;" OwnerUserId="746" LastEditorUserId="746" LastEditorDisplayName="" LastEditDate="2010-12-26T06:59:19.260" LastActivityDate="2010-12-26T06:59:19.260" CommentCount="5" />
  <row Id="15514" PostTypeId="1" AcceptedAnswerId="15517" CreationDate="2010-12-25T20:38:32.623" Score="3" ViewCount="325" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is, how can I solve the following derivative question?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$y = \sin(\arctan x) + \tan(\arcsin x)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance&lt;/p&gt;&#xA;" OwnerUserId="3669" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-26T00:13:27.530" LastActivityDate="2010-12-26T04:03:54.567" Title="How can I find the derivative of $y = \sin(\arctan x) + \tan(\arcsin x)$?" Tags="&lt;calculus&gt;&lt;trigonometry&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="15515" PostTypeId="1" AcceptedAnswerId="15529" CreationDate="2010-12-25T20:44:52.837" Score="2" ViewCount="92" Body="&lt;p&gt;Show that $\binom{2n}{2} = 2 \binom{n}{2}+n^2$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;LHS: This is the number of pairs of $2n$ distinct elements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;RHS: We can rewrite this as $2 \binom{n}{2}+ \binom{n}{1} \binom{n}{1}$. So you can divide up the $2n$ distinct elements into two subsets of $n$ elements. Then pick an element from each of these subsets to form a pair. Or you can just choose $2$ elements from the $n$ element subset (and multiply by $2$ since order matters).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this the correct idea? &lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2010-12-25T23:14:27.703" Title="Combinatorial Identity and Counting" Tags="&lt;combinatorics&gt;&lt;binomial-coefficients&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="0" />
  <row Id="15516" PostTypeId="2" ParentId="15514" CreationDate="2010-12-25T20:53:40.040" Score="0" ViewCount="" Body="&lt;p&gt;What's your intended result?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A derivative, i.e. $y&amp;#39;$? Use the chain rule.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solve for $x$? Use $cos^2(x) = 1 - sin^2(x)$ and $tan(x) = sin(x)/cos(x)$ to simplify&lt;/p&gt;&#xA;" OwnerUserId="917" LastActivityDate="2010-12-25T20:53:40.040" />
  <row Id="15517" PostTypeId="2" ParentId="15514" CreationDate="2010-12-25T20:57:28.200" Score="8" ViewCount="" Body="&lt;p&gt;It's not clear to me where the derivative comes in to your question, but to simplify $\sin(\arctan x)+\tan(\arcsin x)$, I'd suggest the following: since arctangent and arcsine both result in angles in the first or fourth quadrants (between $-\frac{\pi}{2}$ and $\frac{\pi}{2}$), draw two diagrams showing those two quadrants, one showing an angle labeled $\arctan x$, the other showing an angle labeled $\arcsin x$.  Draw right triangles to the $x$-axis in each diagram and label the legs in terms of $x$; for example, if the angle is $\arctan x$, that suggests that the leg opposite the angle is $x$ and the leg adjacent to the angle is $1$ (make sure that the labeling makes sense for positive and negative values of $x$).  Use these diagrams to find simpler algebraic expressions for $\sin(\arctan x)$ and $\tan(\arcsin x)$.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;edit&lt;/em&gt;:  Here's a bit more detail on part of the problem.  If you're still stuck, please comment and let me know.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Start by drawing an angle in the proper quadrant(s).  $\arctan x$ is an angle in the first or fourth quadrant, so:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/HWKLT.png&quot; alt=&quot;diagram 1&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;($\tan^{-1}(x)$ is another notation for $\arctan x$.)  Either the black (1st quadrant) or the gray (4th quadrant) triangle could be the proper triangle, depending on whether $x$ is positive or negative.  Since the angle is $\arctan x$, the tangent of that angle is $x$.  The easiest way to make that happen is for the leg opposite the angle to have length $x$ and the leg adjacent to the angle to have length $1$.  Note that if $x$ is negative, the gray triangle is the correct triangle and in the sense of directed-distances in the coordinate plane, that leg should have &quot;negative&quot; length.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/JyHko.png&quot; alt=&quot;diagram 2&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Knowing the lengths of 2 sides of a right triangle, you can use the Pythagorean theorem to find the length of the third side.  In this case, the unknown side is the hypotenuse and the directed-distance length of the hypotenuse should always be positive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/MjwgA.png&quot; alt=&quot;diagram 3&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With all three sides of the triangle labeled, you can use the triangle to write an expression for $\sin(\arctan x)$, which is the sine of the marked angle, so the ratio of the length of the opposite leg to the length of the hypotenuse, $\frac{x}{\sqrt{x^2+1}}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The procedure for the $\tan(\arcsin x)$ term is similar, but the labeling of the sides of the triangle is different.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you were working with, for example, arccosine, the two triangles would be in the first and second quadrants because the range of the arccosine function is $0\le\arccos x\le\pi$.&lt;/p&gt;&#xA;" OwnerUserId="72" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-12-25T22:23:11.743" LastActivityDate="2010-12-25T22:23:11.743" CommentCount="2" />
  <row Id="15519" PostTypeId="2" ParentId="15162" CreationDate="2010-12-25T21:29:36.747" Score="2" ViewCount="" Body="&lt;p&gt;Asaf's answer explains that there is no set of all cardinals, and that under the axiom of choice, the cardinals are well-ordered, and so it is impossible to have a homomorphism as you want.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What remains is to see whether it is possible, in some models where the axiom of choice fails, to have a homomorphism $p$ as you suggest, with range a collection of sets, no two of which have the same cardinality. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is actually possible. In fact, it holds in models where the &lt;a href=&quot;http://en.wikipedia.org/wiki/Axiom_of_determinacy&quot; rel=&quot;nofollow&quot;&gt;axiom of determinacy&lt;/a&gt; is true, but the argument is sophisticated, and follows from recent work that started with results of Alexander Kechris and Scot Adams on Borel equivalence relations. The original reference is &quot;Linear algebraic groups and countable Borel equivalence relations&quot;, Journal of the American Mathematical Society, Vol 13 (4), (2000) 909-943. This paper is explicitly about Borel equivalence relations, but their arguments hold in more general contexts, assuming that all sets of reals are &quot;sufficiently nice&quot; in ways that the axiom of determinacy guarantees. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Benjamin Miller suggested the following general statement (I believe in this specific form the result is due to him):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If $\le$ is an analytic (i.e., $\Sigma^1_1$) partial order on a Hausdorff space $X$, then there is a sequence of countable analytic equivalence relations $(E_x)_{x \in X}$ such that $$ x &amp;lt; y \Longrightarrow | X/E_x | &amp;lt; | X/ E_x \sqcup X/ E_y | = | X/ E_x \times X/E_y | = | X/ E_y | $$ for all $x, y \in X$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here, &lt;em&gt;analytic&lt;/em&gt; means the continuous image of a Borel set. Without choice, an inequality $|A|&amp;lt;|B|$ simply means that there is an injection from $A$ into $B$ but not one from $B$ into $A$. $A\sqcup B$ is the disjoint union of $A$ and $B$; this is the canonical set associated to the cardinal sum $|A|+|B|$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason why this solves the problem is that the ordering in the space $A/\sim$ is analytic in this sense, and so we can assign to each class $[a]$ the set $p([a])=X/E_{[a]}$ for $(E_{[a]})_{[a]\in A/\sim}$ a sequence of countable equivalence relations as granted by the statement above. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What follows is Ben's (very high level) sketch:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) In all of these results, the key is the use of what we call &quot;rigidity arguments&quot;, which provide us with aperiodic countable Borel equivalence relations $F_x$ equipped with ergodic, invariant Borel probability measures $m_x$ with the property that for all distinct $x, y \in X$, the pair $(F_x, m_x)$ is $F_y$-ergodic (i.e., if $C$ is a set of full $m_x$-measure and $f$ is an $m_x$-measurable homomorphism from $F_x\upharpoonright C$ to $F_y$, then there is a subset $B$ of $C$ with $m_x(C \setminus B) = 0$ such that $f[B]$ is contained in a single $F_y$-class).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) Let $E_x$ denote the disjoint union of the equivalence relations of the form $F_y$, for $y \le x$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(3) We have that $x \le y$ implies that the identity map is a reduction of $E_x$ to $E_y$, i.e., it gives us an injection of $X/E_x$ into $X/E_y$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(4) If $x \le y$ is false, however, then any homomorphism from $E_x$ to $E_y$ would restrict down to a homomorphism from $F_x$ to $E_y$. Ergodicity of $m_x$ would then give a set $C$ of full $m_x$-measure and some $z \le y$ such that the map in question is a homomorphism from $F_x\upharpoonright C$ to $F_z$. As $x\ne z$, the map must therefore concentrate $m_x$-almost everywhere on a single $F_z$-class, and this implies that the original map could not have been a reduction. Thus $X/E_x$ cannot inject into $X/E_y$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(5) From the construction one can directly argue that (for any $x$) $X/E_x \times X/E_x$ is in bijection with $X/E_x$. From this, it is straightforward to check (using &lt;a href=&quot;http://en.wikipedia.org/wiki/Cantor%E2%80%93Bernstein%E2%80%93Schroeder_theorem&quot; rel=&quot;nofollow&quot;&gt;Schroeder-Bernstein&lt;/a&gt;) that if $x\le y$ then $X/E_x\sqcup X/E_y$, $X/E_x\times X/E_y$, and $X/E_y$ are all in bijection with one another. &lt;/p&gt;&#xA;" OwnerUserId="462" LastActivityDate="2010-12-25T21:29:36.747" CommentCount="1" />
  <row Id="15520" PostTypeId="1" AcceptedAnswerId="15733" CreationDate="2010-12-25T21:40:11.670" Score="3" ViewCount="138" Body="&lt;p&gt;First, define a probability distribution with density:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align*}&#xA;\newcommand\bm\boldsymbol&#xA;\newcommand{\mbl}{\mathbf{l}}&#xA;f(\mbl \mid \bm \alpha)&#xA;= \frac{1}{B(\bm \alpha + \bm 1)}\exp\left(\mbl \cdot \bm\alpha\right)&#xA;\end{align*}&#xA;where&#xA;$\mathbf l \in (-\infty,0]^k, \sum_i\exp(l_i) = 1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm having trouble finishing this integral (to calculate the convolution):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align*}&#xA;\newcommand{\mba}{\mathbf{a}}&#xA;\newcommand{\mbb}{\mathbf{b}}&#xA;\newcommand{\mbl}{\mathbf{l}}&#xA;\newcommand{\mbp}{\mathbf{p}}&#xA;\newcommand{\mbv}{\mathbf{v}}&#xA;\newcommand{\mbx}{\mathbf{x}}&#xA;f(\mbl \mid \mba, \mbb)&#xA;&amp;amp;= \int_{l_1}^0 \dotsi \int_{l_k}^0 f(\mbx \mid \mba)f(\mbl-\mbx \mid \mbb)d\mbx \newline&#xA;&amp;amp;\propto \int_{l_1}^0 \dotsi \int_{l_k}^0 \exp(\mbx \cdot \mba + (\mbl-\mbx) \cdot \mbb)d\mbx \newline&#xA;&amp;amp;= \int_{l_1}^0 \dotsi \int_{l_k}^0 \exp(\mbx \cdot (\mba - \mbb) + \mbl \cdot \mbb)d\mbx \newline&#xA;&amp;amp;= \prod_{i=1}^k\frac{\exp(a_il_i) - \exp(b_il_i)}{a_i-b_i} ???&#xA;\end{align*}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After looking at mpiktas's answer, the final line should have been&#xA;\begin{align*}&#xA;&amp;amp;= \prod_{i=1}^k\frac{\exp(b_il_i) - \exp(a_il_i)}{a_i-b_i}&#xA;\end{align*}&#xA;which ensures that the terms are all positive.&lt;/p&gt;&#xA;" OwnerUserId="774" LastEditorUserId="774" LastEditorDisplayName="" LastEditDate="2010-12-29T12:03:57.223" LastActivityDate="2010-12-29T12:03:57.223" Title="Convolution of probability densities. Having trouble integrating over vector" Tags="&lt;probability-theory&gt;&lt;integral&gt;" AnswerCount="1" CommentCount="0" />
  <row Id="15521" PostTypeId="1" AcceptedAnswerId="15540" CreationDate="2010-12-25T21:55:28.620" Score="5" ViewCount="283" Body="&lt;p&gt;I was trying to solve a problem similar to the &quot;how many ways are there to make change for a dollar&quot; problem.  I ran across a &lt;a href=&quot;http://www.maa.org/features/mathchat/mathchat_4_19_01.html&quot; rel=&quot;nofollow&quot;&gt;site&lt;/a&gt; that said I could use a generating function similar to the one quoted below.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The answer to our problem (293) is the&#xA;  coefficient of x^100 in the reciprocal&#xA;  of the following:&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;(1-x)(1-x^5)(1-x^10)(1-x^25)(1-x^50)(1-x^100)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But I must be missing something as I can't figure out how they get from that to 293.  Any help on this would be appreciated.&lt;/p&gt;&#xA;" OwnerUserId="5111" LastActivityDate="2010-12-26T09:59:54.467" Title="Making Change for a Dollar (and other number partitioning problems)" Tags="&lt;generating-functions&gt;" AnswerCount="4" FavoriteCount="1" />
  <row Id="15522" PostTypeId="2" ParentId="15521" CreationDate="2010-12-25T22:09:29.000" Score="2" ViewCount="" Body="&lt;p&gt;You &quot;just&quot; have to follow the prescription:  find the formal power series (no need to think about convergence) that is defined and check the number that multiplies x^100.  There's a reason I put just in quotes.  There is no obvious route to 293 that I can see.  Mathematica can do it with just one command, but I can't get Alpha to do it.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-12-25T22:09:29.000" />
  <row Id="15523" PostTypeId="2" ParentId="15521" CreationDate="2010-12-25T22:10:25.853" Score="1" ViewCount="" Body="&lt;p&gt;I think you calculate $[x^{100}](1-x)^{100}(1-x^5)^{20}(1-x^{10})^{10}(1-x^{25})^4(1-x^{50})^2(1-x^{100})$, but that calculation seems to be brute force.&lt;/p&gt;&#xA;" OwnerUserId="774" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-26T00:16:17.030" LastActivityDate="2010-12-26T00:16:17.030" />
  <row Id="15524" PostTypeId="1" AcceptedAnswerId="15525" CreationDate="2010-12-25T22:36:48.340" Score="4" ViewCount="229" Body="&lt;p&gt;I'm not sure how well known this &quot;riddle&quot; is but here it goes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3 people go to a restaurant, each buy food worth 10.00.  When they're done, they give 30.00 to the waitress.  She gives the money to the manager, manager says they paid too much, gives 5.00 back to the waitress.  She comes back to the table, gives a dollar back to each person and then puts the remaining 2.00 in her pocket for tip.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So now, each person has paid 9.00.  9.00 x 3 = 27.00.  Plus the 2.00 in the waitresses pocket is 29.00.  What happened to the 30th dollar?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what's the issue with this way of calculating (since if you backward it works fine), that doesn't give us the correct result?&lt;/p&gt;&#xA;" OwnerUserId="1630" LastActivityDate="2010-12-25T23:24:29.567" Title="Riddle (simple arithmetic problem/illusion)" Tags="&lt;arithmetic&gt;&lt;riddle&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="15525" PostTypeId="2" ParentId="15524" CreationDate="2010-12-25T22:50:53.427" Score="4" ViewCount="" Body="&lt;p&gt;I am not sure how 'mathematical' the riddle part of this story is. :) Each guest paid 9 because together they paid 30, and got back 27. Of those 27, the manager got 25, and the waitress got the other 2 -- there is no sense/reason in adding 27 to 2 -- though you might have a gut feeling that you are headed in the 'right direction' of getting the initial 30 that way. But take a look -- there really is no extra dollar left, that's all it should be -- 25+2. In other words, each guest paid 9 dollars with the tip included into that. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another, possibly more insightful way to look at things:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The guests initially paid 30 dollars. The waitress returned them 3 dollars. So the guests ended up paying 27 (not thirty!). Of those 27 dollars, 2 dollars were pocketed as a tip by the waitress, so you could also say that they paid 8.333... each and then added a two dollar tip. I think it's a question of order of operations (and using it with respect to appropriate quantities) that could be confusing here. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Punchline&lt;/strong&gt;: the waitress' 2 dollars were part of 27 dollars paid by the customers. They saved the other 3 dollars of the initial 30  because of what the manager said, so if you are wondering 'what happened to 30?' it's more of 27+3, or 25+2+3, where 25 is what they paid without a tip, 2 is the tip, and 3 is the amount they saved.&lt;/p&gt;&#xA;" OwnerUserId="3731" LastEditorUserId="3731" LastEditorDisplayName="" LastEditDate="2010-12-25T23:24:29.567" LastActivityDate="2010-12-25T23:24:29.567" CommentCount="1" />
  <row Id="15526" PostTypeId="2" ParentId="9999" CreationDate="2010-12-25T23:04:14.250" Score="0" ViewCount="" Body="&lt;p&gt;On another note, many people have mentioned the Binet formula. There is a nice way to see how this formula comes about. The Fibonacci sequence is a recurrence relation, so we can apply difference calculus to the sequence. In my advanced discrete mathematics course, we looked at the Fibonacci sequence, naturally. You can see the notes from the day we derived a formula for it[1]. It is on page 1 and page 2 of [1].&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A good text book to take a look at with regards to this is [2].&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me know if I can help you anymore!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[1] &lt;a href=&quot;http://www.tylerclark12.com/blog/wp-content/uploads/2010/10/Math_542_9-24.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.tylerclark12.com/blog/wp-content/uploads/2010/10/Math_542_9-24.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[2] Kelley, W. &amp;amp; Peterson, A. (2001). Difference Equations: An Introduction with Applications (2nd Ed.). San Diego, CA: Academic Press.&lt;/p&gt;&#xA;" OwnerUserId="3791" LastActivityDate="2010-12-25T23:04:14.250" />
  <row Id="15527" PostTypeId="2" ParentId="15515" CreationDate="2010-12-25T23:07:32.113" Score="0" ViewCount="" Body="&lt;p&gt;$2n \choose 2$ is the sum of numbers from $1$ to $2n-1$, which is the sum of even numbers from $2$ to $2(n-1)$, which is $2{n\choose 2}$, and odd numbers from $1$ to $2n-1$. The latter is of course $n^2$.&lt;/p&gt;&#xA;" OwnerUserId="4806" LastActivityDate="2010-12-25T23:07:32.113" />
  <row Id="15528" PostTypeId="2" ParentId="15514" CreationDate="2010-12-25T23:11:01.953" Score="1" ViewCount="" Body="&lt;p&gt;First you want to simplify $\tan (\arcsin x)$ and $\sin(\arctan x)$. Let me show&#xA;you the latter one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose $\theta=\arctan x$. Then by definition $-\pi/2&amp;lt;\theta&amp;lt;\pi/2$ and $\tan\theta=x$.&#xA;Since $\sec \theta&amp;gt;0$ (because $-\pi/2&amp;lt;\theta&amp;lt;\pi/2$), and $$\sec^2\theta=1+\tan^2 \theta =1+x^2,$$ you get&#xA;$\sec \theta=\sqrt{1+x^2}$, so $\cos \theta =1/\sqrt{1+x^2},$ and &#xA;$$\sin\theta=\tan\theta \cos\theta =\frac{x}{\sqrt{1+x^2}},$$&#xA;i.e. $\sin(\arctan x)=\frac{x}{\sqrt{1+x^2}}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, you can prove $\tan (\arcsin x)=\frac{x}{\sqrt{1-x^2}}.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using quotient rule, you can prove that the derivative of $\frac{x}{\sqrt{1+x^2}}$&#xA;and $\frac{x}{\sqrt{1-x^2}}$ are &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{1}{(1+x^2)^{3/2}} \mbox{ and }\frac{1}{(1-x^2)^{3/2}} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;respectively. Hence the derivative of  your function is $$\frac{1}{(1+x^2)^{3/2}}+\frac{1}{(1-x^2)^{3/2}}. $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;${\bf Edit.}$ Or you can use chain rule to get:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\cos(\arctan x)\frac{1}{1+x^2}+\sec^2(\arcsin x)\frac{1}{\sqrt{1-x^2}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now simplify it.&lt;/p&gt;&#xA;" OwnerUserId="3249" LastEditorUserId="3249" LastEditorDisplayName="" LastEditDate="2010-12-26T04:03:54.567" LastActivityDate="2010-12-26T04:03:54.567" />
  <row Id="15529" PostTypeId="2" ParentId="15515" CreationDate="2010-12-25T23:14:27.703" Score="2" ViewCount="" Body="&lt;p&gt;Looks good to me.  Also you could just work numerically: ${{2n} \choose 2}=\frac{1}{2} \times 2n(2n-1)=n(2n-1)=2n^2-n=n(n-1)+n^2=2{n \choose 2}+n^2$.&lt;/p&gt;&#xA;" OwnerUserId="139" LastActivityDate="2010-12-25T23:14:27.703" CommentCount="1" />
  <row Id="15530" PostTypeId="1" CreationDate="2010-12-25T23:23:43.797" Score="-1" ViewCount="160" Body="&lt;p&gt;hi, i am trying to figure a algorithm problem that determines the two largest number of the a series of random 10 numbers. Thanks to your help.&lt;/p&gt;&#xA;" OwnerUserId="5112" LastActivityDate="2010-12-26T19:06:05.093" Title="algorithm of determines the two largest of a series" Tags="&lt;algorithms&gt;" AnswerCount="4" CommentCount="5" />
  <row Id="15531" PostTypeId="2" ParentId="15530" CreationDate="2010-12-25T23:47:44.810" Score="5" ViewCount="" Body="&lt;p&gt;Just go through the list and keep track of the two largest numbers, that is have two variables that are the the two largest elements so far while going through.&lt;/p&gt;&#xA;" OwnerUserId="5113" LastActivityDate="2010-12-25T23:47:44.810" CommentCount="2" />
  <row Id="15532" PostTypeId="2" ParentId="15530" CreationDate="2010-12-26T00:07:43.527" Score="3" ViewCount="" Body="&lt;pre&gt;&lt;code&gt;list = [random, random, random, ..., random]&#xA;largest := (list[0] &amp;gt; list[1]) ? list[0] : list[1]&#xA;secondlargest := (list[0] &amp;lt;= list[1]) ? list[0] : list[1]&#xA;for each num in list:&#xA;    if num &amp;gt;= largest:&#xA;        secondlargest := largest&#xA;        largest := num&#xA;    elseif num &amp;gt;= secondlargest:&#xA;        secondlargest := num&#xA;return {largest, secondlargest}&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;You aren't going to be able to avoid iterating across the entire list because it is not sorted, and therefore you are going to have to search in $O(n)$ time. If the list is sorted, it becomes pretty much a non-problem, but there is no sense in sorting it if all you want is the two largest (if you have to do other things it might be worth it though) &lt;/p&gt;&#xA;" OwnerUserId="5038" LastActivityDate="2010-12-26T00:07:43.527" CommentCount="4" />
  <row Id="15533" PostTypeId="1" AcceptedAnswerId="15551" CreationDate="2010-12-26T00:11:33.337" Score="3" ViewCount="180" Body="&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; What is the definition of a small class?  Is there no such thing as a set which contains other sets such as {{1,2},{1},{2}, 1, 2} (ie, is this really called a &quot;class&quot;)?  What are some prototypical examples of small classes?  What are some prototypical examples of large classes?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Motivation:&lt;/strong&gt; A few things I've been paging through (Rotman's homological algebra, a few papers which talk about categories, etc.) talk about small classes using definitions which are slightly confusing to me.  Most of them note that a &quot;class&quot; is a &quot;collection&quot; of &quot;sets&quot;, and that a small class is a class which is a set.  I'm not exactly sure where &quot;small&quot; ends and &quot;large&quot; begins in this respect.  See the specific questions below for what I mean by this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Related Specific Questions:&lt;/strong&gt; If we use the notation {a,b,c} for a set with the elements 'a', 'b', and 'c', then is {a,b,c} a small class because it IS a set?  Is {{a,b,c}} a large class because it is a &quot;set which contains a set&quot;?  Is the set of real numbers a small class?  Is the set of natural numbers a small class?  Is the set of all functions from the real numbers to the real numbers a small class?  Is the class of all small classes a large class?&lt;/p&gt;&#xA;" OwnerUserId="2959" LastActivityDate="2010-12-26T10:40:33.417" Title="Defining &quot;Small Classes.&quot;" Tags="&lt;set-theory&gt;&lt;examples&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="15534" PostTypeId="2" ParentId="15530" CreationDate="2010-12-26T00:39:11.080" Score="1" ViewCount="" Body="&lt;p&gt;It might be worth noting that the naive way to find the $k$ largest elements requires $O(nk)$ operations, since updating the list of $k$ largest elements could take $O(k)$ operations. However, if these are stored in a heap then we get the better complexity $O(n \log k)$; this method is practical (i.e. actually outperforms the naive algorithm) for moderately sized $k$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: Debanjan's method shows how to solve this using $O(n + k\log k)$ operations, or $O(n)$ if we're not interested in the relative order of the $k$ largest elements. First you find the $k$th largest element using the classical $O(n)$ algorithm, and then you collect all elements at least as large as it is; if required, you then sort them.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2010-12-26T19:06:05.093" LastActivityDate="2010-12-26T19:06:05.093" CommentCount="1" />
  <row Id="15535" PostTypeId="2" ParentId="15533" CreationDate="2010-12-26T01:21:50.793" Score="2" ViewCount="" Body="&lt;p&gt;As I understand it, the terms &quot;set&quot; and &quot;small class&quot; are equivalent.  Certainly a set may have other sets as elements, so {{1,2},{1},{2},1,2} is a set.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Informally, a class is a collection of sets which can be &quot;unambiguously defined by a property that all its members share,&quot; to quote Wikipedia.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To &lt;em&gt;formalize&lt;/em&gt; the notion of a large class -- that is, a class which is not small -- requires a choice of set theory that allows the formulation of such a concept.  The standard framework of the Zermelo-Fraenkel axioms (ZF) does &lt;em&gt;not&lt;/em&gt; have a formal notion of a large class: everything in consideration in that theory is a set.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some standard examples of large classes are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;the class of all sets&lt;/li&gt;&#xA;&lt;li&gt;the class of all groups&lt;/li&gt;&#xA;&lt;li&gt;the class of all topological spaces&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These are large classes because they are not small.  For example, the class of all sets is not itself a set: &lt;a href=&quot;http://en.wikipedia.org/wiki/Russel%27s_paradox&quot; rel=&quot;nofollow&quot;&gt;Russell's paradox&lt;/a&gt; can be thought of as a proof of this statement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To answer your specific questions in light of the above: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;{a,b,c} is a small class, because it is a set.  &lt;/li&gt;&#xA;&lt;li&gt;{{a,b,c}} is a small class, because it is a set.&lt;/li&gt;&#xA;&lt;li&gt;Each of the set of real numbers and the set of natural numbers is a small class.&lt;/li&gt;&#xA;&lt;li&gt;The class of all small classes (i.e. the class of all sets) is a large class, as above.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="5116" LastActivityDate="2010-12-26T01:21:50.793" />
  <row Id="15536" PostTypeId="1" AcceptedAnswerId="15539" CreationDate="2010-12-26T01:33:30.293" Score="2" ViewCount="65" Body="&lt;p&gt;How to input such system &#xA;&lt;img src=&quot;http://i.imgur.com/yFGkB.jpg&quot; alt=&quot;alt text&quot;&gt;&#xA;into maple so that it would solve it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;BTW we some part of system will be given to us,, which we do not know certently - may be some Ai's may be some  Bi's as constants. we do not know how much constants we will get. And we would need from given constants get a soltion for other a's and b's. in real life we will be provided with s as a constant but I hoped for general solution=) &lt;/p&gt;&#xA;" OwnerUserId="2065" LastEditorUserId="2065" LastEditorDisplayName="" LastEditDate="2010-12-26T13:36:44.297" LastActivityDate="2010-12-26T13:36:44.297" Title="Maple: How to input such system so that it would be solved?" Tags="&lt;computer-algebra-systems&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="15537" PostTypeId="2" ParentId="14311" CreationDate="2010-12-26T01:45:17.763" Score="1" ViewCount="" Body="&lt;p&gt;jerr18 ---&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ignoring the fact that my posted &quot;answer&quot; was clearly indicated to be a &quot;comment&quot; ...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps you do not understand what a proof is. As a writer: a proof is what makes me believe a result. As a reader whatever you wish to accept is a proof. A counterexample is what is needed to show a proof is defective. Sometimes a believer will use a comment to provide a counterexample. (I am not asking that you make me disbelieve my comment/proof by providing a counter example.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for your question, I believe that the empty language is poly for any alphabet and it satisfies your requirement that every terminal appears an even number of times. I suppose your question should be reposed to conform to the effort that Yuval Filmus put forth.&lt;/p&gt;&#xA;" OwnerUserId="5055" LastActivityDate="2010-12-26T01:45:17.763" CommentCount="4" />
  <row Id="15538" PostTypeId="2" ParentId="15533" CreationDate="2010-12-26T01:50:07.037" Score="1" ViewCount="" Body="&lt;p&gt;You already know the definition of a &quot;small class&quot;, it is what you gave.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Definition:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let X be a class.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then X is a set if and only there exists a class Y such that X is an element of Y.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Theorem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let X be a class.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then X is a set if and only if either/both of the following equivalent conditions hold:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;There exists an ordinal alpha such that X is an element of V_alpha.&lt;/li&gt;&#xA;&lt;li&gt;There exists an ordinal alpha such that X is a subclass of V_alpha.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;(ordinal is defined at en.wikipedia.org/wiki/Ordinal_number,&#xA;&lt;br&gt;&#xA;V_alpha is defined at en.wikipedia.org/wiki/Von_neumann_universe)&lt;/p&gt;&#xA;" OwnerUserId="5114" LastActivityDate="2010-12-26T01:50:07.037" CommentCount="5" />
  <row Id="15539" PostTypeId="2" ParentId="15536" CreationDate="2010-12-26T03:01:22.477" Score="3" ViewCount="" Body="&lt;p&gt;Apparently you want your solution be parametrized in $k,s$. I don't think Maple can&#xA;do that. What you need to do is to solve this linear system for some particular&#xA;$k, s$ and hopefully you can discern a pattern.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can solve linear systems using the linear algebra package: &gt;with(linalg);&lt;/p&gt;&#xA;" OwnerUserId="3249" LastActivityDate="2010-12-26T03:01:22.477" />
  <row Id="15540" PostTypeId="2" ParentId="15521" CreationDate="2010-12-26T03:25:02.650" Score="3" ViewCount="" Body="&lt;p&gt;You should be able to compute it using a Partial Fraction representation (involving complex numbers). For instance see this previous answer: &lt;a href=&quot;http://math.stackexchange.com/questions/12273/minimum-multi-subset-sum-to-a-target/12284#12284&quot;&gt;Minimum multi-subset sum to a target.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note, this partial fraction expansion needs to be calculated only &lt;em&gt;one&lt;/em&gt; time. Once you have that, you can compute the way to make change for an arbitrary amount pretty quickly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case, I doubt they really did that for finding the coefficient of $x^{100}$. It is probably quicker to just multiply out, ignoring the terms which would not contribute to the coefficient of $x^{100}$. Or you could try computing the partial fraction representation of only some of the terms and then multiply out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note, if you are multiplying out to find the coefficient of $x^{100}$, it would be easier not to go to the reciprocal, which arises from considering an infinite number of terms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You just need to multiply out&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ (\sum_{j=0}^{100} x^j)\ (\sum_{j=0}^{20} x^{5j})\ (\sum_{j=0}^{10} x^{10j})\  (\sum_{j=0}^{4} x^{25j})\ (1 + x^{100})$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which would amount to enumerating the different ways to make the change (and in fact is the the way we come up with the generating function in the first place).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could potentially do other things, like computing the $100^{th}$ derivative at $0$, or computing a contour integral of the generating function divided by $x^{100}$, but I doubt they went that route either.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope that helps.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-12-26T03:25:02.650" CommentCount="1" />
  <row Id="15541" PostTypeId="1" AcceptedAnswerId="15581" CreationDate="2010-12-26T04:52:38.150" Score="4" ViewCount="74" Body="&lt;p&gt;This is a problem from Arnold's book on ODEs I cannot solve.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Prove that if $f:U\to V$ is a diffeomorphism, then the Euclidean spaces with the domains $U$ and $V$ as subsets have the same dimension.&#xA;Hint. Use the implicit function theorem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="5119" LastActivityDate="2010-12-26T21:26:11.467" Title="invariance of dimension under diffeomorphism of real subspaces" Tags="&lt;real-analysis&gt;&lt;multivariable-calculus&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="15542" PostTypeId="1" AcceptedAnswerId="15615" CreationDate="2010-12-26T05:14:51.390" Score="7" ViewCount="239" Body="&lt;p&gt;On page 18 of Rudin's &lt;em&gt;Real and Complex&lt;/em&gt; analysis he defines $0 \cdot \infty = 0$ and says that &quot;with this definition the commutative, associative, and distributive laws hold in $[0,\infty]$ without any restriction&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is not clear to me is whether the quoted statement is a justification of the definition or just a consequence. Wouldn't the commutative, associative, and distributive laws also hold if we define $0 \cdot \infty = \infty$? &lt;/p&gt;&#xA;" OwnerUserId="1195" LastActivityDate="2010-12-27T05:22:21.563" Title="Arithmetic on $[0,\infty]$: is $0 \cdot \infty = 0$ the only reasonable choice?" Tags="&lt;real-analysis&gt;&lt;number-systems&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="15543" PostTypeId="2" ParentId="11763" CreationDate="2010-12-26T06:40:51.440" Score="2" ViewCount="" Body="&lt;p&gt;One way to think about this is to look at the chain level. Recall that the singular chains of a pair $C_n(X, Y)$ are sometimes defined as $C_n(X,Y) = C_n(X)/C_n(Y)$. So one way of thinking about this is that we are taking the space $X$ and attaching a cone to $Y$ so that any chain that lives in $Y$ can be shrunk to a point and forgotten about. Relative homology is then defined to be the relative cycles modulo the relative boundaries, i.e. we look at cycles AND chains living in $Y$, and we mod out not only by things that are boundaries, but things that are boundaries plus maybe a chain in $Y$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This intuition works especially well for nice pairs $(X,Y)$ where $H_n(X,Y) = H_n(X/Y)$. In other words, we really are forgetting about all the stuff happening in $Y$ since we are shrinking it to a point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A wonderful treatment of all of this is given in Hatcher's book- he's very good about providing intuition.&lt;/p&gt;&#xA;" OwnerUserId="423" LastActivityDate="2010-12-26T06:40:51.440" />
  <row Id="15544" PostTypeId="2" ParentId="15503" CreationDate="2010-12-26T08:59:12.163" Score="1" ViewCount="" Body="&lt;p&gt;Using only the things you have written, your question does not have an answer. In general there may not be any way to complete the commutative diagram you drew into a square (and in general there may not be any projection $\pi_S$, but this is a smaller problem). The basic problem is that you have only specified what $G_0$ does on objects, so as it stands $G_0$ is not even a functor (and in particular not a subfunctor).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the example where $F$ is the group of units functor, as above. Let $G_0(R)$ be the whole group of units whenever $R$ is infinite, and the trivial group when $R$ is finite. Then you will see that there is in general no way to fill in the square above for e.g. the map $\mathbb{Z} \to \mathbb{Z}/n\mathbb{Z}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fortunately the condition that $G_0$ should be a subfunctor of $F$ means that you have very little choice in how you define $G_0(f)$, for a morphism $f$. The fact that the diagram above should be commutative &lt;em&gt;forces&lt;/em&gt; you to choose $G_0(f)$ to be equal to the restriction of $F(f)$ to $G_0(R)$. What you then need to check, in order for this to be well defined, is that this restriction always lands inside $G_0(S)$. (This fails in the example of the previous paragraph.) Whenever this condition holds, $G_0$ automatically becomes a functor using only that $F$ is a functor (check this).&lt;/p&gt;&#xA;" OwnerUserId="677" LastActivityDate="2010-12-26T08:59:12.163" CommentCount="1" />
  <row Id="15545" PostTypeId="2" ParentId="15486" CreationDate="2010-12-26T08:59:27.530" Score="4" ViewCount="" Body="&lt;p&gt;For nice enough spaces, e.g. metric spaces, compactness is equivalent to sequential compactness. I like to think of sequential compactness as saying that sequences of points can't run off to &quot;infinity&quot;.&lt;/p&gt;&#xA;" OwnerUserId="263" LastActivityDate="2010-12-26T08:59:27.530" CommentCount="1" />
  <row Id="15546" PostTypeId="2" ParentId="11763" CreationDate="2010-12-26T09:13:19.733" Score="3" ViewCount="" Body="&lt;p&gt;See page 115 of Hatcher: Elements of $H_n(X,A)$ are represented by $n$-chains $\alpha \in C_n(X)$ such that $\partial \alpha \in C_{n-1}(A) \subset C_{n-1}(X)$. So you can think of an element of $H_n(X,A)$ as being an $n$-thing in $X$ whose boundary (an $(n-1)$-thing) lies in $A$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the relative homology groups fit into the long exact sequence&#xA;$$\cdots \to H_n(A) \to H_n(X) \to H_n(X,A) \to H_{n-1}(A) \to \cdots,$$&#xA;you can think of them as giving a crude measurement of, e.g., the non-surjectiveness or non-injectiveness of the maps $H_n(A) \to H_n(X)$.&lt;/p&gt;&#xA;" OwnerUserId="263" LastActivityDate="2010-12-26T09:13:19.733" />
  <row Id="15547" PostTypeId="2" ParentId="9186" CreationDate="2010-12-26T09:31:35.473" Score="3" ViewCount="" Body="&lt;p&gt;I detect that you're really asking about Lie groups and Lie algebras.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Much of the theory of Lie groups and Lie algebras can be done algebraically or formally. Analysis only really comes in when you start to talk about convergence -- such as that of $\operatorname{exp} : \mathfrak{g} \to G$. If you don't care about convergence, you can still treat $\operatorname{exp}$ formally. Or, if $\mathfrak{g}$ is nice enough, say nilpotent, then everything works out -- you get convergence of $\operatorname{exp}$ for free.&lt;/p&gt;&#xA;" OwnerUserId="263" LastActivityDate="2010-12-26T09:31:35.473" CommentCount="2" />
  <row Id="15548" PostTypeId="2" ParentId="15521" CreationDate="2010-12-26T09:59:54.467" Score="1" ViewCount="" Body="&lt;p&gt;We can ease the calculation by noting that the number of ways of changing 100 equals the number of ways of representing the numbers less than or equal to $100$ as the sum of the numbers $5, 10, 25, 50$ and $100$, since the pennies can make up any remaining difference.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Noting that all these number are divisible by $5$ we can conclude that the number of ways of representing $100$ in units of $1, 5, 10, 25, 50$ and $100$ is the sum of the coefficients up to and including the term in $x^{20}$ in the expansion of&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \frac{1}{(1-x)(1-x^2)(1-x^5)(1-x^{10})(1-x^{20})} . $$&lt;/p&gt;&#xA;" OwnerUserId="1301" LastActivityDate="2010-12-26T09:59:54.467" CommentCount="1" />
  <row Id="15549" PostTypeId="2" ParentId="15504" CreationDate="2010-12-26T10:04:38.440" Score="2" ViewCount="" Body="&lt;p&gt;This formula is not meant to be a terribly complicated way to compute $f(a+h)$ (e.g. $\log(1+h)$) $exactly$, but it tells you the following: If you compute the value $T_{f,a,n}(a+h)$ (which is a simple polynomial value) instead you commit an error which is small because (i) $h^{n+1}$ might be very small, (ii) ${1\over(n+1)!}$ is very small and (iii) hopefully the $(n+1)^{\rm st}$ derivative of $f$ is bounded by a constant throughout, so that it doesn't matter where the point $a+\theta h$ guaranteed by the theorem lies.&lt;/p&gt;&#xA;" OwnerUserId="1303" LastActivityDate="2010-12-26T10:04:38.440" />
  <row Id="15550" PostTypeId="2" ParentId="15530" CreationDate="2010-12-26T10:35:54.740" Score="5" ViewCount="" Body="&lt;p&gt;What you are looking for can be solved in $O(n)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is called finding the k-th order statistic.The Wikipedia entry of &lt;a href=&quot;http://en.wikipedia.org/wiki/Selection_algorithm&quot; rel=&quot;nofollow&quot;&gt;Selection algorithm&lt;/a&gt; provides some insight.But probably a better discussion is presented in &lt;strong&gt;Chapter 9, Medians and Other statistics from &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0262032937&quot; rel=&quot;nofollow&quot;&gt;Introduction to Algorithms, 2nd Ed&lt;/strong&gt;&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a well described randomized $O(n)$ approach : &lt;a href=&quot;http://pine.cs.yale.edu/pinewiki/QuickSelect&quot; rel=&quot;nofollow&quot;&gt;&lt;strong&gt;QuickSelect&lt;/strong&gt;&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2109" LastEditorUserId="2109" LastEditorDisplayName="" LastEditDate="2010-12-26T13:30:36.513" LastActivityDate="2010-12-26T13:30:36.513" CommentCount="2" />
  <row Id="15551" PostTypeId="2" ParentId="15533" CreationDate="2010-12-26T10:40:33.417" Score="1" ViewCount="" Body="&lt;p&gt;The sets that we consider are actually &lt;a href=&quot;http://en.wikipedia.org/wiki/Hereditary_set&quot; rel=&quot;nofollow&quot;&gt;hereditary sets&lt;/a&gt; meaning that all of their members are themselves hereditary sets.  What this informally means is that a hereditary set is a set of sets of sets of sets, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, the empty set ($\emptyset$) is a hereditary set because it has no members so all of its members are vacuously hereditary sets.  Then $\{\emptyset\}$ is a hereditary set because its only member is a hereditary set.  Then $\{\emptyset, \{\emptyset\}\}$ is a hereditary set because both of its members are hereditary sets, and we can continue this process of tacking on the previous hereditary set as a member ad infinitum to get an infinite collection of hereditary sets.  This process describes how the Natural numbers are constructed:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$0 = \emptyset$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$1 = \{\emptyset\} = \{0\}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$2 = \{\emptyset, \{\emptyset\}\} = \{0, 1\}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$3 = \{\emptyset, \{\emptyset\}, \{\emptyset, \{\emptyset\}\}\} = \{0, 1, 2\}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\vdots$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$n + 1 = \{0, 1, 2, \ldots, n\}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the axiom of infinity, there exists an inductive set (set that contains $0$ and the successor of any of its members) so the class of Natural numbers is actually a hereditary set.  The integers can then be  constructed from the Natural numbers by considering equivalence classes of ordered pairs of Natural numbers where $(a, b)$ is equivalent to $(c, d)$ exactly when $a + d = b + c$ so $(a, b)$ will correspond to the integer $a - b$.  Ordered pairs of hereditary sets are also hereditary sets because $(a, b)$ can be formalized as $\{\{a\}, \{a, b\}\}$.  Consequently, the class of integers will also be a hereditary set.&#xA;The set of Rational numbers can then be defined as equivalence classes of ordered pairs of integers so it too will be a hereditary set.  The Real numbers, which can be constructed by considering select subsets of Rational numbers (or elements from the powerset of the Rational numbers), is also a hereditary set.  Once we formalize the Real numbers as a hereditary set, we can show that a function from the Reals into the Reals, which is a collection of ordered pairs of Reals, is a hereditary set.  Then we can show that the collection of all functions from the Reals into the Reals is a hereditary set.  In short, all of the natural objects would be &quot;small classes&quot; because they can be formed from the axioms of set theory (ZFC).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The point here is that what determines whether a class is &quot;small&quot; or not is not whether it contains elements that look like sets such as $\{1, 2, 3\}$ or $\{a, b, c\}$ because we mostly restrict ourselves to hereditary sets anyway.  Instead, we have as a minimal requirement for smallness that the objects can either be provably constructible from the axioms of ZFC &lt;em&gt;or&lt;/em&gt; the assumption of their existence as sets does not lead to a contradiction in ZFC.  I include the first possibility here because we cannot prove (within ZFC) that the axioms of ZFC won't themselves lead to a contradiction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an example of what we would consider a large class is the class of all small classes or the class of all sets.  This is because such a class cannot be constructed from the axioms of ZFC, and it provably leads to a contradiction from ZFC (see Russell's Paradox from Brad's post).  In a nutshell, if we called the class of all sets $X$ and $X$ were a set, then $A = \{x \in X| x \notin X\}$ would have to be a set by the comprehension axiom.  But then $X \in X$ since $X$ is the class of all sets so $X \in A$ if and only if $X \notin A$, an immediate contradiction.  Other examples of &quot;large classes&quot; include the class of ordinals or as Ricky mentioned any object that does not fall into the $V$ hierarchy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One more thing I should mention here is that you should always be aware of the context when considering whether an object is a large class (also known as proper class) or not.  Specifically, most of the natural objects that you consider will be small classes with respect to the set-theoretic universe.  However, sometimes classes are considered with respect to individual models.  A Grothendieck Universe for example, is a proper class if there are no inaccessible cardinals and we cannot prove that such a large cardinal is even possible from the ZFC axioms of set theory.  As another example, we sometimes do not allow ourselves use of all of the ZFC axioms, and the set of all even numbers for example would be a large class if we explicitly assumed that all sets were finite.&lt;/p&gt;&#xA;" OwnerUserId="4693" LastActivityDate="2010-12-26T10:40:33.417" CommentCount="1" />
  <row Id="15552" PostTypeId="2" ParentId="15499" CreationDate="2010-12-26T11:25:43.560" Score="3" ViewCount="" Body="&lt;p&gt;Dylan's right that diffeomorphism groups are usually infinite dimensional.  There's a theory of infinite dimensional Lie groups, I've heard, but I think it's a lot harder to work out things like convergence.  Worth noting, though, is that in this analogy, if $Diff(M)$ is a Lie group then its Lie algebra should be $Vect(M)$, the (infinite-dimensional) vector space of smooth vector fields on $M$.  The Lie algebra bracket is just the usual vector field bracket (possibly up to a sign? but if so, then it's only because of stupid clashing conventions).  Heuristically, to look for finite dimensional Lie groups we're going to want to put some more structure on $M$ and require that our diffeomorphisms preserve this.  If you're looking at things that aren't so homogeneous you get into the territory of $G$-bundles, which I think you should definitely look into if you don't already know about -- you don't need to know too much Lie theory to get the basics, and it's really cool (I think you'll like it too).  I don't have a reference offhand (please, anyone, post as a comment if you do) but I took an excellent class on them from Robert Bryant last year and a friend of mine is working on TeXing up the notes, so hopefully those will be available (at least in part) before too long.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the way, $Aut(M)/$isotopy is called the &lt;em&gt;mapping class group&lt;/em&gt;.  This can be written $MCG(M)=Aut(M)/Aut_0(M)$, where $Aut_0(M)$ is the connected component of the identity map, which if you think about it is all isotopies are anyways.  This has been extensively studied.  I don't remember much from a class I took with Jeff Brock a few years ago, but checking wikipedia reminds me that e.g. $MCG(\mathbb{T}^n)\simeq GL(n,\mathbb{Z})$.  It also says that the MCGs of surfaces have been extensively studied.  In any case, you can see that this is just the group $\pi_0(Aut(M))$, so assuming you understand $MCG(M)$ (!) then you only need to figure out $Aut_0(M)$ (!) to have the whole picture.  Which brings us back to the original question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know much, but like I said it's definitely easier (and finite-dimensionaler) if we restrict ourselves to manifolds with structure.  We can start by saying something simple: $Isom(S^n)=O(n+1)$.  One potentially fruitful way to think about this is to just start with a (matrix) group and find an invariant subspace of the (affine) space on which it acts, hopefully freely -- very often this should be a manifold, I think.  Otherwise things will probably be significantly more difficult.  If you put the flat metric on a torus and assume its fundamental domain is a square, then you should be able to look at isometries by what they do to your fundamental square sitting on a latticed plane.  To represent an isometry it'll need to be an isometry of the square, and to be a continuous map you'll need the four corners to be mapped to preimages of the same point under $\mathbb{R}^2\rightarrow \mathbb{T}^2$.  I'm pretty sure this ends up meaning $Isom_0(\mathbb{T}^2)\cong \mathbb{T}^2$.  But maybe that's not so surprising, since it's a group itself.  I guess all it's really telling you is that there are no isometries other than the &quot;multiply-by-$g$&quot; map for $g\in \mathbb{T}^2$.  The group $Conf(\mathbb{T}^2)$ should be pretty different, but I think you can't continuously dilate because that would eventually change the degree of your map (which is a discrete invariant), so a similar analysis should go through but you won't get any interesting &lt;em&gt;Lie&lt;/em&gt;-type behavior.  More interesting might be to try this all with the fundamental domain for an $n$-holed torus ($n\geq 2$) sitting on a latticed hyperbolic plane.  But it's late and I'm tired, so I'm not going to.  You should though!  If you do, let me know what you come up with.&lt;/p&gt;&#xA;" OwnerUserId="2159" LastActivityDate="2010-12-26T11:25:43.560" CommentCount="1" />
  <row Id="15553" PostTypeId="1" AcceptedAnswerId="15554" CreationDate="2010-12-26T12:33:56.790" Score="3" ViewCount="106" Body="&lt;p&gt;&lt;strong&gt;If $A$ and $B$ are two conformable matrices,and $AB = A$ and $BA = B$ then prove that A and B are idempotent matrices.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This may be trivial but I am not sure how to proceed on this.&lt;/p&gt;&#xA;" OwnerUserId="2109" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-26T13:05:44.073" LastActivityDate="2010-12-26T13:05:44.073" Title="An elementary proof in matrix algebra" Tags="&lt;linear-algebra&gt;&lt;matrices&gt;" AnswerCount="1" />
  <row Id="15554" PostTypeId="2" ParentId="15553" CreationDate="2010-12-26T12:47:09.663" Score="6" ViewCount="" Body="&lt;p&gt;$$Ax = ABx = A(BA)x = (AB)Ax = AAx$$&lt;/p&gt;&#xA;" OwnerUserId="1269" LastActivityDate="2010-12-26T12:47:09.663" CommentCount="3" />
  <row Id="15555" PostTypeId="1" CreationDate="2010-12-26T13:16:35.977" Score="3" ViewCount="349" Body="&lt;p&gt;What is $\displaystyle \lim_{x \to 0} \frac{\lceil x \rceil}{x}$ ?  Here, $\lceil x \rceil$ is the ceiling function at $x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For left limit and right limit as $x\to 0$.&lt;/p&gt;&#xA;" OwnerUserId="5122" LastEditorUserId="2736" LastEditorDisplayName="" LastEditDate="2011-06-06T00:36:07.313" LastActivityDate="2011-06-06T00:36:07.313" Title="what is $\lim\limits_{x\to 0} \lceil x\rceil/x$ ?" Tags="&lt;calculus&gt;&lt;limit&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="1" />
  <row Id="15556" PostTypeId="1" CreationDate="2010-12-26T13:36:00.910" Score="16" ViewCount="733" Body="&lt;p&gt;Some books say even numbers start from two but if you consider the number line concept, I think zero should be even because it is in between -1 and +1 (i.e in between 2 odd numbers). What is the real answer?&lt;/p&gt;&#xA;" OwnerUserId="5123" LastEditorUserId="6296" LastEditorDisplayName="" LastEditDate="2011-03-14T03:16:29.657" LastActivityDate="2011-03-14T03:16:29.657" Title="Is zero odd or even?" Tags="&lt;number-theory&gt;&lt;math-history&gt;&lt;parity&gt;" AnswerCount="5" CommentCount="2" FavoriteCount="4" />
  <row Id="15557" PostTypeId="2" ParentId="15556" CreationDate="2010-12-26T13:43:48.847" Score="7" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;YES!&lt;/strong&gt; zero is an even number. &lt;a href=&quot;http://mathforum.org/library/drmath/view/57132.html&quot; rel=&quot;nofollow&quot;&gt;Here&lt;/a&gt; is the Dr. Math's explanation.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;This seems to be a matter of confusion for many others around this planet,you may always like to ask  &lt;a href=&quot;http://www.google.com/search?btnG=1&amp;amp;pws=0&amp;amp;q=0+is+even+or+odd#hl=en&amp;amp;expIds=27642&amp;amp;sugexp=ldymls&amp;amp;xhr=t&amp;amp;q=Is+Zero+Even%2C+Odd%2C+or+Neither%3F&amp;amp;cp=30&amp;amp;qe=SXMgWmVybyBFdmVuLCBPZGQsIG9yIE5laXRoZXI_&amp;amp;qesig=Qog_VPHJ0jgC7-PI4nFBIQ&amp;amp;pkc=AFgZ2tlOVslOr2lA43pd92AsxK23pS2E9t-f3AFrd6jeF3PgpPmpcCSsSliwvaKIxTWjKuoevnHhEMGpNeprD9dQlSr05bBGkQ&amp;amp;pf=p&amp;amp;sclient=psy&amp;amp;source=hp&amp;amp;aq=f&amp;amp;aqi=&amp;amp;aql=&amp;amp;oq=Is+Zero+Even,+Odd,+or+Neither%3F&amp;amp;gs_rfai=&amp;amp;pbx=1&amp;amp;fp=b54c4bac999a8c85&quot; rel=&quot;nofollow&quot;&gt;google&lt;/a&gt; for your confusion.&lt;/p&gt;&#xA;" OwnerUserId="2109" LastEditorUserId="2109" LastEditorDisplayName="" LastEditDate="2010-12-26T20:26:15.257" LastActivityDate="2010-12-26T20:26:15.257" CommentCount="13" />
  <row Id="15558" PostTypeId="2" ParentId="15555" CreationDate="2010-12-26T13:49:42.973" Score="9" ViewCount="" Body="&lt;p&gt;Assuming that [x] is the floor of x, then look at &lt;a href=&quot;http://www.wolframalpha.com/input/?i=f%5bx%5d+%3D%3D+Floor%5bx%5d%2Fx+&quot; rel=&quot;nofollow&quot;&gt;this graph&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming that [x] is the ceiling of x, then look at &lt;a href=&quot;http://www.wolframalpha.com/input/?i=f%5bx%5d+%3D%3D+Ceiling%5bx%5d%2Fx+&quot; rel=&quot;nofollow&quot;&gt;this similar looking graph&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming that [x] is the &quot;nearest integer&quot; function, then consider what the nearest integer is on the interval [-0.49, 0.49].  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If this is something else, please specify.&lt;/p&gt;&#xA;" OwnerUserId="2959" LastEditorUserId="2959" LastEditorDisplayName="" LastEditDate="2010-12-26T17:23:04.557" LastActivityDate="2010-12-26T17:23:04.557" CommentCount="1" />
  <row Id="15559" PostTypeId="1" CreationDate="2010-12-26T14:26:57.530" Score="1" ViewCount="162" Body="&lt;p&gt;From wikipedia,&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Riemann_zeta_function&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Riemann_zeta_function&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Furthermore, the fact that $\zeta(s) = \zeta(s^*)^*$ for all complex s ≠ 1 ($s^*$ indicating complex conjugation) implies that the zeros of the Riemann zeta function are symmetric about the real axis.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know the zeros are symmetrical. But what about the other values of $\zeta(s)$?&#xA;My main aim is to find out:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Is $\zeta(s)$ symmetrical about the real axis for all $\Re(s) &amp;gt; 1$ ?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2320" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-27T01:02:21.370" LastActivityDate="2010-12-27T01:02:21.370" Title="Is Riemann Zeta Function symmetrical about the real axis?" Tags="&lt;special-functions&gt;&lt;analytic-number-theory&gt;&lt;zeta-function&gt;" CommentCount="12" />
  <row Id="15560" PostTypeId="1" AcceptedAnswerId="15565" CreationDate="2010-12-26T14:51:09.213" Score="2" ViewCount="49" Body="&lt;p&gt;So I had a system&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;#for given koefs &#xA;k:=3; n:=3; &#xA;#let us solve system:&#xA;koefSolution:= solve({&#xA; sum(a[i], i = 0 .. k) = 0,&#xA; sum(a[i], i = 0 .. k)-(sum(b[i], i = 0 .. k)) = 0,&#xA; sum(i^n*a[i], i = 0 .. k)-(sum(i^(n-1)*b[i], i = 0 .. k)) = 0&#xA;});&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;So I have a vector like &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;koefSolution := {  a[0] = 7*a[2]+26*a[3]-b[1]-4*b[2]-9*b[3], &#xA;                   a[1] = -8*a[2]-27*a[3]+b[1]+4*b[2]+9*b[3], &#xA;                   a[2] = a[2], &#xA;                   a[3] = a[3], &#xA;                   b[0] = -b[1]-b[2]-b[3], &#xA;                   b[1] = b[1], b[2] = b[2], &#xA;                   b[3] = b[3]}&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I have &lt;code&gt;a[0]&lt;/code&gt; so I try &lt;code&gt;solve({koefSolution, a[0] = 1});&lt;/code&gt; why it does not solve my system for given &lt;code&gt;a[0]&lt;/code&gt;? ( main point here is to fill koefSolution with given a[] and b[] and optimize. So here I am trying to emulate analytic way of solving my system (It shall be capable to work for any given a[], b[], n, k... so that all students in our class would get good marks=))&lt;/p&gt;&#xA;" OwnerUserId="2065" LastEditorUserId="2065" LastEditorDisplayName="" LastEditDate="2011-02-01T13:06:12.707" LastActivityDate="2011-02-01T13:06:12.707" Title="How to solve such system with given parts of it? (maple)" Tags="&lt;computer-science&gt;&lt;computer-algebra-systems&gt;&lt;maple&gt;" AnswerCount="1" />
  <row Id="15561" PostTypeId="2" ParentId="15556" CreationDate="2010-12-26T15:15:56.757" Score="2" ViewCount="" Body="&lt;p&gt;Which numbers are you using? Positive integer? In this case, you don't have to consider zero (that's why, perhaps, some books says that even numbers start from 2). If 0 is in your number set, then yes, it is divisible by 2.&lt;/p&gt;&#xA;" OwnerUserId="354" LastActivityDate="2010-12-26T15:15:56.757" />
  <row Id="15562" PostTypeId="1" CreationDate="2010-12-26T15:49:24.040" Score="7" ViewCount="170" Body="&lt;p&gt;Last month, I was introduced to the card game SET. The game raises several interesting questions eg what's the probability that n randomly drawn cards contain k sets, is it possible to end with 12 (or 15) cards which contain no sets... I would like to present the solution to a question like this to undergrads, and through it introduce them to an area of mathematics. For example, as &lt;a href=&quot;http://math.stackexchange.com/questions/6190/a-question-related-to-the-card-game-set&quot;&gt;mentioned previously&lt;/a&gt;, there is the issue of maximal caps, which &lt;a href=&quot;http://en.wikipedia.org/wiki/Set_%28game%29#Variations&quot; rel=&quot;nofollow&quot;&gt;Davis and MacLagan's paper&lt;/a&gt; relates to projective space and error-correcting codes. Can you give me any similar ideas? Perhaps relating to a variation of SET &lt;a href=&quot;http://en.wikipedia.org/wiki/Set_%28game%29#Variations&quot; rel=&quot;nofollow&quot;&gt;listed here&lt;/a&gt;?&lt;/p&gt;&#xA;" OwnerUserId="782" LastEditorUserId="782" LastEditorDisplayName="" LastEditDate="2010-12-27T15:14:48.523" LastActivityDate="2010-12-27T15:14:48.523" Title="Mathematics related to the card game SET" Tags="&lt;combinatorics&gt;" AnswerCount="1" CommentCount="8" />
  <row Id="15564" PostTypeId="1" AcceptedAnswerId="15569" CreationDate="2010-12-26T17:03:11.593" Score="2" ViewCount="173" Body="&lt;p&gt;I'm reading a book about Calculus on my own and am stuck at a problem, the problem is&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;There are two circles of radius 2 that have centers on the line x = 1 and pass through the origin. Find their equations.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The equation for circle is $(x-h)^2 + (y-k)^2 = r^2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any hints will be really appreciated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt; Here is what I did. I drew a triangle from the origin and applied the pythogoras theorem to find the perpendicular, the hypotenuse being 2 (radius = 2) and base 1 (because x = 1), the value of y-coordinate is $\sqrt3$. Can anyone confirm if this is correct?&lt;/p&gt;&#xA;" OwnerUserId="5125" LastEditorUserId="5125" LastEditorDisplayName="" LastEditDate="2010-12-27T03:38:22.137" LastActivityDate="2010-12-27T03:38:22.137" Title="Equation for a circle" Tags="&lt;geometry&gt;&lt;analytic-geometry&gt;" AnswerCount="4" CommentCount="1" />
  <row Id="15565" PostTypeId="2" ParentId="15560" CreationDate="2010-12-26T17:05:52.103" Score="1" ViewCount="" Body="&lt;p&gt;In your solution, the free variables are $a[2],a[3],b[1],b[2],b[3]$. Since $a[0]$ is not&#xA;a free variable, setting it equal to 1 in your koefSolution does not yield a solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you can do is to set $a[0]=1$ in your original system, and solve for the rest&#xA;of the variables. I did that and I get:&#xA;$$ a[1]=-1-a[2]-a[3]$$&#xA;$$a[2]=a[2]$$&#xA;$$a[3]=a[3]$$&#xA;$$b[0]=1-7a[2]-26a[3]+3b[2]+8b[3]$$&#xA;$$b[1]=-1+7a[2]+26a[3]-4b[2]-9b[3]$$&#xA;$$b[2]=b[2]$$&#xA;$$b[3]=b[3]$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The free variables now are $a[2],a[3],b[2],b[3]$.&lt;/p&gt;&#xA;" OwnerUserId="3249" LastActivityDate="2010-12-26T17:05:52.103" />
  <row Id="15566" PostTypeId="2" ParentId="15564" CreationDate="2010-12-26T17:11:57.003" Score="4" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt;: Find out what points $(x, y)$ with $x = 1$ also have distance $2$ to the origin. What would these points represent?&lt;/p&gt;&#xA;" OwnerUserId="917" LastEditorUserId="917" LastEditorDisplayName="" LastEditDate="2010-12-26T19:51:43.710" LastActivityDate="2010-12-26T19:51:43.710" CommentCount="4" />
  <row Id="15567" PostTypeId="2" ParentId="15564" CreationDate="2010-12-26T17:13:06.330" Score="3" ViewCount="" Body="&lt;p&gt;Note that you can write down the equation of a circle if you know the co-ordinates of the center $(h,k)$ and its radius $r$. You already know that the two circles in question have radius $2$. It remains to figure out where their centers lie. You are told that their centers lie on the line $x=1$ which is a line parallel to the $Y$-axis. So you know the $x$-coordinate of the centers. The only thing that remains to be figured out are the $y$-coordinates of the two circles. To figure this out, you are given an additional information that both circles pass through the origin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would suggest drawing a picture of the Cartesian plane and of the line $x=1$ on it. You know the centers lie on this line and you know the centers have to be at a certain distance from the origin (why?). Given these two constraints, figure out what possible locations the centers can be at. If this isn't clear enough, post your work and indicate where you are getting stuck.&lt;/p&gt;&#xA;" OwnerUserId="618" LastActivityDate="2010-12-26T17:13:06.330" CommentCount="2" />
  <row Id="15568" PostTypeId="2" ParentId="15564" CreationDate="2010-12-26T17:50:20.120" Score="0" ViewCount="" Body="&lt;p&gt;Rephrasing Dario's hint: the center of each circle is a point of distance $2$ from the origin with $x=1$. If the center is the point $(x,y)$ then $x=1$, and so you can solve for $y$ (remember, the distance between $(x,y)$ and $(0,0)$ is $2$).&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-12-26T17:50:20.120" />
  <row Id="15569" PostTypeId="2" ParentId="15564" CreationDate="2010-12-26T18:01:54.617" Score="5" ViewCount="" Body="&lt;p&gt;Hint: The distance between a point $P(h,k)=P(1,k)$ and the origin $O(0,0)$&#xA;is given by $\sqrt{1+k^{2}}$. This distance should be equal to $2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/MrptX.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="752" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2010-12-26T21:17:10.877" LastActivityDate="2010-12-26T21:17:10.877" CommentCount="4" />
  <row Id="15570" PostTypeId="1" AcceptedAnswerId="15577" CreationDate="2010-12-26T18:50:17.767" Score="5" ViewCount="110" Body="&lt;p&gt;Show that $\displaystyle 1+ \sum\limits_{k=1}^{n} k \cdot k! = (n+1)!$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;RHS: This is the number of permutations of an $n+1$ element set. We can rewrite this as $n!(n+1)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;LHS: It seems that the $k \cdot k!$ has a similar form to $(n+1)! = (n+1)n!$ Also we can write $1 = 0!$ I think you use the mulitplication principle is being used here (e.g. the permutations of a $k$ element set multiplied by $k$).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that a combinatorial proof is wanted (not an algebraic one).&lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2010-12-26T20:01:50.983" Title="Permutation Identity and Sum" Tags="&lt;combinatorics&gt;&lt;permutations&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="15571" PostTypeId="1" CreationDate="2010-12-26T19:03:45.670" Score="10" ViewCount="1157" Body="&lt;p&gt;I'm not going to prove something, this is just a question. From the first day which I went to University until now I had some root problems in some basic mathematical assumptions and concepts. Please help me to solve the most important of those problems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;For example:&lt;/strong&gt; &lt;em&gt;&lt;strong&gt;infinity&lt;/em&gt;&lt;/strong&gt; ! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is infinity ? Is that real ? Is infinity concept real ? I mean in real world. your answer is &quot;Yes&quot; ? What about &lt;a href=&quot;http://en.wikipedia.org/wiki/Zeno%27s_paradoxes&quot; rel=&quot;nofollow&quot;&gt;Zeno's paradox&lt;/a&gt; ? I know that there are different solutions for this paradox ... but i don't talk about solutions. I'm saying &quot;this is logical&quot; because we accept some basic assumptions. But I can not understand it and compare it with real world. .... This problem assumes that there are infinite numbers between 2 numbers OR There are infinite points between two point. But in real world there is &quot;&lt;a href=&quot;http://iopscience.iop.org/0264-9381/20/15/101/fulltext&quot; rel=&quot;nofollow&quot;&gt;smallest distance&lt;/a&gt;&quot;. It's about &lt;a href=&quot;http://en.wikipedia.org/wiki/Planck_length&quot; rel=&quot;nofollow&quot;&gt;1.6*10^-35&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's talk more about my root problems: is number &quot;0&quot; real ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is &lt;strong&gt;&quot;zero&quot;&lt;/strong&gt; ? You can say that &quot;there is 3 apple on the desk, if you take all of them there is 0 apple on the desk ? &quot; I still don't see number &quot;0&quot; ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i know that they are tools which help us to solve different kind of &quot;real world&quot; problems ( I'm a software developer and I always use 0 and negative numbers ). But i can't see it in nature!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What about &lt;a href=&quot;http://en.wikipedia.org/wiki/Empty_set&quot; rel=&quot;nofollow&quot;&gt;empty set&lt;/a&gt; ? I mean this --&gt; &lt;strong&gt;{}&lt;/strong&gt; ? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;there is space between 2 bracket ... how can it be empty ? Is there any empty space in world ? for example a box which isn't any air in it ! It's an empty box ! ... no ... the element &quot;nothing&quot; is there ! &quot;nothing&quot; is a member of a set ... so there is no set with size &quot;zero&quot;. There is always a &quot;nothing&quot; or &quot;null&quot; element in there. Am I right ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's talk about other things which are panics in my mind!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What about &lt;strong&gt;R set&lt;/strong&gt; ?!( &lt;strong&gt;real&lt;/strong&gt; numbers )&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are millions of mathematical theorems which based on R set ! ... Somebody show me a set which is not countable in real world ? I mean in nature.  (I know &lt;a href=&quot;http://theoremoftheweek.wordpress.com/2010/03/18/theorem-21-the-real-numbers-are-uncountable/&quot; rel=&quot;nofollow&quot;&gt;Cantor's theorem&lt;/a&gt; , but it's based on &quot;infinity&quot; concept ! Which i can't understand it ).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again, I know that this kind of mathematics help us to solve real world problems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not talking about logical relationships between those theorems or other things. They are logical if we base them on these concepts. I'm talking about concepts. Infinity concept, &quot;nothing&quot; concept. etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm sorry if my question seems ridicules for someone !&lt;/p&gt;&#xA;" OwnerUserId="5103" LastEditorUserId="5103" LastEditorDisplayName="" LastEditDate="2010-12-29T16:43:37.290" LastActivityDate="2011-05-19T20:01:58.767" CommunityOwnedDate="2010-12-29T16:43:37.290" Title="Do infinity and zero really exist ?" Tags="&lt;number-theory&gt;&lt;infinity&gt;&lt;number&gt;&lt;zeros&gt;" AnswerCount="8" CommentCount="21" FavoriteCount="6" />
  <row Id="15572" PostTypeId="2" ParentId="15505" CreationDate="2010-12-26T19:06:39.663" Score="0" ViewCount="" Body="&lt;p&gt;I don't claim to have a slick combinatorial proof, but I've upvoted the question in hopes that someone will provide one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would cop out and show equality as follows.  Dividing the LHS by the RHS and expanding the definition of the choose notation, we have&#xA;$$ (n-r)\frac{(n + r - 1)!}{r!(n-1)!} \frac{n!}{r!(n-r)!} \cdot \frac{1}{n} \frac{(2r)!(n-r-1)!}{(n+r-1)!} \frac{r!r!}{(2r)!}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By cancelling appropriately, it follows that the LHS divided by the RHS is 1, so they're equal.&lt;/p&gt;&#xA;" OwnerUserId="3537" LastActivityDate="2010-12-26T19:06:39.663" />
  <row Id="15573" PostTypeId="2" ParentId="15571" CreationDate="2010-12-26T19:18:01.497" Score="9" ViewCount="" Body="&lt;p&gt;It's not mathematics but rather physics which is based on these &quot;lies&quot;. In mathematics, we assume (if we're Platonists) that objects like the real numbers &quot;really exist&quot;, just not in the physical world, and then everything makes sense. Some people pretend that when they're doing mathematics, they're just combining axioms and derivation rules to prove theorems; these people have to take as an article of faith the fact that their chosen axiom system (e.g. ZFC) is consistent, otherwise all their toil makes no sense.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When mathematics is applied to the real word, e.g. in physics, then often some approximations are made, like the fact that (at least in classical physics) a Cartesian coordinate system describes space, and the real line describes time. These approximations can be explained mathematically: taking discreteness into account in general will only slightly alter the results; but it complicates everything greatly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some mathematics is not like that, for example when a poll is being taken, statisticians will calculate the standard deviation - this makes sense even if you don't believe in the real numbers; they are just a theoretical construct introduced to understand discrete phenomena. At a final count, everything reduces to finitistic reasoning, whose validity however rests on some unfounded belief in the consistency of some axiom system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Last but not least, why do you oppose zero? If you take a ruler and mark a peg every inch, then you ask yourself &quot;how many pegs do I need to jump from 2in to 3in&quot;? The answer is $1$. Then &quot;how many pegs do I need to jump from 2in to 2in&quot;? The answer is zero. You can understand negative numbers this way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, zero apples is the number of apples that remain after you've eaten all of them. And there are many more examples, in fact an entire book was written on the subject.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-12-26T19:18:01.497" CommentCount="5" />
  <row Id="15574" PostTypeId="2" ParentId="15571" CreationDate="2010-12-26T19:28:45.473" Score="9" ViewCount="" Body="&lt;p&gt;It seems like your premise for mathematics being a 'lie' is that its theoretical background cannot be completely applied to the real world (e.g. having the smallest distance in the real world, but not in mathematics). If that's how you define a lie, sure, but mathematics is an abstract science and I am not sure why you think it's supposed to hold certain properties that are true in the real world -- abstraction is actually the whole point. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solutions to Zeno's paradox have been offered, I am not sure what you mean by &quot;but i don't talk about solutions. I'm saying &quot;this is LOGICAL&quot;. But it's not TRUE.&quot; In fact, Zeno's paradox shows the &lt;em&gt;lie&lt;/em&gt; of the real world, as Zeno contested that motion does not exist and is an illusion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;You wanna tell me that &quot;there is 3 apple on the desk, if you take all of them there is 0 apple on the desk ? &quot; I still don't see number &quot;0&quot; ?&quot; -- true, but neither you see number 3 when there are 3 apples on the desk, correct? 0 is used to denote lack of objects, and it's definition is a tad more complicated than definition of other integers, and for a reason (it can be used differently in different branches of mathematics). Also, if you want a real-life definition of zero, think of not moving, not gaining weight, not making any progress -- all of this can essentially be seen as addition of zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An empty set is a set (which, again, is an abstract concept) that has no elements. Careful with the words -- it's not the case that it has 'nothing' as an element, it's rather the case that it has no elements at all. Here, it comes down to semantics of the definition, &#xA;but you once again seem to be drawing parallels to real life and trying to find a corresponding entity. I do not quite see why that should be the approach -- we are talking about mathematics, not physics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With respect to the set of Real numbers, again, same story. I do not know if you can find an uncountable set in the 'real world' (though I suppose you could contest that the set of points of any object is uncountably infinite, because as you keep zooming in you will be getting more and more particles, and it's our limitations that do not allow us to see past atoms). But that is not the point. Mathematics is based on some abstract ideas, and it's true that not all of them are necessarily reflected in nature. That does not, however, make mathematics based on 'lies.'&lt;/p&gt;&#xA;" OwnerUserId="3731" LastEditorUserId="3731" LastEditorDisplayName="" LastEditDate="2010-12-26T23:24:12.527" LastActivityDate="2010-12-26T23:24:12.527" CommentCount="2" />
  <row Id="15575" PostTypeId="1" AcceptedAnswerId="15576" CreationDate="2010-12-26T19:42:43.670" Score="4" ViewCount="267" Body="&lt;p&gt;Could someone tell me what the &quot;square of a graph&quot; $G^2$ is? Thanks.&lt;/p&gt;&#xA;" OwnerUserId="4654" LastActivityDate="2011-01-03T22:04:32.767" Title="Square of a graph" Tags="&lt;graph-theory&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="1" />
  <row Id="15576" PostTypeId="2" ParentId="15575" CreationDate="2010-12-26T19:56:26.160" Score="5" ViewCount="" Body="&lt;p&gt;The square of a graph $G$ is obtained by starting with $G$, and adding edges between any two vertices whose distance in $G$ is $2$.&lt;/p&gt;&#xA;" OwnerUserId="1508" LastEditorUserId="1508" LastEditorDisplayName="" LastEditDate="2011-01-03T22:04:32.767" LastActivityDate="2011-01-03T22:04:32.767" CommentCount="11" />
  <row Id="15577" PostTypeId="2" ParentId="15570" CreationDate="2010-12-26T20:01:50.983" Score="6" ViewCount="" Body="&lt;p&gt;Here is a brief description of one combinatorial way to approach this. Suppose we are permuting $\{1,2,3,\ldots,n+1\}$. One permutation is $P=(1,2,3,\ldots,n+1)$. Any other permutation has a first position from the left which differs from $P$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $S_1$ be the set of permutations that first differs from $P$ in position $n$. Let $S_2$ be the set of permutations that first differs from $P$ in position $n-1$. Continue this until we get to $S_n$, which is the set of permutations that first differs from $P$ in position $1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When we count the size of $S_i$, we can convince ourselves that it is $(i+1)!-i! = i \cdot i!$. This is the part I'm skipping over. Also counting $P$ then gives us the left hand side.&lt;/p&gt;&#xA;" OwnerUserId="5124" LastActivityDate="2010-12-26T20:01:50.983" />
  <row Id="15578" PostTypeId="2" ParentId="15575" CreationDate="2010-12-26T20:11:31.347" Score="4" ViewCount="" Body="&lt;p&gt;I would have thought that $G^2$ would either mean the box product of $G$ with itself, or the cross product of $G$ with itself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The definitions of these are as follows: If $G$ and $H$ are graphs with vertex sets $V_G$ and $V_H$, then the box product of $G$ and $H$ has vertex set $V_G \times V_H$ and has an edge from $(g_1, h_1)$ to $(g_2, h_2)$ if and only if &lt;em&gt;either&lt;/em&gt; (1) $g_1=g_2$ and there is an edge from $h_1$ to $h_2$ in $H$ or (2) $h_1=h_2$ and there is an edge from $g_1$ to $g_2$ in $G$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The cross product of $G$ and $H$ has vertex set $V_G \times V_H$ and has an edge from $(g_1, h_1)$ to $(g_2, h_2)$ if and only if there is an edge from $g_1$ to $g_2$ in $G$ and an edge from $h_1$ to $h_2$ in $H$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since TonyK has found yet another definition, I would say that there is more than one thing $G^2$ can denote.&lt;/p&gt;&#xA;" OwnerUserId="448" LastActivityDate="2010-12-26T20:11:31.347" CommentCount="4" />
  <row Id="15580" PostTypeId="1" AcceptedAnswerId="15582" CreationDate="2010-12-26T21:24:20.987" Score="2" ViewCount="216" Body="&lt;p&gt;How can I calculate average distance from square center to points inside the square?&lt;/p&gt;&#xA;" OwnerUserId="5130" LastActivityDate="2010-12-26T21:40:15.637" Title="What is average distance from square center to some point?" Tags="&lt;geometry&gt;&lt;probability&gt;" AnswerCount="1" />
  <row Id="15581" PostTypeId="2" ParentId="15541" CreationDate="2010-12-26T21:26:11.467" Score="2" ViewCount="" Body="&lt;p&gt;Thanks for the comments. After giving it more thought I was able to solve it on my own. Here's what I did.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose $U\subset \mathbb{R}^{n+m},V\subset \mathbb{R}^m$ and $n&amp;gt;0$ (if $n&amp;lt;0$ just consider $f^{-1}$ instead of $f$ in what follows). We have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f&amp;#39;=\left(&#xD;&#xA;\begin{array}{cccccc}&#xD;&#xA; \frac{\partial f_1}{\partial x_1} &amp;amp; \ldots  &amp;amp; \frac{\partial f_1}{\partial x_n} &amp;amp; \frac{\partial f_1}{\partial y_1} &amp;amp; \ldots  &amp;amp; \frac{\partial f_1}{\partial y_m} \\&#xD;&#xA; \ldots  &amp;amp; \ldots  &amp;amp; \ldots  &amp;amp; \ldots  &amp;amp; \ldots  &amp;amp; \ldots  \\&#xD;&#xA; \frac{\partial f_m}{\partial x_1} &amp;amp; \ldots  &amp;amp; \frac{\partial f_m}{\partial x_n} &amp;amp; \frac{\partial f_m}{\partial y_1} &amp;amp; \ldots  &amp;amp; \frac{\partial f_m}{\partial y_m}&#xD;&#xA;\end{array}&#xD;&#xA;\right)=\left(&#xD;&#xA;\begin{array}{cc}&#xD;&#xA; \frac{\partial \left(f_1,\ldots ,f_m\right)}{\partial \left(x_1,\ldots ,x_n\right)} &amp;amp; \frac{\partial \left(f_1,\ldots ,f_m\right)}{\partial \left(y_1,\ldots ,y_m\right)}&#xD;&#xA;\end{array}&#xD;&#xA;\right)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\left(f^{-1}\right)&amp;#39;=\left(&#xD;&#xA;\begin{array}{ccc}&#xD;&#xA; \frac{\partial x_1}{\partial f_1} &amp;amp; \ldots  &amp;amp; \frac{\partial x_1}{\partial f_m} \\&#xD;&#xA; \ldots  &amp;amp; \ldots  &amp;amp; \ldots  \\&#xD;&#xA; \frac{\partial x_n}{\partial f_1} &amp;amp; \ldots  &amp;amp; \frac{\partial x_n}{\partial f_m} \\&#xD;&#xA; \frac{\partial y_1}{\partial f_1} &amp;amp; \ldots  &amp;amp; \frac{\partial y_1}{\partial f_m} \\&#xD;&#xA; \ldots  &amp;amp; \ldots  &amp;amp; \ldots  \\&#xD;&#xA; \frac{\partial y_m}{\partial f_1} &amp;amp; \ldots  &amp;amp; \frac{\partial y_m}{\partial f_m}&#xD;&#xA;\end{array}&#xD;&#xA;\right)=\left(&#xD;&#xA;\begin{array}{c}&#xD;&#xA; \frac{\partial \left(x_1,\ldots ,x_n\right)}{\partial \left(f_1,\ldots ,f_m\right)} \\&#xD;&#xA; \frac{\partial \left(y_1,\ldots ,y_m\right)}{\partial \left(f_1,\ldots ,f_m\right)}&#xD;&#xA;\end{array}&#xD;&#xA;\right)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\left(f^{-1}\right)&amp;#39;\cdot f&amp;#39;=\left(&#xD;&#xA;\begin{array}{cc}&#xD;&#xA; \frac{\partial \left(x_1,\ldots ,x_n\right)}{\partial \left(f_1,\ldots ,f_m\right)}\cdot \frac{\partial \left(f_1,\ldots ,f_m\right)}{\partial \left(x_1,\ldots ,x_n\right)} &amp;amp; \frac{\partial \left(x_1,\ldots ,x_n\right)}{\partial \left(f_1,\ldots ,f_m\right)}\cdot \frac{\partial \left(f_1,\ldots ,f_m\right)}{\partial \left(y_1,\ldots ,y_m\right)} \\&#xD;&#xA; \frac{\partial \left(y_1,\ldots ,y_m\right)}{\partial \left(f_1,\ldots ,f_m\right)}\cdot \frac{\partial \left(f_1,\ldots ,f_m\right)}{\partial \left(x_1,\ldots ,x_n\right)} &amp;amp; \frac{\partial \left(y_1,\ldots ,y_m\right)}{\partial \left(f_1,\ldots ,f_m\right)}\cdot \frac{\partial \left(f_1,\ldots ,f_m\right)}{\partial \left(y_1,\ldots ,y_m\right)}&#xD;&#xA;\end{array}&#xD;&#xA;\right)=\left(&#xD;&#xA;\begin{array}{cc}&#xD;&#xA; I_{(n\times n)} &amp;amp; 0 \\&#xD;&#xA; 0 &amp;amp; I_{(m\times m)}&#xD;&#xA;\end{array}&#xD;&#xA;\right)=I_{(n+m\times n+m)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It follows that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{\partial \left(y_1,\ldots ,y_m\right)}{\partial \left(f_1,\ldots ,f_m\right)}\cdot \frac{\partial \left(f_1,\ldots ,f_m\right)}{\partial \left(y_1,\ldots ,y_m\right)}=I_{(m\times m)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the Jacobian matrix $\frac{\partial \left(f_1,\ldots ,f_m\right)}{\partial \left(y_1,\ldots ,y_m\right)}$ is not singular, we can apply the implicit function theorem. It follows that there is a function $g:A\to B$, where $A\subset \mathbb{R}^n,B\subset \mathbb{R}^m$ are open subsets and $A\times B\subset U$, such that $f(x,g(x))=\text{const}$ for all $x\in A$. It is then clear that $f$ cannot be bijective.&lt;/p&gt;&#xA;" OwnerUserId="5119" LastActivityDate="2010-12-26T21:26:11.467" CommentCount="1" />
  <row Id="15582" PostTypeId="2" ParentId="15580" CreationDate="2010-12-26T21:33:41.807" Score="3" ViewCount="" Body="&lt;p&gt;For the average distance, what you want is the expected value of the distance from the center over the set of all points in the square, which is the sum of the probability of each point times the distance to that point.  The problem is (more or less) that there are a whole lot of points and each point has infinitesimal probability.  One way to deal with this is to use calculus.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the square with sides of length 1 parallel to the axes and centered at the origin.  This square is the set of points $(x,y)$ with $-\frac{1}{2}\le x\le\frac{1}{2}$ and $-\frac{1}{2}\le y\le\frac{1}{2}$.  The total area of the square is 1.  The sum of the probabilities of each point (not dealing with distance yet) is $\int_{-\frac{1}{2}}^{\frac{1}{2}}\int_{-\frac{1}{2}}^{\frac{1}{2}}\;dx\;dy=1$.  Now, at each point, multiply the probability by the distance to get the average distance: $\int_{-\frac{1}{2}}^{\frac{1}{2}}\int_{-\frac{1}{2}}^{\frac{1}{2}}\sqrt{x^2+y^2}\;dx\;dy$.&lt;/p&gt;&#xA;" OwnerUserId="72" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-12-26T21:40:15.637" LastActivityDate="2010-12-26T21:40:15.637" CommentCount="5" />
  <row Id="15583" PostTypeId="2" ParentId="11429" CreationDate="2010-12-26T21:56:21.903" Score="2" ViewCount="" Body="&lt;p&gt;Conceptually, the way amortization of loan payments works is that the lender should come out even at the end of the loan when you account for future value of money based on the loan's interest rate.  If the lender loaned $A$ dollars at an interest rate per period $r$ over $n$ periods, the future value (at the end of the loan) of the money lent is $A(1+r)^n$.  If the payment is $p$, then the future value of the first payment is $p(1+r)^{n-1}$ (paid at the end of the first period), for the second is $p(1+r)^{n-2}$, for the $k$th payment is $p(1+r)^{n-k}$, up to the last payment which is paid at the end of the loan so its future value is $p(1+r)^0=p$.  If you take the sum of these future values of all the payments, it should be equal to the future value of the original loan.  This equation can be solved for $p$ in terms of the general $A$, $r$, and $n$ to get a generalized amortization formula.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, on a period-to-period basis, the balance is defined by the initial balance $b_0=A$ and the recurrence relation $b_{k}=(1+r)b_{k-1}-p$.  That is, the change from one period to the next is to add the interest due and subtract the payment.  Knowing the parameters of the loan and the payment, you can use this to find the balance after a specific number of periods, as well as the breakdown of interest and principal in each payment.&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-12-26T21:56:21.903" />
  <row Id="15584" PostTypeId="1" CreationDate="2010-12-26T22:17:49.077" Score="1" ViewCount="83" Body="&lt;p&gt;is integral of (1/x dx) the same as (dx/x) ??&lt;/p&gt;&#xA;" OwnerUserId="1755" LastActivityDate="2010-12-26T22:20:08.567" Title="small difference in integrals" Tags="&lt;integral&gt;" AnswerCount="2" />
  <row Id="15585" PostTypeId="2" ParentId="15584" CreationDate="2010-12-26T22:19:58.700" Score="6" ViewCount="" Body="&lt;p&gt;Yes, $\int \frac{dx}{x}$ is simply shorthand for $\int \frac1x \, dx$; they mean precisely the same thing.&lt;/p&gt;&#xA;" OwnerUserId="30" LastActivityDate="2010-12-26T22:19:58.700" />
  <row Id="15586" PostTypeId="2" ParentId="15584" CreationDate="2010-12-26T22:20:08.567" Score="5" ViewCount="" Body="&lt;p&gt;Yes. $\int \frac{1}{x} \mathrm dx = \int \frac{\mathrm dx}{x}$.&lt;/p&gt;&#xA;" OwnerUserId="3787" LastActivityDate="2010-12-26T22:20:08.567" />
  <row Id="15587" PostTypeId="1" AcceptedAnswerId="15589" CreationDate="2010-12-26T22:20:21.037" Score="-6" ViewCount="153" Body="&lt;p&gt;Update:&#xA;This post was originaly about the loose usage of even and odd, where people without providing a definition of what they meant by even or odd asked inaccurate questions like &quot;is 0 even?&quot;. Although I tried to draw the attention to the fact the historically neither 1 or 0 was considered even or odd ( 0 was not even around when the concept of even or oddness was coined ), it missed the target by making the readers conclude that it was about claiming that 0 is not even. After continued downvoting without leaving constructive comments it was time for an update. So here it goes : if one uses equivalance classes to define 2 mutualy exclusive sets by &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$Evens = \{ 2k : k \in \mathbb Z \} , Odds = \{ 2k+1 : k \in \mathbb Z \} $ &#xA;Using this everyone can agree which set 1 or 0 fall in, regarding 1 and 0 being always considered even or odd there are plenty of math history sites.&lt;/p&gt;&#xA;" OwnerUserId="3936" LastEditorUserId="3936" LastEditorDisplayName="" LastEditDate="2011-05-31T00:03:50.100" LastActivityDate="2011-05-31T00:03:50.100" Title="Asking questions without providing a definition" Tags="&lt;logic&gt;" AnswerCount="2" CommentCount="10" />
  <row Id="15588" PostTypeId="2" ParentId="15556" CreationDate="2010-12-26T22:23:22.207" Score="4" ViewCount="" Body="&lt;p&gt;This problem arose e.g. during the &lt;a href=&quot;http://uk.reuters.com/article/idUKNOA03972720080620&quot; rel=&quot;nofollow&quot;&gt;Beijing even-odd car ban for the 2008 Olympics&lt;/a&gt;, where cars with odd numbered licence plates were banned one day, then even the next day.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The choice is between:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;0 is even and not odd,&lt;/li&gt;&#xA;&lt;li&gt;0 is odd and not even,&lt;/li&gt;&#xA;&lt;li&gt;0 is both even and odd,&lt;/li&gt;&#xA;&lt;li&gt;0 is neither even nor odd (like infinity or $\pi$) or&lt;/li&gt;&#xA;&lt;li&gt;0 is assigned a unique title (like how 1 called a &quot;unit&quot; -- neither prime nor composite).&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This is a matter of definition, so while you could define 0 to be any of the above, it's best to choose the definition that will be the &lt;em&gt;most consistent&lt;/em&gt; with the usage of &quot;even&quot; and &quot;odd&quot; for numbers other than 0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $W=\{2,4,6,\ldots\}$, $V=\{1,3,5,\ldots,\}$ and lets look at the properties of even and odd numbers on these sets that we are familiar with.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;A number is either even or odd, and not both.&lt;/li&gt;&#xA;&lt;li&gt;If $w,x \in W$ then $w+x \in W$ (even + even = even).&lt;/li&gt;&#xA;&lt;li&gt;If $w \in W$ and $v \in V$ then $w+v \in V$ (even + odd = odd).&lt;/li&gt;&#xA;&lt;li&gt;If $y,v \in V$ then $y+v \in W$ (odd + odd = even).&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;[and probably many others I've forgotten to write here]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So it would be desirable that whichever definition we choose for 0, it preserves the above properties.  Now lets say we let 0 be odd (the second and third cases listed above).  Then our definition is not consistent with these properties.  So, if we choose to define 0 as odd, we should have some substantial benefits to outweigh the losses.  On the other hand, defining 0 to be even and not odd is consistent with the above properties.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The last two candidate definitions are essentially saying there is no consistent way of defining even or oddness to 0.  But in this case, there is -- 0 is even and not odd.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[Note: We also have the property that elements of W are all divisible by 2, but whether or not 0 is divisible by 2 is another matter of definition, for which we should again apply the &quot;which is the most sensible definition&quot; concept.]&lt;/p&gt;&#xA;" OwnerUserId="139" LastActivityDate="2010-12-26T22:23:22.207" />
  <row Id="15589" PostTypeId="2" ParentId="15587" CreationDate="2010-12-26T22:55:04.750" Score="7" ViewCount="" Body="&lt;p&gt;You seem to be proposing that it is somehow an error to use a word beyond the context in which it was originally defined.  This is a ludicrous &lt;em&gt;linguistic&lt;/em&gt; claim, let alone a mathematical one.  As the world and our understanding of it changes, the words we use to describe it must necessarily adapt to those changes.  To invent a new word every time we need to label a new concept would be endlessly tedious, and it would ignore the significant compressive power of metaphor.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-12-26T22:55:04.750" CommentCount="13" />
  <row Id="15590" PostTypeId="2" ParentId="15587" CreationDate="2010-12-26T22:56:35.307" Score="4" ViewCount="" Body="&lt;p&gt;$\ldots,-4,-2,0,2,4,\ldots$ are even integers, $ 2,4,6, \ldots$ are even positive integers.&#xA;$\ldots,-5,-3,-1,1,3,5,\ldots$ are odd integers, $ 1,3,5, \ldots$ are odd positive integers.&#xA;What can be simpler than this?&lt;/p&gt;&#xA;" OwnerUserId="3249" LastActivityDate="2010-12-26T22:56:35.307" />
  <row Id="15591" PostTypeId="1" AcceptedAnswerId="15593" CreationDate="2010-12-26T23:01:32.640" Score="1" ViewCount="205" Body="&lt;p&gt;Suppose we have the following two identities: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\displaystyle \sum_{k=0}^{n} \binom{n}{k} = 2^n$&lt;/li&gt;&#xA;&lt;li&gt;$\displaystyle \sum_{k=0}^{n} (-1)^{k} \binom{n}{k} = 0$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The first says that the number of subsets of an $n$-set is $2^n$. The second says that the number of subsets of even size equals the number of subsets of odd size (of an $n$-set). Thus there are $2^{n-1}$ subsets of even length and $2^{n-1}$ subsets of odd length?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To combinatorially prove the second identity, let $A$ be a $k$-subset of $[n]$. Then note whether $k$ is odd or even? &lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2010-12-26T23:06:06.070" Title="Number of even and odd subsets" Tags="&lt;combinatorics&gt;&lt;binomial-coefficients&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="15592" PostTypeId="1" AcceptedAnswerId="15594" CreationDate="2010-12-26T23:02:32.377" Score="4" ViewCount="204" Body="&lt;p&gt;I want to prove that the size (cardinality) of $\mathbb{N}$ is the same as the size of the set of all ordered triples of natural numbers ; just to be more clear that $|\mathbb{N}|=|\mathbb{N}\times\mathbb{N}\times\mathbb{N}|$.&lt;/p&gt;&#xA;" OwnerUserId="5131" LastEditorUserId="3731" LastEditorDisplayName="" LastEditDate="2010-12-27T00:29:43.257" LastActivityDate="2010-12-27T11:40:50.750" Title="function from $\mathbb{N}$ to $\mathbb{N}\times\mathbb{N}\times\mathbb{N}$" Tags="&lt;logic&gt;&lt;elementary-set-theory&gt;" AnswerCount="3" CommentCount="6" />
  <row Id="15593" PostTypeId="2" ParentId="15591" CreationDate="2010-12-26T23:06:06.070" Score="3" ViewCount="" Body="&lt;p&gt;The second identity can be proven by an inclusion-exclusion argument, but maybe it's clearer this way: pick an arbitrary element of the $n$-element set.  Given a subset, if it contains the arbitrary element, remove it; otherwise, add it in.  (That is, take the symmetric difference.)  This defines a bijection from the set of subsets of odd size and the subsets of even size.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The generalization with $2$ replaced by a larger integer is less trivial; see &lt;a href=&quot;http://math.stackexchange.com/questions/918/how-do-i-count-the-subsets-of-a-set-whose-number-of-elements-is-divisible-by-3-4&quot;&gt;this question&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as I can tell you are going through identities you know how to prove algebraically and trying to see how they can be proven combinatorially.  If you are interested in learning how to do this systematically you should start with Benjamin and Quinn's &lt;a href=&quot;http://books.google.com/books?id=n8Xx2oSlz4MC&amp;amp;pg=PA194&amp;amp;lpg=PA194&amp;amp;dq=art+benjamin+count&quot; rel=&quot;nofollow&quot;&gt;Proofs that Really Count&lt;/a&gt; and then, if you are really serious, Bergeron, Labelle, and Leroux's &lt;a href=&quot;http://books.google.com/books?id=83odtWY4eogC&amp;amp;printsec=frontcover&amp;amp;dq=bergeron+labelle+leroux&quot; rel=&quot;nofollow&quot;&gt;Combinatorial Species and Tree-like Structures&lt;/a&gt;.  There are also some blog posts and other material I could link you to if you want.  Asking a question on this more general topic certainly seems more efficient than what you're currently doing.  &lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-12-26T23:06:06.070" CommentCount="2" />
  <row Id="15594" PostTypeId="2" ParentId="15592" CreationDate="2010-12-26T23:16:29.387" Score="6" ViewCount="" Body="&lt;p&gt;Here is one way: I assume $0\in{\mathbb N}$. Given a triple $(a,b,c)$, assign to it the number $2^a(2(2^b(2c+1)-1)+1)-1$. The same idea shows that ${\mathbb N}$ has the same size as ${\mathbb N}^k$ for any positive integer $k$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(This can be easily modified and in fact gives a slightly easier formula if your convention is that the naturals begin with 1.)&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;There are of course many other ways. The idea I am using above is to first find a bijection between ${\mathbb N}\times{\mathbb N}$ and ${\mathbb N}$, and then use it to create the one you want. The bijection I used is $(a,b)\mapsto 2^a(2b+1)-1$. Another popular choice uses &lt;a href=&quot;http://en.wikipedia.org/wiki/Cantor_pairing_function#Cantor_pairing_function&quot; rel=&quot;nofollow&quot;&gt;Cantor's pairing function&lt;/a&gt;, $\displaystyle (a,b)\mapsto \frac{(a+b)(a+b+1)}2 +b$. As a technical comment, there is actually an advantage to Cantor's choice, in that it is a polynomial, and it is easier to define (in a first-order way) in the usual structure of the natural numbers $({\mathbb N},+,\times,0,1,&amp;lt;)$. The choice I used requires exponentiation, and although this can also be defined in a first-order way, it is significantly less trivial (that exponentiation is first-order definable is the bulk of the argument for &lt;a href=&quot;http://en.wikipedia.org/wiki/Hilbert%27s_tenth_problem&quot; rel=&quot;nofollow&quot;&gt;Hilbert's 10th problem&lt;/a&gt;).&lt;/p&gt;&#xA;" OwnerUserId="462" LastEditorUserId="462" LastEditorDisplayName="" LastEditDate="2010-12-26T23:24:26.197" LastActivityDate="2010-12-26T23:24:26.197" CommentCount="2" />
  <row Id="15595" PostTypeId="2" ParentId="15592" CreationDate="2010-12-26T23:16:44.393" Score="3" ViewCount="" Body="&lt;p&gt;Do you know a &lt;a href=&quot;http://en.wikipedia.org/wiki/Pairing_function&quot; rel=&quot;nofollow&quot;&gt;Pairing function&lt;/a&gt;?  Do you see how this can help?&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-12-26T23:16:44.393" />
  <row Id="15596" PostTypeId="1" CreationDate="2010-12-26T23:53:24.907" Score="2" ViewCount="167" Body="&lt;p&gt;Say I have a time series, with actual measurements of a variable $a$ in different locations.  If I want to know the average of $p$ from time $t_1$ to $t_n$, I could say that $\overline{a_{p}}=\frac{\sum\limits_{t=1}^n a_{{p}_t}}{n}$, where $p$ is the coordinate of the point of measurement.  So far, so good.  Let's generalize and call this averaging function $m_p=f(a_p,t)$, where $f(a,t)$ is some deterministic function and $m$ is its result.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, say I want to calculate $m$ for a location $o$, where $o$ is a points for which I don't have actual measurements and $a_o$ was estimated through a an interpolating function $g(a,o)$.  My intuition tells me that in this case, we should first estimate every $a_{o_t}$ before applying $f(a,t)$. In other words, considering that $a_o = g(a,o)$, we get $m_o$ by doing:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$m_o=f[g(a,o),t]=f(a_o,t)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I have seen so many papers that do the other way around.  They first calculate $f(a,t)$ for every known $a$ and then interpolate the results using the same g() even though there is no guarantee that $m$ behaves like $a$, or:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$m_o=g[f(a,t),o]=g(m,o)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In some cases, they use the same logic when instead of $t$, there is some other variable $b$, i.e., it's not a time series, but $f(a,b)$ is a deterministic function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are the models in those papers conceptually wrong?  If they are wrong, is there a techcnical term to call this kind of mistake?&lt;/p&gt;&#xA;" OwnerUserId="5133" LastEditorUserId="2548" LastEditorDisplayName="" LastEditDate="2010-12-26T23:59:01.643" LastActivityDate="2010-12-27T01:25:47.230" Title="Mean of interpolated data or interpolation of means in geostatistics" Tags="&lt;statistics&gt;" AnswerCount="1" />
  <row Id="15597" PostTypeId="1" AcceptedAnswerId="15600" CreationDate="2010-12-27T00:25:54.307" Score="5" ViewCount="202" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Infinity&quot; rel=&quot;nofollow&quot;&gt;This article&lt;/a&gt; on Wikipedia states the following:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Cardinal arithmetic can be used to show not only that the number of points in a real number line is equal to the number of points in any segment of that line, but that this is equal to the number of points on a plane and, indeed, in any finite-dimensional space.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I was wondering why is it the case for finite-dimensional spaces, but not for $\mathbb{R}^{\omega}$. Can't a PMI proof of a bijection between $[0;1]$ and &#xA;$\mathbb{R}^{\omega}$ be established, or would that proof be showing that one can establish a bijection between $[0;1]$ and $\mathbb{R}^n$ for any given $n\in \mathbb{R}/\mathbb{N}$ but not $\mathbb{R}^{\omega}$? Is there something special about infinitely-dimensional $R$ that makes it have more points than $\mathbb{R}$ or any segment on $\mathbb{R}$? Thanks a lot.&lt;/p&gt;&#xA;" OwnerUserId="5135" LastEditorUserId="5135" LastEditorDisplayName="" LastEditDate="2010-12-27T01:12:29.210" LastActivityDate="2010-12-27T01:12:29.210" Title="Is $[0;1]$ equinumerous to $\mathbb{R}^{\omega}$?" Tags="&lt;elementary-set-theory&gt;" AnswerCount="2" CommentCount="9" />
  <row Id="15598" PostTypeId="2" ParentId="15571" CreationDate="2010-12-27T00:26:01.600" Score="18" ViewCount="" Body="&lt;p&gt;Dear Michel,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think your questions and struggles with these difficult concepts are completely understandable, but your presentation is terrible. You are unnecessarily confrontational and aggressive. Having difficulties with these concepts is completely all right. Assuming that during hundreds of years mathematicians conspired to build a make-believe world makes you a crackpot. Hence the series of down-votes.  I think the first thing you should do is to learn humility and present your ideas with respect. Don't assume that just because something does not make sense &lt;em&gt;to you&lt;/em&gt; it necessarily does not make sense. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, putting aside my initial discomfort with your attitude, let me try to say a few words towards resolving your issues. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) &lt;em&gt;infinity&lt;/em&gt; --- This is indeed something hard to grasp and there are grown mathematicians who are advocating similar views. They are a bit more sophisticated and a lot less arrogant though. As for &lt;em&gt;smallest distance&lt;/em&gt;, it has already been pointed out that that would be a statement in physics and not in math. On the other hand, if you believe in dividing integers by integers, you must realize that there are infinitely many numbers between any two numbers or infinitely many points between any two points, just divide the distance by $n$ where $n$ runs through $\{1,2,3,\dots\}$. But of course, you could try to argue that it does not make sense to talk about arbitrarily large numbers such as $2^{2011}$ as they have no practical applications. I can't say anything to that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) &lt;em&gt;zero&lt;/em&gt; --- I always thought that whoever first invented zero and the empty set had to be a genius. These are indeed difficult concepts. However, you are confusing a few things here. You say &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;...&quot;nothing&quot; is a member of a set...so there is no set with size &quot;zero&quot;. There is always a &quot;nothing&quot; or &quot;null&quot; element in there. Am I right ?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No, you are not right.  First off, what do you mean by &quot;nothing&quot;? The only sensible mathematical meaning of that is the &lt;em&gt;empty set&lt;/em&gt;. So, you are talking about a set that consist of one element, which is the empty set. That's different from the empty set that has no elements at all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You also say&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;You wanna tell me that &quot;there is 3 apple on the desk, if you take all of them there is 0 apple on the desk ? &quot; I still don't see number &quot;0&quot; ?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes, you &lt;em&gt;are right&lt;/em&gt;! There is $0$ apple on the desk and you don't see the number $0$. There are many other things you don't see. For example you don't see &#xA;the dark side of the moon, but you would not doubt that it is there. You don't see the Sun at night but you know it exists even at night.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What about negative numbers? Those don't exist either? So when you take out a loan and you owe the bank a ton of money, it is not real? If you accept negative numbers, you have to accept &lt;em&gt;zero&lt;/em&gt; as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) &lt;em&gt;uncountable sets&lt;/em&gt; --- I guess this is a more sophisticated version of your problem with &lt;em&gt;infinity&lt;/em&gt;. So, can we accept the existence of (non-zero ;) natural numbers? Say $1,2,3,\dots$? If so, then it is easy to show you an uncountable set: take the set that consists of all the subsets of $\{1,2,3,\dots\}$.&lt;/p&gt;&#xA;" OwnerUserId="4654" LastActivityDate="2010-12-27T00:26:01.600" CommentCount="3" />
  <row Id="15599" PostTypeId="2" ParentId="15505" CreationDate="2010-12-27T00:35:22.657" Score="3" ViewCount="" Body="&lt;p&gt;Let $S$ be a set of $n+r-1$ elements. Both sides count the number of ways to select two disjoint sets $A,B\subseteq S$ of size $r$ and possibly an element $c\in S \setminus B$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We first observe that $(n-r)\binom{n}{r}=n\binom{n-1}{r}$ as both sides count the number of ways to form a team of size $r+1$ with a captain out of $n$ people.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Applying the above on the original LHS we get $n\binom{n-1}{r}\binom{n+r-1}{r}$, which corresponds to selecting $A$ ($r$ out of $n+r-1$), then $B$ ($r$ out of the remaining $n-1$) and $c$ ($n-1$ choices in $S\setminus B$ and one option of not choosing $c$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The RHS argument goes as follows: choose $2r$ elements for both $A$ and $B$, then choose $r$ of them to make $B$. Then, as before, there are $n$ options of choosing $c\in S\setminus B$ or none at all.&lt;/p&gt;&#xA;" OwnerUserId="5132" LastActivityDate="2010-12-27T00:35:22.657" />
  <row Id="15600" PostTypeId="2" ParentId="15597" CreationDate="2010-12-27T00:37:23.547" Score="11" ViewCount="" Body="&lt;p&gt;Whoever told you that $[0, 1]$ is smaller than $\mathbb{R}^{\omega}$ is wrong; they both have the same cardinality as $\{ 0, 1 \}^{\omega} = \{ 0, 1 \}^{\omega \cdot \omega}$.  By the way, something about your post suggests to me that you are not convinced that this set has a rigorous definition, and it does: $\mathbb{R}^{\omega}$ is precisely the set of functions from $\omega$ to $\mathbb{R}$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are several other ways one might imagine putting infinitely many copies of $\mathbb{R}$ together.  The above is the infinite direct product; there is also the infinite direct &lt;em&gt;sum&lt;/em&gt; $\bigoplus_{n=1}^{\infty} \mathbb{R}$, which consists of all sequences of real numbers which are eventually zero; unlike the infinite direct product, the infinite direct sum is countable-dimensional.  And also one can consider the sequences $c_0$ of real numbers which converge to zero; this is not countable-dimensional but has nice topological properties.&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-12-27T00:43:30.750" LastActivityDate="2010-12-27T00:43:30.750" CommentCount="5" />
  <row Id="15601" PostTypeId="2" ParentId="15597" CreationDate="2010-12-27T01:06:33.313" Score="4" ViewCount="" Body="&lt;p&gt;The cleanest way of seeing the bijection from $\mathbb{R}^{\omega}$ to $\mathbb{R}$ is probably by taking advantage of your favorite bijection from $\mathbb{R}$ to $\left[0,1\right]$, then treating the numbers in that range as a sequence of decimals, 'stacking' them to get a $\omega\times\omega$ grid of digits, and using a diagonal sweep to compose them into a single decimal; so far instance, if the first three elements of your sequence of reals were .314159, .271828, and .161803, then take the first digit of the first, then the second digit of the first and the first digit of the second, then the third digit of the first, the second digit of the second, and the first digit of the first, continuing to interleave this way: .312471...  It's easy to convince yourself that this is a one-to-one bijection; you can go the other way by writing the digits of your single number one by one diagonally into a grid, and then taking the result as a series of reals.&lt;/p&gt;&#xA;" OwnerUserId="785" LastActivityDate="2010-12-27T01:06:33.313" CommentCount="4" />
  <row Id="15602" PostTypeId="1" AcceptedAnswerId="15604" CreationDate="2010-12-27T01:17:16.193" Score="18" ViewCount="625" Body="&lt;p&gt;I am trying to understand the notion of an orientable manifold.&lt;br&gt;&#xA;Let M be a smooth n-manifold. We say that M is orientable if and only if there exists an atlas $A = \{(U_{\alpha}, \phi_{\alpha})\}$ such that $\textrm{det}(J(\phi_{\alpha} \circ \phi_{\beta}^{-1}))&amp;gt; 0$ (where defined). My question is:&lt;br&gt;&#xA;Using this definition of orientation, how can one prove that the Möbius strip is not orientable?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you!&lt;/p&gt;&#xA;" OwnerUserId="4457" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-27T01:52:06.367" LastActivityDate="2010-12-28T23:09:31.700" Title="Why is the Möbius strip not orientable?" Tags="&lt;geometry&gt;&lt;topology&gt;&lt;differential-geometry&gt;&lt;algebraic-topology&gt;" AnswerCount="3" CommentCount="5" FavoriteCount="4" />
  <row Id="15603" PostTypeId="2" ParentId="15596" CreationDate="2010-12-27T01:25:47.230" Score="2" ViewCount="" Body="&lt;p&gt;If your interpolation scheme is linear in the input $a$'s (and all interpolation schemes I can think of are), then there is no difference between the mean of interpolated values and the interpolation of the means.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your notation is a little confusing, so let me put it this way. Suppose you know the values $a_{i,j} = a(p_i,t_j)$ for $m$ points $p_i$ at $n$ times $t_j$. The interpolated value at a point $o$ at any time $t_j$ can typically be expressed as a weighted sum of the values at known points, $$a(o,t_j) := w_1 a_{1,j} + w_2 a_{2,j} + \cdots + w_m a_{m,j}.$$ The mean of these interpolated values is $$\bar a(o) = \frac{1}{n}\sum_j a(o,t_j).$$ If you expand this out, you'll find that it's equal to $$\begin{align}\bar a(o) &amp;amp;= w_1 \frac{1}{n}\sum_j a_{1,j} + w_2 \frac{1}{n}\sum_j a_{2,j} + \cdots + w_m \frac{1}{n}\sum_j a_{m,j} \\ &amp;amp;= w_1 \bar a_1 + w_2 \bar a_2 + \cdots + w_m \bar a_m,\end{align}$$ which amounts to interpolating the means. Presumably this is better from a computational perspective because if you want to find $\bar a(o)$ at many different $o$, you can keep the $\bar a_i$'s precomputed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your strategy of taking the mean of the interpolated values would be necessary if the sample points were moving. In that case, there's no sensible way to interpolate the means, because the interpolation weights would have been changing.&lt;/p&gt;&#xA;" OwnerUserId="856" LastActivityDate="2010-12-27T01:25:47.230" CommentCount="4" />
  <row Id="15604" PostTypeId="2" ParentId="15602" CreationDate="2010-12-27T01:38:55.613" Score="17" ViewCount="" Body="&lt;p&gt;If you had an orientation, you'd be able to define at each point $p$ a unit vector $n_p$ normal to the strip at $p$, in a way that the map $p\mapsto n_p$ is continuous. Moreover, this map is completely determined once you fix the value of $n_p$ for some specific $p$. (You have two possibilities, this uses a tangent plane at $p$, which is definable using a $(U_\alpha,\phi_\alpha)$ that covers $p$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The point is that the positivity condition you wrote gives you that the normal at any $p&amp;#39;$ is independent of the specific $(U_{\alpha&amp;#39;},\phi_{\alpha&amp;#39;})$ you may choose to use, and path connectedness gives you the uniqueness of the map. Now you simply check that if you follow a loop around the strip, the value of $n_p$ changes sign when you return to $p$, which of course is a contradiction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(This is just a formalization of the intuitive argument.)&lt;/p&gt;&#xA;" OwnerUserId="462" LastActivityDate="2010-12-27T01:38:55.613" CommentCount="4" />
  <row Id="15605" PostTypeId="1" AcceptedAnswerId="15611" CreationDate="2010-12-27T01:57:47.227" Score="11" ViewCount="355" Body="&lt;p&gt;Let M be a smooth manifold. We say that M is orientable if and only if there exists an atlas $A = \{(U_{\alpha}, \phi_{\alpha})\}$ such that for all $\alpha, \beta$, $\textrm{det}(J(\phi_{\alpha} \circ \phi_{\beta}^{-1})) &amp;gt; 0$ (where defined). I'm struggling to understand the reason this definition is made. My question is: What is the intuitive reason for this definition of an orientable manifold? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you!&lt;/p&gt;&#xA;" OwnerUserId="4457" LastEditorUserId="640" LastEditorDisplayName="" LastEditDate="2010-12-27T03:45:48.947" LastActivityDate="2010-12-31T15:10:07.457" Title="How should I think about what it means for a manifold to be orientable?" Tags="&lt;geometry&gt;&lt;topology&gt;&lt;differential-topology&gt;" AnswerCount="5" CommentCount="4" FavoriteCount="4" />
  <row Id="15606" PostTypeId="2" ParentId="15605" CreationDate="2010-12-27T02:32:20.457" Score="5" ViewCount="" Body="&lt;p&gt;Get two intersecting charts $(U,\phi)=(U, (x^i))$, $(V,\psi)=(V, (y^i))$, with $p\in U\cap V$. Then there are two basis for the tangent space of $M$ at $p$: $\{ \frac{\partial}{\partial x^\alpha}\}$ and $\{\frac{\partial}{\partial y^\beta}\}$ . Intuitively, these two bases have the same orientation (in the vector space $T_pM$) if it is possible to go &quot;continuously&quot; from one to another. More precisely, they have $the$ $same$ orientation if the determinant of the matrix of change of basis from $\{ \frac{\partial}{\partial x^\alpha}\}$ to $\{\frac{\partial}{\partial y^\beta}\}$ is positive. If this determinant is negative, we will have to change the direction of one or more axes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ok. Given that, the question is: &quot;How to extend this concept of orientations from one tangent space to the entire manifold&quot;? That is where we define the $orientability$ of the manifold: a manifold will be $orientable$ if it admits an atlas such that, given two charts in this atlas with a nonempty intersection, at every point of this intersection the determinant of the matrix of change of basis from one coordinate basis to the other is positive (just like above). That is, the coordinate basis $\{ \frac{\partial}{\partial x^\alpha}\}$, $\{\frac{\partial}{\partial y^\beta}\}$ at a given point generated by any two charts of this &quot;orientable atlas&quot; have all the same orientation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The matrix of change of basis between two coordinate basis is exactly the jacobian of the coordinate transformation.&lt;/p&gt;&#xA;" OwnerUserId="1178" LastActivityDate="2010-12-27T02:32:20.457" />
  <row Id="15607" PostTypeId="2" ParentId="15605" CreationDate="2010-12-27T02:34:12.000" Score="5" ViewCount="" Body="&lt;p&gt;I am of the opinion that overthinking about &lt;em&gt;what things mean&lt;/em&gt; at the time one is first learning them can often be counterproductive. I like to quote, and fully agree with, D'Alambert, who famously said&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Allez en avant, et la foi vous viendra. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(&quot;push on and faith will catch up with you&quot;, according to a translation I just googled for) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for your concrete question: playing tetris, playing with your right hand and linear algebra should have convinced you that a real vector space has two orientations. Now, a manifold provides, among other things, a family of vector spaces with a piece of extra information: a way to move from one of them to others &quot;continuously&quot;. We say that the manifold is orientable if we can pick an orientation in each of them in a continuous way. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I promise you that in not a long time, it'll make way more sense and you'll even find it difficult to explain it to others because it'll seem &lt;em&gt;intuitive&lt;/em&gt; to you!&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2010-12-27T02:34:12.000" CommentCount="7" />
  <row Id="15608" PostTypeId="2" ParentId="15555" CreationDate="2010-12-27T02:36:46.320" Score="3" ViewCount="" Body="&lt;p&gt;What is the value of $\lceil x\rceil$ for small positive $x$? Think of $0&amp;lt;x&amp;lt;1$.&lt;/p&gt;&#xA;" OwnerUserId="2473" LastActivityDate="2010-12-27T02:36:46.320" />
  <row Id="15609" PostTypeId="2" ParentId="15555" CreationDate="2010-12-27T02:39:01.813" Score="0" ViewCount="" Body="&lt;p&gt;$\lim \limits_{x \to 0} \frac{\lceil x \rceil}{x}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.wolframalpha.com/input/?i=lim+%28Ceiling%5bx%5d/x%29+as+x-%3E0&quot; rel=&quot;nofollow&quot;&gt;This is what the Alpha Wolf says&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$\lim \limits_{x \to 0^-} \frac{\lceil x \rceil}{x} = 0$&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$\lim \limits_{x \to 0^+} \frac{\lceil x \rceil}{x} = \infty$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The two limits aren't equal. And what could this possible mean?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How did we get this? Look &lt;a href=&quot;http://www.wolframalpha.com/input/?i=lim%20%28Ceiling%5bx%5d%29%20as%20x-%3E0&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$\lim \limits_{x \to 0^-} \lceil x \rceil = 0$&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$\lim \limits_{x \to 0^+} \lceil x \rceil = 1$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$\lim \limits_{x \to 0^-} x = 0$&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$\lim \limits_{x \to 0^+} x = 0$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2736" LastEditorUserId="2736" LastEditorDisplayName="" LastEditDate="2010-12-27T03:23:23.053" LastActivityDate="2010-12-27T03:23:23.053" CommentCount="6" />
  <row Id="15611" PostTypeId="2" ParentId="15605" CreationDate="2010-12-27T03:43:06.353" Score="6" ViewCount="" Body="&lt;p&gt;I just want to flesh out something Ronaldo said &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Intuitively, these two bases have the same orientation (in the vector space $T_pM$) if it is possible to go &quot;continuously&quot; from one to another.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To this end, consider the vector space $\mathbb{R}^n$.  Let $Gl = Gl_n(\mathbb{R})$ be the set of all invertible linear maps from $\mathbb{R}^n$ to itself and let $M = M_n(\mathbb{R})$ be the set of all linear maps (possibly noninvertible) from $\mathbb{R}^n$ to itself.  The space $M$ is naturally a vector space and is naturally isomorphic to $\mathbb{R}^{n^2}$.  Use this identification to topologize $M$.  The important thing about this topology is that polynomials in the entries of a matrix are continuous - this is all we'll use about it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have a map (the determinant) $det:M\rightarrow \mathbb{R}$, which is a continuous homomorphism (it's given as a polynomial in the entries of the matrix).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can identify $Gl$ with $det^{-1}(\mathbb{R}-\{0\})$, i.e., $Gl$ is an open subset of $M = \mathbb{R}^{n^2}$.  Give $Gl$ the subspace topology.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we get to the main claim:  With this topology, $Gl$ has precisely two path components.  I'll prove this in a second.  In the mean time, notice that this makes what Ronaldo said precise:  If we start in the component of $Gl$ containing the identity matrix, then a path of elements in $Gl$, starting at the identity matrix, gives rise to a path of basis vectors starting with whatever choice of basis you've started with - it gives you a way to continuously go from one basis to another.  But you'll never be able to get to things of opposite orientation because that would require a path from one component of $Gl$ to another, which is impossible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I'll establish the claim that $Gl$ has precisely two components.  First notice that is must have at least two components because $det:Gl\rightarrow \mathbb{R}^n-\{0\}$ is continuous, surjective, and the range has two components.  Let $X^+\subseteq Gl$ be those matrices of positive determinant and $X^-$ those of negative determinant, so we have $Gl = X^+\cup X^-$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next, I claim that $X^+$ and $X^-$ are homeomorphic.  To see this, let $A\in X^-$ be anything at all and consider the map $f:X^+\rightarrow X^-$ given by $f(B) = BA$.  Since matrix multiplication is given by polynomially in the entries of the matrices, $f$ is continuous.  It has inverse $f^{-1}(B) = BA^{-1}$, which is continuous for the same reason $f$ is.  Thus, $f$ is a homeomorphism.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All the following maps and paths will be continuous for the same reason: they'll be given as polynomials in the entries of the matrices involved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, I just need to show $X^+$ is path connected.  So, let $A\in X^+$ be arbitrary.  I want to connect it to the identity matrix by a path.  Let $E_{ij}$ denote the matrix of all 0s with a 1 in the $ij$th slot.  Let $F_{ij}(t) = I + tE_{ij}$.  Notice that for $i\neq j$, $F_{ij}$ is triangular, so has determinant 1.  Left multiplying by $F_{ij}(t)$ corresponds to the row operation of adding $t$ times the $j$th row to the $i$th row, which doesn't change the determinant if $i\neq j$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the $F_{ij}(t_0)$ are just row operations, we know we can multiply by several of them (each with a different appropriate $t_0$) to bring $A$ to a diagonal form without changing the determinant.  By multiplying by $F_{ij}(t)$ and letting $t$ go from $0$ to $t_0$, we make a path out of this.  Doing this for all the $F_{ij}(t_0)$ simultaneously gives us a path from $A$ to a diagonal matrix where for all time, the determinant is constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We've now connected $A$ to a diagonal matrix $D$ by a path in $X^+$.  It remains to connect the diagonal matrix to the identity.  We know all the entries of $D$ are nonzero because $D$ has positive determinant.  If an entry $d$ of $D$ is positive, use the path $t + (1-t)d$ in that slot to continuously change $d$ to $1$, all the while keeping the determinant nonzero.  If $d$ is negative, using the same trick, we can change $d$ to $-1$.  At this point, the determinant must be $1$ - it's positive because we're in $X^+$ and it's a product of $\pm 1$s.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We've now reduced $D$ to a new diagonal matrix $E$ whose entries are plus and minus 1s.  If all are positive, we're done, so assume we have a $-1$ somewhere.  Since $E$ is in $X^+$, there must be another entry which is $-1$.  For notation, let's assume one $-1$ is in the $i$th position and the other is in the $j$th position.  Think of the two $-1$s as $cos(t\pi)$ and think of the $ij$th entry and $ji$th entries as $sin(t\pi)$, all with $t=1$.  Now, let t go from 1 to 0.  One can easily check the determinant stays $1$, and this gives a path which changes the two $-1$s to $1$s.  Doing this finitely many times eliminates all the $-1$s and changes $E$ to the identity matrix.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, we've connected $A$ to the identity matrix by a path, so $X^+$ is connected.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Kudos to anyone who reads the whole thing - it ended up being WAY longer than I anticipated, but it was a good exercise for me to go through).&lt;/p&gt;&#xA;" OwnerUserId="331" LastEditorUserId="331" LastEditorDisplayName="" LastEditDate="2010-12-31T15:10:07.457" LastActivityDate="2010-12-31T15:10:07.457" CommentCount="2" />
  <row Id="15612" PostTypeId="1" CreationDate="2010-12-27T04:04:50.973" Score="1" ViewCount="177" Body="&lt;p&gt;Let $a,b$ be roots ($a\ne \pm b$) of a Lie algebra $g$ of type $X$, where $X$ can be classic or exceptional $(A,B,C,D,E,F,G)$. It is well known that the length of an $a$-string through $b$ is at most 4.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are all types of $g$ such that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) $a+b$ is a root and $a+2b$ is not a root?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) $a+2b$ is a root and $a+3b$ is not a root?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) $a+3b$ is a root?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="2764" LastEditorUserId="2764" LastEditorDisplayName="" LastEditDate="2010-12-29T02:21:15.510" LastActivityDate="2010-12-29T02:21:15.510" Title="Length of root strings." Tags="&lt;lie-algebra&gt;" CommentCount="2" FavoriteCount="2" />
  <row Id="15614" PostTypeId="2" ParentId="15605" CreationDate="2010-12-27T04:41:21.350" Score="4" ViewCount="" Body="&lt;p&gt;Suppose the universe has shape $M$.  Then $M$ is orientable if and only if you can write yourself a note, then begin a journey, and come back to always find that you can still read the note.  If $M$ is not orientable, it is possible to return from a journey to find that the note is written backwards and, indeed, that everything around you is written backwards, your friends' faces are all reversed, etc.  More seriously, the proteins in your body would have the wrong chirality and I think this would cause problems involving digestion.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is not known, to the best of my knowledge, whether our universe is orientable.  If you are interested in this more intuitive approach to topology and geometry before attacking its rigorous foundations, I recommend reading Weeks' &lt;a href=&quot;http://books.google.com/books?id=A8WBiUWy3SgC&amp;amp;printsec=frontcover&amp;amp;dq=the+shape+of+space&quot; rel=&quot;nofollow&quot;&gt;The Shape of Space&lt;/a&gt; and being very diligent about doing the exercises.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-12-27T04:41:21.350" CommentCount="4" />
  <row Id="15615" PostTypeId="2" ParentId="15542" CreationDate="2010-12-27T05:22:21.563" Score="7" ViewCount="" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;It may seem strange to define $0\cdot\infty=0$.  However, one verifies without difficulty that with this definition the &lt;em&gt;commutative, associative, and distributive laws hold in $[0,\infty]$ without any restriction&lt;/em&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The way this is worded leads naturally to your question, as though Rudin were implying that this is the main justification for defining $0\cdot\infty$ in this way.  Rather, I see this as a bonus after making the convention consistent with what happens when integrating the $0$ function or integrating over a space of measure $0$, as KCd's comment indicates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you ask yourself what the possibilities are, you can start by supposing that $0\cdot\infty=x$ for some $x\in[0,\infty]$, and apply the distributive law to see that $x=2x$, so that $x=0$ or $x=\infty$.  You can then verify that with either convention the commutative, associative, and distributive laws will hold, so something more is needed to motivate the choice.  Such a choice will always depend on context, and in some cases it won't be a good idea to even define $0\cdot\infty$.  However, I am not aware of a mathematical context in which the convention $0\cdot\infty=\infty$ is useful.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-12-27T05:22:21.563" />
  <row Id="15616" PostTypeId="1" CreationDate="2010-12-27T05:45:41.520" Score="2" ViewCount="167" Body="&lt;p&gt;Suppose c is an isolated point in the domain D of a function f.&#xA;In the delta neighbourhood of c, does the function f have the value f(c).&lt;/p&gt;&#xA;" OwnerUserId="5049" LastActivityDate="2010-12-27T06:18:20.590" Title="Continuity of a function at an isolated point" Tags="&lt;real-analysis&gt;" AnswerCount="2" CommentCount="5" />
  <row Id="15617" PostTypeId="2" ParentId="15616" CreationDate="2010-12-27T06:10:56.420" Score="3" ViewCount="" Body="&lt;p&gt;I see now that the comments above provide essentially the answer, with whatever definition of continuity you have. The following ties everything together.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's use as the definition of continuity $\lim_{x \rightarrow c}\,f(x) = f(c)$. Expand: For all $\varepsilon &amp;gt; 0$ there exists $\delta &amp;gt; 0$ such that whenever $0 &amp;lt; |x-c| &amp;lt; \delta$ it is true that $|f(x)-f(c)| &amp;lt; \varepsilon$. When $\delta$ is small enough, there are no points $x$ which work, so the part after the such that is &lt;a href=&quot;http://en.wikipedia.org/wiki/Vacuous_truth&quot; rel=&quot;nofollow&quot;&gt;vacuously true&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="5124" LastActivityDate="2010-12-27T06:10:56.420" CommentCount="2" />
  <row Id="15618" PostTypeId="2" ParentId="15616" CreationDate="2010-12-27T06:18:20.590" Score="5" ViewCount="" Body="&lt;p&gt;You can also see this is true using the topological definition of continuity at a point: a function is continuous at a point $f(x)$ if for any neighborhood $V$ of $f(x)$ there is a neighborhood $U$ of $x$ such that $f(U)$ is contained in $V$.  For an isolated point, you can take the neighborhood consisting of just the point $c$, so its image $f(c)$ will obviously be contained in $V$, as $V$ is a neighborhood of $f(c)$.&lt;/p&gt;&#xA;" OwnerUserId="1769" LastActivityDate="2010-12-27T06:18:20.590" />
  <row Id="15619" PostTypeId="2" ParentId="15571" CreationDate="2010-12-27T07:06:29.017" Score="4" ViewCount="" Body="&lt;p&gt;First of all please agree my apologies because of my attitude and the way I asked my question. I changed it again. I really didn't aggressive in reality just present my question in a bad way. But believe me or not, whenever I asked this kind of questions from my teachers they down voted me in real world. I should ask my question in more warily way. As Yuan commented in my question &quot;I'm not going to get any upvotes from christians.stackexchange.com if I ask them &quot;is Christianity based on lies?&quot;&quot;. BTW, I didn't sleep last night and thinking about my question and these answers and comments.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I read all these 3 answers many times. And thank you. I think the real answer is combination of these 3 answer:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Mathematics is different from Physics. We can't completely compare them with each other. Mathematics is an abstract science while Physics is natural science. So they are different types.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;We should choose an axiomatic system before talking about a theorem. ( like ZFC ).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;About infinity: This is a part of Mathematics not Physics. We can imagine that there is a number bigger than any number. That number is Infinity ! ... I don't know if it exists in physics or not. For example &quot;is world infinity or not !?&quot;... &quot;Is there a new world inside a black hole or not!&quot; ... But in mathematics we can imagine it. And we can work with it without much problem. In many parts we see unbelievable theorems( in physics ) when we assume infinity is exists. But they are logical. So we can accept them. Infinity is just a tool. Nothing else I think!&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;About Zero: zero is a concept. It's not real. We can't see it. But this concept is exists. Show me the word &quot;lie&quot;. What is a &quot;lie&quot; ? I can't see it. but this is a concept and exists. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Empty set: I can't imagine a set which don't have any element in Physics and reality. But I can imagine it in my mind. I can imagine &quot;nothing&quot;. it's a &lt;strong&gt;concept&lt;/strong&gt; again. so we call it &quot;empty set&quot; in mathematics. A set that there is nothing in there. I think I just playing with words in my question. nothing is not an element. it's a word. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Uncountable sets: If I can accept infinity. I can accept this too. So I have no word to say about.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I think this answer is a combination of these answers. I found my answers I think. Thank you again for helping me get out of this confusion. But please let me know if this answer is wrong !&lt;/p&gt;&#xA;" OwnerUserId="5103" LastEditorUserId="5103" LastEditorDisplayName="" LastEditDate="2010-12-27T17:18:28.257" LastActivityDate="2010-12-27T17:18:28.257" CommentCount="3" />
  <row Id="15620" PostTypeId="1" AcceptedAnswerId="15621" CreationDate="2010-12-27T07:32:31.437" Score="4" ViewCount="186" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;In &lt;a href=&quot;http://en.wikipedia.org/wiki/Well-order&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;, well-order is&#xA;defined as a strict total order on a set S&#xA;with the property that every&#xA;non-empty subset of S has a least&#xA;element in this ordering. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But then &lt;a href=&quot;http://en.wikipedia.org/wiki/Well-order#Equivalent_formulations&quot; rel=&quot;nofollow&quot;&gt;later&lt;/a&gt;, well-order is&#xA;defined as a total order on S with&#xA;the property that every non-empty&#xA;subset of S has a least element in&#xA;this ordering. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as I know, a total order and&#xA;a strict total order are different.&#xA;One is not the other. So I was&#xA;wondering if well-order is defined&#xA;for total order or strict total&#xA;order or both? If for both, are they&#xA;equivalent in the sense that if a&#xA;total order is well-order, then its&#xA;corresponding strict total order is&#xA;also well-order? Vice versa?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;At the same &lt;a href=&quot;http://en.wikipedia.org/wiki/Well-order&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt; page, it&#xA;also says &quot;a well-ordering is a&#xA;well-founded strict total order&quot;. As&#xA;I clicked into the definition of&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Well-founded_relation&quot; rel=&quot;nofollow&quot;&gt;Well-founded_relation&lt;/a&gt;, it says&#xA;&quot;a binary relation, R, is&#xA;well-founded (or wellfounded) on a&#xA;class X if and only if every&#xA;non-empty subset of X has a minimal&#xA;element with respect to R&quot;. As&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Minimal_element&quot; rel=&quot;nofollow&quot;&gt;minimal element&lt;/a&gt; is defined for&#xA;partial order not for strict total order, is it true that&#xA;well-founded order is a partial&#xA;order and not a strict total order? So the&#xA;aforementioned &quot;a well-ordering is a&#xA;well-founded strict total order&quot; is&#xA;not well-stated?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Thanks and regards! &lt;/p&gt;&#xA;" OwnerUserId="1281" LastEditorUserId="1281" LastEditorDisplayName="" LastEditDate="2010-12-27T07:38:59.953" LastActivityDate="2010-12-27T08:13:17.187" Title="questions about well-order" Tags="&lt;elementary-set-theory&gt;&lt;order-theory&gt;" AnswerCount="3" />
  <row Id="15621" PostTypeId="2" ParentId="15620" CreationDate="2010-12-27T08:02:38.707" Score="4" ViewCount="" Body="&lt;p&gt;For the first question, taking strict and non-strict orders to be well-ordering is up to you and the way you use it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, in set theory it is generally easier to use strict ordering in the definition because it saves the trouble with $x\le y\wedge y\le x$, furthermore we want $\in$ to define some relations and it has to be strict by the axiom of foundation (also known as axiom of regularity in some parts of the globe). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for minimal elements, they are not only for partial orders - but for any order. If $R$ is some relation on $A$ then $x$ is $R$-minimal if $\forall y (yRx \rightarrow y=x)$, note by the way that if $R$ is not reflexive then this is still true, but you could phrase it as $\forall y\neg(yRx)$ instead, which is clearer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The best way, in my opinion to understand deeply these choices of definitions is to study some set theory theorems about recursive definitions, transfinite inductions and the needed theorems for those. In these proofs it becomes very clear why one prefer strict relations over non-strict ones.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One final remark, although strict and non-strict total orders are &quot;very different&quot;, they only differ by reflexivity which is some vacuous condition that you want to add when it's easier to have it - and you remove it when you find it easier to handle without it.&lt;/p&gt;&#xA;" OwnerUserId="622" LastActivityDate="2010-12-27T08:02:38.707" />
  <row Id="15622" PostTypeId="2" ParentId="15620" CreationDate="2010-12-27T08:07:42.543" Score="5" ViewCount="" Body="&lt;p&gt;&lt;br&gt;1. If a set is totally ordered, the strict total order is implied. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Definition of total order from Wikipedia:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\forall a, b, c \in X$&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If $a ≤ b$ and $b ≤ a$ then $a = b$ (antisymmetry);&lt;br&gt;&#xA;  If $a ≤ b$ and $b ≤ c$ then $a ≤ c$ (transitivity);&lt;br&gt;&#xA;  $a ≤ b$ or $b ≤ a$ (totality).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Definition of strict total order from Wikipedia:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\forall a, b \in X$&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$a &amp;lt; b$ if and only if $a ≤ b$ and $a ≠ b$&lt;br&gt;&#xA;  $a &amp;lt; b$ if and only if not $b ≤ a$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you compare the definitions, you will see that there are no conditions under which the second set of rules is invalid, assuming the first set of rules. So, saying that a set is well-ordered if it is totally ordered makes sense, because total order implies existence of strict total order (and hence, well-ordering) on the given set.&lt;br&gt;&lt;br&gt;&#xA;2. I am fairly sure that one can use the term 'minimal element' with respect to totally ordered sets also.&lt;/p&gt;&#xA;" OwnerUserId="3731" LastEditorUserId="3731" LastEditorDisplayName="" LastEditDate="2010-12-27T08:13:17.187" LastActivityDate="2010-12-27T08:13:17.187" CommentCount="4" />
  <row Id="15623" PostTypeId="2" ParentId="15620" CreationDate="2010-12-27T08:09:47.487" Score="7" ViewCount="" Body="&lt;p&gt;If $\leq$ is a total order on a set $S$, then the new relation $&amp;lt;$ defined by $x &amp;lt; y$  iff ($x \leq y$ and $x \neq y$) is a strict total order on $S$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $&amp;lt;$ is a strict total order on a set $S$, then the new relation $\leq$ defined by &#xA;$x \leq y$  iff ($x &amp;lt; y$ or $x = y$) is a total order on $S$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, in a fairly evident way one can always exchange a total order for a strict total order and conversely.  So it doesn't really matter which definition is taken.  One can easily check that the definition of a well-order in one setting carries over to the definition of a well-order in the other setting.  &lt;/p&gt;&#xA;" OwnerUserId="299" LastActivityDate="2010-12-27T08:09:47.487" CommentCount="3" />
  <row Id="15624" PostTypeId="1" AcceptedAnswerId="15683" CreationDate="2010-12-27T08:34:15.023" Score="5" ViewCount="266" Body="&lt;p&gt;I have a polygon in 2D (defined by a series of Vertex $V$ with coordinates). The polygon can be convex or concave. I have $n$ number of fix points I can put inside the polygon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is, how can I distribute the fix points as uniformly as possible inside the polygon? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The motivation for this question is I want to create a mesh generator and I want all the triangular elements $E$ ( which is defined by a list of vertices $V$) to look good with no too small or too large angles. In order to control the granularity of the mesh I am thinking about using the number of fix points $n$ as the controlling parameter. The fix points are used to control the vertex of the triangular elements. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any algorithm for this?&lt;/p&gt;&#xA;" OwnerUserId="550" LastEditorUserId="1489" LastEditorDisplayName="" LastEditDate="2010-12-27T23:09:31.777" LastActivityDate="2010-12-28T14:35:23.953" Title="Distribute a Fixed Number of Points &quot;Uniformly&quot; inside a Polygon" Tags="&lt;geometry&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="15625" PostTypeId="2" ParentId="15624" CreationDate="2010-12-27T09:05:53.300" Score="6" ViewCount="" Body="&lt;p&gt;The simplest idea is to generate a sequence of points $(X_i,Y_i)$, where $X_i,Y_i$ are independent uniform random variables, in a rectangle containing the polygon (the minimal rectangle bounding the polygon). Then, you simply drop the points which fall outside the polygon. You continue until you get $n$ unrejected points.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As we could expect, this problem has been considered before by many people. You might find &lt;a href=&quot;http://www.gamedev.net/community/forums/topic.asp?topic_id=585566&quot; rel=&quot;nofollow&quot;&gt;this one&lt;/a&gt; very useful. The basic question is how fast/simple you want to generate the $n$ points. The method I described is most simple, but might be time consuming.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;: To add some mathematics to this post, let us justify the rejection method and triangulation approaches (though they might seem intuitively clear). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rejection method approach. Suppose that $(X,Y)$ is uniformly distributed in a rectangle $R$ bounding the polygon $M$, and that $S$ is an arbitrary square contained in $M$. Then,&#xA;$$&#xD;&#xA;{\rm P}((X,Y) \in S |(X,Y) \in M) = \frac{{{\rm P}((X,Y) \in S )}}{{{\rm P}((X,Y) \in M)}} = &#xD;&#xA;\frac{{{\rm area}(S) /{\rm area}(R)}}{{{\rm area}(M)/{\rm area}(R)}} = \frac{{\rm area}(S) }{{{\rm area}(M)}}.&#xD;&#xA;$$ &#xA;Hence, given that $(X,Y) \in M$, $(X,Y)$ is uniformly distributed in $M$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Triangulation approach. Suppose that the polygon $M$ is partitioned into triangles $T_1,\ldots,T_m$, and set $p_i = {\rm area} (T_i)/ {\rm area}(M)$. Suppose that ${\rm P}((X,Y) \in T_i)=p_i$ and that, given $(X,Y) \in T_i$, $(X,Y)$ is uniformly distributed in $T_i$. Now, for fixed $k$, let $S \subset M$ an arbitrary square contained in $T_k$ (think of arbitrarily small squares). Then,&#xA;$$&#xD;&#xA;{\rm P}((X,Y) \in S) = \sum\limits_{i = 1}^m {{\rm P}((X,Y) \in S|(X,Y) \in T_i )p_i } = {\rm P}((X,Y) \in S|(X,Y) \in T_k )p_k.&#xD;&#xA;$$&#xA;Since, given that $(X,Y) \in T_k$, $(X,Y)$ is uniformly distributed in $T_k$, we have&#xA;$$&#xD;&#xA;{\rm P}((X,Y) \in S|(X,Y) \in T_k )  = \frac{{{\rm area}(S)}}{{{\rm area}(T_k )}}.&#xD;&#xA;$$&#xA;Hence, &#xA;$$&#xD;&#xA;{\rm P}((X,Y) \in S) = \frac{{{\rm area}(S)}}{{{\rm area}(T_k )}}\frac{{{\rm area}(T_k )}}{{{\rm area}(M)}} = \frac{{{\rm area}(S)}}{{{\rm area}(M)}}.&#xD;&#xA;$$&#xA;We conclude that $(X,Y)$ is uniformly distributed in $M$.&lt;/p&gt;&#xA;" OwnerUserId="2810" LastEditorUserId="2810" LastEditorDisplayName="" LastEditDate="2010-12-28T14:35:23.953" LastActivityDate="2010-12-28T14:35:23.953" CommentCount="6" />
  <row Id="15626" PostTypeId="2" ParentId="7867" CreationDate="2010-12-27T09:36:07.397" Score="0" ViewCount="" Body="&lt;p&gt;Say you have $n$ data points $\{x_1,x_2,x_3,\ldots,x_n\}$. You have a range of values with which to compare them $[a,b]$. Then, as Ross suggests, I would also assign a score to this set of data points. I would base it on the standard deviation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Values that fall inside the interval get score zero, values $x_k$ that fall outside the interval get score $\min((x_k-a)^2,(x_k-b)^2)$. But if you prefer to take just absolute deviations like Ross, just adapt the formula. Then simply add all scores for all the data points to obtain one score for the entire set of data points. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The lower the score, the closer the points are to the interval, the higher the farther they lie out. If you dispose of several data sets with different scores, you can define percentage deviations by looking at what percentage of data have a lower score than the data set you are examining.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I made a small R script that illustrates the idea:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;n=1000;&#xA;size=10;&#xA;big&amp;lt;-matrix(rnorm(n,0,1),n%/%size,size);&#xA;liminf=-1;&#xA;limsup=1;&#xA;z=rep(0,length(big[,1]);&#xA;perc=z;&#xA;for (k in 1:length(big[,1]))&#xA;{ z[k]&amp;lt;-sum((big[k, big[k,]&amp;lt;liminf]-liminf)^2)+sum((big[k,big[k,]&amp;gt;limsup]-limsup)^2) ; }&#xA;mat&amp;lt;-matrix(c(1,2),1,2);&#xA;layout(mat);&#xA;hist(z,10,main=&quot;Score histogram&quot;);&#xA;for (k in 1:length(big[,1]))&#xA;{ perc[k]=length(z[z&amp;lt;z[k]])/length(z); }&#xA;plot(sort(z),sort(perc),main=&quot;Cumulative plot&quot;);&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;It makes a big matrix of 100 rows of each ten columns. Each row is to be interpreted as one data set. The data have been sampled from a random normal distribution, but you can adapt to any type of data of course. Then, I calculate the score of each row with respect to a chosen interval, here the interval $[-1,1]$. I plotted a score histogram as a bonus. After that, I compute for each score how many percent of the scores is smaller than that score. That's what the &lt;code&gt;perc&lt;/code&gt; is for. I also plotted the cumulative function for the scores based on the computed function &lt;code&gt;perc&lt;/code&gt;.&lt;/p&gt;&#xA;" OwnerUserId="3567" LastEditorUserId="3567" LastEditorDisplayName="" LastEditDate="2010-12-27T13:09:44.917" LastActivityDate="2010-12-27T13:09:44.917" />
  <row Id="15627" PostTypeId="1" CreationDate="2010-12-27T09:38:34.543" Score="-4" ViewCount="387" Body="&lt;p&gt;Just solve it with your own mind without having a look to other answers !&lt;/p&gt;&#xA;" OwnerUserId="5139" LastActivityDate="2010-12-27T09:44:47.807" ClosedDate="2010-12-27T09:58:55.183" Title="1=5, 2=10, 3=15, 4=20 then 5=?" Tags="&lt;puzzle&gt;" AnswerCount="1" CommentCount="6" />
  <row Id="15628" PostTypeId="2" ParentId="15627" CreationDate="2010-12-27T09:44:47.807" Score="2" ViewCount="" Body="&lt;p&gt;I see two ways you can look at this. One appears to be obvious (and is probably not something you are looking for, or rather something that you might expect us to say) -- $1=5$ means $1*5=5$, $2=10$ means $2*5=10$, and so on. Then, $5=?$ means $5*5=25$. Since this is too obvious, the answer could be $5=1$ (which is the very first equality that you provided). I think one could find some other possible answers, too. :)&lt;/p&gt;&#xA;" OwnerUserId="3731" LastActivityDate="2010-12-27T09:44:47.807" CommentCount="7" />
  <row Id="15629" PostTypeId="1" AcceptedAnswerId="15633" CreationDate="2010-12-27T11:09:31.710" Score="4" ViewCount="501" Body="&lt;p&gt;How do I correctly represent the following pseudocode in math notation?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT1: Formula expanded.&#xA;EDIT2: Clarification.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(a,b) represents a line segment on a 1D line. a &amp;lt;= b for each segment. The division show below is done as per the following T-SQL code (which I suppose could be represented as a function in the formula?):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Input:  @a1 real, @b1 real, @an real, @bn real&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;DECLARE @Result real&#xA;&#xA;if @a1 &amp;lt;= @an begin&#xA;    SET @Result = @an - @b1&#xA;&#xA;    if @Result &amp;lt;= 0 RETURN 0&#xA;&#xA;    RETURN @Result / @an&#xA;end&#xA;&#xA;SET @Result = @a1 - @bn&#xA;&#xA;if @Result &amp;lt;= 0 RETURN 0&#xA;&#xA;RETURN @Result / @a1&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Formula:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;if m = 1 then&#xA;  if (a,b)_1 intersects (a,b)_n then&#xA;    r = 1&#xA;  else if (a,b)_1 &amp;lt; (a,b)_n then&#xA;    r = (a,b)_1 / (a,b)_n&#xA;  else&#xA;    r = (a,b)_n / (a,b)_1&#xA;else if m = 2 then&#xA;  if (a,b)_1 intersects (a,b)_n then&#xA;    r = 1&#xA;  else if (a,b)_1 &amp;lt; (a,b)_n then&#xA;    r = (a,b)_1 / (a,b)_n&#xA;  else&#xA;    r = (a,b)_n / (a,b)_1&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The m = 2 block is shown as being the same as the m = 1 one for simplicity's sake.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The divisions are against the two points that are closets to each other, unless the segments intersect, at which point r = 1.&lt;/p&gt;&#xA;" OwnerUserId="4930" LastEditorUserId="4930" LastEditorDisplayName="" LastEditDate="2010-12-27T11:52:53.543" LastActivityDate="2010-12-28T06:44:31.907" Title="Representing IF ... THEN ... ELSE ... in math notation" Tags="&lt;computer-science&gt;&lt;notation&gt;" AnswerCount="4" CommentCount="12" />
  <row Id="15630" PostTypeId="2" ParentId="15592" CreationDate="2010-12-27T11:40:50.750" Score="3" ViewCount="" Body="&lt;p&gt;I will not give you an answer but a geometrical figure in proving what is equivalent to the naturals. You can think that the natural numbers is an elastic rope (infinite from one side )&#xA;and the natural numbers are marked points on this rope .If you want to prove that the integers have the same cardinality as $N^2$ you can think that $N^2$ is a lattice in $R^2$ and you have to create an 1-1 and surjective function or start passing the &quot;natural&quot; rope through the lattice such that every natural number fits on every vertex of the lattice .You can see then what is the function i.e. 1 goes to (0,1) 2 goes to (1,1) 3 goes to (1,0) 4 to (2,0) and you create a spiral round the (0,0).It is an 1-1 surjective function that sends n to (exercise).The same goes to 3 or n dimensions. Of course you can have different ways to use the elastic natural rope to create bijective functions (of course you must have a way to pass the rope through every vertex so usually you make a spiral in a way that grows biger and biger)  but you can have an image that $N$ have the same cardinality as $N^k$ or $Z^k$ for any k.  &lt;/p&gt;&#xA;" OwnerUserId="5010" LastActivityDate="2010-12-27T11:40:50.750" CommentCount="1" />
  <row Id="15631" PostTypeId="1" AcceptedAnswerId="15695" CreationDate="2010-12-27T11:54:30.867" Score="3" ViewCount="242" Body="&lt;p&gt;Hello, the problem is the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For all $x \in \mathbb{R}$, we assign a finite set $\Phi(x) \subset \mathbb{R} - \{ x \}$. We say that a set $S \subset \mathbb{R}$ is independent if for any $x,y \in S$, $x \notin \Phi(y)$ and $y \notin \Phi(x)$, that is, if $S \cap \Phi(S) = \emptyset$. Prove that there is an independent set $S$ with the cardinality of the continuum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any ideas?&#xA;Thanks in advance.&lt;/p&gt;&#xA;" OwnerUserId="2926" LastEditorUserId="856" LastEditorDisplayName="" LastEditDate="2010-12-27T12:20:19.963" LastActivityDate="2010-12-28T00:31:36.213" Title="Problem in set theory" Tags="&lt;set-theory&gt;" AnswerCount="2" CommentCount="7" />
  <row Id="15632" PostTypeId="1" AcceptedAnswerId="15636" CreationDate="2010-12-27T12:40:31.217" Score="1" ViewCount="154" Body="&lt;p&gt;Given implicit function F(x,y) = 0, how can I find its asymptotes?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT:  Sorry, my calculations were wrong. Here is correct function:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$F(x,y)=\sqrt{(x-a)^2 + (y-b)^2} - \sqrt{(x-c)^2 + (y-d)^2}  - e$&lt;/p&gt;&#xA;" OwnerUserId="3532" LastEditorUserId="3532" LastEditorDisplayName="" LastEditDate="2010-12-27T16:04:28.563" LastActivityDate="2010-12-28T03:34:17.990" Title="finding hyperbola asymptotes" Tags="&lt;geometry&gt;&lt;analytic-geometry&gt;&lt;plane-curves&gt;&lt;conic-sections&gt;" AnswerCount="2" CommentCount="8" />
  <row Id="15633" PostTypeId="2" ParentId="15629" CreationDate="2010-12-27T12:44:26.410" Score="6" ViewCount="" Body="&lt;p&gt;In general, if you have &quot;If $\varphi$ then $\psi$, otherwise $\tau$&quot; you can write this as the following formula (or sentence, depending on $\varphi,\psi,\tau$):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\varphi\rightarrow\psi\wedge\neg\varphi\rightarrow\tau$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have several cases, you can either nest them (i.e. $\tau$ would be &quot;if second condition then, else ...&quot;) or if you can express them as $\varphi_1$ meaning only the first case holds, and none of the others as $(\varphi_1\rightarrow\psi_1)\wedge(\varphi_2\rightarrow\psi_2)\wedge\ldots$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Addendum:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(a=b\rightarrow x=1)\wedge(a&amp;lt;b\rightarrow x=\frac{a}{b})\wedge(a&amp;gt;b\rightarrow x=\frac{b}{a})$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have added $x$ as a variable, because writing just $\rightarrow 1$ seems very meaningless to me, you can of course replace $x$ by anything you'd like.&#xA;The idea, essentially is that you express &quot;IF ... THEN ... ELSE&quot; structures using the $\rightarrow$ (or $\implies$ sometimes) and I gave you in my original post the method of doing so.&lt;/p&gt;&#xA;" OwnerUserId="622" LastEditorUserId="622" LastEditorDisplayName="" LastEditDate="2010-12-27T13:34:22.457" LastActivityDate="2010-12-27T13:34:22.457" CommentCount="9" />
  <row Id="15634" PostTypeId="1" CreationDate="2010-12-27T13:02:08.373" Score="6" ViewCount="234" Body="&lt;p&gt;Let $f$ be a function of period 1, Riemann integrable on [0,1]. Let $\xi_n$ be a sequence&#xA;which is equidistributed in $[0,1)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(a) Is it true that $$\frac{1}{N}\sum_{n=1}^N f(x+\xi_n)$$ converges to the constant $\int_0^1 f(y) dy$&#xA;for each $x\in \mathbb{R}$ as $N\to \infty$ ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(b) If so, is the convergence uniform over all $x$?&lt;/p&gt;&#xA;" OwnerUserId="3249" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-27T22:10:46.240" LastActivityDate="2010-12-28T01:41:03.847" Title="Equidistributed sequence and Riemann integrable function" Tags="&lt;analysis&gt;&lt;sequences-and-series&gt;&lt;integral&gt;&lt;fourier-analysis&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="15635" PostTypeId="2" ParentId="15562" CreationDate="2010-12-27T13:18:12.697" Score="1" ViewCount="" Body="&lt;p&gt;While I personally don't remember the mathematics, I can give you two leads which, if you follow it up, should give you some ideas. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Try contacting &lt;a href=&quot;http://www.math.wisc.edu/~ellenber/&quot; rel=&quot;nofollow&quot;&gt;Jordan Ellenberg&lt;/a&gt;. Several years ago he gave a talk to the Princeton University Undergraduate Math Club on precisely what you described in your question. He will definitely be able to give you some pointers. &lt;/li&gt;&#xA;&lt;li&gt;Also, try asking &lt;a href=&quot;http://www-math.mit.edu/people/profile.php?pid=1272&quot; rel=&quot;nofollow&quot;&gt;Aaron Potechin&lt;/a&gt;. he is an expert Set player and I remember seeing a pre-print of his on the Maximal Cap issue a few years ago, when he was still an undergrad. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1543" LastActivityDate="2010-12-27T13:18:12.697" CommunityOwnedDate="2010-12-27T13:18:12.697" CommentCount="1" />
  <row Id="15636" PostTypeId="2" ParentId="15632" CreationDate="2010-12-27T13:26:35.290" Score="3" ViewCount="" Body="&lt;p&gt;After rationalization, your implicit equation can be expressed in standard quadratic form as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$px^2+qxy+ry^2+sx+ty+u=0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$p=4(a-c)^2-4e^2$$&lt;br&gt;&#xA;$$q=8(a-c)(b-d)$$&lt;br&gt;&#xA;$$r=4(b-d)^2-4e^2$$&lt;br&gt;&#xA;$$s=4(a+c)e^2-4(a-c)\left(a^2+b^2-c^2-d^2\right)$$&lt;br&gt;&#xA;$$t=4(b+d)e^2-4(b-d)\left(a^2+b^2-c^2-d^2\right)$$&lt;br&gt;&#xA;$$u=e^4-2\left(a^2+b^2+c^2+d^2\right)e^2+\left(a^2+b^2-c^2-d^2\right)^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The earlier version of my answer would now apply, but since Isaac has written a clearer way of doing it, I'll demonstrate the alternative &quot;dumb&quot; approach (&quot;dumb&quot; in the sense of not exploiting the structure of the problem).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again, before you go through the trouble of doing these computations, you'd first want to ensure that the discriminant $\Delta=64e^2((a-c)^2+(b-d)^2-e^2)$ is negative.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Having done so, the idea is as follows: replace all instances of $y$ in the implicit equation $F(x,y)=0$ with $mx$, divide by $x^2$ (since the degree of the algebraic equation is 2), and let $x\to\infty$. This yields the quadratic equation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$rm^2+qm+p=0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which should be solved for $m$; the initial check of the discriminant would ensure that both roots of that quadratic in $m$ are real.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Letting $m_i$ be any of the two roots $m_1$ and $m_2$ of that quadratic, one now goes back to the original implicit equation, and replaces $y$ with $m_i x+k_i$. From this, note that the quadratic terms disappear, and only the linear terms remain. Divide by $x$, let $x\to\infty$, equate the remaining terms to 0, and solve for the appropriate value of $k_i$. If you do that, the expression you should get for $k_i$ is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$k_i=-\frac{tm_i+s}{2rm_i+q}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;from which $y=m_1 x+k_1$ and $y=m_2 x+k_2$ are your two asymptotes.&lt;/p&gt;&#xA;" OwnerUserId="498" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-28T03:34:17.990" LastActivityDate="2010-12-28T03:34:17.990" CommentCount="3" />
  <row Id="15637" PostTypeId="2" ParentId="15634" CreationDate="2010-12-27T13:34:51.860" Score="3" ViewCount="" Body="&lt;p&gt;For question (a), see &lt;a href=&quot;http://en.wikipedia.org/wiki/Equidistributed_sequence&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; (the section &quot;Properties&quot;); for a proof, see &lt;a href=&quot;http://individual.utoronto.ca/hannigandaley/equidistribution.pdf&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2810" LastEditorUserId="2810" LastEditorDisplayName="" LastEditDate="2010-12-27T14:05:35.863" LastActivityDate="2010-12-27T14:05:35.863" CommentCount="2" />
  <row Id="15638" PostTypeId="1" AcceptedAnswerId="15639" CreationDate="2010-12-27T13:41:33.467" Score="5" ViewCount="364" Body="&lt;p&gt;Can you make a simple example of an uncountable ordinal? With &lt;i&gt;simple&lt;/i&gt; I mean that it is easy to prove that the ordinal is uncountable. I know that the set of all the countable ordinals is an uncountable ordinal, but the only proof that I know is quite complicated.&lt;/p&gt;&#xA;" OwnerUserId="354" LastEditorUserId="630" LastEditorDisplayName="" LastEditDate="2010-12-28T02:33:55.830" LastActivityDate="2010-12-28T02:33:55.830" Title="Simple example of uncountable ordinal" Tags="&lt;set-theory&gt;&lt;cardinals&gt;&lt;ordinals&gt;" AnswerCount="4" CommentCount="2" />
  <row Id="15639" PostTypeId="2" ParentId="15638" CreationDate="2010-12-27T13:49:05.897" Score="10" ViewCount="" Body="&lt;p&gt;The proof that the set $\Omega$ of all countable ordinals is uncountable is not difficult. First, it's an ordinal. Next, if $\Omega$ were countable, then $\Omega + 1$ would also be a countable ordinal. Finally, it is impossible for $\Omega + 1$ to be in $\Omega$, because that would mean that $\Omega + 1 &amp;lt; \Omega$. &lt;/p&gt;&#xA;" OwnerUserId="630" LastActivityDate="2010-12-27T13:49:05.897" CommentCount="10" />
  <row Id="15640" PostTypeId="2" ParentId="15638" CreationDate="2010-12-27T14:02:18.020" Score="4" ViewCount="" Body="&lt;p&gt;The only two that I think have simple proofs are $\Omega$, as Carl says, and $2^{\aleph_0}$, the set of infinite binary sequences.  &lt;a href=&quot;http://en.wikipedia.org/wiki/Cantor%27s_diagonal_argument&quot; rel=&quot;nofollow&quot;&gt;Cantor's diagonal argument&lt;/a&gt; proves this one uncountable.  Alternately, as all cardinals are ordinals, any cardinal greater than $\aleph_0$ is an uncountable ordinal just from the definition of countability.  That's simple to write, but probably not what you are looking for.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-12-27T14:02:18.020" CommentCount="5" />
  <row Id="15641" PostTypeId="1" CreationDate="2010-12-27T14:07:04.457" Score="2" ViewCount="66" Body="&lt;p&gt;Given an abelian group $X$, let $F_n(X)$ denote the simplicial abelian group defined as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$F_n(X)_j=0$ for all $j&amp;lt;n$ and $F_n(X)_j=X$ for all $j\geq n$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with the appropriate zero and identity maps between them so the normalization (normalized moore complex) gives the appropriate homology (say, $d_i=0$ for every $i\geq 1$ (obviously for the parts higher than $n$) and $d_0=id_X$).  Then by the Hurewicz theorem, this simplicial abelian group seems like it should have homotopy concentrated in degree $n$ with $n$th component isomorphic to $X$ by computing the homology of the normalization.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since this thing is a Kan complex (because it's a simplicial abelian group), isn't this a representative for the Eilenberg-Mac Lane space $\kappa(X,n)$?&lt;/p&gt;&#xA;" OwnerUserId="5142" LastActivityDate="2010-12-27T14:07:04.457" Title="What's stopping me from choosing the nth Eilenberg Mac Lane space to be the following simplicial abelian group?" Tags="&lt;homotopy-theory&gt;" CommentCount="3" />
  <row Id="15642" PostTypeId="2" ParentId="15638" CreationDate="2010-12-27T14:07:34.300" Score="5" ViewCount="" Body="&lt;p&gt;The existence of an uncountable ordinal is established by the argument of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Burali-Forti_paradox&quot; rel=&quot;nofollow&quot;&gt;Burali-Forti &quot;Paradox&quot;&lt;/a&gt;.  See for instance Section 1.4 &lt;a href=&quot;http://math.uga.edu/~pete/settheorypart3.pdf&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.  (Yes, this is the same argument as in Carl Mummert's answer.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is probably the simplest way to show the existence of an uncountable ordinal.  As Section 1.2 of the notes linked to above shows, the usual operations of addition, multiplication and even exponentiation of ordinals do not produce uncountable ordinals from countable ones.  That this is not the case for exponentiation is a big difference between ordinal and cardinal arithmetic.&lt;/p&gt;&#xA;" OwnerUserId="299" LastActivityDate="2010-12-27T14:07:34.300" CommentCount="2" />
  <row Id="15643" PostTypeId="2" ParentId="15631" CreationDate="2010-12-27T14:58:33.393" Score="3" ViewCount="" Body="&lt;p&gt;Here is a sketch of a proof (it uses axiom of choice).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's introduce a equivalence relation on real numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We say that two numbers $x$ and $y$ are &quot;equal&quot; of there exists a finite&#xA;set of reals $a_0 = x$, $a_1$, $a_2$, $\cdots$, $a_{n-1}$, $a_n = y$ such&#xA;that for any $i$ we have $a_i \in \Phi(a_{i+1})$ or $a_{i+1} \in \Phi(a_i)$.&#xA;Note that $n$ could be zero, so $x$ and $x$ are also considered equal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's clear that this equivalence relation is symmetric, reflexive and transitive.&#xA;Hence we can consider equivalence classes under this relation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, every equivalence classes is countable (because $\Phi(x)$ is finite for all $x$).&#xA;Hence we have splitted reals into a set of disjoint countable sets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This implies that number of such sets is a continuum.&#xA;Now we can just take any element from each set to form $S$.&lt;/p&gt;&#xA;" OwnerUserId="334" LastActivityDate="2010-12-27T14:58:33.393" CommentCount="1" />
  <row Id="15644" PostTypeId="1" AcceptedAnswerId="15645" CreationDate="2010-12-27T15:19:25.287" Score="0" ViewCount="201" Body="&lt;p&gt;If we only consider the natural numbers, is 3/2 undefined? If not what is the answer?&lt;/p&gt;&#xA;" OwnerUserId="5141" LastActivityDate="2010-12-28T01:56:53.580" Title="Is 3/2 undefined when only considering the natural numbers?" Tags="&lt;natural-number&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="15645" PostTypeId="2" ParentId="15644" CreationDate="2010-12-27T15:22:27.863" Score="10" ViewCount="" Body="&lt;p&gt;It depends on what you mean by $3/2$, of course.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But in the usual intepretation where $3/2$ denotes the number such that when it is multiplied by $2$ gives you $3$, then, if you are restricting 'number' to mean 'natural number', then there is no such number.&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2010-12-27T15:22:27.863" CommentCount="1" />
  <row Id="15646" PostTypeId="1" AcceptedAnswerId="15652" CreationDate="2010-12-27T15:24:09.367" Score="3" ViewCount="191" Body="&lt;h3&gt;Problem&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Given:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\vec r = r(\sin \theta \cos \phi, \sin \theta \sin \phi, \cos \theta)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$S: | \vec r  | = a$, with $\hat n$ outwards&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\vec r_0 = \frac{3a}{5}(\hat x + \hat y - \hat z) = \frac{3a}{5}(1,1,-1)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\vec F = k \frac{\vec r - \vec r_0}{ {| \vec r - \vec r_0 |}^3 }$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Calculate:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_S \vec F \cdot \mathrm d \vec S $$&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Solution (added after accepted answer)&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Thanks to the accepted answer which confirmed that the explicit calculation on/in $S$ &lt;strong&gt;is&lt;/strong&gt; messy, and the suggestion of using translation invariance, I have noted down the solution below (let me know if you have further suggestions).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An argument is used which concludes that the integrand is zero inside the sphere $S$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using Gauss' theorem: $\int_S \vec F \cdot \mathrm d \vec S = \int_V \nabla \cdot \vec F \mathrm dV $. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the calculation is not easy to explicitly calculate in (or on) $S$, instead it is shown that $\nabla \cdot \vec F = 0$ exterior to $S_{\epsilon}$, which is a sphere containing the singularity, and since $S$ is exterior to $S_{\epsilon}$ the integrand must be $0$ and so $\int_V \nabla \cdot \vec F \mathrm dV = 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$S_{\epsilon}: |\vec r - \vec r_0 | = \epsilon, \epsilon &amp;gt; 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;translate the coordinate system so that $S_{\epsilon}$ is the origin in the translated system:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\vec R = \vec r - \vec r_0 = R (\sin \theta \cos \phi, \sin \theta \sin \phi, \cos \theta)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$F(\vec R) = k \frac{\vec R}{|\vec R|^3} = k \frac{1}{R^2} \hat R$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$S_{\epsilon}: |\vec R| = \epsilon$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\nabla \cdot \vec F (\vec R) = \frac{1}{R^2} \frac{\partial}{\partial R} \left(R^2 \vec F_R \right) = \frac{1}{R^2} \frac{\partial}{\partial R} \left(R^2 k \frac{1}{R^2} \right) = 0$$, ($k$ is a constant)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is true for any value of $R$ except at the singularity ($\epsilon$ can be made as small as required), since $S$ does not contain the singularity, the integrand ($\nabla \cdot \vec F(\vec r)$ is 0 and the integral is therefore zero.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;My question (updated)&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;I am having difficulties in explicitly calculating the value of this integral. &#xA;Specifically, the divergence of the field becomes messy, I am not able to see how I can use the symmetry of $S$ due to $\vec r_0$. (When applying the Gauss' theorem, I am stuck in evaluating the divergence of the field). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am able to argue for that this integral is indeed zero (the field only has one singularity and it is exterior to the sphere $S$). However, I am not able to explicitly show (by calculation) that this integral is zero. Thankful for any help.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please note that this is not homework, I am studying for an exam.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;My question (original)&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;What is an easy way to calculate this integral? Any suggestions on the approaches below?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My apologies if this is due to lack of some basic knowledge (I am back studying after 2.5 years)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Intuitively I understand the integral is zero (the point charge is located outside the sphere, anything flowing into the sphere will also flow out), however, I have issues with the calculation.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Calculate directly&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;$$\int_S \vec F \cdot \mathrm d \vec S = \int_0^{\pi} \mathrm d \theta \int_0^{2\pi} \mathrm d \phi r^2 \sin {\theta} \vec F_r \cdot \hat r = \int_0^{\pi} \mathrm d \theta \int_0^{2\pi} \mathrm d \phi r^2 \sin \theta k \frac{a - \frac{3a}{5} \left(  \sin \theta \cos \phi + \sin \theta \sin \phi - \cos \theta) \right)}{ {\left| \vec r - \vec r_0 \right|}^3 }$$&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Using Gauss theorem&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;I am tempted to use Gauss theorem, $\int_S \vec F \cdot \mathrm d \vec S = \int_V \nabla \cdot \vec F \mathrm dV $, in spherical coordinates. However $\vec F$ has components also in $\hat \theta$ and $\hat \phi$ (due to $\vec r_0$) and ${\left| \vec r - \vec r_0 \right|}^3$ is not that nice to derivate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\vec F_r = k \frac{\vec r - \vec r_0}{ {\left| \vec r - \vec r_0 \right|}^3 } \cdot \hat r = k \frac{r - \frac{3a}{5} \left(  \sin \theta \cos \phi + \sin \theta \sin \phi - \cos \theta) \right)}{ {\left| \vec r - \vec r_0 \right|}^3 }$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\vec F_{\theta} = k \frac{\vec r - \vec r_0}{ {\left| \vec r - \vec r_0 \right|}^3 } \cdot \hat \theta = k \frac{ \frac{3a}{5} \left( \cos \theta \cos \phi + \cos \theta \sin \phi - \sin \theta) \right)}{ {\left| \vec r - \vec r_0 \right|}^3 }$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\vec F_{\phi} = k \frac{\vec r - \vec r_0}{ {\left| \vec r - \vec r_0 \right|}^3 } \cdot \hat \phi = k \frac{ \frac{3a}{5} \left( - \sin \phi + \cos \phi ) \right)}{ {\left| \vec r - \vec r_0 \right|}^3 }$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\nabla \cdot \vec F = \frac{1}{r^2} \frac{\partial}{\partial r} \left(r^2 \vec F_r \right) + \frac{1}{r \sin \theta} \frac{\partial}{\partial \theta} \left( \sin \theta F_{\theta} \right) + \frac{1}{r \sin \theta} \frac{\partial F_{\phi}}{\partial \phi}$$&lt;/p&gt;&#xA;" OwnerUserId="5464" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-28T01:16:24.663" LastActivityDate="2010-12-28T01:29:32.880" Title="Calculate surface integral of point charge located outside the surface" Tags="&lt;calculus&gt;&lt;vector-calculus&gt;" AnswerCount="2" CommentCount="8" FavoriteCount="2" />
  <row Id="15647" PostTypeId="2" ParentId="15644" CreationDate="2010-12-27T15:36:32.723" Score="1" ViewCount="" Body="&lt;p&gt;Kind'a similar analogy... $\sqrt{-1}$ is undefined when considering only real numbers. So mathematicians derived the whole complex number system to give that some meaning.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Its really defining the undefined that has progressed mathematics. So the answer to your question is, 3/2 is defined but only if you extend the natural numbers to a more generalized number system, like positive reals. :)&lt;/p&gt;&#xA;" OwnerUserId="2320" LastActivityDate="2010-12-27T15:36:32.723" CommentCount="2" />
  <row Id="15648" PostTypeId="1" CreationDate="2010-12-27T15:48:02.487" Score="1" ViewCount="180" Body="&lt;p&gt;Show that $f_{n-1} + L_n = 2f_{n}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we need to find a $2$ to $1$ correspondence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Set 1: Tilings an $n$-board.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Set 2: Tiling of an $n-1$-board or tiling of an $n$-bracelet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we need to decompose a tiling of an $n$-board to a tiling of an $n-1$-board or a tiling of an $n-1$-bracelet? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Source: &lt;em&gt;Proofs that Really Count&lt;/em&gt; by Art Benjamin and Jennifer Quinn&lt;/p&gt;&#xA;" OwnerUserId="2548" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-12-28T21:41:51.063" LastActivityDate="2011-05-22T12:00:29.797" Title="Lucas Numbers and Tilings" Tags="&lt;combinatorics&gt;&lt;fibonacci-number&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="15650" PostTypeId="1" CreationDate="2010-12-27T16:12:34.003" Score="2" ViewCount="188" Body="&lt;p&gt;Does f(n+1)/f(n) converge as n-&gt;infinity for f(n) defined by a linear recursion, for all linear recursions?&lt;/p&gt;&#xA;" OwnerUserId="5108" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2011-06-03T13:53:11.673" LastActivityDate="2011-06-03T13:53:11.673" Title="Linear Recursion. Do ratio of consecutive terms converge for ALL linear recursions?" Tags="&lt;linear-algebra&gt;&lt;convergence&gt;&lt;recurrence-relations&gt;" AnswerCount="3" CommentCount="4" />
  <row Id="15651" PostTypeId="2" ParentId="15632" CreationDate="2010-12-27T16:31:18.333" Score="3" ViewCount="" Body="&lt;p&gt;The points $(a,b)$ and $(c,d)$ are the foci of the hyperbola and the hyperbola is the locus of points for which the difference of the distances to the foci is $|e|$, though because your $F(x,y)=0$ is only looking at the difference in one direction, your graph is probably only half of the hyperbola (the other half being $0=\sqrt{(x-a)^2+(y-b)^2}-\sqrt{(x-c)^2+(y-d)^2}+e$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The asymptotes of the hyperbola pass through the center, which is the midpoint of the segment joining the foci, $$\left(\frac{a+c}{2},\frac{b+d}{2}\right).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $f=\sqrt{(a-c)^2+(b-d)^2}$, the distance between the foci, and let $g=\sqrt{f^2-e^2}$.  If the foci had the same $x$-coordinate (that is, if the hyperbola were horizontally-oriented), the slopes of the asymptotes would be $\pm\frac{g}{e}$, so the acute angles the asymptotes make with the line through the foci are $\pm\arctan\frac{g}{e}$.  The acute angle the line through the foci makes with the $x$-axis is $\arctan\frac{b-d}{a-c}$, so the angles from the $x$-axis to the asymptotes are $\arctan\frac{b-d}{a-c}\pm\arctan\frac{g}{e}$, and the slopes of the asymptotes are $$\tan\left(\arctan\frac{b-d}{a-c}\pm\arctan\frac{g}{e}\right)=\frac{be-de+ag-cg}{ae-ce-bg+dg}\text{ or }\frac{be-de-ag+cg}{ae-ce+bg-dg}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With a point and the slopes, you should be able to write equations of the asymptotes.&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-12-27T16:31:18.333" CommentCount="1" />
  <row Id="15652" PostTypeId="2" ParentId="15646" CreationDate="2010-12-27T16:54:54.907" Score="3" ViewCount="" Body="&lt;p&gt;(I don't know what you mean by $\hat x+\hat y -\hat z$, so I disregard the information about $r_0$.) Your field $\vec F$ is the gravitational field produced by a point mass at $r_0$. The divergence of this field is $\equiv0$ away from $r_0$. If the point $r_0$ is in the exterior of the sphere $S$ then it follows by Gauss' theorem that the integral in question is $0$. If $r_0$ lies in the interior of $S$ then you should draw a tiny sphere $S_\epsilon$ with center $r_0$ and apply Gauss' theorem to the region between $S_\epsilon$ and $S$. You are left with an integral over $S_\epsilon$ which you can almost do in your head.&lt;/p&gt;&#xA;" OwnerUserId="1303" LastActivityDate="2010-12-27T16:54:54.907" CommentCount="4" />
  <row Id="15653" PostTypeId="2" ParentId="15249" CreationDate="2010-12-27T17:03:42.800" Score="2" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Note&lt;/strong&gt;: I'm adding my original comment as an answer at the OP's suggestion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe that this notion of infinite set is called a &lt;a href=&quot;http://en.wikipedia.org/wiki/Dedekind-infinite_set&quot; rel=&quot;nofollow&quot;&gt;Dedekind infinite set&lt;/a&gt;. The Wikipedia article states some equivalent conditions under the section &lt;a href=&quot;http://en.wikipedia.org/wiki/Dedekind-infinite_set#Dedekind-infinite_sets_in_ZF&quot; rel=&quot;nofollow&quot;&gt;Dedekind infinite sets in ZF&lt;/a&gt; where it says explicitly that the Axiom of Choice is not required to prove their equivalence. In particular, one of the equivalent conditions for $A$ to be Dedekind infinite is that $A$ has a countably infinite subset.&lt;/p&gt;&#xA;" OwnerUserId="900" LastEditorUserId="900" LastEditorDisplayName="" LastEditDate="2010-12-27T17:59:29.050" LastActivityDate="2010-12-27T17:59:29.050" CommentCount="3" />
  <row Id="15654" PostTypeId="2" ParentId="15605" CreationDate="2010-12-27T17:13:36.763" Score="4" ViewCount="" Body="&lt;p&gt;I think it is very hard to get a better answer than any of the above. But there is a perspective that has not been mentioned, at least not explicitly. First, I would like to say that I happen to agree with Marianos answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The body of your question is much more explicit than your title, so I will address the question in your title since it seems others have addressed the one in the body of your question. I like to think of orientations, or in fact most structures on a manifold, in terms of the tangent bundle on the manifold in question. The tangent bundle on a compact n dimensional manifold $M$ is classified (up to isomorphism) by a homotopy class of map $M \to BO(n)$. If we can lift this map over the map $BSO(n) \to BO(n)$ then the bundle is orientable, and hence so is the manifold. Here, the notion of orientability, our structure is really a fact about the tangent bundle. We can talk about almost complex structures or spin structures in a similar way. The map $BSO(n) \to BO(n)$ is induced by the inclusion $SO(n) \to O(n)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another way is to look at $\Lambda^n TM$, the top exterior power of the tangent bundle of M. Then an orientation is a nowhere zero continuous section of this bundle. Similarly, you can think of an orientation as a choice of generator of the top dimensional integral homology or cohomology. These are all probably less helpful than the above answers, but these are the ways I like to think about what an orientation is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;please let me know if you want me to expand on some of these.&lt;/p&gt;&#xA;" OwnerUserId="627" LastActivityDate="2010-12-27T17:13:36.763" CommentCount="1" />
  <row Id="15655" PostTypeId="2" ParentId="11763" CreationDate="2010-12-27T17:28:46.127" Score="2" ViewCount="" Body="&lt;p&gt;First, let me say why all axiomatic treatments insist on talking about homology of pairs. One of the (most important/useful?) axioms is the Long exact sequence axiom, for any pair $A$ contained in $X$ we get the long exact sequence of homology groups. This is crucial for making computations. The concept of the homology of a pair is amazingly well behaved. Much better behaved than that of a quotient space, for homotopical purposes. So the best formulation of the axioms are in terms of pairs, because they require the least assumptions on the types of quotients you will allow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As Dylan alludes to, we really want to think of $(X,A)$ as the cofiber of the inclusion $A \to X$. This cofiber construction (quotient of the mapping cylinder, or rather the mapping cone) is a much better behaved notion. Its homotopy type depends only on the homotopy type of the inclusion. This is not the case for the standard quotient, unless the subspace is &quot;nice.&quot; For surfaces though, probably every subspace you are looking at is nice, so I think it would probably be okay if you just compute $H_*(X,A)$ as $H_*(X/A)$.&lt;/p&gt;&#xA;" OwnerUserId="627" LastActivityDate="2010-12-27T17:28:46.127" />
  <row Id="15656" PostTypeId="1" CreationDate="2010-12-27T17:39:05.940" Score="34" ViewCount="752" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The volume of an $n$-dimensional ball of radius $1$ is given by the classical formula&#xA;$$V_n=\frac{\pi^{n/2}}{\Gamma(n/2+1)}.$$ &#xA;For small values of $n$, we have&#xA;$$V_1=2\qquad$$&#xA;$$V_2\approx 3.14$$&#xA;$$V_3\approx 4.18$$&#xA;$$V_4\approx 4.93$$&#xA;$$V_5\approx 5.26$$ &#xA;$$V_6\approx 5.16$$&#xA;$$V_7\approx 4.72$$&#xA;It is not difficult to prove that $V_n$ assumes its maximal value when $n=5$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question.&lt;/strong&gt; Is there any non-analytic (i.e. geometric, probabilistic, combinatorial...) demonstration of this fact? What is so special about $n=5$?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I also have a similar question concerning the $n$-dimensional volume $S_n$ (&quot;surface area&quot;) of a unit $n$-sphere. Why is the maximum of $S_n$ attained at $n=7$ from a geometric point of view?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;note&lt;/strong&gt;: the question has also been &lt;a href=&quot;http://mathoverflow.net/questions/53119/volumes-of-n-balls-what-is-so-special-about-n5&quot; rel=&quot;nofollow&quot;&gt;asked on MathOverflow&lt;/a&gt; for those curious to other answers. &lt;/p&gt;&#xA;" OwnerUserId="723" LastEditorUserId="1543" LastEditorDisplayName="" LastEditDate="2011-02-01T12:15:51.373" LastActivityDate="2011-02-01T12:15:51.373" Title="Volumes of n-balls: what is so special about n=5?" Tags="&lt;geometry&gt;" AnswerCount="2" CommentCount="9" FavoriteCount="8" />
  <row Id="15657" PostTypeId="1" AcceptedAnswerId="15659" CreationDate="2010-12-27T17:40:30.437" Score="2" ViewCount="230" Body="&lt;p&gt;Given Kn define sequence S:  writing all vertices in order of traversing it by Euler circle&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How many distinct Hamilton cycles are in this sequence S?&lt;/p&gt;&#xA;" OwnerUserId="5145" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-27T22:12:12.293" LastActivityDate="2010-12-28T02:41:16.123" Title="Euler circle and Hamilton cycles" Tags="&lt;graph-theory&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="15658" PostTypeId="2" ParentId="15638" CreationDate="2010-12-27T17:52:21.253" Score="5" ViewCount="" Body="&lt;p&gt;Here is another way of arguing which is slightly different: Consider the set $X={\mathcal P}({\mathbb N}\times{\mathbb N})$ of all subsets of ${\mathbb N}^2$. Given $E\subseteq {\mathbb N}\times{\mathbb N}$, let $A_E$ be the set of all numbers that appear in either the domain or range of $E$ (this is sometimes called the &lt;em&gt;field&lt;/em&gt; of $E$). If it happens that $E$ is a well-ordering of $A_E$, let $\alpha_E$ be the unique ordinal that is order isomorphic to $(A_E,E)$. Otherwise, let $\alpha_E=0$. Then $\{\alpha_E\mid E\in X\}$ is a set (is the image of $X$ under the map $E\mapsto\alpha_E$) and it is obvious that it consists precisely of the countable ordinals. One easily sees then that it is itself an ordinal, and uncountable (since no ordinal belongs to itself). This is $\omega_1$.   &lt;/p&gt;&#xA;" OwnerUserId="462" LastActivityDate="2010-12-27T17:52:21.253" CommentCount="3" />
  <row Id="15659" PostTypeId="2" ParentId="15657" CreationDate="2010-12-27T18:02:30.173" Score="2" ViewCount="" Body="&lt;p&gt;It can certainly vary.  As an Eulerean path uses all n(n-1)/2 edges and a Hamiltonian path uses n, you can't have more than (n-1)/2.  Can you have that many?  There aren't any Eulerean paths if n is even and &gt;2.  Depending upon your definition, I can define an Eulearean path that has no segment that is Hamiltonian &lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2010-12-28T01:38:01.077" LastActivityDate="2010-12-28T01:38:01.077" CommentCount="4" />
  <row Id="15660" PostTypeId="2" ParentId="15650" CreationDate="2010-12-27T18:10:26.130" Score="4" ViewCount="" Body="&lt;p&gt;How about f(n+1)=f(n-1) started with f(0)=0, f(1)=1?&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-12-27T18:10:26.130" />
  <row Id="15661" PostTypeId="2" ParentId="15656" CreationDate="2010-12-27T18:29:10.347" Score="2" ViewCount="" Body="&lt;p&gt;The Gamma function has a minimum close to $5/2 +1$ for positive reals, and for your &quot;dimensionless quantity&quot; it's all that matters.&lt;/p&gt;&#xA;" OwnerUserId="5105" LastEditorUserId="5105" LastEditorDisplayName="" LastEditDate="2010-12-27T18:36:35.583" LastActivityDate="2010-12-27T18:36:35.583" CommentCount="1" />
  <row Id="15662" PostTypeId="2" ParentId="15249" CreationDate="2010-12-27T18:40:52.170" Score="0" ViewCount="" Body="&lt;p&gt;Surely a Dedekind infinite set trivially has an infinite proper subset: its image under the bijective function that is required for it to be Dedekind infinite.&lt;/p&gt;&#xA;" OwnerUserId="1508" LastActivityDate="2010-12-27T18:40:52.170" CommentCount="6" />
  <row Id="15663" PostTypeId="1" AcceptedAnswerId="15664" CreationDate="2010-12-27T19:04:24.600" Score="2" ViewCount="191" Body="&lt;p&gt;Let $f\colon\mathbb{R}^n\to\mathbb{R}$ be a function with $n\geq 2$, and define $S_{f}=\{(x_1,x_2, \ldots x_n) \in\mathbb{R}^n \mid f(x_1,x_2, \ldots x_n)=1\} = f^{-1}(1)$.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Is $S_f$ path-connected when $f$ is $C^{\infty}$? If not, what about if $f \in \mathbb{Q}[x_1,\ldots,x_n]$?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Context&lt;/strong&gt;: I had to establish that the set of diagonal square matrices with $\det=1$ was path-connected, which I did, but it got me thinking on more general lines. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the way, please answer (if possible) such that one needs as little prerequisites as possible to understand. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added&lt;/strong&gt;: Though Eric has already answered the question. There's a natural next question to ask: For what kinds of functions is this set path-connected? &lt;/p&gt;&#xA;" OwnerUserId="4806" LastEditorUserId="4806" LastEditorDisplayName="" LastEditDate="2010-12-28T11:46:34.913" LastActivityDate="2010-12-28T11:46:34.913" Title="Show set is path-connected" Tags="&lt;topology&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="15664" PostTypeId="2" ParentId="15663" CreationDate="2010-12-27T19:10:16.633" Score="4" ViewCount="" Body="&lt;p&gt;Not in general. Let $f:R \to R, f(x) = x^2 - 1$.  Then $S = \{\pm \sqrt{2}\}$ which is not path-connected.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: for the $n=2$ case you can take an equation defining a hyperbola, e.g. $x^2 - y^2$.  The set where this function is 1 has two connected components.&lt;/p&gt;&#xA;" OwnerUserId="9" LastEditorUserId="9" LastEditorDisplayName="" LastEditDate="2010-12-27T19:22:33.943" LastActivityDate="2010-12-27T19:22:33.943" CommentCount="3" />
  <row Id="15666" PostTypeId="2" ParentId="15663" CreationDate="2010-12-27T19:25:14.877" Score="3" ViewCount="" Body="&lt;p&gt;This properly belongs as a comment to Eric's answer, but I don't have enough reputation to post one.  More generally, defining $f:\mathbb{R}^n\to\mathbb{R}$ by $f(x_1,\ldots,x_n) = x_1^2-1$ gives $S$ with two connected components, the affine hyperplanes $\{\pm \sqrt 2\}\times\mathbb{R}^{n-1}$.&lt;/p&gt;&#xA;" OwnerUserId="5116" LastEditorUserId="5116" LastEditorDisplayName="" LastEditDate="2010-12-27T19:32:47.120" LastActivityDate="2010-12-27T19:32:47.120" CommentCount="2" />
  <row Id="15667" PostTypeId="1" AcceptedAnswerId="15673" CreationDate="2010-12-27T19:55:34.207" Score="2" ViewCount="148" Body="&lt;p&gt;Let $\sigma=(1\ 3),\ \tau=(2\ 4\ 5),\ \pi=(2\ 3\ 4) \in S_{5}$. Find $\pi\circ\tau\circ\sigma$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know the solution is $(1\ 4\ 5\ 3)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What i'm doing now is writing the permutations in the &quot;expanded&quot; form,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for example $\tau=\begin{pmatrix} &#xD;&#xA;  1\ 2\ 3\ 4\ 5\\ &#xD;&#xA;  1\ 5\ 3\ 2\ 4 &#xD;&#xA;\end{pmatrix}$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and then &quot;follow the numbers&quot; over the composition. This way, i find $(1\ 3\ 5\ 2)$, which is incorrect. What am i doing wrong?&lt;/p&gt;&#xA;" OwnerUserId="5148" LastActivityDate="2010-12-27T20:41:27.627" Title="Can someone explain the algorithm for composition of cycles?" Tags="&lt;homework&gt;&lt;algorithms&gt;&lt;permutations&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" />
  <row Id="15668" PostTypeId="1" AcceptedAnswerId="15693" CreationDate="2010-12-27T19:58:36.013" Score="7" ViewCount="328" Body="&lt;p&gt;In the &lt;a href=&quot;http://en.wikipedia.org/wiki/Axiom_of_choice&quot; rel=&quot;nofollow&quot;&gt;wikipedia&lt;/a&gt; article, two examples are given which use/ do not use the axiom of choice. They are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Given an infinite pair of socks, one needs AC to pick one sock out of each pair.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Given an infinite collection of pairs of shoes, one shoe can be specified without AC by choosing the left one.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Aren't these equivalent examples (just with different objects)? Why can't one just choose the left sock in (i) (so that AC is not needed)? &lt;/p&gt;&#xA;" OwnerUserId="2548" LastEditorUserId="2548" LastEditorDisplayName="" LastEditDate="2010-12-27T21:23:02.523" LastActivityDate="2010-12-28T00:08:28.153" Title="Axiom of Choice Examples" Tags="&lt;intuition&gt;&lt;elementary-set-theory&gt;&lt;axiom-of-choice&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="15669" PostTypeId="2" ParentId="15667" CreationDate="2010-12-27T20:00:18.510" Score="2" ViewCount="" Body="&lt;p&gt;You're probably composing from the left to the right, instead of from the right to the left.&lt;/p&gt;&#xA;" OwnerUserId="589" LastActivityDate="2010-12-27T20:00:18.510" CommentCount="2" />
  <row Id="15670" PostTypeId="2" ParentId="15556" CreationDate="2010-12-27T20:13:37.323" Score="7" ViewCount="" Body="&lt;p&gt;Yes, the &quot;classification&quot; of naturals by their &lt;strong&gt;parity&lt;/strong&gt; (remainder modulo $2\:$) extends immediately to integers. Namely, &lt;em&gt;even&lt;/em&gt; integers are those divisible by $2$, i.e. $\rm\: n = 2\:m\equiv 0\ (mod\ 2)\:,\: $ and &lt;em&gt;odd&lt;/em&gt; integers are those leaving remainder $\:1\:$ when divided by $\:2\:,\ $ i.e. $\rm\ n = 2\:m + 1\equiv 1\ (mod\ 2)\:.\: $ The effectiveness of this parity classification arises from the fact that it is &lt;em&gt;compatible&lt;/em&gt; with integer arithmetic operations, i.e. if $\rm\ \bar{a}\ :=\ a\ (mod\ 2)\ $ then $\rm\ \overline{ a+b}\ =\ \bar a + \bar b,\ \ \overline{a\ b}\ =\ \bar a\ \bar b\:.\: $  Iterating, we infer that equalities between expressions composed of these integer operations (i.e. integer polynomial expressions) are &lt;em&gt;preserved&lt;/em&gt; by taking their images modulo $2$ (ditto mod $\rm\:m\:$ for any $\rm\:m\:,$ e.g mod $9$ reduction yields &lt;a href=&quot;http://en.wikipedia.org/wiki/Casting%5Fout%5Fnines&quot; rel=&quot;nofollow&quot;&gt;casting out nines&lt;/a&gt;). In this way we can hope to understand integer equations by studying their images in the simpler (finite!) rings $\rm\: \mathbb Z/m\: $ of integers modulo $\rm\:m\:.\:$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, if an integer coefficient polynomial has an integer root $\rm\ f(n) = 0\ $ then it remains a root modulo $2$, i.e. $\rm\ f(n)\equiv 0\ (mod\ 2)\:.\:$ So, contrapositively, if a polynomial has no roots modulo $2$ then it has no integer roots. This leads to the following simple &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;PARITY ROOT TEST&lt;/strong&gt; $\ $ A polynomial $\rm\:P(x)\:$ with integer coefficients &#xA;has no integer roots if its constant coefficient and coefficient sum are both odd. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof&lt;/strong&gt; $\ $ The test  verifies that $\rm\  P(0) \equiv  P(1)\equiv 1\ \ (mod\ 2)\:,\ $ i.e. &#xA;that $\rm\:P(x)\:$ has no roots modulo $2$, hence no integer roots. $\quad$ &lt;strong&gt;QED&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;E.g. $\rm\:\ a\ X^2 + b\ X + c\ $  has no integer roots &#xA;if  $\rm\:c\:$  is odd and $\rm\:a,\:b\:$  have equal parity $\rm\:a\equiv b\ (mod\ 2)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Compare the conciseness of this test to the messy reformulation that would result if we had to restrict it to positive integers. Then we could no longer represent polynomial equations in the normal form $\rm\:f(x) = 0\:$ but, rather, we would need to consider general equalities $\rm\:f(x) = g(x)\:$ where both polynomials have &lt;em&gt;positive&lt;/em&gt; coefficients. Now the test would be much messier - bifurcating into motley cases. Indeed, historically, before the acceptance of negative integers and zero, the formula for the solution of a quadratic equation was stated in such an analogous obfuscated way - involving many cases. But by extending the naturals to the ring of integers we are able to &lt;em&gt;unify&lt;/em&gt; what were previously motley separate cases into a single &lt;em&gt;universal&lt;/em&gt; method of solving a general quadratic equation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Analogous examples exist throughout history that help serve to motivate the reasons behind various number system enlargements. Studying mathematical history will help provide one with a much better appreciation of the motivations behind the successive enlargements of the notion of &quot;number systems&quot;, e.g. see&#xA;Kleiner: &lt;a href=&quot;http://retro.seals.ch/cntmng?type=pdf&amp;amp;rid=elemat-001:1998:53%3a:22&amp;amp;subp=hires&quot; rel=&quot;nofollow&quot;&gt;From numbers to rings: the early history of ring theory.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Above is but one of many examples where &quot;completing&quot; a structure in some manner serves to simplify its theory. Such ideas motivated many of the extensions of the classical number systems (as well as analogous geometrical and topological completions concept, e.g. adjoining points at $\infty$, projective closure, compactification, model completion, etc). For some interesting expositions on such methods see the &lt;a href=&quot;http://math.stackexchange.com/questions/3203/infinite-limits/3206#3206&quot;&gt;references here.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Note:&lt;/strong&gt; $\: $ Analogous remarks (on the power gained by normalizing equations to the form $\ldots = 0\:$) hold true more generally for any algebraic structure whose congruences are determined by ideals - so-called &lt;em&gt;ideal determined&lt;/em&gt; varieties, e.g. see my &lt;a href=&quot;http://groups.google.com/groups?selm=y8zps1nln5o.fsf@nestle.csail.mit.edu&quot; rel=&quot;nofollow&quot;&gt;post here&lt;/a&gt; and see Gumm and Ursini: &lt;a href=&quot;http://www.springerlink.com/content/h2777414xl217t75/fulltext.pdf&quot; rel=&quot;nofollow&quot;&gt;Ideals in universal algebras&lt;/a&gt;. Without zero and negative numbers (additive inverses) we would not be able to rewrite expressions into such concise normal forms and we would not have available such powerful algorithms such as the Grobner basis algorithm, Hermite/Smith normal forms, etc. &lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-12-27T20:19:47.797" LastActivityDate="2010-12-27T20:19:47.797" />
  <row Id="15672" PostTypeId="2" ParentId="15668" CreationDate="2010-12-27T20:18:00.673" Score="3" ViewCount="" Body="&lt;p&gt;The idea behind the axiom of choice is to tell you how to choose when you can't necessarily distinguish between the items.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you take out a pair of socks from the closet (or drawer, etc) you can't tell which one you had on your left foot and which on your right, or which one is this and which one is that. If you can, well... you need to remember to let someone else make sure you have matching socks :-)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The idea, formally, is that if you take a product of infinitely many non-empty sets then it is not empty, namely there is a function in the product which returns an element in each coordinate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Formally, given a set $I$ such that $\forall i\in I$ we have $A_i\not=\emptyset$ then there is $F\colon I\to\bigcup_{i\in I} A_i$ for which $F(i)\in A_i$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you think about it, this is not a &quot;strange&quot; requirement when you are discussing mathematics, and it might come naturally in many places.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, for any two sets $A,B$ if there is $f\colon A\to B$ which is surjective then there is an injective $g\colon B\to A$. What are we doing? In a sense we choose one representative in each equivalence class of $a_1\sim a_2\iff f(a_1) = f(a_2)$. But if we have infinitely many equivalence classes - then we (usually) need to invoke some choice axiom (perhaps the axiom of choice, or countable choice, or dependent choice, etc...)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, many times the use is not of the axiom of choice, but rather an equivalent principle called &quot;Zorn's Lemma&quot;, it states that if you have a partial order in which every chain is bounded from above - then there is some maximal element (that is no one is strictly above it in the order). Again not something that is unreasonable if you want infinitary processes to &quot;act&quot; similar to finitary ones.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Addendum:&lt;/strong&gt;&lt;br&gt;&#xA;After I gave it some extra thought, I came up with something that might clear up the things. There is a concept which is &quot;a definable element&quot;, that is that you can write some formula $\psi(x)$ such that $a$ is the only element (suppose in $A$) which satisfies the formula. If you want to choose from infinitely many sets then you need to be able and tell which element you have chosen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have at least one definable element in each set, suppose by some uniform formula $\varphi(x)$, then you can clearly choose the one element defined by $\varphi$. However if there are many definable elements in each set, or infinitely many formulae are needed - then you cannot express it simply, and then you must assume that you can do it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And as I remarked above, we simply want infinitary processes to behave well, like finitary ones.&lt;/p&gt;&#xA;" OwnerUserId="622" LastEditorUserId="622" LastEditorDisplayName="" LastEditDate="2010-12-27T21:05:36.000" LastActivityDate="2010-12-27T21:05:36.000" />
  <row Id="15673" PostTypeId="2" ParentId="15667" CreationDate="2010-12-27T20:29:28.160" Score="2" ViewCount="" Body="&lt;p&gt;When you read a composition of functions written in the usual notation for permutations, you must remember to read them from right to left. Thus, when you try to compute the composition you must start by looking successively at what does each permutation in the composition do to each integer from $1$ to $5$ (in this case), but from right to left. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you can start by writing &#xA;$$\pi \circ \tau \circ \sigma = \begin{pmatrix} 2 &amp;amp;3 &amp;amp;4 \end{pmatrix} \circ \begin{pmatrix} 2 &amp;amp;4 &amp;amp;5 \end{pmatrix} \circ \begin{pmatrix} 1 &amp;amp;3 \end{pmatrix}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you start with $1$. The rightmost permutation sends $1 \mapsto 3 $ and then the middle permutation sends $3 \mapsto 3$ and finally the leftmost permutation sends $3 \mapsto 4$. So in the end the total result is that the composition of the three of them sends $1 \mapsto 4$. In the same way you'll proceed for the other integers and in the end you'll get the answer&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \pi \circ \tau \circ \sigma = \begin{pmatrix} 1 &amp;amp;2 &amp;amp;3 &amp;amp;4 &amp;amp;5 \\ 4 &amp;amp;2 &amp;amp;1 &amp;amp;5 &amp;amp;3  \end{pmatrix}$$&#xA;or in cycle notation $\pi \circ \tau \circ \sigma = \begin{pmatrix}  1 &amp;amp;4 &amp;amp;5 &amp;amp;3\end{pmatrix}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I checked doing the computation from left to right and the &quot;answer&quot; I got from doing that was $\pi \circ \tau \circ \sigma = \begin{pmatrix}  1 &amp;amp;3 &amp;amp;5 &amp;amp;2\end{pmatrix}$ which is what you originally had as your answer.&lt;/p&gt;&#xA;" OwnerUserId="900" LastEditorUserId="900" LastEditorDisplayName="" LastEditDate="2010-12-27T20:41:27.627" LastActivityDate="2010-12-27T20:41:27.627" CommentCount="1" />
  <row Id="15674" PostTypeId="2" ParentId="15650" CreationDate="2010-12-27T20:29:36.710" Score="3" ViewCount="" Body="&lt;p&gt;The general solution for a (finite) linear recurrence is a function of the form $$f(n) = \sum_{i=1}^m P_i(n) \lambda_i^n,$$ where the $P_i$ are polynomials with possible complex coefficients, and the $\lambda_i$ can be complex as well (even if the sequence itself is real, for example a cyclic sequence of minimal period $&gt;2$); and vice versa. If the eigenvalue with maximal modulus is unique, say $\lambda$, then $f(n+1)/f(n) \rightarrow \lambda$. If all eigenvalues of maximal modulus are obtained from one of them by multiplying with (integral) roots of unity, then there is a periodic subsequence with this property (e.g. in Ross's example take every second term). Otherwise, the sequence will be pretty chaotic: for example $f(n) = \cos(n)$.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-12-27T20:29:36.710" />
  <row Id="15675" PostTypeId="2" ParentId="15629" CreationDate="2010-12-27T20:48:10.887" Score="2" ViewCount="" Body="&lt;p&gt;Or just $\begin{cases} a &amp;amp; b \\ c &amp;amp; d \end{cases}$&lt;/p&gt;&#xA;" OwnerUserId="5152" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-12-28T01:35:17.797" LastActivityDate="2010-12-28T01:35:17.797" CommentCount="3" />
  <row Id="15676" PostTypeId="1" CreationDate="2010-12-27T20:52:57.740" Score="4" ViewCount="191" Body="&lt;p&gt;Sorry for this uninteresting question but hopefully someone can provide some help.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a way to simplify the following expression?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\binom{n}{m}\sum_{\nu=0}^{n-m}(-1)^{\nu}\binom{n-m}{v}\left(\frac{n-m-v}{n}\right)^{r}\displaystyle \frac{m}{m+v}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$-\binom{n}{m}\sum_{\nu=0}^{n-m}(-1)^{\nu}\binom{n-m}{v}\left(\frac{n-m-v}{n}\right)^{r}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a hint for a problem, but I don't know how to proceed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;UPDATE&lt;/strong&gt;: Would it bother you if I add the original problem?. Maybe some context is needed in order to solve this problem by using this hint.&lt;/p&gt;&#xA;" OwnerUserId="2569" LastEditorUserId="2569" LastEditorDisplayName="" LastEditDate="2010-12-27T23:05:52.667" LastActivityDate="2010-12-27T23:05:52.667" Title="Simplification of sums" Tags="&lt;sequences-and-series&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="15677" PostTypeId="2" ParentId="15676" CreationDate="2010-12-27T20:57:09.370" Score="1" ViewCount="" Body="&lt;p&gt;At a quick glance, it looks like there are some common factors to the two sums and that factoring is likely to help in simplifying.  Also, since the summations have the same indexing and range, they could be combined into a single summation.&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-12-27T20:57:09.370" CommentCount="4" />
  <row Id="15678" PostTypeId="2" ParentId="15629" CreationDate="2010-12-27T21:01:21.120" Score="1" ViewCount="" Body="&lt;p&gt;You could potentially convert it to a mathematical formula too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example say we had the following:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;if (a &amp;lt; b) then c = 100 &#xA;else if (a &amp;gt; b) then c = 200&#xA;else c = 300.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This can be rewritten as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$c  = 300 \ (1 - \text{sgn}^2(a-b)) + \text{sgn}^2(a-b)(50 \ \text{sgn}(a-b) + 150)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $\text{sgn}(x)$ is the sign of $x$, as defined here; &lt;a href=&quot;http://en.wikipedia.org/wiki/Sign_function&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Sign_function&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(It is defined as: 1 for positive, 0 for 0, and -1 for negative)&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-12-27T21:17:00.987" LastActivityDate="2010-12-27T21:17:00.987" />
  <row Id="15679" PostTypeId="2" ParentId="15676" CreationDate="2010-12-27T21:15:32.487" Score="1" ViewCount="" Body="&lt;p&gt;I doubt it, the case $m=0$ would give a simpler formula for &lt;a href=&quot;http://en.wikipedia.org/wiki/Stirling_numbers_of_the_second_kind&quot; rel=&quot;nofollow&quot;&gt;Stirling numbers of the second kind&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$n!\ S(r,n) = \sum_{v = 0}^{n} (-1)^{v} {n \choose v} (n-v)^{r}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps you can write your expression in terms of these numbers...&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-12-27T21:15:32.487" CommentCount="4" />
  <row Id="15680" PostTypeId="2" ParentId="15556" CreationDate="2010-12-27T21:38:08.560" Score="2" ViewCount="" Body="&lt;p&gt;The real answer depends on the definition, because there is math-history tag invoked there was a time that 1 was not considered an odd numbers, 0 and negative numbers for sure where not considered even or odd. Historically the concept was defined only for natural numbers. &#xA;These a days the set of all integers multiplied by 2 is considered the set of even numbers, i.e. $\dots,-4,-2,0,2,4,\dots$ n and all the integers not in that set are defined to be odd.&#xA;There is no real answer, it all depends on the definition, same way that the book is only dealing with natural numbers and not integers.&#xA;The concept was extended from naturals to integers, but there are uncountably many ways to define the even and odds beyond the natural numbers. Just make sure others know what definition you are using to label something even or odd. &lt;/p&gt;&#xA;" OwnerUserId="3936" LastActivityDate="2010-12-27T21:38:08.560" CommentCount="4" />
  <row Id="15681" PostTypeId="1" AcceptedAnswerId="15691" CreationDate="2010-12-27T21:49:49.817" Score="5" ViewCount="160" Body="&lt;p&gt;How many Borel-measurable functions from $\mathbb{R}$ to $\mathbb{R}$ are there? The motivation is from &lt;a href=&quot;http://mathoverflow.net/questions/32126/function-with-range-equal-to-whole-reals-on-every-open-set/49137#49137&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; answer of mine on MathOverflow.&lt;/p&gt;&#xA;" OwnerUserId="3078" LastEditorUserId="3078" LastEditorDisplayName="" LastEditDate="2010-12-27T22:02:43.790" LastActivityDate="2010-12-27T23:59:27.707" Title="How many Borel-measurable functions from R to R are there?" Tags="&lt;measure-theory&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="15682" PostTypeId="2" ParentId="13282" CreationDate="2010-12-27T22:55:51.737" Score="2" ViewCount="" Body="&lt;p&gt;Other possibilities include using the conjugate gradient algorithm to check positive-definiteness. In theory, this method terminates after at most n iterations (n being the dimension of your matrix). In practice, it may have to run a bit longer. It is trivial to implement. You can also use a variant of the Lanczos method to estimate the smallest eigenvalue of your matrix (which is much easier than computing all eigenvalues!) Pick up a book on numerical linear algebra (check the SIAM collection.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At any rate recall that such methods (and the Cholesky decomposition) check &lt;em&gt;numerical&lt;/em&gt; positive-definiteness. It is possible for the smallest eigenvalue of your matrix to be, say, 1.0e-16 and for cancellation errors due to finite-precision arithmetic to cause your Cholesky (or conjugate gradient) to break down.&lt;/p&gt;&#xA;" OwnerUserId="5155" LastActivityDate="2010-12-27T22:55:51.737" CommentCount="1" />
  <row Id="15683" PostTypeId="2" ParentId="15624" CreationDate="2010-12-27T23:06:00.453" Score="8" ViewCount="" Body="&lt;p&gt;Just dropping points into the polygon according to a uniform distribution will not, unfortunately, generate very uniform solutions: with extremely high probability there will be large gaps and tight clusters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obtaining a highly uniform pattern within an arbitrary (convex) polygon is a difficult problem to solve; exact solutions are known only for special polygons and very small numbers of points.  As the number of points $n$ grows, they tend to fall into a hexagonal pattern.  However, like real crystals, these patterns can exhibit faults and fractures imposed by the polygon's boundary.  As a quick and dirty solution you might estimate the average spacing by dividing the polygon's area by $n$ and solving for the dimensions of a hexagon of that area, creating an hexagonal mesh of those dimensions, and performing some trial overlays of that mesh (translating it a bit in each trial) to find something that works well.  You are likely to run into trouble at places just inside the polygon's boundary, though.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One practical way to find solutions that really work is with &lt;a href=&quot;http://cat.inist.fr/?aModele=afficheN&amp;amp;cpsidt=2432316&quot; rel=&quot;nofollow&quot;&gt;spatial simulated annealing&lt;/a&gt;.  Although that's a computationally intensive and purely approximate approach, in practice you can get reasonable solutions for the intended application in an extremely short time (thousands of iterations rather than hundreds of thousands).&lt;/p&gt;&#xA;" OwnerUserId="1489" LastActivityDate="2010-12-27T23:06:00.453" CommentCount="3" />
  <row Id="15684" PostTypeId="1" CreationDate="2010-12-27T23:06:03.683" Score="1" ViewCount="169" Body="&lt;p&gt;Let $\tau \in \mathbb{S}^1$ be such that $\tau$ is not a root of unity. Let $E_\tau$ be the quotient space $S^1/\tau^{\mathbb Z}$. Consider it as a pointed space with basepoint the equivalence class of $1$. This is a path-connected space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the fundamental group of this space?&lt;/p&gt;&#xA;" OwnerUserId="5156" LastActivityDate="2010-12-28T03:31:15.413" Title="Fundamental group of noncommutative torus" Tags="&lt;algebraic-topology&gt;&lt;noncommutative-geometry&gt;" AnswerCount="1" CommentCount="10" />
  <row Id="15685" PostTypeId="2" ParentId="15624" CreationDate="2010-12-27T23:28:55.613" Score="4" ViewCount="" Body="&lt;p&gt;As whuber mentions, using a uniform distribution for sampling will result in &quot;clumpiness&quot; of the points. One way of quantifying this &quot;clumpiness&quot; is through a measure called discrepancy. There are numerous &lt;a href=&quot;http://en.wikipedia.org/wiki/Low-discrepancy_sequence&quot; rel=&quot;nofollow&quot;&gt;low-discrepancy sequences&lt;/a&gt; out there. You may also want to look into the &lt;a href=&quot;http://people.csail.mit.edu/thouis/JonesPoissonPreprint.pdf&quot; rel=&quot;nofollow&quot;&gt;Poisson disk method&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1761" LastActivityDate="2010-12-27T23:28:55.613" CommentCount="1" />
  <row Id="15686" PostTypeId="1" AcceptedAnswerId="15690" CreationDate="2010-12-27T23:29:24.397" Score="7" ViewCount="266" Body="&lt;p&gt;I have to explain the problem I am thinking first:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The solutions of $x^3+6x-20=0$ are $x = 2$, $-1+3i$ and $-1-3i$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The cubic formula for these solutions is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x = \sqrt[3]{10 + \sqrt{108}} + \sqrt[3]{10 + \sqrt{108}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First $\sqrt{108} \approx -10.392304845413$, $10.392304845413$ has two values.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And $\sqrt[3]{10 + -10.392304845413}$ has three values $v_1\approx 0.3660 + 0.6339745i$, $v_2\approx -0.7320508$, $v_3\approx 0.3660254 - 0.6339745i$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And $\sqrt[3]{10 + 10.392304845413}$ has three values $u_1\approx 2.7320508$, $u_2\approx -1.366 + 2.366$, $u_3\approx -1.366 - 2.366i$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So there are nine possible values of $x$:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$v_1 + u_1 = 3.098 + 0.63397i$&lt;/li&gt;&#xA;&lt;li&gt;$v_1 + u_2 = -1 + 3i$&lt;/li&gt;&#xA;&lt;li&gt;$v_1 + u_3 = -1 - 1.732i$&lt;/li&gt;&#xA;&lt;li&gt;$v_2 + u_1 = 2$&lt;/li&gt;&#xA;&lt;li&gt;$v_2 + u_2 = -2.098 + 2.366i$&lt;/li&gt;&#xA;&lt;li&gt;$v_2 + u_3 = -2.098 - 2.366i$&lt;/li&gt;&#xA;&lt;li&gt;$v_3 + u_1 = 3.098 - 0.63397i$&lt;/li&gt;&#xA;&lt;li&gt;$v_3 + u_2 = -1 + 1.732$&lt;/li&gt;&#xA;&lt;li&gt;$v_3 + u_3 = -1 - 3i$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;but the correct ones are $v_1 + u_2$, $v_2 + u_1$, $v_3 + u_3$.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt; Here is a table of 3vu for Arturo Magidin:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$3 v_1 u_1 = 3 + 5.196152i$&lt;/li&gt;&#xA;&lt;li&gt;$3 v_1 u_2 = -6$&lt;/li&gt;&#xA;&lt;li&gt;$3 v_1 u_3 = 3 - 5.196152i$&lt;/li&gt;&#xA;&lt;li&gt;$3 v_2 u_1 = -6$&lt;/li&gt;&#xA;&lt;li&gt;$3 v_2 u_2 = 3 - 5.196152i$&lt;/li&gt;&#xA;&lt;li&gt;$3 v_2 u_3 = 3 + 5.196152i$&lt;/li&gt;&#xA;&lt;li&gt;$3 v_3 u_1 = 3 - 5.196152i$&lt;/li&gt;&#xA;&lt;li&gt;$3 v_3 u_2 = 3 + 5.196152i$&lt;/li&gt;&#xA;&lt;li&gt;$3 v_3 u_3 = -6$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;My questions are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;What are the others values of $x$ (those which are not solutions)?&lt;/li&gt;&#xA;&lt;li&gt;Is there a better notation for taking roots that points out (for example) that the cube roots need to be synchronized?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="4290" LastEditorUserId="4290" LastEditorDisplayName="" LastEditDate="2010-12-30T00:02:37.487" LastActivityDate="2010-12-30T00:38:26.410" Title="Different notations for roots?" Tags="&lt;algebra&gt;" AnswerCount="4" CommentCount="6" />
  <row Id="15687" PostTypeId="1" AcceptedAnswerId="15688" CreationDate="2010-12-27T23:46:48.313" Score="7" ViewCount="252" Body="&lt;p&gt;Does anyone remember a movie about four mathematicians studying infinity, who ended up committing a suicide? It was aired online somewhere, and I can't find it (it's a documentary, if that helps).&lt;/p&gt;&#xA;" OwnerUserId="5157" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-28T02:13:26.547" LastActivityDate="2010-12-28T02:13:26.547" Title="Name of the math movie" Tags="&lt;reference-request&gt;&lt;popular-math&gt;" AnswerCount="2" FavoriteCount="2" />
  <row Id="15688" PostTypeId="2" ParentId="15687" CreationDate="2010-12-27T23:49:16.187" Score="12" ViewCount="" Body="&lt;p&gt;Are you talking about &lt;strong&gt;Dangerous Knowledge&lt;/strong&gt; by BBC? Though it's not quite about four mathematicians (at least one of the people was physicist), and not all of them were studying infinity directly. That's the only one that comes to my mind as I watched it a couple of weeks ago.&lt;/p&gt;&#xA;" OwnerUserId="3731" LastActivityDate="2010-12-27T23:49:16.187" CommentCount="1" />
  <row Id="15689" PostTypeId="2" ParentId="15687" CreationDate="2010-12-27T23:52:09.103" Score="5" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Dangerous Knowledge&lt;/strong&gt; it is!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cantor - Infinity&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Boltzmann - Entropy&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Godel and Turing - Logic&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://video.google.com/videoplay?docid=-5122859998068380459#&quot; rel=&quot;nofollow&quot;&gt;http://video.google.com/videoplay?docid=-5122859998068380459#&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3302" LastActivityDate="2010-12-27T23:52:09.103" />
  <row Id="15690" PostTypeId="2" ParentId="15686" CreationDate="2010-12-27T23:58:25.780" Score="7" ViewCount="" Body="&lt;p&gt;When applying Cardano's formulae, you must take specific branches of the square and cubic root functions. Also, note that the square root of 108 is justs a &lt;em&gt;single&lt;/em&gt; value; while $x^2 - 108=0$ has two (real) solutions, the expression $\sqrt{108}$ is a &lt;em&gt;single&lt;/em&gt; value: the nonnegative solution to $x^2-108=0$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In any case, remember that if you have a depressed cubic, $x^3 + px + q = 0$, we introduce two variables $u$ and $v$ subject to the condition that $u+v = x$ and $3uv + p = 0$. This requires the product of $u$ and $v$ to be a real number. Eventually, you arrive at $u$ and $v$ being the cubic roots in question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But remember that you are assuming that $uv$ is a real number. This is why you cannot just take all possible combinations with all possible roots. In other words, all those &quot;other values&quot; of $u+v$ are values that do not satisfy the system of equations you have:&#xA;\begin{align*}&#xA;(u+v)^3 + p(u+v) + q &amp;amp;=0\\&#xA;3uv + p &amp;amp;=0&#xA;\end{align*}&#xA;There is no general notation that tells you which branches of the complex-valued cubic and square root functions you must take, you just have to remember that the product of the two cubic roots has to be a real number. This is what cuts down from nine to three values.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a standard way of listing all solutions: once you find &lt;strong&gt;one&lt;/strong&gt; value of $u$ and $v$ such that $uv$ is a real number, then the three solutions are given by $u+v$, $\omega u + \omega^2 v$, and $\omega^2u + \omega v$, where $\omega = -\frac{1}{2}+\frac{\sqrt{3}}{2}i$ is a primitive cubic root of unity. (Notice that if $u$ is one value of the cubic root, then the other two values are $\omega u$ and $\omega^2u$, and likewise with $v$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;hr/&gt;&#xA;It seems that you are not quite aware of all the assumptions you are making about $u$ and $v$ when you use Cardano's formula. For one thing, if you review the derivation of the formula, say, in &lt;a href=&quot;http://en.wikipedia.org/wiki/Cubic_function#Cardano.27s_method&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;, you will see that the condition that $3uv+p=0$ is worked into the derivation. Here's the method: remember that $x=u+v$ is supposed to be a solution of&#xA;$$x^3 + px + q = 0.$$&#xA;Plugging in and doing a bit of work, we get:&#xA;\begin{align*}&#xA;0 &amp;amp;= (u+v)^3 + p(u+v) + q\\&#xA;  &amp;amp;= u^3 + 3u^2v + 3uv^2 + v^3 + p(u + v) + q\\&#xA;  &amp;amp;= u^3 + v^3 + 3uv(u+v) + p(u+v) + q\\&#xA;  &amp;amp;= u^3 + v^3 + (3uv+p)(u+v) + q.&#xA;\end{align*}&#xA;It is at this point that the extra condition $3uv+p=0$ is introduced, so that the equation reduces to $u^3+v^3 + q = 0$, Then $u^3+v^3 = -q$, and $(uv)^3 = -\frac{p^3}{27}$. Now remember that if $a$ and $b$ are two numbers, then $(z-a)(z-b) = z^2 - (a+b)z + ab$, so $a$ and $b$ are always the roots of the quadratic that has linear coefficient minus their sum, and constant coefficient their product. Applying this to $u^3$ and $v^3$, since we know their sum and their product, we deduce that $u^3$ and $v^3$ are the two roots of the quadratic:&#xA;$$z^2 +qz - \frac{p^3}{27}=0$$&#xA;and from this, using the quadratic formula, we obtain that&#xA;$$u^3 = -\frac{q}{2} + \sqrt{\frac{q^2}{4} + \frac{p^3}{27}}, \qquad v^3 = -\frac{q}{2} - \sqrt{\frac{q^2}{4} + \frac{p^3}{27}}.$$&#xA;Therefore,&#xA;$$u = \sqrt[3]{-\frac{q}{2} + \sqrt{\frac{q^2}{4} + \frac{p^3}{27}}},\qquad v = \sqrt[3]{-\frac{q}{2} - \sqrt{\frac{q^2}{4} + \frac{p^3}{27}}},$$&#xA;so that the solution $x$ is given by the formula you have:&#xA;$$x = u+v = \sqrt[3]{-\frac{q}{2} + \sqrt{\frac{q^2}{4} + \frac{p^3}{27}}}+\sqrt[3]{-\frac{q}{2} - \sqrt{\frac{q^2}{4} + \frac{p^3}{27}}}.$$&#xA;But the assumption that $3uv+p=0$ was instrumental in obtaining this formula. The very fact that you write that $x$ is equal to the sum of the two cubic roots &lt;strong&gt;already assumes&lt;/strong&gt; that the product of those two (complex-valued) cubic roots will be a real number, specifically, $-\frac{p}{3}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While &lt;em&gt;you&lt;/em&gt; may perhaps not have made that assumption consciously, Cardano certainly did, and by using his formula you are also making the assumption, whether you were aware of it or not.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-30T00:38:26.410" LastActivityDate="2010-12-30T00:38:26.410" CommentCount="8" />
  <row Id="15691" PostTypeId="2" ParentId="15681" CreationDate="2010-12-27T23:59:27.707" Score="6" ViewCount="" Body="&lt;p&gt;Sune, there are as many Borel-measurable functions as there are reals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is easy to see that a function is Borel-measurable iff the preimage of each open interval with rational end-points is a Borel set. Moreover, the function is completely determined by the sequence of pairs $(I,B)$ where $I$ varies over the intervals and $B$ is the preimage of $I$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are countably many possible $I$ and to each corresponds one of ${\mathfrak c}=|{\mathbb R}|$ many possible Borel sets. The total number of Borel functions is then bounded above by $|{\mathbb R}^{\mathbb N}|={\mathfrak c}$. Since each constant function is Borel-measurable, ${\mathfrak c}$ is also a lower bound.&lt;/p&gt;&#xA;" OwnerUserId="462" LastActivityDate="2010-12-27T23:59:27.707" CommentCount="0" />
  <row Id="15692" PostTypeId="2" ParentId="15686" CreationDate="2010-12-28T00:01:20.040" Score="3" ViewCount="" Body="&lt;p&gt;More of an extended comment than an answer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Usually, the way I've seen the radical notation used, it is mean to imply only the &quot;principal&quot; root, for some definition of &quot;principal&quot;—therein lies the trouble.  I have seen principal $n$th root defined as the $n$th root with the least non-negative argument or as the $n$th root with argument in $(-\frac{\pi}{n},\frac{\pi}{n}]$ (and, in fact, the latter is the only definition that I've seen implemented in calculators and computer algebra systems).  Most of the instances where I've seen the cubic formula (or quartic formula), it is intended for the first definition, though it is possible to write the formula (both) in such a way that takes the $n$th root only once and thus avoids the need for synchronizing the choice of principal root across multiple expressions (such as &lt;a href=&quot;http://math.stackexchange.com/questions/785/is-there-a-general-formula-for-solving-4th-degree-equations/786#786&quot;&gt;my answer here&lt;/a&gt; for quartics).&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-12-28T00:01:20.040" CommunityOwnedDate="2010-12-28T00:01:20.040" />
  <row Id="15693" PostTypeId="2" ParentId="15668" CreationDate="2010-12-28T00:08:28.153" Score="3" ViewCount="" Body="&lt;p&gt;In both examples you are given an infinite family of sets of size 2, and a choice function picks an element of each family. In the case of the sets of shoes, each set comes with an ordering (left, right), and so we can &lt;em&gt;define&lt;/em&gt; a choice function explicitly. In the case of pairs of socks, this is not the case: Of course, given any pair, we can assign an ordering to it so we can select one of the two socks. However, there is no obvious way of &lt;em&gt;uniformly&lt;/em&gt; doing this for all pairs at the same time. This means (at least intuitively) that there is way of &lt;em&gt;defining&lt;/em&gt; a choice function. Its existence can only be granted by applying the axiom of choice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are several variants of this example. One that may be useful to think about is the following: One can show explicitly that if $A_n$ is a set of reals and $|A_n|=2$ for each $n\in{\mathbb N}$, then $\bigcup_n A_n$ is a (finite or infinite) countable set. However, it is consistent with the axioms of set theory except choice that there is a sequence $(A_n\mid n\in{\mathbb N})$ of sets, each $|A_n|=2$, and yet $\bigcup_n A_n$ is not countable. Although the construction of the model where this happens is technical, the point is that this formalizes the intuition that there is no &quot;explicit&quot; way of choosing a sock from each pair, simultaneously, and that any way of doing so is essentially non-constructive. &lt;/p&gt;&#xA;" OwnerUserId="462" LastActivityDate="2010-12-28T00:08:28.153" CommentCount="1" />
  <row Id="15694" PostTypeId="2" ParentId="15646" CreationDate="2010-12-28T00:13:51.793" Score="0" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;: This reply assumes that the surface contains the singularity (&quot;source&quot;) at its center, so it does not actually pertain to this problem.  For the sake of clarity, I have now changed the surface from $S$ to $S&amp;#39;$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a lot like an electrostatics problem, and it is probably trying to mimic one due the field's specific form.  I am going to treat this problem like it is so; someone please point out if what follows is incorrect since I am not absolutely sure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The field has rotational symmetry, and the $\vec{r_{0}}$ usually means the location of a &quot;source&quot; of the field.  Thus, no matter the location of the source, this field is still rotationally invariant.  Using that surface for the integral is called a &quot;Gaussian surface&quot; by physicists; it is a tool used to help solve a problem by exploiting its symmetry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, I know you probably just want to solve the integral, so I will get to the point.  Your surface will be a sphere of radius $a$, and so will have a unit normal area vector $\hat{r}&amp;#39;$.  Let us write $\vec{r}&amp;#39;=\vec{r}-\vec{r_{0}}$ and $r=|\vec{r}|$.  If we rewrite the integrand, we get&#xA;$$k\frac{\vec{r}-\vec{r_{0}}}{|\vec{r}-\vec{r_{0}}|^{3}}=k\frac{\vec{r}&amp;#39;}{|\vec{r}&amp;#39;|^{3}}=k\frac{\hat{r}&amp;#39;}{|\vec{r}&amp;#39;|^{2}}~\text{since, for any vector }\vec{v},~\hat{v}=\frac{\vec{v}}{|\vec{v}|}.$$&#xA;The infinitesimal vector area element for the sphere will be&#xA;$$d\vec{S&amp;#39;}=\hat{r}&amp;#39;r&amp;#39;^{2}\sin\phi d\theta d\phi.$$&#xA;Thus, the integral becomes&#xA;$$\begin{aligned}&#xD;&#xA;\int\vec{F}\cdot d\vec{S&amp;#39;}&amp;amp;=\int k\left(\frac{\hat{r}&amp;#39;}{|\vec{r}&amp;#39;|^{2}}\right)\cdot\left(\hat{r}&amp;#39;r&amp;#39;^{2}\sin\phi d\theta d\phi\right) \\&#xD;&#xA;&amp;amp;=\int k\frac{r&amp;#39;^{2}}{r&amp;#39;^{2}}\hat{r}&amp;#39;^{2}\sin\phi d\theta d\phi \\&#xD;&#xA;&amp;amp;=k\int_{0}^{2\pi}d\theta\int_{0}^{\pi}\sin\phi d\phi \\&#xD;&#xA;&amp;amp;=4\pi k.&#xD;&#xA;\end{aligned}$$&#xA;Notice that the radius of the surface is irrelevant.  As long as it contains the &quot;source&quot; point in the center, the value of the integral will always be the same for this field.&lt;/p&gt;&#xA;" OwnerUserId="5153" LastEditorUserId="5153" LastEditorDisplayName="" LastEditDate="2010-12-28T01:29:32.880" LastActivityDate="2010-12-28T01:29:32.880" CommentCount="2" />
  <row Id="15695" PostTypeId="2" ParentId="15631" CreationDate="2010-12-28T00:31:36.213" Score="2" ViewCount="" Body="&lt;p&gt;Charlie, this is a result of Sierpiński. It appears in &quot;    Sur un problème de la théorie des relations&quot;, Fundamenta Mathematicae, vol 28, (1937), 71-74. You can download the paper &lt;a href=&quot;http://matwbn.icm.edu.pl/tresc.php?wyd=1&amp;amp;tom=28&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The general result of which this is a particular case is due to Hajnal (rather than &quot;independent&quot;, one usually talks of &quot;free&quot;). It appears in &quot;Proof of a conjecture of B. Ruziewicz&quot;, Fundamenta Mathematicae, vol 50, (1961), 123-128. It can be downloaded &lt;a href=&quot;http://matwbn.icm.edu.pl/tresc.php?wyd=1&amp;amp;tom=50&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me know if you have difficulty understanding the arguments there, and I'll sketch the proof.&lt;/p&gt;&#xA;" OwnerUserId="462" LastActivityDate="2010-12-28T00:31:36.213" />
  <row Id="15696" PostTypeId="1" AcceptedAnswerId="15699" CreationDate="2010-12-28T00:37:00.967" Score="1" ViewCount="73" Body="&lt;p&gt;I have an input array of line segments (a,b)_1 ... (a,b)_n. Let's call it &quot;S&quot; so (a,b)_S_1 ... (a,b)_S_n.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have another array of the same construct, let's call it &quot;I&quot;, so (a,b)_I_1 ... (a,b)_I_n.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wish to mathematically express the shortest distance of each of I against S. In other words, the matrix formed from the product of I * S. To be very clear, each I is compared to each S. I want the minimum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know ||(a,b)_I_k-(a,b)_S_k|| denotes distance. How do I represent the matrix formed and indicate the Min of the cells?&lt;/p&gt;&#xA;" OwnerUserId="4930" LastActivityDate="2010-12-28T01:28:57.583" Title="Representing shortest distance of a matrix comparison" Tags="&lt;matrices&gt;&lt;elementary-set-theory&gt;&lt;notation&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="15698" PostTypeId="2" ParentId="58" CreationDate="2010-12-28T01:14:13.617" Score="1" ViewCount="" Body="&lt;p&gt;I think that Benford's law is quite intuitive in &quot;real life&quot;, but it would indeed be more suprising if it would often hold in mathematics, and in fact it does, &lt;a href=&quot;http://arxiv.org/abs/1009.0780&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; is a nice related student paper.&lt;/p&gt;&#xA;" OwnerUserId="5105" LastActivityDate="2010-12-28T01:14:13.617" />
  <row Id="15699" PostTypeId="2" ParentId="15696" CreationDate="2010-12-28T01:28:57.583" Score="2" ViewCount="" Body="&lt;p&gt;Denote $(a,b)_{S_i}$ by $S_i$ and $(a,b)_{I_k}$ by $I_k$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Distance between $i^{th}$ line segment of $S$ and $j^{th}$ segment of $I$ is $|| (S_i,\  I_j) ||$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can now denote the minimum by &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \text{min}_{1 \le i \le n,1 \le j \le n} \{|| (S_i,\ I_j)||\}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or you can denote the matrix by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$M_{ij} = || (S_i,\ I_j)||$ and the min by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \text{min}_{1 \le i \le n,1 \le j \le n} \{ M_{ij} \}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does this help?&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-12-28T01:28:57.583" CommentCount="9" />
  <row Id="15700" PostTypeId="2" ParentId="15686" CreationDate="2010-12-28T01:40:58.743" Score="6" ViewCount="" Body="&lt;p&gt;Here's a pretty clean version of Cardano I use (adapted from Numerical Recipes). I assume here that the principal values for both the square and cube root (the result is in the interval $\left(-\frac{\pi}{n},\frac{\pi}{n}\right]$, which as Isaac says applies for most computing systems) are taken.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $x^3+bx+c=0$, letting&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$q=\frac{b}{3}$ and $r=\frac{c}{2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$A=-\mathrm{sign}(r)\sqrt[3]{|r|+\sqrt{r^2+q^3}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$B=\begin{cases}\frac{q}{A}&amp;amp;\text{ if }A\neq0\\0&amp;amp;\text{ if }A=0\end{cases}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the three roots can be expressed as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$A-B$$&lt;br&gt;&#xA;$$-\frac{A-B}{2}\pm i\frac{\sqrt{3}}{2}(A+B)$$&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-12-28T01:40:58.743" />
  <row Id="15701" PostTypeId="2" ParentId="15634" CreationDate="2010-12-28T01:41:03.847" Score="2" ViewCount="" Body="&lt;p&gt;The answer to both questions (a) and (b) is yes, and is a consequence of the fact that Riemann integrable functions can be approximated from above and below by &lt;a href=&quot;http://en.wikipedia.org/w/index.php?title=Step_function&amp;amp;oldid=396614578&quot; rel=&quot;nofollow&quot;&gt;step functions&lt;/a&gt;. The proof follows in a fairly straightforward manner once you realize that this allows you to reduce the problem to that of indicator functions of intervals. To be precise, if $f\colon[0,1]\to\mathbb{R}$ is Riemann integrable then, for each fixed $\epsilon &amp;gt; 0$, there are step functions $g\le f\le h$ with $\int_0^1(h(y)-g(y))\,dy &amp;lt; \epsilon$.&#xA;$$&#xD;&#xA;\begin{align}&#xD;&#xA;\frac1N\sum_{n=1}^Nf(\xi_n)-\int_0^1 f(y)\,dy&amp;amp;\le\frac1N\sum_{n=1}^Nh(\xi_n)-\int_0^1 h(y)\,dy+\int_0^1(h(y)-f(y))\,dy\\&#xD;&#xA;&amp;amp;\le\frac1N\sum_{n=1}^Nh(\xi_n)-\int_0^1 h(y)\,dy+\epsilon.&#xD;&#xA;\end{align}&#xD;&#xA;$$&#xA;Similarly, the reverse inequality holds with $g$ replacing $h$ and $-\epsilon$ replacing $\epsilon$ on the right hand side. So, convergence for step functions implies convergence for Riemann integrable functions. Also, replacing $\xi_n$ by $x+\xi_n$ (mod 1) above, the problem of uniform convergence over all $x$ is also reduced to that of step functions. Linearity further reduces the problem to that of indicator functions of intervals in $(0,1]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if $f=1_A$ is an indicator function of an interval $A\subseteq(0,1]$ then the limit $\frac1N\sum_{n=1}^Nf(x+\xi_n)=\int_0^1f(y)\,dy$ follows by definition of equidistribution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can also obtain uniform convergence (in $x$) as follows: note that $y\mapsto f(x+y)=1_{\{y\in A-x{\rm\ (mod\ 1)}\}}$ is again an indicator function of an interval (taken mod 1). Convergence is uniform simultaneously on all intervals. For any $M &amp;gt; 0$ consider the finite collection of intervals $B_i=((i-1)/M,i/M]$. Then, any interval $A\subseteq(0,1]$ can be sandwiched between unions of the $B_i$. That is, $\cup_{i\in I}B_i\subseteq A\subseteq\cup_{j\in J}B_j$ for $I,J\subseteq\{1,2,\ldots,M\}$ with $I\subseteq J$ and $J\setminus I$ containing no more than two elements.&#xA;$$&#xD;&#xA;\frac1N\sum_{n=1}^N 1_A(\xi_n)-\int_0^1 1_A(y)\,dy\le\sum_{i\in J}\left(\frac1N\sum_{n=1}^N 1_{B_i}(\xi_n)-\int_0^1 1_{B_i}(y)\,dy\right) + 2/M.&#xD;&#xA;$$&#xA;The reverse inequality also holds if we make the sum run over $i\in I$ and replace $2/M$ by $-2/M$ on the right hand side. So,&#xA;$$&#xD;&#xA;\left\vert\frac1N\sum_{n=1}^N 1_A(\xi_n)-\int_0^1 1_A(y)\,dy\right\vert\le\sum_{i=1}^M\left\vert\frac1N\sum_{n=1}^N 1_{B_i}(\xi_n)-\int_0^1 1_{B_i}(y)\,dy\right\vert+2/M.&#xD;&#xA;$$&#xA;For any $\epsilon &amp;gt; 0$ we can choose $2/M &amp;lt; \epsilon/2$ and, using equidistribution, $N$ can be chosen large enough that the first term on the right hand side is less than $\epsilon/2$. So, the left hand side can be made less than $\epsilon$ by choosing $N$ large enough. This choice was independent of $A$ so, in fact, convergence is uniform on the set of all intervals $A$.&lt;/p&gt;&#xA;" OwnerUserId="1321" LastActivityDate="2010-12-28T01:41:03.847" />
  <row Id="15702" PostTypeId="1" AcceptedAnswerId="15706" CreationDate="2010-12-28T01:43:23.033" Score="4" ViewCount="133" Body="&lt;p&gt;One way to prove that a field $K$ has no ideals except the entire field and the trivial ideal is to note the fact that every element $x$ has an inverse. By the definition of an ideal, if $x$ is in the ideal then $x^{-1}x$ is because $x^{-1} \in K$. But now we have that 1 is in the ideal, and so again by the definition of an ideal we have that every element is in the ideal. Therefore it is either the entire field or trivial.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, this works for any would-be ideal that has a unit; hence my question. I don't see how this coheres particularly with the idea that ideals are generalizations of things like &quot;multiple of $n$&quot;, or that we use them to form quotient rings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can someone please explain whether this has a deeper meaning or if it's not really important? I think it might have something to do with what is written in the &quot;motivation&quot; section in the &lt;a href=&quot;http://en.wikipedia.org/wiki/Ideal_%28ring_theory%29&quot; rel=&quot;nofollow&quot;&gt;Wikipedia article for ideals&lt;/a&gt; but I'm not really sure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: I do realize that not all subsets without a unit are ideals. Sorry for the confusion.&lt;/p&gt;&#xA;" OwnerUserId="2954" LastEditorUserId="2954" LastEditorDisplayName="" LastEditDate="2010-12-29T00:25:18.437" LastActivityDate="2010-12-29T00:28:35.997" Title="Interpretation of ideals as unitless subsets" Tags="&lt;ring-theory&gt;" AnswerCount="3" CommentCount="3" />
  <row Id="15703" PostTypeId="2" ParentId="15657" CreationDate="2010-12-28T01:45:40.660" Score="2" ViewCount="" Body="&lt;p&gt;The question does not seem to be very precise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In any case,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is &lt;a href=&quot;http://books.google.com/books?id=PuL2KC_KCj0C&amp;amp;pg=PA21&quot; rel=&quot;nofollow&quot;&gt;well known&lt;/a&gt; that $K_{2k+1}$ can be decomposed into $k$ hamiltonian cycles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we can find an Euler tour with exactly $\dfrac{n-1}{2}$ hamiltonian cycles for odd $n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For even $n$, we can do an &quot;incomplete tour&quot; with  $\dfrac{n}{2}-1$ (See link above) hamiltonian cycles.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-12-28T01:52:45.987" LastActivityDate="2010-12-28T01:52:45.987" CommentCount="2" />
  <row Id="15704" PostTypeId="2" ParentId="15644" CreationDate="2010-12-28T01:56:53.580" Score="1" ViewCount="" Body="&lt;p&gt;One never says that the rational number 3/2 is &quot;undefined&quot; as an integer. Rather one simply says that it is nonintegral, or not an integer. On the other hand if 3/2 denotes the value of the (partial) division function on naturals $\rm\:\mathbb N^2\to \mathbb N$ then one may say that this &lt;em&gt;function&lt;/em&gt; is undefined at (3,2). &lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-12-28T01:56:53.580" />
  <row Id="15705" PostTypeId="2" ParentId="15702" CreationDate="2010-12-28T02:08:33.977" Score="3" ViewCount="" Body="&lt;p&gt;Ideals are not &quot;unitless subsets&quot;, in the sense that there are many unitless subsets that aren't ideals. Proper ideals, indeed, contain no units, but they are also closed under addition, subtraction and multiplication by arbitrary ring elements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure why you're confused about the case of &quot;multiples of n&quot;. In the ring $\mathbb{Z}$ of integers, the proper ideals are precisely the multiples of some fixed integer. Indeed, there are many other &quot;unitless subsets&quot; in $\mathbb{Z}$, but they are not ideals. &lt;/p&gt;&#xA;" OwnerUserId="308" LastActivityDate="2010-12-28T02:08:33.977" />
  <row Id="15706" PostTypeId="2" ParentId="15702" CreationDate="2010-12-28T02:10:48.773" Score="6" ViewCount="" Body="&lt;p&gt;Yes, it's true that any ideal in a ring which contains a unit must be the whole ring. (Also: every left ideal that contains a left-invertible element must be the whole ring).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, there are two items here. One is the notion of generalizing &quot;multiples of $n$&quot;,  and the second is the notion of making quotients.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's start with the first. The notion that ideals generalize things like &quot;all multiples of $n$&quot; in $\mathbb{Z}$ goes back all the way to Dedekind, who in fact came up with the term &quot;ideal&quot; (it was a counterpart to Kummer's &quot;ideal numbers&quot;; at the risk of touting my own horn, check out the paper with David McKinnon, &lt;a href=&quot;http://www.jstor.org/stable/30037491&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;Gauss's Lemma for Number Fields&lt;/em&gt;&lt;/a&gt;, Amer. Math. Monthly &lt;strong&gt;112&lt;/strong&gt; no. 5 (2005) pp. 385-416).  Dedekind was looking at rings like $\mathbb{Z}[\sqrt{-5}]$, and he wanted something along the lines of unique factorization; Kummer had done this by introducing something called &quot;ideal numbers&quot;, which were like the primes for integers but which did not actually occur in the ring in order to explain things like $2\times 3 = (1+\sqrt{-5})(1-\sqrt{-5})$, even though none of $2$, $3$, $1+\sqrt{-5}$ and $1-\sqrt{-5}$ can be written as a product of two numbers (from $\mathbb{Z}[\sqrt{-5}]$), neither of which is $1$ or $-1$. The idea was that there were these &quot;ideal numbers&quot; $\alpha$, $\beta$, and $\gamma$, which had the property that $\alpha\beta=1+\sqrt{-5}$, $\alpha\gamma=1-\sqrt{-5}$, $\alpha^2 = 2$, and $\beta\gamma=3$. Instead of &quot;inventing&quot; these numbers, Dedekind considered a collection of all numbers which would be &quot;the multiples of $\alpha$&quot; &lt;em&gt;if&lt;/em&gt; $\alpha$ was actually an element of $\mathbb{Z}[\sqrt{-5}]$. A necessary and sufficient condition for a collection of elements of $\mathbb{Z}[\sqrt{-5}]$ to qualify as &quot;all multiples of $x$&quot; (where $x$ was either an ideal number of an actual number) was that the collection be nonempty, closed under sums and differences, and that if $a$ was in the collection of $r$ was in $\mathbb{Z}[\sqrt{-5}]$, then $ra$ was in the collection. They were &quot;ideals&quot; because they were playing the role of ideal numbers (here, &quot;ideal&quot; means &lt;em&gt;existing in fancy or imagination&lt;/em&gt;). This generalized to any ring contained in a finite extension of $\mathbb{Q}$ and containing all integral elements (roots of monic polynomials with integer coefficients), called the &quot;number field case&quot;; later Dedekind and Weber developed a similar theory in what is called the &quot;function field&quot; case. Later, when rings were abstracted and generalized from the number field and function field cases by Artin and by Noether, the conditions were extended but they were still based in the notion that the ideals corresponded to &quot;all multiples of&quot; an ideal number in the number field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, there is no problem with the fact that sometimes these collections are &lt;em&gt;everything&lt;/em&gt;: this happens whenever the collection satisfies the conditions and contains a unit. Some collections do, some don't. In fact, that is one way in which one can test to see if an element is a unit: if the ideal it generates is the whole ring, then it is a unit (this works in the case of commutative rings with unit, not in more general rings). This actually makes sense because &lt;em&gt;everything&lt;/em&gt; is a multiple of a unit. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For rings like $\mathbb{Z}$, and to some extent number rings as above, &lt;em&gt;every&lt;/em&gt; ideal consists exactly of &quot;all the multiples of $x$&quot; for some $x$ (though in the number ring, the $x$ may be an &quot;ideal number&quot;). For more general (commutative) rings (with unit), if you have a family of elements $x_1,\ldots,x_n$, then the elements of the ideal &lt;em&gt;generated&lt;/em&gt; by $x_1,\ldots,x_n$ are exactly those elements that are &quot;$R$-linear combinations&quot; of $x_1,\ldots,x_n$; that is, all elements that can be expressed as $r_1x_1+\cdots+r_nx_n$ for some $r_1,\ldots,r_n\in R$; this &lt;em&gt;includes&lt;/em&gt; all multiples of each $x_i$, as well as other elements. If one of the $x_i$ is a unit, then since everything is a multiple of a unit, then everything is in the ideal in question. There are other kinds of ideals (not &quot;finitely generated&quot;, as you start dealing with more and more complicated rings).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, on to the second question. As it happens, Dedekind also noted that the conditions on the set (nonempty, closed under sums and differences, and closed under products by elements of the ring) were exactly the condition one needed in order to be able to do &quot;congruence modulo&quot; the multiples of $x$, so he also called them &quot;modules&quot; (because you could do modular arithmetic with them). This is essentially the same as talking about quotients. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But one can get to ideals differently if you are interested in quotients. This is essentially the same thing I did in &lt;a href=&quot;http://math.stackexchange.com/questions/14282/why-do-we-define-quotient-groups-for-normal-subgroups-only/14315#14315&quot;&gt;this answer&lt;/a&gt; about normal subgroups. Suppose you want to define an equivalence relation $\sim$ on the ring $R$ so that you can make $R/\sim$ into a ring using the operations $[a]+[b]=[a+b]$ and $[a][b]=[ab]$ (where $[a]$ is the equivalence class of $a$ under $\sim$). It turns out that the only way this can work is if $\sim$ is a subring of $R\times R$ (when we consider $\sim$ as a collection of ordered pairs of elements of $R$, hence a subset of $R\times R$), and that if you let $I$ be the collection of all elements of $R$ that are equivalent to $0$, then $I$ is an ideal of $R$ and the equivalence relation $\sim$ is $a\sim b\Longleftrightarrow a-b\in I$. You can try proving the theorems in the abovementioned answer for rings instead of groups, they are essentially the same arguments.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again, this has no problem with regards to having units in the ideal: this is what happens if your equivalence relation is just &quot;everything is related to everything&quot;, or if your quotient is the one element ring; these are perfectly valid (if somewhat boring) equivalence relations and quotients.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But: in a commutative ring with $1$, an ideal is the whole ring &lt;strong&gt;if and only if&lt;/strong&gt; it contains a unit. If there are no units, then it is not the whole ring. A ring is said to be &lt;em&gt;simple&lt;/em&gt; if the only ideals are the zero ideal and the whole ring; for commutative rings, the only simple rings are the fields and the zero ring. In the noncommutative case there are simple rings that are not division rings: for example, the set of $n\times n$ matrices over any simple ring (for instance, any field) is simple for every $n$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note also that while a &lt;em&gt;proper&lt;/em&gt; ideal necessarily does not contain a unit, it is not true that any subset that does not contain a unit is an ideal. The collection of all numbers that are either multiples of $2$ or of $3$ in $\mathbb{Z}$ contains no units, but is not an ideal (it is not even a subgroup). &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-29T00:28:35.997" LastActivityDate="2010-12-29T00:28:35.997" CommentCount="4" />
  <row Id="15707" PostTypeId="1" AcceptedAnswerId="15950" CreationDate="2010-12-28T02:36:26.713" Score="11" ViewCount="324" Body="&lt;p&gt;For which functions $f\colon [0,1] \to [0,1]$ is the function $g(x,y)=f(x)f(y)$ convex over $(x,y) \in [0,1]\times [0,1]$ ? Is there a nice characterization of such functions $f$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The obvious examples are exponentials of the form $e^{ax+b}$ and their convex combinations. Anything else?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: This is a simple observation summarizing the status of this question so far. The class of such functions f includes all log-convex functions, and is included in the class of convex functions. So now, the question becomes: are there any functions $f$ that are not log-convex yet $g(x,y)=f(x)f(y)$ is convex?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Jonas Meyer observed that, by setting $x=y$, the determinant of the hessian of $g(x,y)$ is positive if and only if $f$ is a log-convex. This resolves the problem for twice continuously differentiable $f$. Namely: if $f$ is $C^2$, then $g(x,y)$ is convex if and only if $f$ is log-convex.&lt;/p&gt;&#xA;" OwnerUserId="5159" LastEditorUserId="5159" LastEditorDisplayName="" LastEditDate="2010-12-31T00:56:27.643" LastActivityDate="2010-12-31T01:34:20.630" Title="For what functions $f(x)$ is $f(x)f(y)$ convex?" Tags="&lt;convex-analysis&gt;" AnswerCount="2" CommentCount="5" FavoriteCount="0" />
  <row Id="15708" PostTypeId="2" ParentId="15657" CreationDate="2010-12-28T02:41:16.123" Score="2" ViewCount="" Body="&lt;p&gt;There aren't necessarily &lt;em&gt;any&lt;/em&gt; Hamiltonian cycles in your sequence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As Ross already pointed out, provided $n$ is even and $n &amp;gt; 2$, then all vertices have odd degree and no Eulerian circuit exists.  So, we need only worry about when $n$ is odd and $n &amp;gt; 3$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $K_5$ arranged as in the picture, an Eulerian circuit with no Hamiltonian cycles is given by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/StCdJ.png&quot; alt=&quot;$K_5$&quot;&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$1 \rightarrow 2 \rightarrow 3 \rightarrow 4 \rightarrow 2 \rightarrow 5 \rightarrow 4 \rightarrow 1 \rightarrow 5 \rightarrow 3 \rightarrow 1.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now suppose $K_{n-2}$ has an Eulerian circuit with no Hamiltonian cycles.  First label the vertices of $K_n$ as $1, \ldots, n$.  Our Eulerian circuit with no Hamiltonian cycles for $K_n$ is given by first travelling $1 \rightarrow 2$, then travelling the Eulerian circuit with no Hamiltonian cycles for $K_{n-2}$ across the vertices $2, \ldots, n-1$, arriving back at $2$.  Then for $k = 2, 4, 6, \ldots, n - 3$, we travel $$k \rightarrow n \rightarrow k + 1 \rightarrow 1 \rightarrow k + 2.$$  Finally, we travel $n-1 \rightarrow n \rightarrow 1.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It follows from induction that $K_n$ contains an Eulerian circuit with no Hamiltonian cycles for all odd $n$.&lt;/p&gt;&#xA;" OwnerUserId="3537" LastActivityDate="2010-12-28T02:41:16.123" />
  <row Id="15709" PostTypeId="1" CreationDate="2010-12-28T02:54:33.603" Score="2" ViewCount="281" Body="&lt;p&gt;In &lt;em&gt;Conceptual Mathematics&lt;/em&gt; 1st edition, p. 325-236, there is a sketch of a proof, but I can't carry out the complete proof.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;... This also follows from the appropriate universal mapping properties, which imply that the two composites satisfy properties that only the corresponding identity maps satisfy.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can't figure this out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you give me a clue?&lt;/p&gt;&#xA;" OwnerUserId="5158" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-28T02:56:12.853" LastActivityDate="2011-01-06T01:00:59.687" Title="A question in &quot;Conceptual Mathematics&quot;" Tags="&lt;category-theory&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="15710" PostTypeId="1" CreationDate="2010-12-28T03:20:49.937" Score="16" ViewCount="396" Body="&lt;p&gt;Why is it that $\mathbb{Q}$ cannot be homeomorphic to any complete metric space?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Certainly $\mathbb{Q}$ is not a complete metric space. But completeness is not a topological invariant, so why is the above statement true?&lt;/p&gt;&#xA;" OwnerUserId="5161" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-28T03:54:54.777" LastActivityDate="2010-12-28T16:26:06.617" Title="Why is it that $\mathbb{Q}$ cannot be homeomorphic to _any_ complete metric space?" Tags="&lt;topology&gt;" AnswerCount="4" CommentCount="1" FavoriteCount="1" />
  <row Id="15711" PostTypeId="2" ParentId="15684" CreationDate="2010-12-28T03:31:15.413" Score="1" ViewCount="" Body="&lt;p&gt;Can you see what the topology of the quotient $S^1/(\tau)$ is? Its open sets &quot;are&quot; the open sets in $S^1$ which are invariant under multiplication by $\tau$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you see what the topology is, your question becomes easy.&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2010-12-28T03:31:15.413" />
  <row Id="15712" PostTypeId="2" ParentId="15710" CreationDate="2010-12-28T03:31:16.643" Score="20" ViewCount="" Body="&lt;p&gt;By the Baire Category Theorem, a space that is homeomorphic to a complete metric space must be a Baire Space: the intersection of a countable family of open dense sets must be dense. But $\mathbb{Q}$ does not have this property, because it is countable and no point is isolated: for each $q\in\mathbb{Q}$, let $\mathscr{O}_q = \mathbb{Q}\setminus\{q\}$. This is open (since $\{q\}$ is closed in $\mathbb{R}$, hence in the induced topology of $\mathbb{Q}$) and dense, since every open ball with center in $\mathbb{q}$ intersects $\mathscr{O}_q$. But &#xA;$$\bigcap_{q\in\mathbb{Q}} \mathscr{O}_q = \emptyset$$&#xA;is not dense. Hence $\mathbb{Q}$ is not a Baire space, hence cannot be homeomorphic to a complete metric space (or even to an open subset of a complete pseudometric space).&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-28T06:21:09.703" LastActivityDate="2010-12-28T06:21:09.703" CommentCount="5" />
  <row Id="15713" PostTypeId="1" CreationDate="2010-12-28T03:31:16.987" Score="1" ViewCount="174" Body="&lt;p&gt;I asked I question &lt;a href=&quot;http://math.stackexchange.com/questions/15676/simplification-of-sums&quot;&gt;here&lt;/a&gt; trying to obtain clarification about how to follow a hint. In spite of the fine answers I received there, the hint doesn't look very helpful. I'd like to know a hint for the following problem or a way to use the hint I already have.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The probability $p_{m}(r,n)$ of finding &lt;em&gt;exactly&lt;/em&gt; $m$ cells empty placing $r$ balls into $n$ cells is $$p_{m}(r,n)=\frac{1}{n^{r}}\binom{n}{m}A(r,n-m)=\binom{n}{m}\sum_{\nu=0}^{n-m}(-1)^{\nu}\binom{n-m}{\nu}\left(1-\frac{m+\nu}{n}\right)^{r}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From such probability, conclude that the probability $x_{m}(r,n)$ of finding $m$ &lt;em&gt;or more cells&lt;/em&gt; empty equals&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\binom{n}{m}\sum_{\nu=0}^{n-m}(-1)^{\nu}\binom{n-m}{\nu}\left(1-\frac{m+\nu}{n}\right)^{r}\frac{m}{m+v}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;HINT: Evaluate $x_{m}(r,n)-p_{m}(r,n)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Without using the hint, I tried to find the pattern for $p_{m}$, $p_{m+1}$, etc with the goal of factoring each term and sum over something recognizable, but I couldn't find a way to factor the last term (for example, to factor $(n-m-1-\nu)^{r}$ into some terms including the original expression $(n-m-\nu)^{r}$...I don't think it is possible. Furthermore, I don't think that strategy is going to produce the answer I'm looking for.&lt;/p&gt;&#xA;" OwnerUserId="2569" LastEditorUserId="2569" LastEditorDisplayName="" LastEditDate="2010-12-28T19:14:07.677" LastActivityDate="2011-05-30T15:57:53.790" Title="Probability of finding $m$ or more cells empty" Tags="&lt;probability&gt;&lt;combinatorics&gt;" AnswerCount="3" CommentCount="12" />
  <row Id="15714" PostTypeId="2" ParentId="15710" CreationDate="2010-12-28T03:33:51.197" Score="5" ViewCount="" Body="&lt;p&gt;A complete metric space is not a countable union of nowhere dense sets, but $\mathbb{Q}$ is (each singleton is nowhere dense). Since nowhere denseness is a topological property, $\mathbb{Q}$ cannot be homeomorphic to a complete metric space.&lt;/p&gt;&#xA;" OwnerUserId="3249" LastActivityDate="2010-12-28T03:33:51.197" />
  <row Id="15715" PostTypeId="2" ParentId="15710" CreationDate="2010-12-28T03:57:18.440" Score="2" ViewCount="" Body="&lt;p&gt;A subspace of a separable complete metric space is homeomorphic to a complete metric space if and only if it is a $G_\delta$. (I have a reference that says the direction relevant to your question is proved on page 197 of Bourbaki's &lt;em&gt;&lt;a href=&quot;http://books.google.com/books?id=bQwhdmL6IjUC&amp;amp;source=gbs_navlinks_s&quot; rel=&quot;nofollow&quot;&gt;General topology&lt;/a&gt;&lt;/em&gt;, volume 2, but I don't currently have access to that page.  The other direction is proved on &lt;a href=&quot;http://books.google.com/books?id=bQwhdmL6IjUC&amp;amp;lpg=PP1&amp;amp;pg=PA196#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;page 196&lt;/a&gt;.)  The set of rational numbers is not a $G_\delta$ in the real numbers.  I must admit that the only way I know how to prove that $\mathbb{Q}$ is not a $G_\delta$ is to appeal to Baire's theorem. (Every open set containing the rationals is dense, and the set of irrationals is a countable intersection of dense open sets, so if the set of rationals were a $G_\delta$, then the empty set (obtained by intersecting the sets of rational and irrational numbers) would be a countable intersection of dense open sets, contradicting Baire's theorem.) &lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2010-12-28T04:03:03.253" LastActivityDate="2010-12-28T04:03:03.253" CommentCount="3" />
  <row Id="15716" PostTypeId="1" AcceptedAnswerId="15740" CreationDate="2010-12-28T05:13:27.780" Score="12" ViewCount="279" Body="&lt;p&gt;If $n$ is a natural number, let $\displaystyle \sigma_{11}(n) = \sum_{d \mid n} d^{11}$. &#xA;The modular form $\Delta$ is defined by $\displaystyle \Delta(q) = q \prod_{n=1}^{\infty}(1 - q^n)^{24}$.&#xA;Write $\tau(n)$ for the coefficient of $q^{n}$ in $\Delta(q)$.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to know why $\tau(n) \equiv \sigma_{11}(n) \pmod{691}$. I think the proof may be somewhat difficult, so even just an outline of the argument would be much appreciated. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you!&lt;/p&gt;&#xA;" OwnerUserId="4457" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-28T05:48:22.940" LastActivityDate="2010-12-30T14:48:47.680" Title="Why is $\tau(n) \equiv \sigma_{11}(n) \pmod{691}$?" Tags="&lt;number-theory&gt;&lt;algebraic-number-theory&gt;&lt;modular-forms&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="2" />
  <row Id="15717" PostTypeId="2" ParentId="15713" CreationDate="2010-12-28T06:04:57.103" Score="1" ViewCount="" Body="&lt;p&gt;See the &lt;a href=&quot;http://probabilityandstats.wordpress.com/2010/04/04/a-formula-for-the-occupancy-problem/&quot; rel=&quot;nofollow&quot;&gt;following&lt;/a&gt; (Theorem 2). This gives the probability that exactly $m$ cells are empty. You could probably modify this to get the multiplicative factor $\frac{m}{m+v}$.&lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2010-12-28T06:04:57.103" CommentCount="1" />
  <row Id="15718" PostTypeId="2" ParentId="15629" CreationDate="2010-12-28T06:21:44.337" Score="2" ViewCount="" Body="&lt;p&gt;&#xA;How to implement If-then-else structures in propositional logic:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Example 1&lt;/b&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If P then&lt;br&gt;&#xA;&amp;nbsp; Q&lt;br&gt;&#xA;else&lt;br&gt;&#xA;&amp;nbsp; R&lt;br&gt;&#xA;end if&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(P -&amp;gt; Q) &amp;amp; (~P -&amp;gt; R)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Example 2&lt;/b&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If P then&lt;br&gt;&#xA;&amp;nbsp; Q&lt;br&gt;&#xA;else if R then&lt;br&gt;&#xA;&amp;nbsp; S&lt;br&gt;&#xA;else&lt;br&gt;&#xA;&amp;nbsp; T&lt;br&gt;&#xA;end if&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(P -&amp;gt; Q) &amp;amp; (~P &amp;amp; R -&amp;gt; S) &amp;amp; (~P &amp;amp; ~R -&amp;gt; T)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;/p&gt;&#xA;" OwnerUserId="3515" LastEditorUserId="3515" LastEditorDisplayName="" LastEditDate="2010-12-28T06:44:31.907" LastActivityDate="2010-12-28T06:44:31.907" />
  <row Id="15719" PostTypeId="1" AcceptedAnswerId="15723" CreationDate="2010-12-28T06:46:39.920" Score="17" ViewCount="627" Body="&lt;p&gt;$$\int{\frac{1+x^2}{(1-x^2)\sqrt{1+x^4}}}\mathrm dx$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This was a Calc 2 problem for extra credit (we have done hyperbolic trig functions too, if that helps) and I didn't get it (don't think anyone did) -- how would you go about it?&lt;/p&gt;&#xA;" OwnerUserId="5157" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-28T06:56:57.080" LastActivityDate="2011-01-07T17:45:08.580" Title="How do I integrate the following? $\int{\frac{(1+x^{2})\mathrm dx}{(1-x^{2})\sqrt{1+x^{4}}}}$" Tags="&lt;calculus&gt;&lt;integral&gt;" AnswerCount="4" CommentCount="7" FavoriteCount="5" />
  <row Id="15720" PostTypeId="1" CreationDate="2010-12-28T05:18:23.937" Score="3" ViewCount="171" Body="&lt;p&gt;I have a background in math, but I think the following construction appears often in physics.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a term for a functional which maps one smooth function $f: R^n \rightarrow R$ to another $L(f):R \rightarrow R$ which represents the integral along each level set?  That is, &#xA;$L(f)(a) = \int_{f^{-1}(a)} d\mu $ &#xA;where the measure $d\mu$ might be the induced measure from the metric on $R^n$, or it might also depend on the original $f$ in some intelligent way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, if the initial function is $f:R^2 \rightarrow R$ is defined $f(x)=x^2+y^2$ then $L(f)(a) = 2 \pi \sqrt{a}$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the original function is $f:R \rightarrow R$ is defined $f(x)=x^2$, then we'd have $L(f)(a)=0$ for $a&amp;lt;0$, $1$ for $a=0$, and $2$ for $a&amp;gt;0$.  In both cases it might be more natural to integrate with a term derived from the value of $f&amp;#39;$ normal to the level set. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suppose one could generalize to any smooth map $f:M \rightarrow M&amp;#39;$ from a Riemannian manifold to a another smooth manifold and get a new function that represents a computed volume of the preimage for each point:  $L(f):M&amp;#39; \rightarrow R$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question:  Is there a good way to compute $L(f)$ for some class of function $f$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any thoughts would be appreciated.  It is reminiscent of (but I hope simpler than) stationary phase approximations and path integral calculations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Stationary_phase_approximation&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Stationary_phase_approximation&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="948" LastEditorUserId="1543" LastEditorDisplayName="" LastEditDate="2010-12-30T19:23:13.310" LastActivityDate="2010-12-30T19:51:31.570" Title="Integrating over level sets of a function" Tags="&lt;calculus&gt;&lt;analysis&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" />
  <row Id="15721" PostTypeId="1" CreationDate="2010-12-28T07:36:11.710" Score="8" ViewCount="289" Body="&lt;p&gt;How can I find solution for $x^3 \equiv 1 \pmod p$ ($p$ a prime) efficiently?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Trivial root is $x_1 = 1$. I need to find other roots $x_2, x_3$.&lt;/p&gt;&#xA;" OwnerUserId="5164" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-28T07:44:05.557" LastActivityDate="2010-12-29T04:30:21.000" Title="Solve $x^3 \equiv 1 \pmod p$ for $x$" Tags="&lt;number-theory&gt;" AnswerCount="3" CommentCount="0" />
  <row Id="15722" PostTypeId="2" ParentId="15721" CreationDate="2010-12-28T07:48:12.090" Score="5" ViewCount="" Body="&lt;p&gt;First, note that if $p \equiv 2\text{ mod } 3$, then raising to the third power is a bijection on $\left(\mathbb{Z}/p\mathbb{Z}\right)^\times$, so that 1 is the only root. If $p \equiv 1\text{ mod } 3$, then the only way I can think of to find the roots is to find a primitive root modulo $p$ (i.e. a generator of the cyclic group $\left(\mathbb{Z}/p\mathbb{Z}\right)^\times$) and take it to the powers $(p-1)/3$ and $2(p-1)/3$.&lt;/p&gt;&#xA;" OwnerUserId="3212" LastActivityDate="2010-12-28T07:48:12.090" CommentCount="4" />
  <row Id="15723" PostTypeId="2" ParentId="15719" CreationDate="2010-12-28T07:49:27.237" Score="34" ViewCount="" Body="&lt;p&gt;It might not be elliptic after all... (unless I have made some mistake)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int{\frac{1+x^2}{(1-x^2)\sqrt{1+x^4}}}\mathrm dx$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $\displaystyle u = x -\frac{1}{x}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then $\displaystyle du = (1 + \frac{1}{x^2})dx$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now $$\frac{1+x^2}{(1-x^2)\sqrt{1+x^4}} = -\frac{x^2(1 + 1/x^2)}{x(x-1/x)\sqrt{x^2(x^2 + 1/x^2)}} = -\frac{1 + 1/x^2}{(x-1/x)\sqrt{(x - 1/x)^2 + 2}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int{\frac{1+x^2}{(1-x^2)\sqrt{1+x^4}}}\mathrm dx$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$= -\int \ \frac{\mathrm{du}}{u \sqrt{u^2 + 2}}$$&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-12-28T07:49:27.237" CommentCount="7" />
  <row Id="15724" PostTypeId="2" ParentId="15251" CreationDate="2010-12-28T08:04:38.377" Score="3" ViewCount="" Body="&lt;p&gt;You may read the paper &quot;&lt;a href=&quot;http://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.57.8591&quot; rel=&quot;nofollow&quot;&gt;123 theorem and its extensions&lt;/a&gt;&quot; by Noga Alon&lt;/p&gt;&#xA;" OwnerUserId="2481" LastActivityDate="2010-12-28T08:04:38.377" CommentCount="1" />
  <row Id="15725" PostTypeId="1" CreationDate="2010-12-28T08:58:46.910" Score="5" ViewCount="319" Body="&lt;p&gt;How do we solve the recurrence $T(n) = 2T(n/3) + n\log n$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, is it possible to solve this recurrence by the Master method?&lt;/p&gt;&#xA;" OwnerUserId="3545" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-28T10:22:31.490" LastActivityDate="2010-12-28T21:07:41.663" Title="how to solve the recurrence $T(n) = 2T(n/3) + n\log n$" Tags="&lt;algorithms&gt;&lt;asymptotics&gt;" AnswerCount="2" FavoriteCount="2" />
  <row Id="15726" PostTypeId="2" ParentId="15725" CreationDate="2010-12-28T09:14:04.837" Score="3" ViewCount="" Body="&lt;p&gt;Hint: It is possible to solve by &lt;a href=&quot;http://en.wikipedia.org/wiki/Master_theorem&quot; rel=&quot;nofollow&quot;&gt;Master theorem&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A more generic method is &lt;a href=&quot;http://en.wikipedia.org/wiki/Akra%E2%80%93Bazzi_method&quot; rel=&quot;nofollow&quot;&gt;Akra Bazzi&lt;/a&gt;, but you don't need that for this problem.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-12-28T09:14:04.837" CommentCount="4" />
  <row Id="15727" PostTypeId="2" ParentId="15716" CreationDate="2010-12-28T09:36:07.067" Score="9" ViewCount="" Body="&lt;p&gt;The modular discriminant $\Delta (z) \in M_{12,0},$ the linear space of cusp forms of weight $12.$ Now dimension $M_{12,0} = 1$  (see, for example, &lt;a href=&quot;http://books.google.co.uk/books?id=lyqDSjLY6zEC&amp;amp;pg=PA118&amp;amp;lpg=PA118&amp;amp;dq=dimension+linear+space+cusp+form&amp;amp;source=bl&amp;amp;ots=xxacNSP9vX&amp;amp;sig=B0rUiRlyKk-7SiVYNGn9_KwILa8&amp;amp;hl=en&amp;amp;ei=eqcZTfmWMI-5hAfX79y3Dg&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=5&amp;amp;ved=0CDQQ6AEwBA#v=onepage&amp;amp;q=dimension%20linear%20space%20cusp%20form&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;T.M. Apostol: Moldular Functions and Dirichlet Series in Number Theory&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Every modular form for the full modular group is a polynomial in $E_4$ and $E_6,$ where $E_{2k}$ are the Eisenstein series defined by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$E_{2k}(z) = \frac{1}{2\zeta(2k)} \sum_{(m,n) \ne (0,0) } \frac{1}{(m+nz)^{2k}} \quad \textrm{for } Im(z) &amp;gt;0,$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where the summation extends over all integral $m$ and $n$ not both equal to $0$. So if we use the Eisenstein series to construct cusp forms of weight $12,$ because $\textrm{dim } M_{12,0} = 1$ these will necessarily be constant multiples of $\Delta (z).$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we can show that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$E_{2k}(z) = 1 - \frac{4k}{B_{2k}} \sum_{n=1}^\infty \sigma_{2k-1}(n)e^{2 \pi i n z}&#xD;&#xA;\quad \textrm{for } Im(z) &amp;gt; 0,$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\sigma_k(n) = \sum_{d|n} d^k$ and the $B_{2k}$ are the Bernoulli numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore we can expect identities between the divisors functions $\sigma_{2k-1}(n)$ and $\tau(n)$ to exist. Ramanujan's paper &quot;On Certain Arithmetical Functions&quot; shows how to express&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\phi_{r,s}(x) = \sum_{n=1}^\infty n^r \phi_{s-r}(n)x^n$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for integer $r,s \ge 0$ and $ |x| &amp;lt; 1$ as polynomials in $E_2,E_4,$ and $E_6,$ from which many such identities follow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: To cite a random example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\tau(n) = n^2\sigma_7(n) - 540\sum_{k=1}^{n-1}k(n-k)\sigma_3(k)\sigma_3(n-k)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and so $\tau(n) \equiv n^2\sigma_7(n) \textrm{ mod } 540.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT2: $B_{12} = -691/2730.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT3: For completeness, using Ramanujan's notation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$P = 1 - 24 \sum_{n=1}^\infty \frac{nx^n}{1-x^n}$$&#xA;$$Q = 1 + 240 \sum_{n=1}^\infty \frac{n^3x^n}{1-x^n}$$&#xA;$$R = 1 - 504 \sum_{n=1}^\infty \frac{n^5x^n}{1-x^n},$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the relevant entries from his paper &quot;On Certain Arithmetical Functions&quot; are $(3)$ and $(6)$ from his Table I, given below. (Note that $Q$ is $E_4$ and $R$ is $E_6.$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xD;&#xA;1 - 24\phi_{0,1}(x) &amp;amp;= P \\&#xD;&#xA;1 + 240\phi_{0,3}(x) &amp;amp;= Q \\&#xD;&#xA;1 - 504\phi_{0,5}(x) &amp;amp;= R \quad (3) \\&#xD;&#xA;1 + 480\phi_{0,7}(x) &amp;amp;= Q^2 \\&#xD;&#xA;1 - 264\phi_{0,9}(x) &amp;amp;= QR \\&#xD;&#xA;691 + 65520\phi_{0,11}(x) &amp;amp;= 441Q^3+250R^2 \quad (6) \\&#xD;&#xA;1 - 24\phi_{0,13}(x) &amp;amp;= Q^2R \\&#xD;&#xA;3617 + 16320\phi_{0,15}(x) &amp;amp;= 1617Q^4+2000QR^2 \\&#xD;&#xA;43867 - 28728\phi_{0,17}(x) &amp;amp;= 38367Q^3R+5500R^3 \\&#xD;&#xA;174611 + 13200\phi_{0,19}(x) &amp;amp;= 53361Q^5+121250Q^2R^2 \\&#xD;&#xA;77683 - 552\phi_{0,21}(x) &amp;amp;= 57183Q^4R+20500QR^3 \\&#xD;&#xA;236364091 + 131040\phi_{0,23}(x) &amp;amp;= 49679091Q^6+176400000Q^3R^2 + 10285000R^4 \\&#xD;&#xA;657931 - 24\phi_{0,25}(x) &amp;amp;= 392931Q^5R+265000Q^2R^3 \\&#xD;&#xA;3392780147 + 6960\phi_{0,27}(x) &amp;amp;= 489693897Q^7+2507636250Q^4R^2 + 395450000QR^4 \\&#xD;&#xA;1723168255201 - 171864\phi_{0,29}(x) &amp;amp;= 815806500201Q^6R+88134070500Q^3R^3 + 26021050000R^5 \\&#xD;&#xA;7709321041217 + 32640\phi_{0,31}(x) &amp;amp;= 764412173217Q^8+5323905468000Q^5R^2 \\ &#xD;&#xA;&amp;amp;+ 1621003400000Q^2R^4&#xD;&#xA;&#xD;&#xA;\end{align} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Equations $(3)$ and $(6)$, along with $1728 \sum_{n=1}^\infty \tau(n)x^n = Q^3 - R^2,$ give&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$691+65520 \sum_{n=1}^\infty \sigma_{11}(n) x^n = 441 \times 1728 \sum_{n=1}^\infty \tau(n) x^n + 691R^2.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The result follows by taking this equation modulo $691$ and noting that $566$ and $691$ are coprime. &lt;/p&gt;&#xA;" OwnerUserId="1301" LastEditorUserId="1301" LastEditorDisplayName="" LastEditDate="2010-12-30T14:48:47.680" LastActivityDate="2010-12-30T14:48:47.680" CommentCount="3" />
  <row Id="15728" PostTypeId="2" ParentId="15650" CreationDate="2010-12-28T10:03:44.447" Score="1" ViewCount="" Body="&lt;p&gt;As Ross's counterexample points out the answer, of course, is no in general.  However there are broad special cases where the answer is yes, which all more or less go back to the &lt;a href=&quot;http://en.wikipedia.org/wiki/Perron%E2%80%93Frobenius_theorem&quot; rel=&quot;nofollow&quot;&gt;Perron-Frobenius theorem&lt;/a&gt;.  In particular if the recurrence is combinatorial in the sense that it counts the number of words of length $n$ in a regular language, then subject to some reasonable assumptions about the language the limit you are looking at will exist. &lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-12-28T10:03:44.447" />
  <row Id="15729" PostTypeId="2" ParentId="15719" CreationDate="2010-12-28T10:12:43.353" Score="18" ViewCount="" Body="&lt;p&gt;Somewhat inspired by Moron's wonderful answer, I decided to see if a trigonometric solution would do the job.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Making the substitution $x=\cot\left(\frac{\theta}{2}\right)$, $\mathrm dx=\frac{\mathrm d\theta}{\cos\;\theta-1}$, we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int \frac{\mathrm d\theta}{\cos\;\theta(1-\cos\;\theta)\sqrt{1+\cot^4\frac{\theta}{2}}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$=\int \frac{\mathrm d\theta}{\cos\;\theta(1-\cos\;\theta)\sqrt{1+\left(\frac{1+\cos\;\theta}{1-\cos\;\theta}\right)^2}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$=\frac1{\sqrt{2}}\int \frac{\mathrm d\theta}{\cos\;\theta\sqrt{1+\cos^2\theta}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which integrates to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac1{\sqrt{2}}\tanh^{-1}\frac{\sin\;\theta}{\sqrt{1+\cos^2\theta}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Undoing the substitution, we get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac1{\sqrt{2}}\tanh^{-1}\left(x\sqrt{\frac{2}{x^4+1}}\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and it is easy to verify that the derivative of this last expression gives the original integrand.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-12-28T10:12:43.353" />
  <row Id="15730" PostTypeId="1" CreationDate="2010-12-28T10:30:02.517" Score="3" ViewCount="125" Body="&lt;p&gt;Suppose A is a linear transformation from R^3 to R^3 and |det(A)| = 1. I know that A is volume preserving, but is it also area preserving? For example, if a and b are two vectors in R^3 that span a parallelogram, is the area of this parallelogram equal to the area of the paralellogram spanned by A(a) and A(b)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;THank you!&lt;/p&gt;&#xA;" OwnerUserId="5165" LastActivityDate="2010-12-28T10:45:43.413" Title="Area preserving transformations" Tags="&lt;linear-algebra&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="15731" PostTypeId="2" ParentId="15730" CreationDate="2010-12-28T10:45:43.413" Score="7" ViewCount="" Body="&lt;p&gt;Matrices of the form $\begin{pmatrix}X &amp;amp; 0\\\\0 &amp;amp; \text{det}(X)^{-1}\end{pmatrix}$ with $X$ any invertible 2 by 2 matrix with determinant not equal to $\pm1$ give a host of counter examples: consider the action of such a matrix on a parallelogram in the subspace $\langle(1,0,0),(0,1,0)\rangle$.&lt;/p&gt;&#xA;" OwnerUserId="3212" LastActivityDate="2010-12-28T10:45:43.413" />
  <row Id="15732" PostTypeId="2" ParentId="15707" CreationDate="2010-12-28T11:43:23.683" Score="0" ViewCount="" Body="&lt;p&gt;This may just be a silly observation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, let $\alpha=(a_1,a_2)$ and $\beta=(b_1,b_2)$. If $f$ is convex and the sum of $g$ at the four corners of the square formed by $\alpha$ and $\beta$ is non-positive, then $g(t\alpha + (1-t)\beta) \leq tg(\alpha)+(1-t)g(\beta)$, where $t \in [0,1]$&lt;/p&gt;&#xA;" OwnerUserId="4806" LastActivityDate="2010-12-28T11:43:23.683" />
  <row Id="15733" PostTypeId="2" ParentId="15520" CreationDate="2010-12-28T11:55:21.640" Score="2" ViewCount="" Body="&lt;p&gt;\begin{align}&#xA;\newcommand{\mba}{\mathbf{a}}&#xA;\newcommand{\mbb}{\mathbf{b}}&#xA;\newcommand{\mbl}{\mathbf{l}}&#xA;\newcommand{\mbp}{\mathbf{p}}&#xA;\newcommand{\mbv}{\mathbf{v}}&#xA;\newcommand{\mbx}{\mathbf{x}}&#xA;\int_{l_1}^0 \dotsi \int_{l_k}^0 \exp(\mbx \cdot (\mba - \mbb) + \mbl \cdot \mbb)d\mbx &#xA;&amp;amp;=\exp(\mbl\cdot\mbb)\prod_{i=1}^k\int_{l_i}^0\exp(x_i(a_i-b_i))dx_i\\&#xA;&amp;amp;=\exp(\mbl\cdot\mbb)\prod_{i=1}^k\frac{1-\exp(l_i(a_i-b_i))}{a_i-b_i}&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That should fix the error in your calculations. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However a lot of questions still remain. What is $B(\alpha+1)$? And if your are trying to compute convolution, why isn't your integration domain $(-\infty,0]^k$?&lt;/p&gt;&#xA;" OwnerUserId="4742" LastActivityDate="2010-12-28T11:55:21.640" CommentCount="1" />
  <row Id="15734" PostTypeId="1" AcceptedAnswerId="15741" CreationDate="2010-12-28T12:01:23.910" Score="4" ViewCount="227" Body="&lt;p&gt;In&#xA;&lt;a href=&quot;http://math.stackexchange.com/questions/11945/limit-of-sqrt7-sqrt7-sqrt7-cdots&quot;&gt;Limit of $\sqrt{7+\sqrt{7+\sqrt{7+\cdots}}}$&lt;/a&gt;&#xA;Timothy Wagner gave a correct answer that was questioned for not having shown that the limit exists in the first place. My question is, are there any examples were a value of limit can be derived although the limit does not exist? Please note I am not questioning that limit must exist before it is exhibited but that, how a value for limit can be exhibited if the limit doesn't exist? Is that not a contradiction? On one hand we have a value for the limit and on the other hand the proof that it can't exist! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please give an example were a more general form of convergence does not account for the calculated value. Otherwise it seems as if the value was hinting that the notion of convergence required adjustment and not the calculated value that required justification.&lt;/p&gt;&#xA;" OwnerUserId="3936" LastEditorUserId="3936" LastEditorDisplayName="" LastEditDate="2010-12-28T13:52:48.827" LastActivityDate="2010-12-28T14:35:37.523" Title="Request for example of a value of limit of a sequence where sequence does not converge" Tags="&lt;analysis&gt;&lt;logic&gt;" AnswerCount="4" />
  <row Id="15735" PostTypeId="2" ParentId="15734" CreationDate="2010-12-28T12:06:55.683" Score="14" ViewCount="" Body="&lt;p&gt;Yes. Consider the sequence $a_{n+1} = 2a_n + 1$, where $a_0 = 0$. If this sequence converged to some value $L$, then $(a_{n+1})$ would converge to the same value, so we would get $L = 2L + 1$, which gives $L = - 1$. But the sequence actually diverges to infinity.  I use this sort of example when I teach real analysis to drive home the point that we have to check for convergence before trying to compute the value of a limit. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This method also comes up in with infinite series. Consider the series $1 + -1 + 1 + -1 + \cdots$, which is not convergent. If it &lt;em&gt;were&lt;/em&gt; convergent to a value $s$, then we would have $s = 1 + (-1 + 1 + -1 + \cdots)$ so $s = 1 - s$, which leads to $s = 1/2$. So we can compute a &quot;value&quot; even though the series diverges. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This fact is actually useful in practice when we study generalized notions of convergence. For the moment, define a notion of series convergence to be &quot;nice&quot; if, whenever a series $\sum a_n$ converges, we have $\sum a_n = a_0 + \sum a_{n+1}$ and $\sum -a_n = -\sum a_n$. Then if $1 + -1 + 1 + -1 + \cdots$ converges under any &quot;nice&quot; notion of convergence, its limit will be $1/2$. &lt;/p&gt;&#xA;" OwnerUserId="630" LastEditorUserId="630" LastEditorDisplayName="" LastEditDate="2010-12-28T12:38:56.497" LastActivityDate="2010-12-28T12:38:56.497" CommentCount="17" />
  <row Id="15736" PostTypeId="2" ParentId="15734" CreationDate="2010-12-28T12:10:50.437" Score="2" ViewCount="" Body="&lt;p&gt;What he showed was that &lt;em&gt;if&lt;/em&gt; a limit $L$ exists, then it has to satisfy the equation $L = \sqrt{7+L}$. This is typical of exercises early in a real analysis class where the only convergence theorem you have available is that bounded monotone sequences converge (this is easy to show given that the reals have the least upper bound property).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if we have a sequence $(a_n)$ that is defined recursively by $a_1 = x$, $a_{n+1} = f(a_n)$ and is monotone and unbounded, then any fixed point of $f$ is going to be a candidate for a limit, but $(a_n)$ doesn't converge.&lt;/p&gt;&#xA;" OwnerUserId="1269" LastEditorUserId="1269" LastEditorDisplayName="" LastEditDate="2010-12-28T12:22:38.487" LastActivityDate="2010-12-28T12:22:38.487" CommentCount="4" />
  <row Id="15737" PostTypeId="2" ParentId="15734" CreationDate="2010-12-28T12:13:44.277" Score="0" ViewCount="" Body="&lt;p&gt;The point is often when we try to derive the limit, we invoke algebraic manipulations which would only work if it has already been established that a limit exists (say, by showing the sequence is cauchy). &lt;/p&gt;&#xA;" OwnerUserId="4806" LastActivityDate="2010-12-28T12:13:44.277" CommentCount="7" />
  <row Id="15739" PostTypeId="1" AcceptedAnswerId="16899" CreationDate="2010-12-28T14:16:57.003" Score="0" ViewCount="312" Body="&lt;p&gt;I am referring to the Polyhedron Volume &lt;a href=&quot;http://en.wikipedia.org/wiki/Polyhedron#Volume&quot; rel=&quot;nofollow&quot;&gt;formula here&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$V = \frac{\sum_{i=1}^k \vec{x_{i}} \cdot \hat{n_i} A_i}{3}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is, does the above formula applicable for the Polyhedron below?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://docs.google.com/drawings/pub?id=155vFKEaDL3ZL8sy-zhil60zlL9wA07sJBOuZRHVvrKQ&amp;amp;w=960&amp;amp;h=720&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The given faces for the polyhedron are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f_1: n_1,n_2,n_3, n_4$$&#xA;$$f_2:n_2, n_3,n_8,n_5$$&#xA;$$f_3:n_1, n_4,n_6,n_7$$&#xA;$$f_4:n_6, n_8,n_5,n_7$$&#xA;$$f_5:n_1, n_2,n_5,n_7$$&#xA;$$f_6:n_6, n_8,n_3,n_4$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that for my application, the face $f_6$ may or may not intersect with the face $f_5$. Why is why I would prefer to apply the above formula mechanically, instead of having to determine the intersected line, and then compute the volume separately.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it does, what would be the value of the above calculation?&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$|v_1-v_2|$&lt;/li&gt;&#xA;&lt;li&gt;$|v_1|+|v_2|$&lt;/li&gt;&#xA;&lt;li&gt;None of the above?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Is there a proof showing why the answer is it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: Additional discussion on the validity of the above formula is &lt;a href=&quot;http://math.stackexchange.com/questions/16111/is-this-polyhedron-formula-correct&quot;&gt;shown here&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: Update the formula according to the wiki page. &lt;/p&gt;&#xA;" OwnerUserId="550" LastEditorUserId="550" LastEditorDisplayName="" LastEditDate="2011-01-03T08:55:39.563" LastActivityDate="2011-01-09T16:53:15.887" Title="Does Divergence Theorem Polyhedron Volume Calculation Applicable for a Cut-Through Polygon?" Tags="&lt;geometry&gt;" AnswerCount="2" CommentCount="10" />
  <row Id="15740" PostTypeId="2" ParentId="15716" CreationDate="2010-12-28T14:26:28.670" Score="7" ViewCount="" Body="&lt;p&gt;The proof is not at all obvious if you begin simply with the formula&#xA;$$\Delta(q) = q \prod_{n=1}^{\infty} (1-q^n)^{24}.$$  However, as Derek Jennings explains in his answer, if you use the (absolutely crucial!) fact that $\Delta$ is a cusp form of weight twelve and level one, the proof is actually not very difficult.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As Derek explains, the ring of modular forms of level one is generated by two $q$-expansions, namely&#xA;$$E_4 := 1 + 240 \sum_{n = 1}^{\infty} \sigma_3(n) q^n,$$&#xA;which has weight 4,&#xA;and&#xA;$$E_6 := 1 + 504 \sum_{n=1}^{\infty} \sigma_5(n) q^n,$$&#xA;which has weight 6.  (The coefficients $240$ and $-504$ come from Bernoulli numbers, as Derek explains, but we don't need that at the moment.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we see that we can make two monomials of weight 12 from these, namely&#xA;$E_4^3$ and $E_6^2$.  How do we get $\Delta$?  Well, the constant term of $\Delta$ vanishes, while $E_4^3$ and $E_6^2$ have constant term $1$, so &#xA;$\Delta$ must be proportional to $E_4^3  - E_6^2$.  Since the coefficient&#xA;of $q$ in $\Delta$ is $1$ (i.e. $\tau(1) = 1$) while the coefficient&#xA;of $q$ in $E_4^3 - E_6^2$ is $1728,$ we find that &#xA;$$E_4^3 - E_6^2 = 1728 \Delta.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is useful to note that we can also use $E_4^3$ (say) and $\Delta$ as a basis for the weight $12$ modular forms.  In fact, they are a very convenient basis, because if $f$ is any weight $12$ modular form, with constant term $a_0$, then we can subtract of $a_0 E_4^3$ to get rid of the constant term of $f$, and then&#xA;$f - a_0 E_4^3$ must be a mulitple of $\Delta$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To go further, we have to introduce another fact, also noted by Derek Jennings, namely that&#xA;there is a weight 12 modular form&#xA;$$E_{12} = 1 + \dfrac{65520}{691} \sum_{n=1}^{\infty} \sigma_{11}(n) q^n.$$&#xA;In fact, it is &#xA;(for me)&#xA;easier to work with&#xA;$$691 E_{12} = 691 + 65520\sum_{n=1}^{\infty} \sigma_{11}(n) q^n,$$&#xA;which has integer coefficients.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we apply the above procedure to write $691 E_{12}$ in terms of $E_4^3$ and $\Delta$, to find that&#xA;$$691 E_{12} = 691 E_4^3 + (65520 - 691\cdot 720) \Delta.$$ &#xA;Now all the $q$-expansions in this formula have integral coefficients, and so&#xA;what we find, looking at the coefficient of $q^n$, is that&#xA;$65520\sigma_{11}(n) \equiv 65520\tau(n)$ for each $n \geq 1$.&lt;br&gt;&#xA;Dividing by $65520$ (which is coprime to $691$) gives the desired formula.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The usual way this is summarized is to say that&#xA;$$\dfrac{691}{65520} E_{12} = \dfrac{691}{65520} + \sum_{n=1}^{\infty} \sigma_{11}(n) q^n$$&#xA;is normalized (i.e. has coefficient of $q$ equal to $1$) and is a cuspform modulo $691$ (i.e. its constant term vanishes mod $691$).  This forces it (as we have just seen) to be congruent to $\Delta$ modulo $691$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, let me note that by far the best place to read about the theory of modular forms used here is Serre's beautiful book &lt;em&gt;A course in arithmetic&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Added: Since I was editing this anyway, let me add a cultural remark, namely that the study of congruences between Eisenstein series and cuspforms,&#xA;of which the congruence between $\dfrac{691}{65520} E_{12}$ and $\Delta$ considered above is the first example, is a central topic in modern number theory.  It lies at the heart of Mazur's determination of the possible torsion subgroups of elliptic curves over $\mathbb Q$, and is the basic method via which Ribet proved his &quot;converse to Herbrand's theorem'' result giving a criterion for non-triviality of various $p$-power subgroup of the class group of the cyclotomic field $\mathbb Q(\zeta_p)$.&lt;/p&gt;&#xA;" OwnerUserId="221" LastEditorUserId="221" LastEditorDisplayName="" LastEditDate="2010-12-28T19:50:59.233" LastActivityDate="2010-12-28T19:50:59.233" CommentCount="3" />
  <row Id="15741" PostTypeId="2" ParentId="15734" CreationDate="2010-12-28T14:35:37.523" Score="3" ViewCount="" Body="&lt;p&gt;(This should be a comment under Carl's &lt;a href=&quot;http://math.stackexchange.com/questions/15734/request-for-example-of-a-value-of-limit-of-a-sequence-where-sequence-does-not-con/15735#15735&quot;&gt;answer&lt;/a&gt;.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;@Arjan: One very general way in which to formalize your idea of &quot;possible definitions of convergence&quot; is that of &lt;a href=&quot;http://en.wikipedia.org/wiki/Banach_limit&quot; rel=&quot;nofollow&quot;&gt;Banach limits&lt;/a&gt;. Indeed, a Banach limit is essentially a convergence notion for bounded sequences satisfying a minimal set of reasonable conditions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Banach limits do exist---this was proved by Banach himself as a way to show off his Hahn-Banach theorem---and all Banach limits agree with the usual limit on sequences that converge. More interestingly, the set of bounded sequences on which &lt;em&gt;all&lt;/em&gt; Banach limits agree is strictly larger than the set of convergence sequences, and it was determined explicitly by [Lorentz, G. G. A contribution to the theory of divergent sequences. Acta Math. 80, (1948). 167--190. &lt;a href=&quot;http://www.ams.org/mathscinet-getitem?mr=MR0027868&quot; rel=&quot;nofollow&quot;&gt;MR0027868&lt;/a&gt;]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the result of Lorentz implies that not &lt;em&gt;all&lt;/em&gt; Banach limits agree on all sequences (in other words, that there is more than one Banach limit) It follows from this that there are divergent sequences which can be assigned at least &lt;em&gt;two&lt;/em&gt; different limits under different extensions of the notion of convergence.&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2010-12-28T14:35:37.523" CommentCount="1" />
  <row Id="15742" PostTypeId="1" AcceptedAnswerId="15743" CreationDate="2010-12-28T14:45:32.910" Score="1" ViewCount="135" Body="&lt;p&gt;$$T (n) = 3T (n/5) + \log^2 n$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i am trying to solve this recurrence i have been told that it is master methods case 1 but how do i compare $\log^2n$ with $n^ {\log_5 3}$&lt;/p&gt;&#xA;" OwnerUserId="3545" LastEditorUserId="3249" LastEditorDisplayName="" LastEditDate="2010-12-28T15:13:36.017" LastActivityDate="2010-12-28T16:48:39.820" Title="stuck with master method." Tags="&lt;algorithms&gt;&lt;asymptotics&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="15743" PostTypeId="2" ParentId="15742" CreationDate="2010-12-28T15:34:40.163" Score="3" ViewCount="" Body="&lt;p&gt;A rule of thumb is that polynomial functions always grow faster than logarithmic functions.  E.g. &lt;a href=&quot;http://www.cs.columbia.edu/~mihalis/cs4231/Handouts/Asymptotics.pdf&quot; rel=&quot;nofollow&quot;&gt;this lecture&lt;/a&gt;. So your polynomial function asymptotically dominates your log function, giving you case 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to know why powers grow faster than logs, you can put the functions in a ratio and then evaluate $\lim_{n\to \infty}  \frac{f(n)}{g(n)}$.&lt;/p&gt;&#xA;" OwnerUserId="129" LastActivityDate="2010-12-28T15:34:40.163" CommentCount="6" />
  <row Id="15744" PostTypeId="2" ParentId="15742" CreationDate="2010-12-28T15:42:38.337" Score="2" ViewCount="" Body="&lt;p&gt;So we want to show that $T(n) = \Theta(n^{\log_{b}a})$ and $f(n) = O(n^{\log_{b}(a)-\epsilon})$. So $a=3, b=5, f(n) = \log^{2} n$, and $\log_{5} 3 = 0.6826$. So we check the following:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$f(n) = \log^{2} n = O(n^{0.6826- \epsilon})$. So choose any $\epsilon &amp;gt;0$. In other words, $\log^{2} n \in o(n^{0.6826- \epsilon})$. Note the derivative of $(\log n)^2 = \frac{2 \log n}{n}$ and the derivative of $n^{0.6826}$ is $\frac{0.6826}{n^{0.3174}}$. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The result follows. &lt;/p&gt;&#xA;" OwnerUserId="2548" LastEditorUserId="2548" LastEditorDisplayName="" LastEditDate="2010-12-28T15:53:35.853" LastActivityDate="2010-12-28T15:53:35.853" CommentCount="1" />
  <row Id="15746" PostTypeId="2" ParentId="15720" CreationDate="2010-12-28T15:53:17.547" Score="1" ViewCount="" Body="&lt;p&gt;Using the Fourier transform representation of the Dirac measure on the level set, the volume&#xA;of the level set (which is the required integral according to the examples given in the question)&#xA;is given by (using the same notation as in the question):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$L(f)(a) = \frac{1}{2\pi} \int_{-\infty}^\infty \mathrm{d}k \,exp(-ika)\, \int\,|\nabla f|\,exp(ikf)\,\mathrm{d}\mu$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On a Riemannian manifold, the gradient is given by:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\nabla f = \sum_{ij}g^{ij}\frac{\partial f}{\partial x^i}\frac{\partial}{\partial x^j}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The two examples given in the question can be worked out according to this formula (the second example very easily, the first&#xA;one involves integrals of trigonometric functions of a squared angle, but it can also be &#xA;solved analytically).&lt;/p&gt;&#xA;" OwnerUserId="1430" LastActivityDate="2010-12-28T15:53:17.547" />
  <row Id="15747" PostTypeId="2" ParentId="15710" CreationDate="2010-12-28T16:10:09.870" Score="5" ViewCount="" Body="&lt;p&gt;There is another way to test whether a metric space is homeomorphic to a complete metric space (that is, whether the space is &lt;em&gt;completely metrizable&lt;/em&gt;: there is a complete metric that induces the same topology). This is known as the &lt;em&gt;Choquet game&lt;/em&gt; or the &lt;em&gt;strong Choquet game&lt;/em&gt;. It was introduced by Choquet in his &lt;em&gt;Lectures on analysis&lt;/em&gt; in 1969; a modern treatment can be found in &lt;em&gt;Classical descriptive set theory&lt;/em&gt; by Kechris.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The idea is that each metric space is associated with a two-player game of perfect information, of countable length, such that the metric space is homeomorphic to a complete separable metric space if and only if the second player has a winning strategy in the game.  All the details and proofs can be found in Kechris' book. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is easy to show, once you have all the definitions, that the first player has a winning strategy on this game when it is played on $\mathbb{Q}$, which means the second player does not have a winning strategy, so $\mathbb{Q}$ is not completely metrizable. This gives an alternate way to prove that the rationals are not $G_\delta$ in the reals (cf. the answer by Jonas Meyer). First we prove that the rationals are not completely metrizable, by Choquet's method. Then we prove that a metric space is completely metrizable if and only if it is $G_\delta$ in its completion (actually, this is used in the proof of Choquet's result, so we had to prove it already). Since the completion of $\mathbb{Q}$ is $\mathbb{R}$, this shows that $\mathbb{Q}$ is not $G_\delta$ in $\mathbb{R}$. &lt;/p&gt;&#xA;" OwnerUserId="630" LastEditorUserId="630" LastEditorDisplayName="" LastEditDate="2010-12-28T16:26:06.617" LastActivityDate="2010-12-28T16:26:06.617" />
  <row Id="15748" PostTypeId="2" ParentId="15742" CreationDate="2010-12-28T16:48:39.820" Score="2" ViewCount="" Body="&lt;p&gt;This comment below makes explicit what the first answer already said, and hopefully removes the confusion regarding polylogs that seems to persist in the OP's mind.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To see why, consider $$\lim_{n \to \infty} \frac{(\log n)^k}{n^\epsilon}$$ for any positive $\epsilon$ and any positive integer $k$ (integrality is not really necessary, but suffices for our purposes).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have&#xA;$$\lim_{n \to \infty} \frac{(\log n)^k}{n^\epsilon} = \lim_{n \to \infty} \underbrace{\frac{\log n}{n^{\epsilon/k}}\cdots \frac{\log n}{n^{\epsilon/k}}}_{k-\mathrm{times}}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, it all boils down to proving that $$\lim_{n \to \infty} \frac{\log n}{n^\gamma} = 0,$$&#xA;where $\gamma &amp;gt; 0$, and we already know how to prove this. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, I hope this provides the clarification the OP sought.&lt;/p&gt;&#xA;" OwnerUserId="1709" LastActivityDate="2010-12-28T16:48:39.820" />
  <row Id="15749" PostTypeId="1" AcceptedAnswerId="15762" CreationDate="2010-12-28T16:49:11.707" Score="4" ViewCount="148" Body="&lt;p&gt;My question is on the convergence of the riemann sum, when the value spaces are square-integrable random variables. The convergence does depend on the evaluation point we choose, why is the case. here is some background to make this clearer&#xA;suppose $f\colon \Re \mapsto \Re $ is some continous function on $[a,b]$, the stieltjes integral of $f$ with respect to itself $f$ is $\int^{b}_{a} f(t)df(t)$&#xA;if we take a partition $ \Delta_n = \{t_0, t_1, \cdots, t_n \}$ of $[a,b]$ the riemmans sums is &#xA;$$ L_{n} = \sum^{n}_{i=1} f(t_{i-1})(f(t_{i})-f(t_{i-1})) $$&lt;br&gt;&#xA;Now if the limit exists say $\lim_{n\to\infty}L_{n}= A$ then if we choose the evaluation point $ t_{i}$ then the sum&lt;br&gt;&#xA;$$ R_{n} = \sum^{n}_{i=1} f(t_{i})(f(t_{i})-f(t_{i-1})) $$&lt;br&gt;&#xA;will also converge to $A$ so &#xA;$$\lim_{n\to\infty}L_{n} = \lim_{n\to\infty}R_{n}$$&#xA;Now we apply same idea for a stochastic integral here $W(t)$ is a wiener process and we wish to find&#xA; $$\int^{b}_{a}W(t)dW(t) $$ &#xA;$$ L_{n} = \sum^{n}_{i=1} W(t_{i-1})(W(t_{i})-W(t_{i-1})) $$&lt;br&gt;&#xA;$$ R_{n} = \sum^{n}_{i=1} W(t_{i})(W(t_{i})-W(t_{i-1})) $$&#xA;in $L^2$ norm the limits of  $L_{n}$ and $R_{n}$ exist but are different&#xA;$$\lim_{n\to\infty} \Vert R_{n}-L_{n}\Vert = b-a $$&#xA;can someone explain why the limits are different ? If the limit exists which in this case it does. I would have expected  $\lim_{n\to\infty} \Vert R_{n}-L_{n}\Vert = 0 $ in $L^2$ norm &lt;/p&gt;&#xA;" OwnerUserId="5176" LastActivityDate="2010-12-28T21:19:32.047" Title="stochastic integral &amp; stieltjes-integral" Tags="&lt;real-analysis&gt;&lt;stochastic-processes&gt;&lt;stochastic-integrals&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="1" />
  <row Id="15750" PostTypeId="1" AcceptedAnswerId="15763" CreationDate="2010-12-28T17:02:46.167" Score="14" ViewCount="406" Body="&lt;p&gt;After seeing &lt;a href=&quot;http://math.stackexchange.com/questions/15719/how-do-i-integrate-the-following-int-frac1x2-mathrm-dx1-x2-sqrt&quot;&gt;this&lt;/a&gt; recent question asking how to calculate the following integral&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \int \frac{1 + x^2}{(1 - x^2) \sqrt{1 + x^4}} \, dx $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and some of the comments that suggested that it was an elliptic integral, I tried reading a little bit on the Wikipedia article about &lt;a href=&quot;http://en.wikipedia.org/wiki/Elliptic_integral&quot; rel=&quot;nofollow&quot;&gt;elliptic integrals&lt;/a&gt;.&#xA;It seems that the point is that most elliptic integrals cannot be expressed in terms of elementary functions. The Wikipedia article defines an elliptic integral as an integral of the form&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int R \left( x, \sqrt{ P(x) } \right ) \, dx$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $R(x, y)$ is a rational function and $P(x)$ is a polynomial of degree $3$ or $4$ with no repeated roots.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the article does mention in its introductory section that two exceptions in which the elliptic integrals can be expressed in terms of elementary functions are when the polynomial $P(x)$ has repeated roots or when the rational function $R(x, y)$ does not contain odd powers of $y$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the example in question we have $P(x) = 1 + x^4$ and &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$R(x, y) = \frac{1 + x^2}{(1 - x^2)y}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so certainly it does not correspond to the two exceptions mentioned before. Thus I have a couple of questions about this:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;1) What are the conditions for an elliptic integral (as defined in the Wikipedia article) to be expressible in terms of elementary functions? More specifically, are the two above cited conditions the only exceptions or are there any others which may explain why the above integral is expressible in terms of elementary functions?&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;2) Depending on the answer to my first question, why is it that the above &quot;elliptic integral&quot; can be expressed in terms of elementary functions?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Note: I'm not sure but I suppose that some conditions must be put on the rational function $R(x, y)$ so to avoid trivial cases, but I don't want to speculate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you very much in advance.&lt;/p&gt;&#xA;" OwnerUserId="900" LastActivityDate="2010-12-30T04:47:52.143" Title="When is an elliptic integral expressible in terms of elementary functions?" Tags="&lt;calculus&gt;&lt;integral&gt;&lt;special-functions&gt;" AnswerCount="3" CommentCount="8" FavoriteCount="2" />
  <row Id="15751" PostTypeId="2" ParentId="15749" CreationDate="2010-12-28T17:06:45.380" Score="0" ViewCount="" Body="&lt;p&gt;Hint 1:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The basis of the explanation is in the different behaviour of the increments $f(t_i)-f(t_{i-1})$ and $W(t_i)-W(t_{i-1})$. The increments of the first are $O(t_i-t_{i-1})$ those of the second are $O(\sqrt{t_i-t_{i-1}})$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint 2: $$R_n - L_n =\sum_{i=1}^n \left[ f(t_i)-f(t_{i-1}) \right]^2$$ and likewise for $W(t)$.&lt;/p&gt;&#xA;" OwnerUserId="3567" LastEditorUserId="3567" LastEditorDisplayName="" LastEditDate="2010-12-28T17:21:34.067" LastActivityDate="2010-12-28T17:21:34.067" />
  <row Id="15752" PostTypeId="2" ParentId="15750" CreationDate="2010-12-28T17:11:51.023" Score="1" ViewCount="" Body="&lt;p&gt;I think Hardy's monograph &lt;em&gt;The Integration of Functions of a Single Variable&lt;/em&gt; discusses this but I'm not sure.&lt;/p&gt;&#xA;" OwnerUserId="589" LastActivityDate="2010-12-28T17:11:51.023" CommentCount="1" />
  <row Id="15753" PostTypeId="2" ParentId="15721" CreationDate="2010-12-28T17:40:23.780" Score="5" ViewCount="" Body="&lt;p&gt;Tonelli's algorithm for square-roots can easily be generalized to compute d'th roots in finite fields, e.g. see  Adleman; Manders; Miller: &lt;a href=&quot;http://www.cs.cmu.edu/~glmiller/Publications/AMM77.pdf&quot; rel=&quot;nofollow&quot;&gt;On taking roots in finite fields&lt;/a&gt;, and Bach; Shallit: &lt;a href=&quot;http://books.google.com/books?id=iJx1lP9ZcIkC&amp;amp;pg=PA160&quot; rel=&quot;nofollow&quot;&gt;Algorithmic number theory, section 7.3.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-12-28T17:40:23.780" />
  <row Id="15754" PostTypeId="2" ParentId="15749" CreationDate="2010-12-28T17:48:31.543" Score="1" ViewCount="" Body="&lt;p&gt;First write &#xA;$$&#xD;&#xA;R_n  - L_n  = \sum\limits_{i = 1}^n {[W(t_i ) - W(t_{i - 1} )]^2 },&#xD;&#xA;$$&#xA;then consider &lt;a href=&quot;http://planetmath.org/encyclopedia/QuadraticVariationOfBrownianMotion.html&quot; rel=&quot;nofollow&quot;&gt;Quadratic variation of Brownian motion&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2810" LastEditorUserId="2810" LastEditorDisplayName="" LastEditDate="2010-12-28T21:19:32.047" LastActivityDate="2010-12-28T21:19:32.047" CommentCount="1" />
  <row Id="15755" PostTypeId="2" ParentId="14506" CreationDate="2010-12-28T17:55:43.360" Score="2" ViewCount="" Body="&lt;p&gt;I think there may be a simpler algorithm, which uses straightforward dynamic programming.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I understand you correctly, you're interested in a BDD with $2n+1$ levels, the first and last containing the root and sink, respectively (we just delete all vertices from which $T$ is unreachable). Edges only go between adjacent levels, with the correct annotation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's concentrate on the middle three levels $A,B,C$. Thus all $A \rightarrow B$ and $B \rightarrow C$ are labelled by $\pm x_n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When is a node $c \in C$ reachable from a node $a \in A$? There must be some path going through $B$ with edges labeled by the same polarity of $x_n$. We can delete level $B$ and replace it by arrows going directly from $A$ to $C$; we label each arrow by the allowable assignments for $x_n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose there is some path from the root to a vertex $c \in C$. What is its provenance? It must have passed through some vertex $a \in A$ via some edge $e$ going into $a$. We can shortcut the edge $e$ so that it goes directly to $c$, carrying both its original label and the new descriptive label concerning $x_n$. In this way, we can erase level $A$ as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Having eliminated both levels $A,B$, we have removed $x_n$ from the BDD, and we can iterate our steps; having done this $n$ times, we will end up with a parametrization of &lt;i&gt;all&lt;/i&gt; solutions to the BDD.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Edit: Here is a run of the algorithm on your example. The graph is composed of five levels: $$L_0 = \{x_0\},\; L_1 = \{y_0,y_1\},\; L_2 = \{y_2\},\; L_3 = \{x_2,x_3\},\; L_4 = \{T\}.$$ There are 10 edges, which we list as tuples of from/to/label: $$(x_0,y_0,+x),(x_0,y_1,-x),(y_0,y_2,+y),(y_0,y_2,-y),(y_1,y_2,+y),$$ $$(y_2,x_2,+y),(y_2,x_3,-y),(x_2,T,-x),(x_3,T,-x),(x_3,T,+x).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First iteration: the three middle levels are $L_1,L_2,L_3$. We start by eliminating $L_2$. Edges involving $y_2$ are removed, and the following edges are added: $$(y_0,x_2,+y),(y_0,x_3,-y),(y_1,x_2,+y).$$ Next we eliminate $L_1$. Edges involving $y_0,y_1$ are removed, and the following edges are added: $$(x_0,x_2,+x+y),(x_0,x_3,+x-y),(x_0,x_2,-x+y).$$ We are left with levels $L_0,L_3,L_4$ and the following 6 edges: $$(x_0,x_2,+x+y),(x_0,x_3,+x-y),(x_0,x_2,-x+y),(x_2,T,-x),(x_3,T,-x),(x_3,T,+x).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second iteration: the three middle levels are $L_0,L_3,L_4$. We eliminate $L_3$. Edges involving $x_2,x_3$ are removed, and the following edges are added: $$(x_0,T,-x+y),(x_0,T,+x-y).$$ This gives us the answer.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Naively following the algorithm might give too many edges. However, notice that upon eliminating a variable $x_i$, we need only retain one assignment of the variables $x_{i+1},\ldots,x_n$ for each polarity of $x_i$. This way we won't get &lt;i&gt;all&lt;/i&gt; solutions, but we can decide whether there &lt;i&gt;is&lt;/i&gt; a solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, it is easy to change the algorithm so that it maintains a succinct representation of all solutions. I leave that as an exercise for the reader.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2010-12-29T22:04:43.123" LastActivityDate="2010-12-29T22:04:43.123" CommentCount="2" />
  <row Id="15756" PostTypeId="1" CreationDate="2010-12-28T18:26:03.410" Score="10" ViewCount="578" Body="&lt;p&gt;I am interested in finding out about the current status of the planned book: The Triangle Book by John H. Conway and Steve Sigur. I understand that Steve Sigur died some time back. I got no reply from Prof. Conway; is there someone who knows the fate of this project. Is it abandoned or is it still under preparation? Here is some information about the planned &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/1568811659&quot; rel=&quot;nofollow&quot;&gt;book&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="58" LastActivityDate="2011-04-19T21:05:43.870" Title="Status of The Triangle Book" Tags="&lt;geometry&gt;&lt;books&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" />
  <row Id="15757" PostTypeId="1" AcceptedAnswerId="15761" CreationDate="2010-12-28T19:04:40.723" Score="2" ViewCount="158" Body="&lt;p&gt;Let's consider the sequence space $E =\mathbb R^{\mathbb N}$. If I believe in Choice, I have an isomorphism $E \simeq \mathbb R^{(\mathfrak c)}$ for some cardinal $\mathfrak c$. I further have some inequalities about $\mathfrak c$: first, $\mathfrak c = \dim \mathbb R^{\mathbb N} \leq |\mathbb R^{\mathbb N}| \leq (2^{\aleph_0})^{\aleph_0} = 2^{\aleph_0\times \aleph_0} = 2^{\aleph_0}$. Besides, $\mathbb R^{\mathbb N}$ is the dual space of the infinite-dimensional space $\mathbb R^{(\mathbb N)}$ so we have a strict inequality between their dimensions. To sum up, I get $\aleph_0 &amp;lt; \mathfrak c \leq 2^{\aleph_0}$. So, if I believe in the continuum hypothesis, I get $\mathfrak c = 2^{\aleph_0}$ and an isomorphism $\mathbb R^{\mathbb N} = \mathbb R^{(\mathbb R)}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remark: as we have for a general set $A$ the isomorphism $(\mathbb R^{(A)})&amp;#39; \simeq \mathbb R^A$, an equivalent formulation of the question is: what's the dimension of the dual space of a $\aleph_0$-dimensional space?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm quite ignorant about logic, so this raises a number of questions (if that's too much questions for the general rules of the website, please forgive me!)&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;What about the next step, that is the (algebraic) dual space of $\mathbb R^\mathbb N$? Its dimension should be $&amp;gt;2^{\aleph_0}$ but can we be more precise?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;What if I do not suppose the continuum hypothesis? With Choice, it seems we've defined an increasing function $S$ on cardinals by the equation $\mathbb R^{\mathfrak c} (=(\mathbb R^{(\mathfrak c)})&amp;#39;) = \mathbb R^{(S(\mathfrak c))}$. Do we need (CH) to prove $S(\aleph_0) = 2^{\aleph_0}$? What is known about this function, depending on the various axioms about infinite dimensional sets? (I take it does not depend of the underlying field, but is it true?)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;What if I do not believe in Choice? I know the dimension terminology becomes quite useless, but can we still formulate some theorems about (non)existence of injective/surjective linear maps?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Thanks a lot, and my apologies for the imprecision of the questions. I hope it remains acceptable for MSE.&lt;/p&gt;&#xA;" OwnerUserId="5178" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-28T19:14:03.760" LastActivityDate="2010-12-28T20:00:28.947" Title="Dimension of the sequence space and its dual, depending on status of (AC) and (CH)." Tags="&lt;linear-algebra&gt;&lt;set-theory&gt;&lt;axiom-of-choice&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="15758" PostTypeId="2" ParentId="15721" CreationDate="2010-12-28T19:04:56.743" Score="8" ViewCount="" Body="&lt;p&gt;The integers modulo $p$ form a field. Since $x^3 - 1 = (x-1)(x^2+x+1)$, the problem is equivalent to solving $x^2+x+1\equiv 0 \pmod{p}$. For $p\neq 2$, the usual quadratic formula works; so you would need to find $y$ such that $y^2\equiv -3\pmod{p}$. If $p=2$, then $x^2+x+1=0$ has no solutions, and if $p=3$, then $x^3-1 = (x-1)^3$, so again there is no root other than $x=1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's consider the other primes, $p\gt 3$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using quadratic reciprocity, if $p\equiv 1\pmod{4}$, then $-1$ is a square modulo $p$ and we have:&#xA;$$\left(\frac{-3}{p}\right) = \left(\frac{-1}{p}\right)\left(\frac{3}{p}\right) = \left(\frac{p}{3}\right).$$&#xA;So if $p\equiv 1 \pmod{3}$ (hence $p\equiv 1 \pmod{12}$) then $-3$ is a square modulo $p$; if $p\equiv 2\pmod{3}$ (so $p\equiv 5\pmod{12}$), then $-3$ is not a square modulo $p$, so there is no other solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $p\equiv 3\pmod{4}$, then $-1$ is not a square modulo $p$, and we have:&#xA;$$\left(\frac{-3}{p}\right)=\left(\frac{-1}{p}\right)\left(\frac{3}{p}\right) = \left(\frac{p}{3}\right)$$&#xA;(since $\left(\frac{3}{p}\right) = -\left(\frac{p}{3}\right)$); again, if $p\equiv 1\pmod{3}$ (so $p\equiv 7\pmod{12}$) then $-3$ is a square modulo $p$; and if $p\equiv 2\pmod{3}$ (so $p\equiv 11\pmod{12}$) then $-3$ is not a square modulo $p$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In summary, there are roots of $x^3-1$ modulo $p$ other than $x\equiv 1\pmod{p}$ if and only if $p\equiv 1\pmod{6}$. If $p=6k+1$ is such a prime, then the two roots of $x^3-1$ other than $x=1$ are precisely the roots of $x^2+x+1$, which are&#xA;$$\frac{-1\pm\sqrt{-3}}{2} = -3k(-1\pm a),$$&#xA;where $a$ is an integer such that $a^2\equiv -3\pmod{p}$. This means you still need to figure out the square root, which can be done using any of the methods suggested by Bill Dubuque, such as Tonelli's algorithm (say, in Shanks' version as described &lt;a href=&quot;http://en.wikipedia.org/wiki/Tonelli%E2%80%93Shanks_algorithm#The_algorithm&quot; rel=&quot;nofollow&quot;&gt;in Wikipedia&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added:&lt;/strong&gt; As Alex Bartel notes in comments and in his own answer, one can avoid the use of quadratic reciprocity above. Since the units of $\mathbb{Z}/p\mathbb{Z}$ form a cyclic group of order $p-1$, there are nontrivial solutions to $x^3=1$ if and only if $3$ divides the order, that is, if and only if $p\equiv 1\pmod{3}$. From there, it is an easy jump to solutions if and only if $p\equiv 1 \pmod{6}$, since otherwise we would have $p\equiv 4\pmod{6}$ which is impossible since $p$ is prime. Once one establishes that, the quadratic formula can be applied safe in the knowledge that these primes must have $-3$ as a quadratic residue.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-29T04:30:21.000" LastActivityDate="2010-12-29T04:30:21.000" CommentCount="2" />
  <row Id="15759" PostTypeId="2" ParentId="12996" CreationDate="2010-12-28T19:35:11.343" Score="2" ViewCount="" Body="&lt;p&gt;The answer is NO.&#xA;Take $g=sl(3)$ and $\Delta=\{a_1,a_2\}$. The unique choice for $\Delta&amp;#39;$ is $\{a_2\}$.&#xA;The result is clearly false in this scenery.&lt;/p&gt;&#xA;" OwnerUserId="2764" LastActivityDate="2010-12-28T19:35:11.343" />
  <row Id="15760" PostTypeId="2" ParentId="15602" CreationDate="2010-12-28T19:39:26.457" Score="1" ViewCount="" Body="&lt;p&gt;Let $M:=\{(x,y)|x\in\mathbb R, -1&amp;lt;y&amp;lt;1\}$ be an infinite strip and choose an  $L&amp;gt;0$. The equivalence relation $(x+L,-y)\sim(x,y)$ defines a Möbius strip $\hat M$. Let $\pi: M \to \hat M$ be the projection map. The Möbius strip  $\hat M$ inherits the differentiable structure from ${\mathbb R}^2$. We have to prove that $\hat M$ does not admit an atlas of the described kind which is compatible with the differentiable structure on $\hat M$. Assume that there is such an atlas $(U_\alpha,\phi_\alpha)_{\alpha\in I}$. We then define a function $\sigma:{\mathbb R}\to\{-1,1\}$ as follows: For given $x\in{\mathbb R}$ the point $\pi(x,0)$ is in $\hat M$, so there is an $\alpha\in I$ with $\pi(x,0)\in U_\alpha$. The map $f:=\phi_\alpha$ o $\pi$ is a diffeomorphism in a neighbourhood $V$ of $(x,0)$. Put $\sigma(x):={\rm sgn}\thinspace J_f(x,0)$, where $J_f$ denotes the Jacobian of $f$. One easily checks that $\sigma(\cdot)$ is well defined and is locally constant, whence it is constant on ${\mathbb R}$. On the other hand we have $f(x+L,y)\equiv f(x,-y)$ in $V$ which implies $\sigma(L)=-\sigma(0)$ $-$ a contradiction.&lt;/p&gt;&#xA;" OwnerUserId="1303" LastActivityDate="2010-12-28T19:39:26.457" />
  <row Id="15761" PostTypeId="2" ParentId="15757" CreationDate="2010-12-28T20:00:28.947" Score="3" ViewCount="" Body="&lt;p&gt;(Assuming the Axiom of Choice throughout) Suppose $\mathbf{V}$ is $\aleph_0$-dimensional over the field $F$. Then the cardinality of $\mathbf{V}^*$ is $|F|^{\aleph_0}$. If $|F|=\aleph_{\alpha}$, then as noted in &lt;a href=&quot;http://math.stackexchange.com/questions/14429/whats-the-cardinality-of-all-sequences-with-coefficients-in-an-infinite-set/14434#14434&quot;&gt;this answer&lt;/a&gt;, the cardinality is either $2^{\aleph_0}$, $\aleph_{\alpha}$, or $\aleph_{\gamma}^{\aleph_0}$, where $\aleph_{\gamma}\leq \aleph_{\alpha}$ has cofinality $\aleph_0$ and $\aleph_0\lt\aleph_{\alpha}=|F|$. You need to assume the Generalized Continuum Hypothesis to get more about the cardinality.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given that, suppose the dimension of $\mathbf{V}^*$ is $d$. Then $\mathbf{V}^*$ is bijectable with the set of all almost-null functions $d\to F$; since $d$ is infinite, this is $d|F|=\max\{d,|F|\}=\max\{d,\aleph_{\alpha}\}$. So we must have $\max\{d,\aleph_{\alpha}\}=\aleph_{\alpha}^{\aleph_0}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, suppose $\mathbf{V}$ is an infinite dimensional vector space over $F$, and that $\kappa=\dim(\mathbf{V})\gt|F|=\lambda$. Then the dual has cardinality $\lambda^{\kappa}=2^{\kappa}\gt\kappa\gt|F|$, so the dimension will necessarily be $2^{\kappa}$. Thus your sequence of dimensions becomes $\kappa$, $2^{\kappa}$, $2^{2^{\kappa}}$, etc; that is, the $\beth$ function. I think that if $\dim(\mathbf{V})=|F|$, then you can't say precisely what $\dim(\mathbf{V}^*)$ is, other than that it is strictly larger than $|F|$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if at some point you get to a dimension which is strictly larger than the cardinality of the field, you will get a simple sequence, even without assuming the continuum hypothesis (or GCH). Unless the cardinality of the field is strongly inaccessible (or there is a strongly inaccessible cardinal between the dimension of $\mathbf{V}$ and the cardinality of $F$) you will eventually get there. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't really know what happens without the assumption of the Axiom of Choice; if you assume that the field has a defined cardinality, and the vector space has given basis with a defined cardinality, then you can figure out the cardinality of the dual, but I don't know if you can find a basis for it in that case. The statement &quot;every vector space has a basis&quot; is equivalent to the Axiom of Choice.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-12-28T20:00:28.947" />
  <row Id="15762" PostTypeId="2" ParentId="15749" CreationDate="2010-12-28T20:11:12.890" Score="1" ViewCount="" Body="&lt;p&gt;Limits of $R_n$ and $L_n$ coincide when Stieltjes integral exists. Existence of the Stieltjes integral does not follow from the existence of these limits. In general existence and definition of Stieltjes integral can be messy business as the Figure 2.1 in page 6 (page 10 of the ps file) of &lt;a href=&quot;http://www.google.com/url?sa=t&amp;amp;source=web&amp;amp;cd=4&amp;amp;ved=0CDsQFjAD&amp;amp;url=http%3A%2F%2Fwww.maphysto.dk%2Foldpages%2Fevents%2FLevyProc%2FCAC%2Fnotes%2Fchap1%2B2.ps.gz&amp;amp;rct=j&amp;amp;q=Young%20integral%20Dudley%20p-variation&amp;amp;ei=UkAaTf_HDYyw4Qa1j7mGAg&amp;amp;usg=AFQjCNF-ZJIQMvF_31i5A58zmUmsppJcyQ&quot; rel=&quot;nofollow&quot;&gt;this document can attest&lt;/a&gt;. &lt;/p&gt;&#xA;" OwnerUserId="4742" LastActivityDate="2010-12-28T20:11:12.890" CommentCount="2" />
  <row Id="15763" PostTypeId="2" ParentId="15750" CreationDate="2010-12-28T20:20:58.987" Score="13" ViewCount="" Body="&lt;p&gt;A consideration of &lt;a href=&quot;http://math.stackexchange.com/questions/15719/how-do-i-integrate-the-following-int-frac1x2-mathrm-dx1-x2-sqrt/15723#15723&quot;&gt;Moron's answer&lt;/a&gt; to the linked question&#xA;shows that there is a map from the elliptic curve &#xA;$y^2 = P(x)$ to the conic $v^2 = u^2 + 2$&#xA;given by&#xA;$$(x,y) \mapsto (x - \dfrac{1}{x}, \dfrac{y}{x}),$$&#xA;and the differential &#xA;$$\dfrac{1+x^2}{(1-x^2)\sqrt{1 + x^4}} \, dx$$ &#xA;on the elliptic curve&#xA;is the pull-back of the differential &#xA;$$\dfrac{1}{u v}\,du$$&#xA;on the conic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since a conic has genus zero (i.e. it can be parameterized by a single variable,&#xA;using a classical ``$t$-substitution''), the integral of a differential on a conic can always be expressed via elementary functions.  Thus the same is true&#xA;of the integral of the original differential on the elliptic curve.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer to the general question is the same: if the differential in question can be pulled back from a map to a rational curve (i.e. a genus zero curve),&#xA;then the &quot;elliptic integral&quot; in question can be in fact integrated via elementary functions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, any elliptic curve $y^2 = P(x)$ has a map to the the $x$-line given&#xA;by $(x,y) \mapsto x$.  So if the integral only involves rational functions of $x$ (which will be the case when $y$ appears to even powers, since we can always&#xA;substitute $P(x)$ for $y^2$) then it can be computed in elementary terms.  Also,&#xA;if $P(x)$ has repeated roots, then the curve $y^2 = P(x)$ itself is actually rational (it can be parameterized by a variation of the classical $t$-substiution for conics), and so any &quot;elliptic integral&quot; is actually elementarily integrable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S.  I have used some geomertic terminology here (&lt;em&gt;pull-back&lt;/em&gt;, &lt;em&gt;differential&lt;/em&gt;, &lt;em&gt;elliptic curve&lt;/em&gt;, &lt;em&gt;rational curve&lt;/em&gt;) because the modern point of view on this material is via algebraic geometry.  If some of this is unfamiliar, leave a comment and I (or someone else) can elaborate.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Added: If we have a curve $C$ (which could be our elliptic curve $y^2 = P(x)$,&#xA;or our rational curve $v^2 = u^2 + 2$, or just the $x$-line, or ...) and if $\omega$ is a differential on $C$, then finding the indefinite integral of $\omega$ means finding some function $f$ on $C$ such that $\omega = df$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now if $\varphi: C&amp;#39; \to C$ is a map of curves, then $\varphi^* \omega&#xD;&#xA;= \varphi^* d f  = d (f\circ \varphi).$  So $f\circ \varphi$ is an indefinite&#xA;integral of the pulled back differential $\varphi^*\omega$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, if $f$ is an elementary function of the coordinates on $C$,&#xA;and $\varphi$ is given by expressions which are elementary functions of the&#xA;coordinates, than the composite $f\circ \varphi$ will again be given by&#xA;elementary functions of the coordinates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is what is happening in your example.&#xA;Explicitly, on our curve $v^2 = u^2 + u,$ we had for example the differential&#xA;$$\dfrac{1}{u v} \, du = \frac{1}{2 u^2 v}\, d (u^2 + 2) = \frac{1}{2(v^2-2)v}\, d(v^2) = \dfrac{1}{v^2-2}\, dv = \dfrac{1}{2\sqrt{2}}d\log\bigl( (v-\sqrt{2})/(v+\sqrt{2})\bigr).$$&#xA;Now pulling back this differential via our map $\varphi:(x,y)\mapsto (x-\dfrac{1}{x}, \dfrac{y}{x})$ we obtain&#xA;$$\dfrac{1 + x^2}{(1-x^2)y}\, dx = \dfrac{1}{2\sqrt{2}}d\log\bigl((y -&#xD;&#xA;\sqrt{2}x)/(y+ \sqrt{2}x) \bigr).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As this example shows, &lt;em&gt;pulling back&lt;/em&gt; is just the theoretical terminology&#xA;for  &lt;em&gt;making a substitution&lt;/em&gt;, just as &lt;em&gt;a map of curves&lt;/em&gt; is just theoretical terminology for &lt;em&gt;a change of variables&lt;/em&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If memory serves, Miles Reid's wonderful book &lt;em&gt;Undergraduate algebraic geometry&lt;/em&gt; discusses some of this, and in particular gives some of the history of how the analytic theory of elliptic integrals turned into the &#xA;algebro-geometric theory of elliptic curves.  (If you don't know this book, don't be fooled by the title --- it is a great introduction to the subject for someone at any level, not just undergraduates!)  A much more detailed history can be found in Dieudonne's book on the history of algebraic geometry, but that book is probably not very readable unless you already have some feeling for algebraic geometry as a subject. &lt;/p&gt;&#xA;" OwnerUserId="221" LastEditorUserId="221" LastEditorDisplayName="" LastEditDate="2010-12-28T23:16:20.330" LastActivityDate="2010-12-28T23:16:20.330" CommentCount="3" />
  <row Id="15764" PostTypeId="2" ParentId="15602" CreationDate="2010-12-28T20:26:33.150" Score="7" ViewCount="" Body="&lt;p&gt;A simple answer that IMO is easy to justify using your definition of orientation goes like this. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given &lt;em&gt;any&lt;/em&gt; manifold $M$ and a point $p \in M$ there is a homomorphism $O : \pi_1(M,p) \to \mathbb Z_2$ and the idea is this: if $\phi : [0,1] \to M$ is a path such that $\phi(0)=\phi(1)=p$, given any basis for the tangent space to $M$ at $p$, $T_pM$ you can parallel transport that basis along the path, and you'll get a second basis for the tangent space at $\phi(1)=p$, $T_pM$.  And you can ask, is the change-of-basis map from your 1st to your 2nd basis for $T_pM$ orientation-preserving -- i.e. is the determinant of that linear transformation positive?  If it is, define $O(\phi)=0$, if the determinant is negative, define $O(\phi)=1$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fact: the path-component of $p$ in the manifold $M$ is orientable if and only if $O$ is the zero function, $O=0$. You prove it by cutting your path $\phi$ into small segments and comparing orientations within charts -- the key analytical step is the intermediate value theorem, using that determinant is a continuous function of matrices. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, in this discussion &quot;parallel transport&quot; assumes a Riemann metric but you don't really need a Riemann metric for this argument to work.  The parallel transport of vectors along a path $\phi$ simply means continuously-varying vectors such that the vector corresponding to $t \in [0,1]$ is always tangent to the manifold, i.e. elements of $T_{\phi(t)} M$.  And of course if you're transporting $n$ vectors you demand that these $n$ vectors always make basis for $T_{\phi(t)}M$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And in the case of the Moebius band, given any concrete model of the Moebius band you transport a basis along any path that goes once around the band and $O(\phi)=1$. &lt;/p&gt;&#xA;" OwnerUserId="642" LastEditorUserId="642" LastEditorDisplayName="" LastEditDate="2010-12-28T23:09:31.700" LastActivityDate="2010-12-28T23:09:31.700" />
  <row Id="15765" PostTypeId="2" ParentId="15702" CreationDate="2010-12-28T20:36:43.267" Score="6" ViewCount="" Body="&lt;p&gt;One way to think of an ideal as being the set of multiples of the (possibly non-existent) g.c.d of all the elements that contain it.  (I am thinking here of the case of a commutative ring with $1$, so that distinctions between left, right, and two-sided don't matter.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the integers, for example, any set of elements has a g.c.d. with all the  reasonable properties that you could want, and furthermore if $\{a_i\}_{i \in I}$ is a set of integers,&#xA;then the ideal generated by the $a_i$ is in fact principal, and its generator is a&#xA;g.c.d. of this collection.  In a more general ring, g.c.d.s don't necessarily exist, or even if they do, they don't have all the properties that they do in the integers.  So rather than trying to work with g.c.d.s, we can introduce ideals, which generally have better properties (or, rather, do many of the same jobs in more general rings that g.c.d.s and their set of multiples do in the context of the integers).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With this in mind, one sees (a) why ideals are the natural kernels of quotient maps: think about the case of the integers, working {\em modulo} $n$ means setting all multiples of $n$ equal to zero; (b) why an ideal with a unit will be the trivial ideal (i.e. the whole ring): because the g.c.d. of any set containing a unit will have to be $1$.&lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2010-12-28T20:36:43.267" />
  <row Id="15766" PostTypeId="1" AcceptedAnswerId="15768" CreationDate="2010-12-28T20:52:57.360" Score="4" ViewCount="212" Body="&lt;p&gt;From Rudin's &lt;em&gt;Principles of Mathematical Analysis&lt;/em&gt; (Chapter 2, Exercise 6)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $E&amp;#39;$ be the set of all limit points of a set $E$. Prove that $E&amp;#39;$ is closed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think I got it but my argument is a bit hand wavy:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $x$ is a limit point of $E&amp;#39;$, then every neighborhood of $x$ contains some $y\in E&amp;#39;$, and every neighborhood of $y$ contains some $z\in E$. Therefore every neighborhood of $x$ contains some $z\in E$, and so $x$ is a limit point of $E$. Then $x\in E&amp;#39;$, so $E&amp;#39;$ is closed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The thing that's bugging me is the leap from one neighborhood to another. Is this formally correct? Thanks.&lt;/p&gt;&#xA;" OwnerUserId="1523" LastActivityDate="2010-12-28T22:25:32.960" Title="Proving that the set of limit points of a set is closed." Tags="&lt;topology&gt;" AnswerCount="2" />
  <row Id="15767" PostTypeId="2" ParentId="15725" CreationDate="2010-12-28T20:56:18.700" Score="0" ViewCount="" Body="&lt;p&gt;From my understanding :&#xA;Since $$n^{log_3 2+0.4} = n^0.797$$ .(taking $\epsilon = 0.4$ in this case)&#xA;being smaller then $n \log(n)$  it qualifies for the case 3 . &#xA;Hence we check if ,&#xA;$$2\frac{n}{3} \log{n/3} \leq cn \log(n)$$&#xA;=$$\frac{2}{3} \log{n/3} \leq c  \log(n)$$&#xA;=$$ \frac{n^\frac{2c}{3}}{3} \leq \log(n)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which i guess is true for positive values of c.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Am i doing anything wrong here ? or is everything okay :P&lt;/p&gt;&#xA;" OwnerUserId="3545" LastEditorUserId="3545" LastEditorDisplayName="" LastEditDate="2010-12-28T21:07:41.663" LastActivityDate="2010-12-28T21:07:41.663" CommentCount="1" />
  <row Id="15768" PostTypeId="2" ParentId="15766" CreationDate="2010-12-28T21:01:37.447" Score="5" ViewCount="" Body="&lt;p&gt;Your argument is correct, but incomplete: All you need to finish it is to ensure that you can find a neighborhood of $y$ contained in the neighborhood of $x$ that you began with (any will do, since all contain elements of $E$). Use the triangle inequality to find an appropriate radius for the neighborhood of $y$. &lt;/p&gt;&#xA;" OwnerUserId="462" LastActivityDate="2010-12-28T21:01:37.447" CommentCount="1" />
  <row Id="15769" PostTypeId="1" AcceptedAnswerId="15770" CreationDate="2010-12-28T21:18:49.810" Score="3" ViewCount="247" Body="&lt;p&gt;How can I solve the recurrence &#xA;$$T (n) = T \left(\frac{n}{2} +\sqrt{n}\right) +\sqrt{6046}\ ?$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please don't just write the name of the method, as I just started learning this stuff and things are a little hard for me to figure out on my own. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance.&lt;/p&gt;&#xA;" OwnerUserId="3545" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-28T21:43:34.687" LastActivityDate="2010-12-29T09:03:47.497" Title="how to solve $T (n) = T \left(\frac{n}{2} +\sqrt{n}\right) +\sqrt{6046}$" Tags="&lt;asymptotics&gt;" AnswerCount="3" CommentCount="3" />
  <row Id="15770" PostTypeId="2" ParentId="15769" CreationDate="2010-12-28T21:35:43.473" Score="2" ViewCount="" Body="&lt;p&gt;As you are interested in large $n, \sqrt{n}\lt \frac{n}{a}$ for any $a$.  So you can argue (assuming $T$ is increasing with $n$) that eventually $T(n) \leq T(\frac{2n}{3})+80$.  This fits the method you have been using.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2010-12-28T21:53:15.727" LastActivityDate="2010-12-28T21:53:15.727" CommentCount="8" />
  <row Id="15771" PostTypeId="2" ParentId="15766" CreationDate="2010-12-28T22:04:51.670" Score="0" ViewCount="" Body="&lt;p&gt;An equivalent definition of a neighborhood of $x$ is that it is an open set containing $x$. If you adopt this definition, then your proof is perfect.&lt;/p&gt;&#xA;" OwnerUserId="3249" LastActivityDate="2010-12-28T22:04:51.670" CommentCount="2" />
  <row Id="15772" PostTypeId="2" ParentId="15769" CreationDate="2010-12-28T22:06:47.143" Score="2" ViewCount="" Body="&lt;p&gt;I believe &lt;a href=&quot;http://en.wikipedia.org/wiki/Akra%E2%80%93Bazzi_method&quot; rel=&quot;nofollow&quot;&gt;Akra-Bazzi&lt;/a&gt; works for this and gives you $T(x) = \theta(\log x)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The recurrence you have satisfies the assumptions of Akra-Bazzi and we get $p=0$ (check out wiki for what $p$ is). $p=0$ implies $T(x) = \theta(\log x)$ in this case (as $g(x)$ is constant).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A more elementary proof would involve showing $T(x) = \mathcal{O}(\log x)$ as in Ross's answer, then try to show $T(x) = \Omega(\log x)$.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Here is an explanation of applying Akra-Bazzi to this one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Akra-Bazzi is used to solve recurrences of the form:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ T(x) = g(x) + \sum_{i=1}^{m} a_i T(b_i x + h_i(x)) \ \text{where}\  x &amp;gt; x_0 $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $x$ is the variable, $a_i, b_i$ are constants.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In you case, we have that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$T(x) = \sqrt{6046} + T(0.5 x + \sqrt{x})$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice that, this corresponds to $\displaystyle m=1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The assumptions made by the theorem are&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) $\displaystyle a_i \gt 0$. This is true in your case, as $\displaystyle a_1  = 1$. &lt;br/&gt;&#xA;2) $\displaystyle  0 \lt b_i \lt 1$. This is true in your case, as $\displaystyle b_1 = 0.5$. &lt;br/&gt;&#xA;3) $\displaystyle |g(x)| = \mathcal{O}(x^c)$ for some $c$. In our case $\displaystyle g(x) = \sqrt{6046}$ is constant. &lt;br/&gt;&#xA;4) $h_i(x) = \mathcal{O}(\frac{x}{\log^2 x})$. In our case $h_1(x) = \sqrt{x} = \mathcal{O}(\frac{x}{\log^2 x})$ as $\displaystyle \log^2 x \le K \sqrt{x}$ for sufficiently large $\displaystyle x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are a couple more, but are trivial to verify.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now to apply Akra-Bazzi theorem, you need to find $\displaystyle p$ such that $\displaystyle \sum_{i=1}^{m} a_i (b_i)^p = 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In our case, we get $\displaystyle 0.5^p = 1$ and so $\displaystyle p = 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once we find $\displaystyle p$, by the Akra-Bazzi theorem we have that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$T(x) = \theta\left(x^p \left( 1 + \int_{1}^{x} \frac{g(t)}{t^{p+1}} \ \text{dt}\right)\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $\displaystyle p = 0$ and $\displaystyle g(t)$ is constant, we have that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$T(x) = \theta\left(1 + \int_{1}^{x} \frac{\sqrt{6046}}{t} \ \text{dt}\right) = \theta(\log x)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would also suggest you try to complete the elementary proof that $\displaystyle T(x) = \Omega(\log x)$. The relation $\displaystyle T(x) \leq T(2x/3) + 80$ only shows that $\displaystyle T(x) = \mathcal{O}(\log x)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The statement $\displaystyle T(x) = \theta(\log x)$ is stronger. For instance, $\displaystyle T(x) = 10$ also satisifies $\displaystyle T(x) \leq T(2x/3) + 80$, but it won't satisfy your original recurrence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope that helps.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-12-29T07:21:28.000" LastActivityDate="2010-12-29T07:21:28.000" />
  <row Id="15773" PostTypeId="1" CreationDate="2010-12-28T22:16:41.610" Score="17" ViewCount="362" Body="&lt;p&gt;I have seen two proofs of the simplicity of $A_n,~ n \geq 5$ (Dummit &amp;amp; Foote, Hungerford). But, neither of them are such that they 'stick' to the head (at least my head). In a sense, I still do not have the feeling that I know why they are simple and why should it be 5 and not any other number (perhaps this is only because 3-cycles become conjugate in $A_n$ after $n$ becomes greater than 4).&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What is the most illuminating proof of the simplicity of $A_n,~ n \geq 5$ that you know?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4806" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-29T01:12:54.370" LastActivityDate="2010-12-30T12:43:15.860" Title="Simplicity of $A_n$" Tags="&lt;group-theory&gt;&lt;big-list&gt;" AnswerCount="4" CommentCount="4" FavoriteCount="3" />
  <row Id="15774" PostTypeId="2" ParentId="15713" CreationDate="2010-12-28T22:24:26.420" Score="0" ViewCount="" Body="&lt;p&gt;To use Feller's original notation. Let $E_m(r,n)$ be the number of distributions leaving exactly $m$ cells empty, then we have $$p_m(r,n)=n^{-r}\binom{n}{k}A(r,n-m)$$&#xA;as the corresponding probability ($A$ is to be defined)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, we need compute the probability $x_m(r,n)$ of finding $m$ or more cells empty. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the probability that $m$ or more cells are empty is given by&#xA;$x_m = p_m + p_{m+1} + \cdots + p_{n}$. Now, observe the reason why the &lt;em&gt;hint&lt;/em&gt; was given: $x_{m+1} = x_m - p_m$ (i.e., $m+1$ or more cells being empty is the same as $\ge m$ cells being empty minus &lt;em&gt;exactly&lt;/em&gt; $m$ cells being empty).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So for a sanity test, you could plug in the value of $x_m(r,n)$ given, and verify whether $x_{m}(r,n)-x_{m+1}(r,n)=p_m(r,n)$ holds (with the appropriate boundary cases).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The deeper underlying reason for this hint is that Feller probably wants you to figure out an inclusion-exclusion type argument. If I get time, I will write out the details. But the above should be a sufficiently detailed hint?&lt;/p&gt;&#xA;" OwnerUserId="1709" LastActivityDate="2010-12-28T22:24:26.420" CommentCount="1" />
  <row Id="15776" PostTypeId="2" ParentId="15773" CreationDate="2010-12-28T22:37:53.007" Score="11" ViewCount="" Body="&lt;p&gt;The one I like most goes roughly as follows (my reference is in French [Daniel Perrin, &lt;em&gt;Cours d'algèbre&lt;/em&gt;], but maybe it's the one in Jacobson's &lt;em&gt;Basic Algebra I&lt;/em&gt;) : &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;you prove that A(5) is simple by considering the cardinal of the conjugacy classes and seeing that nothing can be a nontrivial normal subgroup (because no nontrivial union of conjugacy classes including {id} has a cardinal dividing 60). Actually, you don't have to know precisely the conjugacy classes in A(5).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Then, you consider a normal subgroup N in A(n), n &gt; 5 which is strictly larger than {id} and you prove (*) that it contains an element fixing at least n-5 points. The fact that A(5) is simple then gives that N contains every even permutation fixing the same n - 5 points. In particular, it contains a 3-cycle, and therefore contains all of A(n).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;To prove (*), you consider a commutator [x,y], where x is nontrivial in your normal subgroup and y is a 3-cycle: by the very definition, it is the product of the 3-cycle and the conjugate of its inverse. So it's the product of two 3-cycles and has at the very least n-6 fixed points. But it's easy to see that you can chose the 3-cycle so that the commutator has n-5 fixed points (it is enough that the two 3-cycles have overlapping supports).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I like this proof because it keeps the &quot;magical computation&quot; part to a minimum, that simply amounts to the fact that you have automatically knowledge about a commutator if you have knowledge about &lt;em&gt;one&lt;/em&gt; of his factors.&lt;/p&gt;&#xA;" OwnerUserId="5178" LastEditorUserId="5178" LastEditorDisplayName="" LastEditDate="2010-12-29T06:12:02.307" LastActivityDate="2010-12-29T06:12:02.307" />
  <row Id="15777" PostTypeId="1" AcceptedAnswerId="15785" CreationDate="2010-12-28T23:02:41.037" Score="6" ViewCount="130" Body="&lt;p&gt;I am aware that compactness fails on finite models, but the common counter-example uses models of arbitrary big finite size. So if we bound the size what results can we get? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume we have an infinite set of sentences $\Sigma$, of a language of first order logic $\mathcal{L}$. If there is a natural number $n$ such that every finite $S\subset\Sigma$  has a model of size $\leq n$, then does $\Sigma$ have a finite model? How about a model of size $\leq n$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the language is finite or there is a  $k$ such that every function has at most $k$ arguments and every relation is at most $k$-place, we have finite many structures of size $\leq n$ of $\mathcal{L}$, let's call them $N_1,\ldots,N_m$. If for every $i\leq m$ there exists a finite $S_i\subset\Sigma$ such that $N_i$ doesn't satisfy $S_i$, then $\bigcup_{i\leq m}S_i$ wouldn't have a model of size $\leq n$, and thus we get that one of the models satisfy every sentence of $\Sigma$. Is this argument correct? And what about the general case, where we don't restrict the language whatsoever and the structures are infinite many?&lt;/p&gt;&#xA;" OwnerUserId="1772" LastActivityDate="2010-12-29T00:21:45.663" Title="Compactness on models with bounded finite size" Tags="&lt;logic&gt;&lt;model-theory&gt;" AnswerCount="2" />
  <row Id="15778" PostTypeId="1" AcceptedAnswerId="15781" CreationDate="2010-12-28T23:36:55.273" Score="9" ViewCount="159" Body="&lt;p&gt;This is probably just elementary, but I don't know how to do it. I would like to find the minimum value of $a$ such that $x^a \geq \ln(x)$ for all $x &amp;gt; 0$. Numerically, I have found that this minimum value lies between 0.365 and 0.37 (i.e., $x^{0.37} &amp;gt; \ln(x)$ for all $x &amp;gt; 0$, but $x^{0.365}$ is not). Is there any analytical way to find out exactly this minimum value?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Based on the received answers, I finally came up with my own one as follows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the function $f(x) = x^a - \ln(x).$ This function is convex in $x$, and hence, achieves the unique minimum as $x^*$ such that $f&amp;#39;(x^*) = 0.$ Solving that equation yields $$f_{\mathrm{min}} = \min\limits_{x&amp;gt;0} f(x) = \frac{\ln(a)+1}{a}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, by letting $f_{min} = 0$, we get the desired value $a^* = 1/e.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank everyone for the answers!&lt;/p&gt;&#xA;" OwnerUserId="1380" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-29T01:35:53.360" LastActivityDate="2010-12-29T01:35:53.360" Title="What is the minimum value of $a$ such that $x^a \geq \ln(x)$ for all $x &gt; 0$?" Tags="&lt;calculus&gt;&lt;real-analysis&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="15779" PostTypeId="2" ParentId="15773" CreationDate="2010-12-28T23:43:23.773" Score="8" ViewCount="" Body="&lt;p&gt;Here are the ingredients I think about:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;$A_n$ is $n-2$-transitive on $\{1,2,\ldots,n\}$. That is, if $1\leq i_1\lt i_2\lt\cdots \lt i_{n-2}\leq n$ are any $n-2$ integers, and $j_1,\ldots,j_{n-2}$ are any $n-2$ distinct integers between $1$ and $n$, then there is an element of $A_n$ that maps $i_k$ to $j_k$ (in fact, one and only one element of $A_n$ that achieves this).  This is easy: just write out the corresponding permutation. If it is even, this gives you an element of $A_n$. If it is not even, then adding a transposition involving the elements that do not occur among the $j_k$ makes it even and does not change the image of the $i_k$. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;In particular, if $\sigma\in A_n$ fixes at least two elements, then every element with the same cycle structure to $\sigma$ is conjugate to $\sigma$ in $A_n$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If $N$ is a normal subgroup of $A_n$, and $\sigma\in N$ fixes at least two elements, then every element with the same cycle structure as $\sigma$ is in $N$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;In particular, if $N$ is normal, contains a $3$-cycle, and $n\geq 5$, then $N$ contains all $3$-cycles; and since the $3$-cycles generate $A_n$ when $n\geq 3$, then $N=A_n$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If $N$ is nontrivial and normal in $A_n$, and $n\geq 5$, then it contains a $3$-cycle: this involves a bit of manipulation, but the fact that $n\geq 5$ is as essential here as in (4) above: it gives you enough room to maneuver, room that you do not have in $A_4$ ($A_3$ and $A_2$ are also simple, but for silly reasons). &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;But I'm not sure if this particular line qualifies as &quot;illuminating&quot; in terms of giving you great insight into the structure of $A_n$; I feel they give me a good feel for what you can and cannot do in terms of playing with permutations (especially the transitivity, and the actual mechanics of point 5 above), and they remind me why $n\geq 5$ is important. &lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-12-28T23:43:23.773" />
  <row Id="15780" PostTypeId="2" ParentId="15778" CreationDate="2010-12-28T23:49:05.160" Score="5" ViewCount="" Body="&lt;p&gt;Consider the minimum of the function $f(x)=x^{1/e}-\ln(x)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: As a second step, verify that $x^a &amp;lt; \ln(x)$ at $x=(1/a)^{1/a}$, for any $0&amp;lt;a&amp;lt;1/e$.&lt;/p&gt;&#xA;" OwnerUserId="2810" LastEditorUserId="2810" LastEditorDisplayName="" LastEditDate="2010-12-29T00:22:56.967" LastActivityDate="2010-12-29T00:22:56.967" CommentCount="3" />
  <row Id="15781" PostTypeId="2" ParentId="15778" CreationDate="2010-12-28T23:49:39.310" Score="8" ViewCount="" Body="&lt;p&gt;$a = \max_{x&amp;gt;0} \frac{\ln(\ln(x))}{\ln(x)} = 1/e$.&lt;/p&gt;&#xA;" OwnerUserId="589" LastActivityDate="2010-12-28T23:49:39.310" CommentCount="1" />
  <row Id="15783" PostTypeId="1" AcceptedAnswerId="15790" CreationDate="2010-12-28T23:53:13.377" Score="2" ViewCount="198" Body="&lt;p&gt;I just asked this question on ask.metafilter, and it was suggested that I ask here.  Though I'm talking about coding something up, this question is about the math behind it, not the implementation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have done analysis in the past where we've computed an approximation for Gaussian curvature of a surface in Cartesian coordinates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What we've been doing for Cartesian (in MATLAB) is&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;[fu,fv] = gradient(Z)&#xA;&#xA;[fuu, fuv] = gradient(fu)&#xA;&#xA;[fvu,fvv] = gradient(fv)&#xA;&#xA;GC = (fuu*fvv - fuv*fuv)/(1 + fu^2 + fv^2)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;So now I have a surface that I'm modeling in cylindrical coordinates, and I can do the same thing as above for $r$ as a function of $\theta$ and $z$. The problem is that it's only taking into account the change in $r$, not the fact that there is curvature inherent in it being a cylinder.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Looking on &lt;a href=&quot;http://mathworld.wolfram.com/CylindricalCoordinates.html&quot; rel=&quot;nofollow&quot;&gt;Wolfram&lt;/a&gt; (equations 27, 32 and 37 and thereabouts), it seems like there's a centripetal component that I don't know how to apply. Dividing by the (constant?) radius doesn't seem like it would work, so I think I'm missing something.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any help would be appreciated, either explaining how to modify these equations to work correctly, or some other approximation that has worked for you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you.&lt;/p&gt;&#xA;" OwnerUserId="5185" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-29T01:41:39.213" LastActivityDate="2010-12-29T04:39:42.553" Title="How do I compute Gaussian curvature in cylindrical coordinates?" Tags="&lt;differential-geometry&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="15784" PostTypeId="2" ParentId="15777" CreationDate="2010-12-29T00:10:30.420" Score="4" ViewCount="" Body="&lt;p&gt;Your argument is correct. In fact, we can axiomatize any finite model, meaning that if ${\mathcal M}=(M,\dots)$ is a model (in a language of arbitrary size) then there is a set of sentences $\Sigma$ such that any model of $\Sigma$ is isomorphic to ${\mathcal M}$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Say $M=\{a_1,\dots,a_n\}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First you say that the universe has size $|M|=n$, with a sentence $\tau$ such as &quot;there are $x_1,\dots,x_n$ which are pairwise different and such that each $y$ is equal to one of them.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For each relational symbol $R$, consider the formula &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;there are $x_1,\dots,x_n$ pairwise different and such that $\bigwedge_{R^{\mathcal M}(a_{i_1},\dots,a_{i_k})}R(x_{i_1},\dots,x_{i_k})$ and such that $\bigwedge_{R^{\mathcal M}(a_{i_1},\dots,a_{i_k})\mbox{ fails}}\lnot R(x_{i_1},\dots,x_{i_k})$&quot;, &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;where the first big conjunction ranges over all tuples of elements of $M$ that are in the interpretation of $R$, and the second runs over all tuples that are not in that interpretation. Given any model of this formula of size $n$, there is a bijection between this model and $M$ such that the interpretation of $R$ in this model is just the image under the bijection of the interpretation of $R$ in ${\mathcal M}$. Let $\phi_R$ be the formula such that the sentence we just wrote is $\exists \vec x\phi_R$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, there is a sentence $\exists\vec x\psi_f$ describing completely each function $f^{\mathcal M}$, in the same sense as the sentence above completely describes $R^{\mathcal M}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Sigma$ consists of $\tau$, and the following formulas:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For any finitely many constant symbols $c_1,\dots,c_j$, any finitely many relational symbols $R_1,\dots,R_s$, and any finitely many function symbols $f_1,\dots,f_k$, say that $c_l^{\mathcal M}=a_{i_l}$ for each $l$, the sentence &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;$\exists \vec x(\bigwedge_{a=1}^s\phi_{R_a}\land\bigwedge_{b=1}^k\psi_{f_b}\land \bigwedge_{l=1}^j c_l=x_{i_l})$&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Note that if the language is finite, $\Sigma$ is finite as well. But the theory just described uniquely characterizes ${\mathcal M}$ even if the language is infinite.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This result is usually presented as a consequence of Beth’s deﬁnability theorem or of Svenonius’s theorem, see for example W. Hodges, Model theory. Cambridge University Press, (1993), Chapter 10.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I guess this says that there is no useful compactness result for finite models, since the only version that would hold would simply give us a model that we already knew we had.)&lt;/p&gt;&#xA;" OwnerUserId="462" LastActivityDate="2010-12-29T00:10:30.420" CommentCount="1" />
  <row Id="15785" PostTypeId="2" ParentId="15777" CreationDate="2010-12-29T00:15:34.650" Score="4" ViewCount="" Body="&lt;p&gt;For the general case, without any restriction on the language, the Compactness theorem does hold for models having a specified finite size. To see this, argue as follows: suppose that every finite subtheory of $\Sigma$ has a model of finite size at most $n$. Let $\Sigma^+$ be the theory obtained by adding the axiom &quot;there are at most $n$ elements.&quot; Notice that every finite subtheory of $\Sigma^+$ is satisfiable. Thus, by the ordinary Compactness theorem, $\Sigma^+$ is satisfiable. Hence, $\Sigma$ has a model of size at most $n$, as desired.&lt;/p&gt;&#xA;" OwnerUserId="413" LastEditorUserId="413" LastEditorDisplayName="" LastEditDate="2010-12-29T00:21:45.663" LastActivityDate="2010-12-29T00:21:45.663" />
  <row Id="15786" PostTypeId="2" ParentId="11095" CreationDate="2010-12-29T00:46:47.917" Score="0" ViewCount="" Body="&lt;p&gt;How many different ways can &#xA;1000 be written as a sum of&#xA;two positive prime numbers ?&lt;/p&gt;&#xA;" OwnerUserId="5186" LastActivityDate="2010-12-29T00:46:47.917" CommentCount="1" />
  <row Id="15787" PostTypeId="1" AcceptedAnswerId="15789" CreationDate="2010-12-29T01:17:16.550" Score="8" ViewCount="366" Body="&lt;p&gt;I was wondering when in history did people start use the $\sqrt{}$ sign for square root, what did they use before, and why this curious nomenclature is adopted.&lt;/p&gt;&#xA;" OwnerUserId="3301" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-29T01:48:07.350" LastActivityDate="2010-12-29T01:48:07.350" Title="How did the square root get its shape?" Tags="&lt;notation&gt;&lt;math-history&gt;" AnswerCount="2" FavoriteCount="3" />
  <row Id="15788" PostTypeId="2" ParentId="15787" CreationDate="2010-12-29T01:22:28.090" Score="2" ViewCount="" Body="&lt;p&gt;If you're not looking for crazy amounts of information, there's always &lt;a href=&quot;http://en.wikipedia.org/wiki/Nth_root#History&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt;!&lt;/p&gt;&#xA;" OwnerUserId="2959" LastActivityDate="2010-12-29T01:22:28.090" />
  <row Id="15789" PostTypeId="2" ParentId="15787" CreationDate="2010-12-29T01:27:03.443" Score="10" ViewCount="" Body="&lt;p&gt;You can find information about the history of the usage in Jeff Miller's page here: &lt;a href=&quot;http://jeff560.tripod.com/operation.html&quot; rel=&quot;nofollow&quot;&gt;Earliest Uses of Symbols of Operation&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Quoted:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Square root. The first use of a&#xA;  capital R with a diagonal line was in&#xA;  1220 by Leonardo of Pisa in Practica&#xA;  geometriae, where the symbol meant&#xA;  &quot;square root&quot; (Cajori vol. 1, page&#xA;  90).&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The radical symbol first appeared in&#xA;  1525 in Die Coss by Christoff Rudolff&#xA;  (1499-1545). He used the symbol &lt;img src=&quot;http://i.imgur.com/xycFi.gif&quot; alt=&quot;alt text&quot;&gt;&#xA;  (without the vinculum) for square&#xA;  roots. He did not use indices to&#xA;  indicate higher roots, but instead&#xA;  modified the appearance of the radical&#xA;  symbol for higher roots.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;It is often suggested that the origin&#xA;  of the modern radical symbol is that&#xA;  it is an altered letter r, the first&#xA;  letter in the word radix. This is the&#xA;  opinion of Leonhard Euler in his&#xA;  Institutiones calculi differentialis&#xA;  (1775). However, Florian Cajori,&#xA;  author of A History of Mathematical&#xA;  Notations, argues against this theory.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;In 1637 Rene Descartes used &lt;img src=&quot;http://i.imgur.com/JIlEB.gif&quot; alt=&quot;alt text&quot;&gt;, adding&#xA;  the vinculum to the radical symbol La&#xA;  Geometrie (Cajori vol. 1, page 375).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-12-29T01:32:04.923" LastActivityDate="2010-12-29T01:32:04.923" CommentCount="1" />
  <row Id="15790" PostTypeId="2" ParentId="15783" CreationDate="2010-12-29T01:38:51.117" Score="3" ViewCount="" Body="&lt;p&gt;Your cylindrical coordinate surface $r(\theta,z)$ in Cartesian coordinates is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align*}x&amp;amp;=r(\theta,z)\cos\;\theta\\y&amp;amp;=r(\theta,z)\sin\;\theta\\z&amp;amp;=z\end{align*}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which now allows you to apply the usual Gaussian curvature formula. In particular, you should get the expression&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$K=-\frac{r^3\frac{\partial^2 r}{\partial z^2}+r^2\left(\left(\frac{\partial^2 r}{\partial \theta\partial z}\right)^2-\frac{\partial^2 r}{\partial z^2}\frac{\partial^2 r}{\partial \theta^2}\right)+2r\frac{\partial r}{\partial \theta}\left(\frac{\partial^2 r}{\partial z^2}\frac{\partial r}{\partial \theta}-\frac{\partial r}{\partial z}\frac{\partial^2 r}{\partial \theta\partial z}\right)+\left(\frac{\partial r}{\partial \theta}\frac{\partial r}{\partial z}\right)^2}{\left(r^2+\left(r\frac{\partial r}{\partial z}\right)^2+\left(\frac{\partial r}{\partial \theta}\right)^2\right)^2}$$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;For completeness, if you have $z$ as a function of $r$ and $\theta$, your Cartesian parametrization is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align*}x&amp;amp;=r\cos\;\theta\\y&amp;amp;=r\sin\;\theta\\z&amp;amp;=z(r,\theta)\end{align*}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and the corresponding Gaussian curvature expression is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$K=\frac{r^2\frac{\partial^2 z}{\partial r^2}\left(\frac{\partial^2 z}{\partial \theta^2}+r\frac{\partial z}{\partial r}\right)-\left(\frac{\partial z}{\partial \theta}-r\frac{\partial^2 z}{\partial r\partial \theta}\right)^2}{\left(r^2\left(\left(\frac{\partial z}{\partial r}\right)^2+1\right)+\left(\frac{\partial z}{\partial \theta}\right)^2\right)^2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I will leave the derivation of the Gaussian curvature expression for&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align*}r&amp;amp;=f(u,v)\\\theta&amp;amp;=g(u,v)\\z&amp;amp;=h(u,v)\end{align*}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;to the interested reader.&lt;/p&gt;&#xA;" OwnerUserId="498" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-29T04:39:42.553" LastActivityDate="2010-12-29T04:39:42.553" CommentCount="0" />
  <row Id="15791" PostTypeId="5" CreationDate="2010-12-29T01:56:11.260" Score="0" ViewCount="0" Body="&lt;p&gt;Before asking a question on the site, please check if you can find your answer here: &lt;a href=&quot;http://jeff560.tripod.com/mathsym.html&quot; rel=&quot;nofollow&quot;&gt;Earliest Uses of Various Mathematical Symbols&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="-1" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-12-29T01:56:11.260" LastActivityDate="2010-12-29T01:56:11.260" />
  <row Id="15792" PostTypeId="4" CreationDate="2010-12-29T01:56:11.260" Score="0" ViewCount="0" Body="Meaning and history of mathematical symbols and notation." OwnerUserId="-1" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-12-29T01:56:11.260" LastActivityDate="2010-12-29T01:56:11.260" />
  <row Id="15793" PostTypeId="1" AcceptedAnswerId="15799" CreationDate="2010-12-29T04:03:51.913" Score="5" ViewCount="133" Body="&lt;p&gt;Let $\tau: E \to E$ be a measure-preserving transformation of the measure space $(E, \mathcal{E}, \mu)$, i.e. $\mu(\tau^{-1}(A)) = \mu(A)$ for all $A \in \mathcal{E}$. Let $\mathcal{E}_\tau = \{ A \in \mathcal{E} : \tau^{-1}(A) = A \}$. In my lecture notes, it is claimed that a measurable function $f$ is invariant (i.e. $f \circ \tau = f$) if and only if it is measurable with respect to $\mathcal{E}_\tau$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is evidently false: Let $E = \{ 0, 1 \}$, and let $\mathcal{E} = \mathcal{P}(E)$ be the power set. Let $\mu$ be the uniform distribution. Let $\tau$ act on $E$ by transposing 0 and 1. Let $f: (E, \mathcal{E}) \to (E, \mathcal{E}_\tau)$ be the identity map on $E$. Clearly, $f$ is measurable and &lt;em&gt;not&lt;/em&gt; invariant. Yet it's also measurable with respect to $\mathcal{E}_\tau$: the preimage of every measurable set in its codomain is in $\mathcal{E}_\tau$, by construction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Have I misinterpreted the claim? If not, is there a similar claim which &lt;em&gt;is&lt;/em&gt; true, e.g. by fixing the codomain of $f$?&lt;/p&gt;&#xA;" OwnerUserId="5191" LastActivityDate="2010-12-29T06:00:44.263" Title="Measure-preserving transformations and invariant functions" Tags="&lt;probability-theory&gt;&lt;measure-theory&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="15796" PostTypeId="2" ParentId="15571" CreationDate="2010-12-29T05:32:24.127" Score="2" ViewCount="" Body="&lt;p&gt;As an indirect way of answering this question, I'd like to bring up an example from wave mechanics that I was told earlier this semester.  Physically, when an electromagnetic wave hits a perfect conductor, say from the left, the wave disappears to the right of the conductor.  What happens mathematically is the following: the charges emit a wave that is the exact mirror of the original one, cancelling it for all time and space to the right of the conductor.  Now, a natural question in this scenario is, are the two waves to the right of the conductor real, actually extending infinitely, or is there really no wave there any more?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer to this question is: the question does not make physical sense.  Physics is concerned with measurable quantities; you can ask &quot;what is the electric field to the right&quot;, and both interpretations of what's 'really' happening will give you the same (correct) answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the same way, mathematics is built on abstractions.  The question of these abstractions being 'real' or not is not mathematical (nor logical), but philosophical; mathematics is the foundation of our technological society, which works quite well, which is the proof that mathematics is valid in the real world.&lt;/p&gt;&#xA;" OwnerUserId="1769" LastActivityDate="2010-12-29T05:32:24.127" />
  <row Id="15797" PostTypeId="1" AcceptedAnswerId="15800" CreationDate="2010-12-29T05:43:47.080" Score="2" ViewCount="124" Body="&lt;p&gt;Does anyone know how to factorize the following expression: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$4x^4+12x^{10/3} y^{2/3}+33x^{8/3} y^{4/3}+46x^2 y^2+33x^{4/3} y^{8/3}+12x^{2/3} y^{10/3}+4 y^4$$ &#xA;?&lt;/p&gt;&#xA;" OwnerUserId="5193" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-12-29T06:11:09.023" LastActivityDate="2010-12-29T06:15:15.727" Title="How to factorize $4x^4+12x^{10/3} y^{2/3}+ \dots $?" Tags="&lt;algebra-precalculus&gt;&lt;polynomials&gt;&lt;symbolic-computation&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="15798" PostTypeId="2" ParentId="15656" CreationDate="2010-12-29T05:53:49.307" Score="5" ViewCount="" Body="&lt;p&gt;If you compare the volume of the sphere to that of its enclosing hyper-cube you will find that this ratio continually diminishes.   The enclosing hyper-cube is 2 units in length per side if $R=1$.  Then we have:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$V_1/2=1\qquad$$&#xA;$$V_2/4\approx 0.785$$&#xA;$$V_3/8\approx 0.5225$$&#xA;$$V_4/16\approx 0.308$$&#xA;$$V_5/32\approx 0.164$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason for this behavior is how we build hyper-spheres from low dimension to high dimensions.  Think for example extending $S_1$ to $S_2$.  We begin with a segment extending from $-1$ to $+1$ on the $x$ axis.  We build a 2 sphere by sweeping this sphere out along the $y$ axis using the scaling factor $\sqrt{1-y^2}$.  Compare this to the process of sweeping out the respective cube where the scale factor is $1$.  So now we only occupy approximately $3/4$ of the enclosing cube (i.e. square for $n=2$).  Likewise for $n=3$, we sweep the circle along the $z$ axis using the scaling factor, loosing even more volume compared to the cylinder if we had not scaled the circle as it was swept.  So as we extend $S_{n-1}$ to get $S_n$ we start with the diminished volume we have and loose even more as we sweep out into the $n^{th}$ dimension.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It would be easier to explain with figures, however hopefully you can work through how this works for lower dimensions and extend to higher ones.&lt;/p&gt;&#xA;" OwnerUserId="4726" LastEditorUserId="4726" LastEditorDisplayName="" LastEditDate="2011-02-01T03:46:53.590" LastActivityDate="2011-02-01T03:46:53.590" CommentCount="1" />
  <row Id="15799" PostTypeId="2" ParentId="15793" CreationDate="2010-12-29T06:00:44.263" Score="2" ViewCount="" Body="&lt;p&gt;I suspect that &lt;em&gt;measurable function&lt;/em&gt; in the statement should be understood to be&#xA;&lt;em&gt;real-valued measurable function&lt;/em&gt;, or something similar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More generally, if each point $y$ in the codomain $Y$ of $f$ is measurable, and if $f$ is measuarble w.r.t. $\mathcal E_{\tau}$, then&#xA;we see that $f^{-1}(y)$ should be invariant under $\tau$, which is to say that&#xA;$f\circ\tau = f$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem with your counter-example is that the points of $E$ are &lt;em&gt;not&lt;/em&gt; in&#xA;your particular $\mathcal E_{\tau}$, and so the preceding argument breaks down&#xA;(as you implicitly noted!).&lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2010-12-29T06:00:44.263" CommentCount="1" />
  <row Id="15800" PostTypeId="2" ParentId="15797" CreationDate="2010-12-29T06:02:43.200" Score="8" ViewCount="" Body="&lt;p&gt;It is possible, but likely messy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First set $\displaystyle z = \left(\frac{x}{y}\right)^{2/3}$ and divide your expression by $\displaystyle y^4$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$4z^6 + 12 z^5 + 33 z^4 + 46 z^3 + 33 z^2 + 12 z + 4$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now divide this by $\displaystyle z^3$ and set $\displaystyle t = z + 1/z$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$4(z^3 + 1/z^3) + 12(z^2 + 1/z^2) + 33(z+1/z) + 46$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ = 4(t^3 - 3t) + 12(t^2 - 2) + 33t + 46$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ = 4t^3 + 12t^2 + 21t + 22$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which is a messy cubic, according to Wolfram Alpha. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Note it is always possible to factorize a cubic in &quot;closed form&quot;, because of Cardano's method of finding the roots).&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-12-29T06:15:15.727" LastActivityDate="2010-12-29T06:15:15.727" />
  <row Id="15801" PostTypeId="1" AcceptedAnswerId="15991" CreationDate="2010-12-29T06:07:48.687" Score="11" ViewCount="247" Body="&lt;p&gt;The (spherical) loxodrome, or the rhumb line, is the curve of constant bearing on the sphere; that is, it is the spherical curve that cuts the meridians of the sphere at a constant angle. A more picturesque way of putting it is that if one wants to travel from one point of a (spherical) globe to the antipodal point (say, from the North Pole to the South Pole) in a fixed direction, the path one would be taking on the globe would be a loxodrome.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a unit sphere, the loxodrome that cuts the meridians at an angle $\varphi\in\left(0,90^\circ\right]$ is given by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align*}x&amp;amp;=\mathrm{sech}(t\cot\;\varphi)\cos\;t\\y&amp;amp;=\mathrm{sech}(t\cot\;\varphi)\sin\;t\\z&amp;amp;=\tanh(t\cot\;\varphi)\end{align*}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While playing around with loxodromes, I noted that for certain values of $\varphi$, one can orient two identical loxodromes such that they do not intersect (that is, one can position two ships such that if both take similar loxodromic paths, they can never collide). Here for instance are two loxodromes whose constant angle $\varphi$ is 60°, oriented such that they do not cross each other:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/nB2WN.png&quot; alt=&quot;noncrossing 60° loxodromes&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, for the (extreme!) case of $\varphi=90^{\circ}$, the two loxodromes degenerate to great circles, and it is well known that two great circles must always intersect (at two antipodal points).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Less extreme, but seemingly difficult, would be the problem of positioning two 80° loxodromes such that they do not intersect:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/huDYj.png&quot; alt=&quot;two 80° loxodromes&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This brings me to my first question:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;1) For what values of $\varphi$ does it become impossible to orient two&#xA;  loxodromes such that they do not cross each other?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For simplicity, one can of course fix one of the two loxodromes to go from the North Pole to the South Pole, and try to orient the other loxodrome so that it does not cross the fixed loxodrome.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;That's the simpler version of my actual problem. Some experimentation seems to indicate that it is not possible to orient &lt;strong&gt;three&lt;/strong&gt; loxodromes such that they do not cross each other. So...&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;2) Is it true that for all (admissible) values of $\varphi$, one&#xA;  cannot position three loxodromes such that none of them cross each other?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I've tried a bit of searching around to see if the problem has been previously considered, but I have not had any luck. Any pointers to the literature will be appreciated.&lt;/p&gt;&#xA;" OwnerUserId="498" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-30T16:38:30.860" LastActivityDate="2011-01-01T23:46:47.327" Title="On nonintersecting loxodromes" Tags="&lt;geometry&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="1" />
  <row Id="15802" PostTypeId="2" ParentId="15686" CreationDate="2010-12-29T06:34:00.567" Score="0" ViewCount="" Body="&lt;p&gt;Rewrite the equation as $$x = \omega^a \sqrt[3]{10 + \sqrt{108}} - \frac{6}{3 \omega^a \sqrt[3]{10 + \sqrt{108}}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All three values of a = 0,1,2 produce a solution to the equation with no ambiguity.&lt;/p&gt;&#xA;" OwnerUserId="4290" LastEditorUserId="4290" LastEditorDisplayName="" LastEditDate="2010-12-30T00:11:40.473" LastActivityDate="2010-12-30T00:11:40.473" CommentCount="2" />
  <row Id="15803" PostTypeId="2" ParentId="15801" CreationDate="2010-12-29T07:43:22.107" Score="3" ViewCount="" Body="&lt;p&gt;Not an answer, but an approach which might help. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.math.ubc.ca/~israel/m103/mercator/mercator.html&quot; rel=&quot;nofollow&quot;&gt;Mercator projection&lt;/a&gt; is a map such that Loxodromes (corresponding to $0 \lt \varphi \lt 90^{\circ}$) get mapped to straight lines (and vice versa) and angles are preserved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An image of the map:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/2tGEe.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you would need to pick parallel lines from the mercator map. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem would be in choosing the correct antipodal points on the Mercator map and there could also be the problem of antipodal connecting lines possibly wrapping around the mercator map (which probably corresponds to your curvy $80^{\circ}$ loxodromes).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is another page which seems useful: &lt;a href=&quot;http://www.erikdeman.de/html/sail020m.htm&quot; rel=&quot;nofollow&quot;&gt;Notes on Loxodrome Calculations&lt;/a&gt; which I got from here: &lt;a href=&quot;http://www.erikdeman.de/html/sail022d.htm&quot; rel=&quot;nofollow&quot;&gt;http://www.erikdeman.de/html/sail022d.htm&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope that helps.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-12-29T18:27:06.833" LastActivityDate="2010-12-29T18:27:06.833" CommentCount="5" />
  <row Id="15807" PostTypeId="1" AcceptedAnswerId="15809" CreationDate="2010-12-29T09:03:24.013" Score="3" ViewCount="117" Body="&lt;p&gt;How to find the intersection between a set of two ranges of number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;let me explain the question with an example,&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;{2,3} {3,8} would result to 0&#xA;{2,5} {3,8} would result to 2&#xA;{3,6} {3,8} would result to 3&#xA;{4,5} {3,8} would result to 1&#xA;{4,5} {3,8} would result to 1&#xA;{8,9} {3,8} would result to 0&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Note:This could be easily done with series of if in a computer application, I am hungry for a mathematical solution!&lt;/p&gt;&#xA;" OwnerUserId="5197" LastEditorUserId="5197" LastEditorDisplayName="" LastEditDate="2010-12-29T09:39:18.257" LastActivityDate="2010-12-29T09:39:18.257" Title="What is the Shortest possible formula to find the intersection between a set of two ranges of number." Tags="&lt;elementary-set-theory&gt;" AnswerCount="1" CommentCount="6" />
  <row Id="15808" PostTypeId="2" ParentId="15769" CreationDate="2010-12-29T09:03:47.497" Score="1" ViewCount="" Body="&lt;p&gt;You can get precise asymptotics using $f(n)=n/2 \leq n/2 + \sqrt{n} \leq n/2 + \sqrt{n} + 1/2 = g(n)$.  Iterates of $f(n)$ and $g(n)$ have a closed form.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The formulas for $f^k$ and $g^k$ (k'th iterates) imply that the number of iterations of $h(x)=x/2 + \sqrt{x}$ needed to bring $n &amp;gt; 4$ down into any interval $[0,C]$ with $C&amp;gt;4$, is $\log_2{n} + O(1)$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the recurrence, $T(n)$ increases by $\sqrt{6046}$ at each iteration, so the asymptotics are $T(n) = \sqrt{6046} \log_2(n) + O(1)$.&lt;/p&gt;&#xA;" OwnerUserId="467" LastActivityDate="2010-12-29T09:03:47.497" />
  <row Id="15809" PostTypeId="2" ParentId="15807" CreationDate="2010-12-29T09:18:25.430" Score="7" ViewCount="" Body="&lt;p&gt;If I understood you correctly, you have $\{x,x+1,\ldots,x+n\}$ and $\{y,y+1,\ldots,y+k\}$ and granted $x,y$ are integers you want to find out the number of elements in the intersections (i.e. how many numbers appear in both sets).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, take: $$\max\{0,\min\{y+k,x+n\}-\max\{x,y\}\}$$ and that should be it.&lt;/p&gt;&#xA;" OwnerUserId="622" LastActivityDate="2010-12-29T09:18:25.430" CommentCount="3" />
  <row Id="15810" PostTypeId="1" CreationDate="2010-12-29T09:20:13.017" Score="6" ViewCount="206" Body="&lt;p&gt;The following problem is basically from Miranda's &quot;Algebraic Curves and Riemann Surfaces&quot;, which I am reading on my own; if there are any rules against posting textbook problems, my apologies!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $X$ be a smooth projective curve defined by the homogeneous polynomial $F(x,y,z)=0$, with $\deg F = d \geq 3$. Let $f(x,y) = F(x,y,1)$. Show that if $p(u,v)$ is a polynomial of degree at most $d-3$, then $p(u,v) \frac{du}{\partial f/ \partial v}$ defines a holomorphic 1-form on the compact Riemann surface X. If $X$ is not smooth, but has nodes, then this form is a holomorphic 1-form on the resolution. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I see that this is a holomorphic $1$-form on the affine curve defined by $f$, since the charts are just projection to the $x$ or $y$ coordinate; in the former case the form is evidently holomorphic and in the latter case the form transforms to $p(u,v) \frac{dv}{\partial f/ \partial u}$. However, I'm a bit confused as to the computations involved in checking this on the other affine curves, and what extra argument is needed for the nodes case. &lt;/p&gt;&#xA;" OwnerUserId="1491" LastActivityDate="2010-12-29T18:04:34.833" Title="Differential Form on a Riemann Surface" Tags="&lt;riemann-surface&gt;&lt;differential-forms&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="15811" PostTypeId="1" CreationDate="2010-12-29T11:11:54.013" Score="8" ViewCount="223" Body="&lt;p&gt;I've just started learning about rings. Rings are one additive abelian group strung together, through the associative law, with another structured operation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Couldn't we continue stringing together operations in this manner (a multi-operation associative law)? Would what I'm thinking be encompassed by the ring definition through something I'm missing? If not, is this done and, if so, do useful objects come out of it? &lt;/p&gt;&#xA;" OwnerUserId="5198" LastActivityDate="2010-12-30T00:31:33.770" Title="Generalization of a ring?" Tags="&lt;algebra&gt;&lt;ring-theory&gt;&lt;definition&gt;" AnswerCount="3" FavoriteCount="2" />
  <row Id="15812" PostTypeId="2" ParentId="15811" CreationDate="2010-12-29T11:19:41.587" Score="12" ViewCount="" Body="&lt;p&gt;Absolutely!  Such things are studied in a few related disciplines encompassing &lt;a href=&quot;http://en.wikipedia.org/wiki/Universal_algebra&quot; rel=&quot;nofollow&quot;&gt;universal algebra&lt;/a&gt; and parts of &lt;a href=&quot;http://en.wikipedia.org/wiki/Category_theory&quot; rel=&quot;nofollow&quot;&gt;category theory&lt;/a&gt; (including what is called &quot;higher-dimensional algebra&quot;).  On the categorical side some keywords here include (algebras over an) &lt;a href=&quot;http://en.wikipedia.org/wiki/Operad_theory&quot; rel=&quot;nofollow&quot;&gt;operad&lt;/a&gt;, (algebras over a) &lt;a href=&quot;http://en.wikipedia.org/wiki/Monad_(category_theory)&quot; rel=&quot;nofollow&quot;&gt;monad&lt;/a&gt;, and (models of a) &lt;a href=&quot;http://en.wikipedia.org/wiki/Lawvere_theory&quot; rel=&quot;nofollow&quot;&gt;Lawvere theory&lt;/a&gt;.  More concretely there are also &lt;a href=&quot;http://en.wikipedia.org/wiki/Coalgebra&quot; rel=&quot;nofollow&quot;&gt;coalgebras&lt;/a&gt;, &lt;a href=&quot;http://en.wikipedia.org/wiki/Bialgebra&quot; rel=&quot;nofollow&quot;&gt;bialgebras&lt;/a&gt;, &lt;a href=&quot;http://en.wikipedia.org/wiki/Hopf_algebra&quot; rel=&quot;nofollow&quot;&gt;Hopf algebras&lt;/a&gt;, &lt;a href=&quot;http://en.wikipedia.org/wiki/Frobenius_algebra&quot; rel=&quot;nofollow&quot;&gt;Frobenius algebras&lt;/a&gt;, etc.  Ross Street's &lt;a href=&quot;http://books.google.com/books?id=u0slIpdBjYUC&amp;amp;printsec=frontcover&amp;amp;dq=ross+street+quantum+groups&quot; rel=&quot;nofollow&quot;&gt;Quantum Groups: a path to current algebra&lt;/a&gt; was written as an introduction to these more exotic types of algebra that people don't generally bother to tell you about as an undergrad.  Far from being esoteric generalizations of ordinary algebra, these structures turn up all over mathematics in the most unexpected places.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a matter of introduction, however, I don't think that what you've just described is a good way to get a sense of what a ring actually is.  The ring axioms capture the abstract properties that &lt;em&gt;symmetries of abelian groups&lt;/em&gt; satisfy, in the same way that the group axioms capture the abstract properties that &lt;em&gt;symmetries of sets&lt;/em&gt; satisfy.  The point here being that if $A$ is an abelian group and $f : A \to A$ some endomorphism of it, then there are two ways that one can combine such endomorphisms: one can compose them or add them.  This leads to the multiplication and addition operation, respectively, on the ring $\text{End}(A)$ of endomorphisms of $A$.  For example if $A$ is the &lt;em&gt;abelian group&lt;/em&gt; $\mathbb{Z}$, then $\text{End}(A)$ is the &lt;em&gt;commutative ring&lt;/em&gt; $\mathbb{Z}$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then there is a &quot;Cayley's theorem&quot; for rings which says that any abstract ring $R$ can be realized as endomorphisms of some abelian group: in fact $R$ is precisely the ring of endomorphisms of $R$ which respect &lt;em&gt;right&lt;/em&gt; multiplication, that is, which respect the right $R$-&lt;a href=&quot;http://en.wikipedia.org/wiki/Module_(mathematics)&quot; rel=&quot;nofollow&quot;&gt;module&lt;/a&gt; structure on $R$.  &lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-12-29T11:26:23.297" LastActivityDate="2010-12-29T11:26:23.297" />
  <row Id="15813" PostTypeId="2" ParentId="15811" CreationDate="2010-12-29T11:32:29.573" Score="3" ViewCount="" Body="&lt;p&gt;The main point in a ring is not really its additive group or its multiplicative monoid: it's the distributive law, which ties the two operations. When you go on to more complicated structures, it's the laws that tie the operations that matter. See for instance the axioms of vector spaces and algebras. &lt;/p&gt;&#xA;" OwnerUserId="589" LastActivityDate="2010-12-29T11:32:29.573" />
  <row Id="15814" PostTypeId="1" AcceptedAnswerId="15817" CreationDate="2010-12-29T11:37:54.543" Score="0" ViewCount="166" Body="&lt;p&gt;Hey,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am working on a model which predicts the total final score (sum of the scores of both teams) of certain sports matches. I have checked the predictions of the model against the actual scores of matches for the last several years. The error in the predicted total score to the actual toal score is approximately normally distributed with mean = &lt;code&gt;-0.3&lt;/code&gt; and standard deviation = &lt;code&gt;13&lt;/code&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Say I have a predicted total score &lt;code&gt;x&lt;/code&gt;, how can I determine the probability that the actual score will above or below a given score line &lt;code&gt;y&lt;/code&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, say my model predicts that the total score will be &lt;code&gt;8&lt;/code&gt; and the score line is &lt;code&gt;5.5&lt;/code&gt;, what is the probability that the actual score will be &amp;lt;=5 and  what is the probability that the actual score will be &gt;5 (given that the errors in my prediction can be assumed to be normally distributed)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cheers,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pete&lt;/p&gt;&#xA;" OwnerUserId="5199" LastActivityDate="2010-12-29T14:35:37.177" Title="Probability - Calculate chance total score will be over or under a score line" Tags="&lt;probability&gt;&lt;statistics&gt;&lt;normal-distribution&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="15815" PostTypeId="1" AcceptedAnswerId="15822" CreationDate="2010-12-29T12:26:29.443" Score="3" ViewCount="216" Body="&lt;p&gt;I've asked &lt;a href=&quot;http://stackoverflow.com/questions/4526143/construct-polygons-out-of-union-of-many-polygons&quot;&gt;this question at SO&lt;/a&gt;, but only answer I got is a non-answer as far as I can tell, so I would like to try my luck here. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically, I'm looking for a better-than-naive algorithm for the constructions of polygons out of union of many polygons, each with a list of Vertex $V$. The naive algorithm to find the union polygon(s) goes like this:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;First take two polygons, union them,&#xA;  and take another polygon, union it&#xA;  with the union of the two polygons,&#xA;  and repeat this process until every&#xA;  single piece is considered. Then I will run&#xA;  through the union polygon list and&#xA;  check whether there are still some&#xA;  polygons can be combined, and I will&#xA;  repeat this step until a&#xA;  satisfactory result is achieved.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Is there a smarter algorithm?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For this purpose, you can imagine each polygon as a jigsaw puzzle piece, when you complete them you will get a nice picture. But the catch is that a small portion ( say &amp;lt;5%) of the jigsaw is missing, and you are still require to form a picture as complete as possible; that's the polygon ( or polygons)-- maybe with holes-- that I want to form.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note:  I'm &lt;em&gt;not&lt;/em&gt; asking about how to union two polygons, but I am asking about--given that I know how to union two polygons--how to union $n$ number of (where $n&gt;&gt;2$) polygons in an efficient manner. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also,all the polygons can share edges, and some polygon's edge can be shared by one or many other polygon's edge. Polygons can't overlapped among themselves. &lt;/p&gt;&#xA;" OwnerUserId="550" LastEditorUserId="550" LastEditorDisplayName="" LastEditDate="2010-12-29T14:04:39.620" LastActivityDate="2010-12-29T15:47:29.163" Title="How to Union Many Polygons Efficiently" Tags="&lt;computational-geometry&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="1" />
  <row Id="15816" PostTypeId="2" ParentId="15164" CreationDate="2010-12-29T13:09:11.507" Score="1" ViewCount="" Body="&lt;p&gt;The way to get the smaller angle spanned by $\mathbf e_1=(x_1,y_1)$ and $\mathbf e_2=(x_2,y_2)$ is through the expression&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\min(|\arctan(x_1,y_1)-\arctan(x_2,y_2)|,2\pi-|\arctan(x_1,y_1)-\arctan(x_2,y_2)|)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\arctan(x,y)$ is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Atan2&quot; rel=&quot;nofollow&quot;&gt;two-argument arctangent&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-12-29T13:09:11.507" />
  <row Id="15817" PostTypeId="2" ParentId="15814" CreationDate="2010-12-29T13:49:18.407" Score="1" ViewCount="" Body="&lt;p&gt;OK, sorry for misunderstanding your point. So, your model predicts $PV=8$ with errors $\epsilon \sim N(-0.3,13)$. Or, this is equivalent to saying that you expect the actual value to be $AV \sim N(8.3,13)$. ($PV=AV+\epsilon$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, to compute the probability that $X\leq 5.5$ (or $x&gt;5.5$) under the hypothesis that your model is correct. This is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathbb{P}(X\leq 5.5|X \sim N(8.3,13)) \; .$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know what programming language or other computer program you use, but most math or statistics interpretative languages (Matlab, R, etc...) have the necessary commands inbuilt. So in R, you would use the command&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;pnorm(5.5,8.3,13)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;which computes the cumulative normal distribution in point $5.5$ for mean and variance as given. For the probability bigger than $5.5$, you take &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;1-pnorm(5.5,8.3,13)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;If you want the explicit formula for the computation, I can provide that too, but unless you want to make a lot of extra work implenting the functions yourself, I don't think that is really useful.&lt;/p&gt;&#xA;" OwnerUserId="3567" LastEditorUserId="3567" LastEditorDisplayName="" LastEditDate="2010-12-29T14:35:37.177" LastActivityDate="2010-12-29T14:35:37.177" CommentCount="10" />
  <row Id="15818" PostTypeId="2" ParentId="15815" CreationDate="2010-12-29T14:02:10.790" Score="0" ViewCount="" Body="&lt;p&gt;If the polygons either share exactly one edge or are disjoint then create a list of edges and the polygons they belong to and then remove each edge that has two polygons, joining those two polygons.&lt;/p&gt;&#xA;" OwnerUserId="589" LastActivityDate="2010-12-29T14:02:10.790" CommentCount="4" />
  <row Id="15819" PostTypeId="1" AcceptedAnswerId="15821" CreationDate="2010-12-29T14:29:26.103" Score="5" ViewCount="203" Body="&lt;p&gt;From Wikipedia's &lt;a href=&quot;http://en.wikipedia.org/wiki/Weierstrass_factorization_theorem&quot; rel=&quot;nofollow&quot;&gt;Weierstrass Factorization Theorem&lt;/a&gt;, I learned that every entire function can be represented as a product involving its zeroes. Examples are the sine and cosine function. The Riemann zeta function, however, is not entire. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let us assume the Riemann Hypothesis. Can $\zeta(s)$ be represented by an infinite product involving both its trivial zeroes at $\zeta(s)=-2n$ (for $n \in \mathbb{N}$) and its non-trivial zeroes at $\zeta(s)=\frac{1}{2} + i t$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks, Max&lt;/p&gt;&#xA;" OwnerUserId="1055" LastEditorUserId="1055" LastEditorDisplayName="" LastEditDate="2010-12-29T14:46:08.647" LastActivityDate="2010-12-29T17:19:16.650" Title="Infinite product representation of a function in terms of its non-trivial zeroes?" Tags="&lt;functions&gt;&lt;analytic-number-theory&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="15820" PostTypeId="2" ParentId="15819" CreationDate="2010-12-29T14:42:02.723" Score="3" ViewCount="" Body="&lt;p&gt;(Would be a comment, but I don't have the reputation.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Riemann zeta function is certainly not entire: it has a simple pole at s = 1. It is, however, meromorphic.&lt;/p&gt;&#xA;" OwnerUserId="5191" LastActivityDate="2010-12-29T14:42:02.723" CommentCount="2" />
  <row Id="15821" PostTypeId="2" ParentId="15819" CreationDate="2010-12-29T15:19:36.520" Score="10" ViewCount="" Body="&lt;p&gt;The standard approach to this problem is to write&#xA;$\xi(s) = \dfrac{s(s-1)}{2}\pi^{-s/2}\Gamma(s/2)\zeta(s).$  This function is entire, and has zeroes precisely at the non-trivial zeroes of $\zeta(s)$.  It also has a slow enough rate of growth that it can be written as a product over its zeroes:&#xA;$$\xi(s) = \xi(0) \prod_{\rho}(1-\dfrac{s}{\rho}),$$&#xA;where the product is over zeroes of $\xi(s)$, i.e. over non-trivial zeroes of $\zeta(s)$.  (Here I am following more-or-less the notation in Edward's book &lt;em&gt;Riemann's zeta function&lt;/em&gt;, which is a good reference for these sort of things.)&#xA;[Edit: Also, to ensure convergence, the product should be taken over &quot;matching pairs&quot; of zeroes, i.e. the factors for a pair of zeroes of the form $\rho$ and $1-\rho$ should be combined.]&#xA;We can then write&#xA;$$\zeta(s) = \dfrac{2\xi(0)}{s(s-1)}\pi^{s/2}\dfrac{1}{\Gamma(s/2)}\prod_{\rho}(1-\dfrac{s}{\rho}).$$&#xA;If we now replace $\dfrac{1}{\Gamma(s/2)}$ by its Weierstrass product, we get&#xA;a product formula for $\zeta(s)$, namely&#xA;$$\zeta(s) = \dfrac{\xi(0)}{s-1} (\pi e^{\gamma})^{s/2}\prod_{n=1}^{\infty}(1 +\dfrac{s}{2n})e^{-s/2n}&#xD;&#xA;\prod_{\rho}(1-\dfrac{s}{\rho}).$$&#xA;(Here $\gamma$ is Euler's constant.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that we can now compute $\xi(0)$, because we know that the value of $\zeta(s)$ at $s = 0$ is equal to $-1/2$.  We find that $\xi(0) = 1/2$,&#xA;and so &#xA;$$\zeta(s) = \dfrac{1}{2(s-1)}(\pi e^{\gamma})^{s/2}\prod_{n=1}^{\infty}(1+\dfrac{s}{2n})e^{-s/2n}\prod_{\rho}(1-\dfrac{s}{\rho}).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An added cultural remark: Riemann's &lt;a href=&quot;http://en.wikipedia.org/wiki/Explicit_formula&quot; rel=&quot;nofollow&quot;&gt;explicit formula&lt;/a&gt; for the prime counting function is obtained by taking a Fourier transform of the logarithm of this formula.  Combined with the fact that all the non-trivial zeroes $\rho$ have real part $&amp;lt; 1$ (proved by Hadamard and de la Vallee Poussin), this gives the prime number theorem.&lt;/p&gt;&#xA;" OwnerUserId="221" LastEditorUserId="221" LastEditorDisplayName="" LastEditDate="2010-12-29T17:19:16.650" LastActivityDate="2010-12-29T17:19:16.650" CommentCount="7" />
  <row Id="15822" PostTypeId="2" ParentId="15815" CreationDate="2010-12-29T15:47:29.163" Score="3" ViewCount="" Body="&lt;p&gt;Likely the best method is to perform a simultaneous &lt;a href=&quot;http://en.wikipedia.org/wiki/Plane_sweep&quot; rel=&quot;nofollow&quot;&gt;plane sweep&lt;/a&gt; of all the polygons.&#xA;This is discussed in &lt;em&gt;The Algorithms Design Manual&lt;/em&gt; under &quot;&lt;a href=&quot;http://books.google.com/books?id=7XUSn0IKQEgC&amp;amp;pg=PA618&amp;amp;lpg=PA618&amp;amp;dq=plane+sweep+to+union+polygons&amp;amp;source=bl&amp;amp;ots=z6eFQRNNZd&amp;amp;sig=OYm-w6WA0jJQu3jtosuELrsafxA&amp;amp;hl=en&amp;amp;ei=7FUbTd-4CoWKlwed6MXhCw&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=1&amp;amp;ved=0CBMQ6AEwAA#v=onepage&amp;amp;q=plane%20sweep%20to%20union%20polygons&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;Intersection Detection&lt;/a&gt;.&quot;&#xA;(Algorithms to find the intersection can be altered to instead find the union.)&#xA;It is also discussed in my textbook &lt;em&gt;&lt;a href=&quot;http://cs.smith.edu/~orourke/books/compgeom.html&quot; rel=&quot;nofollow&quot;&gt;Computational Geometry in C&lt;/a&gt;&lt;/em&gt;, Chapter 7, &quot;Intersection of NonConvex Polygons.&quot;  The time complexity is $O(n \log n + k)$ for $n$ total vertices and $k$ intersection points between edges of different polygons.&lt;/p&gt;&#xA;" OwnerUserId="237" LastActivityDate="2010-12-29T15:47:29.163" />
  <row Id="15823" PostTypeId="1" AcceptedAnswerId="15826" CreationDate="2010-12-29T15:53:30.183" Score="1" ViewCount="158" Body="&lt;p&gt;Can $$\sin r\pi $$ be rational if $r$ is irrational? Either a direct or existence proof is fine.&lt;/p&gt;&#xA;" OwnerUserId="3249" LastActivityDate="2010-12-29T16:26:15.527" Title="Sine values being rational" Tags="&lt;number-theory&gt;&lt;algebra&gt;&lt;number&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="15824" PostTypeId="1" CreationDate="2010-12-29T16:14:58.213" Score="2" ViewCount="190" Body="&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Broadly I would like to know what is the connection between homogeneous binary quartic forms and elliptic curves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I see that the invariants on the first space under the action of $SL(2,\mathbb{C})$ are precisely the functions $g_2$ and $g_3$. Also given such a quartic form on $\mathbb{C}^2$, if one thinks its 0-set as the intersection of two conics in $\mathbb{C}^3$ then the determinant of the complex symmetric matrix representing the most general conic passing through this intersection is precisely the &quot;RHS&quot; of the elliptic curve equation. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;What does this determinant mean geometrically? &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;More specifically consider the $SL(2,\mathbb{C})$ action on $\mathbb{C}^2$ (spanned by $x$ and $y$) and define $U = x^2$, $V = 2xy$ and $W=y^2$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the action is by the matrix $ \left [ \begin{array}{c  c } &#xD;&#xA;                                     a &amp;amp; b \\&#xD;&#xA;                                     c &amp;amp; d \\&#xD;&#xA;                                                                 &#xD;&#xA;                           &#xD;&#xA;   \end{array}\right ]$ ($ad-bc=1$) of $SL(2,\mathbb{C})$ then the following transformation is affected on $\mathbb{C}^2$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \left [  \begin{array}{c} &#xD;&#xA;                            x\\&#xD;&#xA;                            y\\&#xD;&#xA;                                                                 &#xD;&#xA;                          &#xD;&#xA;   \end{array} \right ]  \rightarrow \left [  \begin{array}{c} &#xD;&#xA;                            ax + by\\&#xD;&#xA;                            cx + dy\\&#xD;&#xA;                                                                 &#xD;&#xA;                           &#xD;&#xA;   \end{array} \right ] $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This in turn implies the following transformation on $\mathbb{C}^3$ (defining the matrices, $X$ and $A$),&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$X = \left [  \begin{array}{c} &#xD;&#xA;                            U\\&#xD;&#xA;                            V\\&#xD;&#xA;                            W&#xD;&#xA;                                                                 &#xD;&#xA;                          &#xD;&#xA;   \end{array} \right ]  \rightarrow  A.X  = \left [  \begin{array}{c c c} &#xD;&#xA;                            a^2 &amp;amp; ab &amp;amp; b^2\\&#xD;&#xA;                            2ac &amp;amp; (ad+bc) &amp;amp; 2bd \\&#xD;&#xA;                            c^2 &amp;amp; cd &amp;amp; d^2\\&#xD;&#xA;                                                                 &#xD;&#xA;                           &#xD;&#xA;   \end{array} \right ] \left [  \begin{array}{c} &#xD;&#xA;                            U\\&#xD;&#xA;                            V\\&#xD;&#xA;                            W&#xD;&#xA;                                                                 &#xD;&#xA;                          &#xD;&#xA;   \end{array} \right ] $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$4UW-V^2$ being identically equal to $0$ is obviously kept invariant by the above transformation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The invariance of $4UW-V^2$ can be thought of as $SL(2,\mathbb{C})$ preserving an inner-product on $\mathbb{C}^3$ with signature, $\left [ \begin{array}{c c c} &#xD;&#xA;                                     &amp;amp;   &amp;amp; 2 \\&#xD;&#xA;                                     &amp;amp; -1 &amp;amp;  \\&#xD;&#xA;                                     2 &amp;amp; &amp;amp;  \\&#xD;&#xA;                              &#xD;&#xA;                           &#xD;&#xA;   \end{array} \right ]$&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;But $A$ does not seem to be an orthogonal matrix though it satisfies the property of $det(A)=1$. Then why is it being said in the books that the $SL(2,\mathbb{C})$ action on $\mathbb{C}^2$ (spanned by $x$ and $y$) induces an action of $SO(3)$ on $\mathbb{C}^3$ (spanned by $U$, $V$ and $W$) with respect to the inner product $ \left [ \begin{array}{c  c c}  &#xD;&#xA;                                     &amp;amp;   &amp;amp; 2 \\&#xD;&#xA;                                     &amp;amp; -1 &amp;amp;  \\&#xD;&#xA;                                     2 &amp;amp; &amp;amp;  \\&#xD;&#xA;                              &#xD;&#xA;                           &#xD;&#xA;   \end{array}\right ]$ ? &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="965" LastActivityDate="2011-01-03T15:42:08.900" Title="Homogeneous binary quartic forms to elliptic curves" Tags="&lt;algebraic-geometry&gt;&lt;lie-groups&gt;" AnswerCount="1" />
  <row Id="15825" PostTypeId="1" CreationDate="2010-12-29T16:16:12.030" Score="2" ViewCount="93" Body="&lt;p&gt;Is there a good online community for discussing optimization models?  The ones I have found don't seem to have a critical mass for active discussions.&lt;/p&gt;&#xA;" OwnerUserId="5205" LastActivityDate="2010-12-29T16:16:12.030" Title="Mathematical Programming Community" Tags="&lt;linear-programming&gt;" CommentCount="2" />
  <row Id="15826" PostTypeId="2" ParentId="15823" CreationDate="2010-12-29T16:21:53.597" Score="7" ViewCount="" Body="&lt;p&gt;As J. M. said, Niven's theorem does it.  There is some $r$ such that $\sin (r\pi) = \frac{1}{3}$  As $\sin (r\pi)$ is rational and not $0, \pm1, \pm \frac{1}{2}$, $r$ is not rational&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-12-29T16:21:53.597" />
  <row Id="15828" PostTypeId="2" ParentId="15811" CreationDate="2010-12-29T16:34:08.090" Score="6" ViewCount="" Body="&lt;p&gt;Yes, any algebraic structure that has multiple operations will have laws like the distributive law that intertwine the operations. For otherwise the operations would not interact in any way and the structure could be studied as two independent structures with the non-interacting operations. For example, if we dropped the distributive law from the ring axioms then we'd simply have a set with a given abelian group structure and given monoid structure with no connection between the two structures. It is the distributive law that ties together these two structures and leads to the rich structure that is unique to rings - structure above and beyond the constituent structure of the additive group and multiplicative monoid. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One can observe the key role played by the distributive law even in the simplest results on rings. For example, consider the proof of the law of signs $\rm\ (-A)\:(-B) = A\:B\:.\ $ One simple proof is to observe that both terms are additive inverses of $\rm\ (-A)\:B\ $ hence they are equal by uniqueness of inverses. But to verify that they are inverse requires applying the distributive law. Similarly, any theorem that is truly ring-theoretic result $\:$ (i.e. $\:$ is not merely a result about abelian groups or monoids) must employ the distributive law in its proof (though perhaps obscured in some remote lemma).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Analogous remarks hold true for any algebraic structure with multiple operations, e.g. lattices with their intertwining absorption law $\rm\ X = X \vee (X \wedge Y)\ $ and it's dual, or distributive lattices (e.g. Boolean algebras) with their distributive law $\rm\ X \vee (Y \wedge Z) = (X\vee Y)\wedge (X\vee Z)\ $ and its dual or, more generally, the important modular law $\rm\ (X\wedge Y)\vee(Y\wedge Z) = Y \wedge ((X\wedge Y)\vee Z)\:.\ $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The common properties of algebraic structures are studied in universal algebra (or general algebra). For example, one major theme is the study of the role played by properties of the lattices of congruences, e.g. congruence lattices of lattices are distributive, and congruence lattices of groups and rings are modular. These properties play  fundamental roles in the theories of these structures.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-12-30T00:31:33.770" LastActivityDate="2010-12-30T00:31:33.770" />
  <row Id="15829" PostTypeId="1" AcceptedAnswerId="15837" CreationDate="2010-12-29T17:27:18.553" Score="3" ViewCount="138" Body="&lt;p&gt;Let $p$ be a prime and $H$ a subgroup of a finite group $G$. Let $P$  be a p-sylow subgroup of G. Prove that there exists $g\in G$ such that $H\cap gPg^{-1}$ is  sylow subgroup of $H$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have no idea how to do this, any hints? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Note:&lt;/em&gt; Originally it was unclear if the problem was for possibly infinite groups or just finite ones. However, since the definition of $p$-Sylow subgroup being used is that it is a $p$-subgroup such that the index and the order are relatively prime, the definition only applies to finite groups.&lt;/p&gt;&#xA;" OwnerUserId="5209" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-29T19:54:41.833" LastActivityDate="2010-12-30T16:59:51.297" Title="sylow subgroup of a subgroup" Tags="&lt;group-theory&gt;" AnswerCount="4" CommentCount="3" FavoriteCount="0" />
  <row Id="15830" PostTypeId="2" ParentId="15829" CreationDate="2010-12-29T17:32:39.393" Score="1" ViewCount="" Body="&lt;p&gt;Well, we have $\vert P \vert =p^n$ where $\vert G \vert = p^n m$ and $(p,m)=1$.  What can be said about $\vert H \cap P \vert$ (or, for that matter, $\vert H \cap Q \vert$ for any Sylow p-subgroup $Q$ of $G$)?&lt;/p&gt;&#xA;" OwnerUserId="5137" LastEditorUserId="5137" LastEditorDisplayName="" LastEditDate="2010-12-29T17:38:37.937" LastActivityDate="2010-12-29T17:38:37.937" CommentCount="4" />
  <row Id="15831" PostTypeId="1" CreationDate="2010-12-29T17:36:27.577" Score="17" ViewCount="669" Body="&lt;p&gt;Some time ago when decomponsing the natural numbers, $\mathbb{N}$, in prime powes I noticed a pattern in their powers.  Taking, for example, the numbers $\lbrace 1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16 \rbrace$ and factorize them, we will get &#xA;$$\begin{align} 1&amp;amp;=2^0\times 3^0\times 5^0\times 7^0\times 11^0\times 13^0\times\ldots \\ 2&amp;amp;=2^1\times 3^0\times 5^0\times 7^0\times 11^0\times 13^0\times\ldots \\3&amp;amp;=2^0\times 3^1\times 5^0\times 7^0\times 11^0\times 13^0\times\ldots \\ 4&amp;amp;=2^2\times 3^0\times 5^0\times 7^0\times 11^0\times 13^0\times\ldots \\ 5&amp;amp;=2^0\times 3^0\times 5^1\times 7^0\times 11^0\times 13^0\times\ldots \\ 6&amp;amp;=2^1\times 3^1\times 5^0\times 7^0\times 11^0\times 13^0\times\ldots \\ 7&amp;amp;=2^0\times 3^0\times 5^0\times 7^1\times 11^0\times 13^0\times\ldots \\ 8&amp;amp;=2^3\times 3^0\times 5^0\times 7^0\times 11^0\times 13^0\times\ldots \\ 9&amp;amp;=2^0\times 3^2\times 5^0\times 7^0\times 11^0\times 13^0\times\ldots \\ 10&amp;amp;=2^1\times 3^0\times 5^1\times 7^0\times 11^0\times 13^0\times\ldots \\ 11&amp;amp;=2^0\times 3^0\times 5^0\times 7^0\times 11^1\times 13^0\times\ldots \\ 12&amp;amp;=2^2\times 3^1\times 5^0\times 7^0\times 11^0\times 13^0\times\ldots \\ 13&amp;amp;=2^0\times 3^0\times 5^0\times 7^0\times 11^0\times 13^1\times\ldots \\ 14&amp;amp;=2^1\times 3^0\times 5^0\times 7^1\times 11^0\times 13^0\times\ldots \\ 15&amp;amp;=2^0\times 3^1\times 5^1\times 7^0\times 11^0\times 13^0\times\ldots \\ 16&amp;amp;=2^4\times 3^0\times 5^0\times 7^0\times 11^0\times 13^0\times\ldots \\\end{align}$$&#xA;Now if we look at the powers of $2$ we will notice that they are $$\lbrace f_2(n)\rbrace=\lbrace 0,1,0,2,0,1,0,3,0,1,0,2,0,1,0,4\rbrace$$ and for the powers of $3$ we have $$\lbrace f_3(n)\rbrace=\lbrace 0,0,1,0,0,1,0,0,2,0,0,1,0,0,1,0\rbrace$$&#xA;This, of course, is a well known fact.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since then I wondered if there was a formula for $f_2(n)$ or $f_3(n)$ or $f_p(n)$, with $p\in \mathbb{P}$. It seemed impossible but I was able to devise the suitable formulas. They are&#xA;$$\displaystyle\begin{align} f_2(n)=\sum_{r=1}^{\infty}\frac{r}{{2^{r+1}}}\sum_{k=0}^{2^{r+1}-1}\cos\left( \frac{2k\pi(n+2^{r})}{2^{r+1}} \right)\end{align}$$&#xA;and for the general case we have&#xA;$$\displaystyle f_p(n)=\sum_{r=1}^{\infty}\frac{r}{p^{r+1}}\sum_{j=1}^{p-1}\left(\sum_{k=0}^{p^{r+1}-1}\cos\left( \frac{2k\pi(n+(p-j)p^{r})}{p^{r+1}} \right)\right)$$&#xA;If one cares to analyse the formula for $f_p(n)$ it can be concluded that it needs not to be restricted to the prime numbers, so that we have $f_m(n), m \in \mathbb{N}$ and similar  patterns for $\lbrace f_m(n)\rbrace$ will result. Now, the wonderfull thing is that we can express the &lt;em&gt;arithmetical divisor functions&lt;/em&gt; $\sigma_k(n)$ in terms of $f_m(n)$ as follows&#xA;$$\displaystyle \sigma_a(n)=1+\sum_{m=2}^{\infty}\sum_{r=1}^{\infty}\frac{m^{a}}{m^{r+1}}\sum_{j=1}^{m-1}\left(\sum_{k=0}^{m^{r+1}-1}\cos\left( \frac{2k\pi(n+(m-j)m^{r})}{m^{r+1}} \right)\right)$$&#xA;And, if we consider the &lt;em&gt;divisor summatory function&lt;/em&gt;, $D(n)$, as&#xA;$$D(n)=\sum_{m \leq n}d(m)$$&#xA;with $$d(n)=\sigma_{0}(n)=\sum_{d|n}1$$&#xA;we can express $D(n)$ as &#xA;$$D(n)=\sum_{m=2}^{\infty}\sum_{r=1}^{\infty}\frac{r}{m^{r+1}}\sum_{j=1}^{m-1}\left(\sum_{k=0}^{m^{r+1}-1}\cos\left( \frac{2k\pi(2^{n}+(m-j)m^{r})}{m^{r+1}} \right)\right)$$&#xA;Now, we know that, $d(n)$ and $D(n)$ are related to the Riemann zeta-function by&#xA;$$\zeta^{2}(z)=\sum_{n=1}^{\infty}\frac{d(n)}{n^{z}}$$&#xA;and &#xA;$$\zeta^{2}(z)=z\int_{1}^{\infty}\frac{D(x)}{x^{z+1}}dx$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, my questions&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;What can we say about the convergence of $f_m(z)$, $\sigma_a(z)$ and $D(z)$ with $z \in \mathbb{C}$? We can see that they converge for $z \in \mathbb{N}$.&lt;/li&gt;&#xA;&lt;li&gt;I think that $\sigma_a(z)$ and $D(z)$ are only &lt;a href=&quot;http://mathoverflow.net/questions/54625/how-important-are-the-formulas-at-number-theory-closed&quot; rel=&quot;nofollow&quot;&gt;curiosities&lt;/a&gt; and aren't interesting in the context of the Riemann zeta-function because they are hard to compute. What do you think? &lt;/li&gt;&#xA;&lt;li&gt;Are formulas $f_m(z)$, $\sigma_a(z)$ and $D(z)$ original. I think they are. I'd like to know if anyone has found something like this before. I've posted this as an answer to &lt;a href=&quot;http://mathoverflow.net/questions/29828/greatest-power-of-two-dividing-an-integer&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; post sometime ago.&lt;/li&gt;&#xA;&lt;li&gt;Finally, is this intereting enough to publish somewhere? I'm just an amateur...&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;To conclude I'd like to apologise for presenting all this formulas without showing how I got them but you can consider &lt;a href=&quot;http://math.stackexchange.com/questions/5574/a-triangular-representation-for-the-divisor-summatory-function-dx&quot;&gt;this&lt;/a&gt; previous post of mine and the question &lt;a href=&quot;http://mathoverflow.net/questions/29828/greatest-power-of-two-dividing-an-integer&quot; rel=&quot;nofollow&quot;&gt;Greatest power of two dividing an integer&lt;/a&gt;, &lt;a href=&quot;http://mathoverflow.net/questions/40427/difficult-infinite-sum&quot; rel=&quot;nofollow&quot;&gt;Difficult Infinite Sum&lt;/a&gt; and &lt;a href=&quot;http://fs.gallup.unm.edu/Atanassov-SomeProblems.pdf&quot; rel=&quot;nofollow&quot;&gt;On the 61-st, the 62-nd, and the 63-rd Smarandache's problem&lt;/a&gt; page 38.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And now a challenge, can you present a formula for the &lt;em&gt;characteristic function of the prime numbers&lt;/em&gt;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt;&#xA;I'm answering my challenge and leaving another. Considering that the &lt;em&gt;characteristic function of the primes&lt;/em&gt;, $u(n)$, is given by &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;\begin{equation}&#xD;&#xA;u(n)=\begin{cases}&#xD;&#xA;&amp;amp;1\;\;\;\text{ if } n \in \mathbb{P} \\&#xD;&#xA;&amp;amp;\\&#xD;&#xA;&amp;amp;\\&#xD;&#xA;&amp;amp;0\;\;\;\text{ if } n \notin \mathbb{P}&#xD;&#xA;\end{cases}&#xD;&#xA;\end{equation}&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have found that $u(n)$ is given by the following formula&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;\begin{equation}&#xD;&#xA;u(n)=\prod_{m=2}^{\infty}\;\;\prod _{r=1}^{\infty} \left\{1-\frac{1}{m^{r+1}} \sum _{j=1}^{m-1}\;\;\;\sum _{k=0}^{m^{r+1}-1} \cos\left(2 k \pi \cdot\frac{n-m+(m-j) m^r }{m^{r+1}}\right)\right\}&#xD;&#xA;\end{equation}&#xD;&#xA;$$&#xA;Now, in the same spirit, what is formula for the &lt;em&gt;prime counting function&lt;/em&gt;, $\pi(x)$?&lt;/p&gt;&#xA;" OwnerUserId="1747" LastEditorUserId="1747" LastEditorDisplayName="" LastEditDate="2011-02-09T08:21:49.367" LastActivityDate="2011-02-09T08:21:49.367" Title="Prime powers, patterns similar to $\lbrace 0,1,0,2,0,1,0,3\ldots \rbrace$ and formulas for $\sigma_k(n)$." Tags="&lt;number-theory&gt;&lt;real-analysis&gt;&lt;riemann-zeta&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="9" />
  <row Id="15832" PostTypeId="2" ParentId="12069" CreationDate="2010-12-29T18:03:44.410" Score="0" ViewCount="" Body="&lt;p&gt;Hi, in searching the solution for a nearly problem, I have been falling in your interesting discussion. But there's something I doubt about Zarrax's answer. If the probablity density function is like your result, we can find out easily there's some values of alpha which makes the distribution function greater than 1, is it physically correct?&lt;/p&gt;&#xA;" OwnerUserId="5210" LastActivityDate="2010-12-29T18:03:44.410" CommentCount="1" />
  <row Id="15833" PostTypeId="2" ParentId="15810" CreationDate="2010-12-29T18:04:34.833" Score="2" ViewCount="" Body="&lt;p&gt;You have two questions: how to change variables, and how to handle nodes.  The case of changing variables, for hyperelliptic curves without nodes, is in chapter III.5.5 of Shafarevich's Basic algebraic geometry.  The general case is on page 105 of Phillip Griffiths' Introduction to algebraic curves (China notes).  The discussion of the nodes there needs supplementing with the argument from page 98, i.e. a proof that the derivative vanishes simply on both the separate branches lying over the node, rather than at the point in the plane, as suggested there in a footnote.  Basically a fraction is holomorphic if the numerator vanishes as much as the denominator.  Here the equation for the curve vanishes twice at the node so the derivative in the denominator vanishes once and is canceled by the vanishing of the adjoint polynomial.  But you still have to finesse the point raised in Griffiths' footnote as mentioned above about the order of vanishing of the pullback to the normalization.  Warm up on some specific examples.&lt;/p&gt;&#xA;" OwnerUserId="5211" LastActivityDate="2010-12-29T18:04:34.833" />
  <row Id="15834" PostTypeId="2" ParentId="15773" CreationDate="2010-12-29T18:15:52.130" Score="3" ViewCount="" Body="&lt;p&gt;Since LIE's answer reduces one to the case of $A_5$, I like identifying $A_5$ with the icosahedral group and using the geometry of the icosahedron to make the calculations of the number of elements conjugate to those leaving vertices, edges, and faces, respectively fixed.  This is on page 20 of the notes for math 843-1 &lt;a href=&quot;http://www.math.uga.edu/~roy/&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To identify Icos with $A_5$, note the 30 edges are partitioned into 5 subsets of size 6, by looking at the 5 sets of orthogonal triples in the 15 lines joining the centers of pairs of opposite edges. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then LIE's answer (and Arturo's) shows why $A_5$ is really the key case.  Of course the apparently ad hoc geometry used here is mirrored abstractly by the conjugacy classes of elements in the group.  I.e. every group acts on itself by conjugation, and sometimes we can represent that action by a geometric figure.&lt;/p&gt;&#xA;" OwnerUserId="5211" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-29T19:27:45.773" LastActivityDate="2010-12-29T19:27:45.773" CommentCount="2" />
  <row Id="15835" PostTypeId="2" ParentId="15829" CreationDate="2010-12-29T18:16:20.763" Score="1" ViewCount="" Body="&lt;p&gt;Let $P&amp;#39;$ be any Sylow $p$-subgroup of $H$. Then there is a maximal $p$-subgroup of $G$ containing it. What is it?&lt;/p&gt;&#xA;" OwnerUserId="5191" LastActivityDate="2010-12-29T18:16:20.763" CommentCount="2" />
  <row Id="15836" PostTypeId="2" ParentId="15571" CreationDate="2010-12-29T18:16:25.023" Score="3" ViewCount="" Body="&lt;p&gt;Concerning infinity, here is my take. When we say, for example, that there is an infinity of natural numbers, we mean only that there is no largest number. Whenever you add 1 to a number, you will get another number that is greater still. Nothing very profound there. To say otherwise would greatly complicate even basic arithmetic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It may even be possible to do away with the use of the word &quot;infinity&quot; (or any equivalent notion) in most if not all of mathematics. It may be nothing more than a convenient shorthand. When we say, for example, the limit of some expression as x tends to infinity, we could as easily have said, the limit as x increases without bound. The latter would have the advantage that there would be no confusion about there being some kind of endpoint called &quot;infinity.&quot;&lt;/p&gt;&#xA;" OwnerUserId="3515" LastActivityDate="2010-12-29T18:16:25.023" CommunityOwnedDate="2010-12-29T18:16:25.023" />
  <row Id="15837" PostTypeId="2" ParentId="15829" CreationDate="2010-12-29T18:32:24.993" Score="3" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Note:&lt;/strong&gt; The following works if $G$ is finite, but may fail in the infinite case; there are infinite groups in which there are $p$-Sylow subgroups $P$ and $P&amp;#39;$ for which no automorphism (inner or outer) of $G$ maps $P$ to $P&amp;#39;$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $K$ be a $p$-Sylow subgroup of $H$ (we know it exists, though it may be trivial). Then $K$ is a $p$-subgroup of $G$ (even if $K={e}$) and by the Sylow Theorems, is contained in &lt;em&gt;some&lt;/em&gt; Sylow $p$-subgroup $Q$ of $G$. By the Sylow Theorems, $Q$ and $P$ are conjugate. Now just verify that $Q\cap H = K$. &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-29T19:12:18.970" LastActivityDate="2010-12-29T19:12:18.970" CommentCount="9" />
  <row Id="15838" PostTypeId="1" CreationDate="2010-12-29T18:40:57.577" Score="0" ViewCount="125" Body="&lt;p&gt;Given the next sets: $A=\{1,2\}$ , $B=\mathbb{N}$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a) Describe all the functions from $A$ to $B$ that are not injective.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;b) Describe all the functions from $B$ to $A$ that are not onto.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;c) Prove or disprove the next:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(i) The next functions exist $f\colon A\to B$ and $g\colon B\to A$ if $g\circ f$ reversal&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(ii) the next functions exist $f\colon A\to B$ and $g\colon B\to A$ if $f\circ g$ reversal&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Need help here. Thanks.&lt;/p&gt;&#xA;" OwnerUserId="5214" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-29T19:53:59.400" LastActivityDate="2010-12-29T19:53:59.400" Title="Sets and functions" Tags="&lt;homework&gt;&lt;functions&gt;" AnswerCount="2" CommentCount="5" />
  <row Id="15839" PostTypeId="2" ParentId="15709" CreationDate="2010-12-29T18:45:34.907" Score="4" ViewCount="" Body="&lt;p&gt;A product $A\times B$ is a triple $(A,B,A\times B)$ equipped with maps $\pi_A\colon A\times B\to A$, and $\pi_B\colon A\times B\to B$ such that given any pair of maps $\psi_A\colon X\to A$ and $\psi_B\colon X\to B$, there is a &lt;strong&gt;unique&lt;/strong&gt; map $\psi\colon X\to A\times B$, such that $\psi_A=\pi_a\circ \psi$ and $\psi_B=\pi_B\circ \psi$. Hence for $X = A\times B$, (and some given maps) if the identity map $A\times B\to A\times B$ fulfills this role, then any other map also fulfilling it equals the identity map.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I.e. to prove that $fg$ is the identity, check that it does satisfy some property that is only satisfied by one unique map, and then check that the identity also satisfies it.  Examples are given in the algebra notes 843-4-5 on my website.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Specifically, look on pages 8-9 of the notes 845-3, for exactly this proof of uniqueness of products.&lt;/p&gt;&#xA;" OwnerUserId="5211" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2011-01-06T01:00:59.687" LastActivityDate="2011-01-06T01:00:59.687" CommentCount="3" />
  <row Id="15840" PostTypeId="2" ParentId="15838" CreationDate="2010-12-29T19:20:06.763" Score="2" ViewCount="" Body="&lt;p&gt;The first two follow from the definitions, so I'm assuming you're having some difficulty with what injective (or one to one) and surjective (or onto) mean.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $f : A \rightarrow B$ be a function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Our function $f$ is injective if, provided there is some $a$ in $A$ with $f(a) = b$, there is only one such $a$.  Put another way, everything in $B$ that is &quot;hit&quot; by $f$ is &quot;hit&quot; only one time.  This is why one-to-one is a good term:  for every output, there is one input.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Our function $f$ is surjective if, for every $b$ in $B$, there is some $a$ with $f(a) = b$.  Essentially, $f$ is surjective if everything in $B$ is &quot;hit&quot; by $f$.  This is why onto is a good term:  you can picture $B$ being covered by $f$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For your part (a), think about functions from ${1,2}$ to $\mathbb{N}$ that do not have the property that &quot;for every output, there is one input.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For your part (b), think about functions from $\mathbb{N}$ to ${1,2}$ that do not have the property that &quot;everything is hit&quot; by the function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You need to reword part (c); it does not make sense as written.&lt;/p&gt;&#xA;" OwnerUserId="3537" LastActivityDate="2010-12-29T19:20:06.763" CommentCount="4" />
  <row Id="15841" PostTypeId="2" ParentId="14921" CreationDate="2010-12-29T19:35:09.690" Score="8" ViewCount="" Body="&lt;p&gt;In John Stillwell's Mathematics and its History it is observed that Euler criticized Newton's classification for lacking a general principle, but that a closer examination of Newton/s work reveals one.  In fact his work gives a general classification into 5 types, depicted on &lt;a href=&quot;http://books.google.com/books?id=WNjRrqTm62QC&amp;amp;pg=PA112&quot; rel=&quot;nofollow&quot;&gt;page 112 of that book&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe the difference in the real and complex cases, is the fact that the &quot;discriminant&quot; locus of singular cubics has real codimension one in the real case, hence separates the space of smooth cubics into distinct connected components, and this may be the types sought for.  By Ehresmann's theorem, a connected family of compact manifolds have the same topological type.  Thus curves on different connected components of the complement of the discriminant locus can have different homeomorphism type.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At least one (two?) of Newton's types is also singular and could represent the general (and special) point of the discriminant locus.  I am not an expert.&lt;/p&gt;&#xA;" OwnerUserId="5211" LastEditorUserId="5211" LastEditorDisplayName="" LastEditDate="2011-01-01T00:52:56.590" LastActivityDate="2011-01-01T00:52:56.590" CommentCount="1" />
  <row Id="15842" PostTypeId="2" ParentId="15838" CreationDate="2010-12-29T19:53:34.627" Score="2" ViewCount="" Body="&lt;p&gt;(a) Since $A$ has only two elements, the &lt;em&gt;only&lt;/em&gt; way for a function $f\colon A\to X$ (for &lt;em&gt;any&lt;/em&gt; set $X$) to &lt;em&gt;not&lt;/em&gt; be injective is if $f(1)=f(2)$. So you want all functions from $A$ to $B$ in which $f(1)=f(2)$. Now, while there are indeed &quot;lots of them&quot; (infinitely many) they all have a very easy description. &lt;em&gt;Hint:&lt;/em&gt; the word that describes these functions starts with &quot;con&quot; and ends with &quot;ants&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(b) Again, since $A$ has only two elements, the &lt;em&gt;only&lt;/em&gt; way for a function $g\colon X\to A$ to &lt;em&gt;not&lt;/em&gt; be onto (surjective) is if either you never get $1$, or you never get $2$ as images of $g$. While there are many, many, functions from $B$ to $A$, there are very few in which this happens. &lt;em&gt;Hint&lt;/em&gt;: the word that describes theses functions is the same as in (a).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(c-i) You want to know if you can define functions $A\stackrel{f}{\to} B\stackrel{g}{\to}A$ such that the composition is both one-to-one and onto. Not too hard, since there are only two functions from $A$ to $A$ that are both one-to-one and onto. See if you can come up with some easy functions $f$ and $g$ that do this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(c-ii) Here you want to know if you can find functions $B\stackrel{g}{\to}A\stackrel{f}{\to}B$ so that the composition is both one-to-one and onto. Again, this is not hard if you think about what these things mean. Can you have a one-to-one map from $B$ to $A$? Can you have an onto map from $A$ to $B$? Think about it. Note that these $f$ and $g$ don't have to be the same as in (c-i), they are only asking you if some such functions exist at all.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-12-29T19:53:34.627" />
  <row Id="15843" PostTypeId="1" CreationDate="2010-12-29T20:27:10.637" Score="4" ViewCount="144" Body="&lt;p&gt;Let $p_n$ be the n'th non-trivial zero of the Riemann zeta function. We define the &lt;em&gt;Zetor&lt;/em&gt; function (acronym of 'zeta' and 'zero') as follows: $$\zeta \rho (s) = \sum_{n=1}^{\infty} \frac{1}{(p_n)^s}. $$ &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;For which values of $s$ does $\zeta \rho (s)$ converge?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Thanks, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Max&lt;/p&gt;&#xA;" OwnerUserId="1055" LastEditorUserId="1055" LastEditorDisplayName="" LastEditDate="2010-12-29T20:53:49.553" LastActivityDate="2010-12-29T20:53:49.553" Title="When does the &quot;Zetor function&quot; converge?" Tags="&lt;functions&gt;&lt;analytic-number-theory&gt;" CommentCount="7" />
  <row Id="15844" PostTypeId="1" AcceptedAnswerId="15845" CreationDate="2010-12-29T20:33:21.877" Score="4" ViewCount="92" Body="&lt;p&gt;I'm looking at &lt;a href=&quot;http://en.wikipedia.org/wiki/Intuitionistic_logic#Hilbert-style_calculus&quot; rel=&quot;nofollow&quot;&gt;Wikipedia's article on Intuitionist logic&lt;/a&gt;, and I can't figure out how it would prove $(p \rightarrow p)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does it prove $(p \rightarrow p)$?  If yes, how?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If no, is there a correct (or reasonable, if there is no &quot;correct&quot;) axiomatization of intuitionistic logic available anywhere online?&lt;/p&gt;&#xA;" OwnerUserId="5114" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-29T20:40:45.007" LastActivityDate="2010-12-29T20:40:45.007" Title="How does (wikipedia's axiomatization of) intuitionistic logic prove $p \rightarrow p$ ?" Tags="&lt;logic&gt;&lt;constructive-mathematics&gt;" AnswerCount="1" />
  <row Id="15845" PostTypeId="2" ParentId="15844" CreationDate="2010-12-29T20:38:48.600" Score="7" ViewCount="" Body="&lt;p&gt;&#xA;&lt;li&gt;p→(p→p) (THEN-1)&lt;/li&gt;&#xA;&lt;li&gt;p→((p→p)→p) (THEN-1)&lt;/li&gt;&#xA;&lt;li&gt;(p→((p→p)→p))→((p→(p→p))→(p→p)) (THEN-2)&lt;/li&gt;&#xA;&lt;li&gt;(p→(p→p))→(p→p) (MP from lines 2,3)&lt;/li&gt;&#xA;&lt;li&gt;(p→p) (MP from lines 1,4)&lt;/li&gt;&lt;/p&gt;&#xA;" OwnerUserId="5203" LastActivityDate="2010-12-29T20:38:48.600" />
  <row Id="15846" PostTypeId="1" AcceptedAnswerId="15848" CreationDate="2010-12-29T20:57:32.247" Score="4" ViewCount="155" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was wondering today about why the word &lt;em&gt;differentiable&lt;/em&gt; is used for describing functions that have a derivative or are differentiable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps because originally one considered &lt;em&gt;finite differences&lt;/em&gt;? But that seems somewhat not right, because &lt;em&gt;roughly speaking&lt;/em&gt; a derivative measures &lt;strong&gt;not&lt;/strong&gt; the difference $f(x+h)-f(x)$, but rather the ratio $(f(x+h)-f(x))/h$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, could people here shed light on why we use &quot;differentiable&quot;? Any pointers to academic / historical / etymological explanations are also welcome. Thanks!&lt;/p&gt;&#xA;" OwnerUserId="1709" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-30T01:37:02.710" LastActivityDate="2010-12-30T01:37:02.710" Title="What is the origin of the term &quot;Differentiable&quot;?" Tags="&lt;reference-request&gt;&lt;terminology&gt;&lt;math-history&gt;" AnswerCount="1" />
  <row Id="15847" PostTypeId="2" ParentId="15773" CreationDate="2010-12-29T20:59:20.873" Score="2" ViewCount="" Body="&lt;p&gt;The way to understand this theorem is to understand the following statement:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;Other than the even permutations, there's no way to choose a collection of cycle structures that's closed under multiplication.&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;To illustrate what I mean, I will prove the following theorem, which is related to but slightly easier than the simplicity of $A_n$:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;Theorem.&lt;/strong&gt; &lt;em&gt;For $n \geq 5$, the only nontrivial normal subgroup of $S_n$ is $A_n$.&lt;/em&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The main idea of the proof is to imagine what a normal subgroup of $S_n$ must look like:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Any normal subgroup of $S_n$ must be a union of conjugacy classes.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;But each conjugacy class of $S_n$ is simply the set of all permutations with a given cycle structure.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Therefore, a normal subgroup of $S_n$ must consist of a collection of possible cycle structures which is closed under multiplication.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;For example, the conjugacy classes in $S_4$ are the following:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;the identity&lt;/li&gt;&#xA;&lt;li&gt;transpositions $(\ast \;\ast)$&lt;/li&gt;&#xA;&lt;li&gt;3-cycles $(\ast\;\ast\;\ast)$&lt;/li&gt;&#xA;&lt;li&gt;4-cycles $(\ast\;\ast\;\ast\;\ast)$&lt;/li&gt;&#xA;&lt;li&gt;Elements of the form $(\ast\;\ast)(\ast\;\ast)$&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Any normal subgroup in $S_4$ would have to be the union of some of these, and would have to be closed under multiplication.  It's easy to check that the only possibilities are $A_4 = (1) \cup (3) \cup (5)$ and the 4-element subgroup $(1) \cup (5)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, the conjugacy classes in $S_5$ are the following:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;the identity&lt;/li&gt;&#xA;&lt;li&gt;transpositions $(\ast \;\ast)$&lt;/li&gt;&#xA;&lt;li&gt;3-cycles $(\ast\;\ast\;\ast)$&lt;/li&gt;&#xA;&lt;li&gt;4-cycles $(\ast\;\ast\;\ast\;\ast)$&lt;/li&gt;&#xA;&lt;li&gt;5-cycles $(\ast\;\ast\;\ast\;\ast\;\ast)$&lt;/li&gt;&#xA;&lt;li&gt;Elements of the form $(\ast\;\ast)(\ast\;\ast)$&lt;/li&gt;&#xA;&lt;li&gt;Elements of the form $(\ast\;\ast\;\ast)(\ast\;\ast)$&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;It's easy to check that the only finite union of these that is closed under multiplication is $A_5$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At this point, the rest of the proof just involves playing around with permutations and cycle structures.  In particular, it's not too hard to prove the following statement:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;For any nonidentity permutation $\alpha \in S_n$, there exists a permutation $\beta  \in S_n$ with the same cycle structure so that $\alpha\beta$ has the form $(\ast\;\ast\;\ast)$ or $(\ast\;\ast)(\ast\;\ast)$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Thus any normal subgroup of $S_n$ must either contain all of the 3-cycles, or all elements of the form $(\ast\;\ast)(\ast\;\ast)$.  For $n \geq 5$, elements of these forms generate  all of $A_n$, and therefore $A_n$ is the only nontrivial normal subgroup of $S_n$.  (Since $A_n$ has index two, any nontrivial subgroup that contains $A_n$ must equal $A_n$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The proof that $A_n$ is simple is fairly similar, but complicated by the fact that the conjugacy classes in $A_n$ aren't precisely the cycle structures.  (Some cycle structures in $A_n$ have two corresponding conjugacy classes.)  Alternatively, it may be possible to prove that any normal subgroup of $A_n$ would also have to be normal in $S_n$, and then apply the above theorem to conclude that $A_n$ is simple.&lt;/p&gt;&#xA;" OwnerUserId="1726" LastActivityDate="2010-12-29T20:59:20.873" CommentCount="1" />
  <row Id="15848" PostTypeId="2" ParentId="15846" CreationDate="2010-12-29T21:03:23.293" Score="9" ViewCount="" Body="&lt;p&gt;From the &lt;a href=&quot;http://jeff560.tripod.com/mathword.html&quot; rel=&quot;nofollow&quot;&gt;Earliest Known Uses of Some of the Words of Mathematics&lt;/a&gt; webpage:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;DIFFERENTIAL CALCULUS.&lt;/strong&gt; The term calculus differentialis was introduced by Leibniz in 1684 in Acta Eruditorum 3. Before introducing this term, he used the expression methodus tangentium directa (Struik, page 271). The OED has a nice quotation from Joseph Raphson’s Mathematical Dictionary of 1702: “A different way....passes....in France under the Name of Leibnitz's Differential Calculus, or Calculus of Differences.” &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="274" LastActivityDate="2010-12-29T21:03:23.293" CommentCount="1" />
  <row Id="15849" PostTypeId="1" AcceptedAnswerId="15851" CreationDate="2010-12-29T22:11:00.407" Score="2" ViewCount="137" Body="&lt;p&gt;I'm attempting to understand the proof of the statement that every finite domain $R$ is a division ring. For a fixed nonzero element $a \in R$, the left-translation map is injective since $R$ is a domain. Since $R$ is finite this map is also surjective which implies that 1 is in its image. Therefore we have the existence of an element $b \in R$ such that $ab=1$, right inverses exist for nonzero elements. By looking at the right-translation map, we get the existence of left inverses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given a finite domain $R$ and $a,b,c \in R$ such that $ab=1$ and $ca=1$ how do we show that $b=c=a^{-1}$? How do we know that the nonzero elements are units?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the answer is as follows: The nonzero elements already form a multiplicative monoid since $R$ is a domain. But to form a group it is enough to demonstrate the existence of left (or right) inverses only by &lt;a href=&quot;http://en.wikipedia.org/wiki/Elementary_group_theory#Alternative_Axioms&quot; rel=&quot;nofollow&quot;&gt;these alternate group axioms&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So maybe we just need left inverses and an identity to get a division ring.&lt;/p&gt;&#xA;" OwnerUserId="1761" LastActivityDate="2010-12-30T17:20:22.103" Title="Equality of left and right inverses in a finite domain." Tags="&lt;abstract-algebra&gt;&lt;ring-theory&gt;" AnswerCount="4" CommentCount="3" />
  <row Id="15850" PostTypeId="2" ParentId="15849" CreationDate="2010-12-29T22:16:53.870" Score="5" ViewCount="" Body="&lt;p&gt;What happens if you mutliply the identity $ab = 1$ on the left by $c$?&lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2010-12-29T22:16:53.870" CommentCount="1" />
  <row Id="15851" PostTypeId="2" ParentId="15849" CreationDate="2010-12-29T22:17:59.247" Score="4" ViewCount="" Body="&lt;p&gt;In &lt;em&gt;any&lt;/em&gt; ring $R$, if $a$ has both a left and a right inverse, then the left and right inverses are the same and the element is a unit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To see this in your case, just write this:&#xA;$$ b = 1\cdot b = (ca)b = c(ab) = c\cdot 1 = c.$$&#xA;Since $b=c$, then $ca = ba = ab = 1$, so $b=c=a^{-1}$, as desired.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the standard argument. It can be used, for example, to show that a function that has left and right inverses must be invertible (and the left and right inverses are the same). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is also true that if you have a ring without zero divisors, which has a left identity $e\neq 0$ and every element has a left inverse, then the ring is a division ring. The argument is the one you give, essentially showing that the nonzero elements form a group under multiplication. But the fact that an element having both left and right inverses is in fact invertible is more basic than that.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-12-29T22:17:59.247" CommentCount="4" />
  <row Id="15852" PostTypeId="1" AcceptedAnswerId="15858" CreationDate="2010-12-29T22:44:00.737" Score="4" ViewCount="185" Body="&lt;p&gt;The word problem for finite groups is decidable. Is it obvious that this is true? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, I'm not entirely sure about what it means for the problem to be decidable (in this case---I think I understand what decidable means in general). I assume it means that we are given a fixed group &lt;em&gt;G&lt;/em&gt; (do we have to assume this), but is the generating set (the letters) also fixed? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To decide the word problem for the group of symmetries of a square, with reflection $r$ and transposition $t$ as the generators, I would first find a canonical form for the group elements $1,r,r^2,r^3,t,tr,tr^2,tr^3$, and then note that using the relations $r^4=1$ and $rt=tr^3$ to show that any word can be reduced to something on my list. However, this seems like a lot of work, and it's not obvious to me what should be done in the case of an arbitrary finite group. &lt;/p&gt;&#xA;" OwnerUserId="3656" LastEditorUserId="3656" LastEditorDisplayName="" LastEditDate="2010-12-30T00:27:59.003" LastActivityDate="2010-12-30T16:44:32.323" Title="The word problem for finite groups" Tags="&lt;group-theory&gt;&lt;computability&gt;" AnswerCount="3" CommentCount="0" />
  <row Id="15853" PostTypeId="2" ParentId="15852" CreationDate="2010-12-29T22:51:25.437" Score="5" ViewCount="" Body="&lt;p&gt;The sense in which I think this statement is usually meant is that you are given the entire multiplication table of the group.  Then it's obvious that the word problem is decidable.  You seem to be thinking about a different question: maybe you are given generators and relations and you are told that they define a finite group, then asked to solve the word problem.  This question, to me, is not obvious.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-12-29T22:51:25.437" CommentCount="10" />
  <row Id="15854" PostTypeId="2" ParentId="15849" CreationDate="2010-12-29T23:00:04.583" Score="2" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\:$ &lt;em&gt;Unify&lt;/em&gt; the rewrite rules $\rm\ ca\to 1,\  ab\to 1\ $ by creating a common generalization of their LHS,$\ $ viz. $\rm\ cab\:.\: $ Now apply the two rules to the unification $\rm\ cab\ $ to deduce that $\rm\ c = b\:.$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same type of unification procedure is at the heart of procedures that generate normal-form rewriting algorithms for various algebraic structures, e.g. google the &lt;a href=&quot;http://en.wikipedia.org/wiki/Knuth%E2%80%93Bendix_completion_algorithm&quot; rel=&quot;nofollow&quot;&gt;Knuth-Bendix completion algorithm.&lt;/a&gt; Many well-known algorithms may be considered as special cases of such, e.g. Gaussian elimination and its nonlinear generalizations such as the &lt;a href=&quot;http://en.wikipedia.org/wiki/Gr%C3%B6bner_basis&quot; rel=&quot;nofollow&quot;&gt;Grobner basis algorithm,&lt;/a&gt; and also the &lt;a href=&quot;http://en.wikipedia.org/wiki/Todd%E2%80%93Coxeter_algorithm&quot; rel=&quot;nofollow&quot;&gt;Todd-Coxeter algorithm&lt;/a&gt;  for enumerating cosets of group presented by generators and relations.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-12-30T17:20:22.103" LastActivityDate="2010-12-30T17:20:22.103" CommentCount="2" />
  <row Id="15855" PostTypeId="1" CreationDate="2010-12-29T23:05:02.523" Score="1" ViewCount="101" Body="&lt;p&gt;Let $X_1,...,X_n$ be some observations, and let $\theta$ be some parameter of the density function we want to estimate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, it is well known that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$l(\theta) = n^{-1} \sum_{i=1}^n \log f(X_i ; \theta)$ is called the average log-likelihood.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is $\mathbb{E}[\log f(X_1 ; \theta)]$ called? Meaning, &quot;the expected value of the likelihood&quot;? It can be called the cross-entropy, or perhaps it has other name, but it seems to me it should have a name relating it to the likelihood, such as &quot;population likelihood&quot; perhaps or something of that sort.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyone knows? Wikipedia did not help much here.&lt;/p&gt;&#xA;" OwnerUserId="7076" LastEditorUserId="7076" LastEditorDisplayName="" LastEditDate="2010-12-30T00:57:02.517" LastActivityDate="2010-12-31T03:56:26.060" Title="What is the &quot;population likelihood&quot; called?" Tags="&lt;statistics&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="15857" PostTypeId="1" AcceptedAnswerId="15864" CreationDate="2010-12-29T23:18:01.923" Score="3" ViewCount="283" Body="&lt;p&gt;Which of these two books should I read first? Is one of them more advanced than the other? I'm planning to use them to self-study geometry. &#xA;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="5119" LastActivityDate="2010-12-30T18:49:15.130" Title="Coxeter's &quot;Geometry Revisited&quot; vs. &quot;Introduction to Geometry&quot;" Tags="&lt;geometry&gt;&lt;books&gt;&lt;self-learning&gt;" AnswerCount="5" />
  <row Id="15858" PostTypeId="2" ParentId="15852" CreationDate="2010-12-29T23:18:48.697" Score="8" ViewCount="" Body="&lt;p&gt;One can prove quite simply a much more general result due to McKinsey, viz.&#xA;&lt;img src=&quot;http://i.imgur.com/lzh3L.jpg&quot; alt=&quot;alt text&quot;&gt;&#xA;&lt;img src=&quot;http://i.imgur.com/aHLrP.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-12-29T23:18:48.697" CommentCount="1" />
  <row Id="15859" PostTypeId="2" ParentId="15857" CreationDate="2010-12-29T23:26:08.530" Score="3" ViewCount="" Body="&lt;p&gt;If you are talking about &quot;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0471504580&quot; rel=&quot;nofollow&quot;&gt;Introduction to Geometry&lt;/a&gt;&quot; by Coxeter and &quot;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0883856190&quot; rel=&quot;nofollow&quot;&gt;Geometry Revisited&lt;/a&gt;&quot; by Coxeter and Greitzer, the consensus seems to be that both of them are pretty advanced, but &quot;Introduction to Geometry&quot; is significantly more so, while &quot;Geometry Revisited&quot; is closer to something 'right after high school geometry class,' so I guess you should start with the latter (assuming you do have some geometry knowledge already).&lt;/p&gt;&#xA;" OwnerUserId="3731" LastActivityDate="2010-12-29T23:26:08.530" />
  <row Id="15860" PostTypeId="1" AcceptedAnswerId="15863" CreationDate="2010-12-29T23:29:18.510" Score="2" ViewCount="103" Body="&lt;p&gt;In a population that grows, would a disparity in having a boy/girl probability cause the ratio of males to females tend to the same? e.g. if the probability of having boys was .49 and having a girl was .51 would that mean a growing population would tend to be 49% male and 51% female?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What about a non growing population?, would that mean finally one gender will be the only gender left after sufficient amount of time?&lt;/p&gt;&#xA;" OwnerUserId="3936" LastActivityDate="2010-12-30T00:28:58.077" Title="Consequences of boy/girl probability disparity on population" Tags="&lt;probability&gt;" AnswerCount="1" />
  <row Id="15862" PostTypeId="2" ParentId="15857" CreationDate="2010-12-29T23:42:15.873" Score="4" ViewCount="" Body="&lt;p&gt;Geometry Revisited is much more elementary, aimed at high-school teachers and bright high-school students or college frosh. For example, K. Strubecker wrote in his AMS Math Review&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The tenor of the translation of Coxeter's beautiful tome Geometry revisited [Random House, New York, 1967] is in keeping with the objectives of the Klett Textbooks in Mathematics series which are intended to convey to freshmen and teachers of mathematics---via interesting representations---an approach to different aspects of mathematics, especially to geometry, that is kept as concrete as possible and so is applicable in schools. The volume contains six chapters which deal with the following topics: (1) Points and lines connected with a triangle; (2) some properties of circles; (3) collinearity and concurrence; (4) transformations; (5) an introduction to inversive geometry; (6) an introduction to projective geometry; The very lucid presentation takes the reader from elementary problems of plane Euclidean geometry to the fundamental concepts of non-Euclidean geometry, whose metric is briefly illustrated by the conformal model. Starting with simple geometric figures (triangle, lines, circle) and their properties, the volume advances to higher problems and figures in a manner that is convenient for the student and also whets his appetite. The always original developments use very simple tools (theorems of Ceva and Menelaos) and soon proceed to higher configurations (theorems of Pascal and Brianchon on the circle). The conics are obtained from circles as polar figures of the circles. The book is rich in remarkable facts and thereby is very effective in promoting the significance and the value of geometry in mathematical teaching, a promotion which is very necessary in view of today's predominance of set theory, analysis and algebra on the school and university level, and which deserves the skillful hand of distinguished scholars. An advantage in this recruiting endeavor is the high degree of visualizability of geometry, the easy comprehensibility of its problems and interesting theorems, and the challenge emanating from these problems to occupy oneself with their solutions. This purpose is also served by the numerous problems contained in the text whose solutions are listed at the end of the book. Many historical remarks are woven into the text.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-12-29T23:42:15.873" />
  <row Id="15863" PostTypeId="2" ParentId="15860" CreationDate="2010-12-29T23:57:57.240" Score="5" ViewCount="" Body="&lt;p&gt;It's not possible for one gender to &quot;go extinct&quot;, as it were. That would happen if there were two different species whose rates of reproduction were different; that is, if girls produced only girls and boys produced only boys.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here, in each generation you expect 49% boys and 51% girls to be born. No matter what the initial population, once the first generation has died you will be left with an expected proportion of 49% males to 51% females. This holds whether or not the population is growing or fixed -- but only assuming the death rates for males and females are the same.&lt;/p&gt;&#xA;" OwnerUserId="856" LastActivityDate="2010-12-29T23:57:57.240" />
  <row Id="15864" PostTypeId="2" ParentId="15857" CreationDate="2010-12-30T00:06:57.403" Score="4" ViewCount="" Body="&lt;p&gt;If you are planning to self study and are just beginning geometry, I would recommend neither at this stage.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Geometry Revisted is from a specific viewpoint: Visiting Elementary Geometry from the viewpoint of transformations. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A snapshot of the backcover of Geometry Revisited:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/KGQC9.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you will miss out on some portions which are not covered by this book.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Introduction to Geometry is an advanced book and presumes fairly high familiarity already.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: The books are excellent. I suggest you revisit Geometry Revisited &lt;em&gt;after&lt;/em&gt; you have done some amount of study and then perhaps take a shot at Introduction to Geometry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Having read it at some point myself, I can vouch for Geometry Revisited. It is an excellent book and ought be read at some point. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If your only choices to self learn were Geometry Revisited and Introduction to Geometry, then I would recommend Geometry Revisited.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-12-30T00:13:58.970" LastActivityDate="2010-12-30T00:13:58.970" />
  <row Id="15865" PostTypeId="1" AcceptedAnswerId="18873" CreationDate="2010-12-30T00:16:08.893" Score="3" ViewCount="282" Body="&lt;p&gt;For the solution of the cubic equation $x^3 + px + q = 0$ Cardano wrote it as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sqrt[3]{-\frac{q}{2} + \sqrt{\frac{q^2}{4} + \frac{p^3}{27}}}+\sqrt[3]{-\frac{q}{2} - \sqrt{\frac{q^2}{4} + \frac{p^3}{27}}}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but this is ambiguous because it does not tell you which cube roots to match up. Why don't people write it this way today:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sqrt[3]{-\frac{q}{2} + \sqrt{\frac{q^2}{4} + \frac{p^3}{27}}}+\frac{-p}{3\sqrt[3]{-\frac{q}{2} + \sqrt{\frac{q^2}{4} + \frac{p^3}{27}}}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is unambiguous.&lt;/p&gt;&#xA;" OwnerUserId="4290" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-30T01:31:25.427" LastActivityDate="2011-01-25T06:52:26.923" Title="Why not write the solutions of a cubic this way?" Tags="&lt;algebra-precalculus&gt;&lt;algebra&gt;&lt;polynomials&gt;" AnswerCount="1" CommentCount="6" FavoriteCount="1" />
  <row Id="15866" PostTypeId="1" AcceptedAnswerId="15873" CreationDate="2010-12-30T00:19:44.237" Score="7" ViewCount="310" Body="&lt;p&gt;I need to compute the intersection(s) between two circular arcs. Each arc is specified by its endpoints and their height. The height is the perpendicular distance from the chord connecting the endpoints to the middle of the arc. I use this representation because it is numerically robust for very slightly bent arcs, as well as straight line segments, for which the height is zero. In these cases, representing an arc using the center of its circle could lead to the center being far far away from the endpoints of the arc, and hence, numerically unstable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question at the highest level is how I would go about computing the intersection points, given that the centers of the circles of the arcs cannot necessarily be computed robustly. At a lower level, I am wondering if there is a parameterization of an arc using only the information I have stated above (which does not include the circle center). Of course, keep in mind numerical robustness is my principal concern here; otherwise I would just do the naive thing and compute the circle center for all non-linear arcs and hope for the best.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;: Formula for computing center of circle of arcs:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose the chord length is $2t$, and the height is $h$. The distance from the chord to the circle center is $c$, so that $r=h+c$. Then it follows that $c=(t^2-h^2)/2h$, which breaks down when $h$ is very small. Computing the location of the circle center is some simple vector arithmetic using the chord vector and its perpendicular.&lt;/p&gt;&#xA;" OwnerUserId="398" LastEditorUserId="398" LastEditorDisplayName="" LastEditDate="2010-12-30T01:26:05.417" LastActivityDate="2010-12-30T22:23:41.340" Title="arc-arc intersection, arcs specified by endpoints and height" Tags="&lt;geometry&gt;&lt;numerical-methods&gt;&lt;analytic-geometry&gt;&lt;computational-geometry&gt;&lt;intersection&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="15867" PostTypeId="1" CreationDate="2010-12-30T00:21:19.133" Score="1" ViewCount="222" Body="&lt;p&gt;I am making a software that deals with employee trust. I am trying to make a math formula (no need to tell that I am bad in math :) ).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is the scenario:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Employee 1 trusts employee 2 = &amp;gt; 20% or 0.2 (average trust of employee 1 with employee 2)&#xA;Employee 2 deals with 10 customers =&amp;gt; 13%  (average - 10 customers trust on Employee 2)&#xA;Employee 1 also deals with 7 customers that customers belongs to Employee 2's 10 customers. =&amp;gt; 37% ( average 7 customers trust on employee 1)&#xA;&#xA;Employee 1 dealt with 10 customers for last 17 years.&#xA;Employee 2 dealt with 7 customers for last 1 years.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Now I want to calculate that how can we say that Employee 1 is much trust worthy than Employee 2? If I just see average than I can say Employee 1 but he only dealt with customers for 1 year. Now I want to make a generic formula to see which one is more trust worthy and I want to use all three values that are ( employee to employee, employee to customer, and number of year) to calculate average percentage. I want to increase or decrease an employee trust based on these values. Please remember I want to decrease as well for some employee I can not say that all employee or 90+ employee are trust worthy etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I did:&#xA;I just simply plus employee to employee and employee to customer trust weight&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Employee 1 = 0.57&#xA;Employee 2 = 0.33&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I don't know what to do for years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Current system just calculate employee to employee trust that is not good enough to trust itself. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not restricted to any programming language because I just want to make a generic formula. All ideas and suggestions are most welcome !&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you think my problem relates to any previously developed trust model then please do let me know (I already know web trust model but I don't know how to fit that in my situation)&lt;/p&gt;&#xA;" OwnerUserId="5223" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-30T01:07:25.573" LastActivityDate="2010-12-30T03:31:56.433" Title="Simple math formula to calculate average" Tags="&lt;mathematical-modeling&gt;" AnswerCount="1" />
  <row Id="15868" PostTypeId="1" CreationDate="2010-12-30T00:26:48.470" Score="2" ViewCount="122" Body="&lt;p&gt;[sorry for the bad english]&#xA;I am fond of astronomy and environment.&#xA;I wanna try to make a &quot;light pollution map&quot; but I haven't my satellites... so I use as approximation of light pollution the cities' population.&#xA;Let say we have for each city C citizens, each one spreads an average of X watt of electricity for lightning ( I have these data ).&#xA;Skip the units ( I need just a rough dimensionless &quot;light power&quot; ) :&#xA;city light power = C * X&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a map, with many cities.&#xA;I know light power is inverse proportional to the distance^2.&#xA;I don't know about sky, air diffraction, clouds reflections.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Start from the simplest model. A flat terrain map.&#xA;N light sources, every one &#xA;   at position X(n), Y(n)&#xA;   has a specific &quot;total light power&quot; = C(n) * X(n) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;At a specific point of coordinates x,y wich is the light power, sum of all the cities light ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried to calculate and plot, but it seems weird ( too far from some real satellite night shot ) and too slow to calculate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please help !&lt;/p&gt;&#xA;" OwnerUserId="5222" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-30T00:54:20.593" LastActivityDate="2010-12-31T18:21:27.527" Title="sum of N light sources" Tags="&lt;geometry&gt;&lt;physics&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="15869" PostTypeId="1" AcceptedAnswerId="15874" CreationDate="2010-12-30T00:27:07.223" Score="7" ViewCount="392" Body="&lt;p&gt;I've been reading On Numbers and Games and I noticed that Conway defines addition in his number system in terms of addition. Similarly in the analysis and logic books that I've read (I'm sure that this is not true of all such books) how addition works is assumed. From what I understand the traditional method of building the number system begins with the natural numbers (and zero)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$0:=|\emptyset|$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$1:=|\{\emptyset\}|$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$2:=|\{\emptyset,\{\emptyset\}\}|$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and so forth. In this construction addition could(?) be defined as the disjoint union of the sets associated with the two numbers. Then the integers could be defined as the additive inverse and so forth. Is this the ideal way to do it though, is there a more elegant method?&lt;/p&gt;&#xA;" OwnerUserId="1930" LastActivityDate="2010-12-30T19:17:53.283" Title="How is addition defined?" Tags="&lt;logic&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="5" />
  <row Id="15870" PostTypeId="2" ParentId="15868" CreationDate="2010-12-30T00:51:30.757" Score="2" ViewCount="" Body="&lt;p&gt;You may be confusing the energy distribution received by a satellite with the energy present at a place on the ground.  The  satellite is above the ground and records the amount of energy received as a function of direction (equivalently as a function of location on the ground) but doesn't add all the cities together.  So it would make a map with light at each city location measured as C*X/r^2.  This should match roughly with the photos you see, though different cities will have very different light/person.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At a location on the ground (in your model) you receive light from all other cities, again in the amount C*X/r^2 from each city, but there is no particular reason to expect this sum to match the satellite photos as the light is traveling along the ground, not up into space.  If you imagine a ring of cities in a circle, there would be a lot of light at the center, but the satellite would see nothing coming from there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is surprising the calculation takes a long time.  How many locations are you calculating the power at?  Even if you have a few hundred cities in the list and are calculating at thousands of locations, that is a few million.  Did you get trapped by the curse of dimension and take a thousand points in each of the x and y directions, which would make hundreds of millions of calculations?&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-12-30T00:51:30.757" />
  <row Id="15871" PostTypeId="2" ParentId="15869" CreationDate="2010-12-30T01:07:36.520" Score="9" ViewCount="" Body="&lt;p&gt;Jacob, there are two ways of defining addition of natural numbers in set theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first is the one you indicate: The sum of two numbers is the size of their disjoint union. In general, you define &lt;em&gt;cardinal addition&lt;/em&gt; this way: If $\kappa$ and $\lambda$ are cardinals (sizes of sets), say $\kappa=|A|$ and $\lambda=|B|$, then $\kappa+\lambda=|A\sqcup B|$, where $\sqcup$ denotes disjoint union. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second way is to define &lt;em&gt;ordinal addition&lt;/em&gt;, considering natural numbers as ordered sets: $0=\emptyset$, $1=\{0\}$, $2=\{0,1\}$, etc, with $n=\{0,\dots,n-1\}$ ordered in the usual way, which actually corresponds to membership: $a&amp;lt;b$ iff $a\in b$. Here we identify two linearly ordered sets if they are order isomorphic. The relevant notion here is that of sum of ordered sets: If $(A,&amp;lt;)$ and $(B,\prec)$ are linear orders, their sum $A+B$ is the set $A\sqcup B$ ordered by $a$ is less than $b$ iff $a,b\in A$ and $a&amp;lt;b$, or $a,b\in B$ and $a\prec b$, or $a\in A$ and $b\in B$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One can check that these two notions coincide when $A$, $B$ are finite sets (i.e., when we look at $n+m$ for $n,m$ finite). One has to be careful, because the notions are different in general for infinite sets. For example, if $\aleph_0=|{\mathbb N}|$, then $\aleph_0+1=1+\aleph_0=\aleph_0$. However, if we think of $\aleph_0$ as the ordered set $\{0&amp;lt;1&amp;lt;\dots\}$, then $1+\aleph_0=\aleph_0$ (remember, we identify sets if they are order isomorphic). However, $\aleph_0+1$ is strictly larger than $\aleph_0$ as ordered sets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a third option in the context of arithmetic: All we need is the notion of successor. The successor of 0 is 1, of 1 is 2, etc. Denote by $S(n)$ the successor of $n$ (in general, $S(n)=n+1$, of course). Note that &quot;successor of $n$&quot; is an order-theoretic notion: The smallest number larger than $n$. Then we define $n+m$ recursively: $n+0=n$, and $n+S(m)=S(n+m)$.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;We can go on, and define, for example, multiplication and exponentiation in these three ways as well. Cardinal multiplication is the size of the Cartesian product. Cardinal exponentiation $|A|^{|B|}$ is the size of the set of functions from $B$ to $A$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ordinal multiplication $\alpha\beta$ means: &quot;$\beta$ copies of $\alpha$&quot;. So $\aleph_02=\aleph_0+\aleph_0$ while $2\aleph_0=\aleph_0$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ordinal exponentiation is a bit trickier to define: For ordered sets $\alpha,\beta$ such that $\alpha$ has a minimum $0$, define $F(\alpha,\beta)$ as the set consisting of those functions $f:\beta\to\alpha$ such that there are only finitely many $\xi$ such that $f(\xi)\ne0$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For functions $f,g$ in $F(\alpha,\beta)$ set $f\triangleleft g$ iff $f(\xi)&amp;lt;g(\xi)$ (in the order of $\alpha$) for $\xi$ largest (in the order of $\beta$) such that $f(\xi)\ne g(\xi)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then $\alpha^\beta$ is defined as the order type of $(F(\alpha,\beta),\triangleleft)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again, these notions coincide for finite numbers, but differ wildly for infinite sets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The recursive definitions are given by $n\cdot 0=0$ and $n\cdot S(m)=nm+n$, and $n^0=1$ (even if $n=0$) and $n^{S(m)}=n^m\cdot n$.&lt;/p&gt;&#xA;" OwnerUserId="462" LastActivityDate="2010-12-30T01:07:36.520" CommentCount="2" />
  <row Id="15873" PostTypeId="2" ParentId="15866" CreationDate="2010-12-30T01:11:11.620" Score="4" ViewCount="" Body="&lt;p&gt;This is an interesting conundrum, and what I will suggest is just one possible approach, by no means a definitive answer.  Perhaps what you could do is compute Bézier curve segment approximating&#xA;your arcs, and then compute the intersection between Bézier segments.&#xA;To compute the Bézier segment for an arc,&#xA;you need the tangent vector at one endpoint (the tangent at the other endpoint is obtained by symmetry).&#xA;So this approach reduces to finding a tangent angle at one endpoint.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let the angle subtended by an arc at the circle center be $\theta$.&#xA;If the arc's chord length is $c$ and its height $h$, then, if I've calculated correctly,&#xA;$$\frac{\theta}{2} = \cos^{-1} \left( \frac{c^2 - 4h^2}{c^2+4h^2} \right) \;.$$&#xA;This avoids calculating the circle radius $r = (c^2 + 4h^2)/(8h)$.&#xA;Now from $\theta$ you could compute the needed tangent angle, and from there&#xA;a Bézier segment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have not analyzed this to see if it is indeed robust.&lt;/p&gt;&#xA;" OwnerUserId="237" LastActivityDate="2010-12-30T01:11:11.620" CommentCount="7" />
  <row Id="15874" PostTypeId="2" ParentId="15869" CreationDate="2010-12-30T01:25:06.027" Score="14" ViewCount="" Body="&lt;p&gt;You don't have an a priori notion of cardinality, so you cannot really say things like &quot;$0:=|\emptyset|$&quot;. In fact, before you can define cardinals you usually define ordinals, and usually the definition of the naturals precedes the definition of the ordinals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The set-theoretic method begins by using the Axiom of Infinity, which states that there exists at least one inductive set; a set $X$ is &lt;em&gt;inductive&lt;/em&gt; if and only if (i) $\emptyset\in X$; and (ii) For all $x$, if $x\in X$, then $s(x) = x\cup\{x\}\in X$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, let $X$ be any inductive set. Then we define&#xA;$$\mathbb{N} = \cap\{ A\subseteq X\mid \text{$A$ is inductive}\}.$$&#xA;One can then prove that $\mathbb{N}$ is well defined (it does not depend on the choice of $X$) and satisfies &quot;Peano's Axioms&quot;:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$\emptyset \in\mathbb{N}$.&lt;/li&gt;&#xA;&lt;li&gt;If $x\in\mathbb{N}$, then $s(x) \in\mathbb{N}$.&lt;/li&gt;&#xA;&lt;li&gt;For all $x\in\mathbb{N}$, $s(x)\neq\emptyset$.&lt;/li&gt;&#xA;&lt;li&gt;For all $x,y\in\mathbb{N}$, if $s(x)=s(y)$, then $x=y$.&lt;/li&gt;&#xA;&lt;li&gt;If $S\subseteq \mathbb{N}$ is inductive, then $S=\mathbb{N}$.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;(In fact, 3 and 4 are just consequences of the definition of $s(x)$). We then define &quot;$0$&quot; to mean $\emptyset$, &quot;$1$&quot; to mean $s(0)$, &quot;$2$&quot; to mean $s(1)=s(s(0))$, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively, you can begin with Peano's Axioms. Here, there is a primitive notion called &quot;natural number&quot;, and a primitive symbol called $0$. We also have a primitive function $s$. The Peano Axioms would be:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$0$ is a natural number.&lt;/li&gt;&#xA;&lt;li&gt;If $n$ is a natural number, then $s(n)$ is a natural number.&lt;/li&gt;&#xA;&lt;li&gt;For all natural numbers $n$, $s(n)\neq 0$.&lt;/li&gt;&#xA;&lt;li&gt;For all natural numbers $n$ and $m$, if $s(n)=s(m)$, then $n=m$.&lt;/li&gt;&#xA;&lt;li&gt;&lt;em&gt;Axiom Schema of Induction.&lt;/em&gt; If $\Phi$ is a predicate such that $\Phi(0)$ is true, and for all $n$, $\Phi(n)\Rightarrow \Phi(s(n))$, then for all natural numbers $k$, $\Phi(k)$.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;(You can also begin with $1$ instead of $0$; I use $0$ because it then parallels the set-theoretic construction). We then define &quot;$1$&quot; to mean $s(0)$; and &quot;$2$&quot; to mean $s(1)=s(s(0))$, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We then need the Recursion Theorem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Recursion Theorem.&lt;/strong&gt; Given a set $X$, an element $a\in X$, and a function $f\colon X\to X$, there exists a unique function function $F\colon\mathbb{N}\to X$ such that $F(0)=a$ and $F(s(n)) = f(F(n))$ for all $n\in\mathbb{N}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once we have these definitions and theorem, we can start defining addition. Fix $n\in\mathbb{N}$. I'm going to define &quot;add $n$&quot;, $+_n\colon \mathbb{N}\to\mathbb{N}$ by letting&#xA;\begin{align*}&#xA;+_n(0) &amp;amp;= n,\\&#xA;+_n(s(m)) &amp;amp;= s(+_n(m)).&#xA;\end{align*}&#xA;Or, in usual notation,&#xA;\begin{align*}&#xA;n+0&amp;amp; = n,\\&#xA;n+s(m) &amp;amp;= s(n+m).&#xA;\end{align*}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With these definitions, we have:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Theorem.&lt;/strong&gt; For all $n\in\mathbb{N}$, $n+0=0+n=n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Proof.&lt;/em&gt; Let $S=\{n\in\mathbb{N}\mid n+0=0+n=n\}$. Note that $0\in S$, since $0+0 = 0$ by the definition of addition. Now assume that $k\in S$; that means that $k+0 = 0+k = k$. Then &#xA;$0+s(k) = s(0+k) = s(k)$ (first equality by the definition of addition with $0$, second by the induction hypothesis). And by the definition of addition with $s(k)$, we have $s(k)+0 = s(k)$. Therefore, $k\in S$ implies $s(k)\in S$. Thus, $S=\mathbb{N}$, as desired. &lt;strong&gt;QED&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Theorem.&lt;/strong&gt; For all $n\in\mathbb{N}$, $s(n)=n+1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Proof.&lt;/em&gt; Let $S=\{n\in\mathbb{N}\mid s(n)=n+1\}$. First, $0\in S$, since $s(0) = 1 = 0+1$, by the previous theorem. Assume that $k\in S$; that means that $s(k)=k+1$. Then $s(s(k)) = s(s(k)+0) = s(k)+s(0) = s(k)+1$. So $k\in S$ implies $s(k)\in S$, hence $S=\mathbb{N}$. &lt;strong&gt;QED&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Theorem.&lt;/strong&gt; For all $\ell,n,m\in\mathbb{N}$, $\ell+(m+n) = (\ell+m)+n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Proof.&lt;/em&gt; Fix $\ell$ and $m$. Let $S=\{n\in\mathbb{N}\mid \ell+(m+n)=(\ell+m)+n\}$. We have $0\in S$, since&#xA;$$\ell + (m+0) = \ell + m = (\ell+m) + 0.$$&#xA;Now assume that $k\in S$; that means that $(\ell+m)+k = \ell+(m+k)$. We prove that $s(k)\in S$. We have:&#xA;$$(\ell+m)+s(k) = s((\ell+m)+k) = s(\ell+(m+k)) = \ell+s(m+k) = \ell+(m+s(k)).$$&#xA;Thus, if $k\in S$ then $s(k)\in S$. Hence, $S=\mathbb{N}$. &lt;strong&gt;QED&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Lemma.&lt;/strong&gt; For all $n\in\mathbb{N}$, $1+n = n+1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Proof.&lt;/em&gt; Let $S=\{n\in\mathbb{N}\mid 1+n=n+1\}$. Then $0\in S$. Suppose that $k\in S$, so that $1+k = k+1 = s(k)$. Then we have:&#xA;$$1+s(k) = s(1+k) = s(k+1) = s(k+s(0)) = s(s(k+0)) = s(s(k)) = s(k)+1.$$&#xA;Thus, $S=\mathbb{N}$. &lt;strong&gt;QED&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Theorem.&lt;/strong&gt; For all $n,m\in\mathbb{N}$, $n+m=m+n$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Proof.&lt;/em&gt; Fix $m$, and let $S=\{n\in\mathbb{N}\mid m+n=n+m\}$. First, $0\in S$, since $m+0=0+m$. Also, $1\in S$ by the previous lemma. Now assume that $k\in S$. Then $m+k=k+m$. To show that $s(k)\in S$, we have:&#xA;\begin{align*}&#xA;m+s(k) &amp;amp;= s(m+k) = s(k+m) = (k+m)+1 = k+(m+1) = k+(1+m)\\&#xA;       &amp;amp;= (k+1)+m = s(k)+m.&#xA;\end{align*}&#xA;Thus, $S=\mathbb{N}$. &lt;strong&gt;QED&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And so on. We can then define multiplication similarly, by fixing $n$ and defining&#xA;\begin{align*}&#xA;n\times 0 &amp;amp;= 0\\&#xA;n\times s(m) &amp;amp;= (n\times m) + n,&#xA;\end{align*}&#xA;and prove the usual properties of multiplication inductively. Then we can define exponentiation also recursively: fix $n$; then&#xA;\begin{align*}&#xA;n^0 &amp;amp; = 1\\&#xA;n^{s(m)} &amp;amp;= n^m\times n.&#xA;\end{align*}&#xA;We later define the order among the natural numbers by &#xA;$$a\leq b\Longleftrightarrow \exists n\in\mathbb{N}(a+n=b),$$&#xA;and prove the usual properties.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Later, we can construct $\mathbb{Z}$ from $\mathbb{N}$, $\mathbb{Q}$ from $\mathbb{Z}$, $\mathbb{R}$ from $\mathbb{Q}$, $\mathbb{C}$ from $\mathbb{R}$, etc. See for example my &lt;a href=&quot;http://math.stackexchange.com/questions/14828/set-theoretic-definition-of-numbers/14842#14842&quot;&gt;answer to this previous question&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-30T19:17:53.283" LastActivityDate="2010-12-30T19:17:53.283" CommentCount="8" />
  <row Id="15875" PostTypeId="1" AcceptedAnswerId="15876" CreationDate="2010-12-30T02:00:23.237" Score="6" ViewCount="284" Body="&lt;p&gt;It is, of course, one of the first results in basic complex analysis that a holomorphic function satisfies the Cauchy-Riemann equations when considered as a differentiable two-variable real function. I have always seen the converse as: if $f$ is &lt;em&gt;continuously&lt;/em&gt; differentiable as a function from $U \subset \mathbb{R}^2$ to $\mathbb{R}^2$ and satisfies the Cauchy-Riemann equations, then it is holomorphic (see e.g. Stein and Shakarchi, or &lt;a href=&quot;http://en.wikipedia.org/wiki/Cauchy%E2%80%93Riemann_equations&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;). Why is the $C^1$ condition necessary? I don't see where this comes in to the proof below. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume that $u(x,y)$ and $v(x,y)$ are continuously differentiable and satisfy the Cauchy-Riemann equations. Let $h=h_1 + h_2i$. Then&lt;br&gt;&#xA;\begin{equation*}&#xA;u(x+h_1, y+h_2) - u(x,y) = \frac{du}{x} h_1 + \frac{du}{dy}h_2 + o(|h|)&#xA;\end{equation*}&#xA;and &#xA;\begin{equation*}&#xA;v(x+h_1, y+h_2) - v(x,y) = \frac{dv}{dx} h_1 + \frac{dv}{dy} h_2 + o(|h|).&#xA;\end{equation*}&#xA;Multiplying the second equation by $i$ and adding the two together gives &#xA;\begin{align*}&#xA;(u+iv)(z+h)-(u+iv)(z) &amp;amp;= \frac{du}{dx} h_1 + i \frac{dv}{dx} h_1 + \frac{du}{dy} h_2 + i \frac{dv}{dy} h_2 + o(|h|)\\&#xA; &amp;amp;= \left( \frac{du}{dx} + i \frac{dv}{dx} \right) (h_1+h_2i) + o(|h|).&#xA;\end{align*} &#xA;Now dividing by $h$ gives us the desired result. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does there exist a differentiable but not $C^1$ function $f: U \rightarrow \mathbb{R}^2$ which satisfies the Cauchy-Riemann equations and does NOT correspond to a complex-differentiable function?  &lt;/p&gt;&#xA;" OwnerUserId="1491" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-30T02:06:13.743" LastActivityDate="2010-12-30T02:37:18.010" Title="When is a function satisfying the Cauchy-Riemann equations holomorphic?" Tags="&lt;complex-analysis&gt;" AnswerCount="2" FavoriteCount="2" />
  <row Id="15876" PostTypeId="2" ParentId="15875" CreationDate="2010-12-30T02:15:39.497" Score="7" ViewCount="" Body="&lt;p&gt;See &lt;a href=&quot;http://www.jstor.org/pss/2321164&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt;. &lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2010-12-30T02:15:39.497" CommentCount="2" />
  <row Id="15877" PostTypeId="2" ParentId="15875" CreationDate="2010-12-30T02:23:21.447" Score="3" ViewCount="" Body="&lt;p&gt;There's also the &lt;a href=&quot;http://en.wikipedia.org/wiki/Looman%E2%80%93Menchoff_theorem&quot; rel=&quot;nofollow&quot;&gt;Looman–Menchoff theorem&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="589" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-12-30T02:37:18.010" LastActivityDate="2010-12-30T02:37:18.010" CommentCount="3" />
  <row Id="15878" PostTypeId="1" AcceptedAnswerId="15879" CreationDate="2010-12-30T02:26:46.793" Score="5" ViewCount="234" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Suppose that $f$ is differentiable at&#xA;  0, and that $f(0) = 0$. Prove that&#xA;  $f(x) = xg(x)$ for some function $g$&#xA;  which is continuous at 0.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is a problem from Spivak's &lt;em&gt;Calculus&lt;/em&gt;, namely problem 27 of Chapter 10. (This is not homework, but rather self-study.) I am not sure how to go about this proof. The hint given in the text is to consider that $g(x)$ can be written as $f(x)/x$, but this puzzles me, because then continuity of $g$ at 0 says that $\lim_{x \to 0} g(x) = g(0) = f(0)/0 = 0/0$. &lt;/p&gt;&#xA;" OwnerUserId="5226" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-12-30T02:32:02.060" LastActivityDate="2010-12-30T02:55:44.050" Title="Spivak's Calculus exercise. Chapter 10, Problem 27." Tags="&lt;calculus&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="15879" PostTypeId="2" ParentId="15878" CreationDate="2010-12-30T02:28:06.137" Score="7" ViewCount="" Body="&lt;p&gt;What is $\displaystyle \lim_{x \to 0} \frac{f(x) - f(0)}{x-0}$ ?&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-12-30T02:28:06.137" CommentCount="4" />
  <row Id="15880" PostTypeId="2" ParentId="15878" CreationDate="2010-12-30T02:37:31.503" Score="0" ViewCount="" Body="&lt;p&gt;Show that $f&amp;#39;(0) = g(0)$ from the the continuity of $g$. Note that $f&amp;#39;(0) = \lim\limits_{h \to 0} \frac{f(0+h)-f(0)}{h}$. &lt;/p&gt;&#xA;" OwnerUserId="2548" LastEditorUserId="2548" LastEditorDisplayName="" LastEditDate="2010-12-30T02:55:44.050" LastActivityDate="2010-12-30T02:55:44.050" />
  <row Id="15881" PostTypeId="1" CreationDate="2010-12-30T02:54:22.773" Score="10" ViewCount="261" Body="&lt;p&gt;(Motivation: I am going to be working with a high school student next week on long division, which is a subject I strongly dislike.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider:  $\frac{1110}{56}=19\frac{46}{56}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is really a super easy problem, since once you realize $56*20=1120$ its trivial to write out $1110=56*19+46$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can work out the long division for yourself if you want; needless to say it makes an otherwise trivial problem into a tedious, multi-step process.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Long division is an &quot;effective procedure&quot;, in the sense that a &lt;a href=&quot;http://en.wikipedia.org/wiki/Turing_machine&quot; rel=&quot;nofollow&quot;&gt;Turing&lt;/a&gt; machine could do any division problem once it's given the instructions for the long division procedure.  To put it another way, an effective procedure is one for which given any problem of a specific type, I can apply this procedure systematically to this type of problem, and always arrive at a correct solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are my questions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Are there other distinct effective procedures for doing division problems besides long division?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Is there a way to measure how &lt;em&gt;efficient&lt;/em&gt; a given effective procedure is for doing division problems?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) Does there exist an optimal effective procedure for division problems, in the sense that this procedure is the most efficient?&lt;/p&gt;&#xA;" OwnerUserId="1417" LastEditorUserId="1417" LastEditorDisplayName="" LastEditDate="2010-12-30T03:13:39.970" LastActivityDate="2011-01-02T20:58:22.513" Title="Is long division the most optimal &quot;effective procedure&quot; for doing division problems?" Tags="&lt;arithmetic&gt;" AnswerCount="2" CommentCount="10" />
  <row Id="15882" PostTypeId="2" ParentId="15881" CreationDate="2010-12-30T03:13:02.873" Score="3" ViewCount="" Body="&lt;p&gt;1) yes, &lt;a href=&quot;http://en.wikipedia.org/wiki/Division_%28digital%29#Fast_division_methods&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Division_%28digital%29#Fast_division_methods&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) yes, &lt;a href=&quot;http://en.wikipedia.org/wiki/Big_O_notation&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Big_O_notation&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) ?, This is an open problem.&lt;/p&gt;&#xA;" OwnerUserId="5114" LastActivityDate="2010-12-30T03:13:02.873" CommentCount="4" />
  <row Id="15883" PostTypeId="2" ParentId="15867" CreationDate="2010-12-30T03:31:56.433" Score="1" ViewCount="" Body="&lt;p&gt;The solution to problems like this always starts with clear and detailed thinking about what data you have and what you want to do with it.  You seem to have 1) cross rating by employees of each other, 2) number of customers served, and 3)years of service.  One approach is to decide what weight you want to give each of these.  Say you want the cross rating to be 50% of the weight-you might scale it to a range of 0-50, scale the number of customers to 0-30 and scale experience to 0-20 and add up the points.  There are several issues here, however.  Maybe all your employees like each other and rate each other between 45 and 50.  Then it only contributes 5% of the grade, not 50.  Maybe one employee knows what you are doing and rates all the rest 0 so s/he will stand out.  Maybe one employee just grades harder than all the rest.  Is experience good, or do you want customers handled/year (in which case experience counts negatively)?  Maybe you want customers handled to &quot;decay&quot;-you get full credit for the ones this year, 80% for ones last year, and decreasing credit further in the past.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are quite right this is independent of programming language-they can all do this, once we know what this is.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-12-30T03:31:56.433" CommentCount="3" />
  <row Id="15884" PostTypeId="1" AcceptedAnswerId="15886" CreationDate="2010-12-30T03:48:08.577" Score="5" ViewCount="91" Body="&lt;p&gt;I could swear I had a formula for this years ago in school, but I'm having trouble tracking it down. The problem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have 3 red balls and 3 black balls in a basket. I draw them out one at a time. How many different sequences of six balls can I get, i.e.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;rbrbrb&#xA;rrrbbb&#xA;bbbrrr&#xA;etc...&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I'm looking for a general formula for n red and m black. This is not a homework, just my aging brain trying to recover an abstract formula.&lt;/p&gt;&#xA;" OwnerUserId="5227" LastActivityDate="2010-12-30T03:54:06.963" Title="Permutation with Duplicates" Tags="&lt;combinatorics&gt;&lt;permutations&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="15885" PostTypeId="2" ParentId="15866" CreationDate="2010-12-30T03:50:53.597" Score="5" ViewCount="" Body="&lt;p&gt;Have you considered finding the intersections using an implicit form for the circles, $$\frac{x^2}{r^2} + \frac{y^2}{r^2} + ax + by + c = 0?$$ This representation doesn't have any coefficients that diverge as the circle approaches a straight line. To find intersections, you'll have to solve a quadratic equation whose leading coefficient could be zero or arbitrarily close to it, but &lt;a href=&quot;http://en.wikipedia.org/wiki/Quadratic_equation#Alternative_formula&quot; rel=&quot;nofollow&quot;&gt;the alternative form of the quadratic formula&lt;/a&gt; should be able to deal with that robustly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You'll then have to do some jiggery-pokery to figure out whether the intersection points lie within the arcs. If the arc's bending angle is smaller than $\pi$, a projection onto the line joining the endpoints will suffice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Disclaimer: While all of this feels like it should work, I haven't analyzed it in any detail. Also, there could still be a problem when the circle is close to a line and you want the &lt;em&gt;longer&lt;/em&gt; arc. But I can't imagine that's a case that would turn up in any practical application.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update:&lt;/strong&gt; For a concrete example, here is the equation for a circular arc passing through the three points $(0,0)$, $(0.5, h)$, and $(1,0)$: $$\kappa^2 x^2 + \kappa^2 y^2 - \kappa^2 x - 2\eta y = 0,$$ where $$\begin{align}\kappa &amp;amp;= \frac{8h}{4h^2 + 1}, \\ \eta &amp;amp;= \frac{8h(4h^2-1)}{(4h^2+1)^2}.\end{align}$$ As you can see, the coefficients remain bounded as $h \to 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update 2:&lt;/strong&gt; Wait, that equation becomes trivial if $h = 0$, which is bad. We really want something like $x^2/r + y^2/r + ax + by + c,$  i.e. multiply the previous expression through by $r$. Then for the same example, our equation becomes $$\kappa x^2 + \kappa y^2 - \kappa x - 2\eta&amp;#39; y = 0,$$ where $\eta&amp;#39; = (4h^2-1)/(4h^2+1)$. Here are some explicit values.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$h = 1/2$: $$2 x^2 + 2 y^2 - 2 x = 0,$$ $h = 0.01$: $$0.07997 x^2 + 0.07997 y^2 - 0.07997 x + 1.998 y = 0,$$ $h = 0$: $$2 y = 0.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the way, in this format, the linear terms will always be simply $-2(x_0/r)x$ and $-2(y_0/r)y$, where the center of the circle is at $(x_0,y_0)$. As the center goes to infinity but the endpoints remain fixed, these coefficients remain bounded and nonzero (i.e. not both zero).&lt;/p&gt;&#xA;" OwnerUserId="856" LastEditorUserId="856" LastEditorDisplayName="" LastEditDate="2010-12-30T22:23:41.340" LastActivityDate="2010-12-30T22:23:41.340" CommentCount="11" />
  <row Id="15886" PostTypeId="2" ParentId="15884" CreationDate="2010-12-30T03:54:06.963" Score="8" ViewCount="" Body="&lt;p&gt;Place the $n+m$ balls in a row; the pick which $m$ you want to be red. You have $\binom{n+m}{m}=\frac{(n+m)!}{n!m!}$ possible ways of doing it, so that is the formula in this case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's an alternative way of thinking about it: place the $n$ red in a row; now you need to decide where to insert the $m$ black. What you want to do is &lt;em&gt;choose&lt;/em&gt; the locations of the $m$ black balls; there are $n+1$ possible locations (before all the red balls, in the $n-1$ spaces between red balls, and after all the red balls). You want to pick those allowing repetitions, and without regard to the order in which you pick them (what matters is how many times you pick each gap).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The number of ways in which you can select $r$ items from $k$ possibilities, allowing repetitions and without regard to order (combinations with repetitions) is $\binom{k+r-1}{r}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So here you want to select $m$ out of $n+1$ possibilities, it gives $\binom{n+m}{m} = \frac{(n+m)!}{m!n!}$, same as before.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To see a derivation of the formula, see for example &lt;a href=&quot;http://en.wikipedia.org/wiki/Combinations#Number_of_combinations_with_repetition&quot; rel=&quot;nofollow&quot;&gt;Wikipedia's article on combinations&lt;/a&gt;. This is sometimes also called the &quot;stars and bars problem&quot;, so you can find a proof in the corresponding &lt;a href=&quot;http://en.wikipedia.org/wiki/Stars_and_bars_%28probability%29&quot; rel=&quot;nofollow&quot;&gt;page&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-12-30T03:54:06.963" CommentCount="2" />
  <row Id="15887" PostTypeId="1" CreationDate="2010-12-30T04:26:28.060" Score="4" ViewCount="53" Body="&lt;p&gt;Can you please let me know if my &quot;proof&quot; is correct: ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $Y$ be a Hausdorff space and assume that each $y \in Y$ has a neighborhood $V$ such that $\overline{V}$ is regular. Prove $Y$ is regular.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proof:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to show that $Y$ is regular we are going to show that for each $y \in Y$ and each open set $U$ which contains $y$ there exists an open neighborhood $W$ of $y$ such that $y \in W \subseteq \overline{W} \subseteq U$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First observation: since regularity is hereditary then $V$ is regular. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let $y \in Y$ and $U$ a neighborhood of $y$. By assumption there exists a neighborhood $V$ of $y$ such that $\overline{V}$ is regular and hence $V$ is regular. Observe $U \cap V$ is a non-empty subset of $V$ and this set is open in $V$ as well. Since $V$ is regular we can find a non-empty subset $S$ of $V$ such that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$S \subseteq \overline{S} \subseteq U \cap V \subseteq U$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But since $V$ is open then $S$ is open in $X$ so the same set $S$ given by regularity of $V$ shows that $U$ is regular as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is the above incorrect?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you.&lt;/p&gt;&#xA;" OwnerUserId="5229" LastActivityDate="2010-12-30T05:08:27.027" Title="Showing a space is regular by knowing each point has a nbd whose closure is regular" Tags="&lt;topology&gt;" AnswerCount="2" />
  <row Id="15888" PostTypeId="2" ParentId="15750" CreationDate="2010-12-30T04:47:52.143" Score="5" ViewCount="" Body="&lt;p&gt;(This was intended to be yet another comment, but it got too big for the comment box.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I finally got a chance to look at the venerable &lt;a href=&quot;http://books.google.com/books?id=W6GV17SP0kQC&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;Handbook of Elliptic Integrals for Engineers and Scientists&lt;/em&gt; by Byrd and Friedman&lt;/a&gt;, and there is indeed a short discussion there on pseudoelliptic integrals, which I shall quote in part:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;One frequently encounters integrals that have all the appearances of being&#xA;  elliptic but which can ultimately be expressed solely in terms of&#xA;  elementary functions. These ... are called &lt;em&gt;pseudoelliptic integrals&lt;/em&gt;, and&#xA;  can always be evaluated as&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$$v_0(t)+\sum_{j=1}^m b_j\ln\;v_j(t)$$&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;where the $b_j$ are constants and the $v_j(t)$ are algebraic functions. &lt;strong&gt;One&#xA;  need not, however, be too concerned at the outset about whether or not his&#xA;  integral at hand is genuinely elliptic, because the substitutions&#xA;  given in this book for the reduction and evaluation of elliptic integrals&#xA;  will lead to the right results (without additional labor) even though&#xA;  the integral might turn out to be elementary.&lt;/strong&gt; We give several important&#xA;  examples.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;...&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;...the more general integral&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$$\int\frac{t\cdot R(t^2)}{\sqrt{at^4+bt^2+c}}\mathrm dt$$&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;($R(u)$ a rational function in $u$) ... reduces to integrals involving odd&#xA;  powers of Jacobi elliptic functions, and can thus always be ultimately&#xA;  evaluated in terms of elementary functions.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;A general case of a less obvious pseudoelliptic integral is&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$$\int \frac{R(t^2)}{\sqrt{(1-t^2)(1-k^2 t^2)}}\mathrm dt=\int \frac{R(\sin^2\theta)}{\sqrt{1-k^2\sin^2\theta}}\mathrm d\theta=\int R(\mathrm{sn}^2(u,k))\mathrm du$$&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;when... for all values of $t$ any one of the following relations holds:&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$$\begin{cases}R(t^2)+R\left(\frac1{k^2 t^2}\right)&amp;amp;=0\\R(t^2)+R\left(\frac{1-k^2 t^2}{k^2(1-t^2)}\right)&amp;amp;=0\\R(t^2)+R\left(\frac{1-t^2}{1-k^2 t^2}\right)&amp;amp;=0\end{cases}$$&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(emphasis mine)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since it's a handbook intended for use by nonspecialists, there's not much discussion of the conditions where an &quot;elliptic integral&quot; is in fact pseudoelliptic; I'd imagine advanced textbooks on the elliptic integrals would discuss the matter, but I have not been able to find any.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-12-30T04:47:52.143" CommentCount="2" />
  <row Id="15889" PostTypeId="2" ParentId="15887" CreationDate="2010-12-30T04:57:25.583" Score="2" ViewCount="" Body="&lt;p&gt;There is a problem with your argument.  The closure of $S$ in $V$ might not be closed in $X$.  I can see why you went to $V$, because you wanted $S$ to be open.  However, if you work instead in $\overline V$, then the closure of $S$ will remain closed in $X$, and since $x$ is in the interior of $\overline V$, you'll still be able to find an open set that does what you want.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have to use the regularity of $\overline V$, and not just of $V$, because there are examples of nonregular spaces in which every point has a regular neighborhood, like the &lt;a href=&quot;http://en.wikipedia.org/wiki/Non-Hausdorff_manifold#Line_with_two_origins&quot; rel=&quot;nofollow&quot;&gt;line with 2 origins&lt;/a&gt;. &lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-12-30T04:57:25.583" />
  <row Id="15890" PostTypeId="1" CreationDate="2010-12-30T05:00:13.047" Score="10" ViewCount="562" Body="&lt;p&gt;What books would you recommend to learn physics, being a a Math major, from classical mechanics, electricity, etc. to modern physics?&lt;/p&gt;&#xA;" OwnerUserId="1971" LastEditorUserId="2548" LastEditorDisplayName="" LastEditDate="2010-12-30T05:50:25.637" LastActivityDate="2011-01-02T19:22:30.967" Title="Books to learn physics, being a math major." Tags="&lt;soft-question&gt;&lt;reference-request&gt;&lt;physics&gt;" AnswerCount="6" CommentCount="5" FavoriteCount="7" />
  <row Id="15891" PostTypeId="2" ParentId="15887" CreationDate="2010-12-30T05:08:27.027" Score="0" ViewCount="" Body="&lt;p&gt;@Jonas Meyer: Thank you. I still have trouble and get confused using subspace topology of $\overline V$. Can you please help some more? (not homeowork, it is for fun).&lt;/p&gt;&#xA;" OwnerUserId="5230" LastActivityDate="2010-12-30T05:08:27.027" CommentCount="1" />
  <row Id="15892" PostTypeId="2" ParentId="15857" CreationDate="2010-12-30T05:28:39.890" Score="1" ViewCount="" Body="&lt;p&gt;Thanks for your answers. Geometry Revisited it is. I already started to read it.&lt;/p&gt;&#xA;" OwnerUserId="5119" LastActivityDate="2010-12-30T05:28:39.890" CommentCount="1" />
  <row Id="15893" PostTypeId="2" ParentId="15890" CreationDate="2010-12-30T05:40:55.440" Score="2" ViewCount="" Body="&lt;ul&gt;&#xA;&lt;li&gt;Kleppner/Kolenkow's  &lt;em&gt;An Introduction to Mechanics&lt;/em&gt;&lt;/li&gt;&#xA;&lt;li&gt;David Morin's &lt;em&gt;Introduction to Classical Mechanics&lt;/em&gt;&lt;/li&gt;&#xA;&lt;li&gt;Purcell's &lt;em&gt;Electricity and Magnetism&lt;/em&gt; &lt;/li&gt;&#xA;&lt;li&gt;A.P. French's &lt;em&gt;Principles of Modern Physics&lt;/em&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;These books have more math than the typical introductory physics textbooks. But they also introduce the &quot;physicists&quot; way of thinking (e.g. how to gain physical intuition). &lt;/p&gt;&#xA;" OwnerUserId="2548" LastEditorUserId="2548" LastEditorDisplayName="" LastEditDate="2010-12-30T05:47:38.430" LastActivityDate="2010-12-30T05:47:38.430" />
  <row Id="15894" PostTypeId="2" ParentId="15890" CreationDate="2010-12-30T05:45:38.117" Score="4" ViewCount="" Body="&lt;p&gt;I will assume you are getting started - once you finish the equivalent of the first year of calculus-based physics, then you should dig in / branch out depending on your interests (optics, dynamics, modern physics - i.e. 20th century - E&amp;amp;M, etc.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For an overview of it all, I recommend &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0201021153&quot; rel=&quot;nofollow&quot;&gt;the Feynman Lectures on Physics&lt;/a&gt;, supplemented liberally with any of the solid textbooks (such as the following) and I suppose some Wikipedia.  I like the Feynman lectures because they are entertaining, conversational, and provide a great insight into physics.  Here are some example introductory texts - I can personally vouch for HLR.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0471216437&quot; rel=&quot;nofollow&quot;&gt;Fundamentals of Physics - Halliday, Resnick, &amp;amp; Walker&lt;/a&gt;. Link is to current edition, but this is the book that many of us have used for decades, including myself at a small liberal arts school.  It's also the book used at Stanford.  I think you can use any edition from the last 10-20 years...&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0534408427&quot; rel=&quot;nofollow&quot;&gt;Physics for Scientists and Engineers - Serway and Jewett&lt;/a&gt; (CalTech's intro book)&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0321500628&quot; rel=&quot;nofollow&quot;&gt;University Physics - Young, Hugh, Ford and Freedman&lt;/a&gt; (one of MIT's intro books)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Frankly, I recommend you go to the syllabus of the introductory, calculus-based physics from a school you respect and select that textbook as a reference.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I did a physics major as an undergraduate, in addition to my studies in mathematics and engineering.  I still find that the first year's material is the most useful content, although I think that the concepts covered in introductory Modern Physics courses are really cool.  If you are interested, in addition to the Feynman lectures that I have read on the side, I was quite impressed with the text we used: &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0534493394&quot; rel=&quot;nofollow&quot;&gt;Modern Physics - Serway, Moses, Moyer&lt;/a&gt;.  While remaining very readable, it provides more of a textbook approach, whereas Feynman is a lecture style.&lt;/p&gt;&#xA;" OwnerUserId="4628" LastEditorUserId="4628" LastEditorDisplayName="" LastEditDate="2011-01-01T20:29:55.250" LastActivityDate="2011-01-01T20:29:55.250" CommentCount="4" />
  <row Id="15895" PostTypeId="2" ParentId="15890" CreationDate="2010-12-30T05:56:20.563" Score="5" ViewCount="" Body="&lt;p&gt;As for mechanics I'd recommend:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Stefan Banach - Mechanics (1951)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Arnold VI - Mathematical Methods of Classical Mechanics&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also recommend Griffiths books:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Griffiths - Introduction to Electrodynamics&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Griffiths - Introduction to Quantum Mechanics&lt;/p&gt;&#xA;" OwnerUserId="5119" LastActivityDate="2010-12-30T05:56:20.563" CommentCount="1" />
  <row Id="15896" PostTypeId="1" CreationDate="2010-12-30T06:06:41.387" Score="4" ViewCount="166" Body="&lt;p&gt;I'm trying to figure out a generic way of determining a series of points on a Bézier curve where all the points are the same distance from their neighboring points. By distance I mean direct distance between the points not distance along the curve. See the image below&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/8PXtm.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've written a program that will solve this in an iterative fashion, but I'm wondering if there is a direct solution I could use.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My program starts by defining a circle of some initial radius (R) centered on the start point of the curve. It then intersects this circle with the curve to find the second point , which is R distance away from the start point). It then continues along the curve in this way finding points until it reaches the end of the curve. In most cases the distance between the last intersection point and the end point of the curve will not equal to R. The program then uses that difference to calculate a new value of R to try, and repeats the process.&lt;/p&gt;&#xA;" OwnerUserId="5029" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-30T06:18:47.933" LastActivityDate="2010-12-30T06:18:47.933" Title="Find points along a Bézier curve that are equal distance from one another" Tags="&lt;geometry&gt;&lt;bezier-curve&gt;" CommentCount="5" />
  <row Id="15897" PostTypeId="1" CreationDate="2010-12-30T06:16:48.620" Score="1" ViewCount="82" Body="&lt;p&gt;If we consider the unit square (i.e $[0,1] \times [0,1]$) with the lexicographic&#xA;order induced by ${\mathbb{R}}^{2}$, what are the limit points of the unit square?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My answer is: the set of limit points is equal to:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ [0,1]^{2} \setminus ((0,1) \times \{1\}))$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is my answer correct? If not, how do you find them? I get confused with this topology.&lt;/p&gt;&#xA;" OwnerUserId="5230" LastActivityDate="2010-12-30T09:28:10.357" Title="Finding limit points in lexicographic order topology" Tags="&lt;topology&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="15898" PostTypeId="1" AcceptedAnswerId="16009" CreationDate="2010-12-30T06:28:40.873" Score="2" ViewCount="213" Body="&lt;p&gt;I define the function $D(\mathbf M)$ for square matrices by taking a maximal set of all triples  $(\mathbf v_i,k_i,n_i)_{i \in I}$ that satisfy $(\mathbf{M}-k\mathbf{I})^n\mathbf{v}_i = \mathbf{0}$ with the $\mathbf{v}_i$ linearly independent, and letting $$D(\mathbf M) = \prod_{i \in I} k_i.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example the matrix&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathbf X=\begin{pmatrix}0&amp;amp;1&amp;amp;1\\1&amp;amp;0&amp;amp;1\\1&amp;amp;1&amp;amp;0\end{pmatrix}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;has the set:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$((-1\quad 0\quad 1)^T,-1,1)$&lt;/li&gt;&#xA;&lt;li&gt;$((-1\quad 1\quad 0)^T,-1,1)$&lt;/li&gt;&#xA;&lt;li&gt;$((1\quad 1\quad 1)^T,2,1)$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;so $D(\mathbf X) = -1 \cdot -1 \cdot 2 = 2$. It is equal to the determinant but I don't want to use that.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;How can it be shown that $D(\mathbf A\mathbf B) = D(\mathbf A)D(\mathbf B)$?&lt;/p&gt;&#xA;" OwnerUserId="4290" LastEditorUserId="4290" LastEditorDisplayName="" LastEditDate="2010-12-31T03:20:07.590" LastActivityDate="2010-12-31T19:45:35.830" Title="Proof of multiplicativity" Tags="&lt;linear-algebra&gt;&lt;matrices&gt;" AnswerCount="1" CommentCount="16" />
  <row Id="15899" PostTypeId="1" AcceptedAnswerId="15900" CreationDate="2010-12-30T06:32:32.243" Score="1" ViewCount="264" Body="&lt;p&gt;What's the general form for $x^y$?&#xA;What's the specialized form for $x^1$ and $x^0$?&lt;/p&gt;&#xA;" OwnerUserId="1080" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-30T06:36:49.960" LastActivityDate="2011-01-03T03:00:47.787" Title="Square for $x^2$, Cube for $x^3$, Quartic for $x^4$, and what's for $x^1$?" Tags="&lt;algebra-precalculus&gt;&lt;terminology&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="15900" PostTypeId="2" ParentId="15899" CreationDate="2010-12-30T07:40:34.513" Score="5" ViewCount="" Body="&lt;p&gt;As J.M. and Rahul Narain pointed out in comments, $x^1$ is often referred to as &quot;linear&quot; and $x^0$ as &quot;constant.&quot;  Commonly, though certainly not universally, $x^y$ is a/the &quot;power&quot; with $x$ as the base and $y$ as the exponent.&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-12-30T07:40:34.513" CommunityOwnedDate="2010-12-30T07:40:34.513" />
  <row Id="15901" PostTypeId="2" ParentId="15897" CreationDate="2010-12-30T09:28:10.357" Score="1" ViewCount="" Body="&lt;p&gt;Every point $P = (a,b)$ on the square lies on some horizontal line segment $y = b$.  If $a \neq 0$ -- i.e., if $P$ is not the leftmost point on its horizontal line segment -- then you can build a sequence of points $(x_n,b)$ to the left of $a$ on that line segment converging to $P$.  If $a \neq 1$ -- i.e., if $P$ is not the rightmost point on its horizontal line segment -- then you can build a sequence of points $(x_n,b)$ to the right of $a$ on that line segment converging to $(a,b)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously $P$ can't be both the leftmost point &lt;em&gt;and&lt;/em&gt; the rightmost point, so the conclusion is...&lt;/p&gt;&#xA;" OwnerUserId="299" LastActivityDate="2010-12-30T09:28:10.357" />
  <row Id="15902" PostTypeId="1" AcceptedAnswerId="15910" CreationDate="2010-12-30T09:30:53.300" Score="15" ViewCount="602" Body="&lt;p&gt;This an interesting problem my friend has been working on for a while now (I just saw it an hour ago, but could not come up with anything substantial besides some PMI attempts).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's the full problem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $x_{1}, x_{2}, x_{3}, \cdots x_{y}$ be all the primes that are less than a given $n&amp;gt;1,n \in \mathbb{N}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Prove that $$x_{1}x_{2}x_{3}\cdots x_{y} &amp;lt; 4^n$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any ideas very much appreciated!&lt;/p&gt;&#xA;" OwnerUserId="3731" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-31T11:55:46.077" LastActivityDate="2011-03-10T15:39:02.550" Title="Show that product of primes, $\prod_{k=1}^{\pi(n)} p_k &lt; 4^n$" Tags="&lt;number-theory&gt;" AnswerCount="3" CommentCount="4" FavoriteCount="5" />
  <row Id="15903" PostTypeId="1" AcceptedAnswerId="15929" CreationDate="2010-12-30T09:44:14.633" Score="5" ViewCount="260" Body="&lt;p&gt;&lt;strong&gt;From &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0136020402&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;Introductory Combinatorics&lt;/em&gt;&lt;/a&gt; by Richard Brualdi&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have a chess master. He has 11 weeks to prepare for a competition so he decides that he will practice everyday by playing at least 1 game a day. To make sure that he's able to take a couple of breaks, he also decides that he wont play more than 12 games per week.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is, how do you prove/dis-prove that there will be a succession of (consecutive) days in which he will play &lt;em&gt;k&lt;/em&gt; games, &lt;em&gt;k&lt;/em&gt; &gt;= 22?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Specifically, I don't quite understand the use of the pigeonhole principle here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance! :D&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll try to clarify what I'm asking:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The goal is to prove that for &lt;strong&gt;every possible practice schedule&lt;/strong&gt; that the chess master makes, there will &lt;strong&gt;&lt;em&gt;always&lt;/em&gt;&lt;/strong&gt; be a consecutive sequence of days such that he plays &lt;strong&gt;exactly&lt;/strong&gt; &lt;em&gt;k&lt;/em&gt; games.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So for example, if &lt;em&gt;k&lt;/em&gt; = 22, how do you prove that no matter what his schedule is like, he will &lt;strong&gt;always&lt;/strong&gt; play &lt;strong&gt;exactly&lt;/strong&gt; &lt;em&gt;k&lt;/em&gt; games in &lt;em&gt;r&lt;/em&gt; days, 1 &amp;lt;= &lt;em&gt;r&lt;/em&gt; &amp;lt;= 77.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My main question is: what is the general method for proving different values of &lt;em&gt;k&lt;/em&gt;? Are there values of &lt;em&gt;k&lt;/em&gt; (from 1 to 132 inclusive) that are impossible to obtain?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that he does not necessarily have to play 132 games within 77 days!&#xA;Also, keep in mind the bounds of &lt;em&gt;k&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="5166" LastEditorUserId="5166" LastEditorDisplayName="" LastEditDate="2010-12-30T17:28:56.323" LastActivityDate="2010-12-30T19:28:23.440" Title="Chess Master Problem" Tags="&lt;combinatorics&gt;&lt;pigeonhole-principle&gt;" AnswerCount="2" CommentCount="9" />
  <row Id="15905" PostTypeId="2" ParentId="15869" CreationDate="2010-12-30T10:02:32.090" Score="1" ViewCount="" Body="&lt;p&gt;An extremely elegant way of defining numbers I want to mention does not use &lt;em&gt;sets&lt;/em&gt; but &lt;a href=&quot;http://en.wikipedia.org/wiki/Lambda-calculus&quot; rel=&quot;nofollow&quot;&gt;lambda calculus&lt;/a&gt;, i.e. functions and function application. The system used there is called &lt;a href=&quot;http://en.wikipedia.org/wiki/Church_encoding&quot; rel=&quot;nofollow&quot;&gt;Church encoding&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The idea goes the following: &lt;em&gt;Two&lt;/em&gt;, for example, means doing something for two times.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More precisely, when we have some operation (a function) and a value, we apply this function twice on this value. In lambda notation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ 2 \equiv \lambda f\,x \mapsto f (f\,x) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So generally, any number $n$ is defined as a function that takes another function and returns it's $n$th &lt;a href=&quot;http://en.wikipedia.org/wiki/Function_iteration&quot; rel=&quot;nofollow&quot;&gt;iterate&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we can simply define addition &lt;strong&gt;as a function composition&lt;/strong&gt;.&#xA;We first apply the function $n$ times, then $m$ times and thus we got a total of $n+m$ applications.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ n + m \equiv \lambda f\,x \mapsto n f (m f x)$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, we end up with&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ 2 + 3 \equiv \lambda f\,x \mapsto f (f (f (f (f\,x)))) \equiv 5 $$&lt;/p&gt;&#xA;" OwnerUserId="917" LastActivityDate="2010-12-30T10:02:32.090" />
  <row Id="15907" PostTypeId="2" ParentId="15902" CreationDate="2010-12-30T10:15:55.980" Score="1" ViewCount="" Body="&lt;p&gt;Note my comment above, also when you look a bit at the net you directly find proofs for this theorem. The easiest ones probably use induction, it is basically like this proof:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://mathrefresher.blogspot.com/2009/11/primorial.html&quot; rel=&quot;nofollow&quot;&gt;http://mathrefresher.blogspot.com/2009/11/primorial.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3123" LastActivityDate="2010-12-30T10:15:55.980" />
  <row Id="15908" PostTypeId="2" ParentId="15890" CreationDate="2010-12-30T11:51:45.237" Score="4" ViewCount="" Body="&lt;p&gt;If you want a solid and consistent approach to physics from the theoretical point of view (I assume that as a mathematician you don't want an experimentalist's point of view) then you can't go past the series by &lt;a href=&quot;http://en.wikipedia.org/wiki/Lev_Landau#Landau_and_Lifshitz_Course_of_Theoretical_Physics&quot; rel=&quot;nofollow&quot;&gt;Landau and Lifshitz&lt;/a&gt; - &lt;a href=&quot;http://en.wikipedia.org/wiki/Course_of_Theoretical_Physics&quot; rel=&quot;nofollow&quot;&gt;&quot;Course of Theoretical Physics&quot;&lt;/a&gt; (&lt;a href=&quot;http://www.amazon.com/Course-Theoretical-Physics-All-Volumes/lm/RMVH4G2YKW3NH&quot; rel=&quot;nofollow&quot;&gt;Amazon&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Mechanics.&lt;/li&gt;&#xA;&lt;li&gt;The Classical Theory of Fields.&lt;/li&gt;&#xA;&lt;li&gt;Quantum Mechanics: Non-Relativistic Theory.&lt;/li&gt;&#xA;&lt;li&gt;Relativistic Quantum Theory.&lt;/li&gt;&#xA;&lt;li&gt;Statistical Physics.&lt;/li&gt;&#xA;&lt;li&gt;Fluid Mechanics.&lt;/li&gt;&#xA;&lt;li&gt;Theory of Elasticity.&lt;/li&gt;&#xA;&lt;li&gt;Electrodynamics of Continuous Media.&lt;/li&gt;&#xA;&lt;li&gt;Statistical Physics, Part 2.&lt;/li&gt;&#xA;&lt;li&gt;Physical Kinetics.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="954" LastActivityDate="2010-12-30T11:51:45.237" CommentCount="6" />
  <row Id="15909" PostTypeId="2" ParentId="15425" CreationDate="2010-12-30T12:01:40.040" Score="2" ViewCount="" Body="&lt;p&gt;I assumed above that all coefficients are positive, but this isn't easiest to compute with, so apply the isomorphism $x \mapsto -x$, and now the coefficients are alternating. The volume is unchanged.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A basis for the differentials is $\{\omega_i : 1\le i \le g\}$, where&#xA;$$\omega_i = x^{i-1}\frac{dx}{2y}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $M$ large, there is a positive root $\alpha_1$ of size $M$, and the other $2g$ roots are small. We now need to figure out a basis for the real homology (loops fixed by complex conjugation). There are $g-1$ independent loops $\gamma_{j+1}$, the $j$-th loop being from $\alpha_{2j}$ to $\alpha_{2j+1}$ and back around the &quot;hole&quot; (for example, if all roots are real, these are simply the paths where the polynomial is positive). One more independent loop is from $\alpha_1$ to $\infty$ and back (being real, this is the easiest to understand).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The volume of the jacobian of the curve is $|det((\int_{\gamma_j} \omega_i)_{ij})|$. So let's approximate these integrals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For any $j&amp;gt;1$, since the roots involved are $O(1)$, the loop $\gamma_j$ is close to the origin. In this vicinity the root $\alpha_1$ contributes to $\int_{\gamma_j} \omega_i$ a factor of $M^{-1/2}$, and if we factor it out (recall that we only want a crude approximation), we are left with an integral close to the origin over a function that is relatively small. Hence $\int_{\gamma_j} \omega_i \approx M^{-1/2}$. Note that this is the same for any $i$ since close to the origin $x^{i-1}$ doesn't add much.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $j=1$, the integral can be approximated as&#xA;$$\int_{\gamma_1} \omega_i = 2\int_M^\infty \frac{x^{i-1} dx}{2y} \approx \int_M^{2M} \frac{x^{i-1} dx}{\sqrt{x^{2g+1}+\cdot a_0}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The small roots give each the denominator a factor of $M^{1/2}$, so we can approximate further to arrive at:&#xA;$$M^{-g} \int_M^{2M} \frac{x^{i-1} dx}{\sqrt{x-M}} \approx M^{1/2+i-1-g}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have approximated all entries of the matrix. The top row's largest coefficient is of size $M^{-1/2}$, and hence the largest product of a generalised diagonal is $(M^{-1/2})^g$, since the entries in the other rows are also of size $M^{-1/2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The crude approximation for the volume of the jacobian of a hyperelliptic curve given by a polynomial as above with positive coefficients of size $M$, a large number, is: $$Vol(J(\mathbb{R})) \approx M^{-g/2}$$&lt;/p&gt;&#xA;" OwnerUserId="4946" LastEditorUserId="4946" LastEditorDisplayName="" LastEditDate="2010-12-30T21:02:23.683" LastActivityDate="2010-12-30T21:02:23.683" />
  <row Id="15910" PostTypeId="2" ParentId="15902" CreationDate="2010-12-30T12:39:44.623" Score="37" ViewCount="" Body="&lt;p&gt;I think the following argument is from Erdős: The binomial coefficient &#xA;$$ {n \choose {\lfloor n/2 \rfloor}} = \frac{n!}{{\lfloor n/2 \rfloor}!{\lceil n/2 \rceil}!}$$&#xA;is an integer. Any prime in the range ${\lceil n/2 \rceil}+1 \le p \le n$ will appear in the  numerator but not in the denominator, as a consequence $n \choose {\lfloor n/2 \rfloor}$ is divisible by the product of all the primes in the range ${\lceil n/2 \rceil}+1 \le p \le n$. On the other hand if $n$ is even then it will be the central term in the binomial expansion of $(1+1)^n$ so &#xA;    $$ {n \choose {\lfloor n/2 \rfloor}} \le 2^n $$ &#xA;and if $n$ is odd then $n \choose \lfloor n/2 \rfloor$ and $n \choose \lceil n/2 \rceil$ will be the two central terms of the binomial expansion of $(1+1)^n$, as they are equal we have&#xA;    $$ {n \choose {\lfloor n/2 \rfloor}} \le 2^{n-1} $$ &#xA;we apply this recursively to $n$, $\lceil n/2 \rceil, \lceil n/4 \rceil, \dots$, but&#xA; $$ { \lceil n/2 \rceil \choose \lfloor \lceil n/2 \rceil/2 \rfloor } =   &#xD;&#xA;    { \lceil n/2 \rceil \choose \lfloor n/4 \rfloor } \le 2^{\lfloor n/2 \rfloor } $$&#xA;using either of the two preceding inequalities depending on the parity of $\lceil n/2 \rceil$, by the same argument we get&#xA;  $${ \lceil n/4 \rceil \choose \lfloor \lceil n/4 \rceil/2 \rfloor } ={ \lceil n/4 \rceil \choose \lfloor n/8 \rfloor } \le 2^{\lfloor n/4 \rfloor }$$&#xA;and so on, so &#xA;$$ \begin{align}{n \choose {\lfloor n/2 \rfloor}}{{\lceil n/2 \rceil} \choose {\lfloor n/4 \rfloor}}{{\lceil n/4 \rceil} \choose {\lfloor n/8 \rfloor}}\cdots  &amp;amp;\le 2^n\cdot 2^{{\lfloor n/2 \rfloor}} \cdot 2^{{\lfloor n/4 \rfloor}} \cdots \\\\&#xD;&#xA; &amp;amp;\le 2^{n + {\lfloor n/2 \rfloor} + {\lfloor n/4 \rfloor} + \cdots } \\\\ &#xD;&#xA; &amp;amp;\le 2^{n(1 + 1/2 + 1/4 + \cdots) } = 2^{2n} = 4^n\end{align}$$&#xA;but the left hand side is divisible by all the primes up to $n$. So the product of any subset of these primes will also be bounded by $4^n$.&lt;/p&gt;&#xA;" OwnerUserId="3274" LastEditorUserId="3274" LastEditorDisplayName="" LastEditDate="2010-12-31T10:44:19.510" LastActivityDate="2010-12-31T10:44:19.510" CommentCount="6" />
  <row Id="15912" PostTypeId="2" ParentId="15849" CreationDate="2010-12-30T13:17:01.220" Score="3" ViewCount="" Body="&lt;p&gt;I thouht I would add an extra answer that goes in a slightly different direction than the rest.&#xA;An alternative proof that a finite domain is a division ring goes as follows:&#xA;Look at different powers of $a$ (ie, $a^n$ for various positive numbers $n$). Now, since the ring is finite, you will at some point get to some element you have already been at before, so you have $a^n = a^m$ with $n &amp;lt; m$. Now we can use that it is a domain, so this leads to $a^{m-n} = 1$ which means that not only does $a$ have an inverse. his inverse is in fact a power of $a$ (and then clearly it commutes with $a$ so it is also an inverse from the other side).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This argument is less elegant than the usual one, but I find it is nice to know that the inverse of an element in a finite ring is actually a power of the element itself.&lt;/p&gt;&#xA;" OwnerUserId="2538" LastActivityDate="2010-12-30T13:17:01.220" />
  <row Id="15913" PostTypeId="1" CreationDate="2010-12-30T13:33:58.993" Score="0" ViewCount="96" Body="&lt;p&gt;What is the general advantage of a real sequence?&lt;/p&gt;&#xA;" OwnerUserId="5241" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-30T13:40:42.217" LastActivityDate="2010-12-30T13:40:42.217" ClosedDate="2010-12-30T17:00:55.820" Title="General advantage of a real sequence" Tags="&lt;real-analysis&gt;" CommentCount="5" />
  <row Id="15914" PostTypeId="2" ParentId="15881" CreationDate="2010-12-30T14:05:21.040" Score="4" ViewCount="" Body="&lt;p&gt;Methods for division and the other basic arithmetic operations are explained in Knuth's ACP vol. 2, Seminumerical Algorithms.  For the sake of humans doing longhand division, perhaps the most relevant point is the discussion of trial divisors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you will recall, when faced with a problem like $1110/56$ (asked to find the quotient and remainder, which amounts to giving the fraction in mixed form), the basic instruction is to use the leading digit of the divisor to &quot;guess&quot; (estimate) the next digit of the quotient.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is often the case that this &quot;guess&quot; will be an overestimate, i.e. that the trial divisor proves to be too big for the divisor as a whole.  For example, here when trying to tell how many times the divisor as a whole $56$ will go into the leading digits $111$ of the dividend, we would &quot;guess&quot; $2$ times, because leading digit $5$ certainly goes into $11$ two times.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This turns out, frustratingly, to be a wrong guess in that $56$ times two gives $112$ (over the limit of $111$), and our trial divisor has to be reduced by one and checked again.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Knuth's explanation covers arbitrary (fixed positive integer) radix &lt;em&gt;b&lt;/em&gt; &amp;gt; 1.  Reading down to the radix ten (decimal) case, the point would be this.  If the divisor's leading digit is &quot;normalized&quot; to be greater than  or equal to half the radix (at least 5 in the decimal case, as was true in the example $1110/56$), then the trial divisor obtained in the usual way (estimating by using the leading digit of the divisor) will be at worst $2$ more than the correct next digit in the quotient.  [It is suggested that the desired normalization can be arranged by repeatedly doubling both numerator and denominator as necessary, until the leading digit of the denominator (divisor) reaches at least $b/2$.]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An alternative method for getting the trial divisor is then explained that has a better worst case behavior.  If the next-to-leading digit of the divisor is at least $b/2$, then for the purpose of the trial divisor we round up the leading digit!  Now using that digit (or possible $b=10$ if the rounding takes us to the next &quot;place&quot; value) we will get an estimate for the next digit of the quotient that is off by at most $1$.  In the case where no rounding up took place, this means the trial divisor is either correct or too big by one, and in the case where we did round up, this means the trial divisor is either correct or too small by one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've seen a monograph where this alternative technique for longhand division was tried with a group of (if memory serves) high school math teachers, as a Master's Thesis in Education that someone did.  As it happens I'm in St. Augustine FL where I saw this monograph years ago in a used bookstore.  I'll go by there today and if it remains on the shelf post the reference details later.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added:&lt;/strong&gt;  I found the book (the bookstore itself had moved about five years ago!).  My memory of it was not perfect but here are some details:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;An Experimental Study of Two Methods of Long Division&lt;/em&gt; by Kenneth G. Fuller&lt;br&gt;&#xA;Bureau of Publications/Teachers College/Columbia University&#xA;New York, 1949 (hardback, 76 pages)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The author refers in his acknowledgments to Prof. Clifford B. Upton, and in an early footnote refers to a paper by the same:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Making Long Division Automatic,&quot; &lt;em&gt;Tenth Yearbook, National Council of Teachers of Mathematics&lt;/em&gt;, 1935  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;where the percentages of correcting estimated quotients using a) a trial divisor that consists simply of a leading digit, and b) a trial divisor that is rounded up just when the following divisor digit is &lt;em&gt;greater than&lt;/em&gt; five.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to this account Upton had shown method a) &quot;gives the correct quotient figure on the first trial in only 66.70 per cent of all cases.&quot;  With method b) one gets &quot;the correct quotient figure on the first trial in 80.43 per cent of all cases, a quotient figure requiring one correction in 19.29 per cent of all cases, and a quotient figure requiring two corrections in .28 per cent of all cases.&quot;  Method b) did not require &quot;more than two corrections&quot; in any circumstance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The study then proceeds to compare method b), the variant of our usual approach to trial divisors, and an &quot;experimental method&quot; that builds a table of multiples 1 through 9 of the actual divisor.  Fuller points out that the history of both methods can be followed back to Roberte Recorde's &lt;em&gt;Arithmetike&lt;/em&gt; (1579, orig. edition 1542).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Building the table of multiples eliminates &quot;trial divisors&quot; and is certainly economical if there are to be several digits in the quotient; table construction can be done by alternate doubling and adding in the divisor.  Fuller's approach entailed checking the table of multiples with casting out 9's.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fuller's comparison of methods was carried out with three fifth grade class during the school year 1946-47 in Connecticut.  If I have read the statistical summary properly, the approach using a table of multiples proved more accurate (less error prone), though not in a statistically significant way except for certain longer problems (more digits in the divisor or quotient).  The table of multiples also took longer for the students to do, though this was in part because of extra checks involved on the answer and in part because of the modest sizes of problems (2-3 digits in the divisor, 2-4 digits in the quotient).&lt;/p&gt;&#xA;" OwnerUserId="3111" LastEditorUserId="3111" LastEditorDisplayName="" LastEditDate="2011-01-02T20:58:22.513" LastActivityDate="2011-01-02T20:58:22.513" CommentCount="2" />
  <row Id="15915" PostTypeId="1" AcceptedAnswerId="15916" CreationDate="2010-12-30T14:35:54.507" Score="1" ViewCount="87" Body="&lt;p&gt;In some course notes on &quot;evolution equations in probability spaces (etc)&quot; I have the following definition:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A point $z \in \mathbb R^n$ is called the &lt;em&gt;approximate limit&lt;/em&gt; of $f$ at $x$ if for every $\epsilon &amp;gt; 0$ the set $\{y \in \Omega : |f(y) - z| \leq \epsilon\}$ has Lebesgue density 1 at $x$. Notation: $\tilde{f}(x) := f(y)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't understand this definition, what exactly does it say intuitively? The given set also does not depend on $x$... is there something wrong?&lt;/p&gt;&#xA;" OwnerUserId="1120" LastActivityDate="2010-12-30T15:32:27.037" Title="Approximate limit." Tags="&lt;measure-theory&gt;" AnswerCount="1" CommentCount="6" />
  <row Id="15916" PostTypeId="2" ParentId="15915" CreationDate="2010-12-30T15:04:37.843" Score="3" ViewCount="" Body="&lt;p&gt;I don't know if this will help your intuition, but unwinding the definition a bit, this says that for all $\epsilon\gt0$, for all $\eta\gt0$, there exists $\delta\gt0$ such that in $B_\delta(x)$ you have $|f(y)-z|\leq\epsilon$ outside of a subset of measure less than $\eta\cdot m(B_\delta(x))$.  For a fixed $\epsilon$, letting $\eta$ go to zero makes the relative size of the set where $|f(y)-z|\gt\epsilon$ go to zero (by taking sufficiently small neighborhoods of $x$).  You get an ordinary limit when you don't have to include the exception of a small subset where $|f(y)-z|$ might be larger than $\epsilon$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the simplest case, the limit may exist if you ignore the values of $f$ off a set of measure zero.  (E.g., take a continuous function but add $1$ to its values on a countable dense subset of the domain.)  In that case you can actually take $\eta=0$.  Perhaps more illustrative is the example of the function $f$ on $\mathbb{R}$ that is $1$ in an open interval of length $2^{-n}$ centered at $\frac{1}{n}$ for each positive integer $n$, and $0$ elsewhere.  Then the approximate limit of $f$ at $0$ is $0$, but for $\epsilon\lt1$ and $\delta\gt0$, the measure of the subset of $B_\delta(0)$ on which $|f(y)-0|\gt\epsilon$ is positive. &lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2010-12-30T15:32:27.037" LastActivityDate="2010-12-30T15:32:27.037" CommentCount="2" />
  <row Id="15917" PostTypeId="1" AcceptedAnswerId="15989" CreationDate="2010-12-30T15:08:38.103" Score="1" ViewCount="117" Body="&lt;p&gt;Hi, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This equation: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\hspace{1in}X=S\times(P_{FC}/P_{DC})$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can be rewritten in terms of percentage change over a time period as follows: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\hspace{1in}x=s+I_{FC}-I_{DC}=s-(I_{DC}-I_{FC})$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I actually rewrite this? I think it has something to do with logarithms because the multiplication becomes an addition and the division becomes a subtraction, but I think taking the log on both sides won't give me the answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example,since $x$ is a percentage change over a time period, I figured its this must be true:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\hspace{1in} x=\frac{X_{2}}{X_{1}}-1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But $\frac{X_{2}}{X_{1}}-1$ is definitely not equal to $\log X$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any hints would be much appreciated.&#xA;Thank you.&lt;/p&gt;&#xA;" OwnerUserId="2973" LastEditorUserId="2973" LastEditorDisplayName="" LastEditDate="2010-12-31T01:58:40.873" LastActivityDate="2010-12-31T12:27:01.150" Title="Rewriting an equation in percentage terms" Tags="&lt;homework&gt;&lt;algebra&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="15918" PostTypeId="1" AcceptedAnswerId="15931" CreationDate="2010-12-30T15:32:04.977" Score="0" ViewCount="175" Body="&lt;p&gt;Hi I am wondering how best to explain this, I am working with WebGL (effectively OpenGL) and I have the ray cast from clicking in 3d space. I have the far and near values of the ray cast, the camera position and the target it is looking at.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do I work out where the ray cast starts and ends, effectively to draw a line to show the ray cast through 3d space. The most confusing part of this for me is the angles, subtracting the angles of the cameras face which are also in radians.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Camera position would obviously be the start point, so I need to use that and using the direction the camera is facing, along with the near and far values of the ray to get the end point! I'm confused, please help.&lt;/p&gt;&#xA;" OwnerUserId="3849" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-12-30T21:41:31.307" LastActivityDate="2010-12-30T21:41:31.307" Title="Help with matrix mathematica" Tags="&lt;geometry&gt;&lt;matrices&gt;&lt;3d&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="15919" PostTypeId="2" ParentId="15917" CreationDate="2010-12-30T16:06:11.530" Score="3" ViewCount="" Body="&lt;p&gt;So $X = S \times (P_{FC}/P_{DC})$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then $\log X = \log S + (\log  P_{FC}- \log P_{DC})$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus $\log X = \log S-(\log P_{DC}- \log P_{FC})$. &lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2010-12-30T16:06:11.530" />
  <row Id="15920" PostTypeId="2" ParentId="15852" CreationDate="2010-12-30T16:44:32.323" Score="5" ViewCount="" Body="&lt;p&gt;I would recommend Rotman's book &quot;An Introduction to the Theory of Groups&quot; for basic material on word problems of groups, and for a reasonably accessible proof of the Novikov-Boone theorem which asserts the existence of finitely presented groups with unsolvable word problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The notion of the word problem of a group is generally applied to groups defined by a specific finite presentation, but it is not hard to prove that, for a group $G$, the solvability of the word problem of $G$ is independent of the given presentation of $G$, so it is customary simply to say that the word problem of $G$ is or is not solvable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lots of familiar classes of groups are known to have solvable word problem, including finite groups, finitely presented residually finite groups (proof given above by Bill Dubuque), automatic groups, finitely generated nilpotent groups.&#xA;But, by  a recent result of Olga Kharlampovich, there exist finitely presented solvable groups of derived length $3$ with unsolvable word problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the word problem is semi-decidable for all groups given by a recursively enumerable presentation, in the sense that if you are given a word in the generators  that does represent the identity element of the group, then it is possible to prove that it does. You can do that naively by systematically trying all possibilities for expressing the word as a product of conjugates of defining relators, but there are more efficient and practical methods of attempting this by using rewriting systems or coset enumeration.&lt;/p&gt;&#xA;" OwnerUserId="2820" LastActivityDate="2010-12-30T16:44:32.323" />
  <row Id="15921" PostTypeId="2" ParentId="15903" CreationDate="2010-12-30T16:50:27.170" Score="2" ViewCount="" Body="&lt;p&gt;He has $11$ weeks of training (or $77$ days). The minimum and maximum number of games for this period is $77$ and $11 \cdot 12=132$, respectively. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have distinct situations:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) At any period of $k$ consecutive days he plays at least $k$ games. As well you asked.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) The smallest period of $n$ days such that he plays $k=22$ games occur when he consider a week with a maximum number of games $12$ plus days after and before this week with a maximum possible number of games. Notice that in one day is possible to play $6$ games. So, take one day with $4$ games, one week with $12$ games, and one more day with $6$ games. You get then $22$ games in $9$ days.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using 1) and 2) we conclude that between $9$ and $k$ days we always can make the choice of $22$ games. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I guess it is very useful!&lt;/p&gt;&#xA;" OwnerUserId="2764" LastActivityDate="2010-12-30T16:50:27.170" />
  <row Id="15922" PostTypeId="2" ParentId="15829" CreationDate="2010-12-30T16:59:51.297" Score="5" ViewCount="" Body="&lt;p&gt;Let $G$ be the direct product of countably many copies of the dihedral group $D$ of order 6 (or, if you prefer, $D$ is the symmetric group $S_3$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can construct a Sylow $2$-subgroup of $G$ by choosing Sylow $2$-subgroups of each of the direct factors of $G$, and taking their direct product. Since $D$ has three Sylow $2$-subgroups, $G$ has uncountably many Sylow $2$-subgroups, so they cannot all be conjugate in the countable group $G$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we let $P$ and $H$ be non-conjugate Sylow $2$-subgroups of $G$, then there is no $g \in G$ such that $H \cap gPg^{-1} \in {\rm Syl}_2(H)$.&lt;/p&gt;&#xA;" OwnerUserId="2820" LastActivityDate="2010-12-30T16:59:51.297" CommentCount="3" />
  <row Id="15923" PostTypeId="1" AcceptedAnswerId="15926" CreationDate="2010-12-30T17:17:04.223" Score="4" ViewCount="162" Body="&lt;p&gt;If $\kappa$ is a measurable cardinal, and $\mathcal{U}$ is a normal ultrafilter which is $\kappa$-complete then $\mathcal{U}$ extends the club filter (i.e. every club is a member of $\mathcal{U}$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One result is that all the sets in the ultrafilter are stationary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;s&gt;Now, given a stationary set $S$ such that its complement in $\kappa$ is also stationary (i.e. there is no club subset in $S$) we can choose either to include $S$ or $\kappa\setminus S$ in an extension of an ultrafilter. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am stuck in showing that I can have a normal ultrafilter with either $S$ or $\kappa\setminus S$ for every stationary $S$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I have to prove something else, but it really reduces to this claim.)&lt;/s&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt;&lt;br&gt;&#xA;In light of how the above statement is false, I will just give the original question:&lt;br&gt;&#xA;I have to show that there can be a normal ultrafilter such that the set of measurable cardinals below $\kappa$ is not in the ultrafilter. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there aren't stationary many of them - then clearly it's true. And since the limit of $\omega$ measurable cardinals is singular (ergo non-measurable); the only nontrivial case is when there is a stationary set of measurable cardinals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I prefer hits over partial solutions, and partial solutions over complete solutions. Thanks.&lt;/p&gt;&#xA;" OwnerUserId="622" LastEditorUserId="622" LastEditorDisplayName="" LastEditDate="2010-12-30T17:43:58.610" LastActivityDate="2011-01-01T17:05:54.090" Title="Normal ultrafilters and Stationary sets" Tags="&lt;homework&gt;&lt;set-theory&gt;&lt;large-cardinals&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="15924" PostTypeId="2" ParentId="15868" CreationDate="2010-12-30T17:20:41.823" Score="2" ViewCount="" Body="&lt;p&gt;I'm not sure what you mean by &quot;light power,&quot; but &lt;em&gt;intensity&lt;/em&gt; seems to be the measure you want (and is inversely proportional to the square of the distance.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The intensity at a point $(x,y)$ of a spherical light source originating at, say, $(x_0, y_0)$ is given by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$I = \mathcal{P} / 4 \pi r^2$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\mathcal{P}$ is the power of the light source (what I think you've called $X$), and $r$ is the distance between $(x,y)$ and $(x_0, y_0)$. This gives the final expression, for all light sources $(x_n, y_n)$ with power $\mathcal{P}_n$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$I = \displaystyle \sum_{n} \frac{\mathcal{P}_n}{4 \pi ((x-x_n)^2 + (y-y_n)^2)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Unless I've forgotten some physics and intensities don't sum like that). For the third dimension, just add the $z$ terms in the distance formula. Hope this helps.&lt;/p&gt;&#xA;" OwnerUserId="5218" LastEditorUserId="5218" LastEditorDisplayName="" LastEditDate="2010-12-31T18:21:27.527" LastActivityDate="2010-12-31T18:21:27.527" CommentCount="2" />
  <row Id="15925" PostTypeId="1" AcceptedAnswerId="15936" CreationDate="2010-12-30T17:33:14.933" Score="3" ViewCount="116" Body="&lt;p&gt;What exactly is the variation of a function ? Is it a distace or an element of some space&#xA;The total of a real valued function $f\colon [0,t] \mapsto \Re $ is as below say $\pi = \{0=t_0,t_1,\cdots , t_n=t\} $ then the $p^{th}$ variation of $f$ is &#xA;$$ \lim_{\Vert \pi\Vert \to 0}V^p(f)=\sum_{i=1}^n \vert f(t_i)-f(t_{i-1})\vert^p $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now say we have $g\colon [0,t] \mapsto \mathcal{X}$ where $\mathcal{x}$ is some linear, normed,Complete space. it seems natural to say variation of $g$ over $[0,t]$ is &#xA;$$ Z_n=\sum_{i=1}^n \Vert g(t_i)-g(t_{i-1})\Vert^p $$ where &#xA;$\Vert.\Vert$ is the norm of $\mathcal{X}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it exists what is the Total variation in this case is it $\lim_{n \to \infty}Z$ ie a value in $[0,\infty]$ or is it some $x$ such that $\Vert Z_n -  x \Vert \to 0 $&#xA;say for wiener process $W(s)$ it quadratic  variation over  $[0,t]$ is $t$ does it mean &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \lim_{\Vert \pi\Vert \to 0}V^2(W)=\sum_{i=1}^n \vert W(t_i)-W(t_{i-1})\vert^2 $$ in $L^2(\Omega) $ norm or does it mean some $X \in L^2(\Omega)$ &lt;/p&gt;&#xA;" OwnerUserId="5176" LastActivityDate="2010-12-31T00:44:44.673" Title="Variation of a function" Tags="&lt;stochastic-processes&gt;" AnswerCount="1" />
  <row Id="15926" PostTypeId="2" ParentId="15923" CreationDate="2010-12-30T17:34:45.947" Score="4" ViewCount="" Body="&lt;p&gt;Asaf, what you are saying is false. In $L[\mu]$, for example, there is a unique normal ultrafilter. (Unless I misunderstand. You are saying that for every $S$ stationary-costationary, you can find a normal $U$ with $S\in U$, right? Obviously, any normal $U$ satisfies that either $S$ or $\kappa\setminus S$ is in $U$, which is what you wrote.)&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit :&lt;/strong&gt; This is a nice homework problem. There are several different solutions, and you may want to study them afterward. Here is a way of thinking about this; it is perhaps not the most efficient, but it is very useful. Suppose $U$ on $\kappa$ is normal and concentrates on measurables. Then $\kappa$ is measurable in $M$, where $j:V\to M$ is given by $U$. So there is a normal $U&amp;#39;$ on $\kappa$ in $M$, and $U&amp;#39;$ really is normal (in $V$, not just in $M$). Let $k:V\to N$ be given by $U&amp;#39;$. what can you say about the sizes of $j(\kappa)$ vs $k(\kappa)$ ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is another approach: Use induction, use that you are at a measurable (so you have a normal $U$) and &quot;integrate&quot; the measures on small cardinals witnessing the claim. Check that the resulting measure is as you want. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Let me add a couple of details to the first approach (I know it is more elaborate than the other one, but the payoff is worth the effort): &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $j:V\to M$ is the ultrapower embedding by a normal measure $U$ on $\kappa$, then ${}^\kappa M\subset M$ and from this it is easy to check that if $M\models$&quot;$U'$ is a normal measure on $\kappa$&quot;, then $U&amp;#39;$ is a normal measure on $\kappa$ in $V$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, suppose that $U$ concentrates on measurables. Since the identity represents $\kappa$ in $M$, it follows that $\kappa$ is measurable in $M$, and there is $U&amp;#39;$ as mentioned. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Essentially, we want to iterate this process: Form the embedding $k:V\to N$ given by $U&amp;#39;$, and if $U&amp;#39;$ concentrates on measurables, then we get $U&amp;#39;&amp;#39;$ in $N$ and form $l:V\to P$, etc. We would like this process to stop after finitely many times. For this, we want to associate to a normal measure $U$ on $\kappa$ an ordinal $\alpha_U$ in such a way that if $U&amp;#39;$ is in the ultrapower by $U$, then $\alpha_{U&amp;#39;}&amp;lt;\alpha_U$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The easiest way to do this is to set $\alpha_U=j(\kappa)$. If $U&amp;#39;\in M$, we can form in $M$ the ultrapower embedding by $U&amp;#39;$, call $k&amp;#39;$ the result, so $k&amp;#39;:M\to M&amp;#39;$, and recall we called $k:V\to N$ the ultrapower of $V$ by $U&amp;#39;$. The point is that $k&amp;#39;(\kappa)=k(\kappa)$, because ${}^\kappa M\subset M$, so all functions $f:\kappa\to\kappa$ are in $M$, and this and $U&amp;#39;$ is all we need to compute the value of the embedding at $\kappa$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But $M\models$&quot;$j(\kappa)$ is inaccessible, while $k'(\kappa)&amp;lt;(2^\kappa)^+$&quot;, so $k&amp;#39;(\kappa)&amp;lt;j(\kappa)$, so $k(\kappa)&amp;lt;j(\kappa)$ or $\alpha_{U&amp;#39;}&amp;lt;\alpha_U$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;A posteriori&lt;/em&gt;, this argument shows that the relation $\prec$ on normal ultrafilters on $\kappa$ given by &quot;$U'\prec U$ iff $U&amp;#39;$ belongs to the ultrapower by $U$&quot; is well-founded, and so we can assign to $U$ the ordinal $o(U)$ given by its rank in this well-founded relation. This is Mitchell's ordering, and it is a very useful tool in large cardinal theory. &lt;/p&gt;&#xA;" OwnerUserId="462" LastEditorUserId="462" LastEditorDisplayName="" LastEditDate="2011-01-01T17:05:54.090" LastActivityDate="2011-01-01T17:05:54.090" CommentCount="6" />
  <row Id="15927" PostTypeId="1" CreationDate="2010-12-30T17:50:56.993" Score="0" ViewCount="78" Body="&lt;p&gt;Suppose that&#xA;$$u_1 = \left(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}},0,0,\ldots,0\right)\quad\text{and}\quad&#xD;&#xA;v_1 = \left(\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}},0,0,\ldots,0\right)$$&#xA;and &#xA;$$u_l(n) = \sum_{k=0}^{2^{l-1}-1} u_1\left(n+ \frac{kN}{2^{l-1}}\right)\quad\text{and}\quad&#xD;&#xA;v_l(n) = \sum_{k=0}^{2^{l-1}-1} v_1\left(n+ \frac{kN}{2^{l-1}}\right).$$&#xA;How to show that &#xA;\begin{align*}&#xA;u_l(0)&amp;amp;=  \frac{1}{\sqrt{2}},\\&#xA;u_l(1)&amp;amp;=  \frac{1}{\sqrt{2}},\\&#xA;u_l(n)&amp;amp;=  0 &amp;amp;\quad&amp;amp;\text{for $2 \leq n \leq \frac{N}{2^{l-1}}-1$,}\\&#xA;v_l(0)&amp;amp;=  \frac{1}{\sqrt{2}},\\&#xA;v_l(1)&amp;amp;=  -\frac{1}{\sqrt{2}},\\&#xA;v_l(n)&amp;amp;=  0&amp;amp;&amp;amp;\text{for $2 \leq n \leq \frac{N}{2^{l-1}}-1$.}&#xA;\end{align*}&lt;/p&gt;&#xA;" OwnerUserId="2723" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-30T18:37:39.293" LastActivityDate="2010-12-30T18:37:39.293" Title="How to show that $u_l(0)= \frac{1}{\sqrt{2}}$, $u_l(1)= \frac{1}{\sqrt{2}}$ and $u_l(n)= 0$ for $2 \leq n \leq (\frac{N}{2^{l-1}}-1$ $\dots$" Tags="&lt;linear-algebra&gt;" CommentCount="5" />
  <row Id="15928" PostTypeId="1" AcceptedAnswerId="15930" CreationDate="2010-12-30T18:05:38.633" Score="6" ViewCount="165" Body="&lt;p&gt;Is there any rule for powers so that i can compare which one is greater without actually calculating? For example  &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;54^53 and 53^54 &#xA;23^26 and 26^23&#xA;3^4 and 4^3 (very simple but how without actually calculating)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="1515" LastEditorUserId="1515" LastEditorDisplayName="" LastEditDate="2010-12-30T18:18:18.330" LastActivityDate="2010-12-30T20:24:22.793" Title="How to Compare powers without calculating?" Tags="&lt;arithmetic&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" />
  <row Id="15929" PostTypeId="2" ParentId="15903" CreationDate="2010-12-30T18:14:23.317" Score="5" ViewCount="" Body="&lt;p&gt;For $\displaystyle 1 \le k \le 24$ you can definitely show that the master must have played exactly $k$ games on some set of consecutive days, using pigeonhole principle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose the total number of games the master has played till the end of day $\displaystyle j$ is $\displaystyle g_j$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now consider the $\displaystyle 154$ numbers: $\displaystyle g_1, g_2, \dots, g_{77}, g_1 + k, g_2 + k, \dots, g_{77} + k$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These are a set of $\displaystyle 154$ numbers between $\displaystyle 1$ and $\displaystyle 132+k$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $\displaystyle k \lt 22$, two of these must be equal. Since $\displaystyle g_i \neq g_j$ (at least one game a day) we must have that $\displaystyle g_j + k = g_i$ for some $\displaystyle i,j$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $\displaystyle k=22$ we must have that the numbers are $\displaystyle 1,2, \dots, 154$, in which case, the first (and last) $\displaystyle 22$ days, the master must have played $\displaystyle 1$ game everyday.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $\displaystyle k=23$, we can assume $\displaystyle g_i \neq 23$, and since $g_i \ge 1$, we have $g_i + k \neq 23$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus by an argument similar to above, we must have have $\displaystyle 154$ numbers taking all values in $\displaystyle 1, 2, \dots, 155$, except $\displaystyle 23$ and the master must have played $\displaystyle 1$ game each of the last $\displaystyle 23$ days.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $\displaystyle k=24$, you can show that the master must have played $\displaystyle 1$ game the first $\displaystyle 23$ days (after eliminating one of the numbers in $\displaystyle 133, 134 \dots$), then a big number of games the next, violating the $\displaystyle 12$ games per week restriction (this is where we actually used that restriction for a &lt;em&gt;specific&lt;/em&gt; week).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We might be able to use this kind of argument to show for $\displaystyle k$ close to $\displaystyle 24$, but for larger $\displaystyle k$, I am guessing that you can find a set of games which will miss that (perhaps a computer search will help there).&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-12-30T19:28:23.440" LastActivityDate="2010-12-30T19:28:23.440" CommentCount="2" />
  <row Id="15930" PostTypeId="2" ParentId="15928" CreationDate="2010-12-30T18:24:07.313" Score="9" ViewCount="" Body="&lt;p&gt;If $a\gt b\gt e , b^a\gt a^b$.  To see this, take logs.  You want to compare $a \ln b$ with $b \ln a$.  $\ln$ rises slowly, so the larger multiplier wins.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-12-30T18:24:07.313" CommentCount="8" />
  <row Id="15931" PostTypeId="2" ParentId="15918" CreationDate="2010-12-30T18:41:45.043" Score="0" ViewCount="" Body="&lt;p&gt;Solved: I obtained the correct values from else where within the WebGL wrapper.&lt;/p&gt;&#xA;" OwnerUserId="3849" LastActivityDate="2010-12-30T18:41:45.043" CommentCount="2" />
  <row Id="15932" PostTypeId="2" ParentId="15857" CreationDate="2010-12-30T18:49:15.130" Score="1" ViewCount="" Body="&lt;p&gt;Geometry Revisited has a much narrower domain of content than an Introduction to Geometry. If your goal is to get a sense of what different kinds of problems, techniques, and concepts geometry has evolved to deal with, Introduction to Geometry is a &quot;dated&quot; but somewhat comprehensive choice. (For example, the whole &quot;world&quot; of computational geometry came on the scene after Coexeter's book.) One typically does not read mathematics books from cover to cover and one can dabble in Introduction to Geometry to see some of geometry's many parts. If one has heavy sledding in places than looking at the wiki article of the topic causing trouble might help.&lt;/p&gt;&#xA;" OwnerUserId="1369" LastActivityDate="2010-12-30T18:49:15.130" />
  <row Id="15933" PostTypeId="2" ParentId="12451" CreationDate="2010-12-30T18:52:01.277" Score="0" ViewCount="" Body="&lt;p&gt;I am not expert, but my own limited view is this.  I said for thirty years I could see no interest in theorems like radon nikodym, having learned them very abstractly.  then once while teaching honors calculus i asked myself what the fundamental theorem of calculus should say for integrals of just Riemann integrable, not necessarily continuous functions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;After originally stating the wrong answer to my class, I learned (with help from an analyst friend who showed me a Cantor function) that the indefinite integral is characterized by being a function which has a derivative equal to the original integrand almost everywhere (wherever that integrand is continuous) and being also not just continuous but Lipschitz continuous.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then I realized at last the radon nikodym theorem is just the fundamental theorem of calculus for more general functions.  We may not realize the analogy from calc 1, because by considering only continuous integrands there we miss out on the singular part.  I.e. we forget to wonder why we are looking only at integrals of derivatives of C^1 (in particular Lipschitz) functions. So any application of FTC type is an application of RN, e.g. it lets you characterize constant functions (a.e?) by weak continuity and differentiability properties.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, Fubini is of course repeated integration, so reduces any integral computation inductively to one of lower dimension (volumes by slicing).  E.g. to show some set has measure zero (as in Sard's theorem) you can do it inductively by showing most of the slices have lower dimensional measure zero.  (See Guillemin and Pollack, appendix, or Milnor's differential topology book.)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of my professors once suggested that virtually all problems in analysis are attacked by either dominated convergence or Fubini.  So if you see an exam problem that dominated convergence won't do, try Fubini.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This answer assumes you retain an interest in the topic even after your exam, unless you are at Harvard, where exams are perhaps still in january.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Following up KCd's comment, you might also peruse old qualifying exams available on the websites at schools like Harvard and UGA.  Harvard's also has a few lists of typical questions of this nature: e.g. if every intersection of a certain subset S of the plane, with a line of slope 1 is countable, what can you say about the Lebesgue measure of S? Gosh, Harvard even has, with login capability, online copies of all exams since 1977, and paper copies in libraries of exams since 1836!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or consider a continuous weakly monotone increasing function f on the interval [0,1].  One knows that f is differentiable a.e. say with derivative g ≥ 0. If  g is integrable and G(x) is the integral of g from 0 to x, then to what extent does G determine f or g, if either?  When does G determine both f and g?  Give an example if possible where G does not determine f, respectively g.&lt;/p&gt;&#xA;" OwnerUserId="5211" LastEditorUserId="5211" LastEditorDisplayName="" LastEditDate="2011-01-06T17:32:20.957" LastActivityDate="2011-01-06T17:32:20.957" CommentCount="3" />
  <row Id="15934" PostTypeId="2" ParentId="15890" CreationDate="2010-12-30T18:57:07.563" Score="5" ViewCount="" Body="&lt;p&gt;Michael Spivak gave me a small volume he wrote on mechanics from the mathematicians perspective, and he is said to be preparing to release a longer book on physics explicitly for mathematicians.  In the meantime you might like Max Born's Atomic Physics.  I myself enjoyed, and learned from, at least looking at J.C.Maxwell's E&amp;amp;M.&lt;/p&gt;&#xA;" OwnerUserId="5211" LastEditorUserId="5211" LastEditorDisplayName="" LastEditDate="2010-12-30T19:16:48.537" LastActivityDate="2010-12-30T19:16:48.537" CommentCount="4" />
  <row Id="15935" PostTypeId="2" ParentId="15720" CreationDate="2010-12-30T19:51:31.570" Score="-1" ViewCount="" Body="&lt;p&gt;If I remember correctly, what you are describing is close to the concept of a &lt;em&gt;distribution function&lt;/em&gt; (it is the name that I am slightly unsure about). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given a measure space $(X,\mathcal{E},\mu)$, and a measurable function $f:X\to \mathbb{R}$, a concept often used in classical and harmonic analysis is the notion of the distribution function $\lambda_f:\mathbb{R}_+\to\mathbb{R}$ defined as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \lambda_f(a) = \mu\left( \{|f| &amp;lt; a\} \right) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;that is, $\lambda_f$ is returns the measure of the set on which $|f|$ is bounded by $a$. One can reformulate the $L^p$ norms of $f$ in terms of the distribution function&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \|f\|_{L^p(X,\mu)} = p \int_0^\infty t^{p-1}\lambda_f(t) dt $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a device that is very useful for analytical estimates. Now, formally speaking the functional $L(f)$ you wrote down is very similar to the derivative $\lambda_f&amp;#39;$. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Another possibility that you may want to look into is the notion of pull-back distributions. If $f:M\to\mathbb{R}$ is a smooth map with non-vanishing gradient, then given any distribution (generalized function) $k$ on $\mathbb{R}$&#xA; you can define the pull-back distribution $f_*k$ on $M$. The definition of $f_*k$ depends closely on the notion of $L(f)$ you wrote down (see e.g. the introduction to distributions by Friedlander and Joshi), and is a generalization of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Coarea_formula&quot; rel=&quot;nofollow&quot;&gt;co-area formula&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I am not exactly sure how to answer your second question of &quot;is there a good way to compute&quot;. If you are asking whether given an analytic expression for the function $f$ whether you can have a &quot;closed form&quot; expression for $L(f)$, the answer is surely no, since the expression also depends on the chosen Riemannian metric/volume form/measure on $M$. If you are asking whether there are any convenient expressions to capture $L(f)$, I think you need to say a bit more about why you care about $L(f)$. You can certainly just write $L(f) = f_*\delta_a$ as the pull-back of the Dirac delta distribution on $\mathbb{R}$; whether that expression is a useful one for your purpose, I don't know.  &lt;/p&gt;&#xA;" OwnerUserId="1543" LastActivityDate="2010-12-30T19:51:31.570" />
  <row Id="15936" PostTypeId="2" ParentId="15925" CreationDate="2010-12-30T20:21:57.850" Score="4" ViewCount="" Body="&lt;p&gt;Consider the case where $W$ is a standard Brownian motion. Given a partition $\pi  = \{ 0 = t_0  &amp;lt; t_1  &amp;lt;  \cdots  &amp;lt; t_n  = t\}$ of the interval $[0,t]$, define&#xA;$$&#xD;&#xA;[W]^\pi   = \sum\limits_{i = 1}^n {[W(t_i ) - W(t_{i - 1} )]^2 }. &#xD;&#xA;$$&#xA;Let $|\pi | = \max _i (t_i  - t_{i - 1} )$ denote the mesh of the partition. As is well known, given a sequence of partitions $\pi_n$  with $|\pi_n| \to 0$ as $n \to \infty$, $[W]^{\pi _n }  \to t$ in the $L^2$ norm, that is ${\rm E}\big[([W]^{\pi _n }  - t)^2 \big] \to 0$. Hence, in particular, $[W]^{\pi _n }  \to t$ in probability, and thus, by definition, the quadratic variation of $W$ is given by $[W]_t = t$. The key point here is that the limit $t$ in the right-hand side of $[W]^{\pi _n }  \to t$ plays a role of a random variable, rather than of a constant. Of course, in this particular example, $t$ is also a constant, but in general the limit is a random variable, which is defined using convergence in probability. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: A much simpler yet instructive example is given by the Poisson process $N$. It is readily seen that the variation of $N$ is given by&#xA;$$&#xD;&#xA;\mathop {\lim }\limits_{|\pi _n | \to 0} \sum\limits_{i = 1}^n {|N(t_{i} ) - N(t_{i-1} )|} = N(t),&#xD;&#xA;$$&#xA;and the quadratic variation of $N$ is given by&#xA;$$&#xD;&#xA;\mathop {\lim }\limits_{|\pi _n | \to 0} \sum\limits_{i = 1}^n {[N(t_{i} ) - N(t_{i-1} )]^2} = N(t).&#xD;&#xA;$$&#xA;You may wish to consider different modes of convergence, as an exercise.&lt;/p&gt;&#xA;" OwnerUserId="2810" LastEditorUserId="2810" LastEditorDisplayName="" LastEditDate="2010-12-31T00:44:44.673" LastActivityDate="2010-12-31T00:44:44.673" CommentCount="4" />
  <row Id="15938" PostTypeId="2" ParentId="329" CreationDate="2010-12-30T21:05:27.587" Score="5" ViewCount="" Body="&lt;p&gt;I recommend &lt;em&gt;Primes of the Form x&lt;sup&gt;2&lt;/sup&gt; + ny&lt;sup&gt;2&lt;/sup&gt;&lt;/em&gt;, by David Cox.  The question of which primes can be written as the sum of two squares was settled by Euler.  The more general question turns out to be much harder, and leads you to more advanced techniques in number theory like &lt;a href=&quot;http://en.wikipedia.org/wiki/Class_field_theory&quot; rel=&quot;nofollow&quot;&gt;class field theory&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/Complex_multiplication&quot; rel=&quot;nofollow&quot;&gt;elliptic curves with complex multiplication&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="4880" LastEditorUserId="4880" LastEditorDisplayName="" LastEditDate="2010-12-30T23:45:33.203" LastActivityDate="2010-12-30T23:45:33.203" CommunityOwnedDate="2010-12-30T21:05:27.587" CommentCount="0" />
  <row Id="15939" PostTypeId="1" CreationDate="2010-12-30T21:06:42.593" Score="1" ViewCount="222" Body="&lt;p&gt;I have an $n\times p$  matrix $Z$ with $p\gt n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a diagonal matrix $A$ with positive entries.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there is a known way to determine the&#xA;MP inverse of $A Z^T Z A$, if I know $A$ and the MP inverse of $Z^T Z$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="5250" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-31T00:36:05.407" LastActivityDate="2011-05-30T21:06:48.707" Title="Moore-Penrose inverse invariance" Tags="&lt;linear-algebra&gt;&lt;matrices&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="15940" PostTypeId="1" CreationDate="2010-12-30T21:20:47.750" Score="3" ViewCount="55" Body="&lt;p&gt;In category theory, you see the idea of a class of epimorphisms being stable under pullback.  For example, in a regular category, the class of regular epimorphisms is closed under pullback.  Every place I've seen the notion of pullback-stability, it's always a part of a bigger definition, such as regular category, or Grothendieck topology.  Is there some bigger significance to the idea?  Is there a theory for pullback-stable classes of epimorphisms?&lt;/p&gt;&#xA;" OwnerUserId="4880" LastActivityDate="2010-12-30T21:20:47.750" Title="Pullback-stability for epimorphisms" Tags="&lt;category-theory&gt;" CommentCount="1" />
  <row Id="15941" PostTypeId="2" ParentId="5879" CreationDate="2010-12-30T21:26:24.213" Score="0" ViewCount="" Body="&lt;p&gt;May I ask, how fast is this canonical labeling algorithm? (linear in n?)&#xA;I need to convert an adjacency matrix to canonical form but I don't know how :(&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;c.wilmer@gmail.com&lt;/p&gt;&#xA;" OwnerUserId="5251" LastActivityDate="2010-12-30T21:26:24.213" />
  <row Id="15942" PostTypeId="1" AcceptedAnswerId="15949" CreationDate="2010-12-30T22:15:16.153" Score="1" ViewCount="145" Body="&lt;p&gt;$f_n(x) :  [0,1] \rightarrow \bf R $,  and $f_n$ is $\frac{1}{n}$-periodic, $\max f_n(x) = n$, $\min f_n(x) = -n$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As $n$ and $v$ goes to infinity simultaneously, prove the convergence of&lt;br&gt;&#xA;$$\lim_{n,v \rightarrow \infty}  \int_0^1 f_n(x) e^{-i2\pi v x}\,dx $$&#xA;if $v,n$ are integers and $v/n$  is not an integer.&lt;/p&gt;&#xA;" OwnerUserId="5181" LastEditorUserId="3249" LastEditorDisplayName="" LastEditDate="2011-01-01T15:57:55.497" LastActivityDate="2011-01-01T15:57:55.497" Title="Convergence of $\lim_{n,v \rightarrow \infty} \int_0^1 f_n (x) e^{-i2\pi v x} \mbox{d} x $" Tags="&lt;fourier-analysis&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="15943" PostTypeId="2" ParentId="12824" CreationDate="2010-12-30T22:53:30.267" Score="2" ViewCount="" Body="&lt;p&gt;You can consider presentations for any algebraic system given by special constants, operations and equations.  For example, groups are given by the special constant 1 (for the multiplicative identity), the unary operation of inverse, x&lt;sup&gt;-1&lt;/sup&gt;, and the binary operation of multiplication, $xy$, the plus the equations that these must satisfy.  In that case, you can specify every object with such operations in terms of a set of generators, and the relations that the generators must satisfy.  The study of these kinds of constructions is known as &lt;a href=&quot;http://en.wikipedia.org/wiki/Universal_algebra&quot; rel=&quot;nofollow&quot;&gt;universal algebra&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the specific case of commutative rings over a field, there is an explicit algorithm for working with presentations, known as the &lt;a href=&quot;http://en.wikipedia.org/wiki/Gr%C3%B6bner_basis&quot; rel=&quot;nofollow&quot;&gt;Groebner basis&lt;/a&gt; algorithm.  It's widely implemented in symbolic algebra packages.  (In the case of commutative rings, you're basically just working with systems of polynomials.)&lt;/p&gt;&#xA;" OwnerUserId="4880" LastEditorUserId="4880" LastEditorDisplayName="" LastEditDate="2010-12-30T23:47:04.843" LastActivityDate="2010-12-30T23:47:04.843" />
  <row Id="15944" PostTypeId="2" ParentId="329" CreationDate="2010-12-30T23:07:44.343" Score="1" ViewCount="" Body="&lt;p&gt;&lt;em&gt;Basic Number Theory&lt;/em&gt; by Andre Weil.  It's hard going and mind-blowing.&lt;/p&gt;&#xA;" OwnerUserId="4253" LastActivityDate="2010-12-30T23:07:44.343" CommunityOwnedDate="2010-12-30T23:07:44.343" CommentCount="5" />
  <row Id="15945" PostTypeId="2" ParentId="329" CreationDate="2010-12-30T23:08:08.963" Score="3" ViewCount="" Body="&lt;p&gt;One of my colleagues, a number theorist, recommended the little book by van den eynden for beginners.  my favorite is by trygve nagell.  (I am a geometer.)  One of my friends, preparing for a PhD in arithmetic geometry?, started with the one recommended by Barry, Basic number theory.  As I recall it's for people who can handle Haar measure popping up on the first page of a &quot;basic&quot; book on number theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also recommend Gauss's Disquisitiones Arithmeticae.&lt;/p&gt;&#xA;" OwnerUserId="5211" LastEditorUserId="5211" LastEditorDisplayName="" LastEditDate="2011-01-05T20:01:58.207" LastActivityDate="2011-01-05T20:01:58.207" CommunityOwnedDate="2010-12-30T23:08:08.963" CommentCount="3" />
  <row Id="15946" PostTypeId="1" AcceptedAnswerId="15947" CreationDate="2010-12-30T23:59:17.993" Score="4" ViewCount="196" Body="&lt;p&gt;Is this series known to converge, and if so, what does it converge to (if known)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $p_n$ is prime number n, and $p_1 = 2$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum\limits_{n=1}^{\infty} \frac{1}{p_n}$$&lt;/p&gt;&#xA;" OwnerUserId="4576" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-12-31T00:31:45.397" LastActivityDate="2010-12-31T00:31:45.397" Title="Does the sum of reciprocals of primes converge?" Tags="&lt;number-theory&gt;&lt;real-analysis&gt;&lt;prime-numbers&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="15947" PostTypeId="2" ParentId="15946" CreationDate="2010-12-31T00:02:46.237" Score="7" ViewCount="" Body="&lt;p&gt;No, it does not converge. See this: &lt;a href=&quot;http://en.wikipedia.org/wiki/Proof_that_the_sum_of_the_reciprocals_of_the_primes_diverges&quot; rel=&quot;nofollow&quot;&gt;Proof of divergence of sum of reciprocals of primes&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact it is known that $$\sum_{p \le x} \frac{1}{p} = \log \log x + A + \mathcal{O}(\frac{1}{\log^2 x})$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Related: &lt;a href=&quot;http://math.stackexchange.com/questions/12566/&quot;&gt;Proving $\sum\limits_{p \leq x} \frac{1}{\sqrt{p}} \geq \frac{1}{2}\log{x} -\log{\log{x}}$&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-12-31T00:12:54.347" LastActivityDate="2010-12-31T00:12:54.347" />
  <row Id="15948" PostTypeId="2" ParentId="15946" CreationDate="2010-12-31T00:08:37.593" Score="4" ViewCount="" Body="&lt;p&gt;I would like to note that this implies that according the &lt;a href=&quot;http://en.wikipedia.org/wiki/M%C3%BCntz%E2%80%93Sz%C3%A1sz_theorem&quot; rel=&quot;nofollow&quot;&gt;Müntz-Szász Theorem&lt;/a&gt; that every continuous function in $[0,1]$ is a uniform limit of polynomials whose exponents are prime numbers!&lt;/p&gt;&#xA;" OwnerUserId="1120" LastEditorUserId="1120" LastEditorDisplayName="" LastEditDate="2010-12-31T00:15:05.557" LastActivityDate="2010-12-31T00:15:05.557" CommentCount="2" />
  <row Id="15949" PostTypeId="2" ParentId="15942" CreationDate="2010-12-31T00:28:29.597" Score="2" ViewCount="" Body="&lt;p&gt;I'll assume you mean to take a limit in some sense as over double sequences indexed by natural numbers $n, \nu$ such that $\frac{\nu}{n}$ is not an integer. For $f_n$ an $\frac{1}{n}$-periodic function such that $\max f_n=-\min f_n =n$, let $g_n$ be defined by $f_n(x)=ng_n(nx)$. Then $g_n$ is $1$-periodic and has max and min equal to 1 and -1. Let $r=\frac{\nu}{n}$. Then you require that $r$ is not an integer.&#xA;Then writing the integral of interest in terms of $g_n$ and $r$ and $n$, you'll be very tempted to change variables and integrate from 0 to n. Do so.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then using the periodicity of $g_n$, you'll want to break the integral into a sum of $n$ integrals. Finally, switch the order of summation and integration, be happy that you can look up or re-derive the formula for geometric sums, and be shocked (or not, but I was pleasantly surprised). &lt;/p&gt;&#xA;" OwnerUserId="5253" LastActivityDate="2010-12-31T00:28:29.597" CommentCount="3" />
  <row Id="15950" PostTypeId="2" ParentId="15707" CreationDate="2010-12-31T00:34:59.247" Score="5" ViewCount="" Body="&lt;p&gt;Suppose $f$ is $C^2$.  First of all, because $g$ is convex in each variable, it follows that $f$ is convex, and hence $f&amp;#39;&amp;#39;\geq0$.  I did not initially have Slowsolver's insight that &lt;a href=&quot;http://en.wikipedia.org/wiki/Logarithmically_convex_function&quot; rel=&quot;nofollow&quot;&gt;log convexity&lt;/a&gt; would be a criterion to look for, but naïvely checking for positivity of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Hessian_matrix&quot; rel=&quot;nofollow&quot;&gt;Hessian&lt;/a&gt; of $g$ leads to the inequalities&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f&amp;#39;&amp;#39;(x)f(y)+f(x)f&amp;#39;&amp;#39;(y)\geq0$$&#xA;and&#xA;$$f&amp;#39;(x)^2f&amp;#39;(y)^2\leq f&amp;#39;&amp;#39;(x)f(x)f&amp;#39;&amp;#39;(y)f(y)$$&#xA;for all $x$ and $y$, coming from the fact that a real symmetric $2$-by-$2$ matrix is positive semidefinite if and only if its trace and determinant are nonnegative.  The first inequality follows from nonnegativity of $f$ and $f&amp;#39;&amp;#39;$.  The second inequality is equivalent to $f&amp;#39;^2\leq f&amp;#39;&amp;#39;f$.  To see the equivalence in one direction, just set $x=y$ and take square roots; in the other direction, multiply the inequalities at $x$ and $y$.  Since $\log(f)&amp;#39;&amp;#39;=\frac{f&amp;#39;&amp;#39;f-f&amp;#39;^2}{f^2}$, this condition is equivalent to $\log(f)&amp;#39;&amp;#39;\geq0$, meaning that $\log(f)$ is convex.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2010-12-31T01:34:20.630" LastActivityDate="2010-12-31T01:34:20.630" />
  <row Id="15952" PostTypeId="2" ParentId="12451" CreationDate="2010-12-31T01:01:22.093" Score="3" ViewCount="" Body="&lt;p&gt;One basic example of what you can do with Fubini's theorem is to represent an integral of a function of a function in terms of its distribution function. For instance, there is the formula (for reasonable $\phi$, $f$ basically arbitrary but nonnegative)&#xA;$$ \int \phi \circ f d\mu = \int_t \mu(\{ f&amp;gt; t\}) \phi&amp;#39;(t) dt$$ which reduces questions about integrals of, for instance, $p$th powers to questions about the distribution function. This is (for example) how the boundedness of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Hardy%E2%80%93Littlewood_maximal_function&quot; rel=&quot;nofollow&quot;&gt;Hardy-Littlewood maximal operator&lt;/a&gt;  on $L^p$ ($p&gt;1$) is proved: you get a bound on the distribution function of the maximal function by general methods and then do an interpolation process. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To prove the formula above, as in Rudin, one can consider the collection $E$ of pairs $(x,t)$ such that $0 \leq t \leq f(x)$. This is a measurable subset of $X \times \mathbb{R}$ if $X$ is the initial measure space on which $f$ is defined and $\mathbb{R}$ has Lebesgue measure. &#xA;Then, one can write the second integral in the displayed equation &#xA;as $\int_t \phi&amp;#39;(t) dt \int_{x \in X} \chi_E(x,t) d \mu$ where $\chi_E$ denotes the characteristic function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now rearranging this integral via Fubini's theorem allows one to integrate with respect to $t$ first, for each $x$; then $t$ goes from $0$ to $f(x)$, and one can see that this integral becomes $\int_x \int_{t=0}^{f(x)} \phi&amp;#39;(t) dt$, which is the right-hand-side of the displayed equation.&lt;/p&gt;&#xA;" OwnerUserId="536" LastActivityDate="2010-12-31T01:01:22.093" CommentCount="1" />
  <row Id="15954" PostTypeId="1" CreationDate="2010-12-31T01:34:02.283" Score="3" ViewCount="170" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Possible Duplicate:&lt;/strong&gt;&lt;br&gt;&#xA;  &lt;a href=&quot;http://math.stackexchange.com/questions/155/how-can-you-prove-that-a-function-has-no-closed-form-integral&quot;&gt;How can you prove that a function has no closed form integral?&lt;/a&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;p&gt;When they say that, e.g., Li(x) has no closed form (for some agreed upon definition of &quot;closed form&quot;), do they mean that it can be &lt;strong&gt;proved&lt;/strong&gt; that there isn't one, or just that none has been found yet.  Most authors seem to just assert that a given integral can't be represented in any simpler way, but I've been wondering if that's because they know this for a fact, or just that it's assumed that if there is one it would have been found by now.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If this is a theorem, how would you go about proving it?&lt;/p&gt;&#xA;" OwnerUserId="5257" LastActivityDate="2010-12-31T01:34:02.283" ClosedDate="2010-12-31T02:09:58.803" Title="How do they know an integral has no closed form solution?" Tags="&lt;terminology&gt;&lt;closed-form&gt;" CommentCount="2" />
  <row Id="15955" PostTypeId="1" AcceptedAnswerId="15956" CreationDate="2010-12-31T02:34:26.537" Score="3" ViewCount="141" Body="&lt;p&gt;Let $f(x) = \sum_{i =0}^\infty a_i x^i$ be a power series which converges for all real $x$. Assume that $f(x)$ is not identically zero. I'm interested in the density of the zeros of $f(x)$. Let $Z$ be the set of zeros of $f(x)$. Which of the following claims about density of $Z$ are true? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Claim 1: $Z$ is nowhere dense.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Claim 2: $Z$ is countable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Claim 3: For any $a,b \in \mathbb{R}$ , $Z \cap [a,b]$ is finite.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe (correct me if I'm wrong) that claim 3 implies the other two. I suspect all three claims are true.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suspect that the answers to these questions are well-known, though I was not able to find an obvious reference. Can anyone suggest a reference with a nice treatment of these questions?&lt;/p&gt;&#xA;" OwnerUserId="5159" LastEditorUserId="3249" LastEditorDisplayName="" LastEditDate="2010-12-31T04:48:20.067" LastActivityDate="2010-12-31T04:48:20.067" Title="Density of zeros of a power series over the reals" Tags="&lt;analysis&gt;&lt;power-series&gt;" AnswerCount="2" CommentCount="0" />
  <row Id="15956" PostTypeId="2" ParentId="15955" CreationDate="2010-12-31T02:58:34.723" Score="7" ViewCount="" Body="&lt;p&gt;Claim 3 is true, and it does imply the other 2.  It is enough to consider analytic functions, i.e. functions that have a power series expansion in some interval centered at each real number, which in particular holds if you have an everywhere convergent power series.  If $[a,b]$ had infinitely many zeros of $f$, then there would be a limit point $c$ of these zeros.  By continuity $f(c)=0$.  Let $n\gt0$ be the smallest positive integer such that $f^{(n)}(c)\neq0$ (using the assumption that $f$ is not identically $0$).  Then $f$ has power series expansion &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align*}&#xD;&#xA;f(x)&amp;amp;=\sum_{k=0}^\infty\frac{f^{(k)}(c)}{k!}(x-c)^k&#xD;&#xA;=\sum_{k=n}^\infty\frac{f^{(k)}(c)}{k!}(x-c)^k\\&#xD;&#xA;    &amp;amp;=(x-c)^n\sum_{k=n}^\infty\frac{f^{(k)}(c)}{k!}(x-c)^{k-n}=(x-c)^ng(x),&#xD;&#xA;\end{align*}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $g(x)$ is a continuous function such that $g(c)\neq0$, and hence $g(x)\neq0$ in some open interval $I$ containing $c$.  So the only zero of $f$ in $I$ is $c$, contradicting the fact that $c$ is a limit point of the set of zeros.  Hence, unless $f$ is identically zero, the limit point $c$ cannot exist.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For reference you can read any good text on complex analysis.  If you prefer to stick to the real case, there is the book &lt;em&gt;&lt;a href=&quot;http://books.google.com/books?id=i4vw2STJl2QC&amp;amp;source=gbs_navlinks_s&quot; rel=&quot;nofollow&quot;&gt;A primer of real analytic functions&lt;/a&gt;&lt;/em&gt; by Krantz and Parks.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2010-12-31T03:09:23.310" LastActivityDate="2010-12-31T03:09:23.310" CommentCount="5" />
  <row Id="15957" PostTypeId="2" ParentId="15955" CreationDate="2010-12-31T02:59:16.647" Score="4" ViewCount="" Body="&lt;p&gt;All three claims are true.  Claim 3 $\Rightarrow$ Claim 2 and Claim 1. The zeros of an analytic function are isolated. This implies Claim 3.&lt;/p&gt;&#xA;" OwnerUserId="3249" LastActivityDate="2010-12-31T02:59:16.647" CommentCount="4" />
  <row Id="15958" PostTypeId="1" AcceptedAnswerId="15959" CreationDate="2010-12-31T03:24:30.947" Score="1" ViewCount="142" Body="&lt;p&gt;On p.33 of Sergei Matveev's &quot;Lectures on Algebraic Topology&quot;, the &lt;em&gt;relative chain group&lt;/em&gt; of a pair $(K,L)$, where $L \subset K$ is a subcomplex, is defined as the free abelian group on simplices with interiors in $K \setminus L$. How is this different from just the chains on $K \setminus L$? How is this the same notion as $C_n(K)/C_n(L)$, i.e. chains in $K$ modded out by chains in $L$, which is another definition I have seen for the relative chain groups? &lt;/p&gt;&#xA;" OwnerUserId="1491" LastActivityDate="2010-12-31T03:52:59.903" Title="The &quot;geometry&quot; behind relative homology" Tags="&lt;algebraic-topology&gt;" AnswerCount="1" />
  <row Id="15959" PostTypeId="2" ParentId="15958" CreationDate="2010-12-31T03:52:59.903" Score="3" ViewCount="" Body="&lt;p&gt;The interior of the boundary of a simplex with interior in $K/L$ need not lie in $K/L$. As a result, the boundary map on the relative chain group $(K,L)$ (in your notation) has to be defined such that the (homological) boundary of a simplex includes only the pieces of the (geometric) boundary that don't lie in $L$. In this manner, we get the same thing as $C_*(K)/C_*(L)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Strictly speaking, $K/L$ is not a simplicial complex, so it doesn't make sense to consider simplicial chains on $K/L$. It does make sense to consider singular chains. In that case $H_*(K,L)$ and $H_*(K-L)$ are generally different. (As a very simple example, take the index to be zero, and consider the case where $L$ disconnects the space $K$.)&lt;/p&gt;&#xA;" OwnerUserId="536" LastActivityDate="2010-12-31T03:52:59.903" CommentCount="1" />
  <row Id="15960" PostTypeId="2" ParentId="15855" CreationDate="2010-12-31T03:56:26.060" Score="4" ViewCount="" Body="&lt;p&gt;slowsolver makes a very good point - the value of the parameter in the expectation is not explicit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;one usually considers log-likelihood-ratios, rather than log-likelihoods - as in the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;let&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$K(\theta;\theta_0) = {\mathrm E}_{\theta_0}{\mathrm log}\frac{f(X|\theta)}{f(X|\theta_0)}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;this is just a kullback-leibler divergence between the pdfs indexed by $\theta$ and $\theta_0$. assuming $\theta_0$ is the [fixed] true value of the parameter, this K-L number differs by a constant from the expected log-likelihood [taken wrt $\theta_0$].&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the significance of this K-L quantity is that it is negative if $\theta\ne\theta_0$ and is zero when $\theta=\theta_0$. [this follows directly from the concavity of log.] thus the K-L quantity, as a function of $\theta$, assumes its max at $\theta=\theta_0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;since the average log-likelihood-ratio for the sample converges [for each $\theta$] to $K(\theta;\theta_0)$, one sees heuristically that the MLE of $\theta$ &lt;em&gt;ought&lt;/em&gt; to be near $\theta_0$ when $n$ is large. [this is a handwaving version of wald's proof that that the MLE is consistent to the true value of the parameter. a rigorous argument requires, &lt;em&gt;inter alia&lt;/em&gt;, continuity of $f$ in $\theta$ and conditions to insure that the K-L number stays bounded away from zero outside of a compact subset of $\theta$ [so that the MLE cannot escape to $\infty$] - as well as assorted other niceties.]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the above behavior of the K-L number as a function of $\theta$  also underlies arguments that [maximum-]likelihood-ratio tests are consistent as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;if the expectation for the 'expected log-likelihood' is wrt some fixed pdf $f(x)$, which need not be one of the $f(\cdot|\theta)$ pdfs, then the value of $\theta$ maximizing the K-L quantity indicates the pdf in the model 'most like' $f(\cdot)$ [it has minimal K-L divergence from $f(\cdot)$]. when $f(\cdot)$ is not in the model [so that the model is incorrect], it is possible that the K-L function attains its max at more than one value of $\theta$. then the MLE can oscillate among the various maximizers of the K-L function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;if the expectation for the 'expected log-likelihood' is wrt $\theta$, rather than a fixed $\theta_0$ [or some other fixed $f(\cdot)$], i don't know offhand what interest there is in that quantity. [it is equivalent to considering $K(\theta,\theta_0)$ as a function of $\theta_0$, rather than as a function of $\theta$. but usually one wants the behavior of the entire average-log-likelihood[-ratio] function under the fixed true value $\theta_0$, not just its behavior at $\theta$ when $\theta$ is the true value.]&lt;/p&gt;&#xA;" OwnerUserId="1547" LastActivityDate="2010-12-31T03:56:26.060" CommentCount="0" />
  <row Id="15962" PostTypeId="2" ParentId="370" CreationDate="2010-12-31T04:28:37.767" Score="1" ViewCount="" Body="&lt;p&gt;as a young student, I enjoyed peter freyd's fun little book on abelian categories.  The nice thing about Freyd's book is it isn't boring, and it has little pieces of wisdom (opinion) such as the remark that categories are not really important, you just define them so you can define functors.  And in fact you just define functors so you can define natural transformations, the really interesting things.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course you may disagree, but blunt debatable assertions (like this one) always make for more interesting reading.  Another provocative remark by this author is the observation that he himself seldom learnt math by reading books, but rather by talking to people.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the nice link above I learned that Goldblatt also quotes a remark (which may have inspired Freyd's) by Eilenberg and Maclane that categories are entirely secondary to functors and natural transformations, on page 194 where he introduces these latter concepts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Leinster's notes linked by Patrick, look nice - a bit like an introduction to Maclane's Categories for the working mathematician, chatty and full of debatable assertions, (many of which I disagree with, but enjoy thinking about).  He does not give much credit, but I believe the adjoint functor theorems he quotes without proof, (GAFT,...) may be due to Freyd.  Leinster's notes are easy reading and informative.&lt;/p&gt;&#xA;" OwnerUserId="5211" LastEditorUserId="5211" LastEditorDisplayName="" LastEditDate="2010-12-31T05:36:55.767" LastActivityDate="2010-12-31T05:36:55.767" CommunityOwnedDate="2010-12-31T04:28:37.767" />
  <row Id="15963" PostTypeId="1" AcceptedAnswerId="15973" CreationDate="2010-12-31T04:47:53.840" Score="8" ViewCount="330" Body="&lt;p&gt;Let $X$ and $Y$ be topological spaces. A function $f: X \rightarrow Y$ is defined as continuous if for each open set $U \subset Y$, $f^{-1}(U)$ is open in $X$. This definition makes sense to me when $X$ and $Y$ are metric spaces- it is equivalent to the usual $\epsilon-\delta$ definition. But why is this a good definition when $X$ and $Y$ are not metric spaces? &#xA;How should we think about this definition intuitively? &lt;/p&gt;&#xA;" OwnerUserId="4457" LastEditorUserId="622" LastEditorDisplayName="" LastEditDate="2010-12-31T15:43:53.757" LastActivityDate="2010-12-31T15:43:53.757" Title="What is the intuition for the point-set topology definition of continuity?" Tags="&lt;topology&gt;&lt;intuition&gt;" AnswerCount="6" CommentCount="1" FavoriteCount="1" />
  <row Id="15964" PostTypeId="2" ParentId="15963" CreationDate="2010-12-31T05:00:07.197" Score="2" ViewCount="" Body="&lt;p&gt;You have it exactly right.  It works well when $X$ and $Y$ are metric spaces, and has proved useful in more general contexts.  When you think of open sets as points &quot;near&quot; another, this is the proper translation of the usual $\epsilon-\delta$ definition.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-12-31T05:00:07.197" />
  <row Id="15965" PostTypeId="2" ParentId="15963" CreationDate="2010-12-31T05:39:36.830" Score="7" ViewCount="" Body="&lt;p&gt;Maybe it's just me, but I've never thought that the usual $\epsilon$-$\delta$ definition of continuity is intuitive at all. Why should a function be continuous at $x$ if every ball of radius $\epsilon$ around $f(x)$ contains the image under $f$ of a ball of radius $\delta$ around $x$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instead, in metric spaces, I think of a function as continuous if it preserves limits, which can be intuitively (and generalizably) be phrased by saying that $f$ is continuous if and only if whenever $x$ is in the closure of a set $A$, then $f(x)$ is in the closure of the set $f(A)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Take a piece of paper and draw out the arguments of the next two paragraphs)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To see that the $\epsilon$-$\delta$ continuity implies 'closure' continuity, suppose that $f$ is not closure continuous at $x$, that is $x$ is in the closure of some set $A$ but $f(x)$ is not in the closure of $f(A)$. Then there exists an $\epsilon$-ball around $f(x)$ that does not intersect $f(A)$ even though every $\delta$-ball intersects $A$. Hence, some $\epsilon$-ball around $f(x)$ contains no image of a $\delta$-ball around $x$, and so $f$ is also not $\epsilon$-$\delta$ continuous.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To see that 'closure' continuity implies $\epsilon$-$\delta$ continuity, suppose that $f$ is not $\epsilon$-$\delta$ continuous. Then there exists an $\epsilon$-ball around $f(x)$ that contains no image of a $\delta$-ball around $x$. In other words, the preimage of the $\epsilon$-ball around $f(x)$ contains no $\delta$-ball around $x$, so let $A$ be the collection of points that are not in the preimage of the $\epsilon$-ball. Then $x$ is in the closure of $A$ since any $\delta$-ball around $x$ has a point outside the preimage of the $\epsilon$-ball and hence in $A$, but $f(x)$ is not in the closure of $f(A)$ since the $\epsilon$-ball around $f(x)$ is disjoint from $f(A)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the cool thing to notice is that the above equivalence of definitions works perfectly fine if you replace $\delta$-balls and $\epsilon$-balls with open sets in the appropriate topological spaces, so really what you should care about is how to make sense of 'closure' continuity in a space that is not a metric space, and the answer is given by the &lt;a href=&quot;http://en.wikipedia.org/wiki/Kuratowski_closure_axioms&quot; rel=&quot;nofollow&quot;&gt;wiki:Kuratowski Closure Axioms&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might also find useful the answers to &lt;a href=&quot;http://mathoverflow.net/questions/19152/why-is-a-topology-made-up-of-open-sets&quot; rel=&quot;nofollow&quot;&gt;this mathoverflow question&lt;/a&gt;, specifically &lt;a href=&quot;http://mathoverflow.net/questions/19152/why-is-a-topology-made-up-of-open-sets/19156#19156&quot; rel=&quot;nofollow&quot;&gt;this one by sigfpe&lt;/a&gt; and &lt;a href=&quot;http://mathoverflow.net/questions/19152/why-is-a-topology-made-up-of-open-sets/19173#19173&quot; rel=&quot;nofollow&quot;&gt;this one by Vectornaught&lt;/a&gt;. The first one talks about how open sets can be thought of as rulers that try to measure imprecisely things in the vector space (but which doesn't explain why continuity it as it is)), while the second phrases the Kuratwoski Closure Axioms in terms of the intuitive notion of 'nearness' of points (which does account for continuity).&lt;/p&gt;&#xA;" OwnerUserId="400" LastActivityDate="2010-12-31T05:39:36.830" />
  <row Id="15966" PostTypeId="1" CreationDate="2010-12-31T05:51:02.327" Score="15" ViewCount="462" Body="&lt;p&gt;A positive integer is &lt;em&gt;$B$-smooth&lt;/em&gt; if and only if all of its prime divisors are less than or equal to a positive real $B$. For example, the $3$-smooth integers are of the form $2^{a} 3^{b}$ with non-negative exponents $a$ and $b$, and those integers less than or equal to $20$ are $\{1,2,3,4,6,8,9,12,16,18\}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In Ramanujan's first letter to G. H. Hardy, Ramanujan emphatically quotes (without proof) his result on the number of $3$-smooth integers less than or equal to $N &amp;gt; 1$,&#xA;\begin{eqnarray}&#xA;\frac{\log 2 N \ \log 3N}{2 \log 2 \ \log 3}.&#xA;\end{eqnarray}&#xA;This is an amazingly accurate approximation, as it differs from the exact value by less than 3 for the first $2^{1000} \approx 1.07 \times 10^{301}$ integers, as shown by Pillai. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Question&lt;/em&gt;: Knowing full well that Ramanujan rarely gave proofs of his own claims, I wonder if a proof of this particular estimate appears somewhere in the literature. Is this problem still open? If not, what is a reference discussing its proof?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="2720" LastEditorUserId="2720" LastEditorDisplayName="" LastEditDate="2011-02-10T20:29:31.177" LastActivityDate="2011-02-25T06:13:26.017" Title="Ramanujan's First Letter to Hardy and the Number of $3$-Smooth Integers" Tags="&lt;number-theory&gt;&lt;real-analysis&gt;&lt;reference-request&gt;&lt;analytic-number-theory&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="1" />
  <row Id="15967" PostTypeId="2" ParentId="15963" CreationDate="2010-12-31T05:58:19.797" Score="1" ViewCount="" Body="&lt;p&gt;It's true that this definition generalizes that for metric spaces, but there are other generalizable definitions (e.g. takes convergent sequences to convergent sequences), and perhaps implicit in the OP's question is: why this particular definition?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course part of the answer is that it turns out to work well, but this is not too satisfying. The following are just a couple of things that just occurred to me. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;More generally, given a class of mathematical objects like a topological space (vector space, group, ring, etc.) it is natural to ask: what are the &quot;structure-preserving maps&quot; between such objects? Vector spaces are sets equipped with a scalar multiplication map; groups are sets equipped with a group operation, etc. and the notions of linear map and homomorphism are precisely defined to preserve this structure. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now with a topological space, of course, the structure comes as a set of &quot;open sets.&quot; Here the generalization from concepts of metric spaces is especially clear. Therefore, a continuous map, a structure-preserving map on a topological space, should be one that &quot;preserves open sets.&quot; At first you might think that such a map should take open sets to open sets (i.e. an open map), but examining the conditions on open sets shows that this is bad. The point is that if $f$ is a set map, then $f^{-1}$ is actually much &quot;nicer&quot; than $f$ in terms of how it commutes with unions, intersections, etc. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, I might make the following observation. A structure-preserving map from $X$ to a set $Y$ should prescribe a natural structure for $Y$. &lt;/p&gt;&#xA;" OwnerUserId="1491" LastActivityDate="2010-12-31T05:58:19.797" CommentCount="1" />
  <row Id="15968" PostTypeId="2" ParentId="15966" CreationDate="2010-12-31T06:29:33.437" Score="9" ViewCount="" Body="&lt;p&gt;If $2^a3^b \lt N, a \ln 2 + b \ln 3 \lt \ln N$ or $a+\frac{b \ln 3}{\ln 2}\lt \frac{\ln N}{\ln 2}$.  For large $N$, we can ignore +1s, so to count the number less than $N$, we have $$\sum_{i=0}^\frac{\ln N}{\ln 3}\frac{\ln N-i\ln 3}{\ln 2}=\frac{(\ln N)^2}{\ln 2 \ln 3}-\frac{(\ln N)^2}{2\ln 2 \ln 3}$$ which is within a constant of the Ramanujan result.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-12-31T06:29:33.437" CommentCount="0" />
  <row Id="15969" PostTypeId="2" ParentId="15966" CreationDate="2010-12-31T06:31:04.923" Score="18" ViewCount="" Body="&lt;p&gt;Taking logs, one finds that $2^a 3^b \leq N$ if and only if &#xA;$a \log 2 + b \log 3 \leq \log N.$&#xA;So estimating the number of 3-smooth integers $\leq N$ is the same as estimating&#xA;the number of integer lattice points in the region&#xA;$a, b \geq 0, a \log 2 + b \log 3 \leq \log N.$&#xA;The standard estimate for the number of integer lattice points in a convex region is the area of the region.  In our case the region is a triangle, of area equal to&#xA;$$\dfrac{1}{2}\dfrac{\log N}{\log 2}\dfrac{\log N}{\log 3},$$&#xA;which is close to the estimate you attribute to Ramanujan.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More precisely, the estimate you wrote down is equal &#xA;to $$\dfrac{1}{2}\dfrac{\log N + \log 2}{\log 2}\dfrac{\log N + \log 3}{\log 3}&#xD;&#xA;= \dfrac{1}{2}(\dfrac{\log N}{\log 2} + 1)(\dfrac{\log N}{\log 3} + 1).$$&#xA;This probably improves the estimate coming from the area, by including correction terms coming from the integer points along the boundary of the region (in particular, the contributions coming from $a = 0$ or $b = 0$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(The fact that these &quot;boundary&quot; contributions come with a multiplicity of $1/2$ seems similar to other contexts in which one approximates a discrete sum by an area, e.g. in Euler--MacLaurin summation, and perhaps thinking from that point of view would let one get a better handle on the precise estimates.)&lt;/p&gt;&#xA;" OwnerUserId="221" LastEditorUserId="221" LastEditorDisplayName="" LastEditDate="2011-02-25T06:13:26.017" LastActivityDate="2011-02-25T06:13:26.017" CommentCount="4" />
  <row Id="15970" PostTypeId="1" AcceptedAnswerId="15972" CreationDate="2010-12-31T06:58:11.373" Score="3" ViewCount="173" Body="&lt;p&gt;I am reading &lt;a href=&quot;http://people.hofstra.edu/Stefan_Waner/diff_geom/Sec1.html&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It says &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Intuitively, an open set is a solid region minus its boundary. If we include the boundary, we get a closed set, which formally is defined as the complement of an open set.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now, question is if a &lt;code&gt;closed set&lt;/code&gt; includes interior points also then how can it be &lt;code&gt;complement&lt;/code&gt;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know basic set theory.&#xA;Enlighten me! :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks! &lt;/p&gt;&#xA;" OwnerUserId="705" LastEditorUserId="723" LastEditorDisplayName="" LastEditDate="2010-12-31T10:24:01.117" LastActivityDate="2010-12-31T10:24:01.117" Title="Does closed set contain only boundary points or interior points also?" Tags="&lt;topology&gt;&lt;elementary-set-theory&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="15972" PostTypeId="2" ParentId="15970" CreationDate="2010-12-31T07:14:15.743" Score="2" ViewCount="" Body="&lt;p&gt;A Closed set is by definition a set whose complement is an open set. Note that this also includes the possibility that a set is both open and closed, for example in a space with two connected components, each component is both open and closed. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, in what you have highlighted the complement of the solid region (inclusive of boundary) i.e. the whole space without the region, is open. Which, means that the solid region (inclusive of boundary) is closed.   &lt;/p&gt;&#xA;" OwnerUserId="4806" LastActivityDate="2010-12-31T07:14:15.743" CommentCount="7" />
  <row Id="15973" PostTypeId="2" ParentId="15963" CreationDate="2010-12-31T07:36:20.263" Score="7" ViewCount="" Body="&lt;p&gt;One abstract way to think about continuity (in the sense that it generalizes to non-metric spaces) is that it is about &lt;em&gt;error.&lt;/em&gt;  A function $f : X \to Y$ is continuous at $x$ precisely when $f(x)$ can be &quot;effectively measured&quot; in the sense that, by measuring $x$ closely enough, we can measure $f(x)$ to any desired precision.  (In other words, the error in our measurement of $f(x)$ can be controlled.  &quot;Precision&quot; here means &quot;to within an arbitrary neighborhood of $f(x)$,&quot; so it does not depend on any metric notions.)  This is an abstract formulation of one of the most basic assumptions of science: that (most of) the quantities we try to measure ($f(x)$) depend continuously on the parameters of our experiments ($x$).  If they didn't, science would be effectively impossible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you like thinking about limits, a function is continuous if and only if it preserves limits of &lt;a href=&quot;http://en.wikipedia.org/wiki/Filter_(mathematics)&quot; rel=&quot;nofollow&quot;&gt;filters&lt;/a&gt; or, equivalently, &lt;a href=&quot;http://en.wikipedia.org/wiki/Net_(mathematics)&quot; rel=&quot;nofollow&quot;&gt;nets&lt;/a&gt;.  These are two ways to generalize converge of sequences to spaces which are not first-countable.  &lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-12-31T10:37:20.663" LastActivityDate="2010-12-31T10:37:20.663" CommentCount="1" />
  <row Id="15974" PostTypeId="2" ParentId="15966" CreationDate="2010-12-31T07:47:10.873" Score="13" ViewCount="" Body="&lt;p&gt;For any reference requirement related to Ramanujan, it is always a good idea to check the series of volumes, titled Ramanujan's notebooks, compiled and annotated by Bruce C. Berndt and others.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a special case of Entry 15 in Ramanujan's second notebook, which is about numbers of the form $\displaystyle a^p b^q$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A reference to this can be found here: &lt;a href=&quot;http://books.google.com/books?id=tX5UUCempn8C&amp;amp;pg=PA66&quot; rel=&quot;nofollow&quot;&gt;Ramanujan's notebooks, Volume 4&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I suggest you read page 66 onwards)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A snapshot (from the google books link itself):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/OI2J1.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This volume talks about the &quot;proof&quot; Ramanujan gave (pages 68 and 69), provides references to Hardy's book (which apparently has a whole chapter on this) and also mentions a paper by Hardy and Littlewood which deals with it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;De Bruijn has considered the number of $y$-smooth numbers in this paper: &lt;a href=&quot;http://alexandria.tue.nl/repository/freearticles/597499.pdf&quot; rel=&quot;nofollow&quot;&gt;On the number of positive integer less than x and free of prime factors greater than y&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The number of $y$-smooth numbers $\le x$ is apparently now known in literature as the DeBruijn function: $\psi(x,y)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A closely related function is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Dickman_function&quot; rel=&quot;nofollow&quot;&gt;DeBruijn-Dickman&lt;/a&gt; function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is also a &lt;a href=&quot;http://archive.numdam.org/ARCHIVE/JTNB/JTNB_1993__5_2/JTNB_1993__5_2_411_0/JTNB_1993__5_2_411_0.pdf&quot; rel=&quot;nofollow&quot;&gt;survey&lt;/a&gt; by Hildebrand and Tenenbaum which should be helpful.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-12-31T23:18:45.257" LastActivityDate="2010-12-31T23:18:45.257" />
  <row Id="15975" PostTypeId="1" CreationDate="2010-12-31T08:08:25.343" Score="1" ViewCount="114" Body="&lt;p&gt;&lt;a href=&quot;http://math.stackexchange.com/questions/13169/what-is-spec-mathbfcx-y-y2-x3-x2&quot;&gt;Here&lt;/a&gt; evgeniamerkulova explains intuition for formal completion very nicely. I was wondering if something could be said if the base ring were $\mathbb C$ instead of $\mathbb Z$.&lt;/p&gt;&#xA;" OwnerUserId="5260" LastActivityDate="2010-12-31T08:08:25.343" Title="Formal completion of a scheme at a point" Tags="&lt;algebraic-geometry&gt;" />
  <row Id="15976" PostTypeId="1" AcceptedAnswerId="16002" CreationDate="2010-12-31T08:33:21.347" Score="7" ViewCount="172" Body="&lt;p&gt;V.I. Arnold put the following question in his Mathematical trivium:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can an asymptotically stable equilibrium position become unstable in the Lyapunov sense under linearization?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It puzzled me for a while, since my experience doesn't include such a pathological case - but I'm still not sure about this one.&lt;/p&gt;&#xA;" OwnerUserId="553" LastActivityDate="2011-01-01T10:34:57.810" Title="Lyapunov stability question (Arnold's trivium)" Tags="&lt;control-theory&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="15977" PostTypeId="1" CreationDate="2010-12-31T08:44:26.170" Score="10" ViewCount="365" Body="&lt;p&gt;I am planning to write a little note detailing several proofs of Lagrange's theorem that every natural number can be written as the sum of four perfect squares. I know of three different proofs so far:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;a completely elementary proof by descent. &lt;/li&gt;&#xA;&lt;li&gt;a proof via Minkowski's theorem and lattices. &lt;/li&gt;&#xA;&lt;li&gt;Jacobi's proof via modular forms. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Can anybody think of any more nice, relatively elementary proofs of this result? Thanks in advance. &lt;/p&gt;&#xA;" OwnerUserId="1491" LastActivityDate="2010-12-31T15:49:01.310" Title="Proofs that every natural number is a sum of four squares" Tags="&lt;number-theory&gt;" AnswerCount="4" CommentCount="1" FavoriteCount="3" />
  <row Id="15978" PostTypeId="2" ParentId="15977" CreationDate="2010-12-31T09:00:26.523" Score="7" ViewCount="" Body="&lt;p&gt;There's a proof in Herstein using quaternions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[See also Hardy and Wright's &lt;em&gt;An Introduction to the Theory of Numbers&lt;/em&gt;.  -- PLC.]&lt;/p&gt;&#xA;" OwnerUserId="1470" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2010-12-31T09:34:04.190" LastActivityDate="2010-12-31T09:34:04.190" />
  <row Id="15979" PostTypeId="2" ParentId="15977" CreationDate="2010-12-31T09:32:49.463" Score="5" ViewCount="" Body="&lt;p&gt;One can easily deduce the four squares theorem from the Gauss-Legendre three squares theorem.  The latter is actually less elementary, but there is a very nice proof using Hasse-Minkowski and the Aubry-Davenport-Cassels Lemma.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the approach taken in Serre's &lt;em&gt;A Course in Arithmetic&lt;/em&gt;, Appendix to Chapter IV.  Quadratic forms which satisfy the conclusion of the Aubry-Davenport-Cassels Lemma (&lt;strong&gt;ADC forms&lt;/strong&gt;) have been a topic of recent research of mine, so it is fair to say that this is currently my favorite proof.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the quaternionic proof of Chris Card's answer (which is also a very important one) can be understood and, arguably, clarified in these terms.  The sum of four squares form does not itself satisfy the hypotheses of the ADC Lemma.  This is related to the fact that it is the norm form of the &lt;em&gt;nonmaximal order&lt;/em&gt; $\mathbb{Z}[i,j,k]$ in the Hamilton quaternion algebra $H = \left(\frac{-1,-1}{\mathbb{Q}} \right)$.  A maximal order containing it is $\mathbb{Z}[i,j,k,\frac{1+i+j+k}{2}]$.  The norm form on this order is the quadratic form &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$q = x^2 + y^2 + z^2 + w^2 + xw + yw + zw$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This form &lt;em&gt;does&lt;/em&gt; satisfy the hypothesis of the ADC Lemma (it is &lt;em&gt;Euclidean&lt;/em&gt; in my terminology), so it follows immediately from Hasse-Minkowski that it is universal, i.e., represents all positive integers.  Moreover the form $q$ is sufficiently closely &#xA;related to the sum of four squares form that a little elementary fiddling around shows that the sum of four squares form is universal as well.  For other examples of this phenomenon, see &lt;a href=&quot;http://opensiuc.lib.siu.edu/cgi/viewcontent.cgi?article=1113&amp;amp;context=math_articles&quot; rel=&quot;nofollow&quot;&gt;this recent preprint of R.W. Fitzgerald&lt;/a&gt; as well as some papers of J.I. Deutsch that are cited there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(If you dare, you can take a look at &lt;a href=&quot;http://www.math.uga.edu/~pete/ADCformsPart1b.pdf&quot; rel=&quot;nofollow&quot;&gt;my magnum opus on Euclidean forms and ADC forms&lt;/a&gt;.  &lt;em&gt;Caveat emptor&lt;/em&gt;: more than being unpolished, there are some unsanded sharp corners here.)  &lt;/p&gt;&#xA;" OwnerUserId="299" LastActivityDate="2010-12-31T09:32:49.463" />
  <row Id="15980" PostTypeId="2" ParentId="15977" CreationDate="2010-12-31T09:50:27.073" Score="4" ViewCount="" Body="&lt;p&gt;Here's one to add to your list by Andrews, Ekhad and Zeilberger.&#xA;&lt;a href=&quot;http://arxiv.org/PS_cache/math/pdf/9206/9206203v1.pdf&quot; rel=&quot;nofollow&quot;&gt;A short proof of Jacobi’s formula for the number of representations of an integer as the&#xA;sum of four squares.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1301" LastActivityDate="2010-12-31T09:50:27.073" />
  <row Id="15981" PostTypeId="1" AcceptedAnswerId="15982" CreationDate="2010-12-31T09:51:12.427" Score="4" ViewCount="167" Body="&lt;p&gt;all examples i have come across so far deal with norm defined on a vector space. Can norm only be defined on vector spaces&lt;/p&gt;&#xA;" OwnerUserId="5261" LastActivityDate="2010-12-31T18:31:19.203" Title="Can a norm only be defined on vector spaces ?" Tags="&lt;real-analysis&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="15982" PostTypeId="2" ParentId="15981" CreationDate="2010-12-31T10:07:33.680" Score="11" ViewCount="" Body="&lt;p&gt;For instance, a norm can be defined on an abelian group: see Section 2.4 &lt;a href=&quot;http://math.uga.edu/~pete/8410Chapter2v2.pdf&quot; rel=&quot;nofollow&quot;&gt;of these notes&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is perhaps less commonly done, but one can define a norm on any group: see Exercise 2.17 in the aforelinked lecture notes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course there are many structures throughout mathematics that are called &quot;norms&quot;, and you can make more if you want!  But the above example is very closely related to the norm on a vector space: there is an induced metric, and so forth.  In fact a norm on a vector space in the usual sense gives a norm on the underlying additive group $(V,+)$.&lt;/p&gt;&#xA;" OwnerUserId="299" LastActivityDate="2010-12-31T10:07:33.680" />
  <row Id="15983" PostTypeId="2" ParentId="15963" CreationDate="2010-12-31T10:17:26.600" Score="0" ViewCount="" Body="&lt;p&gt;Suppose $x$ is a point in some topological space, then we can define $\mathcal{N}_x$ to be the set of (not necessarily open) neighbourhoods of $x$. Then &quot;$f$ is continuous at $x$&quot; is defined as &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\forall V \in \mathcal{N}_{f(x)} \exists U \in \mathcal{N}_x : f(U) \subseteq V.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or more colourfully: Whenever &quot;the enemy&quot; comes with a cleverly chosen and &quot;small&quot; neighbourhood of $f(x)$, we must be able to find a neighbourhood of $x$ that maps into said neighbourhood.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is nice about this is that it is (obviously?) the topological version of the $\varepsilon$-$\delta$ definition from $\mathbb{R}$ and metric spaces that we know (and love(?)), and it's relatively easy to prove that $f: X \to Y$ is continuous at $x$ for all $x \in X$ if and only if $f^{-1}(U)$ is open in $X$ for all open $U \subseteq Y$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I'm trying to say is that I don't have much intuition for the &quot;preimage of open sets is open&quot; definition either, but it's not clear to me that you really need that. We take this as the definition because it's simple, it's entirely written in terms of the topologies of $X$ and $Y$ (i.e. the collections of their respective open sets) and easily shown to be equivalent to something which we do have an intuition about (assuming that one finds $\varepsilon$-$\delta$ intuitive, obviously).&lt;/p&gt;&#xA;" OwnerUserId="1269" LastEditorUserId="1269" LastEditorDisplayName="" LastEditDate="2010-12-31T12:21:57.147" LastActivityDate="2010-12-31T12:21:57.147" />
  <row Id="15984" PostTypeId="1" AcceptedAnswerId="16001" CreationDate="2010-12-31T10:19:10.840" Score="8" ViewCount="341" Body="&lt;p&gt;I've had no luck with this one. None of the convergence tests pop into mind.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried looking at it in this form $\sum \sin n\frac{1+\frac{1}{2}+\cdots+\frac{1}{n}}{n}$ and apply Dirichlets test. I know that $\frac{1+\frac{1}{2}+\cdots+\frac{1}{n}}{n} \to 0$ but not sure if it's decreasing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regarding absolute convergence, I tried:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$|\sin n\frac{1+\frac{1}{2}+\cdots+\frac{1}{n}}{n}|\geq \sin^2 n\frac{1+\frac{1}{2}+\cdots+\frac{1}{n}}{n}=$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$=\frac{1}{2}\frac{1+\frac{1}{2}+\cdots+\frac{1}{n}}{n}-\frac{1}{2}\cos 2n\frac{1+\frac{1}{2}+\cdots+\frac{1}{n}}{n}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But again I'm stuck with $\cos 2n\frac{1+\frac{1}{2}+\cdots+\frac{1}{n}}{n}$.&lt;br&gt;&#xA;Assuming it converges then I've shown that $\sum \sin n\frac{1+\frac{1}{2}+\cdots+\frac{1}{n}}{n}$ doesn't converge absolutely.&lt;/p&gt;&#xA;" OwnerUserId="3237" LastEditorUserId="3237" LastEditorDisplayName="" LastEditDate="2010-12-31T11:50:35.817" LastActivityDate="2010-12-31T18:30:28.143" Title="Does $\sum_{n=1}^\infty\frac{\sin n}{n}(1+\frac{1}{2}+\cdots+\frac{1}{n})$ converge (absolutely)?" Tags="&lt;calculus&gt;&lt;sequences-and-series&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="1" />
  <row Id="15985" PostTypeId="1" AcceptedAnswerId="15986" CreationDate="2010-12-31T10:37:37.283" Score="1" ViewCount="168" Body="&lt;p&gt;Again, I am reading &lt;a href=&quot;http://people.hofstra.edu/Stefan_Waner/diff_geom/Sec2.html&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt;.&#xA;I am finding it a bit difficult to understand the definition of &lt;code&gt;n-dimensional smooth manifold&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\{U_a; x^1_a, x^2_a, ..., x^n_a\}$ ----(1)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is the thing (1) a set? (I think it is not). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it a tuple? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, is $U_a$ a set or a set of sets? What is significance of the subscript $a$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like it very much if someone explains the definition easier to understand. With good examples.  &lt;/p&gt;&#xA;" OwnerUserId="705" LastEditorUserId="705" LastEditorDisplayName="" LastEditDate="2010-12-31T10:44:01.693" LastActivityDate="2011-01-01T10:33:00.080" Title="Help with definition of n-dimensional smooth manifold" Tags="&lt;topology&gt;&lt;differential-geometry&gt;" AnswerCount="2" />
  <row Id="15986" PostTypeId="2" ParentId="15985" CreationDate="2010-12-31T10:49:42.610" Score="4" ViewCount="" Body="&lt;p&gt;This is there in the notes. $\{ U_a \}$ is an open cover of $M$. This is a set where each element $U_a$ is an open subset of $M$. The $a$ is used as an index for this set; so it would help to have put $a\in A$, say, where appropriate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each $x_a^i$ is a coordinate function on the open set $U_a$ (for given $a$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The notes do give an example of the unit sphere which has an open cover consisting of two open subsets.&lt;/p&gt;&#xA;" OwnerUserId="638" LastActivityDate="2010-12-31T10:49:42.610" CommentCount="6" />
  <row Id="15987" PostTypeId="2" ParentId="15985" CreationDate="2010-12-31T10:57:54.193" Score="0" ViewCount="" Body="&lt;p&gt;A n-dim manifold is local homeomorphism like a n-dim Euclidean Space.In the smooth situation,you can put the open set（just a common part） of R^n on the any part of the mainfold with the diffeomorphism,it keeps the full rank.&#xA; sphere is a good and basic example, you can white down the diffeomorphic mapping,but for many other figures that is complicated.&lt;/p&gt;&#xA;" OwnerUserId="4706" LastEditorUserId="4706" LastEditorDisplayName="" LastEditDate="2011-01-01T10:33:00.080" LastActivityDate="2011-01-01T10:33:00.080" />
  <row Id="15988" PostTypeId="2" ParentId="15984" CreationDate="2010-12-31T11:05:51.343" Score="11" ViewCount="" Body="&lt;p&gt;To show that the series does not converge absolutely, it suffices to show that $\sum\nolimits_{n = 1}^\infty  {|\frac{{\sin n}}{n}|}  = \infty $. For this purpose, see &lt;a href=&quot;http://www.math.niu.edu/~rusin/known-math/99/series&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt;: the paragraph starting with &quot;It's not absolutely convergent&quot; as well as the one starting with &quot;This series exhibits rather irregular behaviour&quot; (two different approaches).&lt;/p&gt;&#xA;" OwnerUserId="2810" LastEditorUserId="2810" LastEditorDisplayName="" LastEditDate="2010-12-31T11:15:51.387" LastActivityDate="2010-12-31T11:15:51.387" CommentCount="12" />
  <row Id="15989" PostTypeId="2" ParentId="15917" CreationDate="2010-12-31T12:27:01.150" Score="0" ViewCount="" Body="&lt;p&gt;The confusion stems from the fact that you do not say what quantities change. Say, if you mean that $X$ changes by a small amount $\Delta X$ then &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x=\log (X+\Delta X) - \log X = \log (1+\frac{\Delta X}{X}) \approx \frac{\Delta X}{X} \; .$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So $x$ is really the relative change in the quantity $X$ and likewise for the other quantities.&lt;/p&gt;&#xA;" OwnerUserId="3567" LastActivityDate="2010-12-31T12:27:01.150" />
  <row Id="15990" PostTypeId="1" CreationDate="2010-12-31T12:34:57.667" Score="0" ViewCount="142" Body="&lt;p&gt;In the computational topology literature, the reduction algorithm for computing the Smith normal form of a boundary matrix uses the notation $b_j &amp;gt; 1 \: \text{and}\: b_j|b_{j+1}$ in the context of the diagonal elements of the Smith matrix. Can anyone give me an idea for what it means?&lt;/p&gt;&#xA;" OwnerUserId="4820" LastActivityDate="2010-12-31T12:34:57.667" Title="What does $b_i|b_{i+1}$ mean in this context?" Tags="&lt;linear-algebra&gt;&lt;abstract-algebra&gt;&lt;algebraic-topology&gt;" CommentCount="3" />
  <row Id="15991" PostTypeId="2" ParentId="15801" CreationDate="2010-12-31T12:42:38.370" Score="6" ViewCount="" Body="&lt;p&gt;Updated answer to (2): Three non-intersecting $60^\circ$ loxodromes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/7igYI.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The axes are coplanar and inclined at $120^\circ$ to each other. This image shows that symmetry better:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/ERO5C.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And here's the Mercator projection:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/aQ3W3.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My approach was to plot one loxodrome such that its Mercator projection is a (black) line through the origin. Then, I tilted the spherical loxodrome &quot;toward the camera&quot;; that is, I rotated the sphere about the horizontal axis to get new, curvy (red) projections.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/9Sv5u.gif&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From tilt-angles $108^\circ$ to $143^\circ$, the &quot;curve&quot; lies between parts of the &quot;line&quot;, indicating a range of red loxodromes that don't intersect the black one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/TWfR4.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a certain sub-range ($108^\circ$ to about $125^\circ$), a third (blue) non-intersecting loxodrome can be added by rotating the red one about the Mercator origin. Here's an image from the end of that range, where red and blue are tangent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/ZCKS8.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's the end of the illustrated intro. Now for some equations ...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Starting with your parameterization of the loxodrome, then tilting via angle $\theta$, gives this parameterization of the Mercator projection:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xD;&#xA;u &amp;amp;= \rm{atan}\left( \frac{\sin t}{\cos t \cos \theta + \sinh\left(t \cot\phi\right) \sin \theta }\right) \\&#xD;&#xA;v &amp;amp;= \rm{atanh}\left( \frac{-\cos t \sin\theta + \sinh\left(t \cot\phi\right) \cos\theta }{\cosh\left(t \cot\phi\right)}\right)&#xD;&#xA;\end{align}&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A tilted loxodrome crosses into the range of (possible) non-intersection with the un-tilted loxodrome when the &quot;top&quot; of the outer loop about its tilted north pole meets the Mercator origin. (The nature of loxodromes guarantees that the two loxodromes will be tangent there.) The point on the loop corresponds to $t=\pi$, for which $u$ is already zero; for $v$ to vanish, we must have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$0 = -\cos\pi \sin\theta + \sinh\left(\pi \cot\phi\right) \cos\theta = \sin\theta+\sinh\left(\pi \cot\phi \right) \cos\theta$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\tan\theta = -\sinh\left(\pi\cot\phi\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consequently, appropriately adjusting the &quot;branch&quot; of $\rm{atan}$, the range begins at&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\theta_0 := \pi - \rm{atan}\left(\sinh\left(\pi\cot\phi\right)\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The range of (possible) non-intersection ends when the loop around the tilted loxodrome's south pole brushes against the un-tilted loxodrome. This is when the point corresponding to $t=-\pi$ has $v = \pi\cot\phi$ (matching the upper-right point of the &quot;straight&quot; loxodrome projection). So, the range ends at&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\theta_1 := 2\;\rm{atan}\left( \sinh\left(\pi\cot\phi\right) \right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I write &quot;range of &lt;em&gt;possible&lt;/em&gt; non-intersection&quot;, because that range collapses when $\theta_0 = \theta_1$. This gives us a critical loxodrome angle. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\phi_{*} = \rm{atan}\left(\frac{\pi}{ \rm{asinh}\left( \tan\frac{\pi}{3} \right) } \right) \approx 67.2565^\circ$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/c58Kn.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You cannot arrive at two non-intersecting loxodromes with $\phi &amp;gt; \phi_{*}$ --in particular, with $\phi=80^\circ$-- by tilting one relative to the other in the way I've described.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's $\phi = 80^\circ$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/PEfya.gif&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, &quot;the way I've described&quot; lacks generality. In addition to &quot;vertical&quot; tilts, one should also consider &quot;lateral&quot; spins (horizontal shifts in the Mercator projection). I'll leave that, and a full investigation of the three-loxodrome scenario, as an exercise.&lt;/p&gt;&#xA;" OwnerUserId="409" LastEditorUserId="409" LastEditorDisplayName="" LastEditDate="2011-01-01T23:46:47.327" LastActivityDate="2011-01-01T23:46:47.327" CommentCount="2" />
  <row Id="15993" PostTypeId="2" ParentId="15902" CreationDate="2010-12-31T13:18:41.060" Score="5" ViewCount="" Body="&lt;p&gt;I just like to point out that this argument is in Hardy and Wright, An Introduction to the Theory of Numbers, with the slight differences that they avoid the use of the floor and ceiling functions, and finish off (quite nicely, in my opinion) with induction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll type it here, to save you looking it up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Theorem: $\theta(n) &amp;lt; 2n \log 2$ for all $ n \ge 1,$ where $$\theta(x) = \log \prod_{p \le x} p.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $M = { 2m+1 \choose m},$ an integer, which occurs twice in the expansion of $(1+1)^{2m+1}$ and so $2M &amp;lt; 2^{2m+1}.$ Thus $M&amp;lt; 2^{2m}.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $m+1 &amp;lt; p \le 2m+1,$ $p$ divides $M$ since it divides the numerator, but not the denominator, of $ { 2m+1 \choose m } = \frac{(2m+1)(2m)\ldots(m+2)}{m!}.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left( \prod_{m+1 &amp;lt; p \le 2m+1} p \right) | M $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \theta(2m+1) - \theta(m+1) = \sum_{m+1 &amp;lt; p \le 2m+1} \log p \le \log M &amp;lt; 2m \log 2.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The theorem is clear for $n=1$ and $n=2,$ so suppose that it is true for all $n \le N-1.$ If $N$ is even, we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \theta(N)= \theta(N-1) &amp;lt; 2(N-1) \log 2 &amp;lt; 2N \log 2.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $N$ is odd, $N=2m+1 $ say, we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xD;&#xA;\theta(N)=\theta(2m+1) &amp;amp;=\theta(2m+1)-\theta(m+1)+\theta(m+1) \\&#xD;&#xA;&amp;amp;&amp;lt; 2m \log 2 +2(m+1) \log 2 \\&#xD;&#xA;&amp;amp;=2(2m+1) \log 2 = 2N \log 2,&#xD;&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;since $m+1 &amp;lt; N.$ Hence the theorem is true for $n=N$ and the result follows by induction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: It turns out that this proof was discovered by Erdös and another mathematician, Kalmar, independently and almost simultaneously, in 1939. See &lt;a href=&quot;http://www.ias.ac.in/resonance/Mar1998/pdf/Mar1998Reflections.pdf&quot; rel=&quot;nofollow&quot;&gt;Reflections, Ramanujan and I,&lt;/a&gt; by Paul Erdös.&lt;/p&gt;&#xA;" OwnerUserId="1301" LastEditorUserId="1301" LastEditorDisplayName="" LastEditDate="2011-03-10T15:39:02.550" LastActivityDate="2011-03-10T15:39:02.550" CommentCount="2" />
  <row Id="15994" PostTypeId="2" ParentId="15981" CreationDate="2010-12-31T13:31:29.123" Score="2" ViewCount="" Body="&lt;p&gt;The word &quot;norm&quot; is also used in connection with &lt;a href=&quot;http://en.wikipedia.org/wiki/Euclidean_domain&quot; rel=&quot;nofollow&quot;&gt; Euclidean domains (ring theory)&lt;/a&gt;, although that Wikipedia page tries to avoid confusion by preferring instead &quot;Euclidean function&quot;.  &lt;strong&gt;Example:&lt;/strong&gt;  The degree of a univariate polynomial with coefficients over a field gives a norm for that kind of ring to be a Euclidean domain.&lt;/p&gt;&#xA;" OwnerUserId="3111" LastActivityDate="2010-12-31T13:31:29.123" />
  <row Id="15995" PostTypeId="1" AcceptedAnswerId="16054" CreationDate="2010-12-31T13:48:48.927" Score="9" ViewCount="308" Body="&lt;p&gt;Interesting problem from a 2000 St. Petersburg school olympiad.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;There are 109 soldiers in a camp. Every night three of them go on watch patrol. Prove that it can be arranged so that after a while, every pair of soldiers has shared watch exactly three times.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think I'm missing the key insight here.&lt;/p&gt;&#xA;" OwnerUserId="1313" LastActivityDate="2011-01-01T20:11:45.513" Title="combinatorics olympiad problem - sort out a schedule" Tags="&lt;combinatorics&gt;&lt;contest-math&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="15996" PostTypeId="2" ParentId="15963" CreationDate="2010-12-31T14:00:13.037" Score="1" ViewCount="" Body="&lt;p&gt;Let me also add one little bit (perhaps this may even seem backwards).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Topology is defined using sets satisfying some set of axioms, that we call Open sets. However, the collection of subsets we choose (respecting some set of axioms) to call a Topology on the space may vary with the underlying set. But, we can go from one such collection to another using a special class of maps. But, how do we define such maps?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, properties that are to be 'intrinsic' to the Topology should not depend on this choice, so should be invariant across these maps. But, these properties (whatever they be) are defined via set operations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps we want the assignment to be direct: i.e. $U$ is open and thus $f(U)$ should be open. But, notice that since the map is general: $f(A \cap B)$ is contained and not equal to $f(A) \cap f(B)$ etc. However, with the assignment $f^{-1}$ (which need not even be well-defined) all is fine with unions and intersections, i.e., we have equalities, so, the properties follow through almost trivially (see for example the proofs that continuous maps preserve compactness, connectedness etc.)  &lt;/p&gt;&#xA;" OwnerUserId="4806" LastEditorUserId="4806" LastEditorDisplayName="" LastEditDate="2010-12-31T14:16:30.637" LastActivityDate="2010-12-31T14:16:30.637" CommentCount="1" />
  <row Id="15997" PostTypeId="1" AcceptedAnswerId="15998" CreationDate="2010-12-31T14:09:13.373" Score="2" ViewCount="120" Body="&lt;p&gt;Hi, given a ring $R$, a &lt;em&gt;nilpotent&lt;/em&gt; ideal $I$ and a morphism $\phi$ of $R$-modules $M \to N$, such that $M/IM \to N/IN$ is an isomorphism.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is easy to see that this implies $\phi$ surjective, but what about injectivity?&lt;/p&gt;&#xA;" OwnerUserId="5262" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-31T18:03:56.400" LastActivityDate="2010-12-31T18:03:56.400" Title="$M/IM \to N/IN$ an isomorphism $\Longrightarrow M \to N$ an isomorphism?" Tags="&lt;commutative-algebra&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="15998" PostTypeId="2" ParentId="15997" CreationDate="2010-12-31T14:12:11.817" Score="4" ViewCount="" Body="&lt;p&gt;The answer is no in general. E.g. if $M = R$ and $N = R/I$, then the natural surjection $M \to N$ will induce an isomorphism $M/I \to N/I$, which won't be&#xA;an isomorphism unless $I = 0$.&lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2010-12-31T14:12:11.817" CommentCount="4" />
  <row Id="15999" PostTypeId="2" ParentId="15977" CreationDate="2010-12-31T14:52:48.957" Score="6" ViewCount="" Body="&lt;p&gt;Perhaps the most beautiful solution is by way of Aubry's lemma - which employs a geometric variant of the Euclidean algorithm to turn a rational represention into an integral representation. This is the same technique that leads to the reflective generation of primitive Pythagorean triples and the associated &lt;a href=&quot;http://math.stackexchange.com/questions/5540/k2k12-being-a-perfect-square-for-infinitely-many-k/5567#5567&quot;&gt;ternary tree structure.&lt;/a&gt; Aubry's results are, in fact, very special cases of general results of Wall, Vinberg, Scharlau et al. on reflective lattices, i.e. arithmetic groups of isometries generated by reflections in hyperplanes. Generally reflections generate the orthogonal group of Lorentzian quadratic forms in dim &amp;lt; 10. See my &lt;a href=&quot;http://mathoverflow.net/questions/33697/assistance-with-understanding-parent-child-relationships-in-pythagorean-triples/33726#33726&quot; rel=&quot;nofollow&quot;&gt;MO post here&lt;/a&gt; for further remarks and references.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In my opinion, the results in this area are some of the most beautiful results in elementary number theory. Strangely enough, for all this beauty they appear to be little known. For example, over a decade ago when I mentioned to John Conway the connection between Aubry's work and Cassels and Pfister he was not aware of this (R.K. Guy told me that the presentation of the PPT ternary tree in their &lt;em&gt;Book of Numbers&lt;/em&gt; (1996) is based on a lecture he heard by an undergraduate, Richard Vogeler, at an MAA section at Brigham Young Univ. on 89-04-07.)  Also Pfister apparently was not aware of Aubry's work when he generalized Cassels result to arbitrary quadratic forms, founding the modern algebraic theory of quadratic forms (&quot;Pfister forms&quot;). Someday I hope to write something on the bizarre history of this beautiful circle of ideas so I would be grateful to hear from anyone who may know further details.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-12-31T15:49:01.310" LastActivityDate="2010-12-31T15:49:01.310" />
  <row Id="16000" PostTypeId="1" CreationDate="2010-12-31T15:00:33.767" Score="1" ViewCount="110" Body="&lt;p&gt;Excerpt from Waterhouse, 14.4 Structure of Finite Connected groups.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thm. Let $A$ represent a finite connected group scheme over a perfect field of characteristic $p$. Then $A$ has the form &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$k[X_1, ..., X_n]/(X_1^{p^{e_1}}, \ldots, X_n^{p^{e_n}})$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proof. Let $I_A$ be the augmentation ideal of $A$. By connectedness, $I_A$ is nilpotent. [...] We assume therefore that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;the Hopf subalgebra $B = A^p$ is one of these truncated polynomial algebras, say with generators $x_i$ and relations &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x_i^{q_i} = 0$. Choose $y_i \in A$ with $y_i^p = x_i$, and choose also a set $\{z_j\}$ in $A$ maximal with resprect to &lt;/p&gt;&#xA;&#xA;&lt;p&gt;the requirements that $z_j^p = 0$ and that the $z_j$ be linearly independent in $I_A/I_A^2$. Let $C = k[Y_i, Z_j]/(Y_i^&#xD;&#xA;&#xD;&#xA;{pq_i}, Z_j^p$ which maps in the obvious way to $A$. We claim that this map is an iso.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Embed $A^p$ in $C$ by $x_i \mapsto Y_i^p$. Then $C$ is a free $B$-module. By the main theorem, $A$ is also free over $A^p$. As in (14.2), then, it is enough to show that $C/I_BC \to A/I_BA$ is an iso.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't understand the last sentence.&lt;/p&gt;&#xA;" OwnerUserId="5262" LastEditorUserId="274" LastEditorDisplayName="" LastEditDate="2010-12-31T18:54:57.693" LastActivityDate="2010-12-31T18:54:57.693" Title="question regarding Waterhouse, affine group schemes" Tags="&lt;algebraic-geometry&gt;&lt;commutative-algebra&gt;&lt;algebraic-groups&gt;" />
  <row Id="16001" PostTypeId="2" ParentId="15984" CreationDate="2010-12-31T16:24:41.560" Score="10" ViewCount="" Body="&lt;p&gt;The sequence $$a_n=\frac{1+\frac{1}{2}+\cdots +\frac{1}{n}}{n}$$ is decreasing because proving $a_{n+1}&amp;lt;a_n$ reduces to proving $$\frac{n}{n+1}&amp;lt;1+\frac{1}{2}+\cdots +\frac{1}{n}$$ which is clearly true.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also the sums $$s_k=\sum_{n=1}^k \sin n$$ are bounded. (This can be easily proved by writing $\sin n$ in its complex form and using finite geometric formula; in fact $|s_k|$ are bounded by $\frac{1}{|1-e^i|}+\frac{1}{|1-e^{-i}|}$). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore $a_n\to 0$ as $n\to \infty$ because $1+\frac{1}{2}+\cdots +\frac{1}{n}\approx \log n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the series converges by Dirichlet test. As Shai shows below, the convergence is not absolute.&lt;/p&gt;&#xA;" OwnerUserId="3249" LastEditorUserId="3249" LastEditorDisplayName="" LastEditDate="2010-12-31T18:30:28.143" LastActivityDate="2010-12-31T18:30:28.143" />
  <row Id="16002" PostTypeId="2" ParentId="15976" CreationDate="2010-12-31T16:41:40.403" Score="4" ViewCount="" Body="&lt;p&gt;For the system $dx/dt=y-x^3$, $dy/dt=-y^3$, the origin is an&#xA;asymptotically stable equilibrium. (It is even &lt;em&gt;globally&lt;/em&gt; attracting,&#xA;as can be seen by sketching the phase portrait with the help of the&#xA;nullclines $y=x^3$ and $y=0$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The linearized system at the origin is $dx/dt=y$, $dy/dt=0$,&#xA;which is not Lyapunov stable: the solution starting at $(x_0,y_0)=(0,\epsilon)$&#xA;is $(x(t),y(t)) = (\epsilon t, \epsilon)$,&#xA;which goes to infinity if $\epsilon \neq 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The matrix corresponding to the linearized system is $\begin{pmatrix} 0 &amp;amp; 1 \\ 0 &amp;amp; 0 \end{pmatrix}$,&#xA;which has zero as a double eigenvalue.&#xA;If both eigenvalues have nonzero real part, then the linearized system determines the stability&#xA;of the original system. If the eigenvalues are $\pm c i$ (for some real $c \neq 0$), then&#xA;the linearized system is a neutral center (hence Lyapunov stable).&#xA;So the phenomenon in question can only occur if at least one eigenvalue is zero.&#xA;(I haven't really thought about what can or cannot happen in the case with one&#xA;zero eigenvalue and one real nonzero eigenvalue.)&lt;/p&gt;&#xA;" OwnerUserId="1242" LastEditorUserId="1242" LastEditorDisplayName="" LastEditDate="2011-01-01T10:34:57.810" LastActivityDate="2011-01-01T10:34:57.810" CommentCount="4" />
  <row Id="16003" PostTypeId="1" AcceptedAnswerId="16019" CreationDate="2010-12-31T16:58:59.917" Score="4" ViewCount="274" Body="&lt;p&gt;I may be over-complicating things, but something doesn't seem right (and I swear this isn't homework, I'm friggin 30 years old :P).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to see what it will cost me to make a TWO liter of Coca-Cola using coke syrup and CO2 (assume the water is free).  Assume CO2 costs 0.21usd per liter of water (25usd for 120 liters).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The coke syrup needs to be diluted at a &quot;5.4:1&quot; ratio with the carbonated water.&#xA;So how much will it cost me to make my own 2-liter?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's what I've done so far:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;syrup cost per liter = C / ((G * D) * 3.785)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;C is the cost of G gallons of coke syrup&lt;/li&gt;&#xA;&lt;li&gt;D is the dilution (5.4), and&lt;/li&gt;&#xA;&lt;li&gt;3.785 converts gallons to liters&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;For a 65.00usd box of 5 gallons of coke syrup, I get a cost of 0.64usd per liter.  Along with CO2 cost, that makes it 1.69usd for a 2-liter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That seems very, very costly, given that 2-liters in the store usually cost much less.  If I'm wrong, where am I going wrong?  If I'm right, why don't restaurants use 2-liters instead of fountain machines to increase their profits/lower their costs?  I know they charge a lot for a &lt;em&gt;glass&lt;/em&gt; of coke (and refills), not a &lt;em&gt;2-liter&lt;/em&gt;, but it still seems like it would make them more money.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I put that equation into an Excel sheet, and it seems like the syrup would have to be 30usd in order to get the cost down to 1.00usd/2-liter, and that doesn't even seem like much of a gain over buying 2-liter bottles from a grocery (especially with the equipment overhead costs).  I spoke to a restaurant manager, and &lt;em&gt;his&lt;/em&gt; cost is 56usd for a 5-gallon box, which puts the per 2-liter cost at $1.51 by my calculations, so that's when I suspected I was doing something wrong.&lt;/p&gt;&#xA;" OwnerUserId="5263" LastActivityDate="2010-12-31T23:27:02.890" Title="Ratios make me feel like an idiot - help me mix up some Coca-Cola" Tags="&lt;fractions&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="1" />
  <row Id="16004" PostTypeId="1" AcceptedAnswerId="16005" CreationDate="2010-12-31T17:27:39.497" Score="0" ViewCount="106" Body="&lt;p&gt;Let $(\Omega, \mathcal{F}, \textbf{P})$ be a probability space and $A,B, A_i$ events in $\mathcal{F}$. Prove the following properties of every probability measure: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Monotonicity: If $A \subseteq B$ then $\textbf{P}(A) \leq \textbf{P}(B)$.&lt;/li&gt;&#xA;&lt;li&gt;Sub-additivity: If $A \subseteq \bigcup_{i} A_i$ then $\textbf{P}(A) \leq \sum_{i} \textbf{P}(A_i)$. &lt;/li&gt;&#xA;&lt;li&gt;Continuity from Below: If $A_{i} \uparrow A$, then $\textbf{P}(A_i) \uparrow \textbf{P}(A)$. &lt;/li&gt;&#xA;&lt;li&gt;Continuity from Above: If $A_{i} \downarrow A$, then $\textbf{P}(A_i) \downarrow \textbf{P}(A)$. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;For the first, one could consider $\textbf{P}(B-A)$? For the second, one could consider $\textbf{P}(A_{i}-A)$ over all $i$? It seems like the third and fourth properties follow from the first two. In particular, if $A_{1} \subseteq A_2 \subseteq \dots$ and $\bigcup_{i} A_i = A$, then $\textbf{P}(A_1) \leq \textbf{P}(A_2) \leq \dots$ and $\sum_{i} \textbf{P}(A_i) = \textbf{P}(A)$.  &lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2010-12-31T18:04:06.467" Title="Properties of Probability Measure" Tags="&lt;probability-theory&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="16005" PostTypeId="2" ParentId="16004" CreationDate="2010-12-31T17:41:48.913" Score="5" ViewCount="" Body="&lt;p&gt;Monotonicity follows from the countable additivity of measure. You need to recognize whenever $A \subseteq B$, we have $B = A \cup (B \cap A^{c})$. And note that $A$ and $B \cap A^{c}$ are disjoint sets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sub-additivity is proved as follows. First prove that $\mathbf{P}(\displaystyle \cup_{i=1}^{\infty} A_i) \leq \sum_{i=1}^{\infty} \mathbf{P}(A_i)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This can be proved by re-writing $\displaystyle \cup_{i=1}^{\infty} A_i$ as a disjoint union and then use the countable additivity of a measure to get the desired result. Now use monotonicity proved in the previous part to conclude the desired result.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What, continuity from below means is &quot;If we have a nested sequence of sets, i.e. $A_1 \subseteq A_2 \subseteq \ldots $ and $\cup_{i=1}^{\infty} A_i = A$, then we have continuity from below i.e. $\displaystyle \lim_{n \rightarrow \infty} \mathbf{P}(A_n) = \mathbf{P}(A)$&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The trick here again is to rewrite $\displaystyle \cup_{i=1}^{\infty} A_i$ as a disjoint union and massage around to get the desired result.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, continuity from above means, &quot;If we have a sequence of sets such that $A_{1} \supseteq A_{2} \supseteq \ldots $ and $\displaystyle \cap_{i=1}^{\infty} A_i = A$, then we have continuity from above i.e. $\displaystyle \lim_{n \rightarrow \infty} \mathbf{P}(A_n) = \mathbf{P}(A)$&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This follows from previous part when we consider the sequence of sets $B_i = A_1 \backslash A_i$ and noting that $\mathbb{P}(A_1) \leq 1 &amp;lt; \infty$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Note&lt;/strong&gt;: In general, all these properties except the last one hold true for any measure. The last property holds true if at-least one of the sets in the sequence we consider has a finite measure.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2010-12-31T18:04:06.467" LastActivityDate="2010-12-31T18:04:06.467" />
  <row Id="16007" PostTypeId="2" ParentId="15981" CreationDate="2010-12-31T18:31:19.203" Score="0" ViewCount="" Body="&lt;p&gt;There are other kinds of norms, like field norms and norms of ideals. See &lt;a href=&quot;http://en.wikipedia.org/wiki/Norm#In_mathematics&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Norm#In_mathematics&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="589" LastActivityDate="2010-12-31T18:31:19.203" />
  <row Id="16008" PostTypeId="2" ParentId="15899" CreationDate="2010-12-31T18:51:44.450" Score="5" ViewCount="" Body="&lt;p&gt;$x$ used to be called &lt;em&gt;latus&lt;/em&gt; (latin for &lt;em&gt;side&lt;/em&gt;---of a square, a cube)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are interested in this, you &lt;em&gt;must&lt;/em&gt; read Cajori's &lt;em&gt;History of mathematical notations&lt;/em&gt;, which has a thorough discussion of the history of the notation and nomenclature of powers.&lt;/p&gt;&#xA;" OwnerUserId="274" LastEditorUserId="274" LastEditorDisplayName="" LastEditDate="2011-01-03T03:00:47.787" LastActivityDate="2011-01-03T03:00:47.787" />
  <row Id="16009" PostTypeId="2" ParentId="15898" CreationDate="2010-12-31T18:59:33.047" Score="4" ViewCount="" Body="&lt;p&gt;If I understand you correctly, you want to prove that, for a square matrix with complex coefficients (or more generally, any algebraically closed field) the product of the eigenvalues (with appropriate multiplicity) is a multiplicative function on the matrices, but &lt;em&gt;without&lt;/em&gt; using that the product of the eigenvalues is the determinant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(It is pretty easy to show that the product of the Eigenvalues, with appropriate multiplicity, is the determinant, by using the Jordan canonical form. The determinant is invariant under conjugation, so the determinant of a matrix is the same as the determinant of its Jordan canonical form; this form is upper triangular, and the diagonal contains the eigenvalues, each with the appropriate algebraic multiplicity, yielding that the determinant is indeed the product of the eigenvalues; a Jordan canonical basis yields the triples you have. Once you have that the product is indeed the determinant, the conclusion follows).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are a number of obstacles to the project. First, we need to show that $D(\mathbf{M})$ is well-defined; that is, that it does not depend on the choice of maximal family of vectors. This can be achieved pretty much the way one shows that the generalized eigenspaces yield a decomposition of the vector space (see any book that covers the Jordan canonical form), but the simplest proofs I know will use the Cayley-Hamilton Theorem at some point, which involves the determinant. (The more complicated ones consider the vector space as a module over $\mathbb{C}[x]$ and use the structure theorem for modules over PIDs... don't ask.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But assuming this is taken for granted (that you can find a basis of generalized eigenvectors, so that your maximal family of triples will have $n$ elements, and that the number of times that each $k_i$ occurs is independent of the choice), then we can proceed along similar lines as the way we prove that the determinant is multiplicative.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Lemma.&lt;/strong&gt; If there is vector $\mathbf{v}\neq 0$ such that $\mathbf{M}\mathbf{v}=\mathbf{0}$, then $D(\mathbf{M})=0$. Conversely, if $D(\mathbf{M})=0$, then there exists a vector $\mathbf{v}\neq 0$ such that $\mathbf{M}\mathbf{v}=\mathbf{0}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Proof.&lt;/em&gt; Immediate. &lt;strong&gt;QED&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proposition.&lt;/strong&gt; $D(\mathbf{M})=0$ if and only if $\mathbf{M}$ is not invertible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Proof.&lt;/em&gt; $\mathbf{M}$ is invertible if and only if it is of full rank. By the Rank-Nullity Theorem, this occurs if and only if the nullity of $\mathbf{M}$ is zero, which happens if and only if the nullspace is trivial, if and only if for all $\mathbf{v}$, $\mathbf{M}\mathbf{v}=\mathbf{0}\Rightarrow \mathbf{v}=\mathbf{0}$. The previous lemma then completes the proof. &lt;strong&gt;QED&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proposition.&lt;/strong&gt; If $D(\mathbf{B})=0$, then $D(\mathbf{AB}) = 0 = D(\mathbf{A})D(\mathbf{B})$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Proof.&lt;/em&gt; Let $\mathbf{v}\neq\mathbf{0}$ be such that $\mathbf{B}\mathbf{v}=\mathbf{0}$. Then $(\mathbf{AB})\mathbf{v}=\mathbf{0}$, so by previous result, $D(\mathbf{AB})=0$. &lt;strong&gt;QED&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proposition.&lt;/strong&gt; If $D(\mathbf{A}) = 0$, then $D(\mathbf{AB}) = 0 = D(\mathbf{A})D(\mathbf{B})$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Proof.&lt;/em&gt; If $D(\mathbf{B})=0$, then the result follows by the previous proposition. If $D(\mathbf{B})\neq 0$, then $B$ is invertible (as above). Since $D(\mathbf{A})=0$, there exists $\mathbf{v}\neq \mathbf{0}$ such that $\mathbf{A}\mathbf{v}=\mathbf{0}$. Then &#xA;$$(\mathbf{AB})(\mathbf{B}^{-1}\mathbf{v}) = \mathbf{A}\mathbf{v}=\mathbf{0} = 0(\mathbf{B}^{-1}\mathbf{v}),$$ &#xA;and since $\mathbf{B}^{-1}\mathbf{v}\neq\mathbf{0}$, then $D(\mathbf{AB}) = 0$, as desired. &lt;strong&gt;QED&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Lemma.&lt;/strong&gt; If $\mathbf{B}$ is an elementary matrix, then  &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;If $\mathbf{B}$ is obtained from the identity by transposing two rows, then $D(\mathbf{B}) = -1$.&lt;/li&gt;&#xA;&lt;li&gt;If $\mathbf{B}$ is obtained from the identity by adding a multiple of one row to another row, then $D(\mathbf{B}) = 1$.&lt;/li&gt;&#xA;&lt;li&gt;If $\mathbf{B}$ is obtained from the identity by multiplyig a row by $\lambda\neq 0 $, then $D(\mathbf{B}) = \lambda$.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Proof.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Suppose we exchange rows $i$ and $j$. That means that $\mathbf{B}(e_i) = e_j$ and $\mathbf{B}(e_j) = e_i$, while $\mathbf{B}(e_k) = e_k$ if $k\neq i,j$. We get $n-2$ triples with $(e_k,1,1)$, $k\neq i,j$, and also the triples $(e_i+e_j,1,1)$ and $(e_i-e_j,-1,1)$. Then $D(\mathbf{B}) = 1^{n-1}(-1) = -1$, as claimed.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Suppose we add $r\neq 0$ times the $i$th row to the $j$th row. Then $\mathbf{B}(e_k) = e_k$ if $k\neq i$, and $\mathbf{B}(e_i) = e_i+re_j$. Take the $n-1$ triples $(e_k,1,1)$ with $k\neq i$, and the triple $(e_i,1,2)$. Note that this works, since $(\mathbf{B}-\mathbf{I})e_i = (e_i+re_j) - e_i = re_j$, and $(\mathbf{B}-\mathbf{I})(re_j) = re_j - re_j = \mathbf{0}$, so that $(\mathbf{B}-\mathbf{I})^2e_i = \mathbf{0}$, with $2$ minimal. Thus, $D(\mathbf{B}) = 1^n = 1$. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If we multiply the $i$th row by $\lambda$, then the $n-1$ triples $(e_j,1,1)$ with $j\neq i$, together with $(e_i,\lambda,1)$, make a maximal family and we have $D(\mathbf{B}) = 1^{n-1}\lambda = \lambda$. &lt;strong&gt;QED&lt;/strong&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Lemma.&lt;/strong&gt; If $\mathbf{A}$ and $\mathbf{B}$ are elementary matrices, then $D(\mathbf{AB}) = D(\mathbf{A})D(\mathbf{B})$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Sketch.&lt;/em&gt; There are several cases to consider, depending on which kind of elementary matrices they are and how they interact. For example, if $A$ is obtained from the identity by adding $r$ times the $i$th row to the $j$th row, and $B$ is obtained by multiplying the $\ell$th row by $\lambda$, then we must consider:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;If $\ell\neq i,j$, then we can take the $n-3$ triples $(e_k,1,1)$ with $k\neq i,j,\ell$, the triple $(e_{\ell},\lambda,1)$, the triple $(e_j,1,1)$, and the triple $(e_i,1,2)$. Then $D(\mathbf{AB}) = 1^{n-1}\lambda = \lambda = D(\mathbf{A})D(\mathbf{B})$. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If $\ell = j$, then $\mathbf{AB}(e_k) = e_k$ if $k\neq i,j$; $\mathbf{AB}e_j = \lambda e_j$, and $\mathbf{AB}(e_i) = e_i+re_j$, so we can take the $n-2$ triples $(e_k,1,1)$ with $k\neq i,j$, together with $(e_j,\lambda,1)$ and $(e_i,1,2)$. Then $D(\mathbf{AB}) = 1^{n-1}\lambda = \lambda = D(\mathbf{A})D(\mathbf{B})$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If $\ell=i$, then $\mathbf{AB}(e_k) = e_k$ if $k\neq i$, and $\mathbf{AB}(e_i) = \lambda e_i + \lambda re_j$. We can then take the $n-1$ triples $(e_k,1,1)$, $k\neq i$, and the triple $(e_i,\lambda,2)$, to get $D(\mathbf{AB})=1^{n-1}\lambda = \lambda = D(\mathbf{A})D(\mathbf{B})$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Similar arguments hold for the other combinations of products of elementary matrices. &lt;strong&gt;QED&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Corollary.&lt;/strong&gt; If $E_1,\ldots,E_k$ are elementary $n\times n$ matrices with coefficients in $\mathbb{C}$, then $D(E_1\cdots E_k) = D(E_1)\cdots D(E_k)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Proof.&lt;/em&gt; Induction on $k$. &lt;strong&gt;QED&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;THEOREM.&lt;/strong&gt; If $\mathbf{A}$ and $\mathbf{B}$ are $n\times n$ matrices with coefficients in $\mathbb{C}$, then $D(\mathbf{AB}) = D(\mathbf{A})D(\mathbf{B})$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Proof.&lt;/em&gt; If either $D(\mathbf{A}) = 0$  or $D(\mathbf{B})=0$, then the result follows from previous proposition. If $D(\mathbf{A})\neq 0 \neq D(\mathbf{B})$, then each of $\mathbf{A}$ and $\mathbf{B}$ is invertible, hence a product of elementary matrices. The result now follows from the Corollary above. &lt;strong&gt;QED&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-31T19:45:35.830" LastActivityDate="2010-12-31T19:45:35.830" />
  <row Id="16010" PostTypeId="2" ParentId="15939" CreationDate="2010-12-31T19:01:09.793" Score="0" ViewCount="" Body="&lt;p&gt;Yes, Z has full rank, and A is positive definite, but using the formula I do not&#xA;get how i can compute $(A Z^T Z A)^+ $ as a function of $(Z^T Z)^+ $ and $A$...&#xA;or am I doing something wrong?&lt;/p&gt;&#xA;" OwnerUserId="5267" LastActivityDate="2010-12-31T19:01:09.793" CommentCount="5" />
  <row Id="16011" PostTypeId="1" AcceptedAnswerId="16014" CreationDate="2010-12-31T19:50:40.350" Score="24" ViewCount="1089" Body="&lt;p&gt;I don't know if this is a well know fact but I have observed that every number no matter how large that is equally divided by 9, will equal 9 if you add all the numbers it is made from until there is 1 digit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A quick example of what I mean:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;9*99 = 891&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;8+9+1 = 18&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;1+8 = 9&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This works even with really long numbers like 4376331&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is that? This doesn't work with any other number.&lt;/p&gt;&#xA;" OwnerUserId="5268" LastActivityDate="2011-01-01T16:30:55.763" Title="What makes 9 special?" Tags="&lt;number-theory&gt;" AnswerCount="7" CommentCount="7" FavoriteCount="9" />
  <row Id="16012" PostTypeId="2" ParentId="16011" CreationDate="2010-12-31T19:57:29.690" Score="3" ViewCount="" Body="&lt;p&gt;This property is based on the fact that 9 is one minus 10 which is the base of the number system you are working with because of this any power of 10 will equal 1 modulo 9.&lt;/p&gt;&#xA;" OwnerUserId="771" LastActivityDate="2010-12-31T19:57:29.690" />
  <row Id="16013" PostTypeId="2" ParentId="16011" CreationDate="2010-12-31T19:58:04.493" Score="4" ViewCount="" Body="&lt;p&gt;This follows from the &lt;a href=&quot;http://en.wikipedia.org/wiki/Divisibility_rule&quot; rel=&quot;nofollow&quot;&gt;divisibility test&lt;/a&gt; for $9$ and from the fact that the sum of the digits of a natural number with more than one digit is strictly less than the number itself i.e. if $n$ is a natural number with more than one digit and if $s(n)$ denotes the sum of the digits of the natural number then $s(n)&amp;lt;n$. Your observation can be strengthened to note that if we keep repeatedly adding the digits, the final number we are left with is the remainder when divided by $9$.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastActivityDate="2010-12-31T19:58:04.493" />
  <row Id="16014" PostTypeId="2" ParentId="16011" CreationDate="2010-12-31T19:58:20.250" Score="45" ViewCount="" Body="&lt;p&gt;Not quite right, since $9\times 0 = 0$ and the digits don't add up to $9$; but otherwise correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason it works is that we write numbers in base $10$, and when you divide $10$ by $9$, the remainder is $1$. Take a number, say, $184631$ (I just made it up). Remember what that really means:&#xA;$$184631 = 1 + 3\times 10 + 6\times 10^2 + 4\times 10^3 + 8\times 10^4 + 1\times 10^5.$$&#xA;The remainder when you divide any power of $10$ by $9$ is again just $1$, so adding the digits gives you a number that has the same remainder when dividing by $9$ as the original number does. Keep doing it until you get down to a single digit and you get the remainder of the original number when you divide by $9$, except that you get $9$ instead of $0$ if the number is a nonzero multiple of $9$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But since every multiple of $9$ &lt;em&gt;is&lt;/em&gt; a multiple of $9$, you will always get $9$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that you have a similar phenomenon with $3$ (a divisor of $9$), since adding the digits of a multiple of $3$ will always result in one of the one-digit multiples of $3$: $3$, $6$, or $9$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we wrote in base $8$, instead of base $10$, then $7$ would have the property: if you write a number in base $8$ and add the digits (in base 8) until you get down to a single digit between $1$ and $7$, then the multiples of $7$ will always end yielding $7$, for precisely the same reason. And if we wrote in base $16$, then $15$ (or rather, F) would have the property. In general, if you write in base $b$, then $b-1$ has the property.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a special case of &lt;a href=&quot;http://en.wikipedia.org/wiki/Casting_out_nines&quot; rel=&quot;nofollow&quot;&gt;casting out nines&lt;/a&gt;, which in turn is a special case of &lt;a href=&quot;http://en.wikipedia.org/wiki/Modular_arithmetic&quot; rel=&quot;nofollow&quot;&gt;modular arithmetic&lt;/a&gt;. It is what is behind many &lt;a href=&quot;http://en.wikipedia.org/wiki/Divisibility_tests&quot; rel=&quot;nofollow&quot;&gt;divisibility tests&lt;/a&gt; (e.g., for $2$, $3$, $5$, $9$, and $11$). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Coda.&lt;/strong&gt; This reminds me of an anecdote a professor of mine used to relate: a student once came to him telling him he had discovered a very easy way to test divisibility of any number $N$ by any number $b$: write $N$ in base $b$, and see if the last digit is $0$. I guess, equivalently, you could write $N$ in base $b+1$, and add the digits to see if you get $b$ at the end.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-01T01:50:18.590" LastActivityDate="2011-01-01T01:50:18.590" CommentCount="3" />
  <row Id="16015" PostTypeId="2" ParentId="16011" CreationDate="2010-12-31T20:55:42.317" Score="8" ViewCount="" Body="&lt;p&gt;This is a special case of &lt;a href=&quot;http://en.wikipedia.org/wiki/Casting_out_nines&quot; rel=&quot;nofollow&quot;&gt;casting out nines,&lt;/a&gt; which is best viewed in terms of modular arithmetic:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\quad\quad  mod\ 9:\ \ 10\equiv 1\ \Rightarrow\ P(10)\equiv P(1)\quad $ for all polynomials $\rm\ P(x) \in \mathbb Z[x]$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But radix notation has &lt;em&gt;polynomial&lt;/em&gt; form, e.g. $\rm\ 567 = P(10),\ \ P(x) = 5\: x^2 + 6\: x + 7\:,\ $ so the above says that, mod $9$,  an integer $\rm\ N = P(10)\ $ is congruent to $\rm\ P(1) = $ the sum $\rm\:S_N\:$ of its decimal digits.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yours is the special case $\rm\ N\equiv 0\ \ (mod\ 9)\ $ so the above implies that it remains $\equiv 0\ $ when you map $\rm\:N\:$ to its digit sum $\rm\:S_N\:$.  Since this map is strictly decreasing for $\rm\:N &amp;gt; 9\:,\ $ iterating it must eventually reach some nonzero $\rm\: N&amp;#39; \le 9\:.\ $ But since $\rm\ N&amp;#39;\equiv 0\ \ (mod\ 9)\ $ we conclude that $\rm\: N&amp;#39; = 9\:$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: if modular arithmetic is unfamiliar then you may instead proceed as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Factor Theorem&lt;/strong&gt; $\rm\;\Rightarrow\; X-1\ |\ P(X)-P(1)\ $ so $\rm\ X = 9\ \Rightarrow\ 9\ |\ P(10)-P(1) \;\:$ i.e. $\rm\; 9\ |\ N - S_N$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The analogous result holds true for any radix $\rm\:b\:$, i.e. one can cast $\rm\:b-1$'s in the same way, since $\rm\ P(b) \equiv P(1)\ \ (mod\ b-1)\:.\:$ Thus $\:9\:$ is &quot;special&quot; only in the way it relates to the radix $10\:$. One can similarly cast $11$'s by using $\rm\ P(10)\equiv P(-1)\ \ (mod\ 11)\:.\:$ One can devise countless &quot;divisibility tests&quot; by using modular reduction in this manner, e.g. see my &lt;a href=&quot;http://groups.google.com/groups?selm=y8zwsvvxdmy.fsf@nestle.csail.mit.edu&quot; rel=&quot;nofollow&quot;&gt;post here&lt;/a&gt; for casting out $91$'s.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It deserves to be better known that one may also cast out nines to check rational arithmetic - as long as the fractions have denominator coprime to $3$, e.g. see Hilton; Pedersen: &lt;a href=&quot;http://www.jstor.org/stable/2689631&quot; rel=&quot;nofollow&quot;&gt;Casting out nines revisited&lt;/a&gt;. Analogous remarks hold true for any ring that has $\ \mathbb Z/9\ $ as an image - just as one can apply &lt;strong&gt;parity&lt;/strong&gt; arguments in any ring that has $\ \mathbb Z/2\ $ as an image, e.g. all rationals with odd denominator, or the Gaussian integers $\rm\:\mathbb Z[i]\:$, where the image $\rm\ \mathbb Z[i]/(2,i-1) \cong \mathbb Z/2\ $ yields the natural parity definition that $\rm\ a+b\:i\ $ is even iff $\rm\ a\equiv b\ \ (mod\ 2)\ $, i.e. if $\rm\ a+b\:i\ $ maps to $\:0\:$ via the above isomorphism, which maps $\rm\ 2\to 0,\ i\to 1\:$.  These are but simple prototypical examples of problem-solving by way of &lt;strong&gt;modular reduction&lt;/strong&gt; - one of the keystones of abstract algebra.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See also &lt;a href=&quot;http://math.stackexchange.com/questions/12846/using-fermats-little-theorem/12851#12851&quot;&gt;this post&lt;/a&gt; for &quot;casting out orders&quot; in cyclic groups, and see &lt;a href=&quot;http://groups.google.com/groups?selm=y8zoh73mzp8.fsf@nestle.ai.mit.edu&quot; rel=&quot;nofollow&quot;&gt;this thread&lt;/a&gt; for an in-depth comparison of various inductive proofs of casting nines.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2011-01-01T01:24:38.597" LastActivityDate="2011-01-01T01:24:38.597" />
  <row Id="16016" PostTypeId="1" AcceptedAnswerId="16017" CreationDate="2010-12-31T21:40:57.353" Score="2" ViewCount="162" Body="&lt;p&gt;This may be a simple question, but I can't make it out right now.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Let $M$ be a monoid, and $g_1, \ldots, g_n \in M$ be elements of finite order in $M$.  Is $\langle g_1, \ldots, g_n \rangle$ periodic (i.e, have all its elements of finite order)?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The monoid I'm  interested in is the multiplicative monoid of $n \times n$ matrices over $\mathbb{N}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="5270" LastActivityDate="2011-01-01T05:37:27.103" Title="Is a monoid finitely generated by finite order elements periodic?" Tags="&lt;group-theory&gt;&lt;monoid&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="16017" PostTypeId="2" ParentId="16016" CreationDate="2010-12-31T21:53:33.383" Score="6" ViewCount="" Body="&lt;p&gt;(Edit because I misread the question somewhat.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To show that this is false for an arbitrary monoid it suffices to exhibit a monoid generated by elements of finite order which has elements of infinite order.  Take, for example, $M = \langle a, b | a^2 = b^2 = 1 \rangle$.  In this monoid $ab$ has infinite order.  One can see this geometrically by constructing the following linear representation: let $a$ and $b$ act by reflection across two lines through the origin at an angle $\theta$ with respect to each other which is not a rational multiple of $\pi$.  (For this reason $M$ is known as the infinite dihedral group.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Probably the most important example of this phenomenon &quot;in nature&quot; is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Modular_group&quot; rel=&quot;nofollow&quot;&gt;modular group&lt;/a&gt;, which has group presentation $\langle a, b | a^2 = b^3 = 1 \rangle$ and hence is generated by an element of order $2$ and an element of order $3$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not sure about your particular $M$.  Does $\mathbb{N}$ contain $0$ for you?  (If not I am having trouble finding &lt;em&gt;any&lt;/em&gt; nontrivial elements of finite order.)&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-12-31T22:17:38.750" LastActivityDate="2010-12-31T22:17:38.750" CommentCount="6" />
  <row Id="16018" PostTypeId="1" CreationDate="2010-12-31T22:28:24.550" Score="4" ViewCount="96" Body="&lt;p&gt;Dear all,&#xA;I have systems like &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(A - \lambda B) X = F$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where lambda is being updated inside a loop. I also have a limited number of eigenvectors of the matrix pair (A, B), say 40 eigenpair from a previous analysis step.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I could get the results with direct solution however due to repeated factorization of the operator matrix, I run into the O(n^3) wall. I was wondering if that could be possible to solve this system by some iterative methods. I, myself, tried some iterative methods, cg, minres, and others firstly in MATLAB, however cg like methods do not even converge due to the nature of the operator matrix since that can become indefinite in the loop, minres seems to converge hovewer that also does not converge in reasonable iteration counts. Also in these tries I had to use a PD preconditioner and do a factorization, which also induces other costs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically are there some kind of a mathematical tricks for this kind of problem, where some limited number of eigenpairs are available from a previous analysis step?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a poor engineer I tried all I know, so any further ideas?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;TIA,&#xA;Umut&lt;/p&gt;&#xA;" OwnerUserId="5272" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2010-12-31T22:32:53.890" LastActivityDate="2010-12-31T22:32:53.890" Title="linear system solution, iterative vs direct" Tags="&lt;linear-algebra&gt;" CommentCount="1" />
  <row Id="16019" PostTypeId="2" ParentId="16003" CreationDate="2010-12-31T23:27:02.890" Score="3" ViewCount="" Body="&lt;p&gt;With a $5.4:1$ ratio of carbonated water to syrup, assuming that the quantities are additive (mixing 5.4 units of carbonated water with 1 unit of syrup yields 6.4 units of the mixture), I get: $$5\text{ gallons of syrup}\times\frac{6.4\text{ gallons of mixture}}{1\text{ gallon of syrup}}\times\frac{3.785\text{ liters}}{1\text{ gallon}}\approx 121\text{ liters of mixture}.$$&#xA;At $\$65$ per box, $$\frac{\$65}{121\text{ liters}}\approx\$0.54\text{ per liter}.$$&#xA;This is all ignoring the cost of carbonating the water, because around me, a 2-liter bottle of Coke goes on sale for 99 cents with enough regularity that it already doesn't make sense to mix at home.&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-12-31T23:27:02.890" CommentCount="2" />
  <row Id="16020" PostTypeId="1" AcceptedAnswerId="16021" CreationDate="2011-01-01T01:17:49.603" Score="7" ViewCount="233" Body="&lt;p&gt;Hello, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a little project for myself this winter break, I'm trying to go through as much of Enderton's Elements of Set Theory as I can. I hit a snag trying to show two forms of the Axiom of Choice are equivalent. This is exercise 31 on page 55. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first form is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;For any relation $R$ there is a function $G\subseteq R$ with $\text{dom}\ G=\text{dom}\ R$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and the second form is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;For any set $I$ and any function $H$ with domain $I$, if $H(i)\neq\emptyset$ for all $i\in I$, then $\times_{i\in I}H(i)\neq\emptyset$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here is what I have so far:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume the first form. Take any set $I$ and let $H$ be a function with domain $I$ such that $H(i)\neq\emptyset$ for all $i\in I$. This function $H$ is a relation, so by the Axiom of Choice, there exists a function $G\subseteq H$ such that $\text{dom}\ G=\text{dom}\ H=I$. Since $\text{dom}\ G=I$, for each $i\in I$, there exists some $G(i)$ such that $(i,G(i))\in G$. But since $G\subseteq H$, $(i,G(i))=(j,H(j))$ for some $j\in I$. Since these are ordered pairs, $i=j$ and $G(i)=H(j)$? I suppose I want to be able to show that for all $i\in I$, I can have $G$ &quot;choose&quot; some element $G(i)\in H(i)$, and thus $G\in\times_{i\in I}H(i)$, showing that $\times_{i\in I}H(i)\neq\emptyset$, but I don't see how the first form allows one to do that. Instead, all I see is that $G(i)=H(i)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conversely, I assume the second form. I take any relation $R$, and denote $\text{dom}\ R=I$. Let $H$ be any function with domain $I$. Now if $H(i)\neq\emptyset$ for all $i$, then $\times_{i\in I}H(i)\neq\emptyset$, so then I could take some $f\in\times_{i\in I}H(i)$, so by definition, $\text{dom} f=I$, and for all $i$, $f(i)\in H(i)$. If it is the case that $(i,H(i))\in R$, then $f\subseteq R$, and the first form would be proven. Again, I suppose I want $H$ to be a function that, for each $i\in I$, $H$ takes the value of exactly one $y_i$ such that $iRy_i$, but again, I don't see how the assumed axiom allows one to do this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone explain how to get around these two issues? Thank you.&lt;/p&gt;&#xA;" OwnerUserId="1583" LastEditorUserId="1583" LastEditorDisplayName="" LastEditDate="2011-01-01T05:12:23.443" LastActivityDate="2011-01-01T05:12:23.443" Title="Equivalent statements of the Axiom of Choice" Tags="&lt;set-theory&gt;&lt;axiom-of-choice&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="16021" PostTypeId="2" ParentId="16020" CreationDate="2011-01-01T01:35:33.653" Score="10" ViewCount="" Body="&lt;p&gt;The key in these kinds of proofs is to find a &lt;em&gt;particular&lt;/em&gt; instance of the proposition you are assuming that will yield the one you want. So in order to prove the second form from the first, you start with $I$ and $H$, and want to &quot;cook up&quot; some $R$ for which the $G$ will yield what you want. And conversely. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are sort of on the right track, but the problem is that you are trying to apply each form to the information from the other, instead of cooking up an appropriate set to apply the &quot;other&quot; form in each case. For instance, in your second proof, you shouldn't take $H$ to be &lt;em&gt;any&lt;/em&gt; function, you want to cook up a &lt;em&gt;particular&lt;/em&gt; function. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, before we actually prove they are equivalent, let's think about them. Intuitively, why should these two &quot;be&quot; the Axiom of Choice? In a relation, every element in the domain is associated to many elements in the codomain. Being able to find a function &quot;contained&quot; in the relation is like, for each $x$ in the domain, picking one element from each of the things that are associated to $x$ to be the image of $x$. This is the Axiom of Choice (you are picking one element from each collection of &quot;things related to $i$&quot; for $i\in I$). As for the second, an element of $\mathop{\times}\limits_{i\in I}H(i)$ is a function from $I$ to $\cup H(i)$ such that the image of $i$ is in $H(i)$ for each $i$; again, a choice of elements from a family of sets, again Choice. Knowing how each of them relates to AC, and hence to each other, should help you figure out how to show the equivalence: after all, that equivalence should be one in which the particular &quot;choice&quot; that one form allows you to make is exactly the &quot;choice&quot; that the other form requires you to make. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For more explicit hints:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your attempt to show the first form implies the second, you are focusing on the wrong thing; you don't want to look at $H$ as a relation, because then you will have $G=H$ ($H$ is already a function). Instead, define a relation $R\subseteq I\times \cup H(i)$ by $(i,x)$ if and only if $x\in H(i)$. (That is, $H(i)$ is a nonempty set for each $i$; associate with each $i$ all elements in $H(i)$). Now use the first form with $R$, not with $H$, because the $G$ should give you an element of the product (think of them as described above).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the second, assume that $R$ is a relation, and you want to define a function contained in $R$ with the same domain. Let $I$ be the domain of $R$, and for each $i\in I$, let $H(i) = \{ x\in\mathrm{codom}(R)\mid iRx\}$. Use the fact that the product is nonempty to pick an element in the product (which &quot;picks&quot; an element from each of the sets of &quot;things related to $i$&quot;) and use that to define the function $G$. &lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2011-01-01T01:35:33.653" CommentCount="1" />
  <row Id="16022" PostTypeId="1" AcceptedAnswerId="16024" CreationDate="2011-01-01T03:26:54.403" Score="4" ViewCount="272" Body="&lt;p&gt;It seems to my uneducated mind that if I have $\frac{n}{k}$ subproblems, each of size $\Omega (k \log k)$, that my overall problem must be at least of size $\Omega (n \log k)$, by the reasoning that if I have $\frac{n}{k}$ things, each of which is at least $k \log k$, I should be able to multiply those together to get a lower bound for their sum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, my TA seems to disagree with me, stating that addition is not defined on $\Omega (f(x)) + \Omega (g(x))$, and therefore I can't sum or multiply the subproblems in this fashion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was wondering if someone could explain the error in my reasoning.&lt;/p&gt;&#xA;" OwnerUserId="129" LastActivityDate="2011-01-02T04:04:34.913" Title="Addition on big omega notation" Tags="&lt;computer-science&gt;&lt;discrete-mathematics&gt;" AnswerCount="2" CommentCount="7" />
  <row Id="16023" PostTypeId="2" ParentId="16022" CreationDate="2011-01-01T03:33:36.560" Score="2" ViewCount="" Body="&lt;p&gt;What you need to be talking about is $\displaystyle \Omega(f+g)$ and not $\Omega(f) + \Omega(g)$ as $\Omega(f)$ and $\Omega(g)$ are sets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $\displaystyle |h(x)| \ge c_1 |f(x)|$ and $\displaystyle |h(x)| \ge c_2 |g(x)|$ then we have that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle |h(x)| \ge \frac{c}{2} (|f(x)| + |g(x)|) \ge c&amp;#39; |f(x) + g(x)|$, where $\displaystyle c = \text{min}\{c_1, c_2\}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus if $\displaystyle h \in \Omega(f)$ and $\displaystyle h \in \Omega(g)$, then $h \in \Omega(f+g)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But in your case, even though you have a variable number of sub-problems, I am not really sure I completely agree with Alon. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It really depends on what you consider to be variable when you consider each sub-problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $n,k$ are the variables, and you have a a constant $c$ (independent of $n$ and $k$) such that each of the $\frac{n}{k}$ independent sub-problems is of size at least $c k \log k$, then you &lt;em&gt;can&lt;/em&gt; add them up the way you did, to get the total size. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance consider the problem where an array is almost sorted: each element is within $k$ of its original position. You can sort the array by sorting  $\frac{n}{k}$ chunks of $2k$ elements each. If you wanted to consider the total time taken, say with the algorithm you picked for sorting, each chunk &lt;em&gt;always&lt;/em&gt; takes $\Omega(k\log k)$ time, then the total time is indeed $\Omega(n \log k)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you had the $i^{th}$ subproblem constant, be a function of $i$, then it is not really a constant, as it depends on $\frac{n}{k}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now if you allowed the $i^{th}$ sub-problem constant to be independent of $k$, but not $n$, then adding them up becomes kind of meaningless and could lead to incorrect results, as Alon noted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: If you were trying to prove an $\Omega(n \log k)$ lower bound for sorting in the above problem, by considering $n/k$ sub-problems of size $k$, just adding like that is not a proof of overall lower bounds. In the worst case, each sub-problem could be $\Omega(k \log k)$, but when you hit the worst case for one sub-problem, you might be hitting the best case for another. So even claiming $\Omega(k \log k)$ lower bounds for each sub-problem needs more proof. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could add up for the upper bounds though. You can say that an algorithm would be $\mathcal{O}(n \log k)$ in the worst case (for instance, if you picked heap-sort), even though some sub-problem might take $\mathcal{o}(k \log k)$ comparisons (note: small oh), the total number of comparisons is essentially $\mathcal{O}(n \log k)$.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2011-01-02T04:04:34.913" LastActivityDate="2011-01-02T04:04:34.913" />
  <row Id="16024" PostTypeId="2" ParentId="16022" CreationDate="2011-01-01T03:43:23.877" Score="6" ViewCount="" Body="&lt;p&gt;Your original reasoning is indeed problematic, since you're adding together a variable number of things. Suppose you have $n$ subproblems, each of which takes constant time; can you conclude that the whole thing is $O(n)$, or $\Omega(n)$? No, since the constants may vary: the $i$th subproblem may take time $1/i$, or $i^3$; each is still a constant (independent of $n$), but their sum could be a constant, or grow with $n$ - you can't tell. It depends on the distribution of the constants, and that's something the notation $O$ and $\Omega$ conceals (deliberately, of course). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, as Moron pointed out, adding together just two (or any constant number) of functions is fine. Thus I'm not sure I understand the TA's claim. &lt;/p&gt;&#xA;" OwnerUserId="308" LastActivityDate="2011-01-01T03:43:23.877" CommentCount="15" />
  <row Id="16025" PostTypeId="2" ParentId="16020" CreationDate="2011-01-01T05:01:33.170" Score="6" ViewCount="" Body="&lt;p&gt;Assume the first form. Let $I$ be any set and let $H$ be any function such that $\text{dom}\ H=I$, and $H(i)\neq\emptyset$ for all $i\in I$. Define a relation $R\subseteq I\times\bigcup_{i\in I}H(i)$ by&#xA;$$&#xD;&#xA;\langle i,x\rangle\in R\Leftrightarrow x\in H(i).&#xD;&#xA;$$&#xA;By assumption, there exists a function $G\subseteq R$ with $\text{dom}\ G=\text{dom}\ R=I$, as for each $i\in I$, $i\in\text{dom}\ R$ since $H(i)$ is nonempty. So for all $\langle i,G(i)\rangle\in G$, $\langle i,G(i)\rangle\in R$, and thus by the definition of $R$, $G(i)\in H(i)$. It follows that $G\in\prod_{i\in I}H(i)$, so $\prod_{i\in I}H(i)\neq\emptyset$. Thus the second form follows from the first.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conversely, let $R$ be any relation, and denote $\text{dom}\ R=I$. Define a function &#xA;$$&#xD;&#xA;H\colon I\to\mathscr{P}(\text{ran}\ R)\colon i\mapsto H(i):=\{x\in\text{ran}\ R\ |\ iRx\}.&#xD;&#xA;$$&#xA;In particular, $H$ is a function with domain $I$, and $H(i)\neq\emptyset$ for all $i\in I$. So by the second form, $\prod_{i\in I}H(i)\neq\emptyset$, so take $G\in\prod_{i\in I}H(i)$. Hence $\text{dom}\ G=I$, and for all $i\in I$, $G(i)\in H(i)$. Also, for any $\langle i,G(i)\rangle\in G$, $G(i)\in H(i)\subseteq\text{ran}\ R$, and so $\langle i,G(i)\rangle\in R$, so $G\subseteq R$. Hence the two statements of the Axiom of Choice are equivalent.&lt;/p&gt;&#xA;" OwnerUserId="1583" LastActivityDate="2011-01-01T05:01:33.170" CommentCount="1" />
  <row Id="16026" PostTypeId="2" ParentId="16016" CreationDate="2011-01-01T05:30:59.637" Score="4" ViewCount="" Body="&lt;p&gt;For the case you're interested in it is true, because the set of finite order $n$-by-$n$ matrices over $\mathbb{N}$ is precisely the set of permutation matrices.  In fact, every invertible element of $M_n(\mathbb{N})$ is a permutation matrix.  (Here $\mathbb{N}$ contains $0$, and invertible means invertible in the monoid $M_n(\mathbb{N})$ with multiplication.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose that $B=(b_{ij})\in M_n(\mathbb{N})$ is not a permutation matrix.  Then either one of the entries of $B$ is greater than $1$, one of the rows or columns of $B$ has more than one nonzero entry, or one of the rows or columns of $B$ is zero.   Let $A=(a_{ij})\in M_n(\mathbb{N})$ be invertible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose first that $B$ has an entry greater than $1$, i.e., $b_{i_0j_0}\gt1$ for some $i_0$ and $j_0$.  Since $A$ is invertible, its $i_0^\text{th}$ column is not zero, so there is an $i_1$ such that $a_{i_1i_0}\gt0$.  Then if $AB=(c_{ij})$, we have $c_{i_1j_0}\geq a_{i_1i_0}b_{i_0j_0}\gt1$, so that $AB\neq I$.  Hence, $B$ is not invertible in this case.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now suppose that $B$ has a row with more than one nonzero entry, i.e., $b_{i_0j_0}\gt0$ and $b_{i_0j_1}\gt0$ for some $i_0$ and $j_0\neq j_1$.  Again, there is an $i_1$ such that $a_{i_1i_0}\gt0$, and we have $c_{i_1j_0}\geq a_{i_1i_0}b_{i_0j_0}\gt0$ and $c_{i_1j_1}\geq a_{i_1i_0}b_{i_0j_1}\gt0$, so again $AB\neq I$.  A similar argument applies if $B$ has a column with more than one nonzero entry, either by multiplying in the other order, or by considering the transpose.  Thus in either case $B$ is not invertible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This leaves only the case that $B$ has a zero row or column, in which case it is even a zero divisor.  Thus in any case, $B$ is not invertible.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2011-01-01T05:37:27.103" LastActivityDate="2011-01-01T05:37:27.103" CommentCount="1" />
  <row Id="16027" PostTypeId="2" ParentId="16011" CreationDate="2011-01-01T06:53:53.803" Score="3" ViewCount="" Body="&lt;p&gt;Many ways to say this I suppose - what made sense to me when I first was confronted with this fact is that the number &quot;abcd&quot; is 1000a+100b+10c+d=(999a+99b+9c)+a+b+c+d. The first term is clearly divisible by 9 so if the digit sum is then &quot;abcd&quot; is as well...&lt;/p&gt;&#xA;" OwnerUserId="5275" LastActivityDate="2011-01-01T06:53:53.803" />
  <row Id="16028" PostTypeId="2" ParentId="16011" CreationDate="2011-01-01T07:50:50.950" Score="1" ViewCount="" Body="&lt;p&gt;9 is also special for accounts in finding book keeper errors of transposed digits.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;124+&lt;strong&gt;256&lt;/strong&gt;+345=725&lt;/p&gt;&#xA;&#xA;&lt;p&gt;124+&lt;strong&gt;265&lt;/strong&gt;+345=734&lt;/p&gt;&#xA;&#xA;&lt;p&gt;734-725=9&lt;/p&gt;&#xA;" OwnerUserId="5279" LastActivityDate="2011-01-01T07:50:50.950" />
  <row Id="16029" PostTypeId="1" CreationDate="2011-01-01T08:03:24.060" Score="5" ViewCount="189" Body="&lt;p&gt;If you don't have enough teaching experience, what kind of job you should try to find to increase the teaching experience. Now many jobs have set requirements for teaching experience. It seems contradiction, Without appropriate job we can't increase the teaching experience,but without enough teaching experience we can't find jobs. Thanks for advice!&lt;/p&gt;&#xA;" OwnerUserId="5280" LastActivityDate="2011-01-01T23:41:53.167" CommunityOwnedDate="2011-01-02T14:59:54.620" Title="which job to find to increase teaching experience" Tags="&lt;soft-question&gt;" AnswerCount="2" CommentCount="6" FavoriteCount="1" />
  <row Id="16030" PostTypeId="1" AcceptedAnswerId="16032" CreationDate="2011-01-01T11:25:49.963" Score="1" ViewCount="177" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is, how can I solve the following derivative question?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$y = \ln [\ln(\ln(x^2 +1))]$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance&lt;/p&gt;&#xA;" OwnerUserId="3669" LastEditorUserId="1242" LastEditorDisplayName="" LastEditDate="2011-01-01T13:32:38.983" LastActivityDate="2011-01-01T13:32:38.983" Title="How can I find the derivative of $y = \ln [\ln(\ln(x^2 +1))]$ ?" Tags="&lt;calculus&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="16031" PostTypeId="2" ParentId="16030" CreationDate="2011-01-01T11:38:02.950" Score="2" ViewCount="" Body="&lt;p&gt;Use the chain rule: $$h(x)=f(g(x))\implies h&amp;#39;(x)=f&amp;#39;(g(x))\cdot g&amp;#39;(x)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, the derivative of $\ln x$ is $\frac{1}{x}$, so the derivative of $\ln(\text{something})$ is $\frac{1}{\text{something}}\cdot(\text{derivative of that something})$.  Since you've got several layers of nesting (functions inside functions inside functions...), you should expect to use the chain rule multiple times.&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2011-01-01T11:38:02.950" CommentCount="0" />
  <row Id="16032" PostTypeId="2" ParentId="16030" CreationDate="2011-01-01T11:38:12.653" Score="5" ViewCount="" Body="&lt;p&gt;Assuming you're familiar with the chain rule, recall that for a function $f(x)$, if&#xA;$y=\ln(f(x))$, then $y&amp;#39;=\frac{f&amp;#39;(x)}{f(x)}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in this particular case, let $f(x)=\ln(\ln(x^2+1))$, so&#xA;$$&#xD;&#xA;y&amp;#39;=\frac{f&amp;#39;(x)}{\ln(\ln(x^2+1))}.&#xD;&#xA;$$&#xA;Now you must calculate $f&amp;#39;(x)$. Try writing $f(x)$ as $f(x)=\ln(g(x))$ with $g(x)=\ln(x^2+1)$ and apply the same principle as before.&lt;/p&gt;&#xA;" OwnerUserId="1583" LastActivityDate="2011-01-01T11:38:12.653" />
  <row Id="16033" PostTypeId="1" AcceptedAnswerId="16082" CreationDate="2011-01-01T11:47:23.743" Score="3" ViewCount="126" Body="&lt;p&gt;So I have heard about a lot of fancy numerical methods for solving ODEs. I know that there are methods that give asymptotically a low error like the Runge-Kutta methods. (Assuming sufficient smoothness.) These estimate the solution in a set of points $t_0$, $t_1$, etc. But what if I want to have a function that is close to the correct solution everywhere, not just in a discrete set of points?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can extend the numerical solution to a piecewise linear function. This will be a continuous function and it will converge to the correct solution if the step-size goes to zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the error estimate will be poor in most places unless I use a very small step-size, which rather defeats the purpose of using a high-order method. So how does one go about estimating the solution in practice between the $t_i$?&lt;/p&gt;&#xA;" OwnerUserId="4623" LastActivityDate="2011-01-04T00:51:11.037" Title="Numerically solving ODEs-estimating the solution between the nodes?" Tags="&lt;differential-equations&gt;&lt;numerical-methods&gt;" AnswerCount="3" FavoriteCount="1" />
  <row Id="16034" PostTypeId="1" AcceptedAnswerId="16035" CreationDate="2011-01-01T14:30:52.530" Score="1" ViewCount="206" Body="&lt;p&gt;Hi guys!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have this sum:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{0\le y&amp;lt;k}\sum_{0\le x&amp;lt;k-y}k - \max(x,y)\ ,\qquad k\in\mathbb{N}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a closed form for it? This is no homework, im just a highschool student whose math is too poor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another way to write the formula is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum^{k-1}_{y=0}\sum^{k-y-1}_{x=0}(k-\max(x,y)),$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there is something unclear.&lt;/p&gt;&#xA;" OwnerUserId="5282" LastEditorUserId="5282" LastEditorDisplayName="" LastEditDate="2011-01-01T15:21:34.980" LastActivityDate="2011-01-01T15:48:48.877" Title="How to find a closed form for this fomula" Tags="&lt;closed-form&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="16035" PostTypeId="2" ParentId="16034" CreationDate="2011-01-01T15:20:45.690" Score="9" ViewCount="" Body="&lt;p&gt;Hint: Split the sum into &quot;sum of $k$&quot; minus &quot;sum of max&quot;.&#xA;Then draw the numbers to be added in a coordinate grid according to $x$ and $y$ values; I think you should consider even and odd values of $k$ separately.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, if $k=5$ you have to add the values&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;5&#xA;5 5&#xA;5 5 5&#xA;5 5 5 5&#xA;5 5 5 5 5&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;and from that subtract the sum of the values&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;4&#xA;3 3&#xA;2 2 2&#xA;1 1 2 3&#xA;0 1 2 3 4&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Notice the mirror symmetry here:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;4&#xA;3 3&#xA;2 2     2&#xA;1     1     2 3&#xA;    0     1 2 3 4&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;You have the numbers 0 + 1 + 2 on the diagonal $y=x$, and then there&#xA;are the numbers&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;4&#xA;3 3&#xA;2 2&#xA;1&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;which occur twice; moreover, 1+4=5 and 2+3=5, so they add up to three times 5&#xA;(times two).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now try to generalize these observations to a general pattern. For odd $k$, say $k=2m+1$, I get the expression&#xA;$$&#xD;&#xA;k^2 (k+1)/2 - (1+2k) m(m+1)/2.&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll leave the even case to you (or somebody else).  :)&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2011-01-01T15:20:45.690" CommentCount="2" />
  <row Id="16036" PostTypeId="2" ParentId="16034" CreationDate="2011-01-01T15:48:48.877" Score="6" ViewCount="" Body="&lt;p&gt;I get that for $k$ even, the sum is $$\frac{1}{8}k(2k^2+3k+2),$$ and for $k$ odd, the sum is&#xA;$$\frac{1}{8}(k+1)(2k^2+k+1).$$&#xA;Maybe it is a good exercise to figure the sums yourself.&lt;/p&gt;&#xA;" OwnerUserId="3249" LastActivityDate="2011-01-01T15:48:48.877" />
  <row Id="16037" PostTypeId="2" ParentId="16011" CreationDate="2011-01-01T16:30:55.763" Score="0" ViewCount="" Body="&lt;p&gt;I don't think this number is that much special, I guess studying other numbers in base 12 or 32 or 30 or other bases with P symbols, P being a prime number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wonder how special the base 210 is (2*3*5*7).&lt;/p&gt;&#xA;" OwnerUserId="3709" LastActivityDate="2011-01-01T16:30:55.763" />
  <row Id="16038" PostTypeId="1" AcceptedAnswerId="16055" CreationDate="2011-01-01T16:42:19.163" Score="4" ViewCount="161" Body="&lt;p&gt;This question is somewhat linked to &lt;a href=&quot;http://math.stackexchange.com/questions/16016/is-a-monoid-finitely-generated-by-finite-order-elements-periodic&quot;&gt;this one&lt;/a&gt; -- in fact, I meant to ask the following:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Let $\mathcal{M}_n$ be the multiplicative monoid of $n \times n$ matrices on $\mathbb{N}$ (including $0$).  Let $M_1, \ldots, M_k \in \mathcal{M}_n$ and suppose the monoid generated by each one of them is finite.  Is the monoid generated  by &lt;em&gt;all&lt;/em&gt; of them periodic (i.e. such that any of its elements generates a finite monoid)?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I hope I got it right this time.  Sorry for the noise.&lt;br&gt;&#xA;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="5270" LastEditorUserId="5270" LastEditorDisplayName="" LastEditDate="2011-01-01T17:31:52.753" LastActivityDate="2011-01-02T00:41:51.600" Title="Is a monoid finitely generated by periodic matrices periodic?" Tags="&lt;group-theory&gt;&lt;algebra&gt;&lt;monoid&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="16040" PostTypeId="2" ParentId="16029" CreationDate="2011-01-01T17:01:29.240" Score="5" ViewCount="" Body="&lt;p&gt;Here's my American university perspective.  I've been an adjunct instructor at a community college, a TA at a Research I university, a postdoc at a Research I university, and now I have a permanent position at a small liberal arts college.  Each position has been completely different, and nothing has informed my teaching more than making each of these major steps in my career.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I was a grad student, teaching at community college was an excellent way to get experience.  You only need a Master's degree, which you can usually pick up on the way to a Ph.D., it usually pays better than TAing, and you have complete responsibility for your courses (which is looked at much more positively than merely being an assistant when you look for a permanent job).  It also might expose you to a wider variety of students than you might see at a typical research I school, especially if you teach a course during the normal semester and not a summer course.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many hiring committees also want breadth of teaching experience, not just years.  They will want someone comfortable and capable stepping both into a &quot;developmental math&quot; course (i.e., high-school level stuff) and into a variety of advanced courses, including inter-disciplinary and computer-intensive courses. My advice is:  keep trying to expand your experience when you're starting out --- always try to teach something new whenever possible.  Don't turn down a teaching position to research --- do both!  Some graduate programs let a few select students have complete control over a course.  You usually need to prove yourself to get such a job, so you must put your best into your teaching to prove yourself (if you aren't doing this, you might not want to reconsider a teaching path anyway).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, explore alternative teaching environments.  You might be qualified to try teaching an online course, a course for a test prep company, to do outreach with high-schools in your area, or to start a &quot;math circle&quot;. All of this will show your commitment and your passion for education, which might be what a hiring committee will look for in someone relatively new to teaching rather than just a few semesters of teaching experience.&lt;/p&gt;&#xA;" OwnerUserId="4253" LastEditorUserId="4253" LastEditorDisplayName="" LastEditDate="2011-01-01T18:58:41.380" LastActivityDate="2011-01-01T18:58:41.380" CommunityOwnedDate="2011-01-02T14:59:54.620" />
  <row Id="16041" PostTypeId="2" ParentId="16033" CreationDate="2011-01-01T17:02:29.363" Score="2" ViewCount="" Body="&lt;p&gt;You can then use your favorite interpolation technique.  As you say, you have a bunch of points (t,x).  So you can feed them to a spline routine, or a polynomial fit, or whatever.  If you have some knowledge of the functional form, you can take these data points to fit the form, but it may be better to use your differential equation for that.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-01-01T17:02:29.363" />
  <row Id="16042" PostTypeId="1" AcceptedAnswerId="16043" CreationDate="2011-01-01T17:06:51.103" Score="7" ViewCount="177" Body="&lt;p&gt;$A$ an integral domain, $a,b,c\in A$. If $d$ is a greatest common divisor of $a$ and $b$, is it true that $cd$ is a greatest common divisor of $ca$ and $cb$? I know it is true if $A$ is a UFD, but can't think of a counterexample in general situation.&lt;/p&gt;&#xA;" OwnerUserId="4341" LastActivityDate="2011-03-29T23:37:02.790" Title="About greatest common divisor" Tags="&lt;number-theory&gt;&lt;divisibility&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="16043" PostTypeId="2" ParentId="16042" CreationDate="2011-01-01T17:16:46.240" Score="7" ViewCount="" Body="&lt;p&gt;Here is the best that one can say for arbitrary integral domains:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;LEMMA&lt;/strong&gt; $\rm\ \ (a,b)\ =\ (ac,bc)/c\quad$  if $\rm\ (ac,bc)\ $ exists.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof&lt;/strong&gt; $\rm\quad d\ |\ a,b\ \iff\ dc\ |\ ac,bc\ \iff\ dc\ |\ (ac,bc)\ \iff\ d|(ac,bc)/c$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Generally $\rm\ (ac,bc)\ $ need not exist, as is most insightfully viewed as failure of&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EUCLID'S LEMMA&lt;/strong&gt;  $\rm\quad a\ |\ bc\ $ and $\rm\ (a,b)=1\ \Rightarrow\ a\ |\ c\quad$ if $\rm\ (ac,bc)\ $ exists.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof&lt;/strong&gt; $\ \ $ If $\rm\ (ac,bc)\ $ exists then $\rm\ a\ |\ ac,bc\ \Rightarrow\ a\ |\ (ac,bc) = (a,b)\:c = c\ $  by the Lemma.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore if $\rm\: a,b,c\: $ fail to satisfy the Euclid Lemma $\Rightarrow\:$,&#xA;namely  if  $\rm\ a\ |\ bc\ $  and  $\rm\ (a,b) = 1\ $  but $\rm\ a\nmid c\:$, then one immediately deduces that the gcd $\rm\ (ac,bc)\ $ fails to exist. For the special case $\rm\:a\:$ is an atom (i.e. irreducible), the implication reduces to: atom $\Rightarrow$ prime. So it suffices to find a nonprime atom&#xA;in order to exhibit a pair of elements whose gcd fails to exist. This task is a bit simpler, e.g. for $\rm\ \omega = 1 + \sqrt{-5}\ \in\ \mathbb Z[\sqrt{-5}]\ $ we have that the atom $\rm\: 2\ |\ \omega&amp;#39;\: \omega = 6\:,\:$ but $\rm\ 2\nmid \omega&amp;#39;,\:\omega\:,\:$ so $\rm\:2\:$ is not prime. Therefore we deduce that the gcd $\rm\: (2\:\omega,\ \omega&amp;#39;\:\omega)\ =\ (2+2\sqrt{-5},\:6)\ $ fails to exist in $\rm\ \mathbb Z[\sqrt{-5}]\:$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that if the gcd $\rm\: (ac,bc)\ $ fails to exist then this implies that the ideal $\rm\ (ac,bc)\ $ is not principal. Therefore we've &lt;em&gt;constructively&lt;/em&gt;  deduced that the failure of Euclid's lemma immediately yields both a nonexistent gcd and a nonprincipal ideal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That the $\Rightarrow$ in Euclid's lemma implies that Atoms are Prime $\rm(:= AP)$ is denoted $\rm\ D\ \Rightarrow AP\ $ in the list of domains closely related to GCD domains in my &lt;a href=&quot;http://groups.google.com/groups?selm=y8z8yto8k85.fsf@nestle.ai.mit.edu&quot; rel=&quot;nofollow&quot;&gt;post here.&lt;/a&gt; There you will find links to further literature on domains closely related to GCD domains. See especially the referenced comprehensive survey by D.D. Anderson: &lt;em&gt;GCD domains, Gauss' lemma, and contents of polynomials&lt;/em&gt;, 2000.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See also my &lt;a href=&quot;http://math.stackexchange.com/questions/3118/if-a-m-and-a-1-m-prove-aa-1-m/3149#3149&quot;&gt;post here&lt;/a&gt; for the general &lt;em&gt;universal&lt;/em&gt; definitions of $\rm GCD,\: LCM$ and for further remarks on how such $\iff$ definitions enable slick proofs, and &lt;a href=&quot;http://math.stackexchange.com/questions/3768/upper-and-lower-bounds-of-emptyset/3832#3832&quot;&gt;see here&lt;/a&gt; for another simple example of such. &lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2011-03-29T23:37:02.790" LastActivityDate="2011-03-29T23:37:02.790" CommentCount="7" />
  <row Id="16044" PostTypeId="1" CreationDate="2011-01-01T17:20:18.047" Score="4" ViewCount="422" Body="&lt;p&gt;I'll be finishing &lt;em&gt;Calculus&lt;/em&gt; by Spivak somewhat soon, and want to continue into linear algebra and multivariable calculus afterwards. My current plan for learning the two subjects is just to read and work through Apostol volume II; is this a good idea, or would it be better to get a dedicated Linear Algebra book? Are there better books for multivariable calculus? (I don't think I want to jump right into &lt;em&gt;Calculus on manifolds&lt;/em&gt;.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: I'd like to add, as part of my question, something I mentioned in a comment below. Namely, is it useful to learn multivariable calculus without differential forms and the general results on manifolds before reading something like &lt;em&gt;Calculus on Manifolds&lt;/em&gt; or &lt;em&gt;Analysis on Manifolds&lt;/em&gt;? That is, do I need to learn vector calculus as it is taught in a second semester undergraduate course before approaching differential forms?&lt;/p&gt;&#xA;" OwnerUserId="5226" LastEditorUserId="5226" LastEditorDisplayName="" LastEditDate="2011-01-02T06:55:51.550" LastActivityDate="2011-05-24T09:09:34.530" Title="Path for learning linear algebra and multivariable calculus" Tags="&lt;linear-algebra&gt;&lt;reference-request&gt;&lt;multivariable-calculus&gt;" AnswerCount="4" CommentCount="1" FavoriteCount="0" />
  <row Id="16045" PostTypeId="2" ParentId="16033" CreationDate="2011-01-01T17:26:02.007" Score="2" ViewCount="" Body="&lt;p&gt;Take for example Runge-Kutta methods. Let $t_1$, $t_2$, etc., be the points given by the step-size of the method. Then the Runge-Kutta method not only give you the approximate solution at points $t_1$, $t_2$, etc., but also at intermediate points between, say, $t_1$ and $t_2$, depending on the order of the method. You can use these intermediate points (between $t_1$ and $t_2$, together with the endpoints $t_1$ and $t_2$) to construct an interpolation polynomial between $t_1$ and $t_2$, so that the resulting approximate solution will be a piecewise polynomial function, more general than piecewise linears, and the accuracy will as good as the accuracy of the Runge-Kutta method at the mesh points.&lt;/p&gt;&#xA;" OwnerUserId="2473" LastActivityDate="2011-01-01T17:26:02.007" />
  <row Id="16046" PostTypeId="2" ParentId="16044" CreationDate="2011-01-01T17:30:05.083" Score="0" ViewCount="" Body="&lt;p&gt;If you want to start with abstract linear algebra, try Axler's &lt;em&gt;Linear Algebra Done Right&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="589" LastActivityDate="2011-01-01T17:30:05.083" />
  <row Id="16047" PostTypeId="2" ParentId="16044" CreationDate="2011-01-01T17:46:36.207" Score="1" ViewCount="" Body="&lt;p&gt;Try Theodore Shifrin's &lt;em&gt;Multivariable Mathematics: Linear Algebra, Multivariable Calculus, and Manifolds&lt;/em&gt;. It shows how the two subjects are related.&lt;/p&gt;&#xA;" OwnerUserId="4594" LastEditorUserId="4594" LastEditorDisplayName="" LastEditDate="2011-05-24T09:09:34.530" LastActivityDate="2011-05-24T09:09:34.530" />
  <row Id="16048" PostTypeId="1" AcceptedAnswerId="16058" CreationDate="2011-01-01T18:12:03.577" Score="9" ViewCount="368" Body="&lt;p&gt;Could someone suggest me how to learn some basic theory of schemes? I have two books from algebraic geometry, namely &quot;Diophantine Geometry&quot; from Hindry and Silverman and &quot;Algebraic geometry and arithmetic curves&quot; from Qing Liu. I have had difficulties to prove the equivalence of many definitions. For example Hindry and Silverman defines an affine variety to be an irreducible algebraic subset of $\mathbb{A}^n$ with respect to Zariski topology. On the other hand, Liu defines an affine variety to be the affine scheme associated to a finitely generated algebra over a field.&lt;/p&gt;&#xA;" OwnerUserId="2856" LastActivityDate="2011-01-01T21:38:14.273" Title="Learning schemes" Tags="&lt;algebraic-geometry&gt;&lt;self-learning&gt;&lt;schemes&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="1" />
  <row Id="16049" PostTypeId="1" AcceptedAnswerId="16052" CreationDate="2011-01-01T18:19:09.930" Score="1" ViewCount="90" Body="&lt;p&gt;Hey,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine I am betting on the outcome of flipping $n$ biased coins $C_i$. I know exactly what the probability of each coin landing on heads is $P(H)$ and exactly the probability of each coin landing on tails is $P(T)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For every flip I can place one bet on either (but never both) outcome at odds with 5% value i.e. the probability implied by the odds is 5% lower than $P(H)$ or $P(T)$. Because I am getting value on each bet, I know that in each case my expected value is positive. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I choose to bet on $n$ coin flips, what is the probability that I will make a net loss?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cheers,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pete&lt;/p&gt;&#xA;" OwnerUserId="5199" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2011-01-01T19:10:52.743" LastActivityDate="2011-01-01T19:10:52.743" Title="Probability of Profitability" Tags="&lt;probability&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="16050" PostTypeId="2" ParentId="16048" CreationDate="2011-01-01T18:20:34.050" Score="6" ViewCount="" Body="&lt;p&gt;I learned sheafs and schemes from Hartshorne (as did many people), but I found &lt;strong&gt;Why schemes?&lt;/strong&gt; by David Eisenbud and Joe Harris an invaluable aid. The book was rewritten as &lt;strong&gt;The Geometry of Schemes&lt;/strong&gt; and published in Springer-Verlag's yellow GTM series (here's the &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0387986375&quot; rel=&quot;nofollow&quot;&gt;Amazon link&lt;/a&gt;). &lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2011-01-01T18:20:34.050" CommentCount="0" />
  <row Id="16051" PostTypeId="1" CreationDate="2011-01-01T18:37:06.107" Score="3" ViewCount="192" Body="&lt;p&gt;This is a stronger one related to the question &lt;a href=&quot;http://math.stackexchange.com/questions/15942/convergence-of-lim-n-v-rightarrow-infty-int-01-f-n-x-e-i2-pi-v-x&quot;&gt;Convergence of  $\lim_{n,v \rightarrow \infty}  \int_0^1 f_n (x) e^{-i2\pi v x} \mbox{d} x $&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$F_n(x) :  [0,1] \rightarrow \bf R $, for $1 \leq i \leq n$, $F_n(x)= n\cdot g_{n,i}(x)$ if $x \in [\frac{i-1}{n}, \frac{i}{n})$, with $g_{n,i}$ a series of integrable functions.&#xA;As $n, v \in \bf N$  goes to infinity simultaneously at the same rate, prove the convergence of&lt;br&gt;&#xA;$$\lim_{n,v \rightarrow \infty}  \int_0^1 F_n(x) e^{-i2\pi v x}\,\mbox{d} x $$&#xA;if  $v/n$ is not an integer.&lt;/p&gt;&#xA;" OwnerUserId="5181" LastEditorUserId="5181" LastEditorDisplayName="" LastEditDate="2011-01-09T16:42:57.410" LastActivityDate="2011-05-09T18:56:00.000" Title="Convergence of $\lim_{n,v \rightarrow \infty} \int_0^1 F_n (x) e^{-i2\pi v x} \mbox{d} x $" Tags="&lt;fourier-analysis&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="16052" PostTypeId="2" ParentId="16049" CreationDate="2011-01-01T18:59:11.243" Score="2" ViewCount="" Body="&lt;p&gt;If P(H) varies by coin, it will be hard to do better than adding up all the cases.  If P(H) is the same for all coins and P(T)=1-P(H) you can use the binomial theorem.  First calculate how many coins need to come heads to make a profit.  For example, if P(H) = 0.5 you should win 1/.45 -1= 11/9 for heads (is this what you mean for 5% value?  I could argue for various other figures, but let's use this) and lose 1 for tails.  To break even, then, you need to get tails 11/20 of the time and to lose you need more tails than that.  So the probability of loss in $n$ throws is $$\frac{1}{2^n}\sum_{k=0}^{\lceil \frac{9k}{20} \rceil -1} \binom{n}{k}$$  If the coin is not fair, you will need to bring the P(H) into the sum.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-01-01T18:59:11.243" CommentCount="3" />
  <row Id="16053" PostTypeId="2" ParentId="16044" CreationDate="2011-01-01T19:51:35.860" Score="0" ViewCount="" Body="&lt;p&gt;I don't think Apostol has enough for a first course in Linear Algebra, why don't you try: Linear Algebra by Hoffman and Kunze. The linear algebra bits in Artin's Algebra is also very good.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For multivariable calculus, I found Munkres' Analysis on Manifolds much easier to digest than Spivak.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want something requiring less mathematical maturity then you could try: Hubbard, Vector Calculus, Linear Algebra and Differential Forms. This is a fantastic text. &lt;/p&gt;&#xA;" OwnerUserId="4806" LastActivityDate="2011-01-01T19:51:35.860" CommentCount="2" />
  <row Id="16054" PostTypeId="2" ParentId="15995" CreationDate="2011-01-01T19:51:41.443" Score="5" ViewCount="" Body="&lt;p&gt;I believe the following does the trick and that the $109$ is really a red herring, and that the problem has a solution for each odd &#xA;$n \ge 3.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose $n=2m+1$ and consider the triples $(i,i+j,i+2j)$ where $i=1,2,3,\ldots,(2m+1)$ and $j=1,2,3,\ldots, m.$ We can construct a&#xA;$(2m+1) \times m $ array of triples with $(i,i+j,i+2j)$ in the $i$th row and $j$th column. We work modulo $2m+1.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example here's a solution for $n=9.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{array} {cccc}&#xD;&#xA; (1,2,3) &amp;amp; (1,3,5) &amp;amp; (1,4,7) &amp;amp; (1,5,9) \\&#xD;&#xA; (2,3,4) &amp;amp; (2,4,6) &amp;amp; (2,5,8) &amp;amp; (2,6,1) \\&#xD;&#xA; (3,4,5) &amp;amp; (3,5,7) &amp;amp; (3,6,9) &amp;amp; (3,7,2) \\&#xD;&#xA; (4,5,6) &amp;amp; (4,6,8) &amp;amp; (4,7,1) &amp;amp; (4,8,3) \\&#xD;&#xA; (5,6,7) &amp;amp; (5,7,9) &amp;amp; (5,8,2) &amp;amp; (5,9,4) \\&#xD;&#xA; (6,7,8) &amp;amp; (6,8,1) &amp;amp; (6,9,3) &amp;amp; (6,1,5) \\&#xD;&#xA; (7,8,9) &amp;amp; (7,9,2) &amp;amp; (7,1,4) &amp;amp; (7,2,6) \\&#xD;&#xA; (8,9,1) &amp;amp; (8,1,3) &amp;amp; (8,2,5) &amp;amp; (8,3,7) \\&#xD;&#xA; (9,1,2) &amp;amp; (9,2,4) &amp;amp; (9,3,6) &amp;amp; (9,4,8)&#xD;&#xA;\end{array}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The number of triples = $m(2m+1)= { 2m+1 \choose m } = $ the number of distinct pairs of integers in $S= \{ 1,2,3,\ldots,2m+1 \} $ and all triples are unique, where the order of the elements is taken into account. (For suppose $(x,y,z)$ is in our array, then $x=i,y=i+j$ and $z=i+2j$ for some $i,j$ and thus if $x&amp;lt;y$ this triple is in the $x$th row and $(y-x)$th column, and if $x&amp;gt;y$ it is in the $x$th row and $(y-x + 2m+1)$th column. So its position is uniquely determined by its elements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moreover, if $(x,y,z)=(i,i+j,i+2j)$ is in the array then none of $(x,z,y),(z,y,x)$ and $(y,x,z)$ can be in the array, for if one of the positions of the elements is fixed the latter three triples cannot obey the construction rule, that its elements increase by $j,$ since we know that $x,$ $y$ and $z$ increase by $j.$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now consider a pair of integers $(a,b),$ which appears in a given triple, $(a,b,c)$. There cannot exist another triple in the array $(a,b,d),$ say, with $c \ne d$ since $c$ is uniquely determined by $a$ and $b.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The pair $(a,b)$ cannot appear in more than three triples, in whatever order (we've already noted that we cannot have both $(a,b,c)$ and $(b,a,c)$ hence the three and not six), since the remaining element is uniquely determined by $a$ and $b.$ However, it cannot appear in less than three triples since the number of triples equals the number of pairs of distinct integers in&#xA;$S,$ so this would mean that another pair must appear in more than three triples, a contradiction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus we have a solution for any odd $n \ge 3.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So a solution for $n=11$ is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{array} {ccccc}&#xD;&#xA; (1,2,3) &amp;amp; (1,3,5) &amp;amp; (1,4,7) &amp;amp; (1,5,9) &amp;amp; (1,6,11) \\&#xD;&#xA; (2,3,4) &amp;amp; (2,4,6) &amp;amp; (2,5,8) &amp;amp; (2,6,10) &amp;amp; (2,7,1)\\&#xD;&#xA; (3,4,5) &amp;amp; (3,5,7) &amp;amp; (3,6,9) &amp;amp; (3,7,11) &amp;amp; (3,8,2) \\&#xD;&#xA; (4,5,6) &amp;amp; (4,6,8) &amp;amp; (4,7,10) &amp;amp; (4,8,1) &amp;amp; (4,9,3) \\&#xD;&#xA; (5,6,7) &amp;amp; (5,7,9) &amp;amp; (5,8,11) &amp;amp; (5,9,2) &amp;amp; (5,10,4) \\&#xD;&#xA; (6,7,8) &amp;amp; (6,8,10) &amp;amp; (6,9,1) &amp;amp; (6,10,3) &amp;amp; (6,11,5) \\&#xD;&#xA; (7,8,9) &amp;amp; (7,9,11) &amp;amp; (7,10,2) &amp;amp; (7,11,4) &amp;amp; (7,1,6) \\&#xD;&#xA; (8,9,10) &amp;amp; (8,10,1) &amp;amp; (8,11,3) &amp;amp; (8,1,5) &amp;amp; (8,2,7) \\&#xD;&#xA; (9,10,11) &amp;amp; (9,11,2) &amp;amp; (9,1,4) &amp;amp; (9,2,6) &amp;amp; (9,3,8) \\&#xD;&#xA; (10,11,1) &amp;amp; (10,1,3) &amp;amp; (10,2,5) &amp;amp; (10,3,7) &amp;amp; (10,4,9) \\&#xD;&#xA; (11,1,2) &amp;amp; (11,2,4) &amp;amp; (11,3,6) &amp;amp; (11,4,8) &amp;amp; (11,5,10)&#xD;&#xA;&#xD;&#xA;\end{array}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that I'm not using Steiner triples, as $11 \ne 6k+1$ or $6k+3.$&lt;/p&gt;&#xA;" OwnerUserId="1301" LastEditorUserId="1301" LastEditorDisplayName="" LastEditDate="2011-01-01T20:11:45.513" LastActivityDate="2011-01-01T20:11:45.513" CommentCount="1" />
  <row Id="16055" PostTypeId="2" ParentId="16038" CreationDate="2011-01-01T19:52:49.777" Score="5" ViewCount="" Body="&lt;p&gt;In fact, using the two examples in your comment provides a counterexample.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $A = \begin{bmatrix} 1 &amp;amp; 0\\1&amp;amp;0\end{bmatrix}$ and $B = \begin{bmatrix}0&amp;amp;2\\0&amp;amp;0 \end{bmatrix}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you mention, $A^2 = \begin{bmatrix} 1 &amp;amp; 0 \\ 1 &amp;amp; 0\end{bmatrix}$ so $A$ generates a finite monoid.  Likewise, $B^2$ is the $0$ matrix so this generates a finite monoid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the monoid generated by $A$ and $B$ is not periodic: the element $BA = \begin{bmatrix} 2 &amp;amp; 0 \\ 0 &amp;amp; 0\end{bmatrix}$ doesn't generate a finite monoid since $(BA)^n = \begin{bmatrix} 2^n &amp;amp; 0\\ 0 &amp;amp; 0\end{bmatrix}$, which is distinct for each $n$.&lt;/p&gt;&#xA;" OwnerUserId="331" LastActivityDate="2011-01-01T19:52:49.777" CommentCount="1" />
  <row Id="16056" PostTypeId="2" ParentId="16048" CreationDate="2011-01-01T20:24:58.873" Score="2" ViewCount="" Body="&lt;p&gt;I am also currently learning about sheaves and schemes, and I'm finding Ravi Vakil's notes to be very helpful:&#xA;&lt;a href=&quot;http://math216.wordpress.com/&quot; rel=&quot;nofollow&quot;&gt;http://math216.wordpress.com/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5265" LastActivityDate="2011-01-01T20:24:58.873" />
  <row Id="16057" PostTypeId="1" CreationDate="2011-01-01T20:30:26.747" Score="10" ViewCount="144" Body="&lt;p&gt;I have the following cost function:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\mbox{BSP Cost}=\sum_{i=1}^{\frac{n}{G}}G^{2}\left\lceil \frac{i}{p}\right\rceil +g\left(p\right)\sum_{i=1}^{\frac{n}{G}}Gi+l\left(p\right)\frac{n}{G}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to minimize it by choosing an appropriate G (i.e., G is a function of p and n). I have simplified it to the following form:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\mbox{BSP Cost}=\frac{Gn}{2p}+\frac{n^{2}}{2p}+g\left(p\right)\cdot\left(\frac{n}{2G}+\frac{n^{2}}{2G^{2}}\right)+l\left(p\right)\frac{n}{G}\to\min$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To find the minimization, I used a derivative $\frac{d}{dG}$ on the cost function, and compared to zero. I got this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{nG^{3}-g\left(p\right)npG-g\left(p\right)n^{2}p-2l\left(p\right)npG}{2pG^{3}}=0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$nG^{3}-g\left(p\right)npG-g\left(p\right)np-2l\left(p\right)npG=0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And I'm not sure how to proceed from this point. Can you help me find a function for G(n,p)? If you see any mistakes in the process above, please tell me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;: it might also be important to mention that all the variables and functions (g,n,p,l,G) are positive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT 2&lt;/strong&gt;: an rough approximation formula will do!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT 3&lt;/strong&gt;: Here's an approximation of g,l:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$g\left(p\right)=-0.858p^{3}+12.31p^{2}-47.12p+79.67$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$l\left(p\right)=670.9p^{2}+2815p-2763$&lt;/p&gt;&#xA;" OwnerUserId="922" LastEditorUserId="922" LastEditorDisplayName="" LastEditDate="2011-01-02T10:24:08.113" LastActivityDate="2011-01-02T10:24:08.113" Title="A problem with minimizing a function" Tags="&lt;calculus&gt;&lt;algebra&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="1" />
  <row Id="16058" PostTypeId="2" ParentId="16048" CreationDate="2011-01-01T21:38:14.273" Score="7" ViewCount="" Body="&lt;p&gt;I have found Kenji Ueno's book &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0821808621&quot; rel=&quot;nofollow&quot;&gt;Algebraic Geometry 1: From Algebraic Varieties to Schemes&lt;/a&gt; to be quite satisfying in introducing the basic theory of schemes. Well, to be fair, this is only the first in a series of three books on the subject by the same author. So this first volume basically just develops the definitions of an affine scheme first and then of a scheme in general by &quot;pasting&quot; together affine schemes. It does not go into cohomology and more advanced stuff, which is the subject of the other two books. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, what I really like is that he motivates very carefully the passage from the definition of an affine algebraic variety as an irreducible algebraic set in an affine space $\mathbb{A}_{k}^{n}$ to the definition of an affine variety using schemes, which is where you are having some trouble. What he does is that he starts by doing some algebraic geometry in the classic sense, that is, over an algebraically closed field $k$, in the first chapter of the book. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then the author proves that there is a correspondence between the points in an algebraic set $V$ and the maximal ideals of its associated coordinate ring $k[V]$, where a point $(a_1, \dots , a_n) \in V$ corresponds to the maximal ideal of $k[V]$ determined by the ideal $(x_1 - a_1 , \dots , x_n - a_n) \in k[x_1, \dots, x_n]$, that is, a correspondence between the points in $V$ and the &quot;points&quot; in the &lt;strong&gt;maximal spectrum&lt;/strong&gt; $\text{max-Spec}(k[V])$ of the coordinate ring $k[V \, ]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then Ueno goes on to define an affine algebraic variety as a pair $(V, k[V \, ] )$ where $V$ is an an algebraic set. But he then makes the argument that one can go a little bit further and consider the pair $( \text{max-Spec}(R), R )$ where $R$ is a $k$ algebra. But here Ueno arguments that if the original intention was to study the sets of solutions of polynomial equations, then where is the geometry and where are the equations hidden if an algebraic variety is defined as this pair $( \text{max-Spec}(R), R )$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The interesting thing is that if the $k$ algebra $R$ is finitely generated over $k$ then &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ R \simeq \frac{ k[x_1 , \dots , x_n] }{I}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so that as a consequence&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \text{max-Spec}(R) = V(I)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so that again you'll have some equations (this is all done and explained in detail in the book). So then the author (re)defines an algebraic variety over an algebraically closed field $k$ (remember that he is doing everything in the classic sense) as a pair $( \text{max-Spec}(R), R )$, where $R$ is a finitely generated $k$ algebra.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And then at the end of the first chapter the author motivates the need for a more general theory, for example having in mind the needs of number theory, because since everything was done in the context of an algebraically closed field, then the arguments don't work for the fields (and rings) of interest in number theory. In particular, it is noted how an extension of the definitions to include these cases would need to take into account not only the set of maximal ideals, but the set of all prime ideals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then chapter two develops first some properties of this set of prime ideals, or &lt;strong&gt;prime spectrum&lt;/strong&gt; of a ring, making it into a topological space with the Zariski topology... and then defines the necessary things in order to be able to define an affine scheme and a scheme (I mean, the concepts of a sheaf of rings, a ringed space, etc). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is not a short story of course, but again I prefer this type of approach at first, than having to deal with an unmotivated (and difficult) definition that strives for great generality but I have no idea of where it comes from and what is its purpose.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the book that Arturo recommended is great also but it assumes you already know some algebraic geometry and its level is higher than Ueno's book. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You should take a look at it and see if you like it, the book has a fair amount of examples and some exercises interspersed within the text also. You'll have to study from other sources as well but I believe that this book does a pretty good job at motivating the abstract definitions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope this helps at least a little.&lt;/p&gt;&#xA;" OwnerUserId="900" LastActivityDate="2011-01-01T21:38:14.273" />
  <row Id="16059" PostTypeId="1" AcceptedAnswerId="16079" CreationDate="2011-01-01T21:56:17.463" Score="4" ViewCount="206" Body="&lt;p&gt;I am trying to verify for which $z \in \mathbb{R}$ the series $\sum _{n=1}^{\infty } \left(1-\cos \left(\frac{1}{n}\right)\right)^z$ converges. The only test that was successful for me is the Kummer Test which gave the apparently correct result that it converges if $z &amp;gt; \frac{1}{2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To get there I use the fact that the series $\sum _{n=1}^{\infty } a_n$ converges (when the limit exists) if and only if  $\lim_{n\to \infty } \, n&#xD;&#xA;   \left(\frac{a_n}{a_{n+1}}-1\right) &amp;gt; 1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using Mathematica I get $\lim_{n\to \infty } \, n&#xD;&#xA;   \left(\frac{a_n}{a_{n+1}}-1\right)=\lim_{n\to \infty } \, n \left(\left(1-\cos \left(\frac{1}{n}\right)\right)^z \left(1-\cos&#xD;&#xA;   \left(\frac{1}{n+1}\right)\right)^{-z}-1\right) = 2z$ and therefore $z &amp;gt; \frac{1}{2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Start reading here if you are only interested in the problem and not how i got there:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I try to understand how to get that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\lim_{n\to \infty } \, n \left(\left(1-\cos \left(\frac{1}{n}\right)\right)^z \left(1-\cos&#xD;&#xA;   \left(\frac{1}{n+1}\right)\right)^{-z}-1\right)=\lim_{x\to 0} \, \frac{-1+(1-\cos (x))^z \left(1-\cos&#xD;&#xA;   \left(\frac{x}{1+x}\right)\right)^{-z}}{x} = 2z$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All attempts I did to show this in an elementary way failed and thats why I hope someone has an idea how to verify the result. For me its not obvious why the result to the limit is how it is&lt;/p&gt;&#xA;" OwnerUserId="3123" LastActivityDate="2011-01-02T02:51:27.347" Title="Solving a tough limit" Tags="&lt;real-analysis&gt;&lt;limit&gt;" AnswerCount="2" />
  <row Id="16060" PostTypeId="2" ParentId="16059" CreationDate="2011-01-01T22:11:14.913" Score="7" ViewCount="" Body="&lt;p&gt;Note that ${\displaystyle \lim_{x \rightarrow 0} {1 - \cos(x) \over x^2} = {1 \over 2}}$, so that ${\displaystyle \lim_{n \rightarrow \infty} {\big(1 - \cos({1 \over n})\big)^z \over ({1 \over n^2})^z} = {1 \over 2^z}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus you can use the limit comparison test with ${\displaystyle\sum_{n=1}^{\infty} \bigg({1 \over n^2}\bigg)^z = \sum_{n=1}^{\infty} {1 \over n^{2z}}}$. This converges iff $z &amp;gt; {1 \over 2}$ by the $p$-test.&lt;/p&gt;&#xA;" OwnerUserId="3035" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-02T00:04:59.440" LastActivityDate="2011-01-02T00:04:59.440" CommentCount="1" />
  <row Id="16061" PostTypeId="1" CreationDate="2011-01-01T22:16:25.670" Score="2" ViewCount="114" Body="&lt;p&gt;A dishonest dealer marks his goods $20\%$ above the cost price. He gives a discount of $10\%$ to the customer on the marked price and makes a profit by using a false weight of $900$ gms in place of $1$ kg while buying or selling. Find the percentage profit earned by the shopkeeper.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I do tried this problem but am not quite there,hints are welcome :-)&lt;/p&gt;&#xA;" OwnerUserId="2109" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2011-01-02T19:50:01.357" LastActivityDate="2011-01-02T19:52:38.837" Title="On Profit and loss" Tags="&lt;algebra-precalculus&gt;" AnswerCount="1" />
  <row Id="16062" PostTypeId="1" CreationDate="2011-01-01T19:41:32.217" Score="-2" ViewCount="254" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have this polar function:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;r = A / log(B * tan(t / 2 * N)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where A, B, N are arbitary parameters and t is the angle theta in the polar coordinate system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example graph for A=8, B=0.5, N=4&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/FktGb.png&quot; alt=&quot;Sample graph&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I plot this function onto a cartesian coordinate grid so I get an image like the one above?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;thanks &lt;/p&gt;&#xA;" OwnerUserId="5289" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-05T03:57:41.350" LastActivityDate="2011-01-05T03:57:41.350" Title="How to plot polar function onto cartesian grid?" Tags="&lt;graph&gt;&lt;coordinate-systems&gt;" AnswerCount="2" CommentCount="9" />
  <row Id="16063" PostTypeId="2" ParentId="16062" CreationDate="2011-01-01T19:47:49.380" Score="1" ViewCount="" Body="&lt;p&gt;Incomplete pseudocode sample, but you should get the idea:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;for t in [0, 2pi):&#xA;    r = /* whatever you got depending on t */&#xA;    x = r * cos(t)&#xA;    y = r * sin(t)&#xA;    draw line to (x,y)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" LastActivityDate="2011-01-01T22:24:14.197" CommentCount="3" />
  <row Id="16064" PostTypeId="2" ParentId="16062" CreationDate="2011-01-01T20:56:23.573" Score="0" ViewCount="" Body="&lt;p&gt;Ok, I figured it out. Some example Java code:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;import static java.lang.Math.*;&#xA;&#xA;import java.awt.Color;&#xA;import java.awt.Graphics;&#xA;import java.awt.Point;&#xA;import java.awt.image.BufferedImage;&#xA;&#xA;import javax.swing.ImageIcon;&#xA;import javax.swing.JFrame;&#xA;import javax.swing.JLabel;&#xA;&#xA;&#xA;public class TestPolarPlot {&#xA;    public static void main(String[] args) {&#xA;    final int width = 512;&#xA;    final int height = 512;&#xA;    BufferedImage img = new BufferedImage(width, height, BufferedImage.TYPE_4BYTE_ABGR);&#xA;    Graphics g = img.getGraphics();&#xA;    g.setColor(Color.black);&#xA;    g.fillRect(0, 0, width, height);&#xA;    g.setColor(Color.white);&#xA;    final double A = 8;&#xA;    final double B = 0.5;&#xA;    final double N = 4;&#xA;    final double scale = 128;&#xA;    final double zoom = 50;&#xA;    final double step = 1 / scale;&#xA;    Point last = null;&#xA;    final Point origin = new Point(width/2, height/2);&#xA;&#xA;    for (double t = 0; t &amp;lt;= 2*PI; t+= step) {&#xA;        final double r = zoom * polarFunction(t, A, B, N);&#xA;        final int x = (int)round(r * cos(t));&#xA;        final int y = (int)round(r * sin(t));&#xA;        Point next = new Point(x, y);&#xA;        if (last != null) {&#xA;            g.drawLine(origin.x + last.x, origin.y + last.y,&#xA;                origin.x + next.x, origin.y + next.y);&#xA;        }&#xA;        last = next;&#xA;    }&#xA;&#xA;    JFrame frame = new JFrame(&quot;testit&quot;);&#xA;    frame.setDefaultCloseOperation(JFrame.EXIT_ON_CLOSE);&#xA;    frame.getContentPane().add(new JLabel(new ImageIcon(img)));&#xA;    frame.pack();&#xA;    frame.setLocationRelativeTo(null);&#xA;    frame.setVisible(true);&#xA;}&#xA;&#xA;    public static double polarFunction(double t, double A, double B, double N) {&#xA;        return A / log(B * tan(t / (2 * N)));&#xA;    }&#xA;}&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I didn't expect this to create smooth curves but it works pretty well. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/vMgjo.png&quot; alt=&quot;alt text&quot;&gt;&#xA;&lt;img src=&quot;http://i.imgur.com/aED3T.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="5289" LastActivityDate="2011-01-01T22:24:14.197" />
  <row Id="16065" PostTypeId="2" ParentId="16061" CreationDate="2011-01-01T22:48:30.247" Score="2" ViewCount="" Body="&lt;p&gt;The dealer doesn't actually give his customer a discount.  He charges 90% of the price for 90% of the product.  Thus he still makes a 20% profit over the supplier price. Let $C$ be the supplier price for 1 kg of product. The dealer then gets 1.1kg for the price $C$ or 1kg for $10\over{11}$$C$.  He sells this 1kg for $1.2C$.  The profit margin is $1.2C$-$10\over{11}$$C$ divided by the dealer price, $10\over{11}$$C$, which I calculate to be 32%.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here I have assumed the &quot;cost price&quot; is the supplier price, $C$.  If you assume the cost price is what the dishonest dealer actually pays his supplier, you will get a different answer still.&lt;/p&gt;&#xA;" OwnerUserId="3971" LastEditorUserId="3971" LastEditorDisplayName="" LastEditDate="2011-01-02T19:52:38.837" LastActivityDate="2011-01-02T19:52:38.837" CommentCount="1" />
  <row Id="16066" PostTypeId="1" AcceptedAnswerId="16084" CreationDate="2011-01-01T23:14:26.887" Score="29" ViewCount="706" Body="&lt;p&gt;A friend sent me a link to this item today, which is billed as an &quot;Irrational Numbers Wall Clock.&quot;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/owLvp.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is at least one possible mistake in it, as it is not known whether $\gamma$ is irrational.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, this started me wondering about how to improve the design of this clock from a mathematical standpoint.  Here's my formulation of the problem:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Find 12 numbers that have been proved to be irrational to place around the outside of a clock.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Each of eleven of the numbers must approximate as closely as possible one of the integers from 1 through 11.  The 12th can either be just smaller than 12 or just larger than 0.  &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The numbers must have expressions that are as simple as possible (in the spirit of - or even more simple than - those in the clock given in the picture here).  Thus, for example, no infinite sums, no infinite products, and no continued fractions.  Famous constants and transcendental functions evaluated at small integers encouraged.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Expressions should be as varied as possible.  Better answers would include at least one use of trig functions, logarithms, roots, and famous constants.  &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Obviously, goals 2, 3, and 4 act against each other.  And, as Jonas Meyer points out, &quot;as closely as possible&quot; and &quot;as simple as possible&quot; are not well-defined.  That is intentional.  I am afraid that if I tried to define those precisely I would preclude some answers that I might otherwise consider good.  Thus, in addition to the mathematics, there's a sizable artistic component that goes into what would be considered a good answer.  Hence the &quot;soft-question&quot; tag.  I'm really curious as to what the math.SE community comes up with and then what it judges (via upvoting) to be the best answers, subject to these not-entirely-well-defined constraints.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the designer of the clock given here was not trying to approximate the integers on a clock as closely as possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, it's currently New Year's Day in my time zone.  Perhaps a time-related question is appropriate. :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: There is now a community wiki answer that people can edit if they just want to add a few suggestions rather than all twelve.&lt;/p&gt;&#xA;" OwnerUserId="2370" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2011-01-02T04:31:13.893" LastActivityDate="2011-02-26T12:41:58.520" CommunityOwnedDate="2011-01-02T15:01:17.520" Title="Designing an Irrational Numbers Wall Clock" Tags="&lt;soft-question&gt;&lt;irrational-numbers&gt;" AnswerCount="8" CommentCount="9" FavoriteCount="8" />
  <row Id="16067" PostTypeId="2" ParentId="16029" CreationDate="2011-01-01T23:41:53.167" Score="1" ViewCount="" Body="&lt;p&gt;My experience overlaps Barry Smith's some, but this is too long for a comment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On community colleges: The community college where I taught would hire people with bachelor's degrees in math to teach math courses, but the courses eligible for transfer to a four-year institution (basically, the non-remedial courses) could only be taught by people with master's degrees.  Having a master's degree made you more attractive to the CC folks.  And if you had a PhD, that was even better.  At the CC where I taught for three terms I was the only one on the faculty with a PhD, and so even though I was an adjunct I was handed the most advanced courses they offered (Calculus III and differential equations).  The other thing I will say here is that since I didn't get much teaching experience in graduate school, I don't think I would have been hired for my first tenure-track job at a teaching school without the experience I gained working at the CC.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Depending on the area of math you are studying, don't forget that there are departments outside of math that might be willing to hire you as an adjunct.  My PhD work was in operations research, and because of that I was able to teach some advanced OR-related business classes.  Knowledge of other math-related areas that might get you hired as an adjunct in a business or engineering school are statistics and computer programming.  This is not math, per se, but it would still be teaching experience in a related area.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, there are ways to supplement your experience teaching in the classroom.  Barry Smith mentions some.  If you are a graduate student, here are some more suggestions.  Volunteer to help the undergraduates at your institution prepare for the &lt;a href=&quot;http://math.scu.edu/putnam/&quot; rel=&quot;nofollow&quot;&gt;Putnam&lt;/a&gt; or the &lt;a href=&quot;http://www.comap.com/undergraduate/contests/mcm/&quot; rel=&quot;nofollow&quot;&gt;Mathematical Contest in Modeling&lt;/a&gt;, maybe in a weekly seminar format.  Start a journal club for graduate students, where each week one of you teaches the others the content of some journal article.  Some institutions (such as Colorado - see &lt;a href=&quot;http://gtp.colorado.edu/&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;) have a graduate teacher preparation program.  Take advantage of that.   &lt;/p&gt;&#xA;" OwnerUserId="2370" LastActivityDate="2011-01-01T23:41:53.167" CommunityOwnedDate="2011-01-02T14:59:54.620" />
  <row Id="16068" PostTypeId="2" ParentId="16038" CreationDate="2011-01-01T23:56:29.723" Score="2" ViewCount="" Body="&lt;p&gt;More generally, for any $n\gt 1$, the matrices&#xA;$$A=\left(\begin{array}{cccc}&#xD;&#xA;1 &amp;amp; 0 &amp;amp; \cdots &amp;amp; 0\\&#xD;&#xA;1 &amp;amp; 0 &amp;amp; \cdots &amp;amp; 0\\&#xD;&#xA;\vdots &amp;amp; \vdots &amp;amp; \ddots &amp;amp; \vdots\\&#xD;&#xA;1 &amp;amp; 0 &amp;amp; \cdots &amp;amp; 0&#xD;&#xA;\end{array}\right)\quad\text{and}\quad&#xD;&#xA;B=\left(\begin{array}{cccc}&#xD;&#xA;1 &amp;amp; 1 &amp;amp; \cdots &amp;amp; 1\\&#xD;&#xA;0 &amp;amp; 0 &amp;amp; \cdots &amp;amp; 0\\&#xD;&#xA;\vdots &amp;amp; \vdots &amp;amp; \ddots &amp;amp; \vdots\\&#xD;&#xA;0 &amp;amp; 0 &amp;amp; \cdots &amp;amp; 0&#xD;&#xA;\end{array}\right)$$&#xA;are each idempotent, so they each generate a finite monoid, but neither&#xA;$$AB = \left(\begin{array}{cccc}&#xD;&#xA;1 &amp;amp; 1 &amp;amp; \cdots &amp;amp; 1\\&#xD;&#xA;1 &amp;amp; 1 &amp;amp; \cdots &amp;amp; 1\\&#xD;&#xA;\vdots &amp;amp; \vdots &amp;amp; \ddots &amp;amp; \vdots\\&#xD;&#xA;1 &amp;amp; 1 &amp;amp; \cdots &amp;amp; 1&#xD;&#xA;\end{array}\right)\quad\text{nor}\quad&#xD;&#xA;BA = \left(\begin{array}{cccc}&#xD;&#xA;n &amp;amp; 0 &amp;amp; \cdots &amp;amp; 0\\&#xD;&#xA;0 &amp;amp; 0 &amp;amp; \cdots &amp;amp; 0\\&#xD;&#xA;\vdots &amp;amp; \vdots &amp;amp; \ddots &amp;amp; \vdots\\&#xD;&#xA;0 &amp;amp; 0 &amp;amp; \cdots &amp;amp; 0&#xD;&#xA;\end{array}\right)$$&#xA;are periodic. &lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2011-01-01T23:56:29.723" CommentCount="1" />
  <row Id="16069" PostTypeId="2" ParentId="16057" CreationDate="2011-01-02T00:09:09.077" Score="2" ViewCount="" Body="&lt;p&gt;Assuming $l(p)$ and $g(p)$ are known, I get the same for the derivative. You can simplifiy it by dividing out $n$.  You now have a cubic in $G(n,p)$, though it seems not to be a function of $n$, which can be solved by the usual formula for given n and p, but that is a mess.  Alternately, you can write it as $G^2=g(p)p(1+\frac{1}{G})+2l(p)p$  If a numeric solution is acceptable, this should converge nicely-start by evaluating the RHS with $G=0$, find $G^2$, plug $G$ into the RHS and iterate to convergence.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-01-02T00:09:09.077" />
  <row Id="16070" PostTypeId="2" ParentId="16038" CreationDate="2011-01-02T00:41:51.600" Score="2" ViewCount="" Body="&lt;p&gt;Let $A$ be an element of $\mathcal{M}_n$ such that $A$ is periodic.  Then $A^\mathrm{T}$ is also periodic.  However, $AA^\mathrm{T}$ is periodic if and only if each nonzero entry of $A$ is $1$ and each row and column of $A$ has at most one nonzero entry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose $A=(a_{ij})$ and $a_{i_0j_0}\gt1$ for some $i_0$ and $j_0$.  If $AA^\mathrm{T}=(b_{ij})$, then $b_{i_0i_0}\geq a_{i_0j_0}^2\gt1$.  In particular this implies that the Euclidean norm of $AA^\mathrm{T}e_{i_0}$ is greater than $1$, where $e_{i_0}$ is the $i_0^\text{th}$ standard basis vector.  Since $AA^\mathrm{T}$ is positive semidefinite, this implies that its spectral radius is greater than $1$, and therefore its powers have unbounded spectral radius.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose that $a_{i_0j_0}\gt0$ and $a_{i_0j_1}\gt0$ for some $i_0$ and $j_0\neq j_1$.  Then $b_{i_0i_0}\geq a_{i_0j_0}^2+a_{i_0j_1}^2\gt1$, which again implies that the powers of $AA^\mathrm{T}$ have unbounded spectral radius.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose that $a_{i_0j_0}\gt0$ and $a_{i_1j_0}\gt0$ for some $i_0\neq i_1$ and $j_0$.  Then $b_{i_0i_0}\geq a_{i_0j_0}^2\gt0$ and $b_{i_1i_0}\geq a_{i_1j_0}a_{i_0j_0}\gt0$.  This implies that the Euclidean norm of $AA^\mathrm{T}e_{i_0}$ is at least $\sqrt{2}$.  Thus, again, the powers of $AA^\mathrm{T}$ have unbounded spectral radius.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, if each nonzero entry of $A$ is $1$ and each row and column of $A$ has at most one nonzero entry, then $AA^\mathrm{T}$ is idempotent.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2011-01-02T00:41:51.600" CommentCount="1" />
  <row Id="16071" PostTypeId="1" CreationDate="2011-01-02T01:18:23.640" Score="3" ViewCount="195" Body="&lt;p&gt;I study CS, but on the first semester I have a lot of mathematics. Of course, there is an introduction to set theory and logic. Recently, we had lectures about cardinality, different kinds of infinity, countable sets etc. I know the basics, but I don't understand many ideas as I would like to understand them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, I am looking for a book/article/tutorial/etc. which is a good explanation of cardinality, countable sets, idea of different kinds of infinity and cardinality of sets of different functions. Now, I only need sources, which cover the basics, but I don't mind if You post something on higher level (but &lt;em&gt;introduction stuff&lt;/em&gt; has greater priority).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS: I was looking for similar questions on this site, I've found for exaple [1] and [2], but it isn't what I'm actually looking for.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I woluld appreciate You help.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[1] &lt;a href=&quot;http://math.stackexchange.com/questions/477/cardinality-of-set-of-real-continuous-functions&quot;&gt;Cardinality of set of real continuous functions&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;[2] &lt;a href=&quot;http://math.stackexchange.com/questions/1/different-kinds-of-infinities&quot;&gt;Different kinds of infinities?&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5291" LastEditorUserId="622" LastEditorDisplayName="" LastEditDate="2011-01-12T07:15:38.163" LastActivityDate="2011-01-12T18:08:10.137" Title="Book/article/tutorial as an introduction to Cardinality" Tags="&lt;reference-request&gt;&lt;set-theory&gt;&lt;infinity&gt;" AnswerCount="6" />
  <row Id="16072" PostTypeId="2" ParentId="16066" CreationDate="2011-01-02T01:31:47.993" Score="10" ViewCount="" Body="&lt;p&gt;Here are my meager suggestions: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;For a new &quot;zero&quot;, the symobl $\epsilon$, which although not representing a specific number, is nearly universally used to represent a very small positive quantity. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Alternatively, for zero one could use $\frac{1}{\omega}$, which is one of the most canonical infinitesimals in the surreal numbers.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;For 2, one could use $|1+i|^2$. (I realize this is exact and hence rational, but the expression involves non-rational numbers.) &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;As a mathematician who is also an antique clock collector, I love this question, and I would be interested if after some good answers are submitted, they are collected and made into a suitably attractive pdf image that could be printed and actually used for a clock face. My suggestion would be to adopt an old-style clock typography, if this is possible, since it might also help the clock face look more like a clock face.&lt;/p&gt;&#xA;" OwnerUserId="413" LastActivityDate="2011-01-02T01:31:47.993" CommunityOwnedDate="2011-01-02T15:01:17.520" CommentCount="3" />
  <row Id="16073" PostTypeId="1" AcceptedAnswerId="16076" CreationDate="2011-01-02T01:36:48.700" Score="0" ViewCount="94" Body="&lt;p&gt;3:30am and i was just looking at the law of large numbers and thought about the idea of a series of random variables and got myself very stuck.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;when we consider the series X'_n = X_1 + ... X_n and say it (divided by n) converges to E(X_1), one of the definitions of a.s. convergence is based on the probability space. specifically, the theorem can be stated:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P({ w in omega : 1/n * X'_n(w) --&gt; E(X_1) }) = 1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which then led me to the question, what is X'_n(w)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;indeed w is an element in our sample space. lets say our random variables are i.i.d. (assumption of the theorem) with known distribution (lets say, exponential, even though we wont use this fact). then a single w is but a number on the interval [0,+infinite).&lt;br&gt;&#xA;so X'_n(w) = (X_1 + ... + X_n)(w)&lt;br&gt;&#xA;= X_1(w) + ... X_n(w)&#xA;= n * X_1(w)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;since we have n random variables that are all looking at the same event from the sample space, i.e. the &lt;em&gt;same&lt;/em&gt; event (in this case real number), and since theyre identically distributed theyll all have the same value?!?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;of course thats not the intention of the statement of the law. the theorem says that if we sample n times for large enough n, blah blah blah. which of course means that we sample n different values, different w's, i just dont see how thats expressed in the formulation of the definition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;im missing something very basic - maybe the fact that im way too tired to be thinking about this at 3:30am. staying awake when im too tired to function is one of those mistakes that always seems tempting to make at the time, always turns out to be a mistake afterwards, and is the perpetual lesson that i will never learn from...&lt;/p&gt;&#xA;" OwnerUserId="2881" LastActivityDate="2011-01-02T15:42:46.457" Title="series of random variables" Tags="&lt;sequences-and-series&gt;&lt;probability&gt;" AnswerCount="2" />
  <row Id="16074" PostTypeId="2" ParentId="15193" CreationDate="2011-01-02T01:57:48.957" Score="1" ViewCount="" Body="&lt;p&gt;Let $P\ $ be the number of point that can be placed in $K_n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$P \le n-2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To see this consider three adjacent vertex. Without loss of generality let them be named 1, 2, and 3. Consider the triangle 1-2-3, and notice that it is cut by $n-3$ lines all of which start at vertex 2. (There is $n-1$ lines starting from vertex 2, but two of them are the lines 1-2 and 2-3). Thus the triangle 1-2-3 is divided in $n-2$ smaller triangles. Then notice that the extension of all those small triangles are triangles that have all of their vertices on the perimeter of $K_n$. So all of the space of $K_n$ is divided into $n-2$ triangles. Thus if there is more than $n-2$ points (letter) one of those triangle must contain at least two points (letter).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$P \ge n - 2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To see this take an edge, say 1-2 and consider all the triangles that have this edge and another vertex on the perimeter of $K_n$. There is $n-2$ of them because there is $n-2$ vertices once we remove the vertices 1 and 2. Now each of those triangle contain smaller triangles when $n \ge 5$. For each big triangle place a point in the smaller triangle that is closest to the third vertex (i.e. the vertex that is not 1 or 2). In this way you can place $n-2$ points. It should be clear that each of the smaller triangle that contain a point is not contained in any of the other big triangle because all the big triangles here share the edge 1-2 and a different vertex. Since we know that the formula works for $n \le 4$ we are done.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus $P = n - 2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the proof that $P \ge n - 2$ only shows that there is a one to one map between the points and the big triangle constructed. The proof does not show that each possible big triangle (including those that don't have the edge 1-2) unambiguously determines a point (letter).&lt;/p&gt;&#xA;" OwnerUserId="5292" LastActivityDate="2011-01-02T01:57:48.957" />
  <row Id="16075" PostTypeId="2" ParentId="16071" CreationDate="2011-01-02T02:35:30.983" Score="0" ViewCount="" Body="&lt;p&gt;Perhaps you are looking for something slightly different, but Cantor's &quot;Contributions to the Founding of the Theory of Transfinite Numbers&quot; is a delight to read.&lt;/p&gt;&#xA;" OwnerUserId="5253" LastActivityDate="2011-01-02T02:35:30.983" CommentCount="1" />
  <row Id="16076" PostTypeId="2" ParentId="16073" CreationDate="2011-01-02T02:45:34.183" Score="3" ViewCount="" Body="&lt;p&gt;The variables $X_1,\ldots,X_n$ are intended to be \emph{independent}. That means that given a sample space for one of them $\Omega$, the appropriate sample space for all of them is $\Omega^n$. For example, if $X_i$ is a Bernoulli random variable, then you can take $\Omega = \{0,1\}$ and $\Omega^n$ is just the set of all possible outcomes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, a measure space comes with a $\sigma$-field and a probability measures, so these also have to be extended; when you're dealing with finitely many variables this is straightforward. For infinite (countable) sequences it might be less obvious, but you can look it up in any good introductory book.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2011-01-02T02:45:34.183" CommentCount="2" />
  <row Id="16077" PostTypeId="2" ParentId="16044" CreationDate="2011-01-02T02:49:40.413" Score="0" ViewCount="" Body="&lt;p&gt;I LOVED Advanced Calculus: A Differential Forms Approach by Edwards.  Great book..  It's great to move to the next level in geometry. &lt;/p&gt;&#xA;" OwnerUserId="5293" LastActivityDate="2011-01-02T02:49:40.413" />
  <row Id="16078" PostTypeId="1" CreationDate="2011-01-02T02:49:48.440" Score="1" ViewCount="97" Body="&lt;p&gt;&lt;a href=&quot;http://mathworld.wolfram.com/RenyisParkingConstants.html&quot; rel=&quot;nofollow&quot;&gt;Rényi's Parking Constants&lt;/a&gt; comes up when one puts down unit length cars on a interval, such that the probability of covering any two interval is the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there any published results when the distribution is non-uniform? &lt;/p&gt;&#xA;" OwnerUserId="96" LastActivityDate="2011-02-15T13:10:20.113" Title="Random parking problem on a probability distribution" Tags="&lt;probability&gt;&lt;reference-request&gt;" AnswerCount="0" CommentCount="3" />
  <row Id="16079" PostTypeId="2" ParentId="16059" CreationDate="2011-01-02T02:51:27.347" Score="1" ViewCount="" Body="&lt;p&gt;The key is to use the Taylor series approximation $$\cos x \approx 1 - \frac{x^2}{2},$$ which is good for small $x$. That applies to both $x$ and $x/(1+x)$, and so $$\frac{1-\cos x}{1-\cos \frac{x}{1+x}} \approx (1+x)^2.$$ Substituting this into your expression, we get that the expression is $$\approx \frac{(1+x)^{2z}-1}{x} \approx \frac{(1+2zx)-1}{x} = 2z.$$ This reasoning can be made formal using big-O (or little-O) notation, but I leave that for you to ponder.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2011-01-02T02:51:27.347" />
  <row Id="16080" PostTypeId="1" AcceptedAnswerId="16109" CreationDate="2011-01-02T02:56:32.243" Score="3" ViewCount="194" Body="&lt;p&gt;I'm having trouble solving problem 12 from Section 1.2 in Hatcher's &quot;Algebraic Topology&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's the relevant image for the problem:&#xA;&lt;a href=&quot;http://imgur.com/rxkiT.png&quot; rel=&quot;nofollow&quot;&gt;http://imgur.com/rxkiT.png&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm trying to find $\pi_1(R^3-Z)$, where $Z$ is the graph shown in the first figure. The answer (according to the problem statement) is supposed to be $&amp;lt;a,b,c| aba^{-1}b^{-1}cb^\epsilon c^{-1}=1&amp;gt;$, where $\epsilon=\pm 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I attempted to use Van Kampen's theorem, using a cover of two open sets, depicted in the lower image. The first open set is the area above the bottom horizontal line, minus the graph, and the second open set is the region below the top horizontal line, minus the graph. The intersection is the area in between the two horizontal lines minus the graph. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Call the top set $A$, and the bottom set $B$. With applications of Van Kampen, I got that $\pi_1(A)\cong \pi_1(B)\cong Z*Z*Z$, and that $\pi_1(A \cap B)\cong Z*Z*Z*Z*Z$. After using Van Kampen's theorem again, I got that $\pi_1(R^3-Z)\cong Z*Z$, which is wrong.&lt;/p&gt;&#xA;" OwnerUserId="5290" LastActivityDate="2011-01-03T22:08:34.993" Title="Finding the fundamental group of the complement of a certain graph" Tags="&lt;algebraic-topology&gt;&lt;homotopy-theory&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="16081" PostTypeId="1" CreationDate="2011-01-02T03:21:14.817" Score="0" ViewCount="77" Body="&lt;p&gt;I have a presentation in which I want to point out a difference between two functions.&#xA;Instead of putting the two functions on a slide and pointing to the differences, I want to do something simple.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first function has a form of $f_1(x) = g(x)\log (h(x))$ and the second function has the form $f_2(x) = g(x)\log(r(x)h(x))$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The thing is that saying something along the lines &quot;$f_2(x) - f_1(x)$ have the form of $g(x)\log r(x)$ keeps the $g(x)$ (which is rather complicated) and saying something like $f_2(x)/f_1(x)$ has the form $\log h(x) / \log (r(x)h(x))$ is also cumbersome.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anyone see a clear way to point out the difference between these two functions by only making explicit the additional $r(x)$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would really like to make it simple, ideas are welcome.&lt;/p&gt;&#xA;" OwnerUserId="7076" LastActivityDate="2011-01-02T05:53:37.190" Title="how can I present an idea about the difference between two functions clearly?" Tags="&lt;presentation&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="16082" PostTypeId="2" ParentId="16033" CreationDate="2011-01-02T03:30:20.287" Score="1" ViewCount="" Body="&lt;p&gt;Remember that whatever method you use for solving $y^{\prime}=f(t,y)$, be it Runge-Kutta, Bulirsch-Stoer (extrapolative), Gear/Adams multistep, or fancier methods, one always has a triple of values $(t_i,y_i,y_i^{\prime})$ ($y_i^{\prime}=f(t_i,y_i)$) at each evaluation point. Thus, one can always do cubic Hermite interpolation across the points $(t_i,y_i)$ and $(t_{i+1},y_{i+1})$ (note that I am not assuming that the evaluation points are equispaced, as is often the case when doing adaptive solving). If the underlying method is at most third order accurate, cubic Hermite is a good choice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the upshot is that modern implementations of DE solvers always support so-called &quot;dense output&quot;; briefly, associated with a method with $p$th order &quot;accuracy&quot; (in quotes since &quot;high order doesn't always imply high accuracy&quot; ;) ) has with it an associated $p$th order interpolating function. To use my favorite example, the $(4,5)$ adaptive Runge-Kutta solver based on the Dormand-Prince coefficients has with it an associated fifth order interpolating function. The special properties inherent in the coefficients allow the existence of an associated interpolating function with the same order of &quot;accuracy&quot;; in general, not all Runge-Kutta coefficients will have an associated &quot;nice&quot; interpolating function (but again, one can always do cubic Hermite).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I could say more, but the &lt;a href=&quot;http://www.unige.ch/~hairer/books.html&quot; rel=&quot;nofollow&quot;&gt;books by Hairer/Nørsett/Wanner&lt;/a&gt; have an extensive discussion on dense output (and they say it better than what I can hope to say), as well as usable routines (also available within the site). You would do well to study them.&lt;/p&gt;&#xA;" OwnerUserId="5296" LastEditorUserId="5296" LastEditorDisplayName="" LastEditDate="2011-01-04T00:51:11.037" LastActivityDate="2011-01-04T00:51:11.037" />
  <row Id="16083" PostTypeId="2" ParentId="16081" CreationDate="2011-01-02T03:41:36.037" Score="2" ViewCount="" Body="&lt;p&gt;The short answer to your question is no.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To see this, define $a(x) = g(x) \log(h(x))$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we have $a(x) = f_1(x)$ and $a(x) + \log(r(x)) g(x) = f_2(x)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So here we have two linearly independent equations in $a(x)$ and $g(x)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence at best we can solve for $a(x)$ and $g(x)$ (i.e. we can solve for $h(x)$ and $g(x)$) in terms of $f_1(x)$ and $f_2(x)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, either we need another equation, so that we can plug in what we obtain to get rid of $g(x)$ and $h(x)$ or We need $g(x)$ or $h(x)$ to be some contant so that the dependence of a relation between $f_1(x),f_2(x)$ and $r(x)$ on $g(x)$ or $h(x)$ is hidden.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could however define new functions $\tilde{f_1}$ and $\tilde{f_2}$ depending on $f_1(x)$,$f_2(x)$,$g(x)$ and $h(x)$ and massage to get a relation between $\tilde{f_1}$ and $\tilde{f_2}$.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2011-01-02T03:47:15.760" LastActivityDate="2011-01-02T03:47:15.760" />
  <row Id="16084" PostTypeId="2" ParentId="16066" CreationDate="2011-01-02T03:44:01.783" Score="10" ViewCount="" Body="&lt;p&gt;Here's a list of irrational numbers which almost fulfill your criteria.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each were chosen to be accurate to within &amp;plusmn;0.1, so that no two of them implicitly express the same mathematical approximation, and so that none of them &quot;cheats&quot; in order to fudge an exact result involving integers to obtain a slightly inexact, irrational result. Only the last number fails to meet your criteria, as it is slightly larger than 12.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$\ln(3)$&lt;/li&gt;&#xA;&lt;li&gt;$7\pi/11$&lt;/li&gt;&#xA;&lt;li&gt;$\sqrt 2 + \frac\pi2$&lt;/li&gt;&#xA;&lt;li&gt;$7/{\sqrt[3]5}$&lt;/li&gt;&#xA;&lt;li&gt;$\mathrm e^\phi$&lt;/li&gt;&#xA;&lt;li&gt;$\sec^2(20)$&lt;/li&gt;&#xA;&lt;li&gt;$4\sqrt3$&lt;/li&gt;&#xA;&lt;li&gt;$5\phi$&lt;/li&gt;&#xA;&lt;li&gt;$2\pi+\mathrm e$&lt;/li&gt;&#xA;&lt;li&gt;$\sinh(3)$&lt;/li&gt;&#xA;&lt;li&gt;$\pi^3-20$&lt;/li&gt;&#xA;&lt;li&gt;$\csc^2(16)$&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I would prefer not to use cosecant, integers greater than 12, or more than two additions/subtractions &amp;mdash; it is too easy to get results if you rely on these &amp;mdash; but I think I've spent enough time on this diversion for now. :-)&lt;br&gt;&lt;br&gt;[&lt;strong&gt;EDIT:&lt;/strong&gt; revised the formula for 4 two times now: this first to change the formula for 4 from &amp;radic;3&amp;thinsp;+&amp;thinsp;&amp;radic;5 &amp;mdash; which is too close to &amp;radic;4&amp;thinsp;+&amp;thinsp;&amp;radic;4 &amp;mdash; and the second time to correct the formula as I somehow copied a result which was not approximately 4.]&lt;br&gt;&lt;br&gt;&#xA;&lt;strong&gt;Added:&lt;/strong&gt; Since it seems that I can't sleep tonight, here is a list of approximate values:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\begin{align*}&#xD;&#xA;\ln(3)               &amp;amp; \approx 1.0986 \\&#xD;&#xA;7\pi/11              &amp;amp; \approx 1.9992 \\&#xD;&#xA;\sqrt 2 + \tfrac\pi2 &amp;amp; \approx 2.9850 \\&#xD;&#xA;7 / \sqrt[3]5        &amp;amp; \approx 4.0936 \\&#xD;&#xA;\mathrm e^\phi       &amp;amp; \approx 5.0432 \\&#xD;&#xA;\sec^2(20)           &amp;amp; \approx 6.0049 \\&#xD;&#xA;4\sqrt3              &amp;amp; \approx 6.9282 \\&#xD;&#xA;5\phi                &amp;amp; \approx 8.0902 \\&#xD;&#xA;2\pi+\mathrm e       &amp;amp; \approx 9.0015 \\&#xD;&#xA;\sinh(3)             &amp;amp; \approx 10.018 \\&#xD;&#xA;\pi^3-20             &amp;amp; \approx 11.006 \\&#xD;&#xA;\csc^2(16)           &amp;amp; \approx 12.064&#xD;&#xA;\end{align*}$&lt;/p&gt;&#xA;" OwnerUserId="439" LastEditorUserId="439" LastEditorDisplayName="" LastEditDate="2011-01-02T04:49:36.200" LastActivityDate="2011-01-02T04:49:36.200" CommunityOwnedDate="2011-01-02T15:01:17.520" CommentCount="7" />
  <row Id="16085" PostTypeId="1" AcceptedAnswerId="16090" CreationDate="2011-01-02T04:05:49.763" Score="9" ViewCount="104" Body="&lt;p&gt;Is $\ p_n^{\pi(n)} &amp;lt; 4^n$ where $p_n$ is the largest prime $\leq n$?&#xA;Where $\pi(n)$ is the prime counting function. Using PMT it seems asymptotically $\ p_n^{\pi(n)} \leq x^n$ where $e \leq x$&lt;/p&gt;&#xA;" OwnerUserId="3936" LastActivityDate="2011-01-02T04:35:06.103" Title="Is $\ p_n^{\pi(n)} &lt; 4^n$ where $p_n$ is the largest prime $\leq n$?" Tags="&lt;number-theory&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" />
  <row Id="16086" PostTypeId="1" CreationDate="2011-01-02T04:11:47.577" Score="0" ViewCount="147" Body="&lt;p&gt;I have no idea, how to solve this problem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $\psi: (\mathbb{R} \to \mathbb{N}) \to P(P(\mathbb{R}))$ defined as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\psi (f) = \mathbb{R}/_{ker(f)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) Is this a surjective function? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2)Is this an injective function?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried to do &lt;em&gt;something&lt;/em&gt;, but after hours I can't see any progress... I just don't know, were/how to start solving such problem. And I'm not sure, if I properly understand problem itself; I'm sure, that your explanation woluld be helpful.&lt;/p&gt;&#xA;" OwnerUserId="5298" LastEditorUserId="5298" LastEditorDisplayName="" LastEditDate="2011-01-02T08:38:48.880" LastActivityDate="2011-01-02T08:38:48.880" Title="Function problem (kernel, surjective &amp; injective function)" Tags="&lt;homework&gt;&lt;functions&gt;" CommentCount="6" />
  <row Id="16087" PostTypeId="2" ParentId="16085" CreationDate="2011-01-02T04:13:36.773" Score="4" ViewCount="" Body="&lt;p&gt;Yes, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Asymptotically you have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(p_n)^{\pi(n)} \leq n^{n/\log n} = e^n$$&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2011-01-02T04:13:36.773" CommentCount="2" />
  <row Id="16088" PostTypeId="1" AcceptedAnswerId="16094" CreationDate="2011-01-02T04:22:26.333" Score="4" ViewCount="125" Body="&lt;p&gt;I do not understand the topology of a Lie group clearly. Let $G$ be a Lie group and $T_eG$ be its tangent space at the identity $e \in G$. Why $Aut(T_eG)$ is an open subset of the vector space of endomorphisms of $T_eG$ (i.e. $End(T_eG)$)? What does &quot;open&quot; mean?&lt;/p&gt;&#xA;" LastActivityDate="2011-01-02T05:02:09.923" Title="what is the usual topology on a vector space?" Tags="&lt;lie-groups&gt;&lt;representation-theory&gt;&lt;lie-algebra&gt;" AnswerCount="1" />
  <row Id="16089" PostTypeId="2" ParentId="16066" CreationDate="2011-01-02T04:26:37.080" Score="5" ViewCount="" Body="&lt;p&gt;This is intended to be a community wiki answer that people can edit if they just want to give a few numbers rather than all twelve.  I'll start with suggestions currently in the comments.  Feel free to add more!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For 1: $ -\sin 11$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For 3: $e + \sqrt{\frac{5}{63}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For 7: $22/\pi$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For 12: $\log_{1922}(1782^{12}+1841^{12})$  (Approximately 11.99999999996, &lt;a href=&quot;http://en.wikipedia.org/wiki/Fermat%27s_Last_Theorem_in_fiction&quot; rel=&quot;nofollow&quot;&gt;featured on The Simpsons&lt;/a&gt;.)&lt;/p&gt;&#xA;" OwnerUserId="2370" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2011-01-02T04:43:43.477" LastActivityDate="2011-01-02T04:43:43.477" CommunityOwnedDate="2011-01-02T04:26:37.080" CommentCount="4" />
  <row Id="16090" PostTypeId="2" ParentId="16085" CreationDate="2011-01-02T04:35:06.103" Score="7" ViewCount="" Body="&lt;p&gt;Using $$\pi(x) \le 1.25066 \frac{x}{\log x}$$ for all $x&amp;gt;1$ (from Rosser and Schoenfeld), you have&#xA;$$(p_n)^{\pi(n)} \le e^{1.25066 n}  &amp;lt; 3.5^n$$ for all $n\ge 2$.&lt;/p&gt;&#xA;" OwnerUserId="2937" LastActivityDate="2011-01-02T04:35:06.103" CommentCount="3" />
  <row Id="16091" PostTypeId="2" ParentId="16066" CreationDate="2011-01-02T04:37:42.080" Score="2" ViewCount="" Body="&lt;p&gt;Have a look at [Wikipedia][1] for some interesting ideas:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[1]: &lt;a href=&quot;http://en.wikipedia.org/wiki/Mathematical_coincidence&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Mathematical_coincidence&lt;/a&gt; Wikipedia&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I like $\pi^2 \approx 9.87$. Fooling around with variations of what's there also yielded&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ 7( \pi - e) \approx 2.96 $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ 19 (\pi - e) \approx 8.05 $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore the numbers don't have to be limited to (0, 12], since you could always work in mod 12. For example $ (\pi + 20)^i \approx -1 $ could be used for 11 o'clock.&lt;/p&gt;&#xA;" OwnerUserId="5300" LastActivityDate="2011-01-02T04:37:42.080" CommunityOwnedDate="2011-01-02T15:01:17.520" />
  <row Id="16092" PostTypeId="1" CreationDate="2011-01-02T04:39:44.607" Score="4" ViewCount="119" Body="&lt;p&gt;Let su(2) be the Lie algebra of SU(2), thought of as a representation of SU(2) by conjugation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let S be the rank-2 symmetric tensors over the complex numbers, acted on by SU(2) in the obvious way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have seen it asserted that the complexification of su(2) is isomorphic (as a representation) to S, and presumably there is some simple explicit isomorphism that should be obvious.  Alas, to me it is not obvious.  What is that isomorphism?&lt;/p&gt;&#xA;" OwnerUserId="5302" LastEditorUserId="5302" LastEditorDisplayName="" LastEditDate="2011-01-02T05:12:09.707" LastActivityDate="2011-01-03T00:41:08.360" Title="What is the Isomorphism between the adjoint representation of SU(2) and its representation on rank 2 symmetric tensors?" Tags="&lt;representation-theory&gt;" AnswerCount="1" />
  <row Id="16093" PostTypeId="1" CreationDate="2011-01-02T04:58:42.670" Score="0" ViewCount="44" Body="&lt;p&gt;Hello&#xA;I am trying to model the relationship between two variables, say x and y.  I have a number of subjects - for each subjectm I have a number of x and corresponding y, both of which are always positive.  This data tends to be very sparse.  There are some problem specific constraints:&#xA;1) y(0) = 0 (or very close to it)&#xA;2) y is increasing as a function of x&#xA;3) y' is decreasing as a function of x&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is rather nebulous, but I have a feeling that the most important difference between subjects is in the height of the curve, not in the slope.  Because of the sparsity, I think I can get away with forcing each subject to have the same &quot;slope&quot; (perhaps at a specified x), but allowing the height to vary.  I have been playing around with various sorts of logistic functions, but the asymptote isn't really justifiable.  I have also been looking at things like a*log(x+b), but this doesn't really conform to the intuition delineated above.  Does anyone have any suggestions?&lt;/p&gt;&#xA;" OwnerUserId="5303" LastActivityDate="2011-02-01T05:17:36.600" Title="Finding a good parametric form for a model" Tags="&lt;mathematical-modeling&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="16094" PostTypeId="2" ParentId="16088" CreationDate="2011-01-02T05:02:09.923" Score="4" ViewCount="" Body="&lt;p&gt;First, this has nothing to do with Lie groups or Lie algebras.  The only important part is that $T_eG$ is a vector space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For any vector space $V$ (say, finite dimensional over the reals), the set $End(V)$ is naturally a finite dimensional vector space.  Hence, $End(V)$ is isomorphic to $\mathbb{R}^N$ for some $N$ (in fact, $N = $(dim$V)^2$).  Use any choice of isomorphism to topologize $End(V)$.  This choice of isomorphism is equivalent to choosing a basis of $V$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, one has the determinant $det:End(V)\rightarrow\mathbb{R}$ which is given as a polynomials in the entries of the matrices in $End(V)$ (they are matrices after choosing a basis), and hence is continuous.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $det$ is continuous, $det^{-1}(\mathbb{R}-\{0\})$ is an open subset of $End(V)$.  But this subset is precisely $Aut(V)$, the invertible transformations from $V$ to $V$.&lt;/p&gt;&#xA;" OwnerUserId="331" LastActivityDate="2011-01-02T05:02:09.923" CommentCount="5" />
  <row Id="16095" PostTypeId="2" ParentId="16093" CreationDate="2011-01-02T05:12:31.347" Score="1" ViewCount="" Body="&lt;p&gt;(This is supposed to be a comment.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would say that without knowing the physical process(es) that generated the $y$'s for each corresponding $x$'s, any number of functions would be admissible. Barring that, one usually graphs the data first before even thinking about models...&lt;/p&gt;&#xA;" OwnerUserId="5296" LastActivityDate="2011-01-02T05:12:31.347" CommentCount="2" />
  <row Id="16096" PostTypeId="2" ParentId="16081" CreationDate="2011-01-02T05:53:37.190" Score="1" ViewCount="" Body="&lt;p&gt;instead of $f_1,f_2$ would you rather consider $e^{f_2-f_1}={r(x)^{g(x)}}$? &lt;/p&gt;&#xA;" OwnerUserId="3936" LastActivityDate="2011-01-02T05:53:37.190" />
  <row Id="16097" PostTypeId="2" ParentId="16080" CreationDate="2011-01-02T06:26:04.463" Score="1" ViewCount="" Body="&lt;p&gt;This is admittedly not an answer, but it seems to me like you have already done a lot of the work. So here are my suggestions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Name the generators of each copy of the integers, think about $\pi_1(A \cup B)$ in terms of generators and relations by using van Kampens' theorem as you are doing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Remember what van Kampen says, you are taking the free product of the two over the intersection, so you need to name the generators of $\pi_1(A \cap B)$ in terms of how they include into $\pi_1(A)$ and $\pi_1(B)$ respectively. Think about how you take tensor products.&lt;/p&gt;&#xA;" OwnerUserId="627" LastActivityDate="2011-01-02T06:26:04.463" />
  <row Id="16098" PostTypeId="1" CreationDate="2011-01-02T06:52:08.530" Score="3" ViewCount="201" Body="&lt;p&gt;Consider the following function: $ f(x) = \sum_{i,j&amp;gt;=0} a_{ij} x^i (1-x)^j$ , where the coefficients satisfy $a_{ij} \in [0,1]$. Observe that $f(x)$ converges for $x\in [0,1]$. Is $f$ analytic over $[0,1]$? If not, how about over $[c,d]$ for $c,d \in (0,1)$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think I can show that the answer is yes for the $[c,d]$ case by showing uniform convergence (EDIT: I was wrong... Not sure I can anymore). Is there a simpler way? More importantly, are there nice general conditions that guarantee that the infinite sum of polynomials is analytic over the interval (or a proper sub-interval) of convergence? This motivates the following questions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Question 1: Let $f(x) = \sum_{i,j} a_{ij} x^i (1-x)^j$, where there are no guarantees on $a_{ij}$. Assume $f$ converges on interval $[a,b]$.   Is $f(x)$ analytic over $[a,b]$? What about over $[c,d]$ where $c,d \in (a,b)$? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Question 2: Let $f(x) = \sum_k p_k(x)$, where $p_k$ is a polynomial. Assume $f$ converges on interval $[a,b]$, and that $p_k(x) \geq 0$ for $x \in [a,b]$.   Is $f(x)$ analytic over $[a,b]$? What about over $[c,d]$ where $c,d \in (a,b)$? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Question 3: Same as question 2, but there are no guarantees on the sign of $p_k(x)$ anymore. (EDIT: As Jonas Meyer pointed out, this is false by the Weierstrass approximation theorem)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I guess what I'm trying to get at is understanding which natural conditions, short of manually showing uniform convergence, guarantee that a convergent sum of polynomials is analytic over its region of convergence (or a sub-interval of that). Question 1 starts with a very specific sum of polynomials that I somewhat understand, and Questions 2 and 3 are grasping at a generalization. These questions are my best guesses as to what generalizations may look like, though please feel free to suggest others if I'm off-base here.&lt;/p&gt;&#xA;" OwnerUserId="5159" LastEditorUserId="5159" LastEditorDisplayName="" LastEditDate="2011-01-02T08:16:38.207" LastActivityDate="2011-03-19T07:57:50.493" Title="When is an infinite sum of polynomials analytic?" Tags="&lt;real-analysis&gt;&lt;analysis&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="16099" PostTypeId="2" ParentId="16066" CreationDate="2011-01-02T07:00:49.147" Score="2" ViewCount="" Body="&lt;p&gt;After seeing the &lt;a href=&quot;http://www.thinkgeek.com/homeoffice/lights/ac3f/?cpg=ab&quot; rel=&quot;nofollow&quot;&gt;ThinkGeek :: Pop Quiz Math Clock&lt;/a&gt; a few years ago I asked my friend David a question related to the one here: to use more interesting mathematical approaches to generate each of the integers 1-12.  He provided a great answer at&#xA;&lt;a href=&quot;http://www.astronexus.com/comment/reply/153&quot; rel=&quot;nofollow&quot;&gt;http://www.astronexus.com/comment/reply/153&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5307" LastActivityDate="2011-01-02T07:00:49.147" CommunityOwnedDate="2011-01-02T15:01:17.520" />
  <row Id="16100" PostTypeId="2" ParentId="16098" CreationDate="2011-01-02T07:15:48.307" Score="5" ViewCount="" Body="&lt;p&gt;This answer doesn't answer all of your specific questions.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A uniform limit of polynomial functions on $[0,1]$ need not be analytic.  Every continuous function on $[0,1]$ is a uniform limit of polynomial functions, by the &lt;a href=&quot;http://en.wikipedia.org/wiki/Stone%E2%80%93Weierstrass_theorem#Weierstrass_approximation_theorem&quot; rel=&quot;nofollow&quot;&gt;Weierstrass approximation theorem&lt;/a&gt;.  Of course, if you only have pointwise convergence, things could be worse.  This means that the answer to $2$ and $3$ is no. [Edit after question edit: In the edited version, $2$ is a different question, but the answer is still no, and it can still be answered using Weierstrass's theorem.]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Things work out better if you work over the complex numbers instead of the real numbers.  If a sequence of analytic functions defined on an open subset of $\mathbb{C}$ converges uniformly on compact subsets, then the &lt;a href=&quot;http://en.wikipedia.org/wiki/Analytic_function#Properties_of_analytic_functions&quot; rel=&quot;nofollow&quot;&gt;limit function is analytic&lt;/a&gt;.  There are useful criteria for determining whether a sequence converges uniformly on compact sets, such as &lt;a href=&quot;http://en.wikipedia.org/wiki/Montel%27s_theorem&quot; rel=&quot;nofollow&quot;&gt;Montel's theorem&lt;/a&gt;.  However, there are examples of non-analytic pointwise limits of complex analytic functions.  See for example Davidson's &quot;&lt;a href=&quot;http://www.jstor.org/pss/2975578&quot; rel=&quot;nofollow&quot;&gt;Pointwise limits of analytic functions&lt;/a&gt;.&quot; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are comfortable going to the complex case but start out with functions defined on an interval $[c,d]$ in $\mathbb{R}$, you might want to try extending all of your functions to an open subset of $\mathbb{C}$ that contains $[c,d]$.  If the sequence converges uniformly on some such open set, then the limit function will be complex analytic, and therefore its restriction to $[c,d]$ will be real analytic.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2011-01-02T08:04:25.830" LastActivityDate="2011-01-02T08:04:25.830" CommentCount="3" />
  <row Id="16101" PostTypeId="2" ParentId="16071" CreationDate="2011-01-02T07:23:36.983" Score="1" ViewCount="" Body="&lt;p&gt;If you don't mind relatively high level reading (though it starts from the basics, you just need to be able to even read mathematical texts):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://books.google.com/books?id=wlqSgRJ42-EC&quot; rel=&quot;nofollow&quot;&gt;Introduction to Cardinal Arithmetic&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The preface and first chapter is about introduction to cardinal arithmetic, you probably don't need it all, but there's a good introductory part there.&lt;/p&gt;&#xA;" OwnerUserId="622" LastActivityDate="2011-01-02T07:23:36.983" />
  <row Id="16102" PostTypeId="2" ParentId="16073" CreationDate="2011-01-02T08:13:08.537" Score="3" ViewCount="" Body="&lt;p&gt;First of all, note that a sample point $\omega$ (unlike $X(\omega)$) need not necessarily be a number (or vector). But that's not the point here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As in your example, let the $X_i$ be i.i.d. exponential rv's. For our purposes, let's consider only the finite case of $n$ rv's, $X_1,\ldots,X_n$. Then, to see why a sample point $\omega$ should correspond to a vector $\omega = (\omega_1,\ldots,\omega_n) \in [0,\infty)^n$ (rather than to a number in $[0,\infty)$), consider the following observation:&#xA;$$&#xD;&#xA;{\rm P}\bigg(\sum\limits_{i = 1}^n {X_i }  \in (a,b]\bigg) = {\rm P}\bigg[(X_1 , \ldots ,X_n ) \in \bigg \{ (x_1 , \ldots ,x_n ):\sum\limits_{i = 1}^n {x_i }  \in (a,b]\bigg \}\bigg]. &#xD;&#xA;$$&#xA;Hence, we can actually consider vectors $(X_1 (\omega),\ldots, X_n (\omega))$, and, in turn, vectors $\omega = (\omega_1,\ldots,\omega_n) \in [0,\infty)^n$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Since, in our example, the $X_i$ are i.i.d. exponentials, then given any $n$-dimensional rectangle $R_n=[a_1,b_1] \times \cdots \times [a_n,b_n] \subset [0,\infty)^n$, we have&#xA;$$&#xD;&#xA;{\rm P}[(X_1 , \ldots ,X_n ) \in R_n] = \prod\limits_{i = 1}^n {{\rm P}(X_i  \in [a_i ,b_i ])} = \prod\limits_{i = 1}^n {\int_{a_i }^{b_i } {f(x_i )\,{\rm d}x_i } } ,&#xD;&#xA;$$&#xA;where $f(x_i)=\lambda e^{-\lambda x_i}$, $x_i \geq 0$, is the exponential$(\lambda)$ density function.&#xA;The basic idea here is to consider &quot;product measure space&quot;. Another key concept here is &quot;convolution&quot;.&lt;/p&gt;&#xA;" OwnerUserId="2810" LastEditorUserId="2810" LastEditorDisplayName="" LastEditDate="2011-01-02T15:42:46.457" LastActivityDate="2011-01-02T15:42:46.457" />
  <row Id="16103" PostTypeId="1" AcceptedAnswerId="16106" CreationDate="2011-01-02T08:36:01.333" Score="5" ViewCount="242" Body="&lt;p&gt;Firstly, let's establish what exactly I mean by these symbols. Let $\omega_0 = \{ 0, 1, 2, \ldots \}$, where $0, 1, 2, \ldots$ are the usual von Neumann representations of the natural numbers. Let $n$ be a finite natural number. For each $n$, define $\omega_{n+1} = \sup S$, where $S$ the image of the set $R$ under the function-class mapping well-orders to their von Neumann ordinal, and $R \subset \mathcal{P}(\omega_n \times \omega_n)$ is the set of all well-orders on $\omega_n$. We define $\aleph_n = \omega_n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unless I'm mistaken, this establishes the existence of $\omega_n$ for all $n \in \mathbb{N}$ as sets under the axioms of ZF. It's straightforward to see that there is no injection $\omega_{n+1} \to \omega_n$, as that would establish (via pullback) that $|\omega_{n+1}| \le \aleph_n$, and this is a contradiction as $\omega_{n+1}$ is strictly greater than all ordinals in $S$. This in turn implies, by the axiom of choice, that there is no surjection $\omega_n \to \omega_{n+1}$, and the conclusion that there is no surjection $\aleph_n \to \aleph_{n+1}$ follows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is: Can this be done, using my definitions above, without the axiom of choice? I'm willing to accept reasonable alternative definitions, provided that they don't render the conclusion tautological.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(This is a self-imposed extension to a homework problem: Should I tag with homework?)&lt;/p&gt;&#xA;" OwnerUserId="5191" LastActivityDate="2011-01-02T08:54:18.547" Title="Non-existence of a surjection $\aleph_n \to \aleph_{n+1}$, without the axiom of choice" Tags="&lt;set-theory&gt;&lt;cardinals&gt;&lt;axiom-of-choice&gt;&lt;ordinals&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="16104" PostTypeId="2" ParentId="16066" CreationDate="2011-01-02T08:44:09.193" Score="3" ViewCount="" Body="&lt;p&gt;Despite the large constant, I like tanh(2011) for 1.  It is very close and shows the year.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-01-02T08:44:09.193" CommunityOwnedDate="2011-01-02T15:01:17.520" CommentCount="1" />
  <row Id="16105" PostTypeId="2" ParentId="16103" CreationDate="2011-01-02T08:44:47.820" Score="0" ViewCount="" Body="&lt;p&gt;How are you surpassing AOC in assuming the existence of well orderings?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not quite clear what your definition of \aleph_n is. Is it the nth cardinal number (is that what a Von Neumann ordinal is?)?&lt;/p&gt;&#xA;" OwnerUserId="5310" LastActivityDate="2011-01-02T08:44:47.820" CommentCount="1" />
  <row Id="16106" PostTypeId="2" ParentId="16103" CreationDate="2011-01-02T08:54:18.547" Score="2" ViewCount="" Body="&lt;p&gt;You've made a very big mess, I think. The way you defined cardinal numbers is very awkward, so to say. In my eyes, anyway.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let us review the construction of ordinals:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$0 = \emptyset$&lt;/li&gt;&#xA;&lt;li&gt;$\alpha+1 = \alpha \cup \{\alpha\}$&lt;/li&gt;&#xA;&lt;li&gt;At limit stages, $\delta = \bigcup_{\beta&amp;lt;\delta} \beta$&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Now we define initial ordinals as ordinal numbers which cannot be bijected with any smaller ordinal. For example $\omega$ (the set of natural numbers) is such ordinal, while $\omega+1, \omega+\omega, \omega\cdot\omega$ are not initial ordinals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Under the axiom of choice, every set is well-orderable, and therefore we can choose the initial ordinal out of each equivalence class as a representative. This is the usual notion of $\aleph$ numbers under the axiom of choice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Without assuming choice, the cardinal system is not well-ordered and can behave very strangely.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regardless to that, when you are only dealing with ordinals you don't need choice because there is a canonical choice function (take the minimal element). So even without the axiom of choice it is true that $\omega_\alpha$ has no bijection with $\omega_\beta$ for $\alpha\not=\beta$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The idea behind aleph numbers, as far as I see it, is that it is a well ordering of cardinalities (not necessarily all cardinalities, though) and as such it holds just fine even when not assuming choice. However, in the case you don't have the axiom of choice to help you out, $2^{\aleph_0}$ might not be well-orderable and thus won't be represented by an ordinal, and therefore won't be represented by an $\aleph$ number, same with multiplication. It is equivalent to the axiom of choice that for every set $|X| = |X\times X|$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just last remark, you said that the construction you gave infers the existence of $\aleph_n$ for every natural number $n$ while in fact it gives you $\aleph_\alpha$ for every ordinal $\alpha$ and not just for the natural numbers.&lt;/p&gt;&#xA;" OwnerUserId="622" LastActivityDate="2011-01-02T08:54:18.547" CommentCount="3" />
  <row Id="16107" PostTypeId="2" ParentId="15033" CreationDate="2011-01-02T09:20:04.803" Score="0" ViewCount="" Body="&lt;p&gt;If you are trying to put this into code you may want to consider the function atan2(y,x), it is available in most languages. Note that it takes the y coordinate as the first parameter. Since atan2 takes two parameter it can figure out in which quadrant your vector lies, and it can output the angle in the range $-\pi$ to $\pi$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a perl snippet to demonstrate how it works for your two angles:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;#!/usr/bin/perl&#xA;&#xA;use Math::Trig;&#xA;&#xA;$t_rad = atan2(1/sqrt(2),-1/sqrt(2));&#xA;$t_deg = $t_rad * 180 / pi;&#xA;print &quot;theta_2 = $t_deg\n&quot;;&#xA;&#xA;$t_rad = atan2(-1/sqrt(2),1/sqrt(2));&#xA;$t_deg = $t_rad * 180 / pi;&#xA;print &quot;theta_3 = $t_deg\n&quot;;&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;And here is the output when you run it for the two examples above:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;theta_2 = 135&#xA;theta_3 = -45&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="5292" LastActivityDate="2011-01-02T09:20:04.803" />
  <row Id="16108" PostTypeId="2" ParentId="12397" CreationDate="2011-01-02T09:44:32.633" Score="3" ViewCount="" Body="&lt;p&gt;In section 2.11 on p. 27 of the following there is a description of how to create a bezier curve that exactly represents an arc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.tsplines.com/resources/class_notes/Bezier_curves.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.tsplines.com/resources/class_notes/Bezier_curves.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5292" LastActivityDate="2011-01-02T09:44:32.633" />
  <row Id="16109" PostTypeId="2" ParentId="16080" CreationDate="2011-01-02T10:20:31.847" Score="3" ViewCount="" Body="&lt;p&gt;I remember doing this problem and having similar difficulties, until I realized that it was actually way easier to find $\pi_1(Y)$ than it was to find $\pi_1(\mathbb{R}^3-Z)$.  That these are the same is more or less a proof by picture (remember that fundamental group is an invariant of homotopy type).  I suggest drawing a &quot;fundamental domain&quot; for $Y$ which is modeled on the usual one for $X$, but of course you'll need to modify it to take account of the removed disk.  I can say more if you're still not getting it...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;More:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a (sort of) evident deformation retraction of $\mathbb{R}^3-Z$ onto $Y$, which implies that the two spaces are homotopy equivalent, which implies that they have the same fundamental group.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To find $\pi_1(Y)$, start by drawing the usual square fundamental domain for a Klein bottle:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/crQrV.gif&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To make things easy for ourselves, let's assume (without loss of generality) that the b-circles are attached at the circle which in the projection to $\mathbb{R}^3$ bounds the disk that we'll cut out.  When we do so, we get this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/DD5kY.gif&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It'll be easiest to think of this as a single disk whose boundary is identified to itself.  That way we can consider this as a graph (in fact, a bouquet of circles) with a single disk attached, and it's easy to find the fundamental group of such a thing.  To do this, we add in an edge:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/Nq3ft.gif&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So now we're looking at this space as a bouquet of circles $a,b,c$, and we've sewn on a disk according to the relation $aba^{-1}b^{-1}cb^{-1}c^{-1}$.  If you do this slightly differently you can switch to $\epsilon=1$.&lt;/p&gt;&#xA;" OwnerUserId="2159" LastEditorUserId="2159" LastEditorDisplayName="" LastEditDate="2011-01-03T22:08:34.993" LastActivityDate="2011-01-03T22:08:34.993" CommentCount="3" />
  <row Id="16110" PostTypeId="1" AcceptedAnswerId="16112" CreationDate="2011-01-02T10:50:30.087" Score="1" ViewCount="149" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How to embed a matrix, for example, a $3x3$ singular matrix in to $R^9$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How to compute the induced metric? Is it just the Frobenius norm of the matrix?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many Thanks.&#xA;sam&lt;/p&gt;&#xA;" OwnerUserId="5312" LastEditorUserId="627" LastEditorDisplayName="" LastEditDate="2011-01-03T03:27:42.257" LastActivityDate="2011-01-03T03:27:42.257" Title="Embed an $n\times n$ matrix into $R^{n^2}$" Tags="&lt;linear-algebra&gt;&lt;differential-geometry&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="16111" PostTypeId="1" AcceptedAnswerId="16130" CreationDate="2011-01-02T10:57:45.133" Score="1" ViewCount="130" Body="&lt;p&gt;I am referring to the this &lt;a href=&quot;http://en.wikipedia.org/wiki/Polyhedron#Volume&quot; rel=&quot;nofollow&quot;&gt;wiki page&lt;/a&gt;, which states that the way to find the volume of a Polyhedron can be done via Divergence Theorem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int \nabla \cdot \vec{F} dV=\oint \vec{F} \cdot \hat{n} dS$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By choosing&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \vec{F}=\frac{\vec{x}}{3}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One can then obtain the volume formula for a polyhedron with $k$ faces:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$V = \frac{\sum_{i=1}^k \vec{x_{i}} \cdot \hat{n_i} A_i}{3}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\vec{x_{i}}$,$\hat{n_i}$ and $A_i$ are a point, normal vector and the area of the face $i$. (&lt;del&gt;The wiki page doesn't say $x_i$ is the centroid; it just says that it can be any point on the face, but I believe that it must be centroid, or else the volume obtained not correct&lt;/del&gt;, The $x_i$ can be any point; see comments)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is, is my formula and interpretation correct? And does it hold for all kinds of polyhedron, be it convex or concave? &lt;/p&gt;&#xA;" OwnerUserId="550" LastEditorUserId="550" LastEditorDisplayName="" LastEditDate="2011-01-02T12:44:25.937" LastActivityDate="2011-01-02T16:16:31.550" Title="Is This Polyhedron Formula Correct?" Tags="&lt;geometry&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="16112" PostTypeId="2" ParentId="16110" CreationDate="2011-01-02T11:05:30.983" Score="2" ViewCount="" Body="&lt;p&gt;What restrictions are there on the embedding?  If you just mean a bijection between $3 \times 3$ matrices and points in $\mathbb{R^9}$, there are many.  As there are $9$ entries in a $3 \times 3$ matrix, there are easy bijections taking each entry in the matrix to one component.  The bijection allows an easy metric, too-just use the usual metric in $\mathbb{R^9}$  It is true that the singular matrices have a constraint that reduces them to an eight-dimensional space.  This is the source of the first question-this embedding ignores the impact of singularity.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-01-02T11:05:30.983" CommentCount="5" />
  <row Id="16113" PostTypeId="1" AcceptedAnswerId="16114" CreationDate="2011-01-02T11:19:20.303" Score="0" ViewCount="114" Body="&lt;p&gt;If I have $22x = 23y = 24z$ and $x+y+z = 3865$ how to obtain the values of $x$,$y$ and $z$ ?&lt;/p&gt;&#xA;" OwnerUserId="2109" LastActivityDate="2011-01-05T22:28:05.927" Title="Solving $x$, $y$ and $z$" Tags="&lt;algebra-precalculus&gt;&lt;arithmetic&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="16114" PostTypeId="2" ParentId="16113" CreationDate="2011-01-02T11:24:59.030" Score="3" ViewCount="" Body="&lt;p&gt;use $ y = \frac{22}{23}x$ and $z = \frac{22}{24}x$ in last equation.&lt;/p&gt;&#xA;" OwnerUserId="5312" LastEditorUserId="5312" LastEditorDisplayName="" LastEditDate="2011-01-05T22:28:05.927" LastActivityDate="2011-01-05T22:28:05.927" CommentCount="4" />
  <row Id="16115" PostTypeId="2" ParentId="16113" CreationDate="2011-01-02T11:26:52.633" Score="5" ViewCount="" Body="&lt;p&gt;First, try to express $y$ and $z$ by $x:$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$22x = 23y\Leftrightarrow y = \frac{22}{23}x$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$22x = 24z\Leftrightarrow z = \frac{11}{12}x$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then put this into your term to archieve a value for $x:$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x+y+z=3865\Leftrightarrow x+\frac{22}{23}x+\frac{11}{12}x=3865$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now you have the value of $x.$ To figure out $y$ and $z$ is left as an exercise to you.&lt;/p&gt;&#xA;" OwnerUserId="5282" LastActivityDate="2011-01-02T11:26:52.633" />
  <row Id="16116" PostTypeId="1" CreationDate="2011-01-02T11:32:19.143" Score="1" ViewCount="47" Body="&lt;p&gt;It seems like I am &lt;a href=&quot;http://math.stackexchange.com/questions/16061/on-profit-and-loss&quot;&gt;having troubles&lt;/a&gt; while dealing with dishonest merchants! Here is another problem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A cloth merchant says that due to slump in the market, he sells the cloth at $10\%$ loss but he uses a false meter scale and actually gains $15\%$.Find the actual length of the scale.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any hints, solution, explanation are welcome.&lt;/p&gt;&#xA;" OwnerUserId="2109" LastEditorUserId="2109" LastEditorDisplayName="" LastEditDate="2011-01-02T11:42:34.693" LastActivityDate="2011-01-02T11:44:36.127" Title="On profit loss (II)" Tags="&lt;algebra-precalculus&gt;&lt;arithmetic&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="16117" PostTypeId="1" AcceptedAnswerId="16120" CreationDate="2011-01-02T11:35:53.900" Score="1" ViewCount="102" Body="&lt;p&gt;&lt;strong&gt;$1$.The compound interest on certain sum for $2$ years is $\$410$ and simple interest is $\$400$. Find the rate of interest per-year.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;$2$.A sum of money invested at compound interest amounts in 3 years to $\$ 2400$ an in $4$ years to $ \$ 2520$. Find the rate of interest per-year.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solving them using the traditional compound interest formulas is a bit tedious for me under exam conditions,so please suggest if there is any kind of short-cut procedure to solve them.&lt;/p&gt;&#xA;" OwnerUserId="2109" LastEditorUserId="2109" LastEditorDisplayName="" LastEditDate="2011-01-02T12:01:28.770" LastActivityDate="2011-01-02T12:58:33.990" Title="Suggest quicker method for finding rate percent per annum" Tags="&lt;algebra-precalculus&gt;&lt;arithmetic&gt;" AnswerCount="1" CommentCount="7" FavoriteCount="1" />
  <row Id="16118" PostTypeId="2" ParentId="16116" CreationDate="2011-01-02T11:44:09.227" Score="0" ViewCount="" Body="&lt;p&gt;I think the question itself is a bit unclear, I interpret it like this:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;He buys clothes for $n\frac\$m$ (units are set by me), but he sells for $0.9n\frac\$m$&lt;/li&gt;&#xA;&lt;li&gt;But because his scale is too short, the actually price paid by his customers is $1.15n\frac\$m$. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Thus, if the scale's length is just $xm$ instead of one meter, the equation is like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$0.9n\frac\$m = 1.15n\frac\${xm}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Try to figure out the solution by your self and see whether it holds.&lt;/p&gt;&#xA;" OwnerUserId="5282" LastActivityDate="2011-01-02T11:44:09.227" />
  <row Id="16119" PostTypeId="2" ParentId="16116" CreationDate="2011-01-02T11:44:36.127" Score="3" ViewCount="" Body="&lt;p&gt;So he sells the cloth at 0.9 of &quot;value&quot;.  What is the 15% taken on?  If 15% of the selling price is profit, then he bought at 0.85*0.9.  If 15% of his purchase price is his profit, let the buy price be b. 0.9=1.15b, so b=0.9/1.15  These are very close when the profit is small compared to 1, but try it at 40% and see what happens.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-01-02T11:44:36.127" CommentCount="2" />
  <row Id="16120" PostTypeId="2" ParentId="16117" CreationDate="2011-01-02T11:55:03.517" Score="2" ViewCount="" Body="&lt;p&gt;For 1, the difference between compound interest and simple interest is that you pay interest on the interest.  So if you are compounding annually, you have $\$10$ interest on $\$400$ in one year.  So the interest rate is what?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For 2, again if you are compounding annually, you have $\$120$ interest on $\$2400$.  So the interest rate is what?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are not compounding annually, which magically simplifies things by eliminating the compounding for a one year term, I don't see any easy way.  If you have a spreadsheet that has a goal seek, you can use that.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2011-01-02T12:58:33.990" LastActivityDate="2011-01-02T12:58:33.990" CommentCount="7" />
  <row Id="16121" PostTypeId="1" CreationDate="2011-01-02T13:40:55.983" Score="8" ViewCount="363" Body="&lt;p&gt;This may be a simple question: Find all subrings of $\mathbb{Z}^2$.&lt;/p&gt;&#xA;" OwnerUserId="5315" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-03T01:40:21.903" LastActivityDate="2011-05-26T20:37:18.360" Title="Find all subrings Of $\mathbb{Z}^2$" Tags="&lt;abstract-algebra&gt;&lt;algebra&gt;&lt;ring-theory&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="3" />
  <row Id="16122" PostTypeId="2" ParentId="16071" CreationDate="2011-01-02T13:43:01.643" Score="0" ViewCount="" Body="&lt;p&gt;Goldrei's Classic Set Theory is extremely well written and easy to read, and covers cardinalities&lt;/p&gt;&#xA;" OwnerUserId="5316" LastActivityDate="2011-01-02T13:43:01.643" />
  <row Id="16123" PostTypeId="1" CreationDate="2011-01-02T14:39:58.747" Score="3" ViewCount="594" Body="&lt;p&gt;I was &quot;playing&quot; on &lt;a href=&quot;http://projecteuler.net/&quot; rel=&quot;nofollow&quot;&gt;Project Euler&lt;/a&gt; and passed across the Collatz Problem, and it mentioned it was still unproven.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I immediately noticed two things:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;After the $3n+1$ the result is always even, leading to $\frac{n}{2}$&lt;/li&gt;&#xA;&lt;li&gt;$\frac{n}{2}$ is an trivial action, especially in a binary representation.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The first optimization that can be done is combining the $3n+1$ step with $\frac{n}{2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{3n+1}{2} = 1.5n+0.5 = n+((0.5n)-0.5)+1$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Looking at the binary representation, $0.5n$ is simply $n$ bitshifted 1 position to the right.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We now focus on $n$ being an odd positive integer (as even numbers are always $2^x$ of an odd number).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obtaining $((0.5n)-0.5)$ is easily done with the binary representation of $n$, by bitshifting $n$ to the right, and dropping the least significant bit. (Remember, $n$ is odd.)&lt;/p&gt;&#xA;&#xA;&lt;hr /&gt;&#xA;&#xA;&lt;p&gt;We know full adders have a well known and well defined behaviour:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$S = A \oplus B \oplus C_{i}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$C_{o} = majority(A, B, C_{i})$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Doing $n+((0.5n)-0.5)+1$ is implemented using a full adder where the input $B_{i}$ has been connected to $A_{i+1}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/lxdCz.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;Sorry for the confusion, but the $n$ on the in-/outputs is unrelated to the number $n$; it needs to be $i$. I will fix this in the image.&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr /&gt;&#xA;&#xA;&lt;p&gt;By definition, an odd binary number has always $1$ as the least significant bit, and obviously, a $1$ as the most significant bit. The $A_{i+1}$ input from the adder handling the most significant bit falls outside the $n$, and as a result, $0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$C_{0} = 1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$A_{0} = 1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$A_{i_{max}} = 1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$A_{i_{max+1}} = 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After a complete run has been done, $\frac{n}{2}$ gets done if applicable, and the new $n$ gets fed to the calculator again.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that each time the length of the number grows on the most significant bit side, a &quot;foreign&quot; $0$ gets output to $S_{i_{max}}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$S_{i_{max}} = (1 \oplus 0 \oplus C_{i_{max-1}}) \neq C_{i_{max-1}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$C_{i_{max}} = majority(1, 0, C_{i_{max-1}}) = C_{i_{max-1}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$S_{i_{max}} \neq C_{i_{max}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;(This also implies that $n$ never reaches a value in the form of ${2^x}-1$ after the initial value until it reaches $1$ because of the $0$ passed to the $B$ input for the most significant bit. As the only way not to output a $0$ for $S_{_{[msb]}}$ is to not receive a carry trough $C_{i_{[msb]}}$, which means there is at least an earlier bit with $0$ as value.)&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr /&gt;&#xA;&#xA;&lt;p&gt;But more importantly, when the first $0$ appears in $A_{1}$ will result in an $0$ to $S_{0}$ making the number even, and shrinking the number itself (by applying $\frac{n}{2}$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$S_{0} = (1 \oplus 0 \oplus 1) = 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$C_{0} = majority(1, 0, 1) = 1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if it keeps finding an alternating bits of $1$'s and $0$'s, it keeps outputting $0$'s, and keeps the carry on. (This allows you to do $\frac{n}{2}$ several times in a row.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Only if both $A_{i}$ and $A_{i+1}$ are $1$ again, it will start outputting $1$ on $S$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If however, it encounters a $O$'s on both $A_{i}$ and $A_{i+1}$, the carry $C_{i}$ will will be output to $S_{i}$, and outputs $A_{i}$ to $S_{i}$ until it encounters a $1$'s on both $A_{i}$ and $A_{i+1}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$S = (0 \oplus 0 \oplus C_{i-1}) = C_{i-1}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$C_{i} = majority(0, 0, C_{i-1}) = 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$S = (1 \oplus 1 \oplus C_{i-1}) = C_{i-1}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$C_{i} = majority(1, 1, C_{i-1}) = 1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;hr /&gt;&#xA;&lt;strong&gt;How does this translate to a proof of Collatz conjecture?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;Note that I'm a mere programmer with a hobby interest in math problems, so excuse me for the low quality. However, I still think this is able to prove that Collatz conjecture is valid, as the growth of the number has hard limits (how many carries can reach the end), while the shrink has no hard limit (shrinks until a 1 occurs) and the switching is fair and one-way (full adder).&lt;/sub&gt;&#xA;&lt;hr /&gt;&#xA;&lt;sub&gt;Note: my personal truth table for the MSB and LSB&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&#xA;MSB Ci      MSB&#xA;010-0--- -&gt; 011---&#xA;010-1--- -&gt; 100---&#xA;011-0--- -&gt; 100---&#xA;011-1--- -&gt; 101---&#xA;&#xA;   B LSB      Co LSB&#xA;---0-01  -&gt; ---0-10 (additional $\frac{n}{2}$ once)&#xA;---0-11  -&gt; ---1-01&#xA;---1-01  -&gt; ---1-00 (additional $\frac{n}{2}$ multiple times)&#xA;---1-11  -&gt; ---1-11&#xA;&lt;/pre&gt;&#xA;" OwnerUserId="4360" LastEditorUserId="457" LastEditorDisplayName="" LastEditDate="2011-02-25T12:14:22.600" LastActivityDate="2011-02-25T12:14:22.600" Title="Proof of Collatz conjecture?" Tags="&lt;open-problem&gt;" AnswerCount="1" CommentCount="8" FavoriteCount="1" />
  <row Id="16124" PostTypeId="1" AcceptedAnswerId="16145" CreationDate="2011-01-02T14:50:57.923" Score="2" ViewCount="142" Body="&lt;p&gt;Let $G$ and $H$ be two Lie groups and $\rho: G \to H$ be a homomorphism. How to differentiate $\rho$ to obtain a Lie algebra homomorphism $d\rho_e: T_eG \to T_eH$?&lt;/p&gt;&#xA;" LastActivityDate="2011-02-20T01:11:27.897" Title="How to differentiate a homomorphism between two Lie groups." Tags="&lt;representation-theory&gt;&lt;lie-groups&gt;&lt;lie-algebra&gt;" AnswerCount="3" />
  <row Id="16125" PostTypeId="2" ParentId="16124" CreationDate="2011-01-02T14:59:13.717" Score="4" ViewCount="" Body="&lt;p&gt;Given any two manifolds and a map between them, there is automatically induced a map on the tangent spaces at corresponding points. The only new thing here is to check that the induced map is actually a homomorphism of Lie algebras. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $v_1, v_2 \in T_e(G)$. We can extend them to left-invariant vector fields $V_1, V_2$ on $G$. In fact, we identify $T_e(G)$ with the space of left-invariant vector fields on $G$. The bracket is just given by the usual Lie bracket $[V_1, V_2]$. (One has to check that the Lie bracket preserves left-invariance; this is a general fact about it. In detail, if $X, Y$ are manifolds and $f: X \to Y$ a map, then vector fields $V$ on $X$ and $W$ on $Y$ are said to be $f$-related if $f_*(X) = Y$ wherever both sides are defined. It is a basic fact that if $X_1, Y_1$ and $X_2, Y_2$ are $f$-related, then $[X_1, X_2]$ and $[Y_1, Y_2]$ are $f$-related.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose $v_1, v_2$ get sent to $w_1, w_2 \in T_e(H)$. Extend $w_1, w_2$ to invariant vector fields $W_1, W_2$ on $H$. Then $V_1, V_2$ are $f$-related to $W_1, W_2$ by left-invariance. It follows that $[V_1, V_2]$ is $f$-related to $[W_1, W_2]$. This implies that the values at $e$ match up. This is precisely what you are asking for.&lt;/p&gt;&#xA;" OwnerUserId="536" LastActivityDate="2011-01-02T14:59:13.717" CommentCount="4" />
  <row Id="16126" PostTypeId="2" ParentId="16123" CreationDate="2011-01-02T14:59:28.710" Score="7" ViewCount="" Body="&lt;p&gt;The problem with the Collatz conjecture has to do with the possibility of having long &quot;alternating&quot; chains.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider a sequence of numbers &lt;em&gt;a&lt;/em&gt;&lt;sub&gt;0&lt;/sub&gt;, &lt;em&gt;a&lt;/em&gt;&lt;sub&gt;1&lt;/sub&gt;, &lt;em&gt;a&lt;/em&gt;&lt;sub&gt;2&lt;/sub&gt;, ... obtained by applying the Collatz function, starting with some integer &lt;em&gt;a&lt;/em&gt;&lt;sub&gt;0&lt;/sub&gt;. Consider just some element of this sequence, &lt;em&gt;a&lt;/em&gt;&lt;sub&gt;n&lt;/sub&gt;: as you note, if &lt;em&gt;a&lt;sub&gt;n&lt;/sub&gt;&lt;/em&gt; is odd, then &lt;em&gt;a&lt;/em&gt;&lt;sub&gt;&lt;em&gt;n&lt;/em&gt;+1&lt;/sub&gt;&amp;nbsp;=&amp;nbsp;3&lt;em&gt;a&lt;/em&gt;&lt;sub&gt;&lt;em&gt;n&lt;/em&gt;&lt;/sub&gt;+1 is even, so that &lt;em&gt;a&lt;/em&gt;&lt;sub&gt;&lt;em&gt;n&lt;/em&gt;+2&lt;/sub&gt;&amp;nbsp;=&amp;nbsp;(3&lt;em&gt;a&lt;/em&gt;&lt;sub&gt;&lt;em&gt;n&lt;/em&gt;&lt;/sub&gt;+1)/2. But then what if &lt;em&gt;a&lt;/em&gt;&lt;sub&gt;&lt;em&gt;n&lt;/em&gt;+2&lt;/sub&gt; is again odd? Well, we'd then have &lt;em&gt;a&lt;/em&gt;&lt;sub&gt;&lt;em&gt;n&lt;/em&gt;+4&lt;/sub&gt; = (3(3&lt;em&gt;a&lt;/em&gt;&lt;sub&gt;&lt;em&gt;n&lt;/em&gt;&lt;/sub&gt;+1)/2+1)/2 = (9&lt;em&gt;a&lt;/em&gt;&lt;sub&gt;&lt;em&gt;n&lt;/em&gt;&lt;/sub&gt;+3)/4. And if this number is again odd...?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In an &quot;alternating&quot; chain of elements like this, each two steps yields an increase in size by a fraction of more than 3/2, which of course is an exponential growth rate. One could say that the problem of the Collatz conjecture is to bound how long these alternating chains can be (and how close together they can be).&lt;/p&gt;&#xA;" OwnerUserId="439" LastActivityDate="2011-01-02T14:59:28.710" CommentCount="5" />
  <row Id="16127" PostTypeId="2" ParentId="16071" CreationDate="2011-01-02T15:27:37.800" Score="1" ViewCount="" Body="&lt;p&gt;I like Devlin's &quot;Joy of Sets&quot;.&lt;/p&gt;&#xA;" OwnerUserId="851" LastActivityDate="2011-01-02T15:27:37.800" />
  <row Id="16128" PostTypeId="2" ParentId="16121" CreationDate="2011-01-02T15:57:17.343" Score="13" ViewCount="" Body="&lt;p&gt;The strictest interpretation of the question is the one in which rings have identities and subrings have to share identities.  Then any subring of $\mathbb{Z}^2$ contains the identity $(1, 1)$, hence the subring generated by $(1, 1)$, which is the diagonal copy $(k, k)$ of $\mathbb{Z}$.  If $R$ is a subring, then to describe any extra elements of $R$ it suffices to describe the extra elements of the form $(0, k)$ (since the existence of any particular extra element $(a, b)$ can be deduced from the existence of the extra element $(a, b) - (a, a) = (0, b-a)$.)  If there are any extra elements of the form $(0, k)$, they must form an ideal of the second copy of $\mathbb{Z}$ (since they must be closed under addition and under multiplication by the diagonal copy of $\mathbb{Z}$) hence the only choice is that the extra elements are of the form $(0, nk)$ for some fixed positive integer $n$ and all $k \in \mathbb{Z}$.  These elements generate the ring $R_n$ of all pairs $(a, b)$ such that $a \equiv b \bmod n$, and these are all possible subrings (if one includes $R_{\infty}$, which corresponds to the diagonal case).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are two other interpretations: one in which rings have to have identities but subrings don't have to share them, and one in which rings don't have to have identities.  I'll leave these as an exercise if you're interested; you should think about the subrings of $\mathbb{Z}$ first as a warm-up.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2011-01-02T15:57:17.343" />
  <row Id="16129" PostTypeId="1" AcceptedAnswerId="16132" CreationDate="2011-01-02T16:12:26.630" Score="2" ViewCount="155" Body="&lt;p&gt;Definition of lattice require that any two elements of lattice should have LUB and GLB, while complete lattice extends it to, every subset should have LUB and GLB. But by induction , it is possible to show that if any two elements have LUB and GLB then every subset should have LUB and GLB. I read somewhere that the difference is because of infinite set, in that case it is possible that set along with some partial order is lattice but not complete lattice, can someone please elaborate it with one example?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;regards&lt;/p&gt;&#xA;" OwnerUserId="5318" LastActivityDate="2011-01-30T08:04:24.300" Title="Difference between lattice and complete lattice" Tags="&lt;lattices&gt;" AnswerCount="2" />
  <row Id="16130" PostTypeId="2" ParentId="16111" CreationDate="2011-01-02T16:16:31.550" Score="3" ViewCount="" Body="&lt;p&gt;Yes. The quoted formula holds for any polyhedral surface $S$, or even for a collection of polyhedral surfaces $S_i$, that (together) bound(s) a &quot;volume&quot; $V\subset {\mathbb R}^3$. It is true that a polyhedron has edges and vertices. Therefore, depending on the proof that was originally given for the divergence theorem, one might need an approximation argument to apply the theorem to polyhedra.&lt;/p&gt;&#xA;" OwnerUserId="1303" LastActivityDate="2011-01-02T16:16:31.550" CommentCount="1" />
  <row Id="16131" PostTypeId="2" ParentId="16129" CreationDate="2011-01-02T16:18:10.000" Score="3" ViewCount="" Body="&lt;p&gt;Normal induction doesn't &quot;reach&quot; infinity. So proving something for finite sets doesn't prove it for infinite sets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Explicitly: the integers $\mathbb{Z}$ form a lattice, but it is incomplete, since there is no greatest (or least) integer.&lt;/p&gt;&#xA;" OwnerUserId="5191" LastActivityDate="2011-01-02T16:18:10.000" CommentCount="4" />
  <row Id="16132" PostTypeId="2" ParentId="16129" CreationDate="2011-01-02T16:29:06.560" Score="6" ViewCount="" Body="&lt;p&gt;Regular induction (&quot;holds for $1$&quot; and &quot;if it holds for $k$ then it holds for $k+1$&quot;) only gives you that the result holds for every natural number $n$; it does not let you go beyond the finite numbers. For example, you can prove by induction that there are natural numbers that require $n$ digits to write down in base $10$ for every $n$, but this does not mean that there are natural numbers that require an infinite number of digits to write down in base $10$. &quot;For all $n$&quot; is not the same as &quot;for all sizes, finite or infinite&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(There &lt;em&gt;is&lt;/em&gt; a kind of induction that would allow you to prove something for &lt;em&gt;all&lt;/em&gt; sizes, not just finite. This is called &lt;a href=&quot;http://en.wikipedia.org/wiki/Transfinite_induction&quot; rel=&quot;nofollow&quot;&gt;transfinite induction&lt;/a&gt;. You prove the result holds for $1$, and that whenever it holds for all $m\lt k$, then it also holds for $k$ (or, you prove it holds for $1$, that if it holds for an ordinal/cardinal  $\alpha$ then it holds for $\alpha+1$, and that if it holds for all ordinals/cardinals strictly smaller than $\gamma$, then it holds for $\gamma$). However you would be unable to do such a proof with lattices, because it is false).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, if you have a lattice, then any &lt;em&gt;nonempty finite&lt;/em&gt; subset has a least upper bound and a greatest lower bound, by induction. Even if you have a $0$ and a $1$ (a minimum and a maximum element) so that every set has an upper and a lower bound, you still don't get that every set has a &lt;em&gt;least&lt;/em&gt; upper bound. For example, take $P = \mathbb{Q}\cup\{-\infty,\infty\}$, with the usual order among rationals, $-\infty\leq q\leq \infty$ for all $q\in\mathbb{Q}$. This is a lattice, with operations $a\wedge b = \min\{a,b\}$ and $a\vee b = \max\{a,b\}$ (since it is a totally ordered set). Every &lt;em&gt;finite&lt;/em&gt; subset has a least upper bound (the maximum) and a greatest lower bound (the minimum). But it is precisely the absence of suprema and infima for general sets that stops it from being a complete lattice: the set $\{q\in\mathbb{Q}\mid q^2\lt 2\}$ has no least upper bound and no greatest lower bound.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-02T20:11:16.670" LastActivityDate="2011-01-02T20:11:16.670" />
  <row Id="16133" PostTypeId="1" CreationDate="2011-01-02T16:31:47.627" Score="3" ViewCount="276" Body="&lt;p&gt;What is cardinality of $P(\mathbb{R})$? And $P(P(\mathbb{R}))$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P is a &lt;a href=&quot;http://en.wikipedia.org/wiki/Power_set&quot; rel=&quot;nofollow&quot;&gt;Power Set&lt;/a&gt;, $\mathbb{R}$ is set of real numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general - how can find cardinality of different sets? Is/are there a good method(s)? And is there some kind of handout, that provides list of cardinality of popular sets ($\mathbb{R}$, $\mathbb{Q}$, $P(\mathbb{Q})$) etc.?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And, how to find cardinality of sets of functions? For example, what is the cardinality of set that contains all functions? What about set of functions $f: \mathbb{N} \to \mathbb{R}$?&lt;/p&gt;&#xA;" OwnerUserId="5298" LastEditorUserId="5298" LastEditorDisplayName="" LastEditDate="2011-01-02T22:19:03.787" LastActivityDate="2011-02-08T23:49:22.450" Title="Cardinality of $P(\mathbb{R})$ and $P(P(\mathbb{R}))$" Tags="&lt;elementary-set-theory&gt;&lt;infinity&gt;" AnswerCount="5" CommentCount="2" />
  <row Id="16134" PostTypeId="2" ParentId="16133" CreationDate="2011-01-02T16:50:28.993" Score="4" ViewCount="" Body="&lt;p&gt;Cardinal arithmetic and diagonalization.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The cardinality of the power set is $2^{|X|}$, which is why we write the power set as $2^X$, which also represents the set of characteristic functions of subsets of $X$ (i.e. the set of functions $X\to\{0,1\})$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is very cool when you first realize it:  A function of the form above is determined by its elements.  Given such a function $f: X\to\{0,1\}$, let $S_f:=f^{-1}(1)$ be the subset defined by $f$, and given a subset $S\subseteq X$, let $f_S$ denote the characteristic function of $S$ with respect to $X$.  Then we see immediately that these functions $S_{(-)}:\{0,1\}^X\to \mathcal{P}(X)$ and $f_{(-)}: \mathcal{P}(X)\to \{0,1\}^X$ are inverse to one another.  Therefore, it follows that the power set is precisely a function set as noted above.  Edit: I should note that in abstract nonsense language, this means that that $2=\{0,1\}$ is the &quot;subobject classifier&quot; for the category of sets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To see that $|\mathbf{R}^\mathbf{N}|=|\mathbf{R}|$, notice that this becomes $(2^\mathbf{N})^\mathbf{N}\cong 2^{\mathbf{N}\times \mathbf{N}}\cong 2^\mathbf{N}\cong \mathbf{R}$, where $\cong$ here represents an isomorphism of sets, or a bijective function (baby Rudin calls this equivalence relation &quot;equipotence&quot;). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To find that $|\mathbf{R}|=2^{|\mathbf{N}|}$, this is a matter of using the technique called diagonalization.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To find that the integers and rationals represent the same cardinal as $\mathbf{N}$, you have to do a little bit of tricky encoding, but it is easy to find on the page about cardinality on wikipedia.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To see a worked out proof of why $|2^X|&amp;gt;|X|$ for any set $X$, look at this whimsical &lt;a href=&quot;http://en.wikipedia.org/wiki/Cantor%27s_theorem&quot; rel=&quot;nofollow&quot;&gt;link&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="5319" LastEditorUserId="5319" LastEditorDisplayName="" LastEditDate="2011-01-02T18:13:41.870" LastActivityDate="2011-01-02T18:13:41.870" CommentCount="6" />
  <row Id="16135" PostTypeId="2" ParentId="16133" CreationDate="2011-01-02T16:50:36.917" Score="2" ViewCount="" Body="&lt;p&gt;As for the set of functions $f:\mathbb{N} \rightarrow \mathbb{R}$, consider the functions $f:\mathbb{N} \rightarrow \{0, 1, 2, 3, 4, 5, 6, 7, 8, 9 \}$ (i.e. the functions that map each natural number to a single base-10 digit) and compare that to the closed interval $[0, 1]$.  What conclusions can you then draw about the set of all functions $f:\mathbb{N} \rightarrow A$ where $A$ is a countably infinite set?  And so what conclusions can you draw about the set of all functions $f:\mathbb{N} \rightarrow \mathbb{R}$?&lt;/p&gt;&#xA;" OwnerUserId="506" LastEditorUserId="506" LastEditorDisplayName="" LastEditDate="2011-01-02T17:08:31.380" LastActivityDate="2011-01-02T17:08:31.380" CommentCount="2" />
  <row Id="16136" PostTypeId="2" ParentId="16133" CreationDate="2011-01-02T17:10:36.657" Score="3" ViewCount="" Body="&lt;p&gt;Let $X$ be any set. Then, the cardinality of $\mathcal{P}(X)$ is $\left| 2^X \right| = 2^{|X|}$, since there is a bijection $\mathcal{P}(X) \to 2^X$, where $2^X$ is the set of all functions $X \to \{0, 1\}$. I'm afraid this is best answer that I can give, since, for example, it is not possible to prove that $\aleph_1 = | \mathcal{P}(\mathbb{N}) |$, or that $\aleph_1 \ne | \mathcal{P}(\mathbb{N}) |$, from the usual axioms of mathematics (ZFC).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In terms of aleph numbers, the cardinalities of $\mathbb{R}$, $\mathbb{Q}$, and $\mathcal{P}(\mathbb{Q})$ are $2^{\aleph_0}$, $\aleph_0$, and $2^{\aleph_0}$, respectively. $\aleph_0 = | \mathbb{N} |$ by definition. Here's an interesting fact: the set of all functions $\mathbb{R} \to \mathbb{R}$ has cardinality strictly greater than $| \mathbb{R} |$, but the subset of all &lt;em&gt;continuous&lt;/em&gt; functions has cardinality equal to $| \mathbb{R} |$. (Can you prove this?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, as for the &quot;set&quot; of all functions — this isn't a well-defined set under ZFC. So you can't talk about the cardinality. However, if you fix the domain and codomain, then we do get a set, and we write $Y^X$ for the set of all functions $X \to Y$. The cardinality, as the notation suggests, is $|Y|^{|X|}$. (Actually, this is a small lie. We define the notation $|Y|^{|X|}$ to mean $\left| Y^X \right|$. But it happens to agree with the arithmetic definition of exponentiation in the case where $X$ and $Y$ are finite sets.)&lt;/p&gt;&#xA;" OwnerUserId="5191" LastActivityDate="2011-01-02T17:10:36.657" CommentCount="1" />
  <row Id="16137" PostTypeId="2" ParentId="16133" CreationDate="2011-01-02T17:12:46.163" Score="4" ViewCount="" Body="&lt;p&gt;If you are willing to assume the &lt;a href=&quot;http://en.wikipedia.org/wiki/Generalized_Continuum_Hypothesis#The_generalized_continuum_hypothesis&quot; rel=&quot;nofollow&quot;&gt;Generalized Continuum Hypothesis&lt;/a&gt; (GCH), then the cardinalities in question are simply $\aleph_2$ and $\aleph_3$, i.e., the third and fourth smallest infinite cardinals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I don't know anyone seriously interested in set theory who wants to assume GCH: it's basically assuming away all of the subtleties of cardinal exponentiation.  But without this assumption, I don't think the OP's question has a good answer.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I agree that a &quot;cheatsheet&quot; of cardinalities of commonly encountered infinite sets would be a nice thing, and on that sheet one of the lines would read $\mathbb{2}^{\mathbb{R}}$: i.e., this is the simplest description of a certain cardinality of sets, and one could then rather list other sets which happen to be equinumerous with it: the number of topologies on a countably infinite set,$\ldots$&lt;/p&gt;&#xA;" OwnerUserId="299" LastActivityDate="2011-01-02T17:12:46.163" CommentCount="1" />
  <row Id="16138" PostTypeId="1" CreationDate="2011-01-02T17:19:59.860" Score="3" ViewCount="433" Body="&lt;p&gt;The standard proof of the Prime Number Theorem requires taking into consideration that there are no zeroes of the Riemann Zeta function in which the real part equals one. But consider the following argument:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The probability that a number less than X is prime $\pi(X)/X$ is approximately $\Pi_{p&amp;lt;X} (1-1/p)$ (this is for the same reason that the Sieve of Eratosthenes works), which is approximately $1/\Sigma_{n=1}^X 1/n$, which is approximately $1/\log X$. Hence, $\pi(X)$ is approximately $X/\log X$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This doesn't use complex numbers, but it gives a good reason to believe the Prime Number Theorem. Why do complex numbers (which seem to come from nowhere) make this argument rigorous?&lt;/p&gt;&#xA;" OwnerUserId="3163" LastActivityDate="2011-02-01T12:55:59.083" Title="Why are complex numbers necessary to prove the Prime Number Theorem?" Tags="&lt;analytic-number-theory&gt;" AnswerCount="4" CommentCount="1" />
  <row Id="16139" PostTypeId="1" AcceptedAnswerId="16140" CreationDate="2011-01-02T17:21:15.597" Score="0" ViewCount="83" Body="&lt;p&gt;It's been a &lt;em&gt;really&lt;/em&gt; long time since I've done calculus or any other kind of math beyond tip calculation.  I was given a spreadsheet that calculates and plots a growth curve over time based on a handful of inputs.  It goes out for one year.  I'd like to be able to make educated (though not necessarily precise) predictions of values along the curve for any amount of time, but if I can avoid it, I'd rather not just extend the length of the spreadsheet.  The series is nonlinear, the first values look like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;540 545 561 599 672 780 904 1,016 1,098 1,151 1,193 1,238 1,297 1,367 1,444&lt;/p&gt;&#xA;&#xA;&lt;p&gt;THat's the first ten days of the year, given a 'seed' value of 500 (which of course is one of the variables).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Like I said, I don't need it to be super accurate, just generally describe the slope over time, I just don't know where to even begin figuring out the equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks,&#xA;Paul&lt;/p&gt;&#xA;" OwnerUserId="5321" LastActivityDate="2011-01-02T17:48:24.060" Title="How do you derive a function that describes a series?" Tags="&lt;sequences-and-series&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="16140" PostTypeId="2" ParentId="16139" CreationDate="2011-01-02T17:34:39.860" Score="3" ViewCount="" Body="&lt;p&gt;You are describing a &quot;regression&quot;.  Depending on how the data looks plotted on a graph, you should choose a type of regression that looks like it will fit the data (quadratic, cubic, linear, exponential, logarithmic, logistic, etc.).  However, I would suggest you go to the stackexchange for statistics, since this doesn't really qualify as mathematics.  Thanks!  (Please note that I mean no disrespect.  I am merely pointing out that statisticians will be able to tell you how to avoid common mistakes, like trying to get &quot;too good a fit&quot;. There are infinitely many functions that go through a finite number of points, and there is not a unique function that gives a &quot;best fit&quot; without restricting the class of functions (to some of the classes I mentioned above)).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Point of order (Poo): You do not need to do this by hand.  Most spreadsheet software supports these types of computations (excel, for instance)!&lt;/p&gt;&#xA;" OwnerUserId="5319" LastEditorUserId="5319" LastEditorDisplayName="" LastEditDate="2011-01-02T17:48:24.060" LastActivityDate="2011-01-02T17:48:24.060" CommentCount="4" />
  <row Id="16141" PostTypeId="2" ParentId="16138" CreationDate="2011-01-02T17:38:37.487" Score="13" ViewCount="" Body="&lt;p&gt;They're not!  This is the theorem of Selberg and Erdős!  Look up the &quot;elementary proof of the Prime Number Theorem&quot;. &lt;a href=&quot;http://en.wikipedia.org/wiki/Prime_number_theorem#Elementary_proofs&quot; rel=&quot;nofollow&quot;&gt;This link is clicky and wonderful.&lt;/a&gt; (It is also majestic, so please do click it).&lt;/p&gt;&#xA;" OwnerUserId="5319" LastEditorUserId="5319" LastEditorDisplayName="" LastEditDate="2011-01-02T17:44:46.137" LastActivityDate="2011-01-02T17:44:46.137" />
  <row Id="16142" PostTypeId="2" ParentId="16139" CreationDate="2011-01-02T17:43:04.107" Score="0" ViewCount="" Body="&lt;p&gt;I'm not clear on the boundary between mathematics and statistics in this area.  Any good book on numerical methods will have routines for extrapolation.  My favorite is chapter 3 of &lt;a href=&quot;http://www.nr.com/&quot; rel=&quot;nofollow&quot;&gt;Numerical Recipes&lt;/a&gt;.  Excel has routines for doing this, but a text like NR will have useful cautionary notes.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-01-02T17:43:04.107" />
  <row Id="16143" PostTypeId="2" ParentId="16138" CreationDate="2011-01-02T17:52:56.950" Score="6" ViewCount="" Body="&lt;p&gt;If you try to make the argument you describe rigorous in a naive way, then the error terms rapidly become too big for it to actually work, and this is why nobody uses the &lt;a href=&quot;http://en.wikipedia.org/wiki/Legendre_sieve&quot; rel=&quot;nofollow&quot;&gt;Legendre sieve&lt;/a&gt; to count primes.  This is a well-known problem in sieve theory; see, for example, &lt;a href=&quot;http://terrytao.wordpress.com/2007/06/05/open-question-the-parity-problem-in-sieve-theory/&quot; rel=&quot;nofollow&quot;&gt;this blog post&lt;/a&gt; by Terence Tao.  There are elementary proofs, like the one found by Selberg and Erdős, but they are &lt;em&gt;much&lt;/em&gt; harder than the complex analytic proof; don't be fooled into thinking that &quot;elementary&quot; means &quot;easier.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as the &quot;come from nowhereness&quot; of it all, perhaps you would like to read a &lt;a href=&quot;http://qchu.wordpress.com/2010/11/09/zeta-functions-statistical-mechanics-and-haar-measure/&quot; rel=&quot;nofollow&quot;&gt;blog post&lt;/a&gt; I wrote (and its &lt;a href=&quot;http://qchu.wordpress.com/2010/11/14/a-little-more-about-zeta-functions-and-statistical-mechanics/&quot; rel=&quot;nofollow&quot;&gt;companion&lt;/a&gt;) about how to motivate the definition of the zeta function and, in particular, how to motivate studying its asymptotic behavior near $s = 1$.  It is a nontrivial fact that one can use complex analysis to study this behavior, and complex analysis is always extremely powerful when it applies.&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2011-01-02T18:01:44.987" LastActivityDate="2011-01-02T18:01:44.987" CommentCount="3" />
  <row Id="16144" PostTypeId="2" ParentId="16124" CreationDate="2011-01-02T18:02:41.410" Score="1" ViewCount="" Body="&lt;p&gt;Regarding the comments to Akhil's answer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $x$ is a point in a manifold $M$, the tangent space $T_x M$ can be defined in various ways, one of which is to think of a tangent vector as an equivalence class of curves through $x$ (two curves being equivalent if they pass through the point with the same velocity). If $\gamma$ is such a curve (corresponding to a tangent vector to $M$ at $x$, call it $v$), then a function $F:M \to N$ takes $\gamma$ to a curve $F\circ \gamma$ in $N$, which passes through the point $F(x)$ with a certain velocity, corresponding to a tangent vector to $N$ at $F(x)$,&#xA;and it is this tangent vector that is denoted by $dF(v)$. The mapping from $v \in T_x M$ to $dF(v) \in T_{F(x)} N$ is the induced mapping on the tangent spaces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In local coordinates we can write $F(x) = (F_1(x_1,\dots,x_m), \dots, F_n(x_1,\dots,x_m))$,&#xA;and the linear transformation $dF$ simply becomes the $n \times m$ matrix of derivatives, $(\partial F_i/\partial x_j)$.&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2011-01-02T18:02:41.410" CommentCount="1" />
  <row Id="16145" PostTypeId="2" ParentId="16124" CreationDate="2011-01-02T18:09:49.423" Score="3" ViewCount="" Body="&lt;p&gt;There are a number of equivalent ways to define the derivative of a map of manifolds, since there are a number of equivalent ways to define the tangent space of a manifold at a point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One way which makes the definition of derivative particularly simple is as follows: Let $p$ be a point on a manifold $M$, and let $C^\infty(M)$ be the algebra of smooth real-valued functions on $M$. The tangent space $T_pM$ can be thought of as the space of &lt;em&gt;derivations&lt;/em&gt; at $p$; that is, $T_pM$ is the space of $\mathbb{R}$-linear maps $\delta: C^\infty(M)\to\mathbb{R}$ which satisfy the rule $\delta(fg) = \delta(f)g(p) + f(p)\delta(g)$.  Given a local coordinate system $x_1,\ldots,x_m$ near $p$, a basis for $T_pM$ is given by the operators $\frac{\partial}{\partial x_i}|_p$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It should be noted that with this definition, a &lt;em&gt;vector field&lt;/em&gt; (i.e. a smooth choice of tangent vector for each point) is then the same thing as a derivation of $C^\infty(M)$, i.e. an $\mathbb{R}$-linear map $\Delta:C^\infty(M)\to C^\infty(M)$ satisfying the Leibniz rule $\Delta(fg) = \Delta(f)g + f\Delta(g)$.  Given a derivation $\Delta$, the tangent vector $\delta\in T_p M$ is the map $C^\infty(M)\to\mathbb{R}$ given by composing $\Delta$ with evaluation at $p$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With this description of the tangent space, the derivative $d\varphi_p:T_p M\to T_{\varphi(p)}N$ of a map $\varphi:M\to N$ at a point $p\in M$ can be described as follows: first, $\varphi$ induces a map of algebras $\varphi^*:C^\infty(N)\to C^\infty(M)$ by $\varphi^*(g)=g\circ\varphi$.  The derivative is then given by $d\varphi_p(\delta) = \delta\circ\varphi^*$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the purposes of computation, the derivative can be described in local coordinates as follows: suppose $x = (x_1,\ldots,x_m)$ is a local coordinate system for $M$ at $p$, and $y = (y_1,\ldots,y_n)$ is a local coordinate system for $N$ at $\varphi(p)$.  Then $\varphi$ is given near $p$ in terms of these coordinates by functions $\varphi_1(x),\ldots,\varphi_n(x)$ -- that is, $\varphi_j$ is the $y_j$-component of $\varphi$.  The differential $d\varphi_p$ is then defined on the basis $\frac{\partial}{\partial x_i}|_p$ by $\displaystyle d\varphi_p\left(\frac{\partial}{\partial x_i}|_p\right) = \sum_{j=1}^n \frac{\partial\varphi_j}{\partial x_i}|_p \frac{\partial}{\partial y_j}|_{f(p)}$.&lt;/p&gt;&#xA;" OwnerUserId="5116" LastEditorUserId="5116" LastEditorDisplayName="" LastEditDate="2011-02-20T01:11:27.897" LastActivityDate="2011-02-20T01:11:27.897" CommentCount="6" />
  <row Id="16146" PostTypeId="1" AcceptedAnswerId="16279" CreationDate="2011-01-02T18:38:20.647" Score="2" ViewCount="189" Body="&lt;p&gt;According to Tarski's Fixed Point Theorem, for a complete lattice $L$, and monotone function $f:L \rightarrow L$, the set of fixed points of $f$ forms complete lattice. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Definition of $lfp(f)$ and $gfp(f)$ is given in terms of set $\text{Red}(f)$ and $\text{Ext}(f)$. Where $\text{Red}(f)$ is set of all those points $l&amp;#39; \in L$, which satisfy $f(l&amp;#39;) \le l&amp;#39;$. Similarly $\text{Ext}(f)$ is set of all those points $l&amp;#39;&amp;#39; \in L$, which satisfy $f(l&amp;#39;&amp;#39;) \ge l&amp;#39;&amp;#39;$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$lfp(f) = \bigcap(\text{Red}(f))$ and $gfp(f) = \bigcap(\text{Ext}(f))$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I read somewhere that these definitions are not constructive in nature, so to reach least fixed point we try to find out least upper bound of the sequence $(f^n(\bot))_n$, i.e.&#xA;$f(\bot) \sqcup f(f(\bot)) \sqcup f(f(f(\bot))) \dots$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Here $\bot$ is least element of lattice $L$ ).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can someone please explain it further what do we mean by constructive definition here, and how these two definitions of $lfp(f)$ coincide?&lt;/p&gt;&#xA;" OwnerUserId="5318" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2011-01-03T11:47:58.880" LastActivityDate="2011-01-04T01:05:01.083" Title="Is the Knaster-Tarski Fixed Point Theorem constructive?" Tags="&lt;order-theory&gt;&lt;lattices&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="16147" PostTypeId="2" ParentId="15890" CreationDate="2011-01-02T19:22:30.967" Score="2" ViewCount="" Body="&lt;p&gt;I started in math as an undergrad (topology/geometry) and then went into physics for grad school.  I know your pain... :) Most of the books people have suggested already are excellent works that provide physical intuition and if you're going to do some physics for real then make sure you read 'em.  That said, most of what was mentioned does NOT present stuff the way a math guy likes so -- assuming you are a junior or senior in a math undergrad program -- my recommendations are,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For geometry and dynamical systems as applied to Classical Mechanics:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Classical Dynamics: A Contemporary Approach &#xA;Jorge V. José (Author)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;For (functional) analysis applied to Quantum Mechanics:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Quantum Mechanics in Hilbert Space: Second Edition (Dover Books on Physics) [Paperback]&#xA;Eduard Prugovecki (Author) &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;For Lie Groups and using them like a physicist:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Lie Groups for Pedestrians [Paperback]&#xA;Harry J. Lipkin (Author)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Lie Algebras In Particle Physics: from Isospin To Unified Theories [Paperback]&#xA;Howard Georgi (Author)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;For understanding quantum and path integrals:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Quantum Mechanics and Path Integrals &#xA;Richard P. Feynman (Author), Albert R. Hibbs  (Author), Daniel F. Styer (Author) &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Modern Quantum Mechanics (Revised Edition) [Hardcover]&#xA;J. J. Sakurai (Author)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;You gotta understand rotation stone cold:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Rotations, Quaternions, and Double Groups [Paperback]&#xA;Simon L. Altmann (Author)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The Theory of Spinors by Elie Cartan&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Clifford Algebras and Spinors (London Mathematical Society Lecture Note Series) by Pertti Lounesto&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;And differential forms are required:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Differential Forms with Applications to the Physical Sciences [Paperback]&#xA;Harley Flanders (Author)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;For general relativity:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Gravitation (Physics Series) [Paperback]&#xA;Charles W. Misner (Author)&#xA;Kip S. Thorne (Author)&#xA;John Archibald Wheeler (Author)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Advanced General Relativity (Cambridge Monographs on Mathematical Physics) [Paperback]&#xA;John Stewart (Author) &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;For stat mech:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Equilibrium Statistical Physics (3rd Edition) [Paperback] Michael Plischke (Author)&#xA;(Author), Birger Bergersen (Author) &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;and for a final one-shot book that tries to hit everything under the sun:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The Geometry of Physics: An Introduction, Second Edition [Paperback] Theodore Frankel (Author)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="5293" LastActivityDate="2011-01-02T19:22:30.967" />
  <row Id="16148" PostTypeId="1" CreationDate="2011-01-02T19:45:37.977" Score="-2" ViewCount="191" Body="&lt;p&gt;Wikipedia : In mathematics, an algebraic number is a number that is a root of a non-zero polynomial in one variable with rational (or equivalently, integer) coefficients.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Question : Show that the Set of algebraic number in $\mathbb{C}$ is countable ?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Hint : if $\mathbb{F(z) = a_ 0 + a_1z + .... + a_nz^n}$ we denote the Height of a polynomial as : $\mathbb{h = |a_0| + |a_1| + ...... + |a_n| + n}$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;PS : i have already solve it , i am looking for different way to solve it ;)&lt;/p&gt;&#xA;" OwnerUserId="5315" LastEditorUserId="5315" LastEditorDisplayName="" LastEditDate="2011-01-02T20:51:55.607" LastActivityDate="2011-01-02T20:51:55.607" Title="Set of Algebraic Numbers ." Tags="&lt;abstract-algebra&gt;&lt;elementary-set-theory&gt;" AnswerCount="1" CommentCount="7" />
  <row Id="16149" PostTypeId="2" ParentId="16146" CreationDate="2011-01-02T20:04:53.397" Score="1" ViewCount="" Body="&lt;p&gt;See &lt;a href=&quot;http://projecteuclid.org/DPubS/Repository/1.0/Disseminate?view=body&amp;amp;id=pdf_1&amp;amp;handle=euclid.pjm/1102785059&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2011-01-02T20:04:53.397" CommentCount="1" />
  <row Id="16151" PostTypeId="1" AcceptedAnswerId="16153" CreationDate="2011-01-02T20:15:19.177" Score="1" ViewCount="126" Body="&lt;p&gt;It seems to me that a = x (mod m) can mean either that a is the remainder of x divided by m or that the remainders of a and x are the same when divided by m (eg. with respect to mod m).  For example: 2 = 6 (mod 4) would be the first meaning I discussed, but 6 = 2 (mod 4) would be the second.  It seems that I must be confusing some aspect of modular arithmetic here.  Thanks for the assistance.&lt;/p&gt;&#xA;" OwnerUserId="5323" LastActivityDate="2011-01-02T20:21:34.467" Title="How can a = x (mod m) have multiple meanings in modular arithmetic?" Tags="&lt;modular-arithmetic&gt;" AnswerCount="3" CommentCount="7" />
  <row Id="16152" PostTypeId="2" ParentId="16148" CreationDate="2011-01-02T20:19:28.693" Score="5" ViewCount="" Body="&lt;p&gt;To solve this elegantly I think one needs the following steps:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Define an equivalence relation on the algebraic numbers: $x\sim y$ if and only if they have the same irreducible polynomial, that is the polynomial of minimal degree which has the leading coefficient (the highest power) as $1$. (E.g. $i$ and $-i$ are equivalent, while $\sqrt{2}$ and $\sqrt{3}$ are not)&lt;/li&gt;&#xA;&lt;li&gt;Prove the fact that in each equivalence class there are only finitely many elements&lt;/li&gt;&#xA;&lt;li&gt;Show that there is a simple injection from the equivalence classes into the polynomials over $\mathbb{Q}$ (namely $[x] \mapsto irr(x)$ (the irreducible polynomial of $x$ over the rationals)&lt;/li&gt;&#xA;&lt;li&gt;Show that there are only countably many polynomials (that is, the number of finite sequences from a countable set is countable)&lt;/li&gt;&#xA;&lt;li&gt;Use the fact that finitely many elements in each equivalence class, times countably many classes result in at most countably many elements.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;All in all, you have an injection from algebraic numbers into a countable set, and since all the natural numbers are in particular algebraic it is also clear that there are infinitely many of those.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This proof &lt;em&gt;does not require&lt;/em&gt; the axiom of choice, as Andres remarked well in his comment. All the enumerations mentioned above can be done explicitly (or with the CSB theorem, which also does not require the axiom of choice).&lt;/p&gt;&#xA;" OwnerUserId="622" LastEditorUserId="622" LastEditorDisplayName="" LastEditDate="2011-01-02T20:28:13.357" LastActivityDate="2011-01-02T20:28:13.357" CommentCount="2" />
  <row Id="16153" PostTypeId="2" ParentId="16151" CreationDate="2011-01-02T20:20:33.327" Score="5" ViewCount="" Body="&lt;p&gt;No; $a\equiv b \pmod{m}$ means that $a$ and $b$ are congruent modulo $m$: by definition, the meaning is that $m$ divides $a-b$. As it happens, this is &lt;em&gt;equivalent&lt;/em&gt; to saying that $a$ and $b$ have the same remainder when divided by $m$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, there is also a related notation, in which &quot;mod&quot; acts as a &lt;em&gt;binary operator&lt;/em&gt;: in computer science especially, one often finds expressions like &quot;$a \bmod m$&quot;; in this case, this is interpreted as an operation on $a$ and $m$ that results in the remainder when $a$ is divided by $m$ (usually the one among $0$, $1,\ldots,m-1$, but in some instances instead the ones with smallest absolute value, allowing negative remainder).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But these two are related-but-different. In $a\equiv b\pmod{m}$, what we have is a binary &lt;em&gt;relation&lt;/em&gt; between $a$ and $b$, called &quot;congruence modulo $m$&quot;, and written $\equiv\pmod{m}$. In $a\bmod m$, what we have is a binary &lt;em&gt;operation&lt;/em&gt; (noncommutative) on $a$ and $m$. &lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2011-01-02T20:20:33.327" CommentCount="3" />
  <row Id="16154" PostTypeId="2" ParentId="16151" CreationDate="2011-01-02T20:21:04.193" Score="0" ViewCount="" Body="&lt;p&gt;$a = x (\mathrm{mod} m)$ means $a$ is the remainder when $x$ is divided by $m$.&lt;br&gt;&#xA;$a \equiv x (\mathrm{mod} m)$ means $a$ and $x$ have the same remainder when divided by $m$.   &lt;/p&gt;&#xA;" OwnerUserId="1508" LastActivityDate="2011-01-02T20:21:04.193" />
  <row Id="16155" PostTypeId="2" ParentId="16151" CreationDate="2011-01-02T20:21:34.467" Score="2" ViewCount="" Body="&lt;p&gt;I think you are mixing notation here; perhaps using some superfluous parentheses may help: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a=(x\pmod m)$ has the first meaning you said: $a$ is the remainder of dividing $x$ into $m$; while $(a=x)\pmod m$ would have the second meaning: $a$ and $x$ have the same remainder when divided by $m$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To avoid confusion, it is customary to write the second expression as $a\equiv x\pmod m$. This way, if you see $=$ you know you are using the first meaning, if you see $\equiv$, it is the second. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Of course, every now and then, books mix the two notations, hopefully in contexts where which one is intended is clear.)&lt;/p&gt;&#xA;" OwnerUserId="462" LastActivityDate="2011-01-02T20:21:34.467" CommentCount="2" />
  <row Id="16156" PostTypeId="1" AcceptedAnswerId="16158" CreationDate="2011-01-02T20:23:23.310" Score="1" ViewCount="137" Body="&lt;p&gt;I have been given the following pdf :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;fT (t; B, C) = ( exp(-t/C) - exp(-t/B) ) / ( C - B )  ,  (t&gt;0)&#xA;where the overall mean is B+C.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am unsure as to how to write the log likelihood function of B and C.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The next part of the Q asks me to derive the equations that would have to be solved in order to find the max likelihood estimators of B and C.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would be grateful for any help =)&lt;/p&gt;&#xA;" OwnerUserId="5324" LastActivityDate="2011-01-02T21:34:57.453" Title="How do I write a log likelihood function when I have 2 mean values for my pdf?" Tags="&lt;statistics&gt;" AnswerCount="1" />
  <row Id="16157" PostTypeId="2" ParentId="16121" CreationDate="2011-01-02T20:47:41.817" Score="11" ViewCount="" Body="&lt;p&gt;The argument sketched in Qiaochu's answer is actually a special case of a general relationship between subalgebras and congruences. Generalizing the way that one defines congruences for the ring of integers modulo $\rm\:m\:$, one defines a congruence $\equiv$ as an equivalence relation on an algebra $\rm\:A\:$ that is compatible with the operations on $\rm\:A\:$. For example, for rings such compatibility means that if $\rm a&amp;#39;\equiv a,\ b&amp;#39;\equiv b\:$ then $\rm\: a&amp;#39;+b&amp;#39;\:\equiv\: a+b\ $ and similarly for all other operations. Now, viewing the equivalence relation $\equiv$  as a subset of $\rm\ A\times A\:,\: $ this compatibility condition is a closure condition: if $\rm\: (a&amp;#39;,a),\:(b&amp;#39;,b)\in\;\equiv\:$ then $\rm\: (a&amp;#39;+b&amp;#39;,a+b)\in\; \equiv\:,\: $ i.e. $\:\equiv\:$ is closed under addition in $\rm\: A\times A\:$. Thus an equivalence relation on $\rm\:A\:$ is compatible with the operations of $\rm\:A\:$ iff it forms a &lt;em&gt;subalgebra&lt;/em&gt; of $\rm\:A^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Returning to the case as hand, where $\rm\:A\:$ is the ring $\:\mathbb Z\:,\: $ let $\rm\:S\:$ be a subring of $\rm\:\mathbb Z\times\mathbb Z\:.\ $ By the above, to show that $\rm\:S\:$ is a congruence we need only show that it is an equivalence relation. Firstly, since &lt;em&gt;here&lt;/em&gt; $\rm\ (1,1)\in S\ $ generates the full diagonal $\rm\ (1,1)\ \mathbb Z\:,\ $ we deduce that $\rm\:S\:$ is a &lt;em&gt;reflexive&lt;/em&gt; relation. Secondly, $\rm\:S\:$ is &lt;em&gt;symmetric&lt;/em&gt; since if $\rm\ (a,b)\in S\ $ then also in $\rm\:S\:$ is $\rm\ (a+b,a+b)-(a,b)\ =\ (b,a)\:.\: $ Thirdly,  $\rm\:S\:$ is &lt;em&gt;transitive&lt;/em&gt; since if $\rm\ (a,b),\ (b,c)\in S\ $ then so to is $\rm\ (a,b)\:+\:(c,c)\:-\:(c,b)\:\: =\: (a,c)\:. $ Finally,$\:$ for every ring $\rm\:R\:,\: $ a congruence $\equiv$ is uniquely determined by the congruence class of $\:0\:$, since $\rm\ a\equiv b\iff a-b\equiv 0\:.\ $ But the congruence class of $\:0\:$ has the structure of an &lt;em&gt;ideal&lt;/em&gt; since $\rm\ a,b\equiv 0\ \Rightarrow\ a+b\equiv 0\ $ and $\rm\ c\;a\equiv 0\:,\:$ for all $\rm\:c \in R\:$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This explains - from general principles - the connection observed in Qiaochu's post between subalgebras of $\:\mathbb Z\times\mathbb Z\:$ and ideals of $\rm\mathbb Z\:.\ $ Note that  while the connections between congruences and subalgebras of the square, and congruences and ideals hold true for every ring $\rm\:R\:,\:$ the rest of the above argument doesn't follow since generally  $\rm\:(1,1)\:$ doesn't generate the &lt;em&gt;full&lt;/em&gt; diagonal $\rm\:(1,1)\ R\:$. Indeed, it generates only the diagonal of the characteristic subring (the image of $\:\mathbb Z\:$ in $\rm\:R\:)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The above argument shows that in order to verify that a subalgebra of $\rm\:R\times R\:$ is a ring congruence it suffices to show that the subalgebra contains the diagonal (i.e. it is reflexive), since this implies the other equivalence relation properties (symmetry and transitivity). This leads to the following   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;THEOREM&lt;/strong&gt; $\ $ The following are equivalent for a ring $\rm\:R\:$ and set $\rm\ S\subset R\times R$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm(1)\quad S\ $ is a congruence on $\rm\:R\:$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm(2)\quad S\ $ is a subalgebra of $\rm\:R\times R\:$ and $\rm\:S \supset (1,1)\: R$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm(3)\quad I\: :=\: \{\: r\in R\: :\ (r,0)\in S \:\}\ $ is an ideal in $\rm\:R\:$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof&lt;/strong&gt; $\rm\ (1\Rightarrow 2)\ $ follows the same way as sketched above. $\rm\ (1\iff 3)$ is well-known.&lt;br&gt;&#xA;$\rm (2\Rightarrow 3)\quad\ i,\:j\in I\ \Rightarrow\ (i,0),\:(j,0)\in S\ \Rightarrow\ (i,0)+(j,0)=(i+j,0)\in S\ \Rightarrow\ i+j\in I $&lt;br&gt;&#xA;Also $\rm\; r\in R,\: j\in I\ \Rightarrow\ (r,r),\:(j,0)\in S\ \Rightarrow\ (r,r)\: *\: (j,0) = (r *\: i, 0)\in S\ \Rightarrow\ r\: *\:j \in I$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For further details see any good textbook on universal algebra, e.g. Burris and Sankappanavar's textbook &lt;a href=&quot;http://www.math.uwaterloo.ca/~snburris/htdocs/UALG/univ-algebra.pdf#page=154&quot; rel=&quot;nofollow&quot;&gt;A Course in Universal Algebra&lt;/a&gt; or George Bergman's &lt;a href=&quot;http://www.math.berkeley.edu/~gbergman/245&quot; rel=&quot;nofollow&quot;&gt;An Invitation to General Algebra and Universal Constructions&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2011-05-26T20:37:18.360" LastActivityDate="2011-05-26T20:37:18.360" CommentCount="2" />
  <row Id="16158" PostTypeId="2" ParentId="16156" CreationDate="2011-01-02T20:56:51.243" Score="2" ViewCount="" Body="&lt;p&gt;By definition, the log-likelihood is given by&#xA;$$&#xD;&#xA;\ln \mathcal{L}(B,C|x_1 , \ldots ,x_n ) = \sum\limits_{i = 1}^n {\ln f(x_i|B,C)}.&#xD;&#xA;$$&#xA;Thus, in our example, &#xA;$$&#xD;&#xA;\ln \mathcal{L}(B,C|x_1 , \ldots ,x_n ) = \sum\limits_{i = 1}^n {\ln \bigg[\frac{{e^{ - x_i /C}  - e^{ - x_i /B} }}{{C - B}}\bigg]} .&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: In view of the next part of the question, it may be useful to write&#xA;$$&#xD;&#xA;\ln \mathcal{L}(B,C|x_1 , \ldots ,x_n ) = \sum\limits_{i = 1}^n {\ln [e^{ - x_i /C}  - e^{ - x_i /B} ]}  - n\ln (C - B).&#xD;&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="2810" LastEditorUserId="2810" LastEditorDisplayName="" LastEditDate="2011-01-02T21:34:57.453" LastActivityDate="2011-01-02T21:34:57.453" CommentCount="4" />
  <row Id="16159" PostTypeId="1" AcceptedAnswerId="16166" CreationDate="2011-01-02T21:15:58.833" Score="3" ViewCount="111" Body="&lt;p&gt;I am sorry for the vague title - its because I don't know what I am talking about.&#xA;I have a function whose value is determined by considering the outputs of some other functions. For example,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;F(x) = G(a,b) and H(x) and T(y)&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;As you can see, I am not sure how to write the &quot;and&quot; part. That is, I want to say: In order to calculate F(x), you first have to calculate G(a,b) and H(x) and T(y); and then &quot;mix them up in a certain way&quot;.&lt;/li&gt;&#xA;&lt;li&gt;How can I represent this idea mathematically? Can somone please point me to some resources that further explain this in baby-steps?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Followup:&lt;br&gt;&#xA;Thank you for all your replies. Let me clarify what I want to do.&#xA;Consider a scenario where I want to find the value of a car. This depends on factors such as mileage, model, number of accidents, and also owner-bias. The owner-bias itself is a function of gender, years of driving experience, and age&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So.. how can I come up with a symbolic way to represent this AND also define the interralations between all these factors?&#xA;I am sorry for the newbie question, please feel free to direct me to the relevant literature/terminology&lt;/p&gt;&#xA;" OwnerUserId="5326" LastEditorUserId="5326" LastEditorDisplayName="" LastEditDate="2011-01-03T01:08:22.830" LastActivityDate="2011-01-03T15:01:35.377" Title="Considering a combination of functions" Tags="&lt;functions&gt;" AnswerCount="5" />
  <row Id="16160" PostTypeId="1" CreationDate="2011-01-02T21:26:26.043" Score="4" ViewCount="170" Body="&lt;p&gt;Consider a reversible markov chain Xn whose steady state distribution is known, can we find the expected hitting time to a subset A of the states starting from some state i ? Additionally you can assume the the chain has no self transitions.&lt;/p&gt;&#xA;" OwnerUserId="5327" LastActivityDate="2011-03-16T06:27:14.617" Title="Hitting times of reversible markov chain with known steady state probabilites" Tags="&lt;probability&gt;&lt;probability-theory&gt;&lt;stochastic-processes&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="16161" PostTypeId="2" ParentId="16159" CreationDate="2011-01-02T21:29:59.407" Score="3" ViewCount="" Body="&lt;p&gt;Well, if that's how you calculate $F(x)$, then it's not just a function of $x$, is it?  You should write it as $F(G(a, b), H(x), T(y))$.  Unless you're saying that $a, b, y$ are all themselves functions of $x$, in which case I guess you should write&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$F(x) = S(G(a, b), H(x), T(y)).$$&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2011-01-02T21:29:59.407" CommentCount="1" />
  <row Id="16162" PostTypeId="2" ParentId="16159" CreationDate="2011-01-02T21:32:12.993" Score="4" ViewCount="" Body="&lt;p&gt;The issue appears to be one of &quot;dummy variables&quot; and maybe composition of functions.  When you write, for example, q(x)=x^2, it really doesn't matter that the variable is x.  You also have that q(y)=y^2 and the same for any variable you name.  But we expect the same variables on both sides of the equation.  If we could show F(x)=G(a,b) that would say that F(x) doesn't depend upon a and b, so F(x) and hence G(a,b) must be constant.  If F really depends upon G(a,b), H(x), and T(y) we would expect to see F(a,b,x,y)= some expression involving G(a,b), H(x), and T(y).  An example would be F(a,b,x,y)=G(a,b)*H(x)+T(y).  Sometimes H(x) and T(y) are the inputs to G.  That we would write F(x,y)=G(H(x),T(y))&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Qiaochu Yuan and I are trying to get at the same thing using different language.  I hope at least one is helpful.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-01-02T21:32:12.993" CommentCount="1" />
  <row Id="16163" PostTypeId="1" AcceptedAnswerId="16167" CreationDate="2011-01-02T21:32:34.117" Score="3" ViewCount="179" Body="&lt;p&gt;Hopefully without getting too complicated, how is a norm different from an absolute value?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In context, I am trying to understand relative stability of an algorithim:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the inequality $\frac{|(x_0)- \tilde{f}(\epsilon, x_0)|}{|f(x_0)|} \leq \sigma_{rel} ||\epsilon|| + o(||\epsilon||)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For which $\sigma_{rel}$ describes the relative stability.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And $||\epsilon||$ is the maximum rounding error of an elementary function: $||\epsilon|| = \mbox{max} |\epsilon_i|, i=1,...n$,  $\epsilon = (\epsilon_1,...,\epsilon_n)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After reading some of the wikipedia article on norms, I could not take away much other than a norm is positive and a length. I am confused about the statement that it is a function...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any help trying to understand this would be great, thanks!&lt;/p&gt;&#xA;" OwnerUserId="3711" LastActivityDate="2011-01-02T22:39:48.463" Title="How are norms different from absolute values?" Tags="&lt;norm&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="16164" PostTypeId="1" AcceptedAnswerId="16169" CreationDate="2011-01-02T21:34:38.547" Score="1" ViewCount="70" Body="&lt;p&gt;I am working on my own version of a proof of RSA and have come do a conclusion based on these simplified statements.&lt;br&gt;&#xA;Given: N = pq&lt;br&gt;&#xA;X ≡ 1 (mod p)&lt;br&gt;&#xA;X ≡ 1 (mod q)&lt;br&gt;&#xA;X ≡ 1 (mod N) by chinese remainder theorem&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have the original expression:&#xA;MX ≡ M' (mod N)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When substituting X ≡ 1 (mod N) into this equation, is it correct to say that&#xA;M = M' (removing the modulus) or must I say M ≡ M' (mod N)?&lt;/p&gt;&#xA;" OwnerUserId="5323" LastEditorUserId="5323" LastEditorDisplayName="" LastEditDate="2011-01-02T21:44:16.523" LastActivityDate="2011-01-03T00:09:34.367" Title="Is there a theorem based on substitution to convert a congruency to an equality?" Tags="&lt;modular-arithmetic&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="16165" PostTypeId="1" AcceptedAnswerId="16170" CreationDate="2011-01-02T21:38:17.217" Score="5" ViewCount="95" Body="&lt;p&gt;Given the following sequence $d(n) = x \cdot n - \lfloor x \cdot n \rfloor$ with $n \in \mathbb{N}$, $x \in \mathbb{R} \backslash \mathbb{Q}$ .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to show that for every $z \in [0,1)$ there is a subsequence of $d(n)$ that converges against $z$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notes:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am having serious problems of finding such a subsequence, first I see what the sequence does. If you take for example the subsequence $a(n) = d(10^n)$ you see that it gives you the decimal representation of $x$ starting with the $n&amp;#39;th$ digit.&#xA;Example: let $x=\pi$, then we get $a(0)=0.141592..., a(1)=0.41592..., a(2)=0.1592...$ and so on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also this is a plot of $d(n)$ for $x=\pi$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/VfQSZ.gif&quot; alt=&quot;Plot of d(n)&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You see that for irrational x the height of the function at integers is not constant, nor periodic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also this question I asked several weeks ago seems to be related: &lt;a href=&quot;http://math.stackexchange.com/questions/10280/proving-that-an-equation-is-solvable-floor-function&quot;&gt;Proving that an equation is solvable, Floor function.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3123" LastEditorUserId="3123" LastEditorDisplayName="" LastEditDate="2011-01-02T22:13:44.163" LastActivityDate="2011-01-02T22:18:19.737" Title="Decimal representation series" Tags="&lt;real-analysis&gt;&lt;sequences-and-series&gt;&lt;convergence&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="16166" PostTypeId="2" ParentId="16159" CreationDate="2011-01-02T21:40:47.617" Score="3" ViewCount="" Body="&lt;p&gt;Just some possibilities to get you thinking...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If these are numerical functions, you might have:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;F(x,y,a,b) = G(a,b) + H(x) + T(y)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If a and b are independent variables, you have to list them as arguments for F as above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If, on the other hand, a=H(x) and b=T(y), you might have:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;F(x,y) = G(H(x),T(y))&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The possibilities are, of course, endless. I hope this helps.&lt;/p&gt;&#xA;" OwnerUserId="3515" LastActivityDate="2011-01-02T21:40:47.617" CommentCount="5" />
  <row Id="16167" PostTypeId="2" ParentId="16163" CreationDate="2011-01-02T21:45:42.993" Score="10" ViewCount="" Body="&lt;p&gt;The absolute value is a particular instance of a norm. Or perhaps, you can think of norms as functions $\mathbf{V}\to\mathbb{R}$ where $\mathbf{V}$ is a vector space over a field $\mathbf{F}$, and &quot;absolute values&quot; are &quot;norms on the base field&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The absolute value is a function $|\&amp;gt;|\colon\mathbb{R}\to[0,\infty)$; given any real number $r$, you get a nonnegative real number that we write $|r|$, and which satisfies the following properties:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$|x|\geq 0$ for all $x\in\mathbb{R}$; $|x|=0$ if and only if $x=0$.&lt;/li&gt;&#xA;&lt;li&gt;$|rx| = |r||x|$ for all $r,x$.&lt;/li&gt;&#xA;&lt;li&gt;$|x+y| \leq |x|+|y|$.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;It's a real valued function of real variable, because it takes a real number as an input, and gives a (nonnegative) real number as an output. It's just that instead of calling the function, say, $f$, and writing the output as $f(x)$, we call the function &quot;$|\&gt;|$&quot; and write the output as $|x|$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, we have the modulus function for complex numbers, $|\&amp;gt;|\colon\mathbb{C}\to[0,\infty)$, defined by $|z| = \sqrt{z\overline{z}}$, and which also satisfies the three conditions above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $\mathbf{V}$ is a vector space over $\mathbb{R}$, then a &lt;strong&gt;norm on $\mathbf{V}$&lt;/strong&gt; is a function $||\&amp;gt;||\colon \mathbf{V}\to[0,\infty)$ that generalizes the absolute value; it must satisfy:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$||\mathbf{x}|| \geq 0$ for all $\mathbf{x}\in\mathbf{V}$; $||\mathbf{x}||=0$ if and only if $\mathbf{x}=\mathbf{0}$.&lt;/li&gt;&#xA;&lt;li&gt;$||r\mathbf{x}|| = |r|\,||\mathbf{x}||$ for all $r\in\mathbb{R}$, all $\mathbf{x}\in\mathbf{V}$.&lt;/li&gt;&#xA;&lt;li&gt;$||\mathbf{x}+\mathbf{y}|| \leq ||\mathbf{x}||+\mathbf{y}||$ for all $\mathbf{x},\mathbf{y}\in\mathbf{V}$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Again, this is a function: given any vector $\mathbf{x}$ in the domain, $||\mathbf{x}||$ is the output of the function (a nonnegative real number). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, if you view $\mathbb{R}$ as a vector space over itself, then the absolute value gives a norm on $\mathbb{R}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For vector spaces over $\mathbb{C}$, you replace the absolute value of $r$ in the second property with the modulus of $r$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The norm you describe in your post, $||\epsilon||=\max|\epsilon_i|$ is a particular norm that can be placed on $\mathbb{R}^n$; there are many norms that can be defined on $\mathbb{R}^n$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The notion of norm on a vector space can be done with any field that is contained in $\mathbb{C}$, by restricting the modulus to that field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Added.&lt;/em&gt; One can also extend the notion of norm by starting with any field $\mathbf{F}$, such as the rationals, simply asking for an &quot;absolute value&quot; function $|\&amp;gt;|\colon \mathbf{F}\to [0,\infty)$ that satisfies properties 1, 2, and 3 above. Then you extend the notion of norm for any vector space over that field. It is common to refer to such &quot;absolute values&quot; as &quot;norms&quot; so as not to confuse them with the usual absolute value. One classical example, mentioned by Asaf, are the $p$-adic norms on the rationals. First, fix a prime $p$; given an integer $p$, we define the $p$-order of $a$ to be the largest power of $p$ that divides $a$: that is, $\mathrm{ord}_p(a) = n$ if and only if $p^n$ divides $a$ and $p^{n+1}$ does not divide $a$. We formally set $\mathrm{ord}_p(0)=\infty$. We then extend this to the rationals: given a rational $\frac{a}{b}$, we let $\mathrm{o}_p(\frac{a}{b}) = \mathrm{ord}_p(a) - \mathrm{ord}_p(b)$. Finally, we define the $p$-adic norm on the rationals, $||\&amp;gt;||_p\colon\mathbb{Q}\to[0,\infty)$ by $||\frac{a}{b}||_p = p^{-o_p(a/b)}$. (You can use bases other than $p$; they amount to what are called &quot;equivalent norms&quot;). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here you really want to think of $||\&amp;gt;||_p$ as a kind of &quot;non-standard absolute value&quot; on the rationals; it leads to a lot of interesting mathematics, beginning with the &lt;a href=&quot;http://en.wikipedia.org/wiki/P-adic_numbers&quot; rel=&quot;nofollow&quot;&gt;$p$-adic numbers&lt;/a&gt;, if you use $||\&amp;gt;||_p$ to define &quot;Cauchy-sequences&quot; instead of $|\&amp;gt;|$ and do the same construction that leads to the reals in the latter case. &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-02T22:39:48.463" LastActivityDate="2011-01-02T22:39:48.463" CommentCount="5" />
  <row Id="16168" PostTypeId="1" AcceptedAnswerId="16171" CreationDate="2011-01-02T21:51:17.690" Score="3" ViewCount="278" Body="&lt;p&gt;Many courses and books assume that rings have an identity. They say there is not much loss in generality in doing so as rings studied usually have an identity or can be embedded in a ring with identity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What then are the major applications of rings without identities occurring naturally in mathematics?&lt;/p&gt;&#xA;" OwnerUserId="4594" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-05-23T17:37:44.513" LastActivityDate="2011-05-23T17:37:44.513" Title="Applications of rings without identity" Tags="&lt;ring-theory&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="16169" PostTypeId="2" ParentId="16164" CreationDate="2011-01-02T21:54:15.680" Score="0" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\ $ No,$\ $ e.g.$\ $ let $\rm\ M&amp;#39;\: =\: M + N\:.\ $  But the inference that $\rm\ M&amp;#39;\equiv M\ (mod\ N)\ $ is correct since it's a special case of the &lt;a href=&quot;http://math.stackexchange.com/questions/5366/very-simple-question-but-what-is-the-proof-that-x-y-mod-m-x-mod-m-y-mod-m/5376#5376&quot;&gt;congruence product rule,&lt;/a&gt; viz. $\rm\ X\equiv 1\ \Rightarrow\ M\:X \equiv M\cdot 1 \equiv M\:,\:$ i.e. $\rm\ M&amp;#39;\equiv M\:$.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2011-01-03T00:09:34.367" LastActivityDate="2011-01-03T00:09:34.367" CommentCount="2" />
  <row Id="16170" PostTypeId="2" ParentId="16165" CreationDate="2011-01-02T22:18:19.737" Score="1" ViewCount="" Body="&lt;p&gt;If I have understood your question correctly, I believe this follows from the &lt;a href=&quot;http://en.wikipedia.org/wiki/Equidistribution_theorem&quot; rel=&quot;nofollow&quot;&gt;Equidistribution theorem&lt;/a&gt;. For more info refer: &lt;a href=&quot;http://en.wikipedia.org/wiki/Equidistributed_sequence&quot; rel=&quot;nofollow&quot;&gt;Equidistributed Sequence&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2011-01-02T22:18:19.737" CommentCount="5" />
  <row Id="16171" PostTypeId="2" ParentId="16168" CreationDate="2011-01-02T22:29:22.447" Score="13" ViewCount="" Body="&lt;p&gt;The most common example of rings without identity occurs in functional analysis, when one considers rings of functions. A typical example is to consider the ring of all functions of compact support on a non-compact space. Obviously, as these rings of functions are very important in $C^*$-algebras and in studying the properties of the space, knowledge about rings without identity is very important for studying these spaces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Arbitrary direct sums of rings with unity are not rings with unity, which can also be fairly annoying. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is true that one can always embed a ring (as an ideal, even) into a ring with identity. The most common such embedding is the Dorroh embedding, in which we start with a ring $R$, and consider the ring with underlying set $\mathbb{Z}\times R$ and operations given by $(n,a)+(m,b) = (n+m,a+b)$ and $(n,a)(m,b) = (nm, nb+ma+ab)$. It is not hard to verify that $r\mapsto (0,r)$  embeds $R$ into the Dorroh extension as an ideal. You can preserve the characteristic of $R$ if necessary: if $R$ is of characteristic $n$, then replace $\mathbb{Z}$ with $\mathbb{Z}/n\mathbb{Z}$ in the construction. The extension has other nice properties (ideals of $R$ remain ideals of the extension, for example).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Luckily, I am currently going over a thesis about embedding rings as ideals into rings with identity, so I can give you some other classical results.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the Dorroh extension does not preserve &lt;em&gt;all&lt;/em&gt; ring properties that may be of interest in $R$. For example, a ring is &lt;em&gt;entire&lt;/em&gt; if it has no nonzero zero divisors; a ring is &lt;em&gt;prime&lt;/em&gt; if whenever $A$ and $B$ are ideals and $AB=0$, then either $A=0$ or $B=0$ (that is, &quot;prime&quot; is the ideal version of &quot;entire&quot;; an entire ring is necessarily prime). For example, if you perform the Dorroh extension on $\mathbb{Z}$ itself (perhaps not realizing it already had a $1$) then $(1,-1)(0,r)=(0,0)$ even though $\mathbb{Z}$ is entire. There are nontrivial examples of this situation as well. Another property not necessarily preserved by the Dorroh extension is being semiprime.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are other standard embeddings of rings into rings with identity, such as the Szendrei extension (a quotient of the Dorroh extension). But even so there are ring-theoretic properties that may be very hard to maintain in these kinds of embeddings. Among the more difficult ones are simplicity (if $R$ is simple, can we embed $R$ into a simple ring with identity? Yes; Anne Vakarietis, a student of a colleague, just finished putting together the pieces for this in her dissertation). It's known that every commutative $n$-root ring (rings in which every element has an $n$th root) can be embedded in a commutative $n$-root ring with identity, but it is not known if this is possible for noncommutative rings. Likewise, it is not known if every semiprimary ring can be embedded in a semiprimary ring with identity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And worse, there are some properties that we know &lt;em&gt;cannot&lt;/em&gt; be respected by such embeddings. For example, Fuchs and Rangaswamy proved that not every $\pi$-regular ring can be embedded as an ideal in a $\pi$-regular ring with identity (a ring is $\pi$-regular if every element is $n$-regular for some natural number $n$; an element $x$ is $n$-regular if there exists some $y$ such that $x^nyx^n=x^n$; this is a generalization of von Neumann regularity).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, in summary: yes, rings without identity arise very naturally, and as such they show up when investigating other mathematical objects. And while it is true that one can always embed a ring without identity as an ideal into a ring with identity, this may not be a good thing from the point of view of studying some ring-theoretic properties of these rings.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-02T22:43:18.197" LastActivityDate="2011-01-02T22:43:18.197" CommentCount="3" />
  <row Id="16172" PostTypeId="1" AcceptedAnswerId="16200" CreationDate="2011-01-02T22:56:51.833" Score="3" ViewCount="184" Body="&lt;p&gt;A &lt;strong&gt;paracompact space&lt;/strong&gt; is a space in which every open cover has a locally finite refinement. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A &lt;strong&gt;compact space&lt;/strong&gt; is a space in which every open cover has a finite subcover.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why must the product of a compact and a paracompact space be paracompact?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I really have very little intuition about how to go about this question, so any hints or a proof would be greatly appreciated. &lt;/p&gt;&#xA;" OwnerUserId="4460" LastEditorUserId="4460" LastEditorDisplayName="" LastEditDate="2011-01-03T00:09:55.890" LastActivityDate="2011-01-04T14:00:10.080" Title="The product of a paracompact space and a compact space is paracompact. (Why?)" Tags="&lt;topology&gt;" AnswerCount="3" CommentCount="8" FavoriteCount="2" />
  <row Id="16173" PostTypeId="1" AcceptedAnswerId="16174" CreationDate="2011-01-02T22:58:23.260" Score="2" ViewCount="166" Body="&lt;p&gt;Let $K(s,t)$ be a real-valued function of two real variables, and let $T: L^2(\mathbb{R}) \to L^2(\mathbb{R})$ be defined by $(Tf)(s) = \int_\mathbb{R} K(s,t) f(t) dt$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $||K||_{L^2({\mathbb{R}^2})} &amp;lt; \infty$, can we say that $T$ is a compact operator?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think this is true if we are looking at a bounded domain for $K$ and $f$ (by an application of the Arzela-Ascoli theorem), but I am not sure if it is true in general. &lt;/p&gt;&#xA;" OwnerUserId="1736" LastEditorUserId="1736" LastEditorDisplayName="" LastEditDate="2011-01-03T05:29:25.960" LastActivityDate="2011-01-03T05:29:25.960" Title="Compactness of an Integral operator" Tags="&lt;functional-analysis&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="16174" PostTypeId="2" ParentId="16173" CreationDate="2011-01-02T23:08:51.857" Score="5" ViewCount="" Body="&lt;p&gt;Hint:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Approximate $K(s,t)$ by functions of the form $\sum_{k=1}^n g_k(s)h_k(t)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) A norm limit of finite-rank operators is compact.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For further reference, your operator is called a Hilbert-Schmidt integral operator.  A more general version of this problem (where $\mathbb{R}$ with Lebesgue measure is replaced by an arbitrary $\sigma$-finite measure space) is &lt;a href=&quot;http://books.google.com/books?id=S57XLkgbf0oC&amp;amp;lpg=PP1&amp;amp;pg=PA92#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;Problem 173&lt;/a&gt; in Halmos's &lt;em&gt;Hilbert space problem book&lt;/em&gt;.  The book contains a hint and (in case you give up or solve it but want more information) a solution.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2011-01-02T23:57:42.000" LastActivityDate="2011-01-02T23:57:42.000" CommentCount="5" />
  <row Id="16175" PostTypeId="1" AcceptedAnswerId="16178" CreationDate="2011-01-02T23:15:26.663" Score="4" ViewCount="140" Body="&lt;h2&gt;Background&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;A function $ f: \mathbb{R}^n \rightarrow \mathbb{R} \ $ is linear if it satisfies&#xA;$$ (1)\;\; f(x+y) = f(x) + f(y) \ , \ and $$&#xA;$$ (2)\;\; f(\alpha x) = \alpha f(x) $$&#xA;for all $ x,y \in \mathbb{R}^n $ and all $ \alpha \in \mathbb{R} $.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A function satisfying only (2) is not necessarily linear.  For example* &lt;s&gt; $ f: \mathbb{R}^2 \rightarrow \mathbb{R} \ $ defined by $ f(x) = |x| \ $ (where $ |x| \ $ is the $ L^2 $ norm) satisfies (2) but is not linear&lt;/s&gt;.  However, a function satisfying (1) &lt;em&gt;does&lt;/em&gt; satisfy a weaker version of (2), namely $$ (2b)\;\; f(ax)=af(x) $$ for all $ a \in \mathbb{Q} $.    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;*Edit:  As pointed out in the comments this example doesn't quite work since |ax|=|a||x|.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When $ f $ is continuous it's relatively straight-forward to show that under the extra hypothesis that $ f $ is continuous, (2b) implies (2).  I &lt;em&gt;want&lt;/em&gt; to say that continuity is a &lt;em&gt;necessary&lt;/em&gt; condition for (1) to imply (2), or at least (worst) there is &lt;em&gt;some&lt;/em&gt; extra hypothesis required (possibly weaker than continuity), but I'm not sure how to show it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is therefore two-fold:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;-Is continuity a necessary condition for (1) to imply (2) and how could I go about proving it.&#xA;  -What are some examples (if there are any) of a function satisfying (1) but not (2)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This can be stated in a slightly more general context as follows:&#xA;Suppose $ V\ $ is a vector space over $ \mathbb{R}\ $ and $ f: V \rightarrow \mathbb{R}\ $ satisfies&#xA;$$ (1&amp;#39;) \;\; f(x+y) = f(x)+f(y) $$ for all $ x,y \in V $.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Under what conditions is $ f\ $ a vector space homomorphism?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The reason I believe continuity is necessary is because of the similarity to the fact that $ x^{\alpha} x^{\beta} = x^{\alpha + \beta} $ for all $ \alpha,\beta \in \mathbb{R} $.  Irrational powers can be defined either via continuity (i.e. if $ \alpha \ $ is irrational, then $ x^{\alpha}:= \lim_{q\rightarrow \alpha} x^q \ $ where q takes on only rational values) or by using the exponential and natual log functions, and either way proving the desired identity boils down to continuity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have come up with one example that satisfies (something similar to) (1) and not (2), but it doesn't quite fit the bill:&lt;br&gt;&#xA;$ \ $ Define $ \phi : \mathbb{Q}(\sqrt{2}) \rightarrow \mathbb{Q} \ $ defined by $ \phi(a+b\sqrt{2}) = a+b $.  Then $ \phi(x+y) = \phi(x)+\phi(y) \ $ but if $ \alpha=c+d\sqrt{2} \ $ then $ \phi(\alpha(a+b\sqrt{2})) = ac+2bd + ad+bc \neq \alpha \ \phi(a+b\sqrt{2}) $.&#xA;$ \ $ The problem is that even though $ \mathbb{Q}(\sqrt{2}) \ $ is a vector space over $ \mathbb{Q} $, the $ \alpha \ $ is coming from $ \mathbb{Q}(\sqrt{2}) \ $ instead of the base field $ \mathbb{Q} $.&lt;/p&gt;&#xA;" OwnerUserId="5330" LastEditorUserId="5330" LastEditorDisplayName="" LastEditDate="2011-01-02T23:30:25.117" LastActivityDate="2011-01-02T23:39:27.837" Title="On sort-of-linear functions" Tags="&lt;functions&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="16176" PostTypeId="2" ParentId="16175" CreationDate="2011-01-02T23:22:40.587" Score="3" ViewCount="" Body="&lt;p&gt;Yes, continuity is necessary.  In one variable the counterexamples are known as pathological solutions to &lt;a href=&quot;http://en.wikipedia.org/wiki/Cauchy%27s_functional_equation&quot; rel=&quot;nofollow&quot;&gt;Cauchy's functional equation&lt;/a&gt;.  They require the axiom of choice to construct: given the axiom of choice, we can pick a basis for $\mathbb{R}$ as a vector space over $\mathbb{Q}$, and any function $f : \mathbb{R} \to \mathbb{R}$ which is only required to be $\mathbb{Q}$-linear can be specified by arbitrarily specifying its behavior on this basis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pathological solutions are very strange (for example their graph is dense in the plane) and can be ruled out by almost any &quot;niceness&quot; hypothesis, as described in the wiki article.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2011-01-02T23:22:40.587" CommentCount="1" />
  <row Id="16177" PostTypeId="2" ParentId="16172" CreationDate="2011-01-02T23:23:26.103" Score="2" ViewCount="" Body="&lt;p&gt;You could prove that the product of a paracompact space and a compact space is paracompact by using the &lt;a href=&quot;http://en.wikipedia.org/wiki/Tube_lemma&quot; rel=&quot;nofollow&quot;&gt;tube lemma&lt;/a&gt;. &lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2011-01-02T23:23:26.103" CommentCount="0" />
  <row Id="16178" PostTypeId="2" ParentId="16175" CreationDate="2011-01-02T23:25:07.053" Score="4" ViewCount="" Body="&lt;p&gt;It is not true that $|ax|=a|x|$; the correct identity is $|ax|=|a||x|$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whether or not adding the hypothesis of continuity is necessary for additive functions to be linear depends on the axiom of choice.  Using a Hamel basis $B$ for $\mathbb{R}^n$ over $\mathbb{Q}$ together with one of its countable subsets $A=\{x_1,x_2,\ldots\}$, you can construct a discontinuous $\mathbb{Q}$ linear map from $\mathbb{R}^n$ to $\mathbb{R}$ by taking the unique $\mathbb{Q}$ linear extension of the function $f:B\to\mathbb{R}$ such that $f(x_k)=k|x_k|$ and $f(B\setminus A)=\{0\}$.  Since $\mathbb{R}$ linear maps between finite dimensional real vector spaces are continuous, such a map cannot be linear.  However, it is consistent with ZF that all additive functions from $\mathbb{R}^n$ to $\mathbb{R}$ are continuous (I am however not knowledgeable in the set theoretic background needed to show this).  &lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2011-01-02T23:39:27.837" LastActivityDate="2011-01-02T23:39:27.837" CommentCount="3" />
  <row Id="16179" PostTypeId="1" AcceptedAnswerId="16190" CreationDate="2011-01-02T23:59:53.067" Score="4" ViewCount="181" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm stuck on the following part of a proof: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $\phi: \mathbb R^m \to \mathbb R^n$ be a function such that $\gamma&amp;#39;(t) := \phi(\gamma(t))$ is smooth for every smooth function $\gamma: \mathbb R \to \mathbb R^m$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to show that $\phi$ is smooth under these assumptions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could someone give me a pointer?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;S.L.&lt;/p&gt;&#xA;" OwnerUserId="3208" LastActivityDate="2011-01-03T02:02:51.237" Title="function is smooth iff the composition with any smooth curve is again smooth" Tags="&lt;real-analysis&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" />
  <row Id="16180" PostTypeId="1" AcceptedAnswerId="18621" CreationDate="2011-01-03T00:00:11.450" Score="0" ViewCount="170" Body="&lt;p&gt;My question can be read on many levels and  so I welcome answers to any reading. The general question is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What is the computational complexity of verifying a proof?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;One way of looking at a computational complexity class (for decision problems) is that is is a set of theories (a theory being a set of theorems) whose theorems can be proven with the resources allotted to the class (any computation of a yes/no by a TM can be viewed as a proof that the answer is yes/no). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some complexity classes are defined using provability (at least NP), that is, given a proof (a witness or certificate), it can be checked (verified) by some other (strictly smaller) class's machine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In most proofs of NP-completeness, showing that a problem is actually a member of NP usually is just showing that there is a witness fr any instance that can be verified by a P machine (or, better said, that the verification problem is a member of P). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question comes from my observation that these verification problems (the one's I remember) seem almost always (I can't remember any others) to be very linear (yes, I'm confounding computational complexity with algorithmic running time). &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Are there some verification algorithms that are (provably) polynomial of a higher degree than 1? Or even super-polynomial for EXP, EXP2, etc?&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Or in general is the complexity of verification, applied to -any- complexity class, always linear (in P or some smaller class)?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(even though my question is about complexity classes, I am also curious about real life proofs where people may rely on lemmas that involve, for example, proving the existence of a particular intersection of hyperplanes (presumably (!) in some subclass of P)&lt;/p&gt;&#xA;" OwnerUserId="1919" LastActivityDate="2011-01-23T13:35:19.490" Title="Complexity of verifying proofs" Tags="&lt;computer-science&gt;&lt;proof-theory&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="16181" PostTypeId="1" AcceptedAnswerId="16193" CreationDate="2011-01-03T00:07:05.680" Score="0" ViewCount="90" Body="&lt;p&gt;I am reading the lecture notes of geometric representation theory: &lt;a href=&quot;http://www.math.harvard.edu/~gaitsgde/267y/catO.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.math.harvard.edu/~gaitsgde/267y/catO.pdf&lt;/a&gt;. I have a question on coroot. In general, if we have a root $\alpha$, then the corresponding coroot is $\check{\alpha}=2\alpha/(\alpha, \alpha)$. But on page 1 of the lecture notes, line -14, it is said that $h_{\check{\alpha}}$ is a coroot. I don't understand this. Further, why $[e_{\alpha}, f_{\alpha}] \neq h_{\alpha}$ but $[e_{\alpha}, f_{\alpha}]$ is proportionate to $h_{\check{\alpha}}$? Thank you very much.&lt;/p&gt;&#xA;" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-03T00:18:12.347" LastActivityDate="2011-01-03T02:57:55.403" Title="questions about coroot." Tags="&lt;algebraic-geometry&gt;&lt;representation-theory&gt;&lt;d-modules&gt;" AnswerCount="1" />
  <row Id="16182" PostTypeId="2" ParentId="16092" CreationDate="2011-01-03T00:41:08.360" Score="1" ViewCount="" Body="&lt;p&gt;The map $C^2 \otimes C^2 \ni v \otimes w \mapsto v w^t$ allows us to identify the space of rank-2 symmetric tensors with the space of symmetric $2\times 2$ matrices, which I'll call $Sym$.  The corresponding action of an element $A \in SU(2)$ on such a matrix $B$ is then $ABA^t$.  The complexification of $su(2)$ is $sl(2,C)$, the space of traceless $2\times 2$ complex matrices.  An isomorphism of these two reps is given by the map&#xA;$$&#xD;&#xA;sl(2,C) \to Sym, B \mapsto B \left(\begin{array}{cc} 0 &amp;amp; 1 \\ -1 &amp;amp; 0 \end{array}\right).&#xD;&#xA;$$&#xA;That this is a map of representations follows from the fact that&#xA;$$&#xD;&#xA;A^{-1} \left(\begin{array}{cc} 0 &amp;amp; 1 \\ -1 &amp;amp; 0 \end{array}\right) = \left(\begin{array}{cc} 0 &amp;amp; 1 \\ -1 &amp;amp; 0 \end{array}\right) A^t &#xD;&#xA;$$&#xA;for $A \in SU(2)$.  This fact is easily verified by using the fact that every element in $SU(2)$ is of the form&#xA;$$&#xD;&#xA;\left(\begin{array}{cc} z &amp;amp; w \\ -\bar w &amp;amp; \bar z \end{array}\right)&#xD;&#xA;$$&#xA;with $|z|^2 + |w|^2 = 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found this isomorphism by just playing around, and I think I just got lucky.  There is probably another one that is less ad-hoc.&lt;/p&gt;&#xA;" OwnerUserId="9" LastActivityDate="2011-01-03T00:41:08.360" CommentCount="1" />
  <row Id="16183" PostTypeId="1" AcceptedAnswerId="16185" CreationDate="2011-01-03T00:43:25.940" Score="4" ViewCount="112" Body="&lt;p&gt;Dealing with US Dollars...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming we know the weights of half-dollars, quarters, dimes, nickels, and pennies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also assuming the weights remain constant (don't change from one quarter to another quarter).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we have a bag of these coins (could include any coin), and we weigh them to get a total weight, can we know how much money we have in the bag?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we cannot know exactly, how close of an estimate can we get?&lt;/p&gt;&#xA;" OwnerUserId="1630" LastActivityDate="2011-01-12T09:57:32.793" Title="Calculating Total Amount of Money Based on Weight" Tags="&lt;puzzle&gt;&lt;counting&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="16184" PostTypeId="1" AcceptedAnswerId="16189" CreationDate="2011-01-03T00:55:41.967" Score="2" ViewCount="85" Body="&lt;p&gt;Hello,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm working my way through Enderton's Elements of Set Theory, and one of the theorems from the book has a final note that is giving me some trouble. The theorem is found on page 60, and is as follows:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Theorem 3Q&lt;/strong&gt; Assume that $R$ is an equivalence relation on $A$ and that $F\colon A\to A$. If $F$ is compatible with $R$, then there exists a unique $\hat{F}\colon A/R\to A/R$ such that &#xA;  $$&#xA;  (*)\qquad \hat{F}([x])=[F(x)]\quad\text{for all $x\in A$.}&#xA;  $$&#xA;  If $F$ is not compatible with $R$, then no such $\hat{F}$ exists. Analogous results apply to functions from $A\times A$ into $A$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For $F\colon A\to A$, we say $F$ is compatible with $R$ if for all $x,y\in A$, &#xA;$$&#xD;&#xA;xRy\implies F(x)RF(y).&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm trying to formulate and prove the mentioned analogous results for $F\colon A\times A\to A$. I imagine the corresponding theorem would be &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Theorem 3Q'&lt;/strong&gt; Assume that $R$ is an equivalence relation on $A$ and that $F\colon A\times A\to A$. If $F$ is compatible with $R$, then there exists a unique $\hat{F}\colon A/R\times A/R\to A/R$ such that &#xA;  $$&#xA;  (**)\qquad \hat{F}([x],[y])=[F(x,y)]\quad\text{for all $x,y\in A$.}&#xA;  $$&#xA;  If $F$ is not compatible with $R$, then no such $\hat{F}$ exists.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;but of course I may be wrong! I suppose the definition of compatibility requires some retooling, and the only sensible formulation I could think of was $F$ and $R$ are compatible if for any $x,y\in A$, &#xA;$$&#xD;&#xA;xRy\implies F(x,y)RF(y,x).&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;This is what I have so far, but as a warning, I don't think it really goes anywhere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Based on this, I try to show that if $F$ is compatible with $R$, then such a $\hat{F}$ exists. Since $(**)$ requires that $(([x],[y]), [F(x,y)])\in\hat{F}$, I try to define $\hat{F}$ as&#xA;$$&#xD;&#xA;\hat{F}=\{(([x],[y]), [F(x,y)])\ |\ x,y\in A\}.&#xD;&#xA;$$&#xA;I want $\hat{F}$ to indeed be a function, so I consider the pairs $(([x],[y]), [F(x,y)])$ and $(([u],[v]), [F(u,v)])$ in $\hat{F}$. I would then like to have the implications&#xA;\begin{align*}&#xA;([x],[y])=([u],[v]) &amp;amp;\implies [x]=[u]\wedge [y]=[v] \newline&#xA;&amp;amp;\implies xRu\wedge yRv\newline&#xA;&amp;amp;\implies F(x,u)RF(u,x)\wedge F(y,v)RF(v,y)\newline&#xA;&amp;amp;\implies [F(x,u)]=[F(u,x)]\wedge [F(y,v)]=[F(v,y)]\newline&#xA;&amp;amp;\implies \underline{\hspace{1in}} \newline&#xA;&amp;amp;\implies [F(x,y)]=[F(u,v)]&#xA;\end{align*}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I don't know what to put in the blank to make the desired implication follows, so perhaps my idea of compatibility or something else is wrong. Can someone please explain how to get to the proper formulation of the analogous results? Thanks!&lt;/p&gt;&#xA;" OwnerUserId="1583" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-03T05:21:50.113" LastActivityDate="2011-01-03T05:25:02.530" Title="Defining compatibility between a function $F\colon A\times A\to A$ and an equivalence relation $R$ on $A$." Tags="&lt;functions&gt;&lt;elementary-set-theory&gt;&lt;relations&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" />
  <row Id="16185" PostTypeId="2" ParentId="16183" CreationDate="2011-01-03T01:12:38.877" Score="4" ViewCount="" Body="&lt;p&gt;According to the U.S. Mint, nickels weigh 5.000 grams and pennies 2.500 grams.  So you generally cannot know exactly how much money is in the bag: for example, 10 grams of coins could be two nickels (10 cents) or four pennies (4 cents).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.usmint.gov/about_the_mint/?action=coin_specifications&quot; rel=&quot;nofollow&quot;&gt;http://www.usmint.gov/about_the_mint/?action=coin_specifications&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2937" LastActivityDate="2011-01-03T01:12:38.877" CommentCount="1" />
  <row Id="16186" PostTypeId="2" ParentId="16184" CreationDate="2011-01-03T01:17:23.300" Score="1" ViewCount="" Body="&lt;p&gt;I discussed compatibility of operations with an equivalence relation in a &lt;a href=&quot;http://math.stackexchange.com/questions/16121/find-all-rings-of-mathbbz2/16157#16157&quot;&gt;post only a few hours old.&lt;/a&gt; Perhaps this algebraic perspective (&lt;em&gt;congruences&lt;/em&gt;) may  provide further motivational background.&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2011-01-03T01:17:23.300" CommentCount="1" />
  <row Id="16187" PostTypeId="2" ParentId="16159" CreationDate="2011-01-03T01:23:46.233" Score="1" ViewCount="" Body="&lt;p&gt;To add to what has already been said, it is entirely up to you to ensure that all the &quot;independent&quot; variables (e.g. mileage, model, years of driving experience, age...) are indeed independent; that tends to be a sticky statistical problem. Once you are sure of which independent variables you actually need, then your function $F$ ought to have the same number of arguments.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus your function would be something like&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$F(\text{mileage},\text{number of accidents},B(\text{gender},\text{age},\text{driving experience}))$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $B$ is the function for owner bias... you get the idea.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Another thing: figuring out how to combine these independent variables all together is your burden, in that you have to experiment on/derive how to properly relate all your variables.&lt;/p&gt;&#xA;" OwnerUserId="5296" LastActivityDate="2011-01-03T01:23:46.233" />
  <row Id="16188" PostTypeId="1" AcceptedAnswerId="16191" CreationDate="2011-01-03T01:30:33.250" Score="3" ViewCount="156" Body="&lt;p&gt;Can someone give me, with proof, an example of a Noetherian ring which has Krull dimension one but is not a Dedekind domain? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think it would also be instructive to see other &quot;near misses.&quot; &lt;/p&gt;&#xA;" OwnerUserId="1491" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2011-01-03T02:57:05.140" LastActivityDate="2011-01-03T02:57:05.140" Title="A Noetherian ring with Krull dimension one which is not a Dedekind domain" Tags="&lt;commutative-algebra&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="16189" PostTypeId="2" ParentId="16184" CreationDate="2011-01-03T01:31:34.450" Score="3" ViewCount="" Body="&lt;p&gt;Your Theorem 3Q' is fine, but the condition you want is is not that $xRy\Longrightarrow F(x,y)RF(y,x)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Going back to the original Theorem 3Q, it might help to think about it this way: You have a map $F\colon A\to A$, and also a map $A\to A/R$. So we have a diagram:&#xA;$$\begin{array}{ccc}&#xD;&#xA;A &amp;amp; \stackrel{F}{\longrightarrow} &amp;amp; A\\&#xD;&#xA;\downarrow &amp;amp; &amp;amp; \downarrow\\&#xD;&#xA;A/R &amp;amp; &amp;amp; A/R&#xD;&#xA;\end{array}$$&#xA;The compatibility condition means that you are able to put a function on the bottom row and get a &lt;strong&gt;commutative diagram&lt;/strong&gt;: any two compositions of functions that start in the same place and end in the same place are equal. This is what gives the condition that two elements of $A$ that map to the same thing &quot;going down&quot; ($a$ and $x$ such that $[a]=[x]$) must map to the same thing if we go &quot;forward then down&quot; (so $F(a)RF(x)$ should hold). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now consider the situation in 3Q'. Now you have a map $A\times A \to A$, and corresponding to it two maps: $A\to A/R$, and also $A\times A\to (A/R)\times(A/R)$ (the latter given by $(x,y)\mapsto ([x],[y])$. We now have:&#xA;$$\begin{array}{ccc}&#xD;&#xA;A\times A &amp;amp; \stackrel{F}{\longrightarrow} &amp;amp; A\\&#xD;&#xA;\downarrow &amp;amp; &amp;amp; \downarrow\\&#xD;&#xA;(A/R)\times(A/R) &amp;amp; &amp;amp; A/R&#xD;&#xA;\end{array}$$&#xA;By analogy, the compatibility condition should be that any two pairs $(a,b)$ and $(x,y)$ that map to the same thing under $A\times A\to (A/R)\times (A/R)$ (that is, if $([a],[b])=([x],[y])$) should map to the same thing if we go &quot;forward and down&quot; (that is, $F(a,b)RF(x,y)$). This should suggest what condition to put down: note that $([a],[b]) = ([x],[y])$ if and only if $[a]=[x]$ and $[b]=[y]$. This is clearly &lt;em&gt;necessary&lt;/em&gt;, then you need to show it will be sufficient.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From this, you should be able to write down a Theorem 3Q'' for an $n$-ary function (or even infinite arity). &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-03T05:25:02.530" LastActivityDate="2011-01-03T05:25:02.530" CommentCount="1" />
  <row Id="16190" PostTypeId="2" ParentId="16179" CreationDate="2011-01-03T01:33:27.113" Score="5" ViewCount="" Body="&lt;p&gt;This was proved by Jan Boman in the paper &quot;Differentiability of a function and of its compositions with functions of one variable&quot;, &lt;a href=&quot;http://www.mscand.dk/issue.php?year=1967&amp;amp;volume=20&amp;amp;issue=&quot; rel=&quot;nofollow&quot;&gt;Math. Scand. 20 (1967)&lt;/a&gt;, 249-268.  (The theorem as stated is for the case $n=1$, but that is no problem as Jason DeVito already mentioned in a comment.)  &lt;a href=&quot;http://www.mscand.dk/article.php?id=1825&quot; rel=&quot;nofollow&quot;&gt;Here&lt;/a&gt;'s an online version, and &lt;a href=&quot;http://www.ams.org/mathscinet-getitem?mr=237728&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;'s the MathSciNet link.  According to the article and review, it had been an unpublished conjecture of Rådström.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2011-01-03T02:02:51.237" LastActivityDate="2011-01-03T02:02:51.237" CommentCount="2" />
  <row Id="16191" PostTypeId="2" ParentId="16188" CreationDate="2011-01-03T01:37:00.500" Score="5" ViewCount="" Body="&lt;p&gt;The ring $\mathbb{Z}[2i]$ is an example. It satisfies all the properties of a Dedekind domain except that it is not integrally closed. (To see that it satisfies these properties, note that it is integral over $\mathbb{Z}$, so the Krull dimension is one, as integral extensions preserve dimension. It is also clearly noetherian.)&lt;/p&gt;&#xA;" OwnerUserId="536" LastActivityDate="2011-01-03T01:37:00.500" CommentCount="2" />
  <row Id="16192" PostTypeId="2" ParentId="16138" CreationDate="2011-01-03T01:47:26.553" Score="5" ViewCount="" Body="&lt;p&gt;Here is a heuristic which I find useful for ruling out easy proofs of the PNT. Consider a set of positive integers $P$ with the following properties: Between $2^{2k-1}$ and $2^{2k}$, there are roughly $a \frac{2^{2k-1}}{(2k-1) \log 2}$ elements of $P$ and, between $2^{2k}$ and $2^{2k+1}$, there are roughly $b \frac{2^{2k}}{(2k) \log 2}$ elements of $P$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if $P$ is the primes, then $a=b=1$. Suppose instead that $a=1 + c$ and $b=1 - c$, for some small constant $c$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then $\prod_{p \in P} (1-p^{-s})^{-1}$ has a simple pole at $s=1$, with residue $1$. The sum $\sum_{p \in P,\ p \leq N} 1/p$ grows like $\log \log N$. And, regarding your specific question, $\prod_{p \in P,\ p \leq N} (1-1/p) \approx 1/\log N$.&lt;sup&gt;1&lt;/sup&gt; So these properties can't distinguish $P$ from the set of primes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the PNT does not hold for $P$. Let $\pi_P(N)$ be the number of elements of $P$ which are $\leq N$. Then, if $N=2^{2k}$, then &#xA;$$\frac{\pi_P(N)}{N/\log N} = (2k) \left( a \frac{2^{2k-1}}{2k-1} + b \frac{2^{2k-2}}{2k-2} + a \frac{2^{2k-3}}{2k-3} + \cdots \right)$$&#xA;$$\approx a \left( \frac{1}{2} + \frac{1}{8} + \cdots \right) + b \left( \frac{1}{4} + \frac{1}{16} + \cdots \right) = (2/3) a + (1/3) b. $$&#xA;Similarly, if $N=2^{2k+1}$, then &#xA;$$\frac{\pi_P(N)}{N/\log N} \approx (2/3) b + (1/3) a.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the ratio of $\pi_P(N)/(N/\log N)$ does not approach a well defined ratio. Any proof of the PNT must use facts about the primes which distinguish them from $P$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;1&lt;/sup&gt; There is also a second issue here. It turns out that &#xA;$$\prod_{p\ \mbox{Prime},\ p \leq N} \left(1-\frac{1}{p} \right) \sim \frac{e^{- \gamma}}{\sum_{n \leq N} 1/n},\ \mbox{not}\ \sim \frac{1}{\sum_{n \leq N} 1/n}.$$&#xA;So you would have to explain why that $e^{- \gamma}$ disappears.&lt;/p&gt;&#xA;" OwnerUserId="448" LastEditorUserId="448" LastEditorDisplayName="" LastEditDate="2011-02-01T12:55:59.083" LastActivityDate="2011-02-01T12:55:59.083" CommentCount="1" />
  <row Id="16193" PostTypeId="2" ParentId="16181" CreationDate="2011-01-03T02:26:30.437" Score="1" ViewCount="" Body="&lt;p&gt;You surely will not get $h_{\alpha^\vee}$  for all choices of $e_\alpha\in\mathfrak n_\alpha$ and $f_\alpha\in\mathfrak n_\alpha^-$, both non-zero, but otherwise arbitrary!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You should read through Jim Humphreys's &lt;em&gt;Introduction to Lie Algebras&lt;/em&gt;, which is amazingly readable.&lt;/p&gt;&#xA;" OwnerUserId="274" LastEditorUserId="274" LastEditorDisplayName="" LastEditDate="2011-01-03T02:57:55.403" LastActivityDate="2011-01-03T02:57:55.403" CommentCount="3" />
  <row Id="16194" PostTypeId="2" ParentId="16188" CreationDate="2011-01-03T02:45:03.610" Score="2" ViewCount="" Body="&lt;p&gt;If by &quot;other near misses&quot; you mean other rings that satisfy two of the three conditions for being a Dedekind domain, another such ring is k[x, y], which is Noetherian (by the Hilbert basis theorem), integrally closed (it is a UFD, and UFDs are integrally closed) but not every prime ideal is maximal (for example (x)).&lt;/p&gt;&#xA;" OwnerUserId="5265" LastActivityDate="2011-01-03T02:45:03.610" />
  <row Id="16195" PostTypeId="2" ParentId="16188" CreationDate="2011-01-03T02:53:37.500" Score="2" ViewCount="" Body="&lt;p&gt;The ring $R_1 = \mathbb{Z} \times \mathbb{Z}$ gives a counterexample to your claim (it is not a domain).  However $\operatorname{Spec}(R_1)$ is a Dedekind scheme, so this is a somewhat cheap counterexample.  The counterexample $R_2 = \mathbb{Z}[t]/(t^2)$ is more serious.&lt;/p&gt;&#xA;" OwnerUserId="299" LastActivityDate="2011-01-03T02:53:37.500" />
  <row Id="16196" PostTypeId="2" ParentId="16138" CreationDate="2011-01-03T03:00:44.573" Score="1" ViewCount="" Body="&lt;p&gt;nice and elegant approach... !!!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Answering this question: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Why do complex numbers (which seem to come from nowhere) make this argument rigorous?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What makes an argument rigorous or not, is still very much debatable.&#xA;I think, making the PNT rigorous meant to have a formulation that will be widely accepted as a &quot;true-proof&quot;, and complex numbers did a very nice job at that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I only wonder what would have happened if the elementary proofs of PNT came earlier, and the complex-analytic proof came later. Would that have created the same awe for the complex-analytic proof?&lt;/p&gt;&#xA;" OwnerUserId="2320" LastActivityDate="2011-01-03T03:00:44.573" CommentCount="1" />
  <row Id="16198" PostTypeId="2" ParentId="14200" CreationDate="2011-01-03T03:13:41.813" Score="2" ViewCount="" Body="&lt;p&gt;Since almost all graphs have a trivial automorphism group, there are asymptotically $2^{n \choose 2}/n!=O(2^{n^2})$ &lt;a href=&quot;http://oeis.org/A000088&quot; rel=&quot;nofollow&quot;&gt;non-isomorphic graphs&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose you could encode each graph using 1 bit (of course you can't actually do this, but this will help give a lower bound on the complexity).  If the input string $n$ is encoded using $O(\log n)$ bits, then merely writing a &quot;1&quot; for each non-isomorphic graph would require at least double-exponential time $O(2^{(2^{\log n})^2})=O(2^{2^{2 \log n}})$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On top of this, you want to return a string from which you can reconstruct the graph, and (possibly) require some form of isomorphism detection.  All in all, this is not going to be easy (no matter how clever your algorithm is).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, you might want to play around with &lt;em&gt;geng&lt;/em&gt;, which comes with &lt;a href=&quot;http://cs.anu.edu.au/~bdm/nauty/&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;nauty&lt;/em&gt;&lt;/a&gt;, which can generate a list of such graphs.  There's probably many other programs as well that I'm not familiar with.&lt;/p&gt;&#xA;" OwnerUserId="139" LastActivityDate="2011-01-03T03:13:41.813" />
  <row Id="16199" PostTypeId="1" CreationDate="2011-01-03T04:26:20.390" Score="2" ViewCount="91" Body="&lt;p&gt;Fix $\epsilon &amp;gt;0$. Suppose $p_{n}(z) = z^{n}+a_{n-1}z^{n-1}+ \cdots + a_{0} \in \mathbb{Z}[x]$ is irreducible and has all positive real roots. Show that independently of $n$ except for finitely many explicitly computable exceptions that $|a_{n-1}| \geq (2-\epsilon)n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sum of the roots is $-\frac{a_{n-1}}{a_n}$ which is $-a_{n-1}$. So $-a_{n-1} &amp;gt;0$ which means that $a_{n-1} \leq 0$. So $|a_{n-1}| = -a_{n-1}$. The problem can then be rephrased as follows: Show that the sum of the roots of the above polynomial is $ \geq (2-\epsilon)n$. Could we use a proof by contradiction? I think irreducibility is an important property here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Source: &lt;em&gt;Computational excursions in analysis and number theory&lt;/em&gt; by P Borwein&lt;/p&gt;&#xA;" OwnerUserId="2548" LastEditorUserId="2548" LastEditorDisplayName="" LastEditDate="2011-01-03T04:53:30.923" LastActivityDate="2011-01-03T04:53:30.923" Title="Lower Bound of a Coefficient" Tags="&lt;analysis&gt;&lt;polynomials&gt;" CommentCount="3" />
  <row Id="16200" PostTypeId="2" ParentId="16172" CreationDate="2011-01-03T04:31:00.513" Score="2" ViewCount="" Body="&lt;p&gt;I think Trevor is right. Here are the details (this is an adaptation of the classical proof that the product of two compact spaces is a compact space that you can find in Munkres, for instance).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $X$ be a paracompact space,  $Y$ a compact one and ${\cal U}$ an open cover of $X \times Y$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For any $x \in X$, the slice $\left\{ x \right\} \times Y$ is a compact space and ${\cal U}$ an open cover of it (as a subspace). So it admits a finite subcover $U_{x,1}, \dots , U_{x,n_x} \in {\cal U}$. Let us call $N_x = U_{x,1} \cup \dots \cup U_{x,n_x}$ their union.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So $N_x$ is an open set that contains the slice $\left\{ x \right\} \times Y$. Because of the tube lemma, there exists an open set $W_x \subset X$ such that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;\left\{ x \right\} \times Y \quad \subset \quad W_x \times Y \quad \subset \quad N_x \ .&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, those $W_x$ form an open cover ${\cal W}$ of $X$. By hypothesis, there is a locally finite refinement ${\cal W}&amp;#39; = \left\{ W_{x_i}\right\}$, $i\in I$ for some index set $I$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the following subcover of ${\cal U}$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;{\cal U}&amp;#39; = \left\{ U_{x_i,j}\right\} \ ,&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with $i\in I$ and $j = 1, \dots , n_{x_i}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let us show that ${\cal U}&amp;#39;$ is a locally finite subcover of ${\cal U}$: take any point $(x,y) \in X \times Y$. By hypothesis, there is a neighborhood $V \subset X$ of $x$ such that $V$ interesects only finitely many of the sets of ${\cal W}&amp;#39;$. Then $V\times Y$ is a neighborhood of $(x,y)$ that intersects only finitely many of the sets of ${\cal U}&amp;#39;$.&lt;/p&gt;&#xA;" OwnerUserId="664" LastEditorUserId="664" LastEditorDisplayName="" LastEditDate="2011-01-03T04:36:14.067" LastActivityDate="2011-01-03T04:36:14.067" CommentCount="3" />
  <row Id="16201" PostTypeId="1" CreationDate="2011-01-03T04:47:02.610" Score="3" ViewCount="117" Body="&lt;p&gt;Define $\mathrm{E}(\mathbb{Z}_{i})$ to be the group of invertible elements of the ring with unity $\mathbb{Z}_{i}$.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Show that $\mathrm{E}(\mathbb{Z}_{mn})$ is isomorphic to $\mathrm{E}(\mathbb{Z}_{m}) \times \mathrm{E}(\mathbb{Z}_{n})$ if and only if $m$ and $n$ are relatively prime.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I see that it is relatively easy to prove that $\mathbb{Z}_{mn}$ is isomorphic to $\mathbb{Z}_{m}\times \mathbb{Z}_{n}$ iff $m$ and $n$ are relatively prime using the Chinese Remainder Theorem. However, is it possible to make an extension of this theorem to prove the above statement? &lt;/p&gt;&#xA;" OwnerUserId="4697" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-03T05:25:25.993" LastActivityDate="2011-01-03T05:25:25.993" Title="Show $\mathrm{E}(\mathbb{Z}_{mn})$ is isomorphic to $\mathrm{E}(\mathbb{Z}_{m})\times\mathrm{E}(\mathbb{Z}_{n})$ if and only if $(m,n)=1$" Tags="&lt;homework&gt;&lt;ring-theory&gt;&lt;elementary-number-theory&gt;" CommentCount="4" FavoriteCount="0" />
  <row Id="16202" PostTypeId="2" ParentId="14878" CreationDate="2011-01-03T04:53:28.177" Score="-1" ViewCount="" Body="&lt;p&gt;My son (third grader) was asked questions very related to this thread. &#xA;Partition a square into 4 or 6 or 7 or 8 or 9 squares.&#xA;All these are possible (although he found it difficult for 6,8 which is besides the point) &#xA;I am asking a more general question out of curiosity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For what positive integers n, can you partition a square into n squares ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is easy to see it is possible &#xA;for squares and for even integers and for odd integers of the form 4m-3, m&gt;2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What about 3,5 and the rest ? If not, why not ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks. &lt;/p&gt;&#xA;" OwnerUserId="5335" LastActivityDate="2011-01-03T04:53:28.177" CommentCount="2" />
  <row Id="16203" PostTypeId="1" AcceptedAnswerId="16205" CreationDate="2011-01-03T05:37:46.220" Score="4" ViewCount="224" Body="&lt;p&gt;I have been learning about sheaves and am thinking about the following problem. Let $F$ and $G$ be sheaves, say of abelian groups, on a space $X$. The sheaf $Hom(F, G)$ is defined by $Hom(F, G)(U)=Mor(F|_U, G|_U)$.  Given a point $p \in X$ and an open set $U$ containing $p$, a morphism $\varphi: F|_U \rightarrow G|_U$ induces a homomorphism on stalks $\phi: F_p \rightarrow G_p$, which is an element of $Hom(F_p, G_p)$.  Thus, by the universal property of direct limits, we have a homomorphism from $Hom(F, G)_p$ to $Hom(F_p, G_p)$.  However, this is not in general injective or surjective.  Why not? An example or a hint leading towards an example would be much appreciated.  I have thought about this for some simple sheaves (such as skyscraper sheaves), but it seems to be true in those cases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am also interested in a more general answer if there is one, i.e. something category theoretic about Hom and direct limits.&lt;/p&gt;&#xA;" OwnerUserId="5265" LastActivityDate="2011-01-03T06:53:26.843" Title="Why doesn't Hom commute with taking stalks?" Tags="&lt;algebraic-geometry&gt;&lt;sheaf-theory&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="2" />
  <row Id="16204" PostTypeId="2" ParentId="16203" CreationDate="2011-01-03T05:53:50.663" Score="1" ViewCount="" Body="&lt;p&gt;This bad behavior is due to the righthand coordinate of Hom not commuting with colimits (direct limits are a special type of &lt;em&gt;filtered colimit&lt;/em&gt;, so this addresses your question), and the lefthand coordinate flips them into limits (this is a general phenomenon in category theory and follows from the definition of the limit (resp. colimit)).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I should also note that you're trying to turn two colimits into one, and I don't really see why you would expect this to work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, the following deduction is valid: $Hom(F_p,G_p)\cong Hom(colim_{p\in U}F(U),G_p)\cong lim_{p\in U}Hom(F(U),G_p)$.  However, this looks like a very different mathematical object than the one you mentioned. Assuming, for the sake of argument, that $F(U)$ is finitely presented for all $U$, we can pull the colimit out front as well, but this doesn't really get us any closer to our stated goal.&lt;/p&gt;&#xA;" OwnerUserId="5319" LastEditorUserId="5319" LastEditorDisplayName="" LastEditDate="2011-01-03T06:06:19.797" LastActivityDate="2011-01-03T06:06:19.797" CommentCount="1" />
  <row Id="16205" PostTypeId="2" ParentId="16203" CreationDate="2011-01-03T06:53:26.843" Score="11" ViewCount="" Body="&lt;p&gt;Here's an answer to your call for examples. Take $p$ to be a non-isolated closed point in your favorite topological space $X$. Let $H$ be a non-trivial abelian group.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;$Hom(F,G)_p\to Hom(F_p,G_p)$ fails to be surjective&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Let $G$ be the constant sheaf $H$, and let $F$ be the skyscraper sheaf at $p$ with stalk $H$. Then $Hom(F,G)$ is the zero sheaf: any section of $F$ is trivial away from $p$, so a homomorphism would take it to a section of $G$ which is trivial away from $p$, but any section of $G$ which is trivial away from $p$ must be trivial, so every section of $F$ must be taken to the trivial section of $G$. So $Hom(F,G)_p=0$, but $Hom(F_p,G_p)=Hom(H,H)\neq 0$.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;$Hom(F,G)_p\to Hom(F_p,G_p)$ fails to be injective&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Let $V=X\setminus p$, and let $F=G$ be the extension by zero of the constant sheaf $H$ on $V$ (i.e. it's the constant sheaf $H$ on $V$ and $F(U)=0$ if $U\not\subseteq V$). Then $F_p=G_p=0$, but $Hom(F,G)(U)$ contains a natural copy of $Hom(H,H)\neq 0$ for any $U$, so $Hom(F,G)_p$ contains a natural copy of $Hom(H,H)\neq 0$, so it's not zero.&lt;/p&gt;&#xA;" OwnerUserId="60" LastActivityDate="2011-01-03T06:53:26.843" CommentCount="4" />
  <row Id="16206" PostTypeId="1" CreationDate="2011-01-03T07:46:11.073" Score="2" ViewCount="128" Body="&lt;p&gt;I need to show that for prime numbers of the form p=4n+1 , x=(2n)! solves the congruence x^2 =-1modp.&#xA;I then need to show this implies p isn't a gaussian prime. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have started to solve this using Wilson's theorem that a number z is prime iff (z-1)!=-1modz. Therefore the endpoint of my proof should be that (p-1)!=-1modp.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As p is of the form p=4n+1, I only need to prove that 4n! is congruent to -1modp. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is my working so far: Starting with the congruence x^2 = -1modp :&#xA;x^2 = -1modp &#xA;=&gt; x^2 + 1 = kp&#xA;=&gt; (x-i)(x+i) = kp&#xA;=&gt; ((2n)!-i)((2n!)+i))=kp&#xA;=&gt; 4n!- 1 =kp &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is where I start to run out of any ideas that seem to get me anywhere.&#xA;Any tips would be greatly appreciated! &lt;/p&gt;&#xA;" OwnerUserId="4697" LastEditorUserId="2548" LastEditorDisplayName="" LastEditDate="2011-01-03T08:18:35.770" LastActivityDate="2011-01-03T08:19:31.727" Title="Show for prime numbers of the form p=4n+1, x=(2n)! solves the congruence x^2=-1 mod p. p is therefore not a gaussian prime." Tags="&lt;number-theory&gt;&lt;homework&gt;&lt;prime-numbers&gt;" AnswerCount="1" />
  <row Id="16207" PostTypeId="2" ParentId="16206" CreationDate="2011-01-03T08:05:10.423" Score="4" ViewCount="" Body="&lt;p&gt;If $p=4n+1$, then the nonzero elements of $\mathbb{Z}_p$ are $\{1,2, \dots, 4n \}$. You can write this as $\{1,2, \dots, 2n, -2n, \dots, -2,-1 \}$. The product of all of these elements is $-1$ from Wilson's Theorem. But this is the same thing as $(1 \cdot 2 \cdots 2n)^{2}(-1)^{2n} = (1 \cdot 2 \cdots 2n)^{2} = ((2n!))^{2}$.    &lt;/p&gt;&#xA;" OwnerUserId="2548" LastEditorUserId="2548" LastEditorDisplayName="" LastEditDate="2011-01-03T08:19:31.727" LastActivityDate="2011-01-03T08:19:31.727" CommentCount="1" />
  <row Id="16208" PostTypeId="1" CreationDate="2011-01-03T08:28:11.223" Score="0" ViewCount="112" Body="&lt;p&gt;Does anyone know how to use imperial units (feet and inches) with Excel? I want to do addition and multiplication with Excel using imperial units so that length and area can be calculated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ex: 12' 8&quot; + 5' 11&quot; = 18' 7&quot;&lt;/p&gt;&#xA;" OwnerUserId="5336" LastActivityDate="2011-01-03T08:28:11.223" ClosedDate="2011-01-03T09:43:52.607" Title="How to use imperial units (feet and inches) with Microsoft Excel" Tags="&lt;math-software&gt;" CommentCount="4" FavoriteCount="0" />
  <row Id="16209" PostTypeId="2" ParentId="15739" CreationDate="2011-01-03T08:45:28.950" Score="1" ViewCount="" Body="&lt;p&gt;I would say yes, it is applicable. The formula basically tells you that the volume is the sum of volumes of the two &quot;component&quot; polyhedrons, doesn't it?&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2011-01-03T08:45:28.950" CommentCount="4" />
  <row Id="16210" PostTypeId="1" AcceptedAnswerId="16213" CreationDate="2011-01-03T10:20:01.940" Score="1" ViewCount="184" Body="&lt;p&gt;Are there such numbers $a$ and $b$ that if $a &amp;lt; b$, then $a &amp;gt; b$ ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="5340" LastActivityDate="2011-01-03T12:08:30.753" Title="Conditional existence of real numbers" Tags="&lt;number-theory&gt;" AnswerCount="4" CommentCount="1" />
  <row Id="16211" PostTypeId="2" ParentId="16210" CreationDate="2011-01-03T10:28:18.610" Score="2" ViewCount="" Body="&lt;p&gt;No.  The relation less than (as opposed to less than or equal to) is defined as strict, so $(a \lt b) \implies \neg (b \lt a)$&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-01-03T10:28:18.610" CommentCount="3" />
  <row Id="16212" PostTypeId="2" ParentId="16210" CreationDate="2011-01-03T10:30:54.110" Score="2" ViewCount="" Body="&lt;p&gt;I don't really think so. The condition $a &amp;lt; b$ implies that $a \ngeq b$ because $&amp;lt;$ is a total order on $\mathbb{R}$. If you're not familiar with orderings, they are just relations that are reflexive, anti-simmetric and transitive (see &lt;a href=&quot;http://en.wikipedia.org/wiki/Order_theory&quot; rel=&quot;nofollow&quot;&gt; Wikipedia&lt;/a&gt;. A set is said to be totally ordered if every element of the set can be compared to another with the ordering given. &lt;/p&gt;&#xA;" OwnerUserId="1415" LastEditorUserId="1415" LastEditorDisplayName="" LastEditDate="2011-01-03T10:42:03.597" LastActivityDate="2011-01-03T10:42:03.597" CommentCount="2" />
  <row Id="16213" PostTypeId="2" ParentId="16210" CreationDate="2011-01-03T10:51:42.413" Score="11" ViewCount="" Body="&lt;p&gt;Yes. Take for instance a = 1, b = 0. (Or perhaps this isn't what you meant?)&lt;/p&gt;&#xA;" OwnerUserId="1508" LastActivityDate="2011-01-03T10:51:42.413" CommentCount="4" />
  <row Id="16214" PostTypeId="1" CreationDate="2011-01-03T11:00:41.513" Score="0" ViewCount="113" Body="&lt;p&gt;Hi, how can I divide for example $\frac{x^2+1}{2x+1}$ in $\frac{\mathbb{Z}_3[x]}{x^3+1}$? It's like normal polynomial dividing but here I got in first step $\frac{x}{2} (\frac{x^2}{2x}=\frac{x}{2}$).What is the modulo value of this statement? I need it for solve a linear equation solve, it's part of the Euclidan algorithm..but here I am stuck and I don't know what to do..Right answer is 2x+2 with remainder 2..But I don't know WHY:D&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please help..And please be patience for my poor English and absence of right formatting, today I've an exam so I am a bit hurry:D&lt;/p&gt;&#xA;" OwnerUserId="5342" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-04T17:40:26.150" LastActivityDate="2011-04-03T18:39:15.380" Title="Dividing in $\mathbb{Z}_m[x]/p(x)$" Tags="&lt;polynomials&gt;&lt;discrete-mathematics&gt;&lt;modular-arithmetic&gt;" AnswerCount="1" CommentCount="7" />
  <row Id="16215" PostTypeId="1" AcceptedAnswerId="16218" CreationDate="2011-01-03T12:02:10.640" Score="2" ViewCount="75" Body="&lt;p&gt;I have a simple question on a Lipschitz flow. Recall that a flow on a non-empty set $X$ is a map $\Phi: \mathbb R \times X \to X$ satisfying&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Phi(0,x) = x \text{ and } \Phi(s, \Phi(t, x)) = \Phi(s + t, x)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let $X$ be a Banach space and $F$ a Lipschitz-function on $X$. Now it is well known that there exists for every $x \in X$ a unique function $u_x:\mathbb R \to X$ which is continuously strongly differentiable and satisfies:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$u_x'(t) = F(u_x(t)) \text{ where $x \in \mathbb R$ and } u_x(0) = x.$$&#xA;Where $u&amp;#39;$ is the strong derivative in the Banach space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we can set $\Phi(t,x) = u_x(t)$. This should be a flow, why is that? The first property is easy to verify, but the second one requires us to show that $u_{u_x(t)}(s) = u_x(s + t)$. Can someone give a hint how I obtain this? I'm probably missing something simple.&lt;/p&gt;&#xA;" OwnerUserId="1120" LastEditorUserId="723" LastEditorDisplayName="" LastEditDate="2011-01-03T14:28:14.810" LastActivityDate="2011-01-03T14:28:14.810" Title="Lipschitz flows on Banach spaces." Tags="&lt;differential-equations&gt;&lt;banach-spaces&gt;" AnswerCount="1" />
  <row Id="16216" PostTypeId="1" AcceptedAnswerId="16278" CreationDate="2011-01-03T12:07:13.430" Score="2" ViewCount="184" Body="&lt;p&gt;If $A$ can complete a job in $6$ days. $B$ can complete it in $10$ days.If they work on alternate days, in how many days the job would be completed ? Does it depend on who ($A$ or $B$) starts the job?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The solution suggested in my module is to find the work completed by both of them in $2$ days and if this is not an integer then the work will be completed faster if the faster worker starts the job. But I am not sure how/why this works.&lt;/p&gt;&#xA;" OwnerUserId="2109" LastActivityDate="2011-01-04T00:54:34.193" Title="Problem on working together (Time and Work)" Tags="&lt;algebra-precalculus&gt;" AnswerCount="2" />
  <row Id="16217" PostTypeId="2" ParentId="16210" CreationDate="2011-01-03T12:08:30.753" Score="3" ViewCount="" Body="&lt;p&gt;@TonyK gave the correct answer, but I'll clarify why it's true.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're aware of &lt;a href=&quot;http://en.wikipedia.org/wiki/Logical_implication&quot; rel=&quot;nofollow&quot;&gt;Logical implication&lt;/a&gt;, you should know that a statement is false only in the case the first operand (&lt;em&gt;premise&lt;/em&gt;) is true and the second operand (&lt;em&gt;conclusion&lt;/em&gt;) is false, in other cases the statement is true.&lt;/p&gt;&#xA;" OwnerUserId="5340" LastActivityDate="2011-01-03T12:08:30.753" CommentCount="5" />
  <row Id="16218" PostTypeId="2" ParentId="16215" CreationDate="2011-01-03T12:10:10.553" Score="2" ViewCount="" Body="&lt;p&gt;This follows from uniqueness of solutions to the initial value problem&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$v_x&amp;#39;(t) = F(v_x(t)),\qquad\qquad v_x(0) = u_x(s).$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For every $s\in \mathbb R$, $u_{u_x(t)}(s)$ and $u_x(s + t)$ are both solutions with the same initial data at $t=0$.&lt;/p&gt;&#xA;" OwnerUserId="723" LastActivityDate="2011-01-03T12:10:10.553" CommentCount="1" />
  <row Id="16219" PostTypeId="1" CreationDate="2011-01-03T12:37:43.100" Score="1" ViewCount="94" Body="&lt;p&gt;I am looking for hints on how to solve following problems:&lt;br&gt;&#xA;1. Show that in every simple graph, there is a path of length at least $\frac{2m}{n}$.&lt;br&gt;&#xA;2. In $G$ there is a cycle $C$ and a path of length $k$, that connects two vertices from $C$. Show that there is a cycle of length at least $\sqrt{k}$ in $G$.&lt;br&gt;&#xA;3. For all $k$, give an example of graph $G$ with longest cycle $C$ of length shorter than $4\sqrt{k}$ and which has a path of length $k$ connecting two vertices from $C$.&lt;br&gt;&#xA;Can anyone help? In the first one I tried induction but it didn't work, as for the other two I have no idea how to start.   &lt;/p&gt;&#xA;" OwnerUserId="5343" LastActivityDate="2011-01-03T19:53:42.370" Title="Approximations of cycle/path lengths in graphs" Tags="&lt;graph-theory&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="2" />
  <row Id="16221" PostTypeId="2" ParentId="16216" CreationDate="2011-01-03T12:49:29.083" Score="1" ViewCount="" Body="&lt;p&gt;If a is the part of the total amount of work that A gets done per day (i.e. 1/6=0.1666...) and b is the part of the work that B gets done per day (i.e. 1/10=0.1), then a+b is the amount of work that gets done every two days.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;No matter who starts, the same amount of work will be done at the end of every two days, and the total amount of work done will look the same for every two-day cycle as well so we need only look at the two first days to see how the accumulated amount of work will progress.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the interval between every two days, the accumulated amount of work done depends on who starts. If f(t) is a function describing the work done when A works first - then B, and g(t) is a function descibing the work done when B starts first - then A (t being the number of days), we can for instance write them like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f (t) = \begin{matrix}&#xD;&#xA;a \cdot t &amp;amp; , t \in [0,1] \\&#xD;&#xA;a + b \cdot t &amp;amp; , t \in (1,2]&#xD;&#xA;\end{matrix}$&#xA;&lt;br/&gt;&#xA;&lt;br/&gt;&lt;br/&gt;&#xA;$g (t) = \begin{matrix}&#xD;&#xA;b \cdot t &amp;amp; , t \in [0,1] \\&#xD;&#xA;b + a \cdot t &amp;amp; , t \in (1,2]&#xD;&#xA;\end{matrix}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;($t \in [0,1]$ meaning &quot;for all times between 0 and 1&quot;, and likewise $t \in (1,2]$ &quot;for all times between 1 and 2&quot;.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By inspection, we see that as expected g(0) = f(0), and g(2) = f(2). However, in the interval between 0 and 2, the function beginning with the fastest worker (i.e. f, since A is fastest) will always be greater than the other. So for any given t that isn't a multiple of 2, more work will have been done if the fastest worker works the first day.&lt;/p&gt;&#xA;" OwnerUserId="5344" LastEditorUserId="5344" LastEditorDisplayName="" LastEditDate="2011-01-03T23:51:47.563" LastActivityDate="2011-01-03T23:51:47.563" CommentCount="3" />
  <row Id="16223" PostTypeId="2" ParentId="9302" CreationDate="2011-01-03T13:33:23.740" Score="2" ViewCount="" Body="&lt;p&gt;Here is a simple explanation not necessarily from linear algebra. We have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\|A\|_2=\max_{\|x\|=1}\|Ax\|$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\|\cdot\|$ is simple euclidean norm. This is a constrained optimisation problem with Lagrange function:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$L(x,\lambda)=\|Ax\|^2-\lambda(\|x\|^2-1)=x&amp;#39;A^2x-\lambda(x&amp;#39;x-1)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;here I took squares which do not change anything, but makes the following step easier.&#xA;Taking derivative with respect to $x$ and equating it to zero we get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$A^2x-\lambda x=0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the solution for this problem is the eigenvector of $A^2$. Since $A^2$ is symmetric, all its eigenvalues are real. So $x&amp;#39;A^2x$  will achieve maximum on set $\|x\|^2=1$ with maximal eigenvalue of $A^2$. Now since $A$ is symmetric it admits representation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$A=Q\Lambda Q&amp;#39;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with $Q$ the orthogonal matrix and $\Lambda$ diagonal with eigenvalues in diagonals. For $A^2$ we get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$A^2=Q\Lambda^2 Q&amp;#39;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so the eigenvalues of $A^2$ are squares of eigenvalues of $A$. The norm $\|A\|_2$ is the square root taken from maximum $x&amp;#39;A^2x$ on $x&amp;#39;x=1$, which will be the square root of maximal eigenvalue of $A^2$ which is the maximal absolute eigenvalue of $A$.&lt;/p&gt;&#xA;" OwnerUserId="4742" LastActivityDate="2011-01-03T13:33:23.740" />
  <row Id="16224" PostTypeId="1" AcceptedAnswerId="16226" CreationDate="2011-01-03T13:37:48.860" Score="1" ViewCount="187" Body="&lt;p&gt;Quite simply, I need a function that will produce a graph similar to the one below. Important to note:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Y switches from negative to positive at X=1/3&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Y ranges from -0.1 to +0.1, not to and from infinity&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) Y should be very close to 0 by X=1/6 and X=3/6.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4) It should have curves similar to those in the picture &lt;em&gt;(Mod note: the original image link is dead, for context I'm replacing it with the graph of the function given in the accepted answer.)&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/n73sC.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A massive thank-you to whomever can figure this one out.&lt;/p&gt;&#xA;" OwnerUserId="5345" LastEditorUserId="1543" LastEditorDisplayName="" LastEditDate="2011-05-11T13:26:04.357" LastActivityDate="2011-05-11T13:26:04.357" Title="What function produces the pictured graph?" Tags="&lt;functions&gt;" AnswerCount="3" />
  <row Id="16225" PostTypeId="2" ParentId="16224" CreationDate="2011-01-03T13:43:24.007" Score="3" ViewCount="" Body="&lt;p&gt;Do you need continuity when you go through $x=\frac{1}{3}$?  If not, you can use&#xA;$y=0.1 \exp(-a(x-1/3))$ for $x \gt \frac{1}{3}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$y=-0.1 \exp(a(x-1/3))$ for $x \lt \frac{1}{3}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pick $a$  to make it fall as fast as you want&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-01-03T13:43:24.007" CommentCount="1" />
  <row Id="16226" PostTypeId="2" ParentId="16224" CreationDate="2011-01-03T14:02:46.257" Score="3" ViewCount="" Body="&lt;p&gt;I would suggest&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac1{5\pi}\mathrm{arccot}\left(c\left(x-\frac13\right)\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and adjust the value of $c$ to speed up or slow the rate of decrease.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Yet another alternative:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac1{10}\tanh\left(\frac1{c\left(x-\frac13\right)}\right)$$&lt;/p&gt;&#xA;" OwnerUserId="5296" LastEditorUserId="5296" LastEditorDisplayName="" LastEditDate="2011-01-03T14:36:27.303" LastActivityDate="2011-01-03T14:36:27.303" CommentCount="5" />
  <row Id="16227" PostTypeId="2" ParentId="16224" CreationDate="2011-01-03T14:08:46.477" Score="0" ViewCount="" Body="&lt;p&gt;Using the &lt;a href=&quot;http://en.wikipedia.org/wiki/Heaviside_step_function&quot; rel=&quot;nofollow&quot;&gt;Heaviside step function&lt;/a&gt; $H$, the following seems to be the simplest choice&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$y=H(x-\frac13)(1-H(x-\frac12))(x-\frac12)^2-(1-H(x-\frac13))H(x-\frac16)(x-\frac16)^2$&lt;/p&gt;&#xA;" OwnerUserId="2473" LastEditorUserId="2473" LastEditorDisplayName="" LastEditDate="2011-01-03T14:15:13.977" LastActivityDate="2011-01-03T14:15:13.977" />
  <row Id="16228" PostTypeId="1" CreationDate="2011-01-03T14:18:59.690" Score="1" ViewCount="224" Body="&lt;p&gt;$M_n$ is a $n\times n$ matrix with $M_{n+1}=\begin{pmatrix}M_n &amp;amp; a_n \\ b_n^T &amp;amp; c_n\end{pmatrix}$ and $a_n, b_n, c_n \to 0$ for $n\to \infty$. Is this sufficient to state $$ \lim_{n\to\infty}(M_n^{-1}) = (\lim_{n\to\infty}M_n)^{-1} ?$$&lt;/p&gt;&#xA;" OwnerUserId="163" LastActivityDate="2011-02-25T08:05:01.543" Title="Does $M_n^{-1}$ converge for a series of growing matrices $M_n$?" Tags="&lt;linear-algebra&gt;&lt;matrices&gt;&lt;convergence&gt;&lt;inversion&gt;" CommentCount="12" FavoriteCount="1" />
  <row Id="16229" PostTypeId="1" AcceptedAnswerId="16236" CreationDate="2011-01-03T14:23:03.710" Score="3" ViewCount="158" Body="&lt;p&gt;Hi all&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am trying to calculate the radius of convergence of &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sum_{n=1}^\infty \frac{n^n}{(n!)^2}z^n$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with $z \in \mathbb{C}$. According to WolframAlpha, both the root and the ratio test are conclusive, indicating convergence. However at the root test, I am stuck at: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(For convenience I drop the limes superior in the following equations, but it should be in front of every expression.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sqrt[n]{\frac{n^n}{(n!)^2}}=\frac{n}{\sqrt[n]{n!}\sqrt[n]{n!}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there some estimate of $\sqrt[n]{n!}$ I should know of? For the ratio test, I think I get a reasonable solution:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{\frac{(n+1)^{n+1}}{(n+1)!^2}}{\frac{n^n}{n!^2}}=\frac{(n+1)^{n+1}n!^2}{(n+1)!^2n^n} = \frac{(n+1)^{n-1}}{n^n} = \frac{1}{n}(1+\frac{1}{n})^n = \frac{e}{n}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which results in a radius of convergence of $\infty$. I should get the same result with the root test, shouldn't I? After all the root test is stronger.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hopefully someone can point out of how to continue with the root test.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ftiaronsem&lt;/p&gt;&#xA;" OwnerUserId="3188" LastActivityDate="2011-01-03T15:47:06.103" Title="ratio vs. root test for $\sum_{n=1}^\infty \frac{n^n}{(n!)^2}z^n$" Tags="&lt;analysis&gt;&lt;sequences-and-series&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="16230" PostTypeId="1" CreationDate="2011-01-03T14:23:29.527" Score="0" ViewCount="117" Body="&lt;p&gt;I am sorry that I am an engineering student rather then math student, so my problem statement may not be so specific. Suppose I have a signal $f(x)$, I know the minimum difference between $f(x)$ and the signal which is composed by a series of Gaussian distribution $g(x)$ with known mean and variance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, the problem is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;minimize $t$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;subject to $||f(x) - i(x)*g(x)||_2 &amp;lt; t$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $*$ is convolution operator and $t$ and $i(x)$ are the variables in this problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the problem is not formed so well, if there is any unclear, please ask me. Thanks!&lt;/p&gt;&#xA;" OwnerUserId="5347" LastEditorUserId="5347" LastEditorDisplayName="" LastEditDate="2011-01-05T14:51:58.867" LastActivityDate="2011-06-04T17:11:10.150" Title="Convex optimization for convolution problem" Tags="&lt;convex-analysis&gt;&lt;convex-optimization&gt;&lt;numerical-optimization&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" />
  <row Id="16231" PostTypeId="2" ParentId="16229" CreationDate="2011-01-03T14:26:26.737" Score="3" ViewCount="" Body="&lt;p&gt;I'd use &lt;a href=&quot;http://mathworld.wolfram.com/StirlingsApproximation.html&quot; rel=&quot;nofollow&quot;&gt;Stirling&lt;/a&gt; here.&lt;/p&gt;&#xA;" OwnerUserId="5296" LastActivityDate="2011-01-03T14:26:26.737" CommentCount="6" />
  <row Id="16232" PostTypeId="1" AcceptedAnswerId="16235" CreationDate="2011-01-03T14:30:08.447" Score="6" ViewCount="294" Body="&lt;p&gt;Given a natural number $n$, can we completely characterize when $n$ and $2n$ are each a sum of two squares?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example: $446,382,709=(13010)^{2}+(16647)^{2}$ and $892,765,418=2(446,382,709)=(3637)^{2}+(29657)^{2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do these imply, perhaps, that 446,782,709 is the hypotenuse of a primitive pythagorean triple?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I found this question on a slip of paper while cleaning my office. It turns out that a trivial algebraic identity resolves the question...beyond Euler's characterization of when a natural number is the sum of two squares. Is the question nontrivial if we replace $2n$ by $3n$ above?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: The modification is trivial, too. This is my fault, as I didn't think about my question before posting it. (This was a problem found on a slip of paper in my office. I posted it because it looked like a fun problem for students. This is certainly not the best forum for such things!!!)&lt;/p&gt;&#xA;" OwnerUserId="941" LastEditorUserId="941" LastEditorDisplayName="" LastEditDate="2011-03-09T01:02:36.567" LastActivityDate="2011-03-09T01:02:36.567" Title="When are $n$ and $2n$ both sums of two squares?" Tags="&lt;number-theory&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="16233" PostTypeId="2" ParentId="16159" CreationDate="2011-01-03T14:30:20.393" Score="0" ViewCount="" Body="&lt;p&gt;(Note to webmaster: The Add Comment function doesn't seem to be working, so I am submitting another answer.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Varuman, if you are going to look at several difficult to quantify variables, the statistical approach (my Blue Book example) is widely applicable. If you are determined to get a formula, look up regression analysis. It presents a number of techniques for entering data points and coming up a formula of best fit. Of course, you have to have some idea of what the formula will look like and what the variables are. You can assume, for example, that the function is a linear combination of the independent variables, e.g. y = a0+ a1*x1 + a2*x2 +... where y and the xi are variables for which you enter values and the ai are parameters that the software will calculate. &lt;/p&gt;&#xA;" OwnerUserId="3515" LastEditorUserId="3515" LastEditorDisplayName="" LastEditDate="2011-01-03T15:01:35.377" LastActivityDate="2011-01-03T15:01:35.377" />
  <row Id="16234" PostTypeId="2" ParentId="16232" CreationDate="2011-01-03T14:39:31.130" Score="11" ViewCount="" Body="&lt;p&gt;$2(a^2 + b^2) = (a + b)^2 + (a - b)^2$.  And, of course, the classification of numbers which are the sums of two squares is well-known (it is necessary and sufficient that if $p \equiv 3 \mod 4$, then the largest $k$ such that $p^k | n$ is even).  &lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2011-01-03T18:16:50.460" LastActivityDate="2011-01-03T18:16:50.460" CommentCount="1" />
  <row Id="16235" PostTypeId="2" ParentId="16232" CreationDate="2011-01-03T14:43:14.153" Score="16" ViewCount="" Body="&lt;p&gt;A natural is the sum of two squares iff every prime of the form $\rm\:4\:k+3$ occurs to even power in its prime factorization (iff the same is true for $\rm\:2\:n\:$). Note $\rm\ n = x^2 + y^2\ \Rightarrow\ 2\:n = (x+y)^2 + (x-y)^2$ which arises by compositions of forms, or, in linear form, with the norm $\rm\ N(a+b\ i)\ =\ a^2 + b^2 $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\rm 2\ (x^2+y^2)\ =\ N(1+i)\ N(x-y\ i)\ =\ N((1+i)\ (x-y\ i))\ =\ N(x+y + (x-y)\ i) $$&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2011-01-03T14:58:29.343" LastActivityDate="2011-01-03T14:58:29.343" CommentCount="6" />
  <row Id="16236" PostTypeId="2" ParentId="16229" CreationDate="2011-01-03T14:46:54.040" Score="7" ViewCount="" Body="&lt;p&gt;You can use $n! &amp;gt; (\frac{n}{e})^n$. See &lt;a href=&quot;http://en.wikipedia.org/wiki/Factorial&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; for an elementary proof of this result.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Actually, this result can be easily proved using induction.&lt;/p&gt;&#xA;" OwnerUserId="2810" LastEditorUserId="2810" LastEditorDisplayName="" LastEditDate="2011-01-03T15:47:06.103" LastActivityDate="2011-01-03T15:47:06.103" CommentCount="1" />
  <row Id="16239" PostTypeId="2" ParentId="16214" CreationDate="2011-01-03T15:38:49.403" Score="2" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\rm\ \ 3 \equiv 0\ \Rightarrow\ 1 + 2\ x \:\equiv\: 1-x\:.\ $ Now $\rm\ x^3\:\equiv\:-1\ \Rightarrow\ x^2+1\ \equiv\ x^2 -x^3\ \equiv\ x^2\ (1-x)\ $  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore $\rm\ (x^2+1)/(1-x)\ \equiv\ \ldots$&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2011-01-03T15:57:18.927" LastActivityDate="2011-01-03T15:57:18.927" />
  <row Id="16240" PostTypeId="2" ParentId="15824" CreationDate="2011-01-03T15:42:08.900" Score="1" ViewCount="" Body="&lt;p&gt;It &lt;i&gt;is&lt;/i&gt; orthogonal with respect to the given inner product:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;sage: R.&amp;lt;a,b,c,d&amp;gt; = QQ['a,b,c,d']&lt;br&gt;&#xA;  sage: R2 = R.quotient_ring(a*d-b*c-1)&lt;br&gt;&#xA;  sage: A = Matrix(R2,[[a^2,a*b,b^2],[2*a*c, a*d+b*c, 2*b*d],[c^2,c*d,d^2]])&lt;br&gt;&#xA;  sage: J = Matrix([[0,0,2],[0,-1,0],[2,0,0]])&lt;br&gt;&#xA;  sage: A.transpose()*J*A == J&lt;br&gt;&#xA;  True&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4946" LastActivityDate="2011-01-03T15:42:08.900" CommentCount="3" />
  <row Id="16241" PostTypeId="1" CreationDate="2011-01-03T16:10:11.137" Score="1" ViewCount="194" Body="&lt;p&gt;Hello,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given a 2D picture of a face, how is it possible to measure the distance between two different points on the surface of the face?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Joel&lt;/p&gt;&#xA;" OwnerUserId="5349" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-03T17:29:01.537" LastActivityDate="2011-01-04T19:17:43.280" Title="Distance between points on a face" Tags="&lt;spherical-geometry&gt;" AnswerCount="4" CommentCount="1" />
  <row Id="16243" PostTypeId="2" ParentId="16241" CreationDate="2011-01-03T16:22:08.600" Score="2" ViewCount="" Body="&lt;p&gt;Use a &lt;a href=&quot;http://activitiesforlearning.com/ProductImages/4-in1-Ruler.jpg&quot; rel=&quot;nofollow&quot;&gt;ruler&lt;/a&gt;. If your needs are more specific, perhaps you could explain them?&lt;/p&gt;&#xA;" OwnerUserId="1508" LastActivityDate="2011-01-03T16:22:08.600" CommentCount="3" />
  <row Id="16244" PostTypeId="1" AcceptedAnswerId="16295" CreationDate="2011-01-03T16:22:53.533" Score="10" ViewCount="532" Body="&lt;p&gt;I have been thinking on and off about a problem for some time now. It is inspired by an exam problem which I solved but I wanted to find an alternative solution. The object was to prove that some sequence of functions converges weakly to zero in $L^2$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I managed to show (with some help) that the limit $f$ (of a subsequence) satisfies $\int_0^x f \ dm=0$ for all $x&amp;gt;0 $. From this I want to conclude that $f=0$ a.e. I can do this with the fundamental theorem of calculus in its Lebesgue version but there ought to be a more elementary proof.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can someone here help me out? &lt;/p&gt;&#xA;" OwnerUserId="4623" LastActivityDate="2011-01-09T12:38:51.290" Title="If $\int_0^x f \ dm$ is zero everywhere then $f$ is zero almost everywhere" Tags="&lt;integral&gt;" AnswerCount="5" CommentCount="0" FavoriteCount="1" />
  <row Id="16245" PostTypeId="2" ParentId="16244" CreationDate="2011-01-03T16:28:52.790" Score="1" ViewCount="" Body="&lt;p&gt;If $F(x)=\int_0^x f(t) dt=0$ everywhere, then $F&amp;#39;(x)=0$ for all $x$. Since $f$ is locally integrable, $F&amp;#39;(x)=f(x)$ almost everywhere. Hence $f(x)=0$ almost everywhere.&lt;/p&gt;&#xA;" OwnerUserId="3249" LastActivityDate="2011-01-03T16:28:52.790" CommentCount="3" />
  <row Id="16246" PostTypeId="1" AcceptedAnswerId="16249" CreationDate="2011-01-03T16:32:11.347" Score="8" ViewCount="297" Body="&lt;p&gt;One of my homework questions was to prove, from the axioms of ZF only, that a countable union of countable sets does &lt;em&gt;not&lt;/em&gt; have cardinality $\aleph_2$. My solution shows that it does not have cardinality $\aleph_n$, where $n$ is &lt;em&gt;any&lt;/em&gt; non-zero ordinal (not necessarily finite). I have a sneaking suspicion that my solution is actually invalid, but I can't find any reference which invalidates my conclusion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have read that it is provable in ZF that there are no cardinals $\kappa$ such that $\aleph_0 &amp;lt; \kappa &amp;lt; \aleph_1$, but I believe the conclusion of my proof does not preclude the possibility that the cardinality is incomparable to $\aleph_1$ or some such. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the weakest point in my solution is where I claim that the supremum of a countable set of countable ordinals is again countable. This is true, of course, but it sounds uncomfortably close to the claim &quot;a countable union of countable sets is countable&quot;, which is well-known to be unprovable in ZF. Can anybody confirm that the ordinal version is provable in ZF though? If not, I think I can weaken the claim to &quot;the supremum of any set of countable ordinals is at most $\omega_1$&quot;, and this establishes the weaker result that the cardinality of a countable union of countable sets is not $\aleph_n$ for any ordinal $n \ge 2$.&lt;/p&gt;&#xA;" OwnerUserId="5191" LastActivityDate="2011-01-03T17:15:59.990" Title="The cardinality of a countable union of countable sets, without the axiom of choice" Tags="&lt;homework&gt;&lt;set-theory&gt;&lt;cardinals&gt;&lt;axiom-of-choice&gt;&lt;ordinals&gt;" AnswerCount="1" />
  <row Id="16247" PostTypeId="1" CreationDate="2011-01-03T16:50:37.833" Score="0" ViewCount="216" Body="&lt;p&gt;Given a map $f:B^n \to S^n$, where $B^n$ is the unit ball and $S^n$ is the unit sphere, is it true that the degree of $f|_{S^n}$ is always 0, where $f_{S^n}$ is the restriction of $f$ to $S^n$? If so, why? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="5352" LastActivityDate="2011-01-03T22:43:49.697" Title="Degree of maps on the sphere" Tags="&lt;algebraic-topology&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="16248" PostTypeId="1" AcceptedAnswerId="16250" CreationDate="2011-01-03T16:59:10.037" Score="3" ViewCount="77" Body="&lt;p&gt;Is it 'expected' that PSL(2,p) contains a cyclic subgroup of order $\tfrac{p+1}{2}$? I read such an element exists and generates a &quot;nonsplit torus&quot;, but I'm not sure what that means. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there an easy explanation why such an element should exist and have a cycle structure $\tfrac{p+1}{2},\tfrac{p+1}{2}$ in its action on the projective line?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Edit: On a sidenote, it is on the other hand obvious that PSL(2,p) has cyclic subgroups of order (p-1)/2 and p, generated by &#xA; $(\begin{smallmatrix}r&amp;amp;0\\0&amp;amp;\tfrac{1}{r}\end{smallmatrix})$ and $(\begin{smallmatrix}1&amp;amp;1\\ 0&amp;amp;1\end{smallmatrix})$ &#xA;where r generates $GF(p)^*$ )&lt;/p&gt;&#xA;" OwnerUserId="5189" LastEditorUserId="5189" LastEditorDisplayName="" LastEditDate="2011-01-03T19:32:49.963" LastActivityDate="2011-01-03T19:32:49.963" Title="Cyclic subgroup of PSL" Tags="&lt;group-theory&gt;" AnswerCount="1" />
  <row Id="16249" PostTypeId="2" ParentId="16246" CreationDate="2011-01-03T17:01:06.593" Score="11" ViewCount="" Body="&lt;p&gt;$\omega_1$ can be a countable union of countable ordinals, i.e., the sup of a countable set of countable ordinals. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This consistency result was one of the first found with forcing. It was announced in S. Feferman-A. Levy, &quot;Independence results in set theory by Cohen's method II&quot;, Notices Amer Math Soc., 10, (1963) 593.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The result is that it is consistent with ZF that ${\mathbb R}$ is a countable union of countable sets. From this, it follows easily that $\omega_1$ also has this property. A proof can be found in Jech's book on the Axiom of Choice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is that, as you suspect, &quot;The supremum of a countable set of countable ordinals&quot; cannot be proved without some choice to be countable. The issue is that although we know that each countable ordinal is in bijection with $\omega$, there is no uniform way of picking for each countable ordinal one such bijection. Now, you need this to run the usual proof that a countable union of countable sets is countable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, things can be worse: Gitik showed that it is consistent with ZF that &lt;em&gt;every&lt;/em&gt; infinite (well-ordered) cardinal has cofinality $\omega$. (&quot;All uncountable cardinals can be singular&quot;, Israel J. Math, 35, (1980) 61-88.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, one can check that a countable union of countable sets of ordinals must have size at most $\omega_1$ (which is essentially what your HW is asking to verify). So, in Gitik's model, $\omega_2$ is a countable union of countable unions of countable ordinals, but not a countable union of countable ordinals.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Let me add two comments about other things you say in your question: You write &quot;I have read that it is provable in ZF that there are no cardinals $\kappa$ such that $\aleph_0&amp;lt;\kappa&amp;lt;\aleph_1$&quot;. This is true, but it is stronger than that: By &lt;em&gt;definition&lt;/em&gt; $\aleph_1$ is the first ordinal that is not countable, so of course there are no cardinals in between $\aleph_0$ and $\aleph_1$. Similarly, there are no cardinals between any (well-ordered) cardinal $\kappa$ and its successor $\kappa^+$, &lt;em&gt;by definition&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is true, however, that a countable union of countable sets need not be comparable with $\aleph_1$ without choice. In fact, we can have a non-well-orderable set that can be written as a countable union of sets of size 2.&lt;/p&gt;&#xA;" OwnerUserId="462" LastEditorUserId="462" LastEditorDisplayName="" LastEditDate="2011-01-03T17:15:59.990" LastActivityDate="2011-01-03T17:15:59.990" CommentCount="4" />
  <row Id="16250" PostTypeId="2" ParentId="16248" CreationDate="2011-01-03T17:09:29.967" Score="3" ViewCount="" Body="&lt;p&gt;Elements of $\text{SL}_2(\mathbb{F}_p)$ have eigenvalues in $\mathbb{F}_{p^2}$.  The multiplicative group of $\mathbb{F}_{p^2}$ is cyclic of order $p^2 - 1$ generated by some element $g$, so it follows that $\text{GL}_2(\mathbb{F}_p)$ has a cyclic subgroup of order $p^2 - 1$ (just write out the matrix of multiplication by $g$).  The determinant of this matrix is some element of $\mathbb{F}_p^{\ast}$, so $g^{p-1}$ has determinant $1$ and order $p+1$.  Finally, the image of $g^{p-1}$ in $\text{PSL}_2(\mathbb{F}_p)$ has order $\frac{p+1}{2}$ since $g^{ \frac{p^2 - 1}{2} } = -1$.  You can work out the cycle structure from here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for what a nonsplit torus is, the terminology comes from &lt;a href=&quot;http://en.wikipedia.org/wiki/Algebraic_torus&quot; rel=&quot;nofollow&quot;&gt;algebraic geometry&lt;/a&gt; and I can't help you there.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2011-01-03T17:09:29.967" CommentCount="2" />
  <row Id="16251" PostTypeId="2" ParentId="16244" CreationDate="2011-01-03T17:30:25.420" Score="6" ViewCount="" Body="&lt;p&gt;I think you can use &lt;a href=&quot;http://en.wikipedia.org/wiki/Dynkin%27s_lemma&quot; rel=&quot;nofollow&quot;&gt;Dynkin's lemma&lt;/a&gt; (if you call this &quot;more elementary&quot;). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let D be all the measurable sets $U\subseteq I=[0,1]$ such that $\intop_U f(t) = 0$ (the function $f\mid_I$ is in $L_2$ so it is also in $L_1$, so I assume this from now). $I\in D$ and if $A\subseteq B\subseteq I$ are in $D$ then $B-A \in D$. If $A_i \subseteq I$ is an increasing sequence in D then $\bigcup A_i \subseteq I$ is also in D (by the DCT). This shows that D is a Dynkin system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let P be all the open intervals in I (so $P\subseteq D$). P is not empty and an intersection of two open intervals are open, so P is closed under finite intersection, hence it is a pi system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dynkin's lemma says that if P is a pi system and D a dynkin system such that $P\subseteq D$ then $\sigma(P)\subseteq D$. The sigma algebra generated by P is the Borel algebra.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now look on the set $A=\{x\in I \mid f(x)\geq 0\}$. This is a Lebesgue measurable set, so up to a zero measure set it is Borel measurable set $A&amp;#39;$. Since $\intop_{A&amp;#39;} f(t) = 0$ and f is non negative there, then f is zero almost every where in A'. The same argument work for when f&amp;lt;0, so you get that f is zero almost everywhere in $I$. Now do this for all of $n+I,\;n\in \mathbb{Z}$.&lt;/p&gt;&#xA;" OwnerUserId="2125" LastActivityDate="2011-01-03T17:30:25.420" CommentCount="2" />
  <row Id="16252" PostTypeId="2" ParentId="16241" CreationDate="2011-01-03T17:52:35.920" Score="0" ViewCount="" Body="&lt;p&gt;1) Determine the longitude and lattitude of the points. (Using polar co-ordinates)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) &lt;a href=&quot;http://en.wikipedia.org/wiki/Great-circle_distance&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Great-circle_distance&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1554" LastActivityDate="2011-01-03T17:52:35.920" />
  <row Id="16253" PostTypeId="1" AcceptedAnswerId="16259" CreationDate="2011-01-03T17:52:58.760" Score="1" ViewCount="182" Body="&lt;p&gt;Hello there!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a very noob questions about generators: what algorithm do I have to follow so I can prove that a finite subset is a generator?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is the background story (I'll tell it all because I suck at maths and I might have understood the whole concept wrongly):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to know how to compute the image of a vector space morphism. I'll continue on a concrete example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f\colon \mathbb{R}^2 \to \mathbb{R}^2$, $f(x,y) = (x-y, 2x + y)$ is the morphism.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To find its image, I observe that $f(x,y)$ can be rewritten as $x(1,2) + y(-1,1)$. If I am correct, that means that $\mathrm{Im}(f) = \langle(1,2), (-1,1)\rangle$ (that's the notation we use for set generators). Now the question is: does the $\{(1,2), (-1,1)\}$ set generate $\mathbb{R}^2$ or not?&lt;/p&gt;&#xA;" OwnerUserId="5354" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-03T18:57:21.683" LastActivityDate="2011-01-03T19:19:20.367" Title="How to check if a subset is a generator of a vector space" Tags="&lt;linear-algebra&gt;&lt;vector-space&gt;" AnswerCount="4" CommentCount="2" />
  <row Id="16254" PostTypeId="1" AcceptedAnswerId="17264" CreationDate="2011-01-03T17:58:25.450" Score="12" ViewCount="329" Body="&lt;p&gt;There are many summer &lt;a href=&quot;http://www.nsf.gov/crssprgm/reu/list_result.cfm?unitid=5044&quot; rel=&quot;nofollow&quot;&gt;programs&lt;/a&gt; in the United States, targeted at good, motivated undergraduate students majoring in mathematics. Most programs, however (if not all), demand that the applicant is American or a permenant resident in the United States. Even if it is allowed for students outside the U.S. to apply, these non-American students have to pay the whole &quot;experience&quot; themselves. My question is: are there any such programs outside the United States? A roughly equivalent question was asked at &lt;a href=&quot;http://mathoverflow.net/questions/13847/research-experience-for-undergraduates-summer-programs&quot; rel=&quot;nofollow&quot;&gt;MO&lt;/a&gt;, but perhaps the MSE community knows a bit more on such things.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Max&lt;/p&gt;&#xA;" OwnerUserId="1055" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2011-01-13T03:36:11.353" LastActivityDate="2011-01-13T07:05:33.053" CommunityOwnedDate="2011-01-03T18:04:41.023" Title="Summer programs to gain research experience as an undergraduate, outside the United States?" Tags="&lt;soft-question&gt;&lt;education&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="1" />
  <row Id="16255" PostTypeId="2" ParentId="16253" CreationDate="2011-01-03T18:03:56.067" Score="2" ViewCount="" Body="&lt;p&gt;If I did not misinterpret the question, you want to know if the two vectors span the plane. Use the standard results in this case: they span the plane if and only if they are linearly independent, or the 2-by-2 matrix is invertible, and so on. Alternatively you can just show that for every point in the plane there exists the required $x$ and $y$ by solving a pair of simultaneous equations. &lt;/p&gt;&#xA;" OwnerUserId="4594" LastActivityDate="2011-01-03T18:03:56.067" CommentCount="1" />
  <row Id="16256" PostTypeId="2" ParentId="16253" CreationDate="2011-01-03T18:33:43.903" Score="1" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\ $ Two vectors span $\rm\:\mathbb R^2\:$ iff they are nonzero and have different slopes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Generally you can test if $f$ is onto by computing the &lt;a href=&quot;http://en.wikipedia.org/wiki/Rank_%28linear_algebra%29&quot; rel=&quot;nofollow&quot;&gt;rank&lt;/a&gt; of the matrix of $f$.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2011-01-03T19:19:20.367" LastActivityDate="2011-01-03T19:19:20.367" />
  <row Id="16257" PostTypeId="2" ParentId="16247" CreationDate="2011-01-03T18:37:19.103" Score="2" ViewCount="" Body="&lt;p&gt;The degree is zero because the restriction of $f$ to the boundary of $B^n$ (which most humans write $S^{n-1}$!&amp;nbsp;:)&amp;nbsp;) is homotopic to to a constant map.&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2011-01-03T18:37:19.103" />
  <row Id="16258" PostTypeId="2" ParentId="16253" CreationDate="2011-01-03T19:04:20.290" Score="0" ViewCount="" Body="&lt;p&gt;You compute the rank of the matrix whose entries are the coordinates of the vectors in the set (with respect to some basis, say the canonical basis in $\mathbf{R}^n$). The set generates the space iff the rank equals the dimension of the space. In your example, you want the rank to be 2. The rank of a matrix can be found using Gaussian elimination.&lt;/p&gt;&#xA;" OwnerUserId="589" LastActivityDate="2011-01-03T19:04:20.290" />
  <row Id="16259" PostTypeId="2" ParentId="16253" CreationDate="2011-01-03T19:12:24.670" Score="1" ViewCount="" Body="&lt;p&gt;Others have given you very good answers for this specific question. Let me make a few points for general problems of this type:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have a linear transformation $f\colon\mathbb{R}^n\to\mathbb{R}^m$, since every vector in $\mathbb{R}^n$ can be written as a linear combination of the standard basis vectors:&#xA;$$\begin{array}{rcl}&#xD;&#xA;\mathbf{e}_{1} &amp;amp;=&amp;amp; (1,0,0,\ldots,0)\\&#xD;&#xA;\mathbf{e}_{2} &amp;amp;=&amp;amp; (0,1,0,\ldots,0)\\&#xD;&#xA;&amp;amp;\cdots&amp;amp;\\&#xD;&#xA;\mathbf{e}_n &amp;amp;=&amp;amp; (0,0,0,\ldots,1),&#xD;&#xA;\end{array}$$&#xA;then the span of $f$ equals $\langle f(\mathbf{e}_1,\ldots,f(\mathbf{e}_2)\rangle$. In the example you have, this is essentially what you did, since $f(1,0) = (1,2)$ and $f(0,1)=(-1,1)$. &#xA;Second, to find the span of a set of vectors, you can use Gaussian elimination. Put them as rows of a matrix and perform Gaussian elimination (or Gauss-Jordan elimination). The span of the original vectors equals the span of the rows of the matrix, and elementary row operations does not change the span of the rows. The rank will give you the dimension of the image, and if you only want to know if the image is all of $\mathbb{R}^m$, then you just need to see if the rank of the reduced row echelon form is $m$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But it is more common express the function as an $m\times n$ matrix: in the $i$th column, you write down the image of $\mathbf{e}_i$; the image of $f$ is then the span of the columns of this matrix. In your example, we would have&#xA;$$\left(\begin{array}{rr}&#xD;&#xA;1 &amp;amp; -1\\&#xD;&#xA;2 &amp;amp; 1&#xD;&#xA;\end{array}\right)$$&#xA;(the first column is the image of $(1,0)$; the second column is the image of $(0,1)$). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the dimension of the columnspace is the same as the dimension of the rowspace, you can still use elementary row operations to figure out the rank. However, elementary row operations may change the columnspace (not the dimension of the columnspace, but the columnspace itself yes). So you have to be careful there. Here, it is very easy: subtracting twice the first row from the second row, then dividing the second row by $3$, and adding it to the first row we get:&#xA;$$\left(\begin{array}{rr}&#xD;&#xA;1 &amp;amp; -1\\&#xD;&#xA;2 &amp;amp; 1&#xD;&#xA;\end{array}\right) \to \left(\begin{array}{rr}&#xD;&#xA;1 &amp;amp; -1\\&#xD;&#xA;0 &amp;amp; 3&#xD;&#xA;\end{array}\right)\to\left(\begin{array}{rr}&#xD;&#xA;1 &amp;amp; -1\\&#xD;&#xA;0 &amp;amp; 1&#xD;&#xA;\end{array}\right) \to \left(\begin{array}{rr}&#xD;&#xA;1 &amp;amp; 0\\&#xD;&#xA;0 &amp;amp; 1&#xD;&#xA;\end{array}\right).$$&#xA;Since the reduced row echelon form has rank $2$, the columnspace of the original has dimension $2$, so you conclude that $\dim(\mathrm{Im}(f)) = 2$. Since the only subspace of $\mathbb{R}^2$ that has dimension $2$ is $\mathbb{R}^2$ itself, you are done.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2011-01-03T19:12:24.670" CommentCount="6" />
  <row Id="16260" PostTypeId="2" ParentId="16051" CreationDate="2011-01-03T19:16:46.377" Score="0" ViewCount="" Body="&lt;p&gt;Let $F_n(x)=nx$. We can modify the values of $F_n$ at finitely many points so it satisfies the maximum and minimum conditions in the post. Then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_0^1 F_n(x)e^{-i2\pi vx} dx=\int_0^1 nx e^{-i2\pi vx} dx=\frac{i}{2\pi}\frac{n}{v}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and clearly the limit that you are interested in does not exist.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT. One can also modify $F_n(x)$ at those finitely many points so that $F_n(x)$ is continuous for all $n$. &lt;/p&gt;&#xA;" OwnerUserId="3249" LastEditorUserId="3249" LastEditorDisplayName="" LastEditDate="2011-01-04T01:46:06.093" LastActivityDate="2011-01-04T01:46:06.093" CommentCount="3" />
  <row Id="16261" PostTypeId="1" CreationDate="2011-01-03T19:33:31.817" Score="1" ViewCount="88" Body="&lt;p&gt;Find the values of $m$ in the 2nd degree equation $mx^2-2(m-1)x-m-1=0$ so that it has only one root between $-1$ and $2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Like in this &lt;a href=&quot;http://math.stackexchange.com/questions/14742/find-the-values-of-m-in-the-2nd-degree-equation-3m-2x22mx3m-0-so-that&quot;&gt;almost identical question&lt;/a&gt; there are two ways to solve this, one is acknowledging that $f(-1)f(2) \lt 0$, the other applying the theorems for the two possible scenarios $x_1 \lt -1 \lt x_2 \lt 2$ and $-1 \lt x_1 \lt 2 \lt x_2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Doing $f(-1)f(2)&amp;lt;0$, we have:&#xA;$&#xD;&#xA;\begin{align*}&#xD;&#xA;\left(m(-1)^2-2(m-1)(-1)-m-1\right)\left(m2^2-2(m-1)2-m-1\right)&amp;amp;\lt0\\\&#xD;&#xA;(2m-3)(-m+3) &amp;amp; \lt 0 \\\&#xD;&#xA;-2m^2+9m-9 &amp;amp; \lt 0 \\\&#xD;&#xA; m \lt \frac{3}{2} \quad \vee \quad m \gt 3&#xD;&#xA;\end{align*}&#xD;&#xA;$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The thing is that this answer is &lt;strong&gt;wrong&lt;/strong&gt; since the correct one is $m \lt \frac{3}{2} \quad \wedge \quad m \neq 0 \quad \vee \quad m \gt 3$. How do I get to know and not only verify that $m \neq 0$ what did I do wrong?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the theorems in the scenarios &lt;strong&gt;I&lt;/strong&gt;.$ \quad x_1 \lt -1 \lt x_2 \lt 2$ and  &lt;strong&gt;II&lt;/strong&gt;.$ \quad -1 \lt x_1 \lt 2 \lt x_2$, we have:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;I&lt;/strong&gt;.&#xA;$$x_1 \lt -1 \lt x_2 \lt 2&#xD;&#xA;\implies \Bigl(af(-1) \lt 0\Bigr) \wedge  \Bigl(af(2) \gt 0\Bigr)  \wedge \Bigl(\Delta \gt 0 \Bigr)\wedge \Bigl( \frac{S}{2} \lt 2\Bigr).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;II&lt;/strong&gt;.&#xA;$$&#xD;&#xA;-1 \lt x_1 \lt 2 \lt x_2 \implies&#xD;&#xA;\Bigl( af(-1) \gt 0\Bigr)\wedge\Bigl( af(2) \lt 0\Bigr) \wedge \Bigl(\Delta \gt 0 \Bigr) \wedge\Bigl(\frac{S}{2} \gt -1 \Bigr)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So both &lt;strong&gt;I&lt;/strong&gt; and &lt;strong&gt;II&lt;/strong&gt; have the condition $\Delta \gt 0$ in common.&#xA;\begin{align*}&#xA;\Delta \gt 0 &amp;amp;\implies [-2(m-1)]^2-4m(-m-1) \gt 0\\&#xA;&amp;amp;\implies 8m^2-4m+4 \gt 0 \\&#xA;\end{align*}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is that the $\Delta$ from the equation $8m^2-4m+4=0$ is negative. $\Delta = b^2-4ac = (-4)^2-4(8)(4) \lt 0 \implies \nexists m$ such that $8m^2-4m+4=0&#xD;&#xA;\implies \nexists m$ such that $[-2(m-1)]^2-4m(-m-1) \gt 0$. So doesn't matter what the others conditions says in &lt;strong&gt;I&lt;/strong&gt; (or in &lt;strong&gt;II&lt;/strong&gt;), intersecting them all will give an empty set. And therefore &lt;strong&gt;I&lt;/strong&gt; $\cup$ &lt;strong&gt;II&lt;/strong&gt; will give $\emptyset$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I am doing it wrong both ways. Any thoughts on that?&lt;/p&gt;&#xA;" OwnerUserId="3509" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-03T22:40:34.557" LastActivityDate="2011-01-03T22:40:34.557" Title="Find the values of $m$ in the 2nd degree equation $mx^2-2(m-1)x-m-1=0$ so that it has one root between $-1$ and $2$." Tags="&lt;homework&gt;&lt;algebra-precalculus&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="16262" PostTypeId="2" ParentId="16261" CreationDate="2011-01-03T19:49:46.040" Score="3" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\ $ The case $\rm\: m = 0\:$ should not be excluded since then $\rm\: f = 2\ x - 1\ $ which has only 1 root (between -1 and 2). If, alternatively, you require that there be another root outside the interval then you need to determine the case(s) where the prior argument breaks down due to the quadratic degenerating to a lower degree polynomial. Here that is precisely when the leading coefficient vanishes, i.e. $\rm\: m = 0\:$. So the argument really is no more difficult than the simple proof I gave in the original thread.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2011-01-03T20:03:54.540" LastActivityDate="2011-01-03T20:03:54.540" CommentCount="7" />
  <row Id="16263" PostTypeId="2" ParentId="16219" CreationDate="2011-01-03T19:53:42.370" Score="1" ViewCount="" Body="&lt;p&gt;Here are some hints.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;try to use induction, but instead of taking out one edge (or one vertex), take out a spanning tree (when you can), and if the graph is not connected, show that it has a &quot;big enough&quot; sub graph so that $\frac {|E&amp;#39;|}{|V&amp;#39;|} \geq \frac {|E|}{|V|}$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;write $P=(v_1, ..., v_k)$ and $C=(u_1,...,u_t)$  such that $v_1=u_1$ and $v_k=u_i$ for some i. The problem arises when there are a lot of intersections between P and C. Suppose $i\leq t/2$ , so if $u_2=v_j\in P$ (or generally, the first vertex in C after $u_1$ that is in P), then if $j\geq \sqrt{k}$ we have the cycle $(v_1,...,v_j)$. Other wise , j is small, and we look for the second intersection. So we see that two intersections must be near enough on the one hand, but they must be far enough on the other hand, because the first one is with $v_1$ and the last one is with $v_k$. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Here I think you should use a graph that have a cycle of length $4\sqrt{k} $ and then &quot;square&quot; it, so it has 4 sides of size $\sqrt{k}$ (I think of the vertices as $(i,j),\; 1\leq i,j \leq \sqrt{k})$. Now construct a path the start with (1,1) goes to (2,2) and then right until $(2,\sqrt{k})$. then it goes down again to $(3,\sqrt{k}-1)$ and left until (3,1) etc.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I hope this helps. &lt;/p&gt;&#xA;" OwnerUserId="2125" LastActivityDate="2011-01-03T19:53:42.370" />
  <row Id="16264" PostTypeId="2" ParentId="16261" CreationDate="2011-01-03T19:58:09.593" Score="4" ViewCount="" Body="&lt;p&gt;I'm guessing that the reason for $m\neq 0$ is that if $m=0$, then your original equation,&#xA;$$mx^2-2(m-1)x - m - 1=0$$&#xA;becomes a &lt;strong&gt;linear&lt;/strong&gt; equation, namely $2x - 1 = 0$. Presumably, they are asking that the equation actually have &lt;em&gt;two&lt;/em&gt; roots, of which one and only one lies in the desired interval. While it is true that the equation $2x-1=0$ has one and only one solution between $-1$ and $2$ (namely, $x=\frac{1}{2}$), it has no solutions &lt;em&gt;outside&lt;/em&gt; the interval, which is probably assumed in the statement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that you did far too much work in the first method. From $(2m-3)(-m+3)\lt 0$, you are set: the two roots are $m=\frac{3}{2}$ and $m=3$; that means that $(2m-3)(m-3)$ takes negative values exactly between them; since you have this product times $-1$ (given the factor $(-m+3)$ instead of $(m-3)$), the negative values for $(2m-3)(-m+3) = -(2m-3)(m-3)$ occur exactly outside that interval: on $(-\infty,\frac{3}{2})$ and on $(3,\infty)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the second one, you are not thinking clearly. You want the discriminant positive, yes. The discriminant is $8m^2-4m+4$, yes. And the discriminant of &lt;em&gt;this&lt;/em&gt; is negative, yes. And that means is that $8m^2-4m+4$ is &lt;em&gt;never zero&lt;/em&gt;, yes. But you don't want it to be zero! You want it to be positive. You are looking for the values where it is zero so you can see where it may &lt;em&gt;change&lt;/em&gt; from positive to negative. The fact that it is never zero tells you that it never &lt;strong&gt;changes&lt;/strong&gt; from positive to negative or from negative to positive, so that means that is either always positive or always negative. Plugging in $m=0$ shows you that it is in fact &lt;em&gt;always positive&lt;/em&gt;, so that the condition $\Delta \gt 0$ is &lt;em&gt;always&lt;/em&gt; satsfied. It is the &lt;em&gt;exact opposite&lt;/em&gt; of what you concluded: it's not that the condition is &lt;em&gt;never&lt;/em&gt; satisfied, it is &lt;em&gt;always satisfied&lt;/em&gt;, so you just need to check the other condition; $\Delta\gt 0$ will always hold.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2011-01-03T19:58:09.593" CommentCount="6" />
  <row Id="16265" PostTypeId="1" AcceptedAnswerId="16266" CreationDate="2011-01-03T20:21:51.083" Score="1" ViewCount="195" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Let $\sum_{n=1}^{\infty} a_{n}$ be a&#xA;  convergent series when $\forall n, a_{n}\neq 0$.  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Does $\sum_{n=1}^{\infty} (1-\frac{\sin a_{n}}{a_{n}})$ converge if:&lt;/p&gt;&#xA;  &#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;$\forall n, a_{n}\gt 0$.&lt;/li&gt;&#xA;  &lt;li&gt;$a_{n}\lt 0$ for infinitely many $n$'s.&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For (1) I tried looking at the Maclaurian series for $\sin x$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$|(1-\frac{\sin a_{n}}{a_{n}})|=|(1-\frac{1}{a_{n}}(a_{n}+R_{2}(a_{n}))|=|\frac{1}{6}a_{n}^2 \sin \xi| \leq a_{n}^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure if what I did above is correct but if it is then by the comparison test we get that the series converges. Is there a simpler argument for this? Maybe something that uses the fact that $\sin a_{n} \leq a_{n}$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regarding (2) I think that it can converge or diverge. I tried to guess and it seems that $a_{n}=\frac{(-1)^n}{\sqrt n}$ yields a divergent series according to Mathematica but I'm not sure why. How can I find one that converges or diverges without guessing (maybe again by using a Maclaurian series)?&lt;/p&gt;&#xA;" OwnerUserId="3237" LastEditorUserId="3237" LastEditorDisplayName="" LastEditDate="2011-01-04T20:19:09.110" LastActivityDate="2011-01-04T20:19:09.110" Title="Convergence of $\sum_{n=1}^{\infty} (1-\frac{\sin a_{n}}{a_{n}})$ when $\sum_{n=1}^{\infty} a_{n}$ converges" Tags="&lt;calculus&gt;&lt;homework&gt;&lt;sequences-and-series&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="16266" PostTypeId="2" ParentId="16265" CreationDate="2011-01-03T20:32:50.590" Score="4" ViewCount="" Body="&lt;p&gt;Since $\sum a_n$ converges, $a_n\to0$ as $n\to\infty$. The main thing to notice from the Taylor (or Maclaurin as you mentioned) series is that $\sin x=x-\frac16x^3+O(x^5)$ for small $x$. So we have $1-\frac{\sin a_n}{a_n}=a_n^2/6+O(a_n^4)$ for large $n$. Now one can see that the convergence of the series $\sum(1-\frac{\sin a_n}{a_n})$ is the same as the convergence of the series $\sum a_n^2$.&lt;/p&gt;&#xA;" OwnerUserId="2473" LastActivityDate="2011-01-03T20:32:50.590" CommentCount="4" />
  <row Id="16267" PostTypeId="2" ParentId="16265" CreationDate="2011-01-03T20:51:14.590" Score="2" ViewCount="" Body="&lt;p&gt;For (1), you actually want &#xA;$$&#xD;&#xA;(0 \le )\, 1 - \frac{{\sin x}}{x} \le x&#xD;&#xA;$$&#xA;for all sufficiently small $x&amp;gt;0$ ($x$ plays the role of $a_n \downarrow 0$).&#xA;Indeed, it suffices to show $x - \sin x \leq x^2$, which follows straight by comparing derivatives (twice).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: For divergence in (2), consider a sequence $a_n = (-1)^n b_n$ where $b_n \downarrow 0$ monotonically, and note that $\sin x / x$ is an even function.&#xA;Then, consider $1 - \sin x / x \geq x^3$ for all sufficiently small $x&amp;gt;0$. Finally, assume that $\sum\nolimits_{n = 1}^\infty  {b_n^3 }  = \infty $.&lt;/p&gt;&#xA;" OwnerUserId="2810" LastEditorUserId="2810" LastEditorDisplayName="" LastEditDate="2011-01-03T21:33:49.747" LastActivityDate="2011-01-03T21:33:49.747" CommentCount="1" />
  <row Id="16268" PostTypeId="2" ParentId="16244" CreationDate="2011-01-03T21:34:25.403" Score="7" ViewCount="" Body="&lt;p&gt;It is sufficient to prove that $f$ is zero almost everywhere on any bounded interval. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) By additivity it is easy to see that $$\int_a^bf(x)dx=\int_0^bf(x)dx - \int_0^af(x)dx$$ for all bounded intervals $(a,b)$ (and also for $[a,b)$, $(a,b]$ and $[a,b]$). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) Using (1) it is easy to see that $$\int_Bf(x)dx=0$$ for any bounded Borel measurable set.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(3) Any Lebesgue measurable set $A$ is of the form $A=B\cup Z$ where $B$ is a Borel measurable set and $Z$ is a set of measure zero. Hence, by (2) we acheive&#xA;$$\int_A f(x)dx= 0$$&#xA;for any bounded Lebesgue measurable set $A$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(4) Now look at the sets $A_+(n)=\{x:f(x)&amp;gt;0\}\cap[-n,n]$ and $A_-(n)=[-n,n]\setminus A_+(n)$. Assuming $f$ is measurable these sets are also measurable and by (3) &#xA;$$\int_{A_\pm(n)}f(x)dx=0$$&#xA;&lt;strong&gt;EDIT:&lt;/strong&gt; and hence $f=0$ almost everywhere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please forgive me if I write $dx$ for the Lebesgue measure which I presume is what you refer to as $dm$. &lt;/p&gt;&#xA;" OwnerUserId="1154" LastEditorUserId="1154" LastEditorDisplayName="" LastEditDate="2011-01-04T14:09:07.400" LastActivityDate="2011-01-04T14:09:07.400" CommentCount="7" />
  <row Id="16269" PostTypeId="2" ParentId="16241" CreationDate="2011-01-03T21:38:00.367" Score="3" ViewCount="" Body="&lt;p&gt;First you want to build a 3d-mesh out of your face. To get a general idea of this, the following paper might be interesting for you: &lt;a href=&quot;http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.58.8435&amp;amp;rep=rep1&amp;amp;type=pdf&quot; rel=&quot;nofollow&quot;&gt;http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.58.8435&amp;amp;rep=rep1&amp;amp;type=pdf&lt;/a&gt; . Of course you can improve this algorithm with knowledge, that you already have about the appearance of human faces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then you select your points in the mesh and run a path-finding algorithm to get a good approximation for the distance. When you did everything properly you will get something like this (note that the red line should mark shortest path):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/kbdwa.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope this is what you wanted&lt;/p&gt;&#xA;" OwnerUserId="3123" LastActivityDate="2011-01-03T21:38:00.367" CommentCount="1" />
  <row Id="16270" PostTypeId="1" AcceptedAnswerId="16273" CreationDate="2011-01-03T21:54:44.760" Score="4" ViewCount="243" Body="&lt;p&gt;Hi,  I was trying some old problems and got stuck on this one. Then when I looked at the answer there was a step I could not understand. Perhaps you can explain it to me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;A-3&lt;/strong&gt; Find &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ \displaystyle \lim_{t \to \infty}\left[ e^{-t}\int_0^t \int_0^t \frac{e^x - e^y}{x - y}dx dy \right]$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;or show that the limit does not exist.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Solution&lt;/strong&gt; Let $G(t)$ be the double integral.  Then $\lim\limits_{t \to \infty}\frac{G(t)}{e^t} = \lim\limits_{t \to \infty}\frac{G&amp;#39;(t)}{e^t}$ by L'Hopital.  Then&#xA;$$G&amp;#39;(t) = \int_0^t \frac{e^x-e^t}{x-t}dx+ \int_0^t \frac{e^y-e^t}{y-t}dy$$&#xA;so&#xA;$$G&amp;#39;(t) = 2\int_0^t \frac{e^x-e^t}{x-t}dx.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, the answer continues to show that  $\lim_{t \to \infty}\frac{G&amp;#39;(t)}{e^t}=\infty$ since,&#xA;$$\frac{G&amp;#39;(t)}{2e^t} = \int_0^t \frac{e^{x-t}-1}{x-t}dx = \int_0^t\frac{1-e^{-y}}{y}dy &amp;gt; \int_1^t\frac{1-e^{-y}}{y}dy &amp;gt; \left(1-e^{-1}\right)log\,t.$$&#xA;My question is how $G&amp;#39;(t)$ was found.  I understand the rest of the solution.  I understand differentiating under the integral in the one dimensional case, but I do not understand how it works in the case of a double integral (which I assume is what is being done here), and I couldn't produce the answer's result.&lt;/p&gt;&#xA;" OwnerUserId="565" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-03T22:19:49.103" LastActivityDate="2011-01-03T23:48:16.350" Title="1981 Putnam problem A-3 : Differentiating under a double integral?" Tags="&lt;integral&gt;&lt;putnam&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="16271" PostTypeId="1" AcceptedAnswerId="16307" CreationDate="2011-01-03T22:00:29.603" Score="2" ViewCount="159" Body="&lt;p&gt;Inspired by the recent incident in which a professor was able to [detect that students in his class were cheating][1], I'm curious if there is a standard way to detect if a distribution is bimodal, or a standard measure of the 'bimodality' of a distribution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does such a test exist? What are caveats that I should be aware of in such a test?&lt;/p&gt;&#xA;" OwnerUserId="3917" LastEditorUserId="4742" LastEditorDisplayName="" LastEditDate="2011-01-04T11:18:14.543" LastActivityDate="2011-01-04T11:18:14.543" Title="How to measure if a sequence has a bimodal distribution" Tags="&lt;probability&gt;&lt;statistics&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="16272" PostTypeId="2" ParentId="16271" CreationDate="2011-01-03T22:27:11.880" Score="1" ViewCount="" Body="&lt;p&gt;You want &lt;em&gt;goodness of fit tests&lt;/em&gt;. You can start reading on Wikipedia.&lt;/p&gt;&#xA;" OwnerUserId="3330" LastActivityDate="2011-01-03T22:27:11.880" CommentCount="1" />
  <row Id="16273" PostTypeId="2" ParentId="16270" CreationDate="2011-01-03T22:37:39.560" Score="7" ViewCount="" Body="&lt;p&gt;Let $$F(y,t)=\int_0^t \frac{e^x-e^y}{x-y} dx.$$ Then your $G(t)$ is  $$G(t)=\int_0^t F(y,t) dy.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then  $$G&amp;#39;(t)=F(t,t)+\int_0^t \frac{\partial F}{\partial t} (y,t) dy$$&#xA;$$=\int_0^t \frac{e^x-e^t}{x-t} dx + \int_0^t\frac{e^t -e^y}{t-y} dy  .$$&lt;/p&gt;&#xA;" OwnerUserId="3249" LastEditorUserId="3249" LastEditorDisplayName="" LastEditDate="2011-01-03T23:48:16.350" LastActivityDate="2011-01-03T23:48:16.350" />
  <row Id="16274" PostTypeId="2" ParentId="16247" CreationDate="2011-01-03T22:43:49.697" Score="1" ViewCount="" Body="&lt;p&gt;This is just a bit more of an explanation of Mariano's answer:&#xA;The homotopy class of a map between spheres of the same dimension is completely determined by its degree, that is, two maps $f,g: S^n \to S^n$ are homotopic if and only if they have the same degree. Now you can compute, using which ever definition you like, that the degree of the constant map is 0. A map out of a sphere is null homotopic, ie. homotopic to a constant map, if and only if it can be extended to a map of the whole unit ball. Since the map in question, $f|_{S^n}$, can indeed be extended to a map on the whole unit ball, namely $f$, it must be null homotopic. So by the above it must have the same degree as the constant map.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again, this is just an elaboration of Mariano's answer.&lt;/p&gt;&#xA;" OwnerUserId="627" LastActivityDate="2011-01-03T22:43:49.697" />
  <row Id="16275" PostTypeId="1" CreationDate="2011-01-03T23:40:14.747" Score="16" ViewCount="457" Body="&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Four_color_theorem&quot; rel=&quot;nofollow&quot;&gt;four-colour mapping theorem&lt;/a&gt; states that all maps can be four-coloured (adjacent regions receive distinct colours, and four different colours are used in total).  However, the technical definition of &quot;map&quot; implies that regions must be contiguous, and, in fact, it's easy to construct counter-examples to a proposed &quot;four-colour not-necessarily-contiguous-mapping conjecture&quot;.  For example: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/Qq4QI.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where the two green areas constitute a single non-contiguous region.  This is clearly not four-colourable since each of the five internal regions is adjacent to every other (forming a $K_5$ in the graph-theoretic sense).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some real-world maps have disjoint components (e.g. USA, including Alaska, Hawaii), which leads me to:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Question&lt;/strong&gt;: What is an example of a real-world map (which necessarily must have at least one non-contiguous region) that is not four-colourable?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="139" LastActivityDate="2011-01-04T22:33:20.887" Title="An example of a real-world map that is not 4-colourable?" Tags="&lt;combinatorics&gt;&lt;graph-theory&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="4" />
  <row Id="16276" PostTypeId="2" ParentId="16275" CreationDate="2011-01-04T00:04:23.563" Score="20" ViewCount="" Body="&lt;p&gt;Wikipedia has a list of &lt;a href=&quot;http://en.wikipedia.org/wiki/List_of_sets_of_four_countries_that_border_one_another&quot; rel=&quot;nofollow&quot;&gt;sets of four contries that border one another&lt;/a&gt;; it includes a few divided countries (Russia, in Belarus-Lithuania-Russia-Poland, with Kaliningrad being separated from the rest of Russia; and Azerbajan in Azerbajan-Iran-Armenia-Turkey). There do not seem to be any set of 5 or more currently. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Added:&lt;/em&gt; However, one could argue that the existence of disconnected countries in fact forces certain maps to use more colors, even though they do not involve disconnected countries. For example, a map of Europe would necessarily require different colors for France, Belgium, the Netherlands, Germany, Denmark, Norway, and the United Kingdom (possibly also Ireland), lest it seem that one of those countries is disconnected into two non-adjacent regions. This also shows some of the difficulties that islands produce (as opposed to the idealized maps from the theorem).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There was a time when Britain, Spain, France, Portugal, and the Dutch Republic bordered each other (early 18th century, counting borders lying between colonies); and from the late 19th century until before World War I, so did Belgium, Portugal, Germany, France, the United Kingdom, and the Netherlands (necessitating &lt;em&gt;six&lt;/em&gt; colors; again, this includes colonies that may border one another). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's possible that some smaller divisions (states, provinces, etc.) might include such sets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As I noted in the comment, there is an incorrect assumption in your question that a counterexample &lt;em&gt;must&lt;/em&gt; have disconnected regions: in the four-color map theorem it is assumed that neighboring regions have a border of positive measure. Countries/states meeting at a point (such as Colorado, Utah, New Mexico, and Arizona meeting at the Four State Corner) are required to use different colors in maps, but in the context of the 4-color map theorem regions that only meet at the point could be colored using the same color. &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-04T02:22:04.540" LastActivityDate="2011-01-04T02:22:04.540" CommentCount="3" />
  <row Id="16277" PostTypeId="2" ParentId="16275" CreationDate="2011-01-04T00:41:19.343" Score="6" ViewCount="" Body="&lt;p&gt;Here's a &lt;a href=&quot;http://upload.wikimedia.org/wikipedia/commons/7/7d/Holy_Roman_Empire_1648.svg&quot; rel=&quot;nofollow&quot;&gt;map of the Holy Roman Empire&lt;/a&gt; after the Peace of Westphalia in 1648.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the color scheme the map makers have chosen, the pieces of the map labeled &quot;Bavaria&quot; (green), &quot;Austrian Hapsburg&quot; (light green), &quot;Ecclesiastical&quot; (light purple), &quot;Hohenzollern Franconian&quot; (light brown, including Ansbach and Bayreuth), and &quot;Imperial Cities&quot; (red) all touch each other.  So five colors are required to differentiate them.&lt;/p&gt;&#xA;" OwnerUserId="2370" LastActivityDate="2011-01-04T00:41:19.343" CommentCount="2" />
  <row Id="16278" PostTypeId="2" ParentId="16216" CreationDate="2011-01-04T00:54:34.193" Score="1" ViewCount="" Body="&lt;p&gt;Let's suppose that the job entails 30 (the least common multiple of 6 and 10) equal units of work.  Since $A$ can complete the job in 6 days, $A$ does 5 units of work per day.  Since $B$ can complete the job in 10 days, $B$ does 3 units of work per day.  So, if they alternate, in every 2-day period, they do 8 units of work, which is $\frac{8}{30}=\frac{4}{15}$ of the whole job.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To get at the &quot;how/why&quot; of starting with the faster worker, let's make two tables:&#xA;$$\begin{matrix}&#xD;&#xA;\text{If A goes first...} &amp;amp; &amp;amp; \text{If B goes first...} \\&#xD;&#xA;\begin{matrix}&#xD;&#xA;\text{day} &amp;amp; \text{units today} &amp;amp; \text{units total} \\&#xD;&#xA;1 &amp;amp; 5 &amp;amp; 5 \\&#xD;&#xA;2 &amp;amp; 3 &amp;amp; 8 \\&#xD;&#xA;3 &amp;amp; 5 &amp;amp; 13 \\&#xD;&#xA;4 &amp;amp; 3 &amp;amp; 16 \\&#xD;&#xA;5 &amp;amp; 5 &amp;amp; 21 \\&#xD;&#xA;6 &amp;amp; 3 &amp;amp; 24 \\&#xD;&#xA;7 &amp;amp; 5 &amp;amp; 29 \\&#xD;&#xA;8 &amp;amp; 3 &amp;amp; 32 \\&#xD;&#xA;\end{matrix}&#xD;&#xA;&amp;amp; &amp;amp;&#xD;&#xA;\begin{matrix}&#xD;&#xA;\text{day} &amp;amp; \text{units today} &amp;amp; \text{units total} \\&#xD;&#xA;1 &amp;amp; 3 &amp;amp; 3\\ &#xD;&#xA;2 &amp;amp; 5 &amp;amp; 8\\ &#xD;&#xA;3 &amp;amp; 3 &amp;amp; 11\\ &#xD;&#xA;4 &amp;amp; 5 &amp;amp; 16\\ &#xD;&#xA;5 &amp;amp; 3 &amp;amp; 19\\ &#xD;&#xA;6 &amp;amp; 5 &amp;amp; 24\\ &#xD;&#xA;7 &amp;amp; 3 &amp;amp; 27\\ &#xD;&#xA;8 &amp;amp; 5 &amp;amp; 32&#xD;&#xA;\end{matrix}\end{matrix}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the first case, going into the final day, there is 1 unit of work left to be done at a rate of 3 units per day, so it will take $\frac{1}{3}$ of that last day to finish.  In the second case, going into the final day, there are 3 units of work left to be done at a rate of 5 units per day, so it will take $\frac{3}{5}$ of that last day to finish, which is longer.  As you can see looking at the 7-day totals, if the job entailed 28 or 29 units of work (with the same 3 per day and 5 per day rates), the job could be finished in 7 days in the first case, but not in the second case.&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2011-01-04T00:54:34.193" />
  <row Id="16279" PostTypeId="2" ParentId="16146" CreationDate="2011-01-04T01:05:01.083" Score="3" ViewCount="" Body="&lt;p&gt;The first thing to know about the word &quot;constructive&quot; is that it means too many things. There are so many sorts of &quot;constructive mathematics&quot; that significant context is needed to tell which one is meant. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That being said, the clearest sense in which the definitions you have above are nonconstructive is that they are &quot;&lt;em&gt;impredicative&lt;/em&gt;&quot;. This means that, when you look at the definition $\operatorname{lfp}(f) = \bigcap \operatorname{Red}(f)$, you see that the element $\operatorname{lfp}(f)$ is already a member of $\operatorname{Red}(f)$. In ordinary mathematics as it is practiced, this is just a curiosity, but for a time there was concern about the impredicative definitions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first motivation for this concern was Russell's paradox: if we seek to define &quot;the set of all sets that do not contain themselves&quot;, we see that the set we are defining is already a member of the collection of &quot;all sets&quot; that we quantify over, and moreover this impredicative definition leads to a paradox. This led Russell to be interested in predicativism (the program of developing mathematics without impredicative definitions.  Later Weyl in &lt;em&gt;Das Kontinuum&lt;/em&gt; showed that is is possible to develop a significant amount of mathematics in a predicative way; this work has been further extended and clarified by the reverse mathematics program of Friedman and Simpson. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One way to conceptualize the difference between predicative and impredicative definitions (not formal, but informative) is that impredicative definitions often &quot;identify&quot; something that already exists, while predicative definitions directly &quot;construct&quot; it. For example, you could define the supremum of a nonempty bounded set of real numbers as the least element of the set of all upper bounds - this impredicative definitions identifies which upper bound is the supremum, but it doesn't help you actually construct it. Or you could directly construct a Cauchy sequence whose limit you can prove is both an upper bound for the original set and a lower bound for the set of all upper bounds. This would give you a predicative definition of the supremum.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The way to build the Cauchy sequence is as follows. Suppose $A$ is a nonempty set with an upper bound $x_0 \not \in A$. Pick $y_0 \in A$. Now, by induction, assume $y_i, x_i$ are given with $y_i \in A$, $x_i \not \in A$, $y_i &amp;lt; x_i$. Let $z = (x_i + y_i)/2$. If $z \in A$ let $y_{i+1} = z$ and $x_{i+1} = x_i$. Otherwise let $y_{i+1} = y_i$ and $x_{i+1} = z$. Then the sequence $(x_i)$ is Cauchy, and you can prove its limit is the least upper bound of $A$.  You can see that this took a lot more work than the impredicative definition of the supremum, but it also gives a lot more information: if you knew enough about the set $A$ you could use this procedure to actually approximate the supremum to any precision you like. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, there are two proofs of the Knaster–Tarski theorem. One is impredicative but otherwise easy, while the other is predicative but uses transfinite induction. There are many other mathematical objects that have the same pattern of two definitions. Another common one is the algebra of Borel sets on a topological space. &lt;/p&gt;&#xA;" OwnerUserId="630" LastActivityDate="2011-01-04T01:05:01.083" />
  <row Id="16280" PostTypeId="1" AcceptedAnswerId="16281" CreationDate="2011-01-04T01:52:50.263" Score="0" ViewCount="96" Body="&lt;p&gt;what means by ring counter of a polygon.. and what means by ring counter = 1 of a polygon?&#xA;any one can tell me by drawing an example..&lt;/p&gt;&#xA;" OwnerUserId="4250" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-04T02:26:15.460" LastActivityDate="2011-01-04T04:06:13.333" ClosedDate="2011-01-04T17:35:56.257" Title="ringcounter of polygon" Tags="&lt;geometry&gt;" AnswerCount="2" CommentCount="6" />
  <row Id="16281" PostTypeId="2" ParentId="16280" CreationDate="2011-01-04T02:23:09.827" Score="0" ViewCount="" Body="&lt;p&gt;I think you will find of interest the comments on &quot;rings&quot; and other pseudo-geometric objects constituting the nonrigorous geometrical foundations of various spatial reasoning systems:  &lt;a href=&quot;http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.102.1639&amp;amp;rep=rep1&amp;amp;type=pdf&quot; rel=&quot;nofollow&quot;&gt;Polygons: the unstable foundation of spatial modeling.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For basic background on the 9-intersection model see: Egenhofer, M. and Herring, J. &lt;a href=&quot;http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.94.2585&amp;amp;rep=rep1&amp;amp;type=pdf&quot; rel=&quot;nofollow&quot;&gt;Categorizing binary&#xA;topological relationships between regions, lines and points in&#xA;geographic databases.&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2011-01-04T04:06:13.333" LastActivityDate="2011-01-04T04:06:13.333" CommentCount="8" />
  <row Id="16282" PostTypeId="1" AcceptedAnswerId="16283" CreationDate="2011-01-04T02:46:25.650" Score="2" ViewCount="159" Body="&lt;p&gt;nine points are placed in a square array. if 3 points are randomly selected, what is the probability that they are the vertices of a triangle? &lt;/p&gt;&#xA;" OwnerUserId="5359" LastActivityDate="2011-01-04T02:54:33.213" Title="Probability Question" Tags="&lt;probability&gt;" AnswerCount="1" CommentCount="6" />
  <row Id="16283" PostTypeId="2" ParentId="16282" CreationDate="2011-01-04T02:54:33.213" Score="9" ViewCount="" Body="&lt;p&gt;Any 3 points would form a triangle unless they are collinear. By considering horizontal, vertical and diagonal lines, we see that there are exactly 8 cases of collinearity. Subtract these from the total number of cases and deduce the probability from there.&lt;/p&gt;&#xA;" OwnerUserId="4594" LastActivityDate="2011-01-04T02:54:33.213" CommentCount="5" />
  <row Id="16284" PostTypeId="1" AcceptedAnswerId="16287" CreationDate="2011-01-04T03:08:48.457" Score="0" ViewCount="146" Body="&lt;p&gt;The expression $x^2-4x+5$ is a factor of $ax^3+bx^2+25$. Express the sum a+b as an integer. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please give an explanation of how the answer &lt;/p&gt;&#xA;" OwnerUserId="5359" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-04T03:12:16.577" LastActivityDate="2011-01-04T03:53:11.650" Title="Algebra Problem" Tags="&lt;algebra-precalculus&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="16285" PostTypeId="2" ParentId="16280" CreationDate="2011-01-04T03:10:17.403" Score="1" ViewCount="" Body="&lt;p&gt;Given your link, and also &lt;a href=&quot;http://technet.microsoft.com/en-us/library/bb895267.aspx#bkmkvalidgeographypolygons&quot; rel=&quot;nofollow&quot;&gt;this page&lt;/a&gt;, it seems that &quot;ring&quot; refers to the boundaries of a plane polygon, which are homeomorphic to circles. &quot;Outside ring&quot; and &quot;inside rings&quot; refer to the outside boundary and to any boundaries surrounding holes on the inside of the polygon (making it homeomorphic to punctured discs). The page does not contain any instance of &quot;counter&quot;, but presumably it refers to a &quot;counter&quot; in the sense of computer science: an inner count of how many 'rings' are being specified for the polygon ($n+1$, where $n$ is the number of &quot;holes&quot;). Your question seems to me to be off-topic.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2011-01-04T03:10:17.403" CommentCount="4" />
  <row Id="16286" PostTypeId="2" ParentId="16284" CreationDate="2011-01-04T03:50:40.980" Score="4" ViewCount="" Body="&lt;p&gt;If $x^2-4x+5$ is a factor of $ax^3+bx^2+25$, then $ax^3+bx^2+25=(x^2-4x+5)(\text{something})$.  Since $ax^3+\cdots$ is a polynomial of degree 3 and $x^2-\cdots$ is a polynomial of degree 2, the &quot;something&quot; must be a polynomial of degree 1:&#xA;$$ax^3+bx^2+0x+25=(x^2-4x+5)(\underline{\;\;\;\;\;\;}x+\underline{\;\;\;\;\;\;})$$&#xA;Try to fill in the two blanks based the terms on the left side that don't have $a$ and $b$ in them (for example, how will $+25$ end up in the product?), then finish the multiplication to find the values of $a$ and $b$.&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2011-01-04T03:50:40.980" />
  <row Id="16287" PostTypeId="2" ParentId="16284" CreationDate="2011-01-04T03:53:11.650" Score="3" ViewCount="" Body="&lt;p&gt;$\rm\ 0 = a\ x^3 + b\ x^2 + 25 - (x^2 - 4\ x + 5)\ (a\ x + 5) = (4\ a + b - 5)\ x^2 + (20 - 5\ a)\ x\ \Rightarrow \ a,\: b = \ldots$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: $\rm\ a\ x + 5\ $ comes from comparing leading and constant coefficients.&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2011-01-04T03:53:11.650" />
  <row Id="16289" PostTypeId="1" CreationDate="2011-01-04T04:36:10.863" Score="1" ViewCount="131" Body="&lt;p&gt;May I refer you to:&#xA;&lt;img src=&quot;http://i.imgur.com/DlESY.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;taken from &lt;a href=&quot;http://books.google.com.mx/books?id=4hIq6ExH7NoC&amp;amp;pg=PA92&amp;amp;lpg=PA92&amp;amp;dq=continuous+image+of+a+compact+metrizable+space+is+metrizable&amp;amp;source=bl&amp;amp;ots=p8uRTpf1LC&amp;amp;sig=9JV01MhywqlHkHaEjpBtzaoELbk&amp;amp;hl=es&amp;amp;ei=9ZciTfz_F4fWtQOmv_yICw&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=5&amp;amp;ved=0CD8Q6AEwBA#v=onepageq=continuous%20image%20of%20a%20compact%20metrizable%20space%20is%20metrizable&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;page 92 of &lt;em&gt;Infinite Dimensional Analysis&lt;/em&gt; by Charalambos D. Aliprantis.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why $G \subset f^{-1}(W)$ ? I don't see this inclusion. Can you please help?&lt;/p&gt;&#xA;" OwnerUserId="5361" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-04T17:31:01.080" LastActivityDate="2011-01-10T20:36:36.303" Title="Inclusion in a proof (cts image of a compact metrizable space is metrizable)." Tags="&lt;topology&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="16290" PostTypeId="1" CreationDate="2011-01-04T04:36:36.227" Score="12" ViewCount="432" Body="&lt;p&gt;Hi I am new here and have a calculus question that came up at work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose you have a 4' x 8' piece of plywood. You need 3 circular pieces all equal diameter. What is the maximum size of circles you can cut from this piece of material?&#xA;I would have expected I could write a function for the area of the 3 circles in terms of x and y, then differentiate it, find a maxima/minima point and go from there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My coworker did cut 3 33&quot; circles and that solved the real-world problem. But my passion would be to find the mathematical answer to this. I hope that my new stackexchange.com friends have the same passion, and can help me find the answer to this in general terms. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I mean by that is someone says I have a piece of material Q units by 2Q units, what are the three circles of maximum size?... I hope you understand what I am asking.&#xA;I am looking to be a friend and contributor&#xA;BD&lt;/p&gt;&#xA;" OwnerUserId="5360" LastEditorUserId="237" LastEditorDisplayName="" LastEditDate="2011-01-06T12:20:42.193" LastActivityDate="2011-01-06T12:20:42.193" Title="maximum area of three circles" Tags="&lt;calculus&gt;&lt;geometry&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="1" />
  <row Id="16291" PostTypeId="2" ParentId="16290" CreationDate="2011-01-04T04:55:52.443" Score="11" ViewCount="" Body="&lt;p&gt;I believe the optimal configuration is as shown below, with the circles tangent to the sides of the rectangle and to one another.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/nT3kG.png&quot; alt=&quot;diagram&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let the rectangle have sides of length $Q$ and $2Q$, as you specified, and let the radius of all three circles be $r$.  Consider the right triangle with the segment joining the centers of the left and middle circles as its hypotenuse and with legs parallel to the sides of the rectangle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/4lORT.png&quot; alt=&quot;diagram with triangle&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The hypotenuse has length $2r$, and the legs have lengths $Q-r$ and $Q-2r$, so by the Pythagorean theorem, $(2r)^2=(Q-r)^2+(Q-2r)^2$.  Solving for $r$ in terms of $Q$ gives $r=(3\pm\sqrt{7})Q$, but since $r&amp;lt;Q$, $r=(3-\sqrt{7})Q\approx 0.354249Q$.  For your 4-by-8 board, that gives an optimal radius of approximately $0.354249\cdot 48\approx 17.0039$ inches, so a diameter of just over 34 inches.&lt;/p&gt;&#xA;" OwnerUserId="72" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2011-01-04T05:04:29.063" LastActivityDate="2011-01-04T05:04:29.063" CommentCount="11" />
  <row Id="16292" PostTypeId="1" AcceptedAnswerId="16294" CreationDate="2011-01-04T05:09:56.233" Score="1" ViewCount="142" Body="&lt;p&gt;This is more of a textbook semantics issue.  Is the alternative hypothesis always two-tailed when all that is known is a null hypothesis $H_0$ where $p$ equals some arbitrary figure, where an arbitrary sample proportion and significance level are also given (assuming all requirements are met)? &lt;/p&gt;&#xA;" OwnerUserId="2054" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2011-01-04T05:55:16.627" LastActivityDate="2011-01-04T05:55:16.627" Title="Alternative hypothesis when null hypothesis is ambiguous?" Tags="&lt;statistics&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="16294" PostTypeId="2" ParentId="16292" CreationDate="2011-01-04T05:54:06.287" Score="4" ViewCount="" Body="&lt;p&gt;Yes, although it's more than just a textbook semantics issue.  In the absence of information about the form the alternate hypothesis should take (e.g., in the absence of a research question like &quot;Does the new drug work &lt;em&gt;better&lt;/em&gt; than the best-known drug on the market?&quot;) then the formulation of the alternate should just be the negation of the null.  Thus the form of $H_0$ and $H_a$ should be $p = p_0$ and $p \neq p_0$, respectively.  As non-expert points out in the comments, this then means that you need to perform a two-tailed test.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Two comments:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Remember that the statements of the null and alternate hypotheses don't have anything to do with the sample proportion or with the significance level.  In particular, it's good practice to formulate the hypotheses &lt;em&gt;before&lt;/em&gt; the results from the sample are known.  &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;It is interesting that Ronald Fisher, one of the developers of tests of significance, effectively took the position that one should &lt;em&gt;always&lt;/em&gt; use the two-sided alternate hypothesis.  To quote from the Wikipedia article on the &lt;a href=&quot;http://en.wikipedia.org/wiki/Alternative_hypothesis&quot; rel=&quot;nofollow&quot;&gt;alternative hypothesis&lt;/a&gt;, &quot;[The alternate hypothesis] was not part of Ronald Fisher's formulation of statistical hypothesis testing, and he violently opposed its use.  In Fisher's approach to testing, the central idea is to assess whether the observed dataset could have resulted from chance if the null hypothesis were assumed to hold, notionally without preconceptions about what other model might hold. Modern statistical hypothesis testing accommodates this type of test since the alternative hypothesis can be just the negation of the null hypothesis.&quot;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="2370" LastActivityDate="2011-01-04T05:54:06.287" CommentCount="1" />
  <row Id="16295" PostTypeId="2" ParentId="16244" CreationDate="2011-01-04T06:09:28.267" Score="4" ViewCount="" Body="&lt;p&gt;Indeed, as you expected, a simple proof of the result can be found; see Theorem 2.1 in &lt;a href=&quot;http://www.ualberta.ca/~rjia/Math418/Notes/chap3.pdf&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; useful note on absolutely continuous functions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;: Since this is a quite important result, it is worth giving here the proof in detail. The proof below is essentially the one given in the link above, but somewhat shorter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Theorem&lt;/strong&gt;. If $f$ is integrable on $[a, b]$ and $\int_a^x {f(t){\rm d}t}  = 0$ $\forall x \in [a,b]$, then $f = 0$ a.e. on $[a, b]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proof. An open subset $O$ of $[a,b]$ is a countable union of disjoint open intervals $(c_n, d_n)$; hence,&#xA;$\int_O {f(t){\rm d}t}  = \sum\nolimits_{n = 1}^\infty  {\int_{c_n }^{d_n } {f(t){\rm d}t} }  = 0$.&lt;br&gt;&#xA;If $K$ is a closed subset of $[a,b]$, it thus follows that $\int_K {f(t){\rm }{\rm d}t}  = 0$ (note that $[a,b]$ is the disjoint union of $K$ and $[a,b]\backslash K$, and that $[a,b]\backslash K$ is open). Next let&#xA;$E_ +   = \{ x \in [a,b]:f(x) &amp;gt; 0\}$ and $E_ -   = \{ x \in [a,b]:f(x) &amp;lt; 0\}$. If $\lambda(E_+) &amp;gt; 0$, then there exists some closed set $K \subset E_+$ such that $\lambda(K) &amp;gt; 0$. But $\int_K {f(t){\rm d}t}  = 0$, hence $f=0$ a.e. on $K$. This contradiction shows that $\lambda(E_+) = 0$. Similarly, $\lambda(E_-) = 0$. The theorem is thus established.&lt;/p&gt;&#xA;" OwnerUserId="2810" LastEditorUserId="2810" LastEditorDisplayName="" LastEditDate="2011-01-04T20:11:53.157" LastActivityDate="2011-01-04T20:11:53.157" CommentCount="2" />
  <row Id="16296" PostTypeId="1" AcceptedAnswerId="16297" CreationDate="2011-01-04T06:27:00.083" Score="2" ViewCount="78" Body="&lt;p&gt;Let $c$ be the set of all convergent sequences and $c_0$ the set of sequences tending to zero. How would one find the codimension of $c_0$? In particular, how would one find $\text{dim} &#xD;&#xA;\ c/c_0$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So one needs to choose some arbitrary sequence in $c_0$ and represent it as a linear combination of elements in $c/c_0$?&lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2011-01-04T06:44:54.197" Title="Codimension and sequences" Tags="&lt;linear-algebra&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" />
  <row Id="16297" PostTypeId="2" ParentId="16296" CreationDate="2011-01-04T06:38:40.553" Score="6" ViewCount="" Body="&lt;p&gt;Your final paragraph makes no sense. The elements of $c/c_0$ are &lt;em&gt;equivalence classes&lt;/em&gt; of sequences in $c$: they are elements of the form $[a] = \{b\in c\mid a-b\in c_0\}$ where $a$ is some element of $c$. You cannot express an element of $c_0$ as a linear combination of elements of $c/c_0$, because they are not even the same kind of things: the elements of $c_0$ are sequences, the elements of $c/c_0$ are sets of sequences.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The simplest way to find the codimension is to find an onto linear transformation from $c$ to some vector space $\mathbf{V}$ with kernel exactly $c_0$; the codimension will be dimension of $\mathbf{V}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is easy in this case: you can either work abstractly, directly with $c/c_0$, or try to find the map. Herer the former works: when will $a,b\in c$ be congruent modulo $c_0$? If and only if $a-b\in c_0$, if and only if $\lim(a-b)=0$, if and only if $\lim a = \lim b$. The equivalence classes correspond to the limits of the sequences. This suggests exactly which map to pick: let $T\colon c\to\mathbf{F}$ be the map to the underlying field given by $T(a) = \lim a$. It is straightforward to see that $T$ is linear, and that $\ker(T) = c_0$. Thus, $\mathbf{F} \cong c/c_0$. Since the dimension of $\mathbf{F}$ is $1$, the dimension of $c/c_0$ is $1$, so the codimension of $c_0$ in $c$ is $1$.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-04T06:44:54.197" LastActivityDate="2011-01-04T06:44:54.197" CommentCount="6" />
  <row Id="16298" PostTypeId="2" ParentId="3977" CreationDate="2011-01-04T08:27:26.013" Score="2" ViewCount="" Body="&lt;p&gt;GeoGebra 5.0 beta (with 3D) released:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.geogebra.org/forum/viewtopic.php?f=52&amp;amp;t=19846&quot; rel=&quot;nofollow&quot;&gt;http://www.geogebra.org/forum/viewtopic.php?f=52&amp;amp;t=19846&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5367" LastActivityDate="2011-01-04T08:27:26.013" CommunityOwnedDate="2011-01-04T08:27:26.013" />
  <row Id="16299" PostTypeId="2" ParentId="12815" CreationDate="2011-01-04T09:09:31.087" Score="0" ViewCount="" Body="&lt;p&gt;However, in the theory of dynamical systems appears naturally the value of the zeta function for half-odd numbers like 3/2! (or 5/2). Can someone &quot;explain&quot; this?&lt;/p&gt;&#xA;" OwnerUserId="5368" LastActivityDate="2011-01-04T09:09:31.087" CommentCount="1" />
  <row Id="16300" PostTypeId="1" CreationDate="2011-01-04T10:14:06.997" Score="2" ViewCount="214" Body="&lt;p&gt;How to find the inverse function of the following:&#xA;$f\left( x \right)=\sum_{i=0}^{x}{\frac{2^{i}}{i!}}$&lt;/p&gt;&#xA;" OwnerUserId="953" LastEditorUserId="3971" LastEditorDisplayName="" LastEditDate="2011-01-04T10:34:57.177" LastActivityDate="2011-01-06T15:54:49.970" Title="Inverse function of summation" Tags="&lt;sequences-and-series&gt;&lt;functions&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="1" />
  <row Id="16301" PostTypeId="1" AcceptedAnswerId="16332" CreationDate="2011-01-04T10:45:29.720" Score="2" ViewCount="122" Body="&lt;p&gt;Hi all&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an exercise I was doing a proof about equitconitnuity of a certain function. I noticed that I am always choosing the limits in a way that I finally get:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$Expression &amp;lt; \epsilon$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However it wouldn't hurt showing that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$Expression \leq \epsilon$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;wouldn't it? Since $\epsilon$ is getting infinitesimly small. I have been doing this type of $\epsilon$ proofs quite some time now (Convergence of functions/sequences, limites, etc...), but never asked myself that question. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Am I allowed to write  $Expression \leq \epsilon$ ? (and if so, when)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ftiaronsem&lt;/p&gt;&#xA;" OwnerUserId="3188" LastActivityDate="2011-01-04T15:39:14.000" Title="$Expression&lt;\epsilon$ really necessary for convergence / equicontinuity?" Tags="&lt;analysis&gt;&lt;limit&gt;" AnswerCount="4" />
  <row Id="16302" PostTypeId="1" CreationDate="2011-01-04T10:47:30.727" Score="5" ViewCount="300" Body="&lt;ul&gt;&#xA;&lt;li&gt;Question : Prove that the number &lt;img src=&quot;http://i.imgur.com/bsZAM.gif&quot; alt=&quot;alt text&quot;&gt; is &lt;strong&gt;never&lt;/strong&gt; divisible by 5.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="5315" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2011-01-04T17:36:24.617" LastActivityDate="2011-01-05T13:31:52.963" Title="Prove that $\sum_{i=0}^n2^{3i} \binom {2n+1}{2i+1}$ is never divisible by 5" Tags="&lt;combinatorics&gt;&lt;binomial-coefficients&gt;" AnswerCount="3" CommentCount="5" FavoriteCount="0" />
  <row Id="16303" PostTypeId="2" ParentId="3977" CreationDate="2011-01-04T10:53:55.893" Score="0" ViewCount="" Body="&lt;p&gt;If you're on a mac, theres a plotting software build in.&lt;br/&gt;&#xA;Go to Applications / Utilities / Grapher.&lt;br /&gt;&#xA;It's easy and powerful, can plot 2D and 3D graphs.&lt;/p&gt;&#xA;" OwnerUserId="5370" LastActivityDate="2011-01-04T10:53:55.893" CommunityOwnedDate="2011-01-04T10:53:55.893" />
  <row Id="16304" PostTypeId="2" ParentId="16301" CreationDate="2011-01-04T10:57:31.477" Score="3" ViewCount="" Body="&lt;p&gt;Since $\varepsilon&amp;gt;0$ is arbitrary it does not matter. If you have a non strict inequality for each $\varepsilon&amp;gt;0$ you can get strict inequality by adding arbitrary $\eta&amp;gt;0$.&lt;/p&gt;&#xA;" OwnerUserId="4742" LastActivityDate="2011-01-04T10:57:31.477" CommentCount="3" />
  <row Id="16306" PostTypeId="1" CreationDate="2011-01-04T11:14:50.623" Score="0" ViewCount="89" Body="&lt;p&gt;&lt;strong&gt;Edited Question&lt;/strong&gt;:&#xA;Let $\mu$ be some measure on $\mathbb{R^d}$ and $m$ the Lebesgue measure.&#xA;Define the maximal function M(x) = $sup(\mu(B(x,r)/m(B(x,r)))$ (the supremum is taken over $r&gt;0$).&#xA;Why is M a measurable function?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Old Question&lt;/strong&gt;:&#xA;In class we defined the &quot;maximal function&quot; similar to how it's defined &lt;a href=&quot;http://en.wikipedia.org/wiki/Hardy%E2%80%93Littlewood_maximal_function&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;, but instead of taking the supremum on all balls containing x, we took all balls centered at x.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It was then claimed this function is measurable, but I can't see why.&#xA;I do understand that with the definition given by my wikipedia link, the preimage of (a,$\infty$] is open and thus the maximal function is measurable. But why is the centered maximal function measurable?&#xA;(I missed this lesson, and my friends say the professor said it's obvious, so the answer is probably easy).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;: I meant to gives &lt;a href=&quot;http://en.wikipedia.org/wiki/Maximal_function&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; link to wikipedia, where the maximal function is defined with balls containing, not necessarily centered at, x.&lt;/p&gt;&#xA;" OwnerUserId="3533" LastEditorUserId="3533" LastEditorDisplayName="" LastEditDate="2011-01-04T12:40:41.967" LastActivityDate="2011-01-04T12:40:41.967" Title="Is the centered maximal function measurable?" Tags="&lt;measure-theory&gt;" CommentCount="8" />
  <row Id="16307" PostTypeId="2" ParentId="16271" CreationDate="2011-01-04T11:17:36.337" Score="1" ViewCount="" Body="&lt;p&gt;Well the most simple way is to plot the &lt;a href=&quot;http://en.wikipedia.org/wiki/Kernel_density_estimation&quot; rel=&quot;nofollow&quot;&gt;kernel density function&lt;/a&gt;. If your data has a bimodal distribution then it will certainly show up in the graph. For more complicated analysis you can try to fit a &lt;a href=&quot;http://en.wikipedia.org/wiki/Mixture_model&quot; rel=&quot;nofollow&quot;&gt;mixture model&lt;/a&gt;, in order to determine the cause of bimodality. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Try asking in site &lt;a href=&quot;http://stats.stackexchange.com&quot;&gt;Cross Validated&lt;/a&gt; for more answers.&lt;/p&gt;&#xA;" OwnerUserId="4742" LastActivityDate="2011-01-04T11:17:36.337" />
  <row Id="16308" PostTypeId="1" CreationDate="2011-01-04T11:54:35.107" Score="0" ViewCount="104" Body="&lt;p&gt;What is the general term $(a_n)$ of the alternating sequence $cos(3*pi*n/2)$ from $1$ to $\infty$, $n \in \mathbb{N}$ ?&lt;/p&gt;&#xA;" OwnerUserId="2723" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2011-01-04T14:26:07.927" LastActivityDate="2011-01-04T14:26:07.927" Title="What is the general term $(a_n)$ of the alternating sequence $cos(3*pi*n/2)$?" Tags="&lt;calculus&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="16309" PostTypeId="2" ParentId="16300" CreationDate="2011-01-04T13:14:11.550" Score="2" ViewCount="" Body="&lt;p&gt;In case you actually want to find the inverse cdf of the Poisson distribution (with parameter $\lambda=2$), there seem to be many statistical/mathematical softwares that can do the job for you. For example, see &lt;a href=&quot;http://www.mathworks.com/help/toolbox/stats/poissinv.html&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;: The following refers to the OP's question in a comment below (and above). Let $F$ and $F^{-1}$ denote a CDF and its inverse, respectively. Given $a&amp;gt;0$, define a function $F_a$ by $F_a (x) = aF(x)$. If $y=F_a (x)$, then $y=aF(x)$, hence, $F(x)=y/a$, and so $x=F^{-1}(y/a)$. It thus follows that $F_a^{ - 1} (y) = F^{ - 1} (y/a)$, where $F_a^{-1}$ denotes the inverse of $F_a$. Returning to the original question, take $F(x) = e^{ - \lambda } \sum\nolimits_{i = 0}^{\left\lfloor x \right\rfloor } {\frac{{\lambda ^i }}{{i!}}}$ and $a=e^\lambda$ (where $\lambda=2$). Then, $F_a (x) = \sum\nolimits_{i = 0}^{\left\lfloor x \right\rfloor } {\frac{{\lambda ^i }}{{i!}}}$. Hence we are done, by the relation $F_a^{ - 1} (y) = F^{ - 1} (y/a)$. Note: Here, $F$ is an increasing step function, hence not invertible in the usual sense. However, the inverse can be defined as described in the link above.&lt;/p&gt;&#xA;" OwnerUserId="2810" LastEditorUserId="2810" LastEditorDisplayName="" LastEditDate="2011-01-06T15:54:49.970" LastActivityDate="2011-01-06T15:54:49.970" CommentCount="3" />
  <row Id="16310" PostTypeId="1" AcceptedAnswerId="16318" CreationDate="2011-01-04T13:14:32.447" Score="17" ViewCount="585" Body="&lt;p&gt;I'm doing the exercises in &quot;Introduction to commutive algebra&quot; by Atiyah&amp;amp;MacDonald. In chapter two, exercises 24-26 assume knowledge of the Tor functor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have tried Googling the term, but I don't find any readable sources. Wikipedia's explanation use the the term &quot;take the homology&quot;, which I don't understand (yet).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there any good explanations of what the Tor functor is available online not assuming any knowledge about homology?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first exercise:&#xA;&quot;If M is an A-module, TFAE:&#xA;1) M is flat&#xA;2) $Tor_n^A (M,N)=0$ for all n&gt;0 and all A-modules N.&#xA;3) $Tor_1^A (M,N)=0$ for all A-modules N.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance.&lt;/p&gt;&#xA;" OwnerUserId="4284" LastActivityDate="2011-02-26T22:03:31.510" Title="What is the Tor functor?" Tags="&lt;commutative-algebra&gt;&lt;homological-algebra&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="7" />
  <row Id="16311" PostTypeId="1" AcceptedAnswerId="16319" CreationDate="2011-01-04T13:23:24.060" Score="1" ViewCount="100" Body="&lt;p&gt;Hi!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a point 'a' with known coordinates, from which I have drawn a tangent to a circle with centre 'c' which is also known. What is the best way of finding the coordinates of point 'b', the touching point between the tangent and the circle? Here is a diagram: &lt;a href=&quot;http://i.imgur.com/gcKYn.jpg&quot; rel=&quot;nofollow&quot;&gt;http://i.imgur.com/gcKYn.jpg&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;David&lt;/p&gt;&#xA;" OwnerUserId="5371" LastActivityDate="2011-01-04T16:47:01.667" Title="Find Coordinates of Touching Point of a Tangent on a Circle" Tags="&lt;geometry&gt;&lt;trigonometry&gt;&lt;circle&gt;" AnswerCount="2" />
  <row Id="16312" PostTypeId="1" AcceptedAnswerId="16322" CreationDate="2011-01-04T13:30:46.043" Score="0" ViewCount="67" Body="&lt;p&gt;How to show that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f(x)= \lim_{k \to \infty} \Bigl(\lim_{j \to \infty} \cos^{2j}(k!\cdot \pi \cdot x)\Bigl)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is the &lt;em&gt;Dirichlet's function&lt;/em&gt;.&lt;/p&gt;&#xA;" LastEditorUserId="3249" LastEditorDisplayName="" LastEditDate="2011-01-04T14:49:29.537" LastActivityDate="2011-01-04T14:49:29.537" Title="How to show that $f(x)= \lim_{k \to \infty} \Bigl(\lim_{j \to \infty} \cos^{2j}(k!\cdot \pi \cdot x)\Bigl)$" Tags="&lt;real-analysis&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="16313" PostTypeId="1" CreationDate="2011-01-04T13:38:29.983" Score="0" ViewCount="35" Body="&lt;p&gt;formula for trajectory splitting method in the middle&lt;/p&gt;&#xA;" OwnerUserId="5372" LastActivityDate="2011-01-04T13:38:29.983" ClosedDate="2011-01-05T07:52:25.673" Title="trajectory splitting method" Tags="&lt;physics&gt;" CommentCount="3" />
  <row Id="16314" PostTypeId="2" ParentId="16308" CreationDate="2011-01-04T13:39:03.993" Score="2" ViewCount="" Body="&lt;p&gt;Well, you nearly got it ;-)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lets write down the first terms:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\begin{align}&#xD;&#xA;k&amp;amp; \quad &amp;amp;cos(\frac{3\pi k}{2}) \\&#xD;&#xA;1&amp;amp; \quad &amp;amp; 0 \\&#xD;&#xA;2&amp;amp; \quad &amp;amp;-1 \\&#xD;&#xA;3&amp;amp; \quad &amp;amp; 0 \\&#xD;&#xA;4&amp;amp; \quad &amp;amp; 1 \\&#xD;&#xA;5&amp;amp; \quad &amp;amp; 0 \\&#xD;&#xA;6&amp;amp; \quad &amp;amp;-1 \\&#xD;&#xA;7&amp;amp; \quad &amp;amp; 0 \\&#xD;&#xA;8&amp;amp; \qquad &amp;amp; 1 \\&#xD;&#xA;\end{align}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we are obviously searching for something which is $-1$ every second and $1$ every fourth time. Wat comes to mind? $i^k$ Unfortunatelly every first time we have $i$ and every third $-i$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we have to find a way to cancl out $i$ every first and third time. We are therefore searching a $x$ so that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\begin{align}&#xD;&#xA;k \quad  &amp;amp; i^k   &amp;amp; x^k \\&#xD;&#xA;1 \quad  &amp;amp; i  &amp;amp;-i\\&#xD;&#xA;2 \quad &amp;amp; 1  &amp;amp; 1\\&#xD;&#xA;3 \quad &amp;amp;-i  &amp;amp; i\\&#xD;&#xA;4 \quad  &amp;amp;-1  &amp;amp;-1\\&#xD;&#xA;\end{align}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because if we had that, we simply would sum $x^k$ and $i^k$ , divide it by two and would be finished. After a little thinking&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x=i^k(-1)^k$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;comes to mind, since $(-1)^k$ has exactly the alternating properties we are searching.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So your general term is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a_n = \frac{i^k+i^k(-1)^k}{2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S. If anyone knows a tabular environment for MathJax please leave a comment ;-)&lt;/p&gt;&#xA;" OwnerUserId="3188" LastEditorUserId="3188" LastEditorDisplayName="" LastEditDate="2011-01-04T13:44:22.723" LastActivityDate="2011-01-04T13:44:22.723" CommentCount="3" />
  <row Id="16316" PostTypeId="2" ParentId="12815" CreationDate="2011-01-04T13:40:59.233" Score="0" ViewCount="" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Where do the many many proofs powerful enough to evaluate ζ(2n)  stumble when it comes to evaluating ζ(2n+1)?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Another method of evaluating $ \zeta(2n) \ $ not mentioned yet is by applying Parseval's Theorem to the Fourier series of the &lt;a href=&quot;http://mathworld.wolfram.com/BernoulliPolynomial.html&quot; rel=&quot;nofollow&quot;&gt;Bernoulli Polynomials&lt;/a&gt;.  The Fourier coefficients of $ B_n\ $ have the form $ c_k = (constants)\cdot \frac{1}{k} $.  By Parseval's Theorem,&#xA;$$ \int_0^1 |B_n(t)|^2 dt = \sum_{k\neq0} |c_k|^2 $$&#xA;$$ (constant) = (more\ constants) \sum_{k\neq0} \frac{1}{k^{2n}} $$&#xA;(Note that the constants depend upon n)&lt;br&gt;&#xA;The reason this method falls apart for odd integers is that, since Parseval's Theorem involves squaring the coefficients, so there is no way to get odd powers into the sum.&lt;/p&gt;&#xA;" OwnerUserId="5330" LastActivityDate="2011-01-04T13:40:59.233" CommentCount="2" />
  <row Id="16318" PostTypeId="2" ParentId="16310" CreationDate="2011-01-04T13:54:12.827" Score="19" ViewCount="" Body="&lt;p&gt;The Tor functors are the derived functors of the tensor product. The starting observation is that if $0 \to M&amp;#39; \to M \to M&amp;#39;&amp;#39; \to 0$ is a ses of modules and $N$ is any module (let's work over a fixed commutative ring $R$), then $M&amp;#39; \otimes N \to M \otimes N \to M&amp;#39;&amp;#39; \otimes N \to 0$ is exact, &lt;em&gt;but&lt;/em&gt; you don't necessarily have exactness at the first step. (This is what &quot;the tensor product is right-exact&quot; means.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, whenever you have such a sequence that fails to be short exact at just one step, the general philosophy is that it should be the end of a long exact sequence. The long exact sequence should be like&#xA;$$\dots \to ?_1 \to ?_2 \to ?_3 \to M&amp;#39; \otimes N \to M \otimes N \to M&amp;#39;&amp;#39; \otimes N \to 0$$&#xA;where the question marks indicate that we don't know what goes there yet. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The general approach is provided by the theory of &lt;a href=&quot;http://en.wikipedia.org/wiki/Derived_functor&quot; rel=&quot;nofollow&quot;&gt;derived functors&lt;/a&gt;. Derived functors allow one to construct, associated to a right (or left) exact functor $F$, a collection of functors $L_i F$ for $i \geq 0$ (I'm just going to handle the right-exact case henceforth), such that for each ses &#xA;$0 \to M&amp;#39; \to M \to M&amp;#39;&amp;#39; \to 0$, there &quot;almost ses&quot;&#xA;$$FM&amp;#39; \to FM \to FM&amp;#39;&amp;#39; \to 0$$&#xA;can be completed to a les&#xA;$$\dots \to L_2F(M&amp;#39;&amp;#39;) \to L_1F(M&amp;#39;) \to L_1F(M) \to L_1F(M&amp;#39;&amp;#39;) \to \to FM&amp;#39; \to FM \to FM&amp;#39;&amp;#39; \to 0.$$&#xA;This long exact sequence relies on  &quot;connecting homomorphisms&quot; $L_nF(M&amp;#39;&amp;#39;) \to L_{n-1}F(M&amp;#39;)$, which are required to be functorial in the ses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The exact construction of a derived functor uses projective resolutions of the object $M$ and a reasonable amount of diagram-chasing. I will refer you to books on homological algebra.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So suppose we consider the functor sending $M \to M \otimes N$. Its derived functors are denoted $\mathrm{Tor}_i(M,N)$. This means that the Tor functors provide the missing part of the les that expands out the &quot;almost ses&quot; of the tensor product above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One reason the Tor functors are so useful is that they provide a very effective criterion for flatness. Recall that a module $N$ is flat if tensoring with it is exact. When you derive an exact functor, you just get the initial functor in dimension zero (actually, you always do for this), and then zero elsewhere. So if $N$ is flat, $\mathrm{Tor}_i(M,N) = 0$ for $i&amp;gt;0$. In fact, using the les, one can easily show that the implication is reversible, even if one only has $\mathrm{Tor}_1(M,N)=0$ for all $M$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me give an example where things work out very nicely. If $R$ is a noetherian local ring with residue field $k$, and $M$ is a finite $R$-module, then it turns out $M$ is &lt;em&gt;free&lt;/em&gt; if and only if $\mathrm{Tor}_1(k,M)=0$; in particular, freeness is the same thing for flatness (under these hypotheses).&#xA;(For a proof, see the argument behind Theorem 3.2 in ch. 14 of &lt;a href=&quot;http://www.people.fas.harvard.edu/~amathew/CRing.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.people.fas.harvard.edu/~amathew/CRing.pdf&lt;/a&gt;.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An application of this is provided by the observation (which can be deduced from this) that if the $\mathrm{Tor}$ functors of the pair $(k,k)$ vanish in high dimension, then the &lt;a href=&quot;http://en.wikipedia.org/wiki/Global_dimension&quot; rel=&quot;nofollow&quot;&gt;global dimension&lt;/a&gt; of the ring is finite.  This is fairly interesting because the global dimension is a statement about all modules (and, strictly speaking, involving the functors $\mathrm{Ext}$ that derive Hom, not Tor), while the above reduction is to one involving only the residue field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is, in fact, the key observation behind one half the proof that a noetherian local ring is &lt;a href=&quot;http://en.wikipedia.org/wiki/Regular_local_ring&quot; rel=&quot;nofollow&quot;&gt;regular&lt;/a&gt; iff its global dimension is finite. This argument can be found in EGA 0-IV, sec. 17 (it doesn't presuppose anything beyond dimension theory, which you can read about in sec. 16).&lt;/p&gt;&#xA;" OwnerUserId="536" LastEditorUserId="536" LastEditorDisplayName="" LastEditDate="2011-01-04T23:50:36.493" LastActivityDate="2011-01-04T23:50:36.493" CommentCount="5" />
  <row Id="16319" PostTypeId="2" ParentId="16311" CreationDate="2011-01-04T13:55:56.857" Score="1" ViewCount="" Body="&lt;p&gt;If numeric precision is not a problem you can use trigonometry.  Find ab from the Pythagorean theorem, find the angle at a from bc/ac, calculate the angle of ac relative to the horizontal, add the triangle angle and get the slope of ab.  If the angle gets very small, you may want to use the small angle approximations.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-01-04T13:55:56.857" />
  <row Id="16320" PostTypeId="2" ParentId="16172" CreationDate="2011-01-04T14:00:10.080" Score="0" ViewCount="" Body="&lt;p&gt;Sorry, i have little idea on this particular topic. What i really want is help from any of you please. I am an undergraduate student of Bayero University, Kano in Nigeria studying mathematics. In fulfilment of the requirement of my degree program, i decided to write on ''Paracompactness'' which require me to write a literature review on &#xA;1) The concept of topology&#xA;2) The concept of metrizability&#xA;3) The concept of compactness and &#xA;4) The concept of paracompactness&#xA;and i am totally short of materials. I will be very glad if anyone can help me with materials or references. My email is hafiz201@gmail.com&lt;/p&gt;&#xA;" OwnerUserId="5374" LastActivityDate="2011-01-04T14:00:10.080" CommentCount="1" />
  <row Id="16321" PostTypeId="2" ParentId="16312" CreationDate="2011-01-04T14:01:59.550" Score="0" ViewCount="" Body="&lt;p&gt;If x is rational, once k gets large enough the cosine is 1 independent of j.  If x is irrational, the cosine is less than one so the j limit will be zero.  This is true for all k.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-01-04T14:01:59.550" />
  <row Id="16322" PostTypeId="2" ParentId="16312" CreationDate="2011-01-04T14:02:45.303" Score="2" ViewCount="" Body="&lt;p&gt;Suppose $x = \frac {p}{q} \in \mathbb{Q}$. If $k&amp;gt;q$ then $k! \cdot x \in \mathbb {Z}$ so $ \cos(k!\cdot \pi \cdot x) = \pm 1$&#xA;$$  k&amp;gt;q,\; \; \lim_{j \to \infty} \cos(k!\cdot \pi \cdot x)^{2j} = \lim_{j \to \infty} (\pm 1)^{2j} = 1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so taking the limit of $k \rightarrow \infty$ you get also 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $x \notin \mathbb{Q}$ then for every k we have $k! \cdot x \notin \mathbb {Z}$ so $ |\cos(k!\cdot \pi \cdot x)| &amp;lt; 1$ therefore&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$  \lim_{k \to \infty} \lim_{j \to \infty} \cos(k!\cdot \pi \cdot x)^{2j} = \lim_{k \to \infty} 0 = 0 $$&lt;/p&gt;&#xA;" OwnerUserId="2125" LastActivityDate="2011-01-04T14:02:45.303" />
  <row Id="16323" PostTypeId="1" AcceptedAnswerId="16324" CreationDate="2011-01-04T14:05:34.767" Score="3" ViewCount="127" Body="&lt;p&gt;If $x^3 + \frac{1}{x^3} = 52$, then what is $x^2 + \frac{1}{x^2}$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure which formulae or methods should I used to solve this problem, so could somebody show me a way? what should I be looking for?&lt;/p&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-04T17:14:25.087" LastActivityDate="2011-01-05T00:23:12.593" Title="Need help with a power equation" Tags="&lt;algebra-precalculus&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="16324" PostTypeId="2" ParentId="16323" CreationDate="2011-01-04T14:07:47.233" Score="6" ViewCount="" Body="&lt;p&gt;Hint: Define $y=x+\frac{1}{x}$.  There are frequently problems where this transformation helps.  The equation for $y^3$ has only one real root, which you can probably find by inspection.  Then the quadratic formula will give you the other two.  The problem isn't as clean as I thought originally.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added:&lt;/strong&gt;  this works with symmetric polynomials as well.  Imagine you were given $x^6-52x^3+1=0$.  You might be scared by the sixth degree, but if you notice the coefficients are symmetric around the center, you can divide by $x^3$, define $y=x+\frac{1}{x}$, and cut the degree in half.  Another approach, which doesn't work as slickly on the current problem, is to define $u=x^3$ and rewrite it as $u+\frac{1}{u}=52$.  This is a quadratic which will yield to the usual techniques.  I have seen cases where people give you what looks like a high degree polynomial, but this will tame them.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2011-01-04T17:12:58.723" LastActivityDate="2011-01-04T17:12:58.723" />
  <row Id="16325" PostTypeId="1" CreationDate="2011-01-04T14:09:59.660" Score="4" ViewCount="126" Body="&lt;p&gt;Hello,&#xA;it might be a stupid question but I was discussing with a colleague when the 3rd axiom of probability (sigma additivity) is really needed. &#xA;I argue that in the case of a discrete distribution, say a single die, the first two axioms are sufficient as this distribution has a finite number of events. Is this right?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And then I am stretching it a bit by arguing that for simple/well behaved continuous distributions, such as the uniform distribution, the first two axioms are sufficient to ensure a proper probability distribution. Is that right?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I always had the idea that the 3rd axiom was rather there to ensure that more cumbersome distributions or convolutions of distributions would still be proper probability distributions (although I don't have an example at hand). Or is axiom 3 much more vital than I understand?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for any input!&#xA;Best,&#xA;Stefan&lt;/p&gt;&#xA;" OwnerUserId="5375" LastEditorUserId="367" LastEditorDisplayName="" LastEditDate="2011-01-04T18:35:45.733" LastActivityDate="2011-01-04T18:35:45.733" Title="3rd axiom of probability for discrete distribution" Tags="&lt;probability&gt;&lt;probability-theory&gt;&lt;set-theory&gt;&lt;probability-distributions&gt;" AnswerCount="4" CommentCount="2" />
  <row Id="16326" PostTypeId="2" ParentId="16325" CreationDate="2011-01-04T14:19:16.347" Score="2" ViewCount="" Body="&lt;p&gt;The power of the third axiom is illustrated &lt;a href=&quot;http://en.wikipedia.org/wiki/Probability_axioms#Proofs&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.  &lt;/p&gt;&#xA;" OwnerUserId="2810" LastEditorUserId="2810" LastEditorDisplayName="" LastEditDate="2011-01-04T14:43:26.153" LastActivityDate="2011-01-04T14:43:26.153" CommentCount="2" />
  <row Id="16327" PostTypeId="2" ParentId="16325" CreationDate="2011-01-04T14:23:54.640" Score="1" ViewCount="" Body="&lt;p&gt;Without the third axiom we can define many distributions that don't work like probabilities are supposed to.  For example, the distribution where we assign 1 to all subsets of the universe.  That doesn't look like a probability, but it satisfies the first two axioms, so we need an axiom to rule it out.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-01-04T14:23:54.640" />
  <row Id="16328" PostTypeId="2" ParentId="16325" CreationDate="2011-01-04T14:42:30.280" Score="6" ViewCount="" Body="&lt;p&gt;The first two axioms given on Wikipedia don't imply finite additivity, which you seem to be assuming.  This assumption is enough to do probability with a finite sample space but not with a countable sample space; you need the assumption of $\sigma$-additivity here or else you can't do simple intuitive things like compute the expected number of times you have to roll a die to get a certain result.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you only assume finite additivity, you get a lot of weird measures; for example any &lt;a href=&quot;http://en.wikipedia.org/wiki/Ultrafilter&quot; rel=&quot;nofollow&quot;&gt;non-principal ultrafilter&lt;/a&gt; on an infinite set defines a finitely additive measure.  Whether you want to allow these as probability measures is up to you, but the fact is that it's harder to prove theorems in this generality.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2011-01-04T14:42:30.280" />
  <row Id="16329" PostTypeId="2" ParentId="16301" CreationDate="2011-01-04T14:43:59.287" Score="3" ViewCount="" Body="&lt;p&gt;You are right, it doesn't matter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given an arbitrary $\epsilon &amp;gt; 0$, you can make your expression strictly less than $\epsilon$ by choosing your $\delta$ such that the expression is less than or equal to $\epsilon/2$, for example. (This is possible since you know that you can choose $\delta$ so that your expression is less than or equal to &lt;em&gt;any&lt;/em&gt; positive number that you wish.)&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2011-01-04T14:43:59.287" CommentCount="1" />
  <row Id="16330" PostTypeId="2" ParentId="16325" CreationDate="2011-01-04T14:46:41.100" Score="1" ViewCount="" Body="&lt;p&gt;I am suspecting that you are really asking whether $\sigma$-additivity is really necessary when simple additivity might be enough. If we exclude third axiom as it is defined in &lt;a href=&quot;http://en.wikipedia.org/wiki/Probability_axioms&quot; rel=&quot;nofollow&quot;&gt;wikipedia&lt;/a&gt;, then the probability need not to be additive, which is definitely not something you want. Now for finite sample space $\sigma$-additivity and simple additivity coincides, but for infinite sample space $\sigma$-additivity is necessary as other posters pointed out. Take for example probability space $(\Omega, 2^\Omega, P)$ where $\Omega=\{\omega_1,...,\omega_n,...\}$. Now &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Omega=\bigcup_{i=1}^\infty\{\omega_i\}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\{\omega_i\}\cap\{\omega_j\}=\varnothing,i\neq j$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now $\sigma$-additivity implies &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$P(\Omega)=\sum_{i=1}^\infty P(\omega_i)=1$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;without $\sigma$-additivity this statement need not be true, contradicting axiom 2.&lt;/p&gt;&#xA;" OwnerUserId="4742" LastActivityDate="2011-01-04T14:46:41.100" />
  <row Id="16331" PostTypeId="1" CreationDate="2011-01-04T14:47:56.447" Score="7" ViewCount="157" Body="&lt;p&gt;While trying to answer &lt;a href=&quot;http://math.stackexchange.com/questions/16323&quot;&gt;this question&lt;/a&gt;, I got stuck showing that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sqrt[3]{26+15\sqrt{3}}=2+\sqrt{3}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The identity is easy to show if you already know the $2+\sqrt{3}$ part; just cube the thing. If you don't know this, however, I am unsure how one would proceed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That got me thinking... if you have some quadratic surd $a+b\sqrt{c}$, where $a$, $b$, and $c$ are integers, and $c$ is not a perfect square, how do you find out if that surd is the cube of some other surd?&lt;/p&gt;&#xA;" OwnerUserId="5296" LastActivityDate="2011-01-04T15:45:23.633" Title="How to check if a quadratic surd is a perfect cube?" Tags="&lt;algebra-precalculus&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="16332" PostTypeId="2" ParentId="16301" CreationDate="2011-01-04T14:54:54.740" Score="4" ViewCount="" Body="&lt;p&gt;Suppose for every $\epsilon &amp;gt;0$, there is an $N$ such that $n&amp;gt;N$ implies $x_n\leq\epsilon$. Let $k&amp;gt;0$. Then $\frac{k}{2}&amp;gt;0$, so there is an $M$ such that $n&amp;gt;M$ implies $x_n\leq\frac{k}{2}$ which implies that $x_n&amp;lt;k$. This corresponds to the original definition of continuity, doesn't it?&lt;/p&gt;&#xA;" OwnerUserId="4594" LastActivityDate="2011-01-04T14:54:54.740" CommentCount="1" />
  <row Id="16333" PostTypeId="2" ParentId="16331" CreationDate="2011-01-04T15:01:10.163" Score="2" ViewCount="" Body="&lt;p&gt;If you write $(d+e \sqrt{f})^3=a+b \sqrt{c}$ and collect terms, you see $c=f$, then $d^3+3de^2=a, 3d^2e+e^3c=b$.  For integers, $e$ has to be a factor of $b$, $d$ has to be a factor of $a$ and you can just see if it works pretty easily.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-01-04T15:01:10.163" CommentCount="4" />
  <row Id="16334" PostTypeId="2" ParentId="16301" CreationDate="2011-01-04T15:39:14.000" Score="1" ViewCount="" Body="&lt;p&gt;There are already some good answers here, so I won't bother repeating them. However, I'd like to add a cautionary note that there &lt;em&gt;are&lt;/em&gt; instances where using the wrong inequality gives nonsense — for example, it is &lt;em&gt;essential&lt;/em&gt; in the statement of the Banach fixed point theorem that the Lipschitz constant is strictly less than one. If I remember correctly, even $\dfrac{\| f(x) - f(y) \|}{\| x - y \|} &amp;lt; 1$ for all $x \ne y$ is not good enough; we must have $\sup \left\{ \dfrac{\| f(x) - f(y) \|}{\| x - y \|} : x \ne y \right\} &amp;lt; 1$.&lt;/p&gt;&#xA;" OwnerUserId="5191" LastActivityDate="2011-01-04T15:39:14.000" CommentCount="1" />
  <row Id="16335" PostTypeId="2" ParentId="16331" CreationDate="2011-01-04T15:45:23.633" Score="3" ViewCount="" Body="&lt;p&gt;As has been mentioned this is simply solved since it denests already in the field generated by the radicand. Generally this is not true, but there is a general denesting structure theorem that applies. Here's an extract from my &lt;a href=&quot;http://math.stackexchange.com/questions/4680/simplification-of-expressions-containing-radicals/4697#4697&quot;&gt;Sep 15 post&lt;/a&gt; on denesting radicals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;DENESTING STRUCTURE THEOREM&lt;/strong&gt;$\;\; \;$  Let $\rm\; F \;$ be a real field and &#xA;$\rm\; F&amp;#39; = F(q_1^{1/d1},\ldots,q_k^{1/dk}) \;$ be a real radical extension of $\rm\; F \;$ &#xA;of degree $\rm\; n \;$. By $\rm\; B = \{b_0,\ldots, b_{n-1}\}$ denote the standard &#xA;basis of $\rm\; F&amp;#39; \;$ over $\rm\; F \;$. If $\rm\; r \;$ is in $\rm\; F&amp;#39; \;$ and $\rm\; d \;$ is a positive integer such &#xA;that $\rm\; r^{1/d} \;$ denests over $\rm\; F \;$ using only real radicals, that is, &#xA;$\rm\; r^{1/d} \;$ is in $\rm\; F(a_1^{1/t_1},\ldots,a_m^{1/t_m}) \;$ for some positive integers &#xA;$\rm\; t_i \;$ and positive $\rm\; a_i \in F \;$, then there exists a nonzero $\rm\; q \in F \;$ and a &#xA;$\rm\; b \in B \;$ such that $\rm\; (q b r)^{1/d} \in F&amp;#39; \;$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I.e. multiplying the radicand by a $\rm\; q \;$ in the base field $\rm\; F \;$ &#xA;and a power product $\rm\; b \;=\; q_1^{e_1/d_1}\cdots q_k^{e_k/d_k} \;$ we can &#xA;normalize any denesting so that it denests in the field defined &#xA;by the radicand. E.g. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \sqrt{\sqrt[3]5 - \sqrt[3]4} \;\;=\; \frac{1}3 (\sqrt[3]2 + \sqrt[3]{20} - \sqrt[3]{25})$$&#xA;normalises to $$ \sqrt{18\ (\sqrt[3]10 - 2)} \;\;=\; 2 + 2\ \sqrt[3]{10} - \sqrt[3]{10}^2 $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An example with nontrivial $\rm\:b$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$  \sqrt{12 + 5\ \sqrt 6} \;\;=\; (\sqrt 2 + \sqrt 3)\ 6^{1/4} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;normalises to &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \sqrt{\frac{1}3 \sqrt{6}\: (12 + 5\ \sqrt 6)} \;\;=\; 2 + \sqrt{6} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See &lt;a href=&quot;http://math.stackexchange.com/questions/4680/simplification-of-expressions-containing-radicals/4697#4697&quot;&gt;said post&lt;/a&gt; for further details and references.&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2011-01-04T15:45:23.633" CommentCount="4" />
  <row Id="16336" PostTypeId="1" CreationDate="2011-01-04T15:54:57.870" Score="2" ViewCount="193" Body="&lt;p&gt;I'm writing a software that does simple mathematical calculations for lazy people.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need ideas.&#xA;I already wrote some pieces of code that calculates 'Quadratic formula', but I have no idea what to do next.&lt;/p&gt;&#xA;" OwnerUserId="5381" LastActivityDate="2011-01-04T19:33:17.177" Title="Lazy math program?" Tags="&lt;math-software&gt;" AnswerCount="1" CommentCount="8" />
  <row Id="16337" PostTypeId="2" ParentId="16323" CreationDate="2011-01-04T16:44:40.003" Score="5" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\ \ $ Exploit the symmetry. $\ $ Put $\rm\quad x_k = x^k + x^{-k}\:.\ $ Then we have  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\quad\ x_1^2 = (x+1/x)^2 = x_2 + 2\quad\quad\ $ so $\rm\ \ x_2 = x_1^2 - 2\quad\quad\quad\quad\quad(2)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\quad\ x_1^3 = (x+1/x)^3 = x_3 + 3\ x_1\ $ so $\rm\quad x_3 = x_1^3 - 3\ x_1\quad\quad\quad (3)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, put $\rm\:x_3 = 52\:$ in $(3)$, solve it for $\rm\:x_1\:,\:$ then plug $\rm\:x_1\:$ into $(2)$ to get $\rm\:x_2\:$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Analogous methods work for arbitrary symmetric polynomials, e.g. &lt;a href=&quot;http://math.stackexchange.com/questions/8058/how-to-solve-this/8072#8072&quot;&gt;see here.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2011-01-04T18:56:15.393" LastActivityDate="2011-01-04T18:56:15.393" />
  <row Id="16338" PostTypeId="2" ParentId="16311" CreationDate="2011-01-04T16:47:01.667" Score="0" ViewCount="" Body="&lt;p&gt;You have (x-xc)^2+(y-yc)^2=r^2 for the circle. You also have (y-ya)=m(x-xa). m=(xc-xa)/(yc-ya). Take the second equation, solve for y, plot it into the first equation, and solve for x. &lt;/p&gt;&#xA;" OwnerUserId="5101" LastActivityDate="2011-01-04T16:47:01.667" />
  <row Id="16339" PostTypeId="1" AcceptedAnswerId="16340" CreationDate="2011-01-04T17:38:27.140" Score="4" ViewCount="193" Body="&lt;p&gt;Hello altogether&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the title says,  am searching for a proof of &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If $A,B \in \mathbb{R}^{n\times n}$ and&#xA;  $AB=0$ then $\mathrm{rank}(A)+\mathrm{rank}(B) \leq n$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am doing this as preparation for an upcoming exam and can't figure a way to start. Please just post small hints as answers. I will try to go from there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you &lt;/p&gt;&#xA;&#xA;&lt;p&gt;ftiaronsem&lt;/p&gt;&#xA;" OwnerUserId="3188" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-04T17:40:37.060" LastActivityDate="2011-01-04T17:45:32.033" Title="Proof of: $AB=0 \Rightarrow Rank(A)+Rank(B) \leq n$" Tags="&lt;linear-algebra&gt;&lt;matrices&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" />
  <row Id="16340" PostTypeId="2" ParentId="16339" CreationDate="2011-01-04T17:45:32.033" Score="5" ViewCount="" Body="&lt;p&gt;By the Rank-Nullity Theorem, $\mathrm{rank}(A)+\mathrm{nullity}(A)=n$. The problem would be solved if you could show that $\mathrm{rank}(B)\leq\mathrm{nullity}(A)$. Presumably, $AB=0$ will play a role in that, since the result is false otherwise (if $A$ and $B$ were both invertible, for example, then $AB\neq 0$, and $\mathrm{rank}(A)+\mathrm{rank}(B) = 2n\gt n$). &lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2011-01-04T17:45:32.033" CommentCount="6" />
  <row Id="16342" PostTypeId="1" CreationDate="2011-01-04T17:59:52.247" Score="1" ViewCount="198" Body="&lt;p&gt;Why do you need logarithms?&#xA;In what situations do you use them?&lt;/p&gt;&#xA;" OwnerUserId="5384" LastActivityDate="2011-01-04T23:43:42.360" Title="Point of logarithms?" Tags="&lt;logarithms&gt;" AnswerCount="3" CommentCount="3" />
  <row Id="16343" PostTypeId="2" ParentId="16342" CreationDate="2011-01-04T18:01:38.463" Score="2" ViewCount="" Body="&lt;p&gt;See &lt;a href=&quot;http://www.math.umt.edu/tmme/vol5no2and3/TMME_vol5nos2and3_a14_pp.337_344.pdf&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2011-01-04T18:01:38.463" CommentCount="1" />
  <row Id="16344" PostTypeId="2" ParentId="7295" CreationDate="2011-01-04T18:33:19.823" Score="1" ViewCount="" Body="&lt;p&gt;This question was answered &lt;a href=&quot;http://mathoverflow.net/questions/43069/is-there-a-list-of-all-connected-t-0-spaces-with-5-points&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="367" LastActivityDate="2011-01-04T18:33:19.823" CommunityOwnedDate="2011-01-04T18:33:19.823" />
  <row Id="16345" PostTypeId="2" ParentId="16290" CreationDate="2011-01-04T18:38:59.047" Score="3" ViewCount="" Body="&lt;p&gt;Isaac had the right intuition. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/JEswV.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I used Matlab to globally optimize the radius under the constrait that all three circles have the same radius, are in the rectangle and don't intersect each other. What I got is shown above,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;R = 1.41699, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pos1 = (6.58301,2.58301)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pos2 = (1.41699,2.58301)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pos3 = (4.0,1.41699)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that 1.41699&quot;=17.00388 inches so Isaac found the analytic solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Due to request here the Matlab source (yes Matlab is indeed more powerful than u think):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;circleradius.m:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;%% x = [pos1X,pos1Y,pos2X,pos2Y,pos3X,pos3Y,r]&#xA;function f = circleradius(x)&#xA;f = -x(7); %% minimize the maximum :D&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;constraints.m&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;%% x = [pos1X,pos1Y,pos2X,pos2Y,pos3X,pos3Y,r]&#xA;function [c, ceq] = contraints(x)&#xA;c = [4*x(7)^2 - ((x(1) - x(3))^2 + (x(2) - x(4))^2); %% d(Circle1,Circle2)&amp;lt;=(2r)^2&#xA;4*x(7)^2 - ((x(1) - x(5))^2 + (x(2) - x(6))^2);&#xA;4*x(7)^2 - ((x(3) - x(5))^2 + (x(4) - x(6))^2); &#xA;x(7) - x(1);  %% Circles are in the rectangle&#xA;x(7) - x(3); &#xA;x(7) - x(5); &#xA;x(7) - x(2);&#xA;x(7) - x(4);&#xA;x(7) - x(6);&#xA;x(7) + x(1) - 8; %% Width is 8&#xA;x(7) + x(3) - 8;&#xA;x(7) + x(5) - 8;&#xA;x(7) + x(2) - 4; %% Height is 4&#xA;x(7) + x(4) - 4;&#xA;x(7) + x(6) - 4;&#xA;-x(1); -x(2); -x(3); -x(4); -x(5); -x(6); -x(7)]; %% No negative values&#xA;ceq = [ ];&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Later in main window:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;[x,fval,exitflag] = patternsearch(@circleradius,[0.5000 1 1.5000 1 2.5000 1 0.5000],[],[],[],[],[],[],@contraints)&#xA;output of x &amp;gt;&amp;gt; 4.0000    2.5830    1.4170    1.4171    6.5830    1.4171    1.4170&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Note that the values are truncated and the above is a solution which is symmetric to what I posted before. You can make it use a mesh of size 10e-10, so you are almost sure to get the global maximum, as the function is continuous. Also the result is faster and more reliable than NMinimize of Mathematica.&lt;/p&gt;&#xA;" OwnerUserId="3123" LastEditorUserId="3123" LastEditorDisplayName="" LastEditDate="2011-01-04T22:15:18.960" LastActivityDate="2011-01-04T22:15:18.960" CommentCount="10" />
  <row Id="16346" PostTypeId="2" ParentId="16244" CreationDate="2011-01-04T18:55:40.383" Score="2" ViewCount="" Body="&lt;p&gt;I believe here is an elementary proof (if you are willing to call dominated convergence theorem as elementary).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First a lemma:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Lemma&lt;/strong&gt;: let $\displaystyle A$ be a bounded measurable set and let $\displaystyle  f \in L(A)$. If $A_n \subset A$ is a sequence of measurable sets such that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ A_1 \supset A_2 \supset A_3 \supset \dots$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and $$\lim_{n \to \infty} m(A_n) = 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\lim_{n \to \infty} \int_{A_n} f \ \text{dm} = 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;($\displaystyle m(T)$ is the lebesgue measure of $\displaystyle T$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is well known (and has an elementary proof) that $\displaystyle X  = \bigcap_{n=1}^{\infty} A_n$ is measurable and $\displaystyle m(X) = \lim_{n \to \infty} m(A_n) = 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now define a sequence of (summable) functions&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle f_n(x) = &#xD;&#xA;\begin{cases} 2 f(x) &amp;amp; x \in A_n \\&#xD;&#xA;f(x) &amp;amp; \text{otherwise}&#xD;&#xA;\end{cases}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now $\displaystyle |f_n(x)| \le |2f(x)|$ and $f_n \to f$ almost everywhere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The set of points $\displaystyle S$ where $f_n(x) \to f(x)$ is not true, satisfies $\displaystyle S \subset X$ and hence is measurable and $\displaystyle m(S) = 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the dominated convergence theorem we have that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\lim_{n \to \infty} \int_{A} f_n = \int_{A} f$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But we have that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_{A} f_n = \int_{A} f + \int_{A_n} f$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\lim_{n \to \infty} \int_{A_n} f = 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle \circ$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that if $\displaystyle f$ was bounded, then there is a much simpler proof of the above lemma, which does not make use of the dominated convergence theorem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now back to the original problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $\displaystyle P_n = \{ x : f(x) \ge \frac{1}{n} \}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the set $\displaystyle P = \{x : f(x) \gt 0\} = \bigcup P_n$ is of positive measure, then there is an $\displaystyle n$ for which  $\displaystyle m(P_n) \gt 0$. Now if $\displaystyle P_n$ is unbounded, there is some $\displaystyle M$ for which $\displaystyle m(P_n \cap [M, M+1]) \gt 0$. Call that set $\displaystyle A$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice that $\displaystyle \int_{A} f \ge \frac{m(A)}{n} \gt 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now give an integer $\displaystyle k \gt 0$, there is an open set $\displaystyle G_k \supset A$ such that $\displaystyle m(G_k-A) \lt \frac{1}{k}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that we can choose the $\displaystyle G_i$ such that $\displaystyle G_1 \supset G_2 \supset G_3 \supset \dots$, by taking $\displaystyle G&amp;#39;_k = \bigcap_{i = 1}^{k} G_i$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the sequence of sets $\displaystyle A_k = G&amp;#39;_k -A$ satisfies the conditions of the above lemma,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;we also have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_{G&amp;#39;_k} f = \int_{A} f + \int_{A_k} f$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now since $\displaystyle G&amp;#39;_{k}$ is a countable union of intervals, we have that $\displaystyle \int_{G&amp;#39;_k} f = 0$, since over every interval, the integral of $\displaystyle f$ is $\displaystyle 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_{A} f + \int_{A_k} f = 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Taking limits, and applying above lemma, we get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_{A} f = 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A contradiction. Similarly, we can show that negative set of $\displaystyle f$ is of measure $\displaystyle 0$ (or just consider $\displaystyle -f$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence $\displaystyle f = 0 \ \text{a.e}$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Note: Since this answer almost proves two claims made by other answers, I am including a sketch of proof of those here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Claim 1) For any measurable set $\displaystyle A$, there is a Borel Set $\displaystyle B \supset A$ such that $\displaystyle m(B) = m(A)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a proof of that, consider the $\displaystyle G&amp;#39;_{k}$ above. $\displaystyle B = \bigcap_{k=1}^{\infty} G&amp;#39;_{k}$ is a Borel set such that $\displaystyle m(B) = m(A)$, as $\displaystyle m(B) = \lim_{k \to \infty} m(G&amp;#39;_{k}) = m(A)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Claim 2) For the $\displaystyle f$ in the problem, for any Borel set $\displaystyle B$, $\displaystyle \int_{B} f = 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The proof above actually shows that for &lt;em&gt;any&lt;/em&gt; measurable set $\displaystyle E$, $\displaystyle \int_{E} f = 0$.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2011-01-09T12:38:51.290" LastActivityDate="2011-01-09T12:38:51.290" CommentCount="4" />
  <row Id="16347" PostTypeId="2" ParentId="16302" CreationDate="2011-01-04T18:59:27.960" Score="2" ViewCount="" Body="&lt;p&gt;Since it has already been established in the comments that a linear recurrence will do the job, here's a brief outline of that method that I hope doesn't give away too much.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Working mod $5,$ we set $a_n = \sum_{i=0}^n 3^i { 2n+1 \choose 2i+1},$ so that $a_0=1$ and $a_1=6,$ and show that $a_n$ satisfies the linear recurrence&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$a_n = 8a_{n-1} - 4a_{n-2}, \quad n \ge 2,$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;from which it follows that $a_n,$ and thus the sum in the question, is never congruent to zero mod $5$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: To add some more detail note that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ a_n = \sum_{i=0}^n 3^i { 2n+1 \choose 2i+1} =&#xD;&#xA;\frac{(\sqrt{3}+1)^{2n+1} + (\sqrt{3} - 1)^{2n+1}}{2\sqrt{3}}.$$&lt;/p&gt;&#xA;" OwnerUserId="1301" LastEditorUserId="1301" LastEditorDisplayName="" LastEditDate="2011-01-05T13:31:52.963" LastActivityDate="2011-01-05T13:31:52.963" CommentCount="0" />
  <row Id="16348" PostTypeId="2" ParentId="16241" CreationDate="2011-01-04T19:17:43.280" Score="0" ViewCount="" Body="&lt;p&gt;A 2D image of a face is a certain projection of a three-dimensional object into the plane. The pre-image of any point is a line, whence a lot of information gets lost in the process. It follows that it is ${\it impossible}$ to tell the true distance between two original points by measurements in the 2D image, even if you are able to identify the individual image points with their corresponding points on the face.&lt;/p&gt;&#xA;" OwnerUserId="1303" LastActivityDate="2011-01-04T19:17:43.280" />
  <row Id="16349" PostTypeId="2" ParentId="16336" CreationDate="2011-01-04T19:33:17.177" Score="1" ViewCount="" Body="&lt;p&gt;I'm with Jonas T, I don't know why he didn't make his comment an answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Symbolic integration of polynomials would be an easy next step.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I disagree &lt;em&gt;somewhat&lt;/em&gt; with Raphael.  Coding some simple functions can give you a better appreciation when you do use something like WolframAlpha.&lt;/p&gt;&#xA;" OwnerUserId="4181" LastActivityDate="2011-01-04T19:33:17.177" />
  <row Id="16350" PostTypeId="1" AcceptedAnswerId="16357" CreationDate="2011-01-04T20:02:20.700" Score="2" ViewCount="108" Body="&lt;p&gt;I have some difficulty in understanding the concepts: line bundles, line bundles on a homogeneous space, and  sections of line bundles. These concepts are on page 140 (the first paragraph of section 9.3) of the book &lt;a href=&quot;http://books.google.com/books?id=U9vZal2HCcoC&amp;amp;printsec=frontcover&amp;amp;dq=young+tableaux&amp;amp;source=bl&amp;amp;ots=xxxzyLzSra&amp;amp;sig=KmvmON6e4Xli3Fz2b0g2S5j6XTE&amp;amp;hl=en&amp;amp;ei=vHkjTd_CCoWBlAemzLigDA&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=3&amp;amp;ved=0CDEQ6AEwAg#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;Young Tableaux by Fulton&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that $R \times S^{1}$, where $S^{1}$ is the unit circle and $R$ is the real line, is a line bundle. Mobius strip is also a line bundle. Are there other specific examples of line bundles? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How to understand the sentence in the first paragraph of section 9.3 (page 140) in the book of Fulton clearly: There is a general procedure for producing representations as sections of a line bundle on a homogeneous space?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also have a question in the first paragraph of the book Young Tableaux on page 131. It is said that if $F$ is a subspace of $E$ of co-dimension $d$, then the kernel of the map from $\wedge^{d}(E)$ to $\wedge^{d}(E/F)$ is a hyperplane in $\wedge^{d}(E)$; assigning this hyperplane to $F$ gives a mapping $Gr^{d}E \to P^{*}(\wedge^{d}E)$. What are the maps $\wedge^{d}(E) \to \wedge^{d}(E/F)$ and $Gr^{d}E \to P^{*}(\wedge^{d}E)$ explicitly?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you very much.&lt;/p&gt;&#xA;" LastEditorUserId="1242" LastEditorDisplayName="" LastEditDate="2011-01-04T20:14:58.580" LastActivityDate="2011-01-05T15:59:01.730" Title="Line bundles, line bundles on a homogeneous space, and sections of line bundles" Tags="&lt;algebraic-geometry&gt;&lt;representation-theory&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="16352" PostTypeId="2" ParentId="16342" CreationDate="2011-01-04T20:25:22.333" Score="1" ViewCount="" Body="&lt;p&gt;In the geometric view of real numbers there are two basic forms of &quot;movements&quot;, namely (a) shifts: each point $x\in{\mathbb R}$ is shifted a given amount $a$ to the right and (b) scalings: all distances between points are enlarged by the same factor $b&amp;gt;0$. In some instances (e.g. sizes of adults) the first notion is appropriate for comparison of different sizes, in other instances (e.g. distances between various celestial objects) the second notion. The logarithm provides a natural means to transform one view into the other: The sum of two shifts corresponds to the composition of two scalings.&lt;/p&gt;&#xA;" OwnerUserId="1303" LastActivityDate="2011-01-04T20:25:22.333" />
  <row Id="16353" PostTypeId="2" ParentId="16310" CreationDate="2011-01-04T20:30:39.450" Score="11" ViewCount="" Body="&lt;p&gt;I can sympathize with this question because I am about to teach a first (graduate level) course in commutative algebra.  No homological algebra of any sort is a prerequisite: I'll be happy if all of my students are comfortable with exact sequences.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, just a little bit of Tor is extremely helpful when studying commutative algebra (and homological methods become increasingly important later on in the subject).  Just now I am re/writing my notes on flat modules, including the &quot;Equational Criterion for Flatness&quot;.  The best source I have found so far is Matsumura's &lt;em&gt;Commutative Ring Theory&lt;/em&gt; which does use a little Tor in the proof.  After having asked a question about this on MO and mulling over the answers, I think that avoiding Tor entirely is not a good approach.  So where am I going to tell my students to go to learn about Tor?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you know me, you're probably expecting me to say that I've written up my own notes on Tor.  But I didn't want to do that -- basic homological algebra has a kind of &quot;just do it&quot; aspect to it, such that my own personal take on the subject is only going to be different from someone else's if I happen to screw it up (which is a non-negligible possibility if I try to write up something quick and short).  Since Matsumura uses Tor, I looked to see what he did.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact he has written an essentially optimal $9$-page precis of homological algebra, Tor and Ext as Appendix B of his text: pages 274-282.  (There is also Appendix A on tensor products, direct and inverse limits, which is fantastic as well and should be read first.)  I do not aspire to do any better than his treatment, so I will make it available to my students when the course begins.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If anyone wishes to have a copy of this passage, please let me know.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Perhaps I should add that on the first page of my copy of Matsumura's book it is written &quot;Pete L. Clark, 2/1/98&quot;.  So I've had this one for a while, and my general engagement with the subject of commutative algebra has also been there, off and on but most recently on, for almost exactly $13$ years.  Nevertheless I still find portions of his book to be intimidating: he covers ground so quickly and with little enough explanation that I often have to take a deep breath and gird myself to read something out of his book.  I mention this because I do not feel this way at all about Appendices A and B: they read like a novel.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, I think it is a little weird that Atiyah and Macdonald introduce Tor in their book so as only to use it in a few exercises.  The most significant application that I can recall is the characterization of rings over which every module is flat (i.e., &lt;strong&gt;absolutely flat rings&lt;/strong&gt;), which is interesting but hardly seems essential.  Let me reiterate that on a first reading you can absolutely just ignore all the Tor, and it will go fine.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Added&lt;/b&gt;: As of today there is now a treatment of Tor (and Ext) in &lt;a href=&quot;http://math.uga.edu/~pete/integral.pdf&quot; rel=&quot;nofollow&quot;&gt;my commutative algebra notes&lt;/a&gt;: see $\S 3.10$.  This was intended to fill out a whirlwind lecture I gave on derived functors in my course a couple of weeks ago, and I tried to give enough detail (especially, on definitions) so that the student can see what is really going on, without getting bogged down in any of the nontrivial diagram chases that are so unpleasant to both write up and (I've often found, at least) to read.  I had my copy of Weibel's &lt;em&gt;Homological Algebra&lt;/em&gt; at my side throughout, and a lot of the more technical proofs are farmed out to there.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a side remark, that lecture would probably have been better if I had taken at least part of it verbatim from Akhil Mathew's short answer to this question.  In general I like an expository challenge, but this homological stuff kind of tires me out...&lt;/p&gt;&#xA;" OwnerUserId="299" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2011-02-26T22:03:31.510" LastActivityDate="2011-02-26T22:03:31.510" CommentCount="7" />
  <row Id="16354" PostTypeId="1" AcceptedAnswerId="16355" CreationDate="2011-01-04T20:39:09.750" Score="2" ViewCount="136" Body="&lt;p&gt;Is there a general way to encode if-then-else/ITE in arithmetic, i.e. using the usual mathematical operators?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example: let $f(x) = x^2$ if $x &amp;lt; 10^{30}$ and $log x$ otherwise, shortly written as $f(x) = ITE(x&amp;lt;10^{30},x^2,log x)$. &lt;/p&gt;&#xA;" OwnerUserId="5388" LastActivityDate="2011-01-04T21:36:19.090" Title="Can you encode if-then-else in arithmetic" Tags="&lt;arithmetic&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="16355" PostTypeId="2" ParentId="16354" CreationDate="2011-01-04T20:53:53.180" Score="2" ViewCount="" Body="&lt;p&gt;If you allow the use of &lt;a href=&quot;http://en.wikipedia.org/wiki/Indicator_function&quot; rel=&quot;nofollow&quot;&gt;indicator or characteristic functions&lt;/a&gt;, then the answer is yes.  For your example, if you let $A = \{x \in \mathbb{R}| x &amp;lt; 10^{30}\}$, then $f(x) = x^2 \cdot \chi_{A}(x) + \log x \cdot \chi_{A^{C}}(x)$ where $\chi_A$ and $\chi_A^{C}$ are the characteristic functions of $A$ and $A$ complement, respectively.&lt;/p&gt;&#xA;" OwnerUserId="4693" LastActivityDate="2011-01-04T20:53:53.180" CommentCount="1" />
  <row Id="16356" PostTypeId="1" CreationDate="2011-01-04T21:01:35.657" Score="6" ViewCount="63" Body="&lt;p&gt;I am trying to check why curvature 0 implies that the horizontal distribution is involutive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $\pi:P\to U$ be a principal $G:=GL_n$ bundle. Assume that $P$ is trivial and $\pi$ admits a section. Thus, $P\cong U\times G$. A connection on $P$ is a $G$ equivariant splitting of the short exact sequence of bundles $0\to P\times\mathfrak{g} \to TU\oplus TG\to \pi^*TU\to 0$. Let us denote this splitting by $s$. If $(v,u)$ is a tangent vector at the point $(x,g)$, then $s(v,u)=w_x(v)+(R_{g^{-1}})_*(u)$, where $w\in\Gamma(X,\Omega_U\otimes\mathfrak{g})$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $x_1,x_2,...x_n$ be coordinates on $U$ and consider vector fields $X_i$ corresponding to them. Use these to construct horizontal $G$ invariant vector fields on $P$ which are given by $((X_i)_x, -w_x((X_i)_x))$ at the point $(x,Id)$. This should give an involutive distribution on $P$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using that $[X_1,X_2]=0$, when we compute the Lie bracket, we get &#xA;$[(X_1,-w(X_1)),(X_2,-w(X_2))]=-[w(X_1),X_2]-[X_1,w(X_2)]+[w(X_1),w(X_2)]$.&#xA;=$X_2(w(X_1))-X_1(w(X_2))+[w(x_1),w(X_2)]$. We need to show that this is 0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The condition that curvature is 0 is given by $dw+w\wedge w=0$, i.e. $dw_{ij}+\sum_{k=0}^{n}w_{ik}\wedge w_{kj}=0$. This means that $dw(X_1,X_2)+[w(x_1),w(X_2)]=0$, which is same as saying that $X_1(w(X_2))-X_1(w(X_2))+[w(x_1),w(X_2)]=0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There seems to be some mismatch.&lt;/p&gt;&#xA;" OwnerUserId="5389" LastActivityDate="2011-01-04T21:01:35.657" Title="Curvature 0 and involutive horizontal distributions" Tags="&lt;differential-geometry&gt;" />
  <row Id="16357" PostTypeId="2" ParentId="16350" CreationDate="2011-01-04T21:13:30.340" Score="4" ViewCount="" Body="&lt;p&gt;The formation of exterior powers is a functor, so the quotient map&#xA;$E \to E/F$ induces a map $\wedge^d E \to \wedge^d(E/F)$.  It is not hard to&#xA;see that this latter map is surjective.  Since $E/F$ is of dimension $d$ by assumption, its top exterior power $\wedge^d(E/F)$ is one-dimensional, and so&#xA;the kernel of $\wedge^d E \to \wedge^d(E/F)$ is of codimension one, i.e. is&#xA;a hyperplane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now given $F$, this kernel is a hyperplane in $\wedge^d E,$ and so &#xA;we get a map $Gr^dE \to P^*(\wedge^d E)$ sending each element $F$ of the source&#xA;(i.e. each codimension $d$ subspace $F$ of $E$) to the kernel of the associated&#xA;map $\wedge^d E \to \wedge^d (E/F)$, which as we have just seen is a hyperplane&#xA;in $\wedge^d E$, i.e. is an element of $P^*(\wedge^d E)$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for your question about line bundles, you might want to learn more basic theory before trying to understand the statement about sections of homogeneous line bundles.  For example, the examples you wrote down are line bundles in the smooth category, whereas Fulton is talking about line bundles, and sections thereof, in the algebraic category.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One place to start would be by learning about the tautological bundle over a projective space, since this is an example of the kind of homogeneous line bundle that Fulton is talking about.&lt;/p&gt;&#xA;" OwnerUserId="221" LastEditorUserId="221" LastEditorDisplayName="" LastEditDate="2011-01-05T15:59:01.730" LastActivityDate="2011-01-05T15:59:01.730" CommentCount="5" />
  <row Id="16358" PostTypeId="1" AcceptedAnswerId="16383" CreationDate="2011-01-04T21:18:39.280" Score="12" ViewCount="679" Body="&lt;p&gt;I am trying very hard to understand Godel's Incompleteness Theorem.  I am really interested in what it says about axiomatic languages, but I have some questions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Godel's theorem is proved based on Arithmetic and its four operators: is all mathematics derived from these four operators (×, +, -, ÷) ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Operations such as log(x) and sin(x) are indeed atomic operations, like those of arithmetic, but aren't there infinitely many such operators that have inverses (that is, + and - are &quot;inverse&quot; operations, × and ÷ are inverse operations).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To me it seems as though making a statement about the limitations of provability given 4 arbitrary operators is absurd, but that probably highlights a gap in my understanding, given that he proved this in 1931 and its unlikely that I have found a counter-argument.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a follow-up remark, why the obsession with arithmetic operators?  They probably seem &quot;fundamental&quot; to us as humans, but to me they all seem to be derived from four possible graphical arrangements of numbers (if we consider four sides to a digit), and fundamentally derived from addition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[][] o [][] addition and, its inverse, subtraction&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[][]&lt;br&gt;&#xA;[][] multiplication (iterative addition) and, its inverse, division&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There must be operators that are consistent on the natural numbers that we certainly aren't aware of, no?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please excuse my ignorance, I am hoping I haven't offended any real mathematicians with this posting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;hr&gt;&#xA;edit: I think I am understanding this a lot more, and I think my main difficulty in understanding this was that:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;There are statements that are true that are unprovable.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Seemed like an impossible statement.  It does, however, make sense to me at the moment in the context of an axiomatic language with a limited number of axioms.  Ultimately, suggesting that there are statements that are true and expressible in the language, but are unprovable in the language (because of the limited set of axioms), is what I believe to be the point of the proof -- is this correct?&lt;/p&gt;&#xA;" OwnerUserId="1774" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-05T03:27:00.097" LastActivityDate="2011-02-15T12:51:20.630" Title="Understanding Godel's Incompleteness Theorem" Tags="&lt;logic&gt;&lt;incompleteness&gt;" AnswerCount="5" CommentCount="5" FavoriteCount="8" />
  <row Id="16359" PostTypeId="2" ParentId="16354" CreationDate="2011-01-04T21:36:19.090" Score="2" ViewCount="" Body="&lt;p&gt;This seems related to:  &lt;a href=&quot;http://math.stackexchange.com/questions/15629&quot;&gt;Representing IF ... THEN ... ELSE ... in math notation&lt;/a&gt;  Although, since none of the answers there do a particularly good job at answering the question posted here, I'll just add a little bit more:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On top of what Jason wrote, you may also write:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Let $f:\mathbb{R} \rightarrow \mathbb{R}$ be the function defined by $f(x)=x^2$ for $x&amp;lt;10^{30}$ and $f(x)=\log x$ otherwise.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$f(x) = \begin{cases} x^2 &amp;amp; \text{if } x&amp;lt;10^{30} \\ \log x &amp;amp; \text{otherwise.} \end{cases}$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;These have the advantage that you don't need to introduce an auxiliary function (such as a characteristic function).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The latter is typeset in LaTeX using:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;$f(x) = \begin{cases} x^2 &amp;amp; \text{if } x&amp;lt;10^{30} \\ \log x &amp;amp; \text{otherwise.} \end{cases}$&lt;/code&gt;&lt;/p&gt;&#xA;" OwnerUserId="139" LastActivityDate="2011-01-04T21:36:19.090" />
  <row Id="16360" PostTypeId="2" ParentId="16358" CreationDate="2011-01-04T21:41:35.183" Score="6" ViewCount="" Body="&lt;p&gt;Before you understand Gödel's incompleteness theorem, you need to know what a formal system is. A formal system can be thought of as a very strict language where the grammar completely determines the allowable sentences. So, the incompleteness theorem is a statement that &quot;In a formal system sufficiently strong to represent arithmetic, there are sentences which are true but unprovable&quot; which is, to use your phrase, a &quot;limitation on provability (in a formal system)&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You asked &quot;Godel's theorem is proved based on Arithmetic and its four operators: is all mathematics derived from these four operators (×, +, -, ÷) ?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Answer: No, there is mathematics that does not use those operators. You also are claiming that multiplication and division are inverses on the natural numbers but that is not true. 7/4 is not an allowed division on the natural numbers, it is only when you allow fractions that this is possible. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You asked &quot;There must be operators that are consistent on the natural numbers that we certainly aren't aware of, no?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Answer: Depends on what you mean by &quot;operator&quot;. In mathematics we generally think of an &quot;operator&quot; on natural numbers as a function from a pair of natural numbers to another natural number, so there are in fact infinitely many operators on natural numbers. So yes, there are some we &quot;aren't aware of&quot; because there are more than we could ever count. For example, let's invent the &quot;square product operator&quot; (let's give it the symbol ¤) as taking two numbers to the product of their squares, so 2 ¤ 3 = (2*2)*(3*3) = 36. The possibilities are limitless!&lt;/p&gt;&#xA;" OwnerUserId="1796" LastActivityDate="2011-01-04T21:41:35.183" CommentCount="7" />
  <row Id="16361" PostTypeId="1" CreationDate="2011-01-04T21:43:26.910" Score="3" ViewCount="140" Body="&lt;p&gt;Does anybody know of any techniques to solve the following sum (or a fast way (polynomial in N) to compute it):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{i=0}^{N} \binom{N-i}{A} \cdot \binom{\lfloor iP/Q \rfloor + 1}{B}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $P$ and $Q$ are relatively prime positive integers, and $A$,$B$ are positive integers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When $P = Q = 1$, the solution can be found in Concrete Mathematics, and it's $\binom{N+2}{A+B+1}$. I suspect it's also possible to find a closed form when $Q = 1$, and $P &amp;gt; 1$, but I haven't found it. I'm interested in the general solution, though.&lt;/p&gt;&#xA;" OwnerUserId="5393" LastActivityDate="2011-01-04T21:43:26.910" Title="Solving a certain binomial sum involving the floor function" Tags="&lt;binomial-coefficients&gt;" FavoriteCount="1" />
  <row Id="16362" PostTypeId="1" CreationDate="2011-01-04T21:46:57.660" Score="2" ViewCount="81" Body="&lt;p&gt;Hello Everyone,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Source: Lovasz, Plummer - Matching Theory&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I had a question related to the number of perfect matchings in a graph. While going through the $8^{th}$ Chapter on Determinant and Matchings in the text I stumbled across the first exercise problem. So far, I have been unable to solve either of the directions stated in the problem which I present below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A graph $G$ has an even number of perfect matchings if and only if $\exists S \subseteq V(G); (S \neq \phi)$ such that all vertices in $V(G)$ are adjacent to an even number of vertices in $S$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the moment, all I can see is just that finding determinant of $A(G)$ over $F_2$ reveals the parity of perfect matchings. But, owing to my limited Linear Algebra background, I cannot see anything beyond. I would be real glad if someone can help me with this query.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="5391" LastActivityDate="2011-01-04T21:46:57.660" Title="Parity of Perfect Matchings" Tags="&lt;combinatorics&gt;" CommentCount="3" />
  <row Id="16363" PostTypeId="1" CreationDate="2011-01-04T21:49:38.783" Score="0" ViewCount="118" Body="&lt;p&gt;What are general work habits of people here. Do you do most of your work in the morning or night? &lt;/p&gt;&#xA;" OwnerUserId="5394" LastActivityDate="2011-01-04T21:49:38.783" ClosedDate="2011-01-04T21:57:52.570" Title="Work Habits of people here" Tags="&lt;soft-question&gt;" CommentCount="2" />
  <row Id="16364" PostTypeId="2" ParentId="16358" CreationDate="2011-01-04T21:50:57.743" Score="4" ViewCount="" Body="&lt;p&gt;Gödel's first incompleteness theorem says, roughly, that a consistent formal system must contain unprovable (but true) statements, &lt;em&gt;even if that formal system only covers integer arithmetic&lt;/em&gt;. A stronger system, containing real or complex arithmetic, and sin and cos and so on, must cover integer arithmetic too, so it too must contain unprovable (but true) statements.&lt;/p&gt;&#xA;" OwnerUserId="1508" LastActivityDate="2011-01-04T21:50:57.743" CommentCount="3" />
  <row Id="16365" PostTypeId="2" ParentId="16358" CreationDate="2011-01-04T21:51:04.327" Score="10" ViewCount="" Body="&lt;p&gt;The incompleteness theorem is much more general than &quot;arithmetic and its four operators.&quot;  What it says is that any effectively generated formal system which is sufficiently powerful is either inconsistent or incomplete.  This requires that the system be capable of talking about arithmetic, but it can also talk about much more than arithmetic.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;To me it seems as though making a statement about the limitations of provability given 4 arbitrary operators is absurd&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Then you should study harder!  Mathematics is not beholden to your intuition.  When your intuition clashes with the mathematics, you should assume that your intuition is wrong.  (I am not trying to be harsh here, but this is an attitude you need to digest before you can really learn any mathematics.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You also shouldn't think of arithmetic as just being about the arithmetic operations: it's also about the &lt;em&gt;quantifiers&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;As a follow-up remark, why the obsession with arithmetic operators?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Who claims to have an obsession with arithmetic operators?  Perhaps what you are missing here is historical context.  The historical context is, roughly, this: back in Gödel's day there was a program, initiated by &lt;a href=&quot;http://en.wikipedia.org/wiki/David_Hilbert&quot; rel=&quot;nofollow&quot;&gt;Hilbert&lt;/a&gt;, which sought to give a complete set of axioms for all of mathematics.  That is, Hilbert wanted to write down a set of (consistent) axioms from which all mathematical truths were deducible.  (To really understand why Hilbert wanted to do this you should read about the &lt;a href=&quot;http://en.wikipedia.org/wiki/Foundations_of_mathematics#Foundational_crisis&quot; rel=&quot;nofollow&quot;&gt;foundational crisis in mathematics&lt;/a&gt;).  This was very grand and ambitious and many people had great hopes for Hilbert's program until Gödel destroyed those hopes with the Incompleteness theorem, which shows that Hilbert's program could not possibly succeed: if the axioms are powerful enough to talk about arithmetic, then they are either inconsistent (you can prove false statements) or incomplete (you can't prove some true statements).&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2011-01-04T21:51:04.327" CommentCount="4" />
  <row Id="16366" PostTypeId="1" CreationDate="2011-01-04T22:16:45.807" Score="5" ViewCount="218" Body="&lt;p&gt;I have found interesting problem in Gilbert's Strang book, ,,Introduction to Linear Algebra'' (3rd edition):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;How many corners does a cube have in 4 dimensions? How many faces? How many edges? A typical corner is $(0,0,1,0)$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have found the answer for corners:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;We know, that corner is $(x_1,x_2,x_3,x_4)$. For every $x_i$ we can use either $1$ or $0$. We can do this in $2 \cdot 2 \cdot 2 \cdot 2 = 2^4 = 16$ ways.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The same method can be used for general problem of cube in $n$ dimensions (I suppose):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Let's say, we have $n$-dimensional cube (I assume, that length of edge is $1$, but it can be some $a$, where $a \in \mathbb{R}$ [1]). Here, corner of this cube looks like this: $(x_1,x_2, \ldots , x_n)$. For every $x_i$ there are $2$ possibilities: $x_i = 0$ or $x_i = 1$ ($x_i = a$ in general). So, this cube has $2^n$ corners.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It was pretty simple, I think. But now, there are also faces and edges. To be honest, I do not know, how to find the answer in this cases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know, that solution for this problem is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A four-dimensional cube has $2^4 = 16$ corners and $2 \cdot 4 = 8$ three-dimensional sides and $24$ two-dimensional faces and $32$ on-dimensional edges.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Could You &lt;em&gt;somehow&lt;/em&gt; explain me, how to figure out this solution? I have found solution for corners by myself, using Linear Algebra methods &amp;amp; language. Could You show me, how to find the number of edges and faces, using Linear Algebra methods?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there other method to find these numbers? (I suppose, that answer for this question is positive)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am also interested in articles/textbooks/etc. about space dimensions, if You know some interesting positions about that, share with me (and community).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As I wrote: I am interested in mathematical explanations (in particular using Linear Algebra methods/language but other methods may be also interesting) and some &lt;em&gt;intuitions&lt;/em&gt; (how to find solution using imagination etc. [2]).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank You for help.&lt;/p&gt;&#xA;&#xA;&lt;hr/&gt;&#xA;&#xA;&lt;p&gt;[1] I am not sure of this assumption, because:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(a) I am not sure, how edges (and faces) &lt;em&gt;behave&lt;/em&gt; in $n$ dimensions&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(b) I am not sure, how should I think about the distance in $n$ dimensions. I mean, I know, that my intuition may &lt;em&gt;play tricks&lt;/em&gt; here&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[2] I am not asking, how to imagine $4$ dimensional cube, but I think, that there is a way to find the solution, using reasoning, not only Linear Algebra.&lt;/p&gt;&#xA;&#xA;&lt;hr/&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Addition&lt;/b&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My definition of face (there &lt;em&gt;was&lt;/em&gt; a comment about that) is the same as definition here: &lt;a href=&quot;http://en.wikipedia.org/wiki/Face_(geometry)&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Face_(geometry)&lt;/a&gt;, especially:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In geometry, a face of a polyhedron is any of the polygons that make up its boundaries.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5291" LastEditorUserId="5291" LastEditorDisplayName="" LastEditDate="2011-01-06T02:47:41.783" LastActivityDate="2011-01-06T04:10:11.453" Title="Linear Algebra, cube &amp; dimensions &gt; 3" Tags="&lt;linear-algebra&gt;" AnswerCount="3" CommentCount="3" />
  <row Id="16367" PostTypeId="2" ParentId="16366" CreationDate="2011-01-04T22:25:48.203" Score="2" ViewCount="" Body="&lt;p&gt;As you mentioned, a vertex (or a &lt;em&gt;0-face&lt;/em&gt;) is just a choice of string $(x_1,\ldots, x_n)$, where $x_i\in \{0,1\}$.  If you think back to dimensions 2 or 3 (or even 1!), you'll see that an &lt;em&gt;edge&lt;/em&gt; (i.e., a 1-face) is determined by two vertices which differ in a single slot.  So e.g. on the unit square, the left edge goes between $(0,0)$ and $(0,1)$ -- we may name this edge more succinctly as $(0,*)$.  This method generalizes, too: the bottom square (2-face) of the 3-cube is $(*,*,0)$, etc.  Note that the $n$-cube will have $k$-faces for $0\leq k\leq n$.  See if you can find a formula for how many there are!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This isn't exactly linear algebra, it's more like combinatorics.  Also, this problem is actually a decent introduction to the idea of higher dimensions, if you try to visualize things in 4 dimensions.  Search the internet for some representations of 4-cubes, and try to understand why they look the way they do.&lt;/p&gt;&#xA;" OwnerUserId="2159" LastActivityDate="2011-01-04T22:25:48.203" />
  <row Id="16368" PostTypeId="1" CreationDate="2011-01-04T22:27:28.187" Score="3" ViewCount="57" Body="&lt;p&gt;I am looking for a term used to describe an analog to Voronoi diagrams where instead of a single point defining a cell, a continuous set of points is used.  For example, starting with five triangles on the plane, you could construct a diagram that contained regions where each point in that region would be closer to a certain triangle than any other triangle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is anyone aware of any papers dealing with this type of construction?  I'm having a hard time searching for any since I don't know if there is a name for it.&lt;/p&gt;&#xA;" OwnerUserId="5395" LastActivityDate="2011-01-05T21:59:25.140" Title="Term for Voronoi diagrams with non-point sites" Tags="&lt;geometry&gt;&lt;terminology&gt;" AnswerCount="2" />
  <row Id="16369" PostTypeId="2" ParentId="16275" CreationDate="2011-01-04T22:33:20.887" Score="3" ViewCount="" Body="&lt;p&gt;Google &quot;colonial Africa&quot;. The Belgian, French, German, British, and Portugese colonial areas are all adjacent. The only difficult-to-spot adjacency here is Portuguese-French, which is taken care of by Cabinda and French Equatorial Africa. (This map actually only requires five colors, because the Italian, Belgian, and Spanish colonies are all non-contiguous, and Ethiopia is not contiguous to a German colony). &lt;/p&gt;&#xA;" OwnerUserId="2501" LastActivityDate="2011-01-04T22:33:20.887" />
  <row Id="16370" PostTypeId="2" ParentId="16366" CreationDate="2011-01-04T22:35:34.003" Score="2" ViewCount="" Body="&lt;p&gt;To calculate the number of edges:  as you say there are 2^n corners.  Each one is connected to n other corners.  Adding all of these up counts each edge twice, so there are 2^(n-1)n edges, which equals 32 for n=4.  Another way to count edges is to define E(n) as the number of edges in n dimensions.  If you think of the n+1 dimensional cube as connecting the corresponding corners of two n dimensional cubes, the recurrence is E(n+1)=2E(n)+2^n &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Square faces are made by starting with a square in the one of the two n cubes plus the translation of the edges of the n cube in the new dimension.  So S(n+1)=2S(n)+E(n).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3-cube faces are made by starting with a 3-cube in one of the two n-cubes plus translation of the squares in the new dimension.  So C(n+1)=2C(n)+S(n)&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-01-04T22:35:34.003" CommentCount="2" />
  <row Id="16371" PostTypeId="1" AcceptedAnswerId="16372" CreationDate="2011-01-04T22:50:53.673" Score="2" ViewCount="169" Body="&lt;p&gt;In Rudin's &lt;em&gt;Real and Complex Analysis&lt;/em&gt; (p. 97 in my 3rd edition), the following is stated as a corollary to Baire's Category Theorem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;In a complete metric space, the intersection of any countable collection of dense $G_{\delta}$'s is again a dense $G_{\delta}$.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proof: &quot;This follows from the theorem, since every $G_{\delta}$ is the intersection of a countable collection of open sets, and since the union of countably many countable sets is countable.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't understand the word &lt;em&gt;union&lt;/em&gt; in the argument. It seems like it should be &lt;em&gt;intersection&lt;/em&gt;, as the theorem states that such an intersection is dense. Is this a typo or am I missing something?&lt;/p&gt;&#xA;" OwnerUserId="1761" LastActivityDate="2011-01-05T02:06:26.103" Title="Corollary to Baire's Category Theorem" Tags="&lt;real-analysis&gt;&lt;analysis&gt;&lt;topology&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" />
  <row Id="16372" PostTypeId="2" ParentId="16371" CreationDate="2011-01-04T23:05:30.537" Score="5" ViewCount="" Body="&lt;p&gt;&quot;Union&quot; is the correct word here.  The point is that we have an iterated intersection: we have for each $i$ a countable family $\{U_{ij}\}$ of dense open subsets and we are &#xA;taking the intersection over all $i$: $\bigcap_{i \in I} \bigcap_{j \in J} U_{ij}$.&lt;br&gt;&#xA;Here $I$ and $J$ are both countable index sets.  This intersection can be written as &#xA;$\bigcap_{(i,j) \in I \times J} U_{ij}$.  The passage you quote is giving an argument that $I \times J$ is again a countable index set: it is the countable union of the &#xA;countable sets $\{i\} \times J$ as $i$ ranges over all elements of $I$.  (One could give other arguments for this as well.)&lt;/p&gt;&#xA;" OwnerUserId="299" LastActivityDate="2011-01-04T23:05:30.537" CommentCount="4" />
  <row Id="16373" PostTypeId="2" ParentId="16368" CreationDate="2011-01-04T23:09:35.787" Score="3" ViewCount="" Body="&lt;p&gt;It's still called a Voronoi diagram.&lt;/p&gt;&#xA;" OwnerUserId="589" LastActivityDate="2011-01-04T23:09:35.787" />
  <row Id="16374" PostTypeId="1" AcceptedAnswerId="16377" CreationDate="2011-01-04T23:11:13.663" Score="1" ViewCount="199" Body="&lt;p&gt;Let $f \in C[0,1]$ and $f(0)=f(1)$. Prove that $\exists a \in [0,1/2]$ such that $f(a)=f(a+1/2)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was thinking to use the intermediate value theorem, but it lead me nowhere. I was also thinking maybe it's a periodic function with period 1/2, but didn't manage to prove how (maybe it's a false assumption though).&lt;/p&gt;&#xA;" OwnerUserId="3637" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2011-01-05T04:12:32.257" LastActivityDate="2011-01-05T06:40:15.777" Title="Horizontal chord of length $\frac{1}{2}$ in the graph of a continuous function." Tags="&lt;calculus&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="16375" PostTypeId="1" CreationDate="2011-01-04T23:15:30.910" Score="1" ViewCount="62" Body="&lt;p&gt;We have a problem the arises in asset management.  I think (and hope) it raises interesting enough questions for this forum to consider it.  We made pretty extensive searches of the literature and find things that talk around this question, but nothing directly dealing with the issue.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Background&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have time series data for assets, from which we calculate a correlation matrix.  For 5 assets using Mathematica it might look something like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;m = Correlation[data] &lt;/p&gt;&#xA;&#xA;&lt;p&gt;{{1.0, 0.635562, 0.698852, 0.404792, -0.32746},&#xA; {0.635562, 1.0, 0.410075, 0.314375, -0.0636438},&#xA; {0.698852, 0.410075, 1.0, 0.374416, -0.260137},&#xA; {0.404792, 0.314375, 0.374416, 1.0, 0.293135},&#xA; {-0.32746, -0.0636438, -0.260137, 0.293135, 1.0}}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;m //TableForm&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1.000000,   0.635562,   0.698852,  0.404792,  -0.32746&lt;/p&gt;&#xA;&#xA;&lt;p&gt;0.635562,   1.000000,   0.410075,  0.314375,  -0.0636438&lt;/p&gt;&#xA;&#xA;&lt;p&gt;0.698852,   0.410075,   1.000000,  0.374416,  -0.260137&lt;/p&gt;&#xA;&#xA;&lt;p&gt;0.404792,   0.314375,   0.374416, 1.000000,   0.293135&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-0.32746,  -0.0636438, -0.260137,  0.293135,   1.000000&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In asset management one wants diversification.  If two or more assets in a portfolio correlate too highly it concentrates rather than diversifying the risk.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What we want&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;We want an approach or method to construct a portfolio of assets that minimizes the portfolio’s “concentration” risk.&lt;/em&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I’ll illustrate concentration risk with a couple of examples below, but first...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Why this is an interesting problem or question?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A couple of things make this an interesting and challenging question:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;While related to “efficient frontier”&#xA;we have no assumptions about future&#xA;performance for the individual&#xA;instruments to use.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Minimizing variance gives an answer,&#xA;but not an intuitively satisfying or&#xA;even useful one.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Principal Components Analysis seems a&#xA;natural way to look at this, but also&#xA;doesn’t appear to give us what we&#xA;need.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;We’ve looked at using entropy&#xA;maximization, but while one of our&#xA;guys familiar with discrete entropy&#xA;thought it seemed promising, when we&#xA;tried thinking about this in terms of continuous entropy it proved a dead end.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;A couple of simple examples to illustrate concentration risk&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's easiest to understand what we want to achieve if we look at a portfolio of 3 assets in a thought experiment.  Assume 2 of the instruments have correlations of 1 (100%) and the third at 0, it's correlation matrix would look like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1, 1, 0&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1, 1, 0&lt;/p&gt;&#xA;&#xA;&lt;p&gt;0, 0, 1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From our perspective in this case, it would make sense to put 25% in each of the 2 correlated stocks and 50% in the uncorrelated one. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;25%,   25%,   50%&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This offsets the risk of concentrating in correlated instruments, while recognizing that the 100% correlated assets are in fact still different instruments whose correlation in the future may change.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One might make the case that as the two assets that have 100% correlation move the same, then a wide range of possible allocations could equally serve our purposes e.g:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;50%,    0%,    50%&lt;/p&gt;&#xA;&#xA;&lt;p&gt;0%,     50%,   50%&lt;/p&gt;&#xA;&#xA;&lt;p&gt;10%,    40%,   50%&lt;/p&gt;&#xA;&#xA;&lt;p&gt;... or any such variations on the theme.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, as we don’t know how their future correlation will evolve, we think the best and most intuitive solution remains at:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;25%,  25%,  50% &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another example&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a portfolio of 5 assets with 4 having 100% correlation and 1 having 0% correlation the correlation matrix would look like the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1, 1, 1, 1, 0&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1, 1, 1, 1, 0&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1, 1, 1, 1, 0&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1, 1, 1, 1, 0&lt;/p&gt;&#xA;&#xA;&lt;p&gt;0, 0, 0, 0, 1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and the portfolio allocation we want would have the following proportions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;12.5%,   12.5%,   12.5%,   12.5%,   50%&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course the real world presents us with greater complication.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Things we’ve tried&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Minimizing variance&lt;/em&gt; (promising but doesn’t work)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Someone suggested minimizing variance to do this, but as one can see it doesn’t produce an intuitive solution:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some Mathematica code illustrating this follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For 3 assets:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;m3 = {{1, 1, 0}, {1, 1, 0 }, { 0, 0 , 1}};&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Array[x, Length@m3];&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Minimize[{p.m3.p, Tr@p == 1 &amp;amp;&amp;amp; And @@ Thread[p &gt;= 0]}, p]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;{1/2, {x[1] -&gt; 1/4, x[2] -&gt; 1/4, x[3] -&gt; 1/2}}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This looks good.  It gives us:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;25%,   25%,   50% &lt;/p&gt;&#xA;&#xA;&lt;p&gt;but...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For 5 assets:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;m5 = {{1, 1, 1, 1, 0}, {1 , 1, 1, 1, 0 }, {1 , 1, 1, 1, 0 }, {1 , 1,1, 1, 0 }, { 0, 0 , 0, 0, 1}};&lt;/p&gt;&#xA;&#xA;&lt;p&gt;p = Array[x, Length@m5];&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Minimize[{p.m5.p, Tr@p == 1 &amp;amp;&amp;amp; And @@ Thread[p &gt;= 0]}, p]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;{1/2, {x[1] -&gt; 1/16, x[2] -&gt; 1/4, x[3] -&gt; 1/8, x[4] -&gt; 1/16, x[5] -&gt;1/2}}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Not so good as it gives us:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;6.25%,   25%,   12.50%,   6.25%,   50%&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, minimizing variance doesn’t work even for this simple (if artificial) case let alone something more realistic.&#xA; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;A promising solution&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One contributor to our discussion suggested a promising approach - at least for cases that do not have any negative correlations.  Perhaps it would lead someone to suggest a more complete solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again with Mathematica code:&#xA; &#xA;m = {{1, 1, 0}, {1, 1, 0}, {0, 0, 1}};&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Tr /@ PseudoInverse[m]/Tr[ Tr /@ PseudoInverse[m]]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;{1/4, 1/4, 1/2}&#xA; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Exactly what we would want.&#xA;Note: For those not familiar with Mathematica code the functions: “Tr” finds the trace of a matrix and “/@” maps a function to a list or matrix.  The rest probably makes sense.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another example for four assets:&#xA; &#xA;m = {{1, 1, 1, 0}, {1, 1, 1, 0}, {1, 1, 1, 0}, {0, 0, 0, 1}};&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Tr /@ PseudoInverse[m]/Tr[ Tr /@ PseudoInverse[m]]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;{1/6, 1/6, 1/6, 1/2}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again, exactly what we want.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This works better than minimizing variance, but in a more real world example (the first one described in the post) we get something that doesn’t work:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;m = {{1.0, 0.635562, 0.698852, 0.404792, -0.32746},&#xA; {0.635562, 1.0, 0.410075, 0.314375, -0.0636438},&#xA; {0.698852, 0.410075, 1.0, 0.374416, -0.260137},&#xA; {0.404792, 0.314375, 0.374416, 1.0, 0.293135},&#xA; {-0.32746, -0.0636438, -0.260137, 0.293135, 1.0}}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Tr /@ PseudoInverse[m]/Tr[ Tr /@ PseudoInverse[m]]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;{0.267806, 0.0898877, 0.22403, -0.0541658, 0.472441}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case we have a negative allocation (-0.0541658) for the 4th asset, something that doesn’t make sense for what we want to achieve.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Conclusion&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, we need a kind of &lt;em&gt;MinimizeVarianceIntuitively&lt;/em&gt; function.  I hope all of this describes what we want to achieve clearly enough.  Any suggestions or ideas for attacking this problem in completely different ways or for extending any of the things we’ve tried already much appreciated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many thanks,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Arun Garapata&lt;/p&gt;&#xA;" OwnerUserId="5396" LastEditorUserId="5396" LastEditorDisplayName="" LastEditDate="2011-01-04T23:25:26.397" LastActivityDate="2011-01-04T23:25:26.397" Title="MinimizeVarianceIntuitively function or approach needed" Tags="&lt;analysis&gt;&lt;statistics&gt;&lt;matrices&gt;" CommentCount="2" />
  <row Id="16376" PostTypeId="2" ParentId="16374" CreationDate="2011-01-04T23:16:18.057" Score="5" ViewCount="" Body="&lt;p&gt;Hint:   consider g(x), defined on [0,1/2] by g(x)=f(x+1/2)-f(x)&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-01-04T23:16:18.057" />
  <row Id="16377" PostTypeId="2" ParentId="16374" CreationDate="2011-01-04T23:17:02.080" Score="5" ViewCount="" Body="&lt;p&gt;You want to use the intermediate value theorem, but not applied to $f$ directly. Rather, let $g(x)=f(x)-f(x+1/2)$ for $x\in[0,1/2]$. You want to show that $g(a)=0$ for some $a$. But $g(0)=f(0)-f(1/2)=f(1)-f(1/2)=-(f(1/2)-f(1))=-g(1/2)$. This gives us the result: $g$ is continuous and changes sign, so it must have a zero.&lt;/p&gt;&#xA;" OwnerUserId="462" LastEditorUserId="462" LastEditorDisplayName="" LastEditDate="2011-01-05T06:00:06.843" LastActivityDate="2011-01-05T06:00:06.843" CommentCount="8" />
  <row Id="16378" PostTypeId="2" ParentId="16310" CreationDate="2011-01-04T23:17:43.353" Score="8" ViewCount="" Body="&lt;p&gt;You will be a lot more motivated to learn about Tor once you observe closely how horribly tensor product behaves. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let us look at the simplest example possible. Consider the ring $R=\mathbb C[x,y]$ and the ideal $I=(x,y)$. These are about the most well-understood objects, right? What is the tensor product $I\otimes_RI$? This is quite nasty, it has torsions: the element $u = x\otimes y - y\otimes x$ is non-zero, but $xu=yu=0$!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Tor gives you a black box to understand this kind of things. Take the short exact sequence $0 \to I \to R \to R/I \to 0$ and tensor with $I$ we get:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$0 \to \text{Tor}_1(R/I,I) \to I\otimes I \to I \to I/I^2 \to 0$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;from which you can extract:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$0 \to \text{Tor}_1(R/I,I) \to  I\otimes I \to I^2 \to 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But $\text{Tor}_1(R/I,I) = \text{Tor}_2(R/I, R/I) = \mathbb C$ by standard homological algebra.  So now everything fits nicely: the map from $I\otimes I \to I^2$ takes $f\otimes g$ to $fg$, and the kernel is generated by the element $u$, which is killed by $I$, so it is isomorphic to $R/I \cong \mathbb C$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To summarize: tensor product, despite being a fundamental operation, is actually quite bad, and Tor helps you to understand it.&lt;/p&gt;&#xA;" OwnerUserId="223" LastEditorUserId="223" LastEditorDisplayName="" LastEditDate="2011-01-05T08:27:49.700" LastActivityDate="2011-01-05T08:27:49.700" />
  <row Id="16379" PostTypeId="2" ParentId="16342" CreationDate="2011-01-04T23:43:42.360" Score="1" ViewCount="" Body="&lt;p&gt;Logarithms are defined as the solutions to exponential equations and so are practically useful in any situation where one needs to solve such equations (such as finding how long it will take for a population to double or for a bank balance to reach a given value with compound interest). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Historically, they were also useful because of the fact that the logarithm of a product is the sum of the logarithms and sums are easier to calculate by hand (or to estimate by overlapping rulers as in a slide rule). In addition to providing a computational &quot;trick&quot;, this property is the basis of the mapping property described in Christian Blatter's answer and generalizes to the concept of self-adjoint generators of unitary groups, which has many mathematical applications and relates physical observables to symmetry properties in quantum mechanics.&lt;/p&gt;&#xA;" OwnerUserId="5397" LastActivityDate="2011-01-04T23:43:42.360" />
  <row Id="16380" PostTypeId="1" AcceptedAnswerId="16382" CreationDate="2011-01-05T00:09:43.700" Score="2" ViewCount="52" Body="&lt;p&gt;In &lt;a href=&quot;http://mathcircle.berkeley.edu/BMC6/pdf0607/catalan.pdf&quot; rel=&quot;nofollow&quot;&gt;an article&lt;/a&gt; by Tom Davis on &lt;a href=&quot;https://secure.wikimedia.org/wikipedia/en/wiki/Catalan_number&quot; rel=&quot;nofollow&quot;&gt;Catalan numbers&lt;/a&gt;, Davis mentions several problems that are shown to be equivalent. One of them is the problem of determining the number of strings of $n$ pairs of balanced parentheses. Another is the problem of determining the number of &quot;multiplication orderings&quot; of $n + 1$ numbers. In both cases, the answer to the problem is $C_n$, or the n&lt;sup&gt;th&lt;/sup&gt; Catalan number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, with $n = 3$, there are $C_3 = 5$ different strings of 3 pairs of balanced parentheses:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;()()(), ()(()), (())(), (()()), ((()))&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;With four numbers, there are also $C_3$ multiplication orderings of them:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;(((a·b)·c)·d), ((a·b)·(c·d)), ((a·(b·c))·d), (a·((b·c)·d)), (a·(b·(c·d)))&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Davis shows how to convert a multiplication ordering of $n + 1$ terms to a string of $n$ pairs of balanced parentheses, but I would like to know how to go the other way;  i.e. given a string of $n$ pairs of balanced parentheses, how can I calculate a corresponding multiplication ordering?&lt;/p&gt;&#xA;" OwnerUserId="184" LastEditorUserId="184" LastEditorDisplayName="" LastEditDate="2011-01-05T01:06:42.367" LastActivityDate="2011-01-05T01:06:42.367" Title="What is a convenient way to convert a string of balanced parentheses to a corresponding &quot;multiplication ordering&quot; or rooted binary tree?" Tags="&lt;counting&gt;&lt;combinatorics&gt;&lt;catalan-numbers&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="16381" PostTypeId="2" ParentId="16323" CreationDate="2011-01-05T00:15:02.367" Score="1" ViewCount="" Body="&lt;p&gt;After clearing the fraction (multiplying through by x^3), the given equation is a quadratic in x^3. Solve to get x^3=26+/-15sqrt(3), then just take the cube root(s) to get x, and plug in.&#xA;(I am not sure why Ross Millikan considers this less &quot;slick&quot; than any other method)&lt;/p&gt;&#xA;" OwnerUserId="5397" LastEditorUserId="5397" LastEditorDisplayName="" LastEditDate="2011-01-05T00:23:12.593" LastActivityDate="2011-01-05T00:23:12.593" CommentCount="1" />
  <row Id="16382" PostTypeId="2" ParentId="16380" CreationDate="2011-01-05T00:18:12.863" Score="2" ViewCount="" Body="&lt;p&gt;Knuth covers this in chapter 7.2.1.6 (vol. 4, fascicle 4) of The Art Of Computer Programming, where he talks about generating all trees (which is an essentially equivalent problem) and converting back and forth between representations.  Have a look at &lt;a href=&quot;http://cs.utsa.edu/~wagner/knuth/fasc4a.pdf&quot; rel=&quot;nofollow&quot;&gt;http://cs.utsa.edu/~wagner/knuth/fasc4a.pdf&lt;/a&gt; for more details.&lt;/p&gt;&#xA;" OwnerUserId="785" LastEditorUserId="785" LastEditorDisplayName="" LastEditDate="2011-01-05T01:02:23.103" LastActivityDate="2011-01-05T01:02:23.103" />
  <row Id="16383" PostTypeId="2" ParentId="16358" CreationDate="2011-01-05T00:34:25.097" Score="28" ViewCount="" Body="&lt;p&gt;There's a fair amount of historical context to make some of Gödel's choices in his original proof clear. For a good overview of the proof, Nagel and Newman's &lt;strong&gt;Goedel's Proof&lt;/strong&gt; is pretty good (though not without its detractors). I also highly recommend the late Torkel Franzen's &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/1568812388&quot; rel=&quot;nofollow&quot;&gt;Godel's Theorem: An Incomplete Guide to its Use and Abuse&lt;/a&gt;. I may myself be guilty of some of those abuses below; I'm trying to shy away from a lot of the technical details, and this usually invites imprecision and even abuse in this particular field. I hope those who know better than me will keep me honest via comments and appropriate ear-pulling.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Some history.&lt;/strong&gt; Sometimes I think of the 19th century as the &lt;em&gt;Menagerie century&lt;/em&gt;. A &quot;menagerie&quot; was like a zoo of weird animals. During a lot of the 19th century, people were trying to clean up some of the logical problems that abounded in the foundations of mathematics. Calculus plainly &lt;em&gt;worked&lt;/em&gt;, but a lot of the notions of infinitesimals were simply incompatible with some 'facts' about real numbers (eventually solved by Weierstrass's notion of limits using $\epsilon$s and $\delta$s). A lot of assumptions people had been making implicitly (or explicitly) were shown to be false through the construction of explicit counterexamples (the &quot;weird animals&quot; that lead me to call it the &lt;em&gt;menagerie century&lt;/em&gt;; many mathematicians were like explorers bringing back weird animals nobody had seen before and which challenged people's notions of what was and was not the case): the Dirichlet function to show that you can have functions that are discontinuous everywhere; functions that are continuous everywhere but nowhere differentiable; the failure of Dirichlet's principle for some functions; Peano's curve that filled-up a square; etc. Then, the antinomies (paradoxes and contradictions) in the early set theory. Even some work which today we find completely without problem caused a lot of debate: Hilbert's solution of the problem of a finite basis for invariants in any number of variables was originally derided by Gordan as &quot;theology, not mathematics&quot; (the solution was not constructive and involved the use of an argument by contradiction). Many found a lot of these developments deeply troubling. A foundational crisis arose (see the link in Qiaochu's answer).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hilbert, one of the leading mathematicians of the late 19th century, proposed a way to settle the differences between the two main camps. His proposal was essentially to try to use methods that both camps found unassailably valid to show that mathematics was &lt;em&gt;consistent&lt;/em&gt;: that it was not possible to prove both a proposition and its negation. In fact, his proposal was to use methods that both camps found unassailable to prove that the methods that &lt;em&gt;one&lt;/em&gt; camp found troublesome would not introduce any problems. This was the essence of the &lt;strong&gt;Hilbert Programme.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to be able to accomplish this, however, one needed to have some way to study proofs and mathematics itself. There arose the notion of &quot;formal proof&quot;, the idea of an axiomatization of basic mathematics, etc. There were several competing axiomatic systems for the basics of mathematics: Zermelo attempted to axiomatize set theory (later expanded to Zermelo-Fraenkel set theory); Peano had proposed a collection of axioms for basic arithmetic; and famously Russell and Whitehead had, in their massive book &lt;strong&gt;Principia Mathematica&lt;/strong&gt; attempted to establish an axiomatic and deductive system for all of mathematics (as I recall, it takes hundreds of pages to finally get to $1+1=2$). Some early successes were achieved, with people showing that some parts of such theories were in fact consistent (more on this later). Then came Gödel's work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Consistency and Completeness.&lt;/strong&gt; We say a formal theory is &lt;em&gt;consistent&lt;/em&gt; if you cannot prove both $P$ and $\neg P$ in the theory for some sentence $P$. In fact, because from $P$ and $\neg P$ you can prove anything using classical logic, it is equivalent that a theory is consistent if and only if there is at least one sentence $Q$ such that there is no proof of $Q$ in the theory. By contrast, a theory is said to be &lt;em&gt;complete&lt;/em&gt; if given any sentence $P$, either the has a proof of $P$ or a proof of $\neg P$. (Note that an inconsistent theory is necessarily complete). Hilbert proposed to find a consistent and complete axiomatization of arithmetic, together with a proof (using only the basic mathematics that both camps agreed on) that it was both complete and consistent, and that it would remain so even if some of the tools that his camp used (which the other found unpalatable and doubtful) were used with it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Why arithmetic?&lt;/strong&gt; Arithmetic was a particularly good field to focus on in the early efforts. First, it was the basis of the &quot;other camp&quot;. Kronecker famously said &quot;God gave us the natural numbers, the rest is the work of man.&quot; It was hoped that an axiomatization of the natural numbers and their basic operations (addition, multiplication) and relations (order) would be both relatively easy, and also have a hope of being both consistent a complete. That is, it was a good testing ground, because it contained a lot of interesting and nontrivial mathematics, and yet seemed to be reasonably simple.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Gödel&lt;/strong&gt;. Gödel focused on arithmetic for this reason. As it happens, multiplication is key to the argument (there is something special about multiplication; some theories of the natural numbers that include only addition can be shown to be consistent using only the kinds of tools that Hilbert allowed). To answer one of your questions along the way, Gödel even defined new operations and relations on natural numbers that had little to do with addition and multiplication along the way, so that yes, there are operations other than those (no fetishism about them at all). But in fact, Gödel did not restrict himself &lt;em&gt;solely&lt;/em&gt; to arithmetic. His proof is, on its face, about the entire system of mathematics set forth in Russell and Whitehead's &lt;strong&gt;Principia&lt;/strong&gt;, though as Gödel notes it can easily be adapted to other systems so long as they satisfy certain criteria (that's why Gödel's original paper has a title that explicitly refers to the &lt;em&gt;Principia&lt;/em&gt; &quot;and related systems&quot;). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What Gödel showed was that &lt;em&gt;any&lt;/em&gt; theory, subject to some technical restrictions (for example, you must have a way of recognizing whether a given sentence is or is not an axiom), that is &quot;strong enough&quot; that you can use it to define a certain portion of arithmetic will necessarily be either incomplete or inconsistent (that is, either you can prove &lt;em&gt;everything&lt;/em&gt; in that theory, or else there is at least one sentence $P$ such that neither $P$ nor $\neg P$ can be proven). It's not a limitation based on four operations, or an obsession with those operations: quite the opposite. What it says if that if you want your theory to include &lt;em&gt;at least&lt;/em&gt; some arithmetic, then your theory is going to be so complicated that &lt;em&gt;either&lt;/em&gt; it is inconsistent, or else there are propositions that can neither be proven nor disproven &lt;em&gt;using only the methods that both camps found valid.&lt;/em&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is: what it shows is a limitation of those particular (logically unassailable) methods. If we use other methods, we are able to establish consistency of arithmetic, for example, but if you had your doubts about the consistency of arithmetic in the first place, chances are you will find those methods just as doubtful. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, about you coda, and the statement &quot;statements that are true but unprovable&quot;; this is not very apt. You will find a lot of criticism to this paraphrase in Franzen's book, with good reason. It's best to think that you have statements that are neither provable nor disprovable. In fact, one of the things we know is that if you have such a statement $P$, in a theory $M$, then you can find a &lt;strong&gt;model&lt;/strong&gt; (an interpretation of the axioms of $M$ that makes the axioms true) in which $P$ is true, and a different model in which $P$ is false. So in a sense, $P$ is neither &quot;true&quot; nor &quot;false&quot;, because whether it is true or false will depend on the &lt;em&gt;model&lt;/em&gt; you are using. For example, the Gödel sentence $G$ that proves the First Incompleteness Theorem (that if arithmetic is consistent then it is incomplete, since there can be no proof of the sentence $G$ and no proof of $\neg G$) is often said to be &quot;true but unprovable&quot; because $G$ can be interpreted as saying &quot;There is no proof of $G$ in this theory.&quot; But in fact, you can find a model of arithmetic in which $G$ is &lt;em&gt;false&lt;/em&gt;, so why do we say $G$ is &quot;true&quot;? Well, the point is that $G$ is true in what is called &quot;the standard model.&quot; There is a particular interpretation of arithmetic (of what &quot;natural number&quot; means, of what $0$ means, of what &quot;successor of $n$&quot; means, of what $+$ means, etc) which we usually have in mind; &lt;em&gt;in that model&lt;/em&gt;, $G$ is true but not provable. But we know that there are different models (where 'natural number' may mean something completely different, or perhaps $+$ means something different) where we can &lt;em&gt;show&lt;/em&gt; that $G$ is false &lt;em&gt;under that interpretation&lt;/em&gt; of the axioms. I would stay away from &quot;true&quot; and &quot;false&quot;, and stick with &quot;provable&quot; and &quot;unprovable&quot; when discussing this; it tends to prevent problems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;First Summary.&lt;/strong&gt; So: there were historical reasons why Gödel focused on arithmetic; the limitation is not of arithmetic itself, but rather of the formal methods in question: if your theory is sufficiently &quot;strong&quot; that it can represent part of arithmetic (plus it satisfies the few technical restrictions), then either your theory is inconsistent, or else the finitistic proof methods at issue cannot suffice to settle all questions (there are sentences $P$ which can neither be proven nor disproven).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Can something be rescued?&lt;/strong&gt; Well, ideally of course we would have liked a theory that was complete and consistent, and that we could &lt;em&gt;show&lt;/em&gt; is complete and consistent using only the kinds of logical methods which we, and pretty much everyone else, finds beyond doubt. But perhaps we can at least show that the other methods don't introduce any problems? That is, that the theory is consistent, even if it is not complete? That at least would be somewhat of a victory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, Gödel also proved that this is not the case. He showed that if your theory is sufficiently complex that it can represent the part of arithmetic at issue (and it satisfies the technical conditions alluded to earlier), and the theory &lt;em&gt;is&lt;/em&gt; consistent, then in fact one of the things that it cannot settle is whether the theory is consistent! That is, one can write down a sentence $C$ which makes sense in the theory, and which essentially &quot;means&quot; &quot;This theory is consistent&quot; (much like the Gödel sentence $G$ essentially &quot;means&quot;  that &quot;there is not proof of $G$ in this theory&quot;), and which one can prove that if the theory is consistent then the theory has no proof of $C$ and no proof of $\neg C$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again, this is a limitation of those finitistic methods that everyone finds logically unassailable. In fact, there are proofs of the consistency of arithmetic using transfinite induction, but as I alluded to above, if you harbored doubts about arithmetic in the first place, you are simply not going to be very comfortable with transfinite induction either! Imagine that you are not sure that $X$ is being truthful, and $X$ suggests that you ask $Y$ about $X$s truthfulness; you don't know $Y$, but $X$ assures you that $Y$ is &lt;em&gt;very&lt;/em&gt; trustworthy. Well, that's not going to help you, right?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Key take-away:&lt;/strong&gt; Because the theorem applies to &lt;em&gt;any&lt;/em&gt; theory that is sufficiently complex (and satisfies the technical restrictions), we are not even in a position of enlarging our set of axioms to escape these problems. So long as we restrict ourselves to enlargement methods that we find logically unassailable, the technical restrictions will still be satisfied, so that the new theory, stronger and larger though it will be because it has more axioms, will &lt;em&gt;still&lt;/em&gt; be incomplete (though it will possibly be &lt;em&gt;other&lt;/em&gt; sentences that are now incomplete or unprovable; remember that by adding axioms, we are also potentially expanding the kinds of things about which we can talk). So the theorems are not about shortcomings of &lt;em&gt;particular&lt;/em&gt; axiomatic systems, but rather about those finitistic methods within a very large class of systems. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What about those 'technical restrictions'?&lt;/strong&gt; They are important. Suppose that arithmetic were consistent. That means that there is at least one model for it. We could pick a model $M$, and then say &quot;Let's make a theory whose axioms are exactly those sentences that are true when interpreted in $M$.&quot; This is a &lt;em&gt;complete&lt;/em&gt; and &lt;em&gt;consistent&lt;/em&gt; axiomatic system for arithmetic. Complete, because each sentence is either true in $M$ (and hence an axiom, hence provable in this theory) or else false in $M$, in which case its negation is true (and hence an axiom, and hence provable). And consistent, because it has a model, $M$, and a theory is consistent if and only if it has a model. The problem with this axiomatic theory is that if I give you a sentence, you're going to have a hard time deciding if it is or it is not an axiom! We didn't really achieve anything by taking this axiomatic system. The &quot;technical restrictions&quot; are both in the form of making the system actually usable, and also certain technical issues that arise from the mechanics of the proof. But the restrictions are mild enough that pretty much everyone agrees that most reasonable theories will likely satisfy them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Second summary.&lt;/strong&gt; So: if you have a formal axiomatic system which satisfies certain technical (but mild) restrictions, if the theory is large enough that you can represent (a certain part of) arithmetic in it, then the theory is either inconsistent, or else the finitistic methods that everyone agrees are unassailable are insufficient to prove or disprove every sentence in the theory; worse, one of the things that the finitistic methods cannot prove or disprove is &lt;em&gt;whether&lt;/em&gt; the theory is in fact consistent or not. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope that helps. I really recommend Franzen's book in any case. It will lead you away from potential misinterpretations of what the theorem says (I am likely guilty of a few myself above, which will no doubt be addresssed in comments by those who know better than I do). &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-05T03:21:48.260" LastActivityDate="2011-01-05T03:21:48.260" CommentCount="7" />
  <row Id="16384" PostTypeId="1" AcceptedAnswerId="16387" CreationDate="2011-01-05T01:06:50.017" Score="8" ViewCount="275" Body="&lt;p&gt;Assuming the Axiom of Choice, (it seems that) there is a bijection between $\mathbb{R}$ and $\mathbb{N}$ that follows from any well-ordering of the reals. That is, given a well-ordering of $\mathbb{R}$, the nth real number in the ordering would correspond to the nth natural number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the contrary, if the reals are assumed to be countable, a contradiction can quickly be reached using Cantor's &lt;a href=&quot;http://en.wikipedia.org/wiki/Cantor%27s_diagonal_argument&quot; rel=&quot;nofollow&quot;&gt;Diagonal Argument&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where am I mistaken? Is it my understanding of well-ordering?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance&lt;/p&gt;&#xA;" OwnerUserId="1352" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2011-01-05T01:19:06.353" LastActivityDate="2011-01-23T19:05:22.910" Title="Axiom of Choice and the cardinality of the reals" Tags="&lt;set-theory&gt;&lt;axiom-of-choice&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="1" />
  <row Id="16385" PostTypeId="2" ParentId="16384" CreationDate="2011-01-05T01:12:12.477" Score="5" ViewCount="" Body="&lt;p&gt;Well orderings can be much longer than $|\mathbb{N}|$.  There certainly is an nth real in the well-order, but there are reals with much higher ordinals as well.  One form of AC is that any set can be well-ordered, but that does not imply that all infinite sets have the same cardinality.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-01-05T01:12:12.477" />
  <row Id="16386" PostTypeId="1" CreationDate="2011-01-05T01:13:52.103" Score="1" ViewCount="213" Body="&lt;p&gt;Hi!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am so lost on this question I am not sure even where to start.  I am not looking for an answer but more of a turn in the right direction.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Find all the eigenvalues and eigenfunctions of the following linear transformation:  $T(f(x)) = f(-x)$ from $\mathbf{P}_2$ to $\mathbf{P}_2$.  IS T diagonalizable?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;All I know for sure is that the standard basis is $1$, $x$, $x^2$, and if this is an eigenbasis for $T$ then $T$ is diagonalizable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for any assistance!&lt;/p&gt;&#xA;" OwnerUserId="4179" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-05T01:42:11.637" LastActivityDate="2011-01-05T07:44:09.097" Title="Eigenvalues and &quot;Eigenvectors&quot; of Linear Transformations" Tags="&lt;linear-algebra&gt;&lt;homework&gt;&lt;eigenvectors&gt;&lt;eigenvalues&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="16387" PostTypeId="2" ParentId="16384" CreationDate="2011-01-05T01:16:47.743" Score="13" ViewCount="" Body="&lt;p&gt;Your error is thinking that &quot;well-ordered and infinite&quot; means &quot;bijectable with $\mathbb{N}$&quot;. Your argument is not even enough to give a bijection between $\mathbb{N}$ and the following well-ordering of the integers: order the nonnegative integers in the usual way; make every negative number larger than any nonnegative number, and compare negative numbers by comparing their absolute value. That is, the well ordering &#xA;$$0, 1, 2, 3,\ldots, n,\ldots ; -1, -2, -3, \ldots, -n, \ldots$$&#xA;where &quot;;&quot; means that $-1$ is larger than any nonnegative integer. This type of order is called $\omega+\omega$, because it is essentially two copies of $\mathbb{N}$, one placed after the other ($\omega$ is the ordinal name of the well-order of the natural numbers). This is still countable, of course, but you can probably see already that your argument about well-ordering the reals to get a bijection with $\mathbb{N}$ is already in serious trouble: you have no warrant for assuming that it will actually &quot;hit&quot; every real number (and in fact, it won't). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Added:&lt;/em&gt; Just for completeness: to show this is a well ordering of $\mathbb{Z}$, let $A$ be any nonempty subset of $\mathbb{Z}$. If $A\cap\mathbb{N}$ is nonempty, then the least element of $A$ is the least element $\mathbf{a}$ of $A\cap\mathbb{N}$ (my naturals include $0$, by the by), since given any $a\in A$, if $a\in\mathbb{N}$ then by definition of $\mathbf{a}$ we have $\mathbf{a}\leq a$. And if $a$ is negative, then since $\mathbf{a}$ is nonnegative we have $\mathbf{a}\leq a$. Thus, $\mathbf{a}$ is the least element of $A$. If, on the other hand, we have $A\cap\mathbb{N}=\emptyset$, then that means that $A$ consists only of negative numbers. Let $B=\{ |a|\mid a\in A\}$. Then $B\subseteq\mathbb{N}$ and is nonempty, so it has a least element $\mathbf{b}$. Then $\mathbf{a}=-\mathbf{b}\in A$ is the least element of $A$, since given any $a\in A$, we have that $a$ is negative by assumption and so that $|\mathbf{a}| = \mathbf{b}\leq |a|$; since this is how we compare negative numbers in this order, we have that $\mathbf{a}$ is less than or equal to $a$, hence $\mathbf{a}$ is the least element of $A$, as claimed.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-05T01:29:48.700" LastActivityDate="2011-01-05T01:29:48.700" CommentCount="1" />
  <row Id="16388" PostTypeId="2" ParentId="16386" CreationDate="2011-01-05T01:23:09.097" Score="5" ViewCount="" Body="&lt;p&gt;An eigenvector of a linear tranformation $T\colon\mathbf{V}\to\mathbf{V}$ is a vector $\mathbf{v}$, $\mathbf{v}\neq\mathbf{0}$, such that there is a scalar $\lambda$ for which $T(\mathbf{v})=\lambda \mathbf{v}$. The scalar $\lambda$ is called an eigenvalue. $T$ is diagonalizable if and only if you can find a basis for $\mathbf{V}$ in which every vector in the basis is an eigenvector. This does not mean that &lt;strong&gt;any&lt;/strong&gt; basis will consist of eigenvectors. It means that there is a way to &lt;em&gt;pick&lt;/em&gt; a basis so that every element of the basis is an eigenvector.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, here, $\mathbf{V}=\mathbf{P}_2$. Your function $T$ is defined by the given formula. For example, if $p(x) = x^2 - 3x + 1$, then $p(-x) = (-x)^2 -3(-x) + 1 = x^2 +3x + 1$, so $T(p) = x^2+3x+1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are looking for eigenvalues and eigenvectors. If you already know about the characteristic polynomial, then you should figure out the characteristic polynomial of $T$ (say, using the standard basis to find a coordinate matrix for $T$) and proceed along those lines.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you don't know about the characteristic polynomial, not all is lost. Suppose $p(x) = ax^2+bx+c$ is an eigenvector. That means that $a$, $b$, and $c$ are not all zero, and there is a scalar $\lambda$ such that $T(p)=\lambda p$. Since $T(p(x)) = ax^2 -bx +c$ and $\lambda p = \lambda ax^2 + \lambda bx + \lambda c$, the equation says that&#xA;$$ax^2 - bx + c = T(p) = \lambda p = \lambda ax^2 +\lambda bx + \lambda c.$$&#xA;For this to happen, you must have:&#xA;\begin{align*}&#xA;a &amp;amp;= \lambda a\\&#xA;-b &amp;amp;= \lambda b\\&#xA;c &amp;amp;=\lambda c&#xA;\end{align*}&#xA;and $a$, $b$, and $c$ not all zero. You want to find all values of $\lambda$ for which there are soluctions $a,b,c$ not all zero. This is not terribly hard to do. Try that, and we can take it from there; edit your question if you get stuck to account for how far you have managed to go then.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2011-01-05T01:23:09.097" CommentCount="3" />
  <row Id="16389" PostTypeId="1" AcceptedAnswerId="16390" CreationDate="2011-01-05T01:32:44.903" Score="1" ViewCount="99" Body="&lt;p&gt;Suppose&#xA;$$\frac{d^2y}{dx^2} = \frac{e^{-t}}{1-e^t}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After finding the second derivative, how do I find concavity? It's much easier to solve for $t$ in a problem like $t(2-t) = 0$, but in this case, solving for $t$ seems more difficult. Does it have something to do with $e$? $e$ never becomes $0$, but at what point is the curve upward or downward?&lt;/p&gt;&#xA;" OwnerUserId="3514" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-05T01:53:37.900" LastActivityDate="2011-01-05T09:33:34.843" Title="Finding Concavity in Curves" Tags="&lt;calculus&gt;&lt;homework&gt;" AnswerCount="2" />
  <row Id="16390" PostTypeId="2" ParentId="16389" CreationDate="2011-01-05T01:38:43.670" Score="4" ViewCount="" Body="&lt;p&gt;You want to figure out where the second derivative is positive and where it is negative. Since the second derivative is continuous wherever it is defined (the exponential is continuous, and so is $1-e^t$, and you have a quotient of continuous functions), the second derivative can change from positive to negative or from negative to positive &lt;em&gt;only&lt;/em&gt; at points where it is either equal to $0$, or where it is not defined. (This is the &lt;em&gt;real&lt;/em&gt; reason why you usually &quot;solve&quot; for $\frac{d^2y}{dx^2}=0$: to find the places where it can &lt;em&gt;change&lt;/em&gt; from one sign to the other).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, a fraction is undefined where the denominator is $0$; and is $0$ where the denominator is not zero and the numerator is zero. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you note, the numerator of this second derivative is &lt;em&gt;never&lt;/em&gt; zero. That means the second derivative is never zero. Is the second derivative ever undefined?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you find all the places where the second derivative &lt;em&gt;could&lt;/em&gt; change sign, then you can figure out the sign of the second derivative everywhere else by simply evaluating. For example, say for the sake of argument that you found that the derivative is never zero, and is undefined at $-3$ and at $2$. That means that the only places where it can change signs is at $x=-3$ and at $x=2$. So, it must have one particular sign on $(-\infty,-3)$, one particular sign on $(-3,2)$, and one particular sign on $(2,\infty)$ (some of these may be the same sign, but it cannot change sign anywhere &lt;em&gt;except&lt;/em&gt; at $x=-3$ or at $x=2$). So you could then just plug in, say, $x=-4$, $x=0$, and $x=5$ to determine the sign it has on each of those intervals, and obtain the concavity of the original function that way. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you were to discover that the second derivative is never zero and never undefined, that would mean that it is always positive or always negative, so you could just find the value at a single point to figure out which one and so figure out which concavity the function always has.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope that helps.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-05T01:50:39.853" LastActivityDate="2011-01-05T01:50:39.853" />
  <row Id="16391" PostTypeId="1" AcceptedAnswerId="16393" CreationDate="2011-01-05T02:00:49.053" Score="2" ViewCount="146" Body="&lt;p&gt;Minkowski's inequality says the following: For every sequence of scalars $a = (a_i)$ and $b = (b_i)$, and for $1 \leq p \leq \infty$ we have: $||a+b||_{p} \leq ||a||_{p}+ ||b||_{p}$. Note that $||x||_{p} = \left(\smash{\sum\limits_{i=1}^{\infty}} |x_i|^{p}\right)^{1/p}$. This is how I tried proving it:&#xA;\begin{align*}&#xA;||a+b||^{p} &amp;amp;= \sum |a_k+b_k|^{p}\\&#xA;&amp;amp;\leq \sum(|a_k|+|b_k|)^{p}\\&#xA;&amp;amp;= \sum(|a_k|+|b_k|)^{p-1}|a_k|+ \sum(|a_k|+|b_k|)^{p-1}|b_k|.&#xA;\end{align*}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From here, how would you proceed? I know that you need to use Holders inequality. So maybe we can bound both the sums on the RHS since they are products. &lt;/p&gt;&#xA;" OwnerUserId="2548" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-05T02:03:51.150" LastActivityDate="2011-01-05T02:59:38.153" Title="Minkowski's inequality" Tags="&lt;inequality&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" />
  <row Id="16392" PostTypeId="2" ParentId="16371" CreationDate="2011-01-05T02:06:26.103" Score="0" ViewCount="" Body="&lt;p&gt;Actually, when I first read the quote, I just assumed Rudin was talking about what happens when you take complements.  And that might actually be an easier way to see what's going on.&lt;/p&gt;&#xA;" OwnerUserId="2579" LastActivityDate="2011-01-05T02:06:26.103" />
  <row Id="16393" PostTypeId="2" ParentId="16391" CreationDate="2011-01-05T02:10:45.747" Score="2" ViewCount="" Body="&lt;p&gt;Holder's Inequality would say that&#xA;$$\sum |x_ky_k| \leq \left(\sum |x_k|^r\right)^{1/r}\left(\sum|y_k|^s\right)^{1/s}$$&#xA;where $\frac{1}{r}+\frac{1}{s}=1$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Apply Holder's twice, once to each sum, using $x_k = a_k$, $y_k = (|a_k|+|b_k|)^{p-1}$ in one, and similarly in the other, with $r=p$ and $\frac{1}{s}=1-\frac{1}{p}$. &lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2011-01-05T02:10:45.747" />
  <row Id="16394" PostTypeId="2" ParentId="9524" CreationDate="2011-01-05T02:14:45.260" Score="1" ViewCount="" Body="&lt;p&gt;An example from economics:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose the government gives you a 100 dollar tax rebate. You will save, say, 10, and spend the other 90. So at first glance, it seems that the 100 dollar rebate increased GDP by 90 dollars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the 90 you spent will not just disappear; it will go to other people, who will in turn save 10% of it, and spend the other 90%. The 90% the second round spent will go to a third round of people who spend 90% of that, which goes to a fourth round of people... well, you see where I'm going with this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\begin{align}&#xD;&#xA;\Delta GDP &amp;amp; = 90+81+72.9+\dots \\&#xD;&#xA;&amp;amp; = 100(0.9^1 + 0.9^2 + 0.9^3 \dots) \\&#xD;&#xA;&amp;amp; = 100\left(\lim_{n\to\infty}\sum_{i=1}^{n}0.9^i\right) \\ &#xD;&#xA;&amp;amp; = 100 \left(\frac{1}{1-0.9}\right) \\&#xD;&#xA;&amp;amp; = 100 (10) \\&#xD;&#xA;&amp;amp; = 1000&#xD;&#xA;\end{align}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we can see that the fictional stimulus package of a hundred dollars led to an increase in GDP of a thousand dollars.. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are interested, the &quot;90%&quot; is what's called the &lt;a href=&quot;http://en.wikipedia.org/wiki/Marginal_propensity_to_consume&quot; rel=&quot;nofollow&quot;&gt;marginal propensity to consume&lt;/a&gt; and the sum of the series (10 in this example) is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Fiscal_multiplier&quot; rel=&quot;nofollow&quot;&gt;multiplier&lt;/a&gt;. At my school, this is taught in econ 101 without giving the basis (people are told to just memorize $\frac{1}{1-MPC}$) so some of your students may be interested to learn why this formula is correct.&lt;/p&gt;&#xA;" OwnerUserId="2549" LastEditorUserId="2549" LastEditorDisplayName="" LastEditDate="2011-01-05T21:30:51.157" LastActivityDate="2011-01-05T21:30:51.157" CommentCount="1" />
  <row Id="16395" PostTypeId="2" ParentId="16391" CreationDate="2011-01-05T02:59:38.153" Score="0" ViewCount="" Body="&lt;p&gt;There's a version of the proof using exponential's convexity. I let you try to find it by yourself and will complete my post later to give the solution.&lt;/p&gt;&#xA;" OwnerUserId="1865" LastActivityDate="2011-01-05T02:59:38.153" />
  <row Id="16396" PostTypeId="2" ParentId="9524" CreationDate="2011-01-05T03:07:12.500" Score="1" ViewCount="" Body="&lt;p&gt;I agree with the sentiment that power series is the best way to motivate infinite series.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a natural progression from the elementary operations of addition, subtraction, and multiplication; to the concept of a polynomial.  A polynomial is the easiest way to combine the concept of a variable with addition, subtraction, and multiplication.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the same way, power series are the most natural way to combine the concepts of a polynomial with an infinite limit.  Since any (good) calculus course deals with derivatives as limits, the students at this level already should know about limits; and it's not so hard to combine that with polynomials in the obvious way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think when students first encounter infinite series, it's best to caution them strongly that convergence is something that has to be dealt with in a very careful, rigorous way; and then to introduce the geometric series and use it as an example to discuss the basic ideas of convergence\divergence.&lt;/p&gt;&#xA;" OwnerUserId="1417" LastActivityDate="2011-01-05T03:07:12.500" CommentCount="1" />
  <row Id="16397" PostTypeId="2" ParentId="154" CreationDate="2011-01-05T04:00:07.283" Score="0" ViewCount="" Body="&lt;p&gt;A number of answerers have answered by asking the question of whether the real numbers exist.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are lots of physical examples that a person can give to a lay-person of real numbers, including fractions, zero and negative numbers. Vector addition all by itself provides plenty of examples. For instance, most anyone can develop an intuition for the idea that if a vector pointing north has a positive component, then it can be combined with a vector pointing south, and that the south-pointing vector intuitively and usefully can be assigned a negative component.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the question is, are there intuitive physical examples of imaginary numbers. I've never seen one! For example, I'd like anyone to show me something even resembling or approximating a stick of length 2i, that when lined up perpendicularly with another stick of length 2i outlines a region with area -4. Given this, lay people have a point when they deny complex numbers exist.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Two advanced comments on this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) A lovely and non-obvious fact about sqrt(-1) is that once you have introduced it and called it &quot;i&quot; you don't have to introduce a pile of other numbers. For example, the sqrt(i) = 1 / sqrt(2) + i / sqrt(2). Imaginary numbers would be a lot less compelling to mathematicians if introducing them immediately led to a bunch of other quantities having to be hypothesized and named.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) Quantum mechanics makes heavy use of imaginary numbers. Unlike in electrical engineering, it isn't just a convenience. At the end of the day, in electrical engineering, after having solved a difficult problem, you'll never actually go measure an imaginary voltage or current. The solution to electrical engineering problems always involves &quot;throwing away&quot; the imaginary part of the answer, even though it makes the legwork a lot simpler when coming to an answer. In quantum mechanics, you don't throw the complex part of the wave function away. You work with it at every step of the way, until at the end, you use it's norm-squared. This turns a complex &quot;probability amplitude&quot; into a &quot;probability density.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you might say that quantum mechanics provides an example of the &quot;existence&quot; of complex numbers. However, the problem with this example is that quantum mechanics remains difficult to intuit even for the most experienced physicists. &quot;I think I can safely say that nobody understands quantum mechanics.&quot; &lt;a href=&quot;http://en.wikiquote.org/wiki/Richard_Feynman&quot; rel=&quot;nofollow&quot;&gt;http://en.wikiquote.org/wiki/Richard_Feynman&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5400" LastActivityDate="2011-01-05T04:00:07.283" CommunityOwnedDate="2011-04-15T20:45:29.080" />
  <row Id="16398" PostTypeId="2" ParentId="16374" CreationDate="2011-01-05T04:03:45.883" Score="8" ViewCount="" Body="&lt;p&gt;Interestingly,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The numbers of the form $r = \displaystyle \frac{1}{n} \ \ n \ge 1$ are the only numbers such that for &lt;em&gt;any&lt;/em&gt; continuous function $\displaystyle f:[0,1] \to \mathbb{R}$ such that $\displaystyle f(0) = f(1)$, there is some point $\displaystyle c \in [0,1]$ such that $\displaystyle f(c) = f(c+r)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For any other $r$ we can find a continuous function for which there is no $\displaystyle c$ such that $\displaystyle f(c) = f(c+r)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a proof that $\displaystyle r = \frac{1}{n}$ satisifies this property, let $\displaystyle g(x) = f(x) - f(x+ \frac{1}{n})$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then we have that $\displaystyle  \sum_{k=0}^{n-1} \ g\left(\frac{k}{n}\right) = 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus if none of $\displaystyle g\left(\frac{k}{n}\right)$ are $\displaystyle 0$, then there are some $\displaystyle i,j$ such that $\displaystyle g\left(\frac{i}{n}\right) \gt 0$ and $\displaystyle g\left(\frac{j}{n}\right) \lt 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For any other $\displaystyle r \in (0,1)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the following example, due to Paul Levy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle f(x) = \sin^2\left(\frac{\pi x}{r}\right) - x \ \sin^2\left(\frac{\pi}{r}\right)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we have $\displaystyle f(x) = f(x+r)$, then we have that $\displaystyle r\ \sin^2\left(\frac{\pi}{r}\right) = 0$ and hence $\displaystyle r = \frac{1}{m}$ for some integer $\displaystyle m$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Apparently this is called the &lt;a href=&quot;http://www.jstor.org/pss/2317756&quot; rel=&quot;nofollow&quot;&gt;Universal Chord Theorem&lt;/a&gt; (due to Paul Levy!).&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2011-01-05T06:40:15.777" LastActivityDate="2011-01-05T06:40:15.777" CommentCount="5" />
  <row Id="16399" PostTypeId="1" AcceptedAnswerId="16508" CreationDate="2011-01-05T04:06:02.450" Score="4" ViewCount="160" Body="&lt;ol&gt;&#xA;&lt;li&gt;Given integer $n$ and real $0\leq d&amp;lt;1$, set $x := 0$, set $c := 0$.&lt;/li&gt;&#xA;&lt;li&gt;set $y := n-x$, if $y &amp;lt; 1-d$, we are done&lt;/li&gt;&#xA;&lt;li&gt;Randomly chose $l\in [1-d, \min(1+d,y)]$, set $x := x + l$, $c := c+1$. Go to 2.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;So this is a program that count how many random numbers in $[1-d, 1+d]$ one can add until one reach $n$, except the program won't add a number to go over $n$ when one can add a smaller number. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the expected value of $c$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In one part of my program it output the expected value of $c$ by running the above program many many times. But I don't know if my program is correct, therefore knowing what I should expect will help me test my program's correctness.&lt;/p&gt;&#xA;" OwnerUserId="96" LastActivityDate="2011-01-06T13:34:59.117" Title="Expected amount of number one can add before reach n" Tags="&lt;probability&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="16400" PostTypeId="1" AcceptedAnswerId="16430" CreationDate="2011-01-05T04:14:01.993" Score="1" ViewCount="110" Body="&lt;p&gt;Let's say we have a 2D Polygon ( it can be concaved or convex), and on top of this polygon we are required to build a house roof. What is the algorithm that one can use for this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The conditions:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The roof cannot be 0 degree or 90 degree. &lt;/li&gt;&#xA;&lt;li&gt;The created roof piece must be as smooth as possible ( smooth is defined as first order differentiable), meaning the slope of the roof must not change unnecessarily. &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I'm sorry if the conditions sound vague. If you get lost, just think about this: you have a house, how do you generate the roof profile?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: The roof plan must be uniformly sloping, so no circle, or bump can be used. &lt;/p&gt;&#xA;" OwnerUserId="550" LastEditorUserId="550" LastEditorDisplayName="" LastEditDate="2011-01-05T07:56:32.203" LastActivityDate="2011-01-05T14:54:26.497" Title="Generate Roof from a 2D Polygon" Tags="&lt;geometry&gt;" AnswerCount="3" CommentCount="8" />
  <row Id="16401" PostTypeId="1" CreationDate="2011-01-05T04:24:24.087" Score="1" ViewCount="102" Body="&lt;p&gt;We would like to generate simple mathematical formulas on the fly (e.g. $\sin^2 \theta$). The above function seems to be a MathJaX representation. Most of the solutions depend on writing mathematical equations using MathML or Latex or jsMath or MathJaX. I am assuming MathJaX would be the right solution to use if you are planning write these equations right now and most of these solutions get embedded as javascript directly. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a way to generate these equations on the server side using a language like Java assuming the application logic provides a mechanism to embed them in the HTML. What kind of solutions are available if we have to go down this path.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am a complete novice on this subject, looking for inputs so that we choose the right solution.&lt;/p&gt;&#xA;" OwnerUserId="5401" LastActivityDate="2011-01-05T04:37:02.713" ClosedDate="2011-01-07T01:35:43.173" Title="How do you generate mathematical formulas programmatically using a server side language" Tags="&lt;trigonometry&gt;&lt;math-software&gt;&lt;education&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="16402" PostTypeId="2" ParentId="16401" CreationDate="2011-01-05T04:37:02.713" Score="2" ViewCount="" Body="&lt;p&gt;There are two parts to the problem: how to express the mathematics and how to convert that expression into a visual representation.  For the first part, how to express the mathematics, LaTeX is probably the easiest choice.  For the second part, the benefit of something like MathJax is that it puts the work of doing the displaying on the client side, which is essentially the same as any web page—the server provides marked-up content, the client renders it.  I suspect that it's also easier to work with quirks of different browsers with a client-side solution, but I don't know for sure.  There are ways to render LaTeX on the server side, delivering images files (I know there's an extension for MediaWiki that does this), the results I've seen from it do not match the surrounding text as well as MathJax.&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2011-01-05T04:37:02.713" />
  <row Id="16403" PostTypeId="1" AcceptedAnswerId="16677" CreationDate="2011-01-05T06:10:45.277" Score="2" ViewCount="382" Body="&lt;p&gt;Hello.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have the following issue #26 from &lt;em&gt;What Is the Name of This Book?&lt;/em&gt; of &lt;a href=&quot;http://en.wikipedia.org/wiki/Raymond_Smullyan&quot; rel=&quot;nofollow&quot;&gt;R. Smullyan&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;There is a wide variety of puzzles&#xA;  about an island in which  certain&#xA;  inhabitants called &quot;knights&quot; always&#xA;  tell the truth,  and others called&#xA;  &quot;knaves&quot; always lie. It is assumed&#xA;  that  every inhabitant of the island&#xA;  is either a knight or a knave. I shall&#xA;  start with a well-known puzzle of this&#xA;  type and then  follow it with a&#xA;  variety of puzzles of my own.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;According to this old problem, three&#xA;  of the inhabitants — A, B, and C —&#xA;  were standing together in a garden. A&#xA;  stranger passed by and asked A, &quot;Are&#xA;  you a knight or a knave?&quot; A  answered,&#xA;  but rather indistinctly, so the&#xA;  stranger could not make out what he&#xA;  said. The stranger than asked B, &quot;What&#xA;  did A say?&quot; B replied, &quot;A said that he&#xA;  is a knave.&quot; At this point the third&#xA;  man, C, said, &quot;Don't believe B; he is&#xA;  lying!&quot; The question is, what are B&#xA;  and C?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I supposed that truth tables can be used, and composed the following:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;   |   |   |      F1      |   F2   |    G       &#xA;===|===|===|==============|========|=========&#xA; A | B | C | B ↔ (A ↔ ¬A) | C ↔ ¬B | F1 ^ F2 &#xA;===|===|===|==============|========|=========&#xA; 1 | 1 | 1 |      0       |   0    |    0&#xA; 1 | 1 | 0 |      0       |   1    |    0&#xA; 1 | 0 | 1 |      1       |   1    |    1&#xA; 1 | 0 | 0 |      1       |   0    |    0&#xA; 0 | 1 | 1 |      0       |   0    |    0&#xA; 0 | 1 | 0 |      0       |   1    |    0&#xA; 0 | 0 | 1 |      1       |   1    |    1&#xA; 0 | 0 | 0 |      1       |   0    |    0&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Provided that:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;We use $A$, when A is a knight, and&#xA;$\neg A$, when A is a knave.&lt;/li&gt;&#xA;&lt;li&gt;$F1$ is what B said ($A \leftrightarrow \neg A$), &#xA;i. e. B said that A said he's knave.&#xA;Therefore, B is telling the truth if &#xA;and only if he's a knight ($B$).&lt;/li&gt;&#xA;&lt;li&gt;$F2$ means that C is a knight if and &#xA;only if he's telling the truth, i. e.&#xA;B is a knave ($\neg B$).&lt;/li&gt;&#xA;&lt;li&gt;$G$ allows us to select only those&#xA;claims amongst $F1$ and $F2$&#xA;which are true.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;can I safely say that we have only two cases, when $G$ is true and the following conclusions can be made:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;B is a knave, because there are $0$s (false) in the appropriate rows.&lt;/li&gt;&#xA;&lt;li&gt;C is a knight, because B is telling lies, and there are $1$s (true) in the appropriate rows.&lt;/li&gt;&#xA;&lt;li&gt;We cannot say what is A exactly, because we couldn't make out what he said, and there are two cases in the table with $0$ and $1$ in the appropriate rows, where $G$ is true.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Please tell me if my calculations and the truth table are right, not only the conclusion.&lt;/strong&gt; The best answer is one, which either explains what I'm missing in my truth table, or contains a correct one instead of mine, being supposedly wrong. I'm trying to figure out how they can be used, and I guess this issue is quite simple to play with, after all you have the same reasoning in your mind.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance.&lt;/p&gt;&#xA;" OwnerUserId="5340" LastEditorUserId="5340" LastEditorDisplayName="" LastEditDate="2011-01-07T15:13:37.483" LastActivityDate="2011-01-07T18:02:42.017" Title="Knights and knaves: Who are B and C? (task 26 from &quot;What Is the Name of This Book?&quot;)" Tags="&lt;logic&gt;&lt;puzzle&gt;" AnswerCount="5" CommentCount="1" />
  <row Id="16405" PostTypeId="2" ParentId="16403" CreationDate="2011-01-05T07:19:47.340" Score="7" ViewCount="" Body="&lt;p&gt;A truth table is really the wrong tool for this; and your table contains errors. For example, you assert that A v B v C is always true, even in the case when A, B, and C are all false! Truth tables (when correctly filled out) are useful in some situations, but the proper approach to this problem is to take a more direct deductive route.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, we adopt the rule that knights always speak true statements, and knaves never do, and everyone is either a knight or a knave (and not both, unless they never speak!).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Under these assumptions, no one can say &quot;I am a knave&quot;: a knight cannot say it, because it would not be a true statement; and a knave cannot say it because it would be a true statement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, when B states &quot;A said 'I am a knave'&quot;, then it is immediately the case that B is stating a falsehood, and thus we have proven that B is a knave.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since C states a truth (i.e., that B has stated a falsehood), C cannot be a knave; therefore we have proven C is a knight.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is much simpler and clearer than a logic table... in my humble opinion!&lt;/p&gt;&#xA;" OwnerUserId="4980" LastEditorUserId="4980" LastEditorDisplayName="" LastEditDate="2011-01-05T07:25:54.543" LastActivityDate="2011-01-05T07:25:54.543" CommentCount="2" />
  <row Id="16406" PostTypeId="2" ParentId="16386" CreationDate="2011-01-05T07:44:09.097" Score="2" ViewCount="" Body="&lt;p&gt;Here's a general remark, which may or may not be helpful to you:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The transformation $T$ in your question has the property that if you do it twice you get back to what you started with. Expressed differently, $T$ composed with itself is the identity operator: $T \circ T = I$. (This is because $-(-x)=x$; changing the sign of the variable $x$ twice is the same as doing nothing to $x$.)&#xA;Such a linear transformation is called an &lt;em&gt;involution&lt;/em&gt;, and involutions have the property that the eigenvalues can only be $+1$ and $-1$. This is proved as follows: suppose $v$ is an eigenvector with eigenvalue $\lambda$; this implies that&#xA;$$T(T(v))=T(\lambda v)=\lambda T(v) = \lambda \lambda v = \lambda^2 v,$$ &#xA;and moreover $T(T(v))=v$ since $T$ is an involution. These two expressions for $T(T(v))$ must be equal, so $\lambda^2 v=v$, and since an eigenvector by definition is nonzero, it follows that $\lambda^2 = 1$. Hence $\lambda=\pm 1$. Q.E.D.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $T$ is an involution, the following also holds: for any vector $u$, the vectors $u \pm T(u)$ are eigenvectors (if they are nonzero) with eigenvalues $\pm 1$. (Proving this is a nice little one-line exercise.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://math.stackexchange.com/questions/5274/how-do-i-divide-a-function-into-even-and-odd-sections&quot;&gt;This question&lt;/a&gt; is somewhat related.&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2011-01-05T07:44:09.097" />
  <row Id="16407" PostTypeId="1" AcceptedAnswerId="16413" CreationDate="2011-01-05T07:54:49.300" Score="0" ViewCount="108" Body="&lt;p&gt;I am given a vector space $V$, a basis $e_1,\dots,e_n$ and a vector $v\in V$ and I am asked to find $\lambda_1,\dots\lambda_n$ from the underlying field such that $v=\sum \lambda_i e_i$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I do it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, if I know the basis is orthonormal then I can compute the inner product between $v$ and the $e_i$'s, but let's assume I don't know this, so the question is whether I can solve this problem in this level of abstraction.&lt;/p&gt;&#xA;" OwnerUserId="1818" LastActivityDate="2011-01-05T09:32:34.533" Title="Given a vector and a basis, how can I find the coordinate vector?" Tags="&lt;linear-algebra&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="16408" PostTypeId="2" ParentId="16407" CreationDate="2011-01-05T08:19:16.123" Score="0" ViewCount="" Body="&lt;p&gt;You could use the &lt;a href=&quot;http://en.wikipedia.org/wiki/Gram%E2%80%93Schmidt_process&quot; rel=&quot;nofollow&quot;&gt;Gram-Schmidt process&lt;/a&gt; to create an orthonormal basis.&lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2011-01-05T08:19:16.123" />
  <row Id="16409" PostTypeId="2" ParentId="16302" CreationDate="2011-01-05T08:20:55.010" Score="1" ViewCount="" Body="&lt;p&gt;This problem has appeared in IMO 1974 test. The solution may be found here:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.artofproblemsolving.com/Forum/viewtopic.php?p=358034&amp;amp;sid=4ec9cfb8d7c76cdfe71c058fc2f5c974#p358034&quot; rel=&quot;nofollow&quot;&gt;http://www.artofproblemsolving.com/Forum/viewtopic.php?p=358034&amp;amp;sid=4ec9cfb8d7c76cdfe71c058fc2f5c974#p358034&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" LastActivityDate="2011-01-05T08:20:55.010" CommentCount="1" />
  <row Id="16411" PostTypeId="2" ParentId="16358" CreationDate="2011-01-05T08:48:36.670" Score="4" ViewCount="" Body="&lt;p&gt;This is a personal recommendation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Peter Smith's &quot;An Introduction to Gödel's Theorems&quot; is very readable and self-contained. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe to fully appreciate Gödel's theorems it is a good idea to learn some computability theory before (Smith does include some of this background in his book, but a bit more of perspective doesn't hurt).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Two very good sources for beginners are Cutland's &quot;Computability&quot; and Boolos and Jeffrey &quot;Computability and Logic&quot;. There's a forthcoming book from Enderton (who unfortunately recently passed away), &quot;Computability Theory&quot;, which I have been able to read and highly recommend too. For a quick (and unusually clear) introduction to computability theory you can take a look at Enderton's notes, which, by the way, are an excerpt from his book:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.math.ucla.edu/~hbe/computability.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.math.ucla.edu/~hbe/computability.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope this helps (it did to me).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: Let me also recommend a very nice book by Melvin Fitting, &lt;em&gt;Incompleteness in the Land of Sets&lt;/em&gt;. As the title suggests, the author shows us how to discuss incompleteness in its most natural place: the land of hereditarily finite sets.&lt;/p&gt;&#xA;" OwnerUserId="5405" LastEditorUserId="5405" LastEditorDisplayName="" LastEditDate="2011-02-15T12:51:20.630" LastActivityDate="2011-02-15T12:51:20.630" CommentCount="3" />
  <row Id="16413" PostTypeId="2" ParentId="16407" CreationDate="2011-01-05T09:02:12.543" Score="0" ViewCount="" Body="&lt;p&gt;In the absence of an inner product you could use any n independent linear functionals to get n equations in n unknowns. (eg for a function space of solutions to a Boundary Value Problem you might just use evaluation at n points in the domain)&lt;/p&gt;&#xA;" OwnerUserId="5397" LastActivityDate="2011-01-05T09:02:12.543" />
  <row Id="16414" PostTypeId="2" ParentId="16230" CreationDate="2011-01-05T09:17:04.847" Score="0" ViewCount="" Body="&lt;p&gt;I am sorry that I ask the problem in a wrong way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Actually $g(x)$ is the Gaussian Distribution, &#xA;I am going to minimize over $i(x)$ and $t$ in this problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sorry for the type.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;@Slowsolver, may I know which chapter I can find the answer? Thanks&lt;/p&gt;&#xA;" OwnerUserId="5347" LastActivityDate="2011-01-05T09:17:04.847" CommentCount="2" />
  <row Id="16415" PostTypeId="2" ParentId="16407" CreationDate="2011-01-05T09:32:34.533" Score="0" ViewCount="" Body="&lt;p&gt;If you have another basis in which you can express $v$ and $e_i$, finding $\lambda_i$ reduces to solving system of linear equations.&lt;/p&gt;&#xA;" OwnerUserId="4742" LastActivityDate="2011-01-05T09:32:34.533" />
  <row Id="16416" PostTypeId="2" ParentId="16389" CreationDate="2011-01-05T09:33:34.843" Score="0" ViewCount="" Body="&lt;p&gt;From the t on the RHS I assume that this comes from a pair of parametric equations where you were given x(t) and y(t) and that you got the second derivative by calculating dm/dx=m'(t)/x'(t) where m(t)=dy/dx=y'(t)/x'(t). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In any case, you are right that the curve will be concave up at points where the second derivative is positive and concave down where it is negative. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are also right that the exponentials are never zero. &lt;em&gt;e&lt;/em&gt; itself is just a specific positive constant and so raising it to any (positive or negative) power always gives a positive result. So the numerator in your expression is never zero and the only way the whole thing can change sign is when the denominator does. This happens when &lt;em&gt;e&lt;/em&gt;^t=1, which means that t must be zero. Since &lt;em&gt;e&lt;/em&gt;^t is an increasing function of t it must be bigger than 1 for t&gt;0, and so at those points the second derivative is negative and the curve is concave down. But for the points with negative t, &lt;em&gt;e&lt;/em&gt;^t is less than 1 (eg &lt;em&gt;e&lt;/em&gt;^(-2)=1/&lt;em&gt;e&lt;/em&gt;^2), so at these points the second derivative is positive and the curve is concave up.&lt;/p&gt;&#xA;" OwnerUserId="5397" LastActivityDate="2011-01-05T09:33:34.843" />
  <row Id="16417" PostTypeId="2" ParentId="16302" CreationDate="2011-01-05T09:43:01.727" Score="1" ViewCount="" Body="&lt;p&gt;Set $S(n)=\sum_{i=0}^n 2^{3i} {2n+1\choose 2i+1}$, using Zeilberger's algorithm, we can find the recurrence&#xA;$$S(n+2)=18S(n+1)-49S(n),\quad n\ge0,$$&#xA;with initial value $S(0)=1$ and $S(1)=11$. Mod $5$ both sides, we have &#xA;$$S(n+2) \equiv_5 3S(n+1)+S(n),\quad n\ge0,$$&#xA;with initial value $S(0)\equiv_5 S(1)\equiv_5 1$. It is easy to see that&#xA;$$S(12k)\equiv_5 S(12k+1)\equiv_5 S(12k+8)\equiv_5 1,$$&#xA;$$S(12k+5)\equiv_5 S(12k+9)\equiv_5 S(12k+10)\equiv_5 2,$$&#xA;$$S(12k+3)\equiv_5 S(12k+4)\equiv_5 S(12k+11)\equiv_5 3,$$&#xA;$$S(12k+2)\equiv_5 S(12k+6)\equiv_5 S(12k+7)\equiv_5 4.$$&lt;/p&gt;&#xA;" OwnerUserId="4210" LastActivityDate="2011-01-05T09:43:01.727" />
  <row Id="16418" PostTypeId="2" ParentId="16403" CreationDate="2011-01-05T09:51:39.777" Score="1" ViewCount="" Body="&lt;p&gt;I agree with Chas Brown that the truth table is more work-intensive than required, but it can be used.  What you need to do is (correctly) calculate the truth value of the statements for each possibility of knight/knave assignment.  Then see if the truth value corresponds to the type for each person.  This is a poor selection of a problem to demonstrate the method.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But we will try.  The first column would be A's statement, which as Chas points out, is &quot;I am a knight&quot; regardless of his status.  The second column would be the truth value of B's statement, which as Chas says is always false.  The third column would be the truth value of C's statement, which is always true.  Then the fourth column would be whether the truth value in the second column corresponds to B's type.  This is correct if B is a knave.  Similarly the fifth column would be whether the truth value in the third column corresponds to C's type, which is true if C is a knight.  The sixth column would be the and of the fourth and fifth.  Lines where the sixth column shows true are possible assignments.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-01-05T09:51:39.777" />
  <row Id="16419" PostTypeId="1" AcceptedAnswerId="16480" CreationDate="2011-01-05T10:23:36.327" Score="3" ViewCount="106" Body="&lt;p&gt;Let's $G=(U, V, E)$ be a balanced bipartite graph which $|U|=|V|=n$ and $|E|=n*(n-1)$; All nodes in $U$ are connected to all nodes in $V$ except $u_i$ to $v_i$ for $1\leq i \leq n$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Definition1: Cross edges are two edges in $E$, one with two end points $u_i$, $v_j$ and the other with $u_j$, $v_i$.&lt;br&gt;&#xA;Definition2: Good-perfect matching is a perfect matching with no cross edges.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the complexity of counting the number of Good-perfect matching in $G$?&lt;/p&gt;&#xA;" OwnerUserId="5409" LastActivityDate="2011-01-06T00:34:48.283" Title="Complexity of counting the number of Good-perfect matching in bipartite graph" Tags="&lt;combinatorics&gt;&lt;graph-theory&gt;&lt;permutations&gt;&lt;counting&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="16420" PostTypeId="2" ParentId="16403" CreationDate="2011-01-05T10:31:07.887" Score="1" ViewCount="" Body="&lt;p&gt;To learn about how to use truth tables, and formal logic in general, to solve Smullyan's puzzles on Knights and Knaves you can read Smullyan's own &quot;Logical Labyrinths&quot; esp. chapter 8: Liars, Truth-tellers and propositional logic:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://books.google.es/books?id=8i-FeLk0tCcC&amp;amp;printsec=frontcover&amp;amp;dq=Logical+labyrinths&amp;amp;hl=es&amp;amp;ei=xEckTYiJGY2B5AbknNGoCg&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=1&amp;amp;ved=0CCYQ6AEwAA#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;http://books.google.es/books?id=8i-FeLk0tCcC&amp;amp;printsec=frontcover&amp;amp;dq=Logical+labyrinths&amp;amp;hl=es&amp;amp;ei=xEckTYiJGY2B5AbknNGoCg&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=1&amp;amp;ved=0CCYQ6AEwAA#v=onepage&amp;amp;q&amp;amp;f=false&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I am not mistaken, I think he also explains a bit about this stuff in &quot;Forever undecided&quot;.&lt;/p&gt;&#xA;" OwnerUserId="5405" LastEditorUserId="5405" LastEditorDisplayName="" LastEditDate="2011-01-05T12:29:47.313" LastActivityDate="2011-01-05T12:29:47.313" />
  <row Id="16421" PostTypeId="2" ParentId="2899" CreationDate="2011-01-05T10:45:32.977" Score="3" ViewCount="" Body="&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/jaIuJ.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="5397" LastActivityDate="2011-01-05T10:45:32.977" />
  <row Id="16422" PostTypeId="1" AcceptedAnswerId="16423" CreationDate="2011-01-05T10:46:42.980" Score="2" ViewCount="119" Body="&lt;p&gt;If $a^2 = b^2 $ and $a,b \gt 0$ then we can answer &quot;Is $a &amp;gt; b$?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that $a^2 = b^2 \Rightarrow |a| = |b|$ but still I don't understand why the second condition is absolutely necessary.&lt;/p&gt;&#xA;" OwnerUserId="5410" LastEditorUserId="1543" LastEditorDisplayName="" LastEditDate="2011-05-03T10:09:11.103" LastActivityDate="2011-05-03T10:09:11.103" Title="Why both of the conditions are necessary ?" Tags="&lt;algebra-precalculus&gt;" AnswerCount="4" />
  <row Id="16423" PostTypeId="2" ParentId="16422" CreationDate="2011-01-05T10:50:30.857" Score="4" ViewCount="" Body="&lt;p&gt;If $|a|=|b|$ you don't know whether the signs are the same.  Given, say, $|a|=|b|=2$, you could have a=-2, b=2 in which case b&gt;a, or a=2, b=-2, so a&gt;b, or they could have the same sign so a=b.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-01-05T10:50:30.857" />
  <row Id="16424" PostTypeId="2" ParentId="16422" CreationDate="2011-01-05T10:52:16.190" Score="0" ViewCount="" Body="&lt;p&gt;From |a| = |b| it is impossible to answer to: Is a &gt; b?&#xA;For example, a might be -1 and b be 1, or viceversa. If you know both of them are positive, then you are done.&lt;/p&gt;&#xA;" OwnerUserId="5405" LastActivityDate="2011-01-05T10:52:16.190" />
  <row Id="16425" PostTypeId="2" ParentId="16422" CreationDate="2011-01-05T10:54:44.633" Score="0" ViewCount="" Body="&lt;p&gt;The condition a,b&gt;0 is not the only way to determine whether a&gt;b, but you do need to know something about the signs since without specifying them you only have |a|=|b| and so, depending on which signs apply you could have any of a=b, a &amp;lt; b, or a &gt; b&lt;/p&gt;&#xA;" OwnerUserId="5397" LastActivityDate="2011-01-05T10:54:44.633" />
  <row Id="16426" PostTypeId="1" AcceptedAnswerId="16428" CreationDate="2011-01-05T11:07:13.390" Score="1" ViewCount="74" Body="&lt;p&gt;What is exactly meant by &quot;&lt;strong&gt;angle subtended by all sides of a polygon at its center&lt;/strong&gt;&quot; ? I encountered these while solving this problem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can we determine the number of sides of a polygon,given the angles subtended by all the side at its center are equal to $72 ^\circ$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could somebody please explain (with a figure) what is meant to calculate in this question?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: I tried to Google some help, which took me to this &lt;a href=&quot;http://www.mathopenref.com/polygoncentralangle.html&quot; rel=&quot;nofollow&quot;&gt;page&lt;/a&gt;.I think this is what meant in this question.If so why all sides is mention in this question? and why it is supposed that the polygon will have equal sides?&lt;/p&gt;&#xA;" OwnerUserId="2109" LastEditorUserId="2109" LastEditorDisplayName="" LastEditDate="2011-01-05T11:40:16.917" LastActivityDate="2011-01-05T11:42:14.560" Title="Explain :: Angle subtended by all sides of a polygon at its center" Tags="&lt;geometry&gt;" AnswerCount="2" />
  <row Id="16427" PostTypeId="2" ParentId="16426" CreationDate="2011-01-05T11:13:05.290" Score="1" ViewCount="" Body="&lt;p&gt;If there are 360 degrees in a circle...&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-01-05T11:13:05.290" CommentCount="2" />
  <row Id="16428" PostTypeId="2" ParentId="16426" CreationDate="2011-01-05T11:42:14.560" Score="2" ViewCount="" Body="&lt;p&gt;Here's a rough diagram showing the center of the polygon and some sides and indicating the angle that is subtended by a side.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/dXnqH.png&quot; alt=&quot;diagram&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are given that the measure of all such angles in a particular polygon is 72&amp;deg; and asked to determine the number of sides in the polygon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;edit&lt;/em&gt;: As to why &quot;all sides&quot; is used in the problem, I'd guess it's a translation/language issue and that &quot;each side&quot; is a more appropriate description of what was intended.&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2011-01-05T11:42:14.560" CommentCount="8" />
  <row Id="16429" PostTypeId="2" ParentId="16400" CreationDate="2011-01-05T11:45:23.137" Score="1" ViewCount="" Body="&lt;p&gt;Start at the outside. Step inwards a small amount to generate a first level (polygonal) curve at a small height above the wall. Project &quot;ridge&quot; lines from outward corners of house (or crotch lines for inside corners) through corresponding corners of the new polygon (I guess they'll be bisecting the angles of the original polygon) and whenever two or more ridgelines, or two or more of the edges meet, repeat the process starting with the reduced polygon(s) which give the level curve through the point of intersection.&lt;/p&gt;&#xA;" OwnerUserId="5397" LastActivityDate="2011-01-05T11:45:23.137" />
  <row Id="16430" PostTypeId="2" ParentId="16400" CreationDate="2011-01-05T11:45:57.050" Score="2" ViewCount="" Body="&lt;p&gt;Try computing the &lt;a href=&quot;http://en.wikipedia.org/wiki/Straight_skeleton&quot; rel=&quot;nofollow&quot;&gt;straight skeleton&lt;/a&gt; of the polygon. See also &lt;a href=&quot;http://www.sable.mcgill.ca/~dbelan2/roofs/roofs.html&quot; rel=&quot;nofollow&quot;&gt;http://www.sable.mcgill.ca/~dbelan2/roofs/roofs.html&lt;/a&gt; and &lt;a href=&quot;http://www.vterrain.org/Culture/BldCity/Roof/&quot; rel=&quot;nofollow&quot;&gt;http://www.vterrain.org/Culture/BldCity/Roof/&lt;/a&gt; for instance.&lt;/p&gt;&#xA;" OwnerUserId="589" LastActivityDate="2011-01-05T11:45:57.050" />
  <row Id="16431" PostTypeId="1" AcceptedAnswerId="16433" CreationDate="2011-01-05T12:09:00.430" Score="1" ViewCount="76" Body="&lt;p&gt;I am trying to find the complex roots of:  exp(4z)=i.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However I am confused as how to calculate tan-1(1/0) to find the argument. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any help would be much appreciated. Thanks&lt;/p&gt;&#xA;" OwnerUserId="4522" LastActivityDate="2011-01-06T17:17:19.817" Title="Evaluating the argument of exp(4z)=i." Tags="&lt;complex-analysis&gt;" AnswerCount="2" />
  <row Id="16432" PostTypeId="2" ParentId="16431" CreationDate="2011-01-05T12:14:19.973" Score="3" ViewCount="" Body="&lt;p&gt;So let $z=x+iy$. Then you have $4z=4x+i4y$. Then $$e^{4z}=e^{4x+i4y}=e^{4x}\Bigl[i\sin{4y} + \cos{4y}\Bigr]$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since this is you want $e^{4x}\cos{4y}=0$ and $e^{4x}\sin{4y}=1$.&lt;/p&gt;&#xA;" LastActivityDate="2011-01-06T17:17:19.817" CommentCount="2" />
  <row Id="16433" PostTypeId="2" ParentId="16431" CreationDate="2011-01-05T12:14:59.433" Score="2" ViewCount="" Body="&lt;p&gt;Two suggestions:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;consider for what value(s) of $x$ the expression $\tan x$ is undefined (which could correspond to your $\frac{1}{0}$)&lt;/li&gt;&#xA;&lt;li&gt;rather than applying a formula ($\tan\frac{y}{x}$, or something like that), think about where $i$ is located, specifically at what magnitude of rotation about $0$ from the positive real axis&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="72" LastActivityDate="2011-01-05T12:14:59.433" />
  <row Id="16434" PostTypeId="1" CreationDate="2011-01-05T12:36:46.157" Score="4" ViewCount="138" Body="&lt;p&gt;Assume I have a universe of N elements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is: How many sets of size $X$ are needed to assure that every set of K elements is a subset of (at least) one of these sets (where $K \ll X \lt N$). And also, how can these sets be chosen to obtain this minimum?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, the sizes that interest me are: &#xA;\begin{align*}&#xA;N &amp;amp;= O(2^{n^{c_1}}),\\&#xA;X &amp;amp;= O(2^n),\\&#xA;K &amp;amp;= O(n^{c_2}),&#xA;\end{align*}&#xA;where $n$ is a variable, $c_1$ and $c_2$ are constants.&lt;/p&gt;&#xA;" OwnerUserId="5411" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-05T14:49:17.933" LastActivityDate="2011-01-05T14:49:17.933" Title="Minimum number of subsets needed" Tags="&lt;combinatorics&gt;&lt;set-theory&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="16436" PostTypeId="2" ParentId="16434" CreationDate="2011-01-05T13:38:22.637" Score="1" ViewCount="" Body="&lt;p&gt;Are &lt;a href=&quot;http://www.ccrwest.org/cover.html&quot; rel=&quot;nofollow&quot;&gt;covering numbers&lt;/a&gt; what you had in mind? Their (v,k,t) corresponds to your (N,X,K), if I'm not mistaken. Generally these covering numbers are very hard to compute, even for moderate values of (N,X,K). But perhaps asymptotic results are known that would help with the particular very large sizes that interest you.&lt;/p&gt;&#xA;" OwnerUserId="940" LastActivityDate="2011-01-05T13:38:22.637" />
  <row Id="16437" PostTypeId="2" ParentId="16434" CreationDate="2011-01-05T13:53:40.160" Score="3" ViewCount="" Body="&lt;p&gt;What you are asking about are called &lt;a href=&quot;http://www.ccrwest.org/cover.html&quot; rel=&quot;nofollow&quot;&gt; covering designs&lt;/a&gt;.  A more standard notation in combinatorics defines $(v,k,t)$-covering design to mean:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$v =$ total number of points (your universe N)&lt;br&gt;&#xA;$k =$ size of subsets (blocksize, your set size X)&lt;br&gt;&#xA;$t =$ size of covered combinations (your K elements)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a special case where each t-subset is covered exactly once, and these are known as Steiner systems.  Obviously in those cases the number of blocks used is a minimum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The minimum number of blocks ($k$-subsets) needed to cover all $t$-subsets of a &quot;universe&quot; of size $v$ is termed $C(v,k,t)$, and there is no general formula for it.  In fact not much is known for values of $t$ more than ten.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a well-known general lower bound for $C(v,k,t)$ called the Schönheim inequality:&lt;br&gt;&#xA;$$C(v,k,t) \ge \lceil (v/k) * C(v-1,k-1,t-1) \rceil$$&#xA;which can be applied recursively down to the case $t=1$, where trivially: &#xA;$$C(v,k,1) = \lceil v/k \rceil$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think it is known that this lower bound is fairly tight for sufficiently large $v$.  The La Jolla Covering Repository (linked above) has some good papers on constructing covering designs, but they seems to have gotten hidden when the site underwent a redesign awhile back.  I'll see if I can ferret them out and post those lost links.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added: Lost Links&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.ccrwest.org/cover/cover.pdf&quot; rel=&quot;nofollow&quot;&gt; [New Constructions for Covering Designs -- Gordon et al, 1995]&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.ccrwest.org/gordon/optimal.pdf&quot; rel=&quot;nofollow&quot;&gt; [Asymptotically Optimal Covering Designs -- Gordon et al, 1995]&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.ccrwest.org/gordon/hcd.pdf&quot; rel=&quot;nofollow&quot;&gt; [Handbook of Combinatorial Design, chapter excerpt by Gordon and Stinson]&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3111" LastEditorUserId="3111" LastEditorDisplayName="" LastEditDate="2011-01-05T14:15:45.687" LastActivityDate="2011-01-05T14:15:45.687" CommentCount="1" />
  <row Id="16438" PostTypeId="2" ParentId="16422" CreationDate="2011-01-05T13:58:39.087" Score="0" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\ $ For $\rm\ x = a/b\ $  it is $\rm\ x^2 = 1\:,\ x &amp;gt; 0\ \Rightarrow\ x = 1$ &lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2011-01-05T13:58:39.087" />
  <row Id="16439" PostTypeId="2" ParentId="16400" CreationDate="2011-01-05T14:54:26.497" Score="2" ViewCount="" Body="&lt;p&gt;Here is an image I happen to have prepared for a book.  It shows the &lt;em&gt;medial-axis polyhedron&lt;/em&gt; over a convex polygon, which coincides with the straight skeleton as in lhf's answer, and also with the offset-procedure in alQpr's answer. Sides slope at $45^\circ$.&#xA;&lt;br /&gt;&lt;img src=&quot;http://i.imgur.com/lsZVk.jpg&quot; alt=&quot;MedialAxis Polyhedron&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="237" LastActivityDate="2011-01-05T14:54:26.497" CommentCount="2" />
  <row Id="16440" PostTypeId="2" ParentId="16290" CreationDate="2011-01-05T14:58:47.897" Score="3" ViewCount="" Body="&lt;p&gt;Let $Q:=[-2,2]\times[-1,1]$ be the given rectangle and let $r_0$ be the radius computed by Isaac for this rectangle. I shall prove that 3 circles of radius $r&amp;gt;r_0$ cannot be placed into $Q$ without overlap. The midpoints of these circles would have to lie in the smaller rectangle $Q&amp;#39;:=[-2+r, 2-r]\times[-1+r,1-r]$. The left half $Q&amp;#39;_-$ of $Q&amp;#39;$ has a diameter $&amp;lt;2r_0&amp;lt;2r$ (by definition of $r_0$). It follows that $Q&amp;#39;_-$ can contain at most one center of non-overlapping circles of radius $r$, and the same is true for the right half $Q&amp;#39;_+$.&lt;/p&gt;&#xA;" OwnerUserId="1303" LastEditorUserId="1303" LastEditorDisplayName="" LastEditDate="2011-01-06T08:24:06.597" LastActivityDate="2011-01-06T08:24:06.597" CommentCount="1" />
  <row Id="16441" PostTypeId="1" AcceptedAnswerId="16446" CreationDate="2011-01-05T15:38:30.700" Score="1" ViewCount="75" Body="&lt;p&gt;Let $U$ be unitary and $D$ be diagonal. Can anything be stated about the structure of $$U^\dagger D U$$ then?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, would this change for an infinite matrix, i.e. a discrete linear operator?&lt;/p&gt;&#xA;" OwnerUserId="163" LastActivityDate="2011-01-05T16:27:47.023" Title="Can any statement be made on the structure of a diagonal matrix after an unitary transformation?" Tags="&lt;linear-algebra&gt;&lt;matrices&gt;&lt;transformation&gt;" AnswerCount="2" />
  <row Id="16442" PostTypeId="1" CreationDate="2011-01-05T15:42:52.473" Score="2" ViewCount="166" Body="&lt;p&gt;Consider the triangles with integer sides a, b and c with a ≤ b ≤ c.&#xA;An integer sided triangle (a,b,c) is called primitive if gcd(a,b,c)=1.&#xA;How many primitive integer sided triangles exist with a perimeter not exceeding 10 000 000? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am trying to solve this on euler project. I am wondering what is the best way to go to find the valid triples for constructing a triangle. Of course you can do nested for loops but that is not efficient. Any pointers would help.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="5415" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2011-01-06T05:03:51.717" LastActivityDate="2011-01-06T05:03:51.717" Title="Primitive integer triangles" Tags="&lt;geometry&gt;&lt;diophantine-equations&gt;&lt;project-euler&gt;" AnswerCount="3" CommentCount="0" />
  <row Id="16443" PostTypeId="2" ParentId="16441" CreationDate="2011-01-05T16:01:06.093" Score="4" ViewCount="" Body="&lt;p&gt;The n×n matrices of the form U&lt;sup&gt;*&lt;/sup&gt; D U for D diagonal and U unitary are exactly the &lt;a href=&quot;http://en.wikipedia.org/wiki/Normal_matrix&quot; rel=&quot;nofollow&quot;&gt;Normal matrices&lt;/a&gt;, that is the matrices that commute with their conjugate transpose.  The extension to hilbert-space operators are called normal operators.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If by dagger you mean the plain transpose, then I believe Horn–Johnson's book discusses these matrices, but they are less common.&lt;/p&gt;&#xA;" OwnerUserId="583" LastActivityDate="2011-01-05T16:01:06.093" CommentCount="2" />
  <row Id="16444" PostTypeId="2" ParentId="16442" CreationDate="2011-01-05T16:05:33.727" Score="3" ViewCount="" Body="&lt;p&gt;This is &lt;a href=&quot;http://projecteuler.net/index.php?section=problems&amp;amp;id=276&quot; rel=&quot;nofollow&quot;&gt;problem 276&lt;/a&gt;.  I haven't solved it, but the first thing to realize is you don't have to loop over c-from a and b you should be able to find the number of solutions.  Of course, you don't have time to loop over both a and b, but you may be able to loop over a.  I would start by doing a brute force run up to 100 or 250-you should have time for that.  Keep track of the number of solutions for each a.  Then try to find a formula for the number of triangles that include a.  This will probably be based on the factors of a.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope this helps.  I did a lot of them, but have gotten less energetic now.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-01-05T16:05:33.727" />
  <row Id="16445" PostTypeId="2" ParentId="16442" CreationDate="2011-01-05T16:06:10.167" Score="0" ViewCount="" Body="&lt;p&gt;There is &lt;a href=&quot;http://en.wikipedia.org/wiki/Pythagorean_triple#Generating_a_triple&quot; rel=&quot;nofollow&quot;&gt;a formula&lt;/a&gt; which gives you &lt;em&gt;all&lt;/em&gt; the Pythagorean triples. That should cut down your search space a bit.&lt;/p&gt;&#xA;" OwnerUserId="5191" LastActivityDate="2011-01-05T16:06:10.167" CommentCount="1" />
  <row Id="16446" PostTypeId="2" ParentId="16441" CreationDate="2011-01-05T16:21:43.083" Score="5" ViewCount="" Body="&lt;p&gt;If the entries of $D$ are real, there is a beautiful answer due to Schur. Let the entries of $D$ be $(d_1, d_2, \ldots, d_n)$. Then the possible diagonal entries of $U^* D U$ are the points in the convex hull of the $n!$ permutations of the $d_i$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another way to phrase this is the following: Let $d_1 \geq d_2 \geq \cdots \geq d_n$. Then $(c_1, c_2, \ldots, c_n)$ are possible diagonal entries for $U^* D U$ if and only if&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum c_i = \sum d_i$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and, for any $i_1 &amp;lt; i_2 &amp;lt; \cdots &amp;lt; i_k$, we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$c_{i_1} + c_{i_2} + \cdots + c_{i_k} \leq d_1 + d_2 + \cdots + d_k.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See section 5 of Bhatia's article &lt;a href=&quot;http://math.ucr.edu/~jdolan/schubert2.pdf&quot; rel=&quot;nofollow&quot;&gt;Linear Algebra to Quantum Cohomology: The Story of Alfred Horn's Conjectures&lt;/a&gt; for a proof.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;If $D$ is complex, $D=A+Bi$, then $\mathrm{diag}(U^* D U^*) = \mathrm{diag}(U^* A U) + \mathrm{diag}(U^* B U^*) i$. So the real and imaginary parts of the diagonal separately obey Schur's bounds with respect to $A$ and $B$. I'm not sure if you can do better than this.&lt;/p&gt;&#xA;" OwnerUserId="448" LastEditorUserId="448" LastEditorDisplayName="" LastEditDate="2011-01-05T16:27:47.023" LastActivityDate="2011-01-05T16:27:47.023" CommentCount="4" />
  <row Id="16447" PostTypeId="1" AcceptedAnswerId="16799" CreationDate="2011-01-05T16:24:24.153" Score="17" ViewCount="528" Body="&lt;p&gt;This is an old homework problem of mine that I was never able to solve.  The solution may or may not involve the Baire category theorem, which I am terrible at applying.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $C[0, 1]$ denote the space of continuous functions $[0, 1] \to \mathbb{R}$.  Suppose $|| \cdot ||$ is a norm on $C[0, 1]$ with respect to which the evaluation functions $f \mapsto f(x), x \in [0, 1]$ are continuous.  Show that the topology induced by $|| \cdot ||$ is the same as the usual topology induced by the sup norm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strike&gt;It is straightforward to show that any sequence converging uniformly converges with respect to $|| \cdot ||$.  I am stuck on proving the converse; I cannot seem to figure out how to use the assumption that $|| \cdot ||$ is a norm.&lt;/strike&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit:  Zhen Lin informs me that $|| \cdot ||$ is supposed to be complete.  That should make the problem statement true now!&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2011-01-08T13:21:33.360" LastActivityDate="2011-01-08T22:33:28.427" Title="Norms on C[0, 1] inducing the same topology as the sup norm" Tags="&lt;real-analysis&gt;&lt;homework&gt;&lt;functional-analysis&gt;" AnswerCount="3" CommentCount="10" FavoriteCount="7" />
  <row Id="16448" PostTypeId="1" AcceptedAnswerId="16451" CreationDate="2011-01-05T17:03:06.620" Score="3" ViewCount="76" Body="&lt;p&gt;I'm sure there must be lots of  examples of three (infinite) groups $A$, $B$, $C$ where $A&amp;lt;B&amp;lt;C$ , $A\cong C$, $A\not\cong B$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's one example with free groups: $\mathbf F(a^2,ab) &amp;lt; \mathbf F(a^2,ab,ab^{-1}) &amp;lt; \mathbf F(a,b)$, where A,C are a free group on 2 symbols and B is a free group on 3 symbols (in fact precisely the strings of even length in $\mathbf F(a,b)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But are there any examples that are just as easy but do not involve free groups (or equivalent constructions)?&lt;/p&gt;&#xA;" OwnerUserId="5189" LastActivityDate="2011-01-05T17:07:13.743" Title="Easy example of $A&lt;B&lt;C$ , $A\cong C$, $A\not\cong B$" Tags="&lt;group-theory&gt;" AnswerCount="1" />
  <row Id="16451" PostTypeId="2" ParentId="16448" CreationDate="2011-01-05T17:07:13.743" Score="6" ViewCount="" Body="&lt;p&gt;Take C to be the direct sum of countably many copies of Z/4Z, and A to be the direct summand where you omit one of the Z/4Z summands, call it X.  So A⊕X = C and X≅Z/4Z.  Then B=A⊕2X is not isomorphic to A or C, but A and C are isomorphic.&lt;/p&gt;&#xA;" OwnerUserId="583" LastActivityDate="2011-01-05T17:07:13.743" CommentCount="1" />
  <row Id="16452" PostTypeId="1" AcceptedAnswerId="16454" CreationDate="2011-01-05T17:13:43.270" Score="0" ViewCount="101" Body="&lt;p&gt;How to sum $2^{2m}$ where m varies from 0 t0 n ?&lt;/p&gt;&#xA;" OwnerUserId="5421" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2011-01-05T18:29:57.453" LastActivityDate="2011-01-05T18:29:57.453" Title="Sum of series $2^{2m}$" Tags="&lt;sequences-and-series&gt;" AnswerCount="2" />
  <row Id="16453" PostTypeId="2" ParentId="16452" CreationDate="2011-01-05T17:23:31.397" Score="2" ViewCount="" Body="&lt;p&gt;If you really mean $n$ varies from $0$ to $n$ then the answer is&#xA;$\sum _{n=0}^n 2^{2 m}=2^{2 m} (1 + n)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Otherwise if $m$ goes from $0$ to $n$ its just a geometric series ( &lt;a href=&quot;http://mathworld.wolfram.com/GeometricSeries.html&quot; rel=&quot;nofollow&quot;&gt;http://mathworld.wolfram.com/GeometricSeries.html&lt;/a&gt; )&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sum _{m=0}^n 2^{2 m}=\frac{1}{3} \left(2^{2 n+2}-1\right)$&lt;/p&gt;&#xA;" OwnerUserId="3123" LastActivityDate="2011-01-05T17:23:31.397" CommentCount="2" />
  <row Id="16454" PostTypeId="2" ParentId="16452" CreationDate="2011-01-05T17:27:54.520" Score="3" ViewCount="" Body="&lt;p&gt;Note that you can get: $2^{2m}=4^m$,   now&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle\sum_{m=o}^{n} 2^{2m} = \displaystyle\sum_{m=o}^{n} 4^{m}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The last is easy,( is geometric with r=4), so&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle\sum_{m=o}^{n} 2^{2m} = \displaystyle\sum_{m=o}^{n} 4^{m} = \displaystyle\frac{4-4^{n+1}}{1-4}+1=\displaystyle\frac{4^{n+1}-4}{3}+1$$&lt;/p&gt;&#xA;" OwnerUserId="3421" LastEditorUserId="3421" LastEditorDisplayName="" LastEditDate="2011-01-05T17:58:00.107" LastActivityDate="2011-01-05T17:58:00.107" />
  <row Id="16455" PostTypeId="1" CreationDate="2011-01-05T17:41:40.067" Score="2" ViewCount="109" Body="&lt;p&gt;Here's a problem that I have pondered over many times without ever coming to a satisfactory solution:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's say that we have a series of random events: V(i) for I = 1 to n.  Each of these events will have a result VR(i) of either 0 or 1, and a probability VP(i) (from zero to one) that represents the probability that VR(i) = 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We also have a group of k estimators: E(j) for j = 1 to k, that generate EP(j,i) which are estimates of VP(i).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I want is an evaluation function F(j) that given EP(j,i={1..n})  (that is, E(j)'s estimates of VP(i) for i = 1 to n), and VR(i={1 to n}) (i.e., the actual results), that will return a number that is a &quot;good&quot; valuation of E(j)'s ability to estimate E(i).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Some Notes:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The results, VR(i), are known to the Evaluation function, but (obviously) not to the Estimators, neither before or after each result (so the estimators cannot use the VR(i)'s to adjust their subsequent predictions, if that matters).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The probabilities, VP(i), are not known to the Evaluation function.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The &lt;em&gt;distribution&lt;/em&gt; of the probabilities VP(i) is not know to the Evaluation function.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;EP(j,i) is supposed to be an estimate of the event &lt;em&gt;probabilities&lt;/em&gt;, VP(i), and &lt;strong&gt;not&lt;/strong&gt; an estimate of the &lt;em&gt;results&lt;/em&gt;, VR(i).  Virtually every valuation system that I have seen tends to weight each EP(j,i) solely on its closeness to VR(i), which invariably rewards &quot;polarized&quot; estimators: those that always return 1 if VP(i) &gt; 0.5 and 0 if VP(i) &amp;lt; 0.5.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;One hallmark of the problem in #4 is that if VP(i) = 0.5, then these types of valuation systems will tend to reward all EP(j,i)'s the same, or else give the highest reward to EP's of 0 or 1, and the &lt;em&gt;worst&lt;/em&gt; valuation to the actual correct estimate, 0.5.  What I would like, of course, is just the opposite: for estimates of 0 or 1 for actual probabilities of 0.5 to receive the &lt;em&gt;worst&lt;/em&gt; valuation while the correct estimates of 0.5 to be rated the highest.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This I think is also the essence of the difficulty in this problem: how do I correctly value the probability estimates (and estimators) if I only know the event results, but not the actual probabilities being estimated?&lt;/p&gt;&#xA;" OwnerUserId="5419" LastEditorUserId="5419" LastEditorDisplayName="" LastEditDate="2011-01-05T17:48:21.077" LastActivityDate="2011-04-14T21:37:54.690" Title="How to evaluate probability estimators with only external information?" Tags="&lt;probability&gt;&lt;probability-theory&gt;&lt;approximation&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="1" />
  <row Id="16456" PostTypeId="2" ParentId="6848" CreationDate="2011-01-05T17:51:43.657" Score="1" ViewCount="" Body="&lt;p&gt;$\lim_{x \to 0}\ \cos^2(x)\ \dfrac{\sin(\cos(x))}{\cos(x)}=\lim_{x \to 0}\ \cos^2(x)\ \dfrac{\sin(1)}{1}=\lim_{x \to 0}\ \cos^2(x)\sin(1)=\sin(1)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Simple mistake when simplifying the term $\dfrac{\sin(\cos(x))}{\cos(x)}$&lt;/p&gt;&#xA;" OwnerUserId="3123" LastActivityDate="2011-01-05T17:51:43.657" />
  <row Id="16457" PostTypeId="1" CreationDate="2011-01-05T18:07:16.783" Score="4" ViewCount="176" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I just finished reading the proof that the average height of a random binary of given size $n$ is asymptotically $2\sqrt{\pi n}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm now searching for an intuitive, or geometric, or visual proof of this asymptotic equivalence. In other words, I'm trying to find an intuitive arguments that shows that this number grows like $\sqrt{n}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any ideas?&lt;br&gt;&#xA;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="2964" LastActivityDate="2011-01-05T19:14:58.730" Title="Geometric / Visual explanation that the average height of a random binary tree of given size $n$ is asymptotically $2\sqrt{\pi n}$." Tags="&lt;algorithms&gt;&lt;asymptotics&gt;&lt;trees&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" />
  <row Id="16458" PostTypeId="1" AcceptedAnswerId="16460" CreationDate="2011-01-05T18:12:13.460" Score="1" ViewCount="91" Body="&lt;p&gt;Let M be a finitely generated R-module ,where R is a PID and N a submodule.Is there an epimorphism from M to N?&lt;/p&gt;&#xA;" OwnerUserId="5422" LastActivityDate="2011-01-05T19:31:48.067" Title="Modules and epimorphisms." Tags="&lt;modules&gt;" AnswerCount="1" />
  <row Id="16459" PostTypeId="1" AcceptedAnswerId="16463" CreationDate="2011-01-05T18:20:48.190" Score="2" ViewCount="188" Body="&lt;p&gt;How to find the sum of series, $$\sum_{k=1}^n 2^kC(n,k)$$&lt;/p&gt;&#xA;" OwnerUserId="5421" LastEditorUserId="3936" LastEditorDisplayName="" LastEditDate="2011-05-28T15:47:49.620" LastActivityDate="2011-05-28T15:47:49.620" Title="How to find $\sum_{k=1}^n 2^kC(n,k)$?" Tags="&lt;combinatorics&gt;&lt;sequences-and-series&gt;&lt;binomial-coefficients&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="16460" PostTypeId="2" ParentId="16458" CreationDate="2011-01-05T18:25:08.840" Score="4" ViewCount="" Body="&lt;p&gt;Yes, but the epimorphism might not doing anything nice to N ≤ M.  Just use the invariants.  What do the invariants of submodules look like?  What do the invariants of quotient modules look like?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance R=Z, M=ZxZxZ/4Z has invariants (0,0,4) and its submodules have invariants (0,0,4), (0,0,2), (0,0), (0,4), (0,2), (0), (4), (2), and ().  Its quotient modules have many invariants, but amongst them are all of the submodules' invariants.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could also handle the free and torsion parts separately to get a cleaner proof.  The submodules of the free summand are easy to understand, being free of at most the same rank, and clearly each is isomorphic to a quotient module.  The submodule lattice and quotient module lattice of the torsion summand are isomorphic, so you can do better than just &quot;some random epimorphism&quot;.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Invariants of submodules of (bounded) torsion modules:&lt;/strong&gt; If R is a PID and M is a torsion module, then M is the direct sum of its submodules M&lt;sub&gt;p&lt;/sub&gt; = { m in M : there is an n such that p&lt;sup&gt;n&lt;/sup&gt;⋅m = 0 } where p varies over the set of (nonzero) primes.  If N is a submodule of M, then N&lt;sub&gt;p&lt;/sub&gt; = N∩M&lt;sub&gt;p&lt;/sub&gt; by definition.  To study submodules of torsion modules, it suffices to study the submodules of their p-parts, M&lt;sub&gt;p&lt;/sub&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If M&lt;sub&gt;p&lt;/sub&gt; is finitely generated, then there is a largest n needed to get p&lt;sup&gt;n&lt;/sup&gt;⋅m = 0.  In other words, we are looking at modules M such that p&lt;sup&gt;n&lt;/sup&gt;M = 0, for some prime p in the PID R (or any noetherian, commutative ring where p is a principal maximal ideal).  Such an M decomposes into a direct sum M = (R/pR)&lt;sup&gt;(r&lt;sub&gt;1&lt;/sub&gt;)&lt;/sup&gt; ⊕ (R/p&lt;sup&gt;2&lt;/sup&gt;R)&lt;sup&gt;(r&lt;sub&gt;2&lt;/sub&gt;)&lt;/sup&gt; ⊕⋯⊕ (R/p&lt;sup&gt;n&lt;/sup&gt;R)&lt;sup&gt;(r&lt;sub&gt;n&lt;/sub&gt;)&lt;/sup&gt;.  Call such a module (p; r&lt;sub&gt;1&lt;/sub&gt;, r&lt;sub&gt;2&lt;/sub&gt;, …, r&lt;sub&gt;n&lt;/sub&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;Details: Two such modules are isomorphic iff after removing any &quot;trailing 0s&quot; (that is, with r&lt;sub&gt;n&lt;/sub&gt; ≠ 0) the labels are identical.  Any such labeling represents a bounded module as long as the r&lt;sub&gt;i&lt;/sub&gt; are all cardinal numbers, and the module is finitely generated iff the r&lt;sub&gt;i&lt;/sub&gt; are all finite.&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The isomorphism classes of its submodules are precisely the (p; s&lt;sub&gt;1&lt;/sub&gt;, s&lt;sub&gt;2&lt;/sub&gt;, …, s&lt;sub&gt;n&lt;/sub&gt;) with 0 ≤ s&lt;sub&gt;i&lt;/sub&gt; + s&lt;sub&gt;i+1&lt;/sub&gt; + … s&lt;sub&gt;n&lt;/sub&gt; ≤ r&lt;sub&gt;i&lt;/sub&gt; + r&lt;sub&gt;i+1&lt;/sub&gt; + … r&lt;sub&gt;n&lt;/sub&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example Z/9Z × Z/3Z × Z/3Z has label (3;2,1) and its submodules have labels (3;0,0), (3;0,1), (3;1,0), (3;1,1), (3;2,0), (3;2,1), and (3;3,0).  The submodule Z/9Z × Z/3Z × 1 for instance has label (3;1,1), the submodule Z/9Z × 1 × 1 has label (3;0,1), and the submodule 3Z/9Z &amp;times; Z/3Z &amp;times; Z/3Z has label (3;3,0).&lt;/p&gt;&#xA;" OwnerUserId="583" LastEditorUserId="583" LastEditorDisplayName="" LastEditDate="2011-01-05T19:31:48.067" LastActivityDate="2011-01-05T19:31:48.067" CommentCount="2" />
  <row Id="16461" PostTypeId="2" ParentId="16455" CreationDate="2011-01-05T18:26:55.100" Score="1" ViewCount="" Body="&lt;p&gt;I have thought about this a bit as well.  As you do, let n be the total number of events.  The best I can think of is to bin the EP(j,i) for each j in bins by i.  Say for predictor j you bin the lowest 10% of the predictions.  You can then calculate how many of those events should have happened and a variance on the number.  If the lowest 10% of the predictions were all 5%, say, you should have 5% of these n(10%) events happen with a variance of n(10%)(5%)(95%).  If a higher bin has 30% of its predictions 25% and 70% of its predictions 30%, you would expect 28.5% of those events to happen and can figure the variance.  You can then check the chi-square of all the predictors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think this is in the direction you want.  It can certainly give an answer, but whether it is justified I'm not sure.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-01-05T18:26:55.100" />
  <row Id="16462" PostTypeId="1" AcceptedAnswerId="16471" CreationDate="2011-01-05T18:33:16.260" Score="2" ViewCount="94" Body="&lt;p&gt;I am a beginner in representation theory and algebraic geometry, so that references giving clear explanations of things like the tautological line bundle on $\mathbb P^n$, its dual, and the associated sheaves of sections $\mathcal O(-1)$ and $\mathcal O(1)$ would be especially welcome.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have some difficulty in understanding the sentence in the first paragraph of section 9.3 (page 140) in the book of Fulton clearly: There is a general procedure for producing representations as sections of a line bundle on a homogeneous space on page 140 (the first paragraph of section 9.3) of the book Young Tableaux by Fulton: &lt;a href=&quot;http://books.google.com/books?id=U9vZal2HCcoC&amp;amp;printsec=frontcover&amp;amp;dq=young+tableaux&amp;amp;source=bl&amp;amp;ots=xxxzyLzSra&amp;amp;sig=KmvmON6e4Xli3Fz2b0g2S5j6XTE&amp;amp;hl=en&amp;amp;ei=vHkjTd_CCoWBlAemzLigDA&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=3&amp;amp;ved=0CDEQ6AEwAg#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;http://books.google.com/books?id=U9vZal2HCcoC&amp;amp;printsec=frontcover&amp;amp;dq=young+tableaux&amp;amp;source=bl&amp;amp;ots=xxxzyLzSra&amp;amp;sig=KmvmON6e4Xli3Fz2b0g2S5j6XTE&amp;amp;hl=en&amp;amp;ei=vHkjTd_CCoWBlAemzLigDA&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=3&amp;amp;ved=0CDEQ6AEwAg#v=onepage&amp;amp;q&amp;amp;f=false&lt;/a&gt;. I am sorry that I do not know how to make a good link such that the link is not so long.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you very much.&lt;/p&gt;&#xA;" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2011-01-05T19:02:33.847" LastActivityDate="2011-01-05T19:16:25.843" Title="what are good references for learning about vector bundles and their sheaves of sections?" Tags="&lt;algebraic-geometry&gt;&lt;algebraic-topology&gt;&lt;representation-theory&gt;&lt;manifold&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="16463" PostTypeId="2" ParentId="16459" CreationDate="2011-01-05T18:35:27.107" Score="7" ViewCount="" Body="&lt;p&gt;Substitute $x=2$ into the binomial expansion of $(1+x)^n$ and then rearrange.&lt;/p&gt;&#xA;" OwnerUserId="4594" LastEditorUserId="4594" LastEditorDisplayName="" LastEditDate="2011-05-28T15:14:17.753" LastActivityDate="2011-05-28T15:14:17.753" />
  <row Id="16465" PostTypeId="1" CreationDate="2011-01-05T18:57:51.467" Score="6" ViewCount="553" Body="&lt;p&gt;The solution of the cubic equation is known in terms of a rational function of a cube root of a square root. If we just want to know the value it's easy to approximate it using a numerical method. I would like to know if the actual form (in terms of roots) of the solution of the cubic has any applications in number theory?&lt;/p&gt;&#xA;" OwnerUserId="4290" LastActivityDate="2011-02-21T19:27:50.130" Title="Applications of cubic in number theory?" Tags="&lt;number-theory&gt;&lt;algebra&gt;" AnswerCount="3" />
  <row Id="16466" PostTypeId="1" AcceptedAnswerId="16474" CreationDate="2011-01-05T19:08:24.043" Score="2" ViewCount="126" Body="&lt;p&gt;Hey,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given $f:(0,\infty) \rightarrow \mathbb{R}$ and $f&amp;#39;&amp;#39;(x)&amp;gt;0\forall x \in (0,\infty)$. Is it correct to say that $f&amp;#39;(x)$ is a monotonically increasing function? Can I correctly assume that for any $a,b \in (0,\infty)$ $f&amp;#39;(a) &amp;gt; f(&amp;#39;b)$ if $a &amp;gt; b$?&lt;/p&gt;&#xA;" OwnerUserId="3637" LastActivityDate="2011-01-05T19:33:19.953" Title="Is $f'(x)$ monotonically increasing if $f''(x) &gt;0$?" Tags="&lt;calculus&gt;" AnswerCount="2" CommentCount="0" />
  <row Id="16467" PostTypeId="2" ParentId="16462" CreationDate="2011-01-05T19:11:01.880" Score="0" ViewCount="" Body="&lt;p&gt;A brief introduction to line bundles (and divisors) on schemes is available at &lt;a href=&quot;http://people.fas.harvard.edu/~amathew/linebund.pdf&quot; rel=&quot;nofollow&quot;&gt;http://people.fas.harvard.edu/~amathew/linebund.pdf&lt;/a&gt;, although it is very rough. This covers things like the tautological line bundle on $\mathbb{P}^n$ and the sheaves $\mathcal{O}(n)$. In particular, it gives the correspondence (as in Hartshorne II.6) between Weil and Cartier divisors on regular noetherian integral schemes (such as projective space over a field). I don't know if this is completely relevant to what you need, but it may (or may not) be useful for your initial questions. Having not read Fulton's book, I can't answer your specific question.&lt;/p&gt;&#xA;" OwnerUserId="536" LastActivityDate="2011-01-05T19:11:01.880" CommentCount="1" />
  <row Id="16468" PostTypeId="2" ParentId="16466" CreationDate="2011-01-05T19:14:17.643" Score="2" ViewCount="" Body="&lt;p&gt;Yes. Set $g=f&amp;#39;$. Then $g&amp;#39; = f&amp;#39;&amp;#39; &amp;gt; 0$ and monotonicity of $g=f&amp;#39;$ follows. (By the way: $f$ is convex.)&lt;/p&gt;&#xA;" OwnerUserId="3148" LastActivityDate="2011-01-05T19:14:17.643" />
  <row Id="16469" PostTypeId="2" ParentId="16457" CreationDate="2011-01-05T19:14:58.730" Score="6" ViewCount="" Body="&lt;p&gt;Here's one way to at least guess the growth rate.  A random binary tree of size $n$ is the same thing (plus or minus one) as a random walk on the non-negative integers of length $n$ starting and ending at $0$, with the height of the tree corresponding to the farthest the walk goes from $0$.  The order of magnitude of the height should behave like the order of magnitude of the related problem of looking at the ending position of a random walk on the integers of length $n$ starting at $0$.  (I don't know how to justify this rigorously but it seems pretty plausible to me.)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But this question is much easier, since now it's a sum of independent Bernoulli random variables.  The expected end position is $0$, but it is straightforward to calculate that the variance in the end position is $n$, so the standard deviation is $\sqrt{n}$.  Of course, since we're dealing with a sum of independent Bernoulli random variables, we can be much more precise because the &lt;a href=&quot;http://en.wikipedia.org/wiki/Central_limit_theorem&quot; rel=&quot;nofollow&quot;&gt;central limit theorem&lt;/a&gt; applies.  The point is that this probabilistic reasoning tells us what the right growth rate to expect is: it's also the same rate that crops up in &lt;a href=&quot;http://en.wikipedia.org/wiki/Brownian_motion&quot; rel=&quot;nofollow&quot;&gt;Brownian motion&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2011-01-05T19:14:58.730" CommentCount="6" />
  <row Id="16471" PostTypeId="2" ParentId="16462" CreationDate="2011-01-05T19:16:25.843" Score="1" ViewCount="" Body="&lt;p&gt;The friendliest introduction to complex vector bundles on complex manifolds might be &lt;em&gt;Complex Manifolds, Vector Bundles and Hodge Theory&lt;/em&gt; by Brylinski and Toth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the subject which interests you (link with representation theory), I recommend the article &lt;em&gt;The Borel-Weil theorem for complex projective space&lt;/em&gt; by M.Eastwood and J.Sawon in the book &lt;em&gt;Invitations to Geometry and Topology&lt;/em&gt; (Oxford Graduate Texts In Mathematics #7). The article is self-contained and fairly elementary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could also look at the nice book by Joseph Taylor, &lt;em&gt;Several Complex Variables&lt;/em&gt;, published by the AMS, Graduate  Studies In Mathematics #46. Borel-Weil is carefully handled there, with all the necessary preliminaries.&lt;/p&gt;&#xA;" OwnerUserId="3217" LastActivityDate="2011-01-05T19:16:25.843" CommentCount="1" />
  <row Id="16472" PostTypeId="2" ParentId="16459" CreationDate="2011-01-05T19:17:13.743" Score="6" ViewCount="" Body="&lt;p&gt;While Jasper Loy's answer is the canonical one, I thought folks might like to see a different one.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose you are interested, for some function $f(k)$, in the binomial sum &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$B(n) = \sum_{k=0}^n \binom{n}{k} f(k).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, taking $\Delta f(k) = f(k+1) - f(k)$, denote $A(n)$ by &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$A(n) = \sum_{k=0}^n \binom{n}{k} \Delta f(k).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's fairly easy to prove that $B(n+1) - 2B(n) = A(n)$ with initial condition $B(0) = f(0)$.  See, for example, Theorem 2 in my paper &quot;&lt;a href=&quot;http://math.pugetsound.edu/~mspivey/CombSum.pdf&quot; rel=&quot;nofollow&quot;&gt;Combinatorial Sums and Finite Differences&lt;/a&gt;,&quot; &lt;em&gt;Discrete Mathematics&lt;/em&gt;, 307 (24): 3130-3146, 2007.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the OP's question, $f(k) = 2^k$, and, of course, $\Delta f(k) = 2^k$ as well.  So $A(n) = B(n)$, and thus the sum $B(n) = \sum_{k=0}^n \binom{n}{k} 2^k$ can be found by solving the simple recurrence $B(n+1) = 3B(n)$ with initial condition $B(0) = 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;HR&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Incidentally, this approach can also be used to prove the binomial theorem itself for nonnegative integer values of $n$.  See Identity 1 in the paper.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For two more examples of using finite differences to evaluate binomial sums, see &lt;a href=&quot;http://math.stackexchange.com/questions/7757/how-to-prove-this-binomial-identity-sum-r-0n-r-n-choose-r-n2n-1/7810#7810&quot;&gt;this answer&lt;/a&gt; and &lt;a href=&quot;http://math.stackexchange.com/questions/4175/beautiful-identity-sum-k-mn-1k-m-binomkm-binomnk-delta/6743#6743&quot;&gt;this answer&lt;/a&gt; to previous math.SE questions.&lt;/p&gt;&#xA;" OwnerUserId="2370" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2011-01-05T19:57:27.640" LastActivityDate="2011-01-05T19:57:27.640" />
  <row Id="16473" PostTypeId="1" CreationDate="2011-01-05T19:19:04.310" Score="1" ViewCount="66" Body="&lt;p&gt;$S$ is real symmetric. $T$ is real skew-symmetric. I have shown that $T\pm iS$ is skew-Hermitian. I am further asked to show that $U = (I+T+iS)(I-T-iS)^{-1}$ is unitary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Denoting by $^\dagger$ the conjugate transpose, I have that&lt;br&gt;&#xA;$U^\dagger = [(I+T+iS)(I-T-iS)^{-1}]^\dagger$&lt;br&gt;&#xA;$= ((I-T-iS)^\dagger)^{-1}(I+T+iS)^\dagger$&lt;br&gt;&#xA;$= (I+T+iS)^{-1}(I-T-iS)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But then the products $UU^\dagger$ and $U^\dagger U$ have the factors in the wrong order to cancel out to make identity. I must be missing something obvious here; I'd appreciate a steer!&lt;/p&gt;&#xA;" OwnerUserId="5426" LastActivityDate="2011-01-19T05:55:58.927" Title="Show the matrix product $U = (I+T+iS)(I-T-iS)^{-1}$ is unitary" Tags="&lt;homework&gt;&lt;matrices&gt;" AnswerCount="0" CommentCount="4" />
  <row Id="16474" PostTypeId="2" ParentId="16466" CreationDate="2011-01-05T19:20:11.383" Score="2" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://http://en.wikipedia.org/wiki/Mean_value_theorem&quot; rel=&quot;nofollow&quot;&gt;Mean value theorem&lt;/a&gt; says yes. For each $a&amp;gt;0$, $b&amp;gt;0$ there exists $c\in(a,b)$ with $f&amp;#39;(b)=f&amp;#39;(a)+f&amp;#39;&amp;#39;(c)(b-a)$, since $f&amp;#39;&amp;#39;(x)$ is positive for all $x&amp;gt;0$ we get that $f&amp;#39;(b)\ge f&amp;#39;(a)$ which gives us the required statement. &lt;/p&gt;&#xA;" OwnerUserId="4742" LastEditorUserId="4742" LastEditorDisplayName="" LastEditDate="2011-01-05T19:33:19.953" LastActivityDate="2011-01-05T19:33:19.953" CommentCount="3" />
  <row Id="16475" PostTypeId="2" ParentId="154" CreationDate="2011-01-05T19:40:57.497" Score="-1" ViewCount="" Body="&lt;p&gt;Of course complex numbers do not exist. But that doesn't stop them from being so damn useful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;--B&lt;/p&gt;&#xA;" OwnerUserId="5427" LastActivityDate="2011-01-05T19:40:57.497" CommunityOwnedDate="2011-04-15T20:45:29.080" />
  <row Id="16476" PostTypeId="1" AcceptedAnswerId="16703" CreationDate="2011-01-05T19:55:03.013" Score="1" ViewCount="219" Body="&lt;p&gt;I was reading &quot;Graph Theory&quot; by Diestel and I tried to solve a problem from the chapter 3 (on connectivity). It seems at first sight easy (and really intuitive) but I have to admit that I can't work it out! Here is the problem: prove that every $k$-connected graph of order at least $2k$ contains a cycle of length at least $2k$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anyone have a suggestion?&lt;/p&gt;&#xA;" OwnerUserId="5465" LastEditorUserId="5465" LastEditorDisplayName="" LastEditDate="2011-01-05T22:17:01.573" LastActivityDate="2011-01-08T03:29:06.790" Title="Cycles in $k$-connected graphs" Tags="&lt;graph-theory&gt;" AnswerCount="3" />
  <row Id="16477" PostTypeId="1" AcceptedAnswerId="16494" CreationDate="2011-01-05T20:01:32.207" Score="4" ViewCount="146" Body="&lt;p&gt;A function $t: [a, b] \rightarrow \mathbb{R}$ is called a step function when a $k \in \mathbb{N}$ and numbers $z_0,...,z_k$ with $a = z_0 \leq z_1 \leq ... \leq z_k = b$ exist, such that for all $i \in \{1,2,...k\}$ the restriction $t |_{(z_{i-1},z_{i})}$ is constant. Let $f: [0,1] \rightarrow \mathbb{R}$ be defined by $$f(x) = \left\{ \begin{array}{rcl} 1, &amp;amp; \mbox{if} &amp;amp; x \in \mathbb{Q} \\ 0, &amp;amp; \mbox{if} &amp;amp; x \notin \mathbb{Q} \end{array} \right.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Show:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(i) The function $f$ is a point-wise limit of step functions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(ii) There is no uniform convergent series of step functions, whose limit-function is $f$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the book I'm reading mentions this Dirichlet function all the time. Still I'm having trouble finding a solution to this exercise. All help is very much appreciated!&lt;/p&gt;&#xA;" OwnerUserId="3711" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2011-01-05T22:39:55.540" LastActivityDate="2011-01-05T22:39:55.540" Title="Step functions and the characteristic function of rationals." Tags="&lt;analysis&gt;" AnswerCount="2" />
  <row Id="16478" PostTypeId="1" AcceptedAnswerId="16483" CreationDate="2011-01-05T20:01:51.410" Score="2" ViewCount="82" Body="&lt;p&gt;If $a_{n}$ is a non-negative, decreasing sequence, we know from the integral test that if $f(n)=a_{n}$ is an integrable function, then $\sum_{n=1}^{\infty} a_{n}$ and $\int_{1}^{\infty} f(x)dx$ converge/diverge together.&lt;br&gt;&#xA;When the theorem is proven, it is shown that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\forall k \in \mathbb{N}, a_{k+1} \leq \int_{k}^{k+1} f(x)dx \leq a_{k}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which gives us an upper bound of $a_{n}$. Can I use this fact and conclude the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{n=1}^{\infty} a_{n} \overset{?}{\leq} \int_{1}^{\infty} f(x)dx + a_{1}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I'm trying to do is check if $\sum_{n=1}^{\infty} \frac{a}{a^2+n^2} \lt \frac{\pi}{4}+\frac{1}{2}$ when $0\lt a\lt 1$. Using the above gives:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_{1}^{\infty} \frac{a}{a^2+x^2} dx = \frac{1}{a} \int_{1}^{\infty} \frac{1}{1+(\frac{x}{a})^2} dx=\int_{1/a}^{\infty} \frac{1}{1+t^2} dt=$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$=\lim_{t \to \infty} \arctan t - \arctan 1/a \lt \frac{\pi}{2} - 1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{n=1}^{\infty} \frac{a}{a^2+n^2} \lt \frac{\pi}{2} - 1 + \frac{a}{a^2+1} \lt \frac{\pi}{2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(For some reason the end result is not what I wanted since $\frac{\pi}{2} \gt \frac{\pi}{4} + 1/2$ but I may have made an error along the way).&lt;/p&gt;&#xA;" OwnerUserId="3237" LastActivityDate="2011-01-05T20:26:53.087" Title="Bounding a series from above using the integral test" Tags="&lt;calculus&gt;&lt;sequences-and-series&gt;&lt;integral&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="16479" PostTypeId="1" AcceptedAnswerId="16491" CreationDate="2011-01-05T20:04:37.983" Score="3" ViewCount="140" Body="&lt;p&gt;Let $R$ be a PID,$M$ an $R$-module and $M$ is the direct sum of $M_1, \dots, M_k$ where $M_i \leq M$ for $1 \leq i \leq k$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let $N$ be a submodule of $M$ .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it true that $N$ is the direct sum of $N_1,\dots,N_t$ where $N_i \lt N$ and $N_{k-t+1} \le M_i$?&lt;/p&gt;&#xA;" OwnerUserId="5422" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2011-01-05T21:03:57.113" LastActivityDate="2011-01-05T21:14:45.997" Title="Direct sum of modules" Tags="&lt;modules&gt;" AnswerCount="4" CommentCount="2" />
  <row Id="16480" PostTypeId="2" ParentId="16419" CreationDate="2011-01-05T20:06:57.687" Score="4" ViewCount="" Body="&lt;p&gt;I believe it is linear.  Define P(n) as the number of good-perfect matchings on n points.  We have to match $u_1$ to something, n-1 choices, and it might as well be $v_2$.  Now we are not allowed to connect $u_2$ to $v_1$, so let C(n) be the number of ways of completing a good-perfect matching on n points with one set of mismatched indices that cannot be connected.  Similarly let D(n) be the number of ways of completing a good-perfect matching with one set of mismatched indices that are allowed to be matched.  After we connect $u_2$ to $v_3$ we are allowed to connect $u_3$ to $v_1$, so we have a case of D.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The recurrences are $$P(n)=(n-1)C(n-1)$$  $$C(n)=(n-1)D(n-1)$$  $$D(n)=P(n-1)+(n-1)D(n-1)$$ started with $D(1)=1, C(1)=0, P(1)=0$ which leads to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\begin{array}{rrrr}n&amp;amp;D&amp;amp;C&amp;amp;P\\1&amp;amp;1&amp;amp;0&amp;amp;0\\2&amp;amp;1&amp;amp;1&amp;amp;0\\3&amp;amp;2&amp;amp;2&amp;amp;2\\4&amp;amp;8&amp;amp;6&amp;amp;6\\5&amp;amp;38&amp;amp;32&amp;amp;24\\6&amp;amp;214&amp;amp;190&amp;amp;160\\7&amp;amp;1444&amp;amp;1284&amp;amp;1140\\8&amp;amp;11248&amp;amp;10108&amp;amp;8988\\9&amp;amp;98972&amp;amp;89984&amp;amp;80864\\10&amp;amp;971612&amp;amp;890748&amp;amp;809856\end{array}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is &lt;a href=&quot;http://oeis.org/A038205&quot; rel=&quot;nofollow&quot;&gt;A038205&lt;/a&gt; in OEIS, which does not list a closed form solution.  Note: I corrected an error in the expression for P(n) and the numbers in the table.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2011-01-06T00:34:48.283" LastActivityDate="2011-01-06T00:34:48.283" CommentCount="1" />
  <row Id="16481" PostTypeId="2" ParentId="16465" CreationDate="2011-01-05T20:09:34.753" Score="4" ViewCount="" Body="&lt;p&gt;There are probably more interesting and diverse answers, but one immediate application to Galois theory/algebraic number theory is that knowing the solution of the general cubic equation $x^3+ax^2+bx+c=0$ (where the cubic is irreducible over the rationals) allows us to compute the Galois group of a cubic.  In particular, the discriminant $D=18abc+a^2b^2−4b^3−4a^3c−27c^2$, which appears under the square root that you mentioned, completely determines the Galois group.  In general, the splitting field for the cubic is $K=\mathbb{Q}(r, \sqrt{D})$, where $r$ is any one root of the cubic.  If $\sqrt{D}$ rational, then the Galois group is isomorphic to $A_3$.  Otherwise, the Galois group is isomorphic to $S_3$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: As Qiaochu pointed out, the solution of the cubic is better viewed as an application of Galois theory than the other way around. In general, the particulars of solving polynomial equations (for example, what the solutions look like, i.e. whether we can get them by adjoining roots, etc.) is a motivating problem for a branch of algebra/number theory, and Galois theory provides information with regard to these things. There are a few reasons why the problem of understanding solutions of polynomial equations is interesting.  One of them is that this is one of the simplest interesting questions that we can ask in number theory.  Another is that polynomials tend to come up in a lot of places, so knowing about their solutions is useful to everyone, not just number theorists.  I know I haven't answered your question about applications of the solution of the cubic to number theory (no significant applications come to mind, but maybe there are some), but there is an interesting story here that is motivated by wanting to know about solving polynomial equations.  In particular, you might be interested to read about the insolvability of the general quintic by adjoining radicals and how this fits into the history of Galois theory.&lt;/p&gt;&#xA;" OwnerUserId="5265" LastEditorUserId="5265" LastEditorDisplayName="" LastEditDate="2011-01-11T22:53:19.990" LastActivityDate="2011-01-11T22:53:19.990" CommentCount="5" />
  <row Id="16482" PostTypeId="2" ParentId="329" CreationDate="2011-01-05T20:25:02.063" Score="1" ViewCount="" Body="&lt;p&gt;Elementary Number Theory - by David M. Burton if you want it somewhere halfway between fast and slow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://books.google.com/books?id=HMDHQwAACAAJ&amp;amp;dq=number+theory+burton&amp;amp;hl=en&amp;amp;ei=dtIkTZT3KMKCOuur9JAD&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=1&amp;amp;ved=0CCYQ6AEwAA&quot; rel=&quot;nofollow&quot;&gt;link&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5189" LastActivityDate="2011-01-05T20:25:02.063" CommunityOwnedDate="2011-01-05T20:25:02.063" />
  <row Id="16483" PostTypeId="2" ParentId="16478" CreationDate="2011-01-05T20:26:53.087" Score="3" ViewCount="" Body="&lt;p&gt;Given $\forall k \in \mathbb{N}, a_{k+1} \leq \int_{k}^{k+1} f(x)dx \leq a_{k}$, &#xA;$$\sum_{k=1}^{\infty}a_{k+1} \leq \sum_{k=1}^{\infty}\int_{k}^{k+1} f(x)dx \leq \sum_{k=1}^{\infty}a_{k}$$&#xA;so&#xA;$$\sum_{k=2}^{\infty}a_{k} \leq \int_{1}^{\infty} f(x)dx \leq \sum_{k=1}^{\infty}a_{k}$$&#xA;and, using the left part,&#xA;$$\sum_{k=1}^{\infty}a_{k}=a_1+\sum_{k=2}^{\infty}a_{k} \leq a_1+\int_{1}^{\infty} f(x)dx.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, to $\int_{1}^{\infty} \frac{a}{a^2+x^2} dx=\frac{\pi}{2}-\arctan\frac{1}{a}$.  As in Jack Schmidt's comment, $\arctan\frac{1}{a}&amp;gt;\frac{\pi}{4}$ when $0&amp;lt;a&amp;lt;1$, so $\frac{\pi}{2}-\arctan\frac{1}{a}&amp;lt;\frac{\pi}{4}$.  So,&#xA;$$\sum_{k=1}^{\infty}a_{k}\leq a_1+\int_{1}^{\infty} f(x)dx&amp;lt;\frac{a}{a^2+1}+\frac{\pi}{4}&amp;lt;\frac{1}{2}+\frac{\pi}{4}.$$&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2011-01-05T20:26:53.087" />
  <row Id="16484" PostTypeId="2" ParentId="16476" CreationDate="2011-01-05T20:43:48.760" Score="4" ViewCount="" Body="&lt;p&gt;Consider the longest cycle $C$ in $G$ and suppose its length is less than $2k$. Use Menger's Theorem and the pigeonhole principle to come up with a contradiction. &lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2011-01-05T20:43:48.760" />
  <row Id="16485" PostTypeId="2" ParentId="16479" CreationDate="2011-01-05T20:49:47.120" Score="-2" ViewCount="" Body="&lt;p&gt;Please, post the complete question!&lt;/p&gt;&#xA;" OwnerUserId="2764" LastActivityDate="2011-01-05T20:49:47.120" CommentCount="2" />
  <row Id="16486" PostTypeId="2" ParentId="16442" CreationDate="2011-01-05T21:10:14.500" Score="3" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\rm\ gcd(a,b,c) = gcd(10^7-b-c,b,c) = gcd(10^7,b,c)\:, $ so a triple will be primitive unless $\rm\:b,c\:$ are both divisible by $2$ or $5$. The longest leg $\rm\:c\:$ must be less than half the perimeter $\rm\:P\:$ and at least $\rm\:P/3\:.\: $ Therefore, for each $\rm\:c\:$ in this range, count the values of $\rm\ b = c,\: c-1,\ldots, \lceil (P-c)/2\rceil\ $ such that $\rm\:b,c\:$ have neither $2$ nor $5$ as a common divisor. It depends only on  $\rm\:b,c\:$ modulo $\ldots$&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2011-01-05T21:10:14.500" CommentCount="3" />
  <row Id="16487" PostTypeId="2" ParentId="16477" CreationDate="2011-01-05T21:13:31.737" Score="3" ViewCount="" Body="&lt;p&gt;A hint for the first part would be to recall that (a) the rationals are countable, so it would take a countable number of &quot;steps&quot; to &quot;cover&quot; them (since the irrationals are dense, you'll need to drop to 0 &quot;in between&quot; each rational number), and that (b) one could increase the number of steps (but decrease the size of each step) for each function in your sequence.&lt;/p&gt;&#xA;" OwnerUserId="5429" LastActivityDate="2011-01-05T21:13:31.737" />
  <row Id="16488" PostTypeId="2" ParentId="16479" CreationDate="2011-01-05T21:13:39.867" Score="5" ViewCount="" Body="&lt;p&gt;Take $R=\mathbb{Z}$, $M=\mathbb{Z} \oplus \mathbb{Z}= M_1 \oplus M_2$ and $N=\mathbb{Z}\cdot (1,1)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$N\cap M_1 =N\cap M_2 = 0$ so it can't be a direct sum of submodules of $M_1, M_2$&lt;/p&gt;&#xA;" OwnerUserId="2125" LastActivityDate="2011-01-05T21:13:39.867" CommentCount="0" />
  <row Id="16489" PostTypeId="2" ParentId="16479" CreationDate="2011-01-05T21:14:00.223" Score="5" ViewCount="" Body="&lt;p&gt;If I understand the question correctly then&#xA;$$&#xD;&#xA;\mathbb C\cdot(1,1)\subset \mathbb C\oplus\mathbb C&#xD;&#xA;$$&#xA;is a counterexample (here if have chosen $R=\mathbb C$).&lt;/p&gt;&#xA;" OwnerUserId="367" LastActivityDate="2011-01-05T21:14:00.223" />
  <row Id="16490" PostTypeId="1" AcceptedAnswerId="16493" CreationDate="2011-01-05T21:14:42.560" Score="1" ViewCount="45" Body="&lt;p&gt;Consider the linear space $c^{(3)}$ of all sequences $x = (x_n)_{n=1}^{\infty}$ such that $\{x_{3k+q} \}_{k=0}^{\infty}$ converges for $q = 0,1,2$. Find the dimension and a basis for $c^{(3)}/c_0$. Note that $c_0$ is the linear space of sequences that converge to $0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the dimension is $1$ using the reasoning in the answer to this &lt;a href=&quot;http://math.stackexchange.com/questions/16296/codimension-and-sequences&quot;&gt;question&lt;/a&gt;. We know that any integer can be represented as $3k, 3k+1$ or $3k+2$. So $c^{(3)}$ is equivalent to $c$ (e.g. all convergent sequences)? Is this correct?    &lt;/p&gt;&#xA;" OwnerUserId="2548" LastEditorUserId="2548" LastEditorDisplayName="" LastEditDate="2011-01-05T21:20:27.780" LastActivityDate="2011-01-05T21:24:18.047" Title="Dimension and sequences" Tags="&lt;linear-algebra&gt;" AnswerCount="2" />
  <row Id="16491" PostTypeId="2" ParentId="16479" CreationDate="2011-01-05T21:14:45.997" Score="3" ViewCount="" Body="&lt;p&gt;For a general module M over a PID R, its submodules need not even be isomorphic to direct sums of submodules of its proper direct summands.  For instance, the Z-module Q&lt;sup&gt;10&lt;/sup&gt; is a direct sum of proper submodules, but it has directly indecomposable submodules of rank 10 (so they cannot be written as any non-trivial direct sum).  Since the torsion-free rank of a submodule can only decrease, this means M = Q⊕Q&lt;sup&gt;9&lt;/sup&gt; is a counterexample.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we try to do what the question asks, and get a direct sum decomposition from submodules of the direct summands, then things go wrong even for M=Z/2Z ⊕ Z/2Z and the submodule N={(x,x):x in Z/2Z }.  N is directly indecomposable, and is not a submodule of the specified direct summands of M (though it is itself a summand in a different decomposition).&lt;/p&gt;&#xA;" OwnerUserId="583" LastActivityDate="2011-01-05T21:14:45.997" CommentCount="5" />
  <row Id="16492" PostTypeId="2" ParentId="16490" CreationDate="2011-01-05T21:22:29.570" Score="1" ViewCount="" Body="&lt;p&gt;Aren't the series $\{x_{3k} \}_{k=0}^{\infty}$, $\{x_{3k+1} \}_{k=0}^{\infty}$, and $\{x_{3k+2} \}_{k=0}^{\infty}$ completely unrelated to each other?  They can each have a different limit.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-01-05T21:22:29.570" />
  <row Id="16493" PostTypeId="2" ParentId="16490" CreationDate="2011-01-05T21:24:18.047" Score="4" ViewCount="" Body="&lt;p&gt;Notice that $c^{(3)}\cong c\oplus c \oplus c$ where $c$ is the space of convergent sequences, while $c_0 \cong c_0 \oplus c_0 \oplus c_0$ (both isomorphisms are separating the sequence to its three sub sequences modulo 3).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, in the represetation above of $c^{(3)}$ and $c_0$ we get that each copy of $c_0$ is inside a copy of $c$ so $c^{(3)}/c_0 \cong (c/c_0) \oplus (c/c_0) \oplus (c/c_0)$&lt;/p&gt;&#xA;" OwnerUserId="2125" LastActivityDate="2011-01-05T21:24:18.047" />
  <row Id="16494" PostTypeId="2" ParentId="16477" CreationDate="2011-01-05T21:51:59.463" Score="2" ViewCount="" Body="&lt;p&gt;For part ii). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If a sequence of Riemann integrable functions $f_n$ converges uniformly to $f$, then what can you say about the Riemann integrability of $f$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems like this question is trying to point out a drawback of Riemann integration which, if I remember correctly, motivated the discovery of Lebesgue integration.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2011-01-05T21:51:59.463" />
  <row Id="16495" PostTypeId="2" ParentId="16476" CreationDate="2011-01-05T21:54:39.150" Score="2" ViewCount="" Body="&lt;p&gt;Seems like another application of &lt;a href=&quot;http://mathworld.wolfram.com/DiracsTheorem.html&quot; rel=&quot;nofollow&quot;&gt;Dirac's Theorem&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2011-01-05T21:54:39.150" />
  <row Id="16496" PostTypeId="2" ParentId="16368" CreationDate="2011-01-05T21:59:25.140" Score="3" ViewCount="" Body="&lt;p&gt;Here is one reference, to the &lt;a href=&quot;http://www.cgal.org/Manual/3.1/doc_html/cgal_manual/Segment_Voronoi_diagram_2/Chapter_main.html&quot; rel=&quot;nofollow&quot;&gt;CGAL manual section&lt;/a&gt; explaining its computation&#xA;of the Voronoi diagram of segments, e.g.:&#xA;&lt;br /&gt;&lt;img src=&quot;http://i.imgur.com/iMOGv.jpg&quot; alt=&quot;SegVorDiag&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="237" LastActivityDate="2011-01-05T21:59:25.140" />
  <row Id="16497" PostTypeId="2" ParentId="15719" CreationDate="2011-01-05T22:23:23.450" Score="7" ViewCount="" Body="&lt;p&gt;Moron's and J.M.'s solutions are nice.&#xA;Hopefully this solution is simpler.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Without loss of generality we may assume that $1\gt x\gt 0$.&#xA;Put $x:=\sqrt{y}$, $1\gt y\gt 0$. Then we obtain&#xA;$$&#xD;&#xA;\int{\frac{1+x^2}{(1-x^2)\sqrt{1+x^4}}}\mathrm dx=\int{\frac{1+y}{2(1-y)\sqrt{1+y^2}\sqrt{y}}}\mathrm dy.&#xD;&#xA;$$&#xA;Introduce the new variable &#xA;$$&#xD;&#xA;t:=\frac{1+y}{1-y},\qquad 1\lt t \lt \infty.&#xD;&#xA;$$&#xA;Then we have&#xA;$$&#xD;&#xA;y=\frac{-1+t}{1+t},&#xD;&#xA;$$&#xA;$$&#xD;&#xA;\mathrm dy=\frac{2}{(1+t)^2}\,\mathrm dt.&#xD;&#xA;$$&#xA;Substituting back we obtain&#xA;$$&#xD;&#xA;\int{\frac{1+y}{2(1-y)\sqrt{1+y^2}\sqrt{y}}}\mathrm dy=\int{\frac{t}{2\sqrt{1+\left(\dfrac{-1+t}{1+t} \right)^2}\sqrt{\dfrac{-1+t}{1+t}}}\frac{2}{(1+t)^2}\,\mathrm dt}&#xD;&#xA;$$&#xA;$$&#xD;&#xA;=\frac{1}{\sqrt{2}}\int{\frac{t}{\sqrt{t^4-1}}}\mathrm dt&#xD;&#xA;$$&#xA;$$&#xD;&#xA;=\frac{1}{2\sqrt{2}}\ln(t^2+\sqrt{t^4-1})+C.&#xD;&#xA;$$&#xA;Putting back everything we obtain&#xA;$$&#xD;&#xA;\frac{1}{2\sqrt{2}}\ln\left(\frac{(1+x^2)^2+2\sqrt{2}x\sqrt{1+x^4}}{(1-x^2)^2}\right)+C.&#xD;&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="5418" LastEditorUserId="5418" LastEditorDisplayName="" LastEditDate="2011-01-05T22:29:22.683" LastActivityDate="2011-01-05T22:29:22.683" />
  <row Id="16498" PostTypeId="1" AcceptedAnswerId="16585" CreationDate="2011-01-05T22:32:31.937" Score="3" ViewCount="203" Body="&lt;p&gt;Is there any way to tell Mathematica 7 to use &quot;traditional&quot; axes rather than boxing a three-dimensional graph?  That is, rather than the default view produced by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;Plot3D[Exp[-x^2 - y^2], {x, -2, 2}, {y, -2, 2},Boxed-&amp;gt;False]&lt;/code&gt;,&#xA;&lt;img src=&quot;http://i.imgur.com/XrMLw.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like three &quot;axis arrows&quot; to emanate from the origin.&lt;/p&gt;&#xA;" OwnerUserId="3009" LastActivityDate="2011-01-06T16:56:22.783" Title="Traditional axes in 3d Mathematica plots?" Tags="&lt;mathematica&gt;&lt;computer-algebra-systems&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="16499" PostTypeId="2" ParentId="16498" CreationDate="2011-01-05T22:48:49.047" Score="1" ViewCount="" Body="&lt;p&gt;You need the AxesOrigin Option.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Plot3D[Exp[-x^2 - y^2], {x, -2, 2}, {y, -2, 2},Boxed-&gt;False, &#xA; AxesOrigin-&gt;{0,0,0}]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I misinterpreted your question in an earlier answer and I was suggesting using the &quot;AxesEdge&quot; Option which changes the sides of the bounding box on which the axes are displayed.  However, you might still find that useful:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Plot3D[Exp[-x^2 - y^2], {x, -2, 2}, {y, -2, 2}, Boxed -&gt; False, &#xA;AxesEdge -&gt; {{-1, -1}, {-1, -1}, {-1, -1}}]&lt;/p&gt;&#xA;" OwnerUserId="1004" LastEditorUserId="1004" LastEditorDisplayName="" LastEditDate="2011-01-05T22:55:19.543" LastActivityDate="2011-01-05T22:55:19.543" CommentCount="4" />
  <row Id="16500" PostTypeId="1" CreationDate="2011-01-05T23:06:42.940" Score="2" ViewCount="174" Body="&lt;p&gt;Could someone please help?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question reads:&#xA;For which real numbers $k$ is the zero state a stable equilibrium of the dynamic system &#xA;$x_{t+1} = Ax_t$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$A = \begin{bmatrix} 0.1 &amp;amp;k \\ 0.3 &amp;amp; 0.3 \end{bmatrix}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, my thought is I need to find the eigenvalues.  In  order to do this I calculated the characteristic polynomial as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x^2 - 0.4x + 0.03 - 0.3k = 0$,  with $x$ representing eigenvalues.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the quadratic formula I found that the (real) eigenvalues are &#xA;$$x = \frac{2 \pm \sqrt{1+30k}}{10}$$&#xA;and for the zero state to be in stable equilibrium $\sqrt{1+30k}&amp;lt;8$.  Hence, $k&amp;lt;21/10$ (for stable equilibrium).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is how do I figure out the values for $k$ if the eigenvalues are complex?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do I solve the inequality $\sqrt{-1-30k} &amp;lt; 8$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="5430" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2011-01-06T00:08:38.400" LastActivityDate="2011-01-06T00:08:38.400" Title="Zero State, Stable Equilibrium, Dynamic System" Tags="&lt;linear-algebra&gt;&lt;homework&gt;&lt;dynamical-systems&gt;&lt;eigenvalues&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="16501" PostTypeId="1" AcceptedAnswerId="16503" CreationDate="2011-01-05T23:19:57.770" Score="5" ViewCount="203" Body="&lt;p&gt;One of the very first number theory concepts introduced to students -- even before primeness, divisibility, etc. -- is the idea that a natural number can either be &quot;even&quot; (that is, evenly divisible by 2) or &quot;odd&quot; (all other numbers).  For all intents and purposes, at that time, even and odd numbers were evenly distributed and of the same density in the natural numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There always seemed to be something inherently &quot;special&quot; about a number's evenness or oddness, besides the trivial one that members of one class could be divided equally into two groups and the other cannot.  One would memorize addition and multiplication tables of evenness and oddness (an odd number will remain odd when added to ANY even number).  You could not iterate through all of the even numbers through multiplication alone, while you could for the odds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've heard the chess board being described as possessing evenness and oddness.  For example, dark squares and light squares can represent either even or odd.  A diagonal move can be an &quot;even&quot; move and a side-to-side move is an &quot;odd&quot; one, and moves are represented as additions to a square.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this way, a knight's move is an &quot;odd&quot; move (odd+odd+even), and when added to an odd square will yield an even square; when added to an even square will yield an odd square (odd + odd = even, odd + even = odd)&#xA;A bishop's move can be considered always even, so once a bishop is on an odd square, it can only ever move onto other odd squares.  Likewise for bishops on even squares.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there any more generalizations of this concept to math?  Is it meaningful to talk of even or odd matrices, or even or odd vectors or vector spaces?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've heard the concept applied to functions (even or odd functions), but I don't know if they are related to this by anything other than their name.&lt;/p&gt;&#xA;" OwnerUserId="51" LastEditorUserId="51" LastEditorDisplayName="" LastEditDate="2011-01-06T22:36:44.493" LastActivityDate="2011-01-06T22:36:44.493" Title="Generalizations of the number theory concepts of &quot;even&quot; and &quot;odd&quot;?" Tags="&lt;number-theory&gt;&lt;parity&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="1" />
  <row Id="16502" PostTypeId="1" AcceptedAnswerId="16505" CreationDate="2011-01-05T23:40:58.147" Score="4" ViewCount="251" Body="&lt;p&gt;Given $5x/30x^2$ I was wondering which is the correct equivalent form.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to BEDMAS this expression is equivalent to&#xA;$5*\cfrac{x}{30}*x^2$&#xA;but, intuitively, I believe that it could also look like:&#xA;$\cfrac{5x}{30x^2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I asked this question on MathOverflow (which was &quot;Off-topic&quot; and closed) and was told it was ambiguous. I was wondering what the convention was or if such a convention exists. According to Wikipedia the &lt;a href=&quot;http://en.wikipedia.org/wiki/Order_of_operations#Mnemonics&quot; rel=&quot;nofollow&quot;&gt;order of operations can be different based on the mnemonic used&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="5433" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2011-01-05T23:59:59.980" LastActivityDate="2011-01-06T23:36:12.887" Title="Do values attached to integers have implicit parentheses?" Tags="&lt;algebra-precalculus&gt;" AnswerCount="6" CommentCount="1" FavoriteCount="1" />
  <row Id="16503" PostTypeId="2" ParentId="16501" CreationDate="2011-01-05T23:42:04.607" Score="10" ViewCount="" Body="&lt;p&gt;Yes.  The generalization is provided by &lt;a href=&quot;http://en.wikipedia.org/wiki/Modular_arithmetic&quot; rel=&quot;nofollow&quot;&gt;modular arithmetic&lt;/a&gt;.  The properties you are observing all come from the fact that taking the remainder modulo $n$ respects addition and multiplication, and this generalizes to any $n$.  More generally in abstract algebra we study &lt;a href=&quot;http://en.wikipedia.org/wiki/Ring_(mathematics)&quot; rel=&quot;nofollow&quot;&gt;rings&lt;/a&gt; and their &lt;a href=&quot;http://en.wikipedia.org/wiki/Ideal_(ring_theory)&quot; rel=&quot;nofollow&quot;&gt;ideals&lt;/a&gt; for the same reasons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The notion of evenness and oddness of functions is closely related, but it is somewhat hard to explain exactly why.  The key point is that there is a certain &lt;a href=&quot;http://en.wikipedia.org/wiki/Group_(mathematics)&quot; rel=&quot;nofollow&quot;&gt;group&lt;/a&gt;, the cyclic group $C_2$ of order $2$, which is behind both concepts.  For now, note that the product of two even functions is even, the product of an even and odd function is odd, and the product of two odd functions is odd, so even and odd functions under multiplication behave exactly the same way as even and odd numbers under addition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are also huge generalizations depending on exactly what you're looking at, so it's hard to give a complete list here.  You mentioned chessboards; there is a more general construction here, but it is somewhat hard to explain and there are no good elementary references that I know of.  Once you learn some modular arithmetic, here is the modular arithmetic explanation of the chessboard idea: you can assign integer coordinates $(x, y)$ to each square (for example the coordinate of the lower left corner), and then you partition them into black or white squares depending on whether $x + y$ is even or odd; that is, depending on the value of $x + y \bmod 2$.  Then given two points $(a, b)$ and $(c, d)$ you can consider the difference $c + d - a - b \bmod 2$, and constraints on this difference translate to constraints on the movement of certain pieces.  This idea can be used, for example, to prove that certain chessboards (with pieces cut out of them) cannot be tiled with $1 \times 2$ or $2 \times 1$ tiles because these tiles must cover both a white square and a black square.  Of course there are generalizations with $2$ replaced by a larger modulus and larger tiles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for matrices and vectors, let's just say that there are a lot of things this &lt;em&gt;could&lt;/em&gt; mean, and none of them are straightforward generalizations of the above concept.&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2011-01-05T23:54:21.737" LastActivityDate="2011-01-05T23:54:21.737" />
  <row Id="16504" PostTypeId="2" ParentId="16501" CreationDate="2011-01-05T23:49:29.533" Score="6" ViewCount="" Body="&lt;p&gt;One example that immediately comes to mind is permutations. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a concept of the parity of a permutation, which corresponds to the parity of the number of swaps needed to get it back to the original position. This same concept is sometimes talked in terms of &quot;sign&quot; of a permutation, which is either 1 or -1, depending on whether the permutation is even or odd.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Permutations are related to matrices, as they show up in the definition of the determinant, and in fact, the sign of the permutation is used unlike the definition of the permanent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some configurations of the famous fifteen puzzle are shown to be unsolvable by considering the parity of the permutations involved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Look at this page: &lt;a href=&quot;http://en.wikipedia.org/wiki/Parity_of_a_permutation&quot; rel=&quot;nofollow&quot;&gt;Parity of Permutation&lt;/a&gt; which also talks about generalizations of this concept here: &lt;a href=&quot;http://en.wikipedia.org/wiki/Parity_of_a_permutation#Generalizations&quot; rel=&quot;nofollow&quot;&gt;Generalizations to Coxeter Groups&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2011-01-05T23:49:29.533" />
  <row Id="16505" PostTypeId="2" ParentId="16502" CreationDate="2011-01-05T23:56:58.423" Score="8" ViewCount="" Body="&lt;p&gt;The programming languages I have used give equal priority to multiply and divide, and do them from left to right.  This would support the $5*\cfrac{x}{30}*x^2$ assertion.  As Wikipedia says, the acronym BEDMAS may be misleading as it implies that division precedes multiplication and addition precedes subtraction, while each pair is of equal priority.  However, I suspect most people writing $5x/30x^2$ do mean $\cfrac{5x}{30x^2}$.  I would say there is a tendency in people's minds to make the division slash lower priority than multiplication.  But this is an excellent reason to put in parentheses when you are writing, and to check carefully what the author means when reading. &lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2011-01-06T19:10:46.670" LastActivityDate="2011-01-06T19:10:46.670" CommentCount="4" />
  <row Id="16506" PostTypeId="2" ParentId="16502" CreationDate="2011-01-06T00:03:07.793" Score="7" ViewCount="" Body="&lt;p&gt;It's just sloppy notation.  You should never, ever write an expression like it, and you should complain to whoever gave it to you.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2011-01-06T00:03:07.793" CommentCount="6" />
  <row Id="16507" PostTypeId="2" ParentId="16500" CreationDate="2011-01-06T00:05:20.553" Score="3" ViewCount="" Body="&lt;p&gt;In general, the zero state of a dynamical system of the form $x_{t+1} = Ax_t$ is a stable equilibrium when all of the eigenvalues of $A$ are inside the unit circle in the complex plane.  So if you have an eigenvalue of the form $a+bi$, you get stability when $a^2 + b^2 &amp;lt; 1$.  Since this is a homework question, I'll let you work out the rest of the problem from here.  (This isn't going to be equivalent to solving $\sqrt{-1-30k} &amp;lt; 8$, though.)&lt;/p&gt;&#xA;" OwnerUserId="2370" LastActivityDate="2011-01-06T00:05:20.553" CommentCount="12" />
  <row Id="16508" PostTypeId="2" ParentId="16399" CreationDate="2011-01-06T00:07:37.760" Score="3" ViewCount="" Body="&lt;p&gt;As I commented above, implementation of the OP's algorithm is very easy, noting that a uniform$[a,b]$ variable can be written as $a+(b-a)U$, with $U \sim {\rm uniform}[0,1]$. We now show that, for $0&amp;lt;d \leq 1/3$, the exact expression is quite complicated (and probably the case $d&amp;gt;1/3$ is more complicated).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $0&amp;lt;d \leq 1/3$, one should be able to show that the problem is equivalent to the following one (this is confirmed, as noted in the last paragraph). Suppose that $X_i$ are i.i.d. uniform$[1-d,1+d]$ variables, and let $S_k = \sum\nolimits_{i = 1}^k {X_i }$. Further, let $\tau$ denote the first index $k$ such that $S_k &amp;gt; n - (1-d)$. Find ${\rm E}(\tau)$. (The assumption $d \leq 1/3$ comes from $2(1-d) \geq 1+d$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As I described in some previous post, ${\rm E}(\tau)$ can be calculated as follows. Let $F^{(k)}$ denote the distribution function of $S_k$, that is, $F^{(k)}(x) = {\rm P}(S_k \leq x)$. Set $x = n-(1-d)$. Then, ${\rm E}(\tau) = 1 + \sum\nolimits_{k \ge 1} {F^{(k)} (x)}$ (note that the sum is finite). Now, the $X_i$ can be written as $(1-d)+2dU_i$, where $U_i$ are i.i.d. uniform$[0,1]$. However, the distribution function of the sum of i.i.d. uniform variables is complicated, and hence approximating ${\rm E}(\tau)$ using computer simulations is a good idea. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I implemented the OP's algorithm (as given in the question). Numerical results confirm that, for $d \leq 1/3$, the expectation in the question is indeed given by $1 + \sum\nolimits_{k \ge 1} {F^{(k)} (x)}$ (I approximated $F^{(k)} (x)$ using computer simulations). For example, for $n=3$, $d=0.1$, OP's algorithm gave $\approx 2.87488$, my formula gave $\approx 2.87474$; for $n=6$, $d=0.2$: $\approx 5.77619$ vs. $5.77694$; for $n=10$, $d=0.3$: $9.81563$ vs. $9.81554$. (Of course, we can increase accuracy by using a larger number of simulations.)&lt;/p&gt;&#xA;" OwnerUserId="2810" LastEditorUserId="2810" LastEditorDisplayName="" LastEditDate="2011-01-06T13:34:59.117" LastActivityDate="2011-01-06T13:34:59.117" />
  <row Id="16509" PostTypeId="1" AcceptedAnswerId="16531" CreationDate="2011-01-06T00:10:22.910" Score="0" ViewCount="112" Body="&lt;p&gt;Hi everyone,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I calculated the cubic regression for a set of numbers following the instructions on this site:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://epsstore.ti.com/OA_HTML/csksxvm.jsp?nSetId=71235&quot; rel=&quot;nofollow&quot;&gt;http://epsstore.ti.com/OA_HTML/csksxvm.jsp?nSetId=71235&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am wondering how do I evaluate using the equation provided by the cubic regression without having to write down the equation on paper, and then typing in in the calculator.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ted&lt;/p&gt;&#xA;" OwnerUserId="5434" LastActivityDate="2011-01-06T04:38:05.673" Title="calculate x using regression equation" Tags="&lt;calculator&gt;" AnswerCount="3" CommentCount="0" />
  <row Id="16510" PostTypeId="1" CreationDate="2011-01-06T00:19:29.487" Score="0" ViewCount="111" Body="&lt;p&gt;Recently I have found a need to learn more about p-adic group representations over a p-adic vector space. Generally, this motivates a study of representations $\left( V, \rho \right)$ &#xA;for some group $G$ where $V$ is a vector space over $\mathbb{Q}_p$. Since I'm only familiar&#xA;with the theory for representations acting on a complex vector space, I was hoping for references where I could find more information?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This may seem like a question suited more for google, however, I dont seem to have enough knowledge to prompt google in the correct direction.&lt;/p&gt;&#xA;" OwnerUserId="4859" LastActivityDate="2011-01-06T04:46:37.300" Title="Group representations over p-adic vector spaces" Tags="&lt;representation-theory&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="16511" PostTypeId="1" AcceptedAnswerId="16517" CreationDate="2011-01-06T01:01:28.323" Score="3" ViewCount="81" Body="&lt;p&gt;The affine line $\mathbb{A}^1$ (over $Spec R$) is, as far as I understand, $Spec R[x]$. My guess is that $0\in \mathbb{A}^1$ is given by the (opposite of the) composite homomorphism $R[x] \to R \to F$ where the first map sends a polynomial to its constant term and the second is the canonical map to the field of fractions $F$ of $R$. What ring homomorphism $R[x] \to F$ gives us $1 \in \mathbb{A}^1$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ok, closed points of $\mathbb{A}^1$ (over a field k) correspond to ideals $(x-a) \subset k[x]$ or equivalently the quotient maps $k[x] \to k[x]/(x-a)$. So then 0 corresponds to $k[x] \to k$ (consistent with my question as originally stated) and 1 corresponds (I guess) to $k[x] \to k[x]/(x-1)$. This should extend easily to rings with unit. Is this correct? How do we extend this to rings, bearing in mind I may have it wrong that one passes to fields of fractions.&lt;/p&gt;&#xA;" OwnerUserId="3835" LastEditorUserId="3835" LastEditorDisplayName="" LastEditDate="2011-01-06T01:22:01.377" LastActivityDate="2011-01-06T01:53:29.240" Title="How do we describe the point 1 on the affine line?" Tags="&lt;algebraic-geometry&gt;" AnswerCount="2" />
  <row Id="16512" PostTypeId="1" CreationDate="2011-01-06T01:16:56.063" Score="4" ViewCount="67" Body="&lt;p&gt;Hello!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Definition:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A space $X$ is a Urysohn space iff whenever $x \neq y$ in $X$ there are nhoods of $U$ of $x$ and $V$ of $y$ such that $\overline{U} \cap \overline{V} = \emptyset$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to show that every regular, T_1 space is Urysohn.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My attempt:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $x,y \in X$ be distinct points. Now consider the open set $X \setminus \{y\}$. By regularity we can find an open set $U$ such that $x \in U \subseteq \overline{U} \subseteq X \setminus \{y\}$. Now similarly by considering the open set $X \setminus \{x\}$ and using regularity again we can find an open set $V$ such that $y \in V \subseteq \overline{V} \subseteq X \setminus \{x\}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now from where I don't think we can conclude $U$ and $V$ have disjoint closures. Can we? I'm stuck here, can you please help?&lt;/p&gt;&#xA;" OwnerUserId="5435" LastActivityDate="2011-01-06T01:28:43.090" Title="Every regular , T_1 space is a Urysohn space" Tags="&lt;topology&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="16513" PostTypeId="2" ParentId="16512" CreationDate="2011-01-06T01:28:43.090" Score="3" ViewCount="" Body="&lt;p&gt;Suppose $X$ is $T_1$ and regular.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take $x,y\in X$ two different points. Consider, as you did, $X\setminus\{y\}$ as an open neighbourhood of $x$, there exists $x\in U\subseteq\bar{U}\subseteq X\setminus\{y\}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider now, instead of your choice, $X\setminus\bar{U}$. It is an open neighbourhood of $y$ and from regularity you have $y\in V\subseteq\bar{V}\subseteq X\setminus\bar{U}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\bar{U}\cap\bar{V}=\emptyset$ as needed.&lt;/p&gt;&#xA;" OwnerUserId="622" LastActivityDate="2011-01-06T01:28:43.090" CommentCount="1" />
  <row Id="16514" PostTypeId="1" AcceptedAnswerId="16528" CreationDate="2011-01-06T01:30:30.663" Score="5" ViewCount="245" Body="&lt;p&gt;I am studying for the GRE Mathematics subject exam [1]. I am looking for tips regarding how to more effectively study for it. Does anyone know of any good study materials or have any tips in general?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would really appreciate it!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[1] &lt;a href=&quot;http://www.ets.org/gre/subject/about/content/mathematics&quot; rel=&quot;nofollow&quot;&gt;http://www.ets.org/gre/subject/about/content/mathematics&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3791" LastActivityDate="2011-01-06T07:11:33.160" Title="Math GRE Subject Exam" Tags="&lt;graduate&gt;&lt;studies&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="1" />
  <row Id="16515" PostTypeId="2" ParentId="16514" CreationDate="2011-01-06T01:37:55.427" Score="3" ViewCount="" Body="&lt;p&gt;I wondered the same thing some months ago and the best link I found was &lt;a href=&quot;http://www.physicsgre.com/viewtopic.php?f=13&amp;amp;t=1078&quot; rel=&quot;nofollow&quot;&gt;this one&lt;/a&gt;. Hope it helps.&lt;/p&gt;&#xA;" OwnerUserId="1865" LastActivityDate="2011-01-06T01:37:55.427" />
  <row Id="16516" PostTypeId="2" ParentId="16511" CreationDate="2011-01-06T01:49:32.263" Score="5" ViewCount="" Body="&lt;p&gt;If $r\in R$, there is an $R$-point in $\mathbb A^1$ given by the map $R[x]\to R$ given by evaluation of polynomials at $r$.&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2011-01-06T01:49:32.263" CommentCount="2" />
  <row Id="16517" PostTypeId="2" ParentId="16511" CreationDate="2011-01-06T01:53:29.240" Score="4" ViewCount="" Body="&lt;p&gt;If $R$ is a ring (not necessarily a field) then $0$ is not a point of Spec $\mathbb A^1$,&#xA;it is a section.  Namely, there is a natural map Spec $\mathbb A^1 \to$ Spec $R$, corresponding to the map $R \to R[x]$, and for any $r \in R$ (whether it be $0$, $1$, or some other element) the map $R[x] \to R$ given by mapping $x$ to $r$&#xA;gives a section Spec $R \to \mathbb A^1$ which is a section to the natural map.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Added: As Mariano writes in his answer, such sections are the so-called $R$-valued &#xA;points of $\mathbb A^1$.&lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2011-01-06T01:53:29.240" CommentCount="1" />
  <row Id="16518" PostTypeId="1" AcceptedAnswerId="16519" CreationDate="2011-01-06T02:05:13.417" Score="3" ViewCount="78" Body="&lt;p&gt;I have a question that is: Find a degree 3 polynomial with real coefficients whose leading coefficient is 5 that has -2, 1, and 4 as zeros?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I do not want the question answered for me what I want it explained what it is asking. I answered &lt;code&gt;5(x+2)(x-1)(x-4)&lt;/code&gt;, but I was told I needed to finish the problem. Not exactly sure what there is to finish.&lt;/p&gt;&#xA;" OwnerUserId="2135" LastActivityDate="2011-01-06T02:08:29.280" Title="What is this question asking for?" Tags="&lt;algebra-precalculus&gt;&lt;intuition&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="16519" PostTypeId="2" ParentId="16518" CreationDate="2011-01-06T02:07:24.257" Score="6" ViewCount="" Body="&lt;p&gt;I think it is asking for a polynomial in the form $5x^3 + ax^2 + bx + c$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you would multiply out your answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an aside, if that is what it is, it is a silly question. Your answer is fine, IMO.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2011-01-06T02:07:24.257" CommentCount="4" />
  <row Id="16520" PostTypeId="2" ParentId="16518" CreationDate="2011-01-06T02:08:29.280" Score="1" ViewCount="" Body="&lt;p&gt;After pointing out that this is the polynomial you are suggesting you need to show that it is indeed of the intended degree and have the wanted roots, and that the coefficients are real.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Simplify the given polynomial, it should be enough.&lt;/p&gt;&#xA;" OwnerUserId="622" LastActivityDate="2011-01-06T02:08:29.280" />
  <row Id="16521" PostTypeId="1" AcceptedAnswerId="16821" CreationDate="2011-01-06T02:26:38.873" Score="15" ViewCount="224" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Is there an easy way to tell how many roots f(x) has in Q[x]/(f) given the coefficients of the polynomial f in Q[x]?&lt;/strong&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;Is there an easy way to find the roots as rational expressions in x?&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The easiest example is a pure quadratic: X&lt;sup&gt;2&lt;/sup&gt; + 7 for instance.  If A is a root, then so is −A.  Good ole ±√−7.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the Galois group is abelian (like for any quadratic), then all of the roots can be expressed as polynomials in a given root.  However, I am not sure how to tell by looking at the polynomial if its Galois group is abelian, and even if it is, I am not sure how to find those rational expressions for the other roots.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It might help to see some non-Abelian (non-Galois) examples:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If A is a root of X&lt;sup&gt;6&lt;/sup&gt; + 2X&lt;sup&gt;4&lt;/sup&gt; − 8, then −A is also a root, but its other 4 roots cannot be expressed as rational functions of A (assuming I still understand Galois theory).&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Is there some easy way (not asking a CAS to calculate the Galois group) to see the other 4 roots of of X&lt;sup&gt;6&lt;/sup&gt; + 2X&lt;sup&gt;4&lt;/sup&gt; − 8 cannot be expressed as rational functions of A?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This one had the nice feature that it was a function of X&lt;sup&gt;2&lt;/sup&gt;, so it was easy to find two roots.  For X&lt;sup&gt;6&lt;/sup&gt; − 2X&lt;sup&gt;5&lt;/sup&gt; + 3X&lt;sup&gt;3&lt;/sup&gt; − 2X − 1, I still have not found its other root (even using a CAS).&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If A is a root of X&lt;sup&gt;6&lt;/sup&gt; − 2X&lt;sup&gt;5&lt;/sup&gt; + 3X&lt;sup&gt;3&lt;/sup&gt; − 2X − 1, then what is a rational expression in A for another root?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;This all first came up with the polynomial x&lt;sup&gt;4&lt;/sup&gt;−4x&lt;sup&gt;2&lt;/sup&gt;+2, where several distinct ad hoc arguments each sufficed, but I had no real understanding of how to even tell if my ad hoc arguments were worth trying on other polynomials.  If it helps, the roots are A, −A, A&lt;sup&gt;3&lt;/sup&gt;−3A, and 3A−A&lt;sup&gt;3&lt;/sup&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The context is hand calculations and reasonable degrees (say ≤ 10), though I am not opposed to having a polynomial evaluation oracle that computes f(g(x)) mod f(x) in 1 second (so &quot;try this finite and not too big list of possible roots&quot; is ok).&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;If someone is interested, I am curious what the normalizer of a point stabilizer in the Galois group actually means in terms of Galois theory.  The index of the point stabilizer in its normalizer is the number of roots of f in Q[x]/(f), but I'm not sure if it really means anything useful.&lt;/p&gt;&#xA;" OwnerUserId="583" LastActivityDate="2011-01-10T04:23:48.213" Title="Expressing a root of a polynomial as a rational function of another root" Tags="&lt;abstract-algebra&gt;&lt;polynomials&gt;&lt;galois-theory&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="3" />
  <row Id="16522" PostTypeId="1" AcceptedAnswerId="16524" CreationDate="2011-01-06T02:56:38.810" Score="0" ViewCount="121" Body="&lt;p&gt;Suppose we have the following lemma:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Lemma&lt;/strong&gt; If $E_0 \hookrightarrow E$, and $E_0$ is a closed subspace then $E/E_0$ is a normed space and for $[x] \in E/E_0$ its norm is given by $||[x]|| = \text{inf}_{y \in E_0} ||x-y||$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the intuition of this norm besides that it &quot;works&quot;? Does it just measure how close $x$ and $y$ are to being equivalent? Also does one usually try to introduce some norm on a space $E$ but finds out that it doesn't work? Thus one introduces it as a seminorm and then as a norm on $E/E_0$? In other words, is a seminorm on $E$ just a means to get a norm on $E/E_0$? Why can't one just introduce a norm on $E/E_0$ directly?    &lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2011-01-06T04:13:55.757" Title="Seminorms and norms" Tags="&lt;intuition&gt;&lt;functional-analysis&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="16523" PostTypeId="1" CreationDate="2011-01-06T03:16:09.893" Score="1" ViewCount="216" Body="&lt;p&gt;With positional notation, is there any &lt;em&gt;practical&lt;/em&gt; usage for &lt;a href=&quot;http://en.wikipedia.org/wiki/Negative_base&quot; rel=&quot;nofollow&quot;&gt;negative bases&lt;/a&gt;?  I recently learned that numbers can be represented in positional notation using a negative base, but I don't see any real practical application of this. So, is there any?&lt;/p&gt;&#xA;" OwnerUserId="4724" LastEditorUserId="10557" LastEditorDisplayName="" LastEditDate="2011-05-12T22:49:41.440" LastActivityDate="2011-05-12T22:49:41.440" Title="Usage of Negative bases" Tags="&lt;applications&gt;&lt;number-systems&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="16524" PostTypeId="2" ParentId="16522" CreationDate="2011-01-06T03:31:50.003" Score="3" ViewCount="" Body="&lt;p&gt;Your question seems to be superimposing several concepts/issues at once, and perhaps it will help to separate them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, if $E$ is a semi-normed space and $E_0$ is a subspace then we define&#xA;the semi-norm on $E/E_0$ according to the formula you wrote down.  If you look at what's written, the coset $[x]$ is fixed, while $y$ is varying, so there is no particular $y$ of which it makes sense to ask &quot;how close $x$ and $y$ are to being equivalent&quot;.  Rather, you are measuring the distance of the point $x$ from the subspace $E_0$.  Note that this depends only on $[x]$ (i.e. if we translate $x$ by an element of $E_0$, the distance to $E_0$ doesn't change).  This is the most natural semi-norm to place on the quotient $E/E_0$: the distance of the coset $[x]$ from the $0$ in $E/E_0$ is measured by considering the distance of the representative $x$ from the $0$ coset thought of as a subset of $E$, i.e. the distance of $x$ from $E_0$.  If $E_0$ is in fact closed in $E$ (with respect to the topology defined by the semi-norm on $E$), then one sees that the semi-norm on $E/E_0$ is actually a norm (and conversely).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your second question (which got tangled up with your first, but is really a separate question, I think) is why do we sometimes consider semi-norms rather than norms.  There are some situations in which $E$ naturally comes with a seminorm which is not a norm.  We can then consider the subspace $E_0$ consisting of all elements of semi-norm equal to $0$, which will be a closed subspace (in the topology defined by the semi-norm).  (In fact, $E_0$ is precisely the closure of the point $0$.)   Now we can apply the above discussion to this particular choice of $E_0$, and so we get a norm on the quotient&#xA;$E/E_0$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This procedure is applied for example to construct the $L^p$ spaces in measure theory.  If we define $E$ to the be the space of all measurable real-valued functions on some measure space $X$ for which $\int |f|^p d\mu$ is finite (for some $p \geq 1$), then &#xA;$(\int |f|^p d\mu)^{1/p}$ is a semi-norm on $E$.  One finds that $E_0$ is precisely the subspace of $E$ consisting of functions which vanish a.e., and&#xA;the $L^p(X)$ is defined to be the quotient $E/E_0$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the reason for not introducing $E/E_0$ directly in this case is that one first has to talk about functions, and then the elements of $E/E_0$ are certain equivalence classes of functions which it's not really possible to introduce directly, without first going via the functions themselves and then introducing the equivalence relation.  And when we write down the &quot;$L^p$-norm&quot; on functions, which is what we naturally can write down, it turns out not be a norm, but just a semi-norm.  Passing to $E/E_0$ then gives us an actual normed space, rather than just a semi-normed one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for why we pass from $E$ to $E/E_0$ at all, this is mainly convenience. From the point of view of analysis, there is essentially no difference to working in $E$ and $E/E_0$; if any two functions in $E$ lie in the same $E_0$-coset (i.e. they coincide a.e.) then you can't really tell the difference between them analytically.  So it's just convenient to identify them, and to pass to their common coset in $E/E_0$.  (A general priciple is that it's simpler to work in Hausdorff spaces, like $E/E_0$, than in non-Hausdorff ones, like $E$ itself, and so when nothing is lost by passing from $E$ to $E/E_0$, it's easiest just to do so.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, are seminorms purely an intermediate step on the way to norms, as in the example of $L^p$-spaces?  The answer is &lt;em&gt;no&lt;/em&gt;.  The reason is that in&#xA;some situations one has whole families of semi-norms on a space (which are not necessarily themselves norms), which define an interesting or important topology.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;E.g. consider the space $E$ of all continuous functions on $\mathbb R^n$.  This doesn't have a natural norm: the obvious norm to consider on continuous functions is the sup norm, but since $\mathbb R^n$ is not compact, a continuous function need not be bounded, and so the sup norm is not defined on arbitrary continuous functions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, if $K$ is a compact subset of $\mathbb R^n$, then we can define a semi-norm&#xA;$|| \quad ||_K$ on $E$ as $|| f ||_K =$ the sup of $f$ on $K$.  Note that this is not a norm: there can be continuous functions on $\mathbb R^n$ which are not identically zero, but vanish at every point of the particular compact set $K$.&#xA;But we can use this whole family of semi-norms to define a topology: we define a base of open sets by defining, for each $K$ and each $\epsilon &amp;gt;0$, a n.h.&#xA;$B_{K,\epsilon}(f)$ of the function $f$ to be the set of $g$ such that&#xA;$|| f - g||_K &amp;lt; \epsilon.$ This does form a basis of neighbourhoods of $f$,&#xA;because given $K,\epsilon$ and $K&amp;#39;,\epsilon&amp;#39;,$ if we choose $K&amp;#39;&amp;#39; = K \cup K&amp;#39;$&#xA;and $\epsilon&amp;#39;&amp;#39;$ to be the min of $\epsilon$ and $\epsilon&amp;#39;$,&#xA;then $$B_{K&amp;#39;&amp;#39;,\epsilon&amp;#39;&amp;#39;}(f) \subset B_{K,\epsilon}(f) \cap B_{K&amp;#39;,\epsilon&amp;#39;}(f).$$ &#xA;This is a Hausdorff topology on $E$ (basically because a non-zero function &#xA;must be non-zero when restricted to &lt;em&gt;some&lt;/em&gt; compact set, so not &lt;em&gt;all&lt;/em&gt; these semi-norms can vanish on a non-zero function), making $E$ a so-called Frechet space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So here the semi-norms are appearing in their own right, not just as an intermediate step to getting a normed space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the passage from semi-norms to norms that you ask about is not really about getting rid of semi-norms and replacing them by norms at all: it is rather about getting rid of non-Hausdorff spaces and replacing them by Hausdorff ones.  It just happens that if your topology is defined by a &lt;em&gt;single&lt;/em&gt; semi-norm that is not a norm, then it &lt;em&gt;will be&lt;/em&gt; non-Hausdorff, and when you make it Hausdorff (by taking the quotient by the closure of $0$) the resulting space has its topology defined by a norm.    But in general, there are lots of Hausdorff spaces whose topology is defined not by a norm, but rather by a family of semi-norms.  (In applications, these spaces appear most often in the foundational aspects of distribution theory.)&lt;/p&gt;&#xA;" OwnerUserId="221" LastEditorUserId="221" LastEditorDisplayName="" LastEditDate="2011-01-06T04:13:55.757" LastActivityDate="2011-01-06T04:13:55.757" CommentCount="2" />
  <row Id="16525" PostTypeId="2" ParentId="16509" CreationDate="2011-01-06T03:37:15.700" Score="-1" ViewCount="" Body="&lt;p&gt;You &lt;i&gt;would&lt;/i&gt; expect our friends at Texas Instruments to include a way to actually store the result of the regression, wouldn't you?---but examining my TI-83 and accompanying manual, I can't see how to do it, either.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proposed solution: use a different tool. Wolfram|Alpha (&lt;i&gt;wolframalpha.com&lt;/i&gt;) has a &quot;&lt;a href=&quot;http://www.wolframalpha.com/examples/RegressionAnalysis.html&quot; rel=&quot;nofollow&quot;&gt;cubic fit&lt;/a&gt;&quot; command, and if you mouse over the lower-left corner of the result, there's a &quot;copyable plaintext&quot; link.&lt;/p&gt;&#xA;" OwnerUserId="2773" LastActivityDate="2011-01-06T03:37:15.700" />
  <row Id="16526" PostTypeId="2" ParentId="5" CreationDate="2011-01-06T04:06:46.167" Score="5" ViewCount="" Body="&lt;p&gt;Another method is to use continued fractions (which was used in one of the first proofs irrationality of $\displaystyle \pi$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instead of $\displaystyle \sqrt{2}$, we will consider $\displaystyle 1 + \sqrt{2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now $\displaystyle v = 1 + \sqrt{2}$ satisfies&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$v^2 - 2v - 1 = 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i.e&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$v = 2 + \frac{1}{v}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This leads us to the following continued fraction representation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$1 + \sqrt{2} = 2 + \cfrac{1}{2 + \cfrac{1}{2 + \dots}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any number with an infinite simple continued fraction is irrational and any number with a finite simple continued fraction is rational and has at most two such simple continued fraction representations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus it follows that $\displaystyle 1 + \sqrt{2}$ is irrational, and so $\displaystyle \sqrt{2}$ is irrational.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Exercise&lt;/strong&gt;: Show that the Golden Ratio is irrational.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More information here: &lt;a href=&quot;http://en.wikipedia.org/wiki/Continued_fraction&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Continued_fraction&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2011-01-06T18:48:04.747" LastActivityDate="2011-01-06T18:48:04.747" />
  <row Id="16527" PostTypeId="2" ParentId="16366" CreationDate="2011-01-06T04:10:11.453" Score="1" ViewCount="" Body="&lt;p&gt;First you need to be clear about what you mean by face, edge, vertex, etc.  In my view if we have an $n$ dimensional cube, then $n-1$ dimensional cubes form its faces, and $n-2$ dimensional cubes form its edges.  Vertices as you define above are 0-D points.  We do not have names for $n-3$, $n-4$ dimensional boundaries so I will not attempt to name them here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are looking for an intuitive way to build up the relations you can define inductive relationships using the method of extrusion.  You can start with $n=1$ if you like.  In this case we have one 1-D cube (i.e. a segment) with two 0-D faces (i.e. points). So&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$C_1 = 1$$&#xA;$$F_1 = 2$$&#xA;$$E_1 = 0$$&#xA;$$V_1 = 2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $C_1$ is the number of cubes in 1-D space, $F_1$ is the number of faces, $E_1$ is the number of edges (not defined here), and $V_1$ is the number of vertices.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We create a 2-D cube (i.e. a square) by extruding this 1-D cube in a direction orthogonal direction.  By doing this two things happen:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Every object is copied maintaining its current dimensions.  We had one segment, now we have two, we had two vertices, we how have four.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Every object extruded into the new direction creates one object of one more dimension.  The 1-D segment is extruded over a second dimension to form a square (i.e. 2-D object).  The vertices (0-D) are extruded to become lines.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the number of cubes, faces, edges, vertices becomes &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$C_2 = 1$$&#xA;$$F_2 = 2C_1 + E_1 = 4$$&#xA;$$V_2 = 2V_1 = 4$$&#xA;$$E_2 = V_2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Repeat the process for a 3-D cube.  Again we have two sources for new objects, copying and extrusion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$C_3 = 1$$&#xA;$$F_3 = 2C_2 + E_2 = 6$$&#xA;$$E_3 = 2F_2 + V_2 = 12$$&#xA;$$V_3 = 2V_2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Continue the process to get to higher dimensions.&lt;/p&gt;&#xA;" OwnerUserId="4726" LastActivityDate="2011-01-06T04:10:11.453" />
  <row Id="16528" PostTypeId="2" ParentId="16514" CreationDate="2011-01-06T04:20:27.097" Score="3" ViewCount="" Body="&lt;p&gt;Studying past exams helps. I remember a few of them being somewhat difficult to track down. Here are the ones I have: 8767, 9367, 9768, 0809.  If you have trouble finding them, let me know and I can send them to you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also used the Princeton Review book, but it is somewhat outdated in the sense that there has been content on recent exams that isn't covered in the book.  Also, the older exams seemed easier than some of the more recent ones.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: The three older ones are available here: math.ucsb.edu/mathclub/GRE.  The newer one is available here: &lt;a href=&quot;http://www.ets.org/gre/subject/about/content/mathematics&quot; rel=&quot;nofollow&quot;&gt;http://www.ets.org/gre/subject/about/content/mathematics&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5265" LastEditorUserId="5265" LastEditorDisplayName="" LastEditDate="2011-01-06T04:56:17.110" LastActivityDate="2011-01-06T04:56:17.110" CommentCount="7" />
  <row Id="16529" PostTypeId="2" ParentId="16502" CreationDate="2011-01-06T04:23:43.117" Score="10" ViewCount="" Body="&lt;p&gt;I am going to play Devil's Advocate on this one, because I've given it a little bit of thought after a tangentially related &lt;a href=&quot;http://mathoverflow.net/questions/18593/what-are-the-worst-notations-in-your-opinion/18627#18627&quot; rel=&quot;nofollow&quot;&gt;question on MathOverflow about bad notation&lt;/a&gt;. There is a good lexical argument to be made for the &quot;intuitive&quot; reading of 5&lt;em&gt;x&lt;/em&gt;/30&lt;em&gt;x&lt;/em&gt;&lt;sup&gt;2&lt;/sup&gt; as (5&lt;em&gt;x&lt;/em&gt;)/(30&lt;em&gt;x&lt;/em&gt;&lt;sup&gt;2&lt;/sup&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&lt;strong&gt;Disclaimer.&lt;/em&gt;&lt;/strong&gt; The following argument is not intended to communicate why expressions such as &lt;em&gt;a&lt;/em&gt;/&lt;em&gt;bc&lt;/em&gt; are unambiguous. By definition, if there is no strong socially agreed upon convention, it is therefore ambiguous. However, I &lt;em&gt;will&lt;/em&gt; argue for the hypothesis that almost anyone raised from an early age to read a European language, who quickly reads such an expression without reflecting too much, will probably understand &lt;em&gt;a&lt;/em&gt;/&lt;em&gt;bc&lt;/em&gt;&amp;nbsp;=&amp;nbsp;&lt;em&gt;a&lt;/em&gt;/(&lt;em&gt;bc&lt;/em&gt;). Therefore, there exists a reasonable opportunity to &lt;strong&gt;establish&lt;/strong&gt; such a convention.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of the responders to the question of bad notation &lt;a href=&quot;http://math.stackexchange.com/questions/16502/do-values-attached-to-integers-have-implicit-parentheses&quot;&gt;complained about the usage of juxtaposition for multiplication&lt;/a&gt;: he thought it introduced too much ambiguitiy. (For instance, does &lt;em&gt;f&lt;/em&gt;(&lt;em&gt;x&lt;/em&gt;&amp;thinsp;+&amp;thinsp;&lt;em&gt;y&lt;/em&gt;) refer to the product of a scalar &lt;em&gt;f&lt;/em&gt; with &lt;em&gt;x&lt;/em&gt;&amp;thinsp;+&amp;thinsp;&lt;em&gt;y&lt;/em&gt;, or the value of a function &lt;em&gt;f&lt;/em&gt; evaluated at &lt;em&gt;x&lt;/em&gt;&amp;thinsp;+&amp;thinsp;&lt;em&gt;y&lt;/em&gt;, or something else?) In the comments to that answer, I speculated that the reason for this convention is that it reduces the problem of parsing a mathematical expression such as &amp;nbsp;&lt;em&gt;ax&lt;/em&gt;&lt;sup&gt;2&lt;/sup&gt;&amp;thinsp;+&amp;thinsp;&lt;em&gt;bx&lt;/em&gt;&amp;thinsp;+&amp;thinsp;&lt;em&gt;c&lt;/em&gt;&amp;nbsp; to the previously-solved problem (for Europeans, anyway) of parsing a written sentence in a European language, which breaks sentences into words. Juxtaposed variables form nice little cohesive &quot;words&quot;. This hypothesis can even be extended to account for exponents and subscripts &amp;mdash; more notation in which we don't use operator 'symbols' &amp;mdash; to play the role of intra-word apostrophes or other diacritical marks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Under this hypothesis, I can make a strong case for prefering the reading &lt;em&gt;a&lt;/em&gt;/&lt;em&gt;bc&lt;/em&gt; as &lt;em&gt;a&lt;/em&gt;/(&lt;em&gt;bc&lt;/em&gt;). Unlike multiplication or exponentiation, we introduce an actual symbol &amp;mdash; a prominent punctuation mark, if you will, as intrusive as the + symbol &amp;mdash; into the expression. This divides the expression into two &quot;words&quot;, &lt;em&gt;a&lt;/em&gt; and &lt;em&gt;bc&lt;/em&gt;, upon which we perform the operation of division; thus leading to the reading &lt;em&gt;a&lt;/em&gt;/(&lt;em&gt;bc&lt;/em&gt;). This is especially likely in the case where &lt;em&gt;a&lt;/em&gt; is effectively a place-holder, as in 1/2&lt;em&gt;x&lt;/em&gt;, where for &quot;fluent speakers of mathematics&quot; the 1 is only there as a placeholder to allow the concept of &lt;em&gt;multiplicative inverse&lt;/em&gt; to be expressed. Add to this the fact that it makes no sense at all to write &lt;em&gt;a&lt;/em&gt;/&lt;em&gt;bc&lt;/em&gt; if you wish to convey the meaning &lt;em&gt;ac&lt;/em&gt;/&lt;em&gt;b&lt;/em&gt;, and one can almost safely say that anyone who writes &lt;em&gt;a&lt;/em&gt;/&lt;em&gt;bc&lt;/em&gt; means &lt;em&gt;a&lt;/em&gt;/(&lt;em&gt;bc&lt;/em&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The above is meant as a &lt;em&gt;descriptivist&lt;/em&gt; (as opposed to &lt;em&gt;prescriptivist&lt;/em&gt;) argument for how people read and write mathematics. If this is going to happen anyway, and if (as appears to be the case) it doesn't entail any heavy modification of our existing conventions, why not accept it in order to allow more flexibility in our notational repertoire? We can cut down on parentheses, and get rid of this phony ambiguity between something which can be better expressed anyway and an expression which at present we can only confidently convey with parentheses or with extra vertical space.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Now, of course,&lt;/strong&gt; there is no solid &lt;em&gt;existing&lt;/em&gt; agreement about whether to read &lt;em&gt;a&lt;/em&gt;/&lt;em&gt;bc&lt;/em&gt; as &lt;em&gt;a&lt;/em&gt;/(&lt;em&gt;bc&lt;/em&gt;) or as (&lt;em&gt;a&lt;/em&gt;/&lt;em&gt;b&lt;/em&gt;)&lt;em&gt;c&lt;/em&gt;, except to accept that latter because of BEDMAS or what-have-you. So if your job is to evaluate your student's ability to communicate, then you should conclude that they aren't doing such a good job. But if you are trying to evaluate &lt;em&gt;what they mean&lt;/em&gt; &amp;mdash; and if you believe that they learned to read a European language from a tender age &amp;mdash; then chances that when they write 5&lt;em&gt;x&lt;/em&gt;/30&lt;em&gt;x&lt;/em&gt;&lt;sup&gt;2&lt;/sup&gt;, they mean (5&lt;em&gt;x&lt;/em&gt;)/(30&lt;em&gt;x&lt;/em&gt;&lt;sup&gt;2&lt;/sup&gt;).&lt;/p&gt;&#xA;" OwnerUserId="439" LastEditorUserId="439" LastEditorDisplayName="" LastEditDate="2011-01-06T07:41:01.730" LastActivityDate="2011-01-06T07:41:01.730" />
  <row Id="16530" PostTypeId="2" ParentId="16509" CreationDate="2011-01-06T04:27:27.233" Score="0" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://tibasicdev.wikidot.com/cubicreg&quot; rel=&quot;nofollow&quot;&gt;This&lt;/a&gt; should answer your troubles on how to store the regression equation to an equation variable on the TI 83/84/+/SE calculators. Alternatively, the coefficients a-d are accessible from the [VARS] &gt; Statistics… &gt; EQ menu.&lt;/p&gt;&#xA;" OwnerUserId="5439" LastActivityDate="2011-01-06T04:27:27.233" />
  <row Id="16531" PostTypeId="2" ParentId="16509" CreationDate="2011-01-06T04:38:05.673" Score="2" ViewCount="" Body="&lt;p&gt;On a TI-83/84 series calculator (also an Nspire with the TI-84 keyboard module in place), once you've performed the regression, go to your &quot;y=&quot; screen, to a blank equation, press the &quot;vars&quot; key, go down to &quot;5: Statistics...&quot;, arrow right twice to &quot;EQ&quot;, and press enter to select &quot;RegEQ&quot;—this will insert the full regression equation into the y-variable line you were on in the &quot;y=&quot; screen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/aQsR5.gif&quot; alt=&quot;animated screenshot of the steps&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, you can use the graph and trace to evaluate or use the table to evaluate (or directly evaluate on the home screen using $Y_1(value)$, but that's harder to use).&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2011-01-06T04:38:05.673" />
  <row Id="16532" PostTypeId="2" ParentId="16510" CreationDate="2011-01-06T04:46:37.300" Score="6" ViewCount="" Body="&lt;p&gt;There are a few basic situations to consider that I know about.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first case is when the group $G$ is compact.  Then if we are given a&#xA;continuous rep. $\rho: G \to GL_n(\mathbb Q_p)$, the image is compact,&#xA;so lies in a maximal compact subgroup of $GL_n(\mathbb Q_p)$.  Any such&#xA;is conjugate to $GL_n(\mathbb Z_p)$, so after changing basis, we may&#xA;assume that $\rho: G \to GL_n(\mathbb Z_p)$.  (Another way to phrase this&#xA;is that $G$ must leave some $\mathbb Z_p$-lattice invariant, and there are&#xA;lots of ways to prove this, without having to mention the concept of &quot;maximal&#xA;compact&quot;, if that makes you at all nervous.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At this level of generality, there is not that much more to say.  But there&#xA;are various subcases of interest in which one can say more.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;E.g. if $G$ is a Galois group, then there is an enormous literature about $p$-adic Galois representations of Galois groups.  If this is the case you are interested in, you might want to ask another more specific question about it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are two other basic subcases, but before introducing them, I have&#xA;to mention&#xA;a basic fact about $GL_n(\mathbb Z_p)$, namely that there is a quotient map&#xA;$GL_n(\mathbb Z_p) \to GL_n(\mathbb F_p)$, whose kernel, which I'll denote&#xA;by $K(1)$,  is easily seen to be&#xA;a pro-p-group (i.e. a projective limit of $p$-groups).  (The reason for the&#xA;$(1)$ is that we could also consider $K(n)$, the kernel of the &quot;reduction modulo&#xA;$p^n$'' map, for each $n\geq 1$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here now are two other interesting subcases of the compact case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first is when the group $G$ is profinite, and is virtually pro-prime-to-$p$, i.e. contains an open subgroup that is the projective limit of finite groups of order prime to $p$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Examples of such $G$ are $GL_n(\mathbb Z_{\ell})$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case, let $H$ be the open subgroup that is pro-prime-to-$p$.&#xA;Since $K(1)$ is pro-$p$, we see that $\rho(H)$ and $K(1)$ must have trivial&#xA;intersection, and so $\rho(H)$ injects into $GL_n(\mathbb F_p)$.&lt;br&gt;&#xA;Thus $\rho(H)$ is finite, and hence $\rho(G)$ is finite too.  Thus&#xA;in this case, the continuous $\rho$s all factor through some finite&#xA;quotient of $G$, and we reduce to standard representation theory (i.e. rep'n theory of finite groups over a field of char. $0$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second, and more interesting, case is when $G$ is virtually pro-$p$, i.e. contains &#xA;an open subgroup $H$ which is pro-$p$.  (E.g. if $G$ is itself $GL_n(\mathbb Z_p)$, or a closed subgroup thereof.)  In this case the theory is more genuinely $p$-adic, i.e. it doesn't just reduce to the classical rep'n theory of finite&#xA;groups.  A good place to learn about some aspects of this is Lazard's opus &lt;em&gt;Groupes analytiques $p$-adiques&lt;/em&gt;, in Publ. Math. IHES vol. 26 (1965), where&#xA;among other things he studies the continous cohomology of such representations&#xA;(when the group $G$ is $p$-adic analytic, e.g. a matrix group), explains the relationship to Lie theory and Lie algebra cohomology, and proves various Poincare duality-type results for the cohomology.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;More recent discussions of Lazard's work and related ideas can be found in&#xA;some of the literature surrounding non-abelian Iwasawa theory, e.g. Venjakob's article in J. Eur. Math. Soc. vol. 4 (2002).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, let me mention that&#xA;if $G$ is not compact, but is just locally compact, e.g. $GL_n(\mathbb Q_{\ell})$, then there usually won't be many interesting finite-dimensional representations in which $G$ preserves a lattice, and so it is natural to consider $p$-adic Banach space representations in which $G$ preserves a lattice&#xA;instead. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $G$ contains a profinite open subgroup that is pro-prime-to-$p$,&#xA;then this theory is not so novel --- one can see for example Vigneras's article in Asterisque vol. 330 (2010).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, if $G$ contains a pro-$p$ open subgroup, then the theory is much more involved, and is the subject of a lot of recent work, especially by people thinking about $p$-adic Langlands.  You can see some of the papers of Schneider and Teitelbaum, and of Breuil and Colmez, as well as some of the papers on my web-page.  (I'm Emerton at Northwestern.)  &lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2011-01-06T04:46:37.300" />
  <row Id="16533" PostTypeId="1" CreationDate="2011-01-06T05:03:45.203" Score="0" ViewCount="48" Body="&lt;p&gt;As far as I know, integer set is not a convex set. If my problem has this constraint, how can we relax it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem basically is the optimal input design problem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Minimize  $t$ subject to&#xA;$$||f(x)−i(x)∗g(x)||_2 &amp;lt;= t.$$&#xA;$i(x)$ is integer valued,  $∗$ is the convolution operator, and $t$ and $i(x)$ are the variables in this problem.&lt;/p&gt;&#xA;" OwnerUserId="5347" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-06T05:18:15.153" LastActivityDate="2011-01-06T05:18:15.153" Title="Convex optimization with integer set" Tags="&lt;numerical-methods&gt;&lt;convex-optimization&gt;&lt;numerical-optimization&gt;" />
  <row Id="16534" PostTypeId="1" CreationDate="2011-01-06T05:23:42.900" Score="2" ViewCount="65" Body="&lt;p&gt;Hello!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is to prove Poincare Duality of the form&#xA;$(H^p_c (M))^* = H^{n-p} (M)$&#xA;using direct limits and inverse limits, where $H^p (M)$ denotes the $p$-th de Rham cohomology group of $M$. The outline of the proof has been given, which is to be followed. In particular, no homology theory of any sort can be used. (In fact, I don't know any homology theory; this is why I cannot find any reference that is helpful at all, since most texts (e.g. Hatcher, Greenberg &amp;amp; Harper) build on homology theories. The only text that does not mention homology at all is Madsen &amp;amp; Tornehave, but it bypasses my problem with some trick that is not compatible with the outline.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have figured out most of the question, except the following part:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $M$ be a general manifold, with $M = \bigcup_i U_i$, where $U_i$ is open and $U_i \subset \overline{U_i} \subset U_{i+1}$. If Poincare Duality holds for each $U_i$, show that it holds for $M$ using direct limit and inverse limit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For this my approach is to show that (1) $H^p_c (M) = \lim_{\rightarrow} H^p_c (U_i)$, and that (2) $H^p (M) = \lim_{\leftarrow} H^p (U_i)$, then since the dual of a direct limit is the inverse limit of the duals, the result would follow. (1) is easy to show, but I found (2) to be surprisingly hard.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is, I can see that there is a natural map from $H^p (M)$ to $\lim_{\leftarrow} H^p (U_i)$ induced by inclusion of $U_i$ into $M$, namely the map that restricts $p$-forms to each $U_i$. I believe it is also surjective. However, I can't prove injectivity: It is apparent that if $[\omega] \in H^p (M)$ restricts to $[0]$ for each $U_i$, then it doesn't &quot;see any holes anywhere&quot; on any $U_i$, and hence $[\omega]$ must be $[0]$ itself, but I can't formalize this argument.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I looked up some references, and I have to admit I understood very little. I remember one even saying that this is not true in general, and there is some &quot;derived functor&quot; $\lim^1$ (?) involved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could someone please give me a pointer on this? It doesn't have to be the whole solution; I would appreciate any hint. Thank you!&lt;/p&gt;&#xA;" OwnerUserId="5440" LastEditorUserId="5440" LastEditorDisplayName="" LastEditDate="2011-01-06T07:30:52.977" LastActivityDate="2011-01-06T07:30:52.977" Title="Poincare Duality: Cohom. gp. of $M = \cup U_i$ (increasing) is inverse limit of cohom. gp. of $U_i$?" Tags="&lt;homework&gt;&lt;algebraic-topology&gt;" CommentCount="3" />
  <row Id="16535" PostTypeId="1" CreationDate="2011-01-06T05:24:53.837" Score="4" ViewCount="126" Body="&lt;p&gt;If $R$ is a ring such that $x^5=x$ for all $x\in R$, is $R$ commutative?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the answer to the above question is yes, then what is the least positive integer $k \ge 6$, such that there exists a noncommutative ring $R$ with $x^k=x$ for all $x\in R$?&lt;/p&gt;&#xA;" OwnerUserId="3249" LastActivityDate="2011-01-06T06:06:42.083" Title="Is such ring commutative?" Tags="&lt;abstract-algebra&gt;&lt;ring-theory&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="16536" PostTypeId="2" ParentId="16535" CreationDate="2011-01-06T05:29:32.477" Score="7" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://mathoverflow.net/questions/29590/a-condition-that-implies-commutativity&quot; rel=&quot;nofollow&quot;&gt;http://mathoverflow.net/questions/29590/a-condition-that-implies-commutativity&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://mathoverflow.net/questions/32032/on-a-theorem-of-jacobson&quot; rel=&quot;nofollow&quot;&gt;http://mathoverflow.net/questions/32032/on-a-theorem-of-jacobson&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="223" LastEditorUserId="223" LastEditorDisplayName="" LastEditDate="2011-01-06T05:52:15.310" LastActivityDate="2011-01-06T05:52:15.310" />
  <row Id="16537" PostTypeId="1" CreationDate="2011-01-06T06:02:07.487" Score="0" ViewCount="205" Body="&lt;p&gt;An equation f(x)=0 has 2 roots in the interval between 0 and 8. What is the value or a description for which the Newton Raphson method will fail to converge?&lt;/p&gt;&#xA;" OwnerUserId="5441" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2011-02-05T10:58:24.900" LastActivityDate="2011-02-05T10:58:24.900" Title="At what point does the Newton Raphson method fail to converge?" Tags="&lt;numerical-methods&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="16538" PostTypeId="1" AcceptedAnswerId="16543" CreationDate="2011-01-06T06:02:34.227" Score="1" ViewCount="178" Body="&lt;p&gt;Oftentimes we see functions defined on open sets; I never gave this much thought, but now I'd like to get a better understanding.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Why do we define functions over open subsets of $\mathbb{R}^n$?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&quot;Why&quot; is not meant to be a deep philosophical question; what I mean is, for example,&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;When is it important to define a function on an open set (and when do we not care whether the set is open or not)?&lt;/li&gt;&#xA;&lt;li&gt;What's the motivation for doing so, i.e. what is it that we're going to use about an open set?&lt;/li&gt;&#xA;&lt;li&gt;Does using an open set (as opposed to, say, its closure) make anything more convenient or more manageable?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I realize this is quite a general question, and that I didn't give any context; however, I'm hoping that the principles behind defining functions on open sets are &quot;universal&quot; enough that the generality and lack of specific examples won't be too problematic.&lt;/p&gt;&#xA;" OwnerUserId="4904" LastActivityDate="2011-01-06T10:17:05.543" Title="Why do we define functions over $open$ subsets of $\mathbb{R}^n$?" Tags="&lt;analysis&gt;&lt;multivariable-calculus&gt;" AnswerCount="4" CommentCount="1" />
  <row Id="16540" PostTypeId="2" ParentId="16537" CreationDate="2011-01-06T06:08:07.480" Score="2" ViewCount="" Body="&lt;p&gt;See &lt;a href=&quot;http://en.wikipedia.org/wiki/Newton-Raphson#2._Failure_of_the_method_to_converge_to_the_root%3a&quot; rel=&quot;nofollow&quot;&gt;wikipedia&lt;/a&gt;. Since you're asking about a function with a multiple root, you probably want the answer in section 3.&lt;/p&gt;&#xA;" OwnerUserId="2549" LastActivityDate="2011-01-06T06:08:07.480" />
  <row Id="16541" PostTypeId="2" ParentId="16538" CreationDate="2011-01-06T06:20:13.097" Score="0" ViewCount="" Body="&lt;p&gt;See &lt;a href=&quot;http://mathoverflow.net/questions/19152/why-is-a-topology-made-up-of-open-sets&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2011-01-06T06:20:13.097" CommentCount="1" />
  <row Id="16542" PostTypeId="2" ParentId="16538" CreationDate="2011-01-06T06:22:12.107" Score="4" ViewCount="" Body="&lt;p&gt;When you want to speak about continuity or differentiability at a point x, then by definition you need the function f to be defined in a neighborhood of x, therefore x must be interior.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course you can define one sided (or the equivalent in higher dimension) continuity \ derivatives at a point, so sometimes it is enough if the points x have a &quot;half open&quot; set. For example, Rolle's and Lagrange's theorems need the function to be continuous on $I=[a,b]$ and differentiable on $(a,b)$, so $a$ and $b$ are in half open sets (which are open in the induced topology on I).&lt;/p&gt;&#xA;" OwnerUserId="2125" LastActivityDate="2011-01-06T06:22:12.107" />
  <row Id="16543" PostTypeId="2" ParentId="16538" CreationDate="2011-01-06T06:29:21.863" Score="6" ViewCount="" Body="&lt;p&gt;I'm gonna take a big bunch of random stabs at this question.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;A lot of things we do in calculus won't make a whole lot of sense at boundary points [differentiation and two-sided limits, for example.] and the notion of open set, at least in $R^{n}$, lets us work on a space where there are no boundary points.  We often need conditions in an open set and on the boundary of such a set to apply things like the MVT and Rolle's theorem (we often see let f be differentiable in $(a,b)$ and continuous in $[a,b]$, for example).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;For open subsets of euclidean space, a lot of nice properties transfer: something is connected if and only if it is path connected, a subset is open in the smaller space if and only if its open in $R^{n}$, etc.  It's often nice to use open subsets since they're going to be made up of basis elements of your topology, which are nicer to manipulate than, say, sets which are neither open nor closed like $[0,1)$ or sets which are closed, like [0,1], especially in a space as nice as euclidean space which already has some sweet properties (as a metric space!).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;For many functions, it may not even make sense to take a closure.  For example, $f(x) = \frac{1}{x}$ is nice on $(0,\infty)$, but what happens when we take the closure of that?  Also, if we're given a function on an open interval, can we extend it uniquely to its closure?  Can we extend it uniquely to its closure if we add that it must be continuous?  Can we extend it uniquely to its closure if we add that it must be differentiable?  This latter part may not answer your question, but it's kind of neat to think about.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="2959" LastActivityDate="2011-01-06T06:29:21.863" CommentCount="1" />
  <row Id="16544" PostTypeId="2" ParentId="5" CreationDate="2011-01-06T06:56:00.943" Score="5" ViewCount="" Body="&lt;p&gt;The continued fraction proof in Moron's answer can be recast into an elementary form that requires no knowledge of continued fractions. Below is a variant of such that John Conway (JHC) often mentions, followed by my (WGD) reinterpretation of it to highlight the key role played by the principality of (denominator) ideals in $\:\mathbb Z\:$ (which I call &lt;em&gt;unique fractionization&lt;/em&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;THEOREM (JHC)&lt;/strong&gt; $\ \ \rm r = \sqrt{n}\ \:$ is integral if rational,$\:$ for $\:\rm n\in\mathbb{N}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof&lt;/strong&gt; $\quad$  Put $\ \ \displaystyle\rm r = \frac{A}B ,\;$ least $\rm\; B&amp;gt;0\:.\;$ $\  \displaystyle\rm\sqrt{n}\; = \frac{n}{\sqrt{n}} \ \Rightarrow\ \frac{A}B = \frac{nB}A\:.\ $ Taking fractional parts yields $\rm\displaystyle\ \frac{b}B = \frac{a}A\ $ for $\rm\ 0 \le b &amp;lt; B\:.\ $ But $\rm\displaystyle\ B\nmid A\ \Rightarrow\:\ b\ne 0\ \:\Rightarrow\ \frac{A}B = \frac{a}b\ $ contra leastness of $\rm\:B\quad\quad$ &lt;strong&gt;QED&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Abstracting out the Euclidean descent at the heart of the above proof yields the following&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;THEOREM (WGD)&lt;/strong&gt; $\ \ \rm r = \sqrt{n}\ \:$ is integral if rational,$\:$ for $\:\rm n\in\mathbb{N}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof&lt;/strong&gt; $\quad$  Put $\ \ \displaystyle\rm r = \frac{A}B ,\;$ least $\rm\; B&amp;gt;0\:.\;$ $\  \displaystyle\rm\sqrt{n}\; = \frac{n}{\sqrt{n}} \ \Rightarrow\ \frac{A}B = \frac{nB}A\ \Rightarrow\ B\:|\:A\ $ by this key result:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Unique Fractionization&lt;/strong&gt; $\ $ The least denominator $\rm\:B\:$ of a fraction divides every denominator.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof&lt;/strong&gt; $\rm\displaystyle\ \ \frac{A}B = \frac{C}D\ \Rightarrow\ \frac{D}B = \frac{C}A \:.\ $ Taking fractional parts $\rm\displaystyle\ \frac{b}B = \frac{a}A\ $ where $\rm\ 0 \le b &amp;lt; B\:.\ $ But  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\displaystyle\ \:B\nmid D\ \Rightarrow\ b\ne 0\ \Rightarrow\ \frac{A}B = \frac{a}b\ \ $ contra leastness of $\rm\:B\:.\quad\quad $ &lt;strong&gt;QED&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus JHC's proof essentially &quot;inlines&quot; the above proof - which is better viewed as principality of (denominator) ideals in $\mathbb Z\:,$ cf. &lt;a href=&quot;http://math.stackexchange.com/questions/2883/show-that-3p2-q2-implies-3p-and-3q/2941#2941&quot;&gt;my post here&lt;/a&gt;. See also &lt;a href=&quot;http://groups.google.com/groups?selm=y8ziss6kia5.fsf@nestle.ai.mit.edu&quot; rel=&quot;nofollow&quot;&gt;this discussion&lt;/a&gt; between John Conway and I.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2011-02-28T05:11:06.633" LastActivityDate="2011-02-28T05:11:06.633" />
  <row Id="16545" PostTypeId="2" ParentId="16514" CreationDate="2011-01-06T07:11:33.160" Score="0" ViewCount="" Body="&lt;p&gt;I worked through the Princeton Review book as well as the REA book.  The latter was not great, although I learned a lot of random little things from it (e.g. generating functions, various binomial identities, methods of counting (with and without replacement blah blah blah) etc.).  Beware that there are a few glaring errors in the REA book.  The Princeton Review book was pretty good -- it had short reviews of the important concepts in each of the subject areas, and then mini-tests at the ends of those sections, and then cumulative tests too -- but I'm glad I worked from the two books because they covered each others' weaknesses.  I went through all the problems in both, re-did the ones I got wrong and/or wasn't absolutely sure about, then went through the ones that I was still not totally comfortable with, then did that whole thing over again 2 or 3 more times.  Quite a bit of work, but if you care a lot then I'd recommend it.  The crucial thing (as other people have said) is to be able to work very quickly and not get stumped.&lt;/p&gt;&#xA;" OwnerUserId="2159" LastActivityDate="2011-01-06T07:11:33.160" />
  <row Id="16546" PostTypeId="1" CreationDate="2011-01-06T07:15:46.487" Score="27" ViewCount="1056" Body="&lt;p&gt;As a computer science graduate who had only a basic course in abstract algebra, I want to study some abstract algebra in my free time. I've been looking through some books on the topic, and most seem to 'only' cover groups, rings and fields. Why is this the case? It seems to me you'd want to study simpler  structures like semigroups, too. Especially looking at Wikipedia, there seems to be a huge zoo of different kinds of semigroups.&lt;/p&gt;&#xA;" OwnerUserId="5443" LastActivityDate="2011-05-26T23:57:51.897" Title="Why do books titled &quot;Abstract Algebra&quot; mostly deal with groups/rings/fields?" Tags="&lt;abstract-algebra&gt;&lt;soft-question&gt;&lt;books&gt;" AnswerCount="6" CommentCount="1" FavoriteCount="4" />
  <row Id="16547" PostTypeId="1" AcceptedAnswerId="16549" CreationDate="2011-01-06T07:16:00.667" Score="1" ViewCount="281" Body="&lt;p&gt;This is a problem (Exercise 9.2) from Apostol's &lt;em&gt;Mathematical Analysis&lt;/em&gt; (second edition) which I couldn't solve. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\bullet$ Define two sequences $\{f_{n}\}$ and $\{g_{n}\}$ as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f_{n}(x) = x \Bigl(1 + \frac{1}{n}\Bigr)$ for $x \in \mathbb{R}$, $n \in \mathbb{N}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$g_{n}(x)= \begin{cases}&#xD;&#xA; \frac{1}{n} &amp;amp; x=0 \  \text{or}\  x \in \mathbb{R} - \mathbb{Q} \\&#xD;&#xA; b+\frac{1}{n} &amp;amp; x \in \mathbb{Q},\ \text{say}\  x=\frac{a}{b}, b&amp;gt;0&#xD;&#xA;\end{cases}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\mathbb{R}$ is the set of reals, $\mathbb{Q}$ is the set of rational and $\mathbb{R} - \mathbb{Q}$ is the set of irrationals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have to show that if $h_{n}(x)=f_{n}(x)g_{n}(x)$, then $h_{n}(x)$ does not converge uniformly on any bounded interval. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I show this?&lt;/p&gt;&#xA;" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2011-01-06T08:54:01.943" LastActivityDate="2011-02-14T19:58:59.653" Title="Exercise 9.2 from Apostol's Mathematical Analysis book. Uniform convergence of product." Tags="&lt;real-analysis&gt;&lt;analysis&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="2" />
  <row Id="16548" PostTypeId="2" ParentId="16523" CreationDate="2011-01-06T07:22:56.123" Score="3" ViewCount="" Body="&lt;p&gt;According to the history section of that wikipedia link yes: Negabinary was implemented in the early Polish computer BINEG, built 1957-59, based on ideas by Z. Pawlak and A. Lazarkiewicz from the Mathematical Institute in Warsaw.[2] Implementations since then have been rare.&lt;/p&gt;&#xA;" OwnerUserId="3936" LastActivityDate="2011-01-06T07:22:56.123" />
  <row Id="16549" PostTypeId="2" ParentId="16547" CreationDate="2011-01-06T07:35:49.927" Score="2" ViewCount="" Body="&lt;p&gt;What is the pointwise limit of $f_n(x)g_n(x)$ as $n\rightarrow\infty$? What is the pointwise difference between the limit and the value for a given $n$? Show that, for any $n$, this difference is unbounded on any bounded interval. This follows pretty much immediately from the fact that denominators of rational numbers are unbounded on any bounded interval.&lt;/p&gt;&#xA;" OwnerUserId="3212" LastActivityDate="2011-01-06T07:35:49.927" CommentCount="2" />
  <row Id="16550" PostTypeId="2" ParentId="16546" CreationDate="2011-01-06T07:44:36.880" Score="39" ViewCount="" Body="&lt;p&gt;There is not that much substantial to say about general semigroups at an introductory level, in comparison to groups, say, and what there is to say at this level (e.g. the theory of &lt;a href=&quot;http://en.wikipedia.org/wiki/Greens%27_relations&quot; rel=&quot;nofollow&quot;&gt;Green's relations&lt;/a&gt;) tends to rely on a prior understanding of groups in any case.  (One thing I have in mind here --- but I am speaking as a non-expert --- is that semi-groups are much closer to general universal algebra than groups, and universal algebra is not, I would say, appropriate as a topic for a first course in abstract algebra.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The theory of groups lends itself well to a first course, because the axioms are fairly simple but lead fairly quickly to non-trivial theorems, such as the Sylow theorems.  Groups (especially group actions) are also ubiquitous, and the kind of counting arguments which one can make are very useful to learn.  (It is the inability to make these sorts of counting arguments that causes so many problems when you investigate semi-groups, in comparison to the case of group.)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The theory of fields is particularly useful in number theory, and Galois theory also provides a nice tie-in with the theory of groups, which in fact served as Galois's motivation for introducing groups in the first place. The theory of fields (especially the part to do with finding roots of polynomials) is also the part of abstract algebra that is closest to high school algebra, so it is not surprising that there is a substantial focus on it.     &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The theory of rings normally appears both because it is a precursor to field theory (fields are particular kinds of rings, and the polynomial rings $F[x]$ also play an important role in the study of fields), and because it includes many basic examples from mathematics, such as matrix rings, the integers, quaternions, and so on.  Rings also play an important role in the study of group representations (via the appearance of group rings), which, even if they don't appear in a first course, are just over the horizon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Summary:&#xA;So overall, I think the answer is that groups, rings, and fields are the parts of algebra that are most closely connected to the basic core topics of mathematics, and are also closely integrated with one another.  (Groups not&#xA;immediately obviously so, but because of Galois theory and group rings, &#xA;for example.)   The theory of semigroups, by contrast, doesn't play much role in the rest of mathematics, and the theory that does exist is more complicated than the theory of groups (despite the axioms being simpler).&lt;/p&gt;&#xA;" OwnerUserId="221" LastEditorUserId="221" LastEditorDisplayName="" LastEditDate="2011-05-24T11:35:54.197" LastActivityDate="2011-05-24T11:35:54.197" CommentCount="4" />
  <row Id="16551" PostTypeId="1" AcceptedAnswerId="16552" CreationDate="2011-01-06T07:44:38.690" Score="5" ViewCount="173" Body="&lt;p&gt;I am reading a text related to SVM, and the mathematical language is giving me a little hard time.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Here training vectors x&lt;sub&gt;i&lt;/sub&gt;&#xA;  are mapped into a higher (maybe&#xA;  infinite) dimensional space by the&#xA;  function $\theta$. SVM finds a linear&#xA;  separating hyperplane with the maximal&#xA;  margin in this higher dimensional&#xA;  space.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I do not understand the term &quot;dimensional space&quot; in this case. Drawing on a paper is 2D. We are living in 3D space. In Mathematics, when we say &quot;higher dimensional space&quot;, what are we actually implying in Mathematics?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another term &quot;hyperplane&quot; is also giving me a bit of hard time to understand. Is it simply just a 2D plane? I try to search for its definition, and most of the time, I get a term that leads to many more terms (and more confusing) Frankly, mathematical language is difficult for me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could somebody simplify and relate &quot;dimension&quot; and &quot;hyperplane&quot; in the text above in an easier way to understand?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you very much.&lt;/p&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2011-05-20T22:05:19.230" Title="Could somebody elaborate &quot;dimensional space&quot; and &quot;hyperplane&quot;?" Tags="&lt;machine-learning&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="16552" PostTypeId="2" ParentId="16551" CreationDate="2011-01-06T07:49:55.213" Score="7" ViewCount="" Body="&lt;p&gt;The way I read the quote, &quot;higher (maybe infinite) dimensional space&quot; should be read as &quot;higher-dimensional (maybe infinite-dimensional) space&quot; meaning a space with more dimensions than the $x_i$ had originally (mapping them into a higher-dimensional space).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A hyperplane in $n$-dimensional space is an $(n-1)$-dimensional object that can be described by $\vec{n}\cdot\vec{x}=k$ where $\vec{n}$ is a constant vector orthogonal to the hyperplane, $\vec{x}$ is a variable vector from the origin to a point on the plane, and $k$ is some scalar constant.  A hyperplane in 2-space is a line; a hyperplane in 3-space is a plane.  (&lt;em&gt;edit&lt;/em&gt;: Matt E.'s comment that a hyperplane is a subspace of dimension one less than the whole space is a much nicer definition than mine.)&lt;/p&gt;&#xA;" OwnerUserId="72" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2011-05-17T23:36:08.007" LastActivityDate="2011-05-17T23:36:08.007" CommentCount="1" />
  <row Id="16553" PostTypeId="1" AcceptedAnswerId="16587" CreationDate="2011-01-06T08:28:07.843" Score="5" ViewCount="251" Body="&lt;p&gt;In Titchmarsh's book &quot;The theory of the Riemann Zeta function&quot; pg. 15&#xA;where the functional equation of the zeta function is being derived, &#xA;I couldn't understand this part:&#xA;$$\frac{s}{\pi} \sum_{n=1}^{\infty} \frac{(2n\pi)^s}{n} \int_{0}^{\infty} \frac{\sin y}{y^{s+1}} dy = \frac{s}{\pi} (2\pi)^s \{-\Gamma(-s)\}\sin\frac{1}{2}s\pi\zeta(1-s)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I could not digest Titchmarsh's reasoning.&#xA;Can anyone explain this please?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks,&lt;/p&gt;&#xA;" OwnerUserId="2320" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2011-01-06T18:46:00.553" LastActivityDate="2011-01-06T18:46:00.553" Title="Understanding an integral from page 15 of Titchmarsh's book &quot;The theory of the Riemann Zeta function&quot;." Tags="&lt;complex-analysis&gt;&lt;analytic-number-theory&gt;" AnswerCount="2" />
  <row Id="16554" PostTypeId="1" CreationDate="2011-01-06T10:12:06.933" Score="2" ViewCount="174" Body="&lt;p&gt;What exactly it is meant by &quot;&lt;strong&gt;average increase in percentage&lt;/strong&gt;&quot;? I stumble upon this term in while solving this problem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The average monthly income and expenditure of a person in the year $1995$ is $ \$ 14,000 $ and $ \$ 11,000 $ respectively and that of the year $2000$ is $ \$ 21,000 $ and $ \$ 17,600 $ respectively.Find the average percentage increase in expenditure of the person between $1995$ and $2000$.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;ADDED:&lt;/strong&gt; The solution given in my module goes like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let the cumulative increase in expenditure be $r\%$ then,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$17,600 = 11,000 \times (1+\frac{r}{100})^5 \Rightarrow 1.6 = (1+\frac{r}{100})^5 \Rightarrow r = 10 $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I don't really understand,also how can we possible solve $r$ (by using hand)?For this problem they have utilized the four options given but it's a bit tedious I suppose so I am more interested in any alternative procedure if exists.&lt;/p&gt;&#xA;" OwnerUserId="2109" LastEditorUserId="2109" LastEditorDisplayName="" LastEditDate="2011-01-06T14:33:30.960" LastActivityDate="2011-01-07T09:59:40.757" Title="what is &quot;average increase percentage&quot; and how to calculate it ?" Tags="&lt;algebra-precalculus&gt;&lt;arithmetic&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="16555" PostTypeId="2" ParentId="16538" CreationDate="2011-01-06T10:17:05.543" Score="0" ViewCount="" Body="&lt;p&gt;There's a lemma called &quot;Open Set Criterion for Continuity&quot; which says a map between metric spaces(euclidean space is a metric space) is continuous if and only if the inverse image of every open set is open.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So an open set is required for a function to be continuous.&lt;/p&gt;&#xA;" OwnerUserId="5312" LastActivityDate="2011-01-06T10:17:05.543" CommentCount="1" />
  <row Id="16556" PostTypeId="1" AcceptedAnswerId="16642" CreationDate="2011-01-06T10:53:06.260" Score="1" ViewCount="174" Body="&lt;p&gt;Let (e_n) be the standard unit vector basis for c_0,let x_1=e_1,x_n=e_n-e_(n-1) when n&gt;1. &#xA;Prove that the formal series Σx_n is not weakly subseries convergent.&lt;/p&gt;&#xA;" OwnerUserId="4706" LastActivityDate="2011-01-07T05:13:47.870" Title="A formal series in the Banach space" Tags="&lt;functional-analysis&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="16557" PostTypeId="2" ParentId="16547" CreationDate="2011-01-06T11:00:03.543" Score="0" ViewCount="" Body="&lt;p&gt;Let $I$ be a bounded interval. If $x=0$ or irrational, then $h_n (x) \to 0$. If $x$ is a rational of the form $x=a/b$, $b&amp;gt;0$, then &#xA;$$&#xD;&#xA;h_n (x) = \frac{a}{b}\bigg(1 + \frac{1}{n}\bigg)\bigg(b + \frac{1}{n}\bigg) = a + \frac{a}{b}\frac{1}{n} + \frac{a}{n} + \frac{a}{b}\frac{1}{{n^2 }},&#xD;&#xA;$$&#xA;hence $h_n (x) \to a$. Hence, $h_n$ converges pointwise on $I$; denote the limit by $h$. Since a uniform limit is a pointwise limit, it thus suffices to show that $\sup _{x \in I} |h_n (x) - h(x)| \ge 1$ for infinitely many $n$. Indeed, for any $n \in \mathbb{N}$, there exists a rational number $x_n \in I$ of the form $x_n=a/b$, $b&amp;gt;0$, such that $|a| \geq 2n$. Hence, if we choose $n$ sufficiently large, then&#xA;$$&#xD;&#xA;|h_n (x_n) - h(x_n)| = \Big|\Big(a + \frac{a}{b}\frac{1}{n} + \frac{a}{n} + \frac{a}{b}\frac{1}{{n^2 }}\Big) - a\Big| &amp;gt; \frac{1}{2}\Big|\frac{a}{n}\Big| \geq 1&#xD;&#xA;$$&#xA;(note that $a/b$ is uniformly bounded, since $I$ is a bounded interval).&#xA;The claim is thus established.&lt;/p&gt;&#xA;" OwnerUserId="2810" LastActivityDate="2011-01-06T11:00:03.543" />
  <row Id="16558" PostTypeId="2" ParentId="16551" CreationDate="2011-01-06T11:10:41.370" Score="4" ViewCount="" Body="&lt;p&gt;What that means is that SVMs in order to classify your objects need to put a boundary separating objects of class A from objects of class B, for example. This boundary will be an hyperplane. But picture a circle in 2D and let's say you want two opposite quarters to be classified as A, and the two others as B. You can't draw a line that leaves A on one side and B on another. What SVMs do to solve this is &quot;sending&quot; the objects in $\mathbb{R}^3$ for example and then here put a plane separating the objects (they won't be the same though, so when you &quot;send&quot; them you actually transform them quite much).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More generally, if you have objects in $\mathbb{R}^n$, you will just send them in $\mathbb{R}^p$, with $p &amp;gt; n$, through some transformation, so that you can &quot;draw&quot; an &lt;em&gt;hyperplane&lt;/em&gt; leaving objects of class A on one side, and objects of class B on the other. So here's what is meant by &lt;em&gt;higher dimensional vector space&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, an &lt;em&gt;hyperplane&lt;/em&gt; is a really simple object. That's the generalization of straight lines in $\mathbb{R}^2$, planes in $\mathbb{R}^3$. That is, in $\mathbb{R}^n$, you just forget about one dimension -- what kind of object do you have now ? Well, an &lt;em&gt;hyperplane&lt;/em&gt;. Formally, it's a subspace of dimension $n-1$. Why an hyperplane ? Because you see that in $E = \mathbb{R}^2$, a straight line splits $E$ into two disjoint parts. In $E = \mathbb{R}^3$, a plane splits $E$ in two disjoint parts. And the same holds for higher dimensions (though we have a lot more trouble picturing it). And SVMs will find the hyperplane with the highest margin from both sides, A and B objects. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope this helps.&lt;/p&gt;&#xA;" OwnerUserId="1865" LastEditorUserId="1865" LastEditorDisplayName="" LastEditDate="2011-01-06T11:40:36.893" LastActivityDate="2011-01-06T11:40:36.893" />
  <row Id="16559" PostTypeId="1" AcceptedAnswerId="16671" CreationDate="2011-01-06T11:16:40.673" Score="4" ViewCount="190" Body="&lt;p&gt;Dear community,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;at the moment I am trying to &lt;strong&gt;reproduce&lt;/strong&gt; the &lt;strong&gt;results&lt;/strong&gt; of a &lt;a href=&quot;http://prb.aps.org/abstract/PRB/v74/i15/e153417&quot; rel=&quot;nofollow&quot;&gt;paper&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There, it turns out that a specific physical problem is mapped onto an &lt;strong&gt;integral to be calculated&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$I(\Theta; a, b) = 2\cdot\int_0^\infty \frac{a-u^2}{\left( u^2+\frac{a^2}{b-a}\right) \left(u^2+\frac{b^2}{b-a} \right) \sqrt{a - u^2 -\frac{a}{1-a/b}\sin^2(\Theta) }    }du \equiv \int_0^\infty f(u)du$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where I took the liberty to replace $\epsilon_d\rightarrow a$ and $\epsilon_m \rightarrow -b$ in contrast to the paper and one can &lt;strong&gt;assume&lt;/strong&gt; that both&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a,b &amp;gt; 0$ and $b &amp;gt; a$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Somehow, &lt;em&gt;Mathematica&lt;/em&gt; manages to be able to &lt;strong&gt;calculate&lt;/strong&gt; the &lt;strong&gt;numerical&lt;/strong&gt; values of this integral with some warning messages due to the &lt;strong&gt;pole&lt;/strong&gt; at&#xA;$$u_p = \sqrt{a -\frac{a}{1-a/b}\sin^2(\Theta)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, &lt;em&gt;Mathematica&lt;/em&gt; can calculate the &lt;strong&gt;indefinite integral&lt;/strong&gt;, both for $\Theta = 0$ (which is a special case to compare results) and in the general case. Nevertheless, I am not able to use the result since it is &lt;strong&gt;indefinite in all cases&lt;/strong&gt; for $u\rightarrow \infty$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, I am &lt;strong&gt;asking&lt;/strong&gt; for some advice to calculate the integral at hand with respect to given constant $a$ and $b$.&lt;br&gt;&#xA;The &lt;strong&gt;special case&lt;/strong&gt; of $\Theta = 0$ might already be worth to take a look since it is much easier to calculate it than for the general case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the meantime I tried something like a &lt;a href=&quot;http://en.wikipedia.org/wiki/Cauchy_principal_value&quot; rel=&quot;nofollow&quot;&gt;Cauchy principal value integration&lt;/a&gt; around $u_p$ using the parametrization $u_\delta (\varphi) = u_p -\delta e^{\mathrm{i}\varphi}$ along a half circle $C_\delta$ interpreting $u^2$ as $\bar{u}u$. Then,&lt;br&gt;&#xA;$$I_\delta = \int_0^\pi f(u_\delta(\varphi))\delta d\varphi$$&#xA;is the integral around $C_\delta$ which turned out to vanish for $\delta\rightarrow 0$ such that the whole integral should be given in terms of a principal value one. Noteworthy, I am not sure if my result is correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please, if my question is not stated correctly, or anything is obvious don't hesitate to give me some advice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Thank you&lt;/strong&gt; in advance.&lt;br&gt;&#xA;Sincerely&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Robert&lt;/p&gt;&#xA;" OwnerUserId="2671" LastEditorUserId="2671" LastEditorDisplayName="" LastEditDate="2011-01-06T22:30:10.233" LastActivityDate="2011-01-07T13:31:57.750" Title="An indefinite integral" Tags="&lt;integral&gt;&lt;numerical-methods&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="16561" PostTypeId="2" ParentId="16546" CreationDate="2011-01-06T11:53:11.503" Score="10" ViewCount="" Body="&lt;p&gt;Historical inertia.  A relatively small number of people were responsible for more or less deciding the modern abstract algebra curriculum around the beginning of the 20th century, and their ideas were so influential that their choice of topics is rarely questioned, for better or for worse.  See, for example, Section 9.7 of Reid's &lt;a href=&quot;http://books.google.com/books?id=f78gJ_92XcMC&amp;amp;printsec=frontcover&amp;amp;dq=undergraduate+commutative+algebra&quot; rel=&quot;nofollow&quot;&gt;Undergraduate commutative algebra&lt;/a&gt;.  &lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2011-01-06T11:53:11.503" CommentCount="3" />
  <row Id="16562" PostTypeId="1" AcceptedAnswerId="16563" CreationDate="2011-01-06T12:34:17.333" Score="4" ViewCount="70" Body="&lt;p&gt;This is a question about probability theory. One often sees a probability measure characterised as being additive  &quot;for disjoint $X,Y$ we have: $P(X \cup Y) = P(X) + P(Y)$ &quot; but sometimes as being additive like so: $P(X \cup Y) + P(X \cap Y) = P(X) +P(Y)$ . I guess what I'm asking is, how do you prove the inclusion/exclusion version of additivity from the version for disjoint sets. (The converse is trivial).&lt;/p&gt;&#xA;" OwnerUserId="172" LastActivityDate="2011-01-06T12:50:57.453" Title="Does it follow that for disjoint sets $P(X \cup Y) = P(X) + P(Y)$ that for any sets $P(X \cup Y) + P(X \cap Y) = P(X) +P(Y)$?" Tags="&lt;probability-theory&gt;" AnswerCount="2" />
  <row Id="16563" PostTypeId="2" ParentId="16562" CreationDate="2011-01-06T12:37:17.470" Score="6" ViewCount="" Body="&lt;p&gt;Split $X$ into the disjoint union of $X \cap Y$ and some other set $A$, and split $Y$ into the disjoint union of $X \cap Y$ and some other set $B$.  &lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2011-01-06T12:37:17.470" />
  <row Id="16564" PostTypeId="1" CreationDate="2011-01-06T12:49:52.200" Score="4" ViewCount="186" Body="&lt;p&gt;Hi, I studied fourier series as an undergrad and grad. student in EE but did not fully grasp the concepts. Now that I am involved in medical imaging (MRI) understanding the basics of fourier series and transforms is very important and I am frustrated at my level of understanding. For example, a problem in a MRI physics book asks: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Prove the following fourier series identity: $$\sum_{n=-\infty}^\infty \exp(2i\pi n a) = \sum_{m=-\infty}^\infty \delta(a-m)$$.&#xA;The author gives a hint: &quot;consider integrations over small intervals that either include or exclude the region where the argument of one of the delta functions vanishes&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My approach was to write out the left side of the equation for n = -5 to 5. I end up with the value 11. something's wrong for sure. Any help is greatly appreciated! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you &#xA;-Dave  &lt;/p&gt;&#xA;" OwnerUserId="5448" LastEditorUserId="536" LastEditorDisplayName="" LastEditDate="2011-01-06T17:09:06.113" LastActivityDate="2011-01-06T17:09:06.113" Title="Proving a fourier series identity" Tags="&lt;fourier-analysis&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="16565" PostTypeId="2" ParentId="16562" CreationDate="2011-01-06T12:50:57.453" Score="3" ViewCount="" Body="&lt;p&gt;$X = (X\setminus Y) \cup (X\cap Y)$, therefore $P(X) = P(X\setminus Y) + P(X\cap Y)$.&lt;br&gt;&#xA;On the other hand, $X\cup Y = (X\setminus Y)\cup Y$, and $Y\cap (X\setminus Y) = \emptyset$, therefore:&#xA;$$P(X) + P(Y) = P(X\setminus Y)+P(X\cap Y)+P(Y) = P(X\setminus Y \cup Y) + P(X\cap Y) = P(X\cup Y)+P(X\cap Y)$$&lt;/p&gt;&#xA;" OwnerUserId="622" LastActivityDate="2011-01-06T12:50:57.453" />
  <row Id="16566" PostTypeId="1" CreationDate="2011-01-06T13:24:00.050" Score="3" ViewCount="161" Body="&lt;p&gt;I am a little apprehensive to ask this question because I have a feeling it's a &quot;duh&quot; question but I guess that's the beauty of sites like this (anonymity):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need to find an orthonormal eigenbasis for the $2 \times 2$ matrix $\left(\begin{array}{cc}1&amp;amp;1\\&#xD;&#xA;1&amp;amp;1\end{array}\right)$.  I calculated that the eigenvalues were $x=0$ and $x=2$ and the corresponding eigenvectors were &#xA;$E(0) = \mathrm{span}\left(\begin{array}{r}-1\\1\end{array}\right)$ and $E(2) = \mathrm{span}\left(\begin{array}{c}1\\1\end{array}\right)$. Therefore, an orthonormal eigenbasis would be: &#xA;$$\frac{1}{\sqrt{2}}\left(\begin{array}{r}-1\\1\end{array}\right), &#xD;&#xA;\frac{1}{\sqrt{2}}\left(\begin{array}{c}1\\1\end{array}\right).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here my question: Could the eigenvalues for $E(0)$ been $\mathrm{span}\left(\begin{array}{r}1\\-1\end{array}\right)$??  This would make the final answer $\frac{1}{\sqrt{2}}\left(\begin{array}{r}1\\-1\end{array}\right), \frac{1}{\sqrt{2}}\left(\begin{array}{c}1\\1\end{array}\right)$.  Is one answer more correct than the other (or are they both wrong)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="4179" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-06T17:20:55.947" LastActivityDate="2011-01-06T17:31:45.047" Title="Orthonormal Eigenbasis" Tags="&lt;linear-algebra&gt;&lt;homework&gt;&lt;eigenvectors&gt;&lt;eigenvalues&gt;&lt;orthonormal&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="16567" PostTypeId="2" ParentId="16566" CreationDate="2011-01-06T13:38:02.670" Score="4" ViewCount="" Body="&lt;p&gt;0 and 2 are the correct eigenvalues to your matrix; (1, -1) is one eigenvector, (1, 1) the other. Your solution is correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Span is the set of all linear combinations, so if you consider a vector space over $\mathbb{R}$, it absolutely doesn't matter what scalar in $\mathbb{R}$ you multiply your vectors with inside Span. This does not affect the set Span at all. So the solution is the same.&lt;/p&gt;&#xA;" OwnerUserId="5450" LastEditorUserId="5450" LastEditorDisplayName="" LastEditDate="2011-01-06T16:59:45.823" LastActivityDate="2011-01-06T16:59:45.823" />
  <row Id="16568" PostTypeId="1" CreationDate="2011-01-06T13:38:21.807" Score="2" ViewCount="167" Body="&lt;p&gt;How can we prove that if $R$ is a commutative Noetherian ring, $\mathfrak{m} = (a_1,\ldots,a_n)$ is an ideal, then the completion of $R$ at $\mathfrak{m}$ is isomorphic to $R[[x_1,\ldots,x_n]]/(x_1-a_1,\ldots,x_n-a_n)$?&lt;/p&gt;&#xA;" OwnerUserId="5451" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2011-01-06T18:15:41.573" LastActivityDate="2011-01-07T14:54:47.990" Title="Completion of a Noetherian ring R at $(a_1,\ldots,a_n)$ is isomorphic to $R[[x_1,\ldots,x_n]]/(x_1-a_1,\ldots,x_n-a_n)$" Tags="&lt;ring-theory&gt;&lt;commutative-algebra&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="1" />
  <row Id="16569" PostTypeId="1" AcceptedAnswerId="16576" CreationDate="2011-01-06T14:03:06.943" Score="1" ViewCount="103" Body="&lt;p&gt;This question is about the connection between linear algebra and complex analysis. Coming from a two real dimensional domain a transformation matrix geometrically transforms a set of points (e.g. a circle) in that plane to another set of points in the two real dimensional codomain. The same holds true for a complex valued function which does the transformation from a one complex dimensional domain to the complex dimensional codomain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;I have two questions: &lt;br&gt;&lt;/strong&gt;&#xA;1. Do you know of any software (or add-ons) that let you &quot;paint&quot; some images and have them transformed by an arbitrary transformation matrix and/or complex function? &lt;br&gt;&#xA;2. What is the connection between transformation matrices and complex functions? How can you get from one to the other, i.e. finding corresponding transformation matrices to a given complex function and vice versa to produce the same transformation on the plane?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any answers concerning the above questions will be appreciated! Thank you!&lt;/p&gt;&#xA;" OwnerUserId="346" LastActivityDate="2011-01-06T15:19:01.790" Title="transformation matrices and complex functions as projections" Tags="&lt;geometry&gt;&lt;linear-algebra&gt;&lt;functions&gt;&lt;complex-analysis&gt;&lt;math-software&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" />
  <row Id="16570" PostTypeId="2" ParentId="16568" CreationDate="2011-01-06T14:05:10.480" Score="3" ViewCount="" Body="&lt;p&gt;Use the fact that completing is the same thing as tensoring with the completed ring, and that the completed ring is flat over the original ring.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(All this is true in view of your hypothesis, of course!)&lt;/p&gt;&#xA;" OwnerUserId="274" LastEditorUserId="274" LastEditorDisplayName="" LastEditDate="2011-01-06T14:13:46.007" LastActivityDate="2011-01-06T14:13:46.007" />
  <row Id="16571" PostTypeId="1" CreationDate="2011-01-06T14:06:36.643" Score="0" ViewCount="59" Body="&lt;p&gt;Hi! I read the following problem and I cannot get the right solution:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We consider a system of 12 sensors. The probability of one sensor to detect a signal with a magnitude higher than a is $p=\frac{1}{2\pi }\int_{x=a}^\infty \exp(\frac{-x^2}{2})$. What is the joint probability that 5 sensors detect this signal?  It is stated that if a=2.8, the five signals yield to a joint probability corresponding to 5.7&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="5452" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2011-01-06T14:14:23.213" LastActivityDate="2011-01-06T14:14:23.213" Title="joint probability" Tags="&lt;probability-distributions&gt;" AnswerCount="1" />
  <row Id="16572" PostTypeId="2" ParentId="16571" CreationDate="2011-01-06T14:11:54.730" Score="2" ViewCount="" Body="&lt;p&gt;Use the &lt;a href=&quot;http://en.wikipedia.org/wiki/Binomial_distribution&quot; rel=&quot;nofollow&quot;&gt;binomial distribution&lt;/a&gt;. The probability that exactly 5 sensors of 12 will register is $\binom {12}{5}p^5(1-p)^7$.  Just sum up the chance that 5, 6, ... 12 will register.  The claim (which I have not verified) is that the sum over 5, 6, ... 12 out of 12 for a=2.8 equals the chance of a single unit at a=5.7&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-01-06T14:11:54.730" />
  <row Id="16573" PostTypeId="1" AcceptedAnswerId="16591" CreationDate="2011-01-06T14:28:52.477" Score="0" ViewCount="146" Body="&lt;p&gt;This is actually mentioned in my module,I am not sure how they have got this result but I think this may not be correct,I tried using &lt;a href=&quot;http://www.wolframalpha.com/input/?i=44%2F9.3%20%3D%3D%205.5%2F1.16&amp;amp;t=ff3tb01&quot; rel=&quot;nofollow&quot;&gt;wolfarm alpha&lt;/a&gt; which gives false or am I missing something?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I was missing approximation,it seems like data interpretation problems keeps relying on this, so,$$\frac{44}{9.3} = 4.73118 \text{ and } \frac{5.5}{1.16} =  4.74138 $$ Hence, $44:9.3 \approx 5.5:1.16$&lt;/p&gt;&#xA;" OwnerUserId="2109" LastEditorUserId="2109" LastEditorDisplayName="" LastEditDate="2011-01-06T14:58:28.513" LastActivityDate="2011-01-06T17:15:10.450" Title="How $44:9.3 = 5.5:1.16 $?" Tags="&lt;algebra-precalculus&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="16575" PostTypeId="2" ParentId="16573" CreationDate="2011-01-06T14:43:17.670" Score="0" ViewCount="" Body="&lt;p&gt;A quick computation gives that&#xA;$$ \begin{align}&#xD;&#xA;{440 \over 93} = {550 \over 116} &#xD;&#xA;  &amp;amp;\iff {440 \times 116 \over 93 \times 116} = {550 \times 93 \over 116 \times 93} \\\\&#xD;&#xA;  &amp;amp;\iff 440 \times 116 = 550 \times 93 \\\\&#xD;&#xA;  &amp;amp;\iff 44 \times 116 = 55 \times 93,&#xD;&#xA;\end{align} $$&#xA;and in the last equality the LHS is even and RHS is odd, so they can't possibly be equal.&lt;/p&gt;&#xA;" OwnerUserId="1269" LastEditorUserId="1269" LastEditorDisplayName="" LastEditDate="2011-01-06T14:49:18.473" LastActivityDate="2011-01-06T14:49:18.473" CommentCount="3" />
  <row Id="16576" PostTypeId="2" ParentId="16569" CreationDate="2011-01-06T15:19:01.790" Score="1" ViewCount="" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;For complex functions, try &lt;a href=&quot;http://www.dmat.ufpe.br/~ssc/bombelli/&quot; rel=&quot;nofollow&quot;&gt;Bombelli&lt;/a&gt; or &lt;a href=&quot;http://www.saunalahti.fi/mattpaa/complex/complex.html&quot; rel=&quot;nofollow&quot;&gt;Complex visualizer&lt;/a&gt; (both are Java applets).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;In general, there is no connection. Transformation matrices correspond to &lt;em&gt;linear&lt;/em&gt; mappings, and complex functions are in general &lt;em&gt;nonlinear&lt;/em&gt;. Exception: the function $f(z)=(a+ib)z$ which simply multiplies $z$ by the constant $a+ib$ (where $a$ and $b$ are real) corresponds to the linear transformation given by the matrix $\begin{pmatrix} a &amp;amp; -b \\ b &amp;amp; a \end{pmatrix}$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;(Close to a point $z_0$ where $f&amp;#39;(z_0) \neq 0$, the formula $f(z)-f(z_0) \approx f&amp;#39;(z_0) (z-a)$ shows that the function $f$ can be approximated using a linear function of the above type, but that's a different story.)&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2011-01-06T15:19:01.790" />
  <row Id="16577" PostTypeId="2" ParentId="16564" CreationDate="2011-01-06T15:30:59.967" Score="4" ViewCount="" Body="&lt;p&gt;What you have described is, strictly speaking, an identity of distributions and not functions, as the $\delta$-&quot;function&quot; is a distribution. So the identity means that if you &quot;evaluate&quot; both sides with respect to a &lt;a href=&quot;http://en.wikipedia.org/wiki/Rapidly_decreasing_function&quot; rel=&quot;nofollow&quot;&gt;Schwarz function&lt;/a&gt; (i.e. one which decreases faster than any polynomial asymptotically, and so do all its derivatives), you will obtain the same answer. By definition, &quot;evaluation&quot; means multiplying by a Schwarz function and integrating.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fix a Schwarz function $\phi$. Then the left side (I am assuming that there is a small typo in what you have written, namely that you have omitted an $n$ in the exponent) is &#xA;$$\sum_n \int_{\mathbb{R}} f(x) e^{2 \pi i nx} dx$$&#xA;while the right hand side is &#xA;$$\sum_n f(n)$$&#xA;because, by definition, multiplying a function by the Dirac function and integrating just picks out the function's value at zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the identity you ask for is really  the &lt;a href=&quot;http://en.wikipedia.org/wiki/Poisson_summation_formula&quot; rel=&quot;nofollow&quot;&gt;Poisson summation formula&lt;/a&gt;&#xA;$$\sum_n \hat{f}(n) = \sum f(n).$$&#xA;(Here $\hat{f}$ denotes the Fourier transform.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The proof of the Poisson summation formula is to consider the auxiliary function $g(x) = \sum_{n} f(x +  n x)$. This is $1$-periodic and smooth because $f$ is Schwarz, so we can write by the theory of Fourier series, $$g(0) = \sum_n e^{2\pi i n 0} \left( \int_0^1 g(x) e^{2\pi i nx} dx \right),$$&#xA;which in turn yields the summation formula: $g(0)$ is the sum $\sum_n f(n)$, while one can check by expanding out the definition of $g$ and using the definition of the Fourier transform that the right-hand-side is the sum $\sum_n \hat{f}(n)$.&lt;/p&gt;&#xA;" OwnerUserId="536" LastActivityDate="2011-01-06T15:30:59.967" />
  <row Id="16578" PostTypeId="1" CreationDate="2011-01-06T15:33:08.473" Score="0" ViewCount="69" Body="&lt;p&gt;Hi.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm trying to implement this in matlab: &lt;code&gt;w = R_xx^-1 . S&lt;/code&gt; where &lt;code&gt;R_xx&lt;/code&gt; is a 150x150 matrix and &lt;code&gt;S&lt;/code&gt; is a 5x1 column vector.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand that this can't be multiplied, so how do I implement this? Any tricks?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="5456" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2011-01-06T18:35:53.897" LastActivityDate="2011-01-06T18:35:53.897" Title="Wiener-Hopf solution matrix multiplication help" Tags="&lt;matrices&gt;&lt;math-software&gt;&lt;matlab&gt;" CommentCount="4" />
  <row Id="16579" PostTypeId="2" ParentId="16546" CreationDate="2011-01-06T15:52:41.063" Score="7" ViewCount="" Body="&lt;p&gt;I want to mention that my first (undergraduate!) course in algebra did start with the study of monoids.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Actually even that is not true.  It started with a categorical perspective, especially the &quot;quotient principle&quot;: when one map factors through another.  At some point early on adjoint functors were introduced.  And then, maybe 2-3 weeks into the course, we started talking about monoids.  I especially remember the free monoid on the set $X$, which was referred to as the &lt;strong&gt;James construction&lt;/strong&gt;.  A childhood friend -- who is, by one of life's little coincidences, also a mathematician; at the time he was an undergrad two years ahead of me -- came to visit me in college about a month into the quarter and he was very surprised at what he saw in the course.  Finally we started in on groups about five weeks in.  (The instructor, &lt;strong&gt;Arunas Liulevicius&lt;/strong&gt;, was famous for doing this sort of thing and sort of making it work due in part to his great enthusiasm and friendly manner.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the concept of monoids has been with me for essentially my entire mathematical career.  I do think of this as a (slight) advantage, since a lot of people seem simply not to have this word in their vocabulary and go so for as to snicker when someone else says it.  This is ridiculous because they are a natural structure which comes up all the time, and for instance the group completion of a commutative monoid should be part of every mathematician's toolkit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I agree that the general study of monoids seems not to be as rewarding as that of groups, rings, fields (or modules).  As evidence towards this, I have some &lt;a href=&quot;http://math.uga.edu/~pete/semigroup.pdf&quot; rel=&quot;nofollow&quot;&gt;lecture notes on semigroups and monoids&lt;/a&gt; which I wrote up a couple of years ago.  But they didn't seem to be going anywhere interesting any time soon, so I stopped at $11$ pages and may or may not get back to them later.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;All in all, this may be a situation in which it is good to give the definition, a few easy examples of the definition, and then move on and allow the student an opportunity to see the structure recur in their later mathematical experience.&lt;/p&gt;&#xA;" OwnerUserId="299" LastActivityDate="2011-01-06T15:52:41.063" CommentCount="3" />
  <row Id="16580" PostTypeId="2" ParentId="16573" CreationDate="2011-01-06T15:54:22.950" Score="0" ViewCount="" Body="&lt;p&gt;Probably the approximation arises from solving $\rm\displaystyle\ \frac{44}{x} = \frac{5.5}{1.16}\ \Rightarrow\;\ x = \frac{1.16\cdot 44}{5.5} = 9.28 \approx 9.3$&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2011-01-06T15:54:22.950" />
  <row Id="16581" PostTypeId="2" ParentId="16564" CreationDate="2011-01-06T16:03:34.437" Score="3" ViewCount="" Body="&lt;p&gt;First off, there's a typo in the left hand side.. the $n$th term of the series is given by ${\mathbb\exp (2\pi i n a)}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you're being asked to prove is equivalent to the Poisson summation formula. Since Akhil Mathew already proved this as an answer, I'll just try to give some intuition. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have to be careful when saying ${\mathbb\sum_{n = -\infty}^{\infty}\exp (2\pi i n a) = \lim_{m \rightarrow \infty} \sum_{n = -m}^{m} \exp (2\pi i n a)}$ since the partial sums only converge as distributions and the terms don't go to zero for most $a$. Similarly one has the distributional limit&#xA;$$ \sum_{n = -\infty}^{\infty} \exp (2\pi i n a) = \lim_{r \rightarrow 1-} \sum_{n = -\infty}^{\infty} r^{|n|}\exp (2\pi i n a)$$&#xA;The right-hand sum can be written as the sum two geometric series which can be explicitly summed to $P_r(2\pi a)$, where $P_r(\theta)$ is the famous Poisson kernel. (see &lt;a href=&quot;http://en.wikipedia.org/wiki/Poisson_kernel&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Poisson_kernel&lt;/a&gt;). The explicit formula is&#xA;$$P_r(2\pi a) = {1 - r^2 \over 1 - 2r\cos(2\pi a) + r^2}$$&#xA;Note that $P_r(2\pi a)$ has period $1$. It can be shown that for $r$ near $1$, $P_r(2\pi a)$ is the sum of bump functions surrounding each integer, and as $r$ goes to $1$ from below, this sum of bump functions converges (in the sense of distributions) to the sum of  delta functions at all integers, which is exactly your right-hand side.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Incidentally, if you tried to do the analogous thing with the distributional limit  $\lim_{m \rightarrow \infty} \sum_{n = -m}^{m} \exp (2\pi i n a)$ in place of the above, you'd end out dealing with the Dirichlet kernel instead of the Poisson kernel. One can still show analogous convergence, but it's trickier. This is related to how showing pointwise convergence of Fourier series tends to be harder than showing Abel convergence.   &lt;/p&gt;&#xA;" OwnerUserId="3035" LastEditorUserId="3035" LastEditorDisplayName="" LastEditDate="2011-01-06T16:16:44.737" LastActivityDate="2011-01-06T16:16:44.737" />
  <row Id="16583" PostTypeId="1" AcceptedAnswerId="16594" CreationDate="2011-01-06T16:40:40.833" Score="2" ViewCount="1070" Body="&lt;p&gt;Does the probability of winning the lottery differ between randomly generated numbers vs. selecting the same numbers every time?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Specifically. I'm interested in a breakdown of the odds per number for a given set of numbers that comprise a single US Powerball drawing (five white numbers plus the one powerball number), and how they arrive at the odds seen here: &lt;a href=&quot;http://www.powerball.com/powerball/pb_prizes.asp&quot; rel=&quot;nofollow&quot;&gt;http://www.powerball.com/powerball/pb_prizes.asp&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Tying that back to my original question, I was interested if playing the same numbers every drawing changes those odds.&lt;/p&gt;&#xA;" OwnerUserId="5460" LastEditorUserId="5460" LastEditorDisplayName="" LastEditDate="2011-01-06T17:36:03.593" LastActivityDate="2011-01-18T21:06:24.897" Title="Odds of Winning the Lottery Using the Same Numbers Repeatedly Better/Worse?" Tags="&lt;probability&gt;" AnswerCount="7" CommentCount="8" />
  <row Id="16585" PostTypeId="2" ParentId="16498" CreationDate="2011-01-06T16:56:22.783" Score="2" ViewCount="" Body="&lt;p&gt;In the end, I ended up writing my own arrow routine, which produces scalable arrowheads and scalable labels:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;code&gt;axes[x_, y_, z_, f_, a_] := &#xA;   Graphics3D[&#xA;    Join[{Arrowheads[a]}, &#xA;     Arrow[{{0, 0, 0}, #}] &amp;amp; /@ {{x, 0, 0}, {0, y, 0}, {0, 0, &#xA;        z}}, {&#xA;      Text[Style[&quot;x&quot;, FontSize -&amp;gt; Scaled[f]], {0.9*x, 0.1*y, 0.1*z}], &#xA;      Text[Style[&quot;y&quot;, FontSize -&amp;gt; Scaled[f]], {0.1 x, 0.9*y, 0.1*z}], &#xA;      Text[Style[&quot;z&quot;, FontSize -&amp;gt; Scaled[f]], {0.1*x, 0.1*y, 0.9*z}]}]]&lt;/code&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The arguments are the x, y, and z positions of the x, y, and z arrows, respectively, f is the font scale (try about 0.05), and a is the arrowhead scale (about 0.05 should do it).  This is combined with ordinary 3D graphics using &lt;code&gt;Show[]&lt;/code&gt;, as in &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;code&gt;Show[Plot3D[Exp[-x^2 - y^2], {x, -2, 2}, {y, -2, 2}, Boxed -&amp;gt; False, &#xA;    PlotStyle -&amp;gt; Opacity[0.7], Mesh -&amp;gt; 4, Axes -&amp;gt; None], &#xA;   axes[2.5, 2.5, 1.5, 0.05, 0.02], &#xA;   PlotRange -&amp;gt; {{-3, 3}, {-3, 3}, {0, 1.5}}]&lt;/code&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The resulting plot is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/T2ceT.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="3009" LastActivityDate="2011-01-06T16:56:22.783" CommentCount="0" />
  <row Id="16586" PostTypeId="1" AcceptedAnswerId="16595" CreationDate="2011-01-06T16:57:03.527" Score="2" ViewCount="138" Body="&lt;p&gt;I know the sum of independent Gaussian distributed random variables also has a Gaussian distribution.&#xA;say we have a sequence of Gaussian random variables not necessarily independent&#xA;$ X_0, X_1, X_2, \cdots, X_{n-1}, X_n$ &#xA;and say  $ Z_n = X_n - X_{n-1} $ if we know for all $n \geq 1$ &amp;amp; the sequence &#xA;$$ Z_1,Z_2, \cdots, Z_n $$ are all independent with gaussian distribution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;now let $$ \{Y_n\} = \{(a_1X_1 - a_0 X_0), \cdots, (a_nX_n - a_{n-1} X_{n-1}) \} $$ where $a_n \in [0,\infty) $&#xA;how can be sure $Y_n$ is a sequence of Gaussian random variables too ? &lt;/p&gt;&#xA;" OwnerUserId="5176" LastActivityDate="2011-01-07T07:47:45.433" Title="Gaussian Distributed Random Variables" Tags="&lt;probability&gt;&lt;stochastic-processes&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="16587" PostTypeId="2" ParentId="16553" CreationDate="2011-01-06T16:59:34.713" Score="7" ViewCount="" Body="&lt;p&gt;If I'm reading your question correctly, you'd like to prove the stated equality? If so, perhaps this might orient you a little bit. Write the left hand side as&#xA;\begin{eqnarray}&#xA;\frac{s}{\pi} \left( \int_{0}^{\infty} \frac{\sin y}{y^{s+1}} dy \right) (2\pi)^{s} \left( \sum_{n = 1}^{\infty} n^{s-1} \right)&#xA;\end{eqnarray}&#xA;To prove that this equals the right side you'll need a definition and an identity. The zeta function is defined as $\zeta(s) = \sum_{n = 1}^{\infty} n^{-s}$ with $\mathbf{Re}(s) &amp;gt; 1$, so the sum above is clearly equal to $\zeta(1-s)$ with $\mathbf{Re}(s) &amp;lt; 0$. The hard part is now showing the following gamma function integral representation&#xA;\begin{eqnarray}&#xA;\Gamma(s) = \frac{1}{\sin \frac{\pi s}{2}} \int_{0}^{\infty} \frac{\sin y}{y^{1-s}} dy,&#xA;\end{eqnarray}&#xA;where the integral converges if and only if $-1 &amp;lt; \mathbf{Re}(s) &amp;lt; 1$. Once you've got this in hand, then your equality is true on $-1 &amp;lt; \mathbf{Re}(s) &amp;lt; 0$. Now just analytically continue by individually continuing both the gamma function and the zeta functions to their largest respective domains. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Before I post a proof of the integral representation, I'll give you a few hours to try to work on it yourself. Hint: Use a change of variables on a more common integral representation. More to come)&lt;/p&gt;&#xA;" OwnerUserId="2720" LastActivityDate="2011-01-06T16:59:34.713" CommentCount="3" />
  <row Id="16588" PostTypeId="2" ParentId="16583" CreationDate="2011-01-06T17:03:22.503" Score="8" ViewCount="" Body="&lt;p&gt;The probability of winning does not depend on the specific numbers selected and the numbers drawn for each lottery drawing have no dependence on previous drawings, so there is no benefit to playing the same numbers every time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For most U.S. state lotteries, it is beneficial to choose numbers above 31 when possible as many people play numbers based on dates and so picking numbers above 31 lowers the likelihood of split pots.  But, really, playing the lottery is a losing proposition regardless of the mechanics.&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2011-01-06T17:03:22.503" CommentCount="3" />
  <row Id="16589" PostTypeId="2" ParentId="16583" CreationDate="2011-01-06T17:04:47.277" Score="4" ViewCount="" Body="&lt;p&gt;No. Since the lottery numbers are randomly generated each time independently of the previous draws, the probability of winning is always the same.&lt;/p&gt;&#xA;" OwnerUserId="3249" LastEditorUserId="3249" LastEditorDisplayName="" LastEditDate="2011-01-07T05:11:25.220" LastActivityDate="2011-01-07T05:11:25.220" />
  <row Id="16590" PostTypeId="2" ParentId="16583" CreationDate="2011-01-06T17:07:34.010" Score="0" ViewCount="" Body="&lt;p&gt;No, unless you know the exact probability that each number will come out; in that case, you should always choose the most probable numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another case in which you should choose the same numbers if you have some extra information is in Isaac's answer, but that only applies if you change your goal to maximize your gain (or rather, minimize your loss) rather than maximize the probability of winning.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2011-01-06T17:07:34.010" CommentCount="2" />
  <row Id="16591" PostTypeId="2" ParentId="16573" CreationDate="2011-01-06T17:15:10.450" Score="2" ViewCount="" Body="&lt;p&gt;I eyeballed it this way:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac {44} {9.3} = \frac {11} {2.325}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;while:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac {5.5} {1.16} = \frac {11} {2.32}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So one expects roughly three significant figures of agreement but not equality.&lt;/p&gt;&#xA;" OwnerUserId="3111" LastActivityDate="2011-01-06T17:15:10.450" />
  <row Id="16592" PostTypeId="2" ParentId="16502" CreationDate="2011-01-06T17:15:29.240" Score="-3" ViewCount="" Body="&lt;p&gt;The conventions that a/bc means a/(bc), that ln 2x means ln(2x), and so on were universally established and very useful. Unfortunately, the combined influence of computer-style notation and excessive safe-playing by mathematics examiners has led to the present clutter of parentheses. Just pick up any maths book printed before 1970 (and many more recent ones), and you will see that absolutely no ambiguity arises. We need conventions like these; otherwise we will find ourselves having to write sin(4x) = (2sin(2x))cos(2x) for sin 4x = 2 sin 2x cos 2x, to prove that we don't mean (sin 4)x = 2 sin((2x cos2)x), etc.  &lt;/p&gt;&#xA;" OwnerUserId="875" LastActivityDate="2011-01-06T17:15:29.240" />
  <row Id="16593" PostTypeId="2" ParentId="16566" CreationDate="2011-01-06T17:31:45.047" Score="2" ViewCount="" Body="&lt;p&gt;There is no such thing as &lt;strong&gt;the&lt;/strong&gt; eigenvector of a matrix, or &lt;strong&gt;the&lt;/strong&gt; orthonormal basis of eigenvectors. There are usually many choices.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remember that an eigenvector $\mathbf{v}$ of eigenvalue $\lambda$ is a nonzero vector $\mathbf{v}$ such that $T\mathbf{v}=\lambda\mathbf{v}$. That means that if you take &lt;em&gt;any&lt;/em&gt; nonzero multiple of $\mathbf{v}$, say $\alpha\mathbf{v}$, then we will have&#xA;$$T(\alpha\mathbf{v}) = \alpha T\mathbf{v} = \alpha(\lambda\mathbf{v}) = \alpha\lambda\mathbf{v}=\lambda(\alpha\mathbf{v}),$$&#xA;so $\alpha\mathbf{v}$ is &lt;em&gt;also&lt;/em&gt; an eigenvector corresponding to $\lambda$. More generally, if $\mathbf{v}_1,\ldots,\mathbf{v}_k$ are all eigenvectors of $\lambda$, then any nonzero linear combination $\alpha_1\mathbf{v}_1+\cdots+\alpha_k\mathbf{v}_k\neq \mathbf{0}$ is also an eigenvector corresponding to $\lambda$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, of course, since $\left(\begin{array}{r}-1\\1\end{array}\right)$ is an eigenvector (corresponding to $x=0$), then so is $\alpha\left(\begin{array}{r}-1\\1\end{array}\right)$ for any $\alpha\neq 0$, in particular, for $\alpha=-1$ as you take.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, a set of vectors $\mathbf{w}_1,\ldots,\mathbf{w}_k$ is &lt;strong&gt;orthogonal&lt;/strong&gt; if and only if $\langle \mathbf{w}_i,\mathbf{w}_j\rangle = 0$ if $i\neq j$. If you have an orthogonal set, and you replace, say, $\mathbf{w}_i$ by $\alpha\mathbf{w}_i$ with $\alpha$ any scalar, then the result is &lt;em&gt;still&lt;/em&gt; an orthogonal set: because $\langle\mathbf{w}_k,\mathbf{w}_j\rangle=0$ if $k\neq j$ and neither is equal to $i$, and for $j\neq i$, we have &#xA;$$\langle \alpha\mathbf{w}_i,\mathbf{w}_j\rangle = \alpha\langle\mathbf{w}_i,\mathbf{w}_j\rangle = \alpha 0 = 0$$&#xA;by the properties of the inner product. As a consequence, if you take an orthogonal set, and you take any scalars $\alpha_1,\ldots,\alpha_k$, then $\alpha_1\mathbf{w}_1,\ldots,\alpha_k\mathbf{w}_k$ is also an orthogonal set.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A vector $\mathbf{n}$ is &lt;strong&gt;normal&lt;/strong&gt; if $||\mathbf{n}||=1$. If $\alpha$ is any scalar, then $||\alpha\mathbf{n}|| = |\alpha|\,||\mathbf{n}|| = |\alpha|$. So if you multiply any normal vector $\mathbf{n}$ by a scalar $\alpha$ of absolute value $1$ (or of complex norm $1$), then the vector $\alpha\mathbf{n}$ is also a normal vector.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A set of vectors is orthonormal if it is both orthogonal, and every vector is normal. By the above, if you have a set of orthonormal vectors, and you multiply each vector by a scalar of absolute value $1$, then the resulting set is also orthonormal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In summary: you have an orthonormal set of two eigenvectors. You multiply one of them by $-1$; this does not affect the fact that the two are eigenvectors. The set was orthogonal, so multiplying one of them by a scalar does not affect the fact that the set is orthogonal. And the vectors were normal, and you multiplied one by a scalar of absolute value $1$, so the resulting vectors are still normal. So you &lt;em&gt;still&lt;/em&gt; have an orthonormal set of two eigenvectors. I leave it to you to verify that if you have a linearly independent set, and you multiply each vector by a nonzero scalar, the result is still linearly independent.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2011-01-06T17:31:45.047" />
  <row Id="16594" PostTypeId="2" ParentId="16583" CreationDate="2011-01-06T17:35:53.093" Score="6" ViewCount="" Body="&lt;p&gt;So here is how &lt;a href=&quot;http://en.wikipedia.org/wiki/Powerball&quot; rel=&quot;nofollow&quot;&gt;Powerball&lt;/a&gt; works.  You choose five different numbers between $1$ and $59$ inclusive (the white balls) and one number between $1$ and $39$ inclusive (the red ball).  If the white balls match the winning numbers for the white balls, in &lt;em&gt;any order&lt;/em&gt;, and if the red ball matches the winning number for the red ball, then you win the jackpot.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because you can match the white balls in any order, the Powerball winning numbers are usually presented from smallest to largest, so if you order your numbers from smallest to largest, the two sequences have to match.  The number of ways to pick five different numbers in &lt;em&gt;any&lt;/em&gt; order from $1$ to $59$ is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{59 \cdot 58 \cdot 57 \cdot 56 \cdot 55}{5 \cdot 4 \cdot 3 \cdot 2 \cdot 1} = 5006386$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and every choice of five different numbers in increasing order has the same probability (one over the above number) of being chosen.  One way to get the above number is as follows: first, pretend that order matters.  Then there are $59$ possibilities for the first number.  Since there are $58$ possibilities left, there are $58$ possibilities for the second number.  And so forth.  But since order doesn't matter, you can draw any set of five numbers in $5 \cdot 4 \cdot 3 \cdot 2 \cdot 1$ different ways (five &lt;a href=&quot;http://en.wikipedia.org/wiki/Factorial&quot; rel=&quot;nofollow&quot;&gt;factorial&lt;/a&gt;), so you have to divide by that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Matching the red ball is easy: there are $39$ choices, so you have a $\frac{1}{39}$ chance of doing it.  So, in summary, your odds of winning the jackpot from &lt;em&gt;any&lt;/em&gt; choice of numbers is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{1}{5006386 \cdot 39} = \frac{1}{195249054}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;just as reported on the Powerball website.  In other words, it's one over the total number of possible tickets.  (The other probabilities reported on the website are slightly harder to calculate, but not by much; for that you need to learn about something called the &lt;a href=&quot;http://en.wikipedia.org/wiki/Inclusion%E2%80%93exclusion_principle&quot; rel=&quot;nofollow&quot;&gt;inclusion-exclusion principle&lt;/a&gt;.)&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2011-01-06T17:35:53.093" CommentCount="7" />
  <row Id="16595" PostTypeId="2" ParentId="16586" CreationDate="2011-01-06T17:52:23.083" Score="3" ViewCount="" Body="&lt;p&gt;Verify that&#xA;$$&#xD;&#xA;Y_1  = a_1 Z_1  + (a_1  - a_0 )X_0,&#xD;&#xA;$$&#xA;$$&#xD;&#xA;Y_2  = a_2 Z_2  + (a_2  - a_1 )Z_1  + (a_2  - a_1 )X_0, &#xD;&#xA;$$ &#xA;etc. Following the comments above, if the $Z_i$ are independent of $X_0$, then each $Y_i$ is a linear combination of independent Gaussian random variables, hence Gaussian.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Let's explain why an additional assumption (such as the one we used, namely that the $Z_i$ are independent of $X_0$) is necessary here. Suppose that the following assertion is true: ($\star$) There exist normal random variables $X$ and $Y$, and non-zero constants $a$, $b$, $a&amp;#39;$, and $b&amp;#39;$, such that $aX+bY$ is normal but $a&amp;#39;X+b&amp;#39;Y$ is not. Then, letting $X_0 = -bY$ and $X_n=aX$ for all $n \geq 1$, we have $Z_1 = aX+bY$ and $Z_n = 0$ for all $n \geq 2$. Hence, the $X_i$ are normal and the $Z_i$ are independent normal (using the common convention that constants are normal with variance $0$). However, $(a&amp;#39;/a)X_1-(b&amp;#39;/b)X_0 = a&amp;#39;X + b&amp;#39;Y$, and is hence not normal. So, we only need to explain why we should expect that ($\star$) is true (though it might be difficult to give an example where it is satisfied). For this purpose, consider the fact that $X$ and $Y$ are jointly normal if and only if ANY linear combination of $X$ and $Y$ is univariate normal...&lt;/p&gt;&#xA;" OwnerUserId="2810" LastEditorUserId="2810" LastEditorDisplayName="" LastEditDate="2011-01-07T07:47:45.433" LastActivityDate="2011-01-07T07:47:45.433" CommentCount="1" />
  <row Id="16596" PostTypeId="2" ParentId="5" CreationDate="2011-01-06T17:56:29.533" Score="2" ViewCount="" Body="&lt;p&gt;You can also use the &lt;a href=&quot;http://en.wikipedia.org/wiki/Rational_root_theorem&quot; rel=&quot;nofollow&quot; title=&quot;rational root test&quot;&gt;rational root test&lt;/a&gt; on the polynomial equation $x^2-2=0$ (whose solutions are $\pm \sqrt{2}$).  If this equation were to have a rational solution $\frac{a}{b}$, then $a \vert 2$ and $b \vert 1$, hence $\frac{a}{b}\in \{\pm 1, \pm 2\}$.  However, it's straightforward to check that none of $1,-1,2,-2$ satisfy the equation $x^2-2=0$.  Therefore the equation has no rational roots and $\sqrt{2}$ is irrational.&lt;/p&gt;&#xA;" OwnerUserId="5137" LastActivityDate="2011-01-06T17:56:29.533" />
  <row Id="16597" PostTypeId="1" CreationDate="2011-01-06T18:00:09.550" Score="3" ViewCount="159" Body="&lt;p&gt;A ring $R$ is a Boolean ring if $x^2=x$ for all $x\in R$. By Stone representation theorem&#xA;a Boolean ring is isomorphic to a subring of the ring of power set of some set. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is what is an example of a ring $R$ with $x^3=x$ for all $x\in R$ that is &#xA;not a Boolean ring? (Obviously every Boolean ring satisfies this condition.)&lt;/p&gt;&#xA;" OwnerUserId="3249" LastActivityDate="2011-01-06T18:50:44.043" Title="Example of a ring with $x^3=x$ for all $x$" Tags="&lt;abstract-algebra&gt;&lt;ring-theory&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="16598" PostTypeId="2" ParentId="16597" CreationDate="2011-01-06T18:17:44.833" Score="4" ViewCount="" Body="&lt;p&gt;It is a &lt;a href=&quot;http://mathoverflow.net/questions/30220/abstract-thought-vs-calculation/30273#30273&quot; rel=&quot;nofollow&quot;&gt;theorem of Jacobson&lt;/a&gt; that any ring whose elements satisfy $x^n = x$ for a particular $n$ is commutative.  For $n = 3$, if we restrict ourselves to examples of characteristic $3$ then I suspect all examples are given by the ring of continuous functions $T \to \mathbb{F}_3$ for a &lt;a href=&quot;http://en.wikipedia.org/wiki/Stone%27s_representation_theorem_for_Boolean_algebras#Stone_spaces&quot; rel=&quot;nofollow&quot;&gt;Stone space&lt;/a&gt; $T$ (I think you can more or less go through the proof of Stone's representation theorem).  I do not know whether Mariano's example has characteristic $3$ or not.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: some details.  If $P$ is a prime ideal of a ring $R$ all of whose elements satisfy $x^n = x$, then $R/P$ is an integral domain with the same property.  This implies that $R/P$ is a finite integral domain, hence a finite field $\mathbb{F}_q$.  Hence all prime ideals of $R$ are maximal (so $R$ is zero-dimensional).  Moreover, since $R$ has no nilpotents by assumption, its nilradical is zero, and since its nilradical coincides with its Jacobson radical, its Jacobson radical is zero.  (So elements of $R$ are faithfully represented as functions on $\text{Spec } R$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The elements of any residue field $\mathbb{F}_q$ are all the roots of $x^q = x$, so we have $q - 1 | n - 1$.  If we restrict to $n = p$ prime and require that $R$ has characteristic $p$, then $\mathbb{F}_p$ is the only possible residue field.  It follows that $R$ embeds into the ring of continuous functions $\text{Spec } R \to \mathbb{F}_p$, where $\text{Spec } R$ is a Stone space, and from here (following, for example, the approach used in &lt;a href=&quot;http://qchu.wordpress.com/2010/11/22/boolean-rings-ultrafilters-and-stones-representation-theorem/&quot; rel=&quot;nofollow&quot;&gt;this blog post&lt;/a&gt;) I think we can show that $R$ is isomorphic to this ring.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;More generally, $\text{Spec } R$ can be divided into subspaces according to residue field, and then $R$ embeds into a product of rings of the above form for the various possible $p$ (prime power instead of prime).  I don't know if every example is of this form.&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2011-01-06T18:50:44.043" LastActivityDate="2011-01-06T18:50:44.043" CommentCount="7" />
  <row Id="16599" PostTypeId="2" ParentId="16597" CreationDate="2011-01-06T18:19:34.800" Score="1" ViewCount="" Body="&lt;p&gt;You can always pick a set $X$, consider the free $\mathbb Z$-algebra $A=\mathbb Z\langle X\rangle$, and divide by the ideal generated by all the elements $x^3-x$ for $x\in A$ to que a ring $B_X$, no?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Since the quotient is going to be commutative, you can also start from the polynomial ring...)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To show that the resulting quotient is non-trivial, consider the many maps $B_X\to\mathbb Z_3$, which you get from functions $X\to\mathbb Z_3$. Indeed, for every example of a ring $R$ satisfying the condition, you can pick an $X$ and a surjective map $B_X\to R$. So these examples are universally complicated :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Later:&lt;/strong&gt; Notice that $\mathbb Z_6$ is an example of a ring satisfying the identity, so $B_X$ is not a product of copies of $\mathbb Z_3$.&lt;/p&gt;&#xA;" OwnerUserId="274" LastEditorUserId="274" LastEditorDisplayName="" LastEditDate="2011-01-06T18:44:30.773" LastActivityDate="2011-01-06T18:44:30.773" CommentCount="1" />
  <row Id="16600" PostTypeId="2" ParentId="16553" CreationDate="2011-01-06T18:20:15.143" Score="3" ViewCount="" Body="&lt;p&gt;I do not intend to post it, but for those who are interested the proof of the tough part of this question, namely the classic result that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \Gamma(s) \sin \left( \frac{\pi s}{2} \right) = \int_0^\infty y^{s-1} \sin y \textrm{ d}y $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $-1 &amp;lt; Re(s) &amp;lt; 1 $&#xA;can be found in &quot;Topics in Analytic Number Theory,&quot; by Hans Rademacher (in chapter 6).&lt;/p&gt;&#xA;" OwnerUserId="1301" LastActivityDate="2011-01-06T18:20:15.143" CommentCount="3" />
  <row Id="16601" PostTypeId="2" ParentId="16502" CreationDate="2011-01-06T18:21:28.790" Score="-1" ViewCount="" Body="&lt;p&gt;I think it depends on what is meant by the symbol $/$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the symbol $/$ is solely an indication of the binary division operator, sometimes also written $\div$, then the rule that all multiplications and divisions are performed together, working left-to-right, applies (that is, $a/bc$ should be taken to mean $\frac{a}{b}c$).  If, however, the symbol $/$ is a &lt;a href=&quot;http://en.wikipedia.org/wiki/Vinculum_%28symbol%29&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;vinculum&lt;/em&gt;&lt;/a&gt; (and, despite what Wikipedia says there, I have heard people use &quot;vinculum&quot; in the non-horizontal case) indicating division and grouping simultaneously, then $a/bc$ should be taken to mean $\frac{a}{bc}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if the latter sense is intended, I would expect the vinculum to appear bigger than a standard slash (in fact, at least for me, MathJax typesets &lt;code&gt;$/$&lt;/code&gt; to appear slightly bigger than a standard &lt;code&gt;/&lt;/code&gt;) and I would also expect the &quot;numerator&quot; to be offset higher and the &quot;demoninator&quot; to be offset lower, e.g. $^a/_{bc}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suspect that the predominant interpretation at the moment is the former, that $/$ is equivalent to $\div$.  I also know that on some earlier-model graphing calculators from Texas Instruments (the TI-81, I believe; none from the TI-82 onward), the latter interpretation was used (e.g., the input &lt;code&gt;8/2*2&lt;/code&gt; would return &lt;code&gt;2&lt;/code&gt;).&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2011-01-06T18:21:28.790" CommunityOwnedDate="2011-01-06T18:21:28.790" />
  <row Id="16602" PostTypeId="2" ParentId="16597" CreationDate="2011-01-06T18:21:42.100" Score="1" ViewCount="" Body="&lt;p&gt;Actually, not only would $\mathbb{Z}_3$ work, but it's the only solution that's an integral domain not of characteristic 2 (since, in such a case, $x^3-x=0\,\Rightarrow\,x\in {0,1,-1}$).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another solution would be $R:=\mathbb{Z}_3[\mathbb{Z}_2]$, the group ring of $\mathbb{Z}_3$ over $\mathbb{Z}_2$ (ie the ring of &quot;polynomials&quot; over $\mathbb{Z}_3$, except that exponents are in $\mathbb{Z}_2$).  To see why, given an element $f(x)\in R$ with&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f(x)=a_0+a_1 x^{b_1} + \cdots + a_n x^{b_n}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with $a_i\in \mathbb{Z}_3$ and $b_j \in \mathbb{Z}_2$.  Since $R$ is a ring of characteristic 3, the &lt;a href=&quot;http://en.wikipedia.org/wiki/Freshman%27s_dream&quot; rel=&quot;nofollow&quot;&gt;Freshman's Dream&lt;/a&gt; implies that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(f(x))^3=a_0^3+a_1^3 x^{3b_1} + \cdots + a_n^3 x^{3b_n} = a_0 + a_1 x^{b_1} + \cdots + a_n x^{b_n}=f(x)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, by the same argument, if you're given any ring $T$ of characteristic 3 with $x^3=x$ for all $x\in T$, then $T[\mathbb{Z}_2]$ satisfies this property as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can't think of another class of examples off the top of my head, but I'd be surprised if boolean rings and the class of examples above were the only examples of rings of this type.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt;  Yes, the solution of modding out a free algebra by an appropriate ideal would also work nicely.&lt;/p&gt;&#xA;" OwnerUserId="5137" LastEditorUserId="5137" LastEditorDisplayName="" LastEditDate="2011-01-06T18:29:03.560" LastActivityDate="2011-01-06T18:29:03.560" CommentCount="2" />
  <row Id="16603" PostTypeId="2" ParentId="16546" CreationDate="2011-01-06T18:26:04.870" Score="12" ViewCount="" Body="&lt;p&gt;Historically, the first &quot;modern algebra&quot; textbook was van der Waerden's in 1930, which followed the groups/rings/fields model (in that order). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as I know, the first paper with nontrivial results on semigroups was published in 1928, and the first textbook on semigroups would have to wait until the 1960s.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is also a slight problem with the notion of &quot;simpler&quot;. It is true that semigroups have fewer axioms than groups, and as such should be more &quot;ubiquitous&quot;. However, the theory of semigroups is also in some sense &quot;more complex&quot; than the theory of groups, just as the theory of noncommutative rings is harder than that of commutative rings (even though commutative rings are &quot;more complex&quot; than rings because they have an extra axiom) and the structure theory of fields is simpler than that of rings (fewer ideals, for one thing). Groups have the advantage of being a good balance point between simplicity of structure and yet the ability to obtain a lot of nontrivial and powerful results with relatively little prerequisites: most 1-semester courses, even at the undergraduate level, will likely reach the Sylow theorems, a cornerstone of finite group theory. Semigroups require a lot more machinery even to state the isomorphism theorems (you need to notion of congruences). &lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2011-01-06T18:26:04.870" CommentCount="2" />
  <row Id="16604" PostTypeId="1" AcceptedAnswerId="16644" CreationDate="2011-01-06T18:50:03.773" Score="4" ViewCount="96" Body="&lt;p&gt;What is the definition of a profinite morphism in &lt;a href=&quot;http://www.math.upenn.edu/~pop/Teaching/2010_Math624/2010_Math624PS08.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.math.upenn.edu/~pop/Teaching/2010_Math624/2010_Math624PS08.pdf&lt;/a&gt; problem 5? This is not actually a homework of mine but I was unable to find the definition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Snapshot:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/ILUNQ.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="5462" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2011-01-06T19:02:11.937" LastActivityDate="2011-01-07T05:49:42.643" Title="How to define a profinite morphism." Tags="&lt;algebraic-geometry&gt;&lt;terminology&gt;" AnswerCount="1" CommentCount="7" />
  <row Id="16605" PostTypeId="1" AcceptedAnswerId="16607" CreationDate="2011-01-06T19:35:11.537" Score="2" ViewCount="90" Body="&lt;p&gt;Assume that the $a$'s in the following equations must be limited to a binary values of $\{0,1\}$ and the $c$'s are integer constants: &#xA;\begin{align}&#xA;                      2a_{34}a_{23}a_{24}  &amp;amp;= c_1\\&#xA;                 a_{23} + a_{24} + a_{34}  &amp;amp;= c_2\\&#xA;                     2 a_{34}a_{13}a_{14}  &amp;amp;= c_3\\&#xA;                 a_{13} + a_{14} + a_{34}  &amp;amp;= c_4\\&#xA;                     2 a_{24}a_{12}a_{14}  &amp;amp;= c_5\\&lt;br&gt;&#xA;                 a_{12} + a_{14} + a_{24}  &amp;amp;= c_6&#xA;\end{align}&#xA;I recognize that these are highly restricted form of Diophantine equations, but I've been unable to find what they are called (once I know their names I can lookup proofs on any properties they have). If they don't have a specific name, how could one show that a] a solution exists, b] it is unique c] if not unique, how many solutions there are?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My CS background forces to acknowledge that I could enumerate through all $2^6$ possibilities, but I don't think I'd understand what is going if I did that. Also, I'd like to know how this would scale when there are more then 6 free variables.&lt;/p&gt;&#xA;" OwnerUserId="196" LastEditorUserId="196" LastEditorDisplayName="" LastEditDate="2011-01-06T19:45:11.127" LastActivityDate="2011-01-06T20:06:27.507" Title="What kind of Diophantine equations are these?" Tags="&lt;diophantine-equations&gt;" AnswerCount="2" CommentCount="6" />
  <row Id="16606" PostTypeId="2" ParentId="16605" CreationDate="2011-01-06T19:50:22.067" Score="1" ViewCount="" Body="&lt;p&gt;I don't think that this kind of restriction occurs in diophantine equations to begin with, so I doubt they carry a special name.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are a lot of very simple observations that you can make, though: since each $a$ can only take the value $0$ or $1$, then the only possible values of $c_1$, $c_3$, and $c_5$ are $0$ or $2$; and $c_2$, $c_4$, and $c_6$ just tell you how many of the $a$s are equal to $1$ and how many are equal to $0$, so the only possible values are $0$, $1$, $2$, or $3$. In particular, $c_1 = 2$ if and only if $c_2=3$; and similarly with $c_3$ and $c_4$, and with $c_5$ and $c_6$. Your variables are divided into three non-disjoint subsets: $\{a_{23}, a_{24}, a_{34}\}$, $\{a_{13}, a_{14}, a_{34}\}$, and $\{a_{12}, a_{14}, a_{24}\}$; any two of the sets have exactly one element in common, and the triple intersection is trivial. Since you will know how many in each set are equal to $1$ and how many are equal to $0$, it will be easy to determine any solutions using these facts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can also delete the $2$s from the first, third, and fifth equation (suitably modifying $c_1$, $c_3$, and $c_5$ if necessary), and reduce $c_2$, $c_4$, and $c_6$ modulo $2$ so that you have a system of equations over $\mathbb{F}_2$, the field with two elements; solve it over that field and then check which (if any) of the solutions still works over $\mathbb{Z}$ (which is just a matter of whether $c_2$, $c_4$, and $c_6$ were originally $2$ instead of $0$, or $3$ instead of $1$). &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-06T20:05:21.330" LastActivityDate="2011-01-06T20:05:21.330" />
  <row Id="16607" PostTypeId="2" ParentId="16605" CreationDate="2011-01-06T20:06:27.507" Score="1" ViewCount="" Body="&lt;p&gt;I think this could easily be translated into a boolean problem. We will interpret $x=0$ with $x$ is &lt;em&gt;false&lt;/em&gt; and $x=1$ with $x$ is &lt;em&gt;true&lt;/em&gt;. Now Consider quoations of the form &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a+b+c=z$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $z$ is an integer and $a, b, c \in \{0,1\}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now z gives you the amount of variables that you can set to &quot;true&quot;. So we can observe because of the interval of $a,b,c$ that $z \in [0,3]$ it will give you basically 4 cases. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $z=0$ you get the equoation $a = 0 \land b = 0 \land c = 0$, as all variables are $0$ and therefore &quot;false&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $z=1$, exactly one variable is &quot;true&quot;, so you have multiple cases here, you get the equoations &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(a = 1 \land b = 0 \land c = 0) \lor (a = 0 \land b = 1 \land c = 0) \lor (a = 0 \land b = 0 \land c = 1)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $z=2$ you get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(a = 1 \land b = 1 \land c = 0) \lor (a = 1 \land b = 0 \land c = 1) \lor (a = 0 \land b = 1 \land &#xD;&#xA;  c = 1)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $z=3$ you get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a = 1 \land b = 1 \land c = 1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The $2abc=y$ equations are quite simple, you will either have $y = 0$ or $y = 1$. If you get $y = 0$ the equivalent logical equation is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a = 0 \lor b = 0 \lor c = 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As at least one of the variables is 0 in that case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Else if $y = 1$ you have&#xA;$a = 1 \land b = 1 \land c = 1$ as all variables are &quot;true&quot; in this case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now you are left with a boolean &quot;SAT&quot;-Problem ( &lt;a href=&quot;http://en.wikipedia.org/wiki/Boolean_satisfiability_problem&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Boolean_satisfiability_problem&lt;/a&gt; ) where many algorithms exist that are clever and will work for a huge number of quoations although the problem is NP-complete. &lt;/p&gt;&#xA;" OwnerUserId="3123" LastActivityDate="2011-01-06T20:06:27.507" />
  <row Id="16608" PostTypeId="1" AcceptedAnswerId="16620" CreationDate="2011-01-06T20:28:36.800" Score="3" ViewCount="205" Body="&lt;p&gt;Take $X$ a set and $B_X$ the quotient of $\mathbb Z[X]$ by the ideal generated by all elements $u^3-u$ with $u\in \mathbb Z[X]$, which came up in &lt;a href=&quot;http://math.stackexchange.com/questions/16597/example-of-a-ring-with-x3-x-for-all-x/16599#16599&quot;&gt;this answer&lt;/a&gt; to &lt;a href=&quot;http://math.stackexchange.com/questions/16597/example-of-a-ring-with-x3-x-for-all-x&quot;&gt;this question&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you describe $B_X$ explicitly for small values of $|X|$?&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2011-01-07T01:02:23.703" Title="The free ring where $x^3=x$ identically." Tags="&lt;algebra&gt;" AnswerCount="2" />
  <row Id="16609" PostTypeId="1" AcceptedAnswerId="16612" CreationDate="2011-01-06T20:30:21.573" Score="1" ViewCount="101" Body="&lt;p&gt;What can one say about the set of all $n$-dimensional square matrices $A \in \text{GL}_n(\mathbb{C})$ that have an inverse with entries out of $\mathbb{C}$ with the properties:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;unitary&lt;/strong&gt; $:\Leftrightarrow A^*= A^{-1}$&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;hermitian&lt;/strong&gt; $:\Leftrightarrow A^* = A$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;where $A^*$ is the conjugate transpose of $A$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What obviously follows is $$A^{-1} = A$$&#xA;The most simple matrix that is in this set is the identity matrix. Are there others? How do they look like?&lt;/p&gt;&#xA;" OwnerUserId="5450" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-06T20:43:14.340" LastActivityDate="2011-01-06T20:53:26.667" Title="Orthogonal and symmetric Matrices" Tags="&lt;linear-algebra&gt;&lt;matrices&gt;" AnswerCount="2" CommentCount="6" FavoriteCount="2" />
  <row Id="16611" PostTypeId="1" AcceptedAnswerId="16613" CreationDate="2011-01-06T20:46:34.070" Score="2" ViewCount="126" Body="&lt;p&gt;From Herstein's &lt;em&gt;Topics in Algebra&lt;/em&gt; (exercise 14, section 1.2):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If $S$ is infinite and can be brought into one-to-one correspondence with the set of integers, prove that there is one-to-one correspondence between $S$ and $S \times S$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So far I know there exists some bijection $\sigma : S \rightarrow \mathbb{Z}$. If I can define a bijection $\tau : \mathbb{Z} \rightarrow \mathbb{Z} \times \mathbb{Z}$, then a one-to-one correspondence between $S$ and $S \times S$ is given by $\mu$ such that $s_\mu = (a_{\sigma^{-1}},b_{\sigma^{-1}})$, with $a,b\in \mathbb{Z}$ given by $s_{\sigma \circ \tau} = (a,b)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm having trouble defining $\tau$. I &lt;em&gt;think&lt;/em&gt; a possible way to do so would be to create some kind of spiral (like the Ulam spiral), and assign each point to a different integer. I suppose this would be a one-to-one correspondence but I'm at a loss on how to prove it. Thanks a lot!&lt;/p&gt;&#xA;" OwnerUserId="1523" LastEditorUserId="3731" LastEditorDisplayName="" LastEditDate="2011-01-06T23:18:09.380" LastActivityDate="2011-01-06T23:18:09.380" Title="Bijecting a countably infinite set $S$ and its cartesian product $S \times S$" Tags="&lt;set-theory&gt;&lt;elementary-set-theory&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="16612" PostTypeId="2" ParentId="16609" CreationDate="2011-01-06T20:48:12.100" Score="8" ViewCount="" Body="&lt;p&gt;Such a matrix is such that $A^2=I$, so it is diagonalizable, and its possible eigenvalues are $+1$ and $-1$. Since it is unitary, the eigenspaces corresponding to $1$ and to $-1$ are orthogonal. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conversely, every diagonalizable matrix with eigenvalues contained in $\{+1,-1\}$ and orthogonal eigenspaces is of that form.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It follows that the set of your matrices is in bijection with the set of subspaces of $\mathbb C^n$. Explicitely: If $V$ is one such subspace, there is a unique linear transformation $f:\mathbb C^n\to\mathbb C^n$ such that $V$ and $V^\perp$ are eigenspaces for the eigenvalues $1$ and $-1$. The matrix $A_V$ of $f$ with respect to the standard basis of $\mathbb C^n$ satisfies your conditions. The bijection is $$V\in\{\mathrm{subspaces\;of\;}\mathbb C^n\}\leftrightarrow A_V.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a consequence, considered as a whole, your set is a disjoint union of submanifolds homeomorphic to &lt;a href=&quot;http://en.wikipedia.org/wiki/Grassmannian&quot; rel=&quot;nofollow&quot;&gt;Grassmannian varieties&lt;/a&gt;. Literally books have been written about them.&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2011-01-06T20:48:12.100" CommentCount="1" />
  <row Id="16613" PostTypeId="2" ParentId="16611" CreationDate="2011-01-06T20:52:02.533" Score="6" ViewCount="" Body="&lt;p&gt;It's a little simpler to find a bijection $\mathbb{N}\to\mathbb{N}\times\mathbb{N}$ (e.g., think of the usual proof that $\mathbb{Q}$ is countable). There is also an easy bijection $\mathbb{Z}\to\mathbb{N}$, so you can use that to get a bijection $\mathbb{Z}\to\mathbb{Z}\times\mathbb{Z}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A couple of comments, though: Herstein is only asking you to show that such a bijection exists, not to explicitly construct one. You could do so using any number of tools, including some that are in principle constructive but which you may not want to actually carry out. For example, there are obvious and easy embeddings $\mathbb{Z}\hookrightarrow \mathbb{Z}\times\mathbb{Z}$. If you also have an embedding $\mathbb{Z}\times\mathbb{Z}\hookrightarrow \mathbb{Z}$ you get the existence of a bijection between the two by using Cantor-Bernstein. For instance, you could take:&#xA;$$(a,b)\longmapsto \left\{\begin{array}{ll}&#xA;2^a\times 3^b &amp;amp;\text{if $a,b\geq 0$;}\\&#xA;3^b\times 5^{|a|} &amp;amp;\text{if $a\lt 0$, $b\geq 0$;}\\&#xA;2^a\times 7^{|b|} &amp;amp;\text{if $a\geq 0$, $b\lt 0$;}\\&#xA;5^{|a|}\times 7^{|b|} &amp;amp;\text{if $a,b\lt 0$.}&#xA;\end{array}\right.$$&#xA;So with this you &lt;em&gt;know&lt;/em&gt; there is a bijection $S\to S\times S$, even if you don't go through the rather annoying work of trying to write it out explicitly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, it might be worth noting that if you drop the clause &quot;&lt;em&gt;and can be brought into one-to-one correspondence with the set of integers&lt;/em&gt;&quot;, then the resulting statement is equivalent to the Axiom of Choice (this is a result of Tarski's). &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-06T21:11:47.050" LastActivityDate="2011-01-06T21:11:47.050" CommentCount="13" />
  <row Id="16614" PostTypeId="2" ParentId="16609" CreationDate="2011-01-06T20:53:26.667" Score="1" ViewCount="" Body="&lt;p&gt;(in addition to Mariano's answer), these are sometimes called &lt;a href=&quot;http://en.wikipedia.org/wiki/Householder_transformation&quot; rel=&quot;nofollow&quot;&gt;Householder transformations&lt;/a&gt;. &lt;/p&gt;&#xA;" OwnerUserId="308" LastActivityDate="2011-01-06T20:53:26.667" CommentCount="3" />
  <row Id="16615" PostTypeId="2" ParentId="16611" CreationDate="2011-01-06T20:53:43.060" Score="6" ViewCount="" Body="&lt;p&gt;You can do it the following way:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Define $f\colon\mathbb{Z}\to\mathbb{N}$ to be any bijection (for example, $f(x) = 2x$ if $x\ge 0$ and $f(x) = -2x+1$ otherwise)&lt;/li&gt;&#xA;&lt;li&gt;Define $g\colon\mathbb{N}\times\mathbb{N}\to\mathbb{N}$ your favourite bijection (e.g. $f(a,b) = \frac{(a+b+1)(a+b)}{2}+a$)&lt;/li&gt;&#xA;&lt;li&gt;Assume $\sigma$ is a bijection of $S$ with $\mathbb{Z}$ then define: $\tau(s_1,s_2) = \sigma^{-1}(f^{-1}(g(f(\sigma(s_1)),f(\sigma(s_2)))))$&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The proof that $\tau$ is a bijection is pretty straightforward, I think.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course you can skip many a step in this crazy composition by using Cantor-Schroeder-Bernstein theorem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt;&lt;br&gt;&#xA;I think I should expand a bit on the second part, as Arturo said in the comments to the original question.&#xA;Assume $f\colon\mathbb{Z}\times\mathbb{Z}\to\mathbb{Z}$ a bijection, and $\sigma\colon S\to\mathbb{Z}$ then you have $\tau(s_1,s_2) = f(\sigma(s_1),\sigma(s_2))$ a bijection from $S\times S$ to $\mathbb{Z}$ and therefore you have both $S$ and $S\times S$ equinumerous with the integers, and therefore with each other.&lt;/p&gt;&#xA;" OwnerUserId="622" LastEditorUserId="622" LastEditorDisplayName="" LastEditDate="2011-01-06T21:06:50.900" LastActivityDate="2011-01-06T21:06:50.900" />
  <row Id="16616" PostTypeId="2" ParentId="16611" CreationDate="2011-01-06T20:54:15.090" Score="1" ViewCount="" Body="&lt;p&gt;In addition to Arturo's answer, recall that if you have a bijection between A and B, and one between B and C, you have a bijection between A and C by composition.&lt;/p&gt;&#xA;" OwnerUserId="5429" LastActivityDate="2011-01-06T20:54:15.090" />
  <row Id="16617" PostTypeId="2" ParentId="16608" CreationDate="2011-01-06T21:06:40.823" Score="2" ViewCount="" Body="&lt;p&gt;The arguments I gave in the other question show the following:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Every prime ideal is a maximal ideal, and the Jacobson radical is zero.&lt;/li&gt;&#xA;&lt;li&gt;The residue field of every prime ideal is either $\mathbb{F}_2$ or $\mathbb{F}_3$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;It follows that $B_X$ has characteristic $6$.  A morphism $B_X \to \mathbb{F}_2$ is freely determined by its values on $X$, so the subspace of $\text{Spec } B_X$ with residue fields $\mathbb{F}_2$ is homeomorphic to $\{ 0, 1 \}^X$.  Similarly the subspace of $\text{Spec } B_X$ with residue fields $\mathbb{F}_3$ is homeomorphic to $\{ 0, 1, -1 \}^X$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So $B_X$ embeds into $A \times B$ where $A$ is the ring of continuous functions $\{ 0, 1 \}^X \to \mathbb{F}_2$ and $B$ is the ring of continuous functions $\{ 0, 1, -1 \}^X \to \mathbb{F}_3$.  It is precisely the subring of this ring generated by the images of $X$, where $x \in X$ corresponds to the product of the evaluation $e_x : \{ 0, 1 \}^X \to \mathbb{F}_2$ and the evaluation $e_x&amp;#39;: \{ 0, 1, -1 \}^X \to \mathbb{F}_2$.  So I guess it is a certain fiber product of $A$ and $B$. &lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2011-01-06T21:26:26.180" LastActivityDate="2011-01-06T21:26:26.180" CommentCount="0" />
  <row Id="16618" PostTypeId="1" AcceptedAnswerId="38236" CreationDate="2011-01-06T21:34:03.340" Score="2" ViewCount="134" Body="&lt;p&gt;I'm searching for some good reading material on &lt;a href=&quot;http://en.wikipedia.org/wiki/Multifractal&quot; rel=&quot;nofollow&quot;&gt;multifractal analysis&lt;/a&gt;. Preferably something accessible that doesn't put the stress too much on mathematical proofs but rather on applications. As long as it gives a good review of the status of the field, the interesting results and applications, I would be happy. Also, if anything related to multifractal analysis and statistics or time series comes up, I'll take it as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Books, papers, internetpages, videos, etc... accepted!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;: Since the question has been bumped, I decided to put a bounty on it. But I also want to make a bit more precise what I'm looking for.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have always had the impression when encountering the multifractal techniques that people are able to compute a whole bunch of numbers with some nice and fancy formulas. But I have always missed an &quot;understanding&quot; of what the numbers mean. Why is it useful to do a mutlifractal analysis of fluid flow? Of species abundance distributions? Etc... I feel like the technique is purely descriptive with little theory backing up the connection with some deeper underlying structures. But that may just be due to my limited understanding of the field and that is precisely why I ask for pointers to where I can look for this.&lt;/p&gt;&#xA;" OwnerUserId="3567" LastEditorUserId="3567" LastEditorDisplayName="" LastEditDate="2011-05-10T12:07:40.303" LastActivityDate="2011-05-10T14:23:33.630" Title="Need good material on multifractal analysis." Tags="&lt;books&gt;&lt;education&gt;&lt;fractals&gt;&lt;reviews&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="16619" PostTypeId="1" CreationDate="2011-01-06T21:41:20.150" Score="1" ViewCount="36" Body="&lt;p&gt;Let $f$ be a Dynkin diagram automorphism. Extend $f$ linearly to the root system $\Delta$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is a set of representatives of the orbits of $\Delta$ under $f$ ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks,&lt;/p&gt;&#xA;" OwnerUserId="2764" LastActivityDate="2011-01-06T21:41:20.150" Title="orbit of a Dynkin diagram automorphism" Tags="&lt;lie-algebra&gt;" />
  <row Id="16620" PostTypeId="2" ParentId="16608" CreationDate="2011-01-06T22:07:40.563" Score="1" ViewCount="" Body="&lt;p&gt;Just checking exhaustively:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When |X|=0, B=Z/6Z of size 6.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When |X|=1, B=Z[x]/(6,xxx-x,3xx+3x)={a+bx+c(xx+x):a,b in Z/6Z, c in Z/3Z} of size 108.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When |X|=2, B=Z[x,y]/(6,xxx-x,yyy-y,3xx+3x,3yy+3y) = {a+bx+cy+dxy+e(xx+x)+f(xx+x)y+g(xx+x)(yy+y)+hx(yy+y)+i(yy+y) : a,b,c,d in Z/6Z, e,f,g,h,i in Z/3Z} of size 314928.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Presumably the pattern holds, so that if |X|=n, the size is 6^(2^n)*3^(3^n-2^n) with similar additive invariants, but I didn't check n=3 due to the size.  At any rate, its size is a divisor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For larger exponents (varying the &quot;3&quot;), the characteristic of the ring is related to the denominators of the Bernoulli numbers as in this &lt;a href=&quot;http://oeis.org/A027760&quot; rel=&quot;nofollow&quot;&gt;OEIS entry&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Since 3 + 4 = 1, 3*3 = 3, 4*4 = 4, 3*4 = 0, 3*x = x*3, and 4*x = x*4 in any ring B satisfying the law uuu=u, one has that 3 and 4 are orthogonal central idempotents and B decomposes as (3B) × (4B).  3B has characteristic 2 and 4B has characteristic 3, so it suffices to consider algebras over the fields Z/2Z and Z/3Z (satisfying the law uuu=u) and take their direct product (X finite or not, B free or not).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In characteristic 2, (u+1)^3 - (u+1) = u^2-u, so the algebra 3B is always a Boolean algebra. For B free, 3B is a free Boolean algebra since the relevant homomorphisms all have 4B in their kernel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In characteristic 3, (u+v)^3 - (u+v) = u^3-u + v^3-v, so it suffices to check the law on a basis.  Since the law is clearly multiplicative, it then suffices to check it on an algebra generating set, or more importantly, to impose it there.  This gives the simpler description of 4B = Z[x,y]/(3,xxx-x,yyy-y), etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This gives a reasonably explicit description of the free B as a direct product of a free boolean algebra and the &quot;obvious&quot; characteristic 3 algebra.&lt;/p&gt;&#xA;" OwnerUserId="583" LastEditorUserId="583" LastEditorDisplayName="" LastEditDate="2011-01-07T01:02:23.703" LastActivityDate="2011-01-07T01:02:23.703" CommentCount="8" />
  <row Id="16621" PostTypeId="1" AcceptedAnswerId="16622" CreationDate="2011-01-06T22:34:37.083" Score="1" ViewCount="51" Body="&lt;p&gt;If I want to plot $r = 4\sin3\theta$ what is a good way to pick points? The period is $2\pi/3$ so it will repeat after that, but how do I pick points so that I get a good range of values so that I will be able to tell in which direction or how the curve is moving?&lt;/p&gt;&#xA;" OwnerUserId="3514" LastEditorUserId="3514" LastEditorDisplayName="" LastEditDate="2011-01-06T23:39:53.353" LastActivityDate="2011-01-06T23:39:53.353" Title="What is a good way to pick points for polar equations?" Tags="&lt;homework&gt;&lt;algebra-precalculus&gt;&lt;polar-coordinates&gt;" AnswerCount="2" />
  <row Id="16622" PostTypeId="2" ParentId="16621" CreationDate="2011-01-06T22:42:04.380" Score="1" ViewCount="" Body="&lt;p&gt;First, the period is $\frac{2\pi}{3}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As to picking points, what I'd recommend is not to—that is, instead of picking some specific values for $\theta$ and calculating $r$, sketch a graph of $y=4\sin 3x$ for $0\le x\le 2\pi$.  The maximums, minimums, and zeros on that graph are most likely the points you're looking for, you can probably draw that graph as quickly (or more quickly) than computing specific values for $r$, and the graph also tells you what's happening between those values.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're determined to pick points, note that simple sinusoidal functions (things like $f(x)=a\sin bx+c$) have maximums/minimums/zeros spaced by quarter-periods.&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2011-01-06T22:42:04.380" CommentCount="0" />
  <row Id="16624" PostTypeId="2" ParentId="16621" CreationDate="2011-01-06T23:13:27.133" Score="0" ViewCount="" Body="&lt;p&gt;It depends upon how you plot it.  If I wanted to plot it in Excel, for instance, I would just figure the period $2\pi /3 \approx 2.1$ and plot 100 points at an interval of 0.021.  If you are plotting by hand but calculating by spreadsheet, maybe 20 points evenly spaced is reasonable.  You could start there and see if it meets your needs.  More aggressive would be to figure out where the maximum and minimum r come from ($\pi /6, \pi /2$ and $0, \pi /3, 2\pi /3)$ and plot those, plus halfway in between.  But then we are getting close to 20.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-01-06T23:13:27.133" />
  <row Id="16625" PostTypeId="1" AcceptedAnswerId="16626" CreationDate="2011-01-06T23:16:07.003" Score="4" ViewCount="172" Body="&lt;p&gt;I'm learning for my algorithms exam and I can't derive two logarithm transformations:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$ 3^{log_{4}(n)}=n^{log_{4}(3)} $&lt;/li&gt;&#xA;&lt;li&gt;$ log_{3}(n)=log_{3}(e)*ln(n) $&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I'm not very strong in logarithms, anybody can explain how I get in both cases from the left side of the equation to the right side?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sorry if the terminology is wrong, I'm from Germany.&lt;/p&gt;&#xA;" OwnerUserId="5466" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2011-01-06T23:25:38.000" LastActivityDate="2011-01-07T01:11:21.620" Title="Can anybody explain how these logarithms are transformed?" Tags="&lt;logarithms&gt;&lt;transformation&gt;" AnswerCount="4" CommentCount="2" />
  <row Id="16626" PostTypeId="2" ParentId="16625" CreationDate="2011-01-06T23:20:19.447" Score="6" ViewCount="" Body="&lt;p&gt;Use $\log _a b = \frac{{\ln b}}{{\ln a}}$ and $e^{\ln x}  = x$:&#xA;$$&#xD;&#xA;3^{\log _4 n}  = (e^{\ln 3} )^{^{\ln n/\ln 4} }  = e^{\ln 3\ln n/\ln 4}  = (e^{\ln n} )^{\ln 3/\ln 4}  = n^{\ln 3/\ln 4}  = n^{\log _4 3}, &#xD;&#xA;$$&#xA;$$&#xD;&#xA;\log _3 n = \frac{{\ln n}}{{\ln 3}} = \frac{1}{{\ln 3}}\ln n = \frac{{\ln e}}{{\ln 3}}\ln n = (\log _3 e)\ln n.&#xD;&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="2810" LastActivityDate="2011-01-06T23:20:19.447" />
  <row Id="16627" PostTypeId="2" ParentId="16625" CreationDate="2011-01-06T23:24:32.673" Score="6" ViewCount="" Body="&lt;p&gt;For the first one: Remember that for any number $a$, $4^{\log_4(a)}=a$. Also, $(a^b)^c=a^{bc}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This gives us: $3^{\log_4(n)}=(4^{\log_4(3)})^{\log_4(n)}=4^{\log_4(3)\times\log_4(n)}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, $n^{\log_4(3)}=(4^{\log_4(n)})^{\log_4(3)}=4^{\log_4(n)\times\log_4(3)}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, since $ab=ba$ for any numbers $a,b$, this gives us the result.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the second, it is enough to check that $3^{\log_3(e)\times\ln(n)}=n$, because (by definition) the only number $a$ such that $3^a=n$ is $a=\log_3(n)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now: $3^{\log_3(e)\times\ln(n)}=(3^{\log_3(e)})^{\ln(n)}=e^{\ln(n)}=n$, remembering that $\ln$ is just an abbreviation for $\log_e$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Usually this is presented in a more condensed way, as a &lt;em&gt;change of basis&lt;/em&gt; formula, saying that $$\log_a(b)/\log_a(c)=\log_c(b)$$ for any positive numbers  $a,b,c$. This formula can be proved in exactly the same way as we checked the second of the transformations you asked about. &lt;/p&gt;&#xA;" OwnerUserId="462" LastActivityDate="2011-01-06T23:24:32.673" CommentCount="1" />
  <row Id="16628" PostTypeId="2" ParentId="16502" CreationDate="2011-01-06T23:36:12.887" Score="0" ViewCount="" Body="&lt;p&gt;In hand-written form, a horizontal line is a better delimiter for division. In type-set form, ditto: as in $\frac{5x}{30 x^2}$. Those of us of a certain age learned that a horizontal line was a grouping symbol. The slash creates the apparent ambiguity and should be avoided without parenthetical disambiguation. Observe that in $\TeX$, an over-use of braces is not harmful. The same is true with parentheses. Maturing math students should write mathematical expressions in correct form; otherwise, humans and machines will misinterpret the meaning.&lt;/p&gt;&#xA;" OwnerUserId="722" LastActivityDate="2011-01-06T23:36:12.887" />
  <row Id="16629" PostTypeId="2" ParentId="16625" CreationDate="2011-01-07T00:00:11.323" Score="2" ViewCount="" Body="&lt;p&gt;Let me also give some strategy points with these problems that I think will help.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One rule that should help is the equality $log_b(c^p) = p \cdot \log_b(c)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because the second question is asking for an equality with multiplication, you can try substituting $b = 3$, $c = e$, and $p = ln(n)$ to get:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$log_3(e^{ln(n)}) = ln(n) \cdot \log_3(e)$ and $e^{ln(n)}$ simplifies to $n$ so you're good.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the first problem, you want to change that $3$ to a $4$ because you know that $4^{log_4(n)} = n$.  Thus you want $4^m = 3$ so $m = log_4(3)$.  Expressing $3^{log_4(n)} = (4^{log_4(3)})^{log_4(n)} = (4^{log_4(n)})^{log_4(3)} = n^{log_4(3)}$. &lt;/p&gt;&#xA;" OwnerUserId="4693" LastActivityDate="2011-01-07T00:00:11.323" />
  <row Id="16630" PostTypeId="1" AcceptedAnswerId="16631" CreationDate="2011-01-07T00:32:37.190" Score="5" ViewCount="186" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Burnside%27s_lemma&quot; rel=&quot;nofollow&quot;&gt;Burnside's Lemma&lt;/a&gt;, whose list of names is longer than the proof, says that the number of orbits of a permutation group is the average number of fixed points of its elements. It's a very elegant result, but I'm a bit disappointed by the fact that the examples given in the textbooks always amount to counting some colorings of a symmetric object, up to symmetry (the less original example probably being the cube). My question is then: do you know some funnier (but still rather direct) applications of this result?&lt;/p&gt;&#xA;" OwnerUserId="5178" LastActivityDate="2011-01-07T00:58:05.720" Title="Nice application of the Cauchy?-Frobenius?-Burnside?-Pólya? formula" Tags="&lt;group-theory&gt;&lt;soft-question&gt;&lt;counting&gt;" AnswerCount="1" FavoriteCount="2" />
  <row Id="16631" PostTypeId="2" ParentId="16630" CreationDate="2011-01-07T00:39:02.343" Score="8" ViewCount="" Body="&lt;p&gt;Burnside's lemma can be used to prove the &lt;a href=&quot;http://en.wikipedia.org/wiki/P%C3%B3lya_enumeration_theorem&quot; rel=&quot;nofollow&quot;&gt;Polya enumeration theorem&lt;/a&gt;, which has many applications; see, for example, &lt;a href=&quot;http://qchu.wordpress.com/2009/06/16/gila-iii-the-orbit-counting-lemma-and-baby-polya/&quot; rel=&quot;nofollow&quot;&gt;these&lt;/a&gt; &lt;a href=&quot;http://qchu.wordpress.com/2009/06/21/gila-v-the-polya-enumeration-theorem-and-applications/&quot; rel=&quot;nofollow&quot;&gt;two&lt;/a&gt; blog posts.  The application to the symmetric groups alone is the well-known exponential formula in combinatorics, which has many applications; see &lt;a href=&quot;http://qchu.wordpress.com/2009/06/24/gila-vi-the-cycle-index-polynomials-of-the-symmetric-groups/&quot; rel=&quot;nofollow&quot;&gt;this blog post&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It also has applications to representation theory.  If $X$ is a set on which a group $G$ acts, then the free vector space on $X$ is a representation $V$ of $G$ with character $\text{Fix}(g)$.  Burnside's lemma and the orthogonality relations then tell you that the dimension of the invariant subspace of $V$ is the number of orbits of the action of $G$ on $X$.  They also tell you that if $V$ decomposes as a direct sum $\oplus n_k V_k$ where the $V_k$ are irreducible, then $\sum n_k^2$ is the number of orbits of $G$ acting on $X \times X$.  In particular, if $G$ acts double transitively there are two such orbits, so $V$ is the sum of a trivial representation and an irreducible representation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(This application to representation theory, in turn, has applications to &lt;em&gt;graph theory.&lt;/em&gt;  See &lt;a href=&quot;http://qchu.wordpress.com/2011/01/02/the-schrodinger-equation-on-a-finite-graph/&quot; rel=&quot;nofollow&quot;&gt;this blog post&lt;/a&gt;.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit:  Here are some MO and math.SE answers where I have used Burnside's lemma:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://mathoverflow.net/questions/50253/can-this-nested-sum-be-expressed-in-terms-of-generalized-harmonic-numbers-and-the/50256#50256&quot; rel=&quot;nofollow&quot;&gt;http://mathoverflow.net/questions/50253/can-this-nested-sum-be-expressed-in-terms-of-generalized-harmonic-numbers-and-the/50256#50256&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://mathoverflow.net/questions/30112/elementary-combinatorial-identity-expressing-binomial-coefficients-as-an-alter/30114#30114&quot; rel=&quot;nofollow&quot;&gt;http://mathoverflow.net/questions/30112/elementary-combinatorial-identity-expressing-binomial-coefficients-as-an-alter/30114#30114&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://math.stackexchange.com/questions/5697/coloring-the-faces-of-a-hypercube/5731#5731&quot;&gt;Coloring the faces of a hypercube&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I want to point out one of the applications I mention in one of the above blog posts which I think is particularly &quot;funny&quot;: Fermat's little theorem!  Consider the cyclic group of order $p$ acting on the set of strings of length $p$ from an alphabet of size $a$.  By Burnside's lemma the total number of orbits is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{1}{p} \left( a^p + (p-1)a \right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;since there is one element which fixes every string and $p-1$ elements which only fix strings which repeat one letter $p$ times.  The integrality of this number is equivalent to Fermat's little theorem.  (For a generalization, see &lt;a href=&quot;http://qchu.wordpress.com/2009/08/23/newtons-sums-necklace-congruences-and-zeta-functions/&quot; rel=&quot;nofollow&quot;&gt;these&lt;/a&gt; &lt;a href=&quot;http://qchu.wordpress.com/2009/11/04/newtons-sums-necklace-congruences-and-zeta-functions-ii/&quot; rel=&quot;nofollow&quot;&gt;two&lt;/a&gt; blog posts.)&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2011-01-07T00:58:05.720" LastActivityDate="2011-01-07T00:58:05.720" />
  <row Id="16632" PostTypeId="2" ParentId="16625" CreationDate="2011-01-07T01:11:21.620" Score="0" ViewCount="" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Use $\rm\ x = 4^{\:\ell\ x}\ \ $ in $\rm\ \ (4^{\:\ell\ 3})^{\:\ell\ n}\ =\ (4^{\:\ell\ n})^{\:\ell\ 3}\quad$ where $\rm\quad \ell\ x\ :=\ log_{\:4}\ x$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Use $\rm\ n = e^{ln\ n}\:,\ \ $ take $\rm\ log_{\:3}\ $ of it  &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;See also the closely related prior questions &lt;a href=&quot;http://math.stackexchange.com/questions/5788/simple-dumb-logarithmic-conversion-question/5794#5794&quot;&gt;here&lt;/a&gt; and &lt;a href=&quot;http://math.stackexchange.com/questions/11458/if-x-log-12-27-text-then-what-is-the-the-value-of-log-6-16/11463#11463&quot;&gt;here.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2011-01-07T01:11:21.620" />
  <row Id="16633" PostTypeId="1" CreationDate="2011-01-07T01:13:21.337" Score="1" ViewCount="334" Body="&lt;p&gt;Do many mathematicians find that drinking a bit of alcohol helps them form new ideas? I mean this was true for Von Neumann right?&lt;/p&gt;&#xA;" OwnerUserId="5467" LastActivityDate="2011-01-07T01:13:21.337" ClosedDate="2011-01-07T01:15:04.120" Title="Drinking and Math" Tags="&lt;soft-question&gt;" CommentCount="2" FavoriteCount="1" />
  <row Id="16634" PostTypeId="1" AcceptedAnswerId="16635" CreationDate="2011-01-07T01:36:01.020" Score="2" ViewCount="223" Body="&lt;p&gt;Prove that three points are enough to draw/define one and only one circle, how would this be done?&lt;/p&gt;&#xA;" OwnerUserId="5157" LastActivityDate="2011-01-07T18:26:43.777" Title="Prove that three points are enough to draw/define one and only one circle" Tags="&lt;geometry&gt;" AnswerCount="2" />
  <row Id="16635" PostTypeId="2" ParentId="16634" CreationDate="2011-01-07T01:40:06.887" Score="10" ViewCount="" Body="&lt;p&gt;The set of points that are equidistant from two points $A,B$ is a straight line, right? Precisely, the line that goes through the midpoint of the segment $AB$ and is perpendicular to it. Let's call this line $l(A,B)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given 3 non-collinear points $A,B,C$, the lines $l(A,B)$ and $l(A,C)$ are not parallel, because the lines $AB$, $AC$ are not parallel, and therefore meet in exactly one point. This point is equidistant from $A,B,C$, and is therefore the &lt;em&gt;only&lt;/em&gt; such point. It is the center of the unique circle that goes through these three points. &lt;/p&gt;&#xA;" OwnerUserId="462" LastActivityDate="2011-01-07T01:40:06.887" CommentCount="3" />
  <row Id="16636" PostTypeId="2" ParentId="14789" CreationDate="2011-01-07T02:52:52.200" Score="1" ViewCount="" Body="&lt;p&gt;A proof of the quadratic convergence rate for Newton's method in 2 variables may be found in the book &lt;em&gt;Elements of Numerical Analysis&lt;/em&gt; by P. Henrici (J. Wiley, 1964).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also proved is the condition for this to apply to Bairstow's method.&lt;/p&gt;&#xA;" OwnerUserId="5469" LastActivityDate="2011-01-07T02:52:52.200" />
  <row Id="16637" PostTypeId="1" CreationDate="2011-01-07T02:58:41.930" Score="4" ViewCount="216" Body="&lt;p&gt;Let $X$ be a compact topological space and $Y$ a Hausdorff space. Let $C \subseteq Y$ be closed in $Y$ and $U$ an open set in $X \times Y$ which contains $X \times C$. Prove there exists an open set $V \subseteq Y$ such that $X \times C \subseteq X \times V \subseteq U$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's what I tried.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $b \in C$ fixed. For each $x \in X$ find open sets $U_{x}$ and $V_{x}$ in $X$, $Y$ respectively containing $x$ and $b$. Then the colecction $\{U_{x}: x \in X\}$ covers $X$ so by compactness of $X$ , we have $X \subseteq \bigcup_{i=1}^{n} U_{i}$. From here how to obtain the open set $V$ ?, if we take $V$ as the finite intersection of the V_{i} then this won't satisfy $X \times V \subseteq U$. Also I don't see where to use the hypothesis that $C$ is closed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't think the above approach works. Can you please help? Thanks.&lt;/p&gt;&#xA;" OwnerUserId="5470" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-07T03:29:39.533" LastActivityDate="2011-01-09T05:38:37.953" Title="Existence of open set in product topology" Tags="&lt;topology&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="16638" PostTypeId="1" CreationDate="2011-01-07T04:06:44.630" Score="1" ViewCount="123" Body="&lt;p&gt;I need to solve the following problem:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Suppose $f$ has the intermediate value property, i.e. if $f(a)&amp;lt;c&amp;lt;f(b)$, then there exists a value $d$ between $a$ and $b$ for which $f(d)=c$, and also has the additional property that $f^{-1}(a)$ is closed for every $a$ in a dense subset of $\mathbb{R}$, then $f$ is continuous.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can see plenty of counterexamples when the second property is not added, but I can't seem to bridge the gap between adding the property and proving $f$ is continuous. I can't get there either directly or by contradiction, because the additional property doesn't seem directly relevant to the property of continuity, so could anyone please tell me how to go about doing this? Thanks!&lt;/p&gt;&#xA;" OwnerUserId="5472" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-07T04:27:01.123" LastActivityDate="2011-01-07T04:27:44.917" Title="A converse of sorts to the intermediate value theorem, with an additional property" Tags="&lt;real-analysis&gt;&lt;analysis&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="16639" PostTypeId="2" ParentId="16638" CreationDate="2011-01-07T04:27:44.917" Score="2" ViewCount="" Body="&lt;p&gt;Suppose $x_n$ tends to $x$, but $f(x_n)$ does not tend to $f(x)$. Then there is a neighbourhood $(f(x)-\epsilon, f(x)+\epsilon)$ which $f(x_n)$ misses infinitely often. WLOG $f(x_n)&amp;gt;f(x)+\epsilon$ for infinitely many $n$. Hence $f(y)&amp;gt;f(x)+\epsilon$ for $y$ arbitrarily close to $x$. But then, for any $a$ in $(f(x),f(x)+\epsilon)$, the intermediate value property tells us that $f$ takes the value $a$ at points arbitrarily close to $x$. By our other assumption, at least one such $a$ has closed preimage. Contradiction.&lt;/p&gt;&#xA;" OwnerUserId="5203" LastActivityDate="2011-01-07T04:27:44.917" />
  <row Id="16640" PostTypeId="1" AcceptedAnswerId="16688" CreationDate="2011-01-07T04:33:10.027" Score="4" ViewCount="206" Body="&lt;p&gt;Hello all, I'm wondering about the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $f:D \mapsto D$ be a continuous real-valued bijection from the open unit disc $\{(x,y): x^2 + y^2 &amp;lt;1\}$ to itself. Does f necessarily have a fixed point?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am aware that without the bijective property, it is not necessarily true - indeed, I have constructed a counterexample without any trouble. However, I suspect with bijectivity it may be the case. I'm aware of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Brouwer_fixed_point_theorem&quot; rel=&quot;nofollow&quot;&gt;Brouwer Fixed Point Theorem&lt;/a&gt; and I imagine these two are intricately linked. However, i'm not certain where the bijectivity comes in - I believe we can argue something along the lines f now necessarily maps boundary to boundary - something about how if $x^2+y^2 \to 1$, $\|f(x,y)\| \to 1$ maybe. However, how does this help? Even if we could definitely define a limit to f(x,y) along the whole boundary and apply Brouwer, we can't guarantee the fixed points aren't all on the boundary anyway. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conversely however, I still can't construct a counterexample. Could anyone help me finish this off please? Thanks!&lt;/p&gt;&#xA;" OwnerUserId="5473" LastActivityDate="2011-01-07T15:47:33.150" Title="Continuous bijections from the open unit disc to itself - existence of fixed points" Tags="&lt;real-analysis&gt;&lt;homework&gt;&lt;analysis&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="16641" PostTypeId="1" CreationDate="2011-01-07T04:40:04.217" Score="4" ViewCount="266" Body="&lt;p&gt;This question was posed by my high school geometry teacher, for extra credit: Is it possible, using &#xA;only a pencil, a starightedge (not a ruler) and a compass to trisect an angle of unknown value? If &#xA;so, describe the technique, and mathematically prove its validity. If not, mathematically prove the &#xA;impossibility. I recall wondering at the time, why he was posing this question to a freshman high school geometry class, and did he genuinely expect any of us to come up with an answer. Nothing more was said about it and to my knowledge, nobody submitted an answer. But the question stuck in my mind, and I would often spend hours puzzling over it, trying over and over to come up with a valid solution. I'll be 60 years old this July, and I'm no closer to an answer now than I was that dsy, so many years ago. If anyone knows the answer, I'd appreciate a simple explanation, as I am no mathematician.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ed3&lt;/p&gt;&#xA;" OwnerUserId="4642" LastActivityDate="2011-01-07T05:23:36.423" Title="Trisect unknown angle using pencil, straight edge &amp; compass; Prove validity of technique" Tags="&lt;geometry&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" />
  <row Id="16642" PostTypeId="2" ParentId="16556" CreationDate="2011-01-07T05:13:47.870" Score="3" ViewCount="" Body="&lt;p&gt;Here's an overkill argument: According to what is usually called the Orlicz-Pettis theorem, a series in a normed space is weakly subseries convergent (if and) only if it is subseries convergent in the norm. Your series clearly isn't norm-convergent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More seriously: Note that if there is a weak limit of a sequence in $c_{0}$, it also is its weak$^{\ast}$-limit in $\ell^{\infty} = (c_{0})^{\ast\ast}$ (here $c_{0}$ is viewed as a subspace of $\ell^{\infty}$ via the canonical inclusion). Since the weak$^{\ast}$-topology is Hausdorff, it suffices to exhibit a subseries which is weak$^{\ast}$-convergent to some $s \in \ell^{\infty} \smallsetminus c_{0}$. The subseries $s_{k} = \sum_{n = 1}^{k} x_{2n} = (-1,1,-1,1,\cdots,-1,1,0,0,0,\cdots)$ weak$^{\ast}$-converges to $s = (-1,+1,\cdots)$: For all $(y_{n}) \in \ell^{1}$ and all $\varepsilon &amp;gt; 0$ there is $N$ such that $\sum_{n \geq N} |y_{n}| &amp;lt; \varepsilon$, therefore&#xA;\[&#xA;|\langle s - s_{k}, (y_{n}) \rangle_{\ell^{\infty}, \ell^{1}}|&amp;lt; \varepsilon&#xA;\quad \text{for $k \geq N$}&#xA;\]&#xA;by the Hölder inequality.&lt;/p&gt;&#xA;" OwnerUserId="5363" LastActivityDate="2011-01-07T05:13:47.870" CommentCount="7" />
  <row Id="16643" PostTypeId="2" ParentId="16641" CreationDate="2011-01-07T05:18:34.290" Score="11" ViewCount="" Body="&lt;p&gt;The question, given the &quot;unknown value&quot;, is equivalent to asking whether it is possible to trisect a general angle using only straightedge and compass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is known as the problem of &lt;a href=&quot;http://en.wikipedia.org/wiki/Angle_trisection&quot; rel=&quot;nofollow&quot;&gt;&lt;strong&gt;trisecting the angle&lt;/strong&gt;&lt;/a&gt;, one of the three classic problems of greek geometry, together with the problem of &lt;a href=&quot;http://en.wikipedia.org/wiki/Doubling_the_cube&quot; rel=&quot;nofollow&quot;&gt;&lt;strong&gt;duplicating the cube&lt;/strong&gt;&lt;/a&gt; (given a line segment, construct using only compass and straightedge a second segment such that the cube with edge length the second segment has double the volume of the cube with edge length the original line segment) and &lt;a href=&quot;http://en.wikipedia.org/wiki/Squaring_the_circle&quot; rel=&quot;nofollow&quot;&gt;&lt;strong&gt;squaring the circle&lt;/strong&gt;&lt;/a&gt; (given a line segment, construct a second line segment using only straightedge and compass whose square has the same area as a circle with radius the original line segment). The use of &quot;pencil&quot; is implicit in the statements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The three problems are also known as the &lt;strong&gt;three classical impossibilities&lt;/strong&gt;, because it is impossible to come up with such a technique. This does not mean nobody has been able to find one, but that it has been &lt;strong&gt;proven&lt;/strong&gt; that no such techniques exist. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Below, everytime I say &quot;construct&quot;, I mean &quot;using only pencil, a straightedge, and a compass&quot;.  The straightedge lets you extend any line segment, and construct a line segment through two given points; the compass lets you draw a circle with center a given point and going through another given point. You can determine points by intersecting things you can draw.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Think in terms of cartesian coordinates, place a point in the plane which will be the origin, and assume that you can mark off a distance of $1$ (that is, construct a line segment of length $1$). If you consider all line segments with one end at the origin that you can construct, the collection of all numbers $(a,b)$ that are the &lt;em&gt;other&lt;/em&gt; endpoint of these &quot;constructible line segments&quot; form what are called the &quot;constructible numbers.&quot; It is easy to see that if you have two constructible numbers, you can also construct their sum, difference, product, and quotient, just using a straightedge. A little harder is to show that if you have a constructible number $a\gt 0$, then you can also find its square root $\sqrt{a}$ if you also use the compass (the idea is that the compass will let you construct a curve that is given by a quadratic equation, the straightedge will let you construct a curve that is given by a linear equation, and playing one against the other you can obtain square roots). You can then repeat this process and obtain yet more constructible numbers, and more, and more, using these operations (addition, subtraction, multiplication, division, and taking square roots of nonnegative numbers already obtained).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Turns out that these are the &lt;em&gt;only&lt;/em&gt; numbers you can construct using straightedge and compass. So, for example, one cannot construct the number $\sqrt[3]{2}$, because it cannot be obtained from $0$ and $1$ through a sequence of additions, subtractions, multiplications, divisions, and taking square roots (this is not perhaps obvious; it is nonetheless true; it has to do with the fact that the polynomial with rational coefficients, smallest degree, and leading coefficient $1$ which has $\sqrt[3]{2}$ as a root is $x^3-2$, which is of degree $3$; if it were constructible, the smallest such polynomial would have to have degree a power of $2$). Since this is the length you would need to duplicate the cube if you are given a segment of length $1$, which is constructible, duplicating the cube is in general impossible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Squaring the circle is likewise impossible, because if you are given a line segment of length $1$, then you would need to construct a line segment of length $\sqrt{\pi}$. If you could, then you would also be able to construct $\pi$; but $\pi$ is a transcendental number (proven by von Lindermann in 1882), which means it cannot be constructed by the processes described above. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dealing with the trisection of the angle is a bit more involved, but not much. If you can construct an angle of $\alpha$ radians, then you can construct a right triangle with angle $\theta$, and with that triangle you can construct a segment of length $\cos(\alpha)$. Conversely, if you can construct a segment of length $\cos(\alpha)$, and then you draw a perpendicular to it of length $\sqrt{1-\cos^2(\alpha)}$ (which can be done), then you get a right triangle which has an angle of $\alpha$ radians.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, since we can construct equilateral triangles using straightedge and compass, and the inner angle of an equilateral triangle is $\frac{\pi}{3}$ radians, then the angle of $\frac{\pi}{3}$ is constructible, so someone could possibly &quot;give you&quot; that angle to trisect. In order to be able to trisect this angle, you need to be able to construct an angle of $\frac{\pi}{9}$ radians (20 degrees). So if trisecting the angle were possible, you would be able to construct a line segment of length $\cos(20^{\circ})=\beta$. However, using the triple angle formula&#xA;$$\cos(3\theta) = 4\cos^3(\theta) - 3\cos(\theta)$$&#xA;we have that $4\beta^3 - 3\beta - \frac{1}{2} = 0$ (set $\theta=\frac{\pi}{9}$ in the above), so that the polynomial with integer coefficients, of smallest degree, and leading coefficient $1$, that has $\cos(20^{\circ})$ as a root is $x^3 - \frac{3}{4}x - \frac{1}{8}$.  As with duplication of the cube, because the degree of this polynomial is not a power of $2$, the roots cannot be constructed at all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means that you cannot even construct an angle of $20^{\circ}$, which means that you would never be able to trisect an angle of $60^{\circ}$; in particular, there can be no general method for trisecting a given angle (since you &lt;em&gt;can&lt;/em&gt; be given an angle of $60^{\circ}$, but you cannot even construct the angle of $20^{\circ}$, even &lt;em&gt;knowing&lt;/em&gt; that this is the angle you are supposed to construct). So it is also impossible to trisect the angle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you allow yourself other tools, such as a ruler instead of a straightedge, then duplicating the cube and trisecting the angle become possible; it is also possible to do it with &quot;origami&quot; (by folding the paper you are given in different ways). But if you restrict yourself to the classic tools of straightedge and compass, it cannot be done.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-07T05:23:36.423" LastActivityDate="2011-01-07T05:23:36.423" />
  <row Id="16644" PostTypeId="2" ParentId="16604" CreationDate="2011-01-07T05:49:42.643" Score="5" ViewCount="" Body="&lt;p&gt;A profinite morphism presumably means something like a projective limit of finite morphisms.  Since finite morphisms are affine, we can think what this would mean in terms of maps of affine schemes.  An inductive limit of finite maps&#xA;$R \to S_i$ is the same as a map $R \to S$ where $S$ is integral over $R$,&#xA;so this leads me to guess that a profinite morphism is a morphism $X \to Y$&#xA;which is affine, and so that on affine opens Spec $R$ in $Y$, the preimage&#xA;in $X$ is of the form Spec $S$ with $S$ integral over $R$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is compatible with exercise 1 (b), and exercise 5, on the assignment that you link to.  Note, though, that as others have already pointed out, assuming that my guess is correct, this terminology is not standard, or at least not common.  More typically, such morphisms &#xA;are called &lt;em&gt;integral&lt;/em&gt;.  (This is the terminology used in Ravi Vakil's notes, and presumably also in the stacks project, and I also presume it is the terminology used in EGA.)&lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2011-01-07T05:49:42.643" />
  <row Id="16645" PostTypeId="1" CreationDate="2011-01-07T06:34:26.037" Score="0" ViewCount="67" Body="&lt;p&gt;I see some description of job requirment &quot;has quantative background&quot;, what does it mean?Thanks!&lt;/p&gt;&#xA;" OwnerUserId="1944" LastActivityDate="2011-01-07T06:34:26.037" ClosedDate="2011-01-07T13:01:58.100" Title="What is&quot;quantative background&quot;" Tags="&lt;soft-question&gt;" CommentCount="2" />
  <row Id="16646" PostTypeId="1" AcceptedAnswerId="16735" CreationDate="2011-01-07T07:13:06.143" Score="2" ViewCount="149" Body="&lt;p&gt;Let $X$ be riemann surface (not supposed compact) and $\mathcal D$ be sheaf of divisors on $X$.Remind that this means for $U\subset X$ open then $\mathcal D(U)$ is group of divisors on $U$. How to prove that $H^1(X,\mathcal D)=0$ . This is exercise in §16 of Forster book Riemann Surfaces (but is not homework for me) and he says hint is to use discontinuous partitions of unit but I don't understand. Any other method for proving is for me also satisfying . I suppose $H^2(X,\mathcal D)$ is also zero, no?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Addedd: Problem is in analytic category, not algebraic ( where is trivial : $\mathcal D$ is flabby). So I don't understand Matt E's proof (but thank you very much for answer Matt) because I don't know if cohomology commutes with co-limits in non noetherian case.&lt;/p&gt;&#xA;" OwnerUserId="2872" LastEditorUserId="2872" LastEditorDisplayName="" LastEditDate="2011-01-08T10:36:02.723" LastActivityDate="2011-01-08T10:36:02.723" Title="Why is first cohomology group of divisor sheaf on riemann surface zero?" Tags="&lt;algebraic-geometry&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" />
  <row Id="16647" PostTypeId="2" ParentId="16637" CreationDate="2011-01-07T07:17:19.293" Score="1" ViewCount="" Body="&lt;p&gt;Closedness of $C$ is irrelevant. Here's an argument:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Lemma.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Let $f: X \to Y$ be a (not necessarily continuous) function between topological spaces which is closed, &lt;em&gt;i.e.&lt;/em&gt;, maps closed sets to closed sets. If $A \subset Y$ is arbitrary and $U \subset X$ is open such that $f^{-1}(A) \subset U$ then there is $V \supset A$ open such that $f^{-1}(V) \subset U$.&lt;/li&gt;&#xA;&lt;li&gt;If $X$ is compact and $Y$ is Hausdorff then the projection $\pi: X \times Y \to Y$ is closed.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Since $\pi^{-1}(C) = X \times C \subset U$, combining 1. and 2. with $A = C$ yields $V \supset C$ open such that $\pi^{-1}(V) = X \times V \subset U$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Proof of 1:&lt;/em&gt; Since $f$ is closed, the set $V = Y \smallsetminus f(X \smallsetminus U)$ is open. Since $f^{-1}(A) \subset U$ we have $A \subset V$. Moreover, $X \smallsetminus U \subset f^{-1}(f(X \smallsetminus U))$ yields $f^{-1}(V) = X \smallsetminus f^{-1}(f(X \smallsetminus U)) \subset X \smallsetminus (X \smallsetminus U) = U$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Proof of 2:&lt;/em&gt; Let $F \subset X \times Y$ be closed and let $y \in Y \smallsetminus \pi(F)$ be arbitrary. Observe that $F \cap (X \times \{y\}) = \emptyset$. Since $(X \times Y) \smallsetminus F$ is open and contains $X \times \{y\}$, the definition of the product topology yields for each $x \in X$ open sets $U_{x} \subset X$ and $V_{x} \subset Y$ such that $(x,y) \in U_{x} \times V_{x} \subset (X \times Y) \smallsetminus F$. Since $X \times \{y\}$ is compact, there are $x_{1},\ldots,x_{n}$ such that $X \times \{y\} \subset (U_{x_{1}} \times V_{1}) \cup \cdots \cup (U_{x_{n}} \times V_{x_{n}}) =: W$. By construction $W \subset (X \times Y) \smallsetminus F$ and the open set $V := V_{x_{1}} \cap \cdots \cap V_{x_{n}}$ contains $y$ and satisfies $\pi^{-1}(V) \subset W$, hence $V \cap \pi(F) = \emptyset$, hence $Y \smallsetminus \pi(F)$ is open.&lt;/p&gt;&#xA;" OwnerUserId="5363" LastActivityDate="2011-01-07T07:17:19.293" />
  <row Id="16648" PostTypeId="2" ParentId="815" CreationDate="2011-01-07T07:45:44.773" Score="0" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Grigori_Perelman&quot; rel=&quot;nofollow&quot;&gt;Perelman:&lt;/a&gt;&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;He is a truth seeker and does math for the beauty that lies underneath.&#xA;Its not about Poincare conjecture, but his philosophy and approach towards things in life.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similar other figures I consider as having influenced me are Leibniz, Einstein, Russell, Grothendieck, etc. Although not all of them were mathematicians!!!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are infact many others. But the name that comes to my mind right now, is Perelman.&#xA;As Jesus is to religion, so is Perelman w.r.t to mathematics.&lt;/p&gt;&#xA;" OwnerUserId="2320" LastActivityDate="2011-01-07T07:45:44.773" CommunityOwnedDate="2011-01-07T07:45:44.773" />
  <row Id="16649" PostTypeId="1" CreationDate="2011-01-07T08:24:32.227" Score="3" ViewCount="166" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Possible Duplicate:&lt;/strong&gt;&lt;br&gt;&#xA;  &lt;a href=&quot;http://math.stackexchange.com/questions/8186/when-do-the-multiples-of-two-primes-span-all-large-enough-natural-numbers&quot;&gt;When do the multiples of two primes span all large enough natural numbers? &lt;/a&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;p&gt;We have to generate a number by using only 3's and 5's.&lt;br&gt;&#xA;For ex :   &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;3 = 3&#xA;4 can't be generated&#xA;5 = 5&#xA;6 = 3 + 3&#xA;7 can't be generated&#xA;8 = 3 + 5&#xA;9 = 3 + 3 + 3&#xA;10 = 5 + 5&#xA;11 = 3 + 3 + 5&#xA;and so on...  &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;So from 8, every integer i.e. 9,10,11,12,.. can be generated by using 3's and 5's.&lt;br&gt;&#xA;But I can't prove it. So   &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;    (i) Is there a way to prove that every integer after 7, can be generated  &#xA;        by using only 3's and 5's ?&#xA;    (ii)Are there any other pairs of numbers (as 3 and 5 here), which satisfy  &#xA;        this property that after certain integer, all integers can be generated  &#xA;        by using number only from that pair ?&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="4814" LastActivityDate="2011-01-08T13:59:10.397" ClosedDate="2011-01-07T13:26:44.543" Title="Numbers generated from combination of 3's and 5's" Tags="&lt;number-theory&gt;&lt;combinatorics&gt;" AnswerCount="0" CommentCount="4" />
  <row Id="16650" PostTypeId="1" CreationDate="2011-01-07T08:26:46.787" Score="2" ViewCount="168" Body="&lt;p&gt;Hi everyone,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm working on a problem that requires me to find an upper bound on the probability that the sum of independent draws from a random variable deviates from the expected value of that sum by more than a given constant. Specifically, let $X$ be a random variable and suppose that we draw $m$ values for $X$. Let $S$ be the sum of those draws: $S=\sum_{i=1}^m X_i$, where $X_i$ is the $i$-th draw from $X$. This sum has expected value $E[S]$. If we assume that $X$'s values are always in the interval of $[a, b]$, one could try to find an upper bound on the probability that the sum of the draws deviates from the expected value of the sum by more than $t$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$P(S - E[S] &amp;gt; t)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hoeffding's inequality tells us that an upper bound for this probability is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\exp\bigg(  \frac{-2t^2}{\sum_{i=1}^m (a - b)^2}  \bigg)$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem that I have requires me to find an upper bound on the probability that $k$ times the sum of draws deviates from the expected value by more than $t$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$P(kS - E[S] &amp;gt; t)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $k$ is a constant. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems that it should be easy to find an upper bound for this probability, but I'm kind of stuck: I've tried simple algebraic manipulations in order to try to get rid of the $k$ an transform that probability into something that would allow me to use Hoeffding's bound; I also took a look at other bounds, like the Bernstein inequalities, but nothing seems quite right.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anybody have an idea for a bound for this type of probability? I feel that the answer is right in front of me but I can't see it...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance!&lt;/p&gt;&#xA;" OwnerUserId="2368" LastActivityDate="2011-02-01T11:54:50.550" Title="Bounding probability that sum of random variables deviates from expected value" Tags="&lt;probability&gt;&lt;probability-distributions&gt;" AnswerCount="3" />
  <row Id="16651" PostTypeId="2" ParentId="16403" CreationDate="2011-01-07T08:53:36.863" Score="2" ViewCount="" Body="&lt;p&gt;The problem is this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have three inhabitants A, B and C. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;B said &quot;A said &quot;A is a knave&quot;&quot;, and C said &quot;B is a knave&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the information we have is (if you read the first paragraph of the chapter I referred to in my previous answer and carefully follow the instructions):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$k_B \leftrightarrow (k_A \leftrightarrow \neg k_A)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$k_C \leftrightarrow \neg k_B$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your truth table for both formulas (as it is now written) is correct, you simply used A instead of $k_A$, but that's all. To know which are the possible situations under the givens, you just look for the rows that give simultaneously a value 1 for both formulas. These are the two rows with a 1 in the last column of your table.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What the table is telling you, then, is that there are only two possible situations. In both of them B is a knave and C is a knight. So that's true no matter the situation. Instead, you cannot tell what A is because A can either be a knight or a knave.&lt;/p&gt;&#xA;" OwnerUserId="5405" LastEditorUserId="5405" LastEditorDisplayName="" LastEditDate="2011-01-07T08:59:39.590" LastActivityDate="2011-01-07T08:59:39.590" />
  <row Id="16652" PostTypeId="2" ParentId="16650" CreationDate="2011-01-07T08:57:27.200" Score="2" ViewCount="" Body="&lt;p&gt;Are you looking for something like this:&#xA;$$&#xD;&#xA;{\rm P}[kS - {\rm E}(S) &amp;gt; t] = {\rm P}[kS - {\rm E}(kS) &amp;gt; t - (k - 1){\rm E}(S)].&#xD;&#xA;$$&#xA;Also not that $kS=kX_1 + \cdots + kX_m$ is a sum of $m$ i.i.d. random variables.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;: Assuming, from a practical point of view, that ${\rm E}(X)$ is known (and $X \in [a,b]$), then, by Hoeffding's inequality, the above equation gives&#xA;$$&#xD;&#xA;{\rm P}[kS - {\rm E}(S) \geq t] \leq \exp \bigg( - \frac{{2[t - (k - 1)m{\rm E}(X)]^2 }}{{mk^2 (b - a)^2 }}\bigg),&#xD;&#xA;$$&#xA;provided that $t - (k - 1)m{\rm E}(X) &amp;gt; 0$ (note that $kX$ belongs to the interval $[ka,kb]$, whose length is $k(b-a)$).&#xA;If, on the other hand, ${\rm E}(X)$ is not known, then nothing useful is likely to be achieved.&lt;/p&gt;&#xA;" OwnerUserId="2810" LastEditorUserId="2810" LastEditorDisplayName="" LastEditDate="2011-01-09T21:48:41.750" LastActivityDate="2011-01-09T21:48:41.750" CommentCount="8" />
  <row Id="16653" PostTypeId="1" AcceptedAnswerId="16654" CreationDate="2011-01-07T09:23:09.230" Score="0" ViewCount="193" Body="&lt;p&gt;Somebody divided  (cx+d)/(ax+b) into c/a+(d-bc/a)/(ax+b). For use for integrals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anybody knows how was it done? Could you show me how to do something like that, please?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance.&lt;/p&gt;&#xA;" OwnerUserId="1755" LastActivityDate="2011-01-07T17:33:21.757" Title="How to divide it. (cx+d)/(ax+b)" Tags="&lt;algebra&gt;" AnswerCount="2" />
  <row Id="16654" PostTypeId="2" ParentId="16653" CreationDate="2011-01-07T09:46:18.273" Score="5" ViewCount="" Body="&lt;p&gt;It's &lt;a href=&quot;http://en.wikipedia.org/wiki/Polynomial_long_division&quot; rel=&quot;nofollow&quot;&gt;polynomial long division&lt;/a&gt;. In this relatively simple case (first degree polynomials in both numerator and denominator) you can also do it like this, using an &quot;add-and-subtract&quot; trick:&#xA;$$ \frac{cx+d}{ax+b} = \frac{c(x + d/c)}{a(x+b/a)} = \frac{c}{a} \frac{(x+b/a)+(d/c-b/a)}{x+b/a} = \frac{c}{a} \left( 1 + \frac{d/c-b/a}{x+b/a} \right).$$&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2011-01-07T09:46:18.273" />
  <row Id="16655" PostTypeId="1" AcceptedAnswerId="16662" CreationDate="2011-01-07T09:53:37.817" Score="5" ViewCount="109" Body="&lt;p&gt;$$\frac{{{\varphi ^n} - {{(1 - \varphi )}^n}}}{{\sqrt 5 }} = {2^m} - 1$$&#xA;here $\varphi  = \frac{{1 + \sqrt 5 }}{2}$ . This integer equation has no solution for n&gt;3 and m&gt;2.&#xA;how to prove?&lt;/p&gt;&#xA;" OwnerUserId="1498" LastEditorUserId="1498" LastEditorDisplayName="" LastEditDate="2011-01-07T10:19:34.993" LastActivityDate="2011-01-07T11:03:31.420" Title="Prove: the intersection of fibonacci sequence and Mersenne sequence is just {1,3}" Tags="&lt;number&gt;&lt;fibonacci-number&gt;" AnswerCount="1" FavoriteCount="2" />
  <row Id="16656" PostTypeId="2" ParentId="16554" CreationDate="2011-01-07T09:59:40.757" Score="1" ViewCount="" Body="&lt;p&gt;It is possible to get a good approximate solution to your equation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \left( 1 + \frac{r}{100} \right)^5 = 1.6 $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;by hand if we expand the fifth power and ignore terms of order higher than $(r/100)^2,$ since we know that $r/100$ is small. Thus we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ 1 + \frac{5r}{100} + \frac{10r^2}{10000} = 1.6,$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;from which we get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$r^2 + 50r - 600 = 0 \quad \text{or} \quad (r-10)(r+60)=0,$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and so $r=10.$&lt;/p&gt;&#xA;" OwnerUserId="1301" LastActivityDate="2011-01-07T09:59:40.757" />
  <row Id="16658" PostTypeId="1" AcceptedAnswerId="16660" CreationDate="2011-01-07T10:31:47.640" Score="2" ViewCount="198" Body="&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/9K2A1.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am reading &lt;a href=&quot;http://www.cse.iitk.ac.in/users/sigtacs/references/S02E03-fourier.pdf&quot; rel=&quot;nofollow&quot;&gt;this document&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the meaning of $[n]$ ?&#xA;Is it power set of $\{1,2,3...n\}$?&lt;/p&gt;&#xA;" OwnerUserId="705" LastActivityDate="2011-03-08T13:11:28.303" Title="What does [n] mean here?" Tags="&lt;fourier-analysis&gt;&lt;notation&gt;&lt;boolean-algebra&gt;" AnswerCount="2" />
  <row Id="16659" PostTypeId="2" ParentId="16658" CreationDate="2011-01-07T10:53:44.170" Score="0" ViewCount="" Body="&lt;p&gt;It's probably the set $\{ 1,2,\ldots,n\}$.&lt;/p&gt;&#xA;" OwnerUserId="589" LastActivityDate="2011-01-07T10:53:44.170" />
  <row Id="16660" PostTypeId="2" ParentId="16658" CreationDate="2011-01-07T10:55:18.807" Score="9" ViewCount="" Body="&lt;p&gt;No, it's just the set $\{ 1,2,\dots,n \}$. If you look at the notation carefully, you see that $S$ is written a &lt;em&gt;subset&lt;/em&gt; of that set; if they meant the powerset they would have written $S$ as an &lt;em&gt;element&lt;/em&gt; of the powerset.&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2011-01-07T10:55:18.807" CommentCount="1" />
  <row Id="16662" PostTypeId="2" ParentId="16655" CreationDate="2011-01-07T11:03:31.420" Score="6" ViewCount="" Body="&lt;p&gt;We need to find when $F_n+1$ is a power of 2.  Almost every value of $n$ can be eliminated by considering the &lt;a href=&quot;http://en.wikipedia.org/wiki/Pisano_period&quot; rel=&quot;nofollow&quot;&gt;Pisano period&lt;/a&gt;.  In particular, we can deduce that:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$F_n+1 \equiv 0 \pmod {16}$ if and only if $n \equiv 22 \pmod {24}$ and&lt;/li&gt;&#xA;&lt;li&gt;$F_n+1 \equiv 0 \pmod 9$ if $n \equiv 22 \pmod {24}$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This leaves the few small cases already listed.&lt;/p&gt;&#xA;" OwnerUserId="139" LastActivityDate="2011-01-07T11:03:31.420" CommentCount="6" />
  <row Id="16663" PostTypeId="1" AcceptedAnswerId="16666" CreationDate="2011-01-07T11:42:08.727" Score="5" ViewCount="345" Body="&lt;p&gt;Hello altogether&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Doing an exercise for exam preparation, I stumbled across the following function:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f_n(x)=  n^2x(1-nx),  \quad \text{if }0 \leq x \leq \frac{1}{n} $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f_n(x)=  0, \quad \text{if } \frac{1}{n} &amp;lt; x \leq 1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The task is to find the limit of this function series and to determine whether this function converges uniformly in $[0,1]$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the one hand $\frac{1}{n}$ approaches $0$ for $n \to \infty$. So one would just have to insert $0$ in $n^2x(1-nx)$. Thereby gaining $f_n(x) = 0$ for $n \to \infty$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand the function has a maximum for $x=\frac{n}{2}$. Putting this into $n^2x(1-nx)$ and calculating $f_n(x)$ for $n \to \infty$ afterwards one gets $f_n(x) = \infty$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So whats correct? How does one approach such a problem?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ftiaronsem&lt;/p&gt;&#xA;" OwnerUserId="3188" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2011-01-07T21:57:45.740" LastActivityDate="2011-01-08T20:51:28.160" Title="Convergence of sequence of functions, $f_n(x) = n^2 x(1-nx) \dots $" Tags="&lt;real-analysis&gt;&lt;analysis&gt;&lt;sequences-and-series&gt;" AnswerCount="3" CommentCount="6" />
  <row Id="16665" PostTypeId="1" CreationDate="2011-01-07T12:17:16.490" Score="6" ViewCount="344" Body="&lt;p&gt;I read that Presburger arithmetic is decidable while Peano arithmetic is undecidable, and actually Peano arithmaetic extends Presburger arithmetic just with the addition of the multiplication operator. Can someone please give me 'intutive' idea behind this? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or probably a formula in Peano arithmetic that can not be proved. Does it has something to do with self reference paradox in Goedel's Incompleteness Theorem?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regards&lt;/p&gt;&#xA;" OwnerUserId="5318" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-07T18:58:55.847" LastActivityDate="2011-01-07T20:04:25.203" Title="Why is Peano arithmetic undecidable?" Tags="&lt;number-theory&gt;&lt;logic&gt;" AnswerCount="3" CommentCount="4" FavoriteCount="2" />
  <row Id="16666" PostTypeId="2" ParentId="16663" CreationDate="2011-01-07T12:28:32.917" Score="5" ViewCount="" Body="&lt;p&gt;I'd strongly encourage you to draw a picture of the graph of $f_{n}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your argument that $f_{n} \to 0$ is not quite correct. I'd argue as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have $f_{n}(x) \to 0$ as $(n \to \infty)$ for all $x \in [0,1]$. This is clear for $x = 0$ and for $x &amp;gt; 0$ we have $f_{n}(x) = 0$ for all $n$ so large that $\frac{1}{n} &amp;lt; x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $f_{n} \to f$ uniformly on $[0,1]$ then $f_{n} \to f$ pointwise, hence we must have $f = 0$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, the function $f_{n}$ has a maximum at $\frac{1}{2n}$ (&lt;em&gt;not&lt;/em&gt; $\frac{2}{n}$ as you've written in your question) as can be found by differentiation, for example. Evaluation gives&#xA;\[&#xA;f_{n}(\frac{1}{2n}) = n^{2} \frac{1}{2n} ( 1 - n\frac{1}{2n}) = \frac{n}{4}&#xA;\]&#xA;Therefore&#xA;\[&#xA;\sup_{x \in [0,1]} |f_{n}(x) - f(x)| = \sup_{x \in [0,1]} |f_{n}(x)| = \frac{n}{4} \xrightarrow{n \to \infty} \infty&#xA;\]&#xA;and hence $f_{n}$ does not converge uniformly to $f = 0$.&lt;/p&gt;&#xA;" OwnerUserId="5363" LastActivityDate="2011-01-07T12:28:32.917" CommentCount="14" />
  <row Id="16667" PostTypeId="2" ParentId="16663" CreationDate="2011-01-07T12:29:08.933" Score="3" ViewCount="" Body="&lt;p&gt;The limit function is $f(x) = 0 \quad \forall x \in [0,1]$ because, as $n \rightarrow \infty$ you have that the region where the sequence is $n^2 x(1-nx)$ is always smaller and smaller (this is a nice way of approaching sequences with boundary conditions that involve $n$). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for uniform convergence, you should take the supremum for $x \in [0,1]$ and since the function has a maxmimum inside the interval, you can say that $\sup_{x \in [0,1]}\left| n^2 x(1- n x)\right| = \frac{n}{4}$, and if $n \rightarrow \infty$ it doesn't approach zero, so the convergence is not uniform.&lt;/p&gt;&#xA;" OwnerUserId="1415" LastEditorUserId="1415" LastEditorDisplayName="" LastEditDate="2011-01-07T18:28:04.307" LastActivityDate="2011-01-07T18:28:04.307" CommentCount="4" />
  <row Id="16668" PostTypeId="2" ParentId="16568" CreationDate="2011-01-07T12:31:21.057" Score="0" ViewCount="" Body="&lt;p&gt;I can't see the direction you say Mariano. Can you give me some other hint?&lt;/p&gt;&#xA;" OwnerUserId="5480" LastActivityDate="2011-01-07T12:31:21.057" CommentCount="1" />
  <row Id="16669" PostTypeId="2" ParentId="16640" CreationDate="2011-01-07T13:14:05.170" Score="-2" ViewCount="" Body="&lt;p&gt;The following formulation of Brouwer Fixed Point Theorem comes from &quot;Algebraic Topology&quot; by Allen Hatcher (freely available at www.math.cornell.edu/~hatcher/AT/AT.pdf ) on page 31:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Theorem : Every continuous map $h:D^2 \to D^2$ has a fixed point, that is, a point $x$ with $h(x)=x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can find a proof on page 32 (about 15 lines).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This should answer your question.&lt;/p&gt;&#xA;" OwnerUserId="5482" LastActivityDate="2011-01-07T13:14:05.170" CommentCount="2" />
  <row Id="16670" PostTypeId="1" AcceptedAnswerId="16673" CreationDate="2011-01-07T13:29:33.553" Score="5" ViewCount="180" Body="&lt;p&gt;Do countable Hausdorff connected topological spaces exist?&lt;/p&gt;&#xA;" OwnerUserId="5422" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-07T18:55:31.560" LastActivityDate="2011-01-07T18:55:31.560" Title="Are there any countable Hausdorff connected spaces?" Tags="&lt;general-topology&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" />
  <row Id="16671" PostTypeId="2" ParentId="16559" CreationDate="2011-01-07T13:31:57.750" Score="1" ViewCount="" Body="&lt;p&gt;Dear community,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am answering my own question here since I hope that it can safe some time for somebody.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To compare my results with that of the paper I calculated a complex quantity called &lt;a href=&quot;http://en.wikipedia.org/wiki/Reflection_coefficient&quot; rel=&quot;nofollow&quot;&gt;reflection coefficient&lt;/a&gt; $r = |r|e^{\mathrm{i}\varphi}$. The &lt;strong&gt;absolute value&lt;/strong&gt; $|r|$ tells you roughly how much of some field gets back-reflected in some domain at the boundary to another one, depending on several parameters. Its &lt;strong&gt;phase&lt;/strong&gt; $\varphi$ is also of great importance since it will be useful for effects like &lt;a href=&quot;http://en.wikipedia.org/wiki/Interference_%28wave_propagation%29&quot; rel=&quot;nofollow&quot;&gt;interference&lt;/a&gt; for linear phenomena.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My results were simply different from those presented in the given paper since I did not care for the &lt;strong&gt;convention&lt;/strong&gt; used for the calculation of the phase. Taking the modulus of $-\varphi$ with respect to $\pi$, I was able to reconstruct all given data.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I &lt;strong&gt;apologize&lt;/strong&gt; for any inconvenience. Furthermore, an analytic calculation of the integral at hand is only feasible for the special case of $\Theta = 0$ (using &lt;em&gt;Mathematica&lt;/em&gt;) since the results blow up quickly.&lt;br&gt;&#xA;Greetings&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Robert&lt;/p&gt;&#xA;" OwnerUserId="2671" LastActivityDate="2011-01-07T13:31:57.750" />
  <row Id="16672" PostTypeId="2" ParentId="16670" CreationDate="2011-01-07T13:33:53.263" Score="2" ViewCount="" Body="&lt;p&gt;Trivially yes, a singleton for example ;).  Non-trivial examples abound, for example ''A countable connected Hausdorff space'' by Brown, in Bull. Amer. Math. Soc., 59 (1953) p. 367.&lt;/p&gt;&#xA;" OwnerUserId="4322" LastEditorUserId="4322" LastEditorDisplayName="" LastEditDate="2011-01-07T13:38:55.037" LastActivityDate="2011-01-07T13:38:55.037" CommentCount="6" />
  <row Id="16673" PostTypeId="2" ParentId="16670" CreationDate="2011-01-07T13:34:35.593" Score="4" ViewCount="" Body="&lt;p&gt;Yes, see &lt;a href=&quot;http://mathoverflow.net/questions/46986/countable-connected-hausdorff-space/46990#46990&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; (MathOverflow) for references to some non-trivial examples.&lt;/p&gt;&#xA;" OwnerUserId="5363" LastEditorUserId="1242" LastEditorDisplayName="" LastEditDate="2011-01-07T15:51:13.973" LastActivityDate="2011-01-07T15:51:13.973" CommentCount="3" />
  <row Id="16674" PostTypeId="1" AcceptedAnswerId="16681" CreationDate="2011-01-07T14:07:57.103" Score="3" ViewCount="123" Body="&lt;p&gt;How can I show that:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;There exists a bijective $f: \mathbb{N} \to \mathbb{N}$ such that: $$\sum_{n=1}^{\infty} (-1)^{f(n)}\ln\frac{f(n)+1}{f(n)}=\ln 2010$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have really no idea where to begin... Taylor series doesn't seem related. I thought about the alternating series test which approximates the sum but not sure how to use it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any help will be appreciated.&lt;/p&gt;&#xA;" OwnerUserId="3237" LastActivityDate="2011-01-07T15:10:03.603" Title="Show that there's a bijective $f: \mathbb{N} \to \mathbb{N}$ such that $\sum_{n=1}^{\infty} (-1)^{f(n)}\ln\frac{f(n)+1}{f(n)}=\ln 2010$" Tags="&lt;calculus&gt;&lt;homework&gt;&lt;sequences-and-series&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" />
  <row Id="16676" PostTypeId="1" AcceptedAnswerId="16678" CreationDate="2011-01-07T14:10:14.653" Score="0" ViewCount="114" Body="&lt;p&gt;I have a home made device (think baseball) with an accelerometer and a magnetometer embedded within it and would like to capture some details of an attempted 'throw' of the device.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All I need is the speed of the device at release, and the angle from the horizontal that it was released at. To be clear I'm not talking about the orientation of the device, but the angle of its movement relative to the horizontal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm also not after particularly accurate results there can be a pretty large margin of error here. My data starts at the beginning of the throw and ends at the release, and is composed of x,y,z accelerometer data in m/s/s and azimuth, pitch, yaw data in degrees retrieved from the magnetometer and accelerometer data combined.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've looked into this for a few days now but am afraid my maths is really rusty, and I have a horrible feeling that this isn't possible with just these two devices alone.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Am I right? Is this a lost cause without adding more accelerometers to my device?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is the situation improved if I can guarantee that the device is not rotated by more than a 90 degrees around each axis during movement?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any help you can offer would be much appreciated.&lt;/p&gt;&#xA;" OwnerUserId="5486" LastActivityDate="2011-01-07T14:20:38.580" Title="Is it possible to calculate an angle of movement from the horizontal from accelerometer and magnetometer data?" Tags="&lt;trigonometry&gt;&lt;physics&gt;" AnswerCount="1" />
  <row Id="16677" PostTypeId="2" ParentId="16403" CreationDate="2011-01-07T14:11:43.660" Score="0" ViewCount="" Body="&lt;p&gt;Your table is incorrect in the F1 column in that $A \vee B \vee C$ should evaluate to 0 when all three are  0 (the bottom line).  Otherwise your calculations are OK.  Edit:  this column has been removed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The definition of F1 is not what you want.  F1 is supposed to represent whether A spoke the truth, so should just be A.  Edit:  as this column has been removed, this does not apply.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit:  this is incorrect as I misread the table.  See the paragraph below.  The definition of G is the biggest error.  G should be $(A \leftrightarrow F1) \wedge (B \leftrightarrow F2) \wedge (C \leftrightarrow F3)$.  This is the heart of the matter.  You want A to have spoken the truth if and only if A is a knight, and the same for B and C.  G should always be of this form (maybe more terms if you have more individuals involved) and will pick out the lines where the truth value of the statements matches the type of the individuals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Added:  G is correct.  Your F1 says &quot;B spoke properly for his type&quot; and F2 says &quot;C spoke properly for his type&quot;.  So you want to find the cases both spoke properly.  The fact that the 1's appear opposite B=0, C=1 both times says that B is a knave and C is a knight.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2011-01-07T18:02:42.017" LastActivityDate="2011-01-07T18:02:42.017" CommentCount="5" />
  <row Id="16678" PostTypeId="2" ParentId="16676" CreationDate="2011-01-07T14:20:38.580" Score="0" ViewCount="" Body="&lt;p&gt;Assuming you know the local field where the experiment is taking place, the magnetometer can only orient the ball in two axes.  Rotating the ball around the field line will not change the reading at all.  Imagine the case where z is along the field line-you would be able to know that and integrate the z accelerometer (if the ball doesn't rotate) to get z velocity and position.  Again assuming the ball doesn't rotate, you could combine the x and y accelerometer readings to get perpendicular acceleration and integrate that to perpendicular velocity and position.  But you wouldn't be able to find the azimuthal direction from your data.  What you need is a starting measurement of azimuth and a rotational accelerometer if you want motion in 3 space.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-01-07T14:20:38.580" CommentCount="1" />
  <row Id="16679" PostTypeId="2" ParentId="16568" CreationDate="2011-01-07T14:26:33.310" Score="2" ViewCount="" Body="&lt;p&gt;Here is an argument (maybe the same as the one that Mariano sketched, but with a little more detail):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $S = R[[x_1,\ldots,x_n]]$, and let $T = R[[x_1,\ldots,x_n]]/(x_1-a_1,\ldots,&#xD;&#xA;x_n-a_n).$  The ring $S$ is Noetherian (since $R$ is), and is complete with respect to the ideal $(x_1,\ldots,x_n)$.  By Artin--Rees (for example) the finitely generated&#xA;$S$-module $T$ is also complete with respect to this ideal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus $T = $ proj. limit of $T/(x_1,\ldots,x_n)^i = R/(a_1,\ldots,a_n)^i.$&#xA;Thus $T$ is the completion of $R$ with respect to $(a_1,\ldots,a_n)$, as claimed.&lt;/p&gt;&#xA;" OwnerUserId="221" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2011-01-07T14:54:47.990" LastActivityDate="2011-01-07T14:54:47.990" />
  <row Id="16680" PostTypeId="1" AcceptedAnswerId="16683" CreationDate="2011-01-07T14:39:07.743" Score="0" ViewCount="77" Body="&lt;p&gt;Suppose I have a function $F:[0,1] \times [0,\infty) \rightarrow (0,1]$ of two variables $(s,t)$ that satisfies:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) $F(s,0)=1$ for each $s \in [0,1]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) $\lim_{t \rightarrow \infty} F(s,t)=0$ for all $s \in [0,1]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(3) For each fixed $s \in [0,1]$, the function $t\mapsto F(s,t)$ is strictly decreasing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the function $G:[0,1] \times [0,\infty) \rightarrow \mathbb{R}$ defined by $G(s,t)=\frac{\partial F}{\partial s}(s,t)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it necessarily the case that $\sup_{(s,t) \in [0,1] \times [0,\infty)} \left|G(s,t)\right| \lt \infty$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If not, how would I go about constructing a counter-example?&lt;/p&gt;&#xA;" OwnerUserId="5487" LastActivityDate="2011-01-07T14:54:31.797" Title="Proving the partial derivative of a strictly decreasing function is finite everywhere" Tags="&lt;real-analysis&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="16681" PostTypeId="2" ParentId="16674" CreationDate="2011-01-07T14:39:59.090" Score="9" ViewCount="" Body="&lt;p&gt;Suppose first you take $f(n)=2n$ then&#xA;$$\sum_{n=1}^{\infty} (-1)^{2n}\ln\frac{2n+1}{2n}= \sum_{n=1}^{\infty} \ln(2n+1)-\ln(2n) = \sum_{n=1}^{\infty} \frac {1}{2n+\delta_n} \geq \sum_{n=1}^{\infty} \frac {1}{2n+1} = \infty$$&#xA;where $0&amp;lt;\delta_n&amp;lt;1$ - this is Lagrange's theorem $\frac{f(x) - f(y)}{(x)-y)} = f&amp;#39; (c)$ for some $c\in (x, y)$, and here $x=2n+1,\; y=2n,\; f(t)=\ln(t)$ so $f&amp;#39;(t)=\frac{1}{t}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same works for $f(n)=2n+1$ but with $-\infty$. Now construct another function $f$ as follows : choose enough even numbers until the first time your partial sum is above $\ln 2100$, then choose enough odd numbers until the first time your partial sum is below $\ln 2100$. you can do this since the two sequences I described above converge to $\pm \infty$. you can continue like this to build a bijective map $f:\mathbb{N}\rightarrow \mathbb{N}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the $n-th$ term in the sum converges to zero, show that the sequence converges to $\ln 2100$. &lt;/p&gt;&#xA;" OwnerUserId="2125" LastEditorUserId="2125" LastEditorDisplayName="" LastEditDate="2011-01-07T15:10:03.603" LastActivityDate="2011-01-07T15:10:03.603" CommentCount="6" />
  <row Id="16682" PostTypeId="2" ParentId="815" CreationDate="2011-01-07T14:48:06.590" Score="2" ViewCount="" Body="&lt;h1&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Paul_Erd%C5%91s&quot; rel=&quot;nofollow&quot;&gt;Paul Erdos&lt;/a&gt;&lt;/h1&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&quot;I know numbers are beautiful. If they aren't beautiful, nothing is.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My favorite math teacher as an undergraduate was Hungarian, and he is the person who first turned me on to Erdos.  I remember the first time I checked out volume one of Erdos collected works from the library, it was like a had discovered a lost treasure trove!  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I just love the types of problems he worked on; he was amazingly prolific, and the stories about him and his personality that have survived make him seem like a really special person.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are many funny things about the way Erdos spoke; for example he called children &quot;epsilons&quot;, and he said that anyone who was married was &quot;captured&quot;!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Erdos famously said that God keeps a book of every theorem that will ever be discovered by man, but that there is only one proof in the book for every theorem!  These are the book proofs, and this idea has inspired me ever since I heard about, to always look for book proofs whenever possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Erdos was so prolific, that every mathematician has an &lt;a href=&quot;http://en.wikipedia.org/wiki/Erdos_number&quot; rel=&quot;nofollow&quot;&gt;Erdos number&lt;/a&gt;, which &quot;describes the 'collaborative distance' between a person and mathematician Paul Erdos, as measured by authorship of mathematical papers.&quot;&lt;/p&gt;&#xA;" OwnerUserId="1417" LastActivityDate="2011-01-07T14:48:06.590" CommunityOwnedDate="2011-01-07T14:48:06.590" CommentCount="1" />
  <row Id="16683" PostTypeId="2" ParentId="16680" CreationDate="2011-01-07T14:54:31.797" Score="1" ViewCount="" Body="&lt;p&gt;Hint:  Exponentials are handy for things like this because they are easy to take derivatives of and go nicely to zero if the exponent gets large and negative.  Let $F(s,t)=\exp(g(s,t))$  So you want to construct a function $g(s,t)$ such that $g(s,0)=0, \lim_{t \rightarrow \infty} g(s,t)=-\infty, \lim_{s \rightarrow 0}\frac{\partial g(s,t)}{\partial s}=-\infty$.  Something like $g(s,t)=-\frac{t}{h(s)}$ with $h$ chosen properly will do the trick.  &lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-01-07T14:54:31.797" />
  <row Id="16684" PostTypeId="1" AcceptedAnswerId="16693" CreationDate="2011-01-07T14:55:30.407" Score="0" ViewCount="144" Body="&lt;p&gt;What is the mean euclidean distane between two points on the plane which coordinates are normally distributed?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm assuming this would be&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\int _{-\infty }^{\infty }\int _{-\infty }^{\infty }\int _{-\infty&#xD;&#xA;   }^{\infty }\int _{-\infty }^{\infty }\frac{e^{-\frac{(\text{x1}-\mu&#xD;&#xA;   )^2}{2 \sigma ^2}} e^{-\frac{(\text{x2}-\mu )^2}{2 \sigma ^2}}&#xD;&#xA;   e^{-\frac{(\text{y1}-\mu )^2}{2 \sigma ^2}} e^{-\frac{(\text{y2}-\mu&#xD;&#xA;   )^2}{2 \sigma ^2}}&#xD;&#xA;   \sqrt{(\text{x1}-\text{x2})^2+(\text{y1}-\text{y2})^2}}{\left(\sqrt{&#xD;&#xA;   2 \pi } \sigma \right) \left(\sqrt{2 \pi } \sigma \right)&#xD;&#xA;   \left(\sqrt{2 \pi } \sigma \right) \left(\sqrt{2 \pi } \sigma&#xD;&#xA;   \right)}d\text{y2}d\text{y1}d\text{x2}d\text{x1}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does this integral have a functional representation?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;UPDATE:&lt;/strong&gt;&#xA;After doing some numerical exprimentaion I'm guessing that the answer is $ \sigma \sqrt{\pi } $. But I don't know how to solve this integral.&lt;/p&gt;&#xA;" OwnerUserId="169" LastEditorUserId="169" LastEditorDisplayName="" LastEditDate="2011-01-07T15:30:06.480" LastActivityDate="2011-01-07T16:34:10.347" Title="Mean euclidean distance for normal-random coords" Tags="&lt;geometry&gt;&lt;normal-distribution&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="16685" PostTypeId="1" AcceptedAnswerId="16697" CreationDate="2011-01-07T15:11:50.963" Score="5" ViewCount="139" Body="&lt;p&gt;Does exist open domain  in $\mathbb C$ who has only identity for holomorphic automorphism?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Related question: does exist open domain  in $\mathbb C$ so that every holomorphic automorphism has fixed point?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These questions were inspired by (much easier!) question by mathmos6.&lt;/p&gt;&#xA;" OwnerUserId="2872" LastEditorUserId="2872" LastEditorDisplayName="" LastEditDate="2011-01-07T15:58:41.983" LastActivityDate="2011-01-22T20:42:40.973" Title="Exist domains in complex plane with only trivial automorphisms?" Tags="&lt;complex-analysis&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="2" />
  <row Id="16686" PostTypeId="1" AcceptedAnswerId="16692" CreationDate="2011-01-07T15:12:37.420" Score="0" ViewCount="62" Body="&lt;p&gt;I am currently learning about polar coordinate transformation, especially for integrating over certain regions. Let's say we have to calculate&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\int_{n}{xy \; dx dy}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then I think the correct transformation would be (note the functional determinant $r$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\int_{m}{r cos \phi \cdot r sin \phi \cdot r \; d\phi dr}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;First question: Is that correct?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My second question is: I have some examples a friend of mine once wrote for an exam. The task was for the following region&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ B = \{ (x,y) \in R : 0 \leq x^2 + y^2 \leq 1, x \leq y \}  $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;to calculate the integral&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\int_{B}{x^2+y^2 \; dx dy}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And he began with&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\int_{B}{x^2+y^2 \; dx dy} = \int_{0}^{\frac{5\pi}{4}} \int_{0}^{1} r^2 sin^2 \phi + r^2 cos^2 \phi \; d\phi dr$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Second question: Where is the functional determinant here? Is it missing or am I wrong assuming there should belong one?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd be happy if anyone could help me with some insight on this. Thanks!&lt;/p&gt;&#xA;" OwnerUserId="5488" LastActivityDate="2011-01-07T16:05:28.110" Title="When to use the Functional Determinant in Polar Coordinate Transformation" Tags="&lt;analysis&gt;&lt;integral&gt;&lt;integral-transforms&gt;" AnswerCount="1" />
  <row Id="16688" PostTypeId="2" ParentId="16640" CreationDate="2011-01-07T15:47:33.150" Score="2" ViewCount="" Body="&lt;p&gt;Let $D$ be open disk with center $0$ of radius $1$ in complex plane. Consider holomorphic automorphism of $D$ given by formula &#xA;$z \mapsto \frac{z+a}{1+\bar a z}$ with $a\in D$.&#xA;Does it have fixed point if $a\neq 0$ ? [You must solve $z=\frac{z+a}{1+\bar a z}$]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can also biholomorphically send $D$ to right half plane $H$ of $\mathbb C$ by&#xA;$z \mapsto \frac{1+z}{1-z}$ and then apply Chris Eagle comment to $H$. &lt;/p&gt;&#xA;" OwnerUserId="2872" LastActivityDate="2011-01-07T15:47:33.150" CommentCount="1" />
  <row Id="16689" PostTypeId="2" ParentId="14495" CreationDate="2011-01-07T16:00:35.493" Score="1" ViewCount="" Body="&lt;p&gt;The Delaunay triangulation is a natural and popular way to triangulate the convex hull of a set of points, $P$, in the plane. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conceptually, we can construct it as the dual of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Voronoi_diagram&quot; rel=&quot;nofollow&quot;&gt;Voronoi diagram&lt;/a&gt; of $P$. The Voronoi diagram partitions the plane into Voronoi cells, one for each point in $P$. The Voronoi cell for a point $p \in P$ is defined as that portion of the plane consisting of points  that are at least as close to $p$ as to any other point in $P$. The Voronoi cells are convex polygons, but points on the convex hull of $P$ have unbounded Voronoi cells.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If two points are Voronoi neighbours, i.e., their Voronoi cells share a boundary edge, then they will be connected by an edge in the Delaunay tessellation. A Voronoi vertex is a vertex of the polygon describing a Voronoi cell. A Voronoi vertex belongs to at least three Voronoi cells. If there is no Voronoi vertex that is shared by more than three Voronoi cells, then the Delaunay tessellation will be the unique Delaunay triangulation (in which case we say the points are in general position). Otherwise, the Delaunay tessellation will contain higher order polygons, which may be triangulated arbitrarily to obtain a Delaunay triangulation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Delaunay triangles have the property that the inside of their circumcircles contain no points from $P$ (this is easy to check from the above description). Also, of all possible triangulations of $P$, the Delaunay triangulation maximizes the minimum angle that can be found in a triangle (this does not follow easily from the above description -- it is usually demonstrated via an edge flipping argument).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To construct the Delaunay triangulation in practice, there are several efficient algorithms. There are also many freely available implementations of these algorithms. &lt;a href=&quot;http://www.cgal.org/&quot; rel=&quot;nofollow&quot;&gt;CGAL&lt;/a&gt; is one popular example.&lt;/p&gt;&#xA;" OwnerUserId="1728" LastActivityDate="2011-01-07T16:00:35.493" />
  <row Id="16690" PostTypeId="2" ParentId="16653" CreationDate="2011-01-07T16:01:22.197" Score="2" ViewCount="" Body="&lt;p&gt;It is the linear case of the &lt;strong&gt;polynomial remainder theorem:&lt;/strong&gt; $\rm\ f(x)\ \ mod\ \  a\:x + b\ =\ f(-b/a)\:.\ $ This has a very simple proof. Namely, by the &lt;em&gt;division algorithm&lt;/em&gt; for polynomials over a field,  $\rm\ f(x)\ =\ (a\ x + b)\ q(x) + r\ $ so $\rm\ r = f\:(-b/a)\ $ follows by evaluation at  $\rm\ x = -b/a\:.\ $ In your case $\rm\ f(x)\ =\ c\ x + d\ $ so $\rm\ r = f\:(-b/a) = d - bc/a\:.\ $ Also, comparing degrees shows that the quotient has degree zero $\rm\:q(x) = q\:.\ $ Comparing leading coefficients yields $\rm\ c = a\ q\ $ so $\rm\ q(x) = q = c/a\:.\ $  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence $\rm\displaystyle\quad c\ x + d\ =\ (a\ x + b)\ q + r\ \ \Rightarrow\ \ \frac{c\ x + d}{a\ x + b}\ =\ q + \frac{r}{a\ x + b},\quad q = \frac{c}a,\ \ r = d-\frac{bc}a$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More generally, over any &lt;em&gt;Euclidean domain&lt;/em&gt;, i.e. any integral domain with a &lt;em&gt;division algorithm&lt;/em&gt;, one may use the division algorithm to decompose a fraction into an &lt;em&gt;integral&lt;/em&gt; and &lt;em&gt;fractional&lt;/em&gt; part. This is frequently the inlined $\:$ &lt;em&gt;descent&lt;/em&gt; $\: $ step in proofs that &lt;em&gt;implicitly&lt;/em&gt; invoke &lt;em&gt;unique factorization&lt;/em&gt;.$\ $ For example, see my &lt;a href=&quot;http://math.stackexchange.com/questions/5/how-can-you-prove-that-the-square-root-of-two-is-irrational/16544#16544&quot;&gt;analysis of a proof&lt;/a&gt; of square-root irrationality presented by John Conway.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2011-01-07T17:33:21.757" LastActivityDate="2011-01-07T17:33:21.757" />
  <row Id="16691" PostTypeId="1" CreationDate="2011-01-07T16:01:58.190" Score="2" ViewCount="88" Body="&lt;p&gt;Cn(the nth catalan number) counts the number of expressions containing n pairs of parentheses which are correctly matched&#xA;How to count the possibilities if the maximum nesting level is fixed to d?&lt;/p&gt;&#xA;" OwnerUserId="5421" LastActivityDate="2011-01-07T16:35:00.970" Title="Total no. of balanced parenthesis with maximum nesting of d" Tags="&lt;combinatorics&gt;&lt;sequences-and-series&gt;" AnswerCount="1" />
  <row Id="16692" PostTypeId="2" ParentId="16686" CreationDate="2011-01-07T16:05:28.110" Score="2" ViewCount="" Body="&lt;p&gt;Yes, the determinant is $r$. Your friend made a mistake, there should be an $r$ there too (and the lower limit for $\phi$ should be $\pi/4$).&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2011-01-07T16:05:28.110" CommentCount="1" />
  <row Id="16693" PostTypeId="2" ParentId="16684" CreationDate="2011-01-07T16:34:10.347" Score="2" ViewCount="" Body="&lt;p&gt;If $x_1$ and $x_2$ are normally distributed with mean $\mu$ and variance $\sigma^2$, then $x_1 - x_2$ is normally distributed with mean $0$ and variance $2\sigma^2$. Put $X = x_1-x_2$ and $Y = y_1 - y_2$. Then $(X,Y)$ is a symmetric bivariate normal distribution with mean $(0,0)$ and variance $(2\sigma^2,2\sigma^2)$; the pdf is  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{1}{4\pi\sigma^2} e^{-\frac{1}{4\sigma^2}R^2}$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $R = \sqrt{X^2+Y^2}$. The pdf of the radius $R$ is  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{R}{2\sigma^2} e^{-\frac{1}{4\sigma^2}R^2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The mean of this distribution is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{1}{2\sigma^2} \int_0^\infty R^2e^{-\frac{1}{4\sigma^2}R^2} dR$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is indeed $\sigma\sqrt\pi$.&lt;/p&gt;&#xA;" OwnerUserId="1508" LastActivityDate="2011-01-07T16:34:10.347" CommentCount="1" />
  <row Id="16694" PostTypeId="2" ParentId="16691" CreationDate="2011-01-07T16:35:00.970" Score="3" ViewCount="" Body="&lt;p&gt;The answer is described in several equivalent ways in &lt;a href=&quot;http://qchu.wordpress.com/2009/06/07/the-catalan-numbers-regular-languages-and-orthogonal-polynomials/&quot; rel=&quot;nofollow&quot;&gt;this blog post&lt;/a&gt;.  For fixed $d$ the corresponding sequence $C_{d,n}$ satisfies a linear recurrence, and its generating function is &quot;the $d^{th}$ convergent&quot; of a continued fraction representation of the generating function of the Catalan numbers.  Explicitly, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$C_{d,n} = \frac{1}{d+1} \sum_{i=1}^d 2 \sin^2 \frac{\pi}{d+1} \left( 2 \cos \frac{\pi i}{d+1} \right)^{2n}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(The indexing may be off by one.)&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2011-01-07T16:35:00.970" />
  <row Id="16695" PostTypeId="1" CreationDate="2011-01-07T16:52:26.063" Score="1" ViewCount="64" Body="&lt;p&gt;Definition: A $n$-gon is simple if it has no self intersection and is in general position if no pair of its edges are parallel.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is clear that by extending the edges of each simple $n$-gon in general position we would achieve a set of $n$ lines such that no pair of lines are parallel and no triple of lines intersect in a common point.&#xA;Let’s call the  $n$-gon as inducing $n$-gon and the set of lines as induced arrangement of $n$ lines.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is interesting to consider the reverse, and think about finding such an inducing $n$-gon in a given arrangement of $n$ lines which induces the arrangement, and therefore each line of the arrangement should contain exactly one side of the inducing $n$-gon. It is clear that there are arrangements that cannot be induced by a simple polygon; an arrangement of lines that all intersect in one point, and an arrangement of lines that forms a $3$ \texttimes $2$ parallel grid serve as examples of such arrangements.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It has been proved that every simple arrangement of lines contains at least one inducing $n$-gon. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is clear that an inducing $n$-gon for an simple arrangement is not unique and there are lots of inducing $n$-gons.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any suggestion or idea about complexity of counting the number of inducing $n$-gons in a simple arrangement of $n$ lines?   &lt;/p&gt;&#xA;" OwnerUserId="5409" LastActivityDate="2011-01-07T16:52:26.063" Title="Complexity of Counting the number of inducing $n$-gons" Tags="&lt;combinatorics&gt;&lt;discrete-mathematics&gt;&lt;euclidean-geometry&gt;&lt;discrete-geometry&gt;&lt;computational-geometry&gt;" CommentCount="2" />
  <row Id="16696" PostTypeId="2" ParentId="16665" CreationDate="2011-01-07T17:07:52.287" Score="4" ViewCount="" Body="&lt;p&gt;Essentially, yes, but I believe the undecidability of Peano arithmetic (henceforth, PA) follows from the way the proof of Gödel's incompleteness theorem goes, rather than being a consequence of the fact that PA is incomplete. The proof (outlined below) starts by showing that PA can talk about computable relations, and goes on to show from this how you can construct an unprovable sentence. However, we can take a different approach to show that PA is undecidable: if PA can talk about computable relations, then you can formulate a sentence in the language of PA that is true if and only if a given algorithm halts / does not halt. (Briefly: An algorithm is the same thing as a computable partial function, and an algorithm halts on some input if and only if the corresponding partial function is defined on that input.) So if an algorithm can decide whether arbitrary sentences in PA are provable or not, we have an algorithm which solves the halting problem... but there are nonesuch. So the key point is that PA is rich enough to talk about computation. An essential ingredient for talking about computation is the ability to encode a pair of numbers as a number and the ability to recover the original pair from the encoding. It's not immediately clear to me that $+$ is insufficient to do this, but it's certainly plausible that you need at least two binary operations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a sketch of the proof of Gödel's first incompleteness theorem: First let's select a sufficiently powerful theory $T$, e.g. PA, and assume that it is a consistent theory, i.e. does not prove a contradiction.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;We show that we can encode formulae and proofs in the models of $T$.&lt;/li&gt;&#xA;&lt;li&gt;We show that $T$ is powerful enough to talk about computable relations.&lt;/li&gt;&#xA;&lt;li&gt;We show that there is a computable relation $\mathrm{Prf}(m, n)$ which holds if and only if $m$ encodes a valid proof of the sentence encoded by $n$.&lt;/li&gt;&#xA;&lt;li&gt;The above shows that there is a computable relation $Q(m, n)$ which holds if and only if $n$ encodes a formula $F(-)$ with one free variable and $m$ does &lt;em&gt;not&lt;/em&gt; encode a valid proof of $F(n)$.&lt;/li&gt;&#xA;&lt;li&gt;So we can define a formula $P(x)$ by $\forall m. Q(m, x)$. This means $P(x)$ holds if and only if there is no valid proof of $F(x)$, assuming $x$ encodes a formula $F(-)$ with one free variable.&lt;/li&gt;&#xA;&lt;li&gt;But $P(-)$ is a formula with one free variable, and it can be encoded by some number $n$. So is the sentence $P(n)$ provable or not? Suppose it were. Then, that means $P(n)$ is true, so the theory asserts that there is no valid proof of $P(n)$ — a contradiction. &lt;/li&gt;&#xA;&lt;li&gt;So we are forced to conclude that the sentence $P(n)$ is not provable. This is the Gödel sentence which we wished to prove the existence of, so we are done.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Note I haven't said anything about whether $P(n)$ is actually &lt;em&gt;true&lt;/em&gt;. It turns out there is some subtlety here. Gödel's &lt;em&gt;completeness&lt;/em&gt; theorem tells us that everything that can be proven in a first-order theory is true in every model of that theory and that every sentence which is true in every model of a first-order theory can be proven from the axioms of that theory. With some stronger consistency assumptions, we can also show that $\lnot P(n)$ is also not provable in PA, and this means that there are models of PA where $P(n)$ is true and models where it is false.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The key point is that the phrases &quot;$n$ encodes a formula ...&quot; and &quot;$m$ encodes a valid proof of ...&quot; are strictly &lt;em&gt;outside&lt;/em&gt; the theory. The interpretation of a particular number as a formula or proof is defined externally, and we only define it for &quot;standard&quot; numbers. The upshot of this is that in a model $\mathcal{M}$ of PA where $P(n)$ is false, there is some non-standard number $m \in \mathcal{M}$ which $\mathcal{M}$ &quot;believes&quot; is a valid proof of $P(n)$, but because it's non-standard, we cannot translate it into a real proof.&lt;/p&gt;&#xA;" OwnerUserId="5191" LastActivityDate="2011-01-07T17:07:52.287" CommentCount="1" />
  <row Id="16697" PostTypeId="2" ParentId="16685" CreationDate="2011-01-07T17:08:14.797" Score="5" ViewCount="" Body="&lt;p&gt;Sure. Take $\mathbb{CP}^1$ and delete $5$ points $(0,1,\infty, s, t)$ where $s$ and $t$ are generic. Any automorphism of this domain will extend to an automorphism of $\mathbb{CP}^1$, by Riemann's extension theorem. That automorphism must permute the five points $(0, 1, \infty, s, t)$. For generic $s$ and $t$, there is no nontrivial map in $\mathbb{P}GL_2(\mathbb{C})$ which permutes these points.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I deleted three points, there would be an $S_3$ of such maps; if I deleted four points, the corresponding group is $\mathbb{Z}/2 \times \mathbb{Z}/2$. So, if you are happy with getting a finite symmetry group, then $3$ points are enough.&lt;/p&gt;&#xA;" OwnerUserId="448" LastActivityDate="2011-01-07T17:08:14.797" />
  <row Id="16698" PostTypeId="1" CreationDate="2011-01-07T17:42:58.140" Score="1" ViewCount="159" Body="&lt;p&gt;I'm starting to learn proof and logic, all my knowledge in mathematics is: basic discrete boolean logic, introductory logic, basic Calculus, Introductory Linear Algebra. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've been trying to use what I know to prove this statement, but I simply can't. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If $x$ is real number such that $x^4 + 2x^2  -2x \lt 0$, then $0\lt x\lt 1$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I dont know enough Calculus to find its roots, and if use Newton's method, I am able to find $x=0$ and $x\approx 0.77777$. But I wont know if there are two more real roots.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any tips?&lt;/p&gt;&#xA;" OwnerUserId="5489" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-07T19:15:36.090" LastActivityDate="2011-01-07T19:16:34.247" Title="Newton's method in an elementary proof" Tags="&lt;polynomials&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="16699" PostTypeId="2" ParentId="15719" CreationDate="2011-01-07T17:45:08.580" Score="2" ViewCount="" Body="&lt;p&gt;What a surprise! &#xA;Surfing the net, I found an almost same question on&#xA;&lt;a href=&quot;http://topologicalmusings.wordpress.com/2008/06/27/solution-to-pow-5-a-%E2%80%9Chard%E2%80%9D-integral/&quot; rel=&quot;nofollow&quot;&gt;&quot;hard integral&quot;&lt;/a&gt;&#xA;$$&#xD;&#xA;\displaystyle \int \frac{x^2 - 1}{(x^2 + 1) \sqrt{x^4 + 1}} \, dx&#xD;&#xA;$$&#xA;from June 19, 2008.&lt;/p&gt;&#xA;" OwnerUserId="5418" LastActivityDate="2011-01-07T17:45:08.580" CommentCount="3" />
  <row Id="16700" PostTypeId="2" ParentId="16698" CreationDate="2011-01-07T17:56:11.307" Score="3" ViewCount="" Body="&lt;p&gt;The second derivative is $12x^2+4$ which is positive for all $x$ so the function is convex.  Since you showed the two roots are in $[0,1]$ it follows that $f(x) &amp;lt;0 \implies x\in [0,1]$, since by convexity the function is only negative between its two roots.&lt;/p&gt;&#xA;" OwnerUserId="565" LastActivityDate="2011-01-07T17:56:11.307" CommentCount="1" />
  <row Id="16701" PostTypeId="1" AcceptedAnswerId="16704" CreationDate="2011-01-07T18:01:24.580" Score="2" ViewCount="84" Body="&lt;p&gt;Suppose $A$ and $B$ are two commutative rings, and let $f:A\rightarrow B$ be a ring homomorphism. Suppose that the induced map on the spectra $f^*:Spec B\rightarrow Spec A$ is an open immersion. What does it say about $f$? does it follows that $f$ is a localization?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;edit: Thanks for the answer! What about closed immersions? is this case simpler? Is it true that a closed immersion is always of the form $A\rightarrow A/I$ for some ideal $I$ in $A$?&lt;/p&gt;&#xA;" OwnerUserId="5491" LastEditorUserId="5491" LastEditorDisplayName="" LastEditDate="2011-01-07T18:30:22.223" LastActivityDate="2011-01-07T18:30:22.223" Title="What are open immersions in terms of rings?" Tags="&lt;algebraic-geometry&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="16702" PostTypeId="2" ParentId="16698" CreationDate="2011-01-07T18:04:06.823" Score="2" ViewCount="" Body="&lt;p&gt;The equation can be written as $x(x^3 + 2x -2) = 0$. For a general cubic of the form $x^3 + px + q=0$, you can calculate the discriminant $\Delta$ as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;\Delta = -4p^3 -27q^2&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here, $p=2$ and $q=-2$. Since the discriminant is negative, the cubic equation has only one real root. The other two roots are complex conjugates (http://en.wikipedia.org/wiki/Cubic_function#The_nature_of_the_roots).&lt;/p&gt;&#xA;" OwnerUserId="2641" LastActivityDate="2011-01-07T18:04:06.823" />
  <row Id="16703" PostTypeId="2" ParentId="16476" CreationDate="2011-01-07T18:06:47.237" Score="1" ViewCount="" Body="&lt;p&gt;First suppose for contradiction the length of the longest cycle $C$ is $2k-1$, then consider the set of vertices in $C$, which are connected with vertices outside C. It has cardinality at least k. Otherwise, by deleting these vertices, the graph left is still connected. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then by pigeonhole principle there are two adjacent verticies $A$ and $A&amp;#39;$ in $C$, which are connected with vertices outside $C$, say, $B$ and $B&amp;#39;$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The situation when $B=B&amp;#39;$ automatically makes a longer cycle. When $B$ and $B&amp;#39;$ are different, we delete $C$, and get a connected graph. We connect $B$ and $B&amp;#39;$, using vertices all outside $C$. This makes the cycle longer. &lt;/p&gt;&#xA;" OwnerUserId="5476" LastEditorUserId="5476" LastEditorDisplayName="" LastEditDate="2011-01-08T03:29:06.790" LastActivityDate="2011-01-08T03:29:06.790" CommentCount="2" />
  <row Id="16704" PostTypeId="2" ParentId="16701" CreationDate="2011-01-07T18:10:20.483" Score="1" ViewCount="" Body="&lt;p&gt;See &lt;a href=&quot;http://mathoverflow.net/questions/20782/ring-theoretic-characterization-of-open-affines&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2011-01-07T18:10:20.483" />
  <row Id="16705" PostTypeId="2" ParentId="16698" CreationDate="2011-01-07T18:14:36.383" Score="6" ViewCount="" Body="&lt;p&gt;The highlighted statement can be proved without calculus or root approximations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\ x^4\: +\: 2\:x^2\: -\: 2\:x\ \ge\ 0\ \ $ for $\rm\ \ x \le 0\ $ since all three summands are $\ge 0\ $ there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\ x^4 + 2\:x\: (x-1)\ \ge\ 0\ \ $ for $\rm\ \ x \ge 1\ $ since the $\:$ two summands are $\ge 0\ $ there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, if $\:$  it$\:$ is $\: &amp;lt;\: 0\ $ then $\rm\ 0 &amp;lt; x &amp;lt; 1\:.\quad$ &lt;strong&gt;QED&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The algebra has a vivid geometric interpretation: shifting the quadratic $\rm\: f =  2\ x\ (x-1)$ upward by adding some function $\rm\:g\ge 0\:$ (here $\rm\: g = x^4\:$) &#xA;clearly &lt;em&gt;preserves&lt;/em&gt; the property that  $\rm\ f \ge 0\ $ outside $[0,1]$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/OC3u3.jpg&quot; alt=&quot;alt text&quot;&gt;&#xA;&lt;img src=&quot;http://i.imgur.com/pgxrh.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2011-01-07T19:16:34.247" LastActivityDate="2011-01-07T19:16:34.247" CommentCount="1" />
  <row Id="16706" PostTypeId="1" AcceptedAnswerId="16733" CreationDate="2011-01-07T18:18:36.107" Score="4" ViewCount="82" Body="&lt;p&gt;I'm trying to investigate the notion of primitive spectrum and its so-called Jacobson or hull-kernel topology, but I can only find references which define it for C*-algebras: see the Wikipedia page &quot;Spectrum of a C*-algebra&quot; for the definition I'm talking about. It seems like this definition would make sense for any (let's stick with unital) ring whatsoever, so I suspect the problem is that in full generality we don't actually get a topology.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So here's what I want to ask: can you help me think of an example of a unital ring for which the Jacobson topology on its primitive spectrum is not actually a topology? Or even better, does anyone know of general conditions under which the primitive spectrum has a natural topology? Also, when the Jacobson topology is defined, is the primitive spectrum always quasi-compact?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By way of motivation, certain complex algebras have come up in my research in representation theory. Explaining what they are would take me far afield, but they do have the following nice property. In their paper &quot;Extensions of representations of $p$-adic nilpotent groups,&quot; S. Gelfand and D. Kazhdan call a complex unital algebra $A$ &lt;em&gt;quasi-finite&lt;/em&gt; provided that it has a filtration $A_0 \subset A_1 \subset \cdots \subset A$ by finite-dimensional semisimple subalgebras $A_k$, and simple modules for each $A_k$ are finite-dimensional.&lt;/p&gt;&#xA;" OwnerUserId="4842" LastActivityDate="2011-01-08T00:12:17.270" Title="The primitive spectrum of a unital ring" Tags="&lt;ring-theory&gt;&lt;representation-theory&gt;&lt;noncommutative-geometry&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="16707" PostTypeId="2" ParentId="16665" CreationDate="2011-01-07T18:23:04.610" Score="6" ViewCount="" Body="&lt;p&gt;While this may not be the intuitive idea you're looking for, I'll note that the same ideas behind Godel's proof that Peano Arithmetic is incomplete (that is, that there are true sentences of PA that can't be proven) can also be applied to show that Presburger Arithmetic is 'hard' in a tangible sense; essentially, one can encode multiplication of small enough numbers (specifically, the predicate '$x\times y = z$' with $x$, $y$ and $z$ 'only' doubly-exponential in the length of the formula) into a formula involving only addition.  This means that if sentences in Presburger Arithmetic of length $n$ could be decided without multiplying numbers of size at least (roughly) $2^{2^n}$, then Presburger Arithmetic would suffer from the same Godelian paradox that Peano Arithmetic does and be incomplete.  Paul Young wrote up a much better exposition of this for a conference a couple of decades ago than I can possibly give; have a look at &lt;a href=&quot;http://books.google.com/books?id=2vvg3mRzDtwC&amp;amp;pg=PA503&amp;amp;lpg=PA503&quot; rel=&quot;nofollow&quot;&gt;http://books.google.com/books?id=2vvg3mRzDtwC&amp;amp;pg=PA503&amp;amp;lpg=PA503&lt;/a&gt; for the details.&lt;/p&gt;&#xA;" OwnerUserId="785" LastActivityDate="2011-01-07T18:23:04.610" />
  <row Id="16708" PostTypeId="2" ParentId="16634" CreationDate="2011-01-07T18:26:43.777" Score="0" ViewCount="" Body="&lt;p&gt;An expansion on the other answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More analytically (or explicitly), let $a$,$b$,$c$ be three points.  We want to show there is only one point equidistant to all three.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $a = (a_1,a_2)$, $b = (b_1,b_2)$,$c = (c_1,c_2)$. Points $x = (x_1,x_2)$ equidistant to all three must satisfy,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ (x_1-a_1)^2+(x_2-a_2)^2=(x_1-b_1)^2+(x_2-b_2)^2=(x_1-c_1)^2+(x_2-c_2)^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which is the system of equations&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ (x_1-a_1)^2+(x_2-a_2)^2=(x_1-b_1)^2+(x_2-b_2)^2$$&#xA;$$(x_1-b_1)^2+(x_2-b_2)^2=(x_1-c_1)^2+(x_2-c_2)^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The solution to each equation is a line. Specifically, the solution to the first equation is the line&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ (x_1-a_1)^2+(x_2-a_2)^2=(x_1-b_1)^2+(x_2-b_2)^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ -2a_1x_1+a_1^2 -2a_2x_2+a_2^2=-2b_1x_1+b_1^2 -2b_2x_2+b_2^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ x_1 = \frac{(-2b_2+2a_2)x_2+(b_2^2-a_2^2)}{-2a_1+2a_2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since two lines intersect at at most one place, the solution to the system is either a single point which defines the center of the circle, or there is no solution (the case of $a$,$b$,$c$ collinear).  If we include the point at infinity as Qiaochu mentions then two lines always intersect at exactly one point (with parallel lines intersecting at infinity), and the circle will be through $a$,$b$,$c$ with infinite radius.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Otherwise the radius will be $\sqrt{(x_1-a_1)^2+(x_2-a_2)^2}$ and we have defined a circle.&lt;/p&gt;&#xA;" OwnerUserId="565" LastActivityDate="2011-01-07T18:26:43.777" />
  <row Id="16709" PostTypeId="1" AcceptedAnswerId="16711" CreationDate="2011-01-07T19:21:43.540" Score="7" ViewCount="299" Body="&lt;p&gt;This question was inspired by the Lagrange equation, $\frac{\partial L}{\partial q} - \frac{d}{dt}\frac{\partial L}{\partial \dot{q}} = 0$. What happens if the partial derivatives are replaced by total derivatives, leading to a situation where a function's derivative with respect to one variable is differentiated by the original function?&lt;/p&gt;&#xA;" OwnerUserId="1769" LastActivityDate="2011-01-08T22:19:46.410" Title="What is $\frac{d}{dx}\left(\frac{dx}{dt}\right)$?" Tags="&lt;calculus&gt;" AnswerCount="3" FavoriteCount="1" />
  <row Id="16710" PostTypeId="1" CreationDate="2011-01-07T19:32:08.190" Score="2" ViewCount="148" Body="&lt;p&gt;Hi guys,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does someone know if the function resulting of the sum of two locally bounded functions is also locally bounded?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance!&lt;/p&gt;&#xA;" OwnerUserId="5493" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-07T19:48:20.870" LastActivityDate="2011-01-07T19:48:20.870" Title="Is the the sum of two locally bounded functions also locally bounded?" Tags="&lt;real-analysis&gt;" AnswerCount="2" CommentCount="0" />
  <row Id="16711" PostTypeId="2" ParentId="16709" CreationDate="2011-01-07T19:35:57.203" Score="5" ViewCount="" Body="&lt;p&gt;Write $$\frac{d}{dx}\left(\frac{dx}{dt}\right)=\frac{d}{dx}\left(\frac{1}{\frac{dt}{dx}}\right)$$ and use the chain rule.  $$\frac{d}{dx}\left(\frac{dx}{dt}\right)=\left(\frac{-1}{(\frac{dt}{dx})^2}\right)\frac{d^2t}{dx^2}=-\left (\frac{dx}{dt}\right )^2\frac{d^2t}{dx^2}$$&lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2011-01-07T22:05:43.963" LastActivityDate="2011-01-07T22:05:43.963" />
  <row Id="16712" PostTypeId="2" ParentId="16710" CreationDate="2011-01-07T19:37:41.827" Score="5" ViewCount="" Body="&lt;p&gt;The answer is yes, and it is straightforward enough for you to try to work out for yourself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose that $X$ is a metric space and $f,g: X \rightarrow \mathbb{R}$ are two locally bounded functions.  This means that for each $x \in X$, there exists an open set &#xA;$U_f$ containing $x$ on which $f$ is bounded and an open set $U_g$ containing $x$ on which $g$ is bounded.  We want to use $U_f$ and $U_g$ to construct an open set on which both $f$ and $g$ are bounded, for then $f+g$ will be bounded as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you think of how to do this?&lt;/p&gt;&#xA;" OwnerUserId="299" LastActivityDate="2011-01-07T19:37:41.827" CommentCount="1" />
  <row Id="16713" PostTypeId="2" ParentId="16710" CreationDate="2011-01-07T19:37:49.603" Score="4" ViewCount="" Body="&lt;p&gt;A function $f$ is &lt;em&gt;locally bounded&lt;/em&gt; if for every $x$ there exists an open neighborhood $U_x$ of $x$, and a constant $M_x$ (which may depend on $x$) such that $|f(y)|\leq M_x$ for all $y\in U_x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose $f$ and $g$ are both locally bounded. Let $x$ be an element of the domain. Then there are open neighborhoods $U_x$ and $V_x$ of $x$, and constants $M_x$ and $N_x$ such that $|f(y)|\leq M_x$ for all $y\in U_x$, and $|g(z)|\leq N_x$ for all $z\in V_x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What can you say about $f(y)+g(y)$ if $y\in U_x\cap V_x$?&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2011-01-07T19:37:49.603" CommentCount="2" />
  <row Id="16714" PostTypeId="2" ParentId="16665" CreationDate="2011-01-07T20:04:25.203" Score="3" ViewCount="" Body="&lt;p&gt;Peano arithmetic is strong enough to reason about Turing machines.  Thus, if Peano arithmetic were decidable, the Halting problem would be decidable.  I learned essentially this idea from &lt;a href=&quot;http://www.scottaaronson.com/democritus/lec3.html&quot; rel=&quot;nofollow&quot;&gt;Scott Aaronson&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2011-01-07T20:04:25.203" CommentCount="3" />
  <row Id="16715" PostTypeId="1" AcceptedAnswerId="16722" CreationDate="2011-01-07T21:15:03.917" Score="2" ViewCount="86" Body="&lt;p&gt;I'm currently reading M. Atiyah's &quot;K-Theory,&quot; and I'm a bit confused about the following part.  Let $E = X \times V$ and $F = X\times W$, and let $\phi: E\to F$ be a vector bundle homomorphism.  Why is the induced map $\Phi: X \to Hom(V,W)$ continuous?  $V$ and $W$ are finite dimensional vector spaces (over $\mathbb{C}$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To be a bit more precise, for $(x,v) \in E$, $\phi(x,v) = (x,\Phi(x)v)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think it should be obvious, but I can't come up with a proof.&lt;/p&gt;&#xA;" OwnerUserId="2499" LastActivityDate="2011-01-07T22:09:46.577" Title="Simple Continuity Question concerning Vector Bundles" Tags="&lt;differential-topology&gt;&lt;topological-vector-spaces&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="16716" PostTypeId="1" AcceptedAnswerId="16718" CreationDate="2011-01-07T21:21:01.157" Score="4" ViewCount="102" Body="&lt;p&gt;Let $ A\in {\mathbb{F} }^{n\times n} $ be a fixed matrix. The set of all matrices that commute with A forms a subring of ${\mathbb{F} }^{n\times n}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is any subring of ${\mathbb{F} }^{n\times n }$ (which contains the identity) of the above form?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="766" LastEditorUserId="766" LastEditorDisplayName="" LastEditDate="2011-01-07T21:33:02.507" LastActivityDate="2011-01-07T22:02:19.350" Title="Rings of matrices" Tags="&lt;abstract-algebra&gt;&lt;matrices&gt;&lt;ring-theory&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="16717" PostTypeId="2" ParentId="16716" CreationDate="2011-01-07T21:25:24.047" Score="3" ViewCount="" Body="&lt;p&gt;Let $T$ be subring of $M_2(\mathbb F)$ of upper triangular matrices. Is there an $A\in M_2(\mathbb F)$ such that $T$ is the set of matrices which commute with $A$?&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2011-01-07T21:25:24.047" />
  <row Id="16718" PostTypeId="2" ParentId="16716" CreationDate="2011-01-07T21:26:50.160" Score="5" ViewCount="" Body="&lt;p&gt;Not if $n\gt 1$. Consider the subring of all scalar multiples of the identity matrix (this is isomorphic to $\mathbb{F}$ itself). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the collection of all matrices that commute with $A$ always includes $A$ itself, if the subring of all scalar multiples of the identity were of the form $C(A)$ for some $A$, then $A$ would necessarily be a scalar multiple of the identity; but scalar multiples of the identity are central in $\mathbb{F}^{n\times n}$, so the centralizer of such an $A$ would contain more than just the scalar multiples of the identity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might ask more generally whether every subring of $\mathbb{F}^{n\times n}$ of the form $C(S)$ for some $S\subseteq \mathbb{F}^{n\times n}$, where $C(S) = \{M\in\mathbb{F}^{n\times n}\mid MA=AM\text{ for all }A\in S\}$. But even here the answer is still negative; though my example no longer works in this setting (taking $S=\mathbb{F}^{n\times n}$ gives the scalar multiples of the identity), Mariano's example still does. Note that $S\subseteq C(C(S))$; if $T$ is the set of upper triangular matrices, then considering the matrix $E_{ij}$ that has a $1$ in the $(i,j)$ entry, $i\leq j$, and zeros elsewhere, you have that $E_{ij}A$ is the matrix that has zeros everywhere except the $i$th row, where it has the $j$th row of $A$; while $AE_{ij}$ is the matrix that has zeros everywhere except the $j$th column, where it has the $i$th column of $A$. Thus, if $A\in C(T)$, then $A$ must be a scalar multiple of the identity; this means that if $T=C(S)$ for some $S$, then $S$ must be contained in the scalar multiples of the identity, and once again we obtain a contradiction. &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-07T22:02:19.350" LastActivityDate="2011-01-07T22:02:19.350" />
  <row Id="16719" PostTypeId="1" AcceptedAnswerId="16721" CreationDate="2011-01-07T21:29:11.277" Score="3" ViewCount="171" Body="&lt;p&gt;One can't get too far in abstract algebra before encountering Zorn's Lemma.  For example, it is used in the proof that every nonzero ring has a maximal ideal.  However, it seems that if we restrict our focus to Noetherian rings, we can often avoid Zorn's lemma.  How far could a development of the theory for just Noetherian rings go?  When do non-Noetherian rings come up in an essential way for which there is no Noetherian analog?  For example, Artin's proof that every field has an algebraic closure uses Zorn's lemma.  Is there a proof of this theorem (or some Zorn-less version of this theorem) that avoids it?&lt;/p&gt;&#xA;" OwnerUserId="5265" LastEditorUserId="5265" LastEditorDisplayName="" LastEditDate="2011-01-07T21:56:50.483" LastActivityDate="2011-01-07T21:56:50.483" Title="Algebra without Zorn's lemma" Tags="&lt;abstract-algebra&gt;&lt;axiom-of-choice&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="16720" PostTypeId="2" ParentId="16709" CreationDate="2011-01-07T21:29:51.547" Score="5" ViewCount="" Body="&lt;p&gt;Alternatively, the chain rule $\frac{dy}{dx} = \frac{dy}{dt}\frac{dt}{dx}$ gives $$\frac{d}{dx}(\frac{dx}{dt}) = \frac{d}{dt}(\frac{dx}{dt})\frac{dt}{dx} = \frac{d^2x}{dt^2}\frac{dt}{dx}.$$&#xA;Of course, this is the same as Ross' answer above.  That is, we have the identity $$-\left(\frac{dx}{dt}\right)^2\frac{d^2t}{dx^2} = \frac{d^2x}{dt^2}\frac{dt}{dx},$$&#xA;which follows from differentiating the equation $\frac{dx}{dt}\frac{dt}{dx} = 1$ with respect to $t$.&lt;/p&gt;&#xA;" OwnerUserId="640" LastEditorUserId="640" LastEditorDisplayName="" LastEditDate="2011-01-08T04:42:02.440" LastActivityDate="2011-01-08T04:42:02.440" />
  <row Id="16721" PostTypeId="2" ParentId="16719" CreationDate="2011-01-07T21:34:05.453" Score="3" ViewCount="" Body="&lt;p&gt;The proof of existence and uniqueness of algebraic closures goes through assuming only the ultrafilter lemma, which is strictly weaker than AC; see &lt;a href=&quot;http://mathoverflow.net/questions/46566/is-the-statement-that-every-field-has-an-algebraic-closure-known-to-be-equivalent&quot; rel=&quot;nofollow&quot;&gt;this MO question&lt;/a&gt;.  Exactly how strong this assumption is relative to other well-known forms of AC appears to be unknown.  I don't know what &quot;Noetherian version of this theorem&quot; means, since every field is Noetherian.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2011-01-07T21:34:05.453" CommentCount="3" />
  <row Id="16722" PostTypeId="2" ParentId="16715" CreationDate="2011-01-07T22:09:46.577" Score="4" ViewCount="" Body="&lt;p&gt;The map $\Phi&amp;#39;:X\times V\to W$ such that $\phi(x,v)=(x,\Phi&amp;#39;(x,v))$ is continuous. By properties of the the compact-open topology of function spaces, it follows that the map $\Phi:X\to\mathrm{Map}(V,W)$ such that $\Phi(x)(v)=\Phi&amp;#39;(x,v)$ is also continuous. Now $\Phi$ actually takes values in $\hom(V,W)$, whose usual topology actualy coincides with the restriction of the compact-open topology of $\mathrm{Map}(V,W)$, so the restriction $\Phi:X\to\hom(V,W)$ is continuous.&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2011-01-07T22:09:46.577" />
  <row Id="16723" PostTypeId="1" AcceptedAnswerId="17486" CreationDate="2011-01-07T22:34:35.050" Score="9" ViewCount="556" Body="&lt;p&gt;Let $a, b$ and $c$ be positive integers. Recall that the greatest common divisor (gcd) function has the following representation:&#xA;\begin{eqnarray}&#xA;\textbf{gcd}(b,c) = 2 \sum_{k = 1}^{c- 1} \left \lfloor \frac{kb}{c} \right \rfloor + b + c - bc&#xA;\end{eqnarray}&#xA;as shown by Polezzi, where $\lfloor \cdot \rfloor$ denotes the floor function. In trying to generalize the formula I came across the following summation&#xA;\begin{eqnarray}&#xA;\sum_{k = 0}^{\lfloor a/b \rfloor} \left \lfloor \frac{a - kb}{c} \right \rfloor.&#xA;\end{eqnarray}&#xA;I can prove the following identity for real $x$,&#xA;\begin{eqnarray}&#xA;\sum_{k = 0}^{c-1} \left \lceil \frac{x - kb}{c} \right \rceil = d \left \lceil \frac{x}{d} \right \rceil - \frac{(b-1)(c-1)}{2} - \frac{d-1}{2},&#xA;\end{eqnarray}&#xA;where $\lceil \cdot \rceil$ denotes the ceiling function and $d = \text{gcd}(b,c)$. (Note that in the first summation the upper index is in general independent of $c$.) Ideas or reference suggestions are certainly appreciated. Thanks in advance! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Update&lt;/em&gt; I can prove the identity&#xA;\begin{eqnarray}&#xA;\sum_{k = 0}^{c-1} \left \lfloor \frac{x - kb}{c} \right \rfloor = d \left \lfloor \frac{x}{d} \right \rfloor - \frac{(b+1)(c-1)}{2} + \frac{d-1}{2},&#xA;\end{eqnarray}&#xA;where $d = \text{gcd}(b,c)$. There is another identity which might be useful. If $n = c \ell +r$ with $0 \leq r &amp;lt; c$, then&#xA;\begin{eqnarray}&#xA;\sum_{k = 1}^{n} \left \lfloor \frac{k}{c} \right \rfloor = c \binom{\ell}{2} + (r + 1) \ell.&#xA;\end{eqnarray}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Update 2&lt;/em&gt; Ok, so I can prove that for real $x, y &amp;gt; 0$,&#xA;\begin{eqnarray}&#xA;\sum_{k = 0}^{\lfloor y \rfloor} \left \lfloor x + \frac{k}{y} \right \rfloor = \lfloor xy + (\lceil y \rceil - y) \lfloor x + 1 \rfloor \rfloor + \chi_{\mathbb{N}}(y)(\lfloor x \rfloor + 1),&#xA;\end{eqnarray}&#xA;where $\chi_{\mathbb{N}}$ denotes the characteristic function of the positive integers. My original problem (and a nice generalization of it) will be in hand if I can evaluate the following minor generalization: For real $x, y  &amp;gt; 0$ and $n \in \mathbb{Z}_{\geq 0}$,&#xA;\begin{eqnarray}&#xA;\sum_{k = 0}^{n} \left \lfloor x + \frac{k}{y} \right \rfloor.&#xA;\end{eqnarray}&#xA;Again, any help is certainly appreciated!&lt;/p&gt;&#xA;" OwnerUserId="2720" LastEditorUserId="2720" LastEditorDisplayName="" LastEditDate="2011-01-14T19:50:29.440" LastActivityDate="2011-01-17T08:51:03.823" CommunityOwnedDate="2011-01-09T08:52:56.113" Title="Evaluation of the sum $\sum_{k = 0}^{\lfloor a/b \rfloor} \left \lfloor \frac{a - kb}{c} \right \rfloor$" Tags="&lt;combinatorics&gt;&lt;reference-request&gt;" AnswerCount="3" FavoriteCount="2" />
  <row Id="16724" PostTypeId="1" AcceptedAnswerId="16730" CreationDate="2011-01-07T22:49:11.660" Score="6" ViewCount="241" Body="&lt;p&gt;A few days ago I was recalling some facts about the p-adic numbers, for example the fact that the p-adic metric is an ultrametric implies very strongly that there is no order on $\mathbb{Q}_p$, as any number in the interior of an open ball is in fact its center.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that if you take the completion of the algebraic close of the p-adic completion you get something which is isomorphic to $\mathbb{C}$ (this result was very surprising until I studied model theory, then it became obvious).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore, if the algebraic closure is of an extension of dimension 2 then the field is orderable, or even real closed. Either way, it implies that the p-adic numbers don't have this property.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I was thinking, is there a p-adic number whose square equals 2? 3? 2011? For which prime numbers p? How far down the rabbit hole of algebraic numbers can you go inside the p-adic numbers? Are there general results connecting the choice (or rather properties) of $p$ to the &quot;amount&quot; of algebraic closure it gives?&lt;/p&gt;&#xA;" OwnerUserId="622" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2011-01-07T23:32:46.243" LastActivityDate="2011-01-08T04:08:33.257" Title="How far are the p-adic numbers from being algebraically closed?" Tags="&lt;number-theory&gt;&lt;p-adic-number-theory&gt;" AnswerCount="3" CommentCount="5" FavoriteCount="1" />
  <row Id="16726" PostTypeId="1" CreationDate="2011-01-07T22:55:29.743" Score="3" ViewCount="120" Body="&lt;p&gt;Hello, I am reading Concrete Mathematics and came across an interesting problem, number 37 of chapter 2. The answers to exercises lists no known answer to this problem: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;Will all the 1/k by 1/(k+1) rectangles, for k $\ge$ 1, fit together inside a 1 by 1 square? (Recall that their areas sum to 1)&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;My question is: Has a solution been found in the years after the book's publishing? Mathematics is a very large field, and much of the terminology that might aid in googling I am not yet aware of.&lt;/p&gt;&#xA;" OwnerUserId="5496" LastActivityDate="2011-01-07T23:04:27.837" Title="Proof whether or not 1/k by 1/(k+1) rectangles fit inside a unit square." Tags="&lt;geometry&gt;&lt;sequences-and-series&gt;&lt;discrete-mathematics&gt;&lt;infinity&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" />
  <row Id="16727" PostTypeId="2" ParentId="16726" CreationDate="2011-01-07T23:04:27.837" Score="3" ViewCount="" Body="&lt;p&gt;See &lt;a href=&quot;http://mathoverflow.net/questions/34145/can-we-cover-the-unit-square-by-these-rectangles&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2011-01-07T23:04:27.837" CommentCount="2" />
  <row Id="16728" PostTypeId="1" AcceptedAnswerId="16731" CreationDate="2011-01-07T23:37:15.523" Score="3" ViewCount="80" Body="&lt;p&gt;My class is solving the &lt;a href=&quot;http://en.wikipedia.org/wiki/Cauchy%E2%80%93Euler_equation&quot; rel=&quot;nofollow&quot;&gt;Cauchy-Euler&lt;/a&gt; differential equation $a t^2 y&amp;#39;&amp;#39; + b t y&amp;#39; + c y = 0$.  The solutions are powers of $t$, $y = t^r$ and then you solve for $r$ using the characteristic equation $a r^2 + (b-a) r + c = 0$.  This has two roots $r = \overline{r} \pm \Delta r$ and the general solution is $y = A t^{\overline{r} + \Delta r} + B t^{\overline{r} - \Delta r }$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What happens if we get double roots, i.e. $(b-a)^2 = 4ac$ or $\Delta r = 0$? We guess a solution of $y = t^r \ln t$.  Can derive this guess taking the limit of two distinct roots solution as $\Delta r$ tends to 0?  In some sense &#xA;$$ A t^{\overline{r} + \Delta r} + B t^{\overline{r} - \Delta r }&#xD;&#xA;= t^{\overline{r}} (  A t^{ \Delta r} + B t^{ - \Delta r }) \to  t^{\overline{r}} (C + D \ln t )$$&#xA;for some constants $A,B,C,D$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a way, it is plausible the limit works this way because of the integral formula:&#xA;$$ \int t^{r - 1} dt = \left\{ \begin{array}{cc}  \frac{t^r}{r} &amp;amp; r \neq 0\\ \ln t  &amp;amp; r = 0 \end{array} \right. $$&#xA;And I expect this is coming into play in the Euler-Cauchy equation.  How to make it rigorous, though?&lt;/p&gt;&#xA;" OwnerUserId="4997" LastActivityDate="2011-01-08T00:25:47.517" Title="How log(t) can be the limit of t^r, where r -&gt; 0 ?" Tags="&lt;calculus&gt;&lt;real-analysis&gt;&lt;differential-equations&gt;" AnswerCount="1" />
  <row Id="16729" PostTypeId="2" ParentId="16724" CreationDate="2011-01-07T23:42:12.207" Score="1" ViewCount="" Body="&lt;p&gt;A monic polynomial $f$ with coefficients in $\mathbb{Z}$ has a root in $\mathbb{Z}_p$ if and only if it has a root in $\mathbb{Z}/p^n\mathbb{Z}$ for all $n$; this is because $\mathbb{Z}_p$ can be defined as the limit (&quot;inverse limit&quot;) of the $\mathbb{Z}/p^n\mathbb{Z}$.  Hensel's lemma tells you at what value of $n$ you can stop searching and gives an algorithm for determining whether such a root exists for fixed $f$ and $p$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A necessary condition is that $f$ needs to have a root in $\mathbb{F}_p$, and if $p$ doesn't divide the discriminant of $f$ I think this condition is also sufficient.  For fixed $p$, are you looking for a description of all such $f$?  I don't know that I can be more specific than &quot;the set of all polynomials with a root in $\mathbb{F}_p$.&quot;  You just take all the polynomials in $\mathbb{F}_p[x]$ with roots in $\mathbb{F}_p$ and lift them to $\mathbb{Z}$.  Do you want an algorithm to determine when $f$ has this property?  Just evaluate it at every point.  I'm not really sure what you're looking for here.  &lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2011-01-08T00:33:30.847" LastActivityDate="2011-01-08T00:33:30.847" CommentCount="2" />
  <row Id="16730" PostTypeId="2" ParentId="16724" CreationDate="2011-01-07T23:46:29.813" Score="7" ViewCount="" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;A few days ago I was recalling some facts about the p-adic numbers, for example the fact that the p-adic metric is an ultrametric implies very strongly that there is no order on $\mathbb{Q}_p$, as any number in the interior of an open ball is in fact its center.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This argument is not correct.  For instance why does it not apply to $\mathbb{Q}$ with the $p$-adic metric?  In fact any field which admits an ordering also admits a nontrivial non-Archimedean metric.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is true though that $\mathbb{Q}_p$ cannot be ordered.  By the Artin-Schreier theorem, this is equivalent to the fact that $-1$ is a sum of squares.  Using Hensel's Lemma and a little quadratic form theory it is not hard to show that $-1$ is a sum of four squares in $\mathbb{Q}_p$.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I know that if you take the completion of the algebraic close of the p-adic completion you get something which is isomorphic to $\mathbb{C}$ (this result was very surprising until I studied model theory, then it became obvious).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I don't mean to pick, but I am familiar with basic model theory and I don't see how it helps to establish this result.  Rather it is basic field theory: any two algebraically closed fields of equal characteristic and absolute transcendence degree are isomorphic.  (From this the completeness of the theory of algebraically closed fields of any given characteristic follows easily, by Vaught's test.)&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;So I was thinking, is there a $p$-adic number whose square equals 2? 3? 2011? For which prime numbers $p$? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;All of these answers depend on $p$.  The general situation is as follows: for any odd $p$, the group of &lt;strong&gt;square classes&lt;/strong&gt; $\mathbb{Q}_p^{\times}/\mathbb{Q}_p^{\times 2}$ -- which parameterizes quadratic extensions -- has order $4$, meaning there are exactly three quadratic extensions of $\mathbb{Q}_p$ inside any algebraic closure.  If $u$ is any integer which is not a square modulo $p$, then these three extensions are given by adjoinging $\sqrt{p}$, $\sqrt{u}$ and $\sqrt{up}$.  When $p = 2$ the group of square classes has cardinality $8$, meaning there are $7$ quadratic extensions.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;How far down the rabbit hole of algebraic numbers can you go inside the p-adic numbers? Are there general results connecting the choice (or rather properties) of $p$ to the &quot;amount&quot; of algebraic closure it gives?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I don't know exactly what you are looking for as an answer here.  The absolute Galois group of $\mathbb{Q}_p$ is in some sense rather well understood: it is an infinite profinite group but it is &quot;small&quot; in the technical sense that there are only finitely many open subgroups of any given index.  Also every finite extension of $\mathbb{Q}_p$ is solvable.  All in all it is vague -- but fair -- to say that the fields $\mathbb{Q}_p$ are &quot;much closer to being algebraically closed&quot; than the field $\mathbb{Q}$ but &quot;not as close to being algebraically closed&quot; as the finite field $\mathbb{F}_p$.  This can be made precise in various ways.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are interested in the $p$-adic numbers you should read intermediate level number theory texts on local fields.  For instance &lt;a href=&quot;http://www.math.uga.edu/~pete/MATH8410.html&quot; rel=&quot;nofollow&quot;&gt;this page&lt;/a&gt; collects notes from a course on (in part) local fields that I taught last spring.  I also highly recommend books called &lt;em&gt;Local Fields&lt;/em&gt;: one by Cassels and one by Serre.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Added&lt;/b&gt;: see in particular Sections 5.4 and 5.5 &lt;a href=&quot;http://math.uga.edu/~pete/8410Chapter5.pdf&quot; rel=&quot;nofollow&quot;&gt;of this set of notes&lt;/a&gt; for information about the number of $n$th power classes and the number of field extensions of a given degree.&lt;/p&gt;&#xA;" OwnerUserId="299" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2011-01-07T23:58:30.297" LastActivityDate="2011-01-07T23:58:30.297" CommentCount="6" />
  <row Id="16731" PostTypeId="2" ParentId="16728" CreationDate="2011-01-07T23:56:15.073" Score="4" ViewCount="" Body="&lt;p&gt;Everything works out as long as you are careful about the additive constants in your integrals!  What you should actually have written is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_1^{x} t^{r-1} \, dt = \begin{cases} \frac{x^r - 1}{r} &amp;amp; r \neq 0 \\\ \ln x &amp;amp; r = 0 \end{cases}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and then I leave it as an exercise to verify that indeed $\lim_{r \to 0} \frac{x^r - 1}{r} = \ln x$ for positive $x$ (note that this already follows from the above integral formula since integrals respect uniform limits).  A great example of the importance of keeping track of that $+C$.  This exact limit came up in another math.SE question which I can't track down.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the application to the Euler-Lagrange equation I guess you can take $A = -B = \frac{1}{\Delta r}$; then the limit becomes $2 \ln t$.  Probably you can get this by using the right initial conditions.  &lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2011-01-08T00:25:47.517" LastActivityDate="2011-01-08T00:25:47.517" CommentCount="3" />
  <row Id="16732" PostTypeId="1" CreationDate="2011-01-07T23:59:57.753" Score="4" ViewCount="83" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm trying to find the general solution to &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\text{d}y}{\text{d}x} = \frac{y-x^2}{\sin y-x}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any ideas would be greatly appreciated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="5498" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2011-01-08T00:05:50.443" LastActivityDate="2011-01-08T00:26:43.227" Title="solving a differential equation involving $\frac{y-x^2}{\sin y-x}$" Tags="&lt;differential-equations&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="16733" PostTypeId="2" ParentId="16706" CreationDate="2011-01-08T00:12:17.270" Score="3" ViewCount="" Body="&lt;p&gt;You can do this for an arbitrary ring (with or without unit). Jacobson's original article can be found &lt;a href=&quot;http://www.jstor.org/stable/87790&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; (JSTOR, needs a university subscription).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I cannot do better than to simply quote C. Chevalley's &lt;a href=&quot;http://www.ams.org/mathscinet-getitem?mr=13138&quot; rel=&quot;nofollow&quot;&gt;Math Review&lt;/a&gt; (MathSciNet, needs a university subscription): &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A (two-sided) ideal $\frak J$ in a ring $\frak A$ is called primitive if $0$&#xA;  is the only $x\in\frak A$ such that ${\frak A}x\subset\frak J$. Let $S$ be&#xA;  the set of primitive ideals in $\frak A$; a topology is introduced in $S$&#xA;  by defining the closure of a subset $M$ of $S$ to be the set of primitive&#xA;  ideals which contain the intersection of all ideals in $M$. The space&#xA;  defined in this way is a $T_0$-space, but generally not $T_1$ (except when&#xA;  $\frak A$ is commutative). If $\frak A$ has a unit element, $S$ is compact&#xA;  (that is, bicompact). The space $S$ is not changed if $\frak A$ is replaced&#xA;  by $\frak A/\frak R$, where $\frak R$ is the radical of $\frak A$. If&#xA;  $\frak A$ is semisimple and has a unit element, a necessary and sufficient&#xA;  condition for $S$ to be disconnected is that there should exist a&#xA;  decomposition of $\frak A$ as the direct sum of two two-sided ideals not&#xA;  equal to $\{0\}$. It is shown that, given any totally disconnected compact&#xA;  space $S$, it is possible to construct a ring $\frak A$ whose space of&#xA;  primitive ideals is homeomorphic with $S$: the elements of $S$ are certain&#xA;  mappings of $S$ into a field (not necessarily commutative) which can be&#xA;  selected arbitrarily.&#xA;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5363" LastActivityDate="2011-01-08T00:12:17.270" />
  <row Id="16734" PostTypeId="2" ParentId="16732" CreationDate="2011-01-08T00:26:43.227" Score="7" ViewCount="" Body="&lt;p&gt;Your equation is exact once you write it as $$f(x,y)\,\mathrm d x+g(x,y)\,\mathrm d y=0.$$ Find a potential, and voilà. I'll leave you the fun of doing that; the general solution is implictly defined by the equation $$\frac{x^3}{3}-xy-\cos y=c$$ with $c$ a constant.&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2011-01-08T00:26:43.227" CommentCount="1" />
  <row Id="16735" PostTypeId="2" ParentId="16646" CreationDate="2011-01-08T00:29:10.490" Score="2" ViewCount="" Body="&lt;p&gt;The hint suggests modifying the proof that $H^1$ for $C^\infty$-sheaves vanishes.  This also makes me guess that the problem is about Cech cohomology.  The $C^\infty$ proof appears in many places, one reference is Proposition 4.1 in chapter IX of Miranda's book: Algebraic Curves and Riemann Surfaces.    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's probably a way of doing this more cleanly and in more generally but as a start: on $X$ fix a locally finite covering $\{U_i\}_{i \in I}$ where the index set is linearly ordered: $I \subset \mathbb{Z}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case, choose for each natural number $n$ a set of integers $a_1,..., a_n$ such that $\sum_{1}^n a_i = 1$; e.g. $a_1 = 1$ and all others $0$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then define functions $\phi_i \colon U_i \to \mathbb{Z}$ as follows:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) For $p \in U_i$ let $U_{j_1}, ..., U_{j_n}$ be sets containing $p$ with $j_1 &amp;lt; ... &amp;lt; j_n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) let $k$ be the index such that $j_k = i$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) let $a_1,..., a_n$ be the previously determined integers summing to $1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Define $\phi_i(p) = a_k$.  These are discontinuous functions and by construction $\sum_i \phi_i \equiv 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To show $H^1(X, \mathcal{D}) = H^1(\{U_i\}, \mathcal{D}) = 0$ we have to show that every $1$-cocycle $(f_{ij})$ is a co-boundary. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The proof now proceeds as in the $C^\infty$ case: on $U_i$ define $g_i = -\sum_j \phi_j f_{ij}$.  Then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$g_i - g_j = -\sum_k \phi_kf_{ik} + \sum_k \phi_kf_{jk} = \sum_k \phi_k(f_{jk} - f_{ik})$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now using that $f_{jk} - f_{ik} = f_{ji}$ we get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$g_i - g_j$ = $f_{ji}\sum_k \phi_k = f_{ji}$. &lt;/p&gt;&#xA;" OwnerUserId="403" LastActivityDate="2011-01-08T00:29:10.490" />
  <row Id="16736" PostTypeId="1" CreationDate="2011-01-08T00:45:09.647" Score="2" ViewCount="99" Body="&lt;p&gt;In the question &lt;a href=&quot;http://math.stackexchange.com/questions/16716/rings-of-matrices&quot;&gt;Math.SE #16716&lt;/a&gt;, Natalia asked about representing rings of matrices as centralizers of a matrix.  This is an intriguing question, but had some clear problems as rings of matrices need not contain their own centralizers, which would be pretty bad.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In finite groups, subgroups that contain their own centralizers have an eery importance.  It occurred to me that subalgebras of matrix algebras probably have some similar properties, but I don't know what to search for.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What do you call a subalgebra that contains its own centralizer?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In other words, is there a name for a collection of matrices:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;that contains the scalar matrices,&lt;/li&gt;&#xA;&lt;li&gt;that is closed under addition and multiplication, and&lt;/li&gt;&#xA;&lt;li&gt;that contains every matrix that commutes with every matrix in the collection?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="583" LastActivityDate="2011-01-08T00:45:09.647" Title="What is it called when a subalgebra contains its centralizer?" Tags="&lt;abstract-algebra&gt;&lt;terminology&gt;&lt;ring-theory&gt;" CommentCount="3" />
  <row Id="16737" PostTypeId="1" CreationDate="2011-01-08T00:58:10.067" Score="4" ViewCount="83" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Whitney embedding theorem states that any smooth manifold can be embedded in Euclidean space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the Riemannian setting this naturally leads to the question whether this can be done in such a way that the Riemannian metric is preserved, that is, whether we can find an embedding such that the inherited metric as a submanifold of Euclidean space is the original metric?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I guess there are partial results on this topic (e.g. spaces of constant curvature), but is there anything like the Whitney embedding theorem?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any answers, links, book recommendations would be very much appreciated!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance for your always great responses,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;S.L.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; Since there was a very short definite answer to my original question in form of a Wikipedia link, I would like to ask for recommendations for books on Riemannian geometry (where this fact is proved).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would really like to pursue this topic a little further.&lt;/p&gt;&#xA;" OwnerUserId="3208" LastEditorUserId="3208" LastEditorDisplayName="" LastEditDate="2011-01-08T01:37:05.887" LastActivityDate="2011-01-08T01:37:05.887" Title="Riemannian Immersions into Euclidean Space?" Tags="&lt;differential-geometry&gt;&lt;riemannian-geometry&gt;" CommentCount="5" />
  <row Id="16738" PostTypeId="1" CreationDate="2011-01-08T01:06:56.337" Score="0" ViewCount="136" Body="&lt;p&gt;Looking for the proof of the lemma asserting that the conical surface (envelope) is a closed space.&#xA;Thank you.&lt;/p&gt;&#xA;" OwnerUserId="5499" LastEditorUserId="367" LastEditorDisplayName="" LastEditDate="2011-01-09T10:52:22.233" LastActivityDate="2011-05-09T11:55:55.277" Title="Topology: Proof that Conical Surface are a closed space" Tags="&lt;topology&gt;&lt;topological-vector-spaces&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" />
  <row Id="16739" PostTypeId="1" AcceptedAnswerId="16749" CreationDate="2011-01-08T02:07:42.213" Score="1" ViewCount="142" Body="&lt;p&gt;Hello,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;does there exist the notion of a non-integer power of a matrix? This seems to be accessible via semigroup-theory, yet I have not seen an actual definition so far.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not too firm at this right now, but I am curious. Can you give me a sketch of the definition and provide with some introductory information?&lt;/p&gt;&#xA;" OwnerUserId="3557" LastActivityDate="2011-01-08T03:08:20.793" Title="Arbitrary non-integer power of a matrix" Tags="&lt;analysis&gt;&lt;matrices&gt;" AnswerCount="3" />
  <row Id="16740" PostTypeId="2" ParentId="16739" CreationDate="2011-01-08T02:10:43.207" Score="1" ViewCount="" Body="&lt;p&gt;You can use the &lt;a href=&quot;http://en.wikipedia.org/wiki/Binomial_series&quot; rel=&quot;nofollow&quot;&gt;binomial series&lt;/a&gt; to define powers for appropriate matrices.&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2011-01-08T02:10:43.207" CommentCount="1" />
  <row Id="16741" PostTypeId="2" ParentId="855" CreationDate="2011-01-08T02:10:57.140" Score="4" ViewCount="" Body="&lt;p&gt;One simple, high school level proof: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Every prime number $p&amp;gt;3$ can be written in form $6k \pm 1$. This is easily proved by considering remainders upon dividing by $6$. Using that fact, it suffices to show that any number of that form is going to be divisible by $24$, because that implies that any prime greater than $3$ is going to be divisible by it. Proof uses just a little algebraic manipulation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(6k \pm 1)^2 - 1 \Rightarrow 36k^2 \pm 12k + 1 - 1 \Rightarrow 12k(3k \pm 1)$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We use the fact that for every even number times $12$, resulting number is divisible by $24$. So, if $k$ is even then we are done. However, if $k$ is odd, then $3k \pm 1$ is going to be even. Therefore, $k(3k\pm1)$ is even, so we write:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$k(3k\pm1) = 2m \Rightarrow 12\cdot2m \Rightarrow 24m$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&lt;strong&gt;Addendum&lt;/em&gt;&lt;/strong&gt;: This above result is just a part of generalised result which we will now prove.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $p$ is prime number such that $p&amp;gt;0$, then following holds for all natural numbers $n$: $$ 3 \cdot 2^{2 + n} |\ p^{2^{n}} - 1 $$&lt;br&gt;&#xA;We are going to prove it using the induction on natural numbers.&lt;br&gt;&#xA;The base case, when $n=1$, has already been proven in first part of the post: $3 \cdot 2^{2 + 1} |\ p^{2^{1}} - 1 \Leftrightarrow 24 |\ p^2 - 1$&lt;br&gt;&#xA;Suppose that it is valid for $n$: $3 \cdot 2^{2 + n} |\ p^{2^{n}} - 1$, and let us examine case for $n+1$: $$ p^{2^{n+1}} - 1 = (p^{2^{n}} - 1)(p^{2^{n}} + 1) $$&#xA;By induction hypothesis, we can rewrite this as: $k(3 \cdot 2^{n+2})(p^{2^{n}} + 1)$, for some natural number $k$. Our third term, $(p^{2^{n}} + 1)$, is always going to be even, as power of odd prime will be odd, plus one it will be even, so we can rewrite this as $k(3 \cdot 2^{n+2})2q = 3 \cdot 2^{(n+1) + 2} \cdot kq$, for some natural number $q$. As $3 \cdot 2^{(n+1) + 2} \cdot kq = p^{2^{n+1}} - 1 $ for some natural numbers, $k$ and $q$, it follows that $3 \cdot 2^{(n+1) + 2} |\ p^{2^{n+1}} - 1$ and this completes our proof.&lt;/p&gt;&#xA;" OwnerUserId="5501" LastEditorUserId="5501" LastEditorDisplayName="" LastEditDate="2011-03-27T22:56:38.713" LastActivityDate="2011-03-27T22:56:38.713" />
  <row Id="16742" PostTypeId="1" AcceptedAnswerId="16747" CreationDate="2011-01-08T02:12:28.567" Score="1" ViewCount="157" Body="&lt;p&gt;I have been stuck on this one for hours. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $x$, $y$, $z$ be non-negative real numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also we know $x + z \leq 2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Prove the following:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$(x - 2y + z)^2 \geq 4xz - 8y$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Apparently this can be proven with or without AGM, which is $xy \leq \left(\frac{x + y}{2}\right)^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;This is what I have done so far:&lt;/strong&gt;&#xA;\begin{align}&#xA;((x + z) - 2y)^2 &amp;amp;\geq 4xz - 8y\\&#xA;(x + z)^2 -4y(x + z) + 4y^2 &amp;amp;\geq 4xz - 8y&#xA;&amp;amp;\quad&amp;amp;\text{expanded the squared term keeping }(x+z)\\&#xA;(x + z)^2 -4yx - 4yz + 4y^2 &amp;amp;\geq  4xz - 8y&amp;amp;&amp;amp;\text{now we have AGM}\\&#xA;\left(\frac{x + z}{2}\right)^2 -yx -yz + y^2 &amp;amp;\geq xz -2y&#xA;\end{align}&#xA;Rearranging we have&#xA;$$\left(\frac{x + z}{2}\right)^2 -yx -yz + 2y \geq xz - y^2.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is as far as I got and we also know that the told us this:&#xA;$x + z \leq 2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rearranging we get: $x + z - 2 \leq 0$; then multiply by $y$ to get:&#xA;$yx + yz - 2y \leq 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am new to this proofs and if someone can guide me as to how to attempt these and whats the method to solve these question that would be really great, thanks :)&lt;/p&gt;&#xA;" OwnerUserId="5505" LastEditorUserId="5505" LastEditorDisplayName="" LastEditDate="2011-01-08T02:51:08.330" LastActivityDate="2011-01-08T08:54:01.777" Title="AGM Inequality Proof" Tags="&lt;algebra-precalculus&gt;&lt;inequality&gt;" AnswerCount="4" FavoriteCount="1" />
  <row Id="16743" PostTypeId="1" AcceptedAnswerId="16881" CreationDate="2011-01-08T02:14:54.963" Score="1" ViewCount="133" Body="&lt;p&gt;Suppose I am given the following inductive definition$^1$: Let $T$ be the smallest family of sets $\{T_1,T_2,T_3,\ldots\}$ such that&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$0 \leq k &amp;lt; n \Rightarrow k \in T_n$&lt;/li&gt;&#xA;&lt;li&gt;$t_1 \in T_n \wedge n &amp;gt; 0 \Rightarrow \lambda.t_1 \in T_{n - 1}$&lt;/li&gt;&#xA;&lt;li&gt;$t_1 \in T_n \wedge t_2 \in T_n \Rightarrow (t_1 ~ t_2) \in T_n$&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The elements of $T_n$ are the nameless $\lambda$-terms (with de Bruijn indices) with at most $n$ free variables numbered between 0 and $n - 1$. I want to express $T$ inductively, but am having trouble. In particular, I'm not sure how to express (2) and (3). For instance, suppose I try to define the set $T_{n + 1}$ in terms of $T_n$ for (2): in order to define $T_{n + 1}$, I need to know which elements occur in $T_{n + 2}$. For (3), in order to define $T_{n + 1}$, I need to know which elements occur in $T_{n + 1}$. I am being led in circles!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't understand the underlying principle which would permit me to define the family of sets $T$, because it's not a straightforward inductive definition. I assume the answer lies in defining mutually recursive sets, then finding their union. However, I'm stuck. Can anyone enlighten me?&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Pierce, &lt;em&gt;Types and Programming Languages&lt;/em&gt;, 6.1.2.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="5502" LastActivityDate="2011-01-09T22:56:16.120" Title="Inductively Defined Family of Sets" Tags="&lt;homework&gt;&lt;elementary-set-theory&gt;&lt;lambda-calculus&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" />
  <row Id="16744" PostTypeId="2" ParentId="16739" CreationDate="2011-01-08T02:21:21.913" Score="2" ViewCount="" Body="&lt;p&gt;If your matrix has positive eigenvalues, then one definition is to take non-integer powers of each eigenvalue (but keep the eigenvectors the same).  This is a common definition used to take square roots, for example.  &lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2011-01-08T02:21:21.913" />
  <row Id="16745" PostTypeId="1" CreationDate="2011-01-08T02:22:54.017" Score="6" ViewCount="217" Body="&lt;p&gt;Hi everybody,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As above, I'm trying to find all linear operators $L: C([0,1]) \to  C([0,1])$ which satisfy the following 2 conditions:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;I) $Lf \, \geq \, 0$ for all non-negative $f\in C([0,1])$.&lt;/li&gt;&#xA;&lt;li&gt;II) $Lf = f$ for $f(x)= 1$, $f(x)=x$, and $f(x)=x^2$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I'm honestly not sure where to start here - I'm struggling to use these conditions to pare down the class of linear operators which could satisfy the conditions significantly. Could anyone help me get a result out of this? Thank you! &lt;/p&gt;&#xA;" OwnerUserId="5506" LastEditorUserId="5506" LastEditorDisplayName="" LastEditDate="2011-01-08T14:48:01.670" LastActivityDate="2011-01-08T19:43:26.220" Title="Finding all linear operators $L: C([0,1]) \to C([0,1])$ which satisfy 2 conditions" Tags="&lt;real-analysis&gt;&lt;homework&gt;&lt;analysis&gt;&lt;operator-theory&gt;" AnswerCount="2" CommentCount="13" />
  <row Id="16746" PostTypeId="2" ParentId="11426" CreationDate="2011-01-08T02:39:00.537" Score="1" ViewCount="" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Show that the term ZZ where Z is&#xA;  λz.λx. x(z z x) satisfies the&#xA;  requirement for fixed point&#xA;  combinators that ZZM =β M(ZZM).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You are given a term $Z^2 = ZZ$ where $Z = \lambda z . \lambda x . ~ x ~ (z ~ z ~ x)$ and asked to prove that $Z^2$ satisfies the equality $Z^2M ~ =_{\beta} ~ M(Z^2M)$. In order to demonstrate that $Z^2$ satisfies the equality, it suffices to show that $Z^2M$ $\beta$-reduces to $M(Z^2M)$ (modulo renaming of bound variables):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$Z^2M ~=_{\beta} ~ ((\lambda z .\lambda x.~x~(z~z~x))(\lambda z .\lambda x.~x~(z~z~x)))~ M$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$~~~~~~~~~~=_{\beta}~ (\lambda x.~x~((\lambda z .\lambda x.~x~(z~z~x))~(\lambda z .\lambda x.~x~(z~z~x))~x)~M$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$~~~~~~~~~~=_{\beta}~ M~((\lambda z . \lambda x . ~x~(z~z~x))~(\lambda z.\lambda x.~x(z~z~x)) ~ M)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$~~~~~~~~~=~M~(Z^2 ~ M)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Showing that the combinator satisfies the equality is simply a matter of $\beta$-reduction. For instance, the proof of the fixpoint theorem in $\lambda$-calculus is just the construction of such a combinator.&lt;/p&gt;&#xA;" OwnerUserId="5502" LastActivityDate="2011-01-08T02:39:00.537" />
  <row Id="16747" PostTypeId="2" ParentId="16742" CreationDate="2011-01-08T02:46:27.110" Score="4" ViewCount="" Body="&lt;p&gt;This is the solution to the old version of the problem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The right hand side is $4xy-8y=4y(x-2)$ and hence non-positive, since $y$ is non-negative $x$ is at most $2$. But the left hand side is a square and hence non-negative. So the inequality follows. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;For the revised version:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $u=x+z$, so the LHS is $(u-2y)^2$. Notice that swapping value between $x$ and $z$ does not affect $u$, but by the AGM the quantity $xz$ is at most $(\frac u2)^2$. So the RHS is at most $u^2-8y$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the problem reduces to showing $(u-2y)^2\geq u^2-8y$. This is easy if $y=0$, so we may assume $y\neq 0$. Expand the left, cancel $u^2$ and divide by $4y$ to reduce to $-u+y\geq -2$, or $y\geq u-2$, which is true. So we're done.&lt;/p&gt;&#xA;" OwnerUserId="413" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-08T03:36:54.867" LastActivityDate="2011-01-08T03:36:54.867" CommentCount="4" />
  <row Id="16748" PostTypeId="2" ParentId="16742" CreationDate="2011-01-08T02:57:40.277" Score="1" ViewCount="" Body="&lt;p&gt;EDIT: This solves the earlier version of the question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a more &quot;brute-force&quot; approach which seems silly given JDH's solution, but might help you attack similar questions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First of all, we see that $z$ is just there to confuse us. Suppose $x+z$ is held constant - that determines the LHS. In order to prove the inequality, we should consider the maximal RHS. Since the RHS is monotone increasing with $x$, we might as well put $z=0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now expand the square and simplify $LHS-RHS \geq 0$ to get $$x^2 + 4y^2 - 8xy + 8y \geq 0.$$ What do we do with this unwieldy expression? Suppose there were some &quot;bad&quot; values of $y$, for which some $x \in [0,2]$ would be negative. Since everything is continuous, we could find the &quot;critical&quot; values of $y$ by trying both endpoints. Putting $x = 0$ we get the critical value $y = 0$ (as a solution to the equation $LHS - RHS = 0$). Putting $x = 2$ we get the critical value $y = 1$: $$4 + 4y^2 - 16y + 8y = 4y^2 - 8y + 4 = 4(y-1)^2.$$ This suggests substituting $y = Y + 1$: $$x^2 + 4(Y+1)^2 - 8x(Y+1) + 8(Y+1) = x^2 - 8xY - 8x + 4Y^2 + 16Y + 12.$$ That doesn't look any better, does it? We know that $4Y^2$ should be isolated. We quickly notice the subexpression $-8xY+16Y = 8(2-x)Y \geq 0$. What remains is the factorable $x^2 - 8x + 12 = (x-6)(x-2) \geq 0$, and in total we get $$(x-6)(x-2) + 8(2-x)Y + 4Y^2 \geq 0.$$&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2011-01-08T02:57:40.277" CommentCount="1" />
  <row Id="16749" PostTypeId="2" ParentId="16739" CreationDate="2011-01-08T03:08:20.793" Score="2" ViewCount="" Body="&lt;p&gt;There are a few definitions for functions of matrices. See for instance &lt;a href=&quot;http://en.wikipedia.org/wiki/Matrix_function&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Matrix_function&lt;/a&gt; .&lt;/p&gt;&#xA;" OwnerUserId="589" LastActivityDate="2011-01-08T03:08:20.793" />
  <row Id="16750" PostTypeId="2" ParentId="16742" CreationDate="2011-01-08T03:43:55.670" Score="3" ViewCount="" Body="&lt;p&gt;This way doesn't quite use the AM-GM inequality but is pretty similar: First note that $x$ and $z$ are symmetrical in this problem, so without loss of generality you can assume that $z \leq x$. Since $x + z \leq 2$ this means that $z \leq 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that $(a + b)^2 \geq 4ab$ for all $a$ and $b$. (This is almost the same as the AM-GM inequality). In view of your left hand side this is a natural form to use. The question is what to use as $a$ and what to use as $b$. Looking at what you are trying to show, I use $a = x - 2y$ and $b = z$. I obtain&#xA;$$(x - 2y + z)^2 \geq 4xz - 8yz$$&#xA;Since $z \leq 1$, we have $-8yz \geq -8y$. Substituting this in we get the needed&#xA;$$(x - 2y + z)^2 \geq 4xz - 8y$$&#xA;As to how you'd know to do this.. I got the second to last equation first, then tried to see how to convert this into the form you want. It was clear I needed $z \leq 1$ so then I used the condition $x + z \leq 2$ as above.&lt;/p&gt;&#xA;" OwnerUserId="3035" LastActivityDate="2011-01-08T03:43:55.670" />
  <row Id="16751" PostTypeId="2" ParentId="16724" CreationDate="2011-01-08T04:08:33.257" Score="2" ViewCount="" Body="&lt;p&gt;Suppose that $K$ is an algebraic number field, i.e. a finite extension of $\mathbb Q$.  It has a ring of integers $\mathcal O_K$ (the integral closure of $\mathbb Z$ in $K$).  Suppose that there is a prime ideal $\wp \subset \mathcal O_K$ such that:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;$p \in \wp,$ but $p \not\in \wp^2$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The order of $\mathcal O_K/\wp = p.$  (Note that (1) implies in particular that&#xA;$\wp \cap \mathbb Z = p \mathbb Z$, so that $\mathcal O_K/\wp$ is an extension of $\mathbb Z/p\mathbb Z$.  We are now requiring that it in fact be the trivial extension.)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Then the number field $K$ embeds into $\mathbb Q_p$.  The converse also holds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if you want to know whether you can solve the equation $f(x) = 0$ in $\mathbb Q_p$ (where $f(x)$ is some irreducible polynomial in $\mathbb Q[x]$), then set&#xA;$K = \mathbb Q[x]/f(x)$ and apply this criterion.&#xA;This is easiest to do when&#xA;$f(x)$ has integral coefficients, and remains separable when reduced mod $p$&#xA;(something that you can check by computing the discriminant and seeing whether&#xA;or not it is divisible by $p$),&#xA;because in this case the criterion is equivalent to asking that $f(x)$ have a root mod $p$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Incidentally, there are many $f(x)$ that satisfy this criterion&#xA;(because, among other&#xA;things, the algebraic closure of $\mathbb Q$ in $\mathbb Q_p$ has infinite degree over $\mathbb Q$), but there are also many $f(x)$ that don't.&lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2011-01-08T04:08:33.257" />
  <row Id="16752" PostTypeId="2" ParentId="16646" CreationDate="2011-01-08T04:14:22.910" Score="1" ViewCount="" Body="&lt;p&gt;The sheaf $\mathcal D$ is equal to $\bigoplus_{x \in X} i_{x*} \mathbb Z,$&#xA;where $\mathbb Z$ denotes the constant sheaf on the point $x$, and $i_x:\{x \} \to X$&#xA;is the embedding of the one point space $\{x\}$ into the Riemann surface $X$.&#xA;(So $i_{x*}\mathbb Z$ is the skyscraper sheaf supported at $x$, whose stalk is $\mathbb Z$).&#xA;The vanishing of $H^1$ now follows from the vanishing of higher cohomology for skyscraper sheaves.&lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2011-01-08T04:14:22.910" />
  <row Id="16753" PostTypeId="2" ParentId="16709" CreationDate="2011-01-08T04:27:23.403" Score="2" ViewCount="" Body="&lt;p&gt;$\frac{\partial L}{\partial \dot{q}}$ may be understood as the generalized momentum of the system (e.g. $\frac{d}{dv}\left(\frac{1}{2}mv^2\right) = mv$, the derivative of kinetic energy with respect to velocity is momentum).  Then $\frac{d}{dt}\frac{\partial L}{\partial \dot{q}}$ is the time derivative of the generalized momentum.  Likewise the $\frac{\partial L}{\partial q}$ term behaves like a generalized force. This begins to look like Newton's second law, but for generalized coordinates.  &lt;/p&gt;&#xA;" OwnerUserId="4726" LastEditorUserId="4726" LastEditorDisplayName="" LastEditDate="2011-01-08T22:19:46.410" LastActivityDate="2011-01-08T22:19:46.410" />
  <row Id="16754" PostTypeId="1" CreationDate="2011-01-08T04:54:59.337" Score="3" ViewCount="246" Body="&lt;p&gt;Let $c$ be an integer, not necessarily positive and not a square. Let $R=\mathbb{Z}[\sqrt{c}]$ &#xA;denote the set of numbers of the form $$a+b\sqrt{c}, a,b \in \mathbb{Z}.$$ &#xA;Then $R$ is a subring of $\mathbb{C}$ under the usual addition and multiplication.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is: if $R$ is a UFD (unique factorization domain), does it follow that&#xA;it is also a PID (principal ideal domain)?&lt;/p&gt;&#xA;" OwnerUserId="3249" LastActivityDate="2011-01-08T05:28:56.393" Title="Unique factorization domain that is not a Principal ideal domain" Tags="&lt;abstract-algebra&gt;&lt;ring-theory&gt;" AnswerCount="3" FavoriteCount="2" />
  <row Id="16755" PostTypeId="2" ParentId="16754" CreationDate="2011-01-08T04:59:46.110" Score="3" ViewCount="" Body="&lt;p&gt;Yes. If it is a UFD, then it is integrally closed, hence it is a Dedekind domain because it is of dimension one (being contained in the integral closure of $\mathbb{Z}$ in some finite extension of $\mathbb{Q}$). A Dedekind domain is a UFD iff it is a PID: indeed, this is equivalent to every non-zero prime being principal. (A noetherian domain is a UFD iff every height one prime is principal. So if a Dedekind domain is a UFD, then all its primes are principal, so by factorization of ideals, every ideal is principal.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A simple example of a UFD that is not a PID is the polynomial ring $\mathbb{C}[x,y]$.&lt;/p&gt;&#xA;" OwnerUserId="536" LastActivityDate="2011-01-08T04:59:46.110" CommentCount="0" />
  <row Id="16756" PostTypeId="2" ParentId="16754" CreationDate="2011-01-08T05:04:12.140" Score="4" ViewCount="" Body="&lt;p&gt;The answer is yes.  The argument is as follows: if $R$ is a UFD, then it is necessarily integrally closed in its fraction field $K = \mathbb Q(\sqrt{c})$, and thus is&#xA;equal to the full ring of algebraic integers in $K$.  A general fact about such full rings of algebraic integers is that if they are UFDs then they are PIDs, the reason being that in these rings, one always has the unique factorization of non-zero ideals into prime ideals, and it is not hard to see that the UFD property forces prime ideals (and hence any product of prime ideals) to be principal.&lt;/p&gt;&#xA;" OwnerUserId="221" LastEditorUserId="221" LastEditorDisplayName="" LastEditDate="2011-01-08T05:16:56.487" LastActivityDate="2011-01-08T05:16:56.487" />
  <row Id="16757" PostTypeId="1" AcceptedAnswerId="16758" CreationDate="2011-01-08T05:16:37.343" Score="1" ViewCount="65" Body="&lt;p&gt;Let $X$ be a smooth projective curve of genus $g$. Let $\Omega$ be the sheaf of differentials. Mumford (in &lt;em&gt;Abelian Varieties,&lt;/em&gt; sec. 2.6, in proving the theorem of the cube) asserts that there is an effective divisor of degree $g$ on $X$ such that $H^0(X, \Omega \otimes L(-D))=0$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This should be easy, but I'm missing the argument. Could someone explain?&lt;/p&gt;&#xA;" OwnerUserId="536" LastActivityDate="2011-01-08T05:26:11.503" Title="A divisor on a smooth curve such that $\Omega$ minus it has no nonzero sections" Tags="&lt;algebraic-geometry&gt;" AnswerCount="1" />
  <row Id="16758" PostTypeId="2" ParentId="16757" CreationDate="2011-01-08T05:26:11.503" Score="3" ViewCount="" Body="&lt;p&gt;Dear Akhil,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $\mathcal L$ be an invertible sheaf on the curve $X$, and suppose&#xA;that $H^0(X,\mathcal L)$ has dimension $d$.  (In the case when $\mathcal L&#xD;&#xA;= \Omega$, we have $d = g$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we choose a closed point $x \in X$, then there is a map $H^0(X,\mathcal L) \to &#xD;&#xA;\mathcal L_x/\mathfrak m_x \mathcal L_x$ given by mapping a section to its fibre at $x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To make my life easier, let's assume that $k$ (the field over which $X$ is defined) is algebraically closed, so that $x$ is defined over $k$, and we &#xA;may identify $\mathcal L_x/\mathfrak m_x\mathcal L_x$ with $k$ (since it one&#xA;dimensional, being the fibre of an invertible sheaf).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Evaluation is then a functional $H^0(X,\mathcal L) \to k.$  Now this functional will be identically zero if and only if every section vanishes at $x$.  But a non-zero section has only finitely many zeroes, and $X$ has an infinite number of closed points, so we may certainly choose $x$ so that this functional is not&#xA;identically zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The evaluation map sits in an exact sequence&#xA;$$0 \to H^0(X,L(-x)) \to H^0(X,L) \to k,$$&#xA;and so if we choose $x$ such that evaluation is surjective, we find that&#xA;$H^0(X,L(-x))$ has dimension $d - 1$.  Proceeding by induction, we find&#xA;points $x_1,\ldots,x_d$ such that $H^0(X,L(-x_1-\cdots - x_d)) = 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In summary: we have shown that we may find an effective degree $d$ divisor $D$&#xA;such that $H^0(X,L(-D))$ vanishes.  (And in fact, looking at the proof, we see&#xA;that this vanishing will hold for a generic choice of $D$.)&lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2011-01-08T05:26:11.503" CommentCount="1" />
  <row Id="16759" PostTypeId="2" ParentId="16754" CreationDate="2011-01-08T05:28:56.393" Score="2" ViewCount="" Body="&lt;p&gt;Yes, because such quadratic number rings are easily shown to have dimension at most one (i.e. every nonzero prime ideal is maximal). But $\rm PID\:$s are &lt;em&gt;precisely&lt;/em&gt; the $\rm UFD\:$s having dimension $\le 1\:.\ $ Below is a sketch of a proof of this and closely related results.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;THEOREM&lt;/strong&gt; $\rm\ \ \ TFAE\ $ for a $\rm UFD\ D$ &lt;br/&gt;&#xA;$1)\ $ prime ideals are maximal if nonzero&lt;br&gt;&#xA;$2)\ $ prime ideals are principal&lt;br&gt;&#xA;$3)\ $ maximal ideals are principal&lt;br&gt;&#xA;$4)\ \rm\ gcd(a,b) = 1\ \Rightarrow\ (a,b) = 1$ &lt;br/&gt;&#xA;$5)\ $ $\rm D$ is Bezout&lt;br&gt;&#xA;$6)\ $ $\rm D$ is a $\rm PID$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof&lt;/strong&gt; $\ $ (sketch of $1 \Rightarrow 2 \Rightarrow 3 \Rightarrow 4 \Rightarrow 5 \Rightarrow 6 \Rightarrow 1$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$1\Rightarrow 2)$ $\rm\ \ P\supset (p)\ \Rightarrow\ P = (p)$ &lt;br/&gt;&#xA;$2\Rightarrow 3)$ $\ \: $ Clear.&lt;br&gt;&#xA;$3\Rightarrow 4)$ $\ \ \rm (a,b) \subsetneq P = (p)\ $ so $\rm\ (a,b) = 1$ &lt;br/&gt;&#xA;$4\Rightarrow 5)$ $\ \ \rm c = \gcd(a,b)\ \Rightarrow\ (a,b) = c\ (a/c,b/c) = (c)$ &lt;br/&gt;&#xA;$5\Rightarrow 6)$ $\ \ \rm 0 \ne I \subset D\:$ Bezout is generated by an elt with the least number of prime factors&lt;br&gt;&#xA;$6\Rightarrow 1)$ $\ \ \rm P \supset (p),\ a \not\in (p)\ \Rightarrow\ (a,p) = (1)\ \Rightarrow\ P = (p)$&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2011-01-08T05:28:56.393" CommentCount="2" />
  <row Id="16761" PostTypeId="2" ParentId="16663" CreationDate="2011-01-08T06:11:44.863" Score="3" ViewCount="" Body="&lt;p&gt;Another way to proceed is to look at the area of the function in your domain $[0,1]$. (I usually use this as a first step to check if the function is not uniformly convergent, since it is relatively easy.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The area of the function $f_n(x)$ in the domain is $\frac{1}{2}-\frac{1}{3} = \frac{1}{6}$ irrespective of $n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that $f_n(0) = 0, \forall n$ and $\displaystyle \lim_{n \rightarrow \infty} f_n(x) = 0$, $\forall x \in (0,1]$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This can be seen since for any $x$, $\exists N \in \mathbb{N}$ such that $\forall n &amp;gt; N$, $\frac{1}{n} &amp;lt; x \Rightarrow f_n(x) = 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, $f(x) = \displaystyle \lim_{n \rightarrow \infty} f_n(x) = 0$, $\forall x \in [0,1]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we have $\displaystyle \int_{0}^{1} f_n(x) dx = \frac{1}{6}$, $\forall n \in \mathbb{N}$ and $\displaystyle \int_{0}^{1} f(x) dx = 0$. So we have $$\displaystyle \lim_{n \rightarrow \infty} \int_{0}^{1} f_n(x) dx = \frac{1}{6} \neq 0 = \displaystyle \int_{0}^{1} f(x) dx$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, we have $\displaystyle \lim_{n \rightarrow \infty} \int_{0}^{1} f_n(x) dx \neq \displaystyle \int_{0}^{1} \lim_{n \rightarrow \infty} f_n(x) dx$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And we know that if a sequence of functions converge uniformly, we can swap the limit and the integrals to get the same integral.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, the function is &lt;strong&gt;not&lt;/strong&gt; uniformly convergent.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2011-01-08T20:51:28.160" LastActivityDate="2011-01-08T20:51:28.160" CommentCount="5" />
  <row Id="16762" PostTypeId="2" ParentId="16745" CreationDate="2011-01-08T06:24:21.900" Score="0" ViewCount="" Body="&lt;p&gt;The answer is: $L$ must be the identity of $C([0,1])$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $f: [0,1] \to \mathbb{R}$ be continuous. We want to prove that $Lf = f$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $[0,1]$ is compact, $f$ is uniformly continuous, so for fixed $\varepsilon \gt 0$ we can choose $\delta$ such that $|x - y| \leq \sqrt{\delta}$ implies $|f(x) - f(y)| \lt  \varepsilon$. Observe that&#xA;\[&#xA;|f(x) - f(y)| \leq \varepsilon + \frac{2\|f\|}{\delta}(x - y)^{2}&#xA;\]&#xA;for all $x,y \in [0,1]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;If $|x - y| \leq \sqrt{\delta}$ this is clear and otherwise we have $(x - y)^{2} \gt \delta$, so $\varepsilon + \frac{2\|f\|}{\delta}(x - y)^{2} &amp;gt; \varepsilon + 2\|f\| \gt |f(x) - f(y)|$.&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, for $C = \frac{2\|f\|}{\delta}$ we have&#xA;\[&#xA;-\varepsilon - C(x - y)^{2} \leq f(x) - f(y) \leq \varepsilon + C(x-y)^{2}.&#xA;\]&#xA;Keep $y$ fixed (so we regard $y$ and $f(y)$ as constants), consider the three parts of these inequalities as functions of $x$ and apply $L$. Using that $L$ is monotone and that $L(ax^{2} + bx + c) = ax^{2} + bx + c$ by hypothesis, we get&#xA;\[&#xA;-\varepsilon - C(x - y)^{2} \leq (Lf)(x) - f(y) \leq \varepsilon + C(x-y)^{2} \qquad \text{for all $x,y \in [0,1]$}.&#xA;\]&#xA;In particular, setting $x = y$ yields $|Lf(y) - f(y)| &amp;lt; \varepsilon$. Since $\varepsilon$ and $y$ were arbitrary, we conclude $Lf(y) = f(y)$ for all $y \in [0,1]$.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The argument given here may be strengthened with only little effort:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Korovkin's Theorem.&lt;/strong&gt;&#xA;&lt;em&gt;Let $L_{n}: C([0,1]) \to C([0,1])$ be positive operators such that $\|L_{n}g_{i} - g_{i}\|_{\infty} \xrightarrow{n \to \infty} 0$ for $g_{i}(x) = x^{i}$, $i = 0,1,2$. Then $\|L_{n}f - f\|_{\infty} \to 0$ for all $f \in C([0,1])$.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Its proof (as well as the argument above) is a variant of the usual proof of the Weierstrass approximation theorem using Bernstein polynomials. One may take&#xA;\[&#xA;L_{n}f(x) = \sum_{k = 0}^{n} \begin{pmatrix} n \\ k \end{pmatrix}x^{k}(1-x)^{n-k} f(k/n),&#xA;\]&#xA;in Korovkin's theorem and verify directly that $L_n g_{i} \to g_{i}$ for $i = 0,1,2$, so Korovkin's theorem yields the Weierstrass approximation theorem.&lt;/p&gt;&#xA;" OwnerUserId="5363" LastEditorUserId="5363" LastEditorDisplayName="" LastEditDate="2011-01-08T19:43:26.220" LastActivityDate="2011-01-08T19:43:26.220" CommentCount="0" />
  <row Id="16763" PostTypeId="2" ParentId="16745" CreationDate="2011-01-08T06:39:20.650" Score="3" ViewCount="" Body="&lt;p&gt;Here are some hints:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;You don't have much choice for $L$.&lt;/li&gt;&#xA;&lt;li&gt;If $f: [0,1] \to \mathbb{R}$ is continuous then it is uniformly continuous, so for all $\varepsilon &amp;gt; 0$ you can choose $\delta &amp;gt; 0$ such that $|x - y| &amp;lt; \sqrt{\delta}$ implies $|f(x) - f(y)| &amp;lt; \varepsilon$. This gives&#xA;\[&#xA;|f(x) - f(y)| &amp;lt; \varepsilon + C (x - y)^{2} \quad \text{for ALL $x,y \in [0,1]$}&#xA;\]&#xA;with $C = \frac{2\|f\|_{\infty}}{\delta}$.&lt;/li&gt;&#xA;&lt;li&gt;Use this to estimate $f(x) - f(y)$ from above and below by a quadratic polynomial in $x$ (view $y$ and $f(y)$ as constants).&lt;/li&gt;&#xA;&lt;li&gt;If $p(x) = ax^{2} + bx + c$ then $Lp = p$ and if $g \leq h$ then $Lg \leq Lh$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="5363" LastEditorUserId="5363" LastEditorDisplayName="" LastEditDate="2011-01-08T15:51:40.170" LastActivityDate="2011-01-08T15:51:40.170" CommentCount="8" />
  <row Id="16764" PostTypeId="2" ParentId="16742" CreationDate="2011-01-08T07:05:50.570" Score="1" ViewCount="" Body="&lt;p&gt;Use $\rm\ (x-2y+z)^2 = 4xz-8y + 4y(2-x-z) + (x-z)^2 + 4y^2\:.\:$ But last 3 summands are $\ge 0\:.\ $ Note this proof requires only that $\rm\ \ y(2-x-z)\ge 0\:.$&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2011-01-08T08:54:01.777" LastActivityDate="2011-01-08T08:54:01.777" />
  <row Id="16765" PostTypeId="1" AcceptedAnswerId="16840" CreationDate="2011-01-08T07:42:48.990" Score="4" ViewCount="226" Body="&lt;p&gt;Recently I have started reading about some of the history of mathematics in order to better understand things.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A lot of ideas in algebra come from trying to understand the problem of finding solutions to polynomials in terms of radicals, which is solved by the Abel-Ruffini theorem and Galois theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was wondering if there's a textbook (or history book which emphasizes the mathematics) that goes vaguely in chronological order or explicitly presents theorems and concepts in their historical context.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively, if you think it would be better to attempt to read the original papers (Abel's famous quote about reading the masters comes to mind), such as the works of the mathematicians mentioned in &lt;a href=&quot;http://en.wikipedia.org/wiki/Abstract_algebra#Early_group_theory&quot; rel=&quot;nofollow&quot;&gt;this wikipedia article &lt;/a&gt;, how would I go about doing so?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: while researching some of the recommended books, I found &lt;a href=&quot;http://faculty.umf.maine.edu/michael.molinsky/public.www/cshpm/History_of_Mathematics_References.pdf&quot; rel=&quot;nofollow&quot;&gt; this interesting list (pdf) &lt;/a&gt; of references.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The inspiration for this question came from asking myself &quot;why would someone bother/think of defining a normal subgroup in the first place?&quot; (although I already know the answer) and hence I am asking about Galois theory, but really this question works for any area of mathematics and perhaps someone should open a community wiki question for that.&lt;/p&gt;&#xA;" OwnerUserId="2954" LastEditorUserId="2954" LastEditorDisplayName="" LastEditDate="2011-01-09T00:35:59.197" LastActivityDate="2011-01-09T00:35:59.197" Title="Historical textbook on group theory/algebra" Tags="&lt;algebra&gt;&lt;math-history&gt;" AnswerCount="4" CommentCount="1" />
  <row Id="16766" PostTypeId="1" AcceptedAnswerId="16770" CreationDate="2011-01-08T07:51:59.200" Score="10" ViewCount="444" Body="&lt;p&gt;I'm looking for ideas for a 15-hour mathematical enrichment course in a Chinese high school. What (fairly) elementary subject would you suggest as a topic for such a course?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Background/considerations:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;My students are generally quite good at math, but many of them have not been exposed to rigorous or abstract mathematical reasoning. A good topic would be one that would not be impossibly difficult for students who've never written or read proofs in English.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I've taught this class three times before. (Part of the reason that I'm posting this is that I've used up all my ideas!) The first semester I taught an introductory number theory class (which meandered its way towards a proof of quadratic reciprocity, though I think this was ultimately too advanced/abstract for some of the students). The second semester I taught basic graph theory and applications (with a focus on planarity and coloring). The third semester I taught a class on the Rubik's Cube.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The students' math backgrounds are quite varied: some of them participate in contest math competitions, and so are familiar with IMO-style techniques, but many are not. Some of them may know some calculus, but I can't count on it. All of them are very good at what in America is sometimes termed &quot;pre-calculus&quot;: trigonometry, conic sections, systems of linear equations (though, shockingly, no matrices), and the like. They know what a binomial coefficient is.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So, any ideas? Ideally, I'd like to find something a little &quot;sexy&quot; (like the Rubik's Cube) -- attempts to motivate number theory via cryptography seemed to fall on deaf ears, but being able to &quot;see&quot; group theory on the cube was quite popular.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Responses especially welcome from people who grew up in the PRC -- any mathematical topics you wish had been covered in the high school curriculum?)&lt;/p&gt;&#xA;" OwnerUserId="4743" LastEditorUserId="4743" LastEditorDisplayName="" LastEditDate="2011-01-08T08:06:14.570" LastActivityDate="2011-01-09T11:24:03.927" Title="Topic for a high school-level math elective?" Tags="&lt;soft-question&gt;&lt;teaching&gt;&lt;high-school&gt;" AnswerCount="7" CommentCount="2" FavoriteCount="4" />
  <row Id="16767" PostTypeId="1" AcceptedAnswerId="16778" CreationDate="2011-01-08T07:53:13.400" Score="1" ViewCount="207" Body="&lt;p&gt;Does anyone know a non trivial(that we cannot define on every set) topology defined on the power set of an uncountable set?&lt;/p&gt;&#xA;" OwnerUserId="5422" LastEditorUserId="5422" LastEditorDisplayName="" LastEditDate="2011-01-08T08:06:17.293" LastActivityDate="2011-01-08T10:44:59.017" Title="Topology of the power set" Tags="&lt;general-topology&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="2" />
  <row Id="16768" PostTypeId="2" ParentId="16765" CreationDate="2011-01-08T08:00:12.840" Score="2" ViewCount="" Body="&lt;p&gt;Please allow me to quote from Joseph Rotman's book &quot;Galois Theory&quot;, as follows&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;When writing the definition of Galois group for this text, I asked myself an obvious question: how did such thoughts occur to Galois in the late 1820's? The answer, of course, is that he did not think in such terms; for its first century, 1830-1930, the Galois group was a froup of permutations. In the late 1920's, E.Artin, developing ideas of E.Noether going back at least to Dedekind, recognized that it is both more elegant and more fruitful to describe Galois groups in terms of field automorphisms(Artin's version is isomorphic to the original version). In 1930, van der Waerden incorporated much of Artin's viewpoint into his influential text &quot;Moderne Algebra,&quot; and a decade later Artin published his own lectures. So successful have Artin's ideas proved to be that they have vertually eclipsed earlier expositions. But we have lost the inevitability of the definition; group theory is imposed on the study of polynomials rather arising naturally from it. This appendix is an attempt to remedy this pedagogical problem by telling the story of what happened in the beginning. The reader interested in a more thorough account may read [Edwards] or [Tignol] &quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;H.M.Edwards, Galois Theory,Springer,1984&lt;/p&gt;&#xA;&#xA;&lt;p&gt;J.-P.Tignol, Galois Theory of Algenraic Euqations, Wiley, 1988&lt;/p&gt;&#xA;" OwnerUserId="5476" LastActivityDate="2011-01-08T08:00:12.840" />
  <row Id="16769" PostTypeId="2" ParentId="16767" CreationDate="2011-01-08T08:16:40.450" Score="1" ViewCount="" Body="&lt;p&gt;I don't quite follow your question. But I provide one example anyway. $2^{[0,1]}$ is an example which is compact but not sequentially compact. In fact, we choose a series of points $f_n\in 2^{[0,1]}$ as follows. We definte $f_n(x)$ as the nth digit after writing x in the binary system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, when we choose any subsequence $f_{n_j}$ of $f_n$, we can find a $x\in [0,1]$ such that, after rewriting it in the binary system, it has $n_1$th,$n_2$th, $\cdots$ digits $0,1,0,1,\cdots$, alternatively. Obviously, $f_{n_j}$ has its value in x $0,1,0\cdots$ and thus doesn't have a limit. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But by Tychonoff's theorem, we know this space is compact.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;updated: $2^{[0,1]}$ is endowed with product topology.&lt;/p&gt;&#xA;" OwnerUserId="5476" LastEditorUserId="5476" LastEditorDisplayName="" LastEditDate="2011-01-08T10:44:59.017" LastActivityDate="2011-01-08T10:44:59.017" CommentCount="2" />
  <row Id="16770" PostTypeId="2" ParentId="16766" CreationDate="2011-01-08T08:32:20.163" Score="7" ViewCount="" Body="&lt;p&gt;I grew up in PR.China, and was quite disappointing with the pre-university education in mathematics. I am very happy to see one educator like you posting such a question here. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Combinatorics, graph theory and number theory, in my opinion, are proper fields you can choose materials from. By choosing some topics relating to &quot;big theorems&quot; such as Fermat's last theorem (of course in relatively naive ways) can surly attract young students.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think this could be done topic by topic, instead of stucking in only one small field. I believe one major problem in mathematical education in China is that there are too many restrictions on different branches. There are too many questions such as &quot;what field does this problem belongs to?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A book to recommend is &quot;proofs from the book&quot; written by Martin Aigner and Günter M. Ziegler (with illustrations by Karl H. Hofmann). Although this is wriiten as a graduate level book. One can find materials suiatble for high school students. More importanly, it can greatly enhance the students' taste in modern mathematics. &lt;/p&gt;&#xA;" OwnerUserId="5476" LastEditorUserId="5476" LastEditorDisplayName="" LastEditDate="2011-01-09T11:24:03.927" LastActivityDate="2011-01-09T11:24:03.927" CommentCount="3" />
  <row Id="16771" PostTypeId="2" ParentId="16767" CreationDate="2011-01-08T09:26:55.283" Score="3" ViewCount="" Body="&lt;p&gt;You may find this helpful: &lt;a href=&quot;http://www.au.edu.tw/ox_view/edu/tojms/j_paper/Full_text/Vol-21/No-1/21%281%29-2-0333.pdf&quot; rel=&quot;nofollow&quot;&gt;&quot;A Topology on a Power Set of a Set and Convergence of a Sequence of Sets.&quot;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4743" LastActivityDate="2011-01-08T09:26:55.283" />
  <row Id="16772" PostTypeId="2" ParentId="12825" CreationDate="2011-01-08T09:27:02.240" Score="2" ViewCount="" Body="&lt;p&gt;First note that this family of polynomials is orthogonal, on the interval $[-1,1]$, with the weight factor $(1-x^2)^{-(v+1)}$.  This is not much of a surprise since the definition is very similar to that of the traditional Legendre Polynomials, which are orthogonal.  Next, we use the following deep result involving orthogonal polynomials:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $\{p_n\}$ is a family of orthogonal polynomials with roots in $[-1,1]$ and $N(a,b,n)$ represents the number of roots of $p_n$ in [$\cos(b),\cos(a)$] then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\lim_{n\to \infty}\frac1{n} N(a,b,n)=\frac{b-a}{\pi}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus for any small subinterval [$\cos(b),\cos(a)$], there exists $n$ sufficiently large such that $N(a,b,n)&amp;gt;1$ implying that the roots of these polynomials do form a dense subset of $[-1,1]$.&lt;/p&gt;&#xA;" OwnerUserId="3971" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2011-04-19T17:55:09.233" LastActivityDate="2011-04-19T17:55:09.233" />
  <row Id="16773" PostTypeId="2" ParentId="12160" CreationDate="2011-01-08T09:44:03.503" Score="1" ViewCount="" Body="&lt;p&gt;The density of the roots of any family of orthogonal polynomials follows from this result:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $\{p_n\}$ is a family of orthogonal polynomials with roots in [-1,1] and $N(a,b,n)$ represents the number of roots of $p_n$ in [$cos(b),cos(a)$] then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\lim_{n\to \infty}$$1\over{n}$$N(a,b,n)$=${b-a}\over{\pi}$&lt;/p&gt;&#xA;" OwnerUserId="3971" LastActivityDate="2011-01-08T09:44:03.503" />
  <row Id="16774" PostTypeId="1" AcceptedAnswerId="16786" CreationDate="2011-01-08T09:45:44.017" Score="2" ViewCount="145" Body="&lt;p&gt;Hello &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have following problem:&#xA;We throw a fair dice time after time.What is time expectancy to get first time 66  i.e two six in the row for the first time. I probably should use this &lt;a href=&quot;http://en.wikipedia.org/wiki/Expected_value#Discrete_distribution_taking_only_non-negative_integer_values&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Expected_value#Discrete_distribution_taking_only_non-negative_integer_values&lt;/a&gt; though I don't understand how.                                                                                                                                                                                                                                                                                                                                                                                                     &lt;/p&gt;&#xA;" OwnerUserId="5509" LastEditorUserId="5509" LastEditorDisplayName="" LastEditDate="2011-01-08T11:12:36.253" LastActivityDate="2011-01-08T13:23:03.813" Title="Geometric distrubution and expectancy" Tags="&lt;probability&gt;" AnswerCount="3" CommentCount="6" />
  <row Id="16775" PostTypeId="2" ParentId="16738" CreationDate="2011-01-08T09:59:29.567" Score="0" ViewCount="" Body="&lt;p&gt;The lemma is saying :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The conical (set, envelope,shell) of a finite family(set) of p vectors in R^n is a closed set.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I put many terms between brackets , cause I'm not sure of the terminology (sorry for that), I'm translating from french (enveloppe conique).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To precise I can give the mathematical definition of the conical envelope:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The conical envelope  co{X 1 , · · · , X p } is a family of vectors (arrays) &#xA; {X 1 , . . . , X p } where :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;co{X1 , . . . , X p} = {X | ∃λi ≥ 0, i = 1, . . . , p, v=∑ λi X i, 1&amp;lt;=i&amp;lt;=p }&lt;/p&gt;&#xA;" OwnerUserId="5499" LastActivityDate="2011-01-08T09:59:29.567" CommentCount="2" />
  <row Id="16776" PostTypeId="2" ParentId="16447" CreationDate="2011-01-08T10:01:35.877" Score="0" ViewCount="" Body="&lt;p&gt;OK, assuming completeness, Theo provided a simple proof below. My previous proof also works if one assumes completeness, but it is too convoluted.&lt;/p&gt;&#xA;" OwnerUserId="2093" LastEditorUserId="2093" LastEditorDisplayName="" LastEditDate="2011-01-08T22:33:28.427" LastActivityDate="2011-01-08T22:33:28.427" CommentCount="1" />
  <row Id="16777" PostTypeId="2" ParentId="16765" CreationDate="2011-01-08T10:17:01.117" Score="2" ViewCount="" Body="&lt;p&gt;Algebra is a very large field, so you probably want to be a bit more specific.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In case you are wondering about &lt;em&gt;Galois Theory&lt;/em&gt;, and want to learn its history for the purpose of understanding it, I strongly recommend the following book:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Jörg Bewersdorff. &lt;a href=&quot;http://books.google.com/books?id=RSC2hpPQ0LQC&quot; rel=&quot;nofollow&quot;&gt;Galois Theory for Beginners&lt;/a&gt;. AMS 2006.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It doesn't present the history in a scholarly fashion, but as a means to the end of understanding the modern formulation of Galois Theory in terms of fields and automorphism groups. I don't think I would have understood any of it without this book.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is also a paper&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;John Stillwell. &lt;a href=&quot;http://www.jstor.org/pss/2325119&quot; rel=&quot;nofollow&quot;&gt;Galois Theory for Beginners&lt;/a&gt;. The American Mathematical Monthly. Vol. 101, No. 1 (Jan., 1994), pp. 22-27.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;with a similar goal. I'm mentioning it because the author, John Stillwell, has written another marvelous book about the history of mathematics in general.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are interested in Abel's concurrent proof of the unsolvability of the quintic, have a look at the book&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Peter Pesic. &lt;a href=&quot;http://mitpress.mit.edu/catalog/item/default.asp?ttype=2&amp;amp;tid=10180&quot; rel=&quot;nofollow&quot;&gt;Abel's Proof&lt;/a&gt;. MIT Press 2004.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;although it's written for the general public. However, it does include a modern transcription of Abel's original proof in an appendix, which is the object of interest here.&lt;/p&gt;&#xA;" OwnerUserId="434" LastActivityDate="2011-01-08T10:17:01.117" CommentCount="2" />
  <row Id="16778" PostTypeId="2" ParentId="16767" CreationDate="2011-01-08T10:26:33.877" Score="4" ViewCount="" Body="&lt;p&gt;I think you're asking whether power objects exist in the category of topological spaces. I think the answer is not always: although we have a subobject classifier $\Omega$ (namely the 2-point space with the indiscrete topology), and the exponential object $\Omega^X$ is not guaranteed to exist. The obstruction is the requirement that for every continuous function $f: A \times X \to \Omega$ there must be a unique continuous function $\tilde{f}: A \to \Omega^X$, such that $\tilde{f}(a)(x) = f(a, x)$, and vice-versa. When $X$ is nice enough, e.g. locally compact and Hausdorff, then $\Omega^X$ does exist and has the compact-open topology.&lt;/p&gt;&#xA;" OwnerUserId="5191" LastActivityDate="2011-01-08T10:26:33.877" CommentCount="1" />
  <row Id="16779" PostTypeId="1" AcceptedAnswerId="16820" CreationDate="2011-01-08T11:16:25.390" Score="10" ViewCount="246" Body="&lt;p&gt;As we know, Lie group is a special smooth manifold.  I want to find some  geometric property, which is only satisfied by the Lie group. I only found one property: parallelizability.  Can you show me something more?&lt;/p&gt;&#xA;" OwnerUserId="4706" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2011-01-08T16:50:28.460" LastActivityDate="2011-01-09T22:20:42.700" Title="Conditions for a smooth manifold to admit the structure of a Lie group" Tags="&lt;geometry&gt;&lt;lie-groups&gt;" AnswerCount="3" CommentCount="4" FavoriteCount="3" />
  <row Id="16780" PostTypeId="2" ParentId="16447" CreationDate="2011-01-08T11:26:25.533" Score="2" ViewCount="" Body="&lt;p&gt;I've completely forgotten the solution, but I think something like this might work: The set $\mathcal{F}$ of all evaluation maps is a family of bounded linear operators on $C[0, 1]$. It suffices to show that $\mathcal{F}$ is uniformly bounded, and then the uniform bound gives you the constant needed to establish the equivalence with the uniform norm. I don't remember how to establish uniform boundedness, but I imagine the uniform boundedness principle applies.&lt;/p&gt;&#xA;" OwnerUserId="5191" LastActivityDate="2011-01-08T11:26:25.533" CommentCount="7" />
  <row Id="16781" PostTypeId="1" AcceptedAnswerId="16811" CreationDate="2011-01-08T11:28:37.970" Score="1" ViewCount="63" Body="&lt;p&gt;Is there a classification of all finite indecomposable p-groups?&#xA;(something like: those are exactly the quaternions, primary cylic groups, some dihedral groups, etc.).&lt;/p&gt;&#xA;" OwnerUserId="3533" LastActivityDate="2011-01-08T16:00:38.253" Title="Is there a classification of all finite indecomposable p-groups?" Tags="&lt;group-theory&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="16782" PostTypeId="2" ParentId="16774" CreationDate="2011-01-08T11:40:54.080" Score="1" ViewCount="" Body="&lt;p&gt;The probability that you get a 6 after $n+1$ throws is $$ \frac{5^n}{6^n} \frac{1}{6}$$ that is getting $n$ times something else than a $6$ then getting a $6$. Therefore, the probability to throw $n$ times the dice and then to get two $6$ in a row is $$ \frac{5^n}{6^n} \frac{1}{6^2}.$$ The expectancy is thus $$\frac{1}{6^2} \sum_{n=0}^{\infty}\frac{5^n}{6^n}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does it answer your question?&lt;/p&gt;&#xA;" OwnerUserId="5465" LastActivityDate="2011-01-08T11:40:54.080" CommentCount="2" />
  <row Id="16783" PostTypeId="1" CreationDate="2011-01-08T11:43:25.230" Score="3" ViewCount="135" Body="&lt;p&gt;Hey,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Problem: Given that $f$ is differentiable at $[0,1]$ and $f(0)=f(1)=0$. If $ \forall x\in (0,1)$ $|f&amp;#39;&amp;#39;(x)|\leq A$ show that $\forall x \in [0,1]$ $|f&amp;#39;(x)| \leq (A/2) $.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My attempt was to to develop a Taylor series for $f(x)$ and $f&amp;#39;(x)$ around point $c \in (0,1)$ where $f&amp;#39;(c)=0$.&lt;/p&gt;&#xA;" OwnerUserId="3637" LastEditorUserId="2548" LastEditorDisplayName="" LastEditDate="2011-01-08T11:50:01.197" LastActivityDate="2011-01-08T16:01:44.617" Title="Show $|f'(x)| \leq (A/2) $ if $|f''(x)|\leq A$" Tags="&lt;real-analysis&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" />
  <row Id="16784" PostTypeId="1" CreationDate="2011-01-08T12:00:32.213" Score="0" ViewCount="103" Body="&lt;p&gt;Two tangent vectors at a point on a surface are &#xA;      T1 = 4i + 2j + 3k&#xA;and   T2 = -2i - 3j + 1k&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the property of the dot product of two normal vectors, determine the unit vector normal to the surface at the point&lt;/p&gt;&#xA;" OwnerUserId="5511" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2011-01-08T16:32:25.207" LastActivityDate="2011-01-08T23:40:28.743" Title="vectors dot product solution" Tags="&lt;calculus&gt;&lt;vector-calculus&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="16785" PostTypeId="2" ParentId="11686" CreationDate="2011-01-08T12:14:04.437" Score="0" ViewCount="" Body="&lt;p&gt;From the continued-fraction expansions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{15}{106} = [0;7,15]$ and $\frac{16}{113} = [0;7,16]$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;it is immediate that the simplest fraction between them is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$[0;7,15,2] = \frac{31}{219}$&lt;/p&gt;&#xA;" OwnerUserId="1508" LastActivityDate="2011-01-08T12:14:04.437" />
  <row Id="16786" PostTypeId="2" ParentId="16774" CreationDate="2011-01-08T12:20:15.323" Score="5" ViewCount="" Body="&lt;p&gt;Let $x$ be the expected number of tosses to get one six. Then $x = \frac{5}{6}(1+x)+ \frac{1}{6}(1)$. So $x=6$. Let $y$ be the expected number of throws to two consecutive sixes. Then $y = \frac{1}{6}(x+1)+ \frac{5}{6}(x+1+y)$. So $y = 6(x+1)$. Plugging in $x=6$, we get $y = 42$. &lt;/p&gt;&#xA;" OwnerUserId="2548" LastEditorUserId="2548" LastEditorDisplayName="" LastEditDate="2011-01-08T12:56:40.600" LastActivityDate="2011-01-08T12:56:40.600" CommentCount="8" />
  <row Id="16787" PostTypeId="1" AcceptedAnswerId="16803" CreationDate="2011-01-08T12:27:12.193" Score="2" ViewCount="245" Body="&lt;p&gt;Intuitively what is meant by a left invariant vector field on a manifold?&lt;/p&gt;&#xA;" OwnerUserId="5312" LastEditorUserId="9" LastEditorDisplayName="" LastEditDate="2011-01-08T18:34:06.247" LastActivityDate="2011-01-08T18:34:06.247" Title="Intuitive explanation of Left invariant Vector Field" Tags="&lt;differential-geometry&gt;&lt;intuition&gt;&lt;lie-groups&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="16788" PostTypeId="1" CreationDate="2011-01-08T12:43:16.403" Score="2" ViewCount="113" Body="&lt;p&gt;I have the following constraint inequalities and equalities:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$Ax \leq b$$&#xA;$$A_{eq}x = b_{eq}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is that the objective function, which I am asked to minimized, is defined as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f=\sum\limits_{i=1}^{m}u([\sum\limits_{j=1}^{n}k_{ij}x_j]-c_{i})$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $u(x)$ is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Heaviside_step_function&quot; rel=&quot;nofollow&quot;&gt;Heaviside step function&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Strictly speaking, this is not a linear programming problem ( although quite close!), so it can't be attacked by the standard linear programming techniques.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm aware that I can approximate the step function into a smooth function, but this is not the route I plan to take now. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are the techniques that are available for this kind of problem?&lt;/p&gt;&#xA;" OwnerUserId="550" LastEditorUserId="550" LastEditorDisplayName="" LastEditDate="2011-01-10T07:24:16.180" LastActivityDate="2011-01-10T16:12:07.370" Title="Solution Technique to Optimize Sets of Constraint Functions with Objective Function that is Heaviside Step Function" Tags="&lt;optimization&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="2" />
  <row Id="16790" PostTypeId="2" ParentId="16766" CreationDate="2011-01-08T13:05:46.763" Score="4" ViewCount="" Body="&lt;p&gt;Except for the topics that were already mentioned (mostly those that come from combinatorics, which can be explained relatively easily to students without advanced mathematical background, and also contain many interesting puzzles that can be given to them for solving), I would suggest some &lt;b&gt;basic probability&lt;/b&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could explain different everyday counter-intuitive problems such as the &lt;a href=&quot;http://en.wikipedia.org/wiki/Monty_Hall_problem&quot; rel=&quot;nofollow&quot;&gt;Monty Hall Problem&lt;/a&gt; or the &lt;a href=&quot;http://math.stackexchange.com/questions/4400/boy-born-on-a-tuesday-is-it-just-a-language-trick&quot;&gt;Boy born on a Tuesday&lt;/a&gt; problem, and this topic doesn't require much advanced background.&lt;/p&gt;&#xA;" OwnerUserId="766" LastActivityDate="2011-01-08T13:05:46.763" />
  <row Id="16791" PostTypeId="2" ParentId="16784" CreationDate="2011-01-08T13:07:37.837" Score="4" ViewCount="" Body="&lt;p&gt;First, to find the direction of the normal vector, you want to take the cross product of the vectors for $T_1$ and $T_2$. You get the direction of the normal vector $n$ as such:&#xA;$$&#xD;&#xA;n=T_1\times T_2=11i-10j-8k.&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can verify that this vector is orthogonal to both $T_1$ and $T_2$, by seeing that their dot product with $n$ is $0$ in both cases, and thus normal to the surface. All that remains is to make the vector a unit vector, which you can do by dividing each coordinate by the length of the vector, and then you are done.&lt;/p&gt;&#xA;" OwnerUserId="1583" LastEditorUserId="1583" LastEditorDisplayName="" LastEditDate="2011-01-08T23:40:28.743" LastActivityDate="2011-01-08T23:40:28.743" />
  <row Id="16792" PostTypeId="1" CreationDate="2011-01-08T13:09:29.070" Score="1" ViewCount="53" Body="&lt;p&gt;I've been stuck on this homework question for a couple of days. Any hints would be much appreciated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $G$ be a finite group and $\rho : G \to \mathrm{GL}(V)$ be a faithful representation, with $V$ a vector space over the field $F$. Suppose $Z(G) = \{ e \}$, and $H \le G$ is a subgroup with $Z(H) \ne \{ e \}$. Then, I need to show that the restriction $\rho|_H : H \to \mathrm{GL}(V)$ is reducible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have no idea how to proceed, but apparently Schur's lemma is relevant. I'm not sure if I'm missing some conditions, like say $F = \mathbb{C}$. I can see, for instance, that the image of $Z(H)$ under $\rho$ is a family of commuting automorphisms, and so can be simultaneously diagonalised. If I assume that $\rho|_H$ is irreducible then Schur's lemma implies that in fact $\rho(z) = \lambda_z \mathrm{id}$ for some $\lambda_z \in \mathbb{C}^\times$ for each $z \in Z(H)$. This should lead to a contradiction somewhere, but I'm not seeing it... I also notice I haven't used the fact that the parent group $G$ is centreless.&lt;/p&gt;&#xA;" OwnerUserId="5191" LastActivityDate="2011-01-08T13:09:29.070" Title="Reducibility of a faithful representation of a subgroup of a centreless group" Tags="&lt;homework&gt;&lt;representation-theory&gt;" CommentCount="2" />
  <row Id="16793" PostTypeId="2" ParentId="16774" CreationDate="2011-01-08T13:10:37.553" Score="2" ViewCount="" Body="&lt;p&gt;Define a &lt;em&gt;live n-sequence&lt;/em&gt; to be a sequence of $n$ dice throws in which no double 6 occurs. Define:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$r_n$ = number of live $n$-sequences that don't end in 6&lt;br&gt;&#xA;$s_n$ = number of live $n$-sequences that do end in 6&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then we get the recurrence relations:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$r_{n+1} = 5r_{n} + 5s_{n}$&lt;br&gt;&#xA;$s_{n+1} = r_n$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with initial conditions $r_0 = 1, s_0 = 0$. Substituting the second relation into the first gives  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$s_{n+2} = 5s_{n+1} + 5s_{n}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The solutions of this are of the form $s_n = A\alpha^n + B\beta^n$, where $\alpha$ and $\beta$ are the roots of the equation $x^2 - 5x - 5 = 0$:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\alpha = (5 + 3\sqrt5)/2, \beta = (5 - 3\sqrt5)/2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The probability that we need exactly $n$ throws is $\frac{1}{6} s_{n-1}/6^{n-1} = \frac{s_{n-1}}{6^n}$, so the expected number of throws is  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sum_{n=1}^\infty n\frac{s_{n-1}}{6^n} = \frac{A}{\alpha}\sum_{n=1}^\infty nu^n + \frac{B}{\beta}\sum_{n=1}^\infty nv^n$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $u = \frac{\alpha}{6}$ and $v = \frac{\beta}{6}$. This is  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{A}{\alpha}\frac{u}{(1-u)^2} + \frac{B}{\beta}\frac{v}{(1-v)^2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the initial terms $s_0=0, s_1=1$, we get $A = \frac{1}{3\sqrt 5}, B = -A$. So we get  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{A}{6}(\frac{1}{(1-u)^2} - \frac{1}{(1-v)^2})$, which my shaky algebra simplifies to...wait for it...   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$42$&lt;/p&gt;&#xA;" OwnerUserId="1508" LastEditorUserId="1508" LastEditorDisplayName="" LastEditDate="2011-01-08T13:23:03.813" LastActivityDate="2011-01-08T13:23:03.813" CommentCount="1" />
  <row Id="16794" PostTypeId="1" AcceptedAnswerId="16806" CreationDate="2011-01-08T13:19:35.440" Score="6" ViewCount="178" Body="&lt;p&gt;According to wikipedia the &lt;a href=&quot;http://en.wikipedia.org/wiki/Group_isomorphism_problem&quot; rel=&quot;nofollow&quot;&gt;group isomorphism problem&lt;/a&gt; is an &lt;a href=&quot;http://en.wikipedia.org/wiki/Undecidable_problem&quot; rel=&quot;nofollow&quot;&gt;undecidable problem&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When we restrict to (countable) abelian groups does it become decidable or does it remain undecidable?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In case it becomes decidable I'd love to have a reference to an algorithm deciding it.&lt;/p&gt;&#xA;" OwnerUserId="367" LastEditorUserId="413" LastEditorDisplayName="" LastEditDate="2011-01-08T20:52:56.550" LastActivityDate="2011-01-08T21:11:53.023" Title="Is the group isomorphism problem decidable for abelian groups?" Tags="&lt;group-theory&gt;&lt;computability&gt;" AnswerCount="2" />
  <row Id="16795" PostTypeId="2" ParentId="16766" CreationDate="2011-01-08T13:22:53.913" Score="1" ViewCount="" Body="&lt;p&gt;I think that in the long term the most important thing you have to do is to stimulate their mathematical muscles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to show them mathematical rigor (which is a good idea) I suggest some propositional calculus, or predicate calculus. If you want you could solve some riddles using truth tables or by inference rules (an example in my answer here: &lt;a href=&quot;http://math.stackexchange.com/questions/15199/implication-of-three-statements/15206#15206&quot;&gt;Implication of three statements&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can show the importance of rigor by giving examples to how &quot;All horses are black&quot; or &quot;For every natural number $n$ it holds $n=n+1$&quot;, or find some other example which is simple for them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps you can show them some infinite structures and how infinity behaves differently, start with Hilbert's hotel, and then show them how you can &quot;compress&quot; all the integers into natural numbers, or assign a unique natural number to every rational number - with room to spare.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whatever you choose doing, I think that the most important thing is not to get &lt;strong&gt;too&lt;/strong&gt; deep into technical details and proofs, and keep a bit of the mystery to get them search for it themselves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Addendum&lt;/strong&gt;:&lt;br&gt;&#xA;I just remembered something you might find helpful, &lt;a href=&quot;http://mathoverflow.net/questions/47214/how-to-present-mathematics-to-non-mathematicians&quot; rel=&quot;nofollow&quot;&gt;http://mathoverflow.net/questions/47214/how-to-present-mathematics-to-non-mathematicians&lt;/a&gt; this MO thread which was opened by yours truly, it deals with a different situation altogether (I was giving a 10 minutes presentation to non-mathematicians) but it has a nice list of topics that you can take and further analyze and deconstruct in your time.&lt;/p&gt;&#xA;" OwnerUserId="622" LastEditorUserId="622" LastEditorDisplayName="" LastEditDate="2011-01-08T20:16:01.820" LastActivityDate="2011-01-08T20:16:01.820" />
  <row Id="16796" PostTypeId="1" AcceptedAnswerId="16823" CreationDate="2011-01-08T13:29:34.770" Score="5" ViewCount="238" Body="&lt;p&gt;You have a circular table with $N$ seats.$K$ bellicose guests are going visit your house of-course you don't want them to sit beside each other.As the host, you want to find out how many ways there are to choose $K$ seats such that none of them is adjacent to each other. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I noticed that there is a solution  other than $0$ if ($N \ge 2K $) but I am not sure how to approach for the rest.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT: Only $K$ bellicose guests are visiting,no friendly guest are there the remaining $N-K$ seats will be vacant.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;A possible mathematical translation of this problem:&lt;/strong&gt; Choosing $K$ candidate points from a circle of $N$ indistinguishable points such that there are more than one vacant point between adjacent candidate points.&lt;/p&gt;&#xA;" OwnerUserId="3744" LastEditorUserId="3744" LastEditorDisplayName="" LastEditDate="2011-01-09T15:03:01.833" LastActivityDate="2011-01-09T15:03:01.833" Title="Choosing seats for guests" Tags="&lt;combinatorics&gt;" AnswerCount="2" CommentCount="8" FavoriteCount="0" />
  <row Id="16797" PostTypeId="1" AcceptedAnswerId="16816" CreationDate="2011-01-08T13:30:33.223" Score="5" ViewCount="247" Body="&lt;p&gt;Is there any good way to approximate following integral?&lt;br&gt;&#xA;$$\int_0^{0.5}\frac{x^2}{\sqrt{2\pi}\sigma}\cdot \exp\left(-\frac{(x^2-\mu)^2}{2\sigma^2}\right)\mathrm dx$$&lt;br&gt;&#xA;$\mu$ is between $0$ and $0.25$, the problem is in $\sigma$ which is always positive, but it can be arbitrarily small.&lt;br&gt;&#xA;I was trying to expand it using Taylor series, but terms looks more or less this $\pm a_n\cdot\frac{x^{2n+3}}{\sigma^{2n}}$ and that can be arbitrarily large, so the error is significant.&lt;/p&gt;&#xA;" OwnerUserId="4811" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2011-04-07T05:43:35.887" LastActivityDate="2011-04-07T05:43:35.887" Title="An approximation of an integral" Tags="&lt;calculus&gt;&lt;integral&gt;&lt;approximation&gt;" AnswerCount="4" CommentCount="2" />
  <row Id="16798" PostTypeId="1" CreationDate="2011-01-08T13:45:25.400" Score="0" ViewCount="161" Body="&lt;p&gt;Hi.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a PDE, which I need to calculate the Courant number for. Does anyone have any resources on that which can help I. I do want a general way of calculating the number if that is possible, so I can use it further.&lt;/p&gt;&#xA;" OwnerUserId="3221" LastActivityDate="2011-03-20T01:33:41.603" Title="Calculating the Courant number for a PDE" Tags="&lt;differential-equations&gt;&lt;numerical-methods&gt;&lt;education&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="16799" PostTypeId="2" ParentId="16447" CreationDate="2011-01-08T13:52:26.473" Score="14" ViewCount="" Body="&lt;p&gt;Here's $\DeclareMathOperator{\ev}{ev}$ the answer if $(C[0,1], \|\cdot\|)$ is assumed to be complete. Consider the family $\mathcal{F} = \{\ev_{x}\}_{x \in [0,1]}$ of &lt;em&gt;continuous&lt;/em&gt; linear functionals on $(C[0,1],\|\cdot\|)$. For each $f \in C[0,1]$ we have $\sup_{x \in [0,1]} |\ev_{x}(f)| \leq \|f\|_{\infty}$, so the family $\mathcal{F}$ is &lt;em&gt;pointwise bounded&lt;/em&gt;. By the &lt;a href=&quot;http://en.wikipedia.org/wiki/Uniform_boundedness_principle&quot; rel=&quot;nofollow&quot;&gt;uniform boundedness principle&lt;/a&gt; the family is &lt;em&gt;uniformly bounded&lt;/em&gt;, that is to say $\sup_{x \in [0,1]} \|\ev_{x}\| \leq M$ for some constant $M$. On the other hand $|f(x)| = |\ev_{x}(f)| \leq \|ev_{x}\|\|f\| \leq M \|f\|$ gives $\|f\|_{\infty} = \sup_{x \in [0,1]} |f(x)| \leq M\|f\|$, so the identity $(C[0,1], \|\cdot\|) \to (C[0,1],\|\cdot\|_{\infty})$ has norm at most $M$. Since both spaces are complete, we may apply the &lt;a href=&quot;http://en.wikipedia.org/wiki/Open_mapping_theorem_(functional_analysis)&quot; rel=&quot;nofollow&quot;&gt;open mapping theorem&lt;/a&gt; in order to conclude that its inverse is also continuous. In other words, the norms $\|\cdot\|$ and $\|\cdot\|_{\infty}$ are equivalent.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit.&lt;/strong&gt; Here's an example that shows that completeness of the norm is necessary:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Choose a &lt;em&gt;discontinuous&lt;/em&gt; linear functional $\varphi: (C[0,1],\|\cdot\|_{\infty}) \to \mathbb{R}$ and define a norm on $C[0,1]$ by&#xA;\[&#xA;\|f\| = \|f\|_{\infty} + |\varphi(f)|.&#xA;\]&#xA;Since $\|f\|_{\infty} \leq \|f\|$, we have that the identity $(C[0,1],\|\cdot\|) \to (C[0,1],\|\cdot\|_{\infty})$ is continuous. Since the evaluation functionals are continuous with respect to the sup-norm, they are also continuous with respect to the norm $\|\cdot\|$. But as $\varphi$ is discontinuous, there is a sequence $f_{n}$ with $\|f_{n}\|_{\infty} = 1$ and $|\varphi(f_{n})| \to \infty$, hence the norms cannot be equivalent. Of course, $\|\cdot\|$ cannot be complete because the last sentence would be in contradiction to the open mapping theorem.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit 2.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I forgot to argue why the topologies in the above counterexample are not the same. This is obvious: The functional $\varphi$ is continuous with respect to $\|\cdot\|$ but it isn't continuous with respect to $\|\cdot\|_{\infty}$.&lt;/p&gt;&#xA;" OwnerUserId="5363" LastEditorUserId="5363" LastEditorDisplayName="" LastEditDate="2011-01-08T20:23:56.387" LastActivityDate="2011-01-08T20:23:56.387" CommentCount="3" />
  <row Id="16800" PostTypeId="2" ParentId="2483" CreationDate="2011-01-08T14:06:40.927" Score="14" ViewCount="" Body="&lt;p&gt;The following paper seems to answer your question: P. T. Young, &lt;em&gt;&quot;Quadratic reciprocity via Lucas sequences&quot;&lt;/em&gt;, Fibonacci Quart. &lt;strong&gt;33&lt;/strong&gt; (1995), no. 1, 78–81.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's its &lt;a href=&quot;http://www.ams.org/mathscinet-getitem?mr=1316286&quot; rel=&quot;nofollow&quot;&gt;MathSciNet Review&lt;/a&gt; by A. Grytczuk:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Let $\{\gamma_n\}^\infty_{n=0}$ be a given Lucas sequence defined by $\gamma_0=0$, $\gamma_1=1$, $\gamma_{n+1}=\lambda \gamma_n+\mu \gamma_{n-1}$, $n\geq 1$, $\lambda, \mu\in{\bf Z}$, and let $q$ be an odd prime such that $D=(\frac{-1}q)q=\lambda^2+4\mu$. Then the author proves that there is a unique formal power series $\Phi$ with integer coefficients and constant term zero such that (1) $\sum^\infty_{n=1}\gamma_n\Phi^n(t)/n=\sum^\infty_{n=1}(\frac nq)t^n/n$ holds, where $(\frac nq)$ is the Legendre symbol.&lt;br&gt;&#xA;     From this result follows the Gauss law of quadratic reciprocity in the following form: (2) $(\frac pq)=(\frac Dp)$, where $p$, $q$ are distinct odd primes and $D=(\frac{-1}q) q=\lambda^2+4\mu$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here's the &lt;a href=&quot;http://www.mathstat.dal.ca/FQ/Scanned/33-1/young.pdf&quot; rel=&quot;nofollow&quot;&gt;direct link&lt;/a&gt; to the paper.&lt;/p&gt;&#xA;" OwnerUserId="5363" LastEditorUserId="5363" LastEditorDisplayName="" LastEditDate="2011-01-08T16:59:32.680" LastActivityDate="2011-01-08T16:59:32.680" CommentCount="1" />
  <row Id="16801" PostTypeId="1" AcceptedAnswerId="17093" CreationDate="2011-01-08T14:12:19.607" Score="3" ViewCount="89" Body="&lt;p&gt;I am currently preparing my next physics exam, and I got courious if there may be on the Net some sort of time scale of mathematical discoveries, so that I could compare discoveries and achievements in physics with the mathematic instruments available at the time.&lt;br&gt;&#xA;While there is large abundance of such graphs for computer science and physics, maths related ones are quite hard to find..&lt;br&gt;&#xA;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="5492" LastActivityDate="2011-01-11T14:34:28.070" Title="Where can I find a time scale (or anything similar) listing the main discoveries and achievments in mathematics?" Tags="&lt;math-history&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="16802" PostTypeId="1" AcceptedAnswerId="16842" CreationDate="2011-01-08T14:17:59.117" Score="2" ViewCount="251" Body="&lt;p&gt;Dear,&#xA;I don't understand left coset and Lagrange's theorem... I'm working on some project so please write me anything what could help me understand it better...&#xA;milica from montenegro&lt;/p&gt;&#xA;" OwnerUserId="5513" LastActivityDate="2011-01-09T13:32:12.547" Title="Lagrange's theorem" Tags="&lt;group-theory&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="1" />
  <row Id="16803" PostTypeId="2" ParentId="16787" CreationDate="2011-01-08T14:28:16.007" Score="4" ViewCount="" Body="&lt;p&gt;To talk about left invariance, you probably want to assume your manifold is a Lie group, so that the vector field is left invariant under the (derivative of) the group action. Intuitively, this means that the vector field is entirely determined by the vector at the unit element of the Lie group. Given any other element of the Lie group, say g, the vector at g has to be $l_g(X_0)$, where $l_g$ is the derivative of left-multiplication by $g$ and $X_0$ is the vector at the unit element. So the Lie group action allows you to take a single vector and distribute it out over the manifold in a smooth, nonvanishing way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The simplest example is Euclidean space $\mathbb R^n$ regarded as an abelian Lie group under addition. In this case, a left invariant vector field is simply a constant vector field in the usual calculus sense. All the vectors point in the same direction. The map $l_g$ identifies the tangent spaces at each point in the usual &quot;parallel transport&quot; way.&lt;/p&gt;&#xA;" OwnerUserId="2941" LastActivityDate="2011-01-08T14:28:16.007" CommentCount="2" />
  <row Id="16804" PostTypeId="1" AcceptedAnswerId="16808" CreationDate="2011-01-08T14:34:27.007" Score="1" ViewCount="53" Body="&lt;p&gt;Given a sequence of N numbers, how can we find the minimum number of numbers to be inserted to make this sequence to an Arithmetic progression.(we can insert at any position of this sequence)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example consider a sequence of $4$ numbers :$1,3,7,13$ Here we have to insert at-least $3$ more numbers ($5,9,11$) so that $1,3,5,7,9,11,13$ becomes an A.P with common difference $2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am inquisitive to know an an efficient algorithm for this problem.&lt;/p&gt;&#xA;" OwnerUserId="2109" LastEditorUserId="2109" LastEditorDisplayName="" LastEditDate="2011-01-08T14:51:33.207" LastActivityDate="2011-01-08T15:12:08.453" Title="Minimum number of numbers to be inserted in a sequence to transform it into an A.P" Tags="&lt;number-theory&gt;&lt;sequences-and-series&gt;&lt;algorithms&gt;&lt;computer-science&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="16805" PostTypeId="2" ParentId="16796" CreationDate="2011-01-08T14:52:48.807" Score="1" ViewCount="" Body="&lt;p&gt;Let's call $F=N-K$ number of &quot;friendly&quot; guests. And let's call $S(F,K)$ the count of seating ways assuming that seats and guests are distinguible (rotations are distinct solutions).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know that $F \ge K$. For the limit case $F = K$ we have $S(F,K) = 2 K (K-1)! K! = 2 K!^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the recursion: we count the number of ways when adding a friendly guest:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$S(F+1,K) = (F+K+1) S(F,K)$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From this (if it's correct!) you can get an explicit solution. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Update: this is &lt;strong&gt;not&lt;/strong&gt; correct. See TonyK's answer&lt;/p&gt;&#xA;" OwnerUserId="312" LastEditorUserId="312" LastEditorDisplayName="" LastEditDate="2011-01-09T00:14:40.183" LastActivityDate="2011-01-09T00:14:40.183" CommentCount="3" />
  <row Id="16806" PostTypeId="2" ParentId="16794" CreationDate="2011-01-08T15:05:30.613" Score="3" ViewCount="" Body="&lt;p&gt;This is not a full answer (somehow I cannot comment), but when talking about presentations of abelian groups only the exponentvectors matter. This means: for a relation $x_1^{a_1}...x_n^{a_n}$ the only $[a_1,...,a_n]$ matters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore I presume the question can be decided by creating a matrix where the exponentvectors are its rows (entries are in $\mathbb Z$ ) and bringing it to some normal form (upper triangular) utilizing the euclidean algorithm (note that $\mathbb Z$ is not a field) which will then yield the isomorphism type as stated by the fundamental theorem of abelian groups.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't have a reference on this common knowledge but it might point you in some direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;update2:&lt;/strong&gt; I just noticed this is exactly what the &lt;a href=&quot;http://en.wikipedia.org/wiki/Smith_normal_form&quot; rel=&quot;nofollow&quot;&gt;Smith normal form&lt;/a&gt; computes. But this only applies to finite presentations, as stated in comments.&lt;/p&gt;&#xA;" OwnerUserId="5189" LastEditorUserId="5189" LastEditorDisplayName="" LastEditDate="2011-01-08T19:50:11.303" LastActivityDate="2011-01-08T19:50:11.303" CommentCount="8" />
  <row Id="16807" PostTypeId="2" ParentId="16802" CreationDate="2011-01-08T15:08:12.327" Score="1" ViewCount="" Body="&lt;p&gt;I cannot yet write comments, so I'll write an answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Have you tried reading:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Lagrange%27s_theorem_(group_theory&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Lagrange%27s_theorem_(group_theory&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe what you find confusing is that left cosets are not subgroups, just subsets of the group. You don't give many clues about your actual concerns...&lt;/p&gt;&#xA;" OwnerUserId="5405" LastActivityDate="2011-01-08T15:08:12.327" CommentCount="3" />
  <row Id="16808" PostTypeId="2" ParentId="16804" CreationDate="2011-01-08T15:12:08.453" Score="3" ViewCount="" Body="&lt;p&gt;Take the gcd $d$ of the differences between consecutive numbers in the sequence $a_1, ... a_n$; at worst this should take $O(n \log m)$ time where the numbers have size $O(m)$.  The shortest arithmetic progression containing these numbers in the correct order has common difference $d$ and first and last terms $a_1, a_n$, so the total number of terms is $\frac{a_n - a_1}{d} + 1$ and the number of terms that need to be inserted is $\frac{a_n - a_1}{d} + 1 - n$.  &lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2011-01-08T15:12:08.453" CommentCount="1" />
  <row Id="16809" PostTypeId="1" AcceptedAnswerId="16810" CreationDate="2011-01-08T15:19:08.203" Score="2" ViewCount="177" Body="&lt;p&gt;Hello altogether&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I just had a short look on the definition of a Vector Space and couldn't find any obvious reason why&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\lambda a = a \lambda$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\lambda$ is an element of the field $K$ and $a$ is an Element of the set $M$, should be true. Its kind of intuitive since all the vectorspaces I normally come across have this property (Polynoms, indefinitly diffrentiable functions, the $\mathbb{R}^n$). However since $\lambda \in K$ and and $a \in M$, I see no justification for this product being commutative. Is this dependend on the definition of the product between the field and the set? Or is this generally true? If not it would be great if you could provide some examples. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance.&lt;/p&gt;&#xA;" OwnerUserId="3188" LastActivityDate="2011-01-08T16:27:32.013" Title="Vectorspace: Is $\lambda a = a\lambda$ always true?" Tags="&lt;vector-space&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="16810" PostTypeId="2" ParentId="16809" CreationDate="2011-01-08T15:21:26.823" Score="8" ViewCount="" Body="&lt;p&gt;Scalar multiplication is generally not defined on the right.  You can &lt;em&gt;define&lt;/em&gt; $a \lambda = \lambda a$, and then it will of course be true.  &lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2011-01-08T16:27:32.013" LastActivityDate="2011-01-08T16:27:32.013" CommentCount="5" />
  <row Id="16811" PostTypeId="2" ParentId="16781" CreationDate="2011-01-08T16:00:38.253" Score="4" ViewCount="" Body="&lt;p&gt;I'll give some examples of indecomposable p-groups, a survey of small orders, and an asymptotic description.  As Derek Holt points out, most p-groups are directly indecomposable, and perhaps he has a cleaner way of describing the third section.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Some indecomposable finite p-groups:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;cyclic groups&lt;/li&gt;&#xA;&lt;li&gt;p-groups of maximal class (with nilpotency class n and order p&lt;sup&gt;n+1&lt;/sup&gt;), including:&#xA;&lt;ul&gt;&#xA;&lt;li&gt;dihedral groups (of order at least 8)&lt;/li&gt;&#xA;&lt;li&gt;quaternion groups&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Quasidihedral_group&quot; rel=&quot;nofollow&quot;&gt;semi-dihedral groups&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Extraspecial_group&quot; rel=&quot;nofollow&quot;&gt;extra-special groups&lt;/a&gt; with a center of order p and elementary abelian quotient&lt;/li&gt;&#xA;&lt;li&gt;Any group with a cyclic center&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Every group of order p is directly indecomposable (being cyclic).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One out of two groups of order p&lt;sup&gt;2&lt;/sup&gt; is directly indecomposable (the cyclic one).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Three out of five groups of order p&lt;sup&gt;3&lt;/sup&gt; are directly indecomposable (the cyclic one, and the two non-abelian ones which are both extra-special and maximal class).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Eight out of fourteen groups of order 16 and nine out of fifteen groups of order p&lt;sup&gt;4&lt;/sup&gt; for p odd are directly indecomposable (all but two with cyclic center).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;34 out of 51 of order 32, 49 out of 67 of order 243, 59 out 77 for of order 3125, and then a steady pattern out of 61 + 2p + 2(3,p−1) + (4,p−1).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While the groups are organized into families, even the number of families becomes unmanageable after a point and so a different technique is needed for larger n.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Where do decomposable p-groups come from?  They are direct products of smaller p-groups.  However, there are about p&lt;sup&gt;(2n&lt;sup&gt;3&lt;/sup&gt;/27)&lt;/sup&gt; groups of order p&lt;sup&gt;n&lt;/sup&gt;, and so taking direct products of groups of order p&lt;sup&gt;i&lt;/sup&gt; with groups of order p&lt;sup&gt;n−i&lt;/sup&gt; for 1 ≤ i ≤ n/2 gives a somewhat complicated sum but which is (for large n) less than 1/p&lt;sup&gt;n&lt;/sup&gt; th as big as p&lt;sup&gt;(2n&lt;sup&gt;3&lt;/sup&gt;/27)&lt;/sup&gt;.  In other words, a vanishingly small fraction of p-groups are decomposable, and the rest are indecomposable.&lt;/p&gt;&#xA;" OwnerUserId="583" LastActivityDate="2011-01-08T16:00:38.253" />
  <row Id="16812" PostTypeId="1" CreationDate="2011-01-08T16:01:15.667" Score="4" ViewCount="59" Body="&lt;p&gt;A question goes like this:&#xA;Given that there 2 questions out of 3 questions to pass an exam and follow are details of number of people who attended the questions correctly, what is the &lt;strong&gt;maximum&lt;/strong&gt; number of people who could have qualified the exam.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;1. 2 2 2&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The answer here is 3. Consider ab bc ca. It can also be derived as 6/2 = (total number who cleared / number of questions which are required to clear).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I give another sample data&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;2. 245 272 238&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Even here I can say floor(755/2) = 377 are the maximum number of people who have passed the exam.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But consider this data.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;3. 100 10 1&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Now, the average formula would not work. By reasoning, I can say that 11 people could have passed the exam.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But mathematically, how do I derive this as in for any M tests (and we are given the number of people have cleared those M tests indivually) and minimum requirement is N tests, how we can find the maximum number of people who could have cleared the exam.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope, I explained it properly. &lt;/p&gt;&#xA;" OwnerUserId="5516" LastActivityDate="2011-01-08T16:56:43.177" Title="Probability value in a skewed data." Tags="&lt;probability&gt;&lt;average&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="16813" PostTypeId="2" ParentId="16783" CreationDate="2011-01-08T16:01:44.617" Score="3" ViewCount="" Body="&lt;p&gt;A hint: For a given $c\in[0,1]$ compute $f(0)$ and $f(1)$ by means of a Taylor expansion at $c$ (with Lagrange remainder term) and draw conclusions.&lt;/p&gt;&#xA;" OwnerUserId="1303" LastActivityDate="2011-01-08T16:01:44.617" CommentCount="2" />
  <row Id="16814" PostTypeId="1" AcceptedAnswerId="16817" CreationDate="2011-01-08T16:02:26.043" Score="3" ViewCount="79" Body="&lt;p&gt;Let $A$ be a commutative noetherian ring, and let $M$ be a finitely generated projective $A$-module. It is well known and easy to prove that $A$ is locally free in the sense that for every $p \in Spec A$, the module $M_p$ is a free $A_p$-module.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it true that projectives are also locally free in the following (more geometric?) sense:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are elements $f_1,\dots,f_n \in A$ such that $(f_1,\dots,f_n) = 1$, and such that $M_{f_i}$ is a free $A_{f_i}$-module for all $1\le i \le n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this true? if so, can you provide a reference or explain how to prove it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="5491" LastActivityDate="2011-01-08T16:46:09.713" Title="f.g Projectives are locally free" Tags="&lt;commutative-algebra&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="16815" PostTypeId="2" ParentId="16809" CreationDate="2011-01-08T16:18:52.417" Score="0" ViewCount="" Body="&lt;p&gt;A vector space is a particular ring module. It's a right module of a field over and abelian group. In this case, since it is defined over a field (which is commutative by definition) the left module (where the multiplication by a scalar is defined on the left) and the right module (where the multiplication is defined on the right) coincide (it is a bimodule).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the definition of R-modules you can look at wikipedia.&lt;/p&gt;&#xA;" OwnerUserId="1415" LastActivityDate="2011-01-08T16:18:52.417" CommentCount="3" />
  <row Id="16816" PostTypeId="2" ParentId="16797" CreationDate="2011-01-08T16:29:34.897" Score="4" ViewCount="" Body="&lt;p&gt;If you write y=x^2 and pull the constants out you have $$\frac{1}{2\sqrt{2\pi}\sigma}\int_0^{0.25}\sqrt{y}\cdot \exp(-\frac{(y-\mu )^2}{2\sigma ^2})dy$$  If $\sigma$ is very small, the contribution will all come from a small area in $y$ around $\mu$.  So you can set $\sqrt{y}=\sqrt{\mu}$ and use your error function tables for a close approximation.  A quick search didn't turn up moments of $\sqrt{y}$ against the normal distribution, but maybe they are out there.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2011-01-09T21:35:01.413" LastActivityDate="2011-01-09T21:35:01.413" CommentCount="2" />
  <row Id="16817" PostTypeId="2" ParentId="16814" CreationDate="2011-01-08T16:32:10.137" Score="2" ViewCount="" Body="&lt;p&gt;Yes, this is true.  See &lt;a href=&quot;http://mathoverflow.net/questions/13817/a-finitely-generated-locally-free-module-over-a-domain-which-is-not-projective&quot; rel=&quot;nofollow&quot;&gt;this Math Overflow question&lt;/a&gt; for a precise statement and a reference to its proof in Bourbaki's &lt;em&gt;Commutative Algebra&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This result is also stated in &lt;a href=&quot;http://math.uga.edu/~pete/integral.pdf&quot; rel=&quot;nofollow&quot;&gt;my commutative algebra notes&lt;/a&gt;, but the proof is not unfortunately not yet written up there.  I certainly hope that this will be remedied soon though, as I will be teaching a course out of these notes starting on Monday.  When the proof gets written, I will update this answer with a page number.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Added&lt;/b&gt;: Here is something in the MO answer that I decided was worth a comment here.  For finitely generated modules, this stronger version of local freeness is actually &lt;em&gt;equivalent&lt;/em&gt; to projectivity, whereas the weaker &quot;pointwise local freeness&quot; is subtly weaker in general.&lt;/p&gt;&#xA;" OwnerUserId="299" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2011-01-08T16:46:09.713" LastActivityDate="2011-01-08T16:46:09.713" CommentCount="3" />
  <row Id="16818" PostTypeId="2" ParentId="16812" CreationDate="2011-01-08T16:36:19.140" Score="1" ViewCount="" Body="&lt;p&gt;For $M=3$ and $N=2$, this works:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;floor $((\min(a,b+c) + \min(b,c+a) + \min(c,a+b))/2)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For general $M,N$, I expect the answer to be much more complicated.&lt;/p&gt;&#xA;" OwnerUserId="1508" LastActivityDate="2011-01-08T16:36:19.140" />
  <row Id="16820" PostTypeId="2" ParentId="16779" CreationDate="2011-01-08T17:02:21.503" Score="11" ViewCount="" Body="&lt;p&gt;If $G$ is any topological group, &lt;a href=&quot;http://en.wikipedia.org/wiki/Identity_component&quot; rel=&quot;nofollow&quot;&gt;the connected component $G_0$ containing the identity is a closed normal subgroup&lt;/a&gt; and the other connected components are cosets.  So a necessary condition for a space to admit the structure of a topological group is that all of its connected components be homeomorphic.  In the case of a smooth manifold $G_0$ is also open, so the space of components $G/G_0$ is discrete.  It follows that a smooth manifold $X$ admits the structure of a Lie group iff all of its components are homeomorphic and any one of them admits the structure of a Lie group.  In other words, it is no real loss of generality to assume that $X$ is connected.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are lots of deep necessary conditions on the topology of $X$ for $X$ to admit the structure of a Lie group.  See for instance &lt;a href=&quot;http://www.math.sunysb.edu/~basu/notes/GSS5.pdf&quot; rel=&quot;nofollow&quot;&gt;Theorem 2.7 of these notes&lt;/a&gt; which asserts the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a) $\pi_1(X)$ is a finitely generated abelian group.&lt;br&gt;&#xA;b) $\pi_2(X)$ is the trivial group.&lt;br&gt;&#xA;c) $\pi_3(X)$ is a finitely generated free abelian group.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is much more to say here, although others are more qualified than I to say it.    &lt;/p&gt;&#xA;" OwnerUserId="299" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2011-01-08T19:31:38.207" LastActivityDate="2011-01-08T19:31:38.207" CommentCount="3" />
  <row Id="16821" PostTypeId="2" ParentId="16521" CreationDate="2011-01-08T17:25:18.383" Score="3" ViewCount="" Body="&lt;p&gt;If $f$ has abelian Galois group and you can find an explicit embedding of its splitting field into $\mathbb{Q}(\zeta_n)$, you get a quotient map $(\mathbb{Z}/n\mathbb{Z})^{\ast} \to \text{Gal}(f)$ which makes the action of the Galois group quite explicit.  Applying elements of $(\mathbb{Z}/n\mathbb{Z})^{\ast}$ to a root $a$ of $f$ in $\mathbb{Q}(\zeta_n)$ gives you some explicit expressions in $\mathbb{Q}(\zeta_n)$ which you can then try to express as polynomials in $a$.  I don't know how easy this will be to do, though, but in certain cases it is fairly explicit: for example if $f$ is the minimal polynomial of $2 \cos \frac{2 \pi}{n}$ you get Chebyshev polynomials.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2011-01-08T17:25:18.383" />
  <row Id="16822" PostTypeId="2" ParentId="16779" CreationDate="2011-01-08T19:21:30.990" Score="7" ViewCount="" Body="&lt;p&gt;To add some conditions on the topology, if a compact manifold $M$ admits the structure of an abelian Lie group, then $M$ must be isomorphic to $(S^1)^n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More generally, if $M$ admits the structure of a Lie group, then $M$ must have the same cohomology ring with $\mathbb{Q}$ coefficients as a product of odd spheres.  If $M$ admits the structure of a nonabelian Lie group, then $S^3$ must appear in the product of odd spheres.  Further, such an $M$ must be rationally elliptic, which puts VERY strong constraints on the rational homotopy groups.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The universal cover of $M$ can only have $2,3,$ or $5$ torsion in its cohomology.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A noncompact Lie group must be diffeomorphic to $\mathbb{R}^k\times G$ where $G$ is a compact Lie group.  (I'm not positive about this line.  Certainly a noncompact simply connected group must be diffeomorphic to a product like this, with $G$ simply connected).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, I think it could be VERY difficult to come up with sufficient topological conditions to be a Lie group.  For example, the space $SO(7)\times S^7$ is diffeomorphic to $SO(8)$, and hence carries a Lie group structure.  However, as mentioned by others, $S^7$ itself does not carry a Lie group structure.&lt;/p&gt;&#xA;" OwnerUserId="331" LastActivityDate="2011-01-08T19:21:30.990" />
  <row Id="16823" PostTypeId="2" ParentId="16796" CreationDate="2011-01-08T19:59:20.497" Score="4" ViewCount="" Body="&lt;p&gt;Choose a seat $S$, and a bellicose guest $B$. Sit $B$ in $S$, and tell them not to move, whether they like it or not. That done, there are $(K-1)!$ ways of ordering the remaining bellicose guests clockwise around the table, and $F!$ ways of ordering the friendly guests (here I am using leonbloy's notation $F = N - K$). For each such ordering, we have to choose a &lt;em&gt;pattern&lt;/em&gt; of the form $f...b...f...b...f$. This pattern:&lt;br&gt;&#xA;1. starts and ends with $f$ (so that $B$ is isolated);&lt;br&gt;&#xA;2. contains $(K-1)$ $b$'s and $F$ $f$'s; and&lt;br&gt;&#xA;3. contains no two adjacent $b$'s.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the number of such patterns is the same as the number of patterns that&lt;br&gt;&#xA;1. start with $f$; and&lt;br&gt;&#xA;2. contain $(K-1)$ $b$'s and $(F-K+1)$ $f$'s.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(To see this, just replace each instance of $bf$ in the original pattern by $b$.) The number of such patterns is the binomial coefficient $\binom{F-1}{K-1}$. So we end up with:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(K-1)!F!\binom{F-1}{K-1} = \frac{F!(F-1)!}{(F-K)!}$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the number of seating arrangements with guest $B$ in seat $S$. Multiply by $N$ to get the total number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt; Reading the question more carefully, it asks for the number of (what I call here) patterns, not the number of seatings. For each pattern, there are $K!F!$ seatings, so the answer is  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$N\frac{F!(F-1)!}{(F-K)!}/(K!F!) = \frac{N(F-1)!}{(F-K)!K!}$&lt;/p&gt;&#xA;" OwnerUserId="1508" LastEditorUserId="1508" LastEditorDisplayName="" LastEditDate="2011-01-09T13:32:48.773" LastActivityDate="2011-01-09T13:32:48.773" CommentCount="10" />
  <row Id="16824" PostTypeId="2" ParentId="16766" CreationDate="2011-01-08T20:31:58.010" Score="1" ViewCount="" Body="&lt;p&gt;I'm not sure how well this would work in practice, but I've always been interested in introducing mathematical rigor through an abstract algebra course, rather than through analysis.  As a motivating question, you could have the problem of squaring the circle (for example), though this would involve going into a lot of theory pretty quickly.  (It might also be more appropriate in the west, where squaring the circle and trisecting the angle have been problems of interest since antiquity.)&lt;/p&gt;&#xA;" OwnerUserId="1769" LastActivityDate="2011-01-08T20:31:58.010" CommentCount="1" />
  <row Id="16825" PostTypeId="2" ParentId="16794" CreationDate="2011-01-08T20:50:12.853" Score="6" ViewCount="" Body="&lt;p&gt;If you are considering the notion of decidability in the sense of Turing decidability, as is usual in computability theory, then it doesn't make immediate sense to ask whether the isomorphism problem for arbitrary countable groups is decidable or not, since the &quot;input&quot; for an instance of this problem would consist of two infinite objects, the groups in some form of presentation. This takes one outside the context of decidability with Turing machines, which work with finite input and output. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem does make sense for finitely presented groups, since here one can be given two such presentations. In the abelian case, the isomorphism problem in this case is decidable. In the non-abelian case, it is not decidable---a result following from the non-decidability of the word problem, since one cannot even determine the special case of whether a given presentation is a presentation of the trivial group. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you consider computably presented groups, by taking as inputs two programs that will enumerate the relations, then you will not be able to decide in finite time any nontrivial property of the sets they enumerate. This is a consequence of Rice's theorem. For example, you will not be able to decide whether the programs will enumerate no relations or all relations, and so the isomorphism problem will not be decidable in this context. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nevertheless, by adopting a more set-theoretic rather than computational  perspective, one could inquire what is the descriptive-set-theoretic complexity of the set of pairs of isomorphic countable abelian groups. It clearly has complexity at most $\Sigma^1_1$, that is, lightface analytic, since two groups are isomorphic iff there is an isomorphism, and I believe that it is likely $\Sigma^1_1$-complete, since the countable abelian groups can code quite a bit of information, but I would have to check with my descriptive set-theoretic collegues. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is also the subject of Borel equivalence relation theory, which considers the complexity of equivalence relations under Borel reducibility, and in general, the isomorphism relation for finitely generated groups (not necessarily abelian) is a Borel relation in the relevant space, but quite high in complexity. The isomorphism relation of arbitrary countable groups is much higher still, and as I've said, I think it remains high even just for countable abelian groups.&lt;/p&gt;&#xA;" OwnerUserId="413" LastEditorUserId="413" LastEditorDisplayName="" LastEditDate="2011-01-08T21:11:53.023" LastActivityDate="2011-01-08T21:11:53.023" CommentCount="3" />
  <row Id="16826" PostTypeId="2" ParentId="16779" CreationDate="2011-01-08T21:00:46.073" Score="4" ViewCount="" Body="&lt;p&gt;The cohomology of the manifold has to be a Hopf Algebra.  More generally, the manifold has to be an H-space. This implies the unit sphere bundle is fibrewise homotopy-trivial (see comments below). Earlier I thought manifold + h-space implied triviality of the tangent bundle but that's not clear.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A manifold that's an H-space is called an &quot;H-manifold&quot; in the literature.  Sometimes this is ambiguous since if H is a group, &quot;H-manifold&quot; could mean &quot;manifold with an action of the group H&quot; so beware when searching the literature. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe there is a well-known obstruction theory for H-manifolds to be Lie groups and many algebraic topologists know a pile of examples off the top of their heads.  Unfortunately, if I ever did know such a pile of examples I've forgotten!&lt;/p&gt;&#xA;" OwnerUserId="642" LastEditorUserId="642" LastEditorDisplayName="" LastEditDate="2011-01-09T22:20:42.700" LastActivityDate="2011-01-09T22:20:42.700" CommentCount="5" />
  <row Id="16827" PostTypeId="2" ParentId="16797" CreationDate="2011-01-08T21:19:22.557" Score="1" ViewCount="" Body="&lt;p&gt;Not an answer, but might still be helpful. Using the variable substitution that Ross mentions in his answer, we can treat a simpler case $\mu=0$ more easily. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the following integral, Wolfram Alpha tells us that (I hope I did not make a transcription error here):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_0^\infty \sqrt{y}e^{-y^2/(2\sigma^2)}dy = \frac{\sigma^{3/2}\Gamma(3/4)}{2^{1/4}},$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But your problem goes from $0$ to $0.25$, so some approximations are needed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Link-text: http://www.wolframalpha.com/input/?i=Integrate[Sqrt[x]+Exp[-x^2%2F%282+s^2%29]%2C{x%2C0%2Cinf}]&lt;/p&gt;&#xA;" OwnerUserId="1709" LastActivityDate="2011-01-08T21:19:22.557" CommentCount="1" />
  <row Id="16828" PostTypeId="2" ParentId="16766" CreationDate="2011-01-08T21:23:37.977" Score="2" ViewCount="" Body="&lt;p&gt;Perhaps some non-Euclidean Geometry? What's more fun than pulling the rug out from under everything the students &lt;em&gt;thought&lt;/em&gt; they knew about geometry? :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Marta Sved's &lt;a href=&quot;http://www.maa.org/EbusPPRO/Bookstore/ProductDetail/tabid/170/Default.aspx?ProductId=1014&quot; rel=&quot;nofollow&quot;&gt;Journey into Geometries&lt;/a&gt; provides a readable (if pricey) take on the subject, in the style (and with the characters) of Lewis Carroll's &quot;Alice in Wonderland&quot;. When I first read through the book many years ago, I wanted to share it with students, though I didn't have the opportunity. The entire book may be too much for kids alone, but filtered through a teacher, some notions (such as the &quot;Power of a Point&quot; relative to a circle) can be pretty accessible, as I recall ... but my memory of the details is quite sketchy. (I seem to have lost my copy of the book.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe there are hyperbolic add-ons to software like Geometer's Sketchpad that facilitate playing with circle inversions and such, if that's an option. (No doubt there are many applets on-line as well.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the students are in fact very good at (Euclidean) trigonometry, they may find non-Euclidean trig interesting. If nothing else, students may gain an appreciation for the convenience of our (locally) Euclidean universe and its straightforward way of computing the length of the hypotenuse. (On the other hand, it's hard to beat the formula for the area of a hyperbolic triangle.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And many Escher woodcuts are intriguing visualizations of (approximations of) tesselations of the hyperbolic plane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, I'm in favor of any enrichment exercise that keeps geometry from being &quot;that weird class between Algebra I and Algebra II&quot; that students tend to forget all about.&lt;/p&gt;&#xA;" OwnerUserId="409" LastActivityDate="2011-01-08T21:23:37.977" />
  <row Id="16829" PostTypeId="1" AcceptedAnswerId="17036" CreationDate="2011-01-08T21:34:14.577" Score="11" ViewCount="226" Body="&lt;p&gt;Here is a question that I have had in my head for a little while and was recently reminded of.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Let $X$ be a (nonempty!) topological space.  What are useful (or even nontrivial) sufficient conditions for $X$ to admit a group law compatible with its topology?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There are many necessary conditions: for instance $X$ must be completely regular.  In particular, if it is a Kolmogorov space it must be a Urysohn space.  It must be homogeneous in the sense that the self-homeomorphism group $\operatorname{Aut}(X)$ acts transitively.  In particular $\operatorname{Aut}(X)$ must act transitively on the connected components of $X$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is rather clear though that these conditions are nowhere near sufficient even for very nice topological spaces.  A recent question on this site calls attention to this: even for the class of smooth manifolds there are a sequence of successively more intricate necessary conditions but (apparently) no good sufficient conditions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What about for other classes of spaces?  Here are two examples: it suffices for $X$ to be discrete (i.e., every set is open) or indiscrete (i.e., only $\varnothing$ and $X$ are open).  These are, of course, completely trivial.  Can anyone do any better than this?  For instance:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If $X$ is totally disconnected and compact (for me this implies Hausdorff), must it admit the structure of a topological group?  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Added&lt;/b&gt;: as some have rightly pointed out, the above necessary condition of homogeneity is not implied by my assumptions.  So let us assume this as well.  Note that assuming second-countability as well forces the space to be (finite and discrete or) homeomorphic to the Cantor set, which is well-known to carry a topological group structure, so I don't want to assume that.&lt;/p&gt;&#xA;" OwnerUserId="299" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2011-01-09T00:27:17.793" LastActivityDate="2011-01-10T20:25:11.993" Title="Useful sufficient conditions for a topological space to be the underlying space of a topological group?" Tags="&lt;topology&gt;&lt;topological-groups&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="2" />
  <row Id="16830" PostTypeId="1" AcceptedAnswerId="16833" CreationDate="2011-01-08T21:48:42.417" Score="2" ViewCount="129" Body="&lt;p&gt;The problem is:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;For a prime number p the set of co-primes less than or equal to it is&#xA;given by {1,2,3,4,...p-1} .&#xA;&#xA;We define f(x,p) 0 &amp;lt; x &amp;lt; p = 1 if and only if all the&#xA;numbers from 1 to p-1 can be written as a power of x in modulo-p arithmetic .&#xA;&#xA;Let n be the largest 12-digit prime number . Find the product of&#xA;all integers j less than n such that f(j,n)=1, in modulo-n arithmetic&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Can anyone give me a better explanation?&lt;/p&gt;&#xA;" OwnerUserId="5522" LastActivityDate="2011-01-08T22:09:14.303" Title="Need better explanation of a mathematical/programming problem?" Tags="&lt;prime-numbers&gt;&lt;modular-arithmetic&gt;&lt;powers&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="16831" PostTypeId="1" AcceptedAnswerId="16837" CreationDate="2011-01-08T21:56:36.917" Score="4" ViewCount="401" Body="&lt;p&gt;Hello everybody, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the title says, I'm wondering if there is a continuous function such that f is nonzero on [0, 1], and for which $\int_0^1 f(x)x^n dx = 0 \,\forall n \geq 1$. I am trying to solve a problem proving that if (on C([0, 1])) $\int_0^1 f(x)x^n dx = 0 \,\forall n \geq 0$, then f must be identically zero. I presume then we do require the n=0 case to hold too, otherwise it wouldn't be part of the statement. Is there ay function which is not identically zero which satisfies $\int_0^1 f(x)x^n dx = 0 \,\forall n \geq 1$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The statement i am attempting to prove is homework, but this is just idle curiosity (though i will tag it as homework anyway since it is related). Thank you!&lt;/p&gt;&#xA;" OwnerUserId="5473" LastEditorUserId="5473" LastEditorDisplayName="" LastEditDate="2011-01-08T22:21:51.340" LastActivityDate="2011-01-10T23:23:22.097" Title="A nonzero function in C([0, 1]) for which $\int_0^1 f(x)x^n dx = 0 \,\forall n \geq 1$" Tags="&lt;real-analysis&gt;&lt;homework&gt;&lt;analysis&gt;&lt;integral&gt;" AnswerCount="4" CommentCount="9" FavoriteCount="4" />
  <row Id="16832" PostTypeId="2" ParentId="16797" CreationDate="2011-01-08T22:00:04.307" Score="2" ViewCount="" Body="&lt;p&gt;How about some good old-fashioned trapezoid rule?&lt;/p&gt;&#xA;" OwnerUserId="5523" LastActivityDate="2011-01-08T22:00:04.307" CommentCount="1" />
  <row Id="16833" PostTypeId="2" ParentId="16830" CreationDate="2011-01-08T22:04:00.767" Score="4" ViewCount="" Body="&lt;p&gt;The numbers $x$ you are looking for are called &lt;a href=&quot;http://en.wikipedia.org/wiki/Primitive_root_modulo_n&quot; rel=&quot;nofollow&quot;&gt;primitive roots&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is well known that the product of primitive roots of an odd prime $p \gt 3$ is $1 \mod p$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a proof: If $g$ is a primitive root, what can you say about $g^{-1}$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: $g^{-1}$ is the number $\mod p$ such that $g g^{-1} = 1 \mod p$.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2011-01-08T22:09:14.303" LastActivityDate="2011-01-08T22:09:14.303" CommentCount="2" />
  <row Id="16835" PostTypeId="2" ParentId="16831" CreationDate="2011-01-08T22:11:35.330" Score="3" ViewCount="" Body="&lt;p&gt;The answer is no.  Actually I believe the following is a theorem whose name totally escapes me at the moment: assume that $f$ is continuous and let $a_n$ be a sequence of increasing positive integers such that $\int_0^1 f(x) x^{a_n} \, dx = 0$ for $n \ge 1$.  If $\sum \frac{1}{a_n}$ diverges, then $f$ is identically zero!  (Edit: this is a corollary of the &lt;a href=&quot;http://en.wikipedia.org/wiki/M%C3%BCntz%E2%80%93Sz%C3%A1sz_theorem&quot; rel=&quot;nofollow&quot;&gt;Müntz–Szász&lt;/a&gt; theorem - thanks, Moron!)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, the problem isn't phrased the way it is because stronger statements are false; the stronger versions are just harder to prove.&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2011-01-08T22:16:50.393" LastActivityDate="2011-01-08T22:16:50.393" CommentCount="2" />
  <row Id="16837" PostTypeId="2" ParentId="16831" CreationDate="2011-01-08T22:22:48.200" Score="4" ViewCount="" Body="&lt;p&gt;(I am turning this into Community wiki, since the original version made an obvious mistake). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The result follows, for example, from the &lt;a href=&quot;http://en.wikipedia.org/wiki/Stone%E2%80%93Weierstrass_theorem&quot; rel=&quot;nofollow&quot;&gt;Stone-Weierstrass theorem&lt;/a&gt;, once one justifies that the limit of some integrals is the integral of the limit, which can be done (overkill) using Lebesgue's &lt;a href=&quot;http://en.wikipedia.org/wiki/Dominated_convergence_theorem&quot; rel=&quot;nofollow&quot;&gt;dominated convergence theorem&lt;/a&gt; or (more easily) using simple estimates from the fact that $f$ is bounded, since it is continuous.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Below I give full details, which you should probably not read until after your homework is due, since this also solves your homework.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Spoilers: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a sequence of polynomials $p_n(x)$ that converges uniformly to $xf(x)$ on ${}[0,1]$. We have $\int_0^1xf(x)p_n(x)dx=0$ for all $n$, by assumption, since $xp_n(x)$ is a sum of monomials the integral of whose integral with $f$ is 0. Now take the limit as $n\to\infty$ to conclude that $\int_0^1x(f(x))^2dx=0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This gives us that $f=0$ because if $f(x_0)\ne 0$, continuity ensures a positive $\epsilon&amp;gt;0$ and an interval $(a,b)$ with $a&amp;gt;0$ such that $|f(x)|\ge\epsilon$ for all $x\in(a,b)$. But then $\int_0^1xf(x)^2dx\ge la\epsilon^2&amp;gt;0$, where $l=b-a$ is the length of the interval.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To see that the limit of the integrals is 0 without using dominated convergence, let $M\ge|f(x)|$ for all $x\in[0,1]$. The, for any $\delta&amp;gt;0$, if $n$ is large enough, we have $$\int_0^1f(x)xp_n(x)dx=\int_0^1f\times(p-xf+xf)dx=\int_0^1xf(x)^2dx+\int_0^1f\times(p-xf)dx,$$ and the second integral is bounded by $\int_0^1|f||p-xf|dx\le M(\delta/M)=\delta$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, even this is approach is an overkill. (For example, Müntz's theorem gives a more general fact, as already mentioned in another answer.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Apologies for the original mistake.)&lt;/p&gt;&#xA;" OwnerUserId="462" LastEditorUserId="462" LastEditorDisplayName="" LastEditDate="2011-01-08T23:58:36.823" LastActivityDate="2011-01-08T23:58:36.823" CommunityOwnedDate="2011-01-08T22:37:33.470" CommentCount="3" />
  <row Id="16838" PostTypeId="1" CreationDate="2011-01-08T22:23:48.360" Score="1" ViewCount="423" Body="&lt;p&gt;I need to write a program to solve matrix equations Ax=b where A is an nxn matrix, and b is a vector with n entries using LU decomposition. Unfortunately I'm not allowed to use any prewritten codes in Matlab. I am having problems with the first part of my code where i decompose the matrix in to an upper and lower matrix. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have written the following code, and cannot work out why it is giving me all zeros on the diagonal in my lower matrix. Any improvements would be greatly appreciated.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;function[L R]=LR2(A)&#xA;&#xA;%Decomposition of Matrix AA: A = L R&#xA;&#xA;z=size(A,1);&#xA;&#xA;L=zeros(z,z);&#xA;&#xA;R=zeros(z,z);&#xA;&#xA;for i=1:z&#xA;&#xA;% Finding L&#xA;&#xA;for k=1:i-1&#xA;&#xA;L(i,k)=A(i,k);&#xA;&#xA;for j=1:k-1&#xA;&#xA;L(i,k)= L(i,k)-L(i,j)*R(j,k);&#xA;&#xA;end&#xA;&#xA;L(i,k) = L(i,k)/R(k,k);&#xA;&#xA;end&#xA;&#xA;% Finding R&#xA;&#xA;for k=i:z&#xA;&#xA;R(i,k) = A(i,k);&#xA;&#xA;for j=1:i-1&#xA;&#xA;R(i,k)= R(i,k)-L(i,j)*R(j,k);&#xA;&#xA;end&#xA;&#xA;end&#xA;&#xA;end&#xA;&#xA;R&#xA;&#xA;L&#xA;&#xA;end&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I know that i could simply assign all diagonal components of L to be 1, but would like to understand what the problem is with my program! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am also wondering how to change this program to include pivoting. I understand I need to say that if a diagonal element is equal to zero something needs to be changed. How would I go about this? &#xA;Thanks in advance!&lt;/p&gt;&#xA;" OwnerUserId="4697" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2011-01-09T00:52:18.527" LastActivityDate="2011-05-09T02:55:48.773" Title="Implement a program in Matlab for LU decomposition with pivoting." Tags="&lt;homework&gt;&lt;numerical-methods&gt;&lt;computer-science&gt;&lt;math-software&gt;&lt;matlab&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="16839" PostTypeId="2" ParentId="16784" CreationDate="2011-01-08T22:29:01.400" Score="1" ViewCount="" Body="&lt;p&gt;You can use a null-space method.  Note that if we are trying to find $\mathbf u$ then we require that $\mathbf u \cdot T1 = 0$ and $\mathbf u \cdot T2 = 0$.  We solve this by finding the null space of the matrix $[ T1 \hspace{3pt} T2]^T$.&lt;/p&gt;&#xA;" OwnerUserId="4726" LastActivityDate="2011-01-08T22:29:01.400" />
  <row Id="16840" PostTypeId="2" ParentId="16765" CreationDate="2011-01-08T22:46:10.500" Score="3" ViewCount="" Body="&lt;p&gt;I found the book &lt;em&gt;The genesis of the abstract group concept&lt;/em&gt;, by Hans Wussing, to be very interesting.  It gives a scholarly history of the development of the concept of group, from its roots in number theory, geometry, and the theory of equations prior to the 19th century, through to (more-or-less) the end of the 19th century.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for reading the original papers, while there is a lot to be said for this, you should bear in mind that these are not only typically written in languages other than English, but are also in a mathematical language that is quite different from our modern language.  If you do decide to look at the originals, Wussing's book would be a good guide to the major papers in the initial development of group theory.&lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2011-01-08T22:46:10.500" />
  <row Id="16841" PostTypeId="1" AcceptedAnswerId="16844" CreationDate="2011-01-08T22:49:14.207" Score="4" ViewCount="222" Body="&lt;p&gt;I was discussing with a friend of mine about her research and I came across this problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem essentially boils down to this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f(x)$ is a function defined in $[0,1]$ such that $f(x) + f(1-x) = f(1)$. I want to find a condition on $f(x)$ so that I can conclude $f(x) = f(1)x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clearly, $f \in C^{0}[0,1]$ alone is insufficient to conclude $f(x) = f(1)x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My hunch is if $f(x) \in C^{\infty}[0,1]$, then $f(x) = f(1)x$. However, I am unable to prove it. Further, is there a weaker condition with which I can conclude $f(x) = f(1)x$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This problem closely resembles another problem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $f(x+y) = f(x) + f(y)$, $\forall x,y \in \mathbb{R}$ and if $f(x)$ is continuous at atleast one point, then $f(x) = f(1)x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know how to prove this statement, but I am unable to see whether this will help me with the original problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Though these details might not be of much importance to her, I am curious to know.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As Qiaochu pointed out, I need stronger conditions on $f$ to come up with some reasonable answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is something which I know that $f$ has to satisfy the following: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\forall n \in \mathbb{Z}^{+}\backslash \{1\}$, $f(x_1) + f(x_2) + \cdots + f(x_n) = f(1)$, where $\displaystyle \sum_{k=1}^{n} x_k = 1$, $x_i \geq 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that $n=2$ boils down to what I had written earlier.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2011-01-08T23:30:27.613" LastActivityDate="2011-01-09T02:33:16.543" Title="$f(x) + f(1-x) = f(1)$" Tags="&lt;analysis&gt;&lt;functional-equations&gt;" AnswerCount="2" CommentCount="6" FavoriteCount="1" />
  <row Id="16842" PostTypeId="2" ParentId="16802" CreationDate="2011-01-08T22:50:43.467" Score="8" ViewCount="" Body="&lt;p&gt;Suppose $G$ is a group, and $H$ is any subgroup of $G$. I would like to try to define something similar to &quot;congruence modulo $n$&quot; for integers, but for this arbitrary group $G$ and subgroup $H$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the integers, we say that $a\equiv b\pmod{m}$ if and only if $a-b$ is an element of the subgroup $m\mathbb{Z}$. So we are going to try something similar for $H$ and $G$, taking into account that $G$ need not be abelian.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not going to assume the group is finite until I say so explicitly. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Definition.&lt;/strong&gt; Let $G$ be a group and let $H$ be a subgroup. If $x,y\in G$, we say that $x$ and $y$ are &lt;i&gt;congruent on the right modulo&lt;/i&gt; $H$ if and only if $xy^{-1}\in H$. We write this $x\equiv_H y$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proposition.&lt;/strong&gt; $\equiv_H$ is an equivalence relation on $G$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Proof.&lt;/em&gt; We need to show that the relation is reflexive, symmetric, and transitive. Let $x\in G$. Then $xx^{-1} = e\in H$ (since $H$ is a subgroup, hence contains $e$), so $x\equiv_H x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let $x,y\in G$, and suppose that $x\equiv_H y$; we want to show that $y\equiv_H x$ holds as well. Since $x\equiv_H y$, then $xy^{-1}\in H$. Since $H$ is a subgroup, it is closed under taking inverses, so $(xy^{-1})^{-1} = (y^{-1})^{-1}x^{-1} = yx^{-1}\in H$. By defintion, this means that $y\equiv_H x$, as desired.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, suppose that $x,y,z\in G$ and that $x\equiv_H y$ and $y\equiv_Hz$. We want to show that $x\equiv_H z$. The first congruence implies $xy^{-1}\in H$; the second that $yz^{-1}\in H$. Since $H$ is a subgroup, it is closed under products, so $(xy^{-1})(yz^{-1})=xz^{-1}\in H$, hence $x\equiv_H z$, as desired. &lt;strong&gt;QED&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice that the three basic properties of a subgroup are precisely needed for the three basic properties of an equivalence relation: $H$ contains the identity is used for reflexivity; $H$ is closed under inverses is used for symmetry; and $H$ is closed under products is used for transitivity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, since $\equiv_H$ is an equivalence relation on $G$, by the Fundamental Theorem of Equivalence Relations, $\equiv_H$ induces a partition on $G$; that is, $G$ is broken up into disjoint parts, one for each equivalence class. Our next goal is to figure out if we have some description of the equivalence classes that is independent of the equivalence relation (this is very useful in general). Indeed, we do:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Theorem.&lt;/strong&gt; Let $G$ be a group and let $H$ be a subgroup; let $x\in G$. The equivalence class of $x$ under the relation  $\equiv_H$ is equal to the set $Hx = \{hx\mid h\in H\}$. That is,&#xA;$$[x]_H = \{y\in G\mid x\equiv_Hy\} = \{hx\mid h\in H\} = Hx.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Proof.&lt;/em&gt; We need to show that each element in $Hx$ is in $[x]_H$, and that each element in $[x]_H$ is in $Hx$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $z\in Hx$; that means that $z = hx$ for some $h\in H$. Then &#xA;$$xz^{-1} = x(hx)^{-1} = x(x^{-1}h^{-1}) = h^{-1}\in H,$$&#xA;so by definition we have that $x\equiv_H z$. Thus, if $z\in Hx$, then $z\in[x]_H$. So $Hx\subseteq[x]_H$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conversely, let $y\in [x]_H$. Then $x\equiv_H y$, so $xy^{-1}\in H$. Thus, there exists $h\in H$ such that $xy^{-1}=h$. Multiplying on the right by $y$ we get $x=hy$, and multiplying on the left by $h^{-1}$ we get $h^{-1}x = y$. Since $h^{-1}\in H$, we get $y = h^{-1}x \in Hx$. Thus, if $y\in[x]_H$ then $y\in Hx$, so $[x]_H\subseteq Hx$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Putting the two inclusions together, we conclude that $[x]_H = Hx$ for each $x\in G$. &lt;strong&gt;QED&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Corollary.&lt;/strong&gt; Let $G$ be a group, $H$ a subgroup. Then:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$Hx$ is nonempty for each $x\in G$.&lt;/li&gt;&#xA;&lt;li&gt;$\displaystyle G = \mathop{\cup}_{x\in G} Hx$.&lt;/li&gt;&#xA;&lt;li&gt;For all $x,y\in G$, if $Hx\cap Hy\neq\emptyset$, then $Hx=Hy$.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Proof.&lt;/em&gt; This follows from the fact that since the sets of the form $Hx$ are exactly the equivalence classes of the equivalence relation $\equiv_H$, they form a partition of $G$. &lt;strong&gt;QED&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Corollary.&lt;/strong&gt; Let $G$ be a group, $H$ a subgroup. For all $x,y\in G$, $x\equiv_H y$ if and only if $Hx = Hy$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We give the sets of the form $Hx$ a name:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Definition.&lt;/strong&gt; Let $G$ be a group, $H$ a subgroup, and $x\in G$. The set &#xA;$$Hx = \{ hx\mid h\in H\}$$&#xA;is called the &lt;strong&gt;right coset of&lt;/strong&gt; $x$ &lt;strong&gt;modulo&lt;/strong&gt; $H$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Coset&quot; is short for &quot;congruence set&quot;, because the right coset is exactly the set of all things that are congruent on the right to $x$ modulo $H$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, when you have an equivalence relation, the equivalence classes can have different sizes. But not so with these equivalence relation: because they are obtained by taking all possible products with elements of $H$, all equivalence classes are bijectable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Theorem.&lt;/strong&gt; Let $G$ be a group and $H$ a subgroup. Let $x,y\in G$ be any two elements. Then there is a bijection between the sets $Hx$ and $Hy$, given by &#xA;\begin{align*}&#xA;\psi\colon Hx &amp;amp;\to Hy\\&#xA;hx&amp;amp;\mapsto hy&#xA;\end{align*}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Proof.&lt;/em&gt; To show that $\psi$ is one-to-one, let $hx,h&amp;#39;x\in Hx$ be such that $\psi(hx)=\psi(h&amp;#39;x)$. Then $hy = h&amp;#39;y$. Multiplying on the right by $y^{-1}$, we get $h=h&amp;#39;$, so $hx=h&amp;#39;x$. Thus, $\psi$ is one-to-one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To show $\psi$ is onto, let $hy\in H$. Then $hx\in Hx$, and $\psi(hx)=hy$. Thus, $\psi$ is one-to-one and onto, so it is a bijection. &lt;strong&gt;QED&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Now,&lt;/strong&gt; let's assume that $G$ is finite, and $H$ is a subgroup. Then the equivalence relation $\equiv_H$ partitions $G$ into equivalence classes; each equivalence class is of the form $Hx$ for some $x\in G$, and by the previous theorem, they are all bijectable with one another, so they all have the same size, $k$. If there are $m$ distinct equivalence classes, each of size $k$, and they partition $G$, then the size of $G$ is the sum of the sizes of the distinct equivalence classes; that is, $|G|=mk$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But what is $k$? $k$ is the size of the equivalence classes; they are all the same size. In particular, the equivalence class $He$ has size $k$. But what is $He$? Well,&#xA;$$He = \{ he\mid h\in H\} = \{h\mid h\in H\} = H.$$&#xA;That is, $He = H$, so that $H$ itself has size $k$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $|G|=mk = m|H|$, that means that $|H|$ divides $|G|$. That is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Corollary [Lagrange's Theorem].&lt;/strong&gt; If $G$ is a finite group, and $H$ is a subgroup, then the size of $H$ divides the size of $G$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might ask if you can define a &quot;congruence on the left&quot;; yes, you can. We say $x$ and $y$ are congruent on the left modulo $H$ if $x^{-1}y\in H$, and we write $x{}_H\equiv y$. If we proceed as above, then we get that the equivalence class of $x$ modulo $H$ on the left is $xH$ (instead of $Hx$); the process is completely analogous; these are called &lt;strong&gt;left&lt;/strong&gt; cosets of $H$. Turns out that there is a bijection between left and right cosets. One is tempted to define the bijection by sending the coset $xH$ to the coset $Hx$, but it turns out that this is not well defined (you can have $xH = yH$, but also have $Hx\neq Hy$). However, mapping $xH$ to the coset $Hx^{-1}$ works out (I'll leave it to you to work it out), so that the number of left cosets and the number of right cosets of $H$ in $G$ are the same, and the size of any left coset is the same as the size of any right coset of $H$. In general, the equivalence relation &quot;congruent on the left modulo $H$&quot; is &lt;strong&gt;not the same&lt;/strong&gt; as the equivalence relation &quot;congruent on the right modulo $H$&quot;; the case when they &lt;em&gt;are&lt;/em&gt; the same is interesting in its own right, and corresponds to the case when $H$ is a &lt;em&gt;normal subgroup&lt;/em&gt;, which has a lot of important consequences all of its own that you will no doubt discover soon.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-09T13:32:12.547" LastActivityDate="2011-01-09T13:32:12.547" CommentCount="4" />
  <row Id="16843" PostTypeId="2" ParentId="16841" CreationDate="2011-01-08T22:58:17.937" Score="7" ViewCount="" Body="&lt;p&gt;Let $g(x) = f(x + 1/2)$, which is defined on $[-1/2, 1/2]$ and which satisfies $g(x) + g(-x) = g(1/2)$.  Now, any function $g$ on $[-1/2, 1/2]$ can be written&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$g(x) = \frac{g(x) + g(-x)}{2} + \frac{g(x) - g(-x)}{2} = \frac{g(1/2)}{2} + \frac{g(x) - g(-x)}{2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where the first term is the even part and the second term is the odd part, both of which can be chosen arbitrarily.  The problem conditions stipulate the even part and the odd part $h(x)$ is subject to the condition $h(1/2) = \frac{g(1/2)}{2}$.  So we can pick an arbitrary odd function for $h$ and then everything else is determined, e.g. if $h(x) = x^3$ then $g(x) = \frac{1}{8} + x^3$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the meta-problem is &quot;what conditions on an odd function are strong enough to make it equal to $h(x) = x$?&quot; and I don't think this is a reasonable or interesting question to ask unless you have something very specific in mind.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This problem is not actually much like the Cauchy functional equation because there is only one free parameter instead of two.&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2011-01-08T23:10:09.170" LastActivityDate="2011-01-08T23:10:09.170" CommentCount="9" />
  <row Id="16844" PostTypeId="2" ParentId="16841" CreationDate="2011-01-08T23:00:40.697" Score="7" ViewCount="" Body="&lt;p&gt;I presume $\displaystyle f(1) \neq 0$, in which case we can assume $\displaystyle f(1) = 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $\displaystyle f$ is such a function, then $\displaystyle g(x) = \sin^{2}\left(\frac{\pi f(x)}{2}\right)$ is also such a function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $\displaystyle f(1) = 0$, take $\displaystyle g(x) = \sin(f(x))$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you will really need much stronger restrictions than infinite differentiability etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As to your edit, continuity is enough.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can first show $\displaystyle f(1/q) = f(1)/q$ for integral $\displaystyle q$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This can easily be extended to $\displaystyle f(p/q)$ (for instance $\displaystyle f(2/q) + f(1/q) + \dots + f(1/q) = f(1)$) and by continuity, to the whole of $\displaystyle [0,1]$.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2011-01-08T23:37:36.203" LastActivityDate="2011-01-08T23:37:36.203" CommentCount="4" />
  <row Id="16845" PostTypeId="1" AcceptedAnswerId="16847" CreationDate="2011-01-08T23:55:19.893" Score="1" ViewCount="204" Body="&lt;p&gt;Hi, how do I do the following:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Consider the matrix $\begin{bmatrix}&#xD;&#xA;\cos x &amp;amp; -\sin x\\ &#xD;&#xA;\sin x &amp;amp; \cos x&#xD;&#xA;\end{bmatrix}$&#xA;  and the vector $\begin{bmatrix}&#xD;&#xA;\cos y\\ &#xD;&#xA;\sin y&#xD;&#xA;\end{bmatrix}$. Show that the product of the $2\times 2$ matrix with $2\times 1$ matrix $\begin{bmatrix}&#xD;&#xA;\cos(x+y)\\ &#xD;&#xA;\sin(x+y)&#xD;&#xA;\end{bmatrix}$ graphically, by sketching. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can show this using trig identities, but I am not sure how to graph the $2\times 1$ trig vector in first place, let alone explain the rest of the question graphically. All the help very much appreciated!&lt;/p&gt;&#xA;" OwnerUserId="5157" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-09T02:33:38.817" LastActivityDate="2011-01-09T02:33:38.817" Title="How do I graph a vector [cos x, sin x]?" Tags="&lt;linear-algebra&gt;" AnswerCount="2" />
  <row Id="16846" PostTypeId="2" ParentId="16765" CreationDate="2011-01-08T23:58:46.290" Score="2" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0452288533&quot; rel=&quot;nofollow&quot;&gt;Unknown Quantity&lt;/a&gt; by John Derbyshire is an excellent historical account of the evolution of abstract algebra. The book is aimed at the &quot;mathematically inclined&quot; and covers the history of algebra from birth to present day (or pretty close.)  &lt;/p&gt;&#xA;" OwnerUserId="1352" LastActivityDate="2011-01-08T23:58:46.290" />
  <row Id="16847" PostTypeId="2" ParentId="16845" CreationDate="2011-01-09T00:04:02.373" Score="6" ViewCount="" Body="&lt;p&gt;[cos(y),sin(y)] is the unique point on the unit circle with angle $y$ from the origin. Your $2x2$ matrix rotates the plane about the origin by angle $x$, which is why the result is a point on the same circle with angle $x+y$.&lt;/p&gt;&#xA;" OwnerUserId="5468" LastActivityDate="2011-01-09T00:04:02.373" CommentCount="1" />
  <row Id="16848" PostTypeId="2" ParentId="16845" CreationDate="2011-01-09T00:04:48.467" Score="3" ViewCount="" Body="&lt;p&gt;Every point $(x,y)$ can be thought of as a $ 2 \times 1 $ matrix.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $(x,y)$  is a point in the 2D plane, then for a $2\times 2$ matrix $A$, the product of $A$ and  $[x,y]^{T}$ gives another 2D point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case, you $A$ is a rotation matrix, and the $2 \times 1$ matrix you have corresponds to a point on the unit circle.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2011-01-09T00:04:48.467" CommentCount="4" />
  <row Id="16849" PostTypeId="2" ParentId="16831" CreationDate="2011-01-09T00:26:40.320" Score="1" ViewCount="" Body="&lt;p&gt;Just for fun, here's a proof in non-standard analysis (Nelson-style IST).  I write $a \approx b$ to mean that $a-b$ is infinitesimal.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's enough to prove the result when $f$ is standard.  The Weierstrass approximation theorem gives a polynomial $p$ such that $p(x) \approx x f(x)$ for all $x \in [0,1]$. Note $\int _0 ^1 xfp = 0$, because the integral is linear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now estimate the absolute value of $\int xfp - (xf)^2$.  Factoring the integrand as $xf(x)\cdot  ( p(x) - xf(x) )$, we see that this integral is bounded above by $\operatorname{sup}(xf) \cdot \operatorname{sup}(p-xf)$, where the sup is over all $x \in [0,1]$.  This is infinitesimal, since the sup of a standard continuous function on a standard compact interval is limited.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We therefore have $\int (xf)^2 \approx \int xfp = 0$.  Since $\int (xf)^2$ is standard, it is equal to zero.  Therefore $(xf)^2$ is identically zero, and so is $f$ (being continuous).&lt;/p&gt;&#xA;" OwnerUserId="5316" LastEditorUserId="5316" LastEditorDisplayName="" LastEditDate="2011-01-09T00:49:36.093" LastActivityDate="2011-01-09T00:49:36.093" />
  <row Id="16850" PostTypeId="2" ParentId="16766" CreationDate="2011-01-09T01:03:27.630" Score="1" ViewCount="" Body="&lt;p&gt;As a mathematically-inclined student currently in high school (although not in China, so perhaps my opinion is less relevant than it would appear to be), two subjects I would suggest are basic logic and reasoning (usually introduced in the US as part of geometry, which makes it seem boring and less relevant to &quot;real life&quot; reasoning), and basic probability.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As to how to make these &quot;sexy&quot;, I honestly don't know for the first one. Perhaps you could combine it with some topics from computer science or, more concretely, programming computers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Probability is relatively easy to motivate, since it has applications to a very large number of areas outside of mathematics.&lt;/p&gt;&#xA;" OwnerUserId="2954" LastEditorUserId="2954" LastEditorDisplayName="" LastEditDate="2011-01-09T05:54:07.037" LastActivityDate="2011-01-09T05:54:07.037" />
  <row Id="16851" PostTypeId="1" AcceptedAnswerId="16853" CreationDate="2011-01-09T01:04:35.837" Score="0" ViewCount="59" Body="&lt;p&gt;Two sets&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;x = 1,2,3,4,5...&#xA;y = 1,4,7,10,13...&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I need to write a function&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f(x_n) = y_n$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found that if I &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;take a number from x &lt;/li&gt;&#xA;&lt;li&gt;double it &lt;/li&gt;&#xA;&lt;li&gt;subtract 2&lt;/li&gt;&#xA;&lt;li&gt;add the result to the original number&lt;/li&gt;&#xA;&lt;li&gt;I get the corresponding number from y&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Here's what I have so far (it's in ruby code), it works but is there a better way of doing it.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;def f(x)&#xA;  if x &amp;gt; 1&#xA;    return x + ((2 * x) - 2)&#xA;  else&#xA;    return 1&#xA;  end&#xA;end&#xA;&#xA;y = f(x)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="5525" LastActivityDate="2011-01-09T01:13:27.077" Title="Set x increases by 1, Set y increases by 3. Need help with a function that will take $x_n$ and give me $y_n$" Tags="&lt;algorithms&gt;&lt;functions&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="16853" PostTypeId="2" ParentId="16851" CreationDate="2011-01-09T01:13:27.077" Score="2" ViewCount="" Body="&lt;p&gt;Your values y form what is called an arithmetic progression, namely a sequence where each element is obtained from the previous one just by adding always the same constant. In your case the constant is 3, namely:&#xA;1, 4=1+3, 7=4+3, 10=7+3 and so on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the FIRST time you add 3 corresponds to x=2, the formula is just&lt;/p&gt;&#xA;&#xA;&lt;p&gt;y=1+3*(x-1)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or (equivalently)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;y=3*x-2.&lt;/p&gt;&#xA;" OwnerUserId="688" LastActivityDate="2011-01-09T01:13:27.077" CommentCount="1" />
  <row Id="16854" PostTypeId="2" ParentId="16838" CreationDate="2011-01-09T01:32:15.750" Score="0" ViewCount="" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;function[L R x]=LR2(A,b)&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;% This program will find a solution to&#xA;  Ax=b using first giving the&#xA;  decomposition of the matrix into L and&#xA;  R and then solving.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;% Part 1 - Is this matrix square and&#xA;  nonsingular? give an error if not&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;[z y]=size(A);&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;if (z ~= y )&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;disp ( 'LR2 error: Matrix must be&#xA;  square' );&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;return;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;end;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;if det(A)==0&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;disp('L singular error');&#xA;&#xA;return;&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;  &#xA;  &lt;p&gt;end&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;% Part 2 : Decomposition of matrix&#xA;  into L and R&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;L=zeros(z,z);&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;R=zeros(z,z);&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;for i=1:z&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;% Finding L&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;for k=1:i-1&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;L(i,k)=A(i,k);&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;for j=1:k-1&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;L(i,k)= L(i,k)-L(i,j)*R(j,k);&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;end&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;L(i,k) = L(i,k)/R(k,k);&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;end&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;% Finding R&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;for k=i:z&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;R(i,k) = A(i,k);&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;for j=1:i-1&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;R(i,k)= R(i,k)-L(i,j)*R(j,k);&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;end&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;end&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;end&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;for i=1:z&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;L(i,i)=1;&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;  &#xA;  &lt;p&gt;end&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;% Program shows R and L&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;R&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;L&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;% Now use a vector y to solve 'Ly=b'&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;y=zeros(z,1);&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;y(1)=b(1)/L(1,1);&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;for i=2:z&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;y(i)=-L(i,1)*y(1);&#xA;&#xA;for k=2:i-1&#xA;&#xA;    y(i)=y(i)-L(i,k)*y(k);&#xA;&#xA;    y(i)=(b(i)+y(i))/L(i,i);&#xA;&#xA;end;&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;  &#xA;  &lt;p&gt;end;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;% Now we use this y to solve Rx = y&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;x=zeros(z,1);&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;x(1)=y(1)/R(1,1);&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;for i=2:z&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;x(i)=-R(i,1)*x(1);&#xA;&#xA;for k=i:z&#xA;&#xA;    x(i)=x(i)-R(i,k)*x(k);&#xA;&#xA;    x(i)=(y(i)+x(i))/R(i,i);&#xA;&#xA;end;&#xA;&#xA; end&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;  &#xA;  &lt;p&gt;%print x &#xA;  x &#xA;  end&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have solved the original question. However there are still some problems with my solving of Ly=b and Rx=y. I also need to include pivoting. Any improvements would be greatly appreciated again!&lt;/p&gt;&#xA;" OwnerUserId="4697" LastActivityDate="2011-01-09T01:32:15.750" CommentCount="1" />
  <row Id="16856" PostTypeId="1" AcceptedAnswerId="16857" CreationDate="2011-01-09T03:17:29.340" Score="3" ViewCount="190" Body="&lt;p&gt;Suppose that A is a subset of B. How can we show that B-complement is a subset of A-complement?&lt;/p&gt;&#xA;" OwnerUserId="5530" LastEditorUserId="622" LastEditorDisplayName="" LastEditDate="2011-01-09T06:56:03.047" LastActivityDate="2011-01-09T08:34:36.227" Title="Proving $A \subset B \Rightarrow B' \subset A'$" Tags="&lt;elementary-set-theory&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="16857" PostTypeId="2" ParentId="16856" CreationDate="2011-01-09T03:27:04.090" Score="4" ViewCount="" Body="&lt;p&gt;Suppose $b\in B^c$. Is it possible that $b\in A$? Suppose it is, and derive a contradiction, and you should have your desired containment.&lt;/p&gt;&#xA;" OwnerUserId="1583" LastActivityDate="2011-01-09T03:27:04.090" CommentCount="2" />
  <row Id="16858" PostTypeId="1" AcceptedAnswerId="16859" CreationDate="2011-01-09T04:02:18.077" Score="2" ViewCount="182" Body="&lt;p&gt;$16a + 23b + 12c + 34d -2e -37f + 109g -141h + 139i + 149j + 29k + 12l + 131m = 74608$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$13a + 31b -5c + 17d + 29e -67f -16g -101h -7i -201j + 32k + 17l + 171m = -4194$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$81a -12b -5c + 17d -9e + 17f + 39g + 49h -67i -7j -41k + 13l + 121m = -15793$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$91a -43b + 17c + 19d + 17e + 23f + 97g + 101h -101i + 173j -12k + 137l + 217m = -16661 $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$71a + 47b + 37c + 37d + 41e -37f -67g -17h -2i -6j + 41k -39l -12m = -6862 $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$61a + 13b + 101c + 7d + 117e + 111f -4g -3h -19i -2j + 13k + 17l -91m = 7802 $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$51a + 31b + 37c + 4d + 101e -13f -7g -13h -41i -3j + 31k + 73l -71m = 8846 $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$21a + 12b + 67c -67d -9e -67f + 9g + 98h + 101i + 171j + 13k + 191l -54m = -4544 $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$41a -2b + 5c + 17d + 18e + 19f + 20g + 21h + 22i + 23j -11k + 91l -2m = -4050 $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$31a + 100b + 2c + 4d + 5e + 10f -10g + 20h -50i -4j + 10k -67l -4m = -4459 $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$17a + 67b + 2c -2d -8e -78f -9g -101h + 107i + 145j + 21k -9l -10m = 37104 $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$11a + 3b + 4c + 71d -45e + 87f + 8g + 34h + 45i + 93j + 104k -8l -20m = 4737 $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$19a -3b -9c -12d + 123e + 61f + 71g + 83h + 91i + 10j + 11k -12l + 82m = 623$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How would I go about finding $a, b, c, d, e ... m $?&#xA;I can use programming.&lt;/p&gt;&#xA;" OwnerUserId="5522" LastEditorUserId="5522" LastEditorDisplayName="" LastEditDate="2011-03-30T16:20:15.447" LastActivityDate="2011-03-30T18:48:55.890" Title="Solving 12 linear equations with 13 variables." Tags="&lt;linear-algebra&gt;&lt;math-software&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="16859" PostTypeId="2" ParentId="16858" CreationDate="2011-01-09T04:04:29.370" Score="7" ViewCount="" Body="&lt;p&gt;Have you heard of &lt;a href=&quot;http://en.wikipedia.org/wiki/Gauss%E2%80%93Jordan_elimination&quot; rel=&quot;nofollow&quot;&gt;Gauss-Jordan elimination&lt;/a&gt;? It would be an appropriate algorithm to use for this system of linear equations.&lt;/p&gt;&#xA;" OwnerUserId="3731" LastActivityDate="2011-01-09T04:04:29.370" CommentCount="2" />
  <row Id="16860" PostTypeId="1" CreationDate="2011-01-09T04:14:02.613" Score="1" ViewCount="49" Body="&lt;p&gt;In Milnor and Stasheff, it is taken as part of the first axiom that all Stiefel-Whitney classes of a bundle vanish in dimensions greater than the rank of the bundle.  However, in other sources this is omitted.  The wikipedia article makes it seem like this is actually a result of the axioms.  Is there a simple proof of this (the axioms I am using are those listed here &lt;a href=&quot;http://en.wikipedia.org/wiki/Stiefel%E2%80%93Whitney_class#Axioms&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Stiefel%E2%80%93Whitney_class#Axioms&lt;/a&gt;)?&lt;/p&gt;&#xA;" OwnerUserId="9" LastActivityDate="2011-01-09T04:59:27.067" Title="Differing axioms for Stiefel-Whitney class" Tags="&lt;algebraic-topology&gt;&lt;characteristic-classes&gt;" AnswerCount="1" />
  <row Id="16861" PostTypeId="1" AcceptedAnswerId="16863" CreationDate="2011-01-09T04:15:48.917" Score="0" ViewCount="161" Body="&lt;p&gt;We have been told in our homework to implement the following program:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;function A = Cholesky(A)&#xA;% Cholesky Decomposition of a Matrix A, so that the resulting matrix L gives A=L*L&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;N = size(A,1);   &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;for n=1:N&lt;br&gt;&#xA;     if abs(A(n,n))&#xA;     error('&lt;strong&gt;* ERROR *&lt;/strong&gt; Cholesky Decomp doesn't exist'); &#xA;     end; &#xA;     A(n:N,n) = A(n:N,n) - A(n:N,1:n-1) * A(n,1:n-1)';&lt;br&gt;&#xA;     A(n:N,n) = A(n:N,n) / sqrt(A(n,n));&lt;br&gt;&#xA;  end;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Could anyone please explain what the line &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;&lt;pre&gt;&lt;code&gt;if abs(A(n,n))&amp;lt;eps&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;means. I can sort of see that it is saying that the absolute values of the diagonal values can't be zero however what impact would this have on my decomposition.&#xA;Thanks!!&lt;/p&gt;&#xA;" OwnerUserId="4697" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2011-01-09T04:17:50.593" LastActivityDate="2011-01-09T04:37:55.547" Title="In this Cholesky Algorithm in matlab, what is the abs function doing." Tags="&lt;homework&gt;&lt;computer-science&gt;&lt;math-software&gt;&lt;matlab&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="16863" PostTypeId="2" ParentId="16861" CreationDate="2011-01-09T04:37:55.547" Score="2" ViewCount="" Body="&lt;p&gt;In one of the following lines, one divides by $A_{n,n}$. So $A_{n,n}$ must be non-zero. Now, computations in the computer are not accurate so that the correct way to test $A_{n,n} = 0$ is $|A_{n,n}| &amp;lt; \epsilon$. In fact, when $0&amp;lt;|A_{n,n}| &amp;lt; \epsilon$ the entire procedure will be very sensitive to the exact value of $A_{n,n}$, and so the result of the computation won't be trustworthy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Cholesky decomposition is usually defined only for (symmetric) positive-definite matrices, since in that case (contrary to the non-negative case) the decomposition is unique. The algorithm you describe uses some form of Gaussian elimination, so that $A_{n,n}$ does stand for an eignevalue.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2011-01-09T04:37:55.547" CommentCount="2" />
  <row Id="16864" PostTypeId="1" CreationDate="2011-01-09T04:54:33.657" Score="3" ViewCount="93" Body="&lt;p&gt;The problem is from Marden, the first section (I know, I know):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The polynomial $g(z) = z^n + b_1 z^{n-1} + ... + b_n$ has at least $m$ zeros in an arbitrary neighborhood of a point $z = c$ if $|g^{(k)}(c)| \leq \epsilon$ for $k = 0,1,...,m-1$ and for $\epsilon$ sufficiently small and positive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a hint provided: use Rouché's Theorem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can prove the result in the special case of $c = 0$, because then I can bound each of the relevant $b_j$ by $\epsilon$. Unfortunately I don't see a way to extend this to the general case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would appreciate some help on working toward a solution. I've been stumped since lunch on this one.&lt;/p&gt;&#xA;" OwnerUserId="5531" LastEditorUserId="5531" LastEditorDisplayName="" LastEditDate="2011-01-09T05:02:42.090" LastActivityDate="2011-01-09T05:02:42.090" Title="Trouble with a problem involving Rouché's Theorem" Tags="&lt;complex-analysis&gt;&lt;polynomials&gt;" CommentCount="2" />
  <row Id="16865" PostTypeId="2" ParentId="16860" CreationDate="2011-01-09T04:59:27.067" Score="1" ViewCount="" Body="&lt;p&gt;It is certainly both of constructions of the Stiefel-Whitney classes that I know. The construction via the Thom Isomorphism and the Steenrod operations clearly must vanish seeing as the Thom class of the bundle will be in a dimension that is killed by the Steenrod Operation because of instability. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the construction via universal bundles you look at the universal rank $k$-bundle $\gamma_k$, it has $w_n(\gamma_k)=0$  simply by looking at the cohomology of $BO(k)$, there isn't a polynomial generator in that dimension. The above observation coupled with naturality should give the result, but i don't know if that should count as following from the axioms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can flesh this out a bit later if you like.&lt;/p&gt;&#xA;" OwnerUserId="627" LastActivityDate="2011-01-09T04:59:27.067" CommentCount="3" />
  <row Id="16866" PostTypeId="2" ParentId="16637" CreationDate="2011-01-09T05:38:37.953" Score="0" ViewCount="" Body="&lt;p&gt;@Theo Buehler: Thank you very much.&lt;/p&gt;&#xA;" OwnerUserId="5533" LastActivityDate="2011-01-09T05:38:37.953" CommentCount="1" />
  <row Id="16867" PostTypeId="1" CreationDate="2011-01-09T05:41:09.603" Score="3" ViewCount="135" Body="&lt;p&gt;How can we show, rigorously, that if $f: \mathbb{R} \rightarrow \mathbb{R}$ is a continuous map then its graph and the inverse of the graph (i.e the inverse of the relation) is a nowhere dense set? I'm trying to show that the plane cannot be covered by a countable number of graphs of continuous maps and its inverses, so by showing the above the result follows from the Baire category theorem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance&lt;/p&gt;&#xA;" OwnerUserId="5533" LastActivityDate="2011-01-09T12:45:40.090" Title="Graph of a continuous real valued map is a nowhere dense set" Tags="&lt;topology&gt;" AnswerCount="2" CommentCount="0" />
  <row Id="16868" PostTypeId="2" ParentId="12629" CreationDate="2011-01-09T06:16:48.997" Score="1" ViewCount="" Body="&lt;p&gt;A simple way to split the set of natural numbers.Take the sets $S_n$ of numbers with sum of digits $n$ for each all $n$.&#xA;Obviously no sets overlap and each $S_n$ contain numbers of the form $1111..0000...$ where the string of 1's contain $n$ of them, and as many zeros as you wish  &lt;/p&gt;&#xA;" OwnerUserId="5534" LastActivityDate="2011-01-09T06:16:48.997" />
  <row Id="16869" PostTypeId="2" ParentId="16867" CreationDate="2011-01-09T06:19:48.917" Score="4" ViewCount="" Body="&lt;p&gt;The graph $\Gamma(f)$ of a continuous function $f$ is closed, so we just have to show it has empty interior.  But of course if $\Gamma(f)$ had nonempty interior it could not pass the &quot;vertical line test&quot; and would not be a graph.  For the inverse, just turn your head sideways.&lt;/p&gt;&#xA;" OwnerUserId="822" LastActivityDate="2011-01-09T06:19:48.917" />
  <row Id="16870" PostTypeId="1" CreationDate="2011-01-09T06:37:07.607" Score="3" ViewCount="141" Body="&lt;p&gt;I think people here could guide me in solving a problem related to anomaly detection. The term anomaly here refers to some undesired event occurring in the system like a virus infection.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I could get to know about it from more than one source. For example extracting value from two different data structure and if the value is different it is certain that virus infection is there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to remove the false positive cases, information is gathered from different data structures or mechanisms. In that certain information are less trusted and certain information are more trusted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am looking for a mathematical method, that could easily handle this type of situations. Whether &lt;code&gt;fuzzy/Genetic Algorithm/Neural Net&lt;/code&gt; fits here ? Found in some places they using one &lt;code&gt;normality based approach(using z score)&lt;/code&gt;. Please help.&lt;/p&gt;&#xA;" OwnerUserId="5507" LastEditorUserId="1417" LastEditorDisplayName="" LastEditDate="2011-03-27T17:01:04.437" LastActivityDate="2011-04-26T17:42:42.400" Title="Help needed on method to use for anomaly detection in Computer Science" Tags="&lt;statistics&gt;&lt;computer-science&gt;&lt;fuzzy-logic&gt;&lt;machine-learning&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="2" />
  <row Id="16871" PostTypeId="2" ParentId="16867" CreationDate="2011-01-09T08:02:39.470" Score="0" ViewCount="" Body="&lt;p&gt;The &quot;hard&quot; part is showing that the graph of f, $ \Gamma(f) = \{(x,f(x)) \} \subset \mathbb{R} \times \mathbb{R}$ is closed. It's then clear that no point in this subset can be in the interior of $\Gamma(f)$ for the reason Nate stated above.&lt;/p&gt;&#xA;" OwnerUserId="5523" LastActivityDate="2011-01-09T08:02:39.470" CommentCount="1" />
  <row Id="16872" PostTypeId="2" ParentId="16766" CreationDate="2011-01-09T08:03:58.127" Score="1" ViewCount="" Body="&lt;p&gt;It might be helpful for you to get ahold of some of what I've heard called &quot;liberal arts math&quot; textbooks—more or less, discrete math books aimed at college students not intending to pursue much further study in math.  The two that I'd suggest offhand are COMAP's &lt;a href=&quot;http://www.whfreeman.com/newcatalog.aspx?search=For+All+Practical+Purposes&amp;amp;isbn=1429209003&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;For All Practical Purposes&lt;/em&gt;&lt;/a&gt; (published by W. H. Freeman) and &lt;a href=&quot;http://www.wiley.com/WileyCDA/WileyTitle/productCd-EHEP000304.html&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;The Heart of Mathematics&lt;/em&gt;&lt;/a&gt; by Ed Burger (published by Wiley).  These and similar books have a number of topics that are presented in ways designed to draw in students of varying backgrounds.  Many of these topics are not typically in U.S. secondary school curricula and can readily be extended from a &quot;liberal arts math&quot; level to more rigorous math-major style mathematics (e.g. voting theory, fair division).&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2011-01-09T08:03:58.127" />
  <row Id="16873" PostTypeId="2" ParentId="16856" CreationDate="2011-01-09T08:34:36.227" Score="4" ViewCount="" Body="&lt;p&gt;Instead of using a proof by contradiction, $b\in B^c \implies b \notin A$ since $A \subset B$ hence $b \in A^c$ which implies the result.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The key here being that for all $x$, either $x\in X$ or $x\in X^c$.&lt;/p&gt;&#xA;" OwnerUserId="565" LastActivityDate="2011-01-09T08:34:36.227" />
  <row Id="16874" PostTypeId="1" AcceptedAnswerId="16876" CreationDate="2011-01-09T08:44:52.697" Score="3" ViewCount="147" Body="&lt;p&gt;Let $f : \mathbb R \to \mathbb R$ be a given function with $\lvert f(x) \rvert \le 1$ and $f(0) = 1$. Is there a nice simplified expression for $$\begin{align}F(x) &amp;amp;= f(x) f(x/2) f(x/4) f(x/8) \cdots \\ &amp;amp;= \prod_{i=0}^\infty f(x/2^i)?\end{align}$$ If there isn't a general solution, as seems likely, can anything useful be said about the case when $f(x) = \operatorname{sinc} x = \frac{1}{x} \sin x$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This question arose when idly wondering about the limit of convolving an infinite number of dyadically(?) scaled versions of a kernel $g$ together. Taking the Fourier transform of $g(x) * 2g(2x) * 4g(4x) * 8g(8x) * \cdots$ yields the above expression.&lt;/p&gt;&#xA;" OwnerUserId="856" LastEditorUserId="1709" LastEditorDisplayName="" LastEditDate="2011-01-09T09:07:26.600" LastActivityDate="2011-01-09T09:16:07.960" Title="Evaluating $f(x) f(x/2) f(x/4) f(x/8) \cdots$" Tags="&lt;sequences-and-series&gt;&lt;functional-equations&gt;" AnswerCount="1" CommentCount="6" FavoriteCount="1" />
  <row Id="16875" PostTypeId="2" ParentId="16723" CreationDate="2011-01-09T08:50:07.720" Score="0" ViewCount="" Body="&lt;p&gt;This isn't a complete solution but rather a reformulation for a large number of cases involving the problem above. Write $n = m \lfloor y \rfloor + r$, where $0 \leq r &amp;lt; \lfloor y \rfloor$ and $n, m , r \in \mathbb{Z}_{\geq 0}$. Then for real $x, y &amp;gt; 0$, we have&#xA;\begin{eqnarray}&#xA;\sum_{k = 0}^{n} \left \lfloor x + \frac{k}{y} \right \rfloor &amp;amp; = &amp;amp; \lfloor x \rfloor + \sum_{\ell = 0}^{m-1} \left \lfloor x y + \ell \lfloor y \rfloor  + ( \lceil y \rceil - y) \left \lfloor x + \frac{\ell \lfloor y \rfloor}{y} + 1  \right \rfloor   \right \rfloor +&#xA;\end{eqnarray}&#xA;\begin{eqnarray} &amp;amp; &amp;amp; \sum_{k = 1}^{r}  \left \lfloor   x + \frac{m \lfloor y \rfloor}{y}  + \frac{k}{y} \right \rfloor  +   \chi_{\mathbb{N}}(y) \sum_{\ell = 0}^{m-1} \left \lfloor x + \frac{\ell \lfloor y \rfloor}{y}  + 1 \right \rfloor - \sum_{\ell = 0}^{m-1} \left \lfloor x + \frac{ \ell \lfloor y \rfloor }{y}  \right \rfloor .&#xA;\end{eqnarray}&#xA;This was derived by partitioning the interval $[0,n]$ into subintervals and using the last identity in the description of the problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the original problem, write&#xA;\begin{eqnarray}&#xA;\sum_{k = 0}^{\lfloor a /b \rfloor} \left \lfloor \frac{a - k b}{c} \right \rfloor = \sum_{k = 0}^{\lfloor a /b \rfloor} \left \lfloor \frac{\text{mod}(a,b) + k b}{c} \right \rfloor&#xA;\end{eqnarray}&#xA;where $\text{mod}(a,b) = a - b \lfloor \frac{a}{b} \rfloor$. If $\lfloor \frac{a}{b} \rfloor \gg c$, then use the identity above to vastly reduce the number of terms in the sum.&lt;/p&gt;&#xA;" OwnerUserId="2720" LastEditorUserId="2720" LastEditorDisplayName="" LastEditDate="2011-01-11T09:03:18.463" LastActivityDate="2011-01-11T09:03:18.463" />
  <row Id="16876" PostTypeId="2" ParentId="16874" CreationDate="2011-01-09T09:16:07.960" Score="3" ViewCount="" Body="&lt;p&gt;Assuming analyticity of $f(x)$ it seems more tractable to work with the form involving sums after taking logarithms:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \log F(x) = \sum_{i=0}^\infty \log f(x/2^i)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now if $\log f(x)$ has convergent power series expansion in a neighborhood of zero:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \log f(x) = \sum_{k=1}^\infty a_k x^k$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that since $f(0) = 1$ the constant term $a_0$ of $\log f(x)$ is zero and thus omitted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \log F(x) = \sum_{k=1}^\infty a_k (\sum_{i=0}^\infty 2^{-ik}) x^k$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the inner summations are just $\sum_{i=0}^\infty 2^{-ik} = 1/(1 - 2^{-k})$ and thus uniformly bounded by $2$.&lt;/p&gt;&#xA;" OwnerUserId="3111" LastActivityDate="2011-01-09T09:16:07.960" CommentCount="2" />
  <row Id="16877" PostTypeId="1" AcceptedAnswerId="16884" CreationDate="2011-01-09T10:20:23.327" Score="9" ViewCount="338" Body="&lt;p&gt;In[11]:= Select[Table[Fibonacci[n], {n, 1, 10000}], PrimeQ[# - 1] &amp;amp;]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Out[11]= {3, 8}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: Fibonacci[n]-1 is always composite for n&gt;6. why?&#xA; $$\sum\limits_{i = 0}^n {{F_i}}  = {F_{n + 2}} - 1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In[16]:= Select[Table[Fibonacci[n], {n, 1, 10000}], PrimeQ[# + 1] &amp;amp;]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Out[16]= {1, 1, 2}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fibonacci[n]+1 is always composite for n&gt;3. why?&lt;/p&gt;&#xA;" OwnerUserId="1498" LastEditorUserId="1498" LastEditorDisplayName="" LastEditDate="2011-01-09T10:42:15.297" LastActivityDate="2011-01-09T17:23:58.777" Title="Fibonacci[n]-1 is always composite for n&gt;6. why?" Tags="&lt;fibonacci-number&gt;" AnswerCount="2" CommentCount="6" FavoriteCount="2" />
  <row Id="16879" PostTypeId="1" AcceptedAnswerId="16880" CreationDate="2011-01-09T11:06:43.410" Score="0" ViewCount="45" Body="&lt;p&gt;I'm stuck on something that must be incredible simple to do. I calculate the second derivative of $f(x) = e^{\sin(x)}$ like so in Maple:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;diff(diff(exp(sin(x)),x),x);&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;But now i want to calculate $D^2f(0)$, and can't seem to find a proper way to do this...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried: &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; fsolve(Deriv = 0, x);&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;with &lt;code&gt;Deriv:=diff(diff(...))&lt;/code&gt; but that gives me $1,57...$ but it has to be $1$.&lt;/p&gt;&#xA;" OwnerUserId="5109" LastActivityDate="2011-01-09T11:25:03.673" Title="Derivative of a function for a given x in Maple" Tags="&lt;functions&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="16880" PostTypeId="2" ParentId="16879" CreationDate="2011-01-09T11:25:03.673" Score="2" ViewCount="" Body="&lt;p&gt;Your command of &lt;code&gt;fsolve(Deriv = 0, x);&lt;/code&gt; actually solves the equation $D^2 f(x) = 0$ for $x$, which isn't what you want. Instead, you want to compute $D^2 f(0)$, which is simply the second derivative $D^2 f$ evaluated at $x = 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here, the second derivative $D^2 f$ is &#xA;$$&#xD;&#xA;D^2 f(x) = e^{\sin(x)} \cos^2(x) - e^{\sin(x)} \sin(x), &#xD;&#xA;$$&#xA;so plugging in $x = 0$ yields $D^2 f (0) = 1$ as you wanted.&lt;/p&gt;&#xA;" OwnerUserId="30" LastActivityDate="2011-01-09T11:25:03.673" CommentCount="1" />
  <row Id="16881" PostTypeId="2" ParentId="16743" CreationDate="2011-01-09T11:49:17.227" Score="2" ViewCount="" Body="&lt;p&gt;First note that this problem is given in the context of some larger set $S$ that already satisfies the required properties.  In this case, the &quot;smallest&quot; $T$ (should it exist) would mean a subset $T$ of $S$ satisfying these properties $\mathscr{P}$ such that every set $X \subseteq S$ having $\mathcal{P}$ has $T \subseteq S$.  But the reason such a $T$ will exist is because we are implicitly given a well-defined procedure without choices of how to build up our $T$.  Because of this fact, you can express the properties in a form that's equivalent and adheres to the more formal definition of induction, but this is unnecessary and can inhibit understanding.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To the more practical side of the matter, definitions like these fall into the &quot;closure category&quot; and often involve iterating steps infinitely many times starting with the emptyset and repeating iteratively until you've met all of the conditions (which is usually after an infinite number of steps).  The intuition here is that if you've started with the empyset and added elements only as required by the definitions (&quot;closing under the required properties&quot;), then any elements that you added must always be contained in the &quot;smallest&quot; set.  Many times, stepping through all of the Natural numbers will be sufficient.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's begin the procedure to get an idea.  Here, you will be building up each of the $T_n$'s simultaneously as required by the conditions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Start with $T_n = \emptyset$ for all $n \in \mathbb{N}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Step (1): After condition (1), $T_n = \{0, 1, \ldots, n-1\}$ for all $n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After condition (2), $T_n = \{0, 1, \ldots, n-1, \lambda.0, \lambda.1, \ldots \lambda.n\}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After condition (3), $T_n = \{0, 1, \ldots, n-1, \lambda.0, \lambda.1, \ldots \lambda.n\} \cup$ &#xA;$\{0, 1, \ldots, n-1, \lambda.0, \lambda.1, \ldots \lambda.n\} \times \{0, 1, \ldots, n-1, \lambda.0, \lambda.1, \ldots \lambda.n\}$ (union of result from condition (2) and all possible ordered pairs from condition (2)).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now start with result from Step (1), condition (3): $T_n = \{0, 1, \ldots, n-1, \lambda.0, \lambda.1, \ldots \lambda.n\} \cup$ &#xA;$\{0, 1, \ldots, n-1, \lambda.0, \lambda.1, \ldots \lambda.n\} \times \{0, 1, \ldots, n-1, \lambda.0, \lambda.1, \ldots \lambda.n\}$ and repeat conditions (1) - (3) with this set instead of the emptyset for Step (2).  Then plug in the final result from Step (2), and repeat conditions (1) - (3) with this set for Step (3).  Keep repeating for each Natural number step.  Of course condition (1) doesn't need to be repeated in this process because it won't add anything new.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you continue this process of what is known as &quot;closing under the operations&quot; ad infinitum (all Natural numbers), you will wind up with the desired set.  The reason for this is intuitively simple.  Any terms $t_1$ and $t_2$ in the final set were added by some Natural number steps $m$ and $n$, say $n &amp;gt; m$ here.  But then in step $n+1$, you make sure you've satisfied conditions (2) and (3) for any elements in the set so far constructed, including the elements $t_1$ and $t_2$.&lt;/p&gt;&#xA;" OwnerUserId="4693" LastEditorUserId="4693" LastEditorDisplayName="" LastEditDate="2011-01-09T22:56:16.120" LastActivityDate="2011-01-09T22:56:16.120" CommentCount="2" />
  <row Id="16882" PostTypeId="1" CreationDate="2011-01-09T12:42:34.310" Score="0" ViewCount="59" Body="&lt;p&gt;Hi all,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm trying to code up an implementation of the PQN-NNLS algorithm described on page 10 &lt;a href=&quot;http://www.wfu.edu/~plemmons/papers/nonneg.pdf&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; in C#. I'm having trouble reading the pseudocode though, could anybody give me a hand and write it out in a more intelligible format? I have already seen the matlab implementation that can be found on google but I don't understand matlab code...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance&lt;/p&gt;&#xA;" OwnerUserId="5539" LastActivityDate="2011-01-09T12:42:34.310" Title="PQN-Non Negative Least Squares Algorithm" Tags="&lt;linear-algebra&gt;" CommentCount="2" />
  <row Id="16884" PostTypeId="2" ParentId="16877" CreationDate="2011-01-09T14:06:34.207" Score="13" ViewCount="" Body="&lt;p&gt;mjqxxxx is correct; there is a proof in Honsberger's &lt;em&gt;Mathematical Gems III&lt;/em&gt;, which is short enough to reproduce here.  In fact, it suffices to verify the following eight identities:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$F_{4k} - 1 = F_{2k+1} L_{2k-1}$$&#xA;$$F_{4k+1} - 1 = F_{2k} L_{2k+1}$$&#xA;$$F_{4k+2} - 1 = F_{2k} L_{2k+2}$$&#xA;$$F_{4k+3} - 1 = F_{2k+2} L_{2k+1}$$&#xA;$$F_{4k} + 1 = F_{2k-1} L_{2k+1}$$&#xA;$$F_{4k+1} + 1 = F_{2k+1} L_{2k}$$&#xA;$$F_{4k+2} + 1 = F_{2k+2} L_{2k}$$&#xA;$$F_{4k+3} + 1 = F_{2k+1} L_{2k+2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $L_k$ are the Lucas numbers.  These identities all follow straightforwardly from Binet's formula, although I imagine there are also combinatorial proofs.  &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I suspect at least one of these is an open problem, if not both.  Let me record for now the following observation: modular arithmetic is not enough.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proposition:&lt;/strong&gt;  For any finite set $p_1, ... p_n$ of primes, there exists a Fibonacci number $F_k$ such that $\prod p_i | F_k$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Proof.&lt;/em&gt;  Since $F_n | F_{mn}$ for all $m, n \ge 1$ it suffices to show this for a single prime.  But the Fibonacci sequence is clearly periodic $\bmod p$ for any $p$ (since the Fibonacci recursion is reversible), and $F_0 = 0 \equiv 0 \bmod p$.  (The computation of the exact period $\bmod p$ is a nice exercise.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So both $F_n + 1$ and $F_n - 1$ can avoid any finite set of primes.  &lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2011-01-09T17:23:58.777" LastActivityDate="2011-01-09T17:23:58.777" CommunityOwnedDate="2011-01-09T17:23:58.777" CommentCount="4" />
  <row Id="16885" PostTypeId="1" CreationDate="2011-01-09T14:28:34.483" Score="1" ViewCount="91" Body="&lt;p&gt;Perform a taylor expansion in 3 dimensions in time on the time compontent of of $T^{\alpha \beta}(t - r + n^{i} y_{i})$ given that $r$ is a contstant and $n^{i} y_{i}$ is the scalar product of a normal vector $n$ with a position vector $y$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe the formula is $\phi (\vec{r} + \vec{a}) = \sum^{\infty}_{n=0} \frac{1}{n!} (\vec{a}\cdot \nabla)^{n} \phi(\vec{r})$ in general and that the first term is $T^{\alpha \beta}(t -r)$ but would appreciate some other opinion about what the proceeding terms should be up to order $n=3$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cheers&lt;/p&gt;&#xA;" OwnerUserId="5542" LastActivityDate="2011-01-09T17:51:26.343" Title="Taylor expansion in time of the time component of a stress energy tensor" Tags="&lt;calculus&gt;&lt;taylor-expansion&gt;&lt;vector-calculus&gt;&lt;tensors&gt;" AnswerCount="1" />
  <row Id="16886" PostTypeId="1" CreationDate="2011-01-09T14:41:46.807" Score="0" ViewCount="162" Body="&lt;p&gt;I have the answer of 1. But for the answer they split it up into b_n = n/sqrt(n^2+1) and c_n =sin(n)/sqrt(n^2+1) which is fine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then for b_n they divide evrything through by n^2 incluse everything inside the square root to give 1/sqrt(1+1/n^2) and I thought you couldnt do this and checked with real numbers and they are not equivalent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For c_n they said sin(n)/sqrt(n^2+1) is &amp;lt;=  1/sqrt(n^2+1). Fine. But then they say this is &amp;lt; 1/n by continuity of the square root? and this tends to 0 so sin(n)/sqrt(n^2+1) tends to 0. I thought for this to be true 1/n would have to tend to 0 slower then sin(n)/sqrt(n^2+1)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could you help please&lt;/p&gt;&#xA;" OwnerUserId="4577" LastActivityDate="2011-01-09T15:30:51.730" Title="what does (n+sin(n))/sqrt(n^2+1) converge to?" Tags="&lt;analysis&gt;&lt;sequences-and-series&gt;&lt;limit&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="16887" PostTypeId="1" CreationDate="2011-01-09T14:46:25.903" Score="19" ViewCount="1074" Body="&lt;p&gt;The background of this question is this: Fermat proved that the equation,&#xA;$$x^4+y^4 = z^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;has no solution in the positive integers. If we consider the near-miss,&#xA;$$x^4+y^4-1 = z^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;then this has plenty (in fact, an infinity, as it can be solved by a Pell equation). But J. Cullen, by exhaustive search, found that the other near-miss,&#xA;$$x^4+y^4+1 = z^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;has none with $0 &amp;lt; x,y &amp;lt; 10^6$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does the third equation really have none at all, or are the solutions just enormous?&lt;/p&gt;&#xA;" OwnerUserId="4781" LastEditorUserId="5904" LastEditorDisplayName="" LastEditDate="2011-04-15T10:24:41.517" LastActivityDate="2011-04-16T21:20:59.913" Title="Does the equation $x^4+y^4+1 = z^2$ have a non-trivial solution?" Tags="&lt;number-theory&gt;&lt;algebra&gt;&lt;diophantine-equations&gt;" AnswerCount="2" CommentCount="7" FavoriteCount="4" />
  <row Id="16888" PostTypeId="2" ParentId="16886" CreationDate="2011-01-09T14:48:49.360" Score="2" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Hint:&lt;/strong&gt; The square root $\sqrt{n^2+1}$ behaves just like $n$ for large $n$. Since $\lvert\sin(n)\rvert$ is bounded this summand does not contribute to the limit. To see that $\frac{n}{\sqrt{n^2+1}}$ converges to 1 divide numerator and denominator by $n$.&lt;/p&gt;&#xA;" OwnerUserId="367" LastActivityDate="2011-01-09T14:48:49.360" />
  <row Id="16889" PostTypeId="2" ParentId="16886" CreationDate="2011-01-09T15:30:51.730" Score="1" ViewCount="" Body="&lt;p&gt;This is what was done to the square root: $$\sqrt{n^2+1}=\sqrt{n^2 (1+1/n^2)}=\sqrt{n^2}\sqrt{1+1/n^2}.$$&#xA;For positive $n$ we have $\sqrt{n^2}=n$, and this $n$ then cancels the $n$ in the numerator.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the other estimate, clearly $n^2+1 &amp;gt; n^2$, and since the square root function is increasing this gives $\sqrt{n^2+1} &amp;gt; \sqrt{n^2} = n$. The numbers are positive, so taking reciprocals reverses the inequality:&#xA;$$\frac{1}{\sqrt{n^2+1}} &amp;lt; \frac{1}{n}.$$&#xA;Therefore&#xA;$$0 \le \left| \frac{\sin n}{\sqrt{n^2+1}} \right| &amp;lt; \frac{1}{n},$$&#xA;and because of squeezing the limit of $c_n$ must be zero.&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2011-01-09T15:30:51.730" />
  <row Id="16890" PostTypeId="1" CreationDate="2011-01-09T03:33:10.127" Score="4" ViewCount="183" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Stirling_numbers_of_the_second_kind&quot; rel=&quot;nofollow&quot;&gt;Stirling numbers of the second kind&lt;/a&gt; S(n, k) count the number of ways to partition a set of n elements into k nonempty subsets.What if there were duplicate elements in the set?That is,the set is a &lt;a href=&quot;http://en.wikipedia.org/wiki/Multiset&quot; rel=&quot;nofollow&quot;&gt;multiset&lt;/a&gt;?&lt;/p&gt;&#xA;" OwnerUserId="5555" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2011-05-31T22:59:40.743" LastActivityDate="2011-05-31T22:59:40.743" Title="Stirling numbers of the second kind on Multiset" Tags="&lt;combinatorics&gt;&lt;discrete-mathematics&gt;&lt;stirling-numbers&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="1" />
  <row Id="16891" PostTypeId="2" ParentId="16890" CreationDate="2011-01-09T03:40:19.407" Score="2" ViewCount="" Body="&lt;p&gt;By definition, a &lt;a href=&quot;http://en.wikipedia.org/wiki/Set_%28mathematics%29&quot; rel=&quot;nofollow&quot;&gt;set&lt;/a&gt; only has distinct elements.  If elements are duplicated, then yes, the set in question would be a &lt;a href=&quot;http://en.wikipedia.org/wiki/Multiset&quot; rel=&quot;nofollow&quot;&gt;multiset&lt;/a&gt;.&lt;/p&gt;&#xA;" LastActivityDate="2011-01-09T15:40:54.227" />
  <row Id="16892" PostTypeId="1" CreationDate="2011-01-09T16:02:00.660" Score="2" ViewCount="197" Body="&lt;p&gt;How much faster are various mathematical operations using a Trachtenberg method rather than a conventional method?&lt;/p&gt;&#xA;" OwnerUserId="3157" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2011-01-09T16:12:52.660" LastActivityDate="2011-01-09T18:42:19.117" Title="How much faster is the Trachtenberg system?" Tags="&lt;arithmetic&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="16893" PostTypeId="1" CreationDate="2011-01-09T16:05:00.407" Score="4" ViewCount="60" Body="&lt;p&gt;Let $No$ be the class of surreal numbers and let us be bold and construct the projective line over $No$ as $Proj(No)$ (I don't think a &quot;$No \times No / rel $&quot; construction will work in case of a class). The question is (I know it is a little wide):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a connetcion between the number $\omega \in No$ and the point(s) at infinity of the projective beast (I know that $\omega$ is defined (only?) set-theoretically, and the point at infinity concept is defined (only? scheme theoretically) ? Both of the things are meant to capture infinitely far away.&lt;/p&gt;&#xA;" OwnerUserId="5544" LastActivityDate="2011-01-09T16:05:00.407" Title="The projective line over $No$" Tags="&lt;geometry&gt;&lt;algebraic-geometry&gt;&lt;set-theory&gt;" CommentCount="2" />
  <row Id="16894" PostTypeId="1" AcceptedAnswerId="16896" CreationDate="2011-01-09T16:09:59.570" Score="0" ViewCount="132" Body="&lt;p&gt;If $x$ is &lt;strong&gt;prime&lt;/strong&gt; and $1000000000000 &amp;lt; x &amp;lt; 1000100000000$,&lt;br&gt;&#xA;how can I detect if $1 / x$ will result in a terminal decimal or not?&lt;br&gt;&#xA;I can use programming.&lt;/p&gt;&#xA;" OwnerUserId="5522" LastActivityDate="2011-01-09T16:19:37.223" Title="Detecting if a decimal is terminal or not?" Tags="&lt;prime-numbers&gt;" AnswerCount="2" />
  <row Id="16895" PostTypeId="2" ParentId="16894" CreationDate="2011-01-09T16:19:24.773" Score="6" ViewCount="" Body="&lt;p&gt;If &quot;terminal&quot; means the decimal expansion is eventually zero, then no such number has a terminal expansion. $1/x$ has a terminal expansion iff $x=2^a 5^b$ for some $a$, $b$.&lt;/p&gt;&#xA;" OwnerUserId="5203" LastActivityDate="2011-01-09T16:19:24.773" />
  <row Id="16896" PostTypeId="2" ParentId="16894" CreationDate="2011-01-09T16:19:37.223" Score="7" ViewCount="" Body="&lt;p&gt;Suppose that $p$ is a prime other than $2$ or $5$. Then if $\frac{1}{p}$ was a terminating decimal, we would have that $\frac{10^n}{p} = a$ for some $n,a \in \mathbb{N}$. But this is a contradiction since $p$ clearly doesn't divide $10^n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence $\frac{1}{p}$&amp;#160;is not a terminating decimal for any prime $p$ other than $2$ or $5$.&lt;/p&gt;&#xA;" OwnerUserId="3776" LastActivityDate="2011-01-09T16:19:37.223" CommentCount="1" />
  <row Id="16897" PostTypeId="1" AcceptedAnswerId="16904" CreationDate="2011-01-09T16:28:58.483" Score="1" ViewCount="98" Body="&lt;p&gt;Let $X$ be a variety and $\mathbb{C}$ be the field of complex. Then $L = X \times \mathbb{C}$ is a trivial line bundle. The set of sections of this line bundle is $\Gamma(X, L)$ which consisting all functions $s: X \to L$ such that $\pi \cdot s = id$, where $\pi: L \to X$ is the projection map. Why $\Gamma(X, L)$ is the same as the set of all regular functions on $X$? Thank you very much.&lt;/p&gt;&#xA;" LastEditorUserId="627" LastEditorDisplayName="" LastEditDate="2011-01-09T17:38:00.107" LastActivityDate="2011-01-09T17:38:00.107" Title="Trivial line bundles." Tags="&lt;algebraic-geometry&gt;&lt;bundles&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" />
  <row Id="16898" PostTypeId="1" CreationDate="2011-01-09T16:48:25.293" Score="1" ViewCount="96" Body="&lt;p&gt;Hi, I have a real-life situation that can be solved using Queueing Theory.&lt;br&gt;&#xA;This should be easy for someone in the field.  Any pointers would be appreciated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Scenario:&lt;/strong&gt;&lt;br&gt;&#xA;There is a single Queue and &lt;strong&gt;N&lt;/strong&gt; Servers.&lt;br&gt;&#xA;When a server becomes free, the Task at the front of the queue gets serviced.&lt;br&gt;&#xA;The mean service time is &lt;strong&gt;T&lt;/strong&gt; seconds.&lt;br&gt;&#xA;The mean inter-Task arrival time is &lt;strong&gt;K&lt;/strong&gt; * &lt;strong&gt;T&lt;/strong&gt; (where &lt;strong&gt;K&lt;/strong&gt; is a fraction &amp;lt; 1)&lt;br&gt;&#xA;(assume Poisson or Gaussian distributions, whichever is easier to analyze.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt;&lt;br&gt;&#xA;At steady state, what is the length of the queue?  (in terms of &lt;strong&gt;N&lt;/strong&gt;, &lt;strong&gt;K&lt;/strong&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Related Question:&lt;/strong&gt;&lt;br&gt;&#xA;What is the expected delay for a Task to be completed?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is the real-life situation I am trying to model:&lt;br&gt;&#xA;I have an Apache web server with 25 worker processes.&lt;br&gt;&#xA;At steady-state there are 125 requests in the queue.&lt;br&gt;&#xA;I want to have a theoretical basis to help me optimize resources&#xA;and understand quantitatively how adding more worker processes&#xA;affects the queue length and delay.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know the single queue, single server, Poisson distribution is well analyzed.&lt;br&gt;&#xA;I don't know the more general solution for &lt;strong&gt;N&lt;/strong&gt; servers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;thanks in advance,&lt;br&gt;&#xA;-- David Jones&lt;br&gt;&#xA;dxjones@gmail.com&lt;/p&gt;&#xA;" OwnerUserId="5545" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2011-05-31T17:18:37.573" LastActivityDate="2011-05-31T17:18:37.573" Title="Queueing Theory: How to estimate steady-state queue length for single queue, N servers?" Tags="&lt;dynamical-systems&gt;&lt;simulation&gt;&lt;network-flow&gt;&lt;queueing-theory&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="0" />
  <row Id="16899" PostTypeId="2" ParentId="15739" CreationDate="2011-01-09T16:53:15.887" Score="3" ViewCount="" Body="&lt;p&gt;It helps to observe what the formula is saying geometrically. Each term in the sum represents the signed volume of a pyramid; the cone whose apex is the origin and whose base is the face under consideration. The volume is positive if the normal to the face is pointing away from the origin, otherwise it is negative.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The formula assumes that the polyhedron has been equipped with an outward normal. Thus, for example, if the polyhedron is convex and contains the origin, then the volume is simply given by the sum of the pyramids that compose it, all of which have positive volume. If the origin is not contained in the polyhedron, then all the pyramids with base on the far side of the polyhedron will have positive volume, and the negative volumes corresponding to the faces on the near side will cancel out that portion of the volume that represents parts of the far-side pyramids that are outside of the polyhedron.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To see that it works, it is probably easiest if you play with it yourself. Consider the simpler case of a closed curve in the plane. The same formula works for computing the enclosed area, except now there'll be a 2 instead of a 3 in the denominator of the terms: we're adding the signed areas of triangles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, what happens if your closed curve is not simple? Consider a figure 8. The problem now is that your consistently oriented normals can't always point outside. They will point inside one of the loops and outside of the other.&#xA;The area computed will be the area of the latter minus the area of the former.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the answer to your multiple choice question is: &#xA;3) none of the above&#xA;because you have put absolute value signs on the other answers. Your computed volume will be $\pm (v_1 - v_2)$, where the sign depends on how you chose to orient the normals. As moron mentioned, you need to ensure that the normals are oriented consistently between adjacent faces (so if you unfolded them onto a common plane their normals would agree). &lt;/p&gt;&#xA;" OwnerUserId="1728" LastActivityDate="2011-01-09T16:53:15.887" />
  <row Id="16900" PostTypeId="2" ParentId="16877" CreationDate="2011-01-09T16:54:51.687" Score="9" ViewCount="" Body="&lt;p&gt;Wikipedia cites the following source for the assertion that no sufficiently large Fibonacci number is one greater or one less than a prime:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ross Honsberger, &lt;em&gt;Mathematical Gems III&lt;/em&gt; (AMS Dolciani Mathematical Expositions No. 9), 1985, ISBN 0-88385-318-3, p. 133.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Not sure if that reference has a proof.&lt;/p&gt;&#xA;" OwnerUserId="5546" LastActivityDate="2011-01-09T16:54:51.687" CommentCount="1" />
  <row Id="16901" PostTypeId="2" ParentId="16890" CreationDate="2011-01-09T17:00:53.130" Score="1" ViewCount="" Body="&lt;p&gt;Here are two links to get you started:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Eulerian_number#Eulerian_numbers_of_the_second_kind&quot; rel=&quot;nofollow&quot;&gt;Eulerian numbers of the second kind&lt;/a&gt; may be helpful (for counting ascents, descents, etc., though i think)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Additionally some more useful information may be found &lt;a href=&quot;http://books.google.com/books?id=EvJg1VjIGyMC&amp;amp;lpg=PA26&amp;amp;dq=multiset%20eulerian%20numbers&amp;amp;pg=PA26#v=onepage&amp;amp;q=multiset%20eulerian%20numbers&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;in Stanley's book&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="1709" LastActivityDate="2011-01-09T17:00:53.130" />
  <row Id="16902" PostTypeId="1" CreationDate="2011-01-09T17:05:11.960" Score="0" ViewCount="106" Body="&lt;p&gt;$f(x, y) : (0,1)^2 \rightarrow \mathbb R $, is an integrable function, if  $y=\mbox{mod}(n^2 x,1)$, i.e.,  $y$ is $\frac{1}{n^2}$-periodic over $x$, we can write &#xA;$$ g_n(x) = f(x, \mbox{mod}(n^2 x,1))$$ and&#xA;let $g&amp;#39;_n(x)$ be its derivative, as $n, v \in \mathbb N$  goes to infinity simultaneously at the same rate, prove the convergence of&lt;br&gt;&#xA;$$\lim_{n,v \rightarrow \infty}  \int_0^1 g&amp;#39;_n(x) e^{-i2\pi v x}\,\mbox{d} x $$&lt;/p&gt;&#xA;" OwnerUserId="5181" LastEditorUserId="5181" LastEditorDisplayName="" LastEditDate="2011-01-14T07:16:28.927" LastActivityDate="2011-01-14T07:16:28.927" Title="Convergence of $\lim_{n,v \rightarrow \infty} \int_0^1 g'_n (x) e^{-i2\pi v x} \mbox{d} x $" Tags="&lt;analysis&gt;&lt;fourier-analysis&gt;" />
  <row Id="16903" PostTypeId="2" ParentId="16898" CreationDate="2011-01-09T17:22:50.817" Score="1" ViewCount="" Body="&lt;p&gt;Probably you'll find &lt;a href=&quot;http://www.win.tue.nl/~iadan/queueing.pdf&quot; rel=&quot;nofollow&quot;&gt;this one&lt;/a&gt; very useful. Chapter 5 ($M/M/C$ queue) corresponds to your model, where it is assumed that there are $c$ servers, the service time is exponentially distributed, and so are the interarrival times (of course, with different means). Anyway, the key is $M/M/C$ (or $M/M/N$, etc.) queue.&lt;/p&gt;&#xA;" OwnerUserId="2810" LastActivityDate="2011-01-09T17:22:50.817" CommentCount="2" />
  <row Id="16904" PostTypeId="2" ParentId="16897" CreationDate="2011-01-09T17:27:28.297" Score="8" ViewCount="" Body="&lt;p&gt;This is true for any trivial bundle. By the categorical definition of product we know that all maps $X \to X \times \mathbb{C}^n$ are in bijective correspondence with pairs of maps $X \to X$ and $X \to \mathbb{C}^n$. Now since we know it is a section and that $\pi$ is just the projection onto the $X$ factor the map $X \to X$ is forced to be the identity. So what is left, just our map $X \to \mathbb{C}^n$ which must be a morphism in whatever category you are working in, hence regular. Note that the case you asked about is when $n=1$ and that in fact $\mathbb{C}^n$ could be replace by any object $Y$ in whatever category you are working in. This is a fact about trivial bundles or products rather, not line bundles.&lt;/p&gt;&#xA;" OwnerUserId="627" LastActivityDate="2011-01-09T17:27:28.297" CommentCount="4" />
  <row Id="16905" PostTypeId="2" ParentId="16885" CreationDate="2011-01-09T17:51:26.343" Score="1" ViewCount="" Body="&lt;p&gt;Actually, your tensor depends in a very simple way on the coordinates $y_i$. You can see your tensor as a combination of two functions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f:\mathbb{R}^3 \to \mathbb{R}: y_i \mapsto t-r+n^iy_i$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$T^{\alpha\beta}:\mathbb{R} \to \mathbb{R}^{3\times 3}: t \mapsto T^{\alpha\beta}(t)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All you really need is the basic Taylor expansion formula for the last function combined with the knowledge of the partial derivate of the first function. This will give&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$T^{\alpha\beta}(t-r+n^iy_i) \approx T^{\alpha\beta}(t-r)+(n^iy_i)\dot{T}^{\alpha\beta}(t-r)+\frac{1}{2}(n^iy_i)^2{\ddot{T}}^{\alpha\beta}(t-r)+\frac{1}{6}(n^iy_i)^3{\dddot{T}}^{\alpha\beta}(t-r)$$&lt;/p&gt;&#xA;" OwnerUserId="3567" LastActivityDate="2011-01-09T17:51:26.343" CommentCount="1" />
  <row Id="16907" PostTypeId="1" CreationDate="2011-01-09T18:28:05.187" Score="0" ViewCount="144" Body="&lt;p&gt;I have this circuits and signals course where i have been asked to learn matlab all by myself and that its treated as a basic necessity&lt;/p&gt;&#xA;&#xA;&lt;p&gt;now i wanted some help as to where should i start from as i hardly have around less than a month before my practicals session start&lt;/p&gt;&#xA;&#xA;&lt;p&gt;should i go with video lectures/e books or which one should i prefer over the other?&lt;/p&gt;&#xA;" OwnerUserId="5548" LastActivityDate="2011-05-06T11:28:12.880" Title="Got to learn matlab" Tags="&lt;matlab&gt;" AnswerCount="4" CommentCount="3" FavoriteCount="1" />
  <row Id="16908" PostTypeId="2" ParentId="16892" CreationDate="2011-01-09T18:31:04.077" Score="0" ViewCount="" Body="&lt;p&gt;It depends on what area of arithmetic you are talking about&lt;/p&gt;&#xA;&#xA;&lt;p&gt;some places its really useful where as at times it just involves a lot of by hearting where in the normal conventional methods could have their way much more easily&lt;/p&gt;&#xA;" OwnerUserId="5548" LastActivityDate="2011-01-09T18:31:04.077" CommentCount="1" />
  <row Id="16909" PostTypeId="1" AcceptedAnswerId="17230" CreationDate="2011-01-09T18:35:12.637" Score="3" ViewCount="217" Body="&lt;p&gt;I can't do this exercise of Conway's Book: For $r&amp;gt;0$ let $A=\{w:w=\exp(1/z), 0&amp;lt;|z|&amp;lt;r\}$, determine the set $A$. Any hints?&lt;/p&gt;&#xA;" OwnerUserId="1971" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2011-01-12T14:34:29.997" LastActivityDate="2011-01-12T16:00:39.357" Title="Determine the set $\{w:w=\exp(1/z), 0&lt;|z|&lt;r\}$" Tags="&lt;complex-analysis&gt;&lt;complex-numbers&gt;&lt;complex-variables&gt;" AnswerCount="3" CommentCount="5" />
  <row Id="16910" PostTypeId="2" ParentId="16892" CreationDate="2011-01-09T18:42:19.117" Score="1" ViewCount="" Body="&lt;p&gt;According to the book 'The Trachtenberg Speed System of Basic Mathematics', &quot;A thin, studious-looking boy wearing silver-rimmed spectacles was told to multiply 5132437201 times 452736502783. He blitzed through the problem computing the answer—2323641669144374101785—in seventy seconds. The class was one where the Trachtenberg system of mathematics is taught.&quot; The book also states that &quot;Educators have found that the Trachtenberg system shortens time for mathematical computations by twenty percent.&quot;, although this statement does not seem consistent with the previous example.&lt;/p&gt;&#xA;" OwnerUserId="3157" LastActivityDate="2011-01-09T18:42:19.117" CommentCount="1" />
  <row Id="16911" PostTypeId="1" AcceptedAnswerId="16924" CreationDate="2011-01-09T18:47:35.833" Score="2" ViewCount="166" Body="&lt;p&gt;Let $f : M \rightarrow \mathbb{R}$ be a differentiable function defined on a riemannian manifold. Assume that $| \mathrm{grad}f | = 1$ over all $M$. Show that the integral curves of $\mathrm{grad}f$ are geodesics.&lt;/p&gt;&#xA;" OwnerUserId="5209" LastEditorUserId="5209" LastEditorDisplayName="" LastEditDate="2011-01-11T02:44:41.113" LastActivityDate="2011-01-11T03:38:19.053" Title="Integral curves of the gradient." Tags="&lt;homework&gt;&lt;differential-geometry&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" />
  <row Id="16912" PostTypeId="1" CreationDate="2011-01-09T18:57:49.910" Score="1" ViewCount="146" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Possible Duplicate:&lt;/strong&gt;&lt;br&gt;&#xA;  &lt;a href=&quot;http://math.stackexchange.com/questions/16630/nice-application-of-the-cauchy-frobenius-burnside-polya-formula&quot;&gt;Nice application of the Cauchy?-Frobenius?-Burnside?-P&amp;#243;lya? formula&lt;/a&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;p&gt;Burnside's lemma states that the number of orbits in a group action is equal to the average number of fixed points the elements of the group have. It is well known for its many combinatorial uses, and this is exactly what I'm after - examples of such uses, hopefully as diverse as possible. The most famous example is counting the number of ways to color a cube, so I'm primarily interested in examples of a different nature.&lt;/p&gt;&#xA;" OwnerUserId="1818" LastActivityDate="2011-01-09T18:57:49.910" ClosedDate="2011-01-09T19:39:57.157" Title="Interesting uses of Burnside's lemma?" Tags="&lt;combinatorics&gt;&lt;group-theory&gt;" CommentCount="1" />
  <row Id="16913" PostTypeId="1" CreationDate="2011-01-09T19:31:20.477" Score="1" ViewCount="45" Body="&lt;p&gt;Let $\{e_0, e_1, \dots, e_{N-1} \}$ be the Euclidean basis for $l^2(Z_N)$, and let $\{F_0, F_1, \dots, F_{N−1} \}$ be the Fourier basis( where $F_m(n)= \frac{1}{N} e^{2 \pi i m n/N}$ and $\hat{z}(m) = \sum_{n=0}^{N-1} z(n) e^{-2 \pi i m n/N}$).&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;How to show that $\hat{e}_m(k) = e^{−2πimk/N}$ for all $k$? Notice that $\hat{e}_m$ is very nearly (up to a reflection and a normalization) an element of the Fourier basis. &lt;/li&gt;&#xA;&lt;li&gt;How to show that $\hat{F}_m = e_m$?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="2723" LastEditorUserId="462" LastEditorDisplayName="" LastEditDate="2011-01-09T19:33:15.420" LastActivityDate="2011-01-09T21:55:00.517" Title="How to show that $\hat{e}_m(k) = e^{−2πimk/N}$ for all $k$? How to show that $\hat{F}_m = e_m$?" Tags="&lt;linear-algebra&gt;" AnswerCount="1" />
  <row Id="16914" PostTypeId="1" AcceptedAnswerId="16915" CreationDate="2011-01-09T19:34:05.030" Score="4" ViewCount="101" Body="&lt;p&gt;Hi I have some problem how to get count of elements in $Z_7[x]/3x^2+2x$. I think there belong to only polynomials which are indivisible with $3x^2+2x$ (gcd=1) ..I think it is so as far I know that for example in every $Z_m$, m prime, is the count of belonging elements eqauls to $\phi(m)$..But I really dont know how to get these polynomials in some efficient way..&lt;/p&gt;&#xA;" OwnerUserId="5342" LastEditorUserId="2548" LastEditorDisplayName="" LastEditDate="2011-01-10T00:32:25.360" LastActivityDate="2011-01-10T00:32:25.360" Title="Count of elements in Z_7(x)/3x^2+2x" Tags="&lt;abstract-algebra&gt;&lt;field-theory&gt;&lt;finite-fields&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="1" />
  <row Id="16915" PostTypeId="2" ParentId="16914" CreationDate="2011-01-09T19:51:17.107" Score="4" ViewCount="" Body="&lt;p&gt;The result follows from the following Theorem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Theorem.&lt;/strong&gt; Let $F$ be a finite field of order $q$ and let $p(x)$ be a polynomial in $F[x]$ of degree $n \geq 1$. Then $F[x]/(p(x))$ has order $q^n$.&lt;/p&gt;&#xA;" OwnerUserId="2548" LastEditorUserId="2548" LastEditorDisplayName="" LastEditDate="2011-01-09T21:28:03.247" LastActivityDate="2011-01-09T21:28:03.247" CommentCount="4" />
  <row Id="16916" PostTypeId="2" ParentId="16907" CreationDate="2011-01-09T19:54:29.957" Score="1" ViewCount="" Body="&lt;p&gt;For the elementary syntax, you may look up introduction videos on youtube,&#xA;and start from simple task in this &lt;a href=&quot;http://web.mit.edu/6.094/www/psets.html&quot; rel=&quot;nofollow&quot;&gt;MIT introduction - problem set&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;visit the Mathworks website god idea, it's got very good explanatory videos.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And dont foget 2 most important commands in matlab: &quot;help&quot; and &quot;doc&quot;.&#xA;Matlab got very good documentation with examples.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt; MIT Link is no longer available, you may want to copy and paste &quot;MIT Introduction - problem set&quot; into youtube. I believe the intended video series to view are below: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;MIT 6.00 Introduction to Computer Science and Programming, 2008.&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="5550" LastEditorUserId="8476" LastEditorDisplayName="" LastEditDate="2011-05-06T10:48:18.107" LastActivityDate="2011-05-06T10:48:18.107" />
  <row Id="16917" PostTypeId="1" CreationDate="2011-01-09T20:00:32.123" Score="4" ViewCount="119" Body="&lt;p&gt;I was reading the Wikipedia article for &lt;a href=&quot;http://en.wikipedia.org/wiki/Formal_system#Logical_system&quot; rel=&quot;nofollow&quot;&gt;Mathematical_logic&lt;/a&gt;. When reaching &lt;a href=&quot;http://en.wikipedia.org/wiki/Formal_system&quot; rel=&quot;nofollow&quot;&gt;Formal_logical_systems&lt;/a&gt;, I was curious about its definition and clicked into its own article &lt;a href=&quot;http://en.wikipedia.org/wiki/Formal_system#Logical_system&quot; rel=&quot;nofollow&quot;&gt;Logical_system&lt;/a&gt;, which redirected me to the article for &lt;a href=&quot;http://en.wikipedia.org/wiki/Formal_system&quot; rel=&quot;nofollow&quot;&gt;Formal_system&lt;/a&gt;, where I found &lt;a href=&quot;http://en.wikipedia.org/wiki/Formal_system&quot; rel=&quot;nofollow&quot;&gt;the definition for formal system&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In formal logic, a formal system (also&#xA;  called a logical calculus[citation&#xA;  needed]) consists of a formal language&#xA;  and a set of inference rules, used to&#xA;  derive (to conclude) an expression&#xA;  from one or more other premises that&#xA;  are antecedently supposed (axioms) or&#xA;  derived (theorems). The axioms and&#xA;  rules may be called a deductive&#xA;  apparatus. A formal system may be&#xA;  formulated and studied for its&#xA;  intrinsic properties, or it may be&#xA;  intended as a description (i.e. a&#xA;  model) of external phenomena.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and &lt;a href=&quot;http://en.wikipedia.org/wiki/Formal_system#Logical_system&quot; rel=&quot;nofollow&quot;&gt;the definition for logical system&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A logical system or, for short, logic,&#xA;  is a formal system together with a&#xA;  form of semantics, usually in the form&#xA;  of model-theoretic interpretation,&#xA;  which assigns truth values to&#xA;  sentences of the formal language, that&#xA;  is, formulae that contain no free&#xA;  variables. A logic is sound if all&#xA;  sentences that can be derived are true&#xA;  in the interpretation, and complete&#xA;  if, conversely, all true sentences can&#xA;  be derived.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I thought logical system and formal system were the same concept by self-learning, but here it says logical system is a formal system plus semantics/interpretation. Isn't that in logical system we don't care about the semantics/meaning of propositions and that's the reason that the subject is called formal logic? But why here it says logical system has semantics/interpretation? How to understand &quot;semantics&quot; and whether logical system and formal system are the same?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is logical system a special example of formal system, in the same sense as Lambda calculus is a special example of formal system as in &lt;a href=&quot;http://en.wikipedia.org/wiki/Formal_system#See_also&quot; rel=&quot;nofollow&quot;&gt;examples of formal systems&lt;/a&gt;? Or is logical system a formal system with additional structure called &quot;semantics&quot;, in the same sense as topological vector space is a vector space with additional topological structure?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks and regards!&lt;/p&gt;&#xA;" OwnerUserId="1281" LastActivityDate="2011-05-23T21:42:39.753" Title="Formal System and Formal Logical System" Tags="&lt;logic&gt;&lt;formal-systems&gt;" AnswerCount="1" />
  <row Id="16918" PostTypeId="2" ParentId="16914" CreationDate="2011-01-09T20:06:28.217" Score="4" ViewCount="" Body="&lt;p&gt;Sima: Working with polynomials with coefficients in ${\mathbb Z}_7$ (but the same is true for any field), any polynomial $p$, when divided by a polynomial $q$ of degree larger than $0$, produces a remainder $r$ that is a polynomial of degree strictly less than $q$. If $q(x)=3x^2+2x$, the remainder $r$ is then a polynomial of degree 1 or less, i.e., it has the form $ax+b$ where $a,b$ are elements of ${\mathbb Z}_7$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Two elements of ${\mathbb Z}_7[x]$ are identified in the quotient by $3x^2+2x$ iff they have the same remainder, so the elements of ${\mathbb Z}_7[x]/3x^2+2x$ are in correspondence with the remainders that, by the paragraph above, are precisely the linear polynomials $ax+b$. There are 7 possibilities for $a$ and 7 for $b$, for a total of $7^2=49$ elements.  &lt;/p&gt;&#xA;" OwnerUserId="462" LastEditorUserId="462" LastEditorDisplayName="" LastEditDate="2011-01-09T20:43:10.573" LastActivityDate="2011-01-09T20:43:10.573" />
  <row Id="16919" PostTypeId="2" ParentId="16914" CreationDate="2011-01-09T20:09:50.443" Score="4" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\ $ Use the &lt;em&gt;division algorithm&lt;/em&gt; in $\rm\ \mathbb Z_7[x]\ $ to show that every polynomial $\rm\ f(x)\ \in\ \mathbb Z_7[x]\ $ is congruent $\rm\ mod\ \ 3\ x^2 + 2\ x\ $ to a unique polynomial of degree $\:\le 1\:$, viz. $\rm\ f(x)\ \ mod\ \ 3\ x^2 + 2\ x\:,$ analogous to the fact that every elt of $\rm\ \mathbb{Z}/m\ $ has a unique representative in $\rm\:\{0,1,\cdots,\:m-1\}\:.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The analogous result holds true over any ring if the leading coefficient of the polynomial is a unit, i.e $\rm\ |R[x]/(f(x))|\ =\ |R|^n\ $ for any $\rm\ f(x) \in R[x]\ $ having degree $\rm\:n\:$ and unit leading coefficient. The hypothesis on the leading coefficient guarantees that one can divide by $\rm\:f(x)\:$ with &lt;em&gt;unique&lt;/em&gt; remainder. Indeed, the standard high-school long division algorithm clearly works, and if there were two unequal remainders of degree $\rm &amp;lt; n$ then their difference would be divisible by $\rm\:f\:,$ which is impossible since multiples of $\rm\:f\:$ have degree $\ge n$ (else the leading coefficient of $\rm\:f\:$ is a zero-divisor, not a unit).&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2011-01-09T21:42:05.337" LastActivityDate="2011-01-09T21:42:05.337" />
  <row Id="16920" PostTypeId="1" CreationDate="2011-01-09T20:23:05.813" Score="4" ViewCount="168" Body="&lt;p&gt;Let &lt;em&gt;P&lt;/em&gt; be a point, not the center, in the interior of a (round) disk &lt;em&gt;D&lt;/em&gt;⊂ℝ² and let &lt;em&gt;A&lt;/em&gt; and &lt;em&gt;B&lt;/em&gt; be points on ∂&lt;em&gt;D&lt;/em&gt; such that the line segments &lt;em&gt;AP&lt;/em&gt; and &lt;em&gt;BP&lt;/em&gt; have equal length. Choose an arc &lt;em&gt;AB&lt;/em&gt;. What's shape bounded by the arc and the two segments called?&lt;/p&gt;&#xA;" OwnerUserId="4992" LastActivityDate="2011-01-09T22:44:08.157" Title="name of a shape" Tags="&lt;euclidean-geometry&gt;&lt;circle&gt;" AnswerCount="1" CommentCount="7" FavoriteCount="0" />
  <row Id="16921" PostTypeId="1" AcceptedAnswerId="16968" CreationDate="2011-01-09T20:35:14.813" Score="1" ViewCount="99" Body="&lt;p&gt;Let $\{f_i\}$ be a system of &lt;em&gt;linear&lt;/em&gt; equations in $X_1,...,X_n$ with coefficients in $R = \mathbb Z/p^e \mathbb Z$ (i.e. modulo a primepower).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;there is a unique solution $a_i$.&lt;/li&gt;&#xA;&lt;li&gt;$f_1 = u g_1$ where $u$ is a non-unit, i.e. $p\mid u$.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Is it true that $\{f_i\mid i&amp;gt;1\}$ permits only the same unique solution? Or equivalently: is $f_1$ linearly dependent on the other equations?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Intuitively, it seems obvious. For if we substitute variables such that $g_1 = (Y_1 - a)$ then $Y_1$ is determined by $f_1$ up to some multiple of $p$. But there's no way the other equations could determine &lt;em&gt;what&lt;/em&gt; multiple without determining what $Y_1$ is itself, since the only non-units are all multiples of $p$. But that is, of course, not very solid mathematics.&lt;/p&gt;&#xA;" OwnerUserId="5189" LastEditorUserId="5189" LastEditorDisplayName="" LastEditDate="2011-01-09T21:19:50.440" LastActivityDate="2011-01-10T08:08:54.567" Title="Equations over $\mathbb Z/p^e \mathbb Z$" Tags="&lt;linear-algebra&gt;&lt;abstract-algebra&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="16922" PostTypeId="2" ParentId="16907" CreationDate="2011-01-09T20:40:22.413" Score="2" ViewCount="" Body="&lt;p&gt;I am learning also MatLab, am reading this guide may be usefull for you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The guide has been very usefull for me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Link : &lt;a href=&quot;http://www.phy.ohiou.edu/computer/matlab/techdoc/pdfdocs/getstart.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.phy.ohiou.edu/computer/matlab/techdoc/pdfdocs/getstart.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wait all answers. :)&lt;/p&gt;&#xA;" OwnerUserId="3421" LastActivityDate="2011-01-09T20:40:22.413" CommentCount="3" />
  <row Id="16923" PostTypeId="1" AcceptedAnswerId="16927" CreationDate="2011-01-09T21:08:12.947" Score="0" ViewCount="49" Body="&lt;p&gt;Hi I have another problem..Two polynomials a(x) and b(x) are asociated iff a(x)|b(x) and b(x)|a(x)….Right? And now my problem..And polynomials are indivisible when gcd is asociated with 1..And there it is..In which universe is it for example 2??As I know 1 is indivisible by 2?Is there something I am overlooking? yes and all of this in any field e.g. $Z_3[x]$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Oh sorry I forget to specify which I am asking for:)&#xA;So..I know that polynomials e.g. in $Z_3[x]$ are indivisible iff gcd is 1 or 2??And I am asking why?I know that 1 and 2 has to be mutually asociated..But I dont know why..I understand why these polynomials are indivisible, but I dont know why this asociation is valid..&lt;/p&gt;&#xA;" OwnerUserId="5342" LastEditorUserId="5342" LastEditorDisplayName="" LastEditDate="2011-01-09T21:22:57.793" LastActivityDate="2011-01-09T21:52:17.603" Title="Asociated polynomials" Tags="&lt;polynomials&gt;&lt;modular-arithmetic&gt;&lt;divisibility&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="16924" PostTypeId="2" ParentId="16911" CreationDate="2011-01-09T21:20:09.003" Score="2" ViewCount="" Body="&lt;p&gt;Since this seems to be homework, here is just an outline of the proof.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Show that the map $X\rightarrow \nabla_X \nabla f$ is self adjoint, that is, that $g(\nabla_X \nabla f, Y) = g(\nabla_Y \nabla f, X)$ for any vector fields $X$ and $Y$.  You'll need to use the fact that $\nabla f$ is a gradient field, but you won't need the fact that it has norm 1.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Show that $g(\nabla_{\nabla f} \nabla f, X) = 0$ for all $X$ by using 1. to write it as $g(\nabla_X \nabla f, \nabla f)$ and expanding.  Here, you'll need to use the fact that $\nabla f$ has norm 1.  Once you show this, conclude that $\nabla_{\nabla f} \nabla f = 0$, i.e., that the integral curves are geodesics.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Assuming I remember, or that you send a comment, I can update this in a few days with full solutions to either 1 or 2.&lt;/p&gt;&#xA;" OwnerUserId="331" LastActivityDate="2011-01-09T21:20:09.003" CommentCount="2" />
  <row Id="16925" PostTypeId="2" ParentId="16920" CreationDate="2011-01-09T21:23:39.413" Score="1" ViewCount="" Body="&lt;p&gt;It's called a &lt;a href=&quot;http://en.wikipedia.org/wiki/Circular_segment&quot; rel=&quot;nofollow&quot;&gt;circular segment&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edited to add:&lt;/strong&gt; As Jasper has pointed out in the comments, this is not what the question asked. So I don't know the answer.&lt;/p&gt;&#xA;" OwnerUserId="1508" LastEditorUserId="1508" LastEditorDisplayName="" LastEditDate="2011-01-09T22:44:08.157" LastActivityDate="2011-01-09T22:44:08.157" CommentCount="4" />
  <row Id="16926" PostTypeId="2" ParentId="16907" CreationDate="2011-01-09T21:26:49.130" Score="2" ViewCount="" Body="&lt;p&gt;Dear John,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I recommend you to &lt;strong&gt;learn&lt;/strong&gt; the software &lt;strong&gt;by doing&lt;/strong&gt; some little project. Just learning it for the sake of it doesn't make sense since you will fastly loose track imho.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example you could &lt;strong&gt;ask your tutor&lt;/strong&gt; of the signal course to simulate some nice physical situation for him. You will learn &lt;em&gt;MATLAB&lt;/em&gt; on the fly, get to know how to do some basic research and of course how its done on a personal level.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My best wishes&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Robert &lt;/p&gt;&#xA;" OwnerUserId="2671" LastActivityDate="2011-01-09T21:26:49.130" />
  <row Id="16927" PostTypeId="2" ParentId="16923" CreationDate="2011-01-09T21:28:05.447" Score="1" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\rm\ \ gcd(2\:a,2\:b) = 2\ \iff\ gcd(a,b) = 1\ $ is true in every domain where $\:2 \ne 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The associates of $1$ are the units (invertibles). Therefore in any domain like $\rm\:\mathbb Z_3$ where $2$ is a unit, $\rm\ gcd(a,b)=2\ \iff\ gcd(a,b)=1\:$.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2011-01-09T21:52:17.603" LastActivityDate="2011-01-09T21:52:17.603" />
  <row Id="16928" PostTypeId="1" AcceptedAnswerId="16930" CreationDate="2011-01-09T21:49:25.590" Score="2" ViewCount="197" Body="&lt;p&gt;I must find the area of the region that lies inside the first curve and outside the second curve for:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ r^2 = 8cos2\theta, r = 2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do I find $a$ and $b$? My solution manual tells me that the curves intersect when $cos2\theta = 1/2$ but I don't see where that came from. I tried setting it equal to each other, but keep getting $1/4$.&lt;/p&gt;&#xA;" OwnerUserId="3514" LastActivityDate="2011-01-09T21:55:04.337" Title="Finding the points of intersection in Polar Equations" Tags="&lt;calculus&gt;&lt;homework&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="16929" PostTypeId="2" ParentId="16913" CreationDate="2011-01-09T21:55:00.517" Score="1" ViewCount="" Body="&lt;p&gt;These follow immediately from the definition of the Fourier transform. You may be confusing yourself with &quot;e&quot;s.&lt;/p&gt;&#xA;" OwnerUserId="5523" LastActivityDate="2011-01-09T21:55:00.517" CommentCount="1" />
  <row Id="16930" PostTypeId="2" ParentId="16928" CreationDate="2011-01-09T21:55:04.337" Score="3" ViewCount="" Body="&lt;p&gt;The curves will intersect if $8\cos2\theta$ is positive (since it equals $r^2$) and equals 4 (because $r=2$ so $r^2=2^2=4$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[I emphasize that it must be positive, because for example $r=8\cos2\theta$ and $r=2$ intersect whenever $8\cos2\theta=2$ and also when $8\cos2\theta=-2$.]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, $8\cos2\theta=4$ gives us $\cos2\theta=4/8=1/2$. This gives us the needed values of $\theta$.&lt;/p&gt;&#xA;" OwnerUserId="462" LastActivityDate="2011-01-09T21:55:04.337" CommentCount="2" />
  <row Id="16931" PostTypeId="1" CreationDate="2011-01-09T22:09:53.727" Score="1" ViewCount="41" Body="&lt;p&gt;This is a softmax probability distribution:&lt;br&gt;&#xA;$P(i| w_1, w_2, \ldots, w_n) = \frac{exp(w_i)}{\sum_{i=1}^n exp(w_i)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It known also as Boltzmann distribution. It is used in generalized Bradley-Terry model and in multinomial logistic regression. There are efficient minorization-maximization algorithms for infering $\vec{w}$ from data through Maximal Likelihood principle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm looking for similar distribution but extended with &quot;variance&quot; parameter vector.&#xA;A parameter that would represent the confidence in $\vec{w}$.&#xA;Then I would like to infer both $\vec{w}$ and its confidence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anybody of you happen to know such distribution or a research on the topic?&lt;/p&gt;&#xA;" OwnerUserId="5552" LastActivityDate="2011-01-09T22:28:48.820" Title="Is there a SoftMax distribution with confidence parameters?" Tags="&lt;mathematical-modeling&gt;&lt;probability-distributions&gt;" AnswerCount="0" FavoriteCount="1" />
  <row Id="16933" PostTypeId="1" AcceptedAnswerId="16935" CreationDate="2011-01-09T22:41:08.590" Score="4" ViewCount="128" Body="&lt;p&gt;(this question is about trying to use some combinatorics to simplify an algorithm and save memory)&lt;br&gt;&#xA;Let $K_{2n+1}$ be a complete undirected graph on $2n+1$ vertices.&#xA;I would like to generate a Eulerian circuit of this graph (visit each edge exactly once).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One solution is to run the DFS-based algorithm that can find a Eulerian circuit in any Eulerian graph (a graph with all vertices of even degree).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I'm hoping it's possible to use the nice structure of $K_{2n+1}$ to avoid constructing the graph and running the algorithm. Is there a way to construct one such circuit in O(n) time with O(1) memory?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Motivation&lt;/strong&gt;: The motivation behind this is an algorithm which receives a list L and needs to calculate $B(A(x), A(y))$ for each unordered pair ${x,y}$ of distinct elements of L (I'm only considering unordered pairs since $B(u,w)=B(w,u)$). The result of operation A is very large and takes almost half of the available memory. Such a circuit will allow me to minimize the number of calls to operation $A$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Note&lt;/strong&gt;: This is a sort of &quot;undirected&quot; version of generating a de-Bruijn sequence. I once heard it's possible to generate a de-Bruijn sequence with constant memory, but I don't know how to do that.&lt;/p&gt;&#xA;" OwnerUserId="3533" LastActivityDate="2011-01-10T03:20:36.733" Title="Generating a Eulerian circuit of a complete graph with constant memory" Tags="&lt;combinatorics&gt;&lt;algorithms&gt;&lt;graph-theory&gt;" AnswerCount="1" />
  <row Id="16934" PostTypeId="1" AcceptedAnswerId="16937" CreationDate="2011-01-09T23:37:37.863" Score="1" ViewCount="188" Body="&lt;p&gt;Could somebody check if this is correct?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\lim_{n \to \infty} \frac {\log_{2}(\log_{2}(n))}{\log_{2}(n)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I exponantiate the numerator and the denominator with &lt;strong&gt;2&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac {(\log_{2}(\log_{2}(n)))^2}{(\log_{2}(n))^2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ = \frac {\log_{2}(n)}{n}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I extract the constant from the logarithm&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ = \log_{2}(e) * \lim_{n \to \infty} \frac {\ln(n)}{n}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using de l'Hospital:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ = \log_{2}(e) * \lim_{n \to \infty} \frac {\frac {1}{n}}{1}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ = \log_{2}(e) * \lim_{n \to \infty} \frac {1}{n} = 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is that correct?&lt;/p&gt;&#xA;" OwnerUserId="5466" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2011-01-10T03:29:35.080" LastActivityDate="2011-01-10T07:28:34.597" Title="What is the limit of multiple logarithm quotient $ \frac{\log \log n}{\log n}$" Tags="&lt;calculus&gt;&lt;limit&gt;&lt;logarithms&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="16935" PostTypeId="2" ParentId="16933" CreationDate="2011-01-10T00:15:23.243" Score="5" ViewCount="" Body="&lt;p&gt;Yes it is possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Number the vertices $1,2, \dots 2n+1$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the sequence $1, 2n, 2, 2n-1, 3, 2n-2, \dots, n,n+1$, ignore $2n+1$ at the moment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This gives a hamiltonian path in $K_{2n}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now add $1$ to each vertex $\mod 2n$, to get $2,1,3,2n,4, 2n-1 \dots, n+1, n+2$. This can be visualized as writing the numbers $1,2, \dots, 2n$ in a circle and rotating the path. (see figure below).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Repeat this process of adding $1$, $n$ times. This gives you a set of $n$ edge-disjoint hamiltonian paths in $K_{2n}$. (We can show that by proving that (wlog) $1$ is first adjacent to $2n$, then $2,3$, then $4,5$ etc)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To each of these paths, add $2n+1$ at the end and join back to the first vertex of the path.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This gives you an Euler tour of $K_{2n+1}$ which starts and ends at vertex $2n+1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This can be computed in constant memory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that this basically follows the (classic/folklore) construction that $K_{2n+1}$ can be decomposed into $n$ edge-disjoint Hamiltonian cycles. See this for instance, &lt;a href=&quot;http://books.google.com/books?id=SbZKSZ-1qrwC&amp;amp;pg=PA16&quot; rel=&quot;nofollow&quot;&gt;Modern Graph Theory, page 16&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In case that book is not accessible, here is a snapshot of the relevant part of that page:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/YlFzd.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2011-01-10T03:20:36.733" LastActivityDate="2011-01-10T03:20:36.733" />
  <row Id="16936" PostTypeId="2" ParentId="16934" CreationDate="2011-01-10T00:17:49.213" Score="5" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\ $ Let $\rm\ n = 2^{2^k}\:.\ $ Then the limit becomes that of $\rm\ k/2^k\ $ as $\rm\ k\to \infty$&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2011-01-10T00:17:49.213" />
  <row Id="16937" PostTypeId="2" ParentId="16934" CreationDate="2011-01-10T00:21:29.430" Score="2" ViewCount="" Body="&lt;p&gt;Instead of squaring, it's possible to dive right into L'Hôpital's rule, since as $n\to\infty$, you get the indeterminate form $\infty/\infty$, as you seem to have noticed. It doesn't make much sense to 'extract the constant from the logarithm,' but instead you should have&#xA;$$&#xD;&#xA;\lim_{n\to\infty}\frac{\log_2(\log_2(n))}{\log_2(n)}=\lim_{n\to\infty}\frac{\log_2(\log_2(n))&amp;#39;}{\log_2(n)&amp;#39;}=\lim_{n\to\infty}\frac{\frac{1}{n\ln{2}\ln{n}}}{\frac{1}{n\ln{2}}}=\lim_{n\to\infty}\frac{1}{\ln{n}}.&#xD;&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="1583" LastEditorUserId="1583" LastEditorDisplayName="" LastEditDate="2011-01-10T05:37:27.620" LastActivityDate="2011-01-10T05:37:27.620" CommentCount="1" />
  <row Id="16939" PostTypeId="2" ParentId="9048" CreationDate="2011-01-10T02:21:16.190" Score="0" ViewCount="" Body="&lt;p&gt;For each component (x,y)&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;linp(acc, 6*(a-2*b+c), 6*(b-2*c+d), t)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Obviously &lt;code&gt;linp&lt;/code&gt; is the linear interpolation between two values. The bezier is constructed form a sequence of linear inerpolations as shown &lt;a href=&quot;http://asymptote.sourceforge.net/doc/Bezier-curves.html#Bezier-curves&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&#xA;&lt;img src=&quot;http://i.imgur.com/OOFvQ.png&quot; alt=&quot;Bezier Definition&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The linp module can be made into a well defined function $\mathrm{L}(a,b,t)=a+t\,(b-a)$, with known derivatives. All you have to do is calculate the following&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \mathrm{bez}(a,b,c,d,t)=\mathrm{L}( \mathrm{L}( \mathrm{L}(a,b,t), \mathrm{L}(b,c,t), t), \mathrm{L}( \mathrm{L}(b,c,t), \mathrm{L}(c,d,t), t) ) $$&#xA;$$ \mathrm{acc}(a,b,c,d,t)=\frac{\mathrm{d}^2}{\mathrm{d}t^2}{\mathrm{bez}(a,b,c,d,t)} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With the derivative&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \frac{\mathrm{d}}{\mathrm{d}t}{L(f(t),g(t),t) = (1-t)\frac{\mathrm{d}f}{\mathrm{d}t}+t\frac{\mathrm{d}g}{\mathrm{d}t}-f(t)+g(t)} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rest is just math.&lt;/p&gt;&#xA;" OwnerUserId="3301" LastActivityDate="2011-01-10T02:21:16.190" />
  <row Id="16940" PostTypeId="1" AcceptedAnswerId="17025" CreationDate="2011-01-10T02:24:43.550" Score="5" ViewCount="142" Body="&lt;p&gt;Dear All&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In &lt;a href=&quot;http://math.stackexchange.com/questions/2735/solving-very-large-matrices-in-pieces&quot;&gt;Solving very large matrices in &quot;pieces&quot;&lt;/a&gt; there is a way shown to solve matrix inversion in pieces. Is it possible to apply the method in-place?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am refering to the answer in the referenced question that starts with:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\textbf{A}=\begin{pmatrix}\textbf{E}&amp;amp;\textbf{F}\\\\ \textbf{G}&amp;amp;\textbf{H}\end{pmatrix}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And computes the inverse via:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\textbf{A}^{-1}=\begin{pmatrix}\textbf{E}^{-1}+\textbf{E}^{-1}\textbf{F}\textbf{S}^{-1}\textbf{G}\textbf{E}^{-1}&amp;amp;-\textbf{E}^{-1}\textbf{F}\textbf{S}^{-1}\\\\ -\textbf{S}^{-1}\textbf{G}\textbf{E}^{-1}&amp;amp;\textbf{S}^{-1}\end{pmatrix}$, where $\textbf{S} = \textbf{H}-\textbf{G}\textbf{E}^{-1}\textbf{F}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Best Regards&lt;/p&gt;&#xA;" OwnerUserId="4414" LastEditorUserId="4414" LastEditorDisplayName="" LastEditDate="2011-01-10T14:25:56.753" LastActivityDate="2011-01-29T16:13:52.273" Title="In-place inversion of large matrices" Tags="&lt;linear-algebra&gt;&lt;algorithms&gt;&lt;matrices&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="16941" PostTypeId="1" AcceptedAnswerId="16943" CreationDate="2011-01-10T02:31:57.143" Score="3" ViewCount="144" Body="&lt;p&gt;Hello.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've tried solve this for hours, but I still stuck in the beginning. Here's my problem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $R \subseteq P(\mathbb{N} \times \mathbb{N})^2$ be the following equivalence relation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\langle r,s\rangle \in R \Leftrightarrow (\exists \pi : \mathbb{N} \to \mathbb{N} \forall x,y \in \mathbb{N} (\langle x,y\rangle \in r \leftrightarrow \langle\pi(x),\pi(y)\rangle \in s))$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is cardinality of a set $P(\mathbb{N} \times \mathbb{N})/_R$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notes:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$P(A)$ is a Power Set&lt;/li&gt;&#xA;&lt;li&gt;$\pi$ is a bijection&lt;/li&gt;&#xA;&lt;li&gt;$A/_R$ is a set of equivalence classes, I mean: $A/_R = \{ [a]_R | a \in Dom(R) \}$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="5554" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2011-01-10T06:32:41.303" LastActivityDate="2011-01-10T06:42:38.617" Title="Problem with a strange relation; what is cardinality" Tags="&lt;homework&gt;&lt;elementary-set-theory&gt;&lt;relations&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="16942" PostTypeId="2" ParentId="15756" CreationDate="2011-01-10T02:43:10.150" Score="0" ViewCount="" Body="&lt;p&gt;On a little googling , i find that the book is for sale&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/1568811659&quot; rel=&quot;nofollow&quot;&gt;http://www.amazon.com/Triangle-Book-John-H-Conway/dp/1568811659&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5548" LastActivityDate="2011-01-10T02:43:10.150" CommentCount="3" />
  <row Id="16943" PostTypeId="2" ParentId="16941" CreationDate="2011-01-10T02:48:37.390" Score="3" ViewCount="" Body="&lt;p&gt;Given a bijection $\pi:{\mathbb N}\to{\mathbb N}$, let ${\hat\pi}:{\mathbb N}\times{\mathbb N}\to{\mathbb N}\times{\mathbb N}$ be the function ${\hat\pi}(a,b)=(\pi(a),\pi(b))$. Then ${\hat\pi}$ is a bijection.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can rewrite the condition for being in $R$ as follows: $\langle r,s\rangle\in R$ iff there is a bijection $\pi:{\mathbb N}\to{\mathbb N}$ such that for any $x\in{\mathbb N}\times{\mathbb N}$, we have that $x\in r$ iff ${\hat\pi}(x)\in s$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The advantage of this description is two-fold: In a sense, it is simpler, we do not need to worry about the elements of $r,s$ being pairs of numbers. Second, the condition &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;for any $x\in{\mathbb N}\times{\mathbb N}$, we have that $x\in r$ iff ${\hat\pi}(x)\in s$ &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;can be further simplified: For $r$ a set of pairs of numbers, write ${\hat\pi}&amp;#39;&amp;#39;r$ for the pointwise image of $r$ under ${\hat\pi}$: $${\hat\pi}&amp;#39;&amp;#39;r=\{{\hat\pi}(q)\mid q\in r\}.$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then the condition above is simply $s={\hat\pi}&amp;#39;&amp;#39;r$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the relation $R$ is given by: $(r,s)\in R$ iff there is a bijection $\pi$ such that $s={\hat\pi}&amp;#39;&amp;#39;r$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This should help you figure out how to understand the equivalence classes and how to count them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Below I write some details on how to do this. I suggest that you read the bare minimum for now.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The key observation is that we are saying that $r$ is $R$-related to $s$ iff they are isomorphic. More carefully, let $$f(r)=\{a\mid\exists b\in{\mathbb N}(a,b)\in r\lor(b,a)\in r\}. $$&#xA;This is the &lt;em&gt;field&lt;/em&gt; of $r$. Then $(f(r),r)$ is a binary relation, and if $(r,s)\in R$, then $(f(r),r)$ is isomorphic to $(f(s),s)$, with the isomorphism being the function $\pi$ from before. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Luckily for us, it is enough to consider the case $f(r)={\mathbb N}$, so we do not need to worry about the extra parameter $f(r)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, what we have to do is to figure out how many non-isomorphic countable binary relations there are. The obvious guess is continuum many. In fact, there are continuum many non-isomorphic (infinite) countable linear orders. For any such order $(A,\le)$ there is a binary relation $r\subseteq{\mathbb N}\times{\mathbb N}$ such that $({\mathbb N},r)$ is isomorphic to $(A,\le)$ (this is easy from the fact that $A$ is countably infinite). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, if $(r,s)\in R$, then $({\mathbb N},s)$ is also isomorphic to $(A,\le)$. This gives us that there are at leas continuum many $R$-classes, and since this is also an upper bound, we find that the number of classes is continuum many.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All that remains is to show that there are continuum many countable linear orders. In fact, to each infinite sequence $s$ of 0s and 1s, we can assign an order $&amp;lt;_s$ so that if $s_1$ and $s_2$ are distinct sequences, then $&amp;lt;_{s_1}$ is not isomorphic to $&amp;lt;_{s_2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To do this, let the code of 0, $c(0)$, be a single point, and the code of 1, $c(1)$, be a copy of ${\mathbb Q}$. Then to the sequence $$ a_0,a_1,a_2,\dots $$ where each $a_i$ is 0 or 1, we assign the linear order consisting of: A copy of $c(a_0)$ followed by a copy of ${\mathbb Z}$ followed by a copy of $c(a_1)$ followed by a copy of ${\mathbb Z}$ followed by a copy of $c(a_2)$ ...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is easy to see that these orders are as wanted.&lt;/p&gt;&#xA;" OwnerUserId="462" LastEditorUserId="462" LastEditorDisplayName="" LastEditDate="2011-01-10T06:42:38.617" LastActivityDate="2011-01-10T06:42:38.617" CommentCount="5" />
  <row Id="16944" PostTypeId="1" AcceptedAnswerId="16953" CreationDate="2011-01-10T03:32:43.110" Score="0" ViewCount="55" Body="&lt;p&gt;I need to show that: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;a) v(u) = 1 &amp;lt;=&gt; v is in E(R), where E(R) is the group of elements in R with an inverse&lt;br&gt;&#xA;b) a|b =&gt; v(a)|v(b) =&gt; v(a) ≤ v(b) oder b=0&lt;br&gt;&#xA;c) a~b &amp;lt;=&gt; v(a)=v(b) and a|b  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think I have managed to solve part b. Here is my proof:&lt;br&gt;&#xA;a|b =&gt; there exists x in R s.t. b=ax&lt;br&gt;&#xA;=&gt; v(b)=v(ax)&lt;br&gt;&#xA;(v euclidean) =&gt; v(b) =v(a)v(x) let v(x) = k v(b)=v(a)*k =&gt; v(a)|v(b)&lt;br&gt;&#xA;I just need to solve the last part.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm really not sure where to begin with the other parts. I know that a euclidean ring has no zero divider but apart from that am a bit stumped. Google hasn't thrown up many good explanations of any of these points - so any clues would be greatly appreciated. Thanks&lt;/p&gt;&#xA;" OwnerUserId="4697" LastActivityDate="2011-01-10T04:30:42.497" Title="Let R be a euclidean ring with the multiplicative euclidean norm function v. Show the following." Tags="&lt;homework&gt;&lt;ring-theory&gt;" AnswerCount="1" />
  <row Id="16945" PostTypeId="1" CreationDate="2011-01-10T03:49:00.383" Score="8" ViewCount="278" Body="&lt;p&gt;In &lt;a href=&quot;http://en.wikipedia.org/wiki/Summer_Wars&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;Summer Wars&lt;/em&gt;&lt;/a&gt; the main character (he is a mathematician) calculates the day of the week of someone's birthday (19/07/1992 is Sunday). I know (very) basic modular arithmetic but I can't figure out how to do it. Can someone point me to the right direction? It seems fun to do.&lt;/p&gt;&#xA;" OwnerUserId="1938" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2011-04-20T04:35:29.483" LastActivityDate="2011-05-12T14:39:35.857" Title="Calculate which day of the week a date falls in using modular arithmetic" Tags="&lt;number-theory&gt;&lt;modular-arithmetic&gt;&lt;calendar-computations&gt;" AnswerCount="3" CommentCount="4" FavoriteCount="1" />
  <row Id="16946" PostTypeId="1" CreationDate="2011-01-10T04:00:10.053" Score="13" ViewCount="501" Body="&lt;p&gt;The problem&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Show that the sequence, $49, 4489, 444889, \dots$, gotten by inserting&#xA;  the digits $48$ in the middle of the&#xA;  previous number (all in base $10$), consists only of &#xA;  perfect squares.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;has become a classic. For some reason, I got curious as to who actually discovered this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After doing some research, seems like this problem was used in the Stanford University Competitive Exam in Mathematics for high school seniors, in the year 1964. One pdf which has this is here: &lt;a href=&quot;http://www.computing-wisdom.com/jstor/stanford-exam.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.computing-wisdom.com/jstor/stanford-exam.pdf&lt;/a&gt;. It appears as problem 64.2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The pdf also mentions that there is a booklet which gives references to previous appearances of the problems from the above exam, but that seemed to be a dead-end regarding this particular problem.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Does anyone know who originally&#xA;  discovered this little gem? More&#xA;  interestingly, is it known &lt;em&gt;how&lt;/em&gt; this&#xA;  was discovered?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update 1&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On some more research I found this book from 1903: &lt;a href=&quot;http://books.google.com/books?id=3D04AAAAMAAJ&amp;amp;printsec=frontcover&quot; rel=&quot;nofollow&quot;&gt;Algebra Part II&lt;/a&gt; by E.M. Langley and S.R.N. Bradly which has this as an exercise on &lt;a href=&quot;http://books.google.com/books?id=3D04AAAAMAAJ&amp;amp;pg=PA180&quot; rel=&quot;nofollow&quot;&gt;page 180&lt;/a&gt;. The question seems to have been phrased in such a way as to claim ownership and also tells &lt;em&gt;how&lt;/em&gt; it was discovered. I guess we just need confirmation now.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update 2&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More digging reveals this German book by Dr. Hermann Schubert, &lt;a href=&quot;http://books.google.com/books?id=seEGAAAAYAAJ&amp;amp;pg=PP5#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;Mathematische Mussestunden&lt;/a&gt;, which has this on &lt;a href=&quot;http://books.google.com/books?id=seEGAAAAYAAJ&amp;amp;dq=shubert%20mathematische%2067%20667&amp;amp;pg=PA24#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;page 24&lt;/a&gt;. The book was published in 1900, but the preface seems to be dated earlier. If someone can read German, perhaps maybe they can read and see what the book claims about the origins of this problem. The book seems to have a list of references at the beginning.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2011-01-11T00:41:24.417" LastActivityDate="2011-01-11T00:41:24.417" Title="History of &quot;Show that $44\dots 88 \dots 9$ is a perfect square&quot;." Tags="&lt;reference-request&gt;&lt;math-history&gt;" CommentCount="13" FavoriteCount="2" />
  <row Id="16947" PostTypeId="1" AcceptedAnswerId="16950" CreationDate="2011-01-10T04:03:44.660" Score="3" ViewCount="164" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've run into a small problem while working through Enderton's Elements of Set Theory. I'm doing the following problem:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Call a natural number even if it has the form $2\cdot m$ for some $m$. Call it odd if it has the form $(2\cdot p)+1$ for some $p$. Show that each natural number number is either even or odd, but never both.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I've shown most of this, and along the way I've derived many of the results found in Arturo Magidin's great &lt;a href=&quot;http://math.stackexchange.com/questions/15869/how-is-addition-defined&quot;&gt;post on addition&lt;/a&gt;, so any of the theorems there may be used. It is the 'never both' part with which I'm having trouble. This is some of what I have:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let&#xA;$$&#xD;&#xA;B=\{n\in\omega\ |\neg(\exists m(n=2\cdot m)\wedge\exists p(n=2\cdot p+1))\},&#xD;&#xA;$$&#xA;the set of all natural numbers that are not both even and odd. Since $m\cdot 0=0$, $0$ is even. Also $0$ is not odd, for if $0=2\cdot p+1$, then $0=(2\cdot p)^+=\sigma(2\cdot p)$, but then $0\in\text{ran}\ \sigma$, contrary to the first Peano postulate. Hence $0\in B$. Suppose $k\in B$. Suppose $k$ is odd but not even, so $k=2\cdot p+1$ for some $p$. Earlier work of mine shows that $k^+$ is even. However, $k^+$ is not odd, for if $k^+=2\cdot m+1$ for some $m$, then since the successor function $\sigma$ is injective, we have&#xA;$$&#xD;&#xA;k^+=2\cdot m+1=(2\cdot m)^+\implies k=2\cdot m&#xD;&#xA;$$&#xA;contrary to the fact that $k$ is not even.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now suppose $k$ is even, but not odd. I have been able to show that $k^+$ is odd, but &lt;strong&gt;I can't figure a way to show that $k^+$ is not even.&lt;/strong&gt; I suppose it must be simple, but I'm just not seeing it. Could someone explain this little part? Thank you.&lt;/p&gt;&#xA;" OwnerUserId="1583" LastEditorUserId="1583" LastEditorDisplayName="" LastEditDate="2011-01-10T04:41:21.413" LastActivityDate="2011-01-10T04:41:21.413" Title="Proving there is no natural number which is both even and odd." Tags="&lt;elementary-set-theory&gt;&lt;arithmetic&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="16948" PostTypeId="1" CreationDate="2011-01-10T04:07:32.190" Score="13" ViewCount="359" Body="&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; Show that $\pi_{1}({\mathbb R}^{2} - {\mathbb Q}^{2})$ is uncountable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Motivation:&lt;/strong&gt; This is one of those problems that I saw in Hatcher and felt I should be able to do, but couldn't quite get there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What I Can Do:&lt;/strong&gt;  There are proofs of this being path connected (though, I'm not exactly in love with any of those proofs) and this tells us we can let any point be our base-point.  Now, let $p$ be some point in ${\mathbb R}^{2} - {\mathbb Q}^{2}$ and let's let this be our base-point.  We can take one path from $p$ to $q$ and a second one from $q$ to $p$, and it's not hard to show that if these paths are different then there is at least one rational on the &quot;inside&quot; of it.  Since there are uncountably many $q$, this would seem to imply uncountably many different elements of the fundamental group; the problem I'm having is showing that two loops like we've described are actually different!  For example, a loop starting at $p$ and passing through $q$ should be different from a loop starting at $p$ and passing through $q&amp;#39;$ for none of these points the same, for at least an uncountable number of elements $q&amp;#39;$.  Is there some construction I should be using to show these elements of the fundamental group are different?&lt;/p&gt;&#xA;" OwnerUserId="2959" LastActivityDate="2011-02-01T18:23:02.077" Title="$\pi_{1}({\mathbb R}^{2} - {\mathbb Q}^{2})$ is uncountable." Tags="&lt;algebraic-topology&gt;" AnswerCount="3" CommentCount="8" />
  <row Id="16949" PostTypeId="2" ParentId="16947" CreationDate="2011-01-10T04:10:24.253" Score="0" ViewCount="" Body="&lt;p&gt;Suppose there exists some $n \in \mathbb{N}$ which is both even and odd. Then $n= 2m = 2p+1$. So $2m = 2p+1$ or $2(m-p) = 1$. Contradiction. &lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2011-01-10T04:10:24.253" CommentCount="1" />
  <row Id="16950" PostTypeId="2" ParentId="16947" CreationDate="2011-01-10T04:18:28.360" Score="3" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\ $ Here's the inductive step: $\rm\ 2m \ne 2n+1\ \Rightarrow\ 2m+1 \ne 2(n+1)$&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2011-01-10T04:26:20.253" LastActivityDate="2011-01-10T04:26:20.253" CommentCount="2" />
  <row Id="16951" PostTypeId="1" AcceptedAnswerId="16962" CreationDate="2011-01-10T04:20:00.777" Score="0" ViewCount="131" Body="&lt;p&gt;I know that I need to find an irreducible element that isn't prime.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was wondering whether these examples work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In $\mathbb{Z}[\sqrt{10}]$, $9=(3)(3)=(\sqrt{10} +1)(\sqrt{10} -1)$. &#xA;$9$ is not prime, but is irreducible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In $\mathbb{Z}[\sqrt{-17}]$, $18=(6)(3)=(1-\sqrt{-17})(1+\sqrt{-17})$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is writing the number as a product of 2 units prove its irreducible? Or do I need to do more?&#xA;Thanks&lt;/p&gt;&#xA;" OwnerUserId="4697" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-10T06:08:37.797" LastActivityDate="2011-01-10T06:36:06.447" Title="Are $\mathbb{ Z}[\sqrt{-17}]$ or $\mathbb{Z}[\sqrt{10}]$ factorial rings?" Tags="&lt;homework&gt;&lt;ring-theory&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="16952" PostTypeId="2" ParentId="16521" CreationDate="2011-01-10T04:23:48.213" Score="1" ViewCount="" Body="&lt;p&gt;I haven't found a stellar way to do this by hand, but it is now easy to do with Pari/GP.  The basic idea is you just factor f over Q[x]/(f).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Often this is easy to do: find some prime p in Q such that {x,x^p,x^(p^2),…} has exactly deg(f) distinct residues mod (f,p).  Choose p larger than the twice the largest coefficient of the ones in the (unknown) answer.  Replace a mod p with the integer of smallest absolute value congruent to a mod p for each of the coefficients of x^(p^i) mod (f,p).  Check that the formula works.  I had to take p=31 in the particularly 6th degree case, so this is not exactly great for by hand exams.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are more refined versions of factoring using Hensel lifting or combining several primes both of which allow smaller primes to be used (and for it to work in general).  The details of one (or two) algorithms are in section 3.6.2 of Henri Cohen's textbook for a Course in Computational Algebraic Number Theory, and some others are also in the source code to Pari/GP (with references).&lt;/p&gt;&#xA;" OwnerUserId="583" LastActivityDate="2011-01-10T04:23:48.213" />
  <row Id="16953" PostTypeId="2" ParentId="16944" CreationDate="2011-01-10T04:30:42.497" Score="1" ViewCount="" Body="&lt;p&gt;In (a) one of the implications should be clear (if $a$ is a unit in $R$, there is $b$ in $R$ with $ab = 1$; apply $v$ to both sides and use the defining properties of a norm to conclude that $v(a) = 1$).  For the other, use the &quot;division with remainder&quot; property (i.e., the one thing that makes a &quot;Euclidean ring&quot; special).  If $a$ in $R$ satisfies $v(a) = 1$, by the division with remainder property (dividing $1$ by $a$), either there is $q$ in $R$ with $1 = aq$, in which case $a$ is a unit, or there are $q$ and $r$ in $R$ with $r$ nonzero, $1 = aq + r$ and $v(r) &amp;lt; v(1)$.  But nothing can be smaller than $v(1)$ (because $1$ divides $r$ and (b)) so this case does not occur.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In (c) you do not explain what the relation $\sim$ is but I imagine that the definition of the statement $a \sim b$ is that there is a unit $u$ with $a = bu$ (or something equivalent to this) and that you know that $\sim$ defined in this way is an equivalence relation (so in particular, if $a \sim b$ then $b \sim a$). So if $a \sim b$, there is a unit $u$ with $b = au$, and hence $a$ divides $b$, and the equality $v(b) = v(a)$ follows from applying $v$ to both sides of $b = au$ and using the defining properties of $v$ (and $v(u) = 1$, from (a)).  Conversely, if $a$ divides $b$ and $v(a) = v(b)$, from the first of these hypotheses there is $c$ in $R$ with $b = ca$; applying $v$ to both sides of this equation and using the defining properties of $v$ we conclude that $v(b) = v(c) v(a)$; from the hypothesis that $v(a) = v(b)$ we deduce that $v(c) = 1$ and hence that $c$ is a unit by (a); this shows that $a \sim b$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only trick to proving statements like these is close study of the definitions of the things that appear in the statements.  If you are getting stuck, it is likely because you are not making full use of these definitions.  I do not think it is possible to prove (a), for example, using only the ring axioms and the properties of the norm that do not refer to the &quot;division with remainder&quot; property; by this I mean that it is likely you can find an integral domain $R$, with a function $v$ from it to the integers satisfying $v(ab) = v(a) v(b)$ and other identities of this type, for which the statement (a) is not true.&lt;/p&gt;&#xA;" OwnerUserId="5559" LastActivityDate="2011-01-10T04:30:42.497" CommentCount="1" />
  <row Id="16954" PostTypeId="1" AcceptedAnswerId="17003" CreationDate="2011-01-10T04:58:55.657" Score="4" ViewCount="123" Body="&lt;p&gt;Is it easy to describe them?&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;G = Alt(2p),&lt;/li&gt;&#xA;&lt;li&gt;P a Sylow p-subgroup generated by (1,2,...,p) and (p+1,p+2,...,2p),&lt;/li&gt;&#xA;&lt;li&gt;N = N&lt;sub&gt;G&lt;/sub&gt;(P)&lt;/li&gt;&#xA;&lt;li&gt;M = { g in Alt(2p) : g takes every orbit of P to some orbit of P }&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Is it true that the unique subgroup strictly between N and G is M, unless p=2 or p=3 (where N=M is already maximal)? If not, then what does happen?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I think it is clear that N ≤ M &amp;lt; G.  I am fine with M being maximal in G.  I don't really understand why N would be maximal in M, or why N is contained in a unique maximal subgroup.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the symmetric group is easier to work in, and the answer is close, I am fine with that case as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am trying to understand Bender type ideas in this easy but slightly exotic case, so I'd prefer elementary proofs, but any exposé of Bender's ideas is a welcome (but not necessary) addition.&lt;/p&gt;&#xA;" OwnerUserId="583" LastActivityDate="2011-01-17T17:40:09.900" Title="What are the subgroups of Alt(2p) containing the normalizer of a Sylow p-subgroup?" Tags="&lt;group-theory&gt;" AnswerCount="2" FavoriteCount="2" />
  <row Id="16955" PostTypeId="1" CreationDate="2011-01-10T05:11:23.247" Score="1" ViewCount="76" Body="&lt;p&gt;Hello everyone,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I asked a question a couple days ago so please let me know if I'm asking too frequently on here, I'm not sure what the etiquette generally is. In short, I'm looking to prove that for P, Q distinct points in the complex plane $\mathbb{C}$, $\mathbb{C}-\{P,Q\}$ is not conformally equivalent to either $\mathbb{C}$ or $\mathbb{C}^{*}$ (the punctured plane), and that the same is true for any domain whose complement has more than one point in $\mathbb{C}$, using typical techniques in complex analysis. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only caveat is I hope to use this in order to show using the uniformization theorem that $\mathbb{C}-\{P,Q\}$ is not uniformed by the unit disk, so I would prefer to avoid proving this using the uniformization theorem to avoid a circular argument. Since this is homework, I wouldn't expect too much detail in the answer (unless you care to give it!), but I'm not sure what methods are conventionally used to prove 2 structures are not conformally equivalent. In general, is there any rule on a Riemann surface, compact or otherwise, which determines whether it can be uniformized by the unit disk or the complex plane? (For example, based on genus? This latter question is merely a matter of curiosity.) Could anyone help please? Many thanks in advance. &lt;/p&gt;&#xA;" OwnerUserId="5473" LastEditorUserId="5473" LastEditorDisplayName="" LastEditDate="2011-01-10T05:29:05.000" LastActivityDate="2011-01-10T05:29:05.000" Title="Conformal (in)equivalence of surfaces in $\mathbb{C}$" Tags="&lt;homework&gt;&lt;analysis&gt;&lt;complex-analysis&gt;" CommentCount="2" />
  <row Id="16956" PostTypeId="2" ParentId="14798" CreationDate="2011-01-10T05:16:00.333" Score="0" ViewCount="" Body="&lt;p&gt;All rationals in [0,1] form a countable set. All algebraic numbers form a countable set. So all rationals that have a terminating decimal expansion, being a subset of a countable set, are countable.&lt;/p&gt;&#xA;" OwnerUserId="5561" LastActivityDate="2011-01-10T05:16:00.333" />
  <row Id="16957" PostTypeId="2" ParentId="16948" CreationDate="2011-01-10T05:19:04.177" Score="8" ViewCount="" Body="&lt;p&gt;So now that Jacob has explained why the space is path connected let us finish the problem. In his explanation he points out that there are an uncountable number of lines passing through $p$ and an uncountable number of lines passing through $q$ and since we are removing only countably many points not all of these lines can hit a point in $\mathbb{Q}^2$. So we can construct a path from $p$ to $q$ missing $\mathbb{Q}^2$, in fact more than one path. That gives us a loop, inside this loop will be an element of $\mathbb{Q}^2$ just by looking at the coordinates of $p$ and $q$. This shows that the loop is not contractible, ie it is a nontrivial element of $\pi_1(\mathbb{R}^2-\mathbb{Q}^2)$. Now you can proceed to construct more loops passing through $p$ and $q$ that are not contractible or homotopic to the previously constructed loop by picking different line segments as above. Now just construct another path from $p$ to $q$ and notice that an element of $\mathbb{Q}^2$ is inside each of your new loops (there should be at least 2 new loops, try drawing a picture).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does this help?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is, in my mind, a clearer construction of uncountably many non-homotopic loops. Let $p \in \mathbb{R}^2-\mathbb{Q}^2$ and $L$ a line in $\mathbb{R}^2-\mathbb{Q}^2$ not containing $p$. As established, there is a continuum of lines passing through $p$, and once we remove all of the ones passing through a point in $\mathbb{Q}^2$ we still have uncountably many. All remaining lines, with one possible exception, intersect $L$ at distinct points. We are left with a triangle for each pair of distinct remaining lines passing through $p$. There is a rational point inside each of these triangles (maybe this is something that was unclear, let me know and I think I can make this part more explicit). There is no way to move &quot;homotope&quot; the triangle past one of the rational points so that it would be homotopic to any adjacent triangle. Suppose two &quot;loops&quot; created in this fashion were homotopic, and suppose further that the differ in the second line passing through $p$, then there would have to be a homotopy between two paths which enclose a point of $\mathbb{Q}^2$ which is impossible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want I can draw a picture on my tablet and try and post. (I don't really know the proper way to go about posting such a picture.)&lt;/p&gt;&#xA;" OwnerUserId="627" LastEditorUserId="627" LastEditorDisplayName="" LastEditDate="2011-01-11T19:03:21.560" LastActivityDate="2011-01-11T19:03:21.560" CommentCount="9" />
  <row Id="16958" PostTypeId="1" AcceptedAnswerId="16961" CreationDate="2011-01-10T05:38:25.260" Score="2" ViewCount="140" Body="&lt;p&gt;I was recently working on a piece of software and needed to know the number of different ways that a particular word could be represented in mixed upper-and-lowercase letters.  For example, the word &lt;code&gt;&quot;bar&quot;&lt;/code&gt; could be represented as:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;{bar, Bar, bAr, baR, BAr, bAR, BaR, BAR}&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or, more generally, let 0 = lowercase and 1 = uppercase:&lt;br&gt;&#xA;&lt;code&gt;{000, 100, 010, 001, 110, 011, 101, 111}&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I did a few of these and figured out that:&lt;br&gt;&#xA;&lt;code&gt;f(1) = 2&lt;/code&gt;&lt;br&gt;&#xA;&lt;code&gt;f(2) = 4&lt;/code&gt;&lt;br&gt;&#xA;&lt;code&gt;f(3) = 8&lt;/code&gt;&lt;br&gt;&#xA;&lt;code&gt;f(4) = 16&lt;/code&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which any computer programmer instantly recognizes as  &lt;code&gt;f(n) = 2^n&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But what if I hadn't been that lucky and my data had resulted in something like:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;f(1) = 2&lt;/code&gt;&lt;br&gt;&#xA;&lt;code&gt;f(2) = 7&lt;/code&gt;&lt;br&gt;&#xA;&lt;code&gt;f(3) = 24&lt;/code&gt;&lt;br&gt;&#xA;&lt;code&gt;f(4) = 77&lt;/code&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Is there some sort of process that I could use to discover the function?&lt;/strong&gt;&lt;br&gt;&#xA;( Which, in this example, is &lt;code&gt;f(n) = (3^n) - n&lt;/code&gt;)&lt;/p&gt;&#xA;" OwnerUserId="5560" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2011-01-10T15:04:27.330" LastActivityDate="2011-01-10T18:14:30.480" Title="Given a set, what process is there to trace back to the equation that generates it?" Tags="&lt;sequences-and-series&gt;" AnswerCount="5" CommentCount="1" />
  <row Id="16959" PostTypeId="2" ParentId="16958" CreationDate="2011-01-10T05:52:03.533" Score="4" ViewCount="" Body="&lt;p&gt;In general there isn't one method for matching an equation to a set of numbers and furthermore for any finite sequence there are infinitely many equations that describe the sequence. There are a lot of tools and different ways of looking and sequences to find equations that seem like a &quot;good fit&quot;. I would guess though that you would be best served by searching the &lt;a href=&quot;http://oeis.org/&quot; rel=&quot;nofollow&quot;&gt;encyclopedia of sequences&lt;/a&gt; or just typing your sequence into wolfram alpha. &lt;/p&gt;&#xA;" OwnerUserId="1930" LastActivityDate="2011-01-10T05:52:03.533" CommentCount="1" />
  <row Id="16960" PostTypeId="2" ParentId="16958" CreationDate="2011-01-10T05:55:42.000" Score="3" ViewCount="" Body="&lt;p&gt;In general, you can have any number of functions satisfy a given pattern; if you don't know anything about the function, you won't be able to solve for it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you do know something about the function - for example, that it's an $n$-th degree polynomial - you would be able to solve for it (in this case if you have $n$ distinct values).  But, what you know has to be pretty specific - for example, if you have $n$ points and you know you're looking for a polynomial function, but you don't know its degree, you'll have infinitely many functions to choose from that fit those points.  (Furthermore, even if you do know its degree, but you have fewer points than the degree of the polynomial, you'll still wind up with infinitely many functions; and of course not everything is a polynomial).&lt;/p&gt;&#xA;" OwnerUserId="1769" LastActivityDate="2011-01-10T05:55:42.000" />
  <row Id="16961" PostTypeId="2" ParentId="16958" CreationDate="2011-01-10T05:56:03.923" Score="7" ViewCount="" Body="&lt;p&gt;No, there is no process.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason is simple: given any $k$ numbers $a_1,\ldots,a_k$, I can come up with a formula $f$ with $f(1)=a_1$, $f(2)=a_2,\ldots,f(k)=a_k$; for example, Lagrange Interpolation will produce a polynomial of degree $k-1$ that evaluates to these values; so you can take &lt;em&gt;any&lt;/em&gt; other number $a_{k+1}$, and come up with a polynomial of degree $k$ that produces the same first $k$ values, and with $f(k+1)=a_{k+1}$. So you have at least as many polynomials of degree at most $k$ that generate the list you started with as there are real numbers, all of them distinct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, I could say that you &quot;recognized&quot; your first $f$ &quot;incorrectly&quot;, since it is well known that the next term in the sequence is $31$, not $32$ (there are two ways to obtain $31$: one by using the Lagrange Interpolation Formula; the other is to consider what happens if you place $n+2$ points on a circle and join them all with lines; $f(n)$ is the number of regions into which the circle is divided. See &lt;a href=&quot;http://oeis.org/A000127&quot; rel=&quot;nofollow&quot;&gt;Sequence A000127&lt;/a&gt; in the On-Line Encyclopedia of Integer Sequences. See Carl Linderholm's &lt;strong&gt;Mathematics Made Difficult&lt;/strong&gt;). &lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2011-01-10T05:56:03.923" CommentCount="3" />
  <row Id="16962" PostTypeId="2" ParentId="16951" CreationDate="2011-01-10T06:08:20.620" Score="2" ViewCount="" Body="&lt;p&gt;An element $m$ is &lt;em&gt;irreducible&lt;/em&gt; if and only if it is not a unit, not equal to $0$, and whenever $m=ab$, either $a$ is a unit or $b$ is a unit. An element $p$ is &lt;em&gt;prime&lt;/em&gt; if and only if $p$ is not a unit, is not zero, and whenever $p|ab$, either $p|a$ or $p|b$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are correct that in order to show that these rings are not factorials (unique factorization domain) it suffices to show that there is an irreducible element that is not prime.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, you did &lt;em&gt;not&lt;/em&gt; show that $9$ is irreducible; quite the contrary, you showed that it is &lt;em&gt;reducible&lt;/em&gt;: if $3$ is a unit, then $9=3\times 3$ shows that $9$ is a unit (which it is not, since $\mathbb{Z}[\sqrt{10}]$ contains no noninteger rationals); so $3$ is not a unit, which means that you have shown that $9$ can be expressed as a product of two things that are &lt;em&gt;not&lt;/em&gt; units. You showed that $9$ is &lt;strong&gt;reducible&lt;/strong&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(If $u$ and $v$ are units, then $uv$ is a unit: multiply by $v^{-1}u^{-1}$). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you have not yet produced elements that are irreducible but not prime; neither $9$ nor $18$ are irreducible.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, if you can show that $3$ is irreducible in $\mathbb{Z}[\sqrt{10}]$, you're going to be pretty much done; similarly, if you can show either that $3$ or that $2$ are irreducible in $\mathbb{Z}[\sqrt{-17}]$ you'll be very close to done as well along these lines.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-10T06:36:06.447" LastActivityDate="2011-01-10T06:36:06.447" CommentCount="4" />
  <row Id="16963" PostTypeId="1" CreationDate="2011-01-10T07:28:24.710" Score="1" ViewCount="106" Body="&lt;p&gt;So if $X$ is a vector space, and you define a norm, $x \mapsto \| x \|$, on it, then the bounded subset, $V = \{ x \in X: \|x\| &amp;lt; \infty \}$ is automatically a subspace. This follows from the definition of a norm, so for all $x,y \in V, c \in \mathbb{C}$, $\|x + c y\| \le \|x\| + \|cy\| = \|x\|+ |c| \|y\| &amp;lt; \infty$, so $x+cy \in V$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this the reasoning behind the definition of a norm?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;thanks,&#xA;-nick&lt;/p&gt;&#xA;" OwnerUserId="5564" LastActivityDate="2011-01-10T15:14:24.697" Title="Motivation for normed space definition?" Tags="&lt;norm&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="16964" PostTypeId="2" ParentId="16934" CreationDate="2011-01-10T07:28:34.597" Score="0" ViewCount="" Body="&lt;p&gt;Let $ t = \log_{2}(n) $ then $  \lim \limits_{n \to \infty} \frac {\log_{2}(\log_{2}(n))}{\log_{2}(n)}=\lim \limits_{t \to \infty} \frac {\log_{2}(t)}{t}=0$&lt;/p&gt;&#xA;" OwnerUserId="3936" LastActivityDate="2011-01-10T07:28:34.597" />
  <row Id="16965" PostTypeId="1" AcceptedAnswerId="16967" CreationDate="2011-01-10T07:30:08.183" Score="3" ViewCount="157" Body="&lt;p&gt;Let $Q: \mathbb{R}^3 \rightarrow \mathbb{R}^3$ be an operator that preserves all distances.  Is this condition alone enough for us to say that $Q$ must be a &lt;em&gt;linear&lt;/em&gt; operator?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If not, what are some counterexamples (whether simple or pathological), where an operator preserves all distances but is not a linear operator?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[Edit] Did I even give the correct interpretation of &quot;preserves all distances&quot;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[Edit] I guess not!  What I meant is that $Q$ is an isometry (i.e. $||Qx - Qy|| = ||x - y||$ for any $x$, $y$).&lt;/p&gt;&#xA;" OwnerUserId="4904" LastEditorUserId="4904" LastEditorDisplayName="" LastEditDate="2011-01-10T08:11:53.747" LastActivityDate="2011-01-10T08:16:29.500" Title="A distance preserving operator that's not linear?" Tags="&lt;linear-algebra&gt;" AnswerCount="2" CommentCount="0" />
  <row Id="16966" PostTypeId="2" ParentId="16965" CreationDate="2011-01-10T07:45:44.467" Score="3" ViewCount="" Body="&lt;p&gt;(EDIT: when I wrote this answer, the question asked if &quot;for all $x$, $\|Qx\|=\|x\|$&quot; was the right definition of &quot;$Q$ preserves all distances&quot;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;No, your condition is not equivalent to preserving all distances. For example, the map which maps $(1,0,0)$ to $(-1,0,0)$ and fixes everything else satisfies your condition but does not preserve all distances, as it reduces the distance between $(1,0,0)$ and $(-1,0,0)$ from 2 to 0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The natural notion of preserving all distances is for $Q$ to be an isometry: that is, $\|Q(x)-Q(y)\|=\|x-y\|$ for all $x$ and $y$. A map with this property need not be linear (for example, $Q(x)=x+(1,0,0)$ has this property), but it must be &lt;a href=&quot;http://en.wikipedia.org/wiki/Affine_transformation&quot; rel=&quot;nofollow&quot;&gt;affine&lt;/a&gt;. Thus if you also require $Q$ to fix $0$, then it is linear.&lt;/p&gt;&#xA;" OwnerUserId="5203" LastEditorUserId="5203" LastEditorDisplayName="" LastEditDate="2011-01-10T08:16:29.500" LastActivityDate="2011-01-10T08:16:29.500" CommentCount="8" />
  <row Id="16967" PostTypeId="2" ParentId="16965" CreationDate="2011-01-10T08:07:31.703" Score="8" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Mazur-Ulam_theorem&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Mazur-Ulam_theorem&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2011-01-10T08:07:31.703" CommentCount="1" />
  <row Id="16968" PostTypeId="2" ParentId="16921" CreationDate="2011-01-10T08:08:54.567" Score="1" ViewCount="" Body="&lt;p&gt;Suppose for simplicity that the first equation is $p^tx_1 = 0$, so that $x_1 = kp^{e-t}$. Assume to the contrary that the system without this equation has more than one solution. Look at the projection $P$ of the set of all solutions on $x_1$. It must contain exactly one member of the form $kp^{e-t}$. Since it is linear, this must in fact be $0$. There is also another member $rp^{e-t-s}$ which is not a multiple of $p^{e-t}$ (here $(r,p)=1$ and $s \geq 1$). Again, since the solution space is linear we get that $P$ contains $rp^{e-t} \neq 0$, contradiction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can always transform the first equation to this form by simple substitution. Take the lowest power $p^t$ ($t \geq 1$) dividing any of the coefficients of $f_1$, say $y_1$. So the equation can be written as $p^t(y_1 + \cdots) = 0$. Now let $x_1 = y_1 + \cdots$. Since $y_1 = x_1 - \cdots$, we can replace $y_1$ with $x_1$. This transformation is one-to-one and so preserves all the required properties.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2011-01-10T08:08:54.567" CommentCount="1" />
  <row Id="16969" PostTypeId="1" CreationDate="2011-01-10T08:23:19.973" Score="3" ViewCount="178" Body="&lt;p&gt;I've 2 questions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;let be $W_s$ a standard Brownian motion:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;using Ito's formula show that $\left( W_t,\int_0^t W_sds \right)$ has a normal distribution;&lt;/li&gt;&#xA;&lt;li&gt;and calculate $ E\left[e^{W_t}e^{\int_0^t W_sds}   \right] .$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;For the first part, i know that $W_t$ and $\int_0^t W_sds$ &#xA;have normal distribution with mean and variance respectively $(0,t)$ and $(0, t^3/3)$, but i need help with Ito's formula. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the second part i've tried to solve $E\left[e^{W_t}e^{\int_0^t W_sds}   \right]= \iint e^{W_t}e^{\int_0^t W_sds} \;\phi \left( W_t,\int_0^t W_sds \right)\: dW_t \int_0^t W_sds$... &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is these the only way?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S. sorry for my poor english&lt;/p&gt;&#xA;" OwnerUserId="5565" LastEditorUserId="5565" LastEditorDisplayName="" LastEditDate="2011-01-10T08:29:21.230" LastActivityDate="2011-01-10T23:37:07.397" Title="Expectation of time integral" Tags="&lt;calculus&gt;&lt;probability&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="1" />
  <row Id="16970" PostTypeId="1" AcceptedAnswerId="17045" CreationDate="2011-01-10T08:29:06.520" Score="3" ViewCount="145" Body="&lt;p&gt;I've created a JavaScript-based fractal drawer which you can see here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://jsfiddle.net/xfF3f/12/&quot; rel=&quot;nofollow&quot;&gt;http://jsfiddle.net/xfF3f/12/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you're probably all aware, a Mandelbrot Set is created by iterating over pixels as though they were coordinates on the real and imaginary planes. Each pixel then has a real and an imaginary part (like all complex numbers) which can be fed into an iterative loop of its own:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ z = z^2 + c $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where both &lt;code&gt;z&lt;/code&gt; and &lt;code&gt;c&lt;/code&gt; are complex numbers, &lt;code&gt;z&lt;/code&gt; starts at &lt;code&gt;0&lt;/code&gt;, and &lt;code&gt;c&lt;/code&gt; is the value of our pixel. If you run this a bunch of times, the squared &lt;a href=&quot;http://tutorial.math.lamar.edu/Extras/ComplexPrimer/ConjugateModulus.aspx&quot; rel=&quot;nofollow&quot;&gt;modulus&lt;/a&gt; of &lt;code&gt;z&lt;/code&gt; (&lt;code&gt;|z|&lt;/code&gt;) will either stay below a limit you set (6 in my fractal drawer) or it will go above the limit. If it goes above the limit, you break the iteration and consider that pixel to be outside of the set. When this happens, you color that pixel a certain color depending on how big &lt;code&gt;|z|&lt;/code&gt; has gotten and how many iterations it took to determine that it's outset the set. (See more on the formula used in my drawer here: &lt;a href=&quot;http://math.stackexchange.com/questions/4035/continuous-coloring-of-a-mandelbrot-fractal&quot;&gt;Continuous coloring of a Mandelbrot fractal&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So given all this, we can say that a higher max iteration value will tell you with greater precision whether or not a point is in the set. It will also take more time to run because it's doing more calculations per pixel. There are also other visual factors...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you start at a full zoom out and run the plot with a maxIterations value of 50, 100, and 300, you get this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/VkwU3.png&quot; alt=&quot;mandelbrot full zoom out with max iteration values of 50, 100, and 300&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you can see that while the detail of the edge of the set does get better as you increase the maxIterations value, the pixels outside the set are almost all red. At this zoom level, I'd say something like 50 iterations would be an ideal balance of color variation and edge detail.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now if you zoom in to some arbitrary level keeping 50 maxIterations, you will begin to see something like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/KeXpN.png&quot; alt=&quot;arbitrarily zoomed set at 50 maxiterations&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The detail is horrible and the colors are also a bit homogeneous. So let's see what happens if we keep the same zoom level and change the maxIterations number to 80, 120, 250, 500, 1000, and 2000 (remember, the coordinates and zoom are &lt;strong&gt;exactly&lt;/strong&gt; the same in all images, the only difference is the maxIterations value):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/C8QDW.png&quot; alt=&quot;zoomed in changing only the number of maxIterations&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As usual, increasing the maxIterations value too much leaves most of the points outside of the set red. Here I'd say something between the second (120 maxIterations) and third (250 maxIterations) is more or less ideal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is all relatively simple to do one image at a time with your eye and some tinkering, but this would be very difficult to do if I were to create a zoom like this: &lt;a href=&quot;http://vimeo.com/1908224&quot; rel=&quot;nofollow&quot;&gt;http://vimeo.com/1908224&lt;/a&gt;. I'd need some method of finding something like an ideal maxIterations value depending on the zoom level.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So after all of this, my question is: is there some such method? If not, where might I start to look in order to figure this out for myself? Am I thinking about this wrong? Is there a more obvious solution that I'm missing?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance!&lt;/p&gt;&#xA;" OwnerUserId="1720" LastActivityDate="2011-04-02T21:03:38.653" Title="A way to determine the ideal number of maximum iterations for an arbitrary zoom level in a Mandelbrot fractal" Tags="&lt;geometry&gt;&lt;algebra&gt;&lt;fractals&gt;&lt;graph&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="16971" PostTypeId="1" AcceptedAnswerId="16972" CreationDate="2011-01-10T08:31:14.957" Score="1" ViewCount="254" Body="&lt;p&gt;Can you please give me examples of real analytic functions with finite number of zeros that are not polynomials? I know that there are $\mathrm{e}^x$ and $\operatorname{sh}x$, but I can't think of anything else.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It would be even better if there was a method of constructing a large class of such functions.&lt;/p&gt;&#xA;" OwnerUserId="3793" LastActivityDate="2011-01-10T18:44:01.457" Title="Examples of real analytic functions with finite number of zeros that are not polynomials" Tags="&lt;real-analysis&gt;" AnswerCount="3" />
  <row Id="16972" PostTypeId="2" ParentId="16971" CreationDate="2011-01-10T08:38:32.410" Score="4" ViewCount="" Body="&lt;ol&gt;&#xA;&lt;li&gt;$e^x + $ any polynomial. &lt;/li&gt;&#xA;&lt;li&gt;Any positive power series that whose coefficients decay even faster than the Taylor series of $e^x$.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;You can concoct numerous other examples by combining other analytic functions and perturbing their coefficients. Perhaps you mean to ask something a little different...&lt;/p&gt;&#xA;" OwnerUserId="308" LastActivityDate="2011-01-10T08:38:32.410" CommentCount="3" />
  <row Id="16975" PostTypeId="1" AcceptedAnswerId="16976" CreationDate="2011-01-10T10:18:27.020" Score="0" ViewCount="101" Body="&lt;p&gt;I am trying to solve this problem, and i think i did something, but finally i couldn't get the conclusion. The question is:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Let $(X,d)$ be a metric space and let $A,B \subset X$. If $A$ is an open dense subset, and $B$ is a dense subset, then is $A \cap B$ dense in $X$?.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Well, i think this is true. We have to show that $\overline{A \cap B}=X$. Or in other words, $B(x,r) \cap (A \cap B) \neq \emptyset$ for any $x \in X$. Since $A$ is dense in $X$, so we have $B(x,r) \cap A \neq\emptyset$. That means there is a $y \in B(x,r) \cap A$. Which means there is a $y \in A$. And since $A$ is open we have $B(y,r_{1}) \subseteq A$ for some $r_{1} &amp;gt; 0$. Couldn't get any further. I did try some more from here on, but couldn't get it. Any idea for proving it or giving a counter example.&lt;/p&gt;&#xA;" LastActivityDate="2011-01-10T10:31:01.237" Title="$A,B \subset (X,d)$ and $A$ is open dense subset, $B$ is dense then is $A \cap B$ dense?" Tags="&lt;real-analysis&gt;&lt;analysis&gt;&lt;metric-spaces&gt;&lt;metric&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="16976" PostTypeId="2" ParentId="16975" CreationDate="2011-01-10T10:31:01.237" Score="1" ViewCount="" Body="&lt;p&gt;Good start. Now use that $B(x,r) \cap B(y,r_1)$ is an open set (and nonempty since it contains $y$) and therefore contains some point of $B$.&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2011-01-10T10:31:01.237" CommentCount="1" />
  <row Id="16977" PostTypeId="2" ParentId="13861" CreationDate="2011-01-10T11:18:47.950" Score="1" ViewCount="" Body="&lt;p&gt;From the principle of the &quot;jacobi-rotation&quot; method for obtaining the principal components it is clear, that some ellipsoid (based on the data) is defined by the following idea:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;twodimensional case:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;rotate the cloud of datapoints such that the sum-of-squared x-coordinates (&quot;ssq(x)&quot;) is maximum&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;the sum-of-squared y-coordinates (&quot;ssq(y)&quot;) is then minimal.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;multidimensional case:   &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;you do 1) with all pairs of axes x-y, x-z, x-w,... This gives a &quot;temporary&quot; maximum of ssq(x)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;then you do 1) with all pairs y-z,y-w,... to obtain a &quot;temporary&quot; maximum of ssq(y) and so on with z-w,....&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Then you repeat the whole process until convergence &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;In principle this only defines the &lt;em&gt;direction&lt;/em&gt; of the axes of some ellipsoid. Then, for instance in the monography &quot;Faktorenanalyse&quot; (&quot;factoranalysis&quot;) of K. Ueberla, the size of the ellipsoid is defined by: &lt;em&gt;&quot;length of the axes are the roots of ssq(axis)&quot;&lt;/em&gt; (paraphrased) &#xA;Here no direct reference is made to the property of radial deviation (when representated in terms of polar-coordinates).     &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, the concept of minimizing the radial distances is different by a simple consideration.&lt;br&gt;&#xA;If we have -for instance- only three datapoints which are nearly on one straight line, then the center of the best-fitting-circle tends to infinity and not to the mean. The same may occur with some cloud of datapoints whose overall shape is that of a small but sufficient segment of a circumference of an ellipse. Well, we may &lt;em&gt;force&lt;/em&gt; the origin into the arithmetic center of the cloud - but this introduces a likely artificial restriction. Actually, if you search for &quot;circular regression&quot; you'll arrive at nonlinear, iterative methods much different of the method of principal components.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[update] Here I show the difference between the two concepts in a two-dimensional example. I generated slightly correlated random-data. Rotated to pca-position, such that the x-axis of the plot is the first princ.comp. (eigenvector) and so on.    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The blue ellipse is that ellipse using the axis-lengthes of stddev of the principal components.    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then I computed another ellipse by minimizing the sum-of-squares of radial distances of datapoints to the ellipses circumference (just manual checking), this is the red ellipse.&#xA;Here is the plot:    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/iNBut.png&quot; alt=&quot;ellipsefitting&quot;&gt;     &lt;/p&gt;&#xA;&#xA;&lt;p&gt;[end update]&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;One more comment for the geometric visualization (it is just for the two-dimensional case).    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the cloud of data as endpoints of vectors from the origin. &#xA;Then the common concept of the resultant is the mean of the arithmetic/geometric sum of all vectors. This is also called &quot;centroid&quot; and in early times of factor-analysis, without the help of computers, it was often an accepted approximation for the &quot;central tendency&quot;: the mean of the coordinates are the best fit in the sense of least &quot;squares-of-distances&quot; .    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if some vectors have &quot;opposing&quot; (better word?) direction, their influence neutralizes, although they would nicely define an axis. In the old centroid-method such vectors were &quot;inflected&quot; (sign changed) to correct for that effect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another option is to &lt;em&gt;double the angle of each vector&lt;/em&gt; - then opposing vectors overlay and strengthen each others influence instead of neutralizing. Then take the mean of the new vectors and half its angle. This is then the principal component/axis and thus represents another least-squares estimation for the fit. (The computation of the rotation-criterion agrees perfectly to the minimizing formula of the Jacobi-rotation for PCA)&lt;/p&gt;&#xA;" OwnerUserId="1714" LastEditorUserId="1714" LastEditorDisplayName="" LastEditDate="2011-01-10T13:09:54.847" LastActivityDate="2011-01-10T13:09:54.847" CommentCount="1" />
  <row Id="16978" PostTypeId="2" ParentId="16969" CreationDate="2011-01-10T11:37:03.850" Score="0" ViewCount="" Body="&lt;p&gt;If you have already shown that $W_t$ and $\int_0^tW_sds$ are normal, find their covariance and use &lt;a href=&quot;http://en.wikipedia.org/wiki/Moment-generating_function&quot; rel=&quot;nofollow&quot;&gt;moment generating function&lt;/a&gt; of &lt;a href=&quot;http://en.wikipedia.org/wiki/Bivariate_normal_distribution&quot; rel=&quot;nofollow&quot;&gt;bivariate normal distribution&lt;/a&gt; to solve the second part of your problem.&lt;/p&gt;&#xA;" OwnerUserId="4742" LastActivityDate="2011-01-10T11:37:03.850" />
  <row Id="16979" PostTypeId="1" CreationDate="2011-01-10T11:49:39.683" Score="1" ViewCount="241" Body="&lt;p&gt;Is the P versus NP question asking &quot;P = NP&quot; or &quot;ZFC |- P = NP&quot; (or &quot;|- P = NP&quot; for that matter)? Because if I say P = NP, then I will be asked to prove it. But if the goal is &quot;ZFC |- P = NP&quot; then the result will not be useful because of the set theoretic assumptions of ZFC. So you may say the third choice matches our intuition, but it's not a (complete) question. If P = NP, then we are asked to prove |- P = NP and if P != NP we are not asked ~(|- P = NP) but |- P != NP. So what is the P versus NP question asking?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Actually this question can be asked for any question, it's not about P versus NP alone.&lt;/p&gt;&#xA;" OwnerUserId="5079" LastActivityDate="2011-01-15T00:10:32.880" Title="What is the P versus NP question asking?" Tags="&lt;logic&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="1" />
  <row Id="16980" PostTypeId="1" AcceptedAnswerId="16984" CreationDate="2011-01-10T11:52:51.787" Score="0" ViewCount="86" Body="&lt;p&gt;Suppose $f$ is increasing and there exist numbers $A$ and $B$ such that for all $x$, $A \leq f(x) \leq B$. Show that for each $\epsilon &amp;gt;0$ the number of jumps of size exceeding $\epsilon$ is at most $(B-A)/\epsilon$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $\epsilon &amp;gt;0$. So the jump size is $f(x+)-f(x-)$. So it seems that we divide the maximum possible height $B-A$ by $\epsilon$ to get the maximum number of jumps exceeding $\epsilon$ (e.g. dividing a large number by small number). Is this correct? &lt;/p&gt;&#xA;" OwnerUserId="2548" LastEditorUserId="2548" LastEditorDisplayName="" LastEditDate="2011-01-10T11:59:54.930" LastActivityDate="2011-01-10T14:59:23.540" Title="Number of Jumps" Tags="&lt;real-analysis&gt;" AnswerCount="2" />
  <row Id="16981" PostTypeId="2" ParentId="16979" CreationDate="2011-01-10T12:07:22.387" Score="5" ViewCount="" Body="&lt;p&gt;For the $1 million P vs NP prize, the Clay Mathematics Institute problem description does not even talk about proofs explicitly:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Problem Statement. Does P = NP?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.claymath.org/millennium/P_vs_NP/pvsnp.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.claymath.org/millennium/P_vs_NP/pvsnp.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A proof is any completely convincing argument; set-theoretic foundations are just one tool that helps us study mathematical objects.&lt;/p&gt;&#xA;" OwnerUserId="3643" LastActivityDate="2011-01-10T12:07:22.387" CommentCount="3" />
  <row Id="16982" PostTypeId="1" AcceptedAnswerId="17053" CreationDate="2011-01-10T12:07:30.803" Score="7" ViewCount="189" Body="&lt;p&gt;Dear community,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;one can obtain solutions to the &lt;a href=&quot;http://en.wikipedia.org/wiki/Laplace%27s_equation&quot; rel=&quot;nofollow&quot;&gt;Laplace equation&lt;/a&gt;&#xA;$$\Delta\psi(x) = 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or even for the &lt;a href=&quot;http://en.wikipedia.org/wiki/Poisson%27s_equation&quot; rel=&quot;nofollow&quot;&gt;Poisson equation&lt;/a&gt; $\Delta\psi(x)=\varphi(x)$ in a &lt;a href=&quot;http://en.wikipedia.org/wiki/Dirichlet_problem&quot; rel=&quot;nofollow&quot;&gt;Dirichlet boundary value problem&lt;/a&gt; using a random-walk approach, see e.g. &lt;a href=&quot;http://www.maths.leeds.ac.uk/~veretenn/bmp%28w%29.pdf&quot; rel=&quot;nofollow&quot;&gt;Introduction to Brownian Motion&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, already this fascinating concept is not really clear to me. Nevertheless, it would be worth to get into details if there also exists such a connection to the &lt;a href=&quot;http://en.wikipedia.org/wiki/Helmholtz_equation&quot; rel=&quot;nofollow&quot;&gt;Helmholtz equation&lt;/a&gt;&#xA;$$\Delta\psi(x) +k^2\psi(x)= 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence my question:&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Can we use some random walk to calculate solutions of the Helmholtz equation numerically?&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;It would also be interesting if this is still true for &lt;strong&gt;open systems&lt;/strong&gt; where the boundary conditions are different than in the Dirichlet case and for which $k$ is now a domainwise constant function.&lt;br&gt;&#xA;Thank you in advance&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Robert&lt;/p&gt;&#xA;" OwnerUserId="2671" LastEditorUserId="2671" LastEditorDisplayName="" LastEditDate="2011-01-10T19:06:07.637" LastActivityDate="2011-01-14T22:41:12.147" Title="Relation of Brownian Motion to Helmholtz Equation" Tags="&lt;differential-equations&gt;&lt;numerical-methods&gt;&lt;stochastic-processes&gt;" AnswerCount="2" CommentCount="5" FavoriteCount="4" />
  <row Id="16983" PostTypeId="1" CreationDate="2011-01-10T12:11:40.280" Score="1" ViewCount="172" Body="&lt;p&gt;Why is Godel's first theorem not a proof for the truth of the so called undecidable proposition? You may say it's a proof from the outside, but if not all proofs from the outside be formalized inside the system then the system is really not powerful.&lt;/p&gt;&#xA;" OwnerUserId="5079" LastActivityDate="2011-04-27T04:34:35.770" Title="Why is Godel's first theorem not a proof from the inside?" Tags="&lt;logic&gt;&lt;incompleteness&gt;" AnswerCount="1" CommentCount="6" />
  <row Id="16984" PostTypeId="2" ParentId="16980" CreationDate="2011-01-10T12:29:24.353" Score="4" ViewCount="" Body="&lt;p&gt;Since $f$ is increasing, you can't &quot;undo&quot; a jump upwards. Therefore $n$ jumps of size at least $\epsilon$ contribute an increase of at least $n\epsilon$, and this has to be less than the total increase $B-A$.&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2011-01-10T12:29:24.353" />
  <row Id="16985" PostTypeId="1" AcceptedAnswerId="16986" CreationDate="2011-01-10T13:26:48.717" Score="3" ViewCount="339" Body="&lt;p&gt;Whenever i am not doing anything, i generally, happen to see pages of some good Mathematical Institutes in India, so as to know more about the faculty members and see what they are working on. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;While doing so, in one of the faculty webpage he has written this statement.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The prime number theorem has an one-line proof : The &lt;em&gt;Riemann zeta function&lt;/em&gt; does not vanish on the one line &quot;real part of $s$ equals 1&quot;. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Can anyone explain more about this proof. Is he assuming the &lt;em&gt;Riemann Hypothesis&lt;/em&gt;. I am curious to know.  &lt;/p&gt;&#xA;" LastActivityDate="2011-01-10T17:33:37.490" Title="One line Proof of the Prime Number Theorem" Tags="&lt;number-theory&gt;&lt;analytic-number-theory&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" />
  <row Id="16986" PostTypeId="2" ParentId="16985" CreationDate="2011-01-10T14:08:46.950" Score="7" ViewCount="" Body="&lt;p&gt;The assertion you made about $\zeta(s)$ does not assume the Riemann hypothesis and can, in fact, be proved reasonably quickly using reasonably elementary manipulations of trigonometric functions.  The actual deduction of the prime number theorem from this is not trivial, and can be done (&lt;strong&gt;edit:&lt;/strong&gt; as Matt E observes below, this is not the only -- and, indeed, not the original -- way)  by some sort of &quot;Tauberian&quot;  theorem: in these &lt;a href=&quot;http://www.math.uiuc.edu/~r-ash/CV/CV7.pdf&quot; rel=&quot;nofollow&quot;&gt;notes&lt;/a&gt; by Ash (which are based off a short paper by Donald Newman), the argument is to express $\zeta&amp;#39;(s)/\zeta(s)$ as the &quot;Mellin transform&quot; (basically, a variant of the Laplace transform) of partial sums of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Von_Mangoldt_function&quot; rel=&quot;nofollow&quot;&gt;von Mangoldt function&lt;/a&gt;, denoted $\psi$. The point of the &quot;Tauberian&quot; theorem is that you can deduce properties about the original function from the Mellin transform: since we have some information about the Mellin transform (namely, because $\zeta$ is zero-free on the line $Re(s)=1$), we can deduce some information about $\psi$. This is enough to imply the prime number theorem.&lt;/p&gt;&#xA;" OwnerUserId="536" LastEditorUserId="536" LastEditorDisplayName="" LastEditDate="2011-01-10T17:33:37.490" LastActivityDate="2011-01-10T17:33:37.490" CommentCount="3" />
  <row Id="16987" PostTypeId="2" ParentId="13414" CreationDate="2011-01-10T14:18:49.067" Score="3" ViewCount="" Body="&lt;p&gt;Moron's solution is nice. &#xA;The method of residue is standard in complex function theory.&#xA;Here it is a simple elementary derivation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We may assume that $a\ge 0$.&#xA;$$&#xD;&#xA;\int_0^\infty \frac{x^2\,dx}{(x^2-a^2)^2 + x^2}=\int_0^\infty \frac{1}{1+\left( x-\frac{a^2}{x} \right)^2}\,dx.&#xD;&#xA;$$&#xA;If we had &#xA;$$&#xD;&#xA;\int_0^\infty \frac{1}{1+t^2}\,dt&#xD;&#xA;$$&#xA;then we could calculate it easily.&#xA;This motivates the substitution&#xA;$$&#xD;&#xA;x-\frac{a^2}{x}=:t\, \qquad(1).&#xD;&#xA;$$&#xA;Here&#xA;$$&#xD;&#xA;D_x\left( x-\frac{a^2}{x} \right)=1+\frac{a^2}{x^2}\gt 0, \qquad (x\gt 0).&#xD;&#xA;$$&#xA;From $(1)$ we obtain&#xA;$$&#xD;&#xA;x=\frac{t}{2}+\frac{1}{2}\sqrt{t^2+4 a^2}&#xD;&#xA;$$&#xA;because $x\gt0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From this&#xA;$$&#xD;&#xA;dx=\left( \frac{1}{2}+\frac{1}{2}\cdot\frac{t}{\sqrt{t^2+4a^2}} \right)\,dt.&#xD;&#xA;$$&#xA;Substituting back into the integral we get&#xA;$$&#xD;&#xA;\int_0^\infty \frac{1}{1+\left( x-\frac{a^2}{x} \right)^2}\,dx=&#xD;&#xA;\int_{-\infty}^\infty \left(\frac{1}{2}+\frac{1}{2}\cdot\frac{t}{\sqrt{t^2+4a^2}}\right)\frac{1}{1+t^2}    \,dt&#xD;&#xA;$$&#xA;Here the second integrand is an odd function so the result is&#xA;$$&#xD;&#xA;\int_{-\infty}^\infty \frac{1}{2}\cdot\frac{1}{1+t^2}\,dt=\frac{\pi}{2}.&#xD;&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="5418" LastActivityDate="2011-01-10T14:18:49.067" />
  <row Id="16988" PostTypeId="1" CreationDate="2011-01-10T14:25:24.757" Score="2" ViewCount="103" Body="&lt;p&gt;Dear All&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In &lt;a href=&quot;http://math.stackexchange.com/questions/2735/solving-very-large-matrices-in-pieces&quot;&gt;Solving very large matrices in &quot;pieces&quot;&lt;/a&gt; there is a way shown to solve matrix inversion in pieces. Is it possible to turn it into a true divide and conquer algorithm?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am refering to the answer in the referenced question that starts with:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\textbf{A}=\begin{pmatrix}\textbf{E}&amp;amp;\textbf{F}\\\\ \textbf{G}&amp;amp;\textbf{H}\end{pmatrix}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And computes the inverse via:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\textbf{A}^{-1}=\begin{pmatrix}\textbf{E}^{-1}+\textbf{E}^{-1}\textbf{F}\textbf{S}^{-1}\textbf{G}\textbf{E}^{-1}&amp;amp;-\textbf{E}^{-1}\textbf{F}\textbf{S}^{-1}\\\\ -\textbf{S}^{-1}\textbf{G}\textbf{E}^{-1}&amp;amp;\textbf{S}^{-1}\end{pmatrix}$, where $\textbf{S} = \textbf{H}-\textbf{G}\textbf{E}^{-1}\textbf{F}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this algorithm I will have some steps:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;...&lt;br&gt;&#xA;   sequential&lt;br&gt;&#xA;    1. Compute inverse $\textbf{E}^{-1}$&lt;br&gt;&#xA;    2. Compute inverse $(\textbf{H}-\textbf{G}\textbf{E}^{-1}\textbf{F})^{-1}$&lt;br&gt;&#xA;   ...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would be possible to device an algorithm that has steps:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;...&lt;br&gt;&#xA;   parallel&lt;br&gt;&#xA;    1. Compute inverse $\textbf{X}^{-1}$&lt;br&gt;&#xA;    1. Compute inverse $\textbf{Y}^{-1}$&lt;br&gt;&#xA;   ...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i.e. to identify two smaller matrices $\textbf{X}$ and $\textbf{Y}$ which are independent from their inverse?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Best Regards&lt;/p&gt;&#xA;" OwnerUserId="4414" LastActivityDate="2011-01-10T14:25:24.757" Title="True divide and conquer inversion of large matrices" Tags="&lt;linear-algebra&gt;&lt;algorithms&gt;&lt;matrices&gt;" CommentCount="5" />
  <row Id="16989" PostTypeId="1" AcceptedAnswerId="17039" CreationDate="2011-01-10T14:39:30.667" Score="2" ViewCount="258" Body="&lt;p&gt;Let X be the quotient space of $S^2$ under the identifications x~-x for x in the equator $S^1$. Compute the homology groups $H_i(X)$. Do the same for $S^3$ with antipodal points of the equator $S^2 \subset S^3$ identified.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is probably related to cellular homology. Thanks.&lt;/p&gt;&#xA;" OwnerUserId="5573" LastActivityDate="2011-01-10T21:20:19.100" Title="Homology Group of Quotient Space" Tags="&lt;algebraic-topology&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="16990" PostTypeId="2" ParentId="16963" CreationDate="2011-01-10T14:49:28.520" Score="4" ViewCount="" Body="&lt;p&gt;The short version is that we use norms to help us study convergence in &lt;a href=&quot;http://en.wikipedia.org/wiki/Function_space&quot; rel=&quot;nofollow&quot;&gt;function spaces&lt;/a&gt;; this is the huge subject of &lt;a href=&quot;http://en.wikipedia.org/wiki/Functional_analysis&quot; rel=&quot;nofollow&quot;&gt;functional analysis&lt;/a&gt;, which has applications everywhere in mathematics (PDEs, mathematical physics, probability, representation theory...).  I recommend Dieudonne's &lt;a href=&quot;http://books.google.com/books?id=mg7r4acKgq0C&amp;amp;printsec=frontcover&amp;amp;dq=dieudonne+functional+analysis&quot; rel=&quot;nofollow&quot;&gt;History of Functional Analysis&lt;/a&gt; if you are really interested in the full story.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2011-01-10T14:49:28.520" />
  <row Id="16991" PostTypeId="2" ParentId="16958" CreationDate="2011-01-10T14:57:04.843" Score="2" ViewCount="" Body="&lt;p&gt;Look it up at the &lt;a href=&quot;http://oeis.org/&quot; rel=&quot;nofollow&quot;&gt;OEIS&lt;/a&gt;.  If nothing else, you'll gain a new appreciation for how many terms you need to make a sequence unique (as far as the OEIS is concerned).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the way, many sequences that count lengths of words that you'll run into as a computer programmer are going to be &lt;a href=&quot;http://en.wikipedia.org/wiki/Regular_language&quot; rel=&quot;nofollow&quot;&gt;regular languages&lt;/a&gt;, that is, they are described by a regular expression (equivalently, recognized by a finite state machine).  There is a very well-understood theory of how to count these lengths; I don't think it is covered in computer science textbooks (for example it's not covered in Sipser), but there is some information in the relevant section of Stanley's &lt;a href=&quot;http://www-math.mit.edu/~rstan/ec/&quot; rel=&quot;nofollow&quot;&gt;Enumerative Combinatorics&lt;/a&gt; (rational generating functions), as well as Flajolet and Sedgewick's &lt;a href=&quot;http://algo.inria.fr/flajolet/Publications/books.html&quot; rel=&quot;nofollow&quot;&gt;Analytic Combinatorics&lt;/a&gt;.  I also briefly describe how this works in a few special cases in my &lt;a href=&quot;http://web.mit.edu/~qchu/Public/TopicsInGF.pdf&quot; rel=&quot;nofollow&quot;&gt;notes on generating functions&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, the approach I'm advocating is to &lt;em&gt;figure out&lt;/em&gt; what the sequence is instead of trying to &lt;em&gt;guess&lt;/em&gt; it.  In your example, can you prove that $f(n) = 2^n$?&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2011-01-10T15:07:49.400" LastActivityDate="2011-01-10T15:07:49.400" CommentCount="1" />
  <row Id="16992" PostTypeId="2" ParentId="16980" CreationDate="2011-01-10T14:59:23.540" Score="0" ViewCount="" Body="&lt;p&gt;This question is closely related to &lt;a href=&quot;http://math.stackexchange.com/questions/8979/inequality-on-an-increasing-function&quot;&gt;this one&lt;/a&gt;; in particular, if $x_k \in (a,b)$, $k=1,\ldots,n$, are such that $f(x_k+)-f(x_k-) &amp;gt; \varepsilon$, we have from the latter that&#xA;$$&#xD;&#xA;n \varepsilon &amp;lt; \sum\limits_{k = 1}^n {[f(x_k  + ) - f(x_k  - )]}  \le f(b - ) - f(a + ) \le B - A,&#xD;&#xA;$$ &#xA;hence $n &amp;lt; (B-A) / \varepsilon$.&lt;/p&gt;&#xA;" OwnerUserId="2810" LastActivityDate="2011-01-10T14:59:23.540" />
  <row Id="16993" PostTypeId="1" CreationDate="2011-01-10T14:59:39.590" Score="1" ViewCount="48" Body="&lt;p&gt;I am looking for a way of getting some intuition for some Convection Diffusion Equation. I would like to play around with it, simulate it, change its IC/BCs, having some interactive graphs and stuff.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do you know of any free software that does the job? Thank you!&lt;/p&gt;&#xA;" OwnerUserId="346" LastActivityDate="2011-01-10T14:59:39.590" Title="Software for simulating (partial) differential equations" Tags="&lt;differential-equations&gt;&lt;math-software&gt;" FavoriteCount="1" />
  <row Id="16994" PostTypeId="2" ParentId="16963" CreationDate="2011-01-10T15:14:24.697" Score="1" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://math.stackexchange.com/questions/16522/seminorms-and-norms&quot;&gt;This&lt;/a&gt; might be helpful. &lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2011-01-10T15:14:24.697" />
  <row Id="16995" PostTypeId="1" AcceptedAnswerId="16999" CreationDate="2011-01-10T15:26:21.063" Score="4" ViewCount="193" Body="&lt;p&gt;So I want to find all antiderivaties of&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{x}{x^3-1}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the denominator is of a lesser degree than the numerator, partial fraction is to be used instead of long division.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've started by doing:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{x}{x^3-1} = \frac{A}{x^2+x+1} + \frac{B}{x-1}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x = A(x-1)+B(x^2+x+1)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then setting $x = 1$ to get $3B = 1 =&amp;gt; B = \frac{1}{3}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I have $\int({\frac{A}{x^2+x+1}}+{\frac{1}{3(x-1)}})$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I'm not sure how to continue from here.&lt;/p&gt;&#xA;" OwnerUserId="5574" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-10T17:45:41.747" LastActivityDate="2011-01-10T18:00:03.850" Title="Partial fraction integration" Tags="&lt;calculus&gt;&lt;integral&gt;" AnswerCount="2" CommentCount="12" />
  <row Id="16996" PostTypeId="1" CreationDate="2011-01-10T15:26:45.103" Score="1" ViewCount="97" Body="&lt;p&gt;My pdf for $X$ is&#xA;$$f_X ( x ; \theta ) = (1/\theta) x \exp (-x/\theta).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am unsure as to how to show that the mean value of $X$ is $\theta$?&lt;/p&gt;&#xA;" OwnerUserId="5324" LastEditorUserId="3249" LastEditorDisplayName="" LastEditDate="2011-01-10T15:48:16.573" LastActivityDate="2011-01-10T16:29:36.393" Title="How to show the mean value of a pdf?" Tags="&lt;statistics&gt;" AnswerCount="3" />
  <row Id="16997" PostTypeId="2" ParentId="16996" CreationDate="2011-01-10T15:30:46.257" Score="0" ViewCount="" Body="&lt;p&gt;Well, given a continuous pdf, how do you find the expected value (aka the mean)?&lt;/p&gt;&#xA;" OwnerUserId="5137" LastActivityDate="2011-01-10T15:30:46.257" CommentCount="1" />
  <row Id="16998" PostTypeId="2" ParentId="16971" CreationDate="2011-01-10T15:33:55.767" Score="1" ViewCount="" Body="&lt;p&gt;If I remember correctly, an entire function i.e. function analytic everywhere, has exactly finitely many zeros if and only if it is of the form&#xA;$$ p(z)e^{g(z)}$$ where $p(z)$ is a polynomial and  $g(z)$ is entire.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, $p(z)=A(z-z_1)^a(z-z_2)^b\cdots (z-z_n)^c$ for some constant $A$, if $z_1,z_2,\ldots, z_n$ are zeros of the function with orders $a,b,\ldots, c$ respectively.&lt;/p&gt;&#xA;" OwnerUserId="3249" LastEditorUserId="3249" LastEditorDisplayName="" LastEditDate="2011-01-10T15:43:23.943" LastActivityDate="2011-01-10T15:43:23.943" />
  <row Id="16999" PostTypeId="2" ParentId="16995" CreationDate="2011-01-10T15:41:45.857" Score="7" ViewCount="" Body="&lt;p&gt;You can do partial fractions in this way: $$\frac{x}{x^{3}-1} = \frac{Ax+B}{x^{2}+x+1} + \frac{C}{x-1}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now by equating the coefficients you get:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;$A+C=0$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$B+C-A=1$ and;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$C-B=0$. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Solving you get the values of $A= -\frac{1}{3}$, $B=\frac{1}{3}$ and $C=\frac{1}{3}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the required integral breaks down as $$ \int\frac{x}{x^{3}-1} \ dx = -\frac{1}{3}\int \frac{x -1}{x^{2}+x+1} \ dx + \frac{1}{3} \int \frac{1}{x-1} \ dx$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now $$\int\frac{x-1}{x^{2}+x+1} \ dx = \frac{1}{2} \int \frac{2x-2}{x^{2}+x+1} \ dx =\frac{1}{2}\int\frac{2x+1}{x^{2}+x+1}  -\frac{1}{2} \int \frac{3}{x^{2}+x+1} \ dx$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first part of the integral is easy. For the second part use $$x^{2}+x+1 = \Bigl(x+\frac{1}{2}\Bigr)^{2} + \Bigl(\frac{\sqrt{3}}{2}\Bigr)^{2}$$ and then put $t=x+\frac{1}{2}$ and then the trignometric substitution.&lt;/p&gt;&#xA;" OwnerUserId="9413" LastEditorUserId="9413" LastEditorDisplayName="anonymous" LastEditDate="2011-01-10T18:00:03.850" LastActivityDate="2011-01-10T18:00:03.850" CommentCount="2" />
  <row Id="17000" PostTypeId="2" ParentId="16996" CreationDate="2011-01-10T15:52:56.193" Score="1" ViewCount="" Body="&lt;p&gt;Actually, there is a small error in your expression for the pdf (since it does not integrate to $1$); it should be the pdf of a Gamma$(2,\theta)$ random variable, meaning that &#xA;$$&#xD;&#xA;f(x;\theta ) = \frac{1}{{\theta ^2 }}xe^{ - x/\theta }, \;\; x&amp;gt;0. &#xD;&#xA;$$&#xA;Now, the corresponding mean value is&#xA;$$&#xD;&#xA;\int_0^\infty  {xf(x;\theta )\,dx}  = 2 \theta&#xD;&#xA;$$&#xA;(which can be calculated using integration by parts).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Since a Gamma$(2,\theta)$ random variable is distributed as the sum of two (independnet) exponential$(\theta)$ rv's, having pdf $\theta ^{ - 1} e^{ - x/\theta }$, $x&amp;gt;0$, each, the expectation can be found by $\theta + \theta = 2\theta$.&lt;/p&gt;&#xA;" OwnerUserId="2810" LastEditorUserId="2810" LastEditorDisplayName="" LastEditDate="2011-01-10T16:10:09.963" LastActivityDate="2011-01-10T16:10:09.963" />
  <row Id="17001" PostTypeId="2" ParentId="16788" CreationDate="2011-01-10T16:12:07.370" Score="2" ViewCount="" Body="&lt;p&gt;Your problem can be converted in a &lt;strong&gt;mixed-integer program&lt;/strong&gt; in the following way:&#xA;for each term in the objective function of the form &#xA;$$u\Bigl(\sum_j k_{ij} x_j - c_i\Bigr) ,$$&#xA;introduce an auxiliary &lt;strong&gt;binary&lt;/strong&gt; variable $t_i \in \{0,1\}$ and add the constraint &#xA;$$\sum_j k_{ij} x_j - c_i \le M t_i$$&#xA;where $M$ is a sufficiently large constant (it has to be larger than all possible values for the argument of $u$).&#xA;You can then replace the objective function by the sum of the binary variables, i.e. minimize $\sum_i t_i$ and check that the constraint ensures that the problem is equivalent to the original one (note that this assumes the convention $u(0)=0$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mixed-integer programs are harder to solve than linear programs, but you are guaranteed to find a globally optimal solution exactly.&lt;/p&gt;&#xA;" OwnerUserId="55" LastActivityDate="2011-01-10T16:12:07.370" />
  <row Id="17002" PostTypeId="2" ParentId="16982" CreationDate="2011-01-10T16:13:51.847" Score="5" ViewCount="" Body="&lt;p&gt;It does not seem that anybody has a more complete reply, so I might as well summarize what I have said in the comments. The technique you are refering to in the OP is an application of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Feynman%E2%80%93Kac_formula&quot; rel=&quot;nofollow&quot;&gt;Feynman-Kac formula&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This technique has in fact been applied to solving the Helmholtz equation, as a Google search will reveal:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;http://scitation.aip.org/getabs/servlet/GetabsServlet?prog=normal&amp;amp;id=JASMAN000109000005002260000001&amp;amp;idtype=cvips&amp;amp;gifs=yes&quot; rel=&quot;nofollow&quot;&gt;Probabilistic solutions of the Helmholtz equation&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;http://link.aip.org/link/?JASMAN/114/1733/1&quot; rel=&quot;nofollow&quot;&gt;Random walk approach to wave propagation in wedges and cones&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="3567" LastEditorUserId="3567" LastEditorDisplayName="" LastEditDate="2011-01-10T18:41:15.797" LastActivityDate="2011-01-10T18:41:15.797" CommentCount="1" />
  <row Id="17003" PostTypeId="2" ParentId="16954" CreationDate="2011-01-10T16:21:26.523" Score="3" ViewCount="" Body="&lt;p&gt;I think the maximality of $N$ in $M$ follows from the fact that a sharply 2-transitive Frobenius group $F_p$ of order $p(p-1)$ is maximal in $S_p$, which can be verified using the list of all 2-transitive permutation groups.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $N_0$ be the subgroup of index 2 in $N$ that fixes both orbits of $P$. Then the projection of $N_0$ onto each of these orbits is equal to $F_p$. For any group&#xA;$L$ with $N &amp;lt; L \le M$, the projection of $L_0$ onto these orbits would be larger&#xA;than $F_p$ and hence would equal $S_p$. But now, from the simplicity of $A_p$, we can deduce that $L_0=M_0 = (A_p \times A_p).2$, so $L=M$.&lt;/p&gt;&#xA;" OwnerUserId="2820" LastActivityDate="2011-01-10T16:21:26.523" CommentCount="3" />
  <row Id="17004" PostTypeId="2" ParentId="16996" CreationDate="2011-01-10T16:29:36.393" Score="1" ViewCount="" Body="&lt;p&gt;Actually, since he is supposed to show that the mean is $\theta $, I think the error is the extra x in the function. If the function was $\frac{1}{\theta }{e^{ - \frac{x}{\theta }}}$, you could say that this is exactly the function for an Exponential random variable with $\lambda  = \frac{1}{\theta }$. Thus, since the expectancy of an exponential random variable with parameter $\lambda $ is $\frac{1}{\lambda }$, the expectancy (or mean) is $\frac{1}{{\frac{1}{\theta }}} = \theta $&lt;/p&gt;&#xA;" OwnerUserId="5575" LastActivityDate="2011-01-10T16:29:36.393" />
  <row Id="17005" PostTypeId="1" AcceptedAnswerId="17306" CreationDate="2011-01-10T16:55:46.790" Score="1" ViewCount="98" Body="&lt;pre&gt;&lt;code&gt; - * / 15 - 7 + 1 1 3 + 2 + 1 1 =&#xA; - * / 15 - 7   2   3 + 2 + 1 1 =&#xA; - * / 15     5     3 + 2 + 1 1 =&#xA; - *        3       3 + 2 + 1 1 =&#xA; -          9         + 2 + 1 1 =&#xA; -          9         + 2   2   =&#xA; -          9         4         =&#xA;                5&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I need some guidance. I am trying to approach polish notation. From what I can see here right now is whenever we see two operators together, we could apply the operand recently aforementioned to the two operators right away. This is what I am having in mind. So from above,&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; - * / 15 - 7 + 1 1 3 + 2 + 1 1 =&#xA; - * / 15 - 7   2   3 + 2 + 1 1 =&#xA;&#xA;So + 1 1 = 2, - 7 2 = 5&#xA;&#xA; - * / 15     5     3 + 2 + 1 1 =&#xA; - *        3       3 + 2 + 1 1 =&#xA; -          9         + 2 + 1 1 =&#xA;&#xA;So / 15 5 = 3, * 3 3 = 9, &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I just want to be really sure here. &lt;strong&gt;Is it safe to assume this procedure?&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;starting from the left side of the prefix equation, read the next character one by one. &lt;/li&gt;&#xA;&lt;li&gt;if there are two characters that are operands sequentially, we apply the preceding operator right away. &lt;/li&gt;&#xA;&lt;li&gt;Replace the operation just now with its result. (So + 1 1 becomes 2). Go back to step 2 if there are more possible operation.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Note: Could somebody with higher reputation tag this question for &quot;polish notation&quot;, please?&lt;/p&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="2548" LastEditorDisplayName="" LastEditDate="2011-01-10T17:04:38.897" LastActivityDate="2011-01-12T22:59:18.270" Title="Practicing Polish/Prefix notation" Tags="&lt;polish-notation&gt;" AnswerCount="2" />
  <row Id="17006" PostTypeId="1" AcceptedAnswerId="17008" CreationDate="2011-01-10T17:12:16.917" Score="0" ViewCount="142" Body="&lt;p&gt;In infix notation, (2/2(3+3)) means (2/(2*(3+3)). The multiplication sign is omitted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In polish/prefix notation, is it possible to omit multiplication sign? For instance, is it acceptable to write ( * + 2 3 ( / 12 4)) as ( + 2 3 ( / 12 4))?&lt;/p&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2011-01-10T17:22:44.193" Title="Is it possible to omit multiplication operator in polish/prefix notation?" Tags="&lt;polish-notation&gt;" AnswerCount="1" />
  <row Id="17007" PostTypeId="2" ParentId="16969" CreationDate="2011-01-10T17:16:53.107" Score="2" ViewCount="" Body="&lt;p&gt;Let me expand on mpiktas answer. As you noted, $W_t$ and $\int_0^t {W_s ds}$ are zero mean normal with variance $t$ and $t^3/3$, respectively (the first is trivial). Now, if $(W_t,\int_0^t {W_s ds})$ is bivariate normal, then any linear combination $aW_t + b\int_0^t {W_s ds}$, $a,b \in \mathbb{R}$, is univariate normal. In particular, the sum $W_t + \int_0^t {W_s ds}$ is (zero mean) normal. Since you are given the variance of $W_t$ and $\int_0^t {W_s ds}$, the variance of the sum follows straight from the covariance ${\rm Cov}(W_t,\int_0^t {W_s ds})$, that is from ${\rm E}[W_t \int_0^t {W_s ds}]$. Now, assuming you found that variance, calculating ${\rm E}[\exp (W_t  + \int_0^t {W_s ds} )]$ amounts to calculating the moment-generating function at $1$ of a ${\rm N}(0,\sigma^2)$ variable. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Regarding the first question, let's explain why we should expect $(W_t,\int_0^t {W_s ds})$ to be (bivariate) normal. Since $W$ is continuous on the interval $[0,t]$, $\int_0^t {W_s ds}$ is a standard Riemann integral, and &#xA;$\int_0^t {W_s ds}  \approx \frac{t}{n}\sum\nolimits_{k = 0}^{n - 1} {W_{kt/n} }$. Denote the right-hand side by $Y_t^n$. Then $(W_t,Y_t^n)$ is normal. Indeed, $W$ is a Gaussian process; hence, by definition, $(W_{t_1},\ldots,W_{t_m})$ is normal for any choice of times $t_1,\ldots,t_m$. But $(W_{t_1},\ldots,W_{t_m})$ is normal if and only if any linear combination $\sum\nolimits_{i = 1}^m {a_i W_{t_i } }$, $a_i \in \mathbb{R}$, is (univariate) normal, and the conclusion follows. Finally, since $(W_t,Y_t^n)$ is normal, we should expect that in the limit as $n \to \infty$, $(W_t,\int_0^t {W_s ds})$ is normal.&lt;/p&gt;&#xA;" OwnerUserId="2810" LastEditorUserId="2810" LastEditorDisplayName="" LastEditDate="2011-01-10T21:32:19.060" LastActivityDate="2011-01-10T21:32:19.060" />
  <row Id="17008" PostTypeId="2" ParentId="17006" CreationDate="2011-01-10T17:22:44.193" Score="3" ViewCount="" Body="&lt;p&gt;One of the reasons to have polish notation is to get rid of brackets! Expressions in Polish notation are easy to parse and evaluate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the answer to you question is: No, you cannot get rid of the multiplication by putting in brackets, you are only defeating the purpose of having Polish notation by trying to do that.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2011-01-10T17:22:44.193" />
  <row Id="17009" PostTypeId="1" AcceptedAnswerId="17027" CreationDate="2011-01-10T17:32:39.473" Score="5" ViewCount="195" Body="&lt;p&gt;For example, if we have a conic; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;ax&lt;sup&gt;2&lt;/sup&gt; + 2hxy + by&lt;sup&gt;2&lt;/sup&gt; + 2gx + 2fy + c = 0&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What does homogenising this equation with another line (say ax + by + c = 0 ) actually mean? As in, what are the graphical consequences of this? I've been told that it represents a pair of straight lines passing through the origin and the points of intersection (of the conic and the line), but how did this come about?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update:&lt;/strong&gt; I'm sorry for the lack of context, but I've just started algebraic geometry and I like understanding what things actually mean instead of mugging up a formula.&lt;/p&gt;&#xA;" OwnerUserId="5576" LastEditorUserId="5576" LastEditorDisplayName="" LastEditDate="2011-01-10T18:31:47.883" LastActivityDate="2011-01-11T10:19:01.457" Title="What Does Homogenisation Of An Equation Actually Mean?" Tags="&lt;analytic-geometry&gt;&lt;conic-sections&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="17010" PostTypeId="2" ParentId="17005" CreationDate="2011-01-10T17:33:27.147" Score="1" ViewCount="" Body="&lt;p&gt;The above procedure looks correct. Essentially you work from left to right. &lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2011-01-10T17:33:27.147" />
  <row Id="17012" PostTypeId="2" ParentId="17009" CreationDate="2011-01-10T17:39:41.333" Score="0" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://www.artofproblemsolving.com/Forum/blog.php?b=10559&quot; rel=&quot;nofollow&quot;&gt;This&lt;/a&gt; might help (a post by &lt;a href=&quot;http://math.stackexchange.com/users/232/qiaochu-yuan&quot;&gt;Qiaochu Yuan&lt;/a&gt;). Suppose you have some curve $C$ defined by $x^2+y^2 = 1$ on the plane $z=1$. We want each side of the equation to have the same degree. So we multiply each term on the LHS by $z^0$ and the term on RHS by $z^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Matrix_representation_of_conic_sections&quot; rel=&quot;nofollow&quot;&gt;This&lt;/a&gt; is also a useful reference. &lt;/p&gt;&#xA;" OwnerUserId="2548" LastEditorUserId="2548" LastEditorDisplayName="" LastEditDate="2011-01-10T17:51:25.577" LastActivityDate="2011-01-10T17:51:25.577" CommentCount="1" />
  <row Id="17013" PostTypeId="1" CreationDate="2011-01-10T17:40:33.330" Score="8" ViewCount="193" Body="&lt;p&gt;Here's the setup.  Consider a vector bundle $E$ over a manifold $M$ and let $\Omega^*(M, E)$ denote the space of $E$-valued differential forms (i.e. the space of sections of the vector bundle $\bigwedge^* T^*M \otimes E$).  Let $\nabla$ be an affine connection on $E$, i.e. a map $\Omega^0(M, E) \to \Omega^1(M, E)$ satisfying the Leibniz rule.  For each vector field $X$ there is a contraction operator $\iota(X)$ on $\Omega^*(M,E)$, and we form $\nabla_X = \iota(X) \nabla$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Recall that the curvature of $\nabla$ is defined to be $F(X,Y) = [\nabla_X, \nabla_Y] - \nabla_{[X,Y]}$.  It is well known that $F$ is actually a $End(E)$-valued 2-form.  $\Omega^2(M, End(E))$ acts on $\Omega^*(M, E)$ in the standard way, and it is also well known that $\nabla^2: \Omega^k(M,E) \to \Omega^{k+2}(M, E)$ is given by the action of $F$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is about the proof that &quot;$\nabla^2 = F$&quot; in the sense described above.  The textbook proofs all involve trivializing $E$ and doing a coordinate calculation; I don't mind trivializing $E$, but for my own nefarious (and perhaps misguided) purposes I would like to avoid coordinates.  Here is what I mean.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $E$ is trivial then $\nabla = d + \omega$ where $\omega$ is an $End(E)$-valued 1-form.  So $F(X,Y) = [\iota(X)(d+\omega), \iota(Y)(d + \omega] - \iota([X,Y])(d + \omega)$, and it seems to me that one should be able to carry out the argument using only Cartan's homotopy formulas.  There are complications which arise from the fact that the $d$ operator and the contraction operators are different for $\Omega^*(M, End(E))$ versus $\Omega^*(M,E)$, but nevertheless I can get tantalizingly close.  I can show my progress if it would help you help me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In any event, I've been at it for quite some time now and I've decided it's time to seek assistance.  So I'm wondering if anybody knows either how to do this, a reason why it can't be done, or a place where I could look it up.  Thanks in advance!&lt;/p&gt;&#xA;" OwnerUserId="1509" LastEditorUserId="1509" LastEditorDisplayName="" LastEditDate="2011-01-10T20:07:39.403" LastActivityDate="2011-01-11T13:51:07.623" Title="Coordinate free proof that curvature is the &quot;square&quot; of the connection" Tags="&lt;differential-geometry&gt;&lt;differential-forms&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="2" />
  <row Id="17014" PostTypeId="2" ParentId="16995" CreationDate="2011-01-10T17:41:16.387" Score="7" ViewCount="" Body="&lt;p&gt;Like most techniques of integration, the ideal of partial fractions is to reduce a difficult integral to integrals that are, if not easy, at least doable. The three types of integrals that show up when doing partial fractions are:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;$\displaystyle \int\frac{1}{(ax+b)^n}\,dx$, with $a,b$ constants, $a\neq 0$, and $n\geq 1$. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$\displaystyle \int\frac{x}{(ax^2+bx+c)^n}\,dx$, with $n\geq 1$, $ax^2+bx+c$ irreducible quadratic.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$\displaystyle \int\frac{1}{(ax^2+bx+c)^n}\,dx$ with $n\geq 1$, $ax^2+bx+c$ irreducible quadratic.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The first type is easy to solve: do a substitution $u=ax+b$ and go at it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second type is a bit more involved. By doing a substitution $u=ax^2+bx+c$, we get&#xA;\begin{align*}&#xA;\int\frac{x}{(ax^2+bx+c)^n}\,dx &amp;amp;= \frac{1}{2a}\int \frac{2ax\,dx}{(ax^2+bx+c)^n}\\&#xA; &amp;amp;= \frac{1}{2a}\left(\int\frac{2ax+b}{(ax^2+bx+c)^n}\,dx - \int\frac{b}{(ax^2+bx+c)^n}\,dx\right)\\&#xA;&amp;amp;= \frac{1}{2a}\int\frac{du}{u^n} - \frac{b}{2a}\int\frac{dx}{(ax^2+bx+c)^n}.&#xA;\end{align*}&#xA;The first integral can be done; the second reduces to the third type mentioned above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And so we come down to the third type. When $n=1$, the simplest thing to do is to complete the square; factoring out $a$ we may assume we have $x^2+Bx+C$. Completing the square, you get $(x+\frac{B}{2})^2 + (C - \frac{B^2}{4})$. Because we are assuming that the original quadratic is irreducible, that means that $B^2 - 4C\lt 0$, so that $C-\frac{B^2}{4}\gt 0$. Substituting $u=X+\frac{B}{2}$ turns this into a fraction of the form $\frac{1}{u^2+r^2}$; factor out $r^2$, do another substitution, and you can turn it into a fraction of the form $\frac{1}{w^2+1}$. But $\int\frac{dw}{w^2+1}$ is an easy integral: you have an immediate antiderivative for it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, modulo a bunch of algebra and some substitution, you can solve $\int\frac{1}{ax^2+bx+c}\,dx$ with $ax^2+bx+c$ irreducible quadratic: complete the square, do some substituttions, and turn it into $\int\frac{1}{w^2+1}\,dx$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What if you have $\int\frac{dx}{(ax^2+bx+c)^n}$ with $n\gt 1$, $ax^2+bx+c$ irreducible quadratic? Those are more complicated, but not too bad; you can still complete the square and do a bit of algebra, so that you bring it to the form&#xA;$$\int \frac{du}{(u^2+r^2)^n}$$&#xA;for some positive $r$. Then one can use the reduction formula (obtained by doing integration by parts):&#xA;$$\int\frac{du}{(u^2+r^2)^n} = \frac{1}{2r^2(n-1)}\left(\frac{u}{(u^2+r^2)^{n-1}} + (2n-3)\int\frac{du}{(u^2+r^2)^{n-1}}\right)$$&#xA;and continuing this way you will eventually end in the integral with denominator $u^2+r^2$, which we know how to do.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-10T17:59:46.003" LastActivityDate="2011-01-10T17:59:46.003" />
  <row Id="17015" PostTypeId="1" AcceptedAnswerId="17029" CreationDate="2011-01-10T17:44:11.883" Score="1" ViewCount="112" Body="&lt;p&gt;1)Complete study of the function&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-monotony of the function&#xA;-concavity&#xA;-maximum and minimum&#xA;-Zero of the function&#xA;-if the function is ascending or descending&#xA;-inflection points&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2)make the primitive and find the area “under the curve” in a closed “interval” (break)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-Primitive-make the derivative&#xA;-Calculate the area analytically and numerically in a closed interval&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3)&#xA;-average value [0;1])&#xA;-expected value&#xA;-Variance&#xA;-compare values for the exponential distribuition &lt;/p&gt;&#xA;" OwnerUserId="5578" LastEditorUserId="367" LastEditorDisplayName="" LastEditDate="2011-01-10T18:26:27.570" LastActivityDate="2011-01-10T19:30:50.227" Title="Interesting Study of a function $y = \exp(- x/14) / 14$" Tags="&lt;calculus&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="17016" PostTypeId="1" AcceptedAnswerId="17024" CreationDate="2011-01-10T17:48:51.580" Score="1" ViewCount="222" Body="&lt;p&gt;So $i = \sqrt{-1}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But why is $\sqrt{-x^2}=|x| i$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.google.dk/search?q=square+root+of+-9&quot; rel=&quot;nofollow&quot;&gt;Google&lt;/a&gt; tells me it is.&lt;/p&gt;&#xA;" OwnerUserId="5577" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2011-01-10T18:44:57.973" LastActivityDate="2011-01-10T19:37:02.013" Title="Why is $\sqrt{-x^2}=|x| i$" Tags="&lt;complex-numbers&gt;" AnswerCount="4" CommentCount="4" />
  <row Id="17018" PostTypeId="2" ParentId="14049" CreationDate="2011-01-10T17:52:36.040" Score="0" ViewCount="" Body="&lt;p&gt;You might look to Mandelbrot's own work, a lot of which has to do with boundaries, also the relations of similar shapes on different scales and fractal dimension.&lt;/p&gt;&#xA;" OwnerUserId="1312" LastActivityDate="2011-01-10T17:52:36.040" CommentCount="2" />
  <row Id="17020" PostTypeId="2" ParentId="16971" CreationDate="2011-01-10T18:04:01.117" Score="4" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;: I answered a companion question (for entire functions). I suppose the underlying question (as sort of explained in the comments) is to characterize the real analytic $f$ without zeroes. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Weierstrass_factorization_theorem&quot; rel=&quot;nofollow&quot;&gt;Weierstrass factorization theorem&lt;/a&gt; says the following: Suppose that $f$ is entire. Then $f$ has only countably many zeros (possibly only finitely many), counting multiplicities; if $0$ is a zero of multiplicity $k$, and $z_1,z_2,\dots$ enumerates the distinct non-zero zeros of $f$ (of multiplicities $m_1,m_2,\dots$, respectively), then there is an entire function $g(z)$ and a sequence $p_1,p_2,\dots$ of integers such that &#xA; $$ f(z)=z^k e^{g(z)}\prod_{n\ge 1}E_{p_n}(z/z_n), $$&#xA;where the $E_k(z)$ are the &lt;em&gt;elementary factors&lt;/em&gt;, given by&#xA; $$ E_0(z)=1-z $$ and, for $n&amp;gt;0$, $$E_n(z)=(1-z)exp(z+\frac{z^2}2+\dots+\frac{z^n}n).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(There is also a more general version of this result for functions that are not entire.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that if $f$ has only finitely many zeroes, then it follows that $f$ has the form $p(z)e^{h(z)}$ for some entire $h$ and polynomial $p$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, this indicates the enormous variety of entire functions with given zeroes at prescribed locations, and gives a hint of how to build any. (There are general estimates on the sizes of the elementary factors and general results on convergence of infinite products that can be combined to check in specific cases that a product as above indeed defines an entire function.)&lt;/p&gt;&#xA;" OwnerUserId="462" LastEditorUserId="462" LastEditorDisplayName="" LastEditDate="2011-01-10T18:44:01.457" LastActivityDate="2011-01-10T18:44:01.457" CommunityOwnedDate="2011-01-10T18:44:01.457" CommentCount="3" />
  <row Id="17021" PostTypeId="2" ParentId="17016" CreationDate="2011-01-10T18:09:34.090" Score="4" ViewCount="" Body="&lt;p&gt;First, if $a$ and $b$ are nonnegative real numbers ($a,b\in\mathbb{R}$, $a,b\ge 0$), $\sqrt{a}\sqrt{b}=\sqrt{ab}$.  To deal with a square root of a negative real number, say $-a&amp;lt;0$, $\sqrt{-a}=\sqrt{-1\cdot a}=i\sqrt{a}$.  Note that the first rule required that the two numbers were nonnegative, so $\sqrt{-4}\sqrt{-9}=2i\cdot 3i=$ $2\cdot3\cdot i^2=$ $6\cdot-1=-6$ is not equal to $\sqrt{-4\cdot-9}=\sqrt{36}=6$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second, note that for $x\in\mathbb{R}$, $\sqrt{x^2}=|x|$.  For example, $\sqrt{(-3)^2}=\sqrt{9}=3$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, for $x\in\mathbb{R}$, $x^2\in\mathbb{R}$ and $x^2\ge 0$.  So, $\sqrt{-x^2}=i\sqrt{x^2}=i|x|$.&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2011-01-10T18:09:34.090" CommentCount="6" />
  <row Id="17022" PostTypeId="2" ParentId="16958" CreationDate="2011-01-10T18:14:30.480" Score="2" ViewCount="" Body="&lt;p&gt;As others have said, there can be many different formulas that will generate the terms of the sequence that you already know but give different results for other terms, and the OEIS is a good resource for looking up sequences.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another technique that can be of use is the method of finite differences, &lt;a href=&quot;http://math.stackexchange.com/questions/656/how-can-i-write-an-equation-that-matches-any-sequence/661#661&quot;&gt;briefly described in my answer here&lt;/a&gt;.  In particular, if you have reason to believe that the sequence has a polynomial or exponential formula, finite differences can help determine the specific formula.&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2011-01-10T18:14:30.480" />
  <row Id="17023" PostTypeId="2" ParentId="17016" CreationDate="2011-01-10T18:22:40.240" Score="2" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;: The original question asked why $\sqrt{-x^2}=xi$. This is false in general, as I answered below. Here I just address the new version of the question (using what I wrote below).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that we have, from what I wrote below, that $\sqrt{-x^2}=\pm xi$ for all $x$, with the choice of $+$ or $-$ depending on where $x$ is in the plane. if $x$ happens to be real, $x=r^{e^{i\theta}}$ where $\theta=0$ or $\pi$, and the above is just $\sqrt{-r^2}=ir=i|x|$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $x$ is not real, the equality fails.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;This is essentially a matter of definitions. If $x$ is a complex number, and $x\ne 0$, we can write $x=re^{i\theta}$ for some positive real number $r$ and some real number $\theta$. There are many choices of $\theta$, but the principal argument is defined so $0\le\theta&amp;lt;2\pi$. Here, $e^{i\theta}=\cos(\theta)+i\sin(\theta)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We define $\sqrt x$ as $\sqrt r e^{i\theta/2}$ where $\sqrt r$ is the (real) positive square root of $r$, and $\theta$ is the prinicipal argument. Also, we define $i$ as $\sqrt{-1}$, which according to the conventions just described, gives us: $-1=e^{i\pi}$ so $i=\sqrt{-1}=e^{i\pi/2}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then $\sqrt{-x^2}=\sqrt{r^2e^{i(2\theta+\pi)}}$. Now, if $0\le \theta&amp;lt;\pi/2$ (i.e., if $x$ is in the first quadrant), then $2\theta+\pi$ is the principal argument of $-x^2$, and $$\sqrt{-x^2}=re^{i(\theta+\pi/2)}=ix. $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, this identity does not always hold (Google notwithstanding) if $x$ is not in the first quadrant. For example, if $x=-1$, then $\sqrt{-x^2}=\sqrt{-1}=i$ while $ix=-i$. The reason is that if $\pi/2\le \theta&amp;lt;2\pi$, then $2\theta+\pi$ is &lt;em&gt;not&lt;/em&gt; the principal argument of $-x^2$, the principal argument will be $2\theta-\pi$ or $2\theta-3\pi$, whichever lands us again in the range ${}[0,2\pi)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the first case, we always get a failure of the identity: $$ \sqrt{-x^2}=re^{i(\theta-\pi/2)}=-ix.$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the second, we get another instance where the identity holds:&#xA;$$ \sqrt{-x^2}=re^{i(\theta-\pi-\pi/2)}=x(-1)(-i)=ix. $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, note that we are in the first case above iff $0\le2\theta-\pi&amp;lt;2\pi$, i.e., iff $\pi/2\le\theta&amp;lt;3\pi/2$ (i.e., if $x=a+bi$ where $a,b$ are real, and either $a=0\&amp;amp;b&amp;gt;0$, or $a&amp;lt;0$), while we are in the second case iff $3\pi/2\le\theta&amp;lt;2\pi$.  &lt;/p&gt;&#xA;" OwnerUserId="462" LastEditorUserId="462" LastEditorDisplayName="" LastEditDate="2011-01-10T18:50:50.793" LastActivityDate="2011-01-10T18:50:50.793" />
  <row Id="17024" PostTypeId="2" ParentId="17016" CreationDate="2011-01-10T18:23:13.353" Score="7" ViewCount="" Body="&lt;p&gt;&lt;em&gt;Edit&lt;/em&gt;: Originally, we had &quot;$\sqrt{-x} = i\sqrt{x}$&quot; (or rather, the incorrect $\sqrt{-x}=ix$), and I answered using that; I edited this to account for the correction made to the question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It depends on what kind of animal $x$ is and what you mean by $\sqrt{x}$. (In other words: don't believe everything Google tells you).  For example, say $x=i$; then $|x|=1$, $x^2=-1$, so $\sqrt{-x^2}$ is $\sqrt{1}$, whereas $|x|i=i$. And, no matter how you interpret the square root, it is certainly &lt;strong&gt;not&lt;/strong&gt; true that $\sqrt{1}$ is equal to $i$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, what is going on here?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $x$ is a real number, then $x^2$ is nonnegative, so that $\sqrt{-x^2}$ is asking you to take the square root of a negative real number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But when dealing with complex numbers, we really &lt;em&gt;do&lt;/em&gt; need to keep in mind that the square root function has two &quot;branches&quot;; you may remember that in the real number case, one starts with the function $y=x^2$, which is not one-to-one (does not pass the horizontal line test); in order to get an inverse, we restrict the domain to $x\geq 0$ so that the resulting restricted function &lt;em&gt;is&lt;/em&gt; one-to-one, and $\sqrt{x}$ is defined to be the inverse of this function. In particular, $\sqrt{x}$ only accepts nonnegative numbers as inputs, and only gives nonnegative numbers as outputs. Because $\sqrt{x}$ is always nonnegative, it has some nice properties (chief among them: it is a &lt;em&gt;function&lt;/em&gt;; also, $\sqrt{ab}=\sqrt{a}\sqrt{b}$ if $a$ and $b$ are both nonnegative real numbers).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you go to complex numbers things get more complicated, because there no longer is, in general, a good way of selecting which of the two solutions to $y^2=a$ you want to call &quot;the&quot; square root of $a$. This means that we really do need to keep track of both solutions, otherwise things get really messy and you end up with nonsense (as you will see below). So when dealing with square roots of complex numbers we really have what is sometimes called a &quot;multi-valued function&quot;, or a function with &quot;two branches&quot;: there are really &lt;em&gt;two&lt;/em&gt; possible values for $\sqrt{a}$ when $a$ is a complex number. This includes real numbers when you allow negative ones as inputs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $x$ is a real number, then $x^2$ is nonnegative. Then the two&#xA;the two &lt;strong&gt;complex&lt;/strong&gt; square roots of $-x^2$ are $i|x|$ and $-i|x|$; in fact, you can just take $ix$ and $-ix$, because if $x\geq 0$ then $|x|=x$ and this is what you get, and if $x\leq 0$ then $|x|=-x$, so $i|x|=-ix$ and $-i|x|=ix$. To verify these are the two values of the (multi-valued) square root, square both and you'll see that that they both give $-x^2$. So one branch gives you the value $ix$ (or $i|x|$), and the other branch gives you the value $-ix$ (or $-i|x|$). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Note well:&lt;/strong&gt; You have to be careful with the complex square root function: in particular, it is no longer true that $\sqrt{ab}=\sqrt{a}\sqrt{b}$ always holds: for instance, putting $a=b=-1$, you would get silly things like:&#xA;$$1 = \sqrt{1} = \sqrt{(-1)(-1)} \stackrel{\text{not really}}{=}\sqrt{-1}\sqrt{-1} = i\cdot i = -1.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $x$ is a &lt;strong&gt;nonreal complex number&lt;/strong&gt;, though, you have to be more careful. If you write complex numbers in polar form, then multiplication and square roots are easy. If $z$ and $w$ are complex numbers, you can always find nonnegative real numbers $\rho$ and $\sigma$, and angles $\theta$ and $\phi$ (in radians) such that&#xA;\begin{align*}&#xA;z &amp;amp;= \rho\left(\cos\theta + i\sin\theta\right)\\&#xA;w &amp;amp;= \sigma\left(\cos\phi + i\sin\phi\right).&#xA;\end{align*}&#xA;If you do that, then it is easy to check that&#xA;\begin{align*}&#xA;zw &amp;amp;= (\rho\sigma)\left(\cos(\theta+\phi) + i\sin(\theta+\phi)\right);\\&#xA;\sqrt{z} &amp;amp;= \left\{\begin{array}{l}&#xA;\sqrt{\rho}\left(\cos\frac{\theta}{2} + i\sin\frac{\theta}{2}\right),\\&#xA;\sqrt{\rho}\left(\cos\left(\frac{\theta}{2}+\pi\right) + i\sin\left(\frac{\theta}{2}+\pi\right)\right).&#xA;\end{array}\right.&#xA;\end{align*}&#xA;Above, $\sqrt{\rho}$ is the nonnegative real number whose square is equal to $\rho$ (remember that $\rho$ is a real number greater than or equal to $0$). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, if $x=\rho(\cos\theta+i\sin\theta)$, then &#xA;$$-x^2 = \rho^2\left(\cos(2\theta+\pi) + i\sin(2\theta+\pi)\right)$$&#xA;(I used the fact that $-1 = 1(\cos\pi + i\sin\pi)$ and the formulas above). So the two values of the square root are&#xA;\begin{align*}&#xA;\sqrt{\rho^2}\left(\cos\left(\frac{2\theta+\pi}{2}\right) + i\sin\left(\frac{2\theta+\pi}{2}\right)\right) &amp;amp;= \rho\left(\cos\left(\theta+\frac{\pi}{2}\right) + i\sin\left(\theta+\frac{\pi}{2}\right)\right),\\&#xA;\sqrt{\rho^2}\left(\cos\left(\frac{2\theta+\pi}{2}+\pi\right) + i\sin\left(\frac{2\theta+\pi}{2}+\pi\right)\right) &amp;amp;= \rho\left(\cos\left(\theta+\frac{3\pi}{2}\right) + i\sin\left(\theta+\frac{3\pi}{2}\right)\right).&#xA;\end{align*}&#xA;On the other hand, multiplying $x$ by both $i=\cos\frac{\pi}{2}+i\sin\frac{\pi}{2}$ and $-i=\cos\frac{3\pi}{2}+ i\sin\frac{3\pi}{2}$ gives the two values:&#xA;$$\rho\left(\cos\left(\theta+\frac{\pi}{2}\right)+i\sin\left(\theta+\frac{\pi}{2}\right)\right)\text{ and }\rho\left(\cos\left(\theta+\frac{3\pi}{2}\right)+i\sin\left(\theta+\frac{3\pi}{2}\right)\right)$$&#xA;so that these two are still the correct answers. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How does this relate to the error I noted above? If $x=i$, then the two complex branches of the square root give that the two values of $\sqrt{-x^2}$ are $1$ and $-1$. If we look at the two solutions for the general case, which are $ix$ and $-ix$ (notice that I cannot use the absolute value bars if we don't assume $x$ is real to begin with), then we get $i(i) = -1$ and $-i(i) = 1$, the same tow answers (though in different order). So keeping track of &lt;em&gt;both&lt;/em&gt; branches is paramount in the complex case.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-10T19:37:02.013" LastActivityDate="2011-01-10T19:37:02.013" />
  <row Id="17025" PostTypeId="2" ParentId="16940" CreationDate="2011-01-10T19:08:39.553" Score="3" ViewCount="" Body="&lt;p&gt;We suppose an (m+n) by (m+n) matrix $\textbf{A}$ with block partition:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\textbf{A}=\begin{pmatrix}\textbf{E}&amp;amp;\textbf{F}\\\\ \textbf{G}&amp;amp;\textbf{H}\end{pmatrix}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A sequence of steps can be outlined, invoking recursively an in-place matrix inversion and other steps involving in-place matrix multiplication.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some but not all of these matrix multiplication operations would seem to require additional memory to be &quot;temporarily&quot; allocated in between recursive calls to the in-place matrix inversion.  At bottom we will argue that needed additional memory is linear, e.g. size m+n.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;1.&lt;/code&gt; The first step is recursively to invert in-place matrix $\textbf{E}$:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\begin{pmatrix}\textbf{E}^{-1}&amp;amp;\textbf{F}\\\\ \textbf{G}&amp;amp;\textbf{H}\end{pmatrix}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course to make that work requires the leading principal minor $\textbf{E}$ to be nonsingular.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;2.&lt;/code&gt; The next step is to multiply $\textbf{E}^{-1}$ times block submatrix $\textbf{F}$ and negate the result:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\begin{pmatrix}\textbf{E}^{-1}&amp;amp;-\textbf{E}^{-1}\textbf{F}\\\\ \textbf{G}&amp;amp;\textbf{H}\end{pmatrix}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the matrix multiplication and overwriting in this step can be performed one column of $\textbf{F}$ at a time, something we'll revisit in assessing memory requirements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;3.&lt;/code&gt; Next multiply $\textbf{G}$ times the previous result $-\textbf{E}^{-1}\textbf{F}$ and add it to the existing block submatrix $\textbf{H}$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\begin{pmatrix}\textbf{E}^{-1}&amp;amp;-\textbf{E}^{-1}\textbf{F}\\\\ \textbf{G}&amp;amp;\textbf{H}-\textbf{G}\textbf{E}^{-1}\textbf{F}\end{pmatrix}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This step can also be performed one column at a time, and because the results are accumulated in the existing block submatrix $\textbf{H}$, no additional memory is needed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that what has now overwritten $\textbf{H}$ is $\textbf{S}=\textbf{H}-\textbf{G}\textbf{E}^{-1}\textbf{F}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;4. &amp;amp; 5.&lt;/code&gt; The next two steps can be done in either order.  We want to multiply block submatrix $\textbf{G}$ on the right by $\textbf{E}^{-1}$, and we also want to recursively invert in-place the previous result $\textbf{S}$.  After doing both we have:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\begin{pmatrix}\textbf{E}^{-1}&amp;amp;-\textbf{E}^{-1}\textbf{F}\\\\ \textbf{G}\textbf{E}^{-1}&amp;amp;\textbf{S}^{-1}\end{pmatrix}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the matrix multiplication and overwriting of $\textbf{G}$ can be done one row at a time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;6.&lt;/code&gt; Next we should multiply these last two results, overwriting $\textbf{G}\textbf{E}^{-1}$ with $\textbf{S}^{-1}\textbf{G}\textbf{E}^{-1}$ and negating that block:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\begin{pmatrix}\textbf{E}^{-1}&amp;amp;-\textbf{E}^{-1}\textbf{F}\\\\ -\textbf{S}^{-1}\textbf{G}\textbf{E}^{-1}&amp;amp;\textbf{S}^{-1}\end{pmatrix}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;7.&lt;/code&gt;  We are on the home stretch now!  Multiply the two off-diagonal blocks and add that to the diagonal block containing $\textbf{E}^{-1}$:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\begin{pmatrix}\textbf{E}^{-1}+\textbf{E}^{-1}\textbf{F}\textbf{S}^{-1}\textbf{G}\textbf{E}^{-1}&amp;amp;-\textbf{E}^{-1}\textbf{F}\\\\ -\textbf{S}^{-1}\textbf{G}\textbf{E}^{-1}&amp;amp;\textbf{S}^{-1}\end{pmatrix}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;8.&lt;/code&gt;  Finally we multiply the block containing $-\textbf{E}^{-1}\textbf{F}$ on the right by $\textbf{S}^{-1}$, something that can be done one row at a time:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\textbf{A}^{-1}=\begin{pmatrix}\textbf{E}^{-1}+\textbf{E}^{-1}\textbf{F}\textbf{S}^{-1}\textbf{G}\textbf{E}^{-1}&amp;amp;-\textbf{E}^{-1}\textbf{F}\textbf{S}^{-1}\\\\ -\textbf{S}^{-1}\textbf{G}\textbf{E}^{-1}&amp;amp;\textbf{S}^{-1}\end{pmatrix}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Requirements for additional memory:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A temporary need for additional memory arises when we want to do matrix multiplication and store the result back into the location of one of the two factors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Such a need arises in step 2. when forming $\textbf{E}^{-1}\textbf{F}$, in step 4. (or 5.) when forming $\textbf{G}\textbf{E}^{-1}$, in step 6. when forming $\textbf{S}^{-1}\textbf{G}\textbf{E}^{-1}$, and in step 8. when forming $-\textbf{E}^{-1}\textbf{F}\textbf{S}^{-1}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course other &quot;temporary&quot; allocations are hidden in recursive calls to in-place inversion in step 1. and step 4.&amp;amp;5. when matrices $\textbf{E}$ and $\textbf{S}$ are inverted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In each case the allocated memory can be freed (or reused) after one column or one row of a required matrix multiplication is performed, because the overwriting can be done one column or one row at a time following its computation.  The overhead for such allocations is limited to size m+n, or even max(m,n), i.e. a linear overhead in the size of $\textbf{A}$.&lt;/p&gt;&#xA;" OwnerUserId="3111" LastActivityDate="2011-01-10T19:08:39.553" CommentCount="1" />
  <row Id="17026" PostTypeId="1" AcceptedAnswerId="17040" CreationDate="2011-01-10T19:09:04.477" Score="5" ViewCount="329" Body="&lt;p&gt;This question was motivated by &lt;a href=&quot;http://math.stackexchange.com/questions/16831/a-nonzero-function-in-c0-1-for-which-int-01-fxxn-dx-0-forall-n-g&quot;&gt;another question&lt;/a&gt; in this site.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As explained in that problem (and its answers), if $\displaystyle f$ is continuous on $\displaystyle [0,1]$ and $\displaystyle \int_0^1 f(x)p(x)dx=0$ for all polynomials $\displaystyle p$, then $\displaystyle f$ is zero everywhere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose we remove the restriction that $\displaystyle f$ is continuous. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can we conclude from $\displaystyle f\in L^1([0,1])$ that $\displaystyle f$ is zero almost everywhere? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(This should be terribly standard. My apologies, I am rusty of late.)&lt;/p&gt;&#xA;" OwnerUserId="3340" LastEditorUserId="3340" LastEditorDisplayName="" LastEditDate="2011-01-12T21:58:08.907" LastActivityDate="2011-01-12T21:58:08.907" Title="What can we say about $f$ if $\int_0^1 f(x)p(x)dx=0$ for all polynomials $p$?" Tags="&lt;real-analysis&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="17027" PostTypeId="2" ParentId="17009" CreationDate="2011-01-10T19:09:21.393" Score="5" ViewCount="" Body="&lt;p&gt;A function $f:\mathbb{R}^n \to \mathbb{R}$ is homogeneous of degree $r$ when it has the following property:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f(\lambda x_1,\lambda x_2,\ldots,\lambda x_n) = \lambda^r f(x_1,x_2,\ldots,x_n)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Examples of such functions include:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Linear functions, they are of degree 1. If you scale the graph of the function by a factor $\lambda$, you still get the same graph, except that all points have coordinates scaled up by the factor $\lambda$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The monomials in one variable: $f:\mathbb{R} \to \mathbb{R}: x \mapsto x^n$ is homogeneous of degree $n$, scale $x$ by a factor $\lambda$ and the function will scale by a factor $\lambda^n$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Any polynomial (function) in $n$ variables such that each term appearing in the polynomial is of degree $k$ is homogeneous of degree $k$. For instance $x^2 y + z^3$ is homogeneous of degree $3$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The importance of homogeneity is the scale invariance of the functions. Which implies that the graphs of the functions will be scale invariant. Indeed, imagine a homogeneous function is used to define a geometrical object implicitly:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f(x_1,x_2,\ldots,x_n)=0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;meaning all points with coordinates $(x_1,\ldots,x_n)$ that satisfy this equation will belong to the geometrical figure defined by $f$. If $f$ is homogeneous, it immediately follows that any multiple of these coordinates also satisfies the equations. In other words, any point that satisfies the equation immediately implies the entire ray going through that point and the origin of the space belong to the geometrical object. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Homogenizing an implicit polynomial equation means adding an extra variable $z$ and multiply any term by $z^k$ with $k$ such that the resulting polynomial is homogeneous. Of course, since any $z$-multiple of the polynomial will also be homogeneous, you choose the resulting homogeneous polynomial with smallest possible degree.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your example, this would become&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ax^2 + 2hxy + by^2 + 2gxz + 2fyz + cz^2 = 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you take the intersection of the geometrical figure defined by above equation with the plane $z=1$, you get back the original figure. If you choose another plane parallel with the $z=1$ plane, you get a scaled up or scaled down version of the figure.&lt;/p&gt;&#xA;" OwnerUserId="3567" LastEditorUserId="3567" LastEditorDisplayName="" LastEditDate="2011-01-10T19:19:47.043" LastActivityDate="2011-01-10T19:19:47.043" CommentCount="5" />
  <row Id="17028" PostTypeId="2" ParentId="17016" CreationDate="2011-01-10T19:25:19.170" Score="1" ViewCount="" Body="&lt;p&gt;I think that this should be addressed in a much more fundamental level:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose $x\in\mathbb{R}$ then $\sqrt{x^2} = |x|$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason this is happening, is that we want the $\sqrt{\cdot}$ function to have certain properties, which cannot be well-defined on real numbers otherwise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First we want it to be a function, which means that given $x$ there is exactly one $y$ such that $\sqrt{x} = y$. So we can choose either the positive or the negative root - but not both of them, so we take the positive one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We want to have $\sqrt{x} = x^{\frac{1}{2}}$ so we want to have roots respecting the general laws of exponentiation - for example $(x^m)^n = x^{m\cdot n}$, so we want $(x^{\frac{1}{2}})^{\frac{1}{2}} = \sqrt{\sqrt{x}} = \sqrt[4]{x} = x^{\frac{1}{4}}$, but suppose we had chosen $\sqrt{x^2} = -|x|$ then while we could theoretically define $\sqrt[4]{x^2}$ we could still not define within the real numbers $\sqrt{-|x|}$ and thus the above equality wouldn't hold.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lastly, people often think of the $\sqrt{x}$ as an inverse of the $x^2$ function, but people are often forgetting (or unaware) that inverse can only be defined when the function is injective, namely that $f(x)=f(y)$ happens only if $x=y$ (for example, $f(x)=x+1$ has this property). However within the realm of the real numbers this is not the case, as $(-2)^2 = (2^2) = 4$, and therefore the $x^2$ cannot be inversed unless we bound ourselves to a subset of the real numbers on which squaring a number is indeed injective - e.g. the non-negative real numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is just one of the caveats that one must learn to avoid when doing mathematics properly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now for the sake of completeness, to address your question - based on the aforementioned arguments - we have (for $x\in\mathbb{R}$):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sqrt{-x^2} = \sqrt{-1}\cdot\sqrt{x^2} = i|x|$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem defining the root (in fact more than just square root) over complex numbers is essentially the same, only that in the complex numbers you have $n$ roots for $\sqrt[n]{x}$ and you must define properly which one you're taking.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose we take the least non-negative $\theta$ such that $(\sqrt{r}e^{i\theta})^2$ is $x$ (that is, we choose a root, just like we did with the real numbers by choosing the positive root) then we have:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sqrt{-r^2e^{2i\theta}} = \sqrt{r^2e^{2i(\theta+\frac{\pi}{2})}} = re^{i(\theta+\frac{\pi}{2})}=ire^{i\theta}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However $|re^{i\theta}|=r$, and so the equality above is usually not true for complex numbers (however it can be restated in such way which generalizes it for the above choice of complex root).&lt;/p&gt;&#xA;" OwnerUserId="622" LastActivityDate="2011-01-10T19:25:19.170" />
  <row Id="17029" PostTypeId="2" ParentId="17015" CreationDate="2011-01-10T19:30:50.227" Score="1" ViewCount="" Body="&lt;p&gt;More generally, consider the function $f(x;\theta)=(1/\theta)e^{-x/\theta}$, $x \geq 0$, where $\theta &amp;gt; 0$ is fixed, i.e. the density function of the exponential distribution with mean $\theta$. It is strictly decreasing and convex. $&#xD;&#xA;\int {f(x;\theta )dx}  =  - e^{ - x/\theta }  + C$, and $F(x): = \int_0^x {f(u;\theta )du}  = 1 - e^{ - x/\theta } $, for any $x \geq 0$ (distribution function of exponential distribution); in particular $\int_a^b {f(x;\theta )dx}  = F(b) - F(a) = e^{ - a/\theta }  - e^{ - b/\theta } $. The corresponding expectation is $\int_0^\infty  {xf(x;\theta )dx}  = \theta $, and more generally, the corresponding $n$th moment is given, for all $n \in \mathbb{N}$, by&#xA;$$&#xD;&#xA;\mu&amp;#39;_n := \int_0^\infty  {x^n f(x;\theta )dx}  = \theta ^n \int_0^\infty  {x^n e^{ - x} dx}  = \theta ^n \Gamma (n + 1) = \theta ^n n!,&#xD;&#xA;$$&#xA;where $\Gamma$ denotes the gamma function. In particular, the corresponding variance is given, according to the formula ${\rm Var}(X)={\rm E}(X^2)-{\rm E}^2(X)$, by $\mu&amp;#39;_2 - (\mu&amp;#39;_1)^2 = 2 \theta^2 - \theta^2 = \theta^2$.&lt;/p&gt;&#xA;" OwnerUserId="2810" LastActivityDate="2011-01-10T19:30:50.227" />
  <row Id="17030" PostTypeId="2" ParentId="17026" CreationDate="2011-01-10T19:40:41.417" Score="8" ViewCount="" Body="&lt;ol&gt;&#xA;&lt;li&gt;Yes.  This follows from the fact that polynomials are dense in $L^1$, which follows from (say) Stone-Weierstrass together with the fact that continuous functions are dense in $L^1$.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="232" LastActivityDate="2011-01-10T19:40:41.417" CommentCount="1" />
  <row Id="17031" PostTypeId="1" AcceptedAnswerId="17034" CreationDate="2011-01-10T19:55:54.687" Score="4" ViewCount="161" Body="&lt;p&gt;I think it does but I'm having trouble showing it.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Using the integral test: the function $f(x)=\frac{1}{\ln(x^x+x^2)}$ is decreasing but I'm having trouble integrating $\int_{1}^{\infty} \frac{dx}{\ln(x^x+x^2)}$, trying the substitution $t=x^x$ didn't seem to work.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The root test gives $\sqrt[n]{\left |\frac{1}{\ln(n^n+n^2)}\right |} \to 1$ which doesn't imply anything.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="3237" LastActivityDate="2011-01-10T20:06:05.003" Title="Does $\sum_{n=1}^{\infty} \frac{1}{\ln(n^n+n^2)}$ converge?" Tags="&lt;calculus&gt;&lt;homework&gt;&lt;sequences-and-series&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="17032" PostTypeId="1" AcceptedAnswerId="17044" CreationDate="2011-01-10T19:59:53.553" Score="1" ViewCount="77" Body="&lt;p&gt;I know that there is a first-order sentence $\varphi$  such that &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\varphi$ is written in the vocabulary given by just two binary relation symbols $E_1$, $E_2$ (and hence, without the equality symbol*),&lt;/li&gt;&#xA;&lt;li&gt;$\varphi$ is satisfiable in a model where $E_1$ and $E_2$ are equivalence relations,&lt;/li&gt;&#xA;&lt;li&gt;$\varphi$ is not satisfiable in a finite model where $E_1$ and $E_2$ are equivalence relations.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This has to be true because the first-order theory of two equivalence relations is undecidable (see the paper A. Janiczak, Undecidability of some simple formalized theories, Fundamenta Mathematicae, 1953, 40, 131--139). However, I am not able to find any such formula. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&lt;strong&gt;Does anybody know an example of a formula (or a family) with these three properties?&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Footnote *: The assumption of not allowing the equality symbol is not important. Indeed, if we know a formula with equality where the last two previous properties hold, then the formula obtained replacing all subformulas $x \approx y$ with the formula $E_1(x,y) \land E_2(x,y)$ also satisfies these two properties. &lt;/p&gt;&#xA;" OwnerUserId="5580" LastEditorUserId="5580" LastEditorDisplayName="" LastEditDate="2011-01-10T21:39:37.733" LastActivityDate="2011-01-10T21:52:06.590" Title="Finite Model Property on the First-Order Theory of Two Equivalence Relations" Tags="&lt;logic&gt;&lt;model-theory&gt;" AnswerCount="1" CommentCount="6" />
  <row Id="17034" PostTypeId="2" ParentId="17031" CreationDate="2011-01-10T20:06:05.003" Score="8" ViewCount="" Body="&lt;p&gt;Hint: Compare with $$\sum_{n=1}^{\infty} \frac{K}{n\log n}$$&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2011-01-10T20:06:05.003" CommentCount="7" />
  <row Id="17036" PostTypeId="2" ParentId="16829" CreationDate="2011-01-10T20:25:11.993" Score="4" ViewCount="" Body="&lt;p&gt;In &lt;a href=&quot;http://www.vntl.com/im/pdf/29_2_215_217.pdf&quot; rel=&quot;nofollow&quot;&gt;this short paper&lt;/a&gt; we have a counterexample for an even weaker statement (it admits no left topological group structure). There are also consistent examples (under CH) that are S-spaces etc. Note that any first countable compact zero-dimensional homogeneous space that is not metrisable is a counterexample, as a first countable topological group is metrisable.&lt;/p&gt;&#xA;" OwnerUserId="4280" LastActivityDate="2011-01-10T20:25:11.993" CommentCount="3" />
  <row Id="17037" PostTypeId="2" ParentId="16289" CreationDate="2011-01-10T20:36:36.303" Score="1" ViewCount="" Body="&lt;p&gt;An alternative proof: Of course $Y = f[X]$ is compact and Hausdorff. It has a countable network (which is like a base of a space but without the requirement that its members are open sets), as a continuous image of a space with a countable network. A compact Hausdorff space with a countable network is second-countable (Arhangel'skij's theorem) and we are done by Urysohn. For a proof of the stuff on networks, I have a post &lt;a href=&quot;http://at.yorku.ca/cgi-bin/bbqa?forum=ask_a_topologist_2005;task=show_msg;msg=1755&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; that explains it (at the end it goes into networks). &lt;/p&gt;&#xA;" OwnerUserId="4280" LastActivityDate="2011-01-10T20:36:36.303" />
  <row Id="17038" PostTypeId="2" ParentId="17009" CreationDate="2011-01-10T20:42:28.830" Score="3" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Disclaimer&lt;/strong&gt;: I will try to keep my answer at a very elementary level, as you pointed out that you just started learning algebraic geometry. Also, I will be a bit imprecise at several points of the discussion, as I want to convey some intuition and not mathematical rigor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're studying algebraic geometry, I think a good way to view homogenising an equation is to interpret it as &quot;adding points at infinity&quot; to the zero set of the polynomial.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the two-dimensional case this works as follows: instead of the &quot;normal&quot; affine plane $\mathbb{A}^2$ (which consists just of 2-tuples coming from your ground field $k$) we consider the projective plane $\mathbb{P}^2$, which has some additional points: for each direction &quot;out of the plane&quot;, we introduce a new point. This gives us a whole line of new points, which cannot be drawn in the usual picture of the affine plane, but are thought of as lying &quot;at infinity&quot;. While this seems a bit odd at first it has many advantages. Consider for example two straight lines in the affine plane, then they will intersect in some point of the plane or they will be parallel (in which case they don't intersect). When we look at the same two lines in the projective plane, things get simpler: the lines will always intersect! Either the lines intersect in the &quot;affine part&quot; of our projective plane  or they are parallel. In the second case they will both point &quot;out of the plane&quot; in the same direction, which means that they will intersect at the point of the projective plane that we've previously added for that particular direction. So, all straight lines will intersect in $\mathbb{P}^2$. Isn't that nice?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Starting with this idea, one wants to make all of this precise and more rigorous. The formal way to define curves in the projective plane is to only consider zero sets of homogeneous polynomials, that is polynomials in which each monomial has the same total degree. In order to get the right thing, we also need to introduce an extra variable and change the underlying set of our space (it no longer consists 2-tuples but of 3-tuples, and we mod out by some equivalence relation). For the details of the construction you could read the corresponding Wikipedia article on &lt;a href=&quot;http://en.wikipedia.org/wiki/Projective_space&quot; rel=&quot;nofollow&quot;&gt;Projective space&lt;/a&gt;. I will also try to give a short summary in the following: define the projective plane $\mathbb{P}^2$ over your ground field $k$ as follows: the underlying set is $k^3 \setminus \{(0,0,0)\}/\sim$, where $\sim$ denotes the following equivalence relation: $(a,b,c) \sim (d,e,f)$ iff there is a $\lambda \in k \setminus \{0\}$ such that $(\lambda a, \lambda b, \lambda c) = (d,e,f)$ (in other words, if the two points lie on the same line through the origin in $k^3$). We denote the equivalence class of a point $(a,b,c)$ by $(a:b:c)$, as the equivalence relation we introduced tells us that we only care about the ratios of the coordinates. We also endow this set with a certain topology (the Zariski topology), but I do not want to elablorate on the details here. But also without knowledge of the topology, you can see that, as a set, we have created a space that contains the underlying set of the affine plane: If you look at the subset $\{(a:b:c) \subset \mathbb{P}^2| c \neq 0\}$, then you see that every equivalence class in that set has a unique representation as $(a&amp;#39;:b&amp;#39;:1)$, as multiplying $(a:b:c)$ by $\frac{1}{c}$ will not change the equivalence class of $(a:b:c)$. Now, we can just forget about the 1 in $(a&amp;#39;:b&amp;#39;:1)$ and only look at $(a&amp;#39;,b&amp;#39;)$. Now, I intentionally separate $a&amp;#39;$ and $b&amp;#39;$ by a comma, because the representation $(a&amp;#39;:b&amp;#39;:1)$ was unique, which means that we don't need to consider equivalence classes any further. If you think about it, this means that we have identified $\{(a:b:c) \subset \mathbb{P}^2| c \neq 0\}$ with the subset $k \times k$, which was the underlying subset of our affine plane $\mathbb{A}^2$. Thus $\mathbb{P}^2 = \mathbb{A}^2 \cup \{(a:b:c) \subset \mathbb{P}^2| c = 0\}$. You can check for yourself that $\{(a:b:c) \subset \mathbb{P}^2| c = 0\}$ is a copy of $k$, i.e. a line (plus an additional point, to be precise). This is why we can say that $\mathbb{P}^2$ is the affine plane plus a line at infinity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we also want to define curves in $\mathbb{P}^2$ by zero sets of polynomials. This is more complicated than previously: our points in $\mathbb{P}^2$ have three coordinates , therefore we need an additional variable. Also, we need to cope with the difficulty that we're defining a function on an equivalence class of points. In the end, we solve these problems by only considering homogeneous polynomials in the variables $x,y,z$: for a homogeneous polynomial f(x,y,z), we have that $f(\lambda x, \lambda y, \lambda z) = 0 \Leftrightarrow f(x, y, z) = 0$ for all $\lambda \in k \setminus \{0\}$. Thus, if $f$ is zero on one representative of an equivalence class $(a:b:c)$, it will be on all of them, which is just what we need. The set of equivalence classes $(a:b:c)$, where $f$ is zero will be the curve defined by $f$ in $\mathbb{P}^2$. If you rememeber how we &quot;extracted&quot; $\mathbb{A}^2$ from $\mathbb{P}^2$, you can also check now, what $f$ looks like on the $\mathbb{A}^2$-part of $\mathbb{P}^2$: just set the $z$-variable of $f$ equal to 1 and look at the zero set of $f(x,y,1)$ in the affine plane, as you did before. This process is called de-homogenising.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now you can understand what you're doing when you're homogenising a polynomial in two variables: you reverse the process of de-homogenesing, i.e. you &quot;recover&quot; a curve that is defined on the projective plane, by the zero set of the homogeneous polynomial. On the &quot;affine part&quot; of your projective plane, your curve will just look like the zero set of the polynomial that you started with.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example: look at the line in $\mathbb{A}^2$ defined by $ax +by + c =0$, with $a,b,c \neq 0$. we can homogenize this to $ax +by + cz =0$. This is the equation of a line in $\mathbb{P}^2$, and on the affine part $z=1$, it looks like the line $ax +by + c =0$. On the &quot;infinte part&quot; part $z = 0$, it has additional points: the equation then reads $ax +by =0$. By linear algebra, you know that this has a one-dimensional solution space, i.e. all solutions are of the form $(\lambda d, \lambda e)$ for some fixed $d,e \in k$ and all $\lambda \in k$. Thus on the &quot;infinite part&quot; of $\mathbb{P}^2$ the solutions look like $(\lambda d: \lambda e : 0)$, which is just a single point in $\mathbb{P}^2$ by the defining equivalence relation. Thus, homogenizing the equation $ax +by + c=0$ resulted in obtaining one additional point of the line at infinity.&lt;/p&gt;&#xA;" OwnerUserId="1766" LastEditorUserId="1766" LastEditorDisplayName="" LastEditDate="2011-01-11T10:19:01.457" LastActivityDate="2011-01-11T10:19:01.457" CommentCount="6" />
  <row Id="17039" PostTypeId="2" ParentId="16989" CreationDate="2011-01-10T21:20:19.100" Score="3" ViewCount="" Body="&lt;p&gt;Let $X=S^2/\mathord\sim$ and,letting $S^2_+\subseteq S^2$ be the upper closed hemisphere in the sphere, let $X_+=S^2_+/\mathord\sim$ be the quotient of $S^2_+$ by the restricted equivalence relation. Now consider the long exact sequence in reduced homology for the pair $(X,X_+)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using excision &amp;amp;c, show that the relative homology of $(X,X_+)$ is the same as that of the result of collapsing $X_+$ to a point, so that you get a $2$-sphere. On the other hand, $X_+$ is a projective plane, so you also know its homology. Now use the long exact sequence.&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2011-01-10T21:20:19.100" CommentCount="2" />
  <row Id="17040" PostTypeId="2" ParentId="17026" CreationDate="2011-01-10T21:21:57.990" Score="6" ViewCount="" Body="&lt;p&gt;Assuming you're using Lebesgue integrals, to even make the statement that ${\mathbb \int_0^1 f(x)p(x) = 0}$ for all polynomials $p(x)$, you are forcing $f(x)$ to be in $L^1$. This can be seen by setting $p(x) = 1$; in order for the statement ${\mathbb \int_0^1 f(x) = 0}$ to be well-defined the positive and negative parts $f^+(x)$ and $f^-(x)$ have to integrate to the same finite value.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So suppose $f(x)$ is some $L^1$ function with ${\mathbb \int_0^1 f(x)p(x) = 0}$ for all polynomials $p(x)$. By the Stone-Weierstrass theorem you can uniformly approximate any continuous function by polynomials, so taking limits you also have ${\mathbb \int_0^1 f(x)g(x) = 0}$ for every continuous $g(x)$. In particular it is true for $g(x) = \cos(2\pi nx)$ and $g(x) = \sin(2\pi nx)$ for any $n$. This means each Fourier coefficient of $f(x)$ is zero. By the uniqueness theorem for Fourier coefficients, this means $f(x) = 0$ a.e.&lt;/p&gt;&#xA;" OwnerUserId="3035" LastActivityDate="2011-01-10T21:21:57.990" CommentCount="4" />
  <row Id="17041" PostTypeId="1" AcceptedAnswerId="17042" CreationDate="2011-01-10T21:23:02.560" Score="7" ViewCount="242" Body="&lt;p&gt;Hello all,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, I've been revising for an exam and I came up against the question &quot; prove $4(9^n) + 3(2^n)$ is divisible by 7 for all $n&amp;gt;0$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, I know how to do this. If I assume $n=k$ divisible by $7$, then I need to show that $n=k+1$ is divisible by $7$. The easiest way to do this is to state that if we label $n_k=k$ and $n_{k+1} = k+1$ then if $7|n_{k+1}-n_{k}$ and $7|n_{k} \Rightarrow 7|n_{k+1}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, without further ado, $4(9^k)9 - 4(9) + 3(2^k)2 - 3(2^k) = 8\cdot4(9^k) + 3\cdot2^k = 8(4(9^k) + 3(2^k)) - 7\cdot 3(2^k)$. As required. Now clearly for $n=0$ this expression is $7$ so divisible for all $n\geq 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is, how would I go about proving this via &lt;a href=&quot;http://en.wikipedia.org/wiki/Strong_induction#Complete_induction&quot; rel=&quot;nofollow&quot;&gt;complete induction&lt;/a&gt;? I asked because &quot;proof by strong induction also accepted&quot; was mentioned in the mark scheme. Now according to wikipedia, my first assumption is that not only is $n=k$ true but so is $n=k-1$ and so on down to $n=0$. How do I improve my expression of that and how do I go from there to show a proof using this technique?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;: The build up to the question is on the topic of induction, so that's why I proved it that way, but &lt;a href=&quot;http://math.stackexchange.com/users/1277/yuval-filmus&quot;&gt;Yuval Filmus&lt;/a&gt; has pointed out that if we are simply asked to prove it, the fact that $9 \equiv 2 \mod(7)$ means the proof is trivial.&lt;/p&gt;&#xA;" OwnerUserId="892" LastEditorUserId="892" LastEditorDisplayName="" LastEditDate="2011-01-11T00:47:59.263" LastActivityDate="2011-01-11T00:47:59.263" Title="What is complete induction, by example?" Tags="&lt;induction&gt;" AnswerCount="2" CommentCount="6" />
  <row Id="17042" PostTypeId="2" ParentId="17041" CreationDate="2011-01-10T21:33:40.920" Score="5" ViewCount="" Body="&lt;p&gt;One way you can use complete induction is to notice that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$9^{n+1} - 1 = 8(1 + 9 + 9^2 + \dots + 9^n)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$2^{n+1} - 1 = 1 + 2 + 2^2 + \dots + 2^n$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mutiply the first by $\displaystyle 4$ and second by $\displaystyle 3$ and add them up. Now, notice that $\displaystyle 32(9^k) + 3(2^k) = 28(9^k) + 4(9^k) + 3(2^k)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you assume $\displaystyle 4(9^k) + 3(2^k)$ is divisible by $\displaystyle 7$ for $k=0, 1, 2, \dots, n$, then the above shows that it is also true for $\displaystyle n+1$.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2011-01-10T21:43:03.583" LastActivityDate="2011-01-10T21:43:03.583" CommentCount="1" />
  <row Id="17044" PostTypeId="2" ParentId="17032" CreationDate="2011-01-10T21:52:06.590" Score="1" ViewCount="" Body="&lt;p&gt;Let's call two elements $x$ and $y$ $E_1$-neighbors if $(x \neq y \wedge E_1(x, y))$, and $E_2$-neighbors if $(x \neq y \wedge E_2(x, y))$.  Then the following assertions should suffice to force an infinite model:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Every element has exactly one $E_1$-neighbor.&lt;/li&gt;&#xA;&lt;li&gt;There exists an element with no $E_2$-neighbor; every other element has exactly one $E_2$-neighbor.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;In particular, the single element with no $E_2$-neighbor can be identified with $0$; and the remaining natural numbers are generated by alternating conditions 1) and 2).&lt;/p&gt;&#xA;" OwnerUserId="5546" LastActivityDate="2011-01-10T21:52:06.590" CommentCount="1" />
  <row Id="17045" PostTypeId="2" ParentId="16970" CreationDate="2011-01-10T22:16:18.730" Score="2" ViewCount="" Body="&lt;p&gt;fractal colouring for deep zooms I have rendered such as e214 and &quot;&lt;a href=&quot;http://fractal.co/trip-to-e228/&quot; rel=&quot;nofollow&quot;&gt;last lights on&lt;/a&gt;&quot; have the palette applied in a loop, so it repeats every X number of iterations. can you colour it like that? &#xA;If I find the journey of an animation has areas where contrasting colours are too far apart, then I increase the speed of the palette. &#xA;ie: lower the value of X. - &#xA;Then any change in the overall iteration value will not change the colour placement. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;teamfresh&lt;/p&gt;&#xA;" OwnerUserId="5583" LastActivityDate="2011-01-10T22:16:18.730" CommentCount="2" />
  <row Id="17047" PostTypeId="2" ParentId="16831" CreationDate="2011-01-10T23:23:22.097" Score="7" ViewCount="" Body="&lt;p&gt;As an aside, the answer is &lt;em&gt;yes&lt;/em&gt; if the interval is $(0,\infty)$ instead of $(0,1)$.&#xA;For example the &quot;Stieltjes ghost function&quot;&#xA;$f(x) = \exp(-x^{1/4}) \sin x^{1/4}$ satisfies &#xA;$\int_0^{\infty} f(x) x^n dx = 0$&#xA;for all integers $n \ge 0$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Stieltjes gave this as an example of a case where the&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Moment_problem&quot; rel=&quot;nofollow&quot;&gt;moment problem&lt;/a&gt;&#xA;does not have a unique solution.&#xA;It appears in Section 55 of his famous paper &quot;Recherches sur les fractions continues&quot; from 1894;&#xA;see p. 506 in&#xA;&lt;a href=&quot;http://www.archive.org/details/oeuvresthomasja02stierich&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;Œuvres Complètes&lt;/em&gt;, Vol. II&lt;/a&gt;.&#xA;To compute the moments, use the substitution $x=u^4$ to write&#xA;$I_n = \int_0^{\infty} f(x) x^n dx = 4 \int_0^{\infty} e^{-u} \sin(u) u^{4n+3} du$;&#xA;then integrate by parts four times (differentiating the power of $u$, and integrating the rest) to show that $I_n$ is proportional to $I_{n-1}$,&#xA;and finally check that $I_0=0$.&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2011-01-10T23:23:22.097" />
  <row Id="17048" PostTypeId="2" ParentId="17041" CreationDate="2011-01-10T23:31:09.773" Score="4" ViewCount="" Body="&lt;p&gt;A simple but powerful way to prove by complete induction that $\rm\ f(n) = 4\cdot 9^n + 3\cdot 2^n \equiv 0\ \ (mod\ 7)\:$ is as follows: Put $\rm\ S\ f(n) = f(n+1)\:.\ $ Since $\rm\ S-9\ $ kills $\rm\ 9^n,\ $ and $\rm\ S-2\ $ kills $\rm\:2^n\:,\ $ we see that $\rm (S-9)\ (S-2)\ $ kills $\rm\:f(n)\:,\ $ i.e. $\rm\ f(n)\ $ satisfies  $\rm\ f(n+2) - 11\ f(n+1) + 18\ f(n) = 0\:.\ $ Now since $\rm\ 0\equiv f(0)\equiv f(1)\:,\: $ using the recurrence and complete induction shows that $\rm\ f(n)\equiv 0\ $ for all $\rm\ n \in \mathbb N$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;An analogous complete induction proves that a solution of a monic linear recurrence is determined uniquely by its initial conditions - the &lt;strong&gt;uniqueness theorem for linear difference equations&lt;/strong&gt;. Generally uniqueness theorems provide very powerful tools for proving equalities. See some of my &lt;a href=&quot;http://math.stackexchange.com/search?q=user:242+uniqueness+equalities&quot;&gt;other posts&lt;/a&gt; for further examples of such.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is closely related to inductive proofs of the &lt;strong&gt;recursion theorem&lt;/strong&gt;, which justifies the use of recursive definitions. For a nice introduction see Henkin: &lt;a href=&quot;http://www.phys.uu.nl/~dockum/Leon%20Henkin%20-%20On%20mathematical%20induction.pdf&quot; rel=&quot;nofollow&quot;&gt;On mathematical induction.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2011-01-10T23:49:43.637" LastActivityDate="2011-01-10T23:49:43.637" CommentCount="1" />
  <row Id="17049" PostTypeId="2" ParentId="16969" CreationDate="2011-01-10T23:37:07.397" Score="0" ViewCount="" Body="&lt;p&gt;If i've understood what u say, i've to calculate $E[W_t \int_0^t W_s ds]$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Tell me if i'm right doing this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;as $\int_0^t W_sds=tW_t-\int_0^t s dW_s$, i've to calculate $E[W_t (tW_t-\int_0^t s dW_s)]$, or better $E[t W^2_t]-E[W_t\int_0^t s dW_s)]$, and the first expectation is $t^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;as $W_t=\int_0^t dW_s$ the second expectation become $ E[\int_0^t dW_s\int_0^t s dW_s)]$ that is for Ito-isometry (hope that this is the correct translation) = $ E[\int_0^t s ds)]$, and this expectation is $t^2/2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So covariance is  $t-t^2/2=t^2/2$...  &lt;/p&gt;&#xA;" OwnerUserId="5584" LastActivityDate="2011-01-10T23:37:07.397" CommentCount="3" />
  <row Id="17050" PostTypeId="1" CreationDate="2011-01-11T00:16:16.740" Score="0" ViewCount="120" Body="&lt;p&gt;Hello. I'm trying to solve the following exercise from Munkres book. Can you please check it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $X$ be a second countable space and let $A$ be an uncountable subset of $X$. Show that an uncountable number of points of $A$ are limit points of $A$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Proof&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $C = \{x \in A: \textrm{x is a limit point of A} \}$. We want to show that $C$ is uncountable, so suppose not, then C is countable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $B$ be a countable basis for $X$. Then for each $a \in A \setminus C$ we have&#xA;that a is &lt;em&gt;not&lt;/em&gt; a limit point of $A$ so we can find a basis element $U_{a}$ such that $U_{a} \cap A = \{a\}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The above defines an injection from the set $A \setminus C$ to $B$ by just mapping each $a \in A \setminus C$ to $U_{a}$. Since $B$ is countable (because $X$ is 2nd countable) then $A \setminus C$ is countable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But then $(A \setminus C) \cup C = A$ is countable, contradiction because $A$ is uncountable.&lt;/p&gt;&#xA;" OwnerUserId="5587" LastActivityDate="2011-01-11T00:16:16.740" Title="Uncountably many points of an uncountable set in a second countable are limit points" Tags="&lt;topology&gt;" CommentCount="4" />
  <row Id="17051" PostTypeId="1" AcceptedAnswerId="17052" CreationDate="2011-01-11T00:24:34.353" Score="2" ViewCount="123" Body="&lt;p&gt;It is well known that for a matrix Lie group the Lie exponential map is &lt;strong&gt;$e ^z$&lt;/strong&gt;. This maps a tangent vector $z$ at the &lt;strong&gt;identity&lt;/strong&gt; to a group element. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand the general Riemannian exponential map centered at point $x$ is given by $\exp _x \triangle$  which maps a tangent vector $\triangle$ at point $x$ (not necessarily identity element) to a group element.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a relationship between these two exponential maps?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example is below formula correct? If so, are there any conditions involved?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\exp _x \triangle = xe ^{x^{-1}\triangle}$ &lt;/p&gt;&#xA;" OwnerUserId="5312" LastActivityDate="2011-01-11T01:09:49.743" Title="Relationship between Riemannian Exponential Map and Lie Exponential Map" Tags="&lt;differential-geometry&gt;&lt;lie-algebra&gt;&lt;riemannian-geometry&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="17052" PostTypeId="2" ParentId="17051" CreationDate="2011-01-11T01:09:49.743" Score="4" ViewCount="" Body="&lt;p&gt;Notice that you need to pick a metric on a Lie group for the &quot;general Riemannian exponential map&quot; to be defined.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you happen to pick an invariant metric on a Lie group, then every geodesic is (locally) a translate of a 1-parameter subgroup (so essentially both exponentials are the same thing)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know what happens if there are no invariant metrics.&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2011-01-11T01:09:49.743" CommentCount="6" />
  <row Id="17053" PostTypeId="2" ParentId="16982" CreationDate="2011-01-11T01:35:26.757" Score="7" ViewCount="" Body="&lt;p&gt;The general form for the infinitesimal generator of a continuous diffusion in $\mathbb{R}^n$ is&#xA;$$&#xD;&#xA;Af(x) = \frac12\sum_{ij}a_{ij}\frac{\partial^2 f(x)}{\partial x_i\partial x_j}+\sum_ib_i\frac{\partial f(x)}{\partial x_i}-cf(x).\qquad{\rm(1)}&#xD;&#xA;$$&#xA;Here, $a_{ij}$ is a positive-definite and symmetric nxn matrix, $b_i$ is a vector and $c$ is a non-negative scalar, with the coefficients $a,b,c$ functions of position $x$. Such operators are said to be &lt;em&gt;semi-elliptic second order differential operators&lt;/em&gt;. The case with $c=0$ is the most common - being the generator of a Markov process (or semigroup). However, the $c &amp;gt; 0$ case does occur, and is then a generator of a submarkovian semigroup.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The coefficients $a,b,c$ can be understood as follows: $a$ gives the covariance matrix of the motion over small time intervals (i.e., the level of noise). $b$ gives the mean over small time intervals (the drift) and $c$ is the rate at which the process is &quot;killed&quot;. To be precise, a processes $X_t$ can be modeled formally by adding an additional state $\it\Delta$ called the &lt;em&gt;cemetary&lt;/em&gt;. So, we represent the state space for the killed diffusion as $\mathbb{R}^n\cup\{{\it\Delta}\}$. In a small time interval $\delta t$, the process has probability $c\delta t$ of being killed, in which case it jumps to the cemetary, and stays there. So, $X_t={\it\Delta}$ for all $t\ge\tau$ with $\tau$ being the (random) time when the process is killed. The terminology I am using here is taken from &lt;a href=&quot;http://books.google.co.uk/books?id=1ml95FLM5koC&amp;amp;lpg=PP1&amp;amp;ots=-6lir7APLu&amp;amp;dq=revuz%20yor&quot; rel=&quot;nofollow&quot;&gt;Revuz &amp;amp; Yor (Continuous Martingales and Brownian Motion)&lt;/a&gt;, and will vary between different authors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, getting back to PDEs. Suppose we want to solve the PDE $A\psi(x)=0$ on an open domain $U\subseteq\mathbb{R}^n$ with boundary condition $\psi(x)=\psi_0(x)$ for $x$ on the boundary of $U$ ($\partial U$, say). You can do the following. Simulate the process $X_t$ with initial condition $X_0=x$. Wait until the first time $T$ at which it hits the boundary and, when this occurs (if the process doesn't get killed first. i.e., $T &amp;lt; \tau$), take the expected value.&#xA;$$&#xD;&#xA;\psi(x)=\mathbb{E}_x\left[\psi_0(X_T)1_{\{T &amp;lt; \tau\}}\right].\qquad\qquad{\rm(2)}&#xD;&#xA;$$&#xA;Then $\psi$ satisfies the PDE $A\psi=0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is all very general. Getting back to the Helmholtz equation, we can let $a_{ij}$ be the identity matrix and $b_i=0$, and $c$ is a constant. In that case our generator becomes $A\psi=\frac12\Delta\psi-c\psi$. [&lt;strong&gt;Edit:&lt;/strong&gt; This is not quite the same as the Helmholtz equation, which has $c=-k^2$, because here we have $c &amp;gt; 0$. There is a sign difference which changes the behaviour of the solutions. See below] The process then is the following: run a Brownian motion starting from $x$ until the first time $T$ it hits the boundary. Decide if it has not been killed, which has probability $e^{-cT}$ conditional on $T$. If it hasn't, take the value $\psi_0(X_T)$. Finally, take the average of this process (e.g., using Monte Carlo). There is one practical issue here though. Throwing away all the paths on which the process gets killed is a bit wasteful, so you would simply multiply by the probability of not being killed on each path, rather than actually discarding them. i.e., you simulate a regular Brownian motion, and then calculate&#xA;$$&#xD;&#xA;\psi(x)=\mathbb{E}_x\left[\psi_0(X_T)e^{-cT}\right].\qquad\qquad{\rm(3)}&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can even go the whole hog and solve $A\psi(x)=\varphi(x)$ for a general $x$-dependent coefficients and source term $\varphi$,&#xA;$$&#xD;&#xA;\begin{align}&#xD;&#xA;\psi(x)&amp;amp;=\mathbb{E}_x\left[1_{\{T&amp;lt;\tau\}}\psi_0(X_T)-\int_0^{T\wedge\tau}\varphi(X_t)\,dt\right]\\&#xD;&#xA;&amp;amp;=\mathbb{E}_x\left[e^{-\int_0^Tc(\hat X_s)\,ds}\psi(\hat X_T)-\int_0^Te^{-\int_0^tc(\hat X_s)\,ds}\varphi(\hat X_t)\,dt\right].&#xD;&#xA;\end{align}\qquad{\rm(4)}&#xD;&#xA;$$&#xA;Here, $X$ is the process killed at (state-dependent) rate $c$, and I'm using $\hat X$ for the process without killing, which requires multiplying by the survival probabilities $e^{-\int c(\hat X_s)\,ds}$ instead.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One other area in which you have a '$-cf$' term in the PDE governing diffusions is in finance, and it occurs in two different, but closely related ways. Prices of financial assets are frequently modeled as diffusions (even, jump-diffusions), and the value of a financial derivative would be expressed as the expected value of its future value - under a so-called &quot;risk-neutral measure&quot; or &quot;martingale measure&quot; (which are just a special probability measures). However, you need to take &lt;em&gt;interest rates&lt;/em&gt; into account. If the rate is $r$, then you would multiply the future (time $t$) value by $e^{-rt}$ before taking the expected value, which is effectively the same as adding a $-rf(x)$ term to the generator. And, as in the general case above, $r$ can be a function of the market state. The second main way (which occurs to me) in which such terms appear in finance is due to credit risk. If a counterparty has probability $rdt$ of defaulting in any time $t$, then you would have a $-rf(x)$ term occurring in the diffusion. This is more in line with the &quot;killing&quot; idea discussed above, but behaves in much the same way as interest rates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, I'll mention that the PDE in the more general time-dependent situation is of the form $\partial f/\partial t + Af =0$, where $f$ is the expected value of some function of the process at a future time (and not necessarily the first time it hits the boundary). As mentioned in the other answer this is sometimes known as the &lt;a href=&quot;http://en.wikipedia.org/wiki/Feynman%E2%80%93Kac_formula&quot; rel=&quot;nofollow&quot;&gt;Feynman-Kac formula&lt;/a&gt;, generally by physicists, and also as the &lt;a href=&quot;http://en.wikipedia.org/wiki/Kolmogorov_backward_equations_%28diffusion%29&quot; rel=&quot;nofollow&quot;&gt;Kolmogorov backward equation&lt;/a&gt; by mathematicians. Actually, the backward equation in the Wikipedia link doesn't have the $-cf$ term, but it would in the more general case of diffusions with killing. The adjoint PDE applying to probability densities is known as the &lt;a href=&quot;http://en.wikipedia.org/wiki/Fokker%E2%80%93Planck_equation&quot; rel=&quot;nofollow&quot;&gt;Fokker-Planck equation&lt;/a&gt; by physicists and the Kolmogorov forward equation to mathematicians.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; As mentioned above, what we have here does not quite correspond to the Helmholtz equation, because of the sign of $c$, and the behaviour of the solutions does change depending on whether $c$ is positive or negative. In the probabilistic interpretation, $c &amp;gt; 0$ is the rate at which the process is killed. Looking at (3) and (4), we can see that solutions to the PDE will decay exponentially as we move further from the boundary. Furthermore, if the values on the boundary are non-negative, then $\psi$ has to be non-negative everywhere. The probabilistic method naturally leads to $\psi(x)$ being a positive linear combination of its boundary values (i.e., an integral with respect to a measure on the boundary). On the other hand, the Helmholtz equation has oscillating wavelike solutions. The values of $\psi(x)$ can exceed its values on the boundary and, even if $\psi\vert_{\partial U}$ is positive, it is possible for $\psi(x)$ to go negative inside the domain. So, it is not a positive linear combination of its boundary values. We could just try using a negative $c$ in (3) and (4) but, for the reasons just mentioned, this cannot work in general. What happens is that $e^{\vert c\vert T}$ is not integrable. To get around this, it is possible to transform the Helmholtz equation so that the zeroth order coefficient $-c$ is positive. We can make a substitution such as $\psi(x)=\tilde\psi(x)e^{ikS(x)}$, where $S$ is any solution to $\Vert\nabla S\Vert=1$. Then, the Helmholtz equation becomes,&#xA;$$&#xD;&#xA;\frac12\Delta\tilde\psi + ik\nabla S\cdot\nabla\tilde\psi + \frac{ik}{2}(\Delta S)\tilde\psi=0.&#xD;&#xA;$$&#xA;So we have a zeroth order term of the form $-\tilde c\tilde\psi$ for $\tilde c=-ik\Delta S/2$. This is imaginary, so does not make sense as a &quot;killing&quot; rate any more. However, as its real component is nonnegative (zero here), equations such as (3) and (4) above give bounded and well-defined expressions for $\tilde\psi$. A google search gives the following paper which uses such a substitution: &lt;a href=&quot;http://rspa.royalsocietypublishing.org/content/463/2080/1005.full&quot; rel=&quot;nofollow&quot;&gt;Novel solutions of the Helmholtz equation and their application to diffraction&lt;/a&gt;. I expect that the papers linked to in the other answer use similar techniques to transform the equation into a form which can be handled by the probabilistic method - although I have not had a chance to look at them yet, and are not free access.&lt;/p&gt;&#xA;" OwnerUserId="1321" LastEditorUserId="1321" LastEditorDisplayName="" LastEditDate="2011-01-14T22:41:12.147" LastActivityDate="2011-01-14T22:41:12.147" CommentCount="1" />
  <row Id="17054" PostTypeId="1" AcceptedAnswerId="17057" CreationDate="2011-01-11T01:38:02.723" Score="11" ViewCount="382" Body="&lt;p&gt;Let $G$ be a group where every non-identity element has order 2.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If |G| is finite then $G$ is isomorphic to the direct product $\mathbb{Z}_{2} \times \mathbb{Z}_{2} \times \ldots  \times \mathbb{Z}_{2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is the analogous result &#xA; $G= \mathbb{Z}_{2} \times \mathbb{Z}_{2} \times \ldots $.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;true for the case |G| is infinite? &lt;/p&gt;&#xA;" OwnerUserId="832" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2011-01-11T02:08:42.353" LastActivityDate="2011-01-21T05:06:57.893" Title="Group where every element is order 2." Tags="&lt;group-theory&gt;" AnswerCount="3" FavoriteCount="1" />
  <row Id="17055" PostTypeId="2" ParentId="17054" CreationDate="2011-01-11T01:47:07.453" Score="3" ViewCount="" Body="&lt;p&gt;Not really. The direct sum $\bigoplus_{n\in{\mathbb N}}{\mathbb Z}_2$ is a counterexample. More generally, take $G=\prod_{n=1}^\infty{\mathbb Z}_2$. This set has size $|{\mathbb R}|$. It is easy to see that, given &lt;em&gt;any&lt;/em&gt; element of this group, there is a countable subgroup of $G$ that contains this element. Much more pathological counterexamples are also possible, of course.&lt;/p&gt;&#xA;" OwnerUserId="462" LastActivityDate="2011-01-11T01:47:07.453" CommentCount="2" />
  <row Id="17056" PostTypeId="2" ParentId="17054" CreationDate="2011-01-11T01:58:57.657" Score="9" ViewCount="" Body="&lt;p&gt;If every non-identity element of $G$ has order 2, then the group is abelian. Notationally, it helps to write the group operation additively, with identity $0$. In that case, you can view the group as a vector space over the field with two elements, $F_2=\mathbb{Z}/2\mathbb{Z}$. Every vector space has a basis, so&#xA;$$&#xD;&#xA;G\cong\bigoplus_{i\in I}F_2&#xD;&#xA;$$&#xA;where $I$ has the cardinality of a basis of $G$. The point here is that representing a vector space by a basis corresponds to a &lt;a href=&quot;http://en.wikipedia.org/w/index.php?title=Direct_sum_of_modules&amp;amp;oldid=407170722#Construction_for_an_arbitrary_family_of_modules&quot; rel=&quot;nofollow&quot;&gt;direct sum&lt;/a&gt; rather than the &lt;a href=&quot;http://en.wikipedia.org/w/index.php?title=Direct_product&amp;amp;oldid=407170150#Direct_product_of_modules&quot; rel=&quot;nofollow&quot;&gt;direct product&lt;/a&gt;, because you can only take &lt;em&gt;finite&lt;/em&gt; linear combinations of basis elements. (and, as pointed out in Andres' answer, it is not possible to represent all such groups as direct products).&lt;/p&gt;&#xA;" OwnerUserId="1321" LastActivityDate="2011-01-11T01:58:57.657" />
  <row Id="17057" PostTypeId="2" ParentId="17054" CreationDate="2011-01-11T02:04:54.980" Score="15" ViewCount="" Body="&lt;p&gt;Perhaps the best way to look at the problem is to establish the following more precise result:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a group $G$, the following are equivalent:&lt;br&gt;&#xA;(i) Every non-identity element of $G$ has order $2$.&lt;br&gt;&#xA;(ii) $G$ is commutative, and there is a unique $\mathbb{Z}/2\mathbb{Z}$-vector space structure on $G$ with the group operation as addition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I guess you probably already know how to show that if every nonidentity element has order $2$, $G$ is commutative: for all $x,y \in G$, $e = (xy)^2 = xyxy$.  Multiplying on the left by $x$ and on the right by $y$ gives $xy = yx$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Having established the commutativity, it is convenient to write the group law additively.  Then there is only one possible $\mathbb{Z}/2\mathbb{Z}$-vector space structure on $G$, since it remains to define a scalar multiplication and of course we need $0 \cdot x = 0, \ 1 \cdot x = x$ for all $x \in G$.  But you should check that this actually works: i.e., defines an $\mathbb{Z}/2\mathbb{Z}$-vector space structure, just by checking the axioms: the key point is that for all $x \in G$, $(1+1)x = x + x = 0 = 0x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So now your question is equivalent to: is every $\mathbb{Z}/2\mathbb{Z}$ vector space isomorphic to a product of copies of $\mathbb{Z}/2\mathbb{Z}$?  Well, the only invariant of a vector space is its dimension.  It is clear that every finite-dimensional vector space is of this form.  Every infinite dimensional space is isomorphic to a &lt;em&gt;direct sum&lt;/em&gt; $\bigoplus_{i \in I} \mathbb{Z}/2\mathbb{Z}$, the distinction being that in a direct sum, every element has only finitely many nonzero entries.  (In other words, the allowable linear combinations of basis elements are finite linear combinations.)  Moreover, for any infinite index set $I$, the direct sum $\bigoplus_{i \in I} \mathbb{Z}/2\mathbb{Z}$ has dimension $I$ and also cardinality $I$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, it is not possible for a direct product of two element sets to have countably infinite cardinality: if $I$ is infinite, it is at least countable, and then the infinite direct product has the same cardinality of the real numbers (think of binary expansions).  So the answer to your question is &quot;yes&quot; for direct sums, but &quot;no&quot; for direct products.&lt;/p&gt;&#xA;" OwnerUserId="299" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2011-01-21T05:06:57.893" LastActivityDate="2011-01-21T05:06:57.893" CommentCount="2" />
  <row Id="17058" PostTypeId="2" ParentId="16911" CreationDate="2011-01-11T03:11:24.693" Score="1" ViewCount="" Body="&lt;p&gt;An intuitive method of approaching this is as follows.&#xA;If $t\mapsto x(t)$ is an integral curve then $\vert f(x(t_1))-f(x(t_0))\vert=\vert t_1-t_0\vert$. However, any other curve $t\mapsto y(t)$ joining points $y(s_0)=x(t_0)$ and $y(s_1)=x(t_1)$ satisfies $\vert f(y(s_1))-f(y(s_0))\vert\le\vert s_1-s_0\vert$ when parameterized by arc length. So $x$ is the shortest curve between the points.&lt;/p&gt;&#xA;" OwnerUserId="1321" LastEditorUserId="1321" LastEditorDisplayName="" LastEditDate="2011-01-11T03:38:19.053" LastActivityDate="2011-01-11T03:38:19.053" CommentCount="2" />
  <row Id="17059" PostTypeId="2" ParentId="16970" CreationDate="2011-01-11T03:25:02.017" Score="1" ViewCount="" Body="&lt;p&gt;If I understand you correctly, you are asking &quot;what is the formula that tells you how many iterations to do to get the Mandelbrot set to have a 'nice' resolution at some given level of magnification?&quot;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&quot;I'd need some method of finding something like an ideal max Iterations value depending on the zoom level&quot; is not a well phrased question; for one thing, the concept of &quot;ideal max iterations&quot; is not well defined and could mean different things for different people.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;There is no easy way to make the video like the one you linked to, and I'm sure it took a very long time to make that one because each frame had to be resolved one at a time and then put back together.  I have no way of knowing for sure, but I have always assumed that video was made by determining how many iterations for each frame (zoom level) to do one frame at a time, by trial and error.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;That being said, I believe that the rapidity for which one point on the set converges has little/no effect on the convergence properties of many other points in the set which lie very close in distance; and therefore I don't expect that the formula you are looking for will exist. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is that some regions (and some sub regions...) of the set require more iterations to resolve in a nice way than others.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if you included some additional information such as specifying the path you took as you zoomed in on the set, then it would probably be possible to find a formula that accomplishes the thing you are trying to do.  I say this last point based only from my personal experience playing around with ChaosPro, not for any major theoretical reason.&lt;/p&gt;&#xA;" OwnerUserId="1417" LastEditorUserId="1417" LastEditorDisplayName="" LastEditDate="2011-01-11T03:45:40.023" LastActivityDate="2011-01-11T03:45:40.023" />
  <row Id="17060" PostTypeId="2" ParentId="14183" CreationDate="2011-01-11T04:24:58.707" Score="1" ViewCount="" Body="&lt;p&gt;hi,you can learn calculus from video tuts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;here is the link&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.downloadmela.com/video-watch?sno=15&quot; rel=&quot;nofollow&quot;&gt;calculus video lectures&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5591" LastActivityDate="2011-01-11T04:24:58.707" />
  <row Id="17061" PostTypeId="1" CreationDate="2011-01-11T04:29:29.867" Score="3" ViewCount="117" Body="&lt;p&gt;Do there exist two Lie groups which are diffeomorphic as smooth manifolds, have isomorphic group structures, yet are not isomorphic as Lie groups?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, for this to happen, any diffeomorphism would fail to preserve the group structure, and any group isomorphism would either fail to be smooth, or its inverse would fail to be smooth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have no other reason for asking this other than out of curiosity.  (In particular, this is not a problem I found out of a textbook.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Related Question: Are there topological groups that are homeomorphic and have isomorphic group structures, yet are not isomorphic as topological groups?&lt;/p&gt;&#xA;" OwnerUserId="640" LastActivityDate="2011-01-11T04:29:29.867" Title="Diffeomorphic, group-isomorphic Lie groups that are not isomorphic as Lie groups" Tags="&lt;differential-geometry&gt;&lt;differential-topology&gt;&lt;lie-groups&gt;" CommentCount="2" />
  <row Id="17062" PostTypeId="1" AcceptedAnswerId="17080" CreationDate="2011-01-11T05:56:32.830" Score="7" ViewCount="238" Body="&lt;p&gt;Let $p : [0,1] \to \mathbb{R}^2$ be continuous and injective.&#xA;&lt;br&gt;&#xA;Does there always exists a continuous function $f : [0,1]\times \mathbb{R}^2 \to \mathbb{R}^2$ such that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For all $x\in \mathbb{R}^2$, $f(0,x) = x$&#xA;&lt;br&gt;&#xA;and&#xA;&lt;br&gt;&#xA;For all $t\in [0,1]$, $x\mapsto f(t,x)$ is a homeomorphism&#xA;&lt;br&gt;&#xA;and&#xA;&lt;br&gt;&#xA;For all $s\in [0,1]$, $f(1,p(s)) = \langle s,0\rangle$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;?&lt;/p&gt;&#xA;" OwnerUserId="5114" LastActivityDate="2011-03-08T10:59:10.267" Title="Is there &quot;essentially only 1&quot; Jordan arc in the plane?" Tags="&lt;topology&gt;" AnswerCount="1" />
  <row Id="17063" PostTypeId="1" AcceptedAnswerId="17066" CreationDate="2011-01-11T07:15:47.843" Score="9" ViewCount="305" Body="&lt;p&gt;Hi, everyone. I am afraid that my question is too trivial. But here it is. The Klein four group is the first counterexample to the the statement: &quot;If all proper subgroups of a group are cyclic, then the group is cyclic.&quot; I am looking for other examples, if any. Are there?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance.&lt;/p&gt;&#xA;" OwnerUserId="5595" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-11T17:48:28.300" LastActivityDate="2011-01-11T17:48:28.300" Title="Minimal non-cyclic groups other than the Klein four groups" Tags="&lt;group-theory&gt;" AnswerCount="3" FavoriteCount="2" />
  <row Id="17064" PostTypeId="1" CreationDate="2011-01-11T07:18:49.737" Score="3" ViewCount="75" Body="&lt;ul&gt;&#xA;&lt;li&gt;$G$ is a simple graph&lt;/li&gt;&#xA;&lt;li&gt;$|V(G)|$ is even&lt;/li&gt;&#xA;&lt;li&gt;$\delta(G) \gt \frac{|V(G)|}{2}+3$ where $\delta(G)$ is the minimum degree of a vertex in $G$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The question is to prove there exists a $5$-regular spanning subgraph of $G$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm trying to figure this out for at least 10 hours, but no avail. :(&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only thing I've managed to do is prove that $|V(G)| \ge 8$ and that $G$ is Hamiltonian (via Dirac's theorem)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any ideas?&lt;/p&gt;&#xA;" OwnerUserId="5596" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2011-01-11T17:44:00.240" LastActivityDate="2011-01-11T17:44:00.240" Title="If $\delta(G) \gt \frac{|V(G)|}{2}+3$, there exists a $5$-regular spanning subgraph" Tags="&lt;graph-theory&gt;" CommentCount="2" />
  <row Id="17066" PostTypeId="2" ParentId="17063" CreationDate="2011-01-11T07:26:58.570" Score="13" ViewCount="" Body="&lt;p&gt;The example of the Klein group, $\mathbb{Z}_2 \times \mathbb{Z}_2$, generalizes naturally to $\mathbb{Z}_p \times \mathbb{Z}_p$ for any prime $p$. Other finite examples include any nonabelian group of order $pq$, where $p$ and $q$ are primes (such a group exists whenever $p-1$ is divisible by $q$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But there are even infinite examples. For all sufficiently large primes $p$, there exist infinite groups all of whose proper nontrivial subgroups are cyclic of order $p$. These are the &lt;a href=&quot;http://en.wikipedia.org/wiki/Tarski_monster_group&quot; rel=&quot;nofollow&quot;&gt;Tarski monsters&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="5203" LastActivityDate="2011-01-11T07:26:58.570" CommentCount="8" />
  <row Id="17067" PostTypeId="1" AcceptedAnswerId="17068" CreationDate="2011-01-11T07:42:07.690" Score="4" ViewCount="120" Body="&lt;p&gt;Is there a real function that is differentiable at any point but nowhere monotone?&lt;/p&gt;&#xA;" OwnerUserId="5422" LastActivityDate="2011-01-11T07:48:17.357" Title="Differentiable+Not monotone" Tags="&lt;real-analysis&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="17068" PostTypeId="2" ParentId="17067" CreationDate="2011-01-11T07:48:17.357" Score="7" ViewCount="" Body="&lt;p&gt;Yes. See for example &quot;Everywhere Differentiable, Nowhere Monotone, Functions&quot; by Y. Katznelson and Karl Stromberg. &lt;a href=&quot;http://docs.google.com/viewer?a=v&amp;amp;q=cache%3axCbyLWTvabgJ%3awww.les-mathematiques.net/phorum/file.php?4,file%3D8103,filename%3Dkatznelson-stromberg.pdf&quot; rel=&quot;nofollow&quot;&gt;Google docs link&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5203" LastActivityDate="2011-01-11T07:48:17.357" />
  <row Id="17069" PostTypeId="1" AcceptedAnswerId="17070" CreationDate="2011-01-11T08:12:18.207" Score="3" ViewCount="224" Body="&lt;p&gt;Is there a continuous and monotone function that's nowhere differentiable ?&lt;/p&gt;&#xA;" OwnerUserId="5422" LastActivityDate="2011-01-11T08:23:57.500" Title="Monotone+continuous but not differentiable" Tags="&lt;real-analysis&gt;" AnswerCount="2" />
  <row Id="17070" PostTypeId="2" ParentId="17069" CreationDate="2011-01-11T08:16:56.960" Score="7" ViewCount="" Body="&lt;p&gt;No.  Even without the assumption of continuity, a monotone function on $\mathbb{R}$ is differentiable except on a set of measure $0$ (and it can have only countably many discontinuities).  This is mentioned on &lt;a href=&quot;http://en.wikipedia.org/wiki/Monotonic_function#Some_basic_applications_and_results&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;, and proofs can be found in books on measure theory such as Royden or Wheeden and Zygmund.  You can read the details in the latter book at &lt;a href=&quot;http://books.google.com/books?id=YDkDmQ_hdmcC&amp;amp;lpg=PP1&amp;amp;pg=PA111#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;this link&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2011-01-11T08:16:56.960" />
  <row Id="17071" PostTypeId="2" ParentId="17069" CreationDate="2011-01-11T08:23:57.500" Score="2" ViewCount="" Body="&lt;p&gt;Even any function of bounded variation is differentiable almost everywhere.&lt;/p&gt;&#xA;" OwnerUserId="2810" LastActivityDate="2011-01-11T08:23:57.500" CommentCount="2" />
  <row Id="17072" PostTypeId="1" AcceptedAnswerId="17082" CreationDate="2011-01-11T09:00:07.737" Score="1" ViewCount="97" Body="&lt;p&gt;I have a system described an equation,&#xA;and I want to find an (DE) equation for z(x,t), in the limit as l-&gt;0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First some definitions to simplify it some:&lt;br&gt;&#xA;$Z1=z(x,t)-z(x+l,t)$&lt;br&gt;&#xA;$Z2=z(x,t)-z(x-l,t)$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;These are not constants, just to simplify expr.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;c however, is a constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EQ:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$c*l^2*d^2z(x,t)/dt^2=(1-l/sqrt(Z1^2+l^2))*Z1+(1-l/sqrt(Z2^2+l^2))*Z2$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, I want to find the limit of this expression as l-&gt;0, from the positive side.&#xA;(l is a physical distance, and this is a physical realizable system, hopefully in the limit too).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So z(x,t) initially only makes physical sense for x=0,l,2l,3l.&lt;br&gt;&#xA;And I want to make it into a continuous function for all x.&lt;br&gt;&#xA;I dont know if its relevant but z(x,t) and x is also a distance, and t is time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please tell me if I should clarify something, and whether the problem as stated is well defined mathematically.&lt;/p&gt;&#xA;" OwnerUserId="4929" LastEditorUserId="4929" LastEditorDisplayName="" LastEditDate="2011-01-11T12:19:12.637" LastActivityDate="2011-01-11T12:48:48.247" Title="Continuum limit of (DE) equation" Tags="&lt;calculus&gt;&lt;real-analysis&gt;&lt;physics&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="17074" PostTypeId="1" CreationDate="2011-01-11T10:11:44.917" Score="11" ViewCount="239" Body="&lt;p&gt;Einsteins theory of gravitation, &lt;a href=&quot;http://en.wikipedia.org/wiki/General_relativity&quot; rel=&quot;nofollow&quot;&gt;general relativity&lt;/a&gt;, is a purely &lt;a href=&quot;http://en.wikipedia.org/wiki/Differential_geometry&quot; rel=&quot;nofollow&quot;&gt;geometric theory&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a recent question I wanted to know what the relation of &lt;a href=&quot;http://math.stackexchange.com/questions/16982/relation-of-brownian-motion-to-helmholtz-equation&quot;&gt;Brownian Motion to the Helmholtz equation&lt;/a&gt; is and got a very thorough &lt;a href=&quot;http://math.stackexchange.com/questions/16982/relation-of-brownian-motion-to-helmholtz-equation/17053#17053&quot;&gt;answer&lt;/a&gt; from &lt;a href=&quot;http://math.stackexchange.com/users/1321/george-lowther&quot;&gt;George Lowther&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;He pointed out that there is, roughly speaking, a very general relation of semi-elliptic second order differential operators of the form &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$Af = \frac12 a^{ij}f_{,ij} + b^i f_{,i} - cf = 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;to a &quot;killed&quot; Brownian motion. (I used some &lt;a href=&quot;http://mathworld.wolfram.com/EinsteinSummation.html&quot; rel=&quot;nofollow&quot;&gt;summation convention&lt;/a&gt; and $,i = \frac{\partial}{\partial x^i}$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the &lt;a href=&quot;http://en.wikipedia.org/wiki/Einstein_field_equations&quot; rel=&quot;nofollow&quot;&gt;Einstein field equations&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$R_{\mu\nu}-\frac12 g_{\mu\nu}R = \frac{8\pi G}{c^4}T_{\mu\nu}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;are coupled hyperbolic-elliptic partial differential equations (I dropped the cosmological constant here). Can we somehow &lt;strong&gt;adopt&lt;/strong&gt; the relation of a &lt;strong&gt;random process&lt;/strong&gt; to this kind of equation, or&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Is there a way to interprete the&#xA;  Einstein equations stochastically?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Thank you in advance&lt;br&gt;&#xA;Sincerely&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Robert&lt;/p&gt;&#xA;" OwnerUserId="2671" LastEditorUserId="1543" LastEditorDisplayName="" LastEditDate="2011-04-18T12:14:01.830" LastActivityDate="2011-04-18T12:14:01.830" Title="Stochastic interpretation of Einstein Equations" Tags="&lt;differential-geometry&gt;&lt;differential-equations&gt;&lt;stochastic-processes&gt;&lt;physics&gt;&lt;general-relativity&gt;" CommentCount="3" FavoriteCount="3" />
  <row Id="17075" PostTypeId="1" AcceptedAnswerId="17081" CreationDate="2011-01-11T10:34:10.710" Score="0" ViewCount="74" Body="&lt;p&gt;Can someone help me with this problem?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $D$ be a divisor on an algebraic curve $X$ of genus $g$, such that $\deg D = 2g-2$ and $\dim L(D) = g$. Then $D$ must be a canonical divisor. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;By Riemann-Roch, I see that $\dim L(K-D) = 1$ for any canonical divisor $K$, as must certainly be the case. I don't know if this is too helpful. &lt;/p&gt;&#xA;" OwnerUserId="1491" LastActivityDate="2011-01-11T11:44:47.980" Title="Canonical divisor on algebraic curve" Tags="&lt;riemann-surface&gt;" AnswerCount="1" />
  <row Id="17076" PostTypeId="1" AcceptedAnswerId="17077" CreationDate="2011-01-11T10:37:32.767" Score="3" ViewCount="120" Body="&lt;p&gt;I was working out some problems from a entrance test paper, and this problem looked a bit difficult. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Does there exist an analytic function on $\mathbb{C}$ such that $$f\biggl(\frac{1}{2n}\biggr)=f\biggl(\frac{1}{2n+1}\biggr)=\frac{1}{2n} \quad \ \text{for all} \ n\geq 1?$$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I did apply $f$ repeatedly to get some more insights about the problem, but couldn't get anything from it. So how do i solve this one?&lt;/p&gt;&#xA;" LastActivityDate="2011-01-11T10:46:59.200" Title="analytic function on $\mathbb{C}$ such that $f\Bigl(\frac{1}{2n}\Bigr)=f\Bigl(\frac{1}{2n+1}\Bigr)=\frac{1}{2n}$" Tags="&lt;analysis&gt;&lt;complex-analysis&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="17077" PostTypeId="2" ParentId="17076" CreationDate="2011-01-11T10:46:59.200" Score="8" ViewCount="" Body="&lt;p&gt;Suppose there is such an $f(z)$. Let $g(z) = f(z) - z$. If $f(z)$ were analytic, then $g(z)$ would also be analytic. But $g(z)$ has a limiting sequence of zeros (in particular, $g(1/2n) = 0$), and the zeros of any nonconstant analytic function are discrete. Therefore $g(z) = 0$, which is not the case; contradiction. So there is no such analytic function. &lt;/p&gt;&#xA;" OwnerUserId="1491" LastActivityDate="2011-01-11T10:46:59.200" />
  <row Id="17078" PostTypeId="1" CreationDate="2011-01-11T11:09:43.163" Score="3" ViewCount="95" Body="&lt;p&gt;Let $S_3$ be the symmetric group on the three objects $x_1, x_2, x_3$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We are given a (countably infinite) sequence of permutations in $S_3$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sigma_1 \ , \ \sigma_2 \ , \ \sigma_3 \ , \ \ldots \ , \ \sigma_n \ , \ \ldots \ \, .$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it true that there exists a positive integer $N$ such that the sequence of products&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sigma_{N} \ , \ \sigma_{N+1}\sigma_{N} \ , \ \sigma_{N+2}\sigma_{N+1}\sigma_{N} \ , \ldots \, ,$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;has a(n infinite) subsequence whose terms fix $x_1$? Thanks!!!&lt;/p&gt;&#xA;" OwnerUserId="5599" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2011-01-11T17:35:33.120" LastActivityDate="2011-01-11T17:35:33.120" Title="Fixed points in a sequence of permutations" Tags="&lt;combinatorics&gt;&lt;group-theory&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" />
  <row Id="17079" PostTypeId="1" CreationDate="2011-01-11T11:41:25.813" Score="0" ViewCount="34" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;What are the value of physics constant in Kerr metric, including G, M, c, a, r, or others?&#xA;I expect to simplify Gamma&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;why g_compts[1,4] has element and not [4,1]?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Some book assume G = c = 1, what is the meaning of this setting?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Different material have different metric, are there a metric table for element table?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;What is theta in Kerr metric?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&lt;strong&gt;&lt;/em&gt;**&lt;em&gt;*&lt;/em&gt;**&lt;em&gt;&lt;/strong&gt; Kerr metric &lt;strong&gt;&lt;/em&gt;**&lt;em&gt;*&lt;/em&gt;**&lt;em&gt;*&lt;/em&gt;**&lt;/strong&gt;&#xA;t r theta phi&#xA;t&#xA;r&#xA;theta&#xA;phi&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with(tensor):&#xA;coord := [t, r, theta, Phi]:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;g_compts:=array(sparse,1..4,1..4):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;G := 6.67*10^(-11)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;triangle := r^2 - 2*G*M*r/c^2 + a^2:&#xA;p2 := r^2 + ((cos(theta))^2)*a^2:&#xA;A := (r^2+a^2)^2 - (a^2)&lt;em&gt;triangle&lt;/em&gt;(sin(theta))^2:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;g_compts[1,1]:= (triangle - (a^2)&lt;em&gt;(sin(theta))^2)&lt;/em&gt;(c^2)/p2:&#xA;g_compts[1,4]:= 4*G*M*a*r*(sin(theta))^2/(c*p2):&#xA;g_compts[2,2]:= -p2/triangle:&#xA;g_compts[3,3]:= -p2:&#xA;g_compts[4,4]:= -A*(sin(theta)^2)/(p2):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;g1 := create([-1,-1], eval(g_compts)):&#xA;g1_inv := invert( g1, 'detg' ):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;D1g := d1metric( g1, coord ):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cf1_1 := Christoffel1(D1g):&#xA;Cf2_1 := Christoffel2(g1_inv, Cf1_1):&#xA;displayGR(Christoffel2,Cf2_1): &lt;/p&gt;&#xA;" OwnerUserId="5601" LastActivityDate="2011-01-11T11:41:25.813" Title="What are the value of physics constant in Kerr metric?" Tags="&lt;differential-geometry&gt;" CommentCount="3" />
  <row Id="17080" PostTypeId="2" ParentId="17062" CreationDate="2011-01-11T11:42:23.587" Score="10" ViewCount="" Body="&lt;p&gt;The function $f$ that you want is called an &quot;ambient isotopy,&quot; and your question is whether all arcs in the plane are ambient isotopic. The answer is yes, although I'm not sure of a good reference. I believe that one can construct the ambient isotopy by hand using the 2D Schönflies theorem. The idea is to connect the endpoints of your Jordan arc by an arc, using the fact that the complement is path-connected, and apply the Schönflies theorem to the result.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The analogous result for arcs in $\mathbb R^3$ is false. There are &quot;wild&quot; arcs which are not ambiently isotopic to a standard arc. See for example &lt;a href=&quot;http://projecteuclid.org/DPubS?verb=Display&amp;amp;version=1.0&amp;amp;service=UI&amp;amp;handle=euclid.pjm/1102947540&amp;amp;page=record&quot; rel=&quot;nofollow&quot;&gt;this article.&lt;/a&gt; There are also wild spheres in $\mathbb R^3$, such as &lt;a href=&quot;http://mathworld.wolfram.com/AlexandersHornedSphere.html&quot; rel=&quot;nofollow&quot;&gt;Alexander's horned sphere.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; I asked this question on MathOverflow, which got more definitive answers &lt;a href=&quot;http://mathoverflow.net/questions/57766/why-are-there-no-wild-arcs-in-the-plane/57770#57770&quot; rel=&quot;nofollow&quot;&gt;here.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2941" LastEditorUserId="2941" LastEditorDisplayName="" LastEditDate="2011-03-08T10:59:10.267" LastActivityDate="2011-03-08T10:59:10.267" />
  <row Id="17081" PostTypeId="2" ParentId="17075" CreationDate="2011-01-11T11:44:47.980" Score="1" ViewCount="" Body="&lt;p&gt;A divisor of degree $0$ and dimension $1$ is principal.&#xA;Hence by assumption and Riemann-Roch the divisor $K-D$ is principal,&#xA;so that $D$ is linearly equivalent to the canonical divisor $K$.&lt;/p&gt;&#xA;" OwnerUserId="2479" LastActivityDate="2011-01-11T11:44:47.980" CommentCount="2" />
  <row Id="17082" PostTypeId="2" ParentId="17072" CreationDate="2011-01-11T12:31:04.027" Score="2" ViewCount="" Body="&lt;p&gt;OK, I found out how it works. So, Newton's second law and Hooke's law tell you two things. First, for the displacement in the $z$-direction you have:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$m \frac{d^2z}{dt^2} = -k \Delta l_1 \sin\theta_1 -k \Delta l_2 \sin\theta_2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second, since we assume there is no substantial displacement in the $x$-direction:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$-k \Delta l_1 \cos\theta_1 -k \Delta l_2 \cos\theta_2 \approx 0$$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using this second equation we can simplify the first as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$m \frac{d^2z}{dt^2} = -k \Delta l_1 \sin\theta_1 +k \Delta l_1 \tan\theta_2 \cos\theta_1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$m \frac{d^2z}{dt^2} = k \Delta l_1 \cos\theta_1 (\tan\theta_2 - \tan\theta_1)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, $\Delta l_1 \cos\theta_1=l$ and &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\tan\theta_1 = \frac{z(x,t)-z(x-l,t)}{l}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;while&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\tan\theta_2 = \frac{z(x+l,t)-z(x,t)}{l}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so that we get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$m \frac{d^2z}{dt^2} = k (z(x+l,t)+z(x-l,t)-2z(x,t))$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, rearranging and introducing a factor $l^2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{d^2z}{dt^2} = \frac{k l^2}{m} \frac{z(x+l,t)+z(x-l,t)-2z(x,t)}{l^2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the limit will give the second derivative on the right hand side times a constant factor only if $\lim_{l\to 0}\frac{k l^2}{m}=c$. This can be achieved by letting the parameters scale as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\lim_{l \to 0}\frac{m}{l} = M$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\lim_{l \to 0} \; k l = K$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope this clears things up. (I'm sure I made a load of sign mistakes, but I went pretty fast over the calculations, so I just made it so there was an even number of them. ;P )&lt;/p&gt;&#xA;" OwnerUserId="3567" LastEditorUserId="3567" LastEditorDisplayName="" LastEditDate="2011-01-11T12:48:48.247" LastActivityDate="2011-01-11T12:48:48.247" CommentCount="3" />
  <row Id="17083" PostTypeId="1" AcceptedAnswerId="17108" CreationDate="2011-01-11T12:50:02.833" Score="4" ViewCount="126" Body="&lt;p&gt;Let's say I have numbers each taken in a set $A$ of $n$ consecutive naturals, I ask myself : how can I found what are all the unique multisets, which could be created with $k$ elements of this set $A$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example I've got $A=[1,2,\dots,499]$. If I wanted to create unique multiset of 3 elements, I would search all the multisets $\{a,b,c\}$ such as $a\leq b\leq c$ and then have all the unique multisets. As such for each elements $a$ there is $499-(a-1)=500-a$ possibilities for $b$ and $500-b$ possibilities for $c$. Unfortunately I'm stumped here and can't find the number of possible combination, as I didn't do any maths for years. I know that I should have some kind of product, but I don't know how to find the product anymore.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So first, I would like to know if I am right in my original assumption, or if I am looking in the wrong direction. Second I would like to approach that as if I were doing an homework, as I would like to understand the logic behind it: What would be the formula to give the numbers of multisets of $k$ elements from a set $A$ of $n$ consecutive naturals?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S. I'm totally new to math.SE, as such, I'm not sure I tagged the post appropriately.&lt;/p&gt;&#xA;" OwnerUserId="5600" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2011-01-11T17:34:08.100" LastActivityDate="2011-01-12T18:52:18.363" Title="How to find unique multisets of n naturals of a given domain and their numbers?" Tags="&lt;homework&gt;&lt;combinatorics&gt;&lt;statistics&gt;" AnswerCount="4" CommentCount="3" />
  <row Id="17084" PostTypeId="2" ParentId="17078" CreationDate="2011-01-11T13:16:28.017" Score="5" ViewCount="" Body="&lt;p&gt;For each i=1,2,3 look at the sequence &#xA;$$a^{(i)}=(x_i, \sigma_1 (x_i), \sigma_2 \sigma_1 (x_i), \sigma_3 \sigma_2 \sigma_1 (x_i) ...)$$&#xA;Since $\{x_1,x_2,x_3\} = \{a^{(1)}_j, a^{(2)}_j, a^{(3)}_j\}$, then there is at least one $i$ such that $x_1$ appears in $a^{(i)}$ infinite number of time. Choose N such that $a^{(i)}_{N-1}=x_1$ then this you get&#xA;$$ \sigma_{N+k} \cdots \sigma_{N+1}\sigma_{N} x_1 = \sigma_{N+k} \cdots \sigma_{N+1}\sigma_{N} a^{(i)}_{N-1} = a^{(i)}_{N+k}$$&#xA;so $x_1$ appears infinite number of times in this sequence - meaning it is fixed by an infinite sub sequence.&lt;/p&gt;&#xA;" OwnerUserId="2125" LastActivityDate="2011-01-11T13:16:28.017" />
  <row Id="17085" PostTypeId="2" ParentId="17013" CreationDate="2011-01-11T13:51:07.623" Score="1" ViewCount="" Body="&lt;p&gt;Let $\sigma \in \Omega^n(M,E)$ and $X$, $Y$, $Z_1$, . . ., $Z_n$ be vector fields, then:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(\nabla^2\sigma)(X,Y,Z_1,. . . , Z_n) = \nabla_X(\nabla(\sigma))(Y,Z_1,. . . , Z_n) $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$= \nabla_X(\nabla_Y(\sigma))(Z_1,. . . , Z_n) -\nabla\sigma(d_X(Y))(Z_1,. . . , Z_n) $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$= \nabla_X(\nabla_Y(\sigma))(Z_1,. . . , Z_n) -(\nabla_{d_X(Y)}\sigma)(Z_1,. . . , Z_n) $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$= -\nabla_Y(\nabla_X(\sigma))(Z_1,. . . , Z_n) +(\nabla_{d_Y(X)}\sigma)(Z_1,. . . , Z_n) $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$= \frac{1}{2}(\nabla_X(\nabla_Y(\sigma)) -\nabla_Y(\nabla_X(\sigma)))(Z_1,. . . , Z_n) -\frac{1}{2}(\nabla_{d_X(Y)-d_Y(X)}\sigma)(Z_1,. . . , Z_n) $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$= \frac{1}{2}(([\nabla_X, \nabla_Y])(\sigma))(Z_1,. . . , Z_n) -\frac{1}{2}(\nabla_{[X,Y]}\sigma)(Z_1,. . . , Z_n) $&lt;/p&gt;&#xA;" OwnerUserId="1430" LastActivityDate="2011-01-11T13:51:07.623" />
  <row Id="17086" PostTypeId="2" ParentId="17083" CreationDate="2011-01-11T13:57:56.273" Score="1" ViewCount="" Body="&lt;p&gt;For a three-item multiset, the easiest approach is to start by picking $b$.  As you say, there are $500-b$ choices for $c$.  Similar logic says there are $b$ choices for $a$.  Multiplying gives the number of choices for a given $b$.  Now just sum over $b$.  So you have $$\sum_{b=1}^{499} (500-b)b$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a larger set $A$, just change 499 to the largest number in $A$.  For larger numbers of elements, you can make the formula have an upper limit of $n$ instead of 499.  This will give you the number of 4 element multisets with the upper element $n$.  Now you can sum over $n$ from 1 to the largest number in $A$.   This gets tedious with increasing $k$, but maybe you can find a pattern.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We could have followed the strategy in the second paragraph to get the 3 element sets, but the symmetry around the middle element made it a bit easier.  We would have started by saying there are $n$ one element multisets out of $n$.  Then there are $\sum_{i=1}^{n} i=\frac{n(n+1)}{2}$ two element multisets out of $n$ and $$\sum_{i=1}^n \frac{i(i+1)}{2}=\frac{n(n+1)(2n+1)}{12}+\frac{n(n+1)}{4}$$ three element multisets.  This looks different, but agrees with the prior result.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Added:  For the general case, see the section &quot;Multiset coefficients&quot; in &lt;a href=&quot;http://en.wikipedia.org/wiki/Multiset&quot; rel=&quot;nofollow&quot;&gt;Wikipedia's Mulitset&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2011-01-11T21:37:51.650" LastActivityDate="2011-01-11T21:37:51.650" CommentCount="1" />
  <row Id="17087" PostTypeId="1" AcceptedAnswerId="17141" CreationDate="2011-01-11T14:03:27.033" Score="4" ViewCount="152" Body="&lt;p&gt;Suppose E and F are given Banach spaces. Let A be a continuous surjective map. Why is there a small ball around A in the operator topology, s.t. all elements in this ball are surjective?&lt;/p&gt;&#xA;" OwnerUserId="4520" LastActivityDate="2011-01-12T00:22:48.340" Title="Why is the space of surjective operators open?" Tags="&lt;functional-analysis&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="1" />
  <row Id="17088" PostTypeId="1" CreationDate="2011-01-11T14:09:20.363" Score="0" ViewCount="64" Body="&lt;p&gt;Hi.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I have a row that has 8 squares, at the 6th and 7th square is data I want to get. The grid could contain 1-20 columns of these 8 grid squares. How can I find the 6th and 7th squares in every row no matter how many columns there are?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Jeff&lt;/p&gt;&#xA;" OwnerUserId="5603" LastEditorUserId="1016" LastEditorDisplayName="" LastEditDate="2011-01-11T15:54:18.587" LastActivityDate="2011-01-11T15:54:18.587" Title="Finding the same position in a grid" Tags="&lt;homework&gt;" CommentCount="4" />
  <row Id="17089" PostTypeId="1" AcceptedAnswerId="17256" CreationDate="2011-01-11T14:25:06.613" Score="3" ViewCount="219" Body="&lt;p&gt;In the book &quot;A Course of modern analysis&quot;, examples of expanding functions in terms of inverse factorials was given, I am not sure in today's math what subject would that come under but besides the followings : power series ( Taylor Series, Laurent Series ), expansions in terms of theta functions, expanding a function in terms of another function (powers of, inverse factorial etc.), Fourier series, infinite Products (Complex analysis) and partial fractions (Weisenstein series), what other ways of representing functions have been studied? is there a comprehensive list of representation of functions  and the motivation behind each method?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example , power series are relatively easy to work with and establish the domain of convergence e.g. for $ \sin , e^x \text {etc.}$ but infinite product representation makes it trivial to see all the zeroes of $\sin, \cos \text etc. $ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also if anyone can point out the subject that they are studied under would be great.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you &lt;/p&gt;&#xA;" OwnerUserId="3936" LastEditorUserId="2548" LastEditorDisplayName="" LastEditDate="2011-01-11T15:01:45.647" LastActivityDate="2011-01-13T08:39:04.967" Title="Different ways to represent functions other than Laurent and Fourier series?" Tags="&lt;analysis&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="17090" PostTypeId="1" AcceptedAnswerId="17092" CreationDate="2011-01-11T14:25:06.843" Score="6" ViewCount="303" Body="&lt;p&gt;I learned in different math courses, that $\mathbb{C}$ is a &lt;em&gt;clopen&lt;/em&gt; set. I'm very uncertain if I understand this right, please tell me if these two examples are true:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\mathbb{C} \setminus \{ 0 \}$ is &lt;strong&gt;open&lt;/strong&gt;,&lt;/li&gt;&#xA;&lt;li&gt;$\mathbb{C} \setminus \{ z \in \mathbb{C} : |z| &amp;lt; 1 \}$ is &lt;strong&gt;closed&lt;/strong&gt;.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="5450" LastActivityDate="2011-01-11T19:24:20.113" Title="$\mathbb{C}$ is *clopen*. What?" Tags="&lt;topology&gt;&lt;complex-analysis&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="1" />
  <row Id="17091" PostTypeId="2" ParentId="17090" CreationDate="2011-01-11T14:27:20.167" Score="7" ViewCount="" Body="&lt;p&gt;The whole space is always clopen in any topology.  That is part of the definition of a topology.  Your two examples are correct.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2011-01-11T19:24:20.113" LastActivityDate="2011-01-11T19:24:20.113" CommentCount="2" />
  <row Id="17092" PostTypeId="2" ParentId="17090" CreationDate="2011-01-11T14:33:51.630" Score="14" ViewCount="" Body="&lt;p&gt;A clopen set is a set which is both closed and open.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When talking about a space, it is always clopen in itself. This is because we want both the space and the empty set to be open.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So $\mathbb{C}$ is indeed clopen in $\mathbb{C}$ (in any given topology). However, note that $\mathbb{R}$ is clopen in $\mathbb{R}$, but as a subset of $\mathbb{C}$ it's actually just closed. So when you move from $\mathbb{C}$ to something bigger (for example the &lt;a href=&quot;http://en.wikipedia.org/wiki/Riemann_sphere&quot; rel=&quot;nofollow&quot;&gt;Riemann sphere&lt;/a&gt;) then $\mathbb{C}$ is open, and not closed.&lt;/p&gt;&#xA;" OwnerUserId="622" LastActivityDate="2011-01-11T14:33:51.630" CommentCount="4" />
  <row Id="17093" PostTypeId="2" ParentId="16801" CreationDate="2011-01-11T14:34:28.070" Score="2" ViewCount="" Body="&lt;p&gt;Wikipedia has such a list at &lt;a href=&quot;http://en.wikipedia.org/wiki/Timeline_of_mathematics&quot; rel=&quot;nofollow&quot;&gt;this article&lt;/a&gt;.  &lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2011-01-11T14:34:28.070" CommunityOwnedDate="2011-01-11T14:34:28.070" />
  <row Id="17094" PostTypeId="1" AcceptedAnswerId="17095" CreationDate="2011-01-11T14:43:54.300" Score="2" ViewCount="89" Body="&lt;p&gt;What does this mean: ($f$ a function, $I$ an interval and $R$ the real numbers)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f \in C(I,R)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does it mean $f$ is an element of the collection of continuous functions with domain $I$ and range $R$ ? (literal translation so the terms might be off)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, what would this mean:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f \in C^\infty(I,R)$&#xA;?&lt;/p&gt;&#xA;" OwnerUserId="5109" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2011-01-11T17:27:15.317" LastActivityDate="2011-01-11T17:27:15.317" Title="Meaning of $C(I,\mathbb{R})$ and $C^{\infty}(I, \mathbb{R})$ related to continuous functions." Tags="&lt;analysis&gt;&lt;functions&gt;&lt;notation&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="17095" PostTypeId="2" ParentId="17094" CreationDate="2011-01-11T14:52:47.720" Score="7" ViewCount="" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;$C(I,R)$ is the space of continuous functions from $I$ to $R$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$C^{\infty}(I,R)$ is the space of &lt;a href=&quot;http://en.wikipedia.org/wiki/Smooth_function&quot; rel=&quot;nofollow&quot;&gt;smooth&lt;/a&gt; functions from $I$ to $R$. That is, $f$ has derivatives of all orders.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2011-01-11T14:52:47.720" CommentCount="1" />
  <row Id="17096" PostTypeId="2" ParentId="16723" CreationDate="2011-01-11T14:53:43.013" Score="7" ViewCount="" Body="&lt;p&gt;Here is an observation/partial result. For brevity write $t = \lfloor a/b \rfloor .$ When $\text{gcd}(b,c)=1$ and $c \, | \, (t+1) $ we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ S = \sum_{k=0}^{t} \left \lfloor \frac{a - kb}{c} \right \rfloor =&#xD;&#xA;\frac{t+1}{c} \left \lbrace a - \frac{tb}{2} - \frac{c-1}{2} \right \rbrace . $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proof:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xD;&#xA;&#xD;&#xA;a &amp;amp;= m_0 c + r_0 \\&#xD;&#xA;a- b &amp;amp;= m_1 c + r_1 \\&#xD;&#xA;a - 2b &amp;amp;= m_2 c  + r_2 \\&#xD;&#xA;\cdots &amp;amp;= \cdots \\&#xD;&#xA;a - tb &amp;amp;= m_t c  + r_t&#xD;&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for integer $m_i$ and $r_i$ where $ 0 \le r_i &amp;lt; c $ then, adding the above equations,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(t+1)a - \frac{t(t+1)}{2}b = Sc + \sum_{k=0}^t r_k . \quad (1)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now if $\text{gcd}(b,c)=1$ and $k$ runs over a complete system of residues modulo $c$ then&#xA;$a-kb,$ where $a$ is any integer, also runs over a complete system of residues modulo $c$. So when $ c \, | \, (t+1) $ we have that $a-kb$ runs over $(t+1)/c$ complete residue systems modulo $c$.&#xA;Hence&#xA;$$\sum_{k=0}^t r_k = \frac{(t+1)(c-1)}{2}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Substitute this into $(1)$ to obtain the result.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Here is a generalisation for the case $\text{gcd}(b,c)&amp;gt;1,$ which reduces to the above when $b$ and $c$ are coprime.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Write $d=\text{gcd}(b,c)$ and suppose $ a \equiv \lambda \textrm { mod } d $ where&#xA;$ 0 \le \lambda &amp;lt; d.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now $a-kb$ runs through all the residues modulo $c$ that are congruent to $ \lambda $ modulo $c$ as $k$ runs through $0,1,2,\ldots,u-1$ where $u=c/d.$ When $ u \, | \, (t+1) $ we have that $r_0,r_1,\ldots,r_t$ runs through $(t+1)/u$ such residue systems. Hence&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \sum_{k=0}^t r_k =&#xD;&#xA;\frac{t+1}{u} \left \lbrace \frac{du(u-1)}{2} + \lambda u \right \rbrace &#xD;&#xA;= (t+1) \left \lbrace \frac{c-d}{2} + \lambda \right \rbrace .$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And so&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \sum_{k=0}^{t} \left \lfloor \frac{a - kb}{c} \right \rfloor =&#xD;&#xA;\frac{t+1}{c} \left \lbrace a - \frac{tb}{2} - \frac{c-d}{2} - \lambda \right \rbrace . $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT2:&#xA;Here are a couple of numerical examples. With $a=91,b=15 \textrm{ and } c=21$ we have&#xA;$d=\text{gcd}(15,21)=3,$ $t=\lfloor 91/15 \rfloor = 6,$ $u=c/d=21/3=7$ and&#xA;$91 \equiv 1 \textrm{ mod } 3,$ and so $\lambda=1.$ Note that the condition $ u \, | \, (t+1)$ is satisfied. Our formula gives the sum as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{7}{21} \left \lbrace 91 - \frac{6 \cdot 15}{2} - \frac{21-3}{2} - 1 \right \rbrace = 12.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is small enough to check by hand.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \sum_{k=0}^7 \left \lfloor \frac{91-15k}{21} \right \rfloor = 4+3+2+2+1+0+0=12.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With $a=703,b=35 \textrm{ and } c=49$ we have&#xA;$d=\text{gcd}(35,49)=7,$ $t=\lfloor 703/35 \rfloor = 20,$ $u=c/d=49/7=7$ and&#xA;$703 \equiv 3 \textrm{ mod } 7,$ and so $\lambda=3.$ Note that the condition $ u \, | \, (t+1)$ is satisfied. Our formula gives the sum as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{21}{49} \left \lbrace 703 - \frac{20 \cdot 35}{2} - \frac{49-7}{2} - 3 \right \rbrace = 141.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One can verify with WolframAlpha, or similar, that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \sum_{k=0}^{20} \left \lfloor \frac{703-35k}{49} \right \rfloor = 141.$$&lt;/p&gt;&#xA;" OwnerUserId="1301" LastEditorUserId="1301" LastEditorDisplayName="" LastEditDate="2011-01-12T10:32:21.547" LastActivityDate="2011-01-12T10:32:21.547" CommunityOwnedDate="2011-01-11T14:53:43.013" CommentCount="5" />
  <row Id="17097" PostTypeId="2" ParentId="17063" CreationDate="2011-01-11T15:02:44.403" Score="6" ViewCount="" Body="&lt;p&gt;$\Sigma_3$ (the symmetric group on three elements) is the next smallest example.&lt;/p&gt;&#xA;" OwnerUserId="627" LastActivityDate="2011-01-11T15:02:44.403" CommentCount="6" />
  <row Id="17098" PostTypeId="2" ParentId="17090" CreationDate="2011-01-11T15:30:49.903" Score="1" ViewCount="" Body="&lt;p&gt;$\mathbb{C}$ is indeed clopen since it is both open and closed.  Note that one way to define a connected space in topology is that there are no non-trivial clopen sets (the empty set and the whole set will trivially be clopen regardless of the topology).&lt;/p&gt;&#xA;" OwnerUserId="5048" LastActivityDate="2011-01-11T15:30:49.903" />
  <row Id="17099" PostTypeId="1" AcceptedAnswerId="17107" CreationDate="2011-01-11T15:33:06.377" Score="0" ViewCount="44" Body="&lt;p&gt;I have a 3x3 square matrix. I want to find out how many unique matrices I can create if each of the elements can be either 1 or 0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How does the equation change if have NxN matrix?&lt;/p&gt;&#xA;" OwnerUserId="5604" LastActivityDate="2011-01-11T16:19:43.793" Title="How do I calculate variations of a square matrix if the elements are binary?" Tags="&lt;matrices&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="17101" PostTypeId="1" AcceptedAnswerId="17106" CreationDate="2011-01-11T15:45:33.723" Score="3" ViewCount="118" Body="&lt;ol&gt;&#xA;&lt;li&gt;I was wondering about what is the&#xA;general definition of a &lt;a href=&quot;http://en.wikipedia.org/wiki/Recurrence_relation&quot; rel=&quot;nofollow&quot;&gt;recursive&#xA;mapping&lt;/a&gt; between any two sets.&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Is there a condition for a mapping&#xA;to be able to be written in&#xA;recursive form? Is the following&#xA;claim true:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A mapping can be represented&#xA;  recursively if and only if its domain&#xA;  is &lt;a href=&quot;http://en.wikipedia.org/wiki/Recursive_definition&quot; rel=&quot;nofollow&quot;&gt;a set that can be defined recursively&lt;/a&gt;?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Thanks and regards!&lt;/p&gt;&#xA;" OwnerUserId="1281" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2011-06-03T13:52:57.837" LastActivityDate="2011-06-04T01:58:27.487" Title="Recursive Mapping" Tags="&lt;elementary-set-theory&gt;&lt;recursive-mappings&gt;" AnswerCount="2" />
  <row Id="17102" PostTypeId="1" CreationDate="2011-01-11T15:48:43.747" Score="0" ViewCount="121" Body="&lt;p&gt;Hi I have these equations that I wish to know how to reach the variables&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for example&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;A=6&lt;/code&gt;&#xA;&lt;code&gt;D=A-C&lt;/code&gt;&#xA;&lt;code&gt;B=2C&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All I have is the value of A, and A changes from time to time so its not always 6 it can be 8, 20 or anything else.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I get the values of each B, C, and D?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that, all numbers have to be integer, no floats.&lt;/p&gt;&#xA;" OwnerUserId="5605" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2011-01-11T17:21:04.073" LastActivityDate="2011-01-11T17:24:42.807" Title="Three linear equations in four integer variables." Tags="&lt;algebra-precalculus&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="17103" PostTypeId="2" ParentId="17102" CreationDate="2011-01-11T15:58:37.523" Score="1" ViewCount="" Body="&lt;p&gt;To put it simply, you can't. You have 3 linear equations for 4 variables, so you have an underdetermined linear system which admits infinitely many solutions. To use your example, $D$ and $B$ depend entirely on $A$ and $C$. $A$ is fixed, but $C$ is not. Arbitrary choice of $C$ gets you a different set of solutions in the integers. You need more information if you want a single solution.&lt;/p&gt;&#xA;" OwnerUserId="1016" LastActivityDate="2011-01-11T15:58:37.523" />
  <row Id="17104" PostTypeId="2" ParentId="17102" CreationDate="2011-01-11T15:59:23.160" Score="1" ViewCount="" Body="&lt;p&gt;There is no unique solution to the system. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You essentially have two equations in three unknowns:&#xA;\begin{align*}&#xA;C + D &amp;amp;= A\\&#xA;B - 2C &amp;amp;= 0.&#xA;\end{align*}&#xA;Given a value of $A$, if you &lt;em&gt;pick&lt;/em&gt; any value for $D$, then this will give you a value for $C$, which in turn will give you a value for $B$. Or you can &lt;em&gt;pick&lt;/em&gt; any value for $C$, then solve for $D$ and $B$. (You can also pick a value for $B$, so long as it is even, and get a value for $C$ and $D$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even if you require all values to be &lt;em&gt;positive integers&lt;/em&gt;, there is still no unique solution: $D$ gets restricted to only the values $1$, $2,\ldots,A-1$, but each of those values will give you a valid solution with $C$ and $B$ positive integers; so the only case where you would have a unique solution in positive integers is if $A=2$ (in which case, $B=2$ and $C=D=1$). &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-11T17:24:42.807" LastActivityDate="2011-01-11T17:24:42.807" />
  <row Id="17105" PostTypeId="1" AcceptedAnswerId="17185" CreationDate="2011-01-11T16:02:01.550" Score="-1" ViewCount="261" Body="&lt;p&gt;I am looking for an operator $U$, that can do this to a function:&#xA;$$Uf(x)=f(2x).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular I am happy if there is an $U$ for the general case:&#xA;$Uf(x)=f(kx)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does such an operator exist for all $f(x)$?&#xA;What is the operator for known classes of $f(x)$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have so far 1 condition on $f(x)$:&#xA;$$\int_{-\infty}^{\infty} f(x)^2\,dx\text{ converges.}$$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I dont really know where to look for such an operator, or what keyphrases here are.&lt;/p&gt;&#xA;" OwnerUserId="4929" LastEditorUserId="4929" LastEditorDisplayName="" LastEditDate="2011-01-12T11:08:53.947" LastActivityDate="2011-01-12T11:08:53.947" Title="Scale Operator $Uf(x)=f(kx)$" Tags="&lt;calculus&gt;&lt;linear-algebra&gt;&lt;physics&gt;&lt;operator-theory&gt;" AnswerCount="2" CommentCount="14" />
  <row Id="17106" PostTypeId="2" ParentId="17101" CreationDate="2011-01-11T16:15:45.420" Score="4" ViewCount="" Body="&lt;p&gt;Essentially, for you to be able to give a function recursively the domain must be well-ordered.  And it all depends on what you mean by &quot;recursive mapping&quot;; see below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A partial oder $\preceq$ on a set $X$ is a well-order if and only if every nonempty subset of $X$ has a least element under the relation $\preceq$. Assuming the Axiom of Choice, every set can be well-ordered, but these well-orders may not be explicit or constructive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The following is from Halmos's &lt;strong&gt;Naive Set Theory&lt;/strong&gt; (Chapter 18: Transfinite Recursion):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $(W,\preceq)$ is a well-ordered set, and $X$ is an arbitrary set, a &lt;em&gt;sequence of type $a$ in $X$&lt;/em&gt; means a function from the initial segment of $a$ in $W$, $\{w\in W\mid w\prec a\}$, to $X$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A &lt;em&gt;sequence function of type $W$ in $X$&lt;/em&gt; is a function $f$ whose domain is all sequences of type $a$ in $X$, for all $a\in W$, and whose range is contained in $X$; essentially, a &quot;sequence function&quot; tells you how to lengthen a sequence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Transfinite Recursion Theorem.&lt;/strong&gt; If $(W,\preceq)$ is a well-ordered set, and if $f$ is a sequence function of type $W$ in a set $X$, then there exists a unique function $U\colon W\to X$ such that $U(a) = f(U|_{i(a)})$ for each $a\in W$, where $i(a)=\{ w\in W\mid w\prec a\}$ is the initial segment determined by $a$ in $W$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The theorem tells you that you &lt;em&gt;can&lt;/em&gt; define a function recursively on a well-ordered set (with a given well order).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, when can a function be defined recursively? In a sense, &lt;em&gt;always&lt;/em&gt; if your domain is well-ordered. To see why, consider the case of $\mathbb{N}$ first.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose you have an &lt;em&gt;arbitrary&lt;/em&gt; function $f\colon\mathbb{N}\to X$. Let $g\colon \mathbb{N}\times X\to X$ be a function defined as follows: fix $x_0\in X$; then&#xA;$$g(n,x) = \left\{\begin{array}{ll}&#xA;f(n+1) &amp;amp;\text{if $f(n)=x$;}\\&#xA;x_0 &amp;amp;\text{if $f(n)\neq x$.}&#xA;\end{array}\right.$$&#xA;Then we have $f(n+1) = g(n,f(n))$, so that $f$ is defined &quot;recursively&quot;. This can be done with any function, so every function with domain $\mathbb{N}$ can be defined &quot;recursively&quot; this way. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similar constructions hold for an arbitrary well-ordered set. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might say, &quot;but you're cheating! you &lt;em&gt;start&lt;/em&gt; with $f$ and you determine $g$ from it&quot;. Well, yes; but if you were walking on the street and you found $g$ lying on the corner, and didn't know where it came from, you would still be able to use it to define a function. This tells you that there is always &lt;em&gt;a&lt;/em&gt; way of writing $f$ recursively, even if it is a bit silly in the grand scheme of things. The problem is that we have no good way of distinguishing &quot;silly ways&quot; and &quot;real ways&quot;, other than the way that Justice Potter Stewart was able to tell what was and what was not pornography: we know it when we see it.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2011-01-11T16:15:45.420" CommentCount="20" />
  <row Id="17107" PostTypeId="2" ParentId="17099" CreationDate="2011-01-11T16:19:43.793" Score="3" ViewCount="" Body="&lt;p&gt;If you have 2 choices for each of 9 positions in the matrix, there are 2^9 different possibilities.  And if there are N^2 positions,...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is the matrix just an array of numbers, or is there more structure to the problem?&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-01-11T16:19:43.793" CommentCount="1" />
  <row Id="17108" PostTypeId="2" ParentId="17083" CreationDate="2011-01-11T17:33:03.700" Score="4" ViewCount="" Body="&lt;p&gt;Hint: If $O_i$ is the number of occurrences of $i$ in the multiset, the tuple $(O_1, O_2, \dots, O_n)$ uniquely identifies the multiset. What constraint can you write down with the $O_i$, if the multiset has exactly $k$ elements?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ok, to elaborate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The number of multisets of size $k$ is equal to the number of non-negative integral solutions to the equation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$O_1 + O_2 + \dots + O_n = k$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the stars and bars method was already mentioned, here is a derivation of the formula using a more general approach, which is one of the main building blocks of analytic number theory: &lt;a href=&quot;http://www.math.upenn.edu/~wilf/DownldGF.html&quot; rel=&quot;nofollow&quot;&gt;Generating Functions&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The number of solutions to the above equation is same as the coefficient of $x^k$ in&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(1 + x + x^2 + x^3 + \dots )(1+x + x^2 + x^3 + \dots) \dots (1+ x + x^2 + x^3 + \dots) \ \text{repeated} \ n \ \text{times} \ \text{(why?)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The above is same as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \frac{1}{(1-x)^n} = (1-x)^{-n}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;as $$ (1 + x + x^3 + x^3 + \dots) = \frac{1}{1-x}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we apply binomial theorem, (which is also valid for negative exponents)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The coefficient of $x^k$ is $$-1^{k}\ \frac{-n(-n-1)(-n-2)\dots(-n-(k-1))}{k!}  = \frac{n(n+1)(n+2)\dots(n+k-1)}{k!}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$= \frac{(n-1)!n(n+1)(n+2)\dots(n+k-1)}{(n-1)!k!}= {n+k-1 \choose k} $$&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2011-01-12T18:52:18.363" LastActivityDate="2011-01-12T18:52:18.363" CommentCount="2" />
  <row Id="17109" PostTypeId="2" ParentId="17063" CreationDate="2011-01-11T17:33:44.713" Score="5" ViewCount="" Body="&lt;p&gt;Another collection of examples, not yet mentioned, are the Pr&amp;uuml;fer groups (also known as  &lt;em&gt;quasicyclic groups&lt;/em&gt;); it is sometimes denoted $\mathbb{Z}_{p^{\infty}}$. They are infinite, but every proper subgroup is finite (and every nontrivial quotient is isomorphic to the original group). Here are three descriptions:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;For a fixed prime $p$, let $\mathbb{Z}_{p^{\infty}}$ be the group of all $p^k$-th complex roots of unity for all $k\geq 0$, with the group operation being multiplication. It is not hard to show that every proper subgroup is generated by a $p^n$th primitive complex root of unity for some $n$, hence is cyclic.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;As an alternative description, consider subgroup of the additive group $\mathbb{Q}/\mathbb{Z}$ that consists of all classes represented by a fraction whose denominator is a power of $p$. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;A final description: consider the collection of groups $\{\mathbb{Z}/p^n\mathbb{Z}\}_{n\in\mathbb{N}}$, with injections $i_n\colon\mathbb{Z}/p^n\mathbb{Z}\to\mathbb{Z}/p^{n+1}\mathbb{Z}$ given by $i_n(a+p^n\mathbb{Z}) = pa+p^{n+1}\mathbb{Z}$. Then $\mathbb{Z}_{p^{\infty}}$ is the direct limit $\displaystyle\lim_{\longrightarrow}\mathbb{Z}/p^n\mathbb{Z}$ of this direct system. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The additive group $\mathbb{Q}$ is &lt;em&gt;almost&lt;/em&gt; such a group, in that every &lt;em&gt;finitely generated&lt;/em&gt; subgroup is cyclic, but of course it contains noncyclic proper subgroups (e.g., the pre-image of the Pr&amp;uuml;fer group for some $p$). &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-11T17:47:12.123" LastActivityDate="2011-01-11T17:47:12.123" CommentCount="1" />
  <row Id="17110" PostTypeId="1" AcceptedAnswerId="17116" CreationDate="2011-01-11T17:50:59.813" Score="2" ViewCount="241" Body="&lt;p&gt;Calculate all complex roots of the polynomial: $8t^{4} -20t^{3} -10t^{2}-5t-3$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So thanks to matlab, I can easily find out that the roots are $t = 3, -0.5, \pm 0.5i$.&#xA;Unfortunately, achieving this answer by hand has been more difficult.&#xA;Apparently, one valid method is to try to guess one of the roots and then use it to divide the polynomial.&#xA;I was able to verify that this results in $(x-3)(x+\frac{1}{2})(8x^{2}+2)$. However, even with the rational roots test and sythentic division, the &quot;guess&quot; part of the process is a little unappealing to me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I came across another method which seemed more promising: creating such a matrix $\left( \begin{array}{cccc} -\frac{5}{3} &amp;amp; -\frac{10}{3} &amp;amp; -\frac{20}{3} &amp;amp; \frac{8}{3} \\ 1 &amp;amp; 0 &amp;amp; 0 &amp;amp; 0 \\0 &amp;amp;1 &amp;amp;0 &amp;amp;0 \\ 0&amp;amp;0&amp;amp;1&amp;amp;0 \end{array} \right)$, determining the eigenvalues, and calculating the reciprocal of each eigenvalue. When I attempt to use this method, I get $\lambda^{4} + \frac{5}{3} \lambda^{3} + \frac{10}{3}\lambda^{2} + \frac{20}{3}\lambda -\frac{8}{3}$ (confirmed with matlab). However, it doesn't appear the situation has improved much, as I am facing the same problem of calculating roots of another polynomial with degree $4$, right? Should I expect to eventually solve for $\lambda$ if I apply the method once again (to the new polynomial)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Generally are there obvious problems with my understanding of the two approaches above? Would anyone recommend(considering my approximate level) using another method instead?&lt;/p&gt;&#xA;" OwnerUserId="3711" LastActivityDate="2011-01-11T19:29:32.557" Title="How to calculate complex roots of a polynomial" Tags="&lt;polynomials&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="17111" PostTypeId="1" CreationDate="2011-01-11T17:54:18.523" Score="1" ViewCount="127" Body="&lt;p&gt;Is there an epimorphism $f\colon \mathrm{GL}(2,\mathbb{Z})\to \mathrm{GL}(2,\mathbb{Z})$ which is not injective? Here, $\mathrm{GL}(2,\mathbb{Z})$ is the group of invertible $2\times 2$ matrices with integer entries.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Added.&lt;/em&gt; The answer is no: this group is Hopfian. I know a proof in which it is shown that this group is finitely generated (easy) and residually finite(tricky).&#xA;I suspect that there exists another more &quot;elementary&quot; proof if we use isomorphisms of this group.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd really like to see such a proof and so I would be thankfull if someone wrote it (if there exists of course).&lt;/p&gt;&#xA;" OwnerUserId="5422" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-11T18:15:50.060" LastActivityDate="2011-01-12T22:43:57.390" Title="Epimorphism from GL(2,Z) to GL(2,Z)" Tags="&lt;group-theory&gt;&lt;abstract-algebra&gt;&lt;counter-examples&gt;" AnswerCount="2" CommentCount="7" />
  <row Id="17112" PostTypeId="1" AcceptedAnswerId="17115" CreationDate="2011-01-11T18:02:56.387" Score="8" ViewCount="247" Body="&lt;p&gt;Let $ \chi $ be a character on the field $ F_p $, and fix some $a \in F_p $.&#xA;We define a Gauss sum to be:&#xA;$g_a (\chi) = \sum_{t\in F_p}\chi(t)\zeta^{at}$ where $\zeta$ is a primitive $p^{th}$ root of unity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the intuition behind this definition?&lt;/p&gt;&#xA;" OwnerUserId="766" LastEditorUserId="766" LastEditorDisplayName="" LastEditDate="2011-01-11T18:12:49.030" LastActivityDate="2011-01-12T06:33:57.800" Title="What is the intuition behind Gauss sums?" Tags="&lt;number-theory&gt;&lt;intuition&gt;&lt;algebraic-number-theory&gt;" AnswerCount="3" FavoriteCount="6" />
  <row Id="17113" PostTypeId="2" ParentId="17111" CreationDate="2011-01-11T18:07:11.437" Score="4" ViewCount="" Body="&lt;p&gt;No.  It is a theorem of Maltsev (or Malcev depending on how the name is translated into Latin characters) that every finitely generated subgroup of $GL(n,\mathbb{C})$, i.e. a linear group, is residually finite and, hence, Hopfian, which means that every epimorphism is injective.&lt;/p&gt;&#xA;" OwnerUserId="4009" LastActivityDate="2011-01-11T18:07:11.437" CommentCount="1" />
  <row Id="17114" PostTypeId="2" ParentId="17112" CreationDate="2011-01-11T18:09:00.687" Score="2" ViewCount="" Body="&lt;p&gt;They are the discrete Fourier transform of the character $\chi$ (which presumably is a &lt;em&gt;multiplicative&lt;/em&gt; character -- like the quadratic symbol, no?). This allows one to express $\chi$ as a sum of exponentials.&lt;/p&gt;&#xA;" OwnerUserId="536" LastActivityDate="2011-01-11T18:09:00.687" />
  <row Id="17115" PostTypeId="2" ParentId="17112" CreationDate="2011-01-11T18:24:59.180" Score="8" ViewCount="" Body="&lt;p&gt;One explanation is given &lt;a href=&quot;http://math.stackexchange.com/questions/11675/applications-of-gauss-sums/11692#11692&quot;&gt;in this math.SE answer&lt;/a&gt;.  In the language of that answer, you want to describe the unique quadratic subfield of $\mathbb{Q}(\zeta_p)$.  Since it's quadratic, it's generated by the square root of some rational, so the Galois group acts by multiplication by $-1$ on it.  Therefore you want to find an element of $\mathbb{Q}(\zeta_p)$ such that the Galois group $(\mathbb{Z}/p\mathbb{Z})^{\ast}$ acts by multiplication by $-1$ on it, and up to a constant that element is a Gauss sum (for the quadratic character).  Gauss sums with respect to more general characters have a similar relationship to actions of the Galois group.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2011-01-11T18:24:59.180" />
  <row Id="17116" PostTypeId="2" ParentId="17110" CreationDate="2011-01-11T18:29:50.073" Score="4" ViewCount="" Body="&lt;p&gt;The matrix method does not simplify the matter at all (unless you have a mental block about polynomials but feel really comfortable about matrices; or unless you have some really cool method for finding eigenvalues that does not depend on the characteristic polynomial).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Not only did you still get a polynomial of degree $4$, you got &lt;em&gt;essentially the same polynomial you started with!&lt;/em&gt; To see this, let $p(t)$ be your original polynomial, and let $q(\lambda)$ be the characteristic polynomial you found. Then you have:&#xA;\begin{align*}&#xA;-3q(\lambda) &amp;amp;=&#xA;-3\lambda^4 - 5\lambda^3 -10\lambda^2 -20\lambda + 8\\&#xA;&amp;amp;= \lambda^4\left(-3 -5\left(\frac{1}{\lambda}\right) - 10\left(\frac{1}{\lambda^2}\right) - 20\left(\frac{1}{\lambda^3}\right) + 8\left(\frac{1}{\lambda^4}\right)\right)\\&#xA;&amp;amp;= \lambda^4p\left(\frac{1}{\lambda}\right)&#xA;\end{align*}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That means that if $r$ is a root of $p(t)$, then $\frac{1}{r}$ is a root of $q(\lambda)$ (note that $r=0$ is not a root) and conversely; so finding the roots of the characteristic polynomial of the matrix is the same as finding the roots of your original $p(t)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This will always happen with this method: the matrix you write is essentially just the &lt;a href=&quot;http://en.wikipedia.org/wiki/Companion_matrix&quot; rel=&quot;nofollow&quot;&gt;companion matrix&lt;/a&gt; of the original polynomial, so the characteristic polynomial will essentially be the original polynomial again (or perhaps a reversal of the powers, as above). But for any polynomial $p(x) = a_nx^n + \cdots +a_0$ with $a_na_0\neq 0$, the roots of $p(x)$ and the roots of $x^np(\frac{1}{x}) = a_n + a_{n-1}x + \cdots + a_0x^n$ are just reciprocals of each other, so finding roots for the former is essentially the same as finding roots for the latter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Added:&lt;/em&gt; As Andres mentions and Qiaochu hints, it's possible that the change of perspective will let you more easily apply certain algorithms or theorems by going to the matrix rather than dealing with the polynomial directly. But my point is that the matrix method only affords you a change of &lt;em&gt;perspective&lt;/em&gt; rather than a simplification or an essential change in the problem. You are still dealing with the same polynomial when everything is said and done.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rational root method is pretty good for polynomials with relatively small integer coefficients; for polynomials of degree $3$ or $4$ you can always try the formulas by radicals (though they can be pretty annoying to actually use). There are other algorithms for finding, or at least approximating, roots of polynomials; the link Qiaochu gives in the comments is a good place to start.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-11T19:29:32.557" LastActivityDate="2011-01-11T19:29:32.557" CommentCount="3" />
  <row Id="17117" PostTypeId="1" AcceptedAnswerId="17121" CreationDate="2011-01-11T18:42:09.823" Score="3" ViewCount="214" Body="&lt;p&gt;How can I calculate the probability of shaking 7 of a kind using 10 six sided dice?&lt;/p&gt;&#xA;" OwnerUserId="5608" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2011-01-12T01:46:02.843" LastActivityDate="2011-01-12T03:22:13.517" Title="Chance of 7 of a kind with 10 dice" Tags="&lt;probability&gt;" AnswerCount="4" />
  <row Id="17118" PostTypeId="2" ParentId="17117" CreationDate="2011-01-11T18:50:07.850" Score="2" ViewCount="" Body="&lt;p&gt;Calculate the probability of exactly 7 sixes.&lt;br&gt;&#xA;Calculate the probability of exactly 8 sixes.&lt;br&gt;&#xA;Calculate the probability of exactly 9 sixes.&lt;br&gt;&#xA;Calculate the probability of exactly 10 sixes.&lt;br&gt;&#xA;Add them all up, and multiply by 6.&lt;/p&gt;&#xA;" OwnerUserId="1508" LastActivityDate="2011-01-11T18:50:07.850" CommentCount="2" />
  <row Id="17119" PostTypeId="1" AcceptedAnswerId="17165" CreationDate="2011-01-11T18:52:39.183" Score="10" ViewCount="181" Body="&lt;p&gt;Let $S=\{S_1,S_2,\ldots,S_n\}$ be a set of $n$ distinct subsets with $S_i \subseteq \{1,\ldots,n\}$ for $i=1,\ldots, n$ then $k \in \{1,\ldots,n\}$ exists with $S_i \cup \{k\}$ is distinct for $i=1,\ldots,n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found this in a old problem sheet of mine about sets and graph theory. Is there a elegant solution to this problem?&lt;/p&gt;&#xA;" OwnerUserId="3123" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2011-01-13T03:38:07.543" LastActivityDate="2011-01-13T03:38:07.543" Title="Distinctness is maintained after adding some element to all sets." Tags="&lt;combinatorics&gt;&lt;extremal-combinatorics&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" />
  <row Id="17121" PostTypeId="2" ParentId="17117" CreationDate="2011-01-11T19:26:11.010" Score="2" ViewCount="" Body="&lt;p&gt;To be short: The result is $p\approx \frac{1}{672}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can use the second formula in the Wikipedia article and sum the probabilities for every face of 1-6: &lt;a href=&quot;http://en.wikipedia.org/wiki/Liar%27s_dice#Basic_dice_odds&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Liar%27s_dice#Basic_dice_odds&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that Wolframalpha is also able to get this:&#xA;&lt;a href=&quot;http://www.wolframalpha.com/input/?i=10+dice&quot; rel=&quot;nofollow&quot;&gt;http://www.wolframalpha.com/input/?i=10+dice&lt;/a&gt; (Press on more under probabilities)&lt;/p&gt;&#xA;" OwnerUserId="3123" LastEditorUserId="3123" LastEditorDisplayName="" LastEditDate="2011-01-11T20:32:42.543" LastActivityDate="2011-01-11T20:32:42.543" CommentCount="4" />
  <row Id="17122" PostTypeId="2" ParentId="17117" CreationDate="2011-01-11T19:31:39.680" Score="5" ViewCount="" Body="&lt;p&gt;If you want exactly $7$ of a kind, you use &lt;a href=&quot;http://en.wikipedia.org/wiki/Binomial_distribution&quot; rel=&quot;nofollow&quot;&gt;binomial distribution&lt;/a&gt; to first find the probability that you roll $7$ of the same &quot;good&quot; number in $10$ trials. The binomial distribution shows that this probability is&#xA;$$&#xD;&#xA;\binom{10}{7}\left(\frac{1}{6}\right)^7\left(\frac{5}{6}\right)^3,&#xD;&#xA;$$&#xA;since the probability of success is $1/6$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then there are six numbers to choose one from which could be the &quot;good&quot; kind of which you want $7$. Hence you multiply the above expression by $\binom{6}{1}$:&#xA;$$&#xD;&#xA;\binom{6}{1}\binom{10}{7}\left(\frac{1}{6}\right)^7\left(\frac{5}{6}\right)^3.&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that this gives the exact probability $\frac{625}{419904}\approx\frac{1}{672}$, consistent with the approximate value given by Wolframalpha.&lt;/p&gt;&#xA;" OwnerUserId="1583" LastEditorUserId="1583" LastEditorDisplayName="" LastEditDate="2011-01-11T20:47:38.537" LastActivityDate="2011-01-11T20:47:38.537" CommentCount="4" />
  <row Id="17124" PostTypeId="1" AcceptedAnswerId="17127" CreationDate="2011-01-11T19:45:49.637" Score="1" ViewCount="162" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Is $\sin\frac{1}{x} \lt \frac{1}{x},\ \forall x\geq 1$?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I tried &quot;copying&quot; the proof of $\ln x \lt x, \forall x\geq 1$ but it didn't quite work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's what I did: Let $f(x)=\sin\frac{1}{x}-\frac{1}{x}$, I'd like to show that $f(x)\lt 0$. $f&amp;#39;(x)=\frac{1}{x^2}(1-\cos\frac{1}{x}) \gt 0$. So $f(x)$ is increasing, then $f(x)\gt f(1), \forall x \gt 1$ but $f(1)=\sin 1 - 1\lt 0$. And now I'm stuck, not sure if what I did is in the right direction.&lt;/p&gt;&#xA;" OwnerUserId="5611" LastActivityDate="2011-01-11T20:04:07.133" Title="Is $\sin\frac{1}{x} \lt \frac{1}{x},\ \forall x\geq 1$?" Tags="&lt;trigonometry&gt;&lt;inequality&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="17125" PostTypeId="1" AcceptedAnswerId="17129" CreationDate="2011-01-11T19:46:40.477" Score="8" ViewCount="262" Body="&lt;p&gt;Are Riemann integrals special cases of Haar integrals? Why do we need the invariant property under some actions of groups in the definition of Haar integrals? For example, if we have a group of real matrices $G$ and we have an inner product $\langle\&amp;gt;\ ,\ \rangle: G \times G \to \mathbb{R}$, for any $A, B$ in $G$, $C$ in the subgroup $O:=\{C\in G: C^{T}C=I\}$, $\langle A, B\rangle=\langle O^{T}AO,O^{T}BO\rangle$, then we can define an integral $\int_{G} f d\mu$ over $G$. Why do we need the property: $\langle A, B\rangle =\langle O^{T}AO,O^{T}BO\rangle$? Thank you very much.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: I am having a course about random matrices. The course is related to the lecture notes: &lt;a href=&quot;http://arxiv.org/abs/0801.1858&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/0801.1858&lt;/a&gt;. On page 2, equation (2.4), I think that the integral is a haar integral. I don't know why we need (2.5) in the paper. Thank you very much.&lt;/p&gt;&#xA;" LastActivityDate="2011-01-11T23:34:44.607" Title="What are the differences and relations of Haar integrals, Lebesgue integrals, Riemann integrals?" Tags="&lt;integral&gt;&lt;harmonic-analysis&gt;" AnswerCount="1" CommentCount="9" FavoriteCount="1" />
  <row Id="17126" PostTypeId="2" ParentId="17119" CreationDate="2011-01-11T19:48:29.133" Score="2" ViewCount="" Body="&lt;p&gt;THIS IS INCORRECT&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose that this wasn't the case. Then for each $k \in \{1, \dots, n\}$ you can find $S_a \subset S_b$ such that $S_b \setminus S_a = \{k\}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider a directed graph where the $n$ vertices correspond to the sets and there is an edge from $S_a$&amp;#160;to $S_b$&amp;#160;if and only if $S_b \setminus S_a$ consists of exactly one element. Clearly there won't be any loops so it is a forest with at most $n-1$&amp;#160;edges. Now by pigeonhole principle one of the edges must get two $k$s which is impossible.&lt;/p&gt;&#xA;" OwnerUserId="3776" LastEditorUserId="3776" LastEditorDisplayName="" LastEditDate="2011-01-12T05:24:01.197" LastActivityDate="2011-01-12T05:24:01.197" CommentCount="3" />
  <row Id="17127" PostTypeId="2" ParentId="17124" CreationDate="2011-01-11T19:49:56.043" Score="5" ViewCount="" Body="&lt;p&gt;The function is increasing as you note. Now consider&#xA;$$\lim_{x\to\infty} f(x) = \lim_{x\to\infty}\left(\sin\frac{1}{x} - \frac{1}{x}\right).$$&#xA;The limit is $0$; since $f$ is strictly increasing on $[1,\infty)$, has negative value at $1$, and approaches $0$ at $\infty$, can it ever &lt;em&gt;be&lt;/em&gt; equal to zero or positive on $[1,\infty)$?&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2011-01-11T19:49:56.043" CommentCount="6" />
  <row Id="17128" PostTypeId="2" ParentId="17124" CreationDate="2011-01-11T20:01:16.643" Score="0" ViewCount="" Body="&lt;p&gt;$f&amp;#39;(x)=0$ precisely when $\cos\left(\frac{1}{x}\right)=1$, which happens when $\frac{1}{x}\in \{2\pi n\,\vert\,n\in \mathbb{N}\}$ (since you want $x\geq 1$).  So, you have infinitely many critical points ($\frac{1}{2\pi},\pm\frac{1}{4\pi},\cdots$), none of which are in the interval $[1,\infty)$.  So, for all $x\geq 1$, $f(x)\geq f(1)=\sin(1)-1&amp;lt;0$.  Also, it's not too tough to show that $\lim\limits_{x\rightarrow \infty} \sin\left(\frac{1}{x}\right)-\frac{1}{x}=0$ (just use the fact that $-1\leq\sin\left(\frac{1}{x}\right)\leq 1$).  So, we have a differentiable function that's negative at $x=1$, strictly increasing, and has a limit as $x\rightarrow \infty$ of zero.  Thus, it's negative on $[1,\infty)$.  &lt;/p&gt;&#xA;" OwnerUserId="5137" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-11T20:04:07.133" LastActivityDate="2011-01-11T20:04:07.133" CommentCount="1" />
  <row Id="17129" PostTypeId="2" ParentId="17125" CreationDate="2011-01-11T20:06:00.913" Score="7" ViewCount="" Body="&lt;p&gt;Lebesgue integration, in the sense of integration with respect to Lebesgue measure on $\mathbb{R}^n$, is a special case of Haar integration, because Lebesgue measure is the Haar measure on the abelian group $\mathbb{R}^n$ with addition and the usual topology (normalized so that the measure of a unit $n$-cube is $1$).  Riemann integration is closely related to Lebesgue integration, but a fundamental difference is that the Riemann integral is not directly related to a measure, and as a result there are far fewer general results applicable.  For example, a pointwise limit of uniformly bounded Riemann integrable functions on a bounded interval need not be Riemann integrable, even if the convergence is monotone. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lebesgue integration can be seen as a completion of Riemann integration.  If $(R)\int f$ denotes the Riemann integral of a continuous function with compact support on $\mathbb{R}^n$, then $f\mapsto(R)\int f$ is a positive linear functional on $C_c(\mathbb{R}^n)$, and the Riesz representation theorem yields a positive Borel measure $\mu$ on $\mathbb{R}^n$ such that $(R)\int f=\int fd\mu$ for all $f\in C_c(\mathbb{R}^n)$.  The completion of $\mu$ is Lebesgue measure.  For (proper) Riemann integrable functions, the Riemann and the Lebesgue integral are the same, but there are many functions that are &quot;nice&quot; from the measure theoretic perspective but not Riemann integrable, such as the characteristic function of the rationals on $\mathbb{R}$.  On the other hand, there are functions whose &lt;em&gt;improper&lt;/em&gt; Riemann integral exists but whose Lebesgue integral does not exist due to the positive and negative parts not being integrable, such as $\frac{\sin(x)}{x}$ on $\mathbb{R}$.  In such cases, you could define an &quot;improper&quot; Lebesgue integral by taking limits as the domain increases to obtain the same result as with improper Riemann integrals, so there is no real loss in only considering Lebesgue integrals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I recommend reading &lt;a href=&quot;http://books.google.com/books?id=H3zGTvmtp74C&amp;amp;lpg=PP1&amp;amp;pg=PA195#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;pages 195-197&lt;/a&gt; of Körner's &lt;em&gt;A companion to analysis&lt;/em&gt; for a nice discussion of the motivation of going beyond Riemann integration.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2011-01-11T20:44:28.680" LastActivityDate="2011-01-11T20:44:28.680" CommentCount="4" />
  <row Id="17130" PostTypeId="2" ParentId="17105" CreationDate="2011-01-11T20:42:49.583" Score="4" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Proposition:&lt;/strong&gt;  There exists no function $U : \mathbb{R} \to \mathbb{R}$ such that, for every function $f : \mathbb{R} \to \mathbb{R}$, we have the relation $(U \circ f)(x) = f(2x)$.  (This is the interpretation of the question I have managed to glean from your comments.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Proof.&lt;/em&gt;   Let $f, g$ be two functions such that $f(1) = g(1)$ but such that $f(2) \neq g(2)$.  Then $(U \circ f)(1) = (U \circ g)(1)$ but $f(2) \neq g(2)$; contradiction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you can see the proof is robust under many modifications of the conditions on $f$.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2011-01-11T20:42:49.583" CommentCount="4" />
  <row Id="17132" PostTypeId="1" CreationDate="2011-01-11T21:24:13.463" Score="1" ViewCount="298" Body="&lt;p&gt;Suppose $p$ is irreducible and $M$ is a tosion module over a PID $R$  that  can be&#xA;written as a direct sum of cyclic submodules with annihilators of the form $p^{a_1} | \cdots | p^{a_s}$ and $p^{a_i}|p^{a_i+1}$. Let now $N$ be a submodule&#xA;of $M$. How can i prove that $N$ can be written a direct sum of cyclic modules with&#xA;annihilators of the form $p^{b_1} | \cdots | p^{b_t}, t\leq s$ and $\ p^{b_i}| p^&#xD;&#xA;{a_(s-t+i)}$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've already shown that $t\leq s$ considering the epimorphism from a free module to $M$ and from its submodule to $N$.&lt;/p&gt;&#xA;" OwnerUserId="5422" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-12T17:16:05.517" LastActivityDate="2011-01-18T18:12:35.997" Title="Torsion module over PID" Tags="&lt;commutative-algebra&gt;&lt;modules&gt;" AnswerCount="4" CommentCount="6" />
  <row Id="17133" PostTypeId="2" ParentId="17083" CreationDate="2011-01-11T21:28:06.597" Score="1" ViewCount="" Body="&lt;p&gt;Often a combinatorial counting problem is solved elegantly by relating it to something one already knows how to count.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's imagine a fly crawling from one corner of a grid to another (seriously, bear with me and we'll make it back to multisets shortly!).  The fly's walk is efficient in that each step is either up or to the right.  So if it goes from say $(0,0)$ to $(m,p)$, then the fly will take in all $p$-steps up and $m$-steps to the right.  The challenge is to calculate how many possible such paths there are.  Each path can be considered an arrangement of steps up and steps to the right, with a known number of each.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a way to map your counting of $k$-multisets taken from some set $A$ of $n$ items and recast it as a question about how many paths there are for the fly to take.  Consider the horizontal lines of the grid as being labeled by the items in set $A$, so (careful about the fence posts) there are $p = n-1$ vertical edges along the grid.  Consider the number of a particular item in set $A$ as being the number of horizontal steps along its respective line.  Since there will be $k$ entries in total for our multiset, this means $m = k$ horizontal edges across any path (and across the grid as a whole).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Put those ideas together and you'll have a recipe for counting the $k$-multisets from $A$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you actually wanted to list said multisets, it's more of a programming problem, but I think it would be fair game to discuss the algorithm to produce such a list.&lt;/p&gt;&#xA;" OwnerUserId="3111" LastEditorUserId="3111" LastEditorDisplayName="" LastEditDate="2011-01-11T21:44:06.527" LastActivityDate="2011-01-11T21:44:06.527" CommentCount="2" />
  <row Id="17134" PostTypeId="2" ParentId="17132" CreationDate="2011-01-11T22:11:16.613" Score="0" ViewCount="" Body="&lt;p&gt;See Lemma 9 &lt;a href=&quot;http://math.stanford.edu/~emalm/projects/math210b-w08/math210b-fg-modules-over-pid.pdf&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2011-01-11T22:11:16.613" CommentCount="2" />
  <row Id="17135" PostTypeId="2" ParentId="17089" CreationDate="2011-01-11T22:34:57.563" Score="1" ViewCount="" Body="&lt;p&gt;This may not be exactly what you are looking for, but in physics a common example of this shows up in quantum mechanics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The time-independent schroedinger equation in one dimension is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ - k \frac{\partial^2 \psi}{\partial x^2} + V(x) \psi = E \psi $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is essentially an eigenvalue problem for the differential operator&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ - k \frac{\partial^2}{\partial x^2} + V(x) $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $k$ is a parameter and $V(x)$ is a given function to describe a given physical situation (in physics terms it is the potential of the system).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What happens is that (for suitable $V(x)$) if one imposes a boundary condition on $\psi$, such as $\psi$ must be zero at infinity, then one finds that the eigenvalues of the above operator form a discrete set $\lbrace E_n \rbrace$, and the corresponding eigenfunctions $\lbrace \psi_n(x) \rbrace $ form a complete set for $L^2(\mathbb{R})$. (complete here means that any function can be decomposed as a series $\sum a_n \psi_n$ much like a fourier series).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, if $V(x)$ is taken to be the infinite square well potential ( &lt;a href=&quot;http://en.wikipedia.org/wiki/Infinite_square_well&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Infinite_square_well&lt;/a&gt; ) one gets that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\psi_n(x) = \sin(n \pi x / L)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and that the $\psi_n$ form a complete set is just the statement that functions can be decomposed into a fourier series. At any rate, one way to think of this whole thing is that by choosing different $V(x)$'s you can generate some fairly weird looking complete sets. For example take a look at the eigenstates portion of:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Quantum_harmonic_oscillator&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Quantum_harmonic_oscillator&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here $\psi_n = (-1)^n e^{x^2} \frac{d^n}{dx^n} \left( e^{-x^2} \right) $ forms a complete set.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also I believe the study of this kind of problem falls under Functional Analysis.&lt;/p&gt;&#xA;" OwnerUserId="1089" LastActivityDate="2011-01-11T22:34:57.563" />
  <row Id="17136" PostTypeId="2" ParentId="17083" CreationDate="2011-01-11T22:59:48.833" Score="1" ViewCount="" Body="&lt;p&gt;In parallel to &lt;a href=&quot;http://math.stackexchange.com/questions/17083/how-to-find-unique-multisets-of-n-naturals-of-a-given-domain-and-their-numbers/17108#17108&quot;&gt;Moron's hint&lt;/a&gt;, I'd suggest looking into the &quot;stars and bars&quot; technique, which is explained well in &lt;a href=&quot;http://math.stackexchange.com/questions/686/combinations-of-selecting-n-objects-with-k-different-types/691#691&quot;&gt;this answer by Ben Alpert&lt;/a&gt; (or &lt;a href=&quot;http://math.stackexchange.com/questions/7340/basic-combinatorics-question/7389#7389&quot;&gt;here's an answer of mine explaining it&lt;/a&gt;).&lt;/p&gt;&#xA;" OwnerUserId="72" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2011-01-12T00:51:47.643" LastActivityDate="2011-01-12T00:51:47.643" />
  <row Id="17137" PostTypeId="1" AcceptedAnswerId="17148" CreationDate="2011-01-11T23:06:49.433" Score="4" ViewCount="111" Body="&lt;p&gt;As &lt;a href=&quot;http://en.wikipedia.org/wiki/Set_%28mathematics%29#Describing_sets&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt; says, there are two ways to describe a particular set: &lt;a href=&quot;http://en.wikipedia.org/wiki/Intensional_definition&quot; rel=&quot;nofollow&quot;&gt;intensional definition&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/Extensional_definition&quot; rel=&quot;nofollow&quot;&gt;extensional definition&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;For a set able to be described by&#xA;extensional definition, is it necessary and sufficient to say the set has&#xA;to be countable? Do there exist sets that are not describable by extensional definition?&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;For a set able to be described by&#xA;intensional definition, is this&#xA;equivalent to existence of a predicate to&#xA;specify the membership of the set?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As &quot;there are only countably many&#xA;predicates you can actually write&#xA;down (or even possible at all with a&#xA;countable alphabet)&quot; (from a comment&#xA;after &lt;a href=&quot;http://math.stackexchange.com/questions/17101/recursive-mapping/17106#17106&quot;&gt;this reply&lt;/a&gt;, courtesy of&#xA;&lt;a href=&quot;http://math.stackexchange.com/users/742/arturo-magidin&quot;&gt;Arturo Magidin&lt;/a&gt;), are there only&#xA;countable number of sets that can be&#xA;described by intensional definition?&#xA;Do there exist sets that are not&#xA;describable by intensional&#xA;definition?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Are there other ways for describing&#xA;a set other than intensional&#xA;definition and extensional&#xA;definition?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Do there exist sets that are not&#xA;    describable ? &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Thanks and regards!&lt;/p&gt;&#xA;" OwnerUserId="1281" LastEditorUserId="1281" LastEditorDisplayName="" LastEditDate="2011-01-12T03:17:49.920" LastActivityDate="2011-01-12T03:17:49.920" Title="Describability of a set" Tags="&lt;elementary-set-theory&gt;&lt;definition&gt;" AnswerCount="2" CommentCount="14" />
  <row Id="17138" PostTypeId="2" ParentId="17137" CreationDate="2011-01-11T23:31:41.943" Score="4" ViewCount="" Body="&lt;p&gt;For any reasonable definition of &quot;describable,&quot; the answer to #4 has to be &quot;yes.&quot;  In fact, for any reasonable definition of &quot;describable,&quot; it should be true that almost all sets (in an appropriate sense) are not describable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's take our sets to be sets in ZFC, and let's say that a set in ZFC is describable if its elements are precisely the sets $x$ satisfying a predicate $p(x)$.  The correspondence between describable sets and predicates is not perfect because some (possibly most) predicates describe classes, not sets, but in any case there are at most as many describable sets as predicates in ZFC.  But there are only countably many such predicates, and the class of sets is so large that it is not a set (that is, it does not have a well-defined cardinality in ZFC).  So the vast majority of sets are not describable by such a predicate.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2011-01-11T23:31:41.943" CommentCount="9" />
  <row Id="17139" PostTypeId="1" AcceptedAnswerId="17146" CreationDate="2011-01-12T00:09:09.877" Score="1" ViewCount="185" Body="&lt;p&gt;For no real reason, I did a classification of (associative, with a 1) rings with less than 8 elements (they all happen to be commutative).  Most of the rings I got were of a type I knew - namely: cyclic, field, or Boolean - but one wasn't.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$R \hspace{.04 in} := \hspace{.04 in} \Bigg\langle \bigg\{\begin{pmatrix} 0 &amp;amp; 0 \\ 0 &amp;amp; 0 \end{pmatrix},\begin{pmatrix} 1 &amp;amp; 0 \\ 0 &amp;amp; 1 \end{pmatrix},\begin{pmatrix} 0 &amp;amp; 1 \\ 0 &amp;amp; 0 \end{pmatrix},\begin{pmatrix} 1 &amp;amp; 1 \\ 0 &amp;amp; 1 \end{pmatrix}\bigg\} \text{ , matrix operations over the field } \mathbb{Z}/2\mathbb{Z} \Bigg\rangle$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does $R$ arise in any remotely natural situations?&lt;/p&gt;&#xA;" OwnerUserId="5114" LastEditorUserId="5114" LastEditorDisplayName="" LastEditDate="2011-01-12T00:26:12.150" LastActivityDate="2011-01-12T01:23:08.347" Title="&quot;weird&quot; ring with 4 elements - how does it arise?" Tags="&lt;ring-theory&gt;" AnswerCount="3" CommentCount="5" />
  <row Id="17140" PostTypeId="2" ParentId="16797" CreationDate="2011-01-12T00:09:42.857" Score="9" ViewCount="" Body="&lt;p&gt;A standard way to get a good approximation for integrals that &quot;look&quot; Gaussian is to evaluate the Taylor series of the logarithms of their integrands through second order, expanding around the point of maximum value thus (continuing with @Ross Millikan's substitution):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\eqalign{&#xD;&#xA;  &amp;amp;\log\left(\sqrt{y}\cdot \exp\left(-\frac{(y-\mu )^2}{2\sigma ^2}\right)\right) \cr&#xD;&#xA;= &amp;amp;\frac{-\mu ^2-\sigma ^2+\mu  \sqrt{\mu ^2+2 \sigma ^2}+2 \sigma ^2 \log\left[\frac{1}{2} \left(\mu +\sqrt{\mu ^2+2 \sigma ^2}\right)\right]}{4 \sigma ^2} \cr&#xD;&#xA;+ &amp;amp;\left(-\frac{1}{2 \sigma ^2}-\frac{1}{\left(\mu +\sqrt{\mu ^2+2 \sigma ^2}\right)^2}\right) \left(y-\frac{1}{2} \left(\mu +\sqrt{\mu ^2+2 \sigma ^2}\right)\right)^2 \cr&#xD;&#xA;+ &amp;amp;O\left[y-\frac{1}{2} \left(\mu +\sqrt{\mu ^2+2 \sigma ^2}\right)\right]^3 \cr&#xD;&#xA;\equiv &amp;amp;\log(C) - (y - \nu)^2/(2\tau^2)\text{,}&#xD;&#xA;}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;say, with the parameters $C$, $\nu$, and $\tau$ depending on $\mu$ and $\sigma$ as you can see.  The resulting integral now &lt;em&gt;is&lt;/em&gt; a Gaussian, which can be computed (or approximated or looked up) in the usual ways.  The approximation is superb for small $\sigma$ or large $\mu$ and still ok otherwise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The plot shows the original integrand in red (dashed), this approximation in blue, and the simpler approximation afforded by replacing $\sqrt{y} \to \sqrt{\mu}$ in gold for $\sigma = \mu = 1/20$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/gDLAR.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h3&gt;(Added)&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Mathematica&lt;/em&gt; tells us the integral, when taken to $\infty$, can be expressed as a linear combination of modified Bessel Functions $I_\nu$ of orders $\nu = -1/4, 1/4, 3/4, 5/4$ with common argument $\mu^2/(4 \sigma^2)$.  From the Taylor expansion we can see that when both $\mu$ and $\sigma$ are small w.r.t. $1/2$--specifically, $(1/4-\mu)/\sigma \gg 3$, the error made by including the entire right tail will be very small.  (With a little algebra and some simple estimates we can even get good explicit bounds on the error as a function of $\mu$ and $\sigma$.)  There are many ways to compute or approximate Bessel functions, including &lt;a href=&quot;http://www.sciencedirect.com/science?_ob=ArticleURL&amp;amp;_udi=B6TY8-4KKWVK6-4&amp;amp;_user=10&amp;amp;_coverDate=12%2F15%2F2006&amp;amp;_rdoc=1&amp;amp;_fmt=high&amp;amp;_orig=search&amp;amp;_origin=search&amp;amp;_sort=d&amp;amp;_docanchor=&amp;amp;view=c&amp;amp;_searchStrId=1606463395&amp;amp;_rerunOrigin=google&amp;amp;_acct=C000050221&amp;amp;_version=1&amp;amp;_urlVersion=0&amp;amp;_userid=10&amp;amp;md5=6b1a974efc7b54358be324bde6f4a00c&amp;amp;searchtype=a&quot; rel=&quot;nofollow&quot;&gt;polynomial approximations&lt;/a&gt;.  From looking at graphs of the integrand, it appears that the cases where the Bessel function approximation works extremely well more or less complement the cases where the preceding &quot;saddlepoint approximation&quot; works extremely well.&lt;/p&gt;&#xA;" OwnerUserId="1489" LastEditorUserId="1489" LastEditorDisplayName="" LastEditDate="2011-01-13T21:59:54.277" LastActivityDate="2011-01-13T21:59:54.277" CommentCount="5" />
  <row Id="17141" PostTypeId="2" ParentId="17087" CreationDate="2011-01-12T00:22:48.340" Score="3" ViewCount="" Body="&lt;p&gt;Since this question bugged me, I decided to write down the proof (I don't have access to Lang's book, so I hope my argument is not much more complicated than necessary). The idea is the same as in the proof of the Banach-Schauder theorem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the open mapping theorem we may scale the norm on $E$ in such a way that $A$ maps the unit ball of $E$ onto the unit ball of $F$, that is $B_{\leq 1} F \subset A(B_{\leq 1}E)$. Since $A$ is linear we have $B_{\leq r} F \subset A(B_{\leq r}F)$ for all $r &amp;gt; 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Claim.&lt;/em&gt; If $B: E \to F$ is such that $\alpha := \|A - B\| &amp;lt; 1$ then $B$ is onto.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Proof.&lt;/em&gt; Let $f \in F$. We want to show that there is $e$ such that $f = Be$. For convenience, we put $f_{0} = f$ and assume $\|f_{0}\| \leq 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Choose $e_{0}$ with $\|e_{0}\| \leq 1$ such that $Ae_{0} = f_{0}$. Define $f_{1} = f_{0} - Be_{0}$ and observe $\|f_{1}\| = \|(A - B) e_{0}\| \leq \alpha$, so we may choose $e_{1}$ with $\|e_{1}\| \leq \alpha$ such that $Ae_{1} = f_{1}$. Now $f_{2} = f_{1} - Be_1$ has norm $\|f_{2}\| = \|(A - B)e_{1}\| \leq \alpha^{2}$, so we obtain by induction two sequences $\{f_{n}\}_{n=0}^{\infty}$ and $\{e_{n}\}_{n=0}^{\infty}$ having the following properties:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\|e_{n}\|, \|f_{n}\| \leq \alpha^{n}$ for all $n$.&lt;/li&gt;&#xA;&lt;li&gt;$e_{n}$ is such that $f_{n} = A(e_{n})$,&lt;/li&gt;&#xA;&lt;li&gt;$f_{n+1} = f_{n} - B(e_{n}) = (A-B)(e_{n})$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Finally $e = \sum_{n = 0}^{\infty} e_{n}$ has norm $\|e\| \leq \sum_{n=0}^{\infty} \alpha^{n} = \frac{1}{1-\alpha}$ and, moreover,&#xA;\[&#xA;B(e) = \sum_{n=0}^{\infty} B(e_{n}) = \sum_{n=0}^{\infty} (f_{n} - f_{n+1}) = f_{0} = f,&#xA;\]&#xA;as we wanted to show.&lt;/p&gt;&#xA;" OwnerUserId="5363" LastActivityDate="2011-01-12T00:22:48.340" CommentCount="3" />
  <row Id="17142" PostTypeId="1" AcceptedAnswerId="17153" CreationDate="2011-01-12T00:23:52.823" Score="1" ViewCount="98" Body="&lt;p&gt;After solving a system of equations I get this:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;x + 0.03125z = 0.242813&#xA;y + 0.40625z = 0.456563&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;How can I filter solutions that are only positive for x y and z&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ted&lt;/p&gt;&#xA;" OwnerUserId="5434" LastEditorUserId="5434" LastEditorDisplayName="" LastEditDate="2011-01-14T02:32:25.913" LastActivityDate="2011-01-14T02:32:25.913" Title="Finding positive solutions to two equations" Tags="&lt;algebra-precalculus&gt;" AnswerCount="3" />
  <row Id="17143" PostTypeId="1" CreationDate="2011-01-12T00:26:28.687" Score="3" ViewCount="88" Body="&lt;p&gt;In a finite dimensional (think Euclidean) ambient space, let $S$ a compact, convex set and $x$ not in $S$. The two sets can be (weakly) separated, i.e. there exists a vector (normal) that defines a hyperplane separating the point and the set. In fact, there exists an entire (open) set of normals that separate the point and the set. How can one describe this set?&lt;/p&gt;&#xA;" OwnerUserId="5618" LastActivityDate="2011-01-12T10:36:17.700" Title="Describing a set of normals" Tags="&lt;convex-analysis&gt;&lt;convex-sets&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="17144" PostTypeId="2" ParentId="17139" CreationDate="2011-01-12T00:28:22.403" Score="4" ViewCount="" Body="&lt;p&gt;$R$ is a certain subring of the ring $B$ of upper triangular $2 \times 2$ matrices over $\mathbb{F}_2$.  $B$ arises as the endomorphism ring of a &lt;a href=&quot;http://en.wikipedia.org/wiki/Flag_(linear_algebra)&quot; rel=&quot;nofollow&quot;&gt;complete flag&lt;/a&gt; in $(\mathbb{F}_2)^2$, so in that sense it's a natural geometric object to look at.  The construction generalizes to a  vector space over any field.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure exactly what you mean by &quot;natural situation,&quot; though, or why you should expect that every finite ring arises in such a situation.  For any finite field $\mathbb{F}_q$, every subring of $\mathcal{M}_n(\mathbb{F}_q)$ is a finite ring, for example.  As George Lowther says, for finite commutative rings we can also consider quotients $\mathbb{F}_q[x]/p(x)$ for arbitrary polynomials $p(x)$.  These break up into products of rings of the form $\mathbb{F}_q[x]/r(x)^n$ for irreducible $r$, which is a form of the Chinese remainder theorem and is related to the structure theorem for finitely generated modules over a principal ideal domain (in this case, $\mathbb{F}_q[x]$).  Is this &quot;natural&quot;?&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2011-01-12T00:41:22.060" LastActivityDate="2011-01-12T00:41:22.060" CommentCount="3" />
  <row Id="17146" PostTypeId="2" ParentId="17139" CreationDate="2011-01-12T00:33:38.777" Score="5" ViewCount="" Body="&lt;p&gt;I'm not sure what would be natural in this context, but I would find it natural to denote $0 = \begin{pmatrix}0&amp;amp;0\\0&amp;amp;0\end{pmatrix}$,&#xA;$1 = \begin{pmatrix}1&amp;amp;0\\ 0&amp;amp;1\end{pmatrix}$, $\epsilon = \begin{pmatrix}0&amp;amp;1\\0&amp;amp;0\end{pmatrix}$ , $1+\epsilon = \begin{pmatrix}1&amp;amp;1\\0&amp;amp;1\end{pmatrix}$&#xA;Where 0 is a additive unit, $1$ is a multiplicative unit and  $\epsilon^2 = 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So this ring arises as $\mathbb Z[\epsilon]/(2,\epsilon^2)$.&lt;/p&gt;&#xA;" OwnerUserId="5189" LastActivityDate="2011-01-12T00:33:38.777" CommentCount="1" />
  <row Id="17147" PostTypeId="2" ParentId="17142" CreationDate="2011-01-12T00:37:33.220" Score="1" ViewCount="" Body="&lt;p&gt;Perhaps you could glean something from &lt;a href=&quot;http://www.wolframalpha.com/input/?i=x+%2B+0.03125z+%3D+0.242813+y+%2B+0.40625z+%3D+0.456563&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2011-01-12T00:37:33.220" />
  <row Id="17148" PostTypeId="2" ParentId="17137" CreationDate="2011-01-12T00:40:52.637" Score="5" ViewCount="" Body="&lt;p&gt;The thing is, in common practice within set theory, we use &lt;em&gt;parameters&lt;/em&gt; in our descriptions, so actually &lt;em&gt;any&lt;/em&gt; set can be given intentionally, though not necessarily in an interesting way: $a=\{x\mid x\in a\}$. This is not so useless as it first appears. For example, &lt;em&gt;every&lt;/em&gt; set in Goedel's constructible universe $L$ is definable using as parameters &quot;simpler&quot; sets, and this is an important feature of $L$ that allows us to analyze it in detail, being ultimately responsible for the fact that $L$ satisfies the axiom of choice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As already pointed out, if we only use parameter-free formulas, then we can only describe intentionally countably many sets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for extensional descriptions, by the most strict of interpretations, only countable sets can be described this way and, really, we need some explicit way of mentioning their elements. Of course, in practice, we usually write things like $A=\{a_0,a_1,\dots\}$ and I would say this is as extensional as it gets, though of course it is an intentional definition $A=\{a_i\mid i\in{\mathbb N}\}$ in disguise, and even this only makes sense if we have a function $i\mapsto a_i$ to begin with. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, we usually write sets by listing them as $A$ above, even if the sets are not countable, by considering well-orderings, i.e., we write their elements as a long sequence (indexed by some not necessarily countable ordinal). Under choice, any set can be listed that way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think this idea of &quot;extensional&quot; and &quot;intentional&quot; (or &quot;by comprehension&quot;) definitions of sets is a bit of an unintended consequence of the &quot;new math&quot; idea of teaching set theory in schools. So, rather than talking about the axiom of extensionality, that sets are completely determined by their elements, somehow we ended up writing sets by their extensions (i.e., by listing their elements), and rather than talking about the axiom scheme of comprehension, we ended up writing sets by comprehension. Very strange. (I (still) remember being 9 or so and my teacher asking me to write the empty set intentionally. I don't remember my teacher ever explaining how to do this. That was very confusing to me. I guess at that age, writing $\phi$ was akin to saying that $\phi$ had to hold, so the idea of writing a property $\phi(x)$ that fails for all $x$ was completely foreign.)&lt;/p&gt;&#xA;" OwnerUserId="462" LastActivityDate="2011-01-12T00:40:52.637" CommentCount="5" />
  <row Id="17149" PostTypeId="1" AcceptedAnswerId="18732" CreationDate="2011-01-12T01:16:10.117" Score="0" ViewCount="87" Body="&lt;p&gt;I have a line graph, Given some certain points on the graph, I wish to add some explanation on them, such as this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/yQMc2.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any mathematical algorithm that allows me to overlay my explanation textbox &quot;nicely&quot; on the graphs, avoiding self overlapping as far as possible?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand that &quot;nicely&quot; is quite a subjective term, but all I want is that those text fields must be laid out as evenly as possible.&lt;/p&gt;&#xA;" OwnerUserId="550" LastEditorUserId="550" LastEditorDisplayName="" LastEditDate="2011-01-24T08:07:45.467" LastActivityDate="2011-01-24T11:17:39.960" Title="The Mathematics of Putting TextBox Nicely on a Graph" Tags="&lt;computational-geometry&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="17150" PostTypeId="2" ParentId="17143" CreationDate="2011-01-12T01:22:51.717" Score="1" ViewCount="" Body="&lt;p&gt;Define &#xA;\[&#xA;N = \{n \in S^{n-1}\,:\,\langle n, s-x\rangle &amp;lt; 0\;\text{for all $s \in S$}&#xA;\}.&#xA;\]&#xA;The points $n \in N$ satisfy $\langle n, x \rangle &amp;gt; \langle n, s \rangle$ for all $s \in S$. In words, $x$ lies on the positive side of the hyperplane defined by $n$ and all $s \in S$ on the negative side (after a suitable translation in direction of $n$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you don't care about orientation take all vectors in $N \cup (-N)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's how I like to think about it:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After a translation we may assume that $x = 0$. Form the (closed convex) cone&#xA;\[&#xA;C = \{\lambda s\,:\, \lambda \geq 0\}.&#xA;\]&#xA;Its &lt;a href=&quot;http://en.wikipedia.org/wiki/Dual_cone_and_polar_cone&quot; rel=&quot;nofollow&quot;&gt;polar cone&lt;/a&gt; is given by&#xA;\[&#xA;C^{'} = \{y\,:\,\langle y, c\rangle \leq 0 \; \text{for all $c \in C$}\} = \{y\,:\,\langle y, s\rangle \leq 0 \; \text{for all $s \in S$}\}.&#xA;\]&#xA;The sought set of normals is then $N = S^{n-1} \cap \text{int}\,C&amp;#39;$, the intersection of the unit sphere with the interior of $C&amp;#39;$.&lt;/p&gt;&#xA;" OwnerUserId="5363" LastEditorUserId="5363" LastEditorDisplayName="" LastEditDate="2011-01-12T10:36:17.700" LastActivityDate="2011-01-12T10:36:17.700" CommentCount="3" />
  <row Id="17151" PostTypeId="2" ParentId="17139" CreationDate="2011-01-12T01:23:08.347" Score="4" ViewCount="" Body="&lt;p&gt;It's the case $\rm\ R = \mathbb F_2$ of $\rm R[t]/t^2\:.\ $ This is known as the algebra of dual numbers over the ring $\rm R$.  Such rings and higher order analogs $\:\rm R[t]/t^n \;$  prove quite useful when studying (higher) derivations algebraically since such rings provide very convenient &lt;strong&gt;algebraic models of tangent / jet spaces&lt;/strong&gt;. For example, they permit easy transfer of properties of homomorphisms to derivations -- see for example section 8.15 in Jacobson, Basic Algebra II. See &lt;a href=&quot;http://math.stackexchange.com/questions/3101/derivative-of-a-product-and-derivative-of-quotient-of-functions-theorem-i-dont/3178#3178&quot;&gt;this post&lt;/a&gt; for further discussion and links.&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2011-01-12T01:23:08.347" />
  <row Id="17152" PostTypeId="1" AcceptedAnswerId="17234" CreationDate="2011-01-12T01:28:32.847" Score="95" ViewCount="11800" Body="&lt;p&gt;Of course, we've all heard the colloquialism &quot;If a bunch of monkeys pound on a typewriter, eventually one of them will write Hamlet.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a (not very mathematically intelligent) friend who presented it as if it were a mathematical fact.  Which got me thinking... Is this really true?  Of course, I've learned that dealing with infinity can be tricky, but my intuition says that time is countably infinite while the number of works the monkeys could produce is uncountably infinite.  Therefore, it isn't necessarily given that the monkeys would write Hamlet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could someone who's better at this kind of math than me tell me if this is correct?  Or is there more to it than I'm thinking?&lt;/p&gt;&#xA;" OwnerUserId="2519" LastEditorUserId="2736" LastEditorDisplayName="" LastEditDate="2011-01-14T06:35:31.063" LastActivityDate="2011-01-27T21:23:20.230" CommunityOwnedDate="2011-01-15T06:13:52.160" ClosedDate="2011-01-15T06:14:34.063" Title="Given an infinite number of monkeys and an infinite amount of time, would one of them write Hamlet?" Tags="&lt;probability&gt;&lt;infinity&gt;&lt;famous-theorems&gt;" AnswerCount="20" CommentCount="28" FavoriteCount="30" />
  <row Id="17153" PostTypeId="2" ParentId="17142" CreationDate="2011-01-12T01:30:59.647" Score="1" ViewCount="" Body="&lt;p&gt;Are your constants really reals?  It looks like it may be rounded from $x+\frac{z}{32}=\frac{777y}{3200}+\frac{13z}{32}=\frac{1461}{3200}$  If not, you can use the decimals in the following.  Break it into $3200x+100z=1461$ and $777y+1300z=1461$.  You can see that $x \le \frac {1461}{3200}, y \le \frac {1461}{777}, z\le \frac {1461}{1300}$ and as any one gets larger, it drives down the others.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2011-01-12T22:44:01.920" LastActivityDate="2011-01-12T22:44:01.920" />
  <row Id="17154" PostTypeId="2" ParentId="17152" CreationDate="2011-01-12T01:38:29.920" Score="107" ViewCount="" Body="&lt;p&gt;I found online the claim (which we may as well accept for this purpose) that there are 32241 words in Hamlet.  Figuring 5 characters and one space per word, this is 193446 characters.  If the character set is 60 including capitals and punctuation, a random string of 193446 characters has a chance of 1 in 60^193446 (roughly 1 in 10^344000) of being Hamlet.  While very small, this is greater than zero.  So if you try enough times, and infinity times is certainly enough, you will probably produce Hamlet.  But don't hold your breath.  It doesn't even take an infinite number of monkeys or an infinite number of tries.  Only a product of 10^344001 makes it very likely.  True, this is a very large number, but most numbers are larger.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-01-12T01:38:29.920" CommunityOwnedDate="2011-01-15T06:13:52.160" CommentCount="22" />
  <row Id="17155" PostTypeId="2" ParentId="17152" CreationDate="2011-01-12T01:59:48.957" Score="33" ViewCount="" Body="&lt;p&gt;If you have an infinite number of monkeys, then an infinite subset of them will just sit and type out Hamlet, letter-for-letter, straight away. So after a few hours you will have an infinite number of copies of Hamlet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have a finite number of monkeys then you may have to wait. But given in infinite amount of time (and immortal monkeys, etc.), you'll get your Hamlet. Eventually.&lt;/p&gt;&#xA;" OwnerUserId="5619" LastActivityDate="2011-01-12T01:59:48.957" CommunityOwnedDate="2011-01-12T01:59:48.957" CommentCount="10" />
  <row Id="17156" PostTypeId="2" ParentId="17152" CreationDate="2011-01-12T02:01:20.390" Score="61" ViewCount="" Body="&lt;p&gt;[Note that in my answer I am actually assuming that there are only a finite number of monkeys.  I don't see what is gained by having both the number of monkeys and the time frame be infinite: mathematically speaking $\aleph_0 \times \aleph_0 = \aleph_0$, and it is somewhat confusing to contemplate infinitely many monkeys typing simultaneously: too much is happening at once.  In fact, there might as well be only one monkey, or at any rate only one typewriter.]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me take the unusual (for me) step of considering the practical aspects of this question as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As Ross Millikan has explained, there is a simple mathematical model of monkey keyboard pounding under which it is easy to see that the claim is true: the probability that at least one of the monkeys will type out &lt;em&gt;Hamlet&lt;/em&gt; approaches $1$ as the time $n$ approaches infinity.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However there is an assumption here: namely, that the pounding on the typewriter is random or sufficiently close to random.  One way to formalize this is to say that after typing any $n$ characters, the probability of hitting any given key as the $n+1$st character is at least $P$, where $P$ is positive and independent of $n$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is that for actual typewriter banging, this is a very unlikely assumption.  The issue is similar here to what happens if you ask someone to produce a random sequence of digits, say from $0$ to $9$, or even a random sequence of $H$'s and $T$'s (for &quot;heads&quot; and &quot;tails&quot;). Just closing your eyes and banging away will produce something very far from being random.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the question is meant to apply to &lt;em&gt;actual&lt;/em&gt; monkeys with their nonrandom motor behavior, then it is something else entirely.  I would be tempted to say that the probability of producing Hamlet does not approach $1$ as time approaches infinity, but I'm not sure off the top of my head how to justify this.&lt;/p&gt;&#xA;" OwnerUserId="299" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2011-01-13T07:03:09.403" LastActivityDate="2011-01-13T07:03:09.403" CommunityOwnedDate="2011-01-15T06:13:52.160" CommentCount="10" />
  <row Id="17157" PostTypeId="1" AcceptedAnswerId="17235" CreationDate="2011-01-12T03:21:58.510" Score="3" ViewCount="125" Body="&lt;p&gt;Does anyone know of a nice example of a non-singular complete intersection $X$ (say in $\mathbb{P}^n_k$, maybe even $k=\overline{k}$, char($k$)=0) such that it cannot be written as $E\cap H$ where $E$ is a non-singular complete intersection and $H$ is a hypersurface (or maybe this can always be done)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, it is often the case that one might want to use an induction argument for non-singular complete intersection and you know that you can write $X=E\cap H$, but if the argument needs to use non-singularity then we don't necessarily know that $E$ satisfies this.&lt;/p&gt;&#xA;" OwnerUserId="2000" LastActivityDate="2011-01-12T15:40:02.700" Title="Smooth complete intersection counterexample" Tags="&lt;algebraic-geometry&gt;" AnswerCount="1" />
  <row Id="17158" PostTypeId="2" ParentId="17117" CreationDate="2011-01-12T03:22:13.517" Score="0" ViewCount="" Body="&lt;p&gt;Here's another way to get the same answer (and let's be explicit that we're looking for the probability of &lt;em&gt;exactly&lt;/em&gt; 7 of a kind, not &lt;em&gt;at least&lt;/em&gt; 7 of a kind):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Write the result as a sequence, e.g., the ten rolls might come out $(2,5,4,6,5,4,3,1,2,2)$.  Obviously there are $6^{10}$ such sequences, each equally likely.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How many sequences have 7 of a kind?  There are 6 possibilities for what we have 7 of, and there are $10\choose 7$ ways to pick the 7 places that are the same.  Finally, there are $5^3$ ways to fill the other three places, since there are 5 choices for each place.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the answer is $6\cdot{10\choose 7}\cdot 5^3 / 6^{10} = 625/419904$.&lt;/p&gt;&#xA;" OwnerUserId="2682" LastActivityDate="2011-01-12T03:22:13.517" />
  <row Id="17159" PostTypeId="2" ParentId="17152" CreationDate="2011-01-12T03:30:17.533" Score="5" ViewCount="" Body="&lt;p&gt;The answer is yes, With infinite time and all of the infinite monkeys will produce hamlet and every other works infinitely many times. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;After the first keystroke of the infinitely many monkeys, there will be infinitely many hamlets if you just grabbed the first letter from each one of them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also at the first keystroke, infinitely of monkeys would have types the first letter of the entire hamlet already, this shows that if you have infinite number of monkeys you only each one of them to type just as many letters as the number of letters in hamlet to have already infinitely many copies of hamlet, so there is no need for infinite time to have one of them produced hamlet, infinitely many of the monkeys would produce infinitely many hamlets in finite amount of time as long as the finite amount of time is equal or greater than the time to type a single copy of hamlet. (this was already mentioned in Bennett McElwee's answer)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you rephrase the question to : Would an infinite random sequence of letters in 2 dimensions contain at least one hamlet in every and each one of it's columns/rows? then the answer is yes, but there would be infinite number of hamlets contained in each row/column, As an infinite random sequence will contain all it's possible finite sub sequences, infinitely many times. ( Reference needed ).&lt;/p&gt;&#xA;" OwnerUserId="3936" LastEditorUserId="3936" LastEditorDisplayName="" LastEditDate="2011-01-13T12:13:30.380" LastActivityDate="2011-01-13T12:13:30.380" CommunityOwnedDate="2011-01-15T06:13:52.160" CommentCount="6" />
  <row Id="17160" PostTypeId="1" AcceptedAnswerId="17191" CreationDate="2011-01-12T03:42:38.983" Score="0" ViewCount="171" Body="&lt;p&gt;we have skew normal distribution with location=0, scale =1 and shape =0 then it is same as standard normal distribution with mean 0 and variance 1.but if we change the shape parameter say shape=5 then mean and variance also changes.how can we fix mean and variance with different values of shape parameter.&#xA;if we have 3 equation of mean,variance and skewness then how can we fix location, scale and shape parameter.can u explain about it?&lt;/p&gt;&#xA;" OwnerUserId="5620" LastActivityDate="2011-01-12T08:25:47.117" Title="skew normal distribution" Tags="&lt;statistics&gt;" AnswerCount="2" />
  <row Id="17161" PostTypeId="1" AcceptedAnswerId="17163" CreationDate="2011-01-12T03:50:38.030" Score="6" ViewCount="129" Body="&lt;p&gt;Let $L/K$ a finite extension and $f(x)\in K[x]$ a non-linear irreducible polynomial. Prove that  if $\mathrm{gcd}\left( \mathrm{deg}(f) , \left[ L:K  \right]  \right)=1$ then $f(x)$ has no roots in $L$.&lt;/p&gt;&#xA;&#xA;&lt;hr/&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Added: (Solution based on the answer below)&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose $f(x)$ has a root in $L$, namely $\alpha$ and consider the extension $K(\alpha)/K$. Since $f$ is irreducible we have that $[K(\alpha) : K] = \mathrm{deg}(f) &amp;gt; 1$. On the other hand we have that $[L:K]=[L:K(\alpha)][K(\alpha):K]$. Then $[L:K]=[L:K(\alpha)](\mathrm{deg}(f))$ but this is imposible since $\mathrm{gcd}\left( \mathrm{deg}(f) , \left[ L:K  \right]  \right)=1$ and $\mathrm{deg}(f) &amp;gt; 1$. &lt;/p&gt;&#xA;" OwnerUserId="5209" LastEditorUserId="5209" LastEditorDisplayName="" LastEditDate="2011-01-12T04:32:44.793" LastActivityDate="2011-01-12T04:32:44.793" Title="Field Extensions." Tags="&lt;homework&gt;&lt;field-theory&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" />
  <row Id="17162" PostTypeId="1" AcceptedAnswerId="17173" CreationDate="2011-01-12T03:52:31.340" Score="5" ViewCount="272" Body="&lt;p&gt;What is the property where f(g(x)) = g(f(x))?&lt;/p&gt;&#xA;" OwnerUserId="5622" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2011-04-29T18:54:27.577" LastActivityDate="2011-04-29T18:54:27.577" Title="What is the property where f(g(x)) = g(f(x))?" Tags="&lt;functions&gt;&lt;terminology&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="17163" PostTypeId="2" ParentId="17161" CreationDate="2011-01-12T03:53:15.957" Score="8" ViewCount="" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;What do you know about the degree $[K(\alpha):K]$ of an extension when $\alpha$ is a root of an irreducible polynomial $g(x)\in K[x]$? &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;What do you know about the degrees $[L:K]$, $[K:F]$, and $[L:F]$ of extensions when you have a tower $F\subset K\subset L$ ?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="742" LastActivityDate="2011-01-12T03:53:15.957" CommentCount="5" />
  <row Id="17164" PostTypeId="2" ParentId="17162" CreationDate="2011-01-12T03:54:42.650" Score="14" ViewCount="" Body="&lt;p&gt;We say $f$ and $g$ commute (with respect to composition). The property is called &quot;commutativity&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Associativity&quot; is the property that says that $f\circ (g\circ h)$ is the same as $(f\circ g)\circ h$, where $\circ$ is composition.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2011-01-12T03:54:42.650" />
  <row Id="17165" PostTypeId="2" ParentId="17119" CreationDate="2011-01-12T04:15:02.127" Score="8" ViewCount="" Body="&lt;p&gt;This result follows from &lt;a href=&quot;http://en.wikipedia.org/wiki/Bondy%27s_theorem&quot; rel=&quot;nofollow&quot;&gt;Bondy's theorem&lt;/a&gt; (in fact, it is equivalent) which states that,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given $\displaystyle n$ distinct sets $\displaystyle S_1, S_2, \dots, S_n$, each a subset of $\displaystyle \{1,2, \dots, n\}$, there is a set $\displaystyle A \subset \{1,2, \dots, n\}$, with $\displaystyle |A| \leq n-1$ such that the sets $\displaystyle S_i \cap A$ are all distinct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pick a $\displaystyle k \notin A$. Then we have that if $\displaystyle S_i \cup \{k\} = S_j \cup \{k\}$, then $\displaystyle  (S_i \cup \{k\}) \cap A = (S_j \cup \{k\}) \cap A$. Since $\displaystyle k \notin A$, it follows that $\displaystyle S_i \cap A = S_j \cap A$ contradicting the result of Bondy's theorem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can find a &lt;a href=&quot;http://books.google.com/books?id=g6X2VWuB8qIC&amp;amp;pg=PA143&quot; rel=&quot;nofollow&quot;&gt;short proof&lt;/a&gt; and a &lt;a href=&quot;http://books.google.com/books?id=g6X2VWuB8qIC&amp;amp;pg=PA203&quot; rel=&quot;nofollow&quot;&gt;sketch of an elegant linear algebra proof&lt;/a&gt; (originally due to Babai and Frankl) of Bondy's theorem, in the excellent book, &lt;a href=&quot;http://books.google.com/books?id=g6X2VWuB8qIC&quot; rel=&quot;nofollow&quot;&gt;Extremal Combinatorics&lt;/a&gt; by Stasys Jukna.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2011-01-12T05:27:01.770" LastActivityDate="2011-01-12T05:27:01.770" CommentCount="3" />
  <row Id="17166" PostTypeId="1" CreationDate="2011-01-12T04:35:38.137" Score="2" ViewCount="97" Body="&lt;p&gt;According to my textbook:&#xA;$$\int_{-L}^{L} \cos\frac{n \pi x}{L} \cos\frac{m \pi x}{L} dx =&#xD;&#xA;\begin{cases} 0 &amp;amp; \mbox{if } n \neq m \\&#xD;&#xA;              L &amp;amp; \mbox{if } n = m \neq 0 \\&#xD;&#xA;              2L&amp;amp; \mbox{if } n = m = 0&#xD;&#xA; \end{cases}&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to the trig identity given on &lt;a href=&quot;http://www.google.com/url?sa=t&amp;amp;source=web&amp;amp;cd=1&amp;amp;ved=0CBcQFjAA&amp;amp;url=http%3A%2F%2Ftutorial.math.lamar.edu%2Fpdf%2FTrig_Cheat_Sheet.pdf&amp;amp;ei=2SgtTcD8IMWqlAejs6nzCw&amp;amp;usg=AFQjCNEQ_WVPcXrkrKzSIa7Ta9sPXFW1QA&amp;amp;sig2=u9BxMF-Oo0TVR8kprApmOA&quot; rel=&quot;nofollow&quot;&gt;this cheat sheet&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;\cos{\alpha}\cos{\beta} =&#xD;&#xA;    \frac{1}{2}\left [ \cos \left (\alpha -\beta \right ) + &#xD;&#xA;    \cos \left(\alpha +\beta \right ) \right ]&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Substituting this trig identity in and integrating from $-L \mbox{ to } L$ gives:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_{-L}^{L} \cos\frac{n \pi x}{L} \cos\frac{m \pi x}{L} dx =&#xD;&#xA;\frac{L}{\pi} \left [\frac{\sin \left ( \pi (n - m) \right )}{n - m} + \frac{\sin \left ( \pi (n+m) \right )}{n + m}\right ] $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Evaluating the right side at $n = m$ gives a zero denominator, making the whole expression undefined.  Evaluating the right hand side at $n \neq m$ gives $0$ because the sine function is always $0$ for all integer multiples of $\pi$ as can be clearly seen with the unit circle.  None of these results jive with the first equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could you explain what mistakes I am making with my thinking?&lt;/p&gt;&#xA;" OwnerUserId="5621" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2011-01-12T04:50:53.380" LastActivityDate="2011-01-12T06:14:40.437" Title="Apparent inconsistency between integral table and integration using trigonometric identity" Tags="&lt;calculus&gt;&lt;trigonometry&gt;" AnswerCount="2" />
  <row Id="17167" PostTypeId="1" AcceptedAnswerId="17172" CreationDate="2011-01-12T04:39:36.730" Score="1" ViewCount="113" Body="&lt;p&gt;Let $a&amp;lt;b$ be real numbers. Define a cube $Q$ in $R^n$ to be the $n$-fold Cartesian product of $[a,b]$ with itself. Find a formula of the angle between the longest diagonal of $Q$ and any of its edges. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not sure where to start. Any help would be appreciated.&lt;/p&gt;&#xA;" OwnerUserId="2467" LastEditorUserId="2467" LastEditorDisplayName="" LastEditDate="2011-01-12T14:08:25.103" LastActivityDate="2011-01-12T14:17:11.607" Title="Formula of angle between longest diagonal and any edge of a cube" Tags="&lt;real-analysis&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="17168" PostTypeId="2" ParentId="17166" CreationDate="2011-01-12T04:40:00.847" Score="4" ViewCount="" Body="&lt;p&gt;If $a=0$, then $\cos(ax)$ should be simplified before finding its antiderivative.  This will keep you from dividing by zero, and it will make the answer come out right.  (E.g., if $a=(m-m)\pi/L$ or $a=(0+0)\pi/L$.)&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2011-01-12T04:56:31.083" LastActivityDate="2011-01-12T04:56:31.083" CommentCount="1" />
  <row Id="17170" PostTypeId="2" ParentId="17160" CreationDate="2011-01-12T04:52:32.190" Score="1" ViewCount="" Body="&lt;p&gt;If you follow the notation in the &lt;a href=&quot;http://en.wikipedia.org/wiki/Skew_normal_distribution&quot; rel=&quot;nofollow&quot;&gt;Wikipedia article&lt;/a&gt; you have the mean is $\xi + \omega \delta \sqrt{\frac{2}{\pi}}$ and the variance is $\omega^2(1-\frac{2\delta^2}{\pi})$ where $\delta=\frac{\alpha}{\sqrt{1+\alpha^2}}$.  So if you pick your $\alpha$ you can calculate $\delta$.  Your desired variance gives you $\omega$ and finally your desired mean gives $\xi$&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-01-12T04:52:32.190" CommentCount="2" />
  <row Id="17171" PostTypeId="2" ParentId="17166" CreationDate="2011-01-12T04:56:04.937" Score="2" ViewCount="" Body="&lt;p&gt;The integration is wrong if $n=m$ or if $n=-m$, because it is false that $\int \cos(0\pi x)\,dx = \frac{\sin(0\pi x)}{0}+C$. So the very use of the formula assumes that $|n|\neq|m|$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if $|n|\neq|m|$, then your formula &lt;strong&gt;does&lt;/strong&gt; say that the integral should be $0$, the same thing you get after the substitution. What makes you say that it &quot;does not jive&quot;?&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-12T06:14:40.437" LastActivityDate="2011-01-12T06:14:40.437" CommentCount="2" />
  <row Id="17172" PostTypeId="2" ParentId="17167" CreationDate="2011-01-12T04:56:55.227" Score="2" ViewCount="" Body="&lt;p&gt;To get the length of the diagonal, the &lt;a href=&quot;http://en.wikipedia.org/wiki/Pythagorean_theorem&quot; rel=&quot;nofollow&quot;&gt;Pythagorean theorem&lt;/a&gt; works in any dimension.  See the section &quot;Euclidean distance in various coordinate systems&quot;.  Add up the squares of the length in each dimension, take the square root of the sum, and you are there.  Then, as Trevor says, take the &lt;a href=&quot;http://en.wikipedia.org/wiki/Dot_product&quot; rel=&quot;nofollow&quot;&gt;dot product&lt;/a&gt; of a side with the diagonal to find the angle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To be more explicit, imagine the cube with one corner at the origin and sides going out along the positive directions.  The angle will be the same for all cubes.  The length of the diagonal is $\sqrt{n}$.  The dot product of the diagonal with a side is 1.  So $\theta=\arccos \left (\frac{1}{\sqrt{n}}\right)$&lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2011-01-12T14:17:11.607" LastActivityDate="2011-01-12T14:17:11.607" CommentCount="0" />
  <row Id="17173" PostTypeId="2" ParentId="17162" CreationDate="2011-01-12T04:57:42.977" Score="11" ViewCount="" Body="&lt;p&gt;Besides being called (composition) commutative, it is sometimes also said that such functions are &lt;em&gt;permutable&lt;/em&gt;, e.g. see &lt;a href=&quot;http://www.google.com/search?sourceid=chrome&amp;amp;ie=UTF-8&amp;amp;q=ritt+permutable&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;. As an example, a &lt;a href=&quot;http://math.stackexchange.com/questions/11431/when-functions-commute-under-composition/11442#11442&quot;&gt;classic result&lt;/a&gt; of Ritt shows that permutable polynomials are, up to a linear homeomorphism, either both powers of x, both iterates of the same polynomial, or both Chebychev polynomials.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2011-01-12T05:04:27.353" LastActivityDate="2011-01-12T05:04:27.353" />
  <row Id="17174" PostTypeId="1" AcceptedAnswerId="17175" CreationDate="2011-01-12T05:03:05.163" Score="0" ViewCount="142" Body="&lt;p&gt;A child walks due east on the deck of a ship at 1 miles per hour. &#xA;The ship is moving north at a speed of 9 miles per hour.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Find the speed and direction of the child relative to the surface of the water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Speed = &lt;strong&gt;&lt;em&gt;_&lt;/em&gt;_&lt;/strong&gt; mph&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The angle of the direction from the north =  &lt;strong&gt;&lt;em&gt;_&lt;/em&gt;__&lt;/strong&gt; (radians)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The last time I solved a problem like this was last summer, it looks like a &quot;Related Rates&quot; problem; so, I just need a refresher if anyone could care to help. Thank you very much.&lt;/p&gt;&#xA;" OwnerUserId="3800" LastActivityDate="2011-01-12T22:09:54.817" Title="Need help with a Calculus III / Related Rates Problem -- How to Solve It" Tags="&lt;calculus&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="17175" PostTypeId="2" ParentId="17174" CreationDate="2011-01-12T05:06:07.773" Score="3" ViewCount="" Body="&lt;p&gt;Hint:  You have a right angle triangle, 1 mi/hr in the easterly direction and 9 mi/hr in the northerly direction.  The total speed comes from Pythagoras and the angle from trigonometry. &lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2011-01-12T22:09:54.817" LastActivityDate="2011-01-12T22:09:54.817" />
  <row Id="17176" PostTypeId="1" AcceptedAnswerId="17180" CreationDate="2011-01-12T05:29:33.830" Score="1" ViewCount="115" Body="&lt;p&gt;The modified Euclidean Algorithm has you pick an $a,b\in\mathbb{Z}^+$.  Perform the usual division algorithm with the dividend being $b$ and the divisor being $a$.  This produces $b=aq+r$.  The next step is $bq=rq_2+r_2$.  The algorithm terminates and $\sum_{i=0}^n\frac{(-1)^i}{q_i}=\frac{a}{b}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Results like this come up but I can't find the proof in any number theory textbook I own, or online.  My thinking is I'm using the wrong resources.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question then is: Can any of you provide either i) the proof, ii) where you found it and or iii) another place to look up &quot;classical&quot; results like this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Addendum: There is yet another version of this where the sum adds only positive unit fractions and is more like the traditional egyptian fractions.&lt;/p&gt;&#xA;" OwnerUserId="2632" LastEditorUserId="2632" LastEditorDisplayName="" LastEditDate="2011-01-12T05:44:51.167" LastActivityDate="2011-01-12T07:07:37.340" Title="Find proof Egyptian Fractions and modified Euclidean Algorithm." Tags="&lt;number-theory&gt;" AnswerCount="2" />
  <row Id="17177" PostTypeId="1" AcceptedAnswerId="17181" CreationDate="2011-01-12T05:37:08.820" Score="3" ViewCount="113" Body="&lt;p&gt;Find the positive real number(s) $c$ such that the graphs of $y=x^c$ and $x=y^c$ intersect (somewhere) at an angle $\frac{\pi}{4}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A problem from Mark Krusemeyer. Should one be using the symmetry of the graphs along $x=y$ due the functions being inverses of one another? Just a thought.&lt;/p&gt;&#xA;" OwnerUserId="3731" LastEditorUserId="3731" LastEditorDisplayName="" LastEditDate="2011-05-22T16:37:34.513" LastActivityDate="2011-05-22T16:37:34.513" Title="Find the positive real number(s) $c$ such that the graphs of $y=x^c$ and $x=y^c$ intersect (somewhere) at an angle $\frac{\pi}{4}$." Tags="&lt;graph&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="17179" PostTypeId="2" ParentId="17152" CreationDate="2011-01-12T05:50:18.123" Score="12" ViewCount="" Body="&lt;p&gt;It would probably be faster to apply selective pressure to breed intelligent, literary monkeys.  Our common ancestor with chimps lived only &lt;a href=&quot;http://en.wikipedia.org/wiki/Human_evolution&quot; rel=&quot;nofollow&quot;&gt;4 million years&lt;/a&gt; ago, and by design I'm sure we could make something similar happen faster starting with chimps, bonobos, or whatever.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We could probably get orcas and other whales up to speed &lt;em&gt;very&lt;/em&gt; quickly, too!&lt;/p&gt;&#xA;" OwnerUserId="5636" LastActivityDate="2011-01-12T05:50:18.123" CommunityOwnedDate="2011-01-12T05:50:18.123" CommentCount="5" />
  <row Id="17180" PostTypeId="2" ParentId="17176" CreationDate="2011-01-12T05:59:48.203" Score="3" ViewCount="" Body="&lt;p&gt;You are missing key conditions in the Euclidean Algorithm; you require $0\leq r\lt a$ in the first step, and in general $0\leq r_{i+1}\lt r_i$, and &quot;terminates&quot; means the remainder is $0$. This is what guarantees that the process terminates: you have a strictly decreasing sequence of nonnegative integers, and therefore it must be a finite sequence before you get $r_n = 0$ for some $n$; certainly, no more than $a$ steps, but in fact less than that since you can show that after two steps the remainders drop by at least half. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not in my office, but unless I am much mistaken the fact that the process terminates is in Niven, Zuckerman, and Montgomery's &lt;em&gt;An Introduction to the Theory of Numbers&lt;/em&gt;, among others. I'll check tomorrow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The formula $\sum_{i=0}^n\frac{(-1)^i}{q_i}=\frac{a}{b}$ doesn't work with your notation, since you either don't have $q_0$ or you don't have $q_1$; I suspect you mistyped your &quot;next step&quot;, and that you mean mean $b=aq_0 + r_0$, and then $bq_0 = r_0q_1 + r_1$, and more generally $bq_0\cdots q_k = r_kq_{k+1}+r_{k+1}$ (with the restrictions mentioned above on the $r_j$). Moreover, it requires the extra assumption that $a\lt b$ (otherwise, $q_0 = 0$); this is not usually necessary in the Euclidean algorithm, so you need to specify it if you are going to use that formula.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As to a proof: if the remainder is $0$ the result is immediate, since $b=aq$ with all nonzero implies $\frac{1}{q} = \frac{a}{b}$. Assume the formula holds if the algorithm terminates after $k$ steps, and that you have an application with $k+1$ steps. Using the induction hypothesis applied to $bq_0$ and $r_0$ you have that&#xA;$$\sum_{i=0}^{k} \frac{(-1)^i}{q_{i+1}} = \frac{r_0}{bq_0}.$$&#xA;Replacing $r_0$ with $b-aq_0$ we get &#xA;$$\sum_{i=0}^k \frac{(-1)^i}{q_{i+1}} = \frac{b-aq_0}{bq_0} = \frac{1}{q_0} - \frac{a}{b}.$$&#xA;Now solving for $\frac{a}{b}$ gives&#xA;\begin{align*}&#xA;\frac{a}{b} &amp;amp;= \frac{1}{q_0} - \sum_{i=0}^{k}\frac{(-1)^i}{q_{i+1}}&#xA;= \frac{1}{q_0} + \sum_{i=0}^k\frac{(-1)^{i+1}}{q_{i+1}}\\&#xA;&amp;amp;= \frac{1}{q_0} + \sum_{i=1}^{k+1}\frac{(-1)^i}{q_i}&#xA;= \sum_{i=0}^{k+1}\frac{(-1)^i}{q_i},&#xA;\end{align*}&#xA;which proves the formula by induction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;As to the addendum:&lt;/em&gt; I don't know which version you are thinking of, but I'll wager a proof by induction will work along the same lines as the one above for it.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-12T06:38:38.857" LastActivityDate="2011-01-12T06:38:38.857" CommentCount="1" />
  <row Id="17181" PostTypeId="2" ParentId="17177" CreationDate="2011-01-12T06:00:25.957" Score="8" ViewCount="" Body="&lt;p&gt;Here's a brute force approach. There's probably a more elegant way of doing it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The two curves can be written as $y=x^c$ and $y=x^{1/c}$. They always intersect at $x=0$ and $x=1$ (for $c=1$ the curves coincide). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/YRLDV.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The derivatives of the two curves are $c x^{c-1}$ and $c^{-1}x^{1/c-1}$ respectively.&#xA;For $0&amp;lt;c\neq1$ the angle between them at $x=0$ is clearly always $\pi/2$.&#xA;At $x=1$ we want the angle between them to be $\pi/4$, so&#xA;$$ \tan^{-1}(c) - \tan^{-1}(1/c) = \pm\frac\pi4 \implies c = \pm 1+\sqrt2 $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The two solutions just reflect the fact that you can swap the two curves, or equivalently, set $c\to1/c$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are also two negative solutions to the above equation which are excluded by the question. These are also valid curves and solutions, which can be obtained by the transformation $x \to 1/x$.&lt;/p&gt;&#xA;" OwnerUserId="954" LastEditorUserId="954" LastEditorDisplayName="" LastEditDate="2011-01-12T08:38:17.430" LastActivityDate="2011-01-12T08:38:17.430" />
  <row Id="17182" PostTypeId="1" CreationDate="2011-01-12T06:23:33.323" Score="3" ViewCount="225" Body="&lt;p&gt;We are trying to implement transformations to evaluate the incomplete integral of the third kind $\Pi(n;\phi|m)$ for arbitrary inputs, and I can't find any references for how to calculate this function with phase $\phi$ less than zero or greater than $\pi/2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there periodic identities for this function for various ranges of $m$ and $n$? Plotting it in Mathematica tends to suggest there is, but the &lt;a href=&quot;http://dlmf.nist.gov/19.2#E10&quot; rel=&quot;nofollow&quot;&gt;DLMF&lt;/a&gt; only lists periodicity for elliptic integrals of the first two kinds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Update: Wolfram has a &lt;a href=&quot;http://functions.wolfram.com/EllipticIntegrals/EllipticPi3/04/02/03/&quot; rel=&quot;nofollow&quot;&gt;periodicity equation&lt;/a&gt; for $-1\le n\le 1$ although in practise it seems to work for $n&amp;lt;-1$ as well. But I'm still looking for a similar equation for $n&amp;gt;1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2nd update: mjqxxxx below gives periodicity for $n&amp;gt;1$, but for $m=1$ things are slightly different. (I'm trying to be complete.) Here's the real part for $\Pi(n&amp;gt;1,\phi|m=1)$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/Xv44X.png&quot; alt=&quot;real EllipticPi for &amp;quot;m&amp;quot; equals one&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which is symmetric around $\phi=\pi/2$ but anti-symmetric around $\phi=0$. But here's the strange one, the imaginary part:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/tydy4.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Symmetric around $\phi=0$ but around $\phi=\pi/2$ the symmetry breaks down. Can the function be mapped from the region $\pi/2&amp;lt;\phi&amp;lt;\pi$ back to $0&amp;lt;\phi&amp;lt;\pi/2$? &lt;/p&gt;&#xA;" OwnerUserId="2961" LastEditorUserId="2961" LastEditorDisplayName="" LastEditDate="2011-01-24T06:11:53.097" LastActivityDate="2011-01-24T23:38:27.233" Title="Periodicity of EllipticPi" Tags="&lt;special-functions&gt;" AnswerCount="2" />
  <row Id="17183" PostTypeId="2" ParentId="17112" CreationDate="2011-01-12T06:33:57.800" Score="7" ViewCount="" Body="&lt;p&gt;When you say $\chi$ is a character of the field $F_p$, you really mean it is a character of the group $F_p^\times$. Any (multiplicative) character $\chi$ on $F_p^\times$ can be extended to a function on $F_p$ by setting $\chi(0) = 0$, and with this convention $\chi$ as a function on $F_p$ is totally multiplicative. Fixing a choice of nontrivial $p$th root of unity $\zeta$, any function $f \colon F_p \rightarrow {\mathbf C}$ has a Fourier transform ${\mathcal F}f \colon F \rightarrow {\mathbf C}$ given by &#xA;$({\mathcal F}f)(a) = \sum_{t \in F_p} f(t)\overline{\zeta^{at}}$.  So the Gauss sum  $g_a(\chi)$ is essentially the Fourier transform at $a$ of the function $\chi$ (as a function on $F$). For more on this, see exercise 7 on page 16 at &#xA;&lt;a href=&quot;http://www.math.uconn.edu/~kconrad/blurbs/grouptheory/charthy.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.math.uconn.edu/~kconrad/blurbs/grouptheory/charthy.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another intuition (besides the idea that a Gauss sum of a character is basically the Fourier transform of that character, viewed as a function on the additive group $F_p$) is that a Gauss sum is a discrete analogue of the Gamma function. See pp. 56--58 of Koblitz's book &quot;$p$-adic Analysis: A Short Course on Recent Work&quot; for a table illustrating this analogy (including the idea that a Jacobi sum is like the Beta function).&lt;/p&gt;&#xA;" OwnerUserId="619" LastActivityDate="2011-01-12T06:33:57.800" />
  <row Id="17184" PostTypeId="2" ParentId="17177" CreationDate="2011-01-12T06:40:37.187" Score="5" ViewCount="" Body="&lt;p&gt;It is possible to use the symmetry as a slight shortcut to &lt;a href=&quot;http://math.stackexchange.com/questions/17177/find-the-positive-real-numbers-c-such-that-the-graphs-of-y-xc-and-x-yc/17181#17181&quot;&gt;Simon's answer&lt;/a&gt;: Given the graph and conditions as laid out there, the relevant point of intersection must be at $x=y=1$.  Because of the symmetry, the angles formed by the tangent lines to the curves at that point and the line $y=x$ must be $\frac{\pi}{8}$, so the slope of the tangent line to the blue curve (lower for $0&amp;lt; x&amp;lt;1$) at $x=1$ must be $\tan\frac{3\pi}{8}=1+\sqrt{2}$.  Now, either $cx^{c-1}|_{x=1}=c=1+\sqrt{2}$ or $c^{-1}x^{1/c-1}|_{x=1}=\frac{1}{c}=1+\sqrt{2}$, so $c=1+\sqrt{2}$ or $c=\frac{1}{1+\sqrt{2}}=-1+\sqrt{2}$.&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2011-01-12T06:40:37.187" CommentCount="3" />
  <row Id="17185" PostTypeId="2" ParentId="17105" CreationDate="2011-01-12T06:46:59.410" Score="-2" ViewCount="" Body="&lt;p&gt;Yes, such operator exists and it is linear. The answer above by Qiaochu Yuan seems to confuse function with operator. That said there is no such function $U:\mathbb{R}\to\mathbb{R}$, but there is a linear operator U as said Arturo Magidin. And on the set of analytic functions it has the following infinite (because the set of analytic functions is infinite-dementional) matrix:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$U=\left(&#xD;&#xA;\begin{array}{cccccc}&#xD;&#xA; 1 &amp;amp; 0 &amp;amp; 0 &amp;amp; . &amp;amp; 0 &amp;amp; . \\&#xD;&#xA; 0 &amp;amp; k^1 &amp;amp; 0 &amp;amp; . &amp;amp; 0 &amp;amp; . \\&#xD;&#xA; 0 &amp;amp; 0 &amp;amp; k^2 &amp;amp; . &amp;amp; 0 &amp;amp; . \\&#xD;&#xA; . &amp;amp; . &amp;amp; . &amp;amp; . &amp;amp; . &amp;amp; . \\&#xD;&#xA; 0 &amp;amp; 0 &amp;amp; 0 &amp;amp; . &amp;amp; k^n &amp;amp; . \\&#xD;&#xA; . &amp;amp; . &amp;amp; . &amp;amp; . &amp;amp; . &amp;amp; .&#xD;&#xA;\end{array}&#xD;&#xA;\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To obtain the resulting function you have to multiply this matrix by the argument function:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f(k x)= Uf(x)= f(0)+\frac {f&amp;#39;(0)}{1!} k x+ \frac{f&amp;#39;&amp;#39;(0)}{2!} k^2 x^2+\frac{f^{(3)}(0)}{3!} k^3 x^3+ \cdots+\frac{f^{(n)}(0)}{n!} k^n x^n +\cdots.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The confusion with your question probably arose because you used the function composition sign $\circ$ where it should not be if you meant applying the operator.&lt;/p&gt;&#xA;" OwnerUserId="2513" LastEditorUserId="2513" LastEditorDisplayName="" LastEditDate="2011-01-12T07:16:06.870" LastActivityDate="2011-01-12T07:16:06.870" CommentCount="6" />
  <row Id="17187" PostTypeId="2" ParentId="17176" CreationDate="2011-01-12T07:01:29.203" Score="2" ViewCount="" Body="&lt;p&gt;By the &lt;em&gt;division algorithm&lt;/em&gt; $\rm\ \ b\ =\  a\ q\ +\ a&amp;#39;\ \ $ for  $\rm\ \ 0\: \le\: a&amp;#39;\: &amp;lt;\: a\:. $   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dividing this by $\rm\: b\: q\: $ yields $\rm\displaystyle\ \frac{a}b\ =\ \frac{1}q\: -\: \frac{a&amp;#39;}{b&amp;#39;}\ $ for $\rm\ b&amp;#39; = b\ q\:.$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now recurse similarly on $\rm\ a&amp;#39;/\:b&amp;#39;\:.\ $ Since each step decreases numerators $\rm\ a&amp;#39; &amp;lt; a\:,\ $ eventually one reaches $\rm\ a&amp;#39; = 0\:,\: $ ending the recursion, so representing $\rm\ a/b\ $ as the claimed sum of unit fractions.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2011-01-12T07:07:37.340" LastActivityDate="2011-01-12T07:07:37.340" CommentCount="2" />
  <row Id="17188" PostTypeId="2" ParentId="17152" CreationDate="2011-01-12T08:07:18.127" Score="3" ViewCount="" Body="&lt;p&gt;Let me be the one who says NO.&#xA;It is a bit hard to handle an infinite number of monkeys (because you will need an infinite number of bananas to feed them), but if you agree to have a finite number of monkeys and infinite time, I would guess that considering the time it should take for this to happen, it is likely that the universe would die before it actually does.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To quote a chat of one of my friends with my first year &quot;intro to CS&quot; professor:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;professor: &quot;So there is no chance that a fair coin would fall with the head side up 50 times in a row.&quot;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;my friend: &quot;No!, there is such a chance it's just very small....&quot;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;professor: &quot;Well yes, its like the chance for all the nitrogen molecules in the air to gather around your head and have you die from lack of oxygen.&quot; &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5649" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2011-01-12T22:58:55.453" LastActivityDate="2011-01-12T22:58:55.453" CommunityOwnedDate="2011-01-12T08:07:18.127" CommentCount="4" />
  <row Id="17190" PostTypeId="1" AcceptedAnswerId="17228" CreationDate="2011-01-12T08:25:20.420" Score="0" ViewCount="186" Body="&lt;p&gt;Please forgive me a really simple question: Am I calculating the relative error reduction correctly? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example: Model 1 gets 80/100 examples wrong, model 2 gets only 20/100 examples wrong, so the 80 errors are reduced by 60, which is an error reduction of 75%.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the relative error reduction formula seems to be&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;(m1_error - m2_error) / m1_error * 100.0&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;What if it's vice versa? Model 1 gets 20/100 wrong, but model 2 gets 80/100 wrong. By what amount does model 2 &quot;reduce&quot; the error? The formula gives -300%, so it increases the  error by 300%. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does that make sense?&lt;/p&gt;&#xA;" OwnerUserId="5652" LastActivityDate="2011-01-12T14:20:06.423" Title="Relative error reduction" Tags="&lt;algebra&gt;" AnswerCount="1" />
  <row Id="17191" PostTypeId="2" ParentId="17160" CreationDate="2011-01-12T08:25:47.117" Score="1" ViewCount="" Body="&lt;p&gt;@Ross Millikan is right: you should look after and understand the background. But as I have answered this question on &lt;a href=&quot;http://stackoverflow.com/questions/4656784/skew-normal-distribution&quot;&gt;stackoverflow&lt;/a&gt;, let me answer it here also.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just look after how the mean and variance of a skew normal distribution can be computed and you got the answer! Knowing that the mean looks like:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/L6XUy.png&quot; alt=&quot;alt text&quot;&gt; &amp;nbsp;&amp;nbsp; and &amp;nbsp;&amp;nbsp;  &lt;img src=&quot;http://i.imgur.com/9pyKn.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can see, that with a xi=0 (location), omega=1 (scale) and alpha=0 (shape) you really get a standard normal distribution (with mean=0, standard deviation=1):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/1vra9.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you only change the alpha (shape) to 5, you can except the mean will differ a lot, and will be positive. If you want to hold the mean around zero with a higher alpha (shape), you will have to decrease other parameters, e.g.: the omega (scale). The most obvious solution could be to set it to zero instead of 1. See:&#xA;&lt;img src=&quot;http://i.imgur.com/hXPut.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mean is set, we have to get a variance equal to zero with a omega set to zero and shape set to 5. The formula is known:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/8UUsS.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With our known parameters:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/n2DxD.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which is insane :) That cannot be done this way. You may also go back and alter the value of xi instead of omega to get a mean equal to zero. But that way you might first compute the only possible value of omega with the formula of variance given.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/9D9B6.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then the omega should be around 1.605681 (negative or positive).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Getting back to mean:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/1NKGX.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, with the following parameters you should get a distribution you was intended to:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;location = 1.256269 (negative or positive), scale = 1.605681 (negative or positive - the opposite sign of location) and shape = 5.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5651" LastActivityDate="2011-01-12T08:25:47.117" CommentCount="1" />
  <row Id="17192" PostTypeId="2" ParentId="17152" CreationDate="2011-01-12T08:51:05.103" Score="0" ViewCount="" Body="&lt;p&gt;There's a very amusing short story about this by R A Lafferty, called &lt;a href=&quot;http://www.isfdb.org/cgi-bin/title.cgi?58333&quot; rel=&quot;nofollow&quot;&gt;&quot;Been a Long, Long Time&lt;/a&gt;&quot;. You might be able to find a copy -- it's been in a few anthologies.&lt;/p&gt;&#xA;" OwnerUserId="5656" LastActivityDate="2011-01-12T08:51:05.103" CommunityOwnedDate="2011-01-15T06:13:52.160" CommentCount="2" />
  <row Id="17194" PostTypeId="1" AcceptedAnswerId="17197" CreationDate="2011-01-12T09:10:12.033" Score="0" ViewCount="112" Body="&lt;p&gt;In Tom Apostol's &lt;em&gt;Analytic Number Theory&lt;/em&gt; book there is a problem which states:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;That there do not exists polynomials $P(x)$ and $Q(x)$ such that $$\pi(x) = \frac{P(x)}{Q(x)}$$ for all $x \in \mathbb{N}$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I tried this problem, but couldn't find a solution. Here is what i attempted. Since $\pi(x) \sim x\log{x}$ as $x \to \infty$, i saw what happens to the Right hand side $\frac{P(x)}{Q(x)}$ as $x \to \infty$. I couldn't conclude anything. If there are interesting proof's for this result, i shall be happy to see it.&lt;/p&gt;&#xA;" LastActivityDate="2011-01-12T15:37:27.827" Title="Expressing $\pi(x) = \frac{P(x)}{Q(x)}$, where $P$ and $Q$ are polynomials." Tags="&lt;analytic-number-theory&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="17196" PostTypeId="2" ParentId="17152" CreationDate="2011-01-12T09:14:11.437" Score="2" ViewCount="" Body="&lt;p&gt;Here's a rather interesting &lt;a href=&quot;http://plus.maths.org/content/infinite-monkey-business&quot; rel=&quot;nofollow&quot;&gt; article &lt;/a&gt; discussing the probability of a monkey producing Shakespeare's works and uses a random letter generator to demonstrate some results.&lt;/p&gt;&#xA;" OwnerUserId="5662" LastEditorUserId="5662" LastEditorDisplayName="" LastEditDate="2011-01-12T11:32:00.137" LastActivityDate="2011-01-12T11:32:00.137" CommunityOwnedDate="2011-01-15T06:13:52.160" />
  <row Id="17197" PostTypeId="2" ParentId="17194" CreationDate="2011-01-12T09:24:25.670" Score="1" ViewCount="" Body="&lt;p&gt;It seems like you've already gotten the answer. By asymptotic considerations, we must have $\deg P(x) &amp;gt; \deg Q(x)$. Moreover, if we have $\deg P(x) = \deg Q(x) + 1$, then we have &#xA;$$&#xD;&#xA;\frac{\pi(x) Q(x)}{P(x)} \sim \log x \frac{xQ(x)}{P(x)},&#xD;&#xA;$$&#xA;but the $xQ(x)/P(x)$ is asymptotic to the ratio of the leading coefficients, so this can't be asymptotic to $1$, i.e. $P(x)/Q(x)$ grows too slowly. On the other hand, if $\deg P(x) = \deg Q(x)+2$, then we similarly see that $P(x)/Q(x)$ grows too quickly to be asymptotic to $x \log x$. &lt;/p&gt;&#xA;" OwnerUserId="1491" LastActivityDate="2011-01-12T09:24:25.670" CommentCount="1" />
  <row Id="17198" PostTypeId="2" ParentId="16909" CreationDate="2011-01-12T09:42:09.533" Score="2" ViewCount="" Body="&lt;p&gt;The Big Picard Theorem (http://en.wikipedia.org/wiki/Picard_theorem) is a great generalization of Casorati-Weierstrass which says that in the neighborhood of an essential singularity a holomorphic function assumes all complex values with possibly one exception. In this case, that is clearly $0$. So the set is $\mathbb{C}-\{0\}$. &lt;/p&gt;&#xA;" OwnerUserId="1491" LastActivityDate="2011-01-12T09:42:09.533" CommentCount="1" />
  <row Id="17199" PostTypeId="2" ParentId="16465" CreationDate="2011-01-12T09:47:11.773" Score="5" ViewCount="" Body="&lt;p&gt;Solving equations numerically is usually good enough for engineers, say, and they probably don't need the explicit formula very often. Pure mathematicians are usually more interested in understanding how things work rather than computing specific numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a lot of information about solutions of a cubic that you cannot read off from approximate solutions. The most obvious such property is their field of definition. E.g. you cannot tell whether a cubic has solutions over the rationals by looking at approximations. Finding rational and integral solutions to polynomial equations is actually a huge part of what algebraic number theory is about in the first place. This area is called the theory of &lt;a href=&quot;http://en.wikipedia.org/wiki/Diophantine_equation&quot; rel=&quot;nofollow&quot;&gt;Diophantine equations&lt;/a&gt; and goes back to Diophantus of Alexandria, almost 2000 years back. It is usually not thought of as a means to an end, but a goal in its own right, since it is ultimately a question about how natural and rational numbers behave. The latter are some of the most immediate mathematical entities and our desire to understand them is just a manifestation of our curiosity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To understand Diophantine equations, one often needs to understand not just the field of rational numbers, but more general &lt;a href=&quot;http://en.wikipedia.org/wiki/Algebraic_number_field&quot; rel=&quot;nofollow&quot;&gt;number fields&lt;/a&gt;. So, we are not only interested in whether or not a cubic has a rational solution, but we want to know exactly the field of definition of these solutions, and for that, an algebraic expression for its roots is very handy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This was a long way of saying that you shouldn't think of the algebraic formula&#xA;for the roots of a cubic as a means to an end, but rather as an accomplishment&#xA;on the way to a more general aim: solving Diophantine equations and understanding how integers and the rational numbers work. This more&#xA;general aim is pursued by number theorists because they find it exciting,&#xA;not because it has applications.&lt;/p&gt;&#xA;" OwnerUserId="3212" LastActivityDate="2011-01-12T09:47:11.773" CommentCount="2" />
  <row Id="17200" PostTypeId="1" AcceptedAnswerId="17205" CreationDate="2011-01-12T09:56:14.927" Score="4" ViewCount="135" Body="&lt;p&gt;Hi everyone! This is my first question here. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As I'm not a matematician I thought I'd ask here for advice how to approach something I'm working on as a hobby project.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;A bit of context&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Let's say there is a collection of items with a description of features and a price. Imagine a list of cars and prices. All cars have a list of features, e.g. engine size, color, horse power, model, year etc. For each make, something like this:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Ford:&#xA;V8, green, manual, 200hp, 2007, $200&#xA;V6, red, automatic, 140hp, 2010, $300&#xA;V6, blue, manual, 140hp, 2005, $100&#xA;...&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Going even further, the list of cars with prices is published with some time-interval which means we have access to historical price data. Might not always include exactly the same cars.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Problem&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;I would like to understand how to model prices for any car based on this base information, most importantly cars not in the initial list.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Ford, v6, red, automatic, 130hp, 2009&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;For the above car, it's almost the same as one in the list, just slightly different in horse power and year. To price this, what is needed?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;What I'm looking for is something practical and simple, but I would also like to hear about more complex approaches how to model something like this.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;What I've tried&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Here is what I've been experimenting with so far:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) using historical data to lookup car X. If not found, no price. This  is of course very limited and one can only use this in combination with some time decay to alter prices for known cars over time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) using a car feature weighting scheme together with a priced sample car. Basically that there is a base price and features just alter that with some factor. Based on this any car's price is derived.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first proved to be not enough and the second proved to not be always correct and I might not have had the best approach to using the weights. This also seems to be a bit heavy on  maintaining weights, so that's why I thought maybe there is some way to use the historical data as statistics in some way to get weights or to get something else. I just don't know where to start.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Any suggestions how a problem like this could be approached?&lt;/strong&gt; All ideas are more than welcome. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks a lot in advance and looking forward to reading your suggestions!&lt;/p&gt;&#xA;&#xA;UPDATE:&#xA;&#xA;&lt;p&gt;Thanks a lot for the useful suggestions so far! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I just want to add that as a beginner like me, it would be good to also consider the aspects what it takes to set things up. What about the following aspects:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;integrate into some software project I have. Either by using existing libraries or writing algorithm myself.&lt;/li&gt;&#xA;&lt;li&gt;fast recalculation when new historical data comes in.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="5646" LastEditorUserId="5646" LastEditorDisplayName="" LastEditDate="2011-01-12T12:56:13.643" LastActivityDate="2011-01-16T10:18:54.373" Title="How to model prices?" Tags="&lt;regression&gt;" AnswerCount="4" CommentCount="9" FavoriteCount="1" />
  <row Id="17202" PostTypeId="1" CreationDate="2011-01-12T10:02:02.503" Score="1" ViewCount="48" Body="&lt;p&gt;Given a divisor $D$ on an algebraic curve $X$, there is a corresponding map $\phi_D$ from $X$ to the projective space (of dimension $\dim L(D)-1$). In particular, we know that if $D$ is a very ample divisor, then this is a holomorphic embedding into projective space. When $D$ is not very ample, how does one read off information about the map? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a specific problem from Miranda's textbook to provide context:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $X$ be an algebraic curve of genus $2$. Let $K$ be a canonical divisor on $X$, and let $\phi_{2K}$ be the associated map to projective space. Show that the image of $\phi_{2K}$ is a smooth projective plane conic, and that the map has degree $2$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are my thoughts so far. We know that $\deg(2K) = 4g-4 = 4$. By Riemann-Roch, we also have $\deg L(2K) = \deg(2K) + 1 - g = 3$. This would suggest that the image is in $\mathbb(P)^2$, and by the degree it could be a double covering of a conic, an embedding as a degree $4$ curve, or perhaps a quadruple covering of a line? &lt;/p&gt;&#xA;" OwnerUserId="1491" LastEditorUserId="1491" LastEditorDisplayName="" LastEditDate="2011-01-13T23:22:00.713" LastActivityDate="2011-01-13T23:22:00.713" Title="Deducing characteristics of a map induced by a divisor" Tags="&lt;algebraic-curves&gt;&lt;projective-space&gt;" CommentCount="2" />
  <row Id="17203" PostTypeId="2" ParentId="17200" CreationDate="2011-01-12T10:16:06.240" Score="1" ViewCount="" Body="&lt;p&gt;I'm not sure, just an idea, but it seems that you are talking about interpolation of function F(price, factor1, ..., factorN) = 0 given a set of knots (basically knowing the price for a certain set of configurations define tuples (price, factor1, ..., factorN), which can be used to create an interpolation net). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And as for a concrete interpolation approach – you can try a bunch of those and see which fits in best.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's just an idea, mathematicians, correct me if I'm wrong.&lt;/p&gt;&#xA;" OwnerUserId="5667" LastActivityDate="2011-01-12T10:16:06.240" CommentCount="1" />
  <row Id="17204" PostTypeId="1" CreationDate="2011-01-12T10:56:07.537" Score="2" ViewCount="43" Body="&lt;p&gt;The following is a problem in Miranda's &lt;em&gt;Algebraic Curves and Riemann Surfaces.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given any algebraic curve $X$ and a point $p \in X$, show that there is a meromorphic $1$-form $\omega$ on $X$ whose Laurent series at $p$ looks like $dz/z^n$ for $n &amp;gt; 1$, and which has no other poles on $X$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The point of this is as a step towards the proof that the Mittag-Leffler problem can be solved for $X$. &lt;/p&gt;&#xA;" OwnerUserId="1491" LastEditorUserId="1491" LastEditorDisplayName="" LastEditDate="2011-01-12T16:42:31.740" LastActivityDate="2011-01-12T16:42:31.740" Title="How to prove that this kind of differential form exists on an algebraic curve?" Tags="&lt;algebraic-curves&gt;&lt;differential-forms&gt;" CommentCount="4" FavoriteCount="1" />
  <row Id="17205" PostTypeId="2" ParentId="17200" CreationDate="2011-01-12T11:02:24.987" Score="3" ViewCount="" Body="&lt;p&gt;Probably, you are interested in &lt;a href=&quot;http://www.statsoft.com/textbook/multiple-regression/&quot; rel=&quot;nofollow&quot;&gt;multiple-regression&lt;/a&gt;. It seems that &lt;a href=&quot;http://office.microsoft.com/en-us/excel-help/perform-a-regression-analysis-HA001111963.aspx&quot; rel=&quot;nofollow&quot;&gt;excel&lt;/a&gt; can also help you for this purpose. Quoting from the second link: &quot;For example, suppose you want to project the appropriate price for a house in your area based on square footage, number of bathrooms, lot size, and age. Using a multiple regression formula, you can estimate a price, based on a database of information gathered from existing houses.&quot;&lt;/p&gt;&#xA;" OwnerUserId="2810" LastEditorUserId="2810" LastEditorDisplayName="" LastEditDate="2011-01-12T11:12:31.243" LastActivityDate="2011-01-12T11:12:31.243" CommentCount="6" />
  <row Id="17206" PostTypeId="2" ParentId="17200" CreationDate="2011-01-12T11:10:23.847" Score="3" ViewCount="" Body="&lt;p&gt;This is a typical example of &quot;hedonic pricing&quot;, used in applied empirical economics: &lt;a href=&quot;http://www.jstor.org/pss/1830899&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; is a paper with a good review of the theory (JPE papers are renowned), and &lt;a href=&quot;http://www.bls.gov/cpi/cpicamco.htm&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; is a closely related application to the contribution of individual camcorder characteristics in price.&lt;/p&gt;&#xA;" OwnerUserId="5618" LastActivityDate="2011-01-12T11:10:23.847" CommentCount="5" />
  <row Id="17207" PostTypeId="1" CreationDate="2011-01-12T11:21:53.980" Score="5" ViewCount="326" Body="&lt;p&gt;What is the most recommended quality a mathematician should have?&#xA;Extremely high IQ levels?Passion for what he does?Patience and &quot;stuborness&quot;?Something else?&#xA;Of course all of them are necessary ,but which of them do you believe is  the most needed ?&lt;/p&gt;&#xA;" OwnerUserId="5422" LastActivityDate="2011-01-12T12:57:29.180" CommunityOwnedDate="2011-01-12T14:19:53.757" Title="What is the key to success for a mathematician?" Tags="&lt;soft-question&gt;" AnswerCount="3" CommentCount="9" FavoriteCount="3" />
  <row Id="17208" PostTypeId="1" AcceptedAnswerId="17222" CreationDate="2011-01-12T11:25:58.910" Score="2" ViewCount="136" Body="&lt;p&gt;This question was inspired by &lt;a href=&quot;http://math.stackexchange.com/questions/17117/chance-of-7-of-a-kind-with-10-dice/17122#17122&quot;&gt;this question&lt;/a&gt; dealing with finding the probability of rolling a $7$ of a kind with $10$ dice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a general method for finding the chance of rolling a $k$ of a kind with $n$ $m$-sided dice when $k\leq n/2$? When $k&amp;gt;n/2$ the probability is found relatively easily with the Bernoulli distribution, but when $k\leq n/2$ there's the possibility that one may roll a second, or third, etc., $k$ of a kind with the other $n-k$ di(c)e. I don't see a clear way around this possible over counting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="1583" LastActivityDate="2011-01-12T18:13:31.617" Title="Probability of getting exactly $k$ of a kind in $n$ rolls of $m$-sided dice, where $k\leq n/2$." Tags="&lt;probability&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="17209" PostTypeId="1" CreationDate="2011-01-12T11:43:55.293" Score="3" ViewCount="56" Body="&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Sprouts_%28game%29&quot; rel=&quot;nofollow&quot;&gt;Wikipedia article on sprouts&lt;/a&gt;, seems to suggest that it does not, although I can't see how to prove it.&lt;/p&gt;&#xA;" OwnerUserId="123" LastActivityDate="2011-01-12T14:12:43.437" Title="Do the positions of the initial dots affect the game sprouts?" Tags="&lt;topology&gt;&lt;game-theory&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="17210" PostTypeId="2" ParentId="17208" CreationDate="2011-01-12T11:49:02.697" Score="1" ViewCount="" Body="&lt;p&gt;Isn't this pretty straightforward? You pick the side that you want there are $m$ ways to do this. Then you are rolling $n$ dice and you want k dice to be a certain side of which there is a $(1/m)^k$ chance (if in a row) and then $((m-1)/m)^{(n-k)}$ for the remaining dice. You also need to account for different orders so you have $n$ choose $k$ choices for your dice on side k and these will determine the positions of your non-k sided dice. Putting everything together we get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$m \binom{n}{k} \frac{1}{m^k}\left(\frac{m-1}{m}\right)^{n-k}.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whoops, wasn't paying attention when I wrote this. Then we want to know the probability that none of the remaining (n-k) dice are a k of a kind. So we need to subtract away the probability that there exists a k of a kind in the spare dice. First we have (n-k) choose k pairs to examine and the probability that all k are not the same so once again we have to pick a side, but note that we can only pick (m-1) sides because we've already ensured that they are not one of the sides. Finally we get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$m \binom{n}{k} \frac{1}{m^k}\left(\frac{m-1}{m}\right)^{n-k}-(m-1)\binom{n-k}{k}\frac{1}{(m-1)^k}.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe this is right, it certainly holds for the k&gt;n/2 case. &lt;/p&gt;&#xA;" OwnerUserId="1930" LastEditorUserId="1930" LastEditorDisplayName="" LastEditDate="2011-01-12T14:59:14.153" LastActivityDate="2011-01-12T14:59:14.153" CommentCount="3" />
  <row Id="17211" PostTypeId="2" ParentId="17208" CreationDate="2011-01-12T11:54:14.157" Score="1" ViewCount="" Body="&lt;p&gt;You can avoid the over-counting using the inclusion-exclusion principle.&lt;/p&gt;&#xA;" OwnerUserId="5676" LastActivityDate="2011-01-12T11:54:14.157" />
  <row Id="17212" PostTypeId="1" CreationDate="2011-01-12T12:26:27.703" Score="4" ViewCount="247" Body="&lt;p&gt;We know there are statements that are undecidable/independent of ZFC. Can there be a statement S, such that (ZFC $\not\vdash$ S and ZFC $\not\vdash$ ~S) is undecidable?&lt;/p&gt;&#xA;" OwnerUserId="5079" LastActivityDate="2011-01-12T19:50:38.437" Title="Is there a statement whose undecidability is undecidable?" Tags="&lt;logic&gt;&lt;set-theory&gt;" AnswerCount="3" CommentCount="4" />
  <row Id="17213" PostTypeId="2" ParentId="17207" CreationDate="2011-01-12T12:27:48.607" Score="3" ViewCount="" Body="&lt;p&gt;A mathematician is interested in the Concepts, the &quot;principles&quot; behind things.  It is important (IMO) to focus on the big ideas, on Understanding, as opposed to procedures and memorization.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't think a high IQ is especially important but most mathematicians I know enjoy playing with ideas (any ideas!) and so enjoy intellectual games or intellectual exercises.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most of us are (eventually) quite passionate about mathematics -- but that passion probably comes later?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And many of us hate memorizing -- I got into math partly because I could do it without any memorization.  If I understood the concepts, it all made sense and no memorization is necessary.&lt;/p&gt;&#xA;" OwnerUserId="5680" LastActivityDate="2011-01-12T12:27:48.607" CommunityOwnedDate="2011-01-12T14:19:53.757" />
  <row Id="17214" PostTypeId="2" ParentId="17207" CreationDate="2011-01-12T12:35:01.237" Score="4" ViewCount="" Body="&lt;p&gt;First of all, like what you do. I'd even say love what you do.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second, be monomaniacal. Just to tell a little anecdote, a cousin of Albert Einstein once sent him a letter asking why her husband, a mathematician, was not making great discoveries like Albert. Albert replied that was because her husband was not monomaniacal about mathematics. This story tells us that if you want to attain the summit of your profession, you should also put all your efforts into it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Third, nowadays, it is very rare for mathematicians, and scientists in general, to work in isolation. Most papers that are published have multiple authors, collaboration is getting more and more important. Definitely communication skills and the ability to collaborate have become paramount.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fourth, you'll also have to learn how to sell your product. This is sadly a result of how research is becoming increasingly funded. You'll be up against a whole lot of bureaucracy.&lt;/p&gt;&#xA;" OwnerUserId="3567" LastEditorUserId="3567" LastEditorDisplayName="" LastEditDate="2011-01-12T12:57:29.180" LastActivityDate="2011-01-12T12:57:29.180" CommunityOwnedDate="2011-01-12T14:19:53.757" CommentCount="1" />
  <row Id="17215" PostTypeId="2" ParentId="17207" CreationDate="2011-01-12T12:36:09.687" Score="9" ViewCount="" Body="&lt;p&gt;People asked Gauss: &quot;how come you always see things that we don't?&quot;&#xA;And Gauss replied: &quot;&lt;em&gt;If only you would have thought about it for as long as I have...&lt;/em&gt;&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On another occasion someone asked Gauss about Fermat's last theorem. Gauss said he didn't care much, since he could easily fabricate a number of similar conjectures that one could neither proof nor disproof. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think that's the most vital aspect: tackle problems you think are just on the boundary of your reach and never ever ever give up. (Unless you're getting danderously close to insanity. But that's another story.)&lt;/p&gt;&#xA;" OwnerUserId="5189" LastActivityDate="2011-01-12T12:36:09.687" CommunityOwnedDate="2011-01-12T14:19:53.757" />
  <row Id="17216" PostTypeId="2" ParentId="17089" CreationDate="2011-01-12T12:48:01.053" Score="2" ViewCount="" Body="&lt;p&gt;There are literally dozens of ways to represent &quot;arbitrary&quot; (given or unknown) functions $f$ in terms of &quot;special&quot; functions. Each of these ways responds to the particular geometrical situation at hand, to ways of encoding the available information about $f$, to a-priori-conditions that $f$ must fulfill, etc. The special functions used in each particular case are &quot;taken from a catalogue&quot;, they are well understood and usually have an algebraically describable behavior with respect to the natural operations (differentiation, shifts, rotations, etc.) present in the given environment.&lt;/p&gt;&#xA;" OwnerUserId="1303" LastEditorUserId="1303" LastEditorDisplayName="" LastEditDate="2011-01-13T08:39:04.967" LastActivityDate="2011-01-13T08:39:04.967" />
  <row Id="17217" PostTypeId="1" CreationDate="2011-01-12T13:01:28.630" Score="2" ViewCount="58" Body="&lt;p&gt;I'm wondering if their exists a visual/behavioral reference for the fundamental families of functions. I'm not a mathematician so excuse my language if I'm being overly vague.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to have a reference where I can go look at images of functions and say: that looks like a behavior/character I want. Or maybe even search on the properties of the function, such as it's periodic, or that the output is always in the [-1, 1] range.&lt;/p&gt;&#xA;" OwnerUserId="5684" LastEditorUserId="5684" LastEditorDisplayName="" LastEditDate="2011-01-12T14:42:26.817" LastActivityDate="2011-01-12T14:42:26.817" Title="Function Shape Reference" Tags="&lt;soft-question&gt;&lt;reference-request&gt;&lt;functions&gt;&lt;functional-equations&gt;" CommentCount="5" />
  <row Id="17218" PostTypeId="1" AcceptedAnswerId="17219" CreationDate="2011-01-12T13:02:33.963" Score="2" ViewCount="72" Body="&lt;p&gt;Consider $\mathbb{R}^n$ equipped with the standard scalar product. Let $f(n)$ denote the maximal cardinality of a set of vectors in $\mathbb{R}^n$ with a pairwise negative scalar product. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is $f(n)$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some thoughts:&#xA;WLOG one can assume, that all vectors have length 1. This turns this problem in a space packaging problem in $S^{n-1}$: How many discs of radius $\pi/4$ can be placed inside $S^{n-1}$, so that the do not touch each other.&#xA;If one allows touching, than the $2n$ vectors $\pm e_i$ (where $e_i$ denotes the standard basis) seem to be quite optimal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The vertices of the regular $n$-simplex have pairwise negative scalar products, so we know, that $f(n)\ge n+1$.&lt;/p&gt;&#xA;" OwnerUserId="707" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2011-01-12T13:59:30.227" LastActivityDate="2011-01-12T13:59:30.227" Title="Maximal number of vectors with (pairwise) negative scalar product" Tags="&lt;geometry&gt;&lt;linear-algebra&gt;&lt;combinatorics&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="17219" PostTypeId="2" ParentId="17218" CreationDate="2011-01-12T13:34:39.757" Score="4" ViewCount="" Body="&lt;p&gt;This question been asked on MathOverflow: &lt;a href=&quot;http://mathoverflow.net/questions/31436/largest-number-of-vectors-with-pairwise-negative-dot-product&quot; rel=&quot;nofollow&quot;&gt;http://mathoverflow.net/questions/31436/largest-number-of-vectors-with-pairwise-negative-dot-product&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f(n)=n+1$.&lt;/p&gt;&#xA;" OwnerUserId="425" LastActivityDate="2011-01-12T13:34:39.757" />
  <row Id="17221" PostTypeId="2" ParentId="14162" CreationDate="2011-01-12T13:54:13.600" Score="0" ViewCount="" Body="&lt;p&gt;I would fit a straight line to the bins near but outside the peak and subtract this from the data, claiming that this level is the background.  Then you can do a 2D fit of the Gaussian parameters (mean and sigma) and see how it turns out.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-01-12T13:54:13.600" />
  <row Id="17222" PostTypeId="2" ParentId="17208" CreationDate="2011-01-12T13:55:31.590" Score="3" ViewCount="" Body="&lt;p&gt;Let $A_i$ be the event that we obtain exactly $k$ $i$'s (assuming that the dice are labeled&#xA;$1,2,3,\ldots,m$ on their sides). Now inclusion-exclusion gives&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$P(A_1 \cup A_2 \cup \ldots \cup A_m) = \sum P(A_i) - \sum P( A_i \cap A_j)&#xD;&#xA;+ \sum P(A_i \cap A_j \cap A_k ) - \cdots $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where the summations range over distinct $i,j,k,\ldots .$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$P(A_i) = { n \choose k} \left( \frac{1}{m} \right)^k&#xD;&#xA;\left( 1 - \frac{1}{m} \right)^{n-k} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$P(A_i \cap A_j) = { n \choose k} \left( \frac{1}{m} \right)^k&#xD;&#xA;\left( 1 - \frac{1}{m} \right)^{n-k} \cdot { n-k \choose k} \left( \frac{1}{m} \right)^k&#xD;&#xA;\left( 1 - \frac{1}{m} \right)^{n-2k} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$P(A_i \cap A_j \cap A_k)$$ &#xA;$$ = { n \choose k} \left( \frac{1}{m} \right)^k&#xD;&#xA;\left( 1 - \frac{1}{m} \right)^{n-k} \cdot { n-k \choose k} \left( \frac{1}{m} \right)^k&#xD;&#xA;\left( 1 - \frac{1}{m} \right)^{n-2k} \cdot { n-2k \choose k} \left( \frac{1}{m} \right)^k&#xD;&#xA;\left( 1 - \frac{1}{m} \right)^{n-3k} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and so on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Simplifying these expressions, the probability that we get exactly $k$ of a number is given by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$P(A_1 \cup A_2 \cup \ldots \cup A_m) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ = { m \choose 1 } { n \choose k} \left( \frac{1}{m} \right)^k&#xD;&#xA;\left( 1 - \frac{1}{m} \right)^{n-k} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ - { m \choose 2 } { n \choose k} { n-k \choose k} \left( \frac{1}{m} \right)^{2k}&#xD;&#xA;\left( 1 - \frac{1}{m} \right)^{2n-3k}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ + { m \choose 3} { n \choose k} { n-k \choose k} { n-2k \choose k} \left( \frac{1}{m} \right)^{3k} \left( 1 - \frac{1}{m} \right)^{3n-6k} - \cdots $$&lt;/p&gt;&#xA;" OwnerUserId="1301" LastEditorUserId="1301" LastEditorDisplayName="" LastEditDate="2011-01-12T18:13:31.617" LastActivityDate="2011-01-12T18:13:31.617" CommentCount="4" />
  <row Id="17223" PostTypeId="2" ParentId="17152" CreationDate="2011-01-12T13:56:05.330" Score="-3" ViewCount="" Body="&lt;p&gt;Just to clarify, from a less &quot;mathy&quot; point of view.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I see people answering about how practical it would be so,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Important:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is a &quot;metaphor&quot;, I think it is safe to say no monkey will every randomly write Hamlet (this is obvious because of the above mentioned practical limitations).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, why do we say such things, because what it important is what is behind, that is the concept of random and infinity:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you turn the monkey into a computer that randomly outputs characters, eventually (&lt;a href=&quot;http://math.stackexchange.com/questions/17152/given-an-infinite-number-of-monkeys-and-an-infinite-amount-of-time-would-one-of/17154#17154&quot;&gt;see the math in Ross' answer&lt;/a&gt;) you will get Hamlet, written.&lt;/p&gt;&#xA;" OwnerUserId="4256" LastActivityDate="2011-01-12T13:56:05.330" CommunityOwnedDate="2011-01-15T06:13:52.160" CommentCount="4" />
  <row Id="17224" PostTypeId="2" ParentId="17152" CreationDate="2011-01-12T14:03:53.560" Score="8" ViewCount="" Body="&lt;p&gt;Hamlet was indeed written after a finite number of monkeys (Shakespeare was a a primate from the family Hominidae!)&lt;/p&gt;&#xA;" OwnerUserId="3162" LastActivityDate="2011-01-12T14:03:53.560" CommunityOwnedDate="2011-01-12T14:03:53.560" CommentCount="6" />
  <row Id="17225" PostTypeId="1" AcceptedAnswerId="17227" CreationDate="2011-01-12T14:06:21.907" Score="4" ViewCount="58" Body="&lt;p&gt;Hello everyone,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've been struggling with the following problem for a couple of days and I don't seem to get any further:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $R$ be a commutative ring. I would like to get (something like) a classification of all finitely generated $R$-modules $M$ that satisfy the following condition:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When we look at $M \otimes_R M \otimes_R M$, the permutation (123) induces an automorphism of $M \otimes_R M \otimes_R M$ by sending $a \otimes b \otimes c$ to $c \otimes a \otimes b$. I demand that this automorphism be the identity map. In other words, in $M \otimes_R M \otimes_R M$, the elements $a \otimes b \otimes c, c \otimes a \otimes b, b \otimes c \otimes a$ should all be the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $R$ is a field, it is easy to see that the only non-trivial finitely generated $R$-modules (i.e. finite-dimensional vector spaces) that satisfy this condition are the 1-dimensional ones. Furthermore, one sees more generally that all cyclic modules satisfy the condition, too. Up to now I've  neither come up with an example of a non-cyclic module that satisfies the condition, nor was I able to prove that all modules that satisfy this condtion must be cyclic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can somebody help with this matter?&lt;/p&gt;&#xA;" OwnerUserId="1766" LastEditorUserId="367" LastEditorDisplayName="" LastEditDate="2011-01-12T14:25:14.533" LastActivityDate="2011-01-12T14:25:14.533" Title="Modules $M$ such that the automorphism of $M \otimes M \otimes M$ induced by the permutation $(123)$ is the identity" Tags="&lt;commutative-algebra&gt;" AnswerCount="1" />
  <row Id="17226" PostTypeId="2" ParentId="17209" CreationDate="2011-01-12T14:12:43.437" Score="4" ViewCount="" Body="&lt;p&gt;It does not, it just depends on how many dots you start with.  Informally, you would just argue that for any game you can move the dots around, stretching the lines as necessary to maintain the relations of inside/outside, to any pattern you like and get a corresponding game.  More formally you could set up a homeomorphism between the plane with dots in one position and the plane with dots in another position and argue that the game plays the same either way.  This last approach suggests correctly that you could make correspondences between partially played games that do not entirely respect the inside/outside-think of the plane as a punctured sphere and you can close it up and puncture it anywhere else to get a game with the same topology but with different inside/outside relations.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-01-12T14:12:43.437" />
  <row Id="17227" PostTypeId="2" ParentId="17225" CreationDate="2011-01-12T14:19:38.827" Score="4" ViewCount="" Body="&lt;p&gt;It is not true that a module that satisfies the above condition is cyclic. For instance, it is possible that $M \neq 0$ but $M \otimes M$ (and thus higher tensor powers) are zero. Consider for instance $M = \mathbb{Q}/\mathbb{Z}$ over the integers. Then $$M \otimes M = \mathbb{Q}/\mathbb{Z} \otimes \mathbb{Q}/\mathbb{Z} = \mathbb{Q} \otimes (\mathbb{Q}/\mathbb{Z})/ \mathbb{Z} \otimes \mathbb{Q}/\mathbb{Z} = 0$$&#xA;since tensoring with $\mathbb{Q}$ does nothing for a divisible group (it's a form of localization at the multiplicative set $\mathbb{Z} - 0$). So any two maps on $M \otimes M \otimes M$ are equal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, suppose $M$ is finitely generated. The above example was not (and, in fact, the above  phenomenon can't happen for a finitely generated module). Then it is easy to see that if $M$ satisfies your condition, so does the base-change $M \otimes_R R&amp;#39;$ for $R&amp;#39;$ any $R$-algebra (considered as an $R&amp;#39;$-module, that is!) because base extension is a &lt;a href=&quot;http://en.wikipedia.org/wiki/Monoidal_functor&quot; rel=&quot;nofollow&quot;&gt;monoidal functor&lt;/a&gt; with respect to the tensor product.&#xA;It follows that for any prime ideal $\mathfrak{p}$, $M \otimes k(\mathfrak{p})$ (for $k(\mathfrak{p})$ the residue field, i.e. the quotient field of the residue ring) satisfies this property, so from what you have proved about vector spaces, it has rank at most one. Thus all the fibers of $M$ are of rank at most one.&#xA;So at least $M$ is &quot;close&quot; to being cyclic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More generally, the above observation on base-change reduces the question to the case of a local ring, because to check that two things are equal, it suffices to check on all the localizations. By Nakayama's lemma, it follows that for a local ring, and your observation for a field, any module satisfying your property is in fact cyclic.  For local rings, we thus have an if and only if statement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Putting the above observations all together, we find:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;blockquote&gt;&#xA;    &lt;p&gt;A module satisfies your property if and only if the localization at each prime is cyclic (as a module over the localized ring).&lt;/p&gt;&#xA;  &lt;/blockquote&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="536" LastActivityDate="2011-01-12T14:19:38.827" CommentCount="1" />
  <row Id="17228" PostTypeId="2" ParentId="17190" CreationDate="2011-01-12T14:20:06.423" Score="1" ViewCount="" Body="&lt;p&gt;It does make sense.  Reducing the error by -300% is increasing it by 300%.  And increasing 20 by 300% gives 80.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-01-12T14:20:06.423" />
  <row Id="17229" PostTypeId="1" AcceptedAnswerId="17238" CreationDate="2011-01-12T14:21:14.030" Score="2" ViewCount="193" Body="&lt;p&gt;Hi everyone!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've been referred to this website, hopefully you have the &#xA;background in set theory to help me out here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Got two questions, the first is on number systems arising out of set theory &#xA;and the second is a small one about functions. I've tried to be as clear as &#xA;possible but please bear with me; the only reason I'm posting this is because &#xA;I'm so confused.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm trying to get my bearings on the set-theoretic construction of the &#xA;natural numbers. Just as I was confused looking at the the way the reals &#xA;could be axiomatically defined, or constructed by Cantor or Dedekind, or &#xA;thinking that Cauchy, Weierstrass etc... all had different constructions so &#xA;now am I going through that with the various explanations of the natural &#xA;numbers as originating from set theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Peano axioms can be taken as the starting point, but I found out that &#xA;they could be taken as a theorem in set theory, so I've been trying to do &#xA;that. It seems that one way is to look at the idea of an ordinal which is a &#xA;transitive set whose elements are also transitive. But there's also this &#xA;idea of equivalence classes that apparently lead to contradictions in ZFC. &#xA;Also there's the Axiom of Infinity method in the wiki link I've given below.&#xA;Perhaps there's even more, or these are the same in some way I can't &#xA;see?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A problem arises for me as I've come to believe that rational numbers are &#xA;defined as a set of equivalence classes, but Wikipedia says that &#xA;equivalence classes are ruled out of axiomatic set theory so what replaces &#xA;Q when dealing with ZFC? When you remove equivalence classes from the &#xA;picture what defines the rational numbers in ZFC? I'm sure whatever you &#xA;do will generalize to the reals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the question could best be put as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Starting from the most fundamentals, i.e., the axioms of a specific model of &#xA;set theory, what is the logical and linear progression of topics that leads to &#xA;the construction of  $\mathbb{N}$, then $\mathbb{Z}$, then $\mathbb{Q}$, then $\mathbb{R}$, then $\mathbb{C}$? It seems to me &#xA;that there are at least two possible methods this way, one using Von &#xA;Neumann ordinals (with an axiom of infinity?), while the other uses the ideas &#xA;of relations/equivalence classes etc... If you know anything about how to &#xA;do this &amp;amp; understand the distinctions between the various methods of &#xA;doing this could you just let me know, hopefully with a few analogies!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an aside, I just read earlier that the difference between the Riemann &amp;amp; &#xA;Lebesgue integrals can be explained by the analogy of a shop taking in &#xA;money, the Riemann integral adds the money up as it comes in all through &#xA;the day while the Lebesgue integral separates the coins out at the end of &#xA;the day according to a certain scheme (i.e. the idea of measure).&#xA;I'm sure there exists a nice way to picture the two (or more) schemes for &#xA;constructing the number systems! (I read this earlier &amp;amp; I had to mention &#xA;it as it would just be so helpful!).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think equivalence classes originated with Frege &amp;amp; Russell used them but &#xA;they run into contradictions which ZFC surmounts but Quine's set theory &#xA;removes the contradictions &amp;amp; allows the use of equivalence classes. I'm &#xA;sure there is more than just the use of equivalence classes that &#xA;distinguishes these two seperate constructions. Is it fair to group this &#xA;into Frege/Russell/Quine on one side &amp;amp; ZFC-Neumann on the other?&#xA;What about the axiom of infinity description in the wiki link below? &#xA;Also, I assume that the construction of the number systems in the &#xA;&quot;Elementary Set Theory with a Universal Set&quot; link below is just the &#xA;Frege-Quine idea right, I mean it's not another new one is it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some links might be helpful for reference of what I'm talking about:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Ordinal_number#Definition_of_an_ordinal_as_an_equivalence_class&quot; rel=&quot;nofollow&quot;&gt;Definition of ordinal number&lt;/a&gt; (from Wikipedia).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;http://books.google.ie/books?id=FE5cluQ3CE4C&amp;amp;printsec=frontcover#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;A Tour through Mathematical Logic&lt;/em&gt;&lt;/a&gt; by Robert S. Wolf. On page 77, notice at the top of the page they say that $\in$ is a &quot;well-founded relation&quot;.&#xA;My knowledge of Naive set theory tells me that $\in$ is one of two undefinable constructs that you're supposed to take as given (I could quote the sources of this if needed), but here they call it a relation. Surely they don't mean relation in the subset of the cartesian product way do they? If it is a relation why would naive set theory call it undefinable when NST deals in defining what relations are?&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;http://wikipedia.org/wiki/Set-theoretic_definition_of_natural_numbers&quot; rel=&quot;nofollow&quot;&gt;Wikipedia's article on the set-theoretic definition of the natural numbers.&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;http://wikipedia.org/wiki/Axiom_of_Infinity&quot; rel=&quot;nofollow&quot;&gt;Wikipedia's article on the Axiom of Infinity.&lt;/a&gt; Notice here they've used something totally different (I think?) to &#xA;construct the natural numbers, the axiom of infinity. There is no &#xA;mention of ordinals in this link.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;http://math.boisestate.edu/%7Eholmes/holmes/head.pdf&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;Elementary Set Theory with a Universal Set&lt;/em&gt; (in PDF format)&lt;/a&gt;, by Holmes and Randall, 1988.  (Surely not a different construction, it's the Frege one right?).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So, that is the current state of my thoughts on this topic, it's taken me a &#xA;quite a while to figure this meagre stuff out &amp;amp; to find good sources which &#xA;aren't at the graduate level to learn properly. If I hadn't read all this &#xA;conflicting stuff I'd be happy to just use the following sources as my &#xA;main material:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;http://www2.latech.edu/~schroder/fund_videos.htm&quot; rel=&quot;nofollow&quot;&gt;Videos on the fundamentals of mathematics&lt;/a&gt; by Bernd Schr&amp;ouml;der.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;http://books.google.ie/books?id=IRpABNX4s0EC&amp;amp;pg=PA243&amp;amp;lpg=PA243&amp;amp;dq=truss+foundations+of+mathematical+analysis+contents&amp;amp;source=bl&amp;amp;ots=Vx1MN5B276&amp;amp;sig=i2-PssbKvV48gT5hOwcBqpLs50c&amp;amp;hl=en&amp;amp;ei=kY8tTYDZO4uYhQefrdX3CA&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=3&amp;amp;ved=0CCgQ6AEwAg#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;Foundations of Mathematical Analysis&lt;/a&gt; by J.K. Truss (in Google Books). &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;http://www.math.bilgi.edu.tr/courses/Lecture%20Notes/SetTheoryLectureNotes.pdf&quot; rel=&quot;nofollow&quot;&gt;Ali Nesin's Set Theory Notes&lt;/a&gt;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Basically if you could help me clean up my thoughts as regards the above &#xA;stuff I've written, i.e. 3 different constructions 1) ZFC, 2) Frege &#xA;Equivalence Relations, 3) Axiom of Infinity &amp;amp; the ∈ relation statement &#xA;then I could move on to greener pastures!&lt;/p&gt;&#xA;&#xA;&lt;hr/&gt;&#xA;&#xA;&lt;p&gt;My second question is brief, just on the topic of a function. First a function &#xA;is the idea of $y = f(x)$ in school. Then a function is the same thing with &#xA;domains and ranges. Then you see that a function is really a special case &#xA;of a mapping $f\colon A \to B$ defined by $f\colon a \mapsto f(a)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the latest thing &#xA;I've read is that a function is really a set $(A,B,F)$ which is also written as &#xA;$((A,B),F)$, &amp;amp; I assume it's to indicate order, i.e. a Kuratowski ordered pair, &#xA;and $F$ is really $F \subseteq A \times B$. This is obviously a relation &amp;amp; it's dictated by &#xA;the rule [&#xA;$$\forall(a\in A)\exists(b\in B)\Bigl( (a,b)\in F\wedge \bigl(f(a)=b\bigr)\Rightarrow (a,b)\in F\Bigr).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll admit that's not exactly clear, if it's possible to clean this up &amp;amp; clarify it &#xA;please let me know. For example, where has the little $f$ come from? &#xA;The general idea here makes a lot of sense &amp;amp; seeing as math is always &#xA;employing set theory I wish they'd just use this notation from the &#xA;beginning in books.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the question is, is this the final resting place for the definition of a &#xA;function? Could you clean this up for me? Could you give me some &#xA;references where this definition is explained more clearly, I can't find &#xA;anything other than a mention on wikipedia &amp;amp; the Ali Nesin set theory &#xA;notes in the link I've given above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks so much, I know it's asking a lot of your time!&lt;/p&gt;&#xA;" OwnerUserId="5696" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-13T14:24:11.257" LastActivityDate="2011-01-13T14:24:11.257" Title="Set Theory as it Relates to Number Systems?" Tags="&lt;real-analysis&gt;&lt;set-theory&gt;" AnswerCount="3" CommentCount="11" />
  <row Id="17230" PostTypeId="2" ParentId="16909" CreationDate="2011-01-12T14:28:58.503" Score="2" ViewCount="" Body="&lt;p&gt;$\exp$ maps each horizontal strip of height $2\pi$ onto $\mathbb{C}\setminus \{0\}$, and for all $r\gt0$, $\{w: w=\frac{1}{z},0\lt|z|\lt r\}$ contains infinitely many such strips.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2011-01-12T14:28:58.503" />
  <row Id="17231" PostTypeId="2" ParentId="17152" CreationDate="2011-01-12T14:39:06.217" Score="2" ViewCount="" Body="&lt;p&gt;If there truly was an infinite amount of time and monkey, yes, it could happen. However, we know that time and monkeys are both limited. Let's say that the universe will be gone when there are no more neutrons. According to &lt;a href=&quot;http://en.wikipedia.org/wiki/Future_of_an_expanding_universe#All_nucleons_decay&quot; rel=&quot;nofollow&quot;&gt;this article&lt;/a&gt;, that will be about 10^40 years. There are approximately &lt;a href=&quot;http://www.madsci.org/posts/archives/1998-10/905633072.As.r.html&quot; rel=&quot;nofollow&quot;&gt;4*10^78 atoms&lt;/a&gt;. And let's just say that an atom monkey can type at 10^15 keys per second. Let's also assume that there are 40 keys on a typewritter (26 A-Z, numbers, period, comma, semicolon, and space). That'll give the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There will be 4e78*365*24*3600*1e15 key strokes per atom, giving a total of 1.26e101. There would then be 5.05e177 key strokes. 40^108=1.0531e+173. That means that there would be segments of around 108 characters of Hamlet around, but certainly not the whole work.&lt;/p&gt;&#xA;" OwnerUserId="5101" LastActivityDate="2011-01-12T14:39:06.217" CommunityOwnedDate="2011-01-15T06:13:52.160" CommentCount="2" />
  <row Id="17232" PostTypeId="2" ParentId="17212" CreationDate="2011-01-12T14:52:00.933" Score="0" ViewCount="" Body="&lt;p&gt;I've wondered about this before. Call a statement whose undecidability is undecidable 2-undecidable. If such statements exist, one is naturally led to consider whether there exist statements such that the 2-undecidability of the statement is undecidable. Call this 3-undecidability. Do there exist $\omega$-undecidable statements?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could you elaborate on your answer a little Qiaochu? &lt;/p&gt;&#xA;" OwnerUserId="3162" LastActivityDate="2011-01-12T14:52:00.933" CommentCount="5" />
  <row Id="17233" PostTypeId="2" ParentId="14162" CreationDate="2011-01-12T14:55:28.137" Score="1" ViewCount="" Body="&lt;p&gt;If it wasn't for the spike at 0 and the mode at around seven, it would look roughly lognormal to me&lt;/p&gt;&#xA;" OwnerUserId="3162" LastActivityDate="2011-01-12T14:55:28.137" />
  <row Id="17234" PostTypeId="2" ParentId="17152" CreationDate="2011-01-12T15:01:20.477" Score="36" ViewCount="" Body="&lt;p&gt;Some references (I am mildly surprised that no one has done this yet).  This is called the &lt;a href=&quot;http://en.wikipedia.org/wiki/Infinite_monkey_theorem&quot; rel=&quot;nofollow&quot;&gt;infinite monkey theorem&lt;/a&gt; in the literature.  It follows from the second &lt;a href=&quot;http://en.wikipedia.org/wiki/Borel%E2%80%93Cantelli_lemma&quot; rel=&quot;nofollow&quot;&gt;Borel-Cantelli lemma&lt;/a&gt; and is related to &lt;a href=&quot;http://en.wikipedia.org/wiki/Kolmogorov%27s_zero-one_law&quot; rel=&quot;nofollow&quot;&gt;Kolmogorov's zero-one law&lt;/a&gt;, which is the result that provides the intuition behind general statements like this.  (The zero-one law tells you that the probability of getting Hamlet is either zero or one, but doesn't tell you which.  This is usually the hard part of applying the zero-one law.)  Since others have addressed the practical side, I am telling you what the mathematical idealization looks like.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;my intuition says that time is countably infinite while the number of works the monkeys could produce is uncountably infinite.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is a good idea!  Unfortunately, the number of finite strings from a finite alphabet is countable.  This is a good exercise and worth working out yourself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: also, regarding some ideas which have come up in the discussions on other answers, Jorge Luis Borges' short story &lt;a href=&quot;http://en.wikipedia.org/wiki/The_Library_of_Babel&quot; rel=&quot;nofollow&quot;&gt;The Library of Babel&lt;/a&gt; is an interesting read.&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2011-01-13T03:50:03.363" LastActivityDate="2011-01-13T03:50:03.363" CommunityOwnedDate="2011-01-15T06:13:52.160" CommentCount="8" />
  <row Id="17235" PostTypeId="2" ParentId="17157" CreationDate="2011-01-12T15:10:59.823" Score="4" ViewCount="" Body="&lt;p&gt;(1) If you ask that $E$ and $H$ both be nonsingular, the answer is &lt;strong&gt;no&lt;/strong&gt;. In $\mathbb{P}^3$, let $Q$ be a degree $2$ hypersurface with a singularity at one point $p$ (the cone on a smooth conic). Let $C$ be a generic smooth cubic; so $C$ does not pass through $p$ and is transverse to $Q$. Then $X:=C \cap Q$ is a degree $6$, genus $4$, smooth curve, embedded by its canonical section. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we want to write $X$ as $A \cap B$, for hypersurfaces $A$ and $B$, then we must have $(\deg A)(\deg B) = 6$. Since $X$ does not lie in any plane, one of $A$ and $B$ must have degree $2$. There is only one degree two hypersurface containing $X$, namely $Q$, and $Q$ is not smooth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, I can give a smaller example. In $\mathbb{P}^2$, take a smooth cubic $C$ and intersect it with $Q$, the union of two lines, where the crossing point of the lines is not in $C$. You get $6$ points, and the same logic applies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) If you work in $\mathbb{P}^n$, over a characteristic zero field, and only ask that $E$ be smooth (and not $H$), the answer is &lt;strong&gt;yes&lt;/strong&gt;.  Let $X = \{ f_1 = f_2 = \cdots = f_r =0 \}$ with $\deg f_i = d_i$, and reorder the terms such that $d_1 \geq d_2 \geq \cdots \geq d_r$. Choose $h_{ij}$ a polymonial of degree $d_i - d_j$, and otherwise generic. Define $g_i = f_i + \sum_{j&amp;gt;i} h_{ij} f_j$. Clearly, the $g_i$ and $f_i$ generate the same homogeneous ideal. We claim that, for generic values of the $h_{ij}$, every intersection $\{ g_1 \cap g_2 \cap \cdots \cap g_s =0 \}$ is smooth, for $s \leq r$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof:&lt;/strong&gt; By induction on $s$. When $s=0$, the claim is trivial. By induction, fix $h_{ij}$ for $i&amp;lt;s$ such that $\{ g_1 \cap g_2 \cap \cdots \cap g_{s-1} =0 \}$ is smooth; call this smooth variety $A$. We want to show that for generic $h_{sj}$, the intersection $A \cap \{ f_s + \sum h_{sj} f_j=0 \}$ is smooth. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We now use a variant of Corollary III.10.9 in Hartshorne. (This is where we use the characteristic zero hypothesis.) Look at the linear system on $A$ spanned by $f_s$ and by $f_j \mathcal{O}(d_s-d_j)|_A$. The variant we want is that, over a field of characteristic zero, on a smooth variety, a generic element of a finite dimensional linear system has singularities contained within the basepoints of the linear system. Hartshorne sketches the proof of this in Remark 10.9.2. (Hartshorne fails to mention that the algebraic closure hypothesis can be removed, but I think that's just an oversight on his part.) So, for generic $h_{sj}$, the singularities of $A \cap \{ f_s + \sum h_{sj} f_j=0 \}$ are contained in $X$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose for the sake of contradiction that $x \in X$ is a singular point of $A \cap \{ f_s + \sum h_{sj} f_j=0 \}$. But then $X = A \cap \{ f_s + \sum h_{sj} f_j=0 \} \cap \{ f_{s+1} = f_{s+2} = \cdots = f_r =0 \}$. Slicing a singular point by a regular sequence leaves a singular point behind, so this shows that $X$ is singular at $x$, contradicting the hypothesis that $X$ is smooth. QED&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(3) If you take $X=\mathbb{P}^2 \times \mathbb{P}^2$, the answer is &lt;strong&gt;no&lt;/strong&gt;. Take the intersection of a $(2,1)$ hypersurface and a $(1,2)$ hypersurface, each with an isolated singularity and meeting transversely. The intersection is a smooth surface $S$. The coholomogy ring of $\mathbb{P}^2 \times \mathbb{P}^2$ is $\mathbb{Z}[u,v]/\langle u^3, v^3 \rangle$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose we could write $S$ as the intersection of a hypersurface of degree $(a_1, a_2)$ and one of degree $(b_1, b_2)$. Then, in $H^2(\mathbb{P}^2 \times \mathbb{P}^2)$, we have $(a_1 u + a_2 v)(b_1 u + b_2 v) = (2 u + v)(u + 2 v)$. By unique factorization of polynomials, we must have $(a_1, a_2) = (2,1)$ and $(b_1, b_2) = (1,2)$, or &lt;i&gt;vice versa&lt;/i&gt;. (Note that I had to go up to $\mathbb{P}^2 \times \mathbb{P}^2$, so that the degree $2$ polynomials would inject in $H^*$; this argument doesn't work in $\mathbb{P}^1 \times \mathbb{P}^1$.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the only $(1,2)$ form which vanishes on $S$ is the singular $(1,2)$ form we started with, and the same for $(2,1)$ forms. So $S$ cannot be written as a complete intersection with either factor being smooth. QED&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I couldn't figure out:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(4) Characteristic $p$, in $\mathbb{P}^n$. I tried to build a counter-example based on Hartshorne Exercise III.10.7 and failed. I don't have an intuition for whether that is because I was not clever enough, or because the result is true.&lt;/p&gt;&#xA;" OwnerUserId="448" LastEditorUserId="448" LastEditorDisplayName="" LastEditDate="2011-01-12T15:40:02.700" LastActivityDate="2011-01-12T15:40:02.700" CommentCount="1" />
  <row Id="17236" PostTypeId="1" AcceptedAnswerId="17322" CreationDate="2011-01-12T15:32:40.670" Score="2" ViewCount="175" Body="&lt;p&gt;How many possible combinations are there in Hua Rong Dao?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hua Rong Dao is a Chinese sliding puzzle game, also called Daughter in a Box in Japan.  You can see a picture here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/File:Hakoiri3.jpg&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/File:Hakoiri3.jpg&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And an explanation here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Klotski#Hua_Rong_Dao&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Klotski#Hua_Rong_Dao&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The puzzle is a 4x5 grid with these pieces&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;2x2 square   (1 piece)&#xA;&lt;li&gt;1x2 vertical  (4 pieces)&#xA;&lt;li&gt;2x1 horizontal  (1 piece)&#xA;&lt;li&gt;1x1 square (4 pieces)&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Though traditionally each type of piece will have different pictures, you can treat each of the 1x2's as identical and each of the 1x1's as identical.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The goal is to slide around the pieces (not removing them) until the 2x2 &quot;general&quot; goes from the middle top to the middle bottom (where it may slide out of the border).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not concerned in this question with the solution, but more curious about the number of combinations.  Naively, I can come up with an upper bound like this&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Place each piece on the board, ignoring overlaps.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The 2x2 can go in any of 3x4 = 12 squares&lt;br&gt;&#xA;The 2x1 can go in any of 4x4 = 16 squares&lt;br&gt;&#xA;The 1x2 can go in any of 3x5 = 15 squares&lt;br&gt;&#xA;The 1x1 can go in any of 4x5 = 20 squares&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you place the pieces one at a time, subtracting out the used squares&lt;/p&gt;&#xA;&#xA;&lt;p&gt;place the 2x2 = 12 options&#xA;&lt;br&gt;place each of the 2x1 =  ((16 - 4) * (16 - 6) * (16 - 8) * (16 - 10)) / 4!&#xA;&lt;br&gt;place the 1x2 = 15 - 6&#xA;&lt;br&gt;place the 1x1 = (20-14) choose 4 = 15&lt;/p&gt;&#xA;&#xA;&lt;p&gt;multiplied together this works out to 388,800.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any way I might be able to narrow this down further?  The two obvious things not taken into account are blocked pieces (a 2x1 pieces will not fit into two separated squares) and the fact that not all possibilities might be accessible when sliding from the starting position.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Update: I realized that the puzzle is bilaterally symmetrical, so if you just care about meaningful differences between positions, you can divide by two.&lt;/p&gt;&#xA;" OwnerUserId="5697" LastEditorUserId="5697" LastEditorDisplayName="" LastEditDate="2011-01-12T23:02:26.780" LastActivityDate="2011-01-13T03:20:04.300" Title="How many possible combinations are there in Hua Rong Dao?" Tags="&lt;puzzle&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="17237" PostTypeId="2" ParentId="17194" CreationDate="2011-01-12T15:37:27.827" Score="0" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\ $ A rational function is asymptotic to $\rm\ x^n,\ $ for some $\rm\ n \in \mathbb Z\:.\ $ But $\rm\ x \log(x)\ $ lies strictly between $\rm\ x\ $ and $\rm x^2$ asymptotically, i.e. $\rm\ (x\ log(x))/x = log(x)\to\infty,\ \ x^2/(x log(x)) = x/log(x)\to \infty$&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2011-01-12T15:37:27.827" />
  <row Id="17238" PostTypeId="2" ParentId="17229" CreationDate="2011-01-12T15:38:52.333" Score="2" ViewCount="" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;A problem arises for me as I've come to believe that rational numbers are defined as a set of equivalence classes, but wikipedia says that equivalence classes are ruled out of axiomatic set theory so what replaces Q when dealing with ZFC? When you remove equivalence classes from the picture what defines the rational numbers in ZFC?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If ZFC weren't capable of talking about equivalence classes it would be a pretty terrible theory.  What you probably read is something like the fact that you can't define cardinals as equivalence classes of sets under bijection, and you can't define ordinals as equivalence classes of well-ordered sets under order-preserving bijection.  In both cases the problem is that the thing you're trying to take equivalence classes with respect to is too big to be a set, so you can't apply the usual operations of set theory to it.  But equivalence classes on &lt;em&gt;sets&lt;/em&gt; are perfectly well-defined in the usual way. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;(Page 77, notice at the top of the page they say that ∈ is a &quot;well-founded relation&quot;. My knowledge of Naive set theory tells me that ∈ is one of two undefinable constructs that you're supposed to take as given (I could quote the sources of this if needed), but here they call it a relation. Surely they don't mean relation in the subset of the cartesian product way do they?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The word &quot;relation&quot; is being used externally here rather than internally.  Given a first-order theory $T$ (such as ZFC), models of $T$ are sets equipped with certain functions and relations specified by the theory $T$ satisfying the axioms of the theory, and in ZFC there are no functions and the only relation is $\in$.  So this is not naive set theory; this is just set theory.  In set theory (as opposed to naive set theory) one must carefully distinguish between &quot;internal&quot; and &quot;external&quot; notions, and the word &quot;relation&quot; here is being used externally.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2011-01-12T15:38:52.333" CommentCount="1" />
  <row Id="17240" PostTypeId="2" ParentId="16909" CreationDate="2011-01-12T16:00:39.357" Score="0" ViewCount="" Body="&lt;p&gt;Let's prove that $A = \mathbb{C} - \{0\}$. First, we'll prove that $A \subset \mathbb{C} - \{0\}$, then we'll show that $\mathbb{C} - \{0\} \subset A$.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Let's show that $A \subset \mathbb{C} - \{0\}$. This amounts to showing that $0$ does not belong to A. Given any z such that $0 &amp;lt; |z| &amp;lt; |r|$ then we can write: $1/z = a + ib$ (with $a, b \in \mathbb{R}$). Then $|z| = exp(a) &amp;gt; 0$. Thus, $0$ is not in A.&lt;/li&gt;&#xA;&lt;li&gt;Now, let's show that any non-zero complex number $x$ can be written as $x = \exp(1/z)$ with $0 &amp;lt; |z| &amp;lt; |r|$. This will demonstrate that $A \subset \mathbb{C} - \{0\}$.  For that, we pick $x \neq 0$. We write $x$ in exponential form ($\rho, \theta \in \mathbb{R}$): &#xA;\begin{eqnarray}&#xA;x &amp;amp; = &amp;amp; \rho e^{i\theta}\&#xA;&amp;amp; = &amp;amp; \exp(ln(\rho) + i\theta)&#xA;\end{eqnarray}&#xA;Let's pick $k \in \mathbb{N}$ such that $|ln(\rho) + i\theta + 2k\pi| &amp;gt; 1/|r|$. Then $z = ln(\rho) + i\theta + 2k\pi$ verifies $0 &amp;lt; |z| &amp;lt; |r|$ and we have $x = exp(1/z)$. Thus $x$ belongs to $A$.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;We have shown that A both contains and is a subset of $\mathbb{C} - \{0\}$. Thus $A = \mathbb{C} - \{0\}$&lt;/p&gt;&#xA;" OwnerUserId="5701" LastActivityDate="2011-01-12T16:00:39.357" />
  <row Id="17241" PostTypeId="1" CreationDate="2011-01-12T16:02:51.803" Score="2" ViewCount="145" Body="&lt;p&gt;We have a set of numbers, of size $m$. We are going to pick $a$ numbers with uniform probability from that set, with replacement. Let X be the random variable denoting the probability of having X of those picks distinct (exactly X distinct values are picked).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Motivation: I need to calculate this probability in order to calculate a more advanced distribution regarding Bloom filters, in particular the distribution of the number of bits set to 1 in a Bloom filter. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Letting that aside, I am having trouble formulating the the PMF for X. I've tried to look out for multi-variate binomial distribution but I couldn't relate it to what I want to do.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is whether there is such a probability distribution in the literature, and if now, how can I approach this problem ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Update:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have managed to make a formulation: the probability we pick $x$ distinct values is &#xA;$$&#xD;&#xA;\frac{1}{m} \frac{1}{m-1} \cdots \frac{1}{m-x+1}&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And the probability of picking the rest of our $a-x$ picks in that set of $x$ values is&#xA;$$&#xD;&#xA;\left(\frac{x}{m}\right)^{a-x}&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, the number of such configurations is $\binom{m}{x}$. Multiplying all that together and simplifying gives us a PMF&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;P(X=x;a,m) = \frac{ \left( \frac{m}{x} \right) ^{x-a}}{x!}&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does that seem to make any sense ?&lt;/p&gt;&#xA;" OwnerUserId="558" LastEditorUserId="558" LastEditorDisplayName="" LastEditDate="2011-01-12T16:44:13.420" LastActivityDate="2011-01-12T17:47:26.727" Title="Does this probability distribution have a name ?" Tags="&lt;probability&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="17242" PostTypeId="1" AcceptedAnswerId="17244" CreationDate="2011-01-12T16:45:03.093" Score="1" ViewCount="115" Body="&lt;p&gt;Hello everyone,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I encountered a programming problem that need a little math. Since I took number-theory long time ago, from the top of my head, I could not think of a way to approach this problem. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Give a base 'b' ( 2, 3, 4, 5, 6 ... -&gt; 26 ). So if I'm at base 10, I will have 10 digits: 0, 1, 2, 3...9. The digit of a base 'b' is called interesting if it satisfy this condition:&#xA;3 is an interesting digit because 118*3 = 354 and 3+5+4 = 12. Which means both 354 and 12 divides 3. By look at the result sets, I can see the relation is:&lt;/strong&gt;&lt;br&gt;&#xA;&lt;em&gt;if b mod d = 1 then d is an interesting digit.&lt;/em&gt;&lt;br&gt;&#xA;For example: 10 mod 9 = 10 mod 3 = 1. So both 3, 9 are interesting digits in base 10.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How could I prove this? Any hint?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks,&lt;br&gt;&#xA;Chan&lt;/p&gt;&#xA;" OwnerUserId="5711" LastEditorUserId="5711" LastEditorDisplayName="" LastEditDate="2011-01-26T15:36:39.977" LastActivityDate="2011-01-26T15:36:39.977" Title="How does base of a number relate to modulos of its each individual digit?" Tags="&lt;number-theory&gt;&lt;elementary-number-theory&gt;" AnswerCount="3" />
  <row Id="17243" PostTypeId="2" ParentId="17241" CreationDate="2011-01-12T16:55:30.193" Score="1" ViewCount="" Body="&lt;p&gt;Look up multiset coefficients in &lt;a href=&quot;http://en.wikipedia.org/wiki/Multiset&quot; rel=&quot;nofollow&quot;&gt;Wikipedia's Multiset&lt;/a&gt;.  Your calculation would actually give $\frac{m!}{x!}\left(\frac{x}{m}\right)^{(a-x)}\binom {m}{x}$ the way you are arguing but the approach is not correct and this will not sum to 1.  You are not counting the correct number of ways to get x different items.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-01-12T16:55:30.193" CommentCount="2" />
  <row Id="17244" PostTypeId="2" ParentId="17242" CreationDate="2011-01-12T16:55:50.350" Score="4" ViewCount="" Body="&lt;p&gt;You seem to be asking for what digits $\displaystyle 0 \lt d \lt b$ can we apply the following divisibility test:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A number is in base $\displaystyle b$ is divisible by $\displaystyle d$ if the sum of digits of the number is divisible by $\displaystyle d$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is certainly true for $\displaystyle b = 1 \mod d$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because $\displaystyle \sum_{k=0}^{n} a_k b^{k} - \sum_{k=0}^{n} a_k = \sum_{k=0}^{n} a_k (b^k -1) = 0 \mod d$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus $\displaystyle \sum_{k=0}^{n} a_k b^{k} = \sum_{k=0}^{n} a_k  \mod d$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $\displaystyle b = r \mod d$ where $\displaystyle r \neq 1$, then we have that there is some $\displaystyle k$ for which $r^k - 1 \neq 0 \mod d$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The number with $\displaystyle 1$ as the digit correponding to $\displaystyle b^k$, and $\displaystyle d-1$ in as the &quot;units&quot; digit (corresponding to $\displaystyle b^0$) is a counterexample to the divisibility rule for $\displaystyle d$.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2011-01-12T17:13:32.790" LastActivityDate="2011-01-12T17:13:32.790" CommentCount="2" />
  <row Id="17245" PostTypeId="2" ParentId="16071" CreationDate="2011-01-12T17:05:09.823" Score="1" ViewCount="" Body="&lt;p&gt;This is at least the fifth time I have posted this on this site.  But it is free, so here it comes again:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have some relatively brief notes on introductory set theory available on my webpage.  Please go to &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.math.uga.edu/~pete/expositions.html&quot; rel=&quot;nofollow&quot;&gt;http://www.math.uga.edu/~pete/expositions.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and scroll most of the way down to the bottom.  Or more directly go this page:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://math.uga.edu/~pete/settheorypart1.pdf&quot; rel=&quot;nofollow&quot;&gt;http://math.uga.edu/~pete/settheorypart1.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you get to the end of these notes and are still curious, try replacing the &quot;1&quot; by a &quot;2&quot;, and so forth.  But in particular the first set of notes is intended to be used by undergraduates with absolutely no prior background in set theory.  &lt;/p&gt;&#xA;" OwnerUserId="299" LastActivityDate="2011-01-12T17:05:09.823" />
  <row Id="17246" PostTypeId="1" AcceptedAnswerId="17250" CreationDate="2011-01-12T17:14:05.520" Score="2" ViewCount="171" Body="&lt;p&gt;Assuming I have the graph of a function $f(x)$ is there function $f_1(f(x))$ that will give me a rotated version of the graph of that function?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example if I plot $\sin(x)$ I will get a sine wave which straddles the $x$-axis, can I apply a function to $\sin(x)$ to yield a wave that straddles the line that would result from $y = 2x$?&lt;/p&gt;&#xA;" OwnerUserId="5718" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2011-01-13T03:37:26.920" LastActivityDate="2011-01-13T03:37:26.920" Title="Is there a way to rotate the graph of a function?" Tags="&lt;geometry&gt;&lt;graph&gt;" AnswerCount="3" />
  <row Id="17247" PostTypeId="1" CreationDate="2011-01-12T17:15:30.863" Score="2" ViewCount="185" Body="&lt;p&gt;If you divide an number by another numbers reciprocal&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i.e &lt;/p&gt;&#xA;&#xA;&lt;p&gt;x / (1 / y)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can this be expressed in another way&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is dividing by the reciprocal equivelant to another operation?&lt;/p&gt;&#xA;" OwnerUserId="5720" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2011-01-13T03:35:29.870" LastActivityDate="2011-01-13T03:35:29.870" Title="can x / (1/y) be expressed more simply" Tags="&lt;algebra-precalculus&gt;" AnswerCount="3" />
  <row Id="17249" PostTypeId="2" ParentId="17247" CreationDate="2011-01-12T17:18:24.097" Score="9" ViewCount="" Body="&lt;p&gt;Yes. $\frac{x}{\frac{1}{y}} = xy$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a simple consequence of two things: $x = \frac{x}{1}$, and&#xA;$$\frac{\quad\frac{a}{b}\quad}{\frac{x}{y}} = \frac {ay}{bx}.$$&#xA;The latter is just the usual rule for &lt;a href=&quot;http://en.wikipedia.org/wiki/Fraction_%28mathematics%29#Division&quot; rel=&quot;nofollow&quot;&gt;dividing fractions&lt;/a&gt;, written in fraction form.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2011-01-12T17:18:24.097" />
  <row Id="17250" PostTypeId="2" ParentId="17246" CreationDate="2011-01-12T17:24:00.877" Score="2" ViewCount="" Body="&lt;p&gt;Once you rotate, it need not remain a function (i.e. one $x$ value can have multiple $y$ values corresponding to it).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But you can use the following transformation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x&amp;#39; = x\cos \theta - y \sin \theta$$&#xA;$$y&amp;#39; = x \sin \theta + y \cos \theta$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;to rotate by an angle of $\theta$. Point $(x,y)$ gets rotated to point $(x&amp;#39;,y&amp;#39;)$. Note: this is a rotation about the origin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your case of $y = 2x$, you need to rotate by $\arctan(2)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See this for more info: &lt;a href=&quot;http://en.wikipedia.org/wiki/Rotation_matrix&quot; rel=&quot;nofollow&quot;&gt;Rotation Matrix&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2011-01-12T17:24:00.877" CommentCount="3" />
  <row Id="17251" PostTypeId="2" ParentId="17246" CreationDate="2011-01-12T17:26:34.117" Score="1" ViewCount="" Body="&lt;p&gt;You can do the rotation as Moron says, or you can write $y=2x+\sin(x)$.  This will remain a function, but doesn't have the same shape as a sine wave.  It depends upon what you want.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-01-12T17:26:34.117" CommentCount="1" />
  <row Id="17252" PostTypeId="2" ParentId="17246" CreationDate="2011-01-12T17:27:15.100" Score="2" ViewCount="" Body="&lt;p&gt;In general, the answer is &lt;strong&gt;no&lt;/strong&gt; since the rotated version of the graph might not be the graph of a function. For instance it could happen that your rotated version of the graph contains two different points with the same $x$-value -- this cannot happen for the graph of a function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A way out could be to parametrise your graph. So instead of a map $x\mapsto y(x)$ you look at the map $t\mapsto (t,y(t))$. After rotating the trajectory of this thing (not the graph!) it will still be the trajectory of a map $$t\mapsto (x(t),y(t)).$$&lt;/p&gt;&#xA;" OwnerUserId="367" LastActivityDate="2011-01-12T17:27:15.100" />
  <row Id="17253" PostTypeId="2" ParentId="17242" CreationDate="2011-01-12T17:31:32.473" Score="1" ViewCount="" Body="&lt;p&gt;If $b \ne 1 \pmod d$, then the number $10$ in base $b$ shows that $d$ is not interesting: it represents the number $b$, so it is not equal to $1 \pmod b$; but $1 + 0 = 1$. This, together with Moron's answer, shows that your conjecture is correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt; Taking the question literally, I really need to exhibit a number that is divisible by $d$ but the sum of whose digits is not; or &lt;em&gt;vice versa&lt;/em&gt;. If $d &amp;lt; b$, take the number $1.(d-1)$; the sum of its digits is $d$, but it represents $b+d-1$, which is not divisible by $d$. If $d \ge b$, then the representation of $d$ in base $b$ is two or more digits, so their sum must be less than what they represent (because all but the last digit represent numbers greater than themselves).&lt;/p&gt;&#xA;" OwnerUserId="1508" LastEditorUserId="1508" LastEditorDisplayName="" LastEditDate="2011-01-12T17:45:28.087" LastActivityDate="2011-01-12T17:45:28.087" CommentCount="2" />
  <row Id="17254" PostTypeId="2" ParentId="17229" CreationDate="2011-01-12T17:47:14.637" Score="4" ViewCount="" Body="&lt;p&gt;Your definition of function is incorrect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The idea of a &quot;function-as-an-ordered-triple&quot; is the following: to describe a function, we must describe three things: the domain, the codomain, and the rule by which each element of the domain is assigned an element of the codomain. In this setting, we say two functions are equal if and only if they have the same domain, the same codomain, and the same value at every point in the domain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Note that we don't always abide by these conventions; for example, we often think of the functions $f\colon\mathbb{R}\to\mathbb{R}$ and $g\colon\mathbb{R}\to[0,\infty)$ given by $f(x)=g(x)=x^2$ as &quot;the same&quot;, even though the have different codomain; but in order to be able to talk about &quot;surjective functions&quot;, the codomain is important).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So a function requires you to specify three things: the domain, the codomain, and the rule. The rule is a special kind of relation between elements of the domain and elements of the codomain, so it is a subset of the cartesian product of the domain and the codomain. If the domain is $A$ and the codomain is $B$, then the function will be a subset $F\subseteq A\times B$. Not every relation will do, however. What are the two properties that distinguish a function from an arbitrary relation? They are:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;strong&gt;Every&lt;/strong&gt; element of the domain $A$ must have an image in $B$.&lt;/li&gt;&#xA;&lt;li&gt;Each element of the domain $A$ has &lt;strong&gt;only one&lt;/strong&gt; image in $B$.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Translating this into logical formulas, we have:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$\forall(a\in A)\exists(b\in B)\Bigl( (a,b)\in F\Bigr)$.&lt;/li&gt;&#xA;&lt;li&gt;$\forall(a\in A)\forall(b,b&amp;#39;\in B)\Bigl( \bigl((a,b)\in F\wedge (a,b&amp;#39;)\in F\bigr)\rightarrow (b=b&amp;#39;)\Bigr)$. &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;When we write $F\colon A\to B$, we mean that we are looking at a triple of the form $(A,B,F)$ with $F\subseteq A\times B$ and which satisfies 1 and 2 above. When we write $F(a)=b$, we mean that $(a,b)\in F$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first condition says that every element of $A$ has at least one image; the second says that every element of $A$ has at most one image.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given this, we can define &quot;function from $A$ to $B$&quot; as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Definition.&lt;/strong&gt; Let $A$ and $B$ be sets. A &lt;em&gt;function $F$ from $A$ to $B$&lt;/em&gt; is an ordered triple $(A,B,F)$, with $F\subseteq A\times B$ such that&#xA;$$\forall(a\in A)\exists(b\in B)\Bigl((a,b)\in F\Bigr)\wedge \forall(a\in A)\forall(b,b&amp;#39;\in B)\Bigl( \bigl((a,b)\in F\wedge (a,b&amp;#39;)\in F\bigr)\rightarrow (b=b&amp;#39;)\Bigr).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Under this definition, two functions $(A,B,F)$ and $(X,Y,G)$ are equal if and only if $A=X$ (same domain), $B=Y$ (same codomain), and $F=G$ (same pairs; hence, same value at every element of the domain). &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-12T17:59:05.573" LastActivityDate="2011-01-12T17:59:05.573" CommentCount="4" />
  <row Id="17255" PostTypeId="2" ParentId="17241" CreationDate="2011-01-12T17:47:26.727" Score="1" ViewCount="" Body="&lt;p&gt;The number of multisets from a set of size $m$ with cardinality $a$ is $\binom{m+a-1}{a}$.&lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2011-01-12T17:47:26.727" CommentCount="3" />
  <row Id="17256" PostTypeId="2" ParentId="17089" CreationDate="2011-01-12T17:52:17.727" Score="2" ViewCount="" Body="&lt;p&gt;You can add the following to your list of representations &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Continued fractions. See Jones&amp;amp;Thron or &lt;a href=&quot;http://www.math.ntnu.no/~lisa/publist.html&quot; rel=&quot;nofollow&quot;&gt;Lorentzen&lt;/a&gt;'s books&lt;/li&gt;&#xA;&lt;li&gt;Integral representations (Mellin-Barnes, etc). See the &lt;a href=&quot;http://eom.springer.de/I/i051620.htm&quot; rel=&quot;nofollow&quot;&gt;ECM entry&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;Exponentials.  See Knoebel's paper &quot;Exponentials reiterated&quot; &lt;a href=&quot;http://www.ams.org/mathscinet-getitem?mr=610484&quot; rel=&quot;nofollow&quot;&gt;MR0610484&lt;/a&gt; &lt;/li&gt;&#xA;&lt;li&gt;Nested radicals. See &lt;a href=&quot;http://www.ams.org/mathscinet-getitem?mr=151586&quot; rel=&quot;nofollow&quot;&gt;MR0151586&lt;/a&gt; Schuske's Infinite radicals in the complex plane&lt;/li&gt;&#xA;&lt;li&gt;Rational or polynomial approximation (e.g Pade approximants)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I guess the general topic is studied under Complex Analysis, Asymptotic Analysis, Harmonic Analysis.  AFAIK, there is no single book which covers all representations.&lt;/p&gt;&#xA;" OwnerUserId="1558" LastActivityDate="2011-01-12T17:52:17.727" />
  <row Id="17257" PostTypeId="1" AcceptedAnswerId="17261" CreationDate="2011-01-12T17:59:38.693" Score="0" ViewCount="75" Body="&lt;p&gt;Is the subspace of $\mathbb C^2$ of those vectors with real coordinates a complex submanifold? &lt;/p&gt;&#xA;" OwnerUserId="5727" LastEditorUserId="274" LastEditorDisplayName="" LastEditDate="2011-01-12T19:07:43.250" LastActivityDate="2011-01-12T19:07:43.250" Title="question about a complex manifold" Tags="&lt;complex-analysis&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="17258" PostTypeId="2" ParentId="17229" CreationDate="2011-01-12T18:00:32.080" Score="1" ViewCount="" Body="&lt;p&gt;As regards your second question about the definition of a function: Not only is it true that for every a in A, there exists b in B s.t. (a,b) in F, it is more strongly the case that there exists a /unique/ b in B satisfying that constraint (hopefully this will work, I'm still learning the markup language here):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;∀(a∈A)∃(b∈B) ((a,b)∈F ∧ ∀(x∈B)( (a,x)∈F ⇒ x=b)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This allows us to then use f(a) as well-defined notation for a unique b in B for each a in A.&lt;/p&gt;&#xA;" OwnerUserId="4980" LastActivityDate="2011-01-12T18:00:32.080" />
  <row Id="17259" PostTypeId="2" ParentId="16071" CreationDate="2011-01-12T18:08:10.137" Score="0" ViewCount="" Body="&lt;p&gt;There is Set Theory, by K. Kuratowski and A. Mostowski.&#xA;It's dry as dirt and yet for some reason I loved reading that book.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it's available in your library, I'd take it home and see if it works out.&lt;/p&gt;&#xA;" OwnerUserId="5189" LastActivityDate="2011-01-12T18:08:10.137" />
  <row Id="17261" PostTypeId="2" ParentId="17257" CreationDate="2011-01-12T18:19:44.800" Score="3" ViewCount="" Body="&lt;p&gt;Call that subspace $X$. Then the composition of the inclusion $X\to \mathbb C^2$ with the first projection $\mathbb C^2\to\mathbb C$ is an holomorphic map defined on a $1$-dimensional complex manifold which takes only real values. Such a function is necessarily constant. Now do the same for the second projection. What can you conclude?&lt;/p&gt;&#xA;" OwnerUserId="274" LastEditorUserId="274" LastEditorDisplayName="" LastEditDate="2011-01-12T18:29:34.840" LastActivityDate="2011-01-12T18:29:34.840" CommentCount="2" />
  <row Id="17262" PostTypeId="1" CreationDate="2011-01-12T18:25:55.447" Score="4" ViewCount="220" Body="&lt;p&gt;I vaguely remember the Frobenius matrix norm &#xA;( ${||A||}_F = \sqrt{\sum_{i,j} a_{i,j}^2}$ ) was somehow considered unsuitable for numerical analysis applications. I only remember, however, that it was not a &lt;em&gt;subordinate matrix norm&lt;/em&gt;, but only because it did not take the identity matrix to $1$. It seems this latter problem could be solved with a rescaling, though. I don't remember my numerical analysis text considering this norm any further after introducing this fact, which seemed to be its death-knell for some reason.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question, then: for fixed $n$, when looking at $n \times n$ matrices, are there any weird gotchas, deficiencies, oddities, &lt;em&gt;etc&lt;/em&gt;, when using the (possibly rescaled) Frobenius norm? For example, is there some weird series of matrices $A_i$ such that the Frobenius norm of the $A_i$ approaches zero while the $\ell_2$-subordinate norm does not converge to zero? (It seems like that can &lt;em&gt;not&lt;/em&gt; happen because the $\ell_2$ norm is the square root of the largest eigenvalue of $A^*A$, and thus bounded from above by the Frobenius norm...) &lt;/p&gt;&#xA;" OwnerUserId="610" LastActivityDate="2011-01-12T21:47:57.043" Title="Why should I avoid the Frobenius Norm?" Tags="&lt;matrices&gt;&lt;numerical-methods&gt;&lt;counter-examples&gt;&lt;norm&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="17263" PostTypeId="2" ParentId="17262" CreationDate="2011-01-12T18:33:02.973" Score="0" ViewCount="" Body="&lt;p&gt;See &lt;a href=&quot;http://www.ripublication.com/aama/aamav2n1_6.pdf&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt;. According to the article, if a set of data contains outliers, then an $L_1$ norm might be better to use.&lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2011-01-12T18:33:02.973" />
  <row Id="17264" PostTypeId="2" ParentId="16254" CreationDate="2011-01-12T18:45:32.240" Score="2" ViewCount="" Body="&lt;p&gt;Here are some summer research opportunities in the US that fund non-US citizens. (I realize that you were asking for programs outside the US, but you framed your question in terms of being disappointed about having to fund it yourself so I thought this list might be helpful) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I'm new, so the system won't let me post more than one hyperlink)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Research in Industrial Projects for Students (RIPS) 2011 at UCLA (Also has a program in HongKong)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://math.williams.edu/small/application-information/&quot; rel=&quot;nofollow&quot;&gt;http://math.williams.edu/small/application-information/&lt;/a&gt;  (their website says they have funded non-US students in the past)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;NIMBios 2011 Summer Program&#xA;National Institute for Mathematical and Biological Synthesis (NIMBioS)&#xA;at the University of Tennessee, Knoxville&lt;/p&gt;&#xA;&#xA;&lt;p&gt;DIMACS REU&#xA;DIMACS/DIMATIA REU&#xA;CCICADA REU&#xA;Rutgers Math Department REU (foreign students enrolled at a U.S. university are eligible under the CCICADA program)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mathematical and Theoretical Biology Institute MTBI Summer Research Program&lt;/p&gt;&#xA;" OwnerUserId="5727" LastActivityDate="2011-01-12T18:45:32.240" CommunityOwnedDate="2011-01-12T18:45:32.240" CommentCount="3" />
  <row Id="17265" PostTypeId="1" AcceptedAnswerId="17271" CreationDate="2011-01-12T18:53:00.077" Score="0" ViewCount="96" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;A common characterization of&#xA;  M-matrices are non-singular square&#xA;  matrices with non-positive&#xA;  off-diagonal entries, positive&#xA;  diagonal entries, non-negative row&#xA;  sums, and at least one positive row&#xA;  sum&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It seems that this characterization depends on the use of rows or columns. I don't understand how that is possible, as the transposed of an M-Matrix is also an M-Matrix. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example: &#xA;$\left( \begin{array}{ccc}&#xD;&#xA;2 &amp;amp; -1 &amp;amp;0\\&#xD;&#xA;-2 &amp;amp; 2&amp;amp;-1\\&#xD;&#xA;0&amp;amp;-1&amp;amp;2 \end{array} \right)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This matrix is not diagonally dominant, but its transpose is. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However the matrix is clearly M-Matrix, because all Eigenvalues are positive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for your help!&lt;/p&gt;&#xA;" OwnerUserId="4514" LastActivityDate="2011-01-14T16:26:15.297" Title="M-Matrix characterization for the transpose" Tags="&lt;linear-algebra&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="17266" PostTypeId="2" ParentId="17212" CreationDate="2011-01-12T19:14:04.773" Score="8" ViewCount="" Body="&lt;p&gt;At least one version of this question has a nearly-trivial answer; if you want to know if there's some statement G such that ZFC (or your PA-compatible logic system of choice) doesn't prove G, ZFC doesn't prove ~G, and we can't prove that ZFC doesn't prove G or not G, then any undecideable statement works!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Specifically, it's impossible for ZFC to prove that it can't prove something, because such a statement is tantamount to the consistency of ZFC; if ZFC were inconsistent then it could prove everything, so proving that there's something that can't be proved is equivalent to proving the consistency of ZFC, and of course this is forbidden by the second incompleteness theorem.&lt;/p&gt;&#xA;" OwnerUserId="785" LastEditorUserId="785" LastEditorDisplayName="" LastEditDate="2011-01-12T19:19:22.077" LastActivityDate="2011-01-12T19:19:22.077" CommentCount="1" />
  <row Id="17267" PostTypeId="1" CreationDate="2011-01-12T19:38:38.603" Score="2" ViewCount="114" Body="&lt;p&gt;In this &lt;a href=&quot;http://math.stackexchange.com/questions/15966/ramanujans-first-letter-to-hardy-and-the-number-of-3-smooth-integers&quot;&gt;question&lt;/a&gt; on Math.SE, I asked about Ramanujan's (ridiculously close) approximation for counting the number of 3-smooth integers less than or equal to a given positive integer $N$, namely, &#xA;\begin{eqnarray}&#xA;\frac{\log 2 N \ \log 3 N}{2 \log 2 \ \log 3}.&#xA;\end{eqnarray}&#xA;In his posthumously published &lt;a href=&quot;http://books.google.com/books?id=tX5UUCempn8C&amp;amp;pg=PA66#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;notebook&lt;/a&gt;, Ramanujan generalized this formula to similarly counting numbers of the form $b_1^{r_1} b_{2}^{r_{2}}$, where $b_1, b_2 &amp;gt; 1$ are (not necessarily distinct) natural numbers and $r_1, r_2 \geq 0$:&#xA;\begin{eqnarray}&#xA;\frac{\log b_1 N \ \log b_2 N}{2 \log b_1 \ \log b_2},&#xA;\end{eqnarray}&#xA;where a factor of $\frac{1}{2}$ is to be added if $N$ is of the prescribed form. Both of these problems could be understood by counting the number of non-negative integer solutions of the Diophantine inequality: $(\log b_1) x_1 + (\log b_2) x_2 \leq \log N$, which also counts the number of $\mathbb{Z}$-lattice points in the $\log N$ dilate of the $2$-polytope $\mathcal{P} = \textbf{conv}(\mathbf{0}, (\log b_{1})^{-1} \mathbf{e}_1, (\log b_2)^{-1} \mathbf{e}_{2})$. Here, $\{ \mathbf{e}_{i}\}$ is the standard basis of $\mathbb{R}^{n}$. Unfortunately, counting lattice points in &lt;em&gt;real&lt;/em&gt; polytopes is &lt;em&gt;hard&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question, now, is how important a contribution to mathematics would it be if one had an &lt;em&gt;exact formula&lt;/em&gt; for counting $y$-smooth integers less than or equal to a real $x &amp;gt; 0$, i.e., an exact formula for the &lt;a href=&quot;http://en.wikipedia.org/wiki/Dickman_function&quot; rel=&quot;nofollow&quot;&gt;de Bruijn function&lt;/a&gt; $\Psi(x,y)$? This would, at the same time, translate into an exact formula for counting lattice points of real polytopes of the form $\textbf{conv}(\mathbf{0}, a_{1} \mathbf{e}_{1}, \dots, a_{n} \mathbf{e}_{n})$ with $a_{i} \in \mathbb{R}_{&amp;gt; 0}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm aware of the review articles by &lt;a href=&quot;http://websites.math.leidenuniv.nl/algebra/sieving.pdf&quot; rel=&quot;nofollow&quot;&gt;Pomerance&lt;/a&gt;, &lt;a href=&quot;http://library.msri.org/books/Book44/files/09andrew.pdf&quot; rel=&quot;nofollow&quot;&gt;Granville&lt;/a&gt;, &lt;a href=&quot;http://archive.numdam.org/ARCHIVE/JTNB/JTNB_1993__5_2/JTNB_1993__5_2_411_0/JTNB_1993__5_2_411_0.pdf&quot; rel=&quot;nofollow&quot;&gt;Hildebrand and Tenenbaum&lt;/a&gt;, which each deal with various estimates of the De Bruijn function and its useful in the Quadratic Sieve Method of factorization and applications to cryptography, and even Waring's Problem. However, none of these review articles deal with the consequences of having an exact formula.&lt;/p&gt;&#xA;" OwnerUserId="2720" LastEditorUserId="2720" LastEditorDisplayName="" LastEditDate="2011-01-12T22:31:00.250" LastActivityDate="2011-01-12T22:31:00.250" Title="&quot;On the consequences of an exact de Bruijn Function&quot;, or &quot;If Ramanujan had more time...&quot;" Tags="&lt;number-theory&gt;&lt;real-analysis&gt;&lt;combinatorics&gt;&lt;reference-request&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="17268" PostTypeId="2" ParentId="17265" CreationDate="2011-01-12T19:40:53.083" Score="1" ViewCount="" Body="&lt;p&gt;See &lt;a href=&quot;http://www.sciencedirect.com/science?_ob=ArticleURL&amp;amp;_udi=B6V0R-3Y3YMRY-N&amp;amp;_user=10&amp;amp;_coverDate=03/01/1996&amp;amp;_rdoc=1&amp;amp;_fmt=high&amp;amp;_orig=search&amp;amp;_origin=search&amp;amp;_sort=d&amp;amp;_docanchor=&amp;amp;view=c&amp;amp;_searchStrId=1605089600&amp;amp;_rerunOrigin=google&amp;amp;_acct=C000050221&amp;amp;_version=1&amp;amp;_urlVersion=0&amp;amp;_userid=10&amp;amp;md5=1e4f76614ba7578c4b3ba7803022fc47&amp;amp;searchtype=a&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; and &lt;a href=&quot;http://www.sciencedirect.com/science?_ob=ArticleURL&amp;amp;_udi=B6TYJ-43XG2D6-9&amp;amp;_user=10&amp;amp;_coverDate=10/31/2001&amp;amp;_rdoc=1&amp;amp;_fmt=high&amp;amp;_orig=search&amp;amp;_origin=search&amp;amp;_sort=d&amp;amp;_docanchor=&amp;amp;view=c&amp;amp;_searchStrId=1605091630&amp;amp;_rerunOrigin=google&amp;amp;_acct=C000050221&amp;amp;_version=1&amp;amp;_urlVersion=0&amp;amp;_userid=10&amp;amp;md5=b1902a46c5c6eb8b61fdb2005e477f65&amp;amp;searchtype=a&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt;. I think your characterization of M-matrices is a sufficient one but not necessary.&lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2011-01-12T19:40:53.083" CommentCount="3" />
  <row Id="17269" PostTypeId="2" ParentId="17247" CreationDate="2011-01-12T19:44:04.607" Score="1" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\ $ &quot;Integralize&quot; the denominator, i.e. multiply by $\rm\ y\ $ the denominator (and numerator).&lt;br&gt;&#xA;This is analogous to &quot;rationalizing&quot; a denominator $\rm\:b\:$ by multiplying by its conjugate $\rm\:b&amp;#39;\:$, i.e. $$\rm\displaystyle \frac{a}{b} = \frac{a\:b&amp;#39;}n,\ \ n = b\:b&amp;#39;\in\mathbb Q\:.$$&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2011-01-12T19:44:04.607" CommentCount="0" />
  <row Id="17270" PostTypeId="2" ParentId="17212" CreationDate="2011-01-12T19:50:38.437" Score="5" ViewCount="" Body="&lt;p&gt;(Zhen: Independence is with respect to ZFC.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given any sentence S, either (1) ZFC proves S, in which case it is a theorem of ZFC that ZFC proves S, or (2) ZFC does not prove S. In this case, ZFC is consistent, and ZFC does not know that it does not prove S (or else it would know that it is consistent, and therefore it wouldn't be, but then it would prove S). So, in this case ZFC does not prove that it does not prove S, and does not prove that it proves S. It follows easily that &quot;ZFC proves S&quot; is independent of ZFC, for any S for which ZFC does not prove S. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To ask whether the statement &quot;ZFC proves S&quot; or a variant of this such as (+): &quot;ZFC does not prove S and does not prove not-S&quot; is decidable, on the other hand, is silly, unless you are using the term in a strange fashion. For any S there is a Turing machine M with no input that outputs the truth value of the statement (+). You do not ask for the decidability of a single statement, but of a family of statements, say with S as a parameter. You probably need to clarify what you mean. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only sensible way of understanding what you are asking is to take the set whose sole element is a sentence formalizing the statement (+), and asking whether that set is decidable, but of course it is as &lt;em&gt;any finite set (of natural numbers)&lt;/em&gt; is trivially decidable. Perhaps more interesting is whether, calling the (formalization of the) statement in quotes $(+)_S$, the set $A=${$S\mid(+)_S$} is decidable. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now: Suppose first that ZFC is inconsistent. Then ZFC proves anything, so all the statements $(+)_S$ are false. Hence, the set $A$ is obviously decidable. Suppose now that ZFC is consistent. Let S be undecidable. Then if a machine solves $A$, it would tell us, upon inputting S, that S is independent. Since the set of independent (of ZFC) statements is independent, we are done: $A$ is undecidable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps you want to know whether ZFC proves that A is decidable, or if it proves that A is undecidable. Note that if ZFC proves that A is undecidable, then the argument above (formalized within ZFC) tells us that ZFC knows that ZFC is consistent. In this case, ZFC &lt;em&gt;is&lt;/em&gt; inconsistent, and it proves anything.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose, then, that ZFC proves that A is decidable. This is possible if ZFC is inconsistent. So, suppose that ZFC is consistent. Then the argument above tells us that ZFC proves that ZFC is inconsistent. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is not expected to be the case, as it tells us that ZFC is wrong about arithmetic statements. If ZFC is a &quot;true&quot; theory, meaning, if the arithmetic consequences of ZFC are true about the natural numbers, then ZFC cannot prove that A is decidable, and &quot;A is decidable&quot; is independent. Of course, if ZFC is not a true theory, I do not think it is too interesting whether it proves something or other about A.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can further complicate things by considering models of ZFC, and checking whether the model thinks that ZFC proves that A is decidable, or not, or any of the posisble variants suggested above. We can in fact, assuming mild consistency requirements on ZFC, check that there are models of ZFC that disagree on whether ZFC proves that A is decidable, proves that it is undecidable, or does not prove either. &lt;/p&gt;&#xA;" OwnerUserId="462" LastActivityDate="2011-01-12T19:50:38.437" CommentCount="5" />
  <row Id="17271" PostTypeId="2" ParentId="17265" CreationDate="2011-01-12T19:52:49.900" Score="1" ViewCount="" Body="&lt;p&gt;There are many characterizations of non-singular M-matrices which make it clear that the property is preserved by taking transposes.  For example, Berman and Plemmons book, Nonnegative matrices in the mathematical sciences, devotes Chapt. 6 to M-matrices.  Thm. 2.3 there gives many equivalences to a Z-matrix (that is, a matrix with nonpositive off-diagonal entries) to be a non-singular M-matrix.  But the condition you've cited is not among them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One equivalence given there ($A_1$) is for all the principal minors to be positive.  Clearly this is preserved by transpose.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sufficient condition you described is not preserved by taking transpose, as your example shows.  However it is certainly an M-matrix.  The cited sufficient condition is closely related to weak diagonal dominance, and it seems to be an echo of the result by Minkowski (that a Z-matrix with all positive row sums has positive determinant) which apparently motivated Ostrowski's choice of the term M-matrix (honoring Minkowski).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added:&lt;/strong&gt; I've corrected the Wikipedia article.  Thanks for pointing out this mistake.&lt;/p&gt;&#xA;" OwnerUserId="3111" LastEditorUserId="3111" LastEditorDisplayName="" LastEditDate="2011-01-14T16:26:15.297" LastActivityDate="2011-01-14T16:26:15.297" />
  <row Id="17272" PostTypeId="1" AcceptedAnswerId="17290" CreationDate="2011-01-12T19:53:41.497" Score="0" ViewCount="73" Body="&lt;p&gt;In Maple the command &lt;a href=&quot;http://www.maplesoft.com/support/help/Maple/view.aspx?path=LinearAlgebra/NullSpace&quot; rel=&quot;nofollow&quot;&gt;NullSpace&lt;/a&gt; turns a given matrix into a set of vectors -- a basis for its null space. This is precisely what it should in a mathematical sense. However, for computations using the obtained basis, it would be more convenient to have as output a matrix whose columns are the basis vectors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm having a hard time &lt;em&gt;converting a set of vectors into a matrix whose columns are the given vectors&lt;/em&gt;. The order doesn't matter (of course). Any help greatly appreciated.&lt;/p&gt;&#xA;" OwnerUserId="367" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2011-01-12T20:57:35.233" LastActivityDate="2011-01-12T21:21:35.550" Title="Maple: How to turn a set of vectors into a matrix?" Tags="&lt;math-software&gt;&lt;computer-algebra-systems&gt;&lt;maple&gt;" AnswerCount="1" />
  <row Id="17273" PostTypeId="2" ParentId="17247" CreationDate="2011-01-12T19:58:54.083" Score="6" ViewCount="" Body="&lt;p&gt;The only possible issue with writing $\frac{x}{\frac{1}{y}}$ as $xy$ is that you lose the information that $y\ne0$.  So if you are thinking about them as functions then they aren't quite the same.  But whenever $y\ne 0$, $\frac{x}{\frac{1}{y}}=xy$ as noted in the other answers.&lt;/p&gt;&#xA;" OwnerUserId="4266" LastActivityDate="2011-01-12T19:58:54.083" />
  <row Id="17274" PostTypeId="2" ParentId="17267" CreationDate="2011-01-12T20:00:19.503" Score="2" ViewCount="" Body="&lt;p&gt;I would say it would be important enough to publish it. According to the survey by Hildebrand and Tenenbaum, the approximate estimates have themselves found surprising applications in number theory. An exact formula will only do better!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In any case, one can never expect the surprising ways some results can be used. Consider the pigeon-hole principle for instance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Congratulations, btw.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Probably not the answer you were expecting, though).&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2011-01-12T20:00:19.503" />
  <row Id="17275" PostTypeId="1" AcceptedAnswerId="17283" CreationDate="2011-01-12T20:05:31.883" Score="1" ViewCount="237" Body="&lt;p&gt;Let $\displaystyle f: [0,1] \rightarrow \mathbb{R}$ given by &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f(x) = \begin{cases}&#xD;&#xA;0 &amp;amp; x \notin \mathbb{Q} \\&#xD;&#xA;\\&#xD;&#xA;0 &amp;amp; x = 0 \\&#xD;&#xA;\\&#xD;&#xA;\frac{1}{q_x} &amp;amp; x = \frac{p_x}{q_x} \in \mathbb{Q} \backslash \{0\}, \ p_x \in \mathbb{Z}, \  q_x \in \mathbb{N}, \ \text{gcd}(|p_x|, q_x) = 1&#xD;&#xA;\end{cases}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is $\displaystyle f$ Riemann integrable?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am trying to use the equivalent statements $\displaystyle g:[a,b] \rightarrow \mathbb{R}$ is Riemann integrable and $\displaystyle \forall \epsilon &amp;gt;0 \ \exists$ step functions $\displaystyle \rho, \psi$ with $\displaystyle \rho \leq g \leq \psi$ such that $\displaystyle \int_a^b (\psi - \rho) \leq \epsilon$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I guess that means I would have to somehow show that given $\displaystyle \epsilon$, there exists only a finite amount of $\displaystyle x \in [0,1]$ with $\displaystyle f(x) \geq \epsilon$? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it recommended that I consider something else instead? If not, how should I do this?&lt;/p&gt;&#xA;" OwnerUserId="3711" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2011-01-13T03:26:40.287" LastActivityDate="2011-01-13T03:26:40.287" Title="Is this function Riemann integrable?" Tags="&lt;homework&gt;&lt;analysis&gt;" AnswerCount="4" CommentCount="1" />
  <row Id="17276" PostTypeId="2" ParentId="17275" CreationDate="2011-01-12T20:07:05.547" Score="1" ViewCount="" Body="&lt;p&gt;Since you asked for recommendations: It is Riemann integrable because it is continuous almost everywhere (at the irrationals) and bounded. Of course, that is not how that Riemann integrability of this function was originally proved (this is an important example/counter-example function), but the result about continuity almost everywhere is quite important.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since this is homework: Here is a hint using step functions: &lt;a href=&quot;http://books.google.com/books?id=5hM0DcdEGf4C&amp;amp;pg=PA11#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;http://books.google.com/books?id=5hM0DcdEGf4C&amp;amp;pg=PA11#v=onepage&amp;amp;q&amp;amp;f=false&lt;/a&gt;. Exercise 5.24.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2011-01-12T20:38:45.693" LastActivityDate="2011-01-12T20:38:45.693" CommentCount="3" />
  <row Id="17277" PostTypeId="1" AcceptedAnswerId="17280" CreationDate="2011-01-12T20:07:59.093" Score="5" ViewCount="239" Body="&lt;p&gt;Cantor's Diagonal Argument is the standard proof of this theorem. However there must be other proofs, what are some of these proofs? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am asking this because whenever I think of this question, I immediately think of the Cantor's Argument, which kills the possibility of  other interesting finds. &lt;/p&gt;&#xA;" OwnerUserId="4728" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-12T20:25:19.753" LastActivityDate="2011-01-12T23:14:58.510" Title="Alternatives to show that $|\mathbb{R}|&gt;|\mathbb{Z}|$" Tags="&lt;analysis&gt;&lt;elementary-set-theory&gt;" AnswerCount="4" CommentCount="1" FavoriteCount="2" />
  <row Id="17279" PostTypeId="2" ParentId="17277" CreationDate="2011-01-12T20:13:01.267" Score="1" ViewCount="" Body="&lt;p&gt;Cantor's diagonal argument shows that a given function cannot be surjective.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One can construct the &lt;a href=&quot;http://en.wikipedia.org/wiki/Cantor_set&quot; rel=&quot;nofollow&quot;&gt;Cantor set&lt;/a&gt; and thus show $|P(\mathbb{N})| \le |\mathbb{R}|$ and from &lt;a href=&quot;http://en.wikipedia.org/wiki/Cantor%27s_theorem&quot; rel=&quot;nofollow&quot;&gt;Cantor's theorem&lt;/a&gt; it follows that $|\mathbb{N}| &amp;lt; |P(\mathbb{N})|$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(of course, I've been using the fact $|\mathbb{N}| = |\mathbb{Z}|$ freely here)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit:&lt;br&gt;&#xA;Do note that this way doesn't really use any implicit diagonalization argument. It just uses the two facts that $|\mathbb{N}|&amp;lt;|P(\mathbb{N})|$ and that $|P(\mathbb{N})|\le|\mathbb{R}|$. There's something to be said about the fact that the &quot;natural&quot; injection of $P(\mathbb{N})$ into the Cantor set only works for the infinite subsets and not for most finite subsets, however from Cantor's theorem we have that the &quot;most&quot; of the power set is the infinite subsets anyway, and thus this is not a real problem.&lt;/p&gt;&#xA;" OwnerUserId="622" LastEditorUserId="622" LastEditorDisplayName="" LastEditDate="2011-01-12T22:50:13.830" LastActivityDate="2011-01-12T22:50:13.830" CommentCount="15" />
  <row Id="17280" PostTypeId="2" ParentId="17277" CreationDate="2011-01-12T20:13:21.867" Score="5" ViewCount="" Body="&lt;p&gt;picakhu: This was &lt;a href=&quot;http://mathoverflow.net/questions/46970/proofs-of-the-uncountability-of-the-reals/&quot; rel=&quot;nofollow&quot;&gt;posted&lt;/a&gt; on MO and there were some interesting answers. I believe the answer I gave there avoids any hidden use of diagonalization. Other answers there are interesting and useful to think about even if, at the end, a diagonal argument is lurking.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Below is an abridged version of my answer there. It assumes some knowledge of ordinals.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;In a more general fashion, the question can be restated as: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Is there a proof of Cantor's theorem that ${}|X|&amp;lt;|{\mathcal P}(X)|$ that is not a diagonal argument?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I learned the following in A. Kanamori-D. Pincus. &quot;Does GCH imply AC locally?&quot;, in &lt;strong&gt;Paul Erdős and his mathematics, II (Budapest, 1999)&lt;/strong&gt;, 413-426, Bolyai Soc. Math. Stud., 11, János Bolyai Math. Soc., Budapest, 2002. It dates back to Zermelo's 1904 well-ordering paper. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;a.&lt;/strong&gt; It is enough to show that there is no injection $F:{\mathcal P}(Y)\to Y$ for any set $Y$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;b.&lt;/strong&gt; If there is such an injection $F$ we can (definably from $F$) produce a pair $(A,B)$ with $A\ne B$ and $F(A)=F(B)$. In effect, Zermelo proved that:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;For any $F:{\mathcal P}(Y)\to Y$ there is a unique  a unique well-ordering $(W, \lt)$ with $W\subseteq Y$ such that:&lt;/p&gt;&#xA;  &#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;$\forall x\in W (F (\{y ∈ W \mid y \lt x\}) = x)$, and &lt;/li&gt;&#xA;  &lt;li&gt;$F (W )\in W$.&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We can then take $A=W$ and $B=\{y\in W\mid y\lt F(W)\}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;c.&lt;/strong&gt; &lt;em&gt;Zermelo's theorem&lt;/em&gt; can be proved using the (set theoretic) recursion theorem and &lt;em&gt;Hartogs theorem&lt;/em&gt; that for any set $A$ there is a least ordinal $\alpha$ does not inject into $A$. Neither uses diagonalization.&lt;/p&gt;&#xA;" OwnerUserId="462" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2011-01-12T21:25:31.533" LastActivityDate="2011-01-12T21:25:31.533" CommentCount="3" />
  <row Id="17281" PostTypeId="2" ParentId="17275" CreationDate="2011-01-12T20:13:49.440" Score="2" ViewCount="" Body="&lt;p&gt;See &lt;a href=&quot;http://en.wikipedia.org/wiki/Thomae%27s_function&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2810" LastActivityDate="2011-01-12T20:13:49.440" />
  <row Id="17282" PostTypeId="2" ParentId="17275" CreationDate="2011-01-12T20:18:53.583" Score="1" ViewCount="" Body="&lt;p&gt;For any $\epsilon &amp;gt; 0$, there are finitely many points in $[0,1]$ such that $R(x) \ge \epsilon / 2$ (specifically, the rational numbers $x$ with denominators $q_x \le 2 / \epsilon$).  You can construct a step function $\psi$ that is equal to $1$ when sufficiently near one of these points, equal to $\epsilon / 2$ otherwise, and whose total area is $\le \epsilon$.  That should let you conclude that $R$ is Riemann-integrable.&lt;/p&gt;&#xA;" OwnerUserId="5546" LastActivityDate="2011-01-12T20:18:53.583" />
  <row Id="17283" PostTypeId="2" ParentId="17275" CreationDate="2011-01-12T20:19:51.300" Score="1" ViewCount="" Body="&lt;p&gt;Using that given $\epsilon$, $R(x) &amp;gt; \epsilon$ for only finitely many $x_i\in [0,1]$ is a step in the right direction.  Think about how to make the intervals containing those $x_i$ arbitrarily small, and $\epsilon$ arbitrarily small at the same time (hint: the size of the intervals should depend on [be bounded by] $\epsilon$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a complete version of this proof in Analysis by Steven Lay on page 278.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;edit: I have left out many details and I can give more info about how to go about it if you get stuck.&lt;/p&gt;&#xA;" OwnerUserId="565" LastEditorUserId="565" LastEditorDisplayName="" LastEditDate="2011-01-13T01:18:23.333" LastActivityDate="2011-01-13T01:18:23.333" />
  <row Id="17284" PostTypeId="1" AcceptedAnswerId="19393" CreationDate="2011-01-12T20:31:46.767" Score="0" ViewCount="190" Body="&lt;p&gt;Let's consider linear operators on the set of complex-valued functions to the same set. I wonder to which categories such operators can be classified. All linear operators I encountered so far fall into the following categories:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Differintegral of complex or variable order $L:f\to\mathbb{D}^{g}f$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Multiplication by a variable or constant coefficient $L:f\to g f$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Right composition $L:f\to f\circ g$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Finite differences of complex or variable order $L:f\to\Delta^{g}f$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Convolution with a function $L:f\to f * g$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Successive combination of finite number of the operators belonging to the above classes&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Sum of finite number of the operators belonging to the above classes&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;For example, Fourier transform can be expresses through combination of convolution and some other operators from the list.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am also aware about limits, but they would differ from the abovementioned only in finite amount of points.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So my question is: are there categories of linear operators on the same set that do not belong to the mentioned categories and differ from them more than just in countable number of points?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there any examples of such operators?&lt;/p&gt;&#xA;" OwnerUserId="2513" LastEditorUserId="2513" LastEditorDisplayName="" LastEditDate="2011-01-28T20:20:47.443" LastActivityDate="2011-01-28T22:17:10.560" Title="Linear operator categories" Tags="&lt;calculus&gt;&lt;linear-algebra&gt;&lt;operator-theory&gt;" AnswerCount="1" CommentCount="7" />
  <row Id="17285" PostTypeId="1" CreationDate="2011-01-12T20:41:12.807" Score="3" ViewCount="233" Body="&lt;p&gt;I have problems in understanding few concepts of elementary set theory. I've choosen a couple of problems (from my problems set) which would help me understand this concepts. To be clear: it's not a homework, I'm just trying to understand elementary set's theory concepts by reading solutions.&#xA;&lt;hr/&gt;&#xA;&lt;strong&gt;Problem 1&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I don't understand this; I mean - not at all)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $f: A \to B$, where $A,B$ are non-empty sets, and let $R$ be an equivalence relation in set $B$. We define equivalence relation $S$ in $A$ set by condition:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$aSb \Leftrightarrow f(a)R f(b)$&#xA;Determine, which inclusion is always true:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(a) $f([a]_S) \subseteq [f(a)]_R$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(b) $[f(a)]_R \subseteq f([a]_S)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notes:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$[a]_S$ is an equivalence class&#xA;&lt;hr/&gt;&#xA;&lt;strong&gt;Problem 2&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I suppose, that (a) is true &amp;amp; (b) is false)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which statement is true, and which is false (+ proof):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(a) If $f: A \xrightarrow{1-1} B$ and $f(A) \not= B$ then $|A| &amp;lt; |B|$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(b) If $|A| &amp;lt; |B|$ and $C \not= \emptyset$ then $|A \times C| &amp;lt; |B \times C|$&#xA;&lt;hr/&gt;&#xA;&lt;strong&gt;Problem 3&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I don't know, how to think about $\mathbb{Q}^{\mathbb{N}}$ and $\{0,1\}^∗$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which sets have the same cardinality:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$P(\mathbb{Q}), \mathbb{R}^{\mathbb{N}},\mathbb{Z}, \mathbb{Q}^{\mathbb{N}}, \mathbb{R} \times \mathbb{R}, \{ 0,1 \}^*, \{ 0,1 \}^{\mathbb{N}},P(\mathbb{R})$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\{ 0,1 \}^*$ means all finite sequences/words that contains $1$ and $0$, for example $000101000100$ or $1010101010101$ etc. $P(A)$ is a Power Set.&#xA;&lt;hr/&gt;&#xA;&lt;strong&gt;Problem 4&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I don't understand this; I mean - not at all)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are: maximum/minimum/greatest/lowest elements in set:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\{\{2; 3; 3; 5; 2\}; \{1; 2; 3; 4; 6\}; \{3\}; \{2; 1; 2; 1\}; \{1; 2; 3; 4; 5\}; \{3; 4; 2; 4; 1\}; \{2; 1; 2; 2; 1\}\}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ordered via subset inclusion&#xA;&lt;hr/&gt;&#xA;&lt;strong&gt;Problem 5&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How many equivalence relations there are in $\mathbb{N}$ which also are partial order?&#xA;&lt;hr/&gt;&#xA;These are simple problems, but I really need to understand, how to solve this kind of problems. I would apreciate Your help.&lt;/p&gt;&#xA;" OwnerUserId="5742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-12T21:07:16.767" LastActivityDate="2011-01-12T21:23:29.493" Title="Basic problems in elementary set theory" Tags="&lt;elementary-set-theory&gt;" AnswerCount="3" CommentCount="4" />
  <row Id="17286" PostTypeId="1" CreationDate="2011-01-12T20:49:24.353" Score="0" ViewCount="107" Body="&lt;p&gt;I'm trying to understand a proof in &quot;Groups with a maximal irredundant 6-cover&quot; &#xA;by A. Abdollahi et al, page 3231, Lemma 4.1 itens (2) and (3).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's my problem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lemma 4.1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) The following are $\mathfrak{G}_6$-groups:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(d) $G \cong (C_5 \times C_5) \rtimes C_4$ with $Z(G) = 1$. In this case, the intersection of a maximal irredundant cover which is core-free has size 2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(3) A subdirect product G of three symmetric groups $Sym_3$ is a $\mathfrak{G}_6$-groups if and only if G is isomorphic to one the following groups: $(C_3 \times C_3) \rtimes C_2$, $S_3 \times S_3$, $(C_3)^3 \rtimes C_2$, where the size of the intersection of an arbitrary maximal irredundant 6-cover with core-free intersection is 1, 1, and 2, respectively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remark: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(*)$ G is $\mathfrak{G}_6$-groups mean $G = M_1 \cup ...\cup M_6$ is a maximal irredundant 6-cover with core-free intersection.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(**)$ In outline proof is used the following function written with GAP (2002) program&lt;/p&gt;&#xA;&#xA;&lt;p&gt;f:= function(G) local&#xA;S, M, n, C, i, T, Q, R; &#xA;n:=Size(G); M:=MaximalSubgroups(G); C:=Combinations(M,6); &#xA;S:=[]; for i in [1..Size(C)] do if&#xA;Size(Union(C[i]))=n then Add(S, C[i]); fi; &#xA;od; T := []; for i in&#xA;[1..Size(S)] do if Size(Core(G, Intersection(S[i])))=1 then&#xA;Add(T,S[i]); fi; &#xA;od; R:=[]; for i in [1..Size(T)] do&#xA;Q:=Combinations(T[i],5); if (n in&#xA;List(Q, i-&gt;Size(Union(i))))=false then Add(R,T[i]); fi; od;&#xA;return R; end;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The input of the function is a group G and the outputs are all&#xA;the irredundant 6-covers with core-free intersection for G, and if there is no such a&#xA;cover for G, the output is the empty list.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How to apply to items (2) and (3) in the function f written in the GAP to give a demonstration?&lt;/p&gt;&#xA;" OwnerUserId="5743" LastEditorUserId="5743" LastEditorDisplayName="" LastEditDate="2011-01-13T14:28:38.363" LastActivityDate="2011-01-13T14:28:38.363" Title="Cover groups for 6 subgroups maximais" Tags="&lt;group-theory&gt;" CommentCount="4" />
  <row Id="17287" PostTypeId="2" ParentId="17285" CreationDate="2011-01-12T21:02:24.917" Score="2" ViewCount="" Body="&lt;p&gt;I'll do 1 and 5.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For one, the first inclusion is true.  For let $x\in f([a]_b)$.  We want to show that $x\in [f(a)]_r$, i.e., that $x$ $r$ $f(a)$.  Since $x\in f([a]_s)$, there is a $b\in A$ such that $b$ $s$ $a$ and f(b) = x.  Since $b$ $s$ $a$, we know $f(b)$ $r$ $f(a)$.  Thus, $x$ $r$ $f(a)$ so $x\in [f(a)]_r$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reverse inclusion is false.  Suppose $A = \{a\}$ and $B = \{b,c\}$ with $b$ and $c$ distinct.  Use the relation $r = B\times B$ - all elements are $r$-related to all elements.  Use the function $f:A\rightarrow B$ given by $f(a) = b$.  Then it's not too hard to show that $c\in [f(a)]_r$ but $c\notin f([a]_s)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For 5, notice that the only difference between a partial order and an equivalence relation is in the symmetry/antisymmetry.  In other words, if $r$ is an equivalence relation, then $a$ $r$ $b$ implies $b$ $r$ $a$.  However, if $r$ is a partial order, then $a$ $r$ $b$ and $b$ $r$ $a$ implies $a = b$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(To be clearer, I think there are several competing definitions of &quot;partial order&quot; - some, drop the antisymmetry condition.  If that's the case, ignore my answer to number 5).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, suppose $r$ is an equivalence relation and a partial order.  Given $a$, what are all the elements it can be related to?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, suppose $a$ $r$ $b$.  Then we conclude that $b$ $r$ $a$ since $r$ is an equivalence relation.  Since $r$ is a partial order, we conclue from this that $a = b$.  Said another way, if $a\neq b$, then $a$ $r$ $b$ cannot be true.  Thus, the only possibility to consider is whether or not $a$ $r$ $a$.  But in an equivalence relation, this must be the case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, our relation $r$ is simply the set of all ordered pairs $(a,a)$ - there is a unique relation on $\mathbb{N}$ (or any set!) which is both an equivalence relation and a partial order.&lt;/p&gt;&#xA;" OwnerUserId="331" LastActivityDate="2011-01-12T21:02:24.917" CommentCount="2" />
  <row Id="17288" PostTypeId="2" ParentId="17285" CreationDate="2011-01-12T21:18:00.170" Score="3" ViewCount="" Body="&lt;p&gt;For problem 3, $\mathbb{Q^N}$ means the set of functions from $\mathbb{N}$ into $\mathbb{Q}$.  So you need to decide how many of them there are.  Certainly there are at least as many as functions from $\mathbb{N}$ into 2, which is $2^\mathbb{N}$, as you could just restrict the functions values to 0 and 1.  Have you seen that before?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For problem 4, &quot;ordered by subset inclusion&quot; means $a \lt b \Leftrightarrow a \subset b$.  So, for example, {3} &amp;lt; {1;2;3;4;6} and three others of the given sets.  Now look at your definitions of maximum/minimum/greatest/least and see which sets satisfy them.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2011-01-12T21:23:29.493" LastActivityDate="2011-01-12T21:23:29.493" CommentCount="1" />
  <row Id="17289" PostTypeId="2" ParentId="17285" CreationDate="2011-01-12T21:20:26.767" Score="3" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Problem 1.&lt;/strong&gt; Here's an example: take $A=\mathbb{N}$, $B=\{0,1,2,3\}$, and let $f\colon A\to B$ map the natural numbers to their remainder when divided by $4$ (so $f(3) = 3$, $f(15) = 3$, $f(21) = 1$, etc). Let $R$ be the equivalence relation on $B$ given by $aRb$ if and only if $a^2\equiv b^2 \pmod{4}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The relation $S$ is defined as follows: first map to $B$, then check $R$ for the images. If the images are related, then the original elements are related; if the images are not related, then the originals are not related. For example, to see whether $3R10$ holds, we take $f(3)=3$ and $f(10)=2$, and check whether $3R2$ holds or not; since $3^2 \not\equiv 2^2\pmod{4}$, then $3\not R2$; that is, $f(3)\not Rf(2)$, so $3\not S 2$. On the other hand, $7S9$ holds: because $f(7) = 3$, $f(9) = 1$, and $f(7)^2 = 9 \equiv 1 = f(9)^2\pmod{4}$, so $f(7)Rf(9)$ is true.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's the idea.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, for a general case. First convince yourself that $S$ is in fact an equivalence relation on $A$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you do that, in order to check (a), you want to see if $f([a]_S)\subseteq [f(a)]_R$. Well, take an element in $f([a]_S)$; this is a $b\in B$ such that $b=f(x)$ for some $x\in[a]_S$. that is, $xSa$ is true, and $f(x)=b$. So you want to see whether $b\in[f(a)]_R$; that is, whether $bRf(a)$. So, from assuming $xSa$ holds, you want to see whether you can always conclude that $f(x)Sf(a)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For (b), you proceed similarly; now take $b\in [f(a)]_R$, and you want to see whether $b\in f([a]_S)$. That is, assume that $bRf(a)$; must there exist an $x\in A$ such that $xSa$ and $f(x)=b$? If so, prove it; if not, give a specific counterexample with specific $A$, $B$, $f$, $R$, and $S$, and an exlicit $a$ and $b$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Problem 2.&lt;/strong&gt; (a) If your sets are finite, the your guesses are right; if your sets are not necessarily finite, you're in trouble. Consider $A=B=\mathbb{N}$ and $f(n) = 2n$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For (b); you know there is a $1-1$ function from $A$ to $B$; see if you can construct a $1-1$ function from $A\times C$ to $B\times C$ (and see if you can spot why you need $C\neq\emptyset$). This will show that $|A|\leq|B|$ implies $|A\times C|\leq|B\times C|$. If you also know that there is no surjection between $A$ and $B$, then again you are going to have two different cases, depending on whether $C$ is &quot;really big&quot;  (compared to $A$ and $B$) or not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Problem 3.&lt;/strong&gt;  The set $A^B$ is the set of all functions from $B$ to $A$. So $\mathbb{Q}^{\mathbb{N}}$ is the set of all functions from $\mathbb{N}$ to $\mathbb{Q}$; this is the set of all &lt;em&gt;rational sequences&lt;/em&gt; (sequences, each term a rational number). The set $\{0,1\}^{\mathbb{N}}$ is the set of all &lt;em&gt;binary sequences&lt;/em&gt; (sequences, each term either $0$ or $1$). &lt;em&gt;Hint.&lt;/em&gt; Think about binary expansion of real numbers between $0$ and $1$, or even better, the ternary expansion of the elements of the Cantor set to deal with $\{0,1\}^{\mathbb{N}}$. For $\{0,1\}^*$, can you count how many sequences of length $n$ there are, for each $n$? Add them all up then. For $\mathbb{Q}^{\mathbb{N}}$, can you exhibit at least as many sequences as there are real numbers?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Problem 4.&lt;/strong&gt; You are comparing sets by inclusion. We say set $A$ is &quot;smaller than or equal to&quot; set $B$ if and only if $A\subseteq B$. So, for example, the set $\{3\}$ is smaller than or equal to the set $\{1,2,3,4,5\}$, but not smaller than the set $\{1,2\}$. When using this order, note that it is possible for you to have two sets, neither of which is smaller than or equal to the other (for example, neither of $\{3\}$ and $\{1,2\}$ is smaller than the other; they are &lt;em&gt;incomparable&lt;/em&gt;). The maximum among the sets you are given would be a set that is greater than or equal (contains) all of the sets you are given; a minimum would be a set that is smaller than or equal (is contained in) all of the sets you are given. They may or may not exist. &quot;Greatest&quot; is the same as &quot;maximum&quot; in this context; &quot;smallest&quot; the same as minimum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Problem 5.&lt;/strong&gt; An equivalence relation is a collection of ordered pairs that is reflexive, symmetric, and transitive. A partial order is a collection of ordered pairs that is reflexive, *&lt;em&gt;anti&lt;/em&gt;*symmetric, and transitive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the question is: how many equivalence relations on $\mathbb{N}$ are also antisymmetric? Or: how many relations are all of reflexive, transitive, symmetric, &lt;em&gt;and&lt;/em&gt; antisymmetric? Think about what having both symmetry and antisymmetry means. &lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2011-01-12T21:20:26.767" />
  <row Id="17290" PostTypeId="2" ParentId="17272" CreationDate="2011-01-12T21:21:35.550" Score="1" ViewCount="" Body="&lt;p&gt;Its quite easy:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;kern := NullSpace(A)    &#xA;Test := Matrix(convert(kern, list))&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;should perfectly do the job for you (note that it also depends on your exact version of Maple)&lt;/p&gt;&#xA;" OwnerUserId="3123" LastActivityDate="2011-01-12T21:21:35.550" CommentCount="1" />
  <row Id="17291" PostTypeId="1" AcceptedAnswerId="17292" CreationDate="2011-01-12T21:30:12.400" Score="2" ViewCount="153" Body="&lt;p&gt;I am having trouble with this:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If $4^x + 4^{-x} = 34$, then what is $2^x + 2^{-x}$ equal to? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I managed to find $4^x$ and it is:&#xA;$$4^x = 17 \pm 12\sqrt{2}$$&#xA;so that means that $2^x$ is:&#xA;$$2^x = \pm \sqrt{17 \pm 12\sqrt{2}}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Correct answer is 6 and I am not getting it :(. What am I doing wrong?&lt;/p&gt;&#xA;" OwnerUserId="5745" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2011-01-13T03:16:51.970" LastActivityDate="2011-01-13T03:16:51.970" Title="If $4^x + 4^{-x} = 34$, then $2^x + 2^{-x}$ is equal to...?" Tags="&lt;algebra-precalculus&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="17292" PostTypeId="2" ParentId="17291" CreationDate="2011-01-12T21:35:03.330" Score="9" ViewCount="" Body="&lt;p&gt;You haven't done anything wrong! To complete your answer, one way you can see the answer is $6$ is to guess that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$17 + 12 \sqrt{2} = (a + b\sqrt{2})^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Giving us&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$17 = a^2 + 2b^2, \ \ ab = 6$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Giving us &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$a = 3, \ \ b = 2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus $$ \sqrt{17 + 12 \sqrt{2}} = 3 + 2\sqrt{2}$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;which gives $$2^x + 2^{-x} = 6$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(And similarly for $17 - 12\sqrt{2}$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A simpler way is to notice that $(2^x + 2^{-x})^2 = 4^x + 4^{-x} + 2$.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2011-01-12T21:47:01.367" LastActivityDate="2011-01-12T21:47:01.367" />
  <row Id="17293" PostTypeId="2" ParentId="17152" CreationDate="2011-01-12T21:35:12.040" Score="-1" ViewCount="" Body="&lt;p&gt;If you allow for evolution and substitute apes for monkeys you could say that the probability is 1, since it has already happened.&lt;/p&gt;&#xA;" OwnerUserId="5749" LastActivityDate="2011-01-12T21:35:12.040" CommunityOwnedDate="2011-01-15T06:13:52.160" CommentCount="4" />
  <row Id="17294" PostTypeId="2" ParentId="17291" CreationDate="2011-01-12T21:35:27.990" Score="6" ViewCount="" Body="&lt;p&gt;Let $a=2^x$. Then $2^{-x}=1/a$ and $$\left(a+\frac1a\right)^2=a^2+2+1/a^2=2+(2^x)^2+(2^{-x})^2=2+4^x+4^{-x}=2+34=36.$$ This means that $a+1/a=6$ (it should be 6 or -6, since these are the square roots of $36$ but since $a$ and $1/a$ are positive, it is 6).&lt;/p&gt;&#xA;" OwnerUserId="462" LastActivityDate="2011-01-12T21:35:27.990" />
  <row Id="17295" PostTypeId="2" ParentId="17152" CreationDate="2011-01-12T21:35:53.210" Score="0" ViewCount="" Body="&lt;p&gt;Monkeys don't produce a proper random distribution on keystrokes. Not even a Markov-chain of keystrokes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given infinite time (and an undying support of monkeys) or just infinite monkeys, they will produce infinite text. But that does not need to imply that Hamlet will be part of this infinite long text.&#xA;Therefore &lt;em&gt;possibly&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="3226" LastEditorUserId="3226" LastEditorDisplayName="" LastEditDate="2011-01-27T21:23:20.230" LastActivityDate="2011-01-27T21:23:20.230" CommunityOwnedDate="2011-01-15T06:13:52.160" CommentCount="2" />
  <row Id="17296" PostTypeId="2" ParentId="17291" CreationDate="2011-01-12T21:37:12.620" Score="1" ViewCount="" Body="&lt;p&gt;$4^{x}+4^{-x} = 34$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$2^{2x}+2^{-2x} = 34$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $z= 2^x$.&lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2011-01-12T21:37:12.620" />
  <row Id="17297" PostTypeId="2" ParentId="17262" CreationDate="2011-01-12T21:47:57.043" Score="4" ViewCount="" Body="&lt;p&gt;The Frobenius norm is actually quite nice, and also natural. it is defined by merely $$\|A\|_F^2 = \mbox{trace}(A&amp;#39;A)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and since it is naturally an inner-product norm, it makes optimization, etc. with it much easier (think quadratic programs, instead of semidefinite programs)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Numerical analysis probably like the operator norm perhaps because they often exploit $\|Ax\| \le \|A\| \|x\|$, and if you use the operator-2 norm, you get a tighter inequality (in general). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Otherwise, what norm you use, should be governed by the application where you are trying to use it.&lt;/p&gt;&#xA;" OwnerUserId="1709" LastActivityDate="2011-01-12T21:47:57.043" />
  <row Id="17298" PostTypeId="2" ParentId="17152" CreationDate="2011-01-12T22:05:11.927" Score="-6" ViewCount="" Body="&lt;p&gt;Forgetting the age of the universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Forgetting how many bananas or Chinese typewriters this might take.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a setting where all variables are expanded to contain all values then one particular related value will, at some point, be generated.  So yes, infinite monkeys (or any random generator) with infinite time will create hamlet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, adding a touch of reality in here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If neither time nor our supply of monkeys is unlimited then it stands to reason that it is incredibly unlikely for any intelligible work (hamlet or not) to be generated and much more likely to have wasted everyone's time.&lt;/p&gt;&#xA;" OwnerUserId="5750" LastActivityDate="2011-01-12T22:05:11.927" CommunityOwnedDate="2011-01-15T06:13:52.160" />
  <row Id="17299" PostTypeId="1" AcceptedAnswerId="17302" CreationDate="2011-01-12T22:25:09.393" Score="1" ViewCount="71" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a problem in vector algebra.&#xA;In this &lt;a href=&quot;http://mathworld.wolfram.com/Point-LineDistance3-Dimensional.html&quot; rel=&quot;nofollow&quot;&gt;Wolfram-page&lt;/a&gt; the last two formulas (9) and (10) are:&#xA;$$&#xD;&#xA;\frac{ | (x_2 - x_1) \times (x_1 - x_0) | }{|x_2 - x_1 |} = &#xD;&#xA;\frac{ | (x_0 - x_1) \times (x_0 - x_2) | }{|x_2 - x_1 |}&#xD;&#xA;$$&#xA;I've tried to apply properties founded in this other &lt;a href=&quot;http://mathworld.wolfram.com/CrossProduct.html&quot; rel=&quot;nofollow&quot;&gt;Wolfram-crossproduct-page&lt;/a&gt; but I still don't understand it. How is possible? And why the do this vectors manipulation?&lt;/p&gt;&#xA;" OwnerUserId="5046" LastEditorUserId="856" LastEditorDisplayName="" LastEditDate="2011-01-13T03:04:16.173" LastActivityDate="2011-01-13T03:04:16.173" Title="Problem involving cross products" Tags="&lt;geometry&gt;&lt;linear-algebra&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="17301" PostTypeId="2" ParentId="17111" CreationDate="2011-01-12T22:34:40.280" Score="8" ViewCount="" Body="&lt;p&gt;The Maltsev proof is in fact quite easy.  Let me give the proofs of the two relevant results.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Theorem:&lt;/strong&gt; If $G$ is finitely generated and residually finite then $G$ is Hopfian, ie every epimorphism is an isomorphism.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Proof:&lt;/em&gt; Suppose not; let $f:G\to G$ be an epimorphism with some non-trivial $g$ in its kernel.  For each $n$ let $g_n$ be such that $f^n(g_n)=g$.  Let $q: G\to Q$ be a map to a finite group such that $q(g)\neq 1$.  Now $q\circ f^n$ kills $g_{n+1}$ but not $g_n$, so the homomorphisms $q\circ f^n$ are all distinct.  But there are only finitely many homomorphisms from a finitely generated group to a finite group, so this is a contradiction. &lt;em&gt;QED&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A Mariano indicates, the proof that $GL_2(\mathbb{Z})$ is residually finite is also easy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Theorem:&lt;/strong&gt; $GL_2(\mathbb{Z})$ is residually finite.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Proof:&lt;/em&gt; Let $A\in GL_2(\mathbb{Z})$, thought of as a matrix.  Choose some prime $p$ that does not divide all of the entries of $A-I$.  The reduction homomorphism $GL_2(\mathbb{Z})\to GL_2(\mathbb{Z}/p)$ is a map to a finite group that does not kill $A$. &lt;em&gt;QED&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See - easy!&lt;/p&gt;&#xA;" OwnerUserId="2287" LastEditorUserId="2287" LastEditorDisplayName="" LastEditDate="2011-01-12T22:43:57.390" LastActivityDate="2011-01-12T22:43:57.390" />
  <row Id="17302" PostTypeId="2" ParentId="17299" CreationDate="2011-01-12T22:35:40.870" Score="2" ViewCount="" Body="&lt;p&gt;$\begin{align}|(x_2-x_1)\times(x_1-x_0)|&amp;amp;=|(x_1-x_2)\times(x_0-x_1)|\\&amp;amp;=|(x_0-x_1)\times(x_1-x_2)|\\&amp;amp;=|(x_0-x_1)\times((x_0-x_1)+(x_1-x_2))|\\&amp;amp;=|(x_0-x_1)\times(x_0-x_2)|\end{align}$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;where the second is because you are in magnitude signs and the third is because the cross product of a vector with itself (or any parallel vector) is zero.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2011-01-13T02:59:13.660" LastActivityDate="2011-01-13T02:59:13.660" CommentCount="4" />
  <row Id="17303" PostTypeId="1" CreationDate="2011-01-12T22:37:42.067" Score="1" ViewCount="87" Body="&lt;p&gt;I would like to hear some intuition about difference between subtraction and division. For binary subtraction operator the standard development is introduction of unary operation of taking negative number. For division however it depends on the domain. For &quot;nice&quot; domains (such as rational and complex numbers) we have unary multiplicative inverse, while for integers the division is adjoint operation introduced via Galois connection, e.g.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;2*5 &amp;lt;= 11 &amp;lt;-&amp;gt; 5 &amp;lt;= 11/2&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Can binary subtraction be defined via Galois connection as well?&lt;/p&gt;&#xA;" OwnerUserId="5754" LastActivityDate="2011-01-12T22:37:42.067" Title="Similarity between subtraction and division" Tags="&lt;algebra&gt;" CommentCount="2" />
  <row Id="17304" PostTypeId="2" ParentId="17142" CreationDate="2011-01-12T22:41:06.487" Score="0" ViewCount="" Body="&lt;p&gt;For equations that look complicated, it helps to artificially simplify them by guessing really boring values that you might solve for. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example let y and z be 0 (yes, that's not positive but it's not particularly negative either (unless you're speaking French)..actually just consider x,y,z as non-negative...makes things easier all around). At least y and z are as small as possible. Then you'll notice that x is as big as your RHS (right hand side), which really means it is as big as possible. And notice that x is positive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now do the same for y (set x and z to 0) and separately for z (set x and y to 0). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice now that if you subtract a little from any one of them, you'll be able to add a little more to the others (not the same amount for each). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the smallest value for each could be 0, and the largest value will be what your solutions are for the above simplifications. And by the above 'notice', all values in between will work.&lt;/p&gt;&#xA;" OwnerUserId="1919" LastActivityDate="2011-01-12T22:41:06.487" />
  <row Id="17305" PostTypeId="2" ParentId="17152" CreationDate="2011-01-12T22:47:02.670" Score="5" ViewCount="" Body="&lt;p&gt;For some reason I want to attribute this reasoning to Douglas Hofstadter, though I couldn't tell you which of his books it's from. Here goes:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you could get sufficient randomness from an infinite number of monkeys (this is trivial if you assume that by mere chance, a infinite subset will fit the bill – or you could follow &lt;a href=&quot;http://math.stackexchange.com/questions/17152/given-an-infinite-number-of-monkeys-and-an-infinite-amount-of-time-would-one-of/17159#17159&quot;&gt;Arjang's approach&lt;/a&gt; and aggregate across monkeys for more entropy, which has the benefit of getting results &lt;em&gt;much&lt;/em&gt; faster), you already have every variation of every story ever told. You'll even have every story that ever could be told. Just get an infinite number of monkeys (or a slightly smaller number of computers) and opening a publishing business. Make a million bucks and retire.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But this rings false, especially since modern computing power (relative to the difficulty of the task) is practically infinite, putting the practice of this philosophy within reach. Just imagine trying it yourself. It's not the monkeys or the computers or the printers doing all of the work. Suddenly, you are wading through millions of pages of gibberish text looking for the book that will make you rich. Good luck. (It is the fact that the filtering process is much slower than the production process that makes me say that computer power is practically infinite.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's not the characters on the page that make Hamlet. Hamlet is a synthesis of information, a composition that can only be guided by intelligence. In a sea of random characters, the sequence that maps isomorphically to Hamlet is just more noise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This may sound like a qualified &quot;yes&quot; in response to the question, but in reality it is an empathic &quot;&lt;em&gt;no&lt;/em&gt;.&quot; It is not the act of producing the character sequence of Hamlet in a random string that writes Hamlet – it is the act of &lt;em&gt;finding&lt;/em&gt; Hamlet in a random string that writes Hamlet. The distinction may sound subtle, but the two tasks are profoundly different.&lt;/p&gt;&#xA;" OwnerUserId="5751" LastActivityDate="2011-01-12T22:47:02.670" CommunityOwnedDate="2011-01-15T06:13:52.160" CommentCount="5" />
  <row Id="17306" PostTypeId="2" ParentId="17005" CreationDate="2011-01-12T22:59:18.270" Score="1" ViewCount="" Body="&lt;p&gt;Yes, your procedure is correct. It eliminates the need for parentheses altogether.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another way to thinking about it is to go the other direction from right to left. The procedure is to, whenever you come to an operator, -always- apply it to two operands you currently have, replacing the 3 items (operator and two operands) with the value of the operation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This latter procedure is called Reverse Polish Notation (RPN). It is a good method for computer implementation because all you need is a stack to remember any intermediate computations.&lt;/p&gt;&#xA;" OwnerUserId="1919" LastActivityDate="2011-01-12T22:59:18.270" />
  <row Id="17307" PostTypeId="2" ParentId="17277" CreationDate="2011-01-12T23:00:15.627" Score="1" ViewCount="" Body="&lt;p&gt;I read about this proof in Raymond Smullyan's book &quot;Satan, Cantor and Infinity&quot;. The proof is due to Willaim Zwicker (according to the book) and involves something called a hypergame. I'll first mentioned the hypergame paradox (because it's fun and can give an insight about the proof) and then write the proof.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's call a game with turns a &lt;em&gt;normal&lt;/em&gt; game if it always ends in finite turns. For example tic-tac-toe and checkers are both normal games. Now imagine the following game which I will call &lt;em&gt;hypergame&lt;/em&gt;: In the first turn the first player names a normal game. In the second turn the second player plays the first move of the normal game named and then the players continue playing that game in turns. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The paradox arises from the question: Is the hypergame a normal game? Assume that it is and assume that we try to play hypergame: I go first and say &quot;Let's play hypergame&quot; (I can do that since it's a normal game). Then it's your turn and you can say &quot;Let's play hypergame&quot; and so forth. Thus the hypergame cannot be normal since if it is we can start a hypergame that doesn't end in finite turns. But if the hypergame isn't normal then every hypergame will end in finite turns (since every game I will name will end in finite moves and our game will last that number of turns plus one) which will make the hypergame normal!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the proof: Assume that a function $f:\mathbb{N}\to\mathcal{P}(\mathbb{N})$ and let's assume that $f$ sends $n$ to $X_n$. We define a &lt;em&gt;path&lt;/em&gt; of numbers as a finite or infinite sequence of numbers such that an element $n$ of the path is either the last (and the path is finite) or the successor of $n$ belongs in $X_n$. For example take a number $n$. If $m\in X_n$ then a successor of $n$ in a path can be $m$. If $X_n$ is empty then $n$ is always the last element of a path. Now we call an element  $n\in\mathbb{N}$ &lt;em&gt;normal&lt;/em&gt; if every path that begins with $n$ is finite. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally take the set $Z$ that contains exactly all the normal elements of $\mathbb{N}$. It's easy to see that there doesn't exist an $n_0$ such that $Z=X_{n_0}$, which would mean that $f$ is not onto. The argument is essentially the same as the one I used above: Assume that $Z=X_{n_0}$. If $n_0\in Z$ then there is an infinite path, namely $(n_0,n_0,n_0,\ldots)$, but $Z$ is the set of all normal elements so that's impossible. If $n\notin Z$ then there is an infinite path that begins with $n_0$ (by the definition of $Z$). Let's call it $(n_0,m,k,\ldots)$. But then there is an infinite path that begins with $m$ (take the aforementioned path and remove $n_0$ from the start). This is again impossible since $m\in Z$ (by the definition of a path) and every element of $Z$ is normal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This proof uses self-reference but doesn't seem to use the diagonal lemma. I don't know if it's hidden somewhere in the proof (though it looks to me like it isn't).&lt;/p&gt;&#xA;" OwnerUserId="1772" LastActivityDate="2011-01-12T23:00:15.627" CommentCount="5" />
  <row Id="17308" PostTypeId="2" ParentId="17277" CreationDate="2011-01-12T23:06:40.737" Score="0" ViewCount="" Body="&lt;p&gt;Cantor had several other proofs before coming up with the diagonalization argument.  (See &lt;a href=&quot;http://en.wikipedia.org/wiki/Cantor%27s_first_uncountability_proof&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Cantor%27s_first_uncountability_proof&lt;/a&gt; for one version.  Or &lt;a href=&quot;http://mathoverflow.net/questions/23953/earliest-diagonal-proof-of-the-uncountability-of-the-reals&quot; rel=&quot;nofollow&quot;&gt;http://mathoverflow.net/questions/23953/earliest-diagonal-proof-of-the-uncountability-of-the-reals&lt;/a&gt;.)&#xA;I also read a recent account of this in a Math Monthly article. (I think a version of that article is at &lt;a href=&quot;http://www.math.jhu.edu/~wright/Cantor_Pick_Phi.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.math.jhu.edu/~wright/Cantor_Pick_Phi.pdf&lt;/a&gt;.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &quot;Nested Intervals&quot; proof is really very nice.  Assume the reals are countable and so arrange them as a sequence.  Then use that ordering to create intervals $[a_n, b_n]$ where $a_{n-1}&amp;lt;a_n&amp;lt;b_n &amp;lt;b_{n-1}$.  The intersection of this sequence of nested intervals is nonempty (this is the Nested Intervals theorem) yet no real number in our list can be inside it!  This contradiction proves that we could not have originally arranged the reals in a sequence.&lt;/p&gt;&#xA;" OwnerUserId="5680" LastEditorUserId="5680" LastEditorDisplayName="" LastEditDate="2011-01-12T23:14:58.510" LastActivityDate="2011-01-12T23:14:58.510" CommentCount="3" />
  <row Id="17309" PostTypeId="2" ParentId="17242" CreationDate="2011-01-12T23:07:33.217" Score="0" ViewCount="" Body="&lt;p&gt;This is simply the radix $\rm\:b\:$ generalization of &lt;a href=&quot;http://en.wikipedia.org/wiki/Casting_out_nines&quot; rel=&quot;nofollow&quot;&gt;casting out nines.&lt;/a&gt; Generally, if $\rm\:d\:$ divides $\rm\ b-1\ $ then $\rm\:  mod\ \ d:\:\ \ b\equiv 1\ \Rightarrow\ a_n b^n +\:\cdots\: + a_1 b + a_0 \ \equiv\ a_n +\: \cdots\: + a_1 + a_0\: =\:  $ sum of the radix $\rm\:b\:$ digits. For further remarks on casting out nines see my &lt;a href=&quot;http://math.stackexchange.com/questions/16011/what-makes-9-special/16015#16015&quot;&gt;post here.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2011-01-18T07:46:53.363" LastActivityDate="2011-01-18T07:46:53.363" />
  <row Id="17310" PostTypeId="1" AcceptedAnswerId="17312" CreationDate="2011-01-12T23:16:51.130" Score="3" ViewCount="244" Body="&lt;p&gt;How do we show that if $g \geq 2$ is an integer, then the two series $$\sum\limits_{n=0}^{\infty} \frac{1}{g^{n^{2}}} \quad  \ \text{and} \ \sum\limits_{n=0}^{\infty} \frac{1}{g^{n!}}$$ both converge to irrational numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, i tried to see what happens, if they converge to a rational but couldn't get anything out it.&lt;/p&gt;&#xA;" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2011-01-13T03:18:17.333" LastActivityDate="2011-01-13T06:36:59.853" Title="Proving that a series converges to an irrational number" Tags="&lt;real-analysis&gt;&lt;sequences-and-series&gt;&lt;irrational-numbers&gt;" AnswerCount="3" FavoriteCount="1" />
  <row Id="17311" PostTypeId="1" CreationDate="2011-01-12T23:16:53.533" Score="4" ViewCount="217" Body="&lt;p&gt;Is there a categorification of $\pi$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have to admit that this is a very vague question. Somehow it is motivated by &lt;a href=&quot;http://mathoverflow.net/questions/51853/what-is-the-state-of-our-ignorance-about-the-normality-of-pi&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; recent MO question, which made me stare at some digits and somehow forgot my animosity about this branch of mathematics, wondering if there is a connection to the branches I love so much.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\pi$ is the area of the unit circle, so perhaps we have to categorify the unit circle (using the projective line?) and the area of such an object. Areas are values of integrals, and there are some kind of integrals in category theory (ends), but this is really just a wild guess.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For some great examples of categorification see &lt;a href=&quot;http://mathoverflow.net/questions/43579/examples-of-categorification&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; list on MO, and for the meaning of categorification see &lt;a href=&quot;http://mathoverflow.net/questions/4841/what-precisely-is-categorification&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; MO question or &lt;a href=&quot;http://arxiv.org/abs/math/9802029&quot; rel=&quot;nofollow&quot;&gt;that article&lt;/a&gt; by Baez/Dolan. Inspired by the answers of Todd Trimble, we may consider a categorified sinus function&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sin(X) = \sum_{n \geq 0} (-1)^{\otimes n} X^{\otimes (2n+1)} / (2n+1)!$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;in any complete symmetric monoidal category (if we can make sense of $-1$).&lt;/p&gt;&#xA;" OwnerUserId="1650" LastEditorUserId="1650" LastEditorDisplayName="" LastEditDate="2011-01-12T23:32:07.013" LastActivityDate="2011-01-13T00:53:02.803" Title="Categorification of $\pi$?" Tags="&lt;number-theory&gt;&lt;category-theory&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="17312" PostTypeId="2" ParentId="17310" CreationDate="2011-01-12T23:21:17.413" Score="19" ViewCount="" Body="&lt;p&gt;Both are transcendental numbers. This follows from Roth's theorem on the rate of approximations to irrational numbers by rationals. See also this &lt;a href=&quot;http://math.stackexchange.com/questions/10327/0-246810121416-is-it-a-algebraic-number/&quot;&gt;question&lt;/a&gt;. The key concept is &quot;irrationality measure&quot;, see for example this &lt;a href=&quot;http://mathworld.wolfram.com/IrrationalityMeasure.html&quot; rel=&quot;nofollow&quot;&gt;entry&lt;/a&gt; from MathWorld.&lt;/p&gt;&#xA;" OwnerUserId="462" LastActivityDate="2011-01-12T23:21:17.413" CommentCount="5" />
  <row Id="17313" PostTypeId="2" ParentId="17310" CreationDate="2011-01-12T23:28:27.167" Score="16" ViewCount="" Body="&lt;p&gt;The numbers are irrational, as they are infinite non-repeated &quot;decimal&quot; numbers in base-$g$.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2011-01-12T23:28:27.167" CommentCount="2" />
  <row Id="17314" PostTypeId="2" ParentId="17132" CreationDate="2011-01-12T23:33:03.130" Score="1" ViewCount="" Body="&lt;p&gt;Here's an ugly argument that I know of (which works for abelian groups, and so I assume it works over arbitrary PID). You need to make three observations:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Maximal cyclic $p$-submodules of a $p$-module $M$ are the direct summands of $M$.&lt;/li&gt;&#xA;&lt;li&gt;Every cyclic $p$-submodule of $M$ is contained in a maximal cyclic $p$-submodule.&lt;/li&gt;&#xA;&lt;li&gt;If $N_1\subset M_1$ and $N_2\subset M_2$ are cyclic $p$-submodules of $M$, then $M_1$ and $M_2$ are disjoint if and only if $N_1$ and $N_2$ (this follows from observing that the submodules of a cyclic $p$-module are totally ordered under inclusion, and considering where $M_1\cap M_2$ is relative to $N_1$ and $N_2$ as a submodule of $M_1$ or $M_2$; you may have to use the fact that cyclic $p$-modules are indecomposable).&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Once you have proven those, take a direct sum decomposition of $N$ into cylic $p$-submodules $N_1\oplus N_2\oplus\dots\oplus N_i$ (which exists by structure theorem of finitely generated modules over a PID), and use 1., 2. and 3. to obtain a decomposition $M=M_1\oplus\dots M_n\oplus M&amp;#39;$, where $M_i$ is the maximal cyclic $p$-submodule that contains the cyclic $p$-submodule $N_i$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;*by $p$-module I mean a module annihilated by some power of the prime $p$ in the PID $R$ (it's the primality of $p$ that matters, not irreducibility, for the total ordering of submodules, though the two coincide over PIDs, which is where you need to be anyway for the structure theorem to work)&lt;/p&gt;&#xA;" OwnerUserId="400" LastActivityDate="2011-01-12T23:33:03.130" />
  <row Id="17315" PostTypeId="1" AcceptedAnswerId="17317" CreationDate="2011-01-12T23:58:21.937" Score="0" ViewCount="112" Body="&lt;p&gt;OK, so here's the question (not &lt;em&gt;my&lt;/em&gt; homework -  which is line segments -_-)&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;According to Chebyshev's theorem, how&#xA;  many standard deviations from the mean&#xA;  would make up the central 60% of&#xA;  scores for this class? [What are the&#xA;  corresponding grades? Answer the same&#xA;  questions for central 80%. Do these&#xA;  values capture more than the desired&#xA;  amount? Does this agree with&#xA;  Chebyshev's theorem?]&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(The stuff in brackets doesn't need to be answered. I included it there just in case.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance.&lt;/p&gt;&#xA;" OwnerUserId="2736" LastActivityDate="2011-01-13T00:15:13.553" Title="Chebyshev's theorem" Tags="&lt;homework&gt;&lt;standard-deviation&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="17316" PostTypeId="2" ParentId="17315" CreationDate="2011-01-13T00:06:42.907" Score="2" ViewCount="" Body="&lt;p&gt;This is the one that says the probability of being outside $k$ standard deviations is less than $\frac{1}{k^2}$.  So $1-0.6=\frac{1}{k^2}$.  Solve for $k$.  Then change .6 to .8.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-01-13T00:06:42.907" />
  <row Id="17317" PostTypeId="2" ParentId="17315" CreationDate="2011-01-13T00:08:23.047" Score="3" ViewCount="" Body="&lt;p&gt;So $P(|X-\mu| \geq k \sigma) \leq \frac{1}{k^2}$. The central $60 \%$ is $1-P(|X-\mu| \leq k \sigma) = 0.4$.  &lt;/p&gt;&#xA;" OwnerUserId="2548" LastEditorUserId="2548" LastEditorDisplayName="" LastEditDate="2011-01-13T00:15:13.553" LastActivityDate="2011-01-13T00:15:13.553" />
  <row Id="17318" PostTypeId="1" AcceptedAnswerId="17342" CreationDate="2011-01-13T00:48:04.400" Score="4" ViewCount="350" Body="&lt;p&gt;I am confused when reading Wikipedia's article on &lt;a href=&quot;http://en.wikipedia.org/wiki/Limit_superior_and_limit_inferior#Sequences_of_sets&quot; rel=&quot;nofollow&quot;&gt;limsup and liminf of a sequence of subsets of a set $X$&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;It says there are two different ways&#xA;to define them, but first gives &lt;strong&gt;what is common for the two&lt;/strong&gt;.&#xA; Quoted:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;There are two common ways to define&#xA;  the limit of sequences of set. In both&#xA;  cases:&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The sequence &lt;strong&gt;accumulates&lt;/strong&gt; around sets&#xA;  of points rather than single points&#xA;  themselves. That is, because each&#xA;  element of the sequence is itself a&#xA;  set, there exist &lt;strong&gt;accumulation sets&lt;/strong&gt;&#xA;  that are somehow nearby to infinitely&#xA;  many elements of the sequence.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The supremum/superior/outer limit is a&#xA;  set that joins these &lt;strong&gt;accumulation sets&lt;/strong&gt;&#xA;  together. That is, it is the union of&#xA;  all of the &lt;strong&gt;accumulation sets&lt;/strong&gt;. When&#xA;  ordering by set inclusion, the&#xA;  supremum limit is the least upper&#xA;  bound on the set of accumulation&#xA;  points because it contains each of&#xA;  them. Hence, it is the supremum of the&#xA;  limit points.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The infimum/inferior/inner limit is a&#xA;  set where all of these &lt;strong&gt;accumulation sets&lt;/strong&gt; &#xA;  meet. That is, it is the&#xA;  intersection of all of the&#xA;  &lt;strong&gt;accumulation sets&lt;/strong&gt;. When ordering by&#xA;  set inclusion, the infimum limit is&#xA;  the greatest lower bound on the set of&#xA;  accumulation points because it is&#xA;  contained in each of them. Hence, it&#xA;  is the infimum of the limit points.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The difference between the two&#xA;  definitions involves the topology&#xA;  (i.e., how to quantify separation) is&#xA;  defined. In fact, the second&#xA;  definition is identical to the first&#xA;  when the discrete metric is used to&#xA;  induce the topology on $X$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because it mentions that a sequence&#xA;of subsets of a set $X$ &lt;strong&gt;accumulate&lt;/strong&gt; to&#xA;some &lt;strong&gt;accumulation subsets&lt;/strong&gt; of $X$, are&#xA;there some &lt;strong&gt;topology on the power set&#xA;of the set&lt;/strong&gt; for this &lt;strong&gt;accumulation&lt;/strong&gt; to&#xA;make sense? What kind of topology is&#xA;that? Is it induced from some&#xA;structure on the set $X$? Is it possible&#xA;to use mathematic symbols to&#xA;formalize what it means by&#xA;&quot;supremum/superior/outer limit&quot;  and&#xA;&quot;infimum/inferior/inner limit&quot;?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If I understand correctly, here is&#xA;the first way to define&#xA;limsup/liminf of a sequence of&#xA;subsets. Quoted:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;General set convergence&lt;/strong&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;In this case, a sequence of sets&#xA;  approaches a limiting set when its&#xA;  elements of each member of the&#xA;  sequence approach that elements of the&#xA;  limiting set. In particular, if $\{X_n\}$&#xA;  is a sequence of subsets of $X$, then:&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$\limsup X_n$, which is also called the&#xA;  outer limit, consists of those&#xA;  elements which are limits of points in&#xA;  $X_n$ taken from (countably) infinitely&#xA;  many n. That is, $x \in \limsup X_n$ if and&#xA;  only if there exists a sequence of&#xA;  points $x_k$ and a subsequence $\{X_{n_k}\}$ of&#xA;  $\{X_n\}$ such that $x_k \in X_{n_k}$ and $x_k \rightarrow x$ as&#xA;  $k \rightarrow \infty$.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$\liminf X_n$, which is also called the&#xA;  inner limit, consists of those&#xA;  elements which are limits of points in&#xA;  $X_n$ for all but finitely many n (i.e.,&#xA;  cofinitely many n). That is, $x \in \liminf X_n$&#xA;   if and only if there exists a&#xA;  sequence of points $\{x_k\}$ such that $x_k \in X_k$&#xA;   and $x_k \rightarrow x$ as $k \rightarrow \infty$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So I think for this definition, $X$ is&#xA;required to be a topological space.&#xA;This definition is expressed in&#xA;terms of convergence of a sequence&#xA;of points in $X$ with respect to the&#xA;topology of $X$. If referring back to&#xA;what is common for the two ways of&#xA;definitions, I will be wondering how&#xA;to explain what is a &lt;strong&gt;&quot;accumulation&#xA;set&quot;&lt;/strong&gt; in this definition here and&#xA;what topology the &lt;strong&gt;&quot;accumulation set&quot;&lt;/strong&gt;&#xA;is with respect to? i.e. how can the&#xA;definition here fit into&#xA;aforementioned &lt;strong&gt;what is common for&#xA;the two ways&lt;/strong&gt;?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;It says there are two ways to define&#xA;the limit of a sequence of subsets&#xA;of a set $X$. But there seems to be&#xA;just one in the article, as quoted&#xA;in 2. So I was wondering what is the&#xA;second way it refers to?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you might give your answer, here&#xA;is my thought/guess (which has&#xA;actually been written in the article&#xA;but not in a way saying it is the&#xA;second one). Please correct me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In an arbitrary complete lattice, by&#xA;viewing meet as inf and join as sup,&#xA;the limsup of a sequence of points&#xA;$\{x_n\}$ is defined as: $$\limsup&#xD;&#xA;    \, x_n =  \inf_{n \geq 0}&#xD;&#xA;    \left(\sup_{m \geq n} \, x_m\right)&#xD;&#xA;    = \mathop{\wedge}\limits_{n \geq 0}\left( \mathop{\vee}\limits_{m\&#xD;&#xA;    \geq n} \, x_m\right) $$ similarly&#xA;define liminf.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;The power set of any set is a&#xA;complete lattice with union and&#xA;intersection being join and meet, so&#xA;the liminf and limsup of a sequence&#xA;of subsets can be defined in the&#xA;same way.&lt;/em&gt; I was wondering if this is&#xA;the other way the article tries to&#xA;introduce? If it is, then this&#xA;second way of definition does not&#xA;requires $X$ to be a topological&#xA;space. So how can this second way&#xA;fits to &lt;strong&gt;what is common for the two&#xA;ways&lt;/strong&gt; in Part 1, which seems to requires some&#xA;kind of &lt;strong&gt;topology on the power set of&#xA;$X$&lt;/strong&gt;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand this way of definition&#xA;can be shown to be equivalent to a&#xA;special case of the first way  in my&#xA;part 2         when the topology on&#xA;$X$ is induced by discrete metric.&#xA;This is another reason that let me&#xA;doubt it is the second way, because&#xA;I guess the second way should at&#xA;least not be equivalent to a special&#xA;case of the first way.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Can the two ways of definition fit&#xA;into any definition for &lt;a href=&quot;http://en.wikipedia.org/wiki/Limit_superior_and_limit_inferior#Generalized_definitions&quot; rel=&quot;nofollow&quot;&gt;the general&#xA;cases&lt;/a&gt;? In the general cases,&#xA;limsup/liminf is defined for a&#xA;sequence of points in a set with&#xA;some structure. Can limsup/liminf of&#xA;a sequence of subsets of a set be&#xA;viewed as limsup/liminf of a&#xA;sequence of &quot;points&quot;. If not, so in some&#xA;cases, a sequence of subsets must be&#xA;treated just as a sequence of subsets, but not&#xA;as a sequence of &quot;points&quot;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt;   @Arturo: In the last part of &lt;a href=&quot;http://math.stackexchange.com/questions/4705/limit-inferior-and-superior-for-sets-vs-real-numbers/4747#4747&quot;&gt;your reply&lt;/a&gt; to another question,&#xA;did you try to explain how&#xA;limsup/liminf of a sequence of&#xA;points can be viewed as&#xA;limsup/liminf of a sequence of&#xA;subsets? I actually want to&#xA;understand in the opposite&#xA;direction:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is &lt;a href=&quot;http://math.stackexchange.com/questions/17347/limsup-of-a-sequence-of-points-in-a-set&quot;&gt;a post&lt;/a&gt; with my current&#xA;knowledge about limsup/liminf of a&#xA;sequence of points in a set. For&#xA;limsup/liminf of a sequence of&#xA;subsets of any set $X$, defined in&#xA;terms of union and intersection of&#xA;subsets of $X$ as in part 3, it can&#xA;be viewed as limsup/liminf of a&#xA;sequence of points in a complete&#xA;lattice, by viewing the power set of&#xA;$X$ as a complete lattice. But for&#xA;limsup/liminf of a sequence of&#xA;subsets of any set defined in part 2&#xA;when X is a topological space, I was&#xA;wondering if there is some way to&#xA;view it as limsup/liminf of a&#xA;sequence of points in some set?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;It is also great if you have other approaches to understand all the ways of defining limsup/liminf of a sequence of subsets, other than the approach in Wikipedia.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks and regards!&lt;/p&gt;&#xA;" OwnerUserId="1281" LastEditorUserId="1281" LastEditorDisplayName="" LastEditDate="2011-01-13T18:26:51.220" LastActivityDate="2011-01-13T18:26:51.220" Title="limsup and liminf of a sequence of subsets of a set" Tags="&lt;topology&gt;&lt;elementary-set-theory&gt;&lt;order-theory&gt;&lt;lattices&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="17319" PostTypeId="2" ParentId="17311" CreationDate="2011-01-13T00:53:02.803" Score="3" ViewCount="" Body="&lt;p&gt;One approach to questions of this general nature is to find a groupoid whose &lt;a href=&quot;http://ncatlab.org/nlab/show/groupoid+cardinality&quot; rel=&quot;nofollow&quot;&gt;groupoid cardinality&lt;/a&gt; is equal to the number in question; this is a form of &lt;a href=&quot;http://math.ucr.edu/home/baez/groupoidification/&quot; rel=&quot;nofollow&quot;&gt;groupoidification&lt;/a&gt;.  For example, the groupoid cardinality of the groupoid $\text{FinSet}_0$ of finite sets and bijections is $\sum \frac{1}{n!} = e$, so this is a reasonable categorification of $e$; in fact arguably this is &quot;the&quot; categorification of $e$ and goes a long way towards explaining its prevalence in mathematics.  Similarly, for any finite set $X$, the groupoid cardinality of the groupoid of $X$-colored finite sets and color-preserving bijections is $\sum \frac{|X|^n}{n!} = e^{|X|}$.  Note that this groupoid is the coproduct of $|X|$ copies of $\text{FinSet}_0$; see also &lt;a href=&quot;http://math.stackexchange.com/questions/3507/purely-combinatorial-proof-that-d-dxex-ex&quot;&gt;this math.SE question&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a TWF which I can't find (because John Baez made them &lt;em&gt;very hard&lt;/em&gt; to search through for some reason!) John Baez mentions that he looked into the problem of finding a &quot;natural&quot; groupoid whose cardinality is $\pi$, but that he (and possibly some collaborators) weren't able to come up with any nice examples.  So possibly this is the wrong direction to go in the particular case of $\pi$.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2011-01-13T00:53:02.803" CommentCount="8" />
  <row Id="17320" PostTypeId="1" AcceptedAnswerId="17324" CreationDate="2011-01-13T01:03:28.127" Score="1" ViewCount="300" Body="&lt;p&gt;I'm looking for the method by which the partial derangement formula $D_{n,k}$ was derived. I can determine the values for small values of N empirically, but how the general case formula arose still alludes me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any links/books or an explanation will be appreciated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The formula is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$D_{n,k} = {n \choose k}!(n-k)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Links:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://mathworld.wolfram.com/PartialDerangement.html&quot; rel=&quot;nofollow&quot;&gt;Mathworld&lt;/a&gt; &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="5760" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2011-02-07T13:52:22.890" LastActivityDate="2011-02-07T13:52:22.890" Title="Derivation of the Partial Derangement (Rencontres numbers) formula" Tags="&lt;combinatorics&gt;&lt;permutations&gt;" AnswerCount="2" CommentCount="8" />
  <row Id="17321" PostTypeId="2" ParentId="17318" CreationDate="2011-01-13T01:07:30.820" Score="1" ViewCount="" Body="&lt;p&gt;I always found the following definitions of superior limits and inferior limits intuitive: Let $\{E_n \}$ be a sequence of sets ($n = 1,2, \dots$). The superior limit of $\{E_n \}$ is the set consisting of those points which belong to infinitely many $E_n$. The inferior limit is the set of all those points that belong to all but a finite number of the $E_n$. &lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2011-01-13T01:07:30.820" CommentCount="1" />
  <row Id="17322" PostTypeId="2" ParentId="17236" CreationDate="2011-01-13T01:12:21.267" Score="2" ViewCount="" Body="&lt;p&gt;A straightforward search yields the figure of $4392$ for all but the $1 \times 1$ stones. The former fill $14$ out of $20$ squares, so there are $\binom{6}{4} = 15$ possibilities to place the latter. In total, we get $$4392 \times 15 = 65880.$$ These can all be generated, and one can in principle calculate the number of connected components in the resulting graph, where the edges correspond to movements of pieces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: There are 898 different connected components. There are 25955 configurations reachable from the initial state.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2011-01-13T03:20:04.300" LastActivityDate="2011-01-13T03:20:04.300" CommentCount="7" />
  <row Id="17323" PostTypeId="1" AcceptedAnswerId="17344" CreationDate="2011-01-13T01:23:13.327" Score="2" ViewCount="141" Body="&lt;p&gt;Consider a simple random walk on the integers with a reflecting barrier at $0$. What is the expected time the walk spends in the set $\{0, \ldots, k\}$ in the first $n$ steps? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm curious about how the answer scales with $n,k$, and don't really care about the constants.  &lt;/p&gt;&#xA;" OwnerUserId="2120" LastActivityDate="2011-01-13T05:00:00.373" Title="random walk on the nonnegative integers" Tags="&lt;probability&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="17324" PostTypeId="2" ParentId="17320" CreationDate="2011-01-13T01:33:48.463" Score="4" ViewCount="" Body="&lt;p&gt;Here is a very general solution.  There is a fundamental formula in combinatorics called the exponential formula, and one statement of it is as follows.  Given a finite group $G$ acting on a set $X$, its &lt;a href=&quot;http://en.wikipedia.org/wiki/Cycle_index&quot; rel=&quot;nofollow&quot;&gt;cycle index polynomial&lt;/a&gt; is given by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$Z_G = \frac{1}{|G|} \sum_{g \in G} z_1^{c_1(g)} z_2^{c_2(g)} ... $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $c_i(g)$ is the number of cycles of length $i$ in the action of $g$ on $X$.  In particular, the notation $Z_{S_n}$ will denote the cycle index polynomial of $S_n$ acting on an $n$-element set in the usual way; it is a generating function encoding the relative proportions of different cycle types of permutations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://qchu.wordpress.com/2009/06/24/gila-vi-the-cycle-index-polynomials-of-the-symmetric-groups/&quot; rel=&quot;nofollow&quot;&gt;exponential formula&lt;/a&gt; then states that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{n \ge 0} Z_{S_n} t^n = \exp \left( z_1 t + \frac{z_2 t^2}{2} + \frac{z_3 t^3}{3} + ... \right).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In my opinion this is one of the most beautiful formulas in mathematics and a major reason I became interested in combinatorics was because I stumbled upon this formula while solving a Putnam problem (which is described in the blog post I linked to above).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How does it apply to this problem?  Set $z_2 = z_3 = ... = 1$ and $z_1 = z$.  Then the LHS of the exponential formula is a generating function which counts permutations according to how many fixed points ($1$-cycles) they have.  In other words, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$Z_{S_n}(z, 1, 1, ...) = \frac{1}{n!} \sum_{g \in S_n} z^{c_1(g)} = \frac{1}{n!} \sum_{k=0}^n D_{n,k} z^k.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The RHS of the exponential formula, on the other hand, is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\exp \left( zt + \log \frac{1}{1-t} - t \right) = \frac{e^{-t}}{1 - t} e^{zt}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we obtain the beautifully concise formula&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{n \ge 0} \frac{t^n}{n!} \sum_{k=0}^n D_{n,k} z^k = \frac{e^{-t}}{1 - t} e^{zt}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The coefficients of $\frac{e^{-t}}{1 - t}$ are obtained by setting $z = 0$; they give the usual derangement numbers, e.g. the number of permutations of $n$ elements with no fixed points, and this can also be seen directly from the generating function since&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{e^{-t}}{1 - t} = \sum_{n \ge 0} \left( \sum_{k=0}^n \frac{(-1)^k}{k!} \right) t^n$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is equivalent to the formula $D_{n,0} = n! \sum_{k=0}^n \frac{(-1)^k}{k!} \sim \frac{n!}{e}$.  (In fact you can read this asymptotic directly from the generating function.)  The above then gives&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$D_{n,k} = {n \choose n-k} D_{n-k,0} = \frac{n!}{k!} \sum_{i=0}^{n-k} \frac{(-1)^i}{i!}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, there is a much more direct proof of this: observe that specifying a permutation of $n$ elements with $k$ fixed points is equivalent to specifying the $n-k$ elements which are not fixed points, then specifying a fixed-point-free permutation of these.  This immediately gives $D_{n,k} = {n \choose n-k} D_{n-k,0}$, so it suffices to compute $D_{n,0}$, and this can be done by the standard inclusion-exclusion argument. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(In the interest of completeness, the standard inclusion-exclusion argument is as follows: first we start with all $n!$ permutations.  Then we subtract the ones which fix $1$, and the ones which fix $2$, etc., so we subtract $n \cdot (n-1)!$.  But this is overcounting: we need to add back the ones which fix both $1$ and $2$, or more generally both $i$ and $j$ for distinct $i, j$, so we add back ${n \choose 2} \cdot (n-2)!$.  But this is overcounting: we need to subtract the ones which fix any triplet... and so forth.  This gives each term of the formula $n! \sum_{k=0}^n \frac{(-1)^k}{k!}$ one-by-one.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My point in presenting the generating function argument is not that it is any easier in this case but that it generalizes to far more complicated problems in a way which minimizes mental effort: for example you can use it to count permutations by how many $2$-cycles they have, or by $c_3(g) + 17 c_5(g)$, or whatever, and the generating function is also a convenient way to organize the computation of the expected value and variance of various permutation statistics; see, for example, &lt;a href=&quot;http://math.stackexchange.com/questions/1416/algebra-of-random-variables&quot;&gt;this math.SE answer&lt;/a&gt;.  &lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2011-01-13T01:33:48.463" CommentCount="1" />
  <row Id="17325" PostTypeId="1" AcceptedAnswerId="17332" CreationDate="2011-01-13T01:45:44.097" Score="3" ViewCount="207" Body="&lt;p&gt;Given a function $g(x)$, is it possible to find a function $f(x)$ that satisfies&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ f(x+1) - f(x-1) = g(x) $$&lt;/p&gt;&#xA;" OwnerUserId="1393" LastActivityDate="2011-01-13T05:07:58.220" Title="Solving the functional equation $f(x+1) - f(x-1) = g(x)$" Tags="&lt;functions&gt;&lt;functional-equations&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="1" />
  <row Id="17326" PostTypeId="1" CreationDate="2011-01-13T01:56:45.207" Score="0" ViewCount="92" Body="&lt;p&gt;[1)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm wondering if there are two angles in an&#xA;'angle.'&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, the image below would typically be&#xA;identified as a right angle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But couldn't you also say that it's a right angle&#xA;&lt;strong&gt;and&lt;/strong&gt; a 270-degree reflex angle? There are two&#xA;angles?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://tinyurl.com/4fs85kg&quot; rel=&quot;nofollow&quot;&gt;http://tinyurl.com/4fs85kg&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;[2)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or-- An equiangular triangle is defined (?) as a&#xA;triangle with three 60-degree angles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But couldn't you also say that an equiangular&#xA;triangle has three 300-degree reflex angles? Six&#xA;angles?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;[3)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would the rays that form an angle be considered&#xA;straight angles?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If so: Since a ray consists of infinitely many&#xA;points.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And since any two points form a line segment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And since a line segment can be considered a&#xA;straight angle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does it follow that an angle consists of infinitely&#xA;many angles?&lt;/p&gt;&#xA;" OwnerUserId="5761" LastActivityDate="2011-01-13T02:22:26.727" Title="How Many Angles in an Angle?" Tags="&lt;geometry&gt;" AnswerCount="2" />
  <row Id="17327" PostTypeId="2" ParentId="17326" CreationDate="2011-01-13T02:00:46.777" Score="1" ViewCount="" Body="&lt;p&gt;It depends on how you measure angles. If you measure counterclockwise, you will get the usual angles. If you measure clockwise, you will get the other ones you mention.&lt;/p&gt;&#xA;" OwnerUserId="2548" LastEditorUserId="2548" LastEditorDisplayName="" LastEditDate="2011-01-13T02:10:10.943" LastActivityDate="2011-01-13T02:10:10.943" />
  <row Id="17328" PostTypeId="2" ParentId="17325" CreationDate="2011-01-13T02:03:42.090" Score="4" ViewCount="" Body="&lt;p&gt;Let $f(x)=\frac{1}{2}x g(0)$ for $-1\le x \le 1$, $f(x)=g(x-1) + f(x-2)$ for $x&amp;gt;1$, and $f(x)=f(x+2)-g(x+1)$ for $x&amp;lt;-1$.&lt;/p&gt;&#xA;" OwnerUserId="5546" LastEditorUserId="5546" LastEditorDisplayName="" LastEditDate="2011-01-13T04:16:46.910" LastActivityDate="2011-01-13T04:16:46.910" />
  <row Id="17329" PostTypeId="1" AcceptedAnswerId="17330" CreationDate="2011-01-13T02:04:17.957" Score="0" ViewCount="43" Body="&lt;p&gt;How would I go about solving a problem that goes roughly as such?&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Solution A: .1%&#xA;Solution B: .15%&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;How many milliliters of each A &amp;amp; B are required to create 260 milliliters of a solution, C, that is .14%?&lt;/code&gt;&lt;/p&gt;&#xA;" OwnerUserId="4066" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2011-01-13T03:32:29.220" LastActivityDate="2011-01-13T03:32:29.220" Title="Determine concentration." Tags="&lt;algebra-precalculus&gt;&lt;arithmetic&gt;" AnswerCount="1" />
  <row Id="17330" PostTypeId="2" ParentId="17329" CreationDate="2011-01-13T02:05:54.147" Score="2" ViewCount="" Body="&lt;p&gt;Solve the following system:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$x+y = 260$&lt;/li&gt;&#xA;&lt;li&gt;$0.1x+0.15y = (0.14)(260)$&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="2548" LastEditorUserId="2548" LastEditorDisplayName="" LastEditDate="2011-01-13T02:50:27.633" LastActivityDate="2011-01-13T02:50:27.633" CommentCount="2" />
  <row Id="17331" PostTypeId="2" ParentId="17326" CreationDate="2011-01-13T02:22:26.727" Score="0" ViewCount="" Body="&lt;p&gt;You could also say it's a 90+360 = 450 degree angle, or any number $90+360n$ where n is a natural number. The point is, we say that it's 90 degrees clockwise or 270 mostly by convention, but yes, &quot;an angle consists of infinitely many angles&quot; is true.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly for 2) you called it an &quot;equiangular&quot; triangle and so indeed by definition it must have three equal angles, however you measure them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A &quot;straight angle&quot; would just be an angle of zero (or 360, or  720...) degrees.&lt;/p&gt;&#xA;" OwnerUserId="2954" LastActivityDate="2011-01-13T02:22:26.727" />
  <row Id="17332" PostTypeId="2" ParentId="17325" CreationDate="2011-01-13T02:33:39.790" Score="8" ViewCount="" Body="&lt;p&gt;Pick any choice of $f$ on the interval $[-1,1)$. By using the property $f(x) = g(x-1) + f(x-2)$, you can extend this definition to $[1,3)$, and so on to the rest of the positive real line. Similarly, you can extend $f$ to the negative real line using $f(x) = f(x+2) - g(x+1)$. This is one solution to your equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, a different choice of $f$ on $[-1,1)$ will give you a different result. This corresponds to the fact that adding any periodic function $h$ with period 2 to $f$ still satisfies the original equation. In other words, there are infinitely many solutions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For particular restricted classes of functions, as Qiaochu alludes to, there may be a single canonical solution. For example, if $g$ is a polynomial of degree $n$, there is a unique polynomial solution for $f$. This is a polynomial of degree $n+1$, and you can find it by expanding out the original equation and comparing powers starting with the highest. A similar fact holds for exponentials and trigonometric functions (except of course when the trigonometric functions have period 2, in which case you would have to include multiples of $x \sin \pi x$ and $x \cos \pi x$ in $f$ to get a solution).&lt;/p&gt;&#xA;" OwnerUserId="856" LastEditorUserId="856" LastEditorDisplayName="" LastEditDate="2011-01-13T02:44:55.637" LastActivityDate="2011-01-13T02:44:55.637" />
  <row Id="17333" PostTypeId="1" AcceptedAnswerId="17336" CreationDate="2011-01-13T02:34:58.183" Score="1" ViewCount="122" Body="&lt;p&gt;There's a 30 year home loan for $100,000 at 7%. After 15 years the loan is paid off in order to refinance at a lower rate. The loan has a prepayment penalty of six months interest of 80% of the remaining balance of the loan.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a) How much is the remaining balance of the loan?&#xA;b) If the loan can be refinanced with a 15 year loan at 6% with no other costs, then should it be done?&#xA;c) If the loan can be refinanced over 15 yrs with no other costs, then what interest rate would make it worthwhile?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe I got (a)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;month   payment interest principal remaining&lt;br&gt;&#xA;180 | 665.30 | 433.14 | 232.16 | 74019.66&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Not sure about b or c though.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I attempted b by taking 74019.66 and making that my new loan. Find a new payment across 15 years at 6%, which is 624.62. I figured $665.30-624.62 = 40.68$ in savings per month. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One thing I don't know is does prepayment go into the new loan or do you pay out of pocket. If you pay out of pocket then you would save $\$7322.4 - \$2072.55$ (prepayment penalty).&lt;/p&gt;&#xA;" OwnerUserId="3961" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-13T02:49:41.167" LastActivityDate="2011-01-13T12:27:57.250" Title="mortgage math problem" Tags="&lt;finance&gt;" AnswerCount="2" />
  <row Id="17335" PostTypeId="1" CreationDate="2011-01-13T02:40:06.417" Score="0" ViewCount="61" Body="&lt;p&gt;The problem is: Suppose $\phi_1(x)$ and $\phi_2(x)$ are solutions of the differential equation (1):&#xA; $$ y&amp;#39;&amp;#39;(x)+p(x)y&amp;#39;(x)+q(x)y(x)=0 $$&#xA;on the interval $I$. Show that $\lambda(x)=c_1\phi_1(x)+c_2\phi_2(x)$, where $c_1$ and $c_2$ denote arbitrary constants, is also a solution of (1) on the interval $I$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I did was to take the first and second derivative of $\lambda$ and to plug those in for $y&amp;#39;&amp;#39;$, $y&amp;#39;$, and $y$ respectively: $\lambda&amp;#39;= c_1\phi_1&amp;#39;(x)+c_2\phi_2&amp;#39;(x)$ and &#xA;$\lambda&amp;#39;&amp;#39;= c_1\phi_1&amp;#39;&amp;#39;(x)+c_2\phi_2&amp;#39;&amp;#39;(x)$. Then I got the equation&#xA; $$ c_1(\phi_1&amp;#39;&amp;#39;(x)+p(x)\phi_1&amp;#39;(x)+q(x)\phi_1(x)) = c_2(\phi_2&amp;#39;&amp;#39;(x)+p(x)\phi_2&amp;#39;(x)+q(x)\phi_2(x)). $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am having trouble proving that these two are equal since $\phi_1$ and $\phi_2$ could be different. I am looking for some direction as for where to go from here or if I went wrong on one of my steps. Thanks.&lt;/p&gt;&#xA;" OwnerUserId="5765" LastEditorUserId="856" LastEditorDisplayName="" LastEditDate="2011-01-13T02:50:02.247" LastActivityDate="2011-01-13T03:05:34.463" Title="Differential Equation Help, I'm Stuck and Need Help Figuring out the Next Step" Tags="&lt;differential-equations&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="17336" PostTypeId="2" ParentId="17333" CreationDate="2011-01-13T02:45:13.973" Score="1" ViewCount="" Body="&lt;p&gt;I think they want you to increase the balance by the prepayment penalty (so there is no out of pocket cost), then see if the payment is higher or lower.  So for b) you figure the new payment on 74019.66+2072.55 and see if it is lower.  For c) you find the interest rate on the same balance that keeps the payment the same as today.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure I agree with this criterion for a decision in the real world-there is too much chance you will end up paying the loan off early so having the balance higher is a negative.  But I think it is the way you understand the technique in the absence of other instructions.  It might help to clearly lay out your assumptions.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-01-13T02:45:13.973" CommentCount="1" />
  <row Id="17337" PostTypeId="2" ParentId="17335" CreationDate="2011-01-13T02:49:46.277" Score="1" ViewCount="" Body="&lt;p&gt;Katie: What does it mean that $\phi$ is a solution of an equation? It means that whatever the equation says, it is true if instead of the variable you use $\phi$. In this case the &quot;variable&quot; is the function $y(x)$ in (1), so to say that $\phi$ is a solution means that the following is true:&#xA; $$ \phi&amp;#39;&amp;#39;(x)+p(x)\phi&amp;#39;(x)+q(x)\phi(x)=0. $$&#xA;You are told that this is true when $\phi$ is $\phi_1$ and also when $\phi$ is $\phi_2$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you need to prove is that it is true when $\phi$ is $\lambda$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you have done is correct so far, but not enough. I am not sure why you think the equation at the end is what you need. It happens to be true, but it seems strange to focus on it. A big hint towards what needs to be done is that $0=c_10+c_20$.&lt;/p&gt;&#xA;" OwnerUserId="462" LastActivityDate="2011-01-13T02:49:46.277" />
  <row Id="17338" PostTypeId="2" ParentId="17335" CreationDate="2011-01-13T02:50:34.607" Score="0" ViewCount="" Body="&lt;p&gt;You are (almost) done, but seem to have dropped a minus sign.  You should be looking at the sum of the two sides of your equal sign.  As you are told that $\phi_1$ and $\phi_2$ are solutions to the equation, the expressions in parentheses are zero.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-01-13T02:50:34.607" />
  <row Id="17339" PostTypeId="2" ParentId="17335" CreationDate="2011-01-13T02:55:54.523" Score="1" ViewCount="" Body="&lt;p&gt;Put $\rm\displaystyle\: D = \frac{d}{dx}\:.\: $ Since $\rm\ L = D^2 + p\ D + q\ $ is $\rm\:\mathbb C$-&lt;em&gt;linear&lt;/em&gt;, $\rm\ L\:(c_1 \phi_1 + c_2\: \phi_2)\: =\: c_1\ L(\phi_1) + c_2\ L(\phi_2) = 0\:.$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That $\rm\: L\:$ is $\mathbb C$-linear follows simply from the fact that it is a sum of compositions of $\mathbb C$-linear maps, namely the derivative $\rm\ f \to D(f) = f\:&amp;#39;\:,\ $ and the linear maps $\rm f \to p\ f,\ \ f\to q\ f\:.$ &lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2011-01-13T03:05:34.463" LastActivityDate="2011-01-13T03:05:34.463" />
  <row Id="17341" PostTypeId="2" ParentId="17152" CreationDate="2011-01-13T03:42:07.663" Score="-5" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;This has already happened.&lt;/strong&gt; Millions of years ago we started with a finite number of monkeys (well, primates). It took them only a few million years to produce the works of Shakespeare.&lt;/p&gt;&#xA;" OwnerUserId="5619" LastActivityDate="2011-01-13T03:42:07.663" CommunityOwnedDate="2011-01-15T06:13:52.160" CommentCount="4" />
  <row Id="17342" PostTypeId="2" ParentId="17318" CreationDate="2011-01-13T03:49:13.670" Score="2" ViewCount="" Body="&lt;p&gt;First, you might also want to take a look at &lt;a href=&quot;http://math.stackexchange.com/questions/4705/limit-inferior-and-superior-for-sets-vs-real-numbers/4747#4747&quot;&gt;this answer&lt;/a&gt; to a similar question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Okay: the first description assumes that there is some sort of notion of &quot;accumulation point&quot; at work in the set $X$, as you surmise; this may be derived from a topology. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second description talks about limit points, but you can apply it to any set by endowing the set with the discrete topology (every subset is open, every subset is closed). If you do that, then the definition is the &lt;em&gt;usual&lt;/em&gt; definition of limit superior of a sequence of sets: it is the collection of all points that are in infinitely many of the terms of the sequence, while the limit inferior is the collection of all points that are in all sufficiently large terms of the sequence. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &quot;second way&quot; of defining it is in terms of unions and intersection. If $\{X_n\}_{n\in\mathbb{N}}$ is a family of sets, then&#xA;\begin{align*}&#xA;\limsup_{n\in\mathbb{N}} X_n &amp;amp;= \bigcap_{n=1}^{\infty}\left(\bigcup_{j=n}^{\infty} X_j\right)\\&#xA;\liminf_{n\in\mathbb{N}} X_n &amp;amp;= \bigcup_{n=1}^{\infty}\left(\bigcap_{j=n}^{\infty} X_j\right).&#xA;\end{align*}&#xA;This coincides with the notion of the limit superior being the set of all limit points of infinitely many terms in the sequence, under the discrete topology; and the limit inferior being the set of all limit points of all sufficiently large-indexed terms of the sequence (again, under the discrete topology).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The notion of &quot;accumulation point&quot; in the first description is more informal. If you are working with a topological space, then it is limit points as described above and by &quot;accumulation set&quot; you should read &quot;set of all limit points&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For your third point, in order to be able to talk about joins and meets you need to have some sort of &lt;em&gt;complete&lt;/em&gt; lattice order on your set, so that you can talk about those infinite meets and infinite joins; this is the case, for instance, in the real numbers; appropriately interpreted, you do get essentially the definition you propose, though you need to tweak it a bit in order to actually get what the actual definition is (see the other answer quoted above); you don't actually work with the points themselves, but with a slightly different set determined by the points. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think that the previous answer linked to answers essentially your fourth point, of how to interpret limit superior and limit inferior of a sequence of points as a special case of limit superior and limit inferior of sets; but if this is not the case, point it out and I'll try to answer it de nuovo. &lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2011-01-13T03:49:13.670" CommentCount="2" />
  <row Id="17343" PostTypeId="1" CreationDate="2011-01-13T04:46:32.560" Score="0" ViewCount="145" Body="&lt;p&gt;How to characterize the speed of a divergent series ? I have a divergent series with a parameter $x$ in it. How can i characterize the speed of divergence for different $x$ ?&lt;/p&gt;&#xA;" OwnerUserId="2987" LastActivityDate="2011-01-13T05:51:53.993" Title="what is the speed of a divergent series ?" Tags="&lt;sequences-and-series&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="1" />
  <row Id="17344" PostTypeId="2" ParentId="17323" CreationDate="2011-01-13T05:00:00.373" Score="4" ViewCount="" Body="&lt;p&gt;As a commenter noted, this is the same as the total time spent in $[-k,k]$ by a symmetric random walker in the first $n$ steps.  The probability distribution for large $n$ becomes $P(x) = \frac{2}{\sqrt{2\pi n}}\exp\left(-\frac{x^2}{2n}\right)$ for $x$ with the same parity as $n$, and is always $P(x) = 0$ for $x$ with opposite parity.  The probability of being in a fixed interval $[-k,k]$ after exactly $n$ steps is therefore $\Theta\left(k/\sqrt{n}\right)$ as $n \rightarrow \infty$, as the exponential term becomes irrelevant.  Summing this probability over steps $1, 2, ... , n$ gives the desired result, which is $\Theta\left(k\sqrt{n}\right)$.&lt;/p&gt;&#xA;" OwnerUserId="5546" LastActivityDate="2011-01-13T05:00:00.373" />
  <row Id="17345" PostTypeId="2" ParentId="17325" CreationDate="2011-01-13T05:07:58.220" Score="1" ViewCount="" Body="&lt;p&gt;There might be something to gain by interpreting $g(x)$ as being an approximation to the derivative of $f(x)$, (barring a factor of 2). It's not really the derivative, but the centered finite-difference.&lt;/p&gt;&#xA;" OwnerUserId="2256" LastActivityDate="2011-01-13T05:07:58.220" />
  <row Id="17346" PostTypeId="1" CreationDate="2011-01-13T05:13:15.183" Score="3" ViewCount="84" Body="&lt;p&gt;I know that for any surface, the Gaussian curvature $K$ and mean curvature $H$ satisfy the inequality $H^2 \geq K$ , and the sphere is a surface where that inequality becomes an equation. Thus, the sphere has both constant Gaussian and mean curvature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there other surfaces whose Gaussian and mean curvatures are constant and nonzero?&lt;/p&gt;&#xA;" OwnerUserId="5774" LastActivityDate="2011-01-13T10:38:23.693" Title="How does one characterize surfaces with constant nonzero Gaussian and mean curvature." Tags="&lt;differential-geometry&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="17347" PostTypeId="1" AcceptedAnswerId="17504" CreationDate="2011-01-13T05:13:49.640" Score="3" ViewCount="390" Body="&lt;p&gt;&lt;strong&gt;My ways to define/write limsup and liminf of a sequence of points in a set $X$&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They come from what I have understood. If you have other ways of understanding, really appreciate if you can reply here. &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;When $X$ is a complete lattice, then&#xA;$$\limsup \, x_n :=  \inf_{n \geq 0}&#xD;&#xA;        \left(\sup_{m \geq n} \, x_m\right)&#xD;&#xA;        = \mathop{\wedge}\limits_{n \geq 0}\left( \mathop{\vee}\limits_{m\&#xD;&#xA;        \geq n} \, x_m\right). $$&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;When $X$ is a topological space and&#xA;also a partially ordered set (can be&#xA;strengthened to be complete in some&#xA;sense, such as a complete lattice if&#xA;needed), then there are different ways to&#xA;define/write:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;(1) $\displaystyle\limsup \, x_n :=  \sup&#xD;&#xA;        \{\text{limit of any subsequence of&#xD;&#xA;        }\{x_n\}\text{ that converges}\}$&lt;/li&gt;&#xA;&lt;li&gt;(2) $\displaystyle\limsup \, x_n := \lim_{n&#xD;&#xA;        \rightarrow \infty} ( \sup_{m \geq&#xD;&#xA;        n}x_m )$.&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;$\liminf x_n$ is defined similarly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Questions:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Are there some conditions on $X$&#xA;that can make 2(1) and 2(2) equivalent?&lt;/li&gt;&#xA;&lt;li&gt;Are there some conditions on $X$&#xA;that can make 1 and 2(1) equivalent? For example, when $X$ is both a topological space and a complete lattice, I can define limsup in both 1 and 2(1) ways, so when will they agree?&lt;/li&gt;&#xA;&lt;li&gt;Are there other definitions/forms of&#xA;limsup of a sequence of points in a&#xA;set $X$ possibly with some&#xA;structures? Can they be related/shown to be equivalent to the definitions/forms given above?&lt;/li&gt;&#xA;&lt;li&gt;Is 2(1) exactly the&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Limit_superior_and_limit_inferior#Specialization_for_sequences_and_nets&quot; rel=&quot;nofollow&quot;&gt;specialization&lt;/a&gt; from &lt;a href=&quot;http://en.wikipedia.org/wiki/Limit_superior_and_limit_inferior#Definition_for_filter_bases&quot; rel=&quot;nofollow&quot;&gt;limsup of a&#xA;filter base in $X$&lt;/a&gt; to limsup of a sequence of points in $X$:&#xA; $$\limsup \, x_n :=  \sup \cap \{\bar{B}_0 | B_0 \in B  \}$$&#xA; where $B:=\{ \{x_n | n \geq n_0 \} \forall n_0 \in \mathbb{N}  \}$,&#xA; and $\bar{B}_0$ is the closure of $B_0$?&lt;/li&gt;&#xA;&lt;li&gt;When $X$ is a topological space and&#xA;ordered, if $\limsup \, x_n =&#xD;&#xA;    \liminf \, x_n$, both of which are defined in the same way as 2(1),&#xA;what is the condition for $\limsup&#xD;&#xA;    \, x_n = \liminf \, x_n$ to be the&#xA;same as $\lim \, x_n$ defined with&#xA;respect to the topology?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Thanks and regards!&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Self Answer to Question 4:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As to Question 4, I think it is. $\forall B_0 \in B$, it is a tail sequence of $\{ x_n \}$, so any accumulation point of $B_0$ will be an accumulation point of any other tail sequence, and an accumulation point of the original sequence $\{ x_n \}$. Therefore &#xA;\begin{align*}&#xA;\cap \{\bar{B}_0 | B_0 \in B  \} &amp;amp;= \{\text{accumulation point of }\{ x_n \}  \}\\&#xA; &amp;amp;= \{\text{limit of any subsequence of }\{x_n\}\text{ that converges}\}.&#xA;\end{align*}&#xA;Please correct me if I am wrong.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added Questions:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Arturo, really appreciate your reply! It can take me a while to fully understand what you wrote.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think I will probably put the following into a new post if it is not proper of adding questions here given the length of this post. But anyway, here they are:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;In definition 1, when will $\limsup&#xD;&#xA;    x_n = \liminf x_n$?&lt;/li&gt;&#xA;&lt;li&gt;In definition 2(1), when will&#xA;$\limsup x_n = \liminf x_n$?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Basically, I would like to know when one can define &quot;limit&quot; of a sequence in a set, not solely depending on or even independent of topology.&lt;/p&gt;&#xA;" OwnerUserId="1281" LastEditorUserId="1281" LastEditorDisplayName="" LastEditDate="2011-01-14T22:43:28.467" LastActivityDate="2011-01-15T00:09:25.207" Title="limsup and liminf of a sequence of points in a set" Tags="&lt;topology&gt;&lt;elementary-set-theory&gt;&lt;order-theory&gt;&lt;lattices&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" />
  <row Id="17348" PostTypeId="1" AcceptedAnswerId="17352" CreationDate="2011-01-13T05:19:24.960" Score="1" ViewCount="96" Body="&lt;p&gt;Let $X$ and $Y$ be Riemann surfaces, and $\mathscr{O}_X, \mathscr{O}_Y$ be the sheaves of holomorphic functions on $X$ and $Y$ respectively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is obvious that a holomorphic map $f:X \to Y$ gives rise to a morphism of locally ringed spaces $(X, \mathscr{O}_X) \to (Y, \mathscr{O}_Y)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was always under the impression that a morphism of locally ringed spaces also gave rise to a holomorphic map, but I can't for the life of me prove it. I cant see why the induced stalk maps being local homomorphisms forces the topological map to be holomorphic. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any help would be appreciated!&lt;/p&gt;&#xA;" OwnerUserId="1725" LastActivityDate="2011-01-13T06:31:53.273" Title="Locally ringed spaces and Riemann surfaces" Tags="&lt;algebraic-geometry&gt;&lt;reference-request&gt;&lt;riemann-surface&gt;" AnswerCount="3" />
  <row Id="17349" PostTypeId="1" AcceptedAnswerId="17364" CreationDate="2011-01-13T05:24:47.737" Score="0" ViewCount="163" Body="&lt;p&gt;I am experiencing what I think is really simple confusion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take $y(t) = \sin(2 \cdot \pi \cdot t \cdot\omega(t))$&#xA;and $\omega(t) = a \cdot t+b$ for $t \in [0,p)$ and let $\omega(t)$ have a periodic extension with period $p$. The values $a,b,p$ are parameters&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\omega(t)$ looks like the blue function here&lt;img src=&quot;http://i.imgur.com/sWByh.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with $\omega$ as such, $y(t)$ should be two chirps -- a sine wave whose frequency sweeps. My intuition is that the two chirps should be identical, but they're not. The second chirp is higher in frequency. The argument to $\sin$  , the quantity $2 \cdot \pi \cdot t \cdot \omega(t)$ looks like the blue function here:&lt;img src=&quot;http://i.imgur.com/92UKN.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Two piecewise quadratics. The second quadratic should just be a shifted version of the first one, which is what is depicted in red. I can't argue with the math, but there is something very simple wrong with my intuition. The function $\omega$ should modulate the &quot;instantaneous frequency&quot; of the sinewave. &lt;/p&gt;&#xA;" OwnerUserId="2256" LastActivityDate="2011-01-13T07:04:58.853" Title="Frequency Swept sine wave -- chirp" Tags="&lt;signal-processing&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" />
  <row Id="17350" PostTypeId="2" ParentId="17152" CreationDate="2011-01-13T05:37:57.550" Score="8" ViewCount="" Body="&lt;p&gt;You don’t even need an infinite number of monkeys!  For any $\epsilon &amp;gt; 0$ and $k \geq l(\text{Hamlet})$, there is some number $N$ such that $N$ monkeys at typewriters, each typing for $k$ keystrokes, will produce a copy of Hamlet with probability greater than $1-\epsilon$.  (This holds under some quite weak conditions on our model of monkey typing.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is an example of the general “&lt;a href=&quot;http://terrytao.wordpress.com/2007/05/23/soft-analysis-hard-analysis-and-the-finite-convergence-principle/&quot; rel=&quot;nofollow&quot;&gt;soft analysis to hard analysis&lt;/a&gt;” principle, championed by Terry Tao among others: most any proof in analysis may be transformed into a proof of a quantitative statement such as the one above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This can be made precise in some generality using various rather beautiful proof-theoretic methods, such as variants of Gödel’s Dialectica translation; lovely results along these lines have been obtained by e.g. Avigad, Gerhardy and Towsner.  In this particular case, the bounds we get will of course depend on the model of monkey typing used.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, if we assume that the keystrokes are independently uniformly distributed, if our Hamlet-recognition criterion is case-, punctuation- and whitespace-insensitive, then for the case $k = l(\text{Hamlet})$, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$N = \lceil \frac{\log \epsilon}{\log (1 - \frac{1}{26^{l(\text{Hamlet})}})}\rceil$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;will work.  (The proof is an exercise for the reader.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Project Gutenberg’s copy of &lt;a href=&quot;http://www.gutenberg.org/cache/epub/2265/pg2265.txt&quot; rel=&quot;nofollow&quot;&gt;Hamlet (first folio)&lt;/a&gt; weighs in at 117,496 alphanumeric characters.  So if we want to produce Hamlet (first folio) with probability 1/2, in the minimal number of keystrokes, then by some quick slapdash estimating (rounding up a little to be on the safe side), something like $10^{170,000}$ monkeys should certainly suffice!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I guess empirical testing is out — ethical controls are so tricky.  Anyone want to run some simulations?&lt;/p&gt;&#xA;" OwnerUserId="2439" LastActivityDate="2011-01-13T05:37:57.550" CommunityOwnedDate="2011-01-15T06:13:52.160" CommentCount="1" />
  <row Id="17351" PostTypeId="2" ParentId="17348" CreationDate="2011-01-13T05:45:23.683" Score="3" ViewCount="" Body="&lt;p&gt;take a chart of $Y$, and let $y$ be the coordinate function. First show that the morphism actually comes from a honest map from $f: X \to Y$, and that the sheaf map takes $y$ to $y \circ f$. Now $y \circ f$ lying in the sheaf of holomorphic functions on $X$ means exactly that $f$ is holomorphic locally. Varying the chart shows that $f$ is holomorphic.&lt;/p&gt;&#xA;" OwnerUserId="325" LastActivityDate="2011-01-13T05:45:23.683" />
  <row Id="17352" PostTypeId="2" ParentId="17348" CreationDate="2011-01-13T05:45:34.217" Score="3" ViewCount="" Body="&lt;p&gt;If I am not mistaken, for this you need to consider $(X,\mathcal O_X)$ and $(Y,\mathcal O_Y)$ not as &quot;naked&quot; locally ringed spaces, but as spaces locally ringed by $\mathbb C$-algebras, and thus consider morphisms $(X,\mathcal O_X)&#xD;&#xA;\to (Y,\mathcal O_Y)$ that respect the $\mathbb C$-algebra structure on the&#xA;sheaves of rings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With this assumption, it is not hard to show that for such a morphism&#xA;$(f,f^{\natural})$ (so $f:X \to Y$ is continuous, and $f^{\natural}: \mathcal O_Y &#xD;&#xA;\to f_* \mathcal O_X$ is a map of sheaves of $\mathbb C$-algebras which induces&#xA;local maps on stalks), the map $f^{\natural}$ on sheaves is &lt;em&gt;necessarily&lt;/em&gt; given&#xA;by pull-back of functions.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus $f:X \to Y$ is a continuous map which pulls back holomorphic functions on&#xA;open subsets $U$ of $Y$ to holomorphic functions on the preimage of $U$ in $X$.&#xA;Applying this to the holomorphic coordinate functions on coordinate charts of $Y$, one finds that $f$ is in fact a holomorphic map, as required.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Note: there is nothing special about dimension one here; this will work for complex manifolds of any dimension.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Note also that Soarer gives the same answer more succinctly!)&lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2011-01-13T05:45:34.217" CommentCount="1" />
  <row Id="17353" PostTypeId="2" ParentId="17132" CreationDate="2011-01-13T05:47:50.050" Score="0" ViewCount="" Body="&lt;p&gt;just look at the size of the subgroups annihilated by p^k for various k.&lt;/p&gt;&#xA;" OwnerUserId="5211" LastActivityDate="2011-01-13T05:47:50.050" CommentCount="5" />
  <row Id="17354" PostTypeId="2" ParentId="17343" CreationDate="2011-01-13T05:51:19.530" Score="2" ViewCount="" Body="&lt;p&gt;I guess &lt;a href=&quot;http://en.wikipedia.org/wiki/Big_O_notation&quot; rel=&quot;nofollow&quot;&gt;‘big O notation’&lt;/a&gt; and its relatives are what you’re after, or something like that?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $(a_n)$ is a sequence with $a_n \rightarrow \infty$ as $n \rightarrow \infty$, then this notation defines statements like $(a_n) = O(n^2)$, formalising the idea that “in the long run, the sequence $(a_n)$ grows no faster than the sequence $(n^2)$”.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(The use of ‘=’ in this notation is slightly confusing: the precise statement doesn’t assert that any two things are equal.)&lt;/p&gt;&#xA;" OwnerUserId="2439" LastActivityDate="2011-01-13T05:51:19.530" CommentCount="1" />
  <row Id="17355" PostTypeId="2" ParentId="17343" CreationDate="2011-01-13T05:51:53.993" Score="6" ViewCount="" Body="&lt;p&gt;This is related to Hausdorff's &quot;Pantachie&quot; problem. Suppose $x_i$ is a monotone decreasing sequence whose sum is divergent. We say that a similar sequence $y_i$ diverges &lt;i&gt;slower&lt;/i&gt; if $x_i/y_i \rightarrow \infty$. Example: $\sum n^{-1}$ diverges slower then $\sum n^{-0.5}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, if $x_i$ is a monotone decreasing sequence whose sum is &lt;i&gt;convergent&lt;/i&gt;, a similar sequence $y_i$ converges more slowly if $y_i/x_i \rightarrow \infty$. Example: $\sum n^{-2}$ converges more slowly than $\sum n^{-3}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hausdorff proved the following theorem: For any sequence of divergent (convergent) series, there's a sequence diverging (converging) slower than any of them. That means that there is no &quot;expressible by finite strings&quot; characterization of the speed of divergence (convergence), since such a characterization would not allow any series which is diverging (converging) slower.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For more on the subject, look up Hausdorff gaps.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2011-01-13T05:51:53.993" />
  <row Id="17356" PostTypeId="1" AcceptedAnswerId="17360" CreationDate="2011-01-13T05:56:25.870" Score="2" ViewCount="77" Body="&lt;p&gt;Hello people.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Recently, I have been programming a simple game. Very simple:&#xA;There is a tank, and the cannon will aim at whatever position the mouse is at.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now lets talk about the cannon graphic. The cannon graphic points to the north, with 0 rotation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are the variables I have for my game, and that might be important factors for solving my problem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Tx&lt;/em&gt; = The tank's X position in the world.&#xA;&lt;em&gt;Ty&lt;/em&gt; = The tank's Y position in the world.&#xA;&lt;em&gt;Mx&lt;/em&gt; = The mouse's X position in the world.&#xA;&lt;em&gt;My&lt;/em&gt; = The mouse's Y position in the world.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, in this programming language, the greater the &lt;em&gt;Y&lt;/em&gt; coordinate, the &lt;strong&gt;lower&lt;/strong&gt; you are. And the less the &lt;em&gt;Y&lt;/em&gt; coordinate is, the &lt;strong&gt;higher&lt;/strong&gt; you are. So, &lt;em&gt;Y = 0&lt;/em&gt; means the top.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My problem is, how do calculate the rotation needed for my cannon graphic to &quot;point&quot; to the mouse's position?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you for your time.&lt;/p&gt;&#xA;" OwnerUserId="5775" LastEditorUserId="5775" LastEditorDisplayName="" LastEditDate="2011-01-13T06:32:26.790" LastActivityDate="2011-01-13T06:57:59.713" Title="Angle for pointing at a certain point in 2d space." Tags="&lt;trigonometry&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="17357" PostTypeId="1" AcceptedAnswerId="17361" CreationDate="2011-01-13T06:19:31.687" Score="3" ViewCount="163" Body="&lt;p&gt;I'm confused as to how far to go in terms of getting $s_n$ in a telescoping series. I cannot find an explanation for this at all in my textbook. But from the problems I've seen, perhaps it's just coincidence, does it have anything to do with the numerator?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, the section example gives:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(\sum_{n=1}^\infty) \frac{1}{n(n+1)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$s_n = 1 - \frac{1}{n+1}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In another problem I've noticed:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(\sum_{n=2}^\infty) \frac{2}{n^2-1}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$s_n = 1 + \frac{1}{2} - \frac{1}{n-1} - \frac{1}{n}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In another problem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(\sum_{n=1}^\infty) \frac{3}{n(n+3)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$s_n = 1 + \frac{1}{2} + \frac{1}{3} - \frac{1}{1+n} - \frac{1}{2+n} - \frac{1}{3+n}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fractions with denominator n become 0, so they don't count, but I noticed that in the collapsing series, the fractions that did matter equaled the numerator. Just a coincidence? If not, why does this happen?&lt;/p&gt;&#xA;" OwnerUserId="3514" LastActivityDate="2011-01-13T14:04:53.467" Title="How far to go in a telescoping series" Tags="&lt;calculus&gt;&lt;homework&gt;&lt;sequences-and-series&gt;" AnswerCount="3" CommentCount="3" />
  <row Id="17358" PostTypeId="2" ParentId="17348" CreationDate="2011-01-13T06:31:53.273" Score="1" ViewCount="" Body="&lt;p&gt;What I was actually stuck on was showing that the sheaf map is given by pull back. It is painfully simple...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You need to assume that $f^\sharp $ is a morphism of sheaves of $\mathbb{C}$-algebras as pointed out by Matt E&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume $(f,f^\sharp ):X \to Y$ is a morphism of locally ringed spaces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fix $s \in \mathscr{O}_Y(U)$. Fix $p \in U$. Then $s_{f(p)} - (s \circ f(p))_{f(p)}$ is in the maximal ideal of $\mathscr{O}_{Y,f(p)}$. Therefore $f^\sharp (U)(s)_p - (s \circ f(p))_p$ is in the maximal ideal of $\mathscr{O}_{X,p}$. Therefore $f^\sharp (U)(s)(p) = s \circ f(p)$&lt;/p&gt;&#xA;" OwnerUserId="1725" LastActivityDate="2011-01-13T06:31:53.273" />
  <row Id="17359" PostTypeId="2" ParentId="17310" CreationDate="2011-01-13T06:36:59.853" Score="5" ViewCount="" Body="&lt;p&gt;Just for the record: the usual proof of irrationality is as follows (say for the first sum). Suppose that the sum is $p/q$. Multiply by $qg^{n^2}$, where $n$ is &quot;big enough&quot;. We get that $$\sum_{m &amp;gt; n} \frac{q}{g^{m^2-n^2}}$$ is an integer. However, the latter is bounded by the geometric series $$q \sum_{t &amp;gt; 0} g^{-Nt} = \frac{q}{g^N} \cdot \frac{1}{1-g^{-N}} = \frac{q}{g^N-1},$$ where $N = (n+1)^2 - n^2 = 2n+1$. When $n$ is big enough, $q/(g^N-1) &amp;lt; 1$, contradiction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This proofs works for any series $$\sum_{n=1}^\infty \frac{1}{\prod_{i=1}^n a_i}$$ where the non-zero integers $a_i$ satisfy $|a_i| \rightarrow \infty$.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2011-01-13T06:36:59.853" />
  <row Id="17360" PostTypeId="2" ParentId="17356" CreationDate="2011-01-13T06:51:10.673" Score="2" ViewCount="" Body="&lt;p&gt;Suppose this is the situation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/mRawZ.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle D_x$ is the difference of the $\displaystyle x$-coordinates and $\displaystyle D_y$ is the difference of the $\displaystyle y$-coordinates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then angle $\displaystyle w$ is given by $\displaystyle \tan w = \frac{D_y}{D_x}$ and thus $\displaystyle w = \arctan (\frac{D_y}{D_x})$. The angle you will need to rotate would then be anti-clockwise $\displaystyle \frac{\pi}{2} + w$, if the tank is point &quot;up&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: The above assumes that $\displaystyle w$ is acute. I will leave it to you to try to work out the other cases (for different tank and mouse positions) and come up with a general formula.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would suggest reading up on atan or (to avoid a trap involving division by $0$) &lt;a href=&quot;http://en.wikipedia.org/wiki/Atan2&quot; rel=&quot;nofollow&quot;&gt;atan2&lt;/a&gt;. Most likely the Math package of your programming language will have both.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope that helps.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2011-01-13T06:57:59.713" LastActivityDate="2011-01-13T06:57:59.713" CommentCount="2" />
  <row Id="17361" PostTypeId="2" ParentId="17357" CreationDate="2011-01-13T06:52:04.310" Score="4" ViewCount="" Body="&lt;p&gt;This is just an observation, and I'm not sure how rigorous this is, but consider the difference between the linear terms in the factorization of the denominator of the terms of your series. For the first one, the difference between $n$ and $n+1$ is $1$, and notice the telescoping series when written out begins like this:&#xA;$$&#xD;&#xA;(1-1/2)+(1/2-1/3)+(1/3-1/4)+\cdots&#xD;&#xA;$$&#xA;Notice that the terms that vanish, namely $1/2$ and $1/3$, do so by the next term. That is, you only have to add $1$ more term for it to vanish from the sum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Likewise, the difference of the linear terms $n-1$ and $n+1$ in the second series is $2$. Also, when written out, the telescoping sum is&#xA;$$&#xD;&#xA;(1-1/3)+(1/2-1/4)+(1/3-1/5)+(1/4-1/6)+\cdots&#xD;&#xA;$$&#xA;Now the terms that vanish, $1/3, 1/4$, do not do so until $2$ terms after they first appear. That is, $1/3$ first appears in term $a_1$, and doesn't disappear until term $a_3$, so you must add $2$ terms until it goes away. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same applies for the third series, that you must add $3$ terms until the terms which will not contribute to the partial sum vanish. I believe this is helpful in figuring out how many terms you should sum before deciding on a general formula for the partial sum. Then of course induction would be the best way to make sure of your guess. I think it is just coincidence that the difference of the linear factors of the denominator was equal to the numerator in all three of the problems you listed. &lt;/p&gt;&#xA;" OwnerUserId="1583" LastActivityDate="2011-01-13T06:52:04.310" CommentCount="1" />
  <row Id="17362" PostTypeId="2" ParentId="17356" CreationDate="2011-01-13T06:54:42.437" Score="0" ViewCount="" Body="&lt;p&gt;Okay, in your coordinates &quot;north&quot; presumably means up, i.e. the direction of &lt;em&gt;decreasing&lt;/em&gt; $y$ coordinates.  Let's assume for convenience that all $y$ coordinates are nonnegative, so $y=-1$ is off the display.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From a location $(x_0,y_0)$ to target $(x_1,y_1)$ the angle you want has cosine $(y_0 - y_1)/d$ where $d = \sqrt{(x_1 - x_0)^2 + (y_1 - y_0)^2}$ is the distance location to target.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Taking the arccosine of this gives the absolute value of the angle you want.  For the sign you need to say whether the cannon rotates clockwise or counterclockwise as the angle increases.  If increasing the angle (from zero = up/north) rotates clockwise, then choose a positive sign whenever $x_1 &amp;gt; x_0$ and a negative sign when $x_1 &amp;lt; x_0$.  If $x_1 = x_0$, then the angle is either zero when $y_1 &amp;lt; y_0$ or $\pi$ radians (aka 180 degrees) if $y_1 &amp;gt; y_0$.&lt;/p&gt;&#xA;" OwnerUserId="3111" LastActivityDate="2011-01-13T06:54:42.437" />
  <row Id="17364" PostTypeId="2" ParentId="17349" CreationDate="2011-01-13T07:04:58.853" Score="4" ViewCount="" Body="&lt;p&gt;It seems my comment has answered the question, so I suppose I should turn it into an actual answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a chirp signal of the form $\sin(2\pi\phi(t))$, the instantaneous frequency $\omega(t)$ is the time derivative of the instantaneous phase $\phi(t)$. So for given $\omega(t)$, your signal should be $\sin\left(2\pi\int_0^t\omega(\tau)d\tau\right)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For what it's worth, the incorrect expression $\sin(2\pi t\omega(t))$ gives a signal with instantaneous frequency $\omega(t) + t\omega&amp;#39;(t)$. In your example, $\omega&amp;#39;(t)$ is positive almost everywhere, which is why the signal seems to be too high in frequency.&lt;/p&gt;&#xA;" OwnerUserId="856" LastActivityDate="2011-01-13T07:04:58.853" />
  <row Id="17365" PostTypeId="2" ParentId="16254" CreationDate="2011-01-13T07:05:33.053" Score="1" ViewCount="" Body="&lt;p&gt;As an undergraduate in the UK, I did a summer research project in analytic number theory that was funded by the &lt;a href=&quot;http://www.epsrc.ac.uk/funding/students/Pages/vacationbursaries.aspx&quot; rel=&quot;nofollow&quot;&gt;EPSRC Vacation Bursaries Scheme&lt;/a&gt;.  It involved working under the supervision of an advisor on an individual project for approximately 8 weeks, and was a really good opportunity to gain research experience.  I'm not sure what the constraints for eligibility are regarding nationality/citizenship, but the website I've linked to provides information about participating universities, who might have further information.&lt;/p&gt;&#xA;" OwnerUserId="1179" LastActivityDate="2011-01-13T07:05:33.053" CommunityOwnedDate="2011-01-13T07:05:33.053" CommentCount="2" />
  <row Id="17366" PostTypeId="1" AcceptedAnswerId="17367" CreationDate="2011-01-13T07:12:01.327" Score="11" ViewCount="936" Body="&lt;p&gt;Wait before you dismiss this as a crank question :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A friend of mine teaches school kids, and the book she uses states something to the following effect: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If you divide the circumference of &lt;em&gt;any&lt;/em&gt; circle by its diameter, you get the &lt;em&gt;same&lt;/em&gt; number, and this number is an irrational number which starts off as 3.14159... .&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;One of the smarter kids in class now has the following doubt: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Why is this number equal to 3.14159....? Why is it not some &lt;em&gt;other&lt;/em&gt; irrational number?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My friend is in a fix as to how to answer this in a sensible manner. Could you help us with this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have the following idea about how to answer this: Show that the ratio must be greater than $3$. Now show that it must be less than $3.5$. Then show that it must be greater than $3.1$. And so on ... . &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The trouble with this is that&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I don't know of any easy way of doing this, which would also be accessible to a school student. Could you point me to some approximation argument of this form?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5777" LastActivityDate="2011-01-13T12:33:42.227" Title="Teaching : Why is Pi equal to 3.14159..." Tags="&lt;geometry&gt;&lt;teaching&gt;&lt;pi&gt;" AnswerCount="4" CommentCount="5" />
  <row Id="17367" PostTypeId="2" ParentId="17366" CreationDate="2011-01-13T07:31:53.960" Score="12" ViewCount="" Body="&lt;p&gt;You can try doing what Archimedes did: using polygons inside and outside the circle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a webpage which seems to have a good explanation: &lt;a href=&quot;http://itech.fgcu.edu/faculty/clindsey/mhf4404/archimedes/archimedes.html&quot; rel=&quot;nofollow&quot;&gt;http://itech.fgcu.edu/faculty/clindsey/mhf4404/archimedes/archimedes.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An other method one can try is to use the fact that the area of the circle is $\displaystyle \pi r^2$. Take a square, inscribe a circle. Now randomly select points in the square (perhaps by having many students throw pebbles etc at the figure or using rain or a computer etc). Compute the ratio of the points which are  within the circle to the total. This ratio should be approximately the ratio of the areas $ = \displaystyle \frac{\pi}{4}$. Of course, this is susceptible to experimental errors :-)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or maybe just have them compute the approximate area of circle using graph paper, or the approximate perimeter using a piece of string.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Not sure how convincing the physical experiments might be, though.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2011-01-13T07:39:45.370" LastActivityDate="2011-01-13T07:39:45.370" />
  <row Id="17368" PostTypeId="2" ParentId="17366" CreationDate="2011-01-13T08:29:08.000" Score="10" ViewCount="" Body="&lt;p&gt;Given the age of the children, I think that wheeling a bicycle along the ground and measuring the distance traveled for one wheel revolution would be a good idea. This exercise can be continued by asking for suggestions on how to improve the accuracy (e.g. wheel the bike two revolutions etc)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you obviously have children in the class who pose thoughtful questions, one can ask what do you think would happen to the distance measured on the cycle milometer if pi was suddenly smaller or larger (if one let down or pumped up the tyres). What would happen to the speed registered in such cases. I'm sure there is an interesting exercise for your pupils here. Good luck.&lt;/p&gt;&#xA;" OwnerUserId="1301" LastActivityDate="2011-01-13T08:29:08.000" CommentCount="3" />
  <row Id="17369" PostTypeId="2" ParentId="17357" CreationDate="2011-01-13T08:29:08.530" Score="0" ViewCount="" Body="&lt;p&gt;If your terms are hypergeometric then you can use Gosper's algorithm. &lt;a href=&quot;http://mathworld.wolfram.com/GospersAlgorithm.html&quot; rel=&quot;nofollow&quot;&gt;http://mathworld.wolfram.com/GospersAlgorithm.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5676" LastActivityDate="2011-01-13T08:29:08.530" />
  <row Id="17370" PostTypeId="1" AcceptedAnswerId="17371" CreationDate="2011-01-13T08:36:03.610" Score="1" ViewCount="190" Body="&lt;p&gt;I have trouble understanding this question and its supposed answer.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Let $A$ be a subgroup of $G$. Prove that $A$ is a direct summand of $G$ if and only if there is a homomorphism $p:G \to A$ with $p(a)=a$ for all $a \in A$. Here we talk about internal direct sum.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When I see this, I was thinking, why do homomorphism of that ($p:G \to A$) implies it is a direct summand of $G$? That is strange. Considering simple application of homomorphism can give us whether a subgroup $A$ is a direct summand of $G$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any advice on how to prove this is most appreciated. thanks.&lt;/p&gt;&#xA;" OwnerUserId="1683" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-13T14:29:22.717" LastActivityDate="2011-01-14T03:16:30.103" Title="Question about direct sum of groups ." Tags="&lt;homework&gt;&lt;group-theory&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="17371" PostTypeId="2" ParentId="17370" CreationDate="2011-01-13T08:54:09.633" Score="1" ViewCount="" Body="&lt;p&gt;Hint for $\Rightarrow$:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Prove that the map $$t: G \to A \oplus \text{ker}(p)$$ given by $$g \mapsto (p(g),p(g)g^{-1})$$ is a homomorphism of groups&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Prove that $t$ is injective and surjective (i.e. an isomorphism)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Hint for $\Leftarrow$: Consider the projection $p: G = A \oplus B \to A$&lt;/p&gt;&#xA;" OwnerUserId="1766" LastActivityDate="2011-01-13T08:54:09.633" CommentCount="1" />
  <row Id="17373" PostTypeId="2" ParentId="12203" CreationDate="2011-01-13T09:12:11.213" Score="0" ViewCount="" Body="&lt;p&gt;The more general answer to that is:&#xA;\begin{align}&#xA;\forall x \in \mathbb{R},\  cos(x) &amp;amp; = \frac{e^{ix} + e^{-ix}}{2} \\&#xA;sin(x) &amp;amp; = \frac{e^{ix} - e^{-ix}}{2i}&#xA;\end{align}&#xA;These equations can in turn be obtained by summing and subtracting the following two equations:&#xA;\begin{align}&#xA;\forall x \in \mathbb{R},\  e^{ix} &amp;amp; = cos(x) + isin(x) \\&#xA;e^{-ix} &amp;amp; = cos(x) - isin(x) \\&#xA;\end{align}&lt;/p&gt;&#xA;" OwnerUserId="5701" LastActivityDate="2011-01-13T09:12:11.213" />
  <row Id="17374" PostTypeId="2" ParentId="17366" CreationDate="2011-01-13T09:48:19.197" Score="14" ViewCount="" Body="&lt;p&gt;If the kids are not too old you could visually try the following which is very straight forward. Build a few models of a circle of paperboard and then take a wire and put it straigt around it. Mark the point of the line where it is meets itself again and then measure the length of it. You will get something like 3.14.. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/FX2gW.gif&quot; alt=&quot;Pi unrolled&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let them measure themselves the diameter and circumference of different circles and let them plot them into a graph. Tadaa they will see the that its proportional and this is something they (hopefully) already know.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or use the approximation using the archimedes algorithm. Still its not really great as they will have to handle big numbers and the result is rather disappointing as it doesn't reveal the irrationality of pi and just gives them a more accurate number of $\pi$.&lt;/p&gt;&#xA;" OwnerUserId="3123" LastActivityDate="2011-01-13T09:48:19.197" CommentCount="3" />
  <row Id="17375" PostTypeId="1" CreationDate="2011-01-13T09:48:57.860" Score="0" ViewCount="301" Body="&lt;p&gt;$$ D = \{ z \in \mathbb{C} \ : \ |z| &amp;lt; \frac{1}{2} \}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems a disc is open or closed whether it has a &amp;lt; or &amp;lt;= but I don't know how to prove this. All examples have slightly different proofs so I've got no clarity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for your help.&lt;/p&gt;&#xA;" OwnerUserId="4577" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-13T14:28:13.530" LastActivityDate="2011-05-10T11:35:39.167" Title="How do you prove a disc is open?" Tags="&lt;complex-analysis&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="17377" PostTypeId="2" ParentId="17375" CreationDate="2011-01-13T10:11:14.993" Score="4" ViewCount="" Body="&lt;p&gt;The complex numbers form a metric space, meaning we have a function that say how far two points are from one another. In this case, $D$ is all the points which are not further than $\frac{1}{2}$ from $0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Metric spaces are very well behaved spaces, and many things are simplified in them. For example, to say whether or not a set is closed you can check if every convergent sequence inside the set has a limit inside the set. If you want to see if a set is open, you can verify that by choosing an arbitrary point and show there is a positive distance $d$ such that all points closer than $d$ are within the set, i.e. there is an open ball around your point which is completely contained in the set.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another method is to check the complement, namely in some cases it is simpler to check if the complement is closed, rather than if the set is open; and vice versa.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case, however, it's easier to verify directly. That is, if $z_0$ is a complex number such that $|z_0|=r&amp;lt;\frac{1}{2}$ then we can take $d=\frac{\frac{1}{2}-r}{3}$ and show that if $|z-z_0|&amp;lt;d$ then $|z|&amp;lt;\frac{1}{2}$.&lt;br&gt;&#xA;(which I am leaving as an exercise to the reader)&lt;/p&gt;&#xA;" OwnerUserId="622" LastActivityDate="2011-01-13T10:11:14.993" />
  <row Id="17378" PostTypeId="1" CreationDate="2011-01-13T10:30:22.660" Score="1" ViewCount="159" Body="&lt;p&gt;Hi, &#xA;I am  thinking about following problem, but am not able to find answer.&lt;br&gt;&#xA;In how many ways can a number be formed only using 3 digits (0, 1, 2) given following constraints with K and D:-&lt;br&gt;&#xA;1) Each digit appears exactly K times. The total length of the number is 3*K.&lt;br&gt;&#xA;2) At any given position (staring from left), the number of 0's must be at most D more than the number of 1's and number of 1's must be at most D more than number of 2's.&lt;br&gt;&#xA;i.e. for K =3 and D =1 followings are some valid numbers.&lt;br&gt;&#xA;012012012   010212012&lt;br&gt;&#xA;The number 001201212 will be invalid as there are 2 two's before the first 1. Similarly 010122012 is invalid as there are two 1's are there before first 2. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The number 001201212 will be valid if D=2. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So far I have only found out the total number of numbers which can be formed is $\binom{3K}{K}\binom{2K}{K}$&lt;/p&gt;&#xA;" OwnerUserId="5403" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2011-01-13T13:48:03.690" LastActivityDate="2011-01-13T13:48:03.690" Title="how many ways to construct a number" Tags="&lt;combinatorics&gt;&lt;permutations&gt;" CommentCount="7" />
  <row Id="17379" PostTypeId="2" ParentId="17346" CreationDate="2011-01-13T10:38:23.693" Score="1" ViewCount="" Body="&lt;p&gt;In one codimension one can be convinced of this quite easily.  Consider the tracefree curvature tensor&#xA;$&#xD;&#xA;A^o = A - \frac{1}{n}Hg,&#xD;&#xA;$&#xA;where $g$ is the metric, $A$ the second fundamental form, and $H$ the mean curvature.  For surfaces, the norm squared of $A^o$ satisfies:&#xA;$&#xD;&#xA;|A^o|^2 = (k_1 - k_2)^2,&#xD;&#xA;$&#xA;where $k_1$ and $k_2$ are the principal curvatures.  On the other hand,&#xA;$&#xD;&#xA;2|A^o|^2 = 2|A|^2 - H^2 = -2(H^2 - |A|^2) + H^2 = -4K + H^2 .&#xD;&#xA;$&#xA;So if $K$ and $H$ are constant, then the norm of the tracefree curvature tensor is also constant.  Further, the symmetry of $(\nabla A)$ implies that every component of $A^o$ is constant, and so the principal curvatures are themselves constant functions.  If their difference is zero, then the principal curvatures are equal and it follows that the surface is a sphere or a plane.  In higher codimensions this is more complicated.  If their difference is not zero, this contradicts the compactness of the surface.&lt;/p&gt;&#xA;" OwnerUserId="4322" LastActivityDate="2011-01-13T10:38:23.693" />
  <row Id="17380" PostTypeId="1" CreationDate="2011-01-13T10:54:56.157" Score="1" ViewCount="97" Body="&lt;p&gt;Let $X,Y$ be connected sets and $A,B$ proper subsets of $X$ and $Y$ respectively. Then it can be shown that the space $ (X \times Y) \setminus (A \times B)$ is connected. This is an problem from Munkres book. My question is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is the above result still true when we replace connectedness by path-connectedness instead?&lt;/p&gt;&#xA;" OwnerUserId="5782" LastActivityDate="2011-01-13T11:43:05.177" Title="Product of connected sets, proper subsets" Tags="&lt;topology&gt;" AnswerCount="1" />
  <row Id="17381" PostTypeId="2" ParentId="17380" CreationDate="2011-01-13T11:43:05.177" Score="4" ViewCount="" Body="&lt;p&gt;Yes, and the proof is even easier than the connectedness proof. You just need to construct a path between any two points, which you can do piecewise, staying in subsets of the form $X\times\{y\}$ and $\{x\}\times Y$, which are path connected.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively you can copy the connectedness proof, which consists of covering $X\times Y\setminus A\times B$ in stages by connected sets which have a point in common. This still works for path connectivity.&lt;/p&gt;&#xA;" OwnerUserId="2941" LastActivityDate="2011-01-13T11:43:05.177" />
  <row Id="17382" PostTypeId="2" ParentId="17152" CreationDate="2011-01-13T11:54:32.103" Score="-1" ViewCount="" Body="&lt;h1&gt;Hamlet Regurgitated by SuperMonkey&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/A6fV2.jpg&quot; alt=&quot;the monkey who wroteth Hamlet&quot;&gt;&#xA;Monkeyland, 30 Feb 2011: A group of infinite monkeys working since time immemorial in Monkeyland, have successfully produced a verified copy of Hamlet. The group of monkeys was led by a (alleged) descendant and though most of the content produced was illegible to mere humans, this particular work was deciphered properly due to a similar work having being produced by a human some centuries back. The monkey who has accomplished this task is featured in the picture above...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;The above is an excerpt from a future post of my blog-cum-&lt;em&gt;news&lt;/em&gt; website. All rights reserved(Oh damn you cc by South Africa!)&lt;/sub&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Poster's Comments:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This post is to inform you that the problem has been solved. I personally verified the MD5 hashes of the original Hamlet and that produced by Mr. Monkey. Given the limitations of the hash, I would wait for a better comparison done by better methods and/or mathematicians. Additionally the picture used in the article is an old one. The monkeys now use a PC with a (highly) touch sensitive keyboard(hence the regurgitation part of the title... I hope you don't want more details on that front, Although I'd like to add that these keyboards are a lot easier to clean, especially after all the crap that monkeys eat these days.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;I am expecting downvotes. But before you downvote, please know that&lt;/sub&gt; &lt;strong&gt;I love mathematics&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="2366" LastEditorUserId="2366" LastEditorDisplayName="" LastEditDate="2011-01-13T12:08:26.610" LastActivityDate="2011-01-13T12:08:26.610" CommunityOwnedDate="2011-01-15T06:13:52.160" CommentCount="4" />
  <row Id="17383" PostTypeId="2" ParentId="17333" CreationDate="2011-01-13T12:27:57.250" Score="0" ViewCount="" Body="&lt;p&gt;If you do the calculations you will find that the cut off point for the new loan is 6.55% (on a 15 year loan of 74,019.66+2,072.55 = 76,092.21), which gives a repayment of 664.94 (a rate of 6.56% will give you a repayment of  665.36, slightly more than your current repayment).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if they are no other costs, as you suggest, you must ask yourself what is in it for the mortgage company? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;At 6% interest on 76,092.21 the repayment is about 642.11 giving you a monthly saving of&#xA;665.30 - 642.11 = 23.19, which means it will effectively (ignoring complicating factors such as inflation) take 7.5 years to save back the 2,072.55 that has been added to the amount you owe. So you should view this deal as a kind of lock-in, and make a decision with that in mind.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, be wary of making decisions based purely on mathematical calculations. There could be other factors to consider. You may find more help at &lt;a href=&quot;http://money.stackexchange.com/&quot;&gt;Personal Finance and Money.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1301" LastActivityDate="2011-01-13T12:27:57.250" CommentCount="1" />
  <row Id="17384" PostTypeId="2" ParentId="17366" CreationDate="2011-01-13T12:33:42.227" Score="2" ViewCount="" Body="&lt;p&gt;Why is this number equal to 3.14159....? Why is it not some other irrational number?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Answer is that with the usual Euclidean metric that is the number that one gets, the value of $\pi$ is dependent on the geometry that is being used, so on a sphere the $\pi$ used to obtain the area will be different. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another question to ask is that if $S=\pi_1 r^2$ is the area of the circle and $C=\pi_2 2 r$  is the circumference why $\pi_1=\pi_2 = \pi$  ?   What geometries or metrics will result in $\pi_1 \neq \pi_2$ ?&lt;/p&gt;&#xA;" OwnerUserId="3936" LastActivityDate="2011-01-13T12:33:42.227" CommentCount="10" />
  <row Id="17385" PostTypeId="2" ParentId="17375" CreationDate="2011-01-13T13:16:21.543" Score="3" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Hint:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Try to find some continuous function $f:\mathbb{C}\to\mathbb{R}$ such that the open disc is the inverse image $f^{-1}(U)$ for some open set $U\subset\mathbb{R}$.&lt;/p&gt;&#xA;" OwnerUserId="1154" LastEditorUserId="1154" LastEditorDisplayName="" LastEditDate="2011-01-13T21:24:20.700" LastActivityDate="2011-01-13T21:24:20.700" />
  <row Id="17386" PostTypeId="1" CreationDate="2011-01-13T13:17:53.517" Score="0" ViewCount="58" Body="&lt;p&gt;How to prevent singularity when designing system of differential equations?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;sometimes guessing some initial value will get singularity near some value&lt;/p&gt;&#xA;&#xA;&lt;p&gt;when ploting in maple  &lt;/p&gt;&#xA;" OwnerUserId="5792" LastActivityDate="2011-01-13T13:17:53.517" Title="How to prevent singularity when designing system of differential equations?" Tags="&lt;differential-equations&gt;" CommentCount="1" />
  <row Id="17387" PostTypeId="2" ParentId="17357" CreationDate="2011-01-13T14:04:53.467" Score="1" ViewCount="" Body="&lt;p&gt;It is just a coincidence that the number of terms to keep equals the numerator.  In your second example, if your were summing $\frac{1}{n^2-1}$ you would still keep two terms.    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;All your examples are of the form $\frac{a}{n(n+a)}$ (in the second you need to shift $n$ by 1) and the number of terms to keep is $a$.  This is because $\frac{a}{n(n+a)}=\frac{1}{n}-\frac{1}{n+a}$ so terms will telescope $a$ spaces down the line.  The number of terms to keep depends upon the denominator, not the numerator.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You were given $a$ in the numerator to avoid a fraction when making this transformation.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-01-13T14:04:53.467" />
  <row Id="17388" PostTypeId="1" AcceptedAnswerId="17392" CreationDate="2011-01-13T14:41:40.750" Score="2" ViewCount="194" Body="&lt;p&gt;Please answer as many questions as you can.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are the topics one should know before delving into probability theory? (Please recommend any books you know on those topics too.) I think there is set theory, but set theory is a large topic in itself. Does probability need only a little bit of set theory? Also, there is combinatorics, but again combinatorics is big in itself. And is it a good idea to know all topics such as set theory and combinatorics to fully understand probability theory? Or is it enough just to read those topics on the fly? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="5530" LastActivityDate="2011-01-13T15:15:22.797" Title="Prerequisites on Probability Theory" Tags="&lt;probability-theory&gt;&lt;books&gt;" AnswerCount="4" FavoriteCount="2" />
  <row Id="17389" PostTypeId="2" ParentId="17370" CreationDate="2011-01-13T14:44:04.680" Score="1" ViewCount="" Body="&lt;p&gt;Sebastian has given you hints on how to prove this. Let me give you a bit of intuition for why the relation between direct sums and maps.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you have a direct sum, $G=A\oplus B$, you have two pairs of maps that relate $A$, $B$, and $G$. You have projections $\pi_A\colon G\to A$ and $\pi_B\colon G\to B$ (with $\pi_A(a,b) = a$ and $\pi_B(a,b) = b$) and you have injections $i_A\colon A\to G$ and $i_B\colon B\to G$ (with $i_A(a) = (a,1_B)$ and $i_B(b) = (1_A,b)$, where $1_A$ and $1_B$ are the identities of $A$ and $B$, respectively). The maps are related by the property that $\pi_A\circ i_A = \mathrm{id}_A$ and $\pi_B\circ i_B = \mathrm{id}_B$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you identify $A$ with $i_A(A)$, then this translates exactly to the condition in the problem. So the problem is telling you that these conditions are in fact &lt;em&gt;characterized&lt;/em&gt; by the existence of the maps. Why?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the situation where you have a homomorphism $f\colon G\to A$ that is onto. If $B=\mathrm{ker}(f)$, then for each $a\in A$ you can pick an element $g_a\in G$ such that $f(g_a) = a$ (a pre-image). In general, these will not form a subgroup; sometimes, you can &lt;em&gt;pick&lt;/em&gt; the elements cleverly enough so that they &lt;em&gt;will&lt;/em&gt; form a subgroup, sometimes not. For example, if you map $C_{15}\to C_3$ (cyclic group of order 15 to cyclic group of order 3) by sending the generator $x$ to the generator $y$, then you can pick any of $1,x^3,x^6,x^9,x^{12}$ as $g_{1}$, any of $x,x^4,x^7,x^{10},x^{13}$ as $g_y$, and any of $x^2,x^5,x^8,x^{11},x^{14}$ as $g_{y^2}$. If you were to pick, say, $1$, $x^4$, and $x^{11}$ as you preimages, they don't form a group, but if you pick $1$, $x^{10}$, and $x^{5}$, then they &lt;em&gt;do&lt;/em&gt; form a subgroup that is isomorphic to $C_3$. On the other hand if you take $C_4\to C_2$, again mapping generator to generator, the possible preimages of $1$ are $1,x^2$ and the possible preimages of $y$ are $x,x^3$. There is no way to pick an element of $\{1,x^2\}$ and an element of $\{x,x^3\}$ so that the two are a subgroup of $C_4$. So sometimes you can, sometimes you cannot.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you &lt;em&gt;can&lt;/em&gt;, if you let $\mathcal{A}$ be the subgroup of $G$ that is isomorphic to $A$ via the original map $p\colon G\to A$, then every element of $G$ will be written as an element of $\mathcal{A}$ and an element of $B=\mathrm{ker}(p)$. But in general this will not be a direct product, because you will have $(ab)(a&amp;#39;b&amp;#39;) = a(ba&amp;#39;)b&amp;#39; = aa&amp;#39;(a&amp;#39;^{-1}ba&amp;#39;)b&amp;#39;$; unless $a&amp;#39;^{-1}ba&amp;#39;=b$ for all $b$, you don't get a direct product. So you it's not enough to be able to &quot;retract&quot; the projection, there's more needed. &lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2011-01-13T14:44:04.680" />
  <row Id="17390" PostTypeId="2" ParentId="17388" CreationDate="2011-01-13T14:57:19.127" Score="2" ViewCount="" Body="&lt;p&gt;Combinatorics is a very large subject but one set of counting problems that is very helpful for discrete probability is counting the number of inequivalent ways to place balls in boxes. Fundamental cases include when the balls are considered distinguishable or indistinguishable and the boxes are considered distinguishable and indistinguishable. Fred Roberts' book Applied Combinatorics does a nice job with this material.&lt;/p&gt;&#xA;" OwnerUserId="1369" LastActivityDate="2011-01-13T14:57:19.127" />
  <row Id="17391" PostTypeId="2" ParentId="17388" CreationDate="2011-01-13T14:58:19.770" Score="4" ViewCount="" Body="&lt;p&gt;It depends which kind of probability theory you're interested in. An introductory course on probability theory can either dwell on discrete probability or continuous probability.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Discrete probability, which deals with discrete events (e.g. the probability that if you throw a dice it comes up $6$ ten times in a row), only really needs elementary combinatorics. From set theory you need to know the definitions of basic concepts, and from combinatorics you need to know the likes of the binomial coefficient and its properties.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A little more is needed to understand Poisson random variables, namely Stirling's approximation, which is a topic you don't really learn anywhere; this is why these courses often just give the definition, which requires you to know the Taylor expansion of $e^x$. But this topic in its entirety is not necessarily covered.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Continuous probability deals with things like the normal distribution and the central limit theorem - distributions which may take &quot;continuous&quot; values (e.g. every real value rather than only integral values). Sometimes it is given as an addendum to a discrete probability course. To understand continuous probability you will need to know basic calculus (the kind you get from a first course, and then some). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Introductory courses don't usually cover multivariate Gaussians, but these require some linear algebra.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Summarizing, you will need to be confident about some fairly basic topics. Besides some familiarity with basic concepts, it's also best to have some &quot;mathematical maturity&quot;, although not too much of it is actually needed in an introductory course.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2011-01-13T14:58:19.770" />
  <row Id="17392" PostTypeId="2" ParentId="17388" CreationDate="2011-01-13T15:05:58.133" Score="4" ViewCount="" Body="&lt;p&gt;Dependending on how deeply you want to explore the field, you will need more or less.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want a basic introduction then some basic set theory (what is a set and elementary set operations), combinatorics (knowing different ways of counting, inclusion-exclusion principle) and calculus (knowing derivatives and integrals). This could get you through a basic text in probability.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want more serious stuff, I would study &lt;strong&gt;measure theory&lt;/strong&gt; (which serves as the foundation of probability through Kolmogorov's axioms), a thorough knowledge of &lt;strong&gt;analysis&lt;/strong&gt; that goes beyond just knowing calculus, maybe even some functional analysis, &lt;strong&gt;combinatorics&lt;/strong&gt; and generally some discrete mathematics (like working with difference equations).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This will allow you to follow a solid introductory course on probability. After that, it depends a lot on what related branches you want to explore. If you want to study Markov chains, a good knowledge of linear algebra is a must. If you want to delve deeper into statistics (like hypothesis testing and such) more analysis will do you good, etc... &lt;/p&gt;&#xA;" OwnerUserId="3567" LastActivityDate="2011-01-13T15:05:58.133" CommentCount="1" />
  <row Id="17393" PostTypeId="2" ParentId="17388" CreationDate="2011-01-13T15:08:07.983" Score="4" ViewCount="" Body="&lt;p&gt;For elementary, probability theory you can look into these two books:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;A First Course in  Probability&lt;/em&gt; by &lt;em&gt;Sheldon Ross&lt;/em&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;An Introduction to Probability Theory and Its Applications&lt;/em&gt;, by &lt;em&gt;W.Feller.&lt;/em&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;Introduction to Probability and Measure&lt;/em&gt; by &lt;em&gt;K. R. Parthasarathy&lt;/em&gt;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Both books provide very good introduction to the subject. Moreover, it would be nice if you know some basic calculus and set theory because you may need them when you study about Distribution functions of Various Random variables. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The last book which i have added is a really nice book. It's available in Indian edition but i am not sure about it's sales in foreign.&lt;/p&gt;&#xA;" LastActivityDate="2011-01-13T15:15:22.797" CommentCount="3" />
  <row Id="17394" PostTypeId="2" ParentId="11346" CreationDate="2011-01-13T15:27:28.277" Score="2" ViewCount="" Body="&lt;p&gt;Here is a recursive idea that might work.&#xA;Let $s_1$, $s_2$,..., $s_{n-1}$ the sines of the angles of the first vector on $S^{n-1}$, $c_1$, ..., the cosines, and $s&amp;#39;_{\nu}$ and $c&amp;#39;_{\nu}$ those for the second one.&#xA;With $p$ being the scalar product, we have $p=1$ for $n=1$, $p=c c&amp;#39;+s s&amp;#39;$ for $n=2$ and generically $$p(n)=c_1 c&amp;#39;_1 + s_1 s&amp;#39;_1 p(n-1)$$ where $p(n-1)$ is the scalar product of the vectors given by the angles $[\theta_2,...]$ and $[\theta&amp;#39;_2,...]$ on $S^{n-1}$(i.e. by removing the first angle of each sequence).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This, of course, still requires all sines and cosines, but my idea is to simplify $p(n-2)$ using the haversine formula and hope that this reduces the number of calls to trigoniometric functions.&lt;/p&gt;&#xA;" OwnerUserId="5616" LastEditorUserId="6179" LastEditorDisplayName="" LastEditDate="2011-04-15T15:18:20.030" LastActivityDate="2011-04-15T15:18:20.030" />
  <row Id="17395" PostTypeId="1" CreationDate="2011-01-13T15:43:34.387" Score="4" ViewCount="156" Body="&lt;p&gt;What sorts of mathematical tools, models and methods and theoretical frameworks do people use to simulate the function of the brain's neural networks? What mathematical properties do different brains have?&lt;/p&gt;&#xA;" OwnerUserId="5796" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2011-01-13T18:09:02.490" LastActivityDate="2011-01-13T18:18:42.497" Title="What sort of mathematical methods and models are used to model the brain" Tags="&lt;calculus&gt;&lt;topology&gt;&lt;graph-theory&gt;&lt;mathematical-modeling&gt;&lt;network-flow&gt;" AnswerCount="3" CommentCount="7" FavoriteCount="2" />
  <row Id="17396" PostTypeId="1" CreationDate="2011-01-13T15:43:54.607" Score="0" ViewCount="199" Body="&lt;p&gt;I'm posting this as more of a theory/mathematic how-to followup to a stackoverflow question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The non-iterative method for calculating graycode depends on Log2N bytes, to store position information for the next bit in the iteration sequence.&lt;br&gt;&#xA;Specifically, the goal is to know the next bit to change without having to look at the current code.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, for the 3 bit gray code, there's a iteration sequence  0,1,2,1,0,1,2,1 that can be represented with a much simpler function - maintain  &quot;0,1,2,1&quot; in a register and rotate each time (as an example of a more general permutation).&lt;br&gt;&#xA;To reduce the necessary state, this could be kept in two positions, starting 0,1 and a constant function of &quot;xor, permute&quot; applied:   0,3 =&gt; 1,0 =&gt; 2,1 =&gt; 3,2 =&gt; 0,3    (the bit to change being the first, and 3 handled as 1 since only 0,1,2 are valid)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it possible for graycodes to exist for higher values of N, such that the iteration function can be calculated with just a permutation operation?&lt;br&gt;&#xA;If not, with permute/xor?&lt;br&gt;&#xA;Is there a more general way to think of applying constant operations to define the state change? &lt;/p&gt;&#xA;" OwnerUserId="5795" LastActivityDate="2011-01-15T19:45:28.990" Title="How to create a graycode of N bits with a code-agnostic iteration function" Tags="&lt;combinatorics&gt;&lt;computer-science&gt;" AnswerCount="2" CommentCount="8" />
  <row Id="17397" PostTypeId="2" ParentId="3320" CreationDate="2011-01-13T15:56:37.227" Score="7" ViewCount="" Body="&lt;p&gt;Because of the formula $e^{i\pi}+1=0$ you will find $\pi$ appearing in lots of places where it's not clear there is a circle, e.g in the normal distribution formulae.&lt;/p&gt;&#xA;" OwnerUserId="2809" LastActivityDate="2011-01-13T15:56:37.227" />
  <row Id="17398" PostTypeId="1" AcceptedAnswerId="18476" CreationDate="2011-01-13T16:15:13.663" Score="7" ViewCount="345" Body="&lt;p&gt;Hi&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fubini's theorem states that if you have a double integral over a function f(x,y), then you can compute the integral as an iterated integral, if f(x,y) is in L^1. But to find out if f is in L^1 you need to compute the double integral.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What am I missing? The examples I found all apply Fubini's theorem without checking that f(x,y) is in L^1. Many thanks for any clarification!&lt;/p&gt;&#xA;" OwnerUserId="5798" LastActivityDate="2011-01-21T23:57:37.650" Title="Chicken-Egg problem with Fubini’s theorem" Tags="&lt;real-analysis&gt;&lt;analysis&gt;&lt;measure-theory&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="3" />
  <row Id="17399" PostTypeId="1" AcceptedAnswerId="17401" CreationDate="2011-01-13T16:19:27.283" Score="0" ViewCount="88" Body="&lt;p&gt;I'm trying to re-write $v (u x)$ where $v$ and $u$ are row vectors and $x$ is a column vector as some expression $M x$ (or $\bar{v}x$, etc.). &#xA;The motivation is because I'm trying to compute the gradient (and $\nabla^{&amp;#39;}_x [ v (u x) ] \cdot x$ should yield a row vector). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Intuitively, since the expression $v (ux)$ can be re-written as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$[v_1 (ux), v_2(ux), \cdots, v_n(ux)]$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(where $v_i$ is a scalar), this quantity can be computed.&#xA;However, doing this and substituting into a larger expression is pretty obnoxious and I'd like to do it using standard matrix operations (non element-wise) if possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know this has something to do with tensors, but I don't understand how it all fits together. Any help is appreciated.&lt;/p&gt;&#xA;" OwnerUserId="5799" LastActivityDate="2011-01-13T16:37:49.403" Title="gradient of row vector multiplied by scalar" Tags="&lt;vector-calculus&gt;&lt;tensors&gt;" AnswerCount="1" />
  <row Id="17400" PostTypeId="2" ParentId="17395" CreationDate="2011-01-13T16:34:26.697" Score="2" ViewCount="" Body="&lt;p&gt;Commonly used models in mathematical physics for collections of neurons are so-called &lt;a href=&quot;http://en.wikipedia.org/wiki/Neural_network&quot; rel=&quot;nofollow&quot;&gt;neural networks&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is some &lt;a href=&quot;http://en.wikipedia.org/wiki/Artificial_neural_network#Types_of_models&quot; rel=&quot;nofollow&quot;&gt;general explanation of the models&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some popular models are the &lt;a href=&quot;http://en.wikipedia.org/wiki/Hopfield_net&quot; rel=&quot;nofollow&quot;&gt;Hopfield model&lt;/a&gt;, &lt;a href=&quot;http://en.wikipedia.org/wiki/Ashkin-Teller_model&quot; rel=&quot;nofollow&quot;&gt;Ashkin-Teller neural networks&lt;/a&gt;, Blume-Emery-Griffiths neural networks,... &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Techniques used to solve them involve statistical mechanics which result in partition sums (often handled with path integrals), differential equations to model dynamics, stochastic processes, renormalization theory, etc...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just realize that these models are usually rudementary in the sense that they can't model a fully functioning brain but just a &quot;small&quot; collection of interconnected neurons.&lt;/p&gt;&#xA;" OwnerUserId="3567" LastActivityDate="2011-01-13T16:34:26.697" />
  <row Id="17401" PostTypeId="2" ParentId="17399" CreationDate="2011-01-13T16:37:49.403" Score="2" ViewCount="" Body="&lt;p&gt;If by $Mx$ you mean $M$ is a matrix, then clearly you cannot write $v(ux)$ as $Mx$. Why? A matrix represents a linear operation that sends &lt;em&gt;column vectors&lt;/em&gt; to &lt;em&gt;column vectors&lt;/em&gt;. But your operation sends the column vector $x$ to a row vector. What you can do is write&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ v(ux) = \left((v^T\otimes u)x\right)^T $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $v^T\otimes u$ is the tensor product of the column vector $v^T$ and the row vector $u$, and can be represented by the matrix $M$ with entries $M_{ij} = v_iu_j$. So you multiply $x$ by this matrix and then take the transpose. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then taking the gradient is as simple as taking the gradient of $(Mx)^T$. &lt;/p&gt;&#xA;" OwnerUserId="1543" LastActivityDate="2011-01-13T16:37:49.403" />
  <row Id="17402" PostTypeId="2" ParentId="17398" CreationDate="2011-01-13T16:38:04.133" Score="0" ViewCount="" Body="&lt;p&gt;For Fubini to apply you need $f$ to be in $L^1$ as a function of $x$ for almost every $y$ and as a function of $y$ for almost every $x$.&lt;/p&gt;&#xA;" OwnerUserId="2838" LastActivityDate="2011-01-13T16:38:04.133" CommentCount="2" />
  <row Id="17403" PostTypeId="2" ParentId="17395" CreationDate="2011-01-13T16:39:16.517" Score="6" ViewCount="" Body="&lt;p&gt;From what I understand there are broadly two distinct applications of mathematics to neuroscience.  One uses mathematics to study the biological/chemical/physical aspects of the mechanisms in the brain, such as action potentials and the interactions between neurons.  The type of math used here is differential equations/dynamic systems.  Relevant wikipedia articles are &lt;a href=&quot;http://en.wikipedia.org/wiki/Biological_neuron_models&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Biological_neuron_models&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/Biological_neural_network&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Biological_neural_network&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other application is more abstract and concerns itself more with the computational aspects of neurons instead of the biophysical mechanisms by which they operate.  This area is more discrete and more aligned with subjects like computer science, artificial intelligence, and statistical learning.  Relevant wikipedia articles are &lt;a href=&quot;http://en.wikipedia.org/wiki/Artificial_neural_network&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Artificial_neural_network&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/Machine_learning&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Machine_learning&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="9" LastActivityDate="2011-01-13T16:39:16.517" />
  <row Id="17404" PostTypeId="2" ParentId="9916" CreationDate="2011-01-13T16:54:55.533" Score="0" ViewCount="" Body="&lt;p&gt;If $|G:H| = 2$ and $1 \neq g \in G$, then &#xA;$$G = H \; \dot{\cup} \; gH \;\;\; and \;\;\; G = H \; \dot{\cup} \; Hg$$&#xA;Therefore&#xA;$$gH = Hg \;\;\;\; \forall 1 \neq g \in G$$ &#xA;Of course&#xA;$$gH = Hg \;\;\;\; for \;\; 1 = g$$&#xA;Therefore&#xA;$$gH = Hg \;\;\;\; \forall g \in G$$ &lt;/p&gt;&#xA;" OwnerUserId="5802" LastEditorUserId="5802" LastEditorDisplayName="" LastEditDate="2011-01-14T05:13:37.427" LastActivityDate="2011-01-14T05:13:37.427" />
  <row Id="17405" PostTypeId="1" AcceptedAnswerId="22253" CreationDate="2011-01-13T16:59:57.793" Score="4" ViewCount="209" Body="&lt;p&gt;The coordinates of an arc of a circle of length $\frac{2pi}{p}$ are an algebraic number, and when $p$ is a Fermat prime you can find it in terms of square roots.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gauss said that the method applied to a lot more curves than the circle. Will you please tell if you know any worked examples of this (finding the algebraic points on other curves)?&lt;/p&gt;&#xA;" OwnerUserId="4290" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2011-04-19T17:59:46.360" LastActivityDate="2011-04-19T17:59:46.360" Title="Division of Other curves than circles." Tags="&lt;geometry&gt;&lt;algebra&gt;&lt;plane-curves&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="17406" PostTypeId="2" ParentId="17396" CreationDate="2011-01-13T17:05:10.020" Score="1" ViewCount="" Body="&lt;p&gt;The iteration function at stage $t$ is the highest power of $2$ that divides $t$ (this will result in a somewhat different code than yours, but still goes over all numbers, changing only one bit at a time). In general, since there are $n$ possible bits to flip, you will have to store at least $\log_2 n$ bits in memory (if you're not allowed to look at the actual counter).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some processors have a machine instruction that calculates this &quot;highest power of $2$ dividing a number&quot;, or perhaps something similar like &quot;leading zero&quot;. This is an efficient way to implement Gray counting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you don't have such a machine instruction, another good choice is to unroll the increment loop for some small $2^K$, and use some smart iterative approach (using lookup tables) to handle the updates which are not hard-coded.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2011-01-13T17:05:10.020" CommentCount="2" />
  <row Id="17407" PostTypeId="2" ParentId="17405" CreationDate="2011-01-13T17:26:42.413" Score="2" ViewCount="" Body="&lt;p&gt;See &lt;a href=&quot;http://mtl.math.uiuc.edu/book/export/html/26&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt;. In particular, the Gauss-Wantzel Theorem says the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Theorem.&lt;/strong&gt; A regular $n$-gon can be constructed with compass and straightedge iff&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$n$ is a Fermat prime&lt;/li&gt;&#xA;&lt;li&gt;$n$ is a power of $2$&lt;/li&gt;&#xA;&lt;li&gt;$n$ is a product of of a power of $2$ and distinct Fermat primes&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2011-01-13T17:26:42.413" CommentCount="1" />
  <row Id="17408" PostTypeId="1" AcceptedAnswerId="17410" CreationDate="2011-01-13T17:28:43.630" Score="4" ViewCount="270" Body="&lt;p&gt;A linear system of form $A\vec{x}=\vec{0}$ is called homogeneous. Why are all homogenous systems consistent?&lt;/p&gt;&#xA;" OwnerUserId="5804" LastActivityDate="2011-01-15T01:41:53.457" Title="Why are all homogenous systems consistent?" Tags="&lt;linear-algebra&gt;" AnswerCount="3" CommentCount="0" />
  <row Id="17409" PostTypeId="2" ParentId="17408" CreationDate="2011-01-13T17:31:15.267" Score="12" ViewCount="" Body="&lt;p&gt;There is the all zero solution (i.e. the trivial solution).&lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2011-01-13T17:31:15.267" />
  <row Id="17410" PostTypeId="2" ParentId="17408" CreationDate="2011-01-13T17:35:24.517" Score="4" ViewCount="" Body="&lt;p&gt;A system is defined as inconsistent if its row-reduced echelon form contains a row of form $\begin{bmatrix}&#xD;&#xA;0 &amp;amp; 0 &amp;amp; 0 &amp;amp; ... &amp;amp; 0 &amp;amp; | &amp;amp; k &#xD;&#xA;\end{bmatrix}$ where $k \neq 0$ and | is a separator within augmented matrix. Since your system equals $\vec{0}$, it is impossible to have $k \neq 0$, rendering the system consistent.&lt;/p&gt;&#xA;" OwnerUserId="3731" LastActivityDate="2011-01-13T17:35:24.517" CommentCount="2" />
  <row Id="17411" PostTypeId="2" ParentId="12914" CreationDate="2011-01-13T17:38:18.087" Score="4" ViewCount="" Body="&lt;p&gt;For $n=221$ (this is the example in Wikipedia), the liars are 21, 47, 174 and 200.  Now, 21*47 mod 221 gives 103, which is not in that list.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sequence of those composite $n$ for which the liars don't form a group seems to be 65, 85, 145, 185, 205, 221, 265, 305, 325, 365, 377, ... which seems to coincide with &#xA;&quot;A024409 Hypotenuses of more than one primitive Pythagorean triangle&quot; on OEIS (testet up to 629).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: Of course, I included 1 and -1 before checking for being a group.&lt;/p&gt;&#xA;" OwnerUserId="5616" LastEditorUserId="5616" LastEditorDisplayName="" LastEditDate="2011-01-13T18:01:02.233" LastActivityDate="2011-01-13T18:01:02.233" CommentCount="2" />
  <row Id="17412" PostTypeId="1" AcceptedAnswerId="17471" CreationDate="2011-01-13T17:54:29.750" Score="3" ViewCount="270" Body="&lt;p&gt;I ran across a group with presentation $&amp;lt;a,b | a^2b=ba^2, b^2a=ab^2, a^2=b^4&amp;gt;$.  Does this group have a name?&lt;/p&gt;&#xA;" OwnerUserId="1534" LastActivityDate="2011-01-14T08:21:19.387" Title="Can someone tell me what group this is?" Tags="&lt;group-theory&gt;" AnswerCount="2" CommentCount="7" />
  <row Id="17413" PostTypeId="2" ParentId="17375" CreationDate="2011-01-13T18:01:20.650" Score="3" ViewCount="" Body="&lt;p&gt;Think of $C$ as $R^2$ with the Euclidean metric.&#xA;Then a disc is not open if it &quot;has a &amp;lt;=&quot;! In that case it is closed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Open balls (or discs if you like) in $R^2$ are open if &quot;they have &amp;lt;&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's a rule of thumb. Now you need to prove it. You do so by using the definition, that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(DEF) A set B (in this case a ball) in a metric space is open, if for each point $b \in B$ you can draw a ball $O$ around $b$ such that $b \in O \subset B$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The proof starts by saying &quot;Let $x$ be any point in $D$...&quot;&#xA;Then you notice that $x$ lies on the line that goes through $x$ and the centre of $D$. Consider the distances $d_1 $, the distance between x and the centre and $d_2$, the distance between $x$ and the border of your set D. If you now draw a ball around $x$ of radius $r = min(d_1, d_2)/2$, this ball will be contained in $D$ and it will contain $x$ (obviously, since $x$ is its centre).&#xA;Therefore you proved (DEF) and you're done.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for proving that a ball &quot;&amp;lt;=&quot; is closed: to prove a set is closed you prove that its complement is open.&lt;/p&gt;&#xA;" OwnerUserId="5798" LastEditorUserId="5798" LastEditorDisplayName="" LastEditDate="2011-05-10T11:35:39.167" LastActivityDate="2011-05-10T11:35:39.167" />
  <row Id="17414" PostTypeId="2" ParentId="17395" CreationDate="2011-01-13T18:12:24.920" Score="0" ViewCount="" Body="&lt;p&gt;See for example &lt;a href=&quot;http://jn.physiology.org/content/85/1/448.full.pdf&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt;. A lot of information theory is used. &lt;/p&gt;&#xA;" OwnerUserId="2548" LastEditorUserId="2548" LastEditorDisplayName="" LastEditDate="2011-01-13T18:18:42.497" LastActivityDate="2011-01-13T18:18:42.497" />
  <row Id="17415" PostTypeId="1" CreationDate="2011-01-13T18:28:57.250" Score="3" ViewCount="230" Body="&lt;p&gt;How does someone mathematically round a number to its nearest integer?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example &lt;code&gt;1.2&lt;/code&gt; would round down to &lt;code&gt;1&lt;/code&gt; and &lt;code&gt;1.7&lt;/code&gt; would round up to &lt;code&gt;2&lt;/code&gt;&lt;/p&gt;&#xA;" OwnerUserId="5806" LastEditorUserId="367" LastEditorDisplayName="" LastEditDate="2011-01-13T21:20:09.253" LastActivityDate="2011-01-13T21:20:09.253" Title="How do you mathematically round a number?" Tags="&lt;numerical-methods&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="17416" PostTypeId="2" ParentId="17415" CreationDate="2011-01-13T18:34:24.473" Score="1" ViewCount="" Body="&lt;p&gt;This is equivalent to looking at the floor and ceiling functions. See &lt;a href=&quot;http://en.wikipedia.org/wiki/Rounding&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt;. In particular, rounding a number $y$ to an integer $q$ is the same thing as looking as $q = \text{floor}(y)$ or $q = \text{ceil}(y)$. &lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2011-01-13T18:34:24.473" />
  <row Id="17417" PostTypeId="2" ParentId="17415" CreationDate="2011-01-13T18:36:20.243" Score="5" ViewCount="" Body="&lt;p&gt;You could use the &quot;ceiling-function / floor-function&quot; which always rounds an integer up/down ( &lt;a href=&quot;http://en.wikipedia.org/wiki/Floor_and_ceiling_functions&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Floor_and_ceiling_functions&lt;/a&gt; ). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want the number to be rounded the way you want you could just do&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\text{round}(x)= \lceil x-\frac{1}{2} \rceil$ Note that in this case $\text{round}(1.5)=1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or $\text{round&amp;#39;}(x)= \lfloor x+\frac{1}{2} \rfloor$ Note that in this case $\text{round&amp;#39;}(1.5)=2$&lt;/p&gt;&#xA;" OwnerUserId="3123" LastActivityDate="2011-01-13T18:36:20.243" CommentCount="2" />
  <row Id="17418" PostTypeId="2" ParentId="17408" CreationDate="2011-01-13T18:38:50.143" Score="4" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\ $ Zero is a root of every linear map $\rm\:A\:,\:$ since linear maps must  preserve $\rm\ 0 + 0 = 0\:,\ $ i.e.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\rm\ A\ (0 + 0\ =\ 0)\ \ \to\ \ A(0) + A(0)\ =\ A(0)\ \ \Rightarrow\ \ A(0) = 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More generally: monoid homomorphisms preserve idempotents, but the only idempotent element in a &lt;em&gt;cancellative&lt;/em&gt; monoid is the neutral element $\rm\ a + a = a\ \Rightarrow\ a = 0\:.$&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2011-01-15T01:41:53.457" LastActivityDate="2011-01-15T01:41:53.457" />
  <row Id="17419" PostTypeId="2" ParentId="17412" CreationDate="2011-01-13T19:34:10.797" Score="4" ViewCount="" Body="&lt;p&gt;Letting $W$ be an arbitrary word, using the third relation $a^2 = b^4$, we can assume that $W$ contains alternating $a$ and $b^k$ factors. Using the second relations $ab^2 = b^2a$ you can transform $W$ to the form &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ W = b^{k} (ab)^l \mbox{ or } b^k (ba)^l $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But now note $(ba)(ab) = b^6 \implies ba = b^6(ab)^{-1}$, so every word $W$ is equivalent to &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ W = b^k (ab)^l $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for some pair of integer $k,l$. Let us write out the multiplication table for this group in the presentation $(k,l)$. First, the identity is $(0,0)$. Observe that $(2k,0)$ commutes with all other elements. And $(k,l) = (k,0)(0,l)$, but $(0,l)(1,0) = (1+6l,-l)$. In other words, you can reduce to the group $\langle u,b | u^kb = b^{1+6k}u^{-k}\rangle$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or, you can think of the group as the free abelian group $\mathbb{Z}^2 = \langle u,v | uv = vu\rangle$ plus the additional generator $b$ where $b^2 = v$ and $b^{-1}ub = b^6 u^{-1}$. (Nope, I also don't know what this group is, but maybe someone else can take it from here.)&lt;/p&gt;&#xA;" OwnerUserId="1543" LastActivityDate="2011-01-13T19:34:10.797" CommentCount="2" />
  <row Id="17420" PostTypeId="1" AcceptedAnswerId="17422" CreationDate="2011-01-13T19:50:37.400" Score="1" ViewCount="85" Body="&lt;p&gt;I am trying to make a formula where a &quot;table lookup&quot; is not required. Given these two tables:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;  Days   Discount          Refs         Rent_Cost&#xA;  ------------------       ------------------------&#xA;  15     0.00              0-250        $0.200 &#xA;  30     0.05              251-500      $0.210 &#xA;  60     0.10              501-750      $0.220 &#xA;  90     0.18              751-1000     $0.230 &#xA;  150    0.25              1001-1250    $0.240 &#xA;  240    0.30              1251-1500    $0.250 &#xA;                           1501-1750    $0.260 &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;Total_Cost(Refs,Days) = Refs * Rent_Cost * (1-Discount) * (Days/30)&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where &lt;code&gt;Rent_Cost&lt;/code&gt; and &lt;code&gt;Discount&lt;/code&gt; would have to be looked up on the two tables above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in order to find a formula for Rent_Cost, I used FindGraph (http://www.uniphiz.com/findgraph.htm) and plotted all 0 through 1750 Refs with what their Rent_Cost was. I got inaccurate results but it looks as if I played around enough I could get to a formula that would work if I rounded by 0.01. I came up with:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;Rent_Cost(Refs) = 0.195 + (4E-05 * Refs)&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where Rent_Cost would be rounded to the nearest hundredth. This works 100% with no errors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then I did the same with Days and Discount and came up with&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;Discount(Days) = 0.34 - 0.1 ^ ([Days - 137]/-100)&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This works for all Days except for 60. 60 comes out as 0.12 instead of 0.10. You also have to round to the nearest hundredth on this one as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you were to plugin Discount and Rent_Cost into the original formula:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;Total_Cost(Refs,Days) = Refs * (0.195 + (4E-05 * Refs)) * (1-(0.34 - 0.1 ^ ([Days - 137]/-100))) * (Days/30)&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I am asking how some of you would go about improving the formula and hopefully get rid of the inaccurately of my Discount function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: I do not know how to use MathJaX. Please edit my post if its hard to read?&lt;/p&gt;&#xA;" OwnerUserId="5806" LastActivityDate="2011-01-13T20:11:51.807" Title="Is there a more accurate / simpler formula than this?" Tags="&lt;statistics&gt;&lt;functions&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="17421" PostTypeId="2" ParentId="4626" CreationDate="2011-01-13T20:02:46.983" Score="0" ViewCount="" Body="&lt;p&gt;Using the following relations (in Perron, &lt;em&gt;Die Lehre von den Kettenbrüchen&lt;/em&gt;, Band II, 1957 and Sergey Khrushchev, &lt;em&gt;Orthogonal Polynomials and&#xA;Continued Fractions: From Euler's Point of View&lt;/em&gt;, 2008) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{equation}&#xD;&#xA;A_{n}B_{n-1}-A_{n-1}B_{n}=\left( -1\right) ^{n-1}a_{1}a_{2}\cdots a_{n} &#xD;&#xA;\nonumber&#xD;&#xA;\end{equation}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{equation}&#xD;&#xA;A_{n}B_{n-2}-A_{n-2}B_{n}=\left( -1\right) ^{n}b_{n}a_{1}a_{2}\cdots a_{n-1}&#xD;&#xA;\end{equation}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{equation}&#xD;&#xA;A_{n}B_{n-3}-A_{n-3}B_{n }=(-1)^{n-1}a_{1}a_{2}\cdots a_{n -2}\left(&#xD;&#xA;b_{n}b_{n-1}+a_{n}\right)   &#xD;&#xA;\end{equation}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;we can derive&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{eqnarray}&#xD;&#xA;&amp;amp;&amp;amp;\left( A_{n}+w_{n}A_{n-1}\right) \left( B_{n-1}+w_{n-1}B_{n-2}\right)   \\&#xD;&#xA;&amp;amp;&amp;amp;-\left( A_{n-1}+w_{n-1}A_{n-2}\right) \left( B_{n}+w_{n}B_{n-1}\right)  \\&#xD;&#xA;&amp;amp;=&amp;amp;-\left( -1\right) ^{n }a_{1}a_{2}\cdots a_{n-1}\left(&#xD;&#xA;a_{v}-w_{n-1}\left( b_{n}+w_{n}\right) \right)   &#xD;&#xA;\end{eqnarray}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{eqnarray}&#xD;&#xA;&amp;amp;&amp;amp;\left( A_{n -1}+w_{n-1}A_{n-2}\right) \left(&#xD;&#xA;B_{n-2}+w_{n-2}B_{n-3}\right)   \nonumber \\&#xD;&#xA;&amp;amp;&amp;amp;-\left( A_{n-3}+w_{n-3}A_{\nu -3}\right) \left( B_{n-1}+w_{n-1}B_{n&#xD;&#xA;-2}\right)    \\&#xD;&#xA;&amp;amp;=&amp;amp;-\left( -1\right) ^{n -1}a_{1}a_{2}\cdots a_{n-2}\left(&#xD;&#xA;a_{n-1}-w_{n-2}\left( b_{n-1}+w_{n-1}\right) \right)   &#xD;&#xA;\end{eqnarray}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;as well as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{eqnarray}&#xD;&#xA;&amp;amp;&amp;amp;\left( A_{n}+w_{n}A_{n-1}\right) (B_{n-2}+w_{n-2}B_{n-3})   \\&#xD;&#xA;&amp;amp;&amp;amp;-(A_{n-2}+w_{n-2}A_{n-3})\left( B_{n}+w_{n}B_{n -1}\right)    \\&#xD;&#xA;&amp;amp;=&amp;amp;\left( -1\right) ^{n}b_{n }a_{1}a_{2}\cdots a_{n -1}+w_{n&#xD;&#xA;-2}(-1)^{n -1}a_{1}a_{2}\cdots a_{n -2}\left( b_{n}b_{n-1}+a_{n}\right) &#xD;&#xA; \\&#xD;&#xA;&amp;amp;&amp;amp;+w_{n}\left( -1\right) ^{n -2}a_{1}a_{2}\cdots a_{n-1}+w_{n-2}\left(&#xD;&#xA;-1\right) ^{n-1}b_{n-1}a_{1}a_{2}\cdots a_{n-2} &#xD;&#xA;\end{eqnarray}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{eqnarray*}&#xD;&#xA;d_{n} &amp;amp;=&amp;amp;\frac{\left( A_{n}+w_{n}A_{n-1}\right)&#xD;&#xA;(B_{n-2}+w_{n-2}B_{n-3})-(A_{n-2}+w_{n-2}A_{n-3})\left(&#xD;&#xA;B_{n}+w_{n}B_{n-1}\right) }{%&#xD;&#xA;(A_{n-1}+w_{n-1}A_{n-2})(B_{n-2}+w_{n-2}B_{n-3})-(A_{n-2}+w_{n-2}A_{n-3})(B_{n-1}+w_{n-1}B_{n-2})%&#xD;&#xA;} \\&#xD;&#xA;&amp;amp;=&amp;amp;\frac{\left( -1\right) ^{n}b_{n}a_{1}a_{2}\cdots&#xD;&#xA;a_{n-1}+w_{n-2}(-1)^{n-1}a_{1}a_{2}\cdots a_{n-2}\left(&#xD;&#xA;b_{n}b_{n-1}+a_{n}\right) }{-\left( -1\right) ^{n -1}a_{1}a_{2}\cdots&#xD;&#xA;a_{n-2}\left( a_{n-1}-w_{n-2}\left( b_{n -1}+w_{n-1}\right) \right) } \\&#xD;&#xA;&amp;amp;&amp;amp;+\frac{w_{n }\left( -1\right) ^{n-2}a_{1}a_{2}\cdots&#xD;&#xA;a_{n-1}+w_{n-2}\left( -1\right) ^{n-1}b_{n -1}a_{1}a_{2}\cdots a_{n-2}}{%&#xD;&#xA;-\left( -1\right) ^{n-1}a_{1}a_{2}\cdots a_{n-2}\left( a_{v-1}-w_{n-2}\left(&#xD;&#xA;b_{n-1}+r_{n-1}\right) \right) } \\&#xD;&#xA;&amp;amp;=&amp;amp;\dfrac{\left( b_{n}+w_{n}\right) \left(&#xD;&#xA;a_{n-1}-w_{n-2}(b_{n-1}+w_{n-1})\right) -w_{n-2}\left(&#xD;&#xA;a_{n}-w_{n-1}(b_{n}+w_{n})\right) }{a_{n-1}-w_{n-2}(b_{n-1}+w_{n-1})} \\&#xD;&#xA;&amp;amp;=&amp;amp;b_{n }+w_{n}-w_{n-2}\dfrac{a_{n }-w_{n-1}(b_{n }+w_{n})}{a_{n&#xD;&#xA;-1}-w_{n-2}(b_{n-1}+w_{n-1})}&#xD;&#xA;\end{eqnarray*}$$&lt;/p&gt;&#xA;" OwnerUserId="752" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2011-01-13T20:22:43.467" LastActivityDate="2011-01-13T20:22:43.467" CommunityOwnedDate="2011-01-13T20:02:46.983" />
  <row Id="17422" PostTypeId="2" ParentId="17420" CreationDate="2011-01-13T20:11:51.807" Score="3" ViewCount="" Body="&lt;p&gt;First, I'd rewrite the rent cost function as &lt;code&gt;Rent_Cost(Refs) = max(0.20, 0.19 + 0.01 * ceiling(Refs / 250))&lt;/code&gt; or $$\text{Rent Cost}(\text{Refs})=\max(0.20, 0.19+0.01\cdot\left\lceil\frac{\text{Refs}}{250}\right\rceil)$$ (no rounding needed).  (In fact, I suspect that your current formula might give the wrong result at the boundary values of 250, 500, 750, ...)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, to the discount function.  First, note that $\left\lceil\frac{\text{signum}(x-k)+1}{2}\right\rceil$ (where $\text{signum}(x)$ is $0$ if $x=0$, $1$ if $x&amp;gt;0$, and $-1$ if $x&amp;lt;0$) is $0$ if $x&amp;lt;k$ and $1$ if $x\ge k$.  Now, $$\begin{align}\text{Discount}(\text{Days})&amp;amp;=0.05\cdot\left\lceil\frac{\text{signum}(\text{Days}-30)+1}{2}\right\rceil+0.05\cdot\left\lceil\frac{\text{signum}(\text{Days}-60)+1}{2}\right\rceil\\&amp;amp;+0.08\cdot\left\lceil\frac{\text{signum}(\text{Days}-90)+1}{2}\right\rceil+0.07\cdot\left\lceil\frac{\text{signum}(\text{Days}-150)+1}{2}\right\rceil\\&amp;amp;+0.05\cdot\left\lceil\frac{\text{signum}(\text{Days}-240)+1}{2}\right\rceil.\end{align}$$&#xA;This is also an exact formula, no rounding needed.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;edit&lt;/em&gt;: Were I doing this in Excel, I'd actually use a bunch of nested &lt;code&gt;IF()&lt;/code&gt; statements:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Discount(Days) = IF(Days &amp;lt; 30, 0.00,&#xA;                    IF(Days &amp;lt; 60, 0.05,&#xA;                       IF(Days &amp;lt; 90, 0.10,&#xA;                          IF(Days &amp;lt; 150, 0.18,&#xA;                             IF(Days &amp;lt; 240, 0.25,&#xA;                                0.30)))))&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="72" LastActivityDate="2011-01-13T20:11:51.807" CommentCount="1" />
  <row Id="17423" PostTypeId="1" CreationDate="2011-01-13T20:19:04.093" Score="0" ViewCount="224" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The formula for the &lt;a href=&quot;http://en.wikipedia.org/wiki/Gordon_model&quot; rel=&quot;nofollow&quot;&gt;gordon growth model&lt;/a&gt; is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\hspace{1in}P= \sum_{t=1}^{\infty} D\times\frac{(1+g)^t}{(1+k)^t}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So summing the infinite series we get:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\hspace{1in}P=\frac{D(1+g)}{k-g} \hspace{2in}\mbox{(1)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's my attempt to arrive at equation (1):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Geometric_series#Sum&quot; rel=&quot;nofollow&quot;&gt;sum&lt;/a&gt; of a geometric series is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\hspace{1in}\frac{a}{1-r}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So,plugging in the values:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\hspace{1in}\frac{D}{1-\frac{1+g}{1+k}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\hspace{1in}\frac{D}{\frac{1+k-(1+g)}{1+k}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\hspace{1in}\frac{D}{\frac{k-g}{1+k}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\hspace{1in}P=\frac{D(1+k)}{1+g}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Appreciate if someone could highlight where I made a mistake.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks you.&lt;/p&gt;&#xA;" OwnerUserId="2973" LastActivityDate="2011-01-13T20:19:04.093" Title="How to derive Gordon growth model" Tags="&lt;homework&gt;&lt;sequences-and-series&gt;&lt;algebra&gt;&lt;finance&gt;" CommentCount="3" />
  <row Id="17424" PostTypeId="1" AcceptedAnswerId="17429" CreationDate="2011-01-13T20:22:45.340" Score="2" ViewCount="121" Body="&lt;p&gt;I'm reading a text on ray tracing. There is this section about radiometric quantities where radiance is defined as&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;$L = \frac{d^2\Phi}{dA cos\Theta d\omega}$&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;$\Phi$ is the radiant flux&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Theta$ is the solid angle (sr) subtended by the observation or measurement&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\omega$ is the incidence angle measured from the surface normal&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is just one of many equations using $d$ and $d^2$. I'm pretty sure that $d$ has something to do with differential equations. I already read some texts on differential equations but I still don't understand the meaning of $d$ and $d^2$ in this context.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can someone explain this to me or point me to some reference/resource/book whatever?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Especially the $d^2$ puzzles me.&lt;/p&gt;&#xA;" OwnerUserId="3861" LastActivityDate="2011-01-13T20:41:08.293" Title="What does the little d and d^2 mean in equations?" Tags="&lt;trigonometry&gt;&lt;differential-equations&gt;&lt;physics&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" />
  <row Id="17425" PostTypeId="1" AcceptedAnswerId="17430" CreationDate="2011-01-13T20:26:07.537" Score="1" ViewCount="151" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is ; How can I solve the following inverse of function question?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$y=x^2-6x+4$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance,&lt;/p&gt;&#xA;" OwnerUserId="3669" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-13T20:37:39.700" LastActivityDate="2011-01-13T21:16:14.583" Title="How can I find inverse of function ?" Tags="&lt;algebra-precalculus&gt;&lt;functions&gt;" AnswerCount="5" CommentCount="3" />
  <row Id="17426" PostTypeId="2" ParentId="17425" CreationDate="2011-01-13T20:29:53.637" Score="1" ViewCount="" Body="&lt;p&gt;Write $x = y^2-6y+4$ and solve for $y$. You need to restrict the domain and codomain so that the function is bijective (i.e. the inverse exists). &lt;/p&gt;&#xA;" OwnerUserId="2548" LastEditorUserId="2548" LastEditorDisplayName="" LastEditDate="2011-01-13T20:37:26.957" LastActivityDate="2011-01-13T20:37:26.957" CommentCount="1" />
  <row Id="17427" PostTypeId="2" ParentId="17425" CreationDate="2011-01-13T20:31:14.357" Score="0" ViewCount="" Body="&lt;p&gt;Simply rearrange the equation to solve for $x$ in terms of $y$. This is precisely the definition of the inverse.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint: factorise the quadratic expression.&lt;/p&gt;&#xA;" OwnerUserId="56" LastActivityDate="2011-01-13T20:31:14.357" />
  <row Id="17428" PostTypeId="2" ParentId="17425" CreationDate="2011-01-13T20:31:20.427" Score="1" ViewCount="" Body="&lt;p&gt;If your function is $f(x)=x^2-6x+4$, then its inverse (which is not a function) can be found by using the quadratic formula to find the solutions to the equation $0=x^2-6x+4-f(x)$ (which will give a formula for $x$, which becomes $f^{-1}(x)$, in terms of $f(x)$, which becomes $x$).&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2011-01-13T20:31:20.427" />
  <row Id="17429" PostTypeId="2" ParentId="17424" CreationDate="2011-01-13T20:33:19.517" Score="3" ViewCount="" Body="&lt;p&gt;This is fundamental notation in differential calculus. I suggested you pick up a book on the subject and read up; you won't regret it, as its very useful knowledge whatever you do!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your specific cases, the expression is in fact a &lt;em&gt;second-order partial derivative&lt;/em&gt;. The &lt;code&gt;d&lt;/code&gt;s should be written in the curly style - this may be a fault of wherever you saw the expression from.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are some of the basics of notation, to get you started.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{dy}{dx}$ = the derivative (rate of change) of $y$ with respect to $x$. (1st order derivative)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{d^2y}{dx^2}$ = the derivative (rate of change) of $dy/dx$ with respect to $x$. (2nd order derivative, also called &lt;em&gt;curvature&lt;/em&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: a common elementary mistake is to treat differentials as fractions. They are indirectly related, but do not treat them the same.&lt;/p&gt;&#xA;" OwnerUserId="56" LastEditorUserId="56" LastEditorDisplayName="" LastEditDate="2011-01-13T20:41:08.293" LastActivityDate="2011-01-13T20:41:08.293" CommentCount="1" />
  <row Id="17430" PostTypeId="2" ParentId="17425" CreationDate="2011-01-13T20:34:59.343" Score="6" ViewCount="" Body="&lt;p&gt;This function does not have an inverse, since &#xA;\begin{align*}&#xA;y\left(3+\sqrt{5}\right) &amp;amp;= (3+\sqrt{5})^2 - 6(3+\sqrt{5}) + 4\\&#xA;&amp;amp;= 9 + 6\sqrt{5} + 5 -18 - 6\sqrt{5} + 4 = 0,\\&#xA;y\left(3-\sqrt{5}\right) &amp;amp;= (3-\sqrt{5})^2 - 6(3-\sqrt{5}) + 4\\&#xA;&amp;amp;= 9 - 6\sqrt{5} + 5 - 18 + 6\sqrt{5} + 4 = 0.&#xA;\end{align*}&#xA;Since the function is not one to one, it does not have an inverse.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can, on the other hand, restrict the domain; for instance, since&#xA;$$ y = x^2 -6x + 4 = (x-3)^2 - 5$$&#xA;you can restrict yourself to $[3,\infty)$, where the function &lt;em&gt;is&lt;/em&gt; one-to-one. In that case, note that $y\geq -5$, and you have&#xA;\begin{align*}&#xA;y &amp;amp;= x^2 - 6x + 4\\&#xA;x^2 - 6x + (4-y) &amp;amp;= 0&#xA;\end{align*}&#xA;and now you can use the quadratic formula to find an expression for $x$ in terms of $y$. The function will have domain $[-5,\infty)$, and will have range $[3,\infty)$.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-13T21:16:14.583" LastActivityDate="2011-01-13T21:16:14.583" CommentCount="5" />
  <row Id="17431" PostTypeId="2" ParentId="17425" CreationDate="2011-01-13T20:35:26.157" Score="3" ViewCount="" Body="&lt;p&gt;First you have to be careful, because without restricting your domain, the function is not invertible.  For example $2^2-6\cdot 2+4 = -4 = 4^2 - 6\cdot 4 +4$, which shows that $x\mapsto x^2-6x+4$ is not one-to-one.  If you restrict the domain to $x\geq3$ or $x\leq3$, then the resulting function will be one-to-one.  You can see this by finding the vertex of the parabola $y=x^2-6x+4=(x-3)^2-5$, and noticing that the function is strictly monotone (increasing or decreasing) on either side of the vertex.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you define $f(x)=x^2-6x+4$ on $[3,\infty)$, then you can find the inverse function as indicated in the other answers.  The domain will be the range of the previous function $[-5,\infty)$, and for each $x\in[-5,\infty)$, $f^{-1}(x)$ is the unique solution $y$ to $x=y^2-6y+4$ in the interval $[3,\infty)$, which can be found by completing the square or using the quadratic formula. Something similar can be said for the function $g(x)=x^2-6x+4$ on $(-\infty,3]$, where $g^{-1}$ will now have a different range than $f^{-1}$.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2011-01-13T20:40:40.537" LastActivityDate="2011-01-13T20:40:40.537" CommentCount="1" />
  <row Id="17432" PostTypeId="1" AcceptedAnswerId="17433" CreationDate="2011-01-13T20:45:30.233" Score="6" ViewCount="248" Body="&lt;p&gt;Let $X$ be a set and consider the collection $\mathcal{A}(X)$ of countable or cocountable subsets of $X$, that is, $E \in \mathcal{A}(X)$ if $E$ is countable or $X-E$ is countable. If $X$ is countable, then $\mathcal{A}(X)$ coincides with the power set $\mathcal{P}(X)$ of $X$. Now suppose that $X$ is uncountable. Assuming the axiom of choice, we can conclude that $\mathcal{A}(X) \ne \mathcal{P}(X)$, since $|X| = |X| + |X|$. So the question is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Can we prove in ZF that $\mathcal{A}(X) \ne \mathcal{P}(X)$ for every uncountable set $X$?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm assuming that a set $X$ is uncountable if there is no injective function $f : X \rightarrow \mathbb{N}$.&lt;/p&gt;&#xA;" OwnerUserId="329" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2011-01-13T23:31:28.777" LastActivityDate="2011-02-12T22:32:24.757" Title="Uncountable subset with uncountable complement, without the Axiom of Choice" Tags="&lt;set-theory&gt;&lt;axiom-of-choice&gt;" AnswerCount="1" CommentCount="8" />
  <row Id="17433" PostTypeId="2" ParentId="17432" CreationDate="2011-01-13T21:03:32.383" Score="10" ViewCount="" Body="&lt;p&gt;A &lt;em&gt;Dedekind finite&lt;/em&gt; set is one all of whose subsets have strictly smaller cardinality. If $X$ is infinite Dedekind finite (iDf), then $X$ and ${\mathbb N}$ are incomparable in size. This means that no subset of $X$ is countable unless it is finite. Certain Dedekind finite sets may be &lt;em&gt;amorphous&lt;/em&gt;, this means that any subset is finite or cofinite. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is consistent that there are iDf sets, but no amorphous sets.  It is also consistent that there are infinite amorphous sets, so the answer to your question is no, in general. &#xA;However, if every Dedekind finite set is amorphous, then there are no iDf sets. Hence, if there is an iDf set at all, there is one that can be split into two infinite sets, and neither is countable. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Of course, under choice, there are no iDf sets and every uncountable set admits an uncountable subset with uncountable complement.)&lt;/p&gt;&#xA;" OwnerUserId="462" LastEditorUserId="462" LastEditorDisplayName="" LastEditDate="2011-02-12T22:32:24.757" LastActivityDate="2011-02-12T22:32:24.757" CommentCount="1" />
  <row Id="17434" PostTypeId="1" CreationDate="2011-01-13T21:13:40.357" Score="3" ViewCount="96" Body="&lt;p&gt;Let $M$ be the positive borel measures on a hausdorff topological space $X$, which are finite on compacts sets $--$ i.e. the real cone of radon measures.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am given a definition of a derivative of two radon measures, say $\mu, \eta \in M$, as follows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$D^+_\mu \eta (x)&#xD;&#xA;:= \left\{ \begin{array}{1 1}&#xD;&#xA;\limsup \limits_{r \rightarrow 0} \frac{ \eta(\overline{B_r(0)}) }{ \mu(\overline{B_r(0)}) }&#xD;&#xA;&amp;amp; \mu(\overline{B_r(0)}) &amp;gt; 0 \forall r &amp;gt; 0\\&#xD;&#xA;\infty &amp;amp; \mu(\overline{B_r(0)}) = 0 \forall r &amp;gt; 0&#xD;&#xA;\end{array} \right. $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$D^-_\mu \eta (x)&#xD;&#xA;:= \left\{ \begin{array}{1 1}&#xD;&#xA;\liminf \limits_{r \rightarrow 0} \frac{ \eta(\overline{B_r(0)}) }{ \mu(\overline{B_r(0)}) }&#xD;&#xA;&amp;amp; \mu(\overline{B_r(0)}) &amp;gt; 0 \forall r &amp;gt; 0\\&#xD;&#xA;\infty &amp;amp; \mu(\overline{B_r(0)}) = 0 \forall r &amp;gt; 0&#xD;&#xA;\end{array} \right. $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and we write in case of equality&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$D_\mu \eta (x) := D^+_\mu \eta (x) = D^-_\mu \eta (x)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I could accept this ad-hoc definition &quot;as is&quot;. Nevertheless, I know there is a canonical topology on the linear space of bounded radon measures, so is probably on the cone of positive (not necessarily bounded) radon measures (Furthermore, I guess the thoughs pass over the linear space of signed Radon measures)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, I would like to obtain a good intuition how the definition of derivative of above relates to the canonical definition (if applicable) of derivative on infinite-dimensional vector spaces. (I suppose there is such a connection). Can give a explanation or tell me a good tight resource to look this up?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you.&lt;/p&gt;&#xA;" OwnerUserId="3557" LastEditorUserId="3557" LastEditorDisplayName="" LastEditDate="2011-01-14T20:38:48.743" LastActivityDate="2011-01-14T20:38:48.743" Title="How to define the derivative of Radon measures" Tags="&lt;analysis&gt;&lt;measure-theory&gt;&lt;functional-analysis&gt;&lt;topological-vector-spaces&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="1" />
  <row Id="17435" PostTypeId="2" ParentId="5331" CreationDate="2011-01-13T21:50:07.560" Score="2" ViewCount="" Body="&lt;p&gt;There is an exercise in Rudin's &lt;em&gt;Real and complex analysis&lt;/em&gt; whose solution would answer your question, #14 in Chapter 15:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Prove that there is a sequence $\{\alpha_n\}$ with $0\lt\alpha_n\lt1$, which tends to $1$ so rapidly that the Blaschke product with zeros at the points $\alpha_n$ satisfies the condition &#xA;  $$\limsup_{r\to1}|B(r)|=1.$$&#xA;  Hence this $B$ has no radial limit at $z=1$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(The previous exercise says that the limit is $0$ if $\alpha_n=1-n^{-2}$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instead of trying to solve it (with the guess of something like $\alpha_n=1-4^{-n}$), I found the article &quot;&lt;a href=&quot;http://projecteuclid.org/euclid.rmjm/1250127873&quot; rel=&quot;nofollow&quot;&gt;On functions with Weierstrass boundary points everywhere&lt;/a&gt;&quot; by Campbell and Hwang, which says on page 510 (page 4 of the pdf):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Let $B(z)$ be the Blaschke product with zeros at $z=1-\exp(-n)$, for $n=1,2,\ldots$.  Then $B(z)$ has no radial limit at $z=1$....&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The authors cite page 12 of &quot;Sequential and continuous limits of meromorphic functions&quot; by Bagemihl and Seidel for this fact, but I do not currently have access to that article.  Hopefully you can track it down to get your question answered, or perhaps someone will take up the challenge of solving Rudin's problem.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2011-01-13T22:21:20.940" LastActivityDate="2011-01-13T22:21:20.940" CommentCount="3" />
  <row Id="17436" PostTypeId="2" ParentId="3320" CreationDate="2011-01-13T22:01:03.543" Score="3" ViewCount="" Body="&lt;p&gt;Yes, the ratio $\pi$ of a circle's circumference to its diameter shows up in many, many places where one might not expect it!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One partial explanation (similar in spirit to &quot;circles lurk everywhere&quot;) is that the equation for a circle is a $quadratic$ (eg. $x^2+y^2 = r^2$.)  After nice linear functions, the next most commonly used functions are quadratic functions and everywhere one runs into a quadratic function, a trig substitution (e.g. $x = r \cos \theta; y=r\sin \theta$) may be useful, turning the quadratic function into something involving $\pi.$  This explains the antiderivative $\int \frac{1}{1+x^2} dx$ involving $\pi$, the sum of reciprocals of squares $\sum^\infty\frac{1}{k^2}$ involving $\pi$ and the area under the Gaussian distribution involving $\pi$.  And so on....&lt;/p&gt;&#xA;" OwnerUserId="5680" LastActivityDate="2011-01-13T22:01:03.543" CommentCount="5" />
  <row Id="17437" PostTypeId="1" CreationDate="2011-01-13T22:43:01.857" Score="2" ViewCount="112" Body="&lt;p&gt;Hello! I'm not so sure if my &quot;solution&quot; to the following problem is correct. Can you please let me know if is? or what is wrong? I'm new with this stuff so sometimes I get confused with it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway here it is: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose $Y$ is an arbitrary compactification of $X$, if $\beta X$ is the Stone-Cech compactification, prove there is a continuous, surjective closed map $g: \beta X \rightarrow Y$ that equals the identity on X.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My work:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $Y$ is a compactification of $X$ then by definition it is a Hausdorff compact space which contains a dense copy of $X$, so we can view $X$ as a subspace of $Y$.&#xA;Hence we can consider the inclusion map $i: X \rightarrow Y$. This is certainly continuous so by the universal property of extension there exists a cts extension:$g: \beta X \rightarrow Y$. So all it remains to show is that $g$ is surjective and closed. Now it is closed because $\beta X$ is compact , $Y$ is Hausdorff and any continuous map from a compact space to a Hausdorff space is closed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now to show $g$ is surjective I'm kinda stuck. Here's what I tried. By definition $X$ is dense in $Y$. But $\beta X$ contains $X$ so $\beta X$ is dense in $Y$ as well. Now the continuous image of a dense set is dense hence:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\overline{g(\beta X)} = Y$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But $\beta X$ is compact and the continuous image of a compact set is compact and a compact subset in a Hausdorff space (Y) is closed so $\overline{g(\beta X)} = g(\beta X)$. Thus $g(\beta X) = Y$.&lt;/p&gt;&#xA;" OwnerUserId="5817" LastActivityDate="2011-01-13T23:24:39.650" Title="Existence of closed, surjective map defined in the Stone-Cech compactification" Tags="&lt;topology&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="17438" PostTypeId="1" AcceptedAnswerId="17441" CreationDate="2011-01-13T23:24:29.043" Score="3" ViewCount="134" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am told that when transforming coordinates $\nabla (f(Ax)) = A^T(\nabla f)(Ax)$, however I read both of these as &quot;the gradient of f, where f is a function of Ax, where A is a matrix and x is a vector&quot;, with the second multiplied with $A^T$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is obviously incorrect, so how am I misunderstanding the notation?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks, Ash&lt;/p&gt;&#xA;" OwnerUserId="1763" LastActivityDate="2011-01-13T23:57:35.913" Title="Brackets around gradients" Tags="&lt;vector-calculus&gt;" AnswerCount="1" />
  <row Id="17439" PostTypeId="2" ParentId="17437" CreationDate="2011-01-13T23:24:39.650" Score="1" ViewCount="" Body="&lt;p&gt;We know that $X \mapsto cX$ extends to a continuous map $\beta X \mapsto cX$. And $X \mapsto cX$ has a dense image. It follows that $g: \beta X \mapsto Y$ is surjective. &lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2011-01-13T23:24:39.650" CommentCount="2" />
  <row Id="17440" PostTypeId="1" CreationDate="2011-01-13T23:33:53.760" Score="4" ViewCount="123" Body="&lt;p&gt;This exrcise is from the first section of Marden:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Exercise 12.&lt;/strong&gt; Let the interior of a piecewise regular curve $C$ contain the origin $\cal O$ and be star-shaped with respect to $\cal O$.  If the complex numbers $a_1, a_2, \ldots, a_m$ are given, then $n$ and $b_{m+1}, b_{m+2}, \ldots, b_n$ may be determined so that all of the zeros of&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$F(z) = 1 + a_1 z + \cdots + a_m z^m + b_{m+1} z^{m+1} + \cdots + b_n z^n$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;lie on $C$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint:&lt;/strong&gt; Choose the zeros $\zeta_j$ of $G(z) = z^n F(1/z)$ so that $z_j = 1/\zeta_j$ are points of $C$ and so that the Newton-Girard formulas&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$s_k + s_{k-1} a_1 + \cdots + k a_k = 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;are satisfied by the sums $s_r$ of the $r^{\text{th}}$ powers of the $\zeta_j$.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I understand how the requirements of the hint will imply the result, but I do not know how to establish them.  Indeed, if we can satisfy those requirements, then the remaining $b_j$ will be determined by the further Newton-Girard formulas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a generalization of the previous problem in the book:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Exercise 11:&lt;/strong&gt; If $F(z) = 1 + a_1 z + b_2 z^2 + \cdots + b_n z^n$, the quantities $n, b_2, \ldots, b_n$ may be so determined so that all the zeros of F lie on the unit circle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The solution to this is simpler: we choose $n &amp;gt; -a_1$ and values $\zeta_1, \ldots, \zeta_n$ on the unit circle with centroid $-a_1/n$.  Viète's formulas tell us that these are the zeros of a polynomial $z^n + a_1 z^{n-1} + b_2 z^{n-2} + \cdots + b_n = z^n F(1/z)$, so that $1/\zeta_j$ are the zeros of $F(z)$ and lie on the unit circle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My problem with Exercise 12 lies in ensuring the Newton-Girard formulas are satisfied; it is not as simple as choosing a centroid, and I can't see a way to do it even for the case when $C$ is the unit circle.  How do I know that a solution exists here?  Can I extend this to the general case, or is a separate argument needed?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, how do I center formulas?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: I forgot to mention that Marden gives some citations of this result:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gavrilov, L., &lt;em&gt;On the continuation of polynomials&lt;/em&gt;&lt;br&gt;&#xA;Gavrilov, L., &lt;em&gt;On the K-extension of polynomials&lt;/em&gt;&lt;br&gt;&#xA;Cebotarev, N. G., &lt;em&gt;Über die Fortsetzbarkeit von Polynomen auf geschlassene Kurven&lt;/em&gt;&lt;br&gt;&#xA;Cebotarev, N. G., &lt;em&gt;On Hurwitz's problem for transcendental functions&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately I read neither Russian nor German and I can't seem to locate the last one, so these aren't of direct help to me.&lt;/p&gt;&#xA;" OwnerUserId="5531" LastEditorUserId="5531" LastEditorDisplayName="" LastEditDate="2011-01-13T23:50:38.573" LastActivityDate="2011-01-13T23:50:38.573" Title="On the continuation of a polynomial" Tags="&lt;complex-analysis&gt;&lt;polynomials&gt;" CommentCount="2" />
  <row Id="17441" PostTypeId="2" ParentId="17438" CreationDate="2011-01-13T23:57:35.913" Score="5" ViewCount="" Body="&lt;p&gt;There can be a bit of subtlety whenever the gradient operator appears, because it is a differential operator, but the notation doesn't always make it clear what variable the differentiation is with respect to.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the left, you are differentiating $f(Ax)$ with respect to $x$. This is not the gradient of $f$ per se, but of the transformed function $x \mapsto f(Ax)$. On the right, $\nabla f$ &lt;em&gt;is&lt;/em&gt; the gradient of $f$. So $(\nabla f)(Ax)$ is the gradient of $f$ evaluated at $Ax$; in other words, you are differentiating $f$ with respect to its argument (which here is $Ax$), not with respect to $x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It might help to think of the analogous one-dimensional equation: $\frac{d}{dx} f(ax) = a f&amp;#39;(ax)$.&lt;/p&gt;&#xA;" OwnerUserId="856" LastActivityDate="2011-01-13T23:57:35.913" />
  <row Id="17442" PostTypeId="2" ParentId="12914" CreationDate="2011-01-14T00:22:24.930" Score="3" ViewCount="" Body="&lt;p&gt;The strong liars form a subgroup of $(\mathbb{Z}/n\mathbb{Z})^{\times}$ iff $n$ isn't of the form $n=\prod_{j=1}^{k}{p_{j}}^{\alpha_{j}}$, where $k\geq2$, $p_{1},\ldots,p_{j}$ are pairwise different primes such that $p_{j}\equiv 1\pmod4$ and $\alpha_{j}\in\mathbb{N}$, $j\in\{1,\ldots,k\}$. Here's the proof:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $n=1$ everything's trivial so let's assume $n&amp;gt;1$ and set $n-1=2^{x}y$, where $x,y\in\mathbb{Z}$ and $2\nmid y$. Also let $L(n)$ denote the set of strong liars $\bmod n$, ie&#xA;$$L(n)=\big\{a\in(\mathbb{Z}/n\mathbb{Z})^{\times}\colon(\exists t\in\mathbb{Z},0\leq t&amp;lt;x:a^{2^{t}y}=-1)\vee a^y=1\big\}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $2\mid n$, then&#xA;$$L(n)=\big\{a\in(\mathbb{Z}/n\mathbb{Z})^{\times}\colon a^{n-1}=1\big\},$$&#xA;which is obviously a subgroup of $(\mathbb{Z}/n\mathbb{Z})^{\times}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $q\mid n$ for some prime $q$ such that $q\equiv3\pmod4$, then $-1$ is quadratic nonresidue $\bmod{q}$, hence also $\bmod{n}$, so it can't be $a^{2^{t}y}=-1$ for $t&amp;gt;0$ and&#xA;$$L(n)=\big\{a\in(\mathbb{Z}/n\mathbb{Z})^{\times}\colon a^y=\pm1\big\},$$&#xA;which is again a subgroup of $(\mathbb{Z}/n\mathbb{Z})^{\times}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $n=p^\alpha$, where $p$ is a prime such that $p\equiv1\bmod4$ and $\alpha\in\mathbb{N}$ then we have $a^{n-1}-1=(a^{y}-1)\prod_{j=0}^{x-1}(a^{2^{j}y}+1)$, and since at most one of the factors on the RHS of this equation can be $0\bmod p$, we have&#xA;$$L(n)=\big\{a\in(\mathbb{Z}/n\mathbb{Z})^{\times}\colon a^{n-1}=1\big\},$$&#xA;which is once again a subgroup of $(\mathbb{Z}/n\mathbb{Z})^{\times}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let $n=\prod_{j=1}^{k}{p_{j}}^{\alpha_{j}}$, where $k\geq2$, $p_{1},\ldots,p_{j}$ are pairwise different primes such that $p_{j}\equiv 1\pmod4$ and $\alpha_{j}\in\mathbb{N}$, $j\in\{1,\ldots,k\}$. Let $G_{j}=(\mathbb{Z}/{p_{j}}^{\alpha_{j}}\mathbb{Z})^{\times}$, $H_{j}={G_{j}}^y$ be the subgroup of $y$-th powers in $G_{j}$ and $K_{j}=\{a\in G_{j}\colon a^{y}=1\}$ be the $y$-torsion subgroup of $G_{j}$. Then $|G_{j}|=\varphi({p_{j}}^{\alpha_{j}})=(p_{j}-1){p_{j}}^{\alpha_{j}-1}$, $|K_{j}|=\gcd(\varphi({p_{j}}^{\alpha_{j}}),y)$ (because $G_{j}$ is cyclic) and $H_{j}\cong G_{j}/K_{j}$, from which it follows that $4\mid|H_{j}|$. $H_{j}$ is cyclic (it's a subgroup of $G_{j})$, so it contains an unique cyclic subgroup of order $4$. For every $j\in\{1,\ldots,k\}$ let's fix $r_{j}\in G_{j}$ such that ${r_{j}}^{y}$ has order $4$ (such elements exist by the above reasoning). By CRT we have $(\mathbb{Z}/n\mathbb{Z})^{\times}\cong\prod_{j=1}^{k}G_{j}$, let $f_{j}\colon(\mathbb{Z}/n\mathbb{Z})^{\times}\to G_{j}$ be the natural projection. Then there exist $a,b\in(\mathbb{Z}/n\mathbb{Z})^{\times}$ such that $f_{1}(a)=r_1$, $f_{1}(b)=-r_1$ and $f_{j}(a)=f_{j}(b)=r_j$ for $j\geq2$. Then for every $j$ it's $f_{j}(a^{2y})=f_{j}(b^{2y})=-1$, hence $a^{2y}=b^{2y}=-1$ and $a,b\in L(n)$ (because $x\geq2$). But $(ab)^{2y}=1$ so it can't be $(ab)^{2^{t}y}=-1$ for $t\geq1$, and also $f_{1}((ab)^{y})=-{r_{1}}^{2y}=1$, $f_{2}((ab)^{y})={r_{2}}^{2y}=-1$, so it's $(ab)^{y}\neq\pm1$, hence $ab\notin L(n)$, which means that $L(n)$ isn't a subgroup of $(\mathbb{Z}/n\mathbb{Z})^{\times}$ in this case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Concerning the previous post: If $q\mid n$ for some prime $q$ such that $q\equiv 2\text{ or }3\pmod4$ then it's easy to see that there are no primitive Pythagorean triangles with hypotenuse $n$, and if $n=\prod_{j=1}^{k}{p_{j}}^{\alpha_{j}}$, where $p_{1},\ldots,p_{j}$ are pairwise different primes such that $p_{j}\equiv 1\pmod4$ and $\alpha_{j}\in\mathbb{N}$, then it's not difficult to show that there are $2^{k-1}$ different (up to isomorphism) primitive Pythagorean triangles with hypotenuse $n$, so the &quot;bad&quot; $n$'s are really exactly the ones for which there's more than one such triangle. :)&lt;/p&gt;&#xA;" OwnerUserId="5814" LastActivityDate="2011-01-14T00:22:24.930" />
  <row Id="17444" PostTypeId="2" ParentId="17370" CreationDate="2011-01-14T03:16:30.103" Score="2" ViewCount="" Body="&lt;p&gt;are these abelian groups?  otherwise i do not understand any of this.&lt;/p&gt;&#xA;" OwnerUserId="5211" LastActivityDate="2011-01-14T03:16:30.103" CommentCount="1" />
  <row Id="17448" PostTypeId="2" ParentId="16465" CreationDate="2011-01-14T05:00:29.817" Score="1" ViewCount="" Body="&lt;p&gt;This doesn't answer the question as stated but instead is a related result. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Siegel%27s_theorem_on_integral_points&quot; rel=&quot;nofollow&quot;&gt;Siegel's Theorem&lt;/a&gt; is an example of using the discriminant of the cubic equation in algebraic number theory. It states that a smooth algebraic curve $C$ of genus $g &amp;gt; 0$ defined over a number field $K$, written in a given coordinate system over affine space, contains only &lt;em&gt;finitely&lt;/em&gt; many points with coordinates in the ring of integers of $K$. For example, the elliptic curve $y^{2} = x^{3} + a x + b$ with $a,b,c \in \mathbb{Z}$ and $4 a^{3} + 27 b^{2} \neq 0$ has only finitely many solutions $x , y \in \mathbb{Z}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This particular example can be better understood if one knows one solution of the reduced cubic, $x^{3} + p x = q$, which is&#xA;\begin{eqnarray}&#xA;x = \sqrt[3]{ \sqrt{D} + \frac{q}{2}} - \sqrt[3]{ \sqrt{D} - \frac{q}{2}},&#xA;\end{eqnarray}&#xA;where $D = \frac{q^{2}}{4} + \frac{p^{3}}{27}$ is the &lt;em&gt;discriminant&lt;/em&gt; up to a constant factor (which we define as $\Delta$ below). More generally, the discriminant of the cubic $ax^{3} + bx^{2} + cx + d = 0$ is&#xA;\begin{eqnarray}&#xA;\Delta = b^{2} c^{2} - 4ac^{3} - 4 b^{3} d - 27 a^{2} d^{2} + 18 abcd.&#xA;\end{eqnarray}&#xA;There are ways to calculate the discriminant without solving the cubic, however. Note that if $\Delta = 0$, then the cubic has repeated roots. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Considering the right side of the defining equation of elliptic curve $C$, we have&#xA;\begin{eqnarray}&#xA;\Delta(C) = - (4 a^{3} + 27 b^{2}),&#xA;\end{eqnarray}&#xA;which is up to sign the conditional factor above. A nice exercise is to think about what occurs (or is violated) in the case that the associated discriminant vanishes.&lt;/p&gt;&#xA;" OwnerUserId="2720" LastEditorUserId="2720" LastEditorDisplayName="" LastEditDate="2011-02-21T19:27:50.130" LastActivityDate="2011-02-21T19:27:50.130" CommentCount="6" />
  <row Id="17449" PostTypeId="2" ParentId="9340" CreationDate="2011-01-14T05:05:47.730" Score="0" ViewCount="" Body="&lt;p&gt;Assuming $1\leq n \leq x$, split the sum into two parts, $1\leq m &amp;lt;n$ and $n&amp;lt;m\leq x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To handle the first part, make the variable change $m\to n-k$, and notice that&#xA;$$ \sum_{1\leq m &amp;lt;n } \frac{1}{|\log\frac{m}{n}|} = - \sum_{1\leq k \leq n-1} \frac{1}{\log\frac{n-k}{n}}. $$&#xA;Since $-\log\frac{n-k}{n} &amp;gt; \frac{k}{n}$ for $1\leq k \leq n-1$, this last sum is&#xA;$$ &amp;lt; n \sum_{1\leq k \leq n-1} \frac{1}{k} = O(n\log n) = O(x\log x).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second part can be handled similarly making the variable change $m\to n+k$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remark: Out of curiosity, are you computing moments of the Riemann zeta-function?&lt;/p&gt;&#xA;" OwnerUserId="1151" LastActivityDate="2011-01-14T05:05:47.730" CommentCount="1" />
  <row Id="17450" PostTypeId="1" AcceptedAnswerId="17452" CreationDate="2011-01-14T05:17:55.793" Score="1" ViewCount="210" Body="&lt;p&gt;I am looking at the following:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Show that a torsion-free divisible group $G$ is a vector space over $\mathbb{Q}$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have no problem verifying the axioms of vector spaces after noting that the term divisible group $G$ implies that a solution to $ny=x$ exist where $n$ is an integer and is unique by the fact that the group $G$ is torsion-free.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What does vector spaces over $\mathbb{Q}$ mean? And why does the question say that it is a vector spaces over $\mathbb{Q}$ instead of $\mathbb{Z}$? &lt;/p&gt;&#xA;" OwnerUserId="1683" LastEditorUserId="367" LastEditorDisplayName="" LastEditDate="2011-01-14T07:51:51.567" LastActivityDate="2011-01-14T07:57:33.057" Title="Vector space over $\mathbb{Q}$ or $\mathbb{Z}$?" Tags="&lt;linear-algebra&gt;&lt;group-theory&gt;&lt;vector-space&gt;" AnswerCount="4" CommentCount="0" />
  <row Id="17451" PostTypeId="2" ParentId="17450" CreationDate="2011-01-14T05:21:20.253" Score="2" ViewCount="" Body="&lt;p&gt;Because it is talking about vector spaces over $\mathbb{Q}$ (the rational numbers). This is know as a $\mathbb{Q}$-vector space or more generally an $F$-vector space (where $F$ is an arbitrary field). See &lt;a href=&quot;http://www.math.uchicago.edu/~yskim/M255/w080125.pdf&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; for more details.&lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2011-01-14T05:21:20.253" />
  <row Id="17452" PostTypeId="2" ParentId="17450" CreationDate="2011-01-14T05:23:09.670" Score="6" ViewCount="" Body="&lt;p&gt;A vector space is defined over a &lt;i&gt;field&lt;/i&gt;. If you want to define something over a ring, it's known as a &lt;i&gt;module&lt;/i&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2011-01-14T05:23:09.670" />
  <row Id="17453" PostTypeId="2" ParentId="17450" CreationDate="2011-01-14T05:29:44.333" Score="1" ViewCount="" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;What does vector spaces over $\mathbb{Q}$ means?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Not sure, but are you asking what one actually means by a vector space $V$ over a field $F$? The field is the set of acceptable scalars for your vector space. So if the vector space is over $\mathbb{Q}$, you couldn't have $\pi$ as a scalar, for example, whereas you could have $3/4$.&lt;/p&gt;&#xA;" OwnerUserId="1583" LastActivityDate="2011-01-14T05:29:44.333" CommentCount="3" />
  <row Id="17454" PostTypeId="2" ParentId="17152" CreationDate="2011-01-14T05:45:22.407" Score="-1" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;NOTE:&lt;/strong&gt; By probability, I mean the chance of it happening per iteration, and starting with a new page each time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's take a step back, shall we? (Not too many, because there's a cliff behind you.) Let's think of what the probability is of producing the following randomly:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Assuming there are only 26 characters (A-Z, uppercase), the probability would be $\frac{1}{26}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What about this:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;AA&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It'd be $(\frac{1}{26})^2$. This:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;AAA&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It'd be $(\frac{1}{26})^3$. And this:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;XKCD&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It'd be $(\frac{0}{26})^4$. [Just kidding, it's: $(\frac{1}{26})^4$].&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, for every character we add to the quote, it will be: $(1/26)^c)$, where $c$ represents the number of characters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically, it would be a probability of $(\frac{1}{26*2+12})^c$ since the characters used could be: &lt;code&gt;A-Z&lt;/code&gt;, &lt;code&gt;a-z&lt;/code&gt;, &lt;code&gt;.!?,;: &quot;'/()&lt;/code&gt; Of course, there could be more characters, but that's just an example. :)&lt;/p&gt;&#xA;" OwnerUserId="2736" LastEditorUserId="2736" LastEditorDisplayName="" LastEditDate="2011-01-14T05:54:03.957" LastActivityDate="2011-01-14T05:54:03.957" CommunityOwnedDate="2011-01-15T06:13:52.160" />
  <row Id="17455" PostTypeId="1" CreationDate="2011-01-14T05:47:34.807" Score="2" ViewCount="185" Body="&lt;p&gt;Assume $S_1$ and $S_2$ are two $n \times n$ (positive definite if that helps) matrices, $c_1$ and $c_2$ are two variables taking scalar form, and $u_1$ and $u_2$ are two $n \times 1$ vectors. In addition, $c_1+c_2=1$, but in the more general case of $m$ $S$'s, $u$'s, and $c$'s, the $c$'s also sum to 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the derivative of $(c_1 S_1+c_2 S_2)^{-1}(c_1 u_1+c_2 u_2)$ with respect to both $c_1$ and $c_2$?&lt;/p&gt;&#xA;" OwnerUserId="5855" LastEditorUserId="856" LastEditorDisplayName="" LastEditDate="2011-01-14T05:56:51.377" LastActivityDate="2011-01-14T17:45:04.470" Title="Derivative of Matrix-Vector Product" Tags="&lt;calculus&gt;&lt;matrices&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="17457" PostTypeId="1" AcceptedAnswerId="17458" CreationDate="2011-01-14T06:13:16.650" Score="0" ViewCount="81" Body="&lt;p&gt;Hi, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please help me with this.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Let $G$ be abelian group, and let $A_k$ be a family of subgroups of $G$. Prove that $G=\sum A_k$ (internal) if and only if every non-zero element $g\in G$ has a unique expression of the form $g=a_{k_1}+...+a_{k_n}$, where $a_{k_i} \in A_{k_i}$, the $k_i$ are distinct and each $a_{k_i}\neq 0$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have a hunch that the term &quot;$g$ has a unique expression&quot; here means that $A_k \bigcap A_j =\{0\}$. If so, how do we reason it and proved it to be so? &lt;/p&gt;&#xA;" OwnerUserId="1683" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-14T17:13:08.613" LastActivityDate="2011-01-14T17:13:08.613" Title="The meaning of &quot;unique expression&quot;." Tags="&lt;group-theory&gt;" AnswerCount="1" CommentCount="0" />
  <row Id="17458" PostTypeId="2" ParentId="17457" CreationDate="2011-01-14T06:15:20.060" Score="2" ViewCount="" Body="&lt;p&gt;No, it does not mean that. It means that $\displaystyle A_i\cap(\sum_{j\neq i}A_j)=0$ for all $i$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Later:&lt;/strong&gt; Well, actually, it is &lt;em&gt;equivalent&lt;/em&gt; to that. That «$g$ has a unique expression of the form $a_1+\cdots+a_n$ with $a_i\in A_i$» means exactly that &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;first&lt;/em&gt;, there exist $a_1,\dots,a_n$ with $a_i\in A_i$ for each $i$ such that $g=a_1+\cdots+a_n$, and &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;second&lt;/em&gt;, that whenever you have elements $a_1,\dots,a_n, b_1,\dots,n_n$ with $a_i,b_i\in A_i$ for each $i$ such that $=a_1+\cdots+a_n=b_1+\cdots+b_n$, then $a_i=b_i$ for all $i$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="274" LastEditorUserId="274" LastEditorDisplayName="" LastEditDate="2011-01-14T06:21:09.960" LastActivityDate="2011-01-14T06:21:09.960" CommentCount="4" />
  <row Id="17459" PostTypeId="2" ParentId="17455" CreationDate="2011-01-14T06:31:27.467" Score="2" ViewCount="" Body="&lt;p&gt;The condition that $S_1$ and $S_2$ are positive definite is relevant to the existence of the inverse in the definition of the function.  I assume that it is taken as given that the inverse exists at the relevant values of $c_1$ and $c_2$.  This would be true in particular if $c_1$ and $c_2$ were positive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By symmetry, the same method will apply for $c_1$ and $c_2$, and we're basically differentiating the function $f(t)=(tA+B)^{-1}(tu+v)$, where $A$ and $B$ are matrices and $u$ and $v$ are column vectors.  You can write this as $f(t)=g(t)h(t)$, where $g$ is a matrix valued function and $h$ is a vector valued function.  By the product rule, $f&amp;#39;(t)=g&amp;#39;(t)h(t)+g(t)h&amp;#39;(t)$.  So you just need to be able to determine $g&amp;#39;$ and $h&amp;#39;$.  It is straightforward that $h&amp;#39;(t)=u$.  You can also show that $g&amp;#39;(t)=-(tA+B)^{-1}A(tA+B)^{-1}$.  &lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2011-01-14T07:23:36.807" LastActivityDate="2011-01-14T07:23:36.807" />
  <row Id="17460" PostTypeId="1" CreationDate="2011-01-14T06:35:19.963" Score="1" ViewCount="108" Body="&lt;p&gt;Can someone please let me know if my solution is correct:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Define:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Let $A = \{x \in \mathbb{R}^{2}: \text{all coordinates of x are rational} \}$. Show that&#xA;$\mathbb{R}^{2} \setminus A$ is connected.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My answer: just note that $A = \mathbb{Q} \times \mathbb{Q}$ so countable and there's a standard result that $\mathbb{R}^{2}$ minus a countable set is path-connected so connected, thus the result follows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) $B = \{x \in \mathbb{R}^{2}: \text{at least one coordinate is irrational} \}$. Show that $B$ is connected. Well isn't $B$ just $B = \mathbb{R}^{2} \setminus \mathbb{Q}^{2}$ so it is exactly the same problem as above isn't it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a problem in Dugundji's book.&lt;/p&gt;&#xA;" OwnerUserId="5817" LastActivityDate="2011-01-14T06:35:19.963" Title="Connectedness of sets in the plane with rational coordinates and at least one irrational" Tags="&lt;topology&gt;" CommentCount="4" />
  <row Id="17461" PostTypeId="1" AcceptedAnswerId="17463" CreationDate="2011-01-14T06:51:04.317" Score="2" ViewCount="63" Body="&lt;p&gt;I just read about the construction of the real numbers in Enderton's Elements of Set Theory, and am now trying to go through all the exercises. Enderton chooses to construct the reals with left sides of Dedekind cuts. His definition is $y$ is a Dedekind cut if $\emptyset\neq y\neq\mathbb{Q}$, $y$ has no largest member, and $y$ is downward closed, i.e., $q\in y\wedge r&amp;lt;q\implies r\in y$. The following, question 19 of chapter 5 has me a bit baffled.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Assume that $p$ is a positive rational number. Show that for any real number $x$ there is some rational $q$ in $x$ such that $p+q\not\in x$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have this intuitive argument in my head. I guess I want to take $q\in x$, preferably closer to the right in $x$ on the rational line than not, and add it to $p$. If $p+q\not\in x$, then I am done. Otherwise, since $x$ has no greatest element, I can find $q&amp;#39;\in x$ such that $q&amp;#39;&amp;gt;q$, and add it to $p$ until I finally get a $q$ that pushes $p+q$ out of $x$, supposing of course that this process eventually ends. But this is horribly informal, and I have very little experience with analysis, and can't see a way to formalize it (if this is even the correct approach.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could someone please explain how to make this argument more rigorous, or perhaps point out a better way? Thanks.&lt;/p&gt;&#xA;" OwnerUserId="1583" LastEditorUserId="1583" LastEditorDisplayName="" LastEditDate="2011-01-14T07:05:01.880" LastActivityDate="2011-01-14T07:15:40.460" Title="If $p\in\mathbb{Q}^+$, for any $x\in\mathbb{R}$, there is a rational $q\in x$ such that $p+q\not\in x$." Tags="&lt;real-analysis&gt;&lt;elementary-set-theory&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="17462" PostTypeId="1" CreationDate="2011-01-14T06:55:03.070" Score="4" ViewCount="114" Body="&lt;p&gt;First, this was one of the four problems we had to solve in a project last year and I couldn’t find a suitable algorithm so we handle in a brute force solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Problem: The numbers are in a list that is not sorted and supports only one type of operation. The operation is defined as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given a position i and a position j the operation moves the number at position i to position j without altering the relative order of the other numbers. If i &gt; j, the positions of the numbers between positions j and i - 1 increment by 1, otherwise if i &amp;lt; j the positions of the numbers between positions i+1 and j decreases by 1. This operation requires i steps to find a number to move and j steps to locate the position to which you want to move it. Then the number of steps required to move a number of position i to position j is i+j.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We need to design an algorithm that given a list of numbers, determine the optimal (in terms of cost) sequence of moves to rearrange the sequence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Attempts: Part of our investigation was around NP-Completeness, we make it a decision problem and try to find a suitable transformation to any of the problems listed in Garey and Johnson’s book: Computers and Intractability with no results. There is also no direct reference (from our point of view) to this kind of variation in Donald E. Knuth’s book: The art of Computer Programing Vol. 3 Sorting and Searching. We also analyzed algorithms to sort linked lists but none of them gives a good idea to find de optimal sequence of movements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the idea is not to find an algorithm that orders the sequence, but one to tell me the optimal sequence of movements in terms of cost that organizes the sequence, you can make a copy and sort it to analyze the final position of the elements if you want, in fact we may assume that the list contains the numbers from 1 to n, so we know where we want to put each number, we are just concerned with minimizing the total cost of the steps.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We tested several greedy approaches but all of them failed, divide and conquer sorting algorithms can’t be used because they swap with no cost portions of the list and our dynamic programing approaches had to consider many cases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The brute force recursive algorithm takes all the possible combinations of movements from i to j and then again all the possible moments of the rest of the element’s, at the end it returns the sequence with less total cost that sorted the list, as you can imagine the cost of this algorithm is brutal and makes it impracticable for more than 8 elements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Our observations:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;  n movements is not necessarily cheaper than n+1 movements (unlike swaps in arrays that are O(1)).&#xA;&#xA;  There are basically two ways of moving one element from position i to j: one is to move it directly and the other is to move other elements around i in a way that it reaches the position j.&#xA;&#xA;  At most you make n-1 movements (the untouched element reaches its position alone).&#xA;&#xA;  If it is the optimal sequence of movements then you didn’t move the same element twice.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="5829" LastEditorUserId="5829" LastEditorDisplayName="" LastEditDate="2011-01-15T22:44:09.240" LastActivityDate="2011-01-15T22:44:09.240" Title="Optimal sorting algorithm with modified cost" Tags="&lt;sorting&gt;" CommentCount="8" />
  <row Id="17463" PostTypeId="2" ParentId="17461" CreationDate="2011-01-14T07:05:49.217" Score="2" ViewCount="" Body="&lt;p&gt;There is an integer $n$ such that $np\in x$ but $(n+1)p\notin x$. The reason is that there is some rational $r\notin x$, and there is some natural $m$ such that $mp&amp;gt;r$, in particular $mp\notin x$. But there is some $s\in x$, and there is some integer $k$ such that $kp&amp;lt;s$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It follows that there is a &lt;em&gt;least&lt;/em&gt; $m$ such that $mp\notin x$ (since the set of such integers is bounded below by $k$), which means that $(m-1)p\in x$. Ok, then you can take $q=(m-1)p$.&lt;/p&gt;&#xA;" OwnerUserId="462" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2011-01-14T07:15:40.460" LastActivityDate="2011-01-14T07:15:40.460" CommentCount="1" />
  <row Id="17465" PostTypeId="1" CreationDate="2011-01-14T07:13:08.747" Score="1" ViewCount="142" Body="&lt;p&gt;I'm creating a computer program where I need to calculate the parametrized circumference of an ellipse, like this:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;x = r1 * cos(t), y = r2 * sin(t)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Now, say I want this parametrized ellipse to be tilted at an arbitrary angle. How do I go about this? Any obvious simplifications for i.e 30, 45 or 60 degrees? &lt;/p&gt;&#xA;" OwnerUserId="5830" LastActivityDate="2011-01-24T02:15:33.463" Title="Creating a parametrized ellipse at an angle" Tags="&lt;trigonometry&gt;&lt;parametric&gt;" AnswerCount="3" />
  <row Id="17466" PostTypeId="2" ParentId="17465" CreationDate="2011-01-14T07:20:02.897" Score="0" ViewCount="" Body="&lt;p&gt;Update:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x = r_1 \cos(t)$,&#xA;$y = r_2 \sin(t)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;using the equation of ellipse in polar form we have : (could someone confirm this please?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ r(\theta) = \frac {r_1r_2}{\sqrt {{r_1^2 \sin^2 (\theta)}+{r_2^2 \cos^2 (\theta)}}}  $ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;to rotate by angle $\phi$ :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ r(\theta) = \frac {r_1r_2}{\sqrt {{r_1^2 \sin^2 (\theta-\phi)}+{r_2^2 \cos^2 (\theta-\phi)}}}  $ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The original &lt;strong&gt;incorrect&lt;/strong&gt; answer is below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x = r_1 \cos(t + \phi)$,&#xA;$y = r_2 \sin(t + \phi) \text {, where }\phi$ is your angle of tilt in radians&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This was as pointed out was my mistake on thinking polar while working with parametric. I have tried converting this to polar form but adding angle of $\phi$ and reverting back to parametric form so far has not been possible. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Has anyone seen solution of this problem by converting to polar coordinates and performing an angle addition and converting back to parametric form?&lt;/p&gt;&#xA;" OwnerUserId="3936" LastEditorUserId="3936" LastEditorDisplayName="" LastEditDate="2011-01-24T02:15:33.463" LastActivityDate="2011-01-24T02:15:33.463" CommunityOwnedDate="2011-01-24T02:16:14.820" CommentCount="9" />
  <row Id="17467" PostTypeId="2" ParentId="17465" CreationDate="2011-01-14T07:24:54.693" Score="1" ViewCount="" Body="&lt;p&gt;If you rotate $(x=r_1\cos t,y=r_2\sin t)$ by $\theta$ about $(0,0)$, the resulting curve is given by $(x&amp;#39;=r_1\cos t\cos\theta-r_2\sin t\sin\theta, y&amp;#39;=r_1\cos t\sin\theta+r_2\sin t\cos\theta)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Using the fact that complex multiplication by $e^{i\theta}$ rotates by $\theta$ about $0$, the point $(x,y)=x+yi$ is mapped to $(x+iy)(\cos\theta+i\sin\theta)=$ $x\cos\theta-y\sin\theta+i(x\sin\theta+y\cos\theta)=$ $(x\cos\theta-y\sin\theta,x\sin\theta+y\cos\theta)$.)&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2011-01-14T07:24:54.693" CommentCount="1" />
  <row Id="17468" PostTypeId="2" ParentId="17465" CreationDate="2011-01-14T07:26:37.657" Score="3" ViewCount="" Body="&lt;p&gt;If you want to rotate $\theta$ radians, you should use $$t\mapsto \left(&#xD;&#xA;\begin{array}{c}&#xD;&#xA; a \cos (t) \cos (\theta )-b \sin (t) \sin&#xD;&#xA;   (\theta ) \\&#xD;&#xA; a \cos (t) \sin (\theta )+b \sin (t) \cos&#xD;&#xA;   (\theta )&#xD;&#xA;\end{array}&#xD;&#xA;\right)&#xD;&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2011-01-14T07:26:37.657" />
  <row Id="17469" PostTypeId="1" CreationDate="2011-01-14T07:38:19.597" Score="5" ViewCount="144" Body="&lt;p&gt;&lt;strong&gt;Slight Motivation:&lt;/strong&gt; In Mac Lane and Freyd's books (the latter being a reprint of an older book called &quot;Abelian Categories&quot;) they note that instead of defining any Objects in a category we may define an &quot;arrows only&quot; approach by considering the identity morphism associated to an object to be the object itself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; Is it computationally or syntactically easier in category theory to consider a category as objects and morphisms instead of as just as morphisms?  In short, is there a reason we keep objects around?&lt;/p&gt;&#xA;" OwnerUserId="2959" LastActivityDate="2011-01-14T09:41:17.110" Title="Category Theory with and without Objects." Tags="&lt;soft-question&gt;&lt;category-theory&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" />
  <row Id="17470" PostTypeId="2" ParentId="17450" CreationDate="2011-01-14T07:57:33.057" Score="2" ViewCount="" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;And why does the question say that it is a vector spaces over $\mathbb Q$ instead of $\mathbb Z$?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Notice that an Abelian group is a module over $\mathbb Z$ in a canonical way. The problem you quote says that this $\mathbb Z$-module structure &lt;em&gt;extends&lt;/em&gt; canonically to a $\mathbb Q$-module or $\mathbb Q$-vector space structure provided that your group is also torsion-free and divisible .&lt;/p&gt;&#xA;" OwnerUserId="367" LastActivityDate="2011-01-14T07:57:33.057" />
  <row Id="17471" PostTypeId="2" ParentId="17412" CreationDate="2011-01-14T08:08:39.303" Score="9" ViewCount="" Body="&lt;p&gt;Let $G = \langle a, b\ |\ [a,b^2]=1, a^2=b^4\rangle$ be your group.  Consider the map of $G$ to $\mathbb{Z}$, sending $b$ to $1$ and $a$ to $2$.  This has a kernel which is $D_\infty$ (found via Reidemeister-Schreier rewriting).  That is, the kernel is $\langle x,y\ |\ x^2=y^2=1\rangle$, where $x=ab^{-2}$ and $y=bab^{-3}$.  Thus your group splits as a semidirect product $D_\infty \rtimes\mathbb{Z}$, where if $D_\infty=\langle x,y\ |\ x^2=y^2=1\rangle$ and $\mathbb{Z}=\langle t\rangle$, then the action of $t$ is: $x^t=y$, $y^t=x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: So we've gotten the following presentation of your group:&#xA;$$ G= \langle x,y,t\ |\ x^2=y^2=1, x^t=y, y^t=x\rangle.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $K=\langle xy,t\rangle$ be a subgroup of $G$; it is not hard to see $K$ has index 2, and is isomorphic to the Klein bottle group.  Thus your group $G$ is also a semidirect product $K\rtimes C_2$ (here $C_2$ is the cyclic group of order two), where if $C_2=\langle u\rangle$, and $K=\langle r,s\ |\ s^{-1}rs=r^{-1}\rangle$, then $r^u=r^{-1}$ and $s^u=sr^{-1}$.&lt;/p&gt;&#xA;" OwnerUserId="641" LastEditorUserId="641" LastEditorDisplayName="" LastEditDate="2011-01-14T08:21:19.387" LastActivityDate="2011-01-14T08:21:19.387" CommentCount="1" />
  <row Id="17472" PostTypeId="1" AcceptedAnswerId="17481" CreationDate="2011-01-14T08:30:19.550" Score="5" ViewCount="104" Body="&lt;p&gt;Let $R$ be a commutative ring and $m\subseteq R$ be a maximal ideal. &#xA;Can you describe the set of prime ideals of the  $R/m^2$. Are they all maximal ?&lt;/p&gt;&#xA;" OwnerUserId="5422" LastEditorUserId="5422" LastEditorDisplayName="" LastEditDate="2011-01-14T09:46:24.960" LastActivityDate="2011-01-14T19:32:45.613" Title="Description of the set of prime ideals of the $R/m^2$" Tags="&lt;commutative-algebra&gt;" AnswerCount="2" FavoriteCount="0" />
  <row Id="17473" PostTypeId="1" AcceptedAnswerId="17475" CreationDate="2011-01-14T08:39:19.267" Score="4" ViewCount="244" Body="&lt;p&gt;I was watching Beating Blackjack with Andy Bloch where he runs through the basic strategy charts that outline the best strategy with playing the game. Later he also talks about the methodologies to count, but that is not relevant to my question. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;He states in that video that the basic strategy is an outcome of computer simulations, and that suggests to me that its a rather weak method to go about doing it. I have a two-fold question, one is when is a computer solution considered good enough to be &quot;true&quot;, and the second, surely, this is a much easier problem that does not need computer simulations to solve, are there any &lt;em&gt;known&lt;/em&gt; methods to derive the basic strategy. &lt;/p&gt;&#xA;" OwnerUserId="4728" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2011-01-14T18:23:01.443" LastActivityDate="2011-01-16T12:59:21.983" Title="Question about basic strategy in Blackjack" Tags="&lt;statistics&gt;&lt;recreational-mathematics&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="1" />
  <row Id="17474" PostTypeId="1" AcceptedAnswerId="17480" CreationDate="2011-01-14T08:52:13.050" Score="2" ViewCount="108" Body="&lt;p&gt;Let $R$ be a commutative ring and $I,J$ ideals of R such that $J$ is finitely generated and the rings $R/I$ and $R/J$ are Noetherian. Are the $R-$ modules $R/J$ , $J/IJ$ , $R/IJ$ Noetherian ?Is the ring $R/IJ$ is  Noetherian?&lt;/p&gt;&#xA;" OwnerUserId="5422" LastActivityDate="2011-01-14T09:58:21.550" Title="Noetherian quotient rings" Tags="&lt;commutative-algebra&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="17475" PostTypeId="2" ParentId="17473" CreationDate="2011-01-14T09:21:06.540" Score="2" ViewCount="" Body="&lt;p&gt;I'm not sure I understand your first question. You don't have a &quot;computer model&quot; for blackjack. The rules of the game are given. You can model the dealing of the cards as if they were uniformly random draws from a collection of objects (distinct objects if you are using a single deck of cards) without replacement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since it is a game with randomness built in, it is a good candidate for Monte Carlo simulation. To my knowledge, it is not very simple to derive basic strategy using just pen and paper because you would have to deal with a number of different cases and keeping track of them all can be tedious. I'm not aware of any elegant solutions to this problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you go further and look at card counting strategies, things become even more complicated and the methods and strategies appear to be more and more heuristic.&lt;/p&gt;&#xA;" OwnerUserId="2641" LastActivityDate="2011-01-14T09:21:06.540" CommentCount="1" />
  <row Id="17476" PostTypeId="1" CreationDate="2011-01-14T09:31:27.823" Score="5" ViewCount="97" Body="&lt;p&gt;Dear community,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the &lt;a href=&quot;http://en.wikipedia.org/wiki/Helmholtz_equation&quot; rel=&quot;nofollow&quot;&gt;Helmholtz equation&lt;/a&gt; &#xA;$$\Delta \psi + k^2 \psi = 0$$&#xA;has a lot of fundamental applications in physics since it is a form of the wave equation $\Delta\phi - c^{-2}\partial_{tt}\phi = 0$ with an assumed harmonic time dependence $e^{\pm\mathrm{i}\omega t}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$k$ can be seen as some kind of potential - the equation is analogue to the stationary &lt;a href=&quot;http://en.wikipedia.org/wiki/Schr%C3%B6dinger_equation&quot; rel=&quot;nofollow&quot;&gt;Schrödinger equation&lt;/a&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The existance of solutions is to my knowledge linked to the separability of the Laplacian $\Delta$ in certain coordinate systems. Examples are cartesian, elliptical and cylindrical ones.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For now I am interested in a toroidal geometry, &#xA;$$k(\mathbf{r}) = \begin{cases} k_{to} &amp;amp; \mathbf{r}\in T^2 \\ k_{out} &amp;amp; \text{else}\end{cases}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $T^2 = \left\{ (x,y,z):\, r^2 \geq \left(  \sqrt{x^2 + y^2} - R\right)^2 + z^2  \right\}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence the question:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Are there known solutions (in terms of eigenfunctions) of the Helmholtz equation for the given geometry?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Thank you in advance&lt;br&gt;&#xA;Sincerely&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Robert&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: As &lt;a href=&quot;http://math.stackexchange.com/users/1242/hans-lundmark&quot;&gt;Hans&lt;/a&gt; pointed out, there might not be any solution according to a corresponding &lt;a href=&quot;http://en.wikipedia.org/wiki/Toroidal_coordinates#Standard_separation&quot; rel=&quot;nofollow&quot;&gt;Wikipedia article&lt;/a&gt;. Unfortunately, there is no reference given - does anyone know where I could find the proof?&lt;/p&gt;&#xA;" OwnerUserId="2671" LastEditorUserId="2671" LastEditorDisplayName="" LastEditDate="2011-01-14T14:39:15.427" LastActivityDate="2011-01-14T14:39:15.427" Title="Eigenfunctions of the Helmholtz equation in Toroidal geometry" Tags="&lt;differential-equations&gt;&lt;functional-analysis&gt;&lt;physics&gt;" CommentCount="8" FavoriteCount="1" />
  <row Id="17477" PostTypeId="2" ParentId="17469" CreationDate="2011-01-14T09:41:17.110" Score="2" ViewCount="" Body="&lt;p&gt;Category theory without objects is a little easier to formalise as a first-order theory. For instance, &lt;a href=&quot;http://nlab.mathforge.org/nlab/show/fully%20formal%20ETCS&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; is a first-order formulation of Lawvere's elementary theory of the category of sets, which adopts the strategy of identifying objects with their identity arrows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That said, I'm not sure category theory without objects is any easier to &lt;em&gt;think&lt;/em&gt; about.&lt;/p&gt;&#xA;" OwnerUserId="5191" LastActivityDate="2011-01-14T09:41:17.110" />
  <row Id="17478" PostTypeId="1" AcceptedAnswerId="17490" CreationDate="2011-01-14T09:44:18.753" Score="1" ViewCount="194" Body="&lt;p&gt;If:&lt;br&gt;&#xA;I can make two tools which would serve the same purpose and:&lt;br&gt;&#xA;I need 10 times less time to make first tool but work with this tool is a half time slower than work with the second tool, after what period of time I will get advantage by making the tool with which I will work faster by 2 times (the second tool).&lt;/p&gt;&#xA;" OwnerUserId="4366" LastActivityDate="2011-01-14T18:05:41.230" Title="How quickly with better tool" Tags="&lt;algebra&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="17479" PostTypeId="1" CreationDate="2011-01-14T09:55:47.747" Score="5" ViewCount="364" Body="&lt;p&gt;For a function $f(r, \vartheta, \varphi)$ given in spherical coordinates, how can the Fourier transform be calculated best? Possible ideas:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;express $(r,\vartheta,\varphi)$ in cartesian coordinates, yielding a nonlinear argument of $f$&lt;/li&gt;&#xA;&lt;li&gt;express $\vec k,\vec r$ in the $e^{i\vec k\vec r}$ term in spherical coordinates, yielding a nonlinear exponent in $\vartheta$ and $\varphi$&lt;/li&gt;&#xA;&lt;li&gt;decompose $f$ into &lt;a href=&quot;http://en.wikipedia.org/wiki/Spherical_harmonics&quot; rel=&quot;nofollow&quot;&gt;Spherical Harmonics&lt;/a&gt; and then change base to Fourier space, requiring the Fourier transform of the Spherical Harmonics (&lt;a href=&quot;http://www.google.com/search?q=recursion&amp;amp;hl=en&quot; rel=&quot;nofollow&quot;&gt;it is obviously not possible to calculate them using this very method...&lt;/a&gt;, can that be be found somewhere?)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="163" LastActivityDate="2011-05-30T23:06:51.940" Title="How to perform a Fourier transform in spherical coordinates?" Tags="&lt;analysis&gt;&lt;fourier-analysis&gt;&lt;spherical-coordinates&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="17480" PostTypeId="2" ParentId="17474" CreationDate="2011-01-14T09:58:21.550" Score="4" ViewCount="" Body="&lt;p&gt;Sebastian did the $R/J$ case in the comments.  Now $J/IJ$ is $J\otimes R/I$ which, since $J$ is finitely generated and $R/I$ is Noetherian (as an R-module), is a Noetherian R-module.&#xA;The exact sequence&#xA;$$ 0\rightarrow J/IJ \rightarrow R/IJ\rightarrow R/J\rightarrow 0$$&#xA;and the previous two results show $R/IJ$ is Noetherian.&lt;/p&gt;&#xA;" OwnerUserId="641" LastActivityDate="2011-01-14T09:58:21.550" />
  <row Id="17481" PostTypeId="2" ParentId="17472" CreationDate="2011-01-14T10:09:25.067" Score="10" ViewCount="" Body="&lt;p&gt;$m/m^2$ is a maximal ideal of $R/m^2$, and it is nilpotent; it is therefore the nilradical, and being prime, is the only prime ideal of $R/m^2$.&lt;/p&gt;&#xA;" OwnerUserId="641" LastActivityDate="2011-01-14T10:09:25.067" />
  <row Id="17482" PostTypeId="1" AcceptedAnswerId="26538" CreationDate="2011-01-14T10:33:23.923" Score="8" ViewCount="291" Body="&lt;p&gt;We all know that the complex integral calculus can be useful for computing real integrals. I was wondering if there are any similar example where we can use Green's theorem to compute one-variables integrals. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now it is clear that if we have an integral $\int_a^b f dx$ on the real line we can view this as a curve integral in the plane of $\int P dx + Q dy$ for infinitely many choices of $P$ and $Q$ where we integrate over the line between $a$ and $b$. We could then integrate this vector field over some other curve, $\gamma$, with the same endpoints and try computing the difference between our original integral and the new one with Green's theorem. It is clear that for random choices of $P$, $Q$ and $\gamma$ we will not have simplified our problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But are there any examples where this technique is useful? &lt;/p&gt;&#xA;" OwnerUserId="4623" LastActivityDate="2011-03-12T09:59:39.597" Title="Examples of using Green's theorem to compute one-variable integrals?" Tags="&lt;integral&gt;&lt;vector-calculus&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" />
  <row Id="17483" PostTypeId="1" CreationDate="2011-01-14T10:54:37.050" Score="1" ViewCount="57" Body="&lt;p&gt;If R is semiprime,right principally injective and satisfies ACC on right annihilators of elements,is it self-injective?I only know it is right nonsingular.&lt;/p&gt;&#xA;" OwnerUserId="4706" LastActivityDate="2011-01-14T10:54:37.050" Title="How does the right principally injective ring self-injective?" Tags="&lt;abstract-algebra&gt;&lt;ring-theory&gt;" />
  <row Id="17484" PostTypeId="1" AcceptedAnswerId="17487" CreationDate="2011-01-14T11:16:16.507" Score="2" ViewCount="107" Body="&lt;p&gt;Is it true that if a continuous-time stochastic process $X_t$ is &lt;a href=&quot;http://en.wikipedia.org/wiki/Stationary_process#Weak_or_wide-sense_stationarity&quot; rel=&quot;nofollow&quot;&gt;weakly stationary&lt;/a&gt; then $|X_t|$ is also weakly stationary?&lt;/p&gt;&#xA;" OwnerUserId="5835" LastEditorUserId="5835" LastEditorDisplayName="" LastEditDate="2011-01-14T11:36:13.067" LastActivityDate="2011-01-14T16:24:01.833" Title="Weak stationarity of a process modulus" Tags="&lt;stochastic-processes&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="2" />
  <row Id="17485" PostTypeId="1" AcceptedAnswerId="17495" CreationDate="2011-01-14T11:23:25.763" Score="2" ViewCount="106" Body="&lt;p&gt;hi&#xA;i've found this interesting page: &lt;a href=&quot;http://local.wasp.uwa.edu.au/~pbourke/geometry/superellipse/&quot; rel=&quot;nofollow&quot;&gt;superellipse and superellipsoid&lt;/a&gt;&#xA;and i used the formula for one of my computer graphics applications.&#xA;i used (the most usefull for computer graphics) the parametric formula:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/FHN5D.gif&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but for correclty draw the superellipsod i had to calculate also some absolute values of sine and cosine and deal with signs like in this piece of code (it's ANSI C code, i hope it make sense for you):&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;void EvalSuperEllipse(double t1,double t2,double p1,double p2,XYZ *p)&#xA;{&#xA; double tmp;&#xA; double ct1,ct2,st1,st2;&#xA;&#xA; ct1 = cos(t1);&#xA; ct2 = cos(t2);&#xA; st1 = sin(t1);&#xA; st2 = sin(t2);&#xA;&#xA; // SIGN function return 1 if input is positive, -1 if it is negative&#xA; // fabs function calculate absolute value&#xA;&#xA; tmp  = SIGN(ct1) * pow(fabs(ct1),n1);&#xA; p-&amp;gt;x = tmp * SIGN(ct2) * pow(fabs(ct2),n2);&#xA; p-&amp;gt;y = SIGN(st1) * pow(fabs(st1),n1);&#xA; p-&amp;gt;z = tmp * SIGN(st2) * pow(fabs(st2),n2);&#xA;}&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;the first question is how to write in pure math formalism the parametric formula including those sign change (obtained with SIGN and fabs function), &#xA;and the second question is why i have to make this manipulations and how understand that i had to do that in new geometric adventure if i don't find the code ready-to-use.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i hope i do not make an off-topic with this programming question! (i know it's weird to post ANSI C code here but i think is more attinent here instad of in stackoverflow.com)&lt;/p&gt;&#xA;" OwnerUserId="5046" LastActivityDate="2011-01-14T15:34:46.657" Title="superellipsoid: problem in understanding parametric formula" Tags="&lt;computational-geometry&gt;" AnswerCount="2" />
  <row Id="17486" PostTypeId="2" ParentId="16723" CreationDate="2011-01-14T12:49:21.713" Score="5" ViewCount="" Body="&lt;p&gt;I would like to summarise and extend the results of my previous answer in a new answer as I prefer to keep the original in its current form to prevent it from turning into my magnum opus.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Theorem 1:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For positive integers $a, b$ and $c$ where $ d=\text{gcd}(b,c), $ $u=c/d,$&#xA;$t= \lfloor a/b \rfloor $ and $ a \equiv \lambda \textrm{ mod } d, $ where&#xA;$ 0 \le \lambda &amp;lt; d,$ and $ u \, | \, (t+1) $ we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \sum_{k=0}^{t} \left \lfloor \frac{a - kb}{c} \right \rfloor =&#xD;&#xA;\frac{t+1}{c} \left \lbrace a - \frac{tb}{2} - \frac{c-d}{2} - \lambda \right \rbrace . $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So far, the above theorem and its proof are included in my previous answer. The rest is new.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Theorem 2: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Along with the definitions in theorem 1, let $ a \equiv r \textrm{ mod } c, $ where $ 0 \le r &amp;lt; c $ and $ u \, | \, t $ then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \sum_{k=0}^{t} \left \lfloor \frac{a - kb}{c} \right \rfloor =&#xD;&#xA;\frac{t+1}{c} \left \lbrace a - \frac{tb}{2}  \right \rbrace&#xD;&#xA;- \frac{1}{c} \left \lfloor \frac{t+1}{u} \right \rfloor&#xD;&#xA;\left \lbrace \frac{u(c-d)}{2} + \lambda u \right \rbrace - \frac{r}{c} . $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, with $a=1019,b=33$ and $c=55$ we have $d=\text{gcd}(33,55)=11,$&#xA;$u=c/d=55/11=5,$ $t= \lfloor 1019/33 \rfloor = 30$  and $ 1019 \equiv 29 \textrm{ mod } 55, $ and&#xA;$ 1019 \equiv 7 \textrm{ mod } 11. $ Hence $ r = 29 $ and $\lambda = 7.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the condition $ u \, | \, t $ is satisfied, and so theorem 2 gives&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \sum_{k=0}^{30} \left \lfloor \frac{1019 - 33k}{55} \right \rfloor = &#xD;&#xA;\frac{31}{55} \left \lbrace 1019 - \frac{ 30 \cdot 33}{2} \right \rbrace&#xD;&#xA;- \frac{6}{55} \left \lbrace \frac{5(55-11)}{2} + 7 \cdot 5 \right \rbrace - \frac{29}{55} = 279.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is easily verified with WolframAlpha, or similar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both of these theorems are special cases ($ \mu=0$ and $\mu=1$) of the general result:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Theorem 3: Along with the previous definitions in theorems 1 and 2, let&#xA;$ t+1 \equiv \mu \textrm{ mod } u,$ where $ 0 \le \mu &amp;lt; u $ then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \sum_{k=0}^{t} \left \lfloor \frac{a - kb}{c} \right \rfloor =&#xD;&#xA;\frac{t+1}{c} \left \lbrace a - \frac{tb}{2}  \right \rbrace&#xD;&#xA;- \frac{1}{c} \left \lfloor \frac{t+1}{u} \right \rfloor&#xD;&#xA;\left \lbrace \frac{u(c-d)}{2} + \lambda u \right \rbrace $$&#xA;$$ - \frac{1}{c} \sum_{k=0}^{\mu - 1} \left \lbrace&#xD;&#xA;r+k \left \lbrace \left( \left \lfloor \frac{b}{c} \right \rfloor + 1 \right) c - b&#xD;&#xA;\right \rbrace \textrm{ mod } c \right \rbrace .$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are $ \mu $ terms in the last sum, so this is understood to be zero when $ \mu =0.$ The terms in the final summation of the equation are all $ \ge 0 $ and reduced modulo $c.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A sketch of the proof runs as follows. On adding up the $t+1$ equations at the start of my previous answer, we add up $ \lfloor (t+1)/u \rfloor $ “complete” sets of residues modulo $c$&#xA;(not to be confused with a complete residue system modulo $c$) that are congruent to $ \lambda $ modulo $c.$ The term $ \sum_{k=0}^{\mu - 1} \lbrace r+k \lbrace \cdots \rbrace&#xD;&#xA;\textrm{ mod } c \rbrace $ (equivalent, of course, to $ \sum_{k=0}^{\mu - 1} \lbrace (r-kb) \textrm{ mod } c \rbrace $ ) is the sum of the “left over residues.”&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rearranging the equation obtained from the summation proves the theorem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: In the text $ \lbrace \cdot \rbrace $ is only used for readability and does not indicate fractional part (I find doubled-up curved brackets awkward to read).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just for fun, here is what we get when we put $ \mu = 2 $ into theorem 3.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Theorem 2(b): Along with the previous definitions, since $ \mu = 2 $ our condition here is that $ u \, | \, (t-1) .$ Now for $ r \ge b $ we obtain&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \sum_{k=0}^{t} \left \lfloor \frac{a - kb}{c} \right \rfloor =&#xD;&#xA;\frac{t+1}{c} \left \lbrace a - \frac{tb}{2}  \right \rbrace&#xD;&#xA;- \frac{1}{c} \left \lfloor \frac{t+1}{u} \right \rfloor&#xD;&#xA;\left \lbrace \frac{u(c-d)}{2} + \lambda u \right \rbrace - \frac{2r-b}{c} . \quad (1) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, with $a=1037,b=33$ and $c=55$ we have $d=\text{gcd}(33,55)=11,$&#xA;$u=c/d=55/11=5,$ $t= \lfloor 1037/33 \rfloor = 31$  and $ 1037 \equiv 47 \textrm{ mod } 55, $ and&#xA;$ 1037 \equiv 3 \textrm{ mod } 11. $ Hence $ r = 47 $ and $\lambda = 3.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the condition $ u \, | \, (t-1) $ is satisfied, and so theorem 2(b) gives&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \sum_{k=0}^{31} \left \lfloor \frac{1037 - 33k}{55} \right \rfloor = &#xD;&#xA;\frac{32}{55} \left \lbrace 1037 - \frac{ 31 \cdot 33}{2} \right \rbrace&#xD;&#xA;- \frac{6}{55} \left \lbrace \frac{5(55-11)}{2} + 3 \cdot 5 \right \rbrace - \frac{2 \cdot 47 - 33}{55} = 291.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is easily verified with WolframAlpha, or similar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $ r &amp;lt; b $ the final term in $(1)$ would be $ - \frac{2r+c-b}{c}.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(To be continued... time and enthusiasm permitting :-) )&lt;/p&gt;&#xA;" OwnerUserId="1301" LastEditorUserId="1301" LastEditorDisplayName="" LastEditDate="2011-01-17T08:51:03.823" LastActivityDate="2011-01-17T08:51:03.823" CommunityOwnedDate="2011-01-14T12:49:21.713" CommentCount="4" />
  <row Id="17487" PostTypeId="2" ParentId="17484" CreationDate="2011-01-14T13:24:27.580" Score="3" ViewCount="" Body="&lt;p&gt;For simplicity, assume discrete-time process $\{X_t\}_{t\in\mathbb Z}$. A counterexample is the following. One can easily finds random variables $Y$ and $Z$, such that $E(Y) = E(Z), E|Y|\neq E|Z|$ and $E(Y^2)=E(Z^2)$ (*). Then, let $X_{2n}$ be i.i.d. copies of $Y$, and $X_{2n-1}$ be i.i.d. copies of $Z$, and let $X_{2n}$ and $X_{2m-1}$ for all $n,m\in\mathbb Z$ be independent as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now construct an example such that (*) holds. Consider $Y = \epsilon U$, where $U$ is the uniform random variable on $[0,1]$, $P(\epsilon = 1)=P(\epsilon=-1)=1/2$ and $\epsilon$, $U$ are independent. Then, $E Y^2 = 1/3, EY = 0$ and $E|Y| = 1/2$. Take $Z = \epsilon&amp;#39; \delta_{1/\sqrt 3}$. $EZ^2 = 1/3, EZ = 0$, but $E|Z| = 1/\sqrt{3}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now for the continuous-time counterexample. Let $U$ be uniform on $[0,1]$, and let $\tilde X$ denote the discrete one defined above. Define&#xA;$X_t = \tilde X_{\lfloor t+U \rfloor}$.&#xA;Then, for $|t-s|&amp;lt;1$, $E(|X_tX_s|) = (t-s)E|X_0|^2$.&lt;/p&gt;&#xA;" OwnerUserId="5836" LastEditorUserId="5836" LastEditorDisplayName="" LastEditDate="2011-01-14T14:43:58.137" LastActivityDate="2011-01-14T14:43:58.137" CommentCount="4" />
  <row Id="17488" PostTypeId="1" AcceptedAnswerId="17491" CreationDate="2011-01-14T13:53:56.697" Score="3" ViewCount="76" Body="&lt;p&gt;Imagine 2 sets of 6 wires. How would I find how many possible connections there are?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Every wire must be used to be considered a connection.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;//1st set&#xA;ABCDEF&#xA;GHIJKL&#xA;&#xA;//2nd set&#xA;ABCDEF&#xA;HGIJKL&#xA;&#xA;//visual layout of situation&#xA;0 1&#xA;0 1&#xA;0 1&#xA;0 1&#xA;0 1&#xA;0 1&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="1595" LastActivityDate="2011-01-14T14:07:12.247" Title="How many possible solutions for 6 wires?" Tags="&lt;algebra&gt;&lt;logic&gt;&lt;functions&gt;" AnswerCount="1" />
  <row Id="17489" PostTypeId="2" ParentId="15235" CreationDate="2011-01-14T13:54:27.310" Score="3" ViewCount="" Body="&lt;p&gt;The answer is Yes:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume that $X_{k+1}^TX_{k+1}$ is invertible. One can check that its lowest absolute eigenvalue is given by&#xA;$$\left\vert{\frac1{\lambda_{k+1}}}\right\vert=\min\limits_{y\in{\mathbb R}^{k+1},\,\left\lVert y\right\lVert_2=1}\left\lVert{X_{k+1}y}\right\lVert_2^2\,,$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It holds&#xA;$$\min\limits_{y\in{\mathbb R}^{k+1},\,\left\lVert y\right\lVert_2=1}\left\lVert{X_{k+1}y}\right\lVert_2^2\leq\min\limits_{y\in{\mathbb R}^{k},\,\left\lVert y\right\lVert_2=1}\left\lVert{X_{k}y}\right\lVert_2^2\,.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This completes the proof.&lt;/p&gt;&#xA;" OwnerUserId="5838" LastEditorUserId="5838" LastEditorDisplayName="" LastEditDate="2011-01-14T22:44:34.860" LastActivityDate="2011-01-14T22:44:34.860" CommentCount="4" />
  <row Id="17490" PostTypeId="2" ParentId="17478" CreationDate="2011-01-14T14:00:06.630" Score="5" ViewCount="" Body="&lt;p&gt;If it takes $x$ hours to make the first tool then it will take $10x$ hours to make the second tool.  Also, if a job takes $y$ hours with the first tool it will take $\frac{1}{2}y$ hours with the second tool.  So you are asking when&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;x+y\geq 10x+\frac{1}{2}y.&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rearranging this inequality indicates&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;y\geq 18x.&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, once your job requires at least $18x$ hours, the second tool is better.&lt;/p&gt;&#xA;" OwnerUserId="2838" LastActivityDate="2011-01-14T14:00:06.630" />
  <row Id="17491" PostTypeId="2" ParentId="17488" CreationDate="2011-01-14T14:07:12.247" Score="5" ViewCount="" Body="&lt;p&gt;I am not sure what you meant by the sets of wires, but assuming you have two stacks of six pieces of wires, like:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;1st: A,B,C,D,E,F   &lt;/li&gt;&#xA;&lt;li&gt;2nd: G,H,I,J,K,L&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Then A can be connected to 6 pieces (G,H,I,J,K,L), B to 5 (G,H,I,J,K,L except for one) and so on, so the answer could be 6!=720.&lt;/p&gt;&#xA;" OwnerUserId="5651" LastActivityDate="2011-01-14T14:07:12.247" CommentCount="5" />
  <row Id="17492" PostTypeId="2" ParentId="17485" CreationDate="2011-01-14T15:07:51.337" Score="2" ViewCount="" Body="&lt;p&gt;You are right that in $\mathbb{R}$ non-integral powers are not defined for negative bases.  The easiest way is to restrict the angles $\beta$ and $\phi$ to $[0,\pi/2]$, then recognize that you need all eight combinations of plus and  minus signs for $x ,y$ and $z$.  This cuts the trig function calls by a factor eight.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-01-14T15:07:51.337" />
  <row Id="17493" PostTypeId="2" ParentId="17473" CreationDate="2011-01-14T15:24:45.167" Score="1" ViewCount="" Body="&lt;p&gt;The sort of question you would do with Monte Carlo is to decide &quot;If the dealer shows a 6, what value should I draw to?&quot;  You would do this by simulating many deals with the dealer showing a 6, try out various strategies, and see which has the highest expectation.  Under certain statistical assumptions, you can state that a particular strategy wins a% of the money at stake with a standard deviation of b%.  Raising the number of trials will decrease b%, roughly by the square root of the number of trials.  So if one strategy is clearly better than another, it won't take too many tries to know that.  If they are close, it will take a lot.  If they are really close, you will never know, but then it doesn't matter much.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As svenkatr states, the details of Blackjack make it difficult to do an analytic solution, but you could in particular cases.  Going back to dealer shows a 6, you could say suppose I draw to 15.  After tedious counting, you could exactly calculate your winning percentage.  After more counting, you could compare that with the winning percentage of draw to 14, and you would have a rigorous answer. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Added:  counting just changes the frequency distribution of cards that are dealt.  The same approach works-you just ask questions like &quot;If the proportion of cards that count 10 drops from 4/13 to 2/13, how does that change the results?&quot; and adjust your strategy accordingly. Generally if the high count cards are depleted you draw more because you bust less.  More high count cards are good for the player, because s/he can stop and let the dealer bust.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2011-01-14T18:18:55.393" LastActivityDate="2011-01-14T18:18:55.393" />
  <row Id="17494" PostTypeId="1" AcceptedAnswerId="17525" CreationDate="2011-01-14T15:28:30.827" Score="2" ViewCount="116" Body="&lt;p&gt;I'd like a good way to solve an optimization problem I came across.  It's a constrained knapsack problem: I want to find integers&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$1=a_1\le a_2\le\cdots\le a_t$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$a_1+a_2+\cdots+a_t=N$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with $t$ as small as possible such that (with the $S_i\subseteq S={a_1,a_2,\ldots,a_t}$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{e\in S_i}e=C_i$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for a certain fixed collection $C_1 = 1, C_2 = 4,\ldots,C_k=N.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What software exists for this sort of problem?  I feel like writing my own program from the ground up would be the wrong approach.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've been toying with COIN-OR and such but they seem set up for very different sorts of problems.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Ideally, I'd solve an even more complicated problem where $C_1$ and $C_k$ must match exactly and the other sums must be within a fixed range depending on the values of the neighboring C-values... but I don't imagine this is feasible?&lt;/p&gt;&#xA;" OwnerUserId="1778" LastEditorUserId="1778" LastEditorDisplayName="" LastEditDate="2011-01-14T20:00:59.883" LastActivityDate="2011-01-14T23:54:29.400" Title="Solving a knapsack-type problem" Tags="&lt;optimization&gt;&lt;integer-programming&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="17495" PostTypeId="2" ParentId="17485" CreationDate="2011-01-14T15:34:46.657" Score="3" ViewCount="" Body="&lt;p&gt;The equation for an axis-aligned ellipse is $$\left(\frac{x}{a}\right)^2 + \left(\frac{y}{b}\right)^2 = 1.$$  Because $x^2 = (-x)^2$, the ellipse is symmetric under $x \rightarrow -x$ and $y \rightarrow -y$ reflections.  The superellipse is a generalization where the exponent is some $\alpha \neq 2$.  However, $x^\alpha$ is not well-defined for negative $x$ and fractional $\alpha$; and for odd integer $\alpha$, $x^\alpha \neq (-x)^\alpha$, so the reflection symmetries are broken.  One way to patch the symmetries is to use $|x|$ instead of $x$ in the generalized equation: $$\left|\frac{x}{a}\right|^\alpha + \left|\frac{y}{b}\right|^\alpha = 1.$$  Note that this still reduces to the ellipse for $\alpha = 2$.  Similar considerations apply to the parametric representation of the ellipse:&#xA;$$\left(x(\theta),y(\theta)\right) = (a\cos\theta, b\sin\theta)$$ is symmetric under $(x,y,\theta)\rightarrow(-x,y,\pi-\theta)$ and $(x,y,\theta)\rightarrow(x,-y,-\theta)$, and the expressions for $x$ and $y$ become ill-defined when the $\sin$ and $\cos$ terms are raised to the power of $2/\alpha$.  To preserve the symmetries and still solve the defining equation, the correct generalization is $$\left(x(\theta),y(\theta)\right) = \left(a|\cos\theta|^{2/\alpha} \text{sgn}(\cos\theta), b|\sin\theta|^{2/\alpha} \text{sgn}(\sin\theta)\right).$$&#xA;The motivation for the definition of the superellipsoid is completely analogous, and the same type of corrections need to be made to the parametric representation of its surface.  Specifically, its definition is $$\left|\frac{x}{a}\right|^\alpha + \left|\frac{y}{b}\right|^\alpha + \left|\frac{z}{c}\right|^\beta = 1,$$ and its surface can be parametrized as&#xA;$$&#xD;&#xA;\begin{eqnarray}&#xD;&#xA;x(\theta,\phi) &amp;amp;=&amp;amp; a \left(\cos\theta\right)^{2/\beta} |\cos\phi|^{2/\alpha} \text{sgn}(\cos\phi) \\&#xD;&#xA;y(\theta,\phi) &amp;amp;=&amp;amp; b \left(\cos\theta\right)^{2/\beta} |\sin\phi|^{2/\alpha} \text{sgn}(\sin\phi) \\&#xD;&#xA;z(\theta,\phi) &amp;amp;=&amp;amp; c |\sin\theta|^{2/\beta} \text{sgn}(\sin\theta)&#xD;&#xA;\end{eqnarray}&#xD;&#xA;$$&#xA;for $-\pi/2 \le \theta \le \pi/2$ and $0 \le \phi &amp;lt; 2\pi$.&lt;/p&gt;&#xA;" OwnerUserId="5546" LastActivityDate="2011-01-14T15:34:46.657" />
  <row Id="17496" PostTypeId="1" AcceptedAnswerId="17503" CreationDate="2011-01-14T15:57:17.053" Score="4" ViewCount="311" Body="&lt;p&gt;I'm looking for some help disproving an answer provided on a &lt;a href=&quot;http://stackoverflow.com/questions/4681434/f-facebook-hacker-cup-double-squares&quot;&gt;StackOverflow question&lt;/a&gt; I posted about computing the number of double square combinations for a given integer.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The original question is from the Facebook Hacker Cup  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Source: Facebook Hacker Cup&#xA;  Qualification Round 2011&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;A double-square number is an integer $X$&#xA;  which can be expressed as the sum of&#xA;  two perfect squares. For example, 10&#xA;  is a double-square because $10 = 3^2 + 1^2$.&#xA;  Given $X$, how can we determine the&#xA;  number of ways in which it can be&#xA;  written as the sum of two squares? For&#xA;  example, $10$ can only be written as $3^2 + 1^2$&#xA;  (we don't count $1^2 + 3^2$ as being different). On the other hand,&#xA;  $25$ can be written as $5^2 + 0^2$ or as&#xA;  $4^2 + 3^2$.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;You need to solve this problem for $0 \leq X \leq 2,147,483,647$.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Examples:&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$10 \Rightarrow 1$&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$25 \Rightarrow 2$&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$3 \Rightarrow 0$&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$0 \Rightarrow 1$&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$1 \Rightarrow 1$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In response to my original question about optimizing this for F#, I got the following response which I'm unable to confirm solves the given problem correctly.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Source: &lt;a href=&quot;http://stackoverflow.com/questions/4681434/f-facebook-hacker-cup-double-squares/4681771#4681771&quot;&gt;StackOverflow answer by Alexandre C.&lt;/a&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Again, the number of integer solutions&#xA;  of $x^2 + y^2 = k$ is four times the&#xA;  number of prime divisors of $k$ which&#xA;  are equal to $1 \bmod 4$.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Knowing this, writing a program which&#xA;  gives the number of solutions is easy:&#xA;  compute prime numbers up to $46341$ once&#xA;  and for all.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Given $k$, compute the prime divisors of&#xA;  $k$ by using the above list (test up to&#xA;  $\sqrt{k}$). Count the ones which are&#xA;  equal to $1 \bmod 4$, and sum. Multiply&#xA;  answer by $4$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When I go through this algorithm for $25$, I get $8$ which is not correct.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;For each prime factor (pf) of $25$ (5,5 are prime factors of $25%)&#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;if pf % 4 = 1 (true for both 5's), add 1 to count&lt;/li&gt;&#xA;  &lt;/ul&gt;&lt;/li&gt;&#xA;  &lt;li&gt;return 4 * count (count would be 2 here). &lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;  &#xA;  &lt;p&gt;So for $25$, this would be 8&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5848" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2011-01-14T17:03:14.723" LastActivityDate="2011-01-14T20:02:53.157" Title="Number of integer solutions of $x^2 + y^2 = k$" Tags="&lt;number-theory&gt;" AnswerCount="4" FavoriteCount="1" />
  <row Id="17497" PostTypeId="2" ParentId="17496" CreationDate="2011-01-14T16:03:28.067" Score="6" ViewCount="" Body="&lt;p&gt;$25$ has &lt;em&gt;two&lt;/em&gt; prime divisors: $5$ and $5$, and you have to count them both because they are both congruent to $1$ modulo $4$. Now multiply by $4$. You have $8$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This counts the decompositions $$\begin{array}{cc}&#xD;&#xA;3^2+4^2,&amp;amp;&#xD;&#xA;3^2+(-4)^2, \\&#xD;&#xA;(-3)^2+4^2, &amp;amp;&#xD;&#xA;(-3)^2+(-4)^2, \\&#xD;&#xA;4^2+3^2, &amp;amp;&#xD;&#xA;4^2+(-3)^2, \\&#xD;&#xA;(-4)^2+3^2, &amp;amp;&#xD;&#xA;(-4)^2+(-3)^2&#xD;&#xA;\end{array}&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you can see, multiplying by $4$ at the end is done to cater for signs: if you are only interested in &quot;positive&quot; decompositions, then don't do that.&#xA;Also, this counts $a^2+b^2$ as different from $b^2+a^2$. If you don't want that, well, you have to substract the number $N$ of decompositions of the form $a^2+a^2$ (which is easy to compute...) divide by $2$, and then add $N$ back.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, if you want to know &lt;em&gt;why&lt;/em&gt; this works, well, there are very few better things than to consult an introduction to number theory!&lt;/p&gt;&#xA;" OwnerUserId="274" LastEditorUserId="274" LastEditorDisplayName="" LastEditDate="2011-01-14T16:08:44.630" LastActivityDate="2011-01-14T16:08:44.630" CommentCount="11" />
  <row Id="17498" PostTypeId="1" AcceptedAnswerId="17502" CreationDate="2011-01-14T16:03:42.593" Score="2" ViewCount="80" Body="&lt;p&gt;Hey everyone&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's say I have 3 points, A B and C.&#xA;Each point has a fixed distance to every other point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I want to find the right representation for those distances, I have to place A and B according to their distance, draw two circles respectively with center A and B and d(A, C) radius and d(B, C) radius. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The two intersections of the circles are the possible place where I can put C and have the right distance between every point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pretty simple.&#xA;My problem is the following: How do I do the exact same thing with 4 points A B C and D ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The two options for placing my C point never allow me to place D with the right distances from every other point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I am sorry if this is very badly explained, I'm not a native english speaker !)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks you !&#xA;Gael&lt;/p&gt;&#xA;" OwnerUserId="5849" LastEditorUserId="2548" LastEditorDisplayName="" LastEditDate="2011-01-14T16:07:17.860" LastActivityDate="2011-01-14T20:49:20.663" Title="how to find the right position for representing distance between points when 4 points" Tags="&lt;geometry&gt;" AnswerCount="3" CommentCount="5" />
  <row Id="17499" PostTypeId="2" ParentId="9896" CreationDate="2011-01-14T16:07:08.213" Score="4" ViewCount="" Body="&lt;p&gt;I'm not sure what Arturo has in mind with his link (which is an axiomatization of set theory, not of category theory), but it's well-known that there is the same issues with the category of all categories that there is with the set of all sets.  It doesn't normally arise because the types of constructions that you would want to do don't lead to contradiction, but the same thing is true of set theory.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The existence of a contradiction seems to be folklore, but &lt;a href=&quot;http://cs.nyu.edu/pipermail/fom/1999-May/003117.html&quot; rel=&quot;nofollow&quot;&gt;this message&lt;/a&gt; by Steve Simpson on the Foundation of Mathematics mailing list gives a contradiction in pretty careful detail.  Carsten Butz &lt;a href=&quot;http://cs.nyu.edu/pipermail/fom/1999-May/003119.html&quot; rel=&quot;nofollow&quot;&gt;replies&lt;/a&gt; saying that this type of example is well-known.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;People seem to be increasingly careful about the issue, and emphasizing that they are taking about a &quot;large&quot; category of &quot;small&quot; categories, where &quot;large&quot; and &quot;small&quot; can be made precise in the two different ways that Nuno discusses.&lt;/p&gt;&#xA;" OwnerUserId="4880" LastActivityDate="2011-01-14T16:07:08.213" CommentCount="2" />
  <row Id="17500" PostTypeId="2" ParentId="17498" CreationDate="2011-01-14T16:22:34.487" Score="0" ViewCount="" Body="&lt;p&gt;See &lt;a href=&quot;http://planetmath.org/encyclopedia/CompassAndStraightedgeConstructionOfSquare.html&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt;. This is the construction of a square (a special case). Essentially, we are constructing a square of sides of given length $s$. &lt;em&gt;A priori&lt;/em&gt;, from the Pythogorean Theorem, we know the distances between two given points. &lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2011-01-14T16:22:34.487" />
  <row Id="17501" PostTypeId="2" ParentId="17484" CreationDate="2011-01-14T16:24:01.833" Score="1" ViewCount="" Body="&lt;p&gt;A simple counterexample is the following.  Let $m(t)$ be a non-constant function that satisfies $m(t) \ge 2$ for all $t$, and let $p(t)=1/m(t)^2$.  Now take $X_t$ at each point in time to be $m(t)$ with probability $p(t)$, $-m(t)$ with probability $p(t)$, and $0$ with probability $1-2p(t)$.  Clearly $E[X_t] = 0$ for all $t$.  Also $E[X_t^2] = 2p(t)m(t)^2 = 2$; and $E[X_t X_{t+h}]=0$ for $h \neq 0$.  Therefore $X_t$ is weakly stationary.  On the other hand, $E[|X_t|] = 2p(t)m(t) = 2/m(t)$, which is not constant; so $|X_t|$ is not weakly stationary.&lt;/p&gt;&#xA;" OwnerUserId="5546" LastActivityDate="2011-01-14T16:24:01.833" />
  <row Id="17502" PostTypeId="2" ParentId="17498" CreationDate="2011-01-14T16:28:17.490" Score="3" ViewCount="" Body="&lt;p&gt;In general, you cannot place four points in the plane at a given set of distances.  Even with three points, you need to satisfy the triangle inequality.  If the distances do, your construction works.  But then given distances from D to A, B, and C, it is unlikely there is a point that satisfies the distances.  If the distances do allow it, you can just do what you propose for A, B, and C (using either position for C-it is just a mirror image triangle), then locate D relative to A and B and check each of the two choices for being the correct distance from C.  But it will not work unless the distance are carefully chosen.  If you move into $\mathbb{R^3}$ you will be able to do it in general as long as the triangle inequalities are satisfied, and the construction is the same.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-01-14T16:28:17.490" CommentCount="1" />
  <row Id="17503" PostTypeId="2" ParentId="17496" CreationDate="2011-01-14T16:57:23.717" Score="8" ViewCount="" Body="&lt;p&gt;This is worked out in detail in many an introductory textbook on Number Theory. Here is a summary of the treatment in &lt;strong&gt;An Introduction to the Theory of Numbers&lt;/strong&gt; (5th Edition) by Ivan Niven, Herbert S. Zuckerman, and Hugh L. Montgomery (Wiley, 1991, section 3.6).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Theorem&lt;/strong&gt; (Fermat) Let $n$ be a positive integer, and write $n$ in the form&#xA;$$n = 2^{\alpha}\prod_{p\equiv 1\pmod{4}} p^{\beta} \prod_{q\equiv 3 \pmod{4}} q^{\gamma}.$$&#xA;with $p$ and $q$ primes. Then $n$ can be expressed as a sum of two squares if and only if all the exponents $\gamma$ are even.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, define the following:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$R(n)$: the number of &lt;em&gt;ordered pairs&lt;/em&gt; $(x,y)$ of integers such that $x^2+y^2=n$.&lt;/li&gt;&#xA;&lt;li&gt;$r(n)$: the number of ordered pairs $(x,y)$ such that $\gcd(x,y)=1$ and $x^2+y^2=n$ (called &quot;proper representations of $n$&quot;);&lt;/li&gt;&#xA;&lt;li&gt;$P(n)$: the number of representations of $n$ by the form $x^2+y^2$ with $\gcd(x,y)=1$,  $x\gt 0$ and $y\geq 0$;&lt;/li&gt;&#xA;&lt;li&gt;$N(n)$: the number of solutions to the congruence $s^2\equiv -1\pmod{n}$. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Theorem.&lt;/strong&gt; A positive integer $n$ is properly represented by the form $x^2+y^2$ if and only if $-4$ is a square modulo $4n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, since $-4$ is a square modulo $8$ but not modulo $16$, $n$ may be divisible by $2$ but not by $4$. If $p$ is an odd prime of the form $4k+1$, then $-4$ is a square modulo $p$, and by Hensel's Lemma it lifts to a solution modulo $p^k$ for an y $k$; so $n$ may be divisible by arbitrary powers of primes of the form $4k+1$. On the other hand, if $p$ is a prime that divides $n$ and is congruent to $3$ modulo $4$, then $-4$ is not a square modulo $p$. So we get:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Theorem.&lt;/strong&gt; A positive integer $n$ is properly representable as a sum of two squares if and only if the prime factors of $n$ are all of the form $4k+1$, except for the prime $2$ which may occur to at most the first power.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now suppose that $n$ is positive and $n=x^2+y^2$ is an arbitrary representation. Set $g=\gcd(x,y)$; then $g^2|n$, so $n = g^2m$, $\gcd(\frac{x}{g},\frac{y}{g}) = 1$, so $m$ is properly represented as the sum of the squares of $\frac{x}{g}$ and $\frac{y}{g}$. Note that $g$ may have prime factors congruent to $3$ modulo $4$, which divide $n$ to an even power, and the power of $2$ that divides $n$ may be arbitrary as well. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Theorem.&lt;/strong&gt; Suppose that $n\gt 0$. Then $P(n)=N(n)$, $r(n) = 4N(n)$ and $R(n)=\sum r\left(\frac{n}{d^2}\right)$, where the sum is taken over all positive $d$ such that $d^2|n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Theorem.&lt;/strong&gt; Let $n$ be a positive integer and write&#xA;$$ n = 2^{\alpha}\prod_{p}p^{\beta}\prod_q q^{\gamma}$$&#xA;where $p$ runs over prime divisors of $n$ congruent to $1$ modulo $4$ in the first product, and $q$ runs over prime divisors of $n$ of the form $4k+3$ in the second. If $\alpha=0$ or $1$, and all $\gamma$ are $0$, then $r(n) = 2^{t+2}$, where $t$ is the number of primes $p$ of the form $4k+1$ that divide $n$ (counted with multiplicity); otherwise, $r(n)=0$. If all the $\gamma$ are even, then $R(n) = 4\prod_p(\beta+1)$. Otherwise, $R(n)=0$.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2011-01-14T16:57:23.717" CommentCount="4" />
  <row Id="17504" PostTypeId="2" ParentId="17347" CreationDate="2011-01-14T17:24:27.757" Score="3" ViewCount="" Body="&lt;p&gt;I suspect that if you have a complete lattice $X$ and you endow it with the order topology (where a subbasis of the topology is given by the sets of the form $(a,\infty) = \{x\in X\mid a\lt x\}$ and $(-\infty,b) = \{x\in X\mid x\lt b\}$ with $a,b\in X$ arbitrary), then 1, 2(1), and 2(2) will be equivalent in the sense that one exists if and only if the others exists and in that case they will be equal (the only caveat is that we have to be careful with 2(1) in the case of unbounded sequences; technically, for instance, if you take $1,2,1,3,1,4,1,5,1,6,\ldots$, the only converging subsequences converge to $1$, but the sups are never defined, so you could interpret 2(1) as saying that the limit superior is $1$, and 2(2) as saying it does not exists). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I am having trouble proving it in full generality; the equivalence of 2(1) and 2(2) is not difficult if you assume the set is totally ordered and complete, so that suprema exist for any bounded set (throw in an &amp;uuml;ber-sup and an &amp;uuml;ber-inf like $\infty$ and $-\infty$ to the reals if necessary). I'm somewhat unhappy about this, but I figure something is better than nothing here. I've been toying with this and going down a fair number of silly alleys along the way, so maybe after it is written down I'll see that it works in any lattice; however, right now I am using the fact that any sequence contains a monotone subsequence, and this requires total order, not just lattice order. (I'm also using the fact that the order topology is Hausdorff for totally ordered sets; I don't know if that works for lattices).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that a totally ordered set that has the order topology is necessarily Hausdorff: given $x\neq y$, say $x\lt y$. If there exists $z$ such that $x\lt z\lt y$, then the sets $(-\infty,z)$ and $(z,\infty)$ provide disjoint open neighborhoods of $x$ and $y$, respectively; if there is no element of $X$ strictly between $x$ and $y$, then $(-\infty,y)$ and $(x,\infty)$ provide the necessary disjoint neighborhoods.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Theorem.&lt;/strong&gt; Suppose $X$ is a totally ordered set such that every nonempty subset has a supremum. If we endow $X$ with the order topology, then definitions 2(1) and 2(2) are equivalent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Proof.&lt;/em&gt; Say $\lim\limits_{n\to\infty}\sup\{x_m\mid m\geq n\}$ is equal to $L$. This means that for every open neighborhood $U$ of $L$, there exists a $K$ such that for all $k\geq K$, $\sup\{x_m\mid m\geq k\}\in U$, and in particular for every $k\geq K$ there exists $m\geq k$ such that $x_m\in U$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, I claim that the set of limit points of subsequences of $x_n$ is bounded above by $L$. Indeed, suppose that you have a subsequence that converges to $M$, with $M\gt L$. Let $U$ and $V$ be neighborhoods of $L$ and $M$ that are disjoint; intersect $U$ with $(-\infty,M)$ and $V$ with $(L,\infty)$, if necessary. Then there exists $K$ such that for all $k\geq K$, $\sup\{x_m\mid m\geq k\}\in U$; in particular, for all $k\geq K$, we have $x_k \leq\sup\{x_m\mid m\geq k\}\notin V$; therefore, since the sequence is eventually not in $V$, it is not infinitely often in $V$, and therefore there can be no subsequence that converges to $M$. This contradicts the choice of $M$, so we conclude that $M\leq L$, as claimed. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since every sequence contains a monotone subsequence, and monotone subsequences in complete sets have limits (their sup if they are increasing, their inf if they are decreasing), the set of limit points of subsequences is nonempty; therefore, since it is bounded above by $L$, we know that &#xA;$$\sup\{\text{limit of any subsequence of }\{x_n\}\text{ that converges}\} \leq L=\lim_{n\to\infty}\sup\{x_m\mid m\geq n\}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose that the supremum of the limits is $M$, and is strictly smaller than $L$; let $V$ be an open neighborhood of $M$ contained in $(-\infty,L)$, $U$ an open neighborhood of $L$ contained in $(M,\infty)$, with $U\cap V=\emptyset$. We know that there exists $K$ such that for all $k\geq K$, $\sup\{x_m\mid m\geq k\}\in U$; in particular, for all $k\geq K$ there exists $m\geq k$ such that $x_m\in U$. Thus, we can find a subsequence of $\{x_n\}$ that is contained in $U$; this subsequence contains a monotone subsequence, which therefore coverges to some point in the closure of $V$; the closure of $V$ is contained in the complement of $U$ and of $(-\infty,M)$, and therefore the closure of $V$ does not include $M$ and consists only of points strictly larger than $M$; that is, the limit of this monotone subsequence is strictly larger than $M$, contradicting the choice of $M$. Thus, $M\geq L$, giving equality. &lt;em&gt;QED&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know about 3 and 4, I'll think about 5, though again I expect you'll need the topology to be related to the order.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In any lattice, with definition 1, you always have  $\liminf x_n\leq \limsup x_n$. To see this, note that for every $k,m$, you have &#xA;$$\mathop{\wedge}\limits_{n\geq k} x_n \leq \mathop{\vee}\limits_{n\geq m}x_n$$&#xA;because if $r\geq \max\{k,m\}$, then you have&#xA;$$\mathop{\wedge}\limits_{n\geq k}x_n \leq x_r \leq \mathop{\vee}\limits_{n\geq m}x_n.$$&#xA;Therefore, each $\mathop{\vee}\limits_{n\geq m}x_n$ is an upper bound for all of $\mathop{\wedge}\limits_{n\geq k}x_n$, hence for each $m$ we have&#xA;$$\liminf x_n = \sup\mathop{\wedge}\limits_{n\geq k}x_n \leq \mathop{\vee}\limits_{n\geq m}x_n.$$&#xA;Therefore, $\liminf x_n$ is a lower bound for each of the joins, hence is less than or equal to their inf; that is,&#xA;$$\liminf x_n \leq \inf\mathop{\vee}\limits_{n\geq m}x_n = \limsup x_n.$$&#xA;(The same is also true for definition 2(1), for silly reasons: the limit superior is the supremum of a particular nonempty set, the limit inferior is the infimum of that same set, and since the set is nonempty, then the infimum is less than or equal to the maximum).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $L=\liminf x_n$ and $M=\limsup x_n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose that $\liminf x_n \lt \limsup x_n$, and that you endow $X$ with the order topology inherited from the complete lattice structure. Because $\liminf x_n$ and $\limsup x_n$ are comparable, then we can prove the existence of disjoint open neighborhoods that separate them the same way I did for a totally ordered set: if there is an element $z\in X$ such that $L\lt z\lt M$, then the open sets $(-\infty,z)$ and $(z,\infty)$ are disjoint open neighborhoods of $L$ and $M$, respectively; and if there is no such element, then $(-\infty,M)$ and $(L,\infty)$ are disjoint open neighborhoods that separate them. Taking disjoint neighborhoods of $L$ and $M$ you can obtain subsequences that are eventually not in the corresponding neighborhood, so if $L\lt M$ then there are subsequences that do not converge. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $L=M$, on the other hand, then let $U$ be an open neighborhood of $L=M$ that is an interval; because the sequence $w_k = \wedge\{ x_m\mid m\geq k\}$ is nondecreasing, and $U$ is an interval containing $L$, there exists $K$ such that for all $k\geq K$ you have $w_k\in U$. Likewise, by considering the nonincreasing sequence $z_k\vee\{x_m\mid m\geq k\}$ which converges to $M$, there exists $H$ such that for all $h\geq H$ we have $z_h\in U$. Therefore, for every sufficiently large $k$ we have that both the meet and the join of $\{x_m\mid m\geq k\}$ lie in the interval $U$, then all the terms lie in the interval $U$; that is, the sequence is eventually in $U$. That is, given any open neighborhood of $L=M$, then the sequence is eventually in $U$; this means that it converges to $L=M$; and from here it is easy to verify that every subsequence also converges to $L$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, $\limsup x_n = \liminf x_n$ if and only if there exists $L$ such that every subsequence converges to $L$, if and only if there exists $L$ such that the sequence converges (in the topological sense) to $L$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know if the limit needs to be unique (in a topology, you can have a sequence converge to two different things if the topology does not separate the points enough). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For 2(1), assuming the topology is the order topology, and $X$ is totally ordered, if $L\lt M$, then we can find intervals $(-\infty,a)$ and $(b,\infty)$ that are disjoint, $L\lt a$ and $b\lt M$, as above. Then there is a subsequence that converges to an element of $(-\infty,a)$ (in fact, of $[L,a)$) and a subsequence that converges to an element of $(b,\infty)$ (in fact, of $(b,M]$). Therefore, the original sequence is infinitely often in $(-\infty,a)$, and infinitely often in $(b,\infty)$; since these two are disjoint, the sequence cannot converge (if it converges to $x$, then for every open neighborhood of $x$ the sequence is eventually in the neighborhood; but separate $x$ from either $L$ or $M$ and get that the sequence is eventually not in the corresponding neighborhood of $x$). Thus, if $L\lt M$ then the sequence does not converge. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $L=M$, then every converging subsequence converges to $L$; but if the set is not complete, then you'll run into trouble: take for example $X=\mathbb{Q}$, a sequence of rationals that converges to $\sqrt{2}$, and interweave it with the constant sequence $1$. If you take a subsequence, then either it contains infinitely many terms of the sequence that converges to $\sqrt{2}$, and therefore does not converge in $X$, or else it contains only finitely many terms of the sequence, and therefore converges to $1$; so the set of limits of converging subsequences is $\{1\}$, hence the limits inferior and superior are equal under definition (1), but the sequence does not converge; if the space $X$ is complete relative to the order topology (so that every set has a supremum and every set has an infimum), then we have the equivalence between definition 2(1) and 2(2) by the above, and then we are essentially in the situation of 1 that I just dealt with to show that the sequence converges to $L=M$ in the sense of 2(2), and hence in the sense of 2(1). &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-15T00:09:25.207" LastActivityDate="2011-01-15T00:09:25.207" CommunityOwnedDate="2011-01-14T17:24:27.757" CommentCount="1" />
  <row Id="17505" PostTypeId="1" AcceptedAnswerId="17507" CreationDate="2011-01-14T17:33:37.713" Score="1" ViewCount="374" Body="&lt;p&gt;Problem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \Psi_{xx} - \Psi_{tt} - \Psi = \exp(3t) \cdot \delta(x)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;No boundary conditions specified.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;I solved the homogeneous portion, $\Psi_\mathrm{homogeneous}$, of this equation via separation of variables but my solution is just for some random case of $k^2$ where: $F&amp;#39;&amp;#39;/F = G&amp;#39;&amp;#39;/G + 4 = k^2$. I chose the case where $k^2 = 0$ which gave me solutions for $\Psi_\mathrm{homogeneous}$ like $x\sin(2t)$ and $x\cos((2t)$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;With the guess method for $\Psi_\mathrm{particular}$, I have no idea what to guess on a general form of $\exp(3t)\delta(x)$ to plug back in to the PDE.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I have read about Green's Functions but, man, I'm having a hard time understanding the guides I have seen because they skip so many of the intermediate steps. I understand that these Green's Functions can provide a general solution and that seems like what I'm looking for. I likely spent a lot of time for nothing on my first attempt with separation of variables for $\Psi_\mathrm{homogeneous}$ and then looking for $\Psi_\mathrm{particular}$ using the guess method...&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I'm curious if there is a general set of IC/BCs that I should be assuming as well?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="5853" LastEditorUserId="5853" LastEditorDisplayName="" LastEditDate="2011-01-18T03:34:36.100" LastActivityDate="2011-01-26T19:06:21.930" Title="Inhomogeneous PDE Help?" Tags="&lt;differential-equations&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="17506" PostTypeId="2" ParentId="17455" CreationDate="2011-01-14T17:45:04.470" Score="0" ViewCount="" Body="&lt;p&gt;Since I did this initially unregistered, I can't seem to figure out how to get back in and add a comment instead of an answer. This is a bit long, but I am still having trouble resolving this problem satisfactorily. Also, I apologize for the formatting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;@Jonas Meyer I got similar answers to you the first time I had tried it. The problem I'm actually dealing with is more general and contains perhaps a c3, c4, or any higher number. However, one of these cs will always be defined as c=1-sum(other cs).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the simpler problem, S^-1*(c1*u1+c2*u2), if you assume c1 and c2 are independent, then the derivative wrt c1 is S^-1*u1 as I said above. This answer makes quite a lot of sense with regards to my current problem. If S=c1*S1+c2*S2 and S1=S2, I also found the answer that you have in the reply, S^-1*(u1-c1*u1-c2*u2). The problem is that this isn't consistent with what is above. This is a portfolio management problem. The first answer will give the equivalent of a portfolio with confidence c1=100% in the u1 view, while the second will give the difference between that view and the blended final portfolio. I am trying to address how much the confidences c1 and c2 are responsible for the final portfolio weights. In the first case, it is obvious: Multiply S^-1*u1 by c1 since it is homogeneous and divide by S^-1*(c1*u1+c2*u2). It is not obvious in the case of S^-1*(u1-c1*u1-c2*u2). For instance, if c1=100% and c2=0%, then the first case will show 100% of the weights are driven by c1 and 0% are driven by c2. The second case will say the contribution is 0% for c1. Something is mistaken here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If c2=1-c1, then the derivative wrt c1 is S^-1*(u1-u2). For the more complicated case, where S=c1*S1+c2*S2=S2+c1*(S1-S2), the derivative wrt c1 can be calculated using what is in the answer and there will a term in one part so that if S1=S2 then it falls out and the answer becomes S^-1*(u1-u2). So if you assume c2=1-c1, I think you can get consistent answers between the case with 1 S and the case with 2 Ss. However, it still doesn't resolve my problems when you multiply the derivative wrt c1 by c1 and divide by S^-1*(c1*u1+c2*u2). This number will not provide the relative contributions from c1 to the portfolio weights.&lt;/p&gt;&#xA;" OwnerUserId="5855" LastActivityDate="2011-01-14T17:45:04.470" CommentCount="3" />
  <row Id="17507" PostTypeId="2" ParentId="17505" CreationDate="2011-01-14T18:02:51.653" Score="1" ViewCount="" Body="&lt;p&gt;Taking the Fourier transform in the spatial variable only reduces the PDE to a nice ODE except for your &quot;double&quot; exponential term.  If it were just exp(3it) you could easily solve it, but in the case exp(exp(3it)) I don't think you're going to be able to write down exact solutions.&lt;/p&gt;&#xA;" OwnerUserId="5857" LastActivityDate="2011-01-14T18:02:51.653" CommentCount="6" />
  <row Id="17508" PostTypeId="1" CreationDate="2011-01-14T18:05:08.483" Score="2" ViewCount="87" Body="&lt;p&gt;In Allen Hatcher's book p. 117 (bottom) [download here: http://www.math.cornell.edu/~hatcher/AT/ATpage.html] the following is said: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The boundary map $\partial : H_n (X, A)\rightarrow H_{n−1} (A)$ has a very simple description: If a class $[\alpha] \in H_n(X,A)$ is represented by a relative cycle $\alpha$, then $\partial[\alpha]$ is the class of the cycle $\partial \alpha$ in $H_{n−1} (A)$ . This is immediate from the algebraic definition of the boundary&#xA;  homomorphism in the long exact sequence of homology groups associated to a short&#xA;  exact sequence of chain complexes.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can see that $\partial \alpha \in C_{n-1}(A)$ and that it must be a cycle since $\partial\partial = 0$. But how is the relation $\partial[\alpha] = [\partial \alpha]$ established (Left hand side contains the $\partial : H_n (X, A)\rightarrow H_{n−1} $, Right hand side contains $\partial : C_n(A) \rightarrow C_{n-1}(A)$).&lt;/p&gt;&#xA;" OwnerUserId="5856" LastActivityDate="2011-01-14T18:05:08.483" Title="Interpretation of boundary homomorphism in long exact sequence of homology groups" Tags="&lt;algebraic-topology&gt;" CommentCount="2" FavoriteCount="1" />
  <row Id="17509" PostTypeId="2" ParentId="17478" CreationDate="2011-01-14T18:05:41.230" Score="1" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\:$ If the slow tool takes $\rm\:T\:$ hours to build, then the fast tool takes an extra $\rm\: 9\ T\:$ hours to build. For the double-speed tool to make up for this lost time in a job of $\rm\ J\ $ hours, its saved time of $\rm\  J/2\ $ hours must equal the lost build time $\rm\ 9\ T\:,\ $ i.e. $\rm\ J/2\ =\ 9\ T\:.$&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2011-01-14T18:05:41.230" CommentCount="0" />
  <row Id="17510" PostTypeId="2" ParentId="7514" CreationDate="2011-01-14T18:34:34.470" Score="1" ViewCount="" Body="&lt;p&gt;The link &lt;a href=&quot;http://mathoverflow.net/questions/30511/ebook-readers-for-mathematics&quot; rel=&quot;nofollow&quot;&gt;http://mathoverflow.net/questions/30511/ebook-readers-for-mathematics&lt;/a&gt; really answers my original question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wanted to say that I ended up purchasing the Pocketbook Reader 602, and I've been using it to read math papers and textbooks.  There have been some issues with the firmware and wifi connectivity.  I actually had to send it back because of a broken reset button. The bluetooth file transferring interface and program could be a bit better. But overall, I'm very happy with the device.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.pocketbook-usa.com/support/pocketbook-602/&quot; rel=&quot;nofollow&quot;&gt;http://www.pocketbook-usa.com/support/pocketbook-602/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some of the complains my friends had were that it wasn't touch screen and that the screen was too small.  I actually did not want a touch screen, and I opted for the smaller device for the portability.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I personally have no problems with the screen size. The only times this ever became an issue was when the file had too much white blank space. I used a pdf editor and cut out the white space, and now it's totally readable. I have a lot of djvu and ps files, and so far PBR has had no problems handling them without having to convert them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In short, I recommend this device to mathematicians who are interested in getting an e-reader.&lt;/p&gt;&#xA;" OwnerUserId="2499" LastActivityDate="2011-01-14T18:34:34.470" />
  <row Id="17511" PostTypeId="1" AcceptedAnswerId="17512" CreationDate="2011-01-14T18:58:59.323" Score="3" ViewCount="137" Body="&lt;p&gt;I have some maths homework due in on monday, therefore I can't ask my teacher for help and the three equations I have are a bit more advanced than what sir taught us. Here is the question I'm stuck on:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ y \leq 3y + 2 \leq 8 + 2y $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would appreciate it if someone could give me a solution and/or explanation of it please. That way I don't consider it to be cheating, as I am learning not just copying! &lt;/p&gt;&#xA;" OwnerUserId="5854" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2011-01-15T18:07:14.230" LastActivityDate="2011-01-15T18:07:14.230" Title="Inequalities: $y \leq 3y+2 \le 8 + 2y$." Tags="&lt;homework&gt;&lt;algebra-precalculus&gt;&lt;inequality&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="17512" PostTypeId="2" ParentId="17511" CreationDate="2011-01-14T19:04:57.717" Score="5" ViewCount="" Body="&lt;p&gt;$a\leq b\leq c$ just means $a\leq b$ AND $b\leq c$.  If you solve both of these individual inequalities, then the solution to the original problem is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Intersection_%28set_theory%29&quot; rel=&quot;nofollow&quot;&gt;intersection&lt;/a&gt; of their solution sets .  Therefore I recommend starting by solving each of $y\leq 3y+2$ and $3y+2\leq 8+2y$.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2011-01-14T19:04:57.717" CommentCount="3" />
  <row Id="17516" PostTypeId="2" ParentId="17472" CreationDate="2011-01-14T19:19:27.377" Score="3" ViewCount="" Body="&lt;p&gt;Prime $\rm\ P = I + m^2\supset m^2\ \Rightarrow\ P\supset m\ \Rightarrow\ P = m\ $ since $\rm\ m\ $ is maximal.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2011-01-14T19:32:45.613" LastActivityDate="2011-01-14T19:32:45.613" />
  <row Id="17517" PostTypeId="1" CreationDate="2011-01-14T19:20:12.277" Score="1" ViewCount="132" Body="&lt;p&gt;I can't seem to come up with a simple formula to head-tail adding two vectors in spherical coordinates. So I'd like to know:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Can anybody point out a way to do it in spherical coordinates (without converting back and forth from cartesian coordinates)?&lt;/li&gt;&#xA;&lt;li&gt;For the sake of execution speed in a computer program, is it faster to do it straight in spherical coordinates or converting back and forth from cartesian coordinates?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="5859" LastActivityDate="2011-01-15T00:27:07.580" Title="Vector sum in spherical coordinates" Tags="&lt;vector-calculus&gt;&lt;spherical-coordinates&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="17518" PostTypeId="2" ParentId="17496" CreationDate="2011-01-14T19:22:13.460" Score="2" ViewCount="" Body="&lt;p&gt;A simpler way to look at this would be to consider the factorization of $\displaystyle n$ in &lt;a href=&quot;http://en.wikipedia.org/wiki/Gaussian_integer&quot; rel=&quot;nofollow&quot;&gt;Gaussian Integers&lt;/a&gt;: $\displaystyle \mathbb{Z}[i] = \{a + bi \ \ | \ a \in \mathbb{Z}, \ b \in \mathbb{Z}\}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is well known that that&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Gaussian integers have unique factorization property (upto units $\displaystyle \pm 1, \pm i$).&lt;/li&gt;&#xA;&lt;li&gt;Primes (of $\displaystyle \mathbb{Z}$) of the form $\displaystyle 4k+3$ are also primes in $\displaystyle \mathbb{Z}[i]$.&lt;/li&gt;&#xA;&lt;li&gt;Primes (of $\displaystyle \mathbb{Z}$) of the form $\displaystyle 4k+1$ factorize into $\displaystyle w w&amp;#39;$ for a prime $\displaystyle w$ (in $\displaystyle \mathbb{Z}[i]$) and $\displaystyle w&amp;#39;$ is the conjugate of $\displaystyle w$. In fact, if $\displaystyle w = a+ib$, then the prime is of the form $\displaystyle a^2 + b^2$.&lt;/li&gt;&#xA;&lt;li&gt;$\displaystyle 2 = (1+i)(1-i)$ and $\displaystyle 1+i$ is prime.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Now if $\displaystyle n = x^2 + y^2$ then $\displaystyle n = (x+iy)(x-iy)$, which corresponds to factorization of $\displaystyle n$ in $\displaystyle \mathbb{Z}[i]$, which you can get from the &lt;em&gt;prime&lt;/em&gt; factorization of $\displaystyle n$ in $\mathbb{Z}[i]$ by choosing a subset of primes to multiply, to get $\displaystyle x + iy$. The conjugates of those primes go toward $\displaystyle x-iy$. The remaining need to be a perfect square to distribute once each into $\displaystyle x+iy$ and $\displaystyle x-iy$. Note this implies that primes of the form $\displaystyle 4k+3$ (which are also primes in $\displaystyle \mathbb{Z}[i]$) need to have an even power in the factorization.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So it basically amounts to finding out the different factorizations of the perfect squares which are factors of $\displaystyle n$ (after ignoring primes of the form $\displaystyle 4k+3$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: It is the units which gives the multiple of $4$, if you count $(x,y), (y,x), (-x,y),(y,-x)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Of course, this is not a formal proof...)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope the helps.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2011-01-14T19:58:42.727" LastActivityDate="2011-01-14T19:58:42.727" />
  <row Id="17519" PostTypeId="1" AcceptedAnswerId="17524" CreationDate="2011-01-14T19:27:02.590" Score="3" ViewCount="104" Body="&lt;p&gt;I hope this is the right place for help on this question! I expect this should be easy for this audience (and, no, this isn't homework).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a task that takes $X$ seconds to complete (say, moving a rock up a hill). However, sometimes the task is restarted from the beginning (say, an earthquake rolls the rock back to the bottom of the hill). I know that these restart events happen on average every $Y$ seconds ($Y$ less than $X$). I need to find out how long it will take in total to reach the top of the hill.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To be perfectly clear, suppose I start the task at time $t=0$ and, every time a &quot;restart event&quot; happens, I immediately restart. When I complete the task I record $f$, the final value of $t$. What is the expected value of $f$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In my attempt to solve this, I model the restart as a Poisson event with average $\lambda=Y$. However, I'm not sure how to proceed. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;p.s. the restart events are random, of course, not on a schedule happening every Y seconds (otherwise I would never be able to reach the top of the hill).&lt;/p&gt;&#xA;" OwnerUserId="5858" LastEditorUserId="5858" LastEditorDisplayName="" LastEditDate="2011-01-14T19:46:43.970" LastActivityDate="2011-01-15T01:23:36.263" Title="Time until x successes, given p(failure)?" Tags="&lt;stochastic-processes&gt;" AnswerCount="1" CommentCount="7" />
  <row Id="17520" PostTypeId="2" ParentId="17517" CreationDate="2011-01-14T19:28:56.537" Score="0" ViewCount="" Body="&lt;p&gt;See &lt;a href=&quot;http://www.engin.brown.edu/courses/en3/Notes/Vector_Web2/Vectors6a/Vectors6a.htm&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2011-01-14T19:28:56.537" CommentCount="1" />
  <row Id="17521" PostTypeId="2" ParentId="17517" CreationDate="2011-01-14T19:32:45.863" Score="1" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Added&lt;/strong&gt;: As pointed out by David Zaslavsky in his comment the sum below holds only in the trivial case in which  $\overrightarrow{u}_P$, $\overrightarrow{v}_P$ are applied to the same point $P$. If the vectors $\overrightarrow{u}_P$, $\overrightarrow{v}_Q$ are applied to two different points $P,Q$, then the unit vectors $\overrightarrow{e}_{r},\overrightarrow{e}_{\theta },\overrightarrow{e}_{\varphi }$ have different directions, and consequently the sum $\overrightarrow{u}_P+\overrightarrow{v}_Q$ is given by a more complex formula and I do NOT see how to avoid the usual spherical to Cartesian coordinates conversion and back. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;If you have two vectors in spherical coordinates&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\overrightarrow{u}=u_{r}\overrightarrow{e}_{r}+u_{\theta }\overrightarrow{e}%&#xD;&#xA;_{\theta }+u_{\varphi }\overrightarrow{e}_{\varphi }$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\overrightarrow{v}=v_{r}\overrightarrow{e}_{r}+v_{\theta }\overrightarrow{e}%&#xD;&#xA;_{\theta }+v_{\varphi }\overrightarrow{e}_{\varphi }$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;their sum is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\overrightarrow{u}+\overrightarrow{v}=\left( u_{r}+v_{r}\right) &#xD;&#xA;\overrightarrow{e}_{r}+\left( u_{\theta }+v_{\theta }\right) \overrightarrow{%&#xD;&#xA;e}_{\theta }+\left( u_{\varphi }+v_{\varphi }\right) \overrightarrow{e}%&#xD;&#xA;_{\varphi }$&lt;/p&gt;&#xA;" OwnerUserId="752" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2011-01-15T00:27:07.580" LastActivityDate="2011-01-15T00:27:07.580" CommentCount="2" />
  <row Id="17522" PostTypeId="1" AcceptedAnswerId="17526" CreationDate="2011-01-14T19:41:48.463" Score="1" ViewCount="184" Body="&lt;p&gt;I'm just beginning a course in discrete mathematics and I'm learning a few of the basic laws to prove propositions. I understand how to use propositions that use the logical connectives, AND , OR, NOT. However I'm not sure how to prove a proposition that has an implies operator. For example prove the following is a tautology.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(p\land (p\implies q))\implies q$ &lt;/p&gt;&#xA;" OwnerUserId="1040" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2011-01-14T19:49:02.717" LastActivityDate="2011-01-14T22:07:40.943" Title="Are there rules of Logic dealing with the implies operator?" Tags="&lt;logic&gt;" AnswerCount="4" />
  <row Id="17523" PostTypeId="2" ParentId="17522" CreationDate="2011-01-14T19:45:07.417" Score="1" ViewCount="" Body="&lt;p&gt;Funny that you can use the logical operators and ($\land$), or ($\lor$) and not ($\lnot$) but not the imply operator as its nothing else:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(a \implies b) \Leftrightarrow (\lnot a \lor b)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe this helps? Just replace those imply operators in your equation.&lt;/p&gt;&#xA;" OwnerUserId="3123" LastActivityDate="2011-01-14T19:45:07.417" CommentCount="2" />
  <row Id="17524" PostTypeId="2" ParentId="17519" CreationDate="2011-01-14T19:48:12.437" Score="5" ViewCount="" Body="&lt;p&gt;Let $T$ be the requested time, and $R$ a random variable measuring the time until the next restart. We have $$\mathbb{E}[T] = \Pr[R &amp;gt; X] \cdot X + \Pr[R \leq X]( \mathbb{E}[T] + \mathbb{E}[R|R\leq X]).$$ Simplifying, we get $$\mathbb{E}[T] = X + \frac{\Pr[R \leq X]}{\Pr[R &amp;gt; X]} \mathbb{E}[R|R\leq X].$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: Here's a more explicit formula, assuming that $R$ is a continuous random variable with density $f$:&#xA;$$ \mathbb{E}[T] = X + \frac{\int_0^X rf(r)\, dr}{1 - \int_0^X f(r)\, dr}. $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: Here's another formula, for a discrete random variable with integer values (assuming a restart is valid if it happens at time $X$):&#xA;$$ \mathbb{E}[T] = X + \frac{\sum_{t=0}^X t\Pr[R=t]}{1 - \sum_{t=0}^X \Pr[R=t]}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: plugging a geometric distribution with expectation $\lambda$, Wolfram alpha gets &lt;a href=&quot;http://tinyurl.com/4natt6r&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt;. Note that a geometric distribution has a non-zero probability to be $0$ (in that case the restart occurs immediately).&lt;/p&gt;&#xA;" OwnerUserId="1277" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2011-01-15T01:23:36.263" LastActivityDate="2011-01-15T01:23:36.263" CommentCount="7" />
  <row Id="17525" PostTypeId="2" ParentId="17494" CreationDate="2011-01-14T19:48:25.883" Score="2" ViewCount="" Body="&lt;p&gt;This is not a variation of KNAPSACK since you can choose your weights.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From what you write, I assume that $C_1 &amp;lt; C_2 &amp;lt; \dots C_k$. Let me also explicitly state that you never demand the $S_i$ to be disjoint; the following approach would not work then.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note first that you can safely drop the first restriction $1 = a_1 \leq \dots$. If you find feasible $a_i$ in any order, this can always be achieved by renaming.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; The offered solution is suboptimal as has been pointed out in the comments.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $a_1 = 1$ and for $2 \leq i \leq k$ let $a_i = C_i - C_{i-1}$. These $a_i$ are an optimal solution since&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$C_i = \sum_{j=1}^i a_j$ obviously and&lt;/li&gt;&#xA;&lt;li&gt;$t=k$ is the minimum $t$ possible. Since all $C_i$ are pairwise different, we need at least one summand per $C_i$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; A new idea.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Start with $A = \{N\}$.&#xA;For every $i$ from $N-1$ to $1$ do:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If you can sum up $C_i$ with what is in $A$, continue with $i-1$.&lt;/li&gt;&#xA;&lt;li&gt;Let $m = \min \{ a \in A \mid a \geq C_i\}$&lt;/li&gt;&#xA;&lt;li&gt;If $m \neq C_i$, choose $p = \arg\min_r \{ |q - r| \mid r+q=m, r+a=C_i, a\in A_{&amp;lt; C_i}\cup\{0\}  \}$ and set $A = (A \setminus \{m\}) \cup \{p, m-p\}$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The rather ugly argmin in words: For every $C_i$ take the smallest already chosen summand larger than $C_i$ and break it in as big chunks as possible while making sure that you can build $C_i$ with what you have. That should always work out, i.e. yield a solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have no idea regarding optimality. For your example, the solution provided by you is found. Of course, I use &lt;a href=&quot;http://en.wikipedia.org/wiki/Subset_sum_problem&quot; rel=&quot;nofollow&quot;&gt;SUBSET SUM&lt;/a&gt;, what is probably a bad idea. Doh.&lt;/p&gt;&#xA;" OwnerUserId="3330" LastEditorUserId="3330" LastEditorDisplayName="" LastEditDate="2011-01-14T23:54:29.400" LastActivityDate="2011-01-14T23:54:29.400" CommentCount="4" />
  <row Id="17526" PostTypeId="2" ParentId="17522" CreationDate="2011-01-14T19:54:10.637" Score="6" ViewCount="" Body="&lt;p&gt;There are two possible answers. One, your language doesn't really contain the $\implies$ operator, and it is thought of as a shorthand (as in user3123's answer): $$a \implies b \text{ is a shorthand for } \lnot a \lor b.$$&#xA;Two, your language does include it, and then you have some axioms expressing its &quot;meaning&quot;. For example, you could have a system with only $\lnot$ and $\implies$, having the following axioms&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ A \implies (B \implies A) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ (A \implies B) \implies ((B \implies C) \implies (A \implies C)) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ (A \implies B) \implies (\lnot B \implies \lnot A) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and Modus Ponens as the only inference rule: given $A$ and $A \implies B$, deduce $B$. This system is complete (can prove every true proposition). The other connectives are then shorthands:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ A \lor B \text{ stands for } \lnot A \implies B $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ A \land B \text{ stands for } \lnot(\lnot A \lor \lnot B) $$&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2011-01-14T19:54:10.637" />
  <row Id="17527" PostTypeId="1" AcceptedAnswerId="17530" CreationDate="2011-01-14T19:59:59.767" Score="0" ViewCount="57" Body="&lt;p&gt;&lt;strong&gt;Background&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am trying to do a reproducible scientific analysis. My conclusions are not dependent on the random number generator, but the RNG does change the results ~1% between runs. I would like to define starting points for the RNG algorithm to reproduce the same result each time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am using the &quot;Mersenne-Twister&quot; algorithm only because it is the default in R. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am using the RNG to sample parameter values from their distributions and I need about $10^7$ values for the complete analysis. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there any consequences of the choice of a starting value? Intuitively, R's &lt;code&gt;set.seed(0)&lt;/code&gt;  feels very non-random, but rationally, I can't imagine why this would matter.&lt;/p&gt;&#xA;" OwnerUserId="3733" LastEditorUserId="3733" LastEditorDisplayName="" LastEditDate="2011-01-14T20:17:23.300" LastActivityDate="2011-01-14T20:17:23.300" Title="Consequences of choice of a seed for random number generating algorithm?" Tags="&lt;probability&gt;&lt;statistics&gt;&lt;algorithms&gt;&lt;random&gt;" AnswerCount="1" />
  <row Id="17528" PostTypeId="2" ParentId="17496" CreationDate="2011-01-14T20:02:53.157" Score="0" ViewCount="" Body="&lt;p&gt;For a completely unambiguous counterexample, consider $k=15$: it has one prime factor (5) that is congruent to 1 modulo 4, but it cannot be written as a sum of two squares.  The reason (as mentioned in both Moron's and Arturo's answers) is that a number cannot be written as a sum of two squares unless all its prime factors that are congruent to 3 modulo 4 are raised to even powers.&lt;/p&gt;&#xA;" OwnerUserId="5546" LastActivityDate="2011-01-14T20:02:53.157" />
  <row Id="17529" PostTypeId="2" ParentId="17522" CreationDate="2011-01-14T20:05:02.483" Score="1" ViewCount="" Body="&lt;p&gt;You may draw a truth table, in order to investigate the alternatives (see &lt;a href=&quot;http://en.wikipedia.org/wiki/Logical_implication&quot; rel=&quot;nofollow&quot;&gt;Implication at wikipedia&lt;/a&gt;). In words,&#xA;if $A$ and $B$ are some propositions then $A\Rightarrow B$ is a third proposition which is true if and only if $A$ is false, or both $A$ and $B$ are true.&lt;/p&gt;&#xA;" OwnerUserId="1154" LastEditorUserId="1154" LastEditorDisplayName="" LastEditDate="2011-01-14T22:07:40.943" LastActivityDate="2011-01-14T22:07:40.943" />
  <row Id="17530" PostTypeId="2" ParentId="17527" CreationDate="2011-01-14T20:09:49.670" Score="3" ViewCount="" Body="&lt;p&gt;Depending on the algorithm the seed can indeed affect the quality of your random numbers however due to the design of mersenne twister the choice of the seed shouldn't really affect the length of the periodicity of the random numbers or their quality. Still there might be some implementations of the mersenne twister where the seed does matter as you can read here ( &lt;a href=&quot;http://library.gnome.org/devel/glib/unstable/glib-Random-Numbers.html&quot; rel=&quot;nofollow&quot;&gt;http://library.gnome.org/devel/glib/unstable/glib-Random-Numbers.html&lt;/a&gt; ) Obviously they changed the seeding algorithm because the first one could result in bad seeds. If you want to be on the safe side use a reliable library for your random numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might want to try something like this for getting the seed: &lt;a href=&quot;http://www.stanford.edu/~blp/writings/clc/random-seed.html&quot; rel=&quot;nofollow&quot;&gt;http://www.stanford.edu/~blp/writings/clc/random-seed.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that if you don't need too many numbers you could download real random numbers generated from atmospheric noise here: &lt;a href=&quot;http://www.random.org/&quot; rel=&quot;nofollow&quot;&gt;http://www.random.org/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3123" LastEditorUserId="3123" LastEditorDisplayName="" LastEditDate="2011-01-14T20:15:00.537" LastActivityDate="2011-01-14T20:15:00.537" CommentCount="4" />
  <row Id="17531" PostTypeId="2" ParentId="17522" CreationDate="2011-01-14T20:13:13.897" Score="5" ViewCount="" Body="&lt;p&gt;There's a nice shortcut to proving that something with an implies operator is a tautology.  Assume it's not a tautology: the only way $a \Rightarrow b$ can be false is if $a$ is true and $b$ is false, so you can assume that, etc. and this is often a quick way to reach a contradiction.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2011-01-14T20:13:13.897" />
  <row Id="17532" PostTypeId="2" ParentId="17434" CreationDate="2011-01-14T20:18:12.920" Score="1" ViewCount="" Body="&lt;p&gt;I think there is some misunderstanding here. While it is possible to have derivatives of functions defined &lt;em&gt;between Banach spaces&lt;/em&gt; (say) (e.g. with the concept of Frechet- or Gateaux-derivative), the derivative of a Radon measure is a derivative of an &lt;em&gt;object in such a space&lt;/em&gt;. Hence, the derivative of a Radon measure is somewhat more related to the distributional derivative of a function or the weak derivative.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To get an impression of the derivative you are talking about, the following example may be helpful: Let $\mu$ be the one dimensional Lebesgue measure on $[0,1]$. For a integrable and positive function $f:[0,1] \to \mathbf{R}$ and define a measure $\nu$ by $\nu(A) = \int_A f\,d\mu$. Now, $D_\mu \eta$ can be calculated explicitely and is related to the weak derivative of $f$. (Actually I do not know what precisely is meant by $B_r^-(0)$ and hence, I could not work out the details.)&lt;/p&gt;&#xA;" OwnerUserId="3148" LastActivityDate="2011-01-14T20:18:12.920" CommentCount="1" />
  <row Id="17533" PostTypeId="1" AcceptedAnswerId="17539" CreationDate="2011-01-14T20:36:11.833" Score="1" ViewCount="134" Body="&lt;p&gt;Is it legal to distribute the $\land$  or the $\lor$ operators over the $\implies$ operator? For example is it legal for me to to do the following? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(p\land(p\implies q))\implies q $ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(p\land p) \implies (p\land q) \implies q$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Put more simply: Is the &quot;AND&quot; operator  distributive over the &quot;IMPLIES&quot; operator.&lt;/p&gt;&#xA;" OwnerUserId="1040" LastEditorUserId="1040" LastEditorDisplayName="" LastEditDate="2011-01-14T20:49:37.140" LastActivityDate="2011-01-14T21:44:57.790" Title="Laws of distribution" Tags="&lt;logic&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="17534" PostTypeId="2" ParentId="17533" CreationDate="2011-01-14T20:42:50.270" Score="2" ViewCount="" Body="&lt;p&gt;The &quot;truth table&quot; &lt;a href=&quot;http://en.wikipedia.org/wiki/Truth_table&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Truth_table&lt;/a&gt; is an easy method to verify and even proove such assumptions.&lt;/p&gt;&#xA;" OwnerUserId="3123" LastActivityDate="2011-01-14T20:42:50.270" CommentCount="2" />
  <row Id="17535" PostTypeId="2" ParentId="17498" CreationDate="2011-01-14T20:49:20.663" Score="0" ViewCount="" Body="&lt;p&gt;The general problem of determining the positions of n points given their mutual distances crops up in &lt;a href=&quot;https://computation.llnl.gov/casc/sapphire/pubs/148494.pdf&quot; rel=&quot;nofollow&quot;&gt; metric dimension reduction techniques&lt;/a&gt;, where one asks if the distances between a set of points may be preserved by a mapping into a lower dimensional space.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If a solution exists for n points, it can be embedded in not more than n-1 dimensions.  Obviously any solution can be transformed by reflection, translation, and rotation, so at best one hopes for uniqueness up to an orthogonal transformation.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nor does a solution always exist.  As Ross Millikan points out, if a solution exists the distances must satisfy triangle inequalities.  This is sufficient for three points (to determine a triangle in the plane), but it is not sufficient for placing four points.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider having placed points A,B,C in an equilateral triangle with side length 1.  Then we can try to place a fourth point D.  If the distances from D to each of A,B,C are slightly more than 1/2, then all triangle inequalities are satisfied yet there is no feasible location for D, even in three dimensions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Indeed for four points the question amounts, having placed the first three points, whether a feasible location for the fourth point exists.  The distances to the first three points give spheres embedded in three dimensions, the intersection of any two of which forms a circle, a point, or the empty set.  The answer is easily found by cartesian geometry.&lt;/p&gt;&#xA;" OwnerUserId="3111" LastActivityDate="2011-01-14T20:49:20.663" />
  <row Id="17536" PostTypeId="1" AcceptedAnswerId="17537" CreationDate="2011-01-14T20:54:29.307" Score="1" ViewCount="80" Body="&lt;p&gt;I'm TA-ing an intro class on theoretical CS, and this week class only covered the simplest concepts, such as words and languages.  I wanted to take this chance to present some combinatorics on words, and prove the cube-freeness of the Thue-Morse sequence (which I want to define as $w_0 = a, w_i = w_{i-1}w&amp;#39;_{i-1}$ where $w&amp;#39;$ is $w$ where the $a$'s and the $b$'s are exchanged).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Salomaa's old book (&lt;a href=&quot;http://bit.ly/hHeWbK&quot; rel=&quot;nofollow&quot;&gt;http://bit.ly/hHeWbK&lt;/a&gt;) gives the following program: (1) Show that neither $a^3$ nor $b^3$ can appear in $w_i$, (2) Show that neither $ababa$ nor $babab$ can appear in $w_i$, (3) show than any sequence $aa$ or $bb$ appears in $w_i$ on an even position, (4) Conclude by induction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1-3) are pretty easy by induction, but I can't think of a solution for (4).  I came to the easy conclusion that no odd-length word can appear as a cube, but I've got nothing for the even part (and I really want to avoid going through the overlap-free proof, or use the definition of the sequence using the binary expansion of natural numbers).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any idea?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="5270" LastActivityDate="2011-01-14T23:14:09.403" Title="Thue-Morse sequence cube-freeness proof from the Book." Tags="&lt;combinatorics&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="17537" PostTypeId="2" ParentId="17536" CreationDate="2011-01-14T21:06:43.833" Score="1" ViewCount="" Body="&lt;p&gt;Use the fact that the even positions in $w_i$ are $w_{i-1}$, and the odd positions are $w&amp;#39;_{i-1}$ (if we start counting at zero).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: let's prove this bit by induction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Denote by $E(s)$ every second bit starting with $0$, and $O(s)$ every second bit starting with $1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Want to prove that for $i \geq 1$, $E(w_i) = w_{i-1}$ and $O(w_i) = w&amp;#39;_{i-1}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Base case $i=1$: Just check: $$E(w_1) = E(ab) = a = w_0$$ while $$O(w_1) = O(ab) = b = a&amp;#39; = w&amp;#39;_0.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Induction step: Assume that it holds for $i$, prove it for $i+1$: $$E(w_{i+1}) = E(w_i w&amp;#39;_i) = E(w_i) E(w&amp;#39;_i) = E(w_i) E(w_i)&amp;#39; = w_{i-1} w&amp;#39;_{i-1} = w_i,$$ and similarly $$O(w_{i+1}) = O(w_i w&amp;#39;_i) = O(w_i) O(w&amp;#39;_i) = O(w_i) O(w_i)&amp;#39; = w&amp;#39;_{i-1} w_{i-1} = w_i&amp;#39;.$$ In both steps, we used the fact that $|w_i|$ is even (this is true for $i \geq 1$).&lt;/p&gt;&#xA;" OwnerUserId="1277" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2011-01-14T23:14:09.403" LastActivityDate="2011-01-14T23:14:09.403" CommentCount="3" />
  <row Id="17538" PostTypeId="2" ParentId="17533" CreationDate="2011-01-14T21:09:11.897" Score="0" ViewCount="" Body="&lt;p&gt;Assuming that $\implies$ distributes to the right (so by $A \implies B \implies C$ you mean $A \implies (B \implies C)$), then for your examples it works out, since they're both tautologies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Otherwise, your example doesn't work it, since the second statement is no longer a tautology.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2011-01-14T21:09:11.897" />
  <row Id="17539" PostTypeId="2" ParentId="17533" CreationDate="2011-01-14T21:44:57.790" Score="3" ViewCount="" Body="&lt;p&gt;No, in general you can't replace $a \wedge (b \implies c)$ with $(a \wedge b) \implies (a \wedge c)$.  If $a$ is true, then both expressions have the same truth value.  However, if $a$ is false, then the first expression is false and the second expression is true.&lt;/p&gt;&#xA;" OwnerUserId="5546" LastActivityDate="2011-01-14T21:44:57.790" />
  <row Id="17540" PostTypeId="1" AcceptedAnswerId="17541" CreationDate="2011-01-14T21:45:13.363" Score="2" ViewCount="90" Body="&lt;p&gt;The mean value of the fractional part $\{ x \}$ on $\mathbb{R}_+$ is clearly $\tfrac{1}{2}$. I'd like to know if a similar statement holds for $\{ f(x) \}$ where $f \colon X \to \mathbb{R}_{+}$ is a non-negative $C^{k}(X)$ function for some $k \geq 0$ and assumes all values of an infinite continuum subset of $\mathbb{R}$ at least once. For example, if $f(x) = x + 1$, then the average value on $[-1,\infty)$ remains at $\frac{1}{2}$. Does the same value hold if $f(x) = \log_{b} x$ for some base $b &amp;gt; 0$ on $[b,\infty)$? Is the result always $\frac{1}{2}$? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This Mathematica code may be useful:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Plot[Mean[Table[FractionalPart[Log[2, x]], {x, 2, y, 0.01}]], {y, 2, 1000}]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clarification (and edits), examples, counterexamples and references are welcome! Thanks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Edit&lt;/em&gt;: The nice constructions and counter-examples given below imply that further conditions are needed to ensure the mean value is exactly $\frac{1}{2}$. What are they?&lt;/p&gt;&#xA;" OwnerUserId="2720" LastEditorUserId="2720" LastEditorDisplayName="" LastEditDate="2011-02-21T06:24:04.400" LastActivityDate="2011-02-21T06:24:04.400" Title="Average value of the Fractional Part of a $C^{k}$-function" Tags="&lt;real-analysis&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="1" />
  <row Id="17541" PostTypeId="2" ParentId="17540" CreationDate="2011-01-14T22:08:14.220" Score="5" ViewCount="" Body="&lt;p&gt;Consider the standard construction of the bump function: let $\phi$ be a $C^\infty$ function with the property that $0 \leq \phi \leq 1$, $\phi = 0$ on the negative real axis, and $\phi(x) = 1$ on $[1,\infty)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $\phi_\epsilon(x) = \phi(x/\epsilon)$ the scaling of $\phi$. For each $\epsilon &amp;lt; 1$, define the smoothed step-function&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \Psi(x;\epsilon) = \lfloor x\rfloor + \phi_\epsilon(\{x\}) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is pretty clear that the average of fractional part of $\Psi$ can be made, by construction, as close to 0 as you want. Indeed, the average of $\{\Psi\}$ over $[N,M]$ where $N,M\in\mathbb{Z}$ is the same as its average over $[0,1]$ by periodicity, and by construction $\{\Psi(x,\epsilon)\}$ vanishes for $x\in(\epsilon,1]$ and is bounded by 1 for $x\in[0,\epsilon]$, so we have that the average of the fractional part of $\Psi(x;\epsilon)$ must be bounded above by $\epsilon$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(In fact, using this construction you can, for any $y \in (0,1)$, make a function $F$ such that $F$ is monotonically increasing, smooth, defined on $\mathbb{R}$, such that the &quot;mean&quot; of the fractional part of $F$ over $\mathbb{R}$ converges to the desired $y$.)&lt;/p&gt;&#xA;" OwnerUserId="1543" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2011-01-14T22:53:17.533" LastActivityDate="2011-01-14T22:53:17.533" CommentCount="0" />
  <row Id="17542" PostTypeId="1" AcceptedAnswerId="17547" CreationDate="2011-01-14T22:10:04.993" Score="2" ViewCount="216" Body="&lt;p&gt;Hi&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm having an argument about what the notation of $\lim$ means. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume you have $f_n: X \rightarrow \mathbb{R}$. Are the following two sets equal:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\{ x \ |\ &amp;#160;(f_n(x)) \ \text{converges} \} = \{ x \ |\ -\infty \lt&amp;#160;\lim_{n \rightarrow \infty} f_n(x) \lt \infty\}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can assume the reals are not extended and that convergence means that the limit exists and is finite.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The person I'm arguing with insists that what I wrote on the right makes no sense because the limit might no be defined. My argument is that undefined in this case is ruled out by the notation, since you cannot write &quot;undefined &amp;lt; something&quot; so the &quot;&amp;lt;&quot; implies that the limit is defined and therefore that the two sets are equal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for clarification!&lt;/p&gt;&#xA;" OwnerUserId="5798" LastEditorUserId="5798" LastEditorDisplayName="" LastEditDate="2011-01-15T09:12:10.427" LastActivityDate="2011-01-15T09:12:10.427" Title="A question regarding the meaning of &quot;lim&quot;" Tags="&lt;real-analysis&gt;&lt;analysis&gt;&lt;notation&gt;" AnswerCount="1" CommentCount="9" />
  <row Id="17543" PostTypeId="2" ParentId="17540" CreationDate="2011-01-14T22:17:30.670" Score="4" ViewCount="" Body="&lt;p&gt;Another counterexample is in fact $\log_b x$. Because $\log_b bx = 1 + \log_b x$, the mean of $\{\log_b x\}$ over $\mathbb R_+$ is the same as the mean of $\log_b x$ over $[1,b]$. If you look at the shape of the log function, it's pretty clear that the mean is larger than 1/2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When $b = 10$, a related phenomenon gives rise to &lt;a href=&quot;http://en.wikipedia.org/wiki/Benford%27s_law&quot; rel=&quot;nofollow&quot;&gt;Benford's law&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="856" LastActivityDate="2011-01-14T22:17:30.670" />
  <row Id="17544" PostTypeId="1" CreationDate="2011-01-14T22:24:56.327" Score="2" ViewCount="182" Body="&lt;p&gt;So I got to the infamous &quot;the proof is left to you as an exercise&quot; of the book when I tried to look up how to get the Lagrange form of the remainder for a Taylor polynomial. Is this right?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$R_{n}(x)=\frac{1}{n!}\int_{0}^{x}f^{n+1}(t)(x-t)^{n}dt$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;find out why&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$R_{n}(x)=\frac{1}{(n+1)!}f^{n+1}(c)x^{n+1}$ for some $c\in [0,x]$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to FTC,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\int_{0}^{x}f&amp;#39;(t)dt = f(x) - f(0)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, according to the Mean Value Theorem, there exists a $c$ such that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f&amp;#39;(c)(x-0)=f(x)-f(0)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\int_{0}^{x}f&amp;#39;(t)dt = f&amp;#39;(c)(x-0)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;finding the derivative of both sides with respect to $x$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f&amp;#39;(x) = f&amp;#39;(c)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f^{n+1}(x) = f^{n+1}(c)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Going back to the integral form of the remainder:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$R_{n}(x)=\frac{1}{n!}\int_{0}^{x}f^{n+1}(t)(x-t)^{n}dt$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I replace $f^{n+1}(x)$ with $f^{n+1}(c)$ (This is the step I am most unsure of)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $f&amp;#39;(c)$ is a constant, I pull it out of the integral and integrate what's left under the integral, giving me&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$R_{n}(x)=\frac{1}{(n+1)!}f^{n+1}(c)x^{n+1}$ for some $c\in [0,x]$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If this is right, then does it mean that $f&amp;#39;(c)$ is the average value of $f&amp;#39;(x)$  from $0$ to $x$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sorry if my LaTeX/wording/proof is off. I'd appreciate any corrections/answers to be as simple (notation-wise) as possible please - 1st year undergrad here...&lt;/p&gt;&#xA;" OwnerUserId="1958" LastActivityDate="2011-01-14T23:05:21.393" Title="Proving Lagrange Form of Remainder for Taylor Polynomial" Tags="&lt;calculus&gt;&lt;taylor-expansion&gt;" AnswerCount="1" />
  <row Id="17546" PostTypeId="2" ParentId="17544" CreationDate="2011-01-14T22:31:47.473" Score="1" ViewCount="" Body="&lt;p&gt;No, it is not right.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f&amp;#39;(c) (x-0) = f(x) - f(0)$$ is true, but $c$ depends on $x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So it is something like&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f&amp;#39;(c_x) (x-0) = f(x) - f(0)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So derivative of $f&amp;#39;(c) x $ is not really $f&amp;#39;(c)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And the step from&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f&amp;#39;(x) = f&amp;#39;(c)$ to $f^{n+1}(x) = f^{n+1}(c)$ is also wrong. If $c$ is constant (according to your proof), then the derivative on the right side becomes $0$.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2011-01-14T23:05:21.393" LastActivityDate="2011-01-14T23:05:21.393" CommentCount="2" />
  <row Id="17547" PostTypeId="2" ParentId="17542" CreationDate="2011-01-14T22:45:25.107" Score="4" ViewCount="" Body="&lt;p&gt;If you are using the extended Reals, then the definition of convergence also includes the infinities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are not using extended Reals, then the (usual) definition of convergence is restricted to finite reals, in which case the two sets you have are the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even without extended reals, you could say a sequence converges to infinity, but that is not common I believe (of course, I might be wrong about that). In which case, the sets could be different.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the real issue is: What definition of convergence are you using?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;IMO, you are right. Just writing $\lim... \lt \infty$ carries the implicit assumption that it exists. I believe many authors use this as a convenient way to say the limit exists, in fact (I will try to update with references when I find them).&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2011-01-14T23:15:21.430" LastActivityDate="2011-01-14T23:15:21.430" CommentCount="2" />
  <row Id="17548" PostTypeId="1" AcceptedAnswerId="17785" CreationDate="2011-01-14T23:08:41.763" Score="4" ViewCount="106" Body="&lt;p&gt;Hello, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am currently converting a minimization problem from Matlab to C++, using the Numerical Recipes implementation of the Nelder and Mead Downhill simplex method. The function requires me to define a constant lambda for each variable, which represents the &quot;characteristic length scale&quot; of the variable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried to find a more formal definition of what the authors mean by this, and how to choose one, but couldn't find anything. I'm guessing it's something of a &quot;choose something that looks like a reasonable step size for that variable and hope that it converges fast enough; otherwise, try something else&quot;. Any pointers to something a little bit more scientific?&lt;/p&gt;&#xA;" OwnerUserId="64" LastActivityDate="2011-01-16T22:39:33.657" Title="How to define the characteristic length scale in a downhill simplex method?" Tags="&lt;numerical-methods&gt;&lt;optimization&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="17549" PostTypeId="1" AcceptedAnswerId="17553" CreationDate="2011-01-14T23:35:26.717" Score="3" ViewCount="110" Body="&lt;p&gt;We know that $\text{exp}(-\alpha |x|^2)$ is a fixed point for the unitary Fourier transform if $\text{Re } \alpha &amp;gt; 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know many arguments to show this (contour-integration and differentiation).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a not an elegant way where we can exploit that fact that the Gaussian is rotationally symmetric? A sketch would be fine.&lt;/p&gt;&#xA;" OwnerUserId="1120" LastActivityDate="2011-01-15T01:18:52.443" Title="Gaussian fixed point Fourier transform." Tags="&lt;fourier-analysis&gt;" AnswerCount="1" />
  <row Id="17550" PostTypeId="1" AcceptedAnswerId="17558" CreationDate="2011-01-14T23:37:14.577" Score="6" ViewCount="286" Body="&lt;p&gt;In &lt;a href=&quot;http://en.wikipedia.org/wiki/Cross_product#Definition&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;, a cross product between two &quot;vectors&quot; is defined in terms of the angle between the vectors and their magnitudes.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;As I learned cross product in linear&#xA;algebra, which I understand to be a&#xA;topic about vector space, I now&#xA;wonder if cross product is not&#xA;defined on vector space, but instead&#xA;can only be defined on an inner&#xA;product space so that the angle&#xA;between the vectors and their&#xA;magnitudes can make sense?&lt;/li&gt;&#xA;&lt;li&gt;Or is there other definition of&#xA;cross product on vector space?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Thanks and regards!&lt;/p&gt;&#xA;" OwnerUserId="1281" LastActivityDate="2011-01-15T03:15:21.470" Title="Is Cross Product Defined on Vector Space?" Tags="&lt;vector-space&gt;&lt;inner-product-space&gt;" AnswerCount="3" FavoriteCount="2" />
  <row Id="17551" PostTypeId="2" ParentId="17550" CreationDate="2011-01-14T23:43:20.377" Score="6" ViewCount="" Body="&lt;p&gt;The question was studied by &lt;a href=&quot;http://en.wikipedia.org/wiki/Hermann_Grassmann&quot; rel=&quot;nofollow&quot;&gt;Grassmann&lt;/a&gt;. At the time, people laughed at him, but now you can read all about it in the Wikipedia article on &lt;a href=&quot;http://en.wikipedia.org/wiki/Exterior_algebra&quot; rel=&quot;nofollow&quot;&gt;Exterior Algebra&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2011-01-14T23:43:20.377" />
  <row Id="17552" PostTypeId="1" AcceptedAnswerId="17554" CreationDate="2011-01-14T23:44:02.467" Score="1" ViewCount="120" Body="&lt;p&gt;Usually Jacobian matrix encapsulate partial derivatives of a vector f w.r.t. another vector v.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More generally how to define/get the Jacobian matrix of a matrix F w.r.t another matrix V?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it just taking the elements of F and V and arranging them as vectors and getting the usual Jacobian matrix of 2 vectors?&lt;/p&gt;&#xA;" OwnerUserId="5312" LastActivityDate="2011-01-14T23:50:51.493" Title="Jacobian of a Matrix" Tags="&lt;linear-algebra&gt;&lt;vector-calculus&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="17553" PostTypeId="2" ParentId="17549" CreationDate="2011-01-14T23:48:23.180" Score="4" ViewCount="" Body="&lt;p&gt;Here's a different argument. Take a zero-mean Gaussian random variable $X$. We know that the sum of $n$ copies of $X$, scaled down by a factor of $\sqrt{n}$, is distributed the same as $X$. On the other hand, by the characteristic function argument (which can be used to prove the central limit theorem) we know that the Fourier transform of $(X_1+\cdots+X_n)/\sqrt{n}$ converges to $\exp(- \sigma^2x^2/2)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: an even simpler way. Again take $X$ to be a zero-mean Gaussian. We know that $(X+X)/\sqrt{2}$ is equidistributed with $X$. We immediately deduce that all higher-order cummulants are nil, and since the second cummulant is the variance, we get that the Fourier transform is $\exp(-\sigma^2x^2/2)$.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2011-01-15T01:18:52.443" LastActivityDate="2011-01-15T01:18:52.443" CommentCount="3" />
  <row Id="17554" PostTypeId="2" ParentId="17552" CreationDate="2011-01-14T23:50:51.493" Score="2" ViewCount="" Body="&lt;p&gt;The Jacobian is defined for a real function depending on a vector. You can replace the vector by a matrix by considering the matrix as a vector, or by using a four-dimensional tensor instead of a two-dimensional matrix.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2011-01-14T23:50:51.493" CommentCount="1" />
  <row Id="17555" PostTypeId="2" ParentId="17540" CreationDate="2011-01-14T23:53:17.467" Score="2" ViewCount="" Body="&lt;p&gt;Look up Weyl's equidistribution theorem. Weyl's general method is to compute the Fourier transform to get the distribution of the fractional part, using the fact that $\exp(2\pi i t) = \exp(2\pi i \{t\})$.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2011-01-14T23:53:17.467" />
  <row Id="17556" PostTypeId="2" ParentId="16979" CreationDate="2011-01-15T00:10:32.880" Score="5" ViewCount="" Body="&lt;p&gt;By $\vdash P = NP$, I'm going to assume that you mean you can only use the axioms of first-order logic (i.e., formulas that define the meanings of the logical symbols such as $\forall x(x = x)$ for '=') and that &quot;P = NP&quot; is a shorthand for a formula in the language of arithmetic stating something nontrivial.  In that case, P = NP is independent (of the axioms of first-order logic) but so is $0 = 1$.  For the second formula, note that $0 = 1$ is true in a model interpreting $0$ and $1$ as the same element and false in a model interpreting $0$ and $1$ as different elements.  Therefore, we can neither prove it nor disprove it from the axioms of first-order logic alone.  Specifically without axioms governing the interpretations of these symbols, we cannot derive any meaningful conclusions about whether P = NP.  If you assume nothing, you can prove nothing!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, you can be certain that the P versus NP problem does &lt;B&gt;not&lt;/B&gt; ask for whether we can show $\vdash P = NP$ but instead it asks whether $T \vdash P = NP$ for some theory $T$ enforcing the obvious rules we want to hold for arithmetic such as $0 \neq 1$.  Moreover, it should be noted that the problem of P = NP has already been proven to be independent of certain weak theories of arithmetic.   You can see a survey of the P = NP problem in &lt;a href=&quot;http://www.scottaaronson.com/papers/pnp.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.scottaaronson.com/papers/pnp.pdf&lt;/a&gt;.  The whole point is that independence results are less valuable when we assume less, and we cannot hope to prove $P = NP$ or $P \neq NP$ from a theory adequately representing its meaning without the theory being sufficiently powerful or inconsistent. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, reasonable interpretations of the P = NP problem seem to suggest a proof from PA or ZFC.  If you were to find that the problem were independent of one of these stronger theories, this would indeed be very significant.  However, doing so using known techniques is essentially solving the problem itself as discussed in the following excerpt from Wikipedia:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;BLOCKQUOTE&gt;&#xA;These barriers have also led some computer scientists to suggest that the P versus NP problem may be independent of standard axiom systems like ZFC (cannot be proved or disproved within them). The interpretation of an independence result could be that either no polynomial-time algorithm exists for any NP-complete problem, and such a proof cannot be constructed in (e.g.) ZFC, or that polynomial-time algorithms for NP-complete problems may exist, but it's impossible to prove in ZFC that such algorithms are correct.[16] However, if it can be shown, using techniques of the sort that are currently known to be applicable, that the problem cannot be decided even with much weaker assumptions extending the Peano axioms (PA) for integer arithmetic, then there would necessarily exist nearly-polynomial-time algorithms for every problem in NP.[17] Therefore, if one believes (as most complexity theorists do) that not all problems in NP have efficient algorithms, it would follow that proofs of independence using those techniques cannot be possible. Additionally, this result implies that proving independence from PA or ZFC using currently known techniques is no easier than proving the existence of efficient algorithms for all problems in NP.&#xA;&lt;/BLOCKQUOTE&gt;&lt;/p&gt;&#xA;" OwnerUserId="4693" LastActivityDate="2011-01-15T00:10:32.880" />
  <row Id="17557" PostTypeId="2" ParentId="17473" CreationDate="2011-01-15T00:28:19.280" Score="0" ViewCount="" Body="&lt;p&gt;&lt;em&gt;I do not think that the basic strategy would be an outcome of any Monte Carlo simulation&lt;/em&gt;, as blackjack (and other games in casinos) had been played a lot before enough computer capacities were available for such computations. And these games are made up cleverly (just see Pascal's roulette), there is no need to compute the house's edge, as it has been well calculated for ages.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With probability and game theory any outcome can be expressed, and the probabilities (and by this: wins and losses) could be computed, so the basic strategy could be defined. E.g.: you have 15 and the dealer an A, the probability you will go bust with another card is quite high: $\frac{29}{49}$ (around 59 percent) to risk. This could be even worse with a 5 in hand (~61 %). The odds of the dealer has another card worth 10 is quite high: $\frac{16}{49}$ and other cards also help him (7, 8, 9 and also smaller ones with smaller probability). And so on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is true that with modern computers a simulation could be implemented a lot faster then doing all computations for probabilities, that is why I think these techniques are so popular nowadays.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another issue is that the outcome of a Monte Carlo method could hold an  acceptable accuracy with enough repetitions, so human driven (planned) computations just do not worth the time or energy to deal with.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;After editing&lt;/strong&gt;, I try to answer for the concrete question, which says:&#xA;&quot;&lt;em&gt;assuming that no form of card counting has occurred, is is still true that if you have a 10+5 or 9+6 or 8+7 vs a 6 you stand?&lt;/em&gt;&quot; (OP)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's say you have a 10+5 (hard total: 15) and you can see a 6 at the dealer. The basic strategy says you have to stand. Why?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You know nothing about the other card of the dealer, but you are quite sure about he will ask for another card. This is not 100 percent true, as he may have an ace ($p=\frac{4}{49}$), and could stand, if you have no more than 17 and did not bust.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What if you stand? You will loose if the dealer have an ace ($p=\frac{4}{49}$) or have a smaller card, hit and the overall sum will be between 17 and 21. It could be done with a sum of the 2 unknown card being between 11 and 15. It could be any of the followings: 2-9, 2-10, 2-J, 2-Q, 2-K, 2-A, 3-8, 3-9, 3-10, 3-J, 3-Q, 3-K, 3-A, 4-7, 4-8, 4-9, 4-10, 4-J, 4-Q, 4-K, 4-A, 5-6, 5-7, 5-8, 5-9, 5-10, 5-J, 5-Q, 5-K, 6-5, 6-6, 6-7, 6-8, 6-9, 7-4, 7-5, 7-6, 7-7, 7-8, 8-3, 8-4, 8-5, 8-6, 8-7, 9-2, 9-3, 9-4, 9-5, 9-6, 10-2, 10-3, 10-4, 10-5, J-2, J-3, J-4, J-5, Q-2, Q-3, Q-4, Q-5, K-2, K-3, K-4, K-5, A-2, A-3 and A-4. Note: you have 4 pieces of all cards in the deck except for the 5, 6 and 10, as you have seen 1-1 in your or the dealer's hand. That will also affect the computed probability, but I will not deal with this small problem here. So the prob. that you you will loose against the dealer's 3 card is: 68 hands will be bad for you from the possible 169 ($=13^2$) combinations. This and the above soft 17 gives a $p=\frac{4}{49}+\frac{68}{169}=0.4839995$.This means you have around 51.6 percent chance to win!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you do not stand, you will have a quite high risk of busting: $p=\frac{8*4-1}{49}=\frac{31}{49}$, which is a lot higher chance (around 63.26 percent) to loose. And I did not compute the chance of you would not bust but the dealer would hit a higher sum than you. That's why you should stand.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I might have maiden some mistakes in the above example, but I hope the theory can be seen. And of course I do think the above computation could have been done in a lot simpler way using combinatorics, but that was done such a simple way only using the classical theorem of probability.&lt;/p&gt;&#xA;" OwnerUserId="5651" LastEditorUserId="5651" LastEditorDisplayName="" LastEditDate="2011-01-16T12:59:21.983" LastActivityDate="2011-01-16T12:59:21.983" CommentCount="2" />
  <row Id="17558" PostTypeId="2" ParentId="17550" CreationDate="2011-01-15T01:32:02.903" Score="9" ViewCount="" Body="&lt;p&gt;One point of view is that the cross product is the composition of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Hodge_dual&quot; rel=&quot;nofollow&quot;&gt;Hodge dual&lt;/a&gt; and the exterior product $V \times V \to \Lambda^2 V$ in three dimensions.  The Hodge dual requires additional structure to define: you need not only an inner product, but an &lt;a href=&quot;http://en.wikipedia.org/wiki/Orientation_(mathematics)&quot; rel=&quot;nofollow&quot;&gt;orientation&lt;/a&gt;.  This reflects the fact that there is a choice of handedness in the definition of the cross product.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another point of view is that the cross product is not an operation on spatial vectors in $\mathbb{R}^3$, but the Lie bracket on the &lt;a href=&quot;http://en.wikipedia.org/wiki/Lie_algebra&quot; rel=&quot;nofollow&quot;&gt;Lie algebra&lt;/a&gt; of $\text{SO}(3)$.  This is the definition, for example, which is relevant to the physics of angular momentum.  Of course, if you insist on taking the cross product of spatial vectors you need a way to identify spatial vectors with elements of the Lie algebra of $\text{SO}(3)$.  Writing $\text{SO}(3)$ as $\text{Aut}(V)$ where $V$ is a real oriented 3-dimensional inner product space, the Lie algebra of $\text{SO}(3)$ is naturally isomorphic to $\Lambda^2 V$, so this identification is again the Hodge dual.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2011-01-15T01:32:02.903" CommentCount="9" />
  <row Id="17559" PostTypeId="2" ParentId="17550" CreationDate="2011-01-15T03:15:21.470" Score="4" ViewCount="" Body="&lt;p&gt;You are correct in that you need an inner product to define the cross product.  A geometric picture of it is as follows: take two vectors in three dimensions.  They determine a parallelogram and the cross product is defined to be the vector perpendicular (with respect to the inner product) to the parallelogram and with magnitude equal to its area.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that there is something special in 3-dimensions that allows this definition to work, namely that to any plane there is a unique perpendicular direction.  This is not the case in other dimensions so this definition does not generalize.  Indeed, as the wikipedia article states, the only other dimension that has an analogous cross product is 7 (dimension 1 also has a cross product but it is trivial-- just negative of the regular product of real numbers).  What is special about dimensions 1,3 and 7 is that there are division algebras only in dimensions 2,4, and 8 (the real numbers, the quaternions, and the octonions).  The wikipedia article talks more about this construction.&lt;/p&gt;&#xA;" OwnerUserId="9" LastActivityDate="2011-01-15T03:15:21.470" CommentCount="4" />
  <row Id="17560" PostTypeId="1" AcceptedAnswerId="17571" CreationDate="2011-01-15T03:46:27.657" Score="1" ViewCount="99" Body="&lt;p&gt;Hi everyone,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm looking for a proof that if f is $\mu$-integrable ($\mu(|f|) &amp;lt; \infty$, where $\mu(f)$=sup{$\mu(g):g \leq f,\ g  \text{ simple}$}), and $\tau$ is measure preserving ($\tau^{-1}(A)$ measurable for every measurable A), then $f \circ \tau$ is integrable and $\int f d\mu = \int f \circ \tau \, d\mu$. I have tried proving it myself (I need a proof but I don't need to have proved it myself) but I'm not sure where to start. Could anyone either direct me to a proof (online if possible though if it's a book that's not a disaster) or help me with such a proof? I can't seem to get going on it, so the more help you can give the better!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks very much, Giles. &lt;/p&gt;&#xA;" OwnerUserId="5873" LastEditorUserId="5873" LastEditorDisplayName="" LastEditDate="2011-01-15T06:37:42.063" LastActivityDate="2011-01-15T10:11:44.110" Title="Measure theoretic proof on composition of invariant functions and integrability" Tags="&lt;homework&gt;&lt;integral&gt;&lt;measure-theory&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="17561" PostTypeId="1" AcceptedAnswerId="17565" CreationDate="2011-01-15T04:41:25.407" Score="2" ViewCount="167" Body="&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/D3LeC.png&quot; alt=&quot;alt text&quot;&gt;Given any triangle (vertices are known) and a distance X, how can I compute the triangle that is shrunk by X from the original? By shrink, I mean edges of the shrunk triangle are exactly X away from the original edges. So if X is large enough, the shrunk triangle doesn't exist.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: the resulting triangle needs to be inside the given triangle. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Attaching a picture for clarity.&lt;/p&gt;&#xA;" OwnerUserId="2239" LastEditorUserId="2239" LastEditorDisplayName="" LastEditDate="2011-01-15T05:12:58.090" LastActivityDate="2011-01-16T02:50:45.610" Title="How to “shrink” a triangle" Tags="&lt;geometry&gt;&lt;trigonometry&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="1" />
  <row Id="17564" PostTypeId="2" ParentId="17561" CreationDate="2011-01-15T05:38:08.320" Score="0" ViewCount="" Body="&lt;p&gt;Given the points $T_1, T_2, T_3$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$A = T - {T_1}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$s$ is the value to scale the triangle by:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$s = ?$ // I'm figuring this out&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$B = A \cdot s$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$C = B + T_1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$C$ represents your new triangle.&lt;/p&gt;&#xA;" OwnerUserId="2736" LastActivityDate="2011-01-15T05:38:08.320" CommunityOwnedDate="2011-01-15T05:40:34.870" />
  <row Id="17565" PostTypeId="2" ParentId="17561" CreationDate="2011-01-15T05:52:44.910" Score="4" ViewCount="" Body="&lt;p&gt;Green triangle (result) is obviously a homothety of the given (blue) triangle with respect to a certain point &lt;code&gt;Q&lt;/code&gt; inside the triangle, and a certain coefficient &lt;code&gt;k&lt;/code&gt;. As you wish to find &lt;code&gt;Q&lt;/code&gt; and &lt;code&gt;k&lt;/code&gt; providing the distance &lt;code&gt;X&lt;/code&gt; from the original edges, then the direction of the moveout of the vertices lays on the original triangle's bisectors (proof is trivial for every vertex of the green triangle). The intersection point of triangle bisectors is the center of the inscribed circle. So, the homothety center &lt;code&gt;Q&lt;/code&gt; is the center of the inscribed circle. The homothety coefficient &lt;code&gt;k&lt;/code&gt; is the fraction of &lt;code&gt;X&lt;/code&gt; with respect to inscribed circle radius &lt;code&gt;R&lt;/code&gt;: &lt;code&gt;k=R/(R-X)&lt;/code&gt;. Thus, given the original triangle vertices &lt;code&gt;A, B, C&lt;/code&gt;, find the inscribed circle &lt;code&gt;Q, R&lt;/code&gt;, then find the scale &lt;code&gt;k&lt;/code&gt;, then moveout the vertices: &lt;code&gt;A'=A+(Q-A)*k&lt;/code&gt; - same for &lt;code&gt;B, C&lt;/code&gt; - in vector form.&lt;/p&gt;&#xA;" OwnerUserId="3806" LastActivityDate="2011-01-15T05:52:44.910" CommentCount="3" />
  <row Id="17566" PostTypeId="1" AcceptedAnswerId="17567" CreationDate="2011-01-15T07:33:00.347" Score="0" ViewCount="95" Body="&lt;p&gt;Hello There,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am a computer science student and do not have any background about tensors. Recently, I started learning about tensor spaces and tensor products from the text &quot;Tensor spaces and exterior algebra&quot; by Takeo Yokonuma.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I had a question and I understand that its pretty basic - but I would like some help with that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The author is trying to show that given two vector spaces $V$ and $W$ over a field $k$ and any bilinear map $\phi \in L(V,W,U)$ (&lt;strong&gt;EDIT:&lt;/strong&gt; where $L(V,W,U)$ is the set of all bilinear maps from $V \times W \rightarrow U$ for various vector spaces $U$ over the field $k$.) We consider a particular bilinear mapping $l \colon V \times W \to U_0$&#xA;(i) there exists a $k$-linear mapping $F \ \colon U_0 \to U$ such that $\phi = F \ o \ l$. ()&#xA;(ii) $l$ (in the former part) is the mapping from $V \times W$ to $U_0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was going through the proof of part (i) which is where my troubles begin. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Author has already established that the $dim(U_0) \geq mn$ where $dim(V) = n$ and $dim(W) = m$. He proceeds by picking a $U_0$ of dimension exactly $mn$. Thereafter, he takes a basis $\mathcal{G} = {g_1, g_2 \ldots g_{mn}}$ of $U_0$ and observes that the basis $\mathcal{E}$ of $U$ and $\mathcal{F}$ of $W$ are such that we have a one-to-one correspondence between  the sets $\mathcal{E} \times \mathcal{F}$ and $\mathcal{G}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After this observation, the author lets $\phi \in L(V,W,U)$ for a vector space $U$. Then, he defines $F_{\phi} \colon U_0 \to U$ by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$F_{\phi}(u) = F_{\phi}(\sum_{i,j}\gamma_{i,j} \cdot g(i,j) = \sum_{i,j} \gamma_{i,j} \cdot \phi(i,j))$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and this is what I have trouble believing. I am willing to believe that $F_{\phi} (\gamma_{ij}g(i,j)) = \gamma_{ij} \phi(i,j)$ but I cannot assume that this also distributes linearly - as this is what I want to prove. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe it is easy to see but I would like to have some help here (as I do not see why is this definition for $F_{\phi}$ legitimate)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope my question is clear.&#xA;Please help&#xA;Thanks&#xA;-Akash&lt;/p&gt;&#xA;" OwnerUserId="5391" LastEditorUserId="5391" LastEditorDisplayName="" LastEditDate="2011-01-15T13:51:20.717" LastActivityDate="2011-01-15T13:51:20.717" Title="Universality of tensor product" Tags="&lt;tensors&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="17567" PostTypeId="2" ParentId="17566" CreationDate="2011-01-15T09:12:09.910" Score="2" ViewCount="" Body="&lt;p&gt;Linearity is built into the definition: you first define the map $F_\phi$ on the basis elements, then you extend linearly. I am not sure what you mean by &quot;this is what I have trouble believing&quot;. Clearly, if you only define a map on a basis and then extend linearly to linear combinations of the basis elements, you get a linear map. The only thing that you might doubt is that this is well-defined. But in the context of vector spaces there is no such issue, since every element can be &lt;em&gt;uniquely&lt;/em&gt; written as a linear combination of the basis elements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Think of what you do when you define a linear map on a vector space. You just specify what this map does to a basis (that's the whole point behind matrix notation).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the way, if you tried to do that on an arbitrary generating set of some module, then you would be heading for trouble without appropriate care, so your doubt would be appropriate.&lt;/p&gt;&#xA;" OwnerUserId="3212" LastActivityDate="2011-01-15T09:12:09.910" CommentCount="3" />
  <row Id="17568" PostTypeId="1" AcceptedAnswerId="17572" CreationDate="2011-01-15T09:12:43.367" Score="7" ViewCount="490" Body="&lt;p&gt;While reading about cardinality, I've seen a few examples of bijections from the open unit interval $(0,1)$ to $\mathbb{R}$, one example being the function defined by $f(x)=\tan\pi(2x-1)/2$. Another geometric example is found by bending the unit interval into a semicircle with center $P$, and mapping a point to its projection from $P$ onto the real line.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is, is there a bijection between the open unit interval $(0,1)$ and $\mathbb{R}$ such that rationals are mapped to rationals and irrationals are mapped to irrationals?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I played around with mappings similar to $x\mapsto 1/x$, but found that this never really had the right range, and using google didn't yield any examples, at least none which I could find. Any examples would be most appreciated, thanks!&lt;/p&gt;&#xA;" OwnerUserId="1583" LastActivityDate="2011-01-15T20:47:24.283" Title="Is there a bijection between $(0,1)$ and $\mathbb{R}$ that preserves rationality?" Tags="&lt;functions&gt;&lt;elementary-set-theory&gt;" AnswerCount="3" CommentCount="3" />
  <row Id="17570" PostTypeId="2" ParentId="17568" CreationDate="2011-01-15T09:54:59.187" Score="6" ViewCount="" Body="&lt;p&gt;With the axiom of choice we can find well orderings of $\mathbb{R}$ and of $(0,1)$ such that the first $\omega$ elements are all the rationals of the set, then we can define our map to go from one well ordering to another by preserving the index (that is $a_\alpha\mapsto b_\alpha$, for $\alpha&amp;lt;2^{\aleph_0}$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As discussed in the comments below by Colin and Jason (and myself), one does not need the axiom of choice for that. Using the Cantor-Schroeder-Bernstein theorem one can have two bijections, one from $(0,1)\setminus\mathbb{Q}$ to $\mathbb{R}\setminus\mathbb{Q}$ and one from $(0,1)\cap\mathbb{Q}$ to $\mathbb{Q}$, and define a bijection as needed without the use of the axiom of choice.&lt;/p&gt;&#xA;" OwnerUserId="622" LastEditorUserId="622" LastEditorDisplayName="" LastEditDate="2011-01-15T10:56:09.520" LastActivityDate="2011-01-15T10:56:09.520" CommentCount="7" />
  <row Id="17571" PostTypeId="2" ParentId="17560" CreationDate="2011-01-15T10:11:44.110" Score="1" ViewCount="" Body="&lt;p&gt;There are two issues in the formulation of your question:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;The definition&#xA;\[&#xA;\mu(f) = \sup\{\mu(g)\,:\,\text{$g \leq f$ simple}\}&#xA;\]&#xA;only works for $f \geq 0$. For general $f$ you have to write $f = f_+ - f_{-}$ with $f_+ \geq 0$ and $f_{-} \geq 0$ and use that by definition $\mu(f) = \mu(f_{+}) - \mu(f_{-})$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;You forgot the condition $\mu(A) = \mu(\tau^{-1}A)$ in the definition of measure-preserving.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;As Qiaochu pointed out, the case of simple functions implies the general case. I'll treat the case $f \geq 0$ only.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that for the characteristic function $g = [A]$ of a measurable set $A$ we have $g \circ \tau = [\tau^{-1} A]$ and $\mu(g) = \mu(g \circ \tau)$ by hypothesis. By linearity of the integral $\mu(g) = \mu(g \circ \tau)$ for simple $g$. Now if $f \geq 0$ we may choose a sequence $g_{n}$ of simple functions such that $g_{n} \to f$ pointwise (a.e.) and monotone. By using the monotone convergence theorem twice and using the fact that $\tau$ is measure-preserving in the second equality we have&#xA;\[&#xA;\mu(f) = \lim_{n \to \infty} \mu(g_{n}) = \lim_{n \to \infty} \mu(g_{n} \circ \tau) = \mu(f \circ \tau)&#xA;\]&#xA;as desired.&lt;/p&gt;&#xA;" OwnerUserId="5363" LastActivityDate="2011-01-15T10:11:44.110" CommentCount="2" />
  <row Id="17572" PostTypeId="2" ParentId="17568" CreationDate="2011-01-15T10:27:38.273" Score="15" ViewCount="" Body="&lt;p&gt;$(1/x)-2$ on $(0,1/2]$ and $2-(1/(x-1/2))$ on $(1/2,1)$.&lt;/p&gt;&#xA;" OwnerUserId="5838" LastActivityDate="2011-01-15T10:27:38.273" CommentCount="6" />
  <row Id="17573" PostTypeId="1" AcceptedAnswerId="17576" CreationDate="2011-01-15T10:34:42.400" Score="2" ViewCount="157" Body="&lt;p&gt;Is there a monoidal category $\mathcal C$ whose unit object is $I$ (i.e. $I\otimes A\cong A\cong A\otimes I$ for all $A\in \text{Ob}_\mathcal C$), with an object &quot;$-1$&quot; such that&#xA;$$&#xD;&#xA;(-1)\otimes(-1)\cong I ?&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;(Edit: no one misunderstood, but I'm also asking that $-1\neq I$)&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm struggling with that since I read &lt;a href=&quot;http://math.stackexchange.com/questions/17311/categorification-of-pi&quot;&gt;this&lt;/a&gt; math.SE post... &lt;a href=&quot;http://math.stackexchange.com/users/1650/martin-brandenburg&quot;&gt;Martin&lt;/a&gt;, if you see me, your server rejected every mail I tried to send you. :(&lt;/p&gt;&#xA;" OwnerUserId="685" LastEditorUserId="1650" LastEditorDisplayName="" LastEditDate="2011-03-18T18:38:52.820" LastActivityDate="2011-03-18T18:38:52.820" Title="$(-1)\otimes (-1) \cong I$" Tags="&lt;category-theory&gt;&lt;monoidal-categories&gt;" AnswerCount="2" />
  <row Id="17574" PostTypeId="1" AcceptedAnswerId="17590" CreationDate="2011-01-15T10:54:07.260" Score="0" ViewCount="135" Body="&lt;p&gt;Let $ABC = D$ where $B$ and $D$ are symmetrical matrices. However their [rows x columns] values are not same. For example, $B$ is 2x2 and $D$ is 3x3 a matrix. Clearly, in this case, $A$ has to be a 3x2 matrix and $C$ must be a 2x3 matrix. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Prove or disprove that this holds iff $A^T = C$ i.e. if $A$ and $C$ are transpose to each other then $D$ is symmetrical and if $B$ and $D$ are symmetrical then $A$ and $C$ are transpose to each other. I have tested few cases on computer and it seems to be correct. But I am not sure about 'if and only if' part. Does this hold if $B$ and $D$ are positive definite? Any comment about the nature of $A$ and $C$ is welcome. A chocolate for a correct proof, a cup of coffee otherwise! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT : There was a serious flaw in original problem. Corrected now. Thanks to Hardmath and Willie. &lt;/p&gt;&#xA;" OwnerUserId="1674" LastEditorUserId="1674" LastEditorDisplayName="" LastEditDate="2011-01-15T14:29:51.667" LastActivityDate="2011-01-23T13:31:36.517" Title="Transformation of a symmetrical matrix into another symmetrical matrix" Tags="&lt;linear-algebra&gt;" AnswerCount="2" CommentCount="8" />
  <row Id="17575" PostTypeId="1" AcceptedAnswerId="17579" CreationDate="2011-01-15T11:14:25.337" Score="4" ViewCount="200" Body="&lt;p&gt;I have N sensors which are been sampled M times, so I have an N by M readout matrix. If I want to know the relation and dependencies of these sensors simplest thing is to do a Pearson's correlation which gives me an N by N correlation matrix. &#xA;Now let's say if I have the correlation matrix but I want to explore tho possible readout space that can lead to such correlation matrix what can I do? &#xA;So the question is: given an N by N correlation matrix how you can get to a matrix that would have such correlation matrix?&#xA;Any comment is appreciated. &lt;/p&gt;&#xA;" OwnerUserId="5875" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-16T20:54:40.520" LastActivityDate="2011-01-16T23:19:48.327" Title="Going back from a correlation matrix to the original matrix" Tags="&lt;matrices&gt;&lt;correlation&gt;" AnswerCount="4" CommentCount="5" FavoriteCount="3" />
  <row Id="17576" PostTypeId="2" ParentId="17573" CreationDate="2011-01-15T12:29:45.513" Score="5" ViewCount="" Body="&lt;p&gt;what about $\mathbb{Z}_2$ &lt;a href=&quot;http://en.wikipedia.org/wiki/Super_vector_space&quot; rel=&quot;nofollow&quot;&gt;vector spaces&lt;/a&gt; over a field $k$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(k,0)$ is the unit in this category and $(0,k)\otimes (0,k) = (k,0)$.&lt;/p&gt;&#xA;" OwnerUserId="2125" LastActivityDate="2011-01-15T12:29:45.513" CommentCount="1" />
  <row Id="17577" PostTypeId="1" AcceptedAnswerId="17581" CreationDate="2011-01-15T12:40:32.577" Score="4" ViewCount="97" Body="&lt;p&gt;When showing that a (natural family of) diagram of $R$-algebras for all rings $R$ commutes, why does it suffice to show that it commutes for all $R$ local with algebraically closed residue field?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My idea: Reduce to local rings (fpqc descent), then to strictly henselian rings (strict henselisation is faithfully flat). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this correct?&lt;/p&gt;&#xA;" OwnerUserId="5262" LastActivityDate="2011-01-15T14:48:43.310" Title="showing locally that a diagram commutes" Tags="&lt;algebraic-geometry&gt;&lt;commutative-algebra&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="17578" PostTypeId="1" AcceptedAnswerId="17580" CreationDate="2011-01-15T13:52:48.720" Score="4" ViewCount="248" Body="&lt;p&gt;Let $x = i^{i^{i^{i^{.^{.^{.{^ \infty}}}}}}}$. This is the solution of the equation&#xA;$i^x - x = 0 $ . I used Euler's identity to find a solution. But I haven't yet found the real and imaginary parts of the solution. Are there more solutions? If so why did I miss them?&lt;/p&gt;&#xA;" OwnerUserId="5595" LastEditorUserId="5595" LastEditorDisplayName="" LastEditDate="2011-01-15T14:42:57.497" LastActivityDate="2011-03-27T14:43:12.653" Title="How to compute the infinite tower of the complex number $i$, that is$ ^{\infty}i$" Tags="&lt;complex-analysis&gt;" AnswerCount="2" CommentCount="9" />
  <row Id="17579" PostTypeId="2" ParentId="17575" CreationDate="2011-01-15T14:03:07.747" Score="1" ViewCount="" Body="&lt;p&gt;You can do a cholesky-decomposition, such that if your correlation-matrix is &lt;strong&gt;R&lt;/strong&gt; then if&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ L*transpose(L) = R $   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;then $ L = cholesky(R)$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here &lt;strong&gt;L&lt;/strong&gt; is triangular of size &lt;em&gt;N by N&lt;/em&gt;. Now you could also understand &lt;strong&gt;L&lt;/strong&gt; as a simple rotation of your original readout-matrix where the &lt;em&gt;M&lt;/em&gt; columns are rotated to fit into &lt;em&gt;N&lt;/em&gt; (&lt;em&gt;&amp;lt;=M&lt;/em&gt; ) columns/vectorspace.    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you may apply any arbitrary rotation to the columns of &lt;strong&gt;L&lt;/strong&gt;, which may arbitrarily be extended to some &lt;em&gt;M&lt;/em&gt; columns by appending null-columns.    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You may also apply a &quot;stretching&quot; of the rows in that reconstructed &quot;readout&quot;-matrix by multiplication of each row by some standard-deviation since the &lt;strong&gt;L&lt;/strong&gt;-matrix represents rows with norm 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(If you want you can download my (free) MatMate-program for matrix-visualization and see with a simple script what I mean here/how this works. Here is the  &lt;a href=&quot;http://go.helms-net.de/sw/matmate/index.htm&quot; rel=&quot;nofollow&quot;&gt;download page&lt;/a&gt; .)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[update] Here I give an example. After reading more comments I understand, that the goal was to construct a data-matrix given a correlation-matrix. In the example I constructed a random correlation-matrix 4 by 4. Then I produced data and then I check, whether the data, corrleated, reproduce the correlation-matrix. They do, and because the matrix-algebra is simple we can also see that this is exact.    &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;;===========================================&#xA;;MatMate-Listing vom:16.01.2011 18:33:13&#xA;;============================================&#xA;[1] cor = covtocorr(cov)   // this is the correlation-matrix from a given covariance.&#xA;      cor : &#xA;    1.00       -0.38        0.22       -0.85&#xA;   -0.38        1.00        0.30        0.57&#xA;    0.22        0.30        1.00        0.27&#xA;   -0.85        0.57        0.27        1.00&#xA;&#xA;[2] lad = cholesky(cor)   // this provides a &quot;loadings&quot;-matrix by cholesky-decomposition&#xA;                          // of matrix cor&#xA;      lad : &#xA;    1.00        0.00        0.00        0.00&#xA;   -0.38        0.93        0.00        0.00&#xA;    0.22        0.42        0.88        0.00&#xA;   -0.85        0.27        0.39        0.25&#xA;&#xA;Now we want to construct 200 columns of data. As I wrote earlier, the cholesky&#xA;matrix L of a given correlationmatrix R is just a rotation of the data-matrix&#xA;One should add, that the data must be centered (mean=0) and normed(stddev=1)&#xA;Now I do the inverse: I construct a valid rotation-matrix t by just rotating a&#xA;dummy random-matrix of 200 columns to the first column(rotation method:&quot;triangular&quot;).&#xA;[3] n=200&#xA;[4]     dummy = randomu(1,n)&#xA;[5]     t = gettrans(dummy,&quot;tri&quot;)&#xA;[6]     hide t,dummy&#xA;&#xA;So I've got a valid rotation-matrix t. &#xA;Now I provide columns to rotate the cholesky/loadings-matrix within&#xA;and insert the cholesky matrix in the first 4 columns&#xA;[7] data = (lad || null(v,n-v)) * sqrt(n)&#xA;&#xA;Then I rotate that data-matrix with that random rotation&#xA;[8] data = data * t&#xA;[9] disp = data[*,1..10]   // show the first 10 columns of the created dataset&#xA;      disp : &#xA;    1.71       -0.11       -0.20       -0.05       -0.06       -0.21       -0.16       -0.14       -0.11       -0.21&#xA;    0.23       13.12        0.08        0.02        0.02        0.08        0.06        0.05        0.04        0.08&#xA;    2.22        5.77       12.34       -0.01       -0.01       -0.05       -0.03       -0.03       -0.02       -0.05&#xA;   -0.46        3.83        5.63        3.52        0.05        0.18        0.13        0.12        0.09        0.17&#xA;&#xA;Check whether that data really give the correlation-matrix     &#xA;[10] chk_cor = data * data' /n&#xA;      chk_cor : &#xA;    1.00       -0.38        0.22       -0.85&#xA;   -0.38        1.00        0.30        0.57&#xA;    0.22        0.30        1.00        0.27&#xA;   -0.85        0.57        0.27        1.00&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;We see, that the correlation-matrix is reproduced.&lt;br&gt;&#xA;The &quot;mystery&quot; here is simply, that by the definition of correlation (of normed data, to make the example short)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;R = data * data' / n&lt;/p&gt;&#xA;&#xA;&lt;p&gt;here data may have arbitrary many columns&#xA;and t*t' is identity-matrix, so&lt;br&gt;&#xA;  R = data * t * t' * data' / n&lt;br&gt;&#xA;  R = (data*t) * (data*t)'  / n&lt;br&gt;&#xA;and the cholesky-factorization is just the operation to get L and L' from a correlationmatrix    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;L = cholesky(R) ==&gt; L*L' = R&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Thus the result is also exact)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[end update]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[update 2] In response to the remark of mpiktas I'll add a variant which allows to set the standard-deviation and the means in the generated data explicitely.&lt;br&gt;&#xA;The arguing in the first update was to show, how the data form a m-dimensional vectorspace, and the cholesky-decomposition of the according correlation-matrix can simply be understood as a special rotation of the data in that vectorspace.&lt;br&gt;&#xA;However, means of data generated the way I described above are not guaranteed to be zero. Soe here is a variant which produces means=zero, stddevs=1 and the data are then scalable to other stddevs and translatable to prediefined means.&lt;br&gt;&#xA;Here the idea is to use the L-matrix as &quot;recipe for composition&quot; of uncorrelated &quot;factors&quot; as understood in factor analysis/PCA. Having a centered and normed matrix of data &lt;strong&gt;D&lt;/strong&gt; it is assumed that the cholesky-matrix &lt;strong&gt;L&lt;/strong&gt; describes a linear combination of uncorrelated data of unit-variance (&quot;factors&quot;) in a matrix &lt;strong&gt;F&lt;/strong&gt; with the same number of columns as &lt;strong&gt;D&lt;/strong&gt; such that&lt;br&gt;&#xA;  &lt;strong&gt;L * F  = D&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, some random &lt;strong&gt;F&lt;/strong&gt; can be created by a random-generator providing normal distribution with mean=0 and stddev=1 per row. If the rows in &lt;strong&gt;F&lt;/strong&gt; were truely uncorrelated, we had by the above composition &lt;strong&gt;L * F = D&lt;/strong&gt; a simple solution. After that we could scale the rows of &lt;strong&gt;D&lt;/strong&gt; by standard-deviations and also translate to predefined means.&lt;br&gt;&#xA;Unfortunately randomgenerators do not exactly give uncorrelated data, so the correlations in &lt;strong&gt;F&lt;/strong&gt; must be removed first. This is possible by the inverse of the cholesky-matrix of the correlations of &lt;strong&gt;F&lt;/strong&gt;. Here is a script pseudocode    &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;v = 4 // set number of variables/of rows, make a small example        &#xA;n = 200 // set number of observations/ of columns      &#xA;F = randomn(v,n,0,1)  // generate randomdata in F where rows have mean=0, stddev=1      &#xA;covF  = F * F' /n      &#xA;cholF =  cholesky(covF)     &#xA;&#xA;L = cholesky( R )          // R is the targetted correlation-matrix&#xA;data = L * inv(cholF) * F      &#xA;&#xA;sdevs = diag([1,2,3,4])              // provide four stddevs     &#xA;means = colvector([20,10,40,30])     // provide four means     &#xA;&#xA;simuldata = sdevs*data + means   // here the means-vector must be expanded to    &#xA;                                // be added to all columns in simuldata     &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The simulated data are in the matrix &lt;strong&gt;simuldata&lt;/strong&gt;.      &lt;/p&gt;&#xA;&#xA;&lt;p&gt;While we can set stddevs and means explicitely there is still one problem pertaining: by the leftmultiplication of &lt;strong&gt;inv(cholF&lt;/strong&gt;) and &lt;strong&gt;L&lt;/strong&gt; the normal-distribution in &lt;strong&gt;simuldata&lt;/strong&gt; as originally provided by the randomgenerator in &lt;strong&gt;F&lt;/strong&gt; may be affected/distorted a bit. I don't have an idea yet how to solve this...     &lt;/p&gt;&#xA;" OwnerUserId="1714" LastEditorUserId="1714" LastEditorDisplayName="" LastEditDate="2011-01-16T23:19:48.327" LastActivityDate="2011-01-16T23:19:48.327" CommentCount="9" />
  <row Id="17580" PostTypeId="2" ParentId="17578" CreationDate="2011-01-15T14:15:14.130" Score="4" ViewCount="" Body="&lt;p&gt;In case you really mean $ I^x - x = 0 $ you can find this by the iteration&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; x = &amp;lt;some initial value&amp;gt; &#xA; repeat&#xA;     x = I^x   // where I is the imaginary unit&#xA;   until convergence&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;You'll get approximately $ x = 0.438282936727 + 0.360592471871*I $ (using Pari/GP, for instance)    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, you can also find the value using the lambert-w-function.&lt;br&gt;&#xA;[update] using the lambert-w:     &lt;/p&gt;&#xA;&#xA;&lt;p&gt;let $ \lambda=\ln(i)  $&lt;br&gt;&#xA;then&lt;br&gt;&#xA;$i^x = x$&lt;br&gt;&#xA;$1 = x* \exp( -x \lambda) $&lt;br&gt;&#xA;$-\lambda = -x \lambda * \exp( -x \lambda) $&lt;br&gt;&#xA;$ W(-\lambda) = - x \lambda $&lt;br&gt;&#xA;so&lt;br&gt;&#xA;$ x = \frac{W(-\ln(i))}{-\ln(i)}  $     &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;(Whether this is more &quot;exact&quot; is rather a question &quot;exact in terms-of-what?&quot;)&lt;/em&gt;    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;[end update]&lt;/p&gt;&#xA;" OwnerUserId="1714" LastEditorUserId="1714" LastEditorDisplayName="" LastEditDate="2011-01-15T15:08:24.687" LastActivityDate="2011-01-15T15:08:24.687" CommentCount="1" />
  <row Id="17581" PostTypeId="2" ParentId="17577" CreationDate="2011-01-15T14:18:19.610" Score="6" ViewCount="" Body="&lt;p&gt;You don't need anything as fancy or as hard as the general fpqc descent theorem (or even the affine case, which is easier), but only basic facts about faithfully flat  maps. So first of all it is fairly well-known and easy to prove (i.e., from the definition) that two maps of $R$-modules are equal iff the base-changes to the localizations $R_{\mathfrak{p}}$ for $\mathfrak{p}$ prime are equal, for every prime $\mathfrak{p}$.&#xA;This is why you can reduce to the case of a local ring.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The more interesting part of your question is why we can reduce to the case where the residue field is algebraically closed. Here I quote a lemma from EGA $0_{III}$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Lemma:&lt;/strong&gt; Let $(R, \mathfrak{m})$ be a noetherian local ring with residue field $k$, and let $K$ be an extension of $k$. Then there exists a flat, local noetherian $R$-algebra $R&amp;#39;$ such that $\mathfrak{m} R&amp;#39; $ is the maximal ideal of $R&amp;#39;$ and such that the residue field is $K$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $R$ is not assumed noetherian, then $R&amp;#39;$ will not be assumed noetherian, but the argument still works. Actually, it is not hard if you forget about noetherianness, so let me sketch it. We write $K$ as an inductive (by an ordinal whose cardinality is generally large) colimit of towers of extensions generated by one element. By transfinite induction (as a colimit of flat things is flat), we reduce to the case where $K$ is generated by one element over $k$, say some $\alpha$. If $\alpha$ is transcendental, then we can take $R&amp;#39; = (R[t])_{\mathfrak{m} R[t]}$. If $\alpha$ is algebraic, satisfying some monic polynomial $\overline{P} \in k[X]$ that lifts to $P \in R[X]$, then let $R&amp;#39; = R[X]/P$. One can check that this works (the maximal ideals of $R&amp;#39;$ are in correspondence with those of $R[X]/P \otimes_R k$ by Nakayama, and this last thing is clearly a field.)&#xA;In either of these cases, it is easy to check that $R&amp;#39;$ is flat (even free in the second case).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;OK. So if $R&amp;#39;$ is flat over $R$ (assumed local as above, in the situation of the last paragraph),  then $R&amp;#39;$ is &lt;em&gt;faithfully&lt;/em&gt; flat because we are working with local rings. So if $M \rightrightarrows N$ are two maps of $R$-modules, they are equal if and only if $M \otimes_{R} R&amp;#39; \rightrightarrows N \otimes_R R&amp;#39;$. It follows that if we want to prove something about commutativity of diagrams of $R$-modules, we can reduce to proving something about $R&amp;#39;$-modules.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, as I explained above, any local ring admits a flat local homomorphism into a local ring whose residue field is algebraically closed. So if you want to prove some kind of commutativity for local rings, you can reduce to the case where the residue field is algebraically closed. And, as I explained at the very beginning, if you want to prove something about commuting diagrams of modules over a ring, you can reduce to the case of  local rings.&lt;/p&gt;&#xA;" OwnerUserId="536" LastEditorUserId="536" LastEditorDisplayName="" LastEditDate="2011-01-15T14:48:43.310" LastActivityDate="2011-01-15T14:48:43.310" CommentCount="6" />
  <row Id="17582" PostTypeId="1" CreationDate="2011-01-15T14:29:28.603" Score="2" ViewCount="129" Body="&lt;p&gt;Prove that&#xA;$$\binom{2p}{p} \equiv 2 ( mod p^3),$$&#xA;where $p\ge 5$ is a prime number.&lt;/p&gt;&#xA;" OwnerUserId="4210" LastActivityDate="2011-01-27T04:16:24.453" Title="Is there a combinatorial proof of this congruence identity?" Tags="&lt;number-theory&gt;&lt;combinatorics&gt;&lt;binomial-coefficients&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="17583" PostTypeId="2" ParentId="17582" CreationDate="2011-01-15T14:39:09.217" Score="3" ViewCount="" Body="&lt;p&gt;A combinatorial proof for the congruence $\bmod p^2$ is given at &lt;a href=&quot;http://mathoverflow.net/questions/26137/binomial-supercongruences-is-there-any-reason-for-them&quot; rel=&quot;nofollow&quot;&gt;this MO question&lt;/a&gt; but the answerer suggests the congruence $\bmod p^3$ does not have a natural combinatorial proof.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2011-01-15T14:39:09.217" CommentCount="3" />
  <row Id="17584" PostTypeId="1" CreationDate="2011-01-15T14:49:43.183" Score="2" ViewCount="199" Body="&lt;p&gt;Given $BA = A$ and $AB = B$  and $A$ and $B$ are two square matrices, why does $BA +AB = A+B$ ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please explain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ADDED:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The actual problem is it was given that,$BA = A$ and $AB = B$ hence find the value of $A^2 + B^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I simply utilized the fact that if $BA = A$ and $AB = B$ then $A$ and $B$ are idempotent matrices, hence $A^2 + B^2 = A + B$ but the module solution is something more algebraic they break $A^2 + B^2$ and then performs usual substitution after that they showed the result but I don't understand how $BA +AB = A+B$.&lt;/p&gt;&#xA;" OwnerUserId="3744" LastEditorUserId="3744" LastEditorDisplayName="" LastEditDate="2011-01-15T15:03:53.007" LastActivityDate="2011-01-15T15:03:53.007" Title="Why $BA +AB = A+B$ ?" Tags="&lt;matrices&gt;" AnswerCount="2" CommentCount="7" FavoriteCount="1" />
  <row Id="17586" PostTypeId="2" ParentId="17584" CreationDate="2011-01-15T14:55:20.607" Score="2" ViewCount="" Body="&lt;p&gt;This isn't true in general. For example take $A$ as the zero matrix and $B$ as the identity matrix. Then the left hand side is just the zero matrix and the right hand side is the identity matrix, which are not the same.&lt;/p&gt;&#xA;" OwnerUserId="900" LastActivityDate="2011-01-15T14:55:20.607" CommentCount="1" />
  <row Id="17587" PostTypeId="2" ParentId="17584" CreationDate="2011-01-15T14:55:33.337" Score="2" ViewCount="" Body="&lt;p&gt;This is not true, indeed consider the case when $A = 0 $ is the zero matrix and $B= Id$ is the identity, then LHS is the zero matrix but RHS is the identity.&lt;/p&gt;&#xA;" OwnerUserId="5105" LastActivityDate="2011-01-15T14:55:33.337" />
  <row Id="17588" PostTypeId="1" AcceptedAnswerId="17589" CreationDate="2011-01-15T15:23:14.263" Score="3" ViewCount="157" Body="&lt;p&gt;In a normal problem solving I run into a sum:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sum_{k=1}^n \frac{10}{10+k}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let n=20 or something like that (not huge).  Browsing &lt;a href=&quot;http://en.wikipedia.org/wiki/List_of_mathematical_series&quot; rel=&quot;nofollow&quot;&gt;a list of sums&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/Summation&quot; rel=&quot;nofollow&quot;&gt;another&lt;/a&gt;, I'm not finding a formula for this one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm wondering is it possible to find a formula for this type of sum, or if this type of sum has to be computed by a computer summing it.&lt;/p&gt;&#xA;" OwnerUserId="1069" LastEditorUserId="367" LastEditorDisplayName="" LastEditDate="2011-01-15T22:01:44.587" LastActivityDate="2011-01-15T22:01:44.587" Title="Formula for $\sum_{k=1}^n \frac{10}{10+k}$" Tags="&lt;sequences-and-series&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="17589" PostTypeId="2" ParentId="17588" CreationDate="2011-01-15T15:32:11.510" Score="7" ViewCount="" Body="&lt;p&gt;You can express it using the &lt;a href=&quot;http://en.wikipedia.org/wiki/Harmonic_number&quot; rel=&quot;nofollow&quot;&gt;Harmonic numbers&lt;/a&gt; as $\sum_{k=1}^n \frac{10}{10+k}=10(H_{10+n}-H_{10})\approx 10(\ln(10+n)-ln(10))$  But if you want it exactly you need to add it up.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2011-01-15T15:39:58.417" LastActivityDate="2011-01-15T15:39:58.417" CommentCount="2" />
  <row Id="17590" PostTypeId="2" ParentId="17574" CreationDate="2011-01-15T15:33:43.507" Score="2" ViewCount="" Body="&lt;p&gt;The claim is untrue, even if we restrict $B,D$ to identity matrices.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A simple example is $(1 0) I (1 1)^T = I$.  Clearly $A = (1 0)$ is not the transpose of $C = (1 1)^T$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact $B$ could be any symmetric nxn matrix and $A,C$ any compatible row and column.  The result $D$ as a 1x1 matrix will automatically be symmetric.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added:&lt;/strong&gt;  Examples of larger dimensions are easily constructed as well.  For example:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ A = \begin{pmatrix} 1 &amp;amp; 0 &amp;amp; 0 \\\\ 0 &amp;amp; 0 &amp;amp; 1 \end{pmatrix}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ C = \begin{pmatrix} 1 &amp;amp; 0 \\\\ 1 &amp;amp; 0 \\\\ 0 &amp;amp; 1 \end{pmatrix}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;will again satisfy $AIC = I $ for identity matrices of compatible dimensions.&lt;/p&gt;&#xA;" OwnerUserId="3111" LastEditorUserId="3111" LastEditorDisplayName="" LastEditDate="2011-01-15T16:59:49.113" LastActivityDate="2011-01-15T16:59:49.113" />
  <row Id="17591" PostTypeId="1" AcceptedAnswerId="17593" CreationDate="2011-01-15T15:36:17.703" Score="1" ViewCount="96" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've got the following grammar I'm attempting to convert to CNF.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;S -&amp;gt; T01 | USV | epsilon&#xA;U -&amp;gt; X&#xA;V -&amp;gt; S1S | X | 1&#xA;X -&amp;gt; 0XS | 0&#xA;T -&amp;gt; TV | XT | UTU&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I know that when I eliminate epsilon-productions I get...&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;S -&amp;gt; T01 | USV | UV&#xA;U -&amp;gt; X&#xA;V -&amp;gt; S1S | S1 | 1S | X | 1&#xA;X -&amp;gt; 0XS | 0X | 0&#xA;T -&amp;gt; TV | XT | UTU&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I understand that we're getting rid of the epsilon but I don't really get what we're doing with the other symbols. For example how does S now produce &lt;code&gt;UV&lt;/code&gt;? If someone could explain this to me I'd really appreciate it. &lt;/p&gt;&#xA;" OwnerUserId="5325" LastActivityDate="2011-01-15T16:08:30.947" Title="Grammar to Chomskys Normal Form eliminating epsilon productions." Tags="&lt;computer-science&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="17592" PostTypeId="2" ParentId="17568" CreationDate="2011-01-15T15:58:29.787" Score="6" ViewCount="" Body="&lt;p&gt;$$&#xD;&#xA;f(x) = \frac{2x - 1}{1 - |2x - 1|}.&#xD;&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="5546" LastActivityDate="2011-01-15T15:58:29.787" CommentCount="1" />
  <row Id="17593" PostTypeId="2" ParentId="17591" CreationDate="2011-01-15T16:08:30.947" Score="2" ViewCount="" Body="&lt;p&gt;Here's how $S$ produces $UV$: $S \rightarrow USV \rightarrow UV$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Generalizing the example, whenever a non-terminal $X$ produces $\epsilon$, whenever we see $X$ on the right-hand side we can (optionally) delete it. This is more-or-less what the epsilon-elimination algorithm is trying to do.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suggest that instead of trying to follow the algorithm blindly, you try to understand what it's doing. You need to understand three things:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Why the transformations employed by the algorithm do not change the language.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Why the algorithm eventually gets rid of all epsilon productions.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Why the algorithm terminates.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2011-01-15T16:08:30.947" CommentCount="2" />
  <row Id="17594" PostTypeId="1" AcceptedAnswerId="17597" CreationDate="2011-01-15T16:45:53.410" Score="1" ViewCount="144" Body="&lt;p&gt;how can i predict a chaos with feigenbaum constant in the logistic map?&lt;/p&gt;&#xA;" OwnerUserId="5879" LastEditorUserId="2548" LastEditorDisplayName="" LastEditDate="2011-01-15T17:34:39.013" LastActivityDate="2011-01-15T21:03:49.083" Title="feigenbaum constant" Tags="&lt;chaos-theory&gt;" AnswerCount="1" />
  <row Id="17595" PostTypeId="1" CreationDate="2011-01-15T17:03:16.080" Score="2" ViewCount="209" Body="&lt;p&gt;a vector pq and a eqation of vector pq what is difference b/w them ??&lt;/p&gt;&#xA;" OwnerUserId="5880" LastActivityDate="2011-02-14T20:22:52.440" Title="vectors in mathematics" Tags="&lt;vector-analysis&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" />
  <row Id="17596" PostTypeId="2" ParentId="17595" CreationDate="2011-01-15T17:33:27.843" Score="0" ViewCount="" Body="&lt;p&gt;See &lt;a href=&quot;http://en.wikipedia.org/wiki/Euclidean_vector&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2011-01-15T17:33:27.843" />
  <row Id="17597" PostTypeId="2" ParentId="17594" CreationDate="2011-01-15T17:36:24.783" Score="2" ViewCount="" Body="&lt;p&gt;See &lt;a href=&quot;http://www.physics.mq.edu.au/~wardle/PHYS220/chaos.ppt&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; (&lt;em&gt;Chaos and the Logistic Map Lecture Notes&lt;/em&gt; by Lesa Moore).&lt;/p&gt;&#xA;" OwnerUserId="2548" LastEditorUserId="2548" LastEditorDisplayName="" LastEditDate="2011-01-15T21:03:49.083" LastActivityDate="2011-01-15T21:03:49.083" CommentCount="4" />
  <row Id="17599" PostTypeId="2" ParentId="17595" CreationDate="2011-01-15T19:27:04.123" Score="1" ViewCount="" Body="&lt;p&gt;A vector is an element of a vector space. A vector equation, which is what you might mean, is simply an equation involving vectors. For instance, the vector equation $r=a+\lambda b$ with $\lambda\in R$ describes a line.&lt;/p&gt;&#xA;" OwnerUserId="4594" LastActivityDate="2011-01-15T19:27:04.123" />
  <row Id="17600" PostTypeId="1" AcceptedAnswerId="17613" CreationDate="2011-01-15T19:44:20.637" Score="8" ViewCount="122" Body="&lt;p&gt;Suppose $p$ is a prime, $\chi$ and $\lambda$ are characters on $\mathbb{F}_p$.&#xA;How can I show that $\sum_{t\in\mathbb{F}_p}\chi(1-t^m)=\sum_{\lambda}J(\chi,\lambda)$ where $\lambda$ varies over all characters such that $\lambda^m=id$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;($J$ is the Jacobi sum, defined as $J(\chi,\lambda)=\sum_{a+b=1}\chi(a)\lambda(b)$ )&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The exercise is taken from A Classical Introduction to Modern Number Theory by Ireland and Rosen - page 105, ex. 8.&lt;/p&gt;&#xA;" OwnerUserId="766" LastActivityDate="2011-01-15T21:51:28.750" Title="An exercise involving characters" Tags="&lt;number-theory&gt;&lt;homework&gt;&lt;algebraic-number-theory&gt;&lt;characters&gt;" AnswerCount="1" />
  <row Id="17601" PostTypeId="2" ParentId="17396" CreationDate="2011-01-15T19:45:28.990" Score="0" ViewCount="" Body="&lt;p&gt;Set $a_0=0$. To get $a_{n+1}$ from $a_n$, just change bit $r_n$, where $r_n$ is the number of trailing $1$-bits in the binary representation of $n$. So we get:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(r_n) = \lbrace 0,1,0,2,0,1,0,3,0,1,0,2,0,1,0,4,0,1,0,2,0,1,0,3,0,1,0,2,0,1,0,5,\dots \rbrace$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(a_n) = \lbrace 0,1,3,2,6,7,5,4,12,13,15,14,10,11,9,8,24,25,27,26,30,31,29,28,20,\dots \rbrace$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This construction gives the standard Gray code.&lt;/p&gt;&#xA;" OwnerUserId="1508" LastActivityDate="2011-01-15T19:45:28.990" CommentCount="5" />
  <row Id="17603" PostTypeId="2" ParentId="668" CreationDate="2011-01-15T21:09:16.053" Score="1" ViewCount="" Body="&lt;p&gt;If you have a matrix &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$H$&#xA;then you can calculate the correlationmatrix with &lt;/li&gt;&#xA;&lt;li&gt;$G = H \times H^H$&#xA;(H^H denotes the complex conjugated and transposed version of $H$).&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;If you do a eigenvalue decomposition of $G$ cou get eigenvalues $\lambda$ and eigenvectors $v$, that in combination $\lambda\times v$ describes the same space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now there is the following equation, saying:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Determinante($H*H^H$) = Product of all eigenvalues $\lambda$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Ie. If you have a $3\times3$ matrix $H$ -&gt; $G$ is $3\times3$ too giving us three eigenvalues.&#xA;The product of these eigenvalues give as the volume of a cuboid.&#xA;With every extra dimension/eigenvalue the cuboid gets an extra dimension.&lt;/p&gt;&#xA;" OwnerUserId="5884" LastEditorUserId="5884" LastEditorDisplayName="" LastEditDate="2011-01-17T09:45:07.343" LastActivityDate="2011-01-17T09:45:07.343" />
  <row Id="17604" PostTypeId="1" AcceptedAnswerId="17606" CreationDate="2011-01-15T21:11:00.173" Score="4" ViewCount="239" Body="&lt;p&gt;I am taking a Computer Science class soon that requires a solid knowledge of the basics of probability.  I've only had minimal exposure to probability in classes I've taken in the past, so I need to get up to speed quickly.  Can anyone recommend some good self-study resources (e.g. books, online classes, web sites) that I could use to teach myself the fundamentals?&lt;/p&gt;&#xA;" OwnerUserId="607" LastActivityDate="2011-05-24T09:18:57.733" CommunityOwnedDate="2011-01-15T23:06:40.990" Title="Self-study resources for basic probability?" Tags="&lt;probability&gt;&lt;self-learning&gt;" AnswerCount="4" CommentCount="3" FavoriteCount="4" />
  <row Id="17605" PostTypeId="1" CreationDate="2011-01-15T21:13:47.090" Score="1" ViewCount="151" Body="&lt;p&gt;Here's my problem: I have this function $\psi(\lambda)=\frac{e}{\pi\lambda}Im(e^{-\omega(\lambda-1)^{\frac{1}{4}}})$ (where $\omega=e^{-i\frac{\pi}{4}}$) defined on the interval $[1,\infty]$ and want to compute its integral (it should be 1).&#xA;I've been suggested to consider the complex function $e^{-\omega(z-1)^{\frac{1}{4}}}$ (cutting from the complex plane the interval $[1,\infty]$). Integrating on the curve $\gamma$ that goes from $\infty$ to 1, just above the slit, then doing a half-loop around 1 and going back to $\infty$ under the cut, I easily obtain from the residue formula that&#xA;$-\frac{1}{2\pi i}\int_{\gamma}\frac{1}{z}e^{-\omega(z-1)^{\frac{1}{4}}}dz=e^{-1}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, I guess, I have to take the limit (taking $\gamma$ closer and closer to $[1,\infty]$). However this part is a little bit tricky and I don't have figured out yet how to proceed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="5885" LastEditorUserId="5885" LastEditorDisplayName="" LastEditDate="2011-01-16T11:58:57.060" LastActivityDate="2011-01-16T11:58:57.060" Title="Compute an integral using complex analysis techniques" Tags="&lt;integral&gt;&lt;complex-analysis&gt;" CommentCount="2" />
  <row Id="17606" PostTypeId="2" ParentId="17604" CreationDate="2011-01-15T21:15:00.223" Score="4" ViewCount="" Body="&lt;ul&gt;&#xA;&lt;li&gt;&lt;em&gt;A First Course in Probability&lt;/em&gt; by Sheldon Ross&lt;/li&gt;&#xA;&lt;li&gt;&lt;em&gt;Schaum's Outline of Probability, Random Variables, and Random Processes&lt;/em&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;em&gt;An Introduction to Probability Models&lt;/em&gt; by Sheldon Ross&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2011-01-15T21:15:00.223" CommunityOwnedDate="2011-01-15T23:06:40.990" CommentCount="2" />
  <row Id="17607" PostTypeId="2" ParentId="17604" CreationDate="2011-01-15T21:18:50.153" Score="3" ViewCount="" Body="&lt;p&gt;Try &lt;em&gt;Introduction to Probability&lt;/em&gt; by Charles Grinstead and Laurie Snell. The book is available for purchase and also online reading &lt;a href=&quot;http://www.dartmouth.edu/~chance/teaching_aids/books_articles/probability_book/book.html&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="4594" LastEditorUserId="4594" LastEditorDisplayName="" LastEditDate="2011-05-24T09:18:57.733" LastActivityDate="2011-05-24T09:18:57.733" CommunityOwnedDate="2011-01-15T23:06:40.990" CommentCount="1" />
  <row Id="17608" PostTypeId="2" ParentId="17604" CreationDate="2011-01-15T21:20:49.797" Score="3" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://www.cs.cmu.edu/~odonnell/papers/probability-and-computing-lecture-notes.pdf&quot; rel=&quot;nofollow&quot;&gt;Here&lt;/a&gt; you have great lectures from Ryan O'Donnell's course Probability and Computing. I found it due to &lt;a href=&quot;http://cstheory.stackexchange.com/questions/4074/what-lecture-notes-should-everyone-read&quot;&gt;What Lecture Notes Should Everyone Read?&lt;/a&gt; on TCS.&lt;/p&gt;&#xA;" OwnerUserId="4811" LastActivityDate="2011-01-15T21:20:49.797" CommunityOwnedDate="2011-01-15T23:06:40.990" CommentCount="1" />
  <row Id="17609" PostTypeId="1" AcceptedAnswerId="17612" CreationDate="2011-01-15T21:29:47.310" Score="1" ViewCount="116" Body="&lt;p&gt;What would be the formula to estimate the rate of failure of some test as a percentage chance of failure from the number of runs of the test until the first failure was seen?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, consider 0 to mean failure and 1 to me success, the following are possible samples from which each should have an estimated failure rate:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;0 (failed on first try, I would estimate failure rate to be 100%)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;11110 (failed on firth try, so answer is something less than around 20% failure rate)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1111111110 (failed on tenth try, so answer is something less than around 10% failure rate)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Working the other way around, imagine the answer was a 10% failure rate, on average, what test will the first failure be seen at?  Perhaps it is (9/10)^n &amp;lt; 5/10?  What value of n do we reach a 50% chance of failure? Five tries is 59% chance of success, six tries 53%, and seven tries is 48%. Therefore, on average I think we will fail most likely on the seventh trial.  So what would be formula to work backwards from 7 to get ~10%?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So would the answer be?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1 minus the nth root of .5) where n is the first trial to fail&lt;/p&gt;&#xA;" OwnerUserId="5886" LastEditorUserId="5886" LastEditorDisplayName="" LastEditDate="2011-01-15T22:12:34.887" LastActivityDate="2011-01-17T23:16:20.053" Title="How to estimate failure probability from count until first failure?" Tags="&lt;probability&gt;&lt;statistics&gt;&lt;estimation-theory&gt;" AnswerCount="3" />
  <row Id="17610" PostTypeId="2" ParentId="668" CreationDate="2011-01-15T21:37:11.303" Score="2" ViewCount="" Body="&lt;p&gt;In addition to the answers, above, the determinant is a function from the set of set of square matrices into the real numbers that $preserves$ the operation of multiplication:&#xA;\begin{equation}\det(AB) = \det(A)\det(B) \end{equation}&#xA;and so it carries $some$ information about square matrices into the much more familiar set of real numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some examples: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The determinant function maps the identity matrix $I$ to the identity element of the real numbers ($\det(I) = 1$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which real number does $not$ have a multiplicative inverse?  The number 0.  So which square matrices do $not$ have multiplicative inverses? Those which are mapped to 0 by the determinant function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the determinant of the inverse of a matrix?  The inverse of the determinant, of course.  (Etc.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This &quot;operation preserving&quot; property of the determinant explains some of the value of the determinant function and provides a certain level of &quot;intuition&quot; for me in working with matrices.&lt;/p&gt;&#xA;" OwnerUserId="5680" LastActivityDate="2011-01-15T21:37:11.303" />
  <row Id="17611" PostTypeId="2" ParentId="17609" CreationDate="2011-01-15T21:42:36.883" Score="0" ViewCount="" Body="&lt;p&gt;You could probably use the &lt;a href=&quot;http://en.wikipedia.org/wiki/Nelson%E2%80%93Aalen_estimator&quot; rel=&quot;nofollow&quot;&gt;Nelson-Aelen estimator&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2011-01-15T21:42:36.883" CommentCount="1" />
  <row Id="17612" PostTypeId="2" ParentId="17609" CreationDate="2011-01-15T21:47:02.290" Score="1" ViewCount="" Body="&lt;p&gt;If I understand correctly, what you have is a geometric distribution in your experiment. So what you're looking for might be the maximum likelihood parameter estimation (taken from &lt;a href=&quot;http://en.wikipedia.org/wiki/Geometric_distribution&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Geometric_distribution&lt;/a&gt;):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \hat p = (\frac{1}{n} \sum_{i=1}^{n} k_i)^{-1}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You get the $k_i$'s if you do your experiment n times, $k_i$ being the number of times until your test fails. So $\hat p$ is just one over the average number of trials until your test fails.&lt;/p&gt;&#xA;" OwnerUserId="5798" LastActivityDate="2011-01-15T21:47:02.290" CommentCount="2" />
  <row Id="17613" PostTypeId="2" ParentId="17600" CreationDate="2011-01-15T21:51:28.750" Score="5" ViewCount="" Body="&lt;p&gt;Write $J(\chi,\lambda) = \sum_{t \in \mathbb F_p} \chi(1-t)\lambda(t),$&#xA;so that &#xA;$$\sum_{\lambda} J(\chi,\lambda) = \sum_t \chi(1-t)\sum_{\lambda} &#xD;&#xA;\lambda(t).$$&#xA;Now ask your self: what is the value of $\sum_{\lambda}\lambda(t)$, when $\lambda$ ranges&#xA;over all characters of order dividing $m$?  If you sort this out, you will&#xA;have answered your question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(One way to think about is is that the group of $\lambda$ with $\lambda^m = id$ is a subgroup of the full character group, so it is dual to a &lt;em&gt;quotient&lt;/em&gt; of $\mathbb F_p^{\times}$.  What is this quotient group explicitly?)&lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2011-01-15T21:51:28.750" />
  <row Id="17614" PostTypeId="2" ParentId="12591" CreationDate="2011-01-15T21:54:08.013" Score="0" ViewCount="" Body="&lt;p&gt;You should distinguish white noise and Brownian motion. Let $\xi_t$ be $\mathcal{N}(0,1)$ white noise, then if&#xA;$$&#xD;&#xA;dy_t = \xi_t dt&#xD;&#xA;$$&#xA;then $y_t\equiv$const (from the Law of Large Numbers).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact it's better to obtain ODE from SDE, vice versa can be done in the following way: you write perturbed equation&#xA;$$&#xD;&#xA;dX_t = (aX_t + b)dt+\sigma(t,X_t)dB_t&#xD;&#xA;$$&#xA;where $B_t$ is a Brownian motion ( = Wiener process). If you put $\sigma\equiv 0$ then &#xA;$$&#xD;&#xA;dx_t = (ax_t+b)dt&#xD;&#xA;$$&#xA;Also, for arbitrary $\sigma$ it's true for the expected value of the perturbed system $m_t = \mathrm{E}[x_t]$ we have&#xA;$$&#xD;&#xA;dm_t = (am_t+b)dt.&#xD;&#xA;$$&#xA;You can choose any $\sigma$ you like - for example constant, linear in $X_t$, linear in time and so on. I suggest to put $\sigma$ independent of time to leave the system autonomous.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S. You can read in Oksendal that for the white noise $\xi_t$ it's not interesting to consider process with increments $\xi_t dt$ because of its triviality. In fact it's more &quot;correct&quot; to use agreement $dB_t = \xi_t\sqrt{dt}$.&lt;/p&gt;&#xA;" OwnerUserId="5887" LastActivityDate="2011-01-15T21:54:08.013" />
  <row Id="17615" PostTypeId="1" CreationDate="2011-01-15T21:55:24.480" Score="0" ViewCount="42" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have got discrete time model of the point x movement to point y stated as:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;z = x + s * ( (y - x) / (||(y-x)||) )&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;where &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;z - is new position&#xA;s- is some offset for example 0,04&#xA;|| y - x || - euclidean norm operator&#xA;z,x,y are points in some m-dimensional R space&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;How to use this model (how to calculate) in&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;two dimensional real space where&#xA;point x (1, 1) moves to point y (4,&#xA;2) with s 0,04 &lt;/li&gt;&#xA;&lt;li&gt;three dimensional&#xA;real space where point x(1, 1, 1)&#xA;moves to point y(3, 2, 3) with s&#xA;0,04&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="5888" LastActivityDate="2011-05-16T00:57:58.350" Title="How to use model of point movement" Tags="&lt;geometry&gt;&lt;euclidean-geometry&gt;&lt;norm&gt;" AnswerCount="1" />
  <row Id="17616" PostTypeId="2" ParentId="17573" CreationDate="2011-01-15T21:56:07.723" Score="1" ViewCount="" Body="&lt;p&gt;For any locally compact abelian group $G$, the monoidal category of continuous unitary representations of $G$ is generated by the characters $G \to \mathbb{C}$, e.g. the Pontrjagin dual group $\hat{G}$, with tensor product corresponding to the group operation and duals corresponding to the inverse (and the trivial representation corresponding to the identity).  Prometheus's example is the case $G = \hat{G} = \mathbb{Z}/2\mathbb{Z}$.  &lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2011-01-15T21:56:07.723" CommentCount="2" />
  <row Id="17617" PostTypeId="2" ParentId="9209" CreationDate="2011-01-15T22:00:27.077" Score="1" ViewCount="" Body="&lt;p&gt;It's strange that it's mentioned the same space. In fact random process is a collection $(X_t|t\in \mathcal{T})$ where $X_t\in \mathbb{X}_t$ and state spaces can be different. As I remember sometimes notion &quot;stochastic process&quot; for the cases when all state spaces are the same. You can find a lot of literature on the random process with changing state spaces.&lt;/p&gt;&#xA;" OwnerUserId="5887" LastActivityDate="2011-01-15T22:00:27.077" />
  <row Id="17618" PostTypeId="2" ParentId="16583" CreationDate="2011-01-15T22:15:57.137" Score="-2" ViewCount="" Body="&lt;p&gt;I know the lottery keeps no memory of past games.&#xA;Although, I thought playing the same numbers each time would increase chance of a win. Why because each time the lotto does not pick your set of numbers your chance goes up for the next. At some point your number will come up. Is this theory sound at all? I guess what I'm saying is my numbers are a constant with the lotto being a variable. Keeping it the same is better then two variables always random. Like flipping coin in the air letting it hit the floor. It will change in the air many times (the variable) but it will always hit the floor (the constant). &lt;/p&gt;&#xA;" OwnerUserId="5890" LastEditorUserId="5890" LastEditorDisplayName="" LastEditDate="2011-01-15T22:51:33.583" LastActivityDate="2011-01-15T22:51:33.583" CommentCount="4" />
  <row Id="17619" PostTypeId="2" ParentId="17615" CreationDate="2011-01-15T22:19:05.713" Score="0" ViewCount="" Body="&lt;p&gt;You would plug in the values into the model. For (1), your coordinates are of the form $(x,y)$. For (2), your coordinates are of the form $(x,y,k)$. Suppose $x = (x_1, x_2)$ and $y = (y_1, y_2)$. Then $||y-x|| = \sqrt{(y_1-x_1)^2+ (y_2-x_2)^2}$. Likewise, if $x = (x_1, x_2, x_3)$ and $y = (y_1, y_2, y_3)$, then $||y-x|| = \sqrt{(y_1-x_1)^2+ (y_2-x_2)^2+ (y_3-x_3)^2}$.   &lt;/p&gt;&#xA;" OwnerUserId="2548" LastEditorUserId="2548" LastEditorDisplayName="" LastEditDate="2011-01-15T22:26:03.210" LastActivityDate="2011-01-15T22:26:03.210" CommentCount="2" />
  <row Id="17620" PostTypeId="1" CreationDate="2011-01-15T22:23:20.093" Score="0" ViewCount="85" Body="&lt;p&gt;Is every commutative ring a subdirect product of local rings?  (By which I mean is every commutative ring isomorphic to a subring of a direct product of local rings?)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I feel like this is true and that it's not that hard to prove, but I've never actually seen this stated as a theorem, so I'm worried that I'm confused about something basic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Since MattE mentioned Spec A in comments, let me add some detail to just make my confusion completely specific.  Let A be a ring.  You can define a scheme S that's the disjoint union of points, one point for each prime ideal of A, and then attach to each point the localization of A at that prime ideal.  You can define a surjection from S to A.  So if S = Spec B, where B is the direct product of all of the localizations, then we'd be done.  But it's not &amp;mdash; the direct product has additional points that come from &lt;a href=&quot;http://en.wikipedia.org/wiki/Ultraproduct&quot; rel=&quot;nofollow&quot;&gt;ultraproducts&lt;/a&gt; of the residue fields.&lt;/p&gt;&#xA;" OwnerUserId="4880" LastEditorUserId="4880" LastEditorDisplayName="" LastEditDate="2011-01-16T11:59:22.893" LastActivityDate="2011-01-16T11:59:22.893" Title="Commutative Ring as Subdirect Product" Tags="&lt;abstract-algebra&gt;&lt;commutative-algebra&gt;" CommentCount="4" />
  <row Id="17621" PostTypeId="1" AcceptedAnswerId="17622" CreationDate="2011-01-15T22:34:46.213" Score="1" ViewCount="62" Body="&lt;p&gt;Hello everyone,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I just wanted to clear this up, because i think a question I'm attempting may be somewhat badly written. It talks about a set of functions being 'bounded pointwise by 1' - but isn't that just 'bounded by 1'? I don't understand in what sense it is bounded pointwise by a single value, because surely as 1 is constant, it is uniformly bounded by 1? Perhaps the question is just confusingly phrased.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for the help - M.&lt;/p&gt;&#xA;" OwnerUserId="5892" LastActivityDate="2011-01-15T23:03:34.800" Title="Very easy clarification on a question - 'bounded pointwise'" Tags="&lt;homework&gt;&lt;analysis&gt;" AnswerCount="2" />
  <row Id="17622" PostTypeId="2" ParentId="17621" CreationDate="2011-01-15T22:39:11.170" Score="1" ViewCount="" Body="&lt;p&gt;&quot;Bounded pointwise by 1&quot; and &quot;uniformly bounded by 1&quot; mean exactly the same thing.  Perhaps it is written this way for emphasis.  You would have to ask the author.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2011-01-15T22:39:11.170" CommentCount="1" />
  <row Id="17624" PostTypeId="2" ParentId="14530" CreationDate="2011-01-15T22:52:20.513" Score="3" ViewCount="" Body="&lt;p&gt;The problem with expecting the mean of 90°, −90°, and 40° to be (90°−90°+40°)/3 = 13.33° is that you would then expect the mean of 10° and 350° to be (10°+350°)/2 = 180°, and not 0° which is the more reasonable answer. It only gets worse when you have more than two angles (What is the mean of 340°, 350°, 360°, 10°, and 20°? What about 340°, 350°, 0°, 10°, and 20°?). Essentially, what you're doing there is equivalent to setting $z_n = e^{i\theta_n}$ and computing $$\bar z = (z_1 z_2 \cdots z_N)^{1/N},$$ and the problem is of course that it's not obvious &lt;em&gt;a priori&lt;/em&gt; which of the $N$ possible roots of that equation is the right one, if any.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &quot;circular mean&quot; definition is not so bad. In fact, it corresponds to the point which minimizes the sum of its squared distances to the points corresponding to the data, $$\bar z = \underset{\lvert z \rvert = 1}{\arg\min} \sum_{n=1}^N \lvert z - z_n \rvert^2.$$ So this &lt;em&gt;is&lt;/em&gt; almost the same as the formula you like to use; you only have to define the &quot;distance&quot; between angles as the distance between the corresponding points on the unit circle. That is, $d(\theta, \phi) = \sqrt{2 - 2\cos(\theta-\phi)} = 2 \sin(\lvert\theta-\phi\rvert/2)$. This metric is close to $\lvert\theta-\phi\rvert$ when $\theta$ and $\phi$ are close, and has the advantage of being really easy to find the solution to.&lt;/p&gt;&#xA;" OwnerUserId="856" LastActivityDate="2011-01-15T22:52:20.513" CommentCount="1" />
  <row Id="17625" PostTypeId="1" AcceptedAnswerId="17630" CreationDate="2011-01-15T22:56:23.670" Score="3" ViewCount="249" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have to prove the triangle inequality&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(|x_1 - z_1|^p + |x_2 - z_2|^p)^{1/p} \leq (|x_1 - y_1|^p + |x_2 - y_2|^p)^{1/p} + (|y_1 - z_1|^p + |y_2 - z_2|^p)^{1/p}$ for $p \geq 1$ on $\mathbb{R}^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Two things are confusing me:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) How we manipulate the expression, seeing as we can't use the standard triangle inequality $|x - z| \leq |x - y| + |y - z|$. I am used to using a $|x - z| = |x - z + y - y | &amp;lt; |x - y| + |y - z|$ method, however is this still valid given the powers of $p$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Why this is an invalid statement for $0 &amp;lt; p &amp;lt; 1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sorry if these questions are a little basic, I'm just looking for a little guidance not a solution. Any help would be appreciated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks, Ash.&lt;/p&gt;&#xA;" OwnerUserId="1763" LastEditorUserId="1763" LastEditorDisplayName="" LastEditDate="2011-01-15T23:38:55.973" LastActivityDate="2011-01-16T02:42:01.460" Title="Proof of Triangle Inequality on $(\mathbb{R}^n, d_p)$" Tags="&lt;inequality&gt;&lt;metric-spaces&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="17626" PostTypeId="2" ParentId="17621" CreationDate="2011-01-15T23:03:34.800" Score="0" ViewCount="" Body="&lt;p&gt;If some norm were mentioned, the functions could be bounded in norm (and this does not necessarily mean pointwise bounded). Likely, however, the author is just repeating himself at the risk of being repetitive.  &lt;/p&gt;&#xA;" OwnerUserId="4154" LastActivityDate="2011-01-15T23:03:34.800" />
  <row Id="17627" PostTypeId="1" AcceptedAnswerId="17629" CreationDate="2011-01-15T23:13:46.397" Score="4" ViewCount="200" Body="&lt;p&gt;Essentially what the title says - where to me a Hilbert space is a complete (Hermitian) inner product space, am I safe to assume every such real Hilbert space is of uncountable dimension over $\mathbb{R}$, or is there a countable-dimension example?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks a lot :)&lt;/p&gt;&#xA;" OwnerUserId="5473" LastActivityDate="2011-01-16T01:03:12.607" Title="Does there exist a real Hilbert space with countably infinite dimension as a vector space over R?" Tags="&lt;real-analysis&gt;&lt;analysis&gt;&lt;hilbert-spaces&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="4" />
  <row Id="17628" PostTypeId="2" ParentId="16583" CreationDate="2011-01-15T23:17:16.717" Score="-2" ViewCount="" Body="&lt;p&gt;Your right with the two sided coin under static conditions.&#xA;If you had a seven sided object would that logic still apply.&#xA;Plus this a game of fantasy so until you check your numbers your a winner. So, play For fun not investment. If your green shirt is lucky to you, then put it on before you play. Your chance of winning is better if you play then if you don't.&lt;/p&gt;&#xA;" OwnerUserId="5890" LastActivityDate="2011-01-15T23:17:16.717" CommentCount="1" />
  <row Id="17629" PostTypeId="2" ParentId="17627" CreationDate="2011-01-15T23:17:16.770" Score="10" ViewCount="" Body="&lt;p&gt;An infinite dimensional (real) Hilbert space has dimension at least $\mathfrak{c}=2^{\aleph_0}=|\mathbb{R}|$ as a vector space.  One way to see this is by taking an orthornormal sequence $e_1,e_2,\ldots$, and considering the linearly independent set $\{\sum_{k=1}^\infty t^ke_k:0\lt t\lt 1\}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same fact extends to Banach spaces, but there orthogonality cannot be used to write so short of a proof.  A proof is given in &lt;a href=&quot;http://www.jstor.org/stable/2318458&quot; rel=&quot;nofollow&quot;&gt;this short article&lt;/a&gt; by Lacey.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To just see that the dimension cannot be countable, you could use &lt;a href=&quot;http://eom.springer.de/b/b015070.htm&quot; rel=&quot;nofollow&quot;&gt;Baire's theorem&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For more on this in the Hilbert space case, see &lt;a href=&quot;http://books.google.com/books?id=S57XLkgbf0oC&amp;amp;lpg=PP1&amp;amp;dq=hilbert%20space%20problem%20book%20halmos&amp;amp;pg=PA6#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;Problem 7&lt;/a&gt; of Halmos's &lt;em&gt;Hilbert space problem book&lt;/em&gt;.  (It is assumed there that the Hilbert spaces are complex instead of real, but this does not affect your question.)&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1543" LastEditorDisplayName="" LastEditDate="2011-01-16T01:03:12.607" LastActivityDate="2011-01-16T01:03:12.607" />
  <row Id="17630" PostTypeId="2" ParentId="17625" CreationDate="2011-01-15T23:21:31.220" Score="4" ViewCount="" Body="&lt;p&gt;This is where &lt;a href=&quot;http://en.wikipedia.org/wiki/Minkowski_inequality&quot; rel=&quot;nofollow&quot;&gt;Minkowski's Inequality&lt;/a&gt; comes in handy. See &lt;a href=&quot;http://math.stackexchange.com/questions/16391/minkowskis-inequality&quot;&gt;this&lt;/a&gt; (a question about Minkowski's inequality which uses Holder's inequality in the proof).&lt;/p&gt;&#xA;" OwnerUserId="2548" LastEditorUserId="2548" LastEditorDisplayName="" LastEditDate="2011-01-15T23:41:21.540" LastActivityDate="2011-01-15T23:41:21.540" CommentCount="2" />
  <row Id="17631" PostTypeId="2" ParentId="6734" CreationDate="2011-01-15T23:25:07.550" Score="0" ViewCount="" Body="&lt;p&gt;Suppose you have a vector $v$ for which is nonzero.  Then $v \not= 2v$ but $M(2v) = 2M(v)$ and $M(v)$ are not linearly independent.  Conditions B-E are all equivalent. &lt;/p&gt;&#xA;" OwnerUserId="4154" LastActivityDate="2011-01-15T23:25:07.550" />
  <row Id="17632" PostTypeId="1" CreationDate="2011-01-15T23:31:57.023" Score="0" ViewCount="224" Body="&lt;p&gt;Hello mathematicians of the stack exchange community.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Did any of you self-study extra material during undergrad?&lt;/p&gt;&#xA;" OwnerUserId="5563" LastActivityDate="2011-01-15T23:31:57.023" ClosedDate="2011-01-15T23:41:56.547" Title="Do Mathematicians Self-study graduate level-material while they are in undergrad?" Tags="&lt;soft-question&gt;" CommentCount="3" />
  <row Id="17633" PostTypeId="2" ParentId="17575" CreationDate="2011-01-16T00:15:19.827" Score="0" ViewCount="" Body="&lt;p&gt;May I tell you guys my silly solution to this problem? Only if you won't laugh at me! I thought I need to explore this space so I want to randomly sample it (the space of sensor readings). I know what the correlation matrix should look like and I know that the sensor readings are coming from a Gaussian distribution. So I generated a random N by M matrix and started tweaking the values in small steps. and check if each change moves the matriv toward the target or away and kept the changes toward the target. So I choose a random cell in this matrix and increase it by 10% and calculate the correlation matrix and compare it to the target correlation matrix the difference is smaller than what it was before the 10% increase I keep the change and move to next randomly selected cell and continue until I get close enough to the target correlation matrix.  This method, although it is silly, works well and I can get different samples of the sensor reading space.  What you guys think?! In practice I am working on rather large matrices like N = 8000, M = 1000  &lt;/p&gt;&#xA;" OwnerUserId="5875" LastActivityDate="2011-01-16T00:15:19.827" CommentCount="2" />
  <row Id="17634" PostTypeId="1" CreationDate="2011-01-16T01:27:39.963" Score="5" ViewCount="308" Body="&lt;p&gt;I am attempting to integrate a Bessel function of the first kind multiplied by a linear term: $\int xJ_n(x)\mathrm dx$  The textbooks I have open in front of me are not useful (Boas, Arfken, various Schaum's) for this problem.  I would like to do this by hand.  Is it possible?  I have had no luck with expanding out $J_n(x)$ and integrating term by term, as I cannot collect them into something nice at the end.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If possible and I just need to try harder (i.e. other methods or leaving it alone for a few days and coming back to it) that is useful information.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks to anyone with a clue.&lt;/p&gt;&#xA;" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2011-04-19T12:23:08.227" LastActivityDate="2011-04-26T01:05:28.523" Title="How do you integrate a Bessel function? I don't want to memorize answers or use a computer, is this possible?" Tags="&lt;integral&gt;&lt;special-functions&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="17635" PostTypeId="2" ParentId="17634" CreationDate="2011-01-16T01:37:18.580" Score="2" ViewCount="" Body="&lt;p&gt;I would do it numerically. Also for x not too large, you could use the power series expansion, but this will run into convergence issues before x gets too high. Numerical integration can be done to high accuracy without too much computation using Gaussian quadrature. Good luck&#xA;looking for an analytic solution (although maybe one exists?). Possibly you can use Bessel's equation, and by substituting your integral you can derive a new differential equation, but I think you'd be very lucky if this allowed an analytic solution. But, a numerical solution should be straightforward (at least for fixed N).&lt;/p&gt;&#xA;" LastActivityDate="2011-01-16T01:48:49.197" />
  <row Id="17636" PostTypeId="1" CreationDate="2011-01-16T01:57:46.110" Score="2" ViewCount="125" Body="&lt;p&gt;An object with known initial orientation and angular velocity $\omega_0$ is subjected to constant angular acceleration $\alpha$. Its final orientation at time $t$ is some rotation of its initial orientation. How can I find this rotation?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Angular velocity and acceleration are 3D vectors. Note, $\alpha$ is not necessarily parallel to $\omega_0$, so you can't just use the rotation $\omega_0 t+\alpha t^2$ about a common axis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: That should be $\omega_0 t+(\alpha /2) t^2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="5895" LastEditorUserId="2548" LastEditorDisplayName="" LastEditDate="2011-01-16T02:52:52.457" LastActivityDate="2011-04-16T19:40:32.447" Title="Constant angular acceleration" Tags="&lt;classical-mechanics&gt;" AnswerCount="2" CommentCount="6" />
  <row Id="17637" PostTypeId="2" ParentId="17634" CreationDate="2011-01-16T01:59:07.483" Score="2" ViewCount="" Body="&lt;p&gt;There are analytic answers involving hypergeometric functions (which you could look up e.g. on &lt;a href=&quot;http://functions.wolfram.com&quot; rel=&quot;nofollow&quot;&gt;http://functions.wolfram.com&lt;/a&gt;). I don't find hypergeometric functions very enlightening or intuitive, but at least they would give a closed-form answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At large arguments, Bessel functions are approximately proportional to cosines multiplied by $1/\sqrt{x}$, so maybe you could patch together an approximate answer for a definite integral using the power series expansion for small $x$ and the integrals of expressions involving a cosine at large $x$.&lt;/p&gt;&#xA;" OwnerUserId="5271" LastActivityDate="2011-01-16T01:59:07.483" />
  <row Id="17638" PostTypeId="1" CreationDate="2011-01-16T01:59:46.570" Score="4" ViewCount="201" Body="&lt;p&gt;Here is what our professor showed us in our linear algebra class to introduce the idea of determinants:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose we have an $n$-dimensional vector space $V$. Then we can create a function from $V^n$ to $\mathbb{R}$ called $vol$ (for &quot;volume&quot;) satisfying these properties:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$vol$ is multilinear&lt;br&gt;&#xA;$vol$ is alternating (i.e. if any two of $v_1, \ldots, v_n$ are the same, then $vol(v_1, \ldots, v_n) = 0$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From these two properties, we can see that if $e_1, \ldots, e_n$ is a basis of $V$, then the $vol$ function is completely defined by the value $vol(e_1, \ldots, e_n)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus if $T$ is a linear operator on $V$, the ratio:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\dfrac{vol(Te_1, \ldots, Te_n)}{vol(e_1, \ldots, e_n)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is the same for any (multilinear and alternating) $vol$ function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I am having trouble understanding why the ratio is also independent of the basis $e_1, \ldots, e_n$. This is what I am asking for help with. I can see that this invariance implies that intuitively, every $n$-parallelotope is stretched by the same amount by the operator $T$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Our professor then defined the determinant of $T$ as that ratio.)&lt;/p&gt;&#xA;" OwnerUserId="5894" LastActivityDate="2011-03-17T14:33:35.633" Title="Defining the determinant of linear transformations as multilinear alternating form." Tags="&lt;linear-algebra&gt;" AnswerCount="0" CommentCount="5" />
  <row Id="17642" PostTypeId="2" ParentId="17625" CreationDate="2011-01-16T02:42:01.460" Score="1" ViewCount="" Body="&lt;p&gt;(1) It is not in general true that $|x+y|^p \leq |x|^p + |y|^p$ when $p\gt 1$: for example, take $x=y=1$, $p=2$. Then $|x+y|^p = |1+1|^2 = 4$, whereas $|x|^p + |y|^p = 1^2 + 1^2 = 2$. So, no, you cannot use such a &quot;fact&quot; for the proof, because it's not true. You really do need to prove the inequality that is given to you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $x_1=z_1$ and $x_2=z_2$, there is nothing to do, so we may assume that the left hand side is nonzero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you want to show is essentially Mikowski's inequality, much along the lines of the answer Trevor has linked to.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As to why there is such an essential difference between the case $p\geq 1$ and the case $0\lt p\lt 1$, the reason is that while the function $y=x^p$ is convex when $p\gt 1$ (and technically also for $x=1$) it is not so for $0\lt p\lt 1$. This is what hides behind the problems for $0\lt p\lt 1$. &lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2011-01-16T02:42:01.460" />
  <row Id="17643" PostTypeId="1" AcceptedAnswerId="17660" CreationDate="2011-01-16T02:48:40.160" Score="3" ViewCount="256" Body="&lt;p&gt;The notion that the union of two finite sets is again finite is something I took as intuitively true for quite a while. A proof using arithmetic is relatively straight forward.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose $A$ and $B$ are finite sets, so let $m=|A|$ and $n=|B|$. Now $A\cup B=A\cup(B-A)$, and since $B-A$ is a subset of a finite set, it is also finite (I have already seen this statement proven), so let $k=|B-A|$. Then $|A\cup B|=m+k=m+_\omega k\in\omega$, so $A\cup B\approx m+_\omega k$, and thus finite.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've read that it is possible to prove this without any use of arithmetic, but I don't quite see how, as arithmetic seems too essential in showing a set is equinumerous to a natural number. Does anyone have a reference to or a reproduction of such a proof?, I'm curious to see. Thank you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: I was unaware of other definitions of what it means to be finite. I consider a set $A$ finite if there is a bijection $f\colon A\to n$ for some natural number $n$. For arithmetic, I'm trying to refrain from using cardinal arithmetic or the arithmetic of natural numbers, so no $+_\omega$, as was used in the other proof I gave. Also, my definition of natural number is the recursive definition that $0=\emptyset$, and $n^+=n\cup\{n\}$.&lt;/p&gt;&#xA;" OwnerUserId="1583" LastEditorUserId="1583" LastEditorDisplayName="" LastEditDate="2011-01-16T05:52:47.527" LastActivityDate="2011-01-16T13:40:45.583" Title="How can one prove that the union of two finite sets is again finite, without the use of arithmetic?" Tags="&lt;elementary-set-theory&gt;" AnswerCount="4" CommentCount="7" />
  <row Id="17645" PostTypeId="2" ParentId="17636" CreationDate="2011-01-16T02:59:47.833" Score="1" ViewCount="" Body="&lt;p&gt;I assume you mean $\vec{\omega}_0$ and $\vec{\alpha}$ as vector quantities. Unlike rotations, angular speeds and accelerations are vectors that can be added like&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \mathrm{d}\vec{\omega} = \vec{\alpha}\,\mathrm{d}t $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suppose the end result in speed is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \vec{\omega} = \vec{\omega}_0 + \vec{\alpha}\,t $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now for the orientations things are more complicated. They do &lt;strong&gt;not&lt;/strong&gt; add up as vectors. What you have to do is decompose the orientation to three euler angles $\varphi$, $\psi$ and $\theta$ and integrate their derivatives to find the final orientation. If the 3x3 rotation matrix is $R=R_y(\varphi)\,R_x(\psi)\,R_z(\theta)$ then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \vec{\omega} = \hat{j}\dot{\varphi}+R_y(\varphi)\left(\hat{i}\dot{\psi}+R_x(\psi)\,\hat{k}\dot{\theta}\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is a result of the sequencial rotations about $y$, $x$ and $z$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The above can be expressed as $\vec{\omega}=\mathrm{J}\,\pmatrix{\varphi,\psi,\theta}\cdot\begin{bmatrix}\dot{\varphi} \\ \dot{\psi} \\ \dot{\theta}\end{bmatrix}$ with $\mathrm{J}$ the jacobian matrix.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next the acceleration needs to be decomposed by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\vec{\alpha}=\mathrm{J}\,\pmatrix{\varphi,\psi,\theta}\cdot\begin{bmatrix}\ddot{\varphi} \\ \ddot{\psi} \\ \ddot{\theta}\end{bmatrix}+\mathrm{\dot{J}}\,\pmatrix{\varphi,\psi,\theta}\cdot\begin{bmatrix}\dot{\varphi} \\ \dot{\psi} \\ \dot{\theta}\end{bmatrix}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and solved as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \begin{bmatrix}\ddot{\varphi} \\ \ddot{\psi} \\ \ddot{\theta}\end{bmatrix} = \vec{F}\pmatrix{\varphi,\psi,\theta,\dot{\varphi},\dot{\psi},\dot{\theta}} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and integrated (analytically not likely, but numerically).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Good luck!&lt;/p&gt;&#xA;" OwnerUserId="3301" LastActivityDate="2011-01-16T02:59:47.833" CommentCount="3" />
  <row Id="17646" PostTypeId="2" ParentId="17643" CreationDate="2011-01-16T03:01:31.457" Score="4" ViewCount="" Body="&lt;p&gt;If your definition of finite is based on a cardinality equal to a natural number, then perhaps in that sense you would never be free of arithmetic in any such proof.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However an alternative definition of finite could be used that would avoid any explicit reference to cardinal numbers and thus to arithmetic.  Such a definition is being ordered so that a set is simultaneously well-ordered in both directions, ie. that any nonempty subset has both a smallest and a greatest element (not necessarily different if the subset has only one element).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In ZFC this definition of finite is known to be equivalent to the usual one, ie. that a set has cardinality equal to a natural number.&lt;/p&gt;&#xA;" OwnerUserId="3111" LastActivityDate="2011-01-16T03:01:31.457" CommentCount="2" />
  <row Id="17647" PostTypeId="1" CreationDate="2011-01-16T03:06:50.143" Score="1" ViewCount="75" Body="&lt;p&gt;Today matematics are moving away from understanding basics &amp;amp; core methodology. Level of the difficulties are reducing. Questions are becoming more simpler. Hence I would like to see the difficulty level in years Ranging 1940-1970. These Question papers will help improve students to improve in understanding rather than making students to behave like CLERKS by solving repeatadle the similar typ questions not having intelligance in the Questions itself. &lt;/p&gt;&#xA;" OwnerUserId="5897" LastEditorUserId="2548" LastEditorDisplayName="" LastEditDate="2011-01-16T03:13:26.653" LastActivityDate="2011-01-16T03:13:26.653" Title="Where can I get the OLD (1940-1970) high school scholarship question papers?" Tags="&lt;soft-question&gt;&lt;high-school&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="17648" PostTypeId="2" ParentId="17647" CreationDate="2011-01-16T03:08:58.710" Score="1" ViewCount="" Body="&lt;p&gt;You can look at &lt;a href=&quot;http://www.imo-official.org/problems.aspx&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; (old IMO problems).&lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2011-01-16T03:08:58.710" CommentCount="2" />
  <row Id="17649" PostTypeId="1" AcceptedAnswerId="17659" CreationDate="2011-01-16T03:23:40.237" Score="4" ViewCount="320" Body="&lt;p&gt;I think singleton sets $\{x\}$ where $x$ is a member of $\mathbb{R}$ are both open and closed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Singleton sets are open because $\{x\}$ is a subset of itself. There are no points in the neighborhood of $x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to know singleton sets are closed or not.&lt;/p&gt;&#xA;" OwnerUserId="5049" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2011-01-16T06:04:50.550" LastActivityDate="2011-01-16T06:04:50.550" Title="Are Singleton sets in $\mathbb{R}$ both closed and open" Tags="&lt;topology&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="17650" PostTypeId="1" AcceptedAnswerId="17683" CreationDate="2011-01-16T03:24:36.650" Score="2" ViewCount="231" Body="&lt;p&gt;My professor and I have came to a disagreement. My problem is with question number 8. I am pretty sure that you have to use the Bayes'thm for the question. I have tried ONCE to convince him, but regrettably, I have failed. Here is the question(s). I have included the previous questions because some of the information (from previous questions) was used to solve the question 8. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;start&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For questions 5 and 6 use the following information: &#xA;Diabetes - physicians recommend that children with type-1 diabetes keep up with insulin shots to minimize the chance of long-term complications. in addition, some diabetes researchers have observed that growth rate of weight during adolescence among diabetic patients is affected by level of compliance with insulin therapy. suppose 12 year old type-1 diabetic boys who comply with their insulin shots have a weight gain over 1 year that is normal distributed, with mean=12 lbs and variance =12 lbs. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Q5) what is the probability that compliant type-1 diabetic 12 year old boys will gain at least 15 pounds over 1 year? no continuity correction. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Q6) repeat problem #5 but now use the continuity correction (unit of measure is one pound). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Q7) Conversely, 12 year old type-1 diabetic boys who do not take their insulin shots (non compliant) have a weight gain over 1 year that is normally distributed with a mean of 8 pounds and a variance of 12 pounds. what is the probability that these non compliant boys will gain at least 15 pounds over 1 year? no continuity correction. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Q8) for the following problem, notice how the continuity correction is already built into the question: it is generally assumed that 75% of type-1 diabetics comply with their insulin regimen. suppose that a 12 year old type-1 diabetic boy comes to clinic and shows a 5lb weight gain over 1 year. (actually because of measurement error, assume this is an actual weight gain from 4.5 to 5.5 lbs). the boy claims to be taking his insulin medication. what is the probability that he is telling the truth? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;end&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I will jump right into the question. The question asked what is the P(he is telling the truth) i.e. pr(taking medication | weight gain of 5lbs). To me, this is unmistakably conditional. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;find P(A|B) where &#xA;A= taking medication&#xA;B= weight gain of 5lbs. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bayes' thm states... &#xA;P(A|B) = P(B|A)P(A) / P(B)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Find P(B|A) i.e. what is the probability that he gains weight given he is taking med? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given: mu=12 theta= root 12. Using the continuity correction...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;area: min to 5.5 z = (x-mu)/theta = (5.5-12)/root 12 = -1.88 &amp;lt;-&gt; A1 = 0.0301&lt;/p&gt;&#xA;&#xA;&lt;p&gt;area: min to 4.5 z = (x-mu)/theta = (4.5-12)/root 12 = -2.17 &amp;lt;-&gt; A2 = 0.0150&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A1-A2=0.0151 &lt;/p&gt;&#xA;&#xA;&lt;p&gt;P(B|A)=0.0151*&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Find P(A) i.e. what is the probability that he is taking med. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given: In question 8, this was given as 75% or 0.75. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Find P(B) i.e. what is the probability that he gains weight of 5lbs? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bayes' thm also states... &#xA;P(B) = P(A int B)+(not A int B) = P(B|A)P(A) + P(B|not A)P(not A) i.e. P(weight gain given taking med) + P(weight gain given not taking med) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We already found P(weight gain given taking med) = 0.0151* &lt;/p&gt;&#xA;&#xA;&lt;p&gt;so find P(weight gain given not taking med)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given:mu=8 theta=root 12. Using the continuity correction...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;area: min to 5.5 z = (x-mu)/theta = (5.5-8)/root 12 = -0.72 &amp;lt;-&gt; A1 = 0.2358&lt;/p&gt;&#xA;&#xA;&lt;p&gt;area: min to 4.5 z = (x-mu)/theta = (4.5-8)/root 12 = -1.01 &amp;lt;-&gt; A2 = 0.1562&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A1-A2=0.0796&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so, P(B) = 0.0151+0.0796&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Putting P(B|A), P(A), and P(B) together...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P(A|B) = P(B|A)P(A) / P(B) = (0.0151)(0.75)/(0.0151+0.0796) = 0.1196 = 11.96% &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The following is what my professor said about the question: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;start&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Jeff –&#xA;     As worded, it is not a conditional probability problem, therefore you do not use posterior probability.  I strongly believe that the correct answer is the area from min to 5.5 (answer is then 0.0301).  However most students found the area from 4.5 to 5.5 (answer is 0.015) because I mentioned the continuity correction and they were confused so I let this answer stand.  In reality, from min to 5.5 would be non-compliance.  In fact and specific number, even if compliant (e.g., Pr(X = 17)) would be a SMALL number.  Thus the probability really should be 5 or less or Min to 5.5 with the continuity correction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;end&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I think he said was to find the area under the normal distribution curve (as in question 5) from min to 5.5, but that could not possibly be the final answer because the area under the curve is the probability that the boy gains 5lbs., assuming that he is compliant. I went to his office hour and asked him to explain himself. Unfortunately, I did not understand what he was saying.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I feel like I just took a crazy pill. I don't see any logical mistakes in my part. If there is any algebraic mistakes that I've made, please let me know. Thank you!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;my best, &#xA;Jeff Kwak&lt;/p&gt;&#xA;" OwnerUserId="5898" LastEditorUserId="5898" LastEditorDisplayName="" LastEditDate="2011-01-16T04:35:30.043" LastActivityDate="2011-01-16T10:50:42.097" Title="What is the probability that the boy is telling the truth?" Tags="&lt;probability&gt;&lt;statistics&gt;" AnswerCount="2" CommentCount="5" />
  <row Id="17651" PostTypeId="1" AcceptedAnswerId="17655" CreationDate="2011-01-16T03:27:18.890" Score="2" ViewCount="121" Body="&lt;p&gt;p is a non-zero real number. The forms of &lt;a href=&quot;http://en.wikipedia.org/wiki/Generalized_mean&quot; rel=&quot;nofollow&quot;&gt;generalized mean with exponent p&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/Lp_space#.E2.84.93p_spaces&quot; rel=&quot;nofollow&quot;&gt;$l_p$ norm&lt;/a&gt; seem similar except whether to take mean before taking p-root. I am trying to compare them together.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can generalized mean with exponent p become a norm on the real sequence space?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given a sequence, generalized mean with exponent p is nondecreasing with respect to p. Does $l_p$ norm also non-decrease with p?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can someone explain why? Thanks in advance!&lt;/p&gt;&#xA;" OwnerUserId="3348" LastEditorUserId="3348" LastEditorDisplayName="" LastEditDate="2011-01-16T03:35:29.943" LastActivityDate="2011-01-16T06:14:51.423" Title="Trying to compare generalized mean and $l_p$ norm" Tags="&lt;sequences-and-series&gt;" AnswerCount="3" FavoriteCount="1" />
  <row Id="17652" PostTypeId="2" ParentId="17649" CreationDate="2011-01-16T03:27:55.163" Score="2" ViewCount="" Body="&lt;p&gt;It depends on what topology you are looking at. For $T_1$ spaces, singleton sets are always closed. So for the standard topology on $\mathbb{R}$, singleton sets are always closed.&lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2011-01-16T03:27:55.163" CommentCount="1" />
  <row Id="17653" PostTypeId="2" ParentId="17649" CreationDate="2011-01-16T03:35:17.333" Score="7" ViewCount="" Body="&lt;p&gt;If you are giving $\{x\}$ the subspace topology and asking whether $\{x\}$ is open in $\{x\}$ in this topology, the answer is yes.  There is only one possible topology on a one-point set, and it is discrete (and indiscrete).  However, if you are considering singletons as subsets of a larger topological space, this will depend on the properties of that space.  As Trevor indicates, the condition that &lt;a href=&quot;http://en.wikipedia.org/wiki/T1_space&quot; rel=&quot;nofollow&quot;&gt;points are closed&lt;/a&gt; is (equivalent to) the $T_1$ condition, and in particular is true in every metric space, including $\mathbb{R}$.  When $\{x\}$ is open in a space $X$, then $x$ is called an isolated point of $X$.  If all points are isolated points, then the topology is &lt;a href=&quot;http://en.wikipedia.org/wiki/Discrete_space&quot; rel=&quot;nofollow&quot;&gt;discrete&lt;/a&gt;.  In the real numbers, for example, there are no isolated points; every open set is a union of open intervals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In summary, if you are talking about the usual topology on the real line, then singleton sets are closed but not open.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2011-01-16T03:35:17.333" />
  <row Id="17655" PostTypeId="2" ParentId="17651" CreationDate="2011-01-16T03:57:55.917" Score="2" ViewCount="" Body="&lt;p&gt;If $1\leq p\leq\infty$, then yes, both are norms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The ordinary $\ell^p$ norm can decrease with $p$. E.g., Consider $(1,1)$ in $\mathbb{R}^2$. (And as Theo shows in his answer, the $\ell^p$ norm is always nonincreasing.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both norms can be seen as $L^p$ &lt;a href=&quot;http://en.wikipedia.org/wiki/Lp_space&quot; rel=&quot;nofollow&quot;&gt;norms&lt;/a&gt; on $\mathbb{R}^n$ considered as functions on an $n$-point &lt;a href=&quot;http://en.wikipedia.org/wiki/Measure_%28mathematics%29&quot; rel=&quot;nofollow&quot;&gt;measure space&lt;/a&gt;.  For the generalized mean, each point has measure $\frac{1}{n}$, so that the space is normalized to have measure $1$, making it a &quot;&lt;a href=&quot;http://en.wikipedia.org/wiki/Probability_space&quot; rel=&quot;nofollow&quot;&gt;probability space&lt;/a&gt;&quot;.  For the ordinary $\ell^p$ norm, each point has measure $1$.  It is true in general for probability spaces that the $L^p$ norm increases with $p$.  This almost never holds for spaces of measure greater than $1$, as can be seen by taking a characteristic function of a set of finite measure greater than $1$.  On the other hand, having the $L^p$ norm decrease with $p$ is also atypical, and happens only when there are no sets of positive measure less than $1$ (again consider a characteristic function).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since I now see that you want to consider infinite sequences, I'll say a few words on generalized means for that case.  For each sequence of weights $w_1,w_2,\ldots$ consisting of positive numbers summing to $1$, you can define the generalized mean with exponent $p$ by $(w_1|x_1|^p+w_2|x_2|^p+\cdots)^{1/p}$.  The set of all sequences $(x_1,x_2,\ldots)$ with finite mean forms the $L^p$ space of a countable measure space with total measure $1$, if you like thinking of it that way.  It will contain the ordinary $\ell^p$ space.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2011-01-16T06:14:51.423" LastActivityDate="2011-01-16T06:14:51.423" CommentCount="4" />
  <row Id="17656" PostTypeId="2" ParentId="17651" CreationDate="2011-01-16T03:58:32.363" Score="1" ViewCount="" Body="&lt;p&gt;The power mean is defined as (for $p \geq 1$): $$M_p(a_{1}, \dots, a_n) = \left(\frac{1}{n} \sum_{i=1}^{n} a_{i}^{p} \right)^{1/p}$$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The $\ell_p$ norm is defined as follows (for $p \geq 1$): $$ \left(\sum_{i=1}^{n} |x_i|^{p} \right)^{1/p}$$ If we let $x_i = \frac{a_i}{\sqrt[p]{n}}$ then we can view the generalized mean as a norm. &lt;/p&gt;&#xA;" OwnerUserId="2548" LastEditorUserId="2548" LastEditorDisplayName="" LastEditDate="2011-01-16T05:22:11.387" LastActivityDate="2011-01-16T05:22:11.387" CommentCount="3" />
  <row Id="17657" PostTypeId="2" ParentId="17651" CreationDate="2011-01-16T03:58:58.993" Score="2" ViewCount="" Body="&lt;p&gt;For $p \geq 1$ the generalized mean defines a norm, because it is the $\ell^{p}$-norm only up to a factor $\sqrt[p]{n}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if $p \lt 1$, the generalized mean (and also the $\ell^{p}$-expression) don't define a norm because the set $\|x\|_{p} \leq 1$ is not convex: if $x_{i} \geq 0$ and $y_{i} \geq 0$ for all $i$ then $\|x + y\|_{p} \geq \|x\|_{p} + \|y\|_{p}$!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $p \leq q$ then $\|x\|_{q} \leq \|x\|_{p}$. To see this, note that both sides of the inequality are invariant by multiplication with a positive real number, so we may take without loss of generality an $x$ with $\|x\|_{p} = 1$. Then $\|x\|_{q}^{q} = \sum_{j = 1}^{n} |x_{j}|^{q} \leq \sum_{j = 1}^{n} |x_{j}|^{p} = 1$ this is because for $t \leq 1$ and $p \leq q$ we have $t^{q} \leq  t^{p}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't understand what you ask about the sequence space.&lt;/p&gt;&#xA;" OwnerUserId="5363" LastEditorUserId="5363" LastEditorDisplayName="" LastEditDate="2011-01-16T04:36:42.147" LastActivityDate="2011-01-16T04:36:42.147" CommentCount="5" />
  <row Id="17658" PostTypeId="1" CreationDate="2011-01-16T04:08:23.690" Score="1" ViewCount="175" Body="&lt;p&gt;I'm taking a first course in discreet mathematics and there are  a lot of new rules/laws to remember.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is what is the best way to remember the rules of logic?&#xA;How would you remember the:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Absorption Laws &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Domination Laws&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Identity Laws&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there any well known mnemonics, like SOHCAHTOA in trigonometry?&lt;/p&gt;&#xA;" OwnerUserId="1040" LastActivityDate="2011-01-16T04:08:23.690" Title="How should I remember the rules of logic?" Tags="&lt;logic&gt;" CommentCount="7" />
  <row Id="17659" PostTypeId="2" ParentId="17649" CreationDate="2011-01-16T04:08:49.623" Score="9" ViewCount="" Body="&lt;p&gt;As has been noted, the notion of &quot;open&quot; and &quot;closed&quot; is not absolute, but depends on a &lt;em&gt;topology&lt;/em&gt;. So in order to answer your question one must first ask what topology you are considering.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A topological space is a pair, $(X,\tau)$, where $X$ is a nonempty set, and $\tau$ is a collection of subsets of $X$ such that:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\emptyset$ and $X$ are both elements of $\tau$;&lt;/li&gt;&#xA;&lt;li&gt;If $A$ and $B$ are elements of $\tau$, then $A\cap B$ is an element of $\tau$;&lt;/li&gt;&#xA;&lt;li&gt;If $\{A_i\}_{i\in I}$ is an arbitrary family of elements of $\tau$, then $\mathop{\cup}\limits_{i\in I}A_i$ is an element of $\tau$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The elements of $\tau$ are said to be &quot;open&quot; (in $X$, in the topology $\tau$), and a set $C\subseteq X$ is said to be &quot;closed&quot; if and only if $X-C\in\tau$ (that is, if the complement is open).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In $\mathbb{R}$, we can let $\tau$ be the collection of all subsets that are unions of open intervals; equivalently, a set $\mathcal{O}\subseteq\mathbb{R}$ is &lt;strong&gt;open&lt;/strong&gt; if and only if for every $x\in\mathcal{O}$ there exists $\epsilon\gt 0$ such that $(x-\epsilon,x+\epsilon)\subseteq\mathcal{O}$. You may want to convince yourself that the collection of all such sets satisfies the three conditions above, and hence makes $\mathbb{R}$ a topological space. This topology is what is called the &quot;usual&quot; (or &quot;metric&quot;) topology on $\mathbb{R}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are working inside of $\mathbb{R}$ with this topology, then singletons $\{x\}$ are certainly closed, because their complements are open: given any $a\in \mathbb{R}-\{x\}$, let $\epsilon=|a-x|$. Then $x\notin (a-\epsilon,a+\epsilon)$, so $(a-\epsilon,a+\epsilon)\subseteq \mathbb{R}-\{x\}$; hence $\mathbb{R}-\{x\}$ is open, so $\{x\}$ is closed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason you give for $\{x\}$ to be open does not really make sense. &lt;strong&gt;Every&lt;/strong&gt; set is a subset of itself, so if that argument were valid, every set would always be &quot;open&quot;; but we know this is not the case in &lt;em&gt;every&lt;/em&gt; topological space (certainly not in $\mathbb{R}$ with the &quot;usual topology&quot;). So that argument certainly does not work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So: is $\{x\}$ open in $\mathbb{R}$ in the usual topology? Well, $x\in\{x\}$. Does there exist an $\epsilon\gt 0$ such that $(x-\epsilon,x+\epsilon)\subseteq \{x\}$? If so, then congratulations, you have shown the set is open. If there is no such $\epsilon$, and you prove that, then congratulations, you have shown that $\{x\}$ is not open.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2011-01-16T04:08:49.623" />
  <row Id="17660" PostTypeId="2" ParentId="17643" CreationDate="2011-01-16T04:23:42.813" Score="4" ViewCount="" Body="&lt;p&gt;Suppose to start with that the sets are disjoint. Then prove that if the sets are both finite, and if you remove an element from one set and put it in the other, they are both still finite and their union is unchanged. By induction, you will eventually exhaust one of the finite sets, giving the union of a finite set and the empty set, which is therefore a finite set. Because the union was unchanged in each step, this is equal to the union of the original two sets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The proof has to be done by constructing bijections between the sets and natural numbers, transferring the element mapped to the last natural number in one set, and mapping it to the successor of the last natural number mapped to by the other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the case where the sets are not disjoint you can fix it in several ways, such as deleting them from one set without adding to the other, or proving that removing elements from a finite set always gives a finite result, or using your $B-A$ trick.&lt;/p&gt;&#xA;" OwnerUserId="5895" LastActivityDate="2011-01-16T04:23:42.813" CommentCount="1" />
  <row Id="17661" PostTypeId="1" AcceptedAnswerId="17662" CreationDate="2011-01-16T05:54:51.397" Score="0" ViewCount="260" Body="&lt;p&gt;Hello, I'm trying to help a friend of mine solve this problem. It's been like 10 years since I took calculus please help:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Find min/max values of $y = \sin x + \cos x$ on  $[0, 2\pi)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've got this much:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$f&amp;#39;(x) = \cos x - \sin x$. &lt;/li&gt;&#xA;&lt;li&gt;$0 = \cos x - \sin x$. &lt;/li&gt;&#xA;&lt;li&gt;$\sin x = \cos x$. &lt;/li&gt;&#xA;&lt;li&gt;$\tan x = 1$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Evaulate at end points and at the critical point:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$f(0) = 1$&lt;/li&gt;&#xA;&lt;li&gt;$f(2\pi) = 1$&lt;/li&gt;&#xA;&lt;li&gt;$\sin x = \cos x$ at $\pi/4$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Hence,&#xA;$f(x)$ has a minimum at $\pi/4$ and $f(x)$ has a maximum at $x = 0$ on $[0,2\pi)$ (Extreme Value Theorem) &lt;/p&gt;&#xA;" OwnerUserId="5899" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-16T20:48:02.607" LastActivityDate="2011-01-16T20:48:02.607" Title="min and max of trig function" Tags="&lt;calculus&gt;&lt;trigonometry&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="17662" PostTypeId="2" ParentId="17661" CreationDate="2011-01-16T05:59:12.400" Score="3" ViewCount="" Body="&lt;p&gt;No, you don't get $\tan x = 0$; you would get, at best, $\tan x = 1$ (remember that $\tan x = \frac{\sin x}{\cos x}$. If $\sin x$ and $\cos x$ are equal, then the quotient is equal to $1$, not to $0$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, you remembered that you want to find the critical points (points where the derivative is zero). That's good. What you also need to remember is that if you have a continuous function on a closed interval, then the maximum and the minimum will each be achieved at either a critical point or an endpoint.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here, you might as well work over $[0,2\pi]$ (the value at $2\pi$ is the same as the value at $0$). So the maximum and the minimum of $f(x)$ will be achieved either at $x=0$, or at a point where $f&amp;#39;(x)=0$, that is, a point where $\sin(x) = \cos(x)$ in the interval (better to work with these, since this equality does not depend on $\cos(x)\neq 0$, whereas $\tan(x)=1$ does; of course, it does not really matter here because if $\cos(x)=0$, then $\sin(x)\neq \cos(x)$). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the question is: for what points $x$, $0\leq x\leq 2\pi$, do you have $\sin(x)=\cos(x)$? There are two such points; once you have them, simply evaluate the original function at these points, and at $x=0$ (where it has the same value as at $x=2\pi$, which is why we were able to add $2\pi$ to the interval for simplicity). The largest value you get is the maximum, the smallest value you get is the minimum.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-16T06:50:30.480" LastActivityDate="2011-01-16T06:50:30.480" CommentCount="14" />
  <row Id="17663" PostTypeId="2" ParentId="17575" CreationDate="2011-01-16T06:00:42.463" Score="1" ViewCount="" Body="&lt;p&gt;What you want to do is called sampling from a multivariate distribution.  It's not hard to look up how to do this, although in your case if some of your variables are assumed normal and some Poisson you're going to have to calculate the joint PDF of your model yourself.  Rather than trying to explain the whole process here, I'll give you some things to read:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Multivariate_normal_distribution&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Multivariate_normal_distribution&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.statisticalengineering.com/bivariate_normal.htm&quot; rel=&quot;nofollow&quot;&gt;http://www.statisticalengineering.com/bivariate_normal.htm&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.springerlink.com/content/uq8810j473527617/&quot; rel=&quot;nofollow&quot;&gt;http://www.springerlink.com/content/uq8810j473527617/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3296" LastActivityDate="2011-01-16T06:00:42.463" CommentCount="4" />
  <row Id="17664" PostTypeId="1" CreationDate="2011-01-16T07:15:12.113" Score="2" ViewCount="157" Body="&lt;p&gt;one sphere with radius R,named big sphere,two point on it:a(longitude_a,latitue_a),b(longitude_b,latitude_b), dist(a,b)=r, a as center,r as radius,there is another sphere,named little sphere, now what is the circle equation of the intersect of the two spheres? &#xA;parameter equation of the circle: lon=f(xx),lat=g(xx)&lt;/p&gt;&#xA;" OwnerUserId="5900" LastActivityDate="2011-03-17T08:33:35.430" Title="the equation of a circle on sphere?" Tags="&lt;geometry&gt;&lt;circle&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" />
  <row Id="17666" PostTypeId="1" AcceptedAnswerId="17671" CreationDate="2011-01-16T07:21:29.493" Score="1" ViewCount="100" Body="&lt;p&gt;How does one integrate du / (a^2 + u^2)^(3/2)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The table of integrals here: &lt;a href=&quot;http://teachers.sduhsd.k12.ca.us/abrown/classes/CalculusC/IntegralTablesStewart.pdf&quot; rel=&quot;nofollow&quot;&gt;http://teachers.sduhsd.k12.ca.us/abrown/classes/CalculusC/IntegralTablesStewart.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gives it as: u / (a^2 * ( a^2 + u^2)^(1/2) ).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm getting back into calculus and very rusty. I'd like to be comfortable with some of the proofs behind various fundamental &quot;Table of Integrals&quot; integrals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Looking at it, the substitution rule seems like the method of choice. What is the strategy here for choosing a substitution? It has a form similar to many trigonometric integrals, but the final result seems to suggest that they're not necessary in this case.&lt;/p&gt;&#xA;" OwnerUserId="5901" LastActivityDate="2011-01-16T07:57:46.137" Title="Integrate Form du / (a^2 + u^2)^(3/2)" Tags="&lt;integral&gt;" AnswerCount="2" />
  <row Id="17667" PostTypeId="1" CreationDate="2011-01-16T07:32:48.130" Score="2" ViewCount="86" Body="&lt;p&gt;I understand that &lt;em&gt;Diophantine Analysis&lt;/em&gt; is an enormous field! Without first determining the solution set, suppose I'd like to calculate the number of non-negative integer solutions $(x,y,z)$ of &#xA;\begin{eqnarray}&#xA;ax^{p} + by^{q} \leq cz^{r}&#xA;\end{eqnarray}&#xA;for positive integers $a,b,c,p,q$ and $r$. What general methods are available for such enumeration of this or more general Diophantine inequalities, including more variables or nonlinear forms? (Is our understanding of such enumeration limited to a select set of very specific equation types?) If no general methods exist, I'm sure that there are methods of bounding the number of solutions from below and above with error that can be estimated. Which is the best known?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;References are welcome. Thanks!&lt;/p&gt;&#xA;" OwnerUserId="2720" LastEditorUserId="1778" LastEditorDisplayName="" LastEditDate="2011-02-24T15:47:42.463" LastActivityDate="2011-02-24T15:47:42.463" Title="Counting Solutions of Diophantine Inequalities" Tags="&lt;number-theory&gt;&lt;combinatorics&gt;&lt;reference-request&gt;&lt;diophantine-equations&gt;" CommentCount="2" FavoriteCount="1" />
  <row Id="17668" PostTypeId="2" ParentId="17650" CreationDate="2011-01-16T07:34:49.687" Score="3" ViewCount="" Body="&lt;p&gt;This question is fatally flawed. To answer it correctly, you would need to know the probability that the boy would lie given that he skipped his shots. The question seems to assume that this probability is $1$.&lt;/p&gt;&#xA;" OwnerUserId="1508" LastActivityDate="2011-01-16T07:34:49.687" />
  <row Id="17669" PostTypeId="2" ParentId="17575" CreationDate="2011-01-16T07:44:38.147" Score="0" ViewCount="" Body="&lt;p&gt;You could make your &quot;10%&quot; implementation faster by using gradient descent&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's an example of doing it for covariance matrix because it's easier.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have $k\times k$ covariance matrix C and you want to get $k \times n$ observation matrix $A$ that produces it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The task is to find $A$ such that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$A A&amp;#39; = n^2 C$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could start with a random guess for $A$ and try to minimize the sum of squared errors. Using Frobenius norm, we can write this objective as follows&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$J=\|A A&amp;#39;-n^2 C\|^2_F$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $D$ be our matrix of errors, ie $D=A A&amp;#39;-n^2 C$.&#xA;Checking with the &lt;a href=&quot;http://matrixcookbook.com/&quot; rel=&quot;nofollow&quot;&gt;Matrix Cookbook&lt;/a&gt; I get the following for the gradient&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\partial J}{\partial a_{iy}} \propto a_{iy} \sum_j D_{i,j}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, your update to data matrix for sensor $i$, observation $y$ should be proportional to the sum of errors in $i$th row of covariance matrix multiplied by the value of $i,y$'th observation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gradient descent step would be to update all weights at once. It might be more robust to update just the worst rows, ie calculate sum of errors for every row, update entries corresponding to the worst rows, then recalculate errors.&lt;/p&gt;&#xA;" OwnerUserId="998" LastActivityDate="2011-01-16T07:44:38.147" CommentCount="2" />
  <row Id="17670" PostTypeId="2" ParentId="17666" CreationDate="2011-01-16T07:50:08.220" Score="2" ViewCount="" Body="&lt;p&gt;Let $u = a \tan \theta$ and work from there. See &lt;a href=&quot;http://www.math.hmc.edu/calculus/tutorials/trig_substitution/&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; (page about trig substitutions).&lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2011-01-16T07:50:08.220" />
  <row Id="17671" PostTypeId="2" ParentId="17666" CreationDate="2011-01-16T07:57:46.137" Score="4" ViewCount="" Body="&lt;p&gt;A trigonometric substitution does indeed work. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We want to express $(a^2 + u^2)^{3/2}$ as something without square roots. We want to use some form of the Pythagorean trigonometric identity $\sin^2 x + \cos^2 x = 1$. Multiplying each side by $\frac{a^2}{\cos^2 x}$, we get $a^2 \tan^2 x + a^2 = a^2 \sec^2 x$, which is in the desired form. of (sum of two squares) = (something squared).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This suggests that we should use the substitution $u^2 = a^2 \tan^2 x$. Equivalently, we substitute $u = a \tan x$ and $du = a \sec^2 x dx$. Then&#xA;$$&#xD;&#xA;\int \frac{du}{(a^2 + u^2)^{3/2}} &#xD;&#xA;= \int \frac{a \sec^2 x \, dx}{(a^2 + a^2 \tan^2 x)^{3/2}}.&#xD;&#xA;$$&#xA;Applying the trigonometric identity considered above, this becomes&#xA;$$&#xD;&#xA;\int \frac{a \sec^2 x \, dx}{(a^2 \sec^2 x)^{3/2}}&#xD;&#xA;= \int \frac{dx}{a^2 \sec x} = \frac{1}{a^2} \int \cos x \, dx,&#xD;&#xA;$$&#xA;which can be easily integrated as&#xA;$$&#xD;&#xA;=\frac{1}{a^2} \sin x.&#xD;&#xA;$$&#xA;Since we set $u = a \tan x$, we substitute back $x = \tan^{-1} (\frac ua)$ to get that the answer is&#xA;$$&#xD;&#xA;=\frac{1}{a^2} \sin \tan^{-1} \frac{u}{a}.&#xD;&#xA;$$&#xA;Since $\sin \tan^{-1} z = \frac{z}{\sqrt{z^2 + 1}}$, this yields the desired result of &#xA;$$&#xD;&#xA;=\frac{u/a}{a^2 \sqrt{(u/a)^2 + 1}}&#xD;&#xA;= \frac{u}{a^2 (a^2 + u^2)^{1/2}}.&#xD;&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="30" LastActivityDate="2011-01-16T07:57:46.137" />
  <row Id="17672" PostTypeId="2" ParentId="17661" CreationDate="2011-01-16T08:03:16.350" Score="1" ViewCount="" Body="&lt;p&gt;An other thing you might try is to simplify the original function using trigonometric formulas. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Recall the addition formula for the sine function: &#xA;$$\sin(A+B)=\sin(A)\cdot\cos(B)+\sin(B)\cdot\cos(A).$$&#xA;Using this we see&#xA;$f(A)\cdot C=\sin(A)\cdot C+\cos(A)\cdot C=\sin(A+B)$, provided $\cos B =\sin B=C$. &#xA;Next, we have $$1=\sin^2 B+\cos^2 B=2C^2$$&#xA;hence we get $$f(A)=\sqrt{2}\sin(A+B) \quad \text{ or }\quad  f(A)=-\sqrt{2}\sin(A+B)$$&#xA;which shows that the maximal possible value of $f$ is $\sqrt{2}$ and the minimal possible value is $-\sqrt{2}$. (Looking at the &quot;circle-definition&quot; of sine and cosine it is easy to see where these values are attained -- on the line $y=x$).&lt;/p&gt;&#xA;" OwnerUserId="1154" LastEditorUserId="1154" LastEditorDisplayName="" LastEditDate="2011-01-16T17:03:06.167" LastActivityDate="2011-01-16T17:03:06.167" CommentCount="2" />
  <row Id="17673" PostTypeId="2" ParentId="17664" CreationDate="2011-01-16T08:05:53.543" Score="1" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://mathforum.org/library/drmath/view/64311.html&quot; rel=&quot;nofollow&quot;&gt;This&lt;/a&gt; (calculating the equation of the circle intersection between two spheres) will probably help.&lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2011-01-16T08:05:53.543" />
  <row Id="17674" PostTypeId="1" CreationDate="2011-01-16T08:16:22.223" Score="1" ViewCount="49" Body="&lt;p&gt;Suppose we have invariant of graph $G$ that tells us number of subgraphs with $i$ vertices and $j$ edges for every setting of $i$ and $j$. Is there a name for it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I searched for &quot;subgraph polynomial&quot;, but that comes up as something else.&lt;/p&gt;&#xA;" OwnerUserId="998" LastEditorUserId="998" LastEditorDisplayName="" LastEditDate="2011-01-17T22:11:48.650" LastActivityDate="2011-01-17T22:11:48.650" Title="Graph invariant that encodes number of subgraphs with $i$ vertices and $j$ edges" Tags="&lt;graph-theory&gt;&lt;terminology&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="17675" PostTypeId="2" ParentId="17674" CreationDate="2011-01-16T08:31:12.937" Score="1" ViewCount="" Body="&lt;p&gt;See &lt;a href=&quot;http://mathoverflow.net/questions/12310/counting-subgraphs-of-bipartite-graphs&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; (MO question that is similar). I don't think there is any name for it.&lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2011-01-16T08:31:12.937" />
  <row Id="17676" PostTypeId="2" ParentId="14630" CreationDate="2011-01-16T08:42:04.560" Score="1" ViewCount="" Body="&lt;p&gt;Let me say a couple of things which might help here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) The information you give here does not uniquely determine a joint distribution, and in fact information of this sort will never uniquely determine a joint distribution.  For instance, if I say that I have two variables X ~ N(0, 1) and Y ~ N(0, 1) with a given correlation between X and Y, it does not follow that (X, Y) follows a multivariate normal distribution -- there are infinitely many joint distributions with this property.  (This is in some part a consequence of the fact that covariance only gives us information up to the second order.)  You will have to make up a joint density with the desired properties; it will not be uniquely determined.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) There are different ways that covariance could be introduced into this model, and they will give the model different theoretical properties.  For instance, there could be heavy correlation between the Bernoulli variables that decide which distribution to sample from and no correlation between the actual samples.  Conversely the Bernoulli variables for X and Y could be totally uncorrelated, and some or all of the normal/binomial distributions could be correlated to one another.  One could also envision a mixture of both scenarios.  If you have any more fundamental information about your processes (which I'm assuming you do -- how else would you know that it's a mixture of normal and binomial samples?) then you should use that to decide here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The main problem here is coming up with a joint PDF -- you are in reality building a mathematical model here, and that's as much art as science.  Once you have the joint distribution, there are well-established ways to sample from it.  The most straightforward conceptually is to sample x_1, then take a conditional probability distribution for (x_2, x_3, ... x_n), then repeat.  There are more sophisticated ways to deal with this problem as well, and in any case most of these algorithms are built into MATLAB or whatever you're going to use to do your sampling already.&lt;/p&gt;&#xA;" OwnerUserId="3296" LastActivityDate="2011-01-16T08:42:04.560" />
  <row Id="17677" PostTypeId="1" AcceptedAnswerId="17703" CreationDate="2011-01-16T08:56:04.230" Score="1" ViewCount="125" Body="&lt;p&gt;Is there a way to produce parametric equations for a curve?(If we do  know cartesian coordinates of course)&lt;/p&gt;&#xA;" OwnerUserId="5422" LastEditorUserId="5422" LastEditorDisplayName="" LastEditDate="2011-01-16T12:31:38.193" LastActivityDate="2011-01-16T16:57:40.127" Title="Parametric equations of curves" Tags="&lt;differential-geometry&gt;&lt;plane-curves&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="1" />
  <row Id="17678" PostTypeId="1" AcceptedAnswerId="17680" CreationDate="2011-01-16T09:03:13.270" Score="0" ViewCount="59" Body="&lt;p&gt;What is  the  geometrical &quot;meaning&quot; of gaussian curvature?&lt;/p&gt;&#xA;" OwnerUserId="5422" LastEditorUserId="5422" LastEditorDisplayName="" LastEditDate="2011-01-16T09:18:36.597" LastActivityDate="2011-01-16T09:35:00.250" Title="Optimization of gaussian curvature" Tags="&lt;differential-geometry&gt;" AnswerCount="1" />
  <row Id="17680" PostTypeId="2" ParentId="17678" CreationDate="2011-01-16T09:35:00.250" Score="2" ViewCount="" Body="&lt;p&gt;The most geometrical answer I know of is the Gauss–Bonnet theorem. One particular case of it tells us that the sum of the angles of a geodesic triangle on a surface is equal to $\displaystyle \pi + \int_\Delta K \, dA$, where the integral is over the interior of the triangle and $K$ is the Gaussian curvature. &lt;/p&gt;&#xA;" OwnerUserId="5191" LastActivityDate="2011-01-16T09:35:00.250" />
  <row Id="17681" PostTypeId="2" ParentId="17200" CreationDate="2011-01-16T10:18:54.373" Score="0" ViewCount="" Body="&lt;p&gt;If you want to hold a formula in your hands which relates all the parameters to your (potential) price this is the best tool I can think of:&lt;br&gt;&#xA;&lt;a href=&quot;http://ccsl.mae.cornell.edu/eureqa&quot; rel=&quot;nofollow&quot;&gt;http://ccsl.mae.cornell.edu/eureqa&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are not hindered by some form of the resulting equation, it is easy, fast, stable - and free!&lt;/p&gt;&#xA;" OwnerUserId="346" LastActivityDate="2011-01-16T10:18:54.373" />
  <row Id="17682" PostTypeId="2" ParentId="17643" CreationDate="2011-01-16T10:39:10.127" Score="0" ViewCount="" Body="&lt;p&gt;Let $A$ be an arbitrary finite set, and assume by contradiction there exists a $B$ which is finite such that the union is infinite.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have that $B$ is finite, that is there is a bijection of $B$ to some $\{0,\ldots,n-1\}$. So we can assume that this $n$ is the minimal $n$ to have this property.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Choose an arbitrary $b\in B$ (we can do it because $B$ cannot be empty, otherwise the union $A\cup B=A$ and is still finite) then $B\prime = B\setminus\{b\}$ is a finite set whose cardinality is less than $n$ and therefore $A\prime=A\cup B\prime$ is finite by the minimality of $n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have that $A\prime\cup\{b\}$ is infinite, but (this is a much simpler claim) adding one more element to a finite set is still finite, contradiction.&lt;/p&gt;&#xA;" OwnerUserId="622" LastActivityDate="2011-01-16T10:39:10.127" />
  <row Id="17683" PostTypeId="2" ParentId="17650" CreationDate="2011-01-16T10:45:40.197" Score="1" ViewCount="" Body="&lt;p&gt;You are correct that the problem needs Bayes' Theorem. It's hard to tell what mistake your teacher is making, but it looks like he is calculating the P-value: the probability that the boy's weight is that low, given the null hypothesis that he is taking his medication. This is a method for what is called &quot;significance testing&quot;, but it certainly shouldn't be used in this problem, and definitely not in the way your professor has carried it out. In particular your teacher has made no use at all of the fact that 75% percent of patients do take their medication.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The method you've used to tackle the problem is entirely correct. Unfortunately you've made a small slip when you calculated P(B). Although you wrote at the start that:&lt;br&gt;&#xA;P(B) = P(B|A)P(A) + P(B|not A)P(not A)&lt;br&gt;&#xA;when you came to calculate it you used&lt;br&gt;&#xA;P(B) = P(B|A) + P(B|not A)&lt;br&gt;&#xA;so you got that P(B)=0.0151+0.0796=0.0947 and not P(B)=0.0151*0.75+0.0796*0.25=0.031225&#xA;If you correct this in your calculations your final answer goes up to 36.26% (even further away from your teacher's answer!)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As TonyK points out, the question tells us that the child claims to have taken the medicine. Really we should include this in our calculations as evidence (since the child is more likely to claim that he took the medicine if he did in fact take it). However it is hard to ascribe numerical probabilities to statements like &quot;the boy is lying given that he took the medicine&quot;, so this would make the problem much harder. Of course, neither your teacher nor the original writer of the question expected you to take this piece of information into account.&lt;/p&gt;&#xA;" OwnerUserId="1149" LastEditorUserId="1149" LastEditorDisplayName="" LastEditDate="2011-01-16T10:50:42.097" LastActivityDate="2011-01-16T10:50:42.097" CommentCount="3" />
  <row Id="17684" PostTypeId="2" ParentId="8898" CreationDate="2011-01-16T11:09:24.410" Score="0" ViewCount="" Body="&lt;p&gt;One simple way to prove is to use outer product form for AB write A as [a1 | a2 | a3 ----| an]&#xA;and B as [b1 |-------| bn]* and do the operations.&lt;/p&gt;&#xA;" OwnerUserId="5903" LastActivityDate="2011-01-16T11:09:24.410" />
  <row Id="17685" PostTypeId="1" AcceptedAnswerId="17710" CreationDate="2011-01-16T11:19:28.910" Score="2" ViewCount="133" Body="&lt;p&gt;Let $A$ be a commutative noetherian ring, and suppose that $A$ is $I$-adically complete with respect to some ideal $I\subseteq A$. Is it true that for any ideal $J\subseteq I$, the ring $A$ is also $J$-adically complete?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;edit: recall that a ring $A$ is $I$-adically complete if the canonical morphism $A\to \varprojlim A/I^n$ is an isomorphism.&lt;/p&gt;&#xA;" OwnerUserId="5491" LastEditorUserId="5491" LastEditorDisplayName="" LastEditDate="2011-01-16T15:21:35.407" LastActivityDate="2011-01-16T16:30:26.870" Title="Adic completion" Tags="&lt;commutative-algebra&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" />
  <row Id="17686" PostTypeId="2" ParentId="978" CreationDate="2011-01-16T11:23:07.753" Score="0" ViewCount="" Body="&lt;p&gt;Bx=0 will imply ABx=0 but reverse don't hold true that's why number non zero soln. of B for B greater than for AB use it to solve.&lt;/p&gt;&#xA;" OwnerUserId="5903" LastActivityDate="2011-01-16T11:23:07.753" />
  <row Id="17687" PostTypeId="1" CreationDate="2011-01-16T11:28:44.390" Score="3" ViewCount="152" Body="&lt;p&gt;I'm looking for a concrete example of a complete (in the sense that all Cauchy sequences converge) but non-archimedean ordered field, to see that these two properties are independent (an example of archimedean non-complete ordered field is obviously the rationals).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you in advance.&lt;/p&gt;&#xA;" OwnerUserId="2926" LastEditorUserId="2926" LastEditorDisplayName="" LastEditDate="2011-01-16T11:36:30.430" LastActivityDate="2011-01-16T12:10:10.333" Title="Example of a complete, non-archimedean ordered field" Tags="&lt;field-theory&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="17688" PostTypeId="1" AcceptedAnswerId="17844" CreationDate="2011-01-16T11:50:30.523" Score="8" ViewCount="347" Body="&lt;p&gt;This is a puzzle that was sent to me a while back, I am told it is really hard, but supposedly solvable, I cant solve it, but I am interested in the solution,   or any tips on how to proceed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In front of you is an entity named Adam. Adam is a &#xA;solid block with a single speaker, through which &#xA;he hears and communicates. For all propositions &#xA;(statements that are either true or false) $p$, if &#xA;$p$ is true and logically knowable to Adam, then &#xA;Adam knows that $p$ is true. Adam is confined to his &#xA;physical form, cannot move, and only has the sense &#xA;of hearing. The only sounds Adam can make are to play &#xA;one of two pre-recorded audio messages. One message &#xA;consists of a very high note played for one second, &#xA;and the other one a very low note played for one &#xA;second. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Adam has mentally chosen a specific subset of the &#xA;Universe of ordinary mathematics. The Universe &#xA;of ordinary mathematics is defined as follows: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $S_0$ be the set of natural numbers: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$S_0 = \{1,2,3,\ldots\}$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$S_0$ has cardinality $\aleph_0$, the smallest and only &#xA;countable infinity. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The power set of a set $X$, denoted $2^X$, is the set of all subsets &#xA;of $X$. The power set of a set always has a cardinality &#xA;larger than the set itself, $$|2^X| = 2^{|X|}$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $S_1 = S_0 \cup 2^{S_0}$. $S_1$ has cardinality $2^{\aleph_0} = \beth_1$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $S_2 = S_1 \cup 2^{S_1}$. $S_2$ has cardinality $2^{\beth_1} = \beth_2$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, let $S_{n+1} = S_n \cup 2^{S_n}$. $S_{n+1}$ has cardinality $2^{\beth_n} = \beth_{n+1}$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Universe of ordinary mathematics is defined as $$\bigcup_{i=0}^\infty S_i$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This Universe contains all sets of natural numbers, &#xA;all sets of real numbers, all sets of complex numbers, &#xA;all ordered $n$-tuples for all $n$, all functions, all &#xA;relations, all Euclidean spaces, and virtually &#xA;anything that arises in standard analysis. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Universe of ordinary mathematics has cardinality &#xA;$\beth_\omega$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your goal is to determine the subset Adam is thinking &#xA;of, while Adam is trying to prevent you from doing so. &#xA;You are only allowed to ask Adam yes/no questions in &#xA;trying to accomplish your task. Adam must respond to &#xA;each question, and does so by playing a single note. &#xA;After Adam hears your question, he either chooses the &#xA;low note to mean yes and the high note to mean no, or &#xA;the high note to mean yes and the low note to mean no, &#xA;for that question only. He also decides to either tell &#xA;the truth or lie for each question after hearing it. &#xA;If at any time you ask a question which cannot be &#xA;answered by Adam without him contradicting himself, &#xA;Adam will either play the low note or the high note, &#xA;ignoring the question entirely. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Adam has given you an infinite amount of time to &#xA;accomplish your task. More specifically, the set of &#xA;both questions asked by you and notes played by Adam &#xA;can be of any cardinality. If in your strategy this &#xA;set is uncountably large, for any number of possibilities &#xA;of Adam's chosen subset, you must describe the order that &#xA;the elements of this set take place in as completely as &#xA;possible. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;During your questioning, you are keeping track of &#xA;the following numbers: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$B_1 = $ The number of questions in which Adam had the option &#xA;of truthfully responding in the affirmative. (This number &#xA;and the following numbers can of course be cardinal numbers.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$B_2 = $ The number of questions in which Adam had the option &#xA;of truthfully responding in the negative. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$B_3 = $ The number of questions in which Adam had the option &#xA;of falsely responding in the affirmative. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$B_4 = $ The number of questions in which Adam had the option &#xA;of falsely responding in the negative. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$B_5 = $ The number of questions in which Adam responded &#xA;with the high note. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$B_6 = $ The number of questions in which Adam responded &#xA;with the low note. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$B_7 = $ The number of questions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $C = B_1+B_2+B_3+B_4+B_5+B_6+B_7$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A strategy exists which will eventually allow you to &#xA;determine Adam's chosen subset. Describe such a strategy &#xA;in which $C$ is as small as possible, for all possibilities &#xA;of Adam's chosen subset. &lt;/p&gt;&#xA;" OwnerUserId="5904" LastEditorUserId="5904" LastEditorDisplayName="" LastEditDate="2011-01-16T19:54:12.563" LastActivityDate="2011-01-17T12:15:44.720" Title="Complex logic puzzle" Tags="&lt;logic&gt;&lt;set-theory&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="5" />
  <row Id="17689" PostTypeId="1" AcceptedAnswerId="17723" CreationDate="2011-01-16T12:07:46.267" Score="3" ViewCount="164" Body="&lt;p&gt;I was wondering if you would point me to a book where the theory of second order homogeneous linear difference equation with variable coefficients is discussed. I am having difficulties in getting rigorous methods to solve some equations, see an example below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, given the recurrence relation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$X_{n+2} = \frac{3n-2}{n-1}X_{n+1} - \frac{2n}{n-1}X_n$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;two solutions are &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$X(n)= n$ and $X(n) = 2^n$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there an &quot;elementary&quot; way of arriving at these solutions? (i.e. without using transforms, etc.)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance.&lt;/p&gt;&#xA;" OwnerUserId="5595" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2011-06-03T13:52:13.547" LastActivityDate="2011-06-03T13:52:13.547" Title="Second order homogeneous linear difference equation with variable coefficients" Tags="&lt;reference-request&gt;&lt;recurrence-relations&gt;" AnswerCount="2" />
  <row Id="17690" PostTypeId="2" ParentId="17687" CreationDate="2011-01-16T12:10:10.333" Score="1" ViewCount="" Body="&lt;p&gt;In this context, complete means &quot;order complete&quot; rather than complete in the sense of metric spaces.  A non-archimedian ordered field will necessarily have infinitely small elements, so I suspect that if you try to impose a metric you'll get something pathological like the closure of 0 containing all of the infinitely small elements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can construct an order-complete example by taking the field $\mathbb{R}(x)$ of rational functions over $\mathbb{R}$.  Order it such that $x$ is less than every positive number but bigger than 0, and take its order completion.  &lt;/p&gt;&#xA;" OwnerUserId="4880" LastActivityDate="2011-01-16T12:10:10.333" />
  <row Id="17691" PostTypeId="1" AcceptedAnswerId="17696" CreationDate="2011-01-16T12:23:27.910" Score="1" ViewCount="29" Body="&lt;p&gt;Let $n \geq 1$. $V=Z_2^n \setminus \{0\}$. $A \cup B = V$. $A \cap B = \Phi$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it true that either A or B contains a sub-space of dimension $n-1$ (without the zero element)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I reduced a homework question to this one).&lt;/p&gt;&#xA;" OwnerUserId="5906" LastActivityDate="2011-01-16T12:53:58.727" Title="Partition $Z_2^n \setminus \{0\}$ into disjoint sets. Does one contain a maximal subspace?" Tags="&lt;linear-algebra&gt;&lt;homework&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" />
  <row Id="17692" PostTypeId="2" ParentId="17689" CreationDate="2011-01-16T12:26:30.100" Score="1" ViewCount="" Body="&lt;p&gt;I'm not sure this is what you're looking for but do you know of H. Wilf's book &quot;generatingfunctionology&quot;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;He made it available online for free:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.math.upenn.edu/~wilf/gfologyLinked2.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.math.upenn.edu/~wilf/gfologyLinked2.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5798" LastActivityDate="2011-01-16T12:26:30.100" CommentCount="1" />
  <row Id="17693" PostTypeId="1" AcceptedAnswerId="17697" CreationDate="2011-01-16T12:30:21.097" Score="0" ViewCount="142" Body="&lt;p&gt;I did a homework question, which was choosing the shortest path to the other points from a point using Dijkstra's Algorithm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I ended up with the following, whilst an online applet resulted in something else (upper graph is my try, lower one is the applet's one):&#xA;&lt;img src=&quot;http://i.imgur.com/X6233.png&quot; alt=&quot;graphs&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(The vertices in the second image have different names, but I tried to use the same shape of the graph.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, instead of B-D I did E-D. For D, the routes B-D and B-E-D are of equal length, so I was wondering whether both are indeed correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="5905" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2011-01-16T16:09:49.927" LastActivityDate="2011-01-16T16:09:49.927" Title="Dijkstra's Algorithm - two solutions possible?" Tags="&lt;algorithms&gt;" AnswerCount="1" />
  <row Id="17694" PostTypeId="1" AcceptedAnswerId="17698" CreationDate="2011-01-16T12:34:40.280" Score="2" ViewCount="106" Body="&lt;p&gt;I have the feeling that an engineer should be more comfortable with math, than I'm felling I am right now. I also have a tendency toward learning math not required by the math classes I took, and was always leaning toward more fundamental way of doing engineering (math &amp;amp; physics.) Though I'm &quot;fundamentalist&quot; I feel much more comfortable with practical side of engineering, having no difficulty to learn different tools, methods etc. - while my math background seems shaky. I would like to be able to study math for engineering by myself (maybe even reading math papers in the future; distant one) and would like to know what should I learn first (take that I'm 1st year in university, although I'm not.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What math should EE &amp;amp; CT engineer master and what should he be fond of?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="2115" LastActivityDate="2011-01-16T13:18:00.977" Title="Math for &quot;Electrical Engineering &amp; Communication Technology&quot; recommendation." Tags="&lt;soft-question&gt;" AnswerCount="1" FavoriteCount="0" />
  <row Id="17695" PostTypeId="2" ParentId="13975" CreationDate="2011-01-16T12:38:38.937" Score="3" ViewCount="" Body="&lt;p&gt;$(b_1,b_2,b_3,b_4,\ldots)\mapsto(b_1+3^{b_2}-1,b_3+3^{b_4}-1,\ldots)$ answers the question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;@Andres: &quot;As a slightly more challenging exercise, pick any two positive integers $n&amp;lt;m$, and build a &quot;combinatorial&quot; bijection between $\{0,1,…,n\}^{\mathbb N}$ and $\{0,1,…,m\}^{\mathbb N}$. Combinatorial meaning here something in the same spirit of the explicit bijection above.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Answer(partial). Assume that there exists two integers $p,q$ such that $l=(n+1)^p=(m+1)^q$. The bijection comes from a simple remark : there is a natural bijection between the words of length $p$ on the alphabet $\{0,1,…,n\}$ and the words of length $q$ on $\{0,1,…,m\}$. (one can think of two trees with the same leaves (with cardinalty $l$) to see this, &lt;a href=&quot;http://en.wikipedia.org/wiki/Tree_(set_theory)&quot; rel=&quot;nofollow&quot;&gt;look here&lt;/a&gt;). This yields an explicit bijection in the case $(n+1)^p=(m+1)^q$.&lt;/p&gt;&#xA;" OwnerUserId="5838" LastEditorUserId="5838" LastEditorDisplayName="" LastEditDate="2011-01-16T15:15:26.417" LastActivityDate="2011-01-16T15:15:26.417" />
  <row Id="17696" PostTypeId="2" ParentId="17691" CreationDate="2011-01-16T12:53:58.727" Score="0" ViewCount="" Body="&lt;p&gt;This is not true. For $n=4$ let&#xA;$$A=\{(0,1,0,0), (0,0,1,0), (0,0,0,1), (1,1,0,0), (1,1,1,0), (1,1,0,1), (1,0,1,1), (0,1,1,1)\}$$&#xA;$$B=\{(1,0,0,0), (1,0,1,0), (1,0,0,1), (0,1,1,0), (0,1,0,1), (0,0,1,1), (1,1,1,1) \}$$&#xA;$B \cup \{0\}$ is not a subspace, and its size (with the zero vector) is $2^3$, so it doesn't contain a subspace of dimension 4-1=3.&#xA;Notice that if you remove one vector from A, then it will not be a subspace (with the zero vector) so again $A\cup \{0\}$ doesn't contain a 3 dimensional subspace.&lt;/p&gt;&#xA;" OwnerUserId="2125" LastActivityDate="2011-01-16T12:53:58.727" />
  <row Id="17697" PostTypeId="2" ParentId="17693" CreationDate="2011-01-16T13:05:37.233" Score="3" ViewCount="" Body="&lt;p&gt;The solution to the shortest path problem is not unique. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine you have a 2-node graph, let's call the nodes A and B. You have, say, 10 edges between them, all of equal weight w. Then the shortest path from A to B is any one of these 10 edges, so the solution is not unique. In particular, it depends on the order in which you traverse the nodes in each iteration.&lt;/p&gt;&#xA;" OwnerUserId="5798" LastActivityDate="2011-01-16T13:05:37.233" CommentCount="3" />
  <row Id="17698" PostTypeId="2" ParentId="17694" CreationDate="2011-01-16T13:18:00.977" Score="1" ViewCount="" Body="&lt;p&gt;I'll assume that you know first year undergraduate level calculus. My recommendations for higher level math topics are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Probability and random processes - Learn as much probability theory as you can. This is a very important prerequisite for doing any kind of work in communications. All problems in communications are modeled using random variables and random processes. You will definitely use these math concepts in your work. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Linear algebra - All engineers regardless of major should learn some linear algebra. In fact, it is worth it to try and learn the subject in depth. The kind of insights you get by studying it as well as the tools and techniques you learn are very useful when studying linear systems (which you encounter often in engineering).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) Some basic complex analysis - I'm being a little imprecise here, but you should be comfortable with Laplace and Fourier transforms at the very least. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;4) Real analysis -  To be honest, this isn't absolutely necessary because concepts from real analysis are seldom directly applied in an engineering situation, but a course in real analysis is useful because it will teach you to think rigorously (to think a bit like a mathematician if you will). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Additionally, topics like basic graph theory and discrete math are useful if you plan to study topics like networking. Some Group theory is necessary if you plan to study error correcting codes/coding theory.&lt;/p&gt;&#xA;" OwnerUserId="2641" LastActivityDate="2011-01-16T13:18:00.977" CommentCount="1" />
  <row Id="17699" PostTypeId="2" ParentId="17643" CreationDate="2011-01-16T13:40:45.583" Score="2" ViewCount="" Body="&lt;p&gt;Another notion of finite is: &quot;The set does not contain a proper subset bijective with itself.&quot; This does not use arithmetic, and might be used to show what you want.&lt;/p&gt;&#xA;" OwnerUserId="2941" LastActivityDate="2011-01-16T13:40:45.583" />
  <row Id="17700" PostTypeId="1" AcceptedAnswerId="17708" CreationDate="2011-01-16T13:41:15.473" Score="2" ViewCount="154" Body="&lt;p&gt;Hello! I am studying for an exam and I don't know how to determine a general solution for a differential equation. For example,&#xA;$$&#xD;&#xA;x = t x^\prime + (x^\prime)^2&#xD;&#xA;$$&#xA;Can anyone help?&lt;/p&gt;&#xA;" OwnerUserId="2777" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2011-01-16T23:18:23.270" LastActivityDate="2011-01-16T23:46:40.540" Title="General solution for a differential equation $x = t x^\prime + (x^\prime)^2$" Tags="&lt;differential-equations&gt;" AnswerCount="2" />
  <row Id="17702" PostTypeId="1" AcceptedAnswerId="17706" CreationDate="2011-01-16T14:24:02.980" Score="1" ViewCount="85" Body="&lt;p&gt;Suppose I have a value $a_1$ such that $-L &amp;lt; a_1 &amp;lt; L$. I need to compute $a_2$ from $a_1$ and $b$, while satisfying the following requirements:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$-L &amp;lt; a_2 &amp;lt; L$&lt;/li&gt;&#xA;&lt;li&gt;if $b = 0$ then $a_2 = a_1$&lt;/li&gt;&#xA;&lt;li&gt;if $b &amp;lt; 0$ then $a_2 &amp;lt; a_1$&lt;/li&gt;&#xA;&lt;li&gt;if $b &amp;gt; 0$ then $a_2 &amp;gt; a_1$&lt;/li&gt;&#xA;&lt;li&gt;a &lt;em&gt;chaining&lt;/em&gt; requirement, explained below.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The operation can be written as $a_2 = f(a_1, b)$ (with an implicit $L$ argument which I'll leave out for clarity). In this notation, the &lt;em&gt;chaining&lt;/em&gt; requirement can be expressed as: $f(a, b+c) = f(f(a, b), c)$. In intuitive terms, adjusting $a$ by $b$ and &lt;em&gt;then&lt;/em&gt; by $c$ produces the same result as adjusting $a$ by $b + c$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you recommend a function $f$ that satisfies these requirements? If you constructed the function rather than just happened to know one that fits these, could you outline the steps?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(&lt;strong&gt;edited&lt;/strong&gt; to fix the impossible $-L \le a \le $L requirement)&lt;/p&gt;&#xA;" OwnerUserId="5910" LastEditorUserId="5910" LastEditorDisplayName="" LastEditDate="2011-01-16T17:06:10.540" LastActivityDate="2011-01-16T17:06:10.540" Title="Construct a function to &quot;increment&quot; a value while asymptotically approaching a limit" Tags="&lt;functions&gt;&lt;puzzle&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="17703" PostTypeId="2" ParentId="17677" CreationDate="2011-01-16T14:24:50.467" Score="1" ViewCount="" Body="&lt;p&gt;See Chapter 4 of &lt;a href=&quot;http://samizdat.mines.edu/difgeom/Shr3b.pdf&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; (a textbook on differential geometry). Essentially, a curve $\textbf{r}$ depends on two parameters ($u^1$ and $u^2$) in some arbitrary Cartesian coordinate system.&lt;/p&gt;&#xA;" OwnerUserId="2548" LastEditorUserId="2548" LastEditorDisplayName="" LastEditDate="2011-01-16T14:31:37.653" LastActivityDate="2011-01-16T14:31:37.653" />
  <row Id="17704" PostTypeId="1" AcceptedAnswerId="17707" CreationDate="2011-01-16T14:41:08.710" Score="1" ViewCount="231" Body="&lt;p&gt;So, I've been forwarded a math problem with 5 possible answers, and I'm unable to get my answers to match any of the given answers...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll not cloud the issue giving my working up front (but I will post it once I have some answers, for comparison).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is this:&#xA;&lt;img src=&quot;http://i.imgur.com/6z5Ol.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And the possible answers are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;361,430 - 387,012 - 391,552 - 472,148 - 460,577&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any input (working would be greatly appreciated so I can understand where I went right and wrong!) is appreciated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cheers folks,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dave&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: my approach to this...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Calculate 11.2 percent (0.8 less than 12) of the total amount - 425600.(This will give the value of 11% of the following year)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Divide this by 11, then multiply by 100 - giving about 433,000.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Essentially, 425600 * 0.112 = 47667.2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;47667.2 / 11 * 100 = 433,338.182, or 433,338 people.&lt;/p&gt;&#xA;" OwnerUserId="5911" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2011-01-17T00:56:26.550" LastActivityDate="2011-01-17T00:56:26.550" Title="Percentages problem and bus passengers" Tags="&lt;algebra-precalculus&gt;&lt;arithmetic&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" />
  <row Id="17705" PostTypeId="2" ParentId="17704" CreationDate="2011-01-16T14:51:32.833" Score="2" ViewCount="" Body="&lt;p&gt;Calculate how many under 35's travel by bus, then calculate how many this would be if it decreased by 0.8% and then calculate what number this is 11% of.&lt;/p&gt;&#xA;" OwnerUserId="1149" LastActivityDate="2011-01-16T14:51:32.833" CommentCount="2" />
  <row Id="17706" PostTypeId="2" ParentId="17702" CreationDate="2011-01-16T14:54:24.670" Score="3" ViewCount="" Body="&lt;p&gt;This would be easy, if it weren't impossible :-)&lt;br&gt;&#xA;It's impossible because if (say) $a_1 = L$ and $b &amp;gt; 0$, then a suitable $a_2$ would have to be both $\le L$ and $\gt a_1$.&lt;br&gt;&#xA;But suppose we restate the problem, so that $-L \lt a_1 \lt L$ and we require $-L \lt a_2 \lt L$. Now it's easy. All we need is a strictly increasing, invertible function  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$g: (-L,L) \rightarrow \mathbb{R}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance: $g(x) = \tan(\frac{\pi x}{2L})$, with inverse $g^{-1}(y)=\frac{2L}{\pi}tan^{-1}(y)$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now define  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f(a,b) = g^{-1}(g(a)+b)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and we're done.&lt;/p&gt;&#xA;" OwnerUserId="1508" LastActivityDate="2011-01-16T14:54:24.670" CommentCount="2" />
  <row Id="17707" PostTypeId="2" ParentId="17704" CreationDate="2011-01-16T15:06:53.030" Score="2" ViewCount="" Body="&lt;p&gt;Your main mistake is, you cannot subtract percentages like that. The 15% of 100 is 15. Decrease this number by 50% and you get 7.5. If you subtract 50% from 15%, you'd get -35%, and an F.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;p&gt;Let's admit, for lack of a better hypothesis, that those percentages are completely accurate, and that the absolute number of commuters on Rail, Car, Subway and Others remained constant somehow, at least for the Age 34- group.&lt;/p&gt;&lt;p&gt;We know that 425600 × 0.12 = 51,072 under-35s used the Bus. Next year that'll be 51,072 × 0.992 = 50,663&lt;del&gt;.424&lt;/del&gt; &lt;em&gt;sigh&lt;/em&gt;. If that's the 11% of the total, we have 50,663.424 ÷ 0.11 = &lt;strong&gt;460,576&lt;s&gt;.58(18)&lt;/s&gt;&lt;/strong&gt;. Mark &quot;D&quot; as the correct answer and add an humorous note about person fractions.&lt;/p&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="92" LastActivityDate="2011-01-16T15:06:53.030" CommentCount="2" />
  <row Id="17708" PostTypeId="2" ParentId="17700" CreationDate="2011-01-16T15:11:11.127" Score="3" ViewCount="" Body="&lt;p&gt;One approach is to assume a Taylor series expansion for the solution, substitute and equate coefficients. I haven't worked it out to the end and proved that these are all the solutions, but I can find 2 solutions like this.&#xA;Another, simpler way, is to differentiate the ODE once, and see what you can deduce from the resulting 2nd-order ODE - the solutions drop out directly this way.&lt;/p&gt;&#xA;" OwnerUserId="1470" LastEditorUserId="1470" LastEditorDisplayName="" LastEditDate="2011-01-16T15:32:03.553" LastActivityDate="2011-01-16T15:32:03.553" />
  <row Id="17709" PostTypeId="2" ParentId="17702" CreationDate="2011-01-16T15:19:53.650" Score="1" ViewCount="" Body="&lt;p&gt;Except for the restriction of domain and range to $(-L,L)$, ordinary addition would work: $f(a,b) = a+b$.  So if you take $g : \mathbb{R}\rightarrow(-L,L)$ to be any increasing bijection, then $f(a,b) = g\left(g^{-1}(a) + b\right)$ will satisfy the requirements.  A nice specific example is $g(x) = L\tanh(x)$, whose inverse is $$g^{-1}(x) = \frac{1}{2}\log\left(\frac{L+x}{L-x}\right).$$&#xA;The desired function becomes&#xA;$$&#xD;&#xA;f(a,b) = L \tanh\left(b + \frac{1}{2}\log\left(\frac{L+a}{L-a}\right)\right).&#xD;&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="5546" LastActivityDate="2011-01-16T15:19:53.650" CommentCount="1" />
  <row Id="17710" PostTypeId="2" ParentId="17685" CreationDate="2011-01-16T15:25:35.947" Score="3" ViewCount="" Body="&lt;p&gt;Dear Anonymous, the answer is &quot;yes&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $A$ is Noetherian, for any $m$ the finitely generated $A$-module&#xA;$A/J^m$ is $I$-adically complete, and so $A/J^m$ is the inverse limit over $n$&#xA;of $A/(I^n + J^m)$.   Now $J^m \subset I^m,$ and so $I^n \subset I^n + J^m&#xD;&#xA;\subset I^m$ when $n \geq m$.  Thus the inverse limit (over $m$) of $A/J^m$ is&#xA;the same as the inverse limit (over $n$) of $A/I^n$, and we see that $A$ is $J$-adically complete.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another way to think about it is that $A$ is $I$-adically complete (and separated, which is part of the requirement of &quot;complete&quot;) if and only if any $I$-adic Cauchy sequence of elements of $A$ has a unique $I$-adic limit.  Since a $J$-adic Cauchy sequence is also an $I$-adic sequence, a $J$-adic Cauchy $(a_n)$ sequence also has a unique $I$-adic limit, say $a$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now if we choose $n_0$ so that $a_m - a_n \in J^k$ if $m,n \geq n_0$,&#xA;then we see that $a - a_m = a - a_{n} + a_{n} - a_m \in J^k + I^l,$ where &#xA;$l$ can be made arbitrarily large by choosing $n$ large enough (since $a_n$&#xA;converges to $a$ in the $I$-adic topology).  Thus $a - a_m \in \cap_l J^k + I^l.$  This intersection is equal to $J^k$ (by $I$-adic completeness of $A/J^k$)&#xA;and so $a-a_m \in J^k$.  Thus in fact $(a_n)$ converges to $a$ in the $J$-adic topology, and so $A$ is $J$-adically complete.&lt;/p&gt;&#xA;" OwnerUserId="221" LastEditorUserId="221" LastEditorDisplayName="" LastEditDate="2011-01-16T16:30:26.870" LastActivityDate="2011-01-16T16:30:26.870" CommentCount="8" />
  <row Id="17711" PostTypeId="1" AcceptedAnswerId="17715" CreationDate="2011-01-16T15:44:13.160" Score="10" ViewCount="734" Body="&lt;p&gt;In &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0071508619&quot; rel=&quot;nofollow&quot;&gt;this book I'm using&lt;/a&gt; the author seems to feel a need to prove&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$e^{u+v} = e^ue^v$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ln(e^{u+v}) = u + v = ln(e^u) + ln(e^v) = ln(e^u e^v)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence $e^{u+v} = e^u e^v$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But we know from basic algebra that $x^{a+b} = x^ax^b$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Earlier in the chapter the author says that you should not assume $e^x$ &quot;is an ordinary power of a base &lt;em&gt;e&lt;/em&gt; with exponent &lt;em&gt;x&lt;/em&gt;.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is both a math and pedagogy question then, why does he do that?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So 2 questions really&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;em&gt;Do we need to prove&lt;/em&gt; this for such a basic property?&lt;/li&gt;&#xA;&lt;li&gt;If we don't need to, then why does he do it?  Fun?  To make it memorable?  Establish more neural connections?  A case of wildly uncontrolled OCD?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Also I've always taken forgranted the property that $x^{a+b} = x^a x^b$.  I take it as an axiom, but I actually don't know where that axiom is listed.&lt;/p&gt;&#xA;" OwnerUserId="1069" LastEditorUserId="1069" LastEditorDisplayName="" LastEditDate="2011-01-18T16:09:01.330" LastActivityDate="2011-02-10T23:23:39.047" Title="Why do we need to prove $e^{u+v} = e^ue^v$?" Tags="&lt;teaching&gt;" AnswerCount="6" CommentCount="14" FavoriteCount="2" />
  <row Id="17712" PostTypeId="2" ParentId="17711" CreationDate="2011-01-16T15:57:42.573" Score="8" ViewCount="" Body="&lt;p&gt;This may seem like a pretty pointless proof, at least on the surface, but I suspect there's some subtlety in the way the author's defined things here. (It may even appear circular at first, considering that the logarithm is often introduced as the inverse of the exponential function. Saying that, it can be derived the other way round, and this can sometimes be enlightening.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A rigorous proof for integer exponents is very straightforward indeed, and follows simply from the definition of the exponential function (of arbitrary base). For arbitrary exponents, things get slightly more complicated. I present a more complete proof below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, let us suppose that the author began by defining the (natural) logarithm function,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ln a = \int_1^a \frac{dx}{x} .$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can then prove the addition property of logarithms, $\ln (ab) = \ln a + \ln b$, by considering&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;\ln (ab) &#xD;&#xA;= \int_1^{ab} \frac{1}{x} \; dx &#xD;&#xA;= \int_1^a \frac{1}{x} \; dx \; + \int_a^{ab} \frac{1}{x} \; dx &#xD;&#xA;=\int_1^{a} \frac{1}{x} \; dx \; + \int_1^{b} \frac{1}{t} \; dt &#xD;&#xA;= \ln (a) + \ln (b)&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(See &lt;a href=&quot;http://en.wikipedia.org/wiki/Natural_logarithm#Definitions&quot; rel=&quot;nofollow&quot;&gt;this Wikipedia page&lt;/a&gt; for reference.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The exponential function can of course be defined as the inverse of the logarithm, i.e.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$exp(ln(a)) = a$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, to prove the property of exponentials, $e^{u+v} = e^u e^v$, we start as follows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\text{Let}\ u = \ln a, v = \ln b .$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, using this property of logarithms and the definition of the inverse, consider&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$e^{u+v} = e^{\ln a + \ln b} = e^{\ln (ab)} = ab = e^u e^v .\ \square$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That should hopefuly be straightforward enough to follow. There are of course other equivalent definitions of $exp$ and $ln$. (You can for example define the Taylor series of $exp$, use the Cauchy product, and then simplify, but that's slightly trickier.)&lt;/p&gt;&#xA;" OwnerUserId="56" LastEditorUserId="56" LastEditorDisplayName="" LastEditDate="2011-01-16T16:22:28.497" LastActivityDate="2011-01-16T16:22:28.497" CommentCount="4" />
  <row Id="17713" PostTypeId="1" AcceptedAnswerId="17748" CreationDate="2011-01-16T16:00:52.113" Score="2" ViewCount="156" Body="&lt;p&gt;I am a bit perplexed in trying to find values $a,b,c$ so that the approximation is as precise as possible:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{k=n}^{\infty}\frac{(\ln(k))^{2}}{k^{3}} \approx \frac{1}{n^{2}}[a(\ln (n))^{2}+b \ln(n) + c]$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can see from Wolfram that $\ln(x)$ could be written in many different &lt;a href=&quot;http://www.wolframalpha.com/input/?i=ln%28x%29+series&quot; rel=&quot;nofollow&quot;&gt;ways&lt;/a&gt; and that the sum starting from $1$ can be written with Riemann zeta function here. They are probably not something I am looking here. The decreasing exponent hints me that perhaps the term in the sum can be approximated with its derivations, the first derivate here and the second here but when I try to approximate something goes wrong, perhaps wrong premise. Let $z=\frac{\ln(k)^{2}}{k^{3}}$ then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{k=n}^{\infty} z \approx z&amp;#39; z-z&amp;#39; z^3/2!$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;by the Taylor approximation for the odd function ($\ln(x)$ is odd, $x^3$ is odd and the oddity is preserved after the operations, maybe wrong, so $z$ is odd like $\sin(x)$), I get:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{k=n}^{\infty} z \approx (\frac{1}{x^3}(2\ln(x) - 3(\ln(x))^2)) -\frac{1}{x^2}(6 \ln(x)^2-7 \ln(x)+1)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;something wrong because the first term has $x^3$ instead of $x^2$. I am sorry if this hard to read but stuck here. So is it the correct way to approximate the sum? Is it really with Rieman zeta function or does Taylor work here, as I am trying to do above?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I labeled the question with &lt;code&gt;parity&lt;/code&gt; because I think it is central here. I feel I may be misunderstanding how the oddity works over different operations, hopefully shown in my vague explanations, and hence the error.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="5902" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2011-01-16T19:08:57.207" LastActivityDate="2011-01-16T21:12:11.570" Title="How to approximate $\displaystyle\sum_{k=n}^{\infty}\dfrac{(\ln(k))^{2}}{k^{3}}$?" Tags="&lt;analysis&gt;&lt;sequences-and-series&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="2" />
  <row Id="17714" PostTypeId="1" CreationDate="2011-01-16T16:04:09.447" Score="1" ViewCount="106" Body="&lt;p&gt;I'm trying to understand how you derive a context free grammar (CFG) from a Pushdown Automata (PDA)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have the following PDA.&#xA;&lt;img src=&quot;http://i.imgur.com/Uy3lM.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe the following context-free language can be derived..&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;{0^n1^n|n≥0}&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My problem is I don't understand how you can build the context free grammar from a PDA and vice versa. How would I build a PDA from a CFG? I'd really appreciate if someone could walk me through this process.  &lt;/p&gt;&#xA;" OwnerUserId="5325" LastEditorUserId="5325" LastEditorDisplayName="" LastEditDate="2011-01-16T16:57:45.010" LastActivityDate="2011-01-16T16:57:45.010" Title="Pushdown Automata deriving context free language" Tags="&lt;computer-science&gt;" CommentCount="4" />
  <row Id="17715" PostTypeId="2" ParentId="17711" CreationDate="2011-01-16T16:05:47.930" Score="15" ViewCount="" Body="&lt;p&gt;In fact, such a proof &lt;em&gt;is&lt;/em&gt; often necessary, which is why many authors write the function $e^x$ as $\exp(x)$ until they establish that it's just a &quot;normal&quot; exponent. For instance, if the original definition is given as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\exp(x) = \lim_{n \to \infty} \left(1+\frac{x}{n}\right)^n,$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;then proving that $\exp(x + y) = \exp(x) \exp(y)$ is non-obvious, and certainly necessary.&lt;/p&gt;&#xA;" OwnerUserId="4743" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2011-01-16T22:48:31.800" LastActivityDate="2011-01-16T22:48:31.800" CommentCount="3" />
  <row Id="17716" PostTypeId="1" AcceptedAnswerId="17724" CreationDate="2011-01-16T16:13:45.447" Score="1" ViewCount="98" Body="&lt;p&gt;Below are the steps the teacher took to solve:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$y = \sqrt{3}\sin x + \cos x$ find min and max on $[0, 2\pi)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Step 1: = $2\sin(x + \pi/6))$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How does the teacher get this first step?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: No calculus please.&lt;/p&gt;&#xA;" OwnerUserId="5899" LastEditorUserId="5899" LastEditorDisplayName="" LastEditDate="2011-01-16T17:03:10.543" LastActivityDate="2011-01-16T17:03:10.543" Title="How does teacher get first step?" Tags="&lt;trigonometry&gt;" AnswerCount="3" />
  <row Id="17717" PostTypeId="2" ParentId="17711" CreationDate="2011-01-16T16:15:07.820" Score="16" ViewCount="" Body="&lt;p&gt;Mathematicians have a habit of using the same notation to denote many different concepts.  To justify their overloading, they tend to point to similarities between properties of those concepts.  You may think of the exponential as one concept, but it is actually a large family of related concepts:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The exponentials $x^n$ where $x$ is an element of an arbitrary &lt;a href=&quot;http://en.wikipedia.org/wiki/Monoid&quot; rel=&quot;nofollow&quot;&gt;monoid&lt;/a&gt; and $n$ a non-negative integer,&lt;/li&gt;&#xA;&lt;li&gt;The exponentials $x^n$ where $x$ is an element of an arbitrary &lt;a href=&quot;http://en.wikipedia.org/wiki/Group_(mathematics)&quot; rel=&quot;nofollow&quot;&gt;group&lt;/a&gt; and $n$ an integer,&lt;/li&gt;&#xA;&lt;li&gt;The exponentials $x^n$ where $x$ is an element of an arbitrary group and $n$ a rational (which are not guaranteed to exist in general, and are also not guaranteed to be unique in general),&lt;/li&gt;&#xA;&lt;li&gt;The exponential $x^n$ where $x$ is an element of a &lt;a href=&quot;http://en.wikipedia.org/wiki/Topological_group&quot; rel=&quot;nofollow&quot;&gt;topological group&lt;/a&gt; where $x^n$ for $n$ rational is unique, and $n$ a real number (defined by limits as in Rudin),&lt;/li&gt;&#xA;&lt;li&gt;The exponential $e^x = 1 + x + \frac{x^2}{2} + ... $ where $x$ is an element of a &lt;a href=&quot;http://en.wikipedia.org/wiki/Topological_ring&quot; rel=&quot;nofollow&quot;&gt;topological ring&lt;/a&gt; containing $\mathbb{Q}$ and the series converges,&lt;/li&gt;&#xA;&lt;li&gt;etc.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;It is easy to be fooled into thinking that these are all the same concept because they define essentially the same operation on $\mathbb{R}$, but 1) this is not obvious and &lt;strong&gt;requires proof&lt;/strong&gt;, and 2) they generalize in different directions, so should be regarded as different in full generality.  Note that the condition for $x^n$ to be defined where $x$ is an element of a topological group and $n$ a real number is quite strong and rarely satisfied.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note also that your proof using logarithms runs into unnecessary subtleties when $u, v$ are taken to be complex numbers, since in this setting the logarithm is not uniquely defined.  Nevertheless, the exponential law still holds in this setting; it is one of the basic properties that mathematicians point to to justify calling something an exponential in the first place.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In higher mathematics, the exponential further generalizes to:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The exponential map in &lt;a href=&quot;http://en.wikipedia.org/wiki/Exponential_map#Lie_theory&quot; rel=&quot;nofollow&quot;&gt;Lie theory&lt;/a&gt;,&lt;/li&gt;&#xA;&lt;li&gt;The exponential map in &lt;a href=&quot;http://en.wikipedia.org/wiki/Exponential_map#Riemannian_geometry&quot; rel=&quot;nofollow&quot;&gt;Riemannian geometry&lt;/a&gt;,&lt;/li&gt;&#xA;&lt;li&gt;The exponential map in &lt;a href=&quot;http://en.wikipedia.org/wiki/Cartesian_closed_category&quot; rel=&quot;nofollow&quot;&gt;category theory&lt;/a&gt;,&lt;/li&gt;&#xA;&lt;li&gt;etc.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So it is good to keep in mind that &quot;exponential&quot; does not just refer to one thing.&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2011-01-18T02:46:14.737" LastActivityDate="2011-01-18T02:46:14.737" CommentCount="21" />
  <row Id="17718" PostTypeId="2" ParentId="17716" CreationDate="2011-01-16T16:17:58.767" Score="4" ViewCount="" Body="&lt;p&gt;Instead of telling you how your teacher got that, try doing the opposite and expand out $\sin(a+b)=\sin(a)\cos(b)+\cos(a)\sin(b)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rest should be trivial&lt;/p&gt;&#xA;" OwnerUserId="4728" LastEditorUserId="4728" LastEditorDisplayName="" LastEditDate="2011-01-16T16:48:32.960" LastActivityDate="2011-01-16T16:48:32.960" CommentCount="11" />
  <row Id="17719" PostTypeId="2" ParentId="17711" CreationDate="2011-01-16T16:19:27.577" Score="7" ViewCount="" Body="&lt;p&gt;The necessity of a proof depends very much on how each of those things is defined and the domain over which the variables may vary. For example, if $e^x$ is defined by the power series $\displaystyle \sum_{n=0}^{\infty} \frac{x^n}{n!}$ for all $x \in \mathbb{C}$, then there is a very real need to prove that $e^{x + y} = e^x e^y$, and the proof is non-trivial. On the other hand, if $e$ is some constant defined elsewhere and $x$ and $y$ are positive integers, then there's essentially nothing to prove.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would also like to add that the law doesn't hold everywhere. If $A$ and $B$ are square matrices (or endomorphisms of a vector space) that do not commute, then in general $\exp(A + B) \ne \exp(A) \exp(B)$. This fact allows us to study certain non-commutative groups using the tools of analysis and differential geometry — this is the field of study known as Lie theory.&lt;/p&gt;&#xA;" OwnerUserId="5191" LastActivityDate="2011-01-16T16:19:27.577" CommentCount="2" />
  <row Id="17720" PostTypeId="2" ParentId="17677" CreationDate="2011-01-16T16:26:49.090" Score="2" ViewCount="" Body="&lt;p&gt;I recommend the online &lt;a href=&quot;http://www.math.umn.edu/~nykamp/m2374/readings/topicindex.html?4b#s4b&quot; rel=&quot;nofollow&quot;&gt;book&lt;/a&gt;, intuitive introduction, for more deep stuff read the other reply. They call it multivariable calculus, and it covers some parametrization of curves. Sorry but not sure what you are asking here so just giving you a book to look at.&lt;/p&gt;&#xA;" OwnerUserId="5902" LastEditorUserId="5902" LastEditorDisplayName="" LastEditDate="2011-01-16T16:36:08.410" LastActivityDate="2011-01-16T16:36:08.410" />
  <row Id="17721" PostTypeId="1" CreationDate="2011-01-16T16:29:26.230" Score="0" ViewCount="32" Body="&lt;p&gt;I'm wondering if someone could walk me through the steps of generating a CFL from some CFG...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;e.g    S -&gt; aSb | ab&lt;/p&gt;&#xA;" OwnerUserId="5325" LastActivityDate="2011-01-16T16:29:26.230" Title="generate context free language from a context free grammar" Tags="&lt;computer-science&gt;" CommentCount="1" />
  <row Id="17722" PostTypeId="2" ParentId="17716" CreationDate="2011-01-16T16:31:44.813" Score="3" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://math.stackexchange.com/questions/17716/how-does-teacher-get-first-step/17718#17718&quot;&gt;picakhu's answer&lt;/a&gt; is the simplest way to see how it works having already arrived at $y=2\sin(x+\frac{\pi}{6})$ (use the identity there to expand this form).  In general, given $a\sin x+b\cos x$ (let's say for $a,b&gt;0$), it is possible to arrive at a similar equivalent form:&#xA;$$\begin{align}&#xD;&#xA;a\sin x+b\cos x&#xD;&#xA;&amp;amp;=a\left(\sin x+\frac{b}{a}\cos x\right)&#xD;&#xA;\\&#xD;&#xA;&amp;amp;=a\left(\sin x+\tan\left(\arctan\frac{b}{a}\right)\cos x\right)&#xD;&#xA;\\&#xD;&#xA;&amp;amp;=a\left(\sin x+\frac{\sin\left(\arctan\frac{b}{a}\right)}{\cos\left(\arctan\frac{b}{a}\right)}\cos x\right)&#xD;&#xA;\\&#xD;&#xA;&amp;amp;=\frac{a}{\cos\left(\arctan\frac{b}{a}\right)}\left(\sin x\cos\left(\arctan\frac{b}{a}\right)+\sin\left(\arctan\frac{b}{a}\right)\cos x\right)&#xD;&#xA;\\&#xD;&#xA;&amp;amp;=\sqrt{a^2+b^2}\sin\left(x+\arctan\frac{b}{a}\right).&#xD;&#xA;\end{align}$$&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2011-01-16T16:31:44.813" />
  <row Id="17723" PostTypeId="2" ParentId="17689" CreationDate="2011-01-16T16:33:51.130" Score="2" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\ $ Factor the difference operator. With the shift operator $\rm\ S\ X_n = X_{n+1}\ $ we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\rm\ ((n-1)\ S^2 - (3\ n-2)\ S + 2\ n)\ \ X_n\ =\ ((n-1)\ S - n)\ (S - 2)\ \ X_n$$&#xA;Now put $\rm\ Y_n = (S - 2)\ X_n = X_{n+1} - 2\ X_n\:.\ $ Then the above second-order equation reduces to $\rm\ (n-1)\ Y_{n+1} - n\ Y_n = 0\:.\ $ Solve that for $\rm\:Y_n\:$ and then plug it into the prior equation to obtain a first-order &lt;em&gt;nonhomogeneous&lt;/em&gt; equation for $\rm\: X_n\:.$&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2011-01-16T16:45:55.380" LastActivityDate="2011-01-16T16:45:55.380" CommentCount="1" />
  <row Id="17724" PostTypeId="2" ParentId="17716" CreationDate="2011-01-16T16:49:50.013" Score="7" ViewCount="" Body="&lt;p&gt;I'm posting this answer in response to the comment thread under picakhu's answer; writing comments was getting a bit tedious.  The answer has the same content as Isaac's, but is explained a little differently, which might be useful for the OP.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The general problem, of which this is a special case, is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given an expression of the form  $a \sin x + b \cos x$, for some numbers $a$ and $b$, to rewrite it in the form $c \sin (x + \theta)$, for some appropriate choice &#xA;of $c$ and $\theta$ (which will be related to $a$ and $b$ in some way, of course --- and we have to find out what that way is!).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose first that $a^2 + b^2 = 1$.  Then trigonometry (especially the point of view of the unit circle) tells us that there is an angle $\theta$ such&#xA;that $a = \cos \theta$ and $b = \sin \theta$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus we can write $a \sin x + b \cos x = \cos \theta \sin x + \sin \theta \cos x,$ and via the addition theorem for $\sin$, we recognize this as being $\sin(x + \theta).$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now in most examples, it may not be that $a^2 + b^2 = 1$.  But it equals something (!), so let's call that something $c^2$.  Then we see&#xA;that $(a/c)^2 + (b/c)^2 = 1$, so we can choose $\theta$ so that&#xA;$a/c = \cos \theta$ and $b/c = \sin \theta$.  Then as above&#xA;we find that $(a/c)\sin x + (b/c) \cos x = \sin (x + \theta),$&#xA;and so (finishing finally) we have&#xA;$$a \sin x + b \cos x = c \sin (x + \theta).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the OP's example, $a = \sqrt{3}$ and $b = 1$, so $c = 2$,&#xA;and we are led to the given answer.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that in practice this whole process will be easiest when $a/c$ and $b/c$ are easily recognized trig function special values, like $1/2$ or $\sqrt{2}/2$ or $\sqrt{3}/2$.  If they are just somewhat random numbers, then you won't be able to figure out the correct choice of $\theta$ without using a calculator to&#xA;compute $\theta$.   &lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2011-01-16T16:49:50.013" CommentCount="1" />
  <row Id="17725" PostTypeId="2" ParentId="17677" CreationDate="2011-01-16T16:57:40.127" Score="1" ViewCount="" Body="&lt;p&gt;Yes, you fit a cubic-spline through the points. Actually two cubic splines, one for $x$ and one for $y$ in terms of some independent parameter. You can make the parameter $t=0\cdots1$ or whatever you want.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In camshaft profile development we do that all the time, as we take measured points and fit splines to parametrize the curve and then calculate velocity, acceleration, jerk, snap, crackle and pop (1st,2nd,3rd,4th,5th,6th derivatives). Some smoothing is required sometimes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are free online books for cubic spline interpolation in C (&lt;a href=&quot;http://www.nrbook.com/a/bookcpdf.php&quot; rel=&quot;nofollow&quot;&gt;link&lt;/a&gt;), and Fortran (&lt;a href=&quot;http://www.nrbook.com/a/bookfpdf.php&quot; rel=&quot;nofollow&quot;&gt;link&lt;/a&gt;) which is easier to read and re-code to something more modern. A Fortran 90 version uses vectorized operations which are only useful in Matlab, or Fortran. I recommend reading &lt;strong&gt;chapter 3-3&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have ported this routines in C# if interested.&lt;/p&gt;&#xA;" OwnerUserId="3301" LastActivityDate="2011-01-16T16:57:40.127" />
  <row Id="17726" PostTypeId="2" ParentId="17713" CreationDate="2011-01-16T17:09:27.870" Score="2" ViewCount="" Body="&lt;p&gt;Invoke an &lt;a href=&quot;http://en.wikipedia.org/wiki/Integral_test_for_convergence&quot; rel=&quot;nofollow&quot;&gt;Integral test for convergence&lt;/a&gt;&#xA;then two Integration by parts give you (up to a mistake in my calculus)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a=1/2$, $b=1/2$, and $c=1/4$.&lt;/p&gt;&#xA;" OwnerUserId="5838" LastActivityDate="2011-01-16T17:09:27.870" CommentCount="5" />
  <row Id="17727" PostTypeId="1" AcceptedAnswerId="17729" CreationDate="2011-01-16T17:16:01.573" Score="0" ViewCount="57" Body="&lt;p&gt;Hi. I'm in the process of reading a paper and I believe there's a mistake, but it could also be me not noticing something.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's the deal:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $\bar\Pi_k$ be the set of all polynomials of degree at most $k$ that satisfy $p(1) = 1$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next, let $\mu_i, i = 1...n$ be a finite set of numbers in the open interval $(-1,1)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In one of the proofs in the paper, they appear to make the following assumption:&#xA;If $p \in \bar\Pi$, then&#xA;$$ \max_i\, p^2(\mu_i) \le p^2(\max_i\, | \mu_i |)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I find absolutely no reason for this to be true. Note: The coefficients of the polynomials are not guaranteed to be non-negative, otherwise s.th. like Jensen's inequality might have done the trick.&lt;/p&gt;&#xA;" OwnerUserId="4171" LastActivityDate="2011-01-16T17:33:45.790" Title="Certain polynomials monotone on [-1:1]?" Tags="&lt;polynomials&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="17728" PostTypeId="2" ParentId="1870" CreationDate="2011-01-16T17:21:35.323" Score="0" ViewCount="" Body="&lt;p&gt;Also, Bagatrix has a suite of math software that shows you individual steps for math problems that it could solve. In my experience though, it is quite limited in what it can actually solve.&lt;/p&gt;&#xA;" OwnerUserId="3143" LastActivityDate="2011-01-16T17:21:35.323" CommunityOwnedDate="2011-01-16T17:21:35.323" />
  <row Id="17729" PostTypeId="2" ParentId="17727" CreationDate="2011-01-16T17:33:45.790" Score="2" ViewCount="" Body="&lt;p&gt;Counterexample: $p(X)=2-X^2$ then $p(0)\geq p(x)$ for all $x\in(-1,1)$. The same property is true for $p^2$. One can check that $p(1)=1$. Take $\mu_0=0$ and $\mu_1$ whatever value different from zero to get a couterexample.&lt;/p&gt;&#xA;" OwnerUserId="5838" LastActivityDate="2011-01-16T17:33:45.790" CommentCount="4" />
  <row Id="17730" PostTypeId="2" ParentId="1870" CreationDate="2011-01-16T17:57:04.943" Score="0" ViewCount="" Body="&lt;p&gt;Good educational software allows one to explore on one's own and at the same time provides problem solving tools. A very nice example of this, and its free, is:&#xA;&lt;a href=&quot;http://www.geogebra.org/cms/&quot; rel=&quot;nofollow&quot;&gt;http://www.geogebra.org/cms/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1369" LastActivityDate="2011-01-16T17:57:04.943" CommunityOwnedDate="2011-01-16T17:57:04.943" />
  <row Id="17731" PostTypeId="1" AcceptedAnswerId="17733" CreationDate="2011-01-16T18:03:17.353" Score="15" ViewCount="848" Body="&lt;p&gt;I have a general question about the applications of mathematics.&#xA;What are some applications of mathematics that are not scientific, perhaps maybe literary or philosophical, or political. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am basically asking for the scope of mathematics, and looking for concrete examples. &lt;/p&gt;&#xA;" OwnerUserId="4728" LastEditorUserId="9" LastEditorDisplayName="" LastEditDate="2011-01-16T19:33:38.103" LastActivityDate="2011-05-07T13:32:49.573" CommunityOwnedDate="2011-01-16T19:07:18.387" Title="Non-Scientific questions solved by mathematics" Tags="&lt;philosophy&gt;" AnswerCount="14" CommentCount="1" FavoriteCount="1" />
  <row Id="17732" PostTypeId="1" AcceptedAnswerId="17738" CreationDate="2011-01-16T18:08:55.417" Score="2" ViewCount="88" Body="&lt;p&gt;Hi&#xA;I'm trying to prove some properties of $L^1$, the space of $\mu$-integrable functions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have $$ f_n \in L^1$$ and $$ \sum_{n \geq 1} || f_n ||_{1} \lt \infty$$ and I used it to prove $$ \sum_{n\geq 1} f_n(x) \lt \infty $$ $\mu$-almost-everywhere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The next thing I would like to show is that $$ \sum_{n\geq 1} f_n(x) $$ is measurable. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know how to show that the pointwise (i.e. $\forall x \in X$) limit of $\mu$-measurable functions is measurable, but I don't know how to cope with a sequence that only converges $\mu$-almost everywhere, i.e. if the limit is not even defined in some points inside a $\mu$-negligible set.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many thanks for your help!&lt;/p&gt;&#xA;" OwnerUserId="5798" LastActivityDate="2011-01-16T19:21:45.480" Title="Proof of properties of $L^1(\mu, X)$" Tags="&lt;real-analysis&gt;&lt;measure-theory&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="17733" PostTypeId="2" ParentId="17731" CreationDate="2011-01-16T18:15:48.080" Score="21" ViewCount="" Body="&lt;p&gt;In my opinion, the application of mathematics outside of fields where its use is well-established like physics is dangerous.  Mathematics cannot prove anything about the world: it can only prove things about models of the world.  People have a tendency to take their models too seriously, and that leads to all sorts of trouble outside of mathematics and physics.  For example, the recent financial crisis was caused in part because people working in mathematical finance took their models too seriously.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is as far as philosophical, political, or sociological applications go.  I am not sure what you mean by literary applications - do you mean applications to the &lt;em&gt;study&lt;/em&gt; of literature, or to its &lt;em&gt;creation&lt;/em&gt;?  And does the latter mean inspiration or something else?  Certainly art draws its inspiration from many sources.&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2011-01-16T18:43:24.527" LastActivityDate="2011-01-16T18:43:24.527" CommunityOwnedDate="2011-01-16T19:07:18.387" CommentCount="11" />
  <row Id="17734" PostTypeId="2" ParentId="17636" CreationDate="2011-01-16T18:29:09.470" Score="0" ViewCount="" Body="&lt;p&gt;If being considered numerically, it's going to be easier I think to just find angular velocity vectors $\omega_t=(\omega_0+\alpha t_i)$ for each timestep, construct a matrix to rotate about each vector by the appropriate amount, and multiply them together. But I was hoping for something a bit faster and more elegant. Is an algebraic answer impossible, or just difficult?&lt;/p&gt;&#xA;" OwnerUserId="5916" LastActivityDate="2011-01-16T18:29:09.470" />
  <row Id="17735" PostTypeId="1" AcceptedAnswerId="17765" CreationDate="2011-01-16T18:29:14.090" Score="3" ViewCount="147" Body="&lt;p&gt;What makes a context free grammar ambiguous? &lt;/p&gt;&#xA;" OwnerUserId="5325" LastEditorUserId="6296" LastEditorDisplayName="" LastEditDate="2011-03-04T06:04:29.363" LastActivityDate="2011-03-04T06:04:29.363" Title="What makes a context free grammar ambiguous?" Tags="&lt;computer-science&gt;&lt;formal-systems&gt;" AnswerCount="2" CommentCount="5" />
  <row Id="17736" PostTypeId="1" CreationDate="2011-01-16T18:34:34.743" Score="1" ViewCount="99" Body="&lt;p&gt;My question is as to which is the best method to find extremal eigenvalues of a real symmetric matrix? Currently I am using Lanczos Iterations followed by Bisection Method. Does anyone have a better suggestion?&lt;/p&gt;&#xA;" OwnerUserId="5917" LastEditorUserId="367" LastEditorDisplayName="" LastEditDate="2011-01-16T22:17:46.437" LastActivityDate="2011-01-16T22:17:46.437" Title="Which is the best method to find extremal eigenvalues of a symmetric matrix?" Tags="&lt;linear-algebra&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="17737" PostTypeId="2" ParentId="17731" CreationDate="2011-01-16T18:36:51.400" Score="6" ViewCount="" Body="&lt;p&gt;Interestingly, I saw this video a couple of days back and I think this is relevant to the question though it may not directly answer your question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;&lt;a href=&quot;http://www.youtube.com/view_play_list?p=3641A12A6ADDB4B1&quot; rel=&quot;nofollow&quot;&gt;L'importance des mathématiques&lt;/a&gt;&quot; by Timothy Gowers at The Millennium Meeting&lt;/p&gt;&#xA;" OwnerUserId="3302" LastActivityDate="2011-01-16T18:36:51.400" CommunityOwnedDate="2011-01-16T19:07:18.387" CommentCount="1" />
  <row Id="17738" PostTypeId="2" ParentId="17732" CreationDate="2011-01-16T18:39:47.387" Score="3" ViewCount="" Body="&lt;p&gt;denote by $A$ the set where $\sum f_n$ doesn't converge, then you showed that $\mu(A)=0$. Let $g_n = f_n \chi_{X-A}$, then $X-A$ is a measurable set, hence $g_n$ are measurable functions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sum g_n = \sum f_n$ almost everywhere, by the definition of $g_n$, and $\sum_1^m g_n$ is pointwise convergent, so its limit is measurable. now change it back on the set $A$ to be $\sum f_n$, to get that $\sum f_n$ is also measurable (Since we changed a measurable function on a zero measure set). In other words $h = \sum g_n - \sum f_n$ is almost everywhere zero, so it is measurable, hence $\sum f_n = \sum g_n -h$ is measurable.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Suppose f is 0 almost everywhere, and let $A=f^{-1}(0)$ so $\mu(X-A)=0$, then f is measurable. For every $t\in \mathbb{R}$ we have $B=f^{-1}(t,\infty)$ satisfies $A\subseteq B$ or $B \subseteq X-A$. In the first case you have $\mu(X-B)\leq \mu (X-A)=0$ so $B-X$ is measurable (with measure zero) hence B is measurable. In the second case, $\mu (B) \leq \mu(X-A)=0$ and again B is measurable. (actually, we need to do this with the outer measure of B, X-B).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;any way, now that you know that B is measurable, then f is measurable (one of the equivalent definitions). This shows that a function that is zero almost everywhere is measurable, so you can always add it to another measurable function to get another measurable function, or in other words, you can always change a measurable function in a zero measure set, and still have a measurable function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(regarding the answer in your other question, I assumed that the measure is complete - meaning that if $A \subseteq B$ and $\mu(B)=0$ then A is measurable and then its measure must be zero. This is true for example in the Lebesgue measure)&lt;/p&gt;&#xA;" OwnerUserId="2125" LastEditorUserId="2125" LastEditorDisplayName="" LastEditDate="2011-01-16T19:21:45.480" LastActivityDate="2011-01-16T19:21:45.480" CommentCount="4" />
  <row Id="17739" PostTypeId="1" CreationDate="2011-01-16T18:45:21.543" Score="0" ViewCount="99" Body="&lt;p&gt;Is everything that you can write in math (that makes mathematical sense) an expression? If not, what would be examples of non-expressions? And would all expressions be composed of expressions themselves?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, are operators (like the differential operator) by themselves expressions?&lt;/p&gt;&#xA;" OwnerUserId="3143" LastActivityDate="2011-01-16T20:02:47.580" Title="Is everything an expression?" Tags="&lt;definition&gt;" AnswerCount="0" CommentCount="6" FavoriteCount="0" />
  <row Id="17740" PostTypeId="2" ParentId="418" CreationDate="2011-01-16T18:48:17.327" Score="0" ViewCount="" Body="&lt;p&gt;I'm going to give a constructive procedure that hints at generalizations of the cut-and-paste/split function approaches of Akhil/Simon.  The top-level idea is that given a countable $S \subseteq \mathbb{R}$ to be deleted, we will find a countably infinite subset $T \subseteq \mathbb{R} \setminus S$.  We will then split $T$ into two disjoint sets $A$ and $B$ having size $|S|$ and $|T|$, respectively and identify the deleted $S$ from $\mathbb{R}$ with $A$ and $T$ with $B$.  In particular, this will fix all points in $\mathbb{R} \setminus (S \cup T)$ and will send the elements from $S \cup T$ to $A \cup B = T$.  The point is that we work with well-orderable subsets of $\mathbb{R}$ which admit nice bijections. (e.g., a well-order order isomporphic to $\mathbb{N}$ can be put into bijective correspondence with $\mathbb{N} \times \{0, 1\}$ by sending the evens $2n$ to $(n, 0)$ and the odds $2n+1$ to $(n, 1)$.)  Our desired bijections are thus formed from a cut/paste method where we cut out well-ordered sets, work with bijections of them, and then paste the identity mapping with the induced maps from these bijections as Akhil suggests in his answer and as described below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) First consider any injection $j: \mathbb{N} \rightarrow [0, 1]$ including $0$ and $1$ in the range (e.g., $n \mapsto \frac{1}{n}$ for $n &amp;gt; 0$ and $0 \mapsto 0$).  Then we have a $1:1$ correspondence between $\mathbb{N}$ and $C = \text{range}(j)$.  Because $C$ includes $0$ and $1$, we have $[0, 1] \setminus C = (0, 1) \setminus C$ so given a bijection $b$ between $C$ and $(0, 1) \cap C$, we can paste together the identity map and $b$ on these respective domains to get a bijection between $[0, 1]$ and $(0, 1)$.  Now note that $j$ maps exactly two elements $m, n \in \mathbb{N}$ to the $\text{range}(j)$ that are not in $(0, 1)$, mainly the ones being mapped to $0$ and $1$.  Consequently, letting $g: \mathbb{N} \rightarrow \mathbb{N} \setminus \{m, n\}$ be a bijection, say one simply listing the Natural numbers in order that skips $m$ and $n$, you can verify that $j \circ g \circ j^{-1}: C \rightarrow (0, 1) \cap C$ is a well-defined bijection.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) For this part, extend your countable bijection $s: \omega \rightarrow S$ to a countable injection $k: \omega + \omega \rightarrow \mathbb{R}$ ($\omega = \mathbb{N}$ and $\omega + \omega$ has an ordering isomorphic to one on $\mathbb{N} \times \{0, 1\}$ given by $(m, i) \leq (n, i)$ when $m \leq n$ and $(m, 0) \leq (n, 1)$ always).   The reason such an injection exists is because given a set of Reals that resulted from the subtraction of countably many elements, we can always find a countable number that haven't been removed by iteratively diagonalizing against all the ones in our list, adding each new one obtained by this method to the top of the list so that it too gets diagonalized against in the construction of the next Real not in the list. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we have a $1:1$ correspondence between $\omega + \omega$ and $D = \text{range}(k)$.  Because $D$ includes all elements in $S$ by construction, we have $\mathbb{R} \setminus D = (\mathbb{R} \setminus S) \setminus D$ so given a bijection $b^*$ between $D$ and $(\mathbb{R} \setminus S) \cap D$, we can paste together the identity map and $b^*$ on these respective domains to get a bijection between $\mathbb{R}$ and $\mathbb{R} \setminus S$.  Now note that all elements listed by $k$ at and beyond $\omega$ are in $\mathbb{R} \setminus S$ by the fact that $k$ only lists each element once and already listed all elements in $S$ before $\omega$.  Consequently letting $h: \omega + \omega \rightarrow (\omega + \omega) \setminus \omega$ be a bijection,  you can verify that $k \circ h \circ k^{-1}: D \rightarrow (\mathbb{R} \setminus S) \cap D$ is a well-defined bijection.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[I realize that this can be technical without knowing about ordinals but just think of $\omega + \omega$ as being the set of Natural numbers with a copy of the Natural numbers on top and $(\omega + \omega) \setminus \omega$ as the top set of Natural numbers.  Establishing a bijection between the two sets is similar to establishing a bijection between the set of evens and the set of Natural numbers.]&lt;/p&gt;&#xA;" OwnerUserId="4693" LastEditorUserId="4693" LastEditorDisplayName="" LastEditDate="2011-01-17T06:00:31.107" LastActivityDate="2011-01-17T06:00:31.107" />
  <row Id="17741" PostTypeId="2" ParentId="17736" CreationDate="2011-01-16T18:51:16.293" Score="1" ViewCount="" Body="&lt;p&gt;If you are looking for specific software implementations I suggest looking &lt;a href=&quot;http://en.wikipedia.org/wiki/BLAS&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;. Download the source and look what authors of the software are using. Since these are widely used libraries, they are the fastest for the widest range of the problems. Any faster algorithm would definitely be data specific, meaning that for some range of problems it will be faster, but it will be slower for others. Note that this is a generic response, but the so is the question.&lt;/p&gt;&#xA;" OwnerUserId="4742" LastActivityDate="2011-01-16T18:51:16.293" CommentCount="3" />
  <row Id="17742" PostTypeId="1" CreationDate="2011-01-16T18:56:03.297" Score="-1" ViewCount="165" Body="&lt;p&gt;prove that the sum of all the factors of any natural number is=(a^p+1 -1)/a-1*......&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;i&gt;[YF] Here is one possible interpretation of the above:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Prove the formula for the sum-of-divisors function&#xA;$$\sigma\left(\prod_i p_i^{a_i}\right) = \prod_i \frac{p_i^{a_i+1}-1}{p_i-1}.$$&lt;/i&gt;&lt;/p&gt;&#xA;" OwnerUserId="5918" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2011-01-16T21:18:51.170" LastActivityDate="2011-01-17T00:31:18.880" Title="sum of the factors of any natural number" Tags="&lt;number-theory&gt;" AnswerCount="2" CommentCount="6" />
  <row Id="17743" PostTypeId="1" AcceptedAnswerId="17746" CreationDate="2011-01-16T18:59:19.930" Score="2" ViewCount="67" Body="&lt;p&gt;Hi&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $f$ is a $\mu$-measurable function and I change its values on a $\mu$-negligible set, i.e. on $Y \subset Z$ with $\mu (Z) = 0$, why is $f$ still measurable?&lt;/p&gt;&#xA;" OwnerUserId="5798" LastActivityDate="2011-01-16T19:34:31.973" Title="Change the values of a measurable function on a negligible set" Tags="&lt;measure-theory&gt;" AnswerCount="1" />
  <row Id="17744" PostTypeId="2" ParentId="17731" CreationDate="2011-01-16T19:00:06.527" Score="9" ViewCount="" Body="&lt;p&gt;One example of the influence of mathematics on politics would be &lt;a href=&quot;http://en.wikipedia.org/wiki/Arrow%27s_impossibility_theorem&quot; rel=&quot;nofollow&quot;&gt;Arrow's Impossibility Theorem&lt;/a&gt;, which states something like &quot;no voting system is perfectly fair&quot;, but as Qiaochu notes, people have a tendency to take their models too seriously. Whether Arrow's theorem actually means we can never have a voting system that everyone is &quot;reasonably happy with&quot;, I'm not sure (you would have to define what that means; some sort of utility function, but I'm not particularly familiar with economics)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, &lt;a href=&quot;http://en.wikipedia.org/wiki/Game_theory&quot; rel=&quot;nofollow&quot;&gt;game theory&lt;/a&gt; has applications to economics and politics.&lt;/p&gt;&#xA;" OwnerUserId="2954" LastActivityDate="2011-01-16T19:00:06.527" CommunityOwnedDate="2011-01-16T19:07:18.387" CommentCount="3" />
  <row Id="17745" PostTypeId="1" AcceptedAnswerId="17767" CreationDate="2011-01-16T19:02:11.500" Score="4" ViewCount="123" Body="&lt;p&gt;Suppose $K$ is a $p$-adic field (finite extension of the $p$-adics), and let $n$ be any integer (independent of what $p$ is). Define $U$ to be the set of all $x$ in $K$ such that $|x| = 1$ and such that $x = y^n$ for some $y$ in $K$. I would like to show that $U$ is an open set and that as a multiplicative group $U$ has finite index in the group of elements of $K$ of norm $1$. What's the best way of seeing why this is true (assuming it is)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I pretty much have an idea why this holds.. in the $p$-adic case you can prove the $n$th powers are of bounded index in ${\bf Z}_{p^l}$ for each $l$ and then use an inverse limiting-type argument as $l$ goes to infinity to get this for $K = {\mathbb Q_p}$, and I think an analogous argument using powers of a uniformizer in place of powers of $p$ should work for a general $K$. But I keep thinking that this should be some well-known result or something that follows quickly from a well-known result. So I thought I'd throw this out.&lt;/p&gt;&#xA;" OwnerUserId="3035" LastActivityDate="2011-01-16T21:14:19.100" Title="$n$th powers in the p-adics" Tags="&lt;p-adic-number-theory&gt;" AnswerCount="2" />
  <row Id="17746" PostTypeId="2" ParentId="17743" CreationDate="2011-01-16T19:05:56.880" Score="5" ViewCount="" Body="&lt;p&gt;This is not true if the measure is not complete. Simply take a non-measurable subset $Y$ of a measurable null-set $Z$ and look at the characteristic function of $Y$, which is clearly not measurable but agrees with the (measurable) zero function outside of $Y$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, your modified $f$ will be measurable with respect to the completion $\tilde{\mu}$ of the measure and agree with $f$ outside of a $\tilde{\mu}$-null set, so the $\tilde{\mu}$-integral will remain unchanged by such a modification and that's enough for most purposes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To make this more precise, let $g$ be the modified function and $Y = \{x\,:f(x) \neq g(x)\}$. By hypothesis, $Y \subset Z$ with $Z$ measurable and $\mu(Z) = 0$. Observe that $g = [X \smallsetminus Y] \cdot f + [Y] \cdot (g - f)$. With respect to the completion $Y$ and $X \smallsetminus Y$ are measurable because $\tilde{\mu}(Y) = 0$. Note that on a &lt;em&gt;complete&lt;/em&gt; space every function which is nonzero only inside a null-set is measurable. Therefore $g$ is the sum of two $\tilde{\mu}$-measurable functions and hence measurable.&lt;/p&gt;&#xA;" OwnerUserId="5363" LastEditorUserId="5363" LastEditorDisplayName="" LastEditDate="2011-01-16T19:34:31.973" LastActivityDate="2011-01-16T19:34:31.973" CommentCount="3" />
  <row Id="17747" PostTypeId="1" AcceptedAnswerId="17751" CreationDate="2011-01-16T19:06:19.933" Score="3" ViewCount="163" Body="&lt;p&gt;i'm studing graphs in algorithm and complexity,&#xA;(but i'm not very good at math) as in title, &#xA;why a complete graph has $\frac{n(n-1)}{2}$ edges?&#xA;and how this is related with combinatorics?&lt;/p&gt;&#xA;" OwnerUserId="5046" LastActivityDate="2011-01-16T19:20:57.977" Title="why a complete graph has $\frac{n(n-1)}{2}$ edges?" Tags="&lt;graph-theory&gt;" AnswerCount="4" FavoriteCount="1" />
  <row Id="17748" PostTypeId="2" ParentId="17713" CreationDate="2011-01-16T19:06:58.377" Score="6" ViewCount="" Body="&lt;p&gt;You can try using elementary estimates (upper and lower bounds) with integrals (see the end of the answer), or can use the more general and quite useful &lt;a href=&quot;http://en.wikipedia.org/wiki/Euler%E2%80%93Maclaurin_formula&quot; rel=&quot;nofollow&quot;&gt;Euler Mclaurin Summation Formula&lt;/a&gt;, which gives us&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{k=1}^{n} \frac{\log^2 k}{k^3} = \int_{1}^{n} \frac{\log^2 x}{x^3} \ \text{d}x + \frac{\log ^2 n}{2n^3} + C + \mathcal{O}(\frac{1}{n^{3+e}})$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\displaystyle e &amp;gt; 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now $$\int_{1}^{n} \frac{\log^2 x}{x^3} \ \text{d}x  = - \frac{2\log^2 n + 2 \log n + 1}{4n^2} + 1/4$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{k=1}^{n} \frac{\log^2 k}{k^3} = K  - \frac{2\log^2 n + 2 \log n + 1}{4n^2} + \frac{\log^2 n}{2n^3} + \mathcal{O}\left(\frac{1}{n^{3+e}}\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the LHS converges we have that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{k=1}^{n} \frac{\log^2 k}{k^3} = \sum_{k=1}^{\infty}\frac{\log^2 k}{k^3}   - \frac{2\log^2 n + 2 \log n + 1}{4n^2} + \frac{\log^2 n}{2n^3} + \mathcal{O}\left(\frac{1}{n^{3+e}}\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{k=n+1}^{\infty}\frac{\log^2 k}{k^3} = \frac{2\log^2 n + 2 \log n + 1}{4n^2} - \frac{\log^2 n}{2n^3} + \mathcal{O}\left(\frac{1}{n^{3+e}}\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And so&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{k=n}^{\infty}\frac{\log^2 k}{k^3} = \frac{2\log^2 n + 2 \log n + 1}{4n^2} + \frac{\log^2 n}{2n^3} + \mathcal{O}\left(\frac{1}{n^{3+e}}\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the summation formula allows us to figure out the exact terms which make up the $\displaystyle \mathcal{O}\left(\frac{1}{n^{3+e}}\right)$ and so we can make the formula more accurate by including as many lower order terms we need.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;A more elementary way:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $\displaystyle f(x) \ge 0$ and is monotonically decreasing then we have that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_{n}^{M+1} f(x)\  \text{d}x \le \sum_{k=n}^{M} f(k) \le f(n) + \int_{n}^{M} f(x)\  \text{d}x$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which can be seen by bounding the area under the curve by rectangles of width $1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $\displaystyle f(n) = \frac{\log ^2 n}{n^3}$, we can take limit as $\displaystyle M \to \infty$ to get the asymptotic formula you need.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2011-01-16T21:12:11.570" LastActivityDate="2011-01-16T21:12:11.570" CommentCount="2" />
  <row Id="17749" PostTypeId="2" ParentId="17747" CreationDate="2011-01-16T19:13:44.940" Score="6" ViewCount="" Body="&lt;p&gt;A complete graph has an edge between any two vertices. You can get an edge by picking any two vertices.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if there are $n$ vertices, there are $n$ choose $2$ =  ${n \choose 2} = n(n-1)/2$ edges.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does that help?&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2011-01-16T19:13:44.940" />
  <row Id="17750" PostTypeId="2" ParentId="17747" CreationDate="2011-01-16T19:16:54.020" Score="3" ViewCount="" Body="&lt;p&gt;$\frac{n(n-1)}{2} = \binom{n}{2}$ is the number of ways to choose 2 unordered items from n distinct items.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your case, you actually want to count how many unordered pair of vertices you have, since every such pair can be exactly one edge (in a simple complete graph).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;suppose $(v,u)$ is an edge, then v can be any of the vertices in the graph - you have n options for this. u can be any vertex that is not v, so you have (n-1) options for this. the problem is that you counted each edge twice - one time as $(u,v)$ and one time as $(v,u)$ so you need to divide by two, and then you get that you have $\frac {n(n-1)}{2}$ edges in a complete simple graph.&lt;/p&gt;&#xA;" OwnerUserId="2125" LastActivityDate="2011-01-16T19:16:54.020" />
  <row Id="17751" PostTypeId="2" ParentId="17747" CreationDate="2011-01-16T19:19:49.447" Score="8" ViewCount="" Body="&lt;p&gt;A simpler answer without binomials: A complete graph means that every vertex is connected with every other vertex. If you take one vertex of your graph, you therefore have $n-1$ outgoing edges from that particular vertex. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, you have $n$ vertices in total, so you might be tempted to say that there are $n(n-1)$ edges in total, $n-1$ for every vertex in your graph. But this method counts every edge twice, because every edge going out from one vertex is an edge going into another vertex. Hence, you have to divide your result by 2. This leaves you with $n(n-1)/2$.&lt;/p&gt;&#xA;" OwnerUserId="4171" LastActivityDate="2011-01-16T19:19:49.447" />
  <row Id="17752" PostTypeId="2" ParentId="17747" CreationDate="2011-01-16T19:20:57.977" Score="4" ViewCount="" Body="&lt;p&gt;$\frac{n(n-1)}{2}$ comes from simple counting argument. You could directly say that every edge is obtained by choosing a pair of vertices so the number is $C(n,2) = \frac{n(n-1)}{2}$ or you could take the other way of counting. Label the vertices $1,2, \ldots ,n$. The first vertex is now joined to $n-1$ other vertices. The second vertex has already been joined to vertex $1$ and hence has to be joined to the remaining $n-2$ vertices and in general the $k^{th}$ vertex has already been joined to the previous $k-1$ vertices and hence has to be joined to the remaining $n-k$ vertices. So the total number of edges is given by $(n-1) + (n-2) + \ldots 2 + 1 = \frac{n(n-1)}{2}$.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastActivityDate="2011-01-16T19:20:57.977" />
  <row Id="17753" PostTypeId="1" CreationDate="2011-01-16T19:23:20.253" Score="1" ViewCount="52" Body="&lt;p&gt;I'm always needing to learn more about the GAP-groups.&#xA;Can anyone help me on the following question?&#xA;Question:&#xA;I have a list of 500 sublist where each sublist consists of six groups.&#xA;Is there a command to display or renumber only those sublists with trivial intersection?&#xA;Is there a function in GAP can solve this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This questioning arose when I studied a problem on coverage of finite groups.&lt;/p&gt;&#xA;" OwnerUserId="5802" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-16T20:42:59.510" LastActivityDate="2011-01-16T20:42:59.510" Title="Gap groups commands" Tags="&lt;homework&gt;&lt;computer-science&gt;" CommentCount="4" />
  <row Id="17754" PostTypeId="2" ParentId="17731" CreationDate="2011-01-16T19:23:45.367" Score="4" ViewCount="" Body="&lt;p&gt;Gödel's theorem had a great impact in logic and, hence, in philosophy.&lt;/p&gt;&#xA;" OwnerUserId="1971" LastActivityDate="2011-01-16T19:23:45.367" CommunityOwnedDate="2011-01-16T19:23:45.367" CommentCount="3" />
  <row Id="17755" PostTypeId="2" ParentId="17731" CreationDate="2011-01-16T19:24:32.043" Score="3" ViewCount="" Body="&lt;p&gt;There are some literary works attributed to famous writers but doubtfully so. One way of checking whether they're genuine is by extracting and comparing statistical &quot;fingerprints&quot;. The field is called &quot;computational stylometry&quot;. One of the standard challenges are identifying which works are genuine Shakespeare and which should be attributed to Fletcher (as if it mattered who wrote the piece, rather than its &quot;artistic worth&quot;, if there be any such thing).&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2011-01-16T19:24:32.043" CommunityOwnedDate="2011-01-16T19:24:32.043" CommentCount="4" />
  <row Id="17756" PostTypeId="1" AcceptedAnswerId="17763" CreationDate="2011-01-16T19:25:07.567" Score="3" ViewCount="177" Body="&lt;p&gt;What is the maximum number of times a piece of paper (of non-zero thickness) can be folded in half (mathematically)? &lt;br/&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: &lt;br/&gt;&#xA;I totally forgot to mention the &quot;non-zero thickness&quot; part; I've now included it.&lt;/p&gt;&#xA;" OwnerUserId="5915" LastEditorUserId="5915" LastEditorDisplayName="" LastEditDate="2011-01-16T20:35:42.610" LastActivityDate="2011-01-16T20:35:42.610" Title="How many times can a piece of paper be folded in half?" Tags="&lt;geometry&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="17757" PostTypeId="2" ParentId="17742" CreationDate="2011-01-16T19:28:00.490" Score="5" ViewCount="" Body="&lt;p&gt;I assume you are referring to the identity of the divisor function&#xA;$$\sigma (n) = \displaystyle\prod_{i=1}^r\frac{p_i^{\alpha_i+1}-1}{p_i-1}$$&#xA;where $n = p_1^{\alpha_1} p_2^{\alpha_2} \dots p_r^{\alpha_r}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is derived by showing that $\sigma$ is multiplicative and that the sum of the divisors of a prime power $p^k$ is the sum $1 + p + p^2 + \dots + p^k = \frac{p^{k+1}-1}{p-1}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To show $\sigma$ is multiplicative, let $a, b \in \mathbb{N}$, with $\gcd(a,b)=1$. Since any prime that divides $ab$ must divide either $a$ or $b$, as they share no common factors, the sum of divisors of $ab$ must be exactly the sum of the divisors of $a$ multiplied with the sum of the divisors of $b$, as the product of any divisor of $a$ and any divisor of $b$ must also be a unique divisor of $ab$ (see Arturo's answer for a more complete proof). Thus $\sigma(ab) = \sigma(a)\sigma(b)$.&#xA;Since any two primes are coprime, you can decompose $\sigma(n)$ into $\displaystyle\prod_{i=1}^r \sigma(p_i^{\alpha_i})$, and use the hint I mentioned earlier.&lt;/p&gt;&#xA;" OwnerUserId="1016" LastEditorUserId="1016" LastEditorDisplayName="" LastEditDate="2011-01-17T00:31:18.880" LastActivityDate="2011-01-17T00:31:18.880" CommentCount="2" />
  <row Id="17758" PostTypeId="2" ParentId="17756" CreationDate="2011-01-16T19:30:02.273" Score="4" ViewCount="" Body="&lt;p&gt;Theoretically you can fold the paper into half infinite number of times since $1 - (\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \cdots) = 0$. However, practically the number of times you can fold the paper into half will depend on the dimensions of the paper (length, breadth and thickness), the stiffness of the paper, and more importantly your patience and mechanical strength to keep folding it.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2011-01-16T19:40:10.440" LastActivityDate="2011-01-16T19:40:10.440" CommentCount="1" />
  <row Id="17759" PostTypeId="2" ParentId="17731" CreationDate="2011-01-16T19:32:52.527" Score="2" ViewCount="" Body="&lt;p&gt;Through computer science and finally software develeopment, almost all questions people post these days are solved &quot;by&quot; (rather &quot;with the help of&quot;) mathematics. Take Google as a prominent example.&lt;/p&gt;&#xA;" OwnerUserId="3330" LastActivityDate="2011-01-16T19:32:52.527" CommunityOwnedDate="2011-01-16T19:32:52.527" />
  <row Id="17760" PostTypeId="2" ParentId="17731" CreationDate="2011-01-16T19:33:23.703" Score="2" ViewCount="" Body="&lt;p&gt;Michael Harris's article &lt;a href=&quot;http://www.math.jussieu.fr/~harris/otherarticles.html&quot; rel=&quot;nofollow&quot;&gt;An automorphic reading of Thomas Pynchon's &lt;em&gt;Against the Day&lt;/em&gt;&lt;/a&gt; describes some applications of mathematics to literature.  (On the same page, the article &lt;a href=&quot;http://www.math.jussieu.fr/~harris/otherarticles.html&quot; rel=&quot;nofollow&quot;&gt;Do android's prove theorems in their sleep?&lt;/a&gt; gives some applications of literature to mathematics.)&lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2011-01-16T19:33:23.703" CommunityOwnedDate="2011-01-16T19:33:23.703" CommentCount="1" />
  <row Id="17761" PostTypeId="2" ParentId="17731" CreationDate="2011-01-16T19:39:30.903" Score="15" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Cryptanalysis_of_the_Engima&quot; rel=&quot;nofollow&quot;&gt;Breaking the Enigma code&lt;/a&gt; was an application of mathematics to warfare, which would probably not be regarded as a scientific application.  My understanding is that mathematics continues to play a large role in modern cryptanalysis (such as in the work of the NSA in the United States).&lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2011-01-16T19:39:30.903" CommunityOwnedDate="2011-01-16T19:39:30.903" CommentCount="2" />
  <row Id="17762" PostTypeId="2" ParentId="17735" CreationDate="2011-01-16T19:43:23.677" Score="1" ViewCount="" Body="&lt;p&gt;Intuitively, what makes a language inherently ambiguous is that it is composed of distinct parts which have some intersection; the intersection cannot be &quot;singled out&quot;, and therefore the only way to approach is is through the distinct parts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, the language of words of the form $a^xb^yc^z$, where $x=y$ or $y=z$, is context-free, since it's the union of $a^nb^nc^*$ and $a^*b^nc^n$. However, words of the form $a^nb^nc^n$ cannot be &quot;singled out&quot; since that language is not context-free, so we would expect the language to be inherently ambiguous.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The formal proof (using Ogden's lemma) goes like this. Consider the word $a^nb^nc^{n!}$, where $n$ is &quot;big enough&quot;. By Ogden's lemma, the $a^nb^n$ can be pumped up, and so we can derive $a^{n!}b^{n!}c^{n!}$. Similarly, we can derive the same word from $a^{n!}b^nc^n$. The two derivations overlap (over the middle part) and so must be &quot;inherently&quot; different (i.e. have different derivation trees).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another example, from the world of regular languages, is as follows. Consider the language over the alphabet $\{a,b\}$ of all non-empty words having either an even number of $a$'s or an even number of $b$'s (or both). Does it have a disjoint representation as a union $L_1^+ \cup \cdots L_m^+$ (here $+$ is the Kleene plus)? Intuitively, the problem is that a word with both $a$ and $b$ even is &quot;ambiguous&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The formal proof goes by the usual pumping lemma. Pick an odd $n$ that is greater than the pumping length of all $L_i^+$'s in a given representation. The word $a^n b^{n!}$ is generated by a unique $L_i^+$. By the pumping lemma, $L_i^+$ also generates $a^{n!} b^{n!}$. Similarly, the word $a^{n!} b^n$ is generated by a unique $L_j$, which also generates $a^{n!} b^{n!}$. If $i=j$ then $L_i^+$ generates the concatenation, which has both $a$ and $b$ odd - so $i \neq j$ and we get inherent ambiguity.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2011-01-16T19:43:23.677" CommentCount="2" />
  <row Id="17763" PostTypeId="2" ParentId="17756" CreationDate="2011-01-16T20:02:33.960" Score="7" ViewCount="" Body="&lt;p&gt;The zero thickness model is a bad model.  For nonzero thickness the problem was studied by &lt;a href=&quot;http://pomonahistorical.org/12times.htm&quot; rel=&quot;nofollow&quot;&gt;Britney Gallivan&lt;/a&gt;, who derived an upper bound on the number of folds based on the thickness and dimensions of the paper.  The current world record for actual paper was set by her; it's $12$.  There are some additional details and references at &lt;a href=&quot;http://en.wikipedia.org/wiki/Britney_Gallivan&quot; rel=&quot;nofollow&quot;&gt;her Wikipedia article&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2011-01-16T20:02:33.960" CommentCount="4" />
  <row Id="17765" PostTypeId="2" ParentId="17735" CreationDate="2011-01-16T20:02:58.707" Score="5" ViewCount="" Body="&lt;p&gt;And now, an answer to the actual question...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A grammar is &lt;i&gt;ambiguous&lt;/i&gt; if there's a word which has two different derivation trees. You'll have to look up &lt;i&gt;derivation tree&lt;/i&gt; in your textbook since drawing them is awkward, but the idea that it doesn't matter in which order you're doing the derivations as long as it's basically the same derivation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, consider the grammar $S \rightarrow AS|a, A\rightarrow a$. Let's derive $aa$ twice: $$ S \rightarrow AS \rightarrow Aa \rightarrow aa $$ $$ S \rightarrow AS \rightarrow aS \rightarrow aa $$ These derivations share the same derivation tree. However, consider now the grammar $S\rightarrow SS|a$, generating the same language. There are now two &quot;really&quot; different derivations for $aaa$: $$S \rightarrow SS \rightarrow Sa \rightarrow SSa \rightarrow Saa \rightarrow aaa$$ $$S \rightarrow SS \rightarrow aS \rightarrow aSS \rightarrow aaS \rightarrow aaa$$ In the first derivation, the first two $a$'s are derived together, whereas in the second derivation, it is the second two $a$'s which are derived together. So it's $(aa)a$ vs. $a(aa)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case, the language $a^+$ has an unambiguous grammar (the first one I exhibited), but in other cases, all grammars are ambiguous; such languages are called &lt;i&gt;inherently ambiguous&lt;/i&gt;. See my other answer for more on that.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2011-01-16T21:20:40.513" LastActivityDate="2011-01-16T21:20:40.513" CommentCount="3" />
  <row Id="17766" PostTypeId="2" ParentId="17731" CreationDate="2011-01-16T20:07:31.743" Score="0" ViewCount="" Body="&lt;p&gt;Lacan applied Topology to Psychoanalysis, to great acclaim (at least of his multitudinous followers).&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2011-01-16T20:07:31.743" CommunityOwnedDate="2011-01-16T20:07:31.743" CommentCount="1" />
  <row Id="17767" PostTypeId="2" ParentId="17745" CreationDate="2011-01-16T20:09:21.223" Score="5" ViewCount="" Body="&lt;p&gt;There are lots of ways to see this.  My preferred method is as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you want to show is that if $x \equiv 1 \bmod \pi^N$ for sufficiently large&#xA;$N$ then $x$ is an $n$th power; here $\pi$ is uniformizer.  (This shows that $U$ contains all $x$ that are $\equiv 1 \bmod \pi^N$, and hence is open, as desired.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, just use the classical binomial formula: writing $x = 1 + \pi^N y,$&#xA;we have &#xA;$$x^{1/n} = (1 + \pi^N y)^{1/n} = \sum_{i=0}^{\infty} \frac{1}{n}(\frac{1}{n}&#xD;&#xA;- 1) \cdots (\frac{1}{n} - i + 1) \frac{\pi^{N i} y^i}{i !}.$$&#xA;The denominator of the $i$th term is (bounded above by) $n^i i!$, so provided&#xA;that $N$ is large enough, the ratio $\dfrac{\pi^{N i}}{n^i i!}$ will tend to&#xA;zero $\pi$-adically, and thus so will our series.  It's then easy to argue&#xA;that this series in fact converges to an $n$th root of $x$, as required.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A good example to think about is the case $n = 2$ and $K = \mathbb Q_p$,&#xA;first when $p$ is odd, and then when $p = 2$.  In the former case you should&#xA;find that any $x \equiv 1 \bmod p$ is a square, while in the latter case, you will&#xA;find that the condition $x \equiv 1 \bmod 8$ is required.&lt;/p&gt;&#xA;" OwnerUserId="221" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-16T21:14:19.100" LastActivityDate="2011-01-16T21:14:19.100" CommentCount="1" />
  <row Id="17768" PostTypeId="1" AcceptedAnswerId="17786" CreationDate="2011-01-16T20:21:15.520" Score="6" ViewCount="262" Body="&lt;p&gt;Hello there. I've just spent past 5 days trying to get a solution to this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Find all whole numbers, for whom the fraction&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{n^3 + 2010}{n^2 + 2010}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;yields another whole number.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Before asking, please note that this is indeed an exercise from a math competition. However, this competition is already over, so don't worry; you're not doing my homework.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If anyone would be so kind to provide an explanation, could you formulate it in terms of high-school mathematics? Thanks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;P.S.: Please pardon my crude English. It's not my primary language.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="5919" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2011-01-16T20:42:28.763" LastActivityDate="2011-01-17T01:06:10.200" Title="For what n does $n^2 + 2010$ divide $n^3 + 2010$." Tags="&lt;number-theory&gt;" AnswerCount="4" CommentCount="1" FavoriteCount="2" />
  <row Id="17769" PostTypeId="1" CreationDate="2011-01-16T20:22:11.113" Score="4" ViewCount="97" Body="&lt;p&gt;Suppose I have a &lt;strong&gt;stack&lt;/strong&gt; and I want to find the &lt;strong&gt;permutations&lt;/strong&gt; of numbers 1,2,3,...n. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can push and pop. e.g. if n=2: push,pop,push,pop 1,2 and push,push,pop,pop 2,1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;if &lt;strong&gt;n=4 I can only get 14&lt;/strong&gt; from the 24 permutations using the stack.. Does anyone know any function F(n) that could produce the number of permutations the stack &lt;strong&gt;(only one)&lt;/strong&gt; can produce? &#xA;eg f(1)=1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;f(2)=2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;f(4)=14&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you&lt;/p&gt;&#xA;" OwnerUserId="5920" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-16T20:55:23.593" LastActivityDate="2011-01-16T20:55:23.593" Title="Using one stack to find number of permutations" Tags="&lt;combinatorics&gt;&lt;permutations&gt;" AnswerCount="1" />
  <row Id="17770" PostTypeId="2" ParentId="17731" CreationDate="2011-01-16T20:23:37.047" Score="1" ViewCount="" Body="&lt;p&gt;A professor gave a talk in our seminar about using math to find a proto-language, a language that is an earlier ancestor of many modern languages.  Among other techniques they used the &quot;killing fish&quot; method to try to find correlations between two now distinct languages.  The method is described as such:  If you want to find the deepest part of the ocean you can place a single fish in the water and slowly drain the water.  As the water level lowers, the fist will seek out deeper and deeper places.  Eventually the fish will get stuck in a local minimum.&lt;/p&gt;&#xA;" OwnerUserId="2838" LastActivityDate="2011-01-16T20:23:37.047" CommunityOwnedDate="2011-01-16T20:23:37.047" CommentCount="2" />
  <row Id="17771" PostTypeId="2" ParentId="17768" CreationDate="2011-01-16T20:29:45.577" Score="0" ViewCount="" Body="&lt;p&gt;We need to have $n^2+2010$ divide $n^3+2010$ evenly (without a remainder).  After some polynomial division you find that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;n^3+2010 = n(n^2+2010)+2010-2010n&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where the $2010-2010n$ is the remainder term.  Thus the answer is a whole number only when $2010-2010n=0$, or $n=1$.&lt;/p&gt;&#xA;" OwnerUserId="2838" LastEditorUserId="2838" LastEditorDisplayName="" LastEditDate="2011-01-16T20:39:59.467" LastActivityDate="2011-01-16T20:39:59.467" CommentCount="2" />
  <row Id="17772" PostTypeId="2" ParentId="17742" CreationDate="2011-01-16T20:35:07.340" Score="3" ViewCount="" Body="&lt;p&gt;The proof is in any standard book in number theory; it breaks into four steps, two of which are number theoretic, the other is basic algebra.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;If $p$ is a prime, and $n$ is a positive integer, then $\sigma(p^n) = 1+p+p^2+\cdots+p^n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Indeed, the only divisors of $p^n$ are the powers of $p$ with exponent less than or equal to $n$, that is, $p^0 = 1$, $p$, $p^2,\ldots,p^n$. So $\sigma(p^n)$ is their sum.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If $x$ is any number different from $1$, and $n$ is a positive integer, then $1+x+\cdots+x^n = \frac{x^{n+1}-1}{x-1}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Simply notice that $(1+x+\cdots+x^n)(x-1) = x+\cdots +x^{n+1}-(1+\cdots+x^n) = x^{n+1}-1$. Since $x\neq 1$, then $x-1\neq 0$, so divide both sides by $x-1$ to get the identity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, $\sigma(p^n) = 1+p+\cdots+p^n = \frac{p^{n+1}-1}{p-1}$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If $r$ and $s$ are relatively prime, then $\sigma(rs)=\sigma(r)\sigma(s)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose $k$ divides $rs$. Let $b=\gcd(k,r)$, $c=\gcd(k,s)$. Because $r$ and $s$ are relatively prime, $bc=k$. That is: every divisor of $rs$ is the product of a divisor of $r$ and a divisor of $s$. Conversely, if $b|r$ and $c|s$, then $bc|rs$, so the product of a divisor of $r$ and a divisor of $s$ is a divisor of $rs$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is, every summand in $\sigma(rs)$ is the product of a particular summand in $\sigma(r)$, and a particular summand of $\sigma(s)$. That is, $\sigma(rs)=\sigma(r)\sigma(s)$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Induction: if $n=p_1^{a_1}\cdots p_r^{a_r}$, then $\sigma(n)=\prod\sigma(p_i^{a_i})$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The result holds if $r=1$ trivialy, and if $r=2$ by the previous step. Assume the result holds for $k$ distinct prime factors, and $n=p_1^{a_1}\cdots p_r^{a_r}p_{r+1}^{a_{r+1}}$. Let $n_1=p_1^{a_1}\cdots p_r^{a_r}$, $n_2=p_{r+1}^{a_{r+1}}$. Then $n=n_1n_2$, and $\gcd(n_1,n_2)=1$, so $\sigma(n)=\sigma(n_1)\sigma(n_2)$. Now apply induction to $\sigma(n_1)$, and the formula to $\sigma(n_2)$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;And that's it.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2011-01-16T20:35:07.340" />
  <row Id="17773" PostTypeId="2" ParentId="17736" CreationDate="2011-01-16T20:35:57.370" Score="2" ViewCount="" Body="&lt;p&gt;you might benefit from using &lt;a href=&quot;http://www.caam.rice.edu/software/ARPACK/&quot; rel=&quot;nofollow&quot;&gt;ARPACK&lt;/a&gt;, or if you have really large matrices, by using &lt;a href=&quot;http://www.grycap.upv.es/slepc/&quot; rel=&quot;nofollow&quot;&gt;SLEPc&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you say extremal, do you also mean the smallest eigenvalue? In the latter case you might benefit from looking at: &lt;a href=&quot;http://mathoverflow.net/questions/24287/what-is-the-best-algorithm-to-find-the-smallest-nonzero-eigenvalue-of-a-symmetric&quot; rel=&quot;nofollow&quot;&gt;this question on MO&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1709" LastActivityDate="2011-01-16T20:35:57.370" />
  <row Id="17774" PostTypeId="2" ParentId="17731" CreationDate="2011-01-16T20:38:03.323" Score="-2" ViewCount="" Body="&lt;p&gt;These are some (intuitive?) ideas, by no means I advocate them as they lack any type of rigor/sound philosophical reasoning that might justify them, read with extreme suspicion/scepticism/disapproval/disbelief : (This is a non scientific drawing of analogies/hand waving post between math and non scientific things)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Philosophy : Set theory (when dealing with infinities), shows the limits of human knowledge.&#xA;e.g. even if the universe was infinite and we could use every single particle to store information, then the relationship between every object in the universe is in the power set and Just like the real line is beyond any description in a list/book/linear manner. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Philosophy/Religion Gödel's theorem : Many religions use the selling point : &quot;ultimate/absolute truth and complete set of rules to live by&quot; , where as Gödel's theorem shows mush simpler set of rules are incomplete,inconsistent or both.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Philosophy/Religion : Game of life is a good example of determinism does not imply predictability, give very few simple rules,  there is no way to figure out what will happen by the rules alone.&lt;/p&gt;&#xA;" OwnerUserId="3936" LastEditorUserId="3936" LastEditorDisplayName="" LastEditDate="2011-01-18T03:39:03.523" LastActivityDate="2011-01-18T03:39:03.523" CommunityOwnedDate="2011-01-16T20:38:03.323" CommentCount="2" />
  <row Id="17775" PostTypeId="2" ParentId="17769" CreationDate="2011-01-16T20:45:34.627" Score="4" ViewCount="" Body="&lt;p&gt;The number corresponds to the &lt;a href=&quot;http://en.wikipedia.org/wiki/Catalan_number&quot; rel=&quot;nofollow&quot;&gt;Catalan Numbers&lt;/a&gt; which are of the form $\displaystyle \frac{{2n \choose n}}{n+1}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe this particular problem is an exercise Knuth's Art of Programming Volume I.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a hint of proof: Number of pushes $\ge$ Number of pops in any prefix of the string of pushes and pops.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2011-01-16T20:45:34.627" CommentCount="1" />
  <row Id="17776" PostTypeId="1" AcceptedAnswerId="17780" CreationDate="2011-01-16T20:48:26.543" Score="4" ViewCount="314" Body="&lt;p&gt;I have two square matrices - $A$ and $B$. $A^{-1}$ is known and I want to calculate $(A+B)^{-1}$. Are there theorems that helps with calculating the inverse of the sum of matrices? In general case $B^{-1}$ is not known, but if it is necessary then it can be assumed that $B^{1}$ is also known.&lt;/p&gt;&#xA;" OwnerUserId="4811" LastActivityDate="2011-01-16T22:03:37.630" Title="Inverse of the sum of matrices" Tags="&lt;matrices&gt;" AnswerCount="4" CommentCount="2" FavoriteCount="2" />
  <row Id="17777" PostTypeId="2" ParentId="17776" CreationDate="2011-01-16T20:55:53.967" Score="3" ViewCount="" Body="&lt;p&gt;Assuming everything is nicely invertible, you are probably looking for &lt;a href=&quot;http://en.wikipedia.org/wiki/Woodbury_matrix_identity&quot; rel=&quot;nofollow&quot;&gt;the SMW identity&lt;/a&gt;  (which, i think, can also be generalized to pseudoinverses if needed)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please see caveat in the comments below; in general if $B$ is low-rank, then you'd be happy using SMW.&lt;/p&gt;&#xA;" OwnerUserId="1709" LastEditorUserId="1709" LastEditorDisplayName="" LastEditDate="2011-01-16T22:03:37.630" LastActivityDate="2011-01-16T22:03:37.630" CommentCount="3" />
  <row Id="17778" PostTypeId="2" ParentId="17776" CreationDate="2011-01-16T21:02:55.370" Score="4" ViewCount="" Body="&lt;p&gt;A formal power series expansion is possible:&#xA;$$&#xD;&#xA;\begin{eqnarray}&#xD;&#xA;(A + \epsilon B)^{-1}&#xD;&#xA;&amp;amp;=&amp;amp; \left(A \left(I + \epsilon A^{-1}B\right)\right)^{-1} \\&#xD;&#xA;&amp;amp;=&amp;amp; \left(I + \epsilon A^{-1}B\right)^{-1} A^{-1} \\&#xD;&#xA;&amp;amp;=&amp;amp; \left(I - \epsilon A^{-1}B + \epsilon^2 A^{-1}BA^{-1}B - ...\right) A^{-1} \\&#xD;&#xA;&amp;amp;=&amp;amp; A^{-1} - \epsilon A^{-1} B A^{-1} + \epsilon^2 A^{-1} B A^{-1} B A^{-1} - ...&#xD;&#xA;\end{eqnarray}&#xD;&#xA;$$&#xA;Under appropriate conditions on the eigenvalues of $A$ and $B$ (such that $A$ is sufficiently &quot;large&quot; compared to $B$), this will converge to the correct result at $\epsilon=1$.&lt;/p&gt;&#xA;" OwnerUserId="5546" LastActivityDate="2011-01-16T21:02:55.370" CommentCount="1" />
  <row Id="17779" PostTypeId="2" ParentId="17776" CreationDate="2011-01-16T21:04:27.867" Score="3" ViewCount="" Body="&lt;p&gt;$(A+B)^{-1} = A^{-1} - A^{-1}BA^{-1} + A^{-1}BA^{-1}BA^{-1} - A^{-1}BA^{-1}BA^{-1}BA^{-1} + \cdots$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;provided $||A^{-1}B||&amp;lt;1$ or $||BA^{-1}|| &amp;lt; 1$ here $||\cdot||$ means norm.  This is just the Taylor expansion of the inversion function together with basic information on convergence. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(posted essentially at the same time as mjqxxx)&lt;/p&gt;&#xA;" OwnerUserId="642" LastActivityDate="2011-01-16T21:04:27.867" />
  <row Id="17780" PostTypeId="2" ParentId="17776" CreationDate="2011-01-16T21:07:39.360" Score="6" ViewCount="" Body="&lt;p&gt;In general, $A+B$ need not be invertible, even when $A$ and $B$ are. But one might ask whether you can have a formula under the additional assumption that $A+B$ &lt;em&gt;is&lt;/em&gt; invertible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As noted by Adrián Barquero, there is &lt;a href=&quot;http://www.jstor.org/stable/2690437&quot; rel=&quot;nofollow&quot;&gt;a paper by Ken Miller&lt;/a&gt; published in the &lt;em&gt;Mathematics Magazine&lt;/em&gt; in 1981 that addresses this. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;He proves the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Lemma.&lt;/strong&gt; If $A$ and $A+B$ are invertible, and $B$ has rank $1$, then let $g=\mathrm{trace}(BA^{-1})$. Then $g\neq -1$ and &#xA;$$(A+B)^{-1} = A^{-1} - \frac{1}{1+g}A^{-1}BA^{-1}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From this lemma, we can take a general $A+B$ that is invertible and write it as $A+B = A + B_1+B_2+\cdots+B_r$, where $B_i$ each have rank $1$ and such that each $A+B_1+\cdots+B_k$ is invertible (such a decomposition always exists if $A+B$ is invertible and $\mathrm{rank}(B)=r$). Then you get:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Theorem.&lt;/strong&gt; Let $A$ and $A+B$ be nonsingular matrices, and let $B$ have rank $r\gt 0$. Let $B=B_1+\cdots+B_r$, where each $B_i$ has rank $1$, and each $C_{k+1} = A+B_1+\cdots+B_k$ is nonsingular. Setting $C_1 = A$, then&#xA;$$C_{k+1}^{-1} = C_{k}^{-1} - g_kC_k^{-1}B_kC_k^{-1}$$&#xA;where $g_k = \frac{1}{1 + \mathrm{trace}(C_k^{-1}B_k)}$. In particular,&#xA;$$(A+B)^{-1} = C_r^{-1} - g_rC_r^{-1}B_rC_r^{-1}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(If the rank of $B$ is $0$, then $B=0$, so $(A+B)^{-1}=A^{-1}$). &lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2011-01-16T21:07:39.360" CommentCount="1" />
  <row Id="17781" PostTypeId="2" ParentId="17745" CreationDate="2011-01-16T21:11:18.670" Score="3" ViewCount="" Body="&lt;p&gt;I like Matt E's answer as an elementary and appealing way to see that $U^N$ has finite index in $U$ for all $N$.  (Here I am writing $U$ for the full unit group $\mathcal{O}_K^{\times}$ of $K$ and $U^N = \{x^N \ | \ x \in U\}$.  I just want to remark that it is not &lt;em&gt;so&lt;/em&gt; much harder to give a general formula for the index $[U:U^N]$ in the general case (including local fields of positive characteristic $p$, so long as $p \nmid N$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer is that if $v$ is the normalized (i.e., $\mathbb{Z}$-valued) valuation on $K$ and $q$ is the cardinality of the residue field, then &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$[U:U^N] = q^{v(N)} \ \# \mu_N(K).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is Theorem 12 in &lt;a href=&quot;http://math.uga.edu/~pete/8410Chapter5.pdf&quot; rel=&quot;nofollow&quot;&gt;these notes&lt;/a&gt;.  The treatment follows Lang's &lt;em&gt;Algebraic Number Theory&lt;/em&gt;.  (Perhaps it is worth mentioning that the argument is a bit tricky but completely elementary.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $U_n = \{x \in U \ | \ x \equiv 1 \pmod{\mathfrak{p}^n} \}$, so that the $U_n$'s are a cofinal system of open subgroups of $U$.  In other words, for a subgroup of $U$ to be open, it is necessary and sufficient that it contain $U_n$ for some $n$.  We want to show that $U^N$ is open.  But the proof of the above theorem proceeds by showing that for all sufficiently large $r$, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ U_{r+v(N)} = U_r^N \subset U^N,$$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;so indeed $U^N$ is open.  Moreover, since every subgroup $H$ of $U$ of finite index $N$ contains $U^N$, it follows that every finite index subgroup of $U$ is open. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yet another approach is to develop the theory of the logarithm as in Exercise 5.3 of &lt;a href=&quot;http://math.uga.edu/~pete/8410Chapter5.pdf&quot; rel=&quot;nofollow&quot;&gt;loc. cit.&lt;/a&gt; to give, for all sufficiently large $n$, an isomorphism of topological groups from $U_n$ to the additive group $(\mathcal{O}_K,+)$.  This also implies that each $U^N$ is open and of finite index in $U$.  &lt;/p&gt;&#xA;" OwnerUserId="299" LastActivityDate="2011-01-16T21:11:18.670" />
  <row Id="17782" PostTypeId="2" ParentId="17768" CreationDate="2011-01-16T21:38:09.927" Score="7" ViewCount="" Body="&lt;p&gt;The idea to start with polynomial division is correct, but Raskolnikov is right concerning the end of the argument: the answer is a integer as long as the $2010 \frac{1-n}{n^2+2010}$ is an integer, say equal to $k$; you can then state that this is true whenever $\frac{n-1}{n^2+2010} = \frac{k}{2010}$ for $k \in \mathbb{Z}.$ Now if the function on the left becomes smaller than $1/2010,$ you won't have any more solutions; I hope this suffices to work out the rest of the problem.&lt;/p&gt;&#xA;" OwnerUserId="5923" LastActivityDate="2011-01-16T21:38:09.927" CommentCount="5" />
  <row Id="17783" PostTypeId="2" ParentId="17731" CreationDate="2011-01-16T21:46:17.190" Score="1" ViewCount="" Body="&lt;p&gt;I believe that &lt;a href=&quot;http://en.wikipedia.org/wiki/Tarski%27s_indefinability_theorem&quot; rel=&quot;nofollow&quot;&gt;Tarski's truth undefinability&lt;/a&gt; theorem has uses in philosophy, as the mentioned incompleteness theorems, and that's besides cranks attributing it to proofs that god exists, that mathematics is just a bunch of crap etc etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a friend who's studying cognitive science (as well as a math degree) and he says that there's some good math modeling there, and we all saw the financial Nobel prizes that were given to mathematicians.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I have a quarrel with the idea of using mathematics outside the ideal realm of mathematical objects. This is because mathematics is inherently precise and perfect, you can define notions that capture exactly what you want them to capture, while our &quot;real&quot; world is inherently imperfect and imprecise - and we can never judge what's true in the real world, as Tarski's theorem tells us.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The above argument means that taking mathematics into the real world is to allow imperfect and imprecise definitions and &quot;latitude&quot; for things to change beyond our original meaning. This is not mathematics anymore, in my eyes anyway.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the very first math class I had in the academia, the teacher came in and said &quot;Mathematics is the science of deducing from certain assumptions.&quot; and three and a half years later, I only grow to understand deeper and deeper how true this is. In the real world, i.e. outside of mathematical idealism, you can't prove anything - only find evidence to supposedly support a claim, or disprove it. So the ability to assume things and deduce things with absolute certainty becomes problematic. Which is non-mathematical in my eyes anyway.&lt;/p&gt;&#xA;" OwnerUserId="622" LastActivityDate="2011-01-16T21:46:17.190" CommunityOwnedDate="2011-01-16T21:46:17.190" />
  <row Id="17785" PostTypeId="2" ParentId="17548" CreationDate="2011-01-16T22:39:33.657" Score="3" ViewCount="" Body="&lt;p&gt;For the purposes of the Nelder-Mead algorithm, the characteristic length scale for a particular variable is basically your best guess as to the size of the potential solution space in that variable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Example taken from &lt;a href=&quot;http://www.astro.virginia.edu/class/oconnell/astr511/idl_5.1_html/idl6.htm&quot; rel=&quot;nofollow&quot;&gt;this site&lt;/a&gt;.)  For example, in a 3-dimensional problem, if the initial guess is the point $[0,0,0]$, and you know that the function's minimum value occurs in the interval&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$-10 &amp;lt; x_0 &amp;lt; 10,$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$-100 &amp;lt; x_1 &amp;lt; 100,$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$-200 &amp;lt; x_2 &amp;lt; 200,$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;then you could set $\lambda_0$, $\lambda_1$, and $\lambda_2$ to 10, 100, and 200, respectively. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suppose this quantity can be called a step size, but it's only used in the initialization part of the algorithm.   It's not (which is how I normally think of step sizes) used in each iteration of the algorithm to step from one solution to the next.  Those step sizes are the reflection, expansion, contraction, and shrinking parameters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some additional references:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Nelder and Mead's &lt;a href=&quot;http://comjnl.oxfordjournals.org/content/7/4/308.full.pdf+html&quot; rel=&quot;nofollow&quot;&gt;original paper&lt;/a&gt;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;http://reference.kfupm.edu.sa/content/c/o/convergence_properties_of_the_nelder_mea_72334.pdf&quot; rel=&quot;nofollow&quot;&gt;This paper&lt;/a&gt; by Lagarias, et al, contains a nice presentation of the Nelder-Mead algorithm.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;A &lt;a href=&quot;http://wiki.scilab.org/The_Nelder-Mead_Component?action=AttachFile&amp;amp;do=get&amp;amp;target=neldermead.pdf&quot; rel=&quot;nofollow&quot;&gt;Nelder Mead User's Manual&lt;/a&gt;.  In particular, see the discussion of different ways to form the initial simplex.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="2370" LastActivityDate="2011-01-16T22:39:33.657" />
  <row Id="17786" PostTypeId="2" ParentId="17768" CreationDate="2011-01-16T22:52:21.500" Score="7" ViewCount="" Body="&lt;p&gt;Note $\rm\quad\displaystyle\ \frac{n^3+2010}{n^2+2010}\ =\ n + \frac{2010\ (1-n)}{n^2+2010}\ \in \ \mathbb Z\ \iff\ n^2+2010\ |\ 2010\ (1-n) $ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\ gcd(n^2+2010,\: 1-n) = gcd(2011,\: 1-n)\:.\ $ Since $\:2011\:$ is prime, this $\rm gcd$ is $\:1\:$ if $\rm\ 1 &amp;lt; n &amp;lt; 2012\:,\ $ so if $\rm\ n &amp;lt; 2012\ $ then either $\rm\ n = 1\ $ or $\rm\ n^2+2010\ |\ 2010\ \Rightarrow \ n = 0\:.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But $\rm\ \ n \ge 2012\ \Rightarrow\ n^2+2010\ &amp;gt; |2010\ (1-n)|\:.$&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2011-01-16T23:41:03.383" LastActivityDate="2011-01-16T23:41:03.383" CommentCount="7" />
  <row Id="17787" PostTypeId="2" ParentId="13088" CreationDate="2011-01-16T22:58:10.583" Score="1" ViewCount="" Body="&lt;p&gt;Judging from the upvotes on my comment, it looks like I didn't miss anything, so...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think you have it.  You have expressed $P(X(m+1)=j)$  in terms of the previous state $X(m)$  without reference to any of the prior states $X(m−1),X(m−2), \ldots, X(0)$. That's enough to show that $X(m)$ is a Markov chain. &lt;/p&gt;&#xA;" OwnerUserId="2370" LastActivityDate="2011-01-16T22:58:10.583" />
  <row Id="17788" PostTypeId="2" ParentId="17700" CreationDate="2011-01-16T23:16:15.403" Score="3" ViewCount="" Body="&lt;p&gt;A more straightforward approach is the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(x&amp;#39;)^2 +t(x&amp;#39;) - x = 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus by taking the roots of the quadratic in $\displaystyle x&amp;#39;$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ x&amp;#39; = \frac{-t \pm \sqrt{t^2 + 4x}}{2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$2x&amp;#39; + t = \pm \sqrt{t^2 + 4x}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Setting $\displaystyle y = 4x+t^2$ gives us&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{y&amp;#39;}{2} = \pm \sqrt{y}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This should give some of the solutions and in order to get all, the substitution $y = 4x + t^2$ suggests itself, which I believe gives us the differential equation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ (y&amp;#39;)^2 - 4y = 0$$&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2011-01-16T23:46:40.540" LastActivityDate="2011-01-16T23:46:40.540" CommentCount="1" />
  <row Id="17789" PostTypeId="1" AcceptedAnswerId="17791" CreationDate="2011-01-16T23:29:42.310" Score="0" ViewCount="116" Body="&lt;p&gt;I've got a couple of theorems that I must prove, but I'm completely stumped. Could someone help me out?&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Prove: If P(x) can be divided by Q(x), then it can be divided by cQ(x), where c - any non zero number.&lt;/li&gt;&#xA;&lt;li&gt;Prove: If the degrees of P and Q are the same ( equal), then P(x) can be divided by Q(x) then and only then, if Q(x)=cP(x), where c - any non zero number.&lt;/li&gt;&#xA;&lt;li&gt;Prove: If c - any non zero number. then polynomials P(x) and cP(x) have the same divisors &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I know from simple logic, that they are all true, but I'm bad at writing logic in math proofs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="2078" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-17T00:01:58.717" LastActivityDate="2011-01-17T22:41:42.400" Title="Proofs for polynomial divisons." Tags="&lt;homework&gt;&lt;algebra-precalculus&gt;&lt;polynomials&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="17790" PostTypeId="2" ParentId="17789" CreationDate="2011-01-16T23:36:33.233" Score="0" ViewCount="" Body="&lt;p&gt;Assuming you are working over complex numbers,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint: Try using the fundamental theorem of algebra.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2011-01-16T23:36:33.233" CommentCount="4" />
  <row Id="17791" PostTypeId="2" ParentId="17789" CreationDate="2011-01-16T23:36:46.107" Score="3" ViewCount="" Body="&lt;p&gt;I'm assuming that by &quot;number&quot; you mean either &quot;real number&quot; or &quot;complex number&quot;; in fact, most of these statements are true in much greater generality (other classes of numbers).&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The statement &quot;$P(x)$ can be divided by $Q(x)$&quot; means that there exists a polynomial $R(x)$ such that $Q(x)R(x) = P(x)$. In order to show that $P(x)$ can also be divided by $cQ(x)$, you need to find a polynomial $S(x)$ such that $(cQ(x))S(x) = P(x)$. Try using the $R(x)$ you know exists, and the fact that $c\neq 0$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Again, you know that there is a polynomial $R(x)$ such that $Q(x)R(x)=P(x)$. What do you know about the degree of a polynomial product? That is, if the degree of $f(x)$ is $n$ and the degree of $g(x)$ is $m$, what is the degree of $f(x)g(x)$?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Show that if $Q(x)$ divides $P(x)$, then it also divides $cP(x)$. Then use this to show that anything that divides $cP(x)$ also divides $\frac{1}{c}\left(cP(x)\right)$ to get that $P(x)$ and $cP(x)$ have the same divisors: that is, that $Q(x)$ divides $P(x)$ if and only if it divides $cP(x)$. For the second part (that a divisor of $cP(x)$ is a divisor of $P(x)$) you'll need to use that $c\neq 0$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-16T23:45:00.773" LastActivityDate="2011-01-16T23:45:00.773" />
  <row Id="17792" PostTypeId="1" CreationDate="2011-01-17T00:01:41.837" Score="4" ViewCount="96" Body="&lt;p&gt;Hello everyone - the problem I'm stuck on is the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose that S is a countably infinite subset of $\ell_2$ with the property that the linear span of S′ is dense in $\ell_2$ whenever S\S′ is finite. Show that there is some S′ whose linear span is dense in $\ell_2$ and for which S\S′ is infinite.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have tried repeatedly to solve this in somewhat of a 'bang my head against a wall' manner, by constructing a series of subsets of some arbitrary S, such that the complement is finite and of increasing size, but I haven't had any success. I haven't actually used the fact that we're working in $\ell_2$ here, so it's quite likely that I'm meant to use some property of Hilbert spaces - however, I'm not sure what. Could anyone please help? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thankyou very much; Stephen.&lt;/p&gt;&#xA;" OwnerUserId="5926" LastEditorUserId="2548" LastEditorDisplayName="" LastEditDate="2011-01-17T00:55:04.303" LastActivityDate="2011-01-17T04:26:55.573" Title="Constructing a subset of $\ell_2$ with dense linear span and infinite complement" Tags="&lt;homework&gt;&lt;functional-analysis&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="1" />
  <row Id="17793" PostTypeId="1" AcceptedAnswerId="17797" CreationDate="2011-01-17T00:08:36.320" Score="4" ViewCount="234" Body="&lt;p&gt;In respect to a larger proof I need to prove that $(n-m) | (n^r - m^r) $ (where | means divides, ie $a | b$ means &lt;code&gt;b modulus a = 0&lt;/code&gt;). I have played around with this for a while now but to no avail and would be grateful if you could help me.&lt;/p&gt;&#xA;" OwnerUserId="1973" LastActivityDate="2011-01-17T16:04:24.383" Title="Prove that (n-m) | (n^r - m^r)" Tags="&lt;homework&gt;&lt;algebra&gt;&lt;divisibility&gt;" AnswerCount="6" />
  <row Id="17795" PostTypeId="2" ParentId="17793" CreationDate="2011-01-17T00:15:08.870" Score="5" ViewCount="" Body="&lt;p&gt;If $r$ is a positive integer, simply notice that $x^r - y^r = (x-y)(x^{r-1} + x^{r-2}y + \cdots + xy^{r-2} + y^{r-1})$ as polynomials.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2011-01-17T00:15:08.870" />
  <row Id="17796" PostTypeId="2" ParentId="17793" CreationDate="2011-01-17T00:15:46.510" Score="6" ViewCount="" Body="&lt;p&gt;This is a simple telescopic sum trick:&#xA;\[&#xA;n^{r} - m^{r} = (n-m) \sum_{k=1}^{r} n^{r-k}m^{k-1} = (n - m)(n^{r-1} + n^{r-2}m + \cdots +nm^{r-2} + m^{r-1}).&#xA;\]&lt;/p&gt;&#xA;" OwnerUserId="5363" LastActivityDate="2011-01-17T00:15:46.510" CommentCount="1" />
  <row Id="17797" PostTypeId="2" ParentId="17793" CreationDate="2011-01-17T00:18:59.903" Score="13" ViewCount="" Body="&lt;p&gt;There are several ways of proving this. For example, one can use the binomial theorem $(a+b)^r=\sum_{k=1}^r\binom{k}ra^kb^{n-k}$ with $a=n-m$ and $b=m$. (The change of variables from $n,m$ to $a,b$ is in itself a good idea in problems like this, since it tends to reduce the size of some of the expressions.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another way is to use the identity $$n^r-m^r=(n-m)\sum_{k=0}^{r-1}n^{r-1-k}m^k=(n-m)(n^{r-1}+n^{r-2}m+\dots+nm^{r-2}+m^{r-1}).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both of these identities are proved by induction on $r$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or, one can do directly an inductive argument: The result is clear if $r=0$, since $n^r-m^r=0$. Given the result for $r$, prove it for $r+1$ using, for example, that $n^{r+1}-m^{r+1}=n(n^r-m^r)+m^r(n-m)$.&lt;/p&gt;&#xA;" OwnerUserId="462" LastActivityDate="2011-01-17T00:18:59.903" CommentCount="1" />
  <row Id="17798" PostTypeId="1" CreationDate="2011-01-17T00:24:21.227" Score="2" ViewCount="165" Body="&lt;p&gt;Let $(X, \mathscr{O})$  be a ringed space and $\mathscr{F}, \mathscr{G}$  be sheaves of $\mathscr{O}$-modules on $X$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Define $\mathscr{H}(U) = \mathscr{F}(U) \otimes_{\mathscr{O}(U)} \mathscr{G}(U)$. I am stuck trying to prove that $\mathscr{H}_p \cong \mathscr{F}_p \otimes_{\mathscr{O}_p} \mathscr{G}_p$ as $\mathscr{O}_p$-modules.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that if $X$  is a topological space, $\mathscr{F}, \mathscr{G}$ are presheaves of abelian groups on $X$ and $\mathscr{H}(U) = \mathscr{F}(U) \otimes_{\mathbb{Z}} \mathscr{G}(U)$ then $\mathscr{H}_p \cong \mathscr{F}_p \otimes_{\mathbb{Z}} \mathscr{G}_p$. This is just a consequence of the fact that tensor products commute with direct limits. But I don't know how to deal with the case when the base ring is changing&lt;/p&gt;&#xA;" OwnerUserId="1725" LastEditorUserId="1725" LastEditorDisplayName="" LastEditDate="2011-01-17T01:17:49.057" LastActivityDate="2011-02-21T09:28:31.910" Title="Stalks of the tensor product presheaf of two sheaves" Tags="&lt;algebraic-geometry&gt;&lt;sheaf-theory&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" />
  <row Id="17799" PostTypeId="1" CreationDate="2011-01-17T00:24:48.760" Score="0" ViewCount="50" Body="&lt;p&gt;In dynamics, they talk about Abelian differentials on surfaces, are they the same as holomorphic differentials?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Quadratic differentials are multiple valued and can change sign as you move around a zero. &lt;/p&gt;&#xA;" OwnerUserId="4997" LastActivityDate="2011-01-17T02:11:34.647" Title="Abelian vs Holomorphic Differentials vs Quadratic Differentials" Tags="&lt;dynamical-systems&gt;" AnswerCount="1" />
  <row Id="17800" PostTypeId="2" ParentId="17793" CreationDate="2011-01-17T00:32:29.207" Score="3" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\ $ modulo $\rm\ n-m\ $ one has $\rm\ n\equiv m\ \Rightarrow\ n^r \:\equiv\: m^r\ $ by the &lt;a href=&quot;http://math.stackexchange.com/questions/5366/very-simple-question-but-what-is-the-proof-that-x-y-mod-m-x-mod-m-y-mod-m/5376#5376&quot;&gt;congruence product rule.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2011-01-17T00:32:29.207" CommentCount="1" />
  <row Id="17801" PostTypeId="2" ParentId="17768" CreationDate="2011-01-17T00:33:30.050" Score="4" ViewCount="" Body="&lt;p&gt;You find that (Euclidian algorithm)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{ n^3 + 2010 }{ n^2 + 2010 } = n + \frac{ (1-n) 2010 }{ n^2 + 2010 }.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a whole number division you need&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{ (1-n) 2010 }{ n^2 + 2010 } = k, $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for some integer k. You can solve this last equation for 'n'&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$n = -\frac{1005}{k} \pm \frac{1}{k} \sqrt{1005} \sqrt{1005 - 2 k ( k - 1)}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, noting that 1005 is not a square of an integer, you automatically conclude that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$1005 - 2 k ( k - 1) = 1005 a^2,$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;for some integer 'a'. Solving this for 'k' yields&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$k = \frac{1}{2} \pm \frac{1}{2} \sqrt{2011 - 2010 a^2}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, the only solutions are those generated by a = 1, which are&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$k = 0 \Rightarrow n = 1;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$k = 1 \Rightarrow n = 0 \text{ (positive sign solution) and } n = -2010 \text{ (minus sign solution).}$$&lt;/p&gt;&#xA;" OwnerUserId="5928" LastEditorUserId="5928" LastEditorDisplayName="" LastEditDate="2011-01-17T01:06:10.200" LastActivityDate="2011-01-17T01:06:10.200" />
  <row Id="17803" PostTypeId="2" ParentId="1870" CreationDate="2011-01-17T00:51:39.790" Score="0" ViewCount="" Body="&lt;p&gt;Geogebra at &#xA;&lt;a href=&quot;http://www.geogebra.org/cms/&quot; rel=&quot;nofollow&quot;&gt;http://www.geogebra.org/cms/&lt;/a&gt;&#xA;free and multi-level&lt;/p&gt;&#xA;" OwnerUserId="5929" LastActivityDate="2011-01-17T00:51:39.790" CommunityOwnedDate="2011-01-17T00:51:39.790" />
  <row Id="17805" PostTypeId="1" CreationDate="2011-01-17T01:18:50.117" Score="8" ViewCount="404" Body="&lt;p&gt;I have a terrible hand-writing and I'm very good at typing, so I had an idea about taking my math lecture notes using a computer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've tried using a simple syntax (using purely ASCII) but it's getting harder and harder, so I need something a bit more sophisticated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Friend suggested Latex, but said that I probably won't be able to write it fast enough to use it in &quot;real-time&quot;. It also has a quite learning curve.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm on Mac os X. What would you suggest?&lt;/p&gt;&#xA;" OwnerUserId="1759" LastEditorUserId="1208" LastEditorDisplayName="" LastEditDate="2011-04-27T19:23:32.633" LastActivityDate="2011-04-27T19:23:32.633" Title="What software and/or language to use to take Math lecture notes?" Tags="&lt;notation&gt;&lt;math-software&gt;" AnswerCount="5" CommentCount="1" FavoriteCount="3" />
  <row Id="17806" PostTypeId="2" ParentId="17805" CreationDate="2011-01-17T01:21:20.723" Score="2" ViewCount="" Body="&lt;p&gt;See &lt;a href=&quot;http://drija.com/math/33356/best-software-for-taking-math-notes-at-lectures/&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; (site about software for writing math). Also &lt;a href=&quot;http://forums.macrumors.com/showthread.php?t=551147&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; pertains to math typesetting on Mac OS X.&lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2011-01-17T01:21:20.723" CommunityOwnedDate="2011-01-17T20:28:41.003" />
  <row Id="17807" PostTypeId="2" ParentId="17789" CreationDate="2011-01-17T01:24:36.947" Score="1" ViewCount="" Body="&lt;p&gt;The sought divisibility results are special cases of this simple &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Divides Product Rule&lt;/strong&gt; $\ $ Let $\rm\ a,b,c,d\in R $ any ring.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then $\rm\quad\ \ a|b,\ \ c|d\ $ in $\rm\:R\ \ \Rightarrow\ \ ac|bd\ $ in $\rm\:R\:. $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof&lt;/strong&gt; $\rm\ \ \ b/a,\ d/c \ \in\ R\ \Rightarrow\ (bd)/(ac)\in R\:.\quad$ &lt;strong&gt;QED&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let's apply this Rule to derive the sought results in $\rm\ R = K[x],\ K\ $ a field.&lt;br&gt;&#xA;Below I give the main steps, omitting trivial cases such as when $\rm\:p\:$ or $\rm\:q = 0\:.$&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;$\rm c|1,\ q|p\ \Rightarrow\ cq|p\ $ by the Rule (in a field $\rm\ K,\ c\ne 0\ \Rightarrow\ 1/c\in K\ \Rightarrow\ c|1\ $ in $\rm\:K\:).$ &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If $\rm\ deg\ p = \deg\ q\ $ then $\rm\ q|p \iff\ q = c\: p\ $ for some $\rm\ 0 \ne c\in K\:.$&lt;br&gt;&#xA;$\rm(\Leftarrow)\ \ c|1,\ p|p\ \Rightarrow\ cp|p\ $ by the Rule (or $\rm\ q \to p\ $ in 1).&lt;br&gt;&#xA;$\rm(\Rightarrow)\ \ p = q\:r\ \Rightarrow\ \deg\ r = \deg\ p - \deg\ q = 0\ \Rightarrow\ r\in K\:.$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$\rm\ q|p\ \iff\ q|cp\ $&lt;br&gt;&#xA;$\rm(\Leftarrow)\ \ q|cp|p\ $  by $\rm q \to p$ in 1 and &quot;divides&quot; transitivity ($\rm b = c\ \ne 0\:$ in Rule)&lt;br&gt;&#xA;$\rm(\Rightarrow)\ \  1|c,\ q|p\ \Rightarrow\ q|cp\ $ by the Rule.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2011-01-17T22:41:42.400" LastActivityDate="2011-01-17T22:41:42.400" />
  <row Id="17808" PostTypeId="2" ParentId="17792" CreationDate="2011-01-17T01:52:35.380" Score="2" ViewCount="" Body="&lt;p&gt;Most likely you already realize this, but you have to be careful about how you remove the sets.  If $S_1\subset S_2\subset S_3\subset\ldots\subset S$, with $|S_n|=n$, then you know that each $S\setminus S_n$ spans, but you can't say that $S\setminus\cup_n S_n$ spans.  For example, $S$ could be countable, and $\cup_n S_n$ could be $S$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is &lt;a href=&quot;http://books.google.com/books?id=S57XLkgbf0oC&amp;amp;lpg=PP1&amp;amp;pg=PA7#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;Problem 9&lt;/a&gt; in Halmos's &lt;em&gt;A Hilbert space problem book&lt;/em&gt;, which as usual has a lot of good reading leading up to the problem.  There is a hint: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Omit an infinite subset by omitting one element at a time.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There is also a concise solution in case you aren't too wary of having the answer &quot;given away&quot;.  (Don't forget to cite your source if you use this in work you turn in.)  It is worth checking out even if you solve the problem independently.  The given solution does not depend on Hilbert space geometry, and would be valid in any &lt;strike&gt;separable&lt;/strike&gt; normed space.  (As Qiaochu pointed out to me, the reduction to assuming that $S$ is countable does not require separability.)&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;To elaborate a little on the hint, in a way that will hopefully be helpful without giving too much away, make sure at each step that you can approximate the vectors that have been removed with vectors that will never be removed.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2011-01-17T04:26:55.573" LastActivityDate="2011-01-17T04:26:55.573" CommentCount="3" />
  <row Id="17809" PostTypeId="2" ParentId="17793" CreationDate="2011-01-17T01:53:04.203" Score="2" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\ $ For $\rm\ n\to n+m\ $ it is $\rm\ n\ |\ (n+m)^r - m^r\ $ which is obvious by the binomial theorem.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Note&lt;/strong&gt; $\ $ The above proof arises from the following trivial symmetry-inspired proof of the&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Factor Theorem&lt;/strong&gt; $\rm\ \ \ \ x - m \ |\ P(x) - P(m)\ $ in $\rm\: R[x]\:,\ $ for every ring $\rm\:R\:.$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof&lt;/strong&gt; $\ $ Clear if $\rm\ m = 0\::\ \ x\ |\ P(x)-P(0)\:.\: $ Else reduce to $\rm\ m = 0\ $ by the shift $\rm\ x\to x+m\:.\ $ &lt;strong&gt;QED&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The proof exploits an innate symmetry - the problem is invariant under shift automorphisms. Hence, with luck, we can apply these symmetries to reduce the problem to a much simpler one.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2011-01-17T16:04:24.383" LastActivityDate="2011-01-17T16:04:24.383" />
  <row Id="17810" PostTypeId="2" ParentId="17793" CreationDate="2011-01-17T01:53:56.997" Score="1" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\ $ Put $\rm\ P(x) = x^r,\ \ x = n\ $ in $\rm\ x-m\ |\ P(x)-P(m)\ $ in $\rm\ \mathbb Z[x]\ $ (the &lt;strong&gt;Factor Theorem&lt;/strong&gt;)&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2011-01-17T16:03:36.990" LastActivityDate="2011-01-17T16:03:36.990" CommentCount="0" />
  <row Id="17811" PostTypeId="2" ParentId="17799" CreationDate="2011-01-17T02:11:34.647" Score="1" ViewCount="" Body="&lt;p&gt;An Abelian differential is just a traditional name for a holomorphic or meromorphic differential on a compact Riemann surface.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A quadratic differential is just an element of $S^2(\Omega^1)$, the symmetric square of the sheaf of differentials. I do not really see what you mean when you say «[they] are multiple valued and can change sign as you move around a zero».&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2011-01-17T02:11:34.647" CommentCount="4" />
  <row Id="17812" PostTypeId="2" ParentId="17805" CreationDate="2011-01-17T02:33:44.373" Score="10" ViewCount="" Body="&lt;p&gt;I personally only take math lecture notes in $\LaTeX$; there are several benefits to doing this and I don't think it's as hard to learn as you think it is.  I certainly don't have a problem with keeping up with lecturers.  See &lt;a href=&quot;http://sbseminar.wordpress.com/2010/02/25/chromatic-homotopy-ii-or-how-i-learned-to-stop-worrying-and-love-latexing-in-real-time/&quot; rel=&quot;nofollow&quot;&gt;this blog post&lt;/a&gt; and some of the answers at &lt;a href=&quot;http://mathoverflow.net/questions/12638/taking-lecture-notes-in-lectures&quot; rel=&quot;nofollow&quot;&gt;this MO question&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2011-01-17T02:33:44.373" />
  <row Id="17813" PostTypeId="1" AcceptedAnswerId="17995" CreationDate="2011-01-17T02:39:03.977" Score="1" ViewCount="154" Body="&lt;p&gt;how can we generate random numbers using skew normal distribution in multivariate case&lt;/p&gt;&#xA;" OwnerUserId="5620" LastEditorUserId="4742" LastEditorDisplayName="" LastEditDate="2011-01-17T09:22:27.037" LastActivityDate="2011-01-18T15:48:52.637" Title="how can we generate random numbers using skew normal distribution in multivariate case" Tags="&lt;probability&gt;&lt;statistics&gt;&lt;probability-distributions&gt;" AnswerCount="3" />
  <row Id="17814" PostTypeId="1" CreationDate="2011-01-17T02:39:31.937" Score="0" ViewCount="114" Body="&lt;pre&gt;&lt;code&gt;Circle = 0&#xA;Dot = 1&#xA;Line = 2&#xA;Triangle = 3&#xA;Square = 4&#xA;Pentagon = 5&#xA;Hexagon = 6&#xA;Cross = 7&#xA;Heart = 8&#xA;Smiley Face = 9&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Use the following special rule when answering the problem below:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If ever a square is next to a cross during long multiplication, the square shall be treated as a triangle.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Problem:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Show your work in doing long multiplication of Pentagon Pentagon Nonagon by Line Square Octagon. Show your work using symbols not numbers.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5933" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-17T02:41:28.360" LastActivityDate="2011-01-17T02:41:28.360" ClosedDate="2011-01-17T02:48:12.970" Title="“Test to measure your ability to follow directions and solve complex problems in a neat and orderly manner.”" Tags="&lt;geometry&gt;&lt;logic&gt;" CommentCount="2" />
  <row Id="17815" PostTypeId="2" ParentId="17805" CreationDate="2011-01-17T02:42:14.633" Score="5" ViewCount="" Body="&lt;p&gt;Qiaochu's suggestion of LaTeX is also what I would recommend. In this vein, I find it useful to define large numbers of keyboard shortcuts (this can be easily done in vim, for instance, which can be &lt;a href=&quot;http://macvim.org/OSX/index.php&quot; rel=&quot;nofollow&quot;&gt;downloaded&lt;/a&gt; for Mac OS X). For instance, I map &quot;5thm&quot; to a theorem environment, and I usually have Greek letters expand when about half of them is typed (e.g. \alp -&gt; \alpha).&#xA;I have done this  (cf. my website for examples of notes with source code) and don't generally have a problem keeping up with the lecturer. Commutative diagrams are slightly trickier, but they can be done. If you want to live-TikZ your notes, I'm afraid I can't help; but you should learn LaTeX first!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using a word processing system is likely to be much slower for typesetting symbols. If the learning curve for LaTeX bothers you (and it's not that bad with a little practice), you might try &lt;a href=&quot;http://www.lyx.org/&quot; rel=&quot;nofollow&quot;&gt;LyX&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="536" LastActivityDate="2011-01-17T02:42:14.633" />
  <row Id="17816" PostTypeId="1" AcceptedAnswerId="17817" CreationDate="2011-01-17T03:03:55.387" Score="6" ViewCount="176" Body="&lt;p&gt;Given an equation in the form M(x)dx + N(y)dy = 0&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We test that the partial derivative of M with respect to y is equal to the partial derivative of N with respect to X&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If they are equal, then the equation is exact. What is the geometric interpretation of this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Further more to solve the equation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We may integrate M(x) Dx or N(y)Dy, which ever we like better, and then add a constant as a function in terms of the constant variable and solve this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;E.G.&#xA;let f(x) = 3x^2&#xA;then&#xA;F(x) = x^3 + g(y)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After we have our integral we set it's partial differential with respect to the other variable our other given derivative and solve for g(y)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have done the entire homework assignment correctly, but I have no clue why I am doing these steps. What is the geometric interpretation behind this algorithm, and how does it work?&lt;/p&gt;&#xA;" OwnerUserId="5934" LastActivityDate="2011-01-17T04:22:42.500" Title="What is the geometric interpretation behind the method of exact differential equations?" Tags="&lt;calculus&gt;&lt;differential-equations&gt;" AnswerCount="2" FavoriteCount="4" />
  <row Id="17817" PostTypeId="2" ParentId="17816" CreationDate="2011-01-17T03:42:42.253" Score="6" ViewCount="" Body="&lt;p&gt;Great question.  The idea is that $(M(x), N(y))$ defines a &lt;a href=&quot;http://en.wikipedia.org/wiki/Vector_field&quot; rel=&quot;nofollow&quot;&gt;vector field&lt;/a&gt;, and the condition you're checking is equivalent (on $\mathbb{R}^2$) to the vector field being &lt;a href=&quot;http://en.wikipedia.org/wiki/Conservative_vector_field&quot; rel=&quot;nofollow&quot;&gt;conservative&lt;/a&gt;, i.e. being the &lt;a href=&quot;http://en.wikipedia.org/wiki/Gradient&quot; rel=&quot;nofollow&quot;&gt;gradient&lt;/a&gt; of some scalar function $p$ called the potential.  Common physical examples of conservative vector fields include gravitational and electric fields, where $p$ is the gravitational or electric potential.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Geometrically, being conservative is equivalent to the &lt;a href=&quot;http://en.wikipedia.org/wiki/Curl_(mathematics)&quot; rel=&quot;nofollow&quot;&gt;curl&lt;/a&gt; vanishing.  It is also equivalent to the condition that &lt;a href=&quot;http://en.wikipedia.org/wiki/Line_integral&quot; rel=&quot;nofollow&quot;&gt;line integrals&lt;/a&gt; between two points depend only on the beginning and end points and not only on the path chosen.  (The connection between this and the curl is &lt;a href=&quot;http://en.wikipedia.org/wiki/Green%27s_theorem&quot; rel=&quot;nofollow&quot;&gt;Green's theorem&lt;/a&gt;.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The differential equation $M(x) \, dx + N(y) \, dy = 0$ is then equivalent to the condition that $p$ is a constant, and since this is not a differential equation it is a much easier condition to work with.  The analogous one-variable statement is that $M(x) \, dx = 0$ is equivalent to $\int M(x) \, dx = \text{const}$.  Geometrically, the solutions to $M(x) \, dx + N(y) \, dy = 0$ are therefore the &lt;a href=&quot;http://en.wikipedia.org/wiki/Level_set&quot; rel=&quot;nofollow&quot;&gt;level curves&lt;/a&gt; of the potential, which are always orthogonal to its gradient.  The most well-known example of this is probably the diagram of the electric field and the level curves of the electrostatic potential around a dipole.  This is one way to interpret the expression $M(x) \, dx + N(y) \, dy = 0$; it is precisely equivalent to the &quot;dot product&quot; of $(M(x), N(y)$ and $(dx, dy)$ being zero, where you should think of $(dx, dy)$ as being an infinitesimal displacement along a level curve.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(For those in the know, I am ignoring the distinction between vector fields and 1-forms and also the distinction between closed forms and exact forms.)&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2011-01-17T04:22:42.500" LastActivityDate="2011-01-17T04:22:42.500" CommentCount="2" />
  <row Id="17818" PostTypeId="2" ParentId="17816" CreationDate="2011-01-17T04:01:27.013" Score="2" ViewCount="" Body="&lt;p&gt;You might want to look up the wiki article on &lt;a href=&quot;http://en.wikipedia.org/wiki/Exact_differential&quot; rel=&quot;nofollow&quot;&gt;exact differentials&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/Inexact_differential&quot; rel=&quot;nofollow&quot;&gt;inexact differentials&lt;/a&gt;. The famous physical quantity which you cannot write as an exact differential is heat.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastActivityDate="2011-01-17T04:01:27.013" />
  <row Id="17819" PostTypeId="2" ParentId="17813" CreationDate="2011-01-17T04:50:22.003" Score="-1" ViewCount="" Body="&lt;p&gt;Edit: this answers a different question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Supposing that the mean is $\mu$ and the variance is $\Sigma$, calculate the Cholesky decomposition $\Sigma =LL^*$, generate a vector $v$ of $\operatorname{rank} \Sigma$ independent Gaussians, and output $Lv + \mu$.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2011-01-17T05:21:48.187" LastActivityDate="2011-01-17T05:21:48.187" CommentCount="3" />
  <row Id="17820" PostTypeId="2" ParentId="17813" CreationDate="2011-01-17T04:50:40.787" Score="2" ViewCount="" Body="&lt;p&gt;&lt;A href=&quot;http://azzalini.stat.unipd.it/SN/&quot; rel=&quot;nofollow&quot;&gt;This page&lt;/a&gt; should point you to what you want.  I found it just by following links from Wikipedia and in the future it would be best if you try that first before asking.&lt;/p&gt;&#xA;" OwnerUserId="822" LastActivityDate="2011-01-17T04:50:40.787" />
  <row Id="17821" PostTypeId="2" ParentId="17798" CreationDate="2011-01-17T05:05:32.263" Score="1" ViewCount="" Body="&lt;p&gt;I figured out how to prove it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $(X, \mathscr{O})$ be a ringed space. Let $\mathscr{F}, \mathscr{G}$ be sheaves of $\mathscr{O}$-modules. Define $\mathscr{H}(U) = \mathscr{F}(U) \otimes_{\mathscr{O}(U)} \mathscr{G}(U)$. Fix $p \in X$. Assume $U$ is an open n.h of $p$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The $\mathscr{O}_p$-module structure on $\mathscr{F}_p \otimes_{\mathscr{O}_p} \mathscr{G}_p$ induces a $\mathscr{O}(U)$-module structure on $\mathscr{F}_p \otimes_{\mathscr{O}_p} \mathscr{G}_p$ .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Define $\alpha_U : \mathscr{F}(U) \times \mathscr{G}(U) \to \mathscr{F}_p \otimes_{\mathscr{O}_p} \mathscr{G}_p, \quad (s,t) \mapsto s_p \otimes t_p$. This map is $2$-linear over $\mathscr{O}(U)$. Therefore $\alpha_U$ induces an $\mathscr{O}(U)$-module homomorphism from $\mathscr{F}(U) \otimes_{\mathscr{O}(U)} \mathscr{G}(U)$ to $\mathscr{F}_p \otimes_{\mathscr{O}_p} \mathscr{G}_p$. We shall abuse notation and also call this map $\alpha_U$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now forget about the $\mathscr{O}(U)$ module structure on the sections of $\mathscr{H}$. The $\alpha_U$s form a co-cone over the $\mathscr{H}(U)$s with $p \in U$ (that is they make the appropriate diagrams commute), therefore they induce a homomorphism of abelian groups $h : \mathscr{H}_p \to \mathscr{F}_p \otimes_{\mathscr{O}_p} \mathscr{G}_p$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Define $\psi : \mathscr{F}_p \times \mathscr{G}_p \to \mathscr{H}_p$ by $(s_p, t_p) \mapsto (s|_{U \cap V} \otimes t|_{U \cap V})_p$ where $s$ is a section of $\mathscr{F}$ over $U$ and $t$ is a section of $\mathscr{G}$ over $V$. It is easily verified that $\psi$ is 2-linear over $\mathscr{O}_p$. Therefore $\psi$ induces a $\mathscr{O}_p$ homomorphism from $\mathscr{F}_p \otimes_{p} \mathscr{G}_p$ to $\mathscr{H}_p$. It is easily verified that this map and $h$ are inverses.&lt;/p&gt;&#xA;" OwnerUserId="1725" LastEditorUserId="1725" LastEditorDisplayName="" LastEditDate="2011-01-17T05:10:43.380" LastActivityDate="2011-01-17T05:10:43.380" />
  <row Id="17822" PostTypeId="2" ParentId="17711" CreationDate="2011-01-17T05:06:00.010" Score="6" ViewCount="" Body="&lt;p&gt;One advanced reason for giving a proof is that in some related contexts such a formula breaks down. On the 2-adic numbers, the power series $A(x) = \exp(2x^2 - 2x)$ when expanded out and written in standard form turns out to converge on ${\mathbf Z}_2$, so in particular $A(1)$ is defined. One can prove $A(1)^2 = 1$, but although naively one may expect $A(1) = 1$ by just plugging 1 directly into the original definition I gave for $A(x)$, in fact $A(1) = -1$.  (The point here is that the naive calculation $A(1) = \exp(2 - 2) = \exp(2)\exp(-2) = 1$ is wrong since $\exp(2)$ doesn't make sense 2-adically.)&lt;/p&gt;&#xA;" OwnerUserId="619" LastActivityDate="2011-01-17T05:06:00.010" />
  <row Id="17823" PostTypeId="1" CreationDate="2011-01-17T05:21:15.943" Score="0" ViewCount="104" Body="&lt;p&gt;We have a sequence $ S_N(x) = \sum_{i=1}^{N}a_i(x) $ where $ a_i(x) \in \mathbb{R} \bigwedge a_i(x) &amp;gt; 0 \forall i \in \mathbb{N} and \forall x \in \mathbb{R}$ . We also know that the sequence $S_N(x)$ is diveregent $\forall x \in \mathbb{R}$. Now I want to characterize the function $S_N(x)$ for arbitrarily large $N \in \mathbb{N} $ as we would while talking about uniform and non uniform convergence. Please give some thoughts in this direction and also let me know the pit falls in this line of thinking.&lt;/p&gt;&#xA;" OwnerUserId="2987" LastActivityDate="2011-01-17T05:21:15.943" Title="Characterization of divergence" Tags="&lt;real-analysis&gt;&lt;sequences-and-series&gt;&lt;soft-question&gt;" CommentCount="8" />
  <row Id="17824" PostTypeId="1" CreationDate="2011-01-17T05:24:30.603" Score="1" ViewCount="114" Body="&lt;p&gt;Let $X$ be a completely regular space and let $T$ a topological space such that $X \subseteq T \subseteq \beta X$. Then $\beta T = \beta X$, where $\beta$ denotes the Stone-Cech compactification.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solution: Let $f: T \mapsto [0,1]$. Then it suffices to show that the restriction of f to $X$ can be extended to $\beta X$&quot;.. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is this? &lt;/p&gt;&#xA;" OwnerUserId="5937" LastEditorUserId="5937" LastEditorDisplayName="" LastEditDate="2011-01-17T05:37:43.830" LastActivityDate="2011-01-17T06:01:44.630" Title="Elementary question about extension of continuous functions" Tags="&lt;topology&gt;" AnswerCount="3" CommentCount="4" FavoriteCount="1" />
  <row Id="17825" PostTypeId="1" AcceptedAnswerId="17839" CreationDate="2011-01-17T05:36:48.323" Score="3" ViewCount="159" Body="&lt;p&gt;I have learned trigonometry, in school but never understood clearly what it is..just mugged up formula's and theorems to get clear the exams. But now i want to know what exactly, is trigonometry from the basics so that i can understand.&lt;/p&gt;&#xA;" OwnerUserId="671" LastActivityDate="2011-01-17T14:07:07.013" Title="basics of Trigonometry" Tags="&lt;trigonometry&gt;" AnswerCount="3" FavoriteCount="2" />
  <row Id="17826" PostTypeId="2" ParentId="17824" CreationDate="2011-01-17T05:38:09.087" Score="2" ViewCount="" Body="&lt;p&gt;Here is my answer to the modified question: Because $X$ is dense in $\beta X$, it is also dense in $T$, and therefore every continuous function on $T$ is uniquely determined by its restriction to $X$.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Here is my answer to the original question, which asked when, in order to see that a continuous function $f:D\to [0,1]$ has a continuous extension to $Z$, it suffices to show that the restriction of $f$ to $C$ has a continuous extension to $Z$, where $C\subseteq D\subseteq Z$ and $Z$ is a topological space.:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This would suffice precisely when every continuous function from $D$ to $[0,1]$ is determined by its restriction to $C$.  This will always be true if $C$ is dense in $D$.  If $D$ is &lt;a href=&quot;http://en.wikipedia.org/wiki/Tychonoff_space&quot; rel=&quot;nofollow&quot;&gt;completely regular&lt;/a&gt;, then density of $C$ is also necessary.  To see this, suppose $D$ is completely regular and $C$ is not dense.  Then there is a continuous function $f:D\to [0,1]$ such that $f$ is zero on $\overline C$, but $f(x)=1$ for some $x\in D\setminus\overline C$.  Since the zero function on $D$ is also a continuous extension of the zero function on $C$, the zero function on $C$ has more than one continuous extension to $D$.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2011-01-17T05:57:21.873" LastActivityDate="2011-01-17T05:57:21.873" CommentCount="2" />
  <row Id="17827" PostTypeId="2" ParentId="17825" CreationDate="2011-01-17T05:38:38.397" Score="1" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://www.clarku.edu/~djoyce/trig/&quot; rel=&quot;nofollow&quot;&gt;This&lt;/a&gt; is a good site(basic intro to trigonometry). Also &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0817639144&quot; rel=&quot;nofollow&quot;&gt;Trigonometry&lt;/a&gt; by Gelfand is praised highly. It gives a more intuitive/theoretical approach to the subject (and it also has good problems/exercises).&lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2011-01-17T05:38:38.397" CommentCount="1" />
  <row Id="17828" PostTypeId="2" ParentId="17824" CreationDate="2011-01-17T05:46:39.573" Score="0" ViewCount="" Body="&lt;p&gt;The answer to this &lt;a href=&quot;http://math.stackexchange.com/questions/17437/existence-of-closed-surjective-map-defined-in-the-stone-cech-compactification&quot;&gt;question&lt;/a&gt; (also about Stone-Cech compactification) is similar.&lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2011-01-17T05:46:39.573" />
  <row Id="17829" PostTypeId="2" ParentId="17824" CreationDate="2011-01-17T06:01:44.630" Score="2" ViewCount="" Body="&lt;p&gt;Recall that $\beta T$ is characterized by being (up to homeomorphism) the unique compact (Hausdorff) space of which $T$ is a dense subset and such that any continuous map $f:T\to[0,1]$ can be extended to a continuous map $g:\beta T\to[0,1]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ok, so, given $f:T\to[0,1]$, continuous, we want to show that it extends to a continuous  $g:\beta X\to[0,1]$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, $\beta T$ makes sense since $T$ is completely regular, being a subspace of $\beta X$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second, $X$ is dense in $T$, so any continuous $f:T\to[0,1]$ is completely determined by its restriction to $X$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Third, by definition of $\beta X$, this restriction extends to a continuous $g:\beta X\to[0,1]$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $f\upharpoonright X$ uniquely determined $f$, we have shown that any continuous $f:T\to[0,1]$ extends to a continuous $g:\beta X\to[0,1]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This proves the theorem: Since $X\subseteq T$, then $\beta X\subseteq \beta T$. If $\beta T$ happened to be larger than $\beta X$, then we have that any continuous $h:X\to[0,1]$ can be extended to a continuous $j:\beta T\to[0,1]$, contradicting that $\beta X$ is largest with this property. The reason why this extension exists, is because if $i:\beta X\to[0,1]$ is the continuous extension of $h$, then its restriction to $T$ is continuous, and can therefore be extended.&lt;/p&gt;&#xA;" OwnerUserId="462" LastActivityDate="2011-01-17T06:01:44.630" CommentCount="6" />
  <row Id="17830" PostTypeId="1" AcceptedAnswerId="17832" CreationDate="2011-01-17T06:42:48.480" Score="1" ViewCount="158" Body="&lt;p&gt;I know that following is a tautology because I've checked its truth table. I am now attempting to prove that it is a tautology by using the rules of logic, which is more difficult. How should I proceed?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(p\land(p\implies q))\implies q$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(p\land(\lnot p \lor q))\implies q$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(p\land \lnot p) \lor (p\land q) \implies q$ This step is where I'm getting stuck at. I know that $(p\land \lnot p)$ is false. So it seems to me that the truth value of everything to the left of the $\implies$ operator depends on the truth value of $(p\land q)$ So what I want to do is this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;FALSE $\lor (p\land q) \implies q$ which reduces to &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(p\land q) \implies q$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is my thinking correct so far? If so, then I want to rewrite $(p\land q) \implies q$ as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\lnot(p \land q) \lor (p \land q)$ by using the identity $p\implies q \equiv \lnot p \lor q$&#xA;Am I on the right track?&lt;/p&gt;&#xA;" OwnerUserId="1040" LastActivityDate="2011-01-17T21:13:19.130" Title="How should I proceed in proving this tautology?" Tags="&lt;homework&gt;&lt;logic&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="17831" PostTypeId="1" AcceptedAnswerId="17837" CreationDate="2011-01-17T06:47:36.853" Score="2" ViewCount="177" Body="&lt;p&gt;Sylvester's determinant identity states that if $A$ and $B$ are matrices of sizes $m\times n$ and $n\times m$, then $$ det(I+AB)=det(I+BA), $$ where in the first case $I$ denotes the $m\times m$ identity, and in the second, the $n\times n$ identity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could you sketch a proof for me, or point to an accessible reference?&lt;/p&gt;&#xA;" OwnerUserId="3340" LastActivityDate="2011-01-17T15:29:16.673" Title="Sylvester's determinant identity" Tags="&lt;linear-algebra&gt;" AnswerCount="2" />
  <row Id="17832" PostTypeId="2" ParentId="17830" CreationDate="2011-01-17T06:50:10.747" Score="2" ViewCount="" Body="&lt;p&gt;Use &lt;a href=&quot;http://en.wikipedia.org/wiki/DeMorgan%27s_Law&quot; rel=&quot;nofollow&quot;&gt;De Morgan's law&lt;/a&gt; on $\neg(p\land q)$ near the end. This should give you a disjunction which should easily be seen as tautological by the &lt;a href=&quot;http://en.wikipedia.org/wiki/Law_of_excluded_middle&quot; rel=&quot;nofollow&quot;&gt;law of excluded middle&lt;/a&gt;. Also, remember that $(p\land q)\implies q\equiv \neg(p\land q)\lor q$, not $\neg(p\land q)\lor(p\land q)$ as you wrote.&lt;/p&gt;&#xA;" OwnerUserId="1583" LastEditorUserId="1583" LastEditorDisplayName="" LastEditDate="2011-01-17T07:00:50.043" LastActivityDate="2011-01-17T07:00:50.043" CommentCount="1" />
  <row Id="17833" PostTypeId="2" ParentId="17830" CreationDate="2011-01-17T06:51:10.220" Score="1" ViewCount="" Body="&lt;p&gt;Your reasoning is correct so far. It seems to me that since $a\to b$ is equivalent to $\lnot a\lor b$, then $(p\land q)\to q$ is equivalent to $\lnot(p\land q)\lor q$, which is different from what you wrote in the last line. To continue, I guess you want to use that $\lnot(p\land q)$ is equivalent to $\lnot p\lor \lnot q$. &lt;/p&gt;&#xA;" OwnerUserId="3340" LastActivityDate="2011-01-17T06:51:10.220" />
  <row Id="17835" PostTypeId="2" ParentId="17831" CreationDate="2011-01-17T07:01:57.683" Score="6" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;(1)&lt;/strong&gt; Start, for fun, with a silly proof for square matrices:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $A$ is invertible, then &#xA;$$&#xD;&#xA;  \det(I+AB)=\det A^{-1}\cdot\det(I+AB)\cdot\det A=\det(A^{-1}\cdot(I+AB)\cdot A)=\det(I+BA).&#xD;&#xA;$$ Now, in general, both $\det(I+AB)$ and $\det(I+BA)$ are continuous functions of $A$, and equal on the dense set where $A$ is invertible, so they are everywhere equal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;(1)&lt;/strong&gt; Now, more seriously:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;\det\begin{pmatrix}I&amp;amp;-B\\\\A&amp;amp;I\end{pmatrix}&#xD;&#xA;\det\begin{pmatrix}I&amp;amp;B\\\\0&amp;amp;I\end{pmatrix}&#xD;&#xA;=\det\begin{pmatrix}I&amp;amp;-B\\\\A&amp;amp;I\end{pmatrix}\begin{pmatrix}I&amp;amp;B\\\\0&amp;amp;I\end{pmatrix}&#xD;&#xA;=\det\begin{pmatrix}I&amp;amp;0\\\\A&amp;amp;AB+I\end{pmatrix}&#xD;&#xA;=\det(I+AB)&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;\det\begin{pmatrix}I&amp;amp;B\\\\0&amp;amp;I\end{pmatrix}&#xD;&#xA;\det\begin{pmatrix}I&amp;amp;-B\\\\A&amp;amp;I\end{pmatrix}&#xD;&#xA;=\det\begin{pmatrix}I&amp;amp;B\\\\0&amp;amp;I\end{pmatrix}&#xD;&#xA;\begin{pmatrix}I&amp;amp;-B\\\\A&amp;amp;I\end{pmatrix}&#xD;&#xA;=\det\begin{pmatrix}I+BA&amp;amp;0\\\\A&amp;amp;I\end{pmatrix}&#xD;&#xA;=\det(I+BA)&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the leftmost members of these two equalities are equal, we get the equality you want.&lt;/p&gt;&#xA;" OwnerUserId="274" LastEditorUserId="274" LastEditorDisplayName="" LastEditDate="2011-01-17T07:43:17.570" LastActivityDate="2011-01-17T07:43:17.570" CommentCount="4" />
  <row Id="17836" PostTypeId="1" AcceptedAnswerId="17941" CreationDate="2011-01-17T07:09:02.073" Score="1" ViewCount="187" Body="&lt;p&gt;&lt;a href=&quot;http://tieba.baidu.com/%E6%95%B0%E5%AD%A6/shipin/play/f8265eec832c78333a085c0d/&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; is a good math movie. what is your favorite software to make math movies? I used Mathematica, GeoGebra, Sketchpad. But they are not perfect. Also, could you recommend some interesting math movies?&lt;/p&gt;&#xA;" OwnerUserId="1498" LastActivityDate="2011-01-18T05:00:10.467" Title="what tools are good at making mathematic movies?" Tags="&lt;math-software&gt;&lt;math-movie&gt;" AnswerCount="1" CommentCount="6" />
  <row Id="17837" PostTypeId="2" ParentId="17831" CreationDate="2011-01-17T07:33:42.213" Score="9" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\ \ $ Work &lt;em&gt;universally&lt;/em&gt;, i.e. over the polynomial ring $\rm\ \mathbb Z[A_{ij},B_{ij}]\ $ take the determinant of $\rm\ (1+A\ B)\ A\ =\ A\ (1+B\ A)\ \ $ then cancel $\rm\ det(A)\ \ $ (which is valid since the ring is a domain). $\ \ $ Extend to non-square matrices by padding appropriately with $0$'s and $1$'s to get square matrices. Note that the proof is &lt;em&gt;purely&lt;/em&gt; algebraic - it does not require any topological notions (e.g. density).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively, one may proceed by way of Schur decomposition, namely&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\rm\left[ \begin{array}{ccc}&#xD;&#xA;1 &amp;amp; \rm A \\&#xD;&#xA;\rm B &amp;amp; 1 \end{array} \right]\ =\ \left[ \begin{array}{ccc}&#xD;&#xA;1 &amp;amp; \rm 0 \\&#xD;&#xA;\rm B &amp;amp; 1 \end{array} \right]\ \left[ \begin{array}{ccc}&#xD;&#xA;1 &amp;amp; \rm 0 \\&#xD;&#xA;\rm 0 &amp;amp; \rm 1-BA \end{array} \right]\ \left[ \begin{array}{ccc}&#xD;&#xA;1 &amp;amp; \rm A \\&#xD;&#xA;\rm 0 &amp;amp; 1 \end{array} \right]$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\rm\phantom{\left[ \begin{array}{ccc}&#xD;&#xA;1 &amp;amp; \rm B \\&#xD;&#xA;\rm A &amp;amp; 1 \end{array} \right]}\ =\ \left[ \begin{array}{ccc}&#xD;&#xA;1 &amp;amp; \rm A \\&#xD;&#xA;\rm 0 &amp;amp; 1 \end{array} \right]\ \left[ \begin{array}{ccc}&#xD;&#xA;1-AB &amp;amp; \rm 0 \\&#xD;&#xA;\rm 0 &amp;amp; \rm 1 \end{array} \right]\ \left[ \begin{array}{ccc}&#xD;&#xA;1 &amp;amp; \rm 0 \\&#xD;&#xA;\rm B &amp;amp; 1 \end{array} \right]$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See my posts in &lt;a href=&quot;http://groups.google.com/group/sci.math/browse_frm/thread/200930c67b467abc/5aef1ba8cd148297&quot; rel=&quot;nofollow&quot;&gt;this sci.math thread&lt;/a&gt; on 09 Nov 2007 for further discussion.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2011-01-17T15:29:16.673" LastActivityDate="2011-01-17T15:29:16.673" CommentCount="3" />
  <row Id="17839" PostTypeId="2" ParentId="17825" CreationDate="2011-01-17T08:05:30.797" Score="9" ViewCount="" Body="&lt;p&gt;A conceptual view of trigonometry is this: a simple diagram will convince you that if you know the side lengths of two sides of a triangle and the angle between them, then the length of the third side is uniquely determined (do a sketch and convince yourself of that!). A natural question then is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;How do you compute that third length?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If this angle that you are given happens to be a right angle, then you have Pythagoras. But what if you have an arbitrary angle? This question naturally leads to trigonometry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another way to arrive at the same thing: already the Greeks knew that if you start with a triangle and then scale it by a constant factor (zoom in or out, as it were), then the ratios of the corresponding side lengths don't change. In other words, these ratios only depend on the angles of the triangle (while, of course, each side length by itself cannot be computed just from the angles). So the question arises:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What is this dependence precisely? If I give you three angles, can you compute the ratios of the sides?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes - with trigonometry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for specific uses of sin, cos, and tan, they are just names for the aforementioned ratios of side lengths in the special case of &lt;em&gt;right angled&lt;/em&gt; triangles. Now, you might be tempted to proceed and give names to these ratios in arbitrary triangles, but the great insight behind trigonometry is that you don't need to do that: you can express all those ratios using the already introduced notions of sin, cos, and tan. In other words, you can reduce all triangles to right angled ones (by using heights of the triangle, each of which is perpendicular to a side).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, things like Pythagoras generalise quite nicely. If you inspect the &lt;a href=&quot;http://en.wikipedia.org/wiki/Law_of_cosines&quot; rel=&quot;nofollow&quot;&gt;law of cosines&lt;/a&gt;, you will see that it is like &quot;Pythagoras with an error term&quot;. So another way to express how one might arrive at trigonometry is this: you know how to express the hypotenuse in a right angled triangle. You also know that as you begin to shrink the angle while preserving the two side lengths, that hypotenuse gets shorter. By how much? You guessed it...&lt;/p&gt;&#xA;" OwnerUserId="3212" LastEditorUserId="3212" LastEditorDisplayName="" LastEditDate="2011-01-17T12:07:59.110" LastActivityDate="2011-01-17T12:07:59.110" CommentCount="1" />
  <row Id="17840" PostTypeId="1" CreationDate="2011-01-17T08:53:22.257" Score="1" ViewCount="97" Body="&lt;p&gt;how to prove it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am talking about matrixes which:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;( Ax , x ) &gt; 0 for any x not equals 0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How to prove, that A&lt;sup&gt;T&lt;/sup&gt; is also positive?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;x&lt;sup&gt;T&lt;/sup&gt; * A * x = ( x&lt;sup&gt;T&lt;/sup&gt; * A * x )&lt;sup&gt;T&lt;/sup&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;and what? &lt;/p&gt;&#xA;" OwnerUserId="5940" LastEditorUserId="5940" LastEditorDisplayName="" LastEditDate="2011-01-17T09:02:31.750" LastActivityDate="2011-01-17T09:02:31.750" Title="transpose of positive matrix is positive" Tags="&lt;linear-algebra&gt;&lt;matrices&gt;" AnswerCount="1" />
  <row Id="17841" PostTypeId="2" ParentId="17840" CreationDate="2011-01-17T08:55:49.387" Score="4" ViewCount="" Body="&lt;p&gt;Hint: Write the inner product as $x^T A x$ and use the fact that that expression is its own transpose (since it's a 1-by-1 matrix).&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2011-01-17T08:55:49.387" CommentCount="1" />
  <row Id="17842" PostTypeId="1" AcceptedAnswerId="17895" CreationDate="2011-01-17T09:21:30.787" Score="5" ViewCount="131" Body="&lt;p&gt;Working in $F_q^n$. How many different ways do we have to choose $l$ vectors such that every subset of size $k$ of them is linearly independent.&#xA;(Assume n is large)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My Progress&lt;/strong&gt;: For the first k vectors, just keep out of the subspace spanned so far, so for $1 \leq i \leq k$ we have $q^n-q^{i-1}$ ways to choose the $i^{th}$ vector. But the next ones are harder.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;: Although the assignment asks me to find the exact number, a good lower bound can be helpful.&lt;/p&gt;&#xA;" OwnerUserId="5941" LastEditorUserId="5941" LastEditorDisplayName="" LastEditDate="2011-01-17T19:02:52.167" LastActivityDate="2011-01-19T01:02:59.350" Title="How many ways to choose $l$ vectors in $n$-dimensional space such that every $k$-subset is independent" Tags="&lt;linear-algebra&gt;&lt;homework&gt;" AnswerCount="2" CommentCount="7" />
  <row Id="17843" PostTypeId="1" CreationDate="2011-01-17T10:55:39.730" Score="0" ViewCount="119" Body="&lt;p&gt;Given canonical cone equation, how to find equation of cone's generatrix?&lt;/p&gt;&#xA;" OwnerUserId="3532" LastActivityDate="2011-01-17T21:40:10.383" Title="How to find equation of cone's generatrix?" Tags="&lt;differential-geometry&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="17844" PostTypeId="2" ParentId="17688" CreationDate="2011-01-17T12:15:44.720" Score="3" ViewCount="" Body="&lt;p&gt;There is an answer to this question at MO:&lt;br&gt;&#xA;&lt;a href=&quot;http://mathoverflow.net/questions/52246/seemingly-complex-logic-set-theoretic-puzzle/52303#52303&quot; rel=&quot;nofollow&quot;&gt;http://mathoverflow.net/questions/52246/seemingly-complex-logic-set-theoretic-puzzle/52303#52303&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5904" LastActivityDate="2011-01-17T12:15:44.720" />
  <row Id="17845" PostTypeId="1" CreationDate="2011-01-17T13:01:03.180" Score="0" ViewCount="124" Body="&lt;p&gt;Looking at an earlier post &lt;a href=&quot;http://math.stackexchange.com/questions/4510/finding-the-power-series-of-a-rational-function&quot;&gt;Finding the power series of a rational function&lt;/a&gt;, I am trying to get a closed formula for the n'th coefficient in the Taylor series of the rational function (1-x)/(1-2x-x^3). Is it possible to use any of the tricks in that post to not only obtain specific coefficients, but an expression for the n'th coefficient ?If T(x) is the Taylor polynomial I am looking at the equality (1-x) = (1-2x-x^3)*T(x) and differentiating, but I am not able to see a pattern giving me an explicit formula for the coefficients.&lt;/p&gt;&#xA;" OwnerUserId="5943" LastActivityDate="2011-01-17T13:47:27.693" Title="Coefficients for Taylor series of given rational function" Tags="&lt;taylor-expansion&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="17846" PostTypeId="2" ParentId="17845" CreationDate="2011-01-17T13:15:13.947" Score="2" ViewCount="" Body="&lt;p&gt;Denoting the coefficients of $T(x)$ as $t_i$ (so $T(x) = t_0 + t_1 x + ...$), consider the coefficient of $x^a$ in $(1 - 2x - x^3) T(x)$. It shouldn't be hard to show that it's $t_a - 2t_{a-1} - t_{a-3}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we have $t_0 - 2t_{-1} - t_{-3} = 1$, $t_1 - 2t_0 - t_{-2} = -1$, and $a &amp;gt; 1 \Rightarrow t_a - 2t_{a-1} - t_{a-3} = 0$. You can get $t_0$, $t_1$. Do you know how to solve the discrete recurrence $t_{a+3} = 2t_{a+2} + t_{a}$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(If not, here's a hint: suppose you had a number $\alpha$ such that $t_a = \alpha^a$ satisfied the recurrence. What constraint on $\alpha$ can you prove? Now consider a linear combination of the possible such solutions, $t_a = b_0 \alpha_0^a + ... + b_n \alpha_n^a$, and solve $n$ simultaneous equations from your base cases $t_0$, $t_1$, and use $0 = t_{-1} = t_{-2} = ...$ if necessary).&lt;/p&gt;&#xA;" OwnerUserId="5676" LastEditorUserId="5676" LastEditorDisplayName="" LastEditDate="2011-01-17T13:23:12.250" LastActivityDate="2011-01-17T13:23:12.250" />
  <row Id="17849" PostTypeId="2" ParentId="17825" CreationDate="2011-01-17T14:07:07.013" Score="0" ViewCount="" Body="&lt;p&gt;In mathematics we encounter linear objects (i.e. cartesian coordinates, lengths, heights, distances, etc.) and circular type objects (i.e. angles, radii, rotations, etc).  Often mathematical problems (and practical problems) require that we use both.  Trigonometry is the set of mathematical tools by which we relate the linear and the circular objects.  So sin, cos relate the angle and radius to x and y coordinates.  Likewise, tan relates the x and y coordinates to the angle.&lt;/p&gt;&#xA;" OwnerUserId="4726" LastActivityDate="2011-01-17T14:07:07.013" />
  <row Id="17850" PostTypeId="1" AcceptedAnswerId="17855" CreationDate="2011-01-17T14:57:47.533" Score="1" ViewCount="272" Body="&lt;p&gt;$A     = r \int_{0}^{\pi}\int_{0}^{2\pi} e^{i (\alpha+\theta)} d\alpha d\theta&#xD;&#xA;        = r \int_{0}^{\pi} [\frac{-i}{\alpha+\theta} e^{i(\alpha + \theta)}]_{0}^{2\pi} d\theta&#xD;&#xA;        = ...&#xD;&#xA;        = r \int_{0}^{\pi} e^{2i\alpha} ( \frac{1}{\alpha} - \frac{1}{\alpha+2\pi}  ) d\theta$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and according to wolfram it is wrong, here.&#xA;So how do you deduce the area of a sphere in polar coordinates?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;[Edit]&lt;/strong&gt;  Trevor gave us formula $r^2 \int_{2π}^{0}\int^{π}_{0}sin(\phi) d\phi d\theta$ without deduction, by deduction, I mean things on which it is based on. &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Why is it odd function $sin(\phi)$ and not instead an even function $cos(\phi)$? (I know that it is and I can calculate it but some proof here missing)&lt;/li&gt;&#xA;&lt;li&gt;Why no term with $\theta$ but with $\phi$? &lt;/li&gt;&#xA;&lt;li&gt;Why $r^{2}$? -Okay it is area and by symmetry it sounds logical but why $r^{2}$ and not for example $rd$ where $d$ is some lenght or even a path? &lt;/li&gt;&#xA;&lt;li&gt;What about an arbitrary case? &lt;/li&gt;&#xA;&lt;li&gt;What determines the order in integration? &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Many things open to really deduce it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;[Edit]&lt;/strong&gt; &lt;a href=&quot;http://math.stackexchange.com/questions/17850/how-to-deduce-the-area-of-sphere-in-polar-coordinates/17855#17855&quot;&gt;Zhen Lin&lt;/a&gt; has an excellent answer where s/he noted that &lt;code&gt;2-sphere&lt;/code&gt;, i.e. the sphere in three dimensional euclidean space, is two-dimensional. Next deductions should be on &lt;code&gt;0-sphere&lt;/code&gt; and then incrementally on others. More about &lt;code&gt;n-sphere&lt;/code&gt; &lt;a href=&quot;http://en.wikipedia.org/wiki/N-sphere&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;How do you deduce area in &lt;code&gt;0-sphere&lt;/code&gt;, &lt;code&gt;1-sphere&lt;/code&gt;, &lt;code&gt;2-sphere&lt;/code&gt;,  and &lt;code&gt;n-sphere&lt;/code&gt;?&lt;/li&gt;&#xA;&lt;li&gt;Why does the Jacobian matrix in &lt;code&gt;2-sphere-in-3-dim-euclidean-case&lt;/code&gt; have only two two variables $(\phi, \theta)$?&lt;/li&gt;&#xA;&lt;li&gt;Are $\theta$ and $\phi$ non-free-variables because they are bounded somehow by the manifold (maybe misusing terminology)?&lt;/li&gt;&#xA;&lt;li&gt;What is $r$ in the sphere? For example in the &lt;code&gt;2-sphere&lt;/code&gt;?&lt;/li&gt;&#xA;&lt;li&gt;I can sense that the var $r$ is somehow different to $\phi$ and $\theta$ because what is it if you derivate it, zero so perhaps it is a trivial case (again probably misusing terminology, sorry)?&lt;/li&gt;&#xA;&lt;li&gt;What about if $r$ depends on $t$, time, what is it then? $\frac{\partial r}{\partial t}$ is now something. But would the extra dimension be now $t$ instead of $r$?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Please, use the term &lt;code&gt;dimension&lt;/code&gt; in your replies, I feel it is very important to even understand what a sphere is, let alone a dimension. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;[Conclusion]&lt;/strong&gt; The general solution is still open. The $r$ is a homothetic transformation. The terms $(\theta, \phi)$ are apparently tied to the manifold, (ALERT! perhaps misusing terminology). The acceptance of a question does not mean that all things appeared here are solved. Zhen`s solution is really the solution to look at, big thanks for it!&lt;/p&gt;&#xA;" OwnerUserId="5902" LastEditorUserId="5902" LastEditorDisplayName="" LastEditDate="2011-01-17T18:53:37.240" LastActivityDate="2011-01-17T18:53:37.240" Title="How to deduce the area of sphere in polar coordinates?" Tags="&lt;polar-coordinates&gt;" AnswerCount="4" CommentCount="2" FavoriteCount="1" />
  <row Id="17851" PostTypeId="2" ParentId="17850" CreationDate="2011-01-17T15:09:57.613" Score="1" ViewCount="" Body="&lt;p&gt;The surface area of a sphere would be $r^2 \int_{0}^{2 \pi} \int_{0}^{\pi} \sin(\phi) \ d \phi \ d \theta$. This is in spherical coordinates. &lt;/p&gt;&#xA;" OwnerUserId="2548" LastEditorUserId="2548" LastEditorDisplayName="" LastEditDate="2011-01-17T15:15:06.297" LastActivityDate="2011-01-17T15:15:06.297" CommentCount="4" />
  <row Id="17852" PostTypeId="1" AcceptedAnswerId="17853" CreationDate="2011-01-17T15:22:24.277" Score="3" ViewCount="93" Body="&lt;p&gt;If A and B are two different nonempty sets, how many distinct sets can be formed&#xA;with these sets using as many unions,intersections,complements and parentheses as desired.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Four sets are fundamental:$A$,$B$,$A \cup B$,$A \cap B$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other sets are $A \cup B&amp;#39;$,$A' \cup B$,$A \cap B'$,$A' \cap B$,$A' \cup B'$,$A' \cap B'$.&#xA;Any other sets are possible.&lt;/p&gt;&#xA;" OwnerUserId="5049" LastActivityDate="2011-01-17T15:29:07.207" Title="How many different sets can be formed by two nonempty sets" Tags="&lt;elementary-set-theory&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="17853" PostTypeId="2" ParentId="17852" CreationDate="2011-01-17T15:29:07.207" Score="4" ViewCount="" Body="&lt;p&gt;$16$.  The corresponding Venn diagram has four parts, and you can get any combination of those four parts.  &lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2011-01-17T15:29:07.207" CommentCount="2" />
  <row Id="17854" PostTypeId="2" ParentId="17850" CreationDate="2011-01-17T15:52:07.267" Score="1" ViewCount="" Body="&lt;p&gt;I'll try and answer 1, 2 and 3.. It's probably best to make a drawing, but you can also understand it 'algebraically'. An area element &quot;$d\phi \; d\theta$&quot; close to one of the poles is really small, tending to zero as you approach the North or South pole of the sphere. Close to the equator, the area tends to resemble a flat surface. This corresponds (qualitavely) to the factor $\sin(\phi)$, which goes to zero as $\phi$ goes to 0 or $\pi$, and equals 1 for $\pi / 2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is no $\theta$-dependence, since the area element does not depend on your $\theta$-coordinate (i.e. walking parallel to the equator should not change your area element). Your coordinates are axisymmetrical with respect to the axe that goes through both poles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, why should the result depend on some other length $d$? It doesn't appear in the problem description, so finding it in your final result should be very suspect (all you're doing is some calculus, there's no exotic math involved which could introduce such a parameter).&lt;/p&gt;&#xA;" OwnerUserId="5923" LastEditorUserId="5923" LastEditorDisplayName="" LastEditDate="2011-01-17T15:57:52.530" LastActivityDate="2011-01-17T15:57:52.530" />
  <row Id="17855" PostTypeId="2" ParentId="17850" CreationDate="2011-01-17T15:58:35.933" Score="1" ViewCount="" Body="&lt;p&gt;The easiest proof is to use the formula &#xA;$$A = \int_S \sqrt{\det g} \, du \, dv $$&#xA;from Riemannian geometry, where $g = \begin{pmatrix} E &amp;amp; F \\ F &amp;amp; G \end{pmatrix}$ is the metric tensor / first fundamental form and $(u, v)$ are coordinates for the manifold $S$. Unfortunately this doesn't really explain anything, because now the question is how to justify this formula!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I'll attempt to justify it now. First, we need to restrict to the case where we already know how to calculate areas, namely subsets of the Euclidean plane. We know that the area of a subset $S \subset \mathbb{R}^2$ is just&#xA;$$A = \int_S dx \, dy$$&#xA;where $(x, y)$ are the standard Cartesian coordinates. So, for example, the area of a square with corners $(0, 0), (a, 0), (a, b), (0, b)$ is just&#xA;$$A = \int_0^b \int_0^a dx \, dy = ab$$&#xA;But what about shapes which we can't easily parametrise using Cartesian coordinates? The change of variables formula from multivariable calculus tells us we need to introduce a factor — the Jacobian determinant — when we do this. For example, if we use plane polar coordinates $(r, \theta)$, then&#xA;$$\begin{aligned} x &amp;amp; = r \cos \theta \\ y &amp;amp; = r \sin \theta \end{aligned}$$&#xA;so the Jacobian matrix is&#xA;$$J = \begin{pmatrix} \cos \theta &amp;amp; -r \sin \theta \\ \sin \theta &amp;amp; r \cos \theta \end{pmatrix}$$&#xA;and the Jacobian determinant is $\det J = r$. Hence, the polar-coordinate form of the general formula is&#xA;$$A = \int_S r \, dr \, d\theta$$&#xA;and we can use this to calculate, say, the area of a circle of radius $a$:&#xA;$$A = \int_0^{2\pi} \int_{0}^{a} r \, dr \, d\theta = \pi a^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But what does any of this have to do with $g$? Well, it turns out that $g = J^T J$, so for example&#xA;$$g = \begin{pmatrix} 1 &amp;amp; 0 \\ 0 &amp;amp; r^2 \end{pmatrix}$$&#xA;for polar coordinates. Now, by the various properties of the determinant, it follows that in general $\det g = (\det J)^2$, i.e. $\det J = \sqrt{\det g}$. So this justifies the formula at the very beginning of my post.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let's apply this formula to the sphere. We have spherical polar coordinates $(\phi, \theta)$ such that&#xA;$$\begin{aligned}&#xD;&#xA;x &amp;amp; = r \cos \theta \sin \phi \\&#xD;&#xA;y &amp;amp; = r \sin \theta \sin \phi \\&#xD;&#xA;z &amp;amp; = r \cos \phi&#xD;&#xA;\end{aligned}$$&#xA;and this gives the Jacobian&#xA;$$J = \begin{pmatrix}&#xD;&#xA;r \cos \theta \cos \phi &amp;amp; -r \sin \theta \sin \phi \\&#xD;&#xA;r \sin \theta \cos \phi &amp;amp;  r \cos \theta \sin \phi \\&#xD;&#xA;-r \sin \phi &amp;amp; 0&#xD;&#xA;\end{pmatrix}$$&#xA;so the metric is&#xA;$$g = J^T J = \begin{pmatrix} r^2 &amp;amp; 0 \\ 0 &amp;amp; (r \sin \phi)^2 \end{pmatrix}$$&#xA;hence the area of the sphere is just&#xA;$$A = \int_0^{2\pi} \int_0^\pi r^2 \sin \phi \, d\phi \, d\theta = 4 \pi r^2$$&#xA;as we all know.&lt;/p&gt;&#xA;" OwnerUserId="5191" LastActivityDate="2011-01-17T15:58:35.933" CommentCount="11" />
  <row Id="17856" PostTypeId="1" AcceptedAnswerId="17884" CreationDate="2011-01-17T16:05:36.237" Score="11" ViewCount="311" Body="&lt;p&gt;Recently I became interested in trying to visualize the fact that $\pi_1(\text{SO}(3)) = \mathbb{Z}/2\mathbb{Z}$.  For whatever reason, the &lt;a href=&quot;http://en.wikipedia.org/wiki/Plate_trick&quot; rel=&quot;nofollow&quot;&gt;plate trick&lt;/a&gt; doesn't do it for me, so I've been looking for something else.  I know that $\text{SO}(3)$ is homeomorphic to $P^3(\mathbb{R})$, and one concrete visualization of this space is as the solid unit ball with opposite points on the boundary identified.  This model allows a straightforward visualization of the fundamental group, but what it doesn't seem to easily allow for is a corresponding visualization of the action of $\text{SO}(3)$.  (In other words, it doesn't seem to be easy to visually verify that we are actually working in $\text{SO}(3)$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I tried the following instead: $\text{SO}(3)$ acts simply transitively on the unit tangent bundle of $S^2$, so $UT(S^2)$ ought to be homeomorphic to $\text{SO}(3)$.  This satisfies both my requirements: it is easy to visualize paths in this space, and it is also easy to see the action of $\text{SO}(3)$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I would just appreciate someone confirming that I've got the details right: in this model, a representative of the nontrivial homotopy class of paths in $\pi_1(\text{SO}(3))$ is the path which parallel transports a unit vector once around a great circle, correct?  How should I visualize the homotopy between the square of this path and the trivial path?  (This might boil down to the plate trick or the belt trick, but at least it'll be stated in a language that I can understand.)&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2011-01-17T20:34:25.927" Title="Visualizing the fundamental group of SO(3)" Tags="&lt;algebraic-topology&gt;&lt;lie-groups&gt;" AnswerCount="1" CommentCount="7" FavoriteCount="5" />
  <row Id="17857" PostTypeId="2" ParentId="17711" CreationDate="2011-01-17T16:13:25.813" Score="1" ViewCount="" Body="&lt;p&gt;One definition of $e^x$ is $\displaystyle \lim_{n \rightarrow \infty}(1 + \frac{x}{n})^n$. From this definition, it doesn't automatically follow that $e^x e^y = e^{x+y}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, it doesn't even follow immediately that $e^x = \displaystyle \lim_{n \rightarrow \infty}(1 + \frac{x}{n})^n = (\displaystyle \lim_{n \rightarrow \infty}(1 + \frac{1}{n})^n)^x = (e)^x$. What this means is $e^x$ is just a short hand notation for the limit which after some analysis we realize it as $(e)^x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By limit arguments, we can now show that $\displaystyle \lim_{n \rightarrow \infty}(1 + \frac{x}{n})^n = 1 + \sum_{k=1}^{\infty} \frac{x^k}{k!}$, $\forall x \in \mathbb{R}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now $e^x \times e^y = (1 + \sum_{k=1}^{\infty} \frac{x^k}{k!}) \times (1 + \sum_{k=1}^{\infty} \frac{y^k}{k!})$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we need to realize that we can rearrange the terms in the series and multiply terms of the two series since both of them converge absolutely.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence $e^x \times e^y = (1 + \sum_{k=1}^{\infty} \frac{x^k}{k!}) \times (1 + \sum_{k=1}^{\infty} \frac{y^k}{k!}) = 1 + (x+y) + (\frac{x^2 + 2xy + y^2}{2!}) + (\frac{x^3 + 3x^2y + 3xy^2 + y^3}{3!}) + \cdots$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now make use of the binomial theorem to get &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$e^{x} e^{y} = e^{x+y}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS: Though I have taken care to make sure the line of thought is right, you need to be careful when writing down the argument as to when you can interchange terms in an infinite series, multiply out two infinite series etc etc.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2011-02-10T23:23:39.047" LastActivityDate="2011-02-10T23:23:39.047" CommentCount="4" />
  <row Id="17858" PostTypeId="1" CreationDate="2011-01-17T16:26:23.187" Score="10" ViewCount="435" Body="&lt;p&gt;I'm a computer science major and like many of us we have to take two additional sciences. These two additional science courses are in addition to three semesters of calculus,two semesters of physics, and one semester each of linear algebra and discrete math.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I could take the easy way out and take an algebra based astronomy course, but I want the most bang for my buck as it relates to my major.What other types of math classes should I take? I do enjoy math a lot and have always done well.&lt;/p&gt;&#xA;" OwnerUserId="1040" LastActivityDate="2011-05-20T18:18:40.943" Title="What math should a computer scientist take in college?" Tags="&lt;computer-science&gt;" AnswerCount="6" CommentCount="1" FavoriteCount="5" />
  <row Id="17859" PostTypeId="2" ParentId="17858" CreationDate="2011-01-17T16:30:16.043" Score="5" ViewCount="" Body="&lt;p&gt;According to &lt;a href=&quot;http://www.cs.princeton.edu/courses/archive/spr10/cos433/mathcs.pdf&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; (site about math for computer scientists), graph theory,combinatorics, probability, and number theory would be worth taking.&lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2011-01-17T16:30:16.043" />
  <row Id="17860" PostTypeId="1" CreationDate="2011-01-17T16:37:11.373" Score="0" ViewCount="38" Body="&lt;p&gt;$K$ a field, $A\subset K$ a subring, $M\subset K$ an $A$-submodule. Define $$(A:_{K}M):=\{s\in K|sM\subset A\}$$ &#xA;Then it is easy to see that $$M\subset A\Longleftrightarrow A\subset (A:_{K}M),$$&#xA;$$A\subset M\Longrightarrow (A:_{K}M)\subset A$$&#xA;But I couldn't show the reverse implication of the second. It is true if $M$ is invertible, and I'm guessing that it is true only if $M$ is invertible. Any ideas?&lt;/p&gt;&#xA;" OwnerUserId="4341" LastActivityDate="2011-01-17T16:37:11.373" Title="Related to fractional ideals" Tags="&lt;abstract-algebra&gt;&lt;ring-theory&gt;&lt;commutative-algebra&gt;&lt;modules&gt;" CommentCount="1" />
  <row Id="17861" PostTypeId="2" ParentId="17858" CreationDate="2011-01-17T16:40:27.517" Score="6" ViewCount="" Body="&lt;p&gt;I am assuming you are going to do theoretical research in Computer Science.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since you have already taken Calculus, Linear algebra and Discrete Mathematics (which I presume includes Graph theory and some Combinatorics),&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would suggest&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Abstract Algebra&lt;/strong&gt;: For a motivation of this, try reading the paper &lt;a href=&quot;http://www.cse.iitk.ac.in/users/manindra/algebra/primality_v6.pdf&quot; rel=&quot;nofollow&quot;&gt;Primes is in P&lt;/a&gt; which gives a polynomial time algorithm for prime number detection. Abstract Algebra has practical computer science applications too, for example: &lt;a href=&quot;http://en.wikipedia.org/wiki/Coding_theory&quot; rel=&quot;nofollow&quot;&gt;Coding Theory&lt;/a&gt;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Probability&lt;/strong&gt;: For a motivation of this, try reading some of the hardness of approximation results based on the &lt;a href=&quot;http://en.wikipedia.org/wiki/PCP_theorem&quot; rel=&quot;nofollow&quot;&gt;PCP theorem&lt;/a&gt;. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;You might also want to check out &lt;a href=&quot;http://cstheory.stackexchange.com&quot;&gt;http://cstheory.stackexchange.com&lt;/a&gt;. The following two questions from there seem relevant (but hopefully, won't add to the uncertainty) and could perhaps be used to get ideas for future courses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://cstheory.stackexchange.com/questions/1920/examples-of-unrelated-mathematics-playing-a-fundamental-role-in-tcs&quot;&gt;http://cstheory.stackexchange.com/questions/1920/examples-of-unrelated-mathematics-playing-a-fundamental-role-in-tcs&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://cstheory.stackexchange.com/questions/2571/what-kind-of-mathematical-background-is-needed-for-complexity-theory&quot;&gt;http://cstheory.stackexchange.com/questions/2571/what-kind-of-mathematical-background-is-needed-for-complexity-theory&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope that helps.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2011-01-17T17:05:05.563" LastActivityDate="2011-01-17T17:05:05.563" />
  <row Id="17862" PostTypeId="2" ParentId="17850" CreationDate="2011-01-17T16:45:28.507" Score="5" ViewCount="" Body="&lt;p&gt;Here is a diagram that might help you to visually understand that the element of area $\text{d}A = r^2 \sin \phi \, \text{d} \phi \, \text{d} \theta.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that in the diagram $0 \le \phi \le \pi $ and $ 0 \le \theta  \le 2 \pi.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/cBWvL.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1301" LastActivityDate="2011-01-17T16:45:28.507" />
  <row Id="17863" PostTypeId="2" ParentId="16954" CreationDate="2011-01-17T16:53:31.990" Score="2" ViewCount="" Body="&lt;p&gt;The proof seems quite long, and requires the classification of finite simple groups.  Perhaps the most surprising part to me is that the similar question in prime degree &lt;em&gt;p&lt;/em&gt; (rather than degree 2&lt;i&gt;p&lt;/i&gt; as asked) still requires the classification to handle general primes, and requires quite a bit of modular representation theory to handle primes of special forms.  At any rate, here is what I've had time to write up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Itô's conjecture and Wielandt's problem&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;em&gt;Is the normalizer F of a Sylow p-subgroup of the symmetric group on p points maximal?&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This seemed to be true.  The first results along these lines were Galois's investigations on groups of prime degree, followed by Mathieu's discovery of his sporadic groups (especially of prime degree).  Burnside used ordinary character theory to show that a transitive group of prime degree was either a subgroup of F or insoluble.  Brauer (1943) applied his ideas of modular character theory to give some bounds.  Itô (1958+) began to get detailed information on groups of prime degree when the primes were of special form.  Wielandt's 1964 textbook (Part V, especially §31) gathered together Burnside, Schur, and to some extent Brauer's work and gave representation theoretic proofs of several of these results.  Neumann (1972) used modular representation theory to show that any overgroup is triply transitive (see Huppert–Blackburn XII.10.9).  Chillag (1977) showed that if the overgroup inverted an element of order p−1 in the normalizer, then it was the entire symmetric group. Finally after the CFSG was declared complete, lists of 2-transitive groups became available, a published list being clearly presented in Kantor (1985).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;List of 2-transitive permutation groups&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Such groups divide into 2-sorts by Galois: those with elementary abelian socle and those with simple socle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The ones with simple socle N are almost simple and have degree v as in the following list:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;v ≥ 5, N = Alt(v)&lt;/li&gt;&#xA;&lt;li&gt;v = (q&lt;sup&gt;d&lt;/sup&gt;−1)/(q−1), N = PSL(d,q), q a prime power, (d,q) ≠ (2,2), (2,3), and two representations if d ≥ 3&lt;/li&gt;&#xA;&lt;li&gt;v = q&lt;sup&gt;3&lt;/sup&gt;+1, N = PSU(3,q), q a prime power, q &gt; 2&lt;/li&gt;&#xA;&lt;li&gt;v = q&lt;sup&gt;2&lt;/sup&gt;+1, N = Sz(q), q=2&lt;sup&gt;2e+1&lt;/sup&gt;, e a positive integer&lt;/li&gt;&#xA;&lt;li&gt;v = q&lt;sup&gt;3&lt;/sup&gt;+1, N = Ree(q), q=3&lt;sup&gt;2e+1&lt;/sup&gt;, e a positive integer&lt;/li&gt;&#xA;&lt;li&gt;v = 2&lt;sup&gt;(2n−1)&lt;/sup&gt; ± 2&lt;sup&gt;(n−1)&lt;/sup&gt;, N = Sp(2n,2), n ≥ 3&lt;/li&gt;&#xA;&lt;li&gt;v = 11, N = PSL(2,11), two representations&lt;/li&gt;&#xA;&lt;li&gt;v in { 11, 12, 22, 23, 24 }, N = Mathieu(v), and two representations for v=12&lt;/li&gt;&#xA;&lt;li&gt;v = 12, N = Mathieu(11)&lt;/li&gt;&#xA;&lt;li&gt;v = 15, N = Alt(7), two representations&lt;/li&gt;&#xA;&lt;li&gt;v = 176, N = HS, two representations&lt;/li&gt;&#xA;&lt;li&gt;V = 276, N = Co3&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The ones with abelian socle of order v = p&lt;sup&gt;d&lt;/sup&gt; are all contained in AΓL(d,p).  The subgroup G&lt;sub&gt;0&lt;/sub&gt; of permutations fixing the 0 vector comes from a relatively short list, and is a complement to the regular normal subgroup V of order v.  These should be on the &lt;a href=&quot;http://en.wikipedia.org/wiki/List_of_transitive_finite_linear_groups&quot; rel=&quot;nofollow&quot;&gt;List of transitive finite linear groups&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;G ≤ AΓL(1,v) is solvable, G&lt;sub&gt;0&lt;/sub&gt; is cyclic of order dividing v−1&lt;/li&gt;&#xA;&lt;li&gt;SL(n,q) ⊴ G&lt;sub&gt;0&lt;/sub&gt;, q&lt;sup&gt;n&lt;/sup&gt; = p&lt;sup&gt;d&lt;/sup&gt;&lt;/li&gt;&#xA;&lt;li&gt;Sp(n,q) ⊴ G&lt;sub&gt;0&lt;/sub&gt;, q&lt;sup&gt;n&lt;/sup&gt; = p&lt;sup&gt;d&lt;/sup&gt;&lt;/li&gt;&#xA;&lt;li&gt;G&lt;sub&gt;2&lt;/sub&gt;(q)′ ⊴ G&lt;sub&gt;0&lt;/sub&gt;, q&lt;sup&gt;6&lt;/sup&gt; = p&lt;sup&gt;d&lt;/sup&gt;, q even&lt;/li&gt;&#xA;&lt;li&gt;G&lt;sub&gt;0&lt;/sub&gt; in { Alt(6), Alt(7) }, v = 2&lt;sup&gt;4&lt;/sup&gt;&lt;/li&gt;&#xA;&lt;li&gt;SL(2,q) ⊴ G&lt;sub&gt;0&lt;/sub&gt;, q in {3,5}, and either d=2, p in { 5, 7, 11, 19, 23, 29, or 59 }, or d=4, p=3.&lt;/li&gt;&#xA;&lt;li&gt;five more examples with d=4, p=3 (G&lt;sub&gt;0&lt;/sub&gt; a semi-direct product of the quaternion-type extra-special group of order 32 with one of the five transitive subgroups of Sym(5))&lt;/li&gt;&#xA;&lt;li&gt;SL(2,13) = G&lt;sub&gt;0&lt;/sub&gt;, d=6, p=3.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Sifting the list (not done)&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now one has to check which of these groups actually have degree 2p (or just p).  The second half of the list is easily dealt with: 2p is not a prime power unless it is 4, and p is exactly a prime, so the only possibility is AΓL(1,p) = AGL(1,p), the normalizer of the Sylow p-subgroup.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first half of the list is extremely difficult to analyze as it brings up questions similar to Fermat primes.  However, we can also use that these groups have to contain the normalizer in Alt(2p) of a Sylow p-subgroup.  There are six infinite families (alt, psl, unitary, suzuki, ree, symplectic) and a few sporadics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The alternating group is not a maximal subgroup of itself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case 2.PΓL(d,q) = Aut(PSL(d,q)), one would have (q&lt;sup&gt;d&lt;/sup&gt;−1)/(q−1)=2p.  This number is even, so q must be odd, and d must be even.  If p divides q−1, then p divides |N| at least d−1 times.  If p divides the order of the socle N only once, then p must divide |Out(N)| = (q−1,d)⋅f⋅2.  If p divides f, then p divides q−1 by Fermat's little theorem, however, then p divides |N| at least d−1 times, so d = 2 and 2p = q+1, so q=2p-1, but q is also a p'th power.  Since q&lt;sup&gt;p&lt;/sup&gt;−2p+1 is 0 at p=0 and its derivative in p is positive if q ≥ 9 (and since there are no small counterexamples), this is a contradiction.  If p divides (q−1,d), then d≥p, and so p divides |N| at least p−1 times, which is too many since we can assume p ≥ 3. Hence p divides |N| exactly twice, and [G:N] is coprime to p.  If p ≥ d, then a Sylow p-subgroup of N is abelian and semi-simple, so contained in a torus.  In other words, the order k of q mod p satisfies d/3 &amp;lt; k ≤ d/2.  A permutation matrix effects the orbit swap, so we only need to see if all (p−1)/2 multiplication automorphisms from the normalizer of the Sylow p-subgroup in Alt(p) are contained in G.  Hence we have to look at the normalizer of the torus…&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The unitary and Ree cases are impossible as (q&lt;sup&gt;3&lt;/sup&gt;+1) is divisible by 2, (q+1)/2, and (qq-q+1), so is not twice a prime.  Suzuki has odd degree and sympletic cannot occur, 2 divides v only once, so n=2 is out of bounds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the only sporadic possible is M22, but its automorphism group is only divisible by 11 once, so it cannot occur.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Bibliography&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Brauer (1943) MR&lt;a href=&quot;http://www.ams.org/mathscinet-getitem?mr=8237&quot; rel=&quot;nofollow&quot;&gt;0008237&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;Ito (1960) MR&lt;a href=&quot;http://www.ams.org/mathscinet-getitem?mr=117283&quot; rel=&quot;nofollow&quot;&gt;0117283&lt;/a&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;and MR&lt;a href=&quot;http://www.ams.org/mathscinet-getitem?mr=0124389&quot; rel=&quot;nofollow&quot;&gt;0124389&lt;/a&gt; for larger index&lt;/li&gt;&#xA;&lt;li&gt;and his 1963 MR&lt;a href=&quot;http://www.ams.org/mathscinet-getitem?mr=147535&quot; rel=&quot;nofollow&quot;&gt;0147535&lt;/a&gt; where he raises the bar and begins the attack on his more general conjecture by proving multiple transitivity, which Neumann then simplifies and extends&lt;/li&gt;&#xA;&lt;li&gt;and his 1967 MR&lt;a href=&quot;http://www.ams.org/mathscinet-getitem?mr=224697&quot; rel=&quot;nofollow&quot;&gt;0224697&lt;/a&gt; where he handles Fermat primes&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;Neumann (1972) MR&lt;a href=&quot;http://www.ams.org/mathscinet-getitem?mr=313369&quot; rel=&quot;nofollow&quot;&gt;0313369&lt;/a&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;and its application to Itô's conjecture MR&lt;a href=&quot;http://www.ams.org/mathscinet-getitem?mr=0352227&quot; rel=&quot;nofollow&quot;&gt;0352227&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;Chillag (1977) MR&lt;a href=&quot;http://www.ams.org/mathscinet-getitem?mr=437622&quot; rel=&quot;nofollow&quot;&gt;0437622&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;Kantor (1985) MR&lt;a href=&quot;http://www.ams.org/mathscinet-getitem?mr=773556&quot; rel=&quot;nofollow&quot;&gt;0773556&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="583" LastEditorUserId="583" LastEditorDisplayName="" LastEditDate="2011-01-17T17:40:09.900" LastActivityDate="2011-01-17T17:40:09.900" />
  <row Id="17864" PostTypeId="1" AcceptedAnswerId="17867" CreationDate="2011-01-17T17:23:20.443" Score="10" ViewCount="266" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Is there an odd prime integer $\displaystyle q$ such that $\displaystyle p= \frac{q^3+1}{2}$ is also prime?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A quick search did not find any, nor a pattern in the prime factorization of p.  This is a possible quick solution to the unitary and Ree cases of &lt;a href=&quot;http://math.stackexchange.com/questions/16954/what-are-the-subgroups-of-alt2p-containing-the-normalizer-of-a-sylow-p-subgroup&quot;&gt;ME.16954&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="583" LastEditorUserId="583" LastEditorDisplayName="" LastEditDate="2011-05-12T15:51:38.977" LastActivityDate="2011-05-12T15:51:38.977" Title="German sofa primes: Can both $q$ and $\frac{q^3+1}{2}$ be prime?" Tags="&lt;number-theory&gt;&lt;elementary-number-theory&gt;&lt;prime-numbers&gt;" AnswerCount="1" CommentCount="6" />
  <row Id="17865" PostTypeId="1" AcceptedAnswerId="17868" CreationDate="2011-01-17T17:24:08.740" Score="4" ViewCount="103" Body="&lt;p&gt;I was just trying out some problems, when i couldn't solve this question:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Does there exist $n \times n$ matrices $A$ and $B$ with real entries such that $$ \Bigl(I - (AB-BA)\Bigr)^{n}=0?$$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I really don't know how to proceed for this question. What i did was to find some examples, but that didn't quite work. Any ideas on how one goes about thinking on such problems and how to solve them would be of great help.&lt;/p&gt;&#xA;" LastActivityDate="2011-01-17T17:36:20.303" Title="Matrices with real entries such that $(I -(AB-BA))^{n}=0$." Tags="&lt;linear-algebra&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="17867" PostTypeId="2" ParentId="17864" CreationDate="2011-01-17T17:29:59.530" Score="16" ViewCount="" Body="&lt;p&gt;Isn't this divisble by $\displaystyle \frac{q+1}{2}$?&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2011-01-17T17:29:59.530" CommentCount="1" />
  <row Id="17868" PostTypeId="2" ParentId="17865" CreationDate="2011-01-17T17:36:20.303" Score="8" ViewCount="" Body="&lt;p&gt;Chandru1: Here is a hint to get you started: If the equation $(I-(AB-BA))^n=0$ holds, then for any $v$, if $w=(I-(AB-BA))^{n-1}v$, then $(I-(AB-BA))w=0$, or $(AB-BA)w=w$. If $w=0$ for all $v$, then $(I-(AB-BA))^{n-1}=0$. Proceeding inductively this way, you get that either $AB-BA=I$, or else 1 is an eigenvalue of $AB-BA$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the first option is impossible, because the traces of $AB$ and $BA$ coincide, but the trace of $I$ is $n$ rather than 0.&lt;/p&gt;&#xA;" OwnerUserId="462" LastActivityDate="2011-01-17T17:36:20.303" />
  <row Id="17869" PostTypeId="2" ParentId="17858" CreationDate="2011-01-17T17:58:16.267" Score="3" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Linear_programming&quot; rel=&quot;nofollow&quot;&gt;Linear programming&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are interested in game development you should also take a geometry course too.&lt;/p&gt;&#xA;" OwnerUserId="4215" LastActivityDate="2011-01-17T17:58:16.267" CommentCount="1" />
  <row Id="17870" PostTypeId="1" CreationDate="2011-01-17T18:13:52.857" Score="4" ViewCount="167" Body="&lt;p&gt;Let's suppose there are $n$ real numbers $a_0 &amp;lt; ... &amp;lt; a_n$ uniformly selected from interval [0, 1). If one knows $k$ numbers on consecutive positions $a_i &amp;lt; ... &amp;lt; a_{i+k-1}$ how good is $(k - 1) / (a_{i+k-1} - a_i)$ an estimator for $n$? What other estimators are possible/better?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;NOTE: $n &amp;gt;&amp;gt; k$.&lt;/p&gt;&#xA;" OwnerUserId="958" LastEditorUserId="958" LastEditorDisplayName="" LastEditDate="2011-01-17T21:37:27.983" LastActivityDate="2011-01-17T23:26:58.150" Title="Estimating population size" Tags="&lt;probability&gt;&lt;estimation-theory&gt;" AnswerCount="3" CommentCount="5" FavoriteCount="1" />
  <row Id="17871" PostTypeId="2" ParentId="17870" CreationDate="2011-01-17T18:35:09.213" Score="0" ViewCount="" Body="&lt;p&gt;Let $\hat{n} =  \frac{k-1}{a_{i+k-1}-a_i}$. I think you have to find $\mathbb{E}(n)- \mathbb{E}(\hat{n}) = n- \mathbb{E}(\hat{n})$ (i.e. the bias).  I think the margin of error at $95 \%$ conidence would be of the following form: $$ \mathbb{E}(\hat{n}) \pm \underbrace{l_{0.95}\sqrt{\text{Var}(\hat{n})}}_{\text{margin of error}}$$. &lt;/p&gt;&#xA;" OwnerUserId="2548" LastEditorUserId="2548" LastEditorDisplayName="" LastEditDate="2011-01-17T20:31:27.473" LastActivityDate="2011-01-17T20:31:27.473" CommentCount="1" />
  <row Id="17872" PostTypeId="1" CreationDate="2011-01-17T18:39:33.037" Score="1" ViewCount="113" Body="&lt;p&gt;I had referred to this structure earlier in a &lt;a href=&quot;http://math.stackexchange.com/questions/7219/a-particular-pulling-back-and-lifting-of-metric&quot;&gt;previous question&lt;/a&gt; which went unanswered. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $u$ and $v$ are in $T_pM$ and $n \in (T_pM)^\perp$ then the extrinsic curvature form $K$ be defined as,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$K(u,v) = g(n,\nabla _u V)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(where $V$ is a local extension of the vector $v$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly use say $U$ as a local extension of $u$ for defining $K(v,u)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then one claims that the following equalities hold,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$K(u,v) = K(v,u) = -g(v,\nabla _u N)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(where $N$ is a local extension of $n$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second equality can hold if one has metric compatible connection and $U(g(N,V)) = V(g(N,U)) = 0$. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;I am wondering if $n$ being orthogonal to $u$ and $v$ at $p$ is enough to guarantee the above. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I am unable to understand how the symmetric nature of $K$ can be proven. It doesn't seem to follow just by assuming that the connection is Riemann-Christoffel. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="965" LastActivityDate="2011-01-17T20:02:56.160" Title="Symmetricity of the extrinsic curvature tensor" Tags="&lt;differential-geometry&gt;" AnswerCount="1" />
  <row Id="17873" PostTypeId="1" AcceptedAnswerId="17874" CreationDate="2011-01-17T19:37:43.087" Score="3" ViewCount="88" Body="&lt;p&gt;It is my understanding that one-time pad encryption is the only unbreakable encryption, but suffers from the management of huge keys, and the secure distribution of those keys. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Could one-time pads be used without key distribution to create secure communication?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Example Scenario:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Adam encrypts his message to Bob with a one-time pad (OTP) Adam generated and then sends the ciphertext to Bob.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Bob encrypts Adams ciphertext using a second OTP Bob generated and then sends it back to Adam.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Adam attempts to decrypt the message using the original OTP he generated, then sends the results back to Bob.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Bob decrypts the message using his OTP.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="5950" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2011-01-17T20:09:43.360" LastActivityDate="2011-01-17T20:09:43.360" Title="One-time pad without preshared keys" Tags="&lt;number-theory&gt;&lt;cryptography&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="17874" PostTypeId="2" ParentId="17873" CreationDate="2011-01-17T19:50:39.257" Score="6" ViewCount="" Body="&lt;p&gt;Let $M$ be the plain text, and let Adam's and Bob's one-time pads be $A$ and $B$. If I have understood you correctly, what you suggest is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Adam sends Bob $M \oplus A$&lt;br&gt;&#xA;Bob sends Adam $M \oplus A \oplus B$&lt;br&gt;&#xA;Adam sends Bob $M \oplus B$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This doesn't work, because an eavesdropper can recover the plain text by  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$M = (M \oplus A) \oplus (M \oplus A \oplus B) \oplus (M \oplus B)$&lt;/p&gt;&#xA;" OwnerUserId="1508" LastActivityDate="2011-01-17T19:50:39.257" CommentCount="1" />
  <row Id="17875" PostTypeId="2" ParentId="17872" CreationDate="2011-01-17T19:55:24.797" Score="3" ViewCount="" Body="&lt;p&gt;This works only if $\nabla$ is the Levi-Civita connection of the Riemannian structure $g$. The Levi-Civita connection has the property that it is torsion-free, in addition to being compatible with the metric. Torsion-freeness means $\nabla_U V-\nabla_V U=[U,V]$, and $[U,V]$ is tangential to $M$ and thus orthogonal to $n$, which proves the symmetry of $K$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: If $\nabla$ is not the Levi-Civita connection, your $K$ will not even be a tensor in general (that is, $g(n,\nabla_u V)$ will depend on the choice of the &quot;local extension&quot; $V$ of $v$).&lt;/p&gt;&#xA;" OwnerUserId="1609" LastEditorUserId="1609" LastEditorDisplayName="" LastEditDate="2011-01-17T20:02:56.160" LastActivityDate="2011-01-17T20:02:56.160" CommentCount="3" />
  <row Id="17876" PostTypeId="1" AcceptedAnswerId="17878" CreationDate="2011-01-17T20:09:07.150" Score="1" ViewCount="84" Body="&lt;p&gt;For a problem such as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ \sum_{n=1}^\infty \frac{lnn}{n^3}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do you tell if it is decreasing or increasing? I know that you can compute the derivative and test if it is less than or greater than zero, but the way my solutions manual does it, I'm puzzled by.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The derivative is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{1-3lnx}{x^4} &amp;lt; 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solving for x gives $x &amp;gt; e^\frac{1}{3}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm confused as to how that shows that the function is decreasing.&lt;/p&gt;&#xA;" OwnerUserId="3514" LastActivityDate="2011-01-17T20:14:04.323" Title="How do you tell if a series is decreasing?" Tags="&lt;calculus&gt;&lt;homework&gt;&lt;sequences-and-series&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="17877" PostTypeId="1" AcceptedAnswerId="17880" CreationDate="2011-01-17T20:12:01.000" Score="4" ViewCount="102" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;If &#xA;  $$A = \frac{1}{2} \times \begin{pmatrix}&#xD;&#xA; -1&amp;amp; -\sqrt3  &amp;amp; 0 \\ &#xD;&#xA; -\sqrt3&amp;amp; 1 &amp;amp;0 \\ &#xD;&#xA; 0 &amp;amp; 0 &amp;amp; 0&#xD;&#xA;\end{pmatrix}  \text{ and } E = \frac{1}{2} \times \begin{pmatrix}&#xD;&#xA; 1&amp;amp; \sqrt3  &amp;amp; 0 \\ &#xD;&#xA; -\sqrt3&amp;amp; 1 &amp;amp;0 \\ &#xD;&#xA; 0 &amp;amp; 0 &amp;amp; 1&#xD;&#xA;\end{pmatrix}$$&#xA;  then show that $E^{-1}AE$ is diagonal matrix.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Can we prove this without actually computing the the inverses and the product? The actual problem is consist of four options of matrices and I think it's a bit tedious for objective type to do the actual computation, so I am more interested in using some property for the same. I know about $E^{T}AE$ depend on whether A is symmetric or skew-symmetric but is there anything like this for inverses? &lt;/p&gt;&#xA;" OwnerUserId="2109" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-17T22:27:58.450" LastActivityDate="2011-01-18T14:05:25.253" Title="Suggest a tricky procedure" Tags="&lt;linear-algebra&gt;&lt;matrices&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" />
  <row Id="17878" PostTypeId="2" ParentId="17876" CreationDate="2011-01-17T20:14:04.323" Score="2" ViewCount="" Body="&lt;p&gt;If a differentiable function has a negative derivative on an interval, then it is decreasing on that interval, a consequence of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Mean_value_theorem&quot; rel=&quot;nofollow&quot;&gt;mean value theorem&lt;/a&gt;.  What you stated would therefore imply that the function $x\mapsto \frac{\ln x}{x^3}$ is decreasing on $(\sqrt[3]{e},\infty)$, so in particular it is decreasing on the integers greater than $1$.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2011-01-17T20:14:04.323" />
  <row Id="17879" PostTypeId="1" AcceptedAnswerId="17896" CreationDate="2011-01-17T20:14:56.230" Score="0" ViewCount="79" Body="&lt;p&gt;There is a book about rational points on elliptic curves. What about algebraic points?&lt;/p&gt;&#xA;" OwnerUserId="4290" LastActivityDate="2011-01-17T21:49:33.100" Title="Algebraic points on an elliptic curve" Tags="&lt;elliptic-curves&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="17880" PostTypeId="2" ParentId="17877" CreationDate="2011-01-17T20:21:15.580" Score="4" ViewCount="" Body="&lt;p&gt;$E^{-1} A E$ will be diagonal if the columns of $E$ are eigenvectors of $A$ (since then $AE=ED$).&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2011-01-17T20:21:15.580" CommentCount="10" />
  <row Id="17881" PostTypeId="1" CreationDate="2011-01-17T20:23:04.987" Score="3" ViewCount="74" Body="&lt;p&gt;What exactly does it mean by &quot;half a unit in the nth place&quot;? I am reading numerical analysis and this term is popping out very frequent in the first chapter itself,my first encounter with this is while reading the rounding-off error rules.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From few solved examples I can understand (or probably guess) it mean $5$ but what I still don't understand why such a abstruse term ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks,&lt;/p&gt;&#xA;" OwnerUserId="2109" LastActivityDate="2011-01-17T20:49:28.123" Title="Numerical analysis:: meaning of &quot;half a unit in the nth place&quot;" Tags="&lt;numerical-methods&gt;" AnswerCount="1" CommentCount="0" />
  <row Id="17882" PostTypeId="2" ParentId="17870" CreationDate="2011-01-17T20:30:04.700" Score="2" ViewCount="" Body="&lt;p&gt;Are you constrained to use the estimator that you have indicated? If not, another choice for an estimator would be to form the joint pdf of $f(a_i,\dots,a_{i+k-1} \mid n)$ and use the value of $n$ that maximizes this joint pdf as your population estimator.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The joint pdf of the $k$ consecutive values is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\begin{eqnarray}&#xD;&#xA;f(a_i,\dots,a_{i+k-1}) = k!{n \choose k} {{n-k} \choose {i-1}} a_i^{(i-1)} (1-a_{i+k-1})^{(n-i+1)-k}&#xD;&#xA;\end{eqnarray}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This can be seen as follows - Let us say we have $n$ unordered random variables. We choose $k$ of them with order (in $k! {n \choose k}$ ways). Of the remaining $(n-k)$ variables, we choose $(i-1)$ (in ${{n-k} \choose {i-1}}$ ways) and designate them to be the first $(i-1)$ ordered variables in some order. They all are less than $a_i$ and the probability of this happening is $a_i^{i-1}$. The remaining variables we designate to be larger than the $(i+k-1)$th variable which happens with probability $(1-a_{i+k-1})^{(n-i+1-k)}$. It only remains to compute the probability that the $j$th variable takes the value $a_j$ for $i \leq j \leq i+k-1$. But this is just $F(a_j)d(a_j) = d(a_j)$ since the variables are uniformly selected from $[0,1)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maximizing this as a function of $n$ is a potentially messy business. To get an idea of how good this approach is, I used Stirling's approximation for the factorials and further assumed that $n$ is much larger than $i,k$. Then, the maximizing value is approximately&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{eqnarray}&#xD;&#xA;\tilde{n} = \frac{i+k-1}{\log \frac{1}{1-a_{i+k-1}}}&#xD;&#xA;\end{eqnarray}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Simulations seem to suggest that this estimator is better for some parameter ranges (i.e., when $n$ is much larger than $i,k$ for ex.) while in other regimes, the estimator in the question is better (when $k,i$ can't be ignored in comparison with $n$). Even in the latter regime, it might be possible to optimize the $n$ value that maximizes the joint pdf above more carefully (without using asymptotic stuff like Stirling's) to get a better estimator. &lt;strong&gt;Edit&lt;/strong&gt;: I checked in simulation that numerically maximizing the joint pdf without resorting to Stirling's approximation does give a uniformly better estimator.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another thing I noted from the simulation is that if $k=2$, the estimator $\frac{1}{a_{i+1}-a_i}$ performs poorly. To see this, note that the distribution of $D = a_{i+1}-a_i$ is independent of $i$ and is equal to $f_D(d) = n_t(1-d)^{n_t-1}$ where $n_t$ denotes the true value of $N$. This makes the population estimator $\tilde{N}$ have the density function &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\begin{equation}&#xD;&#xA;f_{\tilde{N}}(\tilde{n}) = \frac{n_t}{\tilde{n}^2} \left(1-\frac{1}{\tilde{n}}\right)^{n_t-1}&#xD;&#xA;\end{equation}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This estimator has an unbounded expectation and thus an unbounded bias. But this is a moot point if for some reason you are constrained to use the estimator you had indicated in the question.&lt;/p&gt;&#xA;" OwnerUserId="618" LastEditorUserId="618" LastEditorDisplayName="" LastEditDate="2011-01-17T23:26:58.150" LastActivityDate="2011-01-17T23:26:58.150" CommentCount="6" />
  <row Id="17883" PostTypeId="1" AcceptedAnswerId="17889" CreationDate="2011-01-17T20:33:23.830" Score="2" ViewCount="106" Body="&lt;p&gt;I'm having some difficulty with the following question:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Let $(f_{n}(x))$ be a uniformly convergent functions sequence in $(a,b)$ (where b can be $\infty$) such that $(f_{n}(x)) \to f(x)$. Suppose that for almost all $n$, the limit $\displaystyle \lim_{x\to b^{-}}f_{n}(x)$ exists.  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Prove that $\displaystyle \lim_{x\to b^{-}}f(x)$ exists and $\displaystyle \lim_{n\to \infty}\left (\lim_{x\to b^{-}}f_{n}(x)\right )=\lim_{x\to b^{-}}f(x)$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I tried a method that I see often in theorems regarding uniform convergence:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, let $\displaystyle \left ( a_{n}=\lim_{x\to b^{-}}f_{n}(x) \right )_{n\geq N}$. Such an $N$ exists.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We want to show that $\forall \epsilon,\ \exists \delta,\ \forall x \in (b-\delta, b),\ |f(x)-L|\lt \epsilon$.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can write $|f(x)-L|=|f(x)-f_{n}(x)+f_{n}(x)-a_{n}+a_{n}-L|$, then:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$|f(x)-L|\lt\overbrace{|f(x)-f_{n}(x)|}^{\lt \epsilon / 3}+\overbrace{|f_{n}(x)-a_{n}|}^{\lt \epsilon \lt 3}+\overbrace{|a_{n}-L|}^{\text{?}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I don't know how to deal with $|a_{n}-L|$. Had $a_{n}\to L$, (which is in fact the second part of the question) then I could make sure it's no more than $\epsilon / 3$, but I don't know if $a_{n}$ converges yet and if it converges to $L$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: couldn't think of a better title, if anyone does, feel free to modify it.&lt;/p&gt;&#xA;" OwnerUserId="3237" LastEditorUserId="2548" LastEditorDisplayName="" LastEditDate="2011-01-17T20:44:36.870" LastActivityDate="2011-01-17T20:56:43.787" Title="One-sided limits of a uniformly convergent function sequence and its limit function" Tags="&lt;real-analysis&gt;&lt;homework&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="17884" PostTypeId="2" ParentId="17856" CreationDate="2011-01-17T20:34:25.927" Score="3" ViewCount="" Body="&lt;p&gt;Here's a bit of an elaboration on my earlier comment.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can see $SO_3$ as the quotient of a unit $3$-ball by the antipodal map on the boundary. This is realized explicitly by the exponential map for $SO_3$.  The tangent space to the identity of $SO_3$ is all $3\times 3$ matrices $H$ such that $H^T=-H$.  Up to conjugation by an element of $SO_3$, such a matrix looks like &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left(\begin{matrix} 0 &amp;amp; \theta &amp;amp; 0 \\ -\theta &amp;amp; 0 &amp;amp; 0 \\ 0 &amp;amp; 0 &amp;amp; 0 \end{matrix}\right)$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;which under exponentiation is converted into rotation by $\theta$ about the $z$-axis.  Every element of $SO_3$ is rotation by some angle about a fixed axis since the characteristic polynomial has $1$ as a root -- so this is a rather explicit description of how the exponential map is onto.  If you choose the length of a matrix $[a_{i j}]$ to be $\sqrt{ \sum_{i j} a_{i j}^2}$, then technically $SO_3$ is the quotient of the ball of radius $\pi/\sqrt{2}$, with antipodal points on the boundary identified.  But that's small potatoes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The 3-ball with antipodal points identified, as you've seen has fundamental group $\Bbb Z_2$.  In particular you can think of the 3-ball with antipodal points identified as being built up from the 1-ball with antipodal points identified, union a 2-cell (this gives a CW-decomposition of the 2-ball with antipodal points identified) then union a 3-cell.  So the fundamental group all &quot;happens&quot; in the 2-ball with antipodal points identified.  The 1-skeleton generates the fundamental group and in our model (under exponentiation) corresponds to one full rotation by $2\pi$ about any chosen axis.  The relation corresponds to the 2-cell attachment and you can &lt;em&gt;literally&lt;/em&gt; interpret it as the plate trick in a &quot;sufficiently nice&quot; model.   Let me explain. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The trouble with the plate trick is it uses a human arm.  Human arms you can think of as carrying a core underlying piece of information, a map $f : [0,1] \to SO_3$. $f(t)$ represents the orientation in space of the part of the arm distance $t$ from the shoulder, so $0$ represents the shoulder distance, and $1$ the palm of the hand.  There are technical issues -- if you choose the framing corresponding to the positions of the appropriate bone in the arm, you don't get a continuous map, just a piecewise-constant map -- but if you consider the orientations of the ligaments between the bones it becomes a continuous map.  So the trouble with the plate trick is the map $f$ isn't the only constraint on the arm. The arm is embedded in space.  Bones are rigid, there's only so much ligaments are capable of doing.  So you could be rightly concerned that the plate trick is a demonstration of a peculiarity of anatomy, rather than something about $SO_3$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alright, but formally what's going on with the plate trick?  As mentioned, a configuration of the arm we're representing by $f$, which is an element of the path space $PSO_3$ of continuous paths in $SO_3$ with $f(0) = I \in SO_3$ fixed. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is the path-loop fibration:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Omega SO_3 \to PSO_3 \to SO_3$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and $PSO_3$ is contractible.  The plate trick &quot;is&quot; the induced map $\pi_1 SO_3 \to \pi_0 \Omega SO_3$ from the homotopy long exact sequence of the above fibration, which is guaranteed to be an isomorphism by the path-loop fibration.  Specifically, elements of $\pi_1 SO_3$ you are thinking of as motions of your hand.  The path-loop fibration is telling you that you can realize the motion as a motion of your arm (or at least, a path in $PSO_3$ but the fact that people can &lt;em&gt;do&lt;/em&gt; the plate trick says you can actually realize it as a motion of your arm).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is of course an element that's a bit misleading in all this.  The arm isn't complicated enough to allow for you to lift the product of two generators of $\pi_1 SO_3$ and witness the null homotopy -- since the arm allows a fixed complexity, the null-homotopy happens as you lift the concatenation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But you can see from the rather explicit exponential map above how the null homotopy actually works, as it's sitting in the image of the 2-ball with antipodal points identified. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How is that? &lt;/p&gt;&#xA;" OwnerUserId="642" LastActivityDate="2011-01-17T20:34:25.927" CommentCount="1" />
  <row Id="17886" PostTypeId="2" ParentId="17881" CreationDate="2011-01-17T20:49:28.123" Score="4" ViewCount="" Body="&lt;p&gt;You need to think in terms of significant figures. The nth place is the nth significant figure. Consider, for example, the number 123.45 (base 10). A unit in the second place is 10, in the 3rd place is 1, in the 4th place is 0.1, etc. So half a unit in the second place is 5, in the 3rd place is 0.5, in the 4th place is 0.05, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, in the context in which the book uses it I expect that n is the number of digits used to represent a number. So if all numbers are represented with 5 significant decimal digits, 123.45 really represents the range 123.445 to 123.455 (probably half-open). Then the maximum absolute error in that representation is half a unit in the nth place.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does that make sense?&lt;/p&gt;&#xA;" OwnerUserId="5676" LastActivityDate="2011-01-17T20:49:28.123" />
  <row Id="17888" PostTypeId="2" ParentId="17870" CreationDate="2011-01-17T20:50:41.217" Score="3" ViewCount="" Body="&lt;p&gt;As in the book &lt;a href=&quot;http://books.google.co.il/books?id=dVP-RTea5wcC&amp;amp;printsec=frontcover&amp;amp;dq=%22A+First+Course+in+Order+Statistics%22&amp;amp;source=bl&amp;amp;ots=G4xgpvAAA4&amp;amp;sig=OP2zEM--iAmYJoZPY4e8qF4T6o0&amp;amp;hl=iw&amp;amp;ei=_6o0TZvbHNOSjAeiiMG9Cg&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=6&amp;amp;ved=0CCwQ6AEwBQ#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;A First Course in Order Statistics&lt;/a&gt; (see, in particular, Section 2.5), let us denote by $W_{i,j:n}$ the spacing $W_{i,j:n} = U_{j:n} - U_{i:n}$, $1 \leq i &amp;lt; j \leq n$, where $U_{1:n} &amp;lt; \cdots &amp;lt; U_{n:n}$ are $n$ order statistics from a ${\rm uniform}(0,1)$ distribution. By equation (2.5.21) of that book, the density function of $W_{i,j:n}$ is given by&#xA;$$&#xD;&#xA;f_{W_{i,j:n} } (w) = \frac{{n!}}{{(j - i - 1)!(n - j + i)!}}w^{j - i - 1} (1 - w)^{n - j + i} ,\;\; 0 &amp;lt; w &amp;lt; 1,&#xD;&#xA;$$&#xA;so that $W_{i,j:n}$ has a ${\rm Beta}(j - i,n - j + i + 1)$ distribution (which depends only on $j-i$ and not on $i$ and $j$ individually). Having the distribution of $W_{i,j:n}$, you can check how good is $\frac{{j - i}}{{U_{j:n}  - U_{i:n} }}$ an estimator for $n$, which is what you asked (in different notation).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Using the density function of $W_{i,j:n}$, one can easily calculate the bias (cf. Trevor's answer). My calculation shows that &#xA;$$&#xD;&#xA;{\rm E}\bigg[\frac{{j - i}}{{U_{j:n}  - U_{i:n} }}\bigg] = \frac{{j - i}}{{j - i - 1}}n,&#xD;&#xA;$$&#xA;hence the estimator gets better and better as $j-i$ grows. (The expectation is infinite when $j-i=1$; cf. Dinesh's answer.)&lt;/p&gt;&#xA;" OwnerUserId="2810" LastEditorUserId="2810" LastEditorDisplayName="" LastEditDate="2011-01-17T22:54:31.390" LastActivityDate="2011-01-17T22:54:31.390" CommentCount="2" />
  <row Id="17889" PostTypeId="2" ParentId="17883" CreationDate="2011-01-17T20:56:43.787" Score="2" ViewCount="" Body="&lt;p&gt;Right, the trouble is that you don't have an expression up front for $L$, so you're going to have a hard time showing that $f(x)$ converges to it (as $x \to b^-$).  A good approach in this case is to use completeness of $\mathbb{R}$; then all you have to do is show that $f(x)$ is Cauchy as $x \to b^-$, and it follows that it converges to something, which you can then call $L$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So try to show the following: for all $\epsilon &amp;gt; 0$, there exists $\delta &amp;gt; 0$ such that for all $x,y \in (b, b+\delta)$ we have $|f(x) - f(y)| &amp;lt; \epsilon$.  By completeness, it will follow that $\lim_{x \to b^-} f(x)$ exists, and you can call that number $L$.  (If you think of completeness in terms of Cauchy sequences, then note that the statement implies that for any sequence $x_n \downarrow b$, we have that $f(x_n)$ is a Cauchy sequence, hence convergent to some number $L$, and that $L$ is the same for all such sequences $x_n$.)  It should not be hard then to show that $a_n \to L$ as well.&lt;/p&gt;&#xA;" OwnerUserId="822" LastActivityDate="2011-01-17T20:56:43.787" CommentCount="4" />
  <row Id="17890" PostTypeId="2" ParentId="17830" CreationDate="2011-01-17T21:13:19.130" Score="0" ViewCount="" Body="&lt;p&gt;Using natural deduction notation &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Modus ponens:&#xA;$$\backslash\!\!\!\!\!{A}$$ $$\overline{B}$$ $$\overline{A\to B}$$&#xA;If from the assumption $A$ can deduce $B$, then $A\to B$ is deducible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Law of Conjunction:&#xA;$$A\wedge B$$ $$\overline{\qquad A \qquad}$$ and &#xA;$$A\wedge B$$ $$\overline{\qquad B \qquad}$$ and &#xA;If $A\wedge B$ holds, then both $A$ and $B$ also holds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To prove the statement using ND, start by breaking up the first implication - and you soon figure out what to do. (If there is any doubt just ask.)&lt;/p&gt;&#xA;" OwnerUserId="1154" LastActivityDate="2011-01-17T21:13:19.130" />
  <row Id="17891" PostTypeId="1" CreationDate="2011-01-17T21:17:10.030" Score="8" ViewCount="148" Body="&lt;p&gt;In reference to Stanley's Enumerative Combinatorics Vol. 2: right after he has defined Kostka numbers (section 7.10), he mentions that it is unlikely that a general formula for $K_{\lambda\mu}$ exists, where $K_{\lambda\mu}$ is the number of semistandard Young tableaux of shape $\lambda$ and type $\mu$ with $\lambda\vdash n$ and $\mu$ a weak composition of $n$. Why? In particular, is this an expression of something rigorous, and if so, what?&lt;/p&gt;&#xA;" OwnerUserId="5954" LastEditorUserId="7676" LastEditorDisplayName="" LastEditDate="2011-03-25T05:59:03.700" LastActivityDate="2011-04-25T16:30:52.427" Title="Why is a general formula for Kostka numbers &quot;unlikely&quot; to exist?" Tags="&lt;combinatorics&gt;" AnswerCount="1" />
  <row Id="17892" PostTypeId="1" CreationDate="2011-01-17T21:18:55.453" Score="1" ViewCount="90" Body="&lt;p&gt;You can use the proof of the Schwarz reflection principle to show that if $D$ is a domain in the complex plane, where $D^+$ denotes the subset of D above the real line, and $D^-$ the subset below, if $f:D\to C$ which is analytic on $D^+$ and $D^-$, and continuous on $D$, then $f$ is analytic on $D$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to generalize this as follows: Let $D$ be a domain in the complex plane, and $D&amp;#39;$ an open dense subset of $D$.  If $f$ is a function analytic on $D&amp;#39;$, and continuous on $D$, then $f$ is analytic on $D$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The proof of the Schwarz reflection principle uses integration along arbitrary triangles, and since the intersection of a triangle with the real line is really simple, you can &quot;cut the real line out&quot; so to speak when you are taking the integral.  But if all you know is that $f$ is analytic on an open dense set what can you do?&lt;/p&gt;&#xA;" OwnerUserId="5952" LastEditorUserId="5363" LastEditorDisplayName="" LastEditDate="2011-02-16T23:26:25.473" LastActivityDate="2011-02-16T23:26:25.473" Title="Question about generalizing Schwarz Reflection" Tags="&lt;complex-analysis&gt;&lt;reflection&gt;&lt;analytic&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="17893" PostTypeId="2" ParentId="17843" CreationDate="2011-01-17T21:21:40.257" Score="1" ViewCount="" Body="&lt;p&gt;Consider a cone with caconical equation $x^2/a^2+y^2/b^2-z^2/c^2=0$. Generatrix satisfy the cone equation. Point $x=y=z=0$ satisfy the cone equation and thus satisfy the generatrix equation. Point $x=a y=b z=c\sqrt 2 $ satisfy both cone and generatrix equations. Thus, the direction vector for a certain generatrix will be $(a,b,c\sqrt2)$. Similarly, for any point $x=p, y=q, z=c/ab\sqrt{p^2 b^2+q^2 a^2}$ on a cone parametrized by $(p,q)$ there will be a generatrix with equation $px+qy+zc/ab\sqrt{p^2 b^2+q^2 a^2}=0$, or better, with equation $abpx+abqy+cz\sqrt{p^2 b^2+q^2 a^2}=0$. This is the canonical equation of a desired generatrix.&lt;/p&gt;&#xA;" OwnerUserId="3806" LastEditorUserId="3806" LastEditorDisplayName="" LastEditDate="2011-01-17T21:40:10.383" LastActivityDate="2011-01-17T21:40:10.383" />
  <row Id="17894" PostTypeId="1" CreationDate="2011-01-17T21:30:25.427" Score="0" ViewCount="224" Body="&lt;p&gt;&lt;strong&gt;Axis Angle to quaternion and quaternion to Axis angle question&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Greetings All (matlab / octave code)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Link to text file in case formatting gets messed up&#xA;&lt;a href=&quot;http://db.tt/nVv8Ivj&quot; rel=&quot;nofollow&quot;&gt;http://db.tt/nVv8Ivj&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I created two functions one to convert axis angle to quaternion&#xA;and another one to convert quaternion to axis angle as far as I can tell the formulas &#xA;are correct the question I have is when I create a quaternion from&#xA;the axis angle format example: x=.3 y=.1 z=.6 and the angle is 45 degrees I get&#xA;(0.962730w   0.119633i   0.039878j   0.239266k) which is correct&#xA;when I check it.  But when I plug this quaternion number back into &#xA;my function I get back angle is 31.384 degrees x=.44233, y=.14744 z=.88465&#xA;I was expecting to get back &#xA;x=.3 y=.1 z=.6 and the angle of 45 degrees the same thing I put in&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anyone know what I'm doing wrong? The reason I'm doing it this way&#xA;is to do rotations on values in arrays and not get gimbal lock. I also plan to plot it and&#xA;animate it, and the axis angle format (x,y,z angle) is easier to understand when it comes to plotting&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Iv'e included the code below with a link above to the text file since the&#xA;formating sometimes gets messed up.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;`%Vector taken in [i j k], angle in degrees&#xA;%quaterions given in Format [w xi yj zk] or [w x y z] &#xA;%&#xA;%Example of using function:  ccc=quat([.3 .1 .6],45)  &#xA;%Answer should be -&amp;gt;&#xA;%0.962730w   0.119633i   0.039878j   0.239266k&#xA;&#xA;function Q= quat(vect,theta)&#xA; Q = zeros(1,4);&#xA; xyz_and_angle=[vect theta]    %displays inputed data and theta angle before deg to rad conversion&#xA; %thetaconvert = (theta*pi/180)./norm(theta*pi/180) %convert deg to rad&#xA; thetaconvert=theta*pi/180;&#xA; px=vect(1,1);py=vect(1,2);pz=vect(1,3); &#xA; Q(1,1) = cos(thetaconvert/2); %w value&#xA; Q(1,2)= px*sin(thetaconvert/2); % x value&#xA; Q(1,3)= py*sin(thetaconvert/2); % y value&#xA; Q(1,4)= pz*sin(thetaconvert/2); % z value&#xA;&#xA; q1w=Q(1,1);%qw value&#xA; q1x=Q(1,2);%qx value&#xA; q1y=Q(1,3);%qy value&#xA; q1z=Q(1,4);%qz value&#xA;&#xA; %Normalize Quaternion due to floating point errors&#xA;      qnorm=sqrt(q1w^2+q1x^2+q1y^2+q1z^2) &#xA;      Q=Q./qnorm %normalize Q array&#xA;&#xA;&#xA;%quaterions taken in Format [w xi yj zk] or [w x y z] &#xA;    %example ccc=[0.962730   0.119633   0.039878   0.239266]&#xA;    %ccc2=quattoaxis(ccc)&#xA;    %Answer comes back as angle is 31.384 degrees x=.44233, y=.14744 z=.88465&#xA;    %not sure if the answer is right I thought it would be&#xA;    %x=.3 y=.1 z=.6 and the angle of 45 degrees the same thing I put in above&#xA;&#xA;function Q= quattoaxis(quat)&#xA; Q = zeros(1,4);&#xA; q1w=quat(1,1);q1x=quat(1,2);q1y=quat(1,3);q1z=quat(1,4);&#xA; %Get angle from quat&#xA;&#xA; %convert Quaternion to Axis&#xA; Q(1,1) = 2*acos(q1w) *180/pi %w / angle, I  also included radians to degrees&#xA; Q(1,2)= q1x/sqrt(1-q1w^2)  %x&#xA; Q(1,3)= q1y/sqrt(1-q1w^2) %y&#xA; Q(1,4)= q1z/sqrt(1-q1w^2)   %z&#xA;&#xA; axisangle=Q(1,1) %w / angle, I  also included radians to degrees&#xA; axisx=Q(1,2) %x&#xA; axisy=Q(1,3) %y&#xA; axisz=Q(1,4)   %z&#xA;&#xA;&#xA;&#xA;%Normalize axis angle  due to floating point errors&#xA;    qnorm=sqrt(axisx^2+axisy^2+axisz^2) &#xA;    axisx=Q(1,2)/qnorm; %x&#xA; axisy=Q(1,3)/qnorm; %y&#xA; axisz=Q(1,4)/qnorm;  %z&#xA;&#xA;    Q=[axisangle axisx axisy axisz]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="5957" LastEditorUserId="5957" LastEditorDisplayName="" LastEditDate="2011-03-22T02:23:57.440" LastActivityDate="2011-05-21T02:59:47.513" Title="Axis Angle to quaternion and quaternion to Axis angle question " Tags="&lt;quaternions&gt;" AnswerCount="1" />
  <row Id="17895" PostTypeId="2" ParentId="17842" CreationDate="2011-01-17T21:33:46.023" Score="0" ViewCount="" Body="&lt;p&gt;This is an open question in general, I think even asymptotically speaking. The question &quot;is the number of ways zero or not&quot; is equivalent to the linear coding question, since the distance of a linear code is equal to the minimal number of linearly dependent columns in the parity check matrix. See the Wikipedia entry for &lt;a href=&quot;http://en.wikipedia.org/wiki/Linear_code#Properties&quot; rel=&quot;nofollow&quot;&gt;linear codes&lt;/a&gt; for more.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2011-01-17T21:33:46.023" CommentCount="2" />
  <row Id="17896" PostTypeId="2" ParentId="17879" CreationDate="2011-01-17T21:37:00.990" Score="5" ViewCount="" Body="&lt;p&gt;It is quite common to study elliptic curves over $K$ or to ask about all $K$-points on an elliptic curve (or on an abelian variety), where $K$ is a subfield of the algebraic numbers; usually a number field (finite extension of $\mathbb{Q}$), but sometimes somewhat bigger (e.g., the extension given by adding all $p^n$th roots of unity for some fixed prime $p$). It is also common to work over finite fields, or over general global fields or over local fields. Most of the &lt;em&gt;basic&lt;/em&gt; theory over number fields is very similar to that over $\mathbb{Q}$; e.g., the Mordell-Weil Theorem holds for any number field (the group of $K$-points on an elliptic curve defined over $K$ is finitely generated for any number field $K$). It's known that for any number field $K$, there is a finite list of possible torsion subgroups of the group of $K$-defined points of any elliptic curve defined over $K$ (that is, there is a version of Mazur's Theorem for each number field). There is also the version of integral points ($\mathcal{O}_K$-points, where $\mathcal{O}_K$ is the group of integers of $K$), etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, most of Silverman's books mentioned by Qiaochu work over arbitrary fields, restricting on occasion to local fields or to global fields. Most of the theory is quite general, and yes, there is interest in understanding elliptic curves over number fields. &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-17T21:49:33.100" LastActivityDate="2011-01-17T21:49:33.100" />
  <row Id="17897" PostTypeId="1" AcceptedAnswerId="17899" CreationDate="2011-01-17T21:51:34.793" Score="1" ViewCount="132" Body="&lt;p&gt;The set of rational numbers $\mathbb{Q}$ is an unbounded subset of $\mathbb{R}$ with Lebesgue outer measure zero. In addition, $\mathbb{R}$ is an unbounded subset of itself with Lebesgue outer measure $+\infty$. Therefore the following question came to my mind: is there an unbounded subset of $\mathbb{R}$ with positive Lebesgue outer measure?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there is, can you give me an example?&lt;/p&gt;&#xA;" OwnerUserId="5956" LastActivityDate="2011-01-17T22:42:13.493" Title="Unbounded subset of $\mathbb{R}$ with positive Lebesgue outer measure" Tags="&lt;measure-theory&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="17898" PostTypeId="1" AcceptedAnswerId="17901" CreationDate="2011-01-17T21:55:07.200" Score="5" ViewCount="110" Body="&lt;p&gt;When one has an expression in terms of i, one can send i to -i and, if the expression remains unchanged, one can conclude that the expression is, in fact, real. Analogous statements hold for expressions involving radicals. Why is this?&#xA;One admittedly trivial example is the expression&#xA;$\frac{1}{x-i}+\frac{1}{x+i}$.&lt;/p&gt;&#xA;" OwnerUserId="5959" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-17T22:54:31.173" LastActivityDate="2011-01-17T22:54:31.173" Title="Invariant under transformation $i\longmapsto -i$ implies real?" Tags="&lt;algebra&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="17899" PostTypeId="2" ParentId="17897" CreationDate="2011-01-17T21:57:16.140" Score="4" ViewCount="" Body="&lt;p&gt;I guess you mean with positive and finite outer measure.  An easy example would be something like $[0,1]\cup\mathbb{Q}$.  But perhaps you also want to have nonzero measure outside of each bounded interval?  In that case, consider $[0,1/2]\cup[1,1+1/4]\cup[2,2+1/8]\cup[3,3+1/16]\cup\cdots$.  If you want the set to have positive measure in each subinterval of $\mathbb{R}$, you could let $x_1,x_2,x_3,\ldots$ be a dense sequence (like the rationals) and take a union of open intervals $I_n$ such that $I_n$ contains $x_n$ and has length $1/2^n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, it is often useful to keep in mind that every set of finite measure is &quot;nearly bounded&quot;.  That is, if $m(E)&amp;lt;\infty$ and $\epsilon&amp;gt;0$, then there is an $M\gt0$ such that $m(E\setminus[-M,M])&amp;lt;\epsilon$.  One way to see this is by proving that the monotone sequence $(m(E\cap[-n,n]))_{n=1}^\infty$ converges to $m(E)$.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2011-01-17T22:42:13.493" LastActivityDate="2011-01-17T22:42:13.493" CommentCount="3" />
  <row Id="17900" PostTypeId="2" ParentId="17898" CreationDate="2011-01-17T21:58:49.293" Score="2" ViewCount="" Body="&lt;p&gt;If $x+iy = x-iy$ then $y=0$ (extra characters).&lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2011-01-17T21:58:49.293" />
  <row Id="17901" PostTypeId="2" ParentId="17898" CreationDate="2011-01-17T22:01:30.613" Score="8" ViewCount="" Body="&lt;p&gt;This is a basic fact in &lt;a href=&quot;http://en.wikipedia.org/wiki/Galois_theory&quot; rel=&quot;nofollow&quot;&gt;Galois theory&lt;/a&gt;, namely that for Galois extensions $F/K$ the fixed field of the Galois group $G = \text{Gal}(F/K)$ is precisely $K$.  The quadratic extension $\mathbb{Q}(i)/\mathbb{Q}$ is an example of such an extension, as is the quadratic extension $\mathbb{C}/\mathbb{R}$.  (To apply this to your example one can do a few things: either one regards the expression as a power series and applies the above to each term, or one regards the expression as a family of elements of $\mathbb{C}$ and applies the above separately to each of them.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is instructive to see how this fails when the extension is not Galois.  A simple example is the extension $\mathbb{Q}(\sqrt[3]{2})/\mathbb{Q}$.  This extension has no nontrivial automorphisms because the other two roots of $x^3 - 2$ are complex, so the fixed field of the &quot;Galois group&quot; is all of $\mathbb{Q}(\sqrt[3]{2})$.  To get an analogous statement to the one above one has to pass to the &lt;a href=&quot;http://en.wikipedia.org/wiki/Splitting_field&quot; rel=&quot;nofollow&quot;&gt;splitting field&lt;/a&gt; $\mathbb{Q}(\sqrt[3]{2}, \omega)$, which is Galois.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2011-01-17T22:01:30.613" />
  <row Id="17902" PostTypeId="1" AcceptedAnswerId="17994" CreationDate="2011-01-17T22:09:03.657" Score="3" ViewCount="85" Body="&lt;p&gt;Which computer algebra system allows me to compute the homology of a complex of finitely presented abelian groups which are not necessarily free?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Sage&lt;/em&gt; and &lt;em&gt;Magma&lt;/em&gt; apparently don't: see &lt;a href=&quot;http://www.sagemath.org/doc/reference/sage/homology/chain_complex.html&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; and &lt;a href=&quot;http://magma.maths.usyd.edu.au/magma/handbook/text/538&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: I should specify that the main obstacle for me of doing this by hand with a CAS is the lack of an algorithm for computing the kernel of a homomorphism between finitely generated abelian groups.&lt;/p&gt;&#xA;" OwnerUserId="367" LastEditorUserId="367" LastEditorDisplayName="" LastEditDate="2011-01-18T15:10:51.787" LastActivityDate="2011-01-18T16:38:13.767" Title="CAS: Computing homology of complex of non-free abelian groups" Tags="&lt;homological-algebra&gt;&lt;computer-algebra-systems&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="17903" PostTypeId="2" ParentId="17892" CreationDate="2011-01-17T22:28:37.420" Score="1" ViewCount="" Body="&lt;p&gt;It is not true in general.  Counterexamples and references to counterexamples of this and related questions can be found in the MathOverflow questions &lt;a href=&quot;http://mathoverflow.net/questions/34763/continuous-holomorphic-on-a-dense-open-holomorphic&quot; rel=&quot;nofollow&quot;&gt;Continuous + holomorphic on a dense open =&gt; holomorphic?&lt;/a&gt; and &lt;a href=&quot;http://mathoverflow.net/questions/41212/the-cauchy-riemann-equations-and-analyticity&quot; rel=&quot;nofollow&quot;&gt;The Cauchy-Riemann equations and analyticity&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2011-01-17T22:28:37.420" CommunityOwnedDate="2011-01-17T22:28:37.420" CommentCount="2" />
  <row Id="17904" PostTypeId="1" AcceptedAnswerId="17949" CreationDate="2011-01-17T22:32:22.970" Score="1" ViewCount="89" Body="&lt;p&gt;What conditions are needed on the matrices A and B so that the following is true:&#xA;$A^{T}BA=B$. Clearly, if B is invertible, the determinant of A must be 1 or -1. What other conditions are needed?&lt;/p&gt;&#xA;" OwnerUserId="5959" LastEditorUserId="5959" LastEditorDisplayName="" LastEditDate="2011-01-17T23:28:12.407" LastActivityDate="2011-01-18T14:54:29.330" Title="Conditions on matrix expression" Tags="&lt;linear-algebra&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="17905" PostTypeId="1" AcceptedAnswerId="17907" CreationDate="2011-01-17T22:46:44.253" Score="1" ViewCount="50" Body="&lt;p&gt;I am looking for a term that describes function that is &quot;localized&quot;. What I mean is a function that is non zero in a bounded range and zero else where, such as the a rectangle pulse function. But would also like that the term cover functions such as gaussian that approach zero as you get away from the peak. If you have the whole definition with a reference would be wonderful. I couldn't find anything that pleased me in a basic googling.&lt;/p&gt;&#xA;" OwnerUserId="3799" LastActivityDate="2011-01-17T23:11:35.620" Title="Does the term localized function exists?" Tags="&lt;functions&gt;&lt;terminology&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="17906" PostTypeId="2" ParentId="17877" CreationDate="2011-01-17T22:48:33.923" Score="3" ViewCount="" Body="&lt;p&gt;This is essentially Hans Lundmark's answer, without explicit mention of eigenvectors. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I will ignore the factors of $\frac{1}{2}$, since in the computation of $E^{-1}AE$, they will simply amount to multiplying by $\frac{1}{2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $E$ is invertible, its columns form a basis for $\mathbb{R}^3$; call this basis $\beta$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Every vector in $\mathbb{R}^3$ can be written as a linear combination of elements of $\beta$. We can keep track of this by the use of &quot;coordinate vectors&quot;: if $\mathbf{v}$ is a vector, and&#xA;$$\mathbf{v} = \alpha_1\left(\begin{array}{r}1\\ -\sqrt{3}\\0\end{array}\right) + \alpha_2\left(\begin{array}{r}\sqrt{3}\\1\\0\end{array}\right) + \alpha_3\left(\begin{array}{r}0\\0\\1\end{array}\right)$$&#xA;then since we know the vectors of $\beta$, in order to tell you who $\mathbf{v}$ is, I only need to tell you the values $\alpha_1$, $\alpha_2$, and $\alpha_3$: you know how to use them to get $\mathbf{v}$. This is called the &quot;coordinte vector&quot; of $\mathbf{v}$ with respect to $\beta$,&#xA;$$[\mathbf{v}]_{\beta} = \left(\begin{array}{c}\alpha_1\\ \alpha_2\\ \alpha_3\end{array}\right).$$&#xA;The usual expression of vectors is in fact the coordinate vector with respect to the standard basis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you compute $E^{-1}AE$, you are computing a matrix that, when multiplied by a coordinate vector $[\mathbf{v}]_{\beta}$, will give you the coordinate vector of $A\mathbf{v}$; to see this, notice that if $[\mathbf{v}]_{\beta}=(\alpha_1,\alpha_2,\alpha_3)^T$, then $E[\mathbf{v}]_{\beta} = \mathbf{v}$:&#xA;$$\left(\begin{array}{rrr}&#xD;&#xA;1 &amp;amp; \sqrt{3} &amp;amp; 0\\&#xD;&#xA;-\sqrt{3} &amp;amp; 1 &amp;amp; 0\\&#xD;&#xA;0 &amp;amp; 0 &amp;amp; 1&#xD;&#xA;\end{array}\right)\left(\begin{array}{c}&#xD;&#xA;\alpha_1\\&#xD;&#xA;\alpha_2\\&#xD;&#xA;\alpha_3&#xD;&#xA;\end{array}\right) = \alpha_1\left(\begin{array}{r}&#xD;&#xA;1 \\ -\sqrt{3} \\ 0\end{array}\right) + \alpha_2\left(\begin{array}{r}&#xD;&#xA;\sqrt{3}\\ 1\\ 0\end{array}\right) + \alpha_3\left(\begin{array}{r}0\\ 0\\ 1\end{array}\right) = \mathbf{v}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is not hard to check that $E^{-1}\mathbf{v}$ will in fact give you the coordinate vector of $\mathbf{v}$ relative to $\beta$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So one way to think about $E^{-1}AE$ is the following: you are given a coordinate vector $[\mathbf{v}]_{\beta}$; multiplying by $E$ gives you $\mathbf{v}$. Multiplying by $A$ gives you $A\mathbf{v}$; multiplying by $E^{-1}$ gives you the coordinate vector of $A\mathbf{v}$, that is $[A\mathbf{v}]_{\beta}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When is a $3\times 3$ matrix $M$ a diagonal matrix? It is a diagonal matrix if and only if $M\mathbf{e}_1 = \lambda_1\mathbf{e}_1$, $M\mathbf{e}_2 = \lambda_2\mathbf{e}_2$, and $M\mathbf{e}_3 = \lambda_3\mathbf{e}_3$, where $\mathbf{e}_1,\mathbf{e}_2,\mathbf{e}_3$ is the standard basis for $\mathbb{R}^3$, and $\lambda_1,\lambda_2,\lambda_3$ are scalars; this because $M\mathbf{e}_i$ is just the $i$th column of $M$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, under what conditions will $E^{-1}AE$ be a diagonal matrix? If the vectors $\mathbf{v}_1$, $\mathbf{v}_2$, and $\mathbf{v}_3$ whose coordinate vectors are $\mathbf{e}_1$, $\mathbf{e}_2$, and $\mathbf{e}_3$ are mapped by $A$ to a scalar multiple of themselves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But what vector $\mathbf{v}_1$ has coordinate vector $[\mathbf{v}_1]_{\beta}=\mathbf{e}_1$? The vector&#xA;$$\mathbf{v}_1 = 1\left(\begin{array}{r}1\\ -\sqrt{3}\\ 0\end{array}\right) + 0\left(\begin{array}{r} \sqrt{3}\\1\\0\end{array}\right) + 0\left(\begin{array}{c}0\\0\\1\end{array}\right) = \left(\begin{array}{r}1\\-\sqrt{3}\\0\end{array}\right).$$&#xA;That is, the first column of $E$. Likewise, the vector $\mathbf{v}_2$ with $[\mathbf{v}_2]_{\beta} = \mathbf{e}_2$ is the second column of $E$, and $\mathbf{v}_3$ is the third column of $E$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, $E^{-1}AE$ is diagonal if and only if $A\mathbf{v}_1=\lambda_1\mathbf{v}_1$, $A\mathbf{v}_2=\lambda_2\mathbf{v}_2$, and $A\mathbf{v}_3 = \lambda_3\mathbf{v}_3$, where $\mathbf{v}_1,\mathbf{v}_2,\mathbf{v}_3$ are the columns of $E$, and $\lambda_1,\lambda_2,\lambda_3$ are some scalars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To check this, simply do the product $AE$ (which are the two matrices you have in any case), and compare the columns of the result to the columns of $E$. If the $i$th column of $AE$ is a scalar multiple of the $i$th column of $E$, for $i=1,2,3$, then $E^{-1}AE$ will be a diagonal matrix. &lt;strong&gt;Note:&lt;/strong&gt; This all assumes that $E$ is in fact invertible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This leads to the definition:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Definition.&lt;/strong&gt; Let $A$ be an $n\times n$ matrix with coefficients in $F$. A vector $\mathbf{v}\in F^n$ is an &lt;strong&gt;eigenvector of&lt;/strong&gt; $A$ if and only if $\mathbf{v}\neq\mathbf{0}$, and there exists a scalar $\lambda$ such that $A\mathbf{v}=\lambda\mathbf{v}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(We ask $\mathbf{v}\neq\mathbf{0}$ because $\mathbf{0}$ always satisfies the equation with every scalar, and it will never show up as a column of an invertible matrix). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, given an $n\times n$ matrix $A$ and an invertible $n\times n$ matrix $E$, $E^{-1}AE$ is a diagonal matrix if and only if each column of $E$ is an eigenvector of $A$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Checking if a given vector is an eigenvector is just a matter of multiplying by $A$. So this is really just a matter of computing $AE$: if the $i$th column of $AE$ is a scalar multiple of the $i$th column of $E$ for $i=1,2,3$, then $E^{-1}AE$ is diagonal.  The factors of $\frac{1}{2}$ don't matter for this, they will only change the scalar if it exists. &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-18T14:05:25.253" LastActivityDate="2011-01-18T14:05:25.253" />
  <row Id="17907" PostTypeId="2" ParentId="17905" CreationDate="2011-01-17T22:51:30.380" Score="5" ViewCount="" Body="&lt;p&gt;Given a function $f\colon X\to \mathbb{R}$, the &lt;strong&gt;support of&lt;/strong&gt; $f$ is defined to be&#xA;$$\mathrm{sup}(f) = \{x\in X\mid f(x)\neq 0\}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in the first part, you are talking about &lt;strong&gt;&quot;real variable functions of bounded support&quot;&lt;/strong&gt;. The &quot;real variable&quot; can be dropped if the domain is understood. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For functions that have limit equal to $0$ at infinity you talk about functions that &lt;strong&gt;&quot;vanish at&lt;/strong&gt; $\infty$&quot;. (Thanks to Qiaochu for mentioning this, as I missed it). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that any function with bounded support will necessarily vanish at infinity, but $y = e^{-x^2}$ shows you can have functions that vanish at $\infty$ but are not of bounded support.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-17T23:11:35.620" LastActivityDate="2011-01-17T23:11:35.620" />
  <row Id="17908" PostTypeId="1" AcceptedAnswerId="17911" CreationDate="2011-01-17T22:54:49.190" Score="2" ViewCount="155" Body="&lt;p&gt;Some background: I was in class today, and the professor was proving that given a matrix $A$ that is $n$ by $m$ and a $B$ such that $AB=I$ where $B$ is obviously $m$ by $n$ and $I$ is $n$ by $n$, if it is also the case that $BA=I$, then $A$ must be a square matrix (that is, $m=n$). We didn't manage to get through however, because we couldn't find a reason why $A$ actually had to be square in that case -- it seemed to me that it would just be $BA=I$ where $I$ is $m$ by $m$. So in both cases, we would have an Identity matrix, but of different dimensions (sounds strange, but as far as I recall, as long as the product yields an Identity, $B$ counts as an inverse, even if dimensions vary depending on whether you multiply it from left or right). I was thinking something along the lines of overdetermined systems -- if $A$ is invertible (and not square) at least from the right, it has to have more equations than variables, but that did not bring me anywhere. Thanks!&lt;/p&gt;&#xA;" OwnerUserId="3731" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-17T23:27:33.517" LastActivityDate="2011-01-18T08:39:05.113" Title="Why is it that if A is an n by m matrix, and both BA and AB are indentity matrices, then A is square and can't be rectangular?" Tags="&lt;linear-algebra&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="1" />
  <row Id="17909" PostTypeId="2" ParentId="17905" CreationDate="2011-01-17T22:55:43.027" Score="5" ViewCount="" Body="&lt;p&gt;The second term you're looking for is &lt;a href=&quot;http://en.wikipedia.org/wiki/Vanish_at_infinity&quot; rel=&quot;nofollow&quot;&gt;vanishing at infinity&lt;/a&gt;.  At least on locally compact spaces a natural generalization of the first term is &lt;a href=&quot;http://en.wikipedia.org/wiki/Support_(mathematics)#Compact_support&quot; rel=&quot;nofollow&quot;&gt;compact support&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2011-01-17T22:55:43.027" />
  <row Id="17910" PostTypeId="2" ParentId="17858" CreationDate="2011-01-17T22:58:06.420" Score="0" ViewCount="" Body="&lt;p&gt;I would go with Digital Signal Processing and Applied Mathematics.&lt;/p&gt;&#xA;" OwnerUserId="5961" LastActivityDate="2011-01-17T22:58:06.420" />
  <row Id="17911" PostTypeId="2" ParentId="17908" CreationDate="2011-01-17T23:07:51.660" Score="7" ViewCount="" Body="&lt;p&gt;Assuming your matrices have real coefficients, you can use the Rank-Nullity Theorem, which is essentially what you are talking about. All you need to know is that a homogeneous system of linear equations with more unknowns than equations always has nontrivial solutions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose that $A$ is $n\times m$, $B$ is $m\times n$, and $AB=I_n$, $BA=I_m$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the system of equation $A\mathbf{x}=\mathbf{0}$, where $\mathbf{x}$ is the column vector with $m$ unknowns. A solution $\mathbf{s}$ must be equal to $\mathbf{0}$, because if $A\mathbf{s}=\mathbf{0}$, then &#xA;$$\mathbf{0} = B\mathbf{0} = B(A\mathbf{s}) = (BA)\mathbf{s} = I_m\mathbf{s}=\mathbf{s}.$$&#xA;So the only solution to $A\mathbf{x}=\mathbf{0}$ is the trivial solution, which tells you that you must have at most as many unknowns as equations. Since $A\mathbf{x}=\mathbf{0}$ has $n$ equations and $m$ unknowns, you know $n\geq m$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, repeat the argument with the system $B\mathbf{x}=\mathbf{0}$; multiplying by $A$ on the left you conclude that the system has at least as many equations as unknowns, because the trivial solution is the only solution. So $m\geq n$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Putting the two together you get $n=m$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Notice that you do not need to assume that it is the same matrix $B$ that works on both sides at the outset. If you have &lt;em&gt;some&lt;/em&gt; $m\times n$ matrices $B$ and $C$ such that $CA=I_m$ and $AB=I_n$, then $C=CI_n = C(AB) = (CA)B = I_mB = B$; or you can just do the argument above using $C$ in the first part instead of $B$ to conclude $m=n$ first, then conclude that $C=B=A^{-1}$). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(The argument actually breaks down if your matrices have coefficients in more general rings; in particular, if you allow the entries to come from noncommutative rings, then you &lt;em&gt;can&lt;/em&gt; have nonsquare matrices $A$ and $B$ such that $AB$ and $BA$ are different-sized identities.)&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-17T23:19:08.390" LastActivityDate="2011-01-17T23:19:08.390" CommentCount="4" />
  <row Id="17913" PostTypeId="2" ParentId="17609" CreationDate="2011-01-17T23:16:20.053" Score="0" ViewCount="" Body="&lt;p&gt;I found the answer is 1/n as explained in section 3.1 over here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.madandmoonly.com/doctormatt/mathematics/dice1.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.madandmoonly.com/doctormatt/mathematics/dice1.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5886" LastActivityDate="2011-01-17T23:16:20.053" />
  <row Id="17914" PostTypeId="1" AcceptedAnswerId="17915" CreationDate="2011-01-17T23:26:40.707" Score="7" ViewCount="246" Body="&lt;p&gt;How does one compute the cardinality of the set of functions $f:\mathbb{R} \to \mathbb{R}$ (not necessarily continuous)?&lt;/p&gt;&#xA;" OwnerUserId="5959" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-18T03:21:26.267" LastActivityDate="2011-01-18T03:21:26.267" Title="Cardinality of the set of all real functions of real variable." Tags="&lt;set-theory&gt;" AnswerCount="2" />
  <row Id="17915" PostTypeId="2" ParentId="17914" CreationDate="2011-01-17T23:32:55.960" Score="10" ViewCount="" Body="&lt;p&gt;All you need is a few basics of cardinal arithmetic: if $\kappa$ and $\lambda$ are cardinals, none of them zero, and at least one of them is infinite, then $\kappa+\lambda = \kappa\lambda = \max\{\kappa,\lambda\}$. And cardinal exponentiation satisfies some of the same laws as regular exponentiation; in particular, $(\kappa^{\lambda})^{\nu} = \kappa^{\lambda\nu}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The cardinality of the set of all real functions is then&#xA;$$|\mathbb{R}|^{|\mathbb{R}|} =\mathfrak{c}^{\mathfrak{c}} = (2^{\aleph_0})^{2^{\aleph_0}} = 2^{\aleph_02^{\aleph_0}} = 2^{2^{\aleph_0}} = 2^{\mathfrak{c}}.$$&#xA;In other words, it is equal to the cardinality of the power set of $\mathbb{R}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With a few extra facts, you can get more. In general, if $\kappa$ is an infinite cardinal, and $2\leq\lambda\leq\kappa$, then $\lambda^{\kappa}=2^{\kappa}$. This follows because:&#xA;$$2^{\kappa} \leq \lambda^{\kappa} \leq (2^{\lambda})^{\kappa} = 2^{\lambda\kappa} = 2^{\kappa},$$&#xA;so you get equality throughout. The extra information you need for this is to know that if $\kappa$, $\lambda$, and $\nu$ are nonzero cardinals, $\kappa\leq\lambda$, then $\kappa^{\nu}\leq \lambda^{\nu}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, for any infinite cardinal $\kappa$ you have $\kappa^{\kappa} = 2^{\kappa}$. &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-17T23:43:35.437" LastActivityDate="2011-01-17T23:43:35.437" />
  <row Id="17916" PostTypeId="1" AcceptedAnswerId="17917" CreationDate="2011-01-17T23:37:28.003" Score="26" ViewCount="635" Body="&lt;p&gt;How come the number N! can terminate in exactly 1,2,3,4, or 6 zeroes but never 5 zeroes?&lt;/p&gt;&#xA;" OwnerUserId="5962" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2011-01-18T23:54:49.880" LastActivityDate="2011-04-30T17:27:57.563" Title="Number of Zeros at the end of a factorial" Tags="&lt;elementary-number-theory&gt;&lt;factorial&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="11" />
  <row Id="17917" PostTypeId="2" ParentId="17916" CreationDate="2011-01-17T23:40:43.087" Score="21" ViewCount="" Body="&lt;p&gt;The number of zeros at the end of $N!$ is given by $$\left \lfloor \frac{N}{5} \right \rfloor + \left \lfloor \frac{N}{5^2} \right \rfloor + \left \lfloor \frac{N}{5^3} \right \rfloor + \cdots$$ where $\left \lfloor \frac{x}{y} \right \rfloor$ is the greatest integer $\leq \frac{x}{y}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To make it clear write $N!$ as a product of primes $N! = 2^{\alpha_2} 3^{\alpha_2} 5^{\alpha_5} 7^{\alpha_7} 11^{\alpha_{11}} \ldots$ where $\alpha_i \in \mathbb{N}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that $\alpha_5 &amp;lt; \alpha_2$, $\forall N$. (Why?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The number of zeros at the end of $N!$ is the maximum power of $10$ dividing $N!$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $10^{\alpha}$ divides $N!$ and since $10 = 2 \times 5$, $2^{\alpha} | N!$ and $5^{\alpha} | N!$. Further since $\alpha_5 &amp;lt; \alpha_2$, the maximum power of $10$ dividing $N!$ is the maximum power of $5$ dividing $N!$ which is $\alpha_5$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if you want to find the maximum power of a prime $p$ dividing $N!$, it is given by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left \lfloor \frac{N}{p} \right \rfloor + \left \lfloor \frac{N}{p^2} \right \rfloor + \left \lfloor \frac{N}{p^3} \right \rfloor + \cdots$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first term appears since you want to count the number of terms less than $N$ and are multiples of $p$ and each of these contribute one $p$ to $N!$. But then when you have multiples of $p^2$ you are not adding just one $p$ but you are adding two of these primes $p$ to the product. So you now count the number of multiple of $p^2$ less than $N$ and add and so on and so forth for higher powers of $p$ less than $N$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you will find that for $N \leq 24$, the number of zeros will be less than or equal to 4. However when $N$ hits $25$ you will get 2 additional zeros courtesy $25$ since $25 \times 2^2 = 100$. Hence, there will be a jump when you go from $24$ to $25$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that there will be &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;A jump of $1$ zero going from $(N-1)!$ to $N!$ if $5 || N$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;A jump of $2$ zero going from $(N-1)!$ to $N!$ if $5^2 || N$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;A jump of $3$ zero going from $(N-1)!$ to $N!$ if $5^3 || N$ and in general&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;A jump of $k$ zero going from $(N-1)!$ to $N!$ if $5^k || N$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;where $a || b$ means $a$ divides $b$ and gcd($a,\frac{b}{a}$) = 1&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2011-04-30T17:27:57.563" LastActivityDate="2011-04-30T17:27:57.563" CommentCount="6" />
  <row Id="17918" PostTypeId="1" AcceptedAnswerId="17922" CreationDate="2011-01-17T23:42:06.433" Score="0" ViewCount="76" Body="&lt;p&gt;How do you demonstrate that a local minimum of a function has its derivative equal to zero with the definition of the derivative.&lt;/p&gt;&#xA;" OwnerUserId="5951" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-17T23:58:15.580" LastActivityDate="2011-01-17T23:58:15.580" Title="Proving that a point is a Stationary point with the definition of the derivative" Tags="&lt;calculus&gt;&lt;definition&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="17919" PostTypeId="2" ParentId="17916" CreationDate="2011-01-17T23:44:29.850" Score="11" ViewCount="" Body="&lt;p&gt;To answer that, you have to find the jump point from 4 to 6 zeros. The first is at 5!, the second at 10!, the third at 15!, and the forth at 20!. The next point is at 25!. What happens there is, you're multiplying by 5*5, which will multiply with 2 even numbers, and add 2 more zeros.&lt;/p&gt;&#xA;" OwnerUserId="5101" LastActivityDate="2011-01-17T23:44:29.850" CommentCount="2" />
  <row Id="17920" PostTypeId="2" ParentId="17914" CreationDate="2011-01-17T23:45:52.293" Score="4" ViewCount="" Body="&lt;p&gt;I guess that you know that $|\mathbb{N}| = |\mathbb{N}\times\mathbb{N}|$ and thus $|\mathbb{R}| = |2^{\mathbb{N}}| = |2^{\mathbb{N}\times\mathbb{N}}| = |2^\mathbb{N}\times 2^\mathbb{N}| = |\mathbb{R}\times\mathbb{R}|$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means that $|P(\mathbb{R})| = |P(\mathbb{R}\times\mathbb{R})|$. Since $f\colon\mathbb{R}\to\mathbb{R}$ is an element of $P(\mathbb{R}\times\mathbb{R})$ you have that $\mathbb{R}^\mathbb{R}$ (all the functions from $\mathbb{R}$ to itself) is of cardinality less or equal to the one of $P(\mathbb{R}\times\mathbb{R})$ which in turn means that $|\mathbb{R}^\mathbb{R}|\le |P(\mathbb{R})|$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, since $|P(\mathbb{R})| = |2^\mathbb{R}|$ which is the set of all functions from $\mathbb{R}$ to $\{0,1\}$, and clearly every function from $\mathbb{R}$ into $\{0,1\}$ is in particular a function from $\mathbb{R}$ into itself, we have: &#xA;$$|P(\mathbb{R})| = |2^\mathbb{R}| \le |\mathbb{R}^\mathbb{R}| \le |P(\mathbb{R}\times\mathbb{R})| = |P(\mathbb{R})|$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So all in all we have that $|\mathbb{R}^\mathbb{R}| = |P(\mathbb{R})| = |2^\mathbb{R}|$.&lt;/p&gt;&#xA;" OwnerUserId="622" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2011-01-18T01:00:57.090" LastActivityDate="2011-01-18T01:00:57.090" CommentCount="1" />
  <row Id="17921" PostTypeId="2" ParentId="17918" CreationDate="2011-01-17T23:48:24.777" Score="1" ViewCount="" Body="&lt;p&gt;Having a local minimum at $x$ implies that the right-hand difference quotients are nonnegative and the left-hand difference quotients are nonpositive (when taken sufficiently close to $x$).  &lt;em&gt;If the derivative exists&lt;/em&gt;, then it must equal the limit of both the left-hand and right-hand difference quotients, so it must be a limit of nonnegative numbers and a limit of nonpositive numbers.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2011-01-17T23:48:24.777" />
  <row Id="17922" PostTypeId="2" ParentId="17918" CreationDate="2011-01-17T23:49:31.047" Score="3" ViewCount="" Body="&lt;p&gt;You want to prove that if $f(x)$ has a local minimum at $a$, and $f$ is differentiable at $a$, then $f&amp;#39;(a)=0$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because $f(x)$ has a local minimum at $a$, that means that there exists a $\delta\gt 0$ such that for all $x$, if $|x-a|\leq \delta$, then $f(a)\leq f(x)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The derivative at $a$ is:&#xA;$$\lim_{x\to a}\frac{f(x)-f(a)}{x-a}.$$&#xA;Consider the limits from the left and from the right. If $x\gt a$ and is sufficiently close to $a$ (within $\delta$), then $f(x)-f(a)\geq 0$ and $x-a\gt 0$. So the fraction $\frac{f(x)-f(a)}{x-a}$ satisfies $\frac{f(x)-f(a)}{x-a}\geq 0$. In particular, the limit must be greater than or equal to $0$ (if it exists). Note that we can certainly restrict ourselves to being within $\delta$ of $a$, because the value of the limit only depends on what happens close to $a$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now consider what happens if $x\lt a$. If we are within $\delta$ of $a$, then $f(x)-f(a)\geq 0$ (because $f$ has a minimum at $a$), but $x-a\lt 0$. That means that the fraction $\frac{f(x)-f(a)}{x-a}$ satisfies $\frac{f(x)-f(a)}{x-a}\leq 0$. In particular, the limit, if it exists, must be less than or equal to $0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since we are assuming that the derivative exists, both one-sided limits exist and are equal. The limit from the right must be greater than or equal to $0$, and the limit from the left must be less than or equal to $0$; the only possibility for them to be equal is for them both to be $0$; that is, for $f&amp;#39;(a)$ to be equal to $0$.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-17T23:56:57.110" LastActivityDate="2011-01-17T23:56:57.110" />
  <row Id="17923" PostTypeId="2" ParentId="17908" CreationDate="2011-01-17T23:52:47.020" Score="11" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\ \ $ $\rm\ n\: =\: $ trace $\rm\: AB\ =\:$ trace $\rm\: BA\: = m$&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2011-01-18T00:02:39.477" LastActivityDate="2011-01-18T00:02:39.477" />
  <row Id="17924" PostTypeId="1" AcceptedAnswerId="17929" CreationDate="2011-01-18T00:11:16.857" Score="2" ViewCount="120" Body="&lt;p&gt;A jar contains some red and some yellow jelly beans. If a child ate 1 red jelly bean, 1/7 of the remaining candies would be red. If instead the child ate 5 yellow jelly beans, 1/6 of the remaining candies would be red. How do I found out the number of jelly beans that are in the jar?&lt;/p&gt;&#xA;" OwnerUserId="5962" LastEditorUserId="462" LastEditorDisplayName="" LastEditDate="2011-01-18T00:20:48.977" LastActivityDate="2011-01-18T00:55:12.223" Title="Algebra word problem (Red and Yellow Jelly Beans)" Tags="&lt;algebra-precalculus&gt;" AnswerCount="2" CommentCount="5" FavoriteCount="1" />
  <row Id="17925" PostTypeId="2" ParentId="17924" CreationDate="2011-01-18T00:17:21.420" Score="4" ViewCount="" Body="&lt;p&gt;Use algebra.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let R be the number of red beans, and Y be the number of yellow beans.  Then the second and third sentences can be expressed as equations in R and Y.  Solve this pair of equations, and you've got it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, the second sentence tells us that $\frac{R-1}{R+Y-1}=\frac{1}{7}$.&lt;/p&gt;&#xA;" OwnerUserId="2937" LastActivityDate="2011-01-18T00:17:21.420" CommentCount="1" />
  <row Id="17926" PostTypeId="2" ParentId="17916" CreationDate="2011-01-18T00:23:39.193" Score="4" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\: $ The power of a &lt;em&gt;prime&lt;/em&gt; $\rm\:p\:$ dividing $\rm\ n!\:$ jumps from $\rm\: p-1\:$ for $\rm\: n = p^2-1\:$ to $\rm\: p+1\: $ for $\rm\: n = p^2\:$ since their are $\rm\:p-1\:$ naturals $\rm &amp;lt; p^2\ $ divisible by $\rm\:p\:,\:$ viz. $\rm\ p,\ 2\:p,\:\cdots\:,\: (p-1)\:p\:.\ $ Now put $\rm\ p = 5\:.$&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2011-01-18T00:44:26.367" LastActivityDate="2011-01-18T00:44:26.367" />
  <row Id="17927" PostTypeId="1" AcceptedAnswerId="17938" CreationDate="2011-01-18T00:48:41.337" Score="0" ViewCount="174" Body="&lt;p&gt;What is the best way to prove the mean value theorem.&lt;/p&gt;&#xA;" OwnerUserId="5951" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-18T05:50:28.573" LastActivityDate="2011-01-18T05:50:28.573" Title="What is the best way to prove the mean value theorem" Tags="&lt;calculus&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="17928" PostTypeId="2" ParentId="17927" CreationDate="2011-01-18T00:49:52.807" Score="2" ViewCount="" Body="&lt;p&gt;Have you seen the &lt;a href=&quot;http://en.wikipedia.org/wiki/Mean_value_theorem&quot; rel=&quot;nofollow&quot;&gt;wikipedia&lt;/a&gt; article?&lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2011-01-18T00:49:52.807" />
  <row Id="17929" PostTypeId="2" ParentId="17924" CreationDate="2011-01-18T00:55:12.223" Score="1" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\ $ Put $\rm R = $ #red beans. Then the total #beans $\rm\: =\ 7\ (R-1) + 1\ =\ 6\ R + 5\:.$&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2011-01-18T00:55:12.223" CommentCount="1" />
  <row Id="17930" PostTypeId="1" AcceptedAnswerId="17937" CreationDate="2011-01-18T01:31:53.340" Score="4" ViewCount="140" Body="&lt;p&gt;When you take the derivative of an expression with respect to &lt;em&gt;x&lt;/em&gt;, does &lt;em&gt;x&lt;/em&gt; have to be a variable, or is it allowed to be a polynomial, a term, a vector, or anything else? It doesn't seem to make sense to me if x is not a variable.&lt;/p&gt;&#xA;" OwnerUserId="3143" LastActivityDate="2011-01-18T02:55:23.657" Title="Does the derivative with respect to something have to be a variable?" Tags="&lt;calculus&gt;" AnswerCount="2" CommentCount="6" FavoriteCount="1" />
  <row Id="17931" PostTypeId="2" ParentId="17930" CreationDate="2011-01-18T01:36:20.767" Score="2" ViewCount="" Body="&lt;p&gt;You could look at &lt;a href=&quot;http://en.wikipedia.org/wiki/Matrix_calculus&quot; rel=&quot;nofollow&quot;&gt;matrix calculus&lt;/a&gt; for example.&lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2011-01-18T01:36:20.767" />
  <row Id="17932" PostTypeId="1" AcceptedAnswerId="17936" CreationDate="2011-01-18T01:48:53.883" Score="5" ViewCount="185" Body="&lt;p&gt;This is my first question in mathSE, hope that it is suitable here!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm currently self-studying complex analysis using the book by Stein &amp;amp; Shakarchi, and this is one of the exercises (p.67, Q14) that I have no idea where to start.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Suppose $f$ is holomorphic in an open set $\Omega$ that contains the closed unit disc, except for a pole at $z_0$ on the unit circle. Show that if $f$ has the power series expansion $\sum_{n=0}^\infty a_n z^n$ in the open unit disc, then&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$\displaystyle \lim_{n \to \infty} \frac{a_n}{a_{n+1}} = z_0$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If the limit is taking on $|{a_n}/{a_{n+1}}|$ and assume the limit exists, by the radius of convergence we know that the answer is $1$. But what can we say about the limit of the coefficient ratio, which is a pure complex number? I've tried to expand the limit directly by definition, with no luck. And I couldn't see how we can apply any of the standard theorems in complex analysis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope to get some initial directions about how we can start thinking on the problem, rather than a full answer. Thank you for the help!&lt;/p&gt;&#xA;" OwnerUserId="2767" LastEditorUserId="2767" LastEditorDisplayName="" LastEditDate="2011-01-18T02:05:20.913" LastActivityDate="2011-01-18T08:27:04.110" Title="About the limit of the coefficient ratio for a power series over complex numbers" Tags="&lt;complex-analysis&gt;&lt;power-series&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" />
  <row Id="17933" PostTypeId="2" ParentId="17858" CreationDate="2011-01-18T01:51:41.440" Score="0" ViewCount="" Body="&lt;p&gt;It depends on what you want to do. If you're interested in security, take a cryptography class. If linear equations are your thing, take linear algebra. If you're going to analyze data, a statistics. I know in college it's hard to know exactly what you want to do after graduation, but in order to answer this question, you really do need to think about it.&lt;/p&gt;&#xA;" OwnerUserId="5101" LastActivityDate="2011-01-18T01:51:41.440" />
  <row Id="17935" PostTypeId="1" CreationDate="2011-01-18T02:19:49.653" Score="8" ViewCount="179" Body="&lt;p&gt;How well accepted are proofs where critical parts depend on software of which we have no access to the source, e.g Mathematica, Maple?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More generally how much can we take on faith with regards to software. Do we have to check the compiler for the Four Color Theorem?&lt;/p&gt;&#xA;" OwnerUserId="2809" LastActivityDate="2011-01-18T16:23:04.637" Title="Proof and software" Tags="&lt;math-software&gt;&lt;proof-strategy&gt;&lt;mathematica&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="4" />
  <row Id="17936" PostTypeId="2" ParentId="17932" CreationDate="2011-01-18T02:38:02.047" Score="6" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Hint:&lt;/strong&gt; Assume that you have a simple pole at $z = z_0$, where $|z_0| = 1$ and try to prove it. In particular, take $f(z) = \frac{g(z)}{z-z_0}$ where $g(z)$ is holomorphic on $\Omega$. Prove the result for this case. (Expand $\frac{1}{z-z_0}$ about $z=0$ and do some manipulations). Now the same idea can be extended for higher order poles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt; For a simple pole, $f(z) = \frac{g(z)}{z-z_0} = \displaystyle \sum_{n=0}^{\infty} a_n z^n$. Since $g(z)$ is holomorphic, $g(z) = \displaystyle \sum_{n=0}^{\infty} b_n z^n$. So $\displaystyle \sum_{n=0}^{\infty} b_n z^n = (z-z_0) \displaystyle \sum_{n=0}^{\infty} a_n z^n \Rightarrow b_{n+1} = a_n - z_0 a_{n+1}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now what can we say about $\displaystyle \lim_{n \rightarrow \infty} b_n$ and $\displaystyle \lim_{n \rightarrow \infty} a_n$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Note: $g(z)$ holomorphic on $\Omega$ whereas $f(z)$ is holomorphic except at $z_0$, a point on the unit disc).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This same idea will work for higher order poles as well.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2011-01-18T02:51:11.887" LastActivityDate="2011-01-18T02:51:11.887" CommentCount="10" />
  <row Id="17937" PostTypeId="2" ParentId="17930" CreationDate="2011-01-18T02:55:23.657" Score="5" ViewCount="" Body="&lt;p&gt;Intuitively, the derivative of $f$ with respect to $u$ is the limit of the change in $f$ as $u$ changes, divided by the change in $u$, as the change in $u$ vanishes. This does not require $u$ to be a &quot;variable&quot; in the usual sense: you can certainly ask for the rate of change of, say, $f(x) = \sin(x^2+1)$ with respect to $u=x^2$. So, no, it does not have to be an &quot;independent variable&quot; in the sense that you seem to be thinking about.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, that's what the Chain Rule is all about! It tells you that if $f$ depends on $g$ and $g$ depends on $x$, then the rate of change of $f$ with respect to $x$ is equal to the rate of change of $f$ with respect to $g$, times the rate of change of $g$ with respect to $x$:&#xA;$$\frac{df}{dx} = \frac{df}{dg}\;\frac{dg}{dx}.$$&#xA;Here, we usually have $g$ a &lt;em&gt;function&lt;/em&gt;, not a &quot;variable&quot;. Yet we can talk about the derivative of $f$ with respect to $g$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Every time you have a function, you can try to talk about the rate of change of the function with respect to something else, provided you have some way of quantifying the change.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2011-01-18T02:55:23.657" CommentCount="4" />
  <row Id="17938" PostTypeId="2" ParentId="17927" CreationDate="2011-01-18T02:57:45.883" Score="8" ViewCount="" Body="&lt;p&gt;The most common way to prove the Mean Value Theorem is to first prove Rolle's Theorem; then, given the arbitrary setting of the Mean Value Theorem, you just &lt;em&gt;&quot;tilt your head and apply Rolle's Theorem&quot;&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is, you draw the line joining $(a,f(a))$ and $(b,f(b))$, and let $g(x)$ be the function whose value is the vertical distance between $f(x)$ and this line. Then $g(x)$ is differentiable, and has $g(a)=g(b)=0$; applying Rolle's Theorem to $g$ and then translating this into what it means for $f(x)$ yields the Mean Value Theorem.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2011-01-18T02:57:45.883" CommentCount="5" />
  <row Id="17939" PostTypeId="1" AcceptedAnswerId="38616" CreationDate="2011-01-18T04:00:44.777" Score="10" ViewCount="256" Body="&lt;p&gt;I'm interested in (unity-preserving) homomorphisms $f: S \to T$ between (commutative, with-unity) rings $S$ and $T$ so that if $f(x)$ is a unit, then $x$ was a unit to start with.  For example, an inclusion of fields has this property, but a nontrivial localization like $\mathbb{Z} \to \mathbb{Q}$ does not; it sends $2 \in \mathbb{Z}$, which is not a unit, to $2 \in \mathbb{Q}$, which is.  I'm interested in answers to either of the following two questions:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Is there a name for such $f$ that do preserve units through preimages?&lt;/li&gt;&#xA;&lt;li&gt;If not, is there a class of maps broadly recognized as useful that enjoy this property?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I'm actually only interested in the case of a surjection $R \twoheadrightarrow R_0$ whose kernel is an ideal $I$ satisfying $I^2 = 0$ (i.e., a square-zero extension of $R_0$ by $I$).  But, if this property has a name in general or if square-zero extensions occur as special types of some broadly-recognized class of maps with this property, I'd like to know so I can chat about this with other people and don't go picking a name nobody recognizes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm aware that this can also be viewed as a lifting property, if that jogs anyone's memory of useful geometry words: what kind of map should $\operatorname{spec} S \to \operatorname{spec} R$ be so that for any pair of maps $\operatorname{spec} S \to \mathbb{G}_m$ and $\operatorname{spec} R \to \mathbb{A}^1$ with commuting square $$\begin{array}{ccc} \operatorname{spec} S &amp;amp; \to &amp;amp; \mathbb{G}_m \\ \downarrow &amp;amp; &amp;amp; \downarrow \\ \operatorname{spec} R &amp;amp; \to &amp;amp; \mathbb{A}^1,\end{array}$$ there exists a lift $\operatorname{spec} R \to \mathbb{G}_m$ making both triangles commute?&lt;/p&gt;&#xA;" OwnerUserId="2574" LastActivityDate="2011-05-16T08:30:34.987" Title="Preimaging units to units" Tags="&lt;algebraic-geometry&gt;&lt;terminology&gt;&lt;commutative-algebra&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="2" />
  <row Id="17940" PostTypeId="2" ParentId="17894" CreationDate="2011-01-18T04:52:39.480" Score="0" ViewCount="" Body="&lt;p&gt;Right off the bat, a reality check shows something amiss here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given ANY axis U = (xi, yj, zk) and a rotational angle alpha, the w (i.e., real) value of the resulting quaternion should be cos(alpha/2). But if alpha is 45 degrees, then w = cos(pi/8.0) should be around 0.924... and not somewhere about 0.963..., as you seem to expect when 'correctly' calculating ccc=quat([.3 .1 .6], 45). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your error is that the axis rotation quaternion given axis U and rotation angle alpha assumes that U is already normalized (has length 1.0). So your formula should (in an ideal world, without any round-off error) be: &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; norm(U)*sin(alpha/2) + cos(alpha/2). &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;where norm(U) is U, 'normalized' to have length 1. What you have coded up is instead &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; norm(U*sin(alpha/2) + cos(alpha/2)).&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;As a test, try using your existing code to evaluate ccc = quat([300, 100, 600], 45). It will be much more 'off' than your example (where U is 'relatively' close to being normalized). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That should make sense: the quaternion that rotates 45 degrees around the axis (.3, .1, .6) should be exactly the same quaternion that rotates 45 degrees around the axis (300, 100, 600). In both caes, it's really the same axis: the latter is just a 'longer version' of the former.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since we are not in an idealized world lacking round-off error, what you want to do is first, normalize the axis of rotation U; then construct the quaternion, and then normalize the quaternion:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; norm(norm(U)*sin(alpha/2) + cos(alpha/2))&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;BTW, while it's quite readable, I don't recognize the syntax of the language your are working in. What language is it?&lt;/p&gt;&#xA;" OwnerUserId="4980" LastActivityDate="2011-01-18T04:52:39.480" />
  <row Id="17941" PostTypeId="2" ParentId="17836" CreationDate="2011-01-18T05:00:10.467" Score="2" ViewCount="" Body="&lt;p&gt;There is no &quot;perfect&quot; software for making math visualization videos. The better movies of this kind were likely made using standard 3D animation software such as Cinema4D, Maya, 3d studio, XSI and such. You'll probably have to export data from a math application such as Mathematica into these suites and do the renderings in there. It's possible to render with just POV-Ray but working with a 3D suite can get you much better results.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some examples:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Not Knot, &lt;a href=&quot;http://www.youtube.com/watch?v=AGLPbSMxSUM&quot; rel=&quot;nofollow&quot;&gt;part 1&lt;/a&gt; and &lt;a href=&quot;http://www.youtube.com/watch?v=MKwAS5omW_w&quot; rel=&quot;nofollow&quot;&gt;part 2&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;http://www.youtube.com/watch?v=JX3VmDgiFnY&quot; rel=&quot;nofollow&quot;&gt;Moebius Transformations&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Sphere Eversion &lt;a href=&quot;http://www.youtube.com/watch?v=BVVfs4zKrgk&quot; rel=&quot;nofollow&quot;&gt;Part 1&lt;/a&gt; and &lt;a href=&quot;http://www.youtube.com/watch?v=x7d13SgqUXg&quot; rel=&quot;nofollow&quot;&gt;Part 2&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="308" LastActivityDate="2011-01-18T05:00:10.467" />
  <row Id="17943" PostTypeId="1" CreationDate="2011-01-18T05:05:02.860" Score="0" ViewCount="126" Body="&lt;p&gt;If $\mathbf{x}$ is a vector and $Q$ a symmetric Matrix;&#xA;Does $Q\mathbf{x} = \mathbf{x}^T Q$  ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have this:&#xA;$$\frac{1}{2}\left(Q\mathbf{x} + \mathbf{x}^TQ\right) + \mathbf{c},$$&#xA;where $\mathbf{c}$ is also a vector, and I want to obtain &#xA;$Q\mathbf{x}+\mathbf{c}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I thought that was the answer to my question, but it doesn't seem to be.  Do you see a way to obtain this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT :&#xA;Well Then I believe I made an error prior that, I can explain what I have done before to see if it's the source of the error.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am trying to find the gradient of the function  $$\frac{1}{2}\left(\mathbf{x}^TQ\mathbf{x}\right) + \mathbf{c}\mathbf{x}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;that's how I got the first answer by finding the gradient.&lt;/p&gt;&#xA;" OwnerUserId="5951" LastEditorUserId="5951" LastEditorDisplayName="" LastEditDate="2011-01-18T22:24:04.750" LastActivityDate="2011-01-18T23:12:33.410" Title="Symmetric Matrix Property" Tags="&lt;matrices&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="17944" PostTypeId="2" ParentId="17935" CreationDate="2011-01-18T05:08:58.173" Score="12" ViewCount="" Body="&lt;p&gt;A few years ago the &lt;em&gt;Notices of the American Mathematical Society&lt;/em&gt; ran a series of articles on the state of the art in using computer proof assistants.  They are at least worth skimming.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;&quot;&lt;a href=&quot;http://www.ams.org/notices/200811/tx081101370p.pdf&quot; rel=&quot;nofollow&quot;&gt;Formal Proof&lt;/a&gt;,&quot; by Thomas Hales&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&quot;&lt;a href=&quot;http://www.ams.org/notices/200811/tx081101382p.pdf&quot; rel=&quot;nofollow&quot;&gt;Formal Proof - The Four-Color Theorem&lt;/a&gt;,&quot; by Georges Gonthier&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&quot;&lt;a href=&quot;http://www.ams.org/notices/200811/tx081101395p.pdf&quot; rel=&quot;nofollow&quot;&gt;Formal Proof - Theory and Practice&lt;/a&gt;,&quot; by John Harrison&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&quot;&lt;a href=&quot;http://www.ams.org/notices/200811/tx081101408p.pdf&quot; rel=&quot;nofollow&quot;&gt;Formal Proof - Getting Started&lt;/a&gt;,&quot; by Freek Wiedijk&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I tend to be a traditionalist when it comes to traditional methods versus computer methods, but I do find arguments like the following compelling.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(This is from a &lt;a href=&quot;http://www.ams.org/news?news_id=416&quot; rel=&quot;nofollow&quot;&gt;news release from the AMS&lt;/a&gt; on their series of articles on formal proof.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;When mathematicians prove theorems in the traditional way, they present the argument in narrative form.  They assume previous results, they gloss over details they think other experts will understand, they take shortcuts to make the presentation less tedious, they appeal to intuition, etc.  The correctness of the arguments is determined by the scrutiny of other mathematicians, in informal discussions, in lectures, or in journals.  &lt;em&gt;It is sobering to realize that the means by which mathematical results are verified is essentially a social process and is thus fallible&lt;/em&gt;.&quot; [emphasis mine]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(And this is from Hales's article.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;The computer code that implements the axioms and rules of inference is referred to as the kernel of the system.  It takes fewer than 500 lines of computer code to implement the kernel of HOL Light.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Verifying correctness on such a small program would be a lot easier than verifying correctness of proofs of many theorems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Also from Hales's article.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Experience from other top-tier theorem-proving systems has been that about three to five bugs have been found in each system over a period of 15-20 years of use.  After decades of use on many different systems, to my knowledge, only one proof has ever had to be retracted as a result of a bug in a theorem-proving system, and this in a system that I do not rank in the top tier... &lt;em&gt;We can assert with utmost confidence that the error rates of top-tier theorem proving systems are orders of magnitude lower than error rates in the most prestigious mathematical journals&lt;/em&gt;.&quot; [emphasis mine again]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a side note, Hales's paper also lists several major theorems (already proved the traditional way, of course) that have been since proved by a computer.  These include quadratic reciprocity, the prime number theorem, the fundamental theorems of algebra and calculus, the Jordan curve theorem, and the Brouwer fixed-point theorem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Besides the Four-Color Theorem, apparently the &lt;a href=&quot;http://en.wikipedia.org/wiki/Robbins_algebra&quot; rel=&quot;nofollow&quot;&gt;Robbins conjecture&lt;/a&gt; was first proved by an automated theorem prover. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And, of course, there's Wikipedia on &lt;a href=&quot;http://en.wikipedia.org/wiki/Computer-assisted_proof&quot; rel=&quot;nofollow&quot;&gt;computer-assisted proof&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2370" LastActivityDate="2011-01-18T05:08:58.173" />
  <row Id="17945" PostTypeId="2" ParentId="17943" CreationDate="2011-01-18T05:09:35.160" Score="3" ViewCount="" Body="&lt;p&gt;No, the dimensions don't match.  Qx is a column vector and $x^TQ$ is a row vector.  But you are close-what can you do to fix this?&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-01-18T05:09:35.160" />
  <row Id="17946" PostTypeId="2" ParentId="17943" CreationDate="2011-01-18T05:10:19.653" Score="1" ViewCount="" Body="&lt;p&gt;We know that $Qx = Q^{T}x$. That is all we know. But $Q^{T}x$ is not necessarily equal to $x^{T}Q$ since the dimensions don't match. &lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2011-01-18T05:10:19.653" />
  <row Id="17948" PostTypeId="2" ParentId="1985" CreationDate="2011-01-18T06:02:31.290" Score="1" ViewCount="" Body="&lt;p&gt;These are my choices:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Interactive software: &lt;a href=&quot;http://ipe7.sourceforge.net/&quot; rel=&quot;nofollow&quot;&gt;Ipe&lt;/a&gt; produces postscript and pdf files, can include LaTeX labels and other text, and has helpful snapping modes. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Non-interactive: &lt;a href=&quot;http://www.eukleides.org/&quot; rel=&quot;nofollow&quot;&gt;Eukleides&lt;/a&gt; lets you design pretty high quality geometry diagrams with a simple language. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="3018" LastActivityDate="2011-01-18T06:02:31.290" CommunityOwnedDate="2011-01-18T06:02:31.290" />
  <row Id="17949" PostTypeId="2" ParentId="17904" CreationDate="2011-01-18T06:10:23.003" Score="2" ViewCount="" Body="&lt;p&gt;This can be seen as a special case of the matrix equation $AXB = C$, which is equivalent to $(B^T \otimes A) \operatorname{vec} X = \operatorname{vec} C$, where $B^T \otimes A$ is a &lt;a href=&quot;http://en.wikipedia.org/wiki/Kronecker_product&quot; rel=&quot;nofollow&quot;&gt;Kronecker product&lt;/a&gt; $\operatorname{vec} X$ is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Vectorization_%28mathematics%29&quot; rel=&quot;nofollow&quot;&gt;vectorization&lt;/a&gt; of $X$ (the columns stacked on top of each other).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The equations $AXB = C$ and $(B^T \otimes A) \operatorname{vec} X = \operatorname{vec} C$ can be seen to be equivalent by looking at the columns of $AXB$. Take the $k$:th column, $(AXB)_k$, let $B$ and $X$ have columns $b_k$ and $x_k$ respectively:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(AXB)_k = AXb_k = A \sum_{i=1}^n x_i b_{i,k} = (A b_{1,k}, \dots, Ab_{n,k}) (x_1, \dots, x_n)^T$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, vectorize this (put the columns of top of each other), and you get the result.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case we have $A^TBA = B$ and we get $(A^T \otimes A^T) \operatorname{vec}B = \operatorname{vec}B$, so $B$ has to be an eigenvector of $(A^T \otimes A^T) = (A \otimes A)^T$ with eigenvalue 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, all the eigenvalues of a Kronecker product are products of the eigenvalues of the constituent matrices. Thus, $A^T$ and therefore $A$ has to have at least two eigenvalues $\lambda, \mu$ such that $\lambda\mu = 1$ ($\lambda = \mu$ is ofc ok).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $(\lambda_i, \mu_i)$ for $i = 1, \dots, k$ be all the ordered pairs of eigenvalues to $A$ such that $\lambda_i \mu_i = 1$ and let $(x_i, y_i)$ be the corresponding eigenvectors to $A^T$. Then the following is true:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\operatorname{vec}B = \sum_{i=1}^k \alpha_i x_i \otimes y_i$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and this gives all the $\operatorname{vec}B$ that satisfies the equation. We can also write this on matrix form, $x_i$ forming columns in a matrix and $y_i$ forming rows in another matrix:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$B = \sum_{i=1}^k \alpha_i x_iy_i^T = (x_1, x_2, \dots, x_k)\operatorname{diag}(\alpha_1, \dots, \alpha_n) (y_1, y_2, \dots, y_k)^T$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can here see that $B$ at most can have the same rank as the number of pairs of eigenvalues $(\lambda_i, \mu_i)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\operatorname{rank} B \leq \operatorname{rank} A$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To guarantee a solution, the existence of an eigenvalue $\lambda$ such that $\lambda^2 = 1$ is enough. For $B$ to be invertible, $A$ has to have eigenvalues whose absolute values are all 1, so the absolute value of the determinant has to be 1 (converse to the result stated in the question!).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The idea seems to be also pretty easy to verify in the original equation $A^TBA = B$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is tempting to compare this with the SVD of $B$, but there is a couple of things in the way. $\alpha_i$ can be taken negatively. In some cases the minus sign can be but in one of the vectors instead, but since it's possible that an eigenvector can repeat, this might prove problematic. Second, the eigenvector-sets $\{x_i\}$ and $\{y_i\}$ are not necessarily orthogonal sets, which they need to be.&lt;/p&gt;&#xA;" OwnerUserId="5966" LastEditorUserId="5966" LastEditorDisplayName="" LastEditDate="2011-01-18T14:54:29.330" LastActivityDate="2011-01-18T14:54:29.330" />
  <row Id="17950" PostTypeId="1" AcceptedAnswerId="17956" CreationDate="2011-01-18T06:18:39.480" Score="1" ViewCount="60" Body="&lt;p&gt;Let $V,U$ vector subspaces of $\mathbb{C}^{n+1}$ of dimension $r+1$, $s+1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $\pi : \mathbb{C}^{n+1}/\{0\} \longrightarrow \mathbb{CP}^n$ the projection map to complex projective space defined by $\pi:(x_0, ... ,x_n)\mapsto(x_0:...:x_n)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a question that came up while doing homework and I wanted to check some reasoning.  It is useful to have that $\pi(V)\cap\pi(U) = \pi(V\cap U)$ for a homework question.  Is this true?  It seemed to be true based on my currently poor intuition about projective space, so I tried to prove it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My reasoning for a simple proof was to prove both inclusions. Specifically, if $u\in \pi(U\cap V)$ then there is some $x\in U\cap V$ st $u=\pi(x)$.  But then $x$ is in both $U$ and $V$ so $u\in \pi(U)\cap\pi(V)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Going in the other direction, if $u\in \pi(U)\cap\pi(V)$ then there is $x\in U$ and $y\in V$ such that $\pi(x)=\pi(y) = u$.  But if this is true then $x$ is a scalar multiple of $y$ and vis versa (by the properties of vector subspaces), so $x\in U\cap V$.  Thus $u\in\pi(U\cap V)$ and we have proved the sets are equivalent.&lt;/p&gt;&#xA;" OwnerUserId="565" LastEditorUserId="565" LastEditorDisplayName="" LastEditDate="2011-01-18T06:30:30.863" LastActivityDate="2011-01-18T06:40:38.290" Title="Projections from $\mathbb{C}^{n+1}$ to $\mathbb{CP}^n$. (check my reasoning)" Tags="&lt;homework&gt;&lt;projective-space&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="17951" PostTypeId="1" AcceptedAnswerId="17958" CreationDate="2011-01-18T06:22:57.803" Score="1" ViewCount="173" Body="&lt;p&gt;Show that if A has a right-inverse, then $Ax = b$ has at least one solution for every choice of b in $R^n$.&lt;/p&gt;&#xA;" OwnerUserId="5157" LastActivityDate="2011-01-18T07:07:43.693" Title="Show that if A has a right-inverse, then $Ax = b$ has at least one solution for every choice of b in $R^n$." Tags="&lt;linear-algebra&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="17953" PostTypeId="2" ParentId="17951" CreationDate="2011-01-18T06:33:39.753" Score="2" ViewCount="" Body="&lt;p&gt;Suppose $A$ has a right inverse, so there is a $B$ such that $AB=I$. Then $ABb=Ib=b$ for any $b\in{\mathbb R}^n$. Letting $x=Bb$ gives us the result.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I guess it is worth mentioning that we really didn't use any linear algebra. A similar argument gives that if $A$ has a left inverse, then the null space of $A$ is trivial (${0}$). In fact, these two facts hold for arbitrary functions, not just linear maps (in the second case, the general statement is that $A$ is injective). &lt;/p&gt;&#xA;" OwnerUserId="462" LastEditorUserId="462" LastEditorDisplayName="" LastEditDate="2011-01-18T07:07:43.693" LastActivityDate="2011-01-18T07:07:43.693" CommentCount="2" />
  <row Id="17954" PostTypeId="2" ParentId="17951" CreationDate="2011-01-18T06:34:05.680" Score="1" ViewCount="" Body="&lt;p&gt;Let $A \in \mathbb{R}^{m \times n}$. $A$ has a right-inverse if $A$ is full rank and is a square or fat matrix ($m \leq n$). Let $A^{\dagger} \in \mathbb{R}^{n \times m}$ be the right inverse of $A$ i.e. $A \times A^{\dagger} = I_{m \times m}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take any vector $b \in \mathbb{R}^{m \times 1}$. The claim is $x = A^{\dagger} b$ is always a solution to the system $Ax = b$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This can be easily seen since $$Ax = A \times (A^{\dagger} b) = (A \times A^{\dagger})b = Ib =b$$ where the intermediate step follows from associativity of matrix multiplication.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that if $A$ is a strictly fat full rank matrix, this is not the unique solution since $x = A^{\dagger}b + z$ where $z \in \text{Nullspace}(A)$ is also solution. And since $A$ is a strictly fat matrix, we are assured to have a non-trivial Nullspace.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, for a strictly fat full rank matrix, there will be more than one right inverse. So in fact $A^{\dagger}$ itself is not unique.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2011-01-18T06:48:23.137" LastActivityDate="2011-01-18T06:48:23.137" CommentCount="1" />
  <row Id="17955" PostTypeId="1" AcceptedAnswerId="17968" CreationDate="2011-01-18T06:35:53.150" Score="6" ViewCount="168" Body="&lt;p&gt;I was telling someone about the smash product and he asked whether it was the categorical product in the category of based spaces and I immediately said yes, but after a moment we realized that that wasn't right.  Rather, the categorical product of $(X,x_0)$ and $(Y,y_0)$ is just $(X\times Y,(x_0,y_0))$.  (It seems like in any concrete category $(\mathcal{C},U)$, if we have a product (does a concrete category always have products?) then it must be that $U(X\times Y)=U(X)\times U(Y)$.  But I couldn't prove it.  I should learn category theory.  Maybe functors commute with products or something.)  Anyways, here's what I'm wondering: is the main reason that we like the smash product just that it gives the right exponential law?  It's easy to see that the product $\times$ I gave above has $F(X\times Y,Z)\not\cong F(X,F(Y,Z))$ just by taking e.g. $X=Y=Z=S^0$.&lt;/p&gt;&#xA;" OwnerUserId="2159" LastActivityDate="2011-03-24T23:14:34.040" Title="Why do we use the smash product in the category of based topological spaces?" Tags="&lt;algebraic-topology&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="17956" PostTypeId="2" ParentId="17950" CreationDate="2011-01-18T06:40:38.290" Score="2" ViewCount="" Body="&lt;p&gt;Yes, this is true, and your reasons are correct.  Both describe the set of one dimensional subspaces minus the origin contained in both $U$ and $V$.  Since fibers of $\pi$ are one dimensional subspaces minus the origin, any fiber that intersects a subspace must be contained in that subspace.  Note that $f(A\cap B)\subseteq f(A)\cap f(B)$ is always true for the same reason you gave that $\pi(U\cap V)\subseteq\pi(U)\cap\pi(V)$.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2011-01-18T06:40:38.290" />
  <row Id="17957" PostTypeId="2" ParentId="17955" CreationDate="2011-01-18T06:59:02.473" Score="2" ViewCount="" Body="&lt;p&gt;From &lt;a href=&quot;http://ncatlab.org/nlab/show/smash+product&quot; rel=&quot;nofollow&quot;&gt;nLab&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The smash product is the tensor product in the closed monoidal category of pointed sets. That is,&#xA;  $$\operatorname{Fun}_*(A\wedge B,C)\cong \operatorname{Fun}_*(A,\operatorname{Fun}_*(B,C))$$&#xA;  Here, $\operatorname{Fun}_*(A,B)$ is the set of basepoint-preserving functions from $A$ to $B$, itself made into a pointed set by taking as basepoint the constant function from all of $A$ to the basepoint in $B$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There's more at the link.  I must admit that I know nothing about this, but I recommend nLab as a good place to look for the categorical place of mathematical constructions.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2011-01-18T06:59:02.473" CommentCount="3" />
  <row Id="17958" PostTypeId="2" ParentId="17951" CreationDate="2011-01-18T07:01:42.223" Score="4" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\ $ Right-invertible maps $\rm\:f\:$ are surjective since $\rm\ f\: (f^{-1}(x)) = x\:.\ $ The converse holds true too - it is &lt;a href=&quot;http://en.wikipedia.org/wiki/Surjective_function#Surjections_as_right_invertible_functions&quot; rel=&quot;nofollow&quot;&gt;equivalent&lt;/a&gt; to the axiom of choice.&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2011-01-18T07:01:42.223" CommentCount="1" />
  <row Id="17959" PostTypeId="2" ParentId="17935" CreationDate="2011-01-18T07:08:18.813" Score="2" ViewCount="" Body="&lt;p&gt;Many of your questions are adressed in this new book of Jon Borwein and Keith Devlin:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://books.google.de/books?id=B8Crn_VF9RoC&amp;amp;lpg=PP1&amp;amp;ots=0FLkjZPQ2T&amp;amp;dq=the%20computer%20as%20crucible&amp;amp;hl=en&amp;amp;pg=PP1#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;http://books.google.de/books?id=B8Crn_VF9RoC&amp;amp;lpg=PP1&amp;amp;ots=0FLkjZPQ2T&amp;amp;dq=the%20computer%20as%20crucible&amp;amp;hl=en&amp;amp;pg=PP1#v=onepage&amp;amp;q&amp;amp;f=false&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="346" LastActivityDate="2011-01-18T07:08:18.813" />
  <row Id="17960" PostTypeId="1" CreationDate="2011-01-18T07:13:33.430" Score="3" ViewCount="125" Body="&lt;p&gt;Suppose I chose two rational functions, say, &#xA;$$u = \frac{t(4+t)^5}{(1+4t)^5}, \qquad v = \frac{t^5 (4+t)}{(1+4t)}.$$&#xA;Then I know that $K(X):= \mathbf{C}(u,v)$ is the function field of the projective line (Proof: If $K(Y):=\mathbf{C}(t)$, then there is an inclusion $K(Y) \subseteq K(X)$, and hence a surjection $Y \simeq \mathbf{P}^1 \rightarrow X$, and so $X$ must have genus $0$.)&#xA;From this follows that $K(X) \simeq \mathbf{C}(s)$ for some $s \in \mathbf{C}(t)$. Is there a practical easy algorithm to explicitly construct such an $s$, and, moreover, write $s$ as a rational function of $X$ and $Y$?&#xA;Is there an easy way at least to determine the degree of the map $Y \rightarrow X$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In case you were wondering, the specific choice of $X$ and $Y$ was motivated by the question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://mathoverflow.net/questions/50804/deciding-whether-a-given-power-series-is-modular-or-not&quot; rel=&quot;nofollow&quot;&gt;http://mathoverflow.net/questions/50804/deciding-whether-a-given-power-series-is-modular-or-not&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5972" LastActivityDate="2011-01-18T07:13:33.430" Title="Function field of the projective line." Tags="&lt;algebraic-geometry&gt;" />
  <row Id="17961" PostTypeId="2" ParentId="1985" CreationDate="2011-01-18T07:25:55.923" Score="2" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://sourceforge.net/projects/pgf/&quot; rel=&quot;nofollow&quot; title=&quot;TikZ&quot;&gt;Tikz&lt;/a&gt; is a nice LaTeX package for easily drawing diagrams. Diagrams are made by putting code directly into the TeX document, eliminating the need for extra image files. The package also is very powerful and versatile; the &lt;a href=&quot;http://www.ctan.org/tex-archive/graphics/pgf/base/doc/generic/pgf/pgfmanual.pdf&quot; rel=&quot;nofollow&quot;&gt;manual&lt;/a&gt; contains a very detailed description of its features.&lt;/p&gt;&#xA;" OwnerUserId="30" LastActivityDate="2011-01-18T07:25:55.923" CommunityOwnedDate="2011-01-18T07:25:55.923" />
  <row Id="17962" PostTypeId="1" AcceptedAnswerId="17963" CreationDate="2011-01-18T08:18:00.277" Score="2" ViewCount="153" Body="&lt;p&gt;Is there a compactly supported and smooth (except at one point) function $f : \mathbb{R} \to \mathbb{R}$ where (non smooth point) it is is exactly k times differentiable ?&lt;/p&gt;&#xA;" OwnerUserId="2987" LastActivityDate="2011-01-18T08:54:03.443" Title="Is there a compactly supported smooth function which is exactly k times differentiable at exactly one point ?" Tags="&lt;real-analysis&gt;" AnswerCount="1" FavoriteCount="2" />
  <row Id="17963" PostTypeId="2" ParentId="17962" CreationDate="2011-01-18T08:23:01.883" Score="7" ViewCount="" Body="&lt;p&gt;Yes. Just take $f(x)=0$ for $x&amp;lt;0$, $f(x)=x^{k+1}$ for $x \ge 0$, and multiply by a &lt;a href=&quot;http://en.wikipedia.org/wiki/Non-analytic_smooth_function#Smooth_transition_functions&quot; rel=&quot;nofollow&quot;&gt;smooth transition function&lt;/a&gt; (say one that is $1$ for $x&amp;lt;1$ and $0$ for $x&gt;2$) to make it compactly supported.&lt;/p&gt;&#xA;" OwnerUserId="5203" LastEditorUserId="5203" LastEditorDisplayName="" LastEditDate="2011-01-18T08:54:03.443" LastActivityDate="2011-01-18T08:54:03.443" CommentCount="13" />
  <row Id="17964" PostTypeId="2" ParentId="17932" CreationDate="2011-01-18T08:27:04.110" Score="1" ViewCount="" Body="&lt;p&gt;Let's try another way to solve this problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Construct a contour, which consists of two parts. The first part is a circle and is a little bit larger than the unit circle, except near the point $z_0$. We call it $C_1$, and make it have absolute value strictly larger than $1+\delta$ for some $delta$. The second part is a small circle with radius $\epsilon$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose $z_0$ as pole have degree k, then $f(\zeta)=\frac{g(z)}{(z-z_0)^k}$, where $g(z)$ is holomorhpic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $C_1$, we have that $|\int_{C_1}\frac{g(\zeta)}{\zeta^n}\frac{1}{(\zeta-z_0)^k}d\zeta|\leq \frac{1}{\epsilon^k}\int_{C_1}\frac{M}{(1+\delta)^n}d\zeta \to 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $C_\epsilon$, we have &#xA;$$\int_{C_\epsilon}\frac{g(\zeta)}{\zeta^{n+1}}\frac{1}{(\zeta-z_0)^k}d\zeta=\int_{-\theta_0}^{-\pi+\theta_0} \frac{g(z_0+\epsilon e^\theta_0)}{(z_0+\epsilon e^\theta_0)^{n+1}}e^{-i\theta k} d\theta$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the same way,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_{C_\epsilon}\frac{g(\zeta)}{\zeta^{n+2}}\frac{1}{(\zeta-z_0)^k}d\zeta=\int_{-\theta_0}^{-\pi+\theta_0} \frac{g(z_0+\epsilon e^\theta_0)}{(z_0+\epsilon e^\theta_0)^{n+2}}e^{-i\theta k} d\theta$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By multiplying the second one with $z_0$, and computing the difference, we get,  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \Delta=\int_{-\theta_0}^{-\pi+\theta_0} \frac{g(z_0+\epsilon e^\theta_0)}{(z_0+\epsilon e^\theta_0)^{n+2}}e^{-i\theta k} \epsilon e^{i\theta}d\theta \to 0$$, as $\epsilon \to 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which means &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \frac{\int_{C_\epsilon}\frac{g(\zeta)}{\zeta^{n+1}}\frac{1}{(\zeta-z_0)^k}d\zeta}{z_0\int_{C_\epsilon}\frac{g(\zeta)}{\zeta^{n+2}}\frac{1}{(\zeta-z_0)^k}d\zeta} \to 1 $$ as $\epsilon \to 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Combining all of these and Cauchy's integral formuals that $a_0=f(0)=\frac{1}{2\pi i}\int_C\frac{f(\zeta)}{\zeta}d\zeta$ and $n!a_n=f^{(n)}(0)=\frac{n!}{2\pi i}\int_C\frac{f(\zeta)}{\zeta^{n+1}}d\zeta$, we split $C$ as $C_1$ and $C_\epsilon$, we onle need to choose carefully the $\epsilon$'s and $\delta$'s to complete our proof.&lt;/p&gt;&#xA;" OwnerUserId="5476" LastActivityDate="2011-01-18T08:27:04.110" />
  <row Id="17965" PostTypeId="2" ParentId="17908" CreationDate="2011-01-18T08:39:05.113" Score="0" ViewCount="" Body="&lt;p&gt;Here is another nice exercise which also results in the desired conclusion:&#xA;Let $A$ be an $n\times m$ matrix and $B$ be an $m \times n$ matrix. If $m &amp;lt; n$ then $\det(AB) = 0$ (note that the argument is pretty much the same as Arturo's answer. All you need is that the image of a linear map has at most the same dimension as the domain) &lt;/p&gt;&#xA;" OwnerUserId="2538" LastActivityDate="2011-01-18T08:39:05.113" />
  <row Id="17966" PostTypeId="1" AcceptedAnswerId="17973" CreationDate="2011-01-18T09:53:50.230" Score="1" ViewCount="129" Body="&lt;p&gt;How can we sum up $\sin$ and $\cos$ series when the angles are in A.P ?For example here is the sum of $\cos$ series:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\large \sum_{k=0}^{n-1}\cos (a+k \cdot d) =\frac{\sin(n \times \frac{d}{2})}{\sin ( \frac{d}{2} )} \times \cos \biggl( \frac{ 2 a + (n-1)\cdot d}{2}\biggr)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a slight difference in case of $\sin$ ,which is: $$\large \sum_{k=0}^{n-1}\sin (a+k \cdot d) =\frac{\sin(n \times \frac{d}{2})}{\sin ( \frac{d}{2} )} \times \sin\biggl( \frac{2  a + (n-1)\cdot d}{2}\biggr)$$ what I want to know is that how do we get these results?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks,&lt;/p&gt;&#xA;" OwnerUserId="2109" LastEditorUserId="2109" LastEditorDisplayName="" LastEditDate="2011-01-18T10:46:51.727" LastActivityDate="2011-01-20T14:02:21.420" Title="How can we sum up $\sin$ and $\cos$ series when the angles are in A.P ?" Tags="&lt;trigonometry&gt;" AnswerCount="2" CommentCount="7" />
  <row Id="17967" PostTypeId="2" ParentId="17966" CreationDate="2011-01-18T10:04:20.913" Score="8" ViewCount="" Body="&lt;p&gt;Writing $\cos x = \frac12 (e^{ix} + e^{-ix})$ will reduce the problem to computing two geometric sums.&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2011-01-18T10:04:20.913" CommentCount="4" />
  <row Id="17968" PostTypeId="2" ParentId="17955" CreationDate="2011-01-18T10:20:04.583" Score="4" ViewCount="" Body="&lt;p&gt;Two reasons why we want the smash product:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;There is a natural homeomorphism $(X\times Y)^+ \cong X^+ \wedge Y^+$ for any spaces $X$ and $Y$. This is one of the places that the smash product naturally arises- you want to describe the compactification of a product in terms of each of the factors, how do you do it? It turns out the smash product is the best way to answer the question. In particular, using slightly more suggestive notation, this gives us the result $S^V \wedge S^W \cong S^{V \oplus W}$ for vector spaces $V$ and $W$... look familiar?&lt;/li&gt;&#xA;&lt;li&gt;As Jonas said, the smash product is the analog of the tensor product. Recall that the tensor product for $R$-modules is not the categorical product, but it &lt;em&gt;is&lt;/em&gt; left adjoint to the Hom functor. This gives it all sorts of nice properties; it's one of the reasons why there is a nice duality between Tor and Ext, and all sorts of other nice stuff. However: I don't think that this analogy is incredibly useful until you move to the stable category... there the smash product is much closer to an honest tensor product, since you're working with an additive category. You can start making sense of what it means to &quot;smash over a ring spectrum&quot; just as one &quot;tensors over a ring&quot; (at least I think you can; you know more about stable stuff than I do.)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;As a side note, in response to your parenthetical statement: If you want to know when some functor preserves products or coproducts or some type of limit, it's usually easiest to first check and see if it has an adjoint. See wikipedia on adjoints and (co)continuous functors.&lt;/p&gt;&#xA;" OwnerUserId="423" LastActivityDate="2011-01-18T10:20:04.583" CommentCount="1" />
  <row Id="17969" PostTypeId="1" CreationDate="2011-01-18T10:32:19.163" Score="4" ViewCount="240" Body="&lt;p&gt;Dear all, next year, I will probably teach a one-semester course of Differential Geomtry of curves and surfaces. Its content must be something along the lines of the first four chapters of Do Carmo's famous book and my plan is indeed to follow it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I would like to know some other, more modern, references. Particularly some including material for Matlab or Maple. I'm already aware of the book of John Oprea: any other references?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would also be greatful to know about some links to webpages with this subject.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any hints, ideas, references, suggestions... are welcome. Thank you in advance.&lt;/p&gt;&#xA;" OwnerUserId="664" LastActivityDate="2011-03-10T09:34:45.963" Title="Differential Geometry of curves and surfaces: bibliography?" Tags="&lt;reference-request&gt;&lt;differential-geometry&gt;&lt;teaching&gt;" AnswerCount="5" CommentCount="2" FavoriteCount="3" />
  <row Id="17970" PostTypeId="1" AcceptedAnswerId="17980" CreationDate="2011-01-18T10:41:22.380" Score="0" ViewCount="157" Body="&lt;p&gt;Numerically solving $y&amp;#39;=f(x,y)$ with $k_1=f(x_n,y_n),&#xD;&#xA;k_2=f(x_n+c_2*h, y_n+c_2*h*k_1),&#xD;&#xA;y_{n+1}=y_n+h(b_1*k_1+b_2*k_2)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;what would the local truncation error be?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, how would we perform a Taylor expansion on this to show that it is second order, provided $b_1+b_2=1$ and $b_2*c_2=\frac{1}{2}$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many thanks.&lt;/p&gt;&#xA;" OwnerUserId="4522" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2011-01-18T14:21:54.130" LastActivityDate="2011-01-20T20:34:31.467" Title="Numerical Solutions of ordinary differential equations" Tags="&lt;numerical-methods&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" />
  <row Id="17971" PostTypeId="1" CreationDate="2011-01-18T11:05:06.703" Score="3" ViewCount="111" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have two sets of ranked data which are ranked on different scales, one fixed and one variable, and I would like to convert one of the data sets ranking values to match the other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Set 1 is ranked 10 - 1 with 10 being the highest rank.&#xA;Set 2 is ranked 1 - x with 1 being the highest rank and x being the number of items in the set.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it possible to convert the ranking of set 2 to fit into set 1?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks,&#xA;Phil&lt;/p&gt;&#xA;" OwnerUserId="5973" LastEditorUserId="462" LastEditorDisplayName="" LastEditDate="2011-01-22T21:12:47.843" LastActivityDate="2011-04-22T21:42:21.687" Title="Merging two differently ranked data sets" Tags="&lt;algorithms&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="17972" PostTypeId="1" AcceptedAnswerId="17974" CreationDate="2011-01-18T11:14:26.833" Score="2" ViewCount="182" Body="&lt;p&gt;On the number line, starting from 0. There is a probability of $p$ of moving 1 unit to the positive direction, and $1-p$ of moving 1 unit to the negative direction. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the probability the walk never reaches $-1$ after $n$ steps?&lt;/p&gt;&#xA;" OwnerUserId="96" LastEditorUserId="96" LastEditorDisplayName="" LastEditDate="2011-01-18T11:22:28.903" LastActivityDate="2011-01-19T07:15:42.027" Title="Probability of never reach -1 for a 1D random walk of n steps" Tags="&lt;probability&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="17973" PostTypeId="2" ParentId="17966" CreationDate="2011-01-18T11:31:35.700" Score="5" ViewCount="" Body="&lt;p&gt;Let $$ S = \sin{(a)} + \sin{(a+d)} + \cdots + \sin{(a+nd)}$$ Now multiply both sides by $\sin\frac{d}{2}$. Then you have $$S \times \sin\Bigl(\frac{d}{2}\Bigr) = \sin{(a)}\sin\Bigl(\frac{d}{2}\Bigr) + \sin{(a+d)}\cdot\sin\Bigl(\frac{d}{2}\Bigr) + \cdots + \sin{(a+nd)}\cdot\sin\Bigl(\frac{d}{2}\Bigr)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, note that $$\sin(a)\sin\Bigl(\frac{d}{2}\Bigr) = \frac{1}{2} \cdot \biggl[ \cos\Bigl(a-\frac{d}{2}\Bigr) - \cos\Bigl(a+\frac{d}{2}\Bigr)\biggr]$$ and $$\sin(a+d) \cdot \sin\Bigl(\frac{d}{2}\Bigr) = \frac{1}{2} \cdot \biggl[ \cos\Bigl(a + d -\frac{d}{2}\Bigr) - \cos\Bigl(a+d+\frac{d}{2}\Bigr) \biggr]$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then by doing the same thing you will have some terms cancelled out. You can easily see which terms are going to get Cancelled. Proceed and you should be able to get the formula. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried this by seeing this post. This has been worked for the case when $d=1$. Just take a look here:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://mathforum.org/library/drmath/view/72777.html&quot; rel=&quot;nofollow&quot;&gt;http://mathforum.org/library/drmath/view/72777.html&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" LastEditorUserId="2109" LastEditorDisplayName="" LastEditDate="2011-01-20T14:02:21.420" LastActivityDate="2011-01-20T14:02:21.420" CommentCount="1" />
  <row Id="17974" PostTypeId="2" ParentId="17972" CreationDate="2011-01-18T12:10:04.430" Score="4" ViewCount="" Body="&lt;p&gt;Let $P_n$ be the probability that the walk has still not visited $-1$ after $n$ steps. If $n$ is even, we can't be at $-1$, so $P_{n} = P_{n-1}$ for even $n$. Assume from now on that $n = 2m+1$ is odd.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $0 \le r \le m$, let $P_{n,r}$ denote the probability that after $n=2m+1$ steps, the walk is at position $2r+1$ and has still not visited $-1$. We can count the number $N_{n,r}$ of such walks, as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$N_{n,r} = T_{n,r} - V_{n,r}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $T_{n,r}$ is the total number of walks ending at $2r+1$ (including those that visit $-1$), and $V_{n,r}$ is the number of walks ending at $2r+1$ that do visit $-1$ on the way. A walk that ends at $2r+1$ consists of $m+r+1$ positive steps and $m-r$ negative steps, so $T_{n,r} = \binom{n}{m-r}$.&#xA;To evaluate $V_{n,r}$, we need the following observation:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a 1-1 correspondence between walks ending at $2r+1$ that visit $-1$ on the way, and walks ending at $-(2r+3)$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For, given a walk ending at $2r+1$ that visits $-1$, we can reflect the portion of it that occurs after its first visit to $-1$, to obtain a walk that ends at $-(2r+3)$. And vice versa.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A walk that ends at $-(2r+3)$ consists of $m-r-1$ positive steps and $m+r+2$ negative steps, so $V_{n,r} = \binom{n}{m-r-1}$, giving $N_{n,r} = \binom{n}{m-r} - \binom{n}{m-r-1}$. (If $r = m$, then we understand the second binomial term as zero.) Thus:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$P_{n,r} = p^{m+r+1}q^{m-r}\left(\binom{n}{m-r} - \binom{n}{m-r-1}\right)$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $q = 1-p$. So the total probability of not visiting $-1$ in $n$ steps is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$P_n = \sum_{r=0}^m P_{n,r}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a modification of the binomial distribution. I don't think it can be simplified any further, unless you want to use &lt;a href=&quot;http://en.wikipedia.org/wiki/Beta_function&quot; rel=&quot;nofollow&quot;&gt;regularised incomplete beta functions&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1508" LastEditorUserId="1508" LastEditorDisplayName="" LastEditDate="2011-01-19T07:15:42.027" LastActivityDate="2011-01-19T07:15:42.027" CommentCount="3" />
  <row Id="17975" PostTypeId="2" ParentId="17969" CreationDate="2011-01-18T12:34:26.407" Score="2" ViewCount="" Body="&lt;p&gt;If you can use Mathematica, see &lt;a href=&quot;http://www.wolfram.com/books/profile.cgi?id=6280&quot; rel=&quot;nofollow&quot;&gt;Modern Differential Geometry of Curves and Surfaces with Mathematica&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="589" LastActivityDate="2011-01-18T12:34:26.407" />
  <row Id="17976" PostTypeId="1" CreationDate="2011-01-18T12:47:29.740" Score="11" ViewCount="233" Body="&lt;p&gt;When searching a number on Wolfram Alpha, one of the results is its &lt;em&gt;representation&lt;/em&gt;.&lt;br&gt;&#xA;For example, for &lt;a href=&quot;http://www.wolframalpha.com/input/?i=8549&quot; rel=&quot;nofollow&quot;&gt;8549&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;8549 has the representation 8549 = $5&amp;#183;2^6&amp;#183;3^3-91$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Similarly for &lt;a href=&quot;http://www.wolframalpha.com/input/?i=75290&quot; rel=&quot;nofollow&quot;&gt;75290&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;75290 has the representation 75290 = $3&amp;#183;2^9&amp;#183;7^2+26$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What is the significance of these representations?&lt;/p&gt;&#xA;" OwnerUserId="547" LastActivityDate="2011-01-18T14:29:31.707" Title="What is the meaning of a number's &quot;representation&quot; on Wolfram Alpha?" Tags="&lt;wolframalpha&gt;" AnswerCount="2" CommentCount="6" FavoriteCount="4" />
  <row Id="17977" PostTypeId="1" CreationDate="2011-01-18T12:57:50.817" Score="0" ViewCount="113" Body="&lt;p&gt;Is it possible to write $\int_{-\infty}^\infty e^{-iqu}\left( \frac{A}{\sqrt{1-iau}}+\frac{B}{\sqrt{1-ibu}}\right)^Ndu$ in this form $\int\frac{e^{-iu}}{(\sqrt{1-iau})^n(\sqrt{1-ibu})^m}du$&lt;/p&gt;&#xA;" OwnerUserId="5978" LastEditorUserId="2548" LastEditorDisplayName="" LastEditDate="2011-01-18T14:40:12.180" LastActivityDate="2011-01-18T15:13:05.397" Title="Writing an Integral in Different Form?" Tags="&lt;integral&gt;&lt;transformation&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="17978" PostTypeId="1" CreationDate="2011-01-18T13:37:51.623" Score="2" ViewCount="128" Body="&lt;p&gt;Can you help me with this? &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;For $\lambda \in \mathbb{C}$, show that there exist $m,n \in \mathbb{Z}$ large enough (depending in $\lambda$) such that the equation $e^z = z+\lambda$ has exactly $m+n$ solution in $\{z\mid -2m \pi  &amp;lt; \Im(z) &amp;lt; 2n \pi\}$ (where $\Im(z)$ is the imaginary part of $z$).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Thanks a lot! Jon&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The idea is to use to use the Argument Theorem and to see what is the change of the arg.&lt;/p&gt;&#xA;" OwnerUserId="5979" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-18T16:07:28.313" LastActivityDate="2011-01-18T16:07:28.313" Title="complex number argument theory" Tags="&lt;complex-analysis&gt;&lt;complex-variables&gt;" CommentCount="1" />
  <row Id="17979" PostTypeId="1" AcceptedAnswerId="17983" CreationDate="2011-01-18T13:43:51.490" Score="1" ViewCount="67" Body="&lt;p&gt;Let V be the vector space of all real-valued bounded sequences. Then for $a,b \in V$ $\langle a,b \rangle :=\sum _{n=1}^{\infty } \frac{a(n) b(n)}{n^2}$ defines a dot product. Find a subspace $U \subset V$ with $U \neq 0, U \neq V, U^\bot=0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I couldn't find anything that works, thank you in advance. $U^\bot$ is the orthogonal complement of $U$, in case the notation is confusing.&lt;/p&gt;&#xA;" OwnerUserId="3123" LastActivityDate="2011-01-18T14:03:45.707" Title="Dot Product and Orthogonal Complement" Tags="&lt;homework&gt;&lt;sequences-and-series&gt;&lt;vector-space&gt;&lt;real-analysis&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="17980" PostTypeId="2" ParentId="17970" CreationDate="2011-01-18T13:45:22.367" Score="3" ViewCount="" Body="&lt;p&gt;The local truncation error of a numerical method is the error the method makes in one iteration. This means, to compute this error you assume that you know the exact solution $y(x_n)$ and you apply your method to compute an approximation for $y(x_{n+1})$. Let's call the approximation $y_{n+1}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can rewrite your exact solution $y(x)$ using a Taylor expansion. Because you're interested in $y(x_{n+1}) = y(x_n + h)$, you expand around $x_n$ and then evaluate at $x_n + h$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The expansion is this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ T_y (x) = y(x_n) + (x - x_n)*f(x_n, y(x_n)) + \frac{(x - x_n)^2}{2}*\frac{\partial}{\partial x}f(x_n, y(x_n)) + ...$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that $x_n$ is a constant. Now evaluate $T_y$ at $x_n + h$ to obtain:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ T_y (x_n + h) = y(x_n) + h*f(x_n, y(x_n)) + \frac{h^2}{2}*\frac{\partial}{\partial x}f(x_n, y(x_n)) + ...$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that $T_y (x_n + h) = y(x_n + h)$, i.e. the Taylor expansion is identical with the exact solution $y$ assuming $h$ is small.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To calculate the local truncation error you just need to compute the difference between $y_{n + 1}$ and $T_y (x_n + h)$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ e := T_y (x_n + h) - y_{n+1} = $$&#xA;$$(y(x_n) + h*f(x_n, y(x_n)) + \frac{h^2}{2}*\frac{\partial}{\partial x}f(x_n, y(x_n)) + ...) - (y_n + h*(b_1 k_1 + b_2 k_2))$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You use $y_n = y(x_n)$, your assumption, to obtain:&#xA;$$ e = h(f(x_n, y_n) - b_1f(x_n, y_n) - b_2 f(x_n + c_2 h, y_n + c_2 h k_1)) + \frac{h^2}{2}*\frac{\partial}{\partial x}f(x_n, y_n) + ... $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $f(x_n, y_n) - b_1f(x_n, y_n) - b_2 f(x_n + c_2 h, y_n + c_2 h k_1)$ is non-zero, your truncation error is $O(h)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With $b_2 = 1 - b_1$ you get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ f(x_n, y_n) - b_1f(x_n, y_n) - b_2 f(x_n + c_2 h, y_n + c_2 h k_1) = b_2f(x_n, y_n) - b_2 f(x_n + c_2 h, y_n + c_2 h k_1)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So $$ e = b_2 h (f(x_n, y_n) - f(x_n + c_2 h, y_n + c_2 h k_1)) + \frac{h^2}{2}*\frac{\partial}{\partial x}f(x_n, y_n) + ... $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To finish this, you do the Taylor expansion for $f$ at $(x_n, y_n)$ and use the last condition $b_2 c_2 = \frac{1}{2}$, then the coefficient of $h$ in $e$ vanishes and you see that it's order 2. You can find the details here:&#xA;&lt;a href=&quot;http://math.stackexchange.com/questions/18190/local-truncation-error-calculation&quot;&gt;Local truncation error calculation&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5798" LastEditorUserId="5798" LastEditorDisplayName="" LastEditDate="2011-01-20T20:34:31.467" LastActivityDate="2011-01-20T20:34:31.467" />
  <row Id="17981" PostTypeId="2" ParentId="17731" CreationDate="2011-01-18T13:46:10.863" Score="1" ViewCount="" Body="&lt;p&gt;As an simple example I would cite the &lt;a href=&quot;http://en.wikipedia.org/wiki/Seven_Bridges_of_K%C3%B6nigsberg&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;Seven Bridges of Königsberg&lt;/em&gt; problem&lt;/a&gt; solved by Euler. At that time, Euler was mailed by a politic of the city of Königsberg asking him to help in a problem of finding the &quot;touristically&quot; best walk trought some bridges of the city. Euler, at first, argued that such a problem was not a mathematical problem. But the politic insisted writing back to Euler saying that none engineer that was consulted had a conclusive answer. So, after reviewing the problem, Euler realized that the problem was asking for a solution that had nothing to do with geometry (I mean, a solution that is not metric related) as everybody was expecting. At the end, Euler proofed topologically the existence of a solution and giving rise to a new branch of mathematics, the topology.&lt;/p&gt;&#xA;" OwnerUserId="370" LastEditorUserId="10557" LastEditorDisplayName="" LastEditDate="2011-05-07T08:13:56.790" LastActivityDate="2011-05-07T08:13:56.790" CommunityOwnedDate="2011-01-18T13:46:10.863" />
  <row Id="17982" PostTypeId="2" ParentId="7221" CreationDate="2011-01-18T14:00:49.563" Score="2" ViewCount="" Body="&lt;p&gt;Dear DarthPickley,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;your question tackles some basic concepts of &lt;a href=&quot;http://en.wikipedia.org/wiki/General_relativity&quot; rel=&quot;nofollow&quot;&gt;general relativity&lt;/a&gt; and hence &lt;a href=&quot;http://en.wikipedia.org/wiki/Pseudo-Riemannian_manifold&quot; rel=&quot;nofollow&quot;&gt;pseudo-Riemannian manifolds&lt;/a&gt;. To answer the several parts your question, we will have to go a little into detail.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Limiting behaviour of general relativity&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;First of all, I have to confess that you somehow got the limiting behaviour of Einstein's theory of gravitation the other way round.&lt;br&gt;&#xA;&lt;strong&gt;Locally&lt;/strong&gt;, you will always be able to construct a &lt;a href=&quot;http://en.wikipedia.org/wiki/Minkowski_space&quot; rel=&quot;nofollow&quot;&gt;Minkowski spacetime&lt;/a&gt; with metric &#xA;$$\eta = -c^2 dt^2 + d\mathbf{r}^2$$&#xA;such that your original spacetime metric&#xA;$$g=\eta + \text{higher order terms}_{ij}dx^i dx^j$$&#xA;is kind of Taylor expanded locally. This linearization works pretty well for &quot;normal life&quot; gravitation and you can explain &lt;a href=&quot;http://en.wikipedia.org/wiki/Gravitational_wave&quot; rel=&quot;nofollow&quot;&gt;gravitational waves&lt;/a&gt; or get &lt;a href=&quot;http://en.wikipedia.org/wiki/Newton%27s_law_of_universal_gravitation&quot; rel=&quot;nofollow&quot;&gt;Newtonian gravitation&lt;/a&gt;. How good a linear approximation can be will tell you the value of the scalar Riemann curvature $R = R^a_{\,a}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For &lt;strong&gt;long-range&lt;/strong&gt; interactions, this picture fails as a Taylor expansion will be more and more inaccurate. Then, you have to take the full theory into account and can, under further assumptions, explain things corresponding to &lt;a href=&quot;http://en.wikipedia.org/wiki/Cosmology&quot; rel=&quot;nofollow&quot;&gt;cosmology&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Measuring angles&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;First of all, we have to make a &lt;strong&gt;simplification&lt;/strong&gt;. If you would assume that you want to measure the angles of your triangle on a dynamic spacetime, you would have some conceptual problems defining how to return to the point where you started, and, maybe worse: what is an angle then?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, it is convenient to assume that you want to do that in a &lt;strong&gt;stationary spacetime&lt;/strong&gt;. That means there exists a local isometry of the metric $\varphi$ and a corresponding &lt;a href=&quot;http://en.wikipedia.org/wiki/Killing_vector_field&quot; rel=&quot;nofollow&quot;&gt;Killing field&lt;/a&gt; $\xi = \dot{\varphi}(\tau)$ which is properly parametrized, say $g\left( \xi,\xi \right) = -c^2\,\forall\tau$, $$\varphi^\star g = g$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we can &lt;strong&gt;define&lt;/strong&gt; something like an ordinary &lt;strong&gt;angle&lt;/strong&gt; which I guess you had in mind. Consider the metric $$\tilde{g}\left(X,Y\right) = g\left(X,Y\right) - \frac{1}{g(\xi,\xi)}g\left(\xi,Y\right)g\left(\xi,X\right)$$&#xA;which only takes the spatial part of the scalar product of the vectors $X$ and $Y$, then you can assign an &lt;strong&gt;angle&lt;/strong&gt; $\theta$ to these quantities in the usual sense as&#xA;$$\cos \theta = \frac{\tilde{g}\left(X,Y\right)}{\sqrt{\tilde{g}\left(X,X\right)\cdot\tilde{g}\left(Y,Y\right)}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now you have to think of what $X$ and $Y$ should be. Since you want to compare these quantities at three different points, you will have to &lt;a href=&quot;http://en.wikipedia.org/wiki/Parallel_transport&quot; rel=&quot;nofollow&quot;&gt;parallel transport&lt;/a&gt; them between these points using a &lt;a href=&quot;http://en.wikipedia.org/wiki/Geodesic&quot; rel=&quot;nofollow&quot;&gt;geodesic motion&lt;/a&gt;, again in the 4D setup.&lt;br&gt;&#xA;Light geodesics will be totally different from e.g. human ones. So, generally, the angles will depend on &lt;strong&gt;how you transport&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, DarthPickley, if you have a light ray travelling on the event horizon of a &lt;a href=&quot;http://en.wikipedia.org/wiki/Schwarzschild_metric&quot; rel=&quot;nofollow&quot;&gt;Schwarzschild black hole&lt;/a&gt;, what would be the sum of the angles of a corresponding &quot;triangle&quot;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sincerely  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Robert&lt;/p&gt;&#xA;" OwnerUserId="2671" LastActivityDate="2011-01-18T14:00:49.563" CommentCount="5" />
  <row Id="17983" PostTypeId="2" ParentId="17979" CreationDate="2011-01-18T14:03:45.707" Score="4" ViewCount="" Body="&lt;p&gt;As per Willie's request I make my comment into an answer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sequences with only finitely many non-zero terms form a proper subspace $U$ of $V$. For $k \in \mathbb{N}$ let $\delta_{k} \in U$ be the sequence for which $\delta_{k}(k) = 1$ is the only non-zero entry. Now compute the scalar product (= dot product) $\langle \delta_{k}, b \rangle$ for all $k$ in order to see that $U^{\perp} = 0$. In other words: if $b$ is orthogonal to all $\delta_{k}$'s then $b$ must be zero.&lt;/p&gt;&#xA;" OwnerUserId="5363" LastActivityDate="2011-01-18T14:03:45.707" CommentCount="2" />
  <row Id="17984" PostTypeId="1" AcceptedAnswerId="17986" CreationDate="2011-01-18T14:03:54.537" Score="5" ViewCount="108" Body="&lt;p&gt;Does there exist a continuous function $f : [0, 1] → [0, 1]$ such that the pre-image $f^{−1}(y)$ of any point $y \in [0, 1]$ is uncountable?&lt;/p&gt;&#xA;" OwnerUserId="5980" LastActivityDate="2011-01-19T00:16:59.773" Title="Existence of a continuous function with per-image of each point uncountable" Tags="&lt;real-analysis&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="2" />
  <row Id="17985" PostTypeId="2" ParentId="17977" CreationDate="2011-01-18T14:19:01.667" Score="1" ViewCount="" Body="&lt;p&gt;Dear sun,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://math.stackexchange.com/users/3567/raskolnikov&quot;&gt;Raskolnikov&lt;/a&gt; is absolutely right:&#xA;$$\int e^{-\mathrm{i} qu} \left( \frac{A}{\sqrt{1-\mathrm{i}au}} + \frac{B}{\sqrt{1-\mathrm{i}bu}} \right)^N du = \sum_{k=0}^N \binom{N}{k} \int e^{-\mathrm{i} qu} \left[ \frac{A}{\sqrt{1-\mathrm{i}au}}\right]^{N-k} \left[ \frac{B}{\sqrt{1-\mathrm{i}bu}} \right]^{k} du$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See &lt;a href=&quot;http://en.wikipedia.org/wiki/Binomial_theorem&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;br&gt;&#xA;Sincerely&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Robert&lt;/p&gt;&#xA;" OwnerUserId="2671" LastEditorUserId="1543" LastEditorDisplayName="" LastEditDate="2011-01-18T14:46:01.263" LastActivityDate="2011-01-18T14:46:01.263" CommentCount="2" />
  <row Id="17986" PostTypeId="2" ParentId="17984" CreationDate="2011-01-18T14:19:17.800" Score="8" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Yes&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One nice way to see that is to take a &lt;a href=&quot;http://en.wikipedia.org/wiki/Space-filling_curve&quot; rel=&quot;nofollow&quot;&gt;Peano curve&lt;/a&gt; $c: [0,1] \to [0,1]^{2}$ (that is, a continuous surjection) and to compose it with the projection $p(x,y) = x$. Then $f = p \circ c$ will have the desired property.&lt;/p&gt;&#xA;" OwnerUserId="5363" LastEditorUserId="5363" LastEditorDisplayName="" LastEditDate="2011-01-19T00:16:59.773" LastActivityDate="2011-01-19T00:16:59.773" />
  <row Id="17987" PostTypeId="2" ParentId="17976" CreationDate="2011-01-18T14:26:10.173" Score="2" ViewCount="" Body="&lt;p&gt;What it seems to do is, when $n$ is your number, that it maximizes the number of prime factors of $q$ within the range $q \in (n-100,n+100)$. And then sets $n=q+(n-q)$. Doing this it can easily find that 513 is for example $513=2^9+1$. However for the numbers you gave it is not really interesting.&lt;/p&gt;&#xA;" OwnerUserId="3123" LastActivityDate="2011-01-18T14:26:10.173" CommentCount="2" />
  <row Id="17988" PostTypeId="1" AcceptedAnswerId="17992" CreationDate="2011-01-18T14:27:02.917" Score="0" ViewCount="86" Body="&lt;p&gt;What is an Improper Random Variable? I know the definition in terms of the CDF like, F(∞) - F(-∞) &amp;lt;1. Could any one explain it more clearly, specifically I am looking for an example of an improper random variable. Is Cauchy distributed Random Variable improper? (Since it has a long tail)&lt;/p&gt;&#xA;" OwnerUserId="5982" LastActivityDate="2011-01-18T15:17:34.300" Title="Improper Random Variables" Tags="&lt;probability&gt;&lt;probability-theory&gt;&lt;probability-distributions&gt;" AnswerCount="2" FavoriteCount="2" />
  <row Id="17989" PostTypeId="2" ParentId="17976" CreationDate="2011-01-18T14:29:31.707" Score="1" ViewCount="" Body="&lt;p&gt;I think it is just a curious fact to know and tell.  They seem to be a product of small primes plus or minus a small correction.  For 2010, besides the &quot;obvious&quot; 2010=2^11-38 it also finds that 2010 divides 29^6-1.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-01-18T14:29:31.707" />
  <row Id="17990" PostTypeId="2" ParentId="17971" CreationDate="2011-01-18T14:35:01.327" Score="1" ViewCount="" Body="&lt;p&gt;You can take the rank of an item of rank $r_1$ in set 1 and set $r_2=\frac{(x-1)(10-r_1)}{9}+1$.  This will linearly map the set 1 ranks into the range of set 2.  Is this what you are looking for?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This presumes that the items in the two sets should be &quot;evenly mixed&quot;, which may not be what you want.  Maybe all the items of set 1 should be ranked above all the items in set 2.  It also runs the risk of having two items with the same rank, particularly 1 and x.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-01-18T14:35:01.327" />
  <row Id="17991" PostTypeId="2" ParentId="17977" CreationDate="2011-01-18T15:13:05.397" Score="1" ViewCount="" Body="&lt;p&gt;In view of your recent question on MathOverflow, this new question is regarding (inversion of) characteristic functions. Let's begin with the characteristic function corresponding to the right integral. Suppose that $W \sim {\rm N}(0,1)$. Then $W^2$ has characteristic function of chi-square with $1$ degree of freedom:&#xA;$$&#xD;&#xA;{\rm E}[e^{iuW^2 } ] = \frac{1}{{\sqrt {1 - 2iu} }}.&#xD;&#xA;$$&#xA;From this it follows straightforwardly that the characteristic function of the sum &#xA;$$&#xD;&#xA;{\frac{a}{2}} \sum\limits_{i = 1}^n  W_i^2  + {\frac{b}{2}} \sum\limits_{i = 1}^m  {\tilde W}_i^2, &#xD;&#xA;$$&#xA;where $W_i$ and ${\tilde W}_i$ are independent copies of $W$, is given by&#xA;$$&#xD;&#xA;{\rm E}\bigg[\exp \bigg(i\frac{{au}}{2}\sum\limits_{i = 1}^n {W_i^2 } \big)\bigg]{\rm E}\bigg[\exp \bigg(i\frac{{bu}}{2}\sum\limits_{i = 1}^m {{\tilde W}_i^2 } \bigg)\bigg] = \frac{1}{{(\sqrt {1 - iau} )^n }}\frac{1}{{(\sqrt {1 - ibu} )^m }}.&#xD;&#xA;$$&#xA;As for the characteristic function corresponding to the left integral, define random variable $T$ as follows. With probability $A$, $T = (a/2)W_1^2$, and with probability $B(=1-A)$, $T = (b/2)W_2^2$, where $W_1$ and $W_2$ are independent ${\rm N}(0,1)$ rv's. Then, the characteristic function of $T$ is given by&#xA;$$&#xD;&#xA;A{\rm E}\bigg[\exp \bigg(i\frac{{au}}{2}W_1^2 \bigg)\bigg] + B{\rm E}\bigg[\exp \bigg(i\frac{{bu}}{2}W_1^2 \bigg)\bigg] = \frac{A}{{\sqrt {1 - iau} }} + \frac{B}{{\sqrt {1 - ibu} }}.&#xD;&#xA;$$&#xA;Thus, the characteristic function of the sum of $N$ independent copies of $T$ is given by&#xA;$$&#xD;&#xA;\bigg(\frac{A}{{\sqrt {1 - iau} }} + \frac{B}{{\sqrt {1 - ibu} }}\bigg)^N.&#xD;&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="2810" LastActivityDate="2011-01-18T15:13:05.397" />
  <row Id="17992" PostTypeId="2" ParentId="17988" CreationDate="2011-01-18T15:14:34.087" Score="3" ViewCount="" Body="&lt;p&gt;You can think of an improper random variable $X$ as one which has a positive probability of taking on the values $+\infty$ and $-\infty$.  It's then not hard to see that $F_X(+\infty) = 1-P(X = +\infty)$, $F_X(-\infty) = P(X = -\infty)$ (where $F(\pm\infty)$ are to be interpreted as limits, as usual).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, if $X_n$ is a constant random variable with $P(X_n = n) = 1$, then as $n \to +\infty$, the $X_n$ are converging to $+\infty$ with probability 1.  So the limiting random variable $X$ is improper, and should have $P(X = +\infty) = 1$.  The distribution function of $X$ should just be $F_X(t) = 0$, which you may observe is also the (pointwise) limit of the distribution functions $F_{X_n}(t)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An improper random variable in this sense never has a finite mean, but the converse is not true.  The Cauchy random variable does not have a finite means but is not improper, since the values it takes on are all finite, though they can be very large.  &lt;/p&gt;&#xA;" OwnerUserId="822" LastActivityDate="2011-01-18T15:14:34.087" />
  <row Id="17993" PostTypeId="2" ParentId="17988" CreationDate="2011-01-18T15:17:34.300" Score="1" ViewCount="" Body="&lt;p&gt;I believe the definition in terms of the cumulative distribution function is the right one: $F(\infty) - F(-\infty) &amp;lt; 1$.  Improper random variables can be used, for instance, to describe the distribution over time of events that might never happen.  For example, the probability $M(t)$ of having been married before age $t$ is a CDF for an improper random variable with PDF $m(t)=M&amp;#39;(t)$, whose integral is less than $1$, because some people never marry.&lt;/p&gt;&#xA;" OwnerUserId="5546" LastActivityDate="2011-01-18T15:17:34.300" CommentCount="1" />
  <row Id="17994" PostTypeId="2" ParentId="17902" CreationDate="2011-01-18T15:43:44.583" Score="2" ViewCount="" Body="&lt;p&gt;Magma &lt;em&gt;can&lt;/em&gt; compute the kernel and the image of a homomorphism between abelian groups. For instance,&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;  A&amp;lt;[x]&amp;gt; := AbelianGroup([0]);&#xA;  B&amp;lt;[y]&amp;gt; := AbelianGroup([2,4]);&#xA;  C&amp;lt;[z]&amp;gt; := AbelianGroup([2]);&#xA;  g := hom&amp;lt; B -&amp;gt; C | [z[1],2*z[1]]&amp;gt;;&#xA;  K := Kernel(g);&#xA;  K;&#xA;  f := hom&amp;lt; A -&amp;gt; K | [4*y[1]-2*y[2]]&amp;gt;;&#xA;  I := Image(f);&#xA;  I;&#xA;  K/I;&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;gives the following result:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; Abelian Group isomorphic to Z/4&#xA; Defined on 1 generator in supergroup B:&#xA;     K.1 = y[2]&#xA; Relations:&#xA;     4*K.1 = 0&#xA; Abelian Group isomorphic to Z/2&#xA; Defined on 1 generator in supergroup K:&#xA;     I.1 = 2*K.1&#xA; Relations:&#xA;2*I.1 = 0&#xA; Abelian Group isomorphic to Z/2&#xA; Defined on 1 generator&#xA; Relations:&#xA;     2*$.1 = 0&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="367" LastEditorUserId="367" LastEditorDisplayName="" LastEditDate="2011-01-18T16:38:13.767" LastActivityDate="2011-01-18T16:38:13.767" />
  <row Id="17995" PostTypeId="2" ParentId="17813" CreationDate="2011-01-18T15:48:52.637" Score="1" ViewCount="" Body="&lt;p&gt;If you are interested in theory, as @Nate Eldredge suggested, follow the &lt;a href=&quot;http://azzalini.stat.unipd.it/SN/faq-r.html&quot; rel=&quot;nofollow&quot;&gt;url&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As written there also, you could use &lt;a href=&quot;http://www.r-project.org/&quot; rel=&quot;nofollow&quot;&gt;R&lt;/a&gt; in practice to generate random numbers using a skew normal distribution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Load the library &lt;a href=&quot;http://cran.r-project.org/package=sn&quot; rel=&quot;nofollow&quot;&gt;sn&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;library(sn)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Or if not installed, first install it via &lt;code&gt;install.packages('sn')&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then you can generate any random number with given parameters with the &lt;code&gt;rsn&lt;/code&gt; function.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;rsn(n=100, location=1.256269, scale=1.605681, shape=5)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Will generate 100 random numbers from the distribution with given parameters. If you would like to plot the histogram of your generated values, use higher sample size (e.g. 1.000 or 10.000), like:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;hist(rsn(n=10000, location=1.256269, scale=1.605681, shape=5))&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/5llDr.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="5651" LastActivityDate="2011-01-18T15:48:52.637" CommentCount="1" />
  <row Id="17997" PostTypeId="2" ParentId="17935" CreationDate="2011-01-18T16:23:04.637" Score="0" ViewCount="" Body="&lt;p&gt;I'd argue for a mixed proof system. Computer proofs should be more accepted that they are now, but the tradition is slow to build. I'd rather trust a proof that has also been checked by a computer, in addition to traditional peer-review. After all, there is only so much amount of abstraction that a human brain can handle, and  at times small details can change everything. As Alain Connes put it in the Princeton Companion, the most difficult step in a mathematical work is checking for mistakes.&lt;/p&gt;&#xA;" OwnerUserId="5986" LastActivityDate="2011-01-18T16:23:04.637" />
  <row Id="17998" PostTypeId="1" AcceptedAnswerId="18000" CreationDate="2011-01-18T16:24:48.980" Score="2" ViewCount="89" Body="&lt;p&gt;I have an equation&#xA;$$x = \csc(\theta) - \cot(\theta).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As $\theta$ approaches zero, $x$ approaches zero.  However, trying to solve the equation at zero yields an undefined result.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do I rewrite the equation to be continuous at 0?&lt;/p&gt;&#xA;" OwnerUserId="5947" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-18T16:51:13.057" LastActivityDate="2011-01-18T16:51:13.057" Title="Proving equation at zero?" Tags="&lt;trigonometry&gt;" AnswerCount="1" />
  <row Id="17999" PostTypeId="1" AcceptedAnswerId="18015" CreationDate="2011-01-18T16:28:51.547" Score="3" ViewCount="393" Body="&lt;p&gt;Hello&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose There is a random walk starting in origin while probability to move right is 1/3 and probability to move left 2/3.What is the probability to return to the origin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you&lt;/p&gt;&#xA;" OwnerUserId="5145" LastActivityDate="2011-01-18T20:44:31.170" Title="1D random walk-probability to go back to origin." Tags="&lt;probability&gt;" AnswerCount="3" CommentCount="5" FavoriteCount="2" />
  <row Id="18000" PostTypeId="2" ParentId="17998" CreationDate="2011-01-18T16:32:51.763" Score="5" ViewCount="" Body="&lt;p&gt;Hint:  Write the cosecant and cotangent in terms of sine and cosine.  You can then combine the two fractions to give an expression that goes to 0/0.  Expanding in a Taylor series or L'Hopital's rule will then be your friend.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-01-18T16:32:51.763" CommentCount="3" />
  <row Id="18001" PostTypeId="2" ParentId="17999" CreationDate="2011-01-18T16:40:31.147" Score="1" ViewCount="" Body="&lt;p&gt;You can find a detailed answer for your question on &lt;a href=&quot;http://mathworld.wolfram.com/RandomWalk1-Dimensional.html&quot; rel=&quot;nofollow&quot;&gt;WolframMathWorld&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To give an exact answer, you should define the number of steps you are interested in returning to the origin. Of course the number of steps should be even.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have made a little &quot;simulation&quot; (rather: test run) in &lt;a href=&quot;http://www.r-project.org/&quot; rel=&quot;nofollow&quot;&gt;R&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;results &amp;lt;- NULL&#xA;for (i in 1:50) {&#xA;    N=i*2&#xA;    n1=N/2&#xA;    p=1/3&#xA;    q=1-p&#xA;    results &amp;lt;- c(results, ((factorial(N)/(factorial(n1)*factorial(N-n1)))*(p^n1)*(q^(N-n1))))&#xA;}&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;And got the following results (sorry for ugly table):&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&#xA;  N| probability&#xA;---+--------------&#xA;  1 0.4444444444&#xA;  2 0.2962962963&#xA;  3 0.2194787380&#xA;  4 0.1707056851&#xA;  5 0.1365645481&#xA;  6 0.1112748170&#xA;  7 0.0918458807&#xA;  8 0.0765382339&#xA;  9 0.0642543198&#xA; 10 0.0542592034&#xA; 11 0.0460381120&#xA; 12 0.0392176509&#xA; 13 0.0335193598&#xA; 14 0.0287308798&#xA; 15 0.0246872745&#xA; 16 0.0212584864&#xA; 17 0.0183406549&#xA; 18 0.0158499487&#xA; 19 0.0137180842&#xA; 20 0.0118890063&#xA; 21 0.0103163865&#xA; 22 0.0089617094&#xA; 23 0.0077927908&#xA; 24 0.0067826142&#xA; 25 0.0059084106&#xA; 26 0.0051509221&#xA; 27 0.0044938086&#xA; 28 0.0039231662&#xA; 29 0.0034271337&#xA; 30 0.0029955687&#xA; 31 0.0026197805&#xA; 32 0.0022923080&#xA; 33 0.0020067342&#xA; 34 0.0017575319&#xA; 35 0.0015399328&#xA; 36 0.0013498176&#xA; 37 0.0011836238&#xA; 38 0.0010382665&#xA; 39 0.0009110715&#xA; 40 0.0007997183&#xA; 41 0.0007021917&#xA; 42 0.0006167398&#xA; 43 0.0005418386&#xA; 44 0.0004761612&#xA; 45 0.0004185515&#xA; 46 0.0003680018&#xA; 47 0.0003236328&#xA; 48 0.0002846770&#xA; 49 0.0002504641&#xA; 50 0.0002204084&#xA;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;As you can see here, I computed the probabilities of getting back to the original point for all possible steps (N), where N&amp;lt;=100. That is done by computing the probability for all even N, where the number of steps to the right is equal to the number of steps to the left with the following formula: $$P(n_{1}|N)=\frac{N!}{n_{1}!^2}p^{n_{1}}(1-p)^{n_{1}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where N is the number of steps, $n_{1}=\frac{N}{2}$ and $p=\frac{1}{3}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: the sum of all above is 1.998366. That is higher than 1, as the probability of arriving back to the original point could also include a lower N's probability.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's all I could do, I like to play with simulations, but not good at all with Markov chains extrapolated to infinite :)&lt;/p&gt;&#xA;" OwnerUserId="5651" LastEditorUserId="5651" LastEditorDisplayName="" LastEditDate="2011-01-18T18:11:15.547" LastActivityDate="2011-01-18T18:11:15.547" CommentCount="2" />
  <row Id="18002" PostTypeId="2" ParentId="13820" CreationDate="2011-01-18T16:47:48.867" Score="1" ViewCount="" Body="&lt;p&gt;This partial answer will list 9 combinatorial types of 5-dimensional pure quadtopes. In the following, ‘quadrilateral’ is shortened to ‘quad’. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Types 2 – 5 are each the convex hull of three quads, with pairwise intersections at a vertex. Thus, the quads form a loop. For a given quad, say Q1, there are two ways it can connect to the other two quads; the common vertices can be adjacent on Q1 (described as crenellated), or they can be diagonally opposite on Q1 (described as diamondback). The term crenellated is suggested by the outline of the same-colored squares on the first two rows of a chessboard. There are four resulting combinatorial types, denoted ccc, dcd, cdc, and ddd, corresponding to whether the arrangement of quads 1, 2, and 3 (not necessarily in order) is crenellated or diamondback.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A specific example of each type 2 – 5 contains vertices V2 through V8 – the vertices of a 4-dimensional pure quadtope subtended by parallelograms Q1 (V3, 6, 5, 7) and Q2 (V2, 4, 3, 8), plus V1, affine to it. Each type 2 – 5 is completed by completing a parallelogram Q3 with V9, containing V1 and vertices from each of Q1 and Q2. Vertices V1 through V8 are as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;V1: (1,3,6,4,8)&#xA;V2: (2,3,6,4,8)&#xA;V3: (2,4,7,5,10)&#xA;V4: (2,4,8,5,10)&#xA;V5: (2,4,8,5,11)&#xA;V6: (2,4,8,6,11)&#xA;V7: (2,4,9,6,12)&#xA;V8: (2,5,9,6,12)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For brevity in listing polytope elements, the 4-dimensional pure quadtope type will be denoted P([2]), a k-fold pyramid over a quad as P([1],k), and an n-simplex as s&lt;sub&gt;n&lt;/sub&gt;. P([2]) has 8 facets – four P([1],1) and four s&lt;sub&gt;3&lt;/sub&gt;, with f-vector (7,17,18,8). P([1],2) has 6 facets – two P([1],1) and four s&lt;sub&gt;3&lt;/sub&gt;, with f-vector (6,13,13,6).&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The free join of two quads is like-faceted – its 8 facets are all P([1],2), with f-vector (8,24,32,24,8).&lt;/li&gt;&#xA;&lt;li&gt;ccc; V9 = V1 + V5 – V2.  This polytope has 10 facets – the cyclic polytope C(4,6), three P([2]), three P([1],2) and three s&lt;sub&gt;4&lt;/sub&gt;, with f-vector (9,30,46,33,10).&lt;/li&gt;&#xA;&lt;li&gt;dcd; V9 = V7 + V8 – V1. This polytope has 13 facets – one P([2]), eight P([1],2) and four s&lt;sub&gt;4&lt;/sub&gt;, with f-vector (9,30,48,38,13).&lt;/li&gt;&#xA;&lt;li&gt;cdc; V9 = V1 + V7 – V8. This polytope has 13 facets – two P([2]), five P([1],2) and six s&lt;sub&gt;4&lt;/sub&gt;, with f-vector (9,30,48,38,13).&lt;/li&gt;&#xA;&lt;li&gt;ddd; V9 = V7 + V4 – V1. This like-faceted polytope has 12 facets of type P([1],2), with f-vector (9,30,47,36,12).&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Starting with one of the types 2 – 5, you can build another 5-dimensional combinatorial type by selecting one free vertex each from Q1 – Q3, adding a vertex V10 by completing another parallelogram Q4. So, each quad intersects at three different vertices with the other three quads. On any quad, there are two pairs of adjacent vertices intersecting with other quads, and one pair of opposite vertices intersecting with other quads. So, in describing the arrangement of quads in each of the four quad triplets (as above), ‘c’ must appear 8 times and ‘d’ 4 times. (The quad triplets subtend proper 5-dimensional subpolytopes). There are therefore at least four more combinatorial types:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Type 6. Quad triplets cdc, cdc, cdc, cdc. Beginning with type cdc, V10 = V2 + V6 – V1. This polytope has 17 facets – the direct sum s&lt;sub&gt;3&lt;/sub&gt; ⊕ s&lt;sub&gt;1&lt;/sub&gt;, two P([2]), eight P([1],2), and six s&lt;sub&gt;4&lt;/sub&gt;, with f-vector (10,36,62,51,17).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Type 7. Quad triplets ddd, cdc, ccc, ccc. Beginning with type cdc, V10 = V2 + V9 – V5. This polytope has 12 facets. Two of the polychoron facets have 9 ridges each: two P([1],1), six s&lt;sub&gt;3&lt;/sub&gt;, and a common ridge which is a triangular bipyramid. The remaining facets are three P([2]), five P([1],2), and two s&lt;sub&gt;4&lt;/sub&gt;. The f-vector is (10,36,57,41,12).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Type 8. Quad triplets dcd, cdc, cdc, ccc. Beginning with type dcd, V10 = V1 + V5 – V2. This polytope has 17 facets – the direct sum s&lt;sub&gt;1&lt;/sub&gt; ⊕ s&lt;sub&gt;1&lt;/sub&gt; ⊕ s&lt;sub&gt;2&lt;/sub&gt;, two P([2]), six P([1],2), and eight s&lt;sub&gt;4&lt;/sub&gt;, with f-vector (10,37,64,52,17). Interestingly, although Q3 is a (set) element of Q, Q3 is not a (polytope) element of P – it’s a non-element within the facet of type s&lt;sub&gt;1&lt;/sub&gt; ⊕ s&lt;sub&gt;1&lt;/sub&gt; ⊕ s&lt;sub&gt;2&lt;/sub&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Type 9. Quad triplets dcd, dcd, ccc, ccc. Beginning with type dcd, V10 = V1+ V2 – V5. This polytope has a fifth quad element Q5, a parallelogram with vertices V4, V6, V9, and V10. Q5 can be included in set Q with the quadtope definition still met. This polytope has 12 facets - five P([2]), five P([1],2), and two s&lt;sub&gt;4&lt;/sub&gt;, with f-vector (10,35,55,40,12). This polytope has a kind of rotational symmetry to it, with automorphism group the dihedral group D&lt;sub&gt;5&lt;/sub&gt;, although with a seemingly unavoidable crimp at V10. The quads in the P([2]) elements are two apart (mod 5).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From types 6 – 8, a fifth quad can be added, so there are more 5-dimensional pure quadtope types. The figure below contains diagrams for types 2 – 9, showing the quad arrangements and vertices. In the diagrams, arcs are used to join vertices to indicate that they are actually the same vertex.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/DTo1f.png&quot; alt=&quot;&quot; title=&quot;Hosted by imgur.com&quot; /&gt;&lt;/p&gt;&#xA;" OwnerUserId="1782" LastActivityDate="2011-01-18T16:47:48.867" />
  <row Id="18005" PostTypeId="1" CreationDate="2011-01-18T17:38:42.897" Score="2" ViewCount="368" Body="&lt;p&gt;Can we find two functions $f$ and $g$ that are reasonably defined nontrivial(not everywhere zero, $f\neq g$, not linear polynomials) functions such that the following condition is satisfied?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ f( \left(\int_{0}^{t} g(x) \ \text{d}x\right))  = g( \left(\int_{0}^{t} f(x) \ \text{d}x\right)) $$&lt;/p&gt;&#xA;" OwnerUserId="5595" LastEditorUserId="5595" LastEditorDisplayName="" LastEditDate="2011-01-22T15:15:54.033" LastActivityDate="2011-01-22T15:15:54.033" Title="Are there functions satisfying the following integral condition?" Tags="&lt;integral&gt;&lt;functions&gt;" AnswerCount="0" CommentCount="23" />
  <row Id="18006" PostTypeId="1" AcceptedAnswerId="18011" CreationDate="2011-01-18T17:54:49.477" Score="4" ViewCount="227" Body="&lt;p&gt;I have to demonstrate this formulae:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Prove A = (A\B) ∪ (A ∩ B)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But it seems to me that it is false.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;(A\B) ∪ (A ∩ B) &lt;/p&gt;&#xA;  &#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;X ∈ A/B =&gt;  { x ∈ A  and x ∉ B }&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;  &#xA;  &lt;p&gt;or                                    &lt;/p&gt;&#xA;  &#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;X ∈ A ∩ B =&gt; { x ∈ A and x ∈ B }&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;so: &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;x ∈ A ∩ B&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;so:&#xA;  A ≠ (A\B) ∪ (A ∩ B)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Did I solve the problem or I am just blind?&lt;/p&gt;&#xA;" OwnerUserId="5526" LastEditorUserId="462" LastEditorDisplayName="" LastEditDate="2011-01-18T17:58:32.897" LastActivityDate="2011-01-19T07:14:40.687" Title="Prove A = (A\B) ∪ (A ∩ B)" Tags="&lt;elementary-set-theory&gt;" AnswerCount="5" CommentCount="5" />
  <row Id="18008" PostTypeId="2" ParentId="18006" CreationDate="2011-01-18T17:58:53.177" Score="1" ViewCount="" Body="&lt;p&gt;Let $x \in A$. Then $x \in A \backslash B$ or $x \in A \cap B$. Likewise, if $x \in A \backslash B$ or $x \in A \cap B$ then $x \in A$. &lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2011-01-18T17:58:53.177" />
  <row Id="18010" PostTypeId="2" ParentId="12536" CreationDate="2011-01-18T18:03:19.497" Score="-2" ViewCount="" Body="&lt;p&gt;Attempting actuarial exams is one of the hardest roads (academically) you can take.......................&lt;/p&gt;&#xA;" OwnerUserId="5990" LastActivityDate="2011-01-18T18:03:19.497" CommentCount="2" />
  <row Id="18011" PostTypeId="2" ParentId="18006" CreationDate="2011-01-18T18:04:16.710" Score="5" ViewCount="" Body="&lt;p&gt;To show that two sets are equal, you show they have the same elements. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose first $x\in A$. There are two cases: Either $x\in B$, or $x\notin B$. In the first case, $x\in A$ and $x\in B$, so $x\in A\cap B$ (by definition of intersection). In the second case, $x\in A$ and $x\notin B$, so $x\in A\setminus B$ (again, by definition).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This shows that if $x\in A$, then $x\in A\cap B$ or $x\in A\setminus B$, i.e., $x\in (A\setminus B)\cup(A\cap B)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we have to show, conversely, that if $x\in (A\setminus B)\cup(A\cap B)$, then $x\in A$. Note that $x\in(A\setminus B)\cup(A\cap B)$ means that either $x\in A\setminus B$ or $x\in A\cap B$. In the first case, $x\in A$ (and also, $x\notin B$). In the second case, $x\in A$ (and also, $x\in B$). In either case, $x\in A$, but this is what we needed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In summary: We have shown both $A\subseteq (A\setminus B)\cup(A\cap B)$ and $(A\setminus B)\cup(A\cap B)\subseteq A$. But this means the two sets are equal.&lt;/p&gt;&#xA;" OwnerUserId="462" LastActivityDate="2011-01-18T18:04:16.710" CommentCount="2" />
  <row Id="18012" PostTypeId="2" ParentId="18006" CreationDate="2011-01-18T18:04:38.517" Score="3" ViewCount="" Body="&lt;p&gt;To show set equality you show $\supset$, $\subset$ respectively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\subset$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $x \in A$. Then $x$ either in $A \cap B$ or in $A \cap B^c = A - B$, so $x \in (A \cap B) \cup (A - B)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\supset$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $x \in (A \cap B) \cup (A - B)$. Then either $x$ in $ A \cap B$ or x in $A \cap B^c$. But in both cases $x \in A$, therefore $x \in A$.&lt;/p&gt;&#xA;" OwnerUserId="5798" LastActivityDate="2011-01-18T18:04:38.517" />
  <row Id="18013" PostTypeId="2" ParentId="17132" CreationDate="2011-01-18T18:12:35.997" Score="0" ViewCount="" Body="&lt;p&gt;Here's a proof :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The result can be easily shown by strong induction to $\sum_{i=1}^{s}{a_i}$ .Suppose that the result is true if the sum above is less or equal to k.Suppose now that the sum is equal to k+1.Then for the induction step we use the invariant factors of pM for which the sum is less than k.&lt;/p&gt;&#xA;" OwnerUserId="5422" LastActivityDate="2011-01-18T18:12:35.997" />
  <row Id="18014" PostTypeId="2" ParentId="17969" CreationDate="2011-01-18T18:24:20.557" Score="2" ViewCount="" Body="&lt;p&gt;Another good book is the one of Barrett O'Neil,titled&quot; Elementary Differential Geometry&quot;.I think it's at the same level as  the Do Carmo's book but it makes a different approach.&lt;/p&gt;&#xA;" OwnerUserId="5422" LastActivityDate="2011-01-18T18:24:20.557" />
  <row Id="18015" PostTypeId="2" ParentId="17999" CreationDate="2011-01-18T18:24:45.067" Score="7" ViewCount="" Body="&lt;p&gt;Let $P_{i\ge 0}$ be the probability of ever reaching position $x+i$ when starting from position $x$ (this is independent of $x$, since the transition probabilities are).  Clearly $P_0=1$, and $P_i\rightarrow 0$ as $i\rightarrow\infty$ provided that the right-hop probability $q &amp;lt; 1/2$ (in this case $q = 1/3$).  Otherwise,&#xA;$$&#xD;&#xA;P_i = qP_{i-1} + (1-q)P_{i+1},&#xD;&#xA;$$&#xA;You can guess the solution to be of the form $P_i = \alpha^i$ for some $0 &amp;lt; \alpha &amp;lt; 1$.  This turns out to satisfy the conditions if&#xA;$$&#xD;&#xA;\alpha = q + (1-q) \alpha^2,&#xD;&#xA;$$&#xA;which has the solution $\alpha = q/(1-q)$ in this case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the problem specified, you want to know the total probability of returning to the origin after the first step.  If the first step is to the right (which happens with probability $q$), then you must return to the origin; if it is to the left (with probability $1-q$), then you will return to the origin with probability $P_1 = \alpha = q/(1-q)$.  So the solution is&#xA;$$&#xD;&#xA;P_{\text{return}} = q + (1-q)\alpha = q + (1-q)\frac{q}{1-q} = 2q&#xD;&#xA;$$&#xA;for general $q&amp;lt;1/2$, and $P_{\text{return}} = 2/3$ in this case.&lt;/p&gt;&#xA;" OwnerUserId="5546" LastActivityDate="2011-01-18T18:24:45.067" CommentCount="3" />
  <row Id="18016" PostTypeId="2" ParentId="18006" CreationDate="2011-01-18T18:55:28.410" Score="2" ViewCount="" Body="&lt;p&gt;$\rm\ A\backslash B\  =\ A\cap\overline B\ \ \:$ so $\rm\ \: (A\backslash B)\cup (A\cap B)\ =\ (A\cap\overline B)\cup(A\cap B)\ =\ A\cap(\overline B\cup B)\ =\ A$&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2011-01-19T07:14:40.687" LastActivityDate="2011-01-19T07:14:40.687" />
  <row Id="18017" PostTypeId="2" ParentId="18006" CreationDate="2011-01-18T19:04:10.060" Score="1" ViewCount="" Body="&lt;p&gt;Working inside a universe $X$: &#xA;$$A = A \cap X = A \cap (B \cup (X \setminus B)) = ( A \cap B ) \cup ( A \cap (X \setminus B)) = (A \cap B) \cup (A \setminus B)$$   &lt;/p&gt;&#xA;" OwnerUserId="4280" LastActivityDate="2011-01-18T19:04:10.060" CommentCount="1" />
  <row Id="18018" PostTypeId="2" ParentId="17969" CreationDate="2011-01-18T19:26:09.230" Score="3" ViewCount="" Body="&lt;p&gt;The book by &lt;a href=&quot;http://books.google.com/books?id=TyqUnlyV4Y4C&amp;amp;printsec=frontcover&amp;amp;dq=differential+geometry+kuhnel&amp;amp;source=bl&amp;amp;ots=WXEUYpkN18&amp;amp;sig=yJ3r1xzGZq-WLa07-zqBxSaMq7k&amp;amp;hl=en&amp;amp;ei=muc1TeCtGIX4sAOKp5C4BQ&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=3&amp;amp;ved=0CDcQ6AEwAg#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;Kuhnel&lt;/a&gt; is a good introduction. I am sitting through an introductory differential geometry class at Stanford which is being taught by Richard Schoen and he follows this book. The first four chapters will be a good introduction to differential geometry and the rest concentrates on Riemannian geometry. I have read the first three chapters so far and the book so far seems to be complete and concise. You can look up some reviews of the book on &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0821839888&quot; rel=&quot;nofollow&quot;&gt;Amazon&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastActivityDate="2011-01-18T19:26:09.230" />
  <row Id="18020" PostTypeId="2" ParentId="17969" CreationDate="2011-01-18T19:39:51.460" Score="2" ViewCount="" Body="&lt;p&gt;I like the book &lt;a href=&quot;http://books.google.com/books?id=bwwRg7I02-4C&amp;amp;printsec=frontcover&amp;amp;dq=toponogov&amp;amp;hl=pt-BR&amp;amp;ei=Xes1TdvTG8T_lgeNqe2dCQ&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=1&amp;amp;ved=0CCUQ6AEwAA#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;Differential Geometry of Curves and Surfaces&lt;/a&gt; by Toponogov.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Besides its conciseness, it is well written and has some cool results on geodesic triangles which don't appear on Manfredo´s book.&lt;/p&gt;&#xA;" OwnerUserId="370" LastEditorUserId="6296" LastEditorDisplayName="" LastEditDate="2011-03-10T09:34:45.963" LastActivityDate="2011-03-10T09:34:45.963" />
  <row Id="18021" PostTypeId="2" ParentId="4513" CreationDate="2011-01-18T20:09:48.453" Score="0" ViewCount="" Body="&lt;p&gt;The radius is the distance between the center (i,j) and one of the end points. Solve for the sum of the squares to get the Rad.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rad = sqrt((i-x1)^2+(j-y1)^2)&lt;/p&gt;&#xA;" OwnerUserId="5995" LastActivityDate="2011-01-18T20:09:48.453" />
  <row Id="18022" PostTypeId="1" CreationDate="2011-01-18T20:42:48.463" Score="0" ViewCount="51" Body="&lt;p&gt;I have come across an artificial, simulated, stock-market type of situation, whose rules, I find, create a rather interesting problem.  I want to know if there is a mathematically optimal solution for &quot;trading&quot; on this simplified market, and if not, what may be a good approximation of this optimal solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are the rules we are aware of for the market:&lt;br&gt;&#xA;1. There are two commodities, gold coins and oil.&lt;br&gt;&#xA;2. The price of a barrel of oil cannot exceed 6.4 coins.&lt;br&gt;&#xA;3. The price of a barrel of oil cannot be less than 4.8 coins.&lt;br&gt;&#xA;4. The price of a barrel of oil is evaluated every 5 minutes.&lt;br&gt;&#xA;5. Trades placed within these 5 minutes are guaranteed at the current price.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is observed that the price of oil changes as a large number of purchases or sells are made, and any individual &quot;trader&quot; cannot make a trade large enough to influence the price of oil.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A graph of a typical &quot;day&quot; of trading on this market is &lt;a href=&quot;http://s1221.photobucket.com/albums/dd471/nevroldenf/?action=view&amp;amp;current=graph.png&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;br&gt;&#xA;Actual numbers of the given graph are available in the first comment below (until I have 10 reputation).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first, very simple, solution that I came up with, was as follows:&lt;br&gt;&#xA;1.  If current oil price is greater than previous oil price, sell 10% of oil owned.&lt;br&gt;&#xA;2.  If current oil price is less than previous oil price, spend 10% of gold coins to buy oil.&lt;br&gt;&#xA;3.  If there is no change in oil price, do nothing.&lt;br&gt;&#xA;However, I feel that this solution does not make good use of the conditions of the market.&lt;/p&gt;&#xA;" OwnerUserId="4212" LastActivityDate="2011-01-18T20:42:48.463" Title="What is the optimal investing solution for the given simulated market?" Tags="&lt;statistics&gt;&lt;algorithms&gt;&lt;puzzle&gt;" CommentCount="5" />
  <row Id="18023" PostTypeId="2" ParentId="17999" CreationDate="2011-01-18T20:44:31.170" Score="2" ViewCount="" Body="&lt;p&gt;The problem you refer to is the study of return probability for a random walk on a lattice. In your case, the lattice is the 1-D lattice of integers. This problem is very well studied on general lattices. A famous theorem by Polya says that while the probability of return is $1$ for symmetric random walks (i.e., all moves are equally likely unlike in this question) in $1$ and $2$ dimensional lattices, it is strictly less than $1$ in all higher dimensions. See &lt;a href=&quot;http://mathworld.wolfram.com/PolyasRandomWalkConstants.html&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; for more details.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The solution posted by mjqxxxx is very clever and is perfectly valid. If you are interested in other solutions, a very systematic way of studying this problem is through the use of generating functions. See &lt;a href=&quot;http://math.berkeley.edu/~chr/teach/Lec13_2006.pdf&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; lecture notes for more information (In fact, these notes have the solution to the problem you posed). &lt;/p&gt;&#xA;" OwnerUserId="618" LastActivityDate="2011-01-18T20:44:31.170" />
  <row Id="18024" PostTypeId="1" AcceptedAnswerId="18026" CreationDate="2011-01-18T21:00:00.487" Score="2" ViewCount="96" Body="&lt;p&gt;A set is a collection of distinct objects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose I sample some parameter of some population and the results are $\{35,35,36,36,37 \}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do I need to say (i) the multiset of the samples is $\{35,35, \dots \}$ instead of (ii) the set of samples is $\{35,35, \dots \}$?&lt;/p&gt;&#xA;" OwnerUserId="5996" LastEditorUserId="2548" LastEditorDisplayName="" LastEditDate="2011-01-18T21:11:20.867" LastActivityDate="2011-01-18T21:11:20.867" Title="Set vs Multiset" Tags="&lt;statistics&gt;&lt;elementary-set-theory&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="18025" PostTypeId="2" ParentId="18024" CreationDate="2011-01-18T21:03:14.233" Score="1" ViewCount="" Body="&lt;p&gt;I think it would be proper to use &quot;multiset&quot;. But most statisticians probably just say &quot;set&quot;.&lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2011-01-18T21:03:14.233" />
  <row Id="18026" PostTypeId="2" ParentId="18024" CreationDate="2011-01-18T21:05:53.557" Score="3" ViewCount="" Body="&lt;p&gt;Yes, I think it would be better to use the word &quot;multi-set&quot; here. Formally, say, $\{35,35,7\}$ and $\{35,7\}$ are the same set, but in your case the number of times each element is mentioned actually matters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One caveat: Not everybody is careful, and some see this distinction as pedantic, so you may find in books or websites mention of &lt;em&gt;the set ${35,35,7}$&lt;/em&gt; when really, it is the multi-set that is meant. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, sometimes people use ordered tuples $(35,35,7)$ to avoid this issue.&lt;/p&gt;&#xA;" OwnerUserId="462" LastActivityDate="2011-01-18T21:05:53.557" />
  <row Id="18027" PostTypeId="2" ParentId="16583" CreationDate="2011-01-18T21:06:24.897" Score="-1" ViewCount="" Body="&lt;p&gt;@Qiaochu Yuan: What about applying the probability theory to the equation? Any random event, if done enough, will start to develop patterns. Those patterns can be used to then predict a probable outcome. Obviously, its a lottery and theres no guarantees. However, do you think you can apply that theory to increase your chances?&lt;/p&gt;&#xA;" OwnerUserId="5997" LastActivityDate="2011-01-18T21:06:24.897" CommentCount="1" />
  <row Id="18028" PostTypeId="2" ParentId="18024" CreationDate="2011-01-18T21:07:30.637" Score="3" ViewCount="" Body="&lt;p&gt;Two sets $A$ and $B$ are the same if and only if they have the same elements; that is, if and only if for every $x$, $x\in A$ is true if and only if $x\in B$ is true.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That means that the set $A=\{1,1\}$ and the set $B=\{1\}$ are &lt;em&gt;really&lt;/em&gt; the same set, because they have the exact same elements. Sets cannot &quot;tell&quot; how many times an element is listed in it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your set, $A=\{35, 35, 36, 36, 37\}$, and the set $B=\{35, 36, 37\}$ are the same set, for every $x$, $x$ is in $A$ if and only if $x$ is in $B$. They are both also the same as the set $C=\{37, 37, 37, 37, 35, 36\}$, and the set $D=\{37,36,35,35,35,35,35,35,35,35,35,35\}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Multisets&lt;/strong&gt; are slightly different; in a multiset, the number of times that an element appears &lt;em&gt;does&lt;/em&gt; matter. Two multisets $A$ and $B$ are equal if and only if for every $x$, the &lt;em&gt;number of times&lt;/em&gt; that $x$ occurs in $A$ is the same as the number of times that $x$ occurs in $B$. However, multisets don't care about order, so the &lt;em&gt;multiset&lt;/em&gt; $A=\{35, 35, 36, 36, 37\}$ and the multiset $B=\{35, 36, 37, 35, 36\}$ are equal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to keep track of both the number of times and the order, the simplest thing is to consider ordered pairs; for example, instead of the set you attempt to give, you could have a set in which the $n$th reading is represented by the ordered pair $(n,a)$, where $a$ is the reading itself. So you could take $A=\{(1,35), (2,35), (3,36), (4,36), (5,37)\}$; this would be &lt;em&gt;different&lt;/em&gt; from the set in which the readings are, in order, $35$, $36$, $37$, $36$, and $35$. So you need to figure out exactly what is it that matter and what does not matter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If only what actually occurs matters, use sets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If what matters is what occurs and the number of times it occurs, use multisets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If what matters is what occurs, the order, and the number of times it occurs, use sets of ordered pairs or some other modification.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Either way, to avoid confusion, use the corresponding nomenclature: if you want multisets, say &quot;multiset&quot;; if you want sets, say &quot;set&quot;. &lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2011-01-18T21:07:30.637" />
  <row Id="18029" PostTypeId="1" AcceptedAnswerId="18035" CreationDate="2011-01-18T21:46:31.227" Score="1" ViewCount="131" Body="&lt;p&gt;Question: Let $A \in M(n,n,\mathbb{R}), a_{ij} = 1$ for all $i,j$. Calculate the real eigenvalues and the affiliated eigenspace of $A$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So first of all what I would be trying to calculate are values of $\lambda$ such that $0=\text{det}\left( \begin{array}{cccc} 1- \lambda &amp;amp; 1 &amp;amp; \cdots &amp;amp; 1 \\ 1&amp;amp; 1- \lambda &amp;amp; &amp;amp; \vdots\\ \vdots&amp;amp; &amp;amp;\ddots&amp;amp; 1\\1&amp;amp;1&amp;amp;\cdots&amp;amp;1-\lambda \end{array}\right)$, right?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've come to the conclusion that in terms of $n$ the eigenvalues will be equal to $(-1)^{n}\lambda^{n} + (-1)^{n-1}n\lambda^{n-1}$. This was the result of trial and error and eventually the comparison with the theorem that for a characteristic polynomial of a square matrix:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) $\alpha_n=(-1)^{n}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) $\alpha_{n-1}=(-1)^{n-1}\text{trace}(A)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) $\alpha_0=\text{det}(A)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is, a matrix full of $1$'s is going to have det$=0$ and trace$(A)=n$... However I am now stuck, how do I justify the observation that the rest of the coefficients will be zero?&lt;/p&gt;&#xA;" OwnerUserId="3711" LastActivityDate="2011-01-18T22:10:21.363" Title="Calculate the real eigenvalues and affiliated eigenspace of a square matrix whose elements are all 1" Tags="&lt;linear-algebra&gt;" AnswerCount="2" CommentCount="5" FavoriteCount="1" />
  <row Id="18030" PostTypeId="1" AcceptedAnswerId="18032" CreationDate="2011-01-18T21:49:36.227" Score="0" ViewCount="167" Body="&lt;p&gt;I have the matrix A which is m x n size m &gt; n.&#xA;Q = transpose ( A ) A&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do I determine if Q is Semi-Positive-definite&lt;/p&gt;&#xA;" OwnerUserId="5951" LastActivityDate="2011-01-18T22:01:23.030" Title="Determine if a matrix is Semi-Positive-definite" Tags="&lt;matrices&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="2" />
  <row Id="18031" PostTypeId="1" CreationDate="2011-01-18T21:57:01.337" Score="4" ViewCount="65" Body="&lt;p&gt;As most of us are aware the cost for solving a linear system (&quot;exactly&quot;) with Gauss Elimination and other similar methods with a few right hand side and where the matrix has no structure is $\mathcal{O}(N^3)$ where $N$ is the system size.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am wondering about the lower bound for solving a linear system. An obvious lower bound is $\mathcal{\Omega}(N^2)$ (since the information content is $\mathcal{O}(N^2)$). Are there better lower bounds other than $\mathcal{\Omega}(N^2)$ for solving the linear system? Is there a way to prove that the lower bound of $\mathcal{\Omega}(N^2)$ can never be hit for a matrix with no special structure? (assume that we are solving a system with only one right hand side).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also are there other algorithm which solve these system &quot;exactly&quot; whose cost is less than $\mathcal{O}(N^3)$?  I am aware of Strassen algorithm which perform matrix multiplications in $\mathcal{O}(N^{\log_27})$. Can this be used to solve a linear system in $\mathcal{O}(N^{\log_27})$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(&lt;strong&gt;Note&lt;/strong&gt;: The system has no special structure. Say the matrix is just made up of entries drawn out of a random number generator. I am not worried about the stability and other numerical intricacies of the method as of now. I would appreciate if someone could point to some work done in this regard.)&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2011-01-18T22:35:46.997" LastActivityDate="2011-01-18T22:35:46.997" Title="Cost of Solving Linear System" Tags="&lt;linear-algebra&gt;" AnswerCount="1" />
  <row Id="18032" PostTypeId="2" ParentId="18030" CreationDate="2011-01-18T21:57:42.677" Score="4" ViewCount="" Body="&lt;p&gt;Recall that $B$ is semi-positive-definite iff $x^T B x\ge 0$ for all vectors $x$, where the superscript $T$ denotes the transpose. In this case, $$x^TQx=x^T A^T A x.$$ Now, recall that $(Bv)^T=v^T B^T$, so $$x^TQ x=(Ax)^T(Ax).$$ Finally, note that the square of the norm $\|v\|$ of a vector $v$ is just $\|v\|^2=v^T v$, so $$x^T Qx=\|Ax\|^2 \ge0,$$ and this gives you that $Q$ is as wanted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that in general this won't be &lt;em&gt;positive&lt;/em&gt; definite, because there may be values of $x\ne0$ such that $Ax=0$. (This would be guaranteed if $m&amp;lt;n$. If $m&amp;gt;n$, it may or may not happen, it depends on $A$.)&lt;/p&gt;&#xA;" OwnerUserId="462" LastActivityDate="2011-01-18T21:57:42.677" CommentCount="3" />
  <row Id="18034" PostTypeId="2" ParentId="18029" CreationDate="2011-01-18T22:03:36.313" Score="2" ViewCount="" Body="&lt;p&gt;First: because each row of $A$ adds up to $n$, then you know that $(1,1,\ldots,1)^T$ is an eigenvector with eigenvalue $n$.  (This is an often useful shortcut: if each row of $A$ adds up to $k$, then $k$ is an eigenvalue with eigenvector $(1,1,\ldots,1)^T$; if every &lt;em&gt;column&lt;/em&gt; of $A$ adds up to $\ell$, then $\ell$ is an eigenvalue, though we don't know a priori what an eigenvector may be). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second, because $A$ is not invertible, you know that $0$ is an eigenvalue of $n$. The dimension of the eigenspace of $0$ is the nullity of $A$, which is easy to check is equal to $n-1$. That means that the multiplicity of $\lambda=0$ as an eigenvalue of $A$ is &lt;em&gt;at least&lt;/em&gt; $n-1$; you also know that the eigenvalue $\lambda=n$ has multiplicity &lt;em&gt;at least&lt;/em&gt; $1$. And you know that the sum of the multiplicities of all the eigenvalues is &lt;em&gt;at most&lt;/em&gt; $n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So that means that you already have &lt;em&gt;all&lt;/em&gt; the eigenvalues with their multiplicities: $\lambda=0$ is an eigenvalue with algebraic multiplicity $n-1$, and $\lambda=n$ is an eigenvalue with algebraic multiplicity $1$. That means that the characteristic polynomial is divisible by $\lambda^{n-1}(\lambda-n)$. Since the characteristic polynomial is of degree $n$ and leading coefficient $(-1)^n$, this means that the characteristic polynomial must be $(-1)^n\lambda^{n-1}(\lambda-n) = (-1)^n\lambda^n - (-1)^nn\lambda$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now you need to find the eigenspaces, of course...&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-18T22:10:21.363" LastActivityDate="2011-01-18T22:10:21.363" CommentCount="2" />
  <row Id="18035" PostTypeId="2" ParentId="18029" CreationDate="2011-01-18T22:06:04.317" Score="2" ViewCount="" Body="&lt;p&gt;There is an easier argument. Let's try to see directly what the eigenvalues and eigenvectors are. (I'll assume $n&amp;gt;1$. If $n=1$, $A=[1]$ and the only eigenvalue is 1 and any non-zero real(=$1\times 1$ matrix) is an eigenvector.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, since any two rows are dependent, the matrix $A$ is not invertible, so $0$ is an eigenvalue. Let $x=(a_1,\dots,a_n)^T$ be an eigenvector ($T$ denotes the transpose, so I do not have to write column vectors).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then $Ax=0$ iff $a_1+\dots+a_n=0$. Note that, given any $a_1,\dots,a_{n-1}$, there is exactly one $a_n$ that makes this equation true. So: The space of eigenvectors corresponding to 0 has dimension $n-1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An easy way of finding a basis for this space is to take the vectors (1): $$(1,0,0,\dots,0,-1)^T,(0,1,0,0,\dots,-1),\dots,(0,0,\dots,0,1,-1)^T.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We are left to find the non-zero eigenvalues. Since the trace of $A$ is $n$, and we already located $n-1$ eigenvalues, the remaining one is $n$. Let $x=(b_1,\dots,b_n)^T$ be an eigenvector. Then $Ax=nx$, or $b_1+\dots+b_n=nb_1=nb_2=\dots=n b_n$, so we can take (2): $$x=(1,1,\dots,1)^T.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any other eigenvector is either a (non-zero) linear combination of the vectors listed in (1), or a non-zero multiple of the vector in (2).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Note that the case $n=1$ is a particular case of what we found.)&lt;/p&gt;&#xA;" OwnerUserId="462" LastActivityDate="2011-01-18T22:06:04.317" />
  <row Id="18036" PostTypeId="1" CreationDate="2011-01-18T22:07:48.933" Score="2" ViewCount="88" Body="&lt;p&gt;Using basic operators, derive the additional operator intersection.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$R \cap S = ?$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I thought it might be: $(R \cup S) - ((R -S) \cup (S -R))$ but I also think there might be an easier way of doing it?&lt;/p&gt;&#xA;" OwnerUserId="5999" LastEditorUserId="2548" LastEditorDisplayName="" LastEditDate="2011-01-18T22:17:22.833" LastActivityDate="2011-04-19T00:42:13.087" Title="Different way of writing intersection in relational algebra." Tags="&lt;homework&gt;&lt;elementary-set-theory&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="18037" PostTypeId="2" ParentId="17969" CreationDate="2011-01-18T22:20:56.180" Score="3" ViewCount="" Body="&lt;p&gt;My favorite is Andrew Pressley's &quot;Elementary Differential Geometry.&quot;  The material in it is very standard, yet the treatment is more modern than do Carmo's.  However, I don't think there's much in the way of material for MatLab or Maple...&lt;/p&gt;&#xA;" OwnerUserId="640" LastActivityDate="2011-01-18T22:20:56.180" />
  <row Id="18038" PostTypeId="2" ParentId="18036" CreationDate="2011-01-18T22:25:08.063" Score="1" ViewCount="" Body="&lt;p&gt;You could write $R \cap S = R-(R-S)$ or $R \cap S = S-(S-R)$. As Arturo mentioned, you are looking at one set and throwing away everything that is just in that set.&lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2011-01-18T22:25:08.063" />
  <row Id="18039" PostTypeId="2" ParentId="18031" CreationDate="2011-01-18T22:27:01.813" Score="4" ViewCount="" Body="&lt;p&gt;I assuming we are talking in terms of number of multiplications.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the de-facto algorithm for solving linear equations is Gaussian Elimination, let us only consider Gaussian Elimination.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that Gaussian Elimination can invert matrices in the same complexity as it solves linear equations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is well known that Matrix Inversion is as easy (or hard, if you will) as Matrix Multiplication (see this: &lt;a href=&quot;http://www.lehigh.edu/~gi02/m242/08linstras.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.lehigh.edu/~gi02/m242/08linstras.pdf&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The best known lower bound for Matrix multiplication, is apparently $\Omega(n^2)$, I believe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I doubt you can find better lower bounds than that. In any case, the lower bounds you seek would very likely be same as the lower bounds for Matrix Multiplication. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, that does not really answer your question exactly, but hope that helps.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2011-01-18T22:27:01.813" CommentCount="1" />
  <row Id="18040" PostTypeId="2" ParentId="17943" CreationDate="2011-01-18T22:55:20.547" Score="4" ViewCount="" Body="&lt;p&gt;What you have is $f(x) = \frac{1}{2}x^TQx + cx$ and you want to evaluate $\frac{df(x)}{dx}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First make sure to get the dimensions of all the vectors and matrices in place. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here $x \in \mathbb{R}^{n \times 1}$, $Q \in \mathbb{R}^{n \times n}$ and $c \in \mathbb{R}^{1 \times n}$, $f(x) \in \mathbb{R}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This will help you to make sure that your arithmetic operations are performed on vectors of appropriate dimensions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now lets move on to the differentiation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;All you need to know are the following rules for vector differentiation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{d(x^Ta)}{dx} = \frac{d(a^Tx)}{dx} = a$$ where $x,a \in \mathbb{R}^{n \times 1}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that $x^Ta = a^Tx$ since it is a scalar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Some people follow a different convention i.e. treating the derivative as a row vector instead of a column vector. Make sure to stick to your convention and you will end up with the same conclusion in the end)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Make use of the above two results to get,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{d(x^TQx)}{dx} = (Q^T + Q)x \text{       (use chain rule here)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{d(cx)}{dx} = c^T$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, $\frac{df(x)}{dx} = \frac{1}{2}((Q^T + Q)x) + c^T$ and since $Q$ is symmetric, we get $Q^T = Q$ and hence we get $$\frac{df(x)}{dx} = Qx + c^T$$&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2011-01-18T23:12:33.410" LastActivityDate="2011-01-18T23:12:33.410" CommentCount="2" />
  <row Id="18041" PostTypeId="1" AcceptedAnswerId="18066" CreationDate="2011-01-18T23:07:04.450" Score="3" ViewCount="232" Body="&lt;p&gt;My questions are about a sequence or function with several variables. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;I vaguely remember some while ago one&#xA;of my teachers said taking limits of&#xA;a sequence or function with respect to&#xA;different variables is not&#xA;exchangeable everywhere, i.e.&#xA;$$ \lim_n \lim_m a_{n,m} \neq \lim_m \lim_n a_{n,m}, \quad \lim_x \lim_y f(x,y) \neq \lim_y \lim_x f(x,y).$$&#xA;So my&#xA;question is what are the cases or&#xA;examples when one can exchange the&#xA;order of taking limits and when one&#xA;cannot, to your knowledge? I would&#xA;like to collect the cases together,&#xA;and be aware of their difference and&#xA;avoid making mistakes. If you could&#xA;provide some general guidelines, that&#xA;will be even nicer!&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;To give you an example of what I am&#xA;asking about, this is a question that&#xA;confuses me: Assume $f: [0, \infty)&#xD;&#xA;   \rightarrow (0, \infty)$ is a&#xA;function, satisfying $$&#xD;&#xA;   \int_0^{\infty} x f(x) \, dx &amp;lt; \infty.&#xD;&#xA;   $$ Determine the convergence of this&#xA;series $\sum_{n=1}^{\infty}&#xD;&#xA;   \int_n^{\infty} f(x) dx$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer I saw is to exchange the&#xA;order of $\sum_{n=1}^{\infty}$ and&#xA;$\int_n^{\infty}$ as follows: $$&#xD;&#xA;   \sum_{n=1}^{\infty} \int_n^{\infty}&#xD;&#xA;   f(x) dx&#xD;&#xA;   = \int_1^{\infty} \sum_{n=1}^{\lfloor x \rfloor}  f(x) dx \leq&#xD;&#xA;   \int_1^{\infty} \lfloor x \rfloor &#xD;&#xA;   f(x) dx $$ where $\lfloor x \rfloor$&#xA;is the greatest integer less than&#xA;$x$. In this way, the answer proves the series converges. I was wondering why the two steps are valid? Is there some special meaning of the first equality? Because it looks similar to the tail sum formula for expectation of a random variable $X$ with possible values $\{ 0,1,2,...,n\}$: $$\sum_{i=0}^n i P(X=i) = \sum_{i=0}^n P(X\geq i).$$ The formula is from &lt;a href=&quot;http://books.google.com/books?id=L6IWgaCuilwC&amp;amp;pg=PA172&amp;amp;lpg=PA172&amp;amp;dq=expectation++sum+of+%22tail+probabilities%22&amp;amp;source=bl&amp;amp;ots=yzHtpkh5PF&amp;amp;sig=OWGy7xmkkI0czpZgE7NDGwXVztw&amp;amp;hl=en&amp;amp;ei=HyA2TdvPCYGC8gbv4aCiCQ&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=2&amp;amp;ved=0CBsQ6AEwAQ&quot; rel=&quot;nofollow&quot;&gt;Page 171 of Probability by Jim Pitman, 1993&lt;/a&gt;. Are they really related?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Really appreciate your help!  &lt;/p&gt;&#xA;" OwnerUserId="4705" LastEditorUserId="4705" LastEditorDisplayName="" LastEditDate="2011-01-19T01:25:16.417" LastActivityDate="2011-01-19T02:38:37.413" Title="when can we exchange order of two limits" Tags="&lt;real-analysis&gt;&lt;integral&gt;&lt;limit&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="2" />
  <row Id="18042" PostTypeId="1" CreationDate="2011-01-18T23:12:10.013" Score="1" ViewCount="70" Body="&lt;p&gt;There are 6 of x and 4 of y. A group of 5 is selected. What is the probability of there being 3x and 2 y? I completely forgot how to do this and my textbook doesn't have any information on it.&lt;/p&gt;&#xA;" OwnerUserId="6000" LastActivityDate="2011-01-19T02:26:48.533" Title="Probability homework question" Tags="&lt;homework&gt;&lt;probability&gt;" AnswerCount="1" CommentCount="6" />
  <row Id="18044" PostTypeId="1" AcceptedAnswerId="18094" CreationDate="2011-01-18T23:41:45.777" Score="1" ViewCount="135" Body="&lt;p&gt;$U$ and $W$ are two subspaces of vector space $V$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $U \oplus W = V$, then $\forall v \in V$, there exist two unique vectors $u \in U$ and $w \in W$ such that $v = u + w$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is the reverse true? That is, if any vector can has such unique decomposition, do we have $U \oplus W = V$? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can the above statements be generalized to finite number of subspaces $U_1,...,U_n$ instead of just two $U$ and $W$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for your help!&lt;/p&gt;&#xA;" OwnerUserId="4705" LastEditorUserId="4705" LastEditorDisplayName="" LastEditDate="2011-01-19T00:06:33.430" LastActivityDate="2011-01-19T06:00:29.603" Title="Decomposition of vector with respect to direct sum of vector subspaces" Tags="&lt;linear-algebra&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="18045" PostTypeId="1" AcceptedAnswerId="18080" CreationDate="2011-01-18T23:42:55.640" Score="3" ViewCount="168" Body="&lt;p&gt;Exercise 1.8.5 of Berenstein-Gay &quot;Complex variables&quot; asks to count the number of zeroes in $(0,2\pi)$ of certain trigonometric polynomial. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Towards the exercise, the book suggests to first show the following:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The roots of $$p(z)=a_0+a_1z+\dots+a_nz^n$$ lie in the unit disk. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Do you see how to solve the above?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(The exercise is used in several places later through the book, so I am very frustrated with myself for not being able to do this on my own. The original version of this posting asked for the whole exercise, which is somewhat unreasonable, as it is really several very different questions together. I am editing the question to address directly the part that has been answered. I may post the rest of 1.8.5 at a later date.)&lt;/p&gt;&#xA;" OwnerUserId="3340" LastEditorUserId="3340" LastEditorDisplayName="" LastEditDate="2011-02-10T21:13:23.507" LastActivityDate="2011-02-10T21:13:23.507" Title="Number of roots of trigonometric polynomial" Tags="&lt;complex-analysis&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="18046" PostTypeId="1" CreationDate="2011-01-18T23:59:29.567" Score="5" ViewCount="120" Body="&lt;p&gt;I was thinking about cardinality of all symmetric relations, for example in $\mathbb{Z}$. I know, that if I have finite set (which contains $n$ elements), there are $2^{\frac{n(n+1)}{2}}$ symmetric relations. But what happens when we are talking about infinite sets ($\mathbb{Z}$ for example)? How many symmetric relations are in $\mathbb{Z}$ or $\mathbb{N}$? What about equivalence relation?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I just want to know, how to think about such problems. If You can, please show me some examples of similar problems (maybe with hints, how to solve them).&lt;/p&gt;&#xA;" OwnerUserId="5291" LastActivityDate="2011-01-19T03:13:54.013" Title="Cardinality of relations set" Tags="&lt;elementary-set-theory&gt;&lt;relations&gt;" AnswerCount="2" />
  <row Id="18047" PostTypeId="1" AcceptedAnswerId="18112" CreationDate="2011-01-19T00:09:23.820" Score="3" ViewCount="156" Body="&lt;p&gt;In the Category of Modules over a Ring, the i-th Homology of a Chain Complex is defined as the Quotient&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ker &lt;em&gt;d&lt;/em&gt; / Im &lt;em&gt;d&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where &lt;em&gt;d&lt;/em&gt; as usual denotes the differentials, indexes skipped for simplicity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can this be generalized to a general Abelian Category? Do I have the notion of a Quotient there?&lt;/p&gt;&#xA;" OwnerUserId="6002" LastActivityDate="2011-01-19T09:40:13.180" Title="How to define Homology Functor in an arbitrary Abelian Category?" Tags="&lt;category-theory&gt;&lt;homological-algebra&gt;&lt;cohomology&gt;&lt;abelian-categories&gt;" AnswerCount="2" />
  <row Id="18048" PostTypeId="1" AcceptedAnswerId="18053" CreationDate="2011-01-19T00:10:51.183" Score="1" ViewCount="98" Body="&lt;p&gt;Let $f \in L^1_{loc}$. We call $x \in \mathbb R^n$ an Lebesgue point, if&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$\lim \limits_{R \rightarrow 0} \frac{1}{m(B_R(x))}\int_{B_R(x)} f \;\;\;\;$ exists&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;or&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$\lim \limits_{R \rightarrow 0} \;\; \sup \limits_{ B, B&amp;#39; \in \mathcal B_R(x)} \vert \dfrac{1}{m(B)}\int_{B} f - \dfrac{1}{m(B&amp;#39;)}\int_{B&amp;#39;} f \vert = 0$, where $\mathcal B_R(x)$ denotes all balls that contain $x$ with radius at most $R$&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Most calculus books state only one of these definitions, and whereas they state several conclusions or variations of the theorem, but they usually stick to one of the above definitions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, although the equivalence is obvious, and indeed trivial from 2. to 1., I do not manage to prove the implication from 1. to 2. - can someone please explain me, how this is done?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: I should clarify, that I am aware both definitions are in general different for, say, a continuous function with a jump. Nevertheless, for example in Wikipedia I have encountered both defintions entitled as Lebesgue point.&lt;/p&gt;&#xA;" OwnerUserId="3557" LastEditorUserId="3557" LastEditorDisplayName="" LastEditDate="2011-01-19T01:28:21.307" LastActivityDate="2011-01-19T01:28:21.307" Title="Different definitions for Lebesgue points" Tags="&lt;measure-theory&gt;&lt;classical-analysis&gt;" AnswerCount="1" />
  <row Id="18049" PostTypeId="2" ParentId="18046" CreationDate="2011-01-19T00:17:33.220" Score="2" ViewCount="" Body="&lt;p&gt;You can think of a symmetric relation as an undirected graph, with vertices being $\mathbb{Z}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since there are a countable number of edges (an infinite subset of $\mathbb{Z} \times \mathbb{Z}$), of which you can pick any subset, the number of symmetric relations is $2^{\mathbb{N}}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An equivalence relation is a subset of $\mathbb{Z}\times\mathbb{Z}$ and hence the number of equivalence relations $\le 2^{\mathbb{N}}$. Since we have the partitions $\{S, \mathbb{Z} - S\}$ for any $S \subset \mathbb{Z}$, the number of equivalence relations is also $2^{\mathbb{N}}$.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2011-01-19T00:58:41.687" LastActivityDate="2011-01-19T00:58:41.687" CommentCount="5" />
  <row Id="18050" PostTypeId="2" ParentId="18047" CreationDate="2011-01-19T00:17:54.730" Score="2" ViewCount="" Body="&lt;p&gt;The notion of a kernel makes sense in an arbitrary abelian category (or really a category with a zero object). By definition, a kernel of a morphism $A \to B$ is an object $C$ that represents the functor $Z \to ker(\hom(Z, A) \to \hom(Z,B))$ (where the latter ker is of abelian groups). This means that homming into $Z$ is the same as homming into $A$ such that the composite to $B$ is zero. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, one can define cokernels via a universal property. The image can be defined as the kernel of the cokernel (think of what that translates to for abelian groups), or equivalently as the cokernel of the kernel (as this is one of the axioms of an abelian category). A quotient is a special case of a cokernel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if you have  a complex in an abelian category (such that the composites of maps is zero), then one can  see from the universal properties that there is a map $Im(d) \to Ker(d)$. If this is an isomorphism, then the complex is called exact. If you'd like a concrete example to try this out, you might consider for instance the category of sheaves on a topological space, for which a bunch of exercises that work things out are in chapter II of Hartshorne.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cf. books on homological algebra and category theory, e.g. MacLane's &lt;em&gt;Categories for the working mathematician.&lt;/em&gt; &lt;/p&gt;&#xA;" OwnerUserId="536" LastActivityDate="2011-01-19T00:17:54.730" CommentCount="2" />
  <row Id="18051" PostTypeId="2" ParentId="16798" CreationDate="2011-01-19T00:19:43.043" Score="1" ViewCount="" Body="&lt;p&gt;The wikipedia article explains it pretty clearly: &lt;a href=&quot;http://en.wikipedia.org/wiki/Courant%E2%80%93Friedrichs%E2%80%93Lewy_condition&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Courant%E2%80%93Friedrichs%E2%80%93Lewy_condition&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's simply the velocity in the given direction multiplied by the time step and divided by the step size in the given direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In 1D that would be: vx * dt / dx.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're talking about the constant that the Courant number needs to be smaller than, that's dependent on your system.&lt;/p&gt;&#xA;" OwnerUserId="6003" LastActivityDate="2011-01-19T00:19:43.043" CommentCount="2" />
  <row Id="18052" PostTypeId="2" ParentId="18044" CreationDate="2011-01-19T00:28:19.823" Score="2" ViewCount="" Body="&lt;p&gt;Yes. In fact it suffices to verify that the uniqueness of the decomposition implies that $U$ and $W$ are disjoint. Together, they also span $V$, since each vector possesses such a decomposition. To make the proof explicit, consider the mapping $V \rightarrow U \times W$, sending each vector $v$ to its decomposition $(u,w)$; then you can verify the axioms for the direct sum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the general case, you can do this by recurrence, using the basic properties of direct sums (in particular associativity).&lt;/p&gt;&#xA;" OwnerUserId="6004" LastEditorUserId="462" LastEditorDisplayName="" LastEditDate="2011-01-19T00:41:35.960" LastActivityDate="2011-01-19T00:41:35.960" CommentCount="1" />
  <row Id="18053" PostTypeId="2" ParentId="18048" CreationDate="2011-01-19T00:42:30.507" Score="1" ViewCount="" Body="&lt;p&gt;It isn't done.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;0 is a Lebesgue_1 point of signum&#xA;&lt;br&gt;&#xA;0 is not a Lebesgue_2 point of signum&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Sign_function&quot; rel=&quot;nofollow&quot;&gt;signum&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5114" LastActivityDate="2011-01-19T00:42:30.507" CommentCount="1" />
  <row Id="18054" PostTypeId="2" ParentId="18046" CreationDate="2011-01-19T01:02:50.107" Score="4" ViewCount="" Body="&lt;p&gt;To see that the number of equivalence relations on an infinite set $A$ of size $\kappa$ is $2^{\kappa}=|{\mathcal P}(A)|$ (i.e., the largest possible size), recall that any set of size $\kappa$ can be split into $\kappa$ sets of size $\kappa$: $\kappa=\kappa\times\kappa$. Say $A$ has size $\kappa$. Fix a partition $A=\bigcup_{i\in I}A_i$, where $I$ is an index set of size $\kappa$, each $A_i$ has size $\kappa$, and $A_i\cap A_j=\emptyset$ if $i\ne j$. Also, for each $A_i$, find a partition $A_i=B_i\cup C_i$ where each $B_i,C_i$ is non-empty.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given $D\subseteq I$, let $E_D$ be the equivalence relation on $A$ defined as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $i\in D$, then $B_i$ and $C_i$ are equivalence classes. If $i\in I$ is not in $D$, then the whole of $A_i$ is an equivalence class. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;More precisely, in case the above description is not clear: $a E_D b$ for $a,b\in A$, iff both $a,b$ are in the same $A_i$ (for some -unique- $i\in I$) and (if $i\notin D$, then both $a,b$ are in $B_i$ or both are in $C_i$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, from $E_D$, we can reconstruct $D$, so the map $f:{\mathcal P}(I)\to {\mathcal E}$, where ${\mathcal E}$ is the set of equivalence relations on $A$, given by $f(D)=E_D$ is injective, and $|{\mathcal E}|\ge 2^\kappa$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(In fact, for this inequality to hold, all we needed is that $2\times\kappa=\kappa$: We could have taken each $A_i$ of size 2 and each $B_i,C_i$ a singleton. However, the argument below uses that $\kappa\times\kappa=\kappa$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since each class is a subset of $A\times A$, and different classes are disjoint, each equivalence relation has at most $\kappa=|A|$ classes (cannot have more classes than elements of $A$), and each class is chosen from ${\mathcal P}(A\times A)$, so there are at most $\kappa\times 2^{|\kappa\times\kappa|}=2^\kappa$ equivalence relations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It follows that the number of equivalence relations is $2^\kappa$, as claimed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since each equivalence relation is symmetric, this also shows that there are precisely $2^\kappa$ symmetric relations on $A$ if $|A|=\kappa$.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Let me close with a technical remark: If $A$ is countable (so $\kappa$ is usually denoted $\aleph_0=|{\mathbb N}|=|{\mathbb Z}|$), the equality $\kappa=\kappa\times\kappa$ and the equality $\kappa 2^\kappa=2^\kappa$ can be verified without using the axiom of choice. Same if $A$ has the size of the reals (so $\kappa$ is usually denoted ${\mathfrak c}$ or $2^{\aleph_0}=|{\mathcal P}({\mathbb N})|$). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, for arbitrary infinite $A$, the equalities require the axiom of choice. I haven't checked what the number of equivalence relations for an arbitrary infinite $A$ is if choice fails. I do not think it is $2^{|A|}$ in general.&lt;/p&gt;&#xA;" OwnerUserId="462" LastEditorUserId="462" LastEditorDisplayName="" LastEditDate="2011-01-19T03:13:54.013" LastActivityDate="2011-01-19T03:13:54.013" CommentCount="2" />
  <row Id="18055" PostTypeId="2" ParentId="17842" CreationDate="2011-01-19T01:02:59.350" Score="0" ViewCount="" Body="&lt;p&gt;Consider some collection of vectors satisfying the requirements. If we apply any regular linear transformation, we get another collection satisfying the requirements. There are $$(q^n-1)\cdots(q^n-q^{n-1}) = q^{n^2} (1-q^{-n}) \cdots (1-q^{-1}) \geq c_q q^{n^2}$$ of these. However, we might get the same collection. Let us call two collections &lt;i&gt;equivalent&lt;/i&gt; if there's a regular linear transformation mapping one to the other. This is an equivalence relation, so if we upper bound the size of an equivalence class, we will get a lower bound on the number of collections.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How many regular linear transformations can take a collection into itself? Choose some arbitrary $k$ vectors from the collection. These are mapped into some other $k$ vectors from the collections (less than $l^k$ possibilities). How many regular matrices map the former to the latter? Assuming that the first $k$ vectors where $e_1,\ldots,e_k$, the first $k$ rows of the matrix are forced, and for the rest we have $$(q^n-q^k)\cdots(q^n-q^{n-1}) \leq q^{n(n-k)}$$ possibilities. Thus an equivalence classes consists of at most $q^{n(n-k)} l^k$ regular matrices, and the resulting lower bound on the number of collections is $$\frac{c_q q^{n^2}}{q^{n(n-k)} l^k} = c_q \left(\frac{q^n}{l}\right)^k.$$&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2011-01-19T01:02:59.350" />
  <row Id="18056" PostTypeId="1" CreationDate="2011-01-19T01:26:20.747" Score="0" ViewCount="72" Body="&lt;p&gt;The aim is to find a formula for the question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For n=2 i get (2^(2^n)=16 possible functions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the solution for a boolean function with 2 attributes:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;01) a     -&amp;gt;  separable&#xA;02) b     -&amp;gt;  separable&#xA;03) not a    -&amp;gt;  separable&#xA;04) not b    -&amp;gt;  separable&#xA;05) a and b   -&amp;gt;  separable&#xA;06) a or b    -&amp;gt;  separable&#xA;07) a xor b   -&amp;gt; not separable&#xA;08) a nand b   -&amp;gt;  separable&#xA;09) a nor b   -&amp;gt;  separable&#xA;10) a xnor b   -&amp;gt; not separable&#xA;11) (not a) and b  -&amp;gt;  separable&#xA;12) a and (not b)  -&amp;gt;  separable&#xA;13) (not a) or b  -&amp;gt;  separable&#xA;14) a or (not b) -&amp;gt;  separable&#xA;15) (not a) xor b  -&amp;gt; not separable&#xA;16) a xor (not b) -&amp;gt; not separable&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;But what is the formular for n? &lt;/p&gt;&#xA;" OwnerUserId="6006" LastEditorUserId="6006" LastEditorDisplayName="" LastEditDate="2011-01-19T01:39:03.243" LastActivityDate="2011-01-19T01:39:03.243" Title="How many presentable boolean functions with n attributes are linear separable?" Tags="&lt;boolean-algebra&gt;&lt;linear-algebra&gt;" CommentCount="6" />
  <row Id="18057" PostTypeId="1" AcceptedAnswerId="18061" CreationDate="2011-01-19T01:27:55.500" Score="4" ViewCount="139" Body="&lt;p&gt;i had a geometry/trignometry problem come up at work today, and i've been out of school too long: i've lost my tools.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i'm starting with a rectangle of known width (&lt;code&gt;w&lt;/code&gt;) and height (&lt;code&gt;h&lt;/code&gt;). For graphical simplification i can convert it into a right-angle triangle:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/5VG2u.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i'm trying to find the coordinates of that point above which is perpendicular to the origin:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/MR8kC.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i've labelled the opposite angle &lt;code&gt;t1&lt;/code&gt; (i.e. &lt;code&gt;theta1&lt;/code&gt;, but Microsoft Paint cannot easily do greek and subscripts), and i deduce that the two triangles are similar (i.e. they have the same shape):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/Vl4Oe.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we come to my problem. Given &lt;code&gt;w&lt;/code&gt; and &lt;code&gt;h&lt;/code&gt;, find &lt;code&gt;x&lt;/code&gt; and &lt;code&gt;y&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now things get very difficult to keep drawing graphically, to explain my attempts so far.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if i call the length of the line segment common to both triangles &lt;code&gt;M&lt;/code&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/Fz0Vw.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;then:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;M = w∙sin(t1)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Now i can focus on the other triangle, which i'll call &lt;code&gt;O&lt;/code&gt;-&lt;code&gt;x&lt;/code&gt;-&lt;code&gt;M&lt;/code&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/lhvvf.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and use trig to break it down, giving:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;x = M∙sin(t1)&#xA;  = w∙sin(t1)∙sin(t1)&#xA;&#xA;y = M∙cos(t1)&#xA;  = w∙sin(t1)∙cos(t1)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;with&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;t1 = atan(h/w)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Now this all works (i think, i've not actually tested it yet), and i'll be giving it to a computer, so speed isn't horribly important.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But my god, there must have been an easier way to get there. i feel like i'm missing something.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;By the way, what this will be used for is drawing a linear gradient in along that perpendicular:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/NAr5m.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="835" LastActivityDate="2011-01-19T21:55:08.323" Title="Simpler solution to this geometry/trig problem?" Tags="&lt;geometry&gt;&lt;trigonometry&gt;" AnswerCount="5" CommentCount="3" />
  <row Id="18058" PostTypeId="2" ParentId="18057" CreationDate="2011-01-19T01:39:15.640" Score="3" ViewCount="" Body="&lt;p&gt;Parametrize the line from $(w,0)$ to $(0,h)$ by $(w,0) + t(-w, h)$. Then you are searching for the point $(x,y)$ on the line such that $(x,y)\cdot (-w,h) = 0$. &lt;/p&gt;&#xA;" OwnerUserId="2900" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2011-01-19T21:17:35.033" LastActivityDate="2011-01-19T21:17:35.033" CommentCount="3" />
  <row Id="18059" PostTypeId="1" AcceptedAnswerId="18062" CreationDate="2011-01-19T01:43:06.740" Score="0" ViewCount="121" Body="&lt;p&gt;Hello.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Im having trouble with this derivative: 3&lt;sup&gt;x/2&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that the derivative of  c&lt;sup&gt;x&lt;/sup&gt; = c&lt;sup&gt;x&lt;/sup&gt;*lnc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is  3&lt;sup&gt;x/2&lt;/sup&gt; =  3&lt;sup&gt;x/2&lt;/sup&gt;Ln3 correct?&lt;/p&gt;&#xA;" OwnerUserId="6007" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2011-01-19T01:54:39.153" LastActivityDate="2011-01-19T01:54:39.153" Title="Derivative of 3^(x/2)" Tags="&lt;calculus&gt;&lt;homework&gt;" AnswerCount="2" />
  <row Id="18060" PostTypeId="2" ParentId="18057" CreationDate="2011-01-19T01:49:30.780" Score="1" ViewCount="" Body="&lt;p&gt;Suppose $h = (h_1,h_2)$ and $w = (w_1, w_2)$. Then you know the slope of the line. Let's call this $m_1$. You also know that the line perpendicular to the point $(x,y)$ has slope $-\frac{1}{m_1}$. You get a system of equations. &lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2011-01-19T01:49:30.780" />
  <row Id="18061" PostTypeId="2" ParentId="18057" CreationDate="2011-01-19T01:50:49.810" Score="3" ViewCount="" Body="&lt;p&gt;An equation for the hypotenuse of the right triangle is $y=-\frac{h}{w}x+h$.  Its slope is $-\frac{h}{w}$ (change in $y$-coordinate over change in $x$-coordinate).  The slope of any line perpendicular to the hypotenuse is $\frac{w}{h}$ (perpendicular lines have slopes with product $-1$, or it is sometimes said that the slope of a perpendicular line is the &quot;opposite reciprocal&quot;), so an equation for the line through the origin and perpendicular to the hypotenuse is $y=\frac{w}{h}x$.  The point you want is the solution to the system of equations: $$\begin{align}y&amp;amp;=-\frac{h}{w}x+h\\y&amp;amp;=\frac{w}{h}x\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These equations give two expressions that are both equal to $y$, so set them equal: $$-\frac{h}{w}x+h=\frac{w}{h}x$$ and solve:$$h=\left(\frac{w}{h}+\frac{h}{w}\right)x=\frac{w^2+h^2}{wh}x$$  $$x=\frac{h^2w}{w^2+h^2}$$&#xA;Use one of the original equations to find $y$: $$y=\frac{w}{h}x=\frac{hw^2}{w^2+h^2}$$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;edit&lt;/em&gt; Working from &lt;a href=&quot;http://math.stackexchange.com/questions/18057/simpler-solution-to-this-geometry-trig-problem/18117#18117&quot;&gt;Hans Lundmark's answer, final method&lt;/a&gt;, a vector from the origin to the desired point is in the same direction as $\langle\frac{1}{w},\frac{1}{h}\rangle$ or equivalently $\langle h,w\rangle$.  The distance from the origin to the desired point is the length of the altitude—call this $d$.  Looking at the area of the triangle, it is both $\frac{1}{2}wh$ and $\frac{1}{2}d\sqrt{w^2+h^2}$, so $d=\frac{wh}{\sqrt{w^2+h^2}}$.  The desired vector is thus &#xA;$$\begin{align}&#xD;&#xA;\frac{\langle h,w\rangle}{|\langle h,w\rangle|}d&#xD;&#xA;&amp;amp;=\frac{\langle h,w\rangle}{\sqrt{w^2+h^2}}\frac{wh}{\sqrt{w^2+h^2}}&#xD;&#xA;\\&#xD;&#xA;&amp;amp;=\frac{\langle h,w\rangle wh}{w^2+h^2}&#xD;&#xA;\\&#xD;&#xA;&amp;amp;=\left\langle\frac{h^2w}{w^2+h^2},\frac{w^2h}{w^2+h^2}\right\rangle&#xD;&#xA;.\end{align}$$&lt;/p&gt;&#xA;" OwnerUserId="72" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2011-01-19T11:21:51.727" LastActivityDate="2011-01-19T11:21:51.727" CommentCount="2" />
  <row Id="18062" PostTypeId="2" ParentId="18059" CreationDate="2011-01-19T01:52:10.200" Score="2" ViewCount="" Body="&lt;p&gt;Well, almost. It's a composite function, so you need to use the chain rule; multiply with the derivative of $x/2$. So the correct solution is &#xA;$$ (3^{x/2})^&amp;#39; = \frac{1}{2} \cdot \text{ln}3 \cdot 3^{x/2}.$$&lt;/p&gt;&#xA;" OwnerUserId="4216" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2011-01-19T01:53:58.593" LastActivityDate="2011-01-19T01:53:58.593" />
  <row Id="18063" PostTypeId="2" ParentId="18059" CreationDate="2011-01-19T01:53:14.490" Score="3" ViewCount="" Body="&lt;p&gt;The equals sign should only be used for equality.  For example, when finding the derivative of $x^2$, please do not write &quot;$x^2 = 2x$&quot;.  It does not make sense.  If $f(x)=x^2$, then $f&amp;#39;(x)=2x$ would be a better way to state the derivative.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;No, $3^{x/2}\ln 3$ is not correct, but your formula for the derivative of $c^x$ is correct, and can be obtained using the chain rule by observing that $c^x=e^{x\ln c}$.  And if you know the derivative of $f(x)=3^x$, then you can find the derivative of $g(x)=f(x/2)$ using the chain rule.  Alternatively, note that $3^{x/2}=(\sqrt{3})^x$.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2011-01-19T01:53:14.490" CommentCount="6" />
  <row Id="18064" PostTypeId="1" AcceptedAnswerId="18077" CreationDate="2011-01-19T02:26:23.617" Score="1" ViewCount="73" Body="&lt;p&gt;$g(n)=n^a$,  $f(n)=n^b$, where $a$ and $b$ are real numbers such that:&#xA;$0 \lt a \lt b$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to use L'Hopital's technique to prove that $g(n)$ belongs to $f(n)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My attempt:&#xA;$$ \lim_{n\to\infty}\frac{n^a}{n^{a+1}}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm using $a+1$ to state that $b \gt a$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is that I don't know if using $a+1$ is correct.&lt;/p&gt;&#xA;" OwnerUserId="6007" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-19T03:04:34.523" LastActivityDate="2011-01-19T03:48:44.313" Title="L'Hopital, $g(n)=n^a$ and $f(n)= n^b$" Tags="&lt;calculus&gt;&lt;homework&gt;" AnswerCount="1" CommentCount="6" />
  <row Id="18065" PostTypeId="2" ParentId="18042" CreationDate="2011-01-19T02:26:48.533" Score="2" ViewCount="" Body="&lt;p&gt;We have $6$ $x$'s and $4$ $y$'s. Hence, the total number of $x$'s and $y$'s is $10$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The total number of ways $5$ of them are selected is $C(10,5)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The number of ways of choosing $3$ $x$'s from a pool of $6$ $x$'s is $C(6,3)$. The number of ways of choosing $2$ $y$'s from a pool of $4$ $y$'s is $C(4,2)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the total number of favorable cases is $C(6,3) \times C(4,2)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, the probability is $\frac{C(6,3) \times C(4,2)}{C(10,5)} = \frac{10}{21}$&lt;/p&gt;&#xA;" OwnerUserId="3302" LastActivityDate="2011-01-19T02:26:48.533" />
  <row Id="18066" PostTypeId="2" ParentId="18041" CreationDate="2011-01-19T02:38:37.413" Score="1" ViewCount="" Body="&lt;p&gt;A simple example of a doubly indexed sequence $a_{m,n}$ for which you cannot exchange limits is given in Rudin's &quot;Principles...&quot; Example 7.2 pg. 144:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $a_{m,n} = \frac{m}{m+n}$, then $$\lim_{n\rightarrow\infty}\lim_{m\rightarrow\infty}a_{m,n} = 1$$ but $$\lim_{m\rightarrow\infty}\lim_{n\rightarrow\infty}a_{m,n} = 0$$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, here is a previous post on this question which seems to throughly answer your other questions: &lt;a href=&quot;http://math.stackexchange.com/questions/15240/when-can-you-switch-the-order-of-limits&quot;&gt;When can you switch the order of limits?&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="1692" LastActivityDate="2011-01-19T02:38:37.413" />
  <row Id="18067" PostTypeId="1" CreationDate="2011-01-19T02:39:02.250" Score="2" ViewCount="85" Body="&lt;p&gt;Hi, Everyone:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am trying to understand the justification of step (*) below,&#xA;which I have seen used to find the integral of the Error Function Er(f); the&#xA;cdf of the standard-normal normal .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is the derivation I know:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We set $I:= \int_{-\infty}^\infty e^{-x^2}dx$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then we use:&#xA;$$I^2= \left( \int_{-\infty}^\infty e^{-x^2} dx\right) \left(\int_{-\infty}^\infty e^{-y^2} dy\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then we end up with: (*)&#xA;$$I^2=\int_{-\infty}^\infty e^{-(x^2+y^2)} dxdy$$ --and we have a nice polar&#xA;integral. question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do we conclude that $\big(\int f(x)dx\big) \big(\int f(y)dy\big) = \int f(x)f(y) dxdy$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;AFAIK , integration is not multiplicative. What gives?&lt;/p&gt;&#xA;" OwnerUserId="6008" LastEditorUserId="856" LastEditorDisplayName="" LastEditDate="2011-01-19T02:55:31.650" LastActivityDate="2011-01-19T03:08:02.103" Title="How to Justify Step in Finding Integral of Erf (Error Function, aka Normal c.d.f.)" Tags="&lt;calculus&gt;&lt;integral&gt;" AnswerCount="2" />
  <row Id="18068" PostTypeId="1" CreationDate="2011-01-19T02:48:02.117" Score="1" ViewCount="116" Body="&lt;p&gt;$$\begin{array}{l}&#xD;&#xA;\varphi  = \frac{{1 + \sqrt 5 }}{2}\\&#xD;&#xA;{\varphi ^n} = F(n)\varphi  + F(n - 1)&#xD;&#xA;\end{array}$$&#xA;This formula is very powerful to prove or discover the identities on Fibonacci number.For example: $$\varphi ({\varphi ^{2k}} + 1)({\varphi ^{2k}} - 1) = {\varphi ^{4k + 1}} - \varphi  = F(4k + 1)\varphi  - \varphi  + F(4k)$$&#xA;Compare the series of ${\sqrt 5 }$ of both sides, we can get &#xA;$$F_{4k+1} - 1 = F_{2k} L_{2k+1}$$&#xA;What identity on Fibonacci sequence can not be reduced from the Ultimate Formula? I have two questions:&#xA;1. Do you discover a new identity with this formula?&#xA;2. Can the Ultimate Formula help to find Fibonacci primes?&lt;/p&gt;&#xA;" OwnerUserId="1498" LastActivityDate="2011-01-19T02:48:02.117" Title="$\varphi^n=F(n)\varphi+F(n-1)$: the Ultimate Formula of Fibonacci numbers" Tags="&lt;fibonacci-number&gt;" CommentCount="5" FavoriteCount="1" />
  <row Id="18069" PostTypeId="1" AcceptedAnswerId="18076" CreationDate="2011-01-19T02:49:47.810" Score="0" ViewCount="178" Body="&lt;p&gt;What is the derivative of $f(x)= \log^{k}x$ where $k$ is a positive integer?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hello I have found that log^kn = (logn)^k &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Thomas H. Cormen, Introduction to Algorithms (2ed) p. 53 )&lt;/p&gt;&#xA;" OwnerUserId="6009" LastEditorUserId="6009" LastEditorDisplayName="" LastEditDate="2011-01-19T03:27:32.757" LastActivityDate="2011-01-30T16:44:50.363" Title="Derivative of $\log^{k}(x)$" Tags="&lt;calculus&gt;" AnswerCount="2" CommentCount="7" />
  <row Id="18071" PostTypeId="2" ParentId="18069" CreationDate="2011-01-19T02:53:41.863" Score="2" ViewCount="" Body="&lt;p&gt;Use the chain rule (extra characters).&lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2011-01-19T02:53:41.863" />
  <row Id="18072" PostTypeId="2" ParentId="18067" CreationDate="2011-01-19T03:00:57.240" Score="0" ViewCount="" Body="&lt;p&gt;This can be justified by &lt;a href=&quot;http://en.wikipedia.org/wiki/Fubini%27s_theorem&quot; rel=&quot;nofollow&quot;&gt;Fubini's Theorem&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2011-01-19T03:00:57.240" />
  <row Id="18073" PostTypeId="2" ParentId="18067" CreationDate="2011-01-19T03:08:02.103" Score="2" ViewCount="" Body="&lt;p&gt;What you really have in the last step is a double integral, call it $J$: $$J = \int_{-\infty}^\infty \int_{-\infty}^\infty f(x) f(y)\,dx\,dy,$$ which really means $$J = \int_{-\infty}^\infty \left( \int_{-\infty}^\infty f(x) f(y)\,dx \right) \,dy.$$ Because the inner integral is over $x$, the factor of $f(y)$ is a constant and can be pulled out, and you get $$\begin{align}J &amp;amp;= \int_{-\infty}^\infty f(y) \left( \int_{-\infty}^\infty f(x)\,dx \right) \,dy \\ &amp;amp;= \int_{-\infty}^\infty f(y) I \,dy. \\ &amp;amp;=  I \int_{-\infty}^\infty f(y) \,dy \\ &amp;amp;=  I \cdot I \\ &amp;amp;= I^{2}.\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that integration is not multiplicative in the sense that $\int f(x) g(x) dx \neq \left(\int f(x) dx\right) \left(\int g(x) dx\right)$. However, it is true that $\iint f(x) g(y) dx dy = \left(\int f(x) dx\right) \left(\int g(y) dy\right)$.&lt;/p&gt;&#xA;" OwnerUserId="856" LastActivityDate="2011-01-19T03:08:02.103" />
  <row Id="18074" PostTypeId="1" AcceptedAnswerId="18086" CreationDate="2011-01-19T03:24:49.780" Score="2" ViewCount="110" Body="&lt;p&gt;Hi there, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question refers to functions with values in Banach spaces and under what conditions the limit of a sequence of measurable functions is also measurable. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But first, let me recall some well-known results from real analysis. Let $(X,\mathcal{F})$ be a measurable space, $\bar{\mathbb{R}}$ the extended real line, and $\mathcal{B}(\bar{\mathbb{R}})$ the standard Borel $\sigma$-algebra generated by the open sets of $\bar{\mathbb{R}}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) If a sequence of measurable functions $f_n:(X,\mathcal{F})\rightarrow(\bar{\mathbb{R}},\mathcal{B}(\bar{\mathbb{R}}))$ converges to a function $f$ pointwise in $X$, then $f$ is also measurable w.r.t. $\mathcal{F}$ and $\mathcal{B}(\bar{\mathbb{R}})$. (See, e.g., Rudin, Real and Complex Analysis, Theorem 1.14.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) This can be generalized to the case of a complete measure space $(X,\mathcal{F},\mu)$ and convergence of $f_n$ to $f$ pointwise-$\mu$-a.e. in $X$. (This is stated without proof in Hunter and Nachtergaele, Applied Analysis, Theorem 12.24. Can anyone point to a proof of this, or is it trivial?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dudley (Real Analysis and Probability, Theorem 4.2.2) generalizes part (1) above to functions with values in metric spaces. Therefore (1) also holds for functions with values in a Banach space $(Y,\Vert\cdot\Vert)$, with corresponding Borel $\sigma$-algebra $\mathcal{B}(Y)$. My question is: can part (2) be also generalized? In other words, does the following result hold:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $(X,\mathcal{F},\mu)$ be a complete measure space and $(Y,\Vert\cdot\Vert)$ be a Banach space. Let $f_n:(X,\mathcal{F})\rightarrow(Y,\mathcal{B}(Y))$  be a sequence of measurable functions that converges to a function $f$ pointwise-$\mu$-a.e. in $X$ (i.e., $\Vert f_n(x) - f(x)\Vert \rightarrow 0$ as $n\rightarrow\infty$, for $x$ $\mu$-a.e. in $X$), then $f$ is also measurable w.r.t. $\mathcal{F}$ and $\mathcal{B}(Y)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note, I use &quot;measurable&quot; in the standard sense: if $U\in\mathcal{B}(Y)$ then $f^{-1}(U)\in\mathcal{F}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems to me that the result should hold but I haven't been able to find a proof in the standard references. Is it trivial? Also, why should the measure space $(X,\mathcal{F},\mu)$ be complete? What fails if it is not complete. Finally, if the result holds, does this mean that standard measurability is equivalent to so-called strong or Bochner measurability? (Maybe this last question should be the topic of another post.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance!&lt;/p&gt;&#xA;" OwnerUserId="6010" LastActivityDate="2011-01-19T05:18:53.417" Title="Measurable functions with values in Banach spaces" Tags="&lt;real-analysis&gt;&lt;measure-theory&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="18075" PostTypeId="1" AcceptedAnswerId="18078" CreationDate="2011-01-19T03:28:25.903" Score="0" ViewCount="42" Body="&lt;p&gt;Can someone help me with the formula needed to obtain the following. I need to discount products on a sliding scale. If a customer purchases 2 products, he must get a 15% discount. If he purchases 10, he must get a 30% discount. I don't want a linear slope, but rather as the number of products increases, the increase in the discount decreases.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;2 =&amp;gt; 15%&#xA;4 =&amp;gt; ≈20%&#xA;10 =&amp;gt; 30%&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I need to formula with this, which should include a bend factor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know how do to this for FV, IV, but not this.&lt;/p&gt;&#xA;" OwnerUserId="4930" LastActivityDate="2011-01-19T05:05:21.900" Title="Bending algorithm needed" Tags="&lt;algorithms&gt;&lt;finance&gt;&lt;powers&gt;" AnswerCount="1" />
  <row Id="18076" PostTypeId="2" ParentId="18069" CreationDate="2011-01-19T03:30:24.887" Score="2" ViewCount="" Body="&lt;p&gt;Assuming $k$ is an exponent, use the chain rule, $f(x) = g(h(x))$, $\frac{d f}{d x} = \frac{d g}{d h} \frac{d h}{d x}$. In this case we have $g(h) = h^k$ and $h(x) = \log x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{d}{dx} (\log x)^k = \frac{d}{d h} \left(h^k\right) \frac{d}{d x} \left(\log x \right) = k h^{k-1} \frac{d}{dx} \log x = k (\log x)^{k-1} \frac{d}{dx} \log x$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if $\log$ is the natural logarithm you get:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{d}{dx} (\log x)^k = \frac{k}{x} (\log x)^{k-1}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $\log$ is the base $b$ logarithm you get $\log x = \frac{\ln x}{\ln b}$ (denoting the natural logarithm by $\ln$):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{d}{dx} \log^k x = k (\log x)^{k-1} \frac{d}{dx} \frac{\ln x}{\ln b} = k \frac{(\ln x)^{k-1}}{(\ln b)^{k-1}} \frac{1}{x \ln b} = k \frac{(\ln x)^{k-1}}{x (\ln b)^k}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Given your source you probably want to use $b = 2$ in the formula above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT2: Changed notation from $\log^k x$ to $(\log x)^k$) for clarity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another way to derive the equation above is to use the product rule $f(x) = f_1(x)f_2(x)$, $f&amp;#39;(x) = f_1&amp;#39;(x)f_2(x) + f_1(x)f_2&amp;#39;(x)$. Since $f(x) = (\log x)^k = \log x \log x \cdots \log x$. Now, when you derive you get one term for every factor in the product, each term containing the derivative once. All the terms are equal so you can simply multiply by the number of terms ($k$) instead:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{aligned}&#xD;&#xA;\frac{d}{dx} \left(\log x \right)^k = \left(\frac{d}{dx}  \log x \right) \log x \cdots \log x \log x + \dots + \log x \log x \cdots \log x \left( \frac{d}{dx} \ \log x \right) = \\ &#xD;&#xA;= k \left( \frac{d}{dx} \log x \right) (\log x)^{k-1}&#xD;&#xA;\end{aligned}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is the same expression as obtained above.&lt;/p&gt;&#xA;" OwnerUserId="5966" LastEditorUserId="5966" LastEditorDisplayName="" LastEditDate="2011-01-30T16:44:50.363" LastActivityDate="2011-01-30T16:44:50.363" />
  <row Id="18077" PostTypeId="2" ParentId="18064" CreationDate="2011-01-19T03:48:44.313" Score="0" ViewCount="" Body="&lt;p&gt;$\frac{g(n)}{f(n)}=\frac{n^a}{n^b}=n^{(a-b)}$ where $a-b&amp;lt;0$  So what is the limit as $n\rightarrow \infty $?&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-01-19T03:48:44.313" CommentCount="4" />
  <row Id="18078" PostTypeId="2" ParentId="18075" CreationDate="2011-01-19T03:56:16.853" Score="3" ViewCount="" Body="&lt;p&gt;You might want your discount function something like $$d(n) = 100 k_1 \times (1 - e^{-k_2 n})$$ where $d(n)$ is the percentage discount and $n$ is the number of products purchased and $k_1, k_2$ are constants with $0 &amp;lt; k_1 \leq 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt;&#xA;Choose $k_1 = 0.31$ and $k_2 = 0.35$.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2011-01-19T05:05:21.900" LastActivityDate="2011-01-19T05:05:21.900" CommentCount="8" />
  <row Id="18080" PostTypeId="2" ParentId="18045" CreationDate="2011-01-19T04:29:38.380" Score="3" ViewCount="" Body="&lt;p&gt;For the common exercise here are two similar proofs, but one is completely elementary:&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof 1&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the elementary proof.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{p(z)(z-1)}{z^{n+1}} = a_n  + \sum_{k=0}^{n-1} \frac{a_k - a_{k+1}}{z^{n-k}} - \frac{a_0}{z^{n+1}} = a_n  - \sum_{k=0}^{n-1} \frac{a_{k+1} - a_k}{z^{n-k}} - \frac{a_0}{z^{n+1}} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus&#xA; $$\displaystyle \left|\frac{p(z)(z-1)}{z^{n+1}}\right| \ge a_n  - \sum_{k=0}^{n-1} \frac{a_{k+1} - a_k}{|z^{n-k}|} - \frac{a_0}{|z^{n+1}|}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;using $\displaystyle |z_1 - z_2| \ge |z_1| - |z_2|$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now if $\displaystyle |z| \gt 1$, then $\displaystyle -\frac{a_{k+1} - a_k}{|z^{n-k}|} \gt -(a_{k+1} - a_k)$ and $\displaystyle -\frac{a_0}{|z^{n+1}|} \gt - a_0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle \left|\frac{p(z)(z-1)}{z^{n+1}}\right| \gt a_n  - \sum_{k=0}^{n-1} (a_{k+1} - a_k) - a_0 = 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus if $\displaystyle |z| \gt 1$, then $\displaystyle |p(z)| \neq 0$ and so all the roots of $\displaystyle p(z)$ satisfy $\displaystyle |z| \le 1$.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof 2&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similar to the above consider the polynomial&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$g(z) = p(z)(1-z) + a_n z^{n+1}  = a_0 + \sum_{k=0}^{n-1} (a_{k+1} - a_k) z^{k+1}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now given any $\displaystyle c &amp;gt; 0$, we have that, for any $\displaystyle z$ on the boundary of the unit circle,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$|g(z)| \le a_0 + \sum_{k=0}^{n-1} (a_{k+1} - a_k) = a_n \lt |(a_n + c)z^{n+1}|$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus if $\displaystyle f(z) = (a_n + c) z^{n+1}$, we have that $\displaystyle |g(z)| \lt |f(z)|$ on the boundary of the unit disk and thus both $\displaystyle g(z)-f(z)$ and $\displaystyle f(z)$ have the same number of roots satisfying $\displaystyle |z| \lt 1$, by using &lt;a href=&quot;http://en.wikipedia.org/wiki/Rouch%C3%A9%27s_theorem&quot; rel=&quot;nofollow&quot;&gt;Rouche's Theorem&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus for any $\displaystyle c &amp;gt; 0$, the polynomial $\displaystyle  p(z)(1-z) - c z^{n+1}$ has all its roots within the unit circle, i.e. the root $\displaystyle w$ with maximum modulus satisfies $\displaystyle |w| \lt 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now it is a well known fact that the eigenvalues of a complex matrix are continuous and thus the function of the root with the maximum modulus is continuous (though I am not completely sure of this). Taking limits as $\displaystyle c \to 0$ we get the result that the roots of $\displaystyle p(z)(1-z)$ all satisfy $\displaystyle |z| \le 1$.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2011-01-19T22:38:50.283" LastActivityDate="2011-01-19T22:38:50.283" CommentCount="1" />
  <row Id="18083" PostTypeId="1" CreationDate="2011-01-19T05:02:24.653" Score="12" ViewCount="386" Body="&lt;p&gt;Hi, everyone:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I read the result that every compact $n$-manifold is&#xA;a compactification of $\mathbb{R}^n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, for surfaces, this seems clear: we take&#xA;an n-gon, whose interior (i.e., everything in&#xA;the n-gon except for the edges) is homeo. to $\mathbb{R}^n$,&#xA;and then we identify the edges to end up with&#xA;a surface that is closed and bounded.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can do something similar with the $S^n$'s ; by&#xA;using a &quot;1-gon&quot; (an n-disk), and identifying&#xA;the boundary to a point. Or we can just use the&#xA;stereo projection to show that $S^n-\{{\rm pt}\}\sim\mathbb{R}^n$;&#xA;$S^n$ being compact (as the Alexandroff 1-pt. -compactification&#xA;of $\mathbb{R}^n$, i.e., usual open sets + complements of compact subsets of&#xA;$\mathbb{R}^n$). And then some messy work helps&#xA;us show that $\mathbb{R}^n$ is densely embedded in $S^n$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I don't see how we can generalize this statement&#xA;for any compact n-manifold. Can someone suggest any&#xA;ideas?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in Advance.&lt;/p&gt;&#xA;" OwnerUserId="6008" LastEditorUserId="1321" LastEditorDisplayName="" LastEditDate="2011-01-22T20:22:50.053" LastActivityDate="2011-01-22T20:22:50.053" Title="Is every Compact $n$-Manifold a Compactification of $\mathbb{R}^n$?" Tags="&lt;differential-geometry&gt;&lt;differential-topology&gt;" AnswerCount="3" CommentCount="5" FavoriteCount="1" />
  <row Id="18084" PostTypeId="1" CreationDate="2011-01-19T05:04:33.363" Score="2" ViewCount="241" Body="&lt;p&gt;Hi,&#xA;I need assistance with the following question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The particle starts from the origin on the number line .&#xA;It makes n steps at total.&#xA;At each step it jumps with the probability 1/2 to the origin and with the uniform probability jumps to the one of its neighbors.I have to prove that the expectation of its maximal distance from the origin is proportional to logn&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="6013" LastEditorUserId="6013" LastEditorDisplayName="" LastEditDate="2011-01-19T07:15:08.937" LastActivityDate="2011-01-19T18:59:56.240" Title="1-D Random walk expectation calculation" Tags="&lt;probability&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" />
  <row Id="18085" PostTypeId="1" AcceptedAnswerId="18087" CreationDate="2011-01-19T05:07:50.093" Score="0" ViewCount="176" Body="&lt;p&gt;In Physics, the Simple Harmonic Oscillator is represented by the equation $d^2x/dt^2=-\omega^2x$ .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By using the characteristic polynomial, you get solutions of the form $x(t)=Ae^{i\omega t} + Be^{-i\omega t}$. I get that you use Euler's formula $e^{i\theta}=cos\theta + isin\theta$, but I can't seem to find my way all the way to the 'traditional form' of $Dcos\omega t + Csin\omega t$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm stuck here: $A(cos\omega t + isin\omega t) + B(cos\omega t - isin\omega t)$. What am I missing in taking it all the way? It doesn't seem valid to me (not sure why or why not) to make $C = iA - iB$.&lt;/p&gt;&#xA;" OwnerUserId="6012" LastActivityDate="2011-01-19T08:52:01.617" Title="Simple Harmonic Oscillator Solution" Tags="&lt;differential-equations&gt;" AnswerCount="3" />
  <row Id="18086" PostTypeId="2" ParentId="18074" CreationDate="2011-01-19T05:18:53.417" Score="1" ViewCount="" Body="&lt;p&gt;Yes, it's rather trivial.  You might note the following fact: if $f$ is measurable, $\mu$ is complete, and $f = g$ $\mu$-a.e., then $g$ is measurable.  This is valid for functions from any complete measure space to any other measurable space.  (Proving it is a good exercise.)  Essentially, when you have a complete measure, you can change a function arbitrarily on a null set without breaking its measurability.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now suppose $f_n \to f$ $\mu$-a.e., with each $f_n$ measurable.  Let $A$ be the set of $x$ such that $f_n(x) \to f(x)$.  Note $A$ is a measurable set since $A^C$ is $\mu$-null and hence measurable (by completeness of $\mu$).  Let $$g_n = \begin{cases}f_n(x), &amp;amp; x \in A \\ 0, &amp;amp; x \notin A\end{cases}$$  Then clearly $g_n$ is measurable.  Now $g_n(x)$ converges for &lt;em&gt;all&lt;/em&gt; $x$, so the limit $g(x) = \lim_{n \to \infty} g_n(x)$ is measurable, as you know.  But $g(x) = f(x)$ on $A$, so $f = g$ a.e.  By the fact above, $f$ is measurable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If your underlying measure space is not complete, certainly things can go wrong.  For a trivial example, let $B$ be any set which is $\mu$-null but not measurable.  If $f_n$ is the zero function for all $n$, we have $f_n \to 1_B$ a.e., but $1_B$ is not a measurable function.&lt;/p&gt;&#xA;" OwnerUserId="822" LastActivityDate="2011-01-19T05:18:53.417" CommentCount="3" />
  <row Id="18087" PostTypeId="2" ParentId="18085" CreationDate="2011-01-19T05:19:05.113" Score="2" ViewCount="" Body="&lt;p&gt;When your write $x(t) = A e^{i \omega t} + B e^{-i \omega t}$, $A$ and $B$ in this expression are complex numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if you assume the function $x(t)$ is real, then they are related as $\bar{A} = B$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This can be seen since $\bar{x(t)} = x(t)$ which gives us $\bar{A} = B$ (since $e^{i \omega t}$ and $e^{- i \omega t}$ are orthonormal)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rewriting the complex exponentials as sine and cosine we get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x(t) = A (\cos(\omega t) + i \sin(\omega t)) + B (\cos(\omega t) - i \sin(\omega t)) = (A+B)\cos(\omega t) + i (A-B) \sin(\omega t)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now as proved before $\bar{A} = B$ and hence $(A+B)$ is real and $(A-B)$ is imaginary and hence we can write $(A+B) = C$ and $i (A-B) = D$ where $C,D$ are real numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And the solution becomes $$x(t) = C \cos(\omega t) + D \sin(\omega t)$$ where $C,D \in \mathbb{R}$&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2011-01-19T05:24:57.773" LastActivityDate="2011-01-19T05:24:57.773" CommentCount="1" />
  <row Id="18088" PostTypeId="1" AcceptedAnswerId="18091" CreationDate="2011-01-19T05:27:29.457" Score="7" ViewCount="299" Body="&lt;p&gt;This may be a stretch, but are there examples of proofs that prove that a proof exists for a theorem. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, if A is a theorem, and it is too tedious to prove that, is it possible to show that a proof exists for theorem A. &lt;/p&gt;&#xA;" OwnerUserId="4728" LastEditorUserId="1543" LastEditorDisplayName="" LastEditDate="2011-05-03T10:08:34.427" LastActivityDate="2011-05-08T04:53:13.180" Title="Existence Proofs" Tags="&lt;logic&gt;&lt;proof-theory&gt;" AnswerCount="4" CommentCount="7" FavoriteCount="1" />
  <row Id="18089" PostTypeId="2" ParentId="18085" CreationDate="2011-01-19T05:27:33.387" Score="0" ViewCount="" Body="&lt;p&gt;Remember that $x(t) = A e^{i\omega t} + B e^{-i\omega t}$ is a solution &lt;em&gt;for any values of $A$ and $B$&lt;/em&gt; which means that choosing $A = B = 1/2$ results in the solution&#xA;$$ x_1(t) = e^{i\omega t} + e^{-i \omega t}.&#xD;&#xA;$$&#xA;Euler's fomula, along with the properties $\cos(-\omega t) = \cos(\omega t)$ and $\sin(-\omega t) = - \sin(\omega t)$, simplifies this solution to&#xA;$$&#xD;&#xA;x_1 = \cos(\omega t).&#xD;&#xA;$$&#xA;Similarly, choosing $A = 1/(2i)$ and $B = -1/(2i)$ gives the solution $x_2(t) = \sin(\omega t)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, by linearity you can conclude that the linear combination $C x_1 + D x_2$ is also a solution.&lt;/p&gt;&#xA;" OwnerUserId="4229" LastActivityDate="2011-01-19T05:27:33.387" />
  <row Id="18090" PostTypeId="1" CreationDate="2011-01-19T05:37:54.310" Score="1" ViewCount="259" Body="&lt;p&gt;I am wondering if anyone could prove the following equivalent definition of recurrent/persistent state for Markov chains:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) $P(X_n=i,n\ge1|X_0=i)=1$&lt;p&gt;&#xA;2) Let $T_{ij}=min\{n: X_n=j|X_0=i\}$, a state $i$ is recurrent if for $\forall j$ such that $P(T_{ij}&amp;lt;\infty)&amp;gt;0$, one has $P(T_{ij}&amp;lt;\infty)=1$&lt;/p&gt;&#xA;" OwnerUserId="6016" LastEditorUserId="4728" LastEditorDisplayName="" LastEditDate="2011-01-19T06:08:26.970" LastActivityDate="2011-04-07T08:15:22.057" Title="proof of equivalent definitions of recurrent states for Markov chains" Tags="&lt;probability&gt;&lt;stochastic-processes&gt;" AnswerCount="3" CommentCount="7" />
  <row Id="18091" PostTypeId="2" ParentId="18088" CreationDate="2011-01-19T05:50:32.383" Score="7" ViewCount="" Body="&lt;p&gt;Sure.  For example, using the &lt;a href=&quot;http://en.wikipedia.org/wiki/G%C3%B6del%27s_completeness_theorem&quot; rel=&quot;nofollow&quot;&gt;completeness theorem&lt;/a&gt;, showing that a statement is semantically true is equivalent to showing that it has a syntactic proof, even though you don't have to actually give such a syntactic proof.  For example (!), you can show that a statement in &lt;a href=&quot;http://en.wikipedia.org/wiki/Propositional_calculus&quot; rel=&quot;nofollow&quot;&gt;zeroth-order logic&lt;/a&gt; is a tautology by arguing by contradiction, and it follows by the completeness theorem that that tautology has a proof from the axioms of zeroth-order logic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can't immediately think of any examples of &quot;real&quot; theorems with this property, but here's one way to generate a class of them: suppose you can reduce an infinitary statement to the checking of finitely many cases.  Then without checking them, you know that the act of checking them corresponds to the proof of &lt;em&gt;some&lt;/em&gt; theorem even though you know neither (all the steps of) the proof nor the theorem!  Maybe you want to know the theorem as well, though.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is also the notion of a &lt;a href=&quot;http://en.wikipedia.org/wiki/Zero-knowledge_proof&quot; rel=&quot;nofollow&quot;&gt;zero-knowledge proof&lt;/a&gt;.  This might not be quite what you had in mind, though, as it is interactive and probabilistic.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Upon further thought, without a clear distinction between the first and second uses of the word &quot;proof&quot; in your question (as is made by the first example above), I'm not sure what the difference between having a proof and having a proof that a proof exists is.  &lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2011-01-19T06:00:18.247" LastActivityDate="2011-01-19T06:00:18.247" CommentCount="4" />
  <row Id="18093" PostTypeId="1" AcceptedAnswerId="18130" CreationDate="2011-01-19T05:59:23.270" Score="4" ViewCount="326" Body="&lt;p&gt;If I toss a fair coin $n$ times, calculate the probability that no pattern HHTHTHH occurs.&lt;/p&gt;&#xA;" OwnerUserId="6016" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-19T06:07:51.090" LastActivityDate="2011-01-20T12:39:48.313" Title="probability question related to pattern in coin tossing" Tags="&lt;probability&gt;" AnswerCount="4" CommentCount="5" FavoriteCount="2" />
  <row Id="18094" PostTypeId="2" ParentId="18044" CreationDate="2011-01-19T06:00:29.603" Score="2" ViewCount="" Body="&lt;p&gt;To add to Gerben's answer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If every vector $v$ in $V$ has &lt;em&gt;at least one&lt;/em&gt; expression of the form $v=u+w$ with $u\in U$ and $w\in W$ ($U$ and $W$ subspaces of $V$), then $V=U+W$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If every vector $v$ in $V$ has &lt;em&gt;at most one&lt;/em&gt; (but possibly none) expression of the form $v=u+w$ with $u\in U$ and $w\in W$, then $U\cap W = \{0\}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if every vector $v$ in $V$ has &lt;em&gt;exactly one&lt;/em&gt; expression of the form $v=u+w$ with $u\in U$ and $w\in W$, then $V=U\oplus W$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For more than two spaces you have to be a bit careful: it is no longer enough for the $U_i$ to be &lt;em&gt;pairwise&lt;/em&gt; disjoint (that is, for $U_i\cap U_j$ to equal $\{0\}$ for $i\neq j$). For an example, take $V=\mathbb{R}^2$, $U_1$ the $x$-axis, $U_2$ the $y$-axis, and $U_3$ the line $x=y$. Then $U_1\cap U_2 = U_1\cap U_3 = U_2\cap U_3 = \{0\}$, but $V$ is not the direct sum of $U_1$ and $U_2$. Instead, you have that given subspaces $U_1,\ldots,U_m$ of $V$,&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Each vector $v\in V$ has &lt;em&gt;at least one&lt;/em&gt; expression of the form $v=u_1+\cdots + u_m$ with $u_i\in U_i$ if and only if $V=U_1+\cdots+U_m$ (the span of $U_1,\ldots,U_m$).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Each vector $v\in V$ has &lt;em&gt;at most one&lt;/em&gt; (but possibly no) expression of the form $v=u_1+\cdots + u_m$ with $u_i\in U_i$ if and only if for each $i\in\{1,\ldots,m\}$, $U_i\cap(\mathop{\sum}_{i\neq j}U_j) = \{0\}$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="742" LastActivityDate="2011-01-19T06:00:29.603" />
  <row Id="18095" PostTypeId="2" ParentId="18093" CreationDate="2011-01-19T06:11:29.337" Score="-1" ViewCount="" Body="&lt;p&gt;Your question is a little bit ambiguous. You say you toss a coin n times but at the same time you give an exact outcome. Maybe I'm not getting what you mean by pattern.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If n = 7 as you've shown then you should consider the sample space. For 1 toss it would be {H, T}; for two {HH, HT,TH, TT}. For n tosses it would have 2^n possible outcomes. Since you are looking (as far as I can tell) for the probability of an event not happening, then you would consider the probability of all of the events that can happen minus the probability of the events you don't want to happen. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This would result in 1 - 1/2^n.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope this helps.&lt;/p&gt;&#xA;" OwnerUserId="6018" LastActivityDate="2011-01-19T06:11:29.337" CommentCount="4" />
  <row Id="18097" PostTypeId="1" AcceptedAnswerId="18101" CreationDate="2011-01-19T06:19:52.833" Score="1" ViewCount="127" Body="&lt;p&gt;I'm reading through the first chapter of Ahlfors's Complex Analysis book, and during the section on stereographic projections, he says that we can map any $z = x+iy \in \mathbb{C}$ onto the unit sphere in three dimensions injectively using the equation $z= \frac{x_1+ix_2}{1-x_3}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;He then says that $x:y:-1= x_1:x_2:x_3-1$, which implies that $(x,y,0)$, $(x_1,x_2,x_3)$, and $(0,0,1)$ all lie on the same line. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is what the colon symbol means in this context. I only remember it being used in the context of sets as a replacement for the &quot;|&quot; symbol. &lt;/p&gt;&#xA;" OwnerUserId="1736" LastActivityDate="2011-01-19T13:44:28.983" Title="Notation question on the colon symbol" Tags="&lt;complex-analysis&gt;&lt;notation&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="18098" PostTypeId="1" CreationDate="2011-01-19T06:20:46.670" Score="0" ViewCount="30" Body="&lt;p&gt;You have 4 machines that are supposed to create 100mg chocolate bars. One of the four machines is not working correctly and is making 90mg chocolate bars. You only have 1 chance to weigh the chocolate bars of 2 of the machines. What is the mathematical equation capable of pulling this off?&lt;/p&gt;&#xA;" OwnerUserId="6019" LastActivityDate="2011-01-19T06:28:13.520" Title="4 machines producing 100mg, one of them is making 90mg, once chance to weigh 2 machines versus each other..." Tags="&lt;algebra&gt;" AnswerCount="1" />
  <row Id="18099" PostTypeId="2" ParentId="18098" CreationDate="2011-01-19T06:23:06.940" Score="2" ViewCount="" Body="&lt;p&gt;I presume you need to find the defective machine?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This seems to be variant of a classic puzzle. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Assume you have a scale which tells&#xA;  you the exact weight and you are&#xA;  allowed to take any amount of bars from each&#xA;  machine. Each machine is supposed to&#xA;  produce 100gm bars, but one of them is&#xA;  defective and produces 90gm bars. In&#xA;  one weighing how do you find the&#xA;  defective machine?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The solution for that is to take one bar from machine one, two from machine two etc and weight the total and consider the difference between the weight you get and the expected total of 1kg.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As to your puzzle, it is not possible to guarantee finding the defective machine in &lt;em&gt;one&lt;/em&gt; weighing, if you are only allowed to take bars from two of the four machines.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2011-01-19T06:28:13.520" LastActivityDate="2011-01-19T06:28:13.520" />
  <row Id="18100" PostTypeId="2" ParentId="18097" CreationDate="2011-01-19T06:25:08.947" Score="1" ViewCount="" Body="&lt;p&gt;The notation $a:b = c:d$ usually means that the proportions between $a,b$ and $c,d$ are the same, e.g. $a/b=c/d$ or more generally $ad = bc$. In this case we have a triple proportion.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2011-01-19T06:25:08.947" />
  <row Id="18101" PostTypeId="2" ParentId="18097" CreationDate="2011-01-19T06:27:05.693" Score="5" ViewCount="" Body="&lt;p&gt;It is a symbol that represents ratios; $A:B$ means &quot;the ratio of $A$ to $B$&quot;. (Shows up all the time in Euclid; e.g., &lt;a href=&quot;http://aleph0.clarku.edu/~djoyce/java/elements/bookV/bookV.html&quot; rel=&quot;nofollow&quot;&gt;Book V&lt;/a&gt;.) When we write &quot;$A:B=C:D$&quot;, we mean the ratio of $A$ to $B$ is the same as the ratio of $C$ to $D$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here he is talking about three ratios; the fact that $x:y:-1 = x_1:x_2:x_3$ means that the vectors determined by $(x,y,-1)$ and by $(x_1,x_2,x_3)$ are parallel. Note that $(x,y,-1) = (x,y,0) - (0,0,1)$ is the vector with starting point at $(0,0,1)$ and end point at $(x,y,0)$; and $(x_1,x_2,x_3-1) = (x_1,x_2,x_3) - (0,0,1)$ is the vector with starting point at $(0,0,1)$ and endpoint at $(x_1,x_2,x_3)$. Since $x:y:-1 = x_1:x_2:x_3$, they are parallel; since they both start at the same point, that means that the line through $(0,0,1)$ and $(x,y,0)$ (determined by the first vector) and the line through $(0,0,1)$ and $(x_1,x_2,x_3)$ (determined by the second vector) are the same, since they are parallel and they both go through $(0,0,1)$. &lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2011-01-19T06:27:05.693" CommentCount="4" />
  <row Id="18102" PostTypeId="2" ParentId="18084" CreationDate="2011-01-19T06:37:42.473" Score="1" ViewCount="" Body="&lt;p&gt;Actually the expectation is constant. Going back from the last step, the time until the first restart is (roughly) distributed $\mathrm{Geom}(1/2)$. If the walk lasts for $n$ steps, then the expected distance is about $\sqrt{n}$. So the expected distance is roughly $\int_0^n \sqrt{x}/2^x \, dx$. As it happens, this integral converges, i.e. $\int_0^\infty \sqrt{x}/2^x \, dx &amp;lt; \infty$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since already $\int_0^\infty x/2^x \, dx &amp;lt; \infty$, we didn't even need to know the expected distance after $n$ steps (it's clearly at most $n$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe the question was really about the expected &lt;i&gt;maximal&lt;/i&gt; distance from the origin?&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2011-01-19T06:37:42.473" CommentCount="8" />
  <row Id="18103" PostTypeId="2" ParentId="18088" CreationDate="2011-01-19T06:41:24.763" Score="5" ViewCount="" Body="&lt;p&gt;I think one example of what you are talking about is the constant applications of the Deduction Theorem that are implicit in most of mathematics; the Deduction Theorem which says that if you can prove $B$ from $\Delta\cup\{A\}$, then from $\Delta$ you can prove $A\rightarrow B$:&#xA;$$\Delta\cup \{A\}\models B \Longrightarrow \Delta\models A\rightarrow B.$$&#xA;Hofstadter, in &lt;strong&gt;Goedel, Escher and Bach&lt;/strong&gt;, calls it &lt;em&gt;&quot;the fantasy rule&quot;&lt;/em&gt;, because in order to prove that $A\rightarrow B$ holds, you get to fantasize that you actually &lt;em&gt;know&lt;/em&gt; $A$ is true and then prove $B$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The proof of the Deduction Theorem actually tells you how to take a proof of $B$ from $\Delta\cup\{A\}$ and use it to produce a proof of $A\rightarrow B$ from $\Delta$. The procedure is mechanical, but somewhat involved; a short proof of $B$ from $\Delta\cup\{A\}$ will usually yield a somewhat lengthy proof of $A\rightarrow B$ from $\Delta$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a sense, almost every time we prove a theorem we are not really proving the theorem; the vast majority of theorems are of the form $\Delta\models A\rightarrow B$, but we almost invariably prove $\Delta\cup\{A\}\models B$ (when a theorem says something like &quot;If $A$ and $B$ are disjoint, then $|A\cup B|=|A|+|B|$&quot;, then the proof will usually start with &lt;em&gt;&quot;Assume that $A$ and $B$ are disjoint...&quot;&lt;/em&gt;). When we do so, what we have done is prove that a proof exists, without actually providing the proof of the proposition we claim to be proving. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are similar &quot;Metatheorems&quot; for other typical rules proof arguments, like proofs by cases, proof by contradiction, etc. &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-19T19:15:58.433" LastActivityDate="2011-01-19T19:15:58.433" />
  <row Id="18104" PostTypeId="2" ParentId="18093" CreationDate="2011-01-19T07:34:43.617" Score="2" ViewCount="" Body="&lt;p&gt;Construct a deterministic finite automaton which detects the regular expression $.*(HHTHTHH).*$ such that the sole accepting state $A$ always transitions to a state $B$, which always transitions to itself. The starting state is $S$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is basically a directed graph and we can find the transition matrix $P$, each row of which gives the probabilities of getting from one state (i.e. vertex) corresponding to that row, to the next states (one vertex corresponding to each column).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You basically need the entries corresponding to starting state $S$ and ending state $A$, from each of $P, P^2,  P^3, \dots , P^n$ and add them up to get the probability that the pattern $HHTHTHH$ occurs (or more simply the $(S,B)$ entry of $P^{n+1}$ will do). Subtract that from $1$ and you are done.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The $(S,A)$ entry of $P^k$ gives the probability that we find the pattern exactly at the end of $k$ tosses, and no tosses before that.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2011-01-19T11:13:55.213" LastActivityDate="2011-01-19T11:13:55.213" CommentCount="2" />
  <row Id="18105" PostTypeId="1" AcceptedAnswerId="18159" CreationDate="2011-01-19T07:40:02.253" Score="7" ViewCount="217" Body="&lt;p&gt;I'm glancing over Milnor's paper on exotic 7-spheres, and one of the first few lines says, `every closed 7-manifold $M^7$ is the boundary of an 8-manifold $B^8$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's what I don't understand: The unoriented cobordism ring is isomorphic to a polynomial ring over $\mathbb{Z}_2$ with a generator in every degree except $2^m -1$ for any $m$. In particular, there are generators $x_2$ and $x_5$ in degrees $2$ and $5$, respectively. So: why does their product not represent a nontrivial element of the group of cobordism classes of 7-manifolds?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Apologies if this is a silly question... I'm new to all of this. &lt;/p&gt;&#xA;" OwnerUserId="423" LastActivityDate="2011-01-19T18:24:20.967" Title="Why does every 7-manifold bound an 8-manifold?" Tags="&lt;algebraic-topology&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="2" />
  <row Id="18106" PostTypeId="2" ParentId="18097" CreationDate="2011-01-19T07:41:53.753" Score="1" ViewCount="" Body="&lt;p&gt;The colon notation $\rm\ (x\::\:y\::\:z)\ $ denotes a point in &lt;em&gt;projective&lt;/em&gt; space (vs. comma notation $\rm\ (x\:,\:y\:,\:z)\ $ for points in affine space).$\ $ The notation highlights the fact that $\rm\  (x\::\:y\::\:z)\ =\ (\lambda\:x\::\:\lambda\:y\::\:\lambda\:z)\:,\ $ which explains the equality stated by Ahlfors. See for example &lt;a href=&quot;http://math.mit.edu/~kedlaya/geometryunbound/geom-projective.pdf&quot; rel=&quot;nofollow&quot;&gt;Kedlaya: Projective Geometry.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2011-01-19T13:44:28.983" LastActivityDate="2011-01-19T13:44:28.983" />
  <row Id="18109" PostTypeId="2" ParentId="18085" CreationDate="2011-01-19T08:52:01.617" Score="0" ViewCount="" Body="&lt;p&gt;Dear tau-neutrino,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the answers given are absolutely fine as they show the way from $e^{\pm \mathrm{i} \omega t}$ to $\sin$ and $\cos$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It might be even shorter if you use that&#xA;$$\sin(x) \equiv \frac{1}{2\mathrm{i}}\left( e^{\mathrm{i}x} - e^{-\mathrm{i}x} \right)$$&#xA;and&#xA;$$\cos(x) \equiv \frac{1}{2}\left( e^{\mathrm{i}x} + e^{-\mathrm{i}x} \right)\, .$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This can be interpreted as a definition for the functions or seen from its &lt;a href=&quot;http://en.wikipedia.org/wiki/Sine#Series_definition&quot; rel=&quot;nofollow&quot;&gt;series definitions&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sincerely&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Robert&lt;/p&gt;&#xA;" OwnerUserId="2671" LastActivityDate="2011-01-19T08:52:01.617" />
  <row Id="18111" PostTypeId="1" CreationDate="2011-01-19T09:29:47.250" Score="1" ViewCount="107" Body="&lt;p&gt;I'm trying to understand Serre's proof (in &lt;strong&gt;Local Fields&lt;/strong&gt;) of the fact that a fractional ideal $\mathfrak{a}$ of a Dedekind domain factors uniquely as $\mathfrak{a} = \prod_{\mathfrak{p}}\mathfrak{p}^{v_{p}(\mathfrak{a})}$.&#xA;The approach he took was that since $v_{\mathfrak{p}}(\mathfrak{a}) = 0$ for all but finitely many $\mathfrak{p}$, say, $v_{\mathfrak{p_{j}}}(\mathfrak{a}) = e_{j} \neq 0$ for $j = 1,\ldots, k$. Let $\mathfrak{b} = \prod_{j=1}^{k}\mathfrak{p_{j}}^{e_{j}}$. Then show $\mathfrak{a} = \mathfrak{b}$ locally, i.e, $\mathfrak{a}A_{\mathfrak{p}} = \mathfrak{b}A_{\mathfrak{p}}$ for all prime $\mathfrak{p}$; hence, they must be the same. I have been able to fill in most details except for the fact that  $\mathfrak{a} = \mathfrak{b}$ locally. If they are integral ideal of $A$, I can prove it, but since they are fractional with possibly negative powers, I could not figure out how to prove that claim. Note that Serre seems to use the following definition for $v_{\mathfrak{p}}(\mathfrak{a})$ for a fractional ideal $\mathfrak{a}$:&lt;a href=&quot;http://mathoverflow.net/questions/52198/what-is-the-definition-of-the-valuation-of-a-fractional-ideal&quot; rel=&quot;nofollow&quot;&gt;http://mathoverflow.net/questions/52198/what-is-the-definition-of-the-valuation-of-a-fractional-ideal&lt;/a&gt; and I have trouble applying that definition here. Any help is greatly appreciated. &lt;/p&gt;&#xA;" OwnerUserId="6028" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-19T18:03:04.580" LastActivityDate="2011-01-19T18:03:04.580" Title="Question about a proof in Serre’s &quot;Local fields&quot; of unique factorization of a fractional ideal of a Dedekind Domain" Tags="&lt;algebraic-number-theory&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" />
  <row Id="18112" PostTypeId="2" ParentId="18047" CreationDate="2011-01-19T09:40:13.180" Score="3" ViewCount="" Body="&lt;p&gt;First an answer your question about quotients: One of the axioms of an abelian category says that every morphism has a cokernel. The quotient $B/A$ of a monomorphism $f:A \to B$ is simply its cokernel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;[Recall that the cokernel $p: B \to B/A$ is defined by the following universal property: Given a morphism $g:B \to X$ whose composition with $A \to B$ is zero there is a unique factorization $h: B/A \to X$ such that $g = hp$.]&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;A morphism $f: A \to B$ in an abelian category has &lt;em&gt;four&lt;/em&gt; associated objects and &lt;em&gt;five&lt;/em&gt; associated morphisms:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/BIP2B.png&quot; alt=&quot;&quot; title=&quot;Hosted by imgur.com&quot; /&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;em&gt;kernel&lt;/em&gt; $\text{ker}\,(f): \text{Ker}\,(f) \to A$, the &lt;em&gt;cokernel&lt;/em&gt; $\text{coker}\,(f): B \to \text{Coker}\,(f)$, the &lt;em&gt;coimage&lt;/em&gt; $\text{Coim}\,{(f)} = \text{Coker}\,(\text{ker}\,(f))$ and the &lt;em&gt;image&lt;/em&gt; $\text{Im}\,(f) = \text{Ker}\,(\text{coker}\,(f))$. The main axiom of abelian categories states that the canonical morphism $\hat{f}:\text{Coim}\,(f) \to \text{Im}\,(f)$ (uniquely determined by requiring that the diagram be commutative) always is an isomorphism.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now given two morphisms $f:A&amp;#39; \to A$ and $g:A \to A&amp;#39;&amp;#39;$ such that $gf = 0$ there are &lt;em&gt;three&lt;/em&gt; ways to define the homology of the &quot;complex&quot; $A&amp;#39; \to A \to A&amp;#39;&amp;#39;$:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$\text{Coker}\,(\text{Im}\,(f) \to \text{Ker}\,(g))$,&lt;/li&gt;&#xA;&lt;li&gt;$\text{Ker}\,(\text{Coker}\,(f) \to \text{Coim}\,(g))$,&lt;/li&gt;&#xA;&lt;li&gt;$\text{Im}\,(\text{Ker}\,(g)\to \text{Coker}\,(f))$.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/Pdd5e.png&quot; alt=&quot;&quot; title=&quot;Hosted by imgur.com&quot; /&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first of these corresponds to the usual $\text{Ker}/\text{Im}$ and it is not very hard to show that all three ways give canonically isomorphic objects in an abelian category. It is essential to require the category to be abelian here, the three possibilities are distinct in a general additive category (with kernels and cokernels).&lt;/p&gt;&#xA;" OwnerUserId="5363" LastActivityDate="2011-01-19T09:40:13.180" CommentCount="4" />
  <row Id="18113" PostTypeId="1" AcceptedAnswerId="18121" CreationDate="2011-01-19T09:41:12.147" Score="2" ViewCount="50" Body="&lt;p&gt;I am working with granular materials (seeds). I am looking for a way to correctly scale the amount of different particles in one batch using weight only. I have worked with the problem a bit and instead of going in detail with the applied problem, I will try to present the problem more generally.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's say I have the following equation:&#xA;$\sum_{i=1}^{n} x_i = 1$, for all $x_i \in \mathbb{R}$ and $x_i &amp;gt; 0$. I would like to solve for all $x_i$. Solving this problem, solves the applied problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have the following constraints:&#xA;$x_i / x_j = a_i b_i / a_j b_j $, for all $i,j = 1, 2, \ldots, n$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, for all $a_i \in \mathbb{R}$, $a_j &amp;gt; 0$, $b_i \in \mathbb{R}$, $b_j &amp;gt; 0$, $\sum_{i=1}^{n} a_i = k$, and $k &amp;gt; 0$ (where usually $k = 1$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All $a_i$, $b_i$ and $k$ are given as input to the the problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $n=1$ and $n=2$ the problem is trivial (or atleast easy). But I am not sure about $n&amp;gt;2$. Is there a general solution?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Kofler&lt;/p&gt;&#xA;" OwnerUserId="6021" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2011-01-19T11:09:48.803" LastActivityDate="2011-01-19T12:07:00.453" Title="A solution for equation with N unknowns with specific constraints?" Tags="&lt;linear-algebra&gt;&lt;algebra&gt;" AnswerCount="1" />
  <row Id="18114" PostTypeId="1" CreationDate="2011-01-19T09:45:47.763" Score="1" ViewCount="94" Body="&lt;p&gt;This is an applied problem, which arises from the problem of reorienting of a sensor axes according to particle displacement directions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider a sensor which is located inside the solid substance. This sensor is capable of detecting the substance oscillations along each of the three axes (usually orthogonal, but generally, any non-degenerated (non-coplanar) basis). This sensor produces a recording of the detected oscillations, called a trace, containing the displacement sampled at high-frequency, capable to capture any oscillation frequency existing in the substance. There are three traces, one for each axis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider that at some period of time the sensor registers an interference of an event, consisting of the compression wave, and two shear waves. Traces now contain a recordings of the event, effectively, a projections of the oscillations on the sensor's axes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do I now virtually re-orient the axes, that is, perform the linear transform of the traces, so that they will be oriented each along a corresponding wave displacement vector, and thus will contain only single wave event recording in each of the traces after the transformation?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT1: Actually, we have a 3D-curve $r(t)$ of sensor motion, which is represented by sensor axial readings $r_1(t), r_2(t), r_3(t)$. The task is to find the &quot;primal&quot; directions of $r(t)$ movements and re-orient axes to those directions $q_1(t), q_2(t), q_3(t)$.&lt;/p&gt;&#xA;" OwnerUserId="3806" LastEditorUserId="3806" LastEditorDisplayName="" LastEditDate="2011-01-19T14:30:08.677" LastActivityDate="2011-05-19T17:59:16.750" Title="Applying a linear transformation to time sequences to separate interfering oscillations" Tags="&lt;physics&gt;&lt;transformation&gt;&lt;linear-algebra&gt;&lt;mathematical-physics&gt;" AnswerCount="3" FavoriteCount="1" />
  <row Id="18115" PostTypeId="2" ParentId="18083" CreationDate="2011-01-19T10:02:58.087" Score="1" ViewCount="" Body="&lt;p&gt;My guess it that you could generalize your example of the n-gon to any triangulation (with higher simplices, instead of only triangles) of your manifold. Any manifold admits a triangulation.&lt;/p&gt;&#xA;" OwnerUserId="5882" LastActivityDate="2011-01-19T10:02:58.087" CommentCount="4" />
  <row Id="18116" PostTypeId="1" AcceptedAnswerId="18118" CreationDate="2011-01-19T10:12:43.780" Score="5" ViewCount="230" Body="&lt;p&gt;1 can be divided by 1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2 can be divided by 1, 2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;6 can be divided by 1, 2, 3&lt;/p&gt;&#xA;&#xA;&lt;p&gt;12 can be divided by 1, 2, 3, 4&lt;/p&gt;&#xA;&#xA;&lt;p&gt;60 can be divided by 1, 2, 3, 4, 5&lt;/p&gt;&#xA;&#xA;&lt;p&gt;60 can be divided by 1, 2, 3, 4, 5, 6&lt;/p&gt;&#xA;&#xA;&lt;p&gt;420 can be divided by 1, 2, 3, 4, 5, 6, 7&lt;/p&gt;&#xA;&#xA;&lt;p&gt;840 can be divided by 1, 2, 3, 4, 5, 6, 7, 8&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I just wonder whether there is a general function to derive this number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="6023" LastActivityDate="2011-02-07T20:46:47.117" Title="Is there a general function to derive the minimum number which can be divided by 1, 2, 3, 4 ..." Tags="&lt;number-theory&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="18117" PostTypeId="2" ParentId="18057" CreationDate="2011-01-19T10:13:06.087" Score="2" ViewCount="" Body="&lt;p&gt;Another way: You have identified some angles as being equal in your last figure. This implies that several of the right triangles in the picture are similar. For example,&#xA;$$ \frac{h}{w} = \frac{x}{y} = \frac{y}{w-x},$$&#xA;from which it is not too difficult to solve for $x$ and $y$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And yet another: The hypotenuse lies on the line with equation $x/w+y/h=1$ (since the two points $(x,y)=(w,0)$ and $(x,y)=(0,h)$ satisfy this equation, and two points uniquely determine a line). The normal vector to a line can be read off from the coefficients of $x$ and $y$; it is $\mathbf{n}=(1/w,1/h)$ in this case. The point that you seek (let me call it $(a,b)$ here) lies on the line which goes from the origin in the direction that $\mathbf{n}$ points, so the point must be of the form $(a,b)=t \mathbf{n} = (t/w,t/h)$ for some number $t$. Substituting $(x,y)=(t/w,t/h)$ into the equation for the hypotenuse gives $t(1/w^2+1/h^2)=1$, from which we immediately find $t$ and hence also&#xA;$$(a,b)=\left( \frac{1/w}{1/w^2+1/h^2}, \frac{1/h}{1/w^2+1/h^2} \right).$$&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2011-01-19T10:13:06.087" />
  <row Id="18118" PostTypeId="2" ParentId="18116" CreationDate="2011-01-19T10:25:38.783" Score="11" ViewCount="" Body="&lt;p&gt;The question rephrased:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How does one determine $m_k$, the smallest number which is divisible by all the positive integers from 1 to $k$?&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;As a product of primes&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Well, if one starts with the product $\tilde m_k = 1 \cdot 2 \cdot \dots \cdot k$, this is divisble by $1, 2, \dots, k$, but it's not the smallest. For example $1 \cdot 2 \cdot 3 \cdot 4 = 24$, but the smallest is 12, since 4 is divisble by 2. If we remove the 2 we get the smallest number. This suggest a general procedure.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take all the prime numbers $p_1, \dots, p_l$ from $\{1, 2, \dots, k\}$ and let $q_i$ be the largest integer such that $p_i^{q_i} \leq k$ and set  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$m_k = p_1^{q_1} p_2^{q_2} \cdots p_l^{q_l}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which should be the smallest number which is divisble by $1, \dots, k$.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Recursively&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Calculate $m_k$, let $m_{k-1}$ be given. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Is $m_{k-1}$ divisble by $k$? If so, set&#xA;$m_k = m_{k-1}$.&lt;/li&gt;&#xA;&lt;li&gt;Otherwise, set $m_k =&#xD;&#xA;   \operatorname{lcm}(m_{k-1}, k)$,&#xA;where $\operatorname{lcm}$ is the&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Least_common_multiple&quot; rel=&quot;nofollow&quot;&gt;least common multiple&lt;/a&gt;.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This gives a nice implementation for computer using the following formula:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\operatorname{lcm}(a,b) = \frac{|a b|}{\operatorname{gcd}(a,b)}$$&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Least common multiple&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Of course, you can just set:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$m_k = \operatorname{lcm}(2, 3, \dots, k)$$.&lt;/p&gt;&#xA;" OwnerUserId="5966" LastEditorUserId="5966" LastEditorDisplayName="" LastEditDate="2011-02-07T20:46:47.117" LastActivityDate="2011-02-07T20:46:47.117" CommentCount="2" />
  <row Id="18119" PostTypeId="1" CreationDate="2011-01-19T10:48:18.057" Score="2" ViewCount="164" Body="&lt;p&gt;I've been playing with Fourier transform a little and discovered the identity quoted in the title. More precisely, writing the matrix for the Fourier transform in ${\mathbb Z} / N {\mathbb Z}$ as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$A = {1 \over \sqrt{N}} \pmatrix{&#xD;&#xA;\zeta^0&amp;amp; \zeta^0 &amp;amp; \cdots &amp;amp; \zeta^0 \cr&#xD;&#xA;\zeta^0&amp;amp; \zeta^1 &amp;amp; \cdots &amp;amp; \zeta^{N-1} \cr&#xD;&#xA;\vdots &amp;amp; \vdots &amp;amp;\ddots &amp;amp;\vdots \cr&#xD;&#xA;\zeta^0&amp;amp; \zeta^{N-1}  &amp;amp; \cdots &amp;amp; \zeta^1 \cr&#xD;&#xA;}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(with $\zeta \equiv \exp({2 \pi i \over N})$) one easily obtains the quoted identity by using the famous determinant of Vandermonde matrix and the fact that the transform is unitary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, I'd be interested whether the identity can be proven directly, and ideally given a nice combinatorial interpretation. I suppose this should be possible because of lot of structure in the problem but I wasn't able to come up with anything yet.&lt;/p&gt;&#xA;" OwnerUserId="3223" LastActivityDate="2011-01-19T19:03:05.990" Title="Combinatorial proof of $\Big|\prod\limits_{0 \leq i &lt; j &lt; N} (\zeta^j -\zeta^i)\Big| = \sqrt {N^N}$ for $\zeta \equiv \exp({2\pi i \over N})$." Tags="&lt;linear-algebra&gt;&lt;combinatorics&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="1" />
  <row Id="18120" PostTypeId="1" CreationDate="2011-01-19T10:51:26.977" Score="3" ViewCount="113" Body="&lt;p&gt;Hello&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose there are vending machine that sells its good for $3$.Its known that 1/3 of the buyers use 3 coins of one (1, 1, 1).1/3 of the buyers use $2$ and $1$.And the last third use $5$.The machine start empty (without money).What it is the probability that it could give change after n buyers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you!&lt;/p&gt;&#xA;" OwnerUserId="5145" LastEditorUserId="5145" LastEditorDisplayName="" LastEditDate="2011-01-19T12:54:28.837" LastActivityDate="2011-01-19T12:54:28.837" Title="Probability question maybe markov chain." Tags="&lt;probability&gt;" CommentCount="6" />
  <row Id="18121" PostTypeId="2" ParentId="18113" CreationDate="2011-01-19T11:07:50.490" Score="1" ViewCount="" Body="&lt;p&gt;Let $$S = \sum_{i=1}^{n} a_i b_i$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;then&#xA;$$x_i =  \frac{a_ib_i}{S}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is a solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that $$\sum_{i=1}^{n} x_i = \sum_{i=1}^{n} \frac{a_i b_i}{S} = \frac{\sum_{i=1}^{n} a_i b_i}{S} = \frac{S}{S} = 1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{x_i}{x_j} = \frac{a_ib_i/S}{a_jb_j/S} = \frac{a_ib_i}{a_jb_j}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also note that $x_i \gt 0$.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2011-01-19T12:07:00.453" LastActivityDate="2011-01-19T12:07:00.453" CommentCount="4" />
  <row Id="18122" PostTypeId="1" AcceptedAnswerId="18167" CreationDate="2011-01-19T11:18:37.097" Score="1" ViewCount="140" Body="&lt;p&gt;Hi all,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is my very first question, so don't be too hard on me :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am a beginner programmer, and I am working on the typical helicopter game, working with a one-button push:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the button is pushed, the helicopter ascends, starting ascending slowly, accelerating upwards.&#xA;When the button is released, the upward acceleration slows down, then the helicopter starts falling.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This calculation would run in a loop.&#xA;As a result, I would have to get a floating point number between 0 and 1, which would correlate with the vertical position of the helicopter on screen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I have so far is very wrong, but I post it here anyway:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;newPosition = (oldPosition * gravity) + (acceleration * sensitivity)&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where gravity and sensitivity are constant. Acceleration is the length of the button push, which becomes 0, as soon as the button is released.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, this is wrong, because this way the helicopter descends quickly first, then slows down as it gets closer to the ground (to value 0).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any help would be greatly appreciated -- as you can see, I am not a math genius. It would also be great to get some help on &lt;strong&gt;&lt;em&gt;how to approach these calculations&lt;/em&gt;&lt;/strong&gt;, so I could come up with similar calculations in the future.&lt;/p&gt;&#xA;" OwnerUserId="6001" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2011-01-19T11:56:17.313" LastActivityDate="2011-01-19T16:43:50.907" Title="calculation to change a helicopter's vertical position, adjusted with gravity" Tags="&lt;physics&gt;" AnswerCount="3" />
  <row Id="18123" PostTypeId="2" ParentId="18119" CreationDate="2011-01-19T11:29:33.900" Score="1" ViewCount="" Body="&lt;p&gt;If you look at &lt;a href=&quot;http://www.math.hmc.edu/~benjamin/papers/VandermondeNote.pdf&quot; rel=&quot;nofollow&quot;&gt;Combinatorial Proof of Vandermonde's determinant&lt;/a&gt; perhaps you can extract the desired combinatorial proof.&lt;/p&gt;&#xA;" OwnerUserId="1709" LastActivityDate="2011-01-19T11:29:33.900" CommentCount="2" />
  <row Id="18124" PostTypeId="1" CreationDate="2011-01-19T11:34:38.967" Score="2" ViewCount="420" Body="&lt;p&gt;Hi All,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the limit as $x\to\infty$ of $\cos x$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance.&lt;/p&gt;&#xA;" OwnerUserId="3669" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-19T19:12:54.317" LastActivityDate="2011-01-21T06:58:41.363" Title="What is the limit as $x\to\infty$ of $\cos x$?" Tags="&lt;trigonometry&gt;&lt;limit&gt;&lt;infinity&gt;" AnswerCount="4" CommentCount="4" FavoriteCount="1" />
  <row Id="18125" PostTypeId="2" ParentId="18124" CreationDate="2011-01-19T11:38:38.143" Score="6" ViewCount="" Body="&lt;p&gt;There is no limit, $\lim_{x \to \infty} \cos x$, since $\cos$ oscillates between -1 and 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A bit more detailed:&#xA;We say that a function $f(x)$ has a limit as $x \to \infty$ if there exists a real number $a$ (called the limit) such that $|f(x)-a|$ can be made arbitrarily small for all $x$ which are &quot;large enough&quot;. &quot;Large enough&quot; and arbitrarily small means that for all $\varepsilon &amp;gt; 0$, we should be able to find a number $N$ such that $|f(x)-a| &amp;lt; \varepsilon$ for all $x &amp;gt; N$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of $f(x) = \cos x$ we can't do this when $\varepsilon$ is small.  Independent of which $a$ we are trying out, we can always find a large enough $x$ such that $|\cos x - a| &amp;gt; \frac{1}{2}$ (for example), and infinitely many of them (since $\cos$ is periodic).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As Roupam Ghosh stated in his answer, the $\limsup$ and $\liminf$ for $\cos x$ as $x \to \infty$ are not equal. This gives that the limit does not exist, since then you can always find sufficiently large $x,y$ such that $|f(x)-f(y)|=\left|\limsup_{x \to \infty}\cos x - \liminf_{n \to \infty} \cos x\right|\geq\varepsilon&amp;gt;0$.&lt;/p&gt;&#xA;" OwnerUserId="5966" LastEditorUserId="5966" LastEditorDisplayName="" LastEditDate="2011-01-21T06:58:41.363" LastActivityDate="2011-01-21T06:58:41.363" />
  <row Id="18126" PostTypeId="2" ParentId="18124" CreationDate="2011-01-19T11:38:54.937" Score="7" ViewCount="" Body="&lt;p&gt;The limit does not exist because $\cos{(2\pi n)} = 1$ for $n \in \mathbb{Z}$ and $\cos{(\pi + 2 \pi n)} = -1$ for $n \in \mathbb{Z}$.&lt;/p&gt;&#xA;" OwnerUserId="5363" LastActivityDate="2011-01-19T11:38:54.937" />
  <row Id="18127" PostTypeId="2" ParentId="18124" CreationDate="2011-01-19T11:40:31.300" Score="10" ViewCount="" Body="&lt;p&gt;The limit does not exist. It oscillates between -1 and 1. Just so that you know, the limit supremum or infimum as $x \to \infty$ is given as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\lim \sup_{x\to\infty} \cos(x) = 1$$&#xA;$$\lim \inf_{x\to\infty} \cos(x) = -1$$&lt;/p&gt;&#xA;" OwnerUserId="2320" LastActivityDate="2011-01-19T11:40:31.300" />
  <row Id="18128" PostTypeId="2" ParentId="18122" CreationDate="2011-01-19T11:54:52.167" Score="2" ViewCount="" Body="&lt;p&gt;Keep in mind the following:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Rate of change of distance (or in you&#xA;  case, position) is &lt;em&gt;velocity&lt;/em&gt;.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Rate of change of velocity is&#xA;  &lt;em&gt;acceleration&lt;/em&gt;.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Acceleration is proportional to the&#xA;  net force acting on the object&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So assuming that the button push gives a constant thrust (or force), we can assume your acceleration is constant, while the button is being pushed. When the button is released, the acceleration is due to gravity, which again gives a constant acceleration (different from previous, though).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The following gives us the relation of distance and constant acceleration (for a derivation, see link at the end of the answer)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x = ut + \frac{at^2}{2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $u$ was the initial velocity and $t$ is the total time elapsed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus if you want to calculate new position from old, you need to use&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Delta x =  (u + at)\times \Delta t$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Keep in mind the appropriate signs)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\Delta t$ is time change (which is assumed to be very small), $\Delta x$ is the position change and $t$ is the total time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note, if the acceleration changes abruptly (like losing thrust because of release of button), you need to reset your $t$ and $u$. To reset your $u$, you would need to keep track of your velocity, whose change is given by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Delta v = a \Delta t$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: It could so happen that your $\Delta t$ is not small enough (a possible problem with game loops), in which case you could try using the general formulae&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ x = ut + \frac{at^2}{2}, \  \ v = u + at$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Again, keep in mind the appropriate signs).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, you can approach these problems by using calculus. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance see here: &lt;a href=&quot;http://www.ugrad.math.ubc.ca/coursedoc/math101/notes/applications/velocity.html&quot; rel=&quot;nofollow&quot;&gt;http://www.ugrad.math.ubc.ca/coursedoc/math101/notes/applications/velocity.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope that helps.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2011-01-19T12:19:28.213" LastActivityDate="2011-01-19T12:19:28.213" />
  <row Id="18129" PostTypeId="2" ParentId="18122" CreationDate="2011-01-19T12:10:47.037" Score="1" ViewCount="" Body="&lt;p&gt;I'm not sure I completely understood your approach, but here is how I would tackle it. A uniform acceleration $a$ in one dimension implies the following type of movement for an object:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x(t)=x_0+v_0 \cdot (t-t_0) + \frac{1}{2} \cdot a \cdot (t-t_0)^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $x_0$ is the initial height, $v_0$ the initial velocity and $t_0$ is the initial time. The equation then gives you the position of the object at each instant $t$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is all for a constant acceleration though, if you want to make the acceleration dependent on a button push, you can proceed as follows: the equation describing the position of the helicopter is a differential equation &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{d^2 x}{dt^2} = a(D)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$D$ represents a variable that depends on the duration of the button push. If $D=0$ then the acceleration is zero as well, if the duration of the push is different from $0$, you take $a(D)=s\cdot D$ where $s$ is the sensitivity. To make things more realistic, I think you should place an upper bound on the acceleration, so that you can't accelerate indefinitely. But I guess that is easy to program. Now, you just solve the equation, keeping in mind that $D$ will actually be an function of time that is in general not known, but has to be computed each time for each situation. So what you do is solve the problem for small time steps during which it is reasonable to approximate the acceleration as being constant. This will not be perfectly true for the sudden changes from pushing the button to letting go of the button, but if the time intervals are chosen small enough, it should not matter much. But assuming that for short time intervals the acceleration is constant (or uniform) we are back to our very first equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the new position $x_n $is given in terms of the old position $x_o$ as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x_n=x_o+v_o \cdot (\Delta t) + \frac{1}{2} \cdot a(D) \cdot (\Delta t)^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you don't want to keep track of velocities, you can also use the fact that the velocity of the previous step is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$v_o = \frac{x_o-x_{oo}}{\Delta t}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but this means that you have to keep track of the old position and the position one time step older $x_{oo}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt; One final comment about gravity. When your helicopter is just hovering in the air, gravity is certainly acting, but the pull is compensated by the lift of the rotor blades. So the upward acceleration due to the rotorblades is equal in magnitude but opposite in sign to the gravitational acceleration. There are several ways you can account for this in your formula. Either you consider $a(D)$ to be the acceleration after gravity has already been substracted, which means that when there is no button push, the helicopter will just hover in place. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Either you want $a(D)$ to represent the acceleration produced by the rotorblades, and then, you have to substract the gravitation from $a(D)$ in the formulas above. You also have to adapt the formula for the link between button push and acceleration. Probably you want the helicopter to hover when nothing happens, so the formula will be $a(D)=g+s\cdot D$. But as you can see, the net effect will be the same as in the former suggestion. It gets more tricky though if we want to add things like starting up the helicopter from the ground, etc... but everything can in principle be computed by adapting the function $a(D)$.&lt;/p&gt;&#xA;" OwnerUserId="3567" LastEditorUserId="3567" LastEditorDisplayName="" LastEditDate="2011-01-19T12:26:58.363" LastActivityDate="2011-01-19T12:26:58.363" CommentCount="3" />
  <row Id="18130" PostTypeId="2" ParentId="18093" CreationDate="2011-01-19T12:49:24.910" Score="2" ViewCount="" Body="&lt;p&gt;I ended doing something very similar to Moron.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Define a Markov chain with states i=0..7 as follows. Given a full sequence of &lt;em&gt;n&lt;/em&gt; coins, if the pattern has not yet appeared, find the maximum length of the last tosses (sequence suffix) that coincide with the begginning of the expected pattern, so that&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;   i =  length of matching suffix , if pattern has not appeared&#xA;   i = 7 otherwise&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;For example:  HTTHTTTTHT : i=0 ,  HTTTTHHT  i=3&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's easy to write down the transition probabilities. (i=7 would be a absorbent state), and the probability of being in state 7 in step &lt;em&gt;n&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So your anwser would be $ P(n) = 1- p_7[n]$ where $p[n] = M^n p[0]$ where $M$ is the 8x8 transition matrix and $p[0] = (1,0,0,0,0,0,0,0)$ is the initial state.&#xA;I doubt you'll get a simpler closed form answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One could get a very coarse approximate solution by assumming that the patterns starting at each position are independent (obviously a false assumpion), and get &#xA;$ P(n) \approx (1-2^{-7})^{n-6}$ ($n \ge 7$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;UPDATE: some Octave/Matlab code to test the approximation (it seems to work better than I'd expected)&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;N=200;&#xA;% approximation&#xA;pa = (1-1/128).^((1:N)-6);&#xA;pa(pa&amp;gt;1)=1;&#xA;% transition matrix:&#xA;M = [ 1,1,0,0,0,0,0,0 ;&#xA;      1,0,1,0,0,0,0,0 ;&#xA;      0,0,1,1,0,0,0,0 ;&#xA;      1,0,0,0,1,0,0,0 ;&#xA;      0,0,1,0,0,1,0,0 ;&#xA;      1,0,0,0,0,0,1,0 ;&#xA;      1,0,0,0,0,0,0,1 ;&#xA;      0,0,0,0,0,0,0,2 ]/2;&#xA;p=[1,0,0,0,0,0,0,0];&#xA;p7=zeros(1,N);&#xA;p7(1)=1;&#xA;for n = 1:N&#xA; p = p * M;&#xA; p7(n) = 1 - p(8);&#xA;endfor&#xA;plot(7:N,p7(7:N),7:N,pa(7:N));&#xA;&#xA;&amp;gt;&amp;gt;&amp;gt; pa(5:9)&#xA;ans =&#xA;&#xA;   1.00000   1.00000   0.99219   0.98444   0.97675&#xA;&#xA;&amp;gt;&amp;gt;&amp;gt; p7(5:9)&#xA;ans =&#xA;&#xA;   1.00000   1.00000   0.99219   0.98438   0.97656&#xA;&#xA;&amp;gt;&amp;gt;&amp;gt; pa(N-5:N)&#xA;ans =&#xA;&#xA;   0.22710   0.22533   0.22357   0.22182   0.22009   0.21837&#xA;&#xA;&amp;gt;&amp;gt;&amp;gt; p7(N-5:N)&#xA;ans =&#xA;   0.22717   0.22540   0.22364   0.22189   0.22016   0.21844&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="312" LastEditorUserId="312" LastEditorDisplayName="" LastEditDate="2011-01-20T00:03:59.620" LastActivityDate="2011-01-20T00:03:59.620" CommentCount="12" />
  <row Id="18131" PostTypeId="1" AcceptedAnswerId="18135" CreationDate="2011-01-19T12:59:06.787" Score="3" ViewCount="112" Body="&lt;p&gt;I have several tasks to solve where a set of inequalities is used to describe a region. I should then calculate the area or volume of that region.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's say we have the following inequality (for $x,y,z \geq 0$):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x+2y+3z \leq 1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I need to find out the boundaries for the triple integral. From a few examples I have here, I wasn't able to derive a way to solve such a problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The example solution suggests that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$0 \leq y \leq \frac{1-x}{2}$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;and &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$0 \leq z \leq \frac{(x - 2y)}{3}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$0 \leq x \leq 1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;leading to the integral:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$&#xD;&#xA;\int_{0}^{1} \int_{0}^{\frac{1-x}{2}} \int_{0}^{\frac{x-2y}{3}} { 1 \; dz dy dx }&#xD;&#xA;$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do I find out these boundaries? (Moreover, how do I find out the boundaries in a general approach)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS: I'm not sure if the solution is correct at all, that's probably why I'm confused.&lt;/p&gt;&#xA;" OwnerUserId="5488" LastEditorUserId="5203" LastEditorDisplayName="" LastEditDate="2011-01-19T13:47:48.077" LastActivityDate="2011-02-17T19:54:09.967" Title="Solving Inequalities for Regions" Tags="&lt;inequality&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="18132" PostTypeId="2" ParentId="18124" CreationDate="2011-01-19T13:02:55.887" Score="1" ViewCount="" Body="&lt;p&gt;I believe that the proofs given are not rigorous enough, and are not necessarily easy to understand to someone who is not familiar with the material, so I'll try a more rigorous approach. I'll assume every variable I'm referring to is a real number in &lt;code&gt;R&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume that there is a limit &lt;code&gt;k&lt;/code&gt;. In that case it must be that for every &lt;code&gt;d&lt;/code&gt; there exist &lt;code&gt;N&amp;gt;0&lt;/code&gt; so that if &lt;code&gt;n&amp;gt;N&lt;/code&gt; we have that &lt;code&gt;|cos(n)-k|&amp;lt;d&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since &lt;code&gt;cos(x+n2pi)=cos(x)&lt;/code&gt;, it is easy to see that for every &lt;code&gt;N&lt;/code&gt; we can find a number &lt;code&gt;n0&amp;gt;N&lt;/code&gt; for which &lt;code&gt;cos(n0)=0&lt;/code&gt;, and another number &lt;code&gt;n1&lt;/code&gt; for which &lt;code&gt;cos(n0)=1&lt;/code&gt;. Try to prove that, it's not so hard.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let &lt;code&gt;d=1/2&lt;/code&gt;, whatever &lt;code&gt;k&lt;/code&gt; would be, we'll have that either &lt;code&gt;|cos(n0)-k|=|k|&amp;gt;d=1/2&lt;/code&gt; or &lt;code&gt;|cos(n1)-k|=|1-k|&amp;gt;d=1/2&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefor no number is the limit of &lt;code&gt;lim&lt;/code&gt;&lt;sub&gt;n→∞&lt;/sub&gt;&lt;code&gt;cos(n)&lt;/code&gt;. Since &lt;code&gt;cos(n)&lt;/code&gt; is bounded it cannot be that the limit is &lt;code&gt;∞&lt;/code&gt; or &lt;code&gt;-∞&lt;/code&gt;.&lt;/p&gt;&#xA;" OwnerUserId="52" LastEditorUserId="52" LastEditorDisplayName="" LastEditDate="2011-01-19T14:14:28.500" LastActivityDate="2011-01-19T14:14:28.500" CommentCount="6" />
  <row Id="18133" PostTypeId="1" CreationDate="2011-01-19T13:26:24.820" Score="0" ViewCount="128" Body="&lt;p&gt;In course of a particular research I ran into these two inequalities which I would like to have some help with.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$r,R&amp;gt;0$ for both the questions.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Is there a function $M(r)$ for which this inequality is satisfied?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;$$\cos^{-1}\left (\frac{4M(r)}{r}-1 \right ) + \sqrt{1- \left ( \frac{4M(r)}{r}-1 \right )^2} &amp;gt; \pi$$&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Let $t_0$ be the minimum positive solution of $a \cosh (\omega t) + b \sinh (\omega t) = 0$.  &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Then for $-B&amp;lt;0$ and $t&amp;gt;0$, is it possible to choose the values of variables $a,b$ and $g&amp;gt;0$ such that the expression multiplying $-B$ is negative for $t \leq t_0$ and $r&amp;gt;gR$? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$-B\frac {(a\sqrt{r^2-g^2R^2} \mp bgr\omega)\cosh (\omega t)+ (b\sqrt{r^2-g^2R^2} \mp agr\omega)\sinh(\omega t)}{r\sqrt{r^2-g^2R^2}(a \cosh(\omega t) + b \sinh (\omega t))^{2g+1}}$$ &lt;/p&gt;&#xA;" OwnerUserId="965" LastEditorUserId="965" LastEditorDisplayName="" LastEditDate="2011-01-19T17:42:07.723" LastActivityDate="2011-01-19T19:34:28.650" Title="Two inequalities" Tags="&lt;real-analysis&gt;&lt;algebra-precalculus&gt;&lt;inequality&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="18134" PostTypeId="1" AcceptedAnswerId="18149" CreationDate="2011-01-19T13:33:31.160" Score="7" ViewCount="294" Body="&lt;p&gt;You are and your friend are given a list of N distinct integers and are told this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Six distinct integers from the list are selected at random and placed one at each side of a cube. The cube is placed in the middle of a rectangular room in front of its only door, with one face touching the floor, its 6 sides parallel to the walls of the room. Your friend must enter the room and is allowed to alter the orientation of the cube, with the restriction that afterwards its in the same place with one face touching the floor and its 6 sides kept parallel to the walls of the room. Your friend will then be sendt away, after which you can enter the room and are allowed to observe the 5 visible sides of the cube. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the largest N that guarantees you to be able to determine the number on the bottom of the cube and what should you instruct your friend to do with the cube for that N?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;N=29 is possible see! still in need of strategy&#xA;&lt;a href=&quot;http://mathoverflow.net/questions/52541/hard-cube-puzzle&quot; rel=&quot;nofollow&quot;&gt;http://mathoverflow.net/questions/52541/hard-cube-puzzle&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5904" LastEditorUserId="5904" LastEditorDisplayName="" LastEditDate="2011-01-19T21:20:03.593" LastActivityDate="2011-03-06T21:59:00.280" Title="Guessing a hidden number on a cube." Tags="&lt;combinatorics&gt;&lt;puzzle&gt;" AnswerCount="1" CommentCount="12" FavoriteCount="3" />
  <row Id="18135" PostTypeId="2" ParentId="18131" CreationDate="2011-01-19T13:36:39.747" Score="5" ViewCount="" Body="&lt;p&gt;Take the variables one by one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For your example, start with $z$ and move all the other variables to the other side:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$0 \leq 3z \leq 1 - (x+2y)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Divide by $3$ and you have your bound.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $y$, disregard $z$, but do the same thing, obtaining:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$0 \leq 2y \leq 1 - x$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Same thing for $x$ immediately gives&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$0 \leq x \leq 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the general approach is pretty evident from this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With these limits, you get the integral&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\int_0^1 \int_0^\frac{1-x}{2} \int_0^\frac{1-x-2y}{3} dzdydx = \frac{1}{36}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: This should be correct, since the volume described can be seen as a pyramid with base area $\frac{1}{12}$ (triangle with height $\frac{1}{2}$, base $\frac{1}{3}$) and height 1. The volume of a pyramid is $\frac{1}{3}Bh$. &lt;/p&gt;&#xA;" OwnerUserId="5966" LastEditorUserId="5966" LastEditorDisplayName="" LastEditDate="2011-02-17T17:04:52.837" LastActivityDate="2011-02-17T17:04:52.837" CommentCount="3" />
  <row Id="18136" PostTypeId="2" ParentId="18114" CreationDate="2011-01-19T13:45:24.263" Score="0" ViewCount="" Body="&lt;p&gt;If you have three unit vectors that describe your current axes $r_1, r_2, r_3$ and you want to know what the linear transform is that will change $a_1 r_1 + a_2 r_2 + a_3 r_3 $ into the coordinates of some new axes $r&amp;#39;_1, r&amp;#39;_2, r&amp;#39;_3$, the matrix of this transform is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ \left( \begin{array}{ccc}&#xD;&#xA;r&amp;#39;_{11} &amp;amp; r&amp;#39;_{21} &amp;amp; r&amp;#39;_{31} \\&#xD;&#xA;r&amp;#39;_{12} &amp;amp; r&amp;#39;_{22} &amp;amp; r&amp;#39;_{32} \\&#xD;&#xA;r&amp;#39;_{13} &amp;amp; r&amp;#39;_{23} &amp;amp; r&amp;#39;_{33}&#xD;&#xA;\end{array} \right) $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $r&amp;#39;_{j} = r_{j1} r_1 + r_{j2} r_2 + r_{j3} r_3$. So once you've figured out what the axes are for the coordinate system that you want to transform into, you can just plug those numbers into that matrix and apply it to your trace-vector to get the trace in the new coordinate system.&lt;/p&gt;&#xA;" OwnerUserId="1089" LastActivityDate="2011-01-19T13:45:24.263" CommentCount="2" />
  <row Id="18138" PostTypeId="1" AcceptedAnswerId="18143" CreationDate="2011-01-19T13:59:01.203" Score="5" ViewCount="120" Body="&lt;p&gt;Let $E = \{E_k\}_{k \in \mathbb{N}}$ be an infinite sequence of sets. Then, the following inclusion holds:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\bigcup_{n=1}^{\infty} \bigcap_{k=n}^{\infty} E_k \quad\subseteq\quad \bigcap_{n=1}^{\infty} \bigcup_{k=n}^{\infty} E_k$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know the left-hand side (LHS) represents elements which belong to all but finitely many sets in the sequence $E$, and the right-hand side (RHS) represents elements which belong to infinitely many sets in the sequence $E$. This concludes the proof that the LHS is contained in the RHS. Yet it is not intuitive for me, since the above interpretation is not intuitive per se.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Is there an intuitive reason why the above inclusion holds? That is, (very informally) why &lt;strong&gt;union of intersections&lt;/strong&gt; is contained in &lt;strong&gt;intersection of unions&lt;/strong&gt;?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2420" LastEditorUserId="2420" LastEditorDisplayName="" LastEditDate="2011-01-19T14:04:25.643" LastActivityDate="2011-01-19T15:38:42.737" Title="An Intuition to An Inclusion: &quot;Union of Intersections&quot; vs &quot;Intersection of Unions&quot;" Tags="&lt;sequences-and-series&gt;&lt;set-theory&gt;" AnswerCount="4" />
  <row Id="18139" PostTypeId="2" ParentId="18119" CreationDate="2011-01-19T14:31:14.123" Score="3" ViewCount="" Body="&lt;p&gt;It can be derived directly by considering the n-roots of one $z_k = exp(\frac{i 2\pi k}{N})$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$F(z) = z^N-1 = (z-z_0) ... (z-z_{N-1})$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, this readily implies (by setting $z=0$) that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ \prod z_i = (-1)^{N+1}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second, by deriving $F(z)$ and evaluating at, say, $z=z_0$ we get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ N z_0^{N-1} = (z_0 - z_1) (z_0 - z_2)  ... (z_0 - z_{N-1}) $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rearranging, writing the same equation for all $z_i$, multiplying all them, and using the previous equation, we get at your result.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ N^N (\prod z_i)^{N-1} =  \prod_{i \ne j} (z_i - z_j) $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ N^N (-1)^{N+1} =  \prod_{i \ne j} (z_i - z_j) $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$N^{N/2} = | \prod_{i&amp;lt;j} (z_i - z_j)|$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;No combinatorial interpretation, though.&lt;/p&gt;&#xA;" OwnerUserId="312" LastEditorUserId="312" LastEditorDisplayName="" LastEditDate="2011-01-19T15:42:21.330" LastActivityDate="2011-01-19T15:42:21.330" CommentCount="4" />
  <row Id="18140" PostTypeId="2" ParentId="18138" CreationDate="2011-01-19T14:31:58.800" Score="0" ViewCount="" Body="&lt;p&gt;So on the left side, an element must belong to &lt;strong&gt;every&lt;/strong&gt; $E_k$ starting from some $k$.  On the right side, an element only has to belong to sets that are arbitrarily large.  So if it belongs to every other one, or every prime number one, or something like that, it will be included in the right side.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-01-19T14:31:58.800" />
  <row Id="18141" PostTypeId="2" ParentId="18138" CreationDate="2011-01-19T14:33:48.720" Score="1" ViewCount="" Body="&lt;p&gt;As you said the LHS is the collection of elements which appear in all but finitely many $E_k$, while the elements on the RHS are those which appear in infinitely many $E_k$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you appear in all but finitely many, then you appear in infinitely many; the converse is not always true as you can appear in all the even indices but not once in the odd indices.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The LHS is called sometimes denoted by $\liminf E$, and the RHS is $\limsup E$.&lt;/p&gt;&#xA;" OwnerUserId="622" LastActivityDate="2011-01-19T14:33:48.720" CommentCount="2" />
  <row Id="18142" PostTypeId="2" ParentId="18114" CreationDate="2011-01-19T14:39:57.110" Score="0" ViewCount="" Body="&lt;p&gt;There are two questions here.  The first, which Eric answered for orthogonal axes, is &quot;given the axes to transform to, how do I transform?&quot;  The second is &quot;given the data, how do I figure out the axes to transform to?.  I don't think that is available from a single sensor.  If you have some more knowledge you might be able to.   If you know the source of the disturbance, one axis might be from the source to the sensor.  If you have a model how the longitudinal and transverse disturbances are related, you could try a fit to your data as a function of the axes.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-01-19T14:39:57.110" CommentCount="3" />
  <row Id="18143" PostTypeId="2" ParentId="18138" CreationDate="2011-01-19T14:45:29.633" Score="3" ViewCount="" Body="&lt;p&gt;If $A$ and $B$ are sets, and $A\subseteq B_n$ for all $n$, then $A\subseteq\bigcap_n B_n$. Conversely, $A\subseteq\bigcap_n B_n$ means that $A\subseteq B_n$ for all $n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This tells us that the inclusion you are interested in reduces then to checking $(+)$:&#xA;$$ \bigcup_n \bigcap_{k\ge n}E_k\subseteq \bigcup_{k\ge m}E_k $$&#xA;for all $m$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(This is progress. It is similar to a common move in analysis: To prove that $a\le b$, it is not unusual to check instead that $a\le c$ for any $c&amp;gt;b$. But these inequalities $a\le c$ tend to be easier than the one you really want, $a\le b$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the same idea gives us that to prove the new inclusion $(+)$, it is enough to prove $(++)$: $$\bigcap_{k\ge n}E_k \subseteq \bigcup_{k\ge m}E_k $$&#xA;for all $n,m$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(The corresponding move in analysis is that to prove $a\le b$, it is enough to show $c\le b$ for all $c\le a$. Unions correspond to suprema, intersections to infima in these analogies.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, $(++)$ is really obvious: If somebody is in the left hand side, then it is in &lt;em&gt;all&lt;/em&gt; sufficiently large $E_k$, but then it is in the right hand side.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The point is: It is useful to train oneself to &quot;decode&quot; certain expressions the way we did, by focusing on their &quot;atomic&quot; or at least &quot;more basic&quot; components.&lt;/p&gt;&#xA;" OwnerUserId="462" LastEditorUserId="462" LastEditorDisplayName="" LastEditDate="2011-01-19T15:38:42.737" LastActivityDate="2011-01-19T15:38:42.737" CommentCount="2" />
  <row Id="18144" PostTypeId="2" ParentId="18133" CreationDate="2011-01-19T14:57:07.360" Score="1" ViewCount="" Body="&lt;p&gt;For the first, let $(\frac{4M(r)}{r}-1)=z$  You want $\cos^{-1}z+\sqrt{1-z^2}&amp;gt;\pi$  Wolfram Alpha says the inverse cosine near -1 is less than $\pi-\sqrt{2(z+1)}$ so you want $\sqrt{1-z^2}-\sqrt{2(z+1)}&amp;gt;0$ or $\sqrt{z+1}(\sqrt{1-z}-\sqrt{2})&amp;gt;0$, which is never true because $z&amp;gt;-1$.  If you pick a different branch of inverse cosine, it can be true easily.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2011-01-19T18:21:13.550" LastActivityDate="2011-01-19T18:21:13.550" CommentCount="3" />
  <row Id="18145" PostTypeId="1" CreationDate="2011-01-19T15:16:26.163" Score="2" ViewCount="63" Body="&lt;p&gt;A box contains n number of coins, m of which are fair and rest are biased. The probability of getting a head when a fair coin is tossed is 1/2, while it is 2/3 when a biased coin is tossed. A coin is drawn from the box at random and is tossed twice. The first time it shows head and second time it shows tail. What is the probability that the coin drawn is fair.&lt;/p&gt;&#xA;" OwnerUserId="6025" LastActivityDate="2011-01-19T21:17:56.600" Title="What is the probability that the coin drawn is fair." Tags="&lt;probability&gt;" AnswerCount="3" />
  <row Id="18146" PostTypeId="2" ParentId="18138" CreationDate="2011-01-19T15:16:46.137" Score="0" ViewCount="" Body="&lt;p&gt;There are good explanations of &lt;a href=&quot;http://en.wikipedia.org/wiki/Limit_superior_and_limit_inferior&quot; rel=&quot;nofollow&quot;&gt;liminf and limsup of sets here&lt;/a&gt;. You will probably see these constructions in measure theory.&lt;/p&gt;&#xA;" OwnerUserId="1154" LastActivityDate="2011-01-19T15:16:46.137" />
  <row Id="18147" PostTypeId="2" ParentId="18111" CreationDate="2011-01-19T15:16:58.257" Score="3" ViewCount="" Body="&lt;p&gt;The local ring $A_\mathfrak{p}$ is a DVR and in particular a principal ideal domain. So the fractional ideal $A_{\mathfrak{p}}\mathfrak{a}$ is a free  $A_{\mathfrak{p}}$-module (since finitely generated and torsion free) of rank 1,&#xA;say with basis element $x$ . Any other basis element $x&amp;#39; \in A_{\mathfrak{p}}\mathfrak{a}$ must have the form $x&amp;#39; = ux$ for a unit $u \in A_{\mathfrak{p}}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the valuation $v_{\mathfrak{p}}$ is a function $K^\times \to \mathbf{Z}$&#xA;where $K$ is the field of fractions of $A$ and also of $A_{\mathfrak{p}}$. &#xA;So $v_{\mathfrak{p}}(\mathfrak{a}) := v_{\mathfrak{p}}(x)$ is well-defined.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And I believe that you will find that this definition of $v_{\mathfrak{p}}(\mathfrak{a})$ should make clear what is required for your question.&lt;/p&gt;&#xA;" OwnerUserId="5020" LastActivityDate="2011-01-19T15:16:58.257" />
  <row Id="18148" PostTypeId="1" CreationDate="2011-01-19T15:21:50.523" Score="1" ViewCount="69" Body="&lt;p&gt;I haven't done very much statistics but in a Finnish programming forum I found a some exercises where answers are integers. In every exercise the best result $r$ gets 100 points. Suppose you get the result $s$. If in a given problem the best score is as high as possible, you will be awarded by $100(s/r)^2$ points rounded to the nearest integer. If in a given problem the best score is as small as possible, you will be awarded by $100(r/s)^2$ points rounded to the nearest integer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I have a table of the best scores where top contestants gets many scores $100$ for single problem, some gets lower scores, some haven't done a problem or send any valid solution in which case the score is empty.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What kind of methods there are to approximate how difficult a single problem have been?&lt;/p&gt;&#xA;" OwnerUserId="5462" LastEditorUserId="5462" LastEditorDisplayName="" LastEditDate="2011-01-19T15:31:31.883" LastActivityDate="2011-01-19T15:31:31.883" Title="How to measure how difficult an exercise in a programming contest is?" Tags="&lt;statistics&gt;" CommentCount="3" />
  <row Id="18149" PostTypeId="2" ParentId="18134" CreationDate="2011-01-19T15:24:57.823" Score="7" ViewCount="" Body="&lt;p&gt;A straightforward upper bound is $N=29$.  For a given $N$, you are trying to select one of the possible assignments of numbers to the six sides (up to rotation of the cube), of which there are&#xA;$$&#xD;&#xA;{N \choose 6}\frac{6!}{24} = \frac{N!}{24(N-6)!}.&#xD;&#xA;$$&#xA;The number of different states you can see when you enter the room, on the other hand, is&#xA;$$&#xD;&#xA;{N \choose 5}5! = \frac{N!}{(N-5)!} = \frac{N!}{(N-5)(N-6)!}.&#xD;&#xA;$$&#xA;The number of outputs (i.e., assignments of numbers to the cube) is greater than the number of inputs (i.e., different states you can see) whenever $N-5 &amp;gt; 24$, and so there can't be a surjective mapping from inputs to outputs unless $N \le 29$.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;A lower bound of $N=18$ is given by the following strategy.  Your friend will place the cube so that the largest face and its largest neighbor are visible.  There are 16 such ways to orient the cube, all distinguishable by the visible faces: 4 where the largest face is on top; 4 where its largest neighbor is on top; and 8 where both the largest face and its largest neighbor are on the side.  Order these 16 arbitrarily in advance, corresponding to the integers $0$ through $15$.  Similarly, rank the 16 numbers that are not equal to the largest face or its largest neighbor, from $0$ to $15$.  Your friend will place the cube in the position corresponding to the sum of the other four faces' ranks, modulo 16.  By subtracting the ranks of the three faces you can see, you can deduce the rank, and hence the value, of the hidden face.&lt;/p&gt;&#xA;" OwnerUserId="5546" LastEditorUserId="5546" LastEditorDisplayName="" LastEditDate="2011-01-19T19:17:14.530" LastActivityDate="2011-01-19T19:17:14.530" CommentCount="3" />
  <row Id="18150" PostTypeId="2" ParentId="18105" CreationDate="2011-01-19T15:32:17.843" Score="1" ViewCount="" Body="&lt;p&gt;EDIT: The first paragraph is nonsense... but I'm keeping this here since the comments might be helpful to people looking at this question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Looks like I figured out the answer to my own question! In the shower just now, I came up with a 7-manifold not bounded by an 8-manifold: $\mathbb{RP}^2 \times S\mathbb{RP}^4$ (Edit: this doesn't quite work yet; don't trust things you find out about in the shower. I just need a 5-manifold with odd Euler characteristic... any ideas?). Indeed, notice that its Euler characteristic is the product of the Euler characteristics of the factors, which are both odd. Boundaries of manifolds have even Euler characteristic, therefore this one is not a boundary, and it is clearly 7-dimensional. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So my confusion stemmed from the fact that the claim `every 7-manifold is a boundary' is false! After a more careful skimming of Milnor's paper, it looks like, even though he uses the unqualified phrase &quot;every 7-manifold is a boundary&quot; he really only uses that &quot;every &lt;em&gt;orientable&lt;/em&gt; 7-manifold is a boundary of an &lt;em&gt;orientable&lt;/em&gt; 8-manifold.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not entirely sure why this is true, but some evidence comes from the fact that if we invert $2$ in the oriented cobordism ring, then it becomes isomorphic to a polynomial algebra over $\mathbb{Z}[1/2]$ with generators in every 4th dimension... so at least we know that there's nothing appearing in dimension 7 except possibly some 2-torsion. Anyone know why there's no 2-torsion there?&lt;/p&gt;&#xA;" OwnerUserId="423" LastEditorUserId="423" LastEditorDisplayName="" LastEditDate="2011-01-19T18:24:20.967" LastActivityDate="2011-01-19T18:24:20.967" CommentCount="6" />
  <row Id="18151" PostTypeId="2" ParentId="18145" CreationDate="2011-01-19T15:36:18.710" Score="1" ViewCount="" Body="&lt;p&gt;Let &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;$A$ = Even that coin selected is fair&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$B$ = Coin selected is biased&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$C$= Even that first results in a head, second results in a tail.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Now use &lt;em&gt;Bayes Theorem&lt;/em&gt; to get the answer. Answer is  $\frac{9m}{8n+m}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;$P(A)=\frac{m}{n}$ and $P(B)=\frac{n-m}{n}$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$P(C/A) = \frac{1}{4}$ and $P(C/B) = \frac{2}{3} \times \frac{1}{3} = \frac{1}{9}$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;By &lt;em&gt;Bayes theorem&lt;/em&gt; $$P(A/C) = \frac{P(A) \times P(C/A)}{P(A) \times P(C/A) + P(B) \times P(C/B)}=\frac{9m}{8n+m}$$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" LastActivityDate="2011-01-19T21:17:56.600" CommentCount="1" />
  <row Id="18152" PostTypeId="2" ParentId="18145" CreationDate="2011-01-19T15:37:07.590" Score="0" ViewCount="" Body="&lt;p&gt;What have you done so far?  Hint:  Without the flips, what is the chance you draw a fair coin?  If you draw a fair coin, what is the chance you get the flips specified?  If you draw a biased coin, what is the chance?  &lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-01-19T15:37:07.590" />
  <row Id="18153" PostTypeId="2" ParentId="18145" CreationDate="2011-01-19T15:44:52.883" Score="0" ViewCount="" Body="&lt;p&gt;I know this isn't really answering your question, but a coin toss, is rarely ever fair.  And I'm not talking a few thousandths of a percent, but more like 52% vs 48%.  Below makes for fascinating reading:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.codingthewheel.com/archives/the-coin-flip-a-fundamentally-unfair-proposition&quot; rel=&quot;nofollow&quot;&gt;http://www.codingthewheel.com/archives/the-coin-flip-a-fundamentally-unfair-proposition&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And remember, always be the tosser!&lt;/p&gt;&#xA;" OwnerUserId="5072" LastActivityDate="2011-01-19T15:44:52.883" />
  <row Id="18154" PostTypeId="1" AcceptedAnswerId="18196" CreationDate="2011-01-19T15:56:08.147" Score="4" ViewCount="254" Body="&lt;p&gt;On Mathworld it says some diophantine equations can be solved using algebraic numbers. I know one example which is factor $x^2 + y^2$ in $\mathbb{Z}[\sqrt{-1}]$ to find the Pythagorean triples.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would be very interested in finding some examples of harder equations (not quadratic) which are also solved easily using algebraic numbers. Thank you!&lt;/p&gt;&#xA;" OwnerUserId="4290" LastEditorUserId="1321" LastEditorDisplayName="" LastEditDate="2011-01-31T23:34:44.340" LastActivityDate="2011-01-31T23:34:44.340" Title="Diophantine equations solved using algebraic numbers?" Tags="&lt;number-theory&gt;&lt;algebra&gt;" AnswerCount="6" />
  <row Id="18155" PostTypeId="2" ParentId="18133" CreationDate="2011-01-19T15:59:14.593" Score="1" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;First inequality&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order for the inequality to make sense, I assume that $M(r)\le r/2$. Let $\theta$ be such that&#xA;$$\cos\theta=\frac{4M(r)}{r}-1.$$&#xA;Then what you want is $\theta+|\sin\theta|&amp;gt;\pi$. This cannot be achieved with the usual branch of $\cos^{-1}(x)$, which takes values in $[0,\pi/2]$ for $x\in[0,1]$ and in $[\pi/2,\pi]$ for $x\in[-1,0]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Edit&lt;/em&gt; You con take for instance $\arccos_{1}x=2\pi+\cos^{-1}x$, where $\cos^{-1}$ is the usual branch. Then&#xA;$$\arccos_{1}x+\sqrt{1-x^2}&amp;gt;\pi\quad\forall x \in[-1,1].$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then any $M(r)$ such that $M(r)\le r/2$ will do.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have also included absolte values around the $\sin\theta$ in the previous paragraph.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Second inequality&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $a,b&amp;gt;0$, then $a\cosh t+b\sinh t\ge a&amp;gt;0$ for all $t\ge0$, so that $t_0$ is not defined.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Edit&lt;/em&gt; The O.P. has edited the question in the sense that $a$ and $b$ are such that $a\cosh t+b\sinh t=0$ has a positive solution. The question remains open.&lt;/p&gt;&#xA;" OwnerUserId="4791" LastEditorUserId="4791" LastEditorDisplayName="" LastEditDate="2011-01-19T19:34:28.650" LastActivityDate="2011-01-19T19:34:28.650" CommentCount="3" />
  <row Id="18156" PostTypeId="1" AcceptedAnswerId="18165" CreationDate="2011-01-19T16:00:19.410" Score="2" ViewCount="85" Body="&lt;p&gt;Dear community,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am a little stuck at the moment with the &lt;a href=&quot;http://en.wikipedia.org/wiki/Fourier_transform&quot; rel=&quot;nofollow&quot;&gt;Fourier transform&lt;/a&gt; of a function $w(z)$ that is defined piecewise and discontinuous at $z = 0$:&#xA;$$w\left(z\right) = \begin{cases} -\gamma_1 e^{-\gamma_{1}z} &amp;amp; z &amp;gt; 0 \\ \gamma_2 e^{\gamma_{2}z} &amp;amp; z &amp;lt; 0\end{cases}$$&#xA;with $\gamma_i \geq 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a naive attempt, I calculated the following:&#xA;$$ W \left( k \right) = \int_{-\infty}^{\infty}w(z) e^{-\mathrm{i}k z}dz$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ =  -\gamma_{1}\int_{0}^{\infty}e^{-\left(\gamma_{1}+\mathrm{i}k\right)z}dz+\gamma_{2}\int_{-\infty}^{0}e^{\left(\gamma_{2}-\mathrm{i}k\right)z}dz  = -\frac{\gamma_{1}}{\left(\gamma_{1}+\mathrm{i}k\right)}+\frac{\gamma_{2}}{\left(\gamma_{2}-\mathrm{i}k\right)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I am almost sure that I miscalculated since there should somewhere occur a $\delta$-contribution proportional to $\left(\gamma_2 - \gamma_1 \right)\cdot\delta(0)$ as in the case of the &lt;a href=&quot;http://mathworld.wolfram.com/FourierTransformHeavisideStepFunction.html&quot; rel=&quot;nofollow&quot;&gt;Heaviside function&lt;/a&gt;.&lt;br&gt;&#xA;Hence:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Where is my brain-overflow in the calculation given? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Thank you in advance&lt;br&gt;&#xA;Sincerely&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Robert&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS.: Even though it is not such a thing, I marked this question as homework since it is on this level. &lt;/p&gt;&#xA;" OwnerUserId="2671" LastActivityDate="2011-01-19T17:04:26.340" Title="Fourier Transform of an unsteady function" Tags="&lt;homework&gt;&lt;fourier-analysis&gt;" AnswerCount="3" />
  <row Id="18157" PostTypeId="1" AcceptedAnswerId="18162" CreationDate="2011-01-19T16:06:17.960" Score="4" ViewCount="97" Body="&lt;p&gt;This was an exam question of ours:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Let $\chi$ be the set, $\chi = \left \{  a, b, c, d \right \}$. Create a topology $\tau$ on $\chi$ so that $\left ( \chi ,\tau  \right )$ is regular but not normal.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I couldn't solve the question but I'm wondering what the answer is. Is it possible at all? Can anyone help me with this?&lt;/p&gt;&#xA;" OwnerUserId="3796" LastActivityDate="2011-01-19T16:19:42.043" Title="Regular but not normal space" Tags="&lt;topology&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="18158" PostTypeId="2" ParentId="18156" CreationDate="2011-01-19T16:16:17.103" Score="1" ViewCount="" Body="&lt;p&gt;It depends on whether you want to compute $\int_{ - \infty }^\infty  {e^{ - ikz} w(z)dz} $ or $\int_{ - \infty }^\infty  {e^{ - ikz} dw(z)} $.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Further clarification: Define $w(0)=-\gamma_1$, so that $w$ is right-continuous with left limits. Since $w$ has a jump discontinuity of size $-\gamma_1 - \gamma_2$ at $z=0$, and since $e^{-ik0}=1$,&#xA;$$&#xD;&#xA;\int_{ - \infty }^\infty  {e^{ - ikz} dw(z)} = \int_{ - \infty }^0  {e^{ - ikz} w(z)dz} + (-\gamma_1 - \gamma_2) + \int_{ 0 }^\infty  {e^{ - ikz} w(z)dz}.&#xD;&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="2810" LastEditorUserId="2810" LastEditorDisplayName="" LastEditDate="2011-01-19T17:04:26.340" LastActivityDate="2011-01-19T17:04:26.340" CommentCount="4" />
  <row Id="18159" PostTypeId="2" ParentId="18105" CreationDate="2011-01-19T16:17:53.763" Score="6" ViewCount="" Body="&lt;p&gt;Milnor is referring to &lt;em&gt;oriented&lt;/em&gt; bordism there, not un-oriented bordism.  Not every 7-manifold bounds an 8-manifold as you've observed from Thom's computation of the unoriented bordism ring, being a polynomial ring and having generators in dimension 2 and 5 (and many other dimensions). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I believe it goes back to Rene Thom, that the bordism group of oriented 7-manifolds is trivial. &lt;/p&gt;&#xA;" OwnerUserId="642" LastActivityDate="2011-01-19T16:17:53.763" />
  <row Id="18160" PostTypeId="2" ParentId="18157" CreationDate="2011-01-19T16:18:04.917" Score="3" ViewCount="" Body="&lt;p&gt;No, it is not possible. For example check Munkres's Topology book, p 200:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Theorem 32.1&lt;/strong&gt; Every regular space with a countable basis is normal.&lt;/p&gt;&#xA;" OwnerUserId="5105" LastActivityDate="2011-01-19T16:18:04.917" CommentCount="1" />
  <row Id="18161" PostTypeId="1" AcceptedAnswerId="18164" CreationDate="2011-01-19T16:18:43.523" Score="2" ViewCount="98" Body="&lt;p&gt;Since computers work off formulas, without greater knowledge one would assume that it would come up with the same answer for a set formula. However, you're able to tell it to generate a random number. IE: php rand(). What's the formula that's used to get a random number? How can there be anything &quot;random&quot; in an algebraic expression?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Been wondering for a while, hopefully it's not blatantly obvious.&lt;/p&gt;&#xA;" OwnerUserId="1595" LastActivityDate="2011-02-08T23:25:54.567" Title="Formula for Random" Tags="&lt;algebra&gt;&lt;functional-equations&gt;" AnswerCount="2" />
  <row Id="18162" PostTypeId="2" ParentId="18157" CreationDate="2011-01-19T16:19:42.043" Score="4" ViewCount="" Body="&lt;p&gt;Suppose $\chi$ is a regular space. Suppose $A,B$ are closed disjoint sets. If one of them is a singleton, then the regularity implies that they can be separated by open sets. Otherwise, since the sets are disjoint, then each of them is of size 2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If this is the case then $A= \chi - B$ is open as a complement of a close set so $A,B$ are open and you can separate A from B, therefore $\chi$ is normal.&lt;/p&gt;&#xA;" OwnerUserId="2125" LastActivityDate="2011-01-19T16:19:42.043" CommentCount="1" />
  <row Id="18163" PostTypeId="2" ParentId="18156" CreationDate="2011-01-19T16:21:45.227" Score="1" ViewCount="" Body="&lt;p&gt;You should ask yourself what happens with the evaluation of the integral primitive at infinity (when applying the Barrow rule) if $\gamma =0$. For example, what happens to $e^{-(\gamma_1 + ik)z}$ evaluated in $z=\infty$ if $\gamma_1 =0$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That case must be treated with some care. But, in your case, it's no problem, because then your function is zero and the result is ok. Then, I think your result is correct (but you must go with care, and treat the cases $\gamma_i =0$ separatedly).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It would be different is, for example, your second piece of function would have been $e^{\gamma_2 z}$ instead of $\gamma_2 e^{\gamma_2 z}$&lt;/p&gt;&#xA;" OwnerUserId="312" LastEditorUserId="312" LastEditorDisplayName="" LastEditDate="2011-01-19T16:48:48.183" LastActivityDate="2011-01-19T16:48:48.183" CommentCount="1" />
  <row Id="18164" PostTypeId="2" ParentId="18161" CreationDate="2011-01-19T16:24:10.063" Score="4" ViewCount="" Body="&lt;p&gt;There is a discussion in &lt;a href=&quot;http://en.wikipedia.org/wiki/Random_number_generation&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt; and many books on the subject.  The numbers are deterministic but scattered.  You hope that in the ways you use them, they &quot;act like&quot; they are truly random.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-01-19T16:24:10.063" CommentCount="3" />
  <row Id="18165" PostTypeId="2" ParentId="18156" CreationDate="2011-01-19T16:32:43.050" Score="3" ViewCount="" Body="&lt;p&gt;There's no brain overflow. The $\delta(x)$ part in the Heaviside calculuation comes from the fact that the Heaviside function does not decay at infinity. As $x\to +\infty$, you have $H(x)$ is bounded from below by a constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fact that your function has an exponentially decreasing weight gets rid of that problem. (When the weight $\gamma_i = 0$, by definition your function is zero also, so it has zero Fourier transform.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Key thing to remember when looking at Fourier transforms: it interchanges decay properties at infinity with smoothness of a function. That is (morally speaking), if your function is smooth, then its Fourier transform will have fast decay near infinity. If your function has exponential decay, then its Fourier transform will be smooth. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That your starting function has a jump discontinuity at the origin is thus reflected in the roughly $1/k$ (slow) decay of the Fourier transform (analogous to the second factor in the Fourier transform of the Heaviside function in your link). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Also see rule 205 in &lt;a href=&quot;http://en.wikipedia.org/wiki/Fourier_transform#Tables_of_important_Fourier_transforms&quot; rel=&quot;nofollow&quot;&gt;Wikipedia's list&lt;/a&gt;.)&lt;/p&gt;&#xA;" OwnerUserId="1543" LastActivityDate="2011-01-19T16:32:43.050" CommentCount="1" />
  <row Id="18166" PostTypeId="1" AcceptedAnswerId="18171" CreationDate="2011-01-19T16:38:18.993" Score="2" ViewCount="68" Body="&lt;p&gt;We have:  &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;U = 150 &lt;/li&gt;&#xA;&lt;li&gt;|M| = 90&lt;/li&gt;&#xA;&lt;li&gt;|E| = 75&lt;/li&gt;&#xA;&lt;li&gt;|F| = 80&lt;/li&gt;&#xA;&lt;li&gt;|M ∩ E| = 45&lt;/li&gt;&#xA;&lt;li&gt;|M ∩ F| = 35&lt;/li&gt;&#xA;&lt;li&gt;|M ∩ F ∩ E| = 10&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;We want:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;|E ∩ F ∩ $M^c$| = ?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I know I have to use the exclusion-inclusion principle. I have to take the last statement and transform it so that it is similar to:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;|A' ∪ B'|&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;But I don't know how transform it; that complementary is problematic. I have tried playing with the complementary, but I get this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$|E ∩ F ∩ M^c| = |(E∪F∪M^c)^c|$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But that is not useful, right?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think this properties coud be useful too:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;A∩U =A &lt;/li&gt;&#xA;&lt;li&gt;A∩∅=∅ &lt;/li&gt;&#xA;&lt;li&gt;A∪U=U&lt;/li&gt;&#xA;&lt;li&gt;A∪∅=A&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I would appreciate some tips.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My Solution:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$U = 150 = |E ∪ F ∪ M|$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/qht9F.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://i.imgur.com/qht9F.jpg&quot; rel=&quot;nofollow&quot;&gt;1&lt;/a&gt; $|M^c ∩ F ∩ E| = |F ∩ E| - |M ∩ F ∩ E|$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$|E ∪ F ∪ M| = |E| + |F| + |M| - (|E ∩ F| + |E ∩ M| + |F ∩ M|) + |E ∩ F ∩ M|$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$150 = 75 + 80 + 90 - |E ∩ F| - 45 - 35 + 10$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$|E ∩ F| = 25$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, back to &lt;a href=&quot;http://i.imgur.com/qht9F.jpg&quot; rel=&quot;nofollow&quot;&gt;1&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$|M^c ∩ F ∩ E| = |F ∩ E| - |M ∩ F ∩ E| = 25 - 10 = 15  $&lt;/p&gt;&#xA;" OwnerUserId="5526" LastEditorUserId="5526" LastEditorDisplayName="" LastEditDate="2011-02-08T10:20:11.753" LastActivityDate="2011-02-08T10:20:11.753" Title="How to transform a set with complementary so that I can apply Inclusion-exclusion principle" Tags="&lt;elementary-set-theory&gt;" AnswerCount="2" CommentCount="11" />
  <row Id="18167" PostTypeId="2" ParentId="18122" CreationDate="2011-01-19T16:43:50.907" Score="2" ViewCount="" Body="&lt;p&gt;Ah, the old helicopter game. No sweat, it's actually very simple. But because you're modelling &lt;em&gt;acceleration&lt;/em&gt;, you really do want a variable which represents velocity. Then you can just do this:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;every frame {&#xA;&#xA;    if(mouse down) {&#xA;        velocity += [gravity constant];&#xA;    } else {&#xA;        velocity -= [thrust constant];&#xA;    }&#xA;&#xA;    position += velocity;&#xA;&#xA;}&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Without a &lt;code&gt;velocity&lt;/code&gt; variable, it's very tricky to get the effect you want.&lt;/p&gt;&#xA;" OwnerUserId="6027" LastActivityDate="2011-01-19T16:43:50.907" />
  <row Id="18168" PostTypeId="1" AcceptedAnswerId="18169" CreationDate="2011-01-19T16:53:29.817" Score="0" ViewCount="92" Body="&lt;p&gt;How do I approach this problem $x[f(x+1)-f(x-1)]=1$.&lt;/p&gt;&#xA;" OwnerUserId="5959" LastEditorUserId="5959" LastEditorDisplayName="" LastEditDate="2011-03-01T10:54:46.990" LastActivityDate="2011-03-01T10:54:46.990" Title="$x[f(x+1)-f(x-1)]=1$" Tags="&lt;functional-equations&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="18169" PostTypeId="2" ParentId="18168" CreationDate="2011-01-19T16:59:46.490" Score="2" ViewCount="" Body="&lt;p&gt;Pick any function $f(x)$ on $(0,2]$.  Then $f(x)$ on $(2,\infty )$ is determined-just step downward by 2's until you get into $(0,2]$.  A similar technique works for $x&amp;lt;0$ but you step up.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-01-19T16:59:46.490" CommentCount="0" />
  <row Id="18170" PostTypeId="2" ParentId="18154" CreationDate="2011-01-19T17:03:51.770" Score="3" ViewCount="" Body="&lt;p&gt;The classic example is certain &lt;a href=&quot;http://mathworld.wolfram.com/MordellCurve.html&quot; rel=&quot;nofollow&quot;&gt;Mordell curves&lt;/a&gt;.  I can never quite remember which ones have this property off the top of my head, but for example I think $y^2 = x^3 - 1$ is one of them.  (Write it as $y^2 + 1 = x^3$ and factor the LHS in $\mathbb{Z}[i]$.)  There might be a list in some books on Diophantine equations and/or algebraic number theory and/or elliptic curves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is also the &lt;a href=&quot;http://en.wikipedia.org/wiki/Regular_prime&quot; rel=&quot;nofollow&quot;&gt;regular case&lt;/a&gt; of Fermat's last theorem.  I don't know if this counts as &quot;easy,&quot; though.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2011-01-19T17:03:51.770" CommentCount="1" />
  <row Id="18171" PostTypeId="2" ParentId="18166" CreationDate="2011-01-19T17:21:18.530" Score="2" ViewCount="" Body="&lt;p&gt;We know that $E \cap F \cap M^{c} = (E \cap F) - (E \cap F \cap M)$. Thus you need to find $E \cap F$. I think a Venn Diagram will help here since you know the size of $U$ (the universe). Or one could proceed by inclusion-exclusion. We know that $$|E \cup F \cup M| = |E|+|F|+|M|- (|E \cap F|+ |E \cap M|+ |F \cap M|) + |E \cap F \cap M|$$&lt;/p&gt;&#xA;" OwnerUserId="2548" LastEditorUserId="2548" LastEditorDisplayName="" LastEditDate="2011-01-19T17:35:37.643" LastActivityDate="2011-01-19T17:35:37.643" CommentCount="1" />
  <row Id="18172" PostTypeId="1" AcceptedAnswerId="18175" CreationDate="2011-01-19T17:24:16.407" Score="0" ViewCount="76" Body="&lt;p&gt;Given that $x^3[f(x+1)-f(x-1)]=1$, determine $\lim_{x\rightarrow \infty}(f(x)-f(x-1))$ explicitly.&lt;/p&gt;&#xA;" OwnerUserId="5959" LastActivityDate="2011-01-19T17:46:31.920" Title="$x^3[f(x+1)-f(x-1)]=1$" Tags="&lt;functional-equations&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="18174" PostTypeId="2" ParentId="18088" CreationDate="2011-01-19T17:37:26.750" Score="1" ViewCount="" Body="&lt;p&gt;Somewhat along the lines of Andres's comment:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose you were to assign every formula in arithmetic (e.g., $0 = 0$) a unique Natural number in such a way that you could mechanically figure out which formula corresponded to which Natural number.  Then for any &lt;B&gt;consistent&lt;/B&gt; (non-contradictory) theory $T$ that includes the 15 basic rules of arithmetic from &lt;a href=&quot;http://en.wikipedia.org/wiki/Peano_axioms#Equivalent_axiomatizations&quot; rel=&quot;nofollow&quot;&gt;PA&lt;SUP&gt;-&lt;/SUP&gt;&lt;/a&gt; and only includes statements &lt;B&gt;true about the Natural numbers&lt;/B&gt;, there will be some formula $Prov_T(x)$ representing provability from $T$ under this coding.  In particular, this means that if $\varphi_n$ is some formula assigned Natural number $n$ under our mechanical coding procedure and $T \vdash Prov_T(n)$ (&quot;T proves that $\varphi_n$ is provable&quot;), then $T \vdash \varphi_n$ (&quot;T proves $\varphi_n$&quot;).  The representability of provability also means that if $T \vdash \lnot Prov_T(n)$ (&quot;$T$ proves that $\varphi_n$ is not provable&quot;), then $T \nvdash \varphi_n$ (&quot;$T$ does not prove $\varphi_n$&quot;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Incidentally, &lt;a href=&quot;http://en.wikipedia.org/wiki/Godel%27s_incompleteness_theorems#First_incompleteness_theorem&quot; rel=&quot;nofollow&quot;&gt;Gödel's first incompleteness theorem&lt;/a&gt; implies that there will be a statement $\varphi_n$ so that any (consistent) set of &quot;identifiable&quot; axioms $T$ true in the Natural numbers that includes all of those found in PA&lt;SUP&gt;-&lt;/SUP&gt; cannot prove $\varphi_n$ nor $\lnot\varphi_n$.  In fact this statement was of the form $\lnot Prov_T(m)$ for some Natural number $m$.  If $T$ also includes all axioms of full Peano Arithmetic, then by Gödel's second incompleteness theorem, we can find such a statement by using any Natural number $m$ coding a statement $\lnot\varphi$ where $\varphi$ is actually provable from $T$.&lt;/p&gt;&#xA;" OwnerUserId="4693" LastActivityDate="2011-01-19T17:37:26.750" CommentCount="1" />
  <row Id="18175" PostTypeId="2" ParentId="18172" CreationDate="2011-01-19T17:46:31.920" Score="3" ViewCount="" Body="&lt;p&gt;if $\lim_{x\to\infty}(f(x)-f(x-1))$ exists and it's equal to $l$, then also $\lim_{x\to\infty}(f(x+1)-f(x))=l$. So &#xA;$$&#xD;&#xA;2l=\lim_{x\to\infty}( f(x+1)-f(x-1))=\lim_{x\to\infty}\frac{1}{x^3}=0,&#xD;&#xA;$$ &#xA;hence $l=0$.&lt;/p&gt;&#xA;" OwnerUserId="2607" LastActivityDate="2011-01-19T17:46:31.920" />
  <row Id="18176" PostTypeId="2" ParentId="18111" CreationDate="2011-01-19T17:54:55.597" Score="0" ViewCount="" Body="&lt;p&gt;@Willie Wong: I logged in with my open ID just now but somehow, my account name is still unregistered and I could not vote up the answer. Could you help me with this?&lt;/p&gt;&#xA;" OwnerUserId="6029" LastActivityDate="2011-01-19T17:54:55.597" />
  <row Id="18177" PostTypeId="2" ParentId="11382" CreationDate="2011-01-19T17:55:07.313" Score="0" ViewCount="" Body="&lt;p&gt;In digital signal processing a signal of finite length may be represented as a polynomial ( its Z transform).    So the convolution of two finite length signals is simply the multiplication of the Z transform polynomials.   &lt;/p&gt;&#xA;" OwnerUserId="832" LastEditorUserId="832" LastEditorDisplayName="" LastEditDate="2011-01-19T18:00:47.290" LastActivityDate="2011-01-19T18:00:47.290" />
  <row Id="18178" PostTypeId="1" AcceptedAnswerId="18185" CreationDate="2011-01-19T18:03:17.653" Score="0" ViewCount="76" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Possible Duplicate:&lt;/strong&gt;&lt;br&gt;&#xA;  &lt;a href=&quot;http://math.stackexchange.com/questions/18172/x3fx1-fx-1-1&quot;&gt;$x^3[f(x+1)-f(x-1)]=1$&lt;/a&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;p&gt;Given that f is continuous and $x^3[f(x+1)-f(x)]=1$, determine $\lim_{x\rightarrow \infty}f(x)$ explicitly.&lt;/p&gt;&#xA;" OwnerUserId="5959" LastEditorUserId="5959" LastEditorDisplayName="" LastEditDate="2011-01-19T18:10:08.380" LastActivityDate="2011-01-19T18:38:04.793" Title="$x^3[f(x+1)-f(x)]=1$" Tags="&lt;functional-equations&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="18179" PostTypeId="1" AcceptedAnswerId="18186" CreationDate="2011-01-19T18:17:09.227" Score="10" ViewCount="373" Body="&lt;p&gt;While trying some problems along with my friends we had difficulty in this question.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;True or False: The value of the infinite product $$\prod\limits_{n=2}^{\infty} \biggl(1-\frac{1}{n^{2}}\biggr)$$ is $1$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I couldn't do it and my friend justified it by saying that since the terms in the product have values less than $1$, so the value of the product can never be $1$. I don't know whether this justification is correct or not. But i referred to &lt;em&gt;Tom Apostol's&lt;/em&gt; Mathematical Analysis book and found a theorem which states, that &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The infinite product $\prod(1-a_{n})$ converges if the series $\sum a_{n}$ converges.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This assures that the above product converges. Could anyone help me in finding out where it converges to? And,&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Does there exist a function $f$ in $\mathbb{N}$ ( like $n^{2}$, $n^{3}$) such that $\displaystyle \prod\limits_{n=1}^{\infty} \Bigl(1-\frac{1}{f(n)}\Bigr)$ has the value $1$?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" LastEditorUserId="2937" LastEditorDisplayName="" LastEditDate="2011-01-19T18:47:58.830" LastActivityDate="2011-05-18T05:01:12.700" Title="Finding Value of the Infinite Product $\prod \Bigl(1-\frac{1}{n^{2}}\Bigr)$" Tags="&lt;real-analysis&gt;&lt;analysis&gt;&lt;products&gt;" AnswerCount="4" />
  <row Id="18180" PostTypeId="2" ParentId="18166" CreationDate="2011-01-19T18:20:48.103" Score="2" ViewCount="" Body="&lt;p&gt;Since you know that $M\cup E\cup F = 150$, the Inclusion-Exclusion Formula gives you&#xA;\begin{align*}&#xA;150 &amp;amp;= |M\cup E\cup F|\\&#xA;&amp;amp;= |M|+|E|+|F|-|M\cap E|-|M\cap F|-|E\cap F|+|M\cap E\cap F|\\&#xA;&amp;amp;= 90 + 75 + 80 - 45 - 35 - |E\cap F| + 10\\&#xA;&amp;amp;= 175 - |E\cap F|,&#xA;\end{align*}&#xA;so you know that $|E\cap F|= 25$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This gives you all the information you need: For example, to divide $M$ into its four disjoint parts ($M\cap E\cap F$, $M\cap E^c\cap F$, $M\cap E\cap F^c$, and $M\cap E^c\cap F^c$), you use inclusion-exclusion: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;You know that $M$ contains $10$ elements that are also in $E$ and in $F$ (that is, $|M\cap E\cap F| = 10$).&lt;/li&gt;&#xA;&lt;li&gt;$M$ contains $35$ elements that are also in $E$ but not in $F$ (because $|M\cap E| = |M\cap E\cap F^c| + |M\cap E\cap F|$, and you know that $|M\cap E| = 45$ and $|M\cap E\cap F|=10$).&lt;/li&gt;&#xA;&lt;li&gt;$M$ contains $25$ elements that are also in $F$ but not in $E$ (because $|M\cap E^c\cap F| = |M\cap F| - |M\cap E\cap F|$).&lt;/li&gt;&#xA;&lt;li&gt;And $M$ contains $20$ elements that are in $M$ but not in either $E$ nor $F$ (because $|M\cap E^c\cap F^c| = |M| - |M\cap E| - |M\cap F| + |M\cap E\cap F|$).&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Similarly, $E$ contains $10$ elements that are also in $M$ and $F$; it contains $35$ that are in $M$ but not in $F$; $15$ that are also in $F$ but not in $M$; and therefore $15$ that in neither $M$ nor $F$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And $F$ contains $10$ elements that are also in $M$ and $F$; $25$ that are in $M$ but not in $E$; and $15$ that are in $E$ but not in $M$; leaving $30$ that are in $F$ but not in $E$ nor $M$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is:&#xA;\begin{align*}&#xA;|M\cap E\cap F\;| &amp;amp;= 10;\\&#xA;|M\cap E\cap F^c\;| &amp;amp;= 35;\\&#xA;|M\cap E^c\cap F\;| &amp;amp;= 25;\\&#xA;|M\cap E^c\cap F^c\;| &amp;amp;= 20;\\&#xA;|M^c\cap E\cap F\;| &amp;amp;= 15;\\&#xA;|M^c\cap E\cap F^c\;| &amp;amp;= 15;\\&#xA;|M^c\cap E^c\cap F\;| &amp;amp;= 30;\\&#xA;|M^c\cap E^c\cap F^c\;| &amp;amp;= 0.&#xA;\end{align*}&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-19T19:11:00.887" LastActivityDate="2011-01-19T19:11:00.887" />
  <row Id="18181" PostTypeId="2" ParentId="18179" CreationDate="2011-01-19T18:24:01.277" Score="2" ViewCount="" Body="&lt;p&gt;If $a_m = \displaystyle \prod_{n=2}^{m}(1 - \frac{1}{n^2})$, then $a_{m+1} = (1 - \frac{1}{(m+1)^2})a_m &amp;lt; a_m$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we have $a_m &amp;lt; a_2 = (1 - \frac{1}{4}) = \frac{3}{4}$, $\forall m &amp;gt; 2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, $\displaystyle \lim_{m \rightarrow \infty} a_m &amp;lt; a_2 = \frac{3}{4}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence the statement is false.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastActivityDate="2011-01-19T18:24:01.277" />
  <row Id="18182" PostTypeId="2" ParentId="18179" CreationDate="2011-01-19T18:27:17.213" Score="13" ViewCount="" Body="&lt;p&gt;The product can never be 1. Recall that the product is defined as the limit of the partial products $$ \prod_{n\le k}\left(1-\frac1{f(n)}\right)=A_k. $$ Now, if $f(n)=1$ then the product is 0. I assume you know how to handle &quot;divergence to 0&quot;? To simplify, let's assume $f(n)\ne 1$ for all $n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then we have that $A_k&amp;gt;A_{k+1}$ for all $k$, since $1-\epsilon&amp;lt;1$ for any positive $\epsilon$ (say, $\epsilon=1/f(k+1)$). Since $A_1=1-1/f(1)&amp;lt;1$, our sequences will converge to a positive number smaller than 1, or diverge to 0, but it most certainly cannot converge to 1.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Here is a hint to evaluate $$\prod_{n=2}^\infty\left(1-\frac1{n^2}\right) :$$&#xA;Note that this is a telescoping product, since $1-1/n^2=(n-1)(n+1)/n^2$. Now play with the first few terms to see the emerging pattern.&lt;/p&gt;&#xA;" OwnerUserId="462" LastEditorUserId="462" LastEditorDisplayName="" LastEditDate="2011-01-19T18:34:58.840" LastActivityDate="2011-01-19T18:34:58.840" />
  <row Id="18184" PostTypeId="2" ParentId="18179" CreationDate="2011-01-19T18:31:27.363" Score="12" ViewCount="" Body="&lt;p&gt;Your friend is right: the partial products are clearly $\le 3/4$, and since the infinite product converges (by the test you showed), its value must also be $\le 3/4$.  To find the actual value, note that $1 - 1/n^2 = (n-1)(n+1)/n^2$, and so the infinite product is&#xA;$$&#xD;&#xA;{\large\Pi}_{n=2}^{\infty}\left(1 - \frac{1}{n^2}\right) = \frac{1}{2}\frac{3}{2}\cdot\frac{2}{3}\frac{4}{3}\cdot\frac{3}{4}\frac{5}{4}\cdot\frac{4}{5}\frac{6}{5}\cdot ...&#xD;&#xA;$$&#xA;All terms cancel in pairs except the first; so the value of the product is $1/2$.&lt;/p&gt;&#xA;" OwnerUserId="5546" LastActivityDate="2011-01-19T18:31:27.363" CommentCount="2" />
  <row Id="18185" PostTypeId="2" ParentId="18178" CreationDate="2011-01-19T18:38:04.793" Score="1" ViewCount="" Body="&lt;p&gt;I will assume you are working with the function only defined on the positive real numbers. As any function that is bounded at 0 and 1 (implied by continuity at 0 and 1) cannot satisfy your expression at $x = 0$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In any case, your limit does not have any fixed value, and in fact does not necessarily exist. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;First notice that if you define the function $g(x) = f(x) + c$, then $x^3[g(x+1) - g(x)] = x^3[f(x+1) + c - f(x) - c] = 1$ is another solution to your functional equation. This implies that should the limits exists, $\lim_{x\to\infty}f(x) + c = \lim_{x\to\infty} g(x)$. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Now if the limit does exists, the limit would be independent of how you approach $x\to\infty$. So in particular choose the sequence $x_k = k$ of integers, from which the functional equation $f(x+1)-f(x) = 1/x^3$ leads to a telescoping sum&#xA;$$ \lim_{x\to\infty} f(x) = f(1) + \sum_{k = 1}^{\infty} \frac{1}{k^3} $$&#xA;the sum in the right hand side evaluates to roughly 1.20206. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;But the limit does not have to exist. For any function $h(x)$ defined on $[1,2]$ satisfying $h(1) = h(2) - 1$, you can use the functional equation to generate a solution of $x^3[f(x+1) - f(x)] = 1$. You simply set $$f(x) = h(1 + \{x\}) + \sum_{k = 1}^{\lfloor x\rfloor - 1} \frac{1}{(k+ \{x\})^3}$$ where $\{x\}$ is the fractional part of $x$ and $\lfloor x\rfloor$ is the integral part of $x$. Now consider your initially chosen $h(x)$ to take the values $h(1) = 1$, $h(2) = 2$, $h(3/2) = 0$. Then we have $$\limsup_{x\to\infty} f(x) \geq h(1) + \sum_{k = 1}^{\infty} \frac{1}{k^3}$$ but also $$\liminf_{x\to\infty} f(x) \leq h(3/2) + \lim_{n\to\infty}\sum_{k = 1}^n \frac{1}{(k + 1/2)^3} &amp;lt; h(1) + \sum_{k=1}^\infty\frac{1}{k^3} \leq \limsup_{x\to\infty}f(x)$$ using term by term comparison of the two absolutely converging series, and that $h(3/2) &amp;lt; h(1)$. This shows that for such a choice, the limit doesn't exist (since the $\limsup$ is strictly larger than the $\liminf$). &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1543" LastActivityDate="2011-01-19T18:38:04.793" CommentCount="1" />
  <row Id="18186" PostTypeId="2" ParentId="18179" CreationDate="2011-01-19T18:41:41.663" Score="22" ViewCount="" Body="&lt;p&gt;We have&#xA;\begin{align*}&#xA;p_{k} &amp;amp;= \prod_{n = 2}^{k} \left( 1 - \frac{1}{n^{2}} \right) = \prod_{n=2}^{k} \frac{(n-1)(n+1)}{n^{2}} \\ &amp;amp; = \frac{1 \cdot 3}{2 \cdot 2} \cdot \frac{2\cdot 4}{3 \cdot 3} \cdot \frac{3\cdot 5}{4 \cdot 4} \cdot \cdots \cdot \frac{(k-2)\cdot k}{(k-1)\cdot(k-1)} \cdot \frac{(k-1)(k+1)}{k\cdot k}\\ &amp;amp; = \frac{1}{2}\left(1 + \frac{1}{k}\right)&#xA;\end{align*}&#xA;because all but the first and last numerators and denominators cancel. Therefore&#xA;\[&#xA;\prod_{n=2}^{\infty} \left( 1 - \frac{1}{n^{2}} \right) =&#xA;\lim_{k\to\infty} p_{k} = \frac{1}{2}.&#xA;\]&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;To put this into a little context: Euler has shown that&#xA;\[&#xA;\sin{(\pi z)} = \pi z \prod_{n=1}^{\infty} \left( 1 - \frac{z^{2}}{n^{2}} \right)&#xA;\]&#xA;for all $z \in \mathbb{C}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We would like to plug in $z = 1$ on both sides. This doesn't work because we get $0 = 0$. But a simple trick yields what we want:&#xA;\[&#xA;\pi \prod_{n=2}^{\infty} \left( 1 - \frac{1}{n^{2}} \right) = \lim_{z \to 1} \frac{\sin{(\pi z)}}{(1- z^2)} = \lim_{z \to 1} \frac{\pi \cos{(\pi z)}}{-2z} = \frac{\pi}{2}&#xA;\]&#xA;and cancelling on both sides with $\pi$ gives $\frac{1}{2}$ for the infinite product.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a lot of fun that you can have with Euler's product. For instance $z = \frac{1}{2}$ yields the Wallis product formula&#xA;\[&#xA;\frac{\pi}{2} = \frac{2 \cdot 2}{1 \cdot 3} \cdot \frac{4 \cdot 4}{3 \cdot 5} \cdot \frac{6 \cdot 6}{5 \cdot 7} \cdot \cdots = \prod_{n=1}^{\infty} \frac{2n}{2n -1}\frac{2n}{2n +1}&#xA;\]&#xA;and $z = i$ yields&#xA;\[&#xA;\frac{\sin{(i\pi)}}{i \pi} = \frac{e^{\pi} - e^{-\pi}}{2\pi} =&#xA;\prod_{n = 1}^{\infty} \left( 1 + \frac{1}{n^{2}} \right).&#xA;\]&#xA;A thorough treatment of these and many other topics involving infinite products can be found in chapters 1 and 2 of Remmert's book on complex analysis.&lt;/p&gt;&#xA;" OwnerUserId="5363" LastEditorUserId="6491" LastEditorDisplayName="" LastEditDate="2011-05-18T05:01:12.700" LastActivityDate="2011-05-18T05:01:12.700" CommentCount="3" />
  <row Id="18187" PostTypeId="2" ParentId="18154" CreationDate="2011-01-19T18:43:27.577" Score="4" ViewCount="" Body="&lt;p&gt;Perhaps the simplest example is the parametrization of primitive Pythagorean triples $\rm x^2 + y^2 = z^2\:.$ The essence of the proof is: $\rm\ x+y\ i,\ \ x-y\ i\ $ are coprime factors of a square in a UFD, so they must themselves be squares (up to unit factors). So $\rm\ x + y\ i\ =\ (m + n\ i)^2 =\ m^2 - n^2 + 2\: m\: n\ i $. Similarly one can solve low degree cases of Fermat's Last Theorem by employing analogous factorizations over certain rings of algebraic integers. For example, Gauss showed there are no solutions for exponent 3 by working in the ring of integers of $\rm\ \mathbb Q(\sqrt{-3})\:,\: $ and Dirichlet did similarly for exponent 5 using $\rm\ \mathbb Q(\sqrt{5})\:.$ Later Kummer generalized these techniques to handle all regular prime exponents by working over rings of cyclotomic integers. For a nice exposition see Ribenboim: &lt;em&gt;13 lectures on Fermat's last theorem&lt;/em&gt;. Weil nicely summarizes the essence of these techniques in his &lt;em&gt;Number Theory&lt;/em&gt;, Ch.IV,S.VI,p.335:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/4GLMB.jpg&quot; alt=&quot;alt text&quot;&gt;&#xA;&lt;img src=&quot;http://i.imgur.com/oEEby.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2011-01-19T19:35:42.827" LastActivityDate="2011-01-19T19:35:42.827" />
  <row Id="18188" PostTypeId="2" ParentId="18119" CreationDate="2011-01-19T18:55:03.410" Score="2" ViewCount="" Body="&lt;p&gt;This is not an answer.  I worked out some of the details relevant to my comment above &lt;a href=&quot;http://qchu.wordpress.com/2010/10/08/update-and-the-combinatorics-of-quintic-equations/&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; and &lt;a href=&quot;http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1520810#p1520810&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.  A rough summary is as follows:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The discriminant of $t^n + pt + q$ turns out to equal $(-1)^{ {n \choose 2} } ((1 - n)^{n-1} p^n + n^n q^{n-1} )$.  This is a generalization of the fact you want to prove ($p = 0, q = -1$).&lt;/li&gt;&#xA;&lt;li&gt;There is a unique power series $t$ in the variable $q$ which satisfies $t = q + t^n$ ($p = -1$) and $t(0) = 0$.  It turns out to be precisely $t = \sum \frac{1}{(n-1)k + 1} {nk \choose k} q^{(n-1)k + 1}$ for combinatorial reasons; $t$ counts what I call &quot;$n$-ary trees&quot; in the blog post above (but you should read the definition carefully).&lt;/li&gt;&#xA;&lt;li&gt;Therefore $\frac{dt}{dq} = \sum {nk \choose k} q^{(n-1)k}$.  It turns out that the relation $t = q + t^n$ implies a corresponding algebraic relation for $\frac{dt}{dq}$, and the leading term of this algebraic relation is, up to sign, the discriminant of $t^n - t + q$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Unfortunately I do not know a conceptual reason why the third bullet point is true.  You can see it directly in the case $n = 2$, where the discriminant of $t^2 - t + q$ is $1 - 4q$ and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{dt}{dq} = \sum {2k \choose k} q^k = \frac{1}{\sqrt{1 - 4q}}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here $t$ is the generating function for the &lt;a href=&quot;http://en.wikipedia.org/wiki/Catalan_number&quot; rel=&quot;nofollow&quot;&gt;Catalan numbers&lt;/a&gt;.  You can also work out what happens when $n = 3$ by the cubic formula, but after that things become troublesome.&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2011-01-19T19:03:05.990" LastActivityDate="2011-01-19T19:03:05.990" CommentCount="2" />
  <row Id="18189" PostTypeId="2" ParentId="18084" CreationDate="2011-01-19T18:59:56.240" Score="2" ViewCount="" Body="&lt;p&gt;Here's a heuristic argument.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Upper bound: The probability that a walk will not be restarted after $2\log n$ steps is $1/n^2$, so we don't expect it ever to happen within the alloted $n$ steps. In particular, we don't expect the random walk to get farther than $2\log n$ away.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lower bound: The probability that we make $(\log n)/4$ straight right steps is $1/\sqrt{n}$, so we expect that to happen within the alloted $n$ steps. Every time it happens, either the starting point or the end point must be at least $(\log n)/8$ away from the origin.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2011-01-19T18:59:56.240" CommentCount="2" />
  <row Id="18190" PostTypeId="1" AcceptedAnswerId="18207" CreationDate="2011-01-19T19:04:30.143" Score="0" ViewCount="160" Body="&lt;p&gt;The other day, I tried to answer someone's question, you can read the question and the answer here: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://math.stackexchange.com/questions/17970/numerical-solutions-of-ordinary-differential-equations&quot;&gt;Numerical Solutions of ordinary differential equations&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is, I got stuck towards the end of the calculation. Now my question is: does anyone know how to get order 2?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many thanks for your help!&lt;/p&gt;&#xA;" OwnerUserId="5798" LastActivityDate="2011-01-19T22:48:23.987" Title="Local truncation error calculation" Tags="&lt;numerical-methods&gt;" AnswerCount="1" />
  <row Id="18191" PostTypeId="1" AcceptedAnswerId="18194" CreationDate="2011-01-19T19:10:27.447" Score="1" ViewCount="74" Body="&lt;pre&gt;&lt;code&gt;Category Observed Expected (O-E)    (O-E)^2     (O-E)^2 / E&#xA;0 6 5 1 1 0.5&#xA;1 6 5 1 1 0.5&#xA;2 3 5 2 4 2&#xA;3 3 5 2 4 2&#xA;4 4 5 1 1 0.5&#xA;5 5 5 0 0 0&#xA;6 3 5 2 4 2&#xA;7 8 5 3 9 4.5&#xA;8 5 5 0 0 0&#xA;9 7 5 2 4 2&#xA; 50    14 &amp;lt;---- ChiSquare Value&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Can anyone confirm that I did this correct? Basically I did it in excel, taking 50 random numbers between 0-9 using = RANDBETWEEN(0,9).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am interested in knowing if I did the process correctly. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Referenced : &lt;a href=&quot;http://math.hws.edu/javamath/ryan/ChiSquare.html&quot; rel=&quot;nofollow&quot;&gt;http://math.hws.edu/javamath/ryan/ChiSquare.html&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="6032" LastActivityDate="2011-01-19T22:03:17.427" Title="Chi-Squared Test Process Validation" Tags="&lt;statistics&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="18192" PostTypeId="1" AcceptedAnswerId="18193" CreationDate="2011-01-19T19:40:37.503" Score="10" ViewCount="211" Body="&lt;p&gt;Hello. I want to show that any two disjoint compact sets in a Hausdorff space $X$ can be separated by disjoint open sets. Can you please let me know if the following is correct? Not for homework, just studying for a midterm. I'm trying to improve my writing too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My work:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $C$,$D$ be disjoint compact sets in a Hausdorff space $X$. Now fix $y \in D$ and for each $x \in C$ we can find (using Hausdorffness) disjoint open sets $U_{x}(y)$ and $V_{x}(y)$ such that $x \in U_{x}(y)$ and $y \in V_{x}(y)$. Now the collection $\{U_{x}: x \in C\}$ covers $C$ so by compactness we can find some natural k such that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$C \subseteq \bigcup_{i=1}^{k} U_{x_{i}}(y)$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now for simplicity let $U = \bigcup_{i=1}^{k} U_{x_{i}}(y)$, then $C \subseteq U$ and let $W(y) = \bigcap_{i=1}^{k} V_{x_{i}}(y)$. Then $W(y)$ is a neighborhood of $y$ and disjoint from $U$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now consider the collection $\{W(y): y \in D\}$, this covers D so by compactness we can find some natural q such that $D \subseteq \bigcup_{j=1}^{q} W_{y_{j}}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally set $V = \bigcup_{j=1}^{q} W_{y_{j}}$, then $U$ and $V$ are disjoint open sets containing $C$ and $D$ respectively. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What do you think?&lt;/p&gt;&#xA;" OwnerUserId="6033" LastActivityDate="2011-04-06T01:04:04.473" Title="Disjoint compact sets in a Hausdorff space can be separated" Tags="&lt;topology&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="18193" PostTypeId="2" ParentId="18192" CreationDate="2011-01-19T19:49:03.967" Score="5" ViewCount="" Body="&lt;p&gt;This is a very good start, but there is a slight problem with your argument: as you change $y$, your $U$ changes as well (since $U$ is constructed in terms of $y$); you should really call it $U(y)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your construction gives you an open neighborhood $W(y)$ of $y$ for each $y$; $W(y)$ is disjoint from $U(y)$. But for all you know, $W(y)$ may fail to be disjoint from $U(y&amp;#39;)$ with $y&amp;#39;\neq y$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you really still have a bit more to go before you are done. &lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2011-01-19T19:49:03.967" CommentCount="5" />
  <row Id="18194" PostTypeId="2" ParentId="18191" CreationDate="2011-01-19T19:50:49.037" Score="3" ViewCount="" Body="&lt;p&gt;I agree with most of your spreadsheet; however, I think the last column needs some attention. I think it should be &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;0.2&#xA;0.2&#xA;0.8&#xA;0.8&#xA;0.2&#xA;0.0&#xA;0.8&#xA;1.8&#xA;0.0&#xA;0.8&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;It looks like you might have divided by 2 instead of dividing by the Expected column. Of course, this would change the sum, so $\chi^2 = 5.6$.&lt;/p&gt;&#xA;" OwnerUserId="5220" LastEditorUserId="5220" LastEditorDisplayName="" LastEditDate="2011-01-19T22:03:17.427" LastActivityDate="2011-01-19T22:03:17.427" />
  <row Id="18195" PostTypeId="2" ParentId="18154" CreationDate="2011-01-19T19:53:29.147" Score="1" ViewCount="" Body="&lt;p&gt;There are many examples, some of which I see you have already gotten :-)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is one example I had come across recently:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Show that the following diophantine equation,&#xA;  for integral $\displaystyle a,b$, has no non-trivial solutions.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$$3b^4 + 3b^2 + 1 = a^2$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I came across this while trying to find an &quot;elementary&quot; solution to the diophantine equation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$y^2 = x^3 - 1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is easily solved using $\displaystyle \mathbb{Z}[i]$, see this here: &lt;a href=&quot;http://math.stackexchange.com/questions/5914/integral-solutions-to-y2-x3-1/&quot;&gt;Integral solutions to $y^{2}=x^{3}-1$.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, there might be an easy elementary proof for $\displaystyle 3b^4 + 3b^2 + 1 = a^2$, but, in fact this is closely related to the Pell's equation $\displaystyle 3x^2 + 1 = y^2$ and amounts to showing that the non-trivial odd values of $\displaystyle x$ are such that $\displaystyle \frac{x-1}{2}$ cannot be a perfect square. This itself can be cast into a problem of the terms of a linear recurrence never being perfect squares. So I do expect this problem to give some resistance to an elementary solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To solve the problem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mutiply by $\displaystyle 9b^2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$3b^2(9b^4 + 9b^2 + 3) = (3ab)^2$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;put $\displaystyle x = 3b^2 + 1$ and $\displaystyle y = 3ab$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(x-1)(x^2 +x + 1) = y^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x^3 - 1 = y^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which we can show has no non-trivial solutions easily, by using $\displaystyle \mathbb{Z}[i]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: If one can prove non-existence of $\displaystyle 3b^4 + 3b^2 + 1 = a^2$ by other means, we will have found a different proof of non-existence of $\displaystyle x^3 - 1= y^2$, because the existence of non-trivial solutions to $\displaystyle x^3 - 1= y^2$ implies the existence of non-trivial solutions to $\displaystyle 3b^4 + 3b^2 + 1 = a^2$. (That is how I came across this).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I will leave that to you.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2011-01-19T21:43:30.453" LastActivityDate="2011-01-19T21:43:30.453" CommentCount="2" />
  <row Id="18196" PostTypeId="2" ParentId="18154" CreationDate="2011-01-19T20:10:12.437" Score="2" ViewCount="" Body="&lt;p&gt;Euler famously used algebraic integers to solve diophantine problems, though not always exactly right. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, in article 182 of Chapter XII, Part II, of his &lt;strong&gt;Elements of Algebra&lt;/strong&gt; (p. 396 in my brand new Cambridge Library Collection edition), he writes:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Let, therefore, the formula $x^2+cy^2$ be proposed, and let it be required to make it a square. As it is composed of the factors $(x+y\sqrt{-c})(x-y\sqrt{-c})$, these factors must either be squares, or squares multiplied by the same number. For, if the product of two numbers, for example, $pq$, must be a square, we must have $p=r^2$ and $q=s^2$; that is to say, each factor is of itself a square; or $p=mr^2$ and $q=ms^2$; and therefore these factors are squares multiplied by the same number. For which reason, let us make $x+y\sqrt{-c} = m(p+q\sqrt{-c})^2$; it will follow that $x-y\sqrt{-c} = m(p-q\sqrt{-c})^2$, and we shall have $x^2+cy^2 = m^2(p^2+cq^2)^2$, which is a square. Farther, in order to determine $x$ and $y$, we have the equations $x+y\sqrt{-c} = mp^2 + 2mpq\sqrt{-c} -mcq^2$, and $x-y\sqrt{-c} = mp^2-2mpq\sqrt{-c} - mcq^2$; in which $x$ is necessarily equal to the rational part, and $y\sqrt{-c}$ to the irrational part; so that $x=mp^2 - mcq^2$ and $y\sqrt{-c} = 2mpq\sqrt{-c}$, or $y=2mpq$; and these are the values of $x$ and $y$ that will transform the expression $x^2+cy^2$ into a square, $m^2(p^2+cq^2)^2$, the root of which is $mp^2+mcq^2$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(Note that Euler in fact finds &lt;em&gt;sufficient&lt;/em&gt; conditions, but not necessary ones, here, and there is an implicit assumption of unique factorization). In Article 187, he proceeds along the same lines to discuss when $ax^2+cy^2$ will be a cube, working with the factorization $ax^2+cy^2 = (x\sqrt{a}+y\sqrt{-c})(x\sqrt{a} - y\sqrt{-c})$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Euler himself knew this did not quite work out in general: in Article 195 he discusses when $2x^2-5$ is a cube; his method yields the answer that it is &lt;em&gt;never&lt;/em&gt; a cube, but yet he notes that $x=4$ gives $2x^2 - 5 = 27 = 3^3$; his &quot;investigation&quot; into the failure involves some rather strenuous hand-waving, concluding that the problem is that we have a difference instead of a sum as before. &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-19T20:49:11.437" LastActivityDate="2011-01-19T20:49:11.437" CommentCount="1" />
  <row Id="18197" PostTypeId="1" CreationDate="2011-01-19T21:22:09.543" Score="0" ViewCount="117" Body="&lt;p&gt;I believe the problem of trying to find the Jacobian of the following function highlights a lack of understanding of some concept on my part. I was hoping someone could either provide specific advice about solving this problem, or computing Jacobians in general. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the mapping $h : \mathbb{R}^n \rightarrow \mathbb{R}^n$ where the domain is length-$n$ column vectors and the range length-$n$ row vectors (or a transposed vector, if you like). The function is&#xA;$$h(x) = \frac{\eta v&amp;#39; + (M x)&amp;#39;}{(\eta + u&amp;#39;x)^2},$$&#xA;where the constants $v$ and $u$ are (column) vectors, $\eta$ is a scalar, and $M$ is a square matrix. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So far as I know, the quotient rule for vectors is&#xA;$$\nabla\left(\frac{f}{g}\right) = \frac{g\nabla f - f \nabla g}{g^2}$$&#xA;and&#xA;\begin{align*}&#xA;\nabla f &amp;amp;= M'\\&#xA;\nabla g &amp;amp;= 2(\eta + u'x) u'&#xA;\end{align*}&#xA;Putting it all together, I get&#xA;$$\nabla h = \frac{(\eta + u&amp;#39;x)^2 M&amp;#39; - [\eta v&amp;#39; + (M x)&amp;#39;] 2(\eta + u&amp;#39;x) u&amp;#39;}{(\eta + u&amp;#39;x)^4}.$$&#xA;This expression is clearly not right, and to see why evaluate the Jacobian at $x = \mathbf{0}$:&#xA;$$\nabla h(0) = \frac{\eta^2 M&amp;#39; - 2\eta^2 v&amp;#39; u&amp;#39;}{\eta^4}$$&#xA;The resulting expression should be a $n \times n$ matrix, but in the second term we have two (row) vectors multiplied by one another. It seems likely there should be some sort of outer product here, but I'm not sure where my math is going wrong. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any help you can provide is greatly appreciated.&lt;/p&gt;&#xA;" OwnerUserId="5799" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-20T02:42:38.533" LastActivityDate="2011-01-20T02:42:38.533" Title="Jacobian of a function mapping vectors to vectors" Tags="&lt;vector-calculus&gt;" CommentCount="10" />
  <row Id="18198" PostTypeId="1" AcceptedAnswerId="18199" CreationDate="2011-01-19T21:30:12.233" Score="3" ViewCount="159" Body="&lt;p&gt;I know this is very basic. But very puzzling too and often missed by learners like myself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When talking about continuous random variables with a particular probability distribution, what are the underlying sample spaces? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second Q: why these sample spaces are omitted often times, and one simply says r.v. $X$ follows a uniform distribution on interval $[0,1]$? Isn't the sample sample space critically important? Many thanks!&lt;/p&gt;&#xA;" OwnerUserId="6016" LastActivityDate="2011-03-03T01:49:43.363" Title="what are the sample spaces when talking about continuous random variables" Tags="&lt;probability&gt;&lt;probability-theory&gt;" AnswerCount="4" />
  <row Id="18199" PostTypeId="2" ParentId="18198" CreationDate="2011-01-19T21:39:50.550" Score="8" ViewCount="" Body="&lt;p&gt;You can take it to be a subset of $\mathbb{R}$ or, more generally, $\mathbb{R}^n$.  A random variable uniformly distributed in $[0, 1]$ can be thought of as a random variable on the sample space $[0, 1]$ with probability density function $1$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sample space is in fact &lt;em&gt;not&lt;/em&gt; critically important.  (You may find it convenient to pick one to do computations in, but it doesn't matter which one you pick.  This is analogous to choosing coordinates to do computations in linear algebra.)  This point is explained very clearly in Terence Tao's notes &lt;a href=&quot;http://terrytao.wordpress.com/2010/01/01/254a-notes-0-a-review-of-probability-theory/&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;At a purely formal level, one could call probability theory the study of measure spaces with total measure one, but that would be like calling number theory the study of strings of digits which terminate. At a practical level, &lt;strong&gt;the opposite is true&lt;/strong&gt; [emphasis added]: just as number theorists study concepts (e.g. primality) that have the same meaning in every numeral system that models the natural numbers, we shall see that probability theorists study concepts (e.g. independence) that have the same meaning in every measure space that models a family of events or random variables. And indeed, just as the natural numbers can be defined abstractly without reference to any numeral system (e.g. by the Peano axioms), core concepts of probability theory, such as random variables, can also be defined abstractly, without explicit mention of a measure space; we will return to this point when we discuss free probability later in this course.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Terence Tao's explanation of free probability is &lt;a href=&quot;http://terrytao.wordpress.com/2010/02/10/245a-notes-5-free-probability/&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.  (I found it very enlightening; the framework Tao describes can be used to describe &lt;a href=&quot;http://en.wikipedia.org/wiki/Quantum_probability&quot; rel=&quot;nofollow&quot;&gt;quantum probability&lt;/a&gt; with very little modification, unlike the measure theory framework.)&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2011-01-19T21:46:21.353" LastActivityDate="2011-01-19T21:46:21.353" CommentCount="3" />
  <row Id="18200" PostTypeId="2" ParentId="18057" CreationDate="2011-01-19T21:55:08.323" Score="1" ViewCount="" Body="&lt;p&gt;For fun, here's another approach ...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The area of your triangle is given by $\frac{1}{2} w h$, but it's &lt;em&gt;also&lt;/em&gt; given by $\frac{1}{2}s m$, where $s$ is the length of the hypotenuse and $m$ is the length of your segment $M$. (To see why, flip the triangle over so that it's sitting on its hypotenuse; then, clearly, $s$ is the &quot;base&quot; and $m$ the &quot;height&quot;.) Therefore,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$w h = s m$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Writing $\theta$ for the angle that segment $M$ makes with the horizontal, we have that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x = m \cos\theta = \frac{wh}{s}\cos\theta \hspace{0.5in} y = m\sin\theta = \frac{wh}{s}\sin\theta$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By similar triangles, $\theta$ is also the angle in the top corner of your triangle, so&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\cos\theta = \frac{h}{s} \hspace{0.5in} \sin\theta = \frac{w}{s}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x = \frac{wh^2}{s^2}=\frac{w h^2}{w^2+h^2} \hspace{0.5in} y = \frac{w^2 h}{s^2}=\frac{w^2h}{w^2+h^2}$$&lt;/p&gt;&#xA;" OwnerUserId="409" LastActivityDate="2011-01-19T21:55:08.323" />
  <row Id="18201" PostTypeId="1" AcceptedAnswerId="18202" CreationDate="2011-01-19T21:56:19.850" Score="5" ViewCount="187" Body="&lt;p&gt;This question is motivated by a little anecdote. I was at home teaching some secondary school math to a relative. At some relax time, he glanced at a book I had over the table - it was some text about analytical number theory that I had recently bought, second hand- and I  explained him that that was an area of mathematics quite arcane to me, that I'd like to learn something about it in some future, but I had little hopes... &lt;/p&gt;&#xA;&#xA;&lt;p&gt;He looked puzzled for a moment, and then asked me: &lt;em&gt;&quot;But, wait a moment... You don't know &lt;strong&gt;all&lt;/strong&gt; the mathematics?&quot;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This happened months ago, and I'm still laughing. But also (and here comes the question) I'm still thinking about how to make some picture of this issue: the big extension of &quot;the land of mathematics&quot;, in diverseness and ranges of depth - and the small regions that one has explored.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was specifically looking for some kind of bidimensional (planar?) chart, perhaps with the most basic/known math kwowledge in the center, and with the main math areas as regions, placed according to its mutual relations or some kind of taxonomy. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I guess this should go in community wiki)&lt;/p&gt;&#xA;" OwnerUserId="312" LastEditorUserId="2548" LastEditorDisplayName="" LastEditDate="2011-01-19T22:02:04.403" LastActivityDate="2011-01-20T01:43:04.507" CommunityOwnedDate="2011-01-19T22:51:14.980" Title="Some maps of the land of mathematics?" Tags="&lt;reference-request&gt;&lt;teaching&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="2" />
  <row Id="18202" PostTypeId="2" ParentId="18201" CreationDate="2011-01-19T21:59:15.517" Score="6" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://press.princeton.edu/titles/8350.html&quot; rel=&quot;nofollow&quot;&gt;The Princeton Companion of Mathematics&lt;/a&gt; is a good resource. The &lt;a href=&quot;http://www.math.niu.edu/~rusin/known-math/index/mathmap.html&quot; rel=&quot;nofollow&quot;&gt;mathematical atlas&lt;/a&gt; is good as well. The size of the bubbles are directly proportional to the amount of research activity in each area. This &lt;a href=&quot;http://mathoverflow.net/questions/2556/real-world-applications-of-mathematics-by-arxiv-subject-area&quot; rel=&quot;nofollow&quot;&gt;MO post&lt;/a&gt; might be useful also (looking at real world applications of arxiv areas).&lt;/p&gt;&#xA;" OwnerUserId="2548" LastEditorUserId="2548" LastEditorDisplayName="" LastEditDate="2011-01-19T22:24:07.747" LastActivityDate="2011-01-19T22:24:07.747" CommunityOwnedDate="2011-01-19T22:51:14.980" CommentCount="1" />
  <row Id="18203" PostTypeId="1" AcceptedAnswerId="18206" CreationDate="2011-01-19T22:30:36.747" Score="1" ViewCount="114" Body="&lt;p&gt;Let&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\mathscr{N}(R)$ denote the set of all &lt;em&gt;Nilpotent elements&lt;/em&gt; in a ring $R$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I have actually done an exercise, which states that if $x \in \mathscr{N}(R)$, then $x$ is contained in every prime ideal of $R$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The converse of this statement is if $x \notin \mathscr{N}(R)$, then there is a prime ideal which does not contains $x$. But i am not able to prove it. Can anyone provide me with a proof of this result.&lt;/p&gt;&#xA;" LastActivityDate="2011-01-20T09:09:05.257" Title="$x$ not nilpotent implies, there is a prime ideal not containing $x$." Tags="&lt;abstract-algebra&gt;&lt;algebra&gt;&lt;ring-theory&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="18204" PostTypeId="2" ParentId="18198" CreationDate="2011-01-19T22:35:45.657" Score="3" ViewCount="" Body="&lt;p&gt;While I certainly cannot do better than Tao (and admit that I haven't read Qiaochu's links) I'd like to make the following point:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I see the reason for essentially restricting to $(\Omega,\Sigma, \mu) = [0,1]$ as a rather technical one. If one would like to have &lt;em&gt;sequences&lt;/em&gt; of i.i.d. random variables, one needs to be able to form the product space $(\Omega,\mu)^{\mathbb{N}}$ - you need some form of Kolmogorov's consistency theorem. If you allow too nasty $\Omega$'s product spaces can be very degenerate. There are examples in Halmos's or Neveu's books where they show that a countable product can be carrying the trivial measure even if the factors themselves are  far from trivial. The point is that $\Omega$ might simply be &quot;too big&quot; to be reasonable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A very flexible and technically useful class of &quot;reasonable&quot; measurable spaces is the class of &lt;em&gt;standard Borel spaces&lt;/em&gt; (for which Kolmogorov's consistency theorem holds, luckily). By definition these are the measurable spaces which are (measurably) isomorphic to the Borel $\sigma$-algebra of a &lt;em&gt;complete&lt;/em&gt; and &lt;em&gt;separable&lt;/em&gt; metric space. Here's the surprise (Hausdorff, von Neumann):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Every uncountable standard Borel space is isomorphic to $[0,1]$ with the Borel $\sigma$-algebra. Moreover, every non-atomic probability measure on a standard Borel space is equivalent to Lebesgue-measure on $[0,1]$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So from this point of view there is essentially no restriction in assuming $\Omega$ to be $[0,1]$ to begin with. Of course, atoms are not really an issue. Since we deal with a probability space, there are at most countably many of them, so we'll just get a union of an interval and some countable set.&lt;/p&gt;&#xA;" OwnerUserId="5363" LastActivityDate="2011-01-19T22:35:45.657" CommentCount="1" />
  <row Id="18205" PostTypeId="2" ParentId="18203" CreationDate="2011-01-19T22:45:43.553" Score="3" ViewCount="" Body="&lt;p&gt;Let $f$ not be nilpotent. Consider the set $\Sigma$ of ideals $I$ such that if $n &amp;gt; 0$ then $f^n \notin I$. It is non-empty because $(0) \in \Sigma$. Order $\Sigma$ by inclusion and by Zorn's lemma choose a maximal element $P$. Show that $P$ is prime: let $x,y \notin P$. Then the ideals $P+(x)$ and $P+(y)$ strictly contain $P$, and so er not in $\Sigma$. Thus $f^m \in P+(x)$ and $f^n \in P+(y)$ for some $m,n$ (by def of $\Sigma$). It follows that $f^{m+n} \in P+(xy)$, hence $P+(xy) \notin \Sigma$, that is $xy \notin P$, so $P$ is prime. Hence we have a prime ideal $P$ not containing $f$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(the proof is really just copy from the proof in Atiyah-MacDonalds' book)&lt;/p&gt;&#xA;" OwnerUserId="4284" LastActivityDate="2011-01-19T22:45:43.553" CommentCount="1" />
  <row Id="18206" PostTypeId="2" ParentId="18203" CreationDate="2011-01-19T22:48:00.773" Score="3" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\ $ If $x$ isn't nilpotent then localize at the monoid generated by $x$ to deduce that that some prime ideal doesn't contain $x$.&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2011-01-19T22:48:00.773" />
  <row Id="18207" PostTypeId="2" ParentId="18190" CreationDate="2011-01-19T22:48:23.987" Score="1" ViewCount="" Body="&lt;p&gt;You haven't used $b_2 c_2 = \frac{1}{2}$. Taking from where you left off, $e = b_2 h (f(x_n, y_n) - f(x_n + c_2 h, y_n + c_2 h k_1)) + \frac{h^2}{2}*f^\prime(x_n, y_n) +\ldots$ and using the Taylor expansion of $f$ around $(x_n, y_n)$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f(x, y) = f(x_0, y_0) + \frac{\partial}{\partial x}f(x_0, y_0)\frac{(x-x_0)}{1!} + \frac{\partial}{\partial y}f(x_0, y_0)\frac{(y-y_0)}{1!} + ... $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;we get $f(x_n + c_2 h, y_n + c_2 h k_1) = f(x_n, y_n) + \frac{\partial}{\partial x}f(x_n, y_n)\frac{(c_2 h)}{1!} + \frac{\partial}{\partial y}f(x_n, y_n)\frac{(c_2 h k_1)}{1!} + O(h^2)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now make the substitution to eliminate $c_2$ from the RHS and move a term across:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f(x_n + c_2 h, y_n + c_2 h k_1) - f(x_n, y_n) = \frac{\partial}{\partial x}f(x_n, y_n)\frac{h}{2 b_2} + \frac{\partial}{\partial y}f(x_n, y_n)\frac{h k_1}{2 b_2} + O(h^2)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$e = \frac{h^2}{2} (-\frac{\partial}{\partial x}f(x_n, y_n) - \frac{\partial}{\partial y}f(x_n, y_n)k_1) + \frac{h^2}{2}*f^\prime(x_n, y_n) + O(h^3)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;NB I think that $f^\prime$ is $\frac{\partial}{\partial x}f$, so that could be tidied up a bit.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="5676" LastActivityDate="2011-01-19T22:48:23.987" CommentCount="1" />
  <row Id="18208" PostTypeId="1" CreationDate="2011-01-19T23:07:53.143" Score="4" ViewCount="109" Body="&lt;p&gt;I don't know much (any?) graph theory, but I'm interested in learning a bit, and doing some calculations for a game. Is there a tool online where I could construct a graph (this one has 30-40 vertices, maybe 100 edges), and play around to explore its properties? Useful things to do would be describing paths, finding related paths, and letting me write formulas to calculate the value of a path.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(By contrast with &lt;a href=&quot;http://math.stackexchange.com/questions/13841/online-tool-for-making-graphs-vertices-and-edges&quot;&gt;Online tool for making graphs (vertices and edges)?&lt;/a&gt;, I'm not interested in the presentation, I'm interested in analysis, playing, exploring, manipulating, sharing...)&lt;/p&gt;&#xA;" OwnerUserId="6034" LastActivityDate="2011-01-19T23:07:53.143" Title="Online graph theory analysis tool?" Tags="&lt;analysis&gt;&lt;graph-theory&gt;" CommentCount="3" />
  <row Id="18209" PostTypeId="1" AcceptedAnswerId="18217" CreationDate="2011-01-19T23:09:13.667" Score="3" ViewCount="108" Body="&lt;p&gt;Given two invertible sheaves $\mathcal{F}$ and $\mathcal{G}$, one can define their tensor product, but in this definition $\mathcal{F} \otimes \mathcal{G} (U)$ is (apparently) not simply equal to $\mathcal{F} (U) \otimes \mathcal{G}(U)$ for an open set $U$. This latter structure is a presheaf; can someone give me an example for when it is not a sheaf? Is there a characterization for when this actually does give a sheaf? &lt;/p&gt;&#xA;" OwnerUserId="1491" LastActivityDate="2011-01-20T05:07:12.657" Title="Tensor product of invertible sheaves" Tags="&lt;sheaf-theory&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="18210" PostTypeId="2" ParentId="18154" CreationDate="2011-01-19T23:13:49.853" Score="1" ViewCount="" Body="&lt;p&gt;Fermat's Last Theorem was originally &quot;proved&quot; by using some elementary arguments and the &quot;fact&quot; that the ring of integers of $\mathbb{Q}(\zeta_p)$ is a UFD (which it is not), where $\zeta_p$ is a primitive $p$th root of unity. These arguments do hold when $\mathcal{O}_p$ is a UFD though; for a thorough account of the argument, see Chapter 1 of Marcus's &lt;em&gt;Number Fields.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="1491" LastActivityDate="2011-01-19T23:13:49.853" />
  <row Id="18211" PostTypeId="2" ParentId="18198" CreationDate="2011-01-19T23:31:38.083" Score="1" ViewCount="" Body="&lt;p&gt;Begin with a probability space $(\Omega,\mathcal{F},P)$ ($\mathcal{F}$ is called a $\sigma$-algebra on $\Omega$, $P$ is called a probability measure). The collection of all Borel sets of $\mathbb{R}$ is denoted by $\mathcal{B}(\mathbb{R})$. A mapping $X:\Omega \to \mathbb{R}$ is a (real-valued) random variable if it is $\mathcal{F}$-measurable, that is, $\{ \omega \in \Omega: X(\omega) \in B \}$ is in $\mathcal{F}$ for each $B \in \mathcal{B}(\mathbb{R})$. Write $P[\{ \omega \in \Omega: X(\omega) \in B \}]$ as $P[X \in B]$. As a mapping of $B$, this is a probability measure on $\mathcal{B}(\mathbb{R})$, which can be denoted as $P_X$ and called the distribution of $X$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, to say that a random variable $X$ on $(\Omega,\mathcal{F},P)$ follows a uniform distribution on the interval $[0,1]$ simply means that $P_X$ is the measure on $\mathcal{B}(\mathbb{R})$ satisfying $P_X (\mathbb{R}-[0,1]) = 0$ and&#xA;$P_X (I) = b-a$ for any interval $I \subset [0,1]$ with endpoints $a&amp;lt;b$. For the simplest example, take the probability space $(\Omega,\mathcal{F},P) = ([0,1],\mathcal{B}([0,1]),P)$, where $P$ is the restriction to $[0,1]$ of the measure just defined above, and define $X:\Omega \to \mathbb{R}$ by $X(\omega) = \omega$. Then, for any $B \in \mathcal{B}(\mathbb{R})$,&#xA;$$&#xD;&#xA;P_X (B) = P[X \in B] = P[\{ \omega  \in [0,1]:X(\omega ) \in B\} ] = P[[0,1] \cap B],&#xD;&#xA;$$&#xA;from which it follows that $X$ is uniform on $[0,1]$. (Note that it is not essential that $\Omega$ be the set $[0,1]$.) &lt;/p&gt;&#xA;" OwnerUserId="2810" LastEditorUserId="2810" LastEditorDisplayName="" LastEditDate="2011-01-20T01:52:59.880" LastActivityDate="2011-01-20T01:52:59.880" />
  <row Id="18212" PostTypeId="2" ParentId="4054" CreationDate="2011-01-19T23:46:09.710" Score="1" ViewCount="" Body="&lt;p&gt;In short, the intuitive reason the Cauchy integral formula is true is that&#xA;(a) it is true in the limit for small circle paths $C_r$ about $z$,&#xA;when it reduces to averaging,&#xA;and&#xA;(b) the integral does not change as the path deforms nicely from&#xA;$C_r$ to $C$:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Integrals of complex differentiable functions&#xA;$\zeta\mapsto F(\zeta)$ over closed paths don't change&#xA;when the paths are deformed nicely in the domain of differentiability.&#xA;For a super proof of this that does not use continuous differentiability,&#xA;see H. Hanche-Olsen, ``On Goursat's proof of the Cauchy integral theorem,''&#xA;Amer. Math. Monthly 115 (2008) 648-652.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Integration on circle paths&#xA;$C_r: t\mapsto \zeta=z+re^{it}$ ($t\in[0,2\pi]$) corresponds to averaging:&#xA;$$&#xD;&#xA;\frac{d\zeta}{\zeta-z} = \frac{i r e^{it} dt}{re^{it}} =i\,dt,&#xD;&#xA;\quad\mbox{hence}\ \frac{1}{2\pi i}\int_{C_r}&#xD;&#xA;\frac{f(\zeta)}{\zeta-z}d\zeta = \frac1{2\pi} \int_0^{2\pi}f(z+re^{it})\,dt.&#xD;&#xA;$$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Given $z\in D$, the function $$\zeta\mapsto F(\zeta)=\frac{f(\zeta)}{\zeta-z}$$&#xA;is differentiable in $D\setminus\{z\}$,&#xA;so for any path, like $C$, that can be&#xA;deformed nicely in $D\setminus\{z\}$&#xA;to a circle path $C_r$ as above, we can further deform by taking $r\to0$ and conclude that for all small positive $r$,&#xA;$$&#xD;&#xA;f(z)&#xD;&#xA;=\frac{1}{2\pi i} \int_{C_r} \frac{f(\zeta)}{\zeta-z}d\zeta&#xD;&#xA;=\frac{1}{2\pi i} \int_C \frac{f(\zeta)}{\zeta-z}d\zeta.&#xD;&#xA;$$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="2947" LastEditorUserId="2947" LastEditorDisplayName="" LastEditDate="2011-02-23T06:58:17.837" LastActivityDate="2011-02-23T06:58:17.837" CommentCount="1" />
  <row Id="18214" PostTypeId="1" AcceptedAnswerId="18215" CreationDate="2011-01-19T23:48:47.303" Score="3" ViewCount="198" Body="&lt;p&gt;If I were to say that A intersect B was the empty set, does that imply that B intersect A is also the empty set?&lt;/p&gt;&#xA;" OwnerUserId="6036" LastEditorUserId="462" LastEditorDisplayName="" LastEditDate="2011-01-22T21:12:08.637" LastActivityDate="2011-01-25T22:37:28.373" Title="Are Disjoint Set Assertions Commutative?" Tags="&lt;elementary-set-theory&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="18215" PostTypeId="2" ParentId="18214" CreationDate="2011-01-19T23:58:59.097" Score="6" ViewCount="" Body="&lt;p&gt;Yes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because&#xA;$A \cap B = \{ x | x \in A \text{ and } x \in B \}$ and surely &quot;and&quot; is commutative.&lt;/p&gt;&#xA;" OwnerUserId="4284" LastActivityDate="2011-01-19T23:58:59.097" />
  <row Id="18216" PostTypeId="1" AcceptedAnswerId="18218" CreationDate="2011-01-20T00:32:11.430" Score="5" ViewCount="94" Body="&lt;p&gt;Let $p$ be a permutation of $\mathbb{N}$. We say that $p$ is bounded if there exists $k$ so that $|p(i)-i| \le k$ for all $i$. If $p$ is bounded, must there exist $M&amp;gt;0$ such that&#xA;$p(\{1,2,\ldots, M\}) = \{1,2,\ldots, M\}$? If so, can we bound $M$ by a function of $k$?&lt;/p&gt;&#xA;" OwnerUserId="857" LastEditorUserId="857" LastEditorDisplayName="" LastEditDate="2011-01-20T00:40:16.877" LastActivityDate="2011-04-12T06:54:28.237" Title="Do bounded permutations of N leave an initial segment invariant?" Tags="&lt;combinatorics&gt;&lt;group-theory&gt;&lt;permutations&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="18217" PostTypeId="2" ParentId="18209" CreationDate="2011-01-20T00:33:16.277" Score="6" ViewCount="" Body="&lt;p&gt;If for example $\mathcal F = \mathcal O_X,$ then $\mathcal F(U)\otimes_{\mathcal O_X(U)}\mathcal G(U) = \mathcal O_X(U)\otimes_{\mathcal O_X(U)} \mathcal G(U) = \mathcal G(U)$, so in this case we see that the presheaf $\mathcal F\otimes \mathcal G$ is equal to $\mathcal G$, and so is a sheaf.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But this is not typical.  A more illustrative example is $\mathcal F = \mathcal O(n)$ for $n &amp;gt; 0$ (on some positive dimensional projective space $\mathbb P^d$) and $\mathcal G = \mathcal O(-n) = \mathcal F^{-1}$.  Then if $U = \mathbb P^d$,&#xA;we have $\mathcal G(\mathbb P^d) = 0$, and so the values of the presheaf &#xA;$\mathcal F\otimes\mathcal G$ on $\mathbb P^d$ is equal to $0$.  On the other hand,&#xA;the actual (sheaf) tensor product $\mathcal F \otimes \mathcal G$ is equal to&#xA;the structure sheaf $\mathcal O_{\mathbb P^d}$ (because $\mathcal F$ and $\mathcal G$ are mutually inverse), which has a one-dimensional space of global sections. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So rather than trying to find situations when the presheaf tensor product is actually a sheaf, you will be better off getting an intuition for the sheafification process that goes into forming the sheaf tensor product from the presheaf version.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $\mathcal F$ is a presheaf and $\overline{\mathcal F}$ the corresponding sheaf, then two key points are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;there is a canonical map $\mathcal F(U) \to \overline{\mathcal F}(U)$  for any open set $U$;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;the natural map on stalks $\mathcal F_x \to \overline{\mathcal F}_x$ is an&#xA;isomorphism at every point $x$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Thus, in the context of tensor products, there is always a canonical map $\mathcal F(U)\otimes \mathcal G(U) \to (\mathcal F\otimes\mathcal G)(U)$ (where the target denotes the sheaf tensor product), and the stalk of $\mathcal F\otimes\mathcal G$ at any point is the tensor product of the corresponding stalks of $\mathcal F$ and $\mathcal G$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, if $X =$ Spec $A$ is affine and $\mathcal F$ and $\mathcal G$ correspond to $A$-modules $M$ and $N$ respectively, then $\mathcal F\otimes\mathcal G$ is the invertible sheaf associated to the product&#xA;$M\otimes_A N$.  In particular, if $U = $ Spec $A_f$ is a distinguished open&#xA;associated to $f \in A$, then the sections of $\mathcal F\otimes \mathcal G$&#xA;are equal to $(M\otimes_A N)_f = M_f\otimes_{A_f}N_f,$ which &lt;em&gt;is&lt;/em&gt; the tensor product&#xA;of the sections of $\mathcal F$ and $\mathcal G$ on Spec $A_f$.  So this is one case when one can work with the naive presheaf picture and get the correct answer, which is often useful in calculations (and also psychologically helpful in keeping a down-to-earth perspective on the general formalism).&lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2011-01-20T00:33:16.277" />
  <row Id="18218" PostTypeId="2" ParentId="18216" CreationDate="2011-01-20T00:45:52.823" Score="8" ViewCount="" Body="&lt;p&gt;Nope.  Define $p$ as follows:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$p(2i-1) = 2i+1$ for $i\ge1$,&lt;/li&gt;&#xA;&lt;li&gt;$p(2) = 1$,&lt;/li&gt;&#xA;&lt;li&gt;$p(2i+2) = 2i$ for $i\ge1$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;For an interval $[1, M]$ either $M$ or $M-1$ is odd, so $p$ preserves no interval.  But $|p(i) - i| \le 2$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that when $k = 1$, either $p(1) = 1$ or $p(1) = 2, p(2) = 1$.  Note also that leaving an initial segment requires that there exist finite cycles, so the example above shows that boundedness is not a strong enough assumption to guarantee that finite cycles exist.  &lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="6179" LastEditorDisplayName="" LastEditDate="2011-03-06T13:21:19.477" LastActivityDate="2011-03-06T13:21:19.477" CommentCount="2" />
  <row Id="18220" PostTypeId="1" CreationDate="2011-01-20T01:50:42.050" Score="1" ViewCount="88" Body="&lt;p&gt;Sorry to ask yet another &quot;how do I do this integral&quot; question! But I've really been having a hard time with this one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_0^\infty \frac{J_0(x)}{|C^2 - x^2|}\mathrm{d}x$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've been through a lot of identities giving results for expressions of the form&#xA;$\int Z_\nu(x) x^p \mathrm{d}x$,&#xA;where $Z_\nu$ is a Bessel function or one of its variants. But I haven't found a way to apply them to the integral I'm trying to do; I don't see any way to change variables such that the denominator becomes a simple power of $x$ without shifting the argument of $J_0$ becomes shifted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For what it's worth, I've tried expanding $J_0$ in a power series (for which every term after the first diverged), partial fraction decomposition on the denominator, integration by parts, etc. etc. I considered contour integration, but there is the slight problem that the integrand isn't an analytic function, and if I break it into domains $x &amp;lt; C$ and $x &amp;gt; C$ I'm not really sure how to set up an appropriate contour. I've also tried various other manipulations but nothing looks promising so far, which suggests that some heavy mathematical machinery that I'm not familiar with is called for. So can anyone here suggest a method or point me to a helpful reference?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the integral is divergent, that'll be nice to know, but I'd also like to have a way of characterizing the divergence, since it will be added to a couple other terms with a similar structure and any divergences may cancel out.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;In case it's useful, the particular expression that prompted me to ask this question is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_0^\infty\mathrm{d}k\ k J_0(kr)\int_0^\infty \frac{\mathrm{d}k&amp;#39;^2}{k&amp;#39;^2}\int_0^\infty \frac{\mathrm{d}r&amp;#39;}{r&amp;#39;}\left[\frac{k&amp;#39;^2 J_0(k&amp;#39;r&amp;#39;)}{|k^2 - k&amp;#39;^2|} - \Biggl(\frac{k^2}{|k^2 - k&amp;#39;^2|} - \frac{k^2}{\sqrt{4 k&amp;#39;^{4} + k^{4}}}\Biggr)J_0(kr&amp;#39;)\right]N(r&amp;#39;)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More generally, I'm trying to invert the Fourier transform of the BFKL equation in the paper &lt;a href=&quot;http://arxiv.org/abs/hep-ph/0110325&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/hep-ph/0110325&lt;/a&gt;, basically going from equation (9) to equation (5), and in various methods I've encountered several of these integrals involving &quot;shifted&quot; Bessel functions.&lt;/p&gt;&#xA;" OwnerUserId="1190" LastEditorUserId="1190" LastEditorDisplayName="" LastEditDate="2011-01-20T19:13:39.417" LastActivityDate="2011-01-20T19:13:39.417" Title="How can I integrate a Bessel function divided by a &quot;shifted&quot; value?" Tags="&lt;special-functions&gt;" CommentCount="5" />
  <row Id="18221" PostTypeId="1" AcceptedAnswerId="18222" CreationDate="2011-01-20T02:00:55.527" Score="1" ViewCount="77" Body="&lt;p&gt;If $b_n = \frac{1}{2n+1}$ does the notation $b_{n+1}$ make $b_n = \frac{1}{2(n+1)+1}$ or $b_n = \frac{1}{2n+2}$?&lt;/p&gt;&#xA;" OwnerUserId="3514" LastEditorUserId="2548" LastEditorDisplayName="" LastEditDate="2011-01-20T03:45:45.430" LastActivityDate="2011-01-20T03:45:45.430" Title="Quick Question about Sequence Notation" Tags="&lt;notation&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="18222" PostTypeId="2" ParentId="18221" CreationDate="2011-01-20T02:03:36.393" Score="3" ViewCount="" Body="&lt;p&gt;The former: $b_{n+1} = \frac{1}{2(n+1)+1}$.&lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2011-01-20T02:03:36.393" />
  <row Id="18224" PostTypeId="1" CreationDate="2011-01-20T02:19:49.217" Score="1" ViewCount="181" Body="&lt;p&gt;I have no idea how to solve this. Could anyone give a hand? Thanks!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2x^5a + 6x^3a - x^2a - 3&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS: I'm guessing this doesn't make sense unless it's also &quot;solve for x&quot; and the equation is = 0, but I'm not certain of that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PPS: My intuition was to add 3 to both sides, then multiply both sides by x, then divide by 3, but I didn't know where to go from there (or even if that was right).&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Edit: To be clear, I'm not the 8th grader. I'm simply unable to help her, and would appreciate assistance. You can view my web dev profile at stackoverflow if you need proof: &lt;a href=&quot;http://stackoverflow.com/users/174621/matrym&quot;&gt;http://stackoverflow.com/users/174621/matrym&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="6039" LastActivityDate="2011-01-20T02:40:50.527" Title="8th Grade Algebra Problem I Can't Solve..." Tags="&lt;algebra&gt;" AnswerCount="2" CommentCount="6" />
  <row Id="18225" PostTypeId="2" ParentId="18224" CreationDate="2011-01-20T02:28:31.577" Score="6" ViewCount="" Body="&lt;p&gt;I am assuming the equation you have written is $2x^{5a} + 6x^{3a} - x^{2a} - 3 = 0$ and you want to solve it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $y = x^a$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The equation becomes $2y^5 + 6y^3 - y^2 - 3 = 0$.&#xA;$$2y^5 + 6y^3 - y^2 - 3 = 2y^3 (y^2 + 3) - (y^2 + 3) = 0$$&#xA;$$(2y^3 - 1)(y^2 + 3) = 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solve for $y$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are interested only in real solutions, then $y = \frac{1}{\sqrt[3]{2}}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, $$x = \frac{1}{\sqrt[3a]{2}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are interested in complex roots as well, then you will get $5$ solutions for $y$, namely, $$\frac{1}{\sqrt[3]{2}}, \frac{\omega}{\sqrt[3]{2}}, \frac{\omega^2}{\sqrt[3]{2}}, i\sqrt{3}, -i\sqrt{3}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\omega$ is the complex cube root of unity and $i^2 = -1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Call them $y_1,y_2,y_3,y_4$ and $y_5$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then $$x = \zeta \sqrt[a]{y_i}$$ where $\zeta$ is one of the $a$ roots of $a^{th}$ roots of unity and hence you will end up with $5a$ roots.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2011-01-20T02:37:54.920" LastActivityDate="2011-01-20T02:37:54.920" CommentCount="2" />
  <row Id="18226" PostTypeId="1" CreationDate="2011-01-20T02:29:04.557" Score="2" ViewCount="241" Body="&lt;p&gt;I tried proving the formula presented &lt;a href=&quot;http://math.about.com/od/formulas/ss/surfaceareavol_2.htm&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; by integrating the circumferences of cross-sections of a right circular cone:&#xA;$$\int_{0}^{h}2\pi sdt, \qquad\qquad s = \frac{r}{h}t$$&#xA;so&#xA;$$\int_{0}^{h}2\pi \frac{r}{h}tdt.$$&#xA;Integrating it got me $\pi h r$, which can't be right because $h$ isn't the &lt;em&gt;slant&lt;/em&gt; height. So adding up the areas of differential-width circular strips doesn't add up to the lateral surface area of a cone?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: I now realize that the integral works if I set the upper limit to the slant height - this works if I think of &quot;unwrapping&quot; the cone and forming a portion of a circle. The question still remains though: why won't the original integral work? Won't the value of the sum of the cylinders' areas reach the area of the cone as the number of partitions approaches infinity?&lt;/p&gt;&#xA;" OwnerUserId="1958" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-20T04:00:11.170" LastActivityDate="2011-01-20T07:44:22.553" Title="Setting Up an Integral to Find A Cone's Surface Area" Tags="&lt;integral&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="18227" PostTypeId="2" ParentId="18224" CreationDate="2011-01-20T02:34:57.473" Score="1" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\ $ It is $\rm\ \ 2\ x^{2a}\ (x^{3a}+3) - (x^{3a} + 3)\:.\ $ Pull out the obvious factor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Generally one can apply ideas like the &lt;a href=&quot;http://en.wikipedia.org/wiki/Rational_root_theorem&quot; rel=&quot;nofollow&quot;&gt;Rational Root Test&lt;/a&gt; to help find such binomial factors.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2011-01-20T02:40:50.527" LastActivityDate="2011-01-20T02:40:50.527" />
  <row Id="18228" PostTypeId="1" AcceptedAnswerId="18252" CreationDate="2011-01-20T02:40:40.673" Score="7" ViewCount="248" Body="&lt;p&gt;I am an undergraduate Math student, i find the courses that I am in are too easy (calculus, linear algebra, Discrete math, logic). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to become a successful mathematician. I want to know what material i should study by myself? Should I just do all the extra problems in the book, or should I start reading about general problem solving, or maybe about a specific topic..&lt;/p&gt;&#xA;" OwnerUserId="5563" LastEditorUserId="5563" LastEditorDisplayName="" LastEditDate="2011-01-20T03:24:43.957" LastActivityDate="2011-01-20T06:39:57.617" Title="What material is good for extra-studying" Tags="&lt;soft-question&gt;" AnswerCount="4" CommentCount="6" FavoriteCount="1" />
  <row Id="18229" PostTypeId="2" ParentId="18228" CreationDate="2011-01-20T02:46:36.747" Score="2" ViewCount="" Body="&lt;p&gt;Here are some possible resources for problem solving:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0387982191&quot; rel=&quot;nofollow&quot;&gt;Problem Solving Strategies&lt;/a&gt; by Arthur Engel&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.imo-official.org/problems.aspx&quot; rel=&quot;nofollow&quot;&gt;Old IMO Problems&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/How_to_Solve_It&quot; rel=&quot;nofollow&quot;&gt;How to Solve It&lt;/a&gt; by Polya&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://mathoverflow.net/questions/29297/undergraduate-approach-to-learning-math&quot; rel=&quot;nofollow&quot;&gt;This&lt;/a&gt; (MO post about how to go about studying undergraduate math) advice seems helpful. &lt;/p&gt;&#xA;" OwnerUserId="2548" LastEditorUserId="2548" LastEditorDisplayName="" LastEditDate="2011-01-20T06:39:57.617" LastActivityDate="2011-01-20T06:39:57.617" CommentCount="2" />
  <row Id="18230" PostTypeId="2" ParentId="18154" CreationDate="2011-01-20T02:46:58.870" Score="2" ViewCount="" Body="&lt;p&gt;An interesting example which appears in section $4.9$ of Stewart and Tall's &lt;a href=&quot;http://books.google.co.cr/books?id=PIibasv45boC&amp;amp;printsec=frontcover&amp;amp;dq=algebraic+number+theory+and+Fermat%27s+Last+Theorem&amp;amp;hl=es&amp;amp;ei=f5s3TdPsK8SBlAehxOXnBg&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=1&amp;amp;ved=0CCUQ6AEwAA#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;Algebraic Number Theory and Fermat's Last Theorem&lt;/a&gt; is the solution of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Ramanujan%E2%80%93Nagell_equation&quot; rel=&quot;nofollow&quot;&gt;Ramanujan-Nagell equation&lt;/a&gt; given by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x^2 + 7 = 2^n$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The idea is to work over the quadratic field $\mathbb{Q}(\sqrt{-7})$ and use the fact that it has unique factorization in order to compare two factorizations at the core of the argument. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The argument needs to take care of different cases separately and does not run as &quot;smoothly&quot; as the arguments in some of the examples mentioned in the other answers, but that's probably in some sense due to the fact that this equation involves an exponential, which only makes the already difficult problem of solving a Diophantine equation worse.&lt;/p&gt;&#xA;" OwnerUserId="900" LastActivityDate="2011-01-20T02:46:58.870" />
  <row Id="18231" PostTypeId="1" CreationDate="2011-01-20T03:02:41.520" Score="0" ViewCount="196" Body="&lt;p&gt;Hi,&#xA;I'm stuck on the following problem :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does there exist a real borel measure $\mu$ on $[0,1]$ such that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_{0}^{1}x^n d\mu = e^{-n^2}$$&#xA;for all $n \geq 1$ ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anyone have any hint?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you&lt;/p&gt;&#xA;" OwnerUserId="6040" LastEditorUserId="2548" LastEditorDisplayName="" LastEditDate="2011-01-20T03:10:23.043" LastActivityDate="2011-03-21T22:33:47.483" Title="On the existence of a certain real measure" Tags="&lt;measure-theory&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="1" />
  <row Id="18233" PostTypeId="1" AcceptedAnswerId="18324" CreationDate="2011-01-20T03:23:50.340" Score="4" ViewCount="138" Body="&lt;p&gt;Hello, My undergraduate University does not offer advanced courses on logic, I know truth tables, Boolean algebra, propositional calculus. However I want to pursue Mathematical Logic on the long term as a mathematician.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&lt;strong&gt;Can anyone suggest a study-map of Mathematical Logic.&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;such as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) Learn The following topics : a,b,c,etc..&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) once you learned topics in (1), advance to these topics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(3) ..&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(4) etc..&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you&lt;/p&gt;&#xA;" OwnerUserId="5563" LastActivityDate="2011-01-20T16:26:18.010" Title="Map of Mathematical Logic." Tags="&lt;soft-question&gt;" AnswerCount="2" />
  <row Id="18234" PostTypeId="2" ParentId="18233" CreationDate="2011-01-20T03:25:55.883" Score="2" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://mathoverflow.net/questions/33096/reading-materials-for-mathematical-logic&quot; rel=&quot;nofollow&quot;&gt;This&lt;/a&gt;(similar question on MO) would be helpful.&lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2011-01-20T03:25:55.883" />
  <row Id="18235" PostTypeId="1" AcceptedAnswerId="18286" CreationDate="2011-01-20T03:28:35.293" Score="1" ViewCount="212" Body="&lt;p&gt;Can you calculate rigorously the limit &lt;p&gt;$\lim\limits_{n \to \infty} {(\sin n)^{\frac{1}{n}}}$&lt;/p&gt;&#xA;" OwnerUserId="6016" LastEditorUserId="6016" LastEditorDisplayName="" LastEditDate="2011-01-20T04:04:54.827" LastActivityDate="2011-01-20T20:08:45.910" Title="How to obtain this limit" Tags="&lt;real-analysis&gt;&lt;complex-analysis&gt;" AnswerCount="1" CommentCount="10" />
  <row Id="18236" PostTypeId="2" ParentId="18221" CreationDate="2011-01-20T03:42:26.197" Score="1" ViewCount="" Body="&lt;p&gt;Ask yourself what you are replacing with $n+1$ to get $b_{n+1}$?  You are replacing $n$, right?  Therefore you also replace $n$ with $n+1$ in the expression on the right hand side of the equation. This gives $b_{n+1} = \frac{1}{2(n+1)+1}$ as PEV said.&lt;/p&gt;&#xA;" OwnerUserId="4266" LastActivityDate="2011-01-20T03:42:26.197" />
  <row Id="18237" PostTypeId="1" AcceptedAnswerId="18240" CreationDate="2011-01-20T03:45:54.693" Score="3" ViewCount="124" Body="&lt;p&gt;Hi. I have the following theorem on my notes of a course on complex analysis I'm taking:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If f(z) is not bounded in a neighborhood of its isolated singular point z0, then z0 is a pole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think this cannot be right, for example sin(1/z) at z0=0 is not bounded, but has an essential singularity, not a pole. Just need some confirmation. Thanks.&lt;/p&gt;&#xA;" OwnerUserId="6043" LastActivityDate="2011-01-20T04:55:09.737" Title="analytic functions and poles" Tags="&lt;complex-analysis&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="0" />
  <row Id="18238" PostTypeId="2" ParentId="18228" CreationDate="2011-01-20T03:48:32.560" Score="5" ViewCount="" Body="&lt;p&gt;In addition to &quot;general problem solving&quot; (I second PEV's recommendation of Polya's book), much along the lines of this &lt;a href=&quot;http://math.stackexchange.com/questions/4862/where-to-go-after-calculus/4865#4865&quot;&gt;previous question about what to do after Calculus&lt;/a&gt;, I would suggest to start with Number Theory and History of Math; you can learn a lot about both on your own. I would suggest sticking to &quot;elementary&quot; number theory to begin with on that side (that is, not &quot;Algebraic Number Theory&quot; and not &quot;Analytic Number Theory&quot;). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are still thirsting for more on your own, you can probably do a fair amount with some basic point-set topology, some graph theory, or maybe even some advanced calculus/introductory analysis. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;While old IMO problems or Putnam problems might be interesting, I would not necessarily go there myself; while there are a lot of &lt;em&gt;really good&lt;/em&gt; mathematicians who excelled at either or both, there are also a lot who did not. Rather, I would perhaps suggest leafing through the &lt;em&gt;Mathematics Magazine&lt;/em&gt; (probably available from your Department); it includes a problems section if you want to try your hand at that kind of thing, but will also have other articles, many of which will likely be accessible.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2011-01-20T03:48:32.560" />
  <row Id="18239" PostTypeId="2" ParentId="18231" CreationDate="2011-01-20T03:49:59.080" Score="1" ViewCount="" Body="&lt;p&gt;If $\mu$ is supposed to be positive, then $\langle f,g\rangle=\int fgd\mu$ defines a pre-&lt;a href=&quot;http://en.wikipedia.org/wiki/Inner_product_space&quot; rel=&quot;nofollow&quot;&gt;inner product&lt;/a&gt; on the real vector space of polynomial functions on $[0,1]$.  &lt;a href=&quot;http://en.wikipedia.org/wiki/Gram_matrix&quot; rel=&quot;nofollow&quot;&gt;Gramian matrices&lt;/a&gt; with respect to this inner product must be positive semidefinite.  For example, &#xA;$\begin{bmatrix}&#xD;&#xA;\langle x,x\rangle &amp;amp; \langle x,x^2\rangle \\&#xD;&#xA;\langle x^2,x\rangle &amp;amp; \langle x^2,x^2\rangle&#xD;&#xA;\end{bmatrix}$ should be positive semidefinite.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2011-01-20T03:49:59.080" CommentCount="3" />
  <row Id="18240" PostTypeId="2" ParentId="18237" CreationDate="2011-01-20T03:54:48.137" Score="4" ViewCount="" Body="&lt;p&gt;Yes you are right. &lt;a href=&quot;http://en.wikipedia.org/wiki/Picard_theorem#Big_Picard&quot; rel=&quot;nofollow&quot;&gt;Great Picard's theorem&lt;/a&gt; states that if $w$ is an essential singularity of the analytic function $f$, then in any open set containing $w$, $f$ takes all possible values (with at most one exception), infinitely often.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2011-01-20T04:01:25.613" LastActivityDate="2011-01-20T04:01:25.613" />
  <row Id="18241" PostTypeId="2" ParentId="18237" CreationDate="2011-01-20T04:00:59.123" Score="1" ViewCount="" Body="&lt;p&gt;Riemann's &lt;a href=&quot;http://en.wikipedia.org/wiki/Removable_singularity#Riemann.27s_theorem&quot; rel=&quot;nofollow&quot;&gt;removable singularity theorem&lt;/a&gt; says that $f$ has a removable singularity at $a$ if and only if $\lim_{z\to a}(z-a)f(z)=0$.  This condition is easily satisfied if $f$ is bounded in a neighborhood of $a$, and therefore boundedness in a neighborhood of $a$ is equivalent to having a removable singularity.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2011-01-20T04:00:59.123" />
  <row Id="18242" PostTypeId="1" AcceptedAnswerId="18243" CreationDate="2011-01-20T04:01:44.657" Score="2" ViewCount="189" Body="&lt;p&gt;$$ \lim_{n\to\infty}\frac{2^n}{3^{n/2}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I used &lt;a href=&quot;http://www.wolframalpha.com/&quot; rel=&quot;nofollow&quot;&gt;wolfram&lt;/a&gt; to get the limit as follows:&#xA;&quot;&lt;a href=&quot;http://www.wolframalpha.com/input/?i=lim+n+tends+infinity+2%5En%2F3%5E%28n%2F2%29&quot; rel=&quot;nofollow&quot;&gt;lim n tends infinity 2^n/3^(n/2)&lt;/a&gt;&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And using L'Hospital's rule the result was:&#xA;$\displaystyle\frac{2^{n+6}}{3^{n/2}}$, which tends to infinity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is why does &#xA;$\displaystyle\frac{2^{n+6}}{3^{n/2}}$ tend to infinity?&lt;/p&gt;&#xA;" OwnerUserId="6007" LastEditorUserId="1242" LastEditorDisplayName="" LastEditDate="2011-01-20T07:27:08.810" LastActivityDate="2011-01-20T07:27:08.810" Title="limit of $2^n/3^{n/2}$ as $n\to\infty$" Tags="&lt;calculus&gt;&lt;wolframalpha&gt;" AnswerCount="2" CommentCount="6" />
  <row Id="18243" PostTypeId="2" ParentId="18242" CreationDate="2011-01-20T04:03:08.950" Score="7" ViewCount="" Body="&lt;p&gt;Hint: $$\frac{2^n}{3^{n/2}} = \left(\frac{2}{\sqrt{3}}\right)^n$$&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2011-01-20T04:03:08.950" CommentCount="6" />
  <row Id="18244" PostTypeId="2" ParentId="18242" CreationDate="2011-01-20T04:05:32.683" Score="6" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $$\rm \frac{2^n}{3^{n/2}}\ =\ \bigg(\frac{4}{3}\bigg)^{n/2}$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is simply the exponents analog of pulling a common factor out of a sum:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\rm a^n\ *\ b^{\:m\ \:n}\ \ = \ (a\ *\ b^{\:m}\:)^{\:n} $$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\rm a\ n +  b\ m\ n\ = \ (a + b\ m)\ n $$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So it's essentially a consequence of the &lt;em&gt;distributive&lt;/em&gt; law for exponents.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The given problem is the special case $\rm\ \ \ a,b,m,n\ \to\ 1/3,\:2,\:2,\:n/2$&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2011-01-20T05:08:04.450" LastActivityDate="2011-01-20T05:08:04.450" CommentCount="2" />
  <row Id="18245" PostTypeId="2" ParentId="18226" CreationDate="2011-01-20T04:29:23.613" Score="1" ViewCount="" Body="&lt;p&gt;See the discussion &lt;a href=&quot;http://math.stackexchange.com/questions/6979/areas-versus-volumes-of-revolution&quot;&gt;to a previous question here&lt;/a&gt; which might help&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-01-20T04:29:23.613" />
  <row Id="18246" PostTypeId="2" ParentId="18237" CreationDate="2011-01-20T04:55:09.737" Score="1" ViewCount="" Body="&lt;p&gt;Let $G$ be an open connected set in $\mathbb{C}$.   If $f$ is analytic on $G-\{w\}$ for some $w \in G$ then $w$ is a pole of $f$ if and only $\lim_{z \rightarrow w} f(z) = \infty$.   So $\sin(1/z)$ does not have a pole at zero, since $\sin(1/z)$ does not approach $\infty$ at $z=0$ &lt;/p&gt;&#xA;" OwnerUserId="832" LastActivityDate="2011-01-20T04:55:09.737" />
  <row Id="18247" PostTypeId="1" AcceptedAnswerId="18250" CreationDate="2011-01-20T05:03:40.880" Score="5" ViewCount="119" Body="&lt;p&gt;For a given analytic germ $f(z)=\sum_{n=0}^{\infty} a_n (z-z_0)^n$, and a simple curve $\gamma: [0,1]\to \mathbb{C}$ such that $\gamma(0)=z_0$, suppose one may analytically continue $f(z)$ along $\gamma$. So for each point $\tilde{z} =\gamma(t)$ there is a corresponding power series $f(z;\tilde{z})=\sum_{n=0}^{\infty} \tilde{a}_n (z-\tilde{z})^n$ with its own radius of convergence $r(\tilde{z})$. My question is, what can be said about this function $r$? How regular is it? Is it continuous?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What about for more than one variable? &lt;/p&gt;&#xA;" OwnerUserId="5253" LastActivityDate="2011-01-20T05:16:42.900" Title="How does the radius of convergence depend on the point about which the series is expanded?" Tags="&lt;complex-analysis&gt;" AnswerCount="1" />
  <row Id="18248" PostTypeId="2" ParentId="18209" CreationDate="2011-01-20T05:07:12.657" Score="2" ViewCount="" Body="&lt;p&gt;I'm not an algebraist and maybe Matt will bail me out, but here is an intuitive take on this question.  An invertible sheaf is usually associated to a divisor, and if D is a divisor, then sections of the invertible sheaf O(D) are equivalent to meromorphic functions with pole divisor dominated by D.  So sections of O(D)tens.O(E) are meromorphic functions with pole divisor dominated by D+E.  Hence any product of a section of O(D) with a section of O(E) will give a section of O(D+E), but there is no reason to expect that every meromorphic function with pole divisor dominated by D+E is such a product.  E.g. on an elliptic curve, there are non constant meromorphic functions with 2 poles at any two points P,Q, but there are only constant functions with pole divisor P.  Thus we cannot factor a global non constant meromorphic function with pole divisor P+Q, as a product of functions each having only one pole.  We can however factor it locally on proper open sets.&lt;/p&gt;&#xA;" OwnerUserId="5211" LastActivityDate="2011-01-20T05:07:12.657" />
  <row Id="18250" PostTypeId="2" ParentId="18247" CreationDate="2011-01-20T05:16:36.563" Score="5" ViewCount="" Body="&lt;p&gt;The radius of convergence should be the distance to the nearest singular point.  So it will be continuous, and it will be differentiable (in fact, smooth) except where its argument is equidistant from two or more singular points.&lt;/p&gt;&#xA;" OwnerUserId="5546" LastActivityDate="2011-01-20T05:16:36.563" CommentCount="2" />
  <row Id="18251" PostTypeId="2" ParentId="18228" CreationDate="2011-01-20T05:20:56.573" Score="3" ViewCount="" Body="&lt;p&gt;Read harder books, like Courant's or Spivak's or Apostol's Calculus, and Van der Waerden or M. Artin on algebra and linear algebra.  If those are too easy try Spivak's Calculus on manifolds or Fleming's Calculus of several variables, or Dieudonne's Foundations of modern analysis and Jacobson's Basic algebra I, or Chi Han Sah's Abstract Algebra, or Fulton's Algebraic curves.  If those are too easy try Hatcher's or 'Spanier's algebraic topology and Lang's Algebra, and Arnol'd's Ordinary differential equations.  Or Spivak's Differential geometry, or Rick Miranda's Algebraic Curves and Riemann surfaces, or Shafarevich's Basic algebraic geometry, or Siegel's lectures on Riemann surfaces, Gauss's Disquisitiones, ..........&lt;/p&gt;&#xA;" OwnerUserId="5211" LastActivityDate="2011-01-20T05:20:56.573" CommentCount="1" />
  <row Id="18252" PostTypeId="2" ParentId="18228" CreationDate="2011-01-20T05:42:28.290" Score="9" ViewCount="" Body="&lt;p&gt;I find it telling that you do not list any intermediate level courses among those you have taken: especially, real analysis, topology or algebra.  These are the courses which provide the theoretical foundation for all later study of mathematics (and are, as in Roy Smith's answer, arbitrarily challenging: one cannot be too good for any of them).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Have you not taken these courses just because you are in the first or second year of your undergraduate program, or for some other reason?  Is it too late in the current term for you to take one of these courses?  I think it might be worth a try to transfer into them.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As with many times when an undergraduate looks for help from the internet masses, I also wonder: are you being properly advised?  Do you have a faculty member in the math department at your institution with whom you are discussing these issues?  If not, you should find one right away.  &quot;I am not being properly challenged in my classes and would like to learn more interesting mathematics&quot; is just about the best thing a faculty member can hear from a student.  It is hard for me to imagine that you will not be received with open arms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Added&lt;/b&gt;: I see from your profile that you are a first year undergraduate at a Canadian university.  That actually explains a lot.  Most North American undergraduates are not ready for the sort of theoretical mathematics I described above as first year students -- but some are.  Accommodating both groups is a serious curricular challenge.  At some universities there is a sort of &quot;honors track&quot; for those who are: e.g. both at UGA and at Chicago there is an honors calculus course taught out of Spivak, and at both Chicago and Harvard they have very challenging analysis courses for exceptionally strong first year students.  But you have to be both well-prepared and well-informed in order to place into these courses.  Moreover, I have taught at two Canadian universities, and my feeling is that they are a bit more sticklers for taking courses in a fixed sequence than American universities of comparable quality.  One of my closest friends started an undergrad degree at UBC at around the age of 15.  He did extremely well in his classes from the very beginning, but nevertheless took a lot of &quot;cookie cutter&quot; math courses (e.g. differential equations without any real analysis) that such an obviously prodigious student might at a top US university be well-advised to skip.  Anyway, he got to the good stuff at the advanced age of 17 or so, and he is now a successful grown-up mathematician.  So be aware that that's the local culture to a certain extent.  But I stand by my previous advice: contact a faculty member and let them know what you're feeling.  At the very least you should be able to find someone to talk to as you work through Spivak, or Little Rudin, or Artin, or whatever.  &lt;/p&gt;&#xA;" OwnerUserId="299" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2011-01-20T05:55:24.347" LastActivityDate="2011-01-20T05:55:24.347" CommentCount="2" />
  <row Id="18253" PostTypeId="1" CreationDate="2011-01-20T06:24:43.137" Score="1" ViewCount="279" Body="&lt;p&gt;&lt;strong&gt;Problem edited since i missed a line from the problem,&lt;/strong&gt;&#xA;A monkey wants to climb up a pole of 50 metre height. he first climbs up 1 m but he falls back by same height. Again he climbs up 2 m but he falls back by 1 m. Again he climbs up 3 m and falls back by 1 m. Again he climbs up 4 metre but falls back by 1 m. In this way he reaches top of the pole. If it is known that the monkey needs 10 secs for 1 metre in upward direction and 5 secs for 1 metre in downward direction. Then find the total time required by monkey to reach at the top of the pole.&#xA;Note that, when he will reach at the top he will not be slipped back hence find the total time required by this monkey to reach on the top of the pole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have solved this question and i have getting over 10 minutes as answer. But book's answer is 9 minutes 10 seconds. I am not getting why the answer differs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;solution:&#xA;(1,1) ,(2,1) ,(3,1),(4,1)the pair (x,y) where x=climbing interval and y=falling interval&#xA;now then.....&#xA;1st time =0m&#xA;2nd time =2-1=1m&#xA;3rd time =3-1=2m&#xA;4th time =4-1=3m&#xA;effective distance covered in 1 cycle=(0+1+2+3)=6m&#xA;total distance covered =climbing distance + falling distance......  (1)&#xA;for each cycle.....&#xA;climmbing distance =(1+2+3+4)=10&#xA;but ......&#xA;falling distance  = (1+1+1+1)=4&lt;/p&gt;&#xA;&#xA;&lt;p&gt;total time for each cycle=time(climbing dist)+time(falling dist)&#xA;                         =10*10+4*5=120sec&#xA;this process repeats for 8 interval&lt;br&gt;&#xA;hence 120*8=960 sec&#xA;after this i am not able to solve the problem&lt;/p&gt;&#xA;" OwnerUserId="5918" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2011-01-20T09:33:01.490" LastActivityDate="2011-02-19T16:23:12.417" Title="Word problem: Monkey climbing up a pole." Tags="&lt;homework&gt;&lt;algebra-precalculus&gt;" AnswerCount="4" CommentCount="4" />
  <row Id="18254" PostTypeId="1" AcceptedAnswerId="18255" CreationDate="2011-01-20T06:25:26.053" Score="6" ViewCount="155" Body="&lt;p&gt;Naively I would have thought that a manifold becomes geodesically incomplete if there are missing points in it or if the geodesics are hitting a boundary. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I am not sure how to think of geodesic completeness or not for manifolds with a boundary. Like how to think for the closed disk on $\mathbb{R}^2$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to know of other generic ways in which geodesic incompleteness can happen. (Like for pseudo-Riemannian manifolds the Hawking-Penrose theorems pin down many generic scenarios) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Like from $\mathbb{R}^2$ (with the standard metric) if one removes the point $(0,0)$ then there is no geodesic from say the point $(-1,0)$ to $(1,0)$. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;But I can't visualize how I can actually put a complete metric on this punctured plane and make it geodesically complete? I can try to make the metric hyperbolic near the deleted point so that the geodesics approaching it never actually reach it but go off to infinity but even then I can't see how it will produce a geodesic which connects $(-1,0)$ to $(1,0)$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Also there are two other ways of producing geodesic incompleteness which I know but don't have a good understanding of,&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Any open subset of a complete connected Riemannian manifold is geodesically incomplete. (I guess this follows from Hopf-Rinow but I can't see it clearly)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;A noncompact surface which is not diffeomorphic to $\mathbb{R}^n$, and if for some metric every point on this surface has positive curvature, then the metric on it must be incomplete. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I would be happy to see explanations about the above things and also examples for the second case.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also is it possible to write down as formulas simple examples of incomplete geodesics? &lt;/p&gt;&#xA;" OwnerUserId="965" LastEditorUserId="965" LastEditorDisplayName="" LastEditDate="2011-01-22T15:11:05.883" LastActivityDate="2011-01-22T15:11:05.883" Title="Ways in which a manifold can be geodesically incomplete" Tags="&lt;topology&gt;&lt;differential-geometry&gt;&lt;general-topology&gt;" AnswerCount="1" FavoriteCount="2" />
  <row Id="18255" PostTypeId="2" ParentId="18254" CreationDate="2011-01-20T06:49:56.563" Score="10" ViewCount="" Body="&lt;p&gt;To your first bullet point, you certainly could make it hyperbolic near the point.  The (nonunique) geodesic connecting (-1,0) to (1,0) will go around the cusp and look semicircularish (but I don't think it will actually be a semicircle, just approximately one).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps something easier to visualize is that $\mathbb{R}^2 - \{pt\}$ is diffeomorphic to $S^1\times\mathbb{R}$.  If you give this space the product metric of the usual metrics, then you can easily see and work out the details.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To you second bullet point, you should modify the statement slightly (and somewhat pedantically).  Any &lt;i&gt;proper&lt;/i&gt; nonempty open subset $U$ of a complete connected manifold $M$ is incomplete.  To see this, let $p\in U$ and $q\in M-U$.  By Hopf-Rinow, since $M$ is complete there is a geodesic $\gamma$ starting at $p$ and ending at $q$.  Since $U$ is an open subset, it is totally geodesic: what $U$ thinks are geodesics are precisely what $M$ does.  Thus $U$ thinks $\gamma$ is a geodesic which doesn't stay in $U$ for all time, hence $U$ is incomplete.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To the third bullet point, (you say &quot;surface&quot; then use $\mathbb{R}^n$), the theorem is true for $\mathbb{R}^n$ from Cheeger and Gromoll's &lt;a href=&quot;http://en.wikipedia.org/wiki/Soul_theorem&quot; rel=&quot;nofollow&quot; title=&quot;Soul Theorem&quot;&gt;Soul Theorem&lt;/a&gt; together with Perelman's proof of the Soul Conjecture.  The soul theorem states that if $M$ is complete and has nonnegative sectional curvature, then $M$ has a compact totally convex totally geodesic submanifold $K$ (called the soul) so that $M$ is diffeomorphic to the normal bundle of $K$.  The Soul conjecture asks:  If $M$ has nonnegative curvature everywhere and a point with all sectional curvatures positive, must $K$ be a point?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perelman proved the answer is yes: under these hypothesis, $K$ is a point.  But a normal bundle of a point in a manifold is diffeomorphic to $\mathbb{R}^n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, I have been told, but I have no idea about references/proofs/etc that every noncompact surface has some metric (necessarily incomplete if the surface isn't $\mathbb{R}^2$ by the above) of positive curvature.  I'll try to dig up a reference tomorrow, if I remember to.&lt;/p&gt;&#xA;" OwnerUserId="331" LastActivityDate="2011-01-20T06:49:56.563" CommentCount="4" />
  <row Id="18256" PostTypeId="1" CreationDate="2011-01-20T06:55:29.717" Score="0" ViewCount="60" Body="&lt;p&gt;Hello. This is a problem I'm confused with:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Definition:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We say two subsets $A$ and $B$ of a topological space $X$ are completely separated if there exists a continuous map $f: X \rightarrow \mathbb{R}$ such that $f(A) \subseteq \{0\}$ and $f(B) \subseteq \{1\}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i.e we are not given that $A=f^{-1}(\{0\})$ but just that $f(A) \subseteq \{0\}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Prove that every pair of completely separated sets in a completely regular space have disjoint closures in their Stone-Cech compactification.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Right so we need to find a bounded continuous map $f: X \rightarrow \mathbb{R}$ and then extend this to $\beta(X)$. The problem is that how can we bound $f$ ? all we know is that there is a continuous map $f: X \rightarrow \mathbb{R}$ such that $f(x)=0$ if $x \in A$ and 1 if $x \in B$. But what if $f$ is unbounded in the points outside of $A$ and $B$. We are not given that $X = A \cup B$ so how can we bound $f$ ? can we simply define a new function say $g: X \rightarrow \mathbb{R}$ by $g(x) = min \{1,f(x)\}$ or something like that?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance.&lt;/p&gt;&#xA;" OwnerUserId="6033" LastActivityDate="2011-01-20T06:55:29.717" Title="Completely separated sets, Stone Cech compactification" Tags="&lt;topology&gt;" CommentCount="5" />
  <row Id="18257" PostTypeId="2" ParentId="18253" CreationDate="2011-01-20T06:55:55.777" Score="1" ViewCount="" Body="&lt;p&gt;If $n$ is the number of times (and hence a total of $n$ meters) the monkey falls back, then when he reaches the top of the pole, the total distance he has moved upwards is $50+n$ meters (why? Hint: Instead of the monkey falling back, assume it was the pole that moved up by 1m).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus the time taken in seconds is $(50+n)\times 10 + n \times 5$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming the book answer is correct, we get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(50+n)\times 10 + n \times 5 = 550$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i.e.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ 500 + 15n = 550$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ n = 50/15 = 10/3$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus the book answer you quote is wrong, irrespective of the steps of upward progression, if he falls back $1$m each time. The answer I get, assuming the upward progression as $1,2,4,8,16,32$ is less than $10$ minutes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are you sure you read the book answer correctly? Is it possible that you converted the answer given in seconds to minutes(and seconds) and might have made a mistake there?&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2011-01-20T07:45:33.987" LastActivityDate="2011-01-20T07:45:33.987" CommentCount="3" />
  <row Id="18258" PostTypeId="2" ParentId="18253" CreationDate="2011-01-20T07:20:44.350" Score="0" ViewCount="" Body="&lt;p&gt;I suspect that your work assumed that the monkey climbed 1, 2, 3, 4, 5, 6, ... meters (for which I get over 10 minutes), but the problem as you've stated it suggests that the monkey climbs 1, 2, 4, 8, 16, 32, ... meters or 1, 2, 4, 7, 11, ... meters or some other pattern—more information would be helpful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In any case, the way that I'd do the problem is to start with a spreadsheet.  Label the columns up-distance, up-time, down-distance, down-time, total-distance, total-time.  In the up-distance column, fill in whichever progression of distances is intended.  In the up-time column, multiply the up-distance column by 10.  In the down-distance column, fill in all 1s.  In the down-time column, multiply the down-distance column by 5.  In the total-distance column, take the previous total-distance, add the up-distance, and subtract the down-distance.  In the total-time column, take the previous total-time, add the up-time, and add the down-time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, go down the spreadsheet and look for the last row where the distance is less than 50.  Use this row to figure out whether or not the monkey already reached 50 meters or how much more time the monkey needs to reach 50 meters.&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2011-01-20T07:20:44.350" />
  <row Id="18259" PostTypeId="2" ParentId="18253" CreationDate="2011-01-20T07:23:48.060" Score="0" ViewCount="" Body="&lt;p&gt;Since the numbers are small, it's pretty easy just to work out by brute force.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Round 1, 15 seconds, 0m gained&#xA;Round 2, 40 seconds, 1m gained&#xA;Round 3, 85 seconds, 4m gained&#xA;Round 4, 170 seconds, 11m gained&#xA;Round 5, 335 seconds, 26m gained&lt;/p&gt;&#xA;&#xA;&lt;p&gt;He's now in striking distance. Add 24m, times 10 seconds gives 575 seconds, or 9:35s.&lt;/p&gt;&#xA;" OwnerUserId="6034" LastActivityDate="2011-01-20T07:23:48.060" />
  <row Id="18260" PostTypeId="2" ParentId="18226" CreationDate="2011-01-20T07:31:01.347" Score="3" ViewCount="" Body="&lt;p&gt;You seem to be ignoring the fact that s and r vary as the segment you consider varies. By using the same variable names it appears that you are confusing them to be constants...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, for a derivation, look at the following figure:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/iFvD4.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a cross-section of the cone.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The area of the strip of width $\displaystyle dh$ that corresponds to $\displaystyle h$ (from the apex) is $\displaystyle 2\pi r \frac{dh}{\cos x}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now $\displaystyle r = h \tan x$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus $\displaystyle dA = 2 \pi h \frac{\tan x}{\cos x} dh$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus the total area&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$= \int_{0}^{H} 2 \pi h \frac{\tan x}{\cos x} dh = \pi H^2 \frac{\tan x}{\cos x} = \pi (H \tan x) \left(\frac{H}{\cos x}\right) = \pi R S$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope that helps.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2011-01-20T07:44:22.553" LastActivityDate="2011-01-20T07:44:22.553" />
  <row Id="18261" PostTypeId="1" CreationDate="2011-01-20T07:46:30.697" Score="6" ViewCount="124" Body="&lt;p&gt;The number of zeros which are not possible at the end of the n! is&#xA;a) 82&#xA;b) 73&#xA;c) 156&#xA;d) none of these&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was trying to solve this problem. I know how to calculate the no of zeros in factorial but have no idea how to work out this problem quickly.&lt;/p&gt;&#xA;" OwnerUserId="5918" LastEditorUserId="2548" LastEditorDisplayName="" LastEditDate="2011-01-20T07:48:39.537" LastActivityDate="2011-01-20T10:19:01.217" Title="Number of zeros not possible in $n!$" Tags="&lt;number-theory&gt;&lt;homework&gt;&lt;algebra-precalculus&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="18263" PostTypeId="1" CreationDate="2011-01-20T07:58:21.110" Score="1" ViewCount="141" Body="&lt;p&gt;I'm aware that there are an infinite number of integrals that do not have a closed-form solution. But how does one know if their particular integral is one of these? For example, a particular integral I have at the moment is&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Integrate[ Cos[x]^2 Log[ a + b Cos[x] + Sqrt[ c + d Cos[x] ] ] , x]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;or&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$    \int \cos^2 x \,\log(a+b\cos x+\sqrt{c+d \cos x)} \,\mathrm d x    $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To me this looks rather daunting but I'm constantly surprised how trivial my own maths problems are to others. What resources are available to me to investigate this sort of problem? Are there any good reference books for this sort of thing? (my library is regretfully bare) — electronic books (esp. &lt;em&gt;for&lt;/em&gt; pay) even better.&lt;/p&gt;&#xA;" OwnerUserId="2961" LastActivityDate="2011-01-20T07:58:21.110" Title="How to approach a symbolic integral that Mathematica cannot solve?" Tags="&lt;integral&gt;&lt;math-software&gt;" CommentCount="4" />
  <row Id="18265" PostTypeId="1" CreationDate="2011-01-20T08:07:00.163" Score="1" ViewCount="219" Body="&lt;p&gt;This is a follow-up question to &lt;a href=&quot;http://math.stackexchange.com/q/17932/2767&quot;&gt;About the limit of the coefficient ratio for a power series over complex numbers&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Cauchy's estimation in complex analysis is a consequence of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Cauchy%27s_integral_formula&quot; rel=&quot;nofollow&quot;&gt;Cauchy's integral formula&lt;/a&gt;, which can be stated as follow:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Theorem.&lt;/strong&gt; If $f$ is holomorphic in an open set $\Omega$ that contains a closed disc $D$ centered at $w$ with radius $R$, then for all $n$&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$\displaystyle{ &#xA;  \frac{|f^{(n)}(w)|}{n!} \leq \frac{\sup_{z \in C} |f(z)|}{R^n} &#xA;  }$,&#xA;  where $C$ is the boundary circle of $D$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Note that if $f$ has power series expansion $\sum b_n (z-w)^n$ inside $D$, then $b_n = |f^{(n)}(w)|/n!$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the following question: &#xA;If $f$ is holomorphic in an open set $\Omega$ that contains the closed unit disc except for a pole at $z_0$ on the unit circle, and $f$ has the power series expansion $\sum a_n z^n$ in the open unit disc. By &lt;a href=&quot;http://math.stackexchange.com/questions/17932/about-the-limit-of-the-coefficient-ratio-for-a-power-series-over-complex-numbers/17936#17936&quot;&gt;the answer of Sivaram&lt;/a&gt; to the previous post, the coefficients $a_n$ of the expansion of $f$ will not converge to zero. I'm wondering if this can be proved by something that is a (partial) converse to the Cauchy's estimation, and apply it backwards we can ensure that $a_n$ will not go to zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to know whether the following version exists:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Assume $f$ is holomorphic in an open set $\Omega$ that contains &#xA;  a closed unit disc except at $z_0$ on the unit circle, and $f$ has power series expansion $\sum a_n z^n$ inside the open unit disc.&#xA;  If for all $n$ large enough and all $R &amp;lt; 1$ we have &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$\displaystyle{ &#xA;  |a_n| \leq \frac{\sup_{z \in C} |f(z)|}{R^n} &#xA;  }$,&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;where $D$ is a closed disc centered at $0$ with radius $R$ and $C$ is the boundary circle of $D$, then $f$ is also holomorphic at $z_0$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This version may not work at all, but any reasonable modifications which can be used to prove that $a_n$ will be bounded away from zero are fine.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Question.&lt;/strong&gt; Is there a reasonable version to the converse of the Cauchy's estimation?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2767" LastEditorUserId="2767" LastEditorDisplayName="" LastEditDate="2011-01-20T10:19:56.363" LastActivityDate="2011-05-20T17:59:35.760" Title="A converse theorem to the Cauchy's estimation" Tags="&lt;complex-analysis&gt;&lt;inequality&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" />
  <row Id="18266" PostTypeId="2" ParentId="18261" CreationDate="2011-01-20T08:15:32.870" Score="3" ViewCount="" Body="&lt;p&gt;If you know how to calculate the number of zeros at the end of $n!$, then you know that there are some values of $n$ for which the number of zeros has just increased by 2 (or more), skipping over number(s).  What numbers are skipped?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Further hint (hidden):&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Find the number of zeros are at the end of (a) $24!$; (b) $25!$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;edit&lt;/em&gt; more explicitly:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; the factorials of 5-9 end in 1 zero, 10-14 end in 2 zeros, 15-19 end in 3 zeros, 20-24 end in 4 zeros, 25-29 end in 6 zeros, so 5 is the first number skipped.  For what $n$ will the number of zeros at the end of $n!$ next skip over an integer and what is that number of zeros?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="72" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2011-01-20T10:19:01.217" LastActivityDate="2011-01-20T10:19:01.217" CommentCount="7" />
  <row Id="18267" PostTypeId="2" ParentId="18261" CreationDate="2011-01-20T08:25:22.220" Score="0" ViewCount="" Body="&lt;p&gt;Assuming there is a single answer, the answer is $73$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider $300!$, it has $300/5 + 60/5 + 10/5 = 74$ zeroes. $299!$ has $59 + 11 + 2 = 72$ zeroes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Got this by trial and error and luck.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note this is an application of Legendre's formula of the highest power of a prime $p$ dividing $n!$ being $\displaystyle \sum_{k=1}^{\infty} \left \lfloor \frac{n}{p^k} \right \rfloor$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We only need to consider power of $5$ to get the number of zeroes.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Since you asked for a method: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For smallish numbers, you could try getting a multiple of $6 = 1+5$ close to your number, find the number of zeroes for $25/6$ times that and try to revise your estimate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example for $156 = 6*26$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So try $26*5*5 =  650$. $650!$ has $26*5 + 26 + 5 + 1 = 162$ zeroes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since you overshot by $6$, try a smaller multiple of $6$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So try $25*5*5$, which gives $25*5 + 25 + 5 + 1 = 156$. Bingo!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $82$, try $78 = 13*6$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So try $13*25$. Which gives $65 + 13 + 2 = 80$ zeroes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So try increasing the estimate, say by adding 10 (since we were short by 2).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$13*25 + 10$ gives us $67 + 13 + 2 = 82$ zeroes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $187$ Try $186 = 6*31$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So try $31*5*5$ this gives us $31*5 + 31 + 6 + 1 = 193$ zeroes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;since we overshot by $7$, try reducing it, say&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$30*5*5$ gives us $30*5 + 30 + 6 + 1 = 187$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For larger numbers instead of multiple of 6, consider multiple of $1+5+25$, $1+5+25+125$ etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am pretty sure there must be a better method, but I don't expect the CAT folks to expect candidates to know that!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope that helps.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2011-01-20T09:14:18.917" LastActivityDate="2011-01-20T09:14:18.917" CommentCount="1" />
  <row Id="18268" PostTypeId="1" AcceptedAnswerId="18363" CreationDate="2011-01-20T08:27:25.043" Score="12" ViewCount="456" Body="&lt;p&gt;Let $n$ be a number of people. At least two of them may be born on the same day of the year with probability: $$1-\prod_{i=0}^{n-1} \frac{365-i}{365}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But what is the probability that at least two of them are born on two consecutive days of the year (considering December 31st and January 1st also consecutive)? It seems a good approximation is: $$1-\prod_{i=0}^{n-1} \frac{365-2 \times i}{365}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, simulating pseudo-random integers with Python, the 99%-confidence intervals may be slightly different. So do you have any closed formula?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Results of the simulation with Python. Here are 99%-confidence intervals below:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Number of people:  1    Lower bound: 0.0        Upper bound: 0.0&#xA;Number of people:  2    Lower bound: 0.00528    Upper bound: 0.00567&#xA;Number of people:  3    Lower bound: 0.01591    Upper bound: 0.01657&#xA;Number of people:  4    Lower bound: 0.03185    Upper bound: 0.03277&#xA;Number of people:  5    Lower bound: 0.0528     Upper bound: 0.05397&#xA;Number of people:  6    Lower bound: 0.07819    Upper bound: 0.07959&#xA;Number of people:  7    Lower bound: 0.10844    Upper bound: 0.11006&#xA;Number of people:  8    Lower bound: 0.14183    Upper bound: 0.14364&#xA;Number of people:  9    Lower bound: 0.17887    Upper bound: 0.18086&#xA;Number of people: 10    Lower bound: 0.21816    Upper bound: 0.2203&#xA;Number of people: 11    Lower bound: 0.25956    Upper bound: 0.26183&#xA;Number of people: 12    Lower bound: 0.30306    Upper bound: 0.30544&#xA;Number of people: 13    Lower bound: 0.34678    Upper bound: 0.34925&#xA;Number of people: 14    Lower bound: 0.39144    Upper bound: 0.39397&#xA;Number of people: 15    Lower bound: 0.43633    Upper bound: 0.4389&#xA;Number of people: 16    Lower bound: 0.48072    Upper bound: 0.48331&#xA;Number of people: 17    Lower bound: 0.52476    Upper bound: 0.52734&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I give here &lt;a href=&quot;http://i.imgur.com/pqyak.gif&quot; rel=&quot;nofollow&quot;&gt;some results&lt;/a&gt; with a tweaked approximation formula, using &lt;a href=&quot;http://goo.gl/EHvzD&quot; rel=&quot;nofollow&quot;&gt;Wolfram Alpha&lt;/a&gt;:&#xA;$$\left( 1 - \frac{n-1}{2 \times 365 + n-1} \right) \times \left( 1-\prod_{i=0}^{n-1} \frac{365-2 \times i}{365} \right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, this is just a tweak, ans is clearly wrong for $n=33$ since:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Number of people: 33    My guess: 0.91407&#xA;Number of people: 33    Lower bound: 0.94328    Upper bound: 0.94447&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Thanks to &lt;strong&gt;Jacopo Notarstefano, leonbloy, and Moron&lt;/strong&gt;, here is the (correct) formula:&#xA;$$ 1-\sum_{k=1}^{n}\frac{1}{365^{n-k}k}\left(\prod_{i=1}^{k-1}\frac{365-\left(k+i\right)}{365\times i}\right)\sum_{j=0}^{k-1}\left(-1\right)^{j}C_{k}^{j}\left(k-j\right)^{n} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And here are the results of the computations using this formula with Python:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Number of people:  1    Probability: 0.0&#xA;Number of people:  2    Probability: 0.005479452&#xA;Number of people:  3    Probability: 0.016348283&#xA;Number of people:  4    Probability: 0.032428609&#xA;Number of people:  5    Probability: 0.053459591&#xA;Number of people:  6    Probability: 0.079104502&#xA;Number of people:  7    Probability: 0.108959718&#xA;Number of people:  8    Probability: 0.14256532&#xA;Number of people:  9    Probability: 0.179416899&#xA;Number of people: 10    Probability: 0.218978144&#xA;Number of people: 11    Probability: 0.260693782&#xA;Number of people: 12    Probability: 0.304002428&#xA;Number of people: 13    Probability: 0.34834893&#xA;Number of people: 14    Probability: 0.393195856&#xA;Number of people: 15    Probability: 0.438033789&#xA;Number of people: 16    Probability: 0.482390182&#xA;Number of people: 17    Probability: 0.525836596&#xA;Number of people: 18    Probability: 0.567994209&#xA;Number of people: 19    Probability: 0.608537602&#xA;Number of people: 20    Probability: 0.647196551&#xA;Number of people: 21    Probability: 0.683756966&#xA;Number of people: 22    Probability: 0.718059191&#xA;Number of people: 23    Probability: 0.749995532&#xA;Number of people: 24    Probability: 0.779509664&#xA;Number of people: 25    Probability: 0.806569056&#xA;Number of people: 26    Probability: 0.831211564&#xA;Number of people: 27    Probability: 0.853561895&#xA;Number of people: 28    Probability: 0.873571839&#xA;Number of people: 29    Probability: 0.892014392&#xA;Number of people: 30    Probability: 0.906106867&#xA;Number of people: 31    Probability: 0.919063161&#xA;Number of people: 32    Probability: 0.928791992&#xA;Number of people: 33    Probability: 0.944659069&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="2380" LastEditorUserId="1543" LastEditorDisplayName="" LastEditDate="2011-04-16T18:33:39.970" LastActivityDate="2011-04-16T18:33:39.970" Title="Consecutive birthdays probability" Tags="&lt;probability&gt;&lt;combinatorics&gt;" AnswerCount="3" CommentCount="6" FavoriteCount="3" />
  <row Id="18269" PostTypeId="1" CreationDate="2011-01-20T08:52:57.783" Score="1" ViewCount="158" Body="&lt;p&gt;what is the 444th term of the series $1,2,3,10,11,12,13,20,21,22,23,30,31, \dots$&lt;/p&gt;&#xA;" OwnerUserId="5918" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2011-01-20T09:33:34.627" LastActivityDate="2011-01-20T11:17:21.487" Title="General term of $1,2,3,10,11,12,13, \dots$" Tags="&lt;homework&gt;&lt;algebra-precalculus&gt;" AnswerCount="4" />
  <row Id="18270" PostTypeId="1" AcceptedAnswerId="18279" CreationDate="2011-01-20T09:01:54.880" Score="4" ViewCount="61" Body="&lt;p&gt;Which conditions must the matrix entries satisfy, and what would be an interpretation of the row and column sums of the matrix?&lt;/p&gt;&#xA;" OwnerUserId="1434" LastActivityDate="2011-01-20T10:14:46.070" Title="What are the linear isometries on $R^n$, equipped with the $l_1$ norm?" Tags="&lt;linear-algebra&gt;" AnswerCount="1" />
  <row Id="18271" PostTypeId="1" AcceptedAnswerId="18273" CreationDate="2011-01-20T09:05:48.733" Score="3" ViewCount="128" Body="&lt;p&gt;The question is :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;6 points are located on a circle and lines are drawn connecting these points, each pair of points connected by a single line. What can be the maximum number of regions into which the circle is divided?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My answer is 32 . But the actual answer is 31 how ??&lt;/p&gt;&#xA;" OwnerUserId="6047" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2011-01-20T09:32:24.863" LastActivityDate="2011-01-20T09:32:24.863" Title="Maximum number of regions formed by points on a circle." Tags="&lt;geometry&gt;&lt;combinatorics&gt;" AnswerCount="2" />
  <row Id="18272" PostTypeId="2" ParentId="18269" CreationDate="2011-01-20T09:09:08.580" Score="1" ViewCount="" Body="&lt;p&gt;What do you think the next 10 terms are? Then how can you express the nth term in terms of n/4 and n mod 4?&lt;/p&gt;&#xA;" OwnerUserId="5676" LastActivityDate="2011-01-20T09:09:08.580" CommentCount="1" />
  <row Id="18273" PostTypeId="2" ParentId="18271" CreationDate="2011-01-20T09:12:03.787" Score="0" ViewCount="" Body="&lt;p&gt;Answer is located here:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://testfunda.com/Answers/ViewQuestion.aspx?QID=0e9eb0f3-8a80-4609-8be4-b022545c8224&quot; rel=&quot;nofollow&quot;&gt;http://testfunda.com/Answers/ViewQuestion.aspx?QID=0e9eb0f3-8a80-4609-8be4-b022545c8224&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" LastActivityDate="2011-01-20T09:12:03.787" />
  <row Id="18275" PostTypeId="2" ParentId="18271" CreationDate="2011-01-20T09:18:00.287" Score="3" ViewCount="" Body="&lt;p&gt;In general the maximum number of regions you can get from $n$ points is given by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$${n \choose 4} + {n \choose 2} + 1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This can be proved using induction (other combinatorial proofs exist too). For more information (including at least two proofs), see this: &lt;a href=&quot;http://en.wikipedia.org/wiki/Dividing_a_circle_into_areas&quot; rel=&quot;nofollow&quot;&gt;Dividing a circle into areas&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is an oft cited puzzle to show the perils of generalizing based on first few values. We get powers of $2$ till $n=5$, after which we get $31$.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2011-01-20T09:25:18.130" LastActivityDate="2011-01-20T09:25:18.130" />
  <row Id="18276" PostTypeId="2" ParentId="18269" CreationDate="2011-01-20T09:25:28.380" Score="1" ViewCount="" Body="&lt;p&gt;Every 4th term you add an additional 10. Find out how many 4's there are in 444 and you get the answer. You should call this a sequence, not a series.&lt;/p&gt;&#xA;" OwnerUserId="4594" LastActivityDate="2011-01-20T09:25:28.380" CommentCount="2" />
  <row Id="18277" PostTypeId="1" AcceptedAnswerId="18278" CreationDate="2011-01-20T09:45:28.270" Score="2" ViewCount="134" Body="&lt;p&gt;I was asked to sum the given series:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\displaystyle \sum\limits_{n=1}^{\infty} \frac{1}{n\cdot (2n-1)}=\frac{1}{1} + \frac{1}{2 \cdot 3} + \frac{1}{3 \cdot 5} + \cdots \infty$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Here i workout the details.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align*}&#xA;\sum\limits_{n=1}^{\infty} \frac{1}{n \cdot (2n-1)} &amp;amp;= \frac{1}{1} + \frac{1}{2 \cdot 3} + \frac{1}{3 \cdot 5} + \cdots \infty \\ &amp;amp;= 1 + \Bigl( \frac{1}{2} - \frac{1}{3}\Bigr) + \frac{1}{2} \Bigl(\frac{1}{3} - \frac{1}{5}\Bigr) + \frac{1}{3} \Bigl(\frac{1}{4}-\frac{1}{7}\Bigr) + \cdots \infty \\ &amp;amp;= 1 + \biggl[ \frac{1}{1 \cdot 2 } + \frac{1}{2 \cdot 3} + \cdots \infty\biggr] - \biggl[ \frac{1}{1 \cdot 3} + \frac{1}{2 \cdot 5} + \frac{1}{3 \cdot 7} + \cdots \infty \biggr] \\ &amp;amp;= 2 - \sum\limits_{n=1}^{\infty} \frac{1}{n \cdot (2n+1)}&#xA;\end{align*}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now for summing that second sum note that $$-\log(1-x^{2}) =-\biggl[x^{2} + \frac{x^{4}}{2} + \frac{x^{6}}{3} + \cdots \infty\biggr]$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Have i done this correctly or is there some error in computing the sum. Now what i need to evaluate is the integral $$\int\limits_{0}^{1} \log(1-x^{2}) \ \rm{dx}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone tell me how to evaluate this integral. I tried by using &lt;em&gt;Integration by parts&lt;/em&gt; but that didn't actually work out. Moreover, i would also be interested to see if there are other ways of solving this sum.&lt;/p&gt;&#xA;" LastActivityDate="2011-01-20T10:15:45.900" Title="Computing the sum $\sum \frac{1}{n (2n-1)}$" Tags="&lt;real-analysis&gt;&lt;analysis&gt;&lt;sequences-and-series&gt;" AnswerCount="1" />
  <row Id="18278" PostTypeId="2" ParentId="18277" CreationDate="2011-01-20T09:55:11.090" Score="6" ViewCount="" Body="&lt;p&gt;We have,&#xA;$$\sum_{n=1}^{\infty} \frac{1}{n(2n-1)} = 2 \sum_{n=1}^{\infty} \left( \frac{1}{2n-1} - \frac{1}{2n} \right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The RHS has the alternating harmonic series, and its value is $\ln 2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So your final answer would be $2 \ln 2$&lt;/p&gt;&#xA;" OwnerUserId="2320" LastEditorUserId="2320" LastEditorDisplayName="" LastEditDate="2011-01-20T10:15:45.900" LastActivityDate="2011-01-20T10:15:45.900" />
  <row Id="18279" PostTypeId="2" ParentId="18270" CreationDate="2011-01-20T10:14:46.070" Score="5" ViewCount="" Body="&lt;p&gt;Hint: Such an isometry would have to map the $l^1$-unit-sphere onto itself. How does this &quot;sphere&quot; look like?&lt;/p&gt;&#xA;" OwnerUserId="1303" LastActivityDate="2011-01-20T10:14:46.070" CommentCount="4" />
  <row Id="18280" PostTypeId="2" ParentId="18269" CreationDate="2011-01-20T10:27:05.767" Score="2" ViewCount="" Body="&lt;p&gt;Based on comments on other answers, you may have guessed that the given information does not determine a unique answer.  As to a hint at what the commenters might be thinking (which would also be my first guess), a number like 12571 is normally taken to mean $1\cdot 10^4+2\cdot 10^3+5\cdot 10^2+7\cdot 10^1+1\cdot 10^0$—that is, we normally interpret strings of digits to be in &quot;base 10&quot; where the digits read from the &lt;em&gt;right&lt;/em&gt; represent coefficients of successively higher powers of 10.  In base 10, we use digits 0-9.  Your sequence only uses digits 0-3.  Supposing that your sequence were not in base 10, what would this suggest?&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2011-01-20T10:27:05.767" />
  <row Id="18281" PostTypeId="2" ParentId="18269" CreationDate="2011-01-20T10:29:25.070" Score="1" ViewCount="" Body="&lt;p&gt;First of all, this is a sequence not a series.&#xA;Ok, so how about this...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;01,02,03,10,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;11,12,13,20,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;21,22,23,30,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;31,…&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you see the pattern? :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you do observe the pattern, the 444 term would be 1110&lt;/p&gt;&#xA;" OwnerUserId="2320" LastActivityDate="2011-01-20T10:29:25.070" CommentCount="1" />
  <row Id="18283" PostTypeId="1" CreationDate="2011-01-20T10:47:39.043" Score="1" ViewCount="203" Body="&lt;p&gt;I am trying to solve the matrix equation of the form &lt;strong&gt;XA = XB&lt;/strong&gt;. A, B and the solution sought X are 4 $\times 4$ homegeneous matrices which are composed of a rotation matrix and translation vector, like in&#xA;$ \begin{bmatrix}  r1 &amp;amp; r2 &amp;amp; r3 &amp;amp; tx\\  r4 &amp;amp; r5 &amp;amp; r6 &amp;amp;ty\\  r7 &amp;amp; r8 &amp;amp; r9 &amp;amp; tz\\ 0 &amp;amp; 0 &amp;amp; 0 &amp;amp; 1  \end{bmatrix}$.&#xA;There are several methods to solve equations of the type &lt;strong&gt;AX = XB&lt;/strong&gt;.  These type of equations occur in domains like robotics, for which solutions have been proposed like the Tsai and Lenz method [IEEE 1987], for example.&#xA;1) I feel solving the equation &lt;strong&gt;XA = XB&lt;/strong&gt; is not the same as solving the known form &lt;strong&gt;AX = XB&lt;/strong&gt;. Am I right? &#xA;2) Neither can it be solved like the &lt;strong&gt;Sylvester equation&lt;/strong&gt;, because even that requires AX + XB = C form. What i have is XA = XB, a redundant set of equations. That is,&lt;br&gt;&#xA;A1.X = X.B1&lt;br&gt;&#xA;A2.X = X.B2&lt;br&gt;&#xA;.&lt;br&gt;&#xA;.&lt;br&gt;&#xA;An.X = X.Bn    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If i am correct, these could be rewritten into another form like AX = BX. Should i rewrite the equation and try to solve this problem using any other existing methods?&lt;/p&gt;&#xA;" OwnerUserId="6050" LastEditorUserId="6050" LastEditorDisplayName="" LastEditDate="2011-01-20T12:48:42.293" LastActivityDate="2011-01-20T14:39:19.727" Title="Solving matrix equations of the form XA = XB" Tags="&lt;linear-algebra&gt;&lt;matrices&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" />
  <row Id="18284" PostTypeId="1" AcceptedAnswerId="18299" CreationDate="2011-01-20T10:52:27.367" Score="14" ViewCount="530" Body="&lt;p&gt;I am trying to find the $n$'th derivative of $f(x)=e^{1/x}$. When looking at the first few derivatives I noticed a pattern and eventually found the following formula&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{d^n}{d x^n}f(x)=(-1)^n e^{1/x} \cdot \sum _{k=0}^{n-1} k! \binom{n}{k} \binom{n-1}{k} x^{-2 n+k}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tested it for the first $20$ derivatives and it got them all. Mathematica says that it is some hypergeometric distribution but I don't want to use that. Now I am trying to verify it by induction but my algebra is not good enough to do the induction step.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is what I tried for the induction (incomplete, maybe incorrect)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{d^{n+1}}{d x^{n+1}}f(x)=\frac{d}{dx}(-1)^n e^{1/x} \cdot \sum _{k=0}^{n-1} k! \binom{n}{k} \binom{n-1}{k} x^{-2 n+k}=$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$=(-1)^n e^{1/x} \cdot (\sum _{k=0}^{n-1} k! \binom{n}{k} \binom{n-1}{k} (-2n+k) x^{-2 n+k-1})-e^{1/x} \cdot \sum _{k=0}^{n-1} k! \binom{n}{k} \binom{n-1}{k} x^{-2 (n+1)+k}=$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$=(-1)^n e^{1/x} \cdot \sum _{k=0}^{n-1} k! \binom{n}{k} \binom{n-1}{k}((-2n+k) x^{-2 n+k-1}-x^{-2 (n+1)+k)})=$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$=(-1)^{n+1} e^{1/x} \cdot \sum _{k=0}^{n-1} k! \binom{n}{k} \binom{n-1}{k}(2n x-k x+1) x^{-2 (n+1)+k}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know how to get on from here.&lt;/p&gt;&#xA;" OwnerUserId="3123" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2011-05-03T13:21:11.727" LastActivityDate="2011-05-04T19:26:57.150" Title="$n$th derivative of $e^{1/x}$" Tags="&lt;calculus&gt;&lt;sequences-and-series&gt;&lt;binomial-coefficients&gt;" AnswerCount="4" CommentCount="7" FavoriteCount="4" />
  <row Id="18286" PostTypeId="2" ParentId="18235" CreationDate="2011-01-20T11:07:18.153" Score="3" ViewCount="" Body="&lt;p&gt;Following PEV's hint we argue as follows: There is a $p$ (e.g. $p=42$, according to this: &lt;a href=&quot;http://mathforum.org/library/drmath/view/69162.html&quot; rel=&quot;nofollow&quot;&gt;http://mathforum.org/library/drmath/view/69162.html&lt;/a&gt;) and a $k_0$ with $|\pi -{n\over k}|\geq 1/ k^p$ for all $k&amp;gt;k_0$ and all $n$. Assume $n&amp;gt;4k_0$ and let $k$ be the nearest integer to ${n\over \pi}$. Then $k&amp;gt;k_0$ and therefore $|n - k\pi|\geq 1/ k^{p-1}\geq C/ n^{p-1}$ for some $C&amp;gt;0$ which does not depend on $n$. As $|\sin(x)|\geq 2|x|/\pi$ $\ (|x|\leq{\pi\over2})$ it follows that $|\sin(n)|\geq C&amp;#39;/ n^{p-1}$. Since this is true for all large $n$ the indicated limit (with $|\sin|$ instead of $\sin$) is indeed $1$.&lt;/p&gt;&#xA;" OwnerUserId="1303" LastEditorUserId="1303" LastEditorDisplayName="" LastEditDate="2011-01-20T20:08:45.910" LastActivityDate="2011-01-20T20:08:45.910" />
  <row Id="18288" PostTypeId="1" AcceptedAnswerId="18971" CreationDate="2011-01-20T11:36:10.073" Score="0" ViewCount="185" Body="&lt;p&gt;$$\limsup_{i\rightarrow \infty}E_{i} = \bigcap_{k=1}^{\infty}\bigcup_{i=k}^{\infty} E_{i}$$&#xA;So&#xA;$$\bigcup_{i=k}^{\infty} E_{i} = \{x \in E_{i} | i \in I\}=S$$&#xA;where $I$ is some index set. When I try try to understand next part, I get confused, seems too trivial:&#xA;$$\bigcap_{k=1}^{\infty}S = S$$&#xA;since there is only one set $S$ as the sets were already evaluated with the or quantifier in the operation $\bigcup$. I am probably wrong so how do you quantify inner structure in $\limsup&#xD;&#xA;\limits_{i\rightarrow \infty}E_{i}$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;[Updated]&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\bigcap_{k=1}^{\infty}S$ = S clearly cannot be true in every cases, only trivially. Thank you for the notice. Perhaps, I am not clear what I mean by &lt;code&gt;quantification&lt;/code&gt;. Let's consider simpler case with finite sets.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Suppose an arbitrary amount of sets. What is the maximun number of different types of sets in  $A = \bigcap_{k=1}^{a}\bigcup_{i=k}^{b} E_{i}$?&lt;/li&gt;&#xA;&lt;li&gt;What is the maximum amount of different sets in $\bigcup_{i=k}^{\infty} E_{i}$?&lt;/li&gt;&#xA;&lt;li&gt;I pretty sure you know better measures to analyze it, how?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;... and so on, I want to break the problem into parts to analyze it.&lt;/p&gt;&#xA;" OwnerUserId="5902" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-25T16:36:02.467" LastActivityDate="2011-02-02T02:59:27.777" Title="How to Quantify $\limsup_{i\rightarrow \infty}E_{i} = \bigcap_{k=1}^{\infty}\bigcup_{i=k}^{\infty} E_{i}$?" Tags="&lt;limit&gt;" AnswerCount="5" CommentCount="12" FavoriteCount="1" />
  <row Id="18289" PostTypeId="2" ParentId="18283" CreationDate="2011-01-20T11:48:31.783" Score="4" ViewCount="" Body="&lt;p&gt;For $XA = XB$, solve $X(A-B) = 0$, basically the rows of $X$ can be any vector that is in the nullspace of $(A-B)^T$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can write $X(A-B) = 0 \Leftrightarrow (A-B)^TX^T = 0$. The columns of $X^T$ are the rows of $X$. Let $x_i$ be row $i$ of $X$. The $i$:th column of the product $(A-B)^TX^T$ is $(A-B)^Tx_i$. For this to be a zero vector, $x_i$ has to be in the null space of $(A-B)^T$. Thus all rows of $X$ have to be in this nullspace.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Ok, title corrected. Rephrased a little. Short explanation added.&lt;/p&gt;&#xA;" OwnerUserId="5966" LastEditorUserId="5966" LastEditorDisplayName="" LastEditDate="2011-01-20T14:39:19.727" LastActivityDate="2011-01-20T14:39:19.727" CommentCount="1" />
  <row Id="18290" PostTypeId="2" ParentId="18288" CreationDate="2011-01-20T12:01:54.717" Score="1" ViewCount="" Body="&lt;p&gt;$$ \cup_{i=k}^{\infty} E_i = \{ x |  \exists i \geq k : x \in E_i\}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Did you mean this by inner structure of $lim sup$?&lt;/p&gt;&#xA;" OwnerUserId="5798" LastActivityDate="2011-01-20T12:01:54.717" />
  <row Id="18292" PostTypeId="2" ParentId="18288" CreationDate="2011-01-20T12:13:32.243" Score="3" ViewCount="" Body="&lt;p&gt;I am not quite sure I understand your question. But in order to understand something like $\bigcap_{k=1}^{\infty}\bigcup_{i=k}^{\infty} E_{i}$, we always translate $\bigcap$ as &quot;For Any&quot; and translate $\bigcup$ as &quot;there exists&quot;. Then $\bigcap_{k=1}^{\infty}\bigcup_{i=k}^{\infty} E_{i}$ means &quot;For any k, there exists some $i\geq k$, such that $x\in E_i$&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, let $S_I = \bigcup _{i\in I} E_{i}$. If $I$ only contains finite number of indices, there is actually not much to say about the relation between $S_I$ and $S$. But if I contain infinite many elements, $S\subset S_I$. In fact, $S$ is the joint of all such $S_I$'s.&lt;/p&gt;&#xA;" OwnerUserId="5476" LastEditorUserId="5476" LastEditorDisplayName="" LastEditDate="2011-01-20T12:36:04.937" LastActivityDate="2011-01-20T12:36:04.937" CommentCount="4" />
  <row Id="18293" PostTypeId="1" CreationDate="2011-01-20T12:19:52.703" Score="1" ViewCount="101" Body="&lt;p&gt;Suppose $F : C \to D$ is a functor between concrete categories.  If $U_D : D \to \text{Set}$ and $U_C : C \to \text{Set}$ are the corresponding forgetful functors to $\text{Set}$, then it's natural to require that $U_C = U_D F$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose $U_C$ is representable.  I would like to say then that $U_D$ is a &lt;em&gt;lift&lt;/em&gt; of the representable functor $U_C$ to $D$, or something like that.  The representing object should then acquire extra structure corresponding to whatever is needed for $F$ to land in $D$.  Is there an established term for this?&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2011-01-20T16:16:24.183" LastActivityDate="2011-01-20T16:16:24.183" Title="What is the name for a &quot;lifted&quot; representable functor?" Tags="&lt;category-theory&gt;" CommentCount="3" />
  <row Id="18294" PostTypeId="1" AcceptedAnswerId="18303" CreationDate="2011-01-20T12:24:05.667" Score="2" ViewCount="134" Body="&lt;p&gt;Does there exist an analytic everywhere, piecewise-defined function $f$ such that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f(x) = g(x)$ for $x &amp;lt; k$&lt;br&gt;&#xA;$f(x) = h(x)$ for $x&amp;gt;k$&lt;br&gt;&#xA;$f(x) = r$ for $x=k$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;With $g \ne h $ ($g$ not the same function as $h$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it exists, what is such an example?&lt;/p&gt;&#xA;" OwnerUserId="5904" LastEditorUserId="822" LastEditorDisplayName="" LastEditDate="2011-01-20T14:03:09.007" LastActivityDate="2011-01-20T14:03:09.007" Title="Example of analytic piecewise-defined function" Tags="&lt;real-analysis&gt;&lt;functions&gt;&lt;functional-analysis&gt;" AnswerCount="1" CommentCount="9" />
  <row Id="18295" PostTypeId="1" CreationDate="2011-01-20T12:34:46.793" Score="2" ViewCount="121" Body="&lt;p&gt;Suppose I have a group $G$ with a finite presentation, and I know that $G$ is polycyclic.  So I input the group into GAP via generators and relators.  Now I want to find a presentation of an infinite index subgroup; I can do this by hand (usually), by doing a Reidemeister-Schreier rewriting, and then simplifying the presentation as best I can.  GAP will, in general, not attempt to find presentations of infinite index subgroups.  But if I tell GAP the group is polycyclic, are there various algorithms out there such that GAP can in fact give me a finite presentation for this infinite index subgroup? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If that is too much, can GAP do it for certain &quot;nice&quot; subgroups (the commutator, etc.)?&lt;/p&gt;&#xA;" OwnerUserId="641" LastActivityDate="2011-01-20T12:34:46.793" Title="Finite Presentation in GAP of infinite index subgroup" Tags="&lt;group-theory&gt;" CommentCount="9" />
  <row Id="18296" PostTypeId="2" ParentId="18093" CreationDate="2011-01-20T12:39:48.313" Score="0" ViewCount="" Body="&lt;p&gt;I don't know about exact solution, but numeric approximation is $P(n)=1-e^{-\frac{n}{125}}$.&#xA;&lt;img src=&quot;http://i.imgur.com/0UNPR.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Matlab\Octave Code&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;function test_script_1()&#xA;x=7:30:600;&#xA;y=tsh([],5000,7:30:600);&#xA;er=@(p)sum(((p(1)+p(2)*exp(p(3)*x/500))-y).^2);&#xA;an=fminsearch(er,[0,0,0]);&#xA;anr=round(an);&#xA;disp(anr);&#xA;plot(x,y,x,anr(1)+anr(2)*exp(anr(3)*x));&#xA;legend('calculated','1-e^{-x/125}',4)&#xA;end&#xA;function result=tsh(a,b,n)&#xA;% HHTHTHH (H -&amp;gt; 1 , T -&amp;gt; 0)&#xA;if isempty(a)&#xA;    a=logical([1 1 0 1 0 1 1]);&#xA;end&#xA;if isempty(b)&#xA;    b=1000;&#xA;end&#xA;if isempty(n)&#xA;    n=100;&#xA;end&#xA;if any(size(n)&amp;gt;1)&#xA;    result=zeros(size(n));&#xA;    for pa=1:length(n)&#xA;        result(pa)=test_script_1(a,b,n(pa));        &#xA;    end&#xA;    return&#xA;end&#xA;la=length(a);&#xA;t=false(1,b);&#xA;for i=1:b&#xA;    v=raspr(0,1,n);&#xA;    for j=1:n&#xA;        if all(a==v(j:j+la-1))&#xA;            t(i)=true;&#xA;            break&#xA;        end&#xA;    end&#xA;end&#xA;result=nnz(t)/b;&#xA;end&#xA;function resu=raspr(a1,b1,s)&#xA;resu=round((b1-a1+1)*rand(1,s)+a1-0.5);&#xA;end&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Somebody know why is that?&lt;/p&gt;&#xA;" OwnerUserId="5550" LastActivityDate="2011-01-20T12:39:48.313" CommentCount="4" />
  <row Id="18297" PostTypeId="2" ParentId="12704" CreationDate="2011-01-20T12:59:37.787" Score="0" ViewCount="" Body="&lt;p&gt;No. You can solve these type of equations irrespective of A being a square matrix or even otherwise. In Ax = 0, consider that A is a general mxn matrix. If m&gt;&gt;n, then you have a redundant system of equations. You can solve these equations using either of normal equations(pseudoinverse), Singular value decomposition or QR decomposition. Look at it as a least squares problem. x is a value which minimizes the difference between b(here it is zero) and Ax. &lt;/p&gt;&#xA;" OwnerUserId="6050" LastActivityDate="2011-01-20T12:59:37.787" />
  <row Id="18299" PostTypeId="2" ParentId="18284" CreationDate="2011-01-20T13:06:05.667" Score="9" ViewCount="" Body="&lt;p&gt;How's this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left(\sum _{k=0}^{n-1} k! \binom{n}{k} \binom{n-1}{k} \left(-2n+k\right) x^{-2 n+k-1}\right) - \left(\sum _{k=0}^{n-1} k! \binom{n}{k} \binom{n-1}{k} x^{-2 \left(n+1\right)+k}\right) =$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$= \left(\sum _{k=0}^{n-1} k! \binom{n}{k} \binom{n-1}{k} \left(-2n+k\right) x^{-2\left(n+1\right)+k+1}\right) - \left(\sum _{k=0}^{n-1} k! \binom{n}{k} \binom{n-1}{k} x^{-2 \left(n+1\right)+k}\right) =$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$= \left(\sum _{k&amp;#39;=1}^{n} \left(k&amp;#39;-1\right)! \binom{n}{k&amp;#39;-1} \binom{n-1}{k&amp;#39;-1} \left(-2n+k&amp;#39;-1\right) x^{-2\left(n+1\right)+k&amp;#39;}\right) - \left(\sum _{k=0}^{n-1} k! \binom{n}{k} \binom{n-1}{k} x^{-2 \left(n+1\right)+k}\right) =$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$= \left(\sum _{k&amp;#39;=0}^{n} \left(k&amp;#39;-1\right)! \binom{n}{k&amp;#39;-1} \binom{n-1}{k&amp;#39;-1} \left(-2n+k&amp;#39;-1\right) x^{-2\left(n+1\right)+k&amp;#39;}\right) - \left(\sum _{k=0}^{n} k! \binom{n}{k} \binom{n-1}{k} x^{-2 \left(n+1\right)+k}\right) =$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$= \sum _{k=0}^{n} \left(\left(k-1\right)! \binom{n}{k-1} \binom{n-1}{k-1} \left(-2n+k-1\right) - k! \binom{n}{k} \binom{n-1}{k}\right) x^{-2 \left(n+1\right)+k}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then&#xA;$$\left(k-1\right)! \binom{n}{k-1} \binom{n-1}{k-1} \left(-2n+k-1\right) - k! \binom{n}{k} \binom{n-1}{k} =$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$= \frac{\left(k-1\right)!n!\left(n-1\right)!\left(-2n+k-1\right)}{\left(n-k+1\right)!\left(k-1\right)!\left(n-k\right)!\left(k-1\right)!} - \frac{k!n!\left(n-1\right)!}{\left(n-k\right)!k!k!\left(n-k-1\right)!} =$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$= \frac{n!\left(n-1\right)!\left(-2n+k-1\right)k}{\left(n-k+1\right)!\left(n-k\right)!k!} - \frac{n!\left(n-1\right)!\left(n-k\right)\left(n-k+1\right)}{\left(n-k\right)!k!\left(n-k+1\right)!} =$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$= \frac{n!\left(n-1\right)!}{\left(n-k+1\right)!\left(n-k\right)!k!} \left(\left(-2n+k-1\right)k - \left(n-k\right)\left(n-k+1\right)\right) =$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$= \frac{-n\left(n+1\right)n!\left(n-1\right)!}{\left(n-k+1\right)!\left(n-k\right)!k!} =$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$= -\frac{\left(n+1\right)!}{\left(n-k+1\right)!} \binom{n}{k} =$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$= -k! \binom{n+1}{k} \binom{n}{k}$$&lt;/p&gt;&#xA;" OwnerUserId="5676" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2011-05-03T16:54:01.637" LastActivityDate="2011-05-03T16:54:01.637" CommentCount="4" />
  <row Id="18300" PostTypeId="1" AcceptedAnswerId="18313" CreationDate="2011-01-20T13:40:04.803" Score="2" ViewCount="119" Body="&lt;p&gt;I've created a type of alphabet cipher in which each character is defined based on how distant it is (in the alphabet) from the preceding character.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, assuming that A=0, B=1, C=2, etc., and you have a text input of &quot;GOOSE,&quot; you would construct the output like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;G (6) is 2 letters in front of E (4), so output is C (2);&lt;br&gt;&#xA;O (14) is 8 letters in front of G (6), so output is I (8);&lt;br&gt;&#xA;O (14) is 0 letters in front of O (14), so output is A (0);&lt;br&gt;&#xA;S (18) is 4 letters in front of O (14), so output is E (4);&lt;br&gt;&#xA;E (4) is 12 letters in front of S (18), so output is M (12).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of the benefits of this type of cipher is that, unlike a Caesar cipher, it can't be broken using normal letter-frequency analysis, since the cipher-letter that represents each actual letter is not constant throughout.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know very little about ciphers, so I'm assuming that this type of thing has been done before.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm curious: Does this type of cipher have a name -- and how easy is it to decipher, compared with other techniques?&lt;/p&gt;&#xA;" OwnerUserId="6058" LastActivityDate="2011-01-20T14:45:19.787" Title="What type of cipher is this?" Tags="&lt;cryptography&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="18301" PostTypeId="1" AcceptedAnswerId="18346" CreationDate="2011-01-20T13:50:27.060" Score="4" ViewCount="222" Body="&lt;p&gt;I am stuck up with this simple problem. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $\alpha \cdot \beta = \pi$, then show that $$\sqrt{\alpha}\int\limits_{0}^{\infty} \frac{e^{-x^{2}}}{\cosh{\alpha{x}}} \ \textrm{dx} = \sqrt{\beta} \int\limits_{0}^{\infty} \frac{e^{-x^{2}}}{\cosh{\beta{x}}} \ \textrm{dx}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried replacing $\cosh{x} = \frac{e^{x}+e^{-x}}{2}$ and tried doing some manipulations, but it's of no use. Seems to be a clever problem. Moreover since we have $\alpha \cdot \beta = \pi$, we get $\sqrt{\alpha} = \frac{\sqrt{\pi}}{\sqrt{\beta}}$, but the &lt;em&gt;Beta&lt;/em&gt; factor is in the numerator, which bewilders me.&lt;/p&gt;&#xA;" LastEditorUserId="2937" LastEditorDisplayName="" LastEditDate="2011-01-20T18:36:45.763" LastActivityDate="2011-01-21T12:54:44.543" Title="Proving that the given two integrals are equal." Tags="&lt;calculus&gt;&lt;real-analysis&gt;&lt;analysis&gt;&lt;integral&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="2" />
  <row Id="18302" PostTypeId="2" ParentId="18268" CreationDate="2011-01-20T13:53:04.443" Score="4" ViewCount="" Body="&lt;p&gt;UPDATE - THIS IS WRONG -- SEE BELOW CORRECT ANSWER ---&#xA;Let's call N1 the number of configurations that have at least one day in between birthdays (this excludes not only consecutive birthdays, but also coincident birthdays).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I get (counting &lt;a href=&quot;http://en.wikipedia.org/wiki/Composition_%28number_theory%29&quot; rel=&quot;nofollow&quot;&gt;weak compositions&lt;/a&gt;) :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ \displaystyle N_1 = 365 \frac{(365-n-1)!}{(365-2n)!}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to include coincident birthdays, we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle N_0 =  \frac{365!}{(365-n)!}$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the probability asked is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle P = \frac{N_0 - N_1}{365^n}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Update: there &lt;s&gt;might be&lt;/s&gt; is some error here, &lt;s&gt;I think&lt;/s&gt; I'm failing to taking into account the configurations that have both coincident and consecutive birthdays, I'll revise this tonight. I suspect that N1 is correct, and that allows to compute the probability of having &lt;strong&gt;consecutive OR coincident&lt;/strong&gt; birthdays. To count consecutive (exclusively) birthdays seems more difficult.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;UPDATE:  Here's the correct (I hope) answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The probability of having at least a pair of consecutive birthday for M (=365) days and n persons is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle&#xD;&#xA;   P(M,n) = 1 - \sum_{k=1}^n Q(M,k) {M \choose k} \frac{S(n,k) k! }{M^n}&#xD;&#xA;   =  1 - \frac{1}{M^{n-1}} \sum_{k=1}^n \frac{(M-k-1)!}{(M-2k)!} S(n,k)&#xD;&#xA;$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where&#xA;$ \displaystyle&#xD;&#xA;Q(M,k) = \frac{{M -k - 1 \choose k -1} }{{M -1  \choose k -1} }&#xD;&#xA;$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and $S(n,k)$ are the &lt;a href=&quot;http://en.wikipedia.org/wiki/Stirling_numbers_of_the_second_kind&quot; rel=&quot;nofollow&quot;&gt;Stirling numbers of the second kind&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some computed values follow&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;M=365 n=1 p=0.00000000&#xA;M=365 n=2 p=0.00547835&#xA;M=365 n=3 p=0.01634745&#xA;M=365 n=4 p=0.03242761&#xA;M=365 n=5 p=0.05345896&#xA;M=365 n=6 p=0.07910314&#xA;M=365 n=7 p=0.10895871&#xA;M=365 n=8 p=0.14256439&#xA;M=365 n=9 p=0.17941667&#xA;M=365 n=10 p=0.21897764&#xA;M=365 n=11 p=0.26069278&#xA;M=365 n=12 p=0.30400167&#xA;M=365 n=13 p=0.34834843&#xA;M=365 n=14 p=0.39319571&#xA;M=365 n=15 p=0.43803357&#xA;M=365 n=16 p=0.48239009&#xA;M=365 n=17 p=0.52583640&#xA;&#xA;M=365 n=30 p=0.90729104&#xA;&#xA;M=365 n=42 p=0.99074145 &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Explanation follows: -------------------------------&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $M$ (=365) number of days, and $n$ = number of persons and $k$ = number of distinct birthdays ($k$ is not fixed, it's a random variable in the range $1..n$). Let $P_{M,n}(S)$ be the probability of NOT having consecutive birthdays (S is the event: &quot;all birthdays are separated&quot;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then $P_{M,n}(S) = \sum_k P_{M,n}(S \; k) = \sum_k P_{M,n}( S | k ) P_{M,n}(k )$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The following steps compute the two factors inside the summation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;STEP 1:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$P_{M,n}( S | k )$ is the probability of having the events separated, given that the number of distinct birthdays is $k$. If we think of birthdays as occupied boxes in a circular list, a little reflection shows that all configurations are equiprobable ($n$ does not matter now) and we can assume without altering the result that the first box of the list is occupied. Now, the total number of possible ocurrences are the number of selection of $k-1$ boxes among $M-1$ (combination). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And the number of &quot;succesfull&quot; arrangements are those that result in non-consecutive occupied boxes. But this is equivalent of specifying a list of $k$ numbers greater than 1 that sum up to M; and this is the same as specifying a list of $k$ numbers greater than 0 than sum up to $M-k$; and this can be counted, by a reasoning similar to &lt;a href=&quot;http://en.wikipedia.org/wiki/Composition_%28number_theory%29#Number_of_compositions&quot; rel=&quot;nofollow&quot;&gt;this one&lt;/a&gt;, as the combination of $k -1$ taken from $M -k - 1$. So, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle P_{M,n}( S | k ) =  \frac{{M -k - 1 \choose k -1} }{{M -1  \choose k -1} } = Q(M,k)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;STEP2:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$P_{M,n}(k)$ : if we place $n$ balls at random in $M$ boxes, what is the probability that exactly $k$ boxes will be non-empty?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The total counting is given by $M^n$. To count the &quot;successful&quot; cases, we multiply: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;all the possible sets of occupied boxes (combinations of $k$ taken from $M$)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;the total numbers of ways of putting $n$ balls in $k$ boxes leaving no empty box: that is given by the Stirling numbers of the second kind.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;and we need to multiply by $k!$ (permutations of boxes) because the Stirling numbers consider nondistinct boxes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle P_{M,n}(k ) = {M \choose k} \frac{S(n,k) k! }{M^n}$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;And from there  follows the formula above.&lt;/p&gt;&#xA;" OwnerUserId="312" LastEditorUserId="312" LastEditorDisplayName="" LastEditDate="2011-01-21T17:59:28.190" LastActivityDate="2011-01-21T17:59:28.190" CommunityOwnedDate="2011-01-21T01:58:43.297" CommentCount="7" />
  <row Id="18303" PostTypeId="2" ParentId="18294" CreationDate="2011-01-20T13:57:33.070" Score="5" ViewCount="" Body="&lt;p&gt;If $g$ and $h$ are not necessarily analytic, then just take an arbitrary analytic function $f$ with $f(k) = r$. Define $g$ and $h$ to be&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ g(x) := f(x) \chi(k-x) \qquad h(x) := f(x) \chi(x-k) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\chi$ is a cut-off function, that is: $\chi(x) = 1$ for $x \geq 0$, $\chi(x) = 0$ for $x \leq -1$ and $0 &amp;lt; \chi(x) &amp;lt; 1$ for $-1 &amp;lt; x &amp;lt; 0$. Then you have all your desired properties. You can choose $\chi$ to be continuous, or even smooth. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;If $g$ and $h$ are analytic, then the answer is no. Real analytic functions have the unique continuation property. That is, if a real analytic function is zero on an open set, it must be zero everywhere. Then since $f$ is everywhere analytic, and $g$ is everywhere analytic, their difference $f-g$ is also analytic, and vanishes for $x &amp;lt; k$. This implies that $f=g$ everywhere. Similarly $f =h$ everywhere and $g$ and $h$ must coincide. &lt;/p&gt;&#xA;" OwnerUserId="1543" LastActivityDate="2011-01-20T13:57:33.070" />
  <row Id="18304" PostTypeId="2" ParentId="15194" CreationDate="2011-01-20T14:05:30.017" Score="0" ViewCount="" Body="&lt;p&gt;You may want to take a look at exercise 6 on page 223 of Borwein and Borwein.  (I am guessing anyone who asks about the AGM is familiar with this book.) There they discus a matrix version of the AGM and asks one to prove a connection between it and an elliptic integral. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;They also give a paper by Stickel from 1985: &quot;Fast compuation of Matrix Exponentials and Logarithms&quot; as a reference. The journal name is Analysis. &lt;/p&gt;&#xA;" OwnerUserId="4623" LastEditorUserId="4623" LastEditorDisplayName="" LastEditDate="2011-01-21T10:47:08.060" LastActivityDate="2011-01-21T10:47:08.060" />
  <row Id="18305" PostTypeId="1" AcceptedAnswerId="18316" CreationDate="2011-01-20T14:08:50.197" Score="6" ViewCount="184" Body="&lt;p&gt;Maybe this is a dumb question, but I have to admit that it is not really clear to me what the theory is, in which incompleteness of PA and stronger theories is proved. The texts I did study so far are often not really explicit about that. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What's the (minimal) theory in which &#xA;  incompleteness of PA and stronger theories is proved?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1792" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2011-01-21T00:03:46.283" LastActivityDate="2011-01-21T00:03:46.283" Title="What's the theory in which incompleteness of PA is proved?" Tags="&lt;number-theory&gt;&lt;logic&gt;&lt;incompleteness&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" />
  <row Id="18306" PostTypeId="1" AcceptedAnswerId="18314" CreationDate="2011-01-20T14:08:58.763" Score="0" ViewCount="97" Body="&lt;p&gt;What is the measure of one of the larger angles of a&#xA;parallelogram in the xy-plane that has vertices with&#xA;coordinates (2,1),(5,1),(3,5) and (6,5)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;(A)&lt;/strong&gt; 93.4&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;(B)&lt;/strong&gt; 96.8&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;(C)&lt;/strong&gt; 104.0&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;(D)&lt;/strong&gt; 108.3&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;(E)&lt;/strong&gt; 119.0&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the answer strip says &lt;strong&gt;C&lt;/strong&gt;, but it doesn't explain how to find the angle.&#xA;So how would I go about solving this question? I've tried drawing in a plane but I couldn't relate my drawing to the angles.&lt;/p&gt;&#xA;" OwnerUserId="6060" LastActivityDate="2011-01-20T14:46:48.073" Title="Largest angle of a parallelogram in a plane" Tags="&lt;geometry&gt;" AnswerCount="2" />
  <row Id="18307" PostTypeId="1" CreationDate="2011-01-20T14:11:21.300" Score="0" ViewCount="143" Body="&lt;p&gt;Prove that the the function $f(x) = |x-1|+|x-2|$ is not differentiable at $x = 1$,and $x = 2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is actually an objective type question where there is $3$ other options  $(\{0\},\{1\},\{3\},\{1,2\})$,now the problem is graph is not really much of a choice,hence I guess I have to break the functions into ranges and then solve, any idea ?&lt;/p&gt;&#xA;" OwnerUserId="2109" LastEditorUserId="2109" LastEditorDisplayName="" LastEditDate="2011-01-20T15:02:50.597" LastActivityDate="2011-01-20T19:58:15.397" Title="How to prove nondifferentiability ?" Tags="&lt;calculus&gt;" AnswerCount="2" />
  <row Id="18308" PostTypeId="2" ParentId="18307" CreationDate="2011-01-20T14:18:24.783" Score="1" ViewCount="" Body="&lt;p&gt;Hint: Look at the function $f(x)=|x|$. What happens to the graph when $x=0$? If we try finding the derivative, we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f^\prime(x)=\lim_{\epsilon \to 0^+} \frac{|x+\epsilon|-|x|}{\epsilon}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This approaches 1, but if we try to compute $\lim_{\epsilon \to 0^-} \frac{|x+\epsilon|-|x|}{\epsilon}$ we get -1. However, for the derivative to exist, the limit must be unique.&lt;/p&gt;&#xA;" OwnerUserId="4284" LastActivityDate="2011-01-20T14:18:24.783" />
  <row Id="18309" PostTypeId="2" ParentId="18307" CreationDate="2011-01-20T14:37:28.403" Score="3" ViewCount="" Body="&lt;p&gt;Show that the limits&#xA;$$\lim_{h\to 0}\frac{f(1+h)-f(1)}{h}$$&#xA;and&#xA;$$\lim_{h\to 0}\frac{f(2+h)-f(2)}{h}$$&#xA;do not exist.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In each case, the easiest thing will be to consider the one sided limits, as $h\to 0^+$ and as $h\to 0^{-}$; if you can show that the one-sided limits are different from each other or at least one does not exist (including the case that they equal $\infty$ or $-\infty$), (each of the two limits separately, of course), then this will prove the function is not differentiable, because $f$ is differentiable at $a$ if and only if the limit $\lim\limits_{h\to 0}(f(a+h)-f(a))/h$ exists. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;After the edit&lt;/em&gt;: There is a very big difference between &quot;How can I choose the right answer from a given list of possibilities quickly?&quot; and &quot;How do I prove that the function is not differentiable at $x=1$ and $x=2$?&quot; Please try to ask the question you mean to ask and not something different, lest you get an answer for the question you ask instead of for the question you &lt;em&gt;mean.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The key here is that you should know by now that $y=|t|$ is differentiable everywhere except at $t=0$ (the standard example of a function that is continuous at a point but not differentiable there). That means that a horizontal translate of the function by $a$ (that is, the function $y=|t-a|$, which is the function $y=|t|$ &quot;translated&quot; $a$ units to the right) will necessarily be differentiable everywhere except at $t=a$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, $y=|x-1|$ is differentiable everywhere except at $x=1$, and $y=|x-2|$ is differentiable everywhere except at $x=2$. The sum of two functions is differentiable everywhere that the two functions are differentiable, and there is a chance that if &lt;em&gt;neither&lt;/em&gt; function is differentiable at a point $a$, then their sum &lt;em&gt;will&lt;/em&gt; be differentiable (for example, take $y=|x|$ and $y=-|x|$; neither is differentiable at $x=0$, but $|x|-|x|$ &lt;em&gt;is&lt;/em&gt;). However, if &lt;strong&gt;exactly one&lt;/strong&gt; of $f(x)$ and $g(x)$ are differentiable at $x=a$, then $f(x)+g(x)$ is &lt;strong&gt;not&lt;/strong&gt; differentiable at $x=a$: say $f$ is differentiable at $x=a$, and so is $f+g$. Then $g= (f+g)-f$ is also differentiable at $x=a$, being the sum of two functions that are differentiable at $x=a$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, here, you have that $y=|x-1|+|x-2|$ is certainly differentiable at every $x$ that is not equal to either $1$ or $2$, because both $|x-1|$ and $|x-2|$ are differentiable there; and it is certainly &lt;strong&gt;not&lt;/strong&gt; differentiable at either $x=1$ or $x=2$, because one and only one of $|x-1|$ and $|x-2|$ is differentiable at each of those points. So that tells you quickly that the right answer has to be $\{1,2\}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, you should know at a glance that $y=|x-a|$ is differentiable everywhere except at $x=a$. &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-20T19:58:15.397" LastActivityDate="2011-01-20T19:58:15.397" CommentCount="5" />
  <row Id="18311" PostTypeId="1" CreationDate="2011-01-20T14:42:50.117" Score="1" ViewCount="89" Body="&lt;p&gt;Dear community,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have the following system of equations:&#xA;$$\alpha f_1(x) = \int_\mathbb{R} g(k)  h_1(k) e^{\mathrm{i}kx} dk$$&#xA;$$\beta f_2(x) = \int_\mathbb{R} g(k)  h_2(k) e^{\mathrm{i}kx} dk$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with known functions $f_i(x)$ and $h_i(x)$ and unknown $\alpha$, $\beta$ and $g(k)$.&lt;br&gt;&#xA;Before outlining my ansatz, here directly the question:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Under which conditions on $f_1(x)$ and $f_2(x)$ can we calculate the ratio $\frac\beta\alpha$ uniquely?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h3&gt;My approach&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;The first step was to integrate one of the equations,&#xA;$$\alpha \int_\mathbb{R} f_1(x) e^{-\mathrm{i}\lambda x}dx = \int_{\mathbb{R}^2} g(k)  h_1(k) e^{\mathrm{i}(k-\lambda)x} dxdk$$&#xA;resulting in&#xA;$$\alpha F_1(k) = 2\pi g(k) h_1(k)$$&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Dirac_delta_function#Fourier_transform&quot; rel=&quot;nofollow&quot;&gt;because of&lt;/a&gt; &#xA;$$\delta\left(k-\lambda\right) = \frac{1}{2\pi}\int_{-\infty}^{\infty}e^{\mathrm{i}\left(k-\lambda\right)x}dx$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;hence $g(k)$ can be seen as calculated. Plugging this into the second equation, one gets&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$2\pi\frac\beta\alpha f_2(x) = \int_\mathbb{R} F_1(k)  \frac{h_2(k)}{h_1(k)} e^{\mathrm{i}kx} dk\, .$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I think the answer to the question relates to a &lt;strong&gt;further integration&lt;/strong&gt; of this equation via some kernel $K(x)$. I would think that this is somehow the dual of the function $f_2(x)$ like in a &lt;a href=&quot;http://en.wikipedia.org/wiki/Hilbert_space&quot; rel=&quot;nofollow&quot;&gt;Hilbert space&lt;/a&gt;. But I don't know if this really gives me a well defined unique result, and, how to calculate the dual of $f_2(x)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would be thankful for any hint&lt;br&gt;&#xA;Sincerely&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Robert&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;A little excuse&lt;/em&gt;&lt;br&gt;&#xA;As some might have noticed, I am at the moment occupied with some questions arising in, say, applied &lt;a href=&quot;http://en.wikipedia.org/wiki/Functional_analysis&quot; rel=&quot;nofollow&quot;&gt;functional analysis&lt;/a&gt;. As a physicist I am often not very sure about the justification of what I am doing so I hope it is ok to ask these questions here even though the answers might be obvious to most of the people in this community.&lt;/p&gt;&#xA;" OwnerUserId="2671" LastActivityDate="2011-01-20T14:42:50.117" Title="A Projection Problem in Functional Analysis - Uniqueness of a Solution" Tags="&lt;linear-algebra&gt;&lt;functional-analysis&gt;&lt;integral-transforms&gt;" CommentCount="3" />
  <row Id="18312" PostTypeId="2" ParentId="18306" CreationDate="2011-01-20T14:43:48.507" Score="0" ViewCount="" Body="&lt;p&gt;One idea is to find the cartesian coordinates of the edges of the parallilogram and think of the slope of them(use the difference of their slopes).&lt;/p&gt;&#xA;" OwnerUserId="5422" LastActivityDate="2011-01-20T14:43:48.507" />
  <row Id="18313" PostTypeId="2" ParentId="18300" CreationDate="2011-01-20T14:45:19.787" Score="2" ViewCount="" Body="&lt;p&gt;The kind of cipher you are thinking about is called an &lt;strong&gt;autokey cipher&lt;/strong&gt;. This is a polyalphabetic cipher (several encryption rules) in which the first (possibly first few) encryptions are based on some agreed-upon rule, and later encryptions are based on the text being encrypted. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The idea was originated by Girolamo Cardano (1501-1576): he would encrypt the first word of the message using a fixed key, and then for each following word he would use the first letter of the previous word as the key for encrypting the next word.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your case, you have a collection of Caesar ciphers (alphabetic shifts), and you are selecting which Caesar cypher to use at any given step on the basis of the previous letter in the message. &lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2011-01-20T14:45:19.787" />
  <row Id="18314" PostTypeId="2" ParentId="18306" CreationDate="2011-01-20T14:46:48.073" Score="0" ViewCount="" Body="&lt;p&gt;If you have vectors at your disposal, the easiest thing to do is: (1) find the vectors from one vertex to the two adjacent vertices, call them $\vec{u}$ and $\vec{v}$; (2) the angle formed by these two vectors is $\arccos(\frac{\vec{u}\cdot\vec{v}}{|\vec{u}||\vec{v}|})$.  If this angle is acute, then its supplement is the measure of the obtuse angle in the parallelogram.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Without vectors, you could use the Law of Cosines: (1) find the lengths of two adjacent sides of the parallelogram (call them $a$ and $b$) and the length of the longer diagonal (call it $c$; if you used the shorter diagonal, you'd get the acute angle); (2) the angle is $\arccos(\frac{a^2+b^2-c^2}{2ab})$.&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2011-01-20T14:46:48.073" />
  <row Id="18315" PostTypeId="1" AcceptedAnswerId="18327" CreationDate="2011-01-20T14:48:17.340" Score="4" ViewCount="258" Body="&lt;p&gt;In this very &lt;a href=&quot;http://books.google.com/books?id=GNPnXoeYw_EC&amp;amp;printsec=frontcover&amp;amp;dq=submodular+functions&amp;amp;hl=en&amp;amp;ei=H0g4TZzmLMfsrQfKgcnJCA&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=2&amp;amp;ved=0CC0Q6AEwAQ#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;nice book&lt;/a&gt;, author has defined vector space as set of functions $f : S \rightarrow F$ where $S$ is a finite set and $F$ is a field. It turns out that this definition is closely resemble definition of &lt;a href=&quot;http://planetmath.org/encyclopedia/FreeVectorSpaceOverASet.html&quot; rel=&quot;nofollow&quot;&gt;Free Vector Space&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are &lt;a href=&quot;http://www.physicsforums.com/showthread.php?s=05b5360680e0ea27b955f08614562ffb&amp;amp;t=228674&quot; rel=&quot;nofollow&quot;&gt;one or two threads&lt;/a&gt; on other online community regarding this. While I try to undersand it, could anyone point out some differences between vector spaces defined this way and defined in standard ways e.g. Hoffman and Kunze.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can one give a small concrete example of constructing a vector space using this definition and its implication. Any tutorial or link will be appreciated.&lt;/p&gt;&#xA;" OwnerUserId="1674" LastEditorUserId="1674" LastEditorDisplayName="" LastEditDate="2011-01-20T15:51:15.213" LastActivityDate="2011-01-20T23:21:44.743" Title="'Free Vector Space' and 'Vector Space'" Tags="&lt;linear-algebra&gt;" AnswerCount="3" CommentCount="4" FavoriteCount="4" />
  <row Id="18316" PostTypeId="2" ParentId="18305" CreationDate="2011-01-20T15:15:19.183" Score="15" ViewCount="" Body="&lt;p&gt;Let PRA be the theory of &lt;a href=&quot;http://en.wikipedia.org/wiki/Primitive_recursive_arithmetic&quot; rel=&quot;nofollow&quot;&gt;Primitive recursive arithmetic&lt;/a&gt;. This is a subtheory of PA, and it suffices to prove the incompleteness theorem. It is perhaps not the easiest theory to work with, but the point is that proving incompleteness can be carried out is a significantly weak system with respect to the theories to which incompleteness actually applies. It is sometimes argued that Hilbert's notion of finitism is (as precisely as possible) captured by PRA, so in that sense, this is the ideal setting for proving the incompleteness of PA or related systems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You may benefit from reading the nice book &lt;a href=&quot;http://projecteuclid.org/DPubS?service=UI&amp;amp;version=1.0&amp;amp;verb=Display&amp;amp;handle=euclid.pl/1235421926&quot; rel=&quot;nofollow&quot;&gt;Metamathematics of first order arithmetic&lt;/a&gt; by Hájek and Pudlák, where these matters are carefully discussed, as well as some of the numerous essays by Harvey Friedman, available at his &lt;a href=&quot;http://www.math.ohio-state.edu/~friedman/manuscripts.html&quot; rel=&quot;nofollow&quot;&gt;homepage&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me point out that what one proves in PRA is that &lt;em&gt;either PA is inconsistent&lt;/em&gt; or else, it cannot prove its own consistency (and therefore it is incomplete). If we actually want to prove that PA is incomplete, we need to reason within a system where Con(PA), the formal statement asserting the consistency of PA, is assumed. (Of course, Con(PA) is not a theorem of PA, so in a sense this is a strong assumption.) Similarly, PRA suffices to formalize the incompleteness of much stronger systems, and even, to formalize the &lt;em&gt;equiconsistency&lt;/em&gt; results of set theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some people prefer to be able to reason with sets (i.e., semantically, rather than with formal proofs, that are syntactic objects) in a less cumbersome fashion than via coding. Then, rather than working with PRA, it is convenient to work with a weak subsystem of second order arithmetic. Typically, WKL${}_0$ (A system that allows us to prove a weak version of König's lemma that &lt;em&gt;finite branching infinite trees&lt;/em&gt; have infinite branches) is the chosen system to carry out the formalization of the incompleteness proofs. A good place to learn about this and second order arithmetic in general is the wonderful &lt;a href=&quot;http://www.math.psu.edu/simpson/sosoa/&quot; rel=&quot;nofollow&quot;&gt;book&lt;/a&gt; by Stephen Simpson.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me mention that subsystems of second order arithmetic are understood to be the natural theories to carry out investigations on &lt;a href=&quot;http://en.wikipedia.org/wiki/Reverse_mathematics&quot; rel=&quot;nofollow&quot;&gt;reverse mathematics&lt;/a&gt;, i.e., they are precisely the setting one wants in order to study questions about what formal systems suffice to prove a theorem (such as the incompleteness of PA, in this case).&lt;/p&gt;&#xA;" OwnerUserId="462" LastActivityDate="2011-01-20T15:15:19.183" CommentCount="5" />
  <row Id="18317" PostTypeId="2" ParentId="18253" CreationDate="2011-01-20T15:43:31.247" Score="0" ViewCount="" Body="&lt;p&gt;In each step the monkey climbs and falls, so after $n+1$ steps he is at the heigth of&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(1-1)+(2-1)+(3-1)+\ldots+(n+1-1)= 1+2+\ldots+n+(n+1)-(n+1)=\frac{n(n+1)}{2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;we must find $max\{n\}$ such that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{n(n+1)}{2}\leq 50$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;from this we conclude that $n=9$, so after $10$ steps the monkey is at the heigth of $45m$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the time spent is then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$10\times(1+2+\ldots+8+10)+5\times(1+1+\ldots+1)=10\times55+5\times 10=600$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but to reach the top we need to add more $5\times10$ &lt;em&gt;s&lt;/em&gt; so the total time is $650$ &lt;em&gt;s&lt;/em&gt; and this corresponds to $10$ &lt;em&gt;min&lt;/em&gt; $50$ &lt;em&gt;sec&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If instead of adding the time spent in falls we had subtracted we would have&#xA;$$10\times(1+2+\ldots+8+10)-5\times(1+1+\ldots+1)=10\times55-5\times 10=500$$ and the total time would be $550$ &lt;em&gt;sec&lt;/em&gt; corresponding to $9$ &lt;em&gt;min&lt;/em&gt; $10$ &lt;em&gt;sec&lt;/em&gt; and &lt;strong&gt;is wrong&lt;/strong&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1747" LastActivityDate="2011-01-20T15:43:31.247" CommentCount="1" />
  <row Id="18318" PostTypeId="2" ParentId="18315" CreationDate="2011-01-20T15:46:09.983" Score="5" ViewCount="" Body="&lt;p&gt;What you've defined is not the free vector space on $S$.  It is &lt;em&gt;isomorphic&lt;/em&gt; to it, but is not the same thing.  The problem is that &quot;free vector space&quot; is not just an operation on sets, it is a &lt;a href=&quot;http://en.wikipedia.org/wiki/Functor&quot; rel=&quot;nofollow&quot;&gt;functor&lt;/a&gt;, so it also operates on functions between sets.  And when you look at the space of functions $S \to F$, the natural way to extend this to a functor is &lt;a href=&quot;http://en.wikipedia.org/wiki/Functor#Covariance_and_contravariance&quot; rel=&quot;nofollow&quot;&gt;contravariant&lt;/a&gt;: given a set function $f : S \to T$, there is a natural map from the space of functions $T \to F$ to the space of functions $S \to F$ given by precomposition.  The free vector space functor, however, is &lt;em&gt;covariant.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is true is that there is a natural dual pairing between the free vector space on $S$ and the space of functions $S \to F$ given by composition, and finite-dimensional vector spaces which are dual have the same dimension.  Using this pairing, the dual vector to an element $s \in S$ in the free vector space on $S$ is the indicator function $1_s : S \to F$ which is equal to $1$ at $s$ and equal to $0$ otherwise.  So authors will sometimes identify $s$ with its indicator function.  (Note that when $S$ is infinite it is highly false that the two spaces have the same dimension, which is another reason to avoid this definition.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One might argue that the distinction is academic.  If you think that, write down what happens to a left group action of a group $G$ on $S$ in each of the two cases.  (If you can't tell what the difference is, replace &quot;group&quot; with &quot;monoid.&quot;)&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2011-01-20T15:53:10.977" LastActivityDate="2011-01-20T15:53:10.977" CommentCount="6" />
  <row Id="18319" PostTypeId="1" AcceptedAnswerId="18351" CreationDate="2011-01-20T15:56:06.230" Score="2" ViewCount="251" Body="&lt;p&gt;I want to solve this limit without using L'Hopital's rule:&#xA;$$\lim_{x\to 5}\frac{2^x-2^5}{x-5}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And thanks.&lt;/p&gt;&#xA;" OwnerUserId="6063" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-20T20:31:20.517" LastActivityDate="2011-01-20T21:25:29.530" Title="I want to find $\lim\limits_{x\to 5}\frac{2^x-2^5}{x-5}$ without using l'Hopital's rule..." Tags="&lt;limit&gt;" AnswerCount="4" CommentCount="9" FavoriteCount="1" />
  <row Id="18320" PostTypeId="2" ParentId="18319" CreationDate="2011-01-20T16:00:20.253" Score="12" ViewCount="" Body="&lt;p&gt;$\lim_{x\to 5}\frac{2^x-2^5}{x-5}=\lim_{x\to 5}\frac{f(x)-f(5)}{x-5}=f&amp;#39;(5)$, where $f(x)=2^x$.  $f&amp;#39;(x)=\cdots$ so $f&amp;#39;(5)=\cdots$&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2011-01-20T16:00:20.253" />
  <row Id="18321" PostTypeId="2" ParentId="18319" CreationDate="2011-01-20T16:05:20.410" Score="3" ViewCount="" Body="&lt;p&gt;First write $2^x$ as $e^{x \ln 2}$. Then let $x=5+t$ so that you get a limit as $t \to 0$. Now you should be able to use the standard limit $(e^z-1)/z \to 1$ as $z\to 0$.&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2011-01-20T16:05:20.410" />
  <row Id="18322" PostTypeId="2" ParentId="18319" CreationDate="2011-01-20T16:09:42.677" Score="8" ViewCount="" Body="&lt;p&gt;In principle, you can use L'Hopital's Rule here because both numerator and denominator are differentiable, they both have a limit of $0$ as $x\to 5$, and the limit of the quotient $(2^x-2^5)&amp;#39;/(x-5)&amp;#39;$ exists as $x\to 5$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Morally,&lt;/strong&gt; however, you should not use L'Hopital's Rule; the reason is that in order to use L'Hopital's Rule, you need to know what the derivative of $2^x$ is. But the derivative of $2^x$ (at least, at $x=5$) is given &lt;strong&gt;precisely&lt;/strong&gt; by the limit you are trying to do. It would be like using L'Hopital's Rule in order to find $\lim\limits_{\theta\to 0}\frac{\sin\theta}{\theta}$; the problem is that to use L'Hopital's Rule you need ot know the derivative of $\sin\theta$, and in order to know the derivative of $\sin\theta$ you most likely had to figure out this limit in the first place.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So it's a good thing that you don't want to use L'Hopital's Rule here. Instead, you need to recognize this limit as the limit that defines $f&amp;#39;(5)$ with $f(x)=2^x$, and solve it accordingly, whether using the Chain Rule (since $2^x = e^{\ln(2)x}$), or by some other similar method.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2011-01-20T16:09:42.677" />
  <row Id="18323" PostTypeId="2" ParentId="18265" CreationDate="2011-01-20T16:13:29.040" Score="0" ViewCount="" Body="&lt;p&gt;Unfortunately nothing like that holds. Consider $f(z) = \sqrt{1 - z}$. $f(z)$ is analytic on ${\mathbb C} - \{(x,0): x &amp;gt; 1\}$ and satisfies Cauchy's estimates on circles of radius $r &amp;lt; 1$ since the function is analytic on the unit disc. However, try as you may, you cannot extend $f(z)$ to a neighborhood of $1$. &#xA;Note that Cauchy's estimates are even satisfied on the circle of radius $1$, by taking limits as $r$ goes to $1$ of the $r &amp;lt; 1$ Cauchy estimates. &lt;/p&gt;&#xA;" OwnerUserId="3035" LastActivityDate="2011-01-20T16:13:29.040" CommentCount="4" />
  <row Id="18324" PostTypeId="2" ParentId="18233" CreationDate="2011-01-20T16:26:18.010" Score="4" ViewCount="" Body="&lt;p&gt;Your first step would be to learn about predicate calculus (first order calculus or propositional calculus with the addition of 'for all' and 'there exist' quantifiers). This will then lead into the major subcategories:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;proof theory (syntactic manipulation of proofs, proof calculi, ordinals)&lt;/li&gt;&#xA;&lt;li&gt;model theory (models and structures of other areas of mathematics)&lt;/li&gt;&#xA;&lt;li&gt;set theory (axioms of sets, ZF, the axiom of choice, the general continuum hypothesis)&lt;/li&gt;&#xA;&lt;li&gt;recursion theory (decidability, recursive functions, lambda calculus, with connections to theoretical computer science subjects like computational complexity)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Of course, as with any body of knowledge, these categories are not mutually exclusive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These are specialty areas, meaning that research is done under one of these headings. To get into one of these areas, there are many 'elementary' concepts leading in that you probably would want familiarity with. An accessible topic like Goedel's theorems (the completeness theorem and two incompleteness theorems (these are not the same 'kind' of completeness)) will help get some of these necessary concepts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also a good area to be familiar with is philosophy of math and foundations; though not exactly mathematics or logic, it informs a lot of the motivations of study of the above categories. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Logic that is not mathematically based can be pursued in philosophy (human language arguments, deductive and inductive reasoning, vagueness), mostly in the area of epistemology.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your mathematics department may not have advanced logic, but other departments may (though not necessarily by name. Consider looking in the philosophy department (some mathematical logicians get hired there) or the computer science department (the research may be more applied but they may give courses that are theoretical/basic math).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Executive summary: start with predicate calculus and Goedel's theorem.&lt;/p&gt;&#xA;" OwnerUserId="1919" LastActivityDate="2011-01-20T16:26:18.010" />
  <row Id="18325" PostTypeId="1" CreationDate="2011-01-20T16:27:39.850" Score="2" ViewCount="102" Body="&lt;p&gt;Hello,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of my former students asked me how to go from one presentation of the Hermite Polynomial to another. And I'm embarassed to say, I've been trying and failing miserably. (I'm guessing this is a homework problem that he is having trouble with.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://functions.wolfram.com/Polynomials/HermiteH/07/ShowAll.html&quot; rel=&quot;nofollow&quot;&gt;http://functions.wolfram.com/Polynomials/HermiteH/07/ShowAll.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So he has to go from the Rodriguez type formula (written as a contour integral) to an integral on the real axis, which is the 3rd formula in the link provided above.  It seems like the hint he was given was to start from the contour integral.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Starting with the contour integral, I tried using different semi-circles (assuming that $z$ was real), but this quickly turned into something weird.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also tried to use a circle as the contour, then map it to the real line. That was a failure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried working backwards, from the integral on the real axis. Didn't have luck.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The last resort was to show that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Both expressions are polynomials.&#xA;2) Show that the corresponding coefficients were equal. (That is, I took both functions and evaluated them and their derivatives at 0.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even 2), I couldn't see a nice way of showing that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\int_C \frac{e^{-z^2}}{z^{n+1}}dz = \int_{-\infty}^{\infty} z^n e^{-z^2} dz$ (Up to some missing multiplicative constants.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I feel like I'm missing something really easy.  If someone could give me some hints without giving away the answer, that would be most appreciated.&lt;/p&gt;&#xA;" OwnerUserId="2499" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2011-04-19T03:46:47.167" LastActivityDate="2011-04-19T03:46:47.167" Title="Integral Representations of Hermite Polynomial?" Tags="&lt;homework&gt;&lt;integral&gt;&lt;complex-analysis&gt;&lt;special-functions&gt;" />
  <row Id="18327" PostTypeId="2" ParentId="18315" CreationDate="2011-01-20T16:54:45.680" Score="4" ViewCount="" Body="&lt;p&gt;Yes, the definition probably looks &quot;like&quot; definitions of free objects you may have encountered. The reason is essentially that this author is defining vector spaces &lt;strong&gt;essentially&lt;/strong&gt; as free objects without saying so. And the reason the author can do this is that, as it turns out, every vector space is a free object in the category of vector spaces (at least, every finite dimensional vector space is; the statement that &lt;em&gt;every&lt;/em&gt; vector space is a free object, that is that every vector space has a basis, is equivalent to the Axiom of Choice).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's consider a vector space you are probably very familiar with: $\mathbb{R}^2$; we usually think of this vector space as being &quot;given&quot; by a collection of things called &lt;em&gt;vectors&lt;/em&gt;, which are all things of the form $(x,y)$ with $x,y\in\mathbb{R}$. Then we define a vector sum by $(x,y)+(z,w) = (x+y,z+w)$ (where the sum on the right is the sum of real numbers), a scalar product given by $\lambda(x,y) = (\lambda x,\lambda y)$, and show that it satisfies the axioms, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How is this way of thinking about $\mathbb{R}^2$ related to the definition given by the author? Well, a simple way to think about it is that an element of $\mathbb{R}^2$ is an ordered pair of reals. But an ordered pair is &quot;essentially&quot; the same thing as a function $\{1,2\}\to\mathbb{R}$. How so? Well, the pair $(x,y)$ corresponds to the function $1\mapsto x,\ 2\mapsto y$, because the &lt;em&gt;first&lt;/em&gt; entry is $x$ and the &lt;em&gt;second&lt;/em&gt; entry is $y$. Conversely, given a function $f\colon\{1,2\}\to\mathbb{R}$, the function uniquely determines an ordered pair by the rule $f\mapsto (f(1),f(2))$. So instead of thinking of the elements of $\mathbb{R}^2$ as ordered pairs, we can think of them as functions with domain $\{1,2\}$.  I'll write a function $f\colon\{1,2\}\to \mathbb{R}$ as $\mathbf{f}$ when I want to think of it as a vector.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we think of the elements of $\mathbb{R}^2$, how do the operations translate into &quot;operations of functions&quot;? Well, if $\mathbf{f}$ and $\mathbf{g}$ are functions from $\{1,2\}$ to $\mathbb{R}$, what should $\mathbf{f}+\mathbf{g}$ be? Since $\mathbf{f}$ corresponds to $(f(1),f(2))$ and $\mathbf{g}$ corresponds to $(g(1),g(2))$, then $\mathbf{f}+\mathbf{g}$ should correspond to $(f(1),f(2))+(g(1),g(2)) = (f(1)+g(1),f(2)+g(2))$. The function whose value at $1$ is $f(1)+g(1)$ and whose value at $2$ is $f(2)+g(2)$ is what we usually call the &quot;sum&quot; of $\mathbf{f}$ and $\mathbf{g}$, so $\mathbf{f}+\mathbf{g}$ just corresponds to the &quot;pointwise sum&quot; of $f$ and $g$: to find the value of $\mathbf{f}+\mathbf{g}$ at $i$, we find the value of $f$ and the value of $g$ and add them. Similarly, $\alpha\mathbf{f}$ should have value $\alpha f(1)$ at $1$ and $\alpha f(2)$ at $2$; that is, the &quot;pointwise scalar product&quot;. So in a very natural way, we can think of $\mathbb{R}^2$ as &quot;the set of all functions $f\colon{1,2}\to\mathbb{R}$&quot; instead of as &quot;the set of all ordered pairs $(x,y)$ with $x,y\in\mathbb{R}$.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It should be fairly easy to see that you can do the same thing with $\mathbb{R}^n$, using functions $f\colon\{1,2,\ldots,n\}\to\mathbb{R}$ to represent $n$-tuples (in fact, that is the &lt;em&gt;set-theoretic&lt;/em&gt; definition of an $n$-tuple!). A function $f\colon\{1,2,\ldots,n\}\to\mathbb{R}$ corresponds to the $n$-tuple $(f(1),f(2),\ldots,f(n))$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, this gives you the &quot;obvious&quot; thing to do with $\mathbb{R}^n$. What about other vector spaces, like, say, $\mathbf{P}_n[x]$, the vector space of all polynomials of degree at most $n$ with coefficients in $\mathbb{R}$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The key to dealing with these vector spaces is the fact that they have &lt;em&gt;bases&lt;/em&gt;. Fix a basis $\beta$ for $\mathbf{P}_n[x]$ (or more generally, for $\mathbf{V}$). If $\beta=\{\mathbf{v}_1,\ldots,\mathbf{v}_n\}$ (assuming finite dimensionality for now), then every vector $\mathbf{v}\in\mathbf{V}$ can be written &lt;strong&gt;uniquely&lt;/strong&gt; as a linear combination of elements of $\beta$:&#xA;$$\mathbf{v}=\alpha_1\mathbf{v}_1+\cdots + \alpha_n\mathbf{v}_n.$$&#xA;Thus, if we agree on $\beta$, then we can describe any vector in $\mathbf{V}$ by specifying the coefficients of the linear combination; this is the &quot;coordinate vector&quot; relative to $\beta$; that is, if $\mathbf{v}$ happens to be the linear combination mentioned above, then the coordinate vector of $\mathbf{v}$ with respect to $\beta$ is $[\mathbf{v}]_{\beta}$, with&#xA;$$[\mathbf{v}]_{\beta} = (\alpha_1,\ldots,\alpha_n).$$&#xA;And so, given $\beta$, there is a natural way to define a bijection $\mathbf{V}\leftrightarrow \mathbb{R}^n$, and hence to thinking of $\mathbf{V}$ as a &quot;set of functions&quot;. But in order to keep track of $\beta$, and make sure we are talking about the same bases, it makes a bit more sense in this case to make the vectors of $\mathbf{V}$ correspond to functions with domain $\beta$ rather than $\{1,2,\ldots,n\}$. that is, we make $\mathbf{v}$ correspond to the function $f_{\mathbf{v}}\colon\beta\to \mathbb{R}$ given by $f_{\mathbf{v}}(\mathbf{v}_i) = \alpha_i$, the coefficient of $\mathbf{v}_i$ in the linear combination that expresses $\mathbf{v}$ in terms of $\beta$. Thus, in a very natural way, we can think of the elements of $\mathbf{V}$ as &quot;functions from $\beta$ to $\mathbb{R}$&quot;; turns out pointwise addition and scalar multiplication corresponds to vector addition and vector scalar product in $\mathbf{V}$, so everything works out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can we do this &lt;em&gt;wholesale&lt;/em&gt;? Yes: fix a set $X$, and consider all functions $f\colon X\to\mathbb{R}$ with pointwise addition and pointwise multiplication. Then you get a vector space; and you can realize every vector space as &lt;em&gt;isomorphic&lt;/em&gt; to one of these vector spaces by picking a basis and doing the stuff we did above. Different bases will yield different &quot;realizations&quot; of the vector space (if $\beta\neq\gamma$ as sets, then thinking of $\mathbf{V}$ as &quot;all functions $\beta\to\mathbb{R}$&quot; makes it a &lt;em&gt;different object&lt;/em&gt; than if we think of $\mathbf{V}$ as &quot;all functions $\gamma\to\mathbb{R}$&quot;); but they will be &lt;em&gt;isomorphic as vector spaces&lt;/em&gt;: there is a natural bijection between them which is linear, so there is a way to &quot;translate&quot; from one to the other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason this all works is that every finite dimensional vector space is a &lt;strong&gt;free object&lt;/strong&gt; in the category of finite dimensional (real) vector spaces. (I'll talk a bit about infinite dimensional later). Take the category of finite dimensional real vector spaces, and let $S$ be a set. A &quot;free vector space on $S$&quot; is a (finite dimensional real) vector space $\mathbf{V}$, together with a set-theoretic function $i\colon S\to \mathbf{V}$ such that for any (finite dimensional real) vector spaces $\mathbf{W}$ and set-theoretic function $j\colon S\to\mathbf{W}$, there exists a unique linear transformation $T\colon \mathbf{V}\to\mathbf{W}$ such that $j = T\circ i$. A vector space $\mathbf{V}$ is &quot;free&quot; if there exists a set $S$ such that $\mathbf{V}$ is &quot;free on $S$&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since a linear transformation is completely determined by the image of a basis, and any selection of images of a basis defines a (unique) linear transformation, if you let $\beta$ be a basis for $\mathbf{V}$, and you let $i\colon\beta\to\mathbf{V}$ be the natural inclusion, then this gives you a free object: given any $\mathbf{W}$ and any $j\colon\beta\to\mathbf{W}$, there is a unique linear $T\colon\mathbf{V}\to\mathbf{W}$ that 'extends' $j$; that is, such that $j = T\circ i$. So $\mathbf{V}$ is free, because it is free on (any of) its basis. And since two sets are &quot;isomorphic&quot; in the category $\mathcal{S}et$ if and only if they are bijectable, and free objects are unique up to unique isomorphism, then two vectors spaces with bases of the same size are &quot;isomorphic up to unique isomorphism&quot; that identifies the bases. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem here is that in order to think of a vector space as &quot;free&quot;, you really need to think about the basis as well. That is, it is really &quot;vector-space-together-with-a-specific-basis&quot; that gives you a free object, and not the vector space by itself. Pick a different basis, you get a &lt;em&gt;different&lt;/em&gt; (though &lt;em&gt;isomorphic&lt;/em&gt;) free object.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Going to infinite dimensional vector spaces presents a bit more trouble; given an infinite basis $\beta$, it is not longer true that every function $f\colon\beta\to\mathbb{R}$ represents a vector (though every vector corresponds to a function), because linear combinations must be &lt;em&gt;finite&lt;/em&gt;; we only get the functions with &quot;finite support&quot; (only finitely many nonzero values). You can restrict yourself to there, and then things work out generally okay, with some caveats (for example, a linear transformation $\mathbf{V}\to\mathbf{W}$ is still completely and uniquely determined by an &lt;em&gt;arbitrary&lt;/em&gt; function $f\colon\beta\to\mathbf{W}$, where $\mathbf{W}$ is a basis for $\mathbf{V}$; this function need not have finite support). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you are looking at with this definition, though, is really the representability of vector spaces as Hom-sets. Given a fixed vector space $\mathbf{W}$, for any vector space $\mathbf{V}$ you can consider the set of &quot;homomorphisms&quot; of $\mathbf{V}$ to $\mathbf{W}$; this is the set $\mathcal{L}(\mathbf{V},\mathbf{W})$ of linear transformations from $\mathbf{V}$ to $\mathbf{W}$. You are probably aware that this set is in fact a vector space itself with pointwise addition and scalar multiplication, and that it has dimension $\dim(\mathbf{V})\dim(\mathbf{W})$. If you set $\mathbf{W}=\mathbb{R}$ (thought of as a vector space), then $\mathcal{L}(\mathbf{V},\mathbb{R})$ is isomorphic to $\mathbf{V}$, because it has the same dimension (this vector space is called the &lt;em&gt;dual&lt;/em&gt; of $\mathbf{V}$). And of course, $\mathcal{L}(\mathbf{V},\mathbb{R})$ is naturally bijectable with the collection of all &lt;em&gt;set-theoretic&lt;/em&gt; functions $\beta\to\mathbb{R}$, where $\beta$ is a fixed basis for $\mathbb{R}$. So the author is kind of looking at the dual spaces instead of looking at the spaces; this leads to some difficulties in defining linear transformations, because passing to the dual is &lt;em&gt;contravariant&lt;/em&gt;, as Qiaochu points out. &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-20T18:19:18.620" LastActivityDate="2011-01-20T18:19:18.620" CommentCount="7" />
  <row Id="18328" PostTypeId="2" ParentId="12931" CreationDate="2011-01-20T17:10:46.293" Score="1" ViewCount="" Body="&lt;p&gt;Using a discrete Fourier transformation seems more appropriate if the $a_i$ are integers.  The number of solutions is the value of the convolution product&#xA;$$&#xD;&#xA;\left(\delta_{a_0} + \delta_{-a_0}\right) * \left(\delta_{a_1} + \delta_{-a_1}\right) * ... * \left(\delta_{a_{n-1}} + \delta_{-a_{n-1}}\right)&#xD;&#xA;$$&#xA;evaluated at 0.  This has the obvious dynamic programming solution, which takes something like $O(n A)$ time, where $A = \sum_i |a_i|$; i.e., it is a quasi-polynomial-time algorithm.  If the $a_i$ are chosen at random from (bounded) distributions of integers instead, then the dynamic programming solution takes $O(n A^2)$ to calculate the expected number of solutions.  The discrete Fourier transform translates that problem to a direct product of $n$ functions over $O(A)$ points, each of which can be computed in $O(A \log A)$ using the FFT, reducing the total time to $O(nA \log A)$.  Moreover, the number of &quot;approximate&quot; partitions, i.e., choices of the $\sigma_i$ such that $|\sum_{i=0}^{n-1}\sigma_i a_i| \le k$ for some small $k$, falls out of this calculation essentially for free.&lt;/p&gt;&#xA;" OwnerUserId="5546" LastActivityDate="2011-01-20T17:10:46.293" CommentCount="1" />
  <row Id="18329" PostTypeId="1" CreationDate="2011-01-20T17:19:21.597" Score="0" ViewCount="164" Body="&lt;p&gt;There are several equivalent ways of defining a function. We know that a differentiable function  $f : \mathbb{R} \to \mathbb{R}$ is uniquely defined when its values are specified at every point in $\mathbb{R}$. Now the question is : Is the derivative of such a function $f$ always unique ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS: Pardon me if its a very trivial question !&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT 1: the definition of the derivative is same as usual...i mean that given in the answer by Jonas Meyer and so is the definition of differentiability.&lt;/p&gt;&#xA;" OwnerUserId="2987" LastEditorUserId="2987" LastEditorDisplayName="" LastEditDate="2011-01-20T17:34:50.297" LastActivityDate="2011-01-20T18:22:15.947" Title="Uniqueness of the derivative of a function $f : \mathbb{R} \to \mathbb{R}$" Tags="&lt;real-analysis&gt;" AnswerCount="2" CommentCount="5" />
  <row Id="18331" PostTypeId="2" ParentId="18329" CreationDate="2011-01-20T17:28:08.497" Score="4" ViewCount="" Body="&lt;p&gt;Given $f:\mathbb{R}\to\mathbb{R}$, &lt;em&gt;the&lt;/em&gt; derivative of $f$ at $x\in\mathbb{R}$, if it exists, is typically defined to be $f&amp;#39;(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}$.  Real limits are unique when they exist, so this unambiguously assigns (at most) one value to $f&amp;#39;(x)$.  Therefore the derivative is unique.  Assuming $f$ is everywhere differentiable, this means that $f&amp;#39;:\mathbb{R}\to\mathbb{R}$ is a (&quot;well-defined&quot;) function.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2011-01-20T17:28:08.497" CommentCount="6" />
  <row Id="18332" PostTypeId="2" ParentId="18329" CreationDate="2011-01-20T17:28:36.030" Score="2" ViewCount="" Body="&lt;p&gt;The answer is, if it exists the derivate is unique (unlike a primitive). Even a more abstract form of the derivate, the weak derivate is unique ( &lt;a href=&quot;http://en.wikipedia.org/wiki/Weak_derivative&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Weak_derivative&lt;/a&gt; ). A reason for this is the uniqueness of limits, as the derivate is basically a limit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However the converse is not true, a derivate doesn't have a unique function associated with it, see the example PEV gave you.&lt;/p&gt;&#xA;" OwnerUserId="3123" LastEditorUserId="3123" LastEditorDisplayName="" LastEditDate="2011-01-20T18:22:15.947" LastActivityDate="2011-01-20T18:22:15.947" CommentCount="3" />
  <row Id="18333" PostTypeId="1" AcceptedAnswerId="18355" CreationDate="2011-01-20T17:39:30.880" Score="2" ViewCount="100" Body="&lt;p&gt;Let $f:X\to Y$ be a finite map of varieties and let $BL_Z(Y)$ be the blow-up of a subscheme $Z\subset Y$. Is there a map $$\phi:BL_{f^{-1}(Z)}(X)\to  BL_{Z}(Y)?$$ If so, what can be said about $\phi$? Is it also finite?&lt;/p&gt;&#xA;" OwnerUserId="6064" LastActivityDate="2011-01-20T21:39:13.407" Title="functoriality of blow-ups" Tags="&lt;algebraic-geometry&gt;" AnswerCount="1" CommentCount="0" />
  <row Id="18334" PostTypeId="1" AcceptedAnswerId="18367" CreationDate="2011-01-20T18:03:46.797" Score="7" ViewCount="1084" Body="&lt;p&gt;Source link: &lt;a href=&quot;http://romvf.wordpress.com/2011/01/19/open-letter/&quot; rel=&quot;nofollow&quot;&gt;http://romvf.wordpress.com/2011/01/19/open-letter/&lt;/a&gt; (Referred from slashdot)&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The fact of existence of the polynomial algorithm for 3-SAT problem leads to a conclusion that P=NP.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Is there validity to this argument that there is a P=NP solution? How long till this can be verified? Will this take years to verify or more likely weeks?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Final update: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://cartesianproduct.wordpress.com/2011/02/27/no-proof-of-pnp-after-all-yet/&quot; rel=&quot;nofollow&quot;&gt;http://cartesianproduct.wordpress.com/2011/02/27/no-proof-of-pnp-after-all-yet/&lt;/a&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Vladimir Romanov has conceded that his published “proof” of P=NP is flawed and requires further work.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="3735" LastEditorUserId="3735" LastEditorDisplayName="" LastEditDate="2011-02-28T16:31:12.943" LastActivityDate="2011-02-28T22:19:33.203" Title="How valid is the proposed P=NP solution for the 3-SAT problem?" Tags="&lt;np-complete&gt;&lt;open-problem&gt;" AnswerCount="5" FavoriteCount="4" />
  <row Id="18335" PostTypeId="1" AcceptedAnswerId="18341" CreationDate="2011-01-20T18:22:15.773" Score="3" ViewCount="138" Body="&lt;p&gt;I found on Mathworld ( &lt;a href=&quot;http://mathworld.wolfram.com/Ellipse.html&quot; rel=&quot;nofollow&quot;&gt;http://mathworld.wolfram.com/Ellipse.html&lt;/a&gt; ) that the quadratic equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ax^2 + 2bxy + cy^2 + 2dx + 2fy + g = 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;represent an ellipse only when, after defining:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Delta = \left|\begin{array}{ccc} a&amp;amp;b&amp;amp;d\\b&amp;amp;c&amp;amp;f\\d&amp;amp;f&amp;amp;g\end{array}\right|$$&#xA;$$J = ac-b^2$$&#xA;$$I = a+c$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the following conditions are met: $\Delta \neq 0$, $J &amp;gt; 0$ and $\Delta/I &amp;lt; 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I don't understand are the constraint other than $J &amp;gt; 0$. I've never seen these constraints before, and I really wonder why they are here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, I noticed that the formulas to find the semi-axes and angle given after on the MathWorld article work only if these constraints are met.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second part of my question is then: is there an other way to find the length and orientation of the axis of an ellipse given its quadratic equation? Or is it always possible to transform the quadratic equation of an ellipse into an equivalent one respecting the constraints described on MathWorld?&lt;/p&gt;&#xA;" OwnerUserId="4709" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2011-04-07T06:23:09.923" LastActivityDate="2011-04-07T06:23:09.923" Title="Quadratic equation of an ellipse and ellipse description" Tags="&lt;analytic-geometry&gt;&lt;conic-sections&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="18336" PostTypeId="2" ParentId="3528" CreationDate="2011-01-20T18:28:57.717" Score="3" ViewCount="" Body="&lt;p&gt;The multinomial generalization mentioned by Qiaochu is conceptually simple but getting the details right is messy. The goal is to compute $$\int_0^1 \int_0^{1-t_1} \ldots \int_0^{1-t_1-\ldots-t_{k-2}} t_1^{n_1} t_2^{n_2} \ldots t_{k-1}^{n_{k-1}} t_k^{n_k} dt_1 \ldots dt_{k-1},$$ where $t_k = 1 - t_1 - \ldots - t_{k-1},$ for nonnegative integers $n_1, \ldots, n_k$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Draw $k-1 + \sum_{i = 1}^{k}n_k$ numbers $X_1, \ldots, X_{k-1 + \sum_{i = 1}^{k}n_k}$ independently from a uniform $[0,1]$ distribution. Define $X_0 = 0$ and $X_{k + \sum_{i = 1}^{k}n_k} = 1$ for convenience. Let $E$ be the event that the numbers $X_1$ through $X_{k-1}$ are in ascending order and that the numbers $X_{j + \sum_{i = 1}^{j-1} n_i}$ through $X_{j + \sum_{i = 1}^{j}n_i - 1}$ are between $X_{j-1}$ and $X_j$ for $j = 1, \ldots, k$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Define a linear transformation from $(X_1, \ldots, X_{k-1}) \to (T_1, \ldots, T_{k-1})$ by $T_i = X_i - X_{i-1}$ for $i = 1, \ldots, k-1$. Note that the determinant of this linear transformation is 1 and it is therefore measure-preserving. Given values of $X_1$ through $X_{k-1}$, the conditional probability of $E$ is &#xA;$$\mathbb{P}[E|(X_1, \ldots, X_{k-1}) = (x_1, \ldots, x_{k-1})] = \prod_{i = 1}^{k}(x_i - x_{i-1})^{n_k} \mathbf{1}_{\{x_i &amp;gt; x_{i-1}\}}.$$ Marginalizing with respect to the distribution of $X_1 \times \ldots \times X_{k-1}$ gives&#xA;$$\begin{aligned}&#xD;&#xA;\mathbb{P}[E] &amp;amp;= \int_{0}^1 \ldots \int_{0}^1 \prod_{i = 1}^{k}(x_i - x_{i-1})^{n_k} \mathbf{1}_{\{x_i &amp;gt; x_{i-1}\}} p_{X_1 \times \ldots \times X_{k-1}}(x_1, \ldots, x_{k-1}) dx_{k-1} \ldots dx_{1} \\&#xD;&#xA;&amp;amp;= \int_{0}^1 \int_{-t_1}^{1-t_1} \ldots \int_{-t_1 - \ldots - t_{k-1}}^{1 -t_1 - \ldots - t_{k-1}} \prod_{i = 1}^{k} t_k^{n_k} \mathbf{1}_{\{t_k &amp;gt; 0\}} p_{T_1 \times \ldots \times T_{k-1}}(t_1, \ldots, t_{k-1}) dt_{k-1} \ldots dt_{1} \\&#xD;&#xA;&amp;amp;= \int_0^1 \int_0^{1-t_1} \ldots \int_0^{1-t_1-\ldots-t_{k-2}} t_1^{n_1} \ldots t_{k-1}^{n_{k-1}} t_k^{n_k} dt_{k-1} \ldots dt_{1},&#xD;&#xA;\end{aligned}$$&#xA;so if we can compute $\mathbb{P}[E]$ combinatorially we will have evaluated the desired intergral. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $\{R_i\}_{i \in \{1, \ldots, k-1 + \sum_{i = 1}^{k}n_k\}}$ be the ranks that the numbers $\{X_i\}_{i \in \{1, \ldots, n+m+1\}}$ would have if sorted in ascending order. (Note that the numbers are all distinct with probability 1). Since the numbers were drawn independently from a uniform distribution, the ranks are a random permutation of the integers $1$ through $k-1 + \sum_{i = 1}^{k}n_k$. Note that $E$ is exactly the event that $R_j = j + \sum_{i = 1}^j n_i$ for $j \in \{1, \ldots, k-1\}$ and that for each $l \in \{1, \ldots, k\}$, $$R_j \in \{l + \sum_{i = 1}^{l-1} n_i, \ldots, l + \sum_{i=1}^{l}n_i - 1\}$$ for $$j \in \{k+\sum_{i = 1}^{l-1}n_i, \ldots, k + \sum_{i = 1}^{l}n_i - 1\}.$$ There are $n_1!\ldots n_k!$ possible permutations which satisfy these conditions out of $(\sum_{i=1}^{k}n_i+k-1)!$ total possible permuations, so $$\mathbb{P}[E] = \frac{n_1!\ldots n_k!}{(\sum_{i=1}^{k}n_i+k-1)!}.$$&lt;/p&gt;&#xA;" OwnerUserId="1789" LastEditorUserId="1789" LastEditorDisplayName="" LastEditDate="2011-01-20T18:35:31.860" LastActivityDate="2011-01-20T18:35:31.860" />
  <row Id="18337" PostTypeId="2" ParentId="18335" CreationDate="2011-01-20T18:33:59.287" Score="0" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://www.maa.org/joma/Volume8/Kalman/QuadForm.html&quot; rel=&quot;nofollow&quot;&gt;This&lt;/a&gt; link will help (a discussion about quadratic forms).&lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2011-01-20T18:33:59.287" />
  <row Id="18338" PostTypeId="2" ParentId="18334" CreationDate="2011-01-20T18:39:38.937" Score="7" ViewCount="" Body="&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;I won't try to address the 'Is there validity to this argument...?' except to say that it is number 65 out of 67 at the moment on &lt;a href=&quot;http://www.win.tue.nl/~gwoegi/P-versus-NP.htm&quot; rel=&quot;nofollow&quot;&gt;The P-versus-NP page&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;My guess is that, if the proof doesn't work, it will take a couple of weeks (if people who are able to judge care enough), but much longer the more substance it has. The last really serious proof attempt (Deolalikar, last August) took about a month before people ( a lot of serious computational complexity researchers) decided there were too many problems to bother continuing.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;But really, no matter what truth lies behind the claims, pretty much everybody will be expecting this to fail.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1919" LastActivityDate="2011-01-20T18:39:38.937" CommentCount="3" />
  <row Id="18339" PostTypeId="1" AcceptedAnswerId="18350" CreationDate="2011-01-20T18:43:33.310" Score="3" ViewCount="112" Body="&lt;p&gt;I have been reading a bit about the Bochner integral and now I'm wondering the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the theory to be &quot;nice&quot; on would expect that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$L^1([0, \tau], L^1([0, \tau])) \cong L^1([0,\tau] \times [0, \tau]).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this the case? How do we prove this (or where can I find more about this), since if we take $u$ in the set in the LHS, then we can evaluate $(u(t))(x)$ and we want to map this to some $u(t,x)$. The problem I see is that $L^1$-functions are equivalence classes modulo sets of measure zero. So for each $t$ we have different sets of measure zero than the ones in $[0,\tau]^2$. How do we get around this?&lt;/p&gt;&#xA;" OwnerUserId="1120" LastActivityDate="2011-01-20T20:16:38.987" Title="$L^1$ space with values in a Banach Space." Tags="&lt;functional-analysis&gt;" AnswerCount="3" />
  <row Id="18340" PostTypeId="1" AcceptedAnswerId="18347" CreationDate="2011-01-20T18:55:17.373" Score="13" ViewCount="522" Body="&lt;p&gt;Edge-attaching many hexagons results in a plane. Edge-attaching pentagons yields a dodecahedron.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there some insight into why the alternation of pentagons and hexagons yields an approximated sphere? Is this special, or are there an arbitrary number of assorted &lt;em&gt;n-gons&lt;/em&gt; sets that may be joined together to create regular sphere-like surfaces?&lt;/p&gt;&#xA;" OwnerUserId="4241" LastActivityDate="2011-01-21T03:11:07.830" Title="Why are there 12 pentagons and 20 hexagons on a soccer ball?" Tags="&lt;topology&gt;&lt;3d&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="5" />
  <row Id="18341" PostTypeId="2" ParentId="18335" CreationDate="2011-01-20T18:55:58.090" Score="3" ViewCount="" Body="&lt;p&gt;The condition $\Delta/I &amp;lt; 0$ is necessary for the solution set to exist at all. For example, consider $x^2 + y^2 + 1 = 0$, which has no solutions; here $\Delta = 1$ and $I = 2$. I think the necessity of this condition can be shown using linear algebra; leave a comment if you'd like to see the proof.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unless I'm missing something, $\Delta \neq 0$ is redundant, as it is implied by $\Delta/I &amp;lt; 0$.&lt;/p&gt;&#xA;" OwnerUserId="856" LastActivityDate="2011-01-20T18:55:58.090" CommentCount="3" />
  <row Id="18342" PostTypeId="2" ParentId="18334" CreationDate="2011-01-20T19:01:25.260" Score="5" ViewCount="" Body="&lt;p&gt;The problem with such a claim is that there are so many such claims made every year, not just for the P vs NP problem but other open problems as well. I am not an expert in this topic, so I won't comment on the content of the paper you have linked to. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to check claims of this sort, an expert will probably have to spend at least a few days (perhaps longer) to give a cursory reading of the paper. If it satisfies them at first glance (not a very easy task in itself), they will read through the arguments in the manuscript in detail and try find any possible flaws. Given the proliferation of such claims and how they have turned out, how will you convince an expert to invest the time and effort to give a basic reading of the manuscript first? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Check out this blog post by Scott Aaronson &lt;a href=&quot;http://www.scottaaronson.com/blog/?p=122&quot; rel=&quot;nofollow&quot;&gt;http://www.scottaaronson.com/blog/?p=122&lt;/a&gt; . Does the manuscript you have linked to have an answer for most if not all the points that he is raising? Such a checklist can be helpful for people to decide whether a claim is worth checking in detail.&lt;/p&gt;&#xA;" OwnerUserId="2641" LastEditorUserId="2641" LastEditorDisplayName="" LastEditDate="2011-01-21T09:41:26.607" LastActivityDate="2011-01-21T09:41:26.607" CommentCount="1" />
  <row Id="18343" PostTypeId="2" ParentId="18339" CreationDate="2011-01-20T19:04:37.483" Score="4" ViewCount="" Body="&lt;p&gt;You can define the obvious isomorphism $T$ on representatives, which is isometric by Fubini's theorem.  Isometric maps defined on representatives are well defined on equivalence classes, because if $u_1$ and $u_2$ are representatives of an element of the LHS, then $\|Tu_1-Tu_2\|=\|T(u_1-u_2)\|=\|u_1-u_2\|=0$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is a useful fact that a linear map initially defined on some type of representative can often be seen to be well defined from the fact that it is continuous.  In this case, all the work is hidden in Fubini's theorem.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2011-01-20T19:04:37.483" CommentCount="1" />
  <row Id="18344" PostTypeId="2" ParentId="18340" CreationDate="2011-01-20T19:05:19.623" Score="8" ViewCount="" Body="&lt;p&gt;The soccer ball shape can be formed by &lt;a href=&quot;http://en.wikipedia.org/wiki/Truncated_icosahedron&quot; rel=&quot;nofollow&quot;&gt;truncating an icosahedron&lt;/a&gt;—if each of the 12 vertices is cut off, it will leave a pentagonal face and cause each of the 20 triangular faces to become hexagonal.&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2011-01-20T19:05:19.623" />
  <row Id="18345" PostTypeId="2" ParentId="15438" CreationDate="2011-01-20T19:06:13.173" Score="1" ViewCount="" Body="&lt;p&gt;This whitepaper on &lt;a href=&quot;http://www.salsburg.com/geod/geodesicmath.pdf&quot; rel=&quot;nofollow&quot;&gt;Geodesic Math&lt;/a&gt; may be helpful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Probably less helpful is this Ruby Quiz I hosted on &lt;a href=&quot;http://www.rubyquiz.com/quiz3.html&quot; rel=&quot;nofollow&quot;&gt;writing a program to calculate Geodesic spheres&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="4241" LastActivityDate="2011-01-20T19:06:13.173" />
  <row Id="18346" PostTypeId="2" ParentId="18301" CreationDate="2011-01-20T19:08:48.980" Score="8" ViewCount="" Body="&lt;p&gt;This was shown by Hardy back in 1903/1904.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A mention of it can be found here: &lt;a href=&quot;http://books.google.com/books?id=rgULAAAAYAAJ&amp;amp;pg=PA203&quot; rel=&quot;nofollow&quot;&gt;Quarterly Journal Of Pure And Applied Mathematics, Volume 35, Page 203&lt;/a&gt;, which is somewhere in the middle of a long paper.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a snapshot in case that link does not work:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/6WQ8w.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note, the integral is slightly different, but I suppose it won't be too hard to convert it into the form you have.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See also Hardy's response to Ramanujan here: &lt;a href=&quot;http://books.google.com/books?id=Of5G0r6DQiEC&amp;amp;pg=PA46&quot; rel=&quot;nofollow&quot;&gt;http://books.google.com/books?id=Of5G0r6DQiEC&amp;amp;pg=PA46&lt;/a&gt;. Note: 1b.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;(Edit:) Since the journal itself has no reliable electronic copies, and the proof is actually somewhat more involved then just the excerpt shown above, I'll give a quick description of the proof that Hardy provided. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;First is the concept of &lt;em&gt;reciprocal functions of the first and second kind&lt;/em&gt; introduced by Cauchy. Two functions $\phi$ and $\psi$ defined on the positive real line is called reciprocal functions of the first kind if $$\phi(y) = \sqrt{\frac{2}{\pi}} \int_0^\infty \cos(y x) \psi(x) dx$$ and also the same formula with $\phi$ and $\psi$ swapped. They are called reciprocal functions of the second kind if the $\cos$ in the formula above is replaced by $\sin$. Cauchy gave several examples of each, and also examples of functions which are their own reciprocal function of the first kind (but not for the second), and proved that those functions have the following property: whenever $\alpha \beta = \pi$ $$\sqrt\alpha \left( \frac12 \phi(0) + \phi(\alpha) + \phi(2\alpha) + \cdots\right) = \sqrt\beta \left(\frac12 \psi(0) + \psi(\beta) + \psi(2\beta) + \cdots \right)$$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;In the article linked above, Hardy proved the following two facts (among others).&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The function $f(x) = e^{x^2/2}\int_x^\infty e^{-t^2/2}dt$ is its own reciprocal function of the second kind. (That proof is about 3 pages long, condensed in the typical Hardy fashion.)&lt;/li&gt;&#xA;&lt;li&gt;If $\phi$ and $\psi$ are reciprocal functions of the second kind, the following summation formula (analogue of the one above for functions of the first kind) holds: when $\lambda \mu = 2\pi$, one has $$ \sqrt\lambda \sum_0^\infty (-1)^n \phi\left( (n + \frac12)\lambda\right) = \sqrt\mu \sum_0^\infty (-1)^n \psi\left( (n+\frac12)\mu\right)$$ This expression being the one termed equation (9) in the screenshot above. &lt;/li&gt;&#xA;&lt;/ol&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Hardy provided two proves of the formula asked about above in the question. The first proof proceeds by giving the series expansion $$\int_0^\infty \frac{e^{-\alpha x^2}}{\cosh \pi x} dx = \frac{2}{\pi} \sum (-1)^n F\left( (n + \frac12)\alpha\right)$$ where $$F(x) = \sqrt\pi e^{x^2}\int_x^\infty e^{-t^2}dt$$ and using equation (9) above. The second proof is shown in section 10 in the image above: he obtained a &lt;em&gt;different&lt;/em&gt; series expansion of the expression we want on the left hand side, which can be shown to be term by term equal to the &lt;em&gt;first&lt;/em&gt; series expansion of the expression on the right hand side, avoiding the need to invoke equation (9). &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1543" LastEditorDisplayName="" LastEditDate="2011-01-21T12:54:44.543" LastActivityDate="2011-01-21T12:54:44.543" CommentCount="5" />
  <row Id="18347" PostTypeId="2" ParentId="18340" CreationDate="2011-01-20T19:10:12.927" Score="19" ViewCount="" Body="&lt;p&gt;The possible ways to put polygons together to form a sphere-like object are constrained by &lt;a href=&quot;http://en.wikipedia.org/wiki/Euler_characteristic&quot; rel=&quot;nofollow&quot;&gt;Euler's formula&lt;/a&gt; $V - E + F = 2$ (where $V$ is the number of vertices, $E$ is the number of edges, and $F$ is the number of faces).  Equivalently you can think of this as a statement about &lt;a href=&quot;http://en.wikipedia.org/wiki/Planar_graph&quot; rel=&quot;nofollow&quot;&gt;planar graphs&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose we use $f_3$ triangles, $f_4$ squares, $f_5$ pentagons, etc.  Every edge meets exactly two faces, and an edge of type $f_n$ meets $n$ faces, so let's double-count the number of pairs of an edge and a face next to it: on the one hand, this is $2E$, and on the other hand, this is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$3f_3 + 4f_4 + 5f_5 + ...$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Plugging this into Euler's formula gives $V - \frac{f_3 + 2f_4 + 3f_5 + ...}{2} = 2$.  If in addition the polyhedron is &lt;em&gt;convex&lt;/em&gt; and the polygons are &lt;em&gt;regular&lt;/em&gt;, there are constraints on the faces that can meet at each vertex coming from the fact that the angles must sum up to less than $360^{\circ}$.  (This is one way to prove the classification of Platonic solids.)  For example, at most $5$ faces can meet at each vertex if we allow arbitrary faces; this means $3f_3 + 4f_4 + ... \le 5V$.  (If you really want to, you can allow six triangles to touch at one point, but I would just count this as a hexagon.)  If we don't allow triangles, &lt;em&gt;exactly&lt;/em&gt; $3$ faces meet at each vertex; this means $4f_4 + 5f_5 + ... = 3V$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is an application in chemistry: a &lt;a href=&quot;http://en.wikipedia.org/wiki/Fullerene&quot; rel=&quot;nofollow&quot;&gt;fullerene&lt;/a&gt; is a certain type of molecule made from carbon atoms.  (One of these, the buckyball, looks just like a soccer ball.)  It gives a convex polyhedron in which each face is either a regular pentagon or hexagon.  This gives $V - \frac{3f_5 + 4f_6}{2} = 2$ on the one hand, and $3V = 5f_5 + 6f_6$ on the other.  Together these equations give $f_5 = 12$ and $V - 2f_6 = 20$; in other words, any fullerene must have &lt;em&gt;exactly twelve pentagons&lt;/em&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Hexagons are special.  One way to interpret this result is that an infinite plane can be tiled with hexagons, so hexagons correspond to zero curvature, whereas since pentagons have a smaller angle at each vertex they correspond to positive curvature.  What the above statement says, roughly, is that the total amount of curvature is a constant.  This is a simple form of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Gauss%E2%80%93Bonnet_theorem&quot; rel=&quot;nofollow&quot;&gt;Gauss-Bonnet theorem&lt;/a&gt;, which is closely related to Euler's formula.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are some other things you can prove, again under the assumptions of convexity and regularity:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If you only use triangles (as opposed to triangles and hexagons), then $f_3 \le 20$ and $4 | f_3$.&lt;/li&gt;&#xA;&lt;li&gt;If you only use squares, then $f_4 = 6$.&lt;/li&gt;&#xA;&lt;li&gt;You cannot use only hexagons or higher.  (In the Gauss-Bonnet picture, attempting to use heptagons corresponds to negative curvature, and negative curvature does not interact well with Euclidean or spherical geometry; the natural setting for putting heptagons together is instead &lt;a href=&quot;http://en.wikipedia.org/wiki/Hyperbolic_geometry&quot; rel=&quot;nofollow&quot;&gt;hyperbolic geometry&lt;/a&gt;.)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This is already most of the way to the classification of Platonic solids.  If you're interested in learning more about Euler's formula, I highly recommend David Richeson's &lt;a href=&quot;http://www.eulersgem.com/&quot; rel=&quot;nofollow&quot;&gt;Euler's Gem&lt;/a&gt;.  Extremely well-written and informative.  You might also enjoy David Eppstein's &lt;a href=&quot;http://www.ics.uci.edu/~eppstein/junkyard/euler/&quot; rel=&quot;nofollow&quot;&gt;Nineteen ways to prove Euler's formula&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2011-01-20T23:10:07.093" LastActivityDate="2011-01-20T23:10:07.093" CommentCount="8" />
  <row Id="18348" PostTypeId="2" ParentId="18284" CreationDate="2011-01-20T19:13:26.997" Score="6" ViewCount="" Body="&lt;p&gt;You can get the same result using the &lt;a href=&quot;http://mathworld.wolfram.com/FaadiBrunosFormula.html&quot; rel=&quot;nofollow&quot;&gt;Fa&amp;agrave; di Bruno formula&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/Bell_polynomials&quot; rel=&quot;nofollow&quot;&gt;Bell polynomials&lt;/a&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Fa&amp;agrave; di Bruno formula says that $$\frac{d^n}{dx^n} f(g(x)) = \sum_{k=0}^n f^{(k)}(g(x)) B_{n,k}\left(g&amp;#39;(x),g&amp;#39;&amp;#39;(x),\dots,g^{(n-k+1)}(x)\right),$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $B_{n,k}(x_1,x_2,\dots,x_{n-k+1})$ is a partial Bell polynomial.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $f(x) = e^x$, and $g(x) = 1/x$, we have &#xA;$$\frac{d^n}{dx^n} e^{1/x} = \sum_{k=0}^n e^{1/x} B_{n,k}\left(-x^{-2},2x^{-3},\dots,(-1)^{n-k+1}(n-k+1)!x^{-n+k-2}\right).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is known that $B_{n,k}(1!,2!,\dots,(n-k+1)!) = \binom{n}{k}\binom{n-1}{k-1} (n-k)!$, a &lt;a href=&quot;http://en.wikipedia.org/wiki/Lah_number&quot; rel=&quot;nofollow&quot;&gt;Lah number&lt;/a&gt; (see Wikipedia on &lt;a href=&quot;http://en.wikipedia.org/wiki/Bell_polynomials#Properties&quot; rel=&quot;nofollow&quot;&gt;Bell polynomials&lt;/a&gt; or Comtet's &lt;a href=&quot;http://books.google.com/books?id=C0HPgWhEssYC&amp;amp;pg=PA135&amp;amp;lpg=PA135&amp;amp;dq=advanced+combinatorics+lah+numbers&amp;amp;source=bl&amp;amp;ots=rEixo-VmD3&amp;amp;sig=Y0XdAFDiZgAheKIsLrHP8Y5MYjI&amp;amp;hl=en&amp;amp;ei=5YY4Tc2iNYHQsAPv8cSLAw&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=4&amp;amp;ved=0CCwQ6AEwAw#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;Advanced Combinatorics&lt;/a&gt;, p. 135).  Thus (keeping track of signs and powers of $x$),&#xA;$$\frac{d^n}{dx^n} e^{1/x} = (-1)^n e^{1/x} \sum_{k=0}^n \binom{n}{k}\binom{n-1}{k-1} (n-k)! x^{-n-k},$$&#xA;which is just your formula with an index change (and where $\binom{n-1}{-1} = 0$).&lt;/p&gt;&#xA;" OwnerUserId="2370" LastActivityDate="2011-01-20T19:13:26.997" CommentCount="1" />
  <row Id="18349" PostTypeId="2" ParentId="18339" CreationDate="2011-01-20T19:27:03.973" Score="1" ViewCount="" Body="&lt;p&gt;Since I am not able to comment yet, I write this as an answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Be prepared to face the fact that $L^\infty([0,T],L^\infty([0,T])) \neq L^\infty([0,T]\times[0,T])$. To see this, observe that the function&#xA;$$&#xD;&#xA;f(x,y) = \begin{cases} 1 &amp;amp; x&amp;gt;y\\ 0 &amp;amp; \text{else}\end{cases}&#xD;&#xA;$$&#xA;is not an element of the first space.&lt;/p&gt;&#xA;" OwnerUserId="3148" LastActivityDate="2011-01-20T19:27:03.973" />
  <row Id="18350" PostTypeId="2" ParentId="18339" CreationDate="2011-01-20T20:11:28.880" Score="4" ViewCount="" Body="&lt;p&gt;There is no need to restrict to intervals, the same holds true for measure spaces $X$ and $Y$ and Banach spaces $E$ in general:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$L^{1}(X \times Y, E) \cong L^{1}(X,L^{1}(Y,E))$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;By definition each Bochner-integrable function can be approximated by simple functions. In other words, the functions of the form $[A]e$ for some measurable subset $A \subset X$ of finite measure and $e \in E$ (or out of some dense subspace of $E$, if you prefer) generate a dense subspace of $L^{1}(X,E)$. Therefore you can reduce to the case of defining a bijection $[A]\cdot ([B] \cdot e) \leftrightarrow [A \times B]\cdot e$ and observe that this is a bijection on sets generating dense subspaces of $L^{1}(X, L^{1}(Y,E))$ and $L^{1}(X \times Y, E)$ and extend it linearly. That it is well-defined and an isometry is a special case of Fubini, that it is surjective in both directions follows from density.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The general fact underlying this is the canonical isomorphism&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$L^{1}(Y,E) \cong L^{1}(Y) \hat{\otimes} E$ (projective tensor product)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and the isomorphism $L^{1}(X \times Y) \cong L^{1}(X) \hat{\otimes} L^{1}(Y)$ for $L^{1}$-spaces + associativity of the tensor product (and all this is proved in exactly the same way).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, a remark on user3148's cautionary counterexample. There is an isomorphism $(L^{1}(X,E))^{\ast} \cong L_{w^{\ast}}^{\infty}(X,E^{\ast})$ where the latter space is defined by weak$^{\ast}$-measurably in the sense of Gelfand-Dunford. So in this sense we have $L_{w^{\ast}}^{\infty}(X \times Y, E^{\ast}) \cong L_{w^{\ast}}^{\infty}(X, L_{w^{\ast}}^{\infty}(Y,E^{\ast}))$ simply by duality theory.&lt;/p&gt;&#xA;" OwnerUserId="5363" LastEditorUserId="5363" LastEditorDisplayName="" LastEditDate="2011-01-20T20:16:38.987" LastActivityDate="2011-01-20T20:16:38.987" CommentCount="4" />
  <row Id="18351" PostTypeId="2" ParentId="18319" CreationDate="2011-01-20T20:12:28.027" Score="1" ViewCount="" Body="&lt;p&gt;You could do this as well. Write $x = 5+h$. So as $x \to 5$, $h \to 0$. Therefore the required limit is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align*}&#xA;\lim_{h \to 0 } \frac{2^{5+h}-2^{5}}{h} &amp;amp;= 2^{5} \cdot \lim_{h \to 0}\Bigl[ \frac{2^{h}-1}{h}\Bigr] \\ &amp;amp;= 2^{5} \cdot M&#xA;\end{align*}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $M$= Some value in $\log$. I forgot the formula. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;ADDED:&lt;/strong&gt; The formula is $$\lim_{x \to 0} \frac{a^{x}-1}{x} = \log{a}$$&lt;/p&gt;&#xA;" LastActivityDate="2011-01-20T21:25:29.530" CommentCount="5" />
  <row Id="18352" PostTypeId="1" CreationDate="2011-01-20T21:23:19.317" Score="10" ViewCount="279" Body="&lt;p&gt;Melvyn Nathanson, in his book &lt;em&gt;Methods in Number Theory&lt;/em&gt; &lt;strong&gt;(Chapter 8: Prime Number's)&lt;/strong&gt; states the following:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A conjecture of Schinzel and Sierpinski asserts that every positive rational number $x$ can be represented as a quotient of shifted primes, that $x=\frac{p+1}{q+1}$ for primes $p$ and $q$. It is known that the set of shifted primes, generates a subgroup of the &lt;em&gt;multiplicative group of rational numbers&lt;/em&gt; of index at most $3$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I would like to know &lt;em&gt;what progress has been made&lt;/em&gt; regarding this problem and why is this &lt;em&gt;conjecture&lt;/em&gt; important. Since it generates a subgroup, does the subgroup which it generates have any special properties? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I had actually posed a problem which asks us to prove that given any interval $(a,b)$ there is a rational of the form $\frac{p}{q}$ ($p,q$ primes) which lies inside $(a,b)$. Does, this problem have any connections with the actual conjecture? &lt;/p&gt;&#xA;" OwnerUserId="9413" LastEditorUserId="9413" LastEditorDisplayName="anonymous" LastEditDate="2011-05-10T06:04:53.587" LastActivityDate="2011-05-10T06:04:53.587" Title="A Conjecture of Schinzel and Sierpinski" Tags="&lt;number-theory&gt;&lt;prime-numbers&gt;&lt;analytic-number-theory&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="2" />
  <row Id="18353" PostTypeId="1" CreationDate="2011-01-20T21:38:21.287" Score="3" ViewCount="99" Body="&lt;p&gt;Hello. Let $\tau$ be the usual topology on the real line $\mathbb{R}$. Does there exists a topology $\tau_{0} \subset \tau$ such that $(\mathbb{R},\tau_{0})$ is homeomorphic to the &#xA;figure eight? Also, is it possible to find a topology $\tau_{0} \subset \tau$ and a quotient space with this topology such that this space is homeomorphic to $\mathbb{R}$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What's the trick for this one?&lt;/p&gt;&#xA;" OwnerUserId="6033" LastActivityDate="2011-01-20T22:23:05.863" Title="Finding coarser topology and homeomorphism between R and two tangent circles" Tags="&lt;topology&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="18354" PostTypeId="2" ParentId="6208" CreationDate="2011-01-20T21:38:36.793" Score="2" ViewCount="" Body="&lt;p&gt;Jonas Meyer's answer goes much deeper, but let me say the following as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even if your (nonzero -- you didn't say that, but of course you meant it!) function does extend continuously to the closed unit disk it need not have isolated zeros.  Indeed, suppose that $f$ has infinitely many zeros in the open unit disk.  Then the zero set of the (putative) extension of $f$ to the closed unit disk is an infinite subset of a compact space, so must have an accumulation point: i.e., there will be at least one point on the boundary which is a nonisolated zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, by the Weierstrass Factorization Theorem, the zero set of an analytic function on the open unit disk can be &lt;em&gt;any&lt;/em&gt; discrete subset without accumulation points in the open disk.  You can choose such a set to have the entire boundary $|z| = 1$ of the disk contained in its closure.  We conclude that there exists a nonzero analytic function $f$ on the open unit disk such that -- &lt;strong&gt;if&lt;/strong&gt; it admits a continuous extension to the closed unit disk -- this extension is identically zero on the boundary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would be interested to know if this construction can be made unconditional, i.e., whether any analytic function with such a zero set can be extended continuously to the boundary.&lt;/p&gt;&#xA;" OwnerUserId="299" LastActivityDate="2011-01-20T21:38:36.793" CommentCount="5" />
  <row Id="18355" PostTypeId="2" ParentId="18333" CreationDate="2011-01-20T21:39:13.407" Score="3" ViewCount="" Body="&lt;p&gt;You need more hypotheses for such a map to exist.  For example, suppose&#xA;that $X = pt$ and $Y = \mathbb A^2$ (over a field $k$, say) and $f$ is the map sending the point to the origin.  Now take $Z$ to be the origin of $Y$.  The preimage of $Z$ is all of $X$, and so the blow-up of $X$ at $f^{-1}(Z)$ is just $Z$ again.  On the other hand, there is no natural map from $X$ to the blow-up of $Y$ along $Z$.  (If there were we could pick out a distinguished point on the exceptional divisor, which we can't.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $f$ is finite and flat, then unless I am blundering the blow-up of $X$ along $f^{-1}(Z)$ will precisely equal the fibre product over $Y$ of the blow-up of $Y$ along $Z$ with $X$.  So in this case, there is a map $\phi$, it is just a base-change, and in particular it is again finite and flat.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you like, the problem is that Proj (which is what appears in the definition of blow-ups) is not functorial in the most naive way.&lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2011-01-20T21:39:13.407" />
  <row Id="18356" PostTypeId="1" AcceptedAnswerId="18373" CreationDate="2011-01-20T21:43:19.567" Score="11" ViewCount="290" Body="&lt;p&gt;As a physics student, I have occasionally run across the gamma function&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Gamma(n) \equiv \int_0^{\infty}t^{n-1}e^{-t} \textrm{d}t = (n-1)!$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;when we want to generalize the concept of a factorial.  Why not define the gamma function so that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Gamma(n) = n!$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;instead?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I realize either definition is equally good, but if someone were going to ask me to choose one, I would choose the second option.  Are there some areas of mathematics where the accepted definition looks more natural?  Are there some formulas that work out more cleanly with the accepted definition?&lt;/p&gt;&#xA;" OwnerUserId="3305" LastActivityDate="2011-01-21T11:17:07.947" Title="Why isn't the gamma function defined so that $\Gamma(n) = n! $ ?" Tags="&lt;analysis&gt;&lt;soft-question&gt;&lt;gamma-function&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="18357" PostTypeId="2" ParentId="18315" CreationDate="2011-01-20T22:04:38.253" Score="3" ViewCount="" Body="&lt;p&gt;This is not the definition of an arbitrary vector space, for (at least) two reasons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) The requirement that $S$ be finite means your vector space -- let me call it $F^{(S)}$ -- is a finite-dimensional vector space.  So infinite-dimensional vector spaces are missing.  For this, one can extend the construction to an arbitrary set $S$, but we need to slightly redefine $F^{(S)}$ to be the set of all functions $f: S \rightarrow F$ such that $f^{-1}(F \setminus \{0\})$ is finite (&lt;strong&gt;finitely nonzero functions&lt;/strong&gt;).  Note that this is precisely the definition in the planetmath article on Free Vector Spaces that you link to.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) This definition captures all vector spaces &lt;em&gt;up to isomorphism&lt;/em&gt; only.  It is much closer to the concept of &quot;vector space with distinguished basis&quot;.  Indeed, if $V$ is an $F$-vector space, then taking $S$ to be a basis of $V$, we get a map from $V$ to $F^{(S)}$ by mapping each basis element $e \in S$ to the characteristic (or &quot;delta&quot;) function $1_e: S \rightarrow F$: namely, for $e&amp;#39; \in S$, $1_e(e&amp;#39;) = 1$ if $e = e&amp;#39;$ and $0$ otherwise.  This is consistent with the universal mapping property underlying the definition of &quot;free vector space&quot;, i.e., every vector space can be viewed as (or more accurately, canonically endowed with the structure of) the free vector space on any one of its bases in this way.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In terms of examples: for finite sets $S$, from a concrete perspective this is close to just considering the vector space $F^n$ of $n$-tuples of elements of $F$.  Indeed, arguably $F^n$ is an example of this with $S = \{1,\ldots,n\}$.  In general, the only difference I can see is that the finite set $S$ does not come with a distinguished ordering, so whereas $F^n$ has a canonical &lt;em&gt;ordered basis&lt;/em&gt; $e_1,\ldots,e_n$, $F^{(S)}$ has a canonical unordered basis $\{1_s \ | \ s \in S\}$.  &lt;/p&gt;&#xA;" OwnerUserId="299" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2011-01-20T23:21:44.743" LastActivityDate="2011-01-20T23:21:44.743" CommentCount="2" />
  <row Id="18358" PostTypeId="2" ParentId="18352" CreationDate="2011-01-20T22:08:13.813" Score="12" ViewCount="" Body="&lt;p&gt;The main progress I know of is in the two papers by Elliot:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Elliott, P. D. T. A. &quot;The multiplicative group of rationals generated by the shifted primes. I.&quot; J. Reine Angew. Math. 463 (1995), 169–216. &lt;/li&gt;&#xA;&lt;li&gt;Elliott, P. D. T. A. &quot;The multiplicative group of rationals generated by the shifted primes. II.&quot; J. Reine Angew. Math. 519 (2000), 59–71. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The result you mention is in the first one. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the review at MathReviews for the first paper: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Let ${\mathbf Q}^*$ be the multiplicative group of positive rationals, let $\Gamma$ be the subgroup generated by shifted primes $p+1$ and set $G:={\mathbf Q}^*/\Gamma$. Denote by $\Gamma_k$ the subset of $\Gamma$ comprising all rationals $r$ that have a representation $r=\prod^k_{j=1}(p_j+1)^{\epsilon_j}$ with $\epsilon_j=0,\pm1$ and all $p_j$ prime with at most $k$ factors. A well-known conjecture of Schinzel and Sierpiński asserts that $\Gamma=\Gamma_2={\mathbf Q}^*$, and a weaker assumption is that $G$ is trivial. The main goal of this paper is to prove that (i) $G$ is finite; (ii) there exists an absolute $k$ such that $r^{|G|}\in\Gamma_k$ for all $r\in{\mathbf Q}^*$; (iii) $|G|\leq 3$.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The author derives (i) and (ii) by a straightforward argument from the following result, stated as Lemma 1: For $m\in{\mathbf Z}^+$, let $F_m$ be the set of positive integers representable in the form$(p+1)/(m(q+1))$ with $p,q$ prime. Then there is an absolute constant $c$ such that the lower asymptotic density $\underline{\bf d}F_m$ of $F_m$ satisfies $\underline{\bf d}F_m\geq c$ uniformly in $m$. Moreover, the same argument yields that $|G|\leq 1/c$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you have access to MathReviews, I recommend that you read the whole review (by Tenenbaum!), as it is very insightful. In particular, Tenenbaum mentions prior work by Meyer and himself that had given $|G|\le4$. To obtain $|G|\le 3$, &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The author develops a novel way of studying $G$ based on principles of harmonic analysis. This rests on properties of multiplicative functions on the set of shifted primes. [...] The proofs rely on difficult, delicate estimates of classical analytic number theory type. The large sieve, Selberg's sieve, the dispersion method and Halász's theorem are the main ingredients, all deeply revisited. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here is Tenenbaum's review of the second paper:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In Part I, the author proved that $G$ has order $|G|\le 3$ and that there is a $k$ such that every positive rational $r$ has a representation $r^{|G|}=\prod_{j=1}^k(p_j+1)^{\epsilon_j}$ with $p_j$ prime, $\epsilon_j=\pm1$. The standard conjecture in this context is that $G$ is trivial and $k=2$, forming a multiplicative analogue of the twin primes conjecture. P. Berrizbeitia and Elliott [Ramanujan J. 2 (1998), no. 1-2, 219--223; MR1642879 (2000a:11122)] were able to obtain the explicit value $k=19$. The author now improves this result by showing that $k=9$ is admissible. This is derived from the following theorem: Given an arbitrary positive rational number $\gamma$, integers of the form $(p+1)/\gamma(q+1)$ with $p,q$ prime have density at least $\frac{11}{40}$. This improves a result of J. Meyer and the reviewer [Bull. Sci. Math. (2) 108 (1984), no. 4, 437--444; MR0784678 (86h:11072)], who showed the lower bound $\frac14$ when $\gamma$ is an integer. An approach to replace $\frac{11}{40}$ by $\frac{3}{10}$ is briefly outlined. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Unfortunately I am not well versed enough in multiplicative number theory to do the results justice by indicating their significance. I will just say that &quot;product bases&quot; results tend to be very difficult, and that the study of large sieve method is turning into an essential tool in modern analytic number theory. Conjectures such as this one, though I find them intrinsically interesting, tend to be valued because of the tools that provide us with. I expect that the reviews above indicate some of this.  &lt;/p&gt;&#xA;" OwnerUserId="462" LastActivityDate="2011-01-20T22:08:13.813" CommentCount="3" />
  <row Id="18359" PostTypeId="2" ParentId="18353" CreationDate="2011-01-20T22:23:05.863" Score="2" ViewCount="" Body="&lt;p&gt;We let $\tau_{0}$ be the subset of open sets of $\tau$ consisting of $U$ in $\tau$ so that if $U$ contains $0$ then $U$ contains both some interval $(a,\infty)$ and some interval $(-\infty,b)$.  This essentially glues $\infty$ and $-\infty$ to $0$ giving a figure eight.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$0$ is the intersection of the circles, $\mathbb{R}^{+}$ is one branch and $\mathbb{R}^{-}$ is the other.&lt;/p&gt;&#xA;" LastActivityDate="2011-01-20T22:23:05.863" CommentCount="3" />
  <row Id="18361" PostTypeId="1" CreationDate="2011-01-20T23:07:59.960" Score="2" ViewCount="78" Body="&lt;p&gt;I have a collections of &quot;systems&quot; that I want to test their performance, $1,\ldots,k$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a small set of samples, for which I can test whether each condition does better or not than the other, so I can get a binary result $I(i,j)$ for each $1 \le i,j \le k$ that tests whether $i$ or $j$ do better on this small sample.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This $I(i,j)$ is calculated by the following way:&#xA;each $i$ is associated with a &quot;system&quot;, like I said. We have a collection of samples $X_1,...,X_n$, and we measure $C(i) = 1/n \sum_{l=1}^n l(X_l,i)$ where $l(X_l,i)$ is a measure of how well system $i$ does on sample $l$. Then we just measure whether $C(i) &amp;gt; C(j)$ or vice versa (to get $I(i,j)$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can I pick from these collections of binary indicators the system which is mostly likely to do best on the expected value of the whole distribution? Are there any ways to have theoretical guarantees for that, and do they depend on $k$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By &quot;do best on the expected value&quot; I am referring to $D(i) = \mathbb{E}[l(X),i]$ - this should be highest for the system I pick.&lt;/p&gt;&#xA;" OwnerUserId="7076" LastActivityDate="2011-05-02T16:44:01.277" Title="is statistical hypothesis testing more reliable when there are less possible tests?" Tags="&lt;probability&gt;&lt;statistics&gt;" AnswerCount="1" />
  <row Id="18362" PostTypeId="1" CreationDate="2011-01-20T23:19:43.247" Score="1" ViewCount="199" Body="&lt;p&gt;Since $\Gamma$ is not one to one over the complex domain, Is it possible to define some principal values ( analogues to  Principal Roots for the Root function ) so we can have a $\Gamma^{-1} (z)$ (inverse $\Gamma$ function)?&lt;/p&gt;&#xA;" OwnerUserId="3936" LastEditorUserId="3936" LastEditorDisplayName="" LastEditDate="2011-01-21T08:10:59.510" LastActivityDate="2011-02-20T08:23:18.997" Title="Is there an Inverse Gamma $\Gamma^{-1} (z) $ function ?" Tags="&lt;analysis&gt;&lt;special-functions&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="18363" PostTypeId="2" ParentId="18268" CreationDate="2011-01-20T23:28:21.477" Score="6" ViewCount="" Body="&lt;p&gt;I believe we can give a formula, but I would not call it &quot;closed form&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have $\displaystyle n$ people, and $\displaystyle k$ possible birthdays to choose from (i.e. $\displaystyle k$ days in a year). Let $\displaystyle M$ be the minimum of the two.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We will try and count the number of birthday assignments in which no two people have consecutive birthdays.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To do this, we will try and count the assignments which use exactly $d$ distinct birthdays, for $\displaystyle d=1, 2 \dots, k$, and then add them up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now suppose we had a set of $\displaystyle d$ distinct birthdays (don't worry about the consecutive part, just yet). How many ways can we assign these $\displaystyle d$ birthdays to $\displaystyle n$ people, so that each birthday is used at least once?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is basically the problem of finding the number of ways to partition a set of size $\displaystyle n$ in exactly $\displaystyle d$ &lt;em&gt;non-empty&lt;/em&gt; parts and assign the $\displaystyle d$ birthdays to each part in the partition, exactly one to each.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The number of ways to partition a set of size $\displaystyle n$ into $\displaystyle d$ non-empty parts is given by a &lt;a href=&quot;http://en.wikipedia.org/wiki/Stirling_number_of_the_second_kind&quot; rel=&quot;nofollow&quot;&gt;Stirling Number of Second Kind&lt;/a&gt;, $S(n,d)$. The number of ways to assign $\displaystyle d$ birthdays is $\displaystyle d!$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus the number we are looking for is $\displaystyle S(n,d) \times d!$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now suppose we managed to count the number of subsets containing $\displaystyle d$ elements (from the $\displaystyle k$ birthdays) such that no two elements of the set are consecutive, then we could multiply that number by  $\displaystyle S(n,d) \times d!$ to give the number of birthday assignments which use exactly $\displaystyle d$ birthdays such that no two are consecutive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the moment, ignore the fact that Jan 1 and Dec 31 are consecutive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We need to select $\displaystyle d$ numbers from $\displaystyle 1,2, \dots, k$ such that no two are consecutive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now if $\displaystyle b_1 \lt  b_2 \lt \dots \lt b_d$ were such numbers, then notice that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle 1 \le b_1 \lt b_2 - 1 \lt b_3 - 2 \dots \lt b_d - (d-1) \le k-(d-1)$ gives us a way to select numbers from $\displaystyle 1, 2, \dots, k-(d-1)$ without having to bother about the consecutive issue.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This can be done in $\displaystyle {k-d+1 \choose d}$ ways.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now since Jan 1 and Dec 31 are consecutive, we need to subtract from this, the number of sets which contain both $\displaystyle 1$ and $\displaystyle k$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This number is same as the number of $\displaystyle d-2$ non-consecutive subsets of $\displaystyle \{3, \dots, k-2\}$ which is $\displaystyle {k-d-1 \choose d-2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus the number of ways of selecting $\displaystyle d$ non-consecutive birthdays (assuming $\displaystyle 1$ and $\displaystyle k$ are consecutive) is $\displaystyle {k-d+1 \choose d} - {k-d-1 \choose d-2}$, with the understanding that the term being subtracted is $\displaystyle 0$ for $\displaystyle d = 1$ and that $\displaystyle {a \choose b} = 0$ if $\displaystyle a \lt b$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, for $\displaystyle M = \text{min}\{n,k\}$, the total number we are looking for is, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \sum_{d=1}^{M} S(n,d)\times d! \times \left({k-d+1 \choose d} - {k-d-1 \choose d-2}\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that $\displaystyle S(n,d) \ d! = \sum_{j=0}^{d} (-1)^j {d \choose j} (d-j)^n$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we could also write the formula as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \sum_{d=1}^{M}  \left(\sum_{j=0}^{d} (-1)^j {d \choose j} (d-j)^n\right) \left({k-d+1 \choose d} - {k-d-1 \choose d-2}\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is a bit ugly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus the probability you are looking for is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$1  - \frac{ \sum_{d=1}^{M}  S(n,d)\times d! \times \left({k-d+1 \choose d} - {k-d-1 \choose d-2}\right)}{k^n}$$&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2011-01-20T23:41:47.877" LastActivityDate="2011-01-20T23:41:47.877" CommentCount="6" />
  <row Id="18364" PostTypeId="1" AcceptedAnswerId="18368" CreationDate="2011-01-20T23:32:39.667" Score="5" ViewCount="142" Body="&lt;p&gt;How are graphs plotted or sketched? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have a graph plotting software for example mathematica or matlab, or you want to see what the graph of $e^{2x}$ looks like, how do you plot/sketch such a graph? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you proceed by using a finite number of points, and &quot;extrapolating&quot; and &quot;interpolating&quot;, then how do you know that that method is reliable? There are infinitely many smooth(or non-smooth) graphs that can be drawn through a finite number of points.&lt;/p&gt;&#xA;" OwnerUserId="4728" LastActivityDate="2011-04-16T02:23:12.017" Title="Sketching graphs" Tags="&lt;math-software&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="2" />
  <row Id="18365" PostTypeId="2" ParentId="18356" CreationDate="2011-01-20T23:37:30.203" Score="4" ViewCount="" Body="&lt;p&gt;A completely simple explanation is provided in the first section &lt;a href=&quot;http://www.sosmath.com/calculus/improper/gamma/gamma.html&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Consider a general gamma density function. That is, a function $f(x;c,\lambda)$, $x&amp;gt;0$, of the form $f(x;c,\lambda) = \lambda^c x^{c-1}e^{-\lambda x} / \Gamma(c)$, where $c$ and $\lambda$ are arbitrary positive constants. The counterpart with respect to the alternative definition, $\tilde \Gamma (p) := \int_0^\infty  {t^p e^{ - t} \,{\rm d}t}$, is a function $f(x;c,\lambda)$, $x&amp;gt;0$, of the form $f(x;c,\lambda) = \lambda^{c+1} x^{c}e^{-\lambda x} / \tilde \Gamma(c)$, where $c &amp;gt; -1$ and $\lambda &amp;gt; 0$. Obviously, the former form is preferable.&lt;/p&gt;&#xA;" OwnerUserId="2810" LastEditorUserId="2810" LastEditorDisplayName="" LastEditDate="2011-01-21T11:17:07.947" LastActivityDate="2011-01-21T11:17:07.947" />
  <row Id="18366" PostTypeId="2" ParentId="13889" CreationDate="2011-01-20T23:45:55.530" Score="1" ViewCount="" Body="&lt;p&gt;This is meant to be a comment below Zach Conn's answer, but for some reason I don't seem to have the option of commenting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is really the same answer -- in short, integrals can be viewed as cohomology classes -- but just to give a very concrete example: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider $\int_C \frac{dz}{z}$ where $C$ is a closed curve in the complex plane that misses the origin.  If you think of $C$ as variable, then the value of the integral depends only on the homology class of $C$ in the punctured plane.  (If $C$ represents $n$ times the usual homology generator, then the integral is $2\pi i n$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words $\int_{\cdot} \frac{dz}{z}$ is a linear functional on the homology, which is to say a cohomology class.  Which is my point:  it is often the case that &quot;cohomology&quot; equals &quot;functionals on homology&quot; (though not always, e.g. in the presence of torsion).  So if one understands homology, then this can be a starting point.&lt;/p&gt;&#xA;" OwnerUserId="6071" LastEditorUserId="6071" LastEditorDisplayName="" LastEditDate="2011-01-21T02:31:22.080" LastActivityDate="2011-01-21T02:31:22.080" />
  <row Id="18367" PostTypeId="2" ParentId="18334" CreationDate="2011-01-21T00:02:06.187" Score="7" ViewCount="" Body="&lt;p&gt;A cursory look at the paper reveals that the author is not a professional mathematician. The algorithm looks very simple, and the bibliography is extremely short. So according to Scott Aaronson's criteria (mentioned by svenkatr), the probability that this particular approach works is small.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is aggravated by the statement in the website that the algorithm was already obtained in 2002, but the author waited for computer simulations - this makes no sense (either the proof works or it doesn't). Now it's true that &lt;i&gt;in practice&lt;/i&gt; people can solve most SAT problems that they want to, probably much larger than the ones mentioned in the paper. But NP-hardness is about worst-case analysis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The real question is whether this new proposed algorithm is faster in practice on real-world instances. The author has done no comparison, and so (since people have worked on optimizing the existing approaches), his method is probably worse than the competition.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2011-01-21T00:02:06.187" CommentCount="1" />
  <row Id="18368" PostTypeId="2" ParentId="18364" CreationDate="2011-01-21T00:09:34.577" Score="1" ViewCount="" Body="&lt;p&gt;Basically (keyword: &quot;basically&quot;), it works like this:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;BEGIN&#xA;    DECLARE x&#xA;    SET x TO x1&#xA;    WHILE x IS LESS THAN TO x2&#xA;        DRAW LINE BETWEEN f(x) and f(x-step)   // &quot;linear interpolation&quot;&#xA;        INCREASE x BY step&#xA;    END WHILE&#xA;END&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Where step is some small number... Something like 0.01, depending on how fast you want it. $step = \lim_{x\to 0^+}$ would be &quot;perfectly smooth&quot;, but we don't have the resources for such measures, so we're stuck with anything between $10^{50}$ and $10^{-50}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They won't be &quot;infinitely smooth&quot;. That would require infinite amounts of RAM and CPU. It only has to &quot;look&quot; smooth. It's only limited by your RAM/CPU, time (unless you feel like waiting a few trillion years), and most importantly, your screen resolution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More complex plotters can determine the vertical asymptotes, but I don't need to get into that.&lt;/p&gt;&#xA;" OwnerUserId="2736" LastEditorUserId="2736" LastEditorDisplayName="" LastEditDate="2011-04-16T02:23:12.017" LastActivityDate="2011-04-16T02:23:12.017" CommentCount="2" />
  <row Id="18369" PostTypeId="1" AcceptedAnswerId="18371" CreationDate="2011-01-21T00:09:57.337" Score="4" ViewCount="209" Body="&lt;p&gt;Given n points drawn randomly on the circumference of a circle, what is the probability they will all be within any common semicircle?&lt;/p&gt;&#xA;" OwnerUserId="5959" LastActivityDate="2011-01-25T02:03:13.643" Title="N points on a Circle..." Tags="&lt;probability&gt;" AnswerCount="3" CommentCount="4" />
  <row Id="18370" PostTypeId="2" ParentId="18268" CreationDate="2011-01-21T00:16:54.570" Score="5" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;NB&lt;/strong&gt;: I worked earlier on this problem and came up with the following solution. The first answer that was posted made me think that I got something wrong, and I discarded my work. Since the new answer by Moron agrees (essentially) with my previous work here it is, a slightly different derivation of the same formula.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $k$ be the number of days in a year. Let $m$ be the distinct number of birthdays among the $n$ friends. Let's assume that $k$ is big enough to have a non-trivial problem (say, $k &gt; n/2$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We're interested in binary strings with these three conditions:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Are of length $k$, with $m$ ones and $k-m$ zeros.&lt;/li&gt;&#xA;&lt;li&gt;There's at least one zero between any two ones.&lt;/li&gt;&#xA;&lt;li&gt;Condition 2 holds when &quot;wrapping around&quot; the string.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Let's count them by constructing them with the following algorithm:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Start with a string of $m$ ones: $11\dots 1$&lt;/li&gt;&#xA;&lt;li&gt;There are now three distinct cases: there's a birthday on the first day of the year, there is a birthday on the last day of the year, there's a birthday on neither.&lt;/li&gt;&#xA;&lt;li&gt;In the first case we have to distribute $k-m$ zeros in $m$ non-empty buckets. Those are called compositions, and one can show that there are $\binom{k-m-1}{m-1}$ such assignments.&lt;/li&gt;&#xA;&lt;li&gt;The second case is analogous, giving another $\binom{k-m-1}{m-1}$ possible strings.&lt;/li&gt;&#xA;&lt;li&gt;The third case is similar, giving instead $\binom{k-m-1}{m}$ strings.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Putting this together we have: $$2\cdot \binom{k-m-1}{m-1}+\binom{k-m-1}{m}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $n$ friends share $m$ birthdays we have to account for that, giving this expression:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$p(n,k) =\frac{\sum_{m=1}^n{ m! \cdot S(n,m)\cdot \Bigg (2\cdot \binom{k-m-1}{m-1}+\binom{k-m-1}{m}}\Bigg )}{k^n}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $S(n,m)$ is the Stirling number of the second kind. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strike&gt;@Moron: Why do we need to multiply by $m!$ here?&lt;/strike&gt; Oh right! Stirling numbers of the second kind count the number of coalitions, but we care about their order, too!&lt;/p&gt;&#xA;" OwnerUserId="4471" LastEditorUserId="4471" LastEditorDisplayName="" LastEditDate="2011-01-21T08:51:36.550" LastActivityDate="2011-01-21T08:51:36.550" CommentCount="5" />
  <row Id="18371" PostTypeId="2" ParentId="18369" CreationDate="2011-01-21T00:34:21.757" Score="4" ViewCount="" Body="&lt;p&gt;Suppose the points are $x_1,\ldots,x_n$. If the points are indeed contained in a semicircle, then we can define a &quot;leftmost&quot; point, i.e. a point $x_i$ such that $x_j \in [x_i, x_i+\pi)$ (addition is modulo $2\pi$). There is only one leftmost point, and the probability that $x_i$ is leftmost is $1/2^{n-1}$, since the probability that $x_j \in [x_i, x_i+\pi)$ is $1/2$. Since there are $n$ possible leftmost points, we get $n/2^{n-1}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now generalize the argument to $d$ dimensions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: The argument generalizes to show that the probability that all points lie in an $\alpha$-section of the circle is $n\alpha^{n-1}$ for $\alpha \leq 1/2$. You can challenge yourself by thinking about the case $\alpha &amp;gt; 1/2$.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2011-01-21T01:02:27.187" LastActivityDate="2011-01-21T01:02:27.187" CommentCount="3" />
  <row Id="18372" PostTypeId="1" AcceptedAnswerId="18374" CreationDate="2011-01-21T00:55:09.433" Score="0" ViewCount="57" Body="&lt;p&gt;Drawing a parallel to linear Algebra.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Premise:&#xA;The maximum of a determinate is when all the columns are orthogonal, given that the vectors are normalized.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conclusion:&#xA;The orthogonal trajectory should have a similar effect maximizing the wronskian. However, this does not seem to be true.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given &lt;code&gt;dy/dx = x&lt;/code&gt; the orthogonal trajectory would seem to be &lt;code&gt;-x&lt;/code&gt;. However, this would show us that they are linearly dependent, and not maximized at all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the definition of an &quot;orthogonal function&quot; as to orthogonal vectors? (functions whose tangent are orthogonal for every &lt;code&gt;x&lt;/code&gt; seemed like a reasonable definition ). Shouldn't such functions maximize the wronskian? How would one maximize the wronskian?&lt;/p&gt;&#xA;" OwnerUserId="5934" LastActivityDate="2011-01-21T01:16:01.633" Title="Why doesn't the orthogonal trajectory maximize the Wronskian?" Tags="&lt;linear-algebra&gt;&lt;differential-equations&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="18373" PostTypeId="2" ParentId="18356" CreationDate="2011-01-21T01:00:06.723" Score="6" ViewCount="" Body="&lt;p&gt;I find it more illuminating to see what that extra $ t^{-1} $ does to the integral.  As a generalization of the factorial, $ \Gamma $ is inherently multiplicative*.  On the other hand, integration, which is essentially a sum, is inherently &lt;em&gt;additive&lt;/em&gt;.  Thinking about it this way it seems a bit odd that an integral could give an appropriate generalization.  However, there is a simple function that takes the (positive) multiplicative reals to the reals under addition: the logarithm.  Thinking of gamma as $$ \Gamma(s) = \int_0^{\infty} t^s e^{-t} \frac{dt}{t} $$&#xA;we see that the natural log arises naturally (what's the first thing to come to mind when you see $ \frac{dt}{t} \: $?) in this context (and there are no pesky $s-1$'s left).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I haven't done much with integration theory (so I'm not sure my terminology is correct), but I believe this intuitive argument may be made rigorous by considering the integral formula for $ \Gamma(s) \: $ as an integral over the positive reals under multiplication (i.e. the interval $ (0,+\infty) $ ) with respect to the multiplicative Haar measure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;*note that by &quot;multiplicative&quot; I just mean that it's realized as a product, not that it's a multiplicative arithmetic function or anything like that&lt;/p&gt;&#xA;" OwnerUserId="5330" LastActivityDate="2011-01-21T01:00:06.723" CommentCount="1" />
  <row Id="18374" PostTypeId="2" ParentId="18372" CreationDate="2011-01-21T01:16:01.633" Score="1" ViewCount="" Body="&lt;p&gt;Vectors can only be orthogonal with respect to some inner-product. &#xA;In that case, u is orthogonal to v when $\left \langle u,v \right \rangle=0$, where $\left \langle , \right \rangle$ denotes the inner-product.&#xA;Typically, &quot;orthogonal functions&quot; refers to functions which are orthogonal with respect to the $L^2$ inner-product, i.e,$\left \langle f,g \right \rangle=\int_Ifgd\mu=0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is evident, that two vectors which are orthogonal with respect to one inner-product need not be orthogonal with respect to another inner-product.&lt;/p&gt;&#xA;" OwnerUserId="5959" LastActivityDate="2011-01-21T01:16:01.633" />
  <row Id="18375" PostTypeId="1" AcceptedAnswerId="18379" CreationDate="2011-01-21T01:28:57.770" Score="0" ViewCount="100" Body="&lt;p&gt;How to simplify this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle\frac{n^{\log m}}{m^{\log n}}$&lt;/p&gt;&#xA;" OwnerUserId="6007" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2011-01-21T06:51:15.120" LastActivityDate="2011-01-21T06:51:15.120" Title="Simplifying this expression" Tags="&lt;homework&gt;&lt;algebra-precalculus&gt;&lt;logarithms&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="18376" PostTypeId="2" ParentId="18375" CreationDate="2011-01-21T01:34:55.347" Score="4" ViewCount="" Body="&lt;p&gt;Hint: Apply $\log$ to the whole thing and use the quotient and powers rules for $\log$.  You should get a very simple result, which you can exponentiate to find your answer.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2011-01-21T01:34:55.347" />
  <row Id="18377" PostTypeId="2" ParentId="18375" CreationDate="2011-01-21T01:35:02.100" Score="2" ViewCount="" Body="&lt;p&gt;Note that $n^{\log m} = m^{\log n}$.&lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2011-01-21T01:35:02.100" CommentCount="3" />
  <row Id="18378" PostTypeId="2" ParentId="18288" CreationDate="2011-01-21T01:36:12.380" Score="1" ViewCount="" Body="&lt;p&gt;Another way to think of this is to notice that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\limsup_{n\to\infty} E_n = \lbrace x| x\in E_n \hbox{for infinitely many $n$}\rbrace.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, $x\in E_n$ occurs &quot;frequently.&quot;&lt;/p&gt;&#xA;" OwnerUserId="4154" LastActivityDate="2011-01-21T01:36:12.380" />
  <row Id="18379" PostTypeId="2" ParentId="18375" CreationDate="2011-01-21T01:49:30.293" Score="3" ViewCount="" Body="&lt;p&gt;First note that $x = a^{\log_a x}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we find that $n^{\log m} = e^{\log (n^{\log m})} = e^{\log m \log n}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, we find that $m^{\log n} = e^{\log (m^{\log n})} = e^{\log n \log m}$&lt;/p&gt;&#xA;" OwnerUserId="3302" LastActivityDate="2011-01-21T01:49:30.293" />
  <row Id="18380" PostTypeId="1" CreationDate="2011-01-21T03:05:02.853" Score="0" ViewCount="59" Body="&lt;p&gt;Hi there,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to model a random process $u(t,\omega)$ as a random element in a Bochner space since these spaces have some nice properties (which I am still learning about). However, I also want to be aware of the limitations of this approach. I have in mind applications in which $u$ is used to model uncertain physical properties such as mass density or permeability.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Specifically, consider the complete probability space $(\Omega,\mathcal{F},\mathbb{P})$ and a strongly measurable random element $u:(\Omega,\mathcal{F})\rightarrow (L^2(D),\mathcal{B}(L^2(D)))$, where $D$ is a bounded domain in $\mathbb{R}$, such that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;\int_{\Omega} \Vert u (\omega) \Vert_{L^2(D)}^2 \mathrm{d}\mathbb{P}(\omega) &amp;lt; \infty&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words $u\in L^2(\Omega,\mathbb{P};L^2(D))$. Because $L^2(D)$ is a separable Banach (i.e., Polish) space, then $u$ is also weakly measurable by Pettis' Theorem. Furthermore, because $L^2(D)$ is reflexive, the Radon-Nikodym property holds so I can investigate things like conditional expectations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One problematic issue, I think, is whether standard finite-dimensional distributions or even marginal distributions are well defined for $u(t,\omega)$. For example, the projection operator $\pi_t:L^2(D)\rightarrow \bar{\mathbb{R}}$ could be used to define $\pi_t(u(\omega))=u(t,\omega)$, for fixed $t\in D$, but because $\pi_t$ is not bounded (because $u(t,\omega)$ itself is not necessarily bounded), then it is not in the dual space so weak measurability doesn't necessarily imply that $\pi_t\circ u$ is $\mathbb{P}$-measurable, which means that a marginal distribution cannot be defined.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Am I correct? Are there ways around this? Are there other important limitations?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance.&lt;/p&gt;&#xA;" OwnerUserId="6010" LastActivityDate="2011-01-21T03:05:02.853" Title="Pros and cons of modeling a random process as a random element in a Bochner space" Tags="&lt;measure-theory&gt;&lt;stochastic-processes&gt;" />
  <row Id="18381" PostTypeId="2" ParentId="18340" CreationDate="2011-01-21T03:11:07.830" Score="2" ViewCount="" Body="&lt;p&gt;You'd probably be interested in the &lt;a href=&quot;http://en.wikipedia.org/wiki/Platonic_solid&quot; rel=&quot;nofollow&quot;&gt;Platonic solids&lt;/a&gt;, &lt;a href=&quot;http://en.wikipedia.org/wiki/Archimedean_solid&quot; rel=&quot;nofollow&quot;&gt;Archimedean solids&lt;/a&gt;, &lt;a href=&quot;http://en.wikipedia.org/wiki/Catalan_solid&quot; rel=&quot;nofollow&quot;&gt;Catalan solids&lt;/a&gt;, and some of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Johnson_solids&quot; rel=&quot;nofollow&quot;&gt;Johnson solids&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1995" LastActivityDate="2011-01-21T03:11:07.830" CommentCount="2" />
  <row Id="18382" PostTypeId="1" AcceptedAnswerId="27820" CreationDate="2011-01-21T03:17:53.737" Score="0" ViewCount="321" Body="&lt;p&gt;Greetings All&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks to James and Chas for helping me get this far&#xA;btw Chas the language I wrote it in is in matlab.  I tried to respond&#xA;to your post but wasn't able to do it..I guess the gremlins&#xA;are acting up on my computer.  Also since I'm only able&#xA;to post one link I'll post a link with to the text file&#xA;which will have the animation link in it.&#xA;Here's a link to the code as a text file / animation /&#xA;other links in case the format&#xA;gets messed up --&gt; &lt;a href=&quot;http://db.tt/4eYMMz7&quot; rel=&quot;nofollow&quot;&gt;http://db.tt/4eYMMz7&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have the output of my quaternions in the format &#xA;[w x y z].  My first vector point is [1x 0y 0z]&lt;br&gt;&#xA;and the arbitrary point I want to rotate about is&#xA;[0.5x 0y 0z].  The thing is when I plot the quaternion &#xA;values for the x and y values they don't rotate about&#xA;the origin [0x 0y 0z] or the arbitrary axis origin of &#xA;[0.5x 0y 0z] can someone tell me &#xA;what I'm doing wrong?  I checked my axis-angle to &#xA;quaternion conversion by checking it with the&#xA;spincalcvi.m file located on matlab's file exchange. his&#xA;format was in [x y z w] and mine is in [w x y z] &#xA;see above link&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I used the Multiplication formula for two quaternions from this&#xA;site see above link&#xA;and I tested it using this online quaternion calculator and it matches up&#xA;see above link&#xA;When I step through theta I don't get the circular rotations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ps: Yes I do want to plot points in 3D eventually but I'm trying to&#xA;understand/get the 2D portion to work first.  By first&#xA;getting a point to rotate around the origin [0x 0y 0z] and a point &#xA;to rotate around an arbitrary axis origin of [0.5x 0y 0z]  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;See code below &#xA;Here's a link to animation/video of plot showing values&#xA;as theta changes --&gt;see above link&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a link to the code as a text file in case the format&#xA;gets messed up --&gt; see above link&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;    %Quaterion rotation - around orgin and arbitrary axis&#xA;%quaterions using Format [w xi yj zk] or [w x y z]&#xA;&#xA;for ii=1:10:720&#xA;p1=[1,0,0] %[x y z] %point to rotate&#xA;%q=zeros(1,4);&#xA;theta=0+ii;&#xA;%normalize -p1- angle axis to pointing vector prior to quaternion construction&#xA;p1norm=p1(1:3)./norm(p1(1:3)) %norm is x^2+y^2+z^2&#xA;%Create Quaternion from p1norm&#xA;qp1 = [cosd(theta/2) sind(theta/2)*p1norm] % format [w, xi, yj, zk]&#xA;%normalize quaternion qp1&#xA;qp1mag=sqrt(qp1(1,1)^2+qp1(1,2)^2+qp1(1,3)^2+qp1(1,4)^2)&#xA;qp1norm=qp1(1:4)./qp1mag&#xA;&#xA;%Arbitrary axis section------------------------&#xA;rp2=[.5 0 0] %[x y z]&#xA;%normalize -p2- angle axis to pointing vector prior to quaternion construction&#xA;rp2norm=rp2(1:3)./norm(rp2(1:3))&#xA;%Create Quaternion from p2norm&#xA;qrp2 = [cosd(theta/2) sind(theta/2)*rp2norm] % format [w, xi, yj, zk]&#xA;%normalize quaternion qrp2&#xA;qrp2mag=sqrt(qrp2(1,1)^2+qrp2(1,2)^2+qrp2(1,3)^2+qrp2(1,4)^2)&#xA;qrp2norm=qrp2(1:4)./qrp2mag&#xA;&#xA;%Conjugate of qp2norm&#xA;qrp2normconj=qrp2norm.*[1 -1 -1 -1] % [w x y z] format&#xA;&#xA;%1st Multiplication  (qrp2norm*qp1norm) full formula q*P*q' or qrp2norm*qp1norm*qrp2normconj&#xA;q1w=qrp2norm(1,1);q1x=qrp2norm(1,2);q1y=qrp2norm(1,3);q1z=qrp2norm(1,4); %rotated vector&#xA;p1w=qp1norm(1,1);p1x=qp1norm(1,2);p1y=qp1norm(1,3);p1z=qp1norm(1,4); %part being rotataed&#xA;&#xA;qqw=(q1w*p1w - q1x*p1x - q1y*p1y - q1z*p1z)&#xA;qqx=(q1w*p1x + q1x*p1w + q1y*p1z - q1z*p1y)&#xA;qqy=(q1w*p1y - q1x*p1z + q1y*p1w + q1z*p1x)&#xA;qqz=(q1w*p1z + q1x*p1y - q1y*p1x + q1z*p1w)&#xA;qm1=[qqw qqx qqy qqz]&#xA;&#xA;%2nd Multiplication  (qm1*qrp2normconj) full formula q*P*q' or qrp2norm*qp1norm*qrp2normconj&#xA;q1w=qm1(1,1);q1x=qm1(1,2);q1y=qm1(1,3);q1z=qm1(1,4); %rotated vector&#xA;p1w=qrp2normconj(1,1);p1x=qrp2normconj(1,2);p1y=qrp2normconj(1,3);p1z=qrp2normconj(1,4); %part being rotataed&#xA;&#xA;qqw=(q1w*p1w - q1x*p1x - q1y*p1y - q1z*p1z)&#xA;qqx=(q1w*p1x + q1x*p1w + q1y*p1z - q1z*p1y)&#xA;qqy=(q1w*p1y - q1x*p1z + q1y*p1w + q1z*p1x)&#xA;qqz=(q1w*p1z + q1x*p1y - q1y*p1x + q1z*p1w)&#xA;qm2=[qqw qqx qqy qqz]&#xA;&#xA;%plotting section&#xA;&#xA;plot(0,0,'b',qp1norm(1,2),qp1norm(1,3),'-ro',qm2(1,2),qm2(1,3),'-bx')&#xA;grid on&#xA;hold on&#xA;legend('not used','ROT around origin','ROT around arbitrary axis');&#xA;title( { 'Quaternion Plot problems', ...&#xA;['theta = ', num2str(theta),' degrees'], ...&#xA;['Rot around orgin [w x y z] = ', num2str(qp1norm)], ...&#xA;['Rot around arbitrary axis [w x y z] = ',num2str(qm2) ] } )&#xA;axis([-1.5 1.5 -1.5 1.5 -1.5 1.5])&#xA;view([26 64]);&#xA;xlabel('x axis');ylabel('y axis');zlabel('z axis');&#xA;&#xA;pause(.5)&#xA;M(ii)=getframe;&#xA;end&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;But I'm at a lost as to why I don't get a circular rotation &lt;/p&gt;&#xA;" OwnerUserId="5957" LastEditorUserId="5957" LastEditorDisplayName="" LastEditDate="2011-03-21T21:36:53.930" LastActivityDate="2011-03-21T21:36:53.930" Title="Quaternion and rotation about an origin and an arbitrary axis origin help" Tags="&lt;quaternions&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="18383" PostTypeId="2" ParentId="18369" CreationDate="2011-01-21T03:43:56.997" Score="1" ViewCount="" Body="&lt;p&gt;Let $dp(\theta_1, \theta_2)$ be the probability density that $[\theta_1, \theta_2]$ is a minimal covering arc (i.e., it contains all $n$ points, and any proper sub-arc excludes at least one point).  This is given by&#xA;$$&#xD;&#xA;dp(\theta_1, \theta_2) = n(n-1)\left(\frac{\theta_2 - \theta_1}{2\pi}\right)^{n-2} \frac{d\theta_1}{2\pi} \frac{d\theta_2}{2\pi},&#xD;&#xA;$$&#xA;since one point must be at each end of the arc and the remaining $n-2$ must be in its interior. The probability density that the length of the unique minimal covering arc is $\theta$ is therefore&#xA;$$&#xD;&#xA;dp(\theta) = n(n-1)\left(\frac{\theta}{2\pi}\right)^{n-2}\frac{d\theta}{2\pi}&#xD;&#xA;$$&#xA;for $\theta \le \pi$.  (Note that for $\theta &amp;gt; \pi$ there may be more than one minimal covering arc; for example, if all the points cluster around $0$, $2\pi/3$, and $4\pi/3$, there will be three distinct minimal covering arcs, each of length approximately $4\pi/3$.) The probability that all $n$ points lie in some arc of length $\Theta \le \pi$ is therefore&#xA;$$&#xD;&#xA;P(\Theta) = \int_{0}^{\theta} dp(\theta) = n\left(\frac{\Theta}{2\pi}\right)^{n-1}.&#xD;&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="5546" LastActivityDate="2011-01-21T03:43:56.997" />
  <row Id="18384" PostTypeId="1" AcceptedAnswerId="18385" CreationDate="2011-01-21T03:46:14.637" Score="1" ViewCount="49" Body="&lt;p&gt;Suppose you want to line up $2^n$ balls of which $x$ are black the rest are white.  Find a general method to do this so that the black balls are as dispersed as possible, assuming that the pattern will repeat itself ad infinitum.  The solution can be in closed form, iterative, or algorithmic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, if $n=3$, where $0$ is a white ball and $1$ is a black ball, a solution is:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;x=0: 00000000...&#xA;x=1: 10000000...&#xA;x=2: 10001000...&#xA;x=3: 10010010...&#xA;x=4: 10101010...&#xA;x=5: 01101101...&#xA;x=6: 01110111...&#xA;x=7: 01111111...&#xA;x=8: 11111111...&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="6076" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-21T03:50:29.673" LastActivityDate="2011-01-21T03:56:46.307" Title="Spread evenly $x$ black balls among a total of $2^n$ balls." Tags="&lt;analysis&gt;&lt;statistics&gt;&lt;algorithms&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="18385" PostTypeId="2" ParentId="18384" CreationDate="2011-01-21T03:56:46.307" Score="2" ViewCount="" Body="&lt;p&gt;Let $\theta = x/2^n$. For each $m$, put a (black) ball at $$\min \{ k \in \mathbb{N} : k\theta \geq m \}.$$ In other words, look at the sequence $\lfloor k \theta \rfloor$, and put a ball in each position where the sequence increases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, if $n=3$ and $x = 3$ then $\theta = 3/8$ and the sequence of floors is $$0, 0, 0, 1, 1, 1, 2, 2, 3, 3, 3, 4, 4, 4, 5, 5, \ldots,$$ and so the sequence of balls is $10010010\ldots$ .&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2011-01-21T03:56:46.307" />
  <row Id="18386" PostTypeId="1" AcceptedAnswerId="18389" CreationDate="2011-01-21T03:58:03.233" Score="0" ViewCount="88" Body="&lt;p&gt;I've been given the task of computing two series programmatically  one is of the form:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ 1 + \frac{1}{2} +\frac{1}{3}+\frac{1}{4}...+ \frac{1}{n} $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other one is an estimation for $\pi$ which is of the form:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\pi = 4( 1 - \frac{1}{3} + \frac{1}{5} -\frac{1}{2i -1})$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can write this program any way I see fit. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is can these series be written recursively? &#xA;If so, my other question is, is the base case the number 1 in both methods?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We are not on the chapter on recursion yet,so I'm not required to solve them recursively - if they can be solved in that way at all.&lt;/p&gt;&#xA;" OwnerUserId="1040" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2011-06-03T13:51:58.083" LastActivityDate="2011-06-03T13:51:58.083" Title="Can I write this series as a recursive function" Tags="&lt;sequences-and-series&gt;&lt;recursive-algorithms&gt;" AnswerCount="1" />
  <row Id="18387" PostTypeId="1" AcceptedAnswerId="18398" CreationDate="2011-01-21T03:58:56.057" Score="4" ViewCount="82" Body="&lt;p&gt;Since semigroups don't need to have an identity element, I was wondering if there's any kind of short exact sequence for semigroups.  &lt;/p&gt;&#xA;" OwnerUserId="6077" LastActivityDate="2011-01-21T20:43:16.647" Title="is there an analogue of short exact sequences for semigroups?" Tags="&lt;group-theory&gt;&lt;semigroup&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="18388" PostTypeId="1" AcceptedAnswerId="18392" CreationDate="2011-01-21T04:07:17.013" Score="6" ViewCount="300" Body="&lt;p&gt;I am wondering if there are some other examples of Riemann non-integrable but Lebesgue integrable, besides the well-known Dirichlet function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="6016" LastEditorUserId="6016" LastEditorDisplayName="" LastEditDate="2011-01-21T04:19:41.490" LastActivityDate="2011-01-21T05:30:02.727" Title="what functions or classes of functions are Riemann non-integrable but Lebesgue integrable" Tags="&lt;real-analysis&gt;" AnswerCount="3" FavoriteCount="1" />
  <row Id="18389" PostTypeId="2" ParentId="18386" CreationDate="2011-01-21T04:11:10.767" Score="6" ViewCount="" Body="&lt;p&gt;The two series can be defined recursively, but they are much better programmed iteratively. As an example, I will give the recursive definition of the first:&#xA;$$ s_1 = 1, \quad s_{n+1} = s_n + \tfrac{1}{n+1}. $$&#xA;Recursion is slower than iteration for low-level reasons (unless you're programming in a functional programming language in which tail-recursion is optimized), so in general avoid recursion if you can use iteration at the same &quot;programming cost&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In some cases (e.g. naive recursive computation of the Fibonacci sequence), recursion is exponentially slower than an iterative solution, although this can be fixed with more advanced techniques (memoization).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are C codes for your first series:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;double s_iterative(int n) {&#xA;  double res = 0;&#xA;  for (int i = 1; i &amp;lt;= n; i++)&#xA;    res += 1./n;&#xA;  return res;&#xA;}&#xA;&#xA;double s_recursive(int n) {&#xA;  if (n == 0)&#xA;    return 0;&#xA;  else&#xA;    return s_recursive(n - 1) + 1./n;&#xA;}&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;So as you can see, in this case recursion is not worth the trouble.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moron asked for a tail recursive implementation, so here it is:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;double s_tailrec_aux(int n, double acc);&#xA;double s_tailrec(int n) {&#xA;  return s_tailrec_aux(n, 0);&#xA;}&#xA;double s_tailrec_aux(int n, double acc) {&#xA;  if (n == 0)&#xA;    return acc;&#xA;  else&#xA;    return s_tailrec_aux(n - 1, acc + 1./n);&#xA;}&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;In a tail recursion, the recursive call &quot;replaces&quot; the current call, so that no actual recursion need take place (at the discretion of the compiler). It's simply another way of writing a loop.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2011-01-21T05:54:07.427" LastActivityDate="2011-01-21T05:54:07.427" CommentCount="1" />
  <row Id="18390" PostTypeId="2" ParentId="18388" CreationDate="2011-01-21T04:15:10.450" Score="2" ViewCount="" Body="&lt;p&gt;Check the Wikipedia article on the &lt;a href=&quot;http://en.wikipedia.org/wiki/Riemann_integral#Integrability&quot; rel=&quot;nofollow&quot;&gt;Riemann integral&lt;/a&gt; for some ideas.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2011-01-21T04:15:10.450" CommentCount="2" />
  <row Id="18392" PostTypeId="2" ParentId="18388" CreationDate="2011-01-21T04:22:01.627" Score="7" ViewCount="" Body="&lt;p&gt;Yuval's link shows the general criterion for bounded functions on a bounded interval, namely continuity almost everywhere.  The Dirichlet function is a good example, but it is almost everywhere zero, and the zero function sure is Riemann integrable.  If you consider the characteristic function of a &lt;a href=&quot;http://en.wikipedia.org/wiki/Smith%E2%80%93Volterra%E2%80%93Cantor_set&quot; rel=&quot;nofollow&quot;&gt;Cantor set of positive measure&lt;/a&gt;, then it is not Riemann integrable because it is discontinuous at each point in this &quot;fat&quot; Cantor set, and you cannot modify the values on a set of measure 0 to fix this defect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same idea would work for any closed nowhere dense subset $E\subset[0,1]$ of positive measure, and you can appeal to the definition to see this.  Because every subinterval of $[0,1]$ has nonempty intersection with $[0,1]\setminus E$, the lower Riemann sums of $\chi_E$ are all $0$.  However, the upper Riemann sums are all bounded below by $m(E)\gt0$.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2011-01-21T05:30:02.727" LastActivityDate="2011-01-21T05:30:02.727" CommentCount="7" />
  <row Id="18393" PostTypeId="1" AcceptedAnswerId="18394" CreationDate="2011-01-21T04:22:56.267" Score="0" ViewCount="207" Body="&lt;p&gt;Hello everyone, I'm currently stuck on an exercise problem from Joseph Gallian's book, &quot;Contemporary Abstract Algebra.&quot; The question is from Chapter 2, Exercise 12. It says: &quot;For n&gt;2, show that there are at least two elements in U(n) that satisfy $\ x^2=1$&quot; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where U(n) is the set of positive integers less than n and co-prime with n. This set is a group under multiplication mod n.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I see that a good way of showing this property is by induction. So I've set up my inductive hypothesis, after a couple of base cases of course (starting at n=3). But I'm stuck and I can't go from the hypothesis to the inductive step. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any help would be appreciated. Thanks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;*Edit: Thanks everyone, I got it.&lt;/p&gt;&#xA;" OwnerUserId="5369" LastEditorUserId="5369" LastEditorDisplayName="" LastEditDate="2011-01-21T04:48:32.777" LastActivityDate="2011-03-07T17:17:28.503" Title="Abstract Algebra exercise" Tags="&lt;homework&gt;&lt;abstract-algebra&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="0" />
  <row Id="18394" PostTypeId="2" ParentId="18393" CreationDate="2011-01-21T04:27:03.887" Score="2" ViewCount="" Body="&lt;p&gt;Induction performs poorly on factorization because the factors of n+1 are not well related to the factors of n.  A better approach is to exhibit two different solutions to $x^2=1$.  If you move the 1 over and factor the equation...&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-01-21T04:27:03.887" />
  <row Id="18395" PostTypeId="1" AcceptedAnswerId="18404" CreationDate="2011-01-21T04:27:26.817" Score="3" ViewCount="209" Body="&lt;p&gt;We know that if a real function is in $L^2$ then it is in $L^1$, but the reverse is not necessarily true. So what are the examples of functions that are $L^1$ but not $L^2$, especially those intrinsically unbounded ones if any? Welcome more examples, or classes of examples. Thanks.&lt;/p&gt;&#xA;" OwnerUserId="6016" LastActivityDate="2011-01-21T12:46:41.857" Title="what functions or classes of functions are $L^1$ but not $L^2$" Tags="&lt;real-analysis&gt;" AnswerCount="5" FavoriteCount="1" />
  <row Id="18396" PostTypeId="2" ParentId="18393" CreationDate="2011-01-21T04:27:53.683" Score="3" ViewCount="" Body="&lt;p&gt;Hint: How do you normally (say in real numbers) solve $x^2 =1$?&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2011-01-21T04:27:53.683" />
  <row Id="18397" PostTypeId="2" ParentId="18388" CreationDate="2011-01-21T04:31:11.513" Score="3" ViewCount="" Body="&lt;p&gt;For me the &quot;canonical&quot; example is the characteristic function of the rational numbers in $[0,1]$. The upper integral is one, the lower integral is zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; Jonas and Moron have informed me that some people call this example Dirichlet function (and I vaguely remember that we might have done so as well in our Analysis class).&lt;/p&gt;&#xA;" OwnerUserId="5363" LastEditorUserId="5363" LastEditorDisplayName="" LastEditDate="2011-01-21T05:29:48.873" LastActivityDate="2011-01-21T05:29:48.873" CommentCount="1" />
  <row Id="18398" PostTypeId="2" ParentId="18387" CreationDate="2011-01-21T04:34:35.123" Score="5" ViewCount="" Body="&lt;p&gt;Suppose that you have a homomorphism of semigroups $f\colon S\to T$. The &lt;em&gt;kernel of&lt;/em&gt; $f$ is defined to be the congruence $\mathrm{ker} f$ on $S$ defined by:&#xA;$$\mathrm{ker} f = \{ (s_1,s_1)\in S\times S\mid f(s_1)=f(s_2)\}.$$&#xA;Note that $S/\mathrm{ker} f \cong f(S)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We also define the &lt;em&gt;image of&lt;/em&gt; $f$, $\mathcal{K}_{\mathrm{Im}f}$ to be the relation on $T$ defined by:&#xA;$$\mathcal{K}_{\mathrm{Im} f} = f(S)\times f(S)\cup\{(t,t)\mid t\in T\}.$$&#xA;&lt;em&gt;Edit:&lt;/em&gt; The image need not be a congruence if $f(S)$ is not an &lt;em&gt;ideal&lt;/em&gt; in $T$, because it need not be a subsemigroup of $T\times T$. But this agrees with the situation in, say, not-necessarily-abelian groups, where the image need not be a normal subgroup (and thus, not &quot;eligible&quot; to be equal to a kernel). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given homomorphism $f\colon S\to T$ and $g\colon T\to U$, we say that&#xA;$$S\stackrel{f}{\longrightarrow} T \stackrel{g}{\longrightarrow} U$$&#xA;is &lt;em&gt;exact&lt;/em&gt; (at $T$) if and only if $\mathrm{ker} g = \mathcal{K}_{\mathrm{Im} f}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we say that $S\stackrel{f}{\longrightarrow} T\stackrel{g}{\longrightarrow} U$ is a &lt;em&gt;short exact sequence&lt;/em&gt; if and only if&#xA;$\mathrm{ker} f = \{(s,s)\mid s\in S\}$, $\mathcal{K}_{\mathrm{Im} f} = \mathrm{ker}g$, and $\mathcal{K}_{\mathrm{Im} g} = U\times U$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can also see it by extending to monoids first. Given any semigroup $S$, you can always adjoin an identity element by taking some $1\notin S$, and defining the operation on $S\cup\{1\}$ by extending the multiplication by the rule $1s=s1=s$ for all $s\in S$; this monoid is denoted $S^1$ (even if $S$ already has an identity, we adjoin a new element). If $f\colon S\to T$ is a semigroup homorphism, then this induces a monoid homomorphism $f^1\colon S^1\to T^1$ by $f^1(s)=f(s)$ for all $s\in S$, and $f(1) = 1$. If we do this, then note that&#xA;$$\mathrm{ker}(f^1) = \mathrm{ker}(f)\cup\{(1,1)\}$$&#xA;is a congruence on $S^1$; and that&#xA;\begin{align*}&#xA;\mathcal{K}_{\mathrm{Im}f^1} &amp;amp;= f^1(S^1)\times f^1(S^1)\cup\{(t,t)\mid t\in T^1\}\\&#xA;&amp;amp;= f(S)\times f(S) \cup\{(t,t)\mid t\in T\} \cup\{(1_T,1_T)\}\\&#xA;&amp;amp;= \mathrm{K}_{\mathrm{Im}f}\cup\{(1_T,1_T)\},&#xA;\end{align*}&#xA;so that if we have $S\stackrel{f}{\to}T\stackrel{g}{\to}U$, with corresponding $S^1\stackrel{f^1}{\to} T^1\stackrel{g^1}{\to} U^1$, then $\mathrm{ker}f = \mathcal{K}_{\mathrm{Im}g}$ if and only if $\mathrm{ker}f^1 = \mathcal{K}_{\mathrm{Im}g^1}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we do this, then &#xA;$$S\stackrel{f}{\longrightarrow} T \stackrel{g}{\longrightarrow} U$$&#xA;is a short exact sequence if and only if&#xA;$$1 \longrightarrow S^1 \stackrel{f^1}{\longrightarrow} T^1\stackrel{g^1}{\longrightarrow} U^1 \longrightarrow 1$$&#xA;is exact at $S^1$, $T^1$, and $U^1$.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-21T20:43:16.647" LastActivityDate="2011-01-21T20:43:16.647" />
  <row Id="18399" PostTypeId="2" ParentId="18395" CreationDate="2011-01-21T04:37:08.473" Score="12" ViewCount="" Body="&lt;p&gt;Neither containment holds in general.  On $\mathbb{R}$, the function $f$ such that $f(x)=1/x$ if $x\geq1$ and $f(x)=0$ otherwise is in $L^2\setminus L^1$, and the function $g$ such that $g(x)=1/\sqrt{x}$ if $0\lt x\leq1$ and $g(x)=0$ otherwise is in $L^1\setminus L^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For bounded domains, $L^2\subset L^1$, because what keeps a function from being integrable on a bounded set is being too large, and squares make large numbers larger.  If $f$ is a bounded function in $L^1$, then $f$ is also in $L^2$, because what keeps a bounded function from being integrable is not going to zero fast enough, and squares make numbers go to zero faster.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2011-01-21T04:37:08.473" />
  <row Id="18400" PostTypeId="2" ParentId="18395" CreationDate="2011-01-21T04:38:40.733" Score="2" ViewCount="" Body="&lt;p&gt;You can get an example from any two sequences of positive reals such that $\sum \alpha_n \beta_n &amp;lt; \infty$ but $\sum \alpha_n^2 \beta_n = \infty$, e.g. $\alpha_n = n$ and $\beta_n = 1/n^3$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the reverse non-containment, take $\alpha_n = 1/n$ and $\beta_n = 1$.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2011-01-21T04:38:40.733" />
  <row Id="18401" PostTypeId="2" ParentId="18395" CreationDate="2011-01-21T04:41:51.380" Score="3" ViewCount="" Body="&lt;p&gt;For an important example,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Check out Kolmogorov's example of a function in $L^1$ whose Fourier series diverges almost everywhere: &lt;a href=&quot;http://books.google.com/books?id=ikN59GkYJKIC&amp;amp;pg=PA1&quot; rel=&quot;nofollow&quot;&gt;http://books.google.com/books?id=ikN59GkYJKIC&amp;amp;pg=PA1&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By &lt;a href=&quot;http://en.wikipedia.org/wiki/Carleson%27s_theorem&quot; rel=&quot;nofollow&quot;&gt;Carleson's theorem&lt;/a&gt; this cannot be in $L^2$.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2011-01-21T04:41:51.380" CommentCount="4" />
  <row Id="18403" PostTypeId="1" AcceptedAnswerId="18416" CreationDate="2011-01-21T05:28:26.927" Score="3" ViewCount="163" Body="&lt;p&gt;This is related to the content in the book by Grimmett and Stirzaker &quot;Probability and random processes&quot; 3rd ed. On page 111, it calculated, as an example, the conditional density function of $X_1+X_2$ given $X_1=X_2$ for two i.i.d. exponential r.v. with parameter $\lambda$. They used two methods, &lt;p&gt;&#xA;1) using $Y_1=X_1+X_2$ and $Y_2=X_1/X_2$ and then obtained the result&#xA;$f_{Y_1\vert Y_3}(y_1\vert y_3)=\lambda^2 y_1 e^{-\lambda y_1}$ for $y_1\ge 0$&lt;p&gt;&#xA;2) using $Y_1=X_1+X_2$ and $Y_3=X_1-X_2$ and then obtained the result&#xA;$f_{Y_1\vert Y_3}(y_1\vert y_3)=\lambda e^{-\lambda y_1}$ for $y_1\ge 0$&lt;p&gt;&#xA;The explanation provided there is that the two results are based on different sets of information.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is: Does this only occurs for 2 r.v.'s when conditioned on an event with zero probability, or it is in general true even for the condition involving only 1 r.v.? I am a little confused at the situation when the the conditional event has zero probability. Thanks!&lt;/p&gt;&#xA;" OwnerUserId="6016" LastEditorUserId="6016" LastEditorDisplayName="" LastEditDate="2011-01-21T06:23:07.560" LastActivityDate="2011-01-24T20:14:35.873" Title="question about conditional probability for continuous r.v." Tags="&lt;probability&gt;" AnswerCount="2" />
  <row Id="18404" PostTypeId="2" ParentId="18395" CreationDate="2011-01-21T06:04:45.723" Score="4" ViewCount="" Body="&lt;p&gt;I assume that you mean integrable functions in $[0,1]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the sense of &lt;a href=&quot;http://en.wikipedia.org/wiki/Baire_category_theorem&quot; rel=&quot;nofollow&quot;&gt;Baire&lt;/a&gt; almost every function in $L^{1}[0,1]$ is not in $L^{2}[0,1]$:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The space $L^{2}[0,1]$ is &lt;strong&gt;meager&lt;/strong&gt; in $L^{1}[0,1]$ (that is to say it is a countable union of sets whose closure has empty interior in $L^{1}$).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The easiest way to see this by using the &lt;a href=&quot;http://en.wikipedia.org/wiki/Open_mapping_theorem_(functional_analysis)#Generalizations&quot; rel=&quot;nofollow&quot;&gt;open mapping theorem&lt;/a&gt;: the inclusion $L^{2}[0,1] \to L^{1}[0,1]$ is continuous but not onto. More explicitly, the set $B_{n} = \{f \in L^{1}\,:\,\int |f|^{2} \leq n\}$ is easily seen to be closed and have empty interior and $L^{2}[0,1] = \bigcup_{n=1}^{\infty} B_{n}$. Similarly, if $1 \leq p &amp;lt; q \leq \infty$ then $L^{q}[0,1] \subset L^{p}[0,1]$ is meager.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This result is analogous to Baire's theorem saying that almost every continuous function on $[0,1]$ is nowhere differentiable, and with the same defect: If you choose a 'generic' function it won't be differentiable (or square-integrable) but from the statement you don't have a clue what such a function looks like.&lt;/p&gt;&#xA;" OwnerUserId="5363" LastEditorUserId="5363" LastEditorDisplayName="" LastEditDate="2011-01-21T06:12:19.553" LastActivityDate="2011-01-21T06:12:19.553" />
  <row Id="18405" PostTypeId="1" AcceptedAnswerId="18503" CreationDate="2011-01-21T06:21:52.837" Score="4" ViewCount="314" Body="&lt;p&gt;If I have a two-sided coin with probability $p$ showing head. I repeatedly toss it until either HTHTH or HTHH appears. Can you calculate &lt;p&gt;&#xA;1) the probability when I got HTHTH, and &lt;p&gt;&#xA;2) the expected value of the number of tosses before I stop?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="6016" LastEditorUserId="6016" LastEditorDisplayName="" LastEditDate="2011-01-22T00:05:17.007" LastActivityDate="2011-02-02T13:52:48.973" Title="a question related to two competing patterns in coin tossing" Tags="&lt;probability&gt;" AnswerCount="3" CommentCount="7" FavoriteCount="2" />
  <row Id="18406" PostTypeId="1" AcceptedAnswerId="18414" CreationDate="2011-01-21T06:56:38.250" Score="2" ViewCount="118" Body="&lt;p&gt;Let us suppose we have slots $n$ slots $1, \ldots, n$ and $k$ pebbles, each of which is initially placed in some slot. Now the pebbles want to space themselves out as evenly as possible, and so they do the following. At each time step $t$, each pebble moves to the slot closest to the halfway point between its neighboring pebbles; if there is a tie, it chooses the slot to the left.  The leftmost and rightmost pebbles apply the same procedure, but imagining that there are slots numbered $0$ and $n+1$ with pebbles in them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Formally, numbering the pebbles &#xA;$1, \ldots, k$ from left to right, and letting $x_i(t)$ be the slot of the $i$'th pebble at time t, we have $$ x_i(t+1) = \lfloor \frac{x_{i-1}(t)+x_{i+1}(t)}{2} \rfloor, i = 2, \ldots, k-1$$ $\lfloor \cdot \rfloor$ rounds down to the closest integer. Similarly, $$ x_1(t+1) = \lfloor (1/2) x_2(t) \rfloor, x_k(t+1) = \lfloor \frac{x_{k-1}(t) + (n+1)}{2} \rfloor.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the fixed point of this procedure is the arrangement in which $x_{i+1} - x_i$ and $x_i - x_{i-1}$ differ by $1$. &lt;strong&gt;My question:&lt;/strong&gt; is it true that this fixed point is reached by the above procedure after sufficiently many iterations?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Why I care&lt;/strong&gt;: no concrete reason really, I am just reading about finite difference methods, and this seemed like a simple problem connected with some of the things which are confusing me. &lt;/p&gt;&#xA;" OwnerUserId="2120" LastEditorUserId="2120" LastEditorDisplayName="" LastEditDate="2011-01-21T12:46:04.253" LastActivityDate="2011-01-21T12:46:04.253" Title="A very simple discrete dynamical system with pebbles" Tags="&lt;discrete-mathematics&gt;&lt;dynamical-systems&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="18407" PostTypeId="2" ParentId="18362" CreationDate="2011-01-21T07:06:37.323" Score="1" ViewCount="" Body="&lt;p&gt;A short sketch besides the request of Andres' for further specifications:&lt;br&gt;&#xA;if you use the inverse-series for  $ \Gamma(1+x)-1 $ (in Pari/GP: &lt;em&gt;serreverse (gamma(1+x)-1)&lt;/em&gt; ) you get something like    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ f(x) = -1.73245471460*x + 5.14288192485*x^2 - 22.3588922658*x^3 + 120.586032684*x^4 $&lt;br&gt;&#xA;$ - 732.743269181*x^5 + 4785.68759665*x^6 - 32793.0682929*x^7 + O(x^8) $     &lt;/p&gt;&#xA;&#xA;&lt;p&gt;By 64 terms of the series it is not clear, whether the series has finite radius of convergence $ \rho $, possibly it is something like $ \rho \sim 1/9 $. If we define $ g(x) = f(x-1)+1 $ and use Euler-/Noerlund-sums for acceleration of convergence it seems we can arrive at meaningful values in a small range; examples&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;g(1.5)                  = 0.595332094501  // Noerlund-sum&#xA;gamma(  0.595332094501) = 1.500000000&#xA;g(1.8)                  = 0.492222531811  // Noerlund-sum&#xA;gamma(  0.492222531811) = 1.800000000&#xA;g(2.0)                  = 0.442877396485  // Noerlund-sum&#xA;gamma(  0.442877396485) = 2.000000000&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="1714" LastActivityDate="2011-01-21T07:06:37.323" />
  <row Id="18408" PostTypeId="2" ParentId="18334" CreationDate="2011-01-21T07:36:53.860" Score="3" ViewCount="" Body="&lt;p&gt;I haven't read the paper, but discussions I've seen of it indicate that it doesn't give an answer for all inputs, meaning that it doesn't really solve the problem.&lt;/p&gt;&#xA;" OwnerUserId="5676" LastActivityDate="2011-01-21T07:36:53.860" />
  <row Id="18409" PostTypeId="1" CreationDate="2011-01-21T07:38:39.800" Score="2" ViewCount="168" Body="&lt;p&gt;Given the positive sequence &#xA;$a_{n+2}$ = $\sqrt{a_{n+1}}$ + $\sqrt{a_{n}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to prove these.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1)$|a_{n+2}| &gt; 1 $ for sufficiently large $n \ge N$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2)Let $b_{n} = |a_{n} - 4|$. Show that $b_{n+2} &amp;lt; (b_{n+1} + b_{n})/3$ for $n \ge N$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3)Prove that the sequence converges.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How should I proceed.&#xA;Is there a recurrence formula for $a_{n}$ like a continued fraction.&lt;/p&gt;&#xA;" OwnerUserId="5049" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2011-01-21T08:24:59.780" LastActivityDate="2011-01-21T08:24:59.780" Title="About the positive sequence $a_{n+2}$ = $\sqrt{a_{n+1}} + \sqrt{a_{n}}$" Tags="&lt;sequences-and-series&gt;&lt;limit&gt;&lt;convergence&gt;" AnswerCount="1" CommentCount="10" />
  <row Id="18410" PostTypeId="2" ParentId="18406" CreationDate="2011-01-21T07:58:46.953" Score="0" ViewCount="" Body="&lt;p&gt;You have a total number of unoccupied spaces $n-k$, divided into $k+1$ intervals between pebbles.  Your procedure looks at adjacent pairs of empty intervals and tries to make them more similar.  If they differ by more than one, the larger of the two intervals will be reduced and the smaller increased so that their difference becomes zero or one.  If they differ by zero, they will be left alone.  If they differ by one, they will be swapped if necessary so that the larger of the two intervals is on the right.  In short, the only steady states will be those where each interval is the same size or one greater than its neighboring interval on the left -- there are many of these.  If the update rules are applied one pebble at a time, a steady state will always be reached, because the total discrepancy, $D \equiv \sum_{i=1}^{k} |x_{i+1} - 2x_i + x_{i-1} - 1/2|$, decreases with each move.  But if the update rules are applied simultaneously, it is less clear that you must reach a steady state.  I think that you still must, but I don't have as simple a proof.&lt;/p&gt;&#xA;" OwnerUserId="5546" LastEditorUserId="5546" LastEditorDisplayName="" LastEditDate="2011-01-21T08:15:22.630" LastActivityDate="2011-01-21T08:15:22.630" />
  <row Id="18411" PostTypeId="2" ParentId="18409" CreationDate="2011-01-21T08:02:52.927" Score="4" ViewCount="" Body="&lt;p&gt;Here are some hints.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) You can show that if $a_{n}$ is ever greater than $1$, then so is $a_{n+1}$.  On the other hand, if $a_n&amp;lt;1$, then you can show that $a_{n+1}\gt a_n$, and $a_{n+2}\gt 2a_n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) You can use the triangle inequality and a factorization trick, $\sqrt{x}-\sqrt{4}=\frac{x-4}{\sqrt{x}+\sqrt{4}}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3)  Show that $b_n\to 0$.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2011-01-21T08:02:52.927" />
  <row Id="18412" PostTypeId="1" AcceptedAnswerId="18415" CreationDate="2011-01-21T08:19:00.990" Score="0" ViewCount="171" Body="&lt;p&gt;Consider a sequence of functions $f_n$ where $f_n : \mathbb{R} \to \mathbb{R}$&#xA; and $f_n$ are all differentiable with derivatives $f^\prime_n$. The sequence  $f_n$ and the sequence $f^\prime_n$ both converge uniformly to functions $f$ and $g$ respectively. According to the definition given in this &lt;a href=&quot;http://en.wikipedia.org/wiki/Uniform_convergence&quot; rel=&quot;nofollow&quot;&gt;wiki page on uniform convergence in section 'To Differentiability&quot;&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;If $ f_n $ converges uniformly to $ f $, and if all the $ f_n $ are differentiable, and if the derivatives $f^\prime_n$ converge uniformly to $g$, then $ f $ is differentiable and its derivative is $g$.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Q.1) Can a similar definition be used for higher order derivatives ?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a different definition of derivative which is not same as the usual definition $f&amp;#39;(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Q.2) In case of both not being in agreement with each other which one should I use for further analysis on $f$ for example in theorems involving the derivative of $f$  ?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="2987" LastActivityDate="2011-01-21T14:18:10.727" Title="On differentiability and definition of derivative in case of uniform convergence." Tags="&lt;real-analysis&gt;" AnswerCount="2" CommentCount="5" />
  <row Id="18413" PostTypeId="1" AcceptedAnswerId="18418" CreationDate="2011-01-21T08:25:08.260" Score="4" ViewCount="139" Body="&lt;p&gt;The thought came from the following problem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $V$ be a Euclidean space.  Let $T$ be an inner product on $V$.  Let $f$ be a linear transformation $f:V \to V$ such that $T(x,f(y))=T(f(x),y)$ for $x,y\in V$.  Let $v_1,\dots,v_n$ be an orthonormal basis, and let $A=(a_{ij})$ be the matrix of $f$ with respect to this basis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The goal here is to prove that the $A$ is symmetric.  I can prove this easily enough by saying:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $T$ is an inner product, $T(v_i,v_j)=\delta_{ij}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align*}&#xA;T(A v_j,v_i)&amp;amp;=T(\sum_{k=1}^n a_{kj} v_k,v_i)\\&#xA;&amp;amp;=T(a_{1j} v_1,v_i) + \dots + T(a_{nj} v_n,v_i)\\&#xA;&amp;amp;=a_{1j} T(v_1,v_i) + \dots + a_{nj} T(v_n,v_i)\tag{bilinearity}\\&#xA;&amp;amp;=a_{ij}\tag{$T(v_i,v_j)=\delta_{ij}$}\\&#xA;\end{align*}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the same logic,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align*}&#xA;T(A v_j,v_i)&amp;amp;=T(v_j,A v_i)\\&#xA;&amp;amp;=T(v_j,\sum_{k=1}^n a_{ki} v_k)\\&#xA;&amp;amp;=T(v_j,a_{1i} v_1)+\dots+T(v_j,a_{ni} v_n)\\&#xA;&amp;amp;=a_{1i} T(v_j,v_1)+\dots+a_{ni} T(v_j,v_n)\\&#xA;&amp;amp;= a_{ji}\\&#xA;\end{align*}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By hypothesis, $T(A v_j,v_i)=T(v_j,A v_i)$, therefore $a_{ij}=T(A v_j,v_i)=T(v_j,T v_i)=a_{ji}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I had this other idea though, that since $T$ is an inner product, its matrix is positive definite.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$T(x,f(y))=T(f(x),y)$ in matrix notation is $x^T T A y=(A x)^T T y$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align*}&#xA;x^T T A y &amp;amp;= (A x)^T T y\\&#xA;&amp;amp;=x^T A^T T y\\&#xA;TA &amp;amp;= A^T T\\&#xA;(TA)^T &amp;amp;= (A^T T)^T\\&#xA;A^T T^T &amp;amp;= T^T A\\&#xA;TA &amp;amp;= A^T T^T\tag{T is symmetric}\\&#xA;&amp;amp;= (TA)^T\tag{transpose of matrix product}\\&#xA;\end{align*}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is where I got stuck.  We know that $T$ and $TA$ are both symmetric matrices.  Clearly $T^{-1}$ is symmetric.  If it can be shown that $T^{-1}$ and $AT$ commute, that would show it.&lt;/p&gt;&#xA;" OwnerUserId="326" LastActivityDate="2011-01-22T10:15:36.133" Title="If the product of an invertible symmetric matrix and some other matrix is symmetric, is that other matrix also symmetric?" Tags="&lt;linear-algebra&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="18414" PostTypeId="2" ParentId="18406" CreationDate="2011-01-21T08:27:36.737" Score="3" ViewCount="" Body="&lt;p&gt;Start with 8 slots, and pebbles at 3 and 5. The pebbles go to slots 2 and 6, and then back to 3 and 5, indefinitely.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update&lt;/strong&gt; Even simpler: 5 slots. $(1,4) \rightarrow (2,3) \rightarrow (1,4)$.&lt;/p&gt;&#xA;" OwnerUserId="1508" LastEditorUserId="1508" LastEditorDisplayName="" LastEditDate="2011-01-21T11:27:34.227" LastActivityDate="2011-01-21T11:27:34.227" />
  <row Id="18415" PostTypeId="2" ParentId="18412" CreationDate="2011-01-21T08:27:40.963" Score="2" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Uniform_convergence#To_differentiability&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Uniform_convergence#To_differentiability&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $f_n$ converges uniformly to $f$, and if all the $f_n$ are differentiable, and if the derivatives $f&amp;#39;_n$ converge uniformly to $g$, then $f$ is differentiable and its derivative is $g$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if you use the theorem and the derivates $f&amp;#39;_n$ are also differentiable you have again that the $f&amp;#39;_n$ converge uniformly to $f&amp;#39;$ and you can recursively use the theorem from above (you will need that $f&amp;#39;&amp;#39;_n$ converges uniformly to some $g'$).&lt;/p&gt;&#xA;" OwnerUserId="3123" LastActivityDate="2011-01-21T08:27:40.963" />
  <row Id="18416" PostTypeId="2" ParentId="18403" CreationDate="2011-01-21T08:58:09.847" Score="2" ViewCount="" Body="&lt;p&gt;The conclusion from this example is that sometimes we need to pay special attention when conditioning on events of zero probability. Suppose, as in the example, that $X_1$ and $X_2$ are i.i.d. exponentials. Fix $a,b &gt; 0$ with $a &amp;lt; b$, and consider the question what is ${\rm P}\big[ X_1+X_2 \in [a,b] \big| X_1=X_2 \big]$? You may (naturally) wish to interpret this as&#xA;$$&#xA;{\rm P}\big[X_1  + X_2  \in [a,b]\big|X_1  - X_2  = 0 \big] = \mathop {\lim }\limits_{h \to 0^ +  } {\rm P}\big[ X_1  + X_2  \in [a,b]\big|0 &amp;lt; X_1  - X_2  &amp;lt; h \big],&#xA;$$&#xA;so in this case&#xA;$$&#xA;{\rm P}\big [ X_1+X_2 \in [a,b] \big| X_1=X_2 \big ] = \mathop {\lim }\limits_{h \to 0^ +  } \frac{{{\rm P} [ X_1  + X_2  \in [a,b],0 &amp;lt; X_1  - X_2  &amp;lt; h]}}{{{\rm P} [0 &amp;lt; X_1  - X_2  &amp;lt; h]}}.&#xA;$$&#xA;On the other hand, you may (less likely) wish to interpret that as&#xA;$$&#xA;{\rm P} \big[ X_1  + X_2  \in [a,b] \big|X_1 / X_2  = 1 \big] = \mathop {\lim }\limits_{h \to 0^ +  } {\rm P}\big[ X_1  + X_2  \in [a,b] \big|1 &amp;lt; X_1 / X_2 &amp;lt; 1+h \big],&#xA;$$&#xA;leading to&#xA;$$&#xA;{\rm P}\big [ X_1+X_2 \in [a,b] \big| X_1=X_2 \big ] = \mathop {\lim }\limits_{h \to 0^ +  } \frac{{{\rm P} [ X_1  + X_2  \in [a,b],0 &amp;lt; X_1  - X_2  &amp;lt; X_2 h]}}{{{\rm P} [0 &amp;lt; X_1  - X_2  &amp;lt; X_2 h]}}.&#xA;$$&#xA;However, it should not be surprising that the two interpretations lead to different probabilities, since we have limits of the form $a_i (h) / b_i (h)$, $i=1,2$, with $a_i(h),b_i(h) \to 0$ as $h \to 0^+$. Clearly, ${\rm P} [0 &amp;lt; X_1  - X_2  &amp;lt; h]$ and ${\rm P} [0 &amp;lt; X_1  - X_2  &amp;lt; X_2 h]$ are not expected to have the same behavior as $h \to 0^+$. So, the only problem was how to interpret conditioning on $X_1 = X_2$, and this is up to you. In general, however, there is no such problem; you just use the formula $f_{Y|X} (y|x) = \frac{{f_{X,Y} (x,y)}}{{f_X (x)}}$, to find the conditional density function of $Y$ given $X=x$ (where ${f_{X,Y}}$ is the joint density function of $X$ and $Y$). The conditional distribution function of $Y$ given $X=x$ is obtained by integrating the conditional density.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;: The conditional density function of $X_1+X_2$ given $X_1-X_2=0$ (where $X_1$ and $X_2$ are independent exponential($\lambda$) rv's), when interpreting the conditioning with respect to the random variable $X_1-X_2$, is, by Eq. (11) in the book, the exponential$(\lambda)$ density function, $\lambda e^{-\lambda y}$, $y \geq 0$. Hence, ${\rm P}\big[X_1  + X_2  \in [a,b]\big|X_1  - X_2  = 0 \big]$ is given by&#xA;$$&#xA;\mathop {\lim }\limits_{h \to 0^ +  } {\rm P}\big[ X_1  + X_2  \in [a,b]\big|0 &amp;lt; X_1  - X_2  &amp;lt; h \big] = \int_a^b {\lambda e^{ - \lambda y} \,{\rm d}y}  = e^{ - \lambda a}  - e^{ - \lambda b}. &#xA;$$&#xA;On the other hand, the conditional density function of $X_1+X_2$ given $X_1/X_2=1$, when interpreting the conditioning with respect to the random variable $X_1/X_2$, is, by Eq. (10) in the book, the ${\rm Gamma}(2,\lambda)$ density function, $\lambda ^2 ye^{ - \lambda y}$, $y \geq 0$ (i.e., the density function of $X_1+X_2$; this is since $X_1+X_2$ and $X_1/X_2$ are independent). Hence, ${\rm P} \big[ X_1  + X_2  \in [a,b] \big|X_1 / X_2  = 1 \big]$ is given by&#xA;$$&#xA;\mathop {\lim }\limits_{h \to 0^ +  } {\rm P}\big[ X_1  + X_2  \in [a,b] \big|1 &amp;lt; X_1 / X_2 &amp;lt; 1+h \big] = \int_a^b {\lambda ^2 ye^{ - \lambda y} \,{\rm d}y} = &#xA;(\lambda a + 1)e^{ - \lambda a}  - (\lambda b + 1)e^{ - \lambda b}.&#xA;$$&#xA;Both results agree with numerical simulations (approximating the probabilities for small values of $h$).&lt;/p&gt;&#xA;" OwnerUserId="2810" LastEditorUserId="2810" LastEditorDisplayName="" LastEditDate="2011-01-24T20:14:35.873" LastActivityDate="2011-01-24T20:14:35.873" CommentCount="0" />
  <row Id="18417" PostTypeId="1" AcceptedAnswerId="18428" CreationDate="2011-01-21T11:20:36.287" Score="1" ViewCount="129" Body="&lt;p&gt;I have some questions about the definition of splines and in particular periodic splines. &#xA;So in non periodic case splines are piecewise polynomials of degree $&amp;lt;=m$  which are $m-1$ times differentiable.(I took $m-1$ time differentiability for simplicity)&#xA;By this definition  we can call linear or quadratic polynomials  splines of degree 3.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For periodic case we need boundary conditions on some interval$(a,b)$ $s^{(q)}(a)=s^{(q)}(b),q=1...m-1$.&#xA;Does it mean that periodic splines  are piecewise polynomials of degree $m$(not smaller then $m$) or constants?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there analogue for truncated polynomials basis in periodic case?&lt;/p&gt;&#xA;" OwnerUserId="6052" LastEditorUserId="6052" LastEditorDisplayName="" LastEditDate="2011-01-21T13:45:26.940" LastActivityDate="2011-01-27T14:29:19.653" Title="Some question about definition of splines and periodic splines" Tags="&lt;numerical-methods&gt;" AnswerCount="1" CommentCount="7" />
  <row Id="18418" PostTypeId="2" ParentId="18413" CreationDate="2011-01-21T11:27:04.327" Score="9" ViewCount="" Body="&lt;p&gt;It's not true in general, e.g. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$A = \begin{pmatrix} 2 &amp;amp; 1 \\ 1 &amp;amp; 1 \end{pmatrix}$&#xA;$B = \begin{pmatrix} 1 &amp;amp; 1 \\ 2 &amp;amp; 1 \end{pmatrix}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$AB = \begin{pmatrix} 4 &amp;amp; 3 \\ 3 &amp;amp; 2 \end{pmatrix}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(with thanks to Rahul for formatting help)&lt;/p&gt;&#xA;" OwnerUserId="1470" LastEditorUserId="1470" LastEditorDisplayName="" LastEditDate="2011-01-21T13:28:05.923" LastActivityDate="2011-01-21T13:28:05.923" CommentCount="0" />
  <row Id="18419" PostTypeId="2" ParentId="18412" CreationDate="2011-01-21T11:47:12.500" Score="0" ViewCount="" Body="&lt;p&gt;The theorem you mention certainly applies to higher order derivatives, exactly as written by user3123 above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For your other question, it seems to me that you are worried about the situation where $f$ may not be differentiable, but there exists a sequence $f_n$ which uniformly converges to $f$, with each $f_n$ differentiable and the derivatives $f_n&amp;#39;$ converging uniformly to some function $g$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do you have an example of such a situation?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To rephrase what Jonas mentioned already in his comments, this is not likely to happen.  When the sequence $f_n$ uniformly converges to $f$, and the derivatives $f_n&amp;#39;$ uniformly converge to $g$, the following holds:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\lim_{h\searrow 0}\frac{f(x+h)-f(x)}{h} = g(x).$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This &lt;em&gt;is&lt;/em&gt; the derivative of $f$.  It is not an alternative definition, it really is a theorem.  I'm sorry!  Perhaps you can give some more details on the exact situation you have in mind.&lt;/p&gt;&#xA;" OwnerUserId="4322" LastEditorUserId="4322" LastEditorDisplayName="" LastEditDate="2011-01-21T14:18:10.727" LastActivityDate="2011-01-21T14:18:10.727" CommentCount="5" />
  <row Id="18420" PostTypeId="1" AcceptedAnswerId="18423" CreationDate="2011-01-21T12:36:36.390" Score="0" ViewCount="71" Body="&lt;p&gt;There is a sequence $a_n$ and $a_n \to 0$ as $n \to \infty$. Let $S_N = \sum_{n=0}^{N} a_n$. Let the sequence $S_N$ be converging to a limit $l$ where $l \in \mathbb{R}$. What is $\lim_{K\to\infty}\lim_{N\to\infty}\sum_{n=K}^{N} a_n $ ?&lt;/p&gt;&#xA;" OwnerUserId="2987" LastActivityDate="2011-01-21T12:53:27.260" Title="double limit of a series summation" Tags="&lt;sequences-and-series&gt;" AnswerCount="1" />
  <row Id="18422" PostTypeId="2" ParentId="18395" CreationDate="2011-01-21T12:46:41.857" Score="0" ViewCount="" Body="&lt;p&gt;If we look at the discrete group $\mathbb{Z}$, we have $L^1(\mathbb{Z})=\ell^1(\mathbb{Z})\subset \ell^2(\mathbb{Z})=L^2(\mathbb{Z})$ - see the thread &lt;a href=&quot;http://math.stackexchange.com/questions/4094/how-do-you-show-that-l-p-subset-l-q-for-p-leq-q&quot;&gt;How do you show that $\ell^p\subset \ell^q$  for $p≤q$?&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we look at the compact group $\mathbb{T}$ (the unit circle), we have $L^1(\mathbb{T})\subset L^2(\mathbb{T})$, since by Hölder's inequality&#xA;$$\|f\|_{L^1}=\int|f|dx=\int1\cdot|f|dx\leq\sqrt{\int1^2dx}\sqrt{\int|f|^2dx}=1\cdot\|f\|_{L^2}=\|f\|_{L^2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we look at the continuous non-compact group $\mathbb{R}$ there is no inclusion.&lt;/p&gt;&#xA;" OwnerUserId="1154" LastActivityDate="2011-01-21T12:46:41.857" />
  <row Id="18423" PostTypeId="2" ParentId="18420" CreationDate="2011-01-21T12:53:27.260" Score="1" ViewCount="" Body="&lt;p&gt;0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$l = \lim_{N\to\infty}\sum_{n=0}^{N} a_n$. Therefore $\lim_{N\to\infty}\sum_{n=K}^{N} a_n = l - \sum_{n=0}^{K-1} a_n = l - S_{K-1}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But since the sequence converges to $l$, $\lim_{K\to\infty} |S_{K} - l| = 0$.&lt;/p&gt;&#xA;" OwnerUserId="5676" LastActivityDate="2011-01-21T12:53:27.260" />
  <row Id="18424" PostTypeId="1" CreationDate="2011-01-21T12:57:28.423" Score="0" ViewCount="139" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;There is a three digit number $abc$ and a two digit number $xy$ such that the product of $abc$ with $xy$ is same as the product of $cba$ and $yx$,if $x$,$y$ can be similar digits, then find number of such pairs of $P$ and $Q$?&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;a) 99&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;b) 900&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;c) 810&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;d) cannot be determined&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5918" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2011-02-21T05:35:17.533" LastActivityDate="2011-02-21T05:35:17.533" Title="Number system: Product of a $3$ digit and a $2$ digit number" Tags="&lt;homework&gt;&lt;arithmetic&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="18425" PostTypeId="1" CreationDate="2011-01-21T13:10:42.230" Score="10" ViewCount="137" Body="&lt;p&gt;The theorem of Seifert-Van-Kampen states that the fundamental group $\pi_1$ commutes with &lt;em&gt;certain&lt;/em&gt; colimits. There is a beautiful and conceptual proof in Peter May's &quot;A Concise Course in Algebraic Topology&quot;, stating the Theorem first for groupoids and then gives a formal argument how to deduce the result for $\pi_1$. However, there we need the assumption that our space is covered by path-connected open subsets, whose finite intersections are path-connected again. This assumption can be weakened using a little &quot;perturbation&quot; trick which is explained in the proof given by Hatcher in his book &quot;Algebraic Topology&quot;. After doing this, we only need that triple intersections are path connected (and we cannot do better).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question 1.&lt;/strong&gt; Is it possible to &quot;conceptualize&quot; the perturbation trick, thus weakening the assumption concerning finite intersections? Maybe this is answered by one of the pure categorical proofs of Seifert-van-Kampen?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question 2.&lt;/strong&gt; In practice (explicit calculations of fundamental groups), do we actually often need or use that this weakening of the assumption is possible?&lt;/p&gt;&#xA;" OwnerUserId="1650" LastActivityDate="2011-01-21T13:10:42.230" Title="Perturbation trick in the proof of Seifert-van-Kampen" Tags="&lt;algebraic-topology&gt;" FavoriteCount="2" />
  <row Id="18427" PostTypeId="2" ParentId="18424" CreationDate="2011-01-21T14:23:46.040" Score="1" ViewCount="" Body="&lt;p&gt;Suggeston:  If you write out abc as 100a+10b+c and similarly with the others, you can write an explicit equation that your numbers need to satisfy.  Then if you impose the conditions that all variables are naturals, $0\lt a\lt 10$ and so on, you can look for a solution.  A computer search would be easy.  Without a computer, looking for divisibility requirements is your friend on problems like this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Added:  None of the proposed answers is correct.  I get 843 possibilities with $x\le y$.  810 of them are of the form xx * aba.  The others are ( 12 , 231 ) ( 12 , 462 ) ( 12 , 693 ) ( 13 , 341 ) ( 13 , 682 ) ( 14 , 451 ) ( 15 , 561 ) ( 16 , 671 ) ( 17 , 781 ) ( 18 , 891 ) ( 23 , 352 ) ( 24 , 231 ) ( 24 , 462 ) ( 24 , 693 ) ( 25 , 572 ) ( 26 , 341 ) ( 26 , 682 ) ( 27 , 792 ) ( 28 , 451 ) ( 34 , 473 ) ( 35 , 583 ) ( 36 , 231 ) ( 36 , 462 ) ( 36 , 693 ) ( 39 , 341 ) ( 39 , 682 ) ( 45 , 594 ) ( 46 , 352 ) ( 48 , 231 ) ( 48 , 462 ) ( 48 , 693 ) ( 68 , 473 ) ( 69 , 352 )&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if we count these both ways the total would be 810+2*33=876&lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2011-02-20T23:31:52.357" LastActivityDate="2011-02-20T23:31:52.357" />
  <row Id="18428" PostTypeId="2" ParentId="18417" CreationDate="2011-01-21T14:47:06.577" Score="1" ViewCount="" Body="&lt;p&gt;Wlog we can take the knot at $x=0$, and then we're looking for two polynomials of degree $\le m$ which join there with $C(m-1)$ continuity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the two polynomials both have degree $m-1$ then, since they have $m$ derivatives (including the 0th derivative) in common they are equal. (Whether you consider a polynomial to be a spline on the basis that it's piecewise polynomial with all the pieces having the same function is up to you).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, it's possible for one to be of degree $m$ and the other of degree $m-1$. Consider the case $m=2$. Our polynomials are $f(x) = ax^2 + bx + c$ and $g(x) = dx + e$. $f(0) = g(0) \Rightarrow c = e$, $f^\prime(0) = g^\prime(0) \Rightarrow b = d$. $a$ is unconstrained.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Example&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;$f(x) = \left\{\begin{array}{l}&#xD;&#xA;  (x-1)^2 &amp;amp; 0 &amp;lt;= x &amp;lt; 2 \\&#xD;&#xA;  2x - 3 &amp;amp; 2 &amp;lt;= x &amp;lt; 3 \\&#xD;&#xA;  4 - (x-4)^2 &amp;amp; 3 &amp;lt;= x &amp;lt; 5 \\&#xD;&#xA;  13 - 2x &amp;amp; 5 &amp;lt;= x &amp;lt; 6 \\&#xD;&#xA;  f(x-6) &amp;amp; x &amp;gt;= 6 \\&#xD;&#xA;  f(x+6) &amp;amp; x &amp;lt; 0&#xD;&#xA;\end{array}\right.$&lt;/p&gt;&#xA;" OwnerUserId="5676" LastEditorUserId="5676" LastEditorDisplayName="" LastEditDate="2011-01-27T14:29:19.653" LastActivityDate="2011-01-27T14:29:19.653" CommentCount="4" />
  <row Id="18429" PostTypeId="1" CreationDate="2011-01-21T15:02:46.100" Score="2" ViewCount="130" Body="&lt;p&gt;Suppose I have an integral:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$F \equiv \int_{a}^{b} f(x) dx$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that $f(x)$ is real-valued, finite and non-negative everywhere on the interval $(a, b)$ where $a$, $b$ are real numbers or $\pm \infty$ and $a \leq b$.  I know that in the obvious, well-behaved cases this implies $F \geq 0$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any obscure pathological case where $F &amp;lt; 0$?  You may generalize to cases that involve double, triple or higher order integrals.  However, for each integral the property that the upper bound is greater than or equal to the lower bound and the bounds are always real numbers or $\pm \infty$ must hold.&lt;/p&gt;&#xA;" OwnerUserId="6088" LastEditorUserId="6088" LastEditorDisplayName="" LastEditDate="2011-01-21T16:56:40.030" LastActivityDate="2011-01-21T17:02:17.100" Title="Integrand non-negative everywhere, integral negative?" Tags="&lt;integral&gt;" AnswerCount="1" CommentCount="6" />
  <row Id="18430" PostTypeId="1" CreationDate="2011-01-21T15:04:21.493" Score="3" ViewCount="207" Body="&lt;p&gt;I know that $\int f(x) \overline{g(x)} dx$ is an inner product on $L^2$. But is it one on $L^1$? I think it isn't, but I am have had difficulty figuring out which defining property is violated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance for any pointers!&lt;/p&gt;&#xA;" OwnerUserId="6089" LastActivityDate="2011-01-22T04:30:49.720" Title="Is this an inner product on $L^1$?" Tags="&lt;functional-analysis&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="18431" PostTypeId="2" ParentId="18430" CreationDate="2011-01-21T15:09:46.950" Score="4" ViewCount="" Body="&lt;p&gt;Take $f$ a function in $L^1$ but not in $L^2$.  We had examples recently-they blow up too much near a point.  Then $\int f(x) \overline{f(x)} dx$ is not defined.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-01-21T15:09:46.950" />
  <row Id="18432" PostTypeId="2" ParentId="18083" CreationDate="2011-01-21T15:19:57.007" Score="4" ViewCount="" Body="&lt;p&gt;Since the question is tagged differential-topology, I assume we are talking about a differentiable manifolds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One way to see that $M$ is a compactification of $\mathbb{R}^n$ is the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take $f$ to be a Morse function on $M$ with one minimum.The flow of $\nabla f$ around the minimum will parametrize almost all of $M$ (the missing part will be the unstable manifold of critical points of smaller index).&lt;/p&gt;&#xA;" OwnerUserId="6090" LastActivityDate="2011-01-21T15:19:57.007" CommentCount="11" />
  <row Id="18433" PostTypeId="1" AcceptedAnswerId="18541" CreationDate="2011-01-21T15:28:11.920" Score="0" ViewCount="118" Body="&lt;p&gt;I have an equation in the form :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;M(x,y) + N(x,y)dy/dx&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know the integrating factor for this equation. My problem is I can not rearrange my equation in to the form :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;dy/dx + p(x)y = Q(x)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do I need to rearrange if the integrating factor is known?&lt;/p&gt;&#xA;" OwnerUserId="6085" LastActivityDate="2011-01-22T18:16:52.180" Title="What format does ODE have to be in, if integrating factor is known?" Tags="&lt;differential-equations&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="18434" PostTypeId="2" ParentId="18403" CreationDate="2011-01-21T15:48:58.253" Score="0" ViewCount="" Body="&lt;p&gt;This is not an answer, but rather a comment. This problem is related to the &lt;a href=&quot;http://en.wikipedia.org/wiki/Borel%E2%80%93Kolmogorov_paradox&quot; rel=&quot;nofollow&quot;&gt;Borel paradox&lt;/a&gt;. In general, to define a conditional probability with respect to a $\sigma$-field (in this case, the one generated by continuous random variables), one needs the notion of &lt;a href=&quot;http://en.wikipedia.org/wiki/Regular_conditional_probability&quot; rel=&quot;nofollow&quot;&gt;regular conditional probability&lt;/a&gt;. In short, the (pointwise) conditional probability given a probability-zero event, is not well defined.&lt;/p&gt;&#xA;" OwnerUserId="5836" LastActivityDate="2011-01-21T15:48:58.253" CommentCount="2" />
  <row Id="18435" PostTypeId="1" AcceptedAnswerId="18442" CreationDate="2011-01-21T16:06:10.887" Score="3" ViewCount="84" Body="&lt;p&gt;Hi everyone,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried balancing this chemical equation with a system of equations, but the answer doesn't seem to map well. I get a negative coefficient which is prohibitive in this equations. How do I interpret the answer?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$Al(OH)_3  + H_2SO_4 \rightarrow Al_2(SO_4)_3 + H_2O $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.wolframalpha.com/input/?i=rref&quot; rel=&quot;nofollow&quot;&gt;http://www.wolframalpha.com/input/?i=rref&lt;/a&gt;{{1,0,2,0},{3,4,12,1},{3,2,0,2},{0,1,3,0}}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ted.&lt;/p&gt;&#xA;" OwnerUserId="5434" LastActivityDate="2011-01-21T16:42:41.397" Title="Trying to balance equation" Tags="&lt;chemistry&gt;" AnswerCount="4" />
  <row Id="18436" PostTypeId="1" CreationDate="2011-01-21T16:19:47.953" Score="3" ViewCount="129" Body="&lt;p&gt;Hello, I'm actually a programmer so I'm sorry if this is a stupid question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am trying to solve for $x$, but I got stuck at this step:&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;$$e^{\frac{q1 + x}{b}} = e^{\frac{p - x\cdot o}{mb}} - e^{\frac{q2}{b}}$$&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;I'm not sure how to get x by itself because it's in the exponent in both places if I do: $\ln(e^{\cdots})$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll end up with only one $x$ being solved for :( ie.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;$$x = b\ln(e^{\frac{p - x\cdot o}{mb}} - e^{\frac{q2}{b}}) - q1$$&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;obviously that doesn't get me anywhere&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I considered that there might be 2 (or more) answers to this problem (maybe one positive one negative?), but $x$ is guaranteed to be a positive number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;maybe I need to represent that in the problem somehow?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;edit: &lt;strong&gt;more info&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;all values are known except for $x$; they are plugged into the equation by a computer program. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it is possible to find x with arbitrary numbers plugged in for all the other values I'd like to see how to do that (I didn't see a way to solve it that way)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;edit 2: &lt;em&gt;realistic values&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;b = 5000&#xA;o = -2495&#xA;m = 3000&#xA;q1 = 90&#xA;q2 = 105&#xA;p = 75&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;h1&gt;$$e^{\frac{90 + x}{5000}} = e^{\frac{75 - x(-2495)}{3000\cdot 5000}} - e^{\frac{105}{5000}}$$&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;reduced...&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;$$e^{\frac{90 + x}{5000}} = e^{\frac{75 + 2495x}{15000000}} - e^{0.021}$$&lt;/h1&gt;&#xA;" OwnerUserId="6093" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-21T16:46:15.107" LastActivityDate="2011-01-21T18:14:53.577" Title="How to solve for x when it is in the exponent of 2 variables" Tags="&lt;algebra&gt;" AnswerCount="1" CommentCount="9" />
  <row Id="18437" PostTypeId="2" ParentId="18435" CreationDate="2011-01-21T16:20:25.877" Score="0" ViewCount="" Body="&lt;p&gt;Unless I'm misunderstanding, the right side dictates that the ratio of Al to S should be $2:3$, so the coefficients on the left side must be in the ratio $2:3$.  Suppose they are 2 and 3.  There are now $2\cdot3+3\cdot4=18$ O and $2\cdot3+3\cdot2=12$ H on the left.  On the right, the coefficient of the first term must be 1 (since we used 2 and 3 on the left, there is enough Al and S for only 1), which uses 12 O, leaving 6 O and 12 H.  Setting the coefficient of the last term to 6 matches up with 6 O and 12 H.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Looking at the results from your row-reduction on WolframAlpha, I'm thinking that the $-1$ is because the matrix form is working as if that term were on the same side as the 2 and 3, but on the opposite side, it would be a positive $1$.&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2011-01-21T16:20:25.877" />
  <row Id="18438" PostTypeId="2" ParentId="18430" CreationDate="2011-01-21T16:20:32.540" Score="6" ViewCount="" Body="&lt;p&gt;More explicitely: Take $L^p([0,1])$ and consider $f(x) = 1/\sqrt{x}$. Then $f\in L^1$ but $\int_0^1 f(x)^2 dx$ is infinte.&lt;/p&gt;&#xA;" OwnerUserId="3148" LastActivityDate="2011-01-21T16:20:32.540" />
  <row Id="18439" PostTypeId="1" CreationDate="2011-01-21T16:21:38.060" Score="3" ViewCount="115" Body="&lt;p&gt;This is a followup to &lt;a href=&quot;http://math.stackexchange.com/questions/18406/a-very-simple-discrete-dynamical-system-with-pebbles&quot;&gt;A very simple discrete dynamical system with pebbles&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The setup: we have slots $n$ slots labeled $1, \ldots, n$ and $k$ pebbles, each of which is initially placed in some slot. Now the pebbles want to space themselves out as evenly as possible, and so they do the following. At each time step $t$, each pebble moves to the slot closest to the halfway point between its neighboring pebbles; if there is a tie, it chooses the slot to the left.  The leftmost and rightmost pebbles apply the same procedure, but because they only have one neighbor, they imagine that there are slots numbered $0$ and $n+1$ with pebbles in them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Formally, numbering the pebbles &#xA;$1, \ldots, k$ from left to right, and letting $x_i(t)$ be the slot of the $i$'th pebble at time t, we have $$ x_i(t+1) = \lfloor \frac{x_{i-1}(t)+x_{i+1}(t)}{2} \rfloor, i = 2, \ldots, k-1$$ $\lfloor \cdot \rfloor$ rounds down to the closest integer. Similarly, $$ x_1(t+1) = \lfloor (1/2) x_2(t) \rfloor, x_k(t+1) = \lfloor \frac{x_{k-1}(t) + (n+1)}{2} \rfloor.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My question:&lt;/strong&gt; I feel that it is true that (i) after sufficiently many iterations (ii) for large $n$ this dynamical system will spend all of its time in states which are approximately equispaced. Can this statement be made precise?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Note:&lt;/strong&gt; My earlier question ( &lt;a href=&quot;http://math.stackexchange.com/questions/18406/a-very-simple-discrete-dynamical-system-with-pebbles&quot;&gt;A very simple discrete dynamical system with pebbles&lt;/a&gt;) asked about one way to make the above statement precise, which turned out not to work. &lt;/p&gt;&#xA;" OwnerUserId="2120" LastEditorUserId="2120" LastEditorDisplayName="" LastEditDate="2011-01-21T17:41:11.013" LastActivityDate="2011-01-22T01:40:42.663" Title="A very simple discrete dynamical system with pebbles, take two" Tags="&lt;discrete-mathematics&gt;&lt;dynamical-systems&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="0" />
  <row Id="18440" PostTypeId="2" ParentId="18435" CreationDate="2011-01-21T16:22:24.547" Score="2" ViewCount="" Body="&lt;p&gt;Why don't you just use the actual chemical equation as input? &lt;a href=&quot;http://www.wolframalpha.com/input/?i=Al%28OH%293%2BH2SO4%E2%86%92Al2%28SO4%293%2BH2O&quot; rel=&quot;nofollow&quot;&gt;http://www.wolframalpha.com/input/?i=Al(OH)3%2BH2SO4%E2%86%92Al2(SO4)3%2BH2O&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1016" LastActivityDate="2011-01-21T16:22:24.547" CommentCount="1" />
  <row Id="18441" PostTypeId="2" ParentId="18435" CreationDate="2011-01-21T16:23:50.973" Score="1" ViewCount="" Body="&lt;p&gt;A chemical balance will always admit the solution with all zero coefficients: if you put nothing in, you get nothing out. Or, if you double the amount you put in, you will get double the output. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In linear algebra language, this means that your solution cannot be unique. The set of admissible solutions must form a linear subspace. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The set of admissible solutions will be the kernel of the matrix you wrote down corresponding to the chemical balance (incidentally, I would put negative signs in front of the first two columns since those represent things on the left hand side of the chemical equation and are things that are &lt;em&gt;consumed&lt;/em&gt;), and you are looking for a vector in this kernel with all positive coefficients. Therefore it is in fact &lt;em&gt;necessary&lt;/em&gt; for the row-reduced matrix to have negative entries: if the row-reduced matrix has only positive value entries, then it is impossible to have a vector with positive entries in its kernel (since a sum of positive numbers must be positive, and not zero). &lt;/p&gt;&#xA;" OwnerUserId="1543" LastEditorUserId="1543" LastEditorDisplayName="" LastEditDate="2011-01-21T16:35:51.453" LastActivityDate="2011-01-21T16:35:51.453" />
  <row Id="18442" PostTypeId="2" ParentId="18435" CreationDate="2011-01-21T16:29:40.440" Score="3" ViewCount="" Body="&lt;p&gt;Let $x$ be the number of $Al(OH)_3$; $y$ the number of $H_2SO_4$; $z$ the number of $AL_2(SO_4)_3$, and $w$ the number of $H_2O$. Looking at the number of $Al$, you get $x = 2z$. Looking at $O$, you get $3x + 4y = 12z + w$. Looking at $H$ you get $3x + 2y = 2w$; and looking at $S$ you get $y = 3z$. That looks like what you are getting from Wolfram, except you have the wrong signs for $z$ and $w$; unless you are interpreting the first two entries to represent the &quot;unknowns&quot;, and the last two to represent the &quot;solutions&quot;. I would translate into equations the usual way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you have is the following system of linear equations:&#xA;$$\begin{array}{rcrcrcrcl}&#xD;&#xA;x &amp;amp; &amp;amp; &amp;amp; &amp;amp; -2z &amp;amp; &amp;amp; &amp;amp; = &amp;amp; 0\\&#xD;&#xA;3x &amp;amp; + &amp;amp; 4y &amp;amp; - &amp;amp; 12z &amp;amp; - &amp;amp; w &amp;amp; = &amp;amp; 0\\&#xD;&#xA;3x &amp;amp; + &amp;amp; 2y &amp;amp;    &amp;amp;    &amp;amp; - &amp;amp; 2w &amp;amp; = &amp;amp; 0\\&#xD;&#xA;   &amp;amp;   &amp;amp; y &amp;amp; - &amp;amp; 3z &amp;amp; &amp;amp; &amp;amp; = &amp;amp; 0&#xD;&#xA;\end{array}$$&#xA;This leads (after either some back-substitution from the first and last equations into the second and third, or some easy row reduction) to $x=2z$, $y=3z$, and $6z=w$. Since you only want positive integer solutions, setting $z=1$ gives $x=2$, $y=3$, and $w=6$,  yielding the smallest solution:&#xA;$$2 Al(OH)_3 + 3H_2SO_4 \longrightarrow Al_2(SO_4)_3 + 6H_2O.$$&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-21T16:42:41.397" LastActivityDate="2011-01-21T16:42:41.397" CommentCount="1" />
  <row Id="18444" PostTypeId="2" ParentId="18429" CreationDate="2011-01-21T17:02:17.100" Score="5" ViewCount="" Body="&lt;p&gt;No, there is no pathological case. If $f(x)$ is integrable, and $m\leq f(x)\leq M$ for all $x\in[a,b]$, then &#xA;$$m(b-a) \leq \int_a^b f(x)\,dx \leq M(b-a).$$&#xA;This follows easily from the definition of the integral as a limit of Riemann sums. If $a=x_0\lt x_1\lt\cdots \lt x_n = b$ is a partition of $[a,b]$, $\Delta x_i = x_i - x_{i-1}$, and $x_i^*$ is a point in $[x_{i-1},x_i]$, then $m\leq f(x_i^*)\leq M$ for each $i$, so the Riemann sum satisfies&#xA;$$ m(b-a) = \sum_{i=1}^n m\Delta x_i \leq \sum_{i=1}^n f(x_i^*)\Delta x_i \leq \sum_{i=1}^n M\Delta x_i = M(b-a),$$&#xA;so taking limits you get&#xA;$$m(b-a) \leq \lim_{||P||\to 0}\sum_{i=1}^n f(x_i^*)\Delta x_i \leq M(b-a)$$&#xA;or&#xA;$$m(b-a) \leq \int_a^b f(x)\,dx \leq M(b-a)$$&#xA;(using the assumption that $f$ is integrable on $[a,b]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, if $f(x)\geq 0$ for all $x\in [a,b]$, then&#xA;$$0 = 0(b-a) \leq \int_a^b f(x)\,dx = F.$$&#xA;Redefining a function at a finite number of points on $[a,b]$ does not change its integrability, nor the value of the integral, so if all you know is that $f(x)\geq 0$ on $(a,b)$, you can always redefine it at $a$ and $b$ so that you get $f(x)\geq 0$ on $[a,b]$ without changing the value of the integral.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(In fact, you only need $f(x)\geq 0$ &quot;almost everywhere on $[a,b]$&quot;, meaning at all $x$ except perhaps for a set of measure $0$ that is contained on $[a,b]$). &lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2011-01-21T17:02:17.100" CommentCount="1" />
  <row Id="18445" PostTypeId="1" CreationDate="2011-01-21T17:25:46.130" Score="6" ViewCount="469" Body="&lt;p&gt;What are other faster methods of calculating e^x &lt;strong&gt;upto any number of decimals&lt;/strong&gt; other than using the taylor series formula?&lt;/p&gt;&#xA;" OwnerUserId="6094" LastEditorUserId="6094" LastEditorDisplayName="" LastEditDate="2011-01-21T17:58:32.247" LastActivityDate="2011-02-02T03:56:58.867" Title="Fastest way to calculate $e^x$ upto arbitrary number of decimals?" Tags="&lt;algorithms&gt;" AnswerCount="6" CommentCount="10" FavoriteCount="2" />
  <row Id="18446" PostTypeId="2" ParentId="18445" CreationDate="2011-01-21T17:37:09.590" Score="4" ViewCount="" Body="&lt;p&gt;A very reasonable way to do this is to use&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$e^x = 1 + x/1! + x^2/2! + x^3/3!...$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;as it converges really fast.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A &quot;trick&quot; I read in some library for calculating $e^1$ is to calculate the series for $x=1/2^k$ where $k$ is the root of the amount of required bits, then you have to square the result k times to get your final number, it avoids that big/small numbers make the result too inaccurate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit:&#xA;Also this is the exact same topic like &lt;a href=&quot;http://math.stackexchange.com/questions/5420/iterative-refinement-algorithm-for-computing-expx-with-arbitrary-precision&quot;&gt;Iterative refinement algorithm for computing exp(x) with arbitrary precision&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe it should be merged / closed?&lt;/p&gt;&#xA;" OwnerUserId="3123" LastEditorUserId="3123" LastEditorDisplayName="" LastEditDate="2011-01-21T18:06:19.987" LastActivityDate="2011-01-21T18:06:19.987" CommentCount="2" />
  <row Id="18447" PostTypeId="1" CreationDate="2011-01-21T17:55:04.997" Score="0" ViewCount="101" Body="&lt;p&gt;I am having trouble following the argument in Massey for the proof of Theorem 8.3, which states that if $f: X \rightarrow Y$ is a homotopy equivalence, then $f_*: \pi(X, x) \rightarrow \pi(Y, f(x))$ is an isomorphism for any $x \in X$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The proof in the book assumes Theorem 8.2., which states that if $\phi_0, \phi_1: X \rightarrow Y$ are continuous maps which are homotopic, then the image of $\pi(X, x_0)$ under $\phi_{0*}$ and $\phi_{1*}$ are isomorphic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They use this to claim that $\pi(X, x) \rightarrow \pi(X, gf(x))$ factors through $\pi(Y, f(x))$ via $f_*$ and $g_*$. I do not see why this is true. It seems that theorem 8.2 guarantees that a different diagram commutes.&lt;/p&gt;&#xA;" OwnerUserId="6095" LastEditorUserId="2548" LastEditorDisplayName="" LastEditDate="2011-01-21T18:03:46.863" LastActivityDate="2011-01-21T18:10:56.277" Title="Homotopy equivalence induces an isomorphism" Tags="&lt;algebraic-topology&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" />
  <row Id="18448" PostTypeId="2" ParentId="18284" CreationDate="2011-01-21T18:03:18.563" Score="8" ViewCount="" Body="&lt;p&gt;Here's a completely elementary argument that I like better than my previous answer that uses the Faà di Bruno formula.  All that is required are the Maclaurin series for $e^x$ and the definition of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Lah_number&quot; rel=&quot;nofollow&quot;&gt;Lah numbers&lt;/a&gt; $L(n,k)$ as the coefficients arising when converting &lt;a href=&quot;http://mathworld.wolfram.com/RisingFactorial.html&quot; rel=&quot;nofollow&quot;&gt;rising factorial powers&lt;/a&gt; to &lt;a href=&quot;http://mathworld.wolfram.com/FallingFactorial.html&quot; rel=&quot;nofollow&quot;&gt;falling factorial powers&lt;/a&gt; via &#xA;$$x^{\overline{n}} = \sum_{k=1}^n L(n,k) x^{\underline{k}}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, substitute $1/x$ for $x$ in the Maclaurin series for $e^x$ and differentiate $n$ times.  This yields&#xA;$$\frac{d^n}{dx^n} e^{1/x} = (-1)^n x^{-n} \sum_{j=0}^{\infty} \frac{j^{\overline{n}} x^{-j}}{j!}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then convert rising powers to falling powers to obtain&lt;br&gt;&#xA;$$\frac{d^n}{dx^n} e^{1/x} = (-1)^n x^{-n} \sum_{j=0}^{\infty} \sum_{k=1}^n L(n,k)  \frac{j^{\underline{k}} x^{-j}}{j!}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Working with the summation in $j$, we have&#xA;$$\sum_{j=0}^{\infty} \frac{j^{\underline{k}} x^{-j}}{j!} = \sum_{j=k}^{\infty} \frac{j^{\underline{k}} x^{-j}}{j!} = \sum_{j=k}^{\infty} \frac{j! x^{-j}}{(j-k)! j!} = \sum_{i=0}^{\infty} \frac{x^{-i-k}}{i!} = x^{-k} e^{1/x},$$ where the last step uses, once again, the Maclaurin series for $e^x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus we have &#xA;$$\frac{d^n}{dx^n} e^{1/x} = (-1)^n e^{1/x} \sum_{k=1}^n L(n,k) x^{-n-k},$$ which is just your formula with an index change, as $$L(n,k) = \binom{n}{k}\binom{n-1}{k-1} (n-k)!.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(From another perspective, this is computing the rising factorial moments of a Poisson $(1/x)$ distribution.  With this in mind, converting to falling factorial powers seems reasonable, as the &lt;a href=&quot;http://en.wikipedia.org/wiki/Factorial_moment&quot; rel=&quot;nofollow&quot;&gt;falling factorial moments of a Poisson distribution&lt;/a&gt; have a known simple expression.  See also this &lt;a href=&quot;http://math.stackexchange.com/questions/6335/bell-numbers-and-moments-of-the-poisson-distribution/6348#6348&quot;&gt;related answer&lt;/a&gt; to a previous math.SE question.)&lt;/p&gt;&#xA;" OwnerUserId="2370" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2011-01-21T18:08:29.260" LastActivityDate="2011-01-21T18:08:29.260" />
  <row Id="18449" PostTypeId="2" ParentId="18447" CreationDate="2011-01-21T18:10:56.277" Score="4" ViewCount="" Body="&lt;p&gt;Apply Theorem II.8.2 with $Y=X$, $\varphi_0 = \mathrm{id}_X$, and $\varphi_1$; &#xA;You get:&#xA;$$\begin{array}{ccc}&#xD;&#xA;\pi(X,x) &amp;amp; \stackrel{\mathrm{\varphi}_{0*}}{\longrightarrow} &amp;amp; \pi(X,x)\\&#xD;&#xA;\varphi_{1*}&amp;amp;\searrow &amp;amp; \downarrow u\\&#xD;&#xA;&amp;amp;&amp;amp;\pi(X,gf(x))&#xD;&#xA;\end{array}$$&#xA;Now, $\varphi_{0*}=\mathrm{id}_*$ is just the identity, and $\varphi_{1*} = (gf)_* = g_*f_*$, so you can rewrite the above as:&#xA;$$\begin{array}{ccc}&#xD;&#xA;\pi(X,x) &amp;amp; \stackrel{\mathrm{id}}{\longrightarrow} &amp;amp; \pi(X,x)\\&#xD;&#xA;g_*f_*&amp;amp;\searrow &amp;amp; \downarrow u\\&#xD;&#xA;&amp;amp;&amp;amp;\pi(X,gf(x))&#xD;&#xA;\end{array}$$&#xA;This means that $u = (gf)_* = g_*f_*$, which is exactly what the diagram at the bottom of page 57 says.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2011-01-21T18:10:56.277" />
  <row Id="18450" PostTypeId="2" ParentId="18436" CreationDate="2011-01-21T18:14:53.577" Score="2" ViewCount="" Body="&lt;p&gt;Set $$y = e^{x/15000000}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This gives us&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ Ay^{3000} = By^{2495} - C$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which is a polynomial, and I suppose can be solved easily using standard numerical methods.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope that helps.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2011-01-21T18:14:53.577" CommentCount="3" />
  <row Id="18451" PostTypeId="2" ParentId="18445" CreationDate="2011-01-21T18:15:16.550" Score="1" ViewCount="" Body="&lt;p&gt;I don't know that this is necessarily the fastest way, but it's fun. If you want to know how it works, Google.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&#xA;package com.akshor.pjt33.math;&#xA;&#xA;import com.akshor.pjt33.math.symbalg.Rational;&#xA;import java.math.BigInteger;&#xA;&#xA;public abstract class Spigot&#xA;{&#xA;    /** The relevant components of the prefix matrix. */&#xA;    Rational h00, h01=Rational.ZERO, h11=Rational.ONE;&#xA;    BigInteger i=BigInteger.ZERO;&#xA;    boolean verify=true;&#xA;    final int base;&#xA;    private StringBuilder sb=new StringBuilder();&#xA;&#xA;    public Spigot(BigInteger multiplier) {&#xA;        this(multiplier, 10);&#xA;    }&#xA;&#xA;    public Spigot(BigInteger multiplier, int base) {&#xA;        h00=new Rational(multiplier, BigInteger.ONE);&#xA;        this.base=base;&#xA;    }&#xA;&#xA;    /**&#xA;     * Advances this spigot algorithm, producing as many digits as each consumption permits.&#xA;     * @param beta An upper bound (positive) on the magnitude of all subsequent term ratios.&#xA;     * @return The number of digits output.&#xA;     */&#xA;    public int advance(Rational beta) {&#xA;        if (verify &amp;&amp; beta.signum()0) throw new IllegalArgumentException(Et_t+&quot; exceeds bound &quot;+beta);&#xA;        h00=h00.mul(Et_t);&#xA;        i=i.add(BigInteger.ONE);&#xA;&#xA;        // Produce (if we can: without a converging bound we can't).&#xA;        int rv=0;&#xA;        if (beta.compareTo(Rational.ONE)0; i++) numdigits-=advance(beta);&#xA;        return i;&#xA;    }&#xA;&#xA;    protected abstract Rational termRatio();&#xA;&#xA;    public String toString() {&#xA;        return sb.toString();&#xA;    }&#xA;&#xA;    /**&#xA;     * A spigot implementation which evaluates any hypergeometric sum for which you can&#xA;     * provide a term bound.&#xA;     * E.g. PI = 3 F(1/2, 1, 1, 8/5 ; 3/5, 4/3, 5/3 | 2/27) with term limit 2/27.&#xA;     */&#xA;    public static class Hypergeometric extends Spigot&#xA;    {&#xA;        private final Rational[] a, b;&#xA;        protected final Rational z;&#xA;&#xA;        public Hypergeometric(Rational[] a, Rational[] b, Rational z) {&#xA;            this(a,b,z,BigInteger.ONE);&#xA;        }&#xA;&#xA;        public Hypergeometric(Rational[] a, Rational[] b, Rational z, BigInteger mul) {&#xA;            super(mul);&#xA;            this.a=a.clone();&#xA;            this.b=new Rational[b.length+1];&#xA;            this.b[0]=new Rational(1,1);&#xA;            System.arraycopy(b,0,this.b,1,b.length);&#xA;            this.z=z;&#xA;        }&#xA;&#xA;        protected Rational termRatio() {&#xA;            Rational rv=z;&#xA;            for (Rational num : a) rv=rv.mul(num.add(i));&#xA;            for (Rational num : b) rv=rv.div(num.add(i));&#xA;            return rv;&#xA;        }&#xA;    }&#xA;&#xA;    public static class Exp extends Hypergeometric&#xA;    {&#xA;        public Exp(Rational z) {&#xA;            super(new Rational[0], new Rational[0], z);&#xA;        }&#xA;&#xA;        // We rely on the fact that the term ratio for Exp always decreases.&#xA;        public int output(int numdigits) {&#xA;            int i=0;&#xA;            for (; numdigits&gt;0; i++) numdigits-=advance(termRatio().abs());&#xA;            return i;&#xA;        }&#xA;    }&#xA;&#xA;    public static void main(String[] args) {&#xA;        Exp e=new Exp(Rational.ONE);&#xA;        int count=e.output(100);&#xA;        System.out.println(e);&#xA;        System.out.println(&quot;Iterations required: &quot;+(count+1));&#xA;&#xA;        Exp e2=new Exp(new Rational(2,1));&#xA;        count=e2.output(100);&#xA;        System.out.println(e2);&#xA;        System.out.println(&quot;Iterations required: &quot;+(count+1));&#xA;&#xA;        Exp em2=new Exp(new Rational(-2,1));&#xA;        count = em2.output(100);&#xA;        System.out.println(em2);&#xA;        System.out.println(&quot;Iterations required: &quot;+(count+1));&#xA;    }&#xA;}&#xA;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Implementing the Rational class is left as an exercise. And yes, this code is an argument for limited operator overloading.&lt;/p&gt;&#xA;" OwnerUserId="5676" LastEditorUserId="5676" LastEditorDisplayName="" LastEditDate="2011-01-21T20:42:35.567" LastActivityDate="2011-01-21T20:42:35.567" CommentCount="2" />
  <row Id="18452" PostTypeId="2" ParentId="18439" CreationDate="2011-01-21T18:22:16.663" Score="1" ViewCount="" Body="&lt;p&gt;First some thoughts and not an answer. It is helpful to re-think the problem in terms of the gaps between the stones. You have $n$ slots with $k$ stones. So in the case with boundaries you have $(k+1)$ stretches of &quot;in between space&quot;, each of which we will denote $a_i = x_i - x_{i-1} - 1$. The sum of all these $a_i$ add up to $n-k$. The dynamical system in terms of $a_i$ can be written as follows: let $a_0 = a_1$ and $a_{k+2} = a_{k+1}$ (translating the Dirichlet conditions of your original system to a Neumann boundary condition), and demand&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ a_i(t+1) - a_i(t) = \lfloor \frac{a_{i+1}(t) - a_i(t)}{2}\rfloor - \lfloor \frac{a_i(t) - a_{i-1}(t)}{2}\rfloor $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is in fact (more-or-less) a &lt;strong&gt;discrete heat equation&lt;/strong&gt; for $a:\mathbb{Z}\to\mathbb{Z}$. (The discrete Laplacian may also be written as $\lfloor (a_{i+1} + a_{i-1})/2\rfloor - a_i$ which differs from the RHS about by at most 1. The version above has instead the nice property that it preserves the $\ell_1$ norm of $a$, that is, the sum of all $a_i$ is preserved under this evolution.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we can apply similar intuition behind the heat equation: namely that one can expect convergence of the solution to a harmonic function. (The discretized setting however suggests that the limiting solution will not necessarily have $a_i(t+1) = a_i(t)$, but small fluctuations like $|a_i(t+1) - a_i(t)| \leq 1$ should be expected.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the harmonic functions on an intervale are the constant functions and the linear functions. The latter is ruled-out by the Neumann boundary condition. However, because of your discretization &lt;strong&gt;and&lt;/strong&gt; the preferential placement of the stones to the left spot when two are equal, the modified Laplacian given above admits a harmonic function as described by mjqxxxx in his comment: since differences of +1 between consequentive $a_i$s are treated as if they don't exist. So you can have solutions which &lt;em&gt;locally looks like it is equidistributed&lt;/em&gt; (left side and right side differs by a small amount) but globally isn't. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The periodic model, however, doesn't necessarily help. Consider the case with 4 stones and 12 slots. Take the initial distribution (O for stone, u for empty slot)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;OOuuuuOOuuuu&lt;/p&gt;&#xA;&#xA;&lt;p&gt;this becomes&lt;/p&gt;&#xA;&#xA;&lt;p&gt;uuuOOuuuuOOu&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and the pattern oscillates, never settling down. This example can be generalized to large $n$ also. This oscillation represents a failure of the intuition for passing from the continuous heat equation to the discrete case. When the spatial wavelength of the oscillation is small enough (on the order of the lattice size), artefacts from the discretisation starts to interfere with the intuition.  We will see below that this artefact can only manifest for when there are even number of stones in the periodic case, can cannot manifest in the interval case. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Now more concrete answers. Using the property of the heat equation, one can be inspired to check for dissipation laws (in other words, apply the maximum principle for parabolic differential equations). And indeed, by the definition of the system, in either the interval case or the periodic case, you can easily check that &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The value $a_{min} = \min a_i$ is non-decreasing, and $a_{max} = \max a_i$ is non-increasing. &lt;/li&gt;&#xA;&lt;li&gt;Suppose $a_i$ attains the minimum at time $t$. Then $a_i(t+1) = a_i(t)$ if and only if $a_{i-1}(t) = a_i(t)$ and either $a_{i+1}(t) = a_i(t)$ or $a_{i+1}(t) = a_i(t) + 1$. Similarly for the maximum.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;strike&gt;Reasoning directly from point 2 you obtain that, in the interval case (your original question), any distribution will converge toward a distribution that looks like the following: there exists $k_m$ and $k_M$ such that for all $i \leq k_m$, you have $a_i = a_{min}$. For all $i \geq k_M$, you have $a_i = a_{max}$. In between $k_m$ and $k_M$, you have $a_{i+1} - a_i = 1$. The values $k_m$ and $k_M$ can be $0$ or $k+1$, reflecting a limiting distribution that is a constant. &lt;/strike&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Parity&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let us re-write the discrete heat equation. A bit of computation shows that we can write it in the following way:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ a_i(t+1) = \lfloor \frac{a_{i+1}(t) + a_{i-1}(t) + \left( a_{i-1}(t) - a_i(t) \mod 2\right)}{2}\rfloor$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which says that the value of $a_i$ at $t+1$ is the average of its neighbors at $t$, rounded up if $a_{i-1}-a_i$ is odd, and rounded down if it is even. So in the periodic case with $k$ even, there are even number of $a_i$s, which can be then partitioned into two (effectively non-interacting) subsets (the evens and the odds) that results in the oscillating behaviour noted above. In the case of the interval by the Neumann boundary condition, and in the periodic case with the odd number of stones, there cannot by such partition of the system into two independent subsystems. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(to be cont.)&lt;/p&gt;&#xA;" OwnerUserId="1543" LastEditorUserId="1543" LastEditorDisplayName="" LastEditDate="2011-01-22T01:40:42.663" LastActivityDate="2011-01-22T01:40:42.663" CommentCount="5" />
  <row Id="18454" PostTypeId="2" ParentId="18445" CreationDate="2011-01-21T18:40:39.680" Score="6" ViewCount="" Body="&lt;p&gt;The theoretically fastest way appears to be to use Newton iteration to reduce the problem to computing the logarithm function and then using an algorithm based on the arithmetic-geometric mean to compute the logarithm. See this wikipedia page. &lt;a href=&quot;http://en.wikipedia.org/wiki/Computational_complexity_of_mathematical_operations&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Computational_complexity_of_mathematical_operations&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In practice the Taylor series should work fine, given a good implementation. The following webpage &lt;a href=&quot;http://numbers.computation.free.fr/Constants/constants.html&quot; rel=&quot;nofollow&quot;&gt;http://numbers.computation.free.fr/Constants/constants.html&lt;/a&gt; has an example impletation of using the taylor series to compute e. They claim it took 0.02 seconds to compute e to a thousand decimals on a PentiumII, 450 MHz processor. &lt;/p&gt;&#xA;" OwnerUserId="4623" LastActivityDate="2011-01-21T18:40:39.680" />
  <row Id="18455" PostTypeId="1" AcceptedAnswerId="18480" CreationDate="2011-01-21T18:42:14.187" Score="16" ViewCount="490" Body="&lt;p&gt;I'm looking for good examples where double induction is necessary. What I mean by double induction is induction on $\omega^2$. These are intended as examples in an &quot;Automatas and Formal Languages&quot; course.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One standard example is the following: in order to cut an $n\times m$ chocolate bar into its constituents, we need $nm-1$ cuts. However, there is a much better proof without using induction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another example: the upper bound $\binom{a+b}{a}$ on Ramsey numbers. The problem with this example is that it can be recast as induction on $a+b$, while I want something which is inherently inducting on $\omega^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lukewarm example: Ackermann's function, which seems to be pulled out of the hat (unless we know about the primitive recursive hierarchy).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Better examples: the proof of other theorems in Ramsey theory (e.g. Van der Waerden or Hales-Jewett). While these can possibly be recast as induction on $\omega$, it's less obvious, and so intuitively we really think of these proofs as double induction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another example: cut elimination in the sequent calculus. In this case induction on $\omega^2$ might actually be necessary (although I'm not sure about that).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem with my positive examples is that they are all quite technical and complicated. So I'm looking for a simple, non-contrived example where induction on $\omega^2$ cannot be easily replaced with regular induction (or with an altogether simpler argument). Any suggestions?&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2011-05-18T22:45:42.060" Title="Good examples of double induction" Tags="&lt;induction&gt;" AnswerCount="9" CommentCount="8" FavoriteCount="4" />
  <row Id="18458" PostTypeId="2" ParentId="18455" CreationDate="2011-01-21T18:50:55.700" Score="0" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;https://www.math.lsu.edu/~verrill/teaching/discrete2020/handouts/answers.pdf&quot; rel=&quot;nofollow&quot;&gt;This&lt;/a&gt; has some information&lt;/p&gt;&#xA;" OwnerUserId="4728" LastActivityDate="2011-01-21T18:50:55.700" CommentCount="1" />
  <row Id="18459" PostTypeId="2" ParentId="18445" CreationDate="2011-01-21T19:01:13.097" Score="1" ViewCount="" Body="&lt;p&gt;How is $x$ presented? If you want $e^x$ to (say) 100 significant figures, then you have to provide $x$ to (more or less) $100 + \log x$ significant figures. Are you using a multi-precision library?&lt;/p&gt;&#xA;" OwnerUserId="1508" LastActivityDate="2011-01-21T19:01:13.097" CommentCount="2" />
  <row Id="18460" PostTypeId="2" ParentId="18455" CreationDate="2011-01-21T19:26:55.557" Score="2" ViewCount="" Body="&lt;p&gt;What about prove $m+n = n+m$ for $m,n \in \mathbb{N}$? In particular, see &lt;a href=&quot;http://people.umass.edu/gmhwww/595t/pdf/MetaTheory-03-Mathematical%20Induction.pdf&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; (site talking about double induction).&lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2011-01-21T19:26:55.557" CommentCount="1" />
  <row Id="18461" PostTypeId="2" ParentId="18455" CreationDate="2011-01-21T19:48:06.733" Score="3" ViewCount="" Body="&lt;p&gt;One could concoct a simple example, like proving that every sequence of the following moves on pairs of natural numbers eventually terminates:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(i,j)\mapsto (i-1,N)$ for any natural number $N$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(i,j)\mapsto (i,j-1)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Edited to make it an inherently $\omega^2$ problem.)&lt;/p&gt;&#xA;" OwnerUserId="2941" LastEditorUserId="2941" LastEditorDisplayName="" LastEditDate="2011-01-21T20:23:10.093" LastActivityDate="2011-01-21T20:23:10.093" CommentCount="4" />
  <row Id="18462" PostTypeId="1" AcceptedAnswerId="18463" CreationDate="2011-01-21T19:49:19.397" Score="2" ViewCount="77" Body="&lt;p&gt;Show that if $V$ is a finite-dimensional vector space with a dot product $\langle-,-\rangle$, and $f: V \rightarrow V$ linear with $\forall v,w \in V: \langle v,w \rangle=0 \Rightarrow \langle f(v),f(w) \rangle=0$ then $\exists C \in \mathbb{R}$ such that $(C\cdot f)$ is a linear isometry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notes &amp;amp; Thoughts:&#xA;$g$ is a linear isometry means $\forall v \in V: \lVert g(v)\rVert=v$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Visually the theorem makes sense, if orthogonal vectors remain orthogonal under $f$ then the angles remain, so the original vector just changes its length. (If I understand this correctly)&lt;/p&gt;&#xA;" OwnerUserId="3123" LastEditorUserId="274" LastEditorDisplayName="" LastEditDate="2011-01-21T19:54:26.773" LastActivityDate="2011-01-21T20:26:51.700" Title="Linear isometry" Tags="&lt;linear-algebra&gt;&lt;homework&gt;&lt;vector-space&gt;" AnswerCount="1" CommentCount="11" />
  <row Id="18463" PostTypeId="2" ParentId="18462" CreationDate="2011-01-21T20:15:00.587" Score="2" ViewCount="" Body="&lt;p&gt;Let $x,y\in V$ with $|x| = |y|=1$.  I claim that $|f(x)| = |f(y)|$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To see this, notice that  $0 = |x|^2 - |y|^2 = \langle x+y,x-y\rangle$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By assumption, this is equal to $\langle f(x+y), f(x-y)\rangle = |f(x)|^2 - |f(y)|^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, if we define $C = 1/|f(x)|$ with $|x| = 1$, then $C$ does not depend on the choice of $x$.  I claim this $C$ solves the problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, let $z\in V$ be arbitrary.  I want to show that $|z| = |f(z)|$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $z = 0$, then $Cf(z) = 0$, so $|z| = |f(z)|$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $z\neq 0$, then $z/|z|$ is a unit vector, and so $1 = |Cf(z/|z|)|$.  Multiplying both sides by $|z|$, w get $|z| = |Cf(z)|$, showing $Cf$ is an isometry.&lt;/p&gt;&#xA;" OwnerUserId="331" LastEditorUserId="331" LastEditorDisplayName="" LastEditDate="2011-01-21T20:26:51.700" LastActivityDate="2011-01-21T20:26:51.700" CommentCount="5" />
  <row Id="18464" PostTypeId="1" AcceptedAnswerId="18466" CreationDate="2011-01-21T21:03:46.187" Score="5" ViewCount="309" Body="&lt;p&gt;I would assume the answer to my question is yes, but I want to make sure because my book uses both terminologies.  Please also indicate where zero falls into the mix.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;UPDATE:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is an excerpt from my book:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The definition of Θ(g(n)) requires&#xA;  that every member f(n) ϵ Θ(g(n)) be&#xA;  asymptotically &lt;strong&gt;non-negative&lt;/strong&gt;, that is,&#xA;  that f(n) be &lt;strong&gt;non-negative&lt;/strong&gt; whenever n&#xA;  is sufficiently large. (An&#xA;  asymptotically &lt;strong&gt;positive&lt;/strong&gt; function is&#xA;  one that is &lt;strong&gt;positive&lt;/strong&gt; for all&#xA;  sufficiently large n.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="6096" LastEditorUserId="6096" LastEditorDisplayName="" LastEditDate="2011-01-21T21:16:33.110" LastActivityDate="2011-01-21T23:17:39.183" Title="Is positive the same as non-negative?" Tags="&lt;calculus&gt;&lt;statistics&gt;&lt;terminology&gt;" AnswerCount="4" CommentCount="11" FavoriteCount="1" />
  <row Id="18465" PostTypeId="1" AcceptedAnswerId="18481" CreationDate="2011-01-21T21:14:44.873" Score="4" ViewCount="102" Body="&lt;p&gt;I am curios about the decidability or undecidability of the following decision problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;INPUT: a first-order formula $\varphi$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;OUTPUT:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Yes, if $\varphi$ is satisfiable but not finitely-satisfiable&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;No, otherwise&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The problem is very close to Church's and Trakhtenbrot's Theorems, but somehow I do not see how to modify the proof of these two results in order to get the undecidability of this other result.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Somehow what I am asking corresponds to the fact of whether it is decidable or undecidable the problem of &quot;deciding the finite model property&quot;.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Does anybody know whether the problem is undecidable or not? Does anybody know any reference where this is discussed?&lt;/p&gt;&#xA;" OwnerUserId="5580" LastEditorUserId="5580" LastEditorDisplayName="" LastEditDate="2011-01-21T21:24:05.950" LastActivityDate="2011-01-22T00:36:09.767" Title="(Un)Decidability of satisfiable but not finitely-satisfiable formulas" Tags="&lt;logic&gt;&lt;model-theory&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="18466" PostTypeId="2" ParentId="18464" CreationDate="2011-01-21T21:35:31.920" Score="9" ViewCount="" Body="&lt;p&gt;The real numbers can be partitioned into the positive real numbers, the negative real numbers, and zero.  A real number is one and only one of those three possibilities.  This is called &quot;trichotomy.&quot;  Non-negative (or, correspondingly, non-positive) means not negative (not positive), so zero or positive (zero or negative).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is, non-negative includes zero whereas positive does not.&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2011-01-21T21:35:31.920" CommentCount="1" />
  <row Id="18467" PostTypeId="1" AcceptedAnswerId="18611" CreationDate="2011-01-21T22:03:14.997" Score="8" ViewCount="98" Body="&lt;p&gt;Is it true that: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;If the intersection of the Sylow p-subgroups is trivial, then the intersection of their normalizers is equal to the intersection of their centralizers?&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I half remember this being true for odd p, but I cannot find the reference.  I have not found a counterexample for p=2 or p=3.&lt;/p&gt;&#xA;" OwnerUserId="583" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-21T22:08:30.487" LastActivityDate="2011-01-23T11:30:45.543" Title="Normalizing every Sylow p-subgroup versus centralizing every Sylow p-subgroup" Tags="&lt;group-theory&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="2" />
  <row Id="18468" PostTypeId="2" ParentId="18464" CreationDate="2011-01-21T22:10:44.760" Score="0" ViewCount="" Body="&lt;p&gt;As non-negative is an adjective, generally its meaning depends on what word comes after it.&lt;/p&gt;&#xA;" OwnerUserId="4929" LastActivityDate="2011-01-21T22:10:44.760" />
  <row Id="18470" PostTypeId="1" CreationDate="2011-01-21T22:39:51.677" Score="7" ViewCount="140" Body="&lt;p&gt;I had made up this problem a while back, and I think I had a tedious, uninsightful proof. Also, I am not able to reconstruct the proof I had.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is the problem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $\displaystyle \pi: \{1,2, \dots, 13\} \to \{1,2,\dots, 13\}$ be a bijection, i.e. it is basically a permutation of the first $\displaystyle 13$ positive integers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For each such $\displaystyle \pi$ construct the polynomial $\displaystyle P_{\pi}: \mathbb{Z} \to \mathbb{Z}$ such that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$P_{\pi}(x) = \sum_{j=1}^{13} \ \ \pi(j) \ x^{j-1}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there some $\displaystyle \pi$ such that for every $\displaystyle n \in \mathbb{Z}$, $\displaystyle P_{\pi}(n)$ is never a perfect square?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe it is true that there is such a $\displaystyle \pi$. In fact, I seem to recollect having &quot;proved&quot; that there are at least $7!$ such permutations. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the questions are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a) Is there a &quot;slick&quot; proof of the existence of at least one such $\displaystyle \pi$? (Any proof welcome, though).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;b) What if $\displaystyle 13$ is replaced by a generic natural number $\displaystyle M$? Can we give a different (and hopefully simpler) characterization of the $\displaystyle M$ for which such a $\displaystyle \pi$ exists?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please consider posting an answer even if you don't have an answer for b).&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2011-01-22T07:46:17.543" Title="Never Perfect Square Permutation Polynomials" Tags="&lt;number-theory&gt;" AnswerCount="1" />
  <row Id="18471" PostTypeId="2" ParentId="18464" CreationDate="2011-01-21T22:46:13.907" Score="3" ViewCount="" Body="&lt;p&gt;In mathematical English, &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;positive is defined to be $&amp;gt; 0$&lt;/li&gt;&#xA;&lt;li&gt;negative is defined to be $&amp;lt; 0$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So non-negative means $\ge 0$, not the same as positive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In mathematical French, it just happens that the word 'positif' is defined to be $\ge 0$, that is, 0 is both 'positif' and 'negatif'. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other languages...who knows.&lt;/p&gt;&#xA;" OwnerUserId="1919" LastActivityDate="2011-01-21T22:46:13.907" CommentCount="2" />
  <row Id="18472" PostTypeId="2" ParentId="18464" CreationDate="2011-01-21T22:56:06.690" Score="2" ViewCount="" Body="&lt;p&gt;If we go by your edits, about the book excerpt, it looks like the book treats non-negative as $\ge 0$, and positive as $\gt 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, from the notation it seems like you are talking about functions whose domain is $\mathbb{N}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For an example of an asymptotically positive function, consider&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ f(n) = 1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For an example of an asymptotically non-negative function, consider&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f(n) = \left|\sin\left(\frac{n\pi}{2}\right)\right|$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For sufficiently large $\displaystyle n$, we have that $\displaystyle f(n) \ge 0$. Note that this function is &lt;em&gt;not&lt;/em&gt; asymptotically positive, because it is zero (for even $\displaystyle n$) infinitely often.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any asymptotically positive function is also asymptotically non-negative, but not vice-versa.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For an example of a function which is neither asymptotically non-negative, nor asymptotically positive,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f(n) = \sin\left(\frac{n\pi}{2}\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This function takes the values $\displaystyle 1,-1 \ \text{and}\  0$ infinitely often.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2011-01-21T23:17:39.183" LastActivityDate="2011-01-21T23:17:39.183" />
  <row Id="18473" PostTypeId="2" ParentId="18455" CreationDate="2011-01-21T23:07:15.363" Score="4" ViewCount="" Body="&lt;p&gt;Though you've already dismissed it as 'lukewarm', Ackermann's function (proving totality I think is what is wanted) is your most accessible option (I think it is a great option). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's not contrived/unnatural because it is motivated by very different concepts. If you want to construct another example (prove $f(x,y) = g(x,y)$), you'd probably want to have x and y very much asymmetric (in the sense that they should be used in syntactically very different ways in the computations). And Ackermann's function does just that.&lt;/p&gt;&#xA;" OwnerUserId="1919" LastActivityDate="2011-01-21T23:07:15.363" />
  <row Id="18474" PostTypeId="1" AcceptedAnswerId="18478" CreationDate="2011-01-21T23:17:23.033" Score="2" ViewCount="269" Body="&lt;p&gt;Can anyone tell me how to calculate the Riemann integral of the characteristic function of the Cantor set? It's probably obvious but I don't see how to write it down.&#xA;Many thanks for your help!&lt;/p&gt;&#xA;" OwnerUserId="5798" LastActivityDate="2011-01-21T23:40:03.653" Title="Riemann integral of characteristic function of Cantor set" Tags="&lt;measure-theory&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="18475" PostTypeId="2" ParentId="18470" CreationDate="2011-01-21T23:18:32.333" Score="6" ViewCount="" Body="&lt;p&gt;Edit: The original solution was wrong, but fortunately could be fixed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given $M$, we find a sufficient condition for a permutation $\pi$ to work. We use the following simple property: $m^2 \pmod{4} \in \{0,1\}$. So if we can force $P_\pi(n) \pmod{4} \in \{2,3\}$, we are done.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are four cases to consider, according to $n \pmod{4}$. If $n \pmod{4} = 0$ then $$P_\pi(n) \equiv \pi(1) \pmod{4},$$ and so we need $\pi(1) \pmod{4} \in \{2,3\}$. If $n \pmod{4} = 2$ then $$P_\pi(n) \equiv \pi(1) + 2\pi(2) \pmod{4},$$ and so we need $\pi(2)$ to be even. If $n \pmod{4} = 1$ then $$P_\pi(n) \equiv \sum_{i=1}^M \pi(i) = \sum_{i=1}^M i = \frac{M(M+1)}{2} \pmod{4};$$ that happens whenever $M \pmod{8} \in \{2,3,4,5\}$. Finally, if $n \pmod{4} = 2$ then $$P_\pi(n) \equiv \sum_{i=1}^M (-1)^{i+1} \pi(i) \equiv \frac{M(M+1)}{2} + 2\sum_{j=1}^{\lfloor M/2 \rfloor} \pi(2j) \pmod{4}.$$ Thus we need $$\sum_{j=1}^{\lfloor M/2 \rfloor} \pi(2j) \equiv 0 \pmod{2}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Claim: Whenever $M\geq 3$ satisfies $M \pmod{8} \in \{2,3,4,5\}$, we can find such a $\pi$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proof: Set $\pi(1) = 3$ and $\pi(2m) = 2m$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This method of proof can probably be generalized by replacing $4$ with other moduli.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2011-01-22T07:46:17.543" LastActivityDate="2011-01-22T07:46:17.543" CommentCount="2" />
  <row Id="18476" PostTypeId="2" ParentId="17398" CreationDate="2011-01-21T23:24:27.787" Score="2" ViewCount="" Body="&lt;p&gt;The comment made by Qiaochu Yuan answered my question but because it's a comment I can't accept it and close this question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So hereby I declare this question as solved. Thanks to all for your help.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is the comment:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;You don't have to compute the double&#xA;  integral; you just have to bound it.&#xA;  For example if both of the underlying&#xA;  measure spaces are finite and f is&#xA;  bounded then this is automatically&#xA;  true. Similarly if you are computing&#xA;  an integral over R^2 it suffices to&#xA;  bound f on a sequence of compact&#xA;  subsets of R^2, say concentric circles&#xA;  or unit squares.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5798" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2011-01-21T23:57:37.650" LastActivityDate="2011-01-21T23:57:37.650" />
  <row Id="18477" PostTypeId="2" ParentId="18474" CreationDate="2011-01-21T23:32:08.037" Score="2" ViewCount="" Body="&lt;p&gt;I am presuming you are talking about the Cantor Set in $[0,1]$, where you remove the middle third.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the Cantor set is of measure zero, the Lebesgue integral of its characteristic function is $0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it were Riemann integrable (which it is, as the points of discontinuity is of measure $0$), then the value of the Riemann integral would equal the Lebesgue integral and so would be $0$.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2011-01-21T23:32:08.037" CommentCount="2" />
  <row Id="18478" PostTypeId="2" ParentId="18474" CreationDate="2011-01-21T23:34:20.620" Score="5" ViewCount="" Body="&lt;p&gt;Let $C$ be the Cantor set, and let $C_n$ be the closed set left after $n$ steps of removing middle thirds from $[0,1]$, so $C_n$ is a disjoint union of $2^n$ closed intervals, and the sum of the lengths of these intervals is $\left(\frac{2}{3}\right)^n$, which converges to zero.  The characteristic function $\chi_{C_n}$ of $C_n$ is a step function that dominates the characteristic function of $C$, so its integral, $\left(\frac{2}{3}\right)^n$, is an upper Riemann sum for $\chi_C$. Thus the infimum of the upper Riemann sums for $\chi_C$ is at most $\inf_n\left(\frac{2}{3}\right)^n=0$.  The lower Riemann sums are all greater than or equal to $0$, so this shows that the Riemann integral exists and equals $0$.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2011-01-21T23:40:03.653" LastActivityDate="2011-01-21T23:40:03.653" CommentCount="14" />
  <row Id="18479" PostTypeId="2" ParentId="18455" CreationDate="2011-01-22T00:07:13.970" Score="7" ViewCount="" Body="&lt;p&gt;Let me begin with an example of an induction of length $\epsilon_0$: The proof that &lt;a href=&quot;http://en.wikipedia.org/wiki/Goodstein%27s_theorem&quot; rel=&quot;nofollow&quot;&gt;Goodstein sequences&lt;/a&gt; terminate. I mention this because when I decided to understand this result, I began to compute the length of these sequences and eventually came to a conjecture for a general formula (!) for the length of the sequence. It turned out that proving the conjecture was easy, because the proof organized itself as an induction of length $\epsilon_0$. I was both very amused and very intrigued by this. The little paper that came out of this adventure is &lt;a href=&quot;http://andrescaicedo.files.wordpress.com/2010/04/goodsteins.pdf&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, I also found once a natural example of an induction of length $\omega^2$ when studying a &quot;Ramsey type&quot; problem: the size of min-homogeneous sets for regressive functions on pairs. What I liked about this example is that Ackermann's function injected itself into the picture and ended up providing me with the right rates of growth. The details are in a paper &lt;a href=&quot;http://andrescaicedo.files.wordpress.com/2010/04/regressivepairs.pdf&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="462" LastActivityDate="2011-01-22T00:07:13.970" CommentCount="2" />
  <row Id="18480" PostTypeId="2" ParentId="18455" CreationDate="2011-01-22T00:25:17.763" Score="7" ViewCount="" Body="&lt;p&gt;A nice example arises by relativizing &lt;a href=&quot;http://math.stackexchange.com/questions/3532/how-to-explain-real-big-numbers/3551#3551&quot;&gt;Goodstein's Theorem&lt;/a&gt; from $\rm\ \epsilon_0 = \omega^{\omega^{\omega^{\cdot^{\cdot^{\cdot}}}}}$ down to $\rm\ \omega^2\:.\ $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;$\rm\ \omega^2\ $ Goodstein's Theorem&lt;/strong&gt; $\  $  Given naturals $\rm\ a,\:b,\:c\ $ and an arbitrary increasing &quot;base-bumping&quot; function $\rm\ g(n)\ $ on $\:\mathbb N\:$ the following iteration eventually reaches $0\ $ (i.e. $\rm\ a = c = 0\:$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\quad\quad\quad\ \ a\ b + c \ \ \to\quad\quad\ \ a\ \ \ \ \ g(b)\ +\ \ \ c\ \ -\ 1\quad if\quad\ c &amp;gt; 0 $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\quad\quad\quad\ \ \phantom{a\ b + c}\ \ \to\ \   (a-1)\ g(b)\  +\  g(b)-1\quad if\quad \ c = 0 $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: $\ $  The above iteration is really on triples $\rm\ (a,b,c)\ $ but  I chose the above notation in order to emphasize the relationship with radix notation and with Cantor Normal form for ordinals &amp;lt; $\epsilon_0$. $\ \ $ For more on Goodstein's Theorem see the link in Andres's post or see my 1995\12\11 &lt;a href=&quot;http://groups.google.com/groups?selm=WGD.95Dec11023450%40martigny.ai.mit.edu&quot; rel=&quot;nofollow&quot;&gt;sci.math post.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2011-01-22T00:45:44.003" LastActivityDate="2011-01-22T00:45:44.003" />
  <row Id="18481" PostTypeId="2" ParentId="18465" CreationDate="2011-01-22T00:36:09.767" Score="2" ViewCount="" Body="&lt;p&gt;The question is not decidable, since the halting problem reduces to it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose that we are given a Turing machine $M$ and we want to determine whether it halts on input 0. A halting computation can be described as a finite sequence of finite TM configurations, such that the initial configuration shows the appropriate set-up for M and each step of the configuration follows the computational rules specified by M and the final configuration shows the machine in a halting configuration. These properties of such a sequence of configurations can be described in a single statement $\varphi$, using a language enabling us to speak of cells of the tape and their contents and the head position of a given configuration and so on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, M halts if and only if $\varphi$ has a finite model. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, we can arrange the details so that $\varphi$ trivially has an infinite model. Let us suppose that $\varphi$ describes the configurations in such a way that once the machine has halted, we can simply repeat the halting configuration at the next stage. That is, $\varphi$ should assert that the successor steps of computation work correctly, if the computation has not yet halted, but if it has, then the next configuration just copies the previous one as an exact duplicate. Now, if M does not halt, we can generate an infinite sequence of configurations corresponding to that, and then add another $\mathbb{Z}$-block of configurations all the same, showing a (fake) halting configuration. This will be an infinite model of $\varphi$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, for this $\varphi$, we see that $M$ halts if and only if the answer to your inquiry is no. So your inquiry is not asking a decidable question.&lt;/p&gt;&#xA;" OwnerUserId="413" LastActivityDate="2011-01-22T00:36:09.767" CommentCount="7" />
  <row Id="18482" PostTypeId="1" CreationDate="2011-01-22T00:49:14.400" Score="1" ViewCount="85" Body="&lt;p&gt;Let $a$ and $b$ be positive integers both greater than $1$. I'd like to compute the average value of the summation &#xA;\begin{eqnarray}&#xA;\sum_{k = 0}^{\lfloor \log_{a} x \rfloor} \left \lbrace \log_{b} a^{-k} x \right \rbrace,&#xA;\end{eqnarray}&#xA;where $\{ \cdot  \}$ denotes the fractional part function and $\lfloor \cdot \rfloor$ denotes the floor function. Plotting some values, it seems to grow &lt;em&gt;really slowly&lt;/em&gt;, either logarithmically or double logarithmically. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is what I understand. The average value of $\{ \log_{b} x \}$ on $[1,\infty)$ is same as that of $\log_{b} x$ on $[1,b]$ by the periodicity of the fractional part, which I calculate to be $\frac{b}{b-1} - \frac{1}{\log b}$. How does this result apply to the terms of the sum since they are of the form $\{\log_b x - k \log_{b} a \}$? Any help would certainly be appreciated!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is an extension of my previous question, &lt;a href=&quot;http://math.stackexchange.com/questions/17540/average-value-of-the-fractional-part-of-a-ck-function&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Edit&lt;/em&gt; If I'm understanding things correctly, it should be the case that $\{ \log_{b} cx \}$ has average value on $[1,\infty)$ equal to that of $\log_{b} x$ on $[1, b \ \log_{b} c]$, which I calculate as&#xA;\begin{eqnarray}&#xA;\frac{1}{b \ \log_{b} c - 1} \int_{1}^{b \ \log_{b} c} \log_{b} x \ dx = -\frac{\log b + (b \log c) (-1 + \log( b \ \log_{b} c ))}{(\log b)( \log b - b \ \log c)},&#xA;\end{eqnarray}&#xA;which reduces to the value I quoted above in the limit as $c \to b$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There has to be an easier way of estimating this finite sum!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I do have the naive bounds,&#xA;\begin{eqnarray}&#xA;0 \leq \sum_{k = 0}^{\lfloor \log_{a} x \rfloor} \left \lbrace \log_{b} \ a^{-k} x \right \rbrace \leq \left \lfloor \log_{a} x \right \rfloor + 1.&#xA;\end{eqnarray}&lt;/p&gt;&#xA;" OwnerUserId="2720" LastEditorUserId="2720" LastEditorDisplayName="" LastEditDate="2011-01-22T21:57:07.060" LastActivityDate="2011-01-22T21:57:07.060" Title="Average value of the sum $\sum_{k = 0}^{\lfloor \log_{a} x \rfloor} \{ \log_{b} a^{-k} x \}$" Tags="&lt;real-analysis&gt;" AnswerCount="0" FavoriteCount="1" />
  <row Id="18484" PostTypeId="2" ParentId="18455" CreationDate="2011-01-22T01:25:29.230" Score="1" ViewCount="" Body="&lt;p&gt;Let [n] denote the ordered set (0, ..., n). Show that there are precisely $\binom{n+m+1}{n+1}$ order-preserving maps $[n] \rightarrow [m]$. Also note that the collection of objects [n], together with order-preserving functions, forms a category... one can show by double induction that every map has an epi-monic factorization. I haven't actually tried doing these without double induction... it just seemed more natural that way, so I don't really know whether this is a good example, or I was just being silly. &lt;/p&gt;&#xA;" OwnerUserId="423" LastActivityDate="2011-01-22T01:25:29.230" CommentCount="1" />
  <row Id="18485" PostTypeId="1" CreationDate="2011-01-22T02:05:58.743" Score="2" ViewCount="134" Body="&lt;p&gt;This is a pretty dumb question, but it's been a while since I had to do math like this and it's escaping me at the moment (actually, I'm not sure I ever knew how to do this. I remember the basic trigonometric identities, but not anything like this).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a simple equation of one unknown, but the unknown occurs twice in different trigonometric functions and I'm not sure how to combine the two. I want to simply solve for $\theta$ in the following equation, where $a$ and $b$ are constants.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a=\tan(\theta) - \frac{b}{\cos^2\theta}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I reduce this into a single expression so that I can solve for $\theta$ given any $a$ and $b$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I'm only interested in real solutions and, in practice (this is used to calculate the incidence angle for a projectile such that it will pass through a certain point), it should always have a real solution, but an elegant method of checking that it doesn't would not go unappreciated.)&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Based on Braindead's hint I reduced the equation to:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$0=(b-a)+\tan(\theta)+b\tan^2(\theta)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can now solve for $\tan(\theta)$ using the quadratic equation, which gets me what I'm after. Is this the solution others were hinting towards? It seems like there would be a way to do it as a single trigonometric operation, but maybe not.&lt;/p&gt;&#xA;" OwnerUserId="6105" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2011-01-22T02:56:29.857" LastActivityDate="2011-01-22T02:56:29.857" Title="How can I solve for a single variable which occurs in multiple trigonometric functions in an equation?" Tags="&lt;algebra-precalculus&gt;&lt;trigonometry&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="18486" PostTypeId="2" ParentId="18485" CreationDate="2011-01-22T02:13:42.693" Score="0" ViewCount="" Body="&lt;p&gt;You can write $\tan(\theta)=\frac{\sin(\theta)}{\cos(\theta)}=\frac{\sqrt{1-\cos^2(\theta)}}{\cos(\theta)}$ which gets everything in terms of $\cos(\theta)$ but you may not like the degree of the result when you get rid of the radical.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-01-22T02:13:42.693" CommentCount="1" />
  <row Id="18487" PostTypeId="2" ParentId="18485" CreationDate="2011-01-22T02:22:20.207" Score="1" ViewCount="" Body="&lt;p&gt;Hint:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you solve $$p = \frac{q\sin 2\theta + r}{s\cos 2\theta + t}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ok, more details.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$a = \frac{\sin \theta \cos \theta}{\cos^2 \theta} - \frac{b}{\cos^2 \theta}  = \frac{\sin 2 \theta }{2\cos^2 \theta} - \frac{b}{\cos^2 \theta} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ = \frac{\sin 2\theta - 2b}{2cos^2 \theta} = \frac{ \sin 2\theta - 2b}{\cos 2\theta + 1}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$a(\cos 2 \theta + 1) = \sin 2 \theta - 2 b$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$   \sin 2\theta - a \cos 2\theta = a + 2b$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The equation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ p \cos \alpha + q \sin \alpha = r$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;is standard.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and can be solved by dividing by $\displaystyle \sqrt{p^2 + q^2}$ and noticing that for some $\displaystyle \beta$ we must have that $\displaystyle \sin \beta = \frac{p}{\sqrt{p^2 + q^2}}$ and $\displaystyle  \cos \beta = \frac{q}{\sqrt{p^2 + q^2}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Giving rise to &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \sin(\alpha + \beta) = \frac{r}{\sqrt{p^2 +q^2}}$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I will leave it to you to solve your original equation.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2011-01-22T02:41:30.207" LastActivityDate="2011-01-22T02:41:30.207" />
  <row Id="18488" PostTypeId="1" AcceptedAnswerId="18491" CreationDate="2011-01-22T02:38:59.723" Score="3" ViewCount="124" Body="&lt;p&gt;Inverse of a sparse matrix could be dense, but what if I'm only interested in the main diagonal of the result? Is there a method that is more efficient than computing the full inverse?&lt;/p&gt;&#xA;" OwnerUserId="998" LastActivityDate="2011-01-22T03:00:28.680" Title="Diagonal of an inverse of a sparse matrix" Tags="&lt;linear-algebra&gt;&lt;numerical-methods&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="18489" PostTypeId="2" ParentId="18405" CreationDate="2011-01-22T02:45:12.170" Score="-2" ViewCount="" Body="&lt;p&gt;Assume that the coin tossing is infinite. Define the state at time $n$ to be the most recent $5$ outcomes when $n \geq 5$ and the $n$ most recent outcomes when $n &amp;lt;5$. This is a Markov chain. Now $\pi_{HTHTH}$(the limiting probability) is equal to the inverse of the mean time to go from state HTHTH to HTHTH. Thus $\pi_{HTHTH} = p^{3}q^2$ (probability of getting HTHTH) where $q=1-p$. So $1/(p^{3}q^2)$ is the mean time to go from HTHTH to HTHTH. Thus $$E(\text{time to get} \ HTHTH) = E(\text{time to get} \ HT)+ \frac{1}{p^{3}q^{2}}$$ $$= E(\text{time to get} \ H)+ \frac{1}{pq}+\frac{1}{p^{3}q^{2}} =  \frac{1}{p}+\frac{1}{pq}+\frac{1}{p^{3}q^{2}}$$  &lt;/p&gt;&#xA;" OwnerUserId="2548" LastEditorUserId="2548" LastEditorDisplayName="" LastEditDate="2011-01-22T03:36:52.017" LastActivityDate="2011-01-22T03:36:52.017" CommentCount="7" />
  <row Id="18490" PostTypeId="2" ParentId="18488" CreationDate="2011-01-22T02:56:37.510" Score="2" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://www-users.cs.umn.edu/~saad/PDF/Sparse_days_06_15_2010.pdf&quot; rel=&quot;nofollow&quot;&gt;This&lt;/a&gt;(explanation of how to find diagonal of an inverse of a sparse matrix) should help.&lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2011-01-22T02:56:37.510" />
  <row Id="18491" PostTypeId="2" ParentId="18488" CreationDate="2011-01-22T03:00:28.680" Score="3" ViewCount="" Body="&lt;p&gt;You may be interested in work of Jok M. Tang and Yousef Saad.  Their recent paper &quot;&lt;a href=&quot;http://www-users.cs.umn.edu/~saad/PDF/umsi-2010-114.pdf&quot; rel=&quot;nofollow&quot;&gt;Domain-decomposition-type methods for computing the diagonal of a matrix inverse&lt;/a&gt;&quot; has the following abstract:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/onhSn.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2011-01-22T03:00:28.680" CommentCount="2" />
  <row Id="18492" PostTypeId="1" AcceptedAnswerId="18496" CreationDate="2011-01-22T03:18:36.117" Score="4" ViewCount="201" Body="&lt;p&gt;What are the &lt;strong&gt;automorphism groups&lt;/strong&gt; of non-abelian groups of order 27? (there are two non-abelian groups of order 27). &lt;/p&gt;&#xA;" OwnerUserId="8186" LastActivityDate="2011-04-16T08:04:47.810" Title="Automorphisms of non-abelian groups of order 27" Tags="&lt;group-theory&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="5" />
  <row Id="18493" PostTypeId="2" ParentId="18083" CreationDate="2011-01-22T03:23:31.580" Score="2" ViewCount="" Body="&lt;p&gt;Assuming $M$ is a &lt;em&gt;differentiable&lt;/em&gt; manifold, this can be shown by choosing normal coordinates about any point $P\in M$. The following is expanding on the method suggested by Jason DeVito in the comments.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Putting a metric on the manifold, we can define the exponential map $\exp_P\colon T_PM\to M$ and then, choosing an orthonormal basis $e_1,\ldots,e_n$ for $T_PM$, we get a map&#xA;$$&#xD;&#xA;\begin{align}&#xD;&#xA;f\colon\mathbb{R}^n&amp;amp;\to M,\\&#xD;&#xA;(x^1,\ldots,x^n)&amp;amp;\mapsto\exp_P(x^ie_i)&#xD;&#xA;\end{align}&#xD;&#xA;$$&#xA;(using the summation convention). As the manifold is compact, this is an onto mapping (only completeness is required). The idea, then, is that there is a &lt;a href=&quot;http://en.wikipedia.org/w/index.php?title=Balanced_set&amp;amp;oldid=392602385&quot; rel=&quot;nofollow&quot;&gt;circled&lt;/a&gt; open subset $U\subseteq\mathbb{R}^n$ on which this becomes a coordinate map with dense image. That $U$ is circled (aka balanced) means that the line segment joining any point $x\in U$ to the origin is in $U$. It can be seen that any circled open set is diffeomorphic to $\mathbb{R}^n$ (also see &lt;a href=&quot;http://en.wikipedia.org/w/index.php?title=Star_domain&amp;amp;oldid=392602248&quot; rel=&quot;nofollow&quot;&gt;star domain&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $D$ be the set of $x\in\mathbb{R}^n$ such that $[0,1]\to M$, $t\mapsto f(tx)$ is a minimizing geodesic. Equivalently, $D$ is the set of $x\in\mathbb{R}^n$ with $d(P,f(x))=\Vert x\Vert$, from which we see that it is closed. Also, let $U$ be the interior of $D$. The set $\partial D\equiv D\setminus U$ is called the &lt;a href=&quot;http://en.wikipedia.org/w/index.php?title=Cut_locus_%28Riemannian_manifold%29&amp;amp;oldid=370670320&quot; rel=&quot;nofollow&quot;&gt;cut locus&lt;/a&gt; (more precisely, the cut locus is the subset of $T_PM$ corresponding to $\partial D$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fixing some $x\in\mathbb{R}^n\setminus\{0\}$, let us consider the line $t\mapsto f(tx)$ for $t\ge0$. Define $r &amp;gt; 0$ to be the maximum real number with $rx\in D$. For any $0 &amp;lt; r_0 &amp;lt; r$, we can show the following.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;On $[0,r_0]$, $t\mapsto f(tx)$ is the &lt;em&gt;unique&lt;/em&gt; minimizing geodesic joining $P$ to $f(r_0x)$.&lt;/li&gt;&#xA;&lt;li&gt;The derivative of $f$ at $r_0x$ is invertible, so $f$ is nonsingular at $r_0x$.&lt;/li&gt;&#xA;&lt;li&gt;$r_0x$ is in $U$.&lt;/li&gt;&#xA;&lt;li&gt;$f(D)=M$ and $f(\partial D)\subseteq M$ has zero measure.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Property (3) implies that $U$ is circled, so it is diffeomorphic to $\mathbb{R}^n$. Property (1) says that $f$ is one-to-one on $U$ and, by (2), it is a diffeomorphism. Then, by (4), $f(U)\supseteq M\setminus f(\partial D)$ is dense, so $f$ gives a diffeomorphism from $U$ to a dense subset of $M$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll give proofs of these statements now, although they do seem quite standard. See also &lt;a href=&quot;http://www.google.co.uk/url?sa=t&amp;amp;source=web&amp;amp;cd=12&amp;amp;ved=0CBsQFjABOAo&amp;amp;url=http%3A%2F%2Fwww.math.zju.edu.cn%2Fswm%2FRG_Section_5.pdf&amp;amp;rct=j&amp;amp;q=exponential%20map%20cut%20locus&amp;amp;ei=vDo6TY_9GMyxhAeyvJHkCg&amp;amp;usg=AFQjCNE8XiGGsLnDCTDP4EUhosaxHSCnNg&amp;amp;cad=rja&quot; rel=&quot;nofollow&quot;&gt;these notes&lt;/a&gt; and, in particular, Lemma 5.3 for the proofs of (1) and (2) and Lemma 5.4 for the proof of (4).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proof of (1): Consider a minimizing geodesic $\gamma\colon[0,r_0]\to M$ joining P to $f(r_0x)$, which will have length no more than $r_0\Vert x\Vert$. Then, we can extend $\gamma(t)$ to $r_0\le t\le r$ by setting $\gamma(t)=f(tx)$. As this curve joins $P$ to $f(rx)$ and is of length no more than $r\Vert x\Vert$, it must be a geodesic. So, $\gamma(t)=f(tx)$ for all $t\le r$, and $t\mapsto f(tx)$ is a unique minimizing geodesic on $[0,r_0]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proof of (2): Choose a nonzero $y\in\mathbb{R}^n$ and consider the vector field $t\mapsto Y_t$ given by $Y_t=t\nabla_y f(tx)=\frac{\partial}{\partial s}f(t(x+sy))\vert_{s=0}$. By geodesic deviation, this is a &lt;a href=&quot;http://en.wikipedia.org/w/index.php?title=Jacobi_field&amp;amp;oldid=385149512&quot; rel=&quot;nofollow&quot;&gt;Jacobi field&lt;/a&gt;. Also, $Y_0=0$ so, if $\nabla_yf(tx)$ was zero, $P$ and $f(r_0x)$ would be conjugate points along $t\mapsto f(tx)$. Then, it is a standard result that $t\mapsto f(tx)$ is &lt;em&gt;not&lt;/em&gt; a minimizing geodesic on $[0,r]$ for any $r &amp;gt; r_0$ (see &lt;a href=&quot;http://en.wikipedia.org/w/index.php?title=Cut_locus_%28Riemannian_manifold%29&amp;amp;oldid=370670320#Characterization&quot; rel=&quot;nofollow&quot;&gt;characterization of the cut locus&lt;/a&gt;), contradicting the choice of $r$. So, $\nabla_yf(r_0x)\not=0$, and $f$ is nonsingular at $r_0x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proof of (3): If not, then there would be a sequence $x_i\not\in D$ tending to $r_0x$. By the definition of $D$, there exist $y_i\in\mathbb{R}^n$ with $f(y_i)=f(x_i)$ and $\Vert y_i\Vert &amp;lt; \Vert x_i\Vert$. Passing to a subsequence, we can suppose that $y_i$ tends to a limit $y$. So, by continuity, $\Vert y\Vert\le r_0\Vert x\Vert$ and $f(y)=f(r_0x)$. By (1), this means that $y=r_0x$. But, then, setting $a_i=\Vert y_i-x_i\Vert$, (2) contradicts the limit $\nabla_{(y_i-x_i)/a_i}f(r_0x)\sim (f(y_i)-f(x_i))/a_i=0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proof of (4): By completeness, for any $Q\in M$, there is at least one minimizing geodesic joining $P$ to $Q$. So, $f(x)=Q$ for some $x\in D$, and $f(D)=M$. Next, (3) implies that, for any $x\in\mathbb{R}^n\setminus\{0\}$, the radial line $t\mapsto tx$ ($t\ge0$) intersects $\partial D$ at a single point. This means that $\partial D$ has zero measure. As $f$ is locally Lipschitz, it maps zero measure sets to zero measure sets, so $f(\partial D)$ has zero measure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, I'll admit that the details are a bit trickier than I initially thought when I commented that the method can be made to work &quot;without too much trouble&quot;.&lt;/p&gt;&#xA;" OwnerUserId="1321" LastEditorUserId="1321" LastEditorDisplayName="" LastEditDate="2011-01-22T16:36:58.580" LastActivityDate="2011-01-22T16:36:58.580" CommentCount="1" />
  <row Id="18494" PostTypeId="1" CreationDate="2011-01-22T03:57:41.163" Score="1" ViewCount="146" Body="&lt;p&gt;Hello everyone,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm trying to prove a result similar to &lt;a href=&quot;http://en.wikipedia.org/wiki/Runge%27s_theorem&quot; rel=&quot;nofollow&quot;&gt;Runge's theorem&lt;/a&gt; and Mergelyan's theorem (link at the bottom of the previous link), but without the condition of analyticity. The problem is as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let γ : [0,1] → $\mathbb{C}$ be a C$^1$ curve and K a compact subset disjoint from Γ = {γ(t) : t ∈ [0,1]}. Let f : Γ→ $\mathbb{C}$ be continuous. Define g : K → $\mathbb{C}$ by g(z)=$\int_{\gamma} \frac{f(w)}{w-z} dw$. By splitting up the interval [0, 1] into subintervals, prove g can be uniformly approximated on K by functions of the form $g_n(z)=\sum_{j=1}^{n} \frac{A_j}{w_j - z}$, where $w_j$ ∈ Γ and A$_j$ ∈ $\mathbb{C}$ for each j = 1, 2, ..., n.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is, we're trying to approximate g which, unless I'm wrong, doesn't appear to be analytic. As the question suggests, I suppose we're essentially trying to split the integral over [0,1] into n subintervals, each of which will correspond to a term in the sum g$_n$, and then simply apply the triangle inequality to show the integral is uniformly approximated by the g$_n$, or something along those lines. However, I don't know how we should split the integral; I feel like I should be using something like Runge's theorem because otherwise I don't see what all the 'compact, disjoint' setup is for, but as I've said, I don't know how to apply it: I suppose presumably f is uniformly continuous since K is compact; does this give us a starting point for where to split up [0,1]? Any guidance would be much appreciated - it's personal revision rather than homework, but I'll mark it as homework anyway. Thank-you! &lt;/p&gt;&#xA;" OwnerUserId="6108" LastActivityDate="2011-01-22T05:11:46.787" Title="Complex-Analytic theorem similar to Runge's theorem" Tags="&lt;homework&gt;&lt;complex-analysis&gt;&lt;approximation&gt;" AnswerCount="1" />
  <row Id="18495" PostTypeId="1" CreationDate="2011-01-22T04:10:52.487" Score="-1" ViewCount="242" Body="&lt;p&gt;What are all groups of order 54?&lt;/p&gt;&#xA;" OwnerUserId="8186" LastActivityDate="2011-01-22T10:59:18.530" Title="Groups of order 54" Tags="&lt;group-theory&gt;" AnswerCount="2" CommentCount="5" FavoriteCount="4" />
  <row Id="18496" PostTypeId="2" ParentId="18492" CreationDate="2011-01-22T04:28:28.773" Score="5" ViewCount="" Body="&lt;p&gt;The non-abelian group of order p&lt;sup&gt;3&lt;/sup&gt; with no elements of order p&lt;sup&gt;2&lt;/sup&gt; is the Sylow p-subgroup of GL(3,p).  Its automorphism group can also be viewed as a group of 3x3 matrices, the affine general linear group, AGL(2,p) = { [ a,b,e ; c,d,f ; 0, 0, 1 ] : a,b,c,d,e,f in Z/pZ, ad-bc ≠ 0 }, which is the semi-direct product of GL(2,p) on its natural module.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This description is reasonably famous, especially when considering non-abelian groups of order p&lt;sup&gt;2n+1&lt;/sup&gt; with no elements of order p&lt;sup&gt;2&lt;/sup&gt; whose center and derived subgroup have order p.  Instead of GL(2,p) you get a variation on Sp(2n,p), that simplifies to GL(2,p) when n=1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The non-abelian group of order p&lt;sup&gt;3&lt;/sup&gt; with an element of order p&lt;sup&gt;2&lt;/sup&gt; and p ≥ 3 has as its automorphism group a semi-direct product of AGL(1,p) with the dual of its natural module, so you get all 3×3 matrices { [ a, b, 0 ; 0, 1, 0 ; c, d, 1 ] : a,b,c,d in Z/pZ, a≠0 }.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In both cases the &quot;module part&quot; of the semi-direct product is the group of inner automorphisms and the quotient ( GL(2,p) and AGL(1,p) ) are the outer automorphism groups.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can read about some of this in section A.20 of Doerk–Hawkes.&lt;/p&gt;&#xA;" OwnerUserId="583" LastActivityDate="2011-01-22T04:28:28.773" CommentCount="2" />
  <row Id="18497" PostTypeId="2" ParentId="18430" CreationDate="2011-01-22T04:30:49.720" Score="1" ViewCount="" Body="&lt;p&gt;Suppose that $g$ is a measurable function on $[0,1]$ such that $\int |fg|&amp;lt;\infty$ for all $f\in L^1[0,1]$.  Then $g$ is essentially bounded.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $g$ is not essentially bounded and $E_n=\{x\in[0,1]:n\leq |g(x)|\lt n+1\}$ for each positive integer $n$, then there is a subsequence $E_{n_1},E_{n_2},\ldots$ with $m(E_{n_k})\gt 0$ for all $k$.  Let $\displaystyle{f=\sum_{k=1}^\infty}\frac{1}{k^2m(E_{n_k})}\chi_{E_{n_k}}$, so $f$ is in $L^1$, and $\displaystyle{\int |gf| \geq \sum_{k=1}^\infty\frac{n_k}{k^2}\geq\sum_{k=1}^\infty}\frac{1}{k}=\infty$.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2011-01-22T04:30:49.720" CommentCount="5" />
  <row Id="18498" PostTypeId="2" ParentId="18495" CreationDate="2011-01-22T04:33:33.350" Score="6" ViewCount="" Body="&lt;p&gt;The $3$-Sylow subgroup is normal and of order $27$. Any element of order $2$ splits the quotient, so it will be a semidirect product of a group of order $27$ by $C_2$. There are three abelian groups of order $27$, and two nonabelian ones. From this, things should be straightforward.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2011-01-22T04:33:33.350" CommentCount="9" />
  <row Id="18499" PostTypeId="1" AcceptedAnswerId="18502" CreationDate="2011-01-22T04:42:26.550" Score="0" ViewCount="95" Body="&lt;p&gt;A little stumped!&#xA;This is probably a very basic probability question, but I am lost.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At work I was asked the probability of a user hitting an outage on the website. I have some following metrics. &#xA;Total system downtime = 500,000 seconds a year.&#xA;Total amount of seconds a year = 31,556,926 seconds.&#xA;Thus, p of system down = 0.159 or 1.59%&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, here is the tricky part. We have a metric for amount of total users attempting to use the service = 16,000,000 during the same time-frame. However, these are subdivided, in the total time spent using the service. So, lets say we have 7,000,000 users that spend between 0 - 30 seconds attempting to use the service. So for these users what is the probability of hitting the system when it is unavailable? (We can assume an average of 15 seconds spent total if this simplifies things)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I looked up odds ratios and risk factors, but I am not sure how to calculate the probability of the event occurring at all. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance!&lt;/p&gt;&#xA;" OwnerUserId="6110" LastActivityDate="2011-01-22T06:30:17.387" Title="Probability calculation, system uptime, likelehood of occurence." Tags="&lt;probability&gt;&lt;statistics&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="18500" PostTypeId="2" ParentId="18494" CreationDate="2011-01-22T05:11:46.787" Score="1" ViewCount="" Body="&lt;p&gt;Apparently I can't comment on your post so I'll have to post as an answer. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$g$ is an analytic function in $\mathbb{C}\setminus \Gamma$, in particular it is analytic in $K$ so Runge's theorem applies. The problem is the specific approximating sequence you're asked to derive and I don't know right now how to handle it.&lt;/p&gt;&#xA;" OwnerUserId="4829" LastActivityDate="2011-01-22T05:11:46.787" CommentCount="2" />
  <row Id="18501" PostTypeId="1" AcceptedAnswerId="18515" CreationDate="2011-01-22T05:51:50.823" Score="4" ViewCount="109" Body="&lt;p&gt;Can someone give me an elementary proof of this fact? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: This is an exercise in Marker's text, right after he defines &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\text{acl}(A)=\{x:x \text{ is algebraic over }A\}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question and the full definition is &lt;a href=&quot;http://books.google.com/books?id=QieAHk--GCcC&amp;amp;lpg=PP1&amp;amp;dq=marker%20model%20theory&amp;amp;pg=PA31#v=onepage&amp;amp;q=acl%28acl&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="3656" LastEditorUserId="3656" LastEditorDisplayName="" LastEditDate="2011-01-22T06:28:26.657" LastActivityDate="2011-01-22T12:16:25.923" Title="Prove that acl(acl(A))=acl(A), in model theory" Tags="&lt;logic&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="0" />
  <row Id="18502" PostTypeId="2" ParentId="18499" CreationDate="2011-01-22T06:30:17.387" Score="0" ViewCount="" Body="&lt;p&gt;Let's look at a single week and assume (based on your comment) that there are 18,000 seconds of continuous downtime somewhere in the middle of the week, say starting after $d$ seconds (I think the arrangement of the downtime is only significant if there are periods of uptime that are shorter than the visit lengths, as these periods might as well be counted as downtime, too).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we have a single visitor attempting a 15-second visit, their visit can start any time from time $0$ to time $d-15$ or any time after $d+18000$ (up to 604,800) and be entirely clear of the downtime.  That is, the only times that 15-second visit cannot start and be clear of the downtime are from $d-15$ to $d+18000$, a length of $18000+15=18015$ seconds out of 604,800 seconds.  So, for each single 15-second visit, the probability that it misses the downtime is $1-\frac{18015}{604800}$.  For $n$ 15-second visits in a 1-week span, for none of them to hit the downtime, the probability is $\left(1-\frac{18015}{604800}\right)^n$.  The expected (average) number of visits hitting the downtime would be $\frac{18015}{604800}\cdot n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the number of visits you gave: 7,000,000 visits per year is roughly 134,500 visits per week; $\left(1-\frac{18015}{604800}\right)^{134500}\approx 4.377\times 10^{-1767}\approx 0$; $\frac{18015}{604800}\cdot 134500\approx 4006.31$.  That is, with a single block of 18,000 seconds of downtime in the middle of a 604,800-second span, with 134,500 15-second visits, on average 4006 of those visits ($\frac{18015}{604800}\approx 2.98\%$) will be during the downtime.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you can see from the above work, as long as the length of the visit is relatively short compared to the downtime and the total timeframe, the portion of visitors hitting it is roughly the same as the portion of the total time that is downtime.&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2011-01-22T06:30:17.387" CommentCount="1" />
  <row Id="18503" PostTypeId="2" ParentId="18405" CreationDate="2011-01-22T06:53:17.600" Score="6" ViewCount="" Body="&lt;p&gt;Let $q$ be the probability of tails.  Then the answers are&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$1) \frac{q}{1+q}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$2) \frac{1+pq+p^2q+p^2q^2}{p^3q(1+q)}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The ideas and notation in the following derivation are courtesy of Section 8.4 (&quot;Flipping Coins&quot;) of &lt;em&gt;Concrete Mathematics&lt;/em&gt;.  See that section for more details and examples.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine this as a contest in which Player A wins if HTHTH appears first and Player B wins if HTHH appears first.  Let $S_A$ and $S_B$ denote the sum of the winning sequences of tosses for HTHTH and HTHH, respectively.  (This is overloading the addition operator, but it works.)  Let $N$ denote the sum of the sequences in which neither HTHTH nor HTHH has occurred yet.  ($N$ includes the empty sequence.)  Thus $$S_A = \text{HTHTH + THTHTH + HHTHTH + TTHTHTH + THHTHTH + HHHTHTH} + \cdots,$$&#xA;$$S_B = \text{HTHH + THTHH + HHTHH + TTHTHH + THHTHH + HHHTHH} + \cdots,$$&#xA;$$N = \text{1 + H + T + HH + HT + TH + TT} + \cdots.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, by substituting $p$ for H and $q$ for T, $S_A$, $S_B$, and $N$ become the probability that $A$ wins, the probability that $B$ wins, and the expected number of tosses until the game ends, respectively.  The first two claims are straightforward.  The third follows from the fact that if $X$ is a nonnegative random variable then $E[X] = \sum_{k=0}^{\infty} P(X &amp;gt; k).$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As sums of sequences, we also have &#xA;$$N \text{ HTHTH} = S_A + S_A \text{ TH} + S_A \text{ THTH} + S_B \text{ THTH},$$&#xA;$$N \text{ HTHH} = S_A \text{ H} + S_A \text{ THH} + S_B + S_B \text{ THH}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The idea behind the first equation is that the sequences of tosses obtained by appending HTHTH to $N$ must be a sequence of winning tosses for $A$ or $B$ plus possibly some extra tosses.  In particular, each sequence in $N$ HTHTH must be exactly one of 1) a winning sequence for $A$, 2) a winning sequence for $A$ followed by TH, 3) a winning sequence for $A$ followed by THTH, or 4) a winning sequence for $B$ followed by THTH.  These are all possibilities because they represent all ways in which part of the sequence HTHTH can begin (i.e., overlap) another occurrence of HTHTH or (in 4) in which HTHH can begin an occurrence of HTHTH.  The idea behind the second equation is similar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, substituting $p$ for H and $q$ for T we have&#xA;$$N p^3q^2 = S_A + S_A pq + S_A p^2q^2 + S_B p^2q^2,$$&#xA;$$N p^3q = S_A p + S_A p^2 q + S_B + S_B p^2q.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Simultaneously solving these two equations with $S_A + S_B = 1$ yields&#xA;$$S_A = \frac{q}{1+q},$$&#xA;$$S_B = \frac{1}{1+q},$$&#xA;$$N = \frac{1+pq+p^2q+p^2q^2}{p^3q(1+q)}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;HR&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Update, in response to OP's question in the comments: Why does substituting $p$ for H and $q$ for T and the fact that $E[X] = \sum_{k=0}^{\infty} P(X &amp;gt; k)$ for a nonnegative random variable all mean that $N$ is the expected number of tosses until the game ends?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $X$ be the number of tosses until the game ends.  $P(X &amp;gt; 0)$ is just 1.  $P(X &amp;gt; 1)$ is the probability that we haven't seen one of the patterns yet with a single toss; i.e., H + T, with $p$ subbed in for H and $q$ for T.  $P(X &amp;gt; 2)$ is the probability that we haven't seen one of the patterns yet with two tosses, which is HH + HT + TH + TT, with $p$ subbed in for H and $q$ for T.  $P(X &amp;gt; 3)$ is..., etc.  This doesn't get interesting until we look at $P(X &amp;gt; 4)$, which would include all four-length sequences of H and T except for HTHH.  Adding all those sequences in which neither HTHH nor HTHTH has occurred yet is exactly $N$, and subbing in $p$ for H and $q$ for T in $N$ thus gives $E[X]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See also the solution to Problem 8.21 in &lt;em&gt;Concrete Mathematics&lt;/em&gt;, p. 578 (second edition).  &lt;/p&gt;&#xA;" OwnerUserId="2370" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2011-02-02T05:43:45.880" LastActivityDate="2011-02-02T05:43:45.880" CommentCount="2" />
  <row Id="18505" PostTypeId="1" CreationDate="2011-01-22T08:07:57.790" Score="7" ViewCount="297" Body="&lt;p&gt;&lt;strong&gt;Question&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am looking for a unique characterization of a convex polygon with $n$ vertices, relative to a feature point $p$ in the interior of the polygon. This characterization would be a vector of numbers. Suppose a polygon is described by its feature point and vertices $P=(p,v_1,\ldots,v_n)$, then the characterization function is $C(P)$ such that $C(P)=C(Q)$ if and only if $P$ is congruent* to $Q$. The most important thing I need is that for two polygons which are &quot;almost&quot; congruent, their characterization vectors should be &quot;close&quot; (like, say, small 2-norm difference). The question then is what is a simple definition of the characterization function $C(\cdot)$ which satisfies my requirements?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;*Here I define congruent as identical up to rotation and translation (reflections are not allowed), cyclic permutation of vertex indices, as well as identical feature point locations. If $P$ is congruent to $Q$, then any cyclic permutation of the vertices of either polygon should still leave $C(P)=C(Q)$ (thus $C$ should be invariant to cyclic permutations of the vertices). If two polygons are congruent, then when they are overlaid, the feature points of both polygons must also match up (this is why I originally stated that the polygon is defined relative to the feature point). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;An illustration of what I mean is shown below. The dots inside are the feature points of the surrounding polygon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/PXLss.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Things that don't work&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most characterizations of polygons usually calculate the 2x2 moment of inertial tensor relative to the center of mass of the polygon. This is not good enough because first of all, the moment tensor is not enough to completely define the shape, and second, the feature point of a polygon must also match for two congruent polygons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Ideas&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;A vector of higher order moments relative to the feature point. (Is this unique?)&lt;/li&gt;&#xA;&lt;li&gt;A vector of displacements from a regular $n$-gon vertex positions. (Does this satisfy the nearness aspect?)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="398" LastActivityDate="2011-02-12T21:09:32.570" Title="Unique characterization of convex polygons" Tags="&lt;geometry&gt;&lt;computational-geometry&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="18506" PostTypeId="2" ParentId="17798" CreationDate="2011-01-22T09:28:27.650" Score="1" ViewCount="" Body="&lt;p&gt;Alternatively you can prove that the tensor product commutes with colimits not only in both factors, but also over the base ring. This follows easily from the adjunction.&lt;/p&gt;&#xA;" OwnerUserId="1650" LastActivityDate="2011-01-22T09:28:27.650" />
  <row Id="18507" PostTypeId="2" ParentId="18413" CreationDate="2011-01-22T09:42:06.270" Score="2" ViewCount="" Body="&lt;p&gt;I did some numerical search for higher dimensions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$n=3:$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\left(&#xD;&#xA;\begin{array}{ccc}&#xD;&#xA; 1 &amp;amp; 1 &amp;amp; 0 \\&#xD;&#xA; 1 &amp;amp; 1 &amp;amp; 1 \\&#xD;&#xA; 0 &amp;amp; 1 &amp;amp; 1&#xD;&#xA;\end{array}&#xD;&#xA;\right).\left(&#xD;&#xA;\begin{array}{ccc}&#xD;&#xA; -383 &amp;amp; 13 &amp;amp; -13 \\&#xD;&#xA; -36 &amp;amp; -445 &amp;amp; -36 \\&#xD;&#xA; -13 &amp;amp; 13 &amp;amp; -383&#xD;&#xA;\end{array}&#xD;&#xA;\right)=\left(&#xD;&#xA;\begin{array}{ccc}&#xD;&#xA; -419 &amp;amp; -432 &amp;amp; -49 \\&#xD;&#xA; -432 &amp;amp; -419 &amp;amp; -432 \\&#xD;&#xA; -49 &amp;amp; -432 &amp;amp; -419&#xD;&#xA;\end{array}&#xD;&#xA;\right)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$n=4:$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\left(&#xD;&#xA;\begin{array}{cccc}&#xD;&#xA; 1 &amp;amp; 1 &amp;amp; 0 &amp;amp; 0 \\&#xD;&#xA; 1 &amp;amp; 1 &amp;amp; 1 &amp;amp; 0 \\&#xD;&#xA; 0 &amp;amp; 1 &amp;amp; 1 &amp;amp; 1 \\&#xD;&#xA; 0 &amp;amp; 0 &amp;amp; 1 &amp;amp; 1&#xD;&#xA;\end{array}&#xD;&#xA;\right).\left(&#xD;&#xA;\begin{array}{cccc}&#xD;&#xA; -383 &amp;amp; 13 &amp;amp; -36 &amp;amp; -23 \\&#xD;&#xA; 85 &amp;amp; -360 &amp;amp; 49 &amp;amp; -49 \\&#xD;&#xA; -49 &amp;amp; 49 &amp;amp; -360 &amp;amp; 85 \\&#xD;&#xA; -23 &amp;amp; -36 &amp;amp; 13 &amp;amp; -383&#xD;&#xA;\end{array}&#xD;&#xA;\right)=\left(&#xD;&#xA;\begin{array}{cccc}&#xD;&#xA; -298 &amp;amp; -347 &amp;amp; 13 &amp;amp; -72 \\&#xD;&#xA; -347 &amp;amp; -298 &amp;amp; -347 &amp;amp; 13 \\&#xD;&#xA; 13 &amp;amp; -347 &amp;amp; -298 &amp;amp; -347 \\&#xD;&#xA; -72 &amp;amp; 13 &amp;amp; -347 &amp;amp; -298&#xD;&#xA;\end{array}&#xD;&#xA;\right)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$n=5:$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\left(&#xD;&#xA;\begin{array}{ccccc}&#xD;&#xA; 1 &amp;amp; 1 &amp;amp; 0 &amp;amp; 0 &amp;amp; 0 \\&#xD;&#xA; 1 &amp;amp; 2 &amp;amp; 1 &amp;amp; 0 &amp;amp; 0 \\&#xD;&#xA; 0 &amp;amp; 1 &amp;amp; 3 &amp;amp; 1 &amp;amp; 0 \\&#xD;&#xA; 0 &amp;amp; 0 &amp;amp; 1 &amp;amp; 4 &amp;amp; 1 \\&#xD;&#xA; 0 &amp;amp; 0 &amp;amp; 0 &amp;amp; 1 &amp;amp; 5&#xD;&#xA;\end{array}&#xD;&#xA;\right).\left(&#xD;&#xA;\begin{array}{ccccc}&#xD;&#xA; -7526 &amp;amp; 3158 &amp;amp; -9379 &amp;amp; 7340 &amp;amp; -8405 \\&#xD;&#xA; 5216 &amp;amp; -3477 &amp;amp; 6079 &amp;amp; -6570 &amp;amp; 6359 \\&#xD;&#xA; -3225 &amp;amp; 1486 &amp;amp; -3098 &amp;amp; 2500 &amp;amp; -3543 \\&#xD;&#xA; 1159 &amp;amp; -1300 &amp;amp; 905 &amp;amp; -1249 &amp;amp; 970 \\&#xD;&#xA; -641 &amp;amp; 414 &amp;amp; -841 &amp;amp; 186 &amp;amp; -656&#xD;&#xA;\end{array}&#xD;&#xA;\right)=$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$=\left(&#xD;&#xA;\begin{array}{ccccc}&#xD;&#xA; -2310 &amp;amp; -319 &amp;amp; -3300 &amp;amp; 770 &amp;amp; -2046 \\&#xD;&#xA; -319 &amp;amp; -2310 &amp;amp; -319 &amp;amp; -3300 &amp;amp; 770 \\&#xD;&#xA; -3300 &amp;amp; -319 &amp;amp; -2310 &amp;amp; -319 &amp;amp; -3300 \\&#xD;&#xA; 770 &amp;amp; -3300 &amp;amp; -319 &amp;amp; -2310 &amp;amp; -319 \\&#xD;&#xA; -2046 &amp;amp; 770 &amp;amp; -3300 &amp;amp; -319 &amp;amp; -2310&#xD;&#xA;\end{array}&#xD;&#xA;\right)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All matrices have full rank. However, for high dimensions they are quite ugly.&lt;/p&gt;&#xA;" OwnerUserId="3123" LastEditorUserId="3123" LastEditorDisplayName="" LastEditDate="2011-01-22T10:15:36.133" LastActivityDate="2011-01-22T10:15:36.133" />
  <row Id="18508" PostTypeId="1" CreationDate="2011-01-22T09:47:01.577" Score="2" ViewCount="88" Body="&lt;p&gt;Which books teach numerical method for system of differential equations?&lt;/p&gt;&#xA;" OwnerUserId="6115" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2011-02-21T18:36:00.740" LastActivityDate="2011-04-07T09:14:46.050" Title="Which books teach numerical method for system of differential equations?" Tags="&lt;reference-request&gt;&lt;differential-equations&gt;&lt;numerical-methods&gt;" AnswerCount="3" FavoriteCount="1" />
  <row Id="18509" PostTypeId="1" CreationDate="2011-01-22T10:34:52.483" Score="2" ViewCount="94" Body="&lt;p&gt;Suppose I have found an irrational number, say &lt;em&gt;e^x&lt;/em&gt; upto &lt;em&gt;m&lt;/em&gt; decimal places....&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I want to find upto say &lt;em&gt;m+n&lt;/em&gt; decimal places. Is there any way that I can use the previous value of &lt;em&gt;e^x&lt;/em&gt; upto &lt;em&gt;m&lt;/em&gt; decimal places to &lt;em&gt;help&lt;/em&gt; me find the remaining &lt;em&gt;n&lt;/em&gt; decimal places? (I can save some number, say, like remainder, when finding upto &lt;em&gt;m&lt;/em&gt; decimal places to help me find the later digits...)&lt;/p&gt;&#xA;" OwnerUserId="6094" LastActivityDate="2011-05-06T23:50:07.600" Title="Finding e^x with more number of significant digits using a previously computed value of the same (having lesser number of significant digits)" Tags="&lt;algorithms&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="18510" PostTypeId="1" AcceptedAnswerId="18511" CreationDate="2011-01-22T10:38:12.930" Score="1" ViewCount="226" Body="&lt;p&gt;I'm trying to prove the following claim:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f_n \in C_c$, $C_c$ being the set of continuous functions with compact support, then $lim_{n \rightarrow \infty} ||&amp;#160;f_n - f||_{\infty} = 0$ implies $f_n(x) \rightarrow f(x)$ uniformly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, according to my understanding, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ lim_{n \rightarrow \infty} ||&amp;#160;f_n - f||_{\infty} = 0 $$ &#xA;$$ \Leftrightarrow $$&#xA;$$ \forall \varepsilon &amp;gt; 0 \exists N: n &amp;gt; N \Rightarrow |f_n(x) - f(x)| \leq ||&amp;#160;f_n - f||_{\infty}&amp;#160;&amp;lt; \varepsilon$$ $\mu$-almost everywhere on X. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now my problem is that I don't see how uniform convergence follows from pointwise convergence $\mu$ almost everywhere. Can someone give me a hint? I guess I have to use the fact that they have compact support but I don't see how to put this together.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for your help!&lt;/p&gt;&#xA;" OwnerUserId="5798" LastActivityDate="2011-01-22T11:19:56.167" Title="Convergence in $L_{\infty}$ norm implies uniform convergence" Tags="&lt;analysis&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" />
  <row Id="18511" PostTypeId="2" ParentId="18510" CreationDate="2011-01-22T10:55:05.460" Score="4" ViewCount="" Body="&lt;p&gt;You were not very specific about your hypotheses - I assume you're working on $\mathbb{R}^{n}$ with Lebesgue measure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose there exists a point $x$ such that $|f_{n}(x) - f(x)| &amp;gt; \varepsilon$. Then there exists an open set $U$ containing $x$ such that for all $y \in U$ you have $|f_{n}(y) - f(y)| &amp;gt; \varepsilon$ by continuity. But this contradicts the a.e. statement you gave.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In case you don't know yet that $f$ is continuous (or rather: has a continuous representative in $L^\infty$), a similar argument shows that $(f_{n})$ is a uniform Cauchy sequence (with the sup-norm, not only the essential sup-norm), hence $f$ will be continuous (in fact, uniformly continuous).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that I haven't used compact support at all, just continuity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're working in a more general setting (like a locally compact space), you'd have to require that the measure gives positive mass to each non-empty open set.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, note that $f$ need not have compact support. It will only have the property that it will be arbitrarily small outside some compact set (&quot;vanish at infinity&quot; is the technical term). For instance, $\frac{1}{1+|x|}$ can easily be uniformly approximated by functions with compact support.&lt;/p&gt;&#xA;" OwnerUserId="5363" LastEditorUserId="5363" LastEditorDisplayName="" LastEditDate="2011-01-22T11:19:56.167" LastActivityDate="2011-01-22T11:19:56.167" CommentCount="5" />
  <row Id="18512" PostTypeId="2" ParentId="18510" CreationDate="2011-01-22T10:56:08.930" Score="3" ViewCount="" Body="&lt;p&gt;Just a hint: The compact support doesn't matter, but the &lt;em&gt;continuity&lt;/em&gt; does matter. It helps you to remove null sets.&lt;/p&gt;&#xA;" OwnerUserId="2575" LastActivityDate="2011-01-22T10:56:08.930" />
  <row Id="18513" PostTypeId="2" ParentId="18495" CreationDate="2011-01-22T10:59:18.530" Score="1" ViewCount="" Body="&lt;p&gt;A simple google search finds this: &lt;a href=&quot;http://pari.math.u-bordeaux.fr/galpol/54/index.html&quot; rel=&quot;nofollow&quot;&gt;http://pari.math.u-bordeaux.fr/galpol/54/index.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="589" LastActivityDate="2011-01-22T10:59:18.530" />
  <row Id="18514" PostTypeId="1" CreationDate="2011-01-22T11:46:18.507" Score="5" ViewCount="410" Body="&lt;p&gt;How do I compute $\text{(irrational)}^{\text{(irrational)}}$ up to a required number of decimals say m, in the fastest way ? (one way is of course compute both the irrational numbers to a precision much larger than m and then solve it... but you never know how much excess of m you will need to calculate the irrationals.. )&lt;/p&gt;&#xA;" OwnerUserId="6094" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2011-01-22T15:21:17.753" LastActivityDate="2011-05-16T17:09:41.427" Title="Irrational$^\text{Irrational}$" Tags="&lt;irrational-numbers&gt;" AnswerCount="2" CommentCount="13" />
  <row Id="18515" PostTypeId="2" ParentId="18501" CreationDate="2011-01-22T12:16:25.923" Score="4" ViewCount="" Body="&lt;p&gt;First, note that if Grandma Helen has only finitely many children, and each of those children have only finitely many children, then Grandma Helen has only finitely many grandchildren. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the problem, however, there is the subtlety that perhaps some but not all of the children of Helen have only finitely many children. But you can still use the same idea: If Grandma Helen has only finitely many children, and one of those children has at most 10 children (perhaps some others have infinitely many), then the collection of grandchildren of Helen whose parents have at most 10 children is finite.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Those grandchildren (whose parents have at most 10 children) satisfy a first-order expressible property, with Helen as a parameter, that is satisfied by only finitely many other people. &lt;/p&gt;&#xA;" OwnerUserId="413" LastActivityDate="2011-01-22T12:16:25.923" CommentCount="1" />
  <row Id="18517" PostTypeId="2" ParentId="18445" CreationDate="2011-01-22T13:27:01.343" Score="3" ViewCount="" Body="&lt;p&gt;If I wanted to implement an arbitrary precision approximation to $e^x$ from &quot;scratch&quot; (building on an arbitrary precision numeric library such as &lt;a href=&quot;http://en.wikipedia.org/wiki/GNU_Multi-Precision_Library&quot; rel=&quot;nofollow&quot;&gt; GMP&lt;/a&gt;) with faster convergence than Taylor series/polynomial approximations, I would design around &lt;a href=&quot;http://functions.wolfram.com/ElementaryFunctions/Exp/10/ShowAll.html&quot; rel=&quot;nofollow&quot;&gt; continued fraction/rational function approximations&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just as the precision given by the well-known Taylor series can be extended as required by using additional terms, so too will the continued fraction expansion give any required precision by using additional &lt;a href=&quot;http://en.wikipedia.org/wiki/Convergent_%28continued_fraction%29&quot; rel=&quot;nofollow&quot;&gt; convergents&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At first it may seem that evaluating the continued fraction approximation entails a significant disadvantage in comparison with the power series.  Additional terms can be added to the partial sums of the power series, to obtain left-to-right sequential approximations.  There is a way to achieve the same effect with continued fractions, namely evaluating the partial numerators and denominators into the convergents &lt;a href=&quot;http://en.wikipedia.org/wiki/Fundamental_recurrence_formulas&quot; rel=&quot;nofollow&quot;&gt; through a recurrence relation&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are a couple of important ideas worth considering even if another method for evaluating the exponential function were to be used on a bounded interval such as $[0,1]$, as for example power series or even lookup tables(!).  One is &lt;em&gt;symmetry&lt;/em&gt; of the exponential function, and the other is a method of &lt;em&gt;range reduction&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Although the exponential function is not even (or odd), it satisfies a relation:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$e^{-x} = 1/e^x$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;that implies it is a simple rational transformation of an even function:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$e^x = (1 + x*f(x^2))/(1 - x*f(x^2))$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which may be solved for a power series or continued fraction approximation of f(z).  This is the idea of &lt;em&gt;symmetry&lt;/em&gt; as it applies to the exponential function.  Not only does it reduce the evaluaton of the exponential function on the whole real line to just the positive (or the negative) half, it also &quot;contracts&quot; the number of terms needed for a given accuracy by half (retaining only $x^2$ terms in the expansion).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Although a continued fraction or Taylor series approximation to the exponential function may converge for arbitrary real (or complex) values), the rate of convergence is not uniform.  The Taylor series converges faster the closer the argument is to zero.  Thus a &lt;em&gt;range reduction&lt;/em&gt; is important for expressing the exponential function at an arbitrary real value in terms of a value in some interval like $[0,1]$.  For example, if the floating point arithmetic is binary (radix 2), it can be especially convenient to use the familiar law of exponents with multiples of $\ln 2$:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$e^x = 2^k * e^r \ \text{where} \ x = r + k*\ln 2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which allows $e^x$ to be evaluated, up to a change in binary exponent, using an approximation that converges rapidly over $[0,\ln 2]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Combining these two ideas (symmetry, range reduction) the speed of convergence can be limited to the interval $[0, \ln 2/2]$.  Limiting the interval of evaluation may allow you to determine in advance how many terms of the continued fraction or Taylor series expansion have to be retained to obtain a desired accuracy.  When this is known the evaluation can be done more efficiently (backwards recursion for continued fractions or Horner's method for truncated Taylor series/polynomials) than if we were forced to continually introduced further terms until the desired accuracy is reached.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &quot;faster&quot; continued fraction I had in mind is formula (11.1.2) here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[Handbook of continued fractions for special functions (Cuyt et al, 2008)]&lt;br&gt;&#xA;&lt;a href=&quot;http://books.google.com/books?id=DQtpJaEs4NIC&amp;amp;dq=exponential+function+continued+fraction+convergence&quot; rel=&quot;nofollow&quot;&gt;http://books.google.com/books?id=DQtpJaEs4NIC&amp;amp;dq=exponential+function+continued+fraction+convergence&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Their reference is to this modern classic:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The application of continued fractions and their generalizations to problems in approximation theory&lt;br&gt;&#xA;A.N. Khovanskii, 1963 (P.Noordhoff)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A neat webMathematic based site was here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[Convergence Acceleration with Canonical Contractions of Continued Fractions:&#xA;Exponential function -- Andreas Lauschke Consulting]&lt;br&gt;&#xA;&lt;a href=&quot;http://216.80.120.13:8080/webMathematica/LC/exponential.jsp&quot; rel=&quot;nofollow&quot;&gt;http://216.80.120.13:8080/webMathematica/LC/exponential.jsp&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately it seems that site has been down for the past 24 hours.  Google's cached version does not provide the graphic comparisons of convergence (since those were Mathematica based) nor even the formulas as they were perhaps images.  I have some notes on his formulas (derived by contraction from the one noted above) if that would be helpful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Computing the constant $\ln 2$&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Generations of math students have been introduced to the concept of conditional versus absolute convergence by the example of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Harmonic_series_%28mathematics%29#Alternating_harmonic_series&quot; rel=&quot;nofollow&quot;&gt; alternating harmonic series&lt;/a&gt;:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ln 2 = 1 - 1/2 + 1/3 - 1/4 + ...$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course this series, derived from a power series expansion of $\ln x$ around $x = 1$, has such slow convergence that even if we combine consecutive pairs of terms:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ln 2 = 1/2 + 1/12 + 1/30 + ...$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the resulting absolutely convergent series is useless for obtaining arbitrary precision values of $\ln 2$.  For convenience the first seven partial sum of this are:  &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;0.50000000...  &#xA;0.58333333...  &#xA;0.61666666...  &#xA;0.63452381...  &#xA;0.64563492...  &#xA;0.65870518...  &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Since $\ln 2$ is 0.69314718..., we have an error of about a third of a unit in the first decimal place.  In other words not much more convergence than one decimal place correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It therefore makes a striking contrast with the nice convergence of &lt;a href=&quot;http://functions.wolfram.com/ElementaryFunctions/Log/10/&quot; rel=&quot;nofollow&quot;&gt; a continued fraction expansion&lt;/a&gt; of the same value:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ln 2 = \cfrac{1}{1 + \cfrac{1}{2 + \cfrac{1}{3 + \cfrac{4}{4 + \cfrac{4}{5 + \cfrac{9}{6 + \cfrac{9}{7 + ...}}}}}}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first seven convergents are:  &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;0.66666666...  &#xA;0.70000000...  &#xA;0.69230769...  &#xA;0.69333333...  &#xA;0.69312169...  &#xA;0.69315245...  &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The error here is about half a unit in the fifth decimal place.&lt;/p&gt;&#xA;" OwnerUserId="3111" LastEditorUserId="3111" LastEditorDisplayName="" LastEditDate="2011-01-22T20:45:44.090" LastActivityDate="2011-01-22T20:45:44.090" />
  <row Id="18518" PostTypeId="1" AcceptedAnswerId="18535" CreationDate="2011-01-22T13:46:43.070" Score="4" ViewCount="197" Body="&lt;p&gt;Hello allt,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As exam preparation we were trying to proof the following task:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $V=\mathbb{R}^2$ and let $\phi$ be an endomorphism of $V$ with $\phi \circ \phi = id$ and $\phi \neq id$ and $\phi \neq -id$. Proof that this implies the existence of a basis $B=(b_1,b_2)$ of $V$ with $\phi(b_1) = b_1$ and $\phi(b_2)=-b_2$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately we aren't able to solve that task and would very much appreciate some proofs and and a &quot;how to&quot; of how to approach such problems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for your help.&lt;/p&gt;&#xA;" OwnerUserId="3188" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-23T05:58:28.723" LastActivityDate="2011-01-23T05:58:28.723" Title="proof that $\phi\circ\phi=id$ implies the existence of a diagonal matrix" Tags="&lt;linear-algebra&gt;" AnswerCount="3" CommentCount="6" FavoriteCount="0" />
  <row Id="18519" PostTypeId="1" CreationDate="2011-01-22T14:39:25.537" Score="1" ViewCount="128" Body="&lt;p&gt;The problem is 10.6a from &lt;a href=&quot;http://www.cs.uu.nl/geobook/&quot; rel=&quot;nofollow&quot;&gt;Computational Geometry: Algorithms and Applications&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;We want to solve the following query&#xA;  problem: Given a set $S$ of $n$&#xA;  disjoint line segments in the plane,&#xA;  determine those segments that&#xA;  intersect a vertical ray running from&#xA;  a point $(q_x, q_y)$ vertically&#xA;  upwards to infinity. Describe a data&#xA;  structure for this problem that uses&#xA;  $O(n \log n)$ storage and has a query&#xA;  time of $O(\log n + k)$, where $k$ is&#xA;  the number of reported answers.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It was a homework question to a CG course I was taking a year ago.&#xA;I solve the problem, but I remember the solution was much more elegant. Now I can't find the solution sheet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope someone can reconstruct the elegant solution. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here I present basic idea of my solution:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;There exist a way to order the segments, so there are &quot;higher&quot; segments (segments $a$ is higher than segment $b$ another if there exist a vertical segment from $a$ to $b$, such that the higher intersection is on $a$). Number the segment from highest to lowest. &lt;/li&gt;&#xA;&lt;li&gt;Project all segments to $x$ axis. Create a interval tree. In the interval tree, instead of using 2 sorted list to contain the left end points and right end points, use 2 2D range tree instead.&lt;/li&gt;&#xA;&lt;li&gt;Each segment can be uniquely mapped into a pair $(x,y)$. It can be sorted by end point in $x$ direction, and the order we defined in $y$ direction. So we can query it with the 2D range tree. Using fractional cascading we can do it in $O(\log n+k)$ time. &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="96" LastActivityDate="2011-01-22T14:39:25.537" Title="Set of segments a vertical ray intersects." Tags="&lt;algorithms&gt;&lt;computational-geometry&gt;" CommentCount="1" />
  <row Id="18520" PostTypeId="2" ParentId="18518" CreationDate="2011-01-22T14:42:10.633" Score="2" ViewCount="" Body="&lt;p&gt;My approach would be to write the eigenvalues-eigenvectors equatio:n it's inmediate that the eigenvalues must be +1 or -1. What remains it to see that the eigenvectors are orthogonal and that from the alternatives (1,1) (1,-1) (-1,-1) only the second is possible.&lt;/p&gt;&#xA;" OwnerUserId="312" LastActivityDate="2011-01-22T14:42:10.633" CommentCount="5" />
  <row Id="18521" PostTypeId="2" ParentId="18518" CreationDate="2011-01-22T14:56:35.750" Score="2" ViewCount="" Body="&lt;p&gt;The polynomial $X^2-1=(X-1)(X+1)$ annihilates $\phi$, and splits over $\mathbb{R}$, and with simple roots. Thus $\phi$ is diagonalizable. However, $\phi$ cannot be identity or minus identity, then there exists a basis $B=(b_1,b_2)$ of $V=\mathbb{R}^2$ with $\phi(b_1)=b_1$ and $\phi(b_2)=−b_2$.&lt;/p&gt;&#xA;" OwnerUserId="2380" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-23T05:54:48.510" LastActivityDate="2011-01-23T05:54:48.510" CommentCount="2" />
  <row Id="18522" PostTypeId="2" ParentId="1953" CreationDate="2011-01-22T15:38:28.153" Score="1" ViewCount="" Body="&lt;p&gt;If $B$ is any set (finite or not), and $R$ is a ring, then I would write $\prod_{x \in B} R$ to denote the product of a number of copies of $R$ indexed by $B$ --- this product is again a ring (with the component-wise operations).  This is precisely the set of functions $f: B \to R$ with the operations defined pointwise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I guess one could also write $R^B$ (set-theory notation for the functions from $B$ to $R$), but in the parts of algebra that I commonly read (commutative algebra, algebraic and arithmetic geometry) the product notation of my first paragraph is standard, and the exponential notation is not.&lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2011-01-22T15:38:28.153" />
  <row Id="18523" PostTypeId="1" CreationDate="2011-01-22T16:16:29.010" Score="2" ViewCount="128" Body="&lt;p&gt;Hi,&#xA;I'm trying to learn something about mathematical logic. I think, I partly understand definitions of free variables, bound variables and quantification, but it's not easy topic for me and it's not enough to solve problems. I mean, for example I have such problem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;mark sets on a graph:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\{ z : \mathbb{R} | \forall{x}\exists{x}(x=1) \}$&lt;/li&gt;&#xA;&lt;li&gt;$\{ z : \mathbb{R} | \exists{x}\forall{x}(x=1) \}$&lt;/li&gt;&#xA;&lt;li&gt;$\{ x : \mathbb{R} | \exists{x}\forall{x}(x=1) \}$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;What is $z$ in this case? How to think about such problems? How to solve them?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not asking You to solve that particular problem; I want You to explain me (or give some links to articles, tutorials, books etc.) this topic (free variables,bound variables and quantification).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would really appreciate You help.&lt;/p&gt;&#xA;" OwnerUserId="6123" LastActivityDate="2011-01-22T16:43:46.113" Title="Free variables, bound variables and quantification - explanation" Tags="&lt;logic&gt;&lt;elementary-set-theory&gt;&lt;quantifiers&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="18524" PostTypeId="1" CreationDate="2011-01-22T16:33:41.100" Score="1" ViewCount="68" Body="&lt;p&gt;I have a question, I want to prove cartesian closed category with finite coproduct is distributive category. but do not know how to prove A*(B+C)~A*B+A*C is isomorphism from the property of exponential object, can anyone help me? Thank you.&lt;/p&gt;&#xA;" OwnerUserId="6125" LastEditorUserId="6125" LastEditorDisplayName="" LastEditDate="2011-01-23T01:03:58.547" LastActivityDate="2011-01-23T01:03:58.547" Title="prove isomorphism in distributive category" Tags="&lt;category-theory&gt;" CommentCount="1" />
  <row Id="18526" PostTypeId="2" ParentId="18523" CreationDate="2011-01-22T16:43:46.113" Score="1" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://www.stanford.edu/class/cs103a/handouts/18%20Intro%20to%20Quantification.pdf&quot; rel=&quot;nofollow&quot;&gt;These&lt;/a&gt; (Stanford notes on quantification) notes seem good.&lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2011-01-22T16:43:46.113" />
  <row Id="18527" PostTypeId="1" AcceptedAnswerId="18534" CreationDate="2011-01-22T17:12:43.327" Score="1" ViewCount="172" Body="&lt;p&gt;I have done some higher math, being an engineer. They usually focus on getting the correct answer as supposed to actually understanding math. Glorified calculators, basically. I know how to do stuff with log x, even calculate it with different bases, yay me. So my next question might come as a bit of surprise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can someone explain log from a real life point of view. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bonus points for explain x log x.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please and thank you. &lt;/p&gt;&#xA;" OwnerUserId="6127" LastEditorUserId="2548" LastEditorDisplayName="" LastEditDate="2011-01-22T18:00:30.557" LastActivityDate="2011-01-22T18:00:30.557" Title="Can someone explain log?" Tags="&lt;limit&gt;&lt;logarithms&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="18528" PostTypeId="2" ParentId="18433" CreationDate="2011-01-22T17:13:14.670" Score="0" ViewCount="" Body="&lt;p&gt;It actually depends on the form of M and N. But basically, you can use any technique other than getting the integrating factor to solve for the ODE, like determining whether such ODE is exact, homogeneous, etc.&lt;/p&gt;&#xA;" OwnerUserId="6126" LastActivityDate="2011-01-22T17:13:14.670" />
  <row Id="18529" PostTypeId="2" ParentId="18527" CreationDate="2011-01-22T17:25:10.697" Score="1" ViewCount="" Body="&lt;p&gt;log is usually used in statistical results. Here are few examples:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) If I have obtained a data with values in orders of 10 (like: 10, 100, 50000, 6000000000), I can use log in my graph to shorten the length of the axis values (see: log-log graph)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) There are results which increases exponentially like decaying system, half-life, and many systems in control systems engineering. As stated in 1, it is hard to illustrate values that increases without bound.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(3) There are many instances where you cannot find on how many times a number is multiplied by the same number to produce a result. In equation form, you need to find x in this problem: 5^x = 81.859.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;xlogx means the same thing. A shortened value multiplied by an amplitude to somehow increase its value. It usually comes out on finding the initial value problem on differential equations (generally from mechatronics). &lt;/p&gt;&#xA;" OwnerUserId="6126" LastActivityDate="2011-01-22T17:25:10.697" />
  <row Id="18530" PostTypeId="1" AcceptedAnswerId="18579" CreationDate="2011-01-22T17:26:02.310" Score="3" ViewCount="112" Body="&lt;p&gt;Sorry about the title, I have no idea how to describe these types of problems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Problem statement:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$A(S)$ is the set of 1-1 mappings of $S$ onto itself.  Let $S \supset T$ and consider the subset $U(T) = $ { $f \in A(S)$ | $f(t) \in T$ for every $t \in T$ }.  $S$ has $n$ elements and $T$ has $m$ elements.  Show that there is a mapping $F:U(T) \rightarrow S_m$ such that $F(fg) = F(f)F(g)$ for $f, g \in U(T)$ and $F$ is onto $S_m$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do I write up this reasoning:  When I look at the sets $S$ = { 1, 2, ..., $n$} and $T$ = { 1, 2, ..., $m$}, I can see that there are a bunch of permutations of the elements of $T$ within $S$.  I can see there are $(n - m)!$ members of $S$ for each permutation of $T$'s elements.  But there needs to be some way to get a handle on the positions of the elements in $S$ and $T$ in order to compare them to each other.  But $S$ isn't any particular set, like a set of integers, so how can I relate the positions of the elements to one another?  Or, is this the wrong way to go about it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$U(T_3 \subset S_6) = \left(&#xD;&#xA;\begin{array}{cccccc}&#xD;&#xA; 1 &amp;amp; 2 &amp;amp; 3 &amp;amp; 4 &amp;amp; 5 &amp;amp; 6 \\&#xD;&#xA; 1 &amp;amp; 2 &amp;amp; 3 &amp;amp; 4 &amp;amp; 6 &amp;amp; 5 \\&#xD;&#xA; 1 &amp;amp; 2 &amp;amp; 3 &amp;amp; 5 &amp;amp; 4 &amp;amp; 6 \\&#xD;&#xA; 1 &amp;amp; 2 &amp;amp; 3 &amp;amp; 5 &amp;amp; 6 &amp;amp; 4 \\&#xD;&#xA; 1 &amp;amp; 2 &amp;amp; 3 &amp;amp; 6 &amp;amp; 4 &amp;amp; 5 \\&#xD;&#xA; 1 &amp;amp; 2 &amp;amp; 3 &amp;amp; 6 &amp;amp; 5 &amp;amp; 4 \\&#xD;&#xA; 1 &amp;amp; 3 &amp;amp; 2 &amp;amp; 4 &amp;amp; 5 &amp;amp; 6 \\&#xD;&#xA; 1 &amp;amp; 3 &amp;amp; 2 &amp;amp; 4 &amp;amp; 6 &amp;amp; 5 \\&#xD;&#xA; 1 &amp;amp; 3 &amp;amp; 2 &amp;amp; 5 &amp;amp; 4 &amp;amp; 6 \\&#xD;&#xA; 1 &amp;amp; 3 &amp;amp; 2 &amp;amp; 5 &amp;amp; 6 &amp;amp; 4 \\&#xD;&#xA; 1 &amp;amp; 3 &amp;amp; 2 &amp;amp; 6 &amp;amp; 4 &amp;amp; 5 \\&#xD;&#xA; 1 &amp;amp; 3 &amp;amp; 2 &amp;amp; 6 &amp;amp; 5 &amp;amp; 4 \\&#xD;&#xA; 2 &amp;amp; 1 &amp;amp; 3 &amp;amp; 4 &amp;amp; 5 &amp;amp; 6 \\&#xD;&#xA; 2 &amp;amp; 1 &amp;amp; 3 &amp;amp; 4 &amp;amp; 6 &amp;amp; 5 \\&#xD;&#xA; 2 &amp;amp; 1 &amp;amp; 3 &amp;amp; 5 &amp;amp; 4 &amp;amp; 6 \\&#xD;&#xA; 2 &amp;amp; 1 &amp;amp; 3 &amp;amp; 5 &amp;amp; 6 &amp;amp; 4 \\&#xD;&#xA; 2 &amp;amp; 1 &amp;amp; 3 &amp;amp; 6 &amp;amp; 4 &amp;amp; 5 \\&#xD;&#xA; 2 &amp;amp; 1 &amp;amp; 3 &amp;amp; 6 &amp;amp; 5 &amp;amp; 4 \\&#xD;&#xA; 2 &amp;amp; 3 &amp;amp; 1 &amp;amp; 4 &amp;amp; 5 &amp;amp; 6 \\&#xD;&#xA; 2 &amp;amp; 3 &amp;amp; 1 &amp;amp; 4 &amp;amp; 6 &amp;amp; 5 \\&#xD;&#xA; 2 &amp;amp; 3 &amp;amp; 1 &amp;amp; 5 &amp;amp; 4 &amp;amp; 6 \\&#xD;&#xA; 2 &amp;amp; 3 &amp;amp; 1 &amp;amp; 5 &amp;amp; 6 &amp;amp; 4 \\&#xD;&#xA; 2 &amp;amp; 3 &amp;amp; 1 &amp;amp; 6 &amp;amp; 4 &amp;amp; 5 \\&#xD;&#xA; 2 &amp;amp; 3 &amp;amp; 1 &amp;amp; 6 &amp;amp; 5 &amp;amp; 4 \\&#xD;&#xA; 3 &amp;amp; 1 &amp;amp; 2 &amp;amp; 4 &amp;amp; 5 &amp;amp; 6 \\&#xD;&#xA; 3 &amp;amp; 1 &amp;amp; 2 &amp;amp; 4 &amp;amp; 6 &amp;amp; 5 \\&#xD;&#xA; 3 &amp;amp; 1 &amp;amp; 2 &amp;amp; 5 &amp;amp; 4 &amp;amp; 6 \\&#xD;&#xA; 3 &amp;amp; 1 &amp;amp; 2 &amp;amp; 5 &amp;amp; 6 &amp;amp; 4 \\&#xD;&#xA; 3 &amp;amp; 1 &amp;amp; 2 &amp;amp; 6 &amp;amp; 4 &amp;amp; 5 \\&#xD;&#xA; 3 &amp;amp; 1 &amp;amp; 2 &amp;amp; 6 &amp;amp; 5 &amp;amp; 4 \\&#xD;&#xA; 3 &amp;amp; 2 &amp;amp; 1 &amp;amp; 4 &amp;amp; 5 &amp;amp; 6 \\&#xD;&#xA; 3 &amp;amp; 2 &amp;amp; 1 &amp;amp; 4 &amp;amp; 6 &amp;amp; 5 \\&#xD;&#xA; 3 &amp;amp; 2 &amp;amp; 1 &amp;amp; 5 &amp;amp; 4 &amp;amp; 6 \\&#xD;&#xA; 3 &amp;amp; 2 &amp;amp; 1 &amp;amp; 5 &amp;amp; 6 &amp;amp; 4 \\&#xD;&#xA; 3 &amp;amp; 2 &amp;amp; 1 &amp;amp; 6 &amp;amp; 4 &amp;amp; 5 \\&#xD;&#xA; 3 &amp;amp; 2 &amp;amp; 1 &amp;amp; 6 &amp;amp; 5 &amp;amp; 4&#xD;&#xA;\end{array}&#xD;&#xA;\right),A(T_3) = \left(&#xD;&#xA;\begin{array}{ccc}&#xD;&#xA; 1 &amp;amp; 2 &amp;amp; 3 \\&#xD;&#xA; 1 &amp;amp; 3 &amp;amp; 2 \\&#xD;&#xA; 2 &amp;amp; 1 &amp;amp; 3 \\&#xD;&#xA; 2 &amp;amp; 3 &amp;amp; 1 \\&#xD;&#xA; 3 &amp;amp; 1 &amp;amp; 2 \\&#xD;&#xA; 3 &amp;amp; 2 &amp;amp; 1&#xD;&#xA;\end{array}&#xD;&#xA;\right)$&lt;/p&gt;&#xA;" OwnerUserId="6059" LastEditorUserId="6059" LastEditorDisplayName="" LastEditDate="2011-01-22T18:23:07.293" LastActivityDate="2011-01-23T02:54:12.817" Title="A subset of permutations on the set S such that the elements that are within T, a subset of S, are not exchanged with elements outside of T" Tags="&lt;abstract-algebra&gt;&lt;functions&gt;&lt;permutations&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="18531" PostTypeId="2" ParentId="18514" CreationDate="2011-01-22T17:32:02.813" Score="-4" ViewCount="" Body="&lt;p&gt;Use matlab. That is the fastest way next to a calculator and next to a manual calculation.&lt;/p&gt;&#xA;" OwnerUserId="6126" LastActivityDate="2011-01-22T17:32:02.813" CommentCount="4" />
  <row Id="18532" PostTypeId="1" AcceptedAnswerId="18533" CreationDate="2011-01-22T17:41:49.350" Score="4" ViewCount="161" Body="&lt;p&gt;I'm trying to find $$f(x)=\sum_{n=2}^{\infty} \frac{x^n}{n(n-1)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found the radius of convergence of the above series which is $R=1$. Checking $x=\pm 1$ also yields a convergent series. Therefore the above series converges for all $x\in [-1, 1]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using differentiation of the series term by term we get: $$f&amp;#39;(x)=\sum_{n=2}^{\infty} \frac{x^{n-1}}{n-1}=\sum_{n=1}^{\infty} \frac{x^{n}}{n}=-\log(1-x)$$ which also has $R=1$, and then, by integrating term by term we get: $$f(x)=\int_{0}^{x} f&amp;#39;(t)dt=-\int_{0}^{x} \log(1-t)dt=x-(x-1)\ln(1-x)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;if I understand the theorems in my textbook correctly, the above formulas are true only for $x \in (-1, 1)$. It seems the above is correct since this is also what WolframAlpha says.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm abit confused though. At first, it seemed the above series converges for all $x\in [-1, 1]$ but in the end I only got $f(x)$ for all $|x|\lt 1$, something seems to be missing. What can I say about $f(-1)$ and $f(1)$?&lt;/p&gt;&#xA;" OwnerUserId="6128" LastActivityDate="2011-01-22T18:12:24.557" Title="Finding the sum $f(x)=\sum_{n=2}^{\infty} \frac{x^n}{n(n-1)}$" Tags="&lt;calculus&gt;&lt;sequences-and-series&gt;" AnswerCount="2" CommentCount="0" />
  <row Id="18533" PostTypeId="2" ParentId="18532" CreationDate="2011-01-22T17:47:48.663" Score="3" ViewCount="" Body="&lt;p&gt;Try using &lt;a href=&quot;http://en.wikipedia.org/wiki/Abel%27s_theorem&quot; rel=&quot;nofollow&quot;&gt;Abel's theorem&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2011-01-22T18:00:26.003" LastActivityDate="2011-01-22T18:00:26.003" CommentCount="8" />
  <row Id="18534" PostTypeId="2" ParentId="18527" CreationDate="2011-01-22T17:48:00.130" Score="6" ViewCount="" Body="&lt;p&gt;A basic practical reason to care about logarithms is that there are many numbers in real life which vary greatly in size, and it is both a pain and misleading to compare their sizes directly; one should instead compare the sizes of their &lt;em&gt;logarithms&lt;/em&gt;, for various reasons.  This is why the &lt;a href=&quot;http://en.wikipedia.org/wiki/Richter_magnitude_scale&quot; rel=&quot;nofollow&quot;&gt;Richter scale&lt;/a&gt; is logarithmic; see &lt;a href=&quot;http://en.wikipedia.org/wiki/Orders_of_magnitude&quot; rel=&quot;nofollow&quot;&gt;these&lt;/a&gt; &lt;a href=&quot;http://en.wikipedia.org/wiki/Orders_of_magnitude_(length)&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt; &lt;a href=&quot;http://en.wikipedia.org/wiki/Orders_of_magnitude_(time)&quot; rel=&quot;nofollow&quot;&gt;articles&lt;/a&gt; for some examples.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Logarithms also appear in the basic mathematical description of &lt;a href=&quot;http://en.wikipedia.org/wiki/Information_theory&quot; rel=&quot;nofollow&quot;&gt;information&lt;/a&gt;.  Suppose I send you a message consisting of zeroes and ones.  If the message has length $n$, we say that it contains $n$ &lt;em&gt;bits&lt;/em&gt; of information.  There are $2^n$ possible such messages, which leads to a general principle: whenever you are in a situation where there are $k$ possibilities and you know that one of them happens, you have gained $\log_2 k$ bits of information.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Information is a fundamental concept.  Consider the following &lt;a href=&quot;http://math.stackexchange.com/questions/639/logic-problem-identifying-poisoned-wines-out-of-a-sample-minimizing-test-subjec&quot;&gt;puzzle&lt;/a&gt;: you have $1000$ bottles of wine, and you know that one of them is poisoned.  You have an indeterminate number of rats to which you can feed various wines; if they are poisoned, they will die in $1$ hour.  How many rats do you need to figure out which bottle is poisoned in $1$ hour?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer is $10$.  This is because you want to figure out which of $1000$ possibilities happens, so you want to gain $\log_2 1000 \approx 10$ bits of information.  If you feed $n$ rats some amount of wine, the amount of information you have after $1$ hour is precisely a list of which rats died and which rats didn't - zeroes and ones - so you have gained at most $n$ bits of information.  (You might not reach this upper bound if some of the information you gain is redundant.)  This requires that $n$ is at least $10$, and in fact this bound can be achieved by the following algorithm:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Label the wines $0, 1, 2, ... 999$ and convert the numbers to binary.  Each of these numbers has at most $10$ binary digits.  Assign each of the rats wines as follows: rat $i$ will drink all the wines with the property that the $i^{th}$ binary digit is $1$.  After $1$ hour, the pattern of which rats die spells out the binary expansion of the poisoned wine. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I really like this problem because the problem statement does not mention logarithms at all, but it is an inevitable consequence of the particular optimization you are trying to accomplish that logarithms appear in the solution.  &lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2011-01-22T17:48:00.130" />
  <row Id="18535" PostTypeId="2" ParentId="18518" CreationDate="2011-01-22T17:51:53.247" Score="4" ViewCount="" Body="&lt;p&gt;An (elementary) algebraic approach that leads to this conclusion might be as follows. Since $\phi \neq id$ there is some $x$ for which $\phi(x) - x \neq 0$. Applying $\phi$ to this vector gives&#xA;$$\phi(\phi(x) - x) = \phi \circ \phi(x) - \phi(x) = x - \phi(x)$$&#xA;So $x - \phi(x)$ is a nonzero vector $\phi$ takes to negative itself. Similarly, since $\phi$ is not $-id$, there is some $y$ for which $\phi(y) + y$ is nonzero, and applying $\phi$ analogously to this vector one gets back itself. These two vectors are not multiples of each other since $\phi$ has opposite behaviors on the two vectors. Hence they are a basis.&lt;/p&gt;&#xA;" OwnerUserId="3035" LastActivityDate="2011-01-22T17:51:53.247" CommentCount="1" />
  <row Id="18536" PostTypeId="2" ParentId="18527" CreationDate="2011-01-22T17:54:29.220" Score="0" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://math.stackexchange.com/questions/16342/point-of-logarithms&quot;&gt;This&lt;/a&gt; question might be of relevance to you. It asks about the purpose of logarithms. &lt;a href=&quot;http://www.math.umt.edu/tmme/vol5no2and3/TMME_vol5nos2and3_a14_pp.337_344.pdf&quot; rel=&quot;nofollow&quot;&gt;This&lt;/a&gt; site gives a brief history of logarithms. As Qiaochu explained, logarithms are very useful in information theory. The &lt;a href=&quot;http://math.stackexchange.com/questions/3419/is-it-possible-to-represent-every-huge-number-in-abreviated-form/3421#3421&quot;&gt;answer&lt;/a&gt; to this question elegantly uses logarithms. &lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2011-01-22T17:54:29.220" />
  <row Id="18537" PostTypeId="1" CreationDate="2011-01-22T18:01:08.850" Score="1" ViewCount="91" Body="&lt;p&gt;Hello.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[ In first, I say &quot;I'm sorry!&quot;, because I am not a Englishman and I don't know your language terms very well. ]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;OK, I have some polynomials (like $a^2 +2ab +b^2$ ). And I can't decompress these (for example $a^2 +2ab +b^2 = (a+b)^2$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you help me? (if you can, please write the name or formula of combination (like $(a+b)^2 = a^2 +2ab +b^2$) of each polynomial.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;$(a^2-b^2)x^2+2(ad-bc)x+d^2-c^2$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$2x^2+y^+2x-2xy-2y+1$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$2x^2-5xy+2y^2-x-y-1$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$x^6-14x^4+49x^2-36$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$(a+b)^4+(a-b)^4+(a^2-b^2)^2$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Thank you! very much ....&lt;/p&gt;&#xA;" OwnerUserId="2861" LastEditorUserId="4728" LastEditorDisplayName="" LastEditDate="2011-01-22T19:02:15.560" LastActivityDate="2011-01-26T17:06:35.347" Title="decompose some polynomials" Tags="&lt;homework&gt;&lt;polynomials&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="18538" PostTypeId="2" ParentId="18508" CreationDate="2011-01-22T18:06:15.123" Score="1" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://faculty.olin.edu/bstorey/Notes/DiffEq.pdf&quot; rel=&quot;nofollow&quot;&gt;This&lt;/a&gt; (site about numerical methods for differential equations) seems to be a good resource.&lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2011-01-22T18:06:15.123" />
  <row Id="18539" PostTypeId="2" ParentId="18532" CreationDate="2011-01-22T18:12:24.557" Score="3" ViewCount="" Body="&lt;p&gt;If you rewrite $\frac{1}{n(n-1)}$ in the form $\frac{1}{n-1}-\frac{1}{n}$, then you can rewrite the series in both cases $x = \pm 1$ and compute their values directly.  You can then confirm that in both cases the value you compute coincides with the value $f(\pm 1)$.  (In other words, rather than appealing to Abel's theorem, as Moron suggests, in this particular case you can verify it.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[Caveat: In the case $x = -1$, you will need to use the familiar series&#xA;for $\log 2$, and maybe the easiest way to prove this is by appealing to Abel's theorem (applied to the series for $\log (1 + x)$).  So my approach probably doesn't really avoid Abel's theorem, at least for $x = -1$.]&lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2011-01-22T18:12:24.557" />
  <row Id="18540" PostTypeId="1" AcceptedAnswerId="18543" CreationDate="2011-01-22T18:15:58.520" Score="1" ViewCount="68" Body="&lt;p&gt;I am trying to describe the edges of an undirected graph that contains loops. &lt;a href=&quot;http://en.wikipedia.org/wiki/Graph_%28mathematics%29#Types_of_graphs&quot; rel=&quot;nofollow&quot;&gt;On Wikipedia&lt;/a&gt; they are characterized as &lt;em&gt;2-multisets&lt;/em&gt;, meaning it has two elements which can be identical, and the order is not important. I never heard of that term, and find no reference to it anywhere else, so I am not sure whether it is OK to use it in my thesis. Can anyone clear that up for me?&lt;/p&gt;&#xA;" OwnerUserId="3470" LastEditorUserId="462" LastEditorDisplayName="" LastEditDate="2011-01-22T21:11:36.237" LastActivityDate="2011-01-22T21:11:36.237" Title="Is 2-multiset a valid term?" Tags="&lt;elementary-set-theory&gt;&lt;notation&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="18541" PostTypeId="2" ParentId="18433" CreationDate="2011-01-22T18:16:52.180" Score="0" ViewCount="" Body="&lt;p&gt;If you have an integrating factor $\mu(x,y)$ such that $\mu(x,y)M(x,y)dx + \mu(x,y)N(x,y)dy  = 0$ is exact, then you can follow the normal procedure for exact equations to solve it. There's a completely different notion of integrating factor for solving 1st order linear equations ${dy \over dx} + P(x)y = Q(x)$; it just happens to be called the same thing since in both cases you multiply through by a factor and eventually integrate something when you're solving the differential equation. &lt;/p&gt;&#xA;" OwnerUserId="3035" LastActivityDate="2011-01-22T18:16:52.180" />
  <row Id="18542" PostTypeId="2" ParentId="18537" CreationDate="2011-01-22T18:18:22.947" Score="2" ViewCount="" Body="&lt;p&gt;For&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) $(a^2-b^2)x^2+2(ad-bc)x+d^2-c^2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;think about rearranging&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(a^2-b^2)x^2+2(ad-bc)x+d^2-c^2=a^2x^2+2adx+d^2-(b^2x^2+2bcx+c^2)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same idea can be applied to all your questions. &lt;/p&gt;&#xA;" OwnerUserId="4728" LastActivityDate="2011-01-22T18:18:22.947" CommentCount="1" />
  <row Id="18543" PostTypeId="2" ParentId="18540" CreationDate="2011-01-22T18:29:48.757" Score="1" ViewCount="" Body="&lt;p&gt;Yes. See the &lt;a href=&quot;http://en.wikipedia.org/wiki/Multiset&quot; rel=&quot;nofollow&quot;&gt;Wikipedia article on multisets&lt;/a&gt; and the many references therein for where it has occurred previously in literature. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Note, in the context, a 2-multiset merely means a multiset containing 2 elements.)&lt;/p&gt;&#xA;" OwnerUserId="1543" LastActivityDate="2011-01-22T18:29:48.757" />
  <row Id="18544" PostTypeId="1" AcceptedAnswerId="18547" CreationDate="2011-01-22T18:35:47.267" Score="4" ViewCount="128" Body="&lt;p&gt;If I understand correctly, Gauss proved that given any oriented Riemannian surface, one can find a complex structure on the surface so that the metric on the charts is just $f|dz|$, where $f&amp;gt;0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've heard these coordinates referred as &quot;conformal coordinates,&quot; which makes sense, but I've also heard of them referred to as &quot;isothermal coordinates.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why are these coordinates called &quot;isothermal&quot;? Who named it &quot;isothermal&quot;? Does it have anything to do with physics, or is it kind of like &quot;&lt;a href=&quot;http://math.stackexchange.com/questions/6906/why-is-it-called-sylvesters-law-of-inertia&quot;&gt;Inertia&lt;/a&gt;&quot;?&lt;/p&gt;&#xA;" OwnerUserId="2499" LastEditorUserId="1242" LastEditorDisplayName="" LastEditDate="2011-01-22T18:52:49.640" LastActivityDate="2011-01-23T00:07:12.410" Title="Why are they called &quot;Isothermal&quot; Coordinates?" Tags="&lt;soft-question&gt;&lt;math-history&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="1" />
  <row Id="18545" PostTypeId="2" ParentId="18505" CreationDate="2011-01-22T18:36:33.677" Score="2" ViewCount="" Body="&lt;p&gt;Perhaps &lt;a href=&quot;http://www.sciencedirect.com/science?_ob=ArticleURL&amp;amp;_udi=B6V7D-489RPB9-5P&amp;amp;_user=10&amp;amp;_coverDate=12/31/1981&amp;amp;_rdoc=1&amp;amp;_fmt=high&amp;amp;_orig=search&amp;amp;_origin=search&amp;amp;_sort=d&amp;amp;_docanchor=&amp;amp;view=c&amp;amp;_searchStrId=1616208380&amp;amp;_rerunOrigin=google&amp;amp;_acct=C000050221&amp;amp;_version=1&amp;amp;_urlVersion=0&amp;amp;_userid=10&amp;amp;md5=d12101ad7141b1f0835bb6ffbd0a1af7&amp;amp;searchtype=a&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt;(article about characterizing convex polygons) will help.&lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2011-01-22T18:36:33.677" CommentCount="1" />
  <row Id="18546" PostTypeId="2" ParentId="18544" CreationDate="2011-01-22T18:38:24.457" Score="0" ViewCount="" Body="&lt;p&gt;It is similar to &lt;a href=&quot;http://en.wikipedia.org/wiki/Isotherm&quot; rel=&quot;nofollow&quot;&gt;isotherms&lt;/a&gt;(countours that connect equal points of temperature). See the wikipedia article on &lt;a href=&quot;http://en.wikipedia.org/wiki/Isothermal_coordinates&quot; rel=&quot;nofollow&quot;&gt;isothermal coordinates&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2548" LastEditorUserId="2548" LastEditorDisplayName="" LastEditDate="2011-01-22T18:47:06.477" LastActivityDate="2011-01-22T18:47:06.477" CommentCount="1" />
  <row Id="18547" PostTypeId="2" ParentId="18544" CreationDate="2011-01-22T18:57:40.167" Score="10" ViewCount="" Body="&lt;p&gt;If $(u,v)$ are isothermal coordinates, then $u$ and $v$ are harmonic functions with respect to the Laplace-Beltrami operator on your Riemannian manifold, that is $\Delta u=\Delta v=0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the equation $\Delta f=0$ characterizes the stationary states for the heat equation. The level curves for a harmonic function are therefore the isothermal curves for some heat distribution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It follows that if $(u,v)$ is an isothermal system of coordinates, then the level curves, &lt;em&gt;i.e.&lt;/em&gt;, the coordinate lines, are isothermal curves.&lt;/p&gt;&#xA;" OwnerUserId="274" LastEditorUserId="274" LastEditorDisplayName="" LastEditDate="2011-01-23T00:07:12.410" LastActivityDate="2011-01-23T00:07:12.410" />
  <row Id="18548" PostTypeId="1" AcceptedAnswerId="18552" CreationDate="2011-01-22T19:01:43.820" Score="6" ViewCount="251" Body="&lt;p&gt;There is the interesting identity:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left ( \sum_{i=1}^n i \right )^2 = \left ( \sum_{i=1}^n i^3 \right ) $$ &#xA;which holds for any positive integer $n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know several was of proving this (finite differences, induction, algebraic tricks etc..), but even so I still find it &quot;weird&quot; that it is even true.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a very nice intuitive way to prove this using some kind of combinatorial argument?  (Like the why the sum of the volume of the first $n$ cubes should be the area of ... not sure here?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have any pretty different proof that could be enlightening I would love to see it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks a lot!&lt;/p&gt;&#xA;" OwnerUserId="6075" LastActivityDate="2011-01-22T19:19:52.357" Title="Combinatorial reasoning for a famous identity" Tags="&lt;combinatorics&gt;&lt;intuition&gt;&lt;counting&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" />
  <row Id="18549" PostTypeId="1" AcceptedAnswerId="18558" CreationDate="2011-01-22T19:01:55.250" Score="4" ViewCount="143" Body="&lt;p&gt;One of my pals asked me to look at this question &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $f: [0,1] \to \mathbb{R}$ be differentiable. Suppose that $f(0) = 0$&#xA;and $0 &amp;lt; f&amp;#39;(x) &amp;lt; 1$ for all $x \in (0, 1)$, where $f&amp;#39;(x)$ is the derivative of $f$. Prove that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \left(\int_{0}^{1}f(x) \ dx \right)^{2} \geq \int_{0}^{1}(f(x))^{3} \ dx$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't really have a clue as to where to start. Any ideas will be appreciated. &lt;/p&gt;&#xA;" OwnerUserId="4728" LastEditorUserId="2548" LastEditorDisplayName="" LastEditDate="2011-01-22T19:10:49.733" LastActivityDate="2011-01-22T20:15:35.510" Title="Trying a Calculus question" Tags="&lt;real-analysis&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="18551" PostTypeId="2" ParentId="18548" CreationDate="2011-01-22T19:12:44.660" Score="7" ViewCount="" Body="&lt;p&gt;The formula is due to Nicomachus of Gerasa.  There is a nice discussion of ways to prove it at &lt;a href=&quot;http://golem.ph.utexas.edu/category/2009/07/nichomachus_of_gerasa.html&quot; rel=&quot;nofollow&quot;&gt;this n-category cafe post&lt;/a&gt;, including a bijective proof and some visual / &quot;geometric&quot; proofs.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2011-01-22T19:12:44.660" />
  <row Id="18552" PostTypeId="2" ParentId="18548" CreationDate="2011-01-22T19:14:12.660" Score="6" ViewCount="" Body="&lt;p&gt;I found &lt;a href=&quot;http://merganser.math.gvsu.edu/calculus/summation/cubes.html&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; but sadly do not understand it, can someone explain the method?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://users.tru.eastlink.ca/~brsears/math/oldprob.htm#s32&quot; rel=&quot;nofollow&quot;&gt;This&lt;/a&gt; has the same proof but much easier to understand. It is also found in the link by QY&lt;/p&gt;&#xA;" OwnerUserId="4728" LastEditorUserId="4728" LastEditorDisplayName="" LastEditDate="2011-01-22T19:19:52.357" LastActivityDate="2011-01-22T19:19:52.357" CommentCount="1" />
  <row Id="18553" PostTypeId="2" ParentId="18505" CreationDate="2011-01-22T19:14:18.900" Score="0" ViewCount="" Body="&lt;p&gt;You might explore the &lt;a href=&quot;http://en.wikipedia.org/wiki/Frechet_distance&quot; rel=&quot;nofollow&quot;&gt;Fréchet Distance&lt;/a&gt;.&#xA;In any case, here is a useful survey paper:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;Shape matching: similarity measures and algorithms,&quot;&#xA;  Remco C. Veltkamp&lt;br&gt;&#xA;  &lt;a href=&quot;http://portal.acm.org/citation.cfm?id=884078&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;SMI '01 Proceedings of the International Conference on Shape Modeling &amp;amp; Applications&lt;/em&gt;, 2001&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And here are two references specifically on convex polygons:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;Optimal matching of convex polygons,&quot;&#xA;  Pedro Coxa, Henri Maitrea, Michel Minouxb and Celso Ribeiro.&#xA;  &lt;em&gt;Pattern Recognition Letters&lt;/em&gt;&#xA;  Volume 9, Issue 5, June 1989, Pages 327-334. &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&quot;A simple algorithm for the unique characterization of convex polygons,&quot;&#xA;  P.K Harveya.&#xA;  &lt;em&gt;Computers &amp;amp; Geosciences&lt;/em&gt;&#xA;  Volume 7, Issue 4, 1981, Pages 387-392.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(I see that the latter paper was already linked by PEV.)&lt;/p&gt;&#xA;" OwnerUserId="237" LastEditorUserId="237" LastEditorDisplayName="" LastEditDate="2011-01-22T19:19:23.980" LastActivityDate="2011-01-22T19:19:23.980" />
  <row Id="18554" PostTypeId="1" CreationDate="2011-01-22T19:43:22.420" Score="3" ViewCount="790" Body="&lt;p&gt;I want to know the existence of a class of functions with the following properties. If they exist, how to construct them ? In case they do not exist, please suggest a proof.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $ f : \mathbb{R} \to \mathbb{R} $  be a differentiable function with compact support in [0,1]. Let the notation $ Cf(x) = k$ represent that $ f $ is exactly $ k $ times differentiable at the point $ x \in \mathbb{R} $ where $ k \in \mathbb{N} $.&#xA;$ f $ should satisfy the following properties.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ Cf(x) &amp;gt;= q $ $\forall x \in \mathbb{Q}\cap (0,1] $ where $ q \in \mathbb{N} $ is denominator of $ x $ when written in simplest form.&#xA;$ Cf(0) &amp;gt;= 1 $.&#xA;$ Cf(x) = \infty \forall x \in (\mathbb{R}\setminus\mathbb{Q})\cap[0,1] $. The last statement means that $ f(x) $ is smooth at every $ x\in (\mathbb{R}\setminus\mathbb{Q})\cap[0,1] $.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to study such functions (in case they exist) as they seem to have some very nice properties.Also they might be the most well behaved after smooth functions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clarification : $f(x)$ is exactly $k$ times differentiable at $x$ means it is not differentiable more than $k$ times at $x$ .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT 2 :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clarification : $Cf(x)$ is always finite for $\forall x \in Q\cap (0,1] $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT 3 :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One approach i tried is as given in answer/coments to this question &lt;a href=&quot;http://math.stackexchange.com/questions/17962/is-there-a-compactly-supported-smooth-function-which-is-exactly-k-times-different&quot;&gt;question&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I added such functions starting at every rational point to make up the function and varying the scale of the bump functions $\epsilon$. Making $\epsilon \to 0$ would give a limit function (by making some restrictions) which does not have the desired properties.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT 4 : I would like to know whether you find it interesting due to any reasons or not much due to some obvious reasons. Please comment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To me it appears very beautiful and interesting concept..if i am allowed to make such statements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT 5&lt;/strong&gt;   edited on Feb 12 2011.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;I have constructed such a class of functions which happen to be nonempty.More over it has many interesting and important properties.Please contact me if you are interested in knowing more details and the proof.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT : 6&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;@chandok : thank you for the clarification. I think i should mention something after having a re look at your answer. When i was asking the question, I wanted to rule out a particular condition which seemed not of interest, but later due to my misinterpretation/mistaken view of Darboux Theorem, I thought such a condition would not arise, but strikingly your answer is exactly utilizing such a condition and only after looking at your answer i came to know that such a condition is indeed possible and hence I realize that i made a mistake by not ruling out such a condition in the question itself. Now i would like to mention that condition in the question, but i want you to note that i give full credit to your answer as it brought to notice that such a condition is indeed possible, even though it is not very interesting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;New Condition&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;when ever $f(x)$ is not differentiable more than $n$ times at a point  is is because for the $(n+1)^{th}$ derivative, the left and right limits exist and are not equal. The left and right limits do not diverge.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Keeping the above condition in mind i have constructed such a class of functions and it indeed is not empty and i think it has very important and interesting properties.If interested please contact me for details and the proof. You are also welcome to make any comments.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I once again thank Chandok for taking patience in providing an answer.&lt;/p&gt;&#xA;" OwnerUserId="2987" LastEditorUserId="2987" LastEditorDisplayName="" LastEditDate="2011-02-12T16:29:23.210" LastActivityDate="2011-04-13T16:40:17.290" Title="Is this class of functions nonempty/nontrivial?" Tags="&lt;real-analysis&gt;" AnswerCount="1" CommentCount="26" FavoriteCount="1" />
  <row Id="18555" PostTypeId="1" AcceptedAnswerId="18557" CreationDate="2011-01-22T20:04:16.440" Score="5" ViewCount="155" Body="&lt;p&gt;Problem: count the number of distinct ways to write number X as the sum of numbers {a, b, c...} with replacement.&#xA;For instance, there are 3 ways to write 11:&lt;br/&gt;&#xA;2+2+7 &lt;br/&gt;&#xA;2+2+2+2+3&lt;br/&gt;&#xA;2+3+3+3&lt;br/&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And if I remember correctly, isn't there a way to solve this using generating functions or polynomials of some sort?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(this question inspired by: &lt;a href=&quot;http://www.topcoder.com/stat?c=problem_statement&amp;amp;pm=42&amp;amp;rd=2001&quot; rel=&quot;nofollow&quot;&gt;http://www.topcoder.com/stat?c=problem_statement&amp;amp;pm=42&amp;amp;rd=2001&lt;/a&gt;)&lt;/p&gt;&#xA;" OwnerUserId="6131" LastEditorUserId="6131" LastEditorDisplayName="" LastEditDate="2011-01-22T20:21:02.667" LastActivityDate="2011-03-05T16:50:20.853" Title="What's the name of this problem?" Tags="&lt;number-theory&gt;&lt;combinatorics&gt;&lt;generating-functions&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="18557" PostTypeId="2" ParentId="18555" CreationDate="2011-01-22T20:11:09.837" Score="7" ViewCount="" Body="&lt;p&gt;I believe the problem is called counting &lt;a href=&quot;http://en.wikipedia.org/wiki/Partition_(number_theory)#Restricted_partitions&quot; rel=&quot;nofollow&quot;&gt;restricted partitions&lt;/a&gt;.  (Warning: combinatorialists use the word &quot;partition&quot; in two different ways, and sometimes it's not completely clear from context which one is meant.  One must distinguish between partitions of a &lt;em&gt;number&lt;/em&gt;, which is the sense meant here, and partitions of a &lt;em&gt;set.&lt;/em&gt;)  With replacement, the answer is that the number $s_n$ of ways has generating function&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_n s_n x^n = (1 + x^a + x^{2a} + ...)(1 + x^b + x^{2b} + ...)(1 + x^c + x^{2c} + ...)...$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or equivalently&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{1}{(1 - x^a)(1 - x^b)(1 - x^c)...}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is because choosing a term from each factor corresponds to choosing how many times you use each number you can use.  For example, with the choices $\{ 1, 2, 5, 10, 25 \}$, this is the problem of counting the number of possible ways to make change for a certain amount of money.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The generating function can be converted into an explicit formula when the set of possibilities is finite, but it is messy in the general case and not really worth caring about.  It leads to a straightforward asymptotic which I could explain if you are interested.&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2011-01-22T20:16:39.893" LastActivityDate="2011-01-22T20:16:39.893" CommentCount="3" />
  <row Id="18558" PostTypeId="2" ParentId="18549" CreationDate="2011-01-22T20:15:35.510" Score="6" ViewCount="" Body="&lt;p&gt;This Problem is given in the book&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;em&gt;Problems in Mathematical Analysis - III&lt;/em&gt; by Kaczor and Nowak &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The solution is as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Set $$F(t)= \biggl(\int\limits_{0}^{t} f(x) \ \text{dx}\biggr)^{2} - \int\limits_{0}^{t} (f(x))^{3} \ \text{dx}, \quad t \in [0,1]$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then $$F&amp;#39;(t)=f(t) \cdot \biggl(2 \int\limits_{0}^{t} f(x) \ \text{dx} - (f(t))^{2}\biggr)$$ and if $$G(t)= 2 \int\limits_{0}^{t} f(x) \ \text{dx} - (f(t))^{2}$$, then $G&amp;#39;(t)=2f(t) \cdot (1-f&amp;#39;(t)) \geq 0$. Consequently, $G(t) \geq G(0)=0$, which gives $F&amp;#39;(t) \geq 0$. So, $F(t) \geq 0$,, and in particular $F(1) \geq 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moreover if$F(1)=0$, then $F(t)=0$ for $t \in [0,1]$ and therfore $F&amp;#39;(t)=f(t)G(t)=0$. This, in turn, implies $G&amp;#39;(t)=2f(t) \cdot (1-f&amp;#39;(t))=0$ and $1-f&amp;#39;(t)=0$ for $t \in (0,1)$.&lt;/p&gt;&#xA;" LastActivityDate="2011-01-22T20:15:35.510" />
  <row Id="18559" PostTypeId="1" CreationDate="2011-01-22T20:16:56.117" Score="0" ViewCount="130" Body="&lt;p&gt;I want to get acquainted with approaches made to solve P vs NP problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) What is already achieved in solving the P vs NP problem?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) What articles are most cited/famous in that field? Please, provide links for them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) How one can divide existing approaches to different classes?&lt;/p&gt;&#xA;" OwnerUserId="6132" LastActivityDate="2011-01-22T20:45:16.850" Title="Approaches for solving P vs NP problem" Tags="&lt;np-complete&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="18560" PostTypeId="1" AcceptedAnswerId="18565" CreationDate="2011-01-22T20:37:02.743" Score="2" ViewCount="145" Body="&lt;p&gt;Given $A = B \times C$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using any $9$ digits form two numbers $B$ and $C$.All digits must be used exactly once. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the maximum possible value of $A$?&lt;/p&gt;&#xA;" OwnerUserId="3744" LastEditorUserId="3744" LastEditorDisplayName="" LastEditDate="2011-01-22T20:58:08.100" LastActivityDate="2011-06-05T14:19:14.357" Title="Maximize :: $A = B \times C$" Tags="&lt;number-theory&gt;&lt;puzzle&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="18561" PostTypeId="1" AcceptedAnswerId="18562" CreationDate="2011-01-22T20:39:25.237" Score="3" ViewCount="133" Body="&lt;p&gt;Let&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;$GL_{n}(\mathbb{R}) = \{ A \in M_{n}(\mathbb{R}) \ | \ \text{det}(A) \neq 0 \}$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$SL_{n}(\mathbb{R}) = \{ A \in M_{n}(\mathbb{R}) \ | \ \text{det}(A) = 1 \}$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;We know that $$GL_{n}(\mathbb{R})/SL_{n}(\mathbb{R}) \cong \mathbb{R}^{\ast}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am interested to know the answer for this question.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Does there exist a subgroup $H$ such that $GL_{n}(\mathbb{R})/H \cong \mathbb{R}$?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" LastActivityDate="2011-01-23T00:45:40.467" Title="Subgroup $H$ such that $GL_{n}(R)/H \cong \mathbb{R}$" Tags="&lt;group-theory&gt;&lt;abstract-algebra&gt;&lt;algebra&gt;" AnswerCount="1" FavoriteCount="0" />
  <row Id="18562" PostTypeId="2" ParentId="18561" CreationDate="2011-01-22T20:41:37.120" Score="8" ViewCount="" Body="&lt;p&gt;$\mathbb{R}^{\ast} \simeq \{ \pm 1 \} \times \mathbb{R}$: the isomorphism is gotten by looking at sign and taking the logarithm of the absolute value.  So $\mathbb{R}^{\ast}$ has a quotient isomorphic to $\mathbb{R}$, and the kernel of the corresponding composition (the group of matrices with determinant $\pm 1$) provides the desired $H$. &lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2011-01-23T00:45:40.467" LastActivityDate="2011-01-23T00:45:40.467" CommentCount="3" />
  <row Id="18563" PostTypeId="2" ParentId="16685" CreationDate="2011-01-22T20:42:40.973" Score="2" ViewCount="" Body="&lt;p&gt;Steven Krantz has studied the automorphism group of domains, and has authored or coauthored some papers and surveys on the subject. In particular, Chapter 12 of his recent book &quot;Geometric Function Theory&quot; deals precisely with this topic and provides additional references. (It is a very nice book, by the way!) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are some general comments, they all come from this Chapter: &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The automorphism group of a domain is a Lie Group (this is proved in Kobayashi's &quot;Hyperbolic manifolds and holomorphic mappings&quot;), so it makes sense to talk about its &lt;em&gt;dimension&lt;/em&gt;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The automorphism group of the disk is 3-dimensional, and 3 is the largest dimension possible.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The only bounded domain whose automorphism group is 1-dimensional is the annulus.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;A domain with &quot;very many holes&quot; is 0-dimensional. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The automorphism group of a domain with at least 2 but only finitely many holes is finite.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;This last case, which includes your question, was studied by M. Heins in two papers,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;A note on a theorem of Radó concerning the $(1,m)$ conformal maps of a multiply connected region into itself&quot;, Bulletin of the American Mathematical Society 47 (1941), 128-130.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&quot;On the number of 1-1 directly conformal maps which a multiply-connected plane region of finite connectivity $p$ ($&gt;2$) admits onto itself&quot;, Bulletin of the American Mathematical Society 52 (1946), 454-457.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Heins found sharp bounds $N(k)$ for the size of the automorphism group of a domain $\Omega_k$ with precisely $k\ge 2$ holes:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$N(k)=2k$ if $k\ne 4,6,8,12,20$; &lt;/li&gt;&#xA;&lt;li&gt;$N(4)=12$;&lt;/li&gt;&#xA;&lt;li&gt;$N(6)=N(8)=24$;&lt;/li&gt;&#xA;&lt;li&gt;$N(12)=N(20)=60$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In an exercise, Krantz describes a domain with trivial automorphism group: Start with the &quot;box&quot; $\{\zeta\in{\mathbb C}\mid|{\rm Re}\zeta|&amp;lt;2,|{\rm Im}\zeta|&amp;lt;2\}$, remove the four closed disks of radius $0.1$ and centered at $\pm1\pm i$, and &quot;perturb one of the holes&quot; by 0.1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, let me quote a paragraph at the end of the chapter:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A very interesting open problem is to determine which finite groups arise as the automorphism groups of planar domains (there are some results for finitely connected regions). It is known that if $G$ is a compact Lie group, then there is some smoothly bounded domain in some ${\mathbb C}^n$ with automorphism group equal to $G$. But it is difficult to say how large $n$ must be in terms of elementary properties of the group $G$. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Krantz closes by mentioning two references for the above:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Bedford-Dadok, &quot;Bounded domains with prescribed group of automorphisms&quot;, Comment. Math. Helv. 62 (1987), 561-572.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Saerens-Zame, &quot;The isometry groups of manifolds and the automorphism groups of domains&quot;, Trans. Amer. Math. Soc. 301 (1987), 413-429.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="462" LastActivityDate="2011-01-22T20:42:40.973" />
  <row Id="18564" PostTypeId="2" ParentId="18559" CreationDate="2011-01-22T20:45:16.850" Score="3" ViewCount="" Body="&lt;p&gt;Someone else posted this link in partial reply to another question recently. I think it answers 1 and 2, at least. &lt;a href=&quot;http://www.win.tue.nl/~gwoegi/P-versus-NP.htm&quot; rel=&quot;nofollow&quot;&gt;http://www.win.tue.nl/~gwoegi/P-versus-NP.htm&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5676" LastActivityDate="2011-01-22T20:45:16.850" CommentCount="3" />
  <row Id="18565" PostTypeId="2" ParentId="18560" CreationDate="2011-01-22T20:52:46.487" Score="2" ViewCount="" Body="&lt;p&gt;By brute force (Mathematica): 843,973,902 = 87,531 • 9,642.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Probably not the most elegant way to do it, but here's my code:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;In[21]:= Sort[&#xA; Table[Sort[&#xA;    Module[{a = FromDigits[#[[1 ;; 9 - d]]], &#xA;        b = FromDigits[#[[10 - d ;; 9]]]}, {a b, {a, b}}] &amp;amp; /@ &#xA;     Permutations[{1, 2, 3, 4, 5, 6, 7, 8, &#xA;       9}], #1[[1]] &amp;gt; #2[[1]] &amp;amp;][[1]], {d, 1, 4}], #1[[1]] &amp;gt; #2[[1]] &amp;amp;]&#xA;&#xA;Out[21]= {{843973902, {87531, 9642}}, {843809444, {875321, &#xA;   964}}, {840414816, {8754321, 96}}, {788888889, {87654321, 9}}}&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="72" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2011-01-22T21:00:16.387" LastActivityDate="2011-01-22T21:00:16.387" CommentCount="4" />
  <row Id="18566" PostTypeId="2" ParentId="18555" CreationDate="2011-01-22T21:04:24.463" Score="1" ViewCount="" Body="&lt;p&gt;If you are indeed looking at prime partitions, this is a &lt;em&gt;very&lt;/em&gt; complicated problem.  The usual counting generating series approach will unfortunately not yield too much without additional powerful analytic techniques. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is the abstract of R.C. Vaughn's paper &quot;On the number of partitions into primes.&quot;:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Let $p(n)$ denote the number of ways to write a positive integer $n$ as a sum of prime numbers. Because of the irregularity in the distribution of prime numbers, it has long been believed that it is not possible to obtain an asymptotic formula for $p(n)$. (Of course, this depends on what one means by an asymptotic formula.) In this paper, the author obtains a formula for $p(n)$ in the form &#xA;  $$ p(n)=M(n)(1+O(n^{-1/5})) \quad (n\to\infty), $$&#xA;  where $M(n)$ is a smooth function whose definition involves the generating function of $p(n)$. Moreover, using the same analytic method, the author obtains the following remarkable result: &#xA;  $$ p(n+1)-p(n)\sim\frac{\pi p(n)}{\sqrt{3n\log n}} $$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Reading this paper will definitely give some insights.  Also, to see the sequence $p(n)$ you can visit &lt;a href=&quot;http://oeis.org/A000607&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; OEIS page.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope that helps,&lt;/p&gt;&#xA;" OwnerUserId="6075" LastEditorUserId="6075" LastEditorDisplayName="" LastEditDate="2011-03-05T16:50:20.853" LastActivityDate="2011-03-05T16:50:20.853" CommentCount="1" />
  <row Id="18567" PostTypeId="1" CreationDate="2011-01-22T21:10:38.147" Score="2" ViewCount="177" Body="&lt;p&gt;Determine in How many ways $N!$ can be expressed as sum of consecutive numbers.For example for $3!$ can be as $1+2+3$ only $1$ ways.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please suggest an elegant approach.&lt;/p&gt;&#xA;" OwnerUserId="3744" LastActivityDate="2011-01-22T22:20:05.990" Title="Determine in How many ways $N!$ can be expressed as sum of consecutive numbers." Tags="&lt;number-theory&gt;&lt;algorithms&gt;&lt;puzzle&gt;" AnswerCount="4" CommentCount="4" />
  <row Id="18569" PostTypeId="1" AcceptedAnswerId="18577" CreationDate="2011-01-22T21:28:24.970" Score="3" ViewCount="79" Body="&lt;p&gt;I have the following group:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$A = \mathbb{Z}/p_1^{r_1} \times \mathbb{Z}/p_2^{r_2} \mathbb{Z} \times \cdots \times \mathbb{Z}/p_n^{r_n} \mathbb{Z}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;each $p_i$ being a prime, but they need not be distinct primes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to solve the following isomorphism of abelian groups,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{X}{\mathbb{Z}/p\mathbb{Z}} \cong A$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think there are only two types of possibility:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Direct product $X = \mathbb{Z}/p\mathbb{Z} \times A$&lt;/li&gt;&#xA;&lt;li&gt;If $p = p_i$ we could have $\mathbb{Z}/p_1^{r_1} \times \mathbb{Z}/p_2^{r_2} \mathbb{Z} \times \cdots \times \mathbb{Z}/p_i^{r_n+1} \mathbb{Z} \times \cdots \times \mathbb{Z}/p_n^{r_n} \mathbb{Z}$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;But how could I prove these are the only possibility? thanks a lot!&lt;/p&gt;&#xA;" OwnerUserId="4290" LastActivityDate="2011-01-22T22:50:53.587" Title="Abelian Group Extension" Tags="&lt;group-theory&gt;" AnswerCount="2" />
  <row Id="18570" PostTypeId="2" ParentId="18567" CreationDate="2011-01-22T21:36:17.553" Score="0" ViewCount="" Body="&lt;p&gt;Basically you have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a+(a+1)+(a+2)+\ldots+b=\sum _{k=a}^b k=-(1/2) (-1 + a - b) (a + b)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you have to find positive nontrivial solutions for  the following diophantine equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$-(1/2) (-1 + a - b) (a + b) = n!$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for $n=3$ the only solution is exactly the one you gave with $a=1$ and $b=3$&lt;/p&gt;&#xA;" OwnerUserId="3123" LastActivityDate="2011-01-22T21:36:17.553" CommentCount="2" />
  <row Id="18571" PostTypeId="2" ParentId="18569" CreationDate="2011-01-22T21:39:07.080" Score="3" ViewCount="" Body="&lt;p&gt;There are various ways to approach this (and more general problems of the same flavour), but maybe the easiest in this context is to use the classification of finite abelian groups:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The abelian group $X$ has order equal to $p \times$ the order of $A$,&#xA;and contains a subgroup of order $p$ such that the quotient is isomorphic to $A$.&#xA;If you think about the possible elementary divisors of $X$ which are compatible&#xA;with these two statements, you will get the result you want.  &lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2011-01-22T21:39:07.080" CommentCount="2" />
  <row Id="18572" PostTypeId="2" ParentId="18567" CreationDate="2011-01-22T21:41:10.433" Score="2" ViewCount="" Body="&lt;p&gt;Let's deal with odd length and even length sequences separately.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A number $N$ is the sum of an odd number of consecutive integers, centered at $m$, whenever $m|N$ and $N/m$ is odd.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A number $N$ is the sum of an even number of consecutive integers, centered at $m+1/2$, whenever $2m+1|N$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let $N = 2^k M$, where $M$ is odd. For each factor $m|M$ we get one odd-length sequence centered at $2^km$, and one even-length sequence centered at $m+1/2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example: $N = 6 = 2^1 3$, so we get two odd-length sequences (centered at $2$ and $6$) and two even-length sequences (centered at $.5$ and $1.5$): $$6 = 1 + 2 + 3 = 6 = -5 -4 -3 -2 -1 + 0 + 1 + 2 + 3 + 4 + 5 + 6 = 0 + 1 + 2 + 3.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to obtain a formula for the number of representations of $n!$, all we need to do is to figure out how many odd divisors $n!$ has. I leave that to the reader.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2011-01-22T21:41:10.433" />
  <row Id="18573" PostTypeId="2" ParentId="18560" CreationDate="2011-01-22T21:45:27.120" Score="0" ViewCount="" Body="&lt;p&gt;Instead of &quot;trying all possibilities&quot;, one can try the following approach (which I haven't). Consider some current assignment for $B,C$, and try to modify it somehow - change the location of a single digit. Find some condition ensuring that $(B,C)$ is a local maximum. Enumerate all local maxima, and take the maximal one (if we're lucky, there's only one local maximum, which must be global).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using this approach, we can verify for example that the digits in each number must come in decreasing order (from MSD to LSD).&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2011-01-22T21:45:27.120" />
  <row Id="18574" PostTypeId="2" ParentId="18567" CreationDate="2011-01-22T21:50:17.990" Score="1" ViewCount="" Body="&lt;p&gt;Since $n+(n+1)+\cdots+(n+d)=(d+1)n+\frac{d(d+1)}{2}=\frac{(2n+d)(d+1)}{2}$, you can write the number $M$ as sum of $d+1$ consecutive integers starting from $n$ if you can decompose&#xA;$$&#xD;&#xA;2M=(2n+d)(d+1).&#xD;&#xA;$$&#xA;Note that the two factors have opposite parity, so the question is: given $M$, how many ways we have to write $2M$ as a product $2M=ab$ in such a way that $a$ and $b$ have opposite parities and $a&amp;gt;b-1$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this way one gets some non-trivial examples: $4!=24=7+8+9$, $5!=120=22+23+24+25+26$, $6!=720=41+42+\cdots+55$, and so on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Some answers arrived while I was typing this)&lt;/p&gt;&#xA;" OwnerUserId="688" LastEditorUserId="688" LastEditorDisplayName="" LastEditDate="2011-01-22T22:06:14.597" LastActivityDate="2011-01-22T22:06:14.597" />
  <row Id="18575" PostTypeId="2" ParentId="18567" CreationDate="2011-01-22T22:14:47.430" Score="1" ViewCount="" Body="&lt;p&gt;I am assuming that you want it to be written as a sum of consecutive natural numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $p \leq N$ be a prime. The highest power of $p$ dividing $N$ is given by $\alpha_p = \displaystyle \sum_{k=1}^{\infty} \lfloor \frac{N}{p^k} \rfloor$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So $N! = 2^{\alpha_2} 3^{\alpha_3} 5^{\alpha_5} \ldots q^{\alpha_q}$, where $q$ is the last prime $\leq N$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We want $N! = a + (a+1) + (a+2) + (a+3) + \cdots b = \frac{(b-a+1)(a+b)}{2}$, where $a,b \in \mathbb{N}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we need to find $a$ and $b$ such that $(b-a+1)(a+b) = 2^{(\alpha_2 + 1)} 3^{\alpha_3} 5^{\alpha_5} \ldots q^{\alpha_q}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now note that $(a+b)$ and $(b-a+1)$ are of opposite parity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So one of them has to be odd. Further $(a+b)&amp;gt;(b-a+1)$ since $a \in \mathbb{N}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume that $(a+b)$ is even. So $(b-a+1)$ is odd.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence &#xA;$$(a+b) = 2^{(\alpha_2 + 1)} 3^{\beta_3} 5^{\beta_5} \ldots q^{\beta_q}$$ $$(b-a+1) = 3^{\alpha_3-\beta_3} 5^{\alpha_5-\beta_5} \ldots q^{\alpha_q-\beta_q}$$&#xA;where $0 \leq \beta_p \leq \alpha_p$ and $(a+b) &amp;gt; (b-a+1)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now assume that $(a+b)$ is odd. So $(b-a+1)$ is even.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence &#xA;$$(a+b) = 3^{\beta_3} 5^{\beta_5} \ldots q^{\beta_q}$$ $$(b-a+1) = 2^{(\alpha_2 + 1)} 3^{\alpha_3-\beta_3} 5^{\alpha_5-\beta_5} \ldots q^{\alpha_q-\beta_q}$$&#xA;where $0 \leq \beta_p \leq \alpha_p$ and $(a+b) &amp;gt; (b-a+1)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we relax the fact that it has to be written as a sum of consecutive natural numbers, and assume that the consecutive numbers belong to integers then we get&#xA;$$2 \times (1+\alpha_3) \times (1+\alpha_5) \times (1+\alpha_7) \cdots \times (1+\alpha_q)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the above also acts as a trivial upper bound if it has to be written as a sum of consecutive natural numbers. This upper bound is obtained by violating the constraint $(a+b) &amp;gt; (b-a+1)$&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2011-01-22T22:20:05.990" LastActivityDate="2011-01-22T22:20:05.990" />
  <row Id="18576" PostTypeId="1" AcceptedAnswerId="18580" CreationDate="2011-01-22T22:42:02.273" Score="3" ViewCount="240" Body="&lt;p&gt;$1 + \displaystyle\sum_{j=1}^{n} j!j$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to find a formula for the above and then prove it by induction. The answer according to Wolfram is $(n+1)!-1$, however I have no idea how to get there. Any hints or ideas on how I should tackle this one?&lt;/p&gt;&#xA;" OwnerUserId="2985" LastEditorUserId="2985" LastEditorDisplayName="" LastEditDate="2011-01-23T00:34:51.177" LastActivityDate="2011-01-23T02:09:15.010" Title="Summation of a factorial" Tags="&lt;factorial&gt;&lt;sequences-and-series&gt;" AnswerCount="3" />
  <row Id="18577" PostTypeId="2" ParentId="18569" CreationDate="2011-01-22T22:50:53.587" Score="3" ViewCount="" Body="&lt;p&gt;If you don't know the fundamental theorem of finitely generated abelian groups (which seems like it might be the case given that you don't know what an &quot;elementary divisor&quot; is, unless this is simply a problem of language which is certanly possible), here is another approach.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose $X$ is such a group. Consider the collection of all elements of order $p$ in $X$; it is not empty, since $p$ divides the order of $X$, so by Cauchy's Theorem there is such an element. If there exists $x\in X$ of order $p$ such that $\langle x\rangle \cap A = \{0\}$, then let $B=\langle x\rangle$. Then $A\cap B=\{0\}$, and $|A+B| = |X|$, hence $A+B=X$, and therefore $X=A\oplus B$, proving that $X\cong A \times C_p$, as desired (I use $C_n$ for the cyclic group of order $n$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only other possibility is that every element of order $p$ in $X$ is contained in $A$. Let $x\in X$ be an element of order a power of $p$ that maps to a generator of $X/A$ (it exists, because if you pick any preimage of one generator, and ther order is not a $p$-power, you can pick an adequate multiple to get another element that maps to perhaps a different generator and &lt;em&gt;is&lt;/em&gt; a prime power. Note also that $x\notin A$ and $px\in A$. So now we are working inside the $p$-parts of $X$ and $A$, and so we may assume that $X$ and $A$ are both abelian $p$-groups. So $A=C_{p^{a_1}}\oplus\cdots\oplus C_{p^{a_k}}$ with $a_1\leq\cdots\leq a_k$. Write $px$ as an element of $A$; by adding suitable elements of $A$ to $x$ you can ensure that any component of $px$ that is a multiple of $p$ is actually equal to $0$: for example, if the $i$th component is $p^{\ell}ky_i$, where $y_i$ is the generator, $\ell\gt 0$ and $\gcd(k,p)=1$, then adding $-p^{\ell-1}ky_i$ to $x$ still keeps it an element that maps to a generator of $X/A$, but now the $i$th component of $px$ is $0$. Then replacing any generators by suitable powers, you may assume that the &quot;vector&quot; that describes $px$ has all components equal to either $0$ or $1$. Doing an easy change of basis for $A$ gives you the description of $X$ that you want in this situation.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2011-01-22T22:50:53.587" />
  <row Id="18578" PostTypeId="2" ParentId="18576" CreationDate="2011-01-22T22:56:33.967" Score="0" ViewCount="" Body="&lt;p&gt;It looks like you have a closed form already. If you want to derive it rather than get it from an oracle, Gosper's algorithm will do the job.&lt;/p&gt;&#xA;" OwnerUserId="5676" LastActivityDate="2011-01-22T22:56:33.967" CommentCount="1" />
  <row Id="18579" PostTypeId="2" ParentId="18530" CreationDate="2011-01-22T22:57:27.430" Score="2" ViewCount="" Body="&lt;p&gt;I honestly don't understand what you mean by &quot;the positions of the elements of $S$ and of $T$&quot;. A one-to-one mapping doesn't really &quot;care&quot; &lt;em&gt;where&lt;/em&gt; you write the elements, it only cares what the elements &lt;em&gt;are&lt;/em&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instead, think about this: if all you care about is what is happening to the elements of $T$, what happens if you just &lt;em&gt;ignore&lt;/em&gt; all the elements of $S-T$? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To give you a different example so you can see what I'm talking about: suppose you are very interested in squaring positive real numbers, but you don't really care about numbers other than those numbers $x$ which satisfy $0\lt x\lt 1$. Although you have a perfectlty good function $f\colon\mathbb{R}\to[0,\infty)$ that maps $x$ to $x^2$, you don't really care what $f$ is doing on $[1,\infty)$, or on $(-\infty,0]$. You only care about what is happening on $(0,1)$. So, for instance, you might want to consider an inverse for the function; the function $f$ does &lt;em&gt;not&lt;/em&gt; have an inverse, so instead, since we are only interested in what is going on inside $(0,1)$, we consider the &lt;em&gt;restriction&lt;/em&gt; $g = f\bigm|_{(0,1)}$ of $f$ to $(0,1)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can think of the idea of &quot;restrict the function to $(0,1)$&quot; as a way of taking &lt;em&gt;any&lt;/em&gt; function defined on the whole real numbers, and getting a function that is defined on &lt;em&gt;only&lt;/em&gt; $(0,1)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And here,, you have maps defined on &lt;em&gt;all&lt;/em&gt; of $S$, but you are only interested in (i) those that map $T$ to itself; and (ii) only what they do to $T$. So...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And, once you have that, prove the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Theorem.&lt;/strong&gt; Let $X$ and $Y$ be sets, and let $g\colon X\to Y$ be a bijection between $X$ and $Y$. Then $A(X)$ is isomorphic to $A(Y)$, via the map that sends $\sigma\colon X\to X$ to $g\circ\sigma\circ g^{-1}\colon Y\to Y$. &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-23T02:35:14.047" LastActivityDate="2011-01-23T02:35:14.047" CommentCount="9" />
  <row Id="18580" PostTypeId="2" ParentId="18576" CreationDate="2011-01-22T22:57:31.767" Score="6" ViewCount="" Body="&lt;p&gt;This is a simple induction. Use $(n+1)! = (n+1) \cdot n! = n\cdot n! + n!$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added&lt;/strong&gt;: I don't see any better way than to play around with the formula. Rearrange this and you have&#xA;$j \cdot j! = (j+1)! - j!$. But then you have&#xA;\begin{align*}&#xA;\sum_{j=1}^{n} j \cdot j! &amp;amp; = \sum_{j=1}^{n} [ (j+1)! - j!] \\&#xA;&amp;amp; = [2! - 1!] + [3! - 2!] + \cdots + [n! - (n-1)!] + [(n+1)! - n!]&#xA;\end{align*}&#xA;and you see that everything cancels, except the terms $- 1!$ from the first summand and $(n+1)!$ from the last summand, hence the sum must be equal to $(n+1)! - 1$.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit 2&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's the argument:&#xA;We want to prove that the following statement $T(n)$ holds for all $n \in \mathbb{N}$:&#xA;\[&#xA;(n+1)! - 1 = \sum_{j=1}^{n} j \cdot j!.&#xA;\]&#xA;For $n = 1$ we have the statement $T(1)$:&#xA;\[&#xA;1 = (1+1)! - 1 = \sum_{j=1}^{1} j \cdot j! = 1\cdot 1! = 1,&#xA;\]&#xA;so this is ok. &lt;strong&gt;Assume&lt;/strong&gt; that $T(n)$ holds. We want to prove $T(n+1)$:&#xA;\[&#xA;(n+2)! - 1 = \sum_{j=1}^{n+1} j \cdot j!.&#xA;\]&#xA;Start with the right hand side:&#xA;\[&#xA;\sum_{j=1}^{n+1} j \cdot j! = (n+1)\cdot (n+1)! + \sum_{j=1}^{n} j \cdot j!&#xA;\]&#xA;But the last sum is equal to $(n+1)! - 1$ by our &lt;strong&gt;assumption&lt;/strong&gt; that $T(n)$ is true, so&#xA;\begin{align*}&#xA;\sum_{j=1}^{n+1} j \cdot j! &amp;amp; = (n+1)\cdot (n+1)! + (n+1)! - 1 \\&#xA;&amp;amp; = [(n+1) + 1]\cdot(n+1)! - 1 = (n+2) \cdot (n+1)! - 1\\&#xA;&amp;amp; = (n+2)! - 1,&#xA;\end{align*}&#xA;so $T(n+1)$ holds as well.&lt;/p&gt;&#xA;" OwnerUserId="5363" LastEditorUserId="5363" LastEditorDisplayName="" LastEditDate="2011-01-23T02:09:15.010" LastActivityDate="2011-01-23T02:09:15.010" CommentCount="11" />
  <row Id="18581" PostTypeId="2" ParentId="18576" CreationDate="2011-01-22T23:38:28.653" Score="4" ViewCount="" Body="&lt;p&gt;There is a nice interpretation of this identity in terms of uniqueness of representation in &lt;a href=&quot;http://en.wikipedia.org/wiki/Factorial_number_system&quot; rel=&quot;nofollow&quot;&gt;factorial base&lt;/a&gt;.  But to answer your comment to Theo Buehler's answer, telescoping sequences are just a thing that you should be aware of and try to look for, and the identity Theo used is actually &lt;em&gt;equivalent&lt;/em&gt; to what Wolfram told you, so... &lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2011-01-22T23:38:28.653" CommentCount="1" />
  <row Id="18582" PostTypeId="1" AcceptedAnswerId="18747" CreationDate="2011-01-23T00:29:31.527" Score="3" ViewCount="200" Body="&lt;p&gt;Given that we shouldn't say that &quot;$f(z)$ is a function&quot;, shouldn't we also not write &quot;$p \in k[X_1, \ldots, X_n]$ is a polynomial&quot;? Along those lines, I usually write $p(X_1, \ldots, X_n) \in k[X_1, \ldots, X_n]$ in order to balance the &quot;free variables&quot; on both sides of the relation, but that gets unwieldy when you start dealing with iterated polynomial rings. My question is: &lt;strong&gt;Is there a notation for polynomial rings which allow us to talk about polynomials without explicitly naming the indeterminates?&lt;/strong&gt; Consider, for an analogy, vector spaces $\mathbb{R}^n$. These have a canonical basis, but the notation $\mathbb{R}^n$ does not commit me to &lt;em&gt;naming&lt;/em&gt; the canonical basis, unlike, say, the notation $\operatorname{span} \{ e_1, \ldots, e_n \}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suppose I should fix a definition for polynomial rings. For simplicity let's work in the category $\mathbf{CRing}$ of commutative rings with 1. Let $U: \mathbf{CRing} \to \mathbf{Set}$ be the forgetful functor taking rings to their underlying sets. A polynomial ring in a set of indeterminates $\mathcal{S}$ over a ring $A$ is a ring $R$ together with an inclusion map $\iota: A \hookrightarrow R$ and a set-map $x: \mathcal{S} \hookrightarrow UR$, and has the universal property that for every ring $B$, homomorphism $\phi: A \to B$, and set-map $b: \mathcal{S} \to B$, there is a homomorphism $\epsilon: R \to B$ such that $\epsilon \circ \iota = \phi$ and $U\epsilon \circ x = b$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we write $A[\mathcal{S}]$ for such a ring $R$, then we could write, for instance, $A[5]$ for the ring of polynomials in 5 variables over $A$, but that would, I imagine, be extremely confusing. Yet, on the other hand, if we have a bijection $\mathcal{S} \to \mathcal{S}&amp;#39;$, then this lifts to an isomorphism of $A[\mathcal{S}] \to A[\mathcal{S}&amp;#39;]$, so it is all the more tempting to write $A[\kappa]$, $\kappa = |\mathcal{S}|$, for the canonical representative of this isomorphism class. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $\mathcal{S} = \{ 1, \ldots, n \} \subset \mathbb{N}$ and $\phi: A \to B$ is given, I write $\phi p(b_1, \ldots, b_n)$ for the image of $p \in A[\mathcal{S}]$ under $\epsilon$ for $b(m) = b_m, m \in \mathcal{S}$. When the choice of homomorphism $\phi$ is clear I'll omit it in writing. This justifies my notation $p(X_1, \ldots, X_n) \in A[X_1, \ldots, X_n]$, since I would like to regard $A[X]$ as being analogous to $\mathbb{Z}[\pi]$, i.e. it's a ring with a transcendental element adjoined so is isomorphic to a polynomial ring, but doesn't come with evaluation maps attached. But following this line of thought, how should I denote the object that $p$ itself belongs to?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I recently started attending an algebraic geometry course and at one point the lecturer wrote $k[\mathbb{A}^n]$ for the ring of polynomials in $n$ indeterminates over $k$. This seems like a reasonable solution, but there are some problems:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;It feels suspiciously like a function ring, but in general the map taking formal polynomials to polynomial functions is neither injective nor surjective.&lt;/li&gt;&#xA;&lt;li&gt;The notation makes it look like a ring with $\mathbb{A}^n$ adjoined, but that doesn't seem to make sense. (Is there a way to make sense of it, e.g. by defining ring operations on $\mathbb{A}^n$?)&lt;/li&gt;&#xA;&lt;li&gt;Is it standard notation? I have seen $k[V]$ in some algebraic geometry textbooks for the coordinate ring of the (affine) variety $V$, but never for $V = \mathbb{A}^n$. (I have similar reservations about the notation $k[V]$, but not as strongly.)&lt;/li&gt;&#xA;&lt;li&gt;Would it make sense to write, say, $\mathbb{Z}[\mathbb{A}^n]$?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;A related problem arises from the following: let $p(X)$ and $q(X)$ be formal polynomials in $k[X]$, with $p(X) = q(X^2)$. It's clear that $\operatorname{deg} p = 2 \operatorname{deg} q$... but this shows that, in a certain sense, the degree depends on the ambient polynomial ring: if $p(X)$ were considered as a formal polynomial in $k[X^2]$, its degree would be the same as $q$, since, after all, $p(X) = q(X^2)$. It is clear that we should have $k[X^2] \subset k[X]$, but if we obviate the indeterminates and reduce polynomials to their bare skeletons, then the &quot;inclusion&quot; map $k[X^2] \hookrightarrow k[X]$ is no longer a set-theoretic inclusion map. &lt;strong&gt;Is there a coherent way of thinking about polynomials and polynomial rings which resolves this ambiguity, and what is the notation that goes with it?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="5191" LastEditorUserId="5191" LastEditorDisplayName="" LastEditDate="2011-01-24T14:15:26.360" LastActivityDate="2011-01-24T15:19:59.393" Title="Notation for a polynomial ring and formal polynomials" Tags="&lt;abstract-algebra&gt;&lt;commutative-algebra&gt;&lt;notation&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" />
  <row Id="18583" PostTypeId="2" ParentId="18582" CreationDate="2011-01-23T00:37:41.363" Score="2" ViewCount="" Body="&lt;p&gt;Last question first: degree is not (depending on how you look at it) a property of elements of polynomial rings.  It's a property of elements of &lt;a href=&quot;http://en.wikipedia.org/wiki/Graded_algebra&quot; rel=&quot;nofollow&quot;&gt;graded&lt;/a&gt; polynomial rings, and the easiest way to choose a grading is to choose a set of generators.  Your example is a little confused: when you consider $k[x^2]$ you are implicitly considering $x^2$ to have degree $1$, but you don't need to adopt this convention; you can declare that it has degree $2$ instead.  (In any case, the notation $k[x^2]$ is misleading: when you write this you are really talking about the entire inclusion map $k[x^2] \to k[x]$, so everything depends on whether you want this to be a map of rings or a map of graded rings.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your point 3 seems to answer your first question.  I'm not sure why you find $k[\mathbb{A}^n]$ objectionable but $k[V]$ not given that I assume you think $\mathbb{A}^n$ is a variety.  I think this is a fine way to refer to a polynomial ring without naming its generators.  Your worry about the distinction between polynomials and the functions they induce can be ignored if $k$ is algebraically closed, and otherwise you should just remind yourself that in the remaining cases the functor from $k$-varieties to $\text{Set}$ isn't faithful, so the answer is not to take the set-theoretic picture too seriously in the first place and work directly with the opposite of the category of finitely-generated reduced $k$-algebras.  In this category I describe exactly how to recover the ring of functions geometrically &lt;a href=&quot;http://qchu.wordpress.com/2011/01/21/structures-on-hom-sets/&quot; rel=&quot;nofollow&quot;&gt;in this blog post&lt;/a&gt;.  (I should be more explicit about what I mean here: if you work in the right category, the polynomial ring in $n$ variables over $k$ &lt;em&gt;is&lt;/em&gt; the space of functions on $\mathbb{A}^n$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also don't understand your first two sentences; they seem to be inconsistent with each other.  (Off-topic: I don't know what you look like, but because of my Gravatar you know what I look like.  I'm the guy sitting in the back of class on his laptop, so if you'd like to introduce yourself that would be cool.)&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2011-01-23T00:54:45.437" LastActivityDate="2011-01-23T00:54:45.437" CommentCount="20" />
  <row Id="18584" PostTypeId="1" AcceptedAnswerId="18587" CreationDate="2011-01-23T00:52:28.947" Score="4" ViewCount="141" Body="&lt;p&gt;Sorry this is a really simple problem but I couldnt figure it out. I'm trying to find the value of $3^{1000}\bmod 7$. I know the actual value is 4, but I used a calculator. How would I simplify this problem to get the correct answer without a calculator? I dont want a direct answer, but maybe a nudge in the right direction, such as tricks I could use or maybe an alternate way of writing it.&lt;/p&gt;&#xA;" OwnerUserId="2985" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-23T01:07:39.527" LastActivityDate="2011-01-23T04:59:21.037" Title="Modulo question" Tags="&lt;modular-arithmetic&gt;" AnswerCount="4" CommentCount="3" FavoriteCount="2" />
  <row Id="18585" PostTypeId="2" ParentId="18584" CreationDate="2011-01-23T00:56:56.680" Score="1" ViewCount="" Body="&lt;p&gt;$3^{1000}$ is hard to compute by hand because of the $1000$.  Can you learn anything from trying smaller exponents instead?&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2011-01-23T00:56:56.680" CommentCount="2" />
  <row Id="18586" PostTypeId="2" ParentId="18584" CreationDate="2011-01-23T00:58:07.043" Score="2" ViewCount="" Body="&lt;p&gt;Try to split the huge number in smaller numbers you know the value of. For example&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$3^{1000} \text{mod} 7= (3^{10})^{100}  \text{mod} 7= (59049)^{100} \text{mod} 7= 4^{100}  \text{mod} 7 = \ldots$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $3^{10}=59049$ was still too big you could try to rewrite it again etc.&lt;/p&gt;&#xA;" OwnerUserId="3123" LastActivityDate="2011-01-23T00:58:07.043" CommentCount="2" />
  <row Id="18587" PostTypeId="2" ParentId="18584" CreationDate="2011-01-23T01:12:50.563" Score="7" ViewCount="" Body="&lt;p&gt;If you know Fermat's Little Theorem, then you know you can reduce the exponent. If you don't, you can simply do a bit of testing:&#xA;\begin{align*}&#xA;3 &amp;amp;\equiv 3 &amp;amp;\pmod{7}\\&#xA;3^2 &amp;amp;\equiv 2&amp;amp;\pmod{7}\\&#xA;3^3 &amp;amp;\equiv 2\times 3&amp;amp;\pmod{7}\\&#xA;&amp;amp;\equiv -1 &amp;amp;\pmod{7}.&#xA;\end{align*}&#xA;At this point, you will also know that $3^6 = (3^3)(3^3)\equiv (-1)^2 \equiv 1 \pmod{7}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That means that the remainders will &quot;cycle&quot; every seven steps, since $3^7 = 3(3^6) \equiv 3(1) = 3 \pmod{7}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So now, notice that since $1000 = 6(166) + 4$, then&#xA;$$3^{1000} = 3^{6(166)+4} = (3^6)^{166}3^4 = (3^6)^{166}3^33^1.$$&#xA;You know what each of the factors is modulo $7$, so you're done.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-23T04:59:21.037" LastActivityDate="2011-01-23T04:59:21.037" CommentCount="5" />
  <row Id="18588" PostTypeId="2" ParentId="17182" CreationDate="2011-01-23T01:52:04.620" Score="4" ViewCount="" Body="&lt;p&gt;One definition given by &lt;a href=&quot;http://en.wikipedia.org/wiki/Elliptic_integral&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt; for the incomplete elliptic integral of the third kind is:&#xA;$$&#xD;&#xA;\Pi(n; \phi \vert m) = \int_0^{\sin\phi} \frac{1}{1-nt^2}\frac{dt}{\sqrt{(1-mt^2)(1-t^2)}}.&#xD;&#xA;$$&#xA;This indicates that the function is periodic with period $2\pi$, is symmetric about $\phi=\pi/2$, and is antisymmetric about $\phi=0$.  So the values for $0\le\phi &amp;lt; \pi/2$ define the function for all $\phi$.&lt;/p&gt;&#xA;" OwnerUserId="5546" LastActivityDate="2011-01-23T01:52:04.620" CommentCount="1" />
  <row Id="18589" PostTypeId="1" AcceptedAnswerId="18660" CreationDate="2011-01-23T01:53:44.473" Score="19" ViewCount="578" Body="&lt;p&gt;The Gaussian $\mathbb{Z}[i]$ and Eisenstein $\mathbb{Z}[\omega]$ integers have been used to solve some diophantine equations. I have never seen any examples of the golden integers $\mathbb{Z}[\varphi]$ used in number theory though. If anyone happens to know some equations we can apply this in and how it's done I would greatly appreciate it!&lt;/p&gt;&#xA;" OwnerUserId="4290" LastEditorUserId="4290" LastEditorDisplayName="" LastEditDate="2011-01-23T01:59:56.257" LastActivityDate="2011-03-02T04:29:33.000" Title="Golden Number Theory" Tags="&lt;number-theory&gt;&lt;algebra&gt;" AnswerCount="3" CommentCount="6" FavoriteCount="7" />
  <row Id="18590" PostTypeId="1" AcceptedAnswerId="18632" CreationDate="2011-01-23T02:09:05.653" Score="3" ViewCount="104" Body="&lt;p&gt;A property of characteristic classes $c$ is that $c(f^* E) = f^* c(E)$ where $E\to M$ is a bundle and $f^* E$ is the pullback of $E$ by some map $f: N\to M$.  In Bott and Tu it is stated (for Chern classes) that this implies that if $E \simeq F$ then $c(E) = c(F)$.  However I am unable to see why this is true (from the naturality fact alone-- I know why its true from the construction of Chern classes).  In other texts they get around this by defining characteristic classes to be defined on equivalence classes of bundles or use a more broader definition of naturality that includes bundle maps. Hence I am wondering how straightforward this statement in Bott and Tu really is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know from all the constructions of characteristic classes that this is a really pedantic question (obviously its isomorphism invariant!), but I am just curious.&lt;/p&gt;&#xA;" OwnerUserId="9" LastActivityDate="2011-01-24T03:43:44.000" Title="Does naturality imply isomorphism invariance for characteristic classes?" Tags="&lt;algebraic-topology&gt;&lt;characteristic-classes&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" />
  <row Id="18591" PostTypeId="2" ParentId="18584" CreationDate="2011-01-23T02:51:37.363" Score="2" ViewCount="" Body="&lt;p&gt;Since that 3&lt;sup&gt;6&lt;/sup&gt;mod7=1,&#xA;then we have 3&lt;sup&gt;1000&lt;/sup&gt;=3&lt;sup&gt;6*166+4&lt;/sup&gt;=3&lt;sup&gt;4&lt;/sup&gt;=4 mod7.&lt;/p&gt;&#xA;" OwnerUserId="5971" LastActivityDate="2011-01-23T02:51:37.363" />
  <row Id="18592" PostTypeId="2" ParentId="18530" CreationDate="2011-01-23T02:54:12.817" Score="2" ViewCount="" Body="&lt;p&gt;You have noticed something that turns out to be quite interesting in a more general context: for each $f \in U(T)$, there are $(n-m)!$ elements $g \in U(T)$ that satisfy for all $t \in T, f(t) = g(t)$. This is because the kernel of the homomorphism $F$ we are looking have has order $(n-m)!$; but that doesn't actually get you any closer to the answer of the question that is being asked!i&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, let me suggest that instead of $A(X)$, we use the notation $Sym(X)$ as being the group of all bijections $X \to X$ with the group operation being composition. The important things to notice (and prove!) are that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Theorem: $U(T)$ is a group (and thus a subgroup of $Sym(S)$; it is usually called the stabilizer of $T$ in $Sym(S)$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Theorem: For each $f \in U(T)$, there exists a unique $g \in Sym(T)$ such that for all $t \in T$, $f(t) = g(t)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This lets us define $F$ to be the function that maps any $f \in U(T)$ to that unique element $g \in Sym(T)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Theorem: For each $g \in Sym(T)$, there exists at least one $f \in U(T)$ such that for all $t \in T, f(t) = g(t)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It follows that $F$ must therefore be onto $Sym(T)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next prove that $F$ satisfies: $F(gh) = F(g)F(h)$; i.e. that $F$ is a homomorphism $U(T) \to Sym(T)$ which is onto.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally as Qiaochu hinted and Arturo more explicitly noted, given two sets $X, Y$ and a bijection $T : X \to Y$, we can prove that $Sym(X)$ is isomorphic to $Sym(Y)$. The set $[m] = \{1, 2, ..., m\}$ is in bijection with $T$; and $Sym([m])$ is what is usually meant by $S_m$. The conclusion follows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To put the original problem in more common group theoretical language: Given $T \subset S$, let $G$ be the stabilizer of $T$ in $Sym(S)$. Show that there is a homomorphism from $G$ onto $S_{|T|}$.&lt;/p&gt;&#xA;" OwnerUserId="4980" LastActivityDate="2011-01-23T02:54:12.817" />
  <row Id="18593" PostTypeId="1" AcceptedAnswerId="18625" CreationDate="2011-01-23T04:02:19.913" Score="5" ViewCount="97" Body="&lt;p&gt;Let $V$ be a vector subspace of $R^N$, and $l:V \to R$ a linear mapping such that $l(V\bigcap R_{+}^N)\subseteq R_{+}$ (i.e., $l$ is positive).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have heard that there exists a separating hyperplane sort of argument that allows us to show that $l$ extends to a positive linear functional on $R^N$. I have my own proof, but it is complicated and uses the Bauer-Namioka condition for extension of positive linear functionals on ordered vector spaces of any dimension.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question&lt;/strong&gt;: What is this simple separating hyperplane argument that shows that $l$ extends to a positive linear functional on $R^N$? (I believe it should be quite straightforward, but I was not able to construct the right problem to invoke a separation argument...) A reference would be okay as well. Thanks.&lt;/p&gt;&#xA;" OwnerUserId="6141" LastEditorUserId="2548" LastEditorDisplayName="" LastEditDate="2011-01-23T04:34:33.280" LastActivityDate="2011-01-23T15:46:53.390" Title="Extending a positive linear functional in finite dimensions" Tags="&lt;linear-algebra&gt;&lt;reference-request&gt;&lt;functional-analysis&gt;" AnswerCount="2" CommentCount="6" />
  <row Id="18594" PostTypeId="2" ParentId="18593" CreationDate="2011-01-23T04:16:06.790" Score="0" ViewCount="" Body="&lt;p&gt;Perhaps &lt;a href=&quot;http://books.google.com/books?id=wjzZCLzx6hUC&amp;amp;pg=PA113&amp;amp;lpg=PA113&amp;amp;dq=Extending+a+positive+linear+functional+hyperplane&amp;amp;source=bl&amp;amp;ots=28FHg5LeJA&amp;amp;sig=J9xw-indU8l1x2zzGwwMFiRn6gQ&amp;amp;hl=en&amp;amp;ei=NK47Tc_9PIT48Ab-9uXfCg&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=9&amp;amp;ved=0CFEQ6AEwCDgK#v=onepage&amp;amp;q=Extending%20a%20positive%20linear%20functional%20hyperplane&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; will help (Exercise 12 seems similar).&lt;/p&gt;&#xA;" OwnerUserId="2548" LastEditorUserId="2548" LastEditorDisplayName="" LastEditDate="2011-01-23T04:31:10.663" LastActivityDate="2011-01-23T04:31:10.663" CommentCount="0" />
  <row Id="18595" PostTypeId="1" AcceptedAnswerId="18628" CreationDate="2011-01-23T04:17:35.387" Score="1" ViewCount="145" Body="&lt;p&gt;Let $V$ be a vector space with basis $e_1, \ldots, e_n$ and $V^*$ be its dual space with dual basis $e_1^*, \ldots, e_n^*$. Let $k$ be an integer between $1$ and $n$. Why $\wedge^{n-k}V=\wedge^{k}V^*$? Thank you very much.&lt;/p&gt;&#xA;" LastEditorUserId="1650" LastEditorDisplayName="" LastEditDate="2011-01-23T09:07:33.480" LastActivityDate="2011-01-23T15:50:44.040" Title="Exterior power of dual space." Tags="&lt;linear-algebra&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="18598" PostTypeId="2" ParentId="4453" CreationDate="2011-01-23T05:26:41.677" Score="2" ViewCount="" Body="&lt;p&gt;This question (does established terminology for a certain situation exist) has been answered in the comments (no).  But for posterity--- ie, for students who may meet &quot;principal curvature&quot; terminology before they learn a lot of differential geometry, and who may run across this question on web searches for related topics--- I can't resist emphasizing a point hinted at in the question itself: having both principal curvatures negative at a point is not a well defined property of a point on a surface; the signs of the principal curvatures depend on the coordinate system you calculate them in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, if you pick a point $p$ on a sphere in $\mathbb{R}^3$ and a coordinate patch around $p$ in which to compute principal curvatures, you find that both principal curvatures are &lt;em&gt;positive&lt;/em&gt; at $p$ if the normal vector field associated to the coordinate patch points inward (ie, toward the center of the sphere), and both &lt;em&gt;negative&lt;/em&gt; if the normal vector field associated to the patch points outward.  The choice of normal is yours to make; it doesn't come with the sphere.  (I suppose this point might be better made with a complicated surface that does not correspond to a well-known shape--- e.g. the surface formed by a tangle of ribbon after you have unwrapped a gift.  If you punch a hole in this ribbon and replace the missing disc with a hemisphere, whether this is a &quot;valley&quot; or a &quot;cup&quot; is &lt;em&gt;entirely up to you&lt;/em&gt;.  In choosing the normal direction, you are deciding whether or not curves in the hemisphere are &quot;curving&quot; &quot;toward&quot; the normal or &quot;away from&quot; it.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, this indeterminancy up to sign is the worst thing that can happen: if $S \subseteq \mathbb{R}^3$ is a smooth surface, and the principal curvatures at a point $p$ in $S$ are found in some coordinate system to be $\kappa_1$ and $\kappa_2$, then in any other coordinate system they will be either found to be exactly the same (that is, $\kappa_1$ and $\kappa_2$) or the same, but with opposite sign (that is, $-\kappa_1$ and $-\kappa_2$).  How you prove this depends on how you define the principal curvatures, but arises e.g. from the fact that they are the maximum and minimum normal components of accelerations of unit-speed curves in the surface at $p$, and there are only two possible choices of unit normal at $p$, differing only by sign.  So &quot;both principal curvatures are negative&quot; is a well-defined property of an &lt;em&gt;oriented&lt;/em&gt; smooth surface in $\mathbb{R}^3$ (a property which, we learned in the comments, apparently does not have a well-established name; for what it's worth, I like &quot;bump point&quot;).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This same discussion shows that the slightly weaker condition that &quot;both principal curvatures at $p$ have the &lt;em&gt;same sign&lt;/em&gt;&quot; &lt;em&gt;is&lt;/em&gt; well-defined independent of a choice of normal.  (As pointed out in the comments, it &lt;em&gt;does&lt;/em&gt; have a well-established name: such a $p$ is said to be an &lt;em&gt;elliptic&lt;/em&gt; point.)  What the above discussion does &lt;em&gt;not&lt;/em&gt; show, but is nevertheless true, is that this weaker condition is even independent of how one &quot;embeds&quot; $S$ in any &quot;ambient space&quot; (what the asker meant when he referred to the Gaussian curvature at $p$ as being &quot;intrinsic&quot;).  For more, open up any book on the differential geometry of surfaces and look for the sections around Gauss's &lt;a href=&quot;http://en.wikipedia.org/wiki/Theorema_egregium&quot; rel=&quot;nofollow&quot;&gt;Theorema Egregium&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="3840" LastEditorUserId="3840" LastEditorDisplayName="" LastEditDate="2011-01-23T05:31:51.747" LastActivityDate="2011-01-23T05:31:51.747" />
  <row Id="18599" PostTypeId="1" AcceptedAnswerId="18643" CreationDate="2011-01-23T06:35:12.997" Score="0" ViewCount="215" Body="&lt;p&gt;I haven't taken a math class in a few years and I have come to realize, as I am taking one again, that I do not know/remember how to factor equations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One example from my textbook is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$3x^3-x^2+3x+1$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Looking for a very detailed description of how to factor this, and just factoring in general.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Isaac gave a very good description of how to factor this equation but I am looking for a &lt;strong&gt;general description of how to factor equations&lt;/strong&gt; not just this equation.&lt;/p&gt;&#xA;" OwnerUserId="6144" LastEditorUserId="6144" LastEditorDisplayName="" LastEditDate="2011-01-23T16:52:08.887" LastActivityDate="2011-01-24T05:35:54.757" Title="Factoring polynomials step by step" Tags="&lt;polynomials&gt;" AnswerCount="4" CommentCount="3" />
  <row Id="18600" PostTypeId="2" ParentId="18599" CreationDate="2011-01-23T06:43:28.870" Score="3" ViewCount="" Body="&lt;p&gt;The polynomial given does not factor (with integer coefficients), but perhaps $3x^3-x^2-3x+1$ was intended.  One technique for factoring polynomials with several terms (typically more than 3) is to group terms together and factor a common monomial factor out of each group.  In this instance: $$\begin{align}&#xD;&#xA;3x^3-x^2-3x+1 &amp;amp;= (3x^3-x^2)+(-3x+1)&#xD;&#xA;\\&#xD;&#xA;&amp;amp;= x^2(3x-1)-1(3x-1)&#xD;&#xA;\\&#xD;&#xA;&amp;amp;=(x^2-1)(3x-1)&#xD;&#xA;\\&#xD;&#xA;&amp;amp;=(x+1)(x-1)(3x-1)&#xD;&#xA;\end{align}$$&#xA;The final step is one of several common factoring patterns—this one is sometimes referred to as a difference of squares: $a^2-b^2=(a+b)(a-b)$.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;edit&lt;/em&gt; As to how to factor in general, my inclination is to say that you don't.  If you pick a polynomial with random integer coefficients, the probability that it factors over the integers is 0.  Typically, in textbooks, the factoring problems are common monomial factors, specific factoring patterns (difference of squares, sum/difference of cubes, perfect-square, perfect-cube, etc.), factoring-by-grouping, quadratics, or some other special-case technique (as below).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On a side note, since two other answers have suggested factoring by looking for roots, there are polynomials with integer coefficients that factor over the integers but have no real roots or no rational roots (though I think they must be of degree 4 or greater).  For example: $$\begin{align}&#xD;&#xA;x^4+2x^2+81&#xD;&#xA;&amp;amp;=x^4+18x^2+81-16x^2&#xD;&#xA;\\&#xD;&#xA;&amp;amp;=(x^2+9)^2-(4x)^2&#xD;&#xA;\\&#xD;&#xA;&amp;amp;=(x^2-4x+9)(x^2+4x+9)&#xD;&#xA;\end{align}$$&#xA;(This is an example of a class of such problems [quartics that are quadratics in $x^2$] that I have seen in multiple textbooks where the technique is a sort of &quot;completing the square,&quot; but applied to the middle term, followed by a difference-of-squares factoring.)&lt;/p&gt;&#xA;" OwnerUserId="72" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2011-01-24T02:08:11.923" LastActivityDate="2011-01-24T02:08:11.923" CommentCount="6" />
  <row Id="18601" PostTypeId="2" ParentId="18599" CreationDate="2011-01-23T07:33:50.760" Score="1" ViewCount="" Body="&lt;p&gt;Factoring over the integers, $(bx + c)$ can only divide into $a_n x^n + ... + a_1 x + a_0$ if $b | a_n$ and $c | a_0$. For small $a_n$ and $a_0$ you can easily try the different cases. Here you have $a_n$ = 3 and $a_0$ = 1, so you can evaluate your polynomial at $x \in \{-1, \frac{-1}{3}, \frac{1}{3}, 1\}$ to get the roots. If you don't find enough roots, divide through by the terms you have found and re-test the roots you found to see whether any are repeated.&lt;/p&gt;&#xA;" OwnerUserId="5676" LastActivityDate="2011-01-23T07:33:50.760" CommentCount="1" />
  <row Id="18602" PostTypeId="1" CreationDate="2011-01-23T08:14:39.980" Score="-2" ViewCount="143" Body="&lt;p&gt;Describe a Mathematics Model( to be prepared for a math exhibition )of any topic on Mensuration or Circles for Grade 9&lt;/p&gt;&#xA;" OwnerUserId="6147" LastActivityDate="2011-01-23T10:30:04.497" Title="Describe a Mathematics Model of any topic on Mensuration or Circles for Grade 9" Tags="&lt;high-school&gt;" AnswerCount="1" />
  <row Id="18603" PostTypeId="1" AcceptedAnswerId="18617" CreationDate="2011-01-23T09:59:36.547" Score="1" ViewCount="131" Body="&lt;p&gt;I have a problem understanding second-order conditions at a critical point in finding critical points of two-variable functions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's consider $f(x,y)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first-order conditions are $\frac{\partial f}{\partial x}=0$ and $\frac{\partial f}{\partial y}=0$. So the rate of change of $f$ in respect to both $x$ and $y$ is naught at a critical point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second-order conditions at a critical point that I have in my book are of the following form:&#xA;A point (a,b) is a maximum if $f_{xx} f_{yy} - f_{xy}^2 &amp;gt; 0 $ and $f_{xx} &amp;lt; 0$.&#xA;(For all my functions, $f_{xy}=f_{yx}$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I don't understand is why the second-order conditions have to be so complicated. I think if $f_{xx} &amp;gt; 0$ and $f_{yy} &amp;gt; 0$ (both second-order derivatives are positive), we have a minimum. If $f_{xx}&amp;lt;0$ and $f_{yy}&amp;lt;0$, we have a maximum. And if the signs of two second-order derivatives at (a,b) are different it is a saddle point. &lt;strong&gt;What am I missing?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="2846" LastActivityDate="2011-01-29T14:22:45.953" Title="Optimisation in two variables - second order conditions" Tags="&lt;calculus&gt;&lt;real-analysis&gt;&lt;optimization&gt;" AnswerCount="4" />
  <row Id="18604" PostTypeId="2" ParentId="18603" CreationDate="2011-01-23T10:15:13.150" Score="1" ViewCount="" Body="&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Second_partial_derivative_test&quot; rel=&quot;nofollow&quot;&gt;wikipedia&lt;/a&gt; article will provide some insight. What you are looking at is actually the determinant of a $2 \times 2$ Hessian matrix.&lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2011-01-23T10:15:13.150" CommentCount="1" />
  <row Id="18605" PostTypeId="1" CreationDate="2011-01-23T10:18:17.937" Score="6" ViewCount="329" Body="&lt;p&gt;What is the difference between max, min and sup, inf?&lt;/p&gt;&#xA;" OwnerUserId="6148" LastEditorUserId="263" LastEditorDisplayName="" LastEditDate="2011-01-23T12:01:28.827" LastActivityDate="2011-01-23T13:02:09.487" Title="max and min versus sup and inf" Tags="&lt;real-analysis&gt;" AnswerCount="4" CommentCount="3" />
  <row Id="18606" PostTypeId="2" ParentId="18605" CreationDate="2011-01-23T10:23:39.057" Score="0" ViewCount="" Body="&lt;p&gt;For a bounded set $A \subset \mathbb{R}$ we have the following:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\max A \ \text{exists and equals} \ \sup A \Longleftrightarrow \sup A \in A$&lt;/li&gt;&#xA;&lt;li&gt;$\min A \ \text{exists and equals} \ \inf A \Longleftrightarrow  \inf A \in A$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="2548" LastEditorUserId="2548" LastEditorDisplayName="" LastEditDate="2011-01-23T12:16:56.323" LastActivityDate="2011-01-23T12:16:56.323" CommentCount="14" />
  <row Id="18607" PostTypeId="2" ParentId="18602" CreationDate="2011-01-23T10:30:04.497" Score="1" ViewCount="" Body="&lt;p&gt;The recommendation &lt;a href=&quot;http://mathforum.org/alejandre/circles.html&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;(site about science projects) seems relevant.&lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2011-01-23T10:30:04.497" />
  <row Id="18608" PostTypeId="2" ParentId="18605" CreationDate="2011-01-23T10:40:59.660" Score="0" ViewCount="" Body="&lt;p&gt;Every bounded subset $S$ of the reals has a supremum and an infimum. The supremum, for instance, is just the smallest number which is greater or equal to every element in $S$. This number, however, may not belong to $S$ -- this happens for example for $S=(0,1)$, as lhf noted. In this case, you would say that &lt;em&gt;$S$ does not have a maximum/maximal element&lt;/em&gt;. The reason is that a maximum of $S$ is &lt;em&gt;an element of $S$&lt;/em&gt; which is greater or equal to every element in $S$. Such an element may or may not exist, but if it exists it is unique, which justifies the use of the definite article.&lt;/p&gt;&#xA;" OwnerUserId="367" LastActivityDate="2011-01-23T10:40:59.660" />
  <row Id="18609" PostTypeId="2" ParentId="18605" CreationDate="2011-01-23T10:41:54.123" Score="5" ViewCount="" Body="&lt;p&gt;The main difference is that the infimum and supremum always exists (if you allow the values $\pm \infty$) but the minimum and maximum doesn't.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the set is finite then it is trivial, as a finite set always has a maximal and minimal element, but if you look at something like:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$A = \{x \in \mathbb{Q}: x &amp;gt; 0\}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then you cannot find a smallest element in the set as when you find some candidate for the smallest element, the half of it is still smaller. And $0 \notin A$, therefore the Minimum doesn't exist. In contrast to this the Infimum is just the biggest lower bound and doesn't have to be in $A$ itself, (not even in $\mathbb{Q}$). You can easily see that $\text{inf}(A)=0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another example is $B = \{x \in \mathbb{Q}: x&amp;gt;0,x^2 &amp;gt; 2\}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Certainly $\sqrt{2} \notin \mathbb{Q}$ therefore the minimum doesn't exist as for every candidate for your minimum you can go closer to $\sqrt{2}$ and find a smaller one ($\mathbb{Q}$ is dense in $\mathbb{R}$), but the infimum is $\text{inf}(B)=\sqrt{2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Otherwise if the minimum or maximum does exist, it is surely equal to the infimum/supremum as it is the biggest lower / least upper bound in this case.&lt;/p&gt;&#xA;" OwnerUserId="3123" LastEditorUserId="3123" LastEditorDisplayName="" LastEditDate="2011-01-23T13:02:09.487" LastActivityDate="2011-01-23T13:02:09.487" CommentCount="2" />
  <row Id="18610" PostTypeId="1" AcceptedAnswerId="18613" CreationDate="2011-01-23T11:29:50.090" Score="0" ViewCount="72" Body="&lt;p&gt;First of all I'm really sorry but I don't know how to format my exercise. I wrote it as I would of wolframalpha.I need help proving this inequality.   $$\int_{1}^{2} \left(\frac{x+1}{2} \right)^{x+1} \ dx &amp;lt; \int_{1}^{2} x^x \ dx$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Normally I would define $f(x) = \left(\frac{x+1}{2} \right)^{x+1}$ and $g(x) = x^x$ and try to compare those two. Any suggestions?Thank you. &lt;/p&gt;&#xA;" OwnerUserId="2227" LastEditorUserId="2548" LastEditorDisplayName="" LastEditDate="2011-01-23T11:34:18.700" LastActivityDate="2011-01-23T12:06:04.533" Title="comparing two definite integrals" Tags="&lt;integral&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="18611" PostTypeId="2" ParentId="18467" CreationDate="2011-01-23T11:30:45.543" Score="4" ViewCount="" Body="&lt;p&gt;I think the answer is yes. Let $K$ be the intersection of the normalizers of the Sylow $p$-subgroups of $G$, and $P$ any Sylow $p$-subgroup. Then $K$ is a normal subgroup of $G$, so $[K,P] \le K \cap P$. If $K \cap P$ is nontrivial, then a nonidentity element $g$ has order a power of $p$ and normalizes all Sylow $p$-subgroups, so it must lie in all Sylow $p$-subgroups, contradicting your assumption. So $[K,P] = 1$ and hence $K$ centralizes all Sylow $p$-subgroups.&lt;/p&gt;&#xA;" OwnerUserId="2820" LastActivityDate="2011-01-23T11:30:45.543" CommentCount="1" />
  <row Id="18612" PostTypeId="2" ParentId="18605" CreationDate="2011-01-23T11:58:46.070" Score="4" ViewCount="" Body="&lt;p&gt;If we have a set $A$ and some partial ordering on $A$, for $B$ a subset of $A$ then we define the $\sup$ of $B$ (denoted $\sup B$) as &lt;strong&gt;the least&lt;/strong&gt; element $a$ such that:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$\forall b\in B\colon b&amp;lt;a$ or $b=a$ ($a$ is an upper bound of $B$)&lt;/li&gt;&#xA;&lt;li&gt;If $c$ is such that the above condition holds for $c$ (that is $c$ is an upper bound of $B$) then $a&amp;lt;c$ or $a=c$ (namely, every upper bound of $B$ is at least $a$).&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;In the case that $a=\sup B$ and $a\in B$ we say that $a$ is the maximum of $B$ and denote it by $\max B$. Note that this is not always the case, as presented in other answers $1=\sup \{x\in\mathbb{Q} : x&amp;lt;1\}$ but it is not a maximal element of the set.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The definitions for $\inf$ and $\min$ are symmetric when replacing upper bound by lower bound, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note, however, that not every order relation has this property of having upper bounds, not even for bounded subsets. It is easy to construct an order in which there is a bounded set which has neither $\sup$ nor $\inf$ (and thus no $\max$ and $\min$ as well).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, consider $\{a,b,c\}$ and define $a&amp;lt;b$ and $a&amp;lt;c$ however $b$ and $c$ are incomparable, that is both $b&amp;lt;c$ and $c&amp;lt;b$ are false.&lt;br&gt;&#xA;The subset $\{a\}$ is bounded, as $b$ and $c$ are both upper bounds, but neither of those is a least upper bound. In this case $a$ is a minimum element of $\{a\}$, but you can add a few more elements in a similar fashion to have that a subset of two incomparable elements has no $\sup$ nor $\inf$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When discussing real numbers then we always have a supremum, this is a result of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Dedekind_cuts&quot; rel=&quot;nofollow&quot;&gt;Dedekind cuts&lt;/a&gt; construction of the real numbers, which guarantee that every bounded subset has a supremum and infimum.&lt;/p&gt;&#xA;" OwnerUserId="622" LastEditorUserId="622" LastEditorDisplayName="" LastEditDate="2011-01-23T12:07:52.097" LastActivityDate="2011-01-23T12:07:52.097" />
  <row Id="18613" PostTypeId="2" ParentId="18610" CreationDate="2011-01-23T12:06:04.533" Score="5" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT:&lt;/strong&gt; In $[1,2]$ we have that $x^x &amp;gt; (\frac{x+1}{2})^{x+1}$. Try to show this by finding the minimum of $h(x) = x\log(x)-(x+1)\log(\frac{x+1}{2})$. &lt;/p&gt;&#xA;" OwnerUserId="4471" LastActivityDate="2011-01-23T12:06:04.533" CommentCount="1" />
  <row Id="18614" PostTypeId="1" AcceptedAnswerId="18629" CreationDate="2011-01-23T12:41:20.187" Score="2" ViewCount="109" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In my differential geometry course we had the following&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Theorem (Cartan-Hadamard)&lt;/strong&gt;: Let $M$ be a connected, simply connected, complete Riemannian manifold. Then the following are equivalent:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$M$ has nonpositive curvature&lt;/li&gt;&#xA;&lt;li&gt;$|d\exp_p(v)\hat v| \ge |\hat v|$ for all $p \in M$, $v, \hat v \in T_pM$&lt;/li&gt;&#xA;&lt;li&gt;$d(\exp_p(v), \exp_p(w)) \ge |v - w|$ for all $p \in M$, $v, w \in T_pM$&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;In addition $\exp_p$ is a diffeomorphism if either of these statements holds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, I understand the proof that was given, but I do not see where we have to use simple connectivity of $M$ (in the proof given by my professor the Cartan-Ambrose-Hicks Theorem was used). What's wrong with the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$1. \Leftrightarrow 2.$ didn't make use of simple connectivity. (at least I don't see where)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But now I think 2., completeness and connectedness imply that $\exp_p$ is a diffeo, and in particular $M$ is simply connected.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;_ 2. implies 3.: Let $v, w \in T_pM$ be given, let $\gamma(t) = \exp_p(v(t))$ be a lenght minimizing geodesic connecting $\exp_p(v)$ and $\exp_p(w)$. Then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$&#xD;&#xA;\begin{eqnarray}&#xD;&#xA; d(\exp_p(v), \exp_p(w)) &amp;amp;=&amp;amp; L(\gamma) \\ &amp;amp;=&amp;amp; \int_0^1 |\dot \gamma(t)| \, dt \\ &amp;amp;\ge&amp;amp;  \int_0^1 |\dot v(t)| \, dt \\ &amp;amp;\ge&amp;amp;  \left| \int_0^1 \dot v(t) \, dt \right| \\ &amp;amp;=&amp;amp; |v-w|&#xD;&#xA;\end{eqnarray}&#xD;&#xA;$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore $\exp_p$ is injective. &#xA;Completeness and connectedness imply that $\exp_p$ is surjective (Hopf-Rinow).&#xA;2 implies that $\exp_p$ is a local diffeo.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in conclusion $\exp_p$ is seen to be a bijective local diffeo, hence a diffeo.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Is there anything wrong with this argument? &lt;/li&gt;&#xA;&lt;li&gt;If yes: Could you give an example of a complete, connected manifold with nonpositive curvature, which is not simply connected?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Thanks a lot!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;S.L.&lt;/p&gt;&#xA;" OwnerUserId="3208" LastActivityDate="2011-01-23T17:25:46.410" Title="Nonpositive curvature, Theorem of Cartan-Hadamard" Tags="&lt;differential-geometry&gt;&lt;curvature&gt;&lt;riemannian-geometry&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="2" />
  <row Id="18615" PostTypeId="2" ParentId="18603" CreationDate="2011-01-23T13:06:23.210" Score="0" ViewCount="" Body="&lt;p&gt;I actually found a very good illustration of how my simple theory breaks down on page 2 of &lt;a href=&quot;http://www.math.washington.edu/~lee/Courses/126-95-96/supp20.ps&quot; rel=&quot;nofollow&quot;&gt;this file&lt;/a&gt;. They have very useful graphs there and after I plotted $f(x,y)=x^2+y^2+2xy$ in Wolfram Alpha, I understood how both second derivatives $f_{xx}$ and $f_{yy}$ can be positive at a critical point yet the function has no unique minimum at that point. I think the Alpha plot shows it very well!&lt;/p&gt;&#xA;" OwnerUserId="2846" LastActivityDate="2011-01-23T13:06:23.210" />
  <row Id="18616" PostTypeId="1" CreationDate="2011-01-23T13:13:58.503" Score="12" ViewCount="253" Body="&lt;p&gt;What are some great math social activities for students? I'm looking for things that bring people together with a &quot;light&quot; mathematical touch. The goal is to create a stronger mathematical community in a small college math department.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Examples I've heard so far:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Math Stack Exchange Party: Students get together and surf questions on this site and answer them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Zome Tool Party: Students get together and make cool things out of Zome tools.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Integration Bee: Spelling bee....but with integrals!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm looking for &quot;soft&quot; ways to attract students to the major. Thanks, in advance!&lt;/p&gt;&#xA;" OwnerUserId="941" LastEditorUserId="941" LastEditorDisplayName="" LastEditDate="2011-01-23T14:47:45.377" LastActivityDate="2011-01-25T03:27:11.073" Title="Fun math outreach/social activities" Tags="&lt;education&gt;&lt;recreational-mathematics&gt;" AnswerCount="5" CommentCount="1" FavoriteCount="8" />
  <row Id="18617" PostTypeId="2" ParentId="18603" CreationDate="2011-01-23T13:16:24.977" Score="1" ViewCount="" Body="&lt;p&gt;For a minimum at $(x_0,y_0)$ it is necessary that $f(x_0,y_0)$ is minimal with respect to the $f$-values at neighbouring points ${\it all\ around}$ $(x_0,y_0)$. The derivatives $f_{xx}$ and $f_{yy}$ contain only information about changes in $f$ for horizontal resp. vertical displacements, so we must definitely look at $f_{xy}$ to make sure that we don't have a saddle point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An example: The origin is a critical point for the function $f(x,y):=x^2-4xy+y^2$, and we have $f_{xx}(0,0)=f_{yy}(0,0)=2&amp;gt;0$; so you would diagnose a minimum. But if you pass the origin in $45^\circ$-direction, i.e., via $x(t)=y(t)=t\ \ (-1&amp;lt;t&amp;lt;1)$, then you measure the temperature $\phi(t)=f(x(t),y(t))=-2t^2$, and $f$ seems to be maximal at $(0,0)$. In reality we have a saddle point there.&lt;/p&gt;&#xA;" OwnerUserId="1303" LastEditorUserId="1303" LastEditorDisplayName="" LastEditDate="2011-01-29T14:22:45.953" LastActivityDate="2011-01-29T14:22:45.953" />
  <row Id="18618" PostTypeId="2" ParentId="18603" CreationDate="2011-01-23T13:18:12.773" Score="0" ViewCount="" Body="&lt;p&gt;To elaborate a bit on PEV's answer: you are considering the Hessian matrix of your function&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$H_f= \left( \begin{array}{cc} &#xD;&#xA;f_{xx} &amp;amp; f_{xy}\\&#xD;&#xA;f_{xy} &amp;amp; f_{yy} \end{array} \right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;then you apply a general theorem for maxima and minima for multi-variable functions which says that if $m$ and $M$ are the smallest and greatest eigevalue of the Hessian and $m &amp;gt; 0$, then the point at which you're evaluating the matrix is a maxmum and if $M &amp;lt; 0$ the point is a minimum and if $mM &amp;lt; 0$ the point is neither a maxmimum nor a minumim.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This condition is complicated because you are considering functions of two variables which behave a lot more complicately than functions of one, i.e. they change in an uncountable number of directions (take for an example directional derivatives). The condition for a stationary point of a single-valued function to be a maximum is that $f&amp;#39;&amp;#39; &amp;lt;0$ but then again, we're moving in only one direction, so that's enough. On the other hand (as Christian points out) for multi-variable functions, you need to have information about a lot more directions than just two (which are given by $f_{xx}$ and $f_{yy}$).&#xA; If you think of it this way this theorem is actually a great simplification of all the work one might have to do without it.&lt;/p&gt;&#xA;" OwnerUserId="1415" LastActivityDate="2011-01-23T13:18:12.773" />
  <row Id="18619" PostTypeId="1" CreationDate="2011-01-23T13:27:44.133" Score="2" ViewCount="182" Body="&lt;p&gt;There is an urn of N balls, each of a unique color. In each step one takes out 2 balls without replacement, changes the color on the 2nd ball to that of the first ball and returns them back to the urn. What is the expected number of steps until all the balls in the urn are of the same color?&lt;/p&gt;&#xA;" OwnerUserId="6151" LastActivityDate="2011-01-24T05:13:07.570" Title="yet another urn problem" Tags="&lt;stochastic-processes&gt;&lt;probability&gt;" AnswerCount="0" CommentCount="1" />
  <row Id="18620" PostTypeId="2" ParentId="17574" CreationDate="2011-01-23T13:31:36.517" Score="0" ViewCount="" Body="&lt;p&gt;if you solve this problem a bit you will get C'BA'=ABC&#xA;one solution for this equation is C'=A but if you consider the original C'BA'=ABC it has dxd equations with 2xdxb unknowns and for C'=A the number of unknowns is same i.e 2xdxb but number of equations now are dxb. if size of matrix B is greater than D.. then it doesn't seem to be a problem but if its other way round then no of eqns seem to decrease implying there are other equations meaning other solutions. this all is unclear and we not to solve it rigorously for this set of eqns.&lt;/p&gt;&#xA;" OwnerUserId="6152" LastActivityDate="2011-01-23T13:31:36.517" />
  <row Id="18621" PostTypeId="2" ParentId="16180" CreationDate="2011-01-23T13:35:19.490" Score="0" ViewCount="" Body="&lt;p&gt;I will answer your questions in order:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1- The complexity of a proof is the length of the proof. Remember that a proof (witness or certificate) is simply a branch of a Turing machine on some given input. This corresponds to a solution to some problem. For example consider SAT, the input to the machine will be a SAT instance (i.e. a propositional formula), and the proof will be an assignment of 0-1 values to the variables. Some problems can have exponentially long proofs, like PSPACE-complete problems, or proofs of non-existence like in coNP.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2- The second question is not very clear for me. I think you are asking if there can be proofs with super-polynomial lenghts? Like I've said, PSPACE-complete and coNP languages have exponentially long proofs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3- I don't understand what you mean with linear. You mean that the verifiers are always P machines? Not necessarily, for example, NEXP is also called Exponential-NP. This is because the proofs are verified by deterministic machines that run in exponential time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In real life, you can consider that all the lines of text that a proof have, can be stated in formal logic, and this can be interpreted by a machine. A field in computational complexity that studies formal proofs is proof complexity. For example, you are given a statement written in some theory (e.g. Peano's arithmetic). Then, you calculated what are the necessary resources a machine needs to verify that the statement is true or not. &lt;/p&gt;&#xA;" OwnerUserId="1330" LastActivityDate="2011-01-23T13:35:19.490" CommentCount="7" />
  <row Id="18622" PostTypeId="1" CreationDate="2011-01-23T13:35:58.040" Score="2" ViewCount="126" Body="&lt;p&gt;Given $R$ an integral domain (commutative ring with no zero divisors), and $\mathfrak P$ a prime ideal in $R$, is there a relation between the field of fractions of $R$ and the field of fractions of $R/\mathfrak P$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's trivial to see that whenever $\mathfrak P$ is also maximal, then $\text{Frac}(R/P)\cong R/\mathfrak P$, but in general it would be nice if thing worked like that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) There exists at least a maximal ideal containing $P$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) There exists a maximal maximal ideal $\mathfrak M$ containing $P$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) the field of fractions of $R/\mathfrak P$ is $R/\mathfrak M$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but I'm not able to prove or disprove this...&lt;/p&gt;&#xA;" OwnerUserId="685" LastActivityDate="2011-01-23T15:44:26.990" Title="Quotient field of a quotient ring" Tags="&lt;ring-theory&gt;&lt;commutative-algebra&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="18623" PostTypeId="2" ParentId="18622" CreationDate="2011-01-23T13:52:50.170" Score="2" ViewCount="" Body="&lt;p&gt;Consider the ring ${\Bbb Z}[x]$ of polynomials with integer coefficients and its prime ideals $I=(2)$, $J=(x)$ and the maximal ideal $M=(2,x)$ containing both.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then $R/I={\Bbb Z}/2{\Bbb Z}[x]$ with quotient field ${\Bbb Z}/2{\Bbb Z}(x)$, $R/J={\Bbb Z}$ with quotient field $\Bbb Q$, and the residue field at $M$ is &#xA;${\Bbb Z}/2{\Bbb Z}$ which is a subfield of $Frac(R/I)$ but has nothing to do with $Frac(R/J)$ .........&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, if $I\subset J$ there's a canonical surjective map $R/I\rightarrow R/J$, but a surjective map of domains does NOT induce a map of fields of fractions.&lt;/p&gt;&#xA;" OwnerUserId="688" LastEditorUserId="688" LastEditorDisplayName="" LastEditDate="2011-01-23T14:06:44.700" LastActivityDate="2011-01-23T14:06:44.700" />
  <row Id="18624" PostTypeId="1" CreationDate="2011-01-23T15:13:03.380" Score="3" ViewCount="140" Body="&lt;p&gt;In Penrose's book &quot;The Road to Reality&quot; page 97 figure 5.9 he shows the values of the complex logarithm in a diagram as the intersection of two logarithmic spirals. Can you please explain how these particular spirals are derived from the definition of the logarithm?&lt;/p&gt;&#xA;" OwnerUserId="6155" LastEditorUserId="6155" LastEditorDisplayName="" LastEditDate="2011-01-23T19:09:46.720" LastActivityDate="2011-01-23T19:09:46.720" Title="Logarithm of a complex number as intersections of two logarithmic spirals" Tags="&lt;complex-analysis&gt;&lt;logarithms&gt;" AnswerCount="0" CommentCount="3" />
  <row Id="18625" PostTypeId="2" ParentId="18593" CreationDate="2011-01-23T15:34:25.747" Score="2" ViewCount="" Body="&lt;p&gt;(Edit: reading the comments left by Kevin above, Riesz extension requires a bit more than he assumed. But I think that the separating hyperplane argument would boil down to essentially the same trick as outlined below.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think you can use the following modified proof of the Riesz extension theorem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Marcel Riesz Extension Theorem&lt;/strong&gt; Let $W$ be a vector space (finite dimensional in the following proof; the actual theorem has no such limitations), and $V$ a subspace. Let $F$ be a convex cone in $W$ with the property that for every $w\in W$ there exists $v_+, v_- \in V$ such that $v_+ - w\in F$ and $w - v_- \in F$. Then for any $\phi$ linear on $V$ and positive on $V\cap F$, there exists an extension $\psi$ on $W$ that is positive on $F$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof (sketch)&lt;/strong&gt; Define $\psi_+(w) = \inf_{v-w\in F, v\in V}\phi(v)$ and $\psi_-(w) = \sup_{w-v\in F, v\in V}\phi(v)$. Check that $\psi_+$ is convex, and $\psi_-(w)$ is concave, and $\psi_+(w) \geq \psi_-(w)$. An application of the separating hyperplane theorem implies that there exists a linear map $\psi$ with $\psi_+ \geq \psi \geq \psi_-$. Since $\psi_+ = \psi_- |_V$, you have that $\psi$ is an extension. By definition $\psi_-|_F &amp;gt; 0$, and so you have the conclusion of the theorem. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(See also &lt;a href=&quot;http://terrytao.wordpress.com/2007/11/30/the-hahn-banach-theorem-mengers-theorem-and-hellys-theorem/&quot; rel=&quot;nofollow&quot;&gt;this post on Terry Tao's blog&lt;/a&gt; for some related concepts.)&lt;/p&gt;&#xA;" OwnerUserId="1543" LastEditorUserId="1543" LastEditorDisplayName="" LastEditDate="2011-01-23T15:46:53.390" LastActivityDate="2011-01-23T15:46:53.390" CommentCount="3" />
  <row Id="18627" PostTypeId="2" ParentId="18622" CreationDate="2011-01-23T15:44:26.990" Score="6" ViewCount="" Body="&lt;p&gt;With regard to the question in your first sentence, you may want to think about the example of $R = \mathbb Z$, $\mathfrak P = p \mathbb Z$ for a prime $p$,&#xA;and ask yourself what relationship (if any) there is between $\mathbb Q$ (the field of fractions of $\mathbb Z$) and $\mathbb F_p = \mathbb Z/p\mathbb Z$ (the finite field of $p$ elements).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, if $\mathfrak P$ is prime but not maximal, then the quotient&#xA;$R_{\mathfrak P}/P R_{\mathfrak P}$ (where $R_{\mathfrak P}$ is the &lt;em&gt;localization&lt;/em&gt;  of $R$ at $\mathfrak P$) is equal to the field of fractions of $R/\mathfrak P$,&#xA;and this is the typical method in commutative algebra for finding a link between&#xA;the field of fractions of $R/\mathfrak p$ and the ring $R$ itself.&lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2011-01-23T15:44:26.990" />
  <row Id="18628" PostTypeId="2" ParentId="18595" CreationDate="2011-01-23T15:50:44.040" Score="5" ViewCount="" Body="&lt;p&gt;This is slightly false.  The two are isomorphic, but not canonically so.  There is a natural pairing $\Lambda^{n-k} V \times \Lambda^k V \to \Lambda^n V$ given by exterior product, but this pairing does not identify $\Lambda^{n-k} V$ with $(\Lambda^k V)^{\ast}$ until you pick an isomorphism $\Lambda^n V \simeq k$; this implies a choice of &lt;a href=&quot;http://en.wikipedia.org/wiki/Orientation_(mathematics)&quot; rel=&quot;nofollow&quot;&gt;orientation&lt;/a&gt;, but is slightly stronger; one might say it implies a choice of &quot;volume form.&quot;  But it does not imply a choice of inner product.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The (canonical) isomorphism between $(\Lambda^k V)^{\ast}$ and $\Lambda^k V^{\ast}$ comes from the way duals commute with tensor products.  It should look pretty straightforward with a specific basis.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2011-01-23T15:50:44.040" CommentCount="1" />
  <row Id="18629" PostTypeId="2" ParentId="18614" CreationDate="2011-01-23T15:55:27.633" Score="4" ViewCount="" Body="&lt;p&gt;Dear S.L.,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regarding your comment above: yes, if $M$ is complete and connected with non-postiive curvature, then the universal cover $\widetilde{M}$ of $M$ inherits &#xA;a complete connected non-positive curvature metric (just pull-back the Riemannian&#xA;metric from $M$), and so satisfies the conditions (and so also the conclusions!) of the Cartan--Hadamard theorem.  One then finds that $\widetilde{M}$ is homeomorphic to the tangent space of any point $\widetilde{m}$ of $\widetilde{M}$ (via $\exp_{\widetilde{m}}$.  If $m$ is the image of $\widetilde{m}$ in $M$, then&#xA;the tangent space to $\widetilde{m}$ is naturally identified with the tangent space to $m$, and the exponential map $\exp_m$ is naturally identified with the composite of  $\exp_{\widetilde{m}}$ and the projection $\widetilde{M} \to M$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus indeed, one finds that $\exp_m$ is a covering map.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This circle of ideas is frequently applied, e.g. in the context of hyperbolic manifolds.  If $M$ is a compact connected hyperbolic manifold, then one finds &#xA;by Cartan--Hadamard that $\widetilde{M}$ is isometric to hyperbolic space of &#xA;the appropriate dimension, and so $M$ is isometric to a quotient of hyperbolic space by a discrete cocompact group of isometries.  (This is why hyperbolic manifold theory interacts so tightly with certain parts of group theory.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, one concludes that if $M$ is compact and connected (and positive dimensional --- i.e. not a single point!) with non-positive curvature then $\pi_1(M)$ is infinite (because, writing $M$ as a quotient of $T_m M$ via&#xA;$\exp_m$, we see that to get something compact, we have to quotient out by an&#xA;infinite group of diffeos).&lt;/p&gt;&#xA;" OwnerUserId="221" LastEditorUserId="221" LastEditorDisplayName="" LastEditDate="2011-01-23T17:25:46.410" LastActivityDate="2011-01-23T17:25:46.410" CommentCount="1" />
  <row Id="18632" PostTypeId="2" ParentId="18590" CreationDate="2011-01-23T17:12:30.360" Score="5" ViewCount="" Body="&lt;p&gt;This is probably just an honest mistake... I think the usual naturality condition says that if you have a bundle homomorphism $E \rightarrow F$ covering a map of the bases, $f:B \rightarrow D$, that is an isomorphism on fibers, then $c(E) = f^*c(F)$. At least that's what it is for the Stiefel-Whitney classes. Of course, this definition of naturality gives, obviously, that the classes are isomorphism invariant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It doesn't seem like the definition given in Bott and Tu will work (even though, in their proof, they use something stronger anyway since they use &quot;functorial properties&quot; of the Chern class, and any such property is an isomorphism invariant.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a sort of heuristic argument... Suppose we had decided to work in the category of pointed spaces, and we have a gadget $d$ that assigns, to each vector bundle, the fiber at the distinguished point of the base space. Given a map $f: X \rightarrow Y$, with a bundle over $Y$, &lt;em&gt;define&lt;/em&gt; the pullback $f^*d(E)$ to be $d(f^*E)$. Then this gadget satisfies the naturality condition, but it is not an isomorphism invariant for very silly reasons: the actual sets $d(E)$ and $d(F)$ might differ in superficial ways like the names of their elements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, I could also be missing some clever way to avoid this issue... One reason to hold onto hope is that, in my silly example, $d$ was assigning sets that lived inside a specific bundle, and could be changed if we chose a different bundle... In the Chern class case, we are assigning elements of the cohomology of the base space, and the cohomology of the base space doesn't really care about what particular bundle we have sitting over it. &lt;/p&gt;&#xA;" OwnerUserId="423" LastEditorUserId="423" LastEditorDisplayName="" LastEditDate="2011-01-24T03:43:44.000" LastActivityDate="2011-01-24T03:43:44.000" CommentCount="4" />
  <row Id="18633" PostTypeId="2" ParentId="18616" CreationDate="2011-01-23T17:41:51.923" Score="2" ViewCount="" Body="&lt;p&gt;Mathematics and juggling (figuring out how &lt;a href=&quot;http://en.wikipedia.org/wiki/Siteswap&quot; rel=&quot;nofollow&quot;&gt;siteswap notation&lt;/a&gt; works, for example).&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2011-01-23T17:41:51.923" CommentCount="1" />
  <row Id="18634" PostTypeId="1" AcceptedAnswerId="18670" CreationDate="2011-01-23T18:18:42.253" Score="10" ViewCount="323" Body="&lt;p&gt;I was told this question by a friend, who said that their friend had thought about it on and off for six months without any luck.  I have then had it for a while without any luck either.  It is in the style of a contest question, but it is unlike any other, as really nothing seems to gain any ground.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's the question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $f:[0,1]\rightarrow \mathbb{R}$ be a continuous function satisfying the following property:  If $ABC$ is the equilateral triangle with side lengths 1, we have for any point $P$ inside $ABC$, $f(\overline{AP})+f(\overline{BP})+f(\overline{CP})=0$ where $\overline{AP}$ is the distance from point P to vertex $A$.  (Example: by taking $P=A$ we see $f(0)=-2f(1)$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Prove or disprove: $f$ must be identically zero on its domain &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice that continuity is critical since otherwise I could just take the function $f(x)=0$ whenever $x\in (0,1)$ and $f(0)=1$, $f(1)=\frac{-1}{2}$.  &lt;/p&gt;&#xA;" OwnerUserId="6075" LastActivityDate="2011-01-23T23:52:33.033" Title="The Notorious Triangle Problem" Tags="&lt;analysis&gt;&lt;contest-math&gt;" AnswerCount="3" FavoriteCount="1" />
  <row Id="18635" PostTypeId="2" ParentId="18634" CreationDate="2011-01-23T18:26:25.453" Score="5" ViewCount="" Body="&lt;p&gt;This is a problem asked in the American Mathematical Monthly. May be this link could be of some help:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.jstor.org/stable/2323932&quot; rel=&quot;nofollow&quot;&gt;http://www.jstor.org/stable/2323932&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" LastActivityDate="2011-01-23T18:26:25.453" CommentCount="4" />
  <row Id="18636" PostTypeId="1" CreationDate="2011-01-23T18:26:47.747" Score="3" ViewCount="97" Body="&lt;p&gt;What study practices have led you to the best success in learning mathematics (or applied math, or theoretical CS) in the classroom (undergraduate/lower graduate level), especially in courses that would otherwise be &quot;too hard&quot; for you?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively, if you teach math, what practices have your best students devised that set them apart.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously, &quot;study hard and constantly push your knowledge on all presented material&quot; will be at the core of a lot of this, but I'm looking for something slightly more specific. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Examples of things I've found work well:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-Finding the simplest possible version of a proof that's still reasonably short, even if imprecise, before moving on to a more full and formal version&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-Studying a few core concepts of the class before entering it; it gives a complementary perspective to the instructor and gives more time for concepts to germinate in the mind&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-Do textbook problems, but only hard ones that push your knowledge. Easy textbook problems, that are basically just doing operations by rote, are a waste of time.&lt;/p&gt;&#xA;" OwnerUserId="6157" LastActivityDate="2011-01-23T18:30:10.893" Title="Best Practices for Learning Mathematics (especially in the classroom)" Tags="&lt;computer-science&gt;&lt;physics&gt;&lt;education&gt;&lt;applications&gt;" AnswerCount="1" FavoriteCount="3" />
  <row Id="18637" PostTypeId="2" ParentId="18636" CreationDate="2011-01-23T18:30:10.893" Score="2" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Moore_method&quot; rel=&quot;nofollow&quot;&gt;The Moore Method&lt;/a&gt; helped me learn various areas of math. It abides by the dictum that teaching is the best way of learning. One could modify it by pretending that there is an imaginary audience in the room and lecturing out loud (i.e. if you are by yourself this is an alternative to the traditional Moore Method).&lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2011-01-23T18:30:10.893" CommentCount="3" />
  <row Id="18638" PostTypeId="2" ParentId="18634" CreationDate="2011-01-23T18:42:13.440" Score="1" ViewCount="" Body="&lt;p&gt;Though Chandru has provided the link to the solution, I was wondering if the following approach could be of any help.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instead of choosing $P$ to coincide with a vertex, let $P$ to fall on one of the sides say $AB$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $AP = x$. Since $ABC$ is an equilateral triangle $BP = 1-x$ and $CP = \sqrt{1+x^2 - x}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So $\forall x \in [0,1]$, the function has to satisfy&#xA;$$f(x) + f(1-x) + f(\sqrt{1+x^2 - x}) = 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can we squeeze a proof that $f(x)$ has to be zero out of this?&lt;/p&gt;&#xA;" OwnerUserId="3302" LastActivityDate="2011-01-23T18:42:13.440" CommentCount="1" />
  <row Id="18640" PostTypeId="1" AcceptedAnswerId="18642" CreationDate="2011-01-23T19:03:51.223" Score="6" ViewCount="137" Body="&lt;p&gt;I'm studying Shelah's proof (actually written by Uri Abraham) that adding one generic real implies the existence of a Suslin tree (available in &lt;a href=&quot;http://www.springerlink.com/content/7u9g4u8r41303538/abstract/&quot; rel=&quot;nofollow&quot;&gt;this link&lt;/a&gt;, I think that freely for everyone.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The notion of forcing is the set finite functions from $\omega$ to $\omega$, with stronger being &quot;extending&quot;, then we construct a tree on $\omega_1$ by defining some functions by using the functions, and ensuring that the result gives us a Suslin tree.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At one point, the claim is that if $X$ is an uncountable anti-chain in the tree (in the generic extension, of course) then there exists $p$ such that $p\Vdash X$ is an uncountable anti-chain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, it says, we can find $q$ stronger than $p$ for which $Y=\{\alpha | q\Vdash\alpha\in X\}$ is uncountable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That last statement is unclear to me. I'm sensing that this is something relatively simple like a pigeonhole argument, but I'm uncertain how to deduce it.&lt;/p&gt;&#xA;" OwnerUserId="622" LastActivityDate="2011-01-23T20:02:09.487" Title="Forcing cardinality of a set" Tags="&lt;set-theory&gt;&lt;forcing&gt;" AnswerCount="1" />
  <row Id="18641" PostTypeId="2" ParentId="16384" CreationDate="2011-01-23T19:05:22.910" Score="0" ViewCount="" Body="&lt;p&gt;Given the set of all subsets of the natural numbers and the axiom of choice it does seem that one should be able to form a set by picking an element from each subset. Yet this set will have cardinality of 2^(aleph0). Given that the set of naturals only extend to omega, how could a well ordering be placed upon this set? It's elements will not be unique. I have read that the axiom of choice only applies to collections of sets that have no elements in common. So it seems then that we can only select up to the cardinality of the set that contains this collection. Given this, it seems that the naturals could not be used to order the reals because such an ordering would imply multiple lowest ranking elements. &lt;/p&gt;&#xA;" OwnerUserId="6160" LastActivityDate="2011-01-23T19:05:22.910" CommentCount="2" />
  <row Id="18642" PostTypeId="2" ParentId="18640" CreationDate="2011-01-23T19:19:33.580" Score="5" ViewCount="" Body="&lt;p&gt;Hi Asaf.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider a fixed generic extension obtained by a generic where $p$ belongs. We have that $X$ is uncountable. For each $\alpha\in X$ there is $q\in G$ that in $V$ forces $\alpha\in X$. We may assume $q\le p$ by further extending if necessary. This shows that $A_\alpha=\{q\mid q\le p\land q$ forces in $V$ that $\alpha\in X\}$ is nonempty for each $\alpha\in X$. Since Cohen forcing is countable, the same $q$ must be in uncountably many of these $A_\alpha$, and we are done.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is another way of presenting the argument that avoids talking about forcing extensions: For each $q\le p$ let $A_q$ be the set of $\alpha$ that $q$ forces to be in $X$. If each $A_q$ is countable, then (again, because Cohen forcing is countable) there is an $\alpha$ such that any ordinal forced to be in $X$ by a condition extending $p$ must be strictly below $\alpha$. But then $p$ itself must force $X$ to be contained in $\alpha$, contradicting that p forces $X$ to be uncountable. &lt;/p&gt;&#xA;" OwnerUserId="462" LastEditorUserId="462" LastEditorDisplayName="" LastEditDate="2011-01-23T20:02:09.487" LastActivityDate="2011-01-23T20:02:09.487" CommentCount="5" />
  <row Id="18643" PostTypeId="2" ParentId="18599" CreationDate="2011-01-23T19:23:29.307" Score="1" ViewCount="" Body="&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;There's general factoring of polynomials which is really not doable by humans. Given a random sequence of integer coefficients, it's just intractable (because the roots could be arbitrary complex numbers involving square roots and fractions).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Fortunately, what's usually given as exercises are polynomials that look pretty random but for teaching purposes are 'easy' (have only nice integer or fractional solutions, no roots or imaginaries). So the way to do these is to (intelligently) guess a solution $r$, then try to divide out $x - r$ from the original polynomial. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If that works then you try the same strategy again on the (necessarily) smaller degree polynomial. So really, what you're being asked to do is practice polynomial division.&lt;/li&gt;&#xA;&lt;li&gt;If that guess didn't work, try another guess. Your initial guesses come from &lt;a href=&quot;http://en.wikipedia.org/wiki/Rational_root_theorem&quot; rel=&quot;nofollow&quot;&gt;the rational root theorem&lt;/a&gt; (the first and last coefficients).&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So for the above (fixed) example $3x^3-x^2-3x+1$, you'd guess maybe $1\over 3$ (from Peter Taylor's list; I picked it because I know that it happens to be a factor). Divide through by $(3x- 1)$ (because its root is $1\over3$) and get $x^2 + 0x - 1= x^2-1$ which you already know how to factor by the quadratic formula.&lt;/p&gt;&#xA;" OwnerUserId="1919" LastActivityDate="2011-01-23T19:23:29.307" CommentCount="4" />
  <row Id="18644" PostTypeId="1" CreationDate="2011-01-23T19:33:57.433" Score="2" ViewCount="160" Body="&lt;p&gt;I could not understand a part of the book: &lt;a href=&quot;http://books.google.com/books?id=_XxZdhbtf1sC&amp;amp;printsec=frontcover&amp;amp;dq=algebraic+geometry+a+first+course&amp;amp;hl=en&amp;amp;ei=eX88TcrHIITWgQe89Y3nCA&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=book-thumbnail&amp;amp;resnum=1&amp;amp;ved=0CEIQ6wEwAA#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;&#xA;&lt;strong&gt;Algebraic geometry: a first course&lt;/strong&gt; by Joe Harris&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On page 64, line - 4, why the kernel of the map $\phi(\omega)$ will be the annihilator of the kernel of $\psi(\omega)$? Why the images of the transpose of the two maps are annihilator of each other? On page 65, line 6, why $\Theta_{\alpha, \beta}$ are quadratic? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could someone explain Plucker embedding more easily and clearly? Thank you very much.&lt;/p&gt;&#xA;" LastActivityDate="2011-01-24T19:37:04.927" Title="A question in the book: algebraic geometry: a first course" Tags="&lt;algebraic-geometry&gt;" FavoriteCount="1" />
  <row Id="18645" PostTypeId="1" CreationDate="2011-01-23T20:11:23.410" Score="1" ViewCount="153" Body="&lt;p&gt;Prove that if a differentiable curve g:[0,1] -&gt; C (complex plane) parametrizes counterclockwise the boundary of an open set O in C, then under suitable conditions Area of O is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1/(2i) * itegral(conjugate of z) dz computed over the boundary.&lt;/p&gt;&#xA;" OwnerUserId="6161" LastActivityDate="2011-01-23T20:32:36.513" Title="Complex Line Integral" Tags="&lt;complex-analysis&gt;" AnswerCount="1" />
  <row Id="18646" PostTypeId="1" AcceptedAnswerId="18648" CreationDate="2011-01-23T20:27:01.977" Score="1" ViewCount="281" Body="&lt;p&gt;I know that the product of two Gaussians is a Gaussian, and I know that the convolution of two Gaussians is also a Gaussian. I guess I was just wondering if there's a proof out there to show that the convolution of two Gasussians is a Gaussian. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="6162" LastActivityDate="2011-01-23T21:03:25.853" Title="Convolution of two gaussians is a gaussian" Tags="&lt;homework&gt;" AnswerCount="2" CommentCount="9" />
  <row Id="18647" PostTypeId="2" ParentId="18645" CreationDate="2011-01-23T20:32:36.513" Score="1" ViewCount="" Body="&lt;p&gt;Write your contour integral out explicitly using the parameterization $g(t)$ and then use Green's theorem. &lt;/p&gt;&#xA;" OwnerUserId="3035" LastActivityDate="2011-01-23T20:32:36.513" CommentCount="5" />
  <row Id="18648" PostTypeId="2" ParentId="18646" CreationDate="2011-01-23T20:33:24.920" Score="1" ViewCount="" Body="&lt;p&gt;Fourier Transform will help you out to conclude that the convolution is also a gaussian. &lt;/p&gt;&#xA;" OwnerUserId="3302" LastActivityDate="2011-01-23T20:33:24.920" CommentCount="3" />
  <row Id="18649" PostTypeId="1" AcceptedAnswerId="18650" CreationDate="2011-01-23T20:35:42.157" Score="0" ViewCount="186" Body="&lt;p&gt;Let F be a σ-algebra of subsets of Ω and B ∈ F. Show that&#xA;E = {A ∩ B : A ∈ F } is a σ-algebra of subsets of B. Is it still true&#xA;when B is a subset of Ω that does not belong to F?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks and regards!&lt;/p&gt;&#xA;" OwnerUserId="2076" LastActivityDate="2011-01-23T23:24:26.310" Title="Sigma algebra proof problem" Tags="&lt;probability&gt;&lt;probability-theory&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="18650" PostTypeId="2" ParentId="18649" CreationDate="2011-01-23T20:42:45.363" Score="1" ViewCount="" Body="&lt;p&gt;Your first question can be done by showing that $E$ obeys the definitions of $\sigma$ algebra.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the second case, take $\mathcal{F}$ to be the the trivial $\sigma$ algebra and $B$ to be a non-empty proper subset of $\Omega$. Then $E$ is not a $\sigma$ algebra since it doesn't contain $\Omega$ nor $B^c$&lt;/p&gt;&#xA;" OwnerUserId="3302" LastActivityDate="2011-01-23T20:42:45.363" CommentCount="6" />
  <row Id="18652" PostTypeId="2" ParentId="18646" CreationDate="2011-01-23T20:57:28.483" Score="1" ViewCount="" Body="&lt;p&gt;See &lt;a href=&quot;http://en.wikipedia.org/wiki/Sum_of_normally_distributed_random_variables&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; for two common alternatives.&lt;/p&gt;&#xA;" OwnerUserId="2810" LastEditorUserId="2810" LastEditorDisplayName="" LastEditDate="2011-01-23T21:03:25.853" LastActivityDate="2011-01-23T21:03:25.853" CommentCount="1" />
  <row Id="18653" PostTypeId="1" CreationDate="2011-01-23T21:13:17.660" Score="0" ViewCount="83" Body="&lt;p&gt;Show that the set of all orthogonal matrices in the set of all n cross n matrices  endowed with any norm topology is complact&lt;/p&gt;&#xA;" OwnerUserId="6165" LastActivityDate="2011-01-23T21:38:32.243" Title="Compactness Of Orthogonal Matrices" Tags="&lt;complex-numbers&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="18654" PostTypeId="1" AcceptedAnswerId="18666" CreationDate="2011-01-23T21:20:20.263" Score="0" ViewCount="168" Body="&lt;p&gt;I have this problem. I want to prove that if I'm choosing a basis for polynomials that their exponent is from $1$ until $n$ and they are all $F(2)=0$, they are linear independent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To be more spefic, let's have a vector space which contains $N$ function, first is with $1$ degree and so on, all are $0$ in $x=2$, why should they be independent??&lt;/p&gt;&#xA;&#xA;&lt;p&gt;thanks!&lt;/p&gt;&#xA;" OwnerUserId="6163" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-11T17:47:19.520" LastActivityDate="2011-02-11T17:47:19.520" Title="linearly independent Functions in linear algebra" Tags="&lt;linear-algebra&gt;&lt;vector-space&gt;" AnswerCount="2" CommentCount="5" />
  <row Id="18655" PostTypeId="2" ParentId="18653" CreationDate="2011-01-23T21:38:32.243" Score="2" ViewCount="" Body="&lt;p&gt;Recall a compact subset of $R^{n \times n}$ is a set that is closed and bounded. One way to show closedness is to observe that the orthogonal matrices are the inverse image of the element $I$ under the continuous map $M \rightarrow MM^T$. Boundedness follows for example from the fact that each column or row is a vector of magnitude $1$. &lt;/p&gt;&#xA;" OwnerUserId="3035" LastActivityDate="2011-01-23T21:38:32.243" />
  <row Id="18656" PostTypeId="2" ParentId="18654" CreationDate="2011-01-23T21:40:14.777" Score="0" ViewCount="" Body="&lt;p&gt;Assuming your answer to my comment ment that your functions are of the form $(x-2),(x-2)^2,(x-2)^3,\ldots$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, pick any finite subset of your set of polynomials. We may assume that this finite subset is of the form $(x-2),(x-2)^2,\ldots,(x-2)^k$ by adding finitely many more functions. So assume we have a dependence relation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a_1(x-2)+a_2(x-2)^2+\ldots+a_k(x-2)^k=0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Differentiating $k-1$ times, we get $(k-1)!a_k(x-2)=0$, and choosing $x=1$ we get $a_k=0$. Continuing this way, we get $a_i=0$ for all $i$. Thus your set of functions is linearly endependent (over the underlying field). &lt;/p&gt;&#xA;" OwnerUserId="4284" LastEditorUserId="4284" LastEditorDisplayName="" LastEditDate="2011-01-23T22:26:35.123" LastActivityDate="2011-01-23T22:26:35.123" CommentCount="3" />
  <row Id="18657" PostTypeId="1" AcceptedAnswerId="18665" CreationDate="2011-01-23T22:01:31.023" Score="2" ViewCount="81" Body="&lt;p&gt;A regular local ring is a domain. Is a regular ring (a ring whose localization at every prime ideal is regular) also a domain? I am unable to find/construct a proof or a counterexample. Any help would be appreciated.&lt;/p&gt;&#xA;" OwnerUserId="6166" LastActivityDate="2011-01-23T22:50:51.420" Title="Is a regular ring a domain" Tags="&lt;commutative-algebra&gt;" AnswerCount="1" />
  <row Id="18658" PostTypeId="2" ParentId="18616" CreationDate="2011-01-23T22:01:42.607" Score="2" ViewCount="" Body="&lt;p&gt;Though not an actual activity, most of the videos by &lt;a href=&quot;http://www.youtube.com/user/Vihart&quot; rel=&quot;nofollow&quot;&gt;Vihart&lt;/a&gt; are entertaining enough to watch and get even more interesting once you know the mathematics behind it.&lt;/p&gt;&#xA;" OwnerUserId="6122" LastActivityDate="2011-01-23T22:01:42.607" />
  <row Id="18659" PostTypeId="1" AcceptedAnswerId="18667" CreationDate="2011-01-23T22:05:47.840" Score="2" ViewCount="240" Body="&lt;p&gt;I'm looking for a fast algorithm for generating all the partitions of an integer up to a certain maximum length; ideally, I don't want to have to generate &lt;em&gt;all&lt;/em&gt; of them and then discard the ones that are too long, as this will take around 5 times longer in my case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Specifically, given $L = N(N+1)$, I need to generate all the partitions of $L$ that have at most $N$ parts. I can't seem to find any algorithms that'll do this directly; all I've found that seems relevant is &lt;a href=&quot;http://www.springerlink.com/content/754j3m2171k76730/&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; paper, which I unfortunately can't seem to access via my institution's subscription. It &lt;a href=&quot;http://www.site.uottawa.ca/~ivan/F49-int-part.pdf&quot; rel=&quot;nofollow&quot;&gt;apparently&lt;/a&gt; documents an algorithm that generates the partitions of each individual length, which could presumably be easily adapted to my needs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anyone know of any such algorithms?&lt;/p&gt;&#xA;" OwnerUserId="67" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2011-01-24T16:37:16.550" LastActivityDate="2011-03-21T21:24:16.277" Title="Algorithm for generating integer partitions" Tags="&lt;combinatorics&gt;&lt;algorithms&gt;&lt;partitions&gt;" AnswerCount="4" CommentCount="3" />
  <row Id="18660" PostTypeId="2" ParentId="18589" CreationDate="2011-01-23T22:11:32.013" Score="32" ViewCount="" Body="&lt;p&gt;Using the fact that $\mathbb{Z}[\varphi]$ is a unique factorization domain in which we can decompose&#xA;$$&#xD;&#xA;x^5+y^5=(x+y)(x^2+y^2-\varphi xy)(x^2+y^2-\bar\varphi xy),\quad\qquad{\rm(1)}&#xD;&#xA;$$&#xA;we can give a proof of Fermat's Last Theorem for the case of exponent 5. Here, I am using a bar over a number to denote conjugation, so $\varphi=(1+\sqrt{5})/2$ and $\bar\varphi=(1-\sqrt{5})/2$.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Theorem&lt;/strong&gt;: There are no solutions to $x^5+y^5=z^5$ for nonzero $x,y,z$ in $\mathbb{Z}[\varphi]$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That is, for exponent 5, FLT holds in the ring $\mathbb{Z}[\varphi]$ and, in particular, it holds in the integers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Before going any further, let's note a few facts about factorization in $\mathbb{Z}[\varphi]$. As is well known, it is norm-Euclidean, so is a unique factorization domain. We have the prime factorizations $5=(\sqrt{5})^2$ and $11=q\bar q$, where I am setting $q=4-\sqrt{5}$ (for the remainder of this post). The identity $\varphi\bar\varphi=-1$ shows that $\varphi$ is a unit. In fact, it is a fundamental unit, so that every unit in $\mathbb{Z}[\varphi]$ is of the form $\pm\varphi^r$ for integer $r$. It will also be useful to use mod-q arithmetic (with $q$ as above). Then, $\varphi=(1+\sqrt{5})/2=8$ (mod q). Therefore every element of the quotent $\mathbb{Z}[\varphi]/(q)$ is equal to a rational integer mod q. As 11 = 0 mod q, this gives $\mathbb{Z}[\varphi]/(q)\cong\mathbb{Z}/(11)$. So, mod-q arithmetic in $\mathbb{Z}[\varphi]$ is exactly the same as mod-11 arithmetic in the integers. In particular, every 5'th power is equal to one of $0,1,-1$ mod q. Applying this to the equation $x^5+y^5=z^5$ shows that at least one of $x,y,z$ must have a factor of q. By dividing through by their highest common factor, we reduce to the case where $x,y,z$ are coprime, so exactly one is a multiple of q. Rearranging as $(-z)^5+y^5=(-x)^5$ if necessary, we can always bring the multiple of q to the right hand side. This reduces the problem to the following.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Theorem 2&lt;/strong&gt;: There are no solutions to $x^5+y^5=uz^5$ for nonzero coprime $x,y,z\in\mathbb{Z}[\varphi]$ with $u\in\mathbb{Z}[\varphi]$ a unit and $q$ dividing $z$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Let's prove this by showing that, if we have one solution, then we can find another solution for which $xyz$ has strictly fewer distinct prime factors. Applied to a minimal solution, this would give a contradiction. This is essentially the method of &lt;a href=&quot;http://en.wikipedia.org/wiki/Infinite_descent&quot; rel=&quot;nofollow&quot;&gt;descent&lt;/a&gt; used by Fermat himself for the case of exponent 4.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, suppose we have one solution. Writing $c_0=x+y$, $c_1=x^2+y^2-\varphi xy$ and $c_2=x^2+y^2-\bar\varphi xy$, (1) gives the decomposition $uz^5=c_0c_1c_2$. Also,&#xA;$$&#xD;&#xA;c_0^2-\bar\varphi c_1-\varphi c_2=0.\qquad\qquad{\rm(2)}&#xD;&#xA;$$&#xA;We would like to show that the factors $c_0,c_1,c_2$ are 5'th powers, which will be easier if they are coprime. Using the fact that $x,y$ are coprime to $z$, the identities&#xA;$$&#xD;&#xA;\begin{align}&#xD;&#xA;&amp;amp;c_0^2-c_1=\sqrt{5}\varphi xy,\\&#xD;&#xA;&amp;amp;c_0^2-c_2=-\sqrt{5}\bar\varphi xy,\\&#xD;&#xA;&amp;amp;c_1-c_2=-\sqrt{5}xy&#xD;&#xA;\end{align}&#xD;&#xA;$$&#xA;show that the highest common factor of $c_0^2,c_1,c_2$ is either 1 or $\sqrt{5}$. Consider the case where $\sqrt{5}$ divides $z$. Then it will also divide at least one of $c_i$, and the identities above show that it divides each $c_i$. In particular, 5 divides $c_0^2$, so the identities above show that $\sqrt{5}$ divides each of $c_1,c_2$ exactly once.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case where $z$ is not a multiple of $\sqrt{5}$, let us set $\tilde c_0=c_0^2,\tilde c_1=c_1,\tilde c_2=c_2$ and, in the case where $\sqrt{5}$ divides $z$, set $\tilde c_0=c_0^2/\sqrt{5},\tilde c_1=c_1/\sqrt{5},\tilde c_2=c_2/\sqrt{5}$. These are coprime and&#xA;$$&#xD;&#xA;\tilde c_0\tilde c_1^2\tilde c_2^2 = u^2\left(z^{2}/\sqrt{5}^{m}\right)^5&#xD;&#xA;$$&#xA;where $m=0$ if $\sqrt{5}$ does not divide $z$ and $m=1$ if it does. As they are coprime, each prime factor of $z$ divides exactly one of the $\tilde c_i$, and its exponent is a multiple of 5. So, considering prime factorizations, each $\tilde c_i$ is equal to a unit multiplied by a fifth power $w_i^5$. So, (2) gives&#xA;$$&#xD;&#xA;u_0w_0^5+u_1w_1^5+u_2w_2^5=0&#xD;&#xA;$$&#xA;for units $u_i$. Without loss of generality, we assume that $q$ divides $w_0$ and, dividing through by $-u_1$ if necessary, we suppose that $u_1=-1$. Then, $u_2=\pm1$ mod q. However, being a unit, we have $u_2=\pm(\varphi)^r=\pm 8^r$ (mod q), and, looking at this mod 11, 5 must divide $r$. So, $u_2$ is a fifth power and, by absorbing $-u_2^{1/5}$ into $w_2$, we can take $u_2=-1$. So we have arrived at&#xA;$$&#xD;&#xA;w_1^5+w_2^5=u_0w_0^5.&#xD;&#xA;$$&#xA;Also, all prime factors of $w_0w_1w_2$ are factors of $z$. So, except in the case where $x,y$ are units, we have a solution with strictly fewer prime factors, and we are done.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So suppose that we have a solution to Theorem 2. Iteratively applying the procedure above will keep generating new solutions and, as the number of prime factors of $xyz$ cannot decrease indefinitely, we must eventually settle on the case where $x,y$ are units, so that $x/y=\pm\varphi^r$. Exchanging $x,y$ if necessary, we suppose that $r &amp;gt; 0$. Then, $q^5$ is a factor of $1\pm\varphi^{5r}$. Using the identity $\varphi^5=-1+\varphi^4q$, and applying the binomial identity, it can be seen that $rq$ must be a multiple of $q^{5}$, so $r$ is a multiple of $11^4$. In particular, $\vert x/y\vert=\vert\varphi\vert^r$ will be very large (note, $\vert\varphi\vert^{11^4} &gt; 10^{3000}$). Then, the definitions above for $c_0^2,c_1,c_2$ are dominated by the $x^2$ terms, so the ratios $\tilde c_i/\tilde c_j$ are close to one. Going through these details bounds the ratios $w_i/w_j$ and, in particular, none of them will be as large as $\vert\varphi\vert^{11^4}$. This means that we cannot have $x,y$ and $w_1,w_2$ all units. So, continuing the induction will generate solutions with ever fewer prime factors, giving the required contradiction.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;This method of approaching FLT for exponent 5 was something I came up with after seeing the exponent 3 case in lectures years ago. It is a bit tedious having to separately deal with the case where $x,y$ are units. Maybe that can be tidied up. Essentially, the reason why this method works is because $\mathbb{Z}[\varphi]$ consists precisely of the real algebraic integers of the cyclotomic field $\mathbb{Q}(\zeta_5)$.&lt;/p&gt;&#xA;" OwnerUserId="1321" LastEditorUserId="1321" LastEditorDisplayName="" LastEditDate="2011-01-24T02:59:20.177" LastActivityDate="2011-01-24T02:59:20.177" CommentCount="7" />
  <row Id="18661" PostTypeId="1" AcceptedAnswerId="18664" CreationDate="2011-01-23T22:36:21.517" Score="2" ViewCount="87" Body="&lt;p&gt;I just wrote a &lt;a href=&quot;http://philosopherdeveloper.wordpress.com/2011/01/23/overwhelmed-by-choice/&quot; rel=&quot;nofollow&quot;&gt;blog post&lt;/a&gt; and wasn't sure how to word a particular sentence. Say I have the following function:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;f(x) = x^2&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then I can say that the value of &lt;em&gt;f(x)&lt;/em&gt; grows &lt;strong&gt;quadratically&lt;/strong&gt; with &lt;em&gt;x&lt;/em&gt;*. Similarly with this function:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;f(x) = e^x&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;...I could say that &lt;em&gt;f(x)&lt;/em&gt; grows &lt;strong&gt;exponentially&lt;/strong&gt; with &lt;em&gt;x&lt;/em&gt;. But what about this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;f(x) = x!&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do I say that &lt;em&gt;f(x)&lt;/em&gt; grows &quot;factorially&quot;? What's the proper term?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;*I worded this wrong at first. It should be right now. &lt;strike&gt;Wait, is that even right? Or would &quot;exponentially&quot; imply &lt;em&gt;f(x) = k&lt;sup&gt;x&lt;/sup&gt;&lt;/em&gt;? Should the first term be &quot;quadratically&quot; instead?&lt;/strike&gt;&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="6167" LastEditorUserId="6167" LastEditorDisplayName="" LastEditDate="2011-01-23T23:03:37.130" LastActivityDate="2011-01-23T23:03:37.130" Title="How do I describe the growth of something that scales by a factorial?" Tags="&lt;terminology&gt;&lt;factorial&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="18662" PostTypeId="1" AcceptedAnswerId="18663" CreationDate="2011-01-23T22:38:37.287" Score="0" ViewCount="135" Body="&lt;p&gt;Heya everyone,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm currently learning the chain, product and quotient rules of differenentiation. Only I'm lacking some exercises. I'm looking for a website, or something that's available online with some more difficult examples of those rules combined. For example things like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Find the derivative of:&#xA;$$\ln\frac{(x+1)^2}{x^2+1}$$&#xA;and stuff like: Find the derivative of:&#xA;$\sin(e^x)\tan(x^2)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm just lacking some of these Exercises, does anyone know a website + answers?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cheers!&lt;/p&gt;&#xA;" OwnerUserId="3605" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2011-01-24T00:29:06.627" LastActivityDate="2011-01-24T00:29:06.627" Title="Chain, product, quotient rule exercises: ($\ln\frac{(x+1)^2}{x^2+1}$ etc)" Tags="&lt;calculus&gt;" AnswerCount="1" />
  <row Id="18663" PostTypeId="2" ParentId="18662" CreationDate="2011-01-23T22:42:08.147" Score="1" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://www.math.ubc.ca/~karstenc/Teaching/M184_Prac2.pdf&quot; rel=&quot;nofollow&quot;&gt;These&lt;/a&gt;(a list of derivatives) look relevant.&lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2011-01-23T22:42:08.147" />
  <row Id="18664" PostTypeId="2" ParentId="18661" CreationDate="2011-01-23T22:42:25.120" Score="3" ViewCount="" Body="&lt;p&gt;Using Stirling's formula,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$n! \times e^n \approx Cn^{n + 1/2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not sure if there is a name for that kind of growth. It is super-exponential and might be enough to get the point across, I suppose.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;btw, $f(x) = x^2$ is said to grow &lt;em&gt;quadratically&lt;/em&gt;, not exponentially.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2011-01-23T22:42:25.120" CommentCount="7" />
  <row Id="18665" PostTypeId="2" ParentId="18657" CreationDate="2011-01-23T22:50:51.420" Score="3" ViewCount="" Body="&lt;p&gt;No.  E.g. choose two regular domains and take their product; this will be regular, but not a domain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is more or less the general case, as I will explain:&#xA;In general, a Noetherian ring, all of whose localizations at its prime ideals are domains, is a finite product of domains (and of course a finite product of domains has this localization property).  So a regular Noetherian ring will be a finite product of regular domains (and conversely any such product will be regular).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Geometrically, one can think of this as follows: regularity of $A$ is a local property on Spec $A$, and Spec $A\times B$ is equal to Spec $A \coprod$ Spec $B$.&#xA;So locally Spec $A\times B$ looks like either Spec $A$ or Spec $B$.  In particular,&#xA;local properties, such as regularity (or the condition that the localization at prime ideals be a domain) can't detect global properties (like $A$ itself &#xA;being a domain).&lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2011-01-23T22:50:51.420" CommentCount="1" />
  <row Id="18666" PostTypeId="2" ParentId="18654" CreationDate="2011-01-23T22:51:22.527" Score="3" ViewCount="" Body="&lt;p&gt;The first thing you need to do is make it clear what vector space you are looking for. You cannot determine if a set is a basis unless you know &lt;em&gt;exactly&lt;/em&gt; what the vector space in question is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, it is not the same to work in the vector space of all polynomials of degree at most $n$ with coefficients in $\mathbb{R}$ than to work in the vector space of &lt;em&gt;all&lt;/em&gt; polynomials with coefficients in $\mathbb{R}$; those two vector spaces are also different from the vector space of all &lt;em&gt;continuous&lt;/em&gt; functions $f\colon\mathbb{R}\to\mathbb{R}$ (with pointwise addition and scalar multiplication), which in turn is different from the vector space of all differentiable functions, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But then it seems like you are not asking for a basis, but rather how does one check if a given collection of functions is linearly independent. The answer may again depend on the vector space (different scalars can change the answer; for instance, if you consider the complex numbers as a vector space over $\mathbb{R}$, then $1$ and $i$ are linearly independent, because the only solution to $a1+bi$ with $a,b\in\mathbb{R}$ is $a=b=0$; but if you consider $\mathbb{C}$ as a vector space over itself, then $1$ and $i$ are &lt;em&gt;not&lt;/em&gt; linearly independent, because if we can pick complex scalars then $a1+bi=0$ has the solution $a=i$ and $b=-1$). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, assuming you know what your vector space is: suppose $\mathbf{f}_1,\ldots,\mathbf{f}_n$ are your functions. You need to know whether there are any nonzero solutions to&#xA;$$a_1\mathbf{f}_1 + \cdots a_n\mathbf{f}_n = 0$$&#xA;&lt;em&gt;as functions&lt;/em&gt;. That is, the value at all $x$ in the domain should be zero. At that point, there are any number of ways of doing it, depending on the precise nature of the functions. You assume you have that equality, and through any valid means (algebra, calculus, etc) you want to show that you must have $a_1=\cdots=a_n=0$; if this is the case, they are linearly independent. If there is at least one way to obtain the zero &lt;em&gt;function&lt;/em&gt; with not all $a_1,\ldots,a_n$ equal to $0$, then they are not linearly independent. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you care to be a bit more precise as to what your problem is, I and others may be able to provide more specific hints as opposed to generalities.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-23T22:59:06.600" LastActivityDate="2011-01-23T22:59:06.600" />
  <row Id="18667" PostTypeId="2" ParentId="18659" CreationDate="2011-01-23T23:10:55.643" Score="3" ViewCount="" Body="&lt;p&gt;You can do it recursively. Let $f(n, maxcount, maxval)$ return the list of partitions of $n$ containing no more than $maxcount$ parts and in which each part is no more than $maxval$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $n = 0$ you return a single list containing the empty partition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $n &amp;gt; maxcount * maxval$ you return the empty list.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $n = maxcount * maxval$ you return a single list consisting of the obvious solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Otherwise you make a series of recursive calls to $f(n - x, maxcount - 1, x)$.&lt;/p&gt;&#xA;" OwnerUserId="5676" LastActivityDate="2011-01-23T23:10:55.643" CommentCount="6" />
  <row Id="18668" PostTypeId="1" CreationDate="2011-01-23T23:13:36.873" Score="3" ViewCount="135" Body="&lt;p&gt;Hi all,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found this question in a book (Topology II: homotopy and homology: classical manifolds)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Show that the quotient space $X = S^2 \times S^2 / [(x_1,x_2) \sim (Rx_1,Rx_2)]$ where R is the reflection in the equatorial plane, is homeomorphic to $S^4$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am still in the process of learning topology and I really don't think I can prove this result (or even understand a proof if someone were generous enough to provide me with one). I apologize in advance if this is trivial, but it will be of great help if someone could give a homeomorphism. I would really like to use the mapping of the space $X$ to $S^4$ in my work. If you are aware of the proof and would provide it in your answer, I will definitely make an effort to understand it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you for your time. &lt;/p&gt;&#xA;" OwnerUserId="3050" LastActivityDate="2011-01-24T18:58:50.203" Title="Mapping from X to $S^4$" Tags="&lt;algebraic-topology&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="4" />
  <row Id="18669" PostTypeId="2" ParentId="18649" CreationDate="2011-01-23T23:24:26.310" Score="1" ViewCount="" Body="&lt;p&gt;Both questions can be checked by the definition of the $\sigma$-algebra. For the example by @Sivaram (when $B\notin{\cal F}$), ${\cal E} = \{\emptyset, B\}$, which is still a (trivial) $\sigma$-algebra.&lt;/p&gt;&#xA;" OwnerUserId="5836" LastActivityDate="2011-01-23T23:24:26.310" />
  <row Id="18670" PostTypeId="2" ParentId="18634" CreationDate="2011-01-23T23:52:33.033" Score="6" ViewCount="" Body="&lt;p&gt;Chandru has provided the link to a complete solution, but as Eric notes, it is somewhat involved. Some months ago, being unaware of that solution, I had found &lt;a href=&quot;http://www.reddit.com/r/CasualMath/comments/cw6uh/the_triangle_problem/c0vsvgv?context=1&quot; rel=&quot;nofollow&quot;&gt;a less involved proof&lt;/a&gt; if you assume that $f$ is not only continuous but also differentiable. I reproduce that proof below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $\triangle$ be the set of points inside or on the triangle, and $a, b, c : \triangle \to [0,1]$ be scalar fields on $\triangle$ mapping any point $P$ to the distances $\overline{AP}$, $\overline{BP}$, $\overline{CP}$ respectively. Define the scalar field $$g := f \circ a + f \circ b + f \circ c = 0.$$ It is easily shown that $\nabla a$ is a unit vector pointing away from $A$, and similarly for $b$ and $c$. Then $$\nabla g = f&amp;#39;(a) \nabla a + f&amp;#39;(b) \nabla b + f&amp;#39;(c) \nabla c,$$ where $f&amp;#39;(x) = df/dx$. But $g$ is constant, so $\nabla g = 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take $P$ to be a point on the side $AB$. Then $\nabla a$ and $\nabla b$ are antiparallel, while $\nabla c$ is linearly independent. For their linear combination to be $0$, the coefficient of $\nabla c$, which is $f&amp;#39;(c)$, must be $0$. Since $P$ is an arbitrary point on $AB$, $c$ is any value between $\sqrt{3}/2$ and $1$, so over that range $f&amp;#39;$ is zero, and $f$ is constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once we've shown that $f&amp;#39;$ is zero between $\sqrt{3}/2$ and $1$, for any other value $a$ it's fairly easy to pick a point inside the triangle that's $a$ units from $A$ and over $\sqrt{3}/2$ units from $C$. Now $\nabla g = f&amp;#39;(a) \nabla a + f&amp;#39;(b) \nabla b + f&amp;#39;(c) \nabla c = 0$. The last term drops out because $f&amp;#39;(c) = 0$. $\nabla a$ and $\nabla b$ are linearly independent in the interior of the triangle, so their coefficients $f&amp;#39;(a)$ and $f&amp;#39;(b)$ must be zero. Thus $f&amp;#39; = 0$ everywhere, so $f$ is constant. In particular, $f(1/\sqrt{3}) = 0$, so $f = 0$ everywhere.&lt;/p&gt;&#xA;" OwnerUserId="856" LastActivityDate="2011-01-23T23:52:33.033" CommentCount="3" />
  <row Id="18671" PostTypeId="1" CreationDate="2011-01-23T23:59:07.370" Score="1" ViewCount="100" Body="&lt;p&gt;Let O be a non-empty subset of C (the complex plane), and let g: O-&gt;C be a holomorphic function that satisfies the differential equation g''(z) - z*g(z) = 0 for every z in O. Prove that integral([g(x)]^2)dx over U has the same value for all differentiable curves U in O with the same endpoints.&lt;/p&gt;&#xA;" OwnerUserId="6161" LastEditorUserId="6161" LastEditorDisplayName="" LastEditDate="2011-01-24T00:23:38.737" LastActivityDate="2011-01-24T00:23:38.737" Title="Complex derivatives of line integrals" Tags="&lt;complex-analysis&gt;" CommentCount="2" />
  <row Id="18673" PostTypeId="2" ParentId="18668" CreationDate="2011-01-24T00:49:02.427" Score="2" ViewCount="" Body="&lt;p&gt;I think it might be helpful to restrict your $x_1$ coordinate to be on or above the equator.  As long as $x_1$ isn't on the equator, this gives a unique representative for every point, so you get $\{x_1\}\times S^2$.  When $x_1$ is on the equator, you can apply $R$ freely to the second coordinate, so you get $\{x_1\}\times D^2$.  The puzzle is how to fit these together...&lt;/p&gt;&#xA;" OwnerUserId="2159" LastActivityDate="2011-01-24T00:49:02.427" CommentCount="2" />
  <row Id="18674" PostTypeId="1" AcceptedAnswerId="18675" CreationDate="2011-01-24T01:07:46.127" Score="1" ViewCount="162" Body="&lt;p&gt;&lt;strong&gt;1&lt;/strong&gt;.Let f be a real-valued differentiable function defined on [0, 1]. If $f(0)=0$ and&#xA;$f(1)=1$, prove that there exists two numbers $x_1,x_2 \in [0, 1]$ such that&#xA;$\frac{1}{f&amp;#39;(x_1)}+\frac{1}{f&amp;#39;(x_2)}=2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;2&lt;/strong&gt;.Let f be a real-valued differentiable function defined on (0,$\infty$). If $\lim_{x\rightarrow0^+}f(x)=\lim_{x\rightarrow\infty}f(x)=1$, prove that there exists $c\in(1,\infty)$ such that $f&amp;#39;(c)=0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;1&lt;/strong&gt;.From the question's statement, I wasn't sure if $x_1$ could equal $x_2$. If it can, is the following proof valid?&#xA;Clearly, it suffices to show $f&amp;#39;(x)=1$ for some $x \in [0,1]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We proceed by contradiction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose $f&amp;#39;(x)\neq1$ for all $x \in [0,1]$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then either $f&amp;#39;(x)&amp;gt;1$ or $f&amp;#39;(x)&amp;lt;1$ for all relevant x (otherwise, $\exists x_1,x_2$ s.t. $f&amp;#39;(x_1)&amp;gt;1, f&amp;#39;(x_2)&amp;lt;1$ which contradicts the IVT for derivatives).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume, wlog, $f&amp;#39;(x)&amp;gt;1$ for all x.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then $1=f(1)=\int_{0}^{1}f&amp;#39;(x)dx&amp;gt;\int_{0}^{1}dx=1,$ contradiction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;2&lt;/strong&gt;. Again we proceed by contradiction; suppose $f&amp;#39;(x)\neq0$ for all x.&#xA;By an identical argument to the one used in &lt;strong&gt;1&lt;/strong&gt;, either $f&amp;#39;(x)&amp;gt;0$ or $f&amp;#39;(x)&amp;lt;0$ for all x.&#xA;Wlog, assume $f&amp;#39;(x)&amp;gt;0$. Then f(x) is strictly increasing $\forall x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, if x&gt;1, $1=\lim_{t\rightarrow0^+}f(t)&amp;lt;f(1)&amp;lt;f(x)$.&#xA;Hence $\lim_{x\rightarrow\infty}f(x)&amp;gt;1$ (if it exists), contradiction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, if my proofs are valid, are there simpler proofs? Second, can someone provide me with a proof of the case where $x_1\neq x_2$ in &lt;strong&gt;1&lt;/strong&gt;?&lt;/p&gt;&#xA;" OwnerUserId="5959" LastActivityDate="2011-01-24T17:16:38.277" Title="$\frac{1}{f'(x_1)}+\frac{1}{f'(x_2)}=2$" Tags="&lt;real-analysis&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="18675" PostTypeId="2" ParentId="18674" CreationDate="2011-01-24T01:12:56.183" Score="2" ViewCount="" Body="&lt;p&gt;For (1) I think you could use the Mean Value Theorem. Namely, there exists an $x_1 \in (0,1)$ such that $f&amp;#39;(x_1) = \frac{f(1)-f(0)}{1-0} = 1$. Then put $x_2 = x_1$. For (2), you could use the &lt;a href=&quot;http://en.wikipedia.org/wiki/Rolle%27s_theorem#Generalization&quot; rel=&quot;nofollow&quot;&gt;generalization of Rolle's Theorem&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2548" LastEditorUserId="2548" LastEditorDisplayName="" LastEditDate="2011-01-24T01:34:54.947" LastActivityDate="2011-01-24T01:34:54.947" CommentCount="2" />
  <row Id="18676" PostTypeId="2" ParentId="18668" CreationDate="2011-01-24T01:15:02.813" Score="1" ViewCount="" Body="&lt;p&gt;I don't have enough rep points to respond directly to Eric, but that should not be the case, since $S^4$ doesn't have a boundary either. I think you're thinking of $S^2 / (x \tilde{} Rx) \times S^2 / (x \tilde{} Rx),$ which is not exactly the same space as $X$.&lt;/p&gt;&#xA;" OwnerUserId="6004" LastEditorUserId="6004" LastEditorDisplayName="" LastEditDate="2011-01-24T01:25:19.170" LastActivityDate="2011-01-24T01:25:19.170" CommentCount="1" />
  <row Id="18677" PostTypeId="1" CreationDate="2011-01-24T01:36:47.583" Score="3" ViewCount="102" Body="&lt;p&gt;I was working on something on Matlab couple of months back and found a tweak to speed up Gauss elimination on Matlab by dividing the original matrix into 4 block matrices and then solving them in a block-wise fashion. I wrote a quick article on it couple of days back and it is now on arxiv &lt;a href=&quot;http://arxiv.org/pdf/1101.4081v1&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am however wondering about couple of things.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;What is the reason for the optimal &quot;split parameter&quot; to be $\approx \frac{2}{3}$? Is there a heuristic or quick argument for this?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Why is the number of additions independent of how we block partition the matrix?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="3302" LastActivityDate="2011-01-24T04:25:21.297" Title="Speeding up Gauss Elimination" Tags="&lt;linear-algebra&gt;" AnswerCount="1" />
  <row Id="18678" PostTypeId="1" AcceptedAnswerId="18679" CreationDate="2011-01-24T02:12:16.767" Score="1" ViewCount="155" Body="&lt;p&gt;If we are given some linearly dependent vectors, would the T of those vectors necessarily be dependent (given a transformation from $R^n$ to $R^p$)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And if we are given some linearly independent vectors, would T of those vectors necessarily be independent (given a transformation from $R^n$ to $R^p$)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answers are presumably no and no, but I am struggling to figure out why. Thanks!&lt;/p&gt;&#xA;" OwnerUserId="5157" LastActivityDate="2011-01-24T02:56:41.193" Title="Linear independency before and after Linear Transformation." Tags="&lt;linear-algebra&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="18679" PostTypeId="2" ParentId="18678" CreationDate="2011-01-24T02:16:48.500" Score="5" ViewCount="" Body="&lt;p&gt;If a set of vectors is &lt;em&gt;dependent&lt;/em&gt;, there is a non-trivial combination of some of them that equals 0: $$a_1v_1+\dots+a_n v_n=0,$$ where not all the scalars $a_i$ are $0$. Linearity of $T$ should give you at once that the $Tv_i$ are also linearly dependent (as witnessed by the same $a_i$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Linear independence, on the other hand, does not need to be preserved. For example, consider the linear transformation that maps all the vectors to 0.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Now, under some additional conditions, a linear transformation may preserve independence. For example, suppose that $T$ is injective (i.e., the only solution to $Tv=0$ is $v=0$). Then $T$ preserves linear independence: Given $Tv_1,\dots,T_n,v_n$ are dependent and scalars $b_1,\dots,b_n$, suppose that $$b_1Tv_1+\dots+b_nTv_n=0.$$ By linearity, you get $T(b_1v_1+\dots+b_nv_n)=0$, and if $T$ is injective, then in fact $b_1v_1+\dots+_nv_n=0$. But, since the $v_i$ are independent, we must have $b_1=\dots=b_n=0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Injectivity of $T$ is the only way to ensure that &lt;em&gt;any&lt;/em&gt; independent set is mapped to an independent set. However, even if $T$ is not injective, there may be some independent sets whose independence is preserved. For example, consider the linear transformation $T:{\mathbb R}^3\to{\mathbb R}^2$ given by $T(x,y,z)=(x,y)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then $T$ maps both $(1,0,0)$ and $(1,0,1)$ to $(1,0)$. The original vectors are independent, the resulting vectors obviously are not (they coincide!). But, on the other hand, $T$ preserves the linear independence of any two (independent) vectors whose last coordinate is 0.&lt;/p&gt;&#xA;" OwnerUserId="462" LastEditorUserId="462" LastEditorDisplayName="" LastEditDate="2011-01-24T02:31:12.200" LastActivityDate="2011-01-24T02:31:12.200" />
  <row Id="18680" PostTypeId="1" AcceptedAnswerId="18681" CreationDate="2011-01-24T02:26:50.833" Score="1" ViewCount="93" Body="&lt;p&gt;This is in reference to page 153 of the notes found here&#xA;&lt;a href=&quot;http://www.math.lsa.umich.edu/~hochster/614F10/614.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.math.lsa.umich.edu/~hochster/614F10/614.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the second paragraph, the author is considering inverse limit of a family of subsets of a certain set where the defining maps of the inverse systems are inclusions. I am not sure why the inverse limit is the intersection of the subsets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For e.g. consider a set $X$ and four subsets $X_1,X_2,X_3,X_4$ s.t. $X_1\subset X_2$ and $X_3\subseteq X_4$, but $X_2\cap X_4=\emptyset$. Now, by the general construction of the inverse limit in the category of sets, we look at the elements of $X_1\times X_2\times X_3\times X_4$ s.t. any element in the second coordinate is also in the first coordinate and any element in the fourth coordinate is also in the third. So, the inverse limit is not empty, but the intersection is empty. What am I missing here?&lt;/p&gt;&#xA;" OwnerUserId="6166" LastActivityDate="2011-01-24T02:38:55.950" Title="Inverse limit in the category of sets" Tags="&lt;category-theory&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="18681" PostTypeId="2" ParentId="18680" CreationDate="2011-01-24T02:38:55.950" Score="1" ViewCount="" Body="&lt;p&gt;In those notes it is assumed that the system is directed.  In your example that means that one of the $X_i$ must be a subset of $X_2\cap X_4=\emptyset$.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2011-01-24T02:38:55.950" CommentCount="1" />
  <row Id="18682" PostTypeId="2" ParentId="18678" CreationDate="2011-01-24T02:39:36.453" Score="0" ViewCount="" Body="&lt;p&gt;Suppose $v_1,\cdots v_n$ are linearly independent and T is a linear transformation.  Suppose ker(T) only intersects linear span $W$ of $v_1, \cdots v_n$ only at $\{0\}$.  Then T preserves the linear independence of $v_1, \cdots v_n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This condition is necessary and sufficient.&lt;/p&gt;&#xA;" OwnerUserId="4154" LastActivityDate="2011-01-24T02:39:36.453" />
  <row Id="18683" PostTypeId="2" ParentId="18677" CreationDate="2011-01-24T02:40:29.243" Score="5" ViewCount="" Body="&lt;p&gt;Volker Strassen amazed the algorithmic world in 1969 by producing an $O(n^{\log_2 7})$ algorithm for matrix multiplication, based on a similar idea but applied recursively. The algorithm can also be adapted to matrix inversion and so to the solving of linear equations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since then, researchers have produced even better algorithms. The best known so far is by Coppersmith and Winograd (1987), who achieve $O(n^{2.376})$. Recent work suggests that matrix multiplication is actually possible in $O_\epsilon(n^{2+\epsilon})$ for any $\epsilon &amp;gt; 0$. Usually this is stated as $\omega = 2$, where $\omega$ is the infimum of all exponents $k$ such that matrix multiplication is possible in time $O(n^k)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Matrix multiplication is actually equivalent to matrix inversion, in the sense that they both have the same $\omega$. The proof can be found in the well-known CLRS.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All these algorithms - even Strassen's - are impractical (the constants hidden inside the big O are huge), and so never actually used. However, &quot;blocking&quot; is a wide-spread technique, since accessing memory &quot;locally&quot; is more efficient (due to the way caches work).&lt;/p&gt;&#xA;" OwnerUserId="1277" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2011-01-24T04:25:21.297" LastActivityDate="2011-01-24T04:25:21.297" CommentCount="4" />
  <row Id="18684" PostTypeId="2" ParentId="18678" CreationDate="2011-01-24T02:56:41.193" Score="0" ViewCount="" Body="&lt;p&gt;You want to be a bit careful with the statements; the main difficulty lies in how you deal with collections of sets that include repetitions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most of the time, when we think about vectors and vector spaces, a &lt;em&gt;list&lt;/em&gt; of vectors that includes repetitions is considered to be linearly dependent, even though as a &lt;em&gt;set&lt;/em&gt; it may technically not be. For example, in $\mathbb{R}^2$, the &lt;em&gt;list&lt;/em&gt; $\mathbf{v}_1=(1,0)$, $\mathbf{v}_2=(1,0)$ is considered linearly dependent, because $\mathbf{v}_1 - \mathbf{v}_2 = \mathbf{0}$. But as sets, $\{\mathbf{v}_1,\mathbf{v}_2\} = \{(1,0)\} = S$, you have that $S$ is linearly independent (it contains a single element and the element is nonzero).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;If&lt;/em&gt; you are considering lists, and repetitions are allowed (and make a list linearly dependent), then Andres's answer is completely correct. If you are considering &lt;em&gt;sets&lt;/em&gt;, and repetitions are an issue, then the problem is much more difficult. I suspect this is not the case, since it is &lt;em&gt;almost never&lt;/em&gt; the case; but you might want to double check (and try to figure how you can have a linearly dependent set map to a set of vectors that is linearly independent when considered as a set, where repetitions are ignored). &lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2011-01-24T02:56:41.193" />
  <row Id="18685" PostTypeId="1" CreationDate="2011-01-24T03:49:35.303" Score="0" ViewCount="76" Body="&lt;p&gt;$$dX = \kappa_x (\theta_x - X)dt + \sigma_x \sqrt{X} \,dW_x$$&#xA;$$dY = \kappa_y (\theta_y - Y)dt + \sigma_y \sqrt{Y} \,dW_y$$&#xA;$$dW_x dW_y = \rho\, dt$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;we know that $X$ and $Y$ are marginally distributed as non-central $\chi^2$. What is the joint distribution of $X$ and $Y$ ?&lt;/p&gt;&#xA;" OwnerUserId="6151" LastActivityDate="2011-01-24T03:49:35.303" Title="square root processes with correlated deriving Brownian motion" Tags="&lt;stochastic-processes&gt;" />
  <row Id="18686" PostTypeId="1" AcceptedAnswerId="18695" CreationDate="2011-01-24T04:01:54.707" Score="3" ViewCount="116" Body="&lt;p&gt;Suppose you have an arbitrary triangle with vertices $A$, $B$, and $C$. &lt;a href=&quot;http://www.cs.princeton.edu/~funk/tog02.pdf&quot; rel=&quot;nofollow&quot;&gt;This paper (section 4.2)&lt;/a&gt; says that you can generate a random point, $P$, uniformly from within triangle $ABC$ by the following convex combination of the vertices:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$P = (1 - \sqrt{r_1}) A + (\sqrt{r_1} (1 - r_2))  B + (r_2 \sqrt{r_1}) C$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $r_1, r_2 \sim U[0, 1]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do you prove that the sampled points are uniformly distributed within triangle $ABC$?&lt;/p&gt;&#xA;" OwnerUserId="4841" LastActivityDate="2011-01-24T13:41:36.827" Title="uniform random point in triangle" Tags="&lt;random&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="18688" PostTypeId="1" AcceptedAnswerId="18781" CreationDate="2011-01-24T04:20:41.007" Score="6" ViewCount="368" Body="&lt;p&gt;I want to ask if there is a slick way to prove:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{j=0}^n{(-1)^j{{n}\choose{j}}\left(1-\frac{j}{n}\right)^n}=\frac{n!}{n^n}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know Yuval has given a proof, but that one is not direct. I am requesting for a direct algebraic proof of this identity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="6016" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2011-01-24T20:29:24.253" LastActivityDate="2011-02-01T12:52:11.427" Title="Proof of a combination identity:$\sum_{j=0}^n{(-1)^j{{n}\choose{j}}\left(1-\frac{j}{n}\right)^n}=\frac{n!}{n^n}$" Tags="&lt;combinatorics&gt;" AnswerCount="6" FavoriteCount="0" />
  <row Id="18689" PostTypeId="2" ParentId="18688" CreationDate="2011-01-24T04:24:36.930" Score="4" ViewCount="" Body="&lt;p&gt;This is inclusion-exclusion for counting the number of onto maps from $n$ to $n$.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2011-01-24T04:24:36.930" CommentCount="1" />
  <row Id="18690" PostTypeId="1" AcceptedAnswerId="18704" CreationDate="2011-01-24T04:26:45.410" Score="5" ViewCount="210" Body="&lt;p&gt;I'm well aware of the combinatorial variant of the proof, i.e. noting that each formula is a different representation for the number of subset of a set of $n$ elements. I'm curious though if there's a series of algebraic manipulations that can lead from &#xA;$\sum_{i=0}^n \binom{n}{i}$ to $2^n$.&lt;/p&gt;&#xA;" OwnerUserId="1930" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2011-01-24T05:38:01.273" LastActivityDate="2011-04-08T21:01:10.630" Title="Algebraic Proof that $\sum_{i=0}^n \binom{n}{i}=2^n$" Tags="&lt;combinatorics&gt;&lt;binomial-coefficients&gt;" AnswerCount="3" CommentCount="5" FavoriteCount="1" />
  <row Id="18691" PostTypeId="1" CreationDate="2011-01-24T04:26:52.980" Score="0" ViewCount="173" Body="&lt;p&gt;I know this is kind of minor. But I want to know if I understand this correctly. In &quot;baby rudin&quot;, on page 42, is the condition&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$3^{-m}\lt \frac{\beta-\alpha}{6}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;tight? I thought &#xA;$$3^{-m}\lt \frac{\beta-\alpha}{4}$$&#xA;is sufficient, am I correct?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit:&#xA;Okay, the background is: the Cantor set (i.e. the union of all intervals after repeatedly removing all the middle third interval) does not contain any point lying in the segment&lt;p&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(\frac{3k+1}{3^m},\frac{3k+2}{3^m})$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And he was saying every segment $(\alpha, \beta)$ contains a segment of the form above, if&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$3^{-m}\lt \frac{\beta-\alpha}{6}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I thought this condition is too loose.&lt;/p&gt;&#xA;" OwnerUserId="6016" LastEditorUserId="6016" LastEditorDisplayName="" LastEditDate="2011-01-24T05:01:29.037" LastActivityDate="2011-01-24T13:42:51.613" Title="a very small question in baby rudin" Tags="&lt;real-analysis&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="18692" PostTypeId="2" ParentId="18690" CreationDate="2011-01-24T04:30:51.667" Score="4" ViewCount="" Body="&lt;p&gt;Simply use the binomial formula.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(a + b)^n = \sum_{k=0}^n {n \choose k} a^k b^{n - k}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With $a = b = 1$ you have your result.&lt;/p&gt;&#xA;" OwnerUserId="6174" LastEditorUserId="1919" LastEditorDisplayName="" LastEditDate="2011-04-08T21:01:10.630" LastActivityDate="2011-04-08T21:01:10.630" />
  <row Id="18693" PostTypeId="2" ParentId="18690" CreationDate="2011-01-24T04:32:37.763" Score="3" ViewCount="" Body="&lt;p&gt;Well, here is one. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{i=0}^n \binom{n}{i}=2^n$$&#xA;$$\sum_{i=0}^n \binom{n}{i}+\sum_{i=0}^n \binom{n}{i}=2^{n+1}$$&#xA;$$\binom{n}{0}+\left [ \binom{n}{0}+\binom{n}{1} \right ]+...+\left [ \binom{n}{n-1}+\binom{n}{n}\right ]+\binom{n}{n}=2^{n+1}$$&#xA;$$\sum_{i=0}^{n+1} \binom{n+1}{i}=2^{n+1}$$&lt;/p&gt;&#xA;" OwnerUserId="4728" LastEditorUserId="4728" LastEditorDisplayName="" LastEditDate="2011-01-24T04:43:40.007" LastActivityDate="2011-01-24T04:43:40.007" CommentCount="2" />
  <row Id="18694" PostTypeId="1" AcceptedAnswerId="18698" CreationDate="2011-01-24T04:34:38.117" Score="1" ViewCount="135" Body="&lt;p&gt;I have been on this one for hours, cant figure out how to write this in the proper format/wording.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $f$ and $g$ be functions from $\mathbb{R}$ to $\mathbb{R}$. For the sum and product of $f$ and $g$, determine which statements are true. If true provide a proof; if false, provide a counter example.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;If $f$ and $g$ are bounded, then $f + g$ is bounded.&lt;/li&gt;&#xA;&lt;li&gt;If $f$ and $g$ are bounded, then $fg$ is bounded.&lt;/li&gt;&#xA;&lt;li&gt;If both $f + g$ and $fg$ are bounded, then $f$ and $g$ are bounded.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Any help is appreciated, thanks :)&lt;/p&gt;&#xA;" OwnerUserId="5505" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-24T04:38:53.650" LastActivityDate="2011-01-24T04:52:28.640" Title="Proof Functions and Sets" Tags="&lt;functions&gt;&lt;set-theory&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="18695" PostTypeId="2" ParentId="18686" CreationDate="2011-01-24T04:37:01.203" Score="3" ViewCount="" Body="&lt;p&gt;I would argue that if it is true for any triangle, it is true for all of them, as we can find an affine transformation between them.  So I would pick my favorite triangle, which is $A=(0,0), B=(1,0), C=(0,1)$.  Then the point is $(\sqrt{r_1}(1-r_2),r_2\sqrt{r_1})$ and we need to prove it is always within the triangle and evenly distributed.  To be in the triangle we need $x,y\ge 0, x+y\le 1$, which is clear.  Then show that the probability to be within an area $(0,x) \times (0,y)$ is $2xy$ by integration.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2011-01-24T13:41:36.827" LastActivityDate="2011-01-24T13:41:36.827" CommentCount="2" />
  <row Id="18696" PostTypeId="2" ParentId="18694" CreationDate="2011-01-24T04:44:59.013" Score="1" ViewCount="" Body="&lt;p&gt;For a and b, you have that $f(x)\le f_{max}$ and similarly for g.  What does that tell you about $f(x)+g(x)$ and $f(x)g(x)$?  c is harder-it is easy for one of the conditions like $f(x)=-g(x)$ for the sum and $f(x)=0$ for the product, but for both together?&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-01-24T04:44:59.013" />
  <row Id="18697" PostTypeId="2" ParentId="18694" CreationDate="2011-01-24T04:46:27.613" Score="2" ViewCount="" Body="&lt;p&gt;Remember the definition of &lt;em&gt;bounded&lt;/em&gt;: A function $h\colon\mathbb{R}\to\mathbb{R}$ is &lt;em&gt;bounded&lt;/em&gt; if and only if there exist number $M,N\gt 0$ such that for all $r\in\mathbb{R}$, $-N \leq f(r) \leq M$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Intuitively, the graph of $f$ goes neither &quot;too high&quot; nor &quot;too low&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, suppose that $f$ and $g$ are both bounded. That means that there the values of $f$ don't get bigger than some $M_1\gt 0$, and the values of $g$ don't get bigger than some $M_2\gt 0$. When you take $(f+g)(r) = f(r)+g(r)$, how big can the number $f(r)+g(r)$ be, given that $f(r)\leq M_1$ and $g(r)\leq M_2$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How about discussing how &lt;em&gt;small&lt;/em&gt; they can be, given that you know there are numbers $N_1$ and $N_2$ such that $f(r)\geq N_1$ and $g(r)\geq N_2$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How about the product? That's a bit harder because of the signs, so try working with $|fg|$, $|f|$, and $|g|$. Notice that if $-N\leq f(r)\leq M$, then $0\leq f(r) \leq \max\{M,N\}$ (verify this!).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Number $3$ is a bit trickier... first question: can the sum of two functions, each of which is &lt;em&gt;not&lt;/em&gt; bounded, be bounded? Yes: consider for instance $f(x) = x$, which is not bounded, and $g(x) = 1-x$, which is also not bounded. What is $f+g$?  Of course, this example does &lt;em&gt;not&lt;/em&gt; satisfy that $fg$ is bounded (here, $fg(x) = x-x^2$, which is unbounded). And it's easy to come up with unbounded examples with their &lt;em&gt;product&lt;/em&gt; bounded (just make sure that whenever $f$ gets out of control, $g$ &quot;cancels it out&quot;). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps you may want to check that if $ff$ (the product of $f$ with itself) is bounded, then so is $f$, and then consider Theo Buehler's answer.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-24T04:51:51.273" LastActivityDate="2011-01-24T04:51:51.273" CommentCount="2" />
  <row Id="18698" PostTypeId="2" ParentId="18694" CreationDate="2011-01-24T04:46:51.380" Score="3" ViewCount="" Body="&lt;p&gt;a) That $f$ is bounded means that $|f(x)| \leq C$ for all $x$. Similarly $|g(x)| \leq D$ for all $x$. Therefore the triangle inequality gives $|f(x) + g(x)| \leq |f(x)| + |g(x)| \leq C + D$ for all $x$ and hence $f+g$ is bounded by $C + D$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;b) is similar but easier&lt;/p&gt;&#xA;&#xA;&lt;p&gt;c) $f^2 + g^2 = |f^2 + g^2| = |(f + g)^{2} - 2fg| \leq |f + g|^2 + 2|fg|$ and the last term is bounded by assumption and a) and b).&lt;/p&gt;&#xA;" OwnerUserId="5363" LastEditorUserId="5363" LastEditorDisplayName="" LastEditDate="2011-01-24T04:52:28.640" LastActivityDate="2011-01-24T04:52:28.640" CommentCount="2" />
  <row Id="18699" PostTypeId="1" CreationDate="2011-01-24T04:55:25.773" Score="2" ViewCount="94" Body="&lt;p&gt;I have the following statement in a paper:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $\Psi$ be the formal power series defined over the alphabet $\Omega$ and the log semiring by: $(\Psi, (a, b)) = -log(c((a,b)))$ for $(a,b) \in \Omega$, and let $S$ be the formal pwer series $S$ over the log semiring defined by: $S=\Omega^*+\Psi+\Omega^*$  (an alphabet is a finite set of symbols and $\Omega$ contains pairs of such symbols.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It continues: $S$ is a rational power series as a +-product and closure of the polynomial power series $\Omega$ and $\Psi$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What exactly is meant here? I know about the automata theoretical aspects, but I haven't heard of the notion &quot;formal power series&quot; over an alphabet and a semiring. How can I think of this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The paper is at &lt;a href=&quot;http://www.cs.nyu.edu/~mohri/pub/&quot; rel=&quot;nofollow&quot;&gt;http://www.cs.nyu.edu/~mohri/pub/&lt;/a&gt; , [93] and [99] (virtually identical), page 15 and page 17 in the 22-page paper. It is a verbatim copy.&lt;/p&gt;&#xA;" OwnerUserId="6176" LastEditorUserId="6176" LastEditorDisplayName="" LastEditDate="2011-01-24T05:17:10.187" LastActivityDate="2011-01-24T05:17:10.187" Title="Formal Power Series -- what's in it?" Tags="&lt;power-series&gt;" AnswerCount="1" CommentCount="6" FavoriteCount="1" />
  <row Id="18701" PostTypeId="2" ParentId="18699" CreationDate="2011-01-24T05:13:31.610" Score="1" ViewCount="" Body="&lt;p&gt;A forum power series is a &quot;Symbolic series&quot; and not necessarily one where the object the series is over is a &quot;number&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The idea comes from group there where you have the operations of multiplication and addition in a ring. you can form the expressions a*x + b*x^2 + c*x^3, etc... e.g., the polynomials or P[x] but note x may not necessarily be in the ring. We can still write such expressions and take x as a sort of place holder or symbol&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Polynomial_ring&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Polynomial_ring&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Formal_power_series&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Formal_power_series&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The idea is an extension from the idea of a &quot;normal&quot; power series but applied to a &quot;formal object&quot;. Those two links should clue you in on the exact meaning.&lt;/p&gt;&#xA;" OwnerUserId="6174" LastActivityDate="2011-01-24T05:13:31.610" CommentCount="2" />
  <row Id="18702" PostTypeId="1" AcceptedAnswerId="18705" CreationDate="2011-01-24T05:35:06.560" Score="2" ViewCount="229" Body="&lt;p&gt;Is there any $\sigma$-algebra that is strictly between the Borel $\sigma$-algebra and the Lebesgue $\sigma$-algebra? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How about not in between the two, but in general, are there any other $\sigma$ algebra(s)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What can be concluded about measure too, e.g. is Lebesgue measure the only measure for Lebesgue $\sigma$ algebra?&lt;/p&gt;&#xA;" OwnerUserId="6016" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-24T05:41:31.097" LastActivityDate="2011-02-04T21:23:43.590" Title="between Borel $\sigma$ algebra and Lebesgue $\sigma$ algebra, are there any other $\sigma$ algebra?" Tags="&lt;real-analysis&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="2" />
  <row Id="18703" PostTypeId="2" ParentId="18599" CreationDate="2011-01-24T05:35:54.757" Score="1" ViewCount="" Body="&lt;p&gt;One thing you may always want to try is the rational root theorem (http://en.wikipedia.org/wiki/Rational_root_theorem), if your polynomial has integer coefficients.  This is already touched on by the above answers.  While this isn't a general trick, it may be able to help you find factors, and potentially then reduce this to something that you can easily recognize and factor.  &lt;/p&gt;&#xA;" OwnerUserId="2546" LastActivityDate="2011-01-24T05:35:54.757" CommentCount="1" />
  <row Id="18704" PostTypeId="2" ParentId="18690" CreationDate="2011-01-24T05:37:43.163" Score="5" ViewCount="" Body="&lt;p&gt;Here's one.  Let $g(n) = \sum_{i=0}^n \binom{n}{i}$.  Then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$g(n+1) - g(n) = \sum_{i=0}^{n+1} \binom{n+1}{i} - \sum_{i=0}^n \binom{n}{i} = \sum_{i=0}^{n+1} \left(\binom{n+1}{i} - \binom{n}{i}\right) = \sum_{i=0}^{n+1} \binom{n}{i-1} $$&#xA;$$= \sum_{i=0}^n \binom{n}{i} = g(n).$$&#xA;Here, we use the fact that $\binom{n}{n+1} = \binom{n}{-1} = 0$, as well as the binomial recurrence $\binom{n+1}{i} = \binom{n}{i} + \binom{n}{i-1}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus we have $g(n+1) = 2g(n)$, with $g(0) = 1$.  Since $g(n)$ doubles each time $n$ is incremented by 1, we must have &#xA;$$g(n) = \sum_{i=0}^n \binom{n}{i} = 2^n.$$&lt;/p&gt;&#xA;" OwnerUserId="2370" LastActivityDate="2011-01-24T05:37:43.163" CommentCount="3" />
  <row Id="18705" PostTypeId="2" ParentId="18702" CreationDate="2011-01-24T05:48:06.140" Score="4" ViewCount="" Body="&lt;p&gt;First question (title): Sure. It is not hard to show that the sets of the form $B \cup S$, where $B$ is Borel and $S$ is a subset of the Cantor-set constitute a $\sigma$-algebra. There are $2^{\mathfrak{c}}$ subsets of the Cantor set but only $\mathfrak{c}$ Borel sets, hence this $\sigma$-algebra lies strictly between the Borel sets and the Lebesgue-measurable sets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's very rare that there is no $\sigma$-algebra strictly between two $\sigma$-algebras, it essentially means that there are atoms (sets that cannot be written non-trivially as union of proper subsets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure I understand the last question. You can take for instance Lebesgue measure plus a Dirac measure if you want something strictly different (i.e. not related via Radon-Nikodym).&lt;/p&gt;&#xA;" OwnerUserId="5363" LastActivityDate="2011-01-24T05:48:06.140" CommentCount="2" />
  <row Id="18706" PostTypeId="2" ParentId="18691" CreationDate="2011-01-24T05:48:22.687" Score="1" ViewCount="" Body="&lt;p&gt;I would agree that if $3^{-m}\lt \frac{\beta-\alpha}{4}$ there will be three intervals in $(\alpha, \beta)$ of length $3^{-m}$, one of which will be of the form $(\frac{3k+1}{3^m},\frac{3k+2}{3^m})$.  But you have tight vs. loose backwards.  Rudin's condition, with 6 in the denominator, may force a larger $m$ than your condition with the 4 (or they may be the same).  So your condition is looser (it may allow more values of $m$) than Rudin's.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2011-01-24T13:42:51.613" LastActivityDate="2011-01-24T13:42:51.613" />
  <row Id="18707" PostTypeId="2" ParentId="18702" CreationDate="2011-01-24T05:55:15.163" Score="2" ViewCount="" Body="&lt;p&gt;Your first question has been answered by Theo Buehler; for your second question, you can always take $\mathcal{P}(\mathbb{R})$ as a $\sigma$-algebra; a nontrivial example is the $\sigma$-algebra of all subsets of $\mathbb{R}$ that are either countable or co-countable (that is, all $X$ for which either $|X|\leq\aleph_0$ or $|\mathbb{R}-X|\leq\aleph_0$; it is straightforward to verify this is a $\sigma$-algebra).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For your last question: if you place no restrictions whatsoever on the measure, then the answer is trivial: yes, there are other measures. For a trivial example, just scale the Lebesgue measure by a factor different from $1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More generally, given a measurable positive function $f$, define the measure $\mu$ on the Lebesgue measurable sets by $\mu(X) = \int_X f d\lambda = \int f\chi_X d\lambda$, where $\lambda$ is the Lebesgue measure and $\chi_X$ is the characteristic function of $X$. This is a measure:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;If $X$ is any measurable set, then $\mu(X) = \int_X f\,d\lambda \geq 0$ because $f(x)\geq 0$ for all $x$, so $\mu$ is nonnegative.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If $\{E_i\}$ is a countable collection of pairwise disjoint sets, then $$\mu(\cup E_i) = \int_{\cup E_i}f\,d\lambda = \sum_{i=1}^{\infty} \int _{E_i} f\,d\lambda = \sum_{i=1}^{\infty}\mu(E_i)$$&#xA;(with measures and sum possibly infinite); and&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$\mu(\emptyset) = \int_{\emptyset}f\,d\lambda = 0$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Thus, $\mu$ is a measure on the Lebesgue $\sigma$-algebra; unless $f(x)=1$ for almost all $x$, you have $\mu\neq\lambda$. It may even be a &lt;em&gt;finite&lt;/em&gt; measure, if $f\in\mathcal{L}^1$. &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-04T21:23:43.590" LastActivityDate="2011-02-04T21:23:43.590" />
  <row Id="18708" PostTypeId="2" ParentId="18691" CreationDate="2011-01-24T06:02:59.287" Score="1" ViewCount="" Body="&lt;p&gt;Yes you're right. The condition $\frac{\beta - \alpha}{6}$ is too strong and $\frac{\beta - \alpha}{4}$ is sufficient.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The length of the interval $I_{k} = (\frac{3k+1}{3^m},\frac{3k+2}{3^m})$ is precisely $\frac{1}{3^{m}}$. Of four consecutive intervals at least one is of this form and entirely contained in an interval $(\alpha,\beta)$ of length $&amp;gt; \frac{4}{3^{m}}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose $\alpha = \frac{1}{3^{m}} + \varepsilon$ with $\varepsilon$ very small. Therfore $(\alpha,\beta)$ does not contain $I_{0}$ The next interval of the form $I_{k}$ is $I_{1} = (\frac{4}{3^{m}},\frac{5}{3^{m}})$. It follows that $\beta$ must be at least $\frac{5}{3^{m}}$, so $\beta - \alpha$ should be at least $\frac{4}{3^{m}}$. And this gives your bound $3^{-m} &amp;lt; \frac{\beta - \alpha}{4}$. All other cases are similar but easier.&lt;/p&gt;&#xA;" OwnerUserId="5363" LastActivityDate="2011-01-24T06:02:59.287" />
  <row Id="18710" PostTypeId="1" AcceptedAnswerId="18711" CreationDate="2011-01-24T07:05:56.483" Score="8" ViewCount="233" Body="&lt;p&gt;The arithmetic mean of $y_i ... y_n$ is: $$\frac{1}{n}\sum_{i=1}^n~y_i $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a smooth function $f(x)$, we can find the arithmetic mean of $f(x)$ from $x_0$ to $x_1$ by taking $n$ samples and using the above formula. As $n$ tends to infinity, it becomes an integration: $$\int_{x_0}^{x_1} f(x)~dx \over x_1 - x_0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, the geometric mean of $y_i ... y_n$ is: $$\left( \prod_{i=1}^n~{y_i}\right) ^{1/n}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, we can find the geometric mean of $f(x)$ by taking $n$ samples.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Here is my question: As $n$ tends to infinity, what do we call the resultant mathematical object? The geometric integration?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The geometric mean and the arithmetic mean, along with the quadratic mean (root mean square), the harmonic mean, etc, are special case of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Generalized_mean&quot; rel=&quot;nofollow&quot;&gt;generalized mean&lt;/a&gt; (with $p=0,1,2,-1$, respectively).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left( \frac{1}{n}\sum_{i=1}^n~x_i^p\right) ^{1/p}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do we have a generalized integration for different values of $p$?&lt;/p&gt;&#xA;" OwnerUserId="4237" LastEditorUserId="4237" LastEditorDisplayName="" LastEditDate="2011-01-24T09:31:12.017" LastActivityDate="2011-01-24T13:46:41.437" Title="What is to geometric mean as integration is to arithmetic mean?" Tags="&lt;integral&gt;&lt;arithmetic&gt;&lt;average&gt;&lt;products&gt;" AnswerCount="3" FavoriteCount="2" />
  <row Id="18711" PostTypeId="2" ParentId="18710" CreationDate="2011-01-24T07:33:02.553" Score="10" ViewCount="" Body="&lt;p&gt;You don't need to introduce a new concept for this. The geometric mean of $y_i$ is nothing but $\exp$ of the arithmetic mean of $\log y_i$, and this generalizes in the straightforward way to integration: $$\exp\left(\frac{\int_{x_0}^{x_1} \log f(x)dx}{\int_{x_0}^{x_1} dx}\right).$$ You can do the same thing with the generalized mean, replacing $\log$ and $\exp$ with raising to the power of $p$ and $1/p$ respectively. I'm not aware of whether this has a specific name, but it is quite similar to the concept of the &lt;a href=&quot;http://mathworld.wolfram.com/Lp-Space.html&quot; rel=&quot;nofollow&quot;&gt;$L^p$ norm&lt;/a&gt; of a function.&lt;/p&gt;&#xA;" OwnerUserId="856" LastActivityDate="2011-01-24T07:33:02.553" CommentCount="1" />
  <row Id="18712" PostTypeId="2" ParentId="18710" CreationDate="2011-01-24T07:37:37.757" Score="0" ViewCount="" Body="&lt;p&gt;I do not know what you mean by generalized integration as integration is simply the concept of adding things up in a special way. It is not much different than normal summation. If I gave you some type of answer in terms of integrals it wouldn't be a generalization because it is in terms of integrals(it wouldn't be a new concept).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now if your asking about some type of analog then you could come up with (1/L*int(f(x)^p))^(1/p) and this is actually used(almost at least)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Lp_space&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Lp_space&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as I know there is no thing that is analogous to products as integrals are to sums. But Lp spaces deal a lot with the sum you have given because of the way the metric is defined. i.e., your sum is simply the distance in these spaces(well, scaled at least).&lt;/p&gt;&#xA;" OwnerUserId="6174" LastActivityDate="2011-01-24T07:37:37.757" />
  <row Id="18713" PostTypeId="1" AcceptedAnswerId="18714" CreationDate="2011-01-24T08:01:27.720" Score="3" ViewCount="215" Body="&lt;p&gt;Here is a quote from Derbyshire's &lt;em&gt;&quot;Unknown Quantity&quot;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The rational numbers are &quot;dense.&quot; This means that between any two of them you can always find another one.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This being a pop. math book, this is not meant as a formal definition of dense sets but it got me wondering whether the dense subsets of $\mathbb{R}$ could be defined in this way. (It is clearly insufficient by itself since the empty set would be dense!)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I wonder, is there a nice way to fill in the following statement:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$S \subset \mathbb{R}$ is dense iff $(\forall x,y \in S, x &amp;lt; y)(\exists z \in S)(x &amp;lt; z &amp;lt; y)$ and ...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4764" LastActivityDate="2011-01-24T18:19:33.443" Title="Defining dense subsets of $\mathbb{R}$" Tags="&lt;real-analysis&gt;" AnswerCount="4" CommentCount="9" FavoriteCount="1" />
  <row Id="18714" PostTypeId="2" ParentId="18713" CreationDate="2011-01-24T08:19:33.163" Score="5" ViewCount="" Body="&lt;p&gt;It seems this line of definition leaves open the possibility of &quot;holes&quot; in $S$, even if one adds the condition that $S$ is unbounded in both directions. Consider for example any union of &lt;strong&gt;open&lt;/strong&gt; intervals. Or the intersection of the set of rationals with such a union.&lt;/p&gt;&#xA;" OwnerUserId="6179" LastActivityDate="2011-01-24T08:19:33.163" />
  <row Id="18715" PostTypeId="2" ParentId="18713" CreationDate="2011-01-24T08:20:54.863" Score="2" ViewCount="" Body="&lt;p&gt;A better definition of denseness would be that &quot;between any two points of the &lt;em&gt;complement&lt;/em&gt; there is a point of the set&quot;. You will have no difficulty writing this down formally with quantifiers:&#xA;$$&#xD;&#xA;\forall x&amp;lt;y\in S^C\;\exists z\in S...&#xD;&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="3212" LastEditorUserId="3212" LastEditorDisplayName="" LastEditDate="2011-01-24T09:24:36.847" LastActivityDate="2011-01-24T09:24:36.847" />
  <row Id="18716" PostTypeId="2" ParentId="18090" CreationDate="2011-01-24T08:23:36.293" Score="3" ViewCount="" Body="&lt;p&gt;See &lt;a href=&quot;http://mathoverflow.net/questions/53011/proof-of-equivalent-definitions-of-recurrent-states-for-markov-chains&quot; rel=&quot;nofollow&quot;&gt;this page&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; Since incorrect statements continue to pop up on this page, let me explain once again the situation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, two notations (standard in the field): for every state $i$, $P_i$ denotes the distribution of the Markov chain $(X_n)_{n\ge0}$ conditionally on the event $[X_0=i]$ and $T_i=\inf\{n\ge1;X_n=i\}$ denotes the first hitting time of $i$. Next, a tautology: the events $[\exists n\ge1,X_n=i]$ and $[T_i&amp;lt;+\infty]$ are equal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second, a statement of the question the OP is interested in. One considers two properties which may or may not be satisfied by a given Markov chain $(X_n)_{n\ge0}$ and a given state $i$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) $P_i(\exists n\ge1,X_n=i)=1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) For every state $j$, either $P_i(T_j&amp;lt;+\infty)=0$ or $P_i(T_j&amp;lt;+\infty)=1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Property (1) is often taken as a definition of the fact that $i$ is a &lt;em&gt;recurrent state&lt;/em&gt; of the Markov chain $(X_n)_{n\ge0}$. Property (2) is more unusual.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, an answer to the question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Property (1) implies property (2). Proof: see Byron's post. Note that if (1) holds, both $P_i(T_j&amp;lt;+\infty)=0$ and $P_i(T_j&amp;lt;+\infty)=1$ can occur for different states $i$ and $j$ of the same Markov chain. Example: state space $\{0,1,2\}$, transitions $0\to1$, $0\to2$, $1\to1$, $1\to2$, $2\to1$ and $2\to2$ all with positive probabilities, the other transitions with probability zero. Then $P_i(T_0&amp;lt;+\infty)=0$ and $P_i(T_j&amp;lt;+\infty)=1$ for every $i$ and every $j\ne0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Property (2) does not imply property (1). Proof by an example: state space $\mathbb{Z}$, transitions $i\to i+1$ with probability $1$. Then $P_i(T_j&amp;lt;+\infty)=0$ if $j\le i$, $P_i(T_j&amp;lt;+\infty)=1$ if $j\ge i+1$, but $P_i(\exists n\ge1,X_n=i)=0$ for every $i$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally (and once again), a congenial reference for this is the first chapter of the book &lt;em&gt;Markov chains&lt;/em&gt; by James Norris, which is freely available &lt;a href=&quot;http://www.statslab.cam.ac.uk/~james/Markov&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="6179" LastEditorUserId="6179" LastEditorDisplayName="" LastEditDate="2011-04-07T08:15:22.057" LastActivityDate="2011-04-07T08:15:22.057" />
  <row Id="18717" PostTypeId="1" CreationDate="2011-01-24T09:02:06.287" Score="3" ViewCount="128" Body="&lt;p&gt;The first part of my question asked : State all the irreducible Polynomials in $\mathbb{Z}_2[x]$ of order 3.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was able to do this and get the following polynomials :   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x^3 + x^2 + x + 1 \Rightarrow$ reducible&lt;br&gt;&#xA;$x^3 + x^2 + 1 \Rightarrow$ irreducible&lt;br&gt;&#xA;$x^3 + x + 1 \Rightarrow$ irreducible&lt;br&gt;&#xA;$x^3 + 1 \Rightarrow$ reducible  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;and the other polynomials in $\mathbb{Z}_2[x]$ are trivially reducible.&lt;br&gt;&#xA;I set $f=x^3 + x^2 + 1$, $g=x^3 + x + 1$.&lt;br&gt;&#xA;I then have to take $\mathbb{Z}_2[x]/f$ and $\mathbb{Z}_2[x]/g$ and construct an isomorphism between the two.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My real problem is understanding what these two fields $\mathbb{Z}_2[x]/f$ and $\mathbb{Z}_2[x]/g$ really are- and how I can build an isomorphism between them.    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If anyone could help I'd be very grateful :) &lt;/p&gt;&#xA;" OwnerUserId="4697" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-24T14:42:33.543" LastActivityDate="2011-01-24T17:43:50.007" Title="Construct an isomorphism between fields." Tags="&lt;homework&gt;&lt;polynomials&gt;&lt;field-theory&gt;" AnswerCount="2" FavoriteCount="2" />
  <row Id="18718" PostTypeId="1" AcceptedAnswerId="18744" CreationDate="2011-01-24T09:04:54.147" Score="1" ViewCount="386" Body="&lt;p&gt;How do I find the solutions of &lt;code&gt;tanx = x&lt;/code&gt; upto any number of decimals? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Of course, there is the graphical method but it just helps in finding the approximate value...)&lt;/p&gt;&#xA;" OwnerUserId="6094" LastActivityDate="2011-01-25T01:39:59.373" Title="Solution of tanx = x?" Tags="&lt;algorithms&gt;" AnswerCount="4" />
  <row Id="18719" PostTypeId="1" AcceptedAnswerId="18735" CreationDate="2011-01-24T09:10:11.973" Score="2" ViewCount="119" Body="&lt;p&gt;Hi, i have learn that intersection of pure subgroup of a group G is not necessarily pure. Can someone show me an example when such a case exists?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm aware that if G is torsion-free, then intersection of pure subgroup of G are necessarily pure. So the example above must involve for which the group G is not torsion-free.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any idea? thanks&lt;/p&gt;&#xA;&#xA;&lt;p&gt;edit: The group G here we are talking about is abelian, of course.&lt;/p&gt;&#xA;" OwnerUserId="1683" LastActivityDate="2011-01-26T10:59:09.390" Title="Example of Intersection of Pure Subgroup which is not Pure." Tags="&lt;group-theory&gt;" AnswerCount="1" />
  <row Id="18720" PostTypeId="2" ParentId="18717" CreationDate="2011-01-24T09:21:29.330" Score="6" ViewCount="" Body="&lt;p&gt;First of all, you can treat $\mathbb{F}_2[x]/(f)$ just as a ring. As you know, a quotient of a ring is a set of cosets with the obvious operations of addition and multiplication inherited from the big ring. It just so happens that quotients of integral domains by maximal ideals are also fields, but you needn't be too worried about that at the moment. Now, to write down an isomorphism between quotient rings, the most naive way of doing it would be to write down an isomorphism between the big rings (i.e. in this case between $\mathbb{F}_2[x]$ and itself) that takes the one ideal bijectively to the other, i.e. $(f(x))$ to $(g(x))$. Again, the most naive way of accomplishing that would be to just send one generator to the other. See if you can do that and convince yourself that this procedure really does always give you an isomorphism of quotient rings. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another Exercise: if two isomorphic rings turn out to be fields, why is an isomorphism of rings also an isomorphism of fields?&lt;/p&gt;&#xA;" OwnerUserId="3212" LastActivityDate="2011-01-24T09:21:29.330" CommentCount="2" />
  <row Id="18721" PostTypeId="1" AcceptedAnswerId="18723" CreationDate="2011-01-24T09:47:24.343" Score="0" ViewCount="71" Body="&lt;p&gt;Hey everyone,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm currently practicing differentiation. The exercise I currently have is the following&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Find the derivative of:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(x + 6)^3 (9 x^3 - 2)^5$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Okay, well I can do that now. When I do this I uses the chain/product rule to get the following result:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$3(x+6)^2 (9x^3 -2)^5 + 45(x+6)^3 (3x^2) (9x^3 - 2)^4$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, when I put this in &lt;a href=&quot;http://www.wolframalpha.com/input/?i=differentiate+%28x+%2B+6%29%5E3%289x%5E3+%E2%88%92+2%29%5E5&quot; rel=&quot;nofollow&quot;&gt;wolfram alpha&lt;/a&gt; I get the following result (and it matches the answer from my exercise):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$6(x+6)^2 (2-9 x^3)^4 (27 x^3+135 x^2-1)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm staring at this for an hour now, but I don't get how they get rid of the addition, and how they'd get rid of (for example) $(9x^3-2)^5$&lt;/p&gt;&#xA;" OwnerUserId="3605" LastActivityDate="2011-01-24T10:01:12.230" Title="How did they simplify this function" Tags="&lt;algebra&gt;" AnswerCount="1" />
  <row Id="18723" PostTypeId="2" ParentId="18721" CreationDate="2011-01-24T10:01:12.230" Score="2" ViewCount="" Body="&lt;p&gt;I don't know how wolfram alpha's algorithms actually did this, but here's how it could have been done:&lt;br/&gt;&#xA;$3(x+6)^2(9x^3-2)^5+45(x+6)^3(3x^2)(9x^3-2)^4$&lt;br/&gt;&#xA;$=3(x+6)^2(9x^3-2)^4(9x^3-2)+3(x+6)^2(9x^3-2)^4(45x^2)(x+6)$&lt;br/&gt;&#xA;$=3(x+6)^2(9x^3-2)^4((9x^3-2)+45x^2(x+6))$&lt;br/&gt;&#xA;$=3(x+6)^2(9x^3-2)^4(9x^3-2+45x^3+270x^2)$&lt;br/&gt;&#xA;$=3(x+6)^2(9x^3-2)^4(54x^3+270x^2-2)$&lt;br/&gt;&#xA;$=6(x+6)^2(9x^3-2)^4(27x^3+135x^2-1)$&lt;/p&gt;&#xA;" OwnerUserId="5203" LastActivityDate="2011-01-24T10:01:12.230" CommentCount="3" />
  <row Id="18724" PostTypeId="2" ParentId="18718" CreationDate="2011-01-24T10:12:09.320" Score="5" ViewCount="" Body="&lt;p&gt;Newton-Raphson solving for zeros of $f(x) = tan x - x$&lt;/p&gt;&#xA;" OwnerUserId="5676" LastActivityDate="2011-01-24T10:12:09.320" CommentCount="2" />
  <row Id="18725" PostTypeId="1" AcceptedAnswerId="18728" CreationDate="2011-01-24T10:23:35.810" Score="1" ViewCount="80" Body="&lt;p&gt;I have been puzzling over this for a few days now.  (It's not homework.) Suppose $f$ is a positive, non-decreasing, continuous, integrable function. Suppose there are two finite sequences of positive real numbers $\{a_i\}$ and $\{b_i\}$ where&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum^n_{i=1}{f(a_i)} \leq \sum^n_{i=1}{f(b_i)}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it true that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum^n_{i=1}{\int_0^{a_i}{f(t)dt}} \leq \sum^n_{i=1}{\int_0^{b_i}{f(t)dt}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the answer is no, does it improve things if $f$ is convex?&lt;/p&gt;&#xA;" OwnerUserId="1004" LastActivityDate="2011-01-24T11:03:24.730" Title="If $\sum^n_{i=1}{f(a_i)} \leq \sum^n_{i=1}{f(b_i)}$. Then $\sum^n_{i=1}{\int_0^{a_i}{f(t)dt}} \leq \sum^n_{i=1}{\int_0^{b_i}{f(t)dt}}$?" Tags="&lt;real-analysis&gt;" AnswerCount="1" />
  <row Id="18726" PostTypeId="2" ParentId="18674" CreationDate="2011-01-24T10:24:11.817" Score="2" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Updated for completeness&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, there are an uncountable number of solutions to  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{1}{f'(x_1)} + \frac{1}{f'(x_2)} = 2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with $x_1 \ne x_2$. First, if the graph of $f$ coincides with the main diaognal -- i.e. if $f(x) = x$ for all $x \in [0,1]$ -- then any pair $(x_1,x_2)$ is a solution. So suppose there exists $t$ with $f(t) \ne t$. Then there must exist a point $(u,f(u))$ off the main diagonal with $f'(u) = 1$. To see this, apply the Mean Value Theorem (MVT) to the interval $(l,r)$, where $l$ is the greatest upper bound of those $x \lt t$ with $f(x)=x$, and $r$ is the least upper bound of those points $x \gt t$ with $f(x)=x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, to fix things, suppose that $f(u) \gt u$. Then we apply the MVT to intervals $[0,u]$ and $[u,1]$, to find points $v \in (0,u)$ and $w \in (u,1)$ with  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f'(v) = f(u)/u &gt; 1$ and $f'(w) = (1 - f(u))/(1-u) &amp;lt; 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, given &lt;em&gt;any&lt;/em&gt; $s \in (1,f'(v))$ and $t \in (f'(w),1)$, there exist $x_1 \in (v,u)$ and $x_2 \in (u,w)$ such that $f'(x_1) = s$ and $f'(x_2) = t$ (because a derivative, although not necessarily continuous, does obey the Intermediate Value property). There are uncountably many such pairs $(s,t)$ with the property $\frac{1}{s} + \frac{1}{t} = 2$, so there are uncountably many solutions $(x_1, x_2)$.&lt;/p&gt;&#xA;" OwnerUserId="1508" LastEditorUserId="1508" LastEditorDisplayName="" LastEditDate="2011-01-24T17:16:38.277" LastActivityDate="2011-01-24T17:16:38.277" />
  <row Id="18727" PostTypeId="2" ParentId="18713" CreationDate="2011-01-24T10:44:25.110" Score="3" ViewCount="" Body="&lt;p&gt;There is another definition, which is more general.&#xA;A subset $F \subset E$ is dense in $E$ if the set of all limits of sequences of elements of $F$ is $E$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where does this come from?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A way to answer this question is from the &lt;a href=&quot;http://en.wikipedia.org/wiki/Closure_%28topology%29&quot; rel=&quot;nofollow&quot;&gt;closure&lt;/a&gt; of $F$. The closure of $F$ can be thought of as what I called &lt;em&gt;the set of all limits of sequences of elements of&lt;/em&gt; $F$. It's generally defined as the smallest closed set containing $F$. &lt;strong&gt;Concretely&lt;/strong&gt;, it's &lt;em&gt;what can be reached&lt;/em&gt; by the elements of $F$ without really going &quot;outside&quot; of $F$. What really matters here is that you'll be able to reach the boundaries (as limits of some sequences).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just picture the real interval $]0,1[$ (often written $(0,1)$ in English litterature). Let's consider the sequence $(a_n)_{n \geq 2}$ defined by $a_n = 1 - \frac{1}{n}$. For any $n \geq 2$, $a_n \in ]0,1[$. However, its limit is $1$, which does not belong to $]0,1[$. Here, 1 is an example of what I meant by &quot;can be reached from elements of $F$&quot; earlier. You can guess that the closure of $]0,1[$ is $[0,1]$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So here we can say for instance that $]0,1[$ is dense in $[0,1]$. This definition holds in much more general contexts than $\mathbb{R}$ but that's really a way to look at what that is all about.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope this helps.&lt;/p&gt;&#xA;" OwnerUserId="1865" LastActivityDate="2011-01-24T10:44:25.110" CommentCount="0" />
  <row Id="18728" PostTypeId="2" ParentId="18725" CreationDate="2011-01-24T11:03:24.730" Score="4" ViewCount="" Body="&lt;p&gt;The answer is no, even for convex functions. Let $f(t)=(t)^+$ (the positive part of $t$). Then $f$ is convex, the first inequality means that $a_1+\cdots+a_n\le b_1+\cdots+b_n$, the second inequality means that $a_1^2+\cdots+a_n^2\le b_1^2+\cdots+b_n^2$. As soon as $n\ge2$, the former cannot imply the latter.&lt;/p&gt;&#xA;" OwnerUserId="6179" LastActivityDate="2011-01-24T11:03:24.730" />
  <row Id="18729" PostTypeId="1" CreationDate="2011-01-24T11:04:26.223" Score="2" ViewCount="106" Body="&lt;p&gt;I picked this problem from here &lt;a href=&quot;http://mathalon.in/?page=show_problem.php&amp;amp;pid=106&quot; rel=&quot;nofollow&quot;&gt;http://mathalon.in/?page=show_problem.php&amp;amp;pid=106&lt;/a&gt; . I don't know how to solve it accurately. The number seems too to be done by any computation method.&lt;/p&gt;&#xA;" OwnerUserId="6182" LastActivityDate="2011-01-24T11:14:52.117" Title="Calculate GCD(Fib(531185354674),Fib(613570636967))" Tags="&lt;arithmetic&gt;" AnswerCount="2" />
  <row Id="18730" PostTypeId="2" ParentId="18729" CreationDate="2011-01-24T11:13:44.333" Score="3" ViewCount="" Body="&lt;p&gt;Try and compute $\gcd(F_i,F_j)$ for some small $i$ and $j$ and see if you can spot a pattern. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the following three facts you can prove by induction that the pattern holds in general:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\gcd(F_n,F_{n-1}) = 1$.&lt;/li&gt;&#xA;&lt;li&gt;$F_{m+n} = F_{m+1}F_n + F_mF_{n-1}$.&lt;/li&gt;&#xA;&lt;li&gt;$m$ divides $n \implies F_m$ divides $F_n$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1269" LastActivityDate="2011-01-24T11:13:44.333" />
  <row Id="18731" PostTypeId="2" ParentId="18729" CreationDate="2011-01-24T11:14:52.117" Score="4" ViewCount="" Body="&lt;p&gt;GCD(Fib(531185354674),Fib(613570636967)) = Fib(GCD(531185354674,613570636967))&lt;/p&gt;&#xA;" OwnerUserId="589" LastActivityDate="2011-01-24T11:14:52.117" CommentCount="1" />
  <row Id="18732" PostTypeId="2" ParentId="17149" CreationDate="2011-01-24T11:17:39.960" Score="1" ViewCount="" Body="&lt;p&gt;See &lt;a href=&quot;http://en.wikipedia.org/wiki/Automatic_label_placement&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Automatic_label_placement&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="589" LastActivityDate="2011-01-24T11:17:39.960" />
  <row Id="18733" PostTypeId="2" ParentId="18718" CreationDate="2011-01-24T11:30:37.027" Score="7" ViewCount="" Body="&lt;p&gt;Apart from the solution $x=0$ there are for each $n&amp;gt;0$ two solutions $\pm x_n$ with $x_n={(2n-1)\pi\over2}-\tau_n$ where $\tau_n&amp;gt;0$ is $O({1\over n})$. It is easy to see that $\tau_n$ satisfies the equation&#xA;$$\tau=\arctan{1\over (2n-1)\pi/2 - \tau}=:f(\tau).$$&#xA;As $|f&amp;#39;(\tau)|\ll1$ in the neighbourhood of $\tau=0$ we can obtain $\tau_n$ numerically by iterating $f$ starting at $\tau=0$.&lt;/p&gt;&#xA;" OwnerUserId="1303" LastActivityDate="2011-01-24T11:30:37.027" />
  <row Id="18734" PostTypeId="1" AcceptedAnswerId="18742" CreationDate="2011-01-24T12:06:08.670" Score="1" ViewCount="58" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm having trouble showing the following implication:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $M$ be a Riemannian manifold, let $L\subset M$ be a submanifold such that the following holds:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If $\gamma: I \to M$ is a geodesic s.t. $\gamma(0) \in L$, $\dot{\gamma}(0) \in T_{\gamma(0)}L$, then there is an $\epsilon &amp;gt;0$ such that for $|t| &amp;lt; \epsilon$ we have $\gamma(t) \in L$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Then for any smooth curve $\eta: I \to L$ we have &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$P_\eta (s,t) T_{\eta(t)}L = T_{\eta(s)}L $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $P_\eta (s,t)$ denotes parallel transport along $\eta$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't really have an approach to this one. The differential equations satisfied by a geodesic looks sort of like the ones satisfied by the parallel transport of a vector, but I didn't get anywhere with this. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any pointers would be greatly appreciated. Thanks in advance!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;S.L.&lt;/p&gt;&#xA;" OwnerUserId="3208" LastActivityDate="2011-01-24T13:55:46.757" Title="Completely Geodesic" Tags="&lt;differential-geometry&gt;&lt;riemannian-geometry&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="18735" PostTypeId="2" ParentId="18719" CreationDate="2011-01-24T12:40:25.600" Score="3" ViewCount="" Body="&lt;p&gt;This is &lt;a href=&quot;http://books.google.com/books?id=lYrsiaHSHKcC&amp;amp;pg=PA330&amp;amp;lpg=PA330&amp;amp;dq=intersection+of+pure+subgroups+is+not+pure&amp;amp;source=bl&amp;amp;ots=c62abGmVvx&amp;amp;sig=fJiOflrf5-xwCJk0De3lxRisl2s&amp;amp;hl=en&amp;amp;ei=H3E9TZmwOcTCceea-YQH&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=6&amp;amp;ved=0CD4Q6AEwBQ#v=onepage&amp;amp;q=intersection%20of%20pure%20subgroups%20is%20not%20pure&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;exercise 10.33(i)&lt;/a&gt; in Rotman's &quot;An Introduction to Group Theory&quot;. You can take $G=\mathbb Z_2 \times \mathbb Z_8$. Take $A$ the subgroup generated by (0,1) and $B$ the subgroup generated by (1,1). Then $A$ and $B$ are pure, but their intersection will be the subgroup generated by (0,2), which is not pure in $G$.&lt;/p&gt;&#xA;" OwnerUserId="808" LastEditorUserId="808" LastEditorDisplayName="" LastEditDate="2011-01-26T10:59:09.390" LastActivityDate="2011-01-26T10:59:09.390" CommentCount="6" />
  <row Id="18737" PostTypeId="2" ParentId="11611" CreationDate="2011-01-24T13:13:04.340" Score="0" ViewCount="" Body="&lt;p&gt;Since I don't think anyone else has covered it and because it can be useful for thinking about inverse functions/relations, consider completing the square:&#xA;$$\begin{align}&#xD;&#xA;f(x)&amp;amp;=x^2+x-6&#xD;&#xA;\\&#xD;&#xA;&amp;amp;=x^2+x+\frac{1}{4}-\frac{25}{4}&#xD;&#xA;\\&#xD;&#xA;&amp;amp;=\left(x+\frac{1}{2}\right)^2-\frac{25}{4}&#xD;&#xA;\end{align}$$&#xA;so $f(x)=a(b(c(x)))$ where $a(x)=x-\frac{25}{4}$, $b(x)=x^2$, and $c(x)=x+\frac{1}{2}$.  The advantage here (though probably not significant for your purpose) is that each of these functions can be thought of as a single simple operation on their input (that the input appears only once in each function also makes it easier to think of this way).&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2011-01-24T13:13:04.340" />
  <row Id="18739" PostTypeId="2" ParentId="18659" CreationDate="2011-01-24T13:15:28.613" Score="0" ViewCount="" Body="&lt;p&gt;This article about Gray codes includes partitions. The idea behind a Gray code is to enumerate a cyclic sequence of some combinatorial collection of objects so that the &quot;distance&quot; between consecutive items in the list are &quot;close.&quot;   &lt;a href=&quot;http://linkinghub.elsevier.com/retrieve/pii/0196677489900072&quot; rel=&quot;nofollow&quot;&gt;http://linkinghub.elsevier.com/retrieve/pii/0196677489900072&lt;/a&gt;&#xA;Savage also has other survey articles about Gray codes that include partitions. &lt;a href=&quot;http://reference.kfupm.edu.sa/content/s/u/a_survey_of_combinatorial_gray_codes__213043.pdf&quot; rel=&quot;nofollow&quot;&gt;http://reference.kfupm.edu.sa/content/s/u/a_survey_of_combinatorial_gray_codes__213043.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1369" LastActivityDate="2011-01-24T13:15:28.613" />
  <row Id="18740" PostTypeId="2" ParentId="18659" CreationDate="2011-01-24T13:41:55.767" Score="0" ViewCount="" Body="&lt;p&gt;If you are only interested in using an actual implementation, you could go for the &lt;code&gt;integer_partitions(n[, length])&lt;/code&gt; in &lt;a href=&quot;http://maxima.sourceforge.net/&quot; rel=&quot;nofollow&quot;&gt;Maxima&lt;/a&gt;. More details can be found &lt;a href=&quot;http://maxima.sourceforge.net/docs/manual/en/maxima_38.html&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="689" LastActivityDate="2011-01-24T13:41:55.767" />
  <row Id="18741" PostTypeId="2" ParentId="18710" CreationDate="2011-01-24T13:46:41.437" Score="6" ViewCount="" Body="&lt;p&gt;There is a term for this (actually, more than one).  It's called the &lt;a href=&quot;http://en.wikipedia.org/wiki/Product_integral&quot; rel=&quot;nofollow&quot;&gt;product integral&lt;/a&gt; or multiplicative integral, and together with the corresponding derivative you get what's called product calculus or &lt;a href=&quot;http://en.wikipedia.org/wiki/Multiplicative_calculus&quot; rel=&quot;nofollow&quot;&gt;multiplicative calculus&lt;/a&gt; or &lt;a href=&quot;http://en.wikipedia.org/wiki/Non-Newtonian_calculus#Multiplicative_calculus&quot; rel=&quot;nofollow&quot;&gt;non-Newtonian calculus&lt;/a&gt;.  In addition, your idea of obtaining different versions of calculus by considering the generalized mean is discussed &lt;a href=&quot;http://en.wikipedia.org/wiki/Non-Newtonian_calculus#An_alternative_approach_to_derivatives&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; in the non-Newtonian calculus article.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The idea of product integration goes back at least to Vito Volterra in the late 1800s, and there are various applications of it.  The standard reference is probably Dollard and Friedman's text in the Encyclopedia of Mathematics series, &lt;em&gt;&lt;a href=&quot;http://www.cambridge.org/gb/knowledge/isbn/item5736928/?site_locale=en_GB&quot; rel=&quot;nofollow&quot;&gt;Product Integration with Application to Differential Equations&lt;/a&gt;&lt;/em&gt;.  (The description says, &quot;This book shows the beautiful simplifications that can be brought to the theory of differential equations by treating such equations from the product integral viewpoint.&quot;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, as Rahul Narain points out, it is easy to express the product integral in terms of usual integration.  For that reason, some people don't believe it is really anything new. For more on that, see &lt;a href=&quot;http://en.wikipedia.org/wiki/Multiplicative_calculus#Reception&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/Non-Newtonian_calculus#Criticism&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And for more on the product integral, see one of the many references in the Wikipedia pages cited or &lt;a href=&quot;http://math.pugetsound.edu/~mspivey/ProdCalc.pdf&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; or &lt;a href=&quot;http://math.pugetsound.edu/~mspivey/ProdInt.pdf&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; survey paper I wrote a few years ago. &lt;/p&gt;&#xA;" OwnerUserId="2370" LastActivityDate="2011-01-24T13:46:41.437" />
  <row Id="18742" PostTypeId="2" ParentId="18734" CreationDate="2011-01-24T13:55:46.757" Score="2" ViewCount="" Body="&lt;p&gt;Let $II$ denote the second fundamental form of $L$. That is, given $\nabla$ the Levi-Civita connection of $M$ and its Riemannian metric, given vector fields $v,w$ tangent to $L$, define $II(v,w)$ to be the normal projection of $\nabla_vw$ (normal projection meaning the projection to the orthogonal complement of $TL$). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$II$ is tensorial and symmetric. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The parallel transport condition you want to show is equivalent to $II = 0$&lt;/li&gt;&#xA;&lt;li&gt;The geodesic condition you are given tells you that $II(v,v) = 0$ for all $v$. Now consider $II(v+w,v+w)$. &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="1543" LastActivityDate="2011-01-24T13:55:46.757" CommentCount="1" />
  <row Id="18743" PostTypeId="1" CreationDate="2011-01-24T14:06:42.663" Score="2" ViewCount="112" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Is there a &lt;strong&gt;unique&lt;/strong&gt; normal form for&#xA;  each formula of predicate calculus?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am aware of prenex normal forms of predicate calculus and of disjunctive and conjunctive normal forms of propositional calculus. But what is the name of the combination of the two, i.e. the formula is in prenex normal form and the (quantifier-free) matrix is in - say - disjunctive normal form? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can uniqueness always be achieved by a optimal choice of variables and sorting and arrangement of the literals of the disjunctive normal form (given that the variables and the relation and function symbols are lexicographically sorted) and elimination of double negation?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Example&lt;/strong&gt;: The intended unique normal form of &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(\forall x_2) \neg R_2 (x_2) \rightarrow \Big((\exists x_1)\neg\neg R_3(x_1) \wedge R_1(x_1,x_2) \Big)$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;should be (not quite sure)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(\forall x_1)(\exists x_2)\big(R_1 (x_2,x_1) \vee R_2(x_1)\big) \wedge \big( R_2(x_1) \vee R_3 (x_2) \big)$&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What is the complexity of finding such&#xA;  normal forms?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1792" LastActivityDate="2011-01-24T15:47:27.967" Title="Unique normal form of formulae of predicate calculus?" Tags="&lt;logic&gt;" AnswerCount="1" />
  <row Id="18744" PostTypeId="2" ParentId="18718" CreationDate="2011-01-24T14:12:57.377" Score="2" ViewCount="" Body="&lt;p&gt;You know that there is a solution $x_k$ in a neighbourhood of $2 \pi k$, for each integer $k$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And the equation can be also written as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x_k = atan(x_k) + 2 k \pi$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;where the arc tangent returns the principal value. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then you can iterate:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x_k[0] = 2 k \pi$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x_k[n] = atan(x_k[n-1]) + 2 k \pi$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It converges very quickly - empirically - proof is left as an exercise :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have an starting approximate value, you pick your $k$ by rounding $x_0/(2 \pi)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/CnNwd.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="312" LastEditorUserId="312" LastEditorDisplayName="" LastEditDate="2011-01-24T14:28:43.540" LastActivityDate="2011-01-24T14:28:43.540" CommentCount="1" />
  <row Id="18745" PostTypeId="1" AcceptedAnswerId="18765" CreationDate="2011-01-24T14:16:11.230" Score="0" ViewCount="100" Body="&lt;p&gt;How do i expand and simplify the vector expression&lt;br&gt;&#xA;$(\vec{a}-\vec{b})^T.(\vec{a}-\vec{b})$ = ?&lt;br&gt;&#xA;And if there are matrices A and B instead of vectors a and b, how do i multiply and simplify this expression?&#xA;$(A-B)^T.(A-B)$ = ?&lt;br&gt;&#xA;Also in the normal equations for solving least squares problems, we have expressions of the type&lt;br&gt;&#xA;$(Ax-b)^T.(Ax-b)$ = ?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that is really basic. Sorry but my basics are rusty. In fact, I need to so some matrix calculus next. Please do point me to some relevant links.&lt;/p&gt;&#xA;" OwnerUserId="6050" LastActivityDate="2011-01-24T17:11:49.650" Title="How to multiply vector and matrix expressions involving transposes" Tags="&lt;linear-algebra&gt;&lt;matrices&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="18746" PostTypeId="2" ParentId="90" CreationDate="2011-01-24T14:45:18.877" Score="0" ViewCount="" Body="&lt;p&gt;J. Milne has publicized his lecture notes for free, complete with exercises in fields such as group theory, field theory, algebraic geometry, algebraic number theory, &quot;and so on&quot;. &lt;a href=&quot;http://www.jmilne.org/math/CourseNotes/index.html&quot; rel=&quot;nofollow&quot;&gt;http://www.jmilne.org/math/CourseNotes/index.html&lt;/a&gt; (probably best suited for upper undergrad or grad students, though)&lt;/p&gt;&#xA;" OwnerUserId="4284" LastActivityDate="2011-01-24T14:45:18.877" CommunityOwnedDate="2011-01-24T14:45:18.877" />
  <row Id="18747" PostTypeId="2" ParentId="18582" CreationDate="2011-01-24T15:09:13.787" Score="2" ViewCount="" Body="&lt;p&gt;The other answer is bogged down with a lot of comments, so let me give a fresh answer to the revised question.  The short answer is as follows: $k[\mathbb{A}^n]$ is the notation I support because $k[V]$ is perfectly standard.  The fact that it looks like adjoining $V$ is irrelevant; what you are adjoining is coordinate functions on $V$.  (That is, you should think of it as analogous to the notation $C(X)$ for the ring of real-valued functions on a topological space $X$, except that $k(V)$ is already taken; it means the &lt;em&gt;field&lt;/em&gt; of functions on $V$ when $V$ is irreducible.)  Finally, the fact that it looks like a function ring is also irrelevant because, in the right category, it &lt;em&gt;is&lt;/em&gt; a function ring.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The long answer is as follows: your concern that&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;It feels suspiciously like a function ring, but in general the map taking formal polynomials to polynomial functions is neither injective nor surjective.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;comes from the fact that the functor $\text{Hom}(\text{Spec } k, -) : k\text{-Alg}^{op} \to \text{Set}$ is not faithful, but you don't have to think about this functor.  You can in fact work directly in $k\text{-Alg}^{op}$, and in this category $k[V]$ is &lt;em&gt;precisely&lt;/em&gt; the ring of functions $\text{Hom}(V, k)$ where $k$ is shorthand for the affine line $\mathbb{A}^1(k) \simeq \text{Spec } k[x]$.  One need not make a distinction between polynomials and the functions they define in this case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's why.  Suppose $V$ is given by an ideal $I$ in $k[x_1, ... x_n]$ for some $n$, so that $k[V]$ denotes the ring $k[x_1, ... x_n]/I$.  (This is by definition.)  Then a morphism $\text{Hom}(V, k)$ (in $k\text{-Alg}^{op}$, which is emphatically &lt;em&gt;not&lt;/em&gt; the naive category of affine varieties over $k$) is precisely a morphism $k[x] \to k[x_1, ... x_n]/I$ of $k$-algebras.  By the universal property of $k[x]$, such a morphism is freely determined by the image of $x$, so $\text{Hom}(k[x], -)$ represents the forgetful functor $k\text{-Alg} \to \text{Set}$.  In particular such morphisms are in one-to-one correspondence with elements of $k[V]$, and this may therefore be taken as a definition of $k[V]$.  (The details of how to get the ring operations are discussed, as I said, in &lt;a href=&quot;http://qchu.wordpress.com/2011/01/21/structures-on-hom-sets/&quot; rel=&quot;nofollow&quot;&gt;this blog post&lt;/a&gt;.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When $k$ is algebraically closed and we restrict to the opposite of the category of finitely-generated reduced $k$-algebras, then the functor $\text{Hom}(\text{Spec } k, -)$ &lt;em&gt;is&lt;/em&gt; faithful by the Nullstellensatz, so we can define the ring of regular functions $k[V]$ in a naive way by regarding $V$ as a set of points in $\mathbb{A}^n(k)$.  The point I am trying to make above is that, as long as we switch to a more sophisticated category, we can still do this for $k$ arbitrary (and not necessarily a field) and $V$ an arbitrary $k$-scheme rather than just a variety.  In particular, if $V$ is defined over $\mathbb{Z}$, then the notation $\mathbb{Z}[V]$ makes perfect sense.&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2011-01-24T15:19:59.393" LastActivityDate="2011-01-24T15:19:59.393" CommentCount="14" />
  <row Id="18748" PostTypeId="1" AcceptedAnswerId="18750" CreationDate="2011-01-24T15:15:08.250" Score="1" ViewCount="190" Body="&lt;p&gt;I have found some pretty complete lists (I think) of mathematical symbols &lt;a href=&quot;http://www.uefap.com/speaking/symbols/symbols.htm&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/Table_of_mathematical_symbols&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;, but I don't see a symbol for the word &quot;and&quot; on either list.  A person could easily just write the word &quot;and&quot; or use an ampersand, but I was wondering if there was an actual mathematical symbol for the word &quot;and&quot;.  Also, if anyone knows any lists that are more complete than the ones I have linked to please provide a link.&lt;/p&gt;&#xA;" OwnerUserId="6096" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2011-01-24T17:09:48.270" LastActivityDate="2011-01-24T18:52:45.783" Title="Mathematical symbol for &quot;and&quot;" Tags="&lt;notation&gt;" AnswerCount="3" CommentCount="4" FavoriteCount="1" />
  <row Id="18749" PostTypeId="2" ParentId="18745" CreationDate="2011-01-24T15:18:10.283" Score="0" ViewCount="" Body="&lt;p&gt;Just think of a vector as a matrix of one column, and all the matrix rules apply. &lt;/p&gt;&#xA;" OwnerUserId="312" LastActivityDate="2011-01-24T15:18:10.283" />
  <row Id="18750" PostTypeId="2" ParentId="18748" CreationDate="2011-01-24T15:19:15.607" Score="8" ViewCount="" Body="&lt;p&gt;The logical &quot;and&quot; is $\wedge$ (and the corresponding &quot;or&quot; is $\vee$).&lt;/p&gt;&#xA;" OwnerUserId="3148" LastActivityDate="2011-01-24T15:19:15.607" CommentCount="9" />
  <row Id="18751" PostTypeId="2" ParentId="18743" CreationDate="2011-01-24T15:47:27.967" Score="2" ViewCount="" Body="&lt;p&gt;I would like to point out that the problem of uniqueness of a formula is non-trivial even in propositional logic. Eg. (p&amp;rarr;q)&amp;and;(q&amp;rarr;r)&amp;and;(r&amp;rarr;p) is equal to (&amp;not;p&amp;or;q)&amp;and;(&amp;not;q&amp;or;r)&amp;and;(&amp;not;r&amp;or;p) and also to (&amp;not;p&amp;or;r)&amp;and;(&amp;not;r&amp;or;q)&amp;and;(&amp;not;q&amp;or;p).&lt;/p&gt;&#xA;" OwnerUserId="6192" LastActivityDate="2011-01-24T15:47:27.967" CommentCount="1" />
  <row Id="18752" PostTypeId="1" AcceptedAnswerId="18756" CreationDate="2011-01-24T15:56:49.280" Score="3" ViewCount="167" Body="&lt;p&gt;If there are two different proofs for one theorem, at some level are the two proofs the same, or can they be fundamentally different?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, if you have two proofs of a theorem, can one show that the two proofs are expressing the same thing in different ways, and then remove the redundancies and generate a &quot;shorter&quot; proof?&lt;/p&gt;&#xA;" OwnerUserId="4728" LastEditorUserId="1543" LastEditorDisplayName="" LastEditDate="2011-05-03T10:08:02.430" LastActivityDate="2011-05-24T14:05:43.917" Title="Ideas about Proofs" Tags="&lt;logic&gt;&lt;philosophy&gt;&lt;proof-theory&gt;" AnswerCount="2" CommentCount="5" />
  <row Id="18753" PostTypeId="1" CreationDate="2011-01-24T16:02:11.963" Score="2" ViewCount="88" Body="&lt;p&gt;So &lt;a href=&quot;http://blogs.plos.org/badphysics/2011/01/20/ono/&quot; rel=&quot;nofollow&quot;&gt;Ken Ono says&lt;/a&gt; that the partition numbers behave like fractals, in which case I'd like to try to find an appropriately illuminating way of visualizing them. But I'm sort of stuck at the moment, so here I'm asking for suggestions for visualization methods.&lt;/p&gt;&#xA;" OwnerUserId="890" LastActivityDate="2011-01-24T16:02:11.963" Title="Visualizing the Partition numbers (suggestions for visualization techniques)" Tags="&lt;fractals&gt;&lt;partitions&gt;&lt;visualization&gt;" CommentCount="1" FavoriteCount="1" />
  <row Id="18754" PostTypeId="1" AcceptedAnswerId="18758" CreationDate="2011-01-24T16:11:26.100" Score="2" ViewCount="174" Body="&lt;p&gt;Suppose I found a preprint on arXiv.org (like this one: &lt;a href=&quot;http://arxiv.org/abs/math/0309146&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/math/0309146&lt;/a&gt; ), and I need to check if what's in there is actually true. Any tips? I don't want to study an article only to later find that it contained an error &amp;lt;_&amp;lt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More generally, should I spend my time on arXiv.org preprints? How serious is the danger that I will find myself holding a bunch of false beliefs and unable to publish my own works in any serious journal because I can't reference arXiv.org?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do I find if an arXiv.org preprint was actually published in a peer-reviewed journal later? What is the average time interval between sending an article for peer review and it being published? If the paper was not published in, like, 5 years, is this a good reason not to trust it?&lt;/p&gt;&#xA;" OwnerUserId="3793" LastActivityDate="2011-01-24T16:51:53.107" CommunityOwnedDate="2011-01-24T16:28:51.283" Title="How to check the veracity of arXiv.org papers?" Tags="&lt;soft-question&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="1" />
  <row Id="18755" PostTypeId="2" ParentId="18748" CreationDate="2011-01-24T16:16:29.593" Score="1" ViewCount="" Body="&lt;p&gt;I'll also add that, perversely, the comma can mean either &quot;and&quot; or &quot;or&quot;, depending on context. For example, in classical sequent calculus, $\{ P, Q \} \vdash \{ R, S \}$ means $P \land Q \vdash R \lor S$. Also, in set-builder notation $\{ \ldots : \ldots \}$, in a certain sense, commas in the left half are disjunctions and commas in the right half are conjunctions... which is the exact opposite of $\vdash$.&lt;/p&gt;&#xA;" OwnerUserId="5191" LastActivityDate="2011-01-24T16:16:29.593" />
  <row Id="18756" PostTypeId="2" ParentId="18752" CreationDate="2011-01-24T16:18:37.080" Score="6" ViewCount="" Body="&lt;p&gt;They can be &quot;completely different&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, some existence results have both &lt;em&gt;indirect&lt;/em&gt; proofs and &lt;em&gt;constructive&lt;/em&gt; proofs. There is often no way to interpret the &quot;indirect proof&quot; as &quot;essentially the same&quot; as the constructive proof. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or you have the many different proofs of Quadratic Reciprocity. Gauss's first proof, in the &lt;em&gt;Disquisitiones Arithmeticae&lt;/em&gt;, is very constructive; it is done by recursion, and for example it shows exactly how to transform a solution of $x^2\equiv p\pmod{q}$ into a solution of $x^2\equiv q\pmod{p}$ when $p$ and $q$ are not both congruent to $3$ modulo $4$; whereas his third proof was purely combinatorial, counting certain objects, and his sixth used Gauss sums, again an essentially different approach. Eisenstein used infinite products for his fifth proof, Kummer used quadratic forms, Zolotarev used permutations; Auslander and Tolimieri used the Fourier transform, Weil used theta functions. These are truly essentially different approaches, with no easy way to pare them down to the same thing (unless you &quot;pare them down&quot; to the statement of Quadratic Reciprocity itself). &lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2011-01-24T16:18:37.080" CommentCount="3" />
  <row Id="18757" PostTypeId="2" ParentId="18717" CreationDate="2011-01-24T16:25:00.587" Score="1" ViewCount="" Body="&lt;p&gt;The first field looks like this:&#xA;\begin{align*}&#xA;\mathbb{Z}_{2}[x]/\langle f(x)\rangle &amp;amp;=&#xA;\{ p(x) + \langle f(x)\rangle\mid p(x)\in\mathbb{Z}_2[x],\ \deg(p)\lt\deg(f)\}\\&#xA;&amp;amp;= \left\{&#xA;\begin{array}{l}&#xA;0 + \langle f(x)\rangle,\quad 1 + \langle f(x)\rangle,\quad x + \langle f(x)\rangle, \quad x^2 + \langle f(x)\rangle,\quad (1+x) + \langle f(x)\rangle,\\&#xA;(1+x^2) + \langle f(x)\rangle,\quad (x + x^2) + \langle f(x)\rangle, \quad (1+x+x^2) + \langle f(x)\rangle&#xA;\end{array}\right\}&#xA;\end{align*}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You know that you have found all elements by counting them as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have $|\mathbb{Z}_{2}| = |{ 0, 1}| = 2$ choices for any coefficient.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In any polynomial, you can have terms of degree $0$, $1$ or $2$, each such term can have any of the two possible coefficients.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So your resulting field has $2 \times 2 \times 2 = 2^3 = 8$ elements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regarding notation:&#xA;$\langle f(x)\rangle $ in the above means the ideal generated by $f$. This means, the set of all elements $p(x) \in \mathbb{Z}_{2}[x]$ that are multiples of $f$.&lt;/p&gt;&#xA;" OwnerUserId="5798" LastEditorUserId="5798" LastEditorDisplayName="" LastEditDate="2011-01-24T17:43:50.007" LastActivityDate="2011-01-24T17:43:50.007" CommentCount="6" />
  <row Id="18758" PostTypeId="2" ParentId="18754" CreationDate="2011-01-24T16:27:13.360" Score="6" ViewCount="" Body="&lt;p&gt;The simplest way of finding out if an arXiv paper was published is to check the arXiv to see if they list a publishing date; many papers do (those of mine that have been subsequently published certainly do). Absent that, contact the author(s) and ask. (As Willie Wong notes, the paper you refer to &lt;em&gt;has&lt;/em&gt; such a listing; that paper has appeared in print already; of course, you'll take into account the quality of the journal if this is the case). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is risk of a paper being wrong &lt;em&gt;even if&lt;/em&gt; it has been published in a peer-reviewed journal, of course. The odds are lower (one hopes) than for papers on the arXiv simply because third parties have taken the time to go over the paper before publication. The same is true of papers in the arXiv, though less formally, so you'll want to check to see if anybody has read the arXiv paper (check in fora such as this, or &lt;code&gt;sci.math.research&lt;/code&gt;, or in specialist mailing lists, such as the GroupPub Forum).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You may want to check the author's track record; and of course, you'll want to check the paper before you start quoting it (this is true of published material too!). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for trusting a paper that has not appeared, there are many possible reasons for a paper not to appear that have nothing to do with its correctness. The results may have been subsumed by other papers (of the same author), or it may turn out that the results were known; or the journals it was submitted to may have considered it &quot;correct-but-not-significant-enough-to-publish&quot;, etc. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The average time between submission of a paper and publication varies wildly. Even assuming that you submit it once, it gets reviewed once and accepted, the time span varies. Time between final acceptance and appearance in print can range from a few months to a couple of years depending on the backlog of the journal (the AMS publishes a yearly list of the backlog of the main journals); time between review and final acceptance depends on the nature of the changes requested and how quickly the authors get to them (anything from a week to several months); time between submissions and initial review can vary as well, though it is considered impolite for a referee to take more than six months for a review. If you end up having to submit a paper to several journals (because the referees deem it &quot;correct-but-not-important-enough&quot;, for example) the process can take several years before it finally makes it into print.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My advice would be: check the authors track record. If there is nothing particularly bad about them (a habit of grandiose announcements that are later taken back, lots of errors, etc), then take the paper with double the amount of grains of salt as you take a published paper (I assume you read published papers skeptically as well; if you don't, then you should start), and review the theorems to &lt;em&gt;your satisfaction&lt;/em&gt; before &quot;believing them&quot; or quoting them. &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-24T16:51:53.107" LastActivityDate="2011-01-24T16:51:53.107" CommunityOwnedDate="2011-01-24T16:28:51.283" CommentCount="1" />
  <row Id="18759" PostTypeId="1" AcceptedAnswerId="18761" CreationDate="2011-01-24T16:34:43.853" Score="1" ViewCount="272" Body="&lt;p&gt;Prove that the set of functions $f: \mathbb{Q} \rightarrow \{1,2,3\}$ is uncountably infinite.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm totally stuck on this one. We have just been shown Georg Cantor's diagonalization argument in class for why $\mathbb{R}$ is uncountably infinite (and I for the most part understand it---assemble an enumeration for the reals, then proceed down the diagonal changing the number in that position to create a number that will not be on the list, since the number will differ from the i-th number in the i-th position), so I'm assuming we're going to have to assemble some sort of similar argument for this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyone able to prod me in the right direction perhaps? This is homework so I'd like to not get an answer (until maybe after the deadline), but maybe a starting point or reference (which I can cite) to get me thinking about it the right way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="6193" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-24T16:41:14.283" LastActivityDate="2011-01-24T21:45:30.503" Title="Prove the set of functions $f : \mathbb{Q} \rightarrow \{1,2,3\}$ uncountably infinite" Tags="&lt;homework&gt;&lt;functions&gt;&lt;elementary-set-theory&gt;&lt;infinity&gt;" AnswerCount="6" CommentCount="2" />
  <row Id="18760" PostTypeId="2" ParentId="18759" CreationDate="2011-01-24T16:40:21.710" Score="4" ViewCount="" Body="&lt;p&gt;Use Cantor's Diagonal argument to show that set of functions $f\colon\mathbb{N}\to\{1,2,3\}$ is uncountably infinite. Then show that you have at least that many functions $\mathbb{Q}\to\{1,2,3\}$ (say, by &quot;extending&quot; functions).&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2011-01-24T16:40:21.710" />
  <row Id="18761" PostTypeId="2" ParentId="18759" CreationDate="2011-01-24T16:40:32.133" Score="5" ViewCount="" Body="&lt;p&gt;Hint: Try to map each function to a real number, written in base-3. &lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2011-01-24T16:40:32.133" />
  <row Id="18762" PostTypeId="2" ParentId="18759" CreationDate="2011-01-24T16:43:51.380" Score="-2" ViewCount="" Body="&lt;p&gt;The same diagonalization argument will work here as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $\mathbb{Q}$ in countably finite, we can have a bijection to natural numbers and hence $\mathbb{Q}$ can be written as $q_1,q_2,q_3,\ldots$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Correspondingly write the functional values as $f(q_1),f(q_2),f(q_3),\ldots$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, $1,3,2,2,2,1,3,\ldots$ is one such string of numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now do the same diagonal argument thing as for the real numbers.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2011-01-24T16:50:16.850" LastActivityDate="2011-01-24T16:50:16.850" CommentCount="2" />
  <row Id="18763" PostTypeId="2" ParentId="18759" CreationDate="2011-01-24T16:46:42.797" Score="4" ViewCount="" Body="&lt;p&gt;The set of subsets of $\mathbf Q$ is uncountable and is (equivalent to) the set of functions $\mathbf Q \to \{1,2\}$, which is a subset of the set in question.&lt;/p&gt;&#xA;" OwnerUserId="589" LastActivityDate="2011-01-24T16:46:42.797" />
  <row Id="18764" PostTypeId="1" AcceptedAnswerId="18768" CreationDate="2011-01-24T17:11:42.880" Score="0" ViewCount="43" Body="&lt;p&gt;The following question was part of a problem sheet in numerical maths:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given the scalar equation $\ddot{y} = - \lambda y$, $\lambda &gt; 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;show that for $h &gt; h_{critical}$, the approximate solution goes to infinity as $t$ goes to infinity. This behaviour is called blow-up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Stormer-Verlet-method is $y_{k+1} = -y_{k-1} + 2y_k - h^2 \lambda y_k$, which can be rewritten as &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \left( \begin{array}{c} y_{k+1} \\ y_k \end{array} \right) = &#xA;\left(  \begin{array}{cc} 2 - h^2 \lambda &amp;amp; -1 \\ 1 &amp;amp; 0  \end{array} \right) \left( \begin{array}{c} y_{k} \\ y_{k-1} \end{array} \right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then in the solution they write that there is no blow-up if and only if the eigenvalues $\mu_i$ of the matrix &#xA;$$ B = \left(  \begin{array}{cc} 2 - h^2 \lambda &amp;amp; -1 \\ 1 &amp;amp; 0  \end{array} \right)$$ are all $\leq 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My question is&lt;/strong&gt;: how do they get to this claim about the relationship between eigenvalues and blow-up?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I read that the 2 norm $|| . ||_2$ is the same as the spectral norm if the matrices are square. Then I was wondering if the reason could be something like&#xA;$$ || Av||_2 \leq || A ||_2 || v ||_2$$ and then using that this is the spectral norm: &#xA;$$ || Av||_2 \leq || A ||_2 || v ||_2 \leq ( max_i \mu_i ) || v ||_2$$ where the max is taken over the eigenvalues of $A$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem with this is, that I need the other direction: $$ ( max_i \mu_i ) || v ||_2 \leq || Av||_2 $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, obviously, if the biggest eigenvalue is bigger than 1, the approximation after $k$ steps is $A^k y_0$ and using the above I get a blow-up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many thanks for your help.&lt;/p&gt;&#xA;" OwnerUserId="5798" LastEditorUserId="5798" LastEditorDisplayName="" LastEditDate="2011-01-24T17:48:30.200" LastActivityDate="2011-01-24T18:09:23.640" Title="Behaviour of a numerical solution of an ODE" Tags="&lt;linear-algebra&gt;&lt;numerical-methods&gt;" AnswerCount="1" FavoriteCount="0" />
  <row Id="18765" PostTypeId="2" ParentId="18745" CreationDate="2011-01-24T17:11:49.650" Score="0" ViewCount="" Body="&lt;p&gt;A useful source of useful identities (most of which aren't proved though) is &lt;a href=&quot;http://matrixcookbook.com/&quot; rel=&quot;nofollow&quot;&gt;the matrix cookbook&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1676" LastActivityDate="2011-01-24T17:11:49.650" />
  <row Id="18766" PostTypeId="2" ParentId="18759" CreationDate="2011-01-24T17:13:35.693" Score="2" ViewCount="" Body="&lt;p&gt;We have obvious injective inclusions $2^\mathbb{N} \hookrightarrow 2^\mathbb{Q} \hookrightarrow 3^\mathbb{Q}$ and the first set - i.e. the set of functions from the positive integers to ${0,1}$, or if you will the indicator functions on $\mathbb{N}$ or the powerset of $\mathbb{N}$ - is uncountably infinite by a standard diagonal argument.&lt;/p&gt;&#xA;" OwnerUserId="1269" LastActivityDate="2011-01-24T17:13:35.693" />
  <row Id="18768" PostTypeId="2" ParentId="18764" CreationDate="2011-01-24T17:30:28.720" Score="3" ViewCount="" Body="&lt;p&gt;Your solution will be $(y_{n+1},y_n)^T=B^n(y_1,y_0)^T$ so if you expand $(y_1,y_0)^T$ in terms of eigenvectors of $B$, any one which corresponds to an eigenvalue greater than 1 in absolute value will blow up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Added in response to the comment:&#xA;You find the eignevectors of $B$.  As long as $B$ is well-behaved, they will span the space, so form a basis.  So if the eigenvectors are $\vec{x_1}$ and $\vec{x_2}$ corresponding to $\lambda_1$ and $\lambda_2$ respectively,you would write $(y_1,y_0)^T=a_1\vec{x_1}+a_2\vec{x_2}$ and get$(y_{n+1},y_n)^T=a_1\lambda_1^n\vec{x_1}+a_2\lambda_2^n\vec{x_2}$&lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2011-01-24T18:09:23.640" LastActivityDate="2011-01-24T18:09:23.640" CommentCount="1" />
  <row Id="18769" PostTypeId="2" ParentId="18090" CreationDate="2011-01-24T17:30:32.080" Score="3" ViewCount="" Body="&lt;p&gt;A recurrent state will satisfy property (2), but not the other way around. Take a two state Markov chain that jumps from $i$ to $j$ with probability one, then stays there forever. Then $i$ is transient, but $P(T_{ij}&amp;lt;\infty)=1$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt;&#xA; Here's a sketch of (1) implies (2) in terms of &quot;taboo probabilities&quot;.&#xA;We assume that $i$ is recurrent, and all of the probabilities below are conditioned on $X_0=i$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Define $T_i=\inf(n\geq 1 : X_n=i)$ and $T_j=\inf(n\geq 1 : X_n=j)$.&#xA;We want to show that $P(T_j&amp;lt;\infty)&gt;0$ implies $P(T_j&amp;lt;\infty)=1$. It suffices to assume $i\not= j$, since $P(T_i&amp;lt;\infty)=1$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $i$ is recurrent, we have $P(T_i &amp;lt; \infty)=1$ so that &#xA;$P(T_j &amp;lt; T_i)+P(T_i &amp;lt; T_j)=1,$ since $T_i=T_j$ implies $T_i=\infty$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, if $P(T_j &amp;lt; \infty)&gt;0$ we must have $P(T_j &amp;lt; T_i)&gt;0$,&#xA;by considering the shortest possible path from $i$ to $j$ with non-zero probability.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, letting $n$ represent the number of times we return to $i$ without &#xA;hitting $j$, the strong Markov property gives &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$P(T_j &amp;lt; \infty)=\sum_{n=0}^\infty P(T_i &amp;lt; T_j)^n P(T_j &amp;lt; T_i)&#xA;  =\sum_{n=0}^\infty P(T_i &amp;lt; T_j)^n [1-P(T_i &amp;lt; T_j)]=1.$$  &lt;/p&gt;&#xA;" OwnerUserId="940" LastEditorUserId="940" LastEditorDisplayName="" LastEditDate="2011-01-24T21:36:09.313" LastActivityDate="2011-01-24T21:36:09.313" CommentCount="1" />
  <row Id="18770" PostTypeId="2" ParentId="18090" CreationDate="2011-01-24T17:37:17.577" Score="0" ViewCount="" Body="&lt;p&gt;I think the first property is wrong. That would be a strictly periodic state. Perhaps you missed something?&lt;/p&gt;&#xA;" OwnerUserId="312" LastActivityDate="2011-01-24T17:37:17.577" CommentCount="5" />
  <row Id="18771" PostTypeId="1" AcceptedAnswerId="18856" CreationDate="2011-01-24T17:40:34.330" Score="5" ViewCount="179" Body="&lt;p&gt;Show that if a manifold $M$ is contractible, then every vector bundle over $M$ is equivalent to the trivial bundle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Got this as homework but I'm kind of lost in the hole vector bundle subject, so if anyone could give some hints to this one as well as to understanding the subject, it would be highly appreciated&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="5209" LastActivityDate="2011-01-30T22:36:55.557" Title="Vector Bundles Over a Manifold." Tags="&lt;homework&gt;&lt;topology&gt;&lt;differential-geometry&gt;" AnswerCount="3" />
  <row Id="18772" PostTypeId="2" ParentId="18713" CreationDate="2011-01-24T17:59:00.277" Score="0" ViewCount="" Body="&lt;p&gt;As many have commented, the notion of $S$ being dense &lt;em&gt;by itself&lt;/em&gt; (i.e., as an ordered set) is different from the notion of $S$ being a dense subset of $\mathbb{R}$.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;For the former notion, the definition should refer only to the set itself: $(S, &amp;lt;)$ is dense if for any $x,y \in S$ such that $x&#xA;&lt;li&gt;For the latter notion, the definition needs to refer to the ambient set: $S \subset \mathbb{R}$ is dense if for any $x, y \in \mathbb{R}$ such that $x&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The latter notion is closely related to the former in the case of real numbers because the open intervals $(x, y)$ form a basis for the usual topology on $\mathbb{R}$.&lt;/p&gt;&#xA;" OwnerUserId="5546" LastActivityDate="2011-01-24T17:59:00.277" />
  <row Id="18773" PostTypeId="2" ParentId="18688" CreationDate="2011-01-24T18:04:35.763" Score="1" ViewCount="" Body="&lt;p&gt;The result follows immediately if you know the Stirling Numbers of the second kind, and it's combinatoric  meaning. Not an algebraic proof, though.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Update:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In table 4 of these &lt;a href=&quot;http://www.math.wvu.edu/~gould/&quot; rel=&quot;nofollow&quot;&gt;Combinatorial Identities&lt;/a&gt; it is stated an identity that -I believe- is equivalent to yours:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ \displaystyle \sum_{k=0}^n (-1)^k {n \choose k} k^j = (-1)^n n!$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;if $j=n$, (zero otherwise). And it just say that this is a &lt;strong&gt;&quot;Statement of Euler's finite difference theorem&quot;&lt;/strong&gt;. I don't know what this means, and I would like to know.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Update 2:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Indeed, the identity follows from the formula of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Finite_difference#n-th_difference&quot; rel=&quot;nofollow&quot;&gt;n-th finite difference&lt;/a&gt;, taking the function $f(x)=x^j$&lt;/p&gt;&#xA;" OwnerUserId="312" LastEditorUserId="312" LastEditorDisplayName="" LastEditDate="2011-01-24T20:47:42.460" LastActivityDate="2011-01-24T20:47:42.460" CommentCount="1" />
  <row Id="18774" PostTypeId="2" ParentId="942" CreationDate="2011-01-24T18:05:56.247" Score="0" ViewCount="" Body="&lt;p&gt;Hey akhil: &#xA;If you find a closed point in F, doesnt that mean by minimality that the point {x} IS F?&lt;/p&gt;&#xA;" OwnerUserId="6196" LastActivityDate="2011-01-24T18:05:56.247" CommentCount="1" />
  <row Id="18775" PostTypeId="1" AcceptedAnswerId="18799" CreationDate="2011-01-24T18:08:04.250" Score="2" ViewCount="96" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Ultrafilter&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt; says that if $a\le b$ and $b\le a$ in Rudin–Keisler order for ultrafilters $a$ and $b$, then $a$ and $b$ are Rudin–Keisler equivalent. How to prove this?&lt;/p&gt;&#xA;" OwnerUserId="4876" LastEditorUserId="413" LastEditorDisplayName="" LastEditDate="2011-01-24T21:29:29.833" LastActivityDate="2011-01-24T21:35:59.343" Title="Rudin–Keisler equivalence" Tags="&lt;set-theory&gt;&lt;order-theory&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="18776" PostTypeId="1" CreationDate="2011-01-24T18:22:41.617" Score="2" ViewCount="155" Body="&lt;p&gt;In 2D or 3D I have a fix point y and Gaussian distribution of a random point x. I am now interested in the mean euclidean distance between x and y:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$E_x [ d(x,y) ] = \int _{-\infty }^{\infty } d(x,y) N(x; \mu, \Lambda )dx$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many thanks in advance&lt;/p&gt;&#xA;" OwnerUserId="6197" LastEditorUserId="6197" LastEditorDisplayName="" LastEditDate="2011-01-24T18:31:16.757" LastActivityDate="2011-02-09T23:26:18.880" Title="Mean distance between a fixed point and a gaussian distributed random variable" Tags="&lt;probability&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="18777" PostTypeId="2" ParentId="18771" CreationDate="2011-01-24T18:24:33.137" Score="2" ViewCount="" Body="&lt;p&gt;Show that if $f,g:X\to Y$ are two maps between manifolds which are homotopic, and if $E$ is a vector bundle on $Y$, then $f^{-1}(E)$ and $g^{-1}(E)$ are isomorphic vector bundles on $X$. (You don't really need that both $X$ and $Y$ be manifolds: you can get by if they are just spaces and $X$ is paracompact...)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(You should really provide information about your background, otherwise it is more or less impossible to come up with a useful answer, you see!)&lt;/p&gt;&#xA;" OwnerUserId="274" LastEditorUserId="274" LastEditorDisplayName="" LastEditDate="2011-01-30T22:36:55.557" LastActivityDate="2011-01-30T22:36:55.557" CommentCount="2" />
  <row Id="18778" PostTypeId="1" AcceptedAnswerId="18787" CreationDate="2011-01-24T18:30:45.510" Score="4" ViewCount="110" Body="&lt;p&gt;Given a sequence of sets, is there some well-defined notion of a limit of a set? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is, given some universe set $U$, is there topology on $2^U$ (powersets of $U$) such that the usual intersection and the union limits converge to what they normally converge to in that topology?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, if I were to take the following sequence of sets for $U=\mathbb{N}$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$S_n = \{x\in \mathbb{N} | n&amp;lt; x \le 2n \}$, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;or even simpler, something like:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$T_n = \{x\in \mathbb{N} | n\le x &amp;lt; n+1 \} =\{n\}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm wondering whether there is a notion of convergence that can say whether the limit of such a sequence makes sense or not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On one hand, it feels like the limit of both sequences above should be the empty set, by the following type of argument:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;S_n &amp;amp;\subset (n,\infty) \\&#xA;\lim_{n\to\infty} S_n &amp;amp;\subset \lim_{n\to\infty} (n,\infty) = \cap_{n\in\mathbb{N}} (n,\infty) = \emptyset &#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And of course, the same for $T_n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, I don't see why I should be able to pass a set inclusion to the limit. (I feel like I'm just declaring that this should be a property of limits of sets...)&lt;/p&gt;&#xA;" OwnerUserId="2499" LastActivityDate="2011-01-24T19:18:26.590" Title="Convergence of Sets? (Topology on a Powerset of a Set?)" Tags="&lt;elementary-set-theory&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="18779" PostTypeId="1" CreationDate="2011-01-24T18:32:56.880" Score="1" ViewCount="73" Body="&lt;p&gt;I'm trying to decipher a differential geometric comment on page 23-24 of Berline, Getzler, and Vergne's &quot;Heat Kernels and Dirac Operators&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take a trivial vector bundle $E \times M$ in a manifold $M$ with connection $\nabla = d + \omega$ where $\omega$ is an $End(E)$-valued 1 form.  Let $g: GL(E) \to End(E)$ be the tautological map sending a linear map in $GL(E)$ to itself as an element of $End(E)$.  The claim is that the connection 1-form on the (trivial) frame bundle for $E \times M$ is given by $g^{-1} \pi^* \omega g + g^{-1} d g$.  In particular, if $\omega = 0$ then we get that the trivial connection on the trivial bundle is the Maurer-Cartan 1-form.  Unfortunately, I don't see how to give a convincing proof of this - can someone help?&lt;/p&gt;&#xA;" OwnerUserId="1509" LastActivityDate="2011-01-24T18:32:56.880" Title="Maurer-Cartan 1- form as a connection 1-form" Tags="&lt;differential-geometry&gt;" FavoriteCount="1" />
  <row Id="18780" PostTypeId="1" AcceptedAnswerId="18786" CreationDate="2011-01-24T18:39:58.843" Score="1" ViewCount="228" Body="&lt;p&gt;&lt;strong&gt;Background&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am trying to extract data from scientific publications. Sometimes an experiment can test two factors, e.g. a standard two-way factorial ANOVA. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Call these factors $A$ and $B$. If factor $A$ has two levels and $B$ has three, there are six total treatments.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the effects of $A$ and $B$ are significant, but there is no interaction, only the 'Main' effects might be presented, e.g. five results, one for each level of $A$ and one for each level of $B$, averaged across all of the levels of the other factor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is an example from &lt;a href=&quot;http://dx.doi.org/10.1016/S0961-9534%2801%2900008-3&quot; rel=&quot;nofollow&quot;&gt;Ma 2001&lt;/a&gt; Table 2, in which $A$ would be the row spacing and $B$ would be the nitrogen rate. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/mvMPw.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$7577 = \frac{X_{A_{20},B_{0}} + X_{A_{20},B_{112}} + X_{A_{20},B_{224}}} {3}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$9186 = \frac{X_{A_{80},B_{0}} + X_{A_{80},B_{112}} + X_{A_{80},B_{224}}} {3}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$3706 = \frac{X_{A_{20},B_{0}} + X_{A_{80},B_{0}}} {2}$$&#xA;$$9402 = \frac{X_{A_{20},B_{112}} + X_{A_{80},B_{112}}} {2}$$&#xA;$$12038 = \frac{X_{A_{20},B_{224}} + X_{A_{80},B_{224}}} {2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it possible to calculate the means of each of the six treatments $X_{A,B}$, for $A\in[20,80]$ by $B\in[0,112,224]$ from these results? &lt;/p&gt;&#xA;" OwnerUserId="3733" LastEditorUserId="3733" LastEditorDisplayName="" LastEditDate="2011-01-25T00:28:40.033" LastActivityDate="2011-01-25T02:33:45.417" Title="Given 5 equations, can I solve for 6 unknowns?" Tags="&lt;linear-algebra&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="18781" PostTypeId="2" ParentId="18688" CreationDate="2011-01-24T18:44:50.833" Score="6" ViewCount="" Body="&lt;p&gt;Start with &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{j=0}^{n} (-1)^j \binom{n}{j} (1+x)^{n-j} = x^n$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which follows readily from the binomial theorem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now differentiate, and multiply by $\displaystyle (1+x)$. Repeat this $\displaystyle n$ times and set $\displaystyle x = 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice that the constant term of the resulting polynomial on the right side is $\displaystyle n!$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can prove by induction that the lowest degree of $\displaystyle x$ that appears on the right side after $\displaystyle k$ steps ($0 \lt \displaystyle k \le n$) is $\displaystyle x^{n-k}$ and has the coefficient $\displaystyle n(n-1)\dots(n-k+1)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This gives us the set of identities&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{j=0}^{n} (-1)^j \binom{n}{j} (n-j)^{k} = 0, \ \ 0 \le k \lt n$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{j=0}^{n} (-1)^j \binom{n}{j} (n-j)^{n} = n!$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also seem to recollect there was a proof involving $\displaystyle \log x$, I will update this answer later if I remember it.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2011-01-24T20:33:29.960" LastActivityDate="2011-01-24T20:33:29.960" CommentCount="4" />
  <row Id="18782" PostTypeId="2" ParentId="18780" CreationDate="2011-01-24T18:50:16.023" Score="2" ViewCount="" Body="&lt;p&gt;The general rule is that $n$ equations allows you to solve for $n$ unknowns. So I don't think you'll be able to recover each of the original 6 data points. The best you can do is produce a set of constraints that, given any one of the 6 data points (or a relationship between them that isn't redundant with what you already have), would allow you to find the remaining 5.&lt;/p&gt;&#xA;" OwnerUserId="1190" LastActivityDate="2011-01-24T18:50:16.023" CommentCount="3" />
  <row Id="18783" PostTypeId="2" ParentId="18748" CreationDate="2011-01-24T18:52:45.783" Score="1" ViewCount="" Body="&lt;p&gt;The ampersand &amp;amp; is unmistakeable and just about right in semi-formal statements where &quot;and&quot; would be too wordy and a comma would be not very clear. The notation $\land$ is appropriate for formal logic, but isn't used much in general mathematics. &lt;/p&gt;&#xA;" OwnerUserId="875" LastActivityDate="2011-01-24T18:52:45.783" />
  <row Id="18784" PostTypeId="2" ParentId="18668" CreationDate="2011-01-24T18:53:49.117" Score="0" ViewCount="" Body="&lt;p&gt;Trying to follow up on Aaron's hints. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So one way to look at $X$ is as the product space $D^2 \times S^2$ with an equivalence relation $E$ on the boundary $\partial D^2 \times S^2 = S^1 \times S^2$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Equivalence relation $E$ on the boundary is : $({ b_1 \in S^1 }, { b_2 \in S^2 }) \sim ({ b_1 \in S^1 }, { Rb_2 \in S^2 })$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;SO $X = (D^2 \times S^1) /E$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now consider the following subspace $Y$ of $X$:&lt;br&gt;&#xA;$Y = { [-1,1] \times S^2 }$  with $ E_1 = (-1;y \in S^2) \sim (-1;Ry \in S^2)$ and &#xA;$E_2 = (1;y \in S^2) \sim (1;Ry \in S^2)$ as equivalence relations on $Y$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$I=[-1,1]$ is a line passing through the origin of $D^2$. I think I could prove $Y$ to be homeomorphic to $S^3$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please check: This is probably a very crude way to show the homeomorphism, but this is the only way I know how to do this. May be there are some algebraic methods to prove this result ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Visualize $Y$ as a combination two solid cylinders of $I \times D^2$. So I have two solid cylinders, $C_1$ and $C_2$, when joined along the cylindrical surfaces form $Y$. So each point on the cylindrical surface of $C_1$ has an equivalent point on the cylindrical surface of $C_2$, since they belong to the same sphere $S^2$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The disks that belong to the top and bottom surfaces of cylinder $C_1$ are identified with the disks of cylinder $C_2$ due to equivalence relations $E_1,E_2$ on the boundaries of subspace $I \times S^2$ that it inherits from $E$. The relations are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$E_1 = (-1;y \in S^2) \sim (-1;Ry \in S^2)$ and $E_2 = (1;y \in S^2) \sim (1;Ry \in S^2)$. \&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence I have two cylinders $C_1$ and $C_2$ where any point on the surface of $C_1$ has one and only one equivalent point on the surface of $C_2$. A solid cylinders is homeomorphic to a solid ball, hence the subspace $Y$ is equivalent to two solid spheres where the surface points of one solid sphere are identified with surface points of the other solid sphere. This space is equivalent to $S^3$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;May be there is some way to get $S^4$ by extending the subspace $(I \times S^2)/[E_1,E_2]$ to the entire $(D^2 \times S^2)/E$ ?? &lt;/p&gt;&#xA;" OwnerUserId="3050" LastEditorUserId="3050" LastEditorDisplayName="" LastEditDate="2011-01-24T18:58:50.203" LastActivityDate="2011-01-24T18:58:50.203" />
  <row Id="18785" PostTypeId="2" ParentId="10824" CreationDate="2011-01-24T18:55:02.410" Score="1" ViewCount="" Body="&lt;p&gt;When I was an electrical engineer in the 1950s we used logarithms (or a slide rule) to calculate numerical answers, in the days before computers and hand held electronic calculators. $Log_{10}$ tables give the logarithms of numbers from 1 to 10, i.e. values between 0 and 1. Let's do the simple calculation of $0.345 \times 3 = 1.035$ using logarithms. We generally used 'four figure tables' to get three significant figures (3 s.f.) in our answers, and, surprisingly, this was accurate enough for most of our needs. We also had five figure tables, and, for really accurate work, we used &quot;Chamber's six figure tables&quot; to get 5 s.f. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Looking up $3.45$ in the tables gives us $0.5378$. Now $0.345$ is $3.45$ times $0.1$ so $log(0.345)$ is $log(3.45) + log(0.1)$ or $0.5378 -1 = -0.4622$, which is what you get if you look up $log(0.345)$ on a calculator. $Log(3)$ is $0.4771$, so $log(3\times0.345)$ is $0.4771-0.4622 = 0.0149$. This is the log of $1.0349$, or $1.035$ to 3 s.f. (which we would look up in &quot;anti-log&quot; tables).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a long calculation it was awkward to have some additions and some subtractions in the process, so we note that we could have written the sum as $(3.45\times0.1)\times3$ and the log of the answer is $log(3.45)+log(0.1)+log(3)$ or $0.5378+(-1)+0.4771$. We did not have to subtract to find $log(0.345)$, instead we have replaced it by the simple addition of $-1$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This was incorporated into the practice of using logarithms, by writing the log as two parts: a value between 1 and 10, which we could look up directly, called the 'Mantissa', and a multiplier, a power of ten, always a whole number and called the 'Characteristic'. So, since $0.345$ is $3.45\times10^{- 1}$, its logarithm could be written as $-1+0.5378$. A second short cut was to write this as $\overline{1}.5378$, the overline being a reminder to treat just this integer part as a negative number when adding up. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, we would have written this sum as: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Number  Log&lt;br&gt;&#xA;$0.345\ \ \ \ \ \ \ \overline{1}.5378$&lt;br&gt;&#xA;$3\ \ \ \ \ \ \ \ \ +\ \underline{0.4771}$&lt;br&gt;&#xA;$\ \ \ \ \ \ \ \ \ \ \ = 0.0149\ \ Ans:\ 1.0349$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We would have left out the column headings, and the plus and equals signs in day-to-day calculations.&lt;/p&gt;&#xA;" OwnerUserId="6155" LastEditorUserId="6155" LastEditorDisplayName="" LastEditDate="2011-01-24T19:30:01.713" LastActivityDate="2011-01-24T19:30:01.713" />
  <row Id="18786" PostTypeId="2" ParentId="18780" CreationDate="2011-01-24T18:58:14.143" Score="3" ViewCount="" Body="&lt;p&gt;You can do it if you make some assumption to reduce the number of unknowns to five.  You are saying you have an array&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{array} {ccc}  &amp;amp; 20 &amp;amp; 80 \\ 0 &amp;amp; a &amp;amp; b \\ 112 &amp;amp; c &amp;amp; d \\ 224 &amp;amp; e &amp;amp; f \end{array}$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $a$ through $f$ are what you want to solve for. If the effects are independent and additive, you would expect $b-a=d-c=f-e$, $e-c=f-d$, and $c-a=d-b$.  These reduce the data to only three values, which you can check for consistency.  But without at least one more relation, you will get a one-dimensional continuum of solutions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It sounds like you believe 3706 is some sort of weighted average of $a$ and $b$ and similarly for the other entries.  Is that right?&lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2011-01-24T19:28:58.797" LastActivityDate="2011-01-24T19:28:58.797" CommentCount="5" />
  <row Id="18787" PostTypeId="2" ParentId="18778" CreationDate="2011-01-24T19:10:39.280" Score="4" ViewCount="" Body="&lt;p&gt;The natural topology on $2^U$ is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Compact-open_topology&quot; rel=&quot;nofollow&quot;&gt;compact-open topology&lt;/a&gt;, which here is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Product_topology&quot; rel=&quot;nofollow&quot;&gt;product topology&lt;/a&gt;.  This is precisely the topology of pointwise convergence of indicator functions $U \to 2$.  Thus a sequence $S_1, S_2, ...$ of sets converges in this topology if and only if, for every $u \in U$, either all but finitely many $S_i$ contain $u$ (so that $u$ is in the limit set) or all but finitely many $S_i$ do not contain $u$ (so that $u$ is not in the limit set).  So both of the sequences you describe have limit the empty set as desired.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Equivalently (I think), one can define a sequence of sets to converge if its liminf and limsup (defined in the usual way) converge to the same set.&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2011-01-24T19:18:26.590" LastActivityDate="2011-01-24T19:18:26.590" CommentCount="11" />
  <row Id="18788" PostTypeId="1" AcceptedAnswerId="18792" CreationDate="2011-01-24T19:26:59.483" Score="3" ViewCount="87" Body="&lt;p&gt;Hello,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for $f \in L^1_{loc}$, let the sharp-function be defined as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f^\sharp(x) := \sup_{B \in x} |B|^{-1} \int_B |f(y) - |B|^{-1} \int_B f(z) dz | dy$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We define the space $BMO \subset L^1_{loc}$ via&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$BMO := \{ f \in L^1_{loc} \;\;\;s.t.\;\;\; f^\sharp \in L^{\infty}  \}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and the BMO-(semi)norm on BMO as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Vert f \Vert_{BMO} = \Vert f^\sharp \Vert_{\infty}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The completeness of BMO can be infered from the fact that it is exactly the dual space of the Hardy space $H^1$ with operator norm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a more direct way to show completeness of BMO?&lt;/p&gt;&#xA;" OwnerUserId="3557" LastEditorUserId="3557" LastEditorDisplayName="" LastEditDate="2011-01-24T23:14:27.520" LastActivityDate="2011-01-24T23:14:27.520" Title="Completeness of BMO without duality to $H^1$" Tags="&lt;harmonic-analysis&gt;&lt;banach-spaces&gt;" AnswerCount="1" />
  <row Id="18790" PostTypeId="1" AcceptedAnswerId="18791" CreationDate="2011-01-24T19:57:02.953" Score="2" ViewCount="56" Body="&lt;p&gt;I have a determinant to expand which is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\triangle = &#xA;\begin{bmatrix}&#xA; p&amp;amp; 1  &amp;amp; \frac{-q}{2}{}\\&#xA; 1&amp;amp; 2 &amp;amp;-q \\&#xA; 2&amp;amp; 2 &amp;amp; 3&#xA;\end{bmatrix} = 0 $$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But when I am expanding the determinant along the first row such as $ p(6+2q) - (3+2q)  = 0 $ but when I am trying to expand along first column I am getting $p(6+2q) - (3+q)  = 0$ but I have been told by my teacher that we can expand the determinant of $3\times 3$ matrix in along any row and any column giving the same result. where lies the error ? &lt;/p&gt;&#xA;" OwnerUserId="6199" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-24T20:11:57.747" LastActivityDate="2011-01-24T20:11:57.747" Title="Help me spot the error ?" Tags="&lt;linear-algebra&gt;&lt;matrices&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="18791" PostTypeId="2" ParentId="18790" CreationDate="2011-01-24T20:01:59.173" Score="2" ViewCount="" Body="&lt;p&gt;Seems you like you forgot the third column ($\frac{-q}{2}$) when expanding using the first row.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the first row expansion, just using the first and second columns gives you $p(6+q) - (3 +2q)$. The third column gives an extra $q$, after adding which it matches the second expression you got, using the first column.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2011-01-24T20:01:59.173" CommentCount="1" />
  <row Id="18792" PostTypeId="2" ParentId="18788" CreationDate="2011-01-24T20:07:22.180" Score="4" ViewCount="" Body="&lt;p&gt;Yes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have the following theorem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $f \in \text{BMO}(\mathbb R^n)$. Then for any $\epsilon &gt; 0$ there exists a ball $B$ with center $x_0$ and radius $R$ such that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$R^\epsilon \int_{\mathbb R^n} \frac{|f(x) - \text{Avg}_B f|}{(R + |x - x_0|)^{n + \epsilon}} \, dx \leq C \|f\|_\text{BMO}$$&#xA;where $C$ only depends on $n$ and $\epsilon$. (this is just estimating stuff)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So from this theorem we can deduce that any $\text{BMO}$-Cauchy sequence is Cauchy in $L^1$ on every compact set. From this we can deduce that every Cauchy sequence converges.&lt;/p&gt;&#xA;" OwnerUserId="1120" LastActivityDate="2011-01-24T20:07:22.180" CommentCount="1" />
  <row Id="18793" PostTypeId="1" CreationDate="2011-01-24T20:15:33.670" Score="3" ViewCount="113" Body="&lt;p&gt;It is clear that cyclic groups have the property that they cannot be written in a non-trivial way as an amalgamated free product or as an HNN-extension.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can someone please provide us examples of torsion-free 2-generated groups having this property? Any comments related to this question are welcome!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[I'm aware of Serre's Theorem 15 in Trees.]&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; The term &quot;in a non-trivial way&quot; means the following.&#xA;For an amalgamated free product $G=H*_C K$ it is required $H\ne C\ne K$. For an HNN-extension $G=HNN(H,A,B,t)$ it is required $A\ne H\ne B$.&lt;/p&gt;&#xA;" OwnerUserId="6200" LastEditorUserId="6200" LastEditorDisplayName="" LastEditDate="2011-01-25T00:36:32.067" LastActivityDate="2011-01-27T16:18:28.117" Title="Groups that are not amalgams" Tags="&lt;group-theory&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="18794" PostTypeId="2" ParentId="18589" CreationDate="2011-01-24T20:33:23.220" Score="6" ViewCount="" Body="&lt;p&gt;While not nearly as impressive as George's amazing answer, it's worth noting that $\mathbb{Q}[\phi]$ (though not quite $\mathbb{Z}[\phi]$, the elements are in $\frac{1}{2}\mathbb{Z}[\phi]$) shows up in the icosians (a subgroup of order 120 of the group of unit quaternions) and the theory of the icosahedral group and the 600-cell (and even, tangentially, $E_8$); check out the Wikipedia page on the icosians for more details.  (It's, loosely, a higher-dimensional version of the description of the vertices of the icosahedron as the corners of the golden rectangle $(0, \pm 1, \pm \phi)$ and the other two rectangles given by cyclic permutations of the coordinates)&lt;/p&gt;&#xA;" OwnerUserId="785" LastActivityDate="2011-01-24T20:33:23.220" />
  <row Id="18797" PostTypeId="2" ParentId="18776" CreationDate="2011-01-24T20:52:11.707" Score="1" ViewCount="" Body="&lt;p&gt;You can define a new random variable $\tilde{X} = X-y$. Then the quantity you are interested in is really the expected distance of this random variable from the origin. Let $\tilde{X}$ be the column vector $[x_1 x_2 \dots x_n]^T$. Then the squared distance from the origin is $\tilde{X}^T\tilde{X}$. For the case when the components of $\tilde{X}$ are independent, this expectation is a &lt;a href=&quot;http://en.wikipedia.org/wiki/Noncentral_chi-square_distribution&quot; rel=&quot;nofollow&quot;&gt;non-central chi-square distribution&lt;/a&gt; with $n$ degrees of freedom. For the case when $\tilde{X}$ has a general (non-diagonal) covariance matrix, the distribution of $\tilde{X}^T \tilde{X}$ will be a &lt;a href=&quot;http://en.wikipedia.org/wiki/Generalized_chi-square_distribution&quot; rel=&quot;nofollow&quot;&gt;generalized chi-square distribution&lt;/a&gt;. If you are interested in the expected value of the absolute distance (i.e., $E(\sqrt{\tilde{X}^T \tilde{X}})$), try looking at the &lt;a href=&quot;http://en.wikipedia.org/wiki/Chi_distribution&quot; rel=&quot;nofollow&quot;&gt;chi distribution&lt;/a&gt;. The chi distribution is the square root of the chi-square distribution and hence will give you the expected absolute distance if the components of $\tilde{X}$ are independent. If they are not, there should be a corresponding generalized chi distribution that is the square root of the generalized chi-square distribution.&lt;/p&gt;&#xA;" OwnerUserId="618" LastActivityDate="2011-01-24T20:52:11.707" CommentCount="2" />
  <row Id="18798" PostTypeId="2" ParentId="18688" CreationDate="2011-01-24T21:05:05.563" Score="5" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Lemma:&lt;/strong&gt; Let $f(j) = \sum_{k=0}^n f_k j^k$ be a degree $n$ polynomial. I claim that $\sum_{j=0}^{n} (-1)^j \binom{n}{j} f(j) = n! f_n$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof:&lt;/strong&gt; For any polynomial $g$, define $\Delta(g)$ to be the polynomial $\Delta(g)(j) = g(j) - g(j+1)$. Observe that, if $g$ is a polynomial with leading term $a x^d + \cdots$, then $\Delta(g)$ is a polynomial with leading term $-d a x^{d-1} + \cdots$. In particular, if $f$ is as in the statement of the lemma, then $\Delta^n(f)$ is the constant $(-1)^n n! f_n$. The sum in question is $\Delta^n(f)$ evaluated at $0$. QED&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, apply the lemma to $f(j) = (1-j/n)^n$.&lt;/p&gt;&#xA;" OwnerUserId="448" LastEditorUserId="448" LastEditorDisplayName="" LastEditDate="2011-02-01T12:52:11.427" LastActivityDate="2011-02-01T12:52:11.427" />
  <row Id="18799" PostTypeId="2" ParentId="18775" CreationDate="2011-01-24T21:08:28.087" Score="2" ViewCount="" Body="&lt;p&gt;By composing the two witness functions, it suffices to prove the following fact, which is due I think to Solovay.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Theorem.&lt;/b&gt; If $\mu$ is an ultrafilter on a set $I$ and $f:I\to I$ has the property that $X\in\mu\leftrightarrow f^{-1}X\in\mu$, then $f$ is the identity function on a set in $\mu$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proof. If we regard the function $f$ as a set of ordered pairs, it makes a directed graph on vertex set $I$. By the Axiom of Choice, let $D$ select exactly one member from each connected component of this graph. The components of this graph are the same as the equivalence classes under the relation $x\sim y\leftrightarrow f^i(x)=f^j(y)$ for some finite iterates $i$ and $j$ of $f$. Thus, for every point $x\in I$ there is a unique $y\in D$ such that $f^i(x)=f^j(y)$ for some finite iterates $i$ and $j$ of $f$. Let $A$ be the set of $x$ for which the minimal such $i+j$ is even. Note that if $f(x)\neq x$, then $x\in A\implies f(x)\notin A$, since there will be exactly one more application of $f$. In other words, if $E$ is the set of fixed points of $f$, then $A-E$ is disjoint from $f[A-E]$. From this, it follows by our assumption that $A-E\notin\mu$, since otherwise we would have $A-E\in\mu$ and hence $f[A-E]\in\mu$, but these are disjoint. Thus, $E\in\mu$, and so $f$ is the identity on a set in $\mu$. QED&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This argument uses AC, and I think I recall hearing that AC is required. Thus, I would be very curious to see an argument appealing to Schroeder-Bernstein, since that argument does not use AC. But I suppose that there is a certain affinity of this argument and the Schroeder-Bernstein proof, and perhaps this is what Andres meant. (In any case, does anyone know a reference for showing that the result can fail without AC?) &lt;/p&gt;&#xA;" OwnerUserId="413" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-24T21:35:59.343" LastActivityDate="2011-01-24T21:35:59.343" CommentCount="6" />
  <row Id="18800" PostTypeId="1" CreationDate="2011-01-24T21:12:52.227" Score="0" ViewCount="119" Body="&lt;p&gt;Hello alltogether&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While looking through our Analysis Skript, i got suddenly aware that we had already learned quite an amount of rules and laws for calculating limits.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We got:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\lim\limits_{x \to a} c = c$&lt;br /&gt;&#xA;$\lim\limits_{x \to a} x = a$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the limits exist, one gets with corresponding limits $u,v$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\lim\limits_{x \to a} f(x) \pm g(x) = u \pm v \quad$ &lt;br /&gt; &#xA;$\lim\limits_{x \to a} cg(x) = cv$ &lt;br /&gt;&#xA;$\lim\limits_{x \to a} f(x) g(x) = uv$ &lt;br /&gt;&#xA;$\lim\limits_{x \to a} \frac{f(x)}{g(x)} = \frac{u}{v} \quad $ if $v \neq 0$ &lt;br /&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the limits exist and if $f(x)$ is continous:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\lim\limits_{x \to a} f(x) = f(a)$ &lt;br /&gt;&#xA;$\lim\limits_{x \to a} f(x) \pm g(x) = f(a) \pm g(a)$ &lt;br /&gt;&#xA;$\lim\limits_{x \to a} cf(x) = c f(a)$ &lt;br /&gt;&#xA;$\lim\limits_{x \to a} f(x)g(x) = f(a)g(a)$ &lt;br /&gt;&#xA;$\lim\limits_{x \to a} \frac{f(x)}{g(x)} = \frac{f(a)}{g(a)} \quad $ if $g(a) \neq 0$ &lt;br /&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $f$ is continuous and $\lim\limits_{x \to a} g(x)$ exists and if $f(x)$ is defined at $\lim\limits_{x \to a}g(x)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\lim\limits_{x \to a} f(g(x)) = f(\lim\limits_{x \to a}g(x))$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As rules we have:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\forall x$ in an open intervall, containing $a$: &lt;br /&gt; $u(x)\leq v(x)\leq w(x)  \Rightarrow \lim\limits_{x \to a} u(x) \leq \lim\limits_{x \to a} v(x) \leq \lim\limits_{x \to a} w(x)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;if $f(x)$ and $g(x)$ is differentiable and if  $\lim\limits_{x \to a} f(x),g(x) = \pm \infty$ or $\lim\limits_{x \to a} f(x),g(x) = 0$ and if $\lim\limits_{x \to a} \frac{f&amp;#39;(x)}{g&amp;#39;(x)}$ exists&lt;br /&gt;&#xA;$\lim\limits_{x \to a}\frac{f(x)}{g(x)} = \lim\limits_{x \to a} \frac{f&amp;#39;(x)}{g&amp;#39;(x)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But there is definitly more to it, isn't there? For example, when am I allowed to pull a $lim$ into an infinte sum? I bet Math is simply not ending here. What other rules and laws for calculating limits are there?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt; Updeted the rules based on the comments.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance&lt;/p&gt;&#xA;" OwnerUserId="3188" LastEditorUserId="3188" LastEditorDisplayName="" LastEditDate="2011-01-25T20:04:54.253" LastActivityDate="2011-01-25T20:04:54.253" Title="List of rules and laws for calculating limits" Tags="&lt;analysis&gt;&lt;limit&gt;" CommentCount="11" />
  <row Id="18801" PostTypeId="1" AcceptedAnswerId="18807" CreationDate="2011-01-24T21:22:28.657" Score="1" ViewCount="71" Body="&lt;p&gt;Hi, I want to find out, how many ways I can produce an average 4.6 of 13 numbers, when I have only numbers {1,2,3,4,5}.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I thought, that it could be done by generating functions, but since my knowledge on them isn't very deep yet, I'm a little lost. Also I have a hard time dealing with the average not being integer.&lt;/p&gt;&#xA;" OwnerUserId="4447" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2011-01-24T21:47:03.267" LastActivityDate="2011-01-24T21:59:35.833" Title="Generating function for an Arithmetic mean" Tags="&lt;generating-functions&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="1" />
  <row Id="18802" PostTypeId="1" AcceptedAnswerId="18840" CreationDate="2011-01-24T21:29:01.830" Score="3" ViewCount="170" Body="&lt;p&gt;I would appreciate it if someone could explain the difference(s) between a&#xA;$C^\infty$ and a $C^\omega$ surface embedded in $\mathbb{R}^3$.&#xA;I ran across these terms in M. Berger's &lt;em&gt;Geometry Revealed&lt;/em&gt; book (p.387).&#xA;The context is: There are examples of two different $C^\infty$ compact surfaces&#xA;that are isometric, but no known examples for &quot;two real analytic (class $C^\omega$)&quot;&#xA;surfaces which are isometric.  Thanks!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Clarification.&lt;/b&gt; Thanks to Mariano and Willie for trying to help---I appreciate that!&#xA;It is difficult to be clear when you are confused :-).&#xA;Let me try two more specific questions: (1) Where does the $\omega$ enter into&#xA;the definition of $C^\omega$? Presumably $\omega$ is the first infinite ordinal.&#xA;(2) What I'm really after is the geometric &quot;shape differences&quot; between $C^\infty$ and $C^\omega$.  The non-analytic but smooth functions I know smoothly join, say, an exponential to a straight line, but geometrically they look just like smooth functions.  I guess I don't understand what the constraints of real-analyticity imply geometrically.  Maybe that's why this isometric question Berger mentioned is unsolved?!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Addendum.&lt;/b&gt; Here are Ryan's two functions:&#xA;&lt;br /&gt;&lt;img src=&quot;http://i.imgur.com/AnhsJ.jpg&quot; alt=&quot;Two functions&quot;&gt;&lt;br /&gt;&#xA;Left: $C^\infty$ but not $C^\omega$. Right: $C^\omega$.&lt;/p&gt;&#xA;" OwnerUserId="237" LastEditorUserId="237" LastEditorDisplayName="" LastEditDate="2011-01-25T00:23:08.933" LastActivityDate="2011-01-25T02:53:12.210" Title="$C^\infty$ vs. $C^\omega$ surfaces" Tags="&lt;geometry&gt;&lt;differential-geometry&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="18803" PostTypeId="2" ParentId="18759" CreationDate="2011-01-24T21:45:30.503" Score="0" ViewCount="" Body="&lt;p&gt;A slightly more direct approach: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Create a function g such that g(x) = 1 for all x in Q &lt;em&gt;except&lt;/em&gt; the &#xA;natural numbers. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Now, for contradiction, assume you can count these functions, and &#xA;have them listed as f1, f2, f3, etc. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Set g(1) to be different from f1(1) [you have exactly two choices for g(1)] &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;In general, set g(i) to be different from fi(i) &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;g now maps Q to {1,2,3}, but differs from all the listed fi, since &#xA;fi(i) != g(i) &lt;/p&gt;&#xA;" OwnerUserId="2469" LastActivityDate="2011-01-24T21:45:30.503" />
  <row Id="18804" PostTypeId="2" ParentId="18801" CreationDate="2011-01-24T21:51:15.447" Score="2" ViewCount="" Body="&lt;p&gt;If you are adding 13 numbers and the average is $4.6$, this is equivalent to their sum being $4.6\times 13$. Unfortunately, that gives $59.8$, which is impossible to achieve if your original numbers were all integers. Which suggests (to me, at least) that you are in fact &lt;em&gt;approximating&lt;/em&gt; the average, and that the actual average you want is $\frac{60}{13} = 4.615384615\ldots$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, you want to know how many ways you can obtain $60$ as a sum of exactly $13$ integers, all taken from ${1,2,3,4,5}$. This is equivalent to counting the number of $5$-partitions of $60$ of length exactly $13$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It can probably be done with generating functions, but here you can do it from first principles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the maximum sum you can possibly get with $13$ numbers taken from ${1,2,3,4,5}$ is $65$. So the partitions you want are in 1-to-1 correspondence with the partitions of $5$ other than the trivial one ($5=5$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, the partition $5=1+4$ gives the partition $60 = 11(5) + 4 + 1$ (subtract $1$ from one summand, $4$ from another summand). The partition $5=2+3$ yields the partition $60=11(5) + 3 + 2$. Etc. So you have:&#xA;\begin{align*}&#xA;5 &amp;amp;= 1+4  &amp;amp;\qquad 60 &amp;amp;= 11(5) + 4 + 1;\\&#xA;5 &amp;amp;= 2+3  &amp;amp;\qquad 60 &amp;amp;= 11(5) + 3 + 2;\\&#xA;5 &amp;amp;= 1+1+3 &amp;amp;\qquad 60 &amp;amp;= 10(5) + 4 + 4 + 2;\\&#xA;5 &amp;amp;= 1+2+2 &amp;amp;\qquad 60 &amp;amp;= 10(5) + 4 + 3 + 3;\\&#xA;5 &amp;amp;= 1+1+1+2 &amp;amp;\qquad 60&amp;amp;= 9(5) + 4 + 4 + 4 + 3;\\&#xA;5 &amp;amp;= 1+1+1+1+1 &amp;amp;\qquad 60&amp;amp;= 8(5) + 4 + 4 +4 + 4 + 4;&#xA;\end{align*}&#xA;where $60 = k(5) +\cdots$ means that you take the number $5$ $k$ times, and then the other numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So there are exactly $6$ ways of doing it.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2011-01-24T21:51:15.447" />
  <row Id="18805" PostTypeId="2" ParentId="18801" CreationDate="2011-01-24T21:51:58.590" Score="1" ViewCount="" Body="&lt;p&gt;The mean, which is what we usually are talking about when we say average, is the sum of the numbers divided by how many there are.  So the mean of 4 and 5 is (4+5)/2=4.5  Getting a fractional mean from integer numbers is no surprise, as in this case.  However, the sum must have been an integer, so you cannot average 13 integers and get a mean of 4.6 because the sum would have to be 4.6*13=59.8  So either you have to be rounding or you have to accept sets smaller than 13.  For example, you could have {4,4,5,5,5} with a total of 23 and a mean of 4.6&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To do this with generating functions, you can observe that the only way to get a mean of 4.6 is with a sum of 23 and five numbers or a sum of 46 and 10 numbers.  For 23 from 5 numbers in {1,2,3,4,5} the generating function is (x+x^2+x^3+x^4+x^5)^5 and take the coefficient of the x^23 term.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-01-24T21:51:58.590" />
  <row Id="18807" PostTypeId="2" ParentId="18801" CreationDate="2011-01-24T21:59:35.833" Score="1" ViewCount="" Body="&lt;p&gt;13 numbers with average 4.6 would need to add up to 59.8; so you can't get&#xA;that by choosing from the set of integers ${1,2,3,4,5}$. If you had solvable instance of this problem, for instance, 15 numbers with an average of 4.6, you would then be asking for the number of ways of choosing 15 numbers from ${1,2,3,4,5}$ with a sum of 69. A question like that can be answered by looking at the generating function $$F(x,t)\equiv\frac{1}{(1-tx)(1-tx^2)(1-tx^3)(1-tx^4)(1-tx^5)}$$ where the coefficient of the term $t^{15}x^{69}$ would be the answer you are looking for: $$ {\rm{number \; of \; ways}} = \left. \frac{1}{15!69!} \frac{\partial^{84} F}{\partial x^{69} \partial t^{15}}\right|_{x=0,t=0}$$&lt;/p&gt;&#xA;" OwnerUserId="6104" LastActivityDate="2011-01-24T21:59:35.833" />
  <row Id="18808" PostTypeId="1" CreationDate="2011-01-24T22:14:37.343" Score="1" ViewCount="67" Body="&lt;p&gt;I have a natural number $m$, and a polynomial $p: \mathbb{Z}^n \rightarrow \mathbb{Z}$, and I need to find a set $(x_1, x_2,...,x_n)$ so that $p(x_1, x_2,...,x_n) \bmod m = 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First: Does every non-constant polynomial have such a set? And more importantly: Does there exist an (efficient) procedure to determine it, in case a set exists?&lt;/p&gt;&#xA;" OwnerUserId="6205" LastEditorUserId="6205" LastEditorDisplayName="" LastEditDate="2011-01-24T22:31:25.723" LastActivityDate="2011-01-24T23:53:35.887" Title="Method to find $(x_1, x_2,...,x_n)$ s.t. polynom $p(x_1, x_2,...,x_n) \bmod m = 0$" Tags="&lt;roots&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="18809" PostTypeId="1" AcceptedAnswerId="18812" CreationDate="2011-01-24T22:14:37.983" Score="4" ViewCount="210" Body="&lt;p&gt;Hello,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In my algebraic topology class we showed how to calculate the homology groups of $S^n$, using the tools of singular homology, however we did not discuss other ways of doing it; my question - is there any relatively simple way of doing this, using simplicial homology? I tried thinking about this for a bit, but couldn't see any obvious direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="5573" LastActivityDate="2011-01-24T22:42:49.973" Title="Homology of the Sphere" Tags="&lt;topology&gt;&lt;algebraic-topology&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="1" />
  <row Id="18810" PostTypeId="1" AcceptedAnswerId="18811" CreationDate="2011-01-24T22:16:57.640" Score="1" ViewCount="88" Body="&lt;p&gt;The hypothesis of theorem 5.5 in page 152 of Hungerford's Algebra requires that the homomorphism maps the identity of R to the identity of S, but the proof &quot;never&quot; use this hypothesis. Is the theorem still true if the stronger hypothesis is removed. I am so sorry that I currently do not have link to the book.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Theorem 5.5&lt;/strong&gt; Let $R$ and $S$ be commutative rings with identity and $\varphi: R \to S$ a homomorphism of rings such that $\varphi(1_R) =1_S$. If $s_1, \dots, s_n \in S$, then there is a unique homomorphism of rings $\overline{\varphi}: R[x_1, \dots , x_n] \to S$ such that $\overline{\varphi}|R = \varphi$ and $\overline{\varphi}(x_i) = s_i$ for $i = 1,2, \dots, n$. This property completely determines $R[x_1, \dots, x_n]$ up to isomorphism.&lt;/p&gt;&#xA;" OwnerUserId="2177" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-24T22:27:22.997" LastActivityDate="2011-01-25T03:49:52.157" Title="Rings of polynomials" Tags="&lt;ring-theory&gt;" AnswerCount="1" CommentCount="6" />
  <row Id="18811" PostTypeId="2" ParentId="18810" CreationDate="2011-01-24T22:25:32.983" Score="3" ViewCount="" Body="&lt;p&gt;The hypothesis is being stealthily used in the assertion that $\overline{\varphi}$ is a homomorphism of rings. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Added and rewritten:&lt;/em&gt; More explicitly: the sketch Hungerford provides is to take an element of the polynomial ring&#xA;$$\sum r_{i}x_1^{k_{1i}}x_2^{k_{2i}}\cdots x_n^{k_{ni}}$$&#xA;(suitably finite) and suggests mapping it to&#xA;$$\sum\varphi(r_i)s_1^{k_{1i}}\cdots x_n^{k_{ni}}.$$&#xA;Then Hungerford asserts that this makes $\overline{\varphi}$ into a ring homomorphism.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You don't actually need the full hypothesis that $\varphi(1_R)=1_S$, but you do need &lt;em&gt;something&lt;/em&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $\varphi$ to be a ring homomorphism, we need $\varphi(1_R)$ to be an identity for $\varphi(R)$; for $\overline{\varphi}$ defined as above to be a ring homomorphism, we also need $\varphi(1_R)s_j = s_j$, because taking $p = 1_R$ and $q=x_j$ we have $\overline{\varphi}(pq) = \overline{\varphi}(x_j) =s_j$, and $\overline{\varphi}(p)\overline{\varphi}(q) = \varphi(1_R)s_j$. Thus, for $\overline{\varphi}$ to be a ring homomorphism, it is necessary that $\varphi(1_R)$ be an identity of the subring of $S$ generated by $\varphi(R)$ and $s_1,\ldots,s_n$ (as $R$ and $S$ are both commutative).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, this is also sufficient for you to be able to get a homomorphism: you must require that $\varphi(1_R)$ be the identity for the subring of $S$ generated by $\varphi(R)$ and $s_1,\ldots,s_n$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But it is not hard to check that the apparently weaker condition in the theorem (weaker, because it refers to a smaller class of homomorphisms and possible selections of $s_1,\ldots,s_n$) is in fact sufficient to characterize the polynomial ring up to unique isomorphism; the theorem suffices to prove the more general case by considering first the map onto $S&amp;#39;=\langle \varphi(R),s_1,\ldots,s_n\rangle$ and then composing with the embedding $S&amp;#39;\hookrightarrow S$ (since we are not requiring maps of rings with identity to send the identity to the identity). &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-25T03:49:52.157" LastActivityDate="2011-01-25T03:49:52.157" CommentCount="2" />
  <row Id="18812" PostTypeId="2" ParentId="18809" CreationDate="2011-01-24T22:37:03.567" Score="5" ViewCount="" Body="&lt;p&gt;Using simplicial homology you can triangulate the sphere as the boundary of an $(n+1)$-simplex, and work out the chain complex by hand. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;With cellular homology it is even easier since $S^n$ is the union of an $n$-cell and a $0$-cell. The chain complex has a single $\mathbb Z$ in degree $0$ and a single $\mathbb Z$ in degree $n$. In all other degrees it is zero.&lt;/p&gt;&#xA;" OwnerUserId="2941" LastActivityDate="2011-01-24T22:37:03.567" />
  <row Id="18813" PostTypeId="2" ParentId="18809" CreationDate="2011-01-24T22:37:19.103" Score="2" ViewCount="" Body="&lt;p&gt;Another rather quick way to compute the groups is with cellular homology.  Here the $n$th chain group is generated by the $n$-cells of your CW-complex.  The boundary map has to do with degree maps; but in your case it is simple.  An $n$-sphere is a 0-cell with an $n$-cell attached by mapping the boundary $S^{n-1}$ to the 0-cell.  If $n&amp;gt;1$ then all the maps in the chain complex must be 0 because the chain groups are trivial except for the $n$-chains and 0-chains.  The homology of such a complex is easy to compute.&lt;/p&gt;&#xA;" OwnerUserId="2838" LastEditorUserId="2838" LastEditorDisplayName="" LastEditDate="2011-01-24T22:42:49.973" LastActivityDate="2011-01-24T22:42:49.973" CommentCount="2" />
  <row Id="18814" PostTypeId="2" ParentId="18808" CreationDate="2011-01-24T22:44:04.137" Score="1" ViewCount="" Body="&lt;p&gt;If I am not mistaken, this is not known.  &lt;strike&gt;When $m = 2$ this problem is equivalent to &lt;a href=&quot;http://en.wikipedia.org/wiki/Boolean_satisfiability_problem#Basic_definitions.2C_terminology_and_applications&quot; rel=&quot;nofollow&quot;&gt;Boolean satisfiability&lt;/a&gt; (SAT for short), which is known to be NP-complete.  This is because one can represent logical operations as polynomial operations on the finite field $\mathbb{F}_2$ (with $0$ being false and $1$ being true) as follows: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\neg x = 1 - x, x \text{ and } y = xy, x \text{ or } y = x + y + xy.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It follows that any Boolean expression in $n$ Boolean variables $x_1, ... x_n$ can be written as a polynomial over $\mathbb{F}_2$, and finding solutions is equivalent to finding a non-satisfying assignment for the Boolean expression (so replacing the expression with its negation we get the usual form of SAT).  So the question of whether there exists a polynomial-time algorithm for solving this problem is equivalent to P vs. NP.&lt;/strike&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was mistaken!  Well, in any case, &lt;a href=&quot;http://arxiv.org/abs/quant-ph/0408129&quot; rel=&quot;nofollow&quot;&gt;this paper&lt;/a&gt; might be relevant.&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2011-01-24T23:53:35.887" LastActivityDate="2011-01-24T23:53:35.887" CommentCount="2" />
  <row Id="18815" PostTypeId="2" ParentId="18793" CreationDate="2011-01-24T23:12:25.560" Score="2" ViewCount="" Body="&lt;p&gt;Actually, your first statement is false for the infinite cyclic group:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\mathbb{Z}=1*_1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the trivial HNN extension of the trivial group.  The point is that, by the ideas described in &lt;em&gt;Trees&lt;/em&gt;, a splitting of a group is the same thing as an action of that group on a tree, and there is an obvious action of $\mathbb{Z}$ on the real line, which is a tree.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A group that does not split is said to have &lt;em&gt;Property FA&lt;/em&gt;.  There are many constructions of such groups, though some criteria involve torsion, so your torsion-free restriction makes the question more interesting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One source of examples is given by &lt;em&gt;random groups&lt;/em&gt;, which can be two-generated.  Dahmani, Guirardel and Przytycki &lt;a href=&quot;http://arxiv.org/abs/0904.3854&quot; rel=&quot;nofollow&quot;&gt;showed&lt;/a&gt; that random groups have Property FA.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ADDED:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The following is missing some details; the gaps can be filled in from Serre's &lt;em&gt;Trees&lt;/em&gt; and from Peter Scott's article `The Geometries of 3-manifolds'.  I'll fill in some more details if you ask for them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's an explicit example.  You probably know that &lt;em&gt;triangle groups&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$T(p,q,r)\cong \langle a,b\mid a^p=b^q=(ab)^r=1)\rangle$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;have Property FA. This is quite  easy to see by thinking about actions on trees.  Because they are all of finite order, $a,b$ and $ab$ each have to fix vertices, and the only way this can happen is if they all fix the same vertex.  These groups are 2-generator, but clearly not torsion-free.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now here's a torsion-free example, which is the fundamental group of a certain Seifert-fibred space.  For details on Seifert-fibred spaces, I recommend Peter Scott's article `The Geometries of 3-manifolds'.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $\mathrm{gcd}(p,p&amp;#39;)=\mathrm{gcd}(q,q&amp;#39;)=\mathrm{gcd}(r,r&amp;#39;)=1$. The group in question is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Gamma=\langle a,b,z\mid a^p=z^{p&amp;#39;}, b^q=z^{q&amp;#39;}, (ab)^r=z^{r&amp;#39;}, [a,z]=[b,z]=1\rangle~.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that if $\mathrm{gcd}(p&amp;#39;,q&amp;#39;,r&amp;#39;)=1$ then $\Gamma$ is 2-generated; $\Gamma$ is also torsion-free. $\Gamma$ can be written as a non-split central extension&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$1\to\langle z\rangle\to \Gamma\to T(p,q,r)\to 1~.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose that $\Gamma$ acts on a tree $T$.  Then, using some standard lemmas about actions on trees and the fact that it is normal, either $Z=\langle z\rangle$  fixes $T$ pointwise (in which case we are done), or $T$ can be taken to be a line which $z$ translates.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this latter case, translation distance gives a homomorphism $f:\Gamma\to\mathbb{Z}$ with $f(z)$ non-trivial.  But the existence of this homomorphism implies that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$p&amp;#39;/p+q&amp;#39;/q=r&amp;#39;/r$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which of course won't be true for most choices of parameters.&lt;/p&gt;&#xA;" OwnerUserId="2287" LastEditorUserId="2287" LastEditorDisplayName="" LastEditDate="2011-01-27T16:18:28.117" LastActivityDate="2011-01-27T16:18:28.117" CommentCount="6" />
  <row Id="18816" PostTypeId="1" AcceptedAnswerId="18821" CreationDate="2011-01-24T23:24:32.540" Score="3" ViewCount="154" Body="&lt;p&gt;Let $v_n$ be the volume of the unit ball in $\mathbb{R}^n$. Show by using Fubini's theorem that $$v_n = 2\,v_{n-1}\,\int_0^1{(1-t^2)^{(n-1)/2}\,dt}\,.$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've been trying to do this inductively, and I've been thinking about the integral as kind of a generalization of the integral formula for the area of (part of) the unit disc in $\mathbb{R}^2$, but when I go into more dimensions, the exponent is tripping me up. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of my thoughts was to just write down an integral for the volume of the n-ball as a big iterated integral, $$4\,\int_0^1{\cdots \int_0^1{(1-x_1^2)^{1/2}\cdots (1-x_n^2)^{1/2}\,dx_n\cdots dx_1}}\,,$$ but the more I look at this, the less sense it makes to me. At that point, I was thinking of, like, isolating one coordinate at a time and integrating, where the other coordinates are &quot;smooshed&quot; together in a &quot;y-axis.&quot; Or something.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please help, hahaha&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: Here's a solution (I believe). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, we have $$v_n = \idotsint\limits_{\sum\limits_{i=1}^n{x_i^2}\le 1}{dx_1\cdots dx_n}\,.$$ So by Fubini's theorem, we have $$\int_{-1}^1{\left[\idotsint\limits_{\sum\limits_{i=1}^{n-1}{x_i^2}\le 1-x_n^2}{dx_1\cdots dx_{n-1}}\right]\,dx_n}\,.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now set $y_i=\dfrac{x_i}{\sqrt{1-x_n^2}}$. So $dy_i=\dfrac{dx_i}{\sqrt{1-x_n^2}}$. Thus, we have $\sum\limits_{i=1}^{n-1}{y_i^2} = \sum\limits_{i=1}^{n-1}{\dfrac{x_i^2}{1-x_n^2}}\le 1$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we make our change of variables:&#xA;$$\int_{-1}^1{\left[\idotsint\limits_{\sum\limits_{i=1}^{n-1}{y_i^2}\le 1}{\left(\sqrt{1-x_n^2}\right)^{n-1}\,dy_1\cdots dy_{n-1}}\right]\,dx_n} = 2\,\int_0^1{\left(\sqrt{1-x_n^2}\right)^{n-1}v_{n-1}\,dx_n}\,.$$ Now we can just call $x_n$ whatever we like -- in this case, $t$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How does this look?&lt;/p&gt;&#xA;" OwnerUserId="2313" LastEditorUserId="2313" LastEditorDisplayName="" LastEditDate="2011-01-25T17:02:28.123" LastActivityDate="2011-01-25T17:02:28.123" Title="An application of Fubini's theorem" Tags="&lt;real-analysis&gt;&lt;homework&gt;&lt;integral&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="2" />
  <row Id="18817" PostTypeId="2" ParentId="18802" CreationDate="2011-01-24T23:26:18.577" Score="6" ViewCount="" Body="&lt;p&gt;$C^\omega$ means analytic and $C^\infty$ means infinitely-differentiable.   So for example, the union of the two graphs:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ z = e^{\frac{1}{x^2+y^2-1}}, \text{ for } x^2+y^2&amp;lt;1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ z=0, \text{ for } x^2+y^2 \geq 1 $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is a $C^\infty$-surface in $\mathbb R^3$, but it's not a $C^\omega$-surface. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, $z = \frac{1}{1+x^2+y^2}$ is an $C^\omega$-surface in $\mathbb R^3$ and the graph &quot;looks&quot; similar, in that they're both isolated hill-tops. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Were you looking for something more specific than that? &lt;/p&gt;&#xA;" OwnerUserId="642" LastActivityDate="2011-01-24T23:26:18.577" CommentCount="1" />
  <row Id="18818" PostTypeId="1" CreationDate="2011-01-24T23:27:51.057" Score="3" ViewCount="86" Body="&lt;p&gt;Can you give examples of combinatorial construction riddles, approachable by gifted high school students?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Examples&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Find a finite set $A$ and $B \subset 2^A$ such that any element of $A$ is covered by at least 100 sets in $B$, but $B$ cannot be partitioned into 2 disjoint sets, each covering $A$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Find the shortest sequence of numbers between 1 and K (K=99) such that each pair of numbers are adjacent somewhere in the sequence (1231 is an answer if K=3).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Find a systematic way to arrange a round robin tournament between N teams (N even or odd).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Find (different methods) to cover the complete graph on 50 vertices by disjoint spanning trees.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Motivation&lt;/strong&gt;: I'm working with gifted high school students. This is part of an excellence program in computer science. We have found these kind of problems adequate for stimulating and identifying the better students, before we teach them algorithmic methods. We also let them find winning strategies for mathematical games at this level of their knowledge.&lt;/p&gt;&#xA;" OwnerUserId="3533" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2011-01-24T23:56:24.027" LastActivityDate="2011-01-25T03:28:29.887" Title="For teaching: Combinatorial Construction Riddles" Tags="&lt;combinatorics&gt;&lt;soft-question&gt;&lt;teaching&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="18819" PostTypeId="2" ParentId="18818" CreationDate="2011-01-24T23:32:15.517" Score="3" ViewCount="" Body="&lt;p&gt;1)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a classic/folklore puzzle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have 12 apostles and a messiah sitting around a circular table every night for supper.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can it be so arranged that every night each person gets to sit with different people immediately to the left and right? Try to give a construction which achieves the maximum possible number of nights. What is the maximum possible number?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This basically boils down to decomposing $K_{13}$ (the complete graph on $13$ vertices) into hamiltonian cycles. For a solution, see: &lt;a href=&quot;http://math.stackexchange.com/questions/16933/generating-a-eulerian-circuit-of-a-complete-graph-with-constant-memory/16935#16935&quot;&gt;Generating a Eulerian circuit of a complete graph with constant memory&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Try to arrange the students of the class into two groups so that at least half a student's friends are in the other group.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Assume friendship in mutual).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This can be done by starting with a random split, and then refining the split by moving around students who don't 'fit' in the group.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The proof it works is by looking at the size of the cut between the groups and by showing that it reduces each time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are given a subset of points on a grid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can we find a colouring of the given points red/blue so that for each horizontal and vertical line, the difference of number of red and blue points is no more than 1?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can form a Bipartite graph with one set of vertices to be the x-coordinates, and the other to be the y-coordinates. An edge exists between x and y, if the point (x,y) is given.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I recollect correctly, I believe this can be solved by finding an Euler tour in the above graph.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2011-01-25T00:01:48.397" LastActivityDate="2011-01-25T00:01:48.397" />
  <row Id="18820" PostTypeId="2" ParentId="17182" CreationDate="2011-01-24T23:38:27.233" Score="2" ViewCount="" Body="&lt;p&gt;(Not a full answer, but only some preliminary comments, since I've no reputation to do so.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The identity in the Wolfram Functions site is, as mentioned, valid for characteristic $n$ in the interval $[-1,1)$ ; for $n=1$, your elliptic integral of the third kind degenerates into a particular combination of the elliptic integrals of the other two kinds and trigonometric functions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Pi(n;\phi|m)=F(\phi|m)+\frac{\tan\;\phi\sqrt{1-m\sin^2\phi}-E(\phi|m)}{1-m}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;such that you can use the properties of the other elliptic integrals you already know.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That leaves the &lt;em&gt;circular&lt;/em&gt; case $n&amp;lt;-1$ and the &lt;em&gt;hyperbolic&lt;/em&gt; case $n&amp;gt;1$; in the sequel I assume that the parameter $m$ is in the usual interval $[0,1]$ (if you need to deal with parameters outside the usual range, I'll have to dig further into my references).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the circular case, the identity&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Pi(n;\phi|m)=\frac{n(1-m)}{(n-1)(m-n)}\Pi\left(\frac{m-n}{1-n};\phi|m\right)+\frac{m}{m-n}F(\phi|m)+\frac1{\sqrt{(1-n)(1-m/n)}}\arctan\left(\frac12\sqrt{\frac{n(m-n)}{n-1}}\frac{\sin\;2\phi}{\sqrt{1-m\sin^2\phi}}\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;can be used to express the circular case of an incomplete elliptic integral of the third kind back into the case where the characteristic is in the interval $(m,1)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the hyperbolic case, the identity you will need is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Pi(n;\phi|m)=F(\phi|m)-\Pi\left(\frac{m}{n};\phi|m\right)+\frac1{\sqrt{(n-1)(1-m/n)}}\mathrm{artanh}\left(\sqrt{(n-1)(1-m/n)}\frac{\tan\;\phi}{\sqrt{1-m\sin^2\phi}}\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;such that only characteristics in the interval $(0,m)$ are evaluated; the function is $2\pi$-periodic in $\phi$ for the hyperbolic case, so that if your amplitude $\phi$ is outside the usual interval $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$, either add/subtract appropriate multiples of $2\pi$ from the amplitude, or add/subtract appropriate multiples of $\Pi(n|m)=K(m)-\Pi\left(\frac{m}{n}|m\right)$ from the right hand side of that identity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively, if you can help it, you might do better to just use the Carlson integral of the third kind instead; there is less worry here of having to split into &quot;circular&quot; and &quot;hyperbolic&quot; cases. The relations to interconvert Carlson and Legendre-Jacobi integrals are in the DLMF.&lt;/p&gt;&#xA;" OwnerUserId="6207" LastActivityDate="2011-01-24T23:38:27.233" />
  <row Id="18821" PostTypeId="2" ParentId="18816" CreationDate="2011-01-24T23:52:07.457" Score="2" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Hint:&lt;/strong&gt; How does the $(n-1)$-dimensional volume of a ball scale with its radius $r$?&lt;/p&gt;&#xA;" OwnerUserId="5363" LastActivityDate="2011-01-24T23:52:07.457" CommentCount="3" />
  <row Id="18822" PostTypeId="1" AcceptedAnswerId="18824" CreationDate="2011-01-24T23:54:05.520" Score="3" ViewCount="163" Body="&lt;p&gt;It's late at night and I'm tired, but I just stumbled across this while doing my homework. Any chance this is new? Or, maybe, did I just somehow transform it and it is still basically the same formula? In that case, forgive me, please.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, here it is. It is a recursive function (thus, of limited use!?) to calculate the sum of factorials:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f(n) = \sum\limits_{i=1}^n i! = \frac{(n+1)!}{n}+f(n-2)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f(0) = 0$&#xA;and&#xA;$f(-1) = -1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this useful at all or did I just waste my time? :)&lt;/p&gt;&#xA;" OwnerUserId="3016" LastEditorUserId="2548" LastEditorDisplayName="" LastEditDate="2011-01-25T00:29:49.117" LastActivityDate="2011-01-25T00:29:49.117" Title="Is this a new formula?" Tags="&lt;factorial&gt;" AnswerCount="1" />
  <row Id="18824" PostTypeId="2" ParentId="18822" CreationDate="2011-01-25T00:09:54.187" Score="5" ViewCount="" Body="&lt;p&gt;The formula certainly is true. Re-arranging what you wrote you get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ f(n) - f(n-2) = n! + (n-1)! = n(n-1)! + (n-1)! $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ = (n+1)\times (n-1)! = \frac{(n+1)\times n\times (n-1)!}{n} = \frac{(n+1)!}{n} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but I don't think it is particularly more useful than the observation of, say, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ f(n) = n! + f(n-1) $$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The definition of your function as a sum means that it naturally can be described recursively. I don't think the formulation you gave offers any particular advantage to the summation formula, unfortunately. &lt;/p&gt;&#xA;" OwnerUserId="1543" LastActivityDate="2011-01-25T00:09:54.187" CommentCount="1" />
  <row Id="18825" PostTypeId="1" AcceptedAnswerId="28380" CreationDate="2011-01-25T00:23:35.860" Score="0" ViewCount="162" Body="&lt;p&gt;Greetings All &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have some test matlab code which can scale and do rotation but the &#xA;translation (tl value) doesn't seem to be working. &#xA;I expected the entire object to be moved over x=2 and y=0 but it &#xA;doesn't move any ideas? &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;clf; clear all &#xA;rp=[0 4 2 -1 0; 0 1 -4 -3 0;0 0 0 0 0] %x-y values &#xA;scalenum=2 %scale factor &#xA;tx=2;ty=0; %translate values &#xA;% &#xA;for theta=0:45:90 &#xA;    tl=[1 0 tx; 0 1 ty;0 0 1] %translation &#xA;    rot=[cosd(theta) -sind(theta) 0; sind(theta) cosd(theta) 0;0 0 1]; &#xA;%rotation &#xA;    scale=[scalenum 0 0;0 scalenum 0;0 0 1] %scale &#xA;    rt=tl*rp %new translation &#xA;    r=rot*rp %new rotation of points &#xA;    rs=scale*r %new scale with rotation &#xA;    hold off &#xA;    plot(rt(1,:),rt(2,:)) &#xA;axis([-8 8 -8 8]) &#xA;grid on &#xA;pause(.5) &#xA;end; &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="5957" LastEditorUserId="5957" LastEditorDisplayName="" LastEditDate="2011-03-21T21:38:57.110" LastActivityDate="2011-03-21T22:57:25.690" Title="Affine transformation matrix doing translation " Tags="&lt;linear-algebra&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="18826" PostTypeId="2" ParentId="18616" CreationDate="2011-01-25T00:32:56.820" Score="4" ViewCount="" Body="&lt;p&gt;Most of the following have been done at at least one of the institutions where I've taught.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Math movie night.  Host a showing of some movie with mathematical content and then a short discussion afterward about the mathematics.  Possible movies: &quot;21,&quot; &quot;Proof,&quot; &quot;Fermat's Last Tango,&quot; &quot;N is a Number&quot; (an Erdős biography), &quot;Fermat's Room&quot; (haven't seen it, but it looks interesting)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Math pictionary.  Students have a couple of minutes to draw a representation of some mathematics term while their team members try to guess what it is.  No words or mathematical symbols allowed.  (This can get really funny.)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Math puzzles and games night.  Possible games: Go, Set, Hey! That's My Fish, Hex&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Pi Day on March 14, complete with edible pies, a pi-digit reciting contest, and Kate Bush's song &quot;Pi&quot; playing in the background.  If the faculty are brave enough they could volunteer to be the victims in a pie-throwing contest.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Host a &lt;a href=&quot;http://sites.google.com/site/awmmath/programs/kovalevsky-days&quot; rel=&quot;nofollow&quot;&gt;Sonia Kovalevsky Day&lt;/a&gt; for local high schools.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Design a math club t-shirt.  For most students, the more math puns, the better.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;A trip to the local science center or to a public lecture with some mathematical content.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="2370" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2011-01-25T03:27:11.073" LastActivityDate="2011-01-25T03:27:11.073" CommentCount="1" />
  <row Id="18827" PostTypeId="1" AcceptedAnswerId="18831" CreationDate="2011-01-25T01:04:36.730" Score="2" ViewCount="176" Body="&lt;p&gt;So I have a test on abstract algebra tomorrow. It is basically on rings and polynomials in a ring/field. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I just have a couple of questions about terms. The definition of an isomorphic ring states that&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;there is a function $f: R \rightarrow S$ such that:&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;i) $f$ is injective; &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;ii) $f$ is surjective; &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;iii) $f(a+b) = f(a)+f(b)$ and $f(ab) = f(a)f(b)$ for all $a, b \in R$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What does it mean to be injective and surjective? The book is not clear enough. If anyone can put it in simple english, it would be appreciated. This is my first abstract algebra course.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Examples (the prof said to have a lot of examples in mind) would be helpful. He said examples of homomorphisms would be good. For example:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The function $f: Z \rightarrow Z_6$ is given by $f(a) = [a]$. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can verify axiom three myself, but the example states that the homomorphism is surjective but not injective.. Why is that? Is it because the function maps an integer to its integer congruence class? And hence $ a \neq b$ like  PEV puts it ?&lt;/p&gt;&#xA;" OwnerUserId="287" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-25T01:26:47.460" LastActivityDate="2011-05-24T12:39:04.810" Title="introduction to isomorphisms and homomorphism." Tags="&lt;abstract-algebra&gt;" AnswerCount="4" CommentCount="3" FavoriteCount="0" />
  <row Id="18828" PostTypeId="2" ParentId="18827" CreationDate="2011-01-25T01:06:45.107" Score="1" ViewCount="" Body="&lt;p&gt;Injective means one-to-one, while surjective means onto. I am assuming you know what one-to-one and onto functions are; you can check out these articles if not:  &lt;a href=&quot;http://en.wikipedia.org/wiki/Injective&quot; rel=&quot;nofollow&quot;&gt;One-to-one functions&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/Surjective&quot; rel=&quot;nofollow&quot;&gt;Onto functions&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="3731" LastActivityDate="2011-01-25T01:06:45.107" />
  <row Id="18829" PostTypeId="1" AcceptedAnswerId="18832" CreationDate="2011-01-25T01:07:22.017" Score="3" ViewCount="243" Body="&lt;p&gt;I'm trying to prove that if $E \subset \mathbb{R}$ has finite measure and $\varepsilon \in \mathbb{R}$ such that $\varepsilon&amp;gt;0$, then $E$ is the disjoint union of a finite number of measurable sets, each of which has measure at most $\varepsilon$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is what I've done:&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Let $\varepsilon \in \mathbb{R}$ such that $\varepsilon&amp;gt;0$. Since $m^{*}(E) &amp;lt; +\infty$, there exists a countable collection $\{I_{k}\}_{k \in \mathbb{Z}^{+}}$ of open and bounded intervals (measurable sets) covering $E$ and such that $\sum \limits_{k=1}^{\infty } \ell ( I_{k})&amp;lt;m^{*}( E) +\varepsilon /2$. Define $G_{1}:=I_{1}$, and for each $k\in \mathbb{Z}^{+}$ such that $k\geq 2$ define $G_{k}:=I_{k}\setminus\cup_{n=1}^{k-1} I_{n}$. Therefore ${\{G_{k}\}}_{k\in \mathbb{Z}^{+}}$ is a countable disjoint collection of bounded and measurable sets such that $G:=\cup _{k\in \mathbb{Z}^{+}}G_{k}=\cup _{k\in \mathbb{Z}^{+}}I_{k}$. So, $m^{*}( G) &amp;lt;m^{*}( E) +\varepsilon /2$ and $E\subseteq G$. Now&#xA;define $F_{k}:=\overline{G}_{k}$ (the closure of $G_k$). Since $G_{k}\subseteq I_{k}$, we know that $F_{k}\subseteq \overline{I_{k}}$; i.e., each $F_{k}$ is a closed and bounded subset of $\mathbb{R}$. By the Heine-Borel Theorem we conclude that each $F_{k}$ is compact.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $x\in \mathbb{R}$, then we define $N( x,\varepsilon) := (x-\varepsilon/2,x+\varepsilon/2)$. For each $k\in \mathbb{Z}^{+}$ the collection $\{N( x,\varepsilon ) \colon x\in F_{k}\}$ is an open covering of $F_{k}$. The compactness of each $F_{k}$ implies that there exist $x_{1}^{k},\ldots ,x_{n_{k}}^{k}\in F_{k}$, with $n_{k}\in \mathbb{Z}^{+}$, such that $F_{k}\subseteq \bigcup _{i=1}^{n_{k}}N(x_{i}^{k},\varepsilon ) $. Thus $E\subseteq \cup _{k\in \mathbb{Z}^{+}}G_{k}\subseteq \cup _{k\in \mathbb{Z}^{+}}F_{k}\subseteq \cup _{k\in \mathbb{Z}^{+}}(\cup_{i=1}^{n_{k}}N( x_{i}^{k},\varepsilon ))$ ...&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I can't figure out how to make this last union over $k$ finite. I suppose I've to use the fact that $m^*(G)&amp;lt;m^*(E)+\varepsilon/2$. I can't see how! If that union is finite, then the problem is solved; one would just have to make some intersections and take some complements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, I think that if we remove the hypothesis of finite outer measure on $E$, then the conclusion is the same but with a countably infinite union. Am I right?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you in advance.&lt;/p&gt;&#xA;" OwnerUserId="5956" LastEditorUserId="5956" LastEditorDisplayName="" LastEditDate="2011-01-25T02:20:22.477" LastActivityDate="2011-01-25T02:20:22.477" Title="Every subset of $\mathbb{R}$ with finite measure is the disjoint union of a finite number of measurable sets" Tags="&lt;measure-theory&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="1" />
  <row Id="18830" PostTypeId="2" ParentId="18827" CreationDate="2011-01-25T01:07:33.840" Score="2" ViewCount="" Body="&lt;p&gt;For injectivity: For $a,b \in R$,  $f(a) = f(b) \implies a = b$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Surjectivity: For each $s \in S$ there exists $r \in R$ such that $f(r) = s$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f: \mathbb{Z} \to \mathbb{Z}_6$ defined by $f(x) = [x]$ is not injective because $f(6) = f(12)$ but $ 6 \neq 12$. Or in other words, the kernel is nontrivial. &lt;/p&gt;&#xA;" OwnerUserId="2548" LastEditorUserId="2548" LastEditorDisplayName="" LastEditDate="2011-01-25T01:24:40.067" LastActivityDate="2011-01-25T01:24:40.067" CommentCount="3" />
  <row Id="18831" PostTypeId="2" ParentId="18827" CreationDate="2011-01-25T01:28:13.780" Score="3" ViewCount="" Body="&lt;p&gt;A function f(x) has &lt;em&gt;one&lt;/em&gt; value for each x you plug into it, right?  And it's valid for every x you put into it (it's defined over the whole domain).  Well, injective and surjective are saying these things about the range.  Injective means that for every value that you can get out of f(x), there is only one x you can use to get that value.  Surjective means that you can take any value in the set where f(x) is defined (the codomain) and get some f(x) that maps to that value (there is some x where f(x) equals that value).  In other words, it says the range spans the whole codomain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope this informal explanation is helpful.&lt;/p&gt;&#xA;" OwnerUserId="6059" LastActivityDate="2011-01-25T01:28:13.780" CommentCount="2" />
  <row Id="18832" PostTypeId="2" ParentId="18829" CreationDate="2011-01-25T01:32:08.857" Score="4" ViewCount="" Body="&lt;p&gt;Note: The problem is false as stated. Either $E$ has to be measurable, or some of the sets in the disjoint union must be nonmeasurable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An easier approach is to first show that there is an $M$ such that $m^*(E\setminus[-M,M])&amp;lt;\varepsilon$, and break up $[-M,M]$ into a finite number of disjoint intervals $I_1,\ldots,I_n$ of length less than $\varepsilon$.  Then $E\setminus[-M,M],E\cap I_1,\ldots,E\cap I_n$ will do the trick.  (This gives sets of outer measure less than $\varepsilon$, which cannot all be measurable unless $E$ is.)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you only want to have $E$ contained in the union rather than equal to it, then you can take the sets to be measurable.  Let $U$ be an open set containing $E\setminus{[-M,M]}$ of measure less than $\varepsilon$, and intersect it with $\mathbb{R}\setminus[-M,M]$ if necessary to make it disjoint with $[-M,M]$.  Then $U,I_1,\ldots,I_n$ will do the trick.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's another approach (for the case of equality of the union, where the sets may not be measurable).  Let $n$ be such that $n\varepsilon\gt m^*(E)$ and $(n-1)\varepsilon\leq m^*(E)$, and proceed by induction on $n$.  The base case is that the outer measure is less than $\varepsilon$ already, and for the inductive step you can apply the intermediate value theorem to the function $f(t)=m^*(E\cap[-t,t])$ to find and remove a subset of outer measure $\varepsilon$.  For what it's worth, this would give you the smallest possible number of sets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer to your question about countable unions is &quot;yes&quot;.  You can intersect $E$ with intervals of the form $[n\varepsilon,(n+1)\varepsilon)$.  (Again, you can't get the sets to be measurable unless $E$ is.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure how to salvage your approach.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: The following approach works if $E$ is measurable. I hastily missed the fact that that is not assumed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $E$ is measurable, then here is an approach more in line with yours.  You could prove the useful fact that there is a finite collection of intervals $I_1,\ldots I_n$ such that the measure of the symmetric difference of $E$ with $\cup_k I_k$ is less than $\varepsilon$.  This is (one version of) one of &lt;a href=&quot;http://en.wikipedia.org/wiki/Littlewood%27s_three_principles_of_real_analysis&quot; rel=&quot;nofollow&quot;&gt;Littlewood's 3 principles&lt;/a&gt;.  One way to start the proof would be to find an $M$ similar to above, and then to use a compactness argument along with the definition of measure.  After you have this, you already have control on the stuff outside of $\cup_k I_k$, and the stuff inside $\cup_k I_k$ could be handled by a method similar to yours.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2011-01-25T02:18:30.700" LastActivityDate="2011-01-25T02:18:30.700" CommentCount="1" />
  <row Id="18833" PostTypeId="2" ParentId="18718" CreationDate="2011-01-25T01:39:59.373" Score="1" ViewCount="" Body="&lt;p&gt;(This should be a comment to leonbloy's answer)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The proper expression for the fixed-point iteration for the $n$th nonzero solution of $x=\tan\;x$ is $x_{k+1}=\arctan(x_k)+n\pi$. In fact, for the more general transcendental equation $\tan\;x=\lambda x$, the fixed-point iteration $x_{k+1}=\arctan(\lambda x_k)+n\pi$ works nicely; one can use the fixed point starting with $x_0=n\pi$ for the first few iterations and then switch to Newton-Raphson as soon as the fixed-point iteration has started converging. A similar fixed-point method works for the related transcendental equation $\cot\;x=\mu x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If one needs to compute large roots, it might be more profitable to construct asymptotic approximations to the roots; details are given in chapter 7 of the book &lt;a href=&quot;http://books.google.com/books?id=0XSBQgAACAAJ&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;Numerical Methods for Special Functions&lt;/em&gt; by Gil, Segura, and Temme&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="6210" LastActivityDate="2011-01-25T01:39:59.373" />
  <row Id="18834" PostTypeId="2" ParentId="18780" CreationDate="2011-01-25T01:59:15.210" Score="1" ViewCount="" Body="&lt;p&gt;&lt;em&gt;[Update Note] I saw the related question at mathoverflow. There it seemed we were dealing with a frequencies-table, so I repeated that scheme here. But now I see the question is focused on means in a two-factor anova. I'll see, whether that two concepts can be interchanged here; for instance a 1:1-reference should only be possible if the coefficients under treatments (means?) are based on the same number of observations. Possibly it is better to delete this answer later [end note]&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a solution. I computed the &quot;expected frequencies&quot; based on your values, where I compensated the * .../3* and the * .../2*-divisions. Also I corrected &lt;em&gt;9186&lt;/em&gt; to &lt;em&gt;9817&lt;/em&gt; to make the totals consistent.&lt;br&gt;&#xA;$ \begin{array} {rrrrrrr}&#xD;&#xA;  &amp;amp; &amp;amp; B:  &amp;#39;0&amp;#39; &amp;amp;    &amp;#39;112&amp;#39; &amp;amp; &amp;#39;224&amp;#39;   &amp;amp;| &amp;amp;      (all) &amp;amp; \\\&#xD;&#xA;---    &amp;amp;+&amp;amp;----&amp;amp;-----&amp;amp;-----&amp;amp;+&amp;amp;---&amp;amp; &amp;amp;\\\&#xD;&#xA;A:&amp;#39;20&amp;#39; &amp;amp;|&amp;amp; 3350.08&amp;amp;	 8499.04 &amp;amp;	10881.88 &amp;amp;|&amp;amp;	22731  &amp;amp;/3 = 7577 \\\   &#xD;&#xA;&amp;#39;80&amp;#39;   &amp;amp;|&amp;amp; 4061.92&amp;amp;	10304.96 &amp;amp;	13194.12 &amp;amp;|&amp;amp;	27561 &amp;amp; /3 = 9187 \\\    &#xD;&#xA;---    &amp;amp;+&amp;amp;----&amp;amp;-----&amp;amp;-----&amp;amp;+&amp;amp;--- &amp;amp; \\\     &#xD;&#xA;(all)   &amp;amp; &amp;amp;  7412&amp;amp;	18804 &amp;amp;	24076 &amp;amp;|&amp;amp;	50292  &amp;amp; \\\ &#xD;&#xA;        &amp;amp; &amp;amp;  /2=3706&amp;amp;   /2=9402  &amp;amp;  /2=12038     &amp;amp; &amp;amp;      &amp;amp; \\\&#xD;&#xA; \end{array}&#xD;&#xA;$ &lt;/p&gt;&#xA;" OwnerUserId="1714" LastEditorUserId="1714" LastEditorDisplayName="" LastEditDate="2011-01-25T02:33:45.417" LastActivityDate="2011-01-25T02:33:45.417" CommentCount="2" />
  <row Id="18835" PostTypeId="1" AcceptedAnswerId="18837" CreationDate="2011-01-25T02:02:41.390" Score="0" ViewCount="93" Body="&lt;p&gt;Given a Noetherian ring $R$ and a proper ideal $I$ of it, is it true that $\cap_n I^n=0$ as $n$ varies over all natural numbers. If not, is it true if $I$ is a maximal ideal? If not, is it true if $I$ is the maximal ideal of a local ring $R$. If not, is it true under additional assumptions on $R$ (like $R$ is regular)&lt;/p&gt;&#xA;" OwnerUserId="6166" LastActivityDate="2011-01-25T03:43:26.583" Title="Intersection of powers of an ideal in a Noetherian ring" Tags="&lt;commutative-algebra&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="18836" PostTypeId="2" ParentId="18369" CreationDate="2011-01-25T02:03:13.643" Score="2" ViewCount="" Body="&lt;p&gt;Since the problem is trivial for $n \leq 2$, suppose that $n \geq 3$.&#xA;Without loss of generality, assume that the circle has circumference one and that the first of $n$ points is located on its top (the next $n-1$ points are drawn independently and uniformly on the circumference).&lt;br&gt;&#xA;Then, it is readily seen that the problem is equivalent to the following one. Let $U_1,\ldots,U_{n-1}$ be i.i.d. uniform rv's on $[0,1]$. What is the probability that $U_{\max} - U_{\min} \leq 1/2$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As is well known (cf. &lt;a href=&quot;http://mathoverflow.net/questions/51243/what-is-the-probability-that-the-range-of-a-set-of-n-randomly-chosen-real-numbers&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt;, for example) and easy to show (simple exercise in order statistics, upon conditioning on $U_{\max}$), &#xA;$$&#xD;&#xA;{\rm P}(U_{\rm max} - U_{\rm min} \leq x) = (n-1)x^{n - 2}  - (n - 2)x^{n-1}, \;\; x \in [0,1].&#xD;&#xA;$$ &#xA;Substituting $x=1/2$ gives the desired result of $\frac{n}{{2^{n - 1} }}$.&lt;/p&gt;&#xA;" OwnerUserId="2810" LastActivityDate="2011-01-25T02:03:13.643" />
  <row Id="18837" PostTypeId="2" ParentId="18835" CreationDate="2011-01-25T02:08:44.340" Score="4" ViewCount="" Body="&lt;p&gt;It is not true in general: the ideal may well be idempotent!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For an example, consider a direct product of two fields: there are two ideals, both maximal and both idempotent.&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2011-01-25T02:08:44.340" CommentCount="1" />
  <row Id="18838" PostTypeId="1" AcceptedAnswerId="18851" CreationDate="2011-01-25T02:15:09.167" Score="1" ViewCount="159" Body="&lt;p&gt;Let $f$ be analytic on the open set $G\subset \mathbb{C}$. What's the best way of showing that the function $\phi:G \times G \to \mathbb{C}$ defined by $\phi(z,w)=[f(z)-f(w)]/(z-w)$ for $z \neq w$ and $\phi(z,z)=f&amp;#39;(z)$ is continuous?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is obvious for $z\neq w$, but I'm having some trouble on the points $(z,z)$. I did it one way but I'm not convinced this is good.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: Rudin's Proof (for the continuity on the diagonal)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $z_0\in G$. Since $f$ is analytic, there exists $r&amp;gt;0$ such that $B(a,r)\subset G$ and $|f&amp;#39;(\zeta)-f&amp;#39;(z_0)|&amp;lt;\epsilon$ for all $\zeta \in B(a,r)$. Getting $z,w\in B(a,r)$, than $\zeta(t)=(1-t)z+tw \in B(a,r)$ for $0\le t\le 1$. Now just use that $\phi(z,w)-\phi(z_0,z_0)=\int_0^1[f&amp;#39;(\zeta(t))-f&amp;#39;(z_0)]dt$ and the $\epsilon$-bound to get the desired continuity.&lt;/p&gt;&#xA;" OwnerUserId="1971" LastEditorUserId="1971" LastEditorDisplayName="" LastEditDate="2011-01-25T04:45:11.650" LastActivityDate="2011-03-23T16:23:07.203" Title="Continuous complex function (2 variables)" Tags="&lt;complex-analysis&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="18839" PostTypeId="2" ParentId="18838" CreationDate="2011-01-25T02:39:21.540" Score="-1" ViewCount="" Body="&lt;p&gt;What happens to $\phi(z,z+h)$ when $h\to 0$? Let's see&#xA;$$&#xD;&#xA;\lim_{h\to0}\phi(z,z+h) = \lim_{h\to0}\frac{f(z+h)-f(z)}{h} = f&amp;#39;(z) = \phi(z,z).&#xD;&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="2473" LastActivityDate="2011-01-25T02:39:21.540" CommentCount="3" />
  <row Id="18840" PostTypeId="2" ParentId="18802" CreationDate="2011-01-25T02:53:12.210" Score="4" ViewCount="" Body="&lt;p&gt;To Clarification (1): I think $\omega$ is there just to indicate that the smoothness is &quot;more than $C^\infty$&quot;, and I don't think it has any substantial connection to the first infinite ordinal. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2): Analytic surfaces are much more rigid than smooth surfaces. For example, you can choose any open set on a smooth surface and do surgery so that only the chosen set is affected. For analytic surfaces you knock on a point, you perturb the entire surface.&lt;/p&gt;&#xA;" OwnerUserId="2473" LastActivityDate="2011-01-25T02:53:12.210" CommentCount="1" />
  <row Id="18841" PostTypeId="2" ParentId="18616" CreationDate="2011-01-25T02:55:16.480" Score="2" ViewCount="" Body="&lt;p&gt;To add to Mike's Math movies/documentaries list: (unable to post links in comments)&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.veoh.com/search/videos/q/story+of+maths#watch%3Dv1716790862S86Xhh&quot; rel=&quot;nofollow&quot;&gt;Story of Maths&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://video.google.com/videoplay?docid=-5122859998068380459#&quot; rel=&quot;nofollow&quot;&gt;Dangerous Knowledge&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.youtube.com/watch?v=SVXB5zuZRcM&amp;amp;feature=related&quot; rel=&quot;nofollow&quot;&gt;Fermat's Last Theorem&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.youtube.com/watch?v=BsIJN4YMZZo&amp;amp;feature=&amp;amp;p=3641A12A6ADDB4B1&amp;amp;index=0&amp;amp;playnext=1&quot; rel=&quot;nofollow&quot;&gt;The Importance of Mathematics&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="3302" LastActivityDate="2011-01-25T02:55:16.480" />
  <row Id="18842" PostTypeId="1" AcceptedAnswerId="18854" CreationDate="2011-01-25T02:57:36.587" Score="2" ViewCount="110" Body="&lt;p&gt;I have discovered that Groebner basis computations may help in a problem I am working on. However, I am having some very specific problems. First, the literature I have discovered on Groebner basis calculations are either extremely advanced or sacrifice completeness for simplicity. Secondly, to come to grips with these methods I have tried working with computer algebra systems (specifically Macaulay 2, version 1.2) to wade through examples.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therein, my problem has a similar flavor to this &lt;a href=&quot;http://math.stackexchange.com/questions/3550/using-grobner-bases-for-solving-polynomial-equations&quot;&gt;post&lt;/a&gt; (I found the answers moderately helpful). One of the answers to this question provides an example of a computation that works out correctly - I cannot get Macaulay to behave the same way on a randomly selected toy problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is my understanding that Groebner basis methods are an infallible way to find solutions to nonlinear systems of equations over any field (barring some issues with numerical stability and computational complexity). That is, it is my understanding that Groebner basis computations will &lt;i&gt;always&lt;/i&gt; reduce a system of equations to an equivalent system in which there exist polynomials of one variable (and that solving the system can proceed from finding roots to this one polynomial and then &quot;steamroll&quot; the rest).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe I have followed the steps necessary to get a system of nonlinear equations into its Groebner basis form. However, I do not see a clear means of finding solutions to my system. Here is my Macaulay 2 history.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;i29 : T = ZZ[f,g,h]&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;o29 = T&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;o29 : PolynomialRing&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;i32 : Woot = gb ideal(f^2*g + g^2*h +&#xA;  2*h^2, f + g + h, f^3*g^2*h) o32 =&#xA;  GroebnerBasis[status: done; S-pairs&#xA;  encountered up to degree 9]&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;o32 : GroebnerBasis&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;i33 : ideal gens Woot&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;o33 = ideal ($f+g+h$,&#xA;               $g^3+3g^2h+g*h^2+2h^2$,&#xA;               $8g*h^5+4h^6-8h^5$,&#xA;               $5g^2h^4+2g*h^5+2g^2h^3+4h^5$,&#xA;               $4h^7+2g^2h^4+4g*h^5+24h^6-12g^2h^3+40h^5$,&#xA;               $g*h^6+g^2h^4+4g*h^5-2h^6+2g^2h^3-8h^5$,&#xA;               $g^2h^5+2g^2h^4+4g*h^5+4h^6$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;None of these are univariate. (This particular system of equations is not important to me - I just used them as an example).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Have I chosen a bad monomial ordering? (Is it possible to compute an &quot;unhelpful&quot; Goebner basis if you have a bad monomial ordering?) Have I run the wrong commands through Macaulay? Is my understanding of Groebner basis misinformed? How would I solve this system of equations with Groebner basis reduction?&lt;/p&gt;&#xA;" OwnerUserId="6213" LastActivityDate="2011-01-25T04:36:18.617" Title="Solving a nonlinear system using Groebner basis computations." Tags="&lt;algebraic-geometry&gt;&lt;computer-science&gt;&lt;computer-algebra-systems&gt;" AnswerCount="1" />
  <row Id="18843" PostTypeId="1" CreationDate="2011-01-25T03:02:46.777" Score="15" ViewCount="670" Body="&lt;p&gt;This is an offshoot of the question on &lt;a href=&quot;http://math.stackexchange.com/questions/18616/fun-math-outreach-social-activities&quot;&gt;Fun math outreach/social activities&lt;/a&gt;. I have listed a few videos/documentaries I have seen. I would appreciate if people could add on to this list.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.veoh.com/search/videos/q/story+of+maths#watch%3Dv1716790862S86Xhh&quot; rel=&quot;nofollow&quot;&gt;Story of Maths&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://video.google.com/videoplay?docid=-5122859998068380459#&quot; rel=&quot;nofollow&quot;&gt;Dangerous Knowledge&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.youtube.com/watch?v=SVXB5zuZRcM&amp;amp;feature=related&quot; rel=&quot;nofollow&quot;&gt;Fermat's Last Theorem&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.youtube.com/watch?v=BsIJN4YMZZo&amp;amp;feature=&amp;amp;p=3641A12A6ADDB4B1&amp;amp;index=0&amp;amp;playnext=1&quot; rel=&quot;nofollow&quot;&gt;The Importance of Mathematics&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="3302" LastActivityDate="2011-01-25T14:27:40.887" CommunityOwnedDate="2011-01-25T12:35:45.310" Title="List of Interesting Math Videos/ Documentaries" Tags="&lt;soft-question&gt;&lt;big-list&gt;" AnswerCount="10" CommentCount="2" FavoriteCount="19" />
  <row Id="18844" PostTypeId="1" CreationDate="2011-01-25T03:17:15.237" Score="3" ViewCount="47" Body="&lt;p&gt;I am using the &lt;a href=&quot;http://mathworld.wolfram.com/MathieuDifferentialEquation.html&quot; rel=&quot;nofollow&quot;&gt;Mathieu equation&lt;/a&gt; &#xA;  $$x&amp;#39;&amp;#39;(t) + a x(t) + 2 q \cos(2 t) x(t) = 0,$$&#xA;as a model of a physical system. The conditions on the parameters $a$ and $q$ for stable solutions to exist are well known and are even implemented as built-in functions in systems such as Mathematica.&lt;br&gt;&#xA;Now, I would like to extend my model to the form &#xA;  $$x&amp;#39;&amp;#39;(t) + c x^2(t)+ a x(t) + 2 q \cos(2 t) x(t) = 0,$$&#xA;and establish the stability criteria for this modified model.&lt;br&gt;&#xA;Is this a known model, or can it be transformed to one?&lt;br&gt;&#xA;Does it even make sense to talk about stability conditions for this case -- as far as I can see &lt;a href=&quot;http://mathworld.wolfram.com/FloquetsTheorem.html&quot; rel=&quot;nofollow&quot;&gt;Floquet's theorem&lt;/a&gt; does not apply?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Although the most general form would be most interesting, it would also be interesting to know the stability criteria for the less general case of $a=0$.&lt;/p&gt;&#xA;" OwnerUserId="2842" LastActivityDate="2011-01-25T15:39:26.263" Title="Stability of a generalized form of the Mathieu equation" Tags="&lt;differential-equations&gt;" AnswerCount="1" />
  <row Id="18845" PostTypeId="2" ParentId="18616" CreationDate="2011-01-25T03:18:56.530" Score="2" ViewCount="" Body="&lt;p&gt;When I was a first year undergrad, we organised a mathematical Call My Bluff evening. A game master prepares questions in the vein of &quot;what is a happy number&quot;, &quot;State one interesting theorem from Ramsey theory&quot;, and so on, questions whose answers a first year undergraduate is unlikely to know but likely to understand. Then, several teams have to invent credible answers and quietly submit them on a piece of paper. The game master reads them all out, along with the correct answer and the teams have to guess which answer is right. There are some points for identifying the right answer, but even more points for every team that fell for yours.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's great fun, and actually quite demanding: when you invent a definition, you must take care to not define the empty set and to not define something that you already know under a different name. Accordingly, when you are evaluating the answers, you are trying to rule out those that seem to define/state something boring or something well-known under a different name. Of course, inventing answers is also very creative and makes you realise how difficult it is to ask interesting questions or introduce interesting concepts.&lt;/p&gt;&#xA;" OwnerUserId="3212" LastActivityDate="2011-01-25T03:18:56.530" CommentCount="1" />
  <row Id="18846" PostTypeId="2" ParentId="18843" CreationDate="2011-01-25T03:22:28.990" Score="3" ViewCount="" Body="&lt;p&gt;Documentary of Paul Erdos &quot;N is a Number&quot; is another good documentary. Also there is a documentary of Srinivasa Ramanujan called &lt;em&gt;God, Zero and Infinity&lt;/em&gt; was released by TIFR in 2009.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.youtube.com/watch?v=plohsYdD9LY&quot; rel=&quot;nofollow&quot;&gt;http://www.youtube.com/watch?v=plohsYdD9LY&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Another documentary named &lt;em&gt;Hard Problems&lt;/em&gt; was released by &lt;strong&gt;MAA&lt;/strong&gt; which shows how US students performed in the IMO.&lt;/p&gt;&#xA;" LastActivityDate="2011-01-25T03:40:48.897" CommunityOwnedDate="2011-01-25T12:35:45.310" CommentCount="1" />
  <row Id="18847" PostTypeId="2" ParentId="18818" CreationDate="2011-01-25T03:28:29.887" Score="2" ViewCount="" Body="&lt;h2&gt; The carromboard puzzle &lt;/h2&gt;&#xA;&#xA;&lt;p&gt;You get inside a room with a square table and a lamp.&#xA;On each side of the table there's a toggle switch, which can be pressed to toggle it (the four switches have two states each).&#xA;The lamp turns on if all switches are in the same state (either on or off).&#xA;When you enter a room, the lamp is not on, so you know that the switches are not all the same state - but that's all you know about their states.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your goal is to turn the lamp on in as few rounds as possible.&#xA;The complication is that after each round of pressing (you can press either one switch or two at a time), you must close your eyes, and a goblin turns the table by some unknown amount (wlog it is either 0, 90, 180 or 270 degrees).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you turn the lamp on at all? How many rounds does it take you? Is the optimal solution (essentially) unique?&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Generalizations&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;The puzzle can be generalized as follows. There are $n$ switches and $m$ states for each. Each round, you're allowed to press any number of switches (i.e. you have as many hands as you want), and moreover, you can press any given switch more than once. In other words, you're adding an arbitrary vector to the state vector of the switches. The goblin still turns the $n$-sided table each round.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For which $n,m$ can you turn the lamp? What is the minimal rounds needed? How do all the optimal solutions look like?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(If you don't like switches with $m$ states, you can stick to the case $m=2$, which is also simpler.)&lt;/p&gt;&#xA;&#xA;&lt;h3&gt; Answers &lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Well, not quite. There is a simple condition on $n,m$ that tells you whether there is a solution. When a solution does exist, the size of the minimal solution has a simple formula. When $m$ is prime (e.g. $m=2$), all minimal solutions have a nice description. For the rest of the solvable cases (other than the trivial ones), there's apparently no nice description (this is an open question, for your more ambitious students).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It turns out that it's better to change the turn-on criterion from all-switches-equal to all-switches-on; there's a simple reduction between the two (when $m$ is prime), and the latter criterion is somewhat more natural.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt; Prison &lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The wardens in a state penitentiary are playing a game with the prisoners. All the prisoners are gathered, and told the following: &quot;There is a special cell in the prison, within it a switch with N states (we do not tell you its initial state). From time to time, we will take a prisoner into the special cell. We promise to take each prisoner indefinitely many times into the cell, but we take the prisoners in any order. At any given time, a prisoner can shout &quot;Liberty!&quot; At that point, if all prisoners have been to the cell, we set you free; otherwise, we double your sentences.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For which N do the prisoners have a winning strategy?&#xA;(I think they win even for $N=2$ but I don't remember; larger N are easier anyhow.)&lt;/p&gt;&#xA;&#xA;&lt;h2&gt; Hats &lt;/h2&gt;&#xA;&#xA;&lt;p&gt;A 100 students are standing in a row, so that each one can see the color of everyone in front of her. They know that the color of their hats is either red or blue. The teacher asks them one by one, from back to front, as to the color of their hat. Anyone who answers correctly gets a trip to Disneyworld.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What should the students do to maximize the number of those students who go to Disneyworld?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the ambitious: What happens if there are countably many students?&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2011-01-25T03:28:29.887" />
  <row Id="18848" PostTypeId="2" ParentId="18843" CreationDate="2011-01-25T03:33:59.203" Score="2" ViewCount="" Body="&lt;p&gt;Should this be a community wiki? I really liked &lt;a href=&quot;http://en.wikipedia.org/wiki/Beautiful_Young_Minds&quot; rel=&quot;nofollow&quot;&gt;Beautiful Young Minds&lt;/a&gt; about the British IMO team.&lt;/p&gt;&#xA;" OwnerUserId="3731" LastActivityDate="2011-01-25T03:33:59.203" CommunityOwnedDate="2011-01-25T12:35:45.310" CommentCount="1" />
  <row Id="18849" PostTypeId="1" CreationDate="2011-01-25T03:38:26.200" Score="1" ViewCount="76" Body="&lt;p&gt;I'm performing a typical matrix vector multiplcation, where the matrix is symmetric, and the vector is sparse. After multiplication I am then only using the absolute maximum value from the resulting vector in further calculations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is, is there some sort of method that will allow me to find out which row of the matrix will be producing the corresponding maximum value in the resulting vector, so that I will only need to perform a single row*column multiplication?&lt;/p&gt;&#xA;" OwnerUserId="1588" LastActivityDate="2011-01-25T03:38:26.200" Title="Finding the Maximum Value from a Matrix-Vector Multiplication" Tags="&lt;linear-algebra&gt;" CommentCount="1" />
  <row Id="18850" PostTypeId="2" ParentId="18835" CreationDate="2011-01-25T03:43:26.583" Score="1" ViewCount="" Body="&lt;p&gt;See the Section on the Krull Intersection Theorem (currently Section 8.7) in &lt;a href=&quot;http://math.uga.edu/~pete/integral.pdf&quot; rel=&quot;nofollow&quot;&gt;these notes&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A version of the theorem valid for any ideal $I$ in a Noetherian ring $R$ is as follows: if there exists $x \in \bigcap_{n=1}^{\infty} I^n$, then $x \in xI$.  From this one easily deduces that $\bigcap_{n=1}^{\infty} I^n = \{0\}$ under &lt;em&gt;either&lt;/em&gt; of the following additional hypotheses:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\bullet$ $R$ is a domain and $I$ is a proper ideal, or&lt;br&gt;&#xA;$\bullet$ $I$ is contained in the Jacobson radical $J(R)$ of $R$ (i.e., the intersection of all maximal ideals).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular the second condition holds for any proper ideal in a Noetherian local ring.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As Mariano remarks, &lt;em&gt;some&lt;/em&gt; hypothesis beyond Noetherianity is needed in order to guarantee $\bigcap_n I^n = \{0\}$.  I should probably add his counterexample to my notes!&lt;/p&gt;&#xA;" OwnerUserId="299" LastActivityDate="2011-01-25T03:43:26.583" CommentCount="5" />
  <row Id="18851" PostTypeId="2" ParentId="18838" CreationDate="2011-01-25T03:47:53.673" Score="1" ViewCount="" Body="&lt;p&gt;Probably not best, but one approach to showing continuity at $(z_0,z_0)$ is to use the power series expansion $f(z)=\sum_{k=0}^\infty a_k(z-z_0)^k$ to get &#xA;$$\frac{f(w)-f(z)}{w-z}-f&amp;#39;(z_0)=\sum_{k=1}^\infty a_k\frac{(w-z_0)^k-(z-z_0)^k}{(w-z_0)-(z-z_0)}-f&amp;#39;(z_0)=\sum_{k=2}^\infty a_k\sum_{j=0}^{k-1}(w-z_0)^j(z-z_0)^{k-1-j},$$ which converges absolutely and uniformly near $(z_0,z_0)$ and goes to $0$ at $(z_0,z_0)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A better candidate for &quot;best&quot; is that given in Rudin's &lt;em&gt;Real and complex analysis&lt;/em&gt;, Lemma 10.29 on page 314 of the 3rd Edition.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2011-01-25T03:47:53.673" CommentCount="3" />
  <row Id="18852" PostTypeId="2" ParentId="18843" CreationDate="2011-01-25T03:55:44.923" Score="6" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://www.youtube.com/watch?v=JX3VmDgiFnY&quot; rel=&quot;nofollow&quot;&gt;Mobius transformations revealed&lt;/a&gt; is pretty amazing.  Also, anything by &lt;a href=&quot;http://vihart.com/&quot; rel=&quot;nofollow&quot;&gt;Vi Hart&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="4266" LastActivityDate="2011-01-25T03:55:44.923" CommunityOwnedDate="2011-01-25T12:35:45.310" CommentCount="1" />
  <row Id="18853" PostTypeId="1" CreationDate="2011-01-25T04:16:48.033" Score="2" ViewCount="69" Body="&lt;p&gt;I'm not much good at even basic number theory, so this is really mysterious to me.  I think the actual question might want to restrict to &lt;em&gt;finite&lt;/em&gt; Laurent series, but I don't think it matters: I'm pretty sure what I want to do is figure out what each homogeneous component is allowed to look like.  Moreover, I'm pretty sure that by plugging in $h=k=1$ I get that all the coefficients in nonnegative degrees need to be integral, so the first place to look is in degree $-1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So suppose we have \begin{equation*} f(u,v)_{-1}=\frac{a_1}{a_2 u}+\frac{b_1}{b_2 v}. \end{equation*}  Then \begin{equation*}f(ht,kt)_{-1}=\frac{a_1}{a_2 ht}+\frac{b_1}{b_2 kt} = \frac{a_1 b_2 k+b_1 a_2 h}{a_2b_2hkt};\end{equation*} for this to lie in $\mathbb{Z}[t^{\pm},1/h,1/k]$ we need $a_2b_2|a_1b_2k+b_1a_2 h$.  I believe it's a basic fact that since this has to be true for all nonzero $h$ and $k$, we need that $a_2b_2|gcd(a_1b_2,b_1a_2)$.  This implies that $a_2b_2|a_1b_2 \Rightarrow a_2|a_1$ and similarly $b_2|b_1$.  But assuming we started with our fractions in reduced form, this implies that $a_2=b_2=1$; that is, $f(u,v)_{-1}\in \mathbb{Z}[t^{\pm}]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since I don't think I gained much insight from that calculation, I just tried to find restrictions on \begin{equation*} f(u,v)_{-2}=\frac{a_1}{a_2u^2}+\frac{b_1}{b_2uv}+\frac{c_1}{c_2v^2}. \end{equation*}  Now we have \begin{equation*} f(ht,kt)_{-2} = \frac{a_1}{a_2 h^2t^2}+\frac{b_1}{b_2 hkt^2} + \frac{c_1}{c_2 k^2t^2} \end{equation*} \begin{equation*} = \frac{a_1b_2hkc_2k^2 + b_1a_2h^2c_2k^2 + c_1a_2h^2 b_2hk}{a_2b_2c_2h^3k^3t^2} \end{equation*} \begin{equation*} = \frac{a_1b_2c_2k^2 + b_1a_2c_2hk + c_1a_2b_2h^2}{a_2b_2c_2h^2k^2t^2},\end{equation*} and what we need is for $a_2b_2c_2|a_1b_2c_2k^2+b_1a_2c_2hk+c_1a_2b_2h^2$ (for all $h$ and $k$, of course).  This is where I'm stuck.  Reducing mod $a_2$ and recalling that $gcd(a_1,a_2)=1$ we should get that $a_2|b_2c_2k^2$, and similarly $b_2|a_2c_2hk$ and $c_2|a_2b_2h^2$.  Putting $h=k=1$ this means that $a_2|b_2c_2$, $b_2|a_2c_2$, and $c_2|a_2b_2$.  I tried breaking down these conditions into conditions on the prime factorizations but didn't get very far.  And in any case, numbers that satisfy these need not satisfy the original relation.  Help please????&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Motivation:&lt;/strong&gt; In case you're still reading and you're wondering why I care about this specific calculation, it's because it's closely related to the &quot;$K$-theoretic homology of $K$-theory&quot;; precisely, it's the image of $K_*K\rightarrow K_*K\otimes \mathbb{Q}$.  This is important because for any (co)homology theory $E$, the $E$-cohomology of $E$ (written $E^*E$) is exactly the algebra of stable cohomology operations, and with some additional assumtions on $E$ (which are satisfied by $K$-theory), the $E$-homology of $E$ is the coalgebra of stable homology cooperations.  Furthermore, the $e$-invariant is a homomorphism from the stable homotopy groups of spheres to a subquotient of $K_*K$!&lt;/p&gt;&#xA;" OwnerUserId="2159" LastEditorUserId="2159" LastEditorDisplayName="" LastEditDate="2011-01-25T20:07:18.993" LastActivityDate="2011-01-25T20:07:18.993" Title="What is $\{f(u,v)\in \mathbb{Q}[u^{\pm},v^{\pm}]:f(ht,kt)\in \mathbb{Z}[t^{\pm},1/h,1/k]\forall h,k\in \mathbb{Z}\backslash\{0\}\}$?" Tags="&lt;number-theory&gt;" />
  <row Id="18854" PostTypeId="2" ParentId="18842" CreationDate="2011-01-25T04:25:53.607" Score="2" ViewCount="" Body="&lt;p&gt;Usually, if you want to solve the system, you use something closer to the lexicographical ordering, which gives triangular Groebner bases:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;i1 : T = ZZ[f,g,h, MonomialOrder =&amp;gt; Lex]&#xA;&#xA;o1 = T&#xA;&#xA;o1 : PolynomialRing&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;After this, the rest is just the same as what you did:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;i2 : Woot = gb ideal(f^2*g + g^2*h + 2*h^2, f + g + h, f^3*g^2*h) &#xA;&#xA;o2 = GroebnerBasis[status: done; S-pairs encountered up to degree 10]&#xA;&#xA;o2 : GroebnerBasis&#xA;&#xA;i3 : ideal gens Woot&#xA;&#xA;              8      7      6      5      5     6     5     7       6       5     7     6     5    2 3       6       5     7  &#xA;o3 = ideal (4h  + 24h  + 48h  + 32h , 8g*h  + 4h  - 8h , g*h  + 4g*h  + 4g*h  - 2h  - 8h  - 8h , 8g h  - 3g*h  + 2g*h  - 4h  -&#xA;     -------------------------------------------------------------------------------------------------------------------------&#xA;        6      5   2 4     2 3      6       5     6     5   3     2       2     2&#xA;     10h  - 28h , g h  + 2g h  + g*h  + 4g*h  - 2h  - 8h , g  + 3g h + g*h  + 2h , f + g + h)&#xA;&#xA;o3 : Ideal of T&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Incidentally, the order in which you order the variables is quite important (but I think no one knows how to pick it, so you have to try a bit...) For example, if I change the order in this example, I get a considerably simpler&#xA;triangular system:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;i12 : T = ZZ[g,h,f, MonomialOrder =&amp;gt; Lex]&#xA;&#xA;o12 = T&#xA;&#xA;o12 : PolynomialRing&#xA;&#xA;i13 : ideal gens gb ideal(f^2*g + g^2*h + 2*h^2, f + g + h, f^3*g^2*h) &#xA;&#xA;               8      5     7     7       6    8    2 3      5    6   2 5      6       5     7   3     2      2    3&#xA;o13 = ideal (4f , 4h*f  + 4f , h*f  + 2h*f  + f , 2h f  - h*f  - f , h f  - h*f  + 2h*f  + 2f , h  + 2h f + 2h  - f , g + h +&#xA;      ------------------------------------------------------------------------------------------------------------------------&#xA;      f)&#xA;&#xA;o13 : Ideal of T&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;One can solve it almost mentally now :)&lt;/p&gt;&#xA;" OwnerUserId="274" LastEditorUserId="274" LastEditorDisplayName="" LastEditDate="2011-01-25T04:36:18.617" LastActivityDate="2011-01-25T04:36:18.617" CommentCount="5" />
  <row Id="18855" PostTypeId="2" ParentId="18843" CreationDate="2011-01-25T04:28:31.667" Score="3" ViewCount="" Body="&lt;p&gt;Nova's Mathematical Mystery Tour is still on youtube (let's hope it stays there):&#xA;(link to first part):&#xA;&lt;a href=&quot;http://www.youtube.com/watch?v=vPSx2CkKE3c&quot; rel=&quot;nofollow&quot;&gt;http://www.youtube.com/watch?v=vPSx2CkKE3c&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="890" LastActivityDate="2011-01-25T04:28:31.667" CommunityOwnedDate="2011-01-25T12:35:45.310" />
  <row Id="18856" PostTypeId="2" ParentId="18771" CreationDate="2011-01-25T04:29:43.430" Score="2" ViewCount="" Body="&lt;p&gt;Here are a couple of things that I found helpful when thinking about vector bundles:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Over nice spaces they they can be classified up to isomorphism. So by nice here we mean compact manifold, or CW complex. The idea is that every finite dimensional vector bundle $\xi: E \to X$ is isomorphic to the pull back of the universal bundle $EG \to BG$ where $G$ is some particular structure group. (for $\xi$ a real $n$ dimensional bundle $G=GL_n(\mathbb{R})$, note though that $O(n)$ is homotopy equivalent to $GL_n(\mathbb{R})$). You can learn a lot about a vector bundle by examining the map that classifies it. For example, $\xi: E \to X$ is orientable if and only if the map that classifies it can be lifted over the fibration $BSL_n(\mathbb{R}) \to BGL_n(\mathbb{R})$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Vector bundles are really nicely behaved families of vector spaces over $X$. We usually use them to keep track of some important data about $X$. In fact, most operations on vector spaces extend to operations on vector bundles over $X$. Things such as tensor products, direct sums, exterior and symmetric powers pop up all the time.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Surprisingly enough every vector bundle $\xi: E \to X$ is a homotopy equivalence since the fibers are contractible. It seems like this makes vector bundles sort of silly, but it illustrates that they are inherently geometric objects. While we use them for certain constructions in homotopy theory, they are really about the geometry of the base space. That being said, homotopy theory can say a lot about vector bundles, consider Chern-Weil Theory which relates cohomology classes (a gadget that can only see homotopy theory) to things like curvature.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;A big help for learning about bundles for me was thinking about them like objects, like when you think about a manifold you want to think about the manifold not its parametrizations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mariano's answer is dead on, I am just adding some helpful facts. Let me know if you want me to clarify something.&lt;/p&gt;&#xA;" OwnerUserId="627" LastActivityDate="2011-01-25T04:29:43.430" />
  <row Id="18857" PostTypeId="1" AcceptedAnswerId="18859" CreationDate="2011-01-25T04:42:43.617" Score="0" ViewCount="238" Body="&lt;p&gt;I have never seen a problem like this before, so I was wondering if anyone could give me help getting started. I'm studying for a quiz on Wednesday.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Find an equation of the sphere that passes through the origin and whose center is (5, 10, -9).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;_&lt;/em&gt;__&lt;/strong&gt; = 0 &#xA;Note that you must put everything on the left hand side of the equation and that we desire the coefficients of the quadratic terms to be 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, how would one approach this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Find an equation of the largest sphere with center (2, 10 , 4) that is contained completely in the first octant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;= 0 &#xA;Note that you must move everything to the left hand side of the equation that we desire the coefficients of the quadratic terms to be 1.&lt;/p&gt;&#xA;" OwnerUserId="3800" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-25T05:02:18.140" LastActivityDate="2011-01-25T05:02:18.140" Title="Find an equation of the sphere that passes through the origin and whose center is (5, 10, -9). Help with Calculus III." Tags="&lt;analytic-geometry&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="18858" PostTypeId="2" ParentId="18857" CreationDate="2011-01-25T04:45:38.167" Score="3" ViewCount="" Body="&lt;p&gt;The radius of a sphere is the distance between the center and any point &lt;em&gt;on&lt;/em&gt; the sphere. You know the center, and you know a point on the sphere, so the distance between the two points gives you the radius $r$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you know the radius and the center, the equation of the sphere is very simple. It's almost like the case of a circle on the plane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To jog your memory: the circle with center $(a,b)$ and radius $r$ has equation &#xA;$$(x-a)^2 + (y-b)^2 = r^2.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do you recall now what the equation of a sphere with radius $r$ and center $(a,b,c)$ is?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And since you &lt;em&gt;know&lt;/em&gt; the center, and now you know how to find the radius, then the equation of the sphere is...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And as far as moving everything to one side, well, in the case of the circle I gave above, you can just as well write $(x-a)^2 + (y-b)^2 = r^2$ as $(x-a)^2 + (y-b)^2 - r^2 = 0$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the second problem, again let's think in terms of the plane. Suppose you want the equation of the circle with center at $(2,10)$ and completely contained in the first quadrant. How would we proceed? Imagine the largest circle: it must &lt;em&gt;just&lt;/em&gt; touch at least one of the axes; how far is $(2,10)$ from the axes? It is $2$ away from the $y$-axis, and $10$ away from the $x$-axis. So the largest the radius can be is $2$, otherwise it would &quot;cut through&quot; the $y$-axis. So you'd want the circle with radius $2$ and center in $(2,10)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How would you go about doing this for the sphere in the first octant?&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-25T05:01:00.327" LastActivityDate="2011-01-25T05:01:00.327" />
  <row Id="18859" PostTypeId="2" ParentId="18857" CreationDate="2011-01-25T04:53:52.367" Score="3" ViewCount="" Body="&lt;p&gt;Let $r$ be the radius of the sphere. Then the equation of the sphere with center $(5,10,-9)$ is $$ (x-5)^{2} + (y-10)^{2} + (z+9)^{2} =r^{2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now since your sphere passes through the origin$=(0,0,0)$, therefore you get $r^{2}=5^{2}+10^{2}+9^{2}=25+100+81=206$. Therefore the equation of the sphere is: $$(x-5)^{2}+(y-10)^{2}+(z-9)^{2}=206$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the next one:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the sphere is in the 1st octant, the largest possible sphere must be tangent to either the $xy, xz$, $yz$ planes. That is, the radius of the sphere must be such that the surface of the sphere just touches one of the planes.&lt;/p&gt;&#xA;" LastActivityDate="2011-01-25T04:53:52.367" CommentCount="2" />
  <row Id="18860" PostTypeId="2" ParentId="18843" CreationDate="2011-01-25T04:56:09.370" Score="3" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://www.dimensions-math.org/Dim_E.htm&quot; rel=&quot;nofollow&quot;&gt;Dimensions&lt;/a&gt; is a series of very good videos on stereographic projection, fourth dimension geometry, complex numbers, fibrations and proofs. These are dubbed in various languages.&lt;/p&gt;&#xA;" OwnerUserId="1938" LastActivityDate="2011-01-25T04:56:09.370" CommunityOwnedDate="2011-01-25T12:35:45.310" />
  <row Id="18861" PostTypeId="1" CreationDate="2011-01-25T05:10:15.177" Score="-1" ViewCount="142" Body="&lt;p&gt;For example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose vector u = (-2,3) and vector v = (-5,3) then:&#xA;(u + v = ?) and&#xA;(u - v = ?) and&#xA;(v - u = ?) and&#xA;(6u = ?) and&#xA;(-1/8v = ?) and&#xA;(3u - 4v = ?)&lt;/p&gt;&#xA;" OwnerUserId="3800" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-25T05:19:19.547" LastActivityDate="2011-01-25T05:19:19.547" Title="How do I add Vectors in Vector Calculus?" Tags="&lt;analytic-geometry&gt;&lt;vector-calculus&gt;" AnswerCount="2" CommentCount="5" />
  <row Id="18862" PostTypeId="2" ParentId="18861" CreationDate="2011-01-25T05:15:06.700" Score="0" ViewCount="" Body="&lt;p&gt;This is the most I'll give - &lt;a href=&quot;http://en.wikipedia.org/wiki/Vector_(geometric)#Vector_addition_and_subtraction&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Vector_(geometric)#Vector_addition_and_subtraction&lt;/a&gt; - I feel like this question is too homework-y.&lt;/p&gt;&#xA;" OwnerUserId="4967" LastActivityDate="2011-01-25T05:15:06.700" CommentCount="1" />
  <row Id="18863" PostTypeId="1" CreationDate="2011-01-25T05:15:53.313" Score="6" ViewCount="81" Body="&lt;p&gt;Suppose one is trying to devise a method to generate random matrices with a certain distribution. How does one verify that the generated matrices follow the desired distribution? In particular, I am interested in uniformly distributed random orthogonal matrices, but I am curious how people test their codes for correctness in general.&lt;/p&gt;&#xA;" OwnerUserId="398" LastActivityDate="2011-03-26T16:38:14.497" Title="Verifying the distribution of an ensemble of random matrices" Tags="&lt;probability-theory&gt;&lt;matrices&gt;&lt;probability&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="18864" PostTypeId="2" ParentId="18861" CreationDate="2011-01-25T05:16:10.927" Score="1" ViewCount="" Body="&lt;p&gt;You add vectors coordinate-wise (coordinate-to-coordinate: the first coordinate of the sum is the sum of the first coordinates of the summands; the second coordinate of the sum is the sum of the second coordinate of the summands, etc). You multiply vectors by scalars coordinate-wise as well (multiply each coordinate by the scalar). You do combinations of vector additions and scalar multiplications by performing the multiplications first and the additions later.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S. You are studying for Calculus III and you don't know how to do basic operations with vectors? I confess I find it very hard to believe that this is not contained in your textbook, or was not covered in lecture.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2011-01-25T05:16:10.927" CommentCount="1" />
  <row Id="18865" PostTypeId="1" CreationDate="2011-01-25T05:29:08.523" Score="2" ViewCount="120" Body="&lt;p&gt;I didn't think it was possible to have a finite area circumscribing an infinite volume but on page 89 of &lt;em&gt;Nonplussed!&lt;/em&gt; by Havil (accessible for me at Google Books) it is claimed that such is the goblet-shaped solid generated by revolving the cissoid y$^2$ = x$^3$/(1-x) about the positive y-axis between this axis and the asymptote x = 1. What do you think?&lt;/p&gt;&#xA;" OwnerUserId="1770" LastActivityDate="2011-01-25T14:26:46.150" Title="Property of a Cissoid?" Tags="&lt;calculus&gt;" AnswerCount="3" CommentCount="7" FavoriteCount="1" />
  <row Id="18866" PostTypeId="2" ParentId="18865" CreationDate="2011-01-25T05:47:04.620" Score="1" ViewCount="" Body="&lt;p&gt;I'm looking at the quote right now, and it must be wrong. The goblet is infinitely tall, with a radius of approximately 1, so it must have infinite area as well. The quote at the end by de Sluze refers to the weight being finite (proportional to the volume under the curve, above the x-axis), while the cup itself can hold an infinite volume. No matter how you play with this, the surface area must be infinite.&lt;/p&gt;&#xA;" OwnerUserId="398" LastActivityDate="2011-01-25T05:47:04.620" />
  <row Id="18867" PostTypeId="1" CreationDate="2011-01-25T05:52:17.863" Score="1" ViewCount="189" Body="&lt;p&gt;I have found formulas online for the the roots of a cubic polynomial with real coefficients, but they all said this they would not work for polynomials with complex coefficients.  I need to solve the following:&#xA;$x^3+cx^2+\alpha$ where $c&amp;gt;0$ and $\alpha\in\mathbb{C}$.  Anyone have the formula&lt;/p&gt;&#xA;" OwnerUserId="5952" LastEditorUserId="5952" LastEditorDisplayName="" LastEditDate="2011-02-06T21:29:34.817" LastActivityDate="2011-02-06T21:29:34.817" Title="How does one solve a cubic polynomial with complex coefficients?" Tags="&lt;complex-analysis&gt;&lt;polynomials&gt;" AnswerCount="2" CommentCount="0" />
  <row Id="18868" PostTypeId="1" AcceptedAnswerId="18869" CreationDate="2011-01-25T05:58:00.547" Score="2" ViewCount="147" Body="&lt;p&gt;For a high school student that has the options of taking &lt;strong&gt;one&lt;/strong&gt; of Statistics, Algebra or Calculus as subjects, which subject would be the best choice if the person is still undecided on a specific career path (in other words would want to keep their options as general as possible)?&lt;/p&gt;&#xA;" OwnerUserId="5642" LastActivityDate="2011-01-25T21:12:52.043" Title="Choosing subjects at high school" Tags="&lt;soft-question&gt;&lt;high-school&gt;" AnswerCount="4" />
  <row Id="18869" PostTypeId="2" ParentId="18868" CreationDate="2011-01-25T06:00:21.133" Score="6" ViewCount="" Body="&lt;p&gt;You would need to take Algebra before taking Calculus (which is applied to a variety of fields), do you actually have to choose just one? Statistics is widely applicable to the real world too, possibly even more so than Calculus, but I would probably not do either Calculus or Statistics at the expense of missing out on Algebra -- it really is fundamental to both.&lt;/p&gt;&#xA;" OwnerUserId="3731" LastEditorUserId="3731" LastEditorDisplayName="" LastEditDate="2011-01-25T16:03:58.580" LastActivityDate="2011-01-25T16:03:58.580" />
  <row Id="18871" PostTypeId="2" ParentId="18865" CreationDate="2011-01-25T06:41:34.633" Score="3" ViewCount="" Body="&lt;p&gt;First, let's have a &lt;a href=&quot;http://books.google.com/books?id=lQ0N9H793GQC&amp;amp;pg=PA89&quot; rel=&quot;nofollow&quot;&gt;proper link to the book&lt;/a&gt;, alright?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, as can be seen, they are talking about the cissoid of Diocles, which has the Cartesian equation $y^2=\frac{x^3}{a-x}$ . One can check that the cissoid encloses a finite area along with its asymptote: $\frac34\pi a^2$ ; thus you can expect the corresponding surface of revolution to have finite volume.&lt;/p&gt;&#xA;" OwnerUserId="6215" LastActivityDate="2011-01-25T06:41:34.633" CommentCount="2" />
  <row Id="18872" PostTypeId="2" ParentId="18867" CreationDate="2011-01-25T06:45:00.087" Score="4" ViewCount="" Body="&lt;p&gt;As Moron says, Cardano works. &lt;a href=&quot;http://books.google.com/books?id=w47cyQLVQowC&amp;amp;pg=PA228&quot; rel=&quot;nofollow&quot;&gt;You merely need to be a bit more careful than usual.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="6215" LastActivityDate="2011-01-25T06:45:00.087" />
  <row Id="18873" PostTypeId="2" ParentId="15865" CreationDate="2011-01-25T06:52:26.923" Score="2" ViewCount="" Body="&lt;p&gt;Since no one has posted an answer and since my comment is a sort of tangential answer (and relevant to another question): &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In UCSMP &lt;em&gt;Precalculus and Discrete Mathematics&lt;/em&gt;, 3rd edition, p553 (in the &quot;exploration&quot; question), the cubic formula is given in a form analogous to what you describe (though it is for the general monic cubic, not the depressed cubic). It is given that way for the reason you describe—in particular, because of the way that most calculators and computer algebra systems define the principal root, the &quot;traditional&quot; way of writing the formula does not always yield correct results when computing blindly with technology.  The formula as printed in the first printing run of &lt;em&gt;PDM&lt;/em&gt; is actually missing a term, though it should be correct in subsequent printing runs.  The correct formula reads:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $$A=\frac{\sqrt[3]{-2p^3+9pq-27r+3\sqrt{3}\sqrt{-p^2q^2+4q^3+4p^3r-18pqr+27r^2}}}{3\sqrt[3]{2}}$$ and $$B=\frac{-p^2+3q}{9A}.$$  Then, $$x_1=-\frac{p}{3}+A-B,$$ $$x_2=-\frac{p}{3}+\frac{-1-i\sqrt{3}}{2}A-\frac{-1+i\sqrt{3}}{2}B,$$ and $$x_3=-\frac{p}{3}+\frac{-1+i\sqrt{3}}{2}A-\frac{-1-i\sqrt{3}}{2}B$$ are the solutions to $$x^3+px^2+qx+r=0.$$&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2011-01-25T06:52:26.923" />
  <row Id="18874" PostTypeId="2" ParentId="18867" CreationDate="2011-01-25T06:53:55.020" Score="5" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://math.stackexchange.com/questions/15865/why-not-write-the-solutions-of-a-cubic-this-way/18873#18873&quot;&gt;In my answer here&lt;/a&gt;, I give a cubic formula that works for complex coefficients and works with the way principal roots are defined on non-real complex numbers in the vast majority of calculators and computer algebra systems.&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2011-01-25T06:53:55.020" />
  <row Id="18875" PostTypeId="1" CreationDate="2011-01-25T07:39:12.157" Score="6" ViewCount="143" Body="&lt;p&gt;Given is an almost-square matrix $A$ with $n$ columns and $n-1$ rows with maximum rank. The solutions of the homogeneous system $Ax = 0$ form a 1-dimensional subspace of $\mathbb{R}^n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've discovered the following which I believe to be true but I can't prove: the components of the vector $x$ that spans the (1D) solution space are given by:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x_i = (-1)^{i-1} |A_i|$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;in which $|A_i|$ is the determinant of the square submatrix of A obtained by removing the i-th column from A. For example, in $\mathbb{R^3}$, $A$ is a 2x3 matrix, and $x$ as defined above turns out to be the crossproduct of the two row vectors of $A$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this true, and if so, how can it be proved?&lt;/p&gt;&#xA;" OwnerUserId="6216" LastActivityDate="2011-01-25T20:03:38.593" Title="1-dimensional solution space of homogeneous system Ax=0?" Tags="&lt;linear-algebra&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="18876" PostTypeId="2" ParentId="18843" CreationDate="2011-01-25T08:14:36.093" Score="3" ViewCount="" Body="&lt;p&gt;Here's a really excellent video, &lt;a href=&quot;http://www.youtube.com/watch?v=BipvGD-LCjU&quot; rel=&quot;nofollow&quot;&gt;Finite Simple Group (of Order Two)&lt;/a&gt;, that's definitely more towards the fun side than educational, although if you're not getting some of the jokes it may inspire you to look up a thing or two. &lt;/p&gt;&#xA;" OwnerUserId="1301" LastActivityDate="2011-01-25T08:14:36.093" CommunityOwnedDate="2011-01-25T12:35:45.310" />
  <row Id="18877" PostTypeId="2" ParentId="18875" CreationDate="2011-01-25T08:23:55.423" Score="5" ViewCount="" Body="&lt;p&gt;You only need to show that the $x$ you defined is orthogonal to every row of $A$. You can see that's the case by using the row-expansion of the determinant: given any row $j$ of $A$, consider the $n \times n$ matrix obtained from $A$ by duplicating that row. This new square matrix $A&amp;#39;$ has determinant $0$, and you can compute its determinant by expanding along the newly created duplicate row: $$ 0={\rm det} (A&amp;#39;) = \sum_i (-1)^{j-1} (-1)^i A_{ji} |A_i | = (-1)^{j-1} \sum_i A_{ji} x_i$$&lt;/p&gt;&#xA;" OwnerUserId="6104" LastActivityDate="2011-01-25T08:23:55.423" />
  <row Id="18878" PostTypeId="2" ParentId="18843" CreationDate="2011-01-25T08:47:57.710" Score="1" ViewCount="" Body="&lt;p&gt;May not be fundamental maths videos, but I like the teaching style of &lt;a href=&quot;http://web.stonehill.edu/compsci/shai.htm&quot; rel=&quot;nofollow&quot;&gt;Shai Simonson&lt;/a&gt; and his lectures in &lt;a href=&quot;http://aduni.org/courses/discrete/index.php?view=cw&quot; rel=&quot;nofollow&quot;&gt;Discrete Mathematics&lt;/a&gt; and &lt;a href=&quot;http://aduni.org/courses/theory/index.php?view=cw&quot; rel=&quot;nofollow&quot;&gt;Theory of Computation&lt;/a&gt; are entertaining as well as enlightening.&lt;/p&gt;&#xA;" OwnerUserId="5516" LastActivityDate="2011-01-25T08:47:57.710" CommunityOwnedDate="2011-01-25T12:35:45.310" />
  <row Id="18879" PostTypeId="1" AcceptedAnswerId="18885" CreationDate="2011-01-25T10:00:17.910" Score="2" ViewCount="138" Body="&lt;p&gt;A first-order sentence is (logically) valid iff it's true in every interpretation. And it's valid iff it can be deduced from the FO axioms alone.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One normal case of showing that a FO sentence is true is deducing it (syntactically).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I guess that indirect proofs have to be interpreted more &quot;semantically&quot;: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Assume that ~F is true in an interpretation. [Deduction of a contradiction] Thus ~F couldn't have been true. Thus F must be true in every interpretation.&quot; (But we are left without a deduction of F from the axioms.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this reading of indirect proofs correct?&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Are there also &lt;em&gt;direct&lt;/em&gt; proofs that&#xA;  work &quot;semantically&quot;, i.e. that show&#xA;  directly that a sentence is true in&#xA;  every interpretation? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Like truth tables do.&lt;/p&gt;&#xA;" OwnerUserId="1792" LastActivityDate="2011-01-25T23:59:11.160" Title="Direct and indirect proof, syntactical and semantical" Tags="&lt;logic&gt;" AnswerCount="2" />
  <row Id="18880" PostTypeId="1" AcceptedAnswerId="18888" CreationDate="2011-01-25T10:33:49.230" Score="1" ViewCount="37" Body="&lt;p&gt;I am reading an article and it is written there: &#xA;If A is a k-linear category (possibly without direct sums) we can embed it in the&#xA;additive category A × N, where a morphism (x,m) → (y, n) is an n × m matrix&#xA;with entries in A(x, y) = HomA(x, y). Of course if A is additive then A ≈ A × N. Could anyone explain it? Why is it additive?&lt;/p&gt;&#xA;" OwnerUserId="6223" LastActivityDate="2011-01-25T11:43:06.363" Title="embedding a k-linear category in an additive category" Tags="&lt;category-theory&gt;" AnswerCount="1" />
  <row Id="18881" PostTypeId="1" AcceptedAnswerId="18893" CreationDate="2011-01-25T10:44:00.983" Score="5" ViewCount="459" Body="&lt;p&gt;Can anyone explain me as to why &lt;em&gt;Tensor Products&lt;/em&gt; are important, and what makes Mathematician's to define them in such a manner. We already have &lt;em&gt;Direct Product&lt;/em&gt;, &lt;em&gt;Semi-direct&lt;/em&gt; products, so after all why do we need Tensor Product?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Definition of &lt;strong&gt;Tensor Product&lt;/strong&gt; at Planet Math is confusing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Definition:&lt;/strong&gt;  Let $R$ be a commutative ring, and let $A,&#xD;&#xA;B$ be $R$-modules. There exists an $R$-module $A\otimes B$, called the&#xA;&lt;em&gt;tensor product&lt;/em&gt; of $A$ and $B$ over $R$, together with a canonical&#xA;bilinear homomorphism&#xA;$$\otimes: A\times B\rightarrow A\otimes B,$$&#xA;distinguished, up to isomorphism, by the following universal&#xA;property.&#xA;Every bilinear $R$-module homomorphism&#xA;$$\phi: A\times B\rightarrow C,$$&#xA;lifts to a unique $R$-module homomorphism&#xA;$$\tilde{\phi}: A\otimes B\rightarrow C,$$&#xA;such that&#xA;$$\phi(a,b) = \tilde{\phi}(a\otimes b)$$&#xA;for all $a\in A,\; b\in B.$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The tensor product $A\otimes B$ can be constructed by taking the free&#xA;$R$-module generated by all formal symbols&#xA;$$a\otimes b,\quad a\in A,\;b\in B,$$&#xA;and quotienting by the obvious bilinear relations:&#xA;\begin{align*}&#xA; (a_1+a_2)\otimes b &amp;amp;= a_1\otimes b + a_2\otimes b,\quad &amp;amp;&amp;amp;a_1,a_2\in&#xA; A,\; b\in B \\&#xA; a\otimes(b_1+b_2) &amp;amp;= a\otimes b_1 + a\otimes b_2,\quad &amp;amp;&amp;amp;a\in A,\;b_1,b_2\in&#xA; B \\&#xA; r(a\otimes b) &amp;amp;= (ra)\otimes b= a\otimes (rb)\quad &amp;amp;&amp;amp;a\in A,\;b\in&#xA; B,\; r\in R&#xA;\end{align*}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also what do is the meaning of this statement:(&lt;strong&gt;Why do we need this?&lt;/strong&gt;)&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Every bilinear $R$-module homomorphism&#xA;$$\phi: A\times B\rightarrow C,$$&#xA;lifts to a unique $R$-module homomorphism&#xA;$$\tilde{\phi}: A\otimes B\rightarrow C,$$&#xA;such that&#xA;$$\phi(a,b) = \tilde{\phi}(a\otimes b)$$&#xA;for all $a\in A,\; b\in B.$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Me and my friends have planned to study this topic today. So i hope i am not wrong in asking such questions.&lt;/p&gt;&#xA;" LastActivityDate="2011-01-26T13:16:44.273" Title="Motivation for Tensor Product" Tags="&lt;abstract-algebra&gt;&lt;commutative-algebra&gt;" AnswerCount="7" CommentCount="3" FavoriteCount="2" />
  <row Id="18882" PostTypeId="1" AcceptedAnswerId="18886" CreationDate="2011-01-25T10:46:18.033" Score="3" ViewCount="66" Body="&lt;p&gt;Summations can really get complicated - esp. when you have convoluted n-fold summations with all kinds of different indices.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My question:&lt;/strong&gt; Is there some software (or add-on) with which you can find and prove (complicated) summation identities?&lt;/p&gt;&#xA;" OwnerUserId="346" LastActivityDate="2011-01-25T11:03:34.853" Title="Software for proving sum identites" Tags="&lt;combinatorics&gt;&lt;computer-algebra-systems&gt;&lt;sequences-and-series&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" />
  <row Id="18883" PostTypeId="2" ParentId="18879" CreationDate="2011-01-25T10:57:37.733" Score="3" ViewCount="" Body="&lt;p&gt;I hope I won't mess this up; been some time since I played around with logics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A key feature of every sound, consistent logic is that you can not prove $false$. If you add $\neg F$ to the assumptions (what is exactly what you do) and can now prove $false$ (by deriving a contradiction) -- syntactically, i.e. for all interpretations! -- the only reason for that can be that $\neg F$ is not true for any interpretation. This implies that $F$ holds for any interpretation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that if you want to enter this in an interactive theorem prover, you might have to take care that all variables in $F$ are free or bound only in $F$.&lt;/p&gt;&#xA;" OwnerUserId="3330" LastActivityDate="2011-01-25T10:57:37.733" />
  <row Id="18884" PostTypeId="2" ParentId="18881" CreationDate="2011-01-25T10:57:59.927" Score="1" ViewCount="" Body="&lt;p&gt;What helped me see the charm of Tensor Products is the following: calculate Z/100Z ⊗ Z/101Z. Direct Product is a solution for one universal problem. Tensor Products is a solution for another one.&lt;/p&gt;&#xA;" OwnerUserId="6223" LastActivityDate="2011-01-25T10:57:59.927" CommentCount="2" />
  <row Id="18885" PostTypeId="2" ParentId="18879" CreationDate="2011-01-25T11:03:07.520" Score="3" ViewCount="" Body="&lt;p&gt;We can easily turn an indirect proof of a theorem (or proof of a theorem by contradiction) into an ordinary direct syntactic proof.  This is a special case of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Deduction_theorem&quot; rel=&quot;nofollow&quot;&gt;Deduction Theorem&lt;/a&gt; which tells us that if $\Sigma \cup \{A\} \vdash B$ for some collection of statements $\Sigma$ and specific statements $A$ and $B$, then $\Sigma \vdash A \rightarrow B$.  Specifically, if we let $A$ be $\lnot F$ and $B$ be some contradiction $C \land \lnot C$ derived from $\Sigma \cup \{\lnot F\}$, then the deduction theorem tells us that we have a proof of $\lnot F \rightarrow (C \land \lnot C)$ from $\Sigma$.  But since $F$ is a tautological consequence of this last statement, we can extend such a proof of this $\Sigma$-theorem to a proof of $F$.  This direct syntactic proof of $F$ then gives us a direct semantic proof by &lt;a href=&quot;http://en.wikipedia.org/wiki/Completeness_theorem&quot; rel=&quot;nofollow&quot;&gt;Gödel's completeness theorem&lt;/a&gt; for first-order logic. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the Deduction Theorem itself is a &lt;a href=&quot;http://en.wikipedia.org/wiki/Metatheorem&quot; rel=&quot;nofollow&quot;&gt;metatheorem&lt;/a&gt; that can be proven by induction on the length of the proof of $B$ from $\Sigma \cup \{A\}$, and its proof provides us with a constructive way of getting the actual proof of $A \rightarrow B$ from $\Sigma$.  &lt;/p&gt;&#xA;" OwnerUserId="4693" LastEditorUserId="4693" LastEditorDisplayName="" LastEditDate="2011-01-25T23:59:11.160" LastActivityDate="2011-01-25T23:59:11.160" CommentCount="5" />
  <row Id="18886" PostTypeId="2" ParentId="18882" CreationDate="2011-01-25T11:03:34.853" Score="2" ViewCount="" Body="&lt;p&gt;Fabrizio Caruso implemented &quot;Gosper's and Zeilberger's algorithm for proving and finding indefinite and definite hypergeometric summation identities&quot;, as a free and open source package for use with &lt;a href=&quot;http://maxima.sourceforge.net/&quot; rel=&quot;nofollow&quot;&gt;Maxima&lt;/a&gt;. You can get it &lt;a href=&quot;http://www.risc.jku.at/research/combinat/software/Zeilberger/index.php&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="689" LastActivityDate="2011-01-25T11:03:34.853" />
  <row Id="18887" PostTypeId="1" CreationDate="2011-01-25T11:23:11.467" Score="1" ViewCount="91" Body="&lt;p&gt;Simplifying a business example, I have the following situation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some objects should be distributed in a graph in most &quot;linear&quot; way possible for a given &quot;thermometer&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Say, a voyager visits some cities. Several cities are visited multiple times.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, we have list of cities in ordinate axis (that may be duplicated), and Time in abscissas one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, for a given path, say &lt;strong&gt;(A =&gt; X =&gt; A =&gt; B =&gt; C)&lt;/strong&gt; we should  display a line, in the &quot;most linear way possible&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/Bnk9R.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By eg. in the image above, the &lt;em&gt;green&lt;/em&gt; line is optimal one&lt;br&gt;&#xA;&lt;strong&gt;(1 &gt; 2 &gt; 3 &gt; 4 &gt; 5)&lt;/strong&gt;   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;but there could be multiple possible outputs  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1 &gt; 2 &gt; 1 &gt; 4 &gt; 5)&lt;br&gt;&#xA;(1 &gt; 2 &gt; 3 &gt; 4 &gt; 5)&lt;br&gt;&#xA;(1 &gt; 2 &gt; 6 &gt; 4 &gt; 5)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(3 &gt; 2 &gt; 1 &gt; 4 &gt; 5)&lt;br&gt;&#xA;(3 &gt; 2 &gt; 3 &gt; 4 &gt; 5)&lt;br&gt;&#xA;(3 &gt; 2 &gt; 6 &gt; 4 &gt; 5)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(6 &gt; 2 &gt; 1 &gt; 4 &gt; 5)&lt;br&gt;&#xA;(6 &gt; 2 &gt; 3 &gt; 4 &gt; 5)&lt;br&gt;&#xA;(6 &gt; 2 &gt; 6 &gt; 4 &gt; 5)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there some algorithms helping in such situations?&lt;/p&gt;&#xA;" OwnerUserId="4050" LastEditorUserId="4050" LastEditorDisplayName="" LastEditDate="2011-01-25T17:27:55.633" LastActivityDate="2011-01-25T17:27:55.633" Title="Sample algorithm to “linearize” a graph" Tags="&lt;algorithms&gt;&lt;graph&gt;" CommentCount="10" />
  <row Id="18888" PostTypeId="2" ParentId="18880" CreationDate="2011-01-25T11:43:06.363" Score="1" ViewCount="" Body="&lt;p&gt;Well, you haven't got the objects quite right. A better description of the objects of the free additive category $\oplus A$ over $A$ is: $n$-tuples of objects $x = (x_{1}, \ldots, x_{n})$ of $A$ where $n \in \mathbb{N}$. Now a morphism in&#xA;$\text{Hom}\,(x,y)$ is a matrix $(f_{ij})$, where $f_{ij} \in \text{Hom}_A\,(x_{j},y_{i})$ and composition is given by the familiar formula for matrix multiplication.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you see why &lt;em&gt;this&lt;/em&gt; category is additive?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also show that it deserves its name and think about the following:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;every additive functor $F: A \to B$ to an additive category $B$ extends uniquely to an additive functor $\oplus F : \oplus A \to B$.&lt;/li&gt;&#xA;&lt;li&gt;If $A$ is already additive then $A$ and $\oplus A$ are equivalent.&lt;/li&gt;&#xA;&lt;li&gt;What is the free $k$-linear category over $A$?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="5363" LastActivityDate="2011-01-25T11:43:06.363" />
  <row Id="18889" PostTypeId="1" CreationDate="2011-01-25T11:54:24.503" Score="2" ViewCount="163" Body="&lt;p&gt;fT(t;B,C) = exp(-t/C)-exp(-t/B) / C-B     where our mean is C+B and t&gt;0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so far i have found my log likelihood functions and differentiated them as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;dl/dB = sum[t*exp(t/C) / (B^2(exp(t/c)-exp(t/B)))] +n/(C-B) = 0&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i have also found a similar dl/dC. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have now been asked to comment what you can find in the way of sufficient statistics for estimating these parameters and why there is no simple way of using Maximum Likelihood for estimation in the problem. I am simply unsure as to what to comment upon. Any help would be appreciated. Thanks, Rachel&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Editor's Note: Given here is the probability density function&#xA;$$&#xD;&#xA;f_T (t;B,C) = \frac{{e^{ - t/C}  - e^{ - t/B} }}{{C - B}}, \;\; t &amp;gt; 0,&#xD;&#xA;$$&#xA;where $B$ and $C$ are positive constants such that $C &amp;gt; B$. The mean is $C+B$.&#xA;For the log likelihood function, see the last equation in my answer to &lt;a href=&quot;http://math.stackexchange.com/questions/16156/how-do-i-write-a-log-likelihood-function-when-i-have-2-mean-values-for-my-pdf&quot;&gt;this related question&lt;/a&gt;, and differentiate accordingly (with respect to $B$ and $C$).&lt;/p&gt;&#xA;" OwnerUserId="4522" LastEditorUserId="2810" LastEditorDisplayName="" LastEditDate="2011-01-25T15:08:18.057" LastActivityDate="2011-05-25T23:01:59.610" Title="Maximum likelihood and sufficient statistics" Tags="&lt;statistics&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="18890" PostTypeId="1" AcceptedAnswerId="18899" CreationDate="2011-01-25T11:55:20.447" Score="1" ViewCount="83" Body="&lt;p&gt;Are there any ways to visualize(plot/draw) points on a high dimensional (ex: dimension = 5) manifold? &lt;/p&gt;&#xA;" OwnerUserId="5312" LastActivityDate="2011-01-25T12:26:23.017" Title="How to Visualize points on a high dimensional (&gt;3) Manifold?" Tags="&lt;topology&gt;&lt;differential-geometry&gt;&lt;visualization&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" />
  <row Id="18891" PostTypeId="2" ParentId="18881" CreationDate="2011-01-25T12:13:47.160" Score="5" ViewCount="" Body="&lt;p&gt;We define tensor products for the same reason we define any other abstract mathematical structure: it's a structure that shows up a lot in mathematics, so it's worth having a name for.  I don't see why this reason applies any less to tensor products than to direct sums.  As the Wikipedia article says,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Tensor products are important in areas of abstract algebra, homological algebra, algebraic topology and algebraic geometry&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and tensor products of vector spaces are also important in differential geometry and physics.  I think it is better to learn about these applications thoroughly than to have someone attempt to summarize them.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Tim Gowers has written a short introduction to tensor products &lt;a href=&quot;http://www.dpmms.cam.ac.uk/~wtg10/tensors3.html&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;, but it does not, in my opinion, give a good sense of the wide range of applicability of this notion.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2011-01-25T12:13:47.160" CommentCount="1" />
  <row Id="18892" PostTypeId="2" ParentId="18881" CreationDate="2011-01-25T12:15:50.593" Score="2" ViewCount="" Body="&lt;p&gt;The tensor product is the categorization of the multiplication in rings, so you can now multiply objects (vector spaces, rings, algebras, etc), and it is &quot;linear&quot; in each of the term.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose for example you have an algebra over a ring, say $M_n(\mathbb{Z})$ the matrices with integer entries (the ring is the integers), but you want to be able to multiply these matrices with polynomials from $\mathbb{Q}[x]$. So you can define formally a multiplication of a polynomial f with matrix A by (f,A).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now you want several nice properties you are used to when multiplying, meaning, it is linear in each of the terms (distributivity)&#xA;$$ (f+g,A)=(f,A)+(g,A) $$&#xA;$$ (f,A+B)=(f,A)+(f,B) $$&lt;br&gt;&#xA;Taking the tensor product gives you these properties among others.&#xA;(actually, a multiplication in a ring $R$ is a function $\mu: R\otimes R \rightarrow R$ that satisfies associativity - $\mu (1\otimes \mu)= \mu (\mu \otimes 1)$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If for example you start with a vector space $V$ over a field $F$ and you tensor it with a field $K$ which is an extension of $F$, then you get a new vector field $K \otimes V$, but over the bigger field $K$ and with the same dimension(and the multiplication is $\alpha(\beta \otimes v)= (\alpha \beta \otimes v)$. This operation is called extension of scalars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course there are other reasons why tensor product is important, but I think that this is a good place to start.&lt;/p&gt;&#xA;" OwnerUserId="2125" LastActivityDate="2011-01-25T12:15:50.593" CommentCount="1" />
  <row Id="18893" PostTypeId="2" ParentId="18881" CreationDate="2011-01-25T12:19:54.663" Score="6" ViewCount="" Body="&lt;p&gt;Tensor products are useful because of two reasons:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;they allow you to study certain non linear maps (bilinear maps) by transforming them first into linear ones, to which you can apply linear algebra;&lt;/li&gt;&#xA;&lt;li&gt;they allow you to &lt;em&gt;change&lt;/em&gt; the ring over which a module is defined.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;There are many, many ways in which these two abilities show up in nature.&lt;/p&gt;&#xA;" OwnerUserId="274" LastEditorUserId="274" LastEditorDisplayName="" LastEditDate="2011-01-25T18:01:06.687" LastActivityDate="2011-01-25T18:01:06.687" />
  <row Id="18894" PostTypeId="1" AcceptedAnswerId="18919" CreationDate="2011-01-25T12:20:35.993" Score="1" ViewCount="117" Body="&lt;p&gt;In Poisson distribution, the probability of inter arrival time to be t or less is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ P(X\leq t)= 1 - P(X&amp;gt;t) = 1 - P(0 \mbox{ arrivals in } t) = 1 - e^{-\lambda  t} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and probability of one arrival in t is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ P(k=1)= \lambda t e^{- \lambda t} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wonder how the exponential distribution can be derived from Poisson to reach:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ P(X\leq t)= \lambda e^{- \lambda t} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks,&lt;/p&gt;&#xA;" OwnerUserId="896" LastEditorUserId="1543" LastEditorDisplayName="" LastEditDate="2011-01-25T12:43:48.947" LastActivityDate="2011-01-25T16:14:09.237" Title="Exponential distribution from Poisson" Tags="&lt;probability-distributions&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="18895" PostTypeId="1" CreationDate="2011-01-21T23:34:27.570" Score="4" ViewCount="118" Body="&lt;p&gt;I know that the answer is C(8,2), but I don't get, why. &#xA;Can anyone, please, explain it?&lt;/p&gt;&#xA;" LastEditorUserId="5676" LastEditorDisplayName="" LastEditDate="2011-01-25T13:22:18.710" LastActivityDate="2011-01-26T03:52:46.613" Title="How many bytes contain exactly two 1's?" Tags="&lt;combinatorics&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="18896" PostTypeId="2" ParentId="18895" CreationDate="2011-01-21T23:38:49.763" Score="3" ViewCount="" Body="&lt;p&gt;C(8,2) is kinda self explanatory:&lt;br&gt;&#xA;you have 8 bits, and each time you choose 2 of them to be ones.&lt;/p&gt;&#xA;" OwnerUserId="5881" LastActivityDate="2011-01-21T23:38:49.763" CommentCount="1" />
  <row Id="18897" PostTypeId="2" ParentId="18895" CreationDate="2011-01-21T23:44:06.687" Score="6" ViewCount="" Body="&lt;p&gt;C(8, 2) means &quot;8 choose 2&quot;, which in this case should be interpreted as &quot;how many ways can you choose 2 items out of 8 possible&quot;. Here the 8 items are the bits, and the 2 comes from the bits you &quot;select to be 1&quot;. That is, there are just as many bytes with exactly two 1s as there are ways to select 2 bits from 8 possible.&lt;/p&gt;&#xA;" LastActivityDate="2011-01-21T23:44:06.687" CommentCount="1" />
  <row Id="18898" PostTypeId="2" ParentId="18895" CreationDate="2011-01-24T20:08:52.883" Score="2" ViewCount="" Body="&lt;p&gt;$C[8, 2] = 8!/((8-2)!2!) = (8!/6!)/2! = (8*7)/2 = 28$. Think of it this way. The 8 is the choice of the first bit, and the 7 is the choice of the second bit (it's only 7 because there are only 7 bits available after the first bit is decided). The 2 represents the number of permutations of the chosen bits. We divide by 2 because we don't want to count each ordering of the chosen bits separatetly.  &lt;/p&gt;&#xA;" OwnerUserId="4735" LastEditorUserId="4735" LastEditorDisplayName="" LastEditDate="2011-01-26T03:52:46.613" LastActivityDate="2011-01-26T03:52:46.613" />
  <row Id="18899" PostTypeId="2" ParentId="18890" CreationDate="2011-01-25T12:26:23.017" Score="3" ViewCount="" Body="&lt;p&gt;A usual way to work is to take slices, CAT-scan style. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Have you seen a CAT-scan? It represents a 3-dimensional object as sequence of two-dimensional slices of the object, in a way that you can choose which directions your slices cut (after computer processing). &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;There are two usual ways of representing a 3-dimensional object on a plane. You can take projections (think a picture of the &quot;outside&quot; of the object) or slices. Similarly, to represent higher ($N$) dimensional objects on a plane (draw it on a piece of paper), you must throw away $N-2$ degrees of freedom. Each degree of freedom you can choose to throw away by projection or by slicing. So a typical way of visualizing a 4-dimensional object is to take a sequence of 3-dimensional slices, and project each slice onto a computer screen, and display this sequence one-by-one over time. This way you get the illusion of a 4 dimensional object &quot;passing through&quot; our three dimensional world. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Higher dimensional objects can be done similarly, you just need to take more slices. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;One fun thing you can do with this is to try to play 4-dimensional tic-tac-toe. To make the game not a trivial win for the first player, you need a $5\times 5\times 5 \times 5$ grid, and first to make 5 in a row wins. So on paper you need to draw a $5\times 5$ matrix of $5\times 5$ grids, each grid represent the intersection of two orthogonal hyper-slices of the full 4-dimensional game grid. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Another fun way to develop some intuition on higher dimension spaces is to spend some time playing &lt;a href=&quot;http://www.superliminal.com/cube/cube.htm&quot; rel=&quot;nofollow&quot;&gt;Magic Cube 4D&lt;/a&gt;, a higher dimensional analogue of Rubik's cube. (It could also get you dizzy pretty quickly). &lt;/p&gt;&#xA;" OwnerUserId="1543" LastActivityDate="2011-01-25T12:26:23.017" />
  <row Id="18900" PostTypeId="2" ParentId="18843" CreationDate="2011-01-25T12:40:45.207" Score="2" ViewCount="" Body="&lt;p&gt;A few years back the NHK made a documentary called &quot;The Spell of Poincare Conjecture&quot;.  (You can find a possibly &lt;a href=&quot;http://www.veoh.com/browse/videos/category/educational_and_howto/watch/v18823385EkKAnWzX&quot; rel=&quot;nofollow&quot;&gt;non permanent link here&lt;/a&gt; to part 1 of 4 [with also links to the rest of the video].)&lt;/p&gt;&#xA;" OwnerUserId="1543" LastActivityDate="2011-01-25T12:40:45.207" CommunityOwnedDate="2011-01-25T12:40:45.207" CommentCount="2" />
  <row Id="18901" PostTypeId="2" ParentId="18881" CreationDate="2011-01-25T12:48:29.247" Score="4" ViewCount="" Body="&lt;p&gt;There are literally dozens of independent reasons to invent the tensor product, and just about every area of mathematics needs the tensor product for its own reasons (often several reasons).  Here are a couple examples.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Suppose $X$ and $Y$ are topological spaces (metric spaces are fine if you like them better) and consider the rings $C(X)$ and $C(Y)$ of continuous real-valued functions.  If you are convinced that products are worthy of consideration, then perhaps you are convinced that it is useful to look at $C(X \times Y)$.  It is natural to ask if this can be expressed in terms of $C(X)$ and $C(Y)$; the answer (modulo largely irrelevant technical details) is that $C(X \times Y) = C(X) \otimes C(Y)$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Let $V$ be a vector space over $\mathbb{R}$.  It is often desirable to construct a complex vector space naturally associated to $V$ (the &quot;complexificiation&quot; of $V$).  Here by &quot;naturally&quot; I mean in a way which is coordinate free and transparently compatible with linear maps.  The solution is to set $V_{\mathbb{C}} = V \otimes \mathbb{C}$ (tensor product over $\mathbb{R}$).  This is a special case of the more general phenomenon of &quot;extension of scalars&quot;. As a fancy example demonstrating that this really is as useful as I claim, you might check out the wikipedia page on &quot;pontryagin classes&quot; (though it might be over your head if you haven't learned much algebraic topology).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;One of the reasons why direct sums are important is that they let you turn strange objects into groups.  For example, if $G$ is a group and $V$ and $W$ are two representations of $G$ (vector spaces on which $G$ acts nicely), then $V \oplus W$ is also a representation of $V$.  So the set of all representations of $G$ has an additive structure, and with a little algebraic magic one can upgrade this structure to a group (don't spend too much time worrying about how you subtract representations).  Groups are nice and have lots of their own invariants, but rings are even nicer and have even more invariants.  So it would be great if we could define a natural product of representations.  You guessed it: the product of $V$ and $W$ is just $V \otimes W$.  The set of all representations of $G$ with this structure is the infamous &quot;representation ring&quot; of $G$.  This product structure is apparently of paramount importance in quantum mechanics (I don't know why).  As another example where the tensor product turns a group into a ring, you might check out the Wikipedia page on &quot;topological K-theory&quot;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;There are many more examples.  If you know about functional analysis, the Schwartz kernel theorem is a tool used to investigate existence questions and regularity properties of partial differential equations, and it can be formulated purely in terms of Grothendeick's theory of topological tensor products.  I can't give you any deep reason why the same algebraic gadget has such a diverse array of applications, but I guess that's the way it is.  You'll undoubtedly learn more as you keep studying math.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ADDED:&#xA;I just noticed the other part of your question, in which you ask about the &quot;lifting&quot; property of the tensor product.  If I were forced to give a one sentence explanation of what the tensor product really is, it would be the following sentence.  Given two $R$-modules $A$ and $B$, we want to convert $R$-bilinear maps on $A \times B$ into linear maps on some other object.  We want to do this because for many purposes it reduces the structure theory of bilinear maps to the (extensive!) structure theory for linear maps.  The lifting property that you describe tells us that the tensor product does the job.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But it more than just &quot;does the job&quot; - it does the job in the absolute best way possible.  When you learn about most mathematical objects, such as the direct sum of two vector spaces, it is typical to define the object as some set equipped with some structure and then prove that it has certain nice properties.  With the tensor product, you should go about it backwards: you should think of the tensor product as an object with certain nice properties and then prove that there actually is an object with all of those properties.  This is because the actual construction of the tensor product of two modules is completely unenlightening and completely irrelevant to how you actually use the idea in practice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll be a little less vague and outline how the tensor product should be developed from scratch.  Given two $R$-modules $A$ and $B$, define a tensor product of $A$ and $B$ to be a pair $T, t$ where $T$ is a $R$-module and $t: A \times B \to T$ is a bilinear map with the property that given any bilinear map $Q: A \times B \to C$ there exists a unique linear map $L: T \to C$ such that $B = L \circ t$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lemma 1: If the tensor product exists, it is unique up to unique isomorphism.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lemma 2: The tensor product exists.&lt;/p&gt;&#xA;" OwnerUserId="1509" LastEditorUserId="1509" LastEditorDisplayName="" LastEditDate="2011-01-26T13:16:44.273" LastActivityDate="2011-01-26T13:16:44.273" CommentCount="4" />
  <row Id="18902" PostTypeId="1" AcceptedAnswerId="18912" CreationDate="2011-01-25T13:16:55.857" Score="4" ViewCount="356" Body="&lt;p&gt;Nakayama's lemma states that given a finitely generated $A$-module $M$, and $J(A)$ the Jacobson radical of $A$, with $I\subseteq J(A)$ some ideal, then if $IM=M$, we have $M=0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've read the proof, and while being relatively simple, it doesn't give much insight on why this lemma should be true, for example - is there some way to see how the fact that $J(A)$ is the intersection of all maximal ideals related to the result?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any intuition on the conditions and the result would be of great help. &lt;/p&gt;&#xA;" OwnerUserId="766" LastActivityDate="2011-04-21T20:51:08.090" Title="Intuitive explanation of Nakayama's Lemma" Tags="&lt;abstract-algebra&gt;&lt;intuition&gt;&lt;commutative-algebra&gt;&lt;modules&gt;" AnswerCount="4" CommentCount="2" FavoriteCount="3" />
  <row Id="18903" PostTypeId="2" ParentId="18881" CreationDate="2011-01-25T13:25:01.837" Score="0" ViewCount="" Body="&lt;p&gt;All it's saying is that we have a special operator that takes the two spaces A and B and gives us a new space C' ($A \bigotimes B$). And that every homomorphism from A x B to C gives us another homomorphism from C' to C. (and it is unique) and it's very nice because calculation in C' correspond to calculations in C.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the common generalization of mappings. What it does is let us construct a sort of special space, C' that directly relates in a unique way to C. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Tensor_product_of_modules&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Tensor_product_of_modules&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See the first diagram. Note how similar it is to many other diagrams found in group theory such as from quotient spaces. What is different is that we are working on Cartesian products. In some sense you can think of the tensor product as reducing the Cartesian product of a space to a new space that has a direct relation to the old. This is useful in that the new space may be isomorphic to a simpler space and/or features of the new space can be used to explore features of the old space. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;See also: &lt;a href=&quot;http://www.dpmms.cam.ac.uk/~wtg10/tensors3.html&quot; rel=&quot;nofollow&quot;&gt;http://www.dpmms.cam.ac.uk/~wtg10/tensors3.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="6174" LastActivityDate="2011-01-25T13:25:01.837" CommentCount="1" />
  <row Id="18904" PostTypeId="2" ParentId="18843" CreationDate="2011-01-25T13:38:07.190" Score="3" ViewCount="" Body="&lt;p&gt;When I was a kid we used to watch &quot;Donald in Mathmagic land&quot; starring Donald Duck.  While in undergrad we all gathered to watch and it was still enjoyable.&lt;/p&gt;&#xA;" OwnerUserId="2838" LastActivityDate="2011-01-25T13:38:07.190" CommunityOwnedDate="2011-01-25T13:38:07.190" CommentCount="1" />
  <row Id="18905" PostTypeId="2" ParentId="6179" CreationDate="2011-01-25T13:49:36.993" Score="0" ViewCount="" Body="&lt;p&gt;Hi I work at a University in the UK as a learning technologist. You may have come across the site already (it's now quite famous) but I've been very impressed by the explanations of a wide range of mathematical concepts at:&#xA;The explanations are in the form of 10 minute videos with screen recording and narration&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.khanacademy.org/&quot; rel=&quot;nofollow&quot;&gt;http://www.khanacademy.org/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A long time ago I studied electronic engineering at undergraduate level - if I'd had access to this site at the time it would have helped me enormously!&lt;/p&gt;&#xA;" OwnerUserId="6231" LastActivityDate="2011-01-25T13:49:36.993" CommunityOwnedDate="2011-01-25T13:49:36.993" />
  <row Id="18906" PostTypeId="1" CreationDate="2011-01-25T13:54:17.620" Score="2" ViewCount="150" Body="&lt;p&gt;I have a quadrilateral whose four sides are given &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2341276, 34374833, 18278172, 17267343 units.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I find out its area? What would be the area?&lt;/p&gt;&#xA;" OwnerUserId="6232" LastActivityDate="2011-01-25T22:22:14.083" Title="Finding the area of a quadrilateral" Tags="&lt;geometry&gt;&lt;euclidean-geometry&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="18907" PostTypeId="2" ParentId="18906" CreationDate="2011-01-25T14:02:07.250" Score="2" ViewCount="" Body="&lt;p&gt;Have you seen &lt;a href=&quot;http://mathworld.wolfram.com/Quadrilateral.html&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt;(mathworld article)? In particular, the area of a planar quadrilateral is given by $\frac{1}{4}(b^2+d^2-a^2-c^2) \tan \theta$ where $a,b,c$ and $d$ are the side lengths.&lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2011-01-25T14:02:07.250" CommentCount="1" />
  <row Id="18908" PostTypeId="2" ParentId="18865" CreationDate="2011-01-25T14:26:46.150" Score="2" ViewCount="" Body="&lt;p&gt;The interpretation is mangled. The volume is finite but the surface area infinite, much like &lt;a href=&quot;http://en.wikipedia.org/wiki/Gabriel%27s_horn&quot; rel=&quot;nofollow&quot;&gt;Gabriel's Horn&lt;/a&gt;. The idea of the quote is much as in Gabriel's Horn: since the volume is finite, you can imagine &quot;filling it up&quot; with a finite amount of paint. But the surface area is infinite, which suggests that you are &quot;painting&quot; an infinite surface with a finite amount of paint, a paradox (of course, quantum mechanics gets in your way, even theoretically). So you get an unbounded area (which de Sluze and Huygens called &quot;infinite&quot;) that can be &quot;covered&quot; with a finite quantity (the amount of &quot;paint&quot;).&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2011-01-25T14:26:46.150" />
  <row Id="18909" PostTypeId="2" ParentId="18843" CreationDate="2011-01-25T14:27:40.887" Score="2" ViewCount="" Body="&lt;p&gt;I enjoyed &lt;a href=&quot;http://www.amazon.com/Independent-Lens-Dr-Erik-Demaine/dp/B002NWRMO0&quot; rel=&quot;nofollow&quot;&gt;Between the Folds&lt;/a&gt;, which gives a rather high-level view of the mathematics behind origami and how they provide means for constructing incredibly complex shapes.&lt;/p&gt;&#xA;" OwnerUserId="689" LastActivityDate="2011-01-25T14:27:40.887" CommunityOwnedDate="2011-01-25T14:27:40.887" />
  <row Id="18910" PostTypeId="2" ParentId="18902" CreationDate="2011-01-25T14:44:03.717" Score="1" ViewCount="" Body="&lt;p&gt;This is probably not the answer you're looking for, but here's another proof:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose $M \neq 0$ and let $u_1,\ldots,u_n$ be a minimal set of generators for $M$. Then $u_n \in IM$ so we may write $u_n=a_1u_1+\ldots+a_nu_n$ with $a_i \in I$. Hence $(1-a_n)u_n=a_1u_1+\ldots+a_{n-1}u_{n-1}$. Since $I$ was contained in the Jacobson radical, it follows that $1-a_n$ is a unit. Hence $u_n$ belongs to the submodule generated by the $u_1,\ldots,u_{n-1}$, contradiction.&lt;/p&gt;&#xA;" OwnerUserId="4284" LastEditorUserId="4284" LastEditorDisplayName="" LastEditDate="2011-01-25T14:53:10.020" LastActivityDate="2011-01-25T14:53:10.020" />
  <row Id="18911" PostTypeId="2" ParentId="18902" CreationDate="2011-01-25T14:46:08.423" Score="3" ViewCount="" Body="&lt;p&gt;Here's something that might or might not make sense to you.  You know that every ideal $I$ of a commutative ring $R$ gives rise to an $R$-module $R/I$; these are precisely the $R$-modules on one generator.  The $R$-modules of the form $R/m$ where $m$ is maximal are special among these.  In this language, the elements of the Jacobson radical are precisely the ones that act trivially on all $R$-modules of the form $R/m$.  It follows, for example, that if $I$ is in the Jacobson radical, then we cannot have $IM = M$ for any module of the form $R/m^k$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nakayama's lemma asserts that a similar statement is true for &lt;em&gt;all&lt;/em&gt; finitely generated $R$-modules.  This should be reasonable if you are aware of, for example, the classification of finitely generated $R$-modules when $R$ is a PID.  The Wikipedia describes a geometric interpretation of this, but I'm not familiar enough with it to say more.  &lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2011-01-25T14:46:08.423" />
  <row Id="18912" PostTypeId="2" ParentId="18902" CreationDate="2011-01-25T14:48:57.817" Score="2" ViewCount="" Body="&lt;p&gt;Suppose your module is of finite length. Then you can consider on it the so called &lt;em&gt;radical filtration&lt;/em&gt;, which organises the module into an onion-like thing, with elements of the maximal ideal pushing elements of the module farther in from their starting layer to one right below and, moreover, each layer obtained from the one above it in this way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the condition $\mathfrak m M=M$ tells you that the outermost layer of the module is actually empty: obviously, then, there is not much in the whole thing and $M=0$. We have just discovered Nakayama's lemma!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If your module is arbitrary, exactly the same happens.&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2011-01-25T14:48:57.817" CommentCount="4" />
  <row Id="18913" PostTypeId="2" ParentId="9378" CreationDate="2011-01-25T14:50:00.327" Score="0" ViewCount="" Body="&lt;p&gt;The paper &quot;What is a Sheaf?&quot; is not bad at all, it even has a picture or two. It predates the arxiv though, but not Jstor. It is by Seebach, Seebach and Steen, I believe. If you can't find it let me know.&lt;/p&gt;&#xA;" OwnerUserId="627" LastActivityDate="2011-01-25T14:50:00.327" CommentCount="1" />
  <row Id="18914" PostTypeId="1" AcceptedAnswerId="18933" CreationDate="2011-01-25T14:54:07.560" Score="1" ViewCount="339" Body="&lt;p&gt;I am trying to find an accurate way of calculating the capacity of an underground tank at a given depth.  The tank manufacturer has provided a strapping table for the tank which tells me the capacity at various depths.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Gauge Depth (cm)  /  Capacity (Liters)&#xA;2cm = 29.8 liters&#xA;12cm = 240.4 liters&#xA;...&#xA;66cm = 2372.8 liters&#xA;118cm = 4008.4 liters&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;However, I would like to find an equation for calculating the volume of liquid in the tank based on the liquid depth level.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They have provided the following dimensions for the tank:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;1219mm Diameter&#xA;3785mm Long&#xA;3978 Liters Capacity&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The tank is horizontal with hemispherical ends.  &lt;strong&gt;It is safe to assume that the radius of the hemispherical ends is 1/2 of the diameter of the tank.&lt;/strong&gt;&#xA;Any help or direction would be greatly appreciated!&lt;/p&gt;&#xA;" OwnerUserId="6233" LastEditorUserId="1543" LastEditorDisplayName="" LastEditDate="2011-01-25T15:09:43.080" LastActivityDate="2011-01-25T16:46:38.523" Title="Horizontal tank with hemispherical ends depth to capacity calculation" Tags="&lt;calculus&gt;&lt;analytic-geometry&gt;" AnswerCount="4" CommentCount="2" FavoriteCount="1" />
  <row Id="18915" PostTypeId="2" ParentId="18881" CreationDate="2011-01-25T15:03:51.040" Score="1" ViewCount="" Body="&lt;p&gt;I won't try to convince you of anything with the following, I just think this reason should be recorded. It is the right &quot;product&quot; in the category of Rings. It gives you the right symmetric monoidal structure. Essentially, it is important and useful because it actually is what we should be thinking about. The reasons essentially involve the universal property it has that is mentioned above: a linear map out of a tensor product is a multilinear map out of the cartesian product, and when doing algebra this is what we should be concerning ourselves with.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please note the comment of elgeorges below, the tensor product is the categorical coproduct in the category of Commutative Rings, hence my use of quotation marks.&lt;/p&gt;&#xA;" OwnerUserId="627" LastEditorUserId="627" LastEditorDisplayName="" LastEditDate="2011-01-25T22:56:48.063" LastActivityDate="2011-01-25T22:56:48.063" CommentCount="4" />
  <row Id="18916" PostTypeId="1" AcceptedAnswerId="18918" CreationDate="2011-01-25T15:38:56.730" Score="3" ViewCount="108" Body="&lt;p&gt;Why is the following tuple of vectors not a basis of $\mathbb{R}^{2}$? $\left ( \left ( \begin{array}{c}  1\\ 0\\0 \end{array} \right ), \left ( \begin{array}{c}  0\\ 1\\0 \end{array} \right ) \right )$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would've thought that just like $\left ( \left ( \begin{array}{c}  1\\ 0\end{array} \right ), \left ( \begin{array}{c}  0\\ 1 \end{array} \right ) \right )$ the 1st tuple would span $\mathbb{R}^{2}$ and the vectors clearly linearly independent...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hopefully I'm not fundamentally misunderstanding something, I'm doing one of those &quot;self-test at the end of the chapter only solutions, no explanations&quot; things. It'd be great if anyone could offer a brief explanation so that I can move on... thanks!&lt;/p&gt;&#xA;" OwnerUserId="3711" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-25T16:12:19.960" LastActivityDate="2011-01-25T16:12:19.960" Title="Why is this tuple of vectors not a basis for $\mathbb{R}^{2}$?" Tags="&lt;linear-algebra&gt;" AnswerCount="1" CommentCount="8" />
  <row Id="18917" PostTypeId="2" ParentId="18844" CreationDate="2011-01-25T15:39:26.263" Score="4" ViewCount="" Body="&lt;p&gt;I am not sure of an answer, but note that your new model is a non-linear ODE. In general (unless there exists a nice algebraic transform that maps it to a linear ODE) the stability analysis for non-linear ODEs depends not only on the coefficients of the equation, but also the initial values prescribed. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an example, for initial data with $|x|$ very small, your equation is a small perturbation of the Mathieu equation, and one might expect behaviour similar to the Mathieu equation. But for initial data with $|x|$ very large, your equation behaves more like $x&amp;#39;&amp;#39;(t) = -cx^2$, then in the case where $x &amp;lt; 0$ and $x&amp;#39; &amp;lt; 0$ initially with large $|x|$, the solution behaves like $(t_0 - t)^{-2}$ and blows up in finite time. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To put it in other words, I am not sure if the notion of stability analysis is even useful. For a linear ODE, the characteristic behaviours are, roughly speaking, limited to exponential decay, exponential blowup, and oscillation (corresponding roughly to the $t\to\infty$ behaviour of $x&amp;#39;&amp;#39; = \pm x$ with increasing or decreasing initial data). For a non-linear ODE, vastly different behaviour can manifest.&lt;/p&gt;&#xA;" OwnerUserId="1543" LastActivityDate="2011-01-25T15:39:26.263" CommentCount="1" />
  <row Id="18918" PostTypeId="2" ParentId="18916" CreationDate="2011-01-25T15:42:38.433" Score="7" ViewCount="" Body="&lt;p&gt;Strictly speaking, it is the basis of a two-dimensional subspace of $\mathbb{R}^3$, which happens to be isomorphic to $\mathbb{R}^2$.&lt;/p&gt;&#xA;" OwnerUserId="1508" LastActivityDate="2011-01-25T15:42:38.433" />
  <row Id="18919" PostTypeId="2" ParentId="18894" CreationDate="2011-01-25T16:14:09.237" Score="2" ViewCount="" Body="&lt;p&gt;Suppose that $N = \{N_t: t \geq 0 \}$ is a Poisson process with rate $\lambda$. Let $X$ be the waiting time until the first occurrence in the process $N$. Then, for any $t &amp;gt; 0$, &#xA;$$&#xD;&#xA;{\rm P}(X \le t) = 1 - {\rm P}(X &amp;gt; t) = 1 - {\rm P}(N_t  = 0) = 1 - e^{ - \lambda t}. &#xD;&#xA;$$&#xA;Hence, $X$ is an exponential random variable with density function $f_X$ given by&#xA;$$&#xD;&#xA;f_X {(t)} = \frac{{\rm d}}{{{\rm d}t}}(1 - e^{ - \lambda t} ) = \lambda e^{-\lambda t},\;\; t &amp;gt; 0.&#xD;&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="2810" LastActivityDate="2011-01-25T16:14:09.237" />
  <row Id="18920" PostTypeId="1" AcceptedAnswerId="18923" CreationDate="2011-01-25T16:15:53.573" Score="2" ViewCount="65" Body="&lt;p&gt;Hi!! I need a little bit of help, I found that theorem, but the book doesn't prove it and gives a reference to another book that I don't have; does anyone have an idea?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $R$ be a semi-local ring, and $M$ a finite projective $R$-module. Show that $M$ is free if the localizations $M_m$ have the same rank for all maximal ideals $m$ of $R$.&lt;/p&gt;&#xA;" OwnerUserId="6236" LastActivityDate="2011-01-25T16:26:26.013" Title="projective modules over a semi-local ring" Tags="&lt;commutative-algebra&gt;" AnswerCount="1" />
  <row Id="18921" PostTypeId="1" AcceptedAnswerId="18928" CreationDate="2011-01-25T16:16:44.040" Score="2" ViewCount="100" Body="&lt;p&gt;Suppose $f(x_1,x_2,\dots,x_n)$ is a multivariable function $f:\mathbb{R}^n\rightarrow \mathbb{R}$.  Suppose that for all partial derivatives, $1\le i \le n$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\partial f}{\partial x_i}(q_1,q_2,\ldots,q_n) \ge 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for all $q_i \ge r_i$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is f a non-decreasing function in the set of all points $q_i \ge r_i$ for all $1\le i \le n$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: By &quot;non-decreasing&quot;, I mean that $f(q_1,q_2,\ldots,q_n) \ge f(r_1,r_2,\ldots,r_n) $ if  $q_i \ge r_i$ for all i.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, is the converse true? Does a function that is 'non-decreasing', by my definition, also have all first partial derivatives positive? &lt;/p&gt;&#xA;" OwnerUserId="1004" LastEditorUserId="1004" LastEditorDisplayName="" LastEditDate="2011-01-25T18:02:38.773" LastActivityDate="2011-01-25T18:02:38.773" Title="Is a function with non-negative first partial derivatives increasing?" Tags="&lt;real-analysis&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" />
  <row Id="18922" PostTypeId="2" ParentId="18863" CreationDate="2011-01-25T16:23:01.893" Score="1" ViewCount="" Body="&lt;p&gt;In the following &lt;a href=&quot;http://stuff.mit.edu/people/raj/Acta05rmt.pdf&quot; rel=&quot;nofollow&quot;&gt;article&lt;/a&gt;, an experiment is devised to show that the distribution of the largest eigenvalue is given by the Tracy-Widom distribution. See section 10. Perhaps you could modify it.&lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2011-01-25T16:23:01.893" CommentCount="1" />
  <row Id="18923" PostTypeId="2" ParentId="18920" CreationDate="2011-01-25T16:26:26.013" Score="1" ViewCount="" Body="&lt;p&gt;See &lt;strong&gt;Lemma 1.4.4&lt;/strong&gt; in the given link:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://books.google.co.in/books?id=ouCysVw20GAC&amp;amp;lpg=PA20&amp;amp;ots=K5Q1H1M0lj&amp;amp;dq=Let%20R%20be%20a%20semi-local%20ring%2C%20and%20M%20a%20finite%20projective%20R-module.%20Show%20that%20M%20is%20free%20if%20the%20localizations%20Mm%20have%20the%20same%20rank%20for%20all%20maximal%20ideals%20m%20of%20R.&amp;amp;pg=PA20#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;http://books.google.co.in/books?id=ouCysVw20GAC&amp;amp;lpg=PA20&amp;amp;ots=K5Q1H1M0lj&amp;amp;dq=Let%20R%20be%20a%20semi-local%20ring%2C%20and%20M%20a%20finite%20projective%20R-module.%20Show%20that%20M%20is%20free%20if%20the%20localizations%20Mm%20have%20the%20same%20rank%20for%20all%20maximal%20ideals%20m%20of%20R.&amp;amp;pg=PA20#v=onepage&amp;amp;q&amp;amp;f=false&lt;/a&gt;&lt;/p&gt;&#xA;" LastActivityDate="2011-01-25T16:26:26.013" CommentCount="1" />
  <row Id="18924" PostTypeId="2" ParentId="18914" CreationDate="2011-01-25T16:27:08.670" Score="1" ViewCount="" Body="&lt;p&gt;You can divide the tank into a sphere (formed from the two  hemispherical ends) and a cylinder.  &lt;a href=&quot;http://en.wikipedia.org/wiki/Spherical_cap&quot; rel=&quot;nofollow&quot;&gt;Wikipedia Spherical cap&lt;/a&gt; gives the volume as $\frac{\pi h^2}{3}(3r-h)$ where $r$ is the radius of the sphere and $h$ the depth.  Then &lt;a href=&quot;http://en.wikipedia.org/wiki/Circular_segment&quot; rel=&quot;nofollow&quot;&gt;Circular Segment&lt;/a&gt; gives the area of the segment as $\frac{r^2}{2}(\theta - \sin \theta )$ where $\theta=\arccos(\frac{r-h}{r})$.  Multiply this last by the length, add the spherical cap, and there you are.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-01-25T16:27:08.670" />
  <row Id="18925" PostTypeId="2" ParentId="18914" CreationDate="2011-01-25T16:27:31.910" Score="0" ViewCount="" Body="&lt;p&gt;Let $R$ be the radius of the tank and let the end-to-end length be $L + 2R$.  At each depth $x$ above the center of the tank (where $-R \le x \le R$), the tank's cross-section is a rectangle of length $L$ and width $2r(x)$, plus semicircular caps of radius $r(x)$.  The radius satisfies $r(x)^2 + x^2 = R^2$.  So the volume at depth $d$ is&#xA;$$&#xD;&#xA;\begin{eqnarray}&#xD;&#xA;V(d) &amp;amp;=&amp;amp; \int_{-R}^{-R+d} V&amp;#39;(x) dx \\&#xD;&#xA;&amp;amp;=&amp;amp; \int_{-R}^{-R+d} \left(2 L r(x) + \pi r(x)^2\right) dx \\&#xD;&#xA;&amp;amp;=&amp;amp; \int_{-R}^{-R+d} \left(2 L \sqrt{R^2 - x^2} + \pi R^2 - \pi x^2\right) dx \\&#xD;&#xA;&amp;amp;=&amp;amp; 2 L R^2 \tan^{-1}\left(\sqrt{\frac{d}{2R-d}}\right) + L(d-R)\sqrt{d(2R-d)} + \pi\left(Rd^2 - \frac{1}{3}d^3\right).&#xD;&#xA;\end{eqnarray}&#xD;&#xA;$$&#xA;This fits your numbers reasonably well for $R=609.5\text{mm}$ and $L=2566.0\text{mm}$; it could be improved by knowing the thickness of the tank walls and whether the shape deviates from a perfect cylinder plus hemispherical caps (e.g., is it flattened at the bottom)?&lt;/p&gt;&#xA;" OwnerUserId="5546" LastActivityDate="2011-01-25T16:27:31.910" CommentCount="1" />
  <row Id="18926" PostTypeId="2" ParentId="18914" CreationDate="2011-01-25T16:32:28.433" Score="0" ViewCount="" Body="&lt;p&gt;One thing is clear: you can't get 4008.4 litres into a tank with a capacity of 3978 litres. So you are right not to trust the manufacturer's dipstick.&lt;/p&gt;&#xA;" OwnerUserId="1508" LastActivityDate="2011-01-25T16:32:28.433" CommentCount="1" />
  <row Id="18927" PostTypeId="1" AcceptedAnswerId="18930" CreationDate="2011-01-25T16:34:08.090" Score="1" ViewCount="118" Body="&lt;p&gt;If I have a circle inscribed inside an equilateral triangle, and I know the radius of the circle, what is the formula to determine what the length of the side of the triangle is?&lt;/p&gt;&#xA;" OwnerUserId="6237" LastActivityDate="2011-01-25T16:49:21.720" Title="How to find length of side of equilateral triangle when have radius of circle inscribed inside?" Tags="&lt;geometry&gt;" AnswerCount="2" />
  <row Id="18928" PostTypeId="2" ParentId="18921" CreationDate="2011-01-25T16:35:46.653" Score="4" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Updated&lt;/strong&gt; to answer the updated question  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If all partial derivatives are non-negative, then yes: $f(q_1,q_2,\ldots,q_n) \ge f(r_1,r_2,\ldots,r_n) $ if  $q_i \ge r_i$ for all i. And the converse is true as well. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the first question, take a walk from $r$ to $q$, one coordinate at a time: $(r_1,r_2,\ldots,r_n)$ to $(q_1,r_2,\ldots,r_n)$ to $\ldots$ to $(q_1,q_2,\ldots,q_n)$. None of these steps can decrease the value of $f$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the second (converse) question, just ignore all coordinates except one, and it's easy to see that the partial derivative with respect to that coordinate must be non-negative.&lt;/p&gt;&#xA;" OwnerUserId="1508" LastEditorUserId="1508" LastEditorDisplayName="" LastEditDate="2011-01-25T16:47:28.680" LastActivityDate="2011-01-25T16:47:28.680" />
  <row Id="18929" PostTypeId="2" ParentId="18927" CreationDate="2011-01-25T16:38:42.297" Score="1" ViewCount="" Body="&lt;p&gt;It would be $\tan 30^{\circ} = \frac{r}{x}$ so that $x = \frac{r}{\tan 30^{\circ}}$ where $x$ is half of a side. &lt;/p&gt;&#xA;" OwnerUserId="2548" LastEditorUserId="2548" LastEditorDisplayName="" LastEditDate="2011-01-25T16:49:21.720" LastActivityDate="2011-01-25T16:49:21.720" CommentCount="0" />
  <row Id="18930" PostTypeId="2" ParentId="18927" CreationDate="2011-01-25T16:40:50.777" Score="1" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Equilateral_triangle&quot; rel=&quot;nofollow&quot;&gt;The Wikipedia page one equilateral triangles&lt;/a&gt; gives the radius of the inscribed circle as $r=a\frac{\sqrt{3}}{6}$ where $a$ is the side.  So $a=r\frac{6}{\sqrt{3}}$.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-01-25T16:40:50.777" CommentCount="1" />
  <row Id="18931" PostTypeId="1" AcceptedAnswerId="18934" CreationDate="2011-01-25T16:41:12.430" Score="2" ViewCount="331" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;(Application of the rank-nullity theorem) Suppose $S,T: V\to V$ are linear transformations of a finite dimensional vector space $V$, and that the composition $ST\colon V\to V$ is invertible. Show that then $T$ is one-to-one, and deduce that it is invertible. Show also that $S$ is invertible. Deduce that $ST$ is invertible if and only if $TS$ is invertible.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm not remotely sure how to apply the rank-nullity theorem here. Also, I thought it was self-evident that T and S would have to be invertible since as far as I know the inverse of a composition involves the composition of inverses (and if a matrix is invertible it's definitely one-to-one).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any help would be greatly appreciated!&lt;/p&gt;&#xA;" OwnerUserId="6239" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-25T16:51:42.297" LastActivityDate="2011-01-25T19:34:44.413" Title="Show that the linear transformation T is invertible." Tags="&lt;algebra&gt;&lt;transformation&gt;&lt;linear-algebra&gt;" AnswerCount="2" />
  <row Id="18932" PostTypeId="1" AcceptedAnswerId="18959" CreationDate="2011-01-25T16:44:13.257" Score="5" ViewCount="88" Body="&lt;p&gt;In contrast to the possibility of taking  an arbitrary sequence of elements of submodules in the definition of direct product, the definition for the direct sum of submodules of a module requires the indexed elements to vanish cofinitely(i.e. except finitely many times).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More precisely, Let $R$ be a ring, and $\{M_i : i ∈ I\}$ a family of left $R-$modules indexed by the set $I$. The direct sum of ${M_i}$ is then defined to be the set of all sequences $(α_i)$ where $\alpha_i \in M_i$ and $α_i = 0$ for cofinitely many indices $i$. (The direct product is analogous but the indices do not need to cofinitely vanish.)(Source Wikipedia: &lt;a href=&quot;http://en.wikipedia.org/wiki/Direct_sum_of_modules#Construction_for_an_arbitrary_family_of_modules&quot; rel=&quot;nofollow&quot;&gt;Direct sum of modules&lt;/a&gt;.)We have similar definition for the sum of submodules.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have not yet understood what pathology would be incurred without assuming or what simplification (if any) would be gained in assuming cofiniteness. Can you please explain this in simple terms (possibly, with examples) ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="3992" LastActivityDate="2011-01-25T20:52:07.503" Title="Why is cofiniteness included in the definition of direct sum of submodules?" Tags="&lt;algebra&gt;&lt;modules&gt;" AnswerCount="4" CommentCount="4" FavoriteCount="2" />
  <row Id="18933" PostTypeId="2" ParentId="18914" CreationDate="2011-01-25T16:46:38.523" Score="1" ViewCount="" Body="&lt;p&gt;There are three parts to the tank: the two hemispherical ends and the middle cylinder. The two hemispherical ends add up to one whole sphere, so we'll just consider a sphere and a cylinder. Just to check the data you've given: at 1219mm diameter, the sphere has volume roughly 948.44 liters. The cylinder has diameter 1219mm and height (3785-1219)=2566mm. The volume is roughly then 2994.7 liters. So I'm guessing that there's either a typo on your data sheet (either the total tank capacity is closer 3948 liters, or the total length of the tank is closer to 3815mm), or a mis-description: the ends are not capped by exact hemispheres. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Never mind that, since we are on a mathematics site and who cares about plugging in real numbers /end_sarcasm &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume your tank is composed of two hemispherical ends of radius $R$, and a central cylinder of radius $R$ and length $L$. The diameter would be $2R$ and total length $L + 2R$. We want to calculate volume as a function of water depth. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since your tank is lying horizontally, the height of water, which we will call $h$, should be the same inside the cyclinder portion and inside the spherical portions. So we will compute the total volume as a sum of the volume inside the sphere (remember: two hemispheres make one whole sphere) and the volume inside the cylinder.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ V_{tot} = V_s + V_c $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you know calculus, you can compute the volume of the spherical cap of height $h$ by evaluating an integral. But the formula is &lt;a href=&quot;http://en.wikipedia.org/wiki/Spherical_cap&quot; rel=&quot;nofollow&quot;&gt;well-known&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ V_s = \frac{\pi h^2}{3}(3R - h) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the cylinder, the volume at height $h$ is equal to the length $L$ multiplied by the area of the &lt;a href=&quot;http://mathworld.wolfram.com/CircularSegment.html&quot; rel=&quot;nofollow&quot;&gt;circular segment&lt;/a&gt; of height $h$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ V_c = L \times A = L \times \left[ R^2 \cos^{-1}\left(\frac{R-h}{R}\right) - (R - h) \sqrt{2Rh - h^2}\right] $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with the formula for $A$ derivable by elementary geometry/trigonometry, or again through the evaluation of an integral. So putting it all together you have that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ V_{tot} = \frac{\pi h^2}{3}(3R - h) + L \times \left[ R^2 \cos^{-1}\left(\frac{R-h}{R}\right) - (R - h) \sqrt{2Rh - h^2}\right] $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which gives &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;2 cm  =&amp;gt; 11.4 L&#xA;5 cm  =&amp;gt; 46.4 L&#xA;10 cm =&amp;gt; 134.6 L &#xA;20 cm =&amp;gt; 388.9 L &#xA;30 cm =&amp;gt; 716.7 L &#xA;40 cm =&amp;gt; 1094.5 L &#xA;50 cm =&amp;gt; 1504.5 L &#xA;60 cm =&amp;gt; 1930.8 L &#xA;61 cm (just a tiny bit over half full) =&amp;gt; 1971.6 L&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="1543" LastActivityDate="2011-01-25T16:46:38.523" CommentCount="3" />
  <row Id="18934" PostTypeId="2" ParentId="18931" CreationDate="2011-01-25T16:51:00.957" Score="1" ViewCount="" Body="&lt;p&gt;In general, it's possible for a composition to be invertible and neither function to be invertible: take $f\colon\{a\}\to\{1,2\}$ given by $f(a)=1$, and $g\colon\{1,2\}\to\{x\}$ given by $g(1)=g(2)=x$. Neither $f$ nor $g$ are invertible ($f$ is not onto, $g$ is not one-to-one) but $gf\colon\{a\}\to\{x\}$ is clearly invertible. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What &lt;em&gt;is&lt;/em&gt; true is that &lt;strong&gt;if&lt;/strong&gt; $f$ and $g$ are invertible and you can compose them, then $gf$ is invertible and $(gf)^{-1} = f^{-1}g^{-1}$. But you cannot conclude that $f$ and $g$ are each invertible from the assumption that $gf$ is invertible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a linear example, take $T\colon\mathbb{R}^2\to\mathbb{R}^3$ given by $T(x,y) = (x,y,x-y)$ (not onto), and $S\colon\mathbb{R}^3\to\mathbb{R}^2$ given by $S(x,y,z) = (x,y)$ (not one-to-one). Neither $T$ nor $S$ are invertible, but $ST\colon\mathbb{R}^2\to\mathbb{R}^2$ is invertible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a linear example with maps from the a vector space to itself, necessarily infinite dimensional in view of the theorem above, take the vector space of all sequences of real numbers, and let $T$ be the &quot;right shift&quot; operator and $S$ the &quot;left shift&quot; operator. that is, $T(a_1,a_2,a_3,\ldots) = (0,a_1,a_2,\ldots)$, and $S(a_1,a_2,a_3,\ldots) = (a_2,a_3,\ldots)$. Check that neither $T$ nor $S$ are invertible, but that $ST$ is the identity map, hence invertible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So: first assume that $ST$ is invertible. Use this to show that $T$ must be one-to-one. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Added hint:&lt;/em&gt; To show that $T$ is one-to-one, you want to show that if $\mathbf{v}\in V$ is such that $T(\mathbf{v})=\mathbf{0}$, then $\mathbf{v}=\mathbf{0}$ (that is, the only vector that goes to $\mathbf{0}$ is the zero vector; we're showing the nullspace of $T$ is trivial, or that the nullity is $0$). So, let $\mathbf{v}$ be a vector in $V$ such that $T(\mathbf{v})=\mathbf{0}$. What is $ST(\mathbf{v})$? What does that tell you about $\mathbf{v}$? What does that tell you about $T$ (and why)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you know that $T$ is one-to-one, &lt;em&gt;then&lt;/em&gt; you can use the Rank-Nullity Theorem to conclude that since $T$ is one-to-one and maps from $V$ to itself ($V$ finite dimensional), then $T$ is invertible. And &lt;em&gt;then&lt;/em&gt; you can use that both $ST$ and $T$ are invertible (and hence both $ST$ and $T^{-1}$ are invertible) to show that $S$ is invertible. And if both $S$ and $T$ are invertible, then show that $TS$ is invertible. This proves that if $ST$ is invertible, then $TS$ is invertible; the converse follows by simply swapping the roles of $S$ and $T$.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-25T19:28:18.560" LastActivityDate="2011-01-25T19:28:18.560" CommentCount="5" />
  <row Id="18935" PostTypeId="2" ParentId="18932" CreationDate="2011-01-25T16:54:59.173" Score="2" ViewCount="" Body="&lt;p&gt;I like to think of this as being the difference between polynomials (direct sum) and power series (direct product).  Sometimes one object is more useful than or shows up in context instead of the other.  For instance the cohomology ring of a space is the direct sum&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;H^*(X)=\bigoplus_{n\geq 0}H^n(X).&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the cohomology ring of $\mathbb{C}P^\infty$ is $\mathbb{Z}[x]$, a polynomial ring in one variable.  We choose the direct sum due to some topology issues with the cup product.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other times we may want the direct product.&lt;/p&gt;&#xA;" OwnerUserId="2838" LastActivityDate="2011-01-25T16:54:59.173" CommentCount="1" />
  <row Id="18936" PostTypeId="1" AcceptedAnswerId="18937" CreationDate="2011-01-25T16:56:39.203" Score="1" ViewCount="80" Body="&lt;p&gt;I was reading this book about medicine and I found this paragraph in it &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;A person's daily risk of getting lung illness ... is 0.5 percent. The daily risk of getting a bad lung illness ... plus treatment is 0.05 percent. So, you are looking at the difference between a 99.5 percent chance of staying well, and a 99.95 percent chance of staying well ... But sum it up over a year and it is the difference between an 83 percent chance of making it through the year healthy, and a 16 percent chance. &quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't understand how this conclusion was reached. Was some additional data assumed?&lt;/p&gt;&#xA;" OwnerUserId="6171" LastEditorUserId="6171" LastEditorDisplayName="" LastEditDate="2011-01-26T00:44:28.920" LastActivityDate="2011-01-26T00:44:28.920" Title="A question in Probability, risk of lung illness." Tags="&lt;probability&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="0" />
  <row Id="18937" PostTypeId="2" ParentId="18936" CreationDate="2011-01-25T17:05:43.767" Score="4" ViewCount="" Body="&lt;p&gt;No, they just assumed each day is independent.  Without treatment, with 0.5 percent chance per day, the chanceof getting through a year is .995^365=.160  With treatment, you have .9995^365=.833&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-01-25T17:05:43.767" />
  <row Id="18938" PostTypeId="1" AcceptedAnswerId="18956" CreationDate="2011-01-25T17:09:05.490" Score="3" ViewCount="92" Body="&lt;p&gt;If $N$ points on the circumference of a circle are chosen at random, what is the probability $F(\theta)$ that the maximum gap between neighboring points is at least $\theta$?  Because the gaps sum to $2\pi$, the maximum must be at least $2\pi/N$, so $$F(\theta)=1 \text{  for } \theta\le\frac{2\pi}{N}.$$  At the other extreme, the solution to &lt;a href=&quot;http://math.stackexchange.com/questions/18369/n-points-on-a-circle&quot;&gt;this problem&lt;/a&gt; shows that $$F(\theta) = N\left(1 - \frac{\theta}{2\pi}\right)^{N-1} \text{  for } \theta\ge\pi.$$ &#xA;Is there a closed-form solution for any other values of $\theta$?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update:&lt;/strong&gt; The general closed-form solution is given in the answer below.  In terms of the notation in the question, it is&#xA;$$&#xD;&#xA;F(\theta) = \sum_{k=1}^{K} (-1)^{k-1} {N\choose k} \left(1 - \frac{k \theta}{2\pi}\right)^{N-1},&#xD;&#xA;$$&#xA;where $K = \min(N, \lfloor{2\pi/\theta}\rfloor)$.  This reduces to the limiting cases given in the question when $\theta \le 2\pi/N$ (in which case $K=N$) and when $\theta \ge \pi$ (in which case $K=1$).&lt;/p&gt;&#xA;" OwnerUserId="5546" LastEditorUserId="5546" LastEditorDisplayName="" LastEditDate="2011-01-25T21:05:34.293" LastActivityDate="2011-01-25T21:05:34.293" Title="Maximum gap among N points on a circle" Tags="&lt;probability&gt;&lt;circle&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" />
  <row Id="18939" PostTypeId="1" CreationDate="2011-01-25T17:12:58.103" Score="1" ViewCount="111" Body="&lt;p&gt;I have an integral $\int_0 ^a\int_0 ^b\int_0 ^a\int_0 ^b \sin(x)\sin(\bar{x})\sin(y)\sin(\bar{y})f(x,\bar{x},y,\bar{y})dxdyd\bar{x}d\bar{y}$ &#xA;where $f= \dfrac{\sin(\sqrt{(x-\bar{x})^2+(y-\bar{y})^2})}{(x-\bar{x})^2+(y-\bar{y})^2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;under the transformation $\bar{x}=x+u$ and $\bar{y}=y+v$ $\Rightarrow$ $f$ becomes $f(u,v)$&#xA;but the limits of integration with respect to $u$ and $v$ change.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Considering only the x part, $\int_{-x} ^{a-x}\int_0 ^a \sin(x)\sin(x+u)f(u,v)dxdu$ &#xA;integrating this with respect to u analytically is difficult while with respect to x is easier. But the $u$ limit have $x$ terms can I by anyway, make the limits of $u$ as $0$ to $a$ but yet carry a simpler integration with respect to $x$, changing the limits of $x$ would cause no harm.&lt;/p&gt;&#xA;" OwnerUserId="3892" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-25T18:31:34.927" LastActivityDate="2011-01-25T18:31:34.927" Title="Integration - changing the limits" Tags="&lt;integral&gt;" CommentCount="2" />
  <row Id="18940" PostTypeId="2" ParentId="18932" CreationDate="2011-01-25T17:15:34.913" Score="3" ViewCount="" Body="&lt;p&gt;I'm going to try to motivate the difference by defining the direct sum and product in terms of their universal properties.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The direct sum of a family of $R$-modules $\{ M_\alpha : \alpha \in A \}$ is defined as a module $M$ together with a family of homomorphisms $\iota_\alpha : M_\alpha \to M$ such that for every family of homomorphisms $\phi_\alpha : M_\alpha \to N$, there is a &lt;em&gt;unique&lt;/em&gt; homomorphism $\phi : M \to N$ such that $\phi \circ \iota_\alpha = \phi_\alpha$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, the direct product is defined in essentially the same way except with all the arrows reversed: so we have a module $P$ together with a family of homomorphisms $\pi_\alpha : P \to M_\alpha$ such that for every family of homomorphisms $\psi_\alpha: N \to M_\alpha$ we have a &lt;em&gt;unique&lt;/em&gt; homomorphism $\psi: N \to M$ such that $\pi_\alpha \circ \psi = \psi_\alpha$. With this definition, we can probe $P$ by means of homomorphisms $R \to P$, where $R$ is regarded as the free $R$-module on one generator. Because every homomorphism into $P$ is determined by its projections, and every homomorphism out of $R$ is determined by the image of $1$, it turns out $P$ must contain everything in the set-theoretic Cartesian product $\displaystyle \prod_{\alpha \in A} M_\alpha$, and nothing else. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, it is not so easy to probe the direct sum $M$. All we know is that it contains isomorphic copies of each $M_\alpha$, and the definition of module only gives us finite sums of elements, so we only know that $M$ contains finite linear combinations of elements from each $M_\alpha$. It turns out this is sufficient to define a module, and it's exactly the direct sum as usually defined.&lt;/p&gt;&#xA;" OwnerUserId="5191" LastActivityDate="2011-01-25T17:15:34.913" CommentCount="2" />
  <row Id="18941" PostTypeId="2" ParentId="4658" CreationDate="2011-01-25T17:17:05.057" Score="0" ViewCount="" Body="&lt;p&gt;The probability for NO run in n flips is the coefficient of x^n of the polynom series of&#xA;G:=(p,r,x)-&gt;(1-p^r*x^r) / (1-x+(1-p)*(p^r)*x^(r+1)&#xA;For p=0.5, r=7  the coefficient of x^150 is 0.558041257197.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The probability for One run or more is 0.441958742803  (44.19%)&lt;/p&gt;&#xA;" OwnerUserId="6240" LastActivityDate="2011-01-25T17:17:05.057" />
  <row Id="18942" PostTypeId="1" AcceptedAnswerId="18946" CreationDate="2011-01-25T17:19:36.627" Score="3" ViewCount="91" Body="&lt;p&gt;So, here's the problem (1.5.12, Probability and Statistics, Degroot and Schervish 3e) :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $ A_1, A_2, ... $ be an arbitrary infinite sequence of events, and let $ B_1, B_2, ... $ be another infinite sequence of events defined as follows: $B_1 = A_1, B_2 = A^c_1\cap A_2, B_3 = A^c_1 \cap A^c_2 \cap A_3 ... $ &#xA;Prove that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$Pr\left(\bigcup_{i=1}^{n} A_i\right) = \sum_{i=1}^{n} Pr\left(B_i\right)$ for $n = 1, 2, ... $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not stuck solving it, yet; I just want to make sure I am understanding the problem. I can show that $\bigcup A_i = \bigcup B_i$ and that $B_i$ are all disjoint, which allows me to show that $ Pr(\bigcup B_i) = \sum Pr(B_i) $ for finite $n$ by induction and hence $Pr(\bigcup B_i) = Pr(\bigcup A_i)$ follows? Is there anything wrong with my argument?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for any advice, and this is homework so please don't post a full solution :)&lt;/p&gt;&#xA;" OwnerUserId="2465" LastEditorUserId="2465" LastEditorDisplayName="" LastEditDate="2011-01-25T17:57:30.260" LastActivityDate="2011-01-25T18:02:46.980" Title="Sequence of events probability proof." Tags="&lt;homework&gt;&lt;probability&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="18943" PostTypeId="1" AcceptedAnswerId="18948" CreationDate="2011-01-25T17:22:51.477" Score="3" ViewCount="63" Body="&lt;p&gt;I want to compute the integral &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\int_{[0,1]^m} \sqrt{\sum_{j=1}^m u_j^2} \ du_1 \dots du_m$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wolfram Alpha gives an exact result for $m = 2$, but fails to do so for higher values. I am particularly interested in the asymptotics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm sure the value is a standard result, but don't know where to find it (no, I don't have a copy of G&amp;amp;R handy).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit/followup:&lt;/strong&gt; I'm also interested in the integral $\int_{p \ge 0, \sum_{j=1}^m p_j = 1} \sqrt{\sum_{j=1}^m p_j^2} \ dp_1\dots dp_{m-1}$, i.e. the expected 2-norm of a probability distribution on $[m]$ chosen uniformly at random.&lt;/p&gt;&#xA;" OwnerUserId="241" LastEditorUserId="241" LastEditorDisplayName="" LastEditDate="2011-01-25T17:41:08.123" LastActivityDate="2011-01-25T21:43:02.290" Title="What is the expected 2-norm of a uniform point in the standard m-cube?" Tags="&lt;calculus&gt;" AnswerCount="2" />
  <row Id="18944" PostTypeId="1" AcceptedAnswerId="19036" CreationDate="2011-01-25T17:33:09.873" Score="3" ViewCount="190" Body="&lt;p&gt;I know how to compute the probability that 0 runs (or one or more) of M&#xA;consecutive successes will occur in a sequence of N trials, each with  probability P of success.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there an explicit formula (mass function) or a method of calculation (for example in Maple) to find the probability that &lt;strong&gt;exactly R runs of M consecutive successes&lt;/strong&gt; will occur in a sequence of N trials, each trial with probability P of success?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have the same question about the cumulative distribution function?&lt;/p&gt;&#xA;" OwnerUserId="6240" LastActivityDate="2011-01-26T04:40:08.803" Title="Probability of exactly R runs of M consecutive success" Tags="&lt;probability&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="2" />
  <row Id="18945" PostTypeId="1" AcceptedAnswerId="18960" CreationDate="2011-01-25T17:50:09.010" Score="3" ViewCount="205" Body="&lt;p&gt;Hello,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am reading the wikipedia article on the Laplacian matrix:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Laplacian_matrix&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Laplacian_matrix&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't understand what is the particular use of it; having the diagonals as the degree and why the negative adjacency elements off the diagonal? What use would this have? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then on reading about its norm, first of all what does a norm really mean? And what is the norm for the Laplacian matrix delivering? This norm does not result in a matrix whose terms cancel out or sum to one. Or that the determinant is equal to any consistent value. Any insight?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Best,&lt;/p&gt;&#xA;" OwnerUserId="1587" LastActivityDate="2011-01-25T21:39:06.427" Title="Understanding the properties and use of the Laplacian matrix (and its norm)" Tags="&lt;linear-algebra&gt;&lt;graph-theory&gt;&lt;norm&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="18946" PostTypeId="2" ParentId="18942" CreationDate="2011-01-25T18:02:46.980" Score="2" ViewCount="" Body="&lt;p&gt;You are right. Informally-intutively, the B sequences selects the first true value of the A sequence. Hence, the right side (probability that at least one event A is true) is equal to the left side (sum of probabitites that the first true A event was i).&lt;/p&gt;&#xA;" OwnerUserId="312" LastActivityDate="2011-01-25T18:02:46.980" />
  <row Id="18947" PostTypeId="2" ParentId="18932" CreationDate="2011-01-25T18:04:44.840" Score="2" ViewCount="" Body="&lt;p&gt;The point is that the direct sum and the direct product both exist and occur widely in nature.  Sometimes you want one, sometimes the other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a simple, basic and important example where you absolutely want the &quot;cofiniteness condition&quot; (i.e., the direct sum).  Let $V$ be an infinite-dimensional vector space over a field $K$.  Then a basis $B$ for $V$ induces an isomorphism of $V$ with the infinite &lt;em&gt;direct&lt;/em&gt; &lt;strong&gt;sum&lt;/strong&gt; $\bigoplus_{b \in B} K$.  Linear combinations of infinite sets of vectors still must have only finitely many nonzero coefficients.  (At least in pure algebra.  If you have a topology, then one can make sense of infinite linear combinations...)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a simple example where you want the direct product (taken from yesterday's lecture in my commutative algebra course).  If you have an infinite indexed family $\{R_i\}_{i \in I}$ of rings, then the direct product $R = \prod_{i \in I} R_i$ is a ring.  The infinite direct sum is &lt;em&gt;not&lt;/em&gt; a ring according to the modern definition if infinitely many of the factors are nonzero rings, because the multiplicative identity &#xA;$1 = (1,1,1,\ldots)$ is not present.&lt;/p&gt;&#xA;" OwnerUserId="299" LastActivityDate="2011-01-25T18:04:44.840" CommentCount="3" />
  <row Id="18948" PostTypeId="2" ParentId="18943" CreationDate="2011-01-25T18:19:07.853" Score="2" ViewCount="" Body="&lt;p&gt;Well, a simple asymptotic is (by probability) $\sqrt{m/3}$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This can be obtained by considering that we want the expectation of &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$w = \sqrt{x_1^2 +... x_m^2} = \sqrt{y_1 +... y_2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with $y_i = x_i^2$, so that $\displaystyle f_y(y) = \frac{y^{1/2}}{2}$, and $E(y)=1/3$, $\sigma^2(y)=4/45$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, the variable $z = (y_1 +... y_2)/m$ will approach a gaussian with $\mu_z = 1/3$, $\sigma^2_z=\frac{4}{m 45}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And $E(w) = \sqrt{m} E(\sqrt{z})$ which in first approximation is $E(w) \approx \sqrt{m} \sqrt{\mu_z} = \sqrt{m/3}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One can get better aproximations with the higher moments, for example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$E(w) \approx \mu_z^{1/2}  - \frac{1}{8}\mu_z^{-3/2}\sigma_z^2 = &#xD;&#xA;\sqrt{\frac{m}{3}} - \frac{1}{30}\sqrt{\frac{3}{m}} = \sqrt{\frac{m}{3}}\Bigl(1 - \frac{1}{10 m}\Bigr)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(errors aside).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are you looking for something more precise?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;UPDATE: By a similar reasoning, the additional problem (if I understand it right) gives $\displaystyle \sqrt{\frac{4}{3 m}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;UPDATE2: About the second order approximation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, when we want to find moments of $Y=g(X)$ with $g(.)$ nonlinear but smooth, we can do a Taylor expansion around its media:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$Y = g(\mu_x) + g&amp;#39;(\mu_x)(X-\mu_x) + \frac{1}{2!} g&amp;#39;&amp;#39;(\mu_x)(X-\mu_x)^2 + ...$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle \mu_y = E(Y) = g(\mu_x) + g&amp;#39;&amp;#39;(\mu_x)\frac{\sigma_x^2}{2} +...$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This, BTW, justifies the intuitive notion that $E(g(.)) \approx g(E(.))$ if the variance is small (equality applies only if $g(.)$ is linear, of course) and allows to estimate (and to some extent correct) the error of the approximation.&lt;/p&gt;&#xA;" OwnerUserId="312" LastEditorUserId="312" LastEditorDisplayName="" LastEditDate="2011-01-25T21:43:02.290" LastActivityDate="2011-01-25T21:43:02.290" CommentCount="4" />
  <row Id="18949" PostTypeId="2" ParentId="636" CreationDate="2011-01-25T18:20:15.820" Score="-1" ViewCount="" Body="&lt;p&gt;Maybe someone has at least figured out how many perfect Knight's tours there are - closed tours that start ONLY from one of the four knight home spaces?&lt;/p&gt;&#xA;" OwnerUserId="6241" LastActivityDate="2011-01-25T18:20:15.820" CommentCount="14" />
  <row Id="18950" PostTypeId="1" AcceptedAnswerId="18953" CreationDate="2011-01-25T18:28:50.790" Score="2" ViewCount="104" Body="&lt;p&gt;If  z = x + D(x + y) and I let g(x,y) = x + y&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would I be right in saying that :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;z = x + f(g(x))&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and &lt;/p&gt;&#xA;&#xA;&lt;p&gt;dz/dx = 1 + df/dg.dg/dx ?&lt;/p&gt;&#xA;" OwnerUserId="6085" LastActivityDate="2011-01-25T18:37:00.420" Title="Chain rule for multi-variable" Tags="&lt;multivariable-calculus&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="18951" PostTypeId="2" ParentId="18943" CreationDate="2011-01-25T18:31:20.647" Score="2" ViewCount="" Body="&lt;p&gt;As for the asymptotics, suppose that $U_i$ are i.i.d. uniform$(0,1)$ rv's. Then the integral can be written as&#xA;$$&#xD;&#xA;I_m = {\rm E}\bigg(\sqrt {U_1^2  +  \cdots  + U_m^2 } \bigg) = \sqrt m {\rm E}\bigg(\sqrt {\frac{{U_1^2  +  \cdots  + U_m^2 }}{m}}\bigg ).&#xD;&#xA;$$&#xA;Since ${\rm E}(U_1^2)=1/3$, the strong law of large numbers suggests that &#xA;$$&#xD;&#xA;I_m \sim \sqrt {\frac{m}{3}}.&#xD;&#xA;$$&#xA;This agrees with the upper bound of $\sqrt {m/3}$ obtained from&#xA;$$&#xD;&#xA;{\rm E}\bigg(\sqrt {U_1^2  +  \cdots  + U_m^2 } \bigg) &amp;lt; \sqrt {{\rm E}(U_1^2  +  \cdots  + U_m^2 )}  = \sqrt {\frac{m}{3}}. &#xD;&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="2810" LastActivityDate="2011-01-25T18:31:20.647" />
  <row Id="18952" PostTypeId="2" ParentId="18950" CreationDate="2011-01-25T18:35:37.170" Score="1" ViewCount="" Body="&lt;p&gt;I think you meant $z=x+D(g(x,y))$. If $D$ is a function of &lt;em&gt;one&lt;/em&gt; variable, then you can talk about the partial with respect to $x$; you would be correct in saying that&#xA;$$\frac{\partial z}{\partial x} = 1 + \frac{dD}{dg}\frac{\partial g}{\partial x}.$$&#xA;That is,&#xA;$$\frac{\partial z}{\partial x} = 1 + D&amp;#39;(x+y)$$&#xA;because $\frac{\partial g}{\partial x} = \frac{\partial}{\partial x}(x+y) = 1$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice that for functions of one variable, we use $d$ and $dx$, but for functions of several variables you are talking about the partial derivative, $\partial$ and $\partial x$. &lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2011-01-25T18:35:37.170" CommentCount="1" />
  <row Id="18953" PostTypeId="2" ParentId="18950" CreationDate="2011-01-25T18:37:00.420" Score="1" ViewCount="" Body="&lt;p&gt;I assume $D(.)$ is a function not a constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then yes: $$ \frac{\partial }{\partial x}z(x,y) = 1 + \frac{\partial }{\partial g}D(g(x,y)) \frac{\partial }{\partial x}g(x,y)$$&lt;/p&gt;&#xA;" OwnerUserId="5798" LastActivityDate="2011-01-25T18:37:00.420" />
  <row Id="18954" PostTypeId="2" ParentId="18945" CreationDate="2011-01-25T18:39:18.477" Score="4" ViewCount="" Body="&lt;p&gt;Laplacian matrices are important objects in the field of &lt;a href=&quot;http://en.wikipedia.org/wiki/Spectral_graph_theory&quot; rel=&quot;nofollow&quot;&gt;Spectral Graph Theory&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many properties of the graph can be read off easily from the properties of the corresponding Laplacian Matrix. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The multiplicity of the eigenvalue zero gives the number of connected components of the graph.&lt;/li&gt;&#xA;&lt;li&gt;The largest eigenvalue is 2 if and only if a connected component of the graph is a non-trivial bipartite graph.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;There are plenty more results like these.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would recommend you try the book by Fan Chung (Ronald Graham's wife, I believe) conveniently titled Spectral Graph Theory. Here is a link to the book page: &lt;a href=&quot;http://www.math.ucsd.edu/~fan/research/revised.html&quot; rel=&quot;nofollow&quot;&gt;http://www.math.ucsd.edu/~fan/research/revised.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An explanation of the possible motivation for using the Laplacian (and it's normalized form, which is what you seem to be talking about, when you say norm) appears in the first chapter of that book: &lt;a href=&quot;http://www.math.ucsd.edu/~fan/research/cb/ch1.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.math.ucsd.edu/~fan/research/cb/ch1.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also search the web for a linear algebra proof of Friendship theorem, which uses matrices. Even though Laplacian matrices are not used, I would still recommend you read that, as it will give you an idea about the power of spectral methods in graph theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope that helps. &lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2011-01-25T19:02:27.410" LastActivityDate="2011-01-25T19:02:27.410" />
  <row Id="18955" PostTypeId="2" ParentId="18868" CreationDate="2011-01-25T19:02:50.243" Score="1" ViewCount="" Body="&lt;p&gt;Assuming that you have had some form of algebra, since its not typically a high school subject anymore, I would advise you to take statistics.  The subject of statistics is used in just about every major in college.  In fact, most majors require the student to take a statistics course before graduation.  Exposure to this subject will benefit you when making your career choice in the long run.&lt;/p&gt;&#xA;" OwnerUserId="6243" LastActivityDate="2011-01-25T19:02:50.243" />
  <row Id="18956" PostTypeId="2" ParentId="18938" CreationDate="2011-01-25T19:06:49.030" Score="2" ViewCount="" Body="&lt;p&gt;As Shai Covo indicates, this is given in my answer to the recent Math Overflow question &quot;&lt;a href=&quot;http://mathoverflow.net/questions/52493/expected-second-moment-for-random-points-on-a-circle/52548#52548&quot; rel=&quot;nofollow&quot;&gt;Expected Second Moment for Random Points on a Circle&lt;/a&gt;.&quot;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a unit circle, the answer is&#xA;$$F(x) = n(1-x)^{n-1} - \binom{n}{2} (1 - 2x)^{n-1} + \cdots + (-1)^{k-1} \binom{n}{k} (1 - kx)^{n-1} + \cdots,$$ &#xA;where the sum continues until $kx &amp;gt; 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the derivation see the MO answer.  It leans heavily on Section 6.4 (&quot;Random Division of an Interval&quot;) in David and Nagaraja's &lt;em&gt;&lt;a href=&quot;http://books.google.com/books?id=bdhzFXg6xFkC&amp;amp;pg=PA133&amp;amp;lpg=PA133&amp;amp;dq=david+and+nagaraja+order+statistics+division+of+random+interval&amp;amp;source=bl&amp;amp;ots=OaK1s7x7Yj&amp;amp;sig=L6JuApcAHkMRTMfHfxXtNKfBvZw&amp;amp;hl=en&amp;amp;ei=VB4_Tbz1JIfCsAOIrpziBA&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=1&amp;amp;ved=0CBMQ6AEwAA#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;Order Statistics&lt;/a&gt;&lt;/em&gt;, pages 133-137, and the corresponding exercises on pages 153-155.    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The result was apparently first published by Ronald Fisher in &quot;&lt;a href=&quot;http://www.jstor.org/stable/95247&quot; rel=&quot;nofollow&quot;&gt;Tests of Significance in Harmonic Analysis&lt;/a&gt;,&quot; &lt;em&gt;Proceedings of the Royal Society of London, Series A&lt;/em&gt;, 1929, pp 54-59.  (My apologies that this is a JSTOR link and not one that is freely available.) &lt;/p&gt;&#xA;" OwnerUserId="2370" LastActivityDate="2011-01-25T19:06:49.030" CommentCount="1" />
  <row Id="18957" PostTypeId="1" AcceptedAnswerId="18962" CreationDate="2011-01-25T19:15:44.187" Score="3" ViewCount="145" Body="&lt;p&gt;Take 13 integers. Prove that if any 12 of them can be partitioned into two sets of six each with equal sums, then all the integers are the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anyone know if the general case with 2n+1 integers and sets of n integers results in the same conclusion?&lt;/p&gt;&#xA;" OwnerUserId="5959" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2011-03-07T03:57:26.680" LastActivityDate="2011-03-07T03:57:26.680" Title="13 integers with each set of 12 integers..." Tags="&lt;combinatorics&gt;&lt;puzzle&gt;&lt;diophantine-equations&gt;" AnswerCount="2" CommentCount="7" FavoriteCount="1" />
  <row Id="18958" PostTypeId="1" AcceptedAnswerId="19078" CreationDate="2011-01-25T19:17:53.050" Score="1" ViewCount="53" Body="&lt;p&gt;To compute the condition &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \kappa_{abs}^\infty := max\{||&amp;#160;W(t,t_0, y_0) ||: t_0 \leq t \leq T \}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;of an initial value problem on $[t_0, T]$, I need to compute the Wronski matrix $W$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The approximation of the solution is denoted by the discrete evolution operator $\Psi^{t_0, t}$ and the exact solution is denoted $\Phi^{t_0, t}$ and the Wronksi matrix is given as &#xA;$ W(t,t_0, y) := \frac{d}{dy} \Phi^{t_0, t}y \Big |_{y=z}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My &lt;strong&gt;first question&lt;/strong&gt;: is the condition really a function of the exact solution? According to my understanding numerical analysis deals with finding good approximation functions so the condition should be a measure of how good (stable?) a given approximation is, therefore the Wronski matrix should be a function of $\Psi$, not of $\Phi$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My &lt;strong&gt;second question&lt;/strong&gt;: how do I explicitly calculate the Wronski matrix for a given problem? Say, I'm using the explicit Euler method $y_{k+1} = y(t_k) + h*f(t_k, y(t_k))$ to solve the ODE $\dot{y} = (\alpha - \beta y)y$ or the system $\dot{u} = (\alpha - \beta v)u, \dot{v} = (\delta u - \gamma)v$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many thanks for your help.&lt;/p&gt;&#xA;" OwnerUserId="5798" LastActivityDate="2011-01-26T14:16:06.233" Title="Wronksi matrix of a discrete evolution and condition" Tags="&lt;numerical-methods&gt;" AnswerCount="1" FavoriteCount="0" />
  <row Id="18959" PostTypeId="2" ParentId="18932" CreationDate="2011-01-25T19:20:52.590" Score="4" ViewCount="" Body="&lt;p&gt;To see an example of some pathology, while the direct sum of free modules is always free (with the obvious free basis), the direct product of free modules may fail to be free.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take $R=\mathbb{Z}$; then $M = \mathop{\oplus}\limits_{n=1}^{\infty}\mathbb{Z}$ is free abelian, with basis given by the &quot;obvious&quot; elements $e_i$ (which have a $1$ in the $i$th coordinate and zeros elsewhere). However, $N = \prod\limits_{n=1}^{\infty} \mathbb{Z}$ is &lt;em&gt;not&lt;/em&gt; free abelian. For example, Specker proved (&lt;em&gt;Additive Gruppen von Folgen ganzer Zahlen&lt;/em&gt;, Portugaliae Math. &lt;strong&gt;9&lt;/strong&gt; (1950) 131-140) that $N$ has only countably many homomorphisms onto $\mathbb{Z}$. But since $N$ is uncountable, if it were free it would be free in uncountably many generators, and hence would have uncountably many homomorphisms onto $\mathbb{Z}$ (at least the projections). In fact, if $X$ is any infinite set, then $\mathop{\oplus}_{x\in X}\mathbb{Z}$ is free abelian of rank $|X|$, but $\prod_{x\in X}\mathbb{Z}$ is never free abelian.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have a related pathology with vector spaces (of course, every vector space is free, so that's not what the problem will be, but rather when you think about &quot;free on what set?&quot;). When working with finitely many vector spaces, you have that $\dim(V_1\times V_2) = \dim(V_1\oplus V_2) = \dim(V_1)+\dim(V_2)$ (in the sense of sum of cardinalities). However, once you have infinitely many vector spaces, the equality breaks down for the product, while it holds for the direct sum:&#xA;$$\dim\left(\bigoplus_{i=1}^{\infty} V_i\right) = \sum_{i=1}^{\infty}\dim(V_i)$$&#xA;but for products it need not hold: for a counterexample, take $V_i = \mathbb{Q}$ as a vector space over itself; the sum of dimensions is $\aleph_0$, but the direct product of denumerably many copies of $\mathbb{Q}$ is uncountable, so the dimension is $2^{\aleph_0}$ (so you have a &quot;jump&quot; in the dimension once you get to infinitely many elements). &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-25T20:52:07.503" LastActivityDate="2011-01-25T20:52:07.503" CommentCount="2" />
  <row Id="18960" PostTypeId="2" ParentId="18945" CreationDate="2011-01-25T19:21:24.197" Score="4" ViewCount="" Body="&lt;p&gt;The Laplacian is a discrete analogue of the Laplacian $\sum \frac{\partial^2 f}{\partial x_i^2}$ in multivariable calculus, and it serves a similar purpose: it measures to what extent a function differs at a point from its values at nearby points.  The Laplacian appears in the analysis of random walks and electrical networks on a graph (the standard reference here being &lt;a href=&quot;http://front.math.ucdavis.edu/math.PR/0001057&quot; rel=&quot;nofollow&quot;&gt;Doyle and Snell&lt;/a&gt;), and so it is not surprising that it encodes some of its structural properties: as I described in &lt;a href=&quot;http://qchu.wordpress.com/2011/01/02/the-schrodinger-equation-on-a-finite-graph/&quot; rel=&quot;nofollow&quot;&gt;this blog post&lt;/a&gt;, it can be used to set up three differential equations on a graph (the wave equation, the heat equation, and the Schrödinger equation).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(To be totally clear, when you're using this interpretation you should think of the Laplacian, not as a matrix, but as an operator acting on functions $f : V \to \mathbb{R}$.  In this setting there is a discrete notion of gradient (which sends a function $f$ to a function $\text{grad } f : E \to \mathbb{R}$) and a discrete notion of divergence (which sends a function $g : E \to \mathbb{R}$ to a function $\text{div } g : V \to \mathbb{R}$), and the divergence of the gradient is the Laplacian - just like in the infinitary case.  So the Laplacian defines a certain analogy between graphs and Riemannian manifolds.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The quadratic form defined by the Laplacian appears, for example, as the power running through a circuit with given voltages at each point and unit resistances on each edge.  It is the discrete analogue of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Dirichlet%27s_energy&quot; rel=&quot;nofollow&quot;&gt;Dirichlet energy&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Laplacian appears in the &lt;a href=&quot;http://en.wikipedia.org/wiki/Kirchhoff%27s_theorem&quot; rel=&quot;nofollow&quot;&gt;matrix-tree theorem&lt;/a&gt;: the determinant of the Laplacian (with a bit removed) counts the number of spanning trees.  This is related to its appearance in the study of electrical networks and is still totally mysterious to me.  The group $\mathbb{Z}^n / L$  where $L$ is the Laplacian has rank $1$, and its torsion subgroup is the &lt;em&gt;critical group&lt;/em&gt; of the graph, which has size the number of spanning trees.  The critical group appears in the description of chip-firing games on the graph (another name for this is the abelian sandpile model), and is an interesting invariant of graphs. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is &lt;a href=&quot;http://arxiv.org/abs/math/0608360&quot; rel=&quot;nofollow&quot;&gt;some evidence&lt;/a&gt; that finite graphs are analogous to curves over finite fields, and in this analogy the critical group appears to be analogous to the ideal class group (that is, the Jacobian).  Its size even appears in a &lt;a href=&quot;http://en.wikipedia.org/wiki/Class_number_formula&quot; rel=&quot;nofollow&quot;&gt;class number formula&lt;/a&gt; for graphs coming from the &lt;a href=&quot;http://en.wikipedia.org/wiki/Ihara_zeta_function&quot; rel=&quot;nofollow&quot;&gt;Ihara zeta function&lt;/a&gt; (the analogue of the Dedekind zeta function).  Again, all of this is totally mysterious to me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a &lt;a href=&quot;http://www.fmf.uni-lj.si/~mohar/Papers/Spec.pdf&quot; rel=&quot;nofollow&quot;&gt;nice survey paper&lt;/a&gt; by Mohar on what graph theorists actually use the Laplacian for.  In the literature there are several different normalizations; they correspond to either using a different preferred basis for the space of functions $f : V \to \mathbb{R}$ or varying physical properties of the graph (e.g. changing resistances, adding a potential).&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2011-01-25T21:39:06.427" LastActivityDate="2011-01-25T21:39:06.427" />
  <row Id="18961" PostTypeId="2" ParentId="18957" CreationDate="2011-01-25T19:21:42.293" Score="3" ViewCount="" Body="&lt;p&gt;They must all be even or all be odd.  If not, you could find a set of 12 with odd sum and could not divide it evenly.  Now subtract 1 if they are odd, then divide by 2, and make the same argument, so they are the same mod 4.  Repeat.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Added based on Moron's answer:  I think this (sort of) works for infinite sets as well.  If you use cardinals and cardinal addition, as long as $\frac {n+3}{2}$ are the same ($\kappa$)and the rest are smaller, the sum of each subset will be $\kappa$ and you are done.  But in ordinals (using ordinal addition) you can go through the same argument with Cantor normal form.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2011-01-29T14:06:48.600" LastActivityDate="2011-01-29T14:06:48.600" />
  <row Id="18962" PostTypeId="2" ParentId="18957" CreationDate="2011-01-25T19:26:42.897" Score="3" ViewCount="" Body="&lt;p&gt;This is true for arbitrary reals and arbitrary splits. A recent answer I posted on cstheory:&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;To answer the more general question,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Given any positive integers $\displaystyle k$ and&#xA;  $\displaystyle n \gt 1$, if there are $\displaystyle kn+1$ real numbers&#xA;  $\displaystyle r_i$, such that for any $\displaystyle i$, all except $\displaystyle r_i$&#xA;  can be split into $\displaystyle n$ groups of $\displaystyle k$&#xA;  each, such that the sum of the reals&#xA;  in any group is the same as any other,&#xA;  then $\displaystyle r_i = r_j$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is true.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instead of reals, start with integers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice that if $\displaystyle S$ is the total sum, then $\displaystyle S = r_i \mod n$ and so $\displaystyle r_i = r_j \mod n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;wlog assume $\displaystyle r_1$ is the smallest.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then we have a new set of weights&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle \frac{r_i - r_1}{n}$ which has lower $\displaystyle \text{max}\{r_i\}$ (note all have become non-negative so we can assume they were non-negative to begin with) and hence we end up with all zeroes. Since this is reversible, the original must have been equal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus if the weights were integers, they all have to be the same. This easily extends to rationals weights.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now since $\displaystyle \mathbb{R}$ is an infinite dimensional vector space over $\displaystyle \mathbb{Q}$, we are done. (Search the web for Hamel Basis).&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2011-01-25T19:26:42.897" CommentCount="0" />
  <row Id="18963" PostTypeId="1" CreationDate="2011-01-25T19:28:55.110" Score="5" ViewCount="115" Body="&lt;p&gt;The class number of $\mathbb{Q}(\sqrt{-23})$ is $3$, and the form&#xA;$$2x^2 + xy + 3y^2 = z^3$$&#xA;is one of the two reduced non-principal forms with discriminant $-23$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are the obvious non-primitive solutions $(a(2a^2+ab+3b^2),b(2a^2+ab+3b^2), (2a^2+ab+3b^2))$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm pretty sure there aren't any primitive solutions, but can't seem to find an easy argument. Are there?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, is it possible for a non-principal form to represent an $h$-th power primitively (where $h$ is the class number of the associated quadratic field)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;hr&gt;&#xA;[EDIT]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think I've solved the first question and have posted an answer below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the proof is very technical I don't see how it can generalize to the greater question above ($h$-th powers).&lt;/p&gt;&#xA;" OwnerUserId="4946" LastEditorUserId="4946" LastEditorDisplayName="" LastEditDate="2011-01-25T21:27:06.680" LastActivityDate="2011-01-25T22:36:53.707" Title="Primitive integer solutions to $2x^2+xy+3y^2=z^3$?" Tags="&lt;number-theory&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="2" />
  <row Id="18964" PostTypeId="1" CreationDate="2011-01-25T19:32:11.310" Score="1" ViewCount="177" Body="&lt;p&gt;what about GL(n,C)? is it open, dense in M(n,C)?&lt;/p&gt;&#xA;" OwnerUserId="6244" LastActivityDate="2011-01-26T00:51:58.687" Title="what about GL(n,C)? is it open, dense in M(n,C)?" Tags="&lt;topological-vector-spaces&gt;" AnswerCount="4" CommentCount="2" />
  <row Id="18965" PostTypeId="2" ParentId="18931" CreationDate="2011-01-25T19:34:44.413" Score="0" ViewCount="" Body="&lt;p&gt;If $A:=S\circ T: V\to V$ is &quot;invertible&quot; (i.e., bijective) then $T$ has to be injective (i.e., one-one), or one would not find a unique pre-image $A^{-1}y$ for some $y\in V$, and $S$ has to be surjective (i.e., onto), or $A^{-1}$ would not be defined on all of $V$. An injective $T:V\to V$ is automatically bijective (this is a fundamental theorem of linear algebra), and similarly a surjective $S:V\to V$ is automatically bijective.  Now that both $T$ and $S$ are verified to be bijective it follows that the same is true for $TS$. Conversely, the bijectivity of $TS$ implies the bijectivity of $ST$. (Actually it would be enough to assume the composition $ST$ injective resp. surjective only.)&lt;/p&gt;&#xA;" OwnerUserId="1303" LastActivityDate="2011-01-25T19:34:44.413" />
  <row Id="18966" PostTypeId="1" CreationDate="2011-01-25T19:34:45.520" Score="0" ViewCount="104" Body="&lt;p&gt;I would be glad if someone can help me understand the argument in the first paragraph of page 4 of &lt;a href=&quot;http://arxiv.org/pdf/cond-mat/0109112&quot; rel=&quot;nofollow&quot;&gt;this paper&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Especially I don't understand their first sentence,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&quot;Using N bosons (fermions) distributed over m states, one can construct completely symmetric (antisymmetric) irreducible representations of the group U(m) associated with Young tableaux with N boxes in a row (column)&quot;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I am quite familiar with the Quantum Statistics concepts being alluded to but not so much with the Young-Tableux technology being used) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;All I can see is that $U(n)$ can act on the space of $c_i$ and keep the operators defined in their equation 2.9 and 2.10 unchanged. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also on page 4 is there a typo in Equation 2.13? &#xA;It doesn't seem to follow from the line previous to it and the line previous to it makes no sense to me. I guess there should have been a &quot;=&quot; between the $\lambda^N$ and $exp$ in the line just before 2.13. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even if I make the above &quot;correction&quot; I don't see how 2.13 follows from it. &lt;/p&gt;&#xA;" OwnerUserId="965" LastActivityDate="2011-01-25T21:06:26.873" Title="Representations of U(n) using bosons and fermions" Tags="&lt;representation-theory&gt;&lt;lie-groups&gt;" AnswerCount="1" CommentCount="6" />
  <row Id="18967" PostTypeId="2" ParentId="18964" CreationDate="2011-01-25T19:42:39.657" Score="4" ViewCount="" Body="&lt;p&gt;Yes, it is both open and dense.  One way to show it is open is by considering the determinant.  Alternatively, the set of invertible elements of any Banach algebra is open (and $M(n,\mathbb{C})$ is a Banach algebra if you give it any submultiplicative norm), which follows from the fact that if $a$ is invertible and $\|a-b\|\lt \frac{1}{\|a^{-1}\|}$, then $b$ is invertible.  One way to show it is dense is to consider the fact that $a\in M(n,\mathbb{C})$ has only a finite set of eigenvalues.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2011-01-25T19:42:39.657" />
  <row Id="18968" PostTypeId="2" ParentId="18868" CreationDate="2011-01-25T19:53:10.473" Score="2" ViewCount="" Body="&lt;p&gt;If the three options are statistics, calculus, and algebra:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;I'd guess that &quot;algebra&quot; refers to some sort of linear algebra or basic group theory&lt;/li&gt;&#xA;&lt;li&gt;If the set of potential career paths is largely scientific, calculus is likely to be a requirement somewhere along the way (even if only as a gatekeeper course), so it would probably be the best choice.&lt;/li&gt;&#xA;&lt;li&gt;Otherwise (and especially if the set of potential career paths contains social sciences, psychology, science, and other fields where data is used—there are a lot of such fields), statistics is the most likely to be useful.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="72" LastActivityDate="2011-01-25T19:53:10.473" />
  <row Id="18969" PostTypeId="1" AcceptedAnswerId="18973" CreationDate="2011-01-25T19:56:41.393" Score="5" ViewCount="415" Body="&lt;p&gt;Let us start by clarifying this a bit.  I am aware of &lt;a href=&quot;http://math.stackexchange.com/questions/732/proof-that-the-irrational-numbers-are-uncountable&quot;&gt;some proofs&lt;/a&gt; &lt;a href=&quot;http://en.wikipedia.org/wiki/Cantor_diagonalization&quot; rel=&quot;nofollow&quot;&gt;that irrationals/reals are uncountable&lt;/a&gt;.  My issue comes by way of some properties of the reals. These issues can be summed up by the combination of the following questions:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Is it true that between any two rationals one may find at least one irrational?&lt;/li&gt;&#xA;&lt;li&gt;Is it true that between any two irrationals one may find at least one rational?&lt;/li&gt;&#xA;&lt;li&gt;Why are the reals uncountable?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I've been talking with a friend about why the answer of these three questions can be the case when they somewhat seem to contradict each other.  I seek clarification on the subject.  Herein lies a summary of the discussion:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Person A:&lt;/strong&gt;  &lt;em&gt;By  way of &lt;a href=&quot;http://en.wikipedia.org/wiki/Cantor_diagonalization&quot; rel=&quot;nofollow&quot;&gt;Cantor Diagonalization&lt;/a&gt; it can be shown that the reals are uncountable.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Person B:&lt;/strong&gt;  &lt;em&gt;But it is not also the case that one may find at least one rational between any two irrationals and vice versa?&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Person A:&lt;/strong&gt;  &lt;em&gt;That seems logical, I can't pose a counterexample... but why does that matter?&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Person B:&lt;/strong&gt;  &lt;em&gt;Wouldn't that imply that for every irrational there is a corresponding rational?  And from this the Reals would be equivalent to 2 elements for every element of the rationals?&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Person A:&lt;/strong&gt;  &lt;em&gt;That implies that the Reals are countable, but we have already shown that they weren't... where is the hole in our reasoning?&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;And so I pose it to you... where is the hole in our reasoning?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="3077" LastEditorUserId="822" LastEditorDisplayName="" LastEditDate="2011-01-25T21:18:07.780" LastActivityDate="2011-01-26T13:04:50.570" Title="Why Are the Reals Uncountable?" Tags="&lt;real-analysis&gt;&lt;elementary-set-theory&gt;" AnswerCount="8" CommentCount="1" FavoriteCount="1" />
  <row Id="18970" PostTypeId="2" ParentId="18964" CreationDate="2011-01-25T19:59:11.013" Score="1" ViewCount="" Body="&lt;p&gt;For the denseness part, one way to look at is this: A matrix is in $GL_n({\mathbb C})$ iff its rows are linearly independent over ${\mathbb C}$. So what you need to show is that given any matrix $M$ and an $\epsilon &amp;gt; 0$, you can perturb each entry by less than $\epsilon$ (in magnitude) and end out with linearly independent rows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or you can appeal to algebraic geometry and say that $det(M) = 0$ defines an algebraic variety which therefore has measure zero. But of course this is pretty advanced :) &lt;/p&gt;&#xA;" OwnerUserId="3035" LastEditorUserId="3035" LastEditorDisplayName="" LastEditDate="2011-01-25T20:09:10.007" LastActivityDate="2011-01-25T20:09:10.007" />
  <row Id="18971" PostTypeId="2" ParentId="18288" CreationDate="2011-01-25T20:02:50.010" Score="1" ViewCount="" Body="&lt;p&gt;I assume your collection of sets is countable (and that this is what you mean by &quot;arbtirary&quot;); that is, you have one set for each natural number (or perhaps only up to a certain bound), $E_1$, $E_2$, $E_3$, $E_4,\ldots$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\limsup E_i$ consists of all elements in $\cup E_i$ that are in infinitely many of the $E_i$. $\liminf E_i$ consist of all elements in $E_i$ that are in all except perhaps for finitely many of the $E_i$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you imagine the $E_i$ lined up in front of you, then $\limsup E_i$ is the collection of all things which, no matter where you stand on the line, there will always be an instance of that thing behind you (I think of $E_{i+1}$ as &quot;behind&quot; $E_i$).  On the other hand, $\liminf E_i$ consists of all thing which are in &lt;em&gt;each position&lt;/em&gt; of the queue after a certain point. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, if you think of $\limsup E_i$ as a &quot;sequence&quot; of sets, namely the sets $F_k = \cup_{i=k}^{\infty} E_i$, which you are then intersecting, then the $F_k$ can be all different, or they can change as often as you want or as little as you want, even if all your sets are finite.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For an example where all the $F_i$ are different, take $E_i = \{i\}$. Then $F_k = \{k,k+1,\ldots\}$. &#xA;Or you can tweak it to change every so often; $E_1 = \{1\}$, $E_2=\{1,2\}$, $E_3=\{1,3\}$, $E_4=\{4\}$, $E_5=\{4,5\}$, $E_6=\{4,6\},\ldots$. Then $F_k$ will be $\{r,r+1,r+2,\ldots\}$ if $3r+1\leq k\leq 3r+r$. You can make the changes happen at distinct intervals. Etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If all your sets are contained in a &lt;em&gt;finite&lt;/em&gt; set $A$ with $n$ elements, then you can set the $F_i$ to be as many as $n+1$ distinct sets (notice that you will have $F_1\supseteq F_2\supseteq F_3\supseteq F_4\supseteq\cdots$, and it is easy to check that if they are all contained in a set with $n+1$ elements, then such a chain can have at most $n$ distinct elements in it, since every time you &quot;change&quot; sets you lose at least one element, and you can go from $n$ elements down to $0$ elements). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Symmetrically with the limit inferior. if you set $G_i = \mathop{\cap}_{i=1}^{\infty} E_i$, then you have $G_1\subseteq G_2\subseteq G_3\subseteq\cdots$; if all sets are contained in an infinite set $A$, then you can have the $G_i$ change at each step, every other step, or you can designate exactly where you want them to change and you can set it up so that they change exactly there. If your $E_i$ are all subsets of a finite set with $n$ elements, then there can be at most $n+1$ distinct $G_i$, for the same reason as there can be at most $n+1$ distinct $F$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you consider pairs $(a,b)$ with $a\leq b$, and you set $H_{a,b} = \cup_{k=1}^a\cap_{i=k}^{b} E_i$, then you have $H_{1,1}\supseteq H_{1,2}\supseteq H_{1,3}\supseteq\cdots$, $H_{2,2}\supseteq H_{2,3}\supseteq \cdots$ (and in general $H_{a,b}\supseteq H_{a,b+1}$ for all $a\leq b$, and $H_{a,b}\subseteq H_{a+1,b}$ for all $a\lt b$). You can have lots of distinct things, or they may be limited depending on where your sets $E_i$ are taken from.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The questions you pose are too vague to give anything but very general answers absent more information. &lt;/p&gt;&#xA;&#xA;&lt;hr/&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added:&lt;/strong&gt; Clearly, we are having an impasse of undertanding. Below you say that &quot;quantify&quot; to you means &quot;break up and construct interesting stuff&quot;. I'll wager the reason people (me included) are so confused about your question is that nobody took that word to mean anything like that (I certainly did not). The dictionary gives two meanings: &quot;to determine or express the quantity of&quot;, and the logic related one of &quot;to limit the variables of a proposition by prefixing an operator such as &lt;em&gt;for all&lt;/em&gt; or &lt;em&gt;some&lt;/em&gt;&quot;.  I took it to mean that you wanted to give it a kind of &quot;quantity&quot; or &quot;size&quot;; perhaps you are taking your queue from &quot;quantum mechanics&quot; when you use &quot;quantify&quot;, but I don't think anybody understood you. I confess to still being rather mystified.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-02T02:59:27.777" LastActivityDate="2011-02-02T02:59:27.777" CommentCount="7" />
  <row Id="18973" PostTypeId="2" ParentId="18969" CreationDate="2011-01-25T20:06:08.463" Score="18" ViewCount="" Body="&lt;p&gt;The problem is here: &quot;Wouldn't that imply that for every irrational there is a corresponding rational? And from this the Reals would be equivalent to 2 elements for every element of the rationals?&quot; The problem is that for &lt;em&gt;many&lt;/em&gt; different pairs of irrationals you would be choosing the same rational in between. If you want to avoid this problem, you'd have to describe a procedure by means of which:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;You uniquely specify for each pair of irrationals $a&amp;lt;b$ a rational $c$ in between.&#xA;(This is easy but non-trivial since there are infinitely many candidates for $c$.)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Different pairs get assigned different rationals.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Of course, task 2 is impossible, so there is no correspondence between irrationals and rationals.&lt;/p&gt;&#xA;" OwnerUserId="462" LastActivityDate="2011-01-25T20:06:08.463" CommentCount="4" />
  <row Id="18974" PostTypeId="2" ParentId="18969" CreationDate="2011-01-25T20:07:33.643" Score="-3" ViewCount="" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;Wouldn't that imply that for every&#xA;  irrational there is a corresponding&#xA;  rational?&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No. &lt;/p&gt;&#xA;" OwnerUserId="312" LastActivityDate="2011-01-25T20:07:33.643" />
  <row Id="18975" PostTypeId="2" ParentId="18969" CreationDate="2011-01-25T20:13:15.877" Score="3" ViewCount="" Body="&lt;p&gt;What you &lt;em&gt;can&lt;/em&gt; conclude (although there are other means) is that there are uncountably many &lt;em&gt;sets&lt;/em&gt; of rational numbers.  If $a\lt b$ and $c\lt d$ are distinct pairs of irrational numbers, then the set of rational numbers in the interval $(a,b)$ is distinct from the set of rational numbers in $(c,d)$.  For example, if $b\lt d$, then there is a rational number $r$ in the interval $(\max(b,c),d)$, and thus $r$ is in $(c,d)$ but not $(a,b)$.  To make this more concrete, the sets of rational numbers in the intervals $(0,b)$ are distinct as $b$ ranges over the positive irrational (or real) numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll note that this is quite a roundabout way to prove this, because proving that the set of real numbers is uncountable is no more basic than proving that in fact the set of subsets of any countable set is uncountable, but it strikes me as the closest positive result to what you were looking for.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2011-01-25T20:28:48.177" LastActivityDate="2011-01-25T20:28:48.177" />
  <row Id="18976" PostTypeId="2" ParentId="18969" CreationDate="2011-01-25T20:13:41.273" Score="1" ViewCount="" Body="&lt;p&gt;Your reasoning shows that there is a mapping from ordered pairs of irrationals to the rationals, and a mapping from ordered pairs of rationals to the irrationals.  Neither of these maps has any other special properties; in particular, the first is not one-to-one (many pairs of irrationals must map to the same rational), and the second is not onto (many irrationals are not mapped to by any ordered pair of rationals).  Both the rationals and the irrationals are &lt;em&gt;dense&lt;/em&gt; in the reals, which is what explains your observations; but this does not imply that there is a bijection between them.&lt;/p&gt;&#xA;" OwnerUserId="5546" LastActivityDate="2011-01-25T20:13:41.273" />
  <row Id="18977" PostTypeId="2" ParentId="18969" CreationDate="2011-01-25T20:17:52.283" Score="1" ViewCount="" Body="&lt;p&gt;In fact, between any two distinct irrational numbers $a$ and $b$, there are countably infinitely many rational numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proof: Without loss of generality, suppose $0 &amp;lt; a &amp;lt; b &amp;lt; 1$.  Each has an infinite decimal expansion: $$a = \sum_{i = 1}^{\infty} a_i 10^{-i} \mbox{ and } b = \sum_{i = 1}^{\infty} b_i 10^{-i}$$&#xA;Note that for $b$ to be irrational, infinitely many $b_i$ must be non-zero.  Because $a \neq b$, there is some least value of $i$ such that $a_i \neq b_i$ (i.e. there is a first decimal place at which $a$ and $b$ disagree); let that value be $N$.  Then $a_N &amp;lt; b_N$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let $c_n$ be a rational number defined as $$c_n = \sum_{i = 1}^{n} b_i 10^{-i}$$ where $n &amp;gt; N$.  The decimal places of $c_n$ agree with each of the decimal places of $b$ up to and including $b_N$, therefore the decimal places of $c_n$ agree with the decimal places of $a$ up to, but &lt;em&gt;not&lt;/em&gt; including $a_N$.  So the $N$th decimal place is the first place where $c_n$ and $a$ disagree, and $c_n$'s $N$the decimal place is greater than $a$'s $N$th decimal place, therefore $c_n &amp;gt; a$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yet because the decimal expansion of $c_n$ terminates while the decimal expansion of $b$ does not, $c_n &amp;lt; b$.  So $a &amp;lt; c_n &amp;lt; b$, i.e. $c_n$ is a rational number in between $a$ and $b$.  There are at least as many distinct $c_n$ as there are $n &amp;gt; N$ such that $b_n \neq 0$, i.e. countably infinitely many, which proves the claim.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Moral&lt;/strong&gt;: Infinity is stranger than you think.&lt;/p&gt;&#xA;" OwnerUserId="506" LastEditorUserId="506" LastEditorDisplayName="" LastEditDate="2011-01-26T11:38:57.220" LastActivityDate="2011-01-26T11:38:57.220" />
  <row Id="18978" PostTypeId="2" ParentId="18969" CreationDate="2011-01-25T20:20:48.920" Score="1" ViewCount="" Body="&lt;p&gt;Yes, between any two irrationals there is always (at least one) rational: let $r$ and $s$ be irrationals, say $r\lt s$. For simplicity we may assume they are both positive (if $r\lt 0$ and $s\gt 0$, then we can take $0$ as a rational between them; if $r\lt s\lt 0$, we can do what I'm about to do to find a rational $q$ with $-s\lt q\lt -r$, and then $-q$ will lie between $r$ and $s$ and be rational).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $n$ be an integer such that $0\lt \frac{1}{n}\lt \frac{r-s}{2}$; we know that $S=\{m\in\mathbb{N}\mid \frac{m}{n}\gt r\}$ is nonempty by the Archimedean property. Being a nonempty set of natural numbers, it has a least element; call it $M$. Then $r\lt \frac{M}{n}$. If $s\lt \frac{M}{n}$, then &#xA;$$\frac{M-1}{n} = \frac{M}{n}-\frac{1}{n} \geq \frac{M}{n}-\frac{r-s}{2} \gt s-\frac{r-s}{2} = \frac{s+r}{2} \gt \frac{r+r}{2} = r.$$&#xA;contradicting that $M$ is the &lt;strong&gt;smallest&lt;/strong&gt; positive integer such that $\frac{M}{n}\gt r$ (notice that we cannot have $M-1=0$, because that would mean $M=1$, so $\frac{1}{n}\gt r \gt \frac{r-s}{2}$). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means that $\frac{M}{n}\leq s$; and since $\frac{M}{n}$ is rational and $s$ irrational, then $\frac{M}{n}\lt s$. Therefore, $r\lt \frac{M}{n}\lt s$, so there is at least one rational between $r$ and $s$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yes, between any two rationals there is always at least one irrational. You can proceed along similar lines: given any positive integer $n$, there is always an irrational $x$ with $0\lt x\lt \frac{1}{n}$. To see this, just take $x=\frac{\sqrt{2}}{n+1}$. Then $x\lt\frac{1}{n}$ if and only if $\sqrt{2}\leq n(n+1)$, and since $n(n+1)\geq 2$, this holds. So there is at least one such irrational.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let $r$ and $s$ be two rationals, with $r\lt s$. Let $x$ be an irrational with $0\lt x \lt \frac{s-r}{2}$. Now take $x+r$; this is irrational (if $x+r$ were rational, then so would $(x+r)-r = x$, a contradiction). It is greater than $r$, because $x\gt 0$. And&#xA;$$r \lt r+x \lt r+\frac{s-r}{2} = \frac{s+r}{2} \lt \frac{s+s}{2} = s,$$&#xA;so $r\lt y \lt s$ with $y=r+x$ an irrational.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But &lt;strong&gt;no&lt;/strong&gt;, this does not mean that &quot;for every irrational there is a rational&quot;; for the reasons others have so well explained. &lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2011-01-25T20:20:48.920" />
  <row Id="18979" PostTypeId="2" ParentId="18964" CreationDate="2011-01-25T20:37:25.737" Score="4" ViewCount="" Body="&lt;p&gt;The amount of algebraic geometry needed to prove the denseness of $GL_n({\Bbb C})$ is indeed very minimal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The topology of $M_n({\Bbb C})$ comes from its obvious identification with ${\Bbb C}^{n^2}$. The complement is the zero set of the determinant which is a polynomial in the $n^2$ entries of the matrix. The zero set of a polynomial cannot have inner points: if a polynomial vanishes on an open ball, its Taylor expansion centered at any point of the ball is identically 0, and by analicity the polynomial vanishes identically. Therefore there are no proper closed subsets of $M_n({\Bbb C})$ containing $GL_n({\Bbb C})$.&lt;/p&gt;&#xA;" OwnerUserId="688" LastActivityDate="2011-01-25T20:37:25.737" CommentCount="2" />
  <row Id="18980" PostTypeId="2" ParentId="18966" CreationDate="2011-01-25T20:59:39.633" Score="1" ViewCount="" Body="&lt;p&gt;I see this paper was written by a physicist.  Here's what he's &lt;em&gt;trying&lt;/em&gt; to say.  Let $T$ be a linear operator $V \to V$ with eigenvalues $\lambda_1, ... \lambda_n$.  $T$ defines linear operators $\Lambda^N T : \Lambda^N V \to \Lambda^N V$ and $S^N T : S^N V \to S^N V$ for all $N$.  Let us suppose that $T$ is diagonalizable with eigenvectors $v_1, ... v_n$.  Then a basis of eigenvectors for $\Lambda^N T$ is given by the set of all exterior products of $N$ distinct eigenvectors, and a basis of eigenvectors for $S^N T$ is given by the set of all symmetric products of $N$ eigenvectors.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It follows that $\text{tr } \Lambda^N T$ is the $N^{th}$ &lt;a href=&quot;http://en.wikipedia.org/wiki/Elementary_symmetric_polynomial&quot; rel=&quot;nofollow&quot;&gt;elementary symmetric polynomial&lt;/a&gt; in the eigenvalues (this determines the character of the corresponding representation of $U(n)$) and $\text{tr } S^N T$ is the $N^{th}$ &lt;a href=&quot;http://en.wikipedia.org/wiki/Complete_homogeneous_symmetric_polynomial&quot; rel=&quot;nofollow&quot;&gt;complete homogeneous symmetric polynomial&lt;/a&gt; in the eigenvalues (same).  If $T$ is the time evolution operator $U(t)$ for a quantum system, then the eigenvalues of $T$ are complex exponentials of the eigenvalues of the Hamiltonian.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can't help you with the rest; I don't understand the notation.  As far as I can tell, he's just using the identity described in &lt;a href=&quot;https://qchu.wordpress.com/2009/08/23/newtons-sums-necklace-congruences-and-zeta-functions/&quot; rel=&quot;nofollow&quot;&gt;this blog post&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{1}{\det(I - Tz)} = \frac{1}{\sum_{n \ge 0} \text{tr } \Lambda^n T z^n} = \sum_{n \ge 0} \text{tr } S^n T z^n = \exp \left( \sum \frac{\text{tr } T^n}{n} z^n \right).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This identity expresses, in a way I don't really understand, boson-fermion duality.&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2011-01-25T21:06:26.873" LastActivityDate="2011-01-25T21:06:26.873" CommentCount="5" />
  <row Id="18981" PostTypeId="1" AcceptedAnswerId="18985" CreationDate="2011-01-25T21:10:22.987" Score="1" ViewCount="87" Body="&lt;p&gt;I am looking for an equation to calculate the volume of liquid in an underground tank based on a depth reading.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The tank is an ellipsoid shape with the following dimensions:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;X-Axis Radius:    30.425 cm&#xA;Y-Axis Radius:    30.425 cm&#xA;Z-Axis Radius:    60.95 cm &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I am trying to find an equation that would provide the volume of liquid in liters at a specific depth:&#xA;For example:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;When the tank is filled up with liquid to:&#xA;30cm (1/2 way) &#xA;the volume of liquid would be around:  &#xA;120 Liters.&#xA;&#xA;And when the tank if filled up to the top:&#xA;60cm&#xA;the volume would be around:  &#xA;240 Liters.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Thank you for your assistance!&lt;/p&gt;&#xA;" OwnerUserId="6233" LastActivityDate="2011-01-25T21:28:07.647" Title="Equation for calculating the volume of liquid in an ellipsoid" Tags="&lt;calculus&gt;&lt;geometry&gt;&lt;analytic-geometry&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="18982" PostTypeId="2" ParentId="18868" CreationDate="2011-01-25T21:12:52.043" Score="1" ViewCount="" Body="&lt;p&gt;I believe that chances are whichever of the three courses you decide to take in high school, there is a likely chance that you may have to take it again in college or university.  Certain exceptions are there, but it is common to make some mathematics course a requirement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That much said, algebra (hereby I am assuming elementary algebra, and not abstract algebra) is quite fundamental to doing both calculus and statistics.  &lt;/p&gt;&#xA;" OwnerUserId="2546" LastActivityDate="2011-01-25T21:12:52.043" />
  <row Id="18983" PostTypeId="1" AcceptedAnswerId="18990" CreationDate="2011-01-25T21:21:12.677" Score="10" ViewCount="407" Body="&lt;p&gt;why is $\sum\limits_{k=1}^{n} k^m$ a polynomial with degree $m+1$  in $n$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know this is well-known. But how to prove it rigorously? Even mathematical induction does not seem so straight-forward.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="6016" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2011-01-25T22:57:21.273" LastActivityDate="2011-01-27T08:05:09.720" Title="why is $\sum\limits_{k=1}^{n} k^m$ a polynomial with degree $m+1$ in $n$" Tags="&lt;combinatorics&gt;&lt;polynomials&gt;" AnswerCount="7" CommentCount="2" FavoriteCount="1" />
  <row Id="18984" PostTypeId="2" ParentId="18963" CreationDate="2011-01-25T21:27:59.393" Score="3" ViewCount="" Body="&lt;p&gt;If $4|y$, then $2|z$, which in turn, by reducing mod $4$, $2|x$, contradicting primitivity. Multiply by $2$ and factor the equation over $\mathbb{Q}(\sqrt{-23})$:&#xA;$$(\frac{4x+y+\sqrt{-23}y}{2})(\frac{4x+y-\sqrt{-23}y}{2}) = 2z^3$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that both fractions are integral. The gcd of the two factors divides $\sqrt{-23}y$ and $4x+y$. If $23|4x+y$ then $23|z$, and reducing the original equation modulo $23^2$ we see that $23|y$, hence also $23|x$, contradicting primitivity. So the gcd divides $y$ and $4x+y$, and by the above argument, it then must be either $1$ or $2$ according to wether $y$ is odd or even, respectively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, assume that $y$ is odd. So the gcd is $1$. Thus, for some ideal $I$:&#xA;$$(2x+y\frac{1+\sqrt{-23}}{2})=(2,\frac{1+\sqrt{-23}}{2})I^3$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which implies that $(2,\frac{1+\sqrt{-23}}{2})$ is principal, leading to a contradiction. Now assume that $y$ is even, and that $x$ is therefore odd and $z$ is even. Put $y=2u$, $z=2v$, $u$ odd, so that:&#xA;$$(2x+u+\sqrt{-23}u)(2x+u-\sqrt{-23}u)=16v^3$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both factors are divisble by $2$, so that:&#xA;$$(x+u\frac{1+\sqrt{-23}}{2})(x+u\frac{1-\sqrt{-23}}{2})=4v^3$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As before, the gcd is 1, and since $x+u\frac{1+\sqrt{-23}}{2}=x+u+u\frac{-1+\sqrt{-23}}{2} \in (2, \frac{-1+\sqrt{-23}}{2})$ we must have for some ideal $I$:&#xA;$$(x+u\frac{1+\sqrt{-23}}{2}) = (2, \frac{-1+\sqrt{-23}}{2})^2I^3$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Contradicting that $(2, \frac{-1+\sqrt{-23}}{2})^2$ is non-principal (the ideal above $2$ appears squared since the product of factors is divisible by $4$). We are done!&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;It actually seems that the above can be generalised, but I have to use a major theorem, which seems like it might be a bit of an overkill. Without further ado:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $aX^2+bXY+cY^2$ be a primitive non-principal quadratic form of discriminant $\Delta=b^2-4ac$, and $h$ be the class number of the associated quadratic field. Assume there is a solution $x,y,z$:&#xA;$$ax^2+bxy+cy^2=z^h$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Recalling the Chebotarev Density Theorem (OVERKILL), there is an equivalent form with $a$ an odd prime that doesn't divide $\Delta$, and since the invertible change of variables preserves primitivity, we reduce to this case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Multiplying by $a$ and factoring over $\mathbb{Q}(\sqrt{\Delta})$:&#xA;$$(ax+\frac{b+\sqrt{\Delta}}{2}y)(ax+\frac{b-\sqrt{\Delta}}{2}y)=az^h$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The gcd of the factors divides $(2ax+by,\sqrt{\Delta}y)$. Say $\Delta |2ax+by$, then since&#xA;$$(2ax+by)^2-\Delta y^2=4az^h$$&#xA;we must have $\Delta |z$, so $\Delta |y$, and finally $\Delta |x$ (unless $\Delta=\pm 2$, which is impossible), contradicting primitivity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence the gcd divides $(2a,y)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) gcd$=1$:&#xA;$$(ax+\frac{b+\sqrt{\Delta}}{2}y) = (a,\frac{b+\sqrt{\Delta}}{2})I^h$$&#xA;contradicting that the form is non-principal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) gcd$=2$ or $2a$: then $y$ is even, so $z$ is too, and since $a$ is odd, $x$ must also be even, contradicting primitivity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) gcd$=a$: thus $a|z$. If $a^2|y$, then reducing modulo $a$ we see that $a|x$. So $a||y$. Dividing this gcd out of the two factors, we must have:&#xA;$$(x+\frac{b+\sqrt{\Delta}}{2}\frac{y}{a}) = (a,\frac{b-\sqrt{\Delta}}{2})^{h-1}I^h$$&#xA;for some ideal $I$. Note the particular ideal above $a$, which appears since its conjugate cannot appear as $x$ isn't divisble by $a$. Since $(a,\frac{b-\sqrt{\Delta}}{2})^{h-1} \sim (a,\frac{b+\sqrt{\Delta}}{2})$, this contradicts that the form is non-principal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We are done!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There must be a way to avoid applying a major theorem such as Chebotarev... Please tell me if you have an idea how :D&lt;/p&gt;&#xA;" OwnerUserId="4946" LastEditorUserId="4946" LastEditorDisplayName="" LastEditDate="2011-01-25T22:36:53.707" LastActivityDate="2011-01-25T22:36:53.707" CommentCount="2" />
  <row Id="18985" PostTypeId="2" ParentId="18981" CreationDate="2011-01-25T21:28:07.647" Score="3" ViewCount="" Body="&lt;p&gt;Let the semi-axes (the 'radii') in each direction be $r_x$, $r_y$, and $r_z$, and let the depth of the liquid be $d$. Squeeze (or stretch) the ellipsoid into a sphere of unit radius, so that all of these are $1$; the depth of the liquid is then $d&amp;#39; = d/r_z$, and its volume is $\pi d&amp;#39;^2(3-d&amp;#39;)/3$, as you know from &lt;a href=&quot;http://math.stackexchange.com/questions/18914/horizontal-tank-with-hemispherical-ends-depth-to-capacity-calculation&quot;&gt;your previous question&lt;/a&gt;. Now stretch (or squeeze) the sphere back into its original shape, thus expanding the volume of the liquid by a factor $r_xr_yr_z$.&lt;/p&gt;&#xA;" OwnerUserId="1508" LastActivityDate="2011-01-25T21:28:07.647" CommentCount="2" />
  <row Id="18986" PostTypeId="2" ParentId="18983" CreationDate="2011-01-25T21:33:49.777" Score="2" ViewCount="" Body="&lt;p&gt;Check &lt;a href=&quot;http://en.wikipedia.org/wiki/Faulhaber%27s_formula&quot; rel=&quot;nofollow&quot;&gt;Faulhaber's formula&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3217" LastActivityDate="2011-01-25T21:33:49.777" />
  <row Id="18987" PostTypeId="2" ParentId="18983" CreationDate="2011-01-25T21:35:08.743" Score="1" ViewCount="" Body="&lt;p&gt;Have you seen &lt;a href=&quot;http://planetmath.org/encyclopedia/SumOfPowers.html&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; (article about  sum of powers)? In particular, one uses telescoping to prove the above result.&lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2011-01-25T21:35:08.743" />
  <row Id="18988" PostTypeId="1" CreationDate="2011-01-25T21:45:28.683" Score="0" ViewCount="20" Body="&lt;p&gt;Let $S$ be a semigroup with $I$ an ideal of $S$.  Then the Rees quotient of $S$ with respect to $I$ is the semigroup obtained by crushing $I$ to zero.  I need a reference for the following:  If $S$ is residually finite then $S/I$ is residually finite.  I looked in Clifford and Preston and didn't see anything.  Thanks!&lt;/p&gt;&#xA;" OwnerUserId="1534" LastActivityDate="2011-01-25T21:45:28.683" Title="need a reference that shows that rees quotients of residually finite semigroups are residually finite" Tags="&lt;semigroup&gt;" />
  <row Id="18989" PostTypeId="2" ParentId="18983" CreationDate="2011-01-25T21:48:22.003" Score="5" ViewCount="" Body="&lt;p&gt;The formula just drops right out if we use the &lt;a href=&quot;http://en.wikipedia.org/wiki/Euler%E2%80%93Maclaurin_formula&quot; rel=&quot;nofollow&quot;&gt;Euler McLaurin Summation Formula.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $\displaystyle f(x) = x^m$ we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \sum_{k=0}^{n} f(k) = \int_{0}^{n} f(x)\ \text{d}x + \frac{n^m}{2} + \sum_{j=0}^{\infty} \frac{B_{2j}}{(2j)!} (f^{2j-1}(n) - f^{2j-1}(0))$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $\displaystyle B_j$ are the Bernoulli numbers and $\displaystyle f^{j}(x)$ is the $\displaystyle j^{th}$ derivative of $\displaystyle f$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $\displaystyle f(x)$ is polynomial, the terms in&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \sum_{j=0}^{\infty} \frac{B_2k}{(2j)!} (f^{2j-1}(n) - f^{2j-1}(0))$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;all are zero after a point ($\displaystyle 2j-1 \gt m$) and thus we get the formula for&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle \sum_{k=0}^{m} k^m$ as a polynomial in $\displaystyle n$, with degree $m+1$.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2011-01-26T22:39:56.027" LastActivityDate="2011-01-26T22:39:56.027" CommentCount="1" />
  <row Id="18990" PostTypeId="2" ParentId="18983" CreationDate="2011-01-25T21:50:16.230" Score="14" ViewCount="" Body="&lt;p&gt;Let $V$ be the space of all polynomials $f : \mathbb{N}_{\ge 0} \to F$ (where $F$ is an irrelevant field).  Define the &lt;em&gt;forward difference operator&lt;/em&gt; $\Delta f(n) = f(n+1) - f(n)$.  It is not hard to see that the forward difference of a polynomial of degree $d$ is a polynomial of degree $d-1$, hence defines a linear operator $V_d \to V_{d-1}$ where $V_d$ is the space of polynomials of degree at most $d$.  Note that $\dim V_d = d+1$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We want to think of $\Delta$ as a discrete analogue of the derivative, so it is natural to define the corresponding discrete analogue of the integral $(\int f)(n) = \sum_{k=0}^{n-1} f(k)$.  But of course we need to prove that this actually sends polynomials to polynomials.  Since $(\int \Delta f)(n) = f(n) - f(0)$ (the &quot;fundamental theorem of discrete calculus&quot;), it suffices to show that the forward difference is surjective as a linear operator $V_d \to V_{d-1}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But by the &quot;fundamental theorem,&quot; the image of the integral is precisely the subspace of $V_d$ of polynomials such that $f(0) = 0$, so the forward difference and integral define an isomorphism between $V_{d-1}$ and this subspace.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;More explicitly, you can observe that $\Delta$ is upper triangular in the standard basis, work by induction, or use the &lt;strong&gt;Newton basis&lt;/strong&gt; $1, n, {n \choose 2}, {n \choose 3}, ...$ for the space of polynomials.  In this basis we have $\Delta {n \choose k} = {n \choose k-1}$, and now the result is &lt;em&gt;really&lt;/em&gt; obvious.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The method of finite differences provides a fairly clean way to derive a formula for $\sum n^m$ for fixed $m$.  In fact, for any polynomial $f(n)$ we have the &quot;discrete Taylor formula&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f(n) = \sum_{k \ge 0} \Delta^k f(0) {n \choose k}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and it's easy to compute the numbers $\Delta^k f(0)$ using a finite difference table and then to replace ${n \choose k}$ by ${n \choose k+1}$.  I wrote a blog post that explains this, but it's getting harder to find; I also explained it in my &lt;a href=&quot;http://web.mit.edu/~qchu/Public/TopicsInGF.pdf&quot; rel=&quot;nofollow&quot;&gt;notes on generating functions&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2011-01-26T00:18:25.323" LastActivityDate="2011-01-26T00:18:25.323" CommentCount="3" />
  <row Id="18991" PostTypeId="2" ParentId="18983" CreationDate="2011-01-25T21:58:46.720" Score="10" ViewCount="" Body="&lt;p&gt;You can set up a recursive formula for $\sum_{k=0}^n k^m $ by noting that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(n+1)^{m+1} = \sum_{k=0}^n (k+1)^{m+1}- \sum_{k=0}^n k^{m+1}$$&#xA;$$ = { m+1 \choose 1} \sum_{k=0}^n k^m&#xD;&#xA;+ { m+1 \choose 2} \sum_{k=0}^n k^{m-1} + \cdots $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;by expanding the first summation on the RHS by the binomial theorem. Then shift all the other summations except $\sum_{k=0}^n k^m $ to the LHS.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$${ m+1 \choose 1} \sum_{k=0}^n k^m = (n+1)^{m+1} - { m+1 \choose 2} \sum_{k=0}^n k^{m-1}&#xD;&#xA;- { m+1 \choose 3} \sum_{k=0}^n k^{m-2} + \cdots $$&lt;/p&gt;&#xA;" OwnerUserId="1301" LastEditorUserId="1301" LastEditorDisplayName="" LastEditDate="2011-01-25T22:04:17.257" LastActivityDate="2011-01-25T22:04:17.257" />
  <row Id="18992" PostTypeId="1" AcceptedAnswerId="19064" CreationDate="2011-01-25T22:07:11.967" Score="0" ViewCount="86" Body="&lt;p&gt;Hoping to run into some mathcadders:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a worksheet which includes the following except (and error)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/B3yNp.png&quot; alt=&quot;Mathcad fail&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any help?&lt;/p&gt;&#xA;" OwnerUserId="6247" LastActivityDate="2011-01-26T11:22:16.353" Title="MathCad Variable Undefined (even though I just defined it)" Tags="&lt;math-software&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="18993" PostTypeId="1" CreationDate="2011-01-25T22:13:07.400" Score="2" ViewCount="169" Body="&lt;p&gt;The basic problem is that I want to be extremely clear about the sets&#xA;that mathematical manipulations and operations are taking place in, I am&#xA;hoping for someone who really understands this to read what I've written&#xA;closely and point out what is getting me all mixed up, though of course&#xA;reading &amp;amp;/or responding isn't mandatory (lol) - but it is a long post even&#xA;though it's dealing with just one idea. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The set-theoretic definition of a function is f = (X,Y,F) where F is a subset&#xA;of ordered pairs of the Cartesian product of X &amp;amp; Y, (i.e. F ⊆ (X x Y) a relation).&#xA;This is Bourbaki's way of defining a function and he (they) call F the graph.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But isn't a function itself a relation and therefore musn't we write (X,Y,f)&#xA;as the set in which the function acts? To expand this out:&#xA;(X,Y,f) = (X,Y,(X,Y,F)).&#xA;I've come across notation that specifies (X,Y,f) as ((X,Y),f).&#xA;&lt;a href=&quot;http://www.math.bilgi.edu.tr/courses/Lecture%20Notes/SetTheoryLectureNotes.pdf&quot; rel=&quot;nofollow&quot;&gt;Here, page 35 of this .pdf file&lt;/a&gt;&#xA;So ((X,Y),f) = ((X,Y),((X,Y),F)) would seem to make sense.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bourbaki calls f a set &amp;amp; F it's graph but the notation in the .pdf file says&#xA;that f would be defined in the way I've explained above, i.e. that F is a&#xA;subset of XxY. The thing is that since a function f is itself a relation&#xA;shouldn't it be a relation in a set, i.e. ((X,Y),f)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming that the above is the way to think about these things, how&#xA;would I think of both F &amp;amp; f? In f = (X,Y,F), F ⊆ (X x Y) so (x,y) ∈ F or xFy,&#xA;where obviously (x∈X) &amp;amp; (y∈Y).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How about f? I think f ⊆ (X x Y) so (x,y) ∈ f or xfy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't understand how this makes sense because for the set f = (X,Y,F)&#xA;Bourbaki writes f : X → Y so for (X,Y,f) I'd have to set g = (X,Y,f) and&#xA;write g : X → Y. This is a weird conclusion but it seems to suggest itself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem of being extremely clear about what sets you are using is&#xA;particularly interesting when doing linear algebra.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The use of set-theoretic notation in linear algebra both clarifies things for&#xA;me and brings up similar questions, for a vector space V I could write&#xA;((V,+),(F,+',°),•) with the clarification that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;in (V,+) we have + : V × V → V,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;in (F,+',°) we have (+' : F × F → F) &amp;amp; ( ° : F × F → F).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In • we have (• : F × V → V) or perhaps [• : (V,+) × (F,+',°) → (V,+)]?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This notation clearly illustrates why the two operations, vector addition&#xA;and scalar multiplication are used on a vector space and the axioms for&#xA;each clearly jump out, i.e. (V,+) is abelian, (F,+',°) is a field and • isn't the&#xA;clearest to me but I think it's similar to the way that + &amp;amp; ° are related in a&#xA;field, i.e. &quot;multiplication distributes over addition&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Relating all of this to the concerns I had above in a clear manner, in the&#xA;set (F,+',°) it would make sense that +' is a set of the form (F,+'') where&#xA;+'' is a subset of the cartesian product of F x F. Similarly with °, and in the&#xA;set (V,+) you'd have something similar, also in • you'd have a crazy set&#xA;((V,+), (F,+',°), •') or including even more brackets (((V,+), (F,+',°)), •')&#xA;with •' being a subset of the cartesian product of (V,+) &amp;amp; (F,+',°).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is another problem when you want to give a vector space a norm,&#xA;would I write ((V,+),(F,+',°),•,⊗) where ⊗ : V x V → F ? Would ⊗ itself&#xA;suggest the subset ((V,+), (F,+',°), ⊗') in the manner explained above?&#xA;I don't think so because ⊗' would be the set of ordered pairs (x,a) with&#xA;x ∈ V and a ∈ F but since V x V → F you've got the map (x,x') ↦ a, it's&#xA;quite confusing tbh and need help with this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All this seems crazy but it also makes a lot of sense, I want to be very&#xA;rigorous about what I'm doing and all of the above seems to suggest itself&#xA;but it could be a lot of nonsense caused by simple confusion of a particular&#xA;issue in the post , I'm thinking that (X,Y,F) implying (X,Y,f) is the&#xA;culprit but again this idea clarifies things. If you read to this point thanks&#xA;so much, hopefully you recognise the issue.&lt;/p&gt;&#xA;" OwnerUserId="6248" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-26T04:41:13.273" LastActivityDate="2011-01-26T04:41:13.273" Title="Clarifying Notation (Sets &amp; Functions)" Tags="&lt;elementary-set-theory&gt;" AnswerCount="1" CommentCount="6" />
  <row Id="18994" PostTypeId="2" ParentId="18889" CreationDate="2011-01-25T22:19:17.877" Score="0" ViewCount="" Body="&lt;p&gt;I was just looking at this question, as I completed the same problem myself. However, I am unsure as to why you found the second moment for the final part?&lt;/p&gt;&#xA;" OwnerUserId="5324" LastActivityDate="2011-01-25T22:19:17.877" CommentCount="1" />
  <row Id="18995" PostTypeId="2" ParentId="18906" CreationDate="2011-01-25T22:22:14.083" Score="0" ViewCount="" Body="&lt;p&gt;I can't comment (not enough reputation), but as this sort of a partial answer anyway, I'll just post it as an answer. Rahula Narain's comment is correct that the area depends on more than just the sidelengths. But assuming the sidelengths are listed in the question in order around the polygon as usual (in fact, assuming any particular order on the sidelengths), the polygon is pretty constricted in terms of its shape (note that the first side is very small compared to the others, and the last two sum to about the same as the second). So it should be possible to get bounds on the area. (Even if we don't assume an order on the sidelengths, there are only three possibilities modulo things that don't affect area. And because one side is so small and one so large, the three areas should all have pretty similar bounds.) Any takers?&lt;/p&gt;&#xA;" OwnerUserId="4992" LastActivityDate="2011-01-25T22:22:14.083" />
  <row Id="18996" PostTypeId="1" AcceptedAnswerId="18999" CreationDate="2011-01-25T22:24:02.833" Score="1" ViewCount="143" Body="&lt;p&gt;After reading the wiki page on Enigma's, I am still confused as to how it works. Are there some sources that run through how the enigma machine works that is easy to follow. Which also gives information about the how codes are encrypted and decrypted? &lt;/p&gt;&#xA;" OwnerUserId="4728" LastActivityDate="2011-01-26T08:21:43.657" Title="Enigma Machine and how it works" Tags="&lt;cryptography&gt;" AnswerCount="3" />
  <row Id="18997" PostTypeId="2" ParentId="18969" CreationDate="2011-01-25T22:24:07.997" Score="0" ViewCount="" Body="&lt;p&gt;You're applying intuitive logic that works for finite sets to infinite sets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To complete your proof, you would need to construct 'consecutive' pairs of irrationals, which is not possible.&lt;/p&gt;&#xA;" OwnerUserId="6168" LastActivityDate="2011-01-25T22:24:07.997" />
  <row Id="18998" PostTypeId="2" ParentId="18214" CreationDate="2011-01-25T22:37:28.373" Score="0" ViewCount="" Body="&lt;p&gt;A ∩ B := {(x ∊ U)|(x ∊ A) ⋀ (x ∊ B)}&#xA;You can see that this is a logical statement of the form P ⋀ Q and any logic textbook will &#xA;tell you that P ⋀ Q = Q ⋀ P, i.e. that ⋀ is commutative, that seems to me to be the &#xA;justification.&lt;/p&gt;&#xA;" OwnerUserId="6248" LastActivityDate="2011-01-25T22:37:28.373" />
  <row Id="18999" PostTypeId="2" ParentId="18996" CreationDate="2011-01-25T22:42:07.370" Score="0" ViewCount="" Body="&lt;p&gt;In this (non mathematical) book &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0385495323&quot; rel=&quot;nofollow&quot;&gt;http://www.amazon.com/Code-Book-Science-Secrecy-Cryptography/dp/0385495323/ref=sr_1_1?ie=UTF8&amp;amp;qid=1295995264&amp;amp;sr=8-1&lt;/a&gt; you will certainly find what you whant.&lt;/p&gt;&#xA;" OwnerUserId="3781" LastActivityDate="2011-01-25T22:42:07.370" CommentCount="2" />
  <row Id="19000" PostTypeId="1" CreationDate="2011-01-25T22:45:22.183" Score="4" ViewCount="134" Body="&lt;p&gt;On the sound of sounding ridiculous, but in the line of &quot;There are no stupid quetsions&quot;: Is there a way to express $\omega_1$ (and in general $\omega_k$ with $k &amp;gt;= 1$ as a Conway game (that is $&amp;lt;L|R&amp;gt;$, with L and R the left and right options). And is there a way to do it such that addition, multiplication, etc. makes sense ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can express $\omega_1 = &amp;lt; {i}_{i\in\mathbb{R}} | &amp;gt;$, which is essential the same as $&amp;lt; f: \mathbb{R} \to \mathbb{R} |&amp;gt;$ where $f$ is increasing (${i}_{i\in\mathbb{R}} = f(i)$), which isn't exactly a Conway game notation. I think this also borders on an idea of Gonshor, to express the surreals as maps from initial segments of ordinals to a two-element set). In general, I have replaced the sequence $&amp;lt;0,1,2, ...|&amp;gt;$ from one of the forms of $\omega_0$ by an increasing &quot;sequence&quot; where the index is a positive real number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Questions are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The above question(s) ?&lt;/li&gt;&#xA;&lt;li&gt;Is my idea correct or at leat in the right direction ?&lt;/li&gt;&#xA;&lt;li&gt;Are there any other ways to do this ? (of course this is an open question if my idea is wrong in the first place).&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;And yes, I realize my question is little bit broad and I haven't an idea which model you should be work in (ZF, ZFC, NBG, etc), nor do I know how the answer varies with the choice of the model).&lt;/p&gt;&#xA;" OwnerUserId="6250" LastEditorUserId="6250" LastEditorDisplayName="" LastEditDate="2011-01-25T23:22:53.167" LastActivityDate="2011-01-26T00:39:06.053" Title="Notation for surreal numbers" Tags="&lt;set-theory&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="19001" PostTypeId="1" AcceptedAnswerId="19057" CreationDate="2011-01-25T22:45:36.647" Score="1" ViewCount="87" Body="&lt;p&gt;Is this possible? I have a 4-connected 17-graph and am wondering whether there are another 3 like it that can be put together to make its complement. My current 4-connected example is &lt;a href=&quot;https://docs.google.com/drawings/pub?id=1bbujUQ4rhXrWV3u30bsA_VmUlhTPNB1ZY3BmjBWfeRE&amp;amp;w=881&amp;amp;h=534&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="6249" LastActivityDate="2011-01-27T17:33:23.020" Title="edge-partitioning a strongly connected 17-graph into four 4-connected 17-graphs" Tags="&lt;graph-theory&gt;" AnswerCount="1" CommentCount="8" />
  <row Id="19002" PostTypeId="2" ParentId="18983" CreationDate="2011-01-25T23:42:10.577" Score="5" ViewCount="" Body="&lt;p&gt;Qiaochu's answer is (as always?) right on the money.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I just wanted to mention that I also wrote up some notes about &quot;discrete calculus&quot; which takes the question of finding a closed form for the sums of $k$th powers as a jumping off point.  This came from a sophomore level undergraduate course I taught on introduction to proofs, in which we were doing suspiciously many induction proofs involving such closed forms, and I came to understand belatedly that the students found these confusing.  Not because they didn't understand the mechanism of induction (at least, not for the most part), but because they couldn't help but fixate on the nontrivial question: how do you know what closed form expression to put on the right hand side?  In the context of their exercises, the answer is: &quot;You know because you are being given this information as part of the problem.&quot;  But of course that's a stupid answer: &lt;em&gt;really, how do you know&lt;/em&gt;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I wrote up &lt;a href=&quot;http://www.math.uga.edu/~pete/finitecalc.pdf&quot; rel=&quot;nofollow&quot;&gt;these notes&lt;/a&gt; on the subject of discrete calculus.  Or at least I started to: they are not quite complete and are a little more complicated than what I was expecting of the students, so I ended up not giving them out.  See especially Section 2, which explores the issue raised in the previous paragraph and also gives the extra little push (an innocuous recursive formula for the power sums) you need in order to prove by induction that $1^k + \ldots + n^k$ is a degree $k+1$ polynomial in $n$.  Note also that Qiaochu's nice linear algebra argument comes later, in Section 4 (&quot;Linear Algebra of the Discrete Derivative&quot;) and especially Theorem 11.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As usual, any suggestions or corrections on these notes are most welcome, although (as usual) I certainly do not claim that there are few enough typos for it to be worth your time to point them out to me one by one.  &lt;/p&gt;&#xA;" OwnerUserId="299" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2011-01-26T00:22:31.370" LastActivityDate="2011-01-26T00:22:31.370" />
  <row Id="19003" PostTypeId="2" ParentId="19000" CreationDate="2011-01-25T23:42:39.940" Score="4" ViewCount="" Body="&lt;p&gt;Every ordinal number $\alpha$ has a canonical copy as a surreal number $\hat\alpha$, whose left set has as members $\hat\beta$ for every $\beta\lt\alpha$, and whose right set is empty. Succinctly, &#xA;$$\hat\alpha=\{\ \{\hat\beta\mathrel{:} \beta\lt\alpha\ \}\mathrel{|} \}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One can prove by induction that $\alpha\mapsto\hat\alpha$ is isomorphism of the ordinals with their usual order and arithmetic  to the corresponding suborder of the surreals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, for the successor case, $\alpha+1$ is the next ordinal after $\alpha$, and $\hat(\alpha+1)=\{\hat\alpha\mid\}$ is the next surreal after $\hat\alpha$, or $\hat\alpha+1$ in the surreals. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, $\omega_1$ would be represented by the surreal number $\hat\omega_1$, which I think is very different from the surreal numbers that you propose.&lt;/p&gt;&#xA;" OwnerUserId="413" LastActivityDate="2011-01-25T23:42:39.940" />
  <row Id="19004" PostTypeId="2" ParentId="19000" CreationDate="2011-01-25T23:52:06.857" Score="2" ViewCount="" Body="&lt;p&gt;I should note that your expression for $\omega_1$ isn't actually correct; in part because the cardinality of the reals isn't necessarily $\aleph_1$ (this is just Cantor's Hypothesis, which of course Paul Cohen proved independent of ZF), but mostly because the surreal number you've given &amp;mdash; $\langle i_{i\in\mathbb{R}}|\rangle$ &amp;mdash; is actually equal to $\omega\stackrel{\mathrm{def}}{=}\omega_0$ (exercise: show that whoever plays first in $\langle i_{i\in\mathbb{R}}|\rangle - \omega$ loses).  As JDH notes above, the standard way of defining the ordinals copies over into the surreals to give surreal definitions of $\omega_1$, $\omega_2,$ etc; On Numbers And Games even mentions this in passing, IIRC.  That said, AFAIK the canonical arithmetic operations ($\sqrt{}$, etc) are relatively boring on ordinals of higher cardinality; in effect they suffer from (as I understand it - this isn't really my field!) predicativity issues.  You might want to have a look at the Wikipedia page on (an) 'Ordinal collapsing function' - there's a pretty close tie between surreal definitions of various ordinal-related 'numbers' and the canonical ordinal notations discussed there.&lt;/p&gt;&#xA;" OwnerUserId="785" LastEditorUserId="785" LastEditorDisplayName="" LastEditDate="2011-01-26T00:39:06.053" LastActivityDate="2011-01-26T00:39:06.053" />
  <row Id="19005" PostTypeId="1" AcceptedAnswerId="19008" CreationDate="2011-01-26T00:20:34.697" Score="0" ViewCount="152" Body="&lt;p&gt;In my stochastics class we were given the following problem (BRL stand for Borel σ-algebra on real line and R stands for real numbers):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let f : Ω → R be a function. Let F = {A ⊂ Ω : there exists B ∈&#xA;BRL with A = f^−1(B)}. Show that F is a σ-algebra on Ω.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure how I should break this down. Apparently inverse function f^-1 has some property where its target space is a σ-algebra when it's starting space is a Borel σ-algebra... or am I going down the wrong path?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've been going at this for a good while, any help is appreciated.&lt;/p&gt;&#xA;" OwnerUserId="2076" LastActivityDate="2011-01-26T01:00:37.690" Title="Sigma algebra/Borel sigma algebra proof problem" Tags="&lt;probability&gt;&lt;probability-theory&gt;&lt;sigma-algebra&gt;" AnswerCount="2" />
  <row Id="19006" PostTypeId="2" ParentId="19005" CreationDate="2011-01-26T00:27:57.170" Score="0" ViewCount="" Body="&lt;p&gt;You have to show that $F$ is (i) non-empty, (ii) closed under complementation, and (iii) closed under countable unions.&lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2011-01-26T00:27:57.170" CommentCount="2" />
  <row Id="19007" PostTypeId="1" AcceptedAnswerId="19013" CreationDate="2011-01-26T00:36:56.850" Score="0" ViewCount="242" Body="&lt;p&gt;arctan(n+1) - arctan(n+2) where n varies from 0 to infinity&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm having trouble figuring out this problem. I've calculated the first 5 terms of the series and ended up with the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-.0.32175 - 0.14189 - 0.07677 - 0.07677 - 0.04758 ......&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To me, it is apparent that the series is converging on some number (perhaps -1?) however, I'm not sure how to prove this.&#xA;Any help is appreciated!&lt;/p&gt;&#xA;" OwnerUserId="6252" LastActivityDate="2011-01-26T01:34:41.637" Title="Sum of series: arctan(n+1) - arctan(n+2) where n varies from 0 to infinity" Tags="&lt;sequences-and-series&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="19008" PostTypeId="2" ParentId="19005" CreationDate="2011-01-26T00:38:50.260" Score="1" ViewCount="" Body="&lt;p&gt;This is rather easy and has nothing to do with the fact that the target is the real line with the Borel $\sigma$-algebra.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider a function $f: X \to Y$ and let $\Sigma$ be a $\sigma$-algebra on $Y$. Let $f^{\ast}\Sigma = \{ A \subset X \,:\, \exists B \in \Sigma \; \text{such that} \; f^{-1}(B) = A\}$. This is a $\sigma$-algebra:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is easy to see that $\emptyset, X \in f^{\ast}\Sigma$ and the rest of the axioms (closure under complementation and countable unions) is a formal verification. For instance, if $A_{n} = f^{-1}B_{n}  \in f^{\ast}\Sigma$ for $n \in \mathbb{N}$ then we have that&#xA;\[&#xA;\bigcup A_{n} = \bigcup f^{-1}B_{n} = f^{-1}\bigcup B_{n} \in f^{\ast}\Sigma&#xA;\]&#xA;because $\bigcup B_{n} \in \Sigma$. Remember that taking preimages commutes with arbitrary unions: to see this, write $f^{-1}B_{1} = \{ x \in X \,: \, f(x) \in B_{1}\}$ and from this description the claim follows easily.&lt;/p&gt;&#xA;" OwnerUserId="5363" LastEditorUserId="5363" LastEditorDisplayName="" LastEditDate="2011-01-26T01:00:37.690" LastActivityDate="2011-01-26T01:00:37.690" CommentCount="2" />
  <row Id="19009" PostTypeId="2" ParentId="18964" CreationDate="2011-01-26T00:43:50.040" Score="4" ViewCount="" Body="&lt;p&gt;It seems like we agree that the openness of $\operatorname{GL}_n(\mathbb{C})$ is most easily proved by observing that it is the preimage of the open set $\mathbb{C} \setminus \{0\}$ under the continuous function $\operatorname{det}: M_n(\mathbb{C}) \rightarrow \mathbb{C}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We are giving different answers for the density, and I have not yet seen the simple linear algebra answer I was expecting to see.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here it is: let $M \in M_n(\mathbb{C})$.  It is sufficient to show that there exists $\epsilon &amp;gt; 0$ such that for all nonzero complex numbers $t$ with $|t| &amp;lt; \epsilon$, the matrix &#xA;$M + t I_n$ is nonsingular (because then $M$ is a limit point of this set of nonsingular matrices).  But viewing $t$ as an indeterminate, the determinant of $M+t I_n$ is nothing else than the characteristic polynomial $P(t)$ of $-M$.  Being a monic polynomial of degree $n$, it has $n$ roots $\alpha_1,\ldots,\alpha_n$ (possibly with multiplicity), and we may take $\epsilon = \min_{i \ | \ \alpha_i \neq 0} |\alpha_i|$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that this argument works verbatim for matrices over $\mathbb{R}$ or any nondiscrete normed field.  In fact, for an arbitrary field $k$, if one works instead with the Zariski topology on $M_n(k)$, then what we really need is for any cofinite subset of $k$ (i.e., the complement in $k$ of some finite set) to be dense in the Zariski topology.  This is true iff $k$ is infinite.  If $k$ is finite, the Zariski topology on $M_n(k)$ is discrete, and the result is false.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Added&lt;/b&gt;: A fun application of this &quot;linear algebraic perturbation perspective&quot; is to show that $\operatorname{GL}_n(\mathbb{C})$ is connected (whereas $\operatorname{GL}_n(\mathbb{R})$ is not).  Hint: by general nonsense, it is equivalent to show path-connectedness, so try that instead.  If I am trying to find a path from point $A$ to point $B$ in the complex plane, requiring that my path avoid a finite set of points is not nearly as problematic as it would be on the real line!&lt;/p&gt;&#xA;" OwnerUserId="299" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2011-01-26T00:51:58.687" LastActivityDate="2011-01-26T00:51:58.687" CommentCount="5" />
  <row Id="19010" PostTypeId="1" AcceptedAnswerId="19011" CreationDate="2011-01-26T00:50:22.133" Score="1" ViewCount="97" Body="&lt;p&gt;For example, how many times (N) do I need to flip a coin (p=0.5) to have a P=90% probability of having observed 20 heads. I empirically found that I need N=20+28=48.&#xA;Is it correct? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there an explicit formula for the Negative binomial inverse cumulative function?&lt;/p&gt;&#xA;" OwnerUserId="6240" LastActivityDate="2011-03-27T01:38:15.870" Title="Formula for the Negative binomial inverse cumulative function" Tags="&lt;probability&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="19011" PostTypeId="2" ParentId="19010" CreationDate="2011-01-26T00:54:25.010" Score="1" ViewCount="" Body="&lt;p&gt;We know that $F(k) = P(X \leq k) = 1-I_{p}(k+1,r)$ where $I_{p}(k+1,r)$ is the &lt;a href=&quot;http://mathworld.wolfram.com/IncompleteBetaFunction.html&quot; rel=&quot;nofollow&quot;&gt;incomplete beta function&lt;/a&gt;. So $0.9 = 1-I_{0.5}(k+1,r)$. From this, I believe you have enough information to get $k$ and $r$ and hence $n$.&lt;/p&gt;&#xA;" OwnerUserId="2548" LastEditorUserId="2548" LastEditorDisplayName="" LastEditDate="2011-01-26T00:59:59.373" LastActivityDate="2011-01-26T00:59:59.373" CommentCount="1" />
  <row Id="19013" PostTypeId="2" ParentId="19007" CreationDate="2011-01-26T01:09:46.010" Score="3" ViewCount="" Body="&lt;p&gt;If you are looking for the sum of the series&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;\sum_{n=0}^\infty (\arctan(n+1)-\arctan(n+2))&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;then what you have is a telescoping series.  This means that terms will cancel.  The first part of the sum is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;\arctan(1)-\arctan(2)+\arctan(2)-\arctan(3)+\arctan(3)-\arctan(4)+\cdots&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The terms in the middle will cancel, for instance:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;-\arctan(2)+\arctan(2)=0&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus your $n$th partial sum is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;S_n=\arctan(1)-\arctan(n+2)&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Taking the limit gives&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;\lim_{n\rightarrow \infty}(\arctan(1)-\arctan(n+2))=\arctan(1)-\frac{\pi}{2}.&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And $\arctan(1)=\frac{\pi}{4}$.&lt;/p&gt;&#xA;" OwnerUserId="2838" LastActivityDate="2011-01-26T01:09:46.010" CommentCount="1" />
  <row Id="19014" PostTypeId="2" ParentId="19007" CreationDate="2011-01-26T01:34:41.637" Score="1" ViewCount="" Body="&lt;p&gt;Of course, what makes this (clearly telescoping) series interesting is that it can be written in not-so-obviously telescoping form; using the formula $\mathrm{arctan}(u)$ + $\mathrm{arctan}(v)$ = $\mathrm{arctan}\bigl({u+v\over 1-uv}\bigr)$, we can write it as $\sum_{n=0}^{\infty}\mathrm{arctan}{1\over (n+1)(n+2)-1}$ &amp;mdash; or, shifting indices by one, $\sum_{n=1}^{\infty}\mathrm{arctan}{1\over n^2+n-1}$.&lt;/p&gt;&#xA;" OwnerUserId="785" LastActivityDate="2011-01-26T01:34:41.637" />
  <row Id="19015" PostTypeId="1" CreationDate="2011-01-26T01:44:38.727" Score="1" ViewCount="81" Body="&lt;p&gt;Hello. I have this question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Prove/disprove: The set of natural numbers (including zero) with usual topology is the continuous image of the Sorgenfrey line.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can't we take the map $g: \mathbb{R}_{l} \rightarrow \mathbb{N}$ given by $g(x) = |[x]|$ ? (i.e the absolute value of the floor function).&lt;/p&gt;&#xA;" OwnerUserId="6253" LastActivityDate="2011-01-26T02:23:29.707" Title="Is N the cts image of the Sorgenfrey line?" Tags="&lt;topology&gt;" AnswerCount="2" />
  <row Id="19016" PostTypeId="1" AcceptedAnswerId="19027" CreationDate="2011-01-26T01:49:40.733" Score="3" ViewCount="69" Body="&lt;p&gt;Hi, if $R=k[t,t^{-1}]$ is a graded ring where $R_0=k$ is a field and $t\in R$ is a homogeneous element of positive degree which is trascendental over $k$, how can I prove that every graded $R$-module is free?&lt;/p&gt;&#xA;" OwnerUserId="6236" LastActivityDate="2011-01-26T21:13:56.490" Title="graded modules over $k[t,t^{-1}]$" Tags="&lt;commutative-algebra&gt;" AnswerCount="2" />
  <row Id="19017" PostTypeId="2" ParentId="19015" CreationDate="2011-01-26T01:56:03.603" Score="1" ViewCount="" Body="&lt;p&gt;Yes we can.  Is your question how to show that it is continuous?  Is the inverse image of $\{k\}$ open in $\mathbb{R}_l$?&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2011-01-26T01:56:03.603" CommentCount="2" />
  <row Id="19018" PostTypeId="1" AcceptedAnswerId="19019" CreationDate="2011-01-26T01:56:37.813" Score="1" ViewCount="96" Body="&lt;p&gt;This is a problem that I'm attempting to solve (NOT homework!):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are 10 trees. Each one is planted in the center of the circle of space that has a diameter of 16. The 10 circles (each with diameter of 16) are placed in the shape of an equilateral triangle (4 circles, then 3 circles on top, 2 circles on top, 1 circle on top) so that each side of the triangle is made up of 4 circles. The circles are as close together as they can possibly be without overlapping.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am drawing a triangle around the &quot;triangle of circles&quot;. My triangle has sides that are as close to the circles as possible without overlapping them. What is the length of the sides of the outer triangle? What is the area of that triangle?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would appreciate also if you explain how you came to your answer. Thank you!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a picture of the word problem I am describing:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/MlHYy.png&quot; alt=&quot;Picture of word problem&quot;&gt; &lt;/p&gt;&#xA;" OwnerUserId="6237" LastEditorUserId="6237" LastEditorDisplayName="" LastEditDate="2011-01-26T02:11:10.233" LastActivityDate="2011-01-26T02:11:10.233" Title="Area of equilateral triangle around circles placed in equilateral triangle" Tags="&lt;geometry&gt;" AnswerCount="1" />
  <row Id="19019" PostTypeId="2" ParentId="19018" CreationDate="2011-01-26T02:08:11.957" Score="2" ViewCount="" Body="&lt;p&gt;If the radius of circle is $R$, then I get the length of the triangle as $(6+2\sqrt{3})R$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Drop a perpendicular from the center of the rightmost bottom circle to the bottom side.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The distance from the bottom right vertex of the triangle to the base of the perpendicular is $D = \sqrt{3}\ R$, as we get a triangle of angle $30^{\circ}$ and we get $\tan(30^{\circ}) = \frac{R}{D}$. Since $\tan(30^{\circ}) = \frac{1}{\sqrt{3}}$, we get $D = \sqrt{3}\ R$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now drop perpendiculars from the centers of the bottom circles to the bottom side and add up the lengths of the segments formed on the bottom side.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Area of an equilateral triangle of side $a$ is $\sqrt{3}\ a^2/4$.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2011-01-26T02:08:11.957" />
  <row Id="19020" PostTypeId="2" ParentId="19015" CreationDate="2011-01-26T02:23:29.707" Score="2" ViewCount="" Body="&lt;p&gt;The OP seems to have a pretty good handle on the question and maybe just needs to think it through to be fully satisfied.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But here is a slightly different approach which makes it even more clear (to me) that the OP's method will work.  This was borne out of the remark I made (to myself) upon reading the OP's clarification that the natural numbers include zero (you're damn right they do, by the way): namely, it certainly doesn't matter, because any two countably infinite discrete spaces are homeomorphic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Taking that one step further, it is clearly enough to realize $\mathbb{Z}$ with the discrete topology as a continuous image of the Sorgenfrey line: having done this, compose with any homeomorphism (i.e., bijection!) from $\mathbb{Z}$ to $\mathbb{N}$.  (Or, in fact, with any surjection, as the OP has done.)  For this, literally take the greatest integer function.  The preimage on any given basis element -- i.e., a singleton set $\{n\}$ -- is the half-open interval $[n,n+1)$.  I don't keep too much information about the Sorgenfrey line in my head, but I'm pretty sure those sets are open!  &lt;/p&gt;&#xA;" OwnerUserId="299" LastActivityDate="2011-01-26T02:23:29.707" />
  <row Id="19021" PostTypeId="1" AcceptedAnswerId="19029" CreationDate="2011-01-26T02:27:19.923" Score="2" ViewCount="159" Body="&lt;p&gt;This is a little exercise found in Robin Hartshorne's Euclid: Geometry and Beyond:&#xA;&lt;img src=&quot;http://i.imgur.com/33IeK.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe I have found a solution, which only exists when $AB$ has length at least that of the chord bisected by $P$, and length at most the diameter of the circle. So suppose that is the case. I do the following construction. $\odot cXrYZ$ is an abbreviation for the circle centered at $X$ with radius $YZ$. (I assume the compass is not collapsible.) Also, it's not necessary to know my solution to answer my question, so feel free to skip.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Pick $C$ at random on circle $O$. Describe the circle with center $C$ and radius $AB$. Get intersection $D$.&lt;/li&gt;&#xA;&lt;li&gt;Describe the circle with center $D$ and radius $DC$. This will intersect with the previous circle at $E$ and $F$, say.&lt;/li&gt;&#xA;&lt;li&gt;Connect $EF$.&lt;/li&gt;&#xA;&lt;li&gt;Connect $CD$, get intersection $G$.&lt;/li&gt;&#xA;&lt;li&gt;Draw circle with center $O$ and radius $OG$. (This gives the circle to which all chords of length $AB$ will be tangent to.)&lt;/li&gt;&#xA;&lt;li&gt;Connect $PO$. Get $H$, the intersection of $PO$ with $\odot cRrOG$. We know this exists, since $P$ must fall outside $\odot cOrOG$.&lt;/li&gt;&#xA;&lt;li&gt;$\odot cHrPH$. Get $J$.&lt;/li&gt;&#xA;&lt;li&gt;$\odot cPrPJ$. &lt;/li&gt;&#xA;&lt;li&gt;$\odot cJrJP$, which will intersect with $\odot cPrPJ$ at $K$ and $L$.&lt;/li&gt;&#xA;&lt;li&gt;$\odot cOrPO$.&lt;/li&gt;&#xA;&lt;li&gt;Extend $LK$ through to $M$, the intersection with $\odot cOrPO$.&lt;/li&gt;&#xA;&lt;li&gt;Connect $MO$, get $N$, the intersection on $\odot cOrOH$.&lt;/li&gt;&#xA;&lt;li&gt;Extend $PN$ through to intersect the original circle to get $Q$ and $R$. This line is congruent to $AB$ and passes through $P$.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I apologize that this construction may be hard to follow without the picture. This takes me $13$ steps, but I see that Hartshorne suggests it's possible in $5$, hence the (par=$5$). Can anyone produce a more efficient solution in only $5$ steps? I've been thinking about it, but don't see how to make this any simpler from a compass and straightedge construction.&lt;/p&gt;&#xA;" OwnerUserId="1583" LastActivityDate="2011-01-26T11:33:36.377" Title="Given a line segment $AB$ and a point $P$ inside a circle, can one construct a chord through $P$ congruent to $AB$?" Tags="&lt;euclidean-geometry&gt;&lt;puzzle&gt;" AnswerCount="3" CommentCount="5" />
  <row Id="19022" PostTypeId="1" CreationDate="2011-01-26T02:30:18.483" Score="0" ViewCount="81" Body="&lt;p&gt;how does one solve &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;2e^x(x-1)=0&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="6254" LastActivityDate="2011-01-26T04:32:33.177" Title="Solving a exponential equation" Tags="&lt;logarithms&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="19023" PostTypeId="2" ParentId="18993" CreationDate="2011-01-26T02:31:41.033" Score="2" ViewCount="" Body="&lt;p&gt;I first restate just the parts pertaining to your first question: if $X$ and $Y$ are sets, and $F \subseteq X \times Y$ is a relation with the properties one wants of a function (namely: for all $x \in X$ there is $y \in Y$ with $(x,y) \in F$, and for all $x \in X$ and $y_1, y_2 \in Y$ we have that $(x,y_1) \in F$ and $(x, y_2) \in F$ implies $y_1 = y_2$), some people do indeed view the triple $f = (X,Y,F)$ as &lt;em&gt;being&lt;/em&gt; the the function, and define a function to be such a triple.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You then ask: &quot;But isn't a function itself a relation and therefore musn't we write $(X,Y,f)$ as the set in which the function acts?&quot; The answer to the first part of this question--- &quot;isn't a function itself a relation&quot;--- meaning, isn't the function itself the set $F$--- is &lt;em&gt;no&lt;/em&gt; (and hence the second half of the question is dispensed with entirely).  In the definition you have stated, a function is &lt;em&gt;not&lt;/em&gt; a relation, but an ordered triple whose entries are a set $X$, then another set $Y$, and then finally a relation between $X$ and $Y$ (ie, a subset of $X \times Y$) with the properties one wants of a function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To get a sense of the purpose of this: think of the notion of &lt;em&gt;equality of functions&lt;/em&gt; that it produces: to say that $(X,Y,F)$ and $(X&amp;#39;,Y&amp;#39;,F&amp;#39;)$ are equal is to say more than just $F = F&amp;#39;$, but also that $X=X&amp;#39;$ and $Y = Y&amp;#39;$.  The purpose of this definition of a function is to make explicit the roles of the sets $X$ and $Y$ in the specification of a function.  Note that if you are given a function $(X,Y,F)$, you can infer what $X$ is from $F$ alone ($F$ is a set of ordered pairs, and by the defining properties of a function, $X$ is nothing more than the set of first entries of elements of $F$) but you &lt;em&gt;cannot&lt;/em&gt; infer what $Y$ is from $F$ alone.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;An example might help: let $X = \{0,1\}$ and let $Y = X$ and let $Z$ denote the subset $\{0\}$ of $X$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $f: X \to Z$ denote the only possible function (namely the one whose rule is given by $f(x) = 0$ for both possible values of $x$).  Let $g: X \to Y$ denote the function given by $g(0) = 0$ and $g(1) = 0$.  If you think about it for a moment you see that the rules of $f$ and $g$ are both encoded by the same set of ordered pairs, namely $F = \{(0,0), (1,0)\}$.  The only difference between the two is that in one case we are regarding $Z$ as the codomain, and in the other we are regarding the larger set $Y$ as the codomain.  With the above set-theoretic definition of &quot;function&quot;, $f$ and $g$ are &lt;em&gt;different&lt;/em&gt; functions, because one is the set $(X,X,F)$ and the other is the set $(X,Z,F)$, and these two ordered triples differ in their second coordinate, so they are different ordered triples, so the functions (by this definition) are different.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you think of the set $F$ as &quot;being&quot; the function, then you cannot distinguish between $f$ and $g$ (and more generally, the concept of &quot;codomain&quot; becomes hazy).  Of course, for many applications of the function concept there is no conceivable reason to distinguish between $f$ and $g$--- but most mathematicians, I think, would at least want the &lt;em&gt;option&lt;/em&gt; of seeing $f$ and $g$ as different, and if you want to formalize functions within set theory, then, the definition above allows you to do so.  (Some people &lt;em&gt;do&lt;/em&gt; take as a definition that a function from $X$ to $Y$ is a subset of $X \times Y$ with the defining properties of a function; it is a perfectly good definition but it does not allow one to distinguish between $f$ and $g$ above, and the concept of codomain is not encoded in this definition, although if you are not formally minded, you might not notice.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to reiterate: people generally do not &lt;em&gt;think of&lt;/em&gt; functions as being ordered triples, consisting of the domain set, the codomain set, and then the relation giving the rule--- it's very common to think of the function as just being the rule.  But if you want to &lt;em&gt;formalize&lt;/em&gt; the notion, you quickly find that equating a function with the relation that determines its rule leaves a useful distinction (namely the concept of codomain, or the set a function is &quot;to&quot;) out.  So the purpose of the Bourbaki definition is to explicitly build it in: a function is not just a relation with the defining properties one wants of a function, but it is &lt;em&gt;an explicit choice of domain and codomain&lt;/em&gt; together with this data.  (As I remarked earlier, the domain could actually be inferred from the relation, so this is somewhat &quot;redundant&quot;--- but the object of this definition is merely to formalize a way of viewing functions within set theory, not to do so in a &quot;minimal&quot; way.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hopefully the rest of your questions are slightly cleared up by this.&lt;/p&gt;&#xA;" OwnerUserId="3840" LastEditorUserId="3840" LastEditorDisplayName="" LastEditDate="2011-01-26T02:39:22.770" LastActivityDate="2011-01-26T02:39:22.770" CommentCount="5" />
  <row Id="19025" PostTypeId="2" ParentId="19022" CreationDate="2011-01-26T02:33:44.627" Score="0" ViewCount="" Body="&lt;p&gt;Note that $e^x$ is always positive. So if your equation is $e^{x(x-1)} = 0$ then it has no solution. (Not over the reals, that is.)&lt;/p&gt;&#xA;" OwnerUserId="5189" LastActivityDate="2011-01-26T02:33:44.627" CommentCount="2" />
  <row Id="19026" PostTypeId="2" ParentId="19016" CreationDate="2011-01-26T02:58:03.753" Score="2" ViewCount="" Body="&lt;p&gt;One way:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A graded $H=k[t^{\pm1}]$-module $M$ is the same thing as Hopf $H$-module (so it is an $H$-module, and $H$-comodule, and both structures are compatible). The &lt;em&gt;Fundamental Theorem of Hopf modules&lt;/em&gt; tells you, then, that such a beast is free as an $H$-module.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The best reference for this is, in my opinion, Sweedler's book [Sweedler, Moss E. Hopf algebras. Mathematics Lecture Note Series W. A. Benjamin, Inc., New York 1969 vii+336 pp. &lt;a href=&quot;http://www.ams.org/mathscinet-getitem?mr=MR0252485&quot; rel=&quot;nofollow&quot;&gt;MR0252485&lt;/a&gt;]&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2011-01-26T02:58:03.753" CommentCount="1" />
  <row Id="19027" PostTypeId="2" ParentId="19016" CreationDate="2011-01-26T03:09:49.603" Score="2" ViewCount="" Body="&lt;p&gt;A low-tech way:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $M$ be a graded $k[t^{\pm1}]$-module, so that in particular $M=\bigoplus_{n\in\mathbb Z}M_i$ as a vector space. For each $n\in\mathbb Z$ the map $M_n\to M_{n+1}$ given by multiplication by $t$ is a linear bijection, with inverse given by multiplication by $t^{-1}$, of course. It follows that for all $n\in\mathbb Z$ we have $M_n=t^nM_0$. Moreover,  it is easy to see that the map $k[t^{\pm1}]\otimes_k M_0\to M$ obtained by restricting the multiplication map $k[t^{\pm1}]\otimes_k M\to M$ to the subspace $k[t^{\pm1}]\otimes_k M_0$ is in fact an isomorphism of $k[t^{\pm1}]$-modules, if we see its domain as a $k[t^{\pm1}]$-module in the obvious way. But this obvious module is free: any basis of $M_0$ gives a basis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;N.B.:&lt;/strong&gt; This is, I guess, what the high-tech proof amounts to in this concrete situation...&lt;/p&gt;&#xA;" OwnerUserId="274" LastEditorUserId="274" LastEditorDisplayName="" LastEditDate="2011-01-26T21:13:56.490" LastActivityDate="2011-01-26T21:13:56.490" CommentCount="1" />
  <row Id="19028" PostTypeId="2" ParentId="19021" CreationDate="2011-01-26T03:11:17.263" Score="2" ViewCount="" Body="&lt;p&gt;Shorter, but not down to par:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;construct the circle centered at A with the same radius as circle O&lt;/li&gt;&#xA;&lt;li&gt;construct the circle centered at B with the same radius as circle O&lt;/li&gt;&#xA;&lt;li&gt;join the intersection points of these two circles; call one of these intersection points C and call the intersection that is the midpoint of segment AB D&lt;/li&gt;&#xA;&lt;li&gt;construct the circle centered at O with radius CD (note that the desired chord is the chord through P that is tangent to this circle)&lt;/li&gt;&#xA;&lt;li&gt;construct the circle centered at O with radius OP&lt;/li&gt;&#xA;&lt;li&gt;construct the circle centered at P with radius OP&lt;/li&gt;&#xA;&lt;li&gt;join the intersection points of these two circles, which will intersect OP at its midpoint&lt;/li&gt;&#xA;&lt;li&gt;construct the circle centered at the midpoint of OP and passing through O and P&lt;/li&gt;&#xA;&lt;li&gt;draw the line through P and one of the two intersection points of the circles in steps 4 and 8; this line contains the desired chord.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;edit&lt;/em&gt; Having seen &lt;a href=&quot;http://math.stackexchange.com/questions/19021/given-a-line-segment-ab-and-a-point-p-inside-a-circle-can-one-construct-a-ch/19029#19029&quot;&gt;Rahul Narain's answer&lt;/a&gt;, I was able to apply a similar idea but starting from my basic idea in steps 1-2 above, which constructed the center of a circle congruent to circle O for which AB is a chord, to get down to par.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;construct the circle centered at A with the same radius as circle O&lt;/li&gt;&#xA;&lt;li&gt;construct the circle centered at B with the same radius as circle O&lt;/li&gt;&#xA;&lt;li&gt;construct the circle centered at an intersection of the circles from steps 1-2 (if AB is too long, this intersection point will not exist) with radius OP; label as C one of its points of intersection with AB (if P is too close to O, this intersection will not exist)&lt;/li&gt;&#xA;&lt;li&gt;construct the circle centered at P with radius AC; label as Q one of its intersections with circle O&lt;/li&gt;&#xA;&lt;li&gt;construct the line through P and Q, which contains the desired chord&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="72" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2011-01-26T11:33:36.377" LastActivityDate="2011-01-26T11:33:36.377" CommentCount="1" />
  <row Id="19029" PostTypeId="2" ParentId="19021" CreationDate="2011-01-26T03:15:29.060" Score="4" ViewCount="" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;1. Pick an arbitrary point &lt;em&gt;C&lt;/em&gt; on the circle. Draw an arc centered at &lt;em&gt;C&lt;/em&gt; and radius &lt;em&gt;AB&lt;/em&gt;, intersecting the original circle at point &lt;em&gt;D&lt;/em&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If there is no intersection, stop: &lt;em&gt;AB&lt;/em&gt; is too long for a solution to exist.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;2. Draw line segment &lt;em&gt;CD&lt;/em&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We now have a chord which is congruent to &lt;em&gt;AB&lt;/em&gt;. We need to find a congruent chord which passes through &lt;em&gt;P&lt;/em&gt;, so the first thing we need to do is figure out what point on &lt;em&gt;CD&lt;/em&gt; will correspond to &lt;em&gt;P&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;3. Draw an arc with center &lt;em&gt;O&lt;/em&gt; and radius &lt;em&gt;OP&lt;/em&gt;. Let this intersect &lt;em&gt;CD&lt;/em&gt; in point &lt;em&gt;E&lt;/em&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If there is no intersection, stop: &lt;em&gt;AB&lt;/em&gt; is too short for a solution to exist. Otherwise, &lt;em&gt;E&lt;/em&gt; is the point we needed. &lt;em&gt;OE&lt;/em&gt; is congruent to &lt;em&gt;OP&lt;/em&gt;, and &lt;em&gt;CED&lt;/em&gt; is congruent to the desired chord. Now if we find the point &lt;em&gt;Q&lt;/em&gt; that &lt;em&gt;C&lt;/em&gt; corresponds to, we'll have the chord we want.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;4. Draw an arc with center &lt;em&gt;P&lt;/em&gt; and radius &lt;em&gt;EC&lt;/em&gt; to intersect the original circle at point &lt;em&gt;Q&lt;/em&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The (undrawn) triangles &lt;em&gt;OPQ&lt;/em&gt; and &lt;em&gt;OEC&lt;/em&gt; are congruent, so &lt;em&gt;PQ&lt;/em&gt; and &lt;em&gt;EC&lt;/em&gt; make the same angle with the radius, and that determines the length of the chord.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;5. Draw line segment &lt;em&gt;PQ&lt;/em&gt;, extended beyond &lt;em&gt;P&lt;/em&gt; to meet the circle on the other side.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is the desired chord. It is congruent to &lt;em&gt;CD&lt;/em&gt;, which is congruent to &lt;em&gt;AB&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="856" LastEditorUserId="856" LastEditorDisplayName="" LastEditDate="2011-01-26T05:20:46.313" LastActivityDate="2011-01-26T05:20:46.313" CommentCount="6" />
  <row Id="19030" PostTypeId="2" ParentId="19021" CreationDate="2011-01-26T03:24:10.980" Score="1" ViewCount="" Body="&lt;ol&gt;&#xA;&lt;li&gt;Draw the circle $cOP$ with radius $OP$ and center $O$.&lt;/li&gt;&#xA;&lt;li&gt;Draw ray $OP$; let it intersect the original circle at $Q$.&lt;/li&gt;&#xA;&lt;li&gt;Draw an arc $cQrAB$; let it intersect the original circle at $R$.  (If it has no intersection with the original circle, stop; $AB$ is longer than any chord of the given circle.)&lt;/li&gt;&#xA;&lt;li&gt;Draw the chord $QR$.  Let it intersect the inner circle $cOP$ at $P&amp;#39;$.  (If it does not intersect the inner circle, stop; $AB$ is shorter than any chord of the given circle that passes through $P$.)  Now the chord $QP&amp;#39;R$ is congruent to $AB$, and $OP&amp;#39;$ is congruent to $OP$.&lt;/li&gt;&#xA;&lt;li&gt;Draw ray $OP&amp;#39;$; let it intersect the original circle at $Q&amp;#39;$.&lt;/li&gt;&#xA;&lt;li&gt;Extend the line $PQ&amp;#39;$ to construct the desired chord.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="5546" LastEditorUserId="5546" LastEditorDisplayName="" LastEditDate="2011-01-26T03:31:16.793" LastActivityDate="2011-01-26T03:31:16.793" CommentCount="1" />
  <row Id="19031" PostTypeId="1" AcceptedAnswerId="19034" CreationDate="2011-01-26T04:20:51.823" Score="3" ViewCount="102" Body="&lt;p&gt;Hi, &#xA;I am looking for information for some extended family members. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A medical procedure has a 4% chance of stroke per year over 5 years. &#xA;One family member has said that that is a 20% chance of stroke over all (5*4%). &#xA;Others argue that it is less because 4% each year is smaller. &#xA;I am not sure how to frame this question in order to get an accurate answer. &#xA;Is the chance of something happening being 4% a year for 5 years the same as say an instantaneous 20% chance of something happening?&lt;/p&gt;&#xA;" OwnerUserId="6256" LastActivityDate="2011-01-26T04:40:35.517" Title="20% in one year vs 4% per year over 5 years" Tags="&lt;probability&gt;&lt;statistics&gt;" AnswerCount="1" />
  <row Id="19032" PostTypeId="1" AcceptedAnswerId="19042" CreationDate="2011-01-26T04:26:06.840" Score="0" ViewCount="163" Body="&lt;p&gt;I am having trouble solving this problem.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Suppose $f(x) \in \mathbb{Z}[x]$, and $f(a) \cdot f(b) = -(a-b)^{2}$ for distinct $a,b \in \mathbb{Z}$, then how do we prove that $f(a)+f(b)=0$?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Don't know how to proceed.&lt;/p&gt;&#xA;" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-26T05:06:40.877" LastActivityDate="2011-01-26T15:16:03.247" Title="Proving if $f(x) \in \mathbb{Z}[x]$ and $f(a) \cdot f(b) = -(a-b)^{2}\neq 0$ then $f(a)+f(b)=0$." Tags="&lt;number-theory&gt;" AnswerCount="2" CommentCount="5" />
  <row Id="19033" PostTypeId="2" ParentId="19022" CreationDate="2011-01-26T04:32:33.177" Score="3" ViewCount="" Body="&lt;p&gt;If $y$ and $z$ are Real (Complex) numbers, then $yz = 0$ exactly when at least one of $y = 0$ or $z = 0$.  Since $x \mapsto 2e^x$ and $x \mapsto x-1$ both are maps from Real numbers to Real numbers, $2e^x(x-1) = 0$ will be true exactly when at least one of the following is true:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) $2e^x = 0$ $\Leftrightarrow$ $2 = 0$ or $e^x = 0$ (both impossible)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) $x - 1 = 0$ $\Leftrightarrow$ $x = 1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so your only solution is $x = 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that this is the same reason we factor the quadratic polynomial in the equation $ax^2 + bx +c = 0$, mainly so that we can break up the equation into two linear ones that are easily solved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also since I've seen this done before, $yz = r$ does &lt;B&gt;NOT necessarily&lt;/B&gt; mean that either $y = r$ or $z = r$ when $r \neq 0$.  The reason $0$ holds a special place here is because if say $y \neq 0$, then we may multiply both sides by $1/y$ to get $z = r/y$, which is of course $0$ exactly when $r = 0$.&lt;/p&gt;&#xA;" OwnerUserId="4693" LastActivityDate="2011-01-26T04:32:33.177" />
  <row Id="19034" PostTypeId="2" ParentId="19031" CreationDate="2011-01-26T04:35:10.560" Score="3" ViewCount="" Body="&lt;p&gt;If I interpret your question correctly, you want to know the probability of a stroke occurring in $5$ years given that there is a constant $4\%$ chance of stroke each year?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using complementary probability is the best approach here. Since there is a $4\%$ chance of a stroke happening each year, there is a $96\%$ chance of a stroke not happening each year. Then there is a $(.96)^5\approx.8153$, or $81.53\%$ chance that no stroke happens in those $5$ years. This means that there is a $100\%-81.53\%=18.47\%$ chance of stroke happening in those $5$ years, not $20\%$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Think about this also, if something has a $4\%$ chance of happening each year, does this mean there is a $25\cdot 4\%=100\%$ chance that it happens in $25$ years? It doesn't make very much intuitive sense that a relatively rare event is guaranteed to happen in a certain time span.&lt;/p&gt;&#xA;" OwnerUserId="1583" LastEditorUserId="1583" LastEditorDisplayName="" LastEditDate="2011-01-26T04:40:35.517" LastActivityDate="2011-01-26T04:40:35.517" CommentCount="3" />
  <row Id="19036" PostTypeId="2" ParentId="18944" CreationDate="2011-01-26T04:40:08.803" Score="3" ViewCount="" Body="&lt;p&gt;Let's first count the probability that exactly $R$ runs of $M$ successes happen in $N$ trials, and the last trial fails. Denote by $p$ the probability of success, and by $q$ the probability of failure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each set of runs can be &quot;tokenized&quot; into parts of the form $S^kF$ (this is because the last trial fails). If $k &amp;lt; M$ then we count no runs, and otherwise we count $k - M + 1$ runs. We're going to form a bivariate generating function, with $x$ being the total number of trials, and $y$ being the number of runs. Each part contributes the following:&#xA;$$P(x,y) = xq(1 + px + \cdots + (px)^{M-1} + (px)^My + (px)^{M+1}y^2 + \cdots).$$&#xA;We can get an explicit formula for $P(x,y)$ using geometric series identities:&#xA;$$P(x,y) = xq\left(\frac{(px)^{M-1}-1}{px-1} + \frac{(px)^{M-1}}{1-pxy}\right).$$&#xA;The required probability is the coefficient of $x^N y^R$ in $1/(1-P(x,y))$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let's do the same when the last trial succeeds. The last piece is then $P(x,y)/xq$, and so we want the coefficient of $x^N y^R$ in $(P(x,y)/xq)/(1-P(x,y))$. In total, the required probability is the coefficient of $x^N y^R$ in $$ \frac{1+P(x,y)/xq}{1-P(x,y)}. $$&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2011-01-26T04:40:08.803" CommentCount="2" />
  <row Id="19037" PostTypeId="1" AcceptedAnswerId="19051" CreationDate="2011-01-26T05:23:57.993" Score="2" ViewCount="79" Body="&lt;p&gt;Let $X = [0,1]^{\omega_1}$ (that is, all functions from the set of countable ordinals to the unit interval) have the product topology. Let $Y$ be all functions in $X$ with countable support (that is, all points $(x_\alpha) \in X$ with $x_\alpha = 0$ for all but countably many $\alpha \in \omega_1$). I am trying to find an example to illustrate a somewhat technical point and, if I am correct in thinking that $Y$ as above is nonmeagre, then my job will be done. Unfortunately, I'm not sure how to approach this. Certainly I don't see any obvious way to express $Y$ as a countable union of nowhere dense subsets of $X$...&lt;/p&gt;&#xA;" OwnerUserId="6257" LastActivityDate="2011-01-26T08:42:06.790" Title="Am I correct in thinking this set is nonmeagre?" Tags="&lt;topology&gt;" AnswerCount="1" />
  <row Id="19038" PostTypeId="1" CreationDate="2011-01-26T05:45:32.730" Score="1" ViewCount="100" Body="&lt;p&gt;C^1[0,1] (space of all real valued continous function which are continously diffrentiable on (0,1) and whose derivative can be continously extended to [0,1]),for f is in that set define norm of f = max{ modulas of f(t), modulas of f'(t)} where t is in [0,1].can you show this space  is complete under this norm?and let a linear operator  T:C^1[0,1]--&gt; C[0,1] defined by T(f)=f',show that T is continous and Norm of T=1  &lt;/p&gt;&#xA;" OwnerUserId="6990" LastActivityDate="2011-01-26T06:05:47.480" Title="space of all continous functions with a norm" Tags="&lt;functional-analysis&gt;" AnswerCount="1" />
  <row Id="19039" PostTypeId="1" AcceptedAnswerId="19043" CreationDate="2011-01-26T05:55:38.550" Score="0" ViewCount="103" Body="&lt;p&gt;This should be pretty simple.  Here is what I want to put into a formula/algorithm:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Divide the size of every object of an array by the number of times that object has been accessed in the last &lt;em&gt;n&lt;/em&gt; seconds and &lt;strong&gt;discarding the highest quotient.&lt;/strong&gt; (You can probably ignore the bold part)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can program this just fine, but I am going to be showing it to a Mathematician (one of my professors) and I would like to impress him a bit.  If the description of what I want is not clear and an example in code would help let me know and I'll whip something up real quick.&lt;/p&gt;&#xA;" OwnerUserId="6096" LastActivityDate="2011-01-26T07:19:25.237" Title="I want to put my words into a formula/algorithm but I lack the mathematical fortitude" Tags="&lt;algorithms&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="19040" PostTypeId="2" ParentId="19038" CreationDate="2011-01-26T06:05:47.480" Score="4" ViewCount="" Body="&lt;p&gt;You probably mean $\|f\| =\max{\{\|f\|_{\infty},\|f&amp;#39;\|_{\infty}\}}$ where $\|f\|_{\infty} = \sup_{t \in [0,1]} |f(t)|$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint:&lt;/strong&gt; Recall the basic fact that if $f_{n}$ is a sequence of $C^{1}$ functions such that $f_{n}$ converges uniformly (or only pointwise) to a continuous function $f$ and $f_{n}&amp;#39;$ converges uniformly to a continuous function $g$ then $f \in C^{1}$ and $f&amp;#39; = g$. It's probably a good idea to try to prove this on your own without consulting your  books.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That $\|T\| \leq 1$ is just the definition of the operator norm. Consider also $f(x) = x$.&lt;/p&gt;&#xA;" OwnerUserId="5363" LastActivityDate="2011-01-26T06:05:47.480" />
  <row Id="19041" PostTypeId="1" CreationDate="2011-01-26T06:06:54.313" Score="1" ViewCount="81" Body="&lt;p&gt;Hi, I'm working through Stanford's computer vision course to refresh some of my math skills (http://vision.stanford.edu/teaching/cs223b/syllabus.html), and I've run into a problem while working with a simple 2d Support Vector Machine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically, the SVM library I am using only returns the vector 'w', which is the normal to the hyperplane (or line in 2d). I have verified that I am getting the correct value. The problem is, it does not contain 'b' in the equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;w.x + b = 0&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do I compute 'b', so that when I draw the line it correctly bisects my data?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, I've noticed that the library I'm using (liblinear for those interested) assumes that the hyperplane bisects the origin (again, it doesn't use 'b'). I believe it assumes that the data is centered about the origin? My question is, can I simply take the average position (avg) and translate all of my data points  by avg? Is there a preferred method to prep the data before running it through a SVM?&lt;/p&gt;&#xA;" OwnerUserId="6261" LastEditorUserId="1543" LastEditorDisplayName="" LastEditDate="2011-04-26T17:12:22.240" LastActivityDate="2011-05-26T18:02:49.300" Title="Finding displacement from origin of SVM hyperplane" Tags="&lt;linear-algebra&gt;&lt;math-software&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="19042" PostTypeId="2" ParentId="19032" CreationDate="2011-01-26T06:12:43.540" Score="7" ViewCount="" Body="&lt;p&gt;By translation we can assume that $b = 0$. Let $f = x \mapsto P(x)x + C$. We are given that $(P(a)a + C)C = -a^2$. I claim that $a|C$. Indeed, suppose that $(a,C) = g$. Then also $(a,P(a)a+C) = g$, so that $g = a$ (otherwise the product won't be divisible by $-a^2$). Since $a$ divides both factors, they must both be equal (up to sign) to $a$. So one is $a$ and the other $-a$.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2011-01-26T06:12:43.540" />
  <row Id="19043" PostTypeId="2" ParentId="19039" CreationDate="2011-01-26T06:33:55.123" Score="2" ViewCount="" Body="&lt;p&gt;The sizes of the objects form a vector $\vec{s}$. The access statistics form a vector $\vec{a}$. The quotients form a vector $\vec{q}$ defined by $$q_i = n_i/a_i.$$ This is your array division. The &quot;mathy&quot; notation is $q_i$ for the C &quot;q[i]&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now you want to take your array and ignore the maximum, so let $$p = \mathrm{argmax}_i q_i,$$ and then you want to ignore coordinate $p$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I think your professor also understands English.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2011-01-26T06:33:55.123" CommentCount="1" />
  <row Id="19044" PostTypeId="1" CreationDate="2011-01-26T06:41:48.407" Score="1" ViewCount="182" Body="&lt;p&gt;How to prove closure of $\mathbb{Q}$ is $\mathbb{R}$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can it be proved easily?&lt;/p&gt;&#xA;" OwnerUserId="5049" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-26T07:23:45.423" LastActivityDate="2011-01-26T08:48:07.330" Title="How to prove closure of $\mathbb{Q}$ is $\mathbb{R}$." Tags="&lt;topology&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="19045" PostTypeId="1" CreationDate="2011-01-26T07:14:26.257" Score="0" ViewCount="261" Body="&lt;p&gt;Let $\{t_1,t_2,\ldots,t_n\}$ be given points in $[0,1]$, let $\{w_1,w_2,\ldots,w_n\}$ be given real numbers. Let $f\in\mathcal{C}([0,1])$. Define $T(f)=w_1f(t_1)+\cdots+w_nf(t_n)$. Show that $T$ is a continous linear functional on $\mathcal{C}([0,1])$ and that $||T||=\sum$&lt;/p&gt;&#xA;" OwnerUserId="6990" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-26T07:43:25.013" LastActivityDate="2011-05-26T11:02:11.293" Title="soft question in analysis" Tags="&lt;functional-analysis&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="19046" PostTypeId="2" ParentId="19039" CreationDate="2011-01-26T07:19:25.237" Score="-1" ViewCount="" Body="&lt;p&gt;Use source code, like C language.&lt;/p&gt;&#xA;" OwnerUserId="6262" LastActivityDate="2011-01-26T07:19:25.237" />
  <row Id="19047" PostTypeId="2" ParentId="19044" CreationDate="2011-01-26T07:27:52.847" Score="5" ViewCount="" Body="&lt;p&gt;Presumably, you mean the closure of $\mathbb{Q}$ in $\mathbb{R}$ under the usual topology.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It suffices to show that for every real number $r$ and every $\epsilon\gt 0$, there is at least one rational $q$ which is &quot;$\epsilon$-close&quot; to $r$ (that is, $|r-q|\leq \epsilon$), since this will show that every open ball around $r$ contains a rational. This shows that the complement of $\mathbb{Q}$ has empty interior, so the closure of $\mathbb{Q}$ is all of $\mathbb{R}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To prove that, let $r$ be a real number, and let $\epsilon\gt 0$. Then there exists a natural number $n$ such that $\frac{1}{10^n}\lt \epsilon$. Write $r$ in decimal expansion, and let $q$ be the rational that has the same decimal expansion as $r$ up to the $n$th digit after the decimal point, and terminates there. This is a rational number, since its decimal expansion terminates. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now note that $|r-q|\leq \frac{1}{10^n} \lt \epsilon$, proving that there is a rational that is $\epsilon$-close to $r$.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-26T07:53:07.197" LastActivityDate="2011-01-26T07:53:07.197" />
  <row Id="19048" PostTypeId="2" ParentId="19044" CreationDate="2011-01-26T07:42:44.313" Score="3" ViewCount="" Body="&lt;p&gt;This question really does not make a whole lot of sense. In topology closures only makes sense in some ambient space. Some examples will probably make this clear.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\mathbb{Q}$'s closure in $\mathbb{Q}$ is $\mathbb{Q}$ itself.&lt;/li&gt;&#xA;&lt;li&gt;$\mathbb{Q}$'s closure in $\mathbb{R}$ is $\mathbb{R}$.&lt;/li&gt;&#xA;&lt;li&gt;$\mathbb{Q}$'s closure in $\mathbb{Q}(\sqrt{2})$ is $\mathbb{Q}(\sqrt{2})$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;And this doesn't even go into embeddings of $\mathbb{Q}$ in e.g. the unit circle. What you probably meant to ask was how to prove that the &lt;em&gt;completion&lt;/em&gt; of $\mathbb{Q}$ is $\mathbb{R}$. For that you should take a look at &lt;a href=&quot;http://math.stackexchange.com/questions/11923/completion-of-rational-numbers-via-cauchy-sequences/&quot;&gt;this question&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1269" LastEditorUserId="1269" LastEditorDisplayName="" LastEditDate="2011-01-26T07:53:45.067" LastActivityDate="2011-01-26T07:53:45.067" CommentCount="1" />
  <row Id="19049" PostTypeId="2" ParentId="18996" CreationDate="2011-01-26T07:51:48.213" Score="1" ViewCount="" Body="&lt;p&gt;Don't know if these will help you: the NSA published two booklets about the Engima. One is essentially &lt;a href=&quot;http://ed-thelen.org/comp-hist/NSA-Enigma.html&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;, and has a nice diagram on how the Enigma encrypts. A PDF of their booklet on the cryptographic mathematics of the Enigma machine can be found &lt;a href=&quot;http://www.nsa.gov/about/_files/cryptologic_heritage/publications/wwii/engima_cryptographic_mathematics.pdf&quot; rel=&quot;nofollow&quot;&gt;in their website.&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2011-01-26T07:51:48.213" />
  <row Id="19050" PostTypeId="2" ParentId="18996" CreationDate="2011-01-26T08:21:43.657" Score="1" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://enigmaco.de/_fs/index-enigma.html&quot; rel=&quot;nofollow&quot;&gt;Here&lt;/a&gt; we find a simulation (step by step) in Flash for an enigma, that you can play with yourself (change rotors and steckers, see what input corresponds to output).&#xA;I think the wheels are non-historical, but the principle of an enigma becomes clearer.&#xA;&lt;a href=&quot;http://users.telenet.be/d.rijmenants/en/enigmasim.htm&quot; rel=&quot;nofollow&quot;&gt;This site&lt;/a&gt; has a Windows program with an exact simulation of an historical machine, which could be nice to play with.&lt;a href=&quot;http://www.cryptomuseum.com/crypto/enigma/&quot; rel=&quot;nofollow&quot;&gt;The cryptomuseum site&lt;/a&gt; has a nice overview of historical machines.&lt;/p&gt;&#xA;" OwnerUserId="4280" LastActivityDate="2011-01-26T08:21:43.657" />
  <row Id="19051" PostTypeId="2" ParentId="19037" CreationDate="2011-01-26T08:42:06.790" Score="3" ViewCount="" Body="&lt;p&gt;This $Y$ is the so-called $\Sigma$-product of $\omega_1$ many copies of $[0,1]$ w.r.t. $0$. It is countably compact and dense in the full product $X$, and so by lemma 2.2.17 of &lt;a href=&quot;http://users.dimi.uniud.it/~dikran.dikranjan/ITG.pdf&quot; rel=&quot;nofollow&quot;&gt;this syllabus&lt;/a&gt; (e.g.) $Y$ intersects every non-empty $G_\delta$ subset of $X$, and this implies that, as $X$ is a Baire space, so is $Y$ (it's exercise 2.2.15 in the same syllabus). So this confirms your suspicion. $Y$ is an interesting space: it's normal countably compact, non-compact, ccc but not separable. &lt;/p&gt;&#xA;" OwnerUserId="4280" LastActivityDate="2011-01-26T08:42:06.790" CommentCount="5" />
  <row Id="19052" PostTypeId="2" ParentId="19044" CreationDate="2011-01-26T08:48:07.330" Score="1" ViewCount="" Body="&lt;p&gt;In some sense, the denseness of $\Bbb Q$ in $\Bbb R$ is implicit in the very same construction of $\Bbb R$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are a few equivalent ways to construct $\Bbb R$. One of them is to endow $\Bbb Q$ with a structure of metric space by defining the distance function $d(a,b)=|a-b|$ and define $\Bbb R$ to be the completion of the metric space $({\Bbb Q},d)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is but an instance of a general construction: given &lt;em&gt;any&lt;/em&gt; matric space $(X,d)$ one can construct its completion $\tilde X$ as the set of certain equivalence classes of Cauchy sequences in $X$. It turns out that one can endow $\tilde X$ with a natural metric space stucture such that: (1) $X$ embeds isometrically into $\tilde X$ and the image is a dense set, (2) $\tilde X$ is complete, and (3) if $\tilde Y$ is some metric space satisfying the just mentioned properties, then there is a canonical isometry of metric spaces between $\tilde X$ and $\tilde Y$ which is the identity on $X$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Incidentally, it is possible to define distances in $\Bbb Q$ which are not equivalent to the one given by the usual absolute value (as above) but depend on the choice of a prime number $p$ and are defined arithmetically. The corresponding completions, the so-called fields of $p$-adic numbers ${\Bbb Q}_p$, are metric spaces &lt;strong&gt;not&lt;/strong&gt; equivalent to $\Bbb R$ in which $\Bbb Q$ sits densely. A famous theorem by Ostrowski says that $\Bbb R$ and the ${\Bbb Q}_p$ are, up to equivalence, the only completions of $\Bbb Q$.&lt;/p&gt;&#xA;" OwnerUserId="688" LastActivityDate="2011-01-26T08:48:07.330" />
  <row Id="19053" PostTypeId="2" ParentId="19045" CreationDate="2011-01-26T08:58:05.490" Score="0" ViewCount="" Body="&lt;p&gt;$\ |(Tf)|\leq|w_1f(t_1)+...+w_nf(t_n)|\leq |w_1f(t_1)|+...+|w_nf(t_n)|\leq\sum$ so the function is continuous.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the other half take $\ f=1$. &lt;/p&gt;&#xA;" OwnerUserId="5422" LastEditorUserId="5422" LastEditorDisplayName="" LastEditDate="2011-01-26T09:03:26.560" LastActivityDate="2011-01-26T09:03:26.560" CommentCount="4" />
  <row Id="19054" PostTypeId="1" AcceptedAnswerId="19189" CreationDate="2011-01-26T09:18:55.200" Score="19" ViewCount="521" Body="&lt;p&gt;Is there a real function over the real numbers with this property $\ \sqrt{|x-y|} \leq |f(x)-f(y)|$ ?&#xA;My guess is no but can anyone tell me why?&#xA;This came up as  a question of one of my collegues and i cant give an answer.&lt;/p&gt;&#xA;" OwnerUserId="5422" LastEditorUserId="5422" LastEditorDisplayName="" LastEditDate="2011-01-26T11:07:13.427" LastActivityDate="2011-01-27T16:43:15.970" Title="Is there a function with this property?" Tags="&lt;analysis&gt;&lt;functions&gt;" AnswerCount="6" CommentCount="2" FavoriteCount="5" />
  <row Id="19055" PostTypeId="1" CreationDate="2011-01-26T09:42:34.413" Score="0" ViewCount="55" Body="&lt;p&gt;Hi, someone knows the name of the 3D volume obtained through a boolean subtraction of two centrally symmetrical spherical sectors (one at the top the other at the bottom) from a sphere? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It can also be explained as a regular revolved circle sector (360° revolution around z-axis of a circle sector that lays in the z-y plane and has the center of the circle in O, and is defined by for example +/- 30°). It should look similar to that one but the shape has to be centered on the horizontal plane (with a &quot;conic hole&quot; at the top and an identical one at the bottom): &lt;a href=&quot;http://etc.usf.edu/clipart/65700/65755/65755_sphercone.htm&quot; rel=&quot;nofollow&quot;&gt;http://etc.usf.edu/clipart/65700/65755/65755_sphercone.htm&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it possible that the shape is not named?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See also in wolfram mathworld:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;SphericalSector&lt;/p&gt;&#xA;&#xA;&lt;p&gt;SphericalWedge&lt;/p&gt;&#xA;&#xA;&lt;p&gt;SphericalSegment&lt;/p&gt;&#xA;&#xA;&lt;p&gt;SphericalCap&lt;/p&gt;&#xA;&#xA;&lt;p&gt;SphericalCone&lt;/p&gt;&#xA;" OwnerUserId="6267" LastActivityDate="2011-01-26T09:42:34.413" Title="Name of the shape obtained trough subtraction of two sperical sectors from a sphere?" Tags="&lt;geometry&gt;" />
  <row Id="19056" PostTypeId="1" AcceptedAnswerId="19197" CreationDate="2011-01-26T10:09:31.547" Score="3" ViewCount="181" Body="&lt;p&gt;$$\sum_{m=2}^{10^6}\ \sum_{n=1}^{\infty}\dfrac{m^{\{n\}}+m^{-{\{n\}}}}{m^n}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $\{n\}=\text{ integer nearest to the } \sqrt{n} $.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Looks like a non-trivial summation. How to properly approach this problem?&lt;/p&gt;&#xA;" OwnerUserId="149" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2011-01-26T13:04:45.843" LastActivityDate="2011-01-27T06:32:56.867" Title="Summation of a non trivial series" Tags="&lt;sequences-and-series&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="2" />
  <row Id="19057" PostTypeId="2" ParentId="19001" CreationDate="2011-01-26T10:10:37.410" Score="0" ViewCount="" Body="&lt;p&gt;Ok, I've been able to decompose K17 into 8 2-regular subgraphs. Here's my solution:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/nFPWI.png&quot; alt=&quot;K17 decomposed into 8 2-connected subgraphs&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My method is as follows: Starting from any vertex, skip n, and draw an edge to the vertex you ended up at. Repeat the process from the new vertex until you return to your origin. Any complete graph of order 2x+1 can be decomposed into n subgraphs by repeating the above process for values of n from 0 to x-1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are many ways to pair the 8 sub-grids so that they become 4 isomorphic 4-regular ones. First choose a type of polygon you'd like to be dominant in your sub-grids. For instance with triangles, you can pair n=1 with n=2 (1/2), then 3/6, 4/8, and 5/7 (since n=7 in a 17-node graph is the same as n=10). Alternatively, you could also pair 2/4, 6/5, 7/3, 8/1. There are also two ways to pair if you're aiming for rectangles and pentagons, and I assume all the way up to octagons, for a total of 12 ways to combine.&lt;/p&gt;&#xA;" OwnerUserId="6249" LastEditorUserId="6249" LastEditorDisplayName="" LastEditDate="2011-01-27T17:33:23.020" LastActivityDate="2011-01-27T17:33:23.020" CommentCount="5" />
  <row Id="19058" PostTypeId="2" ParentId="19054" CreationDate="2011-01-26T10:35:26.977" Score="1" ViewCount="" Body="&lt;p&gt;I think you're right, there are indeed none. It's pretty easy to prove that $ \left |\frac{f(x+\varepsilon) - f(x)}{\varepsilon}  \right | \geq \frac{1}{\sqrt{\left |\varepsilon  \right |}} $, now think about what this means when $\varepsilon$ tends to zero. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not 100% sure about the non-differentiable/non-continuous case though.&lt;/p&gt;&#xA;" OwnerUserId="360" LastActivityDate="2011-01-26T10:35:26.977" CommentCount="2" />
  <row Id="19059" PostTypeId="1" AcceptedAnswerId="19090" CreationDate="2011-01-26T10:54:04.867" Score="0" ViewCount="72" Body="&lt;p&gt;Assume $M=N_1+N_2$ is a module, where $N_1,N_2$ are noetherian modules. How can I show that $M$ is also noetherian?&lt;/p&gt;&#xA;" OwnerUserId="766" LastActivityDate="2011-01-26T15:47:58.473" Title="A sum of noetherian modules is a noetherian module." Tags="&lt;abstract-algebra&gt;&lt;commutative-algebra&gt;&lt;modules&gt;" AnswerCount="2" />
  <row Id="19060" PostTypeId="1" CreationDate="2011-01-26T10:59:10.707" Score="0" ViewCount="180" Body="&lt;p&gt;It can be proved that arbitrary union of open sets is open.&#xA;Suppose $v$ is a family of open sets.&#xA;$\cup_{G \in v}G = A$ is an open set.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now by De Morgan's theorem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(\cup_{G \in v}G)^{c} = \cap_{G \in v}G^{c} = B$.&#xA;$B$ is a closed set since it is the complement of open set $A$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$G$ is an open set, so $G^{c}$ is a closed set. &#xA;$B$ is an infinite &lt;strike&gt;union&lt;/strike&gt; intersection of closed sets $G^{c}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence Infinite intersection of closed sets is closed.&#xA;Is it not a wrong conclusion.&lt;/p&gt;&#xA;" OwnerUserId="5049" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-26T17:09:13.693" LastActivityDate="2011-01-26T17:09:13.693" Title="Arbitrary intersection of closed sets" Tags="&lt;topology&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="19061" PostTypeId="1" AcceptedAnswerId="19245" CreationDate="2011-01-26T10:59:31.803" Score="6" ViewCount="322" Body="&lt;p&gt;There are ten questions on a piece of paper. Your task is to fill in each blank with a positive integer less than 10 such that there is no contradiction. You can reuse any digit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is as follows:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Digit 0 appears &lt;em&gt;_&lt;/em&gt;_ times on this paper.&lt;/li&gt;&#xA;&lt;li&gt;Digit 1 appears &lt;em&gt;_&lt;/em&gt;_ times on this paper.&lt;/li&gt;&#xA;&lt;li&gt;Digit 2 appears &lt;em&gt;_&lt;/em&gt;_ times on this paper.&lt;/li&gt;&#xA;&lt;li&gt;Digit 3 appears &lt;em&gt;_&lt;/em&gt;_ times on this paper.&lt;/li&gt;&#xA;&lt;li&gt;Digit 4 appears &lt;em&gt;_&lt;/em&gt;_ times on this paper.&lt;/li&gt;&#xA;&lt;li&gt;Digit 5 appears &lt;em&gt;_&lt;/em&gt;_ times on this paper.&lt;/li&gt;&#xA;&lt;li&gt;Digit 6 appears &lt;em&gt;_&lt;/em&gt;_ times on this paper.&lt;/li&gt;&#xA;&lt;li&gt;Digit 7 appears &lt;em&gt;_&lt;/em&gt;_ times on this paper.&lt;/li&gt;&#xA;&lt;li&gt;Digit 8 appears &lt;em&gt;_&lt;/em&gt;_ times on this paper.&lt;/li&gt;&#xA;&lt;li&gt;Digit 9 appears &lt;em&gt;_&lt;/em&gt;_ times on this paper.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;How to solve it?&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Edit 1:&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;I am not sure this question can be done without computer.&lt;/p&gt;&#xA;" OwnerUserId="3901" LastEditorUserId="3901" LastEditorDisplayName="" LastEditDate="2011-01-26T11:13:56.313" LastActivityDate="2011-01-27T16:11:05.057" Title="How to solve this puzzle?" Tags="&lt;puzzle&gt;" AnswerCount="2" />
  <row Id="19062" PostTypeId="2" ParentId="19060" CreationDate="2011-01-26T11:02:03.973" Score="3" ViewCount="" Body="&lt;p&gt;This is true, and your reasoning is correct too.&lt;/p&gt;&#xA;" OwnerUserId="808" LastActivityDate="2011-01-26T11:02:03.973" />
  <row Id="19063" PostTypeId="2" ParentId="19061" CreationDate="2011-01-26T11:19:16.347" Score="6" ViewCount="" Body="&lt;p&gt;Here's an answer:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;0 appears 1 time&#xA;1 appears 7 times&#xA;2 appears 3 times&#xA;3 appears 2 times&#xA;4 appears 1 time&#xA;5 appears 1 time&#xA;6 appears 1 time&#xA;7 appears 2 times&#xA;8 appears 1 time&#xA;9 appears 1 time&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The logic is roughly as follows.  Since the numbers in the blanks must be less than 10, 0 must appear only 1 time.  Also, since you can't have most of the large numbers appearing more than once, 1 appears $x$ times, where $x$ is one of the larger numbers.  That means $x$ probably appears twice.  But then 2 can't appear 2 times, as then there would be 3 instances of 2.  So perhaps 2 appears 3 times.  Then 3 would appear 2 times, and from there you can determine that $x = 7$.&lt;/p&gt;&#xA;" OwnerUserId="2370" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2011-01-26T11:24:49.253" LastActivityDate="2011-01-26T11:24:49.253" CommentCount="5" />
  <row Id="19064" PostTypeId="2" ParentId="18992" CreationDate="2011-01-26T11:22:16.353" Score="3" ViewCount="" Body="&lt;p&gt;Mathcad &quot;reads&quot; down and right and you have defined the equation above the variable definition. If you drag the equation down it will work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are going to be using Mathcad regularly, you might be interested in joining the Mathcad User Community:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://communities.ptc.com/index.jspa&quot; rel=&quot;nofollow&quot;&gt;http://communities.ptc.com/index.jspa&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Chris Hartmann, Academic Program Manager, Mathcad&lt;/p&gt;&#xA;" OwnerUserId="6268" LastActivityDate="2011-01-26T11:22:16.353" CommentCount="2" />
  <row Id="19065" PostTypeId="2" ParentId="19059" CreationDate="2011-01-26T11:31:16.950" Score="1" ViewCount="" Body="&lt;p&gt;First you can think about how the submodules of $N_1 \oplus N_2$ look like. Then you can use that a module is noetherian iff all of its submodules are finitely generated.&lt;/p&gt;&#xA;" OwnerUserId="1766" LastActivityDate="2011-01-26T11:31:16.950" />
  <row Id="19066" PostTypeId="1" CreationDate="2011-01-26T11:39:07.240" Score="1" ViewCount="149" Body="&lt;p&gt;Let me prove subset $A$ of $\mathbb{R}$ is open if and only if it is a countable union of open subsets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For all $x \in A$ there is an $\epsilon &amp;gt; 0$ such that $(x-\epsilon,x+\epsilon)$ is an open set contained in $A$. Now find rationals such that $r_{x} \in (x-\epsilon,x)$ and $s_{x} \in (x,x+\epsilon)$ and $A = \bigcup_{x \in A}(r_{x},s_{x})$. Note that the intervals with rational end points is less than $\mathbb{Q} \times \mathbb{Q}$. Obviously $\mathbb{Q} \times \mathbb{Q}$ is countable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My doubt is that instead of selecting rationals can I select &lt;b&gt;irrationals&lt;/b&gt;. &lt;/p&gt;&#xA;" OwnerUserId="5049" LastEditorUserId="1269" LastEditorDisplayName="" LastEditDate="2011-01-26T11:54:35.453" LastActivityDate="2011-01-26T14:29:50.093" Title="Show that a subset $A$ of $R$ is open iff it is countable union of open subsets" Tags="&lt;topology&gt;" AnswerCount="4" CommentCount="8" />
  <row Id="19067" PostTypeId="2" ParentId="19054" CreationDate="2011-01-26T11:51:13.387" Score="12" ViewCount="" Body="&lt;p&gt;There &lt;em&gt;is&lt;/em&gt; such a function on the rationals:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $(q_0,q_1,...)$ be any enumeration of the rationals. For each $i \ge 0$, let $f(q_i) = q_r$, where $r$ is the smallest index that maintains the required property of $f$. That is, $r$ is the smallest index such that $|q_r - f(q_j)| \ge \sqrt{|q_i - q_j|}$ for all $j \lt i$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edited to add:&lt;/strong&gt; What's more, given any infinite subset $X \subset \mathbb{N}$, there is such a function $f: \mathbb{Q} \rightarrow X$. Just let $f(q_i)$ be the smallest $n \in X$ such that $|n - f(q_j)| \ge \sqrt{|q_i - q_j|}$ for all $j \lt i$.&lt;/p&gt;&#xA;" OwnerUserId="1508" LastEditorUserId="1508" LastEditorDisplayName="" LastEditDate="2011-01-27T10:59:39.613" LastActivityDate="2011-01-27T10:59:39.613" CommentCount="1" />
  <row Id="19068" PostTypeId="2" ParentId="19066" CreationDate="2011-01-26T12:18:14.570" Score="0" ViewCount="" Body="&lt;p&gt;I do not think picking a rational number as the upper and lower bound is correct. Let A be your open set. Then define, for each  $x \in A$, $$r_x = inf \lbrace x_0 : (x_0,x) \subset A \rbrace$$ $$s_x = sup \lbrace x_0 : (x,x_0) \subset A \rbrace$$ Then $(r_x,s_x)$ will be the maximal open interval containing $x$ that is a subset of $A$ as well. In particular, you can show that the endpoints to not belong into these intervals, and secondly, that for two points $x,y \in A$, the intervals $(r_x,s_x)$ and $(r_y,s_y)$ are either disjoint or equal (by maximality). Then note that you can only have a countable amount of &lt;em&gt;disjoint&lt;/em&gt; open intervals in $\mathbb{R}$ (select a rational number in each...).  &lt;/p&gt;&#xA;" OwnerUserId="5105" LastActivityDate="2011-01-26T12:18:14.570" />
  <row Id="19069" PostTypeId="2" ParentId="19054" CreationDate="2011-01-26T12:29:42.833" Score="10" ViewCount="" Body="&lt;p&gt;Observe that $f(x) = f(y)$ implies $x = y$, so if $f$ is continuous then it must be monotonic.  Now we are done by a fairly simple pigeonhole argument: if, say, $|f(1) - f(0)| = d$, then at least one of $|f(1) - f((n-1)/n)|, |f((n-1)/n) - f((n-2)/n)|, ...$ is at most $\frac{d}{n}$, and taking $n$ large enough this leads to a contradiction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have no idea what happens if $f$ isn't required to be continuous.  But the idea here is that the condition gets harder to satisfy the closer $x$ and $y$ are.    &lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2011-01-26T12:29:42.833" />
  <row Id="19070" PostTypeId="1" AcceptedAnswerId="19074" CreationDate="2011-01-26T13:02:50.437" Score="4" ViewCount="240" Body="&lt;p&gt;Let $M(n;\mathbb{R})$ denote the set of all $n \times n$ matrices with real entries (identified with $\mathbb{R}^{n^{2}}$ and endowed with its usual topology) and let $GL(n;\mathbb{R})$ denote the group of invertible matrices. Let $G$ be a subgroup of $GL(n;\mathbb{R})$. Define $$H = \biggl\{ A \in G  \ \biggl| \ \exists \ \varphi:[0,1] \to G \ \text{continuous such that} \ \varphi(0)=A , \ \varphi(1)=I\biggr\}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then is $H$ normal in $G$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For proving $H$ normal i should verify two things:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;First $H$ is a subgroup.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;For all $A \in G, B \in H$, we must have $A \cdot B \cdot A^{-1} \in H$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;That's it i  am not able to proceed any further. &lt;/p&gt;&#xA;" LastActivityDate="2011-01-28T01:04:15.580" Title="Showing $H$ is a normal subgroup of $G$" Tags="&lt;linear-algebra&gt;&lt;abstract-algebra&gt;&lt;topology&gt;" AnswerCount="4" CommentCount="6" />
  <row Id="19071" PostTypeId="2" ParentId="18969" CreationDate="2011-01-26T13:04:50.570" Score="0" ViewCount="" Body="&lt;p&gt;Although I'm aware that pointwise limits can do all kinds of freaky stuff, I like to think of how rare the rationals actually are with respect to the reals by analogy of how rare the naturals are (I think I saw this or something similar in the nice introductory book to Analysis I by Harro Heuser):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the family of functions $f_{n,k}(x) = \cos^{2n}(k! \pi x)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $q$ is a rational number iff there is a $k$ such that $k!q \in \mathbb Z$, we get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\lim_{k \to \infty} \, \lim_{n \to \infty} \, f_{n,k}(x) = \chi_{\mathbb Q}(x)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in &lt;em&gt;some&lt;/em&gt; sense, the mental picture of two rationals being separated by irrationals in a similar way as the naturals are by nonnaturals is not completely wrong. At least in certain situations (like why the irrationals are a Baire space, whereas the rationals are not) this might help one get a more intuitive understanding, I think.&lt;/p&gt;&#xA;" OwnerUserId="3208" LastActivityDate="2011-01-26T13:04:50.570" CommentCount="2" />
  <row Id="19072" PostTypeId="2" ParentId="18554" CreationDate="2011-01-26T13:06:04.627" Score="1" ViewCount="" Body="&lt;p&gt;Being $C^n$ is a local property so it's a bit unusual to say $C^n$ &quot;at a point&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pick your favourite family of periodic functions that are not $C^{n+1}$ on $\frac{1}{n}\mathbb{Z}$, but are $C^{n}$ everywhere.&#xA;Mine happens to be the family $f_n(x) = (\sin(n \pi x))^{n+1/3}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take note of how big the derivatives of $f_n$ can get for large n.&#xA;Here, $||f_n^{(k)}||_\infty \leq (n+1)^{2k} $ for $n \ge k$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now sum these functions with an appropriate weight such that their sum as well as the sum of all their derivative converges uniformly :&#xA;$$F(x) = \Sigma_{n \ge 1} f_n(x) e^{-n} $$&#xA;Now you can prove that for all n,$F$ is $C^n$ exactly on the open set $\mathbb{R} - \{\frac{k}{d}, k \in \mathbb{Z}, 1 \le d \lt n\}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is a funny example of a function being $C^n$ for all n around irrational points, but nowhere $C^\infty$.&#xA;That said, I don't expect it to be anymore interesting than from having this property.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit : I should clarify why being $C^k$ should be a local property :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Smooth_function#Differentiability_classes&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Smooth_function#Differentiability_classes&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When talking about differentiability, we want talk about the behaviour of the function around a point. If $f$ is $C^k$ &quot;at&quot; a point x, we really want to say that $f$ is well-behaved around that point. It is a property of the behaviour of $f$ around that point, and not a property of what the value of $f$ is at that point :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $f$ is $C^k$ &quot;at&quot; a point, it means that the $k$th derivative of $f$ exists on a neighboorhood of $x$, or else we can't talk about it being continuous at $k$.&#xA;Then, yes, there is a difference between &quot;being continuous at $x$&quot; and &quot;being continuous on a neighboorhood of $x$&quot;. The definition on that wikipedia page uses the latter, while your $Cf(x)$ may use the former.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here in my construction, the difference doesn't appear (until discussing if $F$ is $C^\infty$ somewhere). Whichever definition you choose, $F$ is $C^n$ &quot;at&quot; the same points. Or rather, $F$ is $C^n$ on the open subset that you wanted it to.&lt;/p&gt;&#xA;" OwnerUserId="6270" LastEditorUserId="6270" LastEditorDisplayName="" LastEditDate="2011-01-27T19:58:26.373" LastActivityDate="2011-01-27T19:58:26.373" CommentCount="13" />
  <row Id="19073" PostTypeId="2" ParentId="19066" CreationDate="2011-01-26T13:49:32.647" Score="0" ViewCount="" Body="&lt;p&gt;Yes you can select irrational numbers instead of rational ones. The reason for this is that $\mathbb{Q}$ is dense in $\mathbb{R}$. This means that between any two $r_1, r_2 \in \mathbb{R}$ you can find a rational $q \in \mathbb{Q}$ between them and vice versa, i.e. for any two $q_1, q_2 \in \mathbb{Q}$ there is an $r \in \mathbb{R} \backslash \mathbb{Q}$ such that $q_1 &amp;lt; r &amp;lt; q_2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a consequence, if you have $a \in (a - \varepsilon , a + \varepsilon) \subset A$ and $a \in \mathbb{Q}$ you can find an irrational $r$ such that $a \in (r - \varepsilon , r + \varepsilon) \subset A$ .&lt;/p&gt;&#xA;" OwnerUserId="5798" LastActivityDate="2011-01-26T13:49:32.647" />
  <row Id="19074" PostTypeId="2" ParentId="19070" CreationDate="2011-01-26T13:49:40.687" Score="1" ViewCount="" Body="&lt;p&gt;actually, the answer jug gave in his comments is easier, but here is another proof (though a bit overkill) which shows that this group is actually the matrices with positive determinant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The function $det:GL_n(\mathbb{R})\rightarrow \mathbb{R}$ is continuous, so if $\varphi:[0,1]\rightarrow \mathbb{R}$  such that $\varphi(0)=A,\; \varphi(1)=I$  is continuous then $det \circ \varphi$ is continuous. Since $det(\varphi(0))=det(A)$ and $det\varphi (1)=1$ then we must have $det(A)&amp;gt;0$ otherwise by the intermediate value theorem there is t such that $det(\varphi (t))=0$ but then $\varphi(t) \notin GL_n(\mathbb{R})$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, suppose that $det(A)&amp;gt;0$ then if $E$ is an elementary matrix (corresponding to a row operation) then define $\varphi(t)=((1-t)I+tE)A$ for $0\leq t \leq 1$. It is continuous and its image is in $GL_n(\mathbb{R})$ if $((1-t)I+tE)$ is always invertible. This is for true if E is a multiplication by a positive scalar or if it is an addition of one row to another, but not true for multiplication by a negative scalar, or switching rows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Show that if $det(A)&amp;gt;0$ then you can use such elementary matrices to go from A to I.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This shows that your H is just all the matrices with positive determinant.&lt;/p&gt;&#xA;" OwnerUserId="2125" LastActivityDate="2011-01-26T13:49:40.687" />
  <row Id="19075" PostTypeId="2" ParentId="19066" CreationDate="2011-01-26T13:55:55.717" Score="3" ViewCount="" Body="&lt;p&gt;$\mathbb{R}$ is locally connected, so the connected components of open sets are open.&#xA;So if $O$ is open, write it as the (disjoint) union of its connected components, which are open (as said) and connected and thus intervals or segments, or $\mathbb{R}$ itself.&#xA;As all of these are (at most countable) unions of intervals. Because every different component must contain a different member of $\mathbb{Q}$, there are at most countably many components, and thus countably many intervals to write $O$ as a union all together.&lt;/p&gt;&#xA;" OwnerUserId="4280" LastEditorUserId="4280" LastEditorDisplayName="" LastEditDate="2011-01-26T14:05:44.313" LastActivityDate="2011-01-26T14:05:44.313" />
  <row Id="19076" PostTypeId="1" AcceptedAnswerId="19082" CreationDate="2011-01-26T13:56:02.633" Score="2" ViewCount="88" Body="&lt;p&gt;Hello! I got a problem. I don't know how to solve this kind of problem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Find an integer $n_0$ that for each $n&amp;gt;n_0$ this inequation would be always true:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\dfrac{n^{4}-n^{2}+1}{n^{3}-n}&amp;gt;10000 $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know how to solve this kind of problems, and I hope you could help me understanding how to proceed step by step. &#xA;Thank you in advance.&lt;/p&gt;&#xA;" OwnerUserId="4356" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2011-01-26T16:02:09.880" LastActivityDate="2011-01-26T16:02:09.880" Title="Determination of an integer" Tags="&lt;analysis&gt;&lt;inequality&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="19078" PostTypeId="2" ParentId="18958" CreationDate="2011-01-26T14:16:06.233" Score="1" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;First question:&lt;/strong&gt; As you say, the condition number is a function of the exact solution. It measures how much the (exact) solution of the differential equation changes if you change the initial condition. The condition number depends only on the problem you are solving, not on the numerical method you are using.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, why are we interested in this condition number in numerical analysis? Because the condition number gives a lower bound on the error committed by any numerical method. The idea is that the initial condition is inaccurate, either because it comes from some measurements which is not perfect or because of rounding error. This inaccuracy in the initial condition leads to an error in the solution as measured by the condition number, regardless of how good the numerical method is. On top of that comes the errors committed by the numerical method.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Second question:&lt;/strong&gt; One idea for computing the condition number is to use that the numerical evolution operator $\Psi$ approximates the exact evolution operator $\Phi$. Thus, to compute the condition number, replace $\Phi$ in your definition by $\Psi$. You can compute the derivative $\frac{\partial}{\partial y} \Psi^{t_0,t}(y)$ by finite differences or by evaluating the derivative by hand. &lt;/p&gt;&#xA;" OwnerUserId="520" LastActivityDate="2011-01-26T14:16:06.233" CommentCount="2" />
  <row Id="19079" PostTypeId="2" ParentId="19076" CreationDate="2011-01-26T14:27:05.930" Score="1" ViewCount="" Body="&lt;p&gt;Can you divide out the fraction to a polynomial and remainder?  That will help.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-01-26T14:27:05.930" />
  <row Id="19080" PostTypeId="2" ParentId="19066" CreationDate="2011-01-26T14:29:50.093" Score="0" ViewCount="" Body="&lt;p&gt;First of all, you should change your statement: instead of &quot;open subsets&quot; you should say &quot;open intervals&quot;. Otherwise, as Pete L. Clark observes, every open set is the union of &lt;em&gt;one&lt;/em&gt; open subset, namely itself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for your proof, I don't see how you can assure the existence of your rational numbers $r_x$ and $s_x$ such that $A = \bigcup_{x\in A} (r_x, s_x)$ and indeed there is no need for doing so.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instead, I would proceed as Henno Brandsma, but let me write it using less results perhaps: if you have an open set $U \subset \mathbb{R}$, for every point $x\in U$ there exists an open interval $I_x$ such that $x\in I_x \subset U$, by definition of open sets of the real line (with the standard topology). Hence $U = \bigcup_{x\in U} I_x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, pick a rational number $r_x \in I_x$ for every $x\in U$. You can do this because rationals are dense in $\mathbb{R}$. Since there is only a countable number of rational numbers, you're done.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(And yes: you could also pick an irrational number for each $x$: so what?)&lt;/p&gt;&#xA;" OwnerUserId="664" LastActivityDate="2011-01-26T14:29:50.093" />
  <row Id="19081" PostTypeId="1" CreationDate="2011-01-26T14:34:03.500" Score="0" ViewCount="93" Body="&lt;p&gt;&lt;a href=&quot;http://arxiv.org/pdf/math/0205003v1&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/pdf/math/0205003v1&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In around equation (1.1) the author says&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;By necessity all authors have been led in one way or another to the natural approximation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$F(n) := \sum_{a=1}^n \mu(a) \rho_a$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which tends to $-\chi$ both a.e. and in L1 norm when restricted to (0,1) but which has been shown to diverge in H.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;How to determine $F$ diverges in $L^2(0,1)$? because $$\int_{0}^{1} |1 + F|^2 dx &amp;lt; \infty$$ where $F \to -1$?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am new to functional analysis, so little confused on this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Roupam&lt;/p&gt;&#xA;" OwnerUserId="2320" LastActivityDate="2011-01-26T14:34:03.500" Title="Convergent in $L^1(0,1)$ but not in $L^2(0,1)$ help understanding a paper from arxiv" Tags="&lt;functional-analysis&gt;" CommentCount="2" />
  <row Id="19082" PostTypeId="2" ParentId="19076" CreationDate="2011-01-26T14:41:02.453" Score="3" ViewCount="" Body="&lt;p&gt;Hints: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\dfrac{n^{4}-n^{2}+1}{n^{3}-n}&amp;gt;10000\Leftrightarrow n+\dfrac{1}{n^{3}-n}&amp;gt;10000&#xD;&#xA;$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$0&amp;lt;\dfrac{1}{n^{3}-n}&amp;lt;1\Leftrightarrow n&amp;gt;1$&lt;/p&gt;&#xA;" OwnerUserId="752" LastActivityDate="2011-01-26T14:41:02.453" CommentCount="4" />
  <row Id="19083" PostTypeId="2" ParentId="16870" CreationDate="2011-01-26T14:51:54.787" Score="1" ViewCount="" Body="&lt;p&gt;You might want to take a look to one-class SVM method.&lt;/p&gt;&#xA;" OwnerUserId="6273" LastActivityDate="2011-01-26T14:51:54.787" CommentCount="2" />
  <row Id="19084" PostTypeId="2" ParentId="19041" CreationDate="2011-01-26T15:07:54.473" Score="0" ViewCount="" Body="&lt;p&gt;You should provide bias parameter for your problem.&#xA;The bias parameter (B) leads to augmenting training samples $(x_i)$ to $(x_i, B)$, so the hyperplane vector will be $(w, b)$ instead $(w)$. For more details see section 2 of LIBLINEAR paper.&lt;/p&gt;&#xA;" OwnerUserId="6273" LastEditorUserId="6273" LastEditorDisplayName="" LastEditDate="2011-01-26T15:18:29.863" LastActivityDate="2011-01-26T15:18:29.863" CommentCount="0" />
  <row Id="19085" PostTypeId="2" ParentId="19032" CreationDate="2011-01-26T15:16:03.247" Score="6" ViewCount="" Body="&lt;p&gt;The problem appears as one of the &quot;Quickies&quot; in the &lt;a href=&quot;http://www.jstor.org/stable/2690479&quot; rel=&quot;nofollow&quot;&gt;problems section of Mathematics Magazine Vol. 66, No. 1, February 1993&lt;/a&gt;.  It was proposed by Bjorn Poonen.  The answers to the Quickies appear in the same issue on page 64.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From page 57:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Suppose $P(x)\in\mathbf{Z}[x]$ and $P(a)P(b)=-(a-b)^2$ for some distinct $a,b\in\mathbf{Z}$.  Prove $P(a)+P(b)=0$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;From page 64:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Since $a-b$ divides $P(a)-P(b)$, the roots $P(a)/(a-b)$ and $-P(b)/(a-b)$ of $x^2-[(P(a)-P(b))/(a-b)]x+1$ are integers with product $1$, by the rational root theorem.  Thus $P(a)/(a-b)=-P(b)/(a-b)=\pm1$, so $P(a)+P(b)=0$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2011-01-26T15:16:03.247" />
  <row Id="19086" PostTypeId="1" CreationDate="2011-01-26T15:19:43.123" Score="5" ViewCount="119" Body="&lt;p&gt;It is well known that the Euler totient function $\varphi$ satisfies the formula $n = \sum_{d | n}\varphi(d)$. This follows for example from the fact that $\mathbb Z / n \mathbb Z$ can be written (as monoid) as the disjoint union $\coprod_{d|n} (\mathbb Z / d \mathbb Z)^\times$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is now if this can be generalized to an arbitrary number field $K$? More precisely, if $\mathcal O _K$ denotes the ring of integers of $K$ and $\mathfrak f$ a non-zero integral ideal of $\mathcal O _K$, do we have $\mathcal O _K / \mathfrak f \cong \coprod_{\mathfrak d | \ \mathfrak f} (\mathcal O _K / \mathfrak d)^\times$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I am not mistakes then my question would have an affirmative answer, if the number of units of $\mathcal O _K / \mathfrak p ^k$, for $\mathfrak p$ a prime ideal and $k \geq 1$ a natural number, is equal to $\mathcal N (\mathfrak p)^{k-1}(\mathcal N (\mathfrak p) - 1)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: Reading a little bit in Zariski/Samuel I, the rings $\mathcal O _K / \mathfrak p ^k$ should be local principle ideal rings, the maximal ideal given by $\mathfrak p / \mathfrak p ^k$, i.e. in order to verify my question one only has to show that the number $\# (\mathfrak p / \mathfrak p ^k)$ equals $N(\mathfrak p ^{k-1})$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit2: If I am not mistaken, then the last assertion follows directly from Noether's third Isomorphism Theorem. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you very much in advance.&lt;/p&gt;&#xA;" OwnerUserId="6275" LastEditorUserId="6275" LastEditorDisplayName="" LastEditDate="2011-01-26T18:28:33.547" LastActivityDate="2011-01-26T18:28:33.547" Title="extension of Euler's totient function to number fields" Tags="&lt;algebraic-number-theory&gt;" FavoriteCount="2" />
  <row Id="19088" PostTypeId="1" AcceptedAnswerId="19102" CreationDate="2011-01-26T15:33:54.020" Score="3" ViewCount="115" Body="&lt;p&gt;Hello everyone,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My problem is proving $x^2 - 13y^2 = 1$ has integers solutions. I can find easily see that (+-1, 0) are trivial solution. My question is: is it sufficient to complete the proof? How can I approach this problem? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks,&lt;br&gt;&#xA;Chan &lt;/p&gt;&#xA;" OwnerUserId="5711" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-26T17:38:35.837" LastActivityDate="2011-01-26T17:59:57.393" Title="How to prove Pell's equation for a specific case?" Tags="&lt;number-theory&gt;" AnswerCount="1" CommentCount="8" />
  <row Id="19089" PostTypeId="1" AcceptedAnswerId="37477" CreationDate="2011-01-26T15:36:13.373" Score="3" ViewCount="144" Body="&lt;p&gt;The negation of :  &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;∃x ∀y (P(x,y) ⇒ Q(x,y))&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;is:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;∀x ∃y ¬(P(x,y) ⇒ Q(x,y))&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;But I am not sure about the last part (¬(...)). Is that negation well done, in the sense that couldn't be done more concisely?&lt;/p&gt;&#xA;" OwnerUserId="5526" LastActivityDate="2011-05-06T19:13:21.607" Title="Elementary logic. Negation" Tags="&lt;logic&gt;" AnswerCount="1" CommentCount="10" />
  <row Id="19090" PostTypeId="2" ParentId="19059" CreationDate="2011-01-26T15:47:58.473" Score="4" ViewCount="" Body="&lt;p&gt;The standard way to prove it is considering the exact secuence:&#xA;$$0\longrightarrow N_1 \overset{i}\longrightarrow N_1\oplus N_2\overset{j}\longrightarrow N_2\longrightarrow 0$$&#xA;Where $i(n)=(n,0)$ and $j(n,m)=m$. As it is an exact secuence, the module in the middle, ie &#xA;$N_1\oplus N_2$, is noetherian iff the two other modules are.&lt;/p&gt;&#xA;" OwnerUserId="6276" LastActivityDate="2011-01-26T15:47:58.473" />
  <row Id="19091" PostTypeId="1" AcceptedAnswerId="19113" CreationDate="2011-01-26T15:58:26.247" Score="4" ViewCount="104" Body="&lt;p&gt;This is a question from my homework for a real analysis course. Please hint only.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $M$ be a compact metric space. Let $\mathbb{K}$ be the class of non-empty compact subsets of $M$. The $r$-neighbourhood of $A \in \mathbb{K}$ is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ M_r A = \lbrace x \in M : \exists a \in A \text{ and } d(x,a) &amp;lt; r \rbrace = \bigcup_{a \in A} M_r a. $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $A$, $B \in \mathbb{K}$ define&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$D(A,B) = \inf \lbrace r &amp;gt; 0 : A \subset M_r B \text{ and } B \subset M_r A \rbrace. $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Show that $\mathbb{K}$ is compact.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Another part of the question is that if $M$ is connected, then so is $\mathbb{K}$, but this is not assigned).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many thanks.&lt;/p&gt;&#xA;" OwnerUserId="5105" LastEditorUserId="1269" LastEditorDisplayName="" LastEditDate="2011-01-26T19:15:22.273" LastActivityDate="2011-01-26T19:24:37.983" Title="The Class of Non-empty Compact Subsets of a Compact Metric Space is Compact" Tags="&lt;real-analysis&gt;&lt;homework&gt;&lt;metric-spaces&gt;" AnswerCount="4" FavoriteCount="0" />
  <row Id="19092" PostTypeId="1" AcceptedAnswerId="19233" CreationDate="2011-01-26T16:26:09.393" Score="5" ViewCount="219" Body="&lt;p&gt;Let $f, g: S^{n-1} \to X$ be a pair of homotopic continuous maps. Let $X \cup_f B^n$ and $X \cup_g B^n$ be the respective adjunction spaces (pushouts of $B^n \hookleftarrow S^{n-1} \rightarrow X$). I need to show that $X \cup_f B^n$ and $X \cup_g B^n$ are homotopy-equivalent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can just about convince myself that it's true by imagining deformations of one into another by embedding in some ambient space, but I guess this isn't possible in general. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also tried proving this by general abstract nonsense but got nowhere, and then I realised I hadn't used any special properties of $B^n$ at all. What properties should I be looking at? A friend mentioned that his solution used the fact that $B^n \setminus U$, where $U \subset S^{n-1}$ is open, is homotopy-equivalent to $B^n$ or something like that, but I haven't been able to make any sense of that remark.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;: After thinking a little more, by more general abstract nonsense, I think I've figured out how to get a sort of deformation map, and it seems to work for arbitrary spaces, so let $S$ be a subspace $B$, and let $h: S \times I \to X$ be a homotopy from $f: S \to X$ to $g: S \to X$. Let $H: S \times I \to X \times I$ be the map $(s, t) \mapsto (h(s, t), t)$. Let $(X \times I) \cup_H (B \times I)$ be the pushout of $B \times I \hookleftarrow S \times I \xrightarrow{H} X \times I$. Then, by the universal property of pushouts, we have a projection $p: (X \times I) \cup_H (B \times I) \to I$ that agrees with the obvious projections, and injections $X \cup_f B \to (X \times I) \cup_H (B \times I)$, and indeed it's even $p^{-1}(\{0\})$. Then, by the universal property of $(X \times I) \amalg (B \times I) \simeq (X \amalg B) \times I$, we have a map $(X \amalg B) \times I \to (X \times I) \cup_H (B \times I)$. So we have a homotopy from the quotient map $X \amalg B \to X \cup_f B$ to the quotient map $X \amalg B \to X \cup_g B$. But I'm not sure this is any help...&lt;/p&gt;&#xA;" OwnerUserId="5191" LastEditorUserId="5191" LastEditorDisplayName="" LastEditDate="2011-01-26T20:17:00.677" LastActivityDate="2011-01-28T15:50:25.650" Title="Homotopy equivalence of two different gluings of $B^n$ and an arbitrary space $X$" Tags="&lt;homework&gt;&lt;algebraic-topology&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="3" />
  <row Id="19093" PostTypeId="2" ParentId="19091" CreationDate="2011-01-26T16:33:22.040" Score="0" ViewCount="" Body="&lt;p&gt;You need to show that every open cover of $\mathbb{K}$ has a finite subcover. So it seems that $M_{r}(A)$ is an open cover of $A$.&lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2011-01-26T16:33:22.040" CommentCount="1" />
  <row Id="19094" PostTypeId="2" ParentId="19070" CreationDate="2011-01-26T17:05:00.497" Score="0" ViewCount="" Body="&lt;p&gt;${\rm GL}_n({\bf R})$ has at least two connected components.  It is disconnected by the continuous function $f: {\rm GL}_n({\bf R})\rightarrow \{-1, 1\}$ defined by $f(T) = \det(T)/|\det(T)|$.  Since ${\rm GL}_n({\bf R})$ is a manifold, it is locally path connected. Therefore, it path components coincide with its connected component.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can find a detailed proof that the matrices of positive determinant are connected in Lee, J., {\it Introduction to Smooth Manfolds}, on page 236.  This requires a little exercise in the analysis of the Graham-Schmidt orthonormalization procedure to produce a path connecting any matrix with positive determinant to the identity.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Switching the first two rows of a matrix defines a homeomorphism of the general linear group onto itself.  It maps the matrices of positive determinant homeomorphically onto those of negative determinant.  Therefore we have exactly two connected components.&lt;/p&gt;&#xA;" OwnerUserId="4154" LastActivityDate="2011-01-26T17:05:00.497" />
  <row Id="19095" PostTypeId="2" ParentId="18537" CreationDate="2011-01-26T17:06:35.347" Score="0" ViewCount="" Body="&lt;p&gt;Looks like most of these can be done through factoring by &quot;grouping&quot;.  There are some ways and more practice problems here: &lt;a href=&quot;http://cnx.org/content/m21901/latest/&quot; rel=&quot;nofollow&quot;&gt;http://cnx.org/content/m21901/latest/&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="2546" LastActivityDate="2011-01-26T17:06:35.347" />
  <row Id="19096" PostTypeId="1" AcceptedAnswerId="19097" CreationDate="2011-01-26T17:16:55.603" Score="2" ViewCount="122" Body="&lt;p&gt;I've got $8n^2 \lt 64n\log(n)$ and I need to find the $n$ range if $n\gt 0$ to satisfy the inequality.&lt;/p&gt;&#xA;" OwnerUserId="4193" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-26T17:35:59.847" LastActivityDate="2011-01-26T17:55:47.240" Title="How to solve simple log inequality?" Tags="&lt;analysis&gt;" AnswerCount="2" />
  <row Id="19097" PostTypeId="2" ParentId="19096" CreationDate="2011-01-26T17:35:23.467" Score="4" ViewCount="" Body="&lt;p&gt;So this is n&amp;lt;8*ln(n).  If you plot it you can find a range of n that works quite easily.  You probably won't be able to solve it exactly, but approximate numerical solutions are available.  I find a range of about 1.2 to 26.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2011-01-26T17:45:33.573" LastActivityDate="2011-01-26T17:45:33.573" CommentCount="4" />
  <row Id="19098" PostTypeId="1" AcceptedAnswerId="19181" CreationDate="2011-01-26T17:43:54.600" Score="0" ViewCount="84" Body="&lt;p&gt;Hey guys, I need to prove the following statment (actually a special case of it).&#xA;Let $R$ be a Noetherian ring, $M$ a finite $R$-module and $I$ an ideal of $R$. &#xA;grade(I,M)$\geq$2 if and only if the homomorphism $M\rightarrow$Hom$_R(I,M)$ given by $m\mapsto(i\mapsto im)$ is an isomorphism.&lt;/p&gt;&#xA;" OwnerUserId="6236" LastActivityDate="2011-01-27T03:08:54.943" Title="On the grade of an ideal" Tags="&lt;abstract-algebra&gt;&lt;commutative-algebra&gt;&lt;modules&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="19099" PostTypeId="1" AcceptedAnswerId="19106" CreationDate="2011-01-26T17:43:58.990" Score="4" ViewCount="73" Body="&lt;p&gt;Problem: Given ${a_n}$ and that 2008 and 2009 are partial limits of ${a_n}$. Also, for all $n$, $|a_{n+1} - a_n| \le 1/2$. Prove that ${a_n}$ has at least 3 partial limits.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I attempted to prove it using the definition of partial limits of a sequence and assuming that there are only two partial limits and show some contradiction, but got nowhere.&lt;/p&gt;&#xA;" OwnerUserId="3637" LastActivityDate="2011-01-26T18:02:02.270" Title="Partial limits of sequence" Tags="&lt;sequences-and-series&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="19100" PostTypeId="1" AcceptedAnswerId="19107" CreationDate="2011-01-26T17:48:41.307" Score="3" ViewCount="131" Body="&lt;p&gt;Playing with integrals on the form $$\int \frac{1}{1-x^n}\,dx$$ I noticed that for odd values of n &gt; 5, it doesn't appear to be possible to express the integral as a closed-form expression. Is this true? How can I prove it?&lt;/p&gt;&#xA;" OwnerUserId="4576" LastActivityDate="2011-01-26T20:12:24.470" Title="Do closed-form expressions exist for these integrals?" Tags="&lt;real-analysis&gt;&lt;integral&gt;" AnswerCount="1" CommentCount="0" />
  <row Id="19101" PostTypeId="2" ParentId="19099" CreationDate="2011-01-26T17:52:06.693" Score="2" ViewCount="" Body="&lt;p&gt;2008? 2009? Sounds like a Putnam Competition question. (They often include the year.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, because of the partial limits, $a_n$ is frequently $&amp;lt;2008\frac14$ and frequently $&amp;gt;2008\frac34$, so must frequently be in $\left[2008\frac14,2008\frac34\right]$ (by the inequality given). Frequently in an interval implies a limit in the interval's closure.&lt;/p&gt;&#xA;" OwnerUserId="4992" LastActivityDate="2011-01-26T17:52:06.693" />
  <row Id="19102" PostTypeId="2" ParentId="19088" CreationDate="2011-01-26T17:54:19.677" Score="5" ViewCount="" Body="&lt;p&gt;Presuming you meant an &lt;em&gt;infinite&lt;/em&gt; number of solutions, Pell's equation can be solved using the theory of continued fractions and the theory shows that $\displaystyle x^2 - dy^2 = 1$ always has an infinite number of solutions for irrational $\displaystyle \sqrt{d}$ ($\displaystyle d$ a positive integer).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The following are well known, for $\displaystyle d$ not a perfect square,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The continued fraction expansion of $\displaystyle \sqrt{d}$ is periodic, say with period $\displaystyle r$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance in your case&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sqrt{13} = (3, \overline{1, 1, 1, 1,6})$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and $\displaystyle r = 5$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $\displaystyle \frac{h_m}{k_m}$ is the $m^{th}$ convergent of $\displaystyle \sqrt{d}$, which is of period $\displaystyle r$, then we have that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(h_{nr-1})^2 - d \ (k_{nr-1})^2 = (-1)^{nr}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus in case of $\displaystyle \sqrt{13}$, we have that for any even $\displaystyle n = 2t$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(h_{10t-1})^2 - d \ (k_{10t-1})^2 = 1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which for instance, gives us the following as a solution:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$649^2 - 13 \times 180^2 = 1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note one can go further and prove that the continued fractions help us generate &lt;em&gt;all&lt;/em&gt; the solutions of the Pell's equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would recommend you read the excellent chapter on Continued Fractions in the book: An Introduction to the Theory of Numbers by Niven and Zuckerman.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2011-01-26T17:59:57.393" LastActivityDate="2011-01-26T17:59:57.393" CommentCount="1" />
  <row Id="19103" PostTypeId="2" ParentId="19096" CreationDate="2011-01-26T17:55:47.240" Score="2" ViewCount="" Body="&lt;p&gt;$n$ suggests that it might be a natural number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore you could use&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$8 n^2 &amp;lt; 64 n log(n)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Longleftrightarrow 0 &amp;lt; 8 log(n) - n$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can easily see that $n$ grows much faster than $log(n)$. Therefore you can see that for $n &amp;gt; 26$ there are no integer solutions. Also the case $n=1$ is problematic as $log(1)=0$. So all in all the numbers $n = \{2,3,\ldots,26\}$ are a solution for your problem. If you need real numbers you will have to use, like already mentioned the Lambert W function ( &lt;a href=&quot;http://en.wikipedia.org/wiki/Lambert_W_function&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Lambert_W_function&lt;/a&gt; )&lt;/p&gt;&#xA;" OwnerUserId="3123" LastActivityDate="2011-01-26T17:55:47.240" CommentCount="1" />
  <row Id="19104" PostTypeId="1" AcceptedAnswerId="19109" CreationDate="2011-01-26T17:58:57.160" Score="3" ViewCount="63" Body="&lt;p&gt;Hello everybody. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How many string exist of length 2M with 9 characters and &lt;strong&gt;without&lt;/strong&gt; repeating a character more than M times? You can suppose that M is greater than 4.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that my english is really bad so i'll give you some examples.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose M=5.&lt;br&gt;&#xA;AABBCCDDEE is a valid string&lt;br&gt;&#xA;AAAAAADEEE is not a valid string (A is repeated more then 5 times).&lt;br&gt;&#xA;AA is not a valid string as it is not of length 2M&lt;br&gt;&#xA;ZZ is not a valid string as you can have 9 different chars :A,B,C,D,E, F, G, H, I.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;p.s. Is not homework: the original question is really different, I'm asking you a simplified model.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you.&lt;/p&gt;&#xA;" OwnerUserId="6277" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-26T19:22:04.997" LastActivityDate="2011-01-26T19:22:04.997" Title="Sort of Combination with repetitions" Tags="&lt;combinatorics&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="19105" PostTypeId="1" AcceptedAnswerId="19141" CreationDate="2011-01-26T18:00:57.793" Score="1" ViewCount="106" Body="&lt;p&gt;Let $S$ be a finite sequence of positive integers of length $n$. Let $G$ be a finite graph containing $n$ not null edgeless subgraphs. Call these subgraphs $G_0,G_1,\ldots,G_{n-1}$ and let each $G_i$ have order $S_i$. Define the edges of $G$ such that, for all $i&amp;gt;0$, $G_i$ forms a complete bipartite graph with $G_{i-1}$. $G$ contains no other edges.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does this graph have a name? I would appreciate any suggestions for a name if none already exists.&lt;/p&gt;&#xA;" OwnerUserId="4735" LastActivityDate="2011-01-26T21:10:30.900" Title="Name this graph" Tags="&lt;graph-theory&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" />
  <row Id="19106" PostTypeId="2" ParentId="19099" CreationDate="2011-01-26T18:02:02.270" Score="4" ViewCount="" Body="&lt;p&gt;Take a subsequence converging to 2008, and one converging to 2009. Argue that you may assume that they have no terms in common, and in fact, the terms in the first subsequence are &quot;very close&quot; to 2008, and those in the second are &quot;very close&quot; to 2009 (if $b_n\to L$ then for $n$ large, $|b_n-L|&amp;lt;1/8$, and we can restrict ourselves to large values of $n$ in both subsequences).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Argue that you may assume the indices of the sequences alternate. This means that if $a_{n_1}, a_{n_2},\dots $ is the sequence converging to 2008 and $a_{m_1},a_{m_2},\dots$ is the sequence converging to 2009, then we may assume that $n_1&amp;lt;m_1&amp;lt;n_2&amp;lt;m_2&amp;lt;n_3&amp;lt;\dots$ (for this, all you need is to &quot;thin&quot; out your subsequences).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ok. Now argue that &lt;em&gt;for all $i$&lt;/em&gt;, between $n_i$ and $m_i$ there is some $k_i$ such that both $a_{k_i}-a_{n_i}$ and $a_{m_i}-a_{k_i}$ are $\ge1/4$. It is here that you use that $|a_{n+1}-a_n|\le1/2$ for all $n$, noting that 2008 and 2009 are 1 unit apart, and $|a_{n_i}-2008|\le 1/8$ and $|a_{m_i}-2009|\le 1/8$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the sequence $(a_{k_i}\mid i=1,2,\dots)$ is completely contained in the interval ${}[2008.25,2008.75]$, so it must have a convergent subsequence, and this subsequence must converge somewhere other than 2008 and 2009.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;On the other hand, note that you cannot conclude that there are more than 3 &quot;partial limits&quot;, by considering the sequence $2008, 2008.5, 2009, 2008.5, 2008, 2008.5,\dots$ &lt;/p&gt;&#xA;" OwnerUserId="462" LastActivityDate="2011-01-26T18:02:02.270" />
  <row Id="19107" PostTypeId="2" ParentId="19100" CreationDate="2011-01-26T18:10:16.527" Score="11" ViewCount="" Body="&lt;p&gt;All of those integrals have closed form expressions, for both even and odd $n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The polynomial $x^n-1$ can be factored over the complex numbers as&#xA;$$x^n - 1 = (x-1)(x-\zeta)(x-\zeta^2)\cdots(x-\zeta^{n-1}),$$ &#xA;and when $n$ is odd, the only real solution is $x-1$ and the complex factors pair up in conjugate pairs to give irreducible real quadratic polynomials. So the integral can be written as&#xA;$$\int\frac{1}{(1-x)Q_1(x)\cdots Q_{r}(x)}\,dx$$&#xA;where $r=\frac{n-1}{2}$ and $Q_i(x)$ is an irreducible quadratic polynomial.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the method of &lt;a href=&quot;http://en.wikipedia.org/wiki/Partial_fractions_in_integration&quot; rel=&quot;nofollow&quot;&gt;partial fractions&lt;/a&gt;, you can then rewrite the integral as a sum of integrals&#xA;$$\int \frac{1}{1-x^n}\,dx = \int\frac{A}{1-x}\,dx + \int\frac{B_1x+C_1}{Q_1(x)}\,dx + \cdots + \int\frac{B_rx + C_r}{Q_r(x)}\,dx$$&#xA;with $A$, $B_i$, and $C_i$ all constants. The problem then reduces to solving these integrals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first integral is easy to do with a change of variable, as is any integral of the form $\int \frac{x}{Q_i(x)}\,dx$ with $Q_i$ irreducible quadratic (see below for how to them). So you only need to know how to solve integrals of the form $\int\frac{1}{Q_i(x)}\,dx$; these can be done by completing the square and doing a change of variable, which will lead to an integral of the form $\int\frac{du}{u^2+1}$, which has a closed form. Again, see below for the general method.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In short, they all have closed forms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When $n$ is even, the only difference is that you have a second linear factor in the factorization, $x+1$, and that the number of irreducible quadratics is then $\frac{n-2}{2}$ instead of $\frac{n-1}{2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To see this in action, let's consider the case of $n=7$. The polynomial $1-x^7$ factors as&#xA;\begin{align*}&#xA;(1-x^7) &amp;amp;= (1-x)(x-\zeta)(x-\zeta^2)(x-\zeta^3)(x-\zeta^4)(x-\zeta^5)(x-\zeta^6)\\&#xA;&amp;amp;= (-x)\Bigl((x-\zeta)(x-\zeta^6)\Bigr)\Bigl((x-\zeta^2)(x-\zeta^5)\Bigr)\Bigl((x-\zeta^3)(x-\zeta^4)\Bigr)&#xA;\end{align*}&#xA;Now, $\zeta=\cos(\frac{2\pi}{7}) + i\sin(\frac{2\pi}{7})$. It is now easy to check that&#xA;\begin{align*}&#xA;(x-\zeta)(x-\zeta^6) &amp;amp;= x^2 -2x\cos\frac{2\pi}{7} + 1\\&#xA;&amp;amp;= x^2 -c_2 x + 1\\&#xA;(x-\zeta^2)(x-\zeta^5) &amp;amp;= x^2 - 2x\cos\frac{4\pi}{7} + 1\\&#xA;&amp;amp;= x^2 -c_4x + 1\\&#xA;(x-\zeta^3)(x-\zeta^4) &amp;amp;= x^2 - 2x\cos\frac{6\pi}{7} + 1\\&#xA;&amp;amp;= x^2 - c_6x + 1&#xA;\end{align*}&#xA;where for simplicity we have $c_j = 2\cos\frac{j\pi}{7}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The discriminant of each of these quadratic polynomials is $4(\cos^2\frac{2k\pi}{7} -1)$, which is always negative, so they are all irreducible quadratics. &#xA;Now use the method of partial fractions to obtain a closed form for the integral. It will be in the form of the sum of natural logs and arctangent functions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, the polynomial will have no repeated quadratic factors, which makes it very easy to solve using partial fractions. The irreducible quadratic factors with $n$ odd will be&#xA;$$x^2 -2x\cos\frac{2k\pi}{n} + 1,\qquad k=1,\ldots,\frac{n-1}{2}.$$&#xA;When $n$ is even, the irreducible quadratic factors will be the same, but $k$ only goes up to $\frac{n-2}{2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just to give you a guide on solving those integrals in general:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;An integral of the form $\int\frac{1}{ax+b}\,dx$ can be solved by doing the change of variable $u=ax+b$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;An integral of the form $\int\frac{1}{x^2+1}\,dx$ is immediate: you get $\arctan(x)+C$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;An integral of the form $\int\frac{1}{(x-a)^2+b^2}\,dx$ can be solved by first factoring out $b^2$ from the denominator, then doing the change of variable $u = \frac{x-a}{b}$, which brings it to the form in 2.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If $Q(x) = x^2+Bx+C$ is irreducible quadratic, it can always be rewritten as $(x-a)^2 + b^2$ for some real numbers $a$ and $b$ (complete the square). &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If $x^2+Bx+C$ is irreducible quadratic, then the integral of the form $\int\frac{x}{x^2+Bx+C}\,dx$ can be solved by rewriting as $\frac{1}{2}\int\frac{2x+B}{x^2+Bx+C}\,dx - \frac{B}{2}\int\frac{1}{x^2+Bx+C}\,dx$. The first integral can be solved by the change of variable $u=x^2+Bx+C$, the second using the method in 3. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Since you can rewrite $\int\frac{1}{1-x^n}\,dx$ into a sum of integrals of the forms described above, there is a closed form for it for all $n$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added:&lt;/strong&gt; In principle, &lt;em&gt;any&lt;/em&gt; integral of a rational function (polynomial divided by polynomial) has a closed form, which can be obtained by factoring the denominator into irreducible factors and using a partial fractions decomposition. This is a consequence of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Fundamental_theorem_of_algebra&quot; rel=&quot;nofollow&quot;&gt;Fundamental Theorem of Algebra&lt;/a&gt;; in fact, it was this particular application, the integration of rational functions, which provided the early impetus to &lt;em&gt;prove&lt;/em&gt; the Fundamental Theorem of Algebra. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only practical obstacle to finding the closed form is our ability to factor the denominator into irreducible factors; repeated irreducible quadratic factors introduce some minor complications that are resolved by using integration by parts to obtain reduction formulas. In the case of the polynomials you were looking at, though, factoring the denominator is straightforward, so that there is no practical difficulty in finding a closed form.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-26T20:12:24.470" LastActivityDate="2011-01-26T20:12:24.470" CommentCount="2" />
  <row Id="19108" PostTypeId="1" AcceptedAnswerId="19111" CreationDate="2011-01-26T18:10:27.513" Score="5" ViewCount="193" Body="&lt;p&gt;The field $\mathbb{C} $ of complex numbers is well-defined by the Hamilton axioms of addition and product between complex numbers, i.e., a complex number $z$ is a ordered pair of real numbers $(x,y)$, which satisfy the following operations $+$ and $\cdot$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ (x_1,y_1) + (x_2,y_2) = (x_1+x_2,y_1+x_2) $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(x_1,y_1)(x_2,y_2) = (x_1x_2-y_1y_2,x_1y_2 + x_2y_1)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The properties of field follow from them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is: Is there a problem in defining complex numbers simply by $z = x+iy$, where $i&amp;#178; = -1$ and $x$, $y$ real numbers and import from $\mathbb{R} $ the operations? Or is just a elegant manner to write the same thing?&lt;/p&gt;&#xA;" OwnerUserId="6279" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-26T19:52:20.807" LastActivityDate="2011-01-26T20:30:27.373" Title="Is there a problem in defining a complex number by $ z = x+iy$?" Tags="&lt;complex-numbers&gt;" AnswerCount="3" FavoriteCount="1" />
  <row Id="19109" PostTypeId="2" ParentId="19104" CreationDate="2011-01-26T18:11:49.613" Score="4" ViewCount="" Body="&lt;p&gt;Assuming order does not matter (i.e. you are looking for sorted strings), you are looking for the number of solutions of&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x_1  + x_2 + \dots +x_9 = 2M$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $0 \le x_i \le M$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is same the coefficient of $x^{2M}$ in&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(1+x+x^2 + \dots + x^{M})^9 = (x^{M+1} - 1)^9 \times (x-1)^{-9}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which gives us (using binomial theorem for $9$ and $-9$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$9 \ (-1)^{M} \ \binom{9+M-1-1}{9} + \binom{9 + 2M -1}{9}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ = 9 \ (-1)^{M} \ \binom{7+M}{9} + \binom{8 + 2M}{9}$$&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2011-01-26T18:14:01.903" LastActivityDate="2011-01-26T18:14:01.903" CommentCount="2" />
  <row Id="19110" PostTypeId="2" ParentId="19108" CreationDate="2011-01-26T18:17:15.847" Score="4" ViewCount="" Body="&lt;p&gt;Just set $i=(0,1)$ and $x=(x,0)$ for any real $x$, and the notation $x+iy$ is just a shorthand for the ordered pairs notation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course you could also choose $i=(0,-1)$ ........&lt;/p&gt;&#xA;" OwnerUserId="688" LastActivityDate="2011-01-26T18:17:15.847" CommentCount="4" />
  <row Id="19111" PostTypeId="2" ParentId="19108" CreationDate="2011-01-26T18:27:22.313" Score="10" ViewCount="" Body="&lt;p&gt;There is no &quot;explicit&quot; problem, but if you are going to define them as formal symbols, then you need to distinguish between the &lt;code&gt;+&lt;/code&gt; in the symbol $a$&lt;code&gt;+&lt;/code&gt;$bi$, the $+$ operation from $\mathbb{R}$, and the sum operation that you will be defining later until you show that they can be &quot;confused&quot;/identified with one another.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is, you define $\mathbb{C}$ to be the set of all symbols of the form $a$&lt;code&gt;+&lt;/code&gt;$bi$ with $a,b\in\mathbb{R}$. Then you define an addition $\oplus$ and a multiplication $\otimes$ by the rules&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(a$&lt;code&gt;+&lt;/code&gt;$bi)\oplus(c$&lt;code&gt;+&lt;/code&gt;$di) = (a+c)$ &lt;code&gt;+&lt;/code&gt; $(c+d)i$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(a$&lt;code&gt;+&lt;/code&gt;$bi)\otimes(c$&lt;code&gt;+&lt;/code&gt;$di) = (ac - bd)$ &lt;code&gt;+&lt;/code&gt; $(ad+bc)i$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $+$ and $-$ are the real number addition and subtraction, and &lt;code&gt;+&lt;/code&gt; is merely a formal symbol.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then you can show that you can identify the real number $a$ with the symbol $a$&lt;code&gt;+&lt;/code&gt;$0i$; and that $(0$&lt;code&gt;+&lt;/code&gt;$i)\otimes(0$&lt;code&gt;+&lt;/code&gt;$i) = (-1)$&lt;code&gt;+&lt;/code&gt;$0i$; etc. At that point you can start &lt;em&gt;abusing notation&lt;/em&gt; and describing it as you do, using the same symbol for $+$, $\oplus$, and &lt;code&gt;+&lt;/code&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So... the method you propose (which was in fact how complex numbers were used at first) is just a bit more notationally abusive, while the method of ordered pairs is much more formal, giving a precise &quot;substance&quot; to complex numbers as &quot;things&quot; (assuming you think the plane is a &quot;thing&quot;) and not just as &quot;formal symbols&quot;. &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-26T20:30:27.373" LastActivityDate="2011-01-26T20:30:27.373" />
  <row Id="19112" PostTypeId="2" ParentId="19108" CreationDate="2011-01-26T18:30:17.367" Score="4" ViewCount="" Body="&lt;p&gt;There is a completely rigorous way to do the construction you allude to in the last paragraph, namely by means of quotient rings. Indeed, $\mathbb{C} \simeq \mathbb{R}[X] / (X^2 + 1)$. This generalises, for example, we can construct a commutative ring with elements of the form $x + y \epsilon$, where $\epsilon^2 = 0$. The ring so constructed is emphatically not a field, but it is sometimes useful for doing symbolic differentiation. &lt;/p&gt;&#xA;" OwnerUserId="5191" LastActivityDate="2011-01-26T18:30:17.367" />
  <row Id="19113" PostTypeId="2" ParentId="19091" CreationDate="2011-01-26T18:32:59.287" Score="2" ViewCount="" Body="&lt;p&gt;It is probably easier to appeal to the Heine-Borel Theorem&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Step (1): Show that $D$ is a metric on $\mathbb{K}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now that $(\mathbb{K},D)$ is a metric space, by the Heine-Borel theorem, it suffices to show that it is complete and totally bounded. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Step (2): Let $A_i$ be a Cauchy sequence in $\mathbb{K}$, show that it converges. (In fact, you can show that as long as $M$ is a complete metric space, then so is $\mathbb{K}$ with the metric you just wrote down.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Step (3): Show that for every $\epsilon &amp;gt; 0$ there exists a finite open cover of $\mathbb{K}$ by $\epsilon$-balls. (In fact, as long as $M$ is totally bounded, so will $\mathbb{K}$.)&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Step (1) is obvious, so I won't give a hint. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For Step (2), let $A_i$ be your Cauchy sequence. Consider the set $A$ of points $a$ such that for any $\epsilon &amp;gt; 0$, $B(a,\epsilon)$ intersects all but finitely many $A_i$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For Step (3), start with a finite cover of $M$ by $\epsilon$ balls, let $S$ be the set of the center points of those balls. Consider the power-set $P(S)$ (the set of all subsets of $S$) as a set of points in $\mathbb{K}$. &lt;/p&gt;&#xA;" OwnerUserId="1543" LastActivityDate="2011-01-26T18:32:59.287" CommentCount="5" />
  <row Id="19114" PostTypeId="1" CreationDate="2011-01-26T18:38:36.590" Score="0" ViewCount="105" Body="&lt;p&gt;Hi all,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have some random integers separated by space read from a text file like&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;99 20 30 1 100 400 5 10&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I have to find a sum from any combination of these integers that is closest to a specifiq number like&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;183&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;what is the fastest and accurate way of doing this?&lt;/p&gt;&#xA;" OwnerUserId="6283" LastActivityDate="2011-01-28T03:55:41.150" Title="Series calculation" Tags="&lt;sequences-and-series&gt;&lt;mathematica&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="19116" PostTypeId="2" ParentId="19091" CreationDate="2011-01-26T18:49:32.473" Score="1" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Hints:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;It suffices to show that $\mathbb{K}$ is complete and totally bounded.&lt;/li&gt;&#xA;&lt;li&gt;For every $\varepsilon &amp;gt; 0$ there is a constant $C(\varepsilon)$ such that every $K \in \mathbb{K}$ can be covered by at most $C(\varepsilon)$ balls of radius $\varepsilon$ around points in $K$. To see this, cover $M$ with finitely many balls of radius $\varepsilon/2$, let $K \in \mathbb{K}$ be arbitrary and pick $x_{n}$ in the intersection of $K$ with those balls that intersect $K$. The balls with radius $\varepsilon$ around these balls will cover $K$.&lt;/li&gt;&#xA;&lt;li&gt;Given a Cauchy sequence in $\mathbb{K}$, show that it converges (this doesn't need compactness!).&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="5363" LastActivityDate="2011-01-26T18:49:32.473" CommentCount="2" />
  <row Id="19118" PostTypeId="2" ParentId="19091" CreationDate="2011-01-26T19:24:37.983" Score="1" ViewCount="" Body="&lt;p&gt;As an aside (I do not suggest you prove it this way): this $\mathbb{K}$ is just the so-called hyperspace of $M$, also denoted $H(M)$ (all non-empty closed sets of $M$ in general), and an alternative, non-metric description of its topology is by describing a subbase for it, consisting of all $[U] = \{A \in \mathbb{K}:\, A \cap U \neq \emptyset \}$ and $&amp;lt;U&amp;gt; = \{ A \in \mathbb{K}:\, A \subset U \}$, where $U$ ranges over all non-empty subsets of $M$. If $M$ is compact so is $H(M)$, by a simple application of the Alexander subbase lemma. This is a more general topological way of viewing this space (it's a theorem that for compact metric spaces $M$ the space $H(M)$ is compact metrizable as well, and one of the metrics is the one descibed in the question). &lt;/p&gt;&#xA;" OwnerUserId="4280" LastActivityDate="2011-01-26T19:24:37.983" CommentCount="1" />
  <row Id="19119" PostTypeId="1" AcceptedAnswerId="19128" CreationDate="2011-01-26T19:27:35.680" Score="1" ViewCount="273" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need to estimate $\pi$ using the following integration:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_{0}^{1} \!\sqrt{1-x^2} \ dx$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;using monte carlo &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any help would be greatly appreciated, please note that I'm a student trying to learn this stuff so if you can please please be more indulging and try to explain in depth..&lt;/p&gt;&#xA;" OwnerUserId="6284" LastEditorUserId="6284" LastEditorDisplayName="" LastEditDate="2011-01-26T20:04:43.540" LastActivityDate="2011-01-28T15:23:44.580" Title="Approximating $\pi$ using Monte Carlo integration" Tags="&lt;integral&gt;&lt;approximation&gt;&lt;pi&gt;&lt;monte-carlo&gt;" AnswerCount="4" CommentCount="3" FavoriteCount="1" />
  <row Id="19120" PostTypeId="2" ParentId="19070" CreationDate="2011-01-26T19:34:38.747" Score="0" ViewCount="" Body="&lt;p&gt;Here's another proof.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $Gl = GL_n(\mathbb{R})$ is an open subset of $\mathbb{R}^{n^2}$, it is naturally a smooth manifold.  Hence, the path components are the same as the components of it.  Further, since multiplication and inverses can be written as polynomials in the entries of the matrix, these operations are continuous (smooth in fact).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$H$ is clearly the path component of $Gl$, by the above, it's actually a component in the usual sense.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, fix $h\in H$ and consider the map $L_h:Gl\rightarrow Gl$ sending $A$ to $hA$.  This is continuous by the above, so it sends components to components.  Where does it send $H$?  Well, since $I\in H$, we see that $L_h I = hI = h$, so $L_h$ sends one point in $H$ to another point in $H$.  It follows that $L_h$ sends all of $H$ into itself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This shows that $H$ is closed under multiplication.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What about inverses?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, first note that $L_h$ is a homeomorphism because it has inverse $L_h^{-1} = L_{h^{-1}}$.  This shows that the ONLY component it sends onto $H$ is $H$.  Let $X$ be the component of $Gl$ containing $h^{-1}$.  Then we know $L(X) \subseteq H$ since $h^{-1}\in X$ and $L_h h^{-1} = hh^{-1} = Id\in H$.  This implies that $X = H$, i.e., that $h^{-1}\in H$.  So, $H$ is closed under inverses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have now shown that $H$ is a subgroup - now we need only show it's normal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, let $\in Gl$.  Consider the map $C_g:Gl\rightarrow Gl$ sending $A$ to $hAh^{-1}$.  As above, this map is continuous.  So, it sends components to components.  Where does it send $H$?  Well, $I\in H$ and $C_g(I) = gIg^{-1} = I$, so it sends $H$ to $H$, that is, $C_g(H)\subseteq H$.  But $C_g(H) = gHg^{-1}$, so $H$ is normal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Incidentally, what I really proved is the following:  Let $G$ be any Lie group.  Let $G_0$ be the identity path component of $G$.  Then $G_0$ is a normal subgroup of $G$.  More generally, if $G$ is a topological group such that components and path components agree, this works.  I'm not sure whether or not $G$ is a topological group implies the components and path components agree, though).&lt;/p&gt;&#xA;" OwnerUserId="331" LastActivityDate="2011-01-26T19:34:38.747" />
  <row Id="19121" PostTypeId="2" ParentId="19119" CreationDate="2011-01-26T19:38:18.477" Score="2" ViewCount="" Body="&lt;p&gt;No longer applicable as the integral has been corrected:  I don't understand how this integral gets you $\pi$ and if you pull the $x$ out you get $\int_0^1x\sqrt{10}dx=\frac{\sqrt{10}}{2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The general idea  of a Monte Carlo integration of $\int_0^1 f(x)dx$ is to take random pairs $(x,y)$ with $0\le x\le 1$ and $0\le y \le y_{max}$ and check for how many of the pairs $f(x)\le y$.  The integral is then the number of pairs under the curve divided by the number of trials and multiplied by the area of the box ($x y_{max}$)&lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2011-01-28T15:23:44.580" LastActivityDate="2011-01-28T15:23:44.580" CommentCount="6" />
  <row Id="19122" PostTypeId="2" ParentId="19054" CreationDate="2011-01-26T19:39:22.190" Score="1" ViewCount="" Body="&lt;p&gt;This is not a complete answer, but a small lemma that may be useful: If $f$ exists, it is unbounded on any interval.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Without loss of generality, let the interval be $[0,\ell]$. For some $n$, consider the $n+1$ values $f(0)$, $f(\ell/n)$, $f(2\ell/n)$, $\ldots,$ $f(\ell)$. Any two of these must differ by at least $\sqrt{\ell/n}$, so the $n+1$ values together must span a range of $n\sqrt{\ell/n} = \sqrt{n\ell}$. Let $n$ go to infinity, and the lemma follows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, there &lt;em&gt;do&lt;/em&gt; exist functions that are unbounded on any interval (for example, $g(x) = q$ if $x = p/q$, and $0$ otherwise), so this doesn't rule out the existence of the proposed function. But if it does exist, it would have to be fairly pathological.&lt;/p&gt;&#xA;" OwnerUserId="856" LastActivityDate="2011-01-26T19:39:22.190" CommentCount="0" />
  <row Id="19123" PostTypeId="1" CreationDate="2011-01-26T19:46:21.467" Score="0" ViewCount="91" Body="&lt;p&gt;I'm trying to assemble a calculation that takes a known number of items (30) and places them into a set number of containers (3) and to calculate that items 1-10 go into container 1, 11-20 go into container 2 and 21-30 go into container 3... over time the number of items will grow infinitely and the number of containers will equal #items/10 (rounded up to the next whole number, so that 35 items would have 4 containers)...&#xA;Known = # items&#xA;Known = # containers&#xA;Items must go into specific containers (items 1-10 must go into container 1, 11-20 into contain 2 and so on)&lt;/p&gt;&#xA;" OwnerUserId="6285" LastActivityDate="2011-05-26T23:02:55.673" Title="Sorting items to containers" Tags="&lt;sorting&gt;" AnswerCount="2" />
  <row Id="19124" PostTypeId="2" ParentId="19123" CreationDate="2011-01-26T19:51:56.173" Score="1" ViewCount="" Body="&lt;p&gt;This is &lt;a href=&quot;http://en.wikipedia.org/wiki/Bucket_sort&quot; rel=&quot;nofollow&quot;&gt;bucket sort&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2011-01-26T19:51:56.173" />
  <row Id="19125" PostTypeId="1" CreationDate="2011-01-26T19:56:11.740" Score="2" ViewCount="138" Body="&lt;p&gt;I've attempted to solve the problem, but I got $\langle \frac{1}{\sqrt{\frac{29}{4}}}, \frac{5}{\sqrt{\frac{29}{4}}}\rangle$, which is incorrect. There is not a similar problem in my textbook that I can reference.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that to find a unit vector, we first find the length/magnitude of the given vector, and multiply 1/sqrt(magnitude) by the original vector.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$L = \sqrt{x^2 + y^2}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone give me any ideas on how to solve this problem?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Find the unit vector that has the same direction as the vector from the point A = (-1,2) to the point B = (3,3).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you in advance.&lt;/p&gt;&#xA;" OwnerUserId="3800" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-26T20:14:32.550" LastActivityDate="2011-02-25T20:28:37.970" Title="Homework Help - Finding a Vector when given two points, and then finding a unit vector in the same direction." Tags="&lt;homework&gt;&lt;vector-calculus&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="19126" PostTypeId="2" ParentId="19125" CreationDate="2011-01-26T20:02:26.767" Score="0" ViewCount="" Body="&lt;p&gt;When you took the vector from A to B it looks like you added instead of subtracting.  It should be (4,1).  &lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-01-26T20:02:26.767" />
  <row Id="19127" PostTypeId="2" ParentId="19125" CreationDate="2011-01-26T20:03:45.057" Score="1" ViewCount="" Body="&lt;p&gt;The vector that goes from $A$ to $B$ is the vector $B-A$: to see this, notice that if you add vectors using the parallelogram rule, then adding the vector $V$ you are looking for to $A$ should give you $B$, so $A+V = B$, giving you $V=B-A$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the vector you are looking for is $V = B-A = (3,3) - (-1,2) = (4,1)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now that you know the vector, finding the unit vector in the same direction is done as you indicate: find the magnitude of $V$, divide by the magnitude. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Looks like you took $A+B$ instead of $B-A$) &lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2011-01-26T20:03:45.057" />
  <row Id="19128" PostTypeId="2" ParentId="19119" CreationDate="2011-01-26T20:09:07.957" Score="5" ViewCount="" Body="&lt;p&gt;Generate a sequence $U_1,U_2,\ldots$ of independent uniform$[0,1]$ random variables. &#xA;Let $Y_i = f(U_i)$, where $f(x)=\sqrt{1-x^2}$, $0 \leq x \leq 1$. Then, for sufficiently large $n$,&#xA;$$&#xD;&#xA;\frac{{\sum\nolimits_{i = 1}^n {Y_i } }}{n} \approx \int_0^1 {f(x)\,{\rm d}x = } \int_0^1 {\sqrt {1 - x^2 } \,{\rm d}x}  = \frac{\pi }{4}.&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;: Elaborating.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose that $U_1,U_2,\ldots$ is a sequence of independent uniform$[0,1]$ random variables, and $f$ is an integrable function on $[0,1]$, that is, $\int_0^1 {|f(x)|\,{\rm d}x}  &amp;lt; \infty$. Then, the (finite) integral $\int_0^1 {f(x)\,{\rm d}x}$ can be approximated as follows. Let $Y_i = f(U_i)$, so the $Y_i$ are independent and identically distributed random variables, with mean (expectation) $\mu$ given by&#xA;$$&#xD;&#xA;\mu = {\rm E}[Y_1] = {\rm E}[f(U_1)] = \int_0^1 {f(x)\,{\rm d}x}.&#xD;&#xA;$$&#xA;By the strong law of large numbers, the average $\bar Y_n  = \frac{{\sum\nolimits_{i = 1}^n {Y_i } }}{n}$ converges, with probability $1$, to the expectation $\mu$ as $n \to \infty$. That is, with probability $1$, $\bar Y_n  \to \int_0^1 {f(x)\,{\rm d}x}$ as $n \to \infty$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To get a probabilistic error bound, suppose further that $f$ is square-integrable on $[0,1]$, that is $&#xD;&#xA;\int_0^1 {f^2 (x)\,{\rm d}x}  &amp;lt; \infty $. Then, the $Y_i$ have finite variance, $\sigma_2$, given by&#xA;$$&#xD;&#xA;\sigma^2 = {\rm Var}[Y_1] = {\rm E}[Y_1^2] - {\rm E}^2{[Y_1]} = {\rm E}[f^2{(U_1)}] - {\rm E}^2{[f(U_1)]} = &#xD;&#xA;\int_0^1 {f^2 (x) \,{\rm d}x}  - \bigg[\int_0^1 {f(x)\,{\rm d}x} \bigg]^2 .&#xD;&#xA;$$ &#xA;By linearity of expectation, the average $\bar Y_n $ has expectation &#xA;$$&#xD;&#xA;{\rm E}[\bar Y_n] = \mu.&#xD;&#xA;$$&#xA;Since the $Y_i$ are independent, $\bar Y_n $ has variance&#xA;$$&#xD;&#xA;{\rm Var}[\bar Y_n] = {\rm Var}\bigg[\frac{{Y_1  +  \cdots  + Y_n }}{n}\bigg] = \frac{1}{{n^2 }}{\rm Var}[Y_1  +  \cdots  + Y_n ] = \frac{n}{{n^2 }}{\rm Var}[Y_1 ] = \frac{{\sigma ^2 }}{n}.&#xD;&#xA;$$ &#xA;By Chebyshev's inequality, for any given $\varepsilon &amp;gt; 0$,&#xA;$$&#xD;&#xA;{\rm P}\big[\big|\bar Y_n - {\rm E}[\bar Y_n]\big| \geq \varepsilon \big] \leq \frac{{{\rm Var}[\bar Y_n]}}{{\varepsilon ^2 }},&#xD;&#xA;$$&#xA;so&#xA;$$&#xD;&#xA;{\rm P}\big[\big|\bar Y_n - \mu \big| \geq \varepsilon \big] \leq \frac{{\sigma^2}}{{n \varepsilon ^2 }},&#xD;&#xA;$$&#xA;and hence&#xA;$$&#xD;&#xA;{\rm P}\bigg[\bigg|\bar Y_n -  \int_0^1 {f(x)\,{\rm d}x} \bigg| \geq \varepsilon \bigg] \leq \frac{1}{{n \varepsilon ^2 }} \bigg \lbrace \int_0^1 {f^2 (x) \,{\rm d}x}  - \bigg[\int_0^1 {f(x)\,{\rm d}x} \bigg]^2 \bigg \rbrace.&#xD;&#xA;$$&#xA;So if $n$ is sufficiently large, with high probability the absolute difference between $\bar Y_n$ and $\int_0^1 {f(x)\,{\rm d}x}$ will be smaller than $\varepsilon$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Returning to your specific question, letting $f(x)=\sqrt{1-x^2}$ thus gives&#xA;$$&#xD;&#xA;{\rm P}\Big[\Big|\bar Y_n - \frac{\pi }{4} \Big| \geq \varepsilon \Big] \leq \frac{1}{{n \varepsilon ^2 }} \bigg \lbrace \int_0^1 {(1 - x^2) \,{\rm d}x}  -  \frac{\pi^2 }{16} \bigg \rbrace  = \frac{1}{{n \varepsilon ^2 }} \bigg \lbrace  \frac{2}{3} -  \frac{\pi^2 }{16} \bigg \rbrace &amp;lt; \frac{1}{{20n\varepsilon ^2 }},&#xD;&#xA;$$&#xA;where $\bar Y_n  = \frac{{\sum\nolimits_{i = 1}^n {\sqrt {1 - U_i^2 } } }}{n}$.&lt;/p&gt;&#xA;" OwnerUserId="2810" LastEditorUserId="2810" LastEditorDisplayName="" LastEditDate="2011-01-27T04:25:17.223" LastActivityDate="2011-01-27T04:25:17.223" CommentCount="7" />
  <row Id="19129" PostTypeId="2" ParentId="19125" CreationDate="2011-01-26T20:11:56.547" Score="0" ViewCount="" Body="&lt;p&gt;Thank you! I got the correct answer. If anyone else can help me further: There is not an example problem in my textbook that I can reference (I have said this on two other questions so far, but it's the truth. The example problems and solutions manual contain physics / engineering examples rather than abstract examples).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For this one, I did not know how to begin. Normally I would give the work I had done, even if it is incorrect.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;For what values of t and so does the&#xA;  equality &amp;lt;4-2t, 2s-4t &gt;= &amp;lt; s+4t, 3+2t&gt;.&#xA;  Find t and s. At these values of t and&#xA;  s, find the resulting vector.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Can anyone point me in the correct direction or give me an idea on how to solve this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you in advance.&lt;/p&gt;&#xA;" OwnerUserId="3800" LastActivityDate="2011-01-26T20:11:56.547" CommentCount="4" />
  <row Id="19130" PostTypeId="2" ParentId="19123" CreationDate="2011-01-26T20:14:20.243" Score="0" ViewCount="" Body="&lt;p&gt;If you have $I$ items and $C$ containers, the number per container is $N=\lceil \frac{I}{C} \rceil$ Then the bin number for item $m$ is $\lfloor \frac{m-1}{N}\rfloor +1$&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-01-26T20:14:20.243" />
  <row Id="19131" PostTypeId="1" CreationDate="2011-01-26T20:39:21.400" Score="4" ViewCount="314" Body="&lt;p&gt;Hello :)&#xA;I hit a problem.&#xA;If $\sin x + \cos x = \frac{\sqrt{3} + 1}{2}$, then how much is $\tan x  + \cot x$?&lt;/p&gt;&#xA;" OwnerUserId="5745" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-26T20:42:55.303" LastActivityDate="2011-01-27T00:06:50.377" Title="If $\sin x + \cos x = \frac{\sqrt{3} + 1}{2}$ then $\tan x + \cot x=?$" Tags="&lt;trigonometry&gt;" AnswerCount="5" CommentCount="4" />
  <row Id="19132" PostTypeId="1" AcceptedAnswerId="19145" CreationDate="2011-01-26T20:40:35.630" Score="0" ViewCount="159" Body="&lt;p&gt;I am trying to prove a result, for which i have got one part, but i am not able to get the converse part.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Theorem.&lt;/strong&gt; Let $R$ be a commutative ring with $1$. Then $f(X)=a_{0}+a_{1}X+a_{2}X^{2} + \cdots + a_{n}X^{n}$ is a unit in $R[X]$ if and only if $a_{0}$ is a unit in $R$ and $a_{1},a_{2},\cdots,a_{n}$ are all &lt;em&gt;nilpotent&lt;/em&gt; in $R$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof.&lt;/strong&gt; Suppose $f(X)=a_{0}+a_{1}X+\cdots +a_{n}X^{n}$ is such that $a_{0}$ is a unit in $R$ and $a_{1},a_{2}, \cdots,a_{r}$ are all nilpotent in $R$. Since $R$ is commutative, we get that $a_{1}X,a_{2}X^{2},\cdots,a_{n}X^{n}$ are all nilpotent and hence also their sum is nilpotent. Let $z = \sum a_{i}X^{i}$ then $a_{0}^{-1}z$ is nilpotent and so $1+a_{0}^{-1}z$ is a unit. Thus $f(X)=a_{0}+z=a_{0} \cdot (1+a_{0}^{-1}z)$ is a unit since product of two units in $R[X]$ is a unit. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Converse part, i have not been able to get. Would like to see the proof for the converse part.&lt;/p&gt;&#xA;" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-26T20:47:19.417" LastActivityDate="2011-01-26T22:40:51.160" Title="Completing the proof of a theorem." Tags="&lt;abstract-algebra&gt;&lt;ring-theory&gt;" AnswerCount="2" CommentCount="8" FavoriteCount="1" />
  <row Id="19133" PostTypeId="2" ParentId="19056" CreationDate="2011-01-26T20:42:13.037" Score="0" ViewCount="" Body="&lt;p&gt;Summing over all $n$ from $1$ to $100$ Mathematica produces&#xA;1.0000277854514457314*10^6.&lt;/p&gt;&#xA;" OwnerUserId="1303" LastActivityDate="2011-01-26T20:42:13.037" />
  <row Id="19134" PostTypeId="2" ParentId="19131" CreationDate="2011-01-26T20:45:33.490" Score="3" ViewCount="" Body="&lt;p&gt;Hint: write $\tan(x) + \cot(x)$ in terms of $\sin(x)$ and $\cos(x)$, with a common denominator. Square $\sin(x) + \cos(x) = {\sqrt{3} + 1 \over 2}$ and compare. &lt;/p&gt;&#xA;" OwnerUserId="3035" LastActivityDate="2011-01-26T20:45:33.490" />
  <row Id="19135" PostTypeId="2" ParentId="19131" CreationDate="2011-01-26T20:46:18.943" Score="8" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;$\tan{x}+\cot{x} = \frac{1}{\sin{x} \cdot \cos{x}}$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$(\sin{x}+\cos{x})^{2} -1 = 2 \sin{x} \cdot \cos{x}$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="9413" LastEditorUserId="9413" LastEditorDisplayName="" LastActivityDate="2011-01-26T20:46:18.943" />
  <row Id="19136" PostTypeId="2" ParentId="19131" CreationDate="2011-01-26T20:46:33.990" Score="18" ViewCount="" Body="&lt;p&gt;We have&#xA;$$\tan x + \cot x = \frac{\sin x}{\cos x} + \frac{\cos x}{\sin x} = \frac{\sin^2 x + \cos^2 x}{\sin x\cos x} = \frac{1}{\sin x\cos x}.$$&#xA;So you want to know how much is $\sin x\cos x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Squaring the equality you know, we have:&#xA;\begin{align*}&#xA;(\sin x + \cos x)^2 &amp;amp;=  \left(\frac{\sqrt{3}+1}{2}\right)^2\\&#xA;\sin^2 x + 2\sin x\cos x + \cos^2 x &amp;amp; = \frac{4 + 2\sqrt{3}}{4}\\&#xA;1 + 2\sin x \cos x &amp;amp;= 1 + \frac{\sqrt{3}}{2}\\&#xA;2\sin x \cos x &amp;amp;= \frac{\sqrt{3}}{2}\\&#xA;\sin x \cos x &amp;amp;= \frac{\sqrt{3}}{4}\\&#xA;\tan x + \cot x = \frac{1}{\sin x\cos x} &amp;amp;= \frac{4\sqrt{3}}{3}.&#xA;\end{align*}&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2011-01-26T20:46:33.990" />
  <row Id="19137" PostTypeId="2" ParentId="19131" CreationDate="2011-01-26T20:48:54.953" Score="5" ViewCount="" Body="&lt;p&gt;Since $\tan x=\frac{\sin x}{\cos x}$ and $\cot x=\cdots$ (similar to tangent), you can rewrite $\tan x+\cot x$ in terms of sine and cosine.  The result can be simplified to an expression involving the product of sine and cosine.  The square of $\sin x+\cos x$ can also be simplified to an expression involving the product of sine and cosine.  Together, this yields an answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;However&lt;/strong&gt;, in this particular instance, $\sin x+\cos x=\frac{\sqrt{3}+1}{2}$ strongly suggests $\sin x=\frac{\sqrt{3}}{2}$ and $\cos x=\frac{1}{2}$ (or vice versa), as that will happen for a particularly nice value of $x$, which will make calculating $\tan x+\cot x$ simple.&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2011-01-26T20:48:54.953" />
  <row Id="19138" PostTypeId="1" AcceptedAnswerId="19210" CreationDate="2011-01-26T20:49:34.663" Score="1" ViewCount="133" Body="&lt;p&gt;Given the Butcher table $$ \begin{array}{ccc} \Theta &amp;amp; \Big | &amp;amp; \Theta \\\ --&amp;amp;--&amp;amp;-- \\\ &amp;amp; | &amp;amp; 1\end{array}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a) write down the corresponding evolution $\Psi$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;b) find the stability function $S(z)$ of the &quot;$\Theta$-method&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Somehow, I end up getting the Euler method but I don't know why. I would be very grateful if someone could point out my mistake to me. So here is what I did:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a) The RK formula is $$ c_i := \sum_{j=1}^s a_{ij}$$&#xA;$$ k_i := f(t_0 + c_i h, y_0 + h \sum_{j=1}^s a_{ij} k_j)$$&#xA;$$ \Psi^{t_0, t_0 + h} y_0 := y_0 + h \sum_{i=1}^s b_i k_i$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Padding the Butcher table with zeros, I get the following Runge Kutta coefficients:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$c_1 = 0$, $c_2 = \Theta$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a_{11}= a_{12} = a_{22} = 0$, $a_{21} = \Theta$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$b_1 = 1$, $b_2 = 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$k_1 = f(t + c_1 h, y_0 + h(a_{11}k_1 + a_{12} k_2) = f(t, y_0)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$k_2 = f(t + \Theta h, y_0 + h(a_{21}k_1 + a_{22}k_2)) = f(t + h\Theta , y_0 + h \Theta f(t, y_0))$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's a two step RK method, i.e. s = 2 and my $\Psi$ is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \Psi^{t_0 , t_0 + h} y_0 = y_0 + h(b_1 k_1 + b_2 k_2) = y_0 + h f(t_0, y_0)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;b) The stability function of the Euler method is $S(z) = 1 + z$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that this is wrong because this is an old exam question and the stability function they give in their solution is $S(z) = \frac{1 + (1 - \Theta)z}{1 - \Theta z}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've done it 4 times, getting the Euler method which does not depend on $\Theta$. I would be very grateful if someone could help me with this. Many thanks in advance.&lt;/p&gt;&#xA;" OwnerUserId="5798" LastActivityDate="2011-01-27T11:59:36.807" Title="2 step Runge Kutta method with parameter $\Theta$" Tags="&lt;numerical-methods&gt;" AnswerCount="1" FavoriteCount="0" />
  <row Id="19139" PostTypeId="2" ParentId="19132" CreationDate="2011-01-26T21:04:32.207" Score="0" ViewCount="" Body="&lt;p&gt;Let $f(x) = x^n a_n + x^{n-1}a_{n-1} + \dots + a_0$ be a unit, i.e. $$\exists g(x) \in R[x] : f(x) g(x) = 1 \forall x \in R$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $g(x) = x^m b_m + x^{m-1}b_{m-1} + \dots + b_0$ then $$f(x)g(x) = x^{m + n}a_n b_m + \dots + a_0 b_0 = 1 \forall x \in R$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, for $x = 0$ it follows that $a_0 b_0 = 1$, so it follows that $a_0$ is a unit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Will try and add the rest of the proof later.&lt;/p&gt;&#xA;" OwnerUserId="5798" LastActivityDate="2011-01-26T21:04:32.207" CommentCount="3" />
  <row Id="19140" PostTypeId="2" ParentId="17939" CreationDate="2011-01-26T21:08:44.920" Score="0" ViewCount="" Body="&lt;p&gt;This isn't an answer to the exact question I asked, but it will do:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The morphism $\mathbb{G}_m \to \mathbb{A}^1$ is an open immersion, hence &amp;eacute;tale, hence formally &amp;eacute;tale.  This exactly means that maps in from morphisms $\operatorname{spec} R/I \to \operatorname{spec} R$, where $I$ is a nilpotent ideal in $R$, lift.&lt;/p&gt;&#xA;" OwnerUserId="2574" LastActivityDate="2011-01-26T21:08:44.920" />
  <row Id="19141" PostTypeId="2" ParentId="19105" CreationDate="2011-01-26T21:10:30.900" Score="1" ViewCount="" Body="&lt;p&gt;This is known as a &quot;layered graph&quot;. If all $S_i$ are equal then it's a &quot;trellis diagram&quot;, an object used by the Viterbi decoding algorithm (which is a fancy name for dynamic programming).&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2011-01-26T21:10:30.900" CommentCount="2" />
  <row Id="19142" PostTypeId="2" ParentId="19114" CreationDate="2011-01-26T21:16:46.257" Score="1" ViewCount="" Body="&lt;p&gt;The non-trivial algorithm takes $2^{n/2}$ time and memory. Divide the list into two parts and calculate all the internal sums. Sort both lists, put one pointer at the low end of one list, and another at the high end of the other list. Advance the second pointer until the sum crosses the target. Then advance the first pointer, and repeat. Maintain the &quot;best sum encountered so far&quot;, and once the first pointer is done traversing the list, you have the optimum in your hands.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This algorithm isn't really practical, and in practice you can probably do better, potentially depending on properties of the problem. Also, there are probably nice approximation algorithms that work well in practice.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2011-01-26T21:16:46.257" />
  <row Id="19143" PostTypeId="2" ParentId="19119" CreationDate="2011-01-26T21:19:54.647" Score="1" ViewCount="" Body="&lt;p&gt;C# implementation. Thanks to all who contributed!! Ross,  Esteban, Shai&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;class Program&#xA;{&#xA;    static void Main(string[] args)&#xA;    {&#xA;        double x = 0;&#xA;        var rd = new Random();&#xA;&#xA;        double sum = 0;&#xA;        for (int i = 0; i &amp;lt; 1000000; i++) // the higher the more precise&#xA;        {&#xA;            x = rd.NextDouble();&#xA;            sum += Math.Sqrt(1 - x*x); &#xA;        }&#xA;&#xA;        Console.WriteLine(&quot;Pi={0}&quot;, (sum/1000000)*4);&#xA;        Console.Read();&#xA;    }&#xA;}&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="6284" LastActivityDate="2011-01-26T21:19:54.647" CommentCount="3" />
  <row Id="19144" PostTypeId="1" CreationDate="2011-01-26T21:26:06.817" Score="1" ViewCount="99" Body="&lt;p&gt;I have a ring $R$, a nilpotent ideal $I$, and two descending sequences of ideals $A_i$ and $B_j$ satisfying:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$\bigcap_i A_i = 0 = \bigcap_j B_j$,&lt;/li&gt;&#xA;&lt;li&gt;$A_i \nsubseteq I$ and $I \nsubseteq A_i$ for all $i$, and similarly for $B_j$ for all $j$,&lt;/li&gt;&#xA;&lt;li&gt;$R$ is complete with respect to the topologies induced by both of the inverse limits $\lim_i R/A_i$ and $\lim_j R/B_j$.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I then have four morphisms:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$\lim_i R/(I, A_i) \to \lim_j R/(I, B_j)$,&lt;/li&gt;&#xA;&lt;li&gt;$\lim_j R/(I, B_j) \to \lim_i R/(I, A_i)$,&lt;/li&gt;&#xA;&lt;li&gt;$\lim_i R/A_i \to \lim_j R/B_j$,&lt;/li&gt;&#xA;&lt;li&gt;$\lim_j R/B_j \to \lim_i R/A_i$.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Morphisms 1 and 2 are mutual inverses, as are 3 and 4.  Morphism 3 (resp. 4) reduces to morphism 1 (resp. 2) after quotienting out $I$.  I also know that morphisms 1, 2, and 3 are continuous with respect to the topologies induced by the various limits.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;My question is:&lt;/strong&gt; Is morphism 4 necessarily also continuous?  If not, what's a counterexample?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have specific $R$, $I$, $A_i$, and $B_j$ in mind for which I want this to be true, and while I don't think their exact values are relevant, even a counterexample in the general case would help me pin down what's special about my situation.&lt;/p&gt;&#xA;" OwnerUserId="2574" LastActivityDate="2011-01-26T21:26:06.817" Title="Continuous isomorphisms of rings" Tags="&lt;algebra&gt;&lt;commutative-algebra&gt;&lt;ring-theory&gt;" />
  <row Id="19145" PostTypeId="2" ParentId="19132" CreationDate="2011-01-26T21:28:34.157" Score="2" ViewCount="" Body="&lt;p&gt;If $\rm\:R\:$ is a domain then easily $\rm\:f(x)\:$ a unit $\rm\Rightarrow\ a_i = 0\:$ for $\rm\:i&amp;gt;0\:.\ $ Now $\rm\ R\to R/P,\ $ for $\rm\:P\:$ prime, reduces to the domain case, yielding that the $\rm\:a_i\:,\ i&amp;gt;0\:$ are in every prime ideal. But the intersection of all prime ideals is the nilradical, the set of all nilpotent elements - as you proved &lt;a href=&quot;http://math.stackexchange.com/questions/18203/x-not-nilpotent-implies-there-is-a-prime-ideal-not-containing-x&quot;&gt;a few days ago.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See also &lt;a href=&quot;http://math.stackexchange.com/questions/120/is-1-a-prime-number/5735#5735&quot;&gt;my post here&lt;/a&gt; on reduction to domains by factoring out prime ideals.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2011-01-26T22:40:51.160" LastActivityDate="2011-01-26T22:40:51.160" />
  <row Id="19146" PostTypeId="1" AcceptedAnswerId="19149" CreationDate="2011-01-26T21:32:09.493" Score="3" ViewCount="147" Body="&lt;p&gt;I've recently been studying the functional analysis and I think I need some help with one exercise in the book.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;We have the &lt;strong&gt;Lusin's&lt;/strong&gt; theorem, which is stated the following way:&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given some measurable $E$ where $\mu(E)&amp;lt;\infty$ and some function $f$, which is &lt;strong&gt;measurable&lt;/strong&gt; and &lt;strong&gt;finite&lt;/strong&gt; almost everywhere in $E$, the following statement is true:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\forall \epsilon&amp;gt;0 : \space  \exists  F_\epsilon \subset E$, that ($F_\epsilon$ has to be closed):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) $\mu(E \setminus F_\epsilon) &amp;lt; \epsilon$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) $f$ is continuous on $F_\epsilon$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Now - the exercise is to tell if the inverse theorem is correct &lt;strong&gt;(and give the proof if it is)&lt;/strong&gt; - so that if there exists that $F_\epsilon$, on which our function is continuous &lt;em&gt;(and I assume finite almost everywhere)&lt;/em&gt;, then it's also measurable on the corresponding set $E$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I assume this statement is correct, but unfortunately I can't come up with a proper proof. Proof of the Lusin's theorem doesn't help here, because we have to prove the &lt;strong&gt;continuity -&gt; measurableness&lt;/strong&gt; on the corresponding sets and the Egorov's theorem is no help in that case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could someone share the proof please or give me some clues on accomplishing it.&lt;/p&gt;&#xA;" OwnerUserId="4971" LastEditorUserId="1543" LastEditorDisplayName="" LastEditDate="2011-04-03T18:05:48.440" LastActivityDate="2011-04-03T18:05:48.440" Title="Thoughts about measurable functions" Tags="&lt;measure-theory&gt;&lt;functional-analysis&gt;" AnswerCount="1" FavoriteCount="2" />
  <row Id="19147" PostTypeId="1" AcceptedAnswerId="19148" CreationDate="2011-01-26T21:36:35.770" Score="3" ViewCount="77" Body="&lt;p&gt;Say X and Y are normally distributed random variables, i.e., X,Y ~ N(0,1). For some fixed t and c I know that Pr[X≥t]=c and Pr[Y≥t]=c. Moreover, I also know that X and Y have covariance 0.999.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I deduce what Pr[X≥t and Y≥t] roughly is?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bonus question: what are good references to learn more on related topics for normally distributed random variables?&lt;/p&gt;&#xA;" OwnerUserId="915" LastActivityDate="2011-01-26T21:57:38.857" Title="Determining probability involving normal random variables and given covariance." Tags="&lt;probability-theory&gt;&lt;normal-distribution&gt;" AnswerCount="1" />
  <row Id="19148" PostTypeId="2" ParentId="19147" CreationDate="2011-01-26T21:46:22.830" Score="2" ViewCount="" Body="&lt;p&gt;This is the &lt;a href=&quot;http://www.athenasc.com/Bivariate-Normal.pdf&quot; rel=&quot;nofollow&quot;&gt;bivariate normal distribution&lt;/a&gt;. Here is a &lt;a href=&quot;http://www.stuart.iit.edu/shared/shared_stuartfaculty/TablesfortheStandardBivariateNormalDistributionwithMontiraJantaravareerat.pdf&quot; rel=&quot;nofollow&quot;&gt;paper&lt;/a&gt; estimating the cdf of the bivariate normal distribution since no known analytic form is known. &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0387970622&quot; rel=&quot;nofollow&quot;&gt;This&lt;/a&gt; is a good book explaining multivariate normal distributions.&lt;/p&gt;&#xA;" OwnerUserId="2548" LastEditorUserId="2548" LastEditorDisplayName="" LastEditDate="2011-01-26T21:57:38.857" LastActivityDate="2011-01-26T21:57:38.857" CommentCount="1" />
  <row Id="19149" PostTypeId="2" ParentId="19146" CreationDate="2011-01-26T21:50:36.577" Score="4" ViewCount="" Body="&lt;p&gt;Restating the problem, suppose that $E$ is a measurable set of finite measure, that $f$ is an extended real valued function defined on $E$ and finite almost everywhere, and that for all $\varepsilon\gt 0$ there exists a closed set $F_{\varepsilon}\subset E$ such that $\mu(E\setminus F_{\varepsilon})&amp;lt;\varepsilon$ and $f\vert_{F_\varepsilon}$ is continuous.  Then $f$ is measurable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proof sketch: For each positive integer $n$, let $F_n$ be a closed subset of $E$ such that $\mu(E\setminus F_n)&amp;lt;\frac{1}{n}$ and $f\vert_{F_n}$ is continuous.  Then $f\vert_{F_n}$ is also measurable, and from this it follows that the restriction of $f$ to $\cup_k F_k$ is measurable.  Since $\mu(E\setminus\cup_k F_k)\leq \mu(E\setminus F_n)\lt \frac{1}{n}$ for each $n$, it follows that $f$ is measurable when restricted to a measurable subset whose complement has measure $0$.  Thus (assuming $\mu$ is complete) $f$ is measurable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fact that $F_\epsilon$ is closed isn't needed.  It just has to be measurable.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2011-01-26T22:06:35.973" LastActivityDate="2011-01-26T22:06:35.973" CommentCount="4" />
  <row Id="19150" PostTypeId="1" AcceptedAnswerId="19151" CreationDate="2011-01-26T21:54:03.633" Score="1" ViewCount="72" Body="&lt;p&gt;Is the order of random variables important in the chain rule? I mean, is this true: P(A,B,C) = P(A)P(B|A)P(C|A,B) = P(C)P(B|C)P(A|B,C) = P(C,B,A)? If it is, what is the meaning of such order? Thank you.&lt;/p&gt;&#xA;" OwnerUserId="6289" LastActivityDate="2011-01-26T22:06:34.467" Title="Is order of variables important in probability chain rule" Tags="&lt;probability&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="19151" PostTypeId="2" ParentId="19150" CreationDate="2011-01-26T22:06:34.467" Score="1" ViewCount="" Body="&lt;p&gt;$P[A \cap B \cap C] = P[(A \cap B) \cap C] = P[(A \cap B)|C]P(C) = P[C|A \cap B]P[A \cap B]$. Then you can rewrite $P(A \cap B) = P(A|B)P(B) = P(B|A)P(A)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;These are all useful. Suppose you want to find $P(A \cap B)$. Well $P(A \cap B) = P(A|B)P(B) = P(B|A)P(A)$. But suppose you only know $P(B|A)$. Then $P(B|A)P(A)$ is more useful.&lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2011-01-26T22:06:34.467" CommentCount="2" />
  <row Id="19152" PostTypeId="1" AcceptedAnswerId="19154" CreationDate="2011-01-26T22:06:39.747" Score="4" ViewCount="94" Body="&lt;p&gt;Is there some easy way to compute number of particular permutations? Namely let $S_i$ be the number of permutations $(a_1,...,a_n)$, $\{a_1,\ldots,a_n\}=\{1,...,n\}$, satisfying the following conditions:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$a_i=n$ for some $i\in \{1,\ldots,n\}$.&lt;/li&gt;&#xA;&lt;li&gt;$a_1&amp;lt;a_2&amp;lt;\ldots &amp;lt;a_i$.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Compute $\sum_{i=1}^n S_i$. &lt;/p&gt;&#xA;" OwnerUserId="5462" LastEditorUserId="5462" LastEditorDisplayName="" LastEditDate="2011-01-26T22:23:48.897" LastActivityDate="2011-01-27T13:59:29.843" Title="Number of permutations" Tags="&lt;combinatorics&gt;" AnswerCount="1" CommentCount="7" />
  <row Id="19153" PostTypeId="1" CreationDate="2011-01-26T22:11:34.123" Score="2" ViewCount="196" Body="&lt;p&gt;I would like to know if there a closed form solution for the sum:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ S_n(t) = \sum_{k=0}^{n} \cos( t \sqrt{k} ) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is obviously an easy answer when the sum is replaced by an integral so this question is really asking for the exact form that $S_n(t)$ takes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any hints or references on how to approach this problem would be welcome.&lt;/p&gt;&#xA;" OwnerUserId="4143" LastActivityDate="2011-05-07T02:50:08.310" Title="Closed form for $\sum_{k=0}^{n} \cos( t \sqrt{k} )$ ?" Tags="&lt;analysis&gt;&lt;sequences-and-series&gt;&lt;closed-form&gt;" AnswerCount="1" CommentCount="8" />
  <row Id="19154" PostTypeId="2" ParentId="19152" CreationDate="2011-01-26T22:27:06.883" Score="4" ViewCount="" Body="&lt;p&gt;&lt;em&gt;Edit:&lt;/em&gt; Slight mistake in the computation which trickled down.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, let's count the number of $n$-tuples with with the property you describe. You know that $a_i$ is going to be. To select the previous $i-1$ elements, all you need is to select arbitrary $i-1$ elements from the remaining $n-1$ elements in your set, and place them in order. The remaining $n-i$ elements may be selected for $a_{i+1},\ldots,a_n$ arbitrarily. This gives $\binom{n-1}{i-1}(n-i)! = \frac{(n-1)!}{(i-1)!}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, suppose you have two distinct $n$-tuples corresponding to the same $i$. Can they represent the same permutation? No: because two $n$-cycles represent the same permutation if and only if one can be obtained from the other by repeated cycling, $(a_1,\ldots,a_n) \to (a_2,\ldots,a_n,a_1) \to $ etc. Since both have $n$ in the same position in the cycle, the two $n$-tuples represent distinct permutations. So $S_i = \frac{(n-1)!}{(i-1)!}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the number you want is&#xA;$$\sum_{i=1}^{n} S_i = \sum_{i=1}^n\frac{(n-1)!}{(i-1)!} = e\Gamma(n,1)-\Gamma(n)$$&#xA;(thanks to PEV for the last equality). &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-27T13:59:29.843" LastActivityDate="2011-01-27T13:59:29.843" CommentCount="9" />
  <row Id="19155" PostTypeId="1" AcceptedAnswerId="19156" CreationDate="2011-01-26T22:43:28.850" Score="1" ViewCount="69" Body="&lt;p&gt;Hi there. It is well known that, if HM denotes the harmonic mean and AM the arithmetic mean, we have &#xA;$$ AM(x) \ge HM(x) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I am dealing with the expression&#xA;$$ \frac{1}{HM(x)} - \frac{1}{AM(x)} $$&#xA;A trivial lower bound for this expression is $0$, but is there also a nice upper bound?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cheers!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Or, if there's no general upper bound, might there be one if all numbers involved are positive?&lt;/p&gt;&#xA;" OwnerUserId="4171" LastEditorUserId="4171" LastEditorDisplayName="" LastEditDate="2011-01-26T22:53:31.010" LastActivityDate="2011-01-26T23:02:25.603" Title="Bound on deviation between arithmetic and harmonic mean?" Tags="&lt;inequality&gt;&lt;average&gt;" AnswerCount="1" />
  <row Id="19156" PostTypeId="2" ParentId="19155" CreationDate="2011-01-26T22:51:09.357" Score="3" ViewCount="" Body="&lt;p&gt;No, consider just two numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{x+y}{2xy} - \frac{2}{x+y} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fix $x \gt 0$ and as $y \to 0+$, this is unbounded.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2011-01-26T23:02:25.603" LastActivityDate="2011-01-26T23:02:25.603" CommentCount="3" />
  <row Id="19157" PostTypeId="1" AcceptedAnswerId="19192" CreationDate="2011-01-26T23:08:08.827" Score="5" ViewCount="158" Body="&lt;p&gt;Find all functions $f:R\rightarrow R$ which satisfy $f(x+f(y))=f(x-f(y))+4xf(y)$ $\forall x,y \in R$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I strongly suspect $0$ and $x^2+C$  to be the only solutions but, as is almost the case with functional equations, finding the set the solutions is easy; it is proving that a certain set of solutions represents ALL solutions is what is difficult.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is trivial that if $f$ is bounded $f\equiv0$.&lt;/p&gt;&#xA;" OwnerUserId="5959" LastEditorUserId="5959" LastEditorDisplayName="" LastEditDate="2011-01-26T23:46:45.893" LastActivityDate="2011-01-27T06:57:00.870" Title="$f(x+f(y))=f(x-f(y))+4xf(y)$" Tags="&lt;functional-equations&gt;" AnswerCount="1" CommentCount="9" />
  <row Id="19158" PostTypeId="1" AcceptedAnswerId="19159" CreationDate="2011-01-26T23:12:03.993" Score="1" ViewCount="70" Body="&lt;p&gt;Inputs: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) A set of M x N matrices, {A,B,C...N} containing only integers. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) A single 1 x N matrix of floats, W (weights). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need to pull one row from each input matrix and sum values for each column, then take the dot product of the resulting 1 x N matrix with W, yielding a score K. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ex: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A = [1 3 5 6], B = [3 4 5 6], W = [0.5 0.1 2.0 3.0]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case there is only one row in each input matrix so we pick the only rows from A and B, sum their columns to get [4 7 10 12] and dot that with W to yield 58.7, our value for K. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: In addition there is a constraint on one of the columns such that the corresponding weight from W drops by some amount when the sum for that column is greater than some fixed value, for example we might specify that 3 is the cap for the first column in the previous example, with the weight dropping to 0.2 after that has been exceeded. This would make the score for [4 7 10 12] = [(3 * 0.5) + (1 * 0.2) + (7 * 0.1) + (10 * 2.0) + (12 * 3.0) = 58.4&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The goal is to find the global maximum K for all combinations of rows from the input matrices (where we take one row from each matrix). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the set of input matrices is large and the number of rows is more than a couple, the number of row combinations blows up.&lt;/p&gt;&#xA;" OwnerUserId="6291" LastEditorUserId="6291" LastEditorDisplayName="" LastEditDate="2011-01-26T23:52:02.570" LastActivityDate="2011-01-27T03:07:46.647" Title="Combinatorial Optimization Problem (can I/how do I solve this with integer programming?)" Tags="&lt;combinatorics&gt;&lt;optimization&gt;&lt;nonlinear-optimization&gt;&lt;integer-programming&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="19159" PostTypeId="2" ParentId="19158" CreationDate="2011-01-26T23:19:19.170" Score="2" ViewCount="" Body="&lt;p&gt;Suppose that the input matrices are $A_1,\ldots, A_L$.  Let $r_1, \ldots, r_L$ be a choice of one row from each of them.  We want to maximize $(r_1 + \cdots + r_L ) \bullet W = r_1\bullet W + \cdots + r_L \bullet W$ over all such choices of $r_i$.  The right hand side shows that we can use the greedy algorithm which chooses $r_i$ to be the row of $A_i$ which makes the biggest dot product with $W$: choosing $r_i$ in this way doesn't restrict our choices on the other $r_j$ in any way.  The algorithm to compute all these dot products and choose the biggest one runs in linear time in the size of the input.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: The new version, formalized as a decision problem which asks whether we can pick a set of the rows to give objective value at least $M$, is NP-complete.  To see that it is in NP, have the algorithm guess the optimal rows.  To see that it is NP-hard, we reduce from set cover, which is a problem where we are given a set $\mathcal{S} = \{S_1,\ldots, S_M\}$ of subsets of the set $\{1,\ldots,N\}$ and we want to know whether there are $K$ sets in $\mathcal{S}$ whose union is all of $\{1,\ldots, N\}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We must turn these into an instance of your problem.  Define the matrix $A$ by $A_{ij} = 1$ if $j\in S_i$ and zero otherwise.  The $i^\text{th}$ row of $A$ is then the indicator function for the set $S_i$.  Let $L = K$ all the matrices $A_1 = \cdots = A_L = A$.  Have the objective $W$ put weight $1$ on each column until a value of $1$ is achieved in that column and weight $0$ afterwards.  Then the objective value for a given choice of rows is the cardinality of the union of the corresponding subsets in $\mathcal{S}$.  In particular the optimal objective value is $\geq N$ if and only if the set cover problem was solvable.&lt;/p&gt;&#xA;" OwnerUserId="884" LastEditorUserId="884" LastEditorDisplayName="" LastEditDate="2011-01-27T03:07:46.647" LastActivityDate="2011-01-27T03:07:46.647" CommentCount="1" />
  <row Id="19160" PostTypeId="1" AcceptedAnswerId="19166" CreationDate="2011-01-26T23:41:47.957" Score="0" ViewCount="101" Body="&lt;p&gt;I'm having some trouble with the formatting here, so here's a screenshot of the problem we were given:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://i.imgur.com/bnzd6.png&quot; rel=&quot;nofollow&quot;&gt;http://i.imgur.com/bnzd6.png&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is, that in the book from which the problem is taken from (Probability Essentials by Jean Jacod) we aren't given much help with understanding the notation. Earlier there is a proof where he defines lim sup (n-&gt;inf) A_n but I have no idea what lim sup (n-&gt;inf) 1_... could mean.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this some common notation?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="2076" LastActivityDate="2011-01-27T05:37:00.040" Title="Probability proof, sequence of sets: help with understanding the notation?" Tags="&lt;probability&gt;&lt;probability-theory&gt;" AnswerCount="2" />
  <row Id="19161" PostTypeId="2" ParentId="19160" CreationDate="2011-01-26T23:50:54.733" Score="1" ViewCount="" Body="&lt;p&gt;$\textbf{1}_{A_n}$ is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Indicator_function&quot; rel=&quot;nofollow&quot;&gt;indicator function&lt;/a&gt;. &lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2011-01-26T23:50:54.733" />
  <row Id="19162" PostTypeId="1" CreationDate="2011-01-26T23:51:16.833" Score="1" ViewCount="52" Body="&lt;p&gt;There is not an example problem in my textbook that I can reference (I have said this on two other questions so far, but it's the truth. The example problems and solutions manual contain physics / engineering examples rather than abstract examples).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For this one, I did not know how to begin. Normally I would give the work I had done, even if it is incorrect.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;For what values of t and so does the&#xA;  equality &amp;lt;4-2t, 2s-4t &gt;= &amp;lt; s+4t, 3+2t&gt;.&#xA;  Find t and s. At these values of t and&#xA;  s, find the resulting vector.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Can anyone point me in the correct direction or give me an idea on how to solve this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you in advance.&lt;/p&gt;&#xA;" OwnerUserId="3800" LastActivityDate="2011-01-27T02:26:00.447" Title="Homework Help - Finding a Vector, and values of s and t, when given an equality." Tags="&lt;calculus&gt;&lt;vector-calculus&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="19163" PostTypeId="2" ParentId="19162" CreationDate="2011-01-26T23:54:44.873" Score="2" ViewCount="" Body="&lt;p&gt;Set the $x$-components equal to each other and the $y$-components equal to each other.&lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2011-01-26T23:54:44.873" />
  <row Id="19165" PostTypeId="2" ParentId="19131" CreationDate="2011-01-26T23:58:56.393" Score="3" ViewCount="" Body="&lt;p&gt;Another more general approach is to solve your equation for $x$. Since it is&#xA;linear in $\sin x$ and $\cos x$ it can be transformed into a quadratic&#xA;equation in $\tan \frac{x}{2}$ (see &lt;a href=&quot;http://math.stackexchange.com/questions/9138/solve-trigonometric-equation-1-m-textcos-alpha-textsin-alpha/9185#9185&quot;&gt;this answer&lt;/a&gt;): &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sin x+\cos x=\frac{1+\sqrt{3}}{2}\Leftrightarrow \frac{2\tan \frac{x}{2}}{%&#xD;&#xA;1+\tan ^{2}\frac{x}{2}}+\frac{1-\tan ^{2}\frac{x}{2}}{1+\tan ^{2}\frac{x}{2}}%&#xD;&#xA;=\frac{1+\sqrt{3}}{2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Leftrightarrow 2\tan \frac{x}{2}+1-\tan ^{2}\frac{x}{2}=\left( 1+\tan ^{2}%&#xD;&#xA;\frac{x}{2}\right) \frac{1+\sqrt{3}}{2}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Set $y=\tan \frac{x}{2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$2y+1-y^{2}=\frac{1+\sqrt{3}}{2}+\frac{1+\sqrt{3}}{2}y^{2}\Leftrightarrow \left( 1+\frac{1+\sqrt{3}}{2}\right) y^{2}-2y-1+\frac{1+%&#xD;&#xA;\sqrt{3}}{2}=0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and solve for $y$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$y_{1}=\frac{1}{3}\sqrt{3},y_{2}=2-\sqrt{3}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x_{1}=2\arctan \frac{1}{3}\sqrt{3}=\frac{1}{3}\pi $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x_{2}=2\arctan \left( 2-\sqrt{3}\right) =\frac{1}{6}\pi .$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And finally&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\tan \frac{1}{3}\pi +\cot \frac{1}{3}\pi =\frac{4}{3}\sqrt{3}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\tan \frac{1}{6}\pi +\cot \frac{1}{6}\pi =\frac{4}{3}\sqrt{3}.$$&lt;/p&gt;&#xA;" OwnerUserId="752" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2011-01-27T00:06:50.377" LastActivityDate="2011-01-27T00:06:50.377" />
  <row Id="19166" PostTypeId="2" ParentId="19160" CreationDate="2011-01-27T00:37:29.090" Score="3" ViewCount="" Body="&lt;p&gt;${\mathbf 1}_{A_n}$ is a function from $\Omega$ to $\{0,1\}$, defined by ${\mathbf 1}_{A_n} (\omega) = 1$ if $\omega \in A_n$, ${\mathbf 1}_{A_n} (\omega) = 0$ if $\omega \notin A_n$ (indicator function). $\lim \sup _{n \to \infty } {\mathbf 1}_{A_n }$ is defined pointwise: for each $\omega \in \Omega$, it is the $\lim \sup$ of the sequence of numbers ${\mathbf 1}_{A_n} (\omega)$, that is, $\lim \sup _{n \to \infty } {\mathbf 1}_{A_n }$ is a function on $\Omega$ defined pointwise by&#xA;$$&#xD;&#xA;(\mathop {\lim \sup }\limits_{n \to \infty } {\mathbf 1}_{A_n } )(\omega ) = \mathop {\lim \sup }\limits_{n \to \infty } {\mathbf 1}_{A_n } (\omega ) \,(\in \{ 0,1 \}).&#xD;&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="2810" LastEditorUserId="2810" LastEditorDisplayName="" LastEditDate="2011-01-27T05:37:00.040" LastActivityDate="2011-01-27T05:37:00.040" />
  <row Id="19167" PostTypeId="1" AcceptedAnswerId="19195" CreationDate="2011-01-27T00:54:56.353" Score="0" ViewCount="87" Body="&lt;p&gt;A and B throw a fair 6-sided die in turn. The die is not regular because the faces are {1,2,2,4,4,6}. The winner is whoever first gets a cumulated sum equal or greater than 10. Ask for the probability that A wins, and the expected number of throws when the game generates a winner.&lt;/p&gt;&#xA;" OwnerUserId="6016" LastActivityDate="2011-01-27T05:46:37.633" Title="a probability question about throwing an irregular fair die" Tags="&lt;probability&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="19168" PostTypeId="1" CreationDate="2011-01-27T00:56:04.793" Score="-10" ViewCount="535" Body="&lt;p&gt;Probably, most here have seen the proof that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1 = .999...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and I am not here to debate that point.  Instead, my question is regarding a notation for describing infinitely large quantities.  First, I reject the idea of a non-relative infinity and replace it with the concept of relative infinities.  If our base infinity is infinity_0 then,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;infinity_-1 = infinity_0 - 1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;also,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;infinity_1 = infinity_0 + 1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and, in general,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;infinity_x = infinity_0 + x .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then we may define the notation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1...{0}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;to mean repeat 1 precisely infinity_0 times.  In general,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;x...{y}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Means repeat x precisely infinity_y times.  We can clearly see that this has several implications that, in my view, completely demolishes Cantor's conceptions of infinity, and at the same time, makes practically useful statements that I will demonstrate.  Here is a short list of the implications of relative infinities that I have found very recently:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The integers and natural numbers do NOT have the same cardinality as the rational numbers IF the numerator and denominator of the rationals are both defined to have the same precision as the integers (which they should, since they are integers).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Depending upon the infinite precision of the set you are describing the reals could have more OR LESS elements than the integers.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Cantor's diagonal argument is wrong.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Cantor's CH is false.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;There exist numbers which I call multi-primes numbers, which are numbers that take on two or more different prime values simultaneously.  (I haven't entirely verified this yet, but I'm pretty sure it's true.)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Limits are expressible in terms of integer part operations.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;and probably many other things as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, I will prove that Cantor's CH is false with a practically trivial proof.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Aleph_0 = 1...{0}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, clearly,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Aleph_0 = 1...{0} &amp;lt; 2^(1...{-1}) &amp;lt; 2^(1...{0}) = Aleph_1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, we have found an infinity between Aleph_0 and Aleph_1, namely, 2^(1...{-1}).  But, in fact, there exist infinite infinities between Aleph_0 and Aleph_1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My general question is, am I reinventing the wheel here?  Is this notation new?  Is it old notation possibly reformatted?  I tend to believe these days there's nothing new under the sun, so it is probably not new, but if it isn't, does anybody know what axioms it corresponds to and whether or not they are widely accepted and/or self-consistent?  Many thanks.&lt;/p&gt;&#xA;" OwnerUserId="6293" LastEditorUserId="6293" LastEditorDisplayName="" LastEditDate="2011-01-27T01:10:46.707" LastActivityDate="2011-01-27T05:18:34.313" ClosedDate="2011-01-27T06:54:50.133" Title="Is this a new infinity notation?" Tags="&lt;notation&gt;&lt;infinity&gt;" AnswerCount="3" CommentCount="12" FavoriteCount="0" />
  <row Id="19169" PostTypeId="1" CreationDate="2011-01-27T01:14:53.497" Score="1" ViewCount="174" Body="&lt;p&gt;I'm stuck in two parts of Ahlfors' proof of Cauchy's theorem in a disk (page 113), that is, if $f$ is holomorphic in an open disc $D$ then $\int_\gamma f(z)dz=0$ over every closed curve $\gamma$ in $D$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First part:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fix $z_0\in D$. We define $F(z)=\int_\sigma f(\zeta)d\zeta$ where $\sigma$ is the path joining $z_0$ with $z$ by taking an horizontal line from $z_0$ and getting to $z$ with a vertical line (hope it is clear).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;It is immediately seen that $\frac{\partial F}{\partial y}(z)=if(z)$&quot;. Not for me. I mean, I'm geometrically and intuitively inclined to understand that when we derive vertically, since it is a constant path what we should get is the value of $f$. I don't see how the $i$ appears, though, and I formally don't understand what's going on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First of all, what does $\frac{\partial F}{\partial y}$ mean? This is what I understand by that: if $F=u+iv$ then $\frac{\partial F}{\partial y}=\frac{\partial u}{\partial y} + i \frac{\partial v}{\partial y}$. Right?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also found the following formula scribbled on my notebook:&#xA;$\int_\gamma f(z)dz = \int_\gamma f(z) dx + i \int_\gamma f(z) dy$. Is this correct? I don't see how it makes sense to integrate with respect to $x$ a complex-valued function: what am I supposed to do with the $y$'s in the integrand? I'm guessing some abuse of notation is going on here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, I'm guessing this is something easy and I'm just confused by notation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second part: we get that $\frac{\partial F}{\partial y}(z)=if(z)$ and $\frac{\partial F}{\partial x}(z)=f(z)$. &quot;Hence $F$ is holomorphic with derivative $f$&quot;. How is that? I mean, we have $\frac{\partial F}{\partial x}= -i \frac{\partial F}{\partial y}$, which does not seem like Cauchy-Riemann to me. I'm guessing this is the same confusion as above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope I was not overly verbose.&lt;/p&gt;&#xA;" OwnerUserId="2614" LastEditorUserId="2614" LastEditorDisplayName="" LastEditDate="2011-01-27T02:40:47.660" LastActivityDate="2011-01-27T02:40:47.660" Title="Ahlfors' proof of Cauchy's theorem in a disk" Tags="&lt;complex-analysis&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="19170" PostTypeId="1" AcceptedAnswerId="19175" CreationDate="2011-01-27T02:19:55.150" Score="2" ViewCount="72" Body="&lt;p&gt;Suppose $G$ is a compact connected Lie group and let $\{X_i\}$ be a basis for its Lie algebra $\mathfrak g$.  We know that the exponential $\exp:\mathfrak g \to G$ is surjective but when is it the case that $G$ is generated by $\{\exp(tX_i) : t\in \mathbb R\}$? &lt;/p&gt;&#xA;" OwnerUserId="9" LastActivityDate="2011-01-27T02:30:53.427" Title="Generators of compact Lie groups" Tags="&lt;lie-groups&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="19171" PostTypeId="1" CreationDate="2011-01-27T02:21:57.930" Score="1" ViewCount="145" Body="&lt;p&gt;The examples given &lt;a href=&quot;http://www4.ncsu.edu/~edbancro/teaching/handouts/taylorseriesexample.pdf&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; for example, show that once you know the form of a taylor polynomial as a function of $x$, you can replace the $x$ with another function. It works when you work out the problems but I don't understand &lt;em&gt;why&lt;/em&gt;. It would seem that the chain rule would make the taylor polynomial something else.&lt;/p&gt;&#xA;" OwnerUserId="1958" LastActivityDate="2011-01-27T02:35:07.570" Title="Why Does Substitution In Taylor Series Work?" Tags="&lt;intuition&gt;&lt;taylor-expansion&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="19172" PostTypeId="2" ParentId="19169" CreationDate="2011-01-27T02:23:57.580" Score="1" ViewCount="" Body="&lt;p&gt;Second part: The Cauchy-Riemann equations are &lt;strong&gt;equivalent&lt;/strong&gt; to $\frac{\partial F}{\partial y}(z) = if(z)$. This form dictates when a function is conformal. In particular, look at the matrix $$ \begin{bmatrix} u_x &amp;amp; -v_x \\ v_x &amp;amp; \ \ \ u_x \end{bmatrix}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is precisely the &lt;a href=&quot;http://en.wikipedia.org/wiki/Complex_number#Matrix_representation_of_complex_numbers&quot; rel=&quot;nofollow&quot;&gt;matrix representation of a complex number&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2548" LastEditorUserId="2548" LastEditorDisplayName="" LastEditDate="2011-01-27T02:30:51.147" LastActivityDate="2011-01-27T02:30:51.147" />
  <row Id="19173" PostTypeId="2" ParentId="19162" CreationDate="2011-01-27T02:26:00.447" Score="1" ViewCount="" Body="&lt;p&gt;Two vectors are equal if and only if they are identical: the first component of each should be equal, the second component of each should be equal. So for the vectors $\langle 4-2t, 2s-4t\rangle$ and $\langle s+4t, 3+2t\rangle$ to be equal, you must have $4-2t = s+4t$ (first components equal) and $2s-4t = 3+2t$ (second components equal). This is a system of two equations in two unknowns; solve them to find (all) value(s) of $s$ and $t$ that make the equality true. Then plug the values to get the &quot;resulting vector&quot;(s). &lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2011-01-27T02:26:00.447" />
  <row Id="19174" PostTypeId="2" ParentId="19168" CreationDate="2011-01-27T02:29:01.810" Score="10" ViewCount="" Body="&lt;p&gt;Dear John, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you describe has nothing to do with the &quot;infinity&quot; in Cantor's theorem or the continuum hypothesis. There &quot;infinity&quot; refers simply to whether two sets can be compared -- by a bijective function -- so that an infinite set is one that is not in bijection with any set $\{1, 2, \dots, n\}$ for any $n$. To say that two &quot;infinities&quot; are  equivalent simply states that two sets of those cardinalities have a bijection between them. The operation &quot;infinity plus one&quot; (and this really depends on a fixed infinite cardinality) is not something that can be prescribed arbitrarily, but depends simply on the existence or non-existence of bijective functions. If $A$ is any infinite set, then $A$ and $A \cup \{ x\}$ admit bijections between them for any $x$, so they have the same cardinality. This is all that is meant by $\kappa+1 = \kappa$ for an infinite cardinal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is easy to confuse the notion of &quot;infinity&quot; in set theory with others (e.g. limits to infinity), so the use of precise definitions should clarify things, and in the present situation, explain the problem with your reasoning. You seem to have defined some kind of set of rules for playing with these objects that you label &quot;infinity,&quot; which is all fine, but it does not imply anything about Cantor's argument, which relies on a technically &lt;em&gt;different&lt;/em&gt; idea altogether.&lt;/p&gt;&#xA;" OwnerUserId="536" LastActivityDate="2011-01-27T02:29:01.810" CommentCount="2" />
  <row Id="19175" PostTypeId="2" ParentId="19170" CreationDate="2011-01-27T02:30:53.427" Score="4" ViewCount="" Body="&lt;p&gt;The map $\mathbb{R}^n \to G$ sending $(t_1, \dots, t_n)$ to $\mathrm{exp}(t_1 X_1) \dots \mathrm{exp}(t_n X_n)$ has nonsingular Jacobian at $0$, so its image contains a neighborhood of the origin. By a standard argument, a neighborhood of the origin in a connected topological group generates the full group.&lt;/p&gt;&#xA;" OwnerUserId="536" LastActivityDate="2011-01-27T02:30:53.427" CommentCount="2" />
  <row Id="19176" PostTypeId="2" ParentId="19171" CreationDate="2011-01-27T02:35:07.570" Score="1" ViewCount="" Body="&lt;p&gt;See &lt;a href=&quot;http://books.google.com/books?id=gTF_l082y1MC&amp;amp;pg=PA433&amp;amp;lpg=PA433&amp;amp;dq=substitution+in+taylor+series&amp;amp;source=bl&amp;amp;ots=1TiRU9WaKp&amp;amp;sig=axeSrh6TdLhB10jt38yzdWjRWXY&amp;amp;hl=en&amp;amp;ei=kNlATaTJI8SRgQfKmpmSAw&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=6&amp;amp;ved=0CDUQ6AEwBTgU#v=onepage&amp;amp;q=substitution%20in%20taylor%20series&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; (theorem concerning substitution in Taylor series).&lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2011-01-27T02:35:07.570" />
  <row Id="19177" PostTypeId="1" AcceptedAnswerId="19251" CreationDate="2011-01-27T02:36:14.543" Score="8" ViewCount="186" Body="&lt;p&gt;Suppose I wanted to solve (for x and y) an equation of the form $x^2-dp^2y^2=1$ where d is squarefree and p is a prime.  Of course I could simply generate the solutions to the Pell equation $x^2-dy^2=1$ and check if the value of y was divisible by $p^2,$ but that would be slow.  Any better ideas?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It would be useful to be able to distinguish cases where the equation is solvable from cases where it is unsolvable, even without finding an explicit solution.&lt;/p&gt;&#xA;" OwnerUserId="1778" LastEditorUserId="1778" LastEditorDisplayName="" LastEditDate="2011-01-28T01:25:01.147" LastActivityDate="2011-01-28T05:30:10.020" Title="Non-squarefree version of Pell's equation?" Tags="&lt;number-theory&gt;&lt;diophantine-equations&gt;" AnswerCount="3" CommentCount="6" />
  <row Id="19178" PostTypeId="2" ParentId="18688" CreationDate="2011-01-27T02:43:36.617" Score="0" ViewCount="" Body="&lt;p&gt;Let n be fixed , and &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;f(x) = (x-n)^n&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The standard result in finite difference state that if f is a polynomial of degree n , then &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Δ^n(f) = n!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now , the lhs of the given identity is exactly (1/n^n)Δ^n(f)(0) , and the result follows.&#xA;Q.E.D.&lt;/p&gt;&#xA;" OwnerUserId="6294" LastActivityDate="2011-01-27T02:43:36.617" CommentCount="1" />
  <row Id="19179" PostTypeId="2" ParentId="19168" CreationDate="2011-01-27T02:55:43.227" Score="-8" ViewCount="" Body="&lt;p&gt;I am not confusing notations.  There's nothing you've written that I didn't already know.  I am aware of Cantor's definition of the cardinality of sets, however, I simply disagree.  This is where a deeper examination of the subject is needed.  My underlying problem with his arguments is that &quot;...&quot; is not rigorously and mathematically well-defined for sets such as the integers or reals.  This &quot;...&quot; concept actually relies on a self-referencing definition, which has no real meaning.  I would like somebody to define for me, without self referencing concepts, what &quot;...&quot; means.  You can talk about unbounded sets.  That's fine.  But it doesn't answer my question, because you still must define the logical universe in which the unbounded set exists, at which point the definition becomes self-referencing.  I am basically saying every mathematically well-defined set must have a definite beginning and a definite ending.  The problem with, say, Cantor's diagonal proof is that the definite ending of the set we call integers we could define to be 1...{0}, at which point, we may simply define reals such that their integer and fractional parts do map to integers.&lt;/p&gt;&#xA;" OwnerUserId="6293" LastActivityDate="2011-01-27T02:55:43.227" CommentCount="4" />
  <row Id="19180" PostTypeId="1" AcceptedAnswerId="19196" CreationDate="2011-01-27T03:00:18.907" Score="0" ViewCount="82" Body="&lt;p&gt;Given a list of &lt;a href=&quot;http://en.wikipedia.org/wiki/Point_cloud&quot; rel=&quot;nofollow&quot;&gt;point cloud&lt;/a&gt; in terms of $(x,y,z)$ how to determine abnormal points? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The motivation is this. We need to reconstruct a terrain surface out from those point cloud, which the surveyors obtain when doing field survey. The surveyors would take an equipment and record a sufficient sample of the $x,y,z$ of a terrain. Those points will be recorded into a CAD program.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is that the CAD file can be corrupted from time to time with the introduction of &quot;abnormal&quot; points. Those points do not fit into the terrain surface generally, and tend to have erroneous $z$ value ( i.e., the $z$ value is outside of the normal range). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am aware that the definition of abnormal points is a bit loose; and I can't come up with a rigorous definition of it. However, I know what is an abnormal point when I see the drawing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given all these constraint, is there any algorithm to detect these kinds of abnormal points? &lt;/p&gt;&#xA;" OwnerUserId="550" LastActivityDate="2011-01-27T06:04:05.487" Title="Detect Abnormal Points in Point Cloud" Tags="&lt;computational-geometry&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="19181" PostTypeId="2" ParentId="19098" CreationDate="2011-01-27T03:08:54.943" Score="0" ViewCount="" Body="&lt;p&gt;Use the long exact sequence in Ext from the ses $$ 0 \to I \to R \to R/I \to 0,$$&#xA;leading to&#xA;$$ 0 \to \hom(R/I, M) \to \hom(R, M) \to\hom(I,M) \to \mathrm{Ext}^1(R/I, M) \to 0,$$&#xA;because $R$ is projective. It is known that depth (or grade) can be measured in terms of Exts, and in particular that it is at least 2 if the two ends of the sequence vanish. But this is equivalent to the condition that $\hom(R, M) \to \hom(I,M)$ be an isomorphism.&lt;/p&gt;&#xA;" OwnerUserId="536" LastActivityDate="2011-01-27T03:08:54.943" />
  <row Id="19183" PostTypeId="2" ParentId="6" CreationDate="2011-01-27T03:14:31.967" Score="3" ViewCount="" Body="&lt;p&gt;I really like &lt;a href=&quot;http://www.geogebra.org&quot; rel=&quot;nofollow&quot;&gt;Geogebra&lt;/a&gt; as a web based graphing tool (requires Java).&lt;/p&gt;&#xA;" OwnerUserId="6268" LastEditorUserId="41" LastEditorDisplayName="" LastEditDate="2011-01-27T16:03:07.880" LastActivityDate="2011-01-27T16:03:07.880" CommunityOwnedDate="2011-01-27T03:14:31.967" />
  <row Id="19186" PostTypeId="2" ParentId="17582" CreationDate="2011-01-27T04:16:24.453" Score="-2" ViewCount="" Body="&lt;p&gt;The proof can be found in the book written by Richard Stanley , Enumerative Combinatorics Volume I&lt;/p&gt;&#xA;" OwnerUserId="6294" LastActivityDate="2011-01-27T04:16:24.453" CommentCount="1" />
  <row Id="19187" PostTypeId="1" AcceptedAnswerId="19194" CreationDate="2011-01-27T04:50:37.403" Score="11" ViewCount="675" Body="&lt;p&gt;Actually I know the answer. Well, I believe I know the answer, which is NO.&#xA;Unfortunately, the professor in a Theory of Computation class said that yes, there is such a subset.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is to settle a discussion with fellow students. A discussion that is going nowhere so we go to the internets for a verdict.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance for weighing in on this question.&lt;/p&gt;&#xA;" OwnerUserId="6299" LastEditorUserId="6299" LastEditorDisplayName="" LastEditDate="2011-01-27T16:59:49.127" LastActivityDate="2011-01-31T18:23:08.890" Title="Is there such a thing as a countable set with an uncountable subset?" Tags="&lt;set-theory&gt;" AnswerCount="6" FavoriteCount="2" />
  <row Id="19189" PostTypeId="2" ParentId="19054" CreationDate="2011-01-27T04:57:51.967" Score="16" ViewCount="" Body="&lt;p&gt;Indeed, no such function $f$ is possible. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $Y$  be the image of $f$, and let $g$ be the inverse of $f$.  Then as pointed out by Eric Nitardy,&#xA; $ |g(x)-g(y)|\le(x-y)^2$ for all $x,y$ in $Y$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From this inequality it now follows that $g(Y)$ has &lt;em&gt;measure zero&lt;/em&gt;,&#xA;which contradicts the fact that $g(Y)$ is the whole real line.&#xA; To prove $g(Y)$ has measure zero, it suffices to show&#xA;that for any integer $n$, $g(Y_n)$ has measure zero where&#xA;$Y_n=Y\cap[n,n+1].$ &#xA;But for any $N$, $Y_n$ is contained in a union of $N$ intervals&#xA;of diameter $1/N$ whose images under $g$ have diameter not exceeding $1/N^2$.&lt;/p&gt;&#xA;" OwnerUserId="2947" LastActivityDate="2011-01-27T04:57:51.967" CommentCount="1" />
  <row Id="19190" PostTypeId="2" ParentId="19187" CreationDate="2011-01-27T04:59:11.943" Score="5" ViewCount="" Body="&lt;p&gt;A set is countably infinite if there is an injective function from it to the natural numbers. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose $A$ is countable and $f\colon A \to \omega$ is an injection witnessing that, then if $B\subseteq A$ the restriction of $f$ to $B$ is an injective function from $B$ to the natural numbers, therefore $B$ is countable. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A side note is that without the axiom of choice there are infinite sets without a countable subset. That's a whole other deal though. &lt;/p&gt;&#xA;" OwnerUserId="622" LastEditorUserId="622" LastEditorDisplayName="" LastEditDate="2011-01-27T05:14:10.017" LastActivityDate="2011-01-27T05:14:10.017" CommentCount="5" />
  <row Id="19191" PostTypeId="2" ParentId="19168" CreationDate="2011-01-27T05:02:24.857" Score="5" ViewCount="" Body="&lt;p&gt;The problem with the current formulation that you present is that it does not (yet) seem to rigorously define the ordering and arithmetical operations on your concept of levels of infinity.  Therefore, I cannot argue for or against its inconsistency.  For example, how do you know that $1...._{0} &amp;lt; 2^{1...._{-1}}$ when you haven't yet presented a complete definition of how these infinities are ordered?  While $\aleph_0 &amp;lt; 2^{\aleph_0}$ when $2^{\aleph_0}$ is interpreted in terms of cardinal exponentiation (i.e., cardinality of the functions from $\mathbb{N}$ into $2 = \{0, 1\}$), $\omega = 2^{\omega}$ ($\omega = \mathbb{N}$ is usually expressed this way to denote ordinal) when $2^{\omega}$ means ordinal exponentiation (i.e., the least upper bound of $2^n$ for all Natural numbers $n$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an analogy with the definition that you know, infinite decimal representations are informal representations of rigorous limits.  For example, $.\overline{9}$ is really just a representation of $\lim_{n \rightarrow \infty}s_n$ where $\langle s_n| n \in \mathbb{N}\rangle$ is the sequence of partial sums $\langle \Sigma_{i=1}^n 9 \cdot 10^{-i}| n \in \mathbb{N}\rangle$.  Therefore, the fact that $.\overline{9} = 1$ is really just a consequence of our definition.  If you want to argue against this possibility, you need to redefine what $.\overline{9}$ means.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, if you want to argue against the current formulations of different levels of infinity, that's fine.  But you need new well-defined definitions to do this.  The uncountability of the set of Real numbers is not rigorously shown by diagonalizing against informal infinite decimal expansions.  In fact, &lt;a href=&quot;http://en.wikipedia.org/wiki/Cantor%27s_first_uncountability_proof&quot; rel=&quot;nofollow&quot;&gt;Cantor's original proof&lt;/a&gt; didn't even reference such expansions.  Nevertheless, the standard diagonalization argument can be made completely rigorous because you are constructing the Real number by induction.  Mainly, you will be constructing a Cauchy sequence of the form $\langle \Sigma_{i=1}^n d_i \cdot 10^{-i}| n \in \mathbb{N}\rangle$, and the Real number will be the limit of this sequence of Rational numbers, which is guaranteed to exist because of the completeness of the Real numbers.  The number will be different than your countable list of Real numbers in $[0, 1]$ because for each stage $m \in \mathbb{N}$, you define $d_m$ to disagree with $d_m^{\prime}$ in a sequence $\langle \Sigma_{i=1}^n d_i^{\prime} \cdot 10^{-i}| i \in \mathbb{N}\rangle$ whose limit is the $m^{th}$ Real you're diagonalizing against.  Note that this argument does not rely on the informality of the $\ldots$, but only makes use of the more rigorous definitions of the Real numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now to the question of something potentially similar, you may want to check out the &lt;a href=&quot;http://en.wikipedia.org/wiki/Surreal_numbers&quot; rel=&quot;nofollow&quot;&gt;surreal numbers&lt;/a&gt;, which actually form an ordered field and treat levels of infinity very differently.  For example, $\omega - 1$ and $\omega/2$ are defined in this system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To tie up some odds and ends here, you may be interested in knowing that there are models of arithmetic that think the same things are true as $\langle \mathbb{N}; 0, 1; +, \cdot; &amp;lt;\rangle$ and yet have nonstandard elements, meaning that they have elements $c$ such that $n &amp;lt; c$ for all $n \in \mathbb{N}$ under their order relation.  This seems to get at your question of the ambiguity of $\ldots$  Nevertheless, this point is irrelevant in the diagonalization argument because you are carrying the argument out in the model's version of $\mathbb{N}$ so to say that the cardinality of $\mathbb{N}$ is less than the cardinality of $\mathbb{R}$ is with respect to what the model thinks is $\mathbb{N}$ and $\mathbb{R}$.  If it has a nonstandard $\mathbb{N}$, it will have a nonstandard $\mathbb{R}$ as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, my answer is meant as general purpose for you and anyone who may be interested in reading this post.  My explanation is not intended to suggest that you are not already aware of any information I presented.  Instead, it is an attempt to address your concerns in such a way that (is hopefully) moderately accessible to some others who may not possess your formal training.&lt;/p&gt;&#xA;" OwnerUserId="4693" LastActivityDate="2011-01-27T05:02:24.857" CommentCount="5" />
  <row Id="19192" PostTypeId="2" ParentId="19157" CreationDate="2011-01-27T05:06:06.487" Score="7" ViewCount="" Body="&lt;p&gt;Ok I finally got it: the equation $f(x + f(y)) = f(x - f(y)) + 4xf(y)$ can be rewritten as&#xA;$$f(x + f(y)) - f(x - f(y)) = (x + f(y))^2 - (x - f(y))^2$$&#xA;Replacing $x$ by $x + f(y)$ and rearranging terms we get&#xA;$$f(x + 2f(y)) - (x + 2f(y))^2 = f(x) - x^2$$&#xA;Thus $f(x) - x^2$ has period $2f(y)$ for all $y$. In particular, it has period $2f(a + f(b))$ as well as $2f(a - f(b))$ for all $a$ and $b$. Since the difference of two periods is a period, it also has period $2f(a + f(b)) - 2f(a - f(b))$ for any $a$ and $b$, which is $8af(b)$ by the conditions given. Assuming $f(x)$ is not the zero function, we may pick $b$ so that $f(b)$ is not zero. Since $a$ can be anything, we conclude $f(x) - x^2$ has every real number as a period; in other words it is constant. So if $f(x)$ is not the zero function, then $f(x) = x^2 + C$ for some constant $C$. &lt;/p&gt;&#xA;" OwnerUserId="3035" LastEditorUserId="3035" LastEditorDisplayName="" LastEditDate="2011-01-27T06:57:00.870" LastActivityDate="2011-01-27T06:57:00.870" />
  <row Id="19194" PostTypeId="2" ParentId="19187" CreationDate="2011-01-27T05:18:49.910" Score="28" ViewCount="" Body="&lt;p&gt;Of course it is easy to see that any subset of a countable set is countable. While enumerating the larger set, just skip over any elements that are not in the subset, and you have an enumeration of the subset. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nevertheless, since you mentioned you were in a computability theory course, there are a few computability-like senses in which something like a positive answer to the question is possible.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Every infinite computably enumerable set has numerous subsets that are not computably enumerable. This is simply because every infinite set has uncountably many subsets, and most of these will not be computably enumerable, since there are only countably many c.e. sets. But from a computability perspective, &lt;em&gt;computably enumerable&lt;/em&gt; is often the right analogue of &lt;em&gt;countable&lt;/em&gt;, since those are the sets with a computable enumeration function, and so this is a very reasonable sense in which the claim is nearly true.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;There can be c.e. sets whose complement is infinite, but contains no infinite c.e. subset. Thus, you can enumerate the set $S$, and infinitely many numbers are not in $S$, but there is no computable way to enumerate an infinite set of numbers not in $S$. These are known as the &lt;em&gt;simple&lt;/em&gt; sets, and were studied from the time of Post.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Without AC, it is consistent that you can have a set $A$ with an equivalence relation $\sim$ on it, such that the number of equivalence classes of $\sim$ on $A$ is strictly larger than the cardinality of $A$. For example, we can make an equivalence relation $\sim$ having exactly $\mathbb{R}+\omega_1$ many equivalence classes, by saying that two reals are equivalent iff they are equal, or else they both code relations on $\omega$ that are well-orders of the same length. But under the Axiom of Determinacy, there is no $\omega_1$-sequence of distinct real numbers, and so this cardinality is strictly larger than $\mathbb{R}$. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="413" LastActivityDate="2011-01-27T05:18:49.910" CommentCount="4" />
  <row Id="19195" PostTypeId="2" ParentId="19167" CreationDate="2011-01-27T05:46:37.633" Score="3" ViewCount="" Body="&lt;p&gt;Define the polynomial $P(x)$ by&#xA;$$ P(x) = \frac{x + 2x^2 + 2x^4 + x^6}{6}. $$&#xA;This polynomial represents one throw of the die. The probability to get a sum of $s$ after $k$ throws is the coefficient of $x^s$ in $P(x)^k$. Let's denote that by $[P(x)^k]_{x^s}$. The probability that the sum reaches $10$ for the first time at time $t$ is $$w_t = \sum_{s \geq 10} [P(x)^t]_{x^s} - \sum_{s \geq 10} [P(x)^{t-1}]_{x^s};$$ in other words, $$w_t = \sum_{s &amp;lt; 10} [P(x)^{t-1}]_{x^s} - \sum_{s &amp;lt; 10} [P(x)^t]_{x^s}.$$ The probability that the sum doesn't reach $10$ at time $t$ is $$l_t = \sum_{s &amp;lt; 10} [P(x)^t]_{x^s}.$$ Notice that $w_t = l_{t-1} - l_t$. The probability that B wins is $$\sum_{t \geq 1} l_t w_t.$$ The probability that A wins is, similarly, $$\sum_{t \geq 1} w_t l_{t-1}.$$ The expected number of throws is thus $$\sum_{t \geq 1} 2t l_t w_t + (2t-1) w_t l_{t-1}.$$ Since $l_t = 0$ for $t \geq 10$, all these sums are finite. You can compute everything with a CAS.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are some results from SAGE calculations:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;A wins w.p. $64601710707175/101559956668416 \approx 0.636$.&lt;/li&gt;&#xA;&lt;li&gt;B wins w.p. $36958245961241/101559956668416 \approx 0.364$.&lt;/li&gt;&#xA;&lt;li&gt;The expected number of throws is $550136643228931/101559956668416 \approx 5.42$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2011-01-27T05:46:37.633" CommentCount="3" />
  <row Id="19196" PostTypeId="2" ParentId="19180" CreationDate="2011-01-27T06:04:05.487" Score="1" ViewCount="" Body="&lt;p&gt;The answer is largely determined by your surface reconstruction technique (simple triangulation, marching cubes, and the characteristics of the noise.  If the noise points are infrequent and uncorrelated, then the problem is easier.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;You can presume the surface is locally planar and fit a plane to some small region.  A simple &quot;averaging&quot; fit is straightforward and computationally efficient. Once the fit plane is known, you can reject points outside of some tolerance.  You can choose to keep the &quot;good&quot; points as they are - or you might consider obtaining new &quot;average&quot; points by projecting the good points onto the fit plane to obtain new samples.&lt;/li&gt;&#xA;&lt;li&gt;An iterative technique (like a re-weighted least squares) is better if you have the time to perform a few iterations on each region.  &lt;/li&gt;&#xA;&lt;li&gt;There are other, more robust, surface reconstruction techniques like marching cubes, etc., that can perform limited artifact rejection.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Regardless, a good characterization of the spurious points is needed.&lt;/p&gt;&#xA;" OwnerUserId="6129" LastActivityDate="2011-01-27T06:04:05.487" CommentCount="4" />
  <row Id="19197" PostTypeId="2" ParentId="19056" CreationDate="2011-01-27T06:32:56.867" Score="8" ViewCount="" Body="&lt;p&gt;Define&#xA;$$&#xD;&#xA;S \left( T \right) := \sum_{m = 2}^T \sum_{n = 1}^{\infty} \frac{m^{\left\{ n \right\}} + m^{- \left\{ n \right\}}}{m^n} .&#xD;&#xA;$$&#xA;First, let's find all numbers n such that $\left\{ n \right\} = t$ for some integer t:&#xA;$$&#xD;&#xA;\sqrt{n} - \frac{1}{2} &amp;lt; t &amp;lt; \sqrt{n} + \frac{1}{2} \Leftrightarrow t^2 - t + 1 \leq n \leq t^2 + t&#xD;&#xA;$$&#xA;and so,&#xA;$$&#xD;&#xA;S \left( T \right) = \sum_{m = 2}^T \sum_{t = 1}^{\infty} \sum_{n = t^2 - t + 1}^{t^2+t} \frac{m^t + m^{-t}}{m^n}&#xD;&#xA;= \sum_{m = 2}^T \sum_{t = 1}^{\infty} \frac{\left(m^t + m^{-t} \right) \left(m^{- t^2 + t - 1} - m^{-t^2 - t - 1} \right) m}{m - 1}&#xD;&#xA;.&#xD;&#xA;$$&#xA;The summand can now be rewritten to yield&#xA;$$&#xD;&#xA;S \left( T \right) = \sum_{m = 2}^T \frac{m}{m-1} \sum_{t = 1}^{\infty} \left( m^{- \left( t - 1 \right)^2} - m^{- \left( t + 1 \right)^2} \right),&#xD;&#xA;$$&#xA;which is a telescopic sum in t. Thus,&#xA;$$&#xD;&#xA;S \left( T \right) = \sum_{m = 2}^T \frac{m}{m-1} \left( 1 + \frac{1}{m} \right) =&#xD;&#xA;\sum_{m = 2}^T \left( 1 + \frac{2}{m - 1} \right) = T - 1 + 2 H(T - 1),&#xD;&#xA;$$&#xA;where $H \left( k \right) := \sum_{i = 1}^k \frac{1}{i}$ denotes the harmonic sum.&lt;/p&gt;&#xA;" OwnerUserId="960" LastActivityDate="2011-01-27T06:32:56.867" />
  <row Id="19198" PostTypeId="1" AcceptedAnswerId="19199" CreationDate="2011-01-27T06:49:39.130" Score="1" ViewCount="121" Body="&lt;p&gt;Hi, I'm in the process of reading my first Linear Algebra textbook, and was just wondering...Is the standard basis of a vector space in n dimensions equivalent to the row space of the n x n identity matrix?&lt;/p&gt;&#xA;" OwnerUserId="6300" LastActivityDate="2011-01-27T06:58:38.123" Title="Question about the standard basis's relationship to the identity matrix" Tags="&lt;linear-algebra&gt;&lt;matrices&gt;&lt;vector-space&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="19199" PostTypeId="2" ParentId="19198" CreationDate="2011-01-27T06:58:38.123" Score="3" ViewCount="" Body="&lt;p&gt;I assume you're talking about $\mathbb{R}^n$? First off, the standard basis $\{e_1,e_2,\dots,e_n\}$ is a linearly independent set of $n$ vectors, which spans the vector space. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the row space of $I_n$ is the set of all linear combinations of the row vectors, which gives a subspace of $n$-dimensional space, which is just the whole space in this case. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The key difference is that the standard basis is just a set of basis vectors, but the row space of $I_n$ is a vector space, not a standard basis, which are two different objects.&lt;/p&gt;&#xA;" OwnerUserId="1583" LastActivityDate="2011-01-27T06:58:38.123" CommentCount="3" />
  <row Id="19200" PostTypeId="2" ParentId="19187" CreationDate="2011-01-27T07:36:59.163" Score="0" ViewCount="" Body="&lt;p&gt;Consider a set which contains a countable collection of uncountable sets.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One example would be the set whose $m^{th}$ member (even though they are not ordered) is the set of real numbers between $m$ and $m+1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then I can say that such a set has a countable subset, and that each of these subsets contains an uncountable number of elements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I could not say that such a set contains an uncountable subset of a countable subset of real numbers, merely that it contains a countable subset of (uncountable) sets.&lt;/p&gt;&#xA;" OwnerUserId="1417" LastActivityDate="2011-01-27T07:36:59.163" CommentCount="2" />
  <row Id="19201" PostTypeId="2" ParentId="18983" CreationDate="2011-01-27T08:05:09.720" Score="0" ViewCount="" Body="&lt;p&gt;It also follows from Gosper's most excellent algorithm.&lt;/p&gt;&#xA;" OwnerUserId="6301" LastActivityDate="2011-01-27T08:05:09.720" />
  <row Id="19202" PostTypeId="2" ParentId="19119" CreationDate="2011-01-27T08:32:20.770" Score="3" ViewCount="" Body="&lt;p&gt;Let's also elaborate on Ross Millikan's answer, adapted to the case $f(x)=\sqrt{1-x^2}$, $0 \leq x \leq 1$. Suppose that $(X_1,Y_1),(X_2,Y_2),\ldots$ is a sequence of independent uniform vectors on $[0,1] \times [0,1]$, so that for each $i$, $X_i$ and $Y_i$ are independent uniform$[0,1]$ random variables. Define $Z_i$ as follows: $Z_i = 1$ if $X_i^2 + Y_i^2 \leq 1$, $Z_i = 0$ if $X_i^2 + Y_i^2 &amp;gt; 1$, so the $Z_i$ are independent and identically distributed random variables, with mean $\mu$ given by&#xA;$$&#xD;&#xA;\mu = {\rm E}[Z_1] = {\rm P}[X_1^2 + Y_1^2 \leq 1] = {\rm P}\big[(X_1,Y_1) \in \lbrace (x,y) \in [0,1]^2 : x^2+y^2 \leq 1\rbrace \big] = \frac{\pi }{4}, &#xD;&#xA;$$&#xA;where the last equality follows from ${\rm P}[(X_1,Y_1) \in A] = {\rm area}A$ ($A \subset [0,1]^2$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the strong law of large numbers, the average $\bar Z_n  = \frac{{\sum\nolimits_{i = 1}^n {Z_i } }}{n}$ converges, with probability $1$, to the expectation $\mu$ as $n \to \infty$. That is, with probability $1$, $\bar Z_n  \to \frac{\pi }{4}$ as $n \to \infty$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To get a probabilistic error bound, note first that the $Z_i$ have variance $\sigma^2$ given by&#xA;$$&#xD;&#xA;\sigma^2 = {\rm Var}[Z_1] = {\rm E}[Z_1^2] - {\rm E}^2{[Z_1]} = \frac{\pi }{4} - \Big(\frac{\pi }{4}\Big)^2 = \frac{\pi }{4} \Big(1 - \frac{\pi }{4}\Big) &amp;lt; \frac{10}{59}.&#xD;&#xA;$$&#xA;The average $\bar Z_n $ has expectation ${\rm E}[\bar Z_n] = \mu$ and variance ${\rm Var}[\bar Z_n] = \sigma^2 / n$; hence, by Chebyshev's inequality, for any given $\varepsilon &amp;gt; 0$,&#xA;$$&#xD;&#xA;{\rm P}\big[\big|\bar Z_n - \mu \big| \geq \varepsilon \big] \leq \frac{{\sigma^2}}{{n \varepsilon ^2 }},&#xD;&#xA;$$&#xA;and so&#xA;$$&#xD;&#xA;{\rm P}\bigg[\bigg|\bar Z_n - \frac{\pi }{4}  \bigg| \geq \varepsilon \bigg] &amp;lt; \frac{{10}}{{59n\varepsilon ^2 }}.&#xD;&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="2810" LastActivityDate="2011-01-27T08:32:20.770" CommentCount="1" />
  <row Id="19203" PostTypeId="1" AcceptedAnswerId="19222" CreationDate="2011-01-27T08:36:41.750" Score="5" ViewCount="191" Body="&lt;p&gt;8 cubes can form a torus by gluing them together face-to-face, with each cube sharing a face with each of two other cubes, by torus I just mean a closed loop with a hole. Is the same possible with the dodecahedron? What is the minimum number of dodecahedrons needed for this?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit:&#xA;Is it possible with the tetrahedron and icosahedron ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given a polytope, is there a general way to decide if it is possible, and to determine the minimum number?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any good programmers around to check tetrahedron case?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Looking for software to check for small N!  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Crosspost: &lt;a href=&quot;http://mathoverflow.net/questions/53601/which-platonic-solids-can-form-a-topological-torus&quot; rel=&quot;nofollow&quot;&gt;http://mathoverflow.net/questions/53601/which-platonic-solids-can-form-a-topological-torus&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5904" LastEditorUserId="5904" LastEditorDisplayName="" LastEditDate="2011-01-28T17:32:04.750" LastActivityDate="2011-01-28T17:32:04.750" Title="Closed polytope loop" Tags="&lt;geometry&gt;&lt;topology&gt;&lt;math-software&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" />
  <row Id="19204" PostTypeId="1" CreationDate="2011-01-27T08:39:00.853" Score="2" ViewCount="146" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have two polygons $P_1 \subseteq P_2$ in the plane, and I would like to determine to which polygon a given point $p\in P_1 \setminus P_2$ is &quot;closest&quot;, by means of the following measure: $$r(p)=d(p,P_1)/d_p(P_1,P_2),$$ where $d(p,P_1)$ is the distance between $p$ and $P_1$ and $d_p(P_1,P_2)$ is the distance between $P_1$ and $P_2$. &#xA;The $p$ subscript in that case indicates that this distance between both polygons will vary according to where $p$ is located, and the idea is that the smaller (resp. larger) $r(p)$ is, the closer it is to $P_1$ (resp. $P_2$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know about the formulas for computing the distance between a point and a line segment, but this does not seem to be enough here. I might be thinking a bit too much in algorithmic terms, but my idea was to construct a circle with center $p$ and let it grow until it &quot;hits&quot; either polygon (or both); the distance between $p$ and $P_1$ (or $P_2$) is then given by the radius of the smallest circle centered at $p$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This answers the question of how to compute $d(p,P_1)$. However, I'm having trouble seeing how I should compute $d_p(P_1,P_2)$. Any ideas? Is this well-known? If that helps, one can assume that $P_1$ and $P_2$ are rectangles whose sides are pairwise parallel.&lt;/p&gt;&#xA;" OwnerUserId="689" LastActivityDate="2011-01-28T02:27:15.483" Title="Distances between points and polygons" Tags="&lt;geometry&gt;" AnswerCount="2" />
  <row Id="19205" PostTypeId="1" AcceptedAnswerId="19249" CreationDate="2011-01-27T08:57:43.450" Score="5" ViewCount="164" Body="&lt;p&gt;Given a polytope of dimension n, is there some general way to determine if it can tile R^n ?&lt;/p&gt;&#xA;" OwnerUserId="5904" LastEditorUserId="5904" LastEditorDisplayName="" LastEditDate="2011-01-27T09:07:19.807" LastActivityDate="2011-01-27T16:32:02.443" Title="Decidability of tiling of R^n" Tags="&lt;geometry&gt;&lt;topology&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="1" />
  <row Id="19206" PostTypeId="2" ParentId="19204" CreationDate="2011-01-27T09:14:11.573" Score="1" ViewCount="" Body="&lt;p&gt;Assuming the obvious definition of the distance between two polygons, there are three (not mutually exclusive) possible cases:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The shortest distance between P1 and P2 is equal to the distance between a vertex v1 of P1 and a vertex v2 of P2.&lt;/li&gt;&#xA;&lt;li&gt;The shortest distance between P1 and P2 is equal to the distance between a vertex v of one of them and an edge e of the other.&lt;/li&gt;&#xA;&lt;li&gt;The shortest distance between P1 and P2 is equal to the distance between an edge e1 of P1 and an edge e2 of P2.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The brute force approach is to enumerate all of the distances between candidate pairs (v1, v2), (v, e), and (e1, e2), and then pick the smallest. If the polygons have many vertices then you will obviously want to look for more efficient approaches based around eliminating most of the vertices and edges as candidates.&lt;/p&gt;&#xA;" OwnerUserId="5676" LastActivityDate="2011-01-27T09:14:11.573" />
  <row Id="19207" PostTypeId="1" AcceptedAnswerId="19224" CreationDate="2011-01-27T10:08:44.533" Score="0" ViewCount="150" Body="&lt;p&gt;It is known that the solution of the linear ODE $\frac{d\vec{y}}{dt}=A\vec{y}$ is the span of the columns of $e^{xA}$. This gives us a formula for solving the ODE by solving $e^{xA}$. However, is the reverse also possible? That is, can I calculate $e^{xA}$ by solving the ODE?&lt;/p&gt;&#xA;" OwnerUserId="26" LastActivityDate="2011-01-27T14:35:07.270" Title="Calculating matrix exponential by solving appropriate ODE" Tags="&lt;differential-equations&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="19208" PostTypeId="2" ParentId="19205" CreationDate="2011-01-27T11:25:33.517" Score="3" ViewCount="" Body="&lt;p&gt;Probably there is not! Indeed I believe there is a single (very complicated) 2D tile for which the problem of whether it tiles the plane is logically undecidable [Edit: meaning independent of the axioms of ZFC set theory]! I don't have the reference handy. Perhaps somebody else can supply it. With a quick Google search, the Wikipedia article on Wang tiles is the closest I could come. Wang tiles are presented as colored tiles where the colors have to match on shared edges, but these can be converted to regular tiles by modifying the boundary so that only like colors can join. According to Wikipedia, there is a set of 13 tiles for which the tiling problem is logically undecidable [Edit: meaning independent of the axioms of ZFC set theory], but I seem to recall hearing that this number has been reduced to 1, but possibly no longer in the Wang tile category.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: As was pointed out to me in the comments, this seems different than asking whether the tiling problem is decidable in the decision problem sense, for which the truth of any finite proposition is assumed to be known, and the question is whether an algorithm exists that successfully runs on the set of all tiles. &lt;/p&gt;&#xA;" OwnerUserId="2941" LastEditorUserId="2941" LastEditorDisplayName="" LastEditDate="2011-01-27T15:03:19.633" LastActivityDate="2011-01-27T15:03:19.633" CommentCount="8" />
  <row Id="19209" PostTypeId="1" AcceptedAnswerId="19214" CreationDate="2011-01-27T11:58:52.877" Score="0" ViewCount="59" Body="&lt;p&gt;I have two equations. In this example I'll make them linear, and later see if the solution generalises to different forms of equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$y_1=-0.1x + 0.65$$&#xA;$$y_2=0.15x - 0.825$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm interesting in small ranges of these graphs, say 0.5-2.5 of $y_1$ and the same range +7 of $y_2$. We assume the value of the equation outside of its range is 0. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to find a value for x such that the summed integration of $y_1$ from that point up to the end of its valid range, and the integration of $y_2$ from $(x+7)$ down to the start of its valid range, is minimal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/2x1jc.jpg&quot; alt=&quot;Labeled graph of both lines&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope that makes sense. I've sort of fried my brain getting the problem into these terms, and now can't see the solution for the trees.&lt;/p&gt;&#xA;" OwnerUserId="6303" LastActivityDate="2011-01-27T12:45:45.250" Title="Finding X to minimise summed integral of two equations" Tags="&lt;calculus&gt;&lt;algebra&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="19210" PostTypeId="2" ParentId="19138" CreationDate="2011-01-27T11:59:36.807" Score="1" ViewCount="" Body="&lt;p&gt;Why do you think this is a two-step method? Take $s=1$, consider the given method as an implicit one-step method (you do not need to pad the Butcher tableau with zeroes), and you should get the stability function you want.&lt;/p&gt;&#xA;" OwnerUserId="520" LastActivityDate="2011-01-27T11:59:36.807" CommentCount="1" />
  <row Id="19211" PostTypeId="1" CreationDate="2011-01-27T12:00:18.647" Score="0" ViewCount="80" Body="&lt;p&gt;Given a vector V of n integers and an integer k, k &amp;lt;= n, you want a subvector (a sequence of consecutive elements of the vector ) of maximum length containing at most k distinct elements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The technique that I use for the resolution of the problem is dynamic programming. The complexity of this algorithm must be O(n*k).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The main problem is how to count distinct elements of the vector. as you would resolve it ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How to write the EQUATION OF RECURRENCE ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks you!!!.&lt;/p&gt;&#xA;" OwnerUserId="4670" LastEditorUserId="4670" LastEditorDisplayName="" LastEditDate="2011-01-28T09:59:49.110" LastActivityDate="2011-01-28T18:16:34.167" Title="Problem k-subvector using dynamic programming" Tags="&lt;homework&gt;&lt;algorithms&gt;&lt;computer-science&gt;" AnswerCount="1" />
  <row Id="19212" PostTypeId="1" CreationDate="2011-01-27T12:01:21.517" Score="2" ViewCount="227" Body="&lt;p&gt;I am wondering how to show that the ideal (x,y) is prime and maximal in $\mathbb{Q}[x,y]$.&lt;/p&gt;&#xA;" OwnerUserId="6308" LastEditorUserId="8575" LastEditorDisplayName="" LastEditDate="2011-04-27T12:46:17.010" LastActivityDate="2011-04-27T15:59:37.810" Title="Prime Ideal of a Polynomial Ring" Tags="&lt;ring-theory&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="19213" PostTypeId="2" ParentId="19212" CreationDate="2011-01-27T12:35:36.500" Score="7" ViewCount="" Body="&lt;p&gt;Look at the quotient ring $\mathbb{Q}[x,y]/(x,y)$. Then use that an ideal $I \subset R$ is maximal iff $R/I$ is a field. Using the fact that $I$ is prime iff $R/I$ is an integral domain, you will also be able to show that $(x,y)$ is prime (and all maximal ideals are).&lt;/p&gt;&#xA;" OwnerUserId="1766" LastActivityDate="2011-01-27T12:35:36.500" />
  <row Id="19214" PostTypeId="2" ParentId="19209" CreationDate="2011-01-27T12:45:45.250" Score="1" ViewCount="" Body="&lt;p&gt;Integrating both equations gives:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int y_1 dx = -0.05x^2 + 0.65x$$&#xA;$$\int y_2 dx = 0.075x^2 - 0.825x$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The summed integrations for some value of x with respect to $y_1$ is then:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$s=\int_x^2.5 y_1 dx + \int_7.5^{x+7} y_2 dx$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The integration for the upper bound of $y_1$ and the lower bound of $y_2$ are constants. We'll call them $u_1$ and $l_2$. Expanding the above equation gives:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$s=-0.025x^2+0.425x+u_1+l_2+2.1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $u_1=1.3125$ and $l_2=-1.96875$, the final equation we need to minimise is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$s=-0.025x^2+0.425x+1.44375$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The minimum value of $s$ for the range 0.5-2.5 is 1.65 at $x=0.5$. &lt;/p&gt;&#xA;" OwnerUserId="6303" LastActivityDate="2011-01-27T12:45:45.250" CommentCount="4" />
  <row Id="19215" PostTypeId="1" AcceptedAnswerId="19223" CreationDate="2011-01-27T12:57:45.247" Score="1" ViewCount="130" Body="&lt;p&gt;Given any quadratic equation of the form $y=ax^2+bx+c$, I want to find the minimum value for a specific range of $x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My programmer brain can do it in a branchy, algorythmic way as follows, but is there a more elegant solution?&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;if the $a$ coefficent is positive,&lt;br&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;and the end of my range is before the lowest point of the quadratic, &#xA;&lt;ul&gt;&#xA;&lt;li&gt;return the end of my range&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;and the start of my range is after the lowest point of the quadratic,&#xA;&lt;ul&gt;&#xA;&lt;li&gt;return the start of my range&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;and the lowest point of the quadratic occurs in the middle of my range,&#xA;&lt;ul&gt;&#xA;&lt;li&gt;return the lowest point of the quadratic&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;if the $a$ coefficent is negative, etc...&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="6303" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2011-01-27T14:19:21.433" LastActivityDate="2011-01-27T17:00:59.920" Title="Finding the minimum value of a quadratic within a range" Tags="&lt;algebra&gt;&lt;polynomials&gt;&lt;optimization&gt;&lt;quadratic-forms&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="19216" PostTypeId="1" AcceptedAnswerId="19218" CreationDate="2011-01-27T13:32:07.970" Score="1" ViewCount="69" Body="&lt;p&gt;I am relatively very new to probability distribution and after understanding the concept of Expected value of a discrete random variable,I am trying to understand the variance of the same here is an extract from my book: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$Var(X) = E[(X - E(X))^2] = \sum_{\text{all }X}  (X - E(X))^2 P_x(X) = E(X^2) - (E(X))^2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What my question is that I don't understand this simplification.Any pointers in this regard will be highly appreciated.&lt;/p&gt;&#xA;" OwnerUserId="2109" LastActivityDate="2011-03-03T01:37:05.467" Title="On variance of a random variable" Tags="&lt;probability&gt;&lt;probability-theory&gt;&lt;probability-distributions&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="19217" PostTypeId="2" ParentId="19216" CreationDate="2011-01-27T13:35:59.553" Score="0" ViewCount="" Body="&lt;p&gt;First equality is a definition of variance. Second is by the definition of expected value. To see the third, expand the square, multiply out and simplify.&lt;/p&gt;&#xA;" OwnerUserId="2710" LastActivityDate="2011-01-27T13:35:59.553" />
  <row Id="19218" PostTypeId="2" ParentId="19216" CreationDate="2011-01-27T13:40:56.610" Score="3" ViewCount="" Body="&lt;p&gt;Forget the middle bit. What you need to know is that $E(X)$ is a linear function, this means that $E(X + Y) = E(X) + E(Y)$ and $E(aX) = aE(X)$ where $a$ is a constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Per definition, $Var(X) = E((X - E(X)^2))$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Multiplying out the argument of $E$ gives you&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ E((X - E(X)^2)) = E(X^2 - 2XE(X) +E^2(X)) = $$&#xA;$$E(X^2) - E(2XE(X)) + E(E^2(X)) = E(X^2) - 2 E(X)E(X) +E^2(X) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where the last equality holds because $E(X)$ is a constant, so $E(E(X)) = E(X)$.&lt;/p&gt;&#xA;" OwnerUserId="5798" LastActivityDate="2011-01-27T13:40:56.610" />
  <row Id="19219" PostTypeId="1" AcceptedAnswerId="19225" CreationDate="2011-01-27T13:43:19.593" Score="3" ViewCount="115" Body="&lt;p&gt;&lt;strong&gt;A box contain $A$ white and $B$ black balls and $C$ balls are drawn, then the expected value of the number of white balls drawn is ?&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer is $\large \frac{ca}{a+b}$. How to approach this one?&lt;/p&gt;&#xA;" OwnerUserId="2109" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2011-05-26T16:10:23.560" LastActivityDate="2011-05-26T16:10:23.560" Title="Find the expectation" Tags="&lt;probability&gt;&lt;probability-distributions&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="1" />
  <row Id="19220" PostTypeId="2" ParentId="19092" CreationDate="2011-01-27T13:46:56.360" Score="2" ViewCount="" Body="&lt;p&gt;There are two ways of trying to build homotopy equivalence maps between $X \cup_f B_n$ and $X \cup_g B_n$.&#xA;One is to try to have $B_n$ stay fixed, and have things happen in $X$ instead.&#xA;If you find a construction which works for an arbitrary space $S$, it would be one of this kind.&#xA;The other is to try to have $X$ stay fixed, and have things happen in $B_n$ instead.&#xA;Since we know what $B_n$ is and we don't know what $X$ is, it should be easier to do it this way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we want a continuous map $F : X \cup_f B_n \rightarrow X \cup_g B_n$, where $\forall x \in X, F(x) = x$.&#xA;This means that we only want to find its restriction $F|_{B_n} : B_n \rightarrow X \cup_g B_n$, such that $\forall x \in S_{n-1}, F(x) = f(x)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now then, if you want to use the homotopy $h : S^{n-1} \times [0;1] \rightarrow X$ between $f$ and $g$, it makes sense to try to decompose $B^n$ and build something like this :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$B_n \approx (S^{n-1} \times [0;1]) \cup B_n \rightarrow X \cup_g B_n$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finding such a decomposition and checking that the glueing works well is straightforward.&#xA;Proving that the maps you obtain are an equivalence of homotopy, when you properly write down what their composition does, should be easy as well.&lt;/p&gt;&#xA;" OwnerUserId="6270" LastEditorUserId="6270" LastEditorDisplayName="" LastEditDate="2011-01-27T18:49:33.050" LastActivityDate="2011-01-27T18:49:33.050" CommentCount="2" />
  <row Id="19221" PostTypeId="2" ParentId="19219" CreationDate="2011-01-27T13:55:36.493" Score="1" ViewCount="" Body="&lt;p&gt;The formula for the expected value of number of white balls is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$E($number of white drawn$) = 1P($1 white ball in C balls drawn $) +&#xA;2P($2 white balls in C balls drawn $) + \dots + C P($ C white balls drawn $)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now you need to find the probabilities. For example&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$P($ C white balls drawn$) = P($first ball is white$) * P($ second ball is white given first ball was white$) * P($ third ball is white given balls 1 and 2 were white$) * \dots * P($last ball white given all C-1 balls were white$) $&lt;/p&gt;&#xA;" OwnerUserId="5798" LastEditorUserId="5798" LastEditorDisplayName="" LastEditDate="2011-01-27T14:01:09.853" LastActivityDate="2011-01-27T14:01:09.853" CommentCount="2" />
  <row Id="19222" PostTypeId="2" ParentId="19203" CreationDate="2011-01-27T14:13:19.987" Score="3" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://mathworld.wolfram.com/Dodecahedron.html&quot; rel=&quot;nofollow&quot;&gt;MathWorld&lt;/a&gt; (about half-way down the page) says it can be done with 8, but does not say that's a minimum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/JfQZU.gif&quot; alt=&quot;diagram of 8 dodecahedrons arranged in a ring&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(diagram from MathWorld page)&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2011-01-27T14:13:19.987" />
  <row Id="19223" PostTypeId="2" ParentId="19215" CreationDate="2011-01-27T14:18:31.543" Score="3" ViewCount="" Body="&lt;p&gt;Given that your range is $x_0$ to $x_2$, let $x_1=-\frac{b}{2a}$.  The minimum value of $y$ occurs at $x_0$, $x_2$, or $x_1$ (if $x_0\le x_1\le x_2$).  So, compute $y_0=ax_0^2+bx_0+c$, $y_1=ax_1^2+bx_1+c$ (if $x_1$ is in the range), and $y_2=ax_2^2+bx_2+c$ and pick whichever is the least of those.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;edit&lt;/em&gt; to be clear, what I'm suggesting is something like:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;def quadMin(a, b, c, xmin, xmax):&#xA;  q0 = (a * xmin + b) * xmin + c&#xA;  q2 = (a * xmax + b) * xmax + c&#xA;  x1 = -b / (2*a)&#xA;  if (xmin &amp;lt; x1) and (x1 &amp;lt; xmax):&#xA;    q1 = (a * x1 + b) * x1 + c&#xA;    return min(q0, q1, q2)&#xA;  else:&#xA;    return min(q0, q2)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="72" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2011-01-27T17:00:59.920" LastActivityDate="2011-01-27T17:00:59.920" CommentCount="7" />
  <row Id="19224" PostTypeId="2" ParentId="19207" CreationDate="2011-01-27T14:35:07.270" Score="2" ViewCount="" Body="&lt;p&gt;Yes, you can use Euler's method (or one of the more sophisticated methods for solving an ODE initial value problem) to solve the $n$ initial value problems&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{dy}{dt}=Ay, y(0)=e_{1}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{dy}{dt}=Ay, y(0)=e_{2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ldots$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{dy}{dt}=Ay, y(0)=e_{n}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $e_{k}$ is the $k$th column of the identity matrix.  At time $x$, you'll get $n$ result vectors, which are the columns of $e^{xA}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are many other methods for computing matrix exponentials.  You should definitely read the paper, &quot;Nineteen Dubious Ways To Compute The Matrix Exponential, 25 Years Later.&quot; by Cleve Moler and Charles Van Loan.  See&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;http://www.cs.cornell.edu/cv/researchpdf/19ways+.pdf&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="6310" LastActivityDate="2011-01-27T14:35:07.270" />
  <row Id="19225" PostTypeId="2" ParentId="19219" CreationDate="2011-01-27T14:39:04.703" Score="5" ViewCount="" Body="&lt;p&gt;These calculations are often made clearer by using indicator random variables. For $1\leq i\leq C$, define $Z_i$ by setting $Z_i=1$ if the $i$th ball drawn is white and $Z_i=0$ otherwise. Then the total number of white balls is $\sum_{i=1}^C Z_i$ and by linearity of expectation we have&#xA;$$ E(\mbox{ white balls })=E\left(\sum_{i=1}^C Z_i\right)=\sum_{i=1}^C E(Z_i).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The expectation of an indicator random variable is just the probability of the event it indicates, so $E(Z_i)=P(i\mbox{th ball is white})=A/(A+B)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore we find that &#xA;$$ E(\mbox{ white balls })=\sum_{i=1}^C E(Z_i)={C A\over A+B}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice that we did not need to calculate the probability of getting one, two, three, etc. white balls. &lt;/p&gt;&#xA;" OwnerUserId="940" LastActivityDate="2011-01-27T14:39:04.703" />
  <row Id="19226" PostTypeId="1" AcceptedAnswerId="19246" CreationDate="2011-01-27T14:49:30.003" Score="1" ViewCount="47" Body="&lt;p&gt;Consider a given vector $a$ and scalar $d$. What is the set $X$ such that for any $x \in X$ their dot product equals $d$ : $\forall x \in X: x \cdot a = d$ ?&lt;/p&gt;&#xA;" OwnerUserId="3806" LastActivityDate="2011-01-27T16:24:03.773" Title="What is the locus such that any vector from it has a given dot product with the given vector?" Tags="&lt;homework&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="19227" PostTypeId="1" AcceptedAnswerId="19228" CreationDate="2011-01-27T14:57:09.640" Score="3" ViewCount="148" Body="&lt;p&gt;Find the sum of the series:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$ \displaystyle \cdots + \frac{1}{z^{3}} + \frac{1}{z^{2}} + \frac{1}{z} + 1 + z + z^{2} + z^{3} \cdots$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This series can be summed in the following way:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\cdots + \frac{1}{z^{3}} + \frac{1}{z^{2}} + \frac{1}{z} + 1 = \frac{z}{z-1}$$ and $$ z + z^{2} + z^{3} + \cdots = \frac{z}{1-z}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the sum equals $0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this correct or wrong? Please let me know if there is any error.&lt;/p&gt;&#xA;" LastActivityDate="2011-01-27T15:45:33.853" Title="Sum of a certain Series: Where is the error?" Tags="&lt;real-analysis&gt;&lt;analysis&gt;&lt;sequences-and-series&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="1" />
  <row Id="19228" PostTypeId="2" ParentId="19227" CreationDate="2011-01-27T15:01:53.320" Score="7" ViewCount="" Body="&lt;p&gt;The error is the following: Your first summation only holds for $\lvert z\rvert &amp;gt;1$ while the second one only holds for $\lvert z\rvert &amp;lt;1$.&lt;/p&gt;&#xA;" OwnerUserId="367" LastActivityDate="2011-01-27T15:01:53.320" CommentCount="3" />
  <row Id="19229" PostTypeId="2" ParentId="19227" CreationDate="2011-01-27T15:02:47.887" Score="7" ViewCount="" Body="&lt;p&gt;This series does not converge for any complex value of $z$.  However, in a certain sense if you set $z = e^{2 \pi i t}$, this is the Fourier series of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Dirac_delta_function&quot; rel=&quot;nofollow&quot;&gt;Dirac delta function&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In any case, you can show that &lt;em&gt;if&lt;/em&gt; there is a consistent way to interpret the sum, then for $z \neq 1$ it must be $0$.  This is because the expression $f$ satisfies $f = zf$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(It is possible to work abstractly with two-sided Laurent series, but you cannot multiply them, and you cannot evaluate them anywhere; the best you can do is consider them as a module over the Laurent polynomials.  This is done, for example, in certain branches of mathematical physics.)&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2011-01-27T15:45:33.853" LastActivityDate="2011-01-27T15:45:33.853" />
  <row Id="19230" PostTypeId="1" CreationDate="2011-01-27T15:07:04.317" Score="3" ViewCount="147" Body="&lt;p&gt;I came across this puzzle at some online contest(now completed)&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Alexander, son of Phillip of Macedon, has ascended the throne of his father following his assassination. In these tumultuous times, he appoints you as head of the architectural division of his mighty army. The general of the army wants all the catapults to be inducted in the artillery which have a n-sided polygon base, and no two or more of them should have the same type of polygon as their base.However, for the structures to be agile and economic, it is required that the &lt;strong&gt;n should be an odd number and the polygon of the base should be constructable with help of a compass and straightedge&lt;/strong&gt;.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Given the above situation, you are required to find out the &lt;strong&gt;total number of logs of woods required - each log of wood for each side of the base of structure&lt;/strong&gt; - for the construction of all such catapults&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I thought the result would be equal to the sum of the first 5 &lt;a href=&quot;http://en.wikipedia.org/wiki/Fermat_number&quot; rel=&quot;nofollow&quot;&gt;Fermat's Prime&lt;/a&gt; but unfortunately that wasn't correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do you think the data provided is not sufficient or am I missing something? What could the possible answer to this question?&lt;/p&gt;&#xA;" OwnerUserId="6313" LastActivityDate="2011-01-27T17:30:17.937" Title="Constructible Polygon and Fermat's Prime" Tags="&lt;number-theory&gt;&lt;geometry&gt;&lt;prime-numbers&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="19231" PostTypeId="2" ParentId="19205" CreationDate="2011-01-27T15:09:15.947" Score="3" ViewCount="" Body="&lt;p&gt;The non-existence of an algorithm (for the plane) can be found in a paper by Adler and Holroyd, Geometriae Dedicata, 1981.  The proof uses a straightforward simulation of Wang tiles.  &lt;/p&gt;&#xA;" OwnerUserId="6312" LastActivityDate="2011-01-27T15:09:15.947" CommentCount="2" />
  <row Id="19232" PostTypeId="2" ParentId="19230" CreationDate="2011-01-27T15:12:00.677" Score="1" ViewCount="" Body="&lt;p&gt;You're missing something. Read what your Wikipedia link has to say about constructible polygons again:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;An n-sided regular polygon can be constructed with compass and straightedge if and only if n is the &lt;b&gt;product&lt;/b&gt; of a power of 2 and &lt;b&gt;distinct Fermat primes&lt;/b&gt;. In other words, if and only if n is of the form $n = 2^kp_1p_2…p_s$, where k is a nonnegative integer and the $p_i$ are distinct Fermat primes.&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(My emphasis)&lt;/p&gt;&#xA;" OwnerUserId="5676" LastEditorUserId="5676" LastEditorDisplayName="" LastEditDate="2011-01-27T17:30:17.937" LastActivityDate="2011-01-27T17:30:17.937" CommentCount="3" />
  <row Id="19233" PostTypeId="2" ParentId="19092" CreationDate="2011-01-27T15:12:23.577" Score="2" ViewCount="" Body="&lt;p&gt;You should be looking at the fact that $S^{n-1} \to B^n$ is a cofibration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;:&#xA;This is a response to your below comment. There is a certain diagram that tells you what a cofibration is, it is annoying and ugly at first, but eventually it will make some amount of sense. The idea this diagram is supposed convey is that $A \to X$ (think of $A$ as a subspace of $X$) is a cofibration if we have a function out of $X$, say $f:X \to Y$, then once we have a homotopy of $f$ restricted to $A$ then we get a homotopy of &quot;all&quot; of $f$. There are many ways to think about cofibrations though, not all of them are easy to understand. Morally, a cofibration is a map $i: A \to X$ s.t. $X/i(A)$ behaves well in the eyes of homotopy theory, that is the homotopy type of $X/i(A)$ depends only on the homotopy class of $i$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A couple of important points about cofibrations is that every map is homotopic to a cofibration, think about the mapping cylinder. Also, if you have the inclusion of a subspace $A$ into $X$ where $X$ is obtained from $A$ by attaching cells then that inclusion is a cofibration, this is an easy way of noticing that map in question is a cofibration. Now since it is a cofibration and since you have a homotopy then the quotients should be the same. That last sentence can be made precise, think about what the pushout is actually the quotient of and how you can fit your attaching map into that picture.&lt;/p&gt;&#xA;" OwnerUserId="627" LastEditorUserId="627" LastEditorDisplayName="" LastEditDate="2011-01-28T04:57:41.380" LastActivityDate="2011-01-28T04:57:41.380" CommentCount="2" />
  <row Id="19234" PostTypeId="2" ParentId="19219" CreationDate="2011-01-27T15:18:04.830" Score="3" ViewCount="" Body="&lt;p&gt;For those interested, just to elucidate on the &quot;non-smart&quot; method of attacking this problem, the probability of drawing exactly $k$ white balls in $C$ draws is&#xA;$$ \frac{ {A \choose k}{B \choose C-k}} { {A+B \choose C} } $$&#xA;and so the expected number of white balls is given by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{k=0}^C \frac{ k {A \choose k}{B \choose C-k}} { {A+B \choose C} }&#xD;&#xA;= \frac{AC}{A+B}.$$ &lt;/p&gt;&#xA;" OwnerUserId="1301" LastEditorUserId="1301" LastEditorDisplayName="" LastEditDate="2011-01-27T15:24:22.073" LastActivityDate="2011-01-27T15:24:22.073" />
  <row Id="19235" PostTypeId="2" ParentId="19092" CreationDate="2011-01-27T15:19:51.317" Score="0" ViewCount="" Body="&lt;p&gt;If everything is a CW complex there is the more general fact that any two homotopic maps&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;f,g:B\longrightarrow Y&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for $B\subseteq X$ produce homotopic spaces $Y\cup_f X\simeq Y\cup_g X$ where we attach $X$ along $B$ by $f$.  If $F:B\times I\longrightarrow Y$ is the homotopy then you can think about $Y\cup_F (X\times I)$.  Then this becomes the fact that for CW pairs $(Z,A)$ you have a deformation retract of $Z\times I$ onto $(Z\times\{0\})\cup (A\times I)$.&lt;/p&gt;&#xA;" OwnerUserId="2838" LastEditorUserId="2838" LastEditorDisplayName="" LastEditDate="2011-01-28T15:50:25.650" LastActivityDate="2011-01-28T15:50:25.650" CommentCount="3" />
  <row Id="19236" PostTypeId="1" AcceptedAnswerId="19241" CreationDate="2011-01-27T15:29:44.570" Score="1" ViewCount="113" Body="&lt;p&gt;The question is simple. I would like to implement the Gamma function in my calculator written in C; however, I have not been able to find an easy way to programmatically compute an approximation to arbitrary precision.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a good algorithm to compute approximations of the Gamma function?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="1502" LastActivityDate="2011-01-27T15:51:06.830" Title="Algorithm to compute Gamma function" Tags="&lt;algorithms&gt;&lt;gamma-function&gt;" AnswerCount="3" />
  <row Id="19237" PostTypeId="1" AcceptedAnswerId="19267" CreationDate="2011-01-27T15:36:12.197" Score="4" ViewCount="102" Body="&lt;p&gt;Let $R$ be a discrete valuation ring with field of fraction $K$ and residue field $k$ and let $K&amp;#39;$ be a finite and separable extension of $K$.&#xA;If $R$ is henselian (&quot;Hensel's lemma holds&quot;, e.g. if it is complete) and $k$ is algebraically closed then the extension $K&amp;#39;|K$ is totally ramified (from Hensel's lemma if follows that there's just one prime lying above the prime of $R$ and since $k$ is algebraically closed there's no residual extension).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What happens if we assume only that $R$ is strictly henselian (i.e. that $k$ is only separably closed)?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Equivalently: does it exist a finite separable extension $K&amp;#39;$ of $K$ such that the residual extension is a nontrivial inseparable extension?&#xA;I believe that such an extension should exist.&#xA;Moreover, can we find a Galois extension with this property?&#xA;Can we do this both in characteristics $(0,p)$ and $(p,p)$?&lt;/p&gt;&#xA;" OwnerUserId="6314" LastActivityDate="2011-01-28T12:15:06.863" Title="Is every finite separable extension of a strictly henselian DVR totally ramified?" Tags="&lt;number-theory&gt;&lt;field-theory&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="19238" PostTypeId="2" ParentId="19236" CreationDate="2011-01-27T15:39:31.910" Score="0" ViewCount="" Body="&lt;p&gt;Do you want an algorithm for its complex domain or just for real numbers?&lt;/p&gt;&#xA;" OwnerUserId="6004" LastActivityDate="2011-01-27T15:39:31.910" CommentCount="1" />
  <row Id="19239" PostTypeId="2" ParentId="19236" CreationDate="2011-01-27T15:44:12.560" Score="3" ViewCount="" Body="&lt;p&gt;OFCOURSE!!!!!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Gamma_function&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Gamma_function&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Gamma_function#Approximations&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Gamma_function#Approximations&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="6174" LastActivityDate="2011-01-27T15:44:12.560" />
  <row Id="19240" PostTypeId="1" CreationDate="2011-01-27T15:48:01.920" Score="0" ViewCount="129" Body="&lt;p&gt;How to determine all the ideals of the ring $\mathbb{Z}[X]/(2,X^3+1)$?&lt;/p&gt;&#xA;" OwnerUserId="6308" LastEditorUserId="8575" LastEditorDisplayName="" LastEditDate="2011-04-27T12:51:38.307" LastActivityDate="2011-04-27T12:51:38.307" Title="Set of Ideals of a Polynomial Ring" Tags="&lt;ring-theory&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="2" />
  <row Id="19241" PostTypeId="2" ParentId="19236" CreationDate="2011-01-27T15:51:06.830" Score="2" ViewCount="" Body="&lt;p&gt;Looks like the Lanczos approximation will suit my needs : &#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Lanczos_approximation&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Lanczos_approximation&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for your help!&lt;/p&gt;&#xA;" OwnerUserId="1502" LastActivityDate="2011-01-27T15:51:06.830" />
  <row Id="19242" PostTypeId="2" ParentId="19240" CreationDate="2011-01-27T16:01:46.930" Score="4" ViewCount="" Body="&lt;p&gt;The ring $A={\Bbb Z}[X]/(2,X^3+1)$ is isomorphic to ${\Bbb F}_2[X]/(X^3+1)$ where ${\Bbb F}_2$ is the field with 2 elements. The ideals of the quotient $R/I$ are in one-to-one correspondence with the ideals in $R$ containing $I$. The correspondence is given by the quotient map.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, it's enough to decompose $X^3+1$ in irreducibles over ${\Bbb F}_2$. It is straightforward to check that the decomposition is $X^3+1=(X+1)(X^2+X+1)$. The two factors generate the only two non-trivial ideals in $A$.&lt;/p&gt;&#xA;" OwnerUserId="688" LastActivityDate="2011-01-27T16:01:46.930" />
  <row Id="19244" PostTypeId="2" ParentId="19240" CreationDate="2011-01-27T16:04:38.453" Score="0" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT:&lt;/strong&gt; Use the fact that $\mathbb{Z}[x]/(2,x^{3}+1) \cong \mathbb{Z}_{2}[x]/(x^{3}+1)$&lt;/p&gt;&#xA;" LastActivityDate="2011-01-27T16:04:38.453" />
  <row Id="19245" PostTypeId="2" ParentId="19061" CreationDate="2011-01-27T16:11:05.057" Score="7" ViewCount="" Body="&lt;p&gt;The question was asked by xport in the comments to Mike's solution as to whether this solution is unique.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are no doubt many ways to case bash an answer to this, and here is one, proving that&#xA;Mike's solution is indeed unique. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $x_i$ be the number of times the digit $ i $ appears on the paper, then we must have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \sum_{i=0}^9 x_i = 20 \quad \text{and} \quad \sum_{i=0}^9 ix_i = 45+20=65,$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;since there are $20$ digits on the completed paper and the first column sums to $45.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All the digits appear at least once therefore $x_i \ge 1 $ for all $ i $ and $x_0=1.$&#xA;Thus from the above equations we obtain&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{i=1}^9 x_i = 19 \quad (1) \quad \text{and} \quad \sum_{i=2}^9 (i-1)x_i = 46. \quad (2)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $\sum_{i=1}^8 i = 36$ it follows from $(2)$ that $(i-1)x_i \le 46 - (36-(i-1)),$ and so&#xA;$$ x_i \le \left \lfloor \frac{i+9}{i-1} \right \rfloor \quad \text{for} \quad i \ge 2.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ x_6 \le 3, \, x_7 \le 2, \, x_8 \le 2 \quad \text{and} \quad x_9 \le 2. \quad (3)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now suppose $ x_i &amp;gt; 1 $ for some $ i &amp;gt; 2 $ then, since $i$ occurs in at least three places, we must have $x_j=i$ for some $j \ne i.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And further suppose $j \ge 2,$ then from $(2)$ we obtain&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(i-1)x_i + (j-1)i + (36-(i-1)-(j-1)) \le 46,$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which gives&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ x_i \le \frac{2i+8}{i-1} - j.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence when $j \ge 2$ we must have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x_6 \le 2 \quad \text{and} \quad x_7=x_8=x_9=1. \quad (4)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We now do some case bashing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From $(3)$ we know $x_9 \le 2,$ so suppose $x_9=2$ then $(4)$&#xA;implies $x_1=9,$ which again by $(4)$ implies $x_7=x_8=1,$ since if $x_1=9,$ then $x_1$ cannot equal $7$ or $8.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Putting $x_1=9, x_7=1, x_8=1$ and $x_9=2$ in $(1)$ and $(2)$ we obtain&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x_2+x_3+ \cdots + x_6=6$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x_2 + 2x_3 + \cdots + 5x_6 = 17.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The former clearly only has solution $\lbrace 1,1,1,1,2 \rbrace, $ which satisfies the latter equation only when $x_3=2$ and the other $x_i$ are $1.$ But $3$ does not appear twice in this &quot;solution.&quot; Hence we have a contradiction and so&#xA;$$x_9=1.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similary, suppose $x_8=2,$ then $(4)$&#xA;implies $x_1=8,$ which again by $(4)$ implies $x_7=x_9=1,$ since if $x_1=8,$ then $x_1$ cannot equal $7$ or $9.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Putting $x_1=8, x_7=1, x_9=1$ and $x_8=2$ in $(1)$ and $(2)$ we obtain&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x_2+x_3+ \cdots + x_6=7$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x_2 + 2x_3 + \cdots + 5x_6 = 18.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The former equation only has solutions $\lbrace 1,1,1,1,3 \rbrace $ (which does not satisfy the latter equation) and $\lbrace 1,1,1,2,2 \rbrace,$ which only satisifes the latter&#xA;equation when $x_2=x_3=2$ and the other $x_i$ are $1.$ But again $3$ does not appear twice in this &quot;solution.&quot; Hence we have a contradiction and so&#xA;$$x_8=1.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similary, suppose $x_7=2,$ then $(4)$&#xA;implies $x_1=7,$ which again by $(4)$ implies $x_8=x_9=1,$ since if $x_1=7,$ then $x_1$ cannot equal $8$ or $9.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Putting $x_1=7, x_8=1, x_9=1$ and $x_7=2$ in $(1)$ and $(2)$ we obtain&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x_2+x_3+ \cdots + x_6=8$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x_2 + 2x_3 + \cdots + 5x_6 = 19.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The former equation only has solutions $\lbrace 1,1,1,1,4 \rbrace $ (which does not satisfy the latter equation) and $\lbrace 1,1,2,2,2 \rbrace$ (which also does not satisfy the latter equation) and $\lbrace 1,1,1,2,3 \rbrace, $ which only satisifes the latter&#xA;equation when $x_2=3, x_3=2,x_4=1,x_5=1$ and $x_6=1,$ and this is the solution given in Mike's answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence we have a unique solution when $x_7=2,$ otherwise we must have&#xA;$x_7=x_8=x_9=1.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To complete the proof we need only show that when $x_7=1$ we must have $x_6=1.$ That would give us at least six $1&amp;#39;s$ on the paper, implying that one of the numbers&#xA;$\lbrace 6,7,8,9 \rbrace$ must be repeated, which is a contradiction since we have&#xA;$x_6=x_7=x_8=x_9=1.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So suppose $x_7=1$ and $x_6=3$ (the maximum possible from $(3)$) then $(4)$ implies&#xA;$x_1=6.$ And since $x_8=x_9=1,$ from $(1)$ and $(2)$ we obtain&#xA;$$x_2+ x_3 + x_4 + x_5=7$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x_2 + 2x_3 + 3x_4 + 4x_5 = 10.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These have no solutions since the latter equation implies $x_2=x_3=x_4=x_5=1,$ which does not satisfy the former equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So suppose $x_7=1$ and $x_6=2,$ then since $x_8=x_9=1,$ from $(1)$ and $(2)$ we obtain&#xA;$$x_1+x_2+ x_3 + x_4 + x_5=14 \quad (5) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x_2 + 2x_3 + 3x_4 + 4x_5 = 15. \quad (6)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now $(6)$ implies $x_5 \le 2,$ but if $x_5=2$ then we must have $x_2 \ge 3$ (since&#xA;$x_5=x_6=2$ there are at least $3$ occurrences of $2$) and so the LHS of $(6)$ must be at least $16,$ a contradiction. Hence $x_5=1.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But now we have $x_0=1,x_5=1,x_6=2,x_7=1,x_8=1$ and $x_9=1,$ and since $x_6=2$ some number must occur 6 times but we do not have enough undetermined $x_i$ for this to occur unless&#xA;$x_1=6.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Substituting $x_1=6$ and $x_5=1$ in $(5)$ and $(6)$ we obtain&lt;br&gt;&#xA;$$x_2+ x_3 + x_4=7 \quad (7) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x_2 + 2x_3 + 3x_4 = 11. \quad (8)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and subtracting $(7)$ from $(8)$ we obtain&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x_3+2x_4=4. \quad (9)$$&#xA;However, since we cannot have any more $1&amp;#39;s$ (as $x_1=6$) we must have $x_3 \ge 2$ and&#xA;$x_4 \ge 2,$ which contradicts $(9).$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus we have shown that when $x_7=1$ we must have $x_6=1,$ which as noted above completes the proof that the unique solution is given by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x_0=1,x_1=7,x_2=3,x_3=2,x_4=1,x_5=1,x_6=1,x_7=2,x_8=1$ and $x_9=1.$&lt;/p&gt;&#xA;" OwnerUserId="1301" LastActivityDate="2011-01-27T16:11:05.057" CommentCount="1" />
  <row Id="19246" PostTypeId="2" ParentId="19226" CreationDate="2011-01-27T16:24:03.773" Score="1" ViewCount="" Body="&lt;p&gt;It's easy to write down one point $p = \frac{a * d}{|a|^2}$ in this set.  For any other point $q$, we have $(p - q) \cdot a = 0$, so the set of vectors $p - q$ is precisely the set of vectors orthogonal to $a$.  &lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2011-01-27T16:24:03.773" />
  <row Id="19247" PostTypeId="2" ParentId="19187" CreationDate="2011-01-27T16:29:38.177" Score="0" ViewCount="" Body="&lt;p&gt;Almost certainly what was meant by the professor was that the collection of all subsets of a countable set is an uncountable collection as clearly any specific subset of a countable set is also countable.&lt;/p&gt;&#xA;" OwnerUserId="5613" LastEditorUserId="5613" LastEditorDisplayName="" LastEditDate="2011-01-27T23:08:13.303" LastActivityDate="2011-01-27T23:08:13.303" CommentCount="1" />
  <row Id="19248" PostTypeId="2" ParentId="19054" CreationDate="2011-01-27T16:31:05.820" Score="3" ViewCount="" Body="&lt;p&gt;Whenever $f(x_0)$ and $f(x_1)$ fall in an interval $[a, b]$, we must have $$|x_0 - x_1| \le (f(x_0) - f(x_1))^2 \le (a - b)^{2}.$$  Therefore, the pre-image of any interval $[a,b]$ under $f$ is contained in some interval of length $(a-b)^2$.  Now let $a_0=0$ and $a_k=\sum_{n=1}^{k}n^{-1}$ for $k=1,2,3...$  Since $a_k$ diverges, we have $\cup_k [a_k, a_{k+1}] = [0, \infty)$.  So we can write&#xA;$$&#xD;&#xA;\begin{eqnarray}&#xD;&#xA;f^{-1}\left([0, \infty)\right) &amp;amp;=&amp;amp; f^{-1}\left(\cup_k [a_k, a_{k+1}]\right) \\&#xD;&#xA;&amp;amp;=&amp;amp; \cup_k f^{-1}\left([a_k, a_{k+1}]\right) \\&#xD;&#xA;&amp;amp;\subseteq&amp;amp; \cup_k I_k,&#xD;&#xA;\end{eqnarray}&#xD;&#xA;$$&#xA;where each $I_k$ is an interval of length $(a_{k+1} - a_{k})^2 = 1/(k+1)^2$.  This countable union of intervals is measurable, and has measure $\le \sum_{k=0}^{\infty} 1/(k+1)^2 = \pi^2/6$.  Similarly, we can show that $f^{-1}\left((-\infty, 0]\right)$ is also contained in a measurable set with measure $\le \pi^2/6$.  Combining these results, we have $f^{-1}(\mathbb{R}) \subseteq Q$, where $Q$ is measurable with measure $\pi^2/3$.  Clearly $Q \neq \mathbb{R}$ (since $\mathbb{R}$ has infinite measure); hence $f^{-1}(\mathbb{R}) \neq \mathbb{R}$, and $f$ cannot be defined over the entire real line.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same argument with $a_k$ replaced by $\sum_{n=1}^{k}(n+\alpha)^{-1}$ can be used to make the measure of $Q$ arbitrarily small (by choosing $\alpha$ sufficiently large), and we conclude that the domain of $f$ cannot contain any set of positive measure.&lt;/p&gt;&#xA;" OwnerUserId="5546" LastEditorUserId="5546" LastEditorDisplayName="" LastEditDate="2011-01-27T16:43:15.970" LastActivityDate="2011-01-27T16:43:15.970" CommentCount="2" />
  <row Id="19249" PostTypeId="2" ParentId="19205" CreationDate="2011-01-27T16:32:02.443" Score="5" ViewCount="" Body="&lt;p&gt;The book Tilings and Patterns by Branko Grünbaum and Geoffrey Shephard (W.H. Freeman) is still, despite its 1987 publication date, an excellent source of information about tiling questions. Wang tiles and decidability problems are treated in Chapter 11.  &lt;/p&gt;&#xA;" OwnerUserId="1369" LastActivityDate="2011-01-27T16:32:02.443" CommentCount="2" />
  <row Id="19250" PostTypeId="1" CreationDate="2011-01-27T16:35:59.650" Score="0" ViewCount="59" Body="&lt;p&gt;I was wondering what properties follow if a numerical method is &quot;nonexpansive&quot;. I googled and found this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://books.google.ch/books?id=rqvYq19qwiwC&amp;amp;pg=PA332&amp;amp;lpg=PA332&amp;amp;dq=nonexpansive+numerical+method&quot; rel=&quot;nofollow&quot;&gt;http://books.google.ch/books?id=rqvYq19qwiwC&amp;amp;pg=PA332&amp;amp;lpg=PA332&amp;amp;dq=nonexpansive+numerical+method&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is a link to google books to a definition of nonexpansive. Many thanks for your help!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;&#xA;I'm still interested in the answer to this question. Here is a guess (I'm not an expert in numerical maths):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could it be that if a method is non-expansive then it's not possible to get neither a blow-up nor a collapse?&lt;/p&gt;&#xA;" OwnerUserId="5798" LastEditorUserId="5798" LastEditorDisplayName="" LastEditDate="2011-05-17T09:56:29.197" LastActivityDate="2011-05-17T09:56:29.197" Title="Nonexpansive numerical method" Tags="&lt;numerical-methods&gt;" CommentCount="5" FavoriteCount="0" />
  <row Id="19251" PostTypeId="2" ParentId="19177" CreationDate="2011-01-27T16:37:01.423" Score="6" ViewCount="" Body="&lt;p&gt;William LeVeque's &lt;strong&gt;Topics in Number Theory, Volume I&lt;/strong&gt; treats Pell's equation with nonsquare $d$ in Chapter 8, without the assumption that $d$ is squarefree.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In Section 8.2 (page 139 in the first half of the Dover edition), he considers $x^2-dy^2 = 1$ as follows:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;If $d$ is negative, then when $d=-1$ the only solutions are $(\pm 1,0)$ and $(0,\pm 1)$. If $d\lt -1$, then the only solutions are $(\pm 1,0)$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If $d$ is a square, $d=(d&amp;#39;)^2$, then we can rewrite the equation as $x^2 - (d&amp;#39;y)^2 = 1$; the only squares that differ by $1$ are $0$ and $1$, so again the only solutions are $(\pm 1,0)$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;For positive nonsquare $d$, we start with a lemma on approximation and then proceed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Lemma.&lt;/strong&gt; If $\xi$ is a real number and $t$ is a positive integer, then there are integers $x$ and $y$ such that&#xA;$$\left|\xi - \frac{x}{y}\right| \leq \frac{1}{y(t+1)},\qquad 1\leq y\leq t.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Proof.&lt;/em&gt; The $t+1$ numbers&#xA;$$0\cdot \xi - \lfloor 0\cdot \xi\rfloor,\quad 1\cdot\xi - \lfloor 1\cdot \xi\rfloor, \quad\cdots\quad t\xi - \lfloor t\xi\rfloor$$&#xA;are all in the interval $0\leq u \lt 1$. In increasing order of magnitude, call them $a_0$, $a_1,\ldots,a_t$. Mark the numbers on a circle of perimeter $1$. Then the $t+1$ differences&#xA;$$a_1-a_0,\quad a_2-a_1,\quad\ldots,\quad a_t-a_{t-1},\quad a_0-a_t+1,$$&#xA;are the lengths of the arcs between successive $a$s, so they are nonnegative, and&#xA;$$(a_1-0) + (a_2-a_1) + \cdots + (1-a_t) = 1.$$&#xA;So at least one of these $t+1$ differences is no more than $\frac{1}{t+1}$. But the each difference is of the form $g_1\xi - g_2\xi - m$, with $m$ an integer, so take $y=|g_1-g_2|$, $x=\pm m$. $\Box$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Theorem.&lt;/strong&gt; For any irrational $\xi$, the inequality&#xA;$$|x - \xi y| \lt \frac{1}{y}$$&#xA;has infinitely many integer solutions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Proof.&lt;/em&gt; For each positive integer $t$, $0\lt |x-\xi y|\lt \frac{1}{t}$, $1\leq y\leq t$ has solutions. Taking $t=1$ gives a solution $(x_1,y_1)$ to the original equation. For sufficiently large $t_1$ we have $|x_1 - \xi y_1|\gt \frac{1}{t_1}$, so taking $t=t_1$ gives a new solution $(x_2,y_2)$ to the original equation. Lather, rinse, repeat. $\Box$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The irrationality of $\xi$ is used to ensure that you have strict inequalities when you apply the lemma, so that you can set up the recursion and get infinitely many solutions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Theorem.&lt;/strong&gt; If $d$ is positive and not a square, then there are infinitely many integer solutions to the equation&#xA;$$x^2 - dy^2 = k$$&#xA;for positive integers $x,y$ for some $k$ with $|k|\lt 1+2\sqrt{d}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Proof.&lt;/em&gt; Pick a solution to $|x - \sqrt{d}y|\lt \frac{1}{y}$. Then&#xA;\begin{align*}&#xA;|x+y\sqrt{d}| &amp;amp;= |x-y\sqrt{d} + 2y\sqrt{d}|\\&#xA;&amp;amp;\lt \frac{1}{y}+2y\sqrt{d}\\&#xA;&amp;amp;\leq (1+2\sqrt{d})y&#xA;\end{align*}&#xA;and so $|x^2-dy^2| \lt \frac{1}{y}(1+2\sqrt{d})y = 1+2\sqrt{d}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since there are infinitely many pairs $(x,y)$ you can use, but only finitely many integers that are positive and smaller than $1+2\sqrt{d}$, infinitely many of the values of $x^2-dy^2$ must coincide, which is the theorem. $\Box$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice that the only requirement here is that $\sqrt{d}$ be real and irrational, i.e., that $d$ is positive and not a perfect square.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Theorem.&lt;/strong&gt; If $d\gt 0$ is not a square, then the equation&#xA;$$x^2 - dy^2 = 1$$&#xA;has at least one solution with $y\neq 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Proof.&lt;/em&gt; Take the infinitely many solutions to $x^2-dy^2 = k$ (for some $k$ with $|k|\lt 1+2\sqrt{d}$) and divide them into $k^2$ equivalence classes, where $(x_1,y_1)\sim (x_2,y_2)$ if and only if $x_1\equiv x_2\pmod{k}$ and $y_1\equiv y_2\pmod{k}$. Some class contains more than one solution, say $(x_1,y_1)$ and $(x_2,y_2)$ with $x_1x_2\gt 0$. Let&#xA;$$x = \frac{x_1x_2 - dy_1y_2}{k},\qquad y=\frac{x_1y_2 - x_2y_1}{k};$$&#xA;then $x$ and $y$ are integers with $y\neq 0$ and $x^2-dy^2 = 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Indeed, $x_1y_2\equiv x_2y_1\pmod{k}$, so $y$ is an integer. Also,&#xA;$$x_1x_2 - dy_1y_2 \equiv x_1^2 - dy_1^2 = k \equiv 0 \pmod{k}$$&#xA;so $x$ is an integer. Also&#xA;\begin{align*}&#xA;x^2 - dy^2 &amp;amp;= \frac{1}{k^2}\left( (x_1x_2 - dy_1y_2)^2 - d(x_1y_2-x_2y_1)^2\right)\\&#xA;&amp;amp;= \frac{1}{k^2}\left(x_1^2x_2^2 - dx_1^2y_2^2 + d^2y_1^2y_2^2 - dx_2^2y_1^2\right)\\&#xA;&amp;amp;= \frac{1}{k^2}\left(x_1^2-dy_1^2\right)\left(x_2^2 - dy_2^2\right) = 1.&#xA;\end{align*}&#xA;Finally, if $y=0$, then $x_1y_2=x_2y_1$, so $x_1=ax_2$ and $y_1=ay_2$ for some $a$; but plugging into $x^2- dy^2 = k$ we get $a=1$, contradicting that $(x_1,y_1)$ and $(x_2,y_2)$ are distinct solutions. $\Box$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Theorem.&lt;/strong&gt; If $(x_1,y_1)$ and $(x_2,y_2)$ are solutions to $x^2-dy^2 = 1$, then so are the integers $x$ and $y$ defined by&#xA;$$(x_1 + y_1\sqrt{d})(x_2+y_2\sqrt{d}) = x+y\sqrt{d}.$$&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-27T16:55:13.217" LastActivityDate="2011-01-27T16:55:13.217" />
  <row Id="19252" PostTypeId="1" CreationDate="2011-01-27T16:45:58.920" Score="1" ViewCount="33" Body="&lt;p&gt;I am not quite sure how to classify this conceptual question and I really hope to get an answer.  No comments like &quot;it is the wrong place to ask it&quot;. I am just trying to get different perspectives to write part of my work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is important to define the extended Kac-Moody algebras? What was the necessity which required this construction?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks,&lt;/p&gt;&#xA;" OwnerUserId="2764" LastActivityDate="2011-01-27T16:45:58.920" Title="extended Kac-Moody algebras" Tags="&lt;lie-algebra&gt;" />
  <row Id="19253" PostTypeId="1" CreationDate="2011-01-27T17:51:20.453" Score="3" ViewCount="68" Body="&lt;p&gt;A naive question from a physicist, so forgive the lack of rigor. Consider a Lie group, acting on its Lie algebra by the adjoint action. Does every orbit go through the Cartan subalgebra? Alternatively, for which Lie groups is this true? (I view the statement as a generalization of the elementary fact that every Hermitian matrix can be diagonalized by a unitary transformation.) It would be nice if it held at least for real, compact, and semi-simple Lie algebras. Thank you for any hint or reference to literature!&lt;/p&gt;&#xA;" OwnerUserId="6321" LastActivityDate="2011-01-27T17:51:20.453" Title="does every adjoint orbit of a Lie group go through the Cartan subalgebra?" Tags="&lt;lie-groups&gt;&lt;lie-algebra&gt;" CommentCount="3" />
  <row Id="19254" PostTypeId="2" ParentId="2601" CreationDate="2011-01-27T18:03:06.607" Score="2" ViewCount="" Body="&lt;p&gt;My brain only works well in Greekspace (that charming region of mathematics where everything has to correspond to a physical analog), so these are quite familiar to me.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;If real numbers represent distance along a number line, imaginary numbers represent rotation. See the incredibly well-constructed explanation here: &lt;a href=&quot;http://betterexplained.com/articles/a-visual-intuitive-guide-to-imaginary-numbers/&quot; rel=&quot;nofollow&quot;&gt;http://betterexplained.com/articles/a-visual-intuitive-guide-to-imaginary-numbers/&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;If you want to push a crate, and you don't want it to spin while you're pushing it, the direction you have to push it in is an eigenvector. The amount it moves in proportion to how hard you push it is an eigenvalue.&lt;/li&gt;&#xA;&lt;li&gt;A matrix is a collection of vectors. If you have a matrix of some dimension less than four, you can visualize the vectors floating in 3D space. If not, you'll have to get more creative.&lt;/li&gt;&#xA;&lt;li&gt;Speaking of dimensions, a &quot;dimension&quot; is &quot;anything that you can vary continuously,&quot; whether it's time, color, temperature, w/e. (I actually set up a 3D grid of different-colored vectors in my head for my linear algebra final. Very &quot;A Beautiful Mind,&quot; in retrospect.)&lt;/li&gt;&#xA;&lt;li&gt;The Heaviside function is the curb that your response-function-car hits.&lt;/li&gt;&#xA;&lt;li&gt;Vector components/projections are shadows of the vector projected on x, and orthogonal vectors don't cast shadows on each other.&lt;/li&gt;&#xA;&lt;li&gt;Derivatives and integrals are getting distance/acceleration from the graph of your speed (which is, of course, drawn by a pen attached to your speedometer on a scrolling sheet of paper).&lt;/li&gt;&#xA;&lt;li&gt;Logarithms &amp;amp; their rules didn't really make sense to me until I saw &amp;amp; learned how to use a slide rule.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="6322" LastActivityDate="2011-01-27T18:03:06.607" CommunityOwnedDate="2011-01-27T18:03:06.607" CommentCount="2" />
  <row Id="19256" PostTypeId="2" ParentId="19177" CreationDate="2011-01-27T18:31:41.497" Score="5" ViewCount="" Body="&lt;p&gt;A different and important way to view this equation is still as a norm equation, only in a non-maximal order of the quadratic field - namely, we are looking at the ring $\mathbb{Z}[p\sqrt{d}]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since we are still in the realm of algebra, we can prove results algebraically. For example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Proposition.&lt;/b&gt; Let $p$ be a prime that doesn't divide $d$, and $e=x+y\sqrt{d}$ be a unit with integer $x,y$. Then $e^{p-(d/p)} \in \mathbb{Z}[p\sqrt{d}]$, where $(d/p)$ is the Legendre symbol.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Proof.&lt;/b&gt; If the Legendre symbol is $-1$, this means that $(p)$ doesn't split in $O$ (which will denote the maximal order from here on), so that $O/(p)$ is isomorphic to the finite field of size $p^2$. Now every number in $\mathbb{F}_{p^2}$ has order dividing $p^2-1$, so $e^{p+1} \pmod{p}$ has order diving $p-1$. But the numbers of order diving $p-1$ are exactly those in the subfield $\mathbb{F}_p$, and that means that $e^{p+1} = z+pw\sqrt{d}$, proving the claim in this case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $(p)$ splits into $\pi_1 \pi_2$, then $O/(p)$ is isomorphic to $O/\pi_1 \pi_2$ which is in turn, by chinese remainder theorem, isomorphic to $\mathbb{F}_p \oplus \mathbb{F}_p$. By inspection we see that the galois action permutes the two coordinates. So, since under the isomorphism $e^{p-1}\pmod{\pi_1}=e^{p-1}\pmod{\pi_2}=1$, we see that $e^{p-1}\pmod{p}$ is invariant. Hence it is of the form $z+pw\sqrt{d}$, proving the claim in this case as well. $\Box$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The above proposition shows how to get solutions to the non-squarefree Pell equation from solutions of the squarefree one - simply power appropriately.&lt;/p&gt;&#xA;" OwnerUserId="4946" LastActivityDate="2011-01-27T18:31:41.497" />
  <row Id="19257" PostTypeId="1" AcceptedAnswerId="19367" CreationDate="2011-01-27T19:27:29.323" Score="7" ViewCount="466" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Problem: Given $f:[0,1] \rightarrow \mathbb{R}$ ($f$ continuous ) and $f(x^2) = f(x)$ $\forall x \in [0,1]$. Show that function $f$ is a const.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="3637" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2011-02-22T23:49:20.593" LastActivityDate="2011-02-22T23:49:20.593" Title="If $f$ continuous and $f(x^2) = f(x)$, then $f$ is a const" Tags="&lt;calculus&gt;" AnswerCount="1" CommentCount="14" FavoriteCount="3" />
  <row Id="19258" PostTypeId="1" AcceptedAnswerId="19259" CreationDate="2011-01-27T19:38:37.430" Score="1" ViewCount="136" Body="&lt;p&gt;Prove that there is no arithmetic progression that consists &lt;em&gt;only&lt;/em&gt; of primes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A question that I've been set; I'm guessing it makes use of primes being written in the form 4k+1 and 4k+3? Not sure where to start.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="6323" LastActivityDate="2011-01-27T19:44:42.407" Title="arithmetic progression of primes" Tags="&lt;number-theory&gt;&lt;prime-numbers&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="19259" PostTypeId="2" ParentId="19258" CreationDate="2011-01-27T19:44:42.407" Score="4" ViewCount="" Body="&lt;p&gt;An arithmetic progression is a sequence of the form&#xA;$$a,\quad a+b,\quad a+2b,\quad a+3b,\quad a+4b,\quad\ldots, a+nb,\quad\ldots$$&#xA;with $b\neq 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $a$ is not prime, you're done. So suppose $a$ is a prime. Can you find a term later in the sequence that is &lt;em&gt;guaranteed&lt;/em&gt; to be a multiple of $a$?&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2011-01-27T19:44:42.407" CommentCount="1" />
  <row Id="19260" PostTypeId="1" AcceptedAnswerId="19262" CreationDate="2011-01-27T19:52:01.897" Score="1" ViewCount="85" Body="&lt;p&gt;Let $G = GL_n(\mathbb{R})$ be the general linear group. Does $G$ have any subgroups of finite index?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="5491" LastActivityDate="2011-01-27T20:00:02.213" Title="Subgroups of finite index of the general linear group" Tags="&lt;group-theory&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="19261" PostTypeId="1" AcceptedAnswerId="19270" CreationDate="2011-01-27T19:53:46.653" Score="4" ViewCount="245" Body="&lt;p&gt;A probability problem was posed to me, and I am terrible at probability. It was first posed to me like this:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I have 5 coins. One of them is a&#xA;  two-headed coin, and the others are&#xA;  normal coins. I pick a coin, and then&#xA;  I flip that coin three times; each&#xA;  time is heads. What is the probability&#xA;  that the coin is two-headed?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My answer was, naturally, 1/5. I am still of the opinion that it's the correct answer to that particular phrasing (do you agree?). But of course I was told this is a wrong answer. After some thought, I came up with the following phrasing which is what was really meant:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;You are given a coin and told that it&#xA;  has a 1/5 chance that it is&#xA;  double-headed, otherwise it is a&#xA;  normal coin. You are allowed three&#xA;  flips of the coin. Upon doing these,&#xA;  you receive three heads. What is the&#xA;  probability that your coin is&#xA;  double-headed?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now I have two questions, first of all does that seem like a proper rewording to you? And secondly... What is the answer and the reasoning to reach that answer? I am terrible at probability, and while I can clearly see the problem and maybe spell out the first step or two, I am at a total loss as to how to arrive at an answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: one more question, actually: is this an instance of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Monty_Hall_problem&quot; rel=&quot;nofollow&quot;&gt;Monty Hall problem&lt;/a&gt;? I obviously see that it is very different, yet it feels somehow similar...&lt;/p&gt;&#xA;" OwnerUserId="137" LastEditorUserId="137" LastEditorDisplayName="" LastEditDate="2011-01-27T20:10:21.000" LastActivityDate="2011-01-28T12:39:19.257" Title="Probability of a two-headed coin given a few sample flips?" Tags="&lt;probability&gt;" AnswerCount="4" CommentCount="4" />
  <row Id="19262" PostTypeId="2" ParentId="19260" CreationDate="2011-01-27T19:59:52.223" Score="4" ViewCount="" Body="&lt;p&gt;Here's one, though you may consider it a &quot;trivial&quot; example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The map $\det\colon G\to\mathbb{R}$ maps $G$ to the nonzero reals under multiplication via the determinant function, which is a group homomorphism since $\det(AB)=\det(A)\det(B)$. The positive real numbers are a subgroup of the real numbers under multiplication, of index $2$ (map the nonzero real numbers to $C_2=\{1,x\}$, the cyclic group of order $2$, by mapping positive numbers to $1$, negative numbers to $x$. The composition $G\to C_2$ is a map from $G$ onto a finite group, hence the kernel is a proper subgroup of finite index.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2011-01-27T19:59:52.223" CommentCount="1" />
  <row Id="19263" PostTypeId="2" ParentId="19260" CreationDate="2011-01-27T20:00:02.213" Score="3" ViewCount="" Body="&lt;p&gt;You can take the preimage under $\det:\mathrm{GL}_n(\mathbb R)\to\mathbb R^\times$ of $\mathbb R^+$.&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2011-01-27T20:00:02.213" />
  <row Id="19264" PostTypeId="1" CreationDate="2011-01-27T20:01:26.990" Score="0" ViewCount="71" Body="&lt;p&gt;I am currently studying for an Algorithms class. In my notes from class and in the my textbook, there is not a similar example I can reference. I have worked on the problem and gone to tutoring, but I can not seem to get the correct answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone give me any tips on how to solve this problem?&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Evaluate the summation, where $a$ is a constant, $a \neq 1$.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;a) $\displaystyle\sum_{i={10}}^n (a^i+n)$, where $a\neq 1$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Thank you in advance.&lt;/p&gt;&#xA;" OwnerUserId="6324" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-27T20:08:05.820" LastActivityDate="2011-01-27T21:22:52.103" Title="Homework Help - Summation with Three Variables" Tags="&lt;analysis&gt;&lt;sequences-and-series&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="19265" PostTypeId="1" AcceptedAnswerId="19332" CreationDate="2011-01-27T20:08:44.747" Score="3" ViewCount="175" Body="&lt;p&gt;A Stirling number of the second kind, $S (r,k)$, is defined to be the number of ways one can partition an $r$-element set into $k$ subsets. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the following problem: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have $r$ &lt;em&gt;distinguishable&lt;/em&gt; balls and $n$ &lt;em&gt;undistinguishable&lt;/em&gt; boxes. If you are allowed to place balls into these boxes with no restrictions, how many different ways of doing so are there?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The anwser is $\sum_{k=1}^{n} S (r,k)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can nicely use this model for many combinatorial problems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to ask if anyone has seen, or knows of a problem where one would sum over $r$ instead, i.e. where something like $\sum_{k=i}^{j} S (k,n)$ would appear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This just a general interest question; it came up in a combinatorial methods class. Many thanks for any anwsers/comments.&lt;/p&gt;&#xA;" OwnerUserId="5105" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2011-05-28T20:22:34.560" LastActivityDate="2011-05-28T20:22:34.560" Title="Stirling numbers of the second kind $S (r,k)$ - does one ever sum over $r$?" Tags="&lt;combinatorics&gt;&lt;stirling-numbers&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="1" />
  <row Id="19266" PostTypeId="2" ParentId="19264" CreationDate="2011-01-27T20:09:31.543" Score="2" ViewCount="" Body="&lt;p&gt;You can divide the sum into a sum of powers of $a$, and a sum of $n$. The latter is trivial.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the former, you are obviously expected to use the &lt;a href=&quot;http://en.wikipedia.org/wiki/Geometric_series#Formula&quot; rel=&quot;nofollow&quot;&gt;formula for the sum of a geometric series&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2011-01-27T20:09:31.543" />
  <row Id="19267" PostTypeId="2" ParentId="19237" CreationDate="2011-01-27T20:19:12.377" Score="2" ViewCount="" Body="&lt;p&gt;Let $k$ a separably closed imperfect field of char. $p$. Fix $t \in k$ for which $X^p - t \in k[X]$ has no root in $k$.&#xA;Set $K = k((s))$, and form the extension $L = K(y) \supset K$ obtained by adjoining to $K$ a root $y$ of the poly. $f(Y) = Y^p + sY - t \in K[Y]$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Write $A = k[[s]]$ for the integers of $K$. Since the image of $f(Y)$ in $k[Y] = (A/sA)[Y]$ is irreducible, $f(Y)$ is irreducible in $A[Y]$ and hence in $K[Y]$. Since $f&amp;#39;(Y) = s \ne 0$, it follows that $L/K$ is separable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now write $B$ for the integral closure of $A$ in $L$; thus $y \in B$. Denoting by $\pi$ a uniformizer for the DVR $B$, note that $s \in \pi B$. Thus the image of $y$ in $\ell = B/\pi B$ is a $p$-th root of $t$, so $\ell$ is a proper (purely) inseparable extension of $k$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT:  In hindsight, it seems that with minor tweaking the same argument gives an answer in the &quot;mixed characteristic&quot; case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Indeed, let $t \in k$ as before, and now let $A$ a complete (or Henselian) DVR with residue field $k$ and with fractions $K$ of char. 0. Now choose an element&#xA;$\tau \in A$ whose image in $k$ is $t$. Let $\sigma \in A$ be a uniformizer,&#xA;and set $f(Y) = Y^p + \sigma Y - \tau \in A[Y]$.  Then as before the extension $L = K(y)$ of $K$ is separable, where $y$ is a root of $f(Y)$. And the residue extension is again inseparable.&lt;/p&gt;&#xA;" OwnerUserId="5020" LastEditorUserId="5020" LastEditorDisplayName="" LastEditDate="2011-01-28T12:15:06.863" LastActivityDate="2011-01-28T12:15:06.863" CommentCount="4" />
  <row Id="19268" PostTypeId="2" ParentId="19261" CreationDate="2011-01-27T20:26:26.437" Score="2" ViewCount="" Body="&lt;p&gt;With the Bayesian interpretation of the question (your second phrasing), and leaving it as an exercise to figure out my notation: $$p(HHH|DH)=1$$ $$p(DH)=\frac{1}{5}$$ $$p(DH|HHH)=\frac{p(DH \wedge HHH)}{p(HHH|DH)p(DH)+p(HHH|N)p(N)}$$ $$=\frac{1/5}{1/5+(1/8)(4/5)}=\frac{2}{3}$$&lt;/p&gt;&#xA;" OwnerUserId="6104" LastActivityDate="2011-01-27T20:26:26.437" />
  <row Id="19269" PostTypeId="1" AcceptedAnswerId="19271" CreationDate="2011-01-27T20:27:04.680" Score="3" ViewCount="113" Body="&lt;p&gt;So I'm trying to solve this problem stated like this:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Using Green's Theorem, find the area of the elipse defined by (where $a,b \gt 0$):&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$$\frac{x^2}{a^2} + \frac{y^2}{b^2} \leqq 1$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm having trouble doing this,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_{FrD} \!F \cdot T = {\int\int}_D \left(\frac{df_2}{dx} - \frac{df_1}{dy}\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $F$ is a vector field. My try at solving this was parametize the elipse as polar coords and doing $F$ as $F(x,y) = (1,1)$. Similiar to how to get the area. But I'm pretty sure that's not the correct way. &lt;/p&gt;&#xA;" OwnerUserId="6326" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-27T20:33:31.897" LastActivityDate="2011-01-27T20:51:47.200" Title="Applying Green's Theorem" Tags="&lt;integral&gt;&lt;vector-calculus&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="19270" PostTypeId="2" ParentId="19261" CreationDate="2011-01-27T20:30:02.613" Score="5" ViewCount="" Body="&lt;p&gt;I think the rewording is fine, however, I believe your anwser is not correct. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Indeed, without throwing coins at all, there is a $1/5$ chance you get the double headed coin, and a $4/5$ chance you get a normal coin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the probability of getting three heads on three throws of a regular coin is $1/8$, as you can surely check. And as there is a $4/5$ chance you &lt;em&gt;do&lt;/em&gt; get a regular coin, so you have a $1/8 \cdot 4/5 = 1/10$ chance you get three heads flipping &lt;em&gt;a regular coin&lt;/em&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So to sum up the outcomes: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;$1/5$ chance of getting a double headed coin&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$1/10$ chance of getting a normal coin and three heads&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$7/10$ chance of getting a normal coin and anything but three heads.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;However, note that you are asking: What is the probability that your coin is double-headed assuming you GET three heads? So the only options you are considering are the first two (as you know the third didn't occur).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now there is a $1/5 + 1/10 = 3/10$ chance you get three heads. So there is a $\frac{1/5}{3/10} = 2/3$ chance of getting a double headed coin.&lt;/p&gt;&#xA;" OwnerUserId="5105" LastActivityDate="2011-01-27T20:30:02.613" />
  <row Id="19271" PostTypeId="2" ParentId="19269" CreationDate="2011-01-27T20:37:59.723" Score="4" ViewCount="" Body="&lt;p&gt;This is a standard application, a way to use Green's Theorem to compute areas by doing line integrals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $D$ be the ellipse, and $C$ its boundary $\frac{x^2}{a^2}+\frac{y^2}{b^2} = 1$. The area you are trying to compute is&#xA;$$\int\!\!\int_D 1\,dA.$$&#xA;According to Green's Theorem, if you write $1 = \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}$, then this integral equals&#xA;$$\oint_C(P\,dx + Q\,dy).$$&#xA;There are many possibilities for $P$ and $Q$. Pick one. Then use the parametrization of the ellipse&#xA;\begin{align*}&#xA;x&amp;amp;=a\cos t\\&#xA;y&amp;amp;=b\sin t&#xA;\end{align*}&#xA;to compute the line integral.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you can probably see, the idea of finding $P$ and $Q$ with $\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = 1$ can be used to compute the area of any region enclosed by a simple closed curve. Of course, the line integral may be more complicated than the area computation, but that's another kettle of fish.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-27T20:51:47.200" LastActivityDate="2011-01-27T20:51:47.200" />
  <row Id="19272" PostTypeId="2" ParentId="19269" CreationDate="2011-01-27T20:38:41.957" Score="2" ViewCount="" Body="&lt;p&gt;Let $A$ be the area of the region $D$ bounded by the ellipse with equation $$\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}}=1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $\partial{D}$ denote the boundary. You can parametrize $\partial{D}$ with counterclockwise orientation, by $$\varphi(t) = (a\cos{t},b\sin{t})$$ Then you have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align*}&#xA;A &amp;amp;=\frac{1}{2} \int\limits_{\partial{D}} xdy - ydx \\ &amp;amp;= \frac{1}{2} \int\limits_{0}^{2\pi} (−b \sin(t), a \cos(t)) \cdot (−a \sin(t), b \cos(t) dt \\ &amp;amp;= \frac{1}{2} \int\limits_{0}^{2\pi} (ab \sin^{2}{t} + ab \cos^{2}{t}) \ dt\\ &amp;amp;= \pi \cdot ab&#xA;\end{align*}&lt;/p&gt;&#xA;" LastActivityDate="2011-01-27T20:38:41.957" />
  <row Id="19274" PostTypeId="1" AcceptedAnswerId="19351" CreationDate="2011-01-27T20:58:27.667" Score="1" ViewCount="126" Body="&lt;p&gt;Let $S_n = \frac{1}{n}\sum_{i=1}^n X_i$, and $T_n = \frac{1}{n}\sum_{j=1}^nY_j$, where &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The $X_i$ are iid, the $Y_i$ are iid (with a different law)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$X_i$, and $Y_j$ are dependent  for $i =j$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $i\neq j$, $X_i$ and $Y_j$ are independent. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a central limit type result for $S_n^2 - T_n^2$?&lt;/p&gt;&#xA;" OwnerUserId="6327" LastEditorUserId="2548" LastEditorDisplayName="" LastEditDate="2011-01-28T06:24:45.520" LastActivityDate="2011-01-28T17:41:14.800" Title="Limiting distribution of a squared sum of random variables" Tags="&lt;probability&gt;&lt;central-limit-theorem&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" />
  <row Id="19275" PostTypeId="2" ParentId="19261" CreationDate="2011-01-27T21:14:24.643" Score="1" ViewCount="" Body="&lt;p&gt;Both your choice of the coin and the outcomes of the subsequent flips are random variables (though they are not independent).  You want to calculate the probability of the coin choice &lt;em&gt;given&lt;/em&gt; specific observed outcomes; i.e., you want a conditional probability.  Here, you want $P(\text{two-head} \vert HHH) \equiv P(\text{two-head} \wedge HHH) / P(HHH)$.  The numerator and the denominator are&#xA;$$&#xD;&#xA;P(\text{two-head} \wedge HHH) = P(\text{two-head})P(HHH \vert \text{two-head})&#xD;&#xA;= \left(\frac{1}{5}\right)(1) = \frac{1}{5}&#xD;&#xA;$$&#xA;and&#xA;$$&#xD;&#xA;\begin{eqnarray}&#xD;&#xA;P(HHH) &amp;amp;=&amp;amp; P(\text{two-head} \wedge HHH) + P(\text{normal} \wedge HHH) \\&#xD;&#xA;&amp;amp;=&amp;amp; P(\text{two-head})P(HHH \vert \text{two-head}) + P(\text{normal})P(HHH \vert \text{normal}) \\&#xD;&#xA;&amp;amp;=&amp;amp; \left(\frac{1}{5}\right)(1) + \left(\frac{4}{5}\right)\left(\frac{1}{8}\right) \\&#xD;&#xA;&amp;amp;=&amp;amp; \frac{1}{5} + \frac{1}{10} = \frac{3}{10}.&#xD;&#xA;\end{eqnarray}&#xD;&#xA;$$&#xA;The desired result is&#xA;$$&#xD;&#xA;P(\text{two-head} \vert HHH) = \frac{P(\text{two-head}\wedge HHH)}{P(HHH)} = \frac{1/5}{3/10} = \frac{2}{3}.&#xD;&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="5546" LastActivityDate="2011-01-27T21:14:24.643" />
  <row Id="19276" PostTypeId="2" ParentId="19264" CreationDate="2011-01-27T21:22:52.103" Score="0" ViewCount="" Body="&lt;p&gt;By three variables do you mean a,n, and i? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You will have to choose one value for a and n.&#xA;i will go from 10 to n.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example if a = 3 and n=20 your pseudocode would be:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;int a =3;&#xA;   int n = 20&#xA;   answer = 0;&#xA;   for(i=10:i==n;i++){answer += a^i+ n}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are more elegant ways, but this is a tiny assignement. Arturo is right that you could use the formula for a geometric series too.&lt;/p&gt;&#xA;" OwnerUserId="6329" LastActivityDate="2011-01-27T21:22:52.103" />
  <row Id="19278" PostTypeId="1" AcceptedAnswerId="19280" CreationDate="2011-01-27T22:11:18.043" Score="1" ViewCount="54" Body="&lt;p&gt;For $\alpha \geq 0$ the transformation $x \mapsto \log(x) - \alpha \log(1-x)$&#xA;maps the unit intervall to the real line (in fact for $\alpha = 0$ the transformation is not surjective). For $\alpha=1$ this is the logit transformation, which has a well-known inverse. It is also not difficult to find an inverse for $\alpha = 0.5$ and I would guess along these lines for some more $\alpha = 1/N$, $N$ positive natural number. But I do not see a closed form for an arbitrary $\alpha &amp;gt; 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finding an inverse numerically is not too difficult because the function is strictly monotone, but this is at least aethetically unpleasing ;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a way to show that for arbitrary $\alpha$ no such solution can exist, or is there maybe one?&lt;/p&gt;&#xA;" OwnerUserId="6330" LastEditorUserId="6330" LastEditorDisplayName="" LastEditDate="2011-01-28T21:23:28.790" LastActivityDate="2011-01-29T08:04:28.880" Title="Can the inverse of this logit-like transformation be stated analytically?" Tags="&lt;analysis&gt;&lt;functions&gt;&lt;transformation&gt;&lt;roots&gt;&lt;inversion&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="19279" PostTypeId="1" AcceptedAnswerId="19285" CreationDate="2011-01-27T22:14:59.057" Score="2" ViewCount="71" Body="&lt;p&gt;I was reading a paper, and I encountered a definition of some concept. The definition was of the form:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(+)  $\ \qquad\qquad (\exists x) \phi(x) \quad \Rightarrow \quad (\exists y) \psi(y)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\phi$ and $\psi$ are two formulae.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was wondering if the author's could write the above formula in the form:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(++) $\qquad\qquad (\forall x)(\exists y) \quad \phi(x) \Rightarrow \psi(y)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is, if (+) and (++) are equivalent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please prove the equivalency, or give a counterexample.&lt;/p&gt;&#xA;" OwnerUserId="2420" LastActivityDate="2011-01-27T22:33:52.043" Title="Equivalency of Two First-Order Formulae" Tags="&lt;logic&gt;" AnswerCount="2" />
  <row Id="19280" PostTypeId="2" ParentId="19278" CreationDate="2011-01-27T22:15:59.047" Score="0" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://www.wolframalpha.com/input/?i=inverse+of+log%28x%29%2Btlog%281-x%29+&quot; rel=&quot;nofollow&quot;&gt;According&lt;/a&gt; to Wolfram Alpha, there is no analytic form for the inverse.&lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2011-01-27T22:15:59.047" CommentCount="1" />
  <row Id="19282" PostTypeId="2" ParentId="2601" CreationDate="2011-01-27T22:18:10.723" Score="0" ViewCount="" Body="&lt;p&gt;A graph (either the drawing of a function or the vertices-and-edges kind).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This might seem like a paltry or too general a metaphor but really, take any kind of algebraic expression (say in arithmetic); yes, one can manipulate it symbolically, but it often doesn't become meaningful until you draw something, something that captures a number of possibilities all at once in a picture (e.g. why is 1/x undefined at x = 0? the graph goes in two different directions there). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The vertices/edges kind is so useful for capturing state/objects and transitions among them either for real world description or for mathematical concepts.&lt;/p&gt;&#xA;" OwnerUserId="1919" LastActivityDate="2011-01-27T22:18:10.723" CommunityOwnedDate="2011-01-27T22:18:10.723" />
  <row Id="19284" PostTypeId="2" ParentId="19279" CreationDate="2011-01-27T22:28:01.253" Score="3" ViewCount="" Body="&lt;p&gt;(+) implies (++): suppose first that there is an $x_0$ for which $\varphi(x_0)$ holds, and that (+) holds. Let $y_0$ be a witness for (+) (must exist, since the antecedent holds); then $y_0$ is a witness for (++) for every $x$. If, on the other hand, $\forall(x)(\neg\varphi(x))$ holds, then (++) holds by vacuity (this includes the case in which the domain is empty). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(++) implies (+): if the domain is empty, then (+) holds by vacuity. If the domain is not empty, and $\forall x(\neg\phi(x))$ holds, then (+) is true by vacuity again. And if there exists $x_0$ such that $\phi(x_0)$ holds, then by (++) there exists $y_0$ such that $\phi(x_0)\Rightarrow \psi(y_0)$ holds, and since $\phi(x_0)$ is true then $\psi(y_0)$ holds, hence (+) holds because the consequent is true.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-27T22:33:52.043" LastActivityDate="2011-01-27T22:33:52.043" />
  <row Id="19285" PostTypeId="2" ParentId="19279" CreationDate="2011-01-27T22:32:12.700" Score="2" ViewCount="" Body="&lt;p&gt;Yes the two formulas say the same. $((\exists x)\phi(x))\to((\exists y)\psi(y))$ is equivalent to $(\lnot(\exists x)\phi(x))\lor((\exists y)\psi(y))$ which is equivalent to $((\forall x)\lnot\phi(x))\lor((\exists y)\psi(y))$ which is equivalent to $(\forall x)(\exists y)(\lnot\phi(x)\lor\psi(y))$ which is what you are looking for, namely $(\forall x)(\exists y)(\phi(x)\to\psi(y))$.&lt;/p&gt;&#xA;" OwnerUserId="1772" LastActivityDate="2011-01-27T22:32:12.700" CommentCount="3" />
  <row Id="19286" PostTypeId="1" AcceptedAnswerId="19287" CreationDate="2011-01-27T22:40:37.340" Score="2" ViewCount="138" Body="&lt;p&gt;Let $R$ be the relation defined on $\mathbb{Z}$ where $a\; R\; b$ means that $a + b^2 \equiv 0\pmod{2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How would I go about finding the equivalence class $[-13]$?&lt;/p&gt;&#xA;" OwnerUserId="6331" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-27T22:46:06.777" LastActivityDate="2011-01-27T23:07:25.717" Title="Equivalence Classes" Tags="&lt;discrete-mathematics&gt;" AnswerCount="3" />
  <row Id="19287" PostTypeId="2" ParentId="19286" CreationDate="2011-01-27T22:42:21.677" Score="2" ViewCount="" Body="&lt;p&gt;You would find all $a \in \mathbb{Z}$ such that $a+(-13)^2 = 0 \pmod{2}$ since $[-13] = \{a \in \mathbb{Z}: aR -13 \}$.&lt;/p&gt;&#xA;" OwnerUserId="2548" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-27T22:46:20.473" LastActivityDate="2011-01-27T22:46:20.473" CommentCount="9" />
  <row Id="19288" PostTypeId="2" ParentId="19286" CreationDate="2011-01-27T22:51:32.117" Score="0" ViewCount="" Body="&lt;p&gt;$a+b^2=0\ (2)$ is the same as &quot;$a$ and $b$ have the same parity&quot;, so the set of odds is $-13$'s class.&lt;/p&gt;&#xA;" OwnerUserId="4992" LastActivityDate="2011-01-27T22:51:32.117" CommentCount="3" />
  <row Id="19289" PostTypeId="2" ParentId="19286" CreationDate="2011-01-27T22:59:20.413" Score="0" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\ \ \rm b^2 \equiv -b\ \ (mod\ 2)\ $ thus $\rm\ a\ R\ b\ \iff\ a\ \equiv\ b\ \ (mod\ 2)\ $ which &lt;em&gt;is&lt;/em&gt; an equivalence relation.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2011-01-27T23:07:25.717" LastActivityDate="2011-01-27T23:07:25.717" />
  <row Id="19290" PostTypeId="2" ParentId="2601" CreationDate="2011-01-27T23:02:21.027" Score="0" ViewCount="" Body="&lt;p&gt;I find the concept of a &lt;a href=&quot;http://en.wikipedia.org/wiki/Digital_sundial&quot; rel=&quot;nofollow&quot;&gt;digital sundial&lt;/a&gt; quite intriguing.  From wikipedia:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Roughly speaking, there exists a set&#xA;  with prescribed projections in almost&#xA;  all directions.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="139" LastActivityDate="2011-01-27T23:02:21.027" CommunityOwnedDate="2011-01-27T23:02:21.027" />
  <row Id="19291" PostTypeId="1" AcceptedAnswerId="19294" CreationDate="2011-01-27T23:09:28.907" Score="3" ViewCount="127" Body="&lt;p&gt;Find all continuous $f:[0,1] \rightarrow [0,1]$ such that $f(1-f(x))=f(x)$.&lt;/p&gt;&#xA;" OwnerUserId="5959" LastEditorUserId="5959" LastEditorDisplayName="" LastEditDate="2011-01-27T23:16:20.273" LastActivityDate="2011-01-27T23:33:38.777" Title="$f(1-f(x))=f(x)$" Tags="&lt;functional-equations&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="1" />
  <row Id="19292" PostTypeId="1" AcceptedAnswerId="19298" CreationDate="2011-01-27T23:29:31.557" Score="2" ViewCount="170" Body="&lt;p&gt;Hello,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am trying to learn graph theory and the linear algebra used to analyse graphs. The texts I have read through have lots of lemmas and theorems proved. The proofs are convincing but I fail to see the intuition behind them. How do they connect to the properties of the graphs?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand eigenvectors very clearly when they are taken from a transition matrix for a Markov chain, because very simply they represent a stationary distribution of the matrix. But for the matrices derived from graphs? From the adjacency matrix? Here I cannot understand what relevance the eigenvectors/eigenvalues have to the graph. I cannot imagine using those matrices in the first place for multiplying anything by them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe I don't completely understand totally the power and depth of eigenvectors/eigenvalues. But I guess this application would reveal more. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Best,&lt;/p&gt;&#xA;" OwnerUserId="1587" LastActivityDate="2011-01-27T23:51:14.847" Title="connection between graphs and the eigenvectors of their matrix representation" Tags="&lt;linear-algebra&gt;&lt;matrices&gt;&lt;graph-theory&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" />
  <row Id="19293" PostTypeId="2" ParentId="19292" CreationDate="2011-01-27T23:31:13.460" Score="1" ViewCount="" Body="&lt;p&gt;See &lt;a href=&quot;http://www.math.tu-clausthal.de/Studium/Vorlesungen/WS0506/HSGraphenTh/graph_ev_intro.pdf&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; (article about eigenvalues and graph theory).&lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2011-01-27T23:31:13.460" />
  <row Id="19294" PostTypeId="2" ParentId="19291" CreationDate="2011-01-27T23:33:38.777" Score="6" ViewCount="" Body="&lt;p&gt;Let $m,M$ be the minimum and maximum $f$ achieves on $[0,1]$ (there are such since f is continuous). From the intermediate value theorem, for each $m\leq y \leq M$ there is an $x\in [0,1]$ such that $f(x)=y$, so $f(1-y)=f(1-f(x))=f(x)=y$. This shows that if $m \leq y\leq M$ then $f(1-y)=y$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for the $ [0,1] - [1-M, 1-m] $ you can extend $f$ any way you want as long as its range is in $[m,M]$.  &lt;/p&gt;&#xA;" OwnerUserId="2125" LastActivityDate="2011-01-27T23:33:38.777" />
  <row Id="19295" PostTypeId="1" AcceptedAnswerId="19297" CreationDate="2011-01-27T23:37:18.573" Score="4" ViewCount="79" Body="&lt;p&gt;I'm trying to figure out a lecture example given on our Analysis course. We are currently going through Riemann integrals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $g:[0,1] \to R, g(x) = 1$ when $x \in [0, \frac{1}{2}]$ and $g(x) = 2$ when $x \in ]\frac{1}{2},1]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is g integrable?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The example goes on to prove that g is indeed integrable by choosing $P_n = \{0, \frac{1}{2}, \frac{1}{2} + \frac{1}{n}, 1\}$ as the partition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't understand why simply $P = \{0, \frac{1}{2}, 1\}$ isn't enough to prove that the lower and upper Darboux integrals are the same, thus $g$ is integrable.&lt;/p&gt;&#xA;" OwnerUserId="1759" LastActivityDate="2011-01-27T23:43:20.983" Title="Why isn't this simpler partition enough?" Tags="&lt;real-analysis&gt;&lt;integral&gt;" AnswerCount="1" />
  <row Id="19297" PostTypeId="2" ParentId="19295" CreationDate="2011-01-27T23:43:20.983" Score="6" ViewCount="" Body="&lt;p&gt;Typically the endpoints are included in defining the lower and upper sums, so the lower sum for your partition is $1$ and the upper sum is the correct value of $\frac{3}{2}$.  On the other hand, by adding $\frac{1}{2}+\frac{1}{n}$, the discrepancy caused by that pesky endpoint shrinks to $\frac{1}{n}$, which goes to $0$.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2011-01-27T23:43:20.983" CommentCount="4" />
  <row Id="19298" PostTypeId="2" ParentId="19292" CreationDate="2011-01-27T23:43:29.033" Score="3" ViewCount="" Body="&lt;p&gt;The abstract point of view is that the graph is characterized by its adjacency matrix up to permutation.  When we view the adjacency matrix as an abstract linear transformation, we are looking at it up to &lt;em&gt;conjugation&lt;/em&gt;, which is stronger.  So we lose some information, but the information that is left is still interesting.  And any time we can apply linear algebra to a situation, that is a good thing because linear algebra is really easy compared to almost anything else.  (This sounds trite, but it is one of the most-used principles in mathematics.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The eigenvalues of the adjacency matrix describe closed walks on the graph.  More precisely, if $A$ is the adjacency matrix, then the total number of closed walks on the graph of length $n$ is given by $\text{tr } A^n$, which is the same as $\sum_i \lambda_i^n$ where $\lambda_i$ are the eigenvalues.  So knowing the eigenvalues says something about how closed walks grow.  Unfortunately this perspective does not provide a direct interpretation of the eigenvectors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Spectral methods apply particularly well to graphs with a lot of structure, such as &lt;a href=&quot;http://en.wikipedia.org/wiki/Strongly_regular_graph&quot; rel=&quot;nofollow&quot;&gt;strongly regular graphs&lt;/a&gt;.  In the best case one can write down a matrix equation the adjacency matrix satisfies, and analyzing what this says about the eigenvectors and eigenvalues puts strong constraints on the graph.  This is the mechanism behind the standard proof of the classification of &lt;a href=&quot;http://en.wikipedia.org/wiki/Moore_graph&quot; rel=&quot;nofollow&quot;&gt;Moore graphs&lt;/a&gt; of girth $5$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There &lt;em&gt;is&lt;/em&gt; a fairly concrete interpretation of the eigenvalues and the eigenvectors of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Laplacian_matrix&quot; rel=&quot;nofollow&quot;&gt;Laplacian matrix&lt;/a&gt;, as I explained in &lt;a href=&quot;http://math.stackexchange.com/questions/18945/understanding-the-properties-and-use-of-the-laplacian-matrix-and-its-norm/18960#18960&quot;&gt;this math.SE&lt;/a&gt; answer.  The short version is that one can set up three natural differential equations using the Laplacian, and these all give concrete interpretations of the eigenvalues and eigenvectors:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The wave equation.  Here the eigenvectors are the &quot;harmonics&quot; of the graph: they describe the standing wave solutions of the wave equation, and the eigenvalues describe the frequencies at which these waves resonate.&lt;/li&gt;&#xA;&lt;li&gt;The heat equation.  Here the eigenvectors are still the &quot;harmonics&quot; of the graph, but the eigenvalues describe the rate at which heat decays for each harmonic.  Morally this is about &quot;Brownian motion&quot; on the graph, so it is related to random walks.&lt;/li&gt;&#xA;&lt;li&gt;The Schrödinger equation.  Here the eigenvectors are energy eigenstates of a &lt;a href=&quot;http://en.wikipedia.org/wiki/Quantum_walk&quot; rel=&quot;nofollow&quot;&gt;continuous-time quantum random walk&lt;/a&gt;, and the eigenvalues are (up to a constant) energy eigenvalues.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;When the graph is regular, the Laplacian and adjacency matrix are related in a very simple way; the eigenvectors are the same, and the eigenvalues can be easily related to each other.  Things are more complicated in the irregular case.&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2011-01-27T23:51:14.847" LastActivityDate="2011-01-27T23:51:14.847" CommentCount="1" />
  <row Id="19299" PostTypeId="1" CreationDate="2011-01-27T23:47:26.503" Score="3" ViewCount="121" Body="&lt;p&gt;In Milnor's famous book &quot;Topology from the Differential Viewpoint&quot; he proves the following on page 11:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $f: M\to N$ is a smooth map between manifolds of dimension $m\geq n$ and if $y\in N$ is a regular value, then the set $f^{-1}(y) \subset M$ is a smooth manifold of dimension $m-n$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Right, so I've read this proof previously in another book (using the rank theorem). I cannot however understand one step he does here. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proof:&#xA;Let $x\in f^{-1}(y)$. Since $y$ is a regular value, the derivative $df_x$ must map $TM_x$ onto $TN_y$. The null space $R \subset TM_x$ of $df_x$ will therefore be an $(m-n)$-dimensional vector space.&#xA;If $M\subset \mathbb{R}^k$, choose a linear map $L : \mathbb{R}^k \to \mathbb{R}^{m-n}$ that is nonsingular on this subspace $R\subset TM_x \subset \mathbb{R}^k$. Now define&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$F: M \to N\times\mathbb{R}^{m-n}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;by $F(\xi) = (f(\xi), L(\xi))$. The derivative $dF_x$ is clearly given by the formula $dF_x(v) = (df_x(v), L(v))$. Thus $dF_x$ is nonsingular. Hence $F$ maps some neighborhood $U$ of $x$ diffeomorpically onto a neighborhood $V$ of $(y, L(x))$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;*&lt;em&gt;Note that $f^{-1}(y)$ corresponds, under $F$, to the hyperplane $y\times \mathbb{R}^{m-n}$. *&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact $F$ maps $f^{-1}(y)\cap U$ diffeomorphically onto $(y\times\mathbb{R}^{m-n})\cap V$. This proves that $f^{-1}(y)$ is a smooth manifold of dimension $m-n$. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;So, I do not understand the part between the *'s. I think I'm sort-of missing something obvious ...&lt;/p&gt;&#xA;" OwnerUserId="2900" LastActivityDate="2011-01-27T23:47:26.503" Title="Inverse of regular value is a submanifold, Milnor's proof." Tags="&lt;differential-topology&gt;" CommentCount="5" FavoriteCount="1" />
  <row Id="19300" PostTypeId="2" ParentId="11778" CreationDate="2011-01-28T00:02:22.903" Score="3" ViewCount="" Body="&lt;p&gt;This is at least as hard as the &lt;a href=&quot;http://en.wikipedia.org/wiki/Partition_problem&quot; rel=&quot;nofollow&quot;&gt;Number Partition Problem&lt;/a&gt; (or the related &lt;a href=&quot;http://en.wikipedia.org/wiki/Subset_sum&quot; rel=&quot;nofollow&quot;&gt;Subset Sum&lt;/a&gt;) and is probably equivalent.  The Number Partition Problem is NP-Complete so the solution to such an integral is, in general, probably not polynomial time solvable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To see that it's at least as hard, I will consider a special case of your integral.  First notice that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ a_k \in \mathbb{N}, \ \ k \in [ 0, 1, \dots, n-1 ]  $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \mathbb{I}_{\sigma, a} = \frac{1}{2 \pi} \int_{-\pi}^{\pi} e^{ i \theta \sum_{k=0}^{n-1} \sigma_k a_k } d\theta $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for $ \sigma_k \in \{-1, 1\}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice that $\mathbb{I}_{\sigma, a}$ will be 1 if the sum $\sum_{k=0}^{n-1} \sigma_k a_k = 0$.  Otherwise $\mathbb{I}_{\sigma, a}$ will be 0.  This gives us an indicator function to test whether the configuration is perfect or not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Define $Z_a$ as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ Z_{ a } = \int_{-\pi}^{\pi} \prod_{k=0}^{n-1} ( e^{i \theta a_k} + e^{- i \theta a_k } ) d\theta = \int_{-\pi}^{\pi} e^{- i \theta \sum_{k=0}^{n-1} a_k } \prod_{k=0}^{n-1} (e^{ 2 i a_k } + 1 ) d\theta $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and notice that $Z_a$ also equals the sum of all indicator functions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ Z_a = \sum_{ &amp;lt;\sigma&amp;gt; } \mathbb{I}_{\sigma, a} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $&amp;lt;\sigma&amp;gt;$ denotes all possible configurations of $ \sigma_k \in \{ -1, 1\}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$Z_a$ counts the number of solutions and is a special case of the integral you presented above.  Were you to produce an algorithm to evaluate your integral in polynomial time, it could be applied to the above integral not just showing that $ P = NP \ $ but that the related counting problem is also in $P\ $ ($ \text{#} P = P $).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While this means that you probably cannot produce a polynomial time algorithm (in general) to solve your integral, it also means that you can probably adapt algorithms used to solve the Number Partition Problem (or Subset Sum) to provide a solution to the original integral.  For the integral you presented there is a (what appears to me) small hurdle of having the denominators being other than one, but perhaps you could simplify by using continued fractions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To my knowledge, the current 'state of the art' algorithm in solving the Number Partition Problem is the Complete Karmarkar Karp algorithm (CKK).  See &lt;a href=&quot;http://www.americanscientist.org/issues/pub/the-easiest-hard-problem/2&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;, &lt;a href=&quot;http://alexandria.tue.nl/extra1/wskrap/publichtml/200217.pdf&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; and &lt;a href=&quot;http://ijcai.org/Past%20Proceedings/IJCAI-95-VOL%201/pdf/035.pdf&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; for discussions and descriptions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want further reference on the Number Partition Problem and its integral representation, see Borgs, Chayes and Pittel's paper &lt;a href=&quot;http://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.89.9577&quot; rel=&quot;nofollow&quot;&gt;&quot;Phase transition and finite-size scaling for the integer partitioning problem&quot;&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="4143" LastActivityDate="2011-01-28T00:02:22.903" CommentCount="1" />
  <row Id="19301" PostTypeId="1" AcceptedAnswerId="19354" CreationDate="2011-01-28T00:26:37.917" Score="7" ViewCount="213" Body="&lt;p&gt;I was working out some problems. This is giving me trouble.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If $p$ is a prime number of the form $4n+1$ then how do i show that:&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;$$ \sum\limits_{i=1}^{p-1} \Biggl( \biggl\lfloor{\frac{2i^{2}}{p}\biggr\rfloor}-2\biggl\lfloor{\frac{i^{2}}{p}\biggr\rfloor}\Biggr)= \frac{p-1}{2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Two things which i know are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;If $p$ is a prime of the form $4n+1$, then $x^{2} \equiv -1 \ (\text{mod} \ p)$ can be solved.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$\lfloor{2x\rfloor}-2\lfloor{x\rfloor}$ is either $0$ or $1$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I think the second one will be of use, but i really can't see how i can apply it here.&lt;/p&gt;&#xA;" OwnerUserId="9413" LastEditorUserId="9413" LastEditorDisplayName="" LastActivityDate="2011-01-28T22:03:30.737" Title="$ \sum\limits_{i=1}^{p-1} \Biggl( \biggl\lfloor{\frac{2i^{2}}{p}\biggr\rfloor}-2\biggl\lfloor{\frac{i^{2}}{p}\biggr\rfloor}\Biggr)= \frac{p-1}{2}$" Tags="&lt;number-theory&gt;&lt;functions&gt;&lt;inequality&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="19302" PostTypeId="1" AcceptedAnswerId="19307" CreationDate="2011-01-28T00:27:11.040" Score="0" ViewCount="126" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;A sigma algebra is closed under&#xA;  countable intersections.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is a property of a sigma algebra, but I don't understand what this means in plain English.  Can someone explain this to me in terms a n00b like me would understand?  The part I don't understand the most is what &quot;closed&quot; means in this quote.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Addendum:&#xA;The collection of subsets:&#xA;{0},{x},{y,z},{x,y,z}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is a sigma algebra of:&#xA;{0},{x},{y},{z},{x,y},{x,z},{y,z},{x,y,z}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The collection of subsets is not empty -&gt; The collection contains subsets.&lt;/li&gt;&#xA;&lt;li&gt;Closed under complementation -&gt; The collection of subsets contains {x} and {x,y,z}/{x} = {y,z}&lt;/li&gt;&#xA;&lt;li&gt;Closed under countable unions -&gt; {x} and {y,z} are contained in our sigma algebra, so is {x} U {y,z} = {x,y,z}&lt;/li&gt;&#xA;&lt;li&gt;Close under countable intersections -&gt; {x} and {y,z} are contained in our sigma algebra, so is {0}&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;For point 2. above do you just take the complement of the set of all elements or {x,y,z} (that's what it seems to be according to the Wikipedia page)?&lt;/p&gt;&#xA;" OwnerUserId="6333" LastEditorUserId="6333" LastEditorDisplayName="" LastEditDate="2011-01-28T01:16:59.530" LastActivityDate="2011-01-28T01:24:16.983" Title="Sigma Algebra Question" Tags="&lt;sigma-algebra&gt;" AnswerCount="3" />
  <row Id="19303" PostTypeId="2" ParentId="19302" CreationDate="2011-01-28T00:29:48.090" Score="0" ViewCount="" Body="&lt;p&gt;It means that if $A_1, A_2, \dots \in \tau$ then $A_1 \cap A_2 \cap  \dots \in \tau$ where $\tau$ is a $\sigma$-algebra on a set $X$.&lt;/p&gt;&#xA;" OwnerUserId="2548" LastEditorUserId="2548" LastEditorDisplayName="" LastEditDate="2011-01-28T00:47:53.690" LastActivityDate="2011-01-28T00:47:53.690" CommentCount="2" />
  <row Id="19304" PostTypeId="2" ParentId="19302" CreationDate="2011-01-28T00:31:33.777" Score="0" ViewCount="" Body="&lt;p&gt;A sigma algebra is closed under Countable Intersections means that, if suppose $\tau$ is a $\sigma$-algebra on some set $X$ and if $A_{1},A_{2},A_{3}, \cdots \in \tau$ then $\cap A_{i}$ will also be in $\tau$. Note that you can only take the countable intersections. For more information please read the wikipedia page:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Sigma-algebra&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Sigma-algebra&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" LastActivityDate="2011-01-28T00:31:33.777" CommentCount="1" />
  <row Id="19305" PostTypeId="1" CreationDate="2011-01-28T00:43:20.527" Score="0" ViewCount="105" Body="&lt;p&gt;If I compute the eigenvalues and eigenvectors using &lt;code&gt;numpy.linalg.eig&lt;/code&gt; (from Python), the eigenvalues returned seem to be all over the place. Using, for example, &lt;a href=&quot;http://archive.ics.uci.edu/ml/machine-learning-databases/iris/iris.data&quot; rel=&quot;nofollow&quot;&gt;the Iris dataset&lt;/a&gt;, the normalized Eigenvalues are &lt;code&gt;[2.9108 0.9212 0.1474 0.0206]&lt;/code&gt;, but the ones I currently have are &lt;code&gt;[9206.53059607 314.10307292 12.03601935 3.53031167]&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem I'm facing is that I want to find out how much percentage of the variance each component brings, but with the current Eigenvalues I don't have the right values.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, how can I transform my eigenvalues so that they can give me the correct proportion of variance?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: Just in case it wasn't clear, I'm computing the &lt;code&gt;eig&lt;/code&gt; of the covariance matrix (The process is called Principal Component Analysis).&lt;/p&gt;&#xA;" OwnerUserId="6334" LastEditorUserId="6334" LastEditorDisplayName="" LastEditDate="2011-01-28T00:55:26.780" LastActivityDate="2011-02-27T15:39:17.437" Title="Normalizing the Eigenvalues so their sum is equal to min(n, p)?" Tags="&lt;matrices&gt;&lt;eigenvectors&gt;&lt;eigenvalues&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="19307" PostTypeId="2" ParentId="19302" CreationDate="2011-01-28T00:48:03.067" Score="3" ViewCount="" Body="&lt;p&gt;In general, &quot;closed under operation * on the set $\Sigma$&quot; means that, if you apply operation * (in this case, taking countable intersections) to elements of the set $\Sigma$ (in this case, the set is the $\sigma$-algebra of subsets of a given set $X$), then the result is still an element of $\Sigma$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another example which may be more familiar to you: a vector space is closed under addition (addition of vectors is still a vector)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More precisely, in this case, if $\Sigma$ is a $\sigma$-algebra of subsets of $X$, then if $\{A_i\}_{i=1}^\infty \subset \Sigma$, i.e. $A_1, A_2,\dots, A_n, \dots \in \Sigma$ then $\cap_{i=1}^\infty A_i = A_1 \cap A_2 \cap \dots \cap A_n \cap \dots \in \Sigma$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As I said in a comment above, by allowing countable intersections you get finite intersections too. If $A_1,\dots,A_n \in \Sigma$, then because $X\in \Sigma$ too by definition of $\sigma$-algebra, then using the fact $\Sigma$ is closed under countable intersections, $A_1\cap \dots \cap A_n = A_1 \cap \dots \cap A_n \cap X \cap X \cap X \cap \dots \in \Sigma$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To answer to your addendum:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There seems to be a bit of confusion. If $X=\{x,y,z\}$, then $\mathcal{P}(X)=\{\emptyset,\{x\},\{y\},\{z\},\{x,y\},\{x,z\},\{y,z\},\{x,y,z\}\}$ is the power set of $X$, and a $\sigma$-algebra is a subset of the power set with some additional properties.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, if $\Sigma=\{\emptyset,\{x\},\{y,z\},\{x,y,z\}\} \subset \mathcal{P}(X)$, then it is a $\sigma$-algebra of subsets of $X$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To check item 2, what &quot;closed under complementation&quot; means is that, whenever $A\in \Sigma$, we have $A^c=X\setminus A \in \Sigma$. So, we check that for all four elements of $\Sigma$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\emptyset \in \Sigma \Rightarrow \{x,y,z\} \in \Sigma$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\{x\} \in \Sigma \Rightarrow \{y,z\} \in \Sigma$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\{y,z\} \in \Sigma \Rightarrow \{x\} \in \Sigma$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\{x,y,z\} \in \Sigma \Rightarrow \emptyset \in \Sigma$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All these sentences are true, so item 2 is checked.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, as a side note, be careful, do not confuse $\emptyset$ with $\{\emptyset\}$.&lt;/p&gt;&#xA;" OwnerUserId="2614" LastEditorUserId="2614" LastEditorDisplayName="" LastEditDate="2011-01-28T01:24:16.983" LastActivityDate="2011-01-28T01:24:16.983" CommentCount="3" />
  <row Id="19308" PostTypeId="1" AcceptedAnswerId="19312" CreationDate="2011-01-28T00:59:55.507" Score="7" ViewCount="112" Body="&lt;p&gt;Shafarevich offers the following theorem-definition:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;At any nonsingular point $P$ of an irreducible algebraic curve, there exists a regular function $t$ that vanishes at $P$ and such that every rational function $u$ that is not identically $0$ on the curve can be written in the form $u = t^k v$, with $v$ regular at $P$ and $v(P) \neq 0$. A function $t$ with this property is called a local parameter on the curve at $P$.&quot;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;I've looked through six other books on algebraic geometry (The Geometry of Schemes by Eisenbud and Harris, Algebraic Curves by Fulton, Principles of Algebraic Geometry by Griffiths and Harris, The Red Book of Varieties by Mumford, and Vakil's online notes Foundations of Algebraic Geometry) and, unless I have made an error, none even contain the phrase &quot;local parameter.&quot; Hartshorne does appear to have the phrase in a few instances, but certainly does not give any definition at all similar to the one above, and besides Hartshorne is above my level right now so I am not in a good position to decide whether his usage agrees with that above or not.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The above theorem appears to me to exist only in Shafarevich and nowhere else in the mathematical literature.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Wikipedia offers the following much simpler characterization: &quot;In the geometry of complex algebraic curves, a local parameter for a curve $C$ at a smooth point $P$ is just a meromorphic function on $C$ that has a simple zero at $P$.&quot;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So my question is this: what exactly are these local parameters, and how should I think of them? How can I reconcile what Wikipedia has written with what Shafarevich writes? The name &quot;local parameter&quot; suggests to me there is some simple characterization of these functions which Shafarevich is keeping a mystery from me (or is Shafarevich's definition more intuitive than I am finding it?). And finally are these really present virtually nowhere in the entire mathematical literature except Shafarevich, or do equivalent ideas go under different names?&lt;/p&gt;&#xA;" OwnerUserId="251" LastEditorUserId="251" LastEditorDisplayName="" LastEditDate="2011-01-28T01:11:14.527" LastActivityDate="2011-01-28T01:18:28.360" Title="What is a local parameter in algebraic geometry?" Tags="&lt;algebraic-geometry&gt;&lt;algebraic-curves&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="19309" PostTypeId="2" ParentId="19070" CreationDate="2011-01-28T01:04:15.580" Score="0" ViewCount="" Body="&lt;p&gt;(This is meant to be a comment but I'm not being allowed to comment.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The group $H$ is not (necessarily) equal to the group of matrices with positive determinant.  This is true if $G=GL_n$, but the problem starts with an arbitrary subgroup $G$ of $GL_n$.  So $H$ is the path component of $G$ containing the identity; for instance if $G$ is discrete then $H$ is trivial.  The problem is to show $H$ is always normal in $G$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That said, jug's hints do lead to a proof.&lt;/p&gt;&#xA;" OwnerUserId="6071" LastActivityDate="2011-01-28T01:04:15.580" />
  <row Id="19310" PostTypeId="2" ParentId="19308" CreationDate="2011-01-28T01:11:02.520" Score="5" ViewCount="" Body="&lt;p&gt;A synonym for this term is &quot;uniformizing parameter&quot; or &quot;uniformizer,&quot; and this term &lt;em&gt;does&lt;/em&gt; appear in other books.  I believe it is supposed to be the algebraic analogue of a chart; in other words, it is a &quot;local coordinate&quot; at the point.  Perhaps to understand the geometry behind the term you should look at textbooks on Riemann surfaces first.&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2011-01-28T01:17:59.937" LastActivityDate="2011-01-28T01:17:59.937" />
  <row Id="19311" PostTypeId="2" ParentId="19308" CreationDate="2011-01-28T01:11:31.017" Score="1" ViewCount="" Body="&lt;p&gt;I think page 146 of &lt;a href=&quot;http://alexandria.tue.nl/repository/freearticles/593601.pdf&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; seems like a clear explanation of local parameters. This is an article about algebraic geometry applied to coding theory. It also uses &quot;uniformizing parameter&quot; as a synonym as Qiaochu mentioned.&lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2011-01-28T01:11:31.017" CommentCount="1" />
  <row Id="19312" PostTypeId="2" ParentId="19308" CreationDate="2011-01-28T01:18:28.360" Score="6" ViewCount="" Body="&lt;p&gt;What Shafarevic calls a local parameter is often called a uniformizing parameter at $P$, and is also the same thing as a uniformizer of the local ring of $C$ at $P$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The point is that if $P$ is a smooth point on a curve, then the local ring at $P$ (i.e. the ring of rational functions on $C$ which are regular at $P$) is a DVR, and hence its maximal ideal is principal; a generator of this ideal is called a uniformizer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $t$ is a uniformizer/local parameter/uniformizing parameter at $P$, and if &#xA;$u$ is any other rational function, then if we write $u = t^k v$ where $v(P) \neq 0$ (i.e. $v$ is a unit in the local ring), then $k$ &lt;em&gt;is&lt;/em&gt; the order of vanishing of $u$ at $P$.  In particular, $u$ vanishes to order one if and only if it is equal to $t$ times a unit in the local ring, if and only if it is also a generator of the maximal ideal of the local ring at $P$, if and only if it is also a uniformizer.  Thus Shafarevic and Wikipedia are reconciled.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One is supposed to think of $t$ as being a &quot;local coordinate at $P$.&quot; In the  complex analytic picture you would choose a small disk around $P$, and consider the coordinate $z$ on this disk; this a local coordinate around the smooth point $P$.  This analogy is very tight: indeed, it is not hard to show (when the ground field is the complex numbers) that a rational function $t$ is a local parameter at $P$ if and only $t(P) = 0$,&#xA;and if there is a small neighbourhood of $P$ (in the complex topology) which is mapped isomorphically to a disk around $0$ by $t$, i.e. if and only if $t$ restricts to a local coordinate on a neighbourhood of $P$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, this concept is ubiquitous.  The fact that the local ring at a point on a smooth algebraic curve is a DVR is fundamental in the algebraic approach to the theory of algebraic curves; see e.g. section 6 of Chapter I of Hartshorne.&lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2011-01-28T01:18:28.360" />
  <row Id="19313" PostTypeId="2" ParentId="19301" CreationDate="2011-01-28T02:14:27.547" Score="2" ViewCount="" Body="&lt;p&gt;Without giving everything away: when is $\lfloor2x\rfloor - 2\lfloor x\rfloor$ equal to $0$, and when is it equal to $1$?  Can you find some bijection between values of $i$ in your sum that fall into the first camp, and those that fall into the second?  (You may find the other fact you gave to be useful for finding this bijection!)&lt;/p&gt;&#xA;" OwnerUserId="785" LastActivityDate="2011-01-28T02:14:27.547" CommentCount="3" />
  <row Id="19314" PostTypeId="2" ParentId="19204" CreationDate="2011-01-28T02:27:15.483" Score="3" ViewCount="" Body="&lt;p&gt;If your two polygons $P_1$ and $P_2$ are (a) disjoint, and (b) convex, then there is a simple linear-time algorithm, known as the &lt;em&gt;rotating calipers&lt;/em&gt; algorithm. &lt;a href=&quot;http://cgm.cs.mcgill.ca/~orm/mind2p.html&quot; rel=&quot;nofollow&quot;&gt;Here&lt;/a&gt; is a&#xA;description.  If your polygons are not convex, then it is more complicated: they could&#xA;spiral around one another and be quite entangled.&#xA;There is an efficient algorithm for finding the closest pair of vertices that&#xA;see one another:&#xA;&quot;Finding the minimum visible vertex distance between two non-intersecting simple polygons,&quot;&#xA;&lt;a href=&quot;http://portal.acm.org/citation.cfm?id=10519&quot; rel=&quot;nofollow&quot;&gt;Proceedings of the 2nd Symposium on Computational Geometry, 1986&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A key search term here is &lt;em&gt;collision detection&lt;/em&gt;, as many applications require&#xA;the minimum distance between two polygons to detect and avoid collisions.&lt;/p&gt;&#xA;" OwnerUserId="237" LastActivityDate="2011-01-28T02:27:15.483" CommentCount="2" />
  <row Id="19315" PostTypeId="1" AcceptedAnswerId="19316" CreationDate="2011-01-28T02:40:37.777" Score="4" ViewCount="103" Body="&lt;p&gt;Maybe this is well-known, but suppose you have an invertible sheaf $\mathcal{L}$ on a scheme $X$. If $X$ is a projective space (or even projective bundle over an integral scheme, by a similar argument...I think), then if $H^0(X, \mathcal{L})\neq 0$ and $H^0(X, \mathcal{L}^\vee)\neq 0$ we get that $\mathcal{L}\simeq \mathcal{O}_X$. This is really simple because $\mathcal{L}\simeq \mathcal{O}_X(n)$, so you get that the two hypothesis mean $n\geq 0$ and $n\leq 0$, so $n=0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My guess is this must be true more generally. Are there conditions on $X$ such that any invertible sheaf with the property that having non-trivial global sections of both it and its dual implies it is trivial?&lt;/p&gt;&#xA;" OwnerUserId="2000" LastActivityDate="2011-01-28T02:58:36.610" Title="Global Sections of Sheaf and Dual" Tags="&lt;algebraic-geometry&gt;&lt;sheaf-theory&gt;" AnswerCount="1" />
  <row Id="19316" PostTypeId="2" ParentId="19315" CreationDate="2011-01-28T02:53:36.163" Score="3" ViewCount="" Body="&lt;p&gt;In general, this fails. Take $X = \mathrm{Spec}A$ for $A$ a Dedekind domain. Then line bundles correspond to equivalence classes of Weil divisors. A line bundle corresponds to some formal sum $\sum n_i \mathfrak{p}_i$. But you can always find something in $A$ whose divisor is bigger than this (just take it highly divisible at those primes). So any line bundle will satisfy your condition, but if $A$ is not a UFD it will not necessarily be trivial: the Picard group is the class group of $A$. (See 8.2 of &lt;a href=&quot;http://people.fas.harvard.edu/~amathew/CRing.pdf&quot; rel=&quot;nofollow&quot;&gt;http://people.fas.harvard.edu/~amathew/CRing.pdf&lt;/a&gt; for an explanation of this isomorphism and a general discussion of line bundles on affine schemes.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is true for $X$ an integral &lt;em&gt;proper&lt;/em&gt; scheme over an algebraically closed field $k$. (It's in Mumford's &lt;em&gt;Abelian Varieties.&lt;/em&gt;). You don't need the isomorphism of the Picard group with $\mathbb{Z}$ which is only anyway true when $X$ is all of projective space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose $\mathcal{L}$ is such a line bundle. Then there is a nonzero morphism $\mathcal{O}_X \to \mathcal{L}$ and a nonzero morphism $\mathcal{L} \to \mathcal{O}_X$. This is what it means for the space of global sections of $\mathcal{L}$ and its dual to not vanish. So we get a composition&#xA;$$\mathcal{L} \to \mathcal{L},$$&#xA;which is given by a global regular function that can't be zero (as it isn't zero at the generic point). This is necessarily an element of the field $k$, so an isomorphism of line bundles. Similarly  the composition $\mathcal{O}_X \to  \mathcal{O}_X$ is an isomorphism. (I suppose the key property here is that a nonzero endomorphism of a line bundle is an isomorphism, and this in turn is clear from $\Gamma(X, \mathcal{O}_X) = k$, which is a consequence of properness.) It follows that $\mathcal{L} \to \mathcal{O}_X$ is an isomorphism and $\mathcal{L}$ is trivial.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This doesn't really need $k $ to be algebraically closed (though then $X$ &lt;em&gt;geometrically&lt;/em&gt; integral over $k$ would be necessary to make the argument work, unless I'm missing something). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is false without integrality hypotheses. (Take a disconnected scheme, and a line bundle trivial on one piece but not the other.) &lt;/p&gt;&#xA;" OwnerUserId="536" LastEditorUserId="536" LastEditorDisplayName="" LastEditDate="2011-01-28T02:58:36.610" LastActivityDate="2011-01-28T02:58:36.610" />
  <row Id="19317" PostTypeId="1" CreationDate="2011-01-28T03:03:40.967" Score="2" ViewCount="207" Body="&lt;p&gt;In the book &quot;Structure of Groups of Prime Power order (Charles Richard Leedham-Green, Susan McKay), there is an exercise (2.1.10), which asks to &lt;strong&gt;show that the automorphism group of&lt;/strong&gt; ($\mathbb{Z}_p \times \mathbb{Z}_p) \rtimes \mathbb{Z}_p$  &lt;strong&gt;is&lt;/strong&gt; ($\mathbb{Z}_p \times \mathbb{Z}_p) \rtimes GL(2,p)$. To prove this, first, it is easy to show that there is a short exact sequence $1 \rightarrow \mathbb{Z}_p \times \mathbb{Z}_p \rightarrow Aut((\mathbb{Z}_p \times \mathbb{Z}_p) \rtimes \mathbb{Z}_p) \rightarrow GL(2,p) \rightarrow 1$. Therefore it remains to show that &quot;this exact sequence is &lt;strong&gt;split&lt;/strong&gt;&quot;. How to show it?&lt;/p&gt;&#xA;" OwnerUserId="8186" LastEditorUserId="8186" LastEditorDisplayName="" LastEditDate="2011-01-28T03:50:41.467" LastActivityDate="2011-03-13T23:55:34.043" Title="Automorphisms of a non-abelian group of order $p^{3}$" Tags="&lt;group-theory&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="4" />
  <row Id="19318" PostTypeId="1" CreationDate="2011-01-28T03:03:42.590" Score="16" ViewCount="1480" Body="&lt;p&gt;Someone asked me today, &quot;Why we should care about groups at all?&quot;  I realized that I have absolutely no idea how to respond.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One way to treat this might be to reduce &quot;why should we care about groups&quot; to &quot;why should we care about pure math&quot;, but I don't think this would be a satisfying approach for many people.  So here's what I'm looking for:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;blockquote&gt;&#xA;    &lt;p&gt;Are there any problems that that (1) don't originate from group theory, (2) have very elegant solutions in the framework of group theory, and (3) are completely intractable (or at the very least, extremely cumbersome) without non-trivial knowledge of groups?&lt;/p&gt;&#xA;  &lt;/blockquote&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A non-example of what I'm looking for is the proof of Euler's theorem (because that can be done without groups).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[Edit] I take back &quot;insolubility of the quintic&quot; as a non-example; I also retract the condition &quot;we're assuming group theory only, and no further knowledge of abstract algebra&quot;.&lt;/p&gt;&#xA;" OwnerUserId="4904" LastEditorUserId="4904" LastEditorDisplayName="" LastEditDate="2011-01-29T06:13:57.107" LastActivityDate="2011-03-29T08:12:00.160" Title="Why should we care about groups at all?" Tags="&lt;group-theory&gt;&lt;soft-question&gt;" AnswerCount="7" CommentCount="5" FavoriteCount="6" />
  <row Id="19319" PostTypeId="2" ParentId="19318" CreationDate="2011-01-28T03:07:03.743" Score="7" ViewCount="" Body="&lt;p&gt;See this &lt;a href=&quot;http://www.math.uconn.edu/~kconrad/math216/whygroups.html&quot; rel=&quot;nofollow&quot;&gt;answer&lt;/a&gt; by Keith Conrad. So it seems that some elementary particles were predicted by group theory before being experimentally discovered. &lt;a href=&quot;http://www.youtube.com/watch?v=Y8q0vDP30YQ#t=13m0s&quot; rel=&quot;nofollow&quot;&gt;Here&lt;/a&gt; is a video of Richard Feynman describing the particle (using &quot;Strangeness minus 3&quot;). &lt;/p&gt;&#xA;" OwnerUserId="2548" LastEditorUserId="2548" LastEditorDisplayName="" LastEditDate="2011-03-29T08:12:00.160" LastActivityDate="2011-03-29T08:12:00.160" CommentCount="1" />
  <row Id="19320" PostTypeId="2" ParentId="19114" CreationDate="2011-01-28T03:55:41.150" Score="1" ViewCount="" Body="&lt;p&gt;You can also have a look at &lt;a href=&quot;http://comeoncodeon.wordpress.com/2009/07/21/counting-problem/&quot; rel=&quot;nofollow&quot;&gt;http://comeoncodeon.wordpress.com/2009/07/21/counting-problem/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Problem 1 describes different approaches to see whether a sum K is poosible. The DP approach for the Problem 1 should give you all the numbers that are possible sum of a subset. Then you can simply do search for the nearest.&lt;/p&gt;&#xA;" OwnerUserId="1707" LastActivityDate="2011-01-28T03:55:41.150" />
  <row Id="19321" PostTypeId="2" ParentId="19317" CreationDate="2011-01-28T04:00:53.933" Score="2" ViewCount="" Body="&lt;p&gt;Here is a new, less coordinated version that motivates some of the correct coordinate changes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let V be a vector space, let W = V∧V be the product, and let G be the &lt;a href=&quot;http://math.stackexchange.com/questions/21640/&quot;&gt;wedgey group&lt;/a&gt; on V⊕W with multiplication (v&lt;sub&gt;1&lt;/sub&gt;,w&lt;sub&gt;1&lt;/sub&gt;)⋅(v&lt;sub&gt;2&lt;/sub&gt;,w&lt;sub&gt;2&lt;/sub&gt;) = (v&lt;sub&gt;1&lt;/sub&gt;+v&lt;sub&gt;2&lt;/sub&gt;, w&lt;sub&gt;1&lt;/sub&gt;+w&lt;sub&gt;2&lt;/sub&gt; + v&lt;sub&gt;1&lt;/sub&gt;∧v&lt;sub&gt;2&lt;/sub&gt;).  Powers obey the rule (v,w)&lt;sup&gt;n&lt;/sup&gt; = (nv,nw).  Commutators obey the rule [(v&lt;sub&gt;1&lt;/sub&gt;,w&lt;sub&gt;1&lt;/sub&gt;),(v&lt;sub&gt;2&lt;/sub&gt;,w&lt;sub&gt;2&lt;/sub&gt;)] = [0,2(v&lt;sub&gt;1&lt;/sub&gt;∧v&lt;sub&gt;2&lt;/sub&gt;)].  In particular, G always has nilpotency class 2 when 0≠2 (and is abelian otherwise), and has exponent p whenever V is defined over a field of characteristic p.  When V is one-dimensional over the field &lt;b&gt;Z&lt;/b&gt;/p&lt;b&gt;Z&lt;/b&gt; of p elements, G is the unique non-abelian group of order p&lt;sup&gt;3&lt;/sup&gt; and exponent p (the group in question).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider a block matrix of the form $$M = \begin{bmatrix} A &amp;amp; b \\ . &amp;amp; \Lambda^2(A) \end{bmatrix}$$ where A in End(V) and b in Hom(V,W), with the formal multiplication (v,w)⋅M = ( vA, vb + w⋅Λ&lt;sup&gt;2&lt;/sup&gt;(A) ).  When V is 2-dimensional, W is 1-dimensional and Λ&lt;sup&gt;2&lt;/sup&gt;(A) is just multiplication by det(A).  In general, (v&lt;sub&gt;1&lt;/sub&gt;∧v&lt;sub&gt;2&lt;/sub&gt;)⋅Λ&lt;sup&gt;2&lt;/sup&gt;(A) = (v&lt;sub&gt;1&lt;/sub&gt;⋅A) ∧ (v&lt;sub&gt;2&lt;/sub&gt;⋅A).  When dim(V)=2, then every element of Hom(V,W) can be written as b = ∧u which takes v in V to v∧u in W.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This matrix M defines a homomorphism of G:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{array}{l}&#xD;&#xA;(v_1, w_1)M \cdot (v_2, w_2)M \\&#xD;&#xA;= (v_1\cdot A, ~w_1\cdot\Lambda^2(A) + v_1\cdot b) \cdot&#xD;&#xA;(v_2\cdot A, ~w_2\cdot\Lambda^2(A) + v_2 \cdot b) \\&#xD;&#xA;= (v_1\cdot A + v_2\cdot A, ~w_1\cdot\Lambda^2(A) + v_1 \cdot b + w_2\cdot\Lambda^2(A) + v_2 \cdot b + (v_1\cdot A)\wedge(v_2\cdot A) ) \\&#xD;&#xA;= ( (v_1+v_2)\cdot A, ~(w_1 + w_2 + v_1 \wedge v_2)\cdot\Lambda^2(A) + (v_1+v_2)\cdot b ) \\&#xD;&#xA;= (v_1+v_2, ~w_1+w_2+v_1\wedge v_2)M \\&#xD;&#xA;= ( (v_1,w_1)\cdot(v_2,w_2) )M&#xD;&#xA;\end{array}&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The composition of homomorphisms M and M′ is given by matrix multiplication:&#xA;$$M\cdot M&amp;#39; = \begin{bmatrix} A &amp;amp; b \\ . &amp;amp; \Lambda^2(A) \end{bmatrix}&#xD;&#xA;\cdot \begin{bmatrix} A&amp;#39; &amp;amp; b&amp;#39; \\ . &amp;amp; \Lambda^2(A&amp;#39;) \end{bmatrix}&#xD;&#xA;= \begin{bmatrix} A\cdot A&amp;#39; &amp;amp; A\cdot b&amp;#39; + b \cdot \Lambda^2(A&amp;#39;) \\ . &amp;amp; \Lambda^2(A\cdot A&amp;#39;)\end{bmatrix}&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When dim(V)=2, it would be nice to use parameters (A,u) to describe M, but the question is how to multiply in these parameters (A,u)⋅(A′,u′) is not simply (A⋅A′, Au′+u⋅det(A′)).  Here A∧u′ is the element of Hom(V,W) which takes v in V to (vA)∧u′ in W.  We can in fact simplify A∧u′ to just ∧Bu′ for some matrix B related to A.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, A ≠ B in general.  Instead, one gets B = A&lt;sup&gt;−T&lt;/sup&gt; ⋅ det(A), I believe.  In other words, up to a determinant, we get the action of A on its dual module rather than on its natural module.  I'd like a clearer description of this action, since I am pretty certain the real answer has the natural module.  Hopefully, there is another dual I am forgetting that makes everything ok.  At any rate, I have no trouble doing it with coordinates as below:&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;It is comforting to put coordinates on the group: [x,y,z] = ( x⋅e&lt;sub&gt;1&lt;/sub&gt; + y⋅e&lt;sub&gt;2&lt;/sub&gt;, z⋅e&lt;sub&gt;1&lt;/sub&gt;∧e&lt;sub&gt;2&lt;/sub&gt; ).  As matrices, this can be realized as:&#xA;$$[x,y,z] = \begin{pmatrix} 1 &amp;amp; x &amp;amp; z + xy \\ . &amp;amp; 1 &amp;amp; 2y \\ . &amp;amp; . &amp;amp; 1 \end{pmatrix}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Multiplication follows the rule [a,b,c]⋅[x,y,z] = [a+x,b+y,c+z+ay-bx].&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Powers obey the very simple rule [x,y,z]&lt;sup&gt;n&lt;/sup&gt; = [ nx, ny, nz ].&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Commutators obey the simple rule [ [a,b,c], [x,y,z] ] = [ 0, 0, 2(ay-bx) ].&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now consider a matrix: $$ A = \begin{pmatrix} a &amp;amp; b &amp;amp; e \\ c &amp;amp; d &amp;amp; f \\ . &amp;amp; . &amp;amp; ad-bc \end{pmatrix}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One can check that in these coordinates A is an automorphism.  In particular, the function that takes the row vector [x,y,z] times the matrix A to get a new row vector [x′, y′, z′ ] defines an automorphism of G.  Composition of automorphisms is matrix multiplication.  In this coordinate system everything is perfect.&lt;/p&gt;&#xA;" OwnerUserId="583" LastEditorUserId="583" LastEditorDisplayName="" LastEditDate="2011-03-13T23:55:34.043" LastActivityDate="2011-03-13T23:55:34.043" CommentCount="5" />
  <row Id="19322" PostTypeId="2" ParentId="19318" CreationDate="2011-01-28T04:19:41.043" Score="5" ViewCount="" Body="&lt;p&gt;If you care about geometry/topology then you should care at least a little bit about groups.  For example, the study of manifolds with a group structure (Lie groups) is a very beautiful and very applicable field.  If you can put a group structure on a manifold, you get some topological information for free: e.g. abelian fundamental group, trivial tangent bundle.  The applications to physics are huge (c.f. PEV's answer).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another way group theory comes up in geometry/topology is by assigning a group to a space or an object on the space as a way of measuring something.  For example, most of algebraic topology is assigning group invariants to spaces ((co)homology groups, homotopy groups) to &quot;measure holes&quot;.  This is useful because groups are often easy to distinguish while it is often difficult to directly show that two spaces are not homeomorphic.  An example from differential geometry is holonomy: if we are on a manifold with a metric,the holonomy groups quantify how the space is curved.&lt;/p&gt;&#xA;" OwnerUserId="9" LastActivityDate="2011-01-28T04:19:41.043" CommentCount="2" />
  <row Id="19323" PostTypeId="1" AcceptedAnswerId="19326" CreationDate="2011-01-28T04:22:06.240" Score="2" ViewCount="128" Body="&lt;p&gt;Hi everyone,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need to demonstrate that Ker(AB)=${\vec{0}}$, just like Ker(A) and Ker(B). I have a not-so-elegant way of showing that (I think): A and B are n * p and p * m, and it has to be the case that n $\geq$ p $\geq$ m, and rref's of A and B are either equal to the identity matrix, or are of form I and row(s) of 0's on the bottom. So, product of rref's of A and B will equal a matrix that is either I or I with row(s) of 0's on the bottom, and is n * m where n $\geq$ m, so it has more equations than variables. Multiplication of matrices of the aforementioned forms can only yield a matrix of the aforementioned form (I or I with row(s) of zeros), so it must also have Ker=${\vec{0}}$. Is there a more simple (though not too advanced, as we are barely mid-way through the course) way of showing that this is true? Also, while I know that Ker(A)=Ker(rref(A)) for any matrix A, I am not sure if I can say that Ker(rref(A) * rref(B))=Ker(AB). Is this statement true?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="3731" LastActivityDate="2011-01-28T05:13:54.943" Title="If Ker(A)=$\{\vec{0}\}$ and Ker(B)=$\{\vec{0}\}$ Ker(AB)=?" Tags="&lt;linear-algebra&gt;&lt;homework&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="1" />
  <row Id="19324" PostTypeId="2" ParentId="19323" CreationDate="2011-01-28T04:27:30.427" Score="3" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Hint&lt;/strong&gt;: Assume that $\text{Kernel}(AB)$ is non-trivial. Prove by contradiction by making use of the fact that the nullspace of $A$ and $B$ are trivial.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastActivityDate="2011-01-28T04:27:30.427" CommentCount="4" />
  <row Id="19325" PostTypeId="2" ParentId="19323" CreationDate="2011-01-28T04:28:23.467" Score="1" ViewCount="" Body="&lt;p&gt;You can use the &lt;a href=&quot;http://en.wikipedia.org/wiki/Rank%E2%80%93nullity_theorem&quot; rel=&quot;nofollow&quot;&gt;rank nullity theorem&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2011-01-28T04:28:23.467" />
  <row Id="19326" PostTypeId="2" ParentId="19323" CreationDate="2011-01-28T04:33:00.650" Score="6" ViewCount="" Body="&lt;p&gt;Composition of one-to-one functions is one-to-one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively and directly: let $\mathbf{v}\in\mathrm{Ker}(AB)$. Then $A(B(\mathbf{v}))=\mathbf{0}$, so $B(\mathbf{v})\in\mathrm{ker}(A)=\{\mathbf{0}\}$. Therefore, $B(\mathbf{v})=\mathbf{0}$, so $\mathbf{v}\in\mathrm{ker}(B) = \{\mathbf{0}\}$. Therefore, $\mathbf{v}=\mathbf{0}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This proves that $\mathrm{ker}(AB)\subseteq\{\mathbf{0}\}$, which suffices. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;No need to muck about with sizes of matrices.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added:&lt;/strong&gt; (as per request in the comment) I assume that &quot;rref&quot; means &quot;row reduced echelon form&quot;. It is true that $\mathrm{Ker}(A) = \mathrm{Ker}(\mathrm{rref}(A))$, because we obtain $\mathrm{rref}(A)$ be left-multiplying $A$ by invertible elementary matrices, and this does not change the nullspace (equivalently, elementary row operations do not change the nullspace). However, it is false that $\mathrm{Ker}(AB) = \mathrm{Ker}(\mathrm{rref}(A)\mathrm{rref}(B))$; if you think about it, you'll see that the problem is that those invertible matrices will now also be &lt;em&gt;between&lt;/em&gt; $A$ and $B$. Here is an explicit example: take&#xA;$$A = \left(\begin{array}{cc}&#xD;&#xA;1 &amp;amp; 0 \\&#xD;&#xA;0 &amp;amp; 0&#xD;&#xA;\end{array}\right),\qquad B = \left(\begin{array}{cc}&#xD;&#xA;0 &amp;amp; 0\\&#xD;&#xA;1 &amp;amp; 0&#xD;&#xA;\end{array}\right).$$&#xA;Then the reduced row echelon form of $B$ is $A$, so $\mathrm{rred}(A)\mathrm{rred}(B) = AA = A$. However, $AB$ is the zero matrix. So $\mathrm{Ker}(AB) = \mathbb{R}^2$, but $\mathrm{ker}(\mathrm{rred}(A)\mathrm{rred}(B)) = \mathrm{ker}(A) = \{(0,y)\mid y\in\mathbb{R}\}\neq \mathrm{ker}(AB)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason it does not work is that if $C$ is invertible, then $\mathrm{ker}(CA) = \mathrm{ker}(A)$ (use the same argument as above). But $\mathrm{ker}(AC) = C^{-1}(\mathrm{ker}(A))$, so this may change the kernel. Since $\mathrm{rred}(A) = CA$ for some invertible matrix $C$, and $\mathrm{rred}(B) = DB$ for some invertible matrix $D$, you get $\mathrm{rred}(A)\mathrm{rred}(B) = CADB$, so that $D$ in the middle can mess you up.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-28T05:13:54.943" LastActivityDate="2011-01-28T05:13:54.943" CommentCount="2" />
  <row Id="19327" PostTypeId="2" ParentId="19318" CreationDate="2011-01-28T05:04:35.660" Score="5" ViewCount="" Body="&lt;p&gt;Check out this link &lt;a href=&quot;http://plus.maths.org/content/power-groups&quot; rel=&quot;nofollow&quot;&gt;http://plus.maths.org/content/power-groups&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the subsection titled &quot;Lonely Pursuits with Groups&quot;, where they have used the Klein 4-group to decide how many possible end locations are there for the final marble in the game of solitaire is particularly interesting. I was quite stunned at the elegance of the whole thing when I first saw it. I think it is a great example of the power of group theory. It also satisfies the three conditions you have listed.&lt;/p&gt;&#xA;" OwnerUserId="2641" LastActivityDate="2011-01-28T05:04:35.660" CommentCount="1" />
  <row Id="19328" PostTypeId="2" ParentId="19318" CreationDate="2011-01-28T05:07:17.063" Score="19" ViewCount="" Body="&lt;p&gt;First of all, even phrasing the question as &quot;I don't need group theory if every question that can be solved with group theory can also be solved without it&quot; is misguided. It just so happens that the definition of a group is a &lt;em&gt;natural&lt;/em&gt; thing. You might be able to circumvent it sometimes, but that doesn't mean that you should. If a concept naturally suggests itself, then why should one fight hard not to introduce it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, why is it natural? Because there are loads of structures out there in the Platonic world that consist of a set and a binary operation: the integers, the rational numbers, the non-zero rational numbers, matrices, vectors, geometric symmetries (closely related to matrices), permutations,... the list is almost endless. So it makes sense to capture this common feature of so many familiar objects in a definition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for applications, traditionally groups were only thought of as symmetries of geometric objects. Even in this narrow context, the abstract framework is useful, e.g. to count solutions to puzzles or ways of colouring a shape. Here is a &lt;a href=&quot;http://mathoverflow.net/questions/45841/what-is-this-subgroup-of-mathfrak-s-12/&quot; rel=&quot;nofollow&quot;&gt;concrete example&lt;/a&gt; of a puzzle that could only be solved and understood in its entirety using abstract groups (since it allowed us to identify a group of symmetries of a certain object with an already familiar symmetry group). It is also mainly in this traditional function, that groups are of paramount importance to physicists.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, the great insight of Galois was that the word &quot;symmetry&quot; shouldn't be understood too narrowly, and since then, groups have completely permeated all of mathematics. Groups describe the complexity of a polynomial, they describe the complexity of a topological space, of an algebraic variety, of a number field, etc. Given a number field, say a Galois extension of $\mathbb{Q}$, pretty much all its important invariants are groups: the Galois group, the class group, the ring of integers, the group of units in the ring of integers, etc. Similarly, given a topological space, you have its fundamental group, higher homotopy groups, homology and cohomology... You can tell your friend that if we didn't have groups, then we wouldn't know how to tell a donut from a ball!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I should probably stop here, since to give a full account of the usefulness of groups, one would have to write a compendium of all of mathematics.&lt;/p&gt;&#xA;" OwnerUserId="3212" LastActivityDate="2011-01-28T05:07:17.063" CommentCount="2" />
  <row Id="19329" PostTypeId="1" AcceptedAnswerId="19411" CreationDate="2011-01-28T05:08:41.160" Score="8" ViewCount="290" Body="&lt;p&gt;If $f : [a,b] \rightarrow [c,d]$ is a bijection, $f\in \mathcal{R}$ and $f^{-1}$ exists, then prove or disprove that $f^{-1} \in \mathcal{R} [c,d]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remark: I tried to use integration by parts to find $\int_{c}^{d} f^{-1}$ and to prove that was the right limit of Riemann sum. but failed. But I think this idea might be useful.&lt;/p&gt;&#xA;" OwnerUserId="5980" LastActivityDate="2011-01-28T23:01:06.047" Title="the Riemann integrability of inverse function" Tags="&lt;real-analysis&gt;&lt;homework&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="19330" PostTypeId="2" ParentId="19177" CreationDate="2011-01-28T05:10:50.080" Score="3" ViewCount="" Body="&lt;p&gt;For what it is worth, &lt;a href=&quot;http://math.uga.edu/~pete/4400pellnotes.pdf&quot; rel=&quot;nofollow&quot;&gt;my lecture notes&lt;/a&gt; on the Pell Equation are in the context of $d$ any positive, nonsquare integer.  Also see &lt;a href=&quot;http://math.uga.edu/~pete/4400hw9.pdf&quot; rel=&quot;nofollow&quot;&gt;problem 9.2 here&lt;/a&gt; where I ask the students to exploit the fact that $d$ does not need to be squarefree to show that one can find solutions $(x,y)$ to the Pell equation satisfying an additional congruence $y \equiv 0 \pmod M$ for any $M \in \mathbb{Z}^+$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the perspective of algebraic number theory, this comes down to a version of the Dirichlet Unit Theorem for nonmaximal orders in a number field $K$.  Recall that a $\mathbb{Z}$-order $R$ in $K$ is a subring of $\mathbb{Z}_K$ such that the additive group $(R,+)$ is finitely generated as a $\mathbb{Z}$-module.  The last condition is equivalent to the finiteness of the index $[\mathbb{Z}_K:R]$.  The usual statement of the Dirichlet Unit Theorem is that the unit group $\mathbb{Z}_K^{\times}$ is a finitely generated abelian group with rank equal to $r_1 + r_2 - 1$, where if $K \cong \mathbb{Q}[t]/(P(t))$, the polynomial $P$ has $r_1$ real roots and $r_2$ pairs of complex conjugate non-real roots.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if I am not mistaken (and please let me know if I am!), the standard proof of the Dirichlet Unit Theorem works to show that exactly the same is true for the unit group $R^{\times}$ of any nonmaximal order.  (Certainly $R^{\times}$ is finitely generated, being a subgroup of the finitely generated abelian group $\mathbb{Z}_K^{\times}$; the claim is that its rank is no less than that of $\mathbb{Z}_K^{\times}$.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the structure theory of finitely generated abelian groups, one easily deduces the following &lt;strong&gt;relative&lt;/strong&gt; version of the Dirichlet Unit Theorem: for any order $R$ in $\mathbb{Z}_K^{\times}$, the quotient group $\mathbb{Z}_K^{\times}/R^{\times}$ is finite.&lt;/p&gt;&#xA;" OwnerUserId="299" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2011-01-28T05:30:10.020" LastActivityDate="2011-01-28T05:30:10.020" />
  <row Id="19332" PostTypeId="2" ParentId="19265" CreationDate="2011-01-28T05:53:57.110" Score="6" ViewCount="" Body="&lt;p&gt;There are actually lots of interesting uses of the column sums of the Stirling numbers of the second kind $\left\{ n \atop k\right\}$.  The ordinary and exponential generating functions both have nice forms (nicer than those for the row sums):&#xA;$$\sum_{n=k}^{\infty} \left\{ n \atop k\right\} x^n = x^k \prod_{j=1}^k \frac{1}{1-jx},$$&#xA;$$\sum_{n=k}^{\infty} \left\{ n \atop k\right\} \frac{x^n}{n!} = \frac{(e^x-1)^k}{k!}.$$&#xA;These can be used to prove various identities involving $\left\{ n \atop k\right\}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Multiplying $\left\{ n \atop k\right\}$ by certain other interesting numbers before summing also yields some nice identities, such as these:&#xA;$$\sum_{r=k}^n \binom{n}{r} \left\{ r \atop k\right\} = \left\{ n+1 \atop k+1\right\},$$&#xA;$$\sum_{r=k}^n k^{n-r} \left\{ r-1 \atop k-1\right\} = \left\{ n \atop k\right\},$$&#xA;$$\sum_{r=k}^n \left[ n \atop r\right] \left\{ r \atop k\right\} = \binom{n}{k} \frac{(n-1)!}{(k-1)!} = L(n,k),$$&#xA;where $L(n,k)$ is a &lt;em&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Lah_number&quot; rel=&quot;nofollow&quot;&gt;Lah number&lt;/a&gt;&lt;/em&gt;.  (Lah numbers also show up in the answer to this recent question: &lt;a href=&quot;http://math.stackexchange.com/questions/18284/nth-derivative-of-e1-x&quot;&gt;$n$th derivative of $e^{1/x}$&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Column $k$ is also used to convert reciprocal falling factorial powers $x^{\underline{-k}}$ to ordinary reciprocal powers $x^{-n}$ via &#xA;$$x^{\underline{-k}} = \sum_{n=k}^{\infty} (-1)^{n-k} \left\{ n \atop k\right\} \frac{1}{x^n},$$&#xA;in a sort of inverse to the fact that row $k$ is used to convert ordinary powers $x^n$ to falling powers $x^{\underline{k}}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A good reference is Chapter 8 of Charalambides' &lt;em&gt;&lt;a href=&quot;http://books.google.com/books?id=PDMGA-v5G54C&amp;amp;printsec=frontcover&amp;amp;dq=charalambides+enumerative+combinatorics&amp;amp;source=bl&amp;amp;ots=aXMrTL7BIs&amp;amp;sig=HxQWnE1u6eqp40bFLgLB1bQ45AM&amp;amp;hl=en&amp;amp;ei=hldCTbaSMoLmsQP5_pCZCg&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=2&amp;amp;ved=0CCcQ6AEwAQ#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;Enumerative Combinatorics&lt;/a&gt;&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2370" LastActivityDate="2011-01-28T05:53:57.110" CommentCount="1" />
  <row Id="19333" PostTypeId="2" ParentId="18688" CreationDate="2011-01-28T06:11:38.220" Score="2" ViewCount="" Body="&lt;p&gt;This answer is similar to Moron's but I think is somewhat simpler.  Expand $(e^x-1)^n$ using the binomial theorem to obtain $$(e^x-1)^n = \sum_{j=0}^n (-1)^{n-j} \binom{n}{j} e^{jx}.$$  Differentiate both sides $n$ times.  Every term on the left-hand side will contain an $e^x-1$ factor except for an $n!e^{nx}$ term, and the right side will be $$\sum_{j=0}^n (-1)^{n-j} \binom{n}{j} j^n e^{jx}.$$  Then substitute $x = 0$ to obtain &#xA;$$n! = \sum_{j=0}^n (-1)^{n-j} \binom{n}{j} j^n.$$  Reindex the sum and divide by $n^n$.  &lt;/p&gt;&#xA;" OwnerUserId="2370" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2011-01-28T19:27:14.993" LastActivityDate="2011-01-28T19:27:14.993" CommentCount="4" />
  <row Id="19334" PostTypeId="1" AcceptedAnswerId="19343" CreationDate="2011-01-28T06:31:00.383" Score="2" ViewCount="238" Body="&lt;p&gt;In neutral plane geometry, Euclidean geometry without the parallel postulate, I want to show that the family of parallel lines all perpendicular to a given line pass through all of the plane, leaving no holes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My present formulation of this idea is as follows: Given two lines $l$ and $m$ and a transversal $t$ perpendicular to $l$ at point $A$ and passing through line $m$ at $B$, and given another point $C$ on line $m$ forming a triangle $\triangle ABC$. Then there exists a line $n$ perpendicular to $l$ (and therefore parallel to $t$) which passes through the interior of side $\overline{AC}$ (and therefore the interior of the triangle).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This claim is easy to show in Euclidean geometry where all parallel lines are equidistant. In neutral geometry, parallel lines may bend away from each other, and this claim is much less obvious. Since it appears to be true in models of hyperbolic geometry, I'm guessing it is true in neutral geometry, but I'm at a loss for how to show it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update:&lt;/strong&gt; It occurs to me that this question seems a bit more esoteric than it really is. So let me point out that neutral geometry is really very familiar territory for many of you. As I mentioned it is just ordinary Euclidean geometry without the axiom that asserts the uniqueness of a line parallel to a given line and through a given point and any consequences that follow from that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the usual suspects (theorems) are present:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Isosceles triangle theorems, SAS, ASA, SSS, AAS.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The exterior angle theorem, triangle inequality, scalene inequality, hinge theorem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Actually, many parallel line theorems hold, namely. those that say things like &quot;If two lines are cut with a transversal are pair of congruent angles, then the lines are parallel.&quot; or &quot; If two line share a common perpendicular, then they are parallel.&quot; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is not true are theorems that assert the converse: &quot;If two lines are parallel, then ...&quot; and also not true &quot;If lines $l$ and $m$ are parallel and lines $m$ and $n$ are parallel, then $l$ and $n$ are parallel,&quot; and that parallel lines are equidistant (but equidistant lines &lt;strong&gt;are&lt;/strong&gt; parallel).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Others theorems that are missing include:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Pythagorean theorem&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The angle sum theorem (instead the sum of the angles in a triangle must be 180 or less)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rectangles may not exist.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And the weird one: there may be no such thing as similarity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update:&lt;/strong&gt; To make the answer below understandable, you need (also in the comments but hidden):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proof that two lines perpendicular to a given line are parallel: If they were not parallel, they would intersect forming a isosceles triangle with two 90 degree angles. the complimentary angle to either angle would also be 90 degrees. But by the exterior angle theorem, that exterior angle must be strictly greater than 90 degrees, which is a contradiction.&lt;/p&gt;&#xA;" OwnerUserId="6272" LastEditorUserId="6272" LastEditorDisplayName="" LastEditDate="2011-01-28T20:04:17.537" LastActivityDate="2011-01-28T20:04:17.537" Title="In neutral geometry, can a family of parallel lines leave holes in the plane?" Tags="&lt;geometry&gt;&lt;euclidean-geometry&gt;" AnswerCount="2" CommentCount="7" FavoriteCount="1" />
  <row Id="19335" PostTypeId="1" AcceptedAnswerId="19342" CreationDate="2011-01-28T07:16:56.740" Score="1" ViewCount="99" Body="&lt;p&gt;A subspace V in $R^4$ is defined by the equation $x_{1}$-$x_{2}$+$2x_{3}$+$4x_{4}$=0. I need to find T such that Ker(T)=zero vector, and Im(T)=V. How do I approach this problem? As I understand, the equation given to me is a set of points that are solutions to Im(T)=V, so in a sense they vectors of that plane are elements of Ker(V), but I can't go any further. Thanks.&lt;/p&gt;&#xA;" OwnerUserId="5157" LastActivityDate="2011-01-28T08:41:52.127" Title="Find a T from $R^3$ to $R^4$ given an equation for a subspace in $R^4$" Tags="&lt;linear-algebra&gt;" AnswerCount="2" />
  <row Id="19336" PostTypeId="1" CreationDate="2011-01-28T07:27:57.863" Score="0" ViewCount="196" Body="&lt;p&gt;&lt;em&gt;Note: I wasn't quite sure which StackExchange site to ask this on, so hopefully this is the right one.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As human beings, we observe the world in which we live in three dimensions. However, it is certainly theoretically possible that more dimensions exist.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any direct or indirect evidence supporting a fourth dimension?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;P.S. I can't seem to find anything to tag this question with, so someone will need to retag it.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="4626" LastEditorUserId="1543" LastEditorDisplayName="" LastEditDate="2011-05-03T10:05:53.447" LastActivityDate="2011-05-03T10:05:53.447" ClosedDate="2011-01-28T09:08:34.880" Title="Proof of a fourth dimension?" Tags="&lt;physics&gt;" CommentCount="3" />
  <row Id="19337" PostTypeId="1" CreationDate="2011-01-28T07:50:16.447" Score="2" ViewCount="165" Body="&lt;p&gt;N leaves a remainder of 4 when divided by 33, what are the possible remainders when N is divided by 55?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My approach :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Number has to be 33k+4&#xA;Possible remainders will be (33k+4)/55 = 33k/55 + 4/55 = remainder of (3k/5 + 4/55) = &#xA;remainder of (3k/5) + Remainder of (4/55) &#xA;Since ,Remainder of 3k/5 will be 1,2,3,4 &#xA;And also remainder of 4/55 will be 4 &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, remainder can be 5,6,7,8&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think there is some fundamental logical flaw with my approach. &#xA;Please let me know what it is&lt;/p&gt;&#xA;" OwnerUserId="6047" LastEditorUserId="1583" LastEditorDisplayName="" LastEditDate="2011-01-28T20:20:20.020" LastActivityDate="2011-01-28T20:20:20.020" Title="Number system Problem" Tags="&lt;elementary-number-theory&gt;" AnswerCount="5" CommentCount="1" />
  <row Id="19338" PostTypeId="2" ParentId="19337" CreationDate="2011-01-28T07:57:54.113" Score="2" ViewCount="" Body="&lt;p&gt;Have you learned about the Chinese Remainder Theorem yet? This problem seems specifically geared towards it. I've never seen remainders calculated the way you have, and it doesn't quite work like that. For example taking $k=1$, $33=\text{remainder}(33/55)\neq \text{remainder}(3/5)=3$, since $33=0\cdot 55+33$ and $3=0\cdot 5+3$ when you use the division algorithm, so you can't just cancel factors. Also, $\text{remainder}(33/55)+\text{remainder}(4/55)\neq \text{remainder}(33/55)$, since otherwise you would have $\text{remainder}(4/55)=0$, but it is in fact $4$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, using the Chinese Remainder Theorem, you have a situation like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $N$ has a remainer of $4$ when divided by $33$, you have $N\equiv 4\pmod{33}$. This implies that $N\equiv 4\equiv 1\pmod{3}$ and more importantly, $N\equiv 4\pmod{11}$. To find the possible remainders of $N$ when divided by $55$, you have to use that fact that $55=5\cdot 11$. Choose possible remainders for $N$ upon division by $5$, and then use the Chinese Remainder Theorem to find the remainders modulo $55$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does this help somewhat?&lt;/p&gt;&#xA;" OwnerUserId="1583" LastEditorUserId="1583" LastEditorDisplayName="" LastEditDate="2011-01-28T08:27:22.293" LastActivityDate="2011-01-28T08:27:22.293" CommentCount="2" />
  <row Id="19339" PostTypeId="2" ParentId="19337" CreationDate="2011-01-28T08:10:46.950" Score="0" ViewCount="" Body="&lt;p&gt;You need to find all values of $x$ satisfying&#xA;                     $33k$ + $4=x$  $mod 55$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for some $k\in N$.  Multiplying both sides by 5 gives us&lt;br&gt;&#xA;                      $20=5x$  $mod 55$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that this is not the same as $4=x$  $mod 55$ as 5 is not a unit in $Z\over{55Z}$.  As a result, there are multiple solutions for $x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for your method, it's probably not best to work with fractions, especially reduced fractions.  The problem you run into is that $3k/5$ only produces remainders of 1,2,3, or 4 while its obvious that the problem could have much larger remainders (for instance, if $k=1$ the remainder is 37).&lt;/p&gt;&#xA;" OwnerUserId="3971" LastEditorUserId="3971" LastEditorDisplayName="" LastEditDate="2011-01-28T08:23:53.607" LastActivityDate="2011-01-28T08:23:53.607" />
  <row Id="19340" PostTypeId="2" ParentId="19335" CreationDate="2011-01-28T08:12:19.727" Score="3" ViewCount="" Body="&lt;p&gt;Hint: Find three linearly independent vectors in $\mathbb{R}^4$ that span $V$.&lt;/p&gt;&#xA;" OwnerUserId="3212" LastActivityDate="2011-01-28T08:12:19.727" />
  <row Id="19341" PostTypeId="2" ParentId="19337" CreationDate="2011-01-28T08:20:46.003" Score="8" ViewCount="" Body="&lt;p&gt;One approach which is similar to yours and does work:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $r$ be the remainder when $N$ is divided by 55. So, $N=33k+4=55j+r$. Subtracting $55j$ from both sides, we get $33k-55j+4=r$ or $11(3k-5j)+4=r$. Since $3k-5j$ could end up being any integer, let it be $0,1,2,3$ or $4$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$r=11(0)+4=4$$&#xA;$$r=11(1)+4=15$$&#xA;$$r=11(2)+4=26$$&#xA;$$r=11(3)+4=37$$&#xA;$$r=11(4)+4=48$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other choices for $3k-5j$ produce remainders that are too large (or too small).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; I should note that being able to choose any integer for $3k-5j$ relies on the observation that $3(2n)-5(n)=n$. So choosing $k=2n$ and $j=n$ allows $3k-5j$ to equal any integer $n$. While $3$ and $5$ are small enough to make this easy to see. For larger integers, you may need the Euclidean algorithm and Bézout's identity.&lt;/p&gt;&#xA;" OwnerUserId="6272" LastEditorUserId="6272" LastEditorDisplayName="" LastEditDate="2011-01-28T16:36:23.300" LastActivityDate="2011-01-28T16:36:23.300" CommentCount="2" />
  <row Id="19342" PostTypeId="2" ParentId="19335" CreationDate="2011-01-28T08:41:52.127" Score="2" ViewCount="" Body="&lt;p&gt;I am going to treat &quot;matrix&quot; and &quot;linear transformation&quot; as synonyms in this answer, because we are working in $\mathbb{R}^n$ and have the standard basis at our disposal.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rephrasing your problem: you are given a subspace $V$ of $\mathbb{R}^4$, defined by a homogeneous linear equation.  The problem is to find a matrix $M$ whose nullspace consists only of the zero vector and whose column space is $V$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are some useful tools to have in your toolbox:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) Given a subspace $W$ of $\mathbb{R}^n$ defined by a system of $m$ homogeneous linear equations, write down a matrix $A$ whose nullspace is precisely $W$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[Solution, making use only of the definitions of the words and matrix multiplication: rewrite the system of homogeneous linear equations $\sum_{j=1}^n a_{ij} x_j = 0$, $1 \leq i \leq m$, as a single matrix equation $A x = 0$, where $A$ is the matrix whose row $i$, column $j$ entry is $a_{ij}$, and where $x$ is the column vector whose $i$th entry is $x_i$.  The matrix $A$ will do.]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) Given an $m \times n$ matrix $A$, find a basis for the nullspace of $A$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[Solution: somewhat hard to express in words if you haven't seen it--- look in a textbook.  The standard way this is done is to solve $Ax = 0$ using Gaussian elimination, to introduce a &quot;free variable&quot; for every non-pivot variable column, and thus to write the general solution to $Ax = 0$ in terms of &quot;free variables&quot;.  If you rewrite this general solution in vector form--- as a linear combination of fixed numerical vectors, with the coefficients the free variables--- a basis falls right out.]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(3) Given a basis $v_1, \dots, v_k$ for a subspace $W$ of $\mathbb{R}^m$, find a matrix $T$ whose nullspace is $\{0\}$ and whose range is $W$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[Solution, based only on the definitions of the words and matrix multiplication: write the $v$s as column vectors and make them columns of a matrix.  In general, if $M$ is any matrix, the nullspace of $M$ is $\{0\}$ precisely when its columns are linearly independent, and the range of $M$ is precisely the span of its columns--- so as the $v_1, \dots, v_k$ are a basis for $W$, the matrix $T$ formed with these vectors as columns has the two properties you want.]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this example, doing (1) I see that your $V$ is the nullspace of the $1 \times 4$ matrix $A = (1,-1,2,4)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Doing (2) I want to find the general solution to $Ax = 0$, with $A = (1,-1,2,4)$ and $x$ the $4 \times 1$ column vector of variables.  The augmented matrix for this system is $(1,-1,2,4,0)$ and it is already in row echelon form: the variables $x_2 = s$, $x_3 = t$ and $x_4 = u$ are &quot;free&quot;, and when they have been assigned values $x_1$ must then satisfy $x_1 - s + 2t + 4u = 0$, so that $x_1 = s - 2t - 4u$.  The general solution to $Ax = 0$ is thus the set of all vectors of the form $(s - 2t - 4u, s, t ,u) = s (1,1,0,0) + t (-2, 0, 1, 0) + u(-4, 0, 0, 1)$; in other words, it is the span of the vectors $v_1 = (1,1,0,0)$, $v_2 = (-2,0,1,0)$, and $v_3 = (-4,0,0,1)$.  By the general magic of this approach, these vectors are linearly independent, so they are a basis for the nullspace of $A$ (and hence a basis of $V$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(3) I form the matrix $T = \begin{pmatrix} 1 &amp;amp; -2 &amp;amp; -4 \\ 1 &amp;amp; 0 &amp;amp; 0 \\ 0 &amp;amp; 1 &amp;amp; 0 \\ 0 &amp;amp; 0 &amp;amp; 1 \end{pmatrix}$ with $v_1$, $v_2$, and $v_3$ as columns, and I see that it has all of the desired properties.  Going back into the language of linear transformations, the linear transformation from $\mathbb{R}^3$ to $ \mathbb{R}^4$ whose matrix with respect to the standard basis is this matrix $T$, will do the job.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is worth pointing out that in step (2) of this procedure (finding a basis for the nullspace of $A$) we are making an arbitrary choice in how we do that (any vector space has many bases; somebody with a different method for calculating the basis of a nullspace of a matrix may well get a different set of vectors).  So there are some degrees of freedom here and this problem does not have a unique solution.  If you think about it a while you may be able to describe the set of all possible solutions to this problem.  &lt;/p&gt;&#xA;" OwnerUserId="3840" LastActivityDate="2011-01-28T08:41:52.127" />
  <row Id="19343" PostTypeId="2" ParentId="19334" CreationDate="2011-01-28T08:45:38.327" Score="3" ViewCount="" Body="&lt;p&gt;According to the Wikipedia article &lt;a href=&quot;http://en.wikipedia.org/wiki/Absolute_geometry&quot; rel=&quot;nofollow&quot;&gt; absolute geometry&lt;/a&gt; the first 28 propositions of Euclid's Elements do not require the parallel postulate and hence may be proven for neutral or &quot;absolute&quot; geometry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Prop. 11 is the &lt;a href=&quot;http://aleph0.clarku.edu/~djoyce/java/elements/bookI/propI11.html&quot; rel=&quot;nofollow&quot;&gt; construction of a perpendicular to a line from a point on the line&lt;/a&gt;.  Prop. 12 is the &lt;a href=&quot;http://aleph0.clarku.edu/~djoyce/java/elements/bookI/propI12.html&quot; rel=&quot;nofollow&quot;&gt; construction of a line perpendicular to a given line through a point not on the line&lt;/a&gt;.  So, combined with OP's proof that all lines perpendicular to a given line are parallel, the result is established:  The family of parallel lines perpendicular to a given line covers all points.&lt;/p&gt;&#xA;" OwnerUserId="3111" LastEditorUserId="3111" LastEditorDisplayName="" LastEditDate="2011-01-28T16:39:08.913" LastActivityDate="2011-01-28T16:39:08.913" CommentCount="6" />
  <row Id="19344" PostTypeId="1" AcceptedAnswerId="19345" CreationDate="2011-01-28T08:55:37.433" Score="5" ViewCount="188" Body="&lt;p&gt;I've been trying to prove that every Polish Space is homeomorphic to a $G_\delta$ subspace of the Hilbert Cube. There is a hint saying that given a countable dense subset of the Polish space $\{x_n : n\in\mathbb{N}\}$ define the function $f(x)=(d(x,x_n))_{n\in\mathbb{N}}$ (where $d$ is a metric on the Polish space). I think I've shown that $f$ is continuous and $1-1$ but I don't know how to prove that the converse of the function is continuous and that the image is a $G_\delta$ set. It's been a really long time since I've dealt with topology so I'm having a hard time coming up with any idea and I'm afraid I've forgotten some well known topological fact (so maybe it's something obvious here I'm not seeing). Any help on how to proceed would be kindly appreciated.&lt;/p&gt;&#xA;" OwnerUserId="1772" LastEditorUserId="1772" LastEditorDisplayName="" LastEditDate="2011-06-02T10:55:37.263" LastActivityDate="2011-06-02T10:55:37.263" Title="Polish Spaces and the Hilbert Cube" Tags="&lt;topology&gt;&lt;set-theory&gt;&lt;descriptive-set-theory&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="19345" PostTypeId="2" ParentId="19344" CreationDate="2011-01-28T09:17:03.940" Score="5" ViewCount="" Body="&lt;p&gt;I take it that you've already normalized $d$ such that $0 \leq d \leq 1$ (otherwise replace $d$ by $\frac{d}{1+d}$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you've said, the function $f: x \mapsto f(x) = (d(x,x_{n}))_{n \in \mathbb{N}}$ is continuous and injective. Let $f(y_{m}) \to f(y)$ be a convergent sequence in $f(X)$. We want to show that $y_{m} \to y$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By definition of the product topology, we have $d(y_{m},x_{n}) \xrightarrow{m \to \infty} d(y,x_{n})$ for all $n$. Let $\varepsilon &amp;gt; 0$ and pick a point $x_{n}$ such that $d(y,x_{n}) &amp;lt; \varepsilon/3$ by density. Since $d(y_{m},x_{n}) \to d(y,x_{n})$, there is $M$ such that $|d(y_{m},x_{n}) - d(y,x_{n})| &amp;lt; \varepsilon /3$ for all $m \geq M$, so $d(y_{m},x_{n}) &amp;lt; 2 \varepsilon /3$. But then $d(y_{m},y) \leq d(y_{m},x_{n}) + d(x_{n},y)&amp;lt; \varepsilon$ and hence $y_{m} \to y$.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Why is the image a $G_{\delta}$-set? This seems to be much more difficult. I don't see any easier way than to essentially reprove two classical results on metric spaces which are much more interesting, so I prefer to explain this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Theorem (Kuratowski)&lt;/strong&gt;&#xA;Let $A \subset X$ be a subset of a metrizable space and let $g: A \to Y$ be a continuous map to a completely metrizable space $Y$. Then $g$ can be continuously extended to a $G_{\delta}$-set containing $A$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fix a bounded and complete metric on $Y$. For the proof we need the notion of &lt;em&gt;oscillation&lt;/em&gt; of $g$ at a point $x \in \overline{A}$ (the closure of $A$ in $X$) defined by&#xA;$$\displaystyle&#xD;&#xA;\operatorname{osc}_{g}(x) = \inf\{\operatorname{diam}g(U \cap A)\,:\, x \in U, \;U\; \text{open}\}.&#xD;&#xA;$$&#xA;The set $B = \{x \in \overline{A}\,:\,\operatorname{osc}_{g}(x) = 0\}$ is a $G_{\delta}$-set. To see this, note that $B_{n} = \{x \in \overline{A} \,:\, \operatorname{osc}_{g}(x) &amp;lt; \frac{1}{n}\}$ is an open subset of the closed set $\overline{A}$ and $B = \bigcap_{n \in \mathbb{N}} B_{n}$. The continuity of $f$ implies that $A \subset B$. Now define $f: B \to Z$ by $f(x) = \lim g(x_{n})$, where $x_{n} \to x$. It is not hard to show that $f$ is well-defined (because $\operatorname{osc}_{g}(x) = 0$ implies that $g(x_{n})$ is a Cauchy-sequence) and clearly $f$ extends $g$ and is continuous.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second ingredient we need is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Theorem (Lavrentiev)&lt;/strong&gt;&#xA;Let $X$ and $Y$ be completely metrizable spaces and let $g: A \to B$ be a homeomorphism from $A \subset X$ onto $B \subset Y$. Then there exist $G_{\delta}$-sets $G \supset A$ and $H \supset B$ and a homeomorphism $f: G \to H$ extending $g$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $h = g^{-1}$. Choose $G_{\delta}$-sets $G&amp;#39; \supset A$ and $H&amp;#39; \supset B$ and continuous extensions $g&amp;#39;: G&amp;#39; \to Y$ and $h&amp;#39;: H&amp;#39; \to X$ by Kuratowski's theorem. Let $Z = \operatorname{graph}(g&amp;#39;) \cap \widetilde{\operatorname{graph}}(h&amp;#39;) \subset X \times Y$ be the intersection of the graphs (the tilde indicates the 'switch' $\widetilde{(y,x)} = (x,y)$ of coordinates) and let $G = \operatorname{pr}_{X} (Z)$ and $H = \operatorname{pr}_{Y}(Z)$. Obviously, $f = g&amp;#39;|_{G}$ is a homeomorphism of $G$ onto $H$. One can check that $H$ (and thus also $G$ by symmetry) is a $G_{\delta}$-set as follows: The graph of $g&amp;#39;$ is closed in $G&amp;#39; \times Y$ and thus it is a $G_{\delta}$-set and $H$ is its preimage under the continuous map $y \mapsto (h&amp;#39;(y),y)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Corollary.&lt;/strong&gt; If $Y$ is a completely metrizable space and $X \subset Y$ a completely metrizable subspace then $X$ is a $G_{\delta}$-set.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By Lavrentiev's theorem, the inclusion $X \subset Y$ extends to a homeomorphism onto its image.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A further corollary of these ideas is that a subset of a Polish space is Polish if and only if it is a $G_{\delta}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;More detailed information can be found in any decent book on descriptive set theory, for instance Kechris, &lt;em&gt;Classical descriptive set theory&lt;/em&gt;, or Srivastava, &lt;em&gt;A course on Borel sets&lt;/em&gt;, both appeared in the Springer Graduate Texts in Mathematics series.&lt;/p&gt;&#xA;" OwnerUserId="5363" LastEditorUserId="5363" LastEditorDisplayName="" LastEditDate="2011-01-28T13:37:02.847" LastActivityDate="2011-01-28T13:37:02.847" CommentCount="0" />
  <row Id="19346" PostTypeId="2" ParentId="19318" CreationDate="2011-01-28T09:18:44.310" Score="18" ViewCount="" Body="&lt;p&gt;Group theory (when physicists say this they mean &lt;a href=&quot;http://en.wikipedia.org/wiki/Representation_theory&quot; rel=&quot;nofollow&quot;&gt;representation theory&lt;/a&gt;) is the basis of modern physics.  Via &lt;a href=&quot;http://en.wikipedia.org/wiki/Noether%27s_theorem&quot; rel=&quot;nofollow&quot;&gt;Noether's theorem&lt;/a&gt; it is the abstract mechanism responsible for conservation laws (e.g. conservation of energy, conservation of momentum) even in classical mechanics.  In quantum mechanics, representations are even more important: the representations of a group called &lt;a href=&quot;http://en.wikipedia.org/wiki/Special_unitary_group&quot; rel=&quot;nofollow&quot;&gt;$\text{SU}(2)$&lt;/a&gt; describe the difference between &lt;a href=&quot;http://en.wikipedia.org/wiki/Boson&quot; rel=&quot;nofollow&quot;&gt;bosons&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/Fermion&quot; rel=&quot;nofollow&quot;&gt;fermions&lt;/a&gt; (the difference being their &lt;a href=&quot;http://en.wikipedia.org/wiki/Spin_(physics)&quot; rel=&quot;nofollow&quot;&gt;spin&lt;/a&gt;, which is the physical property that makes &lt;a href=&quot;http://en.wikipedia.org/wiki/Magnetic_resonance_imaging&quot; rel=&quot;nofollow&quot;&gt;MRI&lt;/a&gt; work), and the representations of $\text{SU}(2) \times \text{SU}(2)$ describe the possible orbits of an &lt;a href=&quot;http://en.wikipedia.org/wiki/Hydrogen_atom&quot; rel=&quot;nofollow&quot;&gt;electron in a hydrogen atom&lt;/a&gt;.  So group theory can be used, among many many other things, to predict the structure of the periodic table.  It is also the foundation of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Standard_Model&quot; rel=&quot;nofollow&quot;&gt;Standard Model of particle physics&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Group theory also happens to be fundamental to many modern branches of mathematics, but I figured an application to things that people provably care about outside of mathematics would look better.)&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2011-01-28T19:46:30.370" LastActivityDate="2011-01-28T19:46:30.370" CommentCount="3" />
  <row Id="19347" PostTypeId="1" CreationDate="2011-01-28T09:58:01.717" Score="0" ViewCount="79" Body="&lt;p&gt;I have a theory that in order to best compress 1D data between 0 and 1 in a lossy way, you should map each data point $i$ to:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\int_0^i \! P(x) \, \mathrm{d}x$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $P(x)$ is the probability density function of the data. The probability density function over the mapped data should be level and being level when you store the values to a finite precision, the amount of redundant information should be minimized.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is: how can I generalize this to more dimensions? Say I have a set of points which fit a given distribution and I want to compress points that fit the same distribution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My initial assessment is that I should incrementally move the points so that they follow a level probability density function, then (with lots of arm waving) it should be possible to traverse the mapping using the proximity to the reference points.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A further constraint is that given data that fits something that looks like a spiral, rather than squashing it together, it should unwind it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone shed any light on this?&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;I assume that I'm not the first person to talk about such a concept. What is its name?&lt;/li&gt;&#xA;&lt;li&gt;Is this a good approach?&lt;/li&gt;&#xA;&lt;li&gt;How do you incrementally move the points so that they fit a level probability density function?&lt;/li&gt;&#xA;&lt;li&gt;How do you traverse the mapping?&lt;/li&gt;&#xA;&lt;li&gt;How do I modify the above approach to fit the added constraint?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;edit: To clarify, essentially what I want to do is take a set of points, then warp the space the points are in so that the points are equally spaced.&lt;/p&gt;&#xA;" OwnerUserId="4738" LastEditorUserId="4738" LastEditorDisplayName="" LastEditDate="2011-01-29T08:41:48.507" LastActivityDate="2011-02-28T09:39:19.727" Title="How do you generalize the compression of data into N dimensional space?" Tags="&lt;probability&gt;&lt;probability-distributions&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="19348" PostTypeId="2" ParentId="19337" CreationDate="2011-01-28T11:13:01.230" Score="0" ViewCount="" Body="&lt;p&gt;N is congruent to 4 mod 33 and the highest common factor of 33 and 55 is 11.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So N is congruent to 4 mod 11 and mod 55 can be anything that has this congruence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore 4, 15, 26, 37 or 48&lt;/p&gt;&#xA;" OwnerUserId="6342" LastActivityDate="2011-01-28T11:13:01.230" />
  <row Id="19349" PostTypeId="1" CreationDate="2011-01-28T11:51:29.000" Score="3" ViewCount="156" Body="&lt;p&gt;Assume that there are n snakes. Any 2 ends (tail or head) of the &quot;2n&quot; available have to be picked up and tied together and this process has to be repeated infinitely.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If p/q (gcd(p,q) = 1) is the probability of you getting a single long “snake” in the end, what would be the sum of (p+q) for all 2 &amp;lt;= n &amp;lt;= 40?&lt;/p&gt;&#xA;" OwnerUserId="6343" LastActivityDate="2011-02-27T16:57:36.377" Title="Snakes and Probabilistic Enigma" Tags="&lt;number-theory&gt;&lt;homework&gt;&lt;probability&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" />
  <row Id="19350" PostTypeId="2" ParentId="19261" CreationDate="2011-01-28T12:26:53.757" Score="1" ViewCount="" Body="&lt;p&gt;Urn models!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Tossing (Head,Head,Head) with one coin is like drawing a black ball out of an urn with 1 black and 7 white balls. The Head/Head coin is then like an urn with 8 black balls, let's imagine them as marked with a white dot.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, we put 5 urns (4 representing a normal coin and 1 representing the Head/Head coin) into a box. Because all urns contain the same number of balls, we can smash them without changing the probabilities: We can draw each of the 40 balls with same probability, and it will be with 1/5 probability from the urn representing the Head/Head coin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we draw any ball out of the box, that ball will be marked with the probability of 8/40, because 8 of 40 balls are marked. It will be black with 12/40. Under the precondition that we draw a black ball, it will be marked &lt;strong&gt;with the probability of 8/12&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;It is confusing, that we had the probability 1/5 before. So, what happened? How could we ignore the white balls?&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;One way to ignore the white balls is this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We state, that we already know the case of tossing something else than (Head,Head,Head): 0 of 28 white balls are marked, so that means that we would know with the probability 28/40 that the coin has the probability 0/28 of being that Head/Head coin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That way we ask: &quot;We know that none of the white balls are marked, but how many black balls are marked?&quot;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The other way lets us rephrase the question:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;You are given a coin and told that it has a 1/5 chance that it is double-headed, otherwise it is a normal coin. You are allowed three flips of the coin. Upon doing these, you receive &lt;strong&gt;either&lt;/strong&gt; &lt;em&gt;three heads&lt;/em&gt; &lt;strong&gt;or you give it back to restart this game&lt;/strong&gt;. What is the probability that your coin is double-headed?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That way we can ignore or remove the white balls out of the box.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="3226" LastEditorUserId="3226" LastEditorDisplayName="" LastEditDate="2011-01-28T12:39:19.257" LastActivityDate="2011-01-28T12:39:19.257" />
  <row Id="19351" PostTypeId="2" ParentId="19274" CreationDate="2011-01-28T12:51:59.770" Score="3" ViewCount="" Body="&lt;p&gt;If $X_i$ and $Y_j$ are dependent for $i=j$, but independent for  $i\neq j$ we have an iid sample from bivariate distribution $Z_i=(X_i,Y_i)$. Then central limit theorem gives us&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;\sqrt{n}\left(\frac{1}{n}\sum_{i=1}^nZ_i -EZ_1\right)\xrightarrow{D}N(0,\Sigma)&#xA;\end{align}&#xA;with $\Sigma=cov(Z_1)$ and $\xrightarrow{D}$ indicating convergence in distribution. Note that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(S_n,T_n)=\frac{1}{n}\sum_{i=1}^nZ_i.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we can use delta method, which states that if $r_n(U_n-\theta)\xrightarrow{D}U$ for numbers $r_n\to\infty$, then $r_n(\phi(U_n)-\phi(\theta))\xrightarrow{D}\phi&amp;#39;_\theta(U)$. This statement can be found &lt;a href=&quot;http://books.google.com/books?id=UEuQEM5RjWgC&amp;amp;lpg=PR1&amp;amp;dq=asymptotic%20statistics%20vaart&amp;amp;hl=fr&amp;amp;pg=PA26#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;. Delta method is also described &lt;a href=&quot;http://en.wikipedia.org/wiki/Delta_method&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In our case now we have $r_n=\sqrt{n}$, $\phi(x,y)=x^2-y^2$, $U_n=(S_n,T_n)$ and $\theta=(EX_1,EY_1)$. We have &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\phi_{\theta}&amp;#39;=(2EX_1,-2EY_1)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$U=(U_1,U_2)\sim N(0,\Sigma)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally we get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;\sqrt{n}\left(S_n^2-T_n^2-(EX_1)^2+(EY_1)^2\right)\xrightarrow{D} 2U_1EX_1-2U_2EY_1\sim N(0,4(EX_1,EY_1)\Sigma(EX_1,EY_1)')&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that I basically redid the example after the theorem in the link. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The final answer then depends on $\Sigma=cov((X_1,Y_1))$, but this should be known to original poster.&lt;/p&gt;&#xA;" OwnerUserId="4742" LastEditorUserId="4742" LastEditorDisplayName="" LastEditDate="2011-01-28T17:41:14.800" LastActivityDate="2011-01-28T17:41:14.800" CommentCount="1" />
  <row Id="19352" PostTypeId="1" AcceptedAnswerId="19383" CreationDate="2011-01-28T13:16:56.267" Score="4" ViewCount="132" Body="&lt;p&gt;Does there exists other structures in math, which are seemingly random, but deterministic, and follow rules similar to the prime numbers, by rules I mean there must be statements similar to goldbach's conjecture or twin-prime-conjecture etc, for instance i dont consider the digits of pi to have this sort of structure. Is there some field in math which deals with distributions which &quot;seem&quot; random, but still follow certain rules ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are some examples of similar structures?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, I want to know what are the necessary axioms to produce the prime-number distribution, and if one ca replace these axioms to get other pseudo-random distributions which are not &quot;isomorphic&quot; to the prime-numbers, and which are not constructable in regular axiomatic systems ?&lt;/p&gt;&#xA;" OwnerUserId="5904" LastActivityDate="2011-01-28T18:48:40.660" Title="Are there other pseudo-random distributions like the prime-numbers?" Tags="&lt;soft-question&gt;&lt;prime-numbers&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="19353" PostTypeId="1" CreationDate="2011-01-28T13:23:53.140" Score="0" ViewCount="32" Body="&lt;p&gt;What is $Aut_{Modk}(A)$ if A is a finite generated algebra, k is a field?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it a group of automorphisms of A as a vector space over k or what?&lt;/p&gt;&#xA;" OwnerUserId="6345" LastActivityDate="2011-01-28T13:23:53.140" Title="notation question $Aut_{Modk}(A)$" Tags="&lt;algebra&gt;&lt;representation-theory&gt;" CommentCount="3" />
  <row Id="19354" PostTypeId="2" ParentId="19301" CreationDate="2011-01-28T13:24:23.693" Score="3" ViewCount="" Body="&lt;p&gt;Here are some more detailed hints.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the value of $\lfloor 2x \rfloor - 2 \lfloor x \rfloor$ where $x=n+ \delta$ for&#xA;$ n \in \mathbb{Z}$ and $0 \le \delta &amp;lt; 1/2.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose $p$ is a prime number of the form $4n+1$ and $a$ is a&#xA;quadratic residue modulo $p$ then why is $(p-a)$ also a quadratic residue?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What does this say about the number of quadratic residues $&amp;lt; p/2$ ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All the quadratic residues are congruent to the numbers&#xA;$$1^2,2^2,\ldots, \left( \frac{p-1}{2} \right)^2,$$&#xA;which are themselves all incongruent to each other, so how many times does the set&#xA;$\lbrace 1^2,2^2,\ldots,(p-1)^2 \rbrace$ run through a complete set of&#xA;$\it{quadratic}$ residues?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose $i^2 \equiv a \textrm{ mod } p$ where $i \in \lbrace 1,2,\ldots,p-1 \rbrace$ and $a$ is a quadratic residue $&amp;lt; p/2$ then what is the value of&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \left \lfloor \frac{2i^2}{p} \right \rfloor -&#xD;&#xA; 2 \left \lfloor \frac{i^2}{p} \right \rfloor \quad \text{?}$$&lt;/p&gt;&#xA;" OwnerUserId="1301" LastEditorUserId="1301" LastEditorDisplayName="" LastEditDate="2011-01-28T22:03:30.737" LastActivityDate="2011-01-28T22:03:30.737" CommentCount="3" />
  <row Id="19355" PostTypeId="2" ParentId="19349" CreationDate="2011-01-28T13:45:05.930" Score="2" ViewCount="" Body="&lt;p&gt;Hint:Let P(n) be the probability of making one loop from n snakes.  When you pick up two ends, they might come from the same snake (in which case you fail immediately) or they don't (in which case you tie them together and have n-1 snakes).  So you should be able to make a recurrence expressing P(n) in terms of P(n-1).  And P(1)=1-you will always succeed with a single snake.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Added:  with two snakes, the chance of success is $\frac{2}{3}$ as you just have to avoid the other end of the first snake you pick up.  For three, the chance of avoiding failure the first time is $\frac{4}{5}$ and of overall success is $\frac{2\cdot 4}{3\cdot 5}$.  For $n$ snakes it is $\frac {2^{n-1}(n-1)!}{(2n-1)!!}$ where the two exclamation points are the double factorial-the product of all odd numbers up to $2n-1$.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2011-02-27T16:57:36.377" LastActivityDate="2011-02-27T16:57:36.377" CommentCount="5" />
  <row Id="19356" PostTypeId="1" CreationDate="2011-01-28T12:48:54.837" Score="0" ViewCount="108" Body="&lt;p&gt;Hello,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a dataset of trajectories. These trajectories are represented in 3D space (x,y,z). All trajectories of this dataset are similar in their shape, but they are not exactly the same, I mean, there is some variation along the points. The trajectories are nonlinear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I need is a kind of regression (polynomial?) on the data to fit the curve along the 3D points, at the end I need a smoothed trajectory, say a generalized one (result of curve fitting/regression). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I just find curve fitting for 2D data (x,y). Can anyone give hints of how to solve it? I heard about local polynomial regression on manifolds, but I dont know how it works, seems to be complex.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;thank you in advance&lt;/p&gt;&#xA;" LastActivityDate="2011-02-01T16:55:06.550" Title="regression (using 3D points cloud dataset)" Tags="&lt;regression&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="19358" PostTypeId="1" AcceptedAnswerId="19362" CreationDate="2011-01-28T14:01:30.400" Score="2" ViewCount="128" Body="&lt;p&gt;I am learning algorithms, and there is a chapter which uses linear optimization methods to solve a matching problem. This is the problem definition: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/mpY0T.png&quot; alt=&quot;problem screenshot&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I find the abbreviations AP for the constraints of the primal problem  and APD for the constraints of the dual problem confusing. Could somebody please explain what they stand for in this context? &lt;/p&gt;&#xA;" OwnerUserId="6346" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2011-01-28T19:12:40.373" LastActivityDate="2011-01-28T19:12:40.373" Title="In linear optimization, what does &quot;AP&quot; stand for?" Tags="&lt;terminology&gt;&lt;linear-programming&gt;" AnswerCount="1" />
  <row Id="19359" PostTypeId="2" ParentId="19356" CreationDate="2011-01-28T14:08:02.023" Score="3" ViewCount="" Body="&lt;p&gt;It is not clear what you are looking for.  One approach to smooth each trajectory is a set of splines.  You can just regard each of x, y, and z as functions of time and calculate (for example) a cubic spline.  Section 3.3 of &lt;a href=&quot;http://apps.nrbook.com/c/index.html&quot; rel=&quot;nofollow&quot;&gt;Numberical Recipes&lt;/a&gt; has a discussion.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to combine the trajectories into one &quot;average&quot; trajectory, you have to decide what you want to do.  You could certainly average the x,y,z points of all your trajectories to get an average, then do splines on that.  You probably want to average the points an equal distance or time from the origin.  Alternately, maybe you have some model of what a trajectory should look like (like an elliptical orbit) and you are trying to remove measurement noise.  Then you are in to multidimensional minimization.  You write the equation your points are supposed to satisfy (with adjustable parameters) and adjust the parameters until the error is minimized.  Chapter 10 of the same book can help.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-01-28T14:08:02.023" CommentCount="1" />
  <row Id="19360" PostTypeId="1" CreationDate="2011-01-28T14:39:33.270" Score="2" ViewCount="150" Body="&lt;p&gt;Considering the construction of a matrix as follows. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The $n$th row in the matrix is filled with the coeffcients of $x^r$ in the expansion of $(1+x)^n$ from the columns $2n$ to $3n$ inclusive and circle all the numbers that are divisible by $n$ in the same row&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How would I find the number of columns for which all the elements in a column are circled in the first j columns given j = 547 ?&lt;/p&gt;&#xA;" OwnerUserId="6349" LastEditorUserId="6349" LastEditorDisplayName="" LastEditDate="2011-01-28T15:12:10.513" LastActivityDate="2011-05-28T20:04:49.600" Title="Matrix and Binomial Coefficients" Tags="&lt;number-theory&gt;&lt;homework&gt;&lt;matrices&gt;&lt;binomial-coefficients&gt;&lt;puzzle&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="19361" PostTypeId="2" ParentId="19360" CreationDate="2011-01-28T14:57:20.500" Score="0" ViewCount="" Body="&lt;p&gt;Hint: Have you tried looking at Pascal's triangle modulo some small primes?  You could see a pattern that would apply.  The &lt;a href=&quot;http://en.wikipedia.org/wiki/Binomial_coefficient&quot; rel=&quot;nofollow&quot;&gt;Divisibility properties&lt;/a&gt; section of Wikipedia's Binomial Coefficient article has some useful information.  &lt;a href=&quot;http://ecademy.agnesscott.edu/~lriddle/ifs/siertri/Pascal.htm&quot; rel=&quot;nofollow&quot;&gt;This page&lt;/a&gt; has some neat images mod 2.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-01-28T14:57:20.500" CommentCount="5" />
  <row Id="19362" PostTypeId="2" ParentId="19358" CreationDate="2011-01-28T15:02:13.633" Score="3" ViewCount="" Body="&lt;p&gt;$\text{AP}$ stands for assignment problem or allocation problem. So $\text{AP}_{D}$ is the dual problem.&lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2011-01-28T15:02:13.633" />
  <row Id="19363" PostTypeId="2" ParentId="19347" CreationDate="2011-01-28T15:26:36.560" Score="1" ViewCount="" Body="&lt;p&gt;If I understand it right, what you are trying to do is conceptually similar to &lt;a href=&quot;http://en.wikipedia.org/wiki/Vector_quantization&quot; rel=&quot;nofollow&quot;&gt;vector quantization&lt;/a&gt;: this tries to partition the N-dimensional space in zones that measure (in probability) the same, so that you can assign to each zone a representative value (which would corespond to the quantized/coded value).&lt;/p&gt;&#xA;" OwnerUserId="312" LastActivityDate="2011-01-28T15:26:36.560" />
  <row Id="19364" PostTypeId="2" ParentId="19305" CreationDate="2011-01-28T15:36:55.193" Score="1" ViewCount="" Body="&lt;p&gt;The sum of the eigenvalues equals  the trace of the matriz. For a $N  \times N$ covariance matriz, this would amount to $N VAR$  - where VAR is the variance of each variable (assuming they are equal - otherwise it would be the mean variance).&#xA;Put in other way, the mean value of the eigenvalues is equal to the mean value of the variances.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And that's pretty much what can be said. Perhaps you are computing a covariance matriz by just multiplying the data matrices? If so, you just should divide by $N.&lt;/p&gt;&#xA;" OwnerUserId="312" LastActivityDate="2011-01-28T15:36:55.193" />
  <row Id="19365" PostTypeId="2" ParentId="19337" CreationDate="2011-01-28T15:50:33.823" Score="2" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\ $ By the &lt;a href=&quot;http://en.wikipedia.org/wiki/Chinese_remainder_theorem&quot; rel=&quot;nofollow&quot;&gt;Chinese Remainder Theorem,&lt;/a&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\quad\quad\quad\quad\quad\quad\quad x\equiv r\ (mod\ 55)\ $  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\quad\quad\quad\quad\quad\quad\quad x\equiv 4\ (mod\ 33)\ $ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;is solvable iff $\rm\ r\equiv 4\ (mod\ 11)\ \ $ since $\rm\ \ 11 = gcd(55,33)$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence since $\rm\ \ r = 4 + 11\ k\:\ $ we deduce $\rm\ r = 4,\:15\:,\:\cdots\ (mod\ 55)$&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2011-01-28T15:50:33.823" />
  <row Id="19366" PostTypeId="1" CreationDate="2011-01-28T15:55:22.367" Score="2" ViewCount="87" Body="&lt;p&gt;how to calculate $a^{(P-1)/2}\pmod{P}$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for example $3^{500001}\pmod{1000003}$ given that $1000003$ is prime&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that if we square the number $3^{500001}$ the result will be either $1$ or $-1$ modulo $1000003$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but my question is how to continue?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have this question as bonus in previous exam.&lt;/p&gt;&#xA;" OwnerUserId="6351" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-28T16:28:50.873" LastActivityDate="2011-01-28T17:27:06.707" Title="calculate $a^{(P-1)/2}\pmod{P}$ for large prime" Tags="&lt;modular-arithmetic&gt;&lt;elementary-number-theory&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="19367" PostTypeId="2" ParentId="19257" CreationDate="2011-01-28T15:57:05.690" Score="12" ViewCount="" Body="&lt;p&gt;Consider the sequence $a,a^2,a^4,a^8,\ldots$ i.e. $x_n = a^{2^{n}}$ where $a \in [0,1)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clearly, we have $\displaystyle \lim_{n \rightarrow \infty} x_n = 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have $f(a) = f(a^2)$, $\forall a \in [0,1]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using this, it is easy to prove by induction that $f(a) = f(a^{2^{n}})$, $\forall a \in [0,1]$ and $\forall n \in \mathbb{N}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Further, every continuous function is sequentially continuous i.e. $\displaystyle \lim_{n \rightarrow \infty} f(x_n) = f(\lim_{n \rightarrow \infty} x_n) $.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, $\displaystyle \lim_{n \rightarrow \infty} f(x_n) = f(\lim_{n \rightarrow \infty} x_n)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the above arguments, we get that $\forall a \in [0,1)$, $$f(a) = \displaystyle \lim_{n \rightarrow \infty} f(a) = \displaystyle \lim_{n \rightarrow \infty} f(a^{2^{n}}) = f(\displaystyle \lim_{n \rightarrow \infty} a^{2^{n}}) = f(0)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, $f(a) = f(0)$, $\forall a \in [0,1)$. Use continuity to conclude that $f(1) = f(0)$ and hence $$f(a) = f(0), \forall a \in [0,1]$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I just want to make this argument symmetric for $0$ and $1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just like we argued out that $f(a) = f(0)$, $\forall a \in [0,1)$, we can argue out that $f(a) = f(1)$, $\forall a \in (0,1]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instead of considering the sequence $a,a^2,a^4,a^8,\ldots$ consider $a, \sqrt{a}, \sqrt[4]{a}, \sqrt[8]{a}, \ldots$ i.e. $x_n = \sqrt[2^n]{a}$ where $a \in (0,1]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clearly, we have $\displaystyle \lim_{n \rightarrow \infty} x_n = 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have $f(a) = f(\sqrt{a})$, $\forall a \in [0,1]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using this, it is easy to prove by induction that $f(a) = f(\sqrt[2^n]{a})$, $\forall a \in [0,1]$ and $\forall n \in \mathbb{N}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Further, every continuous function is sequentially continuous i.e. $\displaystyle \lim_{n \rightarrow \infty} f(x_n) = f(\lim_{n \rightarrow \infty} x_n) $.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, $\displaystyle \lim_{n \rightarrow \infty} f(x_n) = f(\lim_{n \rightarrow \infty} x_n)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the above arguments, we get that $\forall a \in (0,1]$, $$f(a) = \displaystyle \lim_{n \rightarrow \infty} f(a) = \displaystyle \lim_{n \rightarrow \infty} f(\sqrt[{2^{n}}]{a}) = f(\displaystyle \lim_{n \rightarrow \infty} \sqrt[{2^{n}}]{a}) = f(1)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, $f(a) = f(1)$, $\forall a \in (0,1]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we have that $f(0) = f(a) = f(1)$, $\forall a \in [0,1]$.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2011-02-12T00:42:03.170" LastActivityDate="2011-02-12T00:42:03.170" CommentCount="1" />
  <row Id="19368" PostTypeId="2" ParentId="19366" CreationDate="2011-01-28T16:10:13.200" Score="2" ViewCount="" Body="&lt;p&gt;By using repeated squaring one can compute this &lt;a href=&quot;http://en.wikipedia.org/wiki/Legendre_symbol&quot; rel=&quot;nofollow&quot;&gt;Legendre symbol&lt;/a&gt; in $\ O((lg\ p)^3)\ $ bit operations. Even better, by using &lt;a href=&quot;http://en.wikipedia.org/wiki/Jacobi_symbol&quot; rel=&quot;nofollow&quot;&gt;Jacobi symbols,&lt;/a&gt; one can compute $\:(a|n)\:$ in $\ O((lg\ a)\ (lg\ n))\ $ bit operations. See e.g. Bach &amp;amp; Shallit: &lt;a href=&quot;http://books.google.com/books?id=iJx1lP9ZcIkC&amp;amp;pg=PA110&quot; rel=&quot;nofollow&quot;&gt;Algorithmic Number Theory: Efficient algorithms, pp.110-111&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: for hand computations one often exploits tricks such as noticing certain factorizations, e.g. pulling out factors of 2. The above linked Wiki articles have some examples of such.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2011-01-28T17:27:06.707" LastActivityDate="2011-01-28T17:27:06.707" />
  <row Id="19369" PostTypeId="2" ParentId="19366" CreationDate="2011-01-28T16:24:22.843" Score="3" ViewCount="" Body="&lt;p&gt;You seem to have misspoken. You know that if you square $3^{500001}$ you get $3^{1000002}$, which by Fermat's Little Theorem &lt;em&gt;must&lt;/em&gt; be $1$ modulo $1000003$; there is no possibility of the square of $3^{1000002}$ being $-1$ modulo $1000003$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, since there are only two congruence classes whose square modulo $p$ is $1$, namely $1$ and $-1$, you know that $3^{500001}$ &lt;strong&gt;itself&lt;/strong&gt; is either $1$ or $-1$ modulo $p$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The key to this is Euler's Criterion:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Euler's Criterion.&lt;/strong&gt; If $p$ is an odd prime and $\gcd(a,p)=1$, then&#xA;  $$x^2 \equiv a\pmod{p}$$&#xA;  has two solutions if $a^{(p-1)/2}\equiv 1 \pmod{p}$, and no solutions if $a^{(p-1)/2}\equiv -1 \pmod{p}$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So: $3^{500001} \equiv 1 \pmod{p}$ if and and only if $3$ is a square modulo $1000003$, and $3^{500001}\equiv -1\pmod{p}$ if and only if $3$ is &lt;em&gt;not&lt;/em&gt; a square modulo $1000003$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question then transforms into &quot;How do we tell if $3$ is a square modulo $1000003$?&quot;, or more generally:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Given $a$ and a large prime $P$, $1\lt a\lt P$, how do we tell if $a$ is a square modulo $P$?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The answer is that we do it pretty easily using Legendre or Jacobi symbols, as noted by Bill.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, here, since $3$ and $1000003$ are both congruent to $3$ modulo $4$, quadratic reciprocity says that&#xA;$$\left(\frac{3}{1000003}\right) = -\left(\frac{1000003}{3}\right) = -\left(\frac{1}{3}\right) = -1,$$&#xA;so $3$ is &lt;em&gt;not&lt;/em&gt; a square modulo $1000003$, hence $3^{500001}\equiv -1\pmod{1000003}$. &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-28T17:04:00.330" LastActivityDate="2011-01-28T17:04:00.330" />
  <row Id="19370" PostTypeId="1" AcceptedAnswerId="19377" CreationDate="2011-01-28T16:38:11.830" Score="3" ViewCount="249" Body="&lt;p&gt;Demonstrate by induction: $(1+a)^n ≥1+an$ is true, given a real number $a$, for any $n ∈ \mathbb N$. With $a &amp;gt; 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need to demostre this using the induction principle. My doubt is in the second part of the demonstration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is what I have:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to demonstrate the predicate these two points must be true:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$P(1)$ must be True.&lt;/li&gt;&#xA;&lt;li&gt;$∀ n ∈ \mathbb N , P(n) =&amp;gt; P(n+1)$ is true.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;We demonstrate the first point:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(1+a)^1 ≥ 1+a*1$, &#xA;$1+a ≥ 1+a$  So it is true&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the second part is where I have the problem. I do not know what to do. I understand the theory but I don't know how to apply it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(1+a)^{n+1} ≥ 1+a(n+1)$    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I don't see that as useful. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any tips?&lt;/p&gt;&#xA;" OwnerUserId="5526" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-28T17:01:44.777" LastActivityDate="2011-02-01T01:12:58.400" Title="demonstration by induction: $(1+a)^n ≥1+an$" Tags="&lt;induction&gt;" AnswerCount="4" CommentCount="3" />
  <row Id="19372" PostTypeId="2" ParentId="19212" CreationDate="2011-01-28T16:39:39.633" Score="0" ViewCount="" Body="&lt;p&gt;Here's a hint extracted from one of my old posts (which shows even more)&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Even  easier: the ideals $\rm\ \ (x)\:\ &amp;lt;\ (x,y)\ &amp;lt;\ 1\ $ are distinct primes$\ \ \ $  (or $1$)&lt;br&gt;&#xA;  since the residue rings $\rm\ \mathbb Q[y]\ &amp;gt;\ \ \ \mathbb Q\ \ \ \ &amp;gt;\ 0\ $ are distinct domains (or $0$)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="242" LastActivityDate="2011-01-28T16:39:39.633" CommentCount="2" />
  <row Id="19373" PostTypeId="2" ParentId="19370" CreationDate="2011-01-28T16:44:02.483" Score="4" ViewCount="" Body="&lt;p&gt;Hint:  Now you assume that $(1+a)^n \ge 1 + an$ and need to prove that $(1+a)^{(n+1)} \ge 1 + a(n+1)$.  So $(1+a)^{(n+1)}=(1+a)^n*(1+a) \ge $ what?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Added:$(1+a)^{(n+1)}=(1+a)^n*(1+a) \ge (1+an)(1+a)$ by the inductive hypothesis&#xA;$(1+an)(1+a)=1+a(n+1)+a^2n \ge 1+a(n+1)$&#xA;So, as Arturo showed in his example, we have proven that if the relation is true for $n$, it is also true for $n+1$.  Given that you verified the base case of $n=1$, it is true for all $n$&lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2011-01-31T19:42:16.723" LastActivityDate="2011-01-31T19:42:16.723" CommentCount="7" />
  <row Id="19374" PostTypeId="1" CreationDate="2011-01-28T16:54:34.893" Score="0" ViewCount="67" Body="&lt;p&gt;I want to show some users of a piece of software some solutions of one-dimension boundary value problems (can also be initial value problems). I'm after a collection of problems whose solutions are very interesting or the BVP are of a particularly interesting nature.  The problems can be linear or non-linear. Does anyone have any suggests?&lt;/p&gt;&#xA;" OwnerUserId="1794" LastActivityDate="2011-01-28T18:42:28.540" CommunityOwnedDate="2011-01-28T18:31:41.973" Title="One dimensional boundary value problems showing interesting behaviour." Tags="&lt;differential-equations&gt;&lt;examples&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="19375" PostTypeId="2" ParentId="19370" CreationDate="2011-01-28T17:08:19.403" Score="5" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\ $ Put $\rm\ \ x = 1+a\ \ $ in $\rm\displaystyle\ \ \frac{x^n -1}{x-1}\ =\ x^{n-1}+\:\cdots\: +x+1\ \ge\ n\ $ for $\rm\ x\ge 1$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many induction problems are best solved in this manner, i.e. by transforming them into a form that makes the induction completely trivial.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2011-01-28T17:18:41.377" LastActivityDate="2011-01-28T17:18:41.377" CommentCount="2" />
  <row Id="19376" PostTypeId="2" ParentId="19374" CreationDate="2011-01-28T17:26:06.540" Score="1" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://people.uncw.edu/hermanr/mat463/ODEBook/Book/BVPs.pdf&quot; rel=&quot;nofollow&quot;&gt;This&lt;/a&gt;(site about boundary value problems) should be useful. Some of the interesting boundary value problems they list are the following:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Heat Equation&lt;/li&gt;&#xA;&lt;li&gt;Wave Equation&lt;/li&gt;&#xA;&lt;li&gt;Laplace Equation&lt;/li&gt;&#xA;&lt;li&gt;Poisson Equation&lt;/li&gt;&#xA;&lt;li&gt;Schrodigner's Equation&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2011-01-28T17:26:06.540" CommunityOwnedDate="2011-01-28T18:31:41.973" />
  <row Id="19377" PostTypeId="2" ParentId="19370" CreationDate="2011-01-28T17:26:15.363" Score="9" ViewCount="" Body="&lt;p&gt;Ross has pretty much given you all the hints necessary to answer this particular question, so let me instead give you some general advice on proofs by induction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In doing the induction, especially at first and until you get very comfortable with induction, be very explicit. Label the base &lt;strong&gt;Base.&lt;/strong&gt; Label the inductive step as &quot;&lt;strong&gt;Inductive step.&lt;/strong&gt;&quot; State explicitly what the &lt;em&gt;Induction Hypothesis&lt;/em&gt; is, state explicitly what you want to prove. Use this as a way to organize your thoughts and have the information at the ready. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second: The key to most proofs by induction is to take the case you need to prove, and to somehow &lt;em&gt;reduce it&lt;/em&gt; to &quot;the previous case plus something extra&quot;; then one applies the inductive hypothesis to simplify/answer the part that is the previous case, and then deal with the &quot;something extra.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's an example different from the one at hand, so you can see what I mean.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the following:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Prove that for all natural numbers $n$, $$1\cdot 2 + 2\cdot 3 + \cdots + n(n+1) = \frac{n(n+1)(n+2)}{3}.$$&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;em&gt;Proof.&lt;/em&gt; We proceed by indution on $n$.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;Base.&lt;/strong&gt; We prove the statement for $n=1$: indeed, $1\cdot 2 = \frac{1(2)(3)}{3}$. &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;Inductive step.&lt;/strong&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;em&gt;Induction Hypothesis.&lt;/em&gt; We assume the result holds for $k$. That is, we assume that&#xA;  $$ 1\cdot 2 + 2\cdot 3 + \cdots + k(k+1) = \frac{k(k+1)(k+2)}{3}$$&#xA;  is true.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;em&gt;To prove:&lt;/em&gt; We need to show that the result holds for $k+1$, that is, that &#xA;  $$ 1 \cdot 2 + 2\cdot 3 + \cdots + (k+1)(k+2) = \frac{(k+1)\bigl((k+1)+1\bigr)\bigl((k+1)+2\bigr)}{3}.$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(Now comes the point where we take $1\cdot 2 + 2\cdot 3 + \cdots + (k+1)(k+2)$ and try to think of it as &quot;the $k$-case plus something extra&quot;). &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;\begin{align*}&#xA;  &amp;amp;{1\cdot 2 + 2\cdot 3 + \cdots + (k+1)(k+2)}\\&#xA;  \quad &amp;amp;= \Bigl(1\cdot 2+ 2\cdot 3+\cdots + k(k+1)\Bigr) + (k+1)(k+2) &amp;amp;&amp;amp;\mbox{(associativity of $+$)}\\&#xA;  \quad&amp;amp;= \frac{k(k+1)(k+2)}{3} + (k+1)(k+2) &amp;amp;&amp;amp;\mbox{(This is the induction hypothesis!)}\\&#xA;  &amp;amp;= \frac{k(k+1)(k+2)}{3} + \frac{3(k+1)(k+2)}{3} &amp;amp;&amp;amp;\mbox{(Just algebra)}\\&#xA;  &amp;amp;= \frac{(k+1)(k+2)\bigl(k+3\bigr)}{3} &amp;amp;&amp;amp;\mbox{(factor out $(k+1)(k+2)$)}\\&#xA;  &amp;amp;= \frac{(k+1)\bigl( (k+1)+1\bigr)\bigl((k+1)+2\bigr)}{3}.&#xA;  \end{align*}&#xA;  Thus, $1\cdot 2 + 2\cdot 3 + \cdots + (k+1)(k+2) = \frac{(k+1)\bigl((k+1)+1\bigr)\bigl((k+1)+2\bigr)}{3}$. This proves the inductive step.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Since the statement holds for $n=1$, and if it holds for $k$ then it holds for $k+1$, then by mathematical induction we conclude that &#xA;  $$1\cdot 2 + 2\cdot 3 + \cdots + n(n+1) = \frac{n(n+1)(n+2)}{3}$$&#xA;  for all natural numbers $n$. &lt;strong&gt;QED&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So: be very explicit and clear, to help organize your thoughts. As you get more comfortable with induction, you can leave off makign explicit statement of what you need to prove, etc., but until you are &lt;em&gt;very&lt;/em&gt; comfortable, better to be explicit than to be confused.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-28T18:15:35.477" LastActivityDate="2011-01-28T18:15:35.477" CommentCount="6" />
  <row Id="19378" PostTypeId="2" ParentId="19334" CreationDate="2011-01-28T17:51:15.500" Score="0" ViewCount="" Body="&lt;p&gt;There might be a similar proof in an excellent textbook: &quot;Euclidean and Non-Euclidean Geometries&quot; by Marvin J. Greenberg.  The first several chapters are devoted to neutral geometry.&lt;/p&gt;&#xA;" OwnerUserId="6243" LastActivityDate="2011-01-28T17:51:15.500" CommentCount="1" />
  <row Id="19379" PostTypeId="1" AcceptedAnswerId="19437" CreationDate="2011-01-28T18:03:21.163" Score="1" ViewCount="96" Body="&lt;p&gt;What's $\mathrm{Spec}\mathbf{Z}[\mu_{p-1},\frac{1}{(p-1)p}] \cap \mathbf{Z}_p$, where $\mu_{p-1}$ denotes the $(p-1)$-th roots of unity and $\mathbf{Z}_p$ the $p$-adic integers?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It should be $\mathbf{Z}[\mu_{p-1}]$ without (the places above the primes dividing $p-1$ and all but one places above $p$).&lt;/p&gt;&#xA;" OwnerUserId="5262" LastActivityDate="2011-01-29T05:11:13.600" Title="understanding $\mathrm{Spec}\mathbf{Z}[\mu_{p-1},\frac{1}{(p-1)p}] \cap \mathbf{Z}_p$" Tags="&lt;commutative-algebra&gt;" AnswerCount="1" />
  <row Id="19380" PostTypeId="1" CreationDate="2011-01-28T18:13:04.360" Score="2" ViewCount="96" Body="&lt;p&gt;Hi, how can I use prime avoidance to prove that If $R$ is a Noetherian ring and $J\subset R$ an ideal with $\mathrm{height}\;J=n$ then there exists $x_1,\ldots,x_n\in J$ such that $\mathrm{height}(x_1,\ldots,x_i)=i$ for all $i=0,\ldots,n$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that if $R$ is local and Noetherian of dimension $d$ then there exist $x_1,\ldots,x_d$ such that $\mathrm{height}(x_1,\ldots,x_i)\geq i$ for all $i=1\ldots,d$.&lt;/p&gt;&#xA;" OwnerUserId="6236" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-28T18:18:50.403" LastActivityDate="2011-01-28T18:18:50.403" Title="Prime avoidance" Tags="&lt;abstract-algebra&gt;&lt;commutative-algebra&gt;&lt;ring-theory&gt;" FavoriteCount="1" />
  <row Id="19381" PostTypeId="2" ParentId="19211" CreationDate="2011-01-28T18:16:34.167" Score="1" ViewCount="" Body="&lt;p&gt;Using V[i:j] to represent items i through j of the vector (not i though j-1 as in Python), I think you want to find i and j such that j-1 is maximized and the number of distinct entries is k (because if it were less than k, you could increase j or decrease i).  My naive thought would be to start with i=1, j=k and make a table (kx2) with the distinct entries of V in the first column  and the count in the current interval in the second.  If you don't have k distinct entries, keep adding to j, updating the table, until you do.  Record i and j as the best seen yet.  Then increase i to 2, updating your table, and see if you can increase j.  Continue through V, at each step seeing if the gap has increased.  If so, replace the stored best so far i and j.  I don't think this is worse than O(nk)-the original sort is k ln k, and you make only one pass through the n items.  For each new item, you might have to look through k lines in the table once or twice, but it shouldn't be k^2.  Does that work?&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-01-28T18:16:34.167" />
  <row Id="19382" PostTypeId="2" ParentId="19374" CreationDate="2011-01-28T18:42:28.540" Score="1" ViewCount="" Body="&lt;p&gt;In terms of the initial value problem, a large class of interesting 1-dimensional questions arise from studying various approximations to &lt;a href=&quot;http://en.wikipedia.org/wiki/Surface_wave&quot; rel=&quot;nofollow&quot;&gt;surface waves&lt;/a&gt; starting from fundamental equations of fluid dynamics. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Korteweg%E2%80%93de_Vries_equation&quot; rel=&quot;nofollow&quot;&gt;Korteweg-de Vries equation and its cousins gKdV and mKdV&lt;/a&gt; all lead to demonstrations of a competition between cohesive self-interaction that leads to soliton formation, and dispersive wave-like phenomenon. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(What is especially interesting is to look at the purely dispersive part of the KdV equation &#xA;$$ \partial_t \phi + \partial_x^3 \phi = 0 $$&#xA;and the purely transport part of the equation&#xA;$$ \partial_t \phi + \phi\partial_x\phi = 0 $$&#xA;and note their characteristic behaviours. Then compare it against various different initial data for the KdV equation.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can also consider looking at the &lt;a href=&quot;http://en.wikipedia.org/wiki/Benjamin%E2%80%93Bona%E2%80%93Mahony_equation&quot; rel=&quot;nofollow&quot;&gt;Benjamin-Bona-Mahony equation&lt;/a&gt;, which shows off some long wavelength behaviour (in constrast to the KdV solitons, which tends to be more spatially concentrated). &lt;/p&gt;&#xA;" OwnerUserId="1543" LastActivityDate="2011-01-28T18:42:28.540" CommunityOwnedDate="2011-01-28T18:42:28.540" CommentCount="1" />
  <row Id="19383" PostTypeId="2" ParentId="19352" CreationDate="2011-01-28T18:48:40.660" Score="2" ViewCount="" Body="&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Ulam_number&quot; rel=&quot;nofollow&quot;&gt;Ulam numbers&lt;/a&gt; are a good example of a seemingly &quot;random&quot; sequence, which certainly satisfies the Goldbach requirement (every positive integer is the sum of two Ulam numbers) and I believe it is conjectured that there are infinitely many pairs n, n+1 of Ulam numbers (I conjecture it, at least).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is not hard to invent infinitely many sequences with sufficiently high natural density for which analogues to Goldbach and the Twin Prime conjectures can be expected to apply.  Give it a try!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might enjoy Halberstam and Roth's book &lt;em&gt;Sequences&lt;/em&gt; if you are interested in these sorts of features of sequences.&lt;/p&gt;&#xA;" OwnerUserId="2937" LastActivityDate="2011-01-28T18:48:40.660" />
  <row Id="19384" PostTypeId="1" CreationDate="2011-01-28T19:03:10.987" Score="-1" ViewCount="137" Body="&lt;p&gt;What is the modern classification of linear operators?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Especially interesting regarding linear operators on numerical functions.&lt;/p&gt;&#xA;" OwnerUserId="2513" LastActivityDate="2011-01-28T19:52:55.380" ClosedDate="2011-01-29T00:43:58.000" Title="What is the classification of linear operators?" Tags="&lt;linear-algebra&gt;&lt;functional-analysis&gt;" AnswerCount="1" CommentCount="13" />
  <row Id="19385" PostTypeId="2" ParentId="9728" CreationDate="2011-01-28T19:14:02.480" Score="0" ViewCount="" Body="&lt;p&gt;Suppose $g&amp;#39;(1)&amp;lt;1$. Let $f(x)=g(x)-x$. Then if $x\in(0,1), f&amp;#39;(x)=g&amp;#39;(x)-1&amp;lt; g&amp;#39;(1)-1&amp;lt;1-1=0$, since g'(x) is increasing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence $g(x)-x=f(x)&amp;gt;f(1)=g(1)-1&amp;gt;0 \;\;\;\forall x\in(0,1)$, since by the preceding f is strictly decreasing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now assume $f(x)=g(x)-x\neq0 \;\;\;\forall x\in (0,1)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then $g(x)&amp;gt;x \;\;\;\forall x\in(0,1)$ (otherwise $\exists x_0$ such that $f(x_0)&amp;lt;0$ and since $f(0)&amp;gt;0$ we'd have a contradiction to the IVT).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, $\forall x\in(0,1), \frac{g(1)-g(x)}{1-x}&amp;lt;\frac{g(1)-x}{1-x}=1$. It follows $g&amp;#39;(1)\leq1$.&lt;/p&gt;&#xA;" OwnerUserId="5959" LastActivityDate="2011-01-28T19:14:02.480" />
  <row Id="19386" PostTypeId="1" AcceptedAnswerId="19387" CreationDate="2011-01-28T19:18:09.603" Score="4" ViewCount="144" Body="&lt;p&gt;Is it possible to form a closed loop by joining regular (platonic) tetrahedrons together side-to-side, with each tetrahedron having two neighbours? It should be a loop with a hole in, as can be done with 8 cubes, or 8 dodecahedrons as shown below. What is the minimum number of tetrahedrons needed?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/3896K.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit:&#xA;Could it be that it is possible to create such a ring by allowing the tetrahedrons to extend in one more spatial dimension (R^4)?&lt;/p&gt;&#xA;" OwnerUserId="5904" LastEditorUserId="5904" LastEditorDisplayName="" LastEditDate="2011-01-28T21:53:43.450" LastActivityDate="2011-01-29T01:52:59.533" Title="Tetrahedral torus" Tags="&lt;geometry&gt;&lt;topology&gt;" AnswerCount="3" />
  <row Id="19387" PostTypeId="2" ParentId="19386" CreationDate="2011-01-28T19:38:32.660" Score="7" ViewCount="" Body="&lt;p&gt;No, it isn't possible.  Here is the reference, which I haven't found online (maybe someone can find a link):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mason, J. (1972). &quot;Can Regular Tetrahedra Be Glued Together Face to Face to Form a Ring?&quot; Mathematical Gazette 56 (397) p.194-197.&lt;/p&gt;&#xA;" OwnerUserId="5546" LastActivityDate="2011-01-28T19:38:32.660" CommentCount="2" />
  <row Id="19389" PostTypeId="2" ParentId="19384" CreationDate="2011-01-28T19:52:55.380" Score="4" ViewCount="" Body="&lt;p&gt;There are any number of classes of linear operators that people are interested in, depending on what situation you are in. Some examples:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Diagonalizable_matrix&quot; rel=&quot;nofollow&quot;&gt;&lt;strong&gt;Diagonalizable operators,&lt;/strong&gt; &lt;/a&gt; operators for which there is a basis of eigenvalues.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Projection_%28linear_algebra%29&quot; rel=&quot;nofollow&quot;&gt;&lt;strong&gt;Projection operators,&lt;/strong&gt;&lt;/a&gt; operators $T$ such that $T^2 = T$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;If your space comes equipped with an inner product, one has:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Hermitian_operator&quot; rel=&quot;nofollow&quot;&gt;&lt;strong&gt;Hermitian&lt;/strong&gt;&lt;/a&gt; or &lt;strong&gt;Self adjoint&lt;/strong&gt; operators.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Normal_operator&quot; rel=&quot;nofollow&quot;&gt;&lt;strong&gt;Normal operators&lt;/a&gt;.&lt;/strong&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Unitary_operator&quot; rel=&quot;nofollow&quot;&gt;&lt;strong&gt;Unitary operators&lt;/a&gt;.&lt;/strong&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Orthogonal projections.&lt;/strong&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;For normed spaces you have &lt;strong&gt;bounded operators.&lt;/strong&gt; Etc. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this more or less what you are thinking of?&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2011-01-28T19:52:55.380" CommentCount="14" />
  <row Id="19390" PostTypeId="1" AcceptedAnswerId="19394" CreationDate="2011-01-28T20:23:39.513" Score="3" ViewCount="156" Body="&lt;p&gt;Hi, can someone tell me a way of integrating functions like $e^{f(x)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a specific case: $\int e^{-3x}\,dx$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS: I'm not looking for the answer of this, but the way of doing it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for your help&lt;/p&gt;&#xA;" OwnerUserId="6350" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-28T20:26:36.737" LastActivityDate="2011-01-28T20:56:42.220" Title="Integrating $e^{f(x)}$" Tags="&lt;calculus&gt;&lt;integral&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="19391" PostTypeId="2" ParentId="19390" CreationDate="2011-01-28T20:30:44.673" Score="5" ViewCount="" Body="&lt;p&gt;If you mean a way to obtain an anti-derivative in terms of elementary functions, there is no such general algorithm: it is known that for $f(x) = -x^2$, $\int e^{f(x)}$ cannot be written in terms of elementary functions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are some general algorithms for computing anti-derivatives though, for instance: &lt;a href=&quot;http://en.wikipedia.org/wiki/Risch_algorithm&quot; rel=&quot;nofollow&quot;&gt;Risch's algorithm&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your specific case is much easier than what you have generalized your problem too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint: What is the derivative of $e^{-3x}$ ?&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2011-01-28T20:30:44.673" CommentCount="1" />
  <row Id="19392" PostTypeId="2" ParentId="19386" CreationDate="2011-01-28T20:34:05.390" Score="3" ViewCount="" Body="&lt;p&gt;One can get interesting geometrical &quot;rings&quot; of tetrahedra when they are attached edge to edge: &lt;a href=&quot;http://www.mathematische-basteleien.de/kaleidocycles.htm&quot; rel=&quot;nofollow&quot;&gt;http://www.mathematische-basteleien.de/kaleidocycles.htm&lt;/a&gt; On this page some of the examples use regular tetrahedra and others less regular ones.&lt;/p&gt;&#xA;" OwnerUserId="1369" LastActivityDate="2011-01-28T20:34:05.390" />
  <row Id="19393" PostTypeId="2" ParentId="17284" CreationDate="2011-01-28T20:36:40.427" Score="3" ViewCount="" Body="&lt;p&gt;If I understand correctly, your vector space is $\mathbb{C}^{\mathbb{C}}$, the space of all functions $f: \mathbb{C} \rightarrow \mathbb{C}$.  As a $\mathbb{C}$-vector space this has dimension $c^c = 2^c$.  You have also written down $2^c$ different linear operators.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the space $\operatorname{End} \mathbb{C}^{\mathbb{C}}$ has dimension at least $2^{2^c}$.  Thus you are incredibly far from having written down all the linear operators.  To get the right cardinal number, consider all possible permutations of a basis of $\mathbb{C}^{\mathbb{C}}$.  &lt;/p&gt;&#xA;" OwnerUserId="299" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2011-01-28T22:17:10.560" LastActivityDate="2011-01-28T22:17:10.560" CommentCount="6" />
  <row Id="19394" PostTypeId="2" ParentId="19390" CreationDate="2011-01-28T20:38:43.437" Score="1" ViewCount="" Body="&lt;p&gt;For the case where $f(x)$ is linear, a nice $u$-substitution works. I assume you know how to integrate $\int e^xdx$? So in order to integrate a function of the form $e^{f(x)}$, let $u=f(x)$, and thus $du=f&amp;#39;(x)dx$, which allows you to 'solve' for $dx$ in terms of $du$. Then your original integral goes from form&#xA;$$&#xD;&#xA;\int e^{f(x)}dx&#xD;&#xA;$$&#xA;to&#xA;$$&#xD;&#xA;\int \frac{e^u}{f&amp;#39;(x)}du.&#xD;&#xA;$$&#xA;Of course, this is not always so easy to integrate, as Moron points out. When $f(x)$ is linear, you have a nice situation, because $f&amp;#39;(x)$ is just a constant. Other situations may not be so easily handled, as far as I'm aware.&lt;/p&gt;&#xA;" OwnerUserId="1583" LastEditorUserId="1583" LastEditorDisplayName="" LastEditDate="2011-01-28T20:56:42.220" LastActivityDate="2011-01-28T20:56:42.220" CommentCount="3" />
  <row Id="19395" PostTypeId="1" AcceptedAnswerId="19408" CreationDate="2011-01-28T20:49:48.913" Score="6" ViewCount="194" Body="&lt;p&gt;I found an interesting topic on this site with regards to the series I am trying to evaluate:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://math.stackexchange.com/questions/8884/&quot;&gt;Summing $\frac{1}{e^{2\pi}-1} + \frac{2}{e^{4\pi}-1} + \frac{3}{e^{6\pi}-1} + \cdots \text{ad inf}$&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was wondering if there is a closed form for even m when we have:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{n=1}^{\infty}\frac{n^{2m-1}}{e^{2\pi n}-1}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have the series $$\sum_{n=1}^{\infty}\frac{n^{7}}{e^{2\pi n}-1},$$&#xA;I am trying to evaluate. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The aforementioned thread mentions that if m &gt; 1 and odd, then we can use&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{B_{2m}}{4m}$$ to find the sum. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, if m is even, the formula omits and error term.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anyone have info on this error term or how to evaluate my series or others that involve even m?. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I noticed that when m=1, there is an error term of $$\frac{-1}{8\pi}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The error appears to get smaller the larger m, and thus the power of n, becomes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks to all.&lt;/p&gt;&#xA;" OwnerUserId="6358" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-28T21:14:57.350" LastActivityDate="2011-03-28T04:38:26.640" Title="infinite series $n^7/(\exp(2\pi n)-1)$" Tags="&lt;sequences-and-series&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="19396" PostTypeId="2" ParentId="19390" CreationDate="2011-01-28T20:52:49.090" Score="0" ViewCount="" Body="&lt;p&gt;Interesting, I fired my old symbolic algebra program and typed the following&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \int{\exp\left(C_{0}+C_{1}x-C_{2}x^{2}\right)}\,\mathrm{d}x= $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and it gave the answer&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ =\sqrt{\frac{\pi}{4C_{2}}}\exp\left(C_{0}+\frac{C_{1}^{2}}{4C_{2}}\right)\mathrm{erf}\left(\frac{C_{1}}{2\sqrt{C_{2}}}-\sqrt{C_{2}}\, x\right) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or more simply&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \int{\exp\left(-x^2\right)}\,\mathrm{d}x = \frac{\sqrt{\pi}}{2}\,\mathrm{erf}(x) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with $\mathrm{erf}(x)$ the &lt;a href=&quot;http://en.wikipedia.org/wiki/Error_function&quot; rel=&quot;nofollow&quot;&gt;error-function&lt;/a&gt; which is &lt;strong&gt;not&lt;/strong&gt; an elementary function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So although a general solution may not exist, I am sure there are special cases where as solution is feasible.&lt;/p&gt;&#xA;" OwnerUserId="3301" LastActivityDate="2011-01-28T20:52:49.090" />
  <row Id="19397" PostTypeId="1" AcceptedAnswerId="19398" CreationDate="2011-01-28T20:59:39.437" Score="12" ViewCount="167" Body="&lt;p&gt;I was recently wondering where the word `kernel' comes from in mathematics.  I am sure the internet must know.  I did manage to find &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.pballew.net/etyindex.html#k&quot; rel=&quot;nofollow&quot;&gt;http://www.pballew.net/etyindex.html#k&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which contains the origin of many math words but it doesn't have `kernel'; it also only seems to address math words you might run into in at most a 1st or 2nd year college math course.  I'm wondering if anyone is aware of a reference that might address the origin of more modern math words (but not so modern that it doesn't discuss kernel). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: By `math etymology' I mean why a particular word was used for a certain math concept.  Of course such an explanation might involve discussing the actual etymology of the word.&lt;/p&gt;&#xA;" OwnerUserId="403" LastEditorUserId="403" LastEditorDisplayName="" LastEditDate="2011-01-28T21:32:04.737" LastActivityDate="2011-01-28T21:32:04.737" Title="A place to learn about math etymology?" Tags="&lt;reference-request&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" />
  <row Id="19398" PostTypeId="2" ParentId="19397" CreationDate="2011-01-28T21:05:41.300" Score="7" ViewCount="" Body="&lt;p&gt;You'll want to check out &lt;a href=&quot;http://jeff560.tripod.com/mathword.html&quot; rel=&quot;nofollow&quot;&gt;Earliest Known Uses of Some of the Words of Mathematics&lt;/a&gt;, maintained by Jeff Miller.  In particular, an entry for 'kernel' appears &lt;a href=&quot;http://jeff560.tripod.com/k.html&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2011-01-28T21:05:41.300" CommunityOwnedDate="2011-01-28T21:05:41.300" CommentCount="2" />
  <row Id="19399" PostTypeId="2" ParentId="19397" CreationDate="2011-01-28T21:07:54.347" Score="1" ViewCount="" Body="&lt;p&gt;It's unclear if you want to discuss the reason for a particular choice of a word (whether coined or borrowed), or the etymological meaning of the word. For example, &quot;kernel&quot; is a perfectly fine English word, refering to a whole seed of a cereal (the word coming from the Indo-European root &lt;em&gt;greno-&lt;/em&gt;). But the etyomological origin is somewhat different from the reason why it was chosen to refer to the particular mathematical concept it refers to. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One source for the etymology (and sometimes mathematical origin) is &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0883855119&quot; rel=&quot;nofollow&quot;&gt;&lt;strong&gt;The Words of Mathematics:&lt;/strong&gt;&lt;/a&gt; &lt;em&gt;An Etymological Dictionary of Mathematical Terms Used in English&lt;/em&gt;, by Steven Schwartzmann, published by the Mathematical Association of America. It will tell you things like what the Latin origin of &quot;conjecture&quot; is, and so on. It does not provide much in the way of historical mathematical origin (i.e., why the word was chosen for that particular concept). Sometimes this is clear from the origin of the word, but often this is not the case. &lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2011-01-28T21:07:54.347" CommentCount="2" />
  <row Id="19400" PostTypeId="1" AcceptedAnswerId="19421" CreationDate="2011-01-28T21:17:06.273" Score="8" ViewCount="195" Body="&lt;p&gt;I've been giving a public talk about Art and Mathematics for a few years now as part of my University's outreach program. Audience members are usually well-educated but may not have much knowledge of math. I've been concentrating on explaining Richard Taylor's analysis of Jackson Pollock's work using fractal dimension, but I'm looking to revise the talk, and wonder if anyone has some good ideas about what might go in it. M.C. Escher and Helaman Ferguson's work are some obvious possibilities, but I'd like to hear other ideas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: I'd like to thank the community for their suggestions, and report back that Kaplan and Bosch's TSP art was a real crowd pleaser. The audience was fascinated by the idea that the Mona Lisa picture was traced out by a single intricate line. I also mentioned Tony Robbin and George Hart, which were well-received as well.&lt;/p&gt;&#xA;" OwnerUserId="2941" LastEditorUserId="2941" LastEditorDisplayName="" LastEditDate="2011-04-06T22:56:36.697" LastActivityDate="2011-04-06T22:56:36.697" CommunityOwnedDate="2011-01-29T11:39:42.757" Title="Suggestions for topics in a public talk about art and mathematics" Tags="&lt;art&gt;&lt;public-outreach&gt;" AnswerCount="6" CommentCount="3" />
  <row Id="19401" PostTypeId="1" CreationDate="2011-01-28T21:18:59.093" Score="1" ViewCount="100" Body="&lt;p&gt;I would like to gain some intuition regarding the modules of Kahler differentials $\Omega^j_{A/k}$ of an affine algebra $A$ over a (say - algebraically closed) field $k$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let us recall the definition: let $A^e = A\otimes_k A$, let $f:A^e\to A$ be the map defined by $f(a\otimes b) = ab$, and let $I = \ker f$. Then $\Omega^1_{A/k} = I/I^2$. And, $\Omega^j_{A/k} = \wedge^j \Omega^1_{A/k}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An important theorem regarding Kahler differentials says: If $k \to A$ is smooth of relative dimension $n$, then $\Omega^n_{A/k}$ is a projective module of finite rank.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question: I was wondering if anyone could provide some examples of:&#xA;1. How does the module of kahler differentials looks for some singular varieties? for example, what is $\Omega^1_{A/k}$ for $A = k[x,y]/(y^2-x^3)$.&#xA;2. Can anyone provide an example of a nonsingular affine variety with coordinate ring $A$, such that $k \to A$ is smooth of relative dimension $n$, but $\Omega^n_{A/k}$ is projective but not free?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would be happy for any concrete example that will help my intuition on the subject.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="5491" LastActivityDate="2011-01-28T21:45:16.177" Title="Kahler differentials of affine varieties" Tags="&lt;algebraic-geometry&gt;&lt;commutative-algebra&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="2" />
  <row Id="19402" PostTypeId="2" ParentId="19400" CreationDate="2011-01-28T21:20:33.167" Score="2" ViewCount="" Body="&lt;p&gt;I heard &lt;a href=&quot;http://www.math.brown.edu/~banchoff/&quot; rel=&quot;nofollow&quot;&gt;Thomas Banchoff&lt;/a&gt; give a nice talk about Salvador Dali's work a few years ago.  Apparently they were even friends.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a &lt;a href=&quot;http://vimeo.com/10878800&quot; rel=&quot;nofollow&quot;&gt;link&lt;/a&gt; to a lecture by Banchoff on Dali.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It looks like Banchoff wrote a paper in &lt;strike&gt;Spanish&lt;/strike&gt; Catalan on Dali, too: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;La Quarta Dimensio i Salvador Dali,&quot;  Breu Viatge al mon de la Mathematica, 1 (1984), pp. 19-24. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even with my &lt;strike&gt;poor Spanish&lt;/strike&gt; nonexistent Catalan skills I can translate the title as &quot;The Fourth Dimension and (or &quot;in&quot;?) Salvador Dali.&quot; :)&lt;/p&gt;&#xA;" OwnerUserId="2370" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2011-01-28T23:59:06.710" LastActivityDate="2011-01-28T23:59:06.710" CommunityOwnedDate="2011-01-29T11:39:42.757" CommentCount="4" />
  <row Id="19403" PostTypeId="2" ParentId="19400" CreationDate="2011-01-28T21:30:27.063" Score="3" ViewCount="" Body="&lt;p&gt;There is a lot of beautiful mathematics (mostly, if not all geometry) in perspective drawing.  I've heard Annalisa Crannell give beautiful lectures on this.  Unfortunately, her book isn't due out until July 2011 (you can find it on amazon, and I can only include one link).  There is a nice description of a public lecture she gave at the MAA, with audio of the lecture and a few embedded videos:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://maa.org/news/092310Crannell.html&quot; rel=&quot;nofollow&quot;&gt;http://maa.org/news/092310Crannell.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="6361" LastActivityDate="2011-01-28T21:30:27.063" CommunityOwnedDate="2011-01-29T11:39:42.757" />
  <row Id="19404" PostTypeId="2" ParentId="19400" CreationDate="2011-01-28T21:33:59.130" Score="6" ViewCount="" Body="&lt;p&gt;I also can't resist mentioning Kaplan and Bosch's work using the traveling salesman problem to reproduce famous artworks; i.e., TSP art.  See these links:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.cgl.uwaterloo.ca/~csk/projects/tsp/&quot; rel=&quot;nofollow&quot;&gt;http://www.cgl.uwaterloo.ca/~csk/projects/tsp/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.oberlin.edu/math/faculty/bosch/tspart-page.html&quot; rel=&quot;nofollow&quot;&gt;http://www.oberlin.edu/math/faculty/bosch/tspart-page.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.tsp.gatech.edu/data/art/index.html&quot; rel=&quot;nofollow&quot;&gt;http://www.tsp.gatech.edu/data/art/index.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, this reproduction of the Mona Lisa is actually a solution to a traveling salesman problem.  There's a single path that creates the image. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/tl4GQ.gif&quot; alt=&quot;Mona Lisa&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="2370" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2011-01-28T21:39:17.050" LastActivityDate="2011-01-28T21:39:17.050" CommunityOwnedDate="2011-01-29T11:39:42.757" CommentCount="3" />
  <row Id="19405" PostTypeId="2" ParentId="19401" CreationDate="2011-01-28T21:35:10.510" Score="4" ViewCount="" Body="&lt;p&gt;Let's do your example (1).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a short exact sequence $$0\to A\to A\mathrm dx\oplus A\mathrm dy\to\Omega^1\to 0$$ in which the first map maps $1$ to $2x\,\mathrm dx+3y\,\mathrm dy$. It follows that $\Omega^1$ is isomorphic to the module $$\frac{A\oplus A}{(2x,3y)}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the characteristic is $2$ or $3$, this looks like $A\oplus A/(y)$ or $A\oplus A/(y)$. If the characteristic is not one of those two, then it is clear, from this description, that if you localize at a prime not containing $x$ or $y$, then you get a free module. Now localize at one of the other ideals, and see what you get!&lt;/p&gt;&#xA;" OwnerUserId="274" LastEditorUserId="274" LastEditorDisplayName="" LastEditDate="2011-01-28T21:45:16.177" LastActivityDate="2011-01-28T21:45:16.177" CommentCount="1" />
  <row Id="19406" PostTypeId="1" CreationDate="2011-01-28T21:00:28.963" Score="1" ViewCount="38" Body="&lt;p&gt;Taken a break from my personal project and getting some work done. I would like to find some examples, tutorials on relation algebra.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anything with good examples will be very useful&lt;/p&gt;&#xA;" LastActivityDate="2011-02-27T22:39:16.977" Title="Where can I find good information, tutorials and or examples on relational algebra queries?" Tags="&lt;algebra&gt;&lt;relational-algebra&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="19407" PostTypeId="2" ParentId="19406" CreationDate="2011-01-28T21:05:40.900" Score="1" ViewCount="" Body="&lt;p&gt;Here is the one i use:&#xA;&lt;a href=&quot;http://db.grussell.org/section010.html&quot; rel=&quot;nofollow&quot;&gt;Relational Algebra&lt;/a&gt;&lt;/p&gt;&#xA;" LastActivityDate="2011-01-28T21:05:40.900" />
  <row Id="19408" PostTypeId="2" ParentId="19395" CreationDate="2011-01-28T21:55:35.477" Score="0" ViewCount="" Body="&lt;p&gt;Take a look at equation $(6)$ in&#xA;&lt;a href=&quot;http://arxiv.org/abs/math/0505125&quot; rel=&quot;nofollow&quot;&gt;Ramanujan's Formula for the Logarithmic Derivative of the Gamma Function&lt;/a&gt;&#xA;by David Bradley.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that just before the formula it says, &quot;Let $N$ be a positive integer,&quot; but the formula&#xA;is valid for negative $N$ as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: To answer Cody's question in the comments:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When $N$ is a negative even integer, where $N=-2m,$ the sum on the RHS of $(6)$ is taken to be the empty sum, and so is equal to zero. This means that for $N=-4$ we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{k=1}^\infty \frac{k^7}{ e^{2\pi k} -1 } = \frac{\pi}{8}&#xD;&#xA;\sum_{k=1}^\infty \frac{k^8}{ \sinh^2(\pi k) } - \frac{1}{480}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and more generally&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{k=1}^\infty \frac{k^{4m-1}}{ e^{2\pi k} -1 } = \frac{\pi}{4m}&#xD;&#xA;\sum_{k=1}^\infty \frac{k^{4m}}{ \sinh^2(\pi k) } + \frac{B_{4m}}{8m}.$$&lt;/p&gt;&#xA;" OwnerUserId="1301" LastEditorUserId="1301" LastEditorDisplayName="" LastEditDate="2011-01-29T09:37:47.060" LastActivityDate="2011-01-29T09:37:47.060" CommentCount="2" />
  <row Id="19409" PostTypeId="2" ParentId="19400" CreationDate="2011-01-28T22:00:25.757" Score="2" ViewCount="" Body="&lt;p&gt;You might get some inspiration from the articles in the Journal of Mathematics and the Arts (published by Taylor and Francis -  &lt;a href=&quot;http://www.informaworld.com/smpp/title~content=t755420531~db=all&quot; rel=&quot;nofollow&quot;&gt;http://www.informaworld.com/smpp/title~content=t755420531~db=all&lt;/a&gt;) and/or from the proceedings of the Bridges Conferences (which deal with connections between mathematics and the arts) - &lt;a href=&quot;http://www.bridgesmathart.org/&quot; rel=&quot;nofollow&quot;&gt;http://www.bridgesmathart.org/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1369" LastActivityDate="2011-01-28T22:00:25.757" CommunityOwnedDate="2011-01-29T11:39:42.757" />
  <row Id="19410" PostTypeId="1" CreationDate="2011-01-28T22:28:58.007" Score="2" ViewCount="116" Body="&lt;p&gt;Hello, I'm trying to prove that a graph $G$ contains $k$ independent edges iff $q(G-S) \le |S|+|G|-2k$&#xA;for all sets $S \subset V(G)$ (where $q(G-S)$ is the odd components). I'm having trouble with verifying whether the following direction is correct though, so, does this look valid for showing that if a graph contains $k$ independent edges, then $q(G-S) \le |S|+|G|-2k$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's assume that the graph $G$ with order $n$ contains $k$ independent edges, and add $n-2k$ vertices that all are adjacent to the vertices in $G$, call this graph $G&amp;#39;$. Then, we can find a perfect matching in $G&amp;#39;$ by matching each of these $n-2k$ added vertices with the remaining $n-2k$ vertices in $G$ that are unmatched. Then, by Tutte's theorem, we have that $q(G&amp;#39;-S) \le |S|$ for all $S \subset G&amp;#39;$. Take a subset $S$ of $V(G)$, and also include all the $n-2k$ added vertices to $S$, and call this subset $S&amp;#39;$. Then&#xA;$q(G&amp;#39;-S&amp;#39;) = q(G-S) \le |S&amp;#39;| =|S|+|G|-2k$.&lt;/p&gt;&#xA;" OwnerUserId="6363" LastEditorUserId="856" LastEditorDisplayName="" LastEditDate="2011-01-28T23:36:59.610" LastActivityDate="2011-04-29T02:43:11.410" Title="Tutte's theorem and independent edges." Tags="&lt;graph-theory&gt;" AnswerCount="1" />
  <row Id="19411" PostTypeId="2" ParentId="19329" CreationDate="2011-01-28T22:40:46.870" Score="8" ViewCount="" Body="&lt;p&gt;Let $C\subset[0,1]$ be the &lt;a href=&quot;http://en.wikipedia.org/wiki/Cantor_set&quot; rel=&quot;nofollow&quot;&gt;middle thirds Cantor set&lt;/a&gt;, and let $D\subset[0,1]$ be a &lt;a href=&quot;http://en.wikipedia.org/wiki/Smith%E2%80%93Volterra%E2%80%93Cantor_set&quot; rel=&quot;nofollow&quot;&gt;fat Cantor set&lt;/a&gt;.  Define $f:[0,1]\to[0,1]$ such that $f\vert_C$ is an order preserving homeomorphism of $C$ onto $D$ (by mapping to corresponding endpoints of the removed intervals), and $f\vert_{[0,1]\setminus C}$ is an order reversing homeomorphism of $[0,1]\setminus C$ onto $[0,1]\setminus D$ (by mapping linearly to corresponding removed intervals and then composing with $x\mapsto 1-x$).  Then $f$ is discontinuous at each point of $C$ and continuous at each point of $[0,1]\setminus C$.  Since $C$ has measure zero, $f$ &lt;a href=&quot;http://en.wikipedia.org/wiki/Riemann_integral#Integrability&quot; rel=&quot;nofollow&quot;&gt;is Riemann integrable&lt;/a&gt;. On the other hand, $f^{-1}$ is discontinuous at each point of $D$, so it is not Riemann integrable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you don't want to appeal to the Lebesgue criterion of Riemann integrability, you could work explicitly with Riemann sums of $f$ and $f^{-1}$.  In the case of $f$, you can choose partitions such that the contribution of intervals containing points of $C$ is arbitrarily small.  In the case of $f^{-1}$, the values of $f^{-1}$ on $D\cap[\frac{1}{2},1]$ are at least $\frac{2}{3}$ and the values of $f^{-1}$ on $[\frac{1}{2},1]\setminus D$ are at most $\frac{1}{2}$.  Every interval contains points from $[0,1]\setminus D$, so the difference between upper and lower sums will always be at least $(\frac{2}{3}-\frac{1}{2})(\frac{1}{2}m(D))\gt 0$.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2011-01-28T23:01:06.047" LastActivityDate="2011-01-28T23:01:06.047" CommentCount="1" />
  <row Id="19412" PostTypeId="1" AcceptedAnswerId="19413" CreationDate="2011-01-28T23:42:47.203" Score="3" ViewCount="168" Body="&lt;p&gt;I'm wondering, for no particular reason: are there differentiable vector-valued functions $\vec{f}(\vec{x})$ in three dimensions, other than the constant function $\vec{f}(\vec{x}) = \vec{C}$, that have zero divergence and zero curl? If not, how would I prove that one doesn't exist?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I had a vague memory of learning some reason that such a function doesn't exist, but there's a pretty good chance my mind is just making things up to trick me ;-) But I thought about it for a little while and couldn't think of another divergence-free curl-free function off the top of my head, so I'm curious whether I was thinking of real math or not.&lt;/p&gt;&#xA;" OwnerUserId="1190" LastActivityDate="2011-01-28T23:49:00.047" Title="Is any divergence-free curl-free vector field necessarily constant?" Tags="&lt;functions&gt;&lt;vector-calculus&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="19413" PostTypeId="2" ParentId="19412" CreationDate="2011-01-28T23:49:00.047" Score="5" ViewCount="" Body="&lt;p&gt;How about Gradient of a harmonic function?&lt;/p&gt;&#xA;" OwnerUserId="3302" LastActivityDate="2011-01-28T23:49:00.047" />
  <row Id="19414" PostTypeId="1" CreationDate="2011-01-29T00:04:40.880" Score="0" ViewCount="115" Body="&lt;p&gt;Is there a general rule for dealing with such expression?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\prod_{k=0}^{x-1}f(k)^{g(k)}$$&lt;/p&gt;&#xA;" OwnerUserId="2513" LastActivityDate="2011-01-29T00:04:40.880" ClosedDate="2011-01-29T03:41:04.663" Title="Is there a general rule for product of powers?" Tags="&lt;products&gt;" CommentCount="6" />
  <row Id="19415" PostTypeId="1" AcceptedAnswerId="19417" CreationDate="2011-01-29T00:17:16.027" Score="1" ViewCount="107" Body="&lt;p&gt;suppose $A$ is a normed vector space and $a_i \in A$ and we have the following sequence &#xA;      $$ \Vert a_1\Vert \leq \Vert a_2 \Vert \leq \cdots \leq \Vert a_n \Vert $$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;for any $1\leq i&amp;lt;n-1$ how can i show the following is not necessarily true &#xA;     $$\Vert a_{i+1} -a_i \Vert \leq \Vert a_{i+2} - a_i \Vert $$ &#xA;it easy to see this is not true for $\mathbb{R}$ any other ways of seeing this&lt;/p&gt;&#xA;&#xA;&lt;p&gt;thanks&lt;/p&gt;&#xA;" OwnerUserId="5176" LastActivityDate="2011-01-29T02:06:22.897" Title="normed vector spaces" Tags="&lt;real-analysis&gt;" AnswerCount="3" CommentCount="0" />
  <row Id="19416" PostTypeId="1" AcceptedAnswerId="19420" CreationDate="2011-01-29T00:24:42.880" Score="4" ViewCount="137" Body="&lt;p&gt;Is there any relation between the eigenvalues of possibly non-Hermitian matrix A and those of exp(A)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For hermitian matrices, they are just exponentials of the corresponding values. But in general, any relation?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="6016" LastActivityDate="2011-01-29T01:36:59.060" Title="Is there any relation between the eigenvalues of possibly non-Hermitian matrix A and those of exp(A)?" Tags="&lt;linear-algebra&gt;" AnswerCount="3" FavoriteCount="1" />
  <row Id="19417" PostTypeId="2" ParentId="19415" CreationDate="2011-01-29T00:27:04.733" Score="2" ViewCount="" Body="&lt;p&gt;It mostly relies on the triangle inequality.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have &#xA;$$||a_1|| \leq \cdots \leq ||a_n||$$&#xA;then by substracting by $||a_i||$ for a given $i$ gives :&#xA;$$0 \leq ||a_{i+1}|| - ||a_i|| \leq ||a_{i+2}|| - ||a_i||$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the triangle inequality only tells you $||a_{i+1}|| - ||a_i|| \leq ||a_{i+1} - a_i||$ and $||a_{i+2}|| - ||a_i|| \leq ||a_{i+2} - a_i||$ respectively. I don't see any way to go any further.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can &lt;em&gt;visualize&lt;/em&gt; this a bit: imagine $a_i$ and $a_{i+1}$ have close norms but &quot;point to&quot; very different directions, and that $a_i$ and $a_{i+2}$ have very different norms but sensibly the same direction. Then in that case you can see it wouldn't work.&lt;/p&gt;&#xA;" OwnerUserId="1865" LastActivityDate="2011-01-29T00:27:04.733" />
  <row Id="19418" PostTypeId="2" ParentId="19415" CreationDate="2011-01-29T00:34:46.353" Score="3" ViewCount="" Body="&lt;p&gt;Let $A$ be $\mathbb R^2$ with the Euclidean norm and let $n=3$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $a_1=(0,1)$, $a_2=(1,0)$, and $a_3=(0,2)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then $a_1$ and $a_2$ have norm $1$ and $a_3$ has norm $2$. That is, your condition on the norms is fulfilled.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, $\Vert a_2-a_1\Vert=\sqrt{2}&amp;gt;1=\Vert a_3-a_1\Vert$.&lt;/p&gt;&#xA;" OwnerUserId="367" LastActivityDate="2011-01-29T00:34:46.353" CommentCount="1" />
  <row Id="19419" PostTypeId="2" ParentId="19416" CreationDate="2011-01-29T00:35:10.270" Score="5" ViewCount="" Body="&lt;p&gt;The same result holds in general.  This is probably easiest to see using the &lt;a href=&quot;http://en.wikipedia.org/wiki/Jordan_normal_form&quot; rel=&quot;nofollow&quot;&gt;Jordan canonical form&lt;/a&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Results of this form go under the name &quot;spectral mapping theorem&quot;.  If $f$ is any power series whose disk of convergence contains the eigenvalues of $A$, then the eigenvalues of $f(A)$ are precisely the values of $f$ evaluated at the eigenvalues of $A$, as can be seen first with polynomials using Jordan form and then taking limits. You can also apply holomorphic functions using Cauchy integrals in contexts where power series aren't applicable , as seen in a more general context in the Wikipedia article on &lt;a href=&quot;http://en.wikipedia.org/wiki/Holomorphic_functional_calculus&quot; rel=&quot;nofollow&quot;&gt;holomorphic functional calculus&lt;/a&gt;.  The &lt;a href=&quot;http://en.wikipedia.org/wiki/Holomorphic_functional_calculus#Spectral_mapping_theorem&quot; rel=&quot;nofollow&quot;&gt;spectral mapping theorem&lt;/a&gt; holds there, too.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2011-01-29T00:35:10.270" CommentCount="3" />
  <row Id="19420" PostTypeId="2" ParentId="19416" CreationDate="2011-01-29T00:36:06.040" Score="5" ViewCount="" Body="&lt;p&gt;The same relation holds. To see this, bring $A$ in Jordan normal form and observe that $\exp{(TAT^{-1})} = T\exp(A)T^{-1}$. Since powers of an upper triangular matrix are upper triangular, the diagonal entries of the exponential of an upper triangular matrix are the exponentials of the diagonal entries of the original matrix and you're done.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is special about the Hermitian case, though, is that $\exp{(iA)}$ is a unitary matrix because $\exp(iA)^{\ast} = \exp{(-iA)}$ and $\exp(A + B) = \exp(A)\exp(B)$ for &lt;em&gt;commuting&lt;/em&gt; matrices.&lt;/p&gt;&#xA;" OwnerUserId="5363" LastEditorUserId="5363" LastEditorDisplayName="" LastEditDate="2011-01-29T00:58:16.440" LastActivityDate="2011-01-29T00:58:16.440" CommentCount="3" />
  <row Id="19421" PostTypeId="2" ParentId="19400" CreationDate="2011-01-29T00:55:17.227" Score="3" ViewCount="" Body="&lt;p&gt;The influence of the 4th dimension on art. Duchamp is an obvious example. Art historian Linda Henderson believes that Picasso was 4-dimensionally influenced as does &lt;a href=&quot;http://tonyrobbin.net/&quot; rel=&quot;nofollow&quot;&gt; Tony Robbin &lt;/a&gt; who clearly and unabashedly is. Tony's work influenced me to draw a bunch of stuff at  my website,  and these drawings are directly influencing some of the mathematics that I am doing. In addition, you might want to mention Carlos Seguin (sic?) and of course, George Hart. His daughter Vi Hart has been in the news recently. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is not that there is a lack of material, but there is too much. Many great and serious artists are influenced by us, and as I indicated above some of us are influenced by them. You should also look at the work of Radmila Sazdonovic and Slavan Jablan for example.  &lt;/p&gt;&#xA;" OwnerUserId="722" LastActivityDate="2011-01-29T00:55:17.227" CommunityOwnedDate="2011-01-29T11:39:42.757" CommentCount="1" />
  <row Id="19422" PostTypeId="2" ParentId="19410" CreationDate="2011-01-29T01:19:14.117" Score="1" ViewCount="" Body="&lt;p&gt;Yes that looks correct to me. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Essentially, $G&amp;#39;-S&amp;#39; = G-S$.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2011-01-29T01:19:14.117" />
  <row Id="19423" PostTypeId="2" ParentId="19416" CreationDate="2011-01-29T01:36:59.060" Score="3" ViewCount="" Body="&lt;p&gt;If you are willing to forget about problems of convergence for a moment, think first with polynomials: let&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;f(t) = a_0 + a_1 t + a_2 t^2 + \cdots + a_n t^n  &#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and $A$ any square matrix. Then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;f(A) = a_0 I + a_1 A + a_2 A^2 + \cdots + a_n A^n \ .&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume $v$ is an eigenvector of $A$ with eigenvalue $\lambda$: $Av = \lambda v$. Then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;\begin{align}&#xD;&#xA;f(A)v &amp;amp;= a_0 Iv + a_1 Av + a_2 A^2v + \cdots + a_n A^nv \\&#xD;&#xA;      &amp;amp;= a_0 v + a_1 \lambda v + a_2 \lambda^2 v + \cdots + a_n \lambda^n v \\&#xD;&#xA;      &amp;amp;= \left( a_0 + a_1 \lambda + a_2 \lambda^2 + \cdots + a_n \lambda^n \right) v \\&#xD;&#xA;      &amp;amp;= f(\lambda ) v \ .&#xD;&#xA;\end{align}&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is, $v$ is also an eigenvector of $f(A)$ with eigenvalue $f(\lambda )$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same result holds for analytic functions, such $\exp$: if&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;f(t) = \sum_{n=0}^\infty a_n t^n&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;then also $v$ is an eigenvector of $f(A)$ with eigenvalue $f(\lambda )$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, for no matter what kind of square matrix $A$, if $v$ is an eigenvector of $A$ with eigenvalue $\lambda$, $v$ is an eigenvector of $\exp (A)$ with eigenvalue $\exp (\lambda )$ too.&lt;/p&gt;&#xA;" OwnerUserId="664" LastActivityDate="2011-01-29T01:36:59.060" CommentCount="2" />
  <row Id="19424" PostTypeId="1" AcceptedAnswerId="19429" CreationDate="2011-01-29T01:41:17.357" Score="3" ViewCount="60" Body="&lt;p&gt;I had to do some homework problems involving the polynomial ring $R=\mathbb{Z}+x\,\mathbb{Q}[x]\subset\mathbb{Q}[x]$. This is an integral domain but not a UFD. Further, $x$ is not prime in $R$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of the problems was to describe to $R/(x)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $x$ is not a prime element, we know $(x)$ is not a prime ideal. So at the very least, $R/(x)$ is not an integral domain.... but what else can I say? This is perhaps something I should not admit, but problems of this form have always befuddled me. I know there's not any one &quot;answer&quot; they're looking for, but I never quite know what to say.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, this homework has been submitted already, so I am not including a homework tag. I'm just curious how you all would describe this particular quotient. &lt;/p&gt;&#xA;" OwnerUserId="2313" LastActivityDate="2011-01-29T02:00:53.087" Title="Describing a particular quotient ring involving a subring of $\mathbb{Q}[x]$" Tags="&lt;algebra&gt;&lt;ring-theory&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="19425" PostTypeId="2" ParentId="19415" CreationDate="2011-01-29T01:44:28.067" Score="1" ViewCount="" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;it easy to see this is not true for R any other ways of seeing this&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you already know it is not true for $\mathbb{R}$, you can take any counterexample in $\mathbb{R}$ and multiply by a fixed nonzero vector in $A$ (assuming $A$ has a nonzero vector, of course).  For example, multiplying $1,-1$, and $2$ by $x\neq 0$ gives $\|x\|=\|-x\|\lt\|2x\|$ but $\|-x-x\|\gt\|2x-x\|$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But perhaps by &quot;other ways&quot; you meant that you had already considered this method and wanted a different type of example.  In that case, Rasmus's nice example is definitely a better answer.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2011-01-29T02:06:22.897" LastActivityDate="2011-01-29T02:06:22.897" CommentCount="2" />
  <row Id="19426" PostTypeId="1" CreationDate="2011-01-29T01:47:06.510" Score="3" ViewCount="134" Body="&lt;p&gt;In particular, I'm told if $k$ is commutative (ring), $R$ and $S$ are commutative $k$-algebras such that $R$ is noetherian, and $S$ is a finitely generated $k$-algebra, then the tensor product $R\otimes_k S$ of $R$ and $S$ over $k$ is a noetherian ring. &lt;/p&gt;&#xA;" OwnerUserId="6369" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-29T04:51:31.083" LastActivityDate="2011-01-29T04:51:31.083" Title="When is a tensor product of two commutative rings noetherian?" Tags="&lt;abstract-algebra&gt;&lt;ring-theory&gt;" CommentCount="5" />
  <row Id="19427" PostTypeId="1" CreationDate="2011-01-29T01:52:59.390" Score="13" ViewCount="504" Body="&lt;p&gt;What are the major and minor contributions of Galois Theory to Mathematics? I mean direct contributions (like being aplied as it appears in Algebra) or simply by serving as a model to other theories.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have my own answers and point of view to this question, but I think it would be nice to know your opinion too.&lt;/p&gt;&#xA;" OwnerUserId="1971" LastActivityDate="2011-05-02T00:23:55.570" CommunityOwnedDate="2011-01-29T01:54:40.360" Title="Contributions of Galois Theory to Mathematics" Tags="&lt;soft-question&gt;&lt;algebra&gt;&lt;galois-theory&gt;" AnswerCount="7" CommentCount="2" FavoriteCount="6" />
  <row Id="19428" PostTypeId="2" ParentId="19386" CreationDate="2011-01-29T01:52:59.533" Score="3" ViewCount="" Body="&lt;p&gt;To the follow up question about allowing embeddings in $\mathbb{R}^4$: the answer is &lt;strong&gt;yes&lt;/strong&gt;. One simple example is to start with the &lt;a href=&quot;http://en.wikipedia.org/wiki/Hexadecachoron&quot; rel=&quot;nofollow&quot;&gt;hexadecachoron&lt;/a&gt; or &lt;a href=&quot;http://en.wikipedia.org/wiki/600-cell&quot; rel=&quot;nofollow&quot;&gt;600-cell&lt;/a&gt; and remove some of the constituent tetrahedra. &lt;/p&gt;&#xA;" OwnerUserId="1543" LastActivityDate="2011-01-29T01:52:59.533" CommentCount="2" />
  <row Id="19429" PostTypeId="2" ParentId="19424" CreationDate="2011-01-29T01:55:25.917" Score="4" ViewCount="" Body="&lt;p&gt;The key is to understand what that ideal $(x)$ looks like in the first place. So, clearly all polynomials in that ideal have degree at least 1, but what degree 1 polynomials can occur? For $f(x)*x$ to have degree 1, $f(x)$ must have a non-zero constant term. That term must be in $\mathbb{Z}$, and it is easy to see that you get any coefficients for higher degrees. So $(x)=\mathbb{Z}x + \mathbb{Q}x^2+\ldots+\mathbb{Q}x^n+\ldots$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, in the quotient, any coset is represented by some $n + ax$, $n\in \mathbb{Z},a\in \mathbb{Q}$, and another element $m+bx$ represents a different coset if and only if $n\neq m$ or $a-b\notin\mathbb{Z}$. In other words, the quotient is isomorphic to $\left(\mathbb{Z} + \mathbb{Q}/\mathbb{Z}x\right)/(x^2).$&lt;/p&gt;&#xA;" OwnerUserId="3212" LastEditorUserId="3212" LastEditorDisplayName="" LastEditDate="2011-01-29T02:00:53.087" LastActivityDate="2011-01-29T02:00:53.087" CommentCount="1" />
  <row Id="19430" PostTypeId="2" ParentId="19427" CreationDate="2011-01-29T02:10:43.730" Score="10" ViewCount="" Body="&lt;p&gt;The theory of covering spaces is a great example of a Galois correspondence.&lt;/p&gt;&#xA;" OwnerUserId="2159" LastActivityDate="2011-01-29T02:10:43.730" CommunityOwnedDate="2011-01-29T02:10:43.730" CommentCount="4" />
  <row Id="19431" PostTypeId="2" ParentId="19400" CreationDate="2011-01-29T02:14:03.117" Score="0" ViewCount="" Body="&lt;p&gt;The golden ratio in architecture.  I saw an interesting talk on this at Union College.  They're fond of the golden ratio because of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Nott_Memorial&quot; rel=&quot;nofollow&quot;&gt;Nott&lt;/a&gt; &lt;a href=&quot;http://www.union.edu/N/DS/edition_display.php?e=13&amp;amp;s=1509&quot; rel=&quot;nofollow&quot;&gt;Memorial.&lt;/a&gt;  There are a number of ratios, e.g. diameter to height, that approximate the golden ratio.  The ratio is also incorporated into the design of the Gothic arches.&lt;/p&gt;&#xA;" OwnerUserId="4181" LastEditorUserId="4181" LastEditorDisplayName="" LastEditDate="2011-02-02T17:40:27.403" LastActivityDate="2011-02-02T17:40:27.403" CommunityOwnedDate="2011-01-29T11:39:42.757" />
  <row Id="19433" PostTypeId="1" CreationDate="2011-01-29T03:46:45.080" Score="5" ViewCount="314" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I apologize for my total ignorance in the sphere of mathematics and the possibly very silly question I'm about to ask. My mathematical knowledge level is quite limited (pretty much finished with some slightly more advanced stuff then grade 12) so please if possible limit too much terminology to about that level of math. Again I don't mean to offend anyone &amp;amp; I'm sorry if the following sounds like a joke but I am genuinely interested and cannot quite grasp the reason for it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've been curious for quite sometime now, What is the significance for a mathematics to frequently require proof for both finite &amp;amp; infinite cases of theorems ? Why isn't it satisfactory to prove any theorem for a reasonably high finite x (whatever x is - be it set of some numbers) ? The reason why I'm asking is that in real-life applications (not talking about software application but life applications like count a bag of money or something like that) there is likely never need to deal with infinite of anything really - it might be a very high quantity but never infinite. So why does mathematics needs and requires proof for the infinite case as well, instead being satisfied proving only finite case ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for any advise!!&lt;/p&gt;&#xA;" OwnerUserId="6372" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2011-01-29T09:04:37.473" LastActivityDate="2011-01-29T09:25:20.280" Title="A non-mathematician’s (programmer’s) question on infinity ?" Tags="&lt;infinity&gt;" AnswerCount="3" CommentCount="11" />
  <row Id="19435" PostTypeId="2" ParentId="19427" CreationDate="2011-01-29T04:27:33.200" Score="9" ViewCount="" Body="&lt;p&gt;The idea behind the connection between subgroups and subfields in Galois Theory has wide applications; they form an entire subject called &lt;strong&gt;Galois connections&lt;/strong&gt; (same as &quot;Galois correspondence&quot; mentioned by Aaron Mazel-Gee). See for example the section on Galois connections in George Bergman's &lt;a href=&quot;http://math.berkeley.edu/~gbergman/245/&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;An Invitation to General Algebra and Universal Constructions&lt;/em&gt;&lt;/a&gt; (the files are PostScript, the index is very good, so look in the index under &quot;Galois connection&quot;). The structure theorem for modules over PIDs can be seen as a Galois connection, as can the Zariski correspondence between ideals of $k[x_1,\ldots,x_n]$ and varieties in $\mathbb{A}^n(k)$. (The &quot;really interesting&quot; parts of the Fundamental Theorem of Galois Theory, meaning the one that do not follow from general properties, are the characterizations of the point stabilizer of a subfield as a Galois Group, of the fixed sets in the field as the subfields, and of the Galois subextensions as corresponding to normal subgroups; the comparable part in the theory of the correspondence between the ideals and the varietiese is Hilbert's Nullstellensatz, which characterizes the &quot;closed&quot; ideals as the radical ideals).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are several properties that one often proves from scratch in all these situations that in fact follow from the general setting. It is also mirrored to some extent when one studies group actions in general. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Galois Theory itself has lots of applications, of course: algebraic number theory uses it all the time to study rings of integers, to name one.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-29T05:21:45.963" LastActivityDate="2011-01-29T05:21:45.963" CommunityOwnedDate="2011-01-29T04:27:33.200" />
  <row Id="19436" PostTypeId="2" ParentId="19433" CreationDate="2011-01-29T04:41:23.293" Score="0" ViewCount="" Body="&lt;p&gt;I personally think the main reason is to have to deal with the boundary. We cannot exactly fix the boundaries in every case. For your case it is simple when you are considering years but consider a case where you are calculating the time taken by the light to reach the other end of universe (just a real life example). You can't fix a limit in every case so you need a abstract value that works with every possible value. Hence infinity.&lt;/p&gt;&#xA;" OwnerUserId="6335" LastActivityDate="2011-01-29T04:41:23.293" />
  <row Id="19437" PostTypeId="2" ParentId="19379" CreationDate="2011-01-29T05:11:13.600" Score="4" ViewCount="" Body="&lt;p&gt;To answer your question, you first have to choose a map&#xA;$\mathbb Q[\mu_{p-1}] \subset \mathbb Q_p.$&#xA;To this end, note that $p \equiv 1 \bmod p-1,$ hence $p$ splits completely&#xA;in $\mathbb Q[\mu_{p-1}]$, and so choosing such an embedding amounts to&#xA;choosing one prime $\mathfrak p$ above $p$.   The embedding is then characterized by the fact that the maximal ideal of $\mathbb Z_p$ pulls back &#xA;to $\mathfrak p$, while all the other prime ideals dividing $p$ generate the unit ideal of $\mathbb Z_p$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to keep things concrete, it will be easier to think about actual elements (or, if you like, pricipal ideals) rather than prime ideals that are not necessarily principal.&#xA;To this end, choose a power $h$&#xA;such that $\mathfrak p^h$ is principal, say equal to $\pi$.  We may then write&#xA;$p^h = \pi \pi&amp;#39;,$ where $\pi&amp;#39;$ is prime to $\mathfrak p$. (The ideal generated&#xA;by $\pi&amp;#39;$ is the product of the $h$th powers of all the primes &lt;em&gt;other than&lt;/em&gt;&#xA;$\mathfrak p$ which divide $p$.)&#xA;Then we see that $$\mathbb Z[\mu_{p-1},\frac{1}{(p-1)p}] = \mathbb Z[\mu_{p-1},\frac{1}{(p-1)^h\pi\pi&amp;#39;}].$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The elements $p-1$ and $\pi&amp;#39;$ are invertible in $\mathbb Z_p$, while &#xA;$\pi$ is not (it generates the $h$th power of the maximal ideal of $\mathbb Z_p$).    Thus we see that&#xA;$$\mathbb Z[\mu_{p-1},\frac{1}{(p-1)p}] \cap \mathbb Z_p&#xD;&#xA;= \mathbb Z[\mu_{p-1},\frac{1}{(p-1)\pi&amp;#39;}],$$&#xA;and so its Spec coincides with that of $\mathbb Z[\mu_{p-1}]$ with&#xA;the prime above $p-1$ and all the primes dividing $\pi&amp;#39;$ (i.e. all but one&#xA;of the primes over $p$) removed (confirming your claim).&lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2011-01-29T05:11:13.600" />
  <row Id="19438" PostTypeId="2" ParentId="19433" CreationDate="2011-01-29T05:45:23.633" Score="9" ViewCount="" Body="&lt;p&gt;Let me point out a few reasons why, in my opinion, we are often interested in the infinite.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Often, dealing with the infinite case is easier or, at least, leads to simpler and more understandable statements. Picking a well known and celebrated example, the prime number theorem (PNT) says that the number of primes less than a given number $x$ is approximately $x/\log x$. Consider what this is really saying. Writing $\pi(x)$ for the number of primes less than $x$, the PNT is saying that the ratio $\pi(x)/(x/\log x)$ tends to one as $x$ &lt;em&gt;tends to infinity&lt;/em&gt;. This is written as&#xA;$$&#xD;&#xA;\pi(x)\sim \frac{x}{\log x}\qquad\qquad{\rm(1)}&#xD;&#xA;$$&#xA;where the '$\sim$' means that the ratio of the two sides tends to one as $x$ tends to infinity. Precisely, this means the following: if you pick any positive number $\epsilon$, no matter how small, then the ratio of the two sides of (1) will be within a distance $\epsilon$ of 1 when $x$ is large enough. We can express this by saying that there is a number $N_\epsilon$ (depending on your choice of $\epsilon$) such that $1-\epsilon &amp;lt; \pi(x)/(x/\log x) &amp;lt; 1+\epsilon$ for all numbers $x$ greater than $N_\epsilon$. You might ask, what does this say about the number of primes less than any given $x$. Say, the number of primes less than $2^{100}$? The answer is, absolutely nothing! Without specifying explicit bounds on the ratio $\pi(x)/(x/\log x)$ for a given number $x$, all we have is a statement about the infinite. Of course, bounds can be given, and you will need these to apply the PNT to some actual fixed value of $x$. This is besides the point though. The PNT as above is a simply stated and enlightening theorem. Much more so than just writing out some complicated but accurate expression for all values of $x$ up to some huge number. There will also be many such approximations of varying simplicity and accuracy, whereas the PNT is a single elegant statement which every mathematician knows.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;More than the raw statements of theorems, what often interests mathematicians is the insight that they provide. Saying that something holds for &lt;em&gt;every&lt;/em&gt; natural number -- and proving it -- is much more interesting, and provides more insight into the properties of numbers than just saying that you have checked that it is true for the first billion numbers. People have known that the &lt;a href=&quot;http://en.wikipedia.org/wiki/Riemann_hypothesis&quot; rel=&quot;nofollow&quot;&gt;Riemman hypothesis&lt;/a&gt; is true for many zeros of the Zeta function for some time. I'm not sure what the latest count is, but at least 10 trillion zeros have been checked, and do indeed lie on the `critical line' in accordance with the hypothesis. If however, just one zero was found off the critical line, then that would go against many of the properties people expect of such functions, and go against many of the expected properties of the prime numbers. If the Riemann hypothesis was found to be wrong, a lot of other conjectures would be thrown into doubt too.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;When we prove mathematical theorems, they will usually be used to derive further results. This is often done in rather involved ways. Even if you prove a theorem up to 'reasonably high' values of $x$, someone will want to use that theorem as a stepping stone in proving further results. In that case, it will likely be needed for unspecified valued of $x$. Coupled with techniques such as induction, even huge values of $x$ could be important. As mathematical proofs are frequently built up out of a large number of small steps, it is imperative that each individual part is watertight. Otherwise, you could quickly end up with nonsense. So, it is important that results can be stated and proven for &lt;em&gt;all&lt;/em&gt; cases within certain precisely specified parameters. Just checking a finite number of cases would be of limited use.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="1321" LastActivityDate="2011-01-29T05:45:23.633" />
  <row Id="19439" PostTypeId="1" CreationDate="2011-01-29T06:00:45.120" Score="0" ViewCount="87" Body="&lt;p&gt;Given:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$F(F(n)) = n$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$F(F(n + 2) + 2) = n$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$F(0) = 1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where n is a non-negative integer. $$F(129) = ?$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can we solve such kind of functional equations? Is there any simpler approach other than doing iteration over the values manually?&lt;/p&gt;&#xA;" OwnerUserId="6374" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2011-01-29T06:08:09.073" LastActivityDate="2011-01-29T06:41:33.360" Title="Functional Equation Analysis" Tags="&lt;functional-equations&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="19440" PostTypeId="1" AcceptedAnswerId="19442" CreationDate="2011-01-29T06:34:45.053" Score="3" ViewCount="115" Body="&lt;p&gt;If $G$ is a group acting on a set $S$, then the &quot;orbit&quot; of a point $x$ in $S$ is defined as the set of all elements of the form $gx$ where $g \in G$. My question: why was the word &quot;orbit&quot; chosen for this concept? It is not obviously related to previous uses of this word, such as the path of a planet around the Sun.&lt;/p&gt;&#xA;" OwnerUserId="6375" LastActivityDate="2011-01-29T08:04:32.517" Title="Origin of mathematical use of &quot;orbit&quot;" Tags="&lt;terminology&gt;&lt;math-history&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="19441" PostTypeId="2" ParentId="19439" CreationDate="2011-01-29T06:41:33.360" Score="5" ViewCount="" Body="&lt;p&gt;Firstly, I think you had better allow $n$ to be an arbitrary integer, because&#xA;these equations force the value of $F(n)$ to be negative when $n &amp;gt; 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Secondly, to solve this system of equations, apply $F$ to the second equation,&#xA;and then appeal to the first equation, to find that&#xA;$$F(n+2) + 2 = F(n),$$ i.e. that&#xA;$$F(n+2) = F(n) - 2.$$&#xA;This allows us to reduce computing $F$ to the case when either $n = 0$ or $n = 1$.&#xA;The first case is given to us in the question, namely $F(0) = 1$.  For the second&#xA;case, note that the first equation, with $n = 0$, gives $F(F(0)) = 0,$ i.e. $F(1) = 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we can easily compute all values of $F$, and indeed we find that $F$ has a very simple closed form, the derivation of which I leave as an exercise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The basic idea in the above is the that first equation establishes $F$ as&#xA;its own inverse, and so this can be used to simplify the second equation.&lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2011-01-29T06:41:33.360" />
  <row Id="19442" PostTypeId="2" ParentId="19440" CreationDate="2011-01-29T07:10:21.813" Score="6" ViewCount="" Body="&lt;p&gt;It seems the concept goes back further than D.W. Hall (a student of Whyburn) - at least to Kuratowski, cf. the following excerpt from &lt;a href=&quot;http://books.google.com/books?id=IM7cdgtodqUC&amp;amp;pg=PA14&quot; rel=&quot;nofollow&quot;&gt;Kuczma et al. Iterative functional equations, p. 14.&lt;/a&gt; $\quad $ One can probably find further historical details by Googling &quot;Kuratowski-Whyburn orbit&quot; etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/CFu9S.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: for a simple yet enlightening example of the key role that orbits play in the solution of functional equation see &lt;a href=&quot;http://math.stackexchange.com/questions/3888/find-polynomials-such-that-x-16p2x-16x-1px/4150#4150&quot;&gt;this post.&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2011-01-29T08:04:32.517" LastActivityDate="2011-01-29T08:04:32.517" />
  <row Id="19443" PostTypeId="2" ParentId="19427" CreationDate="2011-01-29T07:57:26.063" Score="3" ViewCount="" Body="&lt;p&gt;I'm not an expert on the work of Alain Connes, but he has used Galois theory as a &#xA;Leitmotiv in developing 'motivic Galois theory' to attack problems in quantum theory. Maybe someone more knowledgeable could expand on this. If you read French, you should probably take a look at his (highly informal) article 'La pensée d'Evariste Galois et le formalisme moderne', available online [http://www.alainconnes.org/docs/galoistext.pdf].&lt;/p&gt;&#xA;" OwnerUserId="6004" LastActivityDate="2011-01-29T07:57:26.063" CommunityOwnedDate="2011-01-29T07:57:26.063" />
  <row Id="19444" PostTypeId="2" ParentId="19278" CreationDate="2011-01-29T08:04:28.880" Score="0" ViewCount="" Body="&lt;p&gt;If $y = \log x - \alpha \log (1-x)$, then $e^y = x/(1-x)^\alpha$. If you pick $\alpha = 1/5$ or $\alpha = 5$, you will have to solve a quintic equation for $x$. Some of the coefficients will involve $e^y$ which is almost always transcendental, and (I think) this implies that the quintic will not in general be solvable.&lt;/p&gt;&#xA;" OwnerUserId="856" LastActivityDate="2011-01-29T08:04:28.880" />
  <row Id="19445" PostTypeId="2" ParentId="19433" CreationDate="2011-01-29T09:07:06.020" Score="10" ViewCount="" Body="&lt;p&gt;It is debatable whether any application of mathematics to real life (e.g. solving differential equations) actually implies that real life is described by the infinitary objects we use to model it (e.g. the real numbers).  However, it tends to be extremely &lt;em&gt;convenient&lt;/em&gt; to use infinitary objects over finitary ones because when we pass to the infinite limit, &lt;em&gt;error terms go away&lt;/em&gt; and the main term of various approximations becomes exact.  (For example, it is debatable whether derivatives &quot;exist&quot; in any meaningful sense.  However, it is much easier to work with derivatives than their finitary analogues.)  So using infinite objects is a good way to systematically ignore certain details in order to better focus on the relevant ones.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A basic example is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Central_limit_theorem&quot; rel=&quot;nofollow&quot;&gt;central limit theorem&lt;/a&gt;.  Its statement and proof require infinitary objects, but what comes out of it is a finitary understanding of how a large number of i.i.d. random variables behave together, and we get the Bell curve, which allows us to understand a lot of natural phenomena.&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2011-01-29T09:25:20.280" LastActivityDate="2011-01-29T09:25:20.280" CommentCount="0" />
  <row Id="19446" PostTypeId="1" CreationDate="2011-01-29T09:18:19.177" Score="2" ViewCount="103" Body="&lt;p&gt;The area/arc-length is given by an integral or the integral defines the area/arc-length is one the first things we learn in Calculus, but that is done in the Cartesian coordinates, next one moves to polar coordinates and the area is transformed/redefined by a new integral using the Jacobian. My question is : Since area and arc length are invariant of coordinate system, is there a way to define them other than in Cartesian coordinates and transform them from coordinate system to coordinate system with the help of Jacobian? In other words is there a way that area/arc-length is defined independet of coordinate system and then according to the structure of coordinate system it's integral form is reached, without tranforming between coordinate systems, but for a given coordinate system it is derived. (and without the use of Jacobian to move between coordinate systems)&lt;/p&gt;&#xA;" OwnerUserId="3936" LastActivityDate="2011-01-29T14:21:23.210" Title="Representing area as an integral independent of the coordinate systems" Tags="&lt;calculus&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="19447" PostTypeId="2" ParentId="9728" CreationDate="2011-01-29T10:08:48.443" Score="1" ViewCount="" Body="&lt;p&gt;To put it more concisely: If there is a $\xi\in]0,1[$ with $g(\xi)=\xi$ then by the mean-value theorem there is a $\xi&amp;#39;\in ]\xi,1[$ with $g&amp;#39;(\xi&amp;#39;)=1$, whence $g&amp;#39;(1)&amp;gt;1$, as $g&amp;#39;$ is strictly increasing. Conversely, if $g&amp;#39;(1)&amp;gt;1$ then $g(a)&amp;lt;a$ for some $a$ immediately to the left of $1$. As $g(0)&amp;gt;0$ there must be a $\xi\in]0,a[$ with $g(\xi)=\xi$.&lt;/p&gt;&#xA;" OwnerUserId="1303" LastActivityDate="2011-01-29T10:08:48.443" />
  <row Id="19449" PostTypeId="2" ParentId="19427" CreationDate="2011-01-29T10:54:32.487" Score="6" ViewCount="" Body="&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Risch_algorithm&quot; rel=&quot;nofollow&quot;&gt;Risch algorithm&lt;/a&gt; in Differential Galois Theory can decide if a function has an elementary primitive or not and, if it does, find it.&lt;/p&gt;&#xA;" OwnerUserId="6276" LastActivityDate="2011-01-29T10:54:32.487" CommunityOwnedDate="2011-01-29T10:54:32.487" />
  <row Id="19450" PostTypeId="1" AcceptedAnswerId="19451" CreationDate="2011-01-29T10:55:30.403" Score="6" ViewCount="215" Body="&lt;p&gt;Does there exist for all subsets of the natural numbers, with infinite elements, a function that maps each element in that subset to each of the natural numbers, and the other way around?&lt;/p&gt;&#xA;" OwnerUserId="5904" LastEditorUserId="822" LastEditorDisplayName="" LastEditDate="2011-01-29T13:21:26.403" LastActivityDate="2011-01-29T13:21:26.403" Title="Do all infinite subsets of N have a bijective function to N?" Tags="&lt;set-theory&gt;&lt;elementary-set-theory&gt;" AnswerCount="1" />
  <row Id="19451" PostTypeId="2" ParentId="19450" CreationDate="2011-01-29T11:18:34.860" Score="9" ViewCount="" Body="&lt;p&gt;If $A\subseteq \mathbb{N}$ and $A$ is infinite then since $\aleph _0$ is the smallest infinite cardinality then $\aleph_0 \leq |A|\leq |\mathbb{N}| =\aleph_0$, so A and $\mathbb{N}$ have the same cardinality - meaning there is a bijection between the two.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This can also be seen directly - suppose $A\subseteq \mathbb{N}$ and $A$ is infinite, define a function $\varphi: \mathbb{N}\rightarrow A$ as follows: A has a minimal number $a_1 \in A$ so $\varphi(1)=a_1$. $A-\{a_1\}$ has minimal number $a_2$ so define $\varphi(2)=a_2$. and in general, A has a k-th minimal number $a_k$ so $\varphi(k)=a_k$. This is true for all k since A is infinite.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is easy to see that this is injective map. to show surjectivity notice that if $a\in A$ then there is only finite natural number (denote by k-1) smaller than $a$ that are in $A$, so $\varphi(k)=a$. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;For the case of $\mathbb{Z}$ this is also true, since $|\mathbb{Z}|=2|\mathbb{N}|=2\aleph_0 =\aleph_0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the direct approach, suppose $A\subseteq \mathbb{Z}$ is infinite. Let $B=A\cap \mathbb{N}$ then if $B$ and $A-B$ (the positive, and negative parts of A) are infinite then there is a bijection between B and $\mathbb{N}$ and a bijection between $A-B$ and the negative integers, so join the two bijections to get one from A to $\mathbb{Z}$.&#xA;Suppose now that one of them is finite, wlog $B-A$ is finite, so A has a minimal number. The same proof as before shows that there is a bijection between A and $\mathbb{N}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a bijection between $\mathbb{N}$ and $\mathbb{Z}$ by ordering the integers, for example by $0,1,-1,2,-2,3,-3,...$. Now just take the composition of the maps - from A to $\mathbb{N}$ and then to $\mathbb{Z}$.&lt;/p&gt;&#xA;" OwnerUserId="2125" LastEditorUserId="2125" LastEditorDisplayName="" LastEditDate="2011-01-29T11:38:17.573" LastActivityDate="2011-01-29T11:38:17.573" CommentCount="4" />
  <row Id="19453" PostTypeId="1" AcceptedAnswerId="19469" CreationDate="2011-01-29T12:43:02.083" Score="0" ViewCount="77" Body="&lt;p&gt;Hi all, Does anyone know any available reference to learn how works CORDIC algorithm to implement a division?&#xA;Thank you!&lt;/p&gt;&#xA;" OwnerUserId="4255" LastActivityDate="2011-01-29T22:13:07.117" Title="References with explanations of how works CORDIC algorithm for division" Tags="&lt;algorithms&gt;&lt;reference-request&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="19454" PostTypeId="2" ParentId="19446" CreationDate="2011-01-29T14:21:23.210" Score="2" ViewCount="" Body="&lt;p&gt;Arc length, area and volume are geometrical concepts that make sense in a coordinate-free world. Note that Archimedes computed many interesting areas and volumes without using coordinates!  But now we have this edifice called &quot;calculus&quot; with its primitives, Fubini theorem, change-of-variables formula, etc. In order to make this edifice useful for computing lengths and areas we have to coordinatize our curves and surfaces. We then have to ${\it prove}$ that our intuitive idea of arc length as being the $\sup$ of the lengths of inscribed polygons, leads to the formula $L(\gamma)=\int_a^b |\dot{\bf x}(t)|dt$. This is hard work! What we have gained in this way is access to a machinery that allows to compute an unlimited number of lengths and areas in a more or less automatic way. But, alas, we shall not obtain a simple formula for the circumference of an ellipse $\ldots$&lt;/p&gt;&#xA;" OwnerUserId="1303" LastActivityDate="2011-01-29T14:21:23.210" />
  <row Id="19455" PostTypeId="2" ParentId="19453" CreationDate="2011-01-29T14:47:58.270" Score="1" ViewCount="" Body="&lt;p&gt;See &lt;a href=&quot;http://math.stackexchange.com/questions/14066/calculator-algorithms/14070#14070&quot;&gt;Calculator algorithms&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="589" LastActivityDate="2011-01-29T14:47:58.270" />
  <row Id="19456" PostTypeId="1" CreationDate="2011-01-29T14:51:51.980" Score="0" ViewCount="71" Body="&lt;p&gt;Suppose we have the following situation: $I = \oplus_{i=1}^{p-1}I_1$ is a decomposition of an $R$-ideal in invertible modules of rank $1$, $I_iI_j \subset I_{i+j \pmod{p-1}}$ and $I_1^i = I_{i \pmod{p-1}}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then it is clear that $Sym_RI_1$ surjects onto $R$, but why is the kernel of the natural map $Sym_RI_1 \to R$ generated by $(a-1)\otimes I_1^{\otimes p}$, where $a \in I_1^{\otimes 1-p} = Hom(I_1^{\otimes p}, I_1)$ is the multiplication map $I_1^{\otimes p} \to I_1$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is somewhat clear, but I need a stringent argument.&lt;/p&gt;&#xA;" OwnerUserId="5262" LastActivityDate="2011-01-29T14:51:51.980" Title="symmetric algebra" Tags="&lt;commutative-algebra&gt;" />
  <row Id="19457" PostTypeId="1" AcceptedAnswerId="19546" CreationDate="2011-01-29T15:13:51.413" Score="2" ViewCount="272" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is ; How can I solve the following integral question?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle \int_{\sqrt2/2}^{1}\int_{\sqrt{1-x^2}}^{x}\frac{1}{\sqrt{x^2+y^2}}dydx$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance,&lt;/p&gt;&#xA;" OwnerUserId="3669" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2011-01-29T15:59:29.067" LastActivityDate="2011-01-30T09:49:11.207" Title="How can I find $\int_{\sqrt2/2}^{1}\int_{\sqrt{1-x^2}}^{x}\frac{1}{\sqrt{x^2+y^2}}dydx$ ?" Tags="&lt;integral&gt;" AnswerCount="4" CommentCount="1" />
  <row Id="19458" PostTypeId="1" CreationDate="2011-01-29T15:30:19.877" Score="1" ViewCount="73" Body="&lt;p&gt;I am a computational physics postgrad student, working with libraries like ATLAS and MAGMA.  I have a matrix which is upper-triangular, and is the result of a Cholesky decomposition.  I need to convert the upper-triangular matrix in to a symmetric, square matrix where the elements of the lower triangle are&lt;/p&gt;&#xA;&#xA;&lt;p&gt;mat(j,i) = mat(i,j)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a naive routine in C++ that simply does the above in a loop, but it's incredibly slow.  I believe it's so slow because the CPU cache is poorly utilised - element i,j is quite far from element j,i.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a clever mathematical trick that I can use to optimise this loop?  Or alternatively, does anyone know of any fuctions in BLAS/similar libraries that can do this kind of operation?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many thanks in advance for any advice you can give.&lt;/p&gt;&#xA;" OwnerUserId="6383" LastActivityDate="2011-01-29T15:30:19.877" Title="From a triangular matrix to a symmetric square matrix" Tags="&lt;linear-algebra&gt;&lt;matrices&gt;" CommentCount="6" />
  <row Id="19459" PostTypeId="2" ParentId="19457" CreationDate="2011-01-29T15:42:08.020" Score="3" ViewCount="" Body="&lt;p&gt;Hint:for the y integral, consider x to be a constant.  It is a standard form that can be solved by a trigonometric substitution.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-01-29T15:42:08.020" />
  <row Id="19460" PostTypeId="1" CreationDate="2011-01-29T16:05:02.697" Score="2" ViewCount="89" Body="&lt;p&gt;Dear All&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to know whether a Heyting algebra gives rise to ring in a similar way that a Boolean algebra gives rise to a Boolean ring. In a Boolean algebra $(B,\lor,\land,\lnot,0,1)$ I can define multiplication and addition as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$a * b = a \land b$$&#xA;$$a + b = (a \land \lnot b) \lor (b \land \lnot a)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And I get a Boolean ring $(B,+,*,0,1)$. From the Boolean ring we can reconstruct the Boolean algebra again via the following definitions (right?):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$a \land b = a * b$$&#xA;$$a \lor b = (a + 1) * (b + 1) + 1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a Heyting algebra $(H,\rightarrow,\land,0,1)$ we do only have a pseudo complement. Can we nevertheless define a ring, that in turn would allow us to reconstruct the algebra?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Best Regards&lt;/p&gt;&#xA;" OwnerUserId="4414" LastActivityDate="2011-01-29T21:29:48.373" Title="Is there something like a Heyting Ring?" Tags="&lt;algebra&gt;&lt;ring-theory&gt;&lt;boolean-algebra&gt;" AnswerCount="0" CommentCount="5" />
  <row Id="19461" PostTypeId="2" ParentId="16940" CreationDate="2011-01-29T16:13:52.273" Score="0" ViewCount="" Body="&lt;p&gt;I was considering a similar problem few months back and wrote a small write-up which you can find it &lt;a href=&quot;http://arxiv.org/PS_cache/arxiv/pdf/1101/1101.4081v1.pdf&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;. Hope it is of some use.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastActivityDate="2011-01-29T16:13:52.273" CommentCount="1" />
  <row Id="19462" PostTypeId="2" ParentId="19457" CreationDate="2011-01-29T16:26:45.640" Score="3" ViewCount="" Body="&lt;p&gt;Try converting to polar coordinates, you should end out integrating $\sec(\theta) - 1$.&lt;/p&gt;&#xA;" OwnerUserId="3035" LastActivityDate="2011-01-29T16:26:45.640" CommentCount="1" />
  <row Id="19463" PostTypeId="1" AcceptedAnswerId="19477" CreationDate="2011-01-29T16:41:14.267" Score="5" ViewCount="175" Body="&lt;p&gt;Why are spectral sequences called &quot;spectral&quot;? Is that use of &quot;spectral&quot; related to other uses in math, such as spectra in homotopy theory, the spectrum of a ring in algebraic geometry or the spectrum of an operator? Why are those things called spectral, for that matter?&lt;/p&gt;&#xA;" OwnerUserId="1070" LastActivityDate="2011-01-29T19:42:32.990" Title="Why are spectral sequences called &quot;spectral&quot;?" Tags="&lt;terminology&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="19464" PostTypeId="1" AcceptedAnswerId="19468" CreationDate="2011-01-29T16:45:31.797" Score="2" ViewCount="196" Body="&lt;p&gt;Suppose $A$ and $B$ are two commuting square matrices over an algebraically closed field. It is true that the spectrum of $A+B$ is contained in the set $\{\lambda_1+\lambda_2:\lambda_1 \in \sigma(A), \lambda_2 \in \sigma(B)\}$ where $\sigma(A)$ denotes the spectrum of $A$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If so, can you provide a reference for this? can more be said?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="5491" LastActivityDate="2011-01-29T17:21:41.007" Title="Spectrum of the sum of two commuting matrices" Tags="&lt;linear-algebra&gt;" AnswerCount="4" />
  <row Id="19465" PostTypeId="2" ParentId="19464" CreationDate="2011-01-29T16:53:25.383" Score="1" ViewCount="" Body="&lt;p&gt;Yes see &lt;a href=&quot;http://en.wikipedia.org/wiki/Commuting_matrices&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt;(article about commutating matrices). In particular, if $\alpha_i$ and $\beta_j$ are the eigenvalues of $A$ and $B$, then one can choose an eigenbasis such that the eigenvalues of $A+B$ are $\alpha_i+ \beta_j$. &lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2011-01-29T16:53:25.383" CommentCount="1" />
  <row Id="19466" PostTypeId="2" ParentId="19464" CreationDate="2011-01-29T16:54:43.917" Score="5" ViewCount="" Body="&lt;p&gt;Frobenius has proved that commuting matrices over an algebraically closed field can be simultaneously brought into upper triangular form (this holds more generally for a solvable Lie algebras of matrices by Lie's theorem). To prove this for commuting matrices, show that they have a simultaneous eigenvector and proceed by induction on the dimension.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One way to prove this last fact (again for commuting matrices) is to use the Hilbert Nullstellensatz, see  &lt;a href=&quot;http://en.wikipedia.org/wiki/Hilbert%27s_Nullstellensatz&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;, a nice proof of which you can find online e.g. on p.116 of&#xA;&lt;a href=&quot;http://math.uga.edu/~pete/integral.pdf&quot; rel=&quot;nofollow&quot;&gt;Pete L. Clark's notes&lt;/a&gt;. &lt;/p&gt;&#xA;" OwnerUserId="5363" LastEditorUserId="5363" LastEditorDisplayName="" LastEditDate="2011-01-29T17:21:41.007" LastActivityDate="2011-01-29T17:21:41.007" CommentCount="1" />
  <row Id="19467" PostTypeId="2" ParentId="19464" CreationDate="2011-01-29T17:01:41.507" Score="2" ViewCount="" Body="&lt;p&gt;Yes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Matrix Analysis&lt;/em&gt; by Horn and Johnson has a proof.&lt;/p&gt;&#xA;" OwnerUserId="4266" LastActivityDate="2011-01-29T17:01:41.507" CommentCount="1" />
  <row Id="19468" PostTypeId="2" ParentId="19464" CreationDate="2011-01-29T17:02:37.067" Score="6" ViewCount="" Body="&lt;h2&gt;Diagonalizable matrices&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Let's say that $A$ and $B$ are diagonalizable. Then, since they commute, there exists an invertible matrix $T$ such that both $T^{-1}AT = D_A$ and $T^{-1}BT = D_B$ are diagonal matrices (Horn &amp;amp; Johnson, Matrix Analysis, page 52). The elements of the diagonal matrices are the eigenvalues. Thus we get:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$A + B = TD_AT^{-1} + TD_BT^{-1} = T(D_A + D_B)T^{-1}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and so the eigenvalues of $A+B$ have the desired form.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;General case&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;In the general case you can use simultaneous triangularization. The idea is the same, since $A$ and $B$ commute, there exists a unitary $U$ such that $U^*AU = T_A$ and $U^*BU = T_B$ where $T_A$ and $T_B$ are upper triangular matrices with the eigenvalues on the diagonal (Horn &amp;amp; Johnson, Matrix Analysis, page 81). Now we get:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$A+B = U(T_A+T_B)U^*$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and thus the eigenvalues have the desired form.&lt;/p&gt;&#xA;" OwnerUserId="5966" LastActivityDate="2011-01-29T17:02:37.067" />
  <row Id="19469" PostTypeId="2" ParentId="19453" CreationDate="2011-01-29T17:36:36.197" Score="2" ViewCount="" Body="&lt;p&gt;CORDIC stands for &lt;a href=&quot;http://en.wikipedia.org/wiki/CORDIC&quot; rel=&quot;nofollow&quot;&gt; &quot;coordinate rotation digital computer&quot;&lt;/a&gt; and was described as a method for &lt;a href=&quot;http://www.jacques-laporte.org/Volder_CORDIC.pdf&quot; rel=&quot;nofollow&quot;&gt; computing trigonometric functions [PDF]&lt;/a&gt; by Jack E. Volder in 1959.  Later the method was generalized by John S. Walther at Hewlett-Packard to &lt;a href=&quot;http://www.jacques-laporte.org/Welther-Unified%20Algorithm.pdf&quot; rel=&quot;nofollow&quot;&gt; other elementary transcendental functions&lt;/a&gt; and operations useful in pocket calculators.  The advantage is simplicity of hardware required to perform additions, subtractions, bit-shifts, and table lookups (avoiding the need for hardware multiplication and division).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Applied to performing &lt;a href=&quot;http://www.drdobbs.com/184404244&quot; rel=&quot;nofollow&quot;&gt; a CORDIC division&lt;/a&gt;, these ideas amount to producing the bits (or digits) of the quotient one at a time by aligning the divisor &quot;under&quot; the dividend and subtracting until what remains in the dividend is too small to allow a further subtraction.  It is essentially the same as the long division algorithm taught in elementary school, especially if applied to binary numbers (where at a given position in the quotient, the divisor goes either 1 time or 0 times).  Cases where the divisor is zero must be trapped as exceptions.  If the divisor is negative, its sign and that of the dividend should both be changed (so that the algorithm is applied with a positive divisor).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See the illustrations here for &lt;a href=&quot;http://courses.cs.vt.edu/~cs1104/BuildingBlocks/divide.030.html&quot; rel=&quot;nofollow&quot;&gt; Binary Division by Shift and Subtract&lt;/a&gt;, including a Java applet that demonstrates the steps.&lt;/p&gt;&#xA;" OwnerUserId="3111" LastEditorUserId="3111" LastEditorDisplayName="" LastEditDate="2011-01-29T22:13:07.117" LastActivityDate="2011-01-29T22:13:07.117" />
  <row Id="19470" PostTypeId="1" AcceptedAnswerId="19472" CreationDate="2011-01-29T17:48:35.710" Score="0" ViewCount="55" Body="&lt;p&gt;Hello &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any general methods available for transforming a 2nd order Boundary value problem such as &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$F\left(x,y,\frac{\text{dy}}{\text{dx}},\frac{d^2y}{\text{dx}^2}\right)=0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$y(a)=y_0$ and $y(b)=y_1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;into an initial value problem?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regards &lt;/p&gt;&#xA;" OwnerUserId="6386" LastActivityDate="2011-02-28T18:39:19.820" Title="Are there any general methods available for transforming a BVP into an IVP" Tags="&lt;differential-equations&gt;" AnswerCount="1" />
  <row Id="19471" PostTypeId="1" AcceptedAnswerId="19473" CreationDate="2011-01-29T17:49:37.990" Score="5" ViewCount="132" Body="&lt;p&gt;I would like to know if every sufficiently differentiable function is convex near a local minimum. The background to my question is that I became curious if one could motivate the usefulness of convex optimization techniques by saying that at least locally they work for all continuous functions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately I discovered there are $C^1$-functions which have  minima where they are not locally convex. But my construction involves the use of everywhere non-differentiable continuous functions so I started to wonder if perhaps $C^2$-functions are convex near a local minimum. &lt;/p&gt;&#xA;" OwnerUserId="4623" LastActivityDate="2011-01-29T18:23:57.723" Title="Is a smooth function convex near a local minimum?" Tags="&lt;optimization&gt;&lt;convex-analysis&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="19472" PostTypeId="2" ParentId="19470" CreationDate="2011-01-29T17:54:20.120" Score="0" ViewCount="" Body="&lt;p&gt;For a certain class of BVP, the answer is yes according to &lt;a href=&quot;http://www.jstor.org/pss/2027566&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; article.&lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2011-01-29T17:54:20.120" CommentCount="0" />
  <row Id="19473" PostTypeId="2" ParentId="19471" CreationDate="2011-01-29T18:06:56.990" Score="8" ViewCount="" Body="&lt;p&gt;No, this isn't true. Let $\phi(x)$ be a nonnegative smooth &quot;bump&quot; function that is zero except on $(0,1)$ where it is positive. Then $\psi(x) = \sin^2({1 \over x})\phi(x)$ is a smooth nonnegative function (define $\psi(0) = 0$) which has zeroes at $x = {1 \over k\pi}$ for positive integers $k$ but is positive between these zeroes. $\psi(x)$ will still have a local minimum at zero because $\psi(0) = 0$, but because of the humps in $\psi(x)$ between zeroes, a chord connecting $({1 \over k\pi},0)$ to $({1 \over (k +1)\pi},0)$ will lie below the graph. So $\psi(x)$ is not convex.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I should add that if $x_0$ is a local minimum of $f(x)$ such that $f^{(l)}(x_0)$ is nonzero for some $l &amp;gt; 0$, then it will be convex on some interval centered at $x_0$ as long as $f(x)$ is $C^{l+1}$. To see this, note that without loss of generality we may assume $l$ is minimal. By Taylor expanding one gets&#xA;$$f(x) = {1 \over l!} f^{(l)}(x_0)(x - x_0)^l + O((x - x_0)^{l+1})$$&#xA;If $|x - x_0|$ is sufficiently small the remainder term will be dominated by the ${1 \over l!} f^{(l)}(x_0)(x - x_0)^l$ term. Thus the only way a local minimum can occur is if $l$ is even and $f^{(l)}(x_0) &amp;gt; 0$. Next note that the second derivative of $f$ has Taylor expansion given by&#xA;$${d^2 f \over dx^2} = {1 \over (l-2)!} f^{(l)}(x_0)(x - x_0)^{l-2} + O((x - x_0)^{l-1})$$&#xA;The remainder term is domninated by the first term once again, which is nonnegative on an interval containing $x_0$ since $l$ is even and $f^{(l)}(x_0) &amp;gt; 0$. Thus the second derivative of $f$ is nonnegative on this interval and therefore the function is convex there. &lt;/p&gt;&#xA;" OwnerUserId="3035" LastEditorUserId="3035" LastEditorDisplayName="" LastEditDate="2011-01-29T18:23:57.723" LastActivityDate="2011-01-29T18:23:57.723" CommentCount="1" />
  <row Id="19474" PostTypeId="1" AcceptedAnswerId="19475" CreationDate="2011-01-29T18:54:01.927" Score="2" ViewCount="123" Body="&lt;p&gt;Does every unfolding of an icosahedron has the same number of edges to be glued to construct it back to the solid?&#xA;If yes, what are those numbers for Platonic solids?&#xA;If no, which unfoldings have the least number of edges to be glued for Platonic solids?&lt;/p&gt;&#xA;" OwnerUserId="3806" LastActivityDate="2011-02-04T02:13:22.923" Title="Which unfolding of an icosahedron has the least number of edges to be glued?" Tags="&lt;geometry&gt;" AnswerCount="2" />
  <row Id="19475" PostTypeId="2" ParentId="19474" CreationDate="2011-01-29T19:00:37.280" Score="8" ViewCount="" Body="&lt;p&gt;Every unfolding corresponds to a spanning tree on the faces of the solid, giving $F-1$ edges which are already joined. So, you still have to glue $E-F+1$ edges.&lt;/p&gt;&#xA;" OwnerUserId="856" LastActivityDate="2011-01-29T19:00:37.280" CommentCount="3" />
  <row Id="19476" PostTypeId="2" ParentId="19457" CreationDate="2011-01-29T19:38:17.717" Score="1" ViewCount="" Body="&lt;p&gt;Have you tried matlab?&#xA;If you say:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;int('1/sqrt(x^2+y^2)','y','sqrt(1-x^2)','x')&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;then you'll get: &#xA;&lt;code&gt;log(x + (2*x^2)^(1/2)) - log((1 - x^2)^(1/2) + 1)&lt;/code&gt; as first integral.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then: &lt;code&gt;int('log(x + (2*x^2)^(1/2)) - log((1 - x^2)^(1/2) + 1)', 'x')&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Result:  &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;x*(log(x + (2*x^2)^(1/2)) - 1) - int(log((1 - x^2)^(1/2) + 1), x)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;But the last part can't be integrated symbolically, so, in matlab, you can evaluate it numerically:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;`double(int('log((1 - x^2)^(1/2) + 1)', 'x',sqrt(2)/2,1))` &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;which will give : &lt;code&gt;ans = 0.1143&lt;/code&gt;, and the other part you can evaluate yourself.&#xA;It should be easy to replace x with the 2 numbers.&#xA;Hope this helps!&lt;/p&gt;&#xA;" OwnerUserId="6388" LastActivityDate="2011-01-29T19:38:17.717" CommentCount="2" />
  <row Id="19477" PostTypeId="2" ParentId="19463" CreationDate="2011-01-29T19:42:32.990" Score="2" ViewCount="" Body="&lt;p&gt;As per Adrián Barquero's request, I post my comment as an answer (even if I think that it doesn't answer the question but rather indicates that the answer isn't really known).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See the related discussion on &lt;a href=&quot;http://mathoverflow.net/questions/17357/what-is-so-spectral-about-spectral-sequences&quot; rel=&quot;nofollow&quot;&gt;Math Overflow&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="5363" LastActivityDate="2011-01-29T19:42:32.990" />
  <row Id="19478" PostTypeId="2" ParentId="19427" CreationDate="2011-01-29T20:20:30.223" Score="1" ViewCount="" Body="&lt;p&gt;Galois Theory is probably the first example of a successful application of an abstract theory to solving a concrete problem. &lt;/p&gt;&#xA;" OwnerUserId="589" LastActivityDate="2011-01-29T20:20:30.223" CommunityOwnedDate="2011-01-29T20:20:30.223" CommentCount="4" />
  <row Id="19479" PostTypeId="1" AcceptedAnswerId="19509" CreationDate="2011-01-29T20:36:42.900" Score="1" ViewCount="176" Body="&lt;p&gt;Hello, I'm looking at this second order quasilinear PDE:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\alpha_{xx} - \alpha_{yy} - m\alpha^2 = 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Attempted Strategies&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Fourier Transform resulted in convolution due to the $\alpha^2$ term, I don't think that would be pretty to work on.&lt;/li&gt;&#xA;&lt;li&gt;Separation of Variables resulted in not being able to separate the variables.&lt;/li&gt;&#xA;&lt;li&gt;I tried substituting $u=\alpha_x$ and $v=\alpha_y$ but I'm left with $(\int u_x dx)^2$ or $(\int u_y dy)^2$ in my equation afterwords due to substituting $u$ and $v$ back in, which seems like a dead end.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Do you guys have any suggestions for an analytical solution method?&lt;/p&gt;&#xA;" OwnerUserId="6390" LastEditorUserId="6390" LastEditorDisplayName="" LastEditDate="2011-01-29T20:55:17.107" LastActivityDate="2011-01-30T04:23:01.637" Title="Quasilinear PDE of Second Order - Some Analytical Direction Needed" Tags="&lt;differential-equations&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="19480" PostTypeId="1" AcceptedAnswerId="19489" CreationDate="2011-01-29T20:53:32.443" Score="3" ViewCount="118" Body="&lt;p&gt;I was reading the Wikipedia article of &lt;a href=&quot;http://en.wikipedia.org/wiki/Dynamic_programming&quot; rel=&quot;nofollow&quot;&gt;Dynamic programming&lt;/a&gt;, however, I'm having a hard time understanding the explanation given in the example for &lt;a href=&quot;http://en.wikipedia.org/wiki/Dynamic_programming#A_type_of_balanced_0-1_matrix&quot; rel=&quot;nofollow&quot;&gt;a type of balanced 0-1 matrix&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is stated as:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;blockquote&gt;&#xA;    &lt;p&gt;Consider the problem of assigning values, either zero or one, to the positions of an n × n matrix, n even, so that each row and each column contains exactly n / 2 zeros and n / 2 ones.&lt;/p&gt;&#xA;  &lt;/blockquote&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I don't quote the whole text because I suppose there is no need to do that. However, I don't really understand the explanation after &quot; The process of subproblem creation involves iterating...&quot; in the third paragraph of the subsection.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyone can explain better that procedure? Of course, an example of this at work would be great.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="2569" LastActivityDate="2011-01-29T21:42:20.420" Title="Dynamic programming - A type of balanced 0-1 matrix" Tags="&lt;optimization&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="19481" PostTypeId="1" CreationDate="2011-01-29T20:53:50.233" Score="2" ViewCount="62" Body="&lt;p&gt;At one step of an algorithm (i=1: something big) I have to draw a vector from the normal distribution $X_p^{(i)}\sim N((A^{(i)} + D^{(i)}_p)^{-1}m, (A^{(i)} + D^{(i)}_p)^{-1})$ for $p=1:P$, with a large $P$ (say 5,000). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strike&gt;$A^{(i)}$ is symmetric positive definite and rank $K$,&lt;/strike&gt; $A^{(i)}$ is $K\times K$ symmetric positive definite with $K$ is much smaller than $P$. $(A^{(i)})^{-1}$ is cheap relative to the rest of the calculation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I'd like to efficiently get $(A + D^{(i)}_p)^{-1}$ (or a matrix square root).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've tried using the matrix inversion lemma (aka Woodbury) to get a quick update from $A^{-1}$ or $(A+D_p)^{-1}$ to $(A+D_{p+1})^{-1}$, but it takes me in circles because it's a full rank update.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It &lt;em&gt;feels&lt;/em&gt; like this should have a simple solution since the $D^{(i)}_p$ are all diagonal, but if there is it escapes me...&lt;/p&gt;&#xA;" OwnerUserId="6389" LastEditorUserId="6389" LastEditorDisplayName="" LastEditDate="2011-01-29T22:27:31.360" LastActivityDate="2011-05-30T00:05:48.863" Title="Inverting $(A + D_p)$ efficiently for many diagonal $D_p$ (and $A$ SPD)" Tags="&lt;linear-algebra&gt;&lt;statistics&gt;&lt;matrices&gt;&lt;inversion&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="19482" PostTypeId="2" ParentId="19481" CreationDate="2011-01-29T21:04:47.677" Score="0" ViewCount="" Body="&lt;p&gt;I assume you mean $A$ is symmetric positive semidefinite with rank $k$. If you can do an efficient low rank factorization of $A = U U^T$, then &lt;a href=&quot;http://en.wikipedia.org/wiki/Woodbury_matrix_identity&quot; rel=&quot;nofollow&quot;&gt;Woodbury update&lt;/a&gt; is the way to go for this. Further if you know the rank $k$ before hand then you could actually get a low-rank factorization in $\mathcal{O}(k^2n)$ where $n$ is the size of $A$.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastActivityDate="2011-01-29T21:04:47.677" CommentCount="1" />
  <row Id="19483" PostTypeId="1" AcceptedAnswerId="19490" CreationDate="2011-01-29T21:18:05.220" Score="3" ViewCount="93" Body="&lt;p&gt;I am taking a beginner linear algebra class and would like a hint on a homework question.  I don't need the answer, just a little guidance.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the differential equation:&#xA;$$\frac{d^2x}{dt^2} - \frac{dx}{dt} - x = \cos(t).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We are told that the differential equation has a solution of the form&#xA;$$x(t) = a\sin(t) + b\cos(t).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Find $a$ and $b$, and graph the solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This question comes from the first section in chapter one of Linear Algebra with Applications Fourth Edition by Otto Bretscher.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know how to set up the system to use elimination to solve.&lt;/p&gt;&#xA;" OwnerUserId="6391" LastEditorUserId="6391" LastEditorDisplayName="" LastEditDate="2011-01-29T23:59:55.847" LastActivityDate="2011-01-30T19:07:53.470" Title="Linear Systems: Elimination" Tags="&lt;linear-algebra&gt;&lt;homework&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="19485" PostTypeId="1" AcceptedAnswerId="19488" CreationDate="2011-01-29T21:20:48.133" Score="10" ViewCount="327" Body="&lt;p&gt;I've never really understood why math induction is supposed to work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have these 3 steps:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Prove true for base case (n=0 or 1 or whatever)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Assume true for n=k.  Call this the &lt;em&gt;induction hypothesis&lt;/em&gt;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Prove true for n=k+1, &lt;strong&gt;somewhere using the induction hypothesis&lt;/strong&gt; in your proof.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;In my experience the proof is usually algebraic, and you just manipulate the problem until you get the induction hypothesis to appear.  If you can do that and it works out, then you say the proof holds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's one I just worked out,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Show $\displaystyle\lim_{x\to\infty} \frac{(\ln x)^n}{x} = 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you go:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Use L'Hospital's rule.&#xA;$\displaystyle\lim_{x\to\infty} \frac{\ln x}{x} = 0$.&#xA;Since that's&#xA;$\displaystyle\lim_{x\to\infty} \frac{1}{x} = 0$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Assume true for $n=k$&lt;/strong&gt;.&#xA;$\displaystyle\lim_{x\to\infty} \frac{(\ln x)^k}{x} = 0$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Prove true for $n=k+1$.  You get&#xA;$\displaystyle\lim_{x\to\infty} \frac{(\ln x)^{k+1}}{x} = 0.$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Use L'Hospital again:&#xA;$\displaystyle\lim_{x\to\infty} \frac{(k+1)(\ln x)^{k}}{x} = 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then you see the induction hypothesis appear, and you can say this is equal to $0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I'm not comfortable with is this idea that you can just &lt;em&gt;assume&lt;/em&gt; something to be true (n=k), then based on that assumption, form a proof for $n=k+1$ case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't see how you can use something you've assumed to be true to prove something else to be true.&lt;/p&gt;&#xA;" OwnerUserId="1069" LastEditorUserId="6296" LastEditorDisplayName="" LastEditDate="2011-03-04T05:54:52.067" LastActivityDate="2011-03-04T15:06:16.717" Title="Dominoes and induction, or how does induction work?" Tags="&lt;intuition&gt;&lt;induction&gt;" AnswerCount="5" CommentCount="3" FavoriteCount="2" />
  <row Id="19486" PostTypeId="2" ParentId="19485" CreationDate="2011-01-29T21:27:37.633" Score="4" ViewCount="" Body="&lt;p&gt;You prove the base case for n=1 (or 0, let's use 1 for the sake of simplicity), and then you use inductive reasoning. The way you do it us by arguing that &lt;em&gt;if&lt;/em&gt; the formula holds for n=k, it must also hold for n=k+1. &lt;em&gt;If&lt;/em&gt; is the key here. Once you have proven that, you have effectively proven that the formula will work for all n, and here's why -- you can always start with n=1 (n=k), and prove that n=1+1 (n=k+1) will work. Then you can treat n=1+1 as n=k, and prove than n=1+1+1 (n=k+1) will satisfy the formula too, and so on to the infinity. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, since you have demonstrated that the &quot;base step&quot; works, you only need to demonstrate that the &quot;base step + 1&quot; will work, too. Then the &quot;base step + 1&quot; can be treated as a 'new base step,' crudely put, and the sequence of &quot;base step + 1&quot; can be continued on and on. It does not matter if the statement holds when n=k or not, all that matters is proving the implication -- if it holds for k, it will also hold for k+1. That allows you to move further from the base step.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, maybe, the more correct way of stating the method is &quot;Assuming that statement holds for some n, show that it holds for n+1.&quot; Once you get that, you can go further infinitely as described above. Obviously, this principle does not work for all proofs, as there are problems where you will need to use Principle of Strong Induction (break the problem down into cases, for which n and n+1 actually correspond). &lt;/p&gt;&#xA;" OwnerUserId="3731" LastEditorUserId="3731" LastEditorDisplayName="" LastEditDate="2011-01-29T21:35:42.677" LastActivityDate="2011-01-29T21:35:42.677" />
  <row Id="19488" PostTypeId="2" ParentId="19485" CreationDate="2011-01-29T21:38:08.553" Score="18" ViewCount="" Body="&lt;p&gt;The inductive step is a proof of an implication: you are proving that &lt;strong&gt;if&lt;/strong&gt; the property you want holds for $k$, &lt;strong&gt;then&lt;/strong&gt; it holds for $k+1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is a result of formal logic that if you can prove $P\rightarrow Q$ (that $P$ implies $Q$), then from $P$ you can prove $Q$; and &lt;em&gt;conversely&lt;/em&gt;, that if from assuming that $P$ is true you can prove $Q$, then you can in fact prove $P\rightarrow Q$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We do this pretty much every time we prove something. For example, suppose you want to prove that if $n$ is a natural number, then $n^2$ is a natural number. How do we start? &quot;Let $n$ be a natural number.&quot; Wait! Why are you allowed to just &lt;em&gt;assume&lt;/em&gt; that you already have a natural number? Shouldn't you have to start by &lt;em&gt;proving&lt;/em&gt; it's a natural number? The answer is &lt;strong&gt;no&lt;/strong&gt;, we don't have to, because we are not trying to prove an absolute, we are trying to prove a &lt;em&gt;conditional&lt;/em&gt; statement: that &lt;strong&gt;if&lt;/strong&gt; $n$ is a natural number, &lt;strong&gt;then&lt;/strong&gt; something happens. So we may begin by assuming we are already in the case where the antecedent is true. (Intuitively, this is because &lt;em&gt;if&lt;/em&gt; the antecedent is false, then the implication is necessarily true and there is nothing to be done; formally, it is because the Deduction Theorem, which is what I described above, tells you that if you manage to find a formal proof that ends with &quot;$n^2$ is a natural number&quot; by assuming that &quot;$n$ is a natural number&quot; is true, then you can use that proof to produce a formal proof that establishes the implication &quot;if $n$ is a natural number then $n^2$ is a natural number&quot;; we don't have to go through the exercise of actually producing the latter proof, we know it's &quot;out there&quot;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We do that in Calculus: &quot;if $\lim\limits_{x\to x_0}f(x) = a$ and $\lim\limits_{x\to x_0}g(x) = b$, then $\lim\limits_{x\to x_0}(f(x)+g(x)) = a+b$.&quot; How do we prove this? We begin by &lt;em&gt;assuming&lt;/em&gt; that the limit of $f(x)$ as $x\to x_0$ &lt;em&gt;is&lt;/em&gt; $a$, and that the limit of $g(x)$ as $x\to x_0$ &lt;em&gt;is&lt;/em&gt; $b$. We assume the premise/antecedent, and procede to try to prove the consequent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What this means in the case of induction is that, since the &quot;Inductive Step&quot; is actually a statement that says that an implication holds:&#xA;$$\mbox{&quot;It&quot; holds for $k$}\rightarrow \mbox{&quot;it&quot; holds for $k+1$},$$&#xA;then in order to prove this implication we can begin by assuming that the &lt;em&gt;antecedent&lt;/em&gt; is already true, and then procede to prove the consequent. Assuming that the antecedent is true is precisely the &quot;Induction Hypothesis&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you are done with the inductive step, you have in fact not proven that it holds for any &lt;em&gt;particular&lt;/em&gt; number, you have &lt;em&gt;only&lt;/em&gt; shown that &lt;strong&gt;if&lt;/strong&gt; it holds for a particular number $k$, &lt;strong&gt;then&lt;/strong&gt; it must hold for the next number $k+1$. It is a &lt;em&gt;conditional&lt;/em&gt; statement, not an absolute one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is only when you combine that conditional statement with the base, which &lt;em&gt;is&lt;/em&gt; an absolute statement that says &quot;it&quot; holds for a specific number, that you can conclude that the original statement holds for all natural numbers (greater than or equal to the base). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since you mention dominoes in your title, I assume you are familiar with the standard metaphor of induction like dominoes that are standing all in a row falling. The inductive step is like arguing that all the dominos will fall if you topple the first one (without actually toppling it): first, you argue that each domino is sufficiently close to the &lt;em&gt;next&lt;/em&gt; domino so that &lt;strong&gt;if&lt;/strong&gt; one falls, &lt;strong&gt;then&lt;/strong&gt; the next one falls. You are not tumbling every domino. And when you argue this, you argue along the lines of &quot;suppose this one falls; since it's length is ...&quot;, that is, you &lt;em&gt;assume&lt;/em&gt; it falls in order to argue the next one will &lt;em&gt;then&lt;/em&gt; fall. This is the same with the inductive step.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a sense you are right that if feels like &quot;cheating&quot; to assume what you want; but the point is that you aren't really assuming what you want. Again, the inductive step does not in fact establish that the result holds for &lt;em&gt;any&lt;/em&gt; number, it only establishes a &lt;em&gt;conditional&lt;/em&gt; statement. &lt;strong&gt;If&lt;/strong&gt; the result happens to hold for some $k$, &lt;strong&gt;then&lt;/strong&gt; it would &lt;em&gt;necessarily&lt;/em&gt; have to also hold for $k+1$. But we are completely silent on whether it &lt;em&gt;actually&lt;/em&gt; holds for $k$ or not. We are not saying anything about that at the inductive-step stage. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;hr/&gt;&#xA;&lt;em&gt;Added:&lt;/em&gt; Here's an example to emphasize that the &quot;inductive step&quot; does not make any absolute statement, but only a conditional statement: Suppose you want to prove that for all natural numbers $n$, $n+1 = n$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Inductive step.&lt;/strong&gt; &lt;em&gt;Induction Hypothesis:&lt;/em&gt; The statement holds for $k$; that is, I'm assuming that $k+1 = k$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;To be proven:&lt;/em&gt; The statement holds for $k+1$. Indeed: notice that since $k+1= k$, then adding one to both sides of the equation we have $(k+1)+1 = k+1$; this proves the statement holds for $k+1$. &lt;strong&gt;QED&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a perfectly valid proof! It says that &lt;strong&gt;if&lt;/strong&gt; $k+1=k$, &lt;strong&gt;then&lt;/strong&gt; $(k+1)+1=k+1$.  &lt;em&gt;This is true!&lt;/em&gt; Of course, the antecedent is never true, but the &lt;em&gt;implication&lt;/em&gt; is. The reason this is not a full proof by induction of a false statement is that there is no &quot;base&quot;; the inductive step only proves the conditional, nothing more. &lt;/p&gt;&#xA;&#xA;&lt;hr/&gt;&#xA;&#xA;&lt;p&gt;By the way: Yes, most proofs by induction that one encounters early on involve algebraic manipulations, but not all proofs by induction are of that kind. Consider the following simplified game of Nim: there are a certain number of matchsticks, and players alternate taking $1$, $2$, or $3$ matchsticks every turn. The person who takes the last matchstick wins. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proposition.&lt;/strong&gt; In the simplified game above, the first player has a winning strategy if the number of matchsticks is not divisible by $4$, and the second player has a winning strategy if the number of matchsticks is divisible by 4.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The proof is by (strong) induction, and it involves no algebraic manipulations whatsoever.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-30T04:58:38.250" LastActivityDate="2011-01-30T04:58:38.250" CommentCount="1" />
  <row Id="19489" PostTypeId="2" ParentId="19480" CreationDate="2011-01-29T21:42:20.420" Score="4" ViewCount="" Body="&lt;p&gt;The dynamic programming algoritm uses a memoized recursive function that solves a more flexible subproblem :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The function takes a number k of rows to fill, and for each column, the number of 1s it has to place in those k rows.&#xA;It has to return the number of possible assignments of 1s in the block of k rows such that each row contains n/2 1s, and each column contains the proper number of 1s stated in the argument.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The principle is that :&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;if a column has to contain a negative number of 1s, there is 0 solution.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;if k=0 (and all the columns have to be empty), then there is 1 solution.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;otherwise, an assignment of a block of k rows is an assignment of the 1st row + an assignment of a block (k-1) rows.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So it iterates over the possible assignments of the 1st row. For each candidate assignment, it recursively asks itself for the number of ways of completing the remaining (k-1) rows with the updated information about the number of 1s that have to be in each column.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you don't memoise the function, it is exactly what the backtracking algorithm does, essentially it is n nested extremely big loops.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you memoise the function, you will find that it is called for the same k and for the same constraints on the column a great number of times. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, while doing the backtracking algorithm, at some point you chose the assignment A1 for row 1 and A2 for row 2, and proceed to count every solution starting with (A1,A2) with (n-2) nested loops. Then later in the execution, you choose the assignment A2 for row 1 and A1 for row 2, and proceed to count every solution starting with (A2,A1), again with (n-2) nested loops.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if you look carefully at the subproblem you have to solve, it has the same constraints in both case.&#xA;A solution in the first case is the same as a solution in the 2nd case, only with the first two rows switched. So you are counting the same number twice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you memoize your recursive function, then it will recognize that these two subproblems are equivalent because they ask for the same constraints, and only compute its result once. In all the later calls it will directly return with that same result, and avoid a very expensive recomputation.&lt;/p&gt;&#xA;" OwnerUserId="6270" LastActivityDate="2011-01-29T21:42:20.420" CommentCount="1" />
  <row Id="19490" PostTypeId="2" ParentId="19483" CreationDate="2011-01-29T21:44:20.133" Score="2" ViewCount="" Body="&lt;p&gt;I hope you won't try to solve the differential equation from scratch!${}^{1}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you just plug in the given $x(t)$ into the equation, what do we get?&#xA;\begin{align*}&#xA;x(t) &amp;amp; = a\sin(t) + b\cos (t)\\&#xA;x'(t) &amp;amp;= a\cos(t) - b\sin(t)\\&#xA;x''(t) &amp;amp;= -a\sin(t) -b\cos(t)\\&#xA;\frac{d^2 x}{dt^2} - \frac{dx}{dt} - x &amp;amp;=&#xA;(-a\sin t-b\cos t) - (a\cos t - b\sin t) - (a\sin t + b\cos t)\\&#xA;&amp;amp;= (-2a+b)\sin t + (-2b-a)\cos t.&#xA;\end{align*}&#xA;Since you &lt;em&gt;know&lt;/em&gt; that this is a solution, you &lt;em&gt;should&lt;/em&gt; get $\cos t$. That tells you something about the coefficients of $\sin t$ and $\cos t$ that you got by plugging in. Does it look like a system of linear equations?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;${}^1$: Happens every semester that I teach Calculus II; the book has early mention of Differential Equations as something integrals will be good for, but the students don't have the tools yet to solve even the most simple differential equations. A few problems ask the students to verify that a certain function is a solution, or to find conditions on certain parameters for a given function to &lt;em&gt;be&lt;/em&gt; a solution, and invariably 99% of the students have a pavlovian response and try to &lt;em&gt;solve&lt;/em&gt; the given equation and if successful compare the solution they find with the given solution, instead of simply plugging in the given answer to check if it works. This happens even though I tell them that to &lt;em&gt;check&lt;/em&gt; something is a solution you don't need to solve, even though I do examples on the board before the homework is due, even though I give them quiz problems that specifically say things like &lt;em&gt;&quot;even though you cannot yet find the general solution to this equation&quot;&lt;/em&gt;... Sigh. &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-30T19:07:53.470" LastActivityDate="2011-01-30T19:07:53.470" CommentCount="6" />
  <row Id="19491" PostTypeId="1" CreationDate="2011-01-29T22:50:35.200" Score="0" ViewCount="243" Body="&lt;p&gt;Here's the question:&#xA;Solve the recurrence by obtaining a $\theta$ bound for $T(n)$ given that $T(1) = \theta(1)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$T(n) = n + T(n-3).$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Attempted Solution:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align*}&#xA;T(n) &amp;amp;= T(n-6) + (n-3) + n \\&#xA;&amp;amp;= T(n-9) + (n-6) + (n-3) + n \\ &#xA;&amp;amp;= T(n-(n-1)) + [(n-n) + (n-(n-3)) + (n-(n-6)) + \cdots + n]\\&#xA;&amp;amp;= T(1) + [0 + 3 + 6 + \cdots + n]\\&#xA;&amp;amp;= \theta(1) = 3[1 + 2 + 3 + \cdots + n/3]\\&#xA;&amp;amp;= \theta(1) + \frac{(n/3)(n/3 + 1)}{2}\\&#xA;&amp;amp;= \theta(1) + \frac{n^2+3n}{6}.&#xA;\end{align*}&#xA;When I double check to see if the solution fits the recurrence, it doesn't work. I can solve these fairly easily when it involves the previous term, but this $T(n-3)$ is throwing me off.  Been at this one for a while now; please help T__T&lt;/p&gt;&#xA;" OwnerUserId="6392" LastEditorUserId="1919" LastEditorDisplayName="" LastEditDate="2011-01-29T23:43:53.510" LastActivityDate="2011-01-31T02:30:43.063" Title="Recurrence Relation Homework Struggles" Tags="&lt;homework&gt;&lt;discrete-mathematics&gt;" AnswerCount="4" CommentCount="15" />
  <row Id="19492" PostTypeId="1" CreationDate="2011-01-29T22:51:03.077" Score="2" ViewCount="95" Body="&lt;p&gt;Hi all,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm trying to show that, if we embed $S^m$ in $S^n$ as the subspace {$x_1,x_2,...,x_{m+1},0,0,...0$}, $S^n-S^m$ is homotopy equivalent to $S^{n-m-1}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to show the homotopy equivalence, I tried to use the inclusion map $S^{n-m-1}$ into $S^n - S^m$, including in the last n-m coordinates, and I'm trying to use funcions along the lines of &lt;/p&gt;&#xA;&#xA;&lt;p&gt;f($x_1,x_2,...,x_{n+1}$)=($x_{m+1},x_{m+2},...,x_{n},\sqrt{x_1^2+...x_m^2+x_{n+1}^2}$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then f $\circ$ g is easily shown to be the identity map except with issues of positivity in the last coordinate, so I tried &lt;/p&gt;&#xA;&#xA;&lt;p&gt;f($x_1,x_2,...,x_{n+1}$)=($x_{m+1},x_{m+2},...,x_{n},a(x_{n+1})\sqrt{x_1^2+...x_m^2+x_{n+1}^2}$), &lt;/p&gt;&#xA;&#xA;&lt;p&gt;where a is either $\pm 1$ or 0 depending on the sign of the final coordinate, but then there's an issue in the other direction, i.e. that the final coordinate can only be 0 when x$_{n+1}$=0. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it possible to amend my approach somehow so that g $\circ$ f is also homotopic to the identity map? Thankyou for any thoughts or advice,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;M.&lt;/p&gt;&#xA;" OwnerUserId="6393" LastActivityDate="2011-01-30T00:01:00.380" Title="Homotopy equivalence of Unit Balls" Tags="&lt;homework&gt;&lt;topology&gt;&lt;algebraic-topology&gt;&lt;vector-space&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" />
  <row Id="19493" PostTypeId="2" ParentId="19187" CreationDate="2011-01-29T22:52:31.080" Score="0" ViewCount="" Body="&lt;p&gt;If your course was a TCS-Course, then maybe your professor mixed up countable sets with enumerable sets. (This happens sometimes, especially in german language areas, where it is &quot;abzählbar&quot; vs. &quot;aufzählbar).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Every subset of a countable set is countable or finite (or empty). However, not every subset of an enumerable set is enumerable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, $\mathbb{N}$ is trivially enumerable. But the codomain of the busy bever function, which is a subset of $\mathbb{N}$, is not.&lt;/p&gt;&#xA;" OwnerUserId="1858" LastActivityDate="2011-01-29T22:52:31.080" CommentCount="1" />
  <row Id="19494" PostTypeId="2" ParentId="19479" CreationDate="2011-01-29T22:57:44.490" Score="1" ViewCount="" Body="&lt;p&gt;I think it's pretty hopeless to get an explicit formula for solutions for this equation (although specific examples of solutions might be known). Your equation is called the nonlinear wave equation, and it is an active area of research (to get a glimpse of the variety of research results, find the DispersiveWiki website, although this will not help in your case). In general it is not even true that an initial value problem (that is, prescribing the solution for $x=0$, here I'm interpreting your $x$ as time) has a global solution; often one has local solutions (that is, for small $x$) which can blow up in finite time.&lt;/p&gt;&#xA;" OwnerUserId="1609" LastEditorUserId="1609" LastEditorDisplayName="" LastEditDate="2011-01-29T23:05:38.300" LastActivityDate="2011-01-29T23:05:38.300" CommentCount="6" />
  <row Id="19495" PostTypeId="1" CreationDate="2011-01-29T23:20:33.037" Score="3" ViewCount="133" Body="&lt;p&gt;The ODE: $y&amp;#39; = (1-2x)y^2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Initial Value: $y(0) = -1/6$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've solved the particular solution, which is $1/(x^2-x-6)$. I don't understand what they mean about the &lt;em&gt;solution&lt;/em&gt; is defined, because when I graph $1/(x^2-x-6)$, it's only discontinuous at $x = -2, 3$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What does &quot;Determine the &lt;em&gt;interval&lt;/em&gt; in which the solution is defined&quot; mean?&lt;/p&gt;&#xA;" OwnerUserId="6394" LastEditorUserId="6394" LastEditorDisplayName="" LastEditDate="2011-01-29T23:35:19.830" LastActivityDate="2011-01-31T07:15:39.643" Title="Determine the *interval* in which the solution is defined?" Tags="&lt;differential-equations&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="19496" PostTypeId="2" ParentId="19491" CreationDate="2011-01-29T23:24:07.710" Score="2" ViewCount="" Body="&lt;p&gt;From the initial condition on $T(1)$ and the recurrence relation, there is no way to estimate $T(n)$ unless $n=3k+1$ for some $k\ge0$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For every $k\ge0$, let $U(k)=T(3k+1)$. Then $U(k)=3k+1+U(k-1)$, hence $U(k)=(3k+1)+(3k-2)+\cdots+4+U(0)$, that is, $T(3k+1)=\frac12k(3k+5)+\theta(1)$.&lt;/p&gt;&#xA;" OwnerUserId="6179" LastActivityDate="2011-01-29T23:24:07.710" />
  <row Id="19497" PostTypeId="1" AcceptedAnswerId="19532" CreationDate="2011-01-29T23:25:21.130" Score="6" ViewCount="356" Body="&lt;p&gt;I was reading my daily reddit and came accross this link to a new double major at Oxford, Computer Science and Philosophy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.comlab.ox.ac.uk/admissions/ugrad/Computer_Science_and_Philosophy&quot; rel=&quot;nofollow&quot;&gt;http://www.comlab.ox.ac.uk/admissions/ugrad/Computer_Science_and_Philosophy&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I just wish my university had a double major like this, i would have taken it as an undergrad for sure (I am now doing graduate research in CS). Anyways, i wanted to ask what books are good for beginning and intermediate levels in Logic. I know theres like Deductive Logic, Set Theory, but the actual books i don't know any.&lt;/p&gt;&#xA;" OwnerUserId="43" LastEditorUserId="6622" LastEditorDisplayName="" LastEditDate="2011-04-24T08:14:28.183" LastActivityDate="2011-04-30T21:06:35.270" CommunityOwnedDate="2011-05-05T18:35:43.217" Title="Philosophy (Logic)" Tags="&lt;logic&gt;&lt;books&gt;&lt;philosophy&gt;" AnswerCount="4" CommentCount="1" FavoriteCount="4" />
  <row Id="19498" PostTypeId="2" ParentId="19495" CreationDate="2011-01-29T23:28:36.110" Score="3" ViewCount="" Body="&lt;p&gt;The solution is not only &lt;em&gt;discontinuous&lt;/em&gt; at $x=-2$ and $x=3$, it is also &lt;em&gt;undefined&lt;/em&gt; at these points. The maximal interval containing $0$ on which there exists a solution is the open interval $(-2,3)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit to answer a comment:&lt;/strong&gt; More generally the ODE around a point $x$ such that $y(x)\ne0$ is equivalent to $y&amp;#39;/y^2=1-2x$, that is, $(1/y)&amp;#39;=(x^2-x)&amp;#39;$. Hence, on every interval where $y$ is defined and not identically zero, there exists a constant $c$ such that $y(x)=1/(x^2-x+c)$. The value of $c$ depends on the initial condition one is given, hence the interval on which $y$ is defined also depends on the initial condition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your example, $y(0)=-1/6$, hence $c=-6$, $y(x)=1/(x^2-x-6)$ and the maximal interval of definition around $0$ is bounded by the roots of $x^2-x-6=(x+2)(x-3)$ which are closest to $0$, namely $-2$ and $3$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But consider another example: if $y(-3)=-1/8$, then $c=-20$, $y(x)=1/(x^2-x-20)$ and the maximal interval of definition around $-3$ is bounded by the roots of $x^2-x-20=(x-5)(x+4)$ which are closest to $-3$, hence this interval is $(-4,+5)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The interval can also be semi-infinite or infinite. For instance, if $y(-3)=1/15$, then $c=3$, $y(x)=1/(x^2-x+3)$ and the maximal interval of definition around $-3$ is the real line $(-\infty,+\infty)$. If $y(-3)=1/12$, then $c=0$, $y(x)=1/(x^2-x)$ and the maximal interval of definition around $-3$ is the half line $(-\infty,0)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And so on.&lt;/p&gt;&#xA;" OwnerUserId="6179" LastEditorUserId="6179" LastEditorDisplayName="" LastEditDate="2011-01-31T07:15:39.643" LastActivityDate="2011-01-31T07:15:39.643" CommentCount="3" />
  <row Id="19499" PostTypeId="1" CreationDate="2011-01-29T23:39:59.913" Score="2" ViewCount="266" Body="&lt;p&gt;Let $f(x)$ be a real-valued function on $\mathbb{R}$ such that $x^nf(x), n=0,1,2,\ldots$ are &#xA;Lebesgue integrable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose $$\int_{-\infty}^\infty x^n f(x) dx=0$$ for all $n=0,1,2,\ldots. $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does it follow that  $f(x)=0$ almost everywhere?&lt;/p&gt;&#xA;" OwnerUserId="3249" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2011-01-30T10:43:33.990" LastActivityDate="2011-01-30T10:51:05.907" Title="Does $\int_{\mathbb R} f(x)x^n dx = 0$ for $n=0,1,2,\ldots$ imply $f=0$ a.e.?" Tags="&lt;integral&gt;&lt;measure-theory&gt;" AnswerCount="1" CommentCount="12" FavoriteCount="2" />
  <row Id="19500" PostTypeId="1" AcceptedAnswerId="19512" CreationDate="2011-01-29T23:41:07.607" Score="0" ViewCount="121" Body="&lt;p&gt;Since I often draw wrong conclusions, I need a confirmation.&#xA;My argument is that for nonzero prime $\mathfrak{p}\subset A$, $A/\mathfrak{p}$ is injective being a field (hence divisible), and $A/p^{k}$ is its quotient hence again is injective, and the f.g. torsion module is a finite direct sum of such quotients hence is injective.&#xA;I couldn't find any book that mentions this fact and I'm wondering I might have made a mistake somewhere.&#xA;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="4341" LastActivityDate="2011-01-30T06:10:56.680" Title="Is a finitely generated torsion module over PID an injective module?" Tags="&lt;algebra&gt;&lt;modules&gt;" AnswerCount="2" CommentCount="7" />
  <row Id="19501" PostTypeId="2" ParentId="19492" CreationDate="2011-01-29T23:46:20.067" Score="1" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Hint:&lt;/strong&gt; If $(x_{1}, \ldots, x_{m+1},y_{m+2},\ldots,y_{n+1}) \in S^{n} \smallsetminus S^{m}$ then $\frac{(\lambda x, y)}{\|(\lambda x,y)\|} \in S^{n}$ for all $\lambda$. Use this to retract $S^{n} \smallsetminus S^{m}$ to $S^{n-m-1}$ (embedded as $(0,y)$).&lt;/p&gt;&#xA;" OwnerUserId="5363" LastEditorUserId="5363" LastEditorDisplayName="" LastEditDate="2011-01-30T00:01:00.380" LastActivityDate="2011-01-30T00:01:00.380" CommentCount="1" />
  <row Id="19502" PostTypeId="2" ParentId="19491" CreationDate="2011-01-29T23:55:05.760" Score="1" ViewCount="" Body="&lt;ul&gt;&#xA;&lt;li&gt;if you have no clue about the general manipulation of the formula, try some specifics. For example, let $T(0)=T(1)=T(2)=0$, then $T(3) =$ ??, $T(6) =$ ??, etc. and then guess a solution.&lt;/li&gt;&#xA;&lt;li&gt;you have no problem with recurrences with T(n-1). One trick with recurrences is to convert the weird inscrutable ones (like for you this $T(n-3)$) -somehow- to $T(n-1)$. The previous bullet point will help with that...create a new function to solve for $S(n)$ where $S(n) =  T(3n)$ (you might also have to do it for $T(3n+1)$ and $T(3n+2)$ ). This is called  'change of variables'.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This might seem like more work than it's worth, just to solve this problem, but you've already spent time trying to solve it without getting anywhere so might as well try the extra work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Actually...you seem to have implicitly done these steps in your attempted solution. Have you checked your steps? What is $n/3(n/3+1)$?&lt;/p&gt;&#xA;" OwnerUserId="1919" LastEditorUserId="1919" LastEditorDisplayName="" LastEditDate="2011-01-30T23:04:01.157" LastActivityDate="2011-01-30T23:04:01.157" CommentCount="2" />
  <row Id="19503" PostTypeId="2" ParentId="19497" CreationDate="2011-01-30T00:00:11.990" Score="4" ViewCount="" Body="&lt;p&gt;&lt;em&gt;Lectures in Logic and Set Theory&lt;/em&gt; (Volumes 1 and 2) by George Tourlakis. This is (in my opinion) a good beginner-intermediate book. I flipped through both volumes and liked the writing style and presentation. It's not too verbose or too terse.&lt;/p&gt;&#xA;" OwnerUserId="2548" LastEditorUserId="2548" LastEditorDisplayName="" LastEditDate="2011-01-31T02:50:57.550" LastActivityDate="2011-01-31T02:50:57.550" CommunityOwnedDate="2011-05-05T18:35:43.217" CommentCount="6" />
  <row Id="19504" PostTypeId="2" ParentId="19500" CreationDate="2011-01-30T00:06:57.100" Score="5" ViewCount="" Body="&lt;p&gt;You can have an exact sequence $0 \rightarrow Z/2Z \rightarrow Z/4Z \rightarrow Z/2Z \rightarrow 0$, where the maps are inclusion and projection. Then this sequence does not split, since $Z/4Z$ is not isomorphic to $Z/2Z \oplus Z/2Z$. Therefore $Z/2Z$ is not injective. &lt;/p&gt;&#xA;" OwnerUserId="6369" LastActivityDate="2011-01-30T00:06:57.100" />
  <row Id="19505" PostTypeId="2" ParentId="19495" CreationDate="2011-01-30T00:07:59.253" Score="5" ViewCount="" Body="&lt;p&gt;It makes sense to consider solutions only on intervals that contain the initial time. The &quot;interval where the solution is defined&quot; (I would call it the (maximal) interval of existence) is the maximal interval of all intervals $I$ which contain 0 and there exists a solution on $I$. This maximal interval turns out to be $(-2,3)$ in your case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So why doesn't it make sense to consider solutions defined on other subsets of $\mathbb{R}$ than intervals, e.g. $(-\infty,-2)\cup (-2,3)\cup (3,\infty)$? One reason is that you wouldn't have uniqueness of solutions, for example, $$\begin{cases}1/(x^2-x-6) &amp;amp; x&amp;lt;3 \cr 0 &amp;amp; x&amp;gt;3\end{cases}$$ would be another &quot;solution&quot;. What happens in the other components which don't contain 0 can be rather arbitrary, so one does not allow them.&lt;/p&gt;&#xA;" OwnerUserId="1609" LastActivityDate="2011-01-30T00:07:59.253" />
  <row Id="19506" PostTypeId="1" CreationDate="2011-01-30T00:09:40.250" Score="2" ViewCount="122" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Let $R=\mathbb{Q}[x,y,z]$, then every simple $R$-module $M$ is finite dimensional over $\mathbb{Q}$. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Had this been over $\mathbb{C}$ (complex field), it would have been rather easy. I have tried to use a theorem which says simple modules over $R$ is isomorphic to $R/I$, where I is a maximal regular ideal. But I don't understand regular ideals all that well. (For example, $Q[x,y]/(xy-1) \simeq Q(y)$, but $Q(y)$ is not finite dimensional over $Q$.) &lt;/p&gt;&#xA;" OwnerUserId="6369" LastEditorUserId="6369" LastEditorDisplayName="" LastEditDate="2011-01-30T00:51:41.587" LastActivityDate="2011-01-30T00:51:41.587" Title="about simple modules over $\mathbb{Q}[x,y,z] $" Tags="&lt;abstract-algebra&gt;&lt;ring-theory&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="19507" PostTypeId="2" ParentId="19485" CreationDate="2011-01-30T00:11:30.810" Score="3" ViewCount="" Body="&lt;p&gt;Arturo's lengthy answer hits all the right points in great detail.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More briefly: you are not doing yourself any favors by writing the Principle of Mathematical Induction as you are: the word &quot;assume&quot; need not appear anywhere in it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a statement of POMI (&lt;a href=&quot;http://www.math.uga.edu/~pete/3200induction.pdf&quot; rel=&quot;nofollow&quot;&gt;see here&lt;/a&gt; for variants on this statement and more on mathematical induction; but unfortunately this handout does not explicitly address your question.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $S$ be a subset of the positive integers satisfying both of the following properties:&lt;br&gt;&#xA;(i) $1 \in S$.&lt;br&gt;&#xA;(ii) For all positive integers $n$, $n \in S \implies n+1 \in S$.&lt;br&gt;&#xA;Then $S$ is the set of all positive integers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you need to understand the logic of an &lt;strong&gt;implication&lt;/strong&gt; $A \implies B$.  You seem to be thinking that from $A \implies B$ we can deduce $A$: this is absolutely false.  Rather $A \implies B$ is a logical predicate defined by its truth table: i.e., for each of the four possible combinations of truth/falsity of $A$ and $B$, we define whether $A \implies B$ is true or false, as follows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $A$ is true and $B$ is true, then $A \implies B$ is true.&lt;br&gt;&#xA;If $A$ is true and $B$ is false, then $A \implies B$ is false.&lt;br&gt;&#xA;If $A$ is false and $B$ is true, then $A \implies B$ is true.&lt;br&gt;&#xA;If $A$ is false and $B$ is false, then $A \implies B$ is true.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, the only case in which $A \implies B$ is false is if $A$ is true and $B$ is false.  So, if you are trying to prove that $A \implies B$ is true, it is enough to look at the situation when $A$ is true and show that in those cases $B$ is also true.  So you see that we are &lt;strong&gt;not assuming that A is true&lt;/strong&gt;: rather, we are looking at that case, since the other case is trivial: if $A$ is false, we don't need to know whether $B$ is true to know that the implication is true. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;As an aside, I have to say that I wish I had a better understanding of why students have so much trouble with the logic of induction.  I personally never did.  While that was of some use to me mathematically, at this point I sort of wish that I did have some trouble at first: I would presumably have long since gotten over it but would be in a better situation with regards to my current teaching.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As I alluded to above, I do think that going over some (mildly) formal logic is a good way to be clear about induction (and this is also an argument for covering logic in &quot;transitions&quot; courses, as I wrote about in an answer to a MO question).  Still I confess that I don't completely get the problems: this semester I am teaching an intermediate course on sequences and series which has this other intro to proofs course as a prerequisite.  So the students have all seen induction before and seem to mostly get it.  Mostly but not perfectly: when I did the second induction proof in class [the first one had two variables in it and I was inducting on one of them; in retrospect, that must have been just too complicated for some of the students], several students commented afterwards on the fact that I wrote &quot;Induction step: suppose $n \in \mathbb{N}$.  We will show that $P(n) \implies P(n+1)$.&quot;  (And yes, after that, I said &quot;Assume $P(n)$&quot; and wrote it out: this &quot;assume&quot; business is the language which is used in practice!)  In their previous course, for the induction step the variable name had been switched to $k$.  They wanted to know &quot;if it was okay&quot; to use $n$ instead.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(By the way, I have often been asked &quot;if it is okay&quot; to do various things, things which were different in appearance from what they've learned in previous courses.  I think they are not asking me &quot;Is what you just did correct?&quot; but rather &quot;Will it be acceptable to you and to the grader if we write it the way you just did?&quot; Recently I tried to answer all these questions with the blanket statement: &quot;You can assume that whatever I write on the board is acceptable and sufficient for you to write on an assignment or an exam unless I explicitly say otherwise [which I occasionally do].&quot;  Part of me was tempted to say, &quot;Yes, this is 'okay' because it is a correct and complete argument, as I hope you can see.&quot;  But I guess I am forgetting what it is like to be a student, where sometimes your instructor does want to see things written in a certain way, and it is often easier to humor him.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I tried putting my hands on my hips and asking: is there any logical difference between the statements&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;For all $n \in \mathbb{N}, P(n) \implies P(n+1)$&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;For all $k \in \mathbb{N}, P(k) \implies P(k+1)$&quot;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course not: we can use whatever names we want for the variables!  (In my own &quot;intro to proofs course&quot; I would have stressed that the point that these are quantified variables, so it really can't make a difference, but I am leery of referring to things that other instructors may not have covered.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But this answer reveals a sort of lack of understanding of the cognitive process behind the question.  What are they really confused about?  Can someone suggest a better answer?&lt;/p&gt;&#xA;" OwnerUserId="299" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-30T04:56:58.313" LastActivityDate="2011-01-30T04:56:58.313" />
  <row Id="19508" PostTypeId="1" AcceptedAnswerId="19601" CreationDate="2011-01-30T00:28:16.047" Score="4" ViewCount="170" Body="&lt;p&gt;Application field can vary from biology\biochemistry, to computer science\coding theory, the more unexpected a connection to a field, the better. And paper preferably should be not very large one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you.&lt;/p&gt;&#xA;" OwnerUserId="5550" LastActivityDate="2011-01-30T21:51:18.250" Title="recommend paper on application of group theory" Tags="&lt;group-theory&gt;&lt;reference-request&gt;&lt;applications&gt;" AnswerCount="4" FavoriteCount="2" />
  <row Id="19509" PostTypeId="2" ParentId="19479" CreationDate="2011-01-30T00:28:54.490" Score="1" ViewCount="" Body="&lt;p&gt;Letting $t=x+y$ and $s=x-y$ as in the standard treatment of the 1-D wave equation, the equation $$\alpha_{st}=\frac{m}{4} \alpha^2.$$ If $m&amp;gt;0$, trying the trick of separation of variables does, indeed, lead to an explicit solution (i.e., assume $\alpha(s,t)=A(s)B(t)$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As Willie Wong mentions below, this method had no hope from the start of producing a set of particular solutions which could be used to write a general solution, since a linear combination of particular solutions will not be a solution, in general.&lt;/p&gt;&#xA;" OwnerUserId="5253" LastEditorUserId="5253" LastEditorDisplayName="" LastEditDate="2011-01-30T04:23:01.637" LastActivityDate="2011-01-30T04:23:01.637" CommentCount="3" />
  <row Id="19510" PostTypeId="2" ParentId="19506" CreationDate="2011-01-30T00:29:35.840" Score="3" ViewCount="" Body="&lt;p&gt;This follows easily from a result known as &quot;Zariski's Lemma&quot;: if $k$ is a field, $k&amp;#39;$ an extension which is a finitely generated $k$-algebra, then $k&amp;#39;$ is a finite extension of $k$. So for any maximal ideal $I \subset k[x_1, \dots, x_n]$, the ring $k[x_1, \dots, x_n]/I$ is a finite field extension of $k$. For a short proof of this due to McCare, see section 7.5 of &lt;a href=&quot;http://people.fas.harvard.edu/~amathew/CRing.pdf&quot; rel=&quot;nofollow&quot;&gt;http://people.fas.harvard.edu/~amathew/CRing.pdf&lt;/a&gt; (in particular the exercise at the end for a really quick one, not using the machinery of the Noether normalization theorem).&lt;/p&gt;&#xA;" OwnerUserId="536" LastActivityDate="2011-01-30T00:29:35.840" CommentCount="6" />
  <row Id="19511" PostTypeId="2" ParentId="19508" CreationDate="2011-01-30T00:32:39.340" Score="1" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://cs.ecs.baylor.edu/~maurer/CSI5v93DA/Group.pdf&quot; rel=&quot;nofollow&quot;&gt;Here&lt;/a&gt; is a paper about applications of group theory to symmetric gates.&lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2011-01-30T00:32:39.340" />
  <row Id="19512" PostTypeId="2" ParentId="19500" CreationDate="2011-01-30T01:04:04.737" Score="4" ViewCount="" Body="&lt;p&gt;One way to check statements about injective modules is to see what's happening for $\mathbb{Z}$-modules -- i.e., abelian groups.  (In general, it's a good idea to test out statements of the form &quot;does this property of modules/rings/spaces/... imply this other property?&quot; by considering relatively simple, well-understood examples.  But I don't want to preach too much...)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $R = \mathbb{Z}$, there is the following nice criterion (due, I believe, to R. Baer): a $\mathbb{Z}$-module $M$ is injective iff it is a &lt;strong&gt;divisible&lt;/strong&gt; group: i.e., for all $n \in \mathbb{Z}^+$, the map $x \mapsto nx$ is surjective.  (Here I am writing the group law on $M$ additively, as is traditional.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From this it is easy to see that any nonzero injective $\mathbb{Z}$-module must be infinite.  Indeed, if $M$ is finite, then multiplication by $n = \# M$ on $M$ is identically zero.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For that matter, no finitely generated $\mathbb{Z}$-module is injective.  The simplest example of an injective $\mathbb{Z}$-module is probably $(\mathbb{Q},+)$ (i.e., the additive group of the rational field).  Since quotients of divisible groups are divisible, also $\mathbb{Q}/\mathbb{Z}$ is divisible.  The first example is torsion-free, and the second example is torsion.  These examples give the flavor of the general case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Added&lt;/b&gt;: Actually Baer's Criterion extends to all modules over PIDs as follows: for a ring $R$, an $R$-module $M$ is &lt;strong&gt;divisible&lt;/strong&gt; if for all $r \in R \setminus \{0\}$, the endomorphism $r \cdot$ of $M$ is surjective.  (This is a direct generalization of the notion of divisible $\mathbb{Z}$-module.)  Then it follows from another basic result of Baer that a module over any PID is injective iff it is divisible.  I learned this on wikipedia (which is getting better and better at math, I must say): &lt;a href=&quot;http://en.wikipedia.org/wiki/Injective_module#Baer.27s_criterion&quot; rel=&quot;nofollow&quot;&gt;see here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that applying this criterion of Baer immediately gives the following fact: if $R$ is a PID of characteristic $0$ (equivalently, the natural map $\mathbb{Z} \rightarrow R$ is an injection), then an injective $R$-module must also be an injective $\mathbb{Z}$-module.  In particular no finite $R$-module is injective.  &lt;/p&gt;&#xA;" OwnerUserId="299" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2011-01-30T06:10:56.680" LastActivityDate="2011-01-30T06:10:56.680" />
  <row Id="19513" PostTypeId="2" ParentId="19508" CreationDate="2011-01-30T01:21:11.373" Score="1" ViewCount="" Body="&lt;p&gt;See &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0387964134&quot; rel=&quot;nofollow&quot;&gt;Polya's book&lt;/a&gt; for an application of Burnside's lemma to chemistry.&lt;/p&gt;&#xA;" OwnerUserId="325" LastActivityDate="2011-01-30T01:21:11.373" />
  <row Id="19514" PostTypeId="2" ParentId="19485" CreationDate="2011-01-30T01:24:58.627" Score="1" ViewCount="" Body="&lt;p&gt;Just for another try at the intuition: usually an induction proof is proving something about natural numbers, that is, proving a statement for n, where n goes over all numbers starting from 0 and going on up. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you want to prove the statement for all $n$. You could prove it for a constant, but that doesn't do it for all $n$. The 'inductive' way is to show that, if your statement is true for some $n$ (you don't (yet) know it's true for all $n$) you can show that the statement is true for $n+1$. The $n$ here isn't -all- $n$, just one you don't know yet. And whichever $n$ this is, you can always reach $n+1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The domino image is that the inductive case of a proof shows you can always 'go' from one integer to the next, you -can- always knock over the next domino. The base case just means you can knock over the first domino in a sequence. The intuition is if those two things hold, then -all- the dominos will fall over. Knock over block 0, and it will knock over block 1, and then that will knock over block 2 and...that will knock over block $n$, and then that will knock over $n+1$, and then that will... (go on forever).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Back to the proof itself, you're not assuming the thing you're trying to prove (which is truth for -all- $n$). You're assuming that -if- it is true for some $n$, then it must be true for $n+1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In formal symbols you're trying to prove $${\rm for all}\ n,\ P(n)$$ ($P(n)$ is the formula with $n$ in it).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The inductive part of the proof is $${\rm for all}\ n,\ (P(n) \implies P(n+1) )$$ (any one domino can knock over the next, no matter which domino you're thinking of).&lt;/p&gt;&#xA;" OwnerUserId="1919" LastActivityDate="2011-01-30T01:24:58.627" />
  <row Id="19515" PostTypeId="1" AcceptedAnswerId="19549" CreationDate="2011-01-30T01:39:05.587" Score="1" ViewCount="193" Body="&lt;p&gt;If $f$ is continuous at $(0,0)$, is $g(x) := f(x, \sin x)$ continuous at $0$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Intuitively this seems true because as you approach 0 the $\sin$ function also approaches 0 from both sides.  Any push on how to prove this would be appreciated!&lt;/p&gt;&#xA;" OwnerUserId="2533" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-30T05:08:38.453" LastActivityDate="2011-01-30T10:14:42.957" Title="If $f$ is continuous at $(0,0)$, is $g(x) := f(x, \sin x)$ continuous at $0$?" Tags="&lt;calculus&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="19516" PostTypeId="2" ParentId="19515" CreationDate="2011-01-30T01:48:05.453" Score="4" ViewCount="" Body="&lt;p&gt;Since $f$ is continuous at $(0,0)$ you know that for all $\varepsilon\gt 0$ there exists $\delta\gt 0$ such that $|x|+|\sin x|\lt\delta$ implies $|f(x,\sin x) - f(0,0)|\lt\varepsilon$.  Since $|f(x,\sin x)-f(0,0)|=|g(x)-g(0)|$, it will be enough to show that there is a $\delta&amp;#39;\gt 0$ such that $|x|\lt\delta&amp;#39;$ implies $|x|+|\sin x| &amp;lt;\delta$.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2011-01-30T01:48:05.453" CommentCount="7" />
  <row Id="19517" PostTypeId="1" CreationDate="2011-01-30T01:53:53.660" Score="4" ViewCount="53" Body="&lt;p&gt;I'm looking for a formula or quick trick for computing the branch locus of a hypersurface under projection.  My specific case:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $V\subset\mathbb{P}^3$ be a surface of degree $d$, $p\in\mathbb{P}^3$ (it's actually a node of $V$, if that's relevant), and I'm attempting to compute the branch locus of the projection $V\dashrightarrow \mathbb{P}^2$, which should be a plane curve.&lt;/p&gt;&#xA;" OwnerUserId="176" LastActivityDate="2011-01-30T01:53:53.660" Title="Branch locus of a projection" Tags="&lt;algebraic-geometry&gt;" CommentCount="1" />
  <row Id="19518" PostTypeId="1" AcceptedAnswerId="19520" CreationDate="2011-01-30T01:55:08.107" Score="3" ViewCount="242" Body="&lt;p&gt;I saw on wikipedia that the determinant of a rotation matrix is always one (possibly by definition?), but it doesn't say anything about the determinant of the Jacobian of such a matrix. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since applying a rotation shouldn't change the integral of a rotationally invariant function over $\mathbb{R}^d$, if $(v_1,...,v_d) = \phi(u_1,...,u_d)$ is our rotation, then it should be the case that $d_{v_1}...d_{v_d} = |det(D \phi)(u)| d_{u_1}...d_{u_d} = d_{u_1}...d_{u_d}$ right (here $D$ stands for the Jacobian)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know enough about rotation matrices to make that justify that claim though.&lt;/p&gt;&#xA;" OwnerUserId="1736" LastEditorUserId="1736" LastEditorDisplayName="" LastEditDate="2011-01-30T02:13:34.697" LastActivityDate="2011-01-31T04:49:12.027" Title="Is the determinant of the jacobian of a rotation matrix always equal to 1?" Tags="&lt;calculus&gt;" AnswerCount="2" />
  <row Id="19520" PostTypeId="2" ParentId="19518" CreationDate="2011-01-30T02:07:33.953" Score="4" ViewCount="" Body="&lt;p&gt;The Jacobian matrix of the map on $\mathbb{R}^n$ induced by a matrix $T$ is everywhere equal to $T$.  Keep in mind that the Jacobian gives the best linear approximation of a function, and that this means equality if the function is linear.  So when you are doing a change of variables with a linear map (like a rotation), you just multiply by the determinant of the map, in your case $1$.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2011-01-30T02:07:33.953" CommentCount="8" />
  <row Id="19521" PostTypeId="2" ParentId="19474" CreationDate="2011-01-30T02:11:29.893" Score="5" ViewCount="" Body="&lt;p&gt;The question has been answered, but if you don't mind a bit of additional information&#xA;which I find interesting: An icosahedron may be cut open and refolded to a flat, doubly covered parallelogram.  I wrote &lt;a href=&quot;http://arxiv.org/abs/1010.2450&quot; rel=&quot;nofollow&quot;&gt;a note&lt;/a&gt; on this, from which the figure below is copied.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/mx6WT.jpg&quot; alt=&quot;Zip Pair for Icosahedron&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Addendum.&lt;/b&gt; I was not addressing the issue, just showing a neat unfolding. :-)&#xA;To address the issue more directly: for an icosahedron, $E=30$, $F=20$, and $E-F+1=11$.&#xA;Note that in the illustrated unfolding, the cut tree is a Hamiltonian path of 11 edges.&#xA;Of course, every spanning cut tree must have the same number of edges (as per Rahul's answer).&#xA;So the unfolding is bounded by 22 half-edges, each corresponding to the cutting&#xA;of one of the 11 edges of the path.  But notice that many of the original polyhedron&#xA;edges end up collinear in the unfolding.  As a polygon, the unfolding has 16 edges,&#xA;or 16 &quot;half-edges&quot; if you think of each as deriving from 8 straight slices on&#xA;the icosahedron surface.  So there is a sense in which one need only glue 8 segments&#xA;to fold the unfolding back to the icosahedron.&lt;/p&gt;&#xA;" OwnerUserId="237" LastEditorUserId="237" LastEditorDisplayName="" LastEditDate="2011-02-04T02:13:22.923" LastActivityDate="2011-02-04T02:13:22.923" />
  <row Id="19522" PostTypeId="2" ParentId="19518" CreationDate="2011-01-30T02:46:53.440" Score="1" ViewCount="" Body="&lt;p&gt;Let $A$ be any $n$x$n$ matrix. Let $h(\mathbb{x}) = A \cdot \mathbb{x}$. Now if you take some $S \subset \mathbb{R}^n$, then you have that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$volume(h(S)) = |\det A| \cdot volume(S)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is just a direct application of a change of variables. Rotation preserves volume, hence if $A$ is your rotation matrix, then $|det A| = 1$. In general, the deteminant is negative in the case that the transformation is orientation reversing. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now to consider the Jacobian: clearly $D h(\mathbb{x}) = D (A \cdot \mathbb{x}) = A$, hence $|\det \ D h| = 1$ as well.&lt;/p&gt;&#xA;" OwnerUserId="5105" LastEditorUserId="5105" LastEditorDisplayName="" LastEditDate="2011-01-31T04:49:12.027" LastActivityDate="2011-01-31T04:49:12.027" CommentCount="2" />
  <row Id="19523" PostTypeId="1" AcceptedAnswerId="19525" CreationDate="2011-01-30T02:54:41.807" Score="3" ViewCount="113" Body="&lt;p&gt;If $g(x):=f(x, kx^m)$ is continuous at $0\;\;\;$    $\forall k\in R$, $\;\;\;\forall m\in N$, then $f(x,y)$ is continuous at $(0,0)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not quite sure what is meant here by at 0.  This means $f(x, kx^m) = 0$? Seems there could be an issue in this statement when k = 0?  &lt;/p&gt;&#xA;" OwnerUserId="2533" LastActivityDate="2011-01-30T09:12:05.350" Title="If $g(x):=f(x, kx^m)$ is continuous at $0$, then $f(x,y)$ is continuous at $(0,0)$" Tags="&lt;calculus&gt;&lt;homework&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="19524" PostTypeId="2" ParentId="8937" CreationDate="2011-01-30T02:57:38.867" Score="2" ViewCount="" Body="&lt;p&gt;See Theorem 2.5 in Chapter 6 of the &lt;a href=&quot;http://people.fas.harvard.edu/~amathew/CRing.pdf&quot; rel=&quot;nofollow&quot;&gt;CRing project&lt;/a&gt; for another elementary argument due to Perdry (American Math. Monthly, 2004) using only the Hilbert basis theorem. In fact, it shows more: if $R$ is a noetherian domain, $I \subset R$ a proper ideal, then the intersection of the powers of $I$ is trivial.&lt;/p&gt;&#xA;" OwnerUserId="536" LastActivityDate="2011-01-30T02:57:38.867" CommentCount="6" />
  <row Id="19525" PostTypeId="2" ParentId="19523" CreationDate="2011-01-30T04:19:34.417" Score="5" ViewCount="" Body="&lt;p&gt;It isn't true.  Let $f$ be defined by $f(x,y)=\frac{xy^2}{x^2+y^4}$ if $(x,y)\neq (0,0)$ and $f(0,0)=0$.  I got this example by Googling &quot;continuous on each line&quot; and clicking on &lt;a href=&quot;http://www.math.harvard.edu/archive/21a_fall_07/exhibits/halloween/index.html&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt;.  You can see that $f$ is not continuous at zero by approaching $(0,0)$ along the curve $x=y^2$, but it satisfies the hypotheses of the problem.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;There's a much easier example.  Define $f$ by $f(0,y)=1$ if $y\neq0$ and $f(x,y)=0$ otherwise.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2011-01-30T09:12:05.350" LastActivityDate="2011-01-30T09:12:05.350" />
  <row Id="19526" PostTypeId="1" AcceptedAnswerId="19543" CreationDate="2011-01-30T04:37:47.120" Score="4" ViewCount="154" Body="&lt;p&gt;Let $f,g \in L^2(\mathbb{R^n})$, $\{ f_n \}, \{ g_m \} \subset C^\infty_0(\mathbb{R}^n)$ (infinitely differentiable functions with compact support) where $f_n \to f$ in $L^2$, and $g_n \to g$ in $L^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then $f_n \star g_m(x) \to f \star g(x)$ pointwise as $n,m \to \infty$ (where $\star$ denotes convolution). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it true that $\int_{\mathbb{R}^n} f_n \star g_m dx \to \int_{\mathbb{R}^n} f \star g dx$? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've thought about using the dominated convergence theorem, but I couldn't figure out how, since I don't really know what I can do with the sequences of functions given only the facts at hand.    &lt;/p&gt;&#xA;" OwnerUserId="1736" LastActivityDate="2011-01-30T15:32:31.993" Title="limits and convolution" Tags="&lt;real-analysis&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="19527" PostTypeId="1" CreationDate="2011-01-30T05:20:33.757" Score="0" ViewCount="78" Body="&lt;p&gt;I need to prove the following:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If $a\gt 0$, $n\in \mathbb{N}-\{0\}$, and if $f$ analytic on $(0, 2a)$ with $|f^{(n)}(0)| = \frac{n!}{a^n}*\sup\{|f(z)|\mid : |z|=a\}$, then $f(z) = rz^n$ for some $r\in \mathbb{C}$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have tried to borrow ideas from the proof of Liouville's Theorem and go backwards. However, the problem doesn't quite seem to fit the hypothesis of Liouville's Theorem. Any suggestion would be appreciated. Thanks!&lt;/p&gt;&#xA;" OwnerUserId="6399" LastEditorUserId="6399" LastEditorDisplayName="" LastEditDate="2011-01-30T07:12:22.210" LastActivityDate="2011-01-30T07:12:22.210" Title="analytic class C(n)" Tags="&lt;homework&gt;&lt;complex-analysis&gt;" CommentCount="7" />
  <row Id="19528" PostTypeId="1" AcceptedAnswerId="19529" CreationDate="2011-01-30T05:22:19.880" Score="5" ViewCount="197" Body="&lt;p&gt;We know that if all the entries of an invertible matrix $A$ are rational, then all the entries of $A^{-1}$ are also rational.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now suppose that all entries of an invertible matrix $A$ are integers. Then it's not necessary that entries of $A^{-1}$ are all integers. Question is:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;What are all the invertible matrices with integer entries such that its inverse is also having integer entries?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-30T05:29:35.797" LastActivityDate="2011-01-31T05:29:13.203" Title="Integer Matrices with Inverse Integer Matrix." Tags="&lt;algebra&gt;&lt;matrices&gt;" AnswerCount="3" CommentCount="0" />
  <row Id="19529" PostTypeId="2" ParentId="19528" CreationDate="2011-01-30T05:24:23.590" Score="6" ViewCount="" Body="&lt;p&gt;Exactly those whose determinant is $1$ or $-1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See &lt;a href=&quot;http://math.stackexchange.com/questions/3655/units-of-m-2z/3659#3659&quot;&gt;the previous question about the $2\times 2$ case&lt;/a&gt;. The determinant map gives necessity, the adjugate formula for the inverse gives sufficiency.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-30T05:58:12.600" LastActivityDate="2011-01-30T05:58:12.600" CommentCount="2" />
  <row Id="19530" PostTypeId="2" ParentId="19528" CreationDate="2011-01-30T05:27:30.997" Score="4" ViewCount="" Body="&lt;p&gt;The inverse of an integer matrix is again an integer matrix iff if the determinant of the matrix is $\pm 1$. Integer matrices of determinant $\pm 1$ form the General Linear Group $GL(n,\mathbb{Z})$&lt;/p&gt;&#xA;" OwnerUserId="3302" LastActivityDate="2011-01-30T05:27:30.997" />
  <row Id="19531" PostTypeId="1" CreationDate="2011-01-30T05:37:00.450" Score="1" ViewCount="73" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Possible Duplicate:&lt;/strong&gt;&lt;br&gt;&#xA;  &lt;a href=&quot;http://math.stackexchange.com/questions/8186/when-do-the-multiples-of-two-primes-span-all-large-enough-natural-numbers&quot;&gt;When do the multiples of two primes span all large enough natural numbers? &lt;/a&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;p&gt;If $m,n$ are positive integers and $(m,n)=1$.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What's the largest integer $N$ that cannot be expressed as $N=am+bn$, where $a$ and $b$ are nonnegative integers?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The answer is $N=mn-m-n$. How to prove it?&lt;/p&gt;&#xA;" OwnerUserId="6401" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-30T05:39:28.730" LastActivityDate="2011-01-30T05:39:28.730" ClosedDate="2011-01-30T05:51:11.547" Title="$(m,n)=1$,$N=am+bn$,where $a ,b$ are non-negative integers. what's the max integer $N$,that cannot expressed as $N=am+bn$?" Tags="&lt;number-theory&gt;&lt;number&gt;" CommentCount="4" />
  <row Id="19532" PostTypeId="2" ParentId="19497" CreationDate="2011-01-30T05:37:32.553" Score="6" ViewCount="" Body="&lt;p&gt;My preferred intro logic text is Enderton's &quot;A mathematical introduction to logic.&quot; Smullyan's book on first-order logic is very nice, though it uses proof trees, which is (awesome but) somewhat unusual. I'm also a fan of Peter Smith's &quot;An Introduction to Godel's Theorems.&quot; As far as set theory goes, it really depends how far you'd like to go. I imagine that if you're doing graduate research in CS you know the &quot;basics&quot; that would be covered in Halmos' &quot;Naive Set Theory.&quot; If you're looking for more, I also like Enderton's &quot;Set Theory,&quot; and after that a big standard text is Jech's. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's obviously more. You need a good textbook for modal logic, and I don't know any. Hopefully someone else can help fill in the picture.&lt;/p&gt;&#xA;" OwnerUserId="6402" LastActivityDate="2011-01-30T05:37:32.553" CommunityOwnedDate="2011-05-05T18:35:43.217" CommentCount="1" />
  <row Id="19533" PostTypeId="2" ParentId="13605" CreationDate="2011-01-30T05:46:53.663" Score="0" ViewCount="" Body="&lt;p&gt;A Filtration is a growing sequence of sigma algebras $$\mathcal{F_1}\subseteq \mathcal{F_2}\ldots \subseteq \mathcal{F_n}$$. Now when talking of martingales we need to talk of conditional expectations, and in particular conditional expectations w.r.t $\sigma$ algebra's. So whenever we write $$ E[Y_n|X_1,x_2,\ldots,X_n]$$ we can alternatively write it as $$E[Y_{n+1}| \mathcal{F_{n}}]$$, where $\mathcal{F}_{n}$ is a sigma algebra that makes random variables $$x_1,\ldots,x_n$$ measurable. Finally a flitration $\mathcal{F_1},\ldots \mathcal{F_n}$ is simply an increasing sequence of simga algebras. That is  we are conditioning on growing amounts of information.&lt;/p&gt;&#xA;" OwnerUserId="6398" LastActivityDate="2011-01-30T05:46:53.663" />
  <row Id="19534" PostTypeId="1" AcceptedAnswerId="19535" CreationDate="2011-01-30T06:02:06.293" Score="3" ViewCount="155" Body="&lt;p&gt;from my understanding,every set has at least two subsets; the null set and the original set itself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is, what is the power set of the null set? Shouldn't it be just itself?&lt;/p&gt;&#xA;" OwnerUserId="1040" LastActivityDate="2011-01-30T06:06:12.257" Title="How do I understand the null set" Tags="&lt;elementary-set-theory&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="19535" PostTypeId="2" ParentId="19534" CreationDate="2011-01-30T06:06:12.257" Score="13" ViewCount="" Body="&lt;p&gt;Every &lt;em&gt;nonempty&lt;/em&gt; set has at least two &lt;em&gt;distinct&lt;/em&gt; subsets, namely the empty set and the set itself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the empty set has only one subset: itself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, the power set of the empty set has one element, namely the empty set. That is, $\mathcal{P}(\emptyset) = \{\emptyset\}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice that the set whose only element is the empty set, $\{\emptyset\}$, is &lt;em&gt;not&lt;/em&gt; empty: a bag that has an empty bag inside is not, itself, empty. So the power set of the empty set is &lt;em&gt;not&lt;/em&gt; the empty set. &lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2011-01-30T06:06:12.257" CommentCount="4" />
  <row Id="19536" PostTypeId="1" AcceptedAnswerId="19537" CreationDate="2011-01-30T06:07:59.283" Score="0" ViewCount="129" Body="&lt;p&gt;If $q\colon\mathbb{C}\to\mathbb{C}$ is a polynomial, $f\colon\mathbb{C}\to\mathbb{C}$ is analytic on all of $\mathbb{C}$, and if there exists $a\gt 0$ such that  $|f(z)| \lt  a|q(z)|$ for every $z\in \mathbb{C}$, then $f = bq$  for some $b\in \mathbb{C}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can an arbitrary analytic function (on all of $\mathbb{C}$) replace $q$?&lt;/p&gt;&#xA;" OwnerUserId="6399" LastEditorUserId="6399" LastEditorDisplayName="" LastEditDate="2011-01-30T06:45:06.600" LastActivityDate="2011-01-30T06:45:06.600" Title="analytic functions" Tags="&lt;complex-analysis&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" />
  <row Id="19537" PostTypeId="2" ParentId="19536" CreationDate="2011-01-30T06:34:18.263" Score="2" ViewCount="" Body="&lt;p&gt;If you really mean strict inequality, then this follows from &lt;a href=&quot;http://en.wikipedia.org/wiki/Liouville%27s_theorem_%28complex_analysis%29&quot; rel=&quot;nofollow&quot;&gt;Liouville's theorem&lt;/a&gt; applied to $f/q$. Note that $q$ must be constant too, as $q$ can have no zeros from the condition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is actually true if $&amp;lt;$ is replaced by $\leq$, though. (Maybe you meant this?) First, the Cauchy estimates show that $f$ is a polynomial; indeed, since $f$ grows polynomially, we can just take averages of $f/(z-\alpha)^N$ over larger and larger circles. From this one can see that $f$ is a polynomial. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the question reduces to showing that if a polynomial $p$ is bounded by a constant multiple of another polynomial $q$, then $p$ and $q$ differ by a constant. This is a straightforward consequence of factorization.&lt;/p&gt;&#xA;" OwnerUserId="536" LastActivityDate="2011-01-30T06:34:18.263" CommentCount="2" />
  <row Id="19538" PostTypeId="1" AcceptedAnswerId="19539" CreationDate="2011-01-30T06:34:51.653" Score="10" ViewCount="340" Body="&lt;p&gt;If $A$ is an $n \times n$ matrix such that $A^2=0$, is $A+I_{n}$ invertible?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This question yielded two different proofs from my professors, which managed to get conflicting results (true and false). Could you please weigh in and explain what's happening, and offer a working proof?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proof that it is invertible: Consider matrix $A-I_{n}$. Multiplying $(A+I_{n})$ by $(A-I_{n})$&#xA;we get $A^2-AI_{n}+AI_{n}-I^2_{n}$. This simplifies to $A^2-I^2_{n}$ which is equal to $-I_{n}$, since $A^2=0$. So, the professor argued, since we have shown that there exists a $B$ such that $(A+I_{n})$ times $B$ is equal to $I$, $(A+I_{n})$ must be invertible. I am afraid, though, that she forgot about the negative sign that was leftover in front of the $I$ -- from what I understand, $(A+I_{n})$*$(A-I_{n})$=$-I$ does not mean that $(A+I_{n})$ is invertible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proof that it is not invertible: Assume that $A(x)=0$ has a non-trivial solution. Now, given $(A+I_{n})(x)=\vec{0}$, multiply both sides by $A$. We get $A(A+I_{n})(x)=A(\vec{0})$, which can be written as $(A^2+A)(x)=\vec{0}$, which simplifies to $A(x)=0$, as $A^2=0$. Since we assumed that $A(x)=0$ has a non-trivial solution, we just demonstrated that $(A+I_{n})$ has a non-trivial solution, too. Hence, it is not invertible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not sure if I reproduced the second proof in complete accuracy (I think I did), but the idea was to show that if $A(x)=\vec{0}$ has a non-trivial solution, $A(A+I_{n})$ does too, rendering $A(A+I_{n})$ non-invertible. But regardless of the proofs, I can think of examples that show that at least in some cases, the statement is true; consider matrices $\begin{bmatrix}&#xD;&#xA;0 &amp;amp; 0\\ &#xD;&#xA;0 &amp;amp; 0&#xD;&#xA;\end{bmatrix}$ and $\begin{bmatrix}&#xD;&#xA;0 &amp;amp; 1\\ &#xD;&#xA;0 &amp;amp; 0&#xD;&#xA;\end{bmatrix}$ which, when added $I_{2}$ to, become invertible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks a lot!&lt;/p&gt;&#xA;" OwnerUserId="3731" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-30T06:43:38.470" LastActivityDate="2011-01-30T23:39:35.357" Title="If $A$ is an $n \times n$ matrix such that $A^2=0$, is $A+I_{n}$ invertible?" Tags="&lt;linear-algebra&gt;" AnswerCount="3" CommentCount="5" FavoriteCount="3" />
  <row Id="19539" PostTypeId="2" ParentId="19538" CreationDate="2011-01-30T06:40:17.337" Score="13" ViewCount="" Body="&lt;p&gt;The minus sign is not an obstacle: If $AB = -I$, then $A(-B) = -(AB) = -(-I) = I$. So in fact, if $A^2 = 0$, then $(A+I)(I-A) = A - A^2 + I - A = I$, so $A+I$ is invertible, as your first professor noted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The error in the second argument is the following: It is true that if $B\mathbf{x}=\mathbf{0}$ has a nontrivial solution, then $CB\mathbf{x}=\mathbf{0}$ has a nontrivial solution. Thus, if $B$ is not invertible, then $CB$ is not invertible. But that is not what was argued. What was argued instead was that since $CB\mathbf{x}=\mathbf{0}$ has a nontrivial solution, then it follows that $B\mathbf{x}=\mathbf{0}$ has a nontrivial solution (with $B=A+I$ and $C=A$). This argument is incorrect: you can always take $C=0$, and that would mean that no matrix is invertible. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is certainly true that if $A$ is not invertible, then no multiple of $A$ is invertible (so for every $C$, neither $CA$ nor $AC$ are invertible); so you &lt;em&gt;can&lt;/em&gt; deduce that $A(A+I)$ is not invertible. This does &lt;em&gt;not&lt;/em&gt; prove that $A+I$ is not invertible, however, which is what you wanted to show.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, for bonus points, show that if $A$ is an $n\times n$ matrix and $A^k=0$ for some positive integer $k$, then $A+\lambda I_n$ is invertible for any nonzero $\lambda$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Added:&lt;/em&gt; For bonus bonus points, explain why the argument would break down if we replace $\lambda I_n$ with an arbitrary invertible matrix $B$. &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-30T07:23:31.977" LastActivityDate="2011-01-30T07:23:31.977" CommentCount="1" />
  <row Id="19540" PostTypeId="2" ParentId="19538" CreationDate="2011-01-30T07:46:28.867" Score="9" ViewCount="" Body="&lt;p&gt;For what it's worth I just want to mention that what is happening here is actually an instance of a more general result about rings. If $R$ is a ring then an element $a \in R$ is said to be &lt;strong&gt;nilpotent&lt;/strong&gt; if there is $n \in \mathbb{N}$ such that $a^n = 0$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your question, the condition on your matrix that $A^2 = 0$ just means that it is nilpotent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now if $R$ is a ring with unity, then if $a \in R$ is nilpotent it can be proved that $1 - a$ is an &lt;strong&gt;invertible&lt;/strong&gt; element (or unit) in the ring $R$, meaning that there is a $b \in R$ such that $(1 - a)b = b(1 - a) = 1$. From this you can then just change $a$ with $-a$ to also see that $1 + a$ is invertible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm pretty sure that this is what Arturo had in mind by adding that exercise for bonus points for you, so I will not give the argument here. You can find it in &lt;a href=&quot;http://planetmath.org/encyclopedia/UnityPlusNilpotentIsUnit.html&quot; rel=&quot;nofollow&quot;&gt;this planetmath entry&lt;/a&gt; if you want to look at it. But I would suggest to you to first try it for yourself, for matrices at least.&lt;/p&gt;&#xA;" OwnerUserId="900" LastEditorUserId="900" LastEditorDisplayName="" LastEditDate="2011-01-30T23:39:35.357" LastActivityDate="2011-01-30T23:39:35.357" />
  <row Id="19541" PostTypeId="2" ParentId="19538" CreationDate="2011-01-30T07:55:16.107" Score="5" ViewCount="" Body="&lt;p&gt;I suggest thinking of the problem in terms of eigenvalues.  Try proving the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $A$ is an $n \times n$ matrix (over any field) which is &lt;strong&gt;nilpotent&lt;/strong&gt; -- i.e., $A^k = 0$ for some positive integer $k$, then $-1$ is not an eigenvalue of $A$ (or equivalently, $1$ is not an eigenvalue of $-A$).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you can prove this, you can prove a stronger statement and collect bonus points from Arturo Magidin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(&lt;b&gt;Added&lt;/b&gt;: Adrian's answer -- which appeared while I was writing mine -- is similar, and probably better: simpler and more general.  But I claim it is always a good idea to keep eigenvalues in mind when thinking about matrices!)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Added&lt;/b&gt;: here's a hint for a solution that has nothing to do with eigenvalues (or, as Adrian rightly points out, really nothing to do with matrices either.)  Recall the formula for the sum of an infinite geometric series:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{1}{1-x} = 1 + x + x^2 + \ldots + x^n + \ldots$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As written, this is an analytic statement, so issues of convergence must be considered.  (For instance, if $x$ is a real number, we need $|x| &amp;lt; 1$.)  But if it happens that some power of $x$ is equal zero, then so are all higher powers and the series is not infinite after all...With only a little work, one can make purely algebraic sense out of this.&lt;/p&gt;&#xA;" OwnerUserId="299" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2011-01-30T08:13:30.380" LastActivityDate="2011-01-30T08:13:30.380" CommentCount="1" />
  <row Id="19543" PostTypeId="2" ParentId="19526" CreationDate="2011-01-30T09:23:11.830" Score="5" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;No&lt;/strong&gt; it is not true.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First note that we can improve pointwise convergence to &lt;em&gt;uniform convergence&lt;/em&gt; of $f_{n} \ast g_{m} \to f \ast g$. This is an easy consequence of H&amp;ouml;lder's inequality (markup doesn't like my subscripts, so I'll use $\|\cdot\|$ for the $L^{2}$-norm):&#xA;\[&#xA;|(f \ast g)(x)| \leq \int |f(x - y) g(y)|\,dy \leq \|f\| \|g\|.&#xA;\]&#xA;As $f_{n} \ast g_{m} - f \ast g = f_{n} \ast (g_{m} - g) + (f_{n} - f) \ast g$ we get &#xA;\[&#xA;|(f_{n} \ast g_{m})(x) - (f \ast g)(x)| \leq \|f_{n}\| \|g_{m} - g\| + \|f_{n} - f\| \|g\|. &#xA;\]&#xA;Since $\|f_{n}\| \to \|f\|$ and both $\|f_{n} - f\| \to 0$ and $\|g_{m} - g\| \to 0$, the right hand side can be made arbitrarily small independently of $x$. It follows in particular that $f \ast g$ is continuous and vanishes at infinity (that's one reason why I prefer writing $C_{c}$ for functions with compact support and $C_{0}$ for functions vanishing at infinity).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But we cannot do better, i.e., get $L^1$-convergence without further hypotheses. To see what happens, I prefer to ignore the condition that $f_{n} \in C^{\infty}$ but rather look at the function $f_{n} = \frac{1}{n}[-n,n]$. Then $f_{n} \to 0$ in $L^{2}$ but not in $L^{1}$. By the above argument we have $f_{n} \ast f_{n} \to 0$ uniformly on $\mathbb{R}$. On the other hand it is easy to see that for $x \in [-\frac{n}{2},\frac{n}{2}]$&#xA;\[&#xA;(f_{n} \ast f_{n})(x) = \int f_{n}(y) f_{n}(x-y)\,dy \geq \frac{1}{n^{2}} \frac{n}{2} = \frac{1}{2n}&#xA;\]&#xA;(the intervals $[-n,n]$ and $[-n-x,n-x]$ have an overlap of length at least $\frac{n}{2}$)&#xA;and therefore we have the estimate $\int (f_{n} \ast f_{n})  \geq \frac{1}{2}$. Thus we cannot have $\int f_{n} \ast f_{n} \to \int f \ast f = 0$. It is now easy to cook up an example along these lines with smooth functions.&lt;/p&gt;&#xA;" OwnerUserId="5363" LastEditorUserId="5363" LastEditorDisplayName="" LastEditDate="2011-01-30T09:40:54.903" LastActivityDate="2011-01-30T09:40:54.903" CommentCount="2" />
  <row Id="19544" PostTypeId="1" CreationDate="2011-01-30T09:39:58.073" Score="1" ViewCount="92" Body="&lt;p&gt;I needed to find the derivative of $\cos^3(x)$ and find it in a list of given derivatives, but I could not find $-3\sin(x)\cos^2(x)$, so I waited until I did more problems and the only one left was $-3\cos^3(x)\tan(x)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If anyone could show me the steps to convert $-3\sin(x)\cos^2(x)$ into $-3\cos^3(x)\tan(x)$, that would be awesome.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="6406" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2011-01-30T17:41:56.713" LastActivityDate="2011-01-30T17:41:56.713" Title="Help with converting the derivative of $\cos^3(x)$" Tags="&lt;calculus&gt;&lt;homework&gt;" AnswerCount="1" />
  <row Id="19545" PostTypeId="2" ParentId="19544" CreationDate="2011-01-30T09:48:44.773" Score="2" ViewCount="" Body="&lt;p&gt;Recall that $\tan{x}=\frac{\sin{x}}{\cos{x}}$. So&#xA;$$&#xD;&#xA;-3\cos^3{x}\tan{x}=-3\cos^3{x}\cdot\frac{\sin{x}}{\cos{x}}=-3\cos^2{x}\sin{x}.&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if you want to convert the derivative to that form, you multiply by $\frac{\cos{x}}{\cos{x}}$. Personally, I think the answer you got should have been the one on the list, as I don't see how anyone would arrive to this other form instead without some unnecessary manipulation. &lt;/p&gt;&#xA;" OwnerUserId="1583" LastEditorUserId="1583" LastEditorDisplayName="" LastEditDate="2011-01-30T10:03:18.800" LastActivityDate="2011-01-30T10:03:18.800" CommentCount="1" />
  <row Id="19546" PostTypeId="2" ParentId="19457" CreationDate="2011-01-30T09:49:11.207" Score="5" ViewCount="" Body="&lt;p&gt;You've had some time to study this, so let's look closer at the two evident approaches:  (a) conversion to polar coordinates, (b) integrate directly with a standard hyperbolic substitution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(a) Conversion to polar coordinates: Let&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$I = \int_{1/\sqrt{2}}^1 \, \int_{\sqrt{1-x^2}}^x \frac{1}{\sqrt{x^2+y^2}} \text{d}y \, \text{d}x.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you draw a diagram of the situation you will see that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$I = \int_0^{\pi/4} \, \int_{r=1}^{r=\sec \theta} \frac{1}{r} r \text{d}r \, \text{d} \theta = \int_0^{\pi/4} \sec \theta \, - 1 \textrm{ d} \theta $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$= \left[ \frac{1}{2} \log \left| \frac{1+ \sin \theta}{1- \sin \theta} \right| - \theta  \right]_0^{\pi/4} = \log(1+\sqrt{2}) - \frac{\pi}{4}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(b) Standard hyperbolic substitution: To evaluate the inner integral set $y= x \sinh u$ and noting that $\sinh^{-1} u = \log(u+ \sqrt{1+u^2})$ we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_{\sqrt{1-x^2}}^x \frac{1}{\sqrt{x^2+y^2}} \text{d}y  =&#xD;&#xA;\left[ \log \left( \frac{y}{x} + &#xD;&#xA;\sqrt{ 1+ \frac{y^2}{x^2} }\right) \right]_{\sqrt{1-x^2}}^x $$&#xA;$$= \log(1+ \sqrt{2}) + \log x - \log( 1+ \sqrt{1-x^2}). \quad (1)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both the logs can be integrated by parts. The first is standard, $\int \log x \textrm{ d}x = x \log x - x + C$ and the second&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \int \log( 1+ \sqrt{1-x^2} ) \textrm{ d}x = x \log( 1+ \sqrt{1-x^2}) -x + \sin^{-1} x + C.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You will need (do the integration) to note that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{1}{\sqrt{1-x^2}} - 1 = \frac{x^2}{(1-x^2) + \sqrt{1-x^2}}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And so upon integrating $(1)$ between $1/\sqrt{2}$ and $1$ you obtain&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$I = \left(1 - \frac{1}{\sqrt{2}} \right) \log(1+ \sqrt{2}) +&#xD;&#xA;\left[ x \log x - x\log( 1+ \sqrt{1-x^2}) - \sin^{-1} x \right]_{1/\sqrt{2}}^1$$&#xA;$$= \log(1+\sqrt{2}) - \frac{\pi}{4}.$$&lt;/p&gt;&#xA;" OwnerUserId="1301" LastActivityDate="2011-01-30T09:49:11.207" />
  <row Id="19547" PostTypeId="1" AcceptedAnswerId="19633" CreationDate="2011-01-30T09:50:11.027" Score="0" ViewCount="129" Body="&lt;p&gt;So I used the split step method on the Schrodinger equation and have produced the following equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Psi(x,t+dt)=F^{-1} \left\{ e^{-i\frac{\hbar^2k^2}{2m}\frac{dt}{\hbar}}F\left\{e^{-iV(x)\frac{dt}{\hbar}}\Psi(x,t) \right\}\right\}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which when scaled to dimensionless the time evolution step can be written as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Psi(x,t+dt)=F^{-1}\left\{e^{-ik^2dt}F\left\{e^{-iV(x)dt}\Psi(x,t)\right\}\right\}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;F above represents the Fourier transform operator acting on the equations.&#xA;Now my problem is that I don't know how to implement the above in Fast Fourier Transform.&#xA;For example: If I were to take the exponential factor with V(x), do I multiply -iV(x) by dt?&#xA;Not sure I understand what is going on here....&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="4789" LastActivityDate="2011-02-03T18:56:54.263" Title="Quantum wave packet propagation, how to use it in FFT?" Tags="&lt;fourier-analysis&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="19548" PostTypeId="1" CreationDate="2011-01-30T10:09:22.133" Score="0" ViewCount="159" Body="&lt;p&gt;If we have &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;An equation $E$ for which the conditions for the existence of a solution are satisfied but we can't prove the uniqueness of the solutions.&lt;/li&gt;&#xA;&lt;li&gt;A perturbed equation $E_p$ of $E$ which the existence and the  uniqueness of solution are satisfied.&lt;/li&gt;&#xA;&lt;li&gt;The solutions $S_p$ of $E_p$ converge strongly to the solutions $S$ of $E$.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Question:&#xA;Do (2) and (3) implies the uniqueness of solution in (1)?&lt;/p&gt;&#xA;" OwnerUserId="6407" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-30T20:20:50.667" LastActivityDate="2011-01-30T21:53:51.680" Title="Uniqueness of solutions" Tags="&lt;real-analysis&gt;&lt;homework&gt;&lt;probability&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="19549" PostTypeId="2" ParentId="19515" CreationDate="2011-01-30T10:14:42.957" Score="7" ViewCount="" Body="&lt;p&gt;You can also use sequences: If $(x_n)$ is a sequence with $x_n\to0$ then also $\sin x_n\to0$, so the continuity of $f$ at $(0,0)$ implies that $g(x_n) = f(x_n,\sin x_n) \to f(0,0) = g(0)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a special case of a more general theorem: If $h$ is continuous at $a$ and $f$ is continuous at $h(a)$, then $g = f\circ h$ is continuous at $a$. In the above example one has $h(x) = (x,\sin x)$ and $a=0$.&lt;/p&gt;&#xA;" OwnerUserId="2575" LastActivityDate="2011-01-30T10:14:42.957" />
  <row Id="19550" PostTypeId="1" CreationDate="2011-01-30T10:43:35.150" Score="2" ViewCount="103" Body="&lt;p&gt;As per the title, I'm looking for the name and for a way to construct a ${\rm GF}(2)$ square matrix of size $N$ with the following properties:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;All rows/columns should be linearly independent&lt;/li&gt;&#xA;&lt;li&gt;On each row and each column there should be $N/2$ ones and zeros&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Can anybody provide some pointers? It doesn't have to be a general solution (i.e. for all $N$) as long as it works for some $N$ it will be ok.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;edit: It has been pointed out that if N is a multiple of 4 the two conditions above are incompatible. Let me relax number 2) by allowing, when N is a multiple of 4, the rows and columns to be (minimally) unbalanced.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;edit2: Bonus points for pointing out a way to build matrices as the ones above, with the added constraint that the conditions above should be valid also for their inverse (obviously number 1 is implied).&lt;/p&gt;&#xA;" OwnerUserId="5859" LastEditorUserId="5859" LastEditorDisplayName="" LastEditDate="2011-02-01T14:58:53.183" LastActivityDate="2011-02-01T14:58:53.183" Title="Invertible NxN matrix over GF(2) having on each row and column N/2 ones" Tags="&lt;matrices&gt;&lt;random&gt;&lt;inversion&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="19551" PostTypeId="2" ParentId="19499" CreationDate="2011-01-30T10:51:05.907" Score="5" ViewCount="" Body="&lt;p&gt;No. This can be seen via Fourier transform. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $f$ be such that for the Fourier transform $\hat f$ it holds that for all $n$&#xA;\begin{equation*}&#xA;\hat f^{(n)}(0) = 0.&#xA;\end{equation*}&#xA; Note that this does not imply that $f$ itself is zero since there are &lt;a href=&quot;http://en.wikipedia.org/wiki/Flat_function&quot; rel=&quot;nofollow&quot;&gt;infinitely flat functions&lt;/a&gt;. By using the rules for Fourier transforms and derivatives you see that $\int x^n f(x) dx = 0$ for all $n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One application of such functions with infinitely many vanishing moment is the construction of infinitely regualel orthonomal wavelet bases (see, e.g., the Meyer wavelet).&lt;/p&gt;&#xA;" OwnerUserId="3148" LastActivityDate="2011-01-30T10:51:05.907" />
  <row Id="19553" PostTypeId="2" ParentId="19550" CreationDate="2011-01-30T11:40:12.520" Score="3" ViewCount="" Body="&lt;p&gt;It's not possible for $N$ divisible by 4, because the vectors with an even number of 1's form a proper subspace of ${\rm GF}(2)^N$. For $N=6$ there are many examples, such as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[1 1 1 0 0 0]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[1 1 0 1 0 0]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[0 0 1 1 0 1]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[1 0 0 0 1 1]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[0 1 0 0 1 1]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[0 0 1 1 1 0]&lt;/p&gt;&#xA;" OwnerUserId="2820" LastActivityDate="2011-01-30T11:40:12.520" CommentCount="1" />
  <row Id="19554" PostTypeId="1" AcceptedAnswerId="19556" CreationDate="2011-01-30T12:08:47.027" Score="2" ViewCount="277" Body="&lt;p&gt;Hi all, I've decided to learn the basics of proofs and here is my first attempt. Could I improve or simplify my proof in any way? Is my formal language correct? Thanks!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $n$ be any integer. $$n(n+2) = (n+1)^2 - 1$$ I will prove this identify by induction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, check with $n=1$;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$1 \times 3 = 2^2 + 1 \equiv&#xD;&#xA;3 = 3$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Inductive step: Assume that the identity is true for n = k;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$k(k+2) = (k+1)^2 - 1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When $n = k + 1$;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(k+1)((k+1)+2) = ((k+1)+1)^2 - 1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\equiv k+1(k+3) = (k+2)^2 - 1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $n = k + 1$;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$n(n+2) = (n+1)^2 - 1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;□&lt;/p&gt;&#xA;" OwnerUserId="6409" LastEditorUserId="1543" LastEditorDisplayName="" LastEditDate="2011-05-03T10:05:34.460" LastActivityDate="2011-05-03T10:05:34.460" Title="Can my proof be simplified and is it valid? $n(n+2) = (n+1)^2 - 1$ for all integers" Tags="&lt;algebra-precalculus&gt;&lt;induction&gt;" AnswerCount="3" CommentCount="8" />
  <row Id="19555" PostTypeId="2" ParentId="19554" CreationDate="2011-01-30T12:13:03.120" Score="17" ViewCount="" Body="&lt;p&gt;Let $n$ be an integer.&#xA;$$n(n+2) = n^2 + 2n = n^2 + 2n + 1 - 1 = (n+1)^2 - 1$$&lt;/p&gt;&#xA;" OwnerUserId="1865" LastActivityDate="2011-01-30T12:13:03.120" />
  <row Id="19556" PostTypeId="2" ParentId="19554" CreationDate="2011-01-30T12:25:17.287" Score="4" ViewCount="" Body="&lt;p&gt;Your proof is correct (except for parentheses missing around $k+1$ somewhere in the induction step), but since you're looking for advice: you could be clearer halfway through the induction step, like: &quot;the l.h.s. yields ..., while the r.h.s. yields ...&quot;, making it easier to read.&#xA;Finally Alo Mestanogullari's comment is spot on; if you'd present this proof more formally, you could use terms like &quot;because of the distributive law...&quot; etc.&lt;/p&gt;&#xA;" OwnerUserId="6004" LastActivityDate="2011-01-30T12:25:17.287" CommentCount="1" />
  <row Id="19557" PostTypeId="1" AcceptedAnswerId="19583" CreationDate="2011-01-30T13:06:43.093" Score="1" ViewCount="103" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know $(\{0, 1\}, \oplus)$ (where $\oplus$ is the XOR operator) is a well-known abelian group.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But what if i &quot;extend&quot; the underlying set?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, i can get $(\{0, 1, 2, 3\}, \oplus)$, or in general $(\{0, 1, ... , 2^n-1\}, \oplus)$ and they are groups as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Have these groups been studied? Have they any application in CS?&lt;/p&gt;&#xA;" OwnerUserId="5672" LastActivityDate="2011-01-30T19:25:24.663" Title="$(\{0, 1, ... , 2^n-1\}, \oplus)$ groups" Tags="&lt;group-theory&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="19558" PostTypeId="2" ParentId="19548" CreationDate="2011-01-30T13:28:59.523" Score="6" ViewCount="" Body="&lt;p&gt;I don't know the exact context, but if I understood the problem correctly I'd say that this is clearly false.&#xA;As a counterexample consider the equation $E$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x = x$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and the perturbed equation $E_p$ for $p &amp;gt; 0$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(1+p)x = x+p$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$E_p$ of course has a unique solution that for $p \rightarrow 0$ converges to 1; actually it's constantly 1 for every $p \ne 0$, so no matter what is your definition of &quot;strong&quot; this convergence is strong:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(1+p)x-x = p$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x(1+p-1) = p$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x=p/p=1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now $x=1$ is of course also a solution of $E$... but I think $E$ has multiple solutions :-)&lt;/p&gt;&#xA;" OwnerUserId="715" LastEditorUserId="715" LastEditorDisplayName="" LastEditDate="2011-01-30T14:55:38.620" LastActivityDate="2011-01-30T14:55:38.620" CommentCount="6" />
  <row Id="19560" PostTypeId="1" AcceptedAnswerId="19606" CreationDate="2011-01-30T14:13:29.670" Score="2" ViewCount="155" Body="&lt;p&gt;I am a student of Computer Science and currently learning about optimization. We've been introduced to the Fritz-John and Karush-Kuhn-Tucker conditions for convex optimizations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think I understand what they mean, in the sense that these conditions have to be fulfilled in order for a point to be an optimum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I don't really understand the &lt;strong&gt;difference between both&lt;/strong&gt; and my lecture notes are not very good at making a clear point of this difference. In fact, I am allergic to mathematical notation, so I would appreciate a concise and short answer in plain English, maybe with an example.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you!&lt;/p&gt;&#xA;" OwnerUserId="5488" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2011-01-30T18:48:05.983" LastActivityDate="2011-01-30T21:58:05.943" Title="Difference between Fritz John and Karush-Kuhn-Tucker conditions" Tags="&lt;optimization&gt;&lt;convex-optimization&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="19561" PostTypeId="2" ParentId="19526" CreationDate="2011-01-30T14:51:59.360" Score="3" ViewCount="" Body="&lt;p&gt;For any integrable $f$ and $g$ (so in particular for functions in $C_O({\mathbb R}^n)$ one has $\int_{{\mathbb R}^n} f \ast g = \int_{{\mathbb R}^n} f \int_{{\mathbb R}^n} g$.&#xA;So your question is asking if $\int_{{\mathbb R}^n} f_n \int_{{\mathbb R}^n} g_n$ necessarily converges to $\int_{{\mathbb R}^n} f \int_{{\mathbb R}^n} g$. A counterexample will be provided by letting $f_n = g_n$ such that $f_n$ converges to $f$ in $L^2$ but such that $\int_{{\mathbb R}^n} f_n$ does not converge to $\int_{{\mathbb R}^n} f$. To do this, one can let $\phi(x)$ be any smooth function with compact support and integral one, and then let $f_n(x) = {1 \over n} \phi({x \over n})$. Each $f_n$ will have integral $1$, but $f_n(x)$ converges in $L^2$ to $f(x) = 0$ which has integral zero.&lt;/p&gt;&#xA;" OwnerUserId="3035" LastEditorUserId="3035" LastEditorDisplayName="" LastEditDate="2011-01-30T15:32:31.993" LastActivityDate="2011-01-30T15:32:31.993" CommentCount="2" />
  <row Id="19562" PostTypeId="1" AcceptedAnswerId="19563" CreationDate="2011-01-30T15:26:27.990" Score="0" ViewCount="129" Body="&lt;p&gt;As in &lt;a href=&quot;http://en.wikipedia.org/wiki/Polynomial&quot; rel=&quot;nofollow&quot;&gt;wikipedia&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In mathematics, a polynomial is an&#xA;  expression of finite length&#xA;  constructed from variables (also known&#xA;  as indeterminates) and constants&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So, it's only considered a polynomial if it have both variables and constans?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;x² + x&lt;/code&gt; or &lt;code&gt;x + x&lt;/code&gt; are not a polynomials? (as they have only variables)&lt;/p&gt;&#xA;" OwnerUserId="4501" LastEditorUserId="4501" LastEditorDisplayName="" LastEditDate="2011-02-06T13:19:25.927" LastActivityDate="2011-02-16T01:53:10.810" Title="The Polynomial concept needs to include both variables and contants?" Tags="&lt;algebra&gt;&lt;polynomials&gt;&lt;definition&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="19563" PostTypeId="2" ParentId="19562" CreationDate="2011-01-30T15:39:13.907" Score="4" ViewCount="" Body="&lt;p&gt;As Qiaochu pointed out, the coefficients of the polynomials are the constants. Hence, in a polynomial like &#xA;$x^2 + x + 5$, the constants are:&#xA;$1$ for $x^2$,&#xA;$1$ for $x^1 = x$&#xA;and $5$ for $x^0 = 1$. $3$ constants for the $3$ terms.&lt;/p&gt;&#xA;" OwnerUserId="6399" LastEditorUserId="664" LastEditorDisplayName="" LastEditDate="2011-01-30T18:54:11.147" LastActivityDate="2011-01-30T18:54:11.147" />
  <row Id="19564" PostTypeId="2" ParentId="19508" CreationDate="2011-01-30T15:48:45.550" Score="1" ViewCount="" Body="&lt;p&gt;There are many applications of group theory to chemistry, in particular to fullerenes. If you do a &quot;google search&quot; with the strings: +chemistry +&quot;group theory&quot; or +&quot;group theory&quot; +fullerenes you will find many papers that deal with these topics.&lt;/p&gt;&#xA;" OwnerUserId="1369" LastActivityDate="2011-01-30T15:48:45.550" />
  <row Id="19565" PostTypeId="1" AcceptedAnswerId="19571" CreationDate="2011-01-30T17:22:59.867" Score="2" ViewCount="127" Body="&lt;p&gt;How does one prove this result:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If $x \in \mathbb{R}\setminus\mathbb{Q}$, the terms of the sequence $(nx - \lfloor{nx\rfloor})$ constitute a dense subset of $[0,1]$?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Many results in analysis like denseness of $\mathbb{Q}$ in $\mathbb{R}$ are very much useful and applicable when we solve problems. Can anyone tell me if the above result is useful anywhere.&lt;/p&gt;&#xA;" LastActivityDate="2011-01-30T18:03:38.743" Title="Proving that the terms of the sequence $(nx-[nx])$ is dense in $[0,1]$." Tags="&lt;real-analysis&gt;&lt;analysis&gt;" AnswerCount="2" CommentCount="6" FavoriteCount="1" />
  <row Id="19566" PostTypeId="2" ParentId="19560" CreationDate="2011-01-30T17:28:49.363" Score="1" ViewCount="" Body="&lt;p&gt;The conditions are mathematically the same, stated in different ways. The way KKT state the conditions makes it easier to find the a set of points which satisfy necessary optimality conditions and are potentially the local optimum points of the problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if for a particular point one wants to see if it can be a local optimum then it is easy to verify Fritz-John conditions. The reason for that is that you know which constraints are active or inactive.&lt;/p&gt;&#xA;" OwnerUserId="6151" LastActivityDate="2011-01-30T17:28:49.363" />
  <row Id="19567" PostTypeId="2" ParentId="12423" CreationDate="2011-01-30T17:35:43.570" Score="3" ViewCount="" Body="&lt;p&gt;This is true, and follows from a theorem of Fary.  Fary's theorem says that for closed curves inside a ball, the average curvature of the curve is at least as large as the curvature of the boundary of the ball.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your curve isn't closed (or finite in length), so you have to think about what the theorem says in this context.  But it will work:  the average curvature is zero, and so it cannot fit in any (finite) ball.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I learned about Fary's theorem from Serge Tabachnikov; there are several proofs (of the 2-dimensional case, though most generalize to higher dimensions, as he describes at the end) in his beautiful short paper:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.math.psu.edu/tabachni/prints/dna-mass2.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.math.psu.edu/tabachni/prints/dna-mass2.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="6071" LastActivityDate="2011-01-30T17:35:43.570" CommentCount="3" />
  <row Id="19568" PostTypeId="1" CreationDate="2011-01-30T17:39:25.503" Score="0" ViewCount="180" Body="&lt;p&gt;What is the difference between a &quot;time series&quot; and a (discrete-time) stochastic process?&lt;/p&gt;&#xA;" OwnerUserId="6417" LastActivityDate="2011-01-31T04:20:43.243" Title="Difference between time series and stochastic process?" Tags="&lt;statistics&gt;&lt;stochastic-processes&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="19569" PostTypeId="1" AcceptedAnswerId="19596" CreationDate="2011-01-30T17:46:23.853" Score="5" ViewCount="149" Body="&lt;p&gt;Let $T_1$ and $T_2$ be two first-order logical theories (over the same signature) such that $T_1 \subseteq T_2$ and both are recursively axiomatized.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is the following: is it possible that $T_1$ is finitely axiomatizable and decidable, while its extension $T_2$ is undecidable?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to know whether there is or not a pair of logical theories with the previous requirements, but I have not been able to find any answer surfing the web. There is a &lt;a href=&quot;http://www.springerlink.com/content/l26l8r7245445727/&quot; rel=&quot;nofollow&quot;&gt;paper&lt;/a&gt; by Verena Huber Dyson with the title &quot;On the decision problem for theories of finite models&quot; where there is an example for all these conditions except for requiring finitary axiomatizability. This paper is quite old (1964), so maybe someone here (if not, I will try mathoverflow) can provide me a better answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Addendum: By @JDH comments it is clear that&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;$T_1$ cannot be a complete theory (because then there are no consistent extensions).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$T_2$ cannot be finitely axiomatizable (because finite extensions of decidable theories are decidable)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="5580" LastEditorUserId="630" LastEditorDisplayName="" LastEditDate="2011-05-04T18:34:41.083" LastActivityDate="2011-05-04T18:34:41.083" Title="Example of an UnDecidable Logical Theory which is an extension of a Logical Decidable Theory?" Tags="&lt;logic&gt;&lt;computability&gt;&lt;model-theory&gt;" AnswerCount="2" />
  <row Id="19570" PostTypeId="2" ParentId="19565" CreationDate="2011-01-30T17:49:11.477" Score="3" ViewCount="" Body="&lt;p&gt;A hint on one way to prove it (not necessarily shortest) without using much theory: what is the difference between successive terms in the sequence?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(More explicit hint: Let $\{x\} = x - \lfloor x \rfloor$ be the fractional part of $x$. Try proving that $$(n+1)x-\lfloor(n+1)x\rfloor \equiv (nx - \lfloor nx \rfloor) + \{x\} \mod 1 .$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you take it from here, using the fact that $\{x\}$ is not rational?)&lt;/p&gt;&#xA;" OwnerUserId="205" LastActivityDate="2011-01-30T17:49:11.477" CommentCount="1" />
  <row Id="19571" PostTypeId="2" ParentId="19565" CreationDate="2011-01-30T18:03:38.743" Score="3" ViewCount="" Body="&lt;p&gt;It is better than dense, it is equidistributed (in $\mathbb{R}/\mathbb{Z}$), meaning that if $f: \mathbb{R}/\mathbb{Z} \rightarrow \mathbb{C}$ is continuous (it is equivalent to give $f$ or a $1$-periodic continuous function from $\mathbb{R}$ to $\mathbb{C}$), $\lim_{n \rightarrow + \infty} \frac{1}{n} \sum_{k=1}^n f(kx)= \int_0^1 f$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This can be easily proven using the fact that the polynomials in $t \mapsto e^{2 i \pi t}$ and $t \mapsto e^{2 i \pi t}$ (so, sums of the form $\sum_{l=-n}^n \lambda_k e^{2 i \pi l t}$) are dense (for the supremum norm) in our space of $1$-periodic continuous functions (Weierstrass theorem), and it is easy to compute $\lim_{n \rightarrow + \infty} \frac{1}{n} \sum_{k=1}^n e^{2 i \pi l k x} = \delta_{l,0}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The density of your sequence follows: if an open subset of $\mathbb{R}/\mathbb{Z}$ (or $]0,1[$) did not contain any element of the sequence, using a &quot;test function&quot; $f$, non-negative, not identically zero and whose support is contained in our given open subset, would contradict the equidistribution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: if you don't you the density theorem of Weierstrass above, you could also use the fact that $x \mathbb{Z} + \mathbb{Z}$ is dense in $\mathbb{R}$ (since it has no smallest positive element, otherwise $x$ would be rational), but this way you only get density, not equidistribution.&lt;/p&gt;&#xA;" OwnerUserId="2660" LastActivityDate="2011-01-30T18:03:38.743" CommentCount="2" />
  <row Id="19572" PostTypeId="1" AcceptedAnswerId="19573" CreationDate="2011-01-30T18:12:18.890" Score="2" ViewCount="139" Body="&lt;p&gt;For an assignment, I've been asked to evaluate $7^{10507} \bmod 13$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know it's possible to do this using binary fast exponentiation - in fact the question refers to a previous question where I calculated $7^{12} \bmod 13$ to help prove $13$ is prime using Fermat's Little Theorem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I could simply perform the exponentiation, but I feel like I'm missing some way to simplify the question here - why bother referencing the $7^{12}$ question if I couldn't somehow use it to simplify my task.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there some sort of simplification I can perform here? Am I missing something?&lt;/p&gt;&#xA;" OwnerUserId="6418" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-30T21:03:25.470" LastActivityDate="2011-01-30T21:06:05.340" Title="Simplifying or easily evaluating a large exponent in a mod number system." Tags="&lt;number-theory&gt;&lt;homework&gt;&lt;modular-arithmetic&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="19573" PostTypeId="2" ParentId="19572" CreationDate="2011-01-30T18:15:40.340" Score="5" ViewCount="" Body="&lt;p&gt;If you can prove $7^a \equiv 1 \pmod{13}$ for some $a$, then $7^{ka} \equiv 1$ for any natural $k$.  Then $7^{(ka+b)} \equiv 7^b$, and you can reduce the exponent mod $a$.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-30T21:06:05.340" LastActivityDate="2011-01-30T21:06:05.340" CommentCount="1" />
  <row Id="19574" PostTypeId="2" ParentId="19569" CreationDate="2011-01-30T18:30:13.017" Score="4" ViewCount="" Body="&lt;p&gt;Since you aren't requiring that $T_2$ is recursively axiomatized, you can do this in a silly way. Choose a nonrecursive subset $A$ of $\mathbb{N}$, and let the signature consist of countably many constants $c_n$. Let $T_1$ be the empty theory, and $T_2$ be the theory whose axioms are $c_i=c_j$ whenever $i$ and $j$ are both in $A$. Then $T_1$ is decidable but $T_2$ is not.&lt;/p&gt;&#xA;" OwnerUserId="5203" LastActivityDate="2011-01-30T18:30:13.017" CommentCount="3" />
  <row Id="19575" PostTypeId="1" CreationDate="2011-01-30T18:33:32.403" Score="2" ViewCount="108" Body="&lt;p&gt;I'm having a little difficulty wrapping my head around Lie theory (I'm a computer scientist, so perhaps that's to be expected).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Specifically, considering the following definition from Wikipedia for the exponential map. What is the significance of the identity element? I think that this could be to do with the derivative of exp being equal to the function itself, but I'm not quite sure if I'm on the right track here. Any nudges in the right direction would be highly appreciated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Let G be a Lie group and $\mathfrak{g}$ be its Lie algebra &lt;strong&gt;(thought of as the  tangent space to the identity element of G).&lt;/strong&gt; The exponential map is a map $\exp\colon \mathfrak g \to G$ which can be defined in several different ways.&quot;&lt;/p&gt;&#xA;" OwnerUserId="4820" LastActivityDate="2011-01-30T19:54:42.480" Title="Reasoning about Lie theory and the Exponential Map" Tags="&lt;lie-groups&gt;&lt;lie-algebra&gt;" AnswerCount="3" FavoriteCount="1" />
  <row Id="19576" PostTypeId="1" CreationDate="2011-01-30T18:39:39.910" Score="3" ViewCount="296" Body="&lt;p&gt;Let $A$ and $B$ be two operators on a Banach space X.&#xA;I am interested in the relationship between the spectra&#xA;of $A$, $B$ and $AB$.&#xA;In particular, are there any set theoretic inclusions&#xA;or everything can happen in general like:&#xA;$\sigma(A) \subset \sigma(AB)$, and conversely,&#xA;$\sigma(B) \subset \sigma(AB)$, and conversely?&lt;br&gt;&#xA;If we know the spectra $\sigma(A)$, $\sigma(B)$ of&#xA;$A$ and $B$, can we determine the spectrum of $AB$?&#xA;I would appreciate any comment or reference.&lt;/p&gt;&#xA;" OwnerUserId="6379" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2011-01-31T15:49:42.430" LastActivityDate="2011-02-01T09:28:15.547" Title="Spectrum of a product of operators on a Banach space" Tags="&lt;analysis&gt;&lt;functional-analysis&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="2" />
  <row Id="19577" PostTypeId="1" CreationDate="2011-01-30T18:46:21.110" Score="1" ViewCount="80" Body="&lt;p&gt;Hello! &#xA;I am looking for a book in projective geometry, using the apparatus of linear algebra, complex analysis, and, perhaps, modern algebra, in full. The counterexample is the Hartshorne's book on projective geometry that starts out with a list of axioms, as in high school geometry book. I'm looking for something in more Needham-like style.&lt;/p&gt;&#xA;" OwnerUserId="6419" LastEditorUserId="6419" LastEditorDisplayName="" LastEditDate="2011-01-30T19:09:35.177" LastActivityDate="2011-01-30T20:19:55.363" Title="Analytically flavored book in projective geometry" Tags="&lt;books&gt;&lt;projective-geometry&gt;" AnswerCount="1" />
  <row Id="19579" PostTypeId="2" ParentId="19575" CreationDate="2011-01-30T18:56:57.527" Score="2" ViewCount="" Body="&lt;p&gt;I am novice in Lie theory myself, but I'll try to answer your question about the significance of identity element.&#xA;There is, in fact, no significance. The identity element is as good as any other element in the connected component of Lie group, and the tangent to the identity element can be moved anywhere, as long as we stay in the connected component. However, it is easier to perform calculations involving the identity element than those that involve other elements. This is why the identity element is used.&lt;/p&gt;&#xA;" OwnerUserId="6419" LastActivityDate="2011-01-30T18:56:57.527" />
  <row Id="19580" PostTypeId="2" ParentId="19575" CreationDate="2011-01-30T19:05:09.813" Score="2" ViewCount="" Body="&lt;p&gt;The elements of the Lie algebra can be thought of as &quot;infinitesimal elements,&quot; so it's natural to think of them as being infinitesimally close to the identity; if we denote by $\epsilon$ an infinitesimal that squares to zero, then you can informally think of the Lie algebra as the set of elements of the form $I + \epsilon X$ where $I$ is the identity.  For example, the elements of the orthogonal group satisfy $A \cdot A^T = I$, so the &lt;em&gt;infinitesimal&lt;/em&gt; elements of the orthogonal group satisfy $(I + \epsilon X)(I + \epsilon X^T) = I + \epsilon (X + X^T) = 0$, or $X + X^T = 0$.  Hence the Lie algebra of the orthogonal group is precisely the space of skew-symmetric matrices.  (This argument can be made completely formal if we work with algebraic groups instead of Lie groups.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a precise sense in which the Lie algebra consists of &quot;infinitesimal elements&quot;: whenever a Lie group $G$ acts on a manifold $M$ by symmetries, the Lie algebra $\mathfrak{g}$ acts by differential operators on the smooth functions $M \to \mathbb{R}$.  So they &quot;infinitesimally generate&quot; symmetries of $M$ (and the precise sense in which this is true is the exponential map).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you prefer, there is an equivalent definition which does not privilege the identity element: the elements of the Lie algebra are the left-invariant vector fields on the Lie group.  (Since a Lie group acts transitively on itself, a left-invariant vector field is determined by its value at any group element, and again, it's natural to look at the identity element.)&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2011-01-30T19:05:09.813" />
  <row Id="19581" PostTypeId="2" ParentId="19576" CreationDate="2011-01-30T19:06:17.903" Score="10" ViewCount="" Body="&lt;p&gt;If $A$ and $B$ commute, then $\sigma(AB)\subseteq \sigma(A)\sigma(B)$.  This can be seen by applying the Gelfand transform to the commutative Banach algebra consisting of all operators that commute with all operators that commute with $A$ and $B$.  (The reason to take the double commutant rather than simply the Banach algebra generated by $A$, $B$, and the identity is that the spectra of $A$, $B$, and $AB$ may be larger relative to the latter algebra.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $A$ and $B$ do not commute, things aren't as nice.  E.g., if $A$ is a nonzero nilpotent (or quasinilpotent) operator on a Hilbert space and $B=A^*$, then $\sigma(A)=\sigma(B)=\{0\}$, but the spectral radius of $AB$ is $\|A\|^2\neq0$.  For a very concrete example consider $A=\begin{bmatrix}0 &amp;amp; 1 \\ 0 &amp;amp; 0 \end{bmatrix}$ and $B=\begin{bmatrix}0 &amp;amp; 0 \\ 1 &amp;amp; 0 \end{bmatrix}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The spectra of $A$ and $B$ can also be much larger than that of $AB$, even in the commutative case.  The spectrum of an invertible operator $A$ can be an arbitrary compact subset of the complex plane not containing $0$, but $\sigma(AA^{-1})=\{1\}$.  For a noncommutative example, the spectrum of a nonunitary isometry $S$ on Hilbert space (like the unilateral shift) is the closed unit disk, while $\sigma(S^*S)=\{1\}$ and $\sigma(SS^*)=\{0,1\}$.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Here's a little more detail on the first paragraph. Let $a$ and $b$ be commuting elements of a unital Banach algebra $\mathcal{A}$ (for instance, $\mathcal{A}$ could be the algebra of all bounded operators on a Banach space).  In order to compare the spectrum of $ab$ to that of $a$ and $b$, we want to apply the Gelfand transform to a commutative Banach algebra $\mathcal{B}$ containing $a$ and $b$.  One must be careful, however, because if $a,b,1_\mathcal{A}\in\mathcal{B}\subset\mathcal{A}$, there may be elements of $\mathcal{B}$ that are invertible in $\mathcal{A}$ but whose inverses lie in $\mathcal{A}\setminus\mathcal{B}$.  To ensure that this doesn't affect the spectra of $a$, $b$, or $ab$, one can either just stipulate that $\mathcal{B}$ also contains all elements of the form $(a-\lambda)^{-1}$, $(b-\lambda)^{-1}$, and $(ab-\lambda)^{-1}$, as $\lambda$ ranges over the resolvent sets of $a$, $b$, and $ab$ respectively, or just take the double commutant of $\{a,b\}$ in $\mathcal{A}$, which will automatically contain all of these elements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once the problem is reduced to the case of elements $a$ and $b$ of a commutative unital Banach algebra $\mathcal{B}$, the Gelfand transform gives you a homomorphism $\Gamma:\mathcal{B}\to C(X)$, where $C(X)$ is the (Banach) algebra of continuous complex-valued functions on the maximal ideal space of $\mathcal{B}$ with pointwise operations (and $\sup$ norm, but that isn't very important here).  One of the most fundamental properties of the Gelfand transform is that it preserves spectra; that is, for all $c\in\mathcal{B}$, $\sigma(c)=\Gamma(c)(X)$.  Using this fact, the inclusion $\sigma(ab)\subseteq\sigma(a)\sigma(b)$ is translated to a tautological fact about the range of a pointwise product of functions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To prove that $\sigma(c)=\Gamma(c)(X)$, remember that $X$ can be identified with (or is defined to be, depending on your perspective) the set of nonzero homomorphisms from $\mathcal{B}$ to $\mathbb{C}$.  If $\phi\in X$, then $\phi(c-\phi(c))=0$, which, since unital homomorphisms send invertible elements to invertible elements, shows that $c-\phi(c)$ is not invertible.  This shows that $\Gamma(c)(\phi)$ is in the spectrum of $c$, so $\Gamma(c)(X)\subseteq\sigma(c)$.  Conversely, if $\lambda\in\sigma(c)$, then there is a maximal ideal $M$ of $\mathcal{B}$ containing $c-\lambda$, and by the Mazur-Gelfand theorem there is a unique isomorphism $\mathcal{B}/M\cong \mathbb{C}$.  Composing with the quotient map gives a $\phi\in X$ with kernel $M$, so in particular $\phi(c-\lambda)=\phi(c)-\lambda = 0$, showing that $\lambda=\Gamma(c)(\phi)$, so that $\sigma(c)\subseteq\Gamma(c)(X)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To learn more about this, I recommend consulting any good textbook on functional analysis or Banach algebras.  For example, &lt;a href=&quot;http://books.google.com/books?id=Sh_vAAAAMAAJ&amp;amp;q&quot; rel=&quot;nofollow&quot;&gt;Rudin&lt;/a&gt;, &lt;a href=&quot;http://books.google.com/books?id=-OdfXeNmrT0C&amp;amp;dq&quot; rel=&quot;nofollow&quot;&gt;Douglas&lt;/a&gt;, and &lt;a href=&quot;http://books.google.com/books?id=PVrvAAAAMAAJ&amp;amp;q&quot; rel=&quot;nofollow&quot;&gt;Rickart&lt;/a&gt; are good sources for the Gelfand theory that vary widely in emphasis and additional topics covered.  As for your question about whether $\sigma(T^{-1}ATB)\subseteq\sigma(A)\sigma(B)$ for commuting $A$ and $B$, the answer is that this need not hold if $T^{-1}AT$ and $B$ do not commute.  For example, let $A=B=\begin{bmatrix} 0 &amp;amp; 1\\ 0 &amp;amp; 0\end{bmatrix}$ and $T=\begin{bmatrix}0 &amp;amp; 1 \\ 1 &amp;amp; 0 \end{bmatrix}$.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2011-01-31T15:31:51.040" LastActivityDate="2011-01-31T15:31:51.040" CommentCount="5" />
  <row Id="19582" PostTypeId="2" ParentId="17479" CreationDate="2011-01-30T19:15:26.280" Score="0" ViewCount="" Body="&lt;p&gt;Tobias, your notation makes it look like your function $f:\mathbb{R}^3\to\mathbb{R}$, in other words, it takes in points in three-dimensional space and spits out real numbers.  In that case, as you note, it can be written as a function of $(x,y,z)$.  So there doesn't seem to be any reason not to go with your first option.  Maybe you could write down the function so I can see the difficulty.  If on the other hand your function takes in points on the sphere $\{(x,y,z):\, x^2+y^2+z^2=1\}$, then it makes sense to use spherical harmonics.  Your second option doesn't seem reasonable--if you want translation to correspond to phase shifts, then you need to integrate along lines in $\mathbb{R}^3$, and then after a change of coordinates you would be back in your first situation.&lt;/p&gt;&#xA;" OwnerUserId="6327" LastActivityDate="2011-01-30T19:15:26.280" CommentCount="2" />
  <row Id="19583" PostTypeId="2" ParentId="19557" CreationDate="2011-01-30T19:25:24.663" Score="3" ViewCount="" Body="&lt;p&gt;As Sebastian mentioned, the groups are not mysterious at all. In fact, they are quite &quot;boring&quot; since they're Abelian.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are some example applications, most of them practical, and some theoretical.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt; CPUs &lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Your CPU supports an operation called XOR (Exclusive OR) which views the registers as members of the group $\mathbb{Z}_2^n$ (where $n=32$ or $n=64$ for modern CPUs), and applies the operation of the group. The CPU also supports addition in the group $\mathbb{Z}_{2^n}$, i.e. addition modulo $2^n$, as well as multiplication modulo $2^n$.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt; Coding Theory &lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Lots of codes are defined using these groups. CRCs are linear, so they can be viewed as a linear mapping from $\mathbb{Z}_2^n$ to $\mathbb{Z}_2^m$ (where $m=16$ or $m=32$). All error-correcting codes used in practice are also linear.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt; Walsh-Hadamard Transform &lt;/h3&gt;&#xA;&#xA;&lt;p&gt;According to &lt;a href=&quot;http://en.wikipedia.org/wiki/Hadamard_transform&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;, the Walsh-Hadamard transform, which is just the Fourier transform on your groups, has some applications to encryption and encoding (usually real Fourier transforms are used).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This transform is very important in complexity theory (see also the next section), where it is used to construct and analyze PCPs, and to prove important results such as the parallel repetition theorem.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt; Complexity Theory &lt;/h3&gt;&#xA;&#xA;&lt;p&gt;These groups are used to prove lower bounds on constant-depth circuits using the polynomials method in its finite field variety (the so-called Razborov-Smolensky method). The idea is to associate with each gate in the circuit a low-degree polynomial approximating the function computed by the gate (with a small error), and then proving that the target function doesn't have such nice approximations.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2011-01-30T19:25:24.663" CommentCount="1" />
  <row Id="19584" PostTypeId="2" ParentId="19568" CreationDate="2011-01-30T19:36:12.783" Score="1" ViewCount="" Body="&lt;p&gt;A time series is a sequence of actual, fixed, values, like:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;61, 63, 58, 64, 56, 48, 39, 42, ...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A stochastic process is a sequence of random variables that have some kind of specified correlation or other distributional relationship between them.  Stochastic processes are often used in modeling time series data- we assume that the time series we have was produced by a stochastic process, find the parameters of a stochastic process that would be likely to produce that time series, and then use that stochastic process as a model in predicting future values of the time series.  &lt;/p&gt;&#xA;" OwnerUserId="6310" LastActivityDate="2011-01-30T19:36:12.783" />
  <row Id="19585" PostTypeId="2" ParentId="19575" CreationDate="2011-01-30T19:54:42.480" Score="4" ViewCount="" Body="&lt;p&gt;The answers given are both really good; I just want to add an additional reason why the identity element is preferred in consideration of the exponential map.  The reason is because the exponential map gives a one to one correspondence between tangent vectors at the identity and one parameter subgroups of $G$.  A one parameter subgroup of $G$ is a homomorphism $\mathbb R \to G$, or equivalently a curve $\gamma(t)\in G$ such that $\gamma(t+s) = \gamma(t)\gamma(s)$.  Since $\gamma$ is a group homomorphism, we need $\gamma(0)$ to be the identity $e\in G$ (so here we are using the special property of $e$) and any such one parameter subgroup is determined by $\gamma&amp;#39;(0) \in T_e G$.  Conversely, given $X \in T_e G$, the curve $t\mapsto \exp(tX)$ is a one-parameter subgroup with derivative $X$ at time 0.&lt;/p&gt;&#xA;" OwnerUserId="9" LastActivityDate="2011-01-30T19:54:42.480" CommentCount="0" />
  <row Id="19587" PostTypeId="2" ParentId="19577" CreationDate="2011-01-30T20:19:55.363" Score="1" ViewCount="" Body="&lt;p&gt;You could look at Chapters 4, 5, 6 of Berger's stupendous  &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0387116583&quot; rel=&quot;nofollow&quot;&gt;Geometry 1&lt;/a&gt;. There are plenty of calculations there, beautiful illustrations (including one by Escher), applications (to photography, optics) and links to algebraic topology: Hopf fibration, Boy's surface, simple connctedness,  Veronese surface,...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;He also carefully analyzes the canonical projectivization of an affine space and relates affine to projective groups. There is even  a short discussion of projective space over the quaternions and over finite fields.&lt;/p&gt;&#xA;" OwnerUserId="3217" LastActivityDate="2011-01-30T20:19:55.363" />
  <row Id="19588" PostTypeId="1" CreationDate="2011-01-30T20:32:48.590" Score="3" ViewCount="87" Body="&lt;p&gt;Hi, Everyone:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am curious as to what changes do we need to make to the hypotheses of the&#xA;inverse function theorem in order to be able to find the &lt;em&gt;global&lt;/em&gt; differentiable inverse to a differentiable function. We obviously need $f$ to be a bijection, and $f&amp;#39;$ to be non-zero. Is this sufficient for the existence of a &lt;em&gt;global&lt;/em&gt; differentiable inverse?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For functions $f\colon\mathbb{R}\to\mathbb{R}$, we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Motivation:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f^{-1}(f(x))=x$, so $(f&amp;#39;)^{-1}(f(x))f&amp;#39;(x)=1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, we could define $(f&amp;#39;)^{^-1}(f(x))$ to be $1/f&amp;#39;(x)$   ( this is the special case of the formula for the differentiable inverse -- when it exists -- in the IFT)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(and we are assumming $f'(x)\neq 0$) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of $\mathbb{R}^2$, I guess we could think of all the branches of $\log z$ and $\exp z$, and we do have at least a branch-wise global inverse , i.e., if/when $\exp z$ is 1-1 (and it is , of course onto $\mathbb{C}-{0}$), then we have a differentiable inverse.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I guess my question would be: once the conditions of the IFT are satisfied:&#xA;in how big of a neighborhood of $x$ can we define this local diffeomorphism,&#xA;and, in which case would this neighborhood be the entire domain of definition of $f$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I guess the case for manifolds would be a generalization of the case of&#xA;$\mathbb{R}^n$, but it seems like we would need for the manifolds to have a single chart.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, are the conditions of f being a bijective, differentiable map sufficient&#xA;for the existence of a global differentiable inverse? And, if $f$ is differentiable, but not bijective,  does the IFT hold in the largest subset of the domain of definition of $f$ where $f$ is a bijection?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="6421" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-30T21:23:05.080" LastActivityDate="2011-01-30T21:38:18.877" Title="extension/&quot;globalization&quot; of inverse function theorem" Tags="&lt;differential-geometry&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="19589" PostTypeId="1" AcceptedAnswerId="19795" CreationDate="2011-01-30T20:33:35.890" Score="1" ViewCount="102" Body="&lt;p&gt;Given a differential matrix equation, ie $X&amp;#39;=A(z)X+B(z)$ where both $A$ and $B$ are matrix of size $n\times n$ with coefficients that are holomorfic functions in a convex open set $\Omega$ and continuous on the closure $\bar \Omega$,  and an initial data: $X(z_0)=u$, I know there exists a solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I haven't been able to find on the internet a proof of the existence. So the question is how to prove it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I already know it when $A(z)$ has constant coefficients, but it cannot be extended to this case.&#xA;Also I've read about Magnus Series. Although I don't fully understand them, I'd prefer a easier proof of the existence, as I'm not really interested in a generic formula.&lt;/p&gt;&#xA;" OwnerUserId="6276" LastActivityDate="2011-02-01T01:57:21.077" Title="Solution of a differential matrix equation" Tags="&lt;complex-analysis&gt;&lt;differential-equations&gt;" AnswerCount="1" />
  <row Id="19590" PostTypeId="2" ParentId="18771" CreationDate="2011-01-30T20:43:34.910" Score="0" ViewCount="" Body="&lt;p&gt;This may be a repetition of what Mariano and Sean were suggesting: you will most likely&#xA;   ultimately need to either: produce a bundle morphism between your bundle and&#xA;   the trivial bundle, or produce a global trivialization of your top space.&lt;/p&gt;&#xA;" OwnerUserId="6421" LastActivityDate="2011-01-30T20:43:34.910" />
  <row Id="19591" PostTypeId="1" AcceptedAnswerId="19595" CreationDate="2011-01-30T20:51:22.177" Score="4" ViewCount="61" Body="&lt;p&gt;In a number of real analysis texts (I am thinking of Folland in particular), three different kinds of measures are defined.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Positive measures: Take values in $[0, +\infty]$&lt;/li&gt;&#xA;&lt;li&gt;Signed measures: Take values in either $(-\infty, \infty]$ or $[-\infty, \infty)$, but cannot assume both $+\infty$ and $-\infty$.&lt;/li&gt;&#xA;&lt;li&gt;Complex measures: Take values in $\mathbb{C}$.  Any kind of infinity is not allowed.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;My question is: why is this?  Is this because of how we set up integrals with respect to these measures, or does it have to do with adding and subtracting measures to make new ones?&lt;/p&gt;&#xA;" OwnerUserId="640" LastActivityDate="2011-01-30T21:07:06.487" Title="Why are signed and complex measures typically not allowed to assume infinite values?" Tags="&lt;real-analysis&gt;&lt;measure-theory&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="19592" PostTypeId="2" ParentId="19591" CreationDate="2011-01-30T21:01:21.127" Score="0" ViewCount="" Body="&lt;p&gt;For signed measures assuming both $\infty$ and $-\infty$ wouldn't a right thing, for $\infty-\infty$ is undefined.&#xA;With $\mathbb{C}$ the problem is that, I think, there is no ordering on it.&lt;/p&gt;&#xA;" OwnerUserId="6419" LastActivityDate="2011-01-30T21:01:21.127" />
  <row Id="19593" PostTypeId="2" ParentId="12022" CreationDate="2011-01-30T21:03:33.763" Score="0" ViewCount="" Body="&lt;p&gt;Just to add that I think that the value of the integral (i.e., 2Pi times the&#xA;  winding number) depends only on the homotopy-type of the curve, i.e., if a curve&#xA;  C is homotopic to the circle L that winds around the origin, then the integral over&#xA;  C will have the same value as the integral over L.&lt;/p&gt;&#xA;" OwnerUserId="6421" LastActivityDate="2011-01-30T21:03:33.763" />
  <row Id="19594" PostTypeId="1" CreationDate="2011-01-30T21:06:37.597" Score="4" ViewCount="142" Body="&lt;p&gt;As a newbie to vector bundles, it seems like all vector bundles I have  run into ( not that many, I admit) need only two charts to be trivialized; one of these&#xA;  charts will contain the &quot;trouble point&quot; (the point that prevents the top space from&#xA;  being a global product), and one chart that will not contain the trouble point ( I am thinking of the Mobius band over the circle here ).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It also seems like the trivializations ultimately consist of unwrapping the twist&#xA;  in the trouble point in the top( again, in here, in the case of the Mobius bundle, &#xA;  I am thinking of the lift of (1,0) in S^1 ) , unwrapping , and flattenning.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this an at least informally correct generalization of vector bundles ( or, at&#xA;  least of plane bundles, i.e., 1-bundles? )&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="6421" LastActivityDate="2011-02-04T00:50:21.167" Title="Is this an Accurate General description of Line Bundles?" Tags="&lt;differential-geometry&gt;" AnswerCount="5" CommentCount="3" />
  <row Id="19595" PostTypeId="2" ParentId="19591" CreationDate="2011-01-30T21:07:06.487" Score="3" ViewCount="" Body="&lt;p&gt;It has to do with adding measures to satisfy countable summability.  Notice that in the complex plane $+\infty$ and $-\infty$ are the same value.  So we can't allow one without the other, and who knows what it should mean to add them?  On the other hand if we stick to finite values then the summability condition is clear.&lt;/p&gt;&#xA;" OwnerUserId="3111" LastActivityDate="2011-01-30T21:07:06.487" CommentCount="2" />
  <row Id="19596" PostTypeId="2" ParentId="19569" CreationDate="2011-01-30T21:13:32.063" Score="6" ViewCount="" Body="&lt;p&gt;Here are a few observations:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Any finite extension of a decidable theory is decidable. That is, if $T_1$ is decidable and $T_2$ is obtained by adding finitely many additional axioms $\varphi$, then $T_2$ is also decidable, since $T_2\vdash\psi$ is the same as $T_1\vdash\varphi\to\psi$. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If $T_2$ is c.e. and complete, then it is decidable, since to decide whether $T_2\vdash\varphi$ one searches either for $\varphi$ or for $\neg\varphi$ in $T_2$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;As I point out in the comment to Chris's answer, a theory is c.e. if and only if it is computably axiomatizable. The reverse implication is just enumerating proofs. For the forward direction, replace each axiom $\varphi$ with an enormous conjunction $\varphi\wedge\cdots\wedge\varphi$ whose length is the time it takes to appear in the computable enumeration. The set of these conjunctions (although a silly trick) axiomatizes the same theory and is decidable.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;There are finitely axiomatizable theories in the language of arithmetic that have no decidable extensions. For example, Robinson's $Q$ is an example of this. One reason is that if $Q\subset T$ and $T$ were decidable and consistent, then we could find a computable separation of the two sets $A$ and $B$ consisting of the TM programs that halt with output $0$ and $1$, respectively. The reason is that any such halting behavior would be provable by $Q$ and hence also by $T$, and so we would find a computable set $C$ containing $A$ and disjoint from $B$. But this contradicts that these sets are computably inseparable---by the recursion theorem we could create a program that halts without output $0$ iff it is not in $C$, a contradiction.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;There are numerous examples of finitely axiomatizable $T_1$ with c.e. $T_2$ that are not decidable. Chris Eagle gave one in his answer, if you take his set $A$ to be c.e. but not decidable. You can do his trick with any decidable theory $T_1$, provided that it has models with at least two elements. Just add new constant symbols $c_i$ and add the axioms $c_i=c_j$ whenever $i,j\in A$, for some fixed c.e. non-decidable set $A$. We can enumerate these axioms, so the resulting theory $T_2$ is c.e., but not decidable, since from a decision procedure for $T_2$ we could decide membership in $A$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="413" LastActivityDate="2011-01-30T21:13:32.063" CommentCount="9" />
  <row Id="19597" PostTypeId="1" CreationDate="2011-01-30T21:19:44.770" Score="0" ViewCount="83" Body="&lt;p&gt;an arbitrary point P is chosen on side BC of triangle ABC and perpendiculars PU and PV are drawn from P to other two sides of the triangle.  (It may be that U or V lies on an extension of AB or AC and not on the actual side of the triangle.  This can happen, for instance, if angle A is obtuse and point P is very near B or C.)  Show that the sum PU+PV of the lengths of the two perpendiculars is constant as P moves along BC.  In other words, this quantity is independent of the choice of P.&lt;/p&gt;&#xA;" OwnerUserId="6422" LastEditorUserId="5105" LastEditorDisplayName="" LastEditDate="2011-01-30T23:21:29.807" LastActivityDate="2011-01-30T23:21:29.807" Title="Math problem- geometry- arbitrary points" Tags="&lt;homework&gt;&lt;euclidean-geometry&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="19598" PostTypeId="2" ParentId="19597" CreationDate="2011-01-30T21:30:17.520" Score="2" ViewCount="" Body="&lt;p&gt;Taking $P$ to be $B$ and $C$, we get that you actually need to assume that the triangle is isosceles ($AB = AC$). Given that, draw the segment $PA$ and express the area of the triangle in terms of $PU+PV$.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2011-01-30T21:30:17.520" />
  <row Id="19599" PostTypeId="2" ParentId="19597" CreationDate="2011-01-30T21:34:41.893" Score="1" ViewCount="" Body="&lt;p&gt;I don't see how this is true: if you have a triangle such that $AB \neq AC$ then taking $ P = B$ and then $P = C$ gives you a contradiction. I think you at least need an isosceles triangle for this to work. &lt;/p&gt;&#xA;" OwnerUserId="5105" LastActivityDate="2011-01-30T21:34:41.893" />
  <row Id="19600" PostTypeId="2" ParentId="19588" CreationDate="2011-01-30T21:38:18.877" Score="3" ViewCount="" Body="&lt;p&gt;There is a theorem (&quot;Introduction to Smooth Manifolds,&quot; Lee, Thm 7.15) for differentiable manifolds which says that:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If $F: M \to N$ is a differentiable bijective map of constant rank, then $F$ is a diffeomorphism -- so in particular $F^{-1}$ is differentiable.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here, the &lt;em&gt;rank&lt;/em&gt; of differentiable map $F\colon M \to N$ at a point $p \in M$ is defined to be the rank of its pushforward $F_*\colon T_pM \to T_{F(p)}N$.  (Some authors use the word &quot;differential&quot; for &quot;pushforward,&quot; and use the notation $d_pF$ for $F_*$.)&lt;/p&gt;&#xA;" OwnerUserId="640" LastActivityDate="2011-01-30T21:38:18.877" CommentCount="3" />
  <row Id="19601" PostTypeId="2" ParentId="19508" CreationDate="2011-01-30T21:51:18.250" Score="2" ViewCount="" Body="&lt;p&gt;Tom Fiore and his coauthors have used group theory to study various pieces of music.  You might enjoy the article &quot;Musical actions of dihedral groups ( &lt;a href=&quot;http://www-personal.umd.umich.edu/~tmfiore/1/CransFioreSatyendra.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www-personal.umd.umich.edu/~tmfiore/1/CransFioreSatyendra.pdf&lt;/a&gt; ), or take a look at Tom's music page: &lt;a href=&quot;http://www-personal.umd.umich.edu/~tmfiore/1/music.html&quot; rel=&quot;nofollow&quot;&gt;http://www-personal.umd.umich.edu/~tmfiore/1/music.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1392" LastActivityDate="2011-01-30T21:51:18.250" />
  <row Id="19602" PostTypeId="1" AcceptedAnswerId="19608" CreationDate="2011-01-30T21:53:56.733" Score="1" ViewCount="114" Body="&lt;p&gt;I have just started out proofs so I am not fully grasping this concept yet. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Question:&#xA;Is the set $\mathbb{R}^2$, with addition and multiplication defined below a field? Explain.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;$(a, b) + (c, d) = (a + c, b + d)$ &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$(a, b) (c, d) = (ac, bd)$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Attempted solution:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know: $f\colon F \times F \rightarrow F$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f(x, y) = x + y$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$g\colon F \times F \rightarrow F$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$g(x, y) = xy$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So solving 1:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(a, b) + (c, d) = (a + c, b + d)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;\begin{align*}&#xD;&#xA;f(a, b) + (c, d) &amp;amp;= (a + b) + (c + d) &amp;amp;&amp;amp;\text{definition of field} \\&#xD;&#xA;                &amp;amp;= (a + c) + (b + d) \\&#xD;&#xA;                &amp;amp;= (a + c, b + d)&#xD;&#xA;\end{align*}&#xD;&#xA;$$&#xA;The second part I did almost the same thing using $g(x, y) = xy$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sorry for the formatting, I still don't know how to use MathJaX but any help is appreciated. If you can also explain for a beginner that would be great, thanks.&lt;/p&gt;&#xA;" OwnerUserId="5505" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-31T00:05:29.587" LastActivityDate="2011-01-31T01:16:21.610" Title="Addition and Multiplication of Field Proof" Tags="&lt;abstract-algebra&gt;&lt;field-theory&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="19603" PostTypeId="2" ParentId="19594" CreationDate="2011-01-30T21:55:35.790" Score="2" ViewCount="" Body="&lt;p&gt;Despite appearances, you should not think that a bundle is not trivial because there is a &quot;trouble point&quot;. For example in the simplest case, the Möbius bundle (= tautological bundle) over $S^1$, all points of $S^1$ play exactly the same role: the bundle is homogeneous and there is no &quot;trouble point&quot;. Yet this bundle is not trivial, as you already know.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is also false that all bundles can be trivialized by using only two charts. For example the tautological bundle on $\mathbb P^{\infty} (\mathbb R) $ cannot even be trivialized by finitely many charts!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But don't let these oddities discourage you: I'm sure you' ll soon master these subtle points...&lt;/p&gt;&#xA;" OwnerUserId="3217" LastActivityDate="2011-01-30T21:55:35.790" />
  <row Id="19604" PostTypeId="1" AcceptedAnswerId="19639" CreationDate="2011-01-30T21:56:47.783" Score="1" ViewCount="112" Body="&lt;p&gt;Hey,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am stuck on this question from a problem set. The question itself has two parts so I post them under the same topic (they're pretty much related to each other too - at least in the book). I hope it's ok.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Q1: Let $f:[a,\infty] \rightarrow \mathbb R$ derivative function. $\lim_{x\rightarrow\infty} f&amp;#39;(x) = \infty$. Show that $f$ &lt;strong&gt;is not&lt;/strong&gt; uniformly continuous on $[a,\infty]$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Attempts:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;I tried to show that if $f&amp;#39;(x)$ isn't blocked, the function cannot be uniformly continuous, but it only works in the opposite direction. &lt;/li&gt;&#xA;&lt;li&gt;Showing that $\lim_{x\rightarrow\infty} f(x)$ doesn't exist doesn't help me either.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Q2: Let $f:\mathbb R \rightarrow \mathbb R$ continuous function. $\lim_{x\rightarrow\infty} f(x) = \lim_{x\rightarrow-\infty} f(x) = L$. Show that $f$ has either minimum or maximum in $\mathbb R$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Attempts:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Exists $M_1$ (for $\lim_{x\rightarrow\infty} f(x)$) and $M_2$ (for $\lim_{x\rightarrow-\infty} f(x)$) and if function gets max or min in $[M_1, M_2]$ which is equal to L, then we're done.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance!&lt;/p&gt;&#xA;" OwnerUserId="3637" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2011-01-31T02:17:12.850" LastActivityDate="2011-01-31T02:17:12.850" Title="Let $f:[a,\infty] \rightarrow \mathbb R$ and $\lim_{x\rightarrow\infty} f'(x) = \infty$, show that $f$ is not uniformly continuous" Tags="&lt;calculus&gt;&lt;real-analysis&gt;&lt;homework&gt;" AnswerCount="3" />
  <row Id="19605" PostTypeId="2" ParentId="19594" CreationDate="2011-01-30T21:57:58.500" Score="1" ViewCount="" Body="&lt;p&gt;I don't think you mean to ask about any &quot;generalization&quot; (as in your last sentence) but rather &quot;general description&quot; (as in your heading), but the answer is No. More complicated vector bundles over more complicated spaces than the circle will generally not admit triviailizations with just two charts. I'm not sure what &quot;unwrapping the twist in the trouble point&quot; might mean in higher dimensional cases, but this doesn't seem like a generally accurate description either. How would you fit, say, the tangent bundle over a torus into that framework?&lt;/p&gt;&#xA;" OwnerUserId="308" LastActivityDate="2011-01-30T21:57:58.500" />
  <row Id="19606" PostTypeId="2" ParentId="19560" CreationDate="2011-01-30T21:58:05.943" Score="3" ViewCount="" Body="&lt;p&gt;Both sets of conditions are necessary conditions for a point to be optimal, but they're not quite the same mathematically.  The KKT conditions are more restrictive and thus shrink the class of points (from those satisfying the Fritz John conditions) that must be tested for optimality.  The additional restriction with KKT is that the Lagrange multiplier on the gradient of the objective function cannot be zero.  One of the most important resulting differences is that KKT points for linear programs must be optimal, whereas Fritz John points for linear programs don't have to be.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The section on &lt;a href=&quot;http://books.google.com/books?id=qIRDEgYesVUC&amp;amp;pg=PA188&amp;amp;lpg=PA188&amp;amp;dq=we+have+observed+above+that+a+point+x+is+a+FJ+point&amp;amp;source=bl&amp;amp;ots=lpWTuDSrZM&amp;amp;sig=nd5wbPn2GqYAyE4SCQ7TygvoBWk&amp;amp;hl=en&amp;amp;ei=INtFTfawHZHmsQP4m5zWCg&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=1&amp;amp;ved=0CBMQ6AEwAA#v=onepage&amp;amp;q=we%20have%20observed%20above%20that%20a%20point%20x%20is%20a%20FJ%20point&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;KKT conditions in Bazarra, Sherali, and Shetty's &lt;em&gt;Nonlinear Programming: Theory and Algorithms&lt;/a&gt;&lt;/em&gt; (second edition) has a nice discussion of the issues.  There are several good examples, especially one in which a Fritz John point for a linear program is shown not to be optimal.  That can't happen with KKT points for linear programs.&lt;/p&gt;&#xA;" OwnerUserId="2370" LastActivityDate="2011-01-30T21:58:05.943" />
  <row Id="19607" PostTypeId="2" ParentId="19594" CreationDate="2011-01-30T21:58:23.227" Score="2" ViewCount="" Body="&lt;p&gt;No. In principal one should be able to construct a bundle (even a 1-bundle) over a space that requires arbitrarily many charts to trivialize. Rather than think of an example of how to do this, I'll explain what I think you're seeing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a theorem that says any vector bundle over a contractible space is trivial. You just happen to be thinking of spaces (like $S^1$ or $S^n$ for that matter) that are contractible after removing a point. Thus any vector bundle over these spaces you can trivialize with the open set of removing a point and then for the other open set you just take any small contractible neighborhood of that point.&lt;/p&gt;&#xA;" OwnerUserId="2000" LastActivityDate="2011-01-30T21:58:23.227" />
  <row Id="19608" PostTypeId="2" ParentId="19602" CreationDate="2011-01-30T22:00:29.987" Score="4" ViewCount="" Body="&lt;p&gt;Your attempted solution apparently amounts to substituting the definition of addition and multiplication, but these are given to you so there's no need to &quot;find them&quot;. In fact, there are lots of different ways to define addition and multiplication. Most of them would not result in a field (think of $(a,b)+(c,d) = (0,0)$). On the other hand, there might be more than one which will result in a field. So the question really depends on the definition of addition and multiplication.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to check whether this object is a field, you have to verify all the field axioms. For example, in a field $x + y = y + x$. In your case, we need to verify that $$(x_1,x_2) + (y_1,y_2) = (y_1,y_2) + (x_1,x_2).$$ All we need to do is substitute the definition of addition: $$(x_1,x_2) + (y_1,y_2) = (x_1+y_1,x_2+y_2)$$ whereas $$(y_1,y_2) + (x_1,x_2) = (y_1+x_1,y_2+x_2),$$ and both expressions are equal by the commutativity of real addition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some of the field axioms require you to identify a zero element and a unit element, i.e. the $0$ and $1$ of the field. What should these be? Do they satisfy all the required axioms? For example, is every non-zero element invertible?&lt;/p&gt;&#xA;" OwnerUserId="1277" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2011-01-30T22:06:31.913" LastActivityDate="2011-01-30T22:06:31.913" CommentCount="2" />
  <row Id="19609" PostTypeId="2" ParentId="19594" CreationDate="2011-01-30T22:03:25.170" Score="5" ViewCount="" Body="&lt;p&gt;One way to see that a given bundle is nontrivial (or requires a lot of sets in a trivialization) is to use characteristic classes. Consider the tautological bundle over $\mathbb{RP}^n$. The total Stiefel-Whitney class of this bundle is $1+\gamma$ for $\gamma \in H^1(\mathbb{RP}^n, \mathbb{Z}/2)$ nonzero (this is one of the axioms in the axiomatic approach as in Milnor-Stasheff for instance). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I claim that this implies at least $n+1$ open sets are required for a trivialization. (This implies the assertion made in elgeorges's answer.) To see this, suppose $U_1, \dots, U_m$ is an open cover on which the tautological line bundle becomes trivial. Then by naturality of the Stiefel-Whitney classes $\gamma$ becomes trivial on $H^1(U_i)$, so lies in the image of $H^1(\mathbb{RP}^n, U_i)$. But it follows that the total cup-product $\gamma \cup \dots \cup \gamma$ ($m$ times) lies in the image of $H^1(\mathbb{RP}^n, \mathbb{RP}^n) = 0$. However, since it takes a high power to kill $\gamma$ (in fact, we know the cohomology ring is $\mathbb{Z}[\gamma]/(\gamma^{n+1})$, we find that $m \geq n+1$. (This argument for the Hopf bundle, which is a fiber bundle, is explained &lt;a href=&quot;http://amathew.wordpress.com/2010/11/24/how-nontrivial-is-the-hopf-bundle/&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.) &lt;/p&gt;&#xA;" OwnerUserId="536" LastEditorUserId="536" LastEditorDisplayName="" LastEditDate="2011-01-30T22:54:16.943" LastActivityDate="2011-01-30T22:54:16.943" CommentCount="2" />
  <row Id="19610" PostTypeId="1" AcceptedAnswerId="19611" CreationDate="2011-01-30T22:08:27.497" Score="3" ViewCount="73" Body="&lt;p&gt;Need some help with the steps in converting the derivatives of the following functions.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;derivative of  $\cos(\tan(x))$ to  $\frac{-\sin(\tan (x))}{\cos^2(x)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can get $-\sec^2(x) \cdot (\sin(\tan(x))$ using chain rule, but then I am stuck.  I guess I just need help on understanding how $\sec^2(x) = \frac{1}{\cos^2(x)}$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;derivative of $\sin(x)\tan(x)$ to $\sin(x) + \tan(x) \cdot \sec(x)$  &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I can get $\cos(x) \cdot \sec^2(x)$ but then I am unsure what to do.  Thanks for any help!&lt;/p&gt;&#xA;" OwnerUserId="6423" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2011-01-30T22:33:16.977" LastActivityDate="2011-01-30T22:54:04.587" Title="Help with using some trig identities" Tags="&lt;calculus&gt;&lt;homework&gt;&lt;trigonometry&gt;" AnswerCount="2" />
  <row Id="19611" PostTypeId="2" ParentId="19610" CreationDate="2011-01-30T22:12:37.837" Score="2" ViewCount="" Body="&lt;p&gt;For your first question, remember that $\sec$ is the reciprocal of $\cos$ by &lt;a href=&quot;http://en.wikipedia.org/wiki/Trigonometric_functions#Reciprocal_functions&quot; rel=&quot;nofollow&quot;&gt;definition&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the second, you should use the &lt;a href=&quot;http://en.wikipedia.org/wiki/Product_rule&quot; rel=&quot;nofollow&quot;&gt;product rule&lt;/a&gt; to find the derivative, don't simply take the derivative of each factor. Then use the fact that $\tan{x}=\frac{\sin{x}}{\cos{x}}$ to simplify your expression and get the desired form. Try rewriting every function in terms of $\sin$ and $\cos$ to most easily simplify.&lt;/p&gt;&#xA;" OwnerUserId="1583" LastActivityDate="2011-01-30T22:12:37.837" CommentCount="2" />
  <row Id="19612" PostTypeId="2" ParentId="19604" CreationDate="2011-01-30T22:14:48.457" Score="1" ViewCount="" Body="&lt;p&gt;A function is uniformly continuous if for every small &quot;range error&quot; $\Delta y$, there is some small \emph{non-zero} &quot;domain error&quot; $\Delta x$ such that a change of $\pm \Delta x$ around \emph{any} element $x$ in the domain will result in a change of only $\pm \Delta y$ in the range.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the derivative $dy/dx$ is the ratio between &quot;change in range&quot; and &quot;change in domain&quot;. Going back to the definition of uniform continuity, we fixed $\Delta y$ and wanted $\Delta x$ to be bounded \emph{from below} (think why we want a bound from below rather than from above). So the derivative needs to be bounded from above.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;For the second question, suppose that the function is not constant. Take some point which has a value different from $L$, say $M &amp;gt; L$. Now $M$ is not necessarily the maximum, but perhaps there is a way to identify the maximum, as some kind of supremum? &lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2011-01-30T22:14:48.457" />
  <row Id="19613" PostTypeId="2" ParentId="19604" CreationDate="2011-01-30T22:15:26.930" Score="1" ViewCount="" Body="&lt;p&gt;Q1: Try proving by contradiction.  If $f$ is uniformly continuous, there exists some $\delta$ such that $|f(x) - f(y)| \le 1$ whenever $|x-y| \le \delta$.  Now find an interval $[a,b]$ of length longer than $\delta$ such that $f&amp;#39; &amp;gt; 1/\delta$ on $[a,b]$.  If we pick $x=a$, $y=a+\delta$, what does the mean value theorem tell you about $|f(x)-f(y)|$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Q2: If $f$ is identically equal to $L$ we are done.  Otherwise there exists $x_0$ with, say, $f(x_0) &amp;gt; L$.  Now find a closed interval $[a,b]$ with $f(x) &amp;lt; f(x_0)$ outside $[a,b]$, and consider the maximum of $f$ on $[a,b]$...&lt;/p&gt;&#xA;" OwnerUserId="822" LastActivityDate="2011-01-30T22:15:26.930" CommentCount="2" />
  <row Id="19614" PostTypeId="2" ParentId="19554" CreationDate="2011-01-30T22:17:54.740" Score="5" ViewCount="" Body="&lt;p&gt;Using $a^2 - b^2 = (a-b)(a+b)$ we have $$(n+1)^2 - 1 = (n+1)^2 - 1^2 = (n+1-1)(n+1+1) = n(n+2).$$&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2011-01-30T22:17:54.740" />
  <row Id="19615" PostTypeId="2" ParentId="19610" CreationDate="2011-01-30T22:18:33.693" Score="3" ViewCount="" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;$\sec(x)$ is defined to be $\frac{1}{\cos(x)}$, that is the ratio of the length of the hypotenuse and the length of the adjacent side.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Note that $\tan(x)\cdot \sec(x) = \frac{\sin(x)}{\cos(x)} \cdot \frac{1}{\cos(x)} =  \sin(x) \cdot \frac{1}{\cos^2(x)}$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="5105" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-30T22:54:04.587" LastActivityDate="2011-01-30T22:54:04.587" />
  <row Id="19617" PostTypeId="1" AcceptedAnswerId="19618" CreationDate="2011-01-30T22:53:35.693" Score="2" ViewCount="96" Body="&lt;p&gt;Need some clarification on this one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is the set [0, infinity) with the usual &lt;strong&gt;addition and multiplication&lt;/strong&gt; axioms, a field?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Attempted solution:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yes it is a field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;let 0 &amp;lt;= r, s &amp;lt; infinity&lt;/p&gt;&#xA;&#xA;&lt;p&gt;then by commutative addition:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;r + s = s + r&lt;/p&gt;&#xA;&#xA;&lt;p&gt;s + r = r + s&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and by commutative multiplication:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;rs = sr&lt;/p&gt;&#xA;&#xA;&lt;p&gt;sr = rs&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is pretty much restating the axioms so I'm unclear as to whether this is the proof or maybe I'm completely wrong. Any help is appreciated thanks.&lt;/p&gt;&#xA;" OwnerUserId="5505" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-30T23:05:06.083" LastActivityDate="2011-01-30T23:06:31.323" Title="Field and Sets Proof" Tags="&lt;abstract-algebra&gt;&lt;field-theory&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="19618" PostTypeId="2" ParentId="19617" CreationDate="2011-01-30T22:57:59.167" Score="6" ViewCount="" Body="&lt;p&gt;You forgot a lot of axioms!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A &lt;em&gt;field&lt;/em&gt; is a set $F$ together with two binary operations, $\oplus$ and $\odot$, such that:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$\oplus$ is associative: for every $r,s,t\in F$, $r\oplus(s\oplus t)=(r\oplus s)\oplus t$.&lt;/li&gt;&#xA;&lt;li&gt;$\oplus$ is commutative: for every $r,s\in F$, $r\oplus s=s\oplus r$.&lt;/li&gt;&#xA;&lt;li&gt;There exists an element $z\in F$ such that for all $x\in F$, $z\oplus x=x$ (additive identity).&lt;/li&gt;&#xA;&lt;li&gt;For every $x\in F$ there exists $y\in F$ such that $x\oplus y = z$ (every element has an additive inverse).&lt;/li&gt;&#xA;&lt;li&gt;$\odot$ is associative: for every $r,s,t\in F$, $r\odot(s\odot t) = (r\odot s)\odot t$. &lt;/li&gt;&#xA;&lt;li&gt;$\odot$ is commutative: for every $r,s\in F$, $r\odot s=s\odot r$.&lt;/li&gt;&#xA;&lt;li&gt;$\odot$ distributes over $\oplus$: for every $r,s,t\in F$, $r\odot(s\oplus t) = (r\odot s)\oplus(r\odot t)$. &lt;/li&gt;&#xA;&lt;li&gt;There exists an element $e\in F$, $e\neq z$ ($z$ is the element from item 3) such that for all $r\in F4, $e\odot r=r$ (multiplicative identity).&lt;/li&gt;&#xA;&lt;li&gt;For every $r\in F$, if $r\neq z$ ($z$ the element from item 3) there exists $s\in F$ such that $r\odot s = e$ ($e$ the element from item 8) (multiplicative inverses for all nonzero elements).&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;You only checked (2) and (6) (twice each, for some reason). So, what are you still missing?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, you would need to show that addition and multiplication is &lt;em&gt;associative&lt;/em&gt; (yes, both are, so this is not a problem).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then you need to show that there is an identity element for the sum (yes, there is one in your set: namely, $0$, but you need to &lt;em&gt;show it&lt;/em&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Then&lt;/em&gt; you need to show that every element in your set has an &lt;em&gt;additive inverse&lt;/em&gt;: for every $r$, there exists $s$ such that $r+s = 0$. (This might prove difficult here).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Then&lt;/em&gt; you need to show that multiplication distributes over the sum: that for all $r,s,t$ in you set, $r(s+t) = rs + rt$ (this holds in this case). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Then&lt;/em&gt; you need to show that there is an element different from the additive identity, which is a multiplicative identity: that is, an element $e$ such that $er = r$ for all $r$ (there is one, but you need to show it).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And you also need to show that every element other than the additive identity has a multiplicative inverse (for every $r$, $r\neq 0$, there exists $s$ such that $rs=e$, where $e$ is the multiplicative identity from the previous part). (This is not a problem here either).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, you are missing quite a bit there!&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-30T23:04:14.507" LastActivityDate="2011-01-30T23:04:14.507" CommentCount="0" />
  <row Id="19619" PostTypeId="2" ParentId="19617" CreationDate="2011-01-30T23:06:31.323" Score="3" ViewCount="" Body="&lt;p&gt;There are one or two numbers that do not have additive inverses.&lt;/p&gt;&#xA;" OwnerUserId="5105" LastActivityDate="2011-01-30T23:06:31.323" CommentCount="1" />
  <row Id="19620" PostTypeId="2" ParentId="12026" CreationDate="2011-01-30T23:32:09.720" Score="1" ViewCount="" Body="&lt;p&gt;I have heard the term &quot;smooth&quot; being used in the context of the &lt;a href=&quot;http://en.wikipedia.org/wiki/H%C3%B6lder_condition&quot; rel=&quot;nofollow&quot;&gt;Hölder condition&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="3330" LastActivityDate="2011-01-30T23:32:09.720" />
  <row Id="19621" PostTypeId="1" AcceptedAnswerId="19622" CreationDate="2011-01-31T00:11:18.887" Score="4" ViewCount="90" Body="&lt;p&gt;Need some help proving the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Question:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Let F = {0, 1, a, b} be a field with four elements. Prove that a^2 = b&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;You can use a(0) = 0 without proving it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Attempted solution:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a^2 = b&lt;/p&gt;&#xA;&#xA;&lt;p&gt;aa = b&lt;/p&gt;&#xA;&#xA;&lt;p&gt;aa + 0 = b + 0&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know a*0 = 0 so we can substitute for zero&lt;/p&gt;&#xA;&#xA;&lt;p&gt;aa + a*0 = b &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the additive inverse of aa we get:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(aa) + (-aa) + a*0 = b + (-aa)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a*0 = b + (-aa)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Im still not getting the concept of how to prove things, maybe a little insight into what are possible steps to approach problems like these.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thats as far as i got, any help is appreciated.&lt;/p&gt;&#xA;" OwnerUserId="5505" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-31T01:11:43.037" LastActivityDate="2011-01-31T01:11:43.037" Title="Field Proof using Axioms" Tags="&lt;homework&gt;&lt;abstract-algebra&gt;" AnswerCount="2" CommentCount="8" FavoriteCount="1" />
  <row Id="19622" PostTypeId="2" ParentId="19621" CreationDate="2011-01-31T00:16:45.713" Score="3" ViewCount="" Body="&lt;p&gt;Note that $0$ and $1$ are distinguished elements, so none of the others should be equal to them.&#xA;Now, $b \cdot a$ cannot be equal to $a$ or $b$ becaus then $b$ or $a$ would be $1$, respectively, but that cannot happen as they are distinct elements. Similarly, $b \cdot a$ cannot be $0$. Hence, $b \cdot a = 1$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Use this type of reasoning to think what $a^2$ &lt;em&gt;cannot&lt;/em&gt; be, and you'll find the anwser: what is $a$ if $a^2 = 0$?&lt;/p&gt;&#xA;" OwnerUserId="5105" LastActivityDate="2011-01-31T00:16:45.713" CommentCount="10" />
  <row Id="19623" PostTypeId="2" ParentId="19621" CreationDate="2011-01-31T00:21:22.713" Score="5" ViewCount="" Body="&lt;p&gt;Consider that the map $x \mapsto ax$ must be bijective (because the field is finite -- should be easy to show). We know,&#xA;$$a \cdot 0 = 0$$&#xA;$$a \cdot 1 = a$$&#xA;This means $a\cdot a = b$ or $a \cdot a = 1.$ But in the latter case we get $a \cdot b = b,$ which implies $a = 1,$ which is not true (by uniqueness of 1).&lt;/p&gt;&#xA;" OwnerUserId="5218" LastActivityDate="2011-01-31T00:21:22.713" />
  <row Id="19624" PostTypeId="1" AcceptedAnswerId="19635" CreationDate="2011-01-31T00:32:46.250" Score="10" ViewCount="138" Body="&lt;p&gt;I'm taking classes in homological algebra now, and the book (together with the lecturer) seem to assume more category theory than I already know.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A &quot;fact&quot; that is used freely in the book (&quot;Homological algebra&quot; by Weibel, by the way), is that &lt;strong&gt;every abelian group is (isomorphic to) the direct limit of its finitely generated subgroups&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I do not find this particularly obvious. How does one go about prove such things?&lt;/p&gt;&#xA;" OwnerUserId="4284" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2011-01-31T00:56:05.040" LastActivityDate="2011-01-31T01:44:23.753" Title="Why is every abelian group the direct limit of its finitely generated subgroups?" Tags="&lt;category-theory&gt;&lt;abelian-groups&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="3" />
  <row Id="19626" PostTypeId="1" AcceptedAnswerId="19627" CreationDate="2011-01-31T00:36:13.070" Score="5" ViewCount="78" Body="&lt;p&gt;Given this recurrence equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$u_1 = 0, u_2 = 1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$u_n = 4u_{n−1} + 4u_{n−2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is the correct characteristic equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$4x+4 = 4$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Complete solve:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The characteristic equation is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x^2-4x-4=0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We solve the quadratic equation...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\alpha = 5$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\beta=-1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$u_n = c_1 \alpha^n + c_2 \beta^n$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We solve the equation...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$c_1 = 1/30$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$c_2 = 1/6$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$u_n = \dfrac{5^n}{30} + \dfrac{(-1)^n}{6}$&lt;/p&gt;&#xA;" OwnerUserId="5526" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2011-06-03T13:51:29.800" LastActivityDate="2011-06-03T13:51:29.800" Title="Recurrence equation: $u_n = 4u_{n−1} + 4u_{n−2}$ ; is $4x+4 = 4$ the characteristic equation?" Tags="&lt;homework&gt;&lt;recurrence-relations&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="19627" PostTypeId="2" ParentId="19626" CreationDate="2011-01-31T00:46:31.697" Score="4" ViewCount="" Body="&lt;p&gt;When I am working with recurrence relations and trying to simplify them, I like to move all of the &quot;recurring&quot; terms to one side. Doing this here will give you a homogeneous difference equation. So here you would get,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$u_{n} - 4u_{n-1} - 4u_{n-2} = 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now to find the characteristic equation, you may want to shift the subscripts to get,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$u_{n+2} - 4u_{n+1} - 4u_{n} = 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, we can see that the characteristic equation will be,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\lambda^{2} - 4 \lambda - 4 = 0$.&lt;/p&gt;&#xA;" OwnerUserId="3791" LastActivityDate="2011-01-31T00:46:31.697" />
  <row Id="19628" PostTypeId="2" ParentId="19626" CreationDate="2011-01-31T00:47:16.613" Score="3" ViewCount="" Body="&lt;p&gt;Your characteristic equation is incorrect. The way you obtain the characteristic equation is to assume $u_n = m^n$. Plug it in to get a quadratic in $m$. Solve for $m$. Get the two roots say $m_1$ and $m_2$. Now the general solution is given by the linear combination namely $u_n = a_1 m_1^n + a_2 m_2^n$. Solve for $a_1$ and $a_2$ using the initial conditions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This methodology is analogous to plugging in $y=e^{mx}$ when you want to solve a linear ODE. Here you have a difference equation instead of a differential equation.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastActivityDate="2011-01-31T00:47:16.613" CommentCount="3" />
  <row Id="19629" PostTypeId="2" ParentId="19626" CreationDate="2011-01-31T00:48:43.880" Score="0" ViewCount="" Body="&lt;p&gt;The characteristic equation is the equation satisfied by $\lambda$ if $u_n = \lambda^n$ be a solution. For example, if the recurrence is $a_n = a_{n-1} + a_{n-2}$ then a solution $\lambda^n$ must satisfy $$\lambda^n = \lambda^{n-1} + \lambda^{n-2}.$$ The point is that all these equations reduce to the one equation $$\lambda^2 = \lambda + 1.$$ Back to your question, when I follow the same steps I get something else.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2011-01-31T00:48:43.880" />
  <row Id="19630" PostTypeId="1" AcceptedAnswerId="19634" CreationDate="2011-01-31T00:51:16.810" Score="7" ViewCount="158" Body="&lt;p&gt;Most students come to calculus with an intuitive sense of what a tangent line should be for a curve. It is easy enough to give a definition of a tangent to a circle that is both elementary and rigorous. (A line that intersects a circle exactly once.) Yet, when we talk about a curve, such as a polynomial, I think one must talk about infinity to give a rigorous definition. This is not a bad thing... It can help motivate a lesson on the formal definition of a tangent line at a point... But, what intermediate definition could I use to help those students who have no intuition about the matter get an idea of what we are going for before we talk about secants and such? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have looked at some textbooks and online and I find saying a tangent line is one that &quot;just touches&quot; problematic-- it is the kind of phrase that is quite meaningless... Even in the setting of un-rigorous definitions. Yes, I will give examples, but I would like to do better ... I think saying that a tangent line is an arrow that points in the direction that the curve is going at that instant might make sense... Though it makes everything 'directional' and that might confuse them later.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the best intermediate definition you have seen?&lt;/p&gt;&#xA;" OwnerUserId="2563" LastEditorUserId="2736" LastEditorDisplayName="" LastEditDate="2011-01-31T01:02:27.910" LastActivityDate="2011-05-31T16:41:48.987" Title="What is an intermediate definition for a tangent to a curve?" Tags="&lt;calculus&gt;&lt;teaching&gt;" AnswerCount="4" FavoriteCount="1" />
  <row Id="19631" PostTypeId="1" AcceptedAnswerId="19637" CreationDate="2011-01-31T00:52:07.307" Score="3" ViewCount="56" Body="&lt;p&gt;Hi everyone,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a problem I have. Consider having 4 vectors in $R^5$ such that $3\vec{v_{1}}+\vec{v_{2}}=4\vec{v_{4}}$. Let $T$ be a linear transformation from $R^4$ to $R^5$ such that $T(e_{j})=v_{j}$ where $e$ is a normal vector.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(a) Find a nonzero vector in kernel of $T$ -- this I solved as $\begin{bmatrix}&#xD;&#xA;3\\ &#xD;&#xA;1\\ &#xD;&#xA;0\\ &#xD;&#xA;-4&#xD;&#xA;\end{bmatrix}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(b) What is the largest possible dimension for $Im(T)$? -- This is the question I am having trouble with. So the matrix for t is clearly 5 by 4, and it has four vectors, relationship between three of which is given to me. By rank-nullity theorem, the dimension of the image is 4 minus the number of dependent columns/free variables. However, I am struggling to understand how many dependent columns/free variables I have from $3\vec{v_{1}}+\vec{v_{2}}-4\vec{v_{4}}=0$ -- clearly, I am given a relationship between three vectors, but does it mean that one of them is dependent on the other two, hence two free variables, or 1 dependent column?! I guess I am just confused by the connection -- I know the number of free variables and dependent columns should be the same, so I would appreciate if someone could clarify how to get both from the equation I am given (I realize that the third vector is not defined, so it could or could not be dependent).&lt;/p&gt;&#xA;" OwnerUserId="5157" LastActivityDate="2011-01-31T01:09:52.993" Title="What is the largest value for Dimension Im(T)?" Tags="&lt;linear-algebra&gt;" AnswerCount="1" />
  <row Id="19632" PostTypeId="2" ParentId="19624" CreationDate="2011-01-31T00:52:56.560" Score="8" ViewCount="" Body="&lt;p&gt;In nice algebraic categories like $\text{Ab}$, the colimit (direct limit) over a (edit: &lt;em&gt;directed&lt;/em&gt;) diagram of inclusions behaves like a &quot;union&quot; operation ought to, so this result shouldn't come as a surprise.  In any case, let's verify the universal property.  Given an abelian group $G$, let $G_i, i \in I$ be an indexing of its finitely generated subgroups, which fit together into the obvious diagram of inclusions.  We want to verify the following universal property: given an abelian group $H$ and a compatible system of morphisms $\phi_i : G_i \to H$, these morphisms factor through a unique morphism $\phi : G \to H$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is not actually a very deep fact.  A compatible system of morphisms $\phi_i$ is precisely a list of where every element of $G$ goes (since every element of $G$ is contained in a finitely generated subgroup, namely the one it generates; this already proves uniqueness) together with a guarantee that this list respects all the relations of $G$ (since relations involve a finite number of elements, so can be verified in the subgroup they generate).  In other words, it defines a unique morphism $G \to H$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again, in nice algebraic categories the same is true: for example, it's true of groups, rings, commutative rings, ....  This has the following nice application: since a commutative ring is the colimit of its finitely-generated subrings, an affine scheme is the limit of Noetherian schemes (and this is still true relative to a Noetherian base).  It also implies that the spectrum of a Boolean ring is a limit of finite sets, hence a profinite set (that is, a Stone space); see, for example, &lt;a href=&quot;http://qchu.wordpress.com/2010/11/22/boolean-rings-ultrafilters-and-stones-representation-theorem/&quot; rel=&quot;nofollow&quot;&gt;this blog post&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2011-01-31T01:44:23.753" LastActivityDate="2011-01-31T01:44:23.753" CommentCount="5" />
  <row Id="19633" PostTypeId="2" ParentId="19547" CreationDate="2011-01-31T01:00:21.513" Score="2" ViewCount="" Body="&lt;p&gt;So, given the dimensionless Hamiltonian&#xA;$$&#xD;&#xA;\hat{\cal{H}} = \frac{1}{2}\hat{p}^2 + V(\hat{x}),&#xD;&#xA;$$&#xA;the wave function evolves as&#xA;$$&#xD;&#xA;\vert\Psi(t+dt)\rangle = e^{-i\hat{\cal{H}}dt}\vert\Psi(t)\rangle = e^{-\frac{1}{2}i\hat{p}^2 dt}e^{-iV(\hat{x})dt}e^{O(dt^2)}\vert\Psi(t)\rangle.&#xD;&#xA;$$&#xA;The idea is to apply the position-space part of the evolution operator ($e^{-iV(\hat{x})dt}$) to the position-space wave function $\Psi(x, t)$, where it is just a multiplication by $e^{-iV(x)dt}$, and then to apply the momentum-space part of the evolution operator ($e^{-\frac{1}{2}i\hat{p}^2 dt}$) to the momentum-space wave function $\tilde{\Psi}(k, t+dt)$, where it is just a multiplication by $e^{-\frac{1}{2}ik^2 dt}$.  To evolve the wave function in this approximation (errors are introduced because terms of order $dt^2$ arising from the noncommutativity of $\hat{x}$ and $\hat{p}$ are being omitted), you would typically choose a small value for $dt$ and a discretization for $x$ and $k$, and then repeatedly apply the four operations shown in your expression above: multiply by the potential energy term, apply the FFT to convert to momentum space, multiply by the kinetic energy term, and apply the inverse FFT to convert back to position space.&lt;/p&gt;&#xA;" OwnerUserId="5546" LastActivityDate="2011-01-31T01:00:21.513" CommentCount="3" />
  <row Id="19634" PostTypeId="2" ParentId="19630" CreationDate="2011-01-31T01:01:09.487" Score="3" ViewCount="" Body="&lt;p&gt;One possible way to introduce a tangent to a curve at a point is to do as follows. Take a point on the curve where we want to define a tangent to the curve. Consider all set of lines passing through the point. There will be a unique line such that in a neighborhood of the curve the entire curve in the neighborhood lies on only one side of the line. But of course this way of motivating will fail if we are near a point of inflexion.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2011-01-31T01:45:37.903" LastActivityDate="2011-01-31T01:45:37.903" />
  <row Id="19635" PostTypeId="2" ParentId="19624" CreationDate="2011-01-31T01:03:37.557" Score="9" ViewCount="" Body="&lt;p&gt;The fact that a  group is the direct limit of its finitely generated subgroups is true for any group, not just abelian ones. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remember what the direct limit of groups is. You start with a directed set $I$ (directed means that it is a partially ordered set, and if $i,j\in I$, then there exists $k\in I$ such that $i\leq k$ and $j\leq k$). For each $i\in I$, you have a group $H_i$, and if $i\leq k$, then you have a morphism $f_{ik}\colon H_i\to H_k$. The system $\{H_i, f_{ij}\}_{i\in I}$ must also satisfy:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$f_{ii} = \mathrm{id}_{H_i}$ for each $i\in I$; and&lt;/li&gt;&#xA;&lt;li&gt;If $i\leq j\leq k$, then $f_{ik} = f_{jk}\circ f_{ij}$.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The &lt;strong&gt;direct limit&lt;/strong&gt; is then constructed as follows: we let $X$ be the disjoint union of the sets $H_i\times\{i\}$, and we define the equivalence relation $\sim$ on $X$ as follows: given $(h,i)$ and $(h&amp;#39;,j)$, we let $(h,i)\sim (h&amp;#39;,j)$ if and only if there exists $k\in I$ such that $i\leq k$, $j\leq k$, and $f_{ik}(h) = f_{jk}(h&amp;#39;)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We then define a group operation on $X/\sim$ as follows: given $[(h,i)]$ and $[(h&amp;#39;,j)]$ in $X$, let $k\in I$ such that $i\leq k$ and $j\leq k$. We define $$[(h,i)][(h&amp;#39;,j)] = [(f_{ik}(h)f_{jk}(h&amp;#39;),k)].$$&#xA;Note that this makes sense, since $f_{ik}(h)$ and $f_{jk}(h)$ are both in $H_k$. It is then an exercise to show that this gives you a group with the appropriate universal property.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Intuition.&lt;/strong&gt; The idea is just that two elements in $X$ are &quot;equal&quot; if and only if they &quot;eventually&quot; map to the same thing.  And we define the product by first mapping both elements sufficiently &quot;ahead&quot; that they both lie in the same group, and &lt;em&gt;then&lt;/em&gt; multiply them. The direct limit is then a way of &quot;gluing&quot; all of $H_i$ in a compatible way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason we expect the direct limit of the finitely generated subgroups to &quot;be&quot; the group (that is, isomorphic to the group) is that any particular operation we want to do with group elements always occurs in a finitely generated subgroup: if you are performing an operation inside the group, the operation uses only finitely many elements, so it all happens inside a finitely generated group. Thus, everything that &quot;determines&quot; what the group &lt;em&gt;is&lt;/em&gt; is captured if you look at &lt;em&gt;all&lt;/em&gt; finitely generated subgroups: to know how to multiply $x$ by $y$, you can multiply them in the finitely generated subgroup $\langle x,y\rangle$, after all. The direct limit is just a way of putting together all of these subgroups which we should be able to do since they are all &quot;really&quot; already glued together inside of $G$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Sketch of proof.&lt;/strong&gt; &#xA;Now, let $G$ be a group. Let $I$ be the collection of all finitely generated subgroups of $G$, and order $I$ by inclusion of subgroups. For each element $i\in I$, let $H_i$ be the subgroup $i$ itself (remember that $i$ is a subgroup of $G$ by definition of $I$). If $i\leq j$, then $i\subseteq j$ as sets, so we let $f_{ij}\colon H_i \to H_j$ be the inclusion map. Note that if $i=j$, then $f_{ii} = \mathrm{id}_{H_i}$, as required, and if $i\leq j\leq k$, then $H_i\subseteq H_j\subseteq H_k$, and the inclusion $H_i\hookrightarrow H_k$ is the composition of the inclusion $H_i\hookrightarrow H_j$ and $H_J\hookrightarrow H_k$. So $\{H_i,f_{ij}\}_{i\in I}$ is a directed system of groups. Let $H$ be the direct limit, and let $f_i\colon H_i\to H$ be the structure maps into the direct limit. The structure maps are one-to-one, because $[(h,i)] = [(e,j)]$ if and only if there exists $k\geq i$ such that $f_{ik}(h) = e$, but all our $f_{ik}$ are one-to-one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To see that the direct limit is (isomorphic to) a subgroup of $G$, note that you have embeddings $\varphi_i\colon H_i\hookrightarrow G$ from $H_i$ to $G$ for each $i$, and this embeddings commute with the structure maps $f_{ij}$. This means that by the universal property there is a homomorphism $\phi\colon H \to G$ such that $\varphi_i = \phi\circ f_i$ for all $i$. In particular, $\phi$ is an embedding, so $H$ is (isomorphic to) a subgroup of $G$. To see that the map is onto, let $g\in G$. Then letting $i=\langle g\rangle$, we have that $g$ is in the image of $\varphi_i$, hence in the image of $\phi$. Thus, $\phi$ is an isomorphism.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added.&lt;/strong&gt; The same argument holds for any collection $\mathcal{C}$ of subgroups of $G$ such that (i) every element of $G$ lies in at least one subgroup in $\mathcal{C}$; (ii) given any two subgroups $H$ and $K$ in $\mathcal{C}$, there is a subgroup $M$ in $\mathcal{C}$ that contains both $H$ and $K$.  So it works for $\mathcal{C}$ being &quot;all finitely generated subgroups&quot;; &quot;all subgroups&quot;; for infinite cardinal $\kappa$, &quot;all subgroups of cardinality less than or equal to $\kappa$&quot;; and other classes.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-31T01:36:06.570" LastActivityDate="2011-01-31T01:36:06.570" CommentCount="8" />
  <row Id="19636" PostTypeId="2" ParentId="19630" CreationDate="2011-01-31T01:04:06.190" Score="5" ViewCount="" Body="&lt;p&gt;A tangent is a line that intersects the curve once, at least if it is made short enough, but there's a direction such that if you rotate the line about the intersection point in that direction, no matter how little you rotate, it'll hit the curve again. This definition apparently goes back to Euclid in some form, and it works at inflection points, but fails in dimensions higher than $2$.  (I assume you're only talking about plane curves.)  I think we need the derivative to be continuous for it to work, and we also need it not to be constant in a neighborhood of the intersection point. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't see what's wrong with giving a physical definition though: it's where a particle moving along the curve would go if there were suddenly no forces acting on it (Newton's first law).  Or, even more physically, it's the line a ball would begin to travel in if you threw the ball in an arc corresponding to the curve and let it go at the point you're interested in.  This definition has the advantage that it does not depend on the background mathematics: it is (to a first approximation) an empirical fact about the universe that this concept is consistent.&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2011-05-31T16:41:48.987" LastActivityDate="2011-05-31T16:41:48.987" CommentCount="4" />
  <row Id="19637" PostTypeId="2" ParentId="19631" CreationDate="2011-01-31T01:09:52.993" Score="1" ViewCount="" Body="&lt;p&gt;The Rank-Nullity Theorem tells you that if you have a linear transformation $T\colon \mathbf{V}\to\mathbf{W}$, then the dimension of the kernel of $T$ plus the dimension of the image of $T$ equals the dimension of $\mathbf{V}$. In this case, since your $T$ maps $\mathbb{R}^4$ to $\mathbb{R}^5$, you know that the dimension of the kernel of $T$ plus the dimension of $\mathrm{Im}(T)$ equals $4$ (the dimension of $\mathbb{R}^4$). Since you know that $\dim(\mathrm{ker}(T))\geq 1$ (you already produced at least one nonzero element in the kernel), that tells you that $\dim(\mathrm{Im}(T))\leq \cdots$.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2011-01-31T01:09:52.993" CommentCount="2" />
  <row Id="19638" PostTypeId="2" ParentId="19602" CreationDate="2011-01-31T01:16:21.610" Score="0" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\rm\ \ \ 0\ =\ x^2 - x\ =\ (x_1^2-x_1,\ x_2^2-x_2)\ $ has more than two roots $\rm\ x = (x_1,x_2)\in \mathbb R^2\:$.&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2011-01-31T01:16:21.610" />
  <row Id="19639" PostTypeId="2" ParentId="19604" CreationDate="2011-01-31T01:24:33.540" Score="1" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;1&lt;/strong&gt;.Fix $\epsilon,\delta&amp;gt;0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since lim$_{x\rightarrow\infty}f'(x)=\infty,\exists N$ such that $f&amp;#39;(x)&amp;gt;\frac{\epsilon}{\delta}$, if $x&amp;gt;N$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, by MVT, we have that$\;\;\;\forall x&gt;N\;\;\;\exists c\in(x,x+\delta)$ such that $f(x+\delta)-f(x)=f&amp;#39;(c)\delta&amp;gt;\epsilon$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $\epsilon$ and $\delta$ were arbitrary, this shows f can not be uniformly continuous.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;2&lt;/strong&gt;.Suppose f is not constant. Then $\exists x_0$ such that, say, $f(x_0)&amp;gt;L$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $lim_{x\rightarrow\infty}f(x)=lim_{x\rightarrow-\infty}f(x)=L, \exists N$ such that $f(x_0)&amp;gt;f(x)$ if $|x|&amp;gt;N$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now since $f$ is continuous it obtains a maximum on $[-N,N]$, say $f(x_1)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clearly, $f(x_1)\geq f(x_0)$. It follows that $f(x_1)\geq f(x) \;\;\;\forall x\in R$.&lt;/p&gt;&#xA;" OwnerUserId="5959" LastEditorUserId="5959" LastEditorDisplayName="" LastEditDate="2011-01-31T02:04:50.383" LastActivityDate="2011-01-31T02:04:50.383" />
  <row Id="19640" PostTypeId="2" ParentId="19491" CreationDate="2011-01-31T01:37:07.267" Score="0" ViewCount="" Body="&lt;p&gt;OK, let me pretend this is a math problem and solve it that way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose we have a function $T: \mathbb{Z}^+ \rightarrow \mathbb{R}$ such that for all &#xA;$n \in \mathbb{Z}^+$, $T(n) = T(n-3) + n$.  We are asked to find a function $f(n)$ such that $T(n) = \Theta(f(n))$, i.e., that the quantity $\frac{T}{f}$ is bounded above and below by positive constants as $n$ tends to infinity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The point here is that we are missing initial conditions on $T$.  Since we have a (somewhat degenerate) three term recurrence, so specify a unique function we need to know the three smallest values $T(1)$, $T(2)$, $T(3)$.  It turns out that what these values are does not affect the asymptotic behavior of the function, so we can solve the problem without needing explicit values.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We may take advantage of the &lt;em&gt;degeneracy&lt;/em&gt; of the recurrence relation -- i.e., $T(n)$ actually only depends on the value $T(n-3)$, not on $T(n-1)$ and $T(n-2)$ to essentially reduce to a one-term recurrence, but with three cases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Case 1: $n = 3k+1$.  Then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$T(3k+1) = 3k+1 + T(3(k-1)+1) = \ldots = (\sum_{i=1}^k 3i+1) + T(1)$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$= 3(\frac{k(k+1)}{2}) + k + T(1) = 3/2 k^2 + o(k^2) \sim 3/2 k^2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sim  \frac{1}{6} (3k+1)^2  \sim \frac{1}{6} n^2 = \Theta(n^2)$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Case 2: $n = 3k+2$.  Then &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$T(n) = T(3k+2) = 3k+2 + T(3(k-1)+2)) = \ldots = (\sum_{i=1}^k 3i+2) + T(2)$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$= 3(\frac{k(k+1)}{2}) + 2k + T(1) \sim 3/2 k^2 \sim  \frac{1}{6} (3k+2)^2$&#xA;$ \sim \frac{1}{6} n^2 = \Theta(n^2)$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Case 3: $n = 3k+3$.  Then &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$T(3k+3) = 3k+3 = T(3(k-1) + 3) = \ldots = (\sum_{i=1}^k 3i+3) + T(3) $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$= 3(\frac{k(k+1)}{2}) + 3k + T(1) \sim 3/2 k^2 \sim  \frac{1}{6} (3k+3)^2$ &#xA;$\sim \frac{1}{6} n^2 = \Theta(n^2)$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, if $T: \mathbb{Z}^+ \rightarrow \mathbb{R}$ is such that we can split the domain up into three different pieces and show that on each piece $T = \Theta(f)$ for &#xA;some function $f$, then it follows that $T = \Theta(f)$ on its entire domain.  (This comes down to the fact that the supremum and infimum of a finite set of positive numbers are positive and finite.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Stated as a math problem like this, it seems strange that we are asked only a $\Theta$ bound for $T$ rather than an asymptotic bound: we can make the more accurate statement &#xA;$T(n) \sim \frac{n^2}{6}$.  But based on the comments, I am guessing that somehow the statement $T(1) = \Theta(1)$ is supposed to mean that all values of $T$ are given only up to a quantity which is uniformly bounded above and below, so the answer should also be recorded in that form.&lt;/p&gt;&#xA;" OwnerUserId="299" LastActivityDate="2011-01-31T01:37:07.267" CommentCount="5" />
  <row Id="19641" PostTypeId="2" ParentId="19630" CreationDate="2011-01-31T01:39:41.853" Score="3" ViewCount="" Body="&lt;p&gt;My favorite calculus-level definition (which is fairly rigorous, not hard to motivate, and does not rely on pictures) is that the tangent line to $y=f(x)$ at $x_0$ is the (unique) line that goes through $(x_0,f(x_0))$ and affords the best linear approximation to $y=f(x)$ near $x_0$. That is, if you let $g(x)$ be the point on the line with coordinate $x$, then $f(x)-g(x)$ goes to zero faster than $x-x_0$ goes to $0$; that is, $\frac{f(x)-g(x)}{x-x_0} \to 0$ as $x\to x_0$. This captures the idea that the tangent is the line that &quot;best approaches&quot; the graph when you are near the point $x_0$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;See about halfway through &lt;a href=&quot;http://math.stackexchange.com/questions/12287/approaching-to-zero-but-not-equal-to-zero-then-why-do-the-points-get-overlapped/12290#12290&quot;&gt;this previous answer&lt;/a&gt; (starting in paragraph 7, where it says &lt;em&gt;&quot;Now, does the line that join $A$ and $B$ really have a slope that approaches the slope of the tangent?&quot;&lt;/em&gt;).&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-31T01:45:53.213" LastActivityDate="2011-01-31T01:45:53.213" />
  <row Id="19642" PostTypeId="1" AcceptedAnswerId="19643" CreationDate="2011-01-31T01:51:56.450" Score="2" ViewCount="128" Body="&lt;p&gt;I'm trying to teach math to my 7 year old daughter. I'm teaching following type of equations. &#xA;$$\cdots  - x = y$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm able to explain her the rule that: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;when $\cdots- x = y$, we can always take $x$ (value on the left of equation) to the other side of $=$ sign and flip( $-$ to $+$ and vice versa), $-$ to $+$ and get the answer. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Meaning when $\cdots - x = y$, we can always do $\cdots =  y + x$ and get the answer. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This rule works for &#xA;\begin{align*}&#xA;x + \cdots &amp;amp;= y \\&#xA;\cdots + x &amp;amp;= y\\ &#xA;\cdots - x &amp;amp;= y \\&#xA;\end{align*}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But it doesn't work for $x - \cdots = y$. Because if you apply the rule, you get - (answer) and not just ( answer ) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is given that I'm trying to teach this to 7 year old, is there any better method where one rule would cover all 4 cases? Any ideas, thoughts...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align*}&#xA; - x + \cdots &amp;amp;= y\\&#xA; -\cdots + x &amp;amp;= y\\&lt;br&gt;&#xA; - \cdots  - x &amp;amp;= y\\&lt;br&gt;&#xA; - x  - \cdots &amp;amp;= y&#xA;\end{align*}&lt;/p&gt;&#xA;" OwnerUserId="6426" LastEditorUserId="856" LastEditorDisplayName="" LastEditDate="2011-01-31T02:41:06.603" LastActivityDate="2011-01-31T02:41:06.603" Title="Can a rule be formulated to explain this to 7 year old ?" Tags="&lt;arithmetic&gt;&lt;teaching&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="19643" PostTypeId="2" ParentId="19642" CreationDate="2011-01-31T01:57:01.340" Score="4" ViewCount="" Body="&lt;p&gt;You shouldn't be using &quot;rules&quot; at all; that is not what mathematics is about, at any level.  (This is an enormous pet peeve of mine.  There is a commercial floating around Hulu about some kind of online tutoring program where a woman describes to a girl the rule for computing the area of a triangle given its base and height, and then completely fails to draw the diagram that explains why this rule works.  It annoys me to no end.  (This particular example is also brought up in the infamous &lt;a href=&quot;http://www.maa.org/devlin/LockhartsLament.pdf&quot; rel=&quot;nofollow&quot;&gt;Lockhart's lament&lt;/a&gt;.))  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have some amount of money in my bank account.  When I withdraw $x$ dollars, I have $y$ dollars left.  How much money did I have originally?  $x + y$.  How much money do I have now?  $(x + y) - x = y$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have some amount of money in my bank account.  When I deposit $x$ dollars, I now have $y$ dollars.  How much money did I have originally?  $y - x$.  How much money do I have now?  $(y - x) + x = y$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At some point it is probably a good idea to mention that $x + y$ is the same as $y + x$ (that is, depositing $x$ dollars and then depositing $y$ dollars is the same as depositing $y$ dollars and then depositing $x$ dollars).  Then you've covered all of the &quot;cases.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternately, a physical analogy ought to work well.  I am some distance away from a wall.  When I move $x$ feet towards the wall, I am $y$ feet away from the wall.  How far was I originally away from the wall?  $x + y$.  How far am I away from the wall now?  $(x + y) - x = y$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am some distance away from a wall.  When I move $x$ feet away from the wall, I am $y$ feet away from the wall.  How far was I originally away from the wall?  $y - x$.  How far am I away from the wall now?  $(y - x) + x = y$.&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2011-01-31T02:10:08.943" LastActivityDate="2011-01-31T02:10:08.943" CommentCount="5" />
  <row Id="19644" PostTypeId="2" ParentId="19642" CreationDate="2011-01-31T02:03:54.433" Score="2" ViewCount="" Body="&lt;p&gt;I'm not clear on what the &quot;rule&quot; you say &quot;doesn't work&quot; is... Still...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As Qiaochu says, don't do &quot;rules&quot;. The key to all of these manipulations is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If two things are equal, and you do the same thing to each of them, the results will also be equal.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So, if $A$ is equal to $B$, then adding $2$ to $A$ will result in the same thing as adding $2$ to $B$: if $A=B$, then $A+2 = B+2$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have $\cdots - x = y$, then you have two things that are equal. Adding $x$ to both will still give you equal things, so&#xA;$$(\cdots - x) + x = y + x.$$&#xA;Then using the fact that $-x+x = 0$, you get $\cdots = y+x$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;All of the manipulations you propose are instances of this: if you have two equal things, and you do the same thing to both, the results are still equal. &lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2011-01-31T02:03:54.433" CommentCount="3" />
  <row Id="19645" PostTypeId="1" AcceptedAnswerId="19652" CreationDate="2011-01-31T02:19:32.940" Score="3" ViewCount="60" Body="&lt;p&gt;I knew this at one point, and in fact it is embarassing that I have forgotten it. I am wondering what filtartion of the total space of a fibration we use  to get the Serre SS. I feel very comfortable with the Serre SS, I am just essentially looking for a one line answer. I checked have Hatcher, Mosher and Tangora and Stricklands note on Spectral Sequences.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think it has something to do with looking at cells where the bundle is trivializable..&lt;/p&gt;&#xA;" OwnerUserId="627" LastActivityDate="2011-01-31T02:58:21.783" Title="Filtration in the Serre SS" Tags="&lt;algebraic-topology&gt;&lt;homological-algebra&gt;&lt;homotopy-theory&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="19646" PostTypeId="1" AcceptedAnswerId="19655" CreationDate="2011-01-31T02:22:35.370" Score="4" ViewCount="60" Body="&lt;p&gt;I have a solution to a logic problem involving propositions that I don't undersand how a particular step was carried out.The professor called the step I'm having trouble with reverse distribution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Prove: $(p \lor q ) \land\lnot (p \lor \lnot q)\leftrightarrow p$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(p \lor q ) \land (p \lor \lnot q) $ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$p\lor(q \land \lnot q)$ This is the step I don't understand. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$p\lor FALSE $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;p&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second step is throwing me for a loop. What am I not seeing?&lt;/p&gt;&#xA;" OwnerUserId="1040" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-31T03:21:43.213" LastActivityDate="2011-01-31T03:21:43.213" Title="What is the reverse distributive technique ?" Tags="&lt;logic&gt;&lt;propositional-calculus&gt;" AnswerCount="1" />
  <row Id="19647" PostTypeId="1" CreationDate="2011-01-31T02:26:02.557" Score="3" ViewCount="100" Body="&lt;p&gt;Assume $\alpha + \beta = 1$, and one is trying to find a lower bound on the $k$th term in the binomial expansion of $(\alpha + \beta)^m$. The terms are of the form $\binom{m}{k} \alpha^k \beta ^{m-k}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume that $k \ll m \ll 1/\alpha$. More formally, and in &lt;a href=&quot;http://en.wikipedia.org/wiki/Big_O_notation&quot; rel=&quot;nofollow&quot;&gt;big-O notation&lt;/a&gt;, $k = o(m)$ and $m = o(1/\alpha)$. Can we derive good lower bounds for the $k$th term in the binomial expansion? (When $m$ is large enough.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS 1: I read &lt;a href=&quot;http://planetmath.org/encyclopedia/UpperAndLowerBoundsToBinomialCoefficient.html&quot; rel=&quot;nofollow&quot;&gt;upper and lower bounds to binomial coefficient&lt;/a&gt; on PlanetMath, but couldn't figure out what to do.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS 2: In fact, this question stem from the following (rather tight) inequality for large enough $n$'s:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\binom{n^4}{n} \left(\frac{1}{n^3}\right)^n \left(1-\frac{1}{n^3}\right)^{n^4-n} &amp;lt; \frac{1}{4 \sqrt n}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;in which $m = n^4$, $\alpha=1/n^3$, and $k=n$.&lt;/p&gt;&#xA;" OwnerUserId="2420" LastActivityDate="2011-01-31T02:26:02.557" Title="Deriving Inequalities for Binomial Terms" Tags="&lt;inequality&gt;" CommentCount="2" />
  <row Id="19648" PostTypeId="1" AcceptedAnswerId="19653" CreationDate="2011-01-31T02:30:01.827" Score="2" ViewCount="107" Body="&lt;p&gt;I started off substituting $u = e^x$ so $du = e^x\, dx$, giving me &#xA;$$\int \frac{u^8}{u^{16}+36}\,du.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So then I substituted $v = u^8$ and $dv = 8u^7du$, which gives me &#xA;$$\int \frac{1}{v^2 + 36}\,dv.$$&#xA;At this point I am unsure what to do.&lt;/p&gt;&#xA;" OwnerUserId="6423" LastActivityDate="2011-01-31T03:23:17.830" Title="Need help integrating $e^{8x}/(e^{16x}+36)$." Tags="&lt;calculus&gt;&lt;homework&gt;&lt;integral&gt;" AnswerCount="2" />
  <row Id="19649" PostTypeId="2" ParentId="19491" CreationDate="2011-01-31T02:30:43.063" Score="0" ViewCount="" Body="&lt;p&gt;The reasoning given at &lt;a href=&quot;http://stackoverflow.com/questions/4839849/recurrence-relation-homework-struggles&quot;&gt;StackOverflow&lt;/a&gt; works (assuming $n \equiv 1 \pmod 3$) except that there was a slight error in the calculation of the summation $4 + 7 + \ldots + n/3$, which is equal to $(4 + n)(n - 1)/6$ and &lt;em&gt;not&lt;/em&gt; $(4 + n)(n - 1)/3$.  Letting $m = (n-1)/3$, you can express this summation as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$[\sum_{i = 1}^{m}(3i + 1)]$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;= $m + \sum_{i=1}^{m}3i$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;= $m + 3 \cdot \sum_{i=1}^{m}i$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;= $m + 3[ m \cdot (m + 1)/2]$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;= $m + (3m^2 + 3m)/2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;= $(3m^2 + 5m)/2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;= $[3((n-1)/3)^2 + 5((n - 1)/3)]/2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;= $(n^2 - 2n + 1 + 5n - 5)/6$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;= $(n^2 + 3n - 4)/6$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;= $(n + 4)(n - 1)/6$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;= $(n-1)/3 + [(n-1)^2 + 3(n-1)]/6$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;= $(5(n-1) + (n-1)^2)/6$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;= $(5n - 5 + n^2 - 2n + 1)/6$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;= $(n^2 + 3n - 4)/6$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;= $(n + 4)(n-1)/6$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;...so $T(n) = \Theta(1) + (n+4)(n-1)/6$, as you pointed out on StackOverflow.&lt;/p&gt;&#xA;" OwnerUserId="4693" LastActivityDate="2011-01-31T02:30:43.063" CommentCount="9" />
  <row Id="19650" PostTypeId="1" AcceptedAnswerId="19656" CreationDate="2011-01-31T02:36:25.473" Score="3" ViewCount="154" Body="&lt;p&gt;I've been studying some linear algebra for a test and I've stumbled across this particular issue.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm given the transformation matrix $M_{BB&amp;#39;}$, corresponding to some $f \colon \mathbb{R}^n \to \mathbb{R}^m$ with respect to the (ordered) bases $B=(B_1, \dotsc, B_n)$ and $B&amp;#39;=(B&amp;#39;_1, \dotsc, B&amp;#39;_n)$.&#xA;As I understand, if I have a vector $v$ expressed with respect to the basis $B$, then I can find $f(v)$ by calculating $M_{BB&amp;#39;}v$, obtaining a result expressed in the basis $B&amp;#39;$. Also, if I want to find $f(v)$ for any $v$ expressed in the canonical basis $E$, I can calculate $C_{B&amp;#39;E}M_{BB&amp;#39;}C_{EB}$, where $C_{B&amp;#39;E}$ and $C_{EB}$ are the matrices for the change of basis from $B&amp;#39;$ to $E$ and from $E$ to $B$, respectively, and apply it to the vector $v$, obtaining $f(v)$ with respect to the canonical basis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as finding the image of a particular vector goes, I don't think much else can be said. Nonetheless, when I'm asked to find the image of $f$, given $M_{EE} = C_{B&amp;#39;E}M_{BB&amp;#39;}C_{EB}$, I reason that, since $C_{EB}$ is invertible, instead of computing $C_{B&amp;#39;E}M_{BB&amp;#39;}C_{EB}v$ for a generic $v$, I can assume that one and only one $w = C_{EB}^{-1}v$ exists for each $v$, so it suffices to find the image of $C_{B&amp;#39;E}M_{BB&amp;#39;}v$, i.e., to find the image of $M_{BB&amp;#39;}$, find a basis for this space and transform the basis vectors to the canonical basis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the null space of $f$, I'm trying to find all $v$ such that&#xA;$M_{EE}v = C_{B&amp;#39;E}M_{BB&amp;#39;}C_{EB}v = 0$. And since $C_{B&amp;#39;E}$ is invertible, all I need to solve is: $M_{BB&amp;#39;}C_{EB}v = 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, I've tried these &quot;simplified&quot; methods and I get different answers, probably due to a mistake in the calculation, but my questions are the following. Are these methods valid? Is there anything I'm missing? Are these the most efficient ways to compute $Im f$ and $Ker f$? (I'm not really worried about working with large matrices or computational complexity, but I would simply like not to perform unnecessary computations.) Are there other procedures for these goals that I'm not considering? Also, is there any significance as to the &quot;disappearance&quot; of one of the two change of basis matrix in each equation?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any help or insight into this matter is greatly appreciated.&lt;/p&gt;&#xA;" OwnerUserId="6427" LastEditorUserId="6427" LastEditorDisplayName="" LastEditDate="2011-01-31T04:36:47.180" LastActivityDate="2011-01-31T04:36:47.180" Title="How can I find the image and the null space of a linear transformation, given its matrix with respect to non-canonical coordinates?" Tags="&lt;linear-algebra&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="19651" PostTypeId="2" ParentId="19648" CreationDate="2011-01-31T02:42:09.253" Score="2" ViewCount="" Body="&lt;p&gt;Ok. I shall start from the first. Put $t =e^{8x}$ So you have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align*}&#xA;\int \frac{e^{8x}}{e^{16x} + 36} \ \text{dx} &amp;amp; = \frac{1}{8}\int \frac{1}{t^{2} + 36} \ \text{dt} \\ &amp;amp;= \frac{1}{8}\int \frac{1}{36 \tan^{2}{\theta} + 36} \times 6 \sec^{2}{\theta} \ d{\theta} \ \Bigl[ \text{ Putting} \ t=6\tan{\theta}\Bigr]\\ &amp;amp;= \frac{6}{8} \int \frac{1}{36 \cdot (1+\tan^{2}{\theta})} \cdot \sec^{2}{\theta} \ d\theta&#xA;\end{align*}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am sure you can proceed from here. The substitution which you needed was $\tan{\theta}$. Remember the following:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Denominator of the form: $a^{2}- x^{2}$ substitute $x = a\sin{\theta}$ or $a \cos{\theta}$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Denominator of the form : $a^{2} + x^{2}$ substitute $x = a\tan{\theta}$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" LastActivityDate="2011-01-31T03:02:25.850" CommentCount="11" />
  <row Id="19652" PostTypeId="2" ParentId="19645" CreationDate="2011-01-31T02:58:21.783" Score="3" ViewCount="" Body="&lt;p&gt;Yup. I win the bet :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See J. M. McCleary's &lt;em&gt;User's guide to spectral sequences&lt;/em&gt;, Chapter 5.&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2011-01-31T02:58:21.783" CommentCount="1" />
  <row Id="19653" PostTypeId="2" ParentId="19648" CreationDate="2011-01-31T03:05:59.807" Score="3" ViewCount="" Body="&lt;p&gt;It's nice to recognize that the denominator is $(e^{8x})^2 + 6^2$. This suggests trying to get the integral into the form $\int \frac{1}{u^2+a^2}\,du$, which can be solved by elementary means.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, let's try that: setting $w=e^{8x}$, we have $dw = 8e^{8x}\,dx$, so the integral becomes&#xA;$$\int\frac{e^{8x}\,dx}{(e^{8x})^2 + 36} = \frac{1}{8}\int \frac{dw}{w^2+36}.$$&#xA;Now, factor out $6^2$ from the denominator, to get&#xA;$$\frac{1}{8}\int\frac{dw}{w^2 + 36} = \frac{1}{288}\int\frac{{dw}}{\left(\frac{w}{6}\right)^2 + 1}.$$&#xA;Now set $u=\frac{w}{6}$, so $6du = dw$. We have:&#xA;$$\frac{1}{288}\int\frac{dw}{\left(\frac{w}{6}\right)^2 + 1}= \frac{1}{48}\int\frac{du}{u^2+1}.$$&#xA;The last integral should be immediate (&lt;strong&gt;Hint:&lt;/strong&gt; $\frac{1}{1+t^2}$ is the derivative of a standard [inverse] function, so its integral is direct). Then taking $u=\frac{w}{6} = \frac{1}{6}e^{8x}$ gets you the integral; you can always differentiate to check.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-31T03:23:17.830" LastActivityDate="2011-01-31T03:23:17.830" CommentCount="4" />
  <row Id="19655" PostTypeId="2" ParentId="19646" CreationDate="2011-01-31T03:15:23.287" Score="7" ViewCount="" Body="&lt;p&gt;Since $\vee$ distributes over $\wedge$, you know that $A\vee(B\wedge C)$ is equivalent to $(A\vee B)\wedge(A\vee C)$. So if you have the former, you can replace it with the latter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, likewise, if you have the &lt;em&gt;latter&lt;/em&gt;, you can replace it with the former. Your second line, $(p\wedge q)\vee(p\wedge\neg q)$ is of the form $(A\vee B)\wedge(A\vee C)$ (with $A=p$, $B=q$, and $C=\neg q$), so it is equivalent to $A\vee(B\wedge C)$, which is exactly the third line.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words: instead of using the &quot;distributive property&quot; as usual, you use it &quot;in reverse&quot;. It's much like going from $5\times 3 + 5\times 7$ to $5\times(3+7)$, instead of the other way around. You can think of it as the analogue of &quot;factoring out&quot; instead of &quot;distributing through&quot;. &lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2011-01-31T03:15:23.287" CommentCount="1" />
  <row Id="19656" PostTypeId="2" ParentId="19650" CreationDate="2011-01-31T03:36:51.380" Score="2" ViewCount="" Body="&lt;p&gt;Think of the matrices $C_{EB}$ as &quot;dictionaries&quot; or &quot;translators&quot;. $C_{EB}$ is a translator which, if you talk to it in the $E$-language (describing vectors in terms of the basis $E$) will &quot;reply&quot; by telling you the same thing, but in $B$-language (in terms of the basis $B$). Likewise, $C_{B&amp;#39;E}$ is the translator that takes something spoken in $B&amp;#39;$-language, and translates it into $E$-language.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What does $C_{B&amp;#39;E}M_{BB&amp;#39;}C_{EB}$ give, then? You &quot;speak&quot; to it in $E$-language; $C_{EB}$ translates it into $B$ language. $M_{BB&amp;#39;}$ will apply $M$ and give the answer in $B&amp;#39;$-language; and finally $C_{B&amp;#39;E}$ will translate this into $E$-language. So $C_{B&amp;#39;E}M_{BB;}C_{EB} = M_{EE}$, where $M_{EE}$ is the matrix of the &lt;em&gt;same&lt;/em&gt; linear transformation as $M_{BB&amp;#39;}$, but relative to the standard basis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, what is $C_{B&amp;#39;E}M_{BB&amp;#39;}$? It is the matrix $M_{BE}$: you give it a vector described in $B$-language, and it will answer with its image under $M$, but described in $E$-language.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The key to the manipulations/computations you are doing, then, is to keep straight in what language you are speaking and in what language you are getting the answers. I suspect that the reason you are getting &quot;different answers&quot; is that you are obtaining the 'same' answers, but in different languages.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you try to find the nullspace of $f$ by looking for all vectors $\mathbf{v}$ for which $M_{BB&amp;#39;}\mathbf{v}=\mathbf{0}$, what you will find is the nullspace of $f$, &lt;em&gt;described in terms of $B$&lt;/em&gt;. So if $B=[\mathbf{x}_1,\ldots,\mathbf{x}_n]$, and you get, say, that $\mathbf{v}=(1,0,0,\ldots,0)^T$ is one of the vectors in the nullspace, what you &lt;em&gt;really&lt;/em&gt; have is that the vector whose coordinates relative to $B$ are $\mathbf{v}$ is in the nullspace; that is, $\mathbf{v}$ &quot;really&quot; describes the vector $\mathbf{x}_1$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;By contrast, if you try to find the nullspace of $f$ by trying to find all vectors $\mathbf{w}$ for which $C_{B&amp;#39;E}M_{BB&amp;#39;}C_{EB}\mathbf{w}=\mathbf{0}$ you will find the vectors in the nullspace described in terms of $E$, not in terms of $B$. You can &quot;translate&quot; form one to the other by using $C_{EB}$ (which translates from $E$-speak to $B$-speak), or $C_{BE} = (C_{EB})^{-1}$ (which translates from $B$-speak to $E$-speak). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(You get the same answer if you use $M_{BB&amp;#39;}C_{EB}$ as if you use $C_{B&amp;#39;E}M_{BB&amp;#39;}C_{EB}$ because you are still dealing with vectors in the domain described in $E$-speak; the fact that you are translating the results into $B&amp;#39;$-speak at the end is immaterial, since the only vector that is described as $\mathbf{0}$ in &lt;em&gt;any&lt;/em&gt; speak is $\mathbf{0}$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the image, you want to keep track of the same things: if you try to find the image using the image of $M_{BB&amp;#39;}$, you will get the answer described in $B&amp;#39;$-speak; but if you use $C_{B&amp;#39;E}M_{BB&amp;#39;}$, you'll get the answer described in $E$-speak. You can translate from one to the other by using $C_{B&amp;#39;E}$ and $C_{EB&amp;#39;}=(C_{EB&amp;#39;})^{-1}$.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2011-01-31T03:36:51.380" CommentCount="3" />
  <row Id="19657" PostTypeId="1" AcceptedAnswerId="19658" CreationDate="2011-01-31T04:05:06.363" Score="4" ViewCount="298" Body="&lt;p&gt;I needed some help integrating this:&#xA;$$\int\frac{2\,dx}{x\ln(6x)}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have never seen the dx within the problem like that, I am assuming I can't just move it to the outside can I?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can I start by factoring out the 2 and then setting $u = \ln(6x)$ and $du = (1/x)\; dx$.&lt;/p&gt;&#xA;" OwnerUserId="6423" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-31T04:26:39.397" LastActivityDate="2011-04-11T09:56:47.950" Title="Integrating $\int\frac{2dx}{x\ln(6x)}$." Tags="&lt;calculus&gt;&lt;homework&gt;&lt;integral&gt;" AnswerCount="3" CommentCount="7" FavoriteCount="1" />
  <row Id="19658" PostTypeId="2" ParentId="19657" CreationDate="2011-01-31T04:11:28.010" Score="3" ViewCount="" Body="&lt;p&gt;Yes, you can.  In this, $dx$ is a term you can move wherever following the commutativity and associativity of multiplication.  Your approach is quite reasonable.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-01-31T04:11:28.010" />
  <row Id="19659" PostTypeId="1" CreationDate="2011-01-31T04:15:20.723" Score="3" ViewCount="106" Body="&lt;p&gt;The &lt;a href=&quot;http://www.comap.com/undergraduate/contests/mcm/&quot; rel=&quot;nofollow&quot;&gt;Mathematical Contest in Modeling&lt;/a&gt; (MCM) is coming up in a couple of weeks.  With the idea that the competition is ultimately about teaching students to become better at mathematical modeling, I was wondering if anyone else on the site who has participated in the MCM or who coaches would be willing to share strategies they have for preparing for the competition.  Since I'm bringing up the question, I'll start.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For those who aren't familiar with the MCM, the &lt;a href=&quot;http://en.wikipedia.org/wiki/Mathematical_Contest_in_Modeling&quot; rel=&quot;nofollow&quot;&gt;Wikipedia page&lt;/a&gt; also gives a brief overview.  &lt;/p&gt;&#xA;" OwnerUserId="2370" LastActivityDate="2011-01-31T15:46:44.260" CommunityOwnedDate="2011-01-31T10:12:06.957" Title="Preparing for the Mathematical Contest in Modeling" Tags="&lt;education&gt;&lt;contest-math&gt;&lt;mathematical-modeling&gt;" AnswerCount="4" CommentCount="4" />
  <row Id="19660" PostTypeId="2" ParentId="19659" CreationDate="2011-01-31T04:16:09.380" Score="1" ViewCount="" Body="&lt;p&gt;The core of the preparation at my university (where I have coached the MCM teams the past few years) is a zero-credit, pass-fail, weekly seminar each fall semester.  Students spend about three weeks each on three or four old MCM problems, and then we look at an outstanding solution and possibly the judges' commentary on each problem.  (These are published in one of the &lt;a href=&quot;http://www.comap.com/product/periodicals/index.html&quot; rel=&quot;nofollow&quot;&gt;UMAP Journal&lt;/a&gt; issues from that year.)  We also spend a little time on software that may be useful during the competition, such as LaTeX and Mathematica.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also go through this paper, &lt;a href=&quot;http://www.math.ucla.edu/~getreuer/mcmguide.pdf&quot; rel=&quot;nofollow&quot;&gt;Quest for the MCM&lt;/a&gt;, written by three students from the University of Colorado who received the top designation in the competition three years in a row.  It's got a lot of good advice.&lt;/p&gt;&#xA;" OwnerUserId="2370" LastActivityDate="2011-01-31T04:16:09.380" CommunityOwnedDate="2011-01-31T04:16:45.463" />
  <row Id="19661" PostTypeId="2" ParentId="19568" CreationDate="2011-01-31T04:20:43.243" Score="0" ViewCount="" Body="&lt;p&gt;Basically, a stochastic process is to a time series what a random variable is to a number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;em&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Realization_%28probability%29&quot; rel=&quot;nofollow&quot;&gt;realization&lt;/a&gt;&lt;/em&gt; (the &quot;result&quot;, the observed value) of a random variable (say, a dice roll) is a number - (but, as it's a random variable, we know that the number can take values from a given set according to some probability law). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same applies to stochastic process, but now the realization instead of being a single number is a sequence (if the process is discrete) or a function (if it's continuos). Basically, a time series.&lt;/p&gt;&#xA;" OwnerUserId="312" LastActivityDate="2011-01-31T04:20:43.243" />
  <row Id="19662" PostTypeId="1" AcceptedAnswerId="19665" CreationDate="2011-01-31T04:48:32.417" Score="2" ViewCount="217" Body="&lt;p&gt;I have several problems like this to do, but I am having a hard time understanding how they get from one step to the next in the example (see note in picture). If I can understand this example I think I'll be good.  Also want to point out that this is just an example, not the actual homework problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/Ckk45.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="6096" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2011-06-03T13:50:38.567" LastActivityDate="2011-06-03T13:50:38.567" Title="Solving recurrences with the Substitution Method" Tags="&lt;homework&gt;&lt;algorithms&gt;&lt;recurrence-relations&gt;" AnswerCount="2" CommentCount="0" />
  <row Id="19663" PostTypeId="1" CreationDate="2011-01-31T04:55:27.260" Score="2" ViewCount="110" Body="&lt;p&gt;I am working on a problem which involves showing Cauchy's integral formula holds for each $z\in D(0,1)$ and for each f defined on closed unit disc and analytic on $D(0,1)$. Also, I have to show the formula holds under the above conditions but this time f is continuous on the closed unit disc, not necessarily differentiable. &#xA;Any suggestion is appreciated. Thanks.&lt;/p&gt;&#xA;" OwnerUserId="6399" LastActivityDate="2011-01-31T04:55:27.260" Title="Cauchy's Integral" Tags="&lt;homework&gt;&lt;complex-analysis&gt;" CommentCount="4" />
  <row Id="19665" PostTypeId="2" ParentId="19662" CreationDate="2011-01-31T04:59:19.753" Score="2" ViewCount="" Body="&lt;p&gt;To give a direct answer to your question it comes from the guess in step $(a)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$T(n) = n \log_2 (n) + n$ is the guess you made in step $(a)$. Using that replace $n = \frac{n}{2}$, $T(\frac{n}{2}) = \frac{n}{2} \log_2 (\frac{n}{2}) + \frac{n}{2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$T(\frac{n}{2}) = \frac{n}{2} (\log_2 (n) - \log_2(2)) + \frac{n}{2} = \frac{n}{2} \log_2 (n) - \frac{n}{2} + \frac{n}{2} = \frac{n}{2} \log_2 (n)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, $T(n) = 2 T(\frac{n}{2}) + n = 2 \frac{n}{2} \log_2 (n) + n = n \log_2 (n) + n$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While solving some recurrences it is good to recognize some nice things about the recurrence you are actually solving. For instance in this recurrence, notice that at each step you are dividing $n$ by $2$. So strictly speaking through the recurrence you can only obtain $T(2^m)$ where $m \in \mathbb{N}$. For instance, $T(3)$ cannot be obtained using this recurrence. However, such recurrences come up when you do certain things, when $n$ is really large, recursively using say a binary tree. When $n$ is really large you are interested mainly in how the solution behaves in an asymptotic sense and not the precise expression for the solution. In such cases, you assume that $n=2^m$ for the simplicity of calculations and the cost for all $n$ (even when $n$ is not a power of $2$) can be shown to obey the solution, obtained by the assuming $n=2^m$, in an asymptotic sense.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you were to rewrite the recurrence in terms of $m$, and call $T(2^m) = S(m)$, then we would get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$S(m) = 2 S(m-1) + 2^m$ since $\frac{n}{2} = 2^{m-1}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$S(m-1) = 2 S(m-2) + 2^{m-1}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$S(m) = 4 S(m-2) + 2^m + 2^m$ and so on (hidden induction here) to get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$S(m) = 2^m S(0) + 2^m + 2^m + \cdots + 2^m + 2^m = m 2^m + 2^m$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence $T(n) = n \log_2(n) + n$ since $n = 2^m$ i.e. $m = \log_2(n)$&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2011-01-31T05:18:32.093" LastActivityDate="2011-01-31T05:18:32.093" CommentCount="1" />
  <row Id="19666" PostTypeId="2" ParentId="19662" CreationDate="2011-01-31T05:01:52.437" Score="2" ViewCount="" Body="&lt;p&gt;As they say, it is a guess.  If (as in a) $T(n)=n lg n + n$ you just substitute to get  the one you ask &quot;To here&quot;.  So the real question is whether induction is justified.  The point is that if the expression for T is valid up to n it is valid up to 2n (and you need to explain in this case what happens for odd $n$) then it is valid for all $n$.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2011-01-31T05:13:44.377" LastActivityDate="2011-01-31T05:13:44.377" />
  <row Id="19667" PostTypeId="1" CreationDate="2011-01-31T05:28:09.133" Score="2" ViewCount="111" Body="&lt;p&gt;I am wondering if there are some conclusions as to when a series solution using the separate variable method to a PDE exists; i.e. for what requirements on the PDE, what requirements on the initial and boundary conditions so that one can assume the solution as products of terms involving only a single dependent variable, and then proceed to solve the equation in Fourier's way.&lt;/p&gt;&#xA;" OwnerUserId="6016" LastActivityDate="2011-01-31T11:29:56.563" Title="when does a separate-variable series solution exist for a PDE" Tags="&lt;differential-equations&gt;&lt;mathematical-physics&gt;&lt;pde&gt;" AnswerCount="1" />
  <row Id="19668" PostTypeId="2" ParentId="19528" CreationDate="2011-01-31T05:29:13.203" Score="1" ViewCount="" Body="&lt;p&gt;Arturo and Sivaram have already given the general condition for integer matrices with integer inverses; here I only note this particular example due to &lt;a href=&quot;http://dx.doi.org/10.1137/0717041&quot; rel=&quot;nofollow&quot;&gt;Ericksen&lt;/a&gt; that the matrix $\mathbf A$ with entries&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$a_{ij}=\binom{n+j-1}{i-1}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $n$ is an arbitrary nonnegative integer has an integer inverse.&lt;/p&gt;&#xA;" OwnerUserId="6429" LastActivityDate="2011-01-31T05:29:13.203" />
  <row Id="19669" PostTypeId="1" AcceptedAnswerId="19670" CreationDate="2011-01-31T05:46:06.827" Score="2" ViewCount="127" Body="&lt;p&gt;Given that f(x) = x / (x+ (2/x))&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The derivative function is given by f'(x) = (Ax^2 + Bx + C) / (x^2+D)^2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A=&lt;/p&gt;&#xA;&#xA;&lt;p&gt;B=&lt;/p&gt;&#xA;&#xA;&lt;p&gt;C=&lt;/p&gt;&#xA;&#xA;&lt;p&gt;D=&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From rearranging the original equation, I can get D=2 C=0 and A=0, but I cant seem to find B.  Maybe someone could help?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: I found the answer to be 4, but don't see how this makes sense, since I can only clearly see how -2 would work.&lt;/p&gt;&#xA;" OwnerUserId="6423" LastEditorUserId="6423" LastEditorDisplayName="" LastEditDate="2011-02-08T08:45:18.423" LastActivityDate="2011-02-08T08:45:18.423" Title="Help rearranging a derivative." Tags="&lt;calculus&gt;&lt;homework&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="19670" PostTypeId="2" ParentId="19669" CreationDate="2011-01-31T05:52:04.780" Score="2" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT:&lt;/strong&gt; Rearrange $f(x) = \frac{x^2}{x^2 + 2} = 1 - \frac{2}{x^2 + 2}$. Use that the derivative of a constant is $0$ and the derivative of $\frac{1}{x^2 + a}$ is $\frac{-2x}{(x^2+a)^2}$.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastActivityDate="2011-01-31T05:52:04.780" CommentCount="7" />
  <row Id="19671" PostTypeId="2" ParentId="19659" CreationDate="2011-01-31T06:44:31.977" Score="1" ViewCount="" Body="&lt;p&gt;I have participated once and I think the single most useful skill is for at least one member of the team to be fluent in some powerful programming language, like C++. We used it for numerical simulations, for drawing pretty pictures, for testing hypotheses and assumptions and so on. If somebody on the team can code almost any modelling situations very quickly, that makes the competition much easier and also much more fun, since it is then easier to concentrate on the mathematical aspects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, LaTeX is a must and is also worth practising, since many undergraduates will spend inordinate amounts of time writing up their submission.&lt;/p&gt;&#xA;" OwnerUserId="3212" LastActivityDate="2011-01-31T06:44:31.977" CommunityOwnedDate="2011-01-31T06:44:39.903" CommentCount="1" />
  <row Id="19672" PostTypeId="1" AcceptedAnswerId="19806" CreationDate="2011-01-31T06:51:10.123" Score="4" ViewCount="156" Body="&lt;p&gt;A search on wikipedia shows:&#xA;$$\mu(n) = \sum_{k=1,gcd(k,n)=1}^{n} e^{2\pi i \frac{k}{n}}$$&#xA;But that uses complex numbers... and requires finding out the gcd... &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How useful will be a method, if that could find the exact value for $\mu(n)$ function, by just knowing all the values from $\mu(1)$ till $\mu(n-1)$, which does not require factoring any integer and which uses only elementary methods?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Has this been done before? My question is more precisely:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Does any such formula exist?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2320" LastActivityDate="2011-02-01T03:22:58.533" Title="On the Mobius Mu function" Tags="&lt;number-theory&gt;&lt;soft-question&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="19673" PostTypeId="1" AcceptedAnswerId="19675" CreationDate="2011-01-31T06:59:59.057" Score="2" ViewCount="271" Body="&lt;p&gt;If we have a function, say, $y=x^2$, then $dy/dx=2x$.&lt;br&gt;&#xA;Now $y=1,4,9,16...$ for $x=1,2,3,4...$&lt;br&gt;&#xA;and $dy/dx=2,4,6,8...$ for $x=1,2,3,4...$&lt;br&gt;&#xA;Now as $dy/dx$ represents rate of change of y w.r.t x but I can't understand how these particluar&lt;br&gt;&#xA;values of $dy/dx$ represent change ?&lt;br&gt;&#xA;I mean for x=3, the rate of change is 6. What does it mean ?&lt;/p&gt;&#xA;" OwnerUserId="4814" LastEditorUserId="4814" LastEditorDisplayName="" LastEditDate="2011-01-31T07:15:45.643" LastActivityDate="2011-01-31T14:22:01.460" Title="What does rate of change actually mean ?" Tags="&lt;calculus&gt;&lt;intuition&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="19675" PostTypeId="2" ParentId="19673" CreationDate="2011-01-31T07:16:07.303" Score="5" ViewCount="" Body="&lt;p&gt;You could think of it this way: taking the value of $\frac{dy}{dx}$ at some point $x_0$, tells you approximately how much the value of $y$ changes for values very close to $x_0$. So in your example, if $x_0 = 3$, the change is $6$: if you consider the value of $x^2$ at a point $x_0 + \epsilon$ for some small number $\epsilon$, the &lt;em&gt;change&lt;/em&gt; in the values of the function from $x_0$ to $x_0 + \epsilon$ will increase by approximately $6\cdot \epsilon$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You should try this with some numbers. Take  $x_0 = 3$ and $\epsilon = 1$. Then $3^2 = 9$, while $(3+1)^2 = 16$. The change is $7$, which is close to $6\cdot \epsilon = 6$. If you take $\epsilon = \frac{1}{2}$, you can check that the change will be $3\frac{1}{4}$, which is close to $6\cdot \epsilon = 3$. Notice that as you make $\epsilon$ smaller, the error becomes smaller as well. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You should check &lt;a href=&quot;http://en.wikipedia.org/wiki/Derivative&quot; rel=&quot;nofollow&quot;&gt;wikipedia&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="5105" LastEditorUserId="5105" LastEditorDisplayName="" LastEditDate="2011-01-31T07:26:57.443" LastActivityDate="2011-01-31T07:26:57.443" CommentCount="1" />
  <row Id="19676" PostTypeId="2" ParentId="19673" CreationDate="2011-01-31T07:24:55.823" Score="1" ViewCount="" Body="&lt;p&gt;When you let $x$ take the values 1, 2, 3, etc., you are taking fairly large steps, and then you don't see very well what the derivative $dy/dx$ means. Think of $dy/dx$ as measuring how much $y$ changes when $x$ changes by &lt;em&gt;a very small amount&lt;/em&gt;, like when changing the value of $x$ from 3 to 3.00001; then $y$ changes by &lt;em&gt;about 6 times as much&lt;/em&gt;, from 9 to approximately 9.00006 (the exaxt value is 9.0000600001).&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2011-01-31T07:24:55.823" />
  <row Id="19677" PostTypeId="1" AcceptedAnswerId="19703" CreationDate="2011-01-31T07:28:10.967" Score="2" ViewCount="222" Body="&lt;p&gt;My book, professor, and friends make this theorem look very significant:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If &lt;em&gt;A&lt;/em&gt; is an &lt;em&gt;m&lt;/em&gt; by &lt;em&gt;n&lt;/em&gt; matrix, then the following are either all true or all false:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Each vector &lt;strong&gt;b&lt;/strong&gt; in R^m is a linear combination of the columns of &lt;em&gt;A&lt;/em&gt;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;For each &lt;strong&gt;b&lt;/strong&gt; in R^m, the equation &lt;em&gt;A&lt;/em&gt;&lt;strong&gt;x&lt;/strong&gt; = &lt;strong&gt;b&lt;/strong&gt; has a solution.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The columns of &lt;em&gt;A&lt;/em&gt; span R^m.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;A&lt;/em&gt; has a pivot position in every row.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;My question is: isn't that a fairly obvious tautology? I mean, the &lt;em&gt;definition&lt;/em&gt; of matrix multiplication simply expands the second equation into a linear combination, so why do people get so excited about this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sometimes I get the feeling that, in linear algebra, we're just finding fifty ways to state the same thing in different words, and getting excited even though they all stem from the same definition. :\&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Apologies in advance about the formatting, I'm not sure how the math formatting works here.)&lt;/p&gt;&#xA;" OwnerUserId="4890" LastEditorUserId="4890" LastEditorDisplayName="" LastEditDate="2011-01-31T07:50:58.980" LastActivityDate="2011-01-31T14:14:52.393" Title="Linear Algebra -- Why is this theorem significant?" Tags="&lt;linear-algebra&gt;" AnswerCount="1" CommentCount="14" FavoriteCount="1" />
  <row Id="19678" PostTypeId="1" AcceptedAnswerId="19688" CreationDate="2011-01-31T07:39:12.400" Score="2" ViewCount="122" Body="&lt;p&gt;Is $(\sin \phi)^2$ is equal to $\sin^2\phi$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can any one tell what is  the ans for the below expression&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sin^260$ + $\cos^260$ + $\tan^245$ + $\sec^260$ - $\csc^260$&lt;/p&gt;&#xA;" OwnerUserId="6434" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2011-01-31T21:00:57.013" LastActivityDate="2011-01-31T21:19:40.680" Title="Trigonometry Expression" Tags="&lt;trigonometry&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="19679" PostTypeId="2" ParentId="19678" CreationDate="2011-01-31T07:48:48.623" Score="2" ViewCount="" Body="&lt;p&gt;Yes, $sin^2\phi$ is by definition $(sin\phi)^2$. The answer to the last question is: you can calculate sin(60), cos(60) and tan(45) just using Pythagoras. A hint for the first two of these: draw a right angled triangle with one angle of 60 degrees. What is the third angle? Can you exploit the symmetry?&lt;/p&gt;&#xA;" OwnerUserId="3212" LastActivityDate="2011-01-31T07:48:48.623" CommentCount="1" />
  <row Id="19680" PostTypeId="2" ParentId="11923" CreationDate="2011-01-31T08:44:22.073" Score="1" ViewCount="" Body="&lt;p&gt;I just wanted to let people know that I finally got a chance to incorporate the work of Kahen's answer into my field theory notes.  It now appears as Section 12.7 of &lt;a href=&quot;http://www.math.uga.edu/~pete/FieldTheory.pdf&quot; rel=&quot;nofollow&quot;&gt;these notes&lt;/a&gt;, pp. 67-72.  (Yes, it takes five pages!)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I followed Kahen's suggestions mostly, but not completely.  In particular my treatment is a little &quot;nettier&quot;.  I wanted one more little result to be completely convinced that convergence of Cauchy sequences implies convergence of Cauchy nets in an Archimedean ordered field (this is certainly no big deal). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then in the proof of $2 \implies 1$ in Kahen's Proposition 12 (um, his &lt;em&gt;first&lt;/em&gt; Proposition 12), I exploited the fact that I could use Cauchy nets more general than sequences to give a different proof that convergence of Cauchy nets implies Dedekind completeness: it is enough to show that any set which is bounded below has a greatest lower bound, and I actually show that the set of all lower bounds has a maximal element by applying Zorn's Lemma.  So that was amusing.  (I have no illusions that this proof could be both original and correct...)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again, comments are welcome.  I wrote this all up pretty rapidly, so it is even more possible than usual that I made a mistake: please let me know.  If you have any other suggestions please let me know.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once again I would prefer to acknowledge Kahen's contribution, if at all possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Added&lt;/b&gt;: Let me mention that my notes have much less to say than Kahen's answer about the &lt;em&gt;construction&lt;/em&gt; of $\mathbb{R}$ using Cauchy sequences, this being something that was not the primary goal of these field theory notes (and in fact is described in other notes of mine in the context of completion of a normed field).  So this is not worth much as an answer to the current question (hence CW).  I decided to leave it as an answer and not as a comment so that it will be easier for others to comment on it, if they care to.&lt;/p&gt;&#xA;" OwnerUserId="299" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2011-01-31T08:53:30.733" LastActivityDate="2011-01-31T08:53:30.733" CommunityOwnedDate="2011-01-31T08:44:22.073" CommentCount="2" />
  <row Id="19681" PostTypeId="1" AcceptedAnswerId="37643" CreationDate="2011-01-31T09:00:20.940" Score="10" ViewCount="361" Body="&lt;p&gt;I'm interested in sufficient conditions for a notion of sequential convergence to be induced by a topology. More precisely: Let $V$ be a vector space over $\mathbb{C}$ endowed with a notion $\tau$ of sequential convergence. When is there a topology $\mathcal{O}$ on $V$ that makes $V$ a topological vector space such that &quot;sequences suffice&quot; in $(V,\mathcal{O})$, e.g. $(V,\mathcal{O})$ is first countable, and convergence in $(V,\mathcal{O})$ coincides with $\tau$-convergence? Is the topology $\mathcal{O}$ uniquely determined?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By a notion $\tau$ of sequential convergence on a vector space $V$ I mean a &quot;rule&quot; $\tau$ which assigns to certain sequences $(v_n)_{n\in\mathbb{N}}\subset V$ (which one would call &lt;em&gt;convergent&lt;/em&gt; sequences) an element $v\in V$ (a &lt;em&gt;limit&lt;/em&gt; of $(v_n)_n$). One could write $v_n\stackrel{\tau}{\to}v$ in this case. This process of &quot;assigning a limit&quot; should satisfy at least that any constant sequence $(v,v,v,...)$ is convergent and is assigned the limit $v$. Also, given a convergent sequence $(v_n)_n$ with limit $v$ any subsequence $(v_{n_k})_k$ should have $v$ as a limit.&lt;br&gt;&#xA;I would also like this concept of assigning a limit to be compatible with addition in $V$ and multiplication by a scalar.&lt;br&gt;&#xA;Maybe one should include further restrictions. In fact I would like to know which further assumptions on this &quot;limiting process&quot; one has to assume in order to ensure that this limiting procedure corresponds to an actual topology on $V$ which makes $V$ a topological vector space in which a sequence converges if and only if it $\tau$-converges.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me give two examples. If we take for instance a topological vector space $(V,\mathcal{O})$ then we have a notion of convergence in $V$ based upon the set $\mathcal{O}$ of open sets of $V$. This notion of convergence clearly satisfies the above assumptions on $\tau$.&lt;br&gt;&#xA;If on the other hand we consider $L^\infty([0,1])$ equipped with the notion of &lt;em&gt;pointwise convergence almost everywhere&lt;/em&gt;, then there is no topology on $L^\infty([0,1])$ which makes $L^\infty([0,1])$ a TVS in which a sequence converges if and only if it converges pointwise almost everywhere. Still, convergence almost everywhere also satisfies the above assumptions on $\tau$.&lt;br&gt;&#xA;So the above assumptions on this concept of convergence are necessary but not sufficient conditions for what I mean by a notion of convergence to correspond to an actual topology. The question is: Which further assumptions do I have to make?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On a less general level I'm particularly interested in the following case: Let $G\subset\mathbb{C}^d$ be a domain, $X$ a (complex) Banach space and let $H^\infty(G;X)$ denote the space of bounded holomorphic functions $f\colon G\to X$. Now consider the following notion $\tau$ of sequential convergence on $H^\infty(G;X)$: We say that a sequence $(f_n)_{n\in\mathbb{N}}\subset H^\infty(G;X)$ $\tau$-converges to $f\in H^\infty(G;X)$ if $\sup_{n\in\mathbb{N}}\sup_{z\in G} \|f_n(z)\|_X$ is finite and $f_n(z)$ converges in $X$ to $f(z)$ for every $z\in G$. Is there a topology $\mathcal{O}$ on $H^\infty(G;X)$ such that &quot;sequences suffice&quot; in $(H^\infty(G;X),\mathcal{O})$ and a sequence $(f_n)_{n\in\mathbb{N}}\subset H^\infty(G;X)$ converges w.r.t. the topology $\mathcal{O}$ if and only if it $\tau$-converges? Is this topology $\mathcal{O}$ unique if existent? What if we drop the &quot;sequences suffice&quot;-restriction? Is $(H^\infty,\mathcal{O})$ locally convex? Metrizable? What if we replace $X$ by a more general space like a LCTVS or a Frechet space?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you in advance for any suggestions, ideas or references.&lt;/p&gt;&#xA;" OwnerUserId="1403" LastEditorUserId="1403" LastEditorDisplayName="" LastEditDate="2011-01-31T16:21:34.533" LastActivityDate="2011-05-07T19:47:21.257" Title="When is a notion of convergence induced by a topology?" Tags="&lt;topology&gt;&lt;functional-analysis&gt;&lt;topological-vector-spaces&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="3" />
  <row Id="19682" PostTypeId="1" AcceptedAnswerId="19685" CreationDate="2011-01-31T09:11:22.450" Score="2" ViewCount="129" Body="&lt;p&gt;So I'm revising my counting, and I looked at this problem and after a little bit of brainstorming I came up with the following solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Number of 8 digit numbers containing both a 5 and a 6&lt;br&gt;&#xA;= Number of 8 digit numbers - Number of 8 digit numbers lacking both a 5 and a 6&lt;br&gt;&#xA;= 90000000 - $7 * 8^7$&lt;br&gt;&#xA;= 75319936&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does this seem right?&#xA;Can anyone point out an easier way to solve a problem like this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you&lt;/p&gt;&#xA;" OwnerUserId="6436" LastActivityDate="2011-01-31T09:29:54.753" Title="How many eight digit numbers are there that contain both a 5 and a 6." Tags="&lt;number-theory&gt;&lt;combinatorics&gt;&lt;counting&gt;" AnswerCount="1" CommentCount="0" />
  <row Id="19684" PostTypeId="2" ParentId="19672" CreationDate="2011-01-31T09:22:07.883" Score="2" ViewCount="" Body="&lt;p&gt;(This should be a comment.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;FWIW, you can derive from the series you gave the alternative representation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mu(n)=\sum_{j=1}^n [\gcd(n,k)=1]\cos\left(\frac{2\pi k}{n}\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $[p]$ is an Iversonian bracket for the condition $p$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since checking if a number is squarefree is conjectured to be at least as hard as straight factorization, I don't have much hope of an algorithm of the sort you want existing. (If there is, I too would be interested in hearing about it. ;) )&lt;/p&gt;&#xA;" OwnerUserId="6438" LastActivityDate="2011-01-31T09:22:07.883" CommentCount="2" />
  <row Id="19685" PostTypeId="2" ParentId="19682" CreationDate="2011-01-31T09:29:54.753" Score="5" ViewCount="" Body="&lt;p&gt;A number not containing both 5 and 6, either doesn't contain 5 or doesn't contain 6 (or both). For example, the number $5$ doesn't contain both $5$ and $6$. However, it does contain $5$. Always try your ideas on things you can count on your own!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You need to use inclusion-exclusion:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Start with all $9\cdot 10^7$ numbers with exactly $8$ digits.&lt;/li&gt;&#xA;&lt;li&gt;Remove all $8\cdot 9^7$ numbers not containing $5$.&lt;/li&gt;&#xA;&lt;li&gt;Remove all $8\cdot 9^7$ numbers not containing $6$.&lt;/li&gt;&#xA;&lt;li&gt;Numbers containing neither $5$ nor $6$ were removed twice, so we add them once to compensate: $7\cdot 8^7$.&lt;/li&gt;&#xA;&lt;li&gt;In total, the result is $$ 9\cdot 10^7 - 2\cdot 8\cdot 9^7 + 7\cdot 8^7. $$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Now let's check ourselves by seeing what happens for one-digit numbers and two-digit numbers. For one-digit numbers we get $$ 9 - 16 + 7 = 0, $$ and indeed no single-digit number can contain both $5$ and $6$. For two-digit numbers we get $$ 90 - 144 + 56 = 2, $$ and the two numbers in question are $56$ and $65$.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;A different (more complicated) way to solve this is to enumerate on the possible first locations of $5$ and $6$, calculate explicitly how many numbers there are of this form (easy), and then summing everything up (hard). This is simpler to do if you don't care about &quot;exactly $8$ digits&quot; (put otherwise, it's simpler to calculate the number for &quot;up to $8$ digits&quot; and then subtract the analogous count for &quot;up to $7$ digits&quot;).&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2011-01-31T09:29:54.753" />
  <row Id="19686" PostTypeId="1" AcceptedAnswerId="19693" CreationDate="2011-01-31T09:32:51.310" Score="4" ViewCount="266" Body="&lt;p&gt;I thought you have to say a mapping is onto &lt;em&gt;something&lt;/em&gt;... like, you don't say, &quot;the book is on the top of&quot;...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Our book starts out by saying &quot;a mapping is said to be onto R^m&quot;, but thereafter, it just says &quot;the mapping is onto&quot;, without saying onto &lt;em&gt;what&lt;/em&gt;. Is that simply the author's version of being too lazy to write the codomain (sorry for saying something negative, but that's what it looks like to me at the moment), or does it have a different meaning?&lt;/p&gt;&#xA;" OwnerUserId="4890" LastActivityDate="2011-02-01T08:43:11.357" Title="Transformation T is... &quot;onto&quot;?" Tags="&lt;linear-algebra&gt;&lt;transformation&gt;" AnswerCount="2" CommentCount="9" />
  <row Id="19687" PostTypeId="1" AcceptedAnswerId="19689" CreationDate="2011-01-31T09:34:50.323" Score="6" ViewCount="136" Body="&lt;p&gt;To solve $y^2 + 2 = x^3$ you can factor $(y - \sqrt{-2})(y + \sqrt{-2}) = x^3$ and then check that they are relatively prime and by unique factorization both must be cubes then you can solve it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What about $y^2 + 5 = x^3$ which does not have unique factoring?&lt;/p&gt;&#xA;" OwnerUserId="4290" LastEditorUserId="1321" LastEditorDisplayName="" LastEditDate="2011-01-31T23:37:29.280" LastActivityDate="2011-02-01T05:33:39.130" Title="Solving the equation by going into a non-UFD" Tags="&lt;number-theory&gt;&lt;algebraic-number-theory&gt;" AnswerCount="2" CommentCount="0" />
  <row Id="19688" PostTypeId="2" ParentId="19678" CreationDate="2011-01-31T09:36:00.097" Score="1" ViewCount="" Body="&lt;p&gt;($sin \phi$)^2 is equal to $sin^2\phi$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$sin^260$ + $cos^260$ + $tan^245$ + $sec^260$ - $cosec^260$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The trick to that is to use a few trigonometric identities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sin^2\theta + \cos^2\theta = 1$&lt;br&gt;&#xA;$\tan 45 = 1$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The value of $\cos 60$ is $\frac{1}{2}$, so $\sec^260$ will evaluate to 4.&#xA;The value of $\sin 60$ is $\sqrt{\frac{3}{4}}$, so $\csc^260$ will evaluate to $\frac{4}{3}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Adding it all up you have $1 + 1 + 4 - \frac{4}{3} = \frac{14}{3}$&lt;/p&gt;&#xA;" OwnerUserId="6436" LastEditorUserId="6436" LastEditorDisplayName="" LastEditDate="2011-01-31T21:19:40.680" LastActivityDate="2011-01-31T21:19:40.680" CommentCount="2" />
  <row Id="19689" PostTypeId="2" ParentId="19687" CreationDate="2011-01-31T09:55:01.677" Score="6" ViewCount="" Body="&lt;p&gt;Dear Quanta,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Often what matters is not whether the number ring in question is a UFD but whether its class number is relatively prime to a certain key number.  In the case of Fermat's Last Theorem, these factoring methods work well for $x^p + y^p = z^p$ for &lt;strong&gt;regular primes&lt;/strong&gt;, i.e., primes for which $p$ does not divide the class number of the ring of integers of &#xA;$\mathbb{Q}(\zeta_p)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the Bachet-Mordell equation $y^2 + k = x^3$ (I suppose $k$ is squarefree), the favorable case is $k \equiv 1,2 \pmod 4$ (so that $\mathbb{Z}[\sqrt{-k}]$ is the full ring of integers of $\mathbb{Q}(\sqrt{-k})$) and that the class number of $\mathbb{Q}(\sqrt{-k})$ is prime to $3$.  In the particular case $k = 5$, you're in luck: $5 \equiv 1 \pmod 4$ and the class number is $2$, which is as good as $1$!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What exactly do we win when this happens?  By some happy coincidence I wrote up some notes on precisely this: please see Theorem 7 &lt;a href=&quot;http://math.uga.edu/~pete/4400MordellEquation.pdf&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.  (Also see the &quot;tables&quot; on page 9.)&lt;/p&gt;&#xA;" OwnerUserId="299" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2011-01-31T10:20:01.997" LastActivityDate="2011-01-31T10:20:01.997" CommentCount="3" />
  <row Id="19691" PostTypeId="1" AcceptedAnswerId="19727" CreationDate="2011-01-31T09:55:38.557" Score="1" ViewCount="132" Body="&lt;p&gt;Can a polynomial size Context free grammar describe the finite language {$w \pi(w)$ : $\pi(w)$ is fixed string permutation, $|w|=n$ is fixed} over alphabet of {0,1}?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One case this is possible is when $\pi(w)$ is the string reverse permutation (CFG is describing palindromes).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there other choices for $\pi(w)$ with polynomial size CFGs?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Explanation&lt;/strong&gt;: $n$ and $\pi$ &lt;strong&gt;are fixed per language&lt;/strong&gt;. $\pi$ depends on $n$ (the indices of the letters of $\pi(w)$ are a permutation of {1..n} corresponding to indices {1..n} in $w$). How is $\pi$ fixed - two sub-cases:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The answerer can choose $\pi$ (or a class of $\pi$s), having in mind the string reverse case is solved.&lt;/li&gt;&#xA;&lt;li&gt;$\pi$ can vary over all permutations. Again, $\pi$ is fixed per language.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Any references to what languages can be described by polynomial size CFGs (I found only several papers of Asveld, P.R.J. dealing with permutations and circular shifts over large alphabets).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Polynomial size is defined as the size of the Chomsky Normal Form. I included CNF on purpose. In case several answers are given, the size of the CNF will show the ``smaller'' solution.&lt;/p&gt;&#xA;" OwnerUserId="4750" LastEditorUserId="4750" LastEditorDisplayName="" LastEditDate="2011-02-01T07:46:03.200" LastActivityDate="2011-02-01T11:16:29.933" Title="Can a polynomial size CFG describe the finite language \{$w \pi(w)$ : $\pi(w)$ is fixed string permutation, $|w|=n$ is fixed\} over alphabet \{0,1\}?" Tags="&lt;computer-science&gt;&lt;computability&gt;&lt;formal-systems&gt;" AnswerCount="1" CommentCount="12" />
  <row Id="19692" PostTypeId="1" AcceptedAnswerId="19769" CreationDate="2011-01-31T10:07:31.080" Score="3" ViewCount="163" Body="&lt;p&gt;Hello,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm going through Hartshorne's Geometry, and one of the exercises has stumped me for a good few hours. The problem is a version of one of Pappus's theorems:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Let $A$, $B$, $C$, be points on a line $l$, and let $A&amp;#39;$, $B&amp;#39;$, $C&amp;#39;$ be points on a line $m$. Assume $AC&amp;#39;\parallel A&amp;#39;C$ and $B&amp;#39;C\parallel BC&amp;#39;$. Show that $AB&amp;#39;\parallel A&amp;#39;B$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A hint is given to draw a circle through $A$, $B&amp;#39;$, $C&amp;#39;$ meeting $l$ in $D$, and to use cyclic quadrilaterals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've included a picture for clarity:&#xA;&lt;img src=&quot;http://i.imgur.com/Ckdvu.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To the right, I've drawn lines $DB&amp;#39;$ and $DC&amp;#39;$. From the cyclic quadrilateral, $ADB&amp;#39;C&amp;#39;$, I can see that $\angle B&amp;#39;AC&amp;#39;\cong\angle B&amp;#39;DC&amp;#39;$, as they are on the same circumference. I call this angle $\gamma$. Similarly, $\angle C&amp;#39;B&amp;#39;D\cong\angle C&amp;#39;AD$. This allowed me to prove that $\triangle AEC,\triangle CFB,\triangle B&amp;#39;C&amp;#39;E,\triangle C&amp;#39;A&amp;#39;F$ are all similar. I attempted to show $\triangle AEB&amp;#39;$ and $\triangle A&amp;#39;FB$ are similar, but was unsuccessful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My strategy was to show that $\angle BA&amp;#39;C=\gamma$, and this would suffice to show $AB&amp;#39;\parallel BA&amp;#39;$ since $AC&amp;#39;\parallel CA&amp;#39;$, but I haven't quite managed it.  Perhaps someone sees the way to finish off this theorem? Thank you for your help.&lt;/p&gt;&#xA;" OwnerUserId="1583" LastEditorUserId="1583" LastEditorDisplayName="" LastEditDate="2011-01-31T20:17:44.617" LastActivityDate="2011-01-31T22:18:28.150" Title="Euclidean Version of Pappus's theorem" Tags="&lt;euclidean-geometry&gt;" AnswerCount="2" CommentCount="0" />
  <row Id="19693" PostTypeId="2" ParentId="19686" CreationDate="2011-01-31T10:17:30.783" Score="5" ViewCount="" Body="&lt;p&gt;As I mentioned in my comment, the word &quot;onto&quot; is often used as a synonym for the word &quot;surjective&quot;. In the same spirit, you can use &quot;one-to-one&quot; instead of &quot;injective&quot;. See for example &lt;a href=&quot;http://en.wikipedia.org/wiki/Bijection,_injection_and_surjection&quot; rel=&quot;nofollow&quot;&gt;the corresponding Wikipedia article&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;: I agree with the comments by Qiaochu and Jonas that &quot;one-to-one&quot; is a little ambiguous and could refer to a bijection. So it is probably best to stick to the unambiguous terms &quot;injective&quot; and &quot;surjective&quot;.&lt;/p&gt;&#xA;" OwnerUserId="1766" LastEditorUserId="1766" LastEditorDisplayName="" LastEditDate="2011-02-01T08:43:11.357" LastActivityDate="2011-02-01T08:43:11.357" CommentCount="2" />
  <row Id="19694" PostTypeId="1" AcceptedAnswerId="19696" CreationDate="2011-01-31T10:55:51.983" Score="4" ViewCount="67" Body="&lt;p&gt;Consider a large set of points with coordinates that are uniformly distributed within a unit-length segment. Consider a Voronoi diagram built on these points. If we consider only non-infinite cells, what would be (if any) the typical (most frequent) number of edges (that is, neighbors) for those cells? Is there a limit for this number when number of points goes to infinity? Does it have anything common with &lt;code&gt;kissing number&lt;/code&gt;? If so, doest it generalize to higher dimensions, that is, 6 for 2D, 13 for 3D etc?&lt;/p&gt;&#xA;" OwnerUserId="3806" LastActivityDate="2011-03-02T11:39:19.910" Title="What is the most frequent number of edges of Voronoi cells of a large set of random points?" Tags="&lt;topology&gt;&lt;graph&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="19695" PostTypeId="2" ParentId="19667" CreationDate="2011-01-31T11:29:56.563" Score="1" ViewCount="" Body="&lt;p&gt;For starters, the equation should be linear (if you want a term-by-term series solution). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For geometric partial differential equations, separation of variables is often tied intimately to the symmetry properties of the underlying (pseudo-)Riemannian metric. To be more precise, consider the ultrahyperbolic/wave/Laplace equation &#xA;$$ \Box_g \phi = 0 $$&#xA;on some pseudo-Riemannian manifold $(M,g)$. As it turns out, the separation of variables for this equation is closely related to the separability of the Hamilton-Jacobi equations associated to the geodesic flow. For the latter, you need to have enough &quot;conserved quantities&quot; as the degree of freedom. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For an $n$-dimensional manifold, in general you will need to have $n$ independent conserved quantities. For example, you can perform a separation of variable where $M$ is Euclidean space using the rectangular coordinate system because of the conservation of energy-momentum, which gives $n$ conserved scalar quantities. For the linear wave equation, you can also perform a separation of variables in spherical coordinates: for this coordinate system you need the conserved quantities total energy and angular momenta, which adds up to, again, $n$ conserved quantities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(For Hamilton-Jacobi type equations [and similarly ordinary differential equations of dynamical systems] the method of separation of variables [the finding of action-angle coordinates] is fairly well-developed. You should probably consult textbooks in those subjects for treatments on what is known.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While the rule of thumb, as listed above for geometric equations, is to try to find $n$ conserved quantities, in practice, there are also cases where fewer conserved scalar quantities are available, yet a separation of variables can proceed. This is sometimes called &quot;hidden symmetry&quot;. A prime example is the wave equation (and geodesic equations) on the Kerr-Newman space-times of general relativity. The geometry only admits two obvious symmetries: time translation and rotation about the $z$ axis. What saves the day is the rather mysterious &quot;Carter tensor&quot; which gives rise to a higher-order symmetry. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For more details in some of the above, you may want to consult &lt;a href=&quot;http://www.ima.umn.edu/~miller/separability.ps&quot; rel=&quot;nofollow&quot;&gt;this article&lt;/a&gt; and the references therein. (Or &lt;a href=&quot;http://www.jstor.org/stable/54002&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt;.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As in the case of most facts about partial differential equations, there isn't a well developed one-size-fits-all theory. But if you search, say, MathSciNet, for the phrase &quot;separation of variables&quot;, you'll see that it is a field of active research. A large part of the established practice, however, can be summarised as the search for continuous groups of symmetries for solutions of differential equations. A standard reference in that direction is P. Olver's &lt;em&gt;Applications of Lie Groups to Differential Equations&lt;/em&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lastly, one thing I am somewhat fond of whenever talking about separation of variables is the following paper by Eisenhart, &quot;Enumeration of potentials for which one-particle Schrodinger equations are separable,&quot; Phys. Rev. 74, 87-89. &lt;/p&gt;&#xA;" OwnerUserId="1543" LastActivityDate="2011-01-31T11:29:56.563" CommentCount="4" />
  <row Id="19696" PostTypeId="2" ParentId="19694" CreationDate="2011-01-31T11:36:38.960" Score="0" ViewCount="" Body="&lt;p&gt;I think the book &lt;a href=&quot;http://ua.t.u-tokyo.ac.jp/okabelab/Voronoi/&quot; rel=&quot;nofollow&quot;&gt;Spatial tessellations : concepts and applications of Voronoi diagrams&lt;/a&gt; by&#xA;OKABE, BOOTS,and SUGIHARA discusses this.&lt;/p&gt;&#xA;" OwnerUserId="589" LastActivityDate="2011-01-31T11:36:38.960" CommentCount="6" />
  <row Id="19697" PostTypeId="2" ParentId="19681" CreationDate="2011-01-31T11:42:35.693" Score="0" ViewCount="" Body="&lt;p&gt;First note that your notion of $\tau$-convergence imposes to consider only bounded holomorphic functions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $X$ is finite-dimensional, &lt;a href=&quot;http://en.wikipedia.org/wiki/Montel%27s_theorem&quot; rel=&quot;nofollow&quot; title=&quot;Montel's theorem&quot;&gt;Montel's theorem&lt;/a&gt; tells you that $f_n$ $\tau$ converges to $f$ iff $f_n$ converges uniformly to $f$ on any compact subset, and this &quot;notion&quot; is metrizable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $X&amp;#39;$ is separable, the &quot;notion of convergence&quot; is still metrizable (compose $f$ and the $f_n$ with $\Lambda$ for all $\Lambda$ in an enumerable and dense family in $X'$).&lt;/p&gt;&#xA;" OwnerUserId="2660" LastActivityDate="2011-01-31T11:42:35.693" CommentCount="2" />
  <row Id="19698" PostTypeId="1" AcceptedAnswerId="19700" CreationDate="2011-01-31T12:36:58.030" Score="1" ViewCount="63" Body="&lt;p&gt;Hi all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anybody know a simple and differentiable function that converts a 3D vector &lt;code&gt;u = (x, y, z)&lt;/code&gt; to another vector that is orthogonal to &lt;code&gt;u&lt;/code&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To be more precise, I am looking for three differentiable functions &lt;code&gt;{f, g, h}&lt;/code&gt; such that the vector &lt;code&gt;u = (x, y, z)&lt;/code&gt; is orthogonal to &lt;code&gt;v = (f(x,y,z), g(x,y,z), h(x,y,z))&lt;/code&gt; and &lt;code&gt;v&lt;/code&gt; is zero only if &lt;code&gt;u&lt;/code&gt; is zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The functions &lt;code&gt;{f, g, h}&lt;/code&gt; should be as simple as possible. I prefer them linear, but I think no such linear functions exist. Low degree polynomials are also good.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S. I found such functions, but they are not polynomials. For example:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;f(x, y, z) = y*(exp(x) + 3) - z*(exp(x) + 2)&#xA;g(x, y, z) = z*(exp(x) + 1) - x*(exp(x) + 3)&#xA;h(x, y, z) = x*(exp(x) + 2) - y*(exp(x) + 1)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;It's simply the cross product of (x,y,z) with (exp(x)+1, exp(x)+2, exp(x)+3). It satisfies all requirements except for being polynomials. But they are quite simple.&lt;/p&gt;&#xA;" OwnerUserId="5073" LastActivityDate="2011-01-31T13:06:58.990" Title="Differentiable orthogonal 3D vector" Tags="&lt;vector-calculus&gt;&lt;3d&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="19700" PostTypeId="2" ParentId="19698" CreationDate="2011-01-31T13:06:58.990" Score="4" ViewCount="" Body="&lt;p&gt;I don't believe you can find such functions, because if you could you could make a continuous, nowhere 0, vector field on the sphere. See, e.g. &lt;a href=&quot;http://en.wikipedia.org/wiki/Hairy_ball_theorem&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your example doesn't work because if you take x=1, y=(e+2)/(e+1), z=(e+3)/(e+1) then (x,y,z) is parallel to (exp(1)+1, exp(1)+2, exp(1)+3), and so the cross product is 0.&lt;/p&gt;&#xA;" OwnerUserId="4830" LastActivityDate="2011-01-31T13:06:58.990" />
  <row Id="19701" PostTypeId="1" AcceptedAnswerId="19708" CreationDate="2011-01-31T13:26:18.230" Score="1" ViewCount="122" Body="&lt;p&gt;Asymmetric graph is a graph that has only trivial automorphism. Asymptotically, almost all finite graphs are asymmetric.  I'm looking for upper bounds and lower bounds on the growth rate of the number of asymmetric graphs on n nodes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, I'm looking for a function $f(n)$ that exactly counts the number of asymmetric graph on n nodes? Is this function efficiently computable?&lt;/p&gt;&#xA;" OwnerUserId="2459" LastEditorUserId="2459" LastEditorDisplayName="" LastEditDate="2011-02-05T01:11:00.393" LastActivityDate="2011-02-05T01:11:00.393" Title="Counting the number of asymmetric graphs on n nodes?" Tags="&lt;graph-theory&gt;&lt;computer-science&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="19702" PostTypeId="2" ParentId="19659" CreationDate="2011-01-31T14:03:54.613" Score="1" ViewCount="" Body="&lt;p&gt;COMAP (Consortium for Mathematics and Its Applications: &lt;a href=&quot;http://www.comap.com/&quot; rel=&quot;nofollow&quot;&gt;http://www.comap.com/&lt;/a&gt;) which runs MCM has lots of materials dealing with the mathematical tools that often are of use in modeling problems. It also has a collection of modeling problems available at: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.mathmodels.org/&quot; rel=&quot;nofollow&quot;&gt;http://www.mathmodels.org/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1369" LastActivityDate="2011-01-31T14:03:54.613" CommunityOwnedDate="2011-01-31T14:03:54.613" />
  <row Id="19703" PostTypeId="2" ParentId="19677" CreationDate="2011-01-31T14:14:52.393" Score="2" ViewCount="" Body="&lt;p&gt;Hello Mehrdad,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have seen many, many students (and you have probably also seen them yourself) who would answer a definitive 'No' to your first question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course it is obvious to most of us, and if it is obvious to you as well, then fantastic.  As Hans said, you can move on to bigger and better things.  If you are looking for more difficult and deeper results in linear algebra (I notice that you are worried linear algebra is a study in the invariants under change of notation) then perhaps you can look in to eigenvalue problems, and if you have studied some calculus, how they relate to the solutions of ODEs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Linear algebra is one of those fields which crops up all the time.  For example, a standard method of solving a very non-linear system of equations involves 'linearising' the system, whereupon certain nice theorems from linear theory apply, and then returning to the non-linear situation.  While I'm not saying that linear algebra is all one needs to solve linear PDEs (there is a fair bit of analysis after all!), I am trying to say that linear algebra plays a role, and thus attempt to convey the far-reaching nature of linear algebra as a discipline.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I do not know what you are looking for as an answer to this question after the excellent comments you have received already.  But I think that one thing to keep in mind is what Willie already mentioned: mathematics is an ever-changing discipline, where 'good' definitions are those which allow one to prove 'obvious' facts are true.  If you find this hard to believe, I can give one model example, again from PDEs.  A large class of solutions to second order parabolic equations remain positive, if initially so.  However for fourth (and higher) order equations this is not true.  The unsatisfactory point is the &lt;em&gt;extent&lt;/em&gt; to which it is not true.  The negative part of the solution is very 'small' when compared with the magnitude of the positive part.  In certain situations one can make this precise, by speaking about the 'average positivity' or 'eventual positivity'.  Both of these are variations on the definition of 'positive' which allow us to quantify the behaviour we see from these equations.  Before discovery, it was very hard to show this.  (In fact, the opposite was a conjecture of Hadamard.)  But now, working backward from a modern definition of 'almost positive' or 'eventually positive', the proof appears quite simple---almost elementary!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope this answer complements the other comments and instills a little bit of that faith Mariano mentioned ;).&lt;/p&gt;&#xA;" OwnerUserId="4322" LastActivityDate="2011-01-31T14:14:52.393" />
  <row Id="19704" PostTypeId="2" ParentId="19673" CreationDate="2011-01-31T14:22:01.460" Score="0" ViewCount="" Body="&lt;p&gt;I find it helps to consider $x$ as a time variable. So let's write $t$ instead of $x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose you've got a good racing car, and at time $t=0$ you're at the position $y=0$. You're planning to race along a straight road. Your car can accelerate in such a way that after $t$ seconds, the &lt;em&gt;speed&lt;/em&gt; is $2t$ metres per second (that won't work very long ...). Then you can show that after $t$ seconds you'll be $y=t^2$ metres away from where you started.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let's check if that makes sense in terms of a rate of change: Your position at time $t$ is $y=t^2$, and the rate of change is $dy/dt = 2t$. OK, so the rate of change $dy/dt$ is just the &lt;em&gt;current&lt;/em&gt; speed (wheras $y/t = t^2/t = t$ would be the &lt;em&gt;avarage&lt;/em&gt; speed between time 0 and $t$).&lt;/p&gt;&#xA;" OwnerUserId="2575" LastActivityDate="2011-01-31T14:22:01.460" />
  <row Id="19705" PostTypeId="1" CreationDate="2011-01-31T14:46:38.833" Score="2" ViewCount="95" Body="&lt;p&gt;Let $p$ be an odd prime number and $T_p$ is the following set of $2 \times 2$ matrices:&#xA;$$ T_p =  \left\{ A = \begin{bmatrix}&#xD;&#xA; a &amp;amp; b \\ &#xD;&#xA; c &amp;amp; a&#xD;&#xA;\end{bmatrix}\ \Biggm|a,b,c \in \{0,1,2,...(p-1)\} \right\}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;then how to prove that the number $k$ of $A$ in $T_p$ such that the trace of A is not divisible by $p$ but the $det(A)$ is divisible by $p$, is $k=(p-1)^2$ ?&lt;/p&gt;&#xA;" OwnerUserId="2109" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-31T14:53:08.673" LastActivityDate="2011-01-31T16:24:57.657" Title="Counting matrices based on determinant and trace" Tags="&lt;linear-algebra&gt;&lt;combinatorics&gt;&lt;matrices&gt;" AnswerCount="3" CommentCount="6" />
  <row Id="19706" PostTypeId="2" ParentId="19705" CreationDate="2011-01-31T15:00:54.283" Score="0" ViewCount="" Body="&lt;p&gt;This essentially amounts to counting the cardinality of the following set: $T = \{a,b,c \in \mathbb{Z}_p: p \not | 2a \ \text{but} \ p|(a^2-bc) \ \text{for} \ a,b,c \in \mathbb{Z}_p \}$.&lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2011-01-31T15:00:54.283" CommentCount="3" />
  <row Id="19707" PostTypeId="2" ParentId="19705" CreationDate="2011-01-31T15:10:40.187" Score="3" ViewCount="" Body="&lt;p&gt;There are p-1 choices for $a$ (not $0$).  Given $a$, the determinant is $a^2-bc$ and since $\mathbb{Z}_p$ is a field you can solve $a^2-bc=0$ for $c$ unless $b=0$, so $p-1$ choices for $b$.  That makes $(p-1)^2$ total.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2011-01-31T15:24:52.623" LastActivityDate="2011-01-31T15:24:52.623" CommentCount="2" />
  <row Id="19708" PostTypeId="2" ParentId="19701" CreationDate="2011-01-31T15:17:19.930" Score="1" ViewCount="" Body="&lt;p&gt;You might look at the references in &lt;a href=&quot;http://oeis.org/A003400&quot; rel=&quot;nofollow&quot;&gt;A003400&lt;/a&gt; for possibly disconnected graphs and &lt;a href=&quot;http://oeis.org/A124059&quot; rel=&quot;nofollow&quot;&gt;A124059&lt;/a&gt; for connected ones.  Neither gives a formula, even asymptotic.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-01-31T15:17:19.930" CommentCount="3" />
  <row Id="19709" PostTypeId="1" AcceptedAnswerId="19711" CreationDate="2011-01-31T15:42:03.147" Score="0" ViewCount="30" Body="&lt;p&gt;Let G(S) denote the sum of the elements of set S and F(n) be the sum of G(s) for all subsets of the set consisting of the first n natural numbers. For example, F(3) = (1) + (2) + (3) + (1 + 2) + (1 + 3) + (2 + 3) + (1 + 2 + 3) = 24. Given n, calculate F(1) + F(2) + ... + F(n).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Input:&lt;br&gt;&#xA;1&lt;br&gt;&#xA;2&lt;br&gt;&#xA;3&lt;br&gt;&#xA;Output:&lt;br&gt;&#xA;1&lt;br&gt;&#xA;7&lt;br&gt;&#xA;31  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do i go about solving for an equation for sum of F(n) n = 1 to n.&lt;br&gt;&#xA;I know the answer - but no idea how to arrive at it&lt;/p&gt;&#xA;" OwnerUserId="6442" LastEditorUserId="6442" LastEditorDisplayName="" LastEditDate="2011-01-31T15:51:17.623" LastActivityDate="2011-01-31T15:51:48.077" Title="Equation for Summing Subsets" Tags="&lt;sequences-and-series&gt;" AnswerCount="1" FavoriteCount="0" />
  <row Id="19710" PostTypeId="2" ParentId="19659" CreationDate="2011-01-31T15:46:44.260" Score="2" ViewCount="" Body="&lt;p&gt;In my personal experience(Meritorious both years I participated), one of the more obvious but important aspects is overlooked by many teams:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Delegate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Our team began both competitions by selecting a &quot;Coder, Writer, and Researcher&quot;. This should play off your strengths, but essentially, the code was substantial enough that our coder did only that. The Researcher spent the first half of the competition finding out what had been done and communicating with the Writer about the introduction and direction of the report. The writer spent the first half of the competition helping write the algorithm with the Coder, and working out the mathematical models to be used. Also the Writer attempted to synthesize the work of the Researcher into the start of the report. The second half of the project was a bit more hectic, the researcher obtained as much information as possible that might be available and useful. This was particularly important for important real world data into our models. The writer spend the second half... writing. Literally just writing non-stop.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This worked fairly well for us. I also recommend spending the first 1-3 hours letting the members think individually about the problem, then spending about 30-60 minutes explaining to one another their attack. Sometimes people aren't forthcoming with really great ideas. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another recommendation is SLEEP! The first year we hardly slept at all for the entire competition. We drank insane amounts of caffeine and worked through the weekend. The second competition, we slept about 6 hours each night. We got the same ranking both times, but we did better the second year. Looking back, the quality of the writing was better. The code had less bugs, and we managed to produce more mathematics. Don't underestimate the power of getting a good nights rest.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another general reference is &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/093570258X&quot; rel=&quot;nofollow&quot;&gt;consider a spherical cow&lt;/a&gt; which is a fairly standard reference. It is fun to read, and somewhat enlightening. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps I will think of more later, but for now this is all.&lt;/p&gt;&#xA;" OwnerUserId="16" LastActivityDate="2011-01-31T15:46:44.260" CommunityOwnedDate="2011-01-31T15:46:44.260" CommentCount="1" />
  <row Id="19711" PostTypeId="2" ParentId="19709" CreationDate="2011-01-31T15:51:48.077" Score="0" ViewCount="" Body="&lt;p&gt;Each element of $\{1,2,3, \ldots,n\}$ is part of $2^{(n-1)}$ subsets.  So you are looking for $$F(n)=\sum_{i=1}^n i2^{(n-1)}=2^{(n-1)}\frac{n(n+1)}{2}=2^{n-2}n(n+1)$$  Then you need to sum this over $n$ for your final answer.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-01-31T15:51:48.077" CommentCount="5" />
  <row Id="19712" PostTypeId="1" AcceptedAnswerId="19723" CreationDate="2011-01-31T15:53:42.473" Score="9" ViewCount="199" Body="&lt;p&gt;One way to get the famous double cover $\text{SU}(2) \to \text{SO}(3)$ is to note that $\text{SU}(2)$ is isomorphic to the group of unit quaternions and to let unit quaternions $q$ act on the subspace $V$ of $\mathbb{H}$ spanned by $i, j, k$ via conjugation $t \mapsto qtq^{-1}$; this preserves the norm.  (Alternately, this is the adjoint action on the Lie algebra, which preserves the Killing form.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another way to do this is to let $\text{SU}(2)$ act on $\mathbb{P}^1(\mathbb{C})$, which is diffeomorphic to the sphere.  This gives an action of $\text{SU}(2)$ by conformal automorphisms.  However, I don't know how to prove that $\text{SU}(2)$ actually acts by &lt;em&gt;rotations&lt;/em&gt; (at least, not without some explicit and unenlightening calculations).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To be more precise, if we fix an inner product on $\mathbb{C}^2$, then the space of lines in $\mathbb{C}^2$ can be given the &lt;a href=&quot;http://en.wikipedia.org/wiki/Fubini%E2%80%93Study_metric&quot; rel=&quot;nofollow&quot;&gt;Fubini-Study metric&lt;/a&gt;, which $\text{SU}(2)$ preserves.  But how can I prove that the Fubini-Study metric agrees with the natural metric on the sphere (up to a constant)?&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2011-01-31T20:31:08.490" LastActivityDate="2011-01-31T20:42:35.860" Title="The action of SU(2) on the Riemann sphere" Tags="&lt;complex-analysis&gt;&lt;lie-groups&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" />
  <row Id="19713" PostTypeId="1" CreationDate="2011-01-31T15:58:23.510" Score="0" ViewCount="52" Body="&lt;p&gt;For linear fractional transformations, it suffices to specify where three points are mapped to. I am wondering if some analogue of this holds for general conformal mappings.&lt;/p&gt;&#xA;" OwnerUserId="6444" LastActivityDate="2011-01-31T16:24:27.710" Title="Are conformal maps determined by a finite number of points?" Tags="&lt;complex-analysis&gt;" AnswerCount="2" />
  <row Id="19714" PostTypeId="2" ParentId="19713" CreationDate="2011-01-31T16:14:13.123" Score="2" ViewCount="" Body="&lt;p&gt;No. For any finite collection of points $\zeta_1$, $\zeta_2$, ..., $\zeta_N$ in the disc of radius  $1$, consider the map &#xA;$$z \mapsto z+\epsilon \prod(z-\zeta_i).$$&#xA;This is a holomorphic map fixing all the $\zeta_i$ and, if $\epsilon$ is small enough, it is conformal.&lt;/p&gt;&#xA;" OwnerUserId="448" LastActivityDate="2011-01-31T16:14:13.123" />
  <row Id="19715" PostTypeId="2" ParentId="19705" CreationDate="2011-01-31T16:15:31.317" Score="0" ViewCount="" Body="&lt;p&gt;Here's a different point of view that leads to the same answer as Ross's.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Think of the matrices as being matrices over the field of $p$ elements by reducing the entries modulo $p$. The determinant is divisible by $p$ (that is, $0$ modulo $p$) if and only if the rows of the matrix do not form a basis for $\mathbf{F}_p^2$, if ando nly if the rows are not linearly independent, if and only if one row is a multiple of the other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To get trace that is not zero, you just need to avoid $a=0$, since $2a\equiv 0\pmod{p}$ if and only if $a\equiv 0 \pmod{p}$ (since $p$ is odd). Having chosen $a$ and an arbitrary $b$, the only way for the determinant to be a multiple of $p$ is if $(c,a) = k(a,b)$ for some $k\in\mathbf{F}_p$ (operation done modulo $p$). If $b=0$, then no choice of $c$ will do, since $a\neq 0$. So $b\neq 0$. If $b\neq 0$, then the only $k$ that can possibly work is $k\equiv ba^{-1}\pmod{p}$, which forces the value of $c$. So you have one and only one such matrix for each nonzero choice of $b$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In summary: $p-1$ choices for $a$; $p-1$ choices for $b$; once $a$ and $b$ are fixed (both necessarily nonzero), $c$ is forced by the condition that $(c,a)$ must be a scalar multiple of $(a,b)$. &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-31T16:24:57.657" LastActivityDate="2011-01-31T16:24:57.657" />
  <row Id="19716" PostTypeId="1" AcceptedAnswerId="19717" CreationDate="2011-01-31T16:18:34.897" Score="7" ViewCount="259" Body="&lt;p&gt;Hi all, I am attempting to self-study proof techniques and your criticism of my following proof would be greatly appreciated. Feel free to nitpick minor/trivial things; that's how I'll learn!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit: I have appended a revised proof with the criticisms received so that other rookies can learn from my progress too.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit 2: And another, third time lucky?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Theorem: Let $n$ be an integer. If $3n$ is odd then $n$ is odd.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $P$ be the sentence &quot;$3n$ is odd&quot; and $Q$ be the sentence &quot;$n$ is odd.&quot; We therefore have;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$P \rightarrow Q$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the law of the contrapositive, we may obtain;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\lnot Q \rightarrow \lnot P$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which is translates to &quot;If $n$ is not odd then $3n$ is not odd&quot;, or put another way; &quot;If $n$ is even then $3n$ is even.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If an integer $n$ is even then there exists some integer $m$ such that;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$n = 2m$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By multiplying this by $3$ we may obtain;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$3n = 6m \equiv n = 2m$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It has therefore been shown that if $n$ is even so is $3n$ and that this is equivalent to showing that if $3n$ is odd then so is $n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;■&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Attempt 2, taking into consideration previous criticism&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Theorem: Let $n$ be an integer. If $3n$ is odd then $n$ is odd.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $P$ be the sentence &quot;$3n$ is odd&quot; and $Q$ be the sentence &quot;$n$ is odd.&quot; We want to show that $$P \rightarrow Q$$ and that by taking contrapositives this is equivalent to showing $$\lnot Q \rightarrow \lnot P$$ which translates to;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;If $n$ is not odd then $3n$ is not odd&quot;, or put another way; &quot;If $n$ is even then $3n$ is even.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If an integer $n$ is even then there exists some integer $m$ such that;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$n = 2m$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By multiplying this by $3$ we may obtain;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$3n = 6m$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This must still be even as an even integer multiplied by an odd integer produces an even one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We must now show that $3n$ is even. If this is so, then $6m = 2k$ for some integer $k$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As $3n=6m=2k$ we have;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$3n=2k$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It has therefore been shown that if $n$ is even so is $3n$ and that this is equivalent to showing that if $3n$ is odd then so is $n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;■&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Attempt 3&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Theorem: Let $n$ be an integer. If $3n$ is odd then $n$ is odd.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $P$ be the sentence &quot;$3n$ is odd&quot; and $Q$ be the sentence &quot;$n$ is odd.&quot; We want to show that $$P \rightarrow Q$$ and that by taking contrapositives this is equivalent to showing $$\lnot Q \rightarrow \lnot P$$ which translates to;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;If $n$ is not odd then $3n$ is not odd&quot;, or put another way; &quot;If $n$ is even then $3n$ is even.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If an integer $n$ is even then there exists some integer $m$ such that;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$n = 2m$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By multiplying this by $3$ we may obtain;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$3n = 6m$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which can be rewritten as;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$3n = 2(3m)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus we have shown that $3n$ is even, as it is equal to $2(3m)$ which, as an integer multiplied by 2, must be even.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It has therefore been shown that if $n$ is even so is $3n$ and that this is equivalent to showing that if $3n$ is odd then so is $n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;■&lt;/p&gt;&#xA;" OwnerUserId="6409" LastEditorUserId="1543" LastEditorDisplayName="" LastEditDate="2011-05-03T10:04:51.620" LastActivityDate="2011-05-03T10:04:51.620" Title="How can my proof be improved? &quot;Let $n$ be an integer. If $3n$ is odd then so is $n$.&quot;" Tags="&lt;logic&gt;&lt;elementary-number-theory&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="19717" PostTypeId="2" ParentId="19716" CreationDate="2011-01-31T16:23:15.210" Score="8" ViewCount="" Body="&lt;p&gt;The very end of the proof is not very clear. First, there is no reason to go from $3n=6m$ back to $n=2m$. Instead, in order to show that $3n$ is even, you must show that it can be written as $2k$ for &lt;em&gt;some&lt;/em&gt; integer $k$. So what you want is to take $3n=6m$, and express $6m$ as $2k$ for some $k$. &lt;em&gt;That&lt;/em&gt; will prove that $3n$ is even (which you haven't as yet, at least not explicitly). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know what kind of course this is for; if it is for an &quot;introduction to proofs&quot; course, chances are you are required to be very explicit about what you are doing, so the opening of your argument is likely required and good. Later, you should be free to summarize the entire first three paragraphs of your proof into: &lt;em&gt;&quot;We proceed by contrapositive: suppose $n$ is even. Then...&quot;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;hr/&gt; &#xA;&lt;em&gt;Comment on the addition:&lt;/em&gt; The claim that &lt;em&gt;&quot;this must be even because an even integer multiplied by an odd integer produces an even one&quot;&lt;/em&gt; relies on a statement that, for all I know, has not been proven, and which is far more general than is needed here. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Simply write $3n$ as &lt;code&gt;2 x &amp;lt;some integer&amp;gt;&lt;/code&gt; &lt;em&gt;explicitly&lt;/em&gt;. Given that you know that $3n$ is equal to $6m$, it should be a trivial matter to express it that way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second: it's not the equation that &quot;must be even&quot;. Equations (like statements) are neither even nor odd. It is &lt;em&gt;integers&lt;/em&gt; that are even or odd (in this context); so what is the &lt;em&gt;&quot;This&quot;&lt;/em&gt; it refers to? If &quot;This&quot; refers to $3n$, then the next line is unnecessary, but then this sentence amounts to asserting what you are trying to prove (instead of proving it). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Third: Given that you say you have to show that $3n$ is even, you &lt;strong&gt;cannot&lt;/strong&gt; simply say &lt;em&gt;&quot;If this is so...&quot;&lt;/em&gt; and proceed. That would mean that you are assuming what you want to prove in order to prove it, which of course is a circular argument. Of course you have concluded that $3n$ is even in that paragraph: you began by &lt;em&gt;assuming&lt;/em&gt; it's even. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think that you have convinced yourself that there is a clever trick or some deep thing that needs to go into showing that $3n$ is even. There isn't. There is a simple, nay, trivial algebraic manipulation that you need to do from your equation $3n=6m$ that will show that $3n$ is indeed even, and that is all you need to do.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-31T20:35:45.413" LastActivityDate="2011-01-31T20:35:45.413" CommentCount="7" />
  <row Id="19718" PostTypeId="2" ParentId="19713" CreationDate="2011-01-31T16:24:27.710" Score="1" ViewCount="" Body="&lt;p&gt;This depends on the group of conformal automorphisms that your region admits. E.g., the Riemann sphere admits a triply transitive group of conformal automorphisms, so you can prescribe the images of three arbitrary chosen points. If your region is the unit disc $D$ then the group of conformal automorphisms has &quot;three real degrees of freedom&quot;. This means that you can prescribe, e.g., $f(0)\in D$ and $\arg(f&amp;#39;(0))$, but not $f(0)$ and $f({1\over2})$ independently. &lt;/p&gt;&#xA;" OwnerUserId="1303" LastActivityDate="2011-01-31T16:24:27.710" />
  <row Id="19719" PostTypeId="2" ParentId="19716" CreationDate="2011-01-31T16:34:53.670" Score="5" ViewCount="" Body="&lt;p&gt;Your argument is structured as if you were arguing from your conclusion.  The correct way to state this argument is &quot;we want to show that $P \Rightarrow Q$, and by taking contrapositives this is &lt;em&gt;equivalent&lt;/em&gt; (you also did not say this) to showing that $\neg Q \Rightarrow \neg P$.&quot;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I guess you are using $\equiv$ to mean &quot;is equivalent to,&quot; but this is usually reserved for an equivalence relation on mathematical &lt;em&gt;objects&lt;/em&gt;, not on mathematical &lt;em&gt;statements.&lt;/em&gt;  For statements you should use $\Leftrightarrow$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And, of course, Arturo's comments are spot-on.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2011-01-31T16:34:53.670" CommentCount="1" />
  <row Id="19720" PostTypeId="1" AcceptedAnswerId="19746" CreationDate="2011-01-31T16:47:58.327" Score="2" ViewCount="126" Body="&lt;p&gt;I have a set $S$ of $n$ elements. I select $r$ elements from the set $E_1, E_2, E_3, \ldots, E_r$ that may not be all distinct. I select again $r$ elements from the set $F_1,F_2,F_3,\ldots, F_r$ in a similar manner.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the probability $P$ that $E_i = F_j$ for some $i$ and $j$? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have to prove that this probability is at least one half of the probability that $2r$ randomly chosen elements selected from $S$ are not all distinct.&lt;/p&gt;&#xA;" OwnerUserId="6445" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2011-01-31T18:39:27.590" LastActivityDate="2011-01-31T18:52:07.937" Title="variation of Birthday paradox" Tags="&lt;probability&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="19722" PostTypeId="2" ParentId="12674" CreationDate="2011-01-31T16:52:55.837" Score="0" ViewCount="" Body="&lt;p&gt;Take a look at dynamic time warping. I think it's just the solution you need. I've used the R package 'dtw' which is described here. &lt;a href=&quot;http://cran.r-project.org/web/packages/dtw/dtw.pdf&quot; rel=&quot;nofollow&quot;&gt;http://cran.r-project.org/web/packages/dtw/dtw.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="6446" LastActivityDate="2011-01-31T16:52:55.837" />
  <row Id="19723" PostTypeId="2" ParentId="19712" CreationDate="2011-01-31T17:06:47.993" Score="5" ViewCount="" Body="&lt;p&gt;With the F-S metric, $\mathbf P^1(\mathbb C)$ is a Riemannian surface upon which a group acts transitively. That implies that the curvature is constant. Now, the classification of space forms, for example, shows that such a thing is covered locally isometrically by $S^2$, the round sphere, $E^2$, the flat plane, or $H^2$, the hyperbolic plane. Since $\mathbf P^1(\mathbb C)$ is simply connected and compact, its only covering is the identity, and since it is compact, it must be a round sphere.&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2011-01-31T17:06:47.993" CommentCount="8" />
  <row Id="19724" PostTypeId="1" AcceptedAnswerId="19728" CreationDate="2011-01-31T17:12:28.873" Score="2" ViewCount="75" Body="&lt;p&gt;Question: Given is the following pipe system:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/eCm5r.jpg&quot; alt=&quot;pipediagram&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the pressure in a stationary state is at every end and cross-point (with exception of $K_i, i ={1,...,4}$) equal to the average pressure of the neighboring points. On $K_i$ with $i = \{1,...,4\}$ are valves with fixed pressure of $K_i = 10i$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a) Create a system of linear equations to represent the values of the pressure on the points.&#xA;b) Turn this system of linear equations into a matrix-vector representation.&#xA;c) Solve this system of linear equations with matlab.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm having a lot of difficulty for some reason. I'm not sure if I'm on the right track by numbering the points from top left to bottom right as one does a matrix ($a_{ij}$). I believe the resulting equations would be somthing like:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a_{11} = \frac{1}{3}(40 + a_{12}+a_{21})$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a_{21} = \frac{1}{3}(a_{11} + a_{22} +a_{31})$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a_{31} = \frac{1}{3}(a_{21} + a_{32} + a_{41})$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a_{41} = \frac{1}{3}(10 +a_{31} +a_{42})$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and the 6 points in the middle would be of the form:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a_{22} = \frac{1}{4}(a_{12} + a_{21} + a_{23} + a_{32})$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't see how I can put this into a matrix-vector form. Everytime that I try to use substitution(say $a_{11}$ into the equation for $a_{21}$ I am not sure which direction to go and I am thinking that I must be misunderstanding something.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At first I was visualising such a matrix equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\left ( \begin{array}{ccccc} a_{11} &amp;amp; a_{12} &amp;amp;a_{13} &amp;amp;a_{14} &amp;amp;a_{15} \\a_{21} &amp;amp; a_{22} &amp;amp; a_{23} &amp;amp; a_{24} &amp;amp; a_{25} \\ a_{31}&amp;amp; a_{32}&amp;amp; a_{33}&amp;amp; a_{34}&amp;amp; a_{35} \\ a_{41}&amp;amp; a_{42}&amp;amp; a_{43}&amp;amp; a_{44}&amp;amp; a_{45}\end{array}\right ) \left( \begin{array}{c}x_1\\ x_2 \\ x_3 \\ x_4 \\ x_5 \end{array}\right ) = \left( \begin{array}{c} 40 \\ 30 \\ 20 \\ 10\end{array} \right)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;however, I think this must be nonsense, there are 5 unknowns and only 4 'equations'. I am assuming though, that I will be using the matlab &quot;linsolve(A,b)&quot; command, right? I am not even sure at this point what the $A$ and $b$ will be in this simple equation... If anyone could help me start getting this straight I would really appreciate it!&lt;/p&gt;&#xA;" OwnerUserId="3711" LastActivityDate="2011-01-31T17:28:16.370" Title="How do you convert this diagram into a linear system of equations/matrix form?" Tags="&lt;homework&gt;&lt;puzzle&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="19726" PostTypeId="2" ParentId="19657" CreationDate="2011-01-31T17:25:50.560" Score="0" ViewCount="" Body="&lt;p&gt;Here is an other useful approach. By the chain rule&#xA;$$f(g(x))&amp;#39;= f&amp;#39;(g(x))g&amp;#39;(x)$$&#xA;and hence, $$\frac{d}{dx}\log(f(x))=\frac{f&amp;#39;(x)}{f(x)}$$&#xA;(this is called the logarithmic derivative of $f$).&#xA;In your case put $f(x)=\log(6x)$ to get&#xA;$$\frac{d}{dx}\log(f(x))=\frac{f&amp;#39;(x)}{f(x)}=\frac{1/x}{\log(6x)}=\frac{1}{x\log(6x)}.$$&lt;/p&gt;&#xA;" OwnerUserId="1154" LastActivityDate="2011-01-31T17:25:50.560" />
  <row Id="19727" PostTypeId="2" ParentId="19691" CreationDate="2011-01-31T17:28:14.780" Score="2" ViewCount="" Body="&lt;p&gt;I think a more precise question would be :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given a permutation $\pi$ of {1...n}, what is the size of a grammar describing $L_\pi = \{w\pi (w), |w|=n\}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, there is not a lot of ways to build such a grammar : if you use a symbol that can develop into different words, i.e. that can put either a 0 or a 1 in a certain position of the final word, then that symbol has to take care simultaneously of the corresponding position in the other half of the word.&#xA;This means that the only way to have a nontrivial rule is when the permutation $\pi$ has a property &#xA;$$\pi(\{n-k+1...n\}) = \{1...k\}$$&#xA;If this is the case we can split the permutation into two components $\sigma$ and $\tau$ such that :&#xA;$$\forall i \in \{1..k\}, \pi(n-k+i) = \sigma(i) \quad \text{and} \quad \forall j \in \{1...n-k\}, \pi(j) = k+\tau(j)$$&#xA;Then, we can use the fact that $L_\pi = \bigcup_{|w|=n-k} w L_\sigma \tau(w)$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, with $rev_n$, the reverse permutation on {1...n}, we do have $rev_n(\{2...n\}) = \{1...n-1\}$, thus we get the equality $L_{rev_n} = 0L_{rev_{n-1}}0 + 1L_{rev_{n-1}}1$, which translates into a linear sized grammar for $L_{rev_n}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the general case, you get exponentially bigger grammar rules as you have to decide on a greater number of letters of $w$ at the same time. The worst case (which happens pretty often) is when you can't write any non-trivial rule, and then have to enumerate all the words of $L_\pi$, for example if $\pi = id_n$. The better cases are the permutation that are close to the reverse permutation for example if n is even and $\pi(i) = (n-i+1+(-1)^i))$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit : Let me give a more precise estimation of the size of such a grammar.&#xA;Define S to be the subset of {0...n} consisting of those integers k such that $\pi(\{n-k+1...n\}) = \{1...k\}$. S = {k1=0, k2, ..., kp = n} (0 and n give trivial decompositions and are always in this set).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then the shortest grammar uses (p-1) symbols S_0 ... S_{p-1}, and each rule is of the described format, where to expand Si we have to choose $\delta_i = k_{i+1}-k_i$ letters before expanding the next symbol $S_{i+1}$. Thus the &quot;size&quot; of the grammar is about&#xA;$\Sigma_{i=0}^{p-1} \delta_i 2^{\delta_i}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you choose a family of partitions that keep the gaps $\delta_i$ small (for example bounded), you get a linear-sized grammar (there are a linear number of symbols and each rule is about the same size). If you allow them to grow slowly (for example around log(n)), you can still get a polynomial-sized grammar. Since it's not clear what kind of families of permutations you're thinking about, I can't get more precise than that.&lt;/p&gt;&#xA;" OwnerUserId="6270" LastEditorUserId="6270" LastEditorDisplayName="" LastEditDate="2011-02-01T11:16:29.933" LastActivityDate="2011-02-01T11:16:29.933" CommentCount="8" />
  <row Id="19728" PostTypeId="2" ParentId="19724" CreationDate="2011-01-31T17:28:16.370" Score="1" ViewCount="" Body="&lt;p&gt;You are close, but not quite visualising the matrix correctly. The five equations you write should be translated into usual form, as follows:&#xA;\begin{align*}&#xA;a_{11} -\frac{1}{3}a_{12} -\frac{1}{3}a_{21} &amp;amp;= \frac{40}{3}\\&#xA;a_{21} -\frac{1}{3}a_{11} -\frac{1}{3}a_{22} -\frac{1}{3}a_{31} &amp;amp;=0\\&#xA;a_{31} -\frac{1}{3}a_{21} -\frac{1}{3}a_{32} -\frac{1}{3}a_{41} &amp;amp;= 0\\&#xA;a_{41} -\frac{1}{3}a_{31} -\frac{1}{3}a_{42} &amp;amp;=\frac{10}{3}\\&#xA;a_{22} -\frac{1}{4}a_{21} -\frac{1}{4}a_{21} -\frac{1}{4}a_{23} -\frac{1}{4}a_{32} &amp;amp;=0&#xA;\end{align*}&#xA;You get one equation for every node, so you should have a system with 20 equations and 20 unknowns (the unknowns being the pressure on each node). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is, the unknowns are $a_{11},\ldots,a_{45}$; these are &lt;em&gt;not&lt;/em&gt; the coefficients of the matrix of the system. &lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2011-01-31T17:28:16.370" CommentCount="2" />
  <row Id="19729" PostTypeId="1" AcceptedAnswerId="19732" CreationDate="2011-01-31T17:39:57.913" Score="1" ViewCount="93" Body="&lt;p&gt;At the moment, I can't find a similar example in my textbook or in my class notes, and I have looked for one and contemplated this problem for the past hour or so. Can anyone give me an idea of what to do to solve this problem?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A constant force F = -3i -2j -1k acts upon an object moving along a straight line from point (-8,0,-10) to point (-9,9,9). Find the work done if the distance is measured in meters and the magnitude of the force is measured in newtons.&lt;/p&gt;&#xA;" OwnerUserId="3800" LastActivityDate="2011-01-31T17:50:45.573" Title="Homework Help - Calculus III / Physics / Force-Work Problem" Tags="&lt;calculus&gt;&lt;physics&gt;" AnswerCount="2" />
  <row Id="19731" PostTypeId="2" ParentId="19729" CreationDate="2011-01-31T17:50:39.410" Score="0" ViewCount="" Body="&lt;p&gt;Work is the dot product of the force and the displacement.  You have the force as a vector.  What is the displacement?&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-01-31T17:50:39.410" CommentCount="3" />
  <row Id="19732" PostTypeId="2" ParentId="19729" CreationDate="2011-01-31T17:50:45.573" Score="1" ViewCount="" Body="&lt;p&gt;The work is defined to be $W = \vec{F} \cdot \vec{d}$ (look here: &lt;a href=&quot;http://en.wikipedia.org/wiki/Work_(physics&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Work_(physics&lt;/a&gt;) )&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I assume in your lecture notes the unit vectors $\vec{e}_1 = (1, 0, 0), \vec{e}_2 = (0, 1, 0), \vec{e}_3 = (0, 0, 1)$ are denored $i,j,k$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then $\vec{F} = (-3, -2, -1)$  and $\vec{d} = (-9,9,9)-(-8,0,-10)$. Now all you have to do to finish this homework question is to compute the scalar product $\vec{F} \cdot \vec{d}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me know if you can do this question now or if you need more help.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the way: if you are satisfied with an answer you should click on the tick to the left of the answer to accept it. If you don't do that, people here will eventually stop answering your questions.&lt;/p&gt;&#xA;" OwnerUserId="5798" LastActivityDate="2011-01-31T17:50:45.573" CommentCount="2" />
  <row Id="19733" PostTypeId="1" AcceptedAnswerId="19736" CreationDate="2011-01-31T17:58:26.790" Score="5" ViewCount="238" Body="&lt;p&gt;Let A be a noetherian ring. How can I show that Spec(A) is noetherian?&#xA;Also, is there a way to show this by showing directly that the closed sets in Spec(A) satisfy the descending chain condition?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(This is exercise 6.8 from Atiyah and Macdonald.)&lt;/p&gt;&#xA;" OwnerUserId="766" LastActivityDate="2011-01-31T18:08:11.877" Title="If A is noetherian, then Spec(A) is noetherian" Tags="&lt;homework&gt;&lt;abstract-algebra&gt;&lt;commutative-algebra&gt;&lt;ring-theory&gt;" CommentCount="2" />
  <row Id="19734" PostTypeId="1" AcceptedAnswerId="19737" CreationDate="2011-01-31T17:59:17.810" Score="2" ViewCount="122" Body="&lt;p&gt;Hello,&#xA;I have some questions as to some good combinatorial interpretations for the sum of elementary symmetric polynomials. I know that for example, for n =2 we have that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$e_0  = 1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$e_1 = x_1+x_2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$e_2 = x_1x_2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And each of these can clearly been seen as the coefficient of $t^k$ in $(1+x_1t)(1+x_2t)$. Now, in general, what combinatorial interpreations are there for say:&#xA;$\sum_{i=0}^n e_i(x)$ for some $x = (x_1,...,x_n)$ ? &lt;/p&gt;&#xA;" OwnerUserId="2394" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2011-01-31T18:09:17.927" LastActivityDate="2011-01-31T19:44:40.560" Title="Combinatorial interpretations of elementary symmetric polynomials?" Tags="&lt;combinatorics&gt;" AnswerCount="4" CommentCount="3" FavoriteCount="1" />
  <row Id="19735" PostTypeId="2" ParentId="19734" CreationDate="2011-01-31T18:04:59.793" Score="2" ViewCount="" Body="&lt;p&gt;You nearly wrote the solution yourself: the sum of $e_i(x)$ for every $i$ is the sum of all the coefficients of the powers of $t$ in $(1+tx_1)\cdots(1+tx_n)$, that is, the value of this polynomial function at $t=1$, that is, $(1+x_1)\cdots(1+x_n)$. (I dunno if this counts as a &lt;em&gt;combinatorial&lt;/em&gt; interpretation.)&lt;/p&gt;&#xA;" OwnerUserId="6179" LastActivityDate="2011-01-31T18:04:59.793" />
  <row Id="19737" PostTypeId="2" ParentId="19734" CreationDate="2011-01-31T18:16:36.190" Score="4" ViewCount="" Body="&lt;p&gt;$e_k(x_1,\dots,x_n)$ is the number of ways of picking a set of $k$ things from a set of $x_1+\cdots+x_n$ things, of which $x_1$ are of color $c_1$, $x_2$ are of color $c_2$, &amp;amp;c, in such a way that colors are not repeated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore $\sum_{k=0}^ne_k(x_1,\dots,x_n)$ is the number of ways of picking some elements out of those $x_1+\cdots+x_n$ things in such a way that their colors are different.&lt;/p&gt;&#xA;" OwnerUserId="274" LastEditorUserId="274" LastEditorDisplayName="" LastEditDate="2011-01-31T18:37:56.787" LastActivityDate="2011-01-31T18:37:56.787" CommentCount="3" />
  <row Id="19738" PostTypeId="1" CreationDate="2011-01-31T18:17:32.970" Score="1" ViewCount="199" Body="&lt;p&gt;I'm in a similar scenario as with my other question. I have tried to answer the problem correctly, and I have found similar (and even an identical) problem in my textbook, yet I still can't seem to yield the correct answer with this specific problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The example problem in my textbook:&#xA;A crate is hauled 8m up a ramp under a constant force of 200 N applied at an angle of 25 degrees to the ramp. Find the work done. W = F * D cos(25) = (200)(8)cos(25) = 1450J.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone give me any advice on how to solve it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A woman exerts a horizontal force of 4 pounds on a box as she pushes it up a ramp that is 5 feet long and inclined at an angle of 30 degrees above the horizontal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Find the work done on the box.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would think that the work would simply be (4)(5)cos(30), but that is not the correct answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I've figured out how to accept answers now, so I'll accept the most helpful answer.)&lt;/p&gt;&#xA;" OwnerUserId="3800" LastActivityDate="2011-05-14T08:57:28.493" Title="Homework Help - Calculus III / Physics / Work-Force Problem #2" Tags="&lt;calculus&gt;&lt;homework&gt;&lt;physics&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="1" />
  <row Id="19739" PostTypeId="2" ParentId="19187" CreationDate="2011-01-31T18:23:08.890" Score="0" ViewCount="" Body="&lt;p&gt;Ask your professor to give you change for a $10 bill. Ask for that change to include a couple of thousand dollar bills. :)&lt;/p&gt;&#xA;" OwnerUserId="5613" LastActivityDate="2011-01-31T18:23:08.890" />
  <row Id="19740" PostTypeId="1" CreationDate="2011-01-31T18:24:33.340" Score="8" ViewCount="221" Body="&lt;p&gt;How can we compute the value of $\mathrm{gcd}(2^n-1,n!)$ efficiently where $n$ is very large? I couldn't think of any fast and efficient method.&lt;/p&gt;&#xA;" OwnerUserId="149" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2011-01-31T18:33:53.037" LastActivityDate="2011-03-28T08:38:27.380" Title="Efficient method to compute $\mathrm{gcd}(2^n-1,n!)$" Tags="&lt;number-theory&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="19741" PostTypeId="1" AcceptedAnswerId="19747" CreationDate="2011-01-31T18:30:05.167" Score="1" ViewCount="60" Body="&lt;p&gt;This is an equality that I am gleaning out of some papers that I have been reading. I am not sure I am reading it right. Hopefully people will correct it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $U$ be a group element and $R$ be a representation of the group. Let $sym^n (R)$ be the $n^{th}$ symmetric power of the representation and similarly $anti^n (R)$ is the $n^{th}$ anti-symmetric power of the representation . And $\chi$  be the character of the representations. Then apparently the following hold,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \sum _{n=0} ^{\infty} t^n \chi _{sym ^n (R)} (U) = e^{\sum _{m =1} ^\infty \frac{t^m}{m} \chi _R (U^m)} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \sum _{n=0} ^{\infty} t^n \chi _{anti ^n (R)} (U) = e^{\sum _{m =1} ^\infty \frac{(-1)^{m+1}t^m}{m} \chi _R (U^m)} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can someone tell me its derivation or let me know if it is known by some name which I can track down in the books?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also does any special thing happen if $U$ is in the adjoint representation? &lt;/p&gt;&#xA;" OwnerUserId="965" LastActivityDate="2011-01-31T18:52:25.957" Title="An equality about characters of representations" Tags="&lt;lie-groups&gt;&lt;representation-theory&gt;&lt;lie-algebra&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="19742" PostTypeId="2" ParentId="19740" CreationDate="2011-01-31T18:32:58.157" Score="1" ViewCount="" Body="&lt;p&gt;One crude way uses &lt;a href=&quot;http://en.wikipedia.org/wiki/Fermat%27s_little_theorem&quot; rel=&quot;nofollow&quot;&gt;Fermat's little theorem&lt;/a&gt;.  $2^{p-1} \equiv 1 \mod p$ so that $p$ divides $2^p - 1$ only when $p-1$ divides $n$.  For general integers $m$, Fermat's little theorem says $2^{\phi(m)} - 1 \equiv 0 \mod m$ where $\phi(\cdot)$ is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Euler%27s_totient_function&quot; rel=&quot;nofollow&quot;&gt;Euler phi function&lt;/a&gt;.  So maybe the problem boils down to finding $m$ such that $\phi(m)$ divides $n$.&lt;/p&gt;&#xA;" OwnerUserId="4997" LastActivityDate="2011-01-31T18:32:58.157" CommentCount="3" />
  <row Id="19743" PostTypeId="2" ParentId="19741" CreationDate="2011-01-31T18:37:03.620" Score="1" ViewCount="" Body="&lt;p&gt;Notice that neither identity involves the whole group: you only need a vector space $R$ and a linea map $U:R\to R$. You can pick a basis of $R$ so that $U$ is in its Jordan canonical form, so that it is easy to compute its trace and that of its powers; if the field is not algebraically closed, you can actually extend scalars, for nothing breaks by doing it: your identities are invariant under field extensions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now compute the trace of the action of $U$ on the symmetric and the antysimmetric powers of $R$ &lt;em&gt;explicitely&lt;/em&gt;, and you are left with an identity between formal series (they actually converge, in fact, over the complex numbers, say) Now prove &lt;em&gt;that&lt;/em&gt; identity is true.&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2011-01-31T18:37:03.620" />
  <row Id="19744" PostTypeId="2" ParentId="19720" CreationDate="2011-01-31T18:43:39.973" Score="3" ViewCount="" Body="&lt;p&gt;The probability $P$ that you're looking for is equal to the probability that among $2r$ randomly chosen elements, at least two &lt;i&gt;whose indices have different parity&lt;/i&gt; are equal. Now consider what happens when, instead of just drawing $2r$ elements, you first draw them and then reorder them randomly (using a permutation from $S_{2r}$). Suppose among the first $2r$ elements there was a pair of equal elements; what is the probability that under the new ordering, these find themselves in positions with opposite parity?&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2011-01-31T18:43:39.973" CommentCount="2" />
  <row Id="19745" PostTypeId="2" ParentId="19734" CreationDate="2011-01-31T18:51:00.077" Score="0" ViewCount="" Body="&lt;p&gt;A term in $e_k$ is the set of combinations of size $i$ from a set of size $n$. So the coefficient of a term in $\sum_{k=0} e_k$ is the number of combinations of -all- sizes, or rather, there is a one-to-one correspondence with the terms in the sum with the subsets of a set of size $n$. &lt;/p&gt;&#xA;" OwnerUserId="1919" LastActivityDate="2011-01-31T18:51:00.077" CommentCount="1" />
  <row Id="19746" PostTypeId="2" ParentId="19720" CreationDate="2011-01-31T18:52:07.937" Score="1" ViewCount="" Body="&lt;p&gt;Take any list of $2r$ elements and randomly permute their order.  Suppose that elements $i$ and $j$ were equal before the shuffle.  After the shuffle, the probability that these elements are in opposite halves of the list (i.e., one is in the first $r$ positions and the other is in the last $r$ positions) is equal to $r/(2r-1)$.  Therefore, given that the original $2r$ elements were not all distinct, the probability that (after shuffling) one of the first $r$ elements is equal to one of the last $r$ elements is at least $r/(2r-1)$ (it is strictly greater if more than one pair of elements is equal).  But this conditional probability is just the ratio that is asked for:&#xA;$$&#xD;&#xA;\frac{P\left[\exists_{i,j}E_i=F_j\right]}{P\left[2r\text{ not distinct}\right]} = P\left[\exists_{i,j}E_i=F_j \vert E, F\text{ not all distinct}\right] \ge \frac{r}{2r-1} &amp;gt; \frac{1}{2}.&#xD;&#xA;$$&#xA;Note that the size of set $S$ is not used, and neither is the fact that all elements were equally likely.  The inequality holds whenever the elements are independent and identically distributed.&lt;/p&gt;&#xA;" OwnerUserId="5546" LastActivityDate="2011-01-31T18:52:07.937" CommentCount="1" />
  <row Id="19747" PostTypeId="2" ParentId="19741" CreationDate="2011-01-31T18:52:25.957" Score="2" ViewCount="" Body="&lt;p&gt;I explain a combinatorial proof of this identity in &lt;a href=&quot;https://qchu.wordpress.com/2009/08/23/newtons-sums-necklace-congruences-and-zeta-functions/&quot; rel=&quot;nofollow&quot;&gt;this blog post&lt;/a&gt; and &lt;a href=&quot;https://qchu.wordpress.com/2009/11/04/newtons-sums-necklace-congruences-and-zeta-functions-ii/&quot; rel=&quot;nofollow&quot;&gt;this one&lt;/a&gt;; I'll sketch the second proof.  First, this is really an identity of symmetric functions in the eigenvalues: the first says that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{n \ge 0} h_n t^n = \exp \left( \sum_{n \ge 1} \frac{p_n}{n} t^n \right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $h_n$ are the &lt;a href=&quot;http://en.wikipedia.org/wiki/Complete_homogeneous_symmetric_polynomial&quot; rel=&quot;nofollow&quot;&gt;complete homogeneous symmetric polynomials&lt;/a&gt; and $p_n$ the &lt;a href=&quot;http://en.wikipedia.org/wiki/Power_sum_symmetric_polynomial&quot; rel=&quot;nofollow&quot;&gt;power symmetric functions&lt;/a&gt; in the eigenvalues of $U$.  The second says that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{n \ge 0} e_n t^n = \exp \left( \sum_{n \ge 1} (-1)^{n+1} \frac{p_n}{n} t^n \right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $e_n$ are the &lt;a href=&quot;http://en.wikipedia.org/wiki/Elementary_symmetric_polynomial&quot; rel=&quot;nofollow&quot;&gt;elementary symmetric functions&lt;/a&gt; in the eigenvalues.  It is not hard to see that $\sum h_n t^n = \frac{1}{\sum e_n t^n}$, so actually these identities are equivalent; we will thus concentrate on the first one, which has no signs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, here is the standard proof (with analytic details omitted, since this is an equality of formal power series anyway), which I find unsatisfactory.  Consider the operator $\frac{1}{1 - Ut} = \sum_{n \ge 0} U^n t^n.$  A standard matrix identity asserts that $\det \exp M = \exp \text{tr } M$ for any matrix $M$.  Applying this identity to the logarithm of the above, we conclude that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\det \frac{1}{1 - Ut} = \exp \text{tr } \log \frac{1}{1 - Ut} = \exp \left( \sum_{n \ge 1} \frac{p_n}{n} t^n \right).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, $\det \frac{1}{1 - Ut} = \frac{1}{\prod (1 - \lambda_i t)}$ where $\lambda_i$ are the eigenvalues, and this is precisely $\sum h_n t^n$.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Now here is a sketch of the combinatorial proof.  First we assume that $U$ is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Adjacency_matrix&quot; rel=&quot;nofollow&quot;&gt;adjacency matrix&lt;/a&gt; of a finite graph $G$.  Then $\text{tr } U^n$ is the number of closed walks of length $n$ on $G$.  On the other hand, $\text{tr } \text{Sym}^n(U)$ describes something more complicated: the graph $G$ has a symmetric product $\text{Sym}^n(G)$ whose vertices are the unordered $n$-tuples of vertices of $G$ and whose edges are chosen so that $\text{Sym}^n(U)$ is its adjacency matrix.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now $\text{tr } \text{Sym}^n(U)$ describes the number of closed walks of length $1$ on the symmetric product; that is, the number of loops.  A loop on $\text{Sym}^n(U)$ is (roughly speaking) an unordered collection of closed walks on $U$ with total length $n$.  This turns out to imply, via the well-known exponential formula in combinatorics, the desired identity.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, since the adjacency matrices of finite graphs are Zariski dense in $\text{GL}_n(\mathbb{C})$, the result follows for all matrices.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;This identity has hidden depths.  For a finite graph $G$ with adjacency matrix $A$, the function $\frac{1}{\det(I - At)}$ is a kind of zeta function for $G$, and the argument above is closely related to the Euler product of this zeta function.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2011-01-31T18:52:25.957" CommentCount="2" />
  <row Id="19748" PostTypeId="2" ParentId="19738" CreationDate="2011-01-31T18:53:37.140" Score="0" ViewCount="" Body="&lt;p&gt;Work done is weight times vertical displacement of the center of gravity. The rest of the problem is playing with the units to get to a consistent set. Well, also friction could add to the work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another source of error, is the &lt;code&gt;sin()&lt;/code&gt; function, in case it expects a number in radians and you supply it in degrees.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You answer in &lt;code&gt;lbs&lt;/code&gt; $\times$ &lt;code&gt;ft&lt;/code&gt; is &lt;em&gt;not&lt;/em&gt; in Joules. Multiply by 1.35581795 to convert to Joules.&lt;/p&gt;&#xA;" OwnerUserId="3301" LastActivityDate="2011-01-31T18:53:37.140" CommentCount="1" />
  <row Id="19749" PostTypeId="1" CreationDate="2011-01-31T19:07:23.030" Score="2" ViewCount="325" Body="&lt;p&gt;Given two vectors, a and b, how do I find the Orthogonal Projection?&#xA;I've already found the Scalar and Vector Projections.&#xA;\begin{align*}&#xA;\text{Scalar}&amp;amp;:\quad  \frac{-90 + -25 + 24}{\sqrt{9^2+5^2+8^2}};\\&#xA;\text{Vector}&amp;amp;:\quad \left(\left(\frac{-91}{\sqrt{170}}\right)\left(\frac{-9}{\sqrt{170}}\right), \left(\frac{-91}{\sqrt{170}}\right)\left(\frac{-5}{\sqrt{170}}\right), \left(\frac{-91}{\sqrt{170}}\right)\left(\frac{8}{\sqrt{170}}\right)\right).&#xA;\end{align*}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's the original question: Let a = (-4, -8, -4) and b = (-3, -2, 0) be vectors. Find the scalar, vector, and orthogonal projections of b onto a.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you for your time and help.&lt;/p&gt;&#xA;" OwnerUserId="3800" LastEditorUserId="3800" LastEditorDisplayName="" LastEditDate="2011-02-01T01:40:16.040" LastActivityDate="2011-02-01T08:41:36.007" Title="How do I find Orthogonal Projection given two Vectors?" Tags="&lt;vector-calculus&gt;" AnswerCount="1" CommentCount="10" />
  <row Id="19750" PostTypeId="2" ParentId="19734" CreationDate="2011-01-31T19:44:40.560" Score="1" ViewCount="" Body="&lt;p&gt;Here are two specific interesting cases to go with Mariano Suárez-Alvarez's general explanation.  On $n$ variables, $e_k(1,1,\ldots,1) = \binom{n}{k}$ and $e_k(1,2,\ldots,n) = \left[ n+1 \atop n-k+1 \right]$, where the latter is a &lt;a href=&quot;http://en.wikipedia.org/wiki/Stirling_numbers_of_the_first_kind&quot; rel=&quot;nofollow&quot;&gt;Stirling number of the first kind&lt;/a&gt;.  (See Comtet's &lt;a href=&quot;http://books.google.com/books?id=C0HPgWhEssYC&amp;amp;printsec=frontcover&amp;amp;dq=advanced+combinatorics&amp;amp;source=bl&amp;amp;ots=rEiyoWYlz2&amp;amp;sig=CKGVhgX5hcxIa_U0TyLsCdrcQ1M&amp;amp;hl=en&amp;amp;ei=ARFHTe7kJ5T6sAPS16jaCg&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=3&amp;amp;ved=0CC8Q6AEwAg#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;Advanced Combinatorics&lt;/em&gt;&lt;/a&gt;, pp. 213-214.)  So summing the former over $k$ gives $2^n$, and summing the latter over $k$ gives $(n+1)!$.&lt;/p&gt;&#xA;" OwnerUserId="2370" LastActivityDate="2011-01-31T19:44:40.560" CommentCount="1" />
  <row Id="19751" PostTypeId="2" ParentId="19692" CreationDate="2011-01-31T19:51:25.217" Score="1" ViewCount="" Body="&lt;p&gt;Without looking in too much detail, I suspect that you have not yet used the property of cyclic quadrilaterals that opposite angles are supplementary.  For example, considering the cyclic quadrilateral $ADB&amp;#39;C&amp;#39;$, $\angle B&amp;#39;AD$ and $\angle B&amp;#39;C&amp;#39;D$ are supplementary, so $\angle B&amp;#39;AD\cong\angle DC&amp;#39;A&amp;#39;$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a further suggestion, though I don't immediately see how to prove it, the circle through $A$, $B&amp;#39;$, and $C&amp;#39;$, the circle through $A&amp;#39;$, $B$, and $C&amp;#39;$, and the circle through $A&amp;#39;$, $B&amp;#39;$, and $C$ all pass through that same point $D$.&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2011-01-31T19:51:25.217" CommentCount="3" />
  <row Id="19752" PostTypeId="2" ParentId="19370" CreationDate="2011-01-31T19:52:29.443" Score="1" ViewCount="" Body="&lt;p&gt;Before you do anything else, ask yourself: &lt;em&gt;do you even believe that this is true?&lt;/em&gt;  Have you tried any examples with specific numbers?  Fixing the value of $a$ (especially taking $a$ to be very small) and varying the value of $n$, perhaps?  Does the result seem plausible after doing this?  Do your experiments suggest anything about what's going on?&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2011-01-31T19:52:29.443" />
  <row Id="19753" PostTypeId="1" AcceptedAnswerId="19758" CreationDate="2011-01-31T20:26:26.543" Score="1" ViewCount="115" Body="&lt;p&gt;While trying to understand artificial neural networks, I came upon an equation for finding the net input of an artificial neuron. Can someone explain this to me and what it means? &lt;a href=&quot;http://www.willamette.edu/~gorr/classes/cs449/ann-overview.html&quot; rel=&quot;nofollow&quot; title=&quot;Here&quot;&gt;Here&lt;/a&gt; is the original article&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$y_i=f\left(\sum_j w_{ij}y_j\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance, ell.&lt;/p&gt;&#xA;" OwnerUserId="6451" LastEditorUserId="6451" LastEditorDisplayName="" LastEditDate="2011-02-06T13:04:59.610" LastActivityDate="2011-02-06T13:04:59.610" Title="What does $y_i=f(\sum_j w_{ij}y_j)$ mean (in an artificial neuron model)?" Tags="&lt;algebra&gt;&lt;notation&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="19754" PostTypeId="1" CreationDate="2011-01-31T20:28:18.187" Score="2" ViewCount="62" Body="&lt;p&gt;I am interested in alternatives to Zeigler's lectures on polytopes, which is the suggested textbook for a course I am attending. I find the conversational style of the book jarring. &lt;/p&gt;&#xA;" OwnerUserId="404" LastActivityDate="2011-02-01T14:38:44.767" Title="Alternative to Zeigler's lectures on polytopes" Tags="&lt;reference-request&gt;&lt;polytopes&gt;" AnswerCount="2" />
  <row Id="19755" PostTypeId="2" ParentId="19754" CreationDate="2011-01-31T20:33:00.487" Score="3" ViewCount="" Body="&lt;p&gt;I think &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0486614808&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; (&lt;em&gt;Regular Polytopes&lt;/em&gt; by Coxeter) would be a good book. I liked his book on geometry. Also the reviews seem to indicate that is is pretty good.&lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2011-01-31T20:33:00.487" />
  <row Id="19756" PostTypeId="2" ParentId="19712" CreationDate="2011-01-31T20:36:24.617" Score="5" ViewCount="" Body="&lt;p&gt;Okay, forget the Fubini-Study metric.  I think I have an alternate solution.  Let's rephrase the problem as follows.  We'll pick a local coordinate $z$ and think of elements of $\text{PSL}_2(\mathbb{C})$ as fractional linear transformations $z \mapsto \frac{az + b}{cz + d}$.  Adding an inner product on the underlying copy of $\mathbb{C}^2$ lets us associate to any $z$ its &quot;orthogonal complement&quot; $- \frac{1}{\bar{z}}$, and the right choice of stereographic projection sends orthogonal complements to &lt;em&gt;antipodes.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, an element of $\text{PSL}_2(\mathbb{C})$ respects antipodes (equivalently, orthogonal complements) if and only if it lies in $\text{PSU}(2)$, so it remains to show that any conformal orientation-preserving antipode-preserving automorphism of the Riemann sphere is a rotation.  Certainly a fractional linear transformation $g$ preserving antipodes must have two antipodal fixed points of the same type.  They can't both be attractive or both repelling (that contradicts the fact that the product of the eigenvalues is $1$), so they are both neither.  This ought to already be enough to conclude that $g$ preserves distances to its fixed points, which should only be possible if it's a rotation (and expanding at a local coordinate at the fixed points shows this if nothing else does).&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2011-01-31T20:42:35.860" LastActivityDate="2011-01-31T20:42:35.860" />
  <row Id="19758" PostTypeId="2" ParentId="19753" CreationDate="2011-01-31T20:48:49.633" Score="2" ViewCount="" Body="&lt;p&gt;So hopefully you understand what the $\sum$ symbol means, it's just a summation across all of those indices. In this case we have our weight vectors&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$w_1,w_2,\dots,w_n$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and our vector $y$. The neuron then is going to compute the new value of y_i, based on a dot product of w_i with y. Hence &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sum_{j} w_{ij}y_j$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is just dot product of w_i and y, where both are n-dimensional vectors. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then f is an activation function, it could be a preceptron or it could be a linear node. So it may output the value of the net, or it may output a 0 or a 1 based on the value of the summation. Or it could have some non-linear neuron it all depends on the network.&lt;/p&gt;&#xA;" OwnerUserId="1930" LastActivityDate="2011-01-31T20:48:49.633" CommentCount="1" />
  <row Id="19759" PostTypeId="2" ParentId="19753" CreationDate="2011-01-31T20:50:38.630" Score="0" ViewCount="" Body="&lt;p&gt;What you are looking at is a simplistic model of the neuron. In particular, the inputs are &lt;a href=&quot;http://en.wikipedia.org/wiki/Dendrite&quot; rel=&quot;nofollow&quot;&gt;dendrites&lt;/a&gt; and the outputs are the &lt;a href=&quot;http://en.wikipedia.org/wiki/Synapse&quot; rel=&quot;nofollow&quot;&gt;synapse&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2011-01-31T20:50:38.630" />
  <row Id="19760" PostTypeId="2" ParentId="19686" CreationDate="2011-01-31T20:52:04.653" Score="2" ViewCount="" Body="&lt;p&gt;This confused me in my first linear algebra class, too. The psychological difference between &quot;onto&quot; and &quot;surjective&quot; is that the latter is only ever introduced as an adjective, whereas prior experience makes us want to read &quot;onto&quot; as a preposition.  I don't think this problem arises for &quot;one-to-one&quot;, because again we first learn this phrase as an adjective, so there's nothing to confuse it with.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Oxford English dictionary has numerous definitions of the preposition &quot;onto&quot;, but the only instance it gives for usage as an adjective is in mathematics.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;B. adj.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Math. In form onto. Designating a mapping of one set on to another.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The following is the earliest quotation given there for this usage:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;1942    S. Lefschetz Algebraic Topol. i. 7   If a transformation is ‘onto’, the inverse image of the complement of a set is the complement of the inverse image of that set.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am confused by this quotation, as the result is true for maps that are not onto.  However, a quick search of &lt;a href=&quot;http://books.google.com/books?id=3IsMJW4WYvsC&amp;amp;vq=onto&amp;amp;source=gbs_navlinks_s&quot; rel=&quot;nofollow&quot;&gt;the book&lt;/a&gt; shows other uses of the adjective &quot;onto&quot; in the modern sense. The next is more apt:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;1951    N. Jacobson Lect. Abstr. Algebra I. 4   If α is a mapping of S into T, and β is a mapping of T into S such that αβ = $1_S$ and βα = $1_T$, then α and β are 1−1, onto mappings and β = α$^{−1}$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2011-01-31T21:05:01.853" LastActivityDate="2011-01-31T21:05:01.853" CommentCount="1" />
  <row Id="19761" PostTypeId="1" AcceptedAnswerId="19762" CreationDate="2011-01-31T21:27:20.477" Score="2" ViewCount="184" Body="&lt;p&gt;I've just had to do a homework on binomial expansion for approximation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$1.07^9$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(1+0.07)^9$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To do binomial expansion you need a calculator for the combinations button (nCr), so why would use a more complicated method, which only gives an approximation be used over just typing 1.07^9 into a calculator?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(or is this never done in real life, and it's just a homework?)&lt;/p&gt;&#xA;" OwnerUserId="2075" LastActivityDate="2011-01-31T22:30:30.963" Title="Binomial expansion for approximation" Tags="&lt;sequences-and-series&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="19762" PostTypeId="2" ParentId="19761" CreationDate="2011-01-31T21:38:15.580" Score="3" ViewCount="" Body="&lt;p&gt;Expanding the whole thing using Binomial Theorem gives you an &lt;em&gt;exact&lt;/em&gt; value. Not an approximation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To get an approximation you can consider a few terms from the expansion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, for &quot;small&quot; $x$, $1+nx$ is a &quot;reasonable&quot; &lt;em&gt;approximation&lt;/em&gt; for $(1+x)^n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice that this corresponds to picking the first two terms from the binomial theorem expansion $(1+x)^n = 1 + \binom{n}{1} \ x + \binom{n}{2}\  x^2 + \dots + x^n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$1.0007^9 \approx 1 + 9\times 0.0007 = 1.0063$ which agrees with $1.0007^9 = 1.0063176688422737867054812736724$ upto $4$ decimal places.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Depending on how accurate you want it, you could consider more terms from the binomial expansion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is based on the fact that for small $x$, as the power $r$ of $x$ gets larger, the term $x^r$ becomes small quite fast.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2011-01-31T22:30:30.963" LastActivityDate="2011-01-31T22:30:30.963" CommentCount="2" />
  <row Id="19763" PostTypeId="2" ParentId="19761" CreationDate="2011-01-31T21:40:37.777" Score="-1" ViewCount="" Body="&lt;p&gt;This looks like a badly thought-out problem to me. The first-order approximation is 1.63, but the true value is greater than 1.83. Can you give us the exact wording of the question?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the practicality of this, I would say that being able to estimate such expressions is a useful skill in many branches of mathematics and statistics. Not essential, but useful.&lt;/p&gt;&#xA;" OwnerUserId="1508" LastActivityDate="2011-01-31T21:40:37.777" />
  <row Id="19764" PostTypeId="1" AcceptedAnswerId="22021" CreationDate="2011-01-31T21:58:05.077" Score="16" ViewCount="255" Body="&lt;p&gt;Let $\phi(n)$ be Euler's totient function. For $n=4m$ what is the smallest $n$ for which &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\phi(n) \ne \phi(k) \textrm{ for any } k&amp;lt;n \textrm{ ?} \quad (1)$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When $n=4m+1$ and $n&amp;gt;1$ the smallest is $n=5$ and when $n=4m+3$ it's $n=3.$ However, when&#xA;$n=4m+2$ the above condition can never be satisfied since&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\phi(4m+2) = (4m+2) \prod_{p | 4m+2} \left( 1 - \frac{1}{p} \right)$$&#xA;$$ = (2m+1) \prod_{p | 2m+1} \left( 1 - \frac{1}{p} \right) = \phi(2m+1).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case $n=4m,$ $n=2^{33}$ is a candidate and $\phi(2^{33})=2^{32}.$ This value satisfies $(1)$ because $\phi(n)$ is a power of $2$ precisely when $n$ is the product of a power of $2$ and any number of distinct &lt;a href=&quot;http://mathworld.wolfram.com/FermatPrime.html&quot; rel=&quot;nofollow&quot;&gt;Fermat primes:&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$2^1+1,2^2+1,2^4+1,2^8+1 \textrm{ and } 2^{16}+1.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note the $n=2^{32}$ does not satisfy condition $(1)$ because the product of the above Fermat primes is $2^{32}-1$ and so $\phi(2^{32})=2^{31}=\phi(2^{32}-1)$ and&#xA;$2^{32}-1 &amp;lt; 2^{32}.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only solutions to $\phi(n)=2^{32}$ are given by numbers of the form&#xA;$n=2^a \prod (2^{x_i}+1)$ where $x_i \in \lbrace 1,2,4,8,16 \rbrace $ and&#xA;$a+ \sum x_i = 33$ (note that the product could be empty),&#xA;so all these numbers are necessarily $ \ge 2^{33}.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why don't many &quot;small&quot; multiples of $4$ satisfy condition $(1)$? Well, note that for $n=2^a(2m+1)$ we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\phi(2^a(2m+1))= 2^a(2m+1) \prod_{p | 2^a(2m+1)} \left( 1 - \frac{1}{p} \right)$$&#xA;$$ = 2^{a-1}(2m+1) \prod_{p | 2m+1} \left( 1 - \frac{1}{p} \right) = 2^{a-1}\phi(2m+1),$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and so, for $a \ge 2,$ if $2^{a-1}\phi(2m+1)+1$ is prime we can take this as our value of $k&amp;lt;n$ and we have $\phi(n)=\phi(k).$ This, together with the existence of the Fermat primes, seems to be why it's difficult to satisfy when $n=4m.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have only made hand calculations so far, so I would not be too surprised if the answer is much smaller than my suggestion. The problem is well within the&#xA;reach of a computer, and possibly further analysis without the aid of a computer. But, anyway, I've decided to ask here as many of you have ready access to good mathematical&#xA;software and I'm very intrigued to know whether there is a smaller solution than $2^{33}.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some background information:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This question arose in my search to bound the function $\Phi(n)$ defined as follows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $\Phi(n)$ be the number of distinct values taken on by $\phi(k)$ for $1 \le k \le n.$&#xA;For example, $\Phi(13)=6$ since $\phi(k)$ takes on the values $\lbrace 1,2,4,6,10,12 \rbrace$ for $1 \le k \le 13.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is clear that $\Phi(n)$ is increasing and increases by $1$ at each prime value of $n,$ except $n=2,$ but it also increases at other values as well. For example, $\Phi(14)=6$ and $\Phi(15)=7.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Currently, for an upper bound, I'm hoping to do better than $\Phi(n) \le \lfloor (n+1)/2 \rfloor .$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But this this not the issue at the moment, although it may well become a separate question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This work originates from &lt;a href=&quot;http://math.stackexchange.com/questions/11398/the-order-of-the-number-of-integer-pairs-satisfying-certain-arithmetical-function&quot;&gt;this stackexchange problem.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1301" LastActivityDate="2011-02-14T16:44:26.013" Title="What is the smallest value of $n$ for which $\phi(n) \ne \phi(k)$ for any $k&lt;n,$ where $n=4m$?" Tags="&lt;number-theory&gt;&lt;totient-function&gt;" AnswerCount="1" CommentCount="10" FavoriteCount="3" />
  <row Id="19765" PostTypeId="1" CreationDate="2011-01-31T21:58:59.250" Score="2" ViewCount="72" Body="&lt;p&gt;If I have a grid that I know is RxCxD in size, and I have number that corresponds to an element in that grid, and the grid is mapped in Row major format, how can I find out the ordered triplet that corresponds to the index that I know? that is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(i,j,k) = index&lt;/p&gt;&#xA;" OwnerUserId="6453" LastActivityDate="2011-01-31T23:50:13.373" Title="calculating coordinates from a flattened 3D array when you know the size, index and ordering" Tags="&lt;linear-algebra&gt;&lt;algebra&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="19766" PostTypeId="1" AcceptedAnswerId="19767" CreationDate="2011-01-31T22:00:00.627" Score="2" ViewCount="149" Body="&lt;p&gt;I know that for a series of non-negative, continuous functions $u_{n}(x)$, a sufficient condition for uniform convergence of $\sum u_{n}(x)$ to $u(x)$ is for $u(x)$ to be continuous in $I\subset \mathbb{R}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I can't think of an example where $\sum f_{n}(x)\to f(x)$ uniformly in $I\subset \mathbb{R}$ but $f(x)$ is discontinuous in $I\subset \mathbb{R}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of $(f_{n}(x))$ must obviously be discontinuous in the interval because (as others have written below) a uniform limit of continuous functions is continuous.&lt;/p&gt;&#xA;" OwnerUserId="3237" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2011-01-31T22:37:55.463" LastActivityDate="2011-01-31T22:52:00.133" Title="Example of discontinuous $u(x)$ where $\sum u_{n}(x)\to u(x)$ uniformly in $I \subset \mathbb{R}$" Tags="&lt;calculus&gt;&lt;sequences-and-series&gt;" AnswerCount="4" CommentCount="3" />
  <row Id="19767" PostTypeId="2" ParentId="19766" CreationDate="2011-01-31T22:12:15.987" Score="3" ViewCount="" Body="&lt;p&gt;One way to get an example for the clarified question is to let $u_1$ be any discontinuous function on $[0,1]$, like $u_1(x)=0$ if $0\leq x\leq \frac{1}{2}$ and $u_1(x)=1$ if $\frac{1}{2}\lt x\leq 1$, and let $u_n(x)=0$ for $n\geq 2$.  Then the series $\sum u_n(x)$ converges uniformly to $u_1(x)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the $u_n$'s are continuous, you won't be able to find such an example, because &lt;a href=&quot;http://en.wikipedia.org/wiki/Uniform_convergence#To_continuity&quot; rel=&quot;nofollow&quot;&gt;a uniform limit of continuous functions is continuous&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Regarding your first statement, it isn't necessarily true if the interval is not compact. For an example of a continuous function that is the nonuniform limit of an increasing sequence of nonnegative continuous functions on an interval, consider $f(x)=1$ on $(0,1)$, and $f_n(x)=1-x^n$.  To phrase this in terms of a series of nonnegative functions, take $u_1=f_1$ and $u_n=f_n-f_{n-1}$ for $n\geq2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if the interval is compact, then it is true that if a monotonically increasing sequence of continuous functions converges pointwise to a continuous function, then the convergence is uniform.  This is called &lt;a href=&quot;http://en.wikipedia.org/wiki/Dini%27s_theorem&quot; rel=&quot;nofollow&quot;&gt;Dini's theorem&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2011-01-31T22:29:39.157" LastActivityDate="2011-01-31T22:29:39.157" CommentCount="6" />
  <row Id="19768" PostTypeId="2" ParentId="19766" CreationDate="2011-01-31T22:13:18.157" Score="3" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt; If you want &lt;em&gt;discontinuous&lt;/em&gt; functions whose sum converges uniformly to a &lt;em&gt;discontinuous&lt;/em&gt; function, consider the characteristic function of the rationals: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;   \lambda(x) = \left\{&#xD;&#xA;     \begin{array}{lr}&#xD;&#xA;       1 &amp;amp; : x \in \mathbb{Q}\\&#xD;&#xA;       0 &amp;amp; : x \notin \mathbb{Q}&#xD;&#xA;     \end{array}&#xD;&#xA;   \right.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then let $\lambda_n = \frac{1}{2^{n+1}} \cdot \lambda$. The sum will indeed converge uniformly to $\lambda$ but will clearly be discontinuous. In fact if $\lambda$ is any discontinuous function, this construction will give you your desired result.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;However, if the $u_n$'s are all continuous, this cannot happen. Let &#xA;$$g_n = \sum_{k=1}^{n} u_k$$&#xA;Then each $g_i$ is countinuous being the sum of continuous functions. But since $g_n \rightarrow u$ uniformly, it must be the case that $u$ is continuous as well. Here's a quick proof:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose we want to show that $u$ is continuous at some $x_0 \in I$. Fix some $\epsilon &amp;gt; 0$. Find $N$ large enough such that $|g_n (x) - u(x) | &amp;lt; \frac{\epsilon}{3}$ for all $x \in I$ and $n \ge N$. Choose some $\delta$ such that $| g_N (x) - g_N (x_0) | &amp;lt; \frac{\epsilon}{3}$ whenever $|x - x_0 | &amp;lt; \delta$, which we can do as $g_N$ is continuous. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we have that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$|u(x) - u(x_0)| \le |u(x) - g_N(x)| + |g_N (x) - g_N (x_0) | + |g_N(x_0) - u(x_0)| &amp;lt; \frac{\epsilon}{3} + \frac{\epsilon}{3} + \frac{\epsilon}{3} = \epsilon $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;by applying the triangular inequality. Hence $u$ is continuous at $x_0$. &lt;/p&gt;&#xA;" OwnerUserId="5105" LastEditorUserId="5105" LastEditorDisplayName="" LastEditDate="2011-01-31T22:41:23.333" LastActivityDate="2011-01-31T22:41:23.333" />
  <row Id="19769" PostTypeId="2" ParentId="19692" CreationDate="2011-01-31T22:13:26.550" Score="2" ViewCount="" Body="&lt;p&gt;I'm not sure what Hartshorne has in mind, but Pappus' theorem is a simple consequence of similarity of Euclidean triangles (in guise of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Intercept_theorem&quot; rel=&quot;nofollow&quot;&gt;intercept theorem&lt;/a&gt;) and there's no need of introducing the circle:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $P$ be the point of intersection of $l$ and $m$. All we need to do is to show that $\frac{PA&amp;#39;}{PB} = \frac{PB&amp;#39;}{PA}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But by assumption we have $\frac{PA&amp;#39;}{PC} = \frac{PC&amp;#39;}{PA}$ and $\frac{PC}{PB} = \frac{PB&amp;#39;}{PC&amp;#39;}$. Multiplying the left hand sides and the right hand sides together, we get what we want.&lt;/p&gt;&#xA;" OwnerUserId="5363" LastEditorUserId="5363" LastEditorDisplayName="" LastEditDate="2011-01-31T22:18:28.150" LastActivityDate="2011-01-31T22:18:28.150" CommentCount="2" />
  <row Id="19770" PostTypeId="2" ParentId="19766" CreationDate="2011-01-31T22:13:32.880" Score="0" ViewCount="" Body="&lt;p&gt;This follows from the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Theorem.&lt;/strong&gt; If the limit function $f(x)$ of a pointwise convergent sequence of continuous functions $\{f_{n}(x) \}$ is discontinuous, then the convergence of the sequence $\{f_{n}(x) \}$ is nonuniform.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the contrapositive is: $\{f_{n}(x) \} \to f \ \text{uniformly} \Longrightarrow f(x) \ \text{is continuous}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence this is impossible.&lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2011-01-31T22:13:32.880" CommentCount="2" />
  <row Id="19772" PostTypeId="2" ParentId="19766" CreationDate="2011-01-31T22:35:27.057" Score="3" ViewCount="" Body="&lt;p&gt;Suppose your interval $I = [-1,1]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $u_n(x) = 2^{-n}$ for $x \in [-1,0)$ and $u_n(x) = 2^{-n} + 1$ for $x \in [0,1]$.  Then $\sum u_n(x) \to u(x)$ uniformly, where $u(x) = 1$ on $[-1,0)$ and $u(x) = 2$ on $[0,1]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For general intervals $I$... well, just change the bounds.&lt;/p&gt;&#xA;" OwnerUserId="640" LastActivityDate="2011-01-31T22:35:27.057" CommentCount="1" />
  <row Id="19773" PostTypeId="1" AcceptedAnswerId="20787" CreationDate="2011-01-31T22:45:19.390" Score="4" ViewCount="145" Body="&lt;p&gt;In &lt;a href=&quot;http://www.springerlink.com/content/y19u81675243r237/fulltext.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.springerlink.com/content/y19u81675243r237/fulltext.pdf&lt;/a&gt;, the author states the following without proof (equation 3.1):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider a random permutation $\pi$ of $\mathbb{Z}_n$. What is the probability that $\pi(i+1)-\pi(i) \pmod{n} &amp;lt;n/2$ for all $i$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The claim is that this is $(2+o(1))^{-n}$, which makes sense and seems like it should be a standard argument. Does anyone have a formal proof?&lt;/p&gt;&#xA;" OwnerUserId="275" LastEditorUserId="275" LastEditorDisplayName="" LastEditDate="2011-02-01T06:21:01.447" LastActivityDate="2011-02-07T06:09:21.917" Title="Random permutations of Z_n" Tags="&lt;probability&gt;&lt;combinatorics&gt;&lt;discrete-mathematics&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="19774" PostTypeId="1" AcceptedAnswerId="19778" CreationDate="2011-01-31T22:50:25.787" Score="6" ViewCount="170" Body="&lt;p&gt;Just wondering about this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it true that there are no prime numbers in Pascal's triangle, with the exception of $\binom{n}{1}$ and $\binom{n}{n-1}$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the first 18 lines it appears that this is true, but I haven't looked beyond that.&#xA;Is this a coincidence or is there a reason for it?&lt;/p&gt;&#xA;" OwnerUserId="2075" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2011-02-01T02:46:07.587" LastActivityDate="2011-02-01T02:55:05.603" Title="prime numbers in Pascal's triangle" Tags="&lt;prime-numbers&gt;&lt;binomial-coefficients&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="19776" PostTypeId="1" AcceptedAnswerId="20163" CreationDate="2011-01-31T22:54:34.483" Score="8" ViewCount="282" Body="&lt;p&gt;Hi,&#xA;I'm trying to prove that if we have the elementary symmetric polynomials that the following identity holds:(where $x = (x_1,..,x_n)$ is a vector of n variables)&#xA;$$\sum_{k=0}^n e_k(x)^2 = x_1\cdots x_n  \sum_{j=0}^{\lfloor n/2 \rfloor}  {2j \choose j} e_{n-2j}(x_1+1/x_1,\cdots x_n+1/x_n).$$ I'm trying to prove this combinatorially, but I'm stuck so I'd love some hints or tips. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Update:&#xA;I guess we can get some interpretation with strictly increasing functions, but that thought isn't fully developed yet.&lt;/p&gt;&#xA;" OwnerUserId="2394" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2011-02-03T00:40:23.733" LastActivityDate="2011-02-03T01:10:45.820" Title="On a combinatorial identity." Tags="&lt;homework&gt;&lt;combinatorics&gt;" AnswerCount="3" CommentCount="5" FavoriteCount="1" />
  <row Id="19777" PostTypeId="1" CreationDate="2011-01-31T22:55:52.907" Score="1" ViewCount="54" Body="&lt;p&gt;What is the best intuition for Calderon-Zygmund operators? Why are they so important in singular integrals, and complementary, which singular integrals don't they cover?&lt;/p&gt;&#xA;" OwnerUserId="3557" LastActivityDate="2011-01-31T22:55:52.907" Title="Intuition for Calderon-Zygmund operator?" Tags="&lt;intuition&gt;&lt;harmonic-analysis&gt;" CommentCount="2" />
  <row Id="19778" PostTypeId="2" ParentId="19774" CreationDate="2011-01-31T23:20:46.597" Score="9" ViewCount="" Body="&lt;p&gt;Yes, it's true.  The identity ${m \choose n} = \frac{m}{n} {m-1 \choose n-1}$ can be rearranged as $n {m \choose n} = m {m-1 \choose n-1}$.  If ${m \choose n}$ is prime it follows that it must divide either $m$ or ${m-1 \choose n-1}$.  In the first case we can only have $n = 1, n = m-1$, as you have already observed, and in the second case, the quotient $\frac{n}{m}$ cannot be an integer unless $n = m$ or $n = 0$, and neither of these cases gives a prime.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2011-01-31T23:20:46.597" />
  <row Id="19779" PostTypeId="2" ParentId="19774" CreationDate="2011-01-31T23:21:57.017" Score="6" ViewCount="" Body="&lt;p&gt;I believe it is true.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $\displaystyle \binom{n}{r} = p$ for some prime $\displaystyle p$ and $\displaystyle 1 \lt r \lt n-1$, then,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $\displaystyle n \ge p$, then $\displaystyle \binom{n}{r} \gt n \ge p$, as the binomial coefficients increase and then decrease as $\displaystyle r$ varies from $\displaystyle 0$ to $\displaystyle n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $\displaystyle n \lt p$ then $\displaystyle \binom{n}{r}$ can never be divisible by $\displaystyle p$, as $\displaystyle n!$ is not divisible by $\displaystyle p$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;OR as&#xA;hardmath succintly put it:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$p \mid \binom{n}{r} \Rightarrow p \le n \lt \binom{n}{r}$&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2011-02-01T02:55:05.603" LastActivityDate="2011-02-01T02:55:05.603" CommentCount="8" />
  <row Id="19780" PostTypeId="2" ParentId="19765" CreationDate="2011-01-31T23:26:43.300" Score="1" ViewCount="" Body="&lt;p&gt;If $i$ increases most quickly, then $j$, then $k$ (so a 2x2x2 array would be 0,0,0 then 1,0,0, then 0,1,0 etc) and you start counting from 0, it is $i+Rj+RCk$.  If you count from 1, it is $(i-1)+R(j-1)+RC(k-1)+1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Going the other direction, if the element number is $N$, then $k=N/RC, j=(N-kRC)/R, i=N-kRC-jR$ where the divisions are integer.  This assumes you start from 0.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2011-01-31T23:50:13.373" LastActivityDate="2011-01-31T23:50:13.373" CommentCount="8" />
  <row Id="19782" PostTypeId="1" AcceptedAnswerId="19786" CreationDate="2011-01-31T23:59:07.330" Score="5" ViewCount="241" Body="&lt;p&gt;A student randomly asked me to compute&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int\sqrt{\cos(2\theta)}\, \mathrm d\theta.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was unable to do so, as were several other instructors.  I typed the integral into Wolfram and it says that it is an elliptic integral of type 2.  I am not familiar with elliptic integrals and I don't have any idea why this student asked me to do this since we are not even doing derivatives yet.  Is this integral important?  Does it have a closed form answer?&lt;/p&gt;&#xA;" OwnerUserId="2838" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2011-04-30T03:48:25.933" LastActivityDate="2011-04-30T03:48:25.933" Title="What's With The Integral $\int\sqrt{\cos(2\theta)}\, \mathrm d\theta$?" Tags="&lt;calculus&gt;&lt;integral&gt;" AnswerCount="1" CommentCount="6" FavoriteCount="1" />
  <row Id="19784" PostTypeId="1" AcceptedAnswerId="19785" CreationDate="2011-02-01T00:54:19.327" Score="-2" ViewCount="60" Body="&lt;p&gt;Could someone please help me out on this question? Many thanks&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/9ww9g.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="6457" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2011-02-01T04:11:10.933" LastActivityDate="2011-02-01T04:11:10.933" Title="Purchase Credit Plan" Tags="&lt;arithmetic&gt;" AnswerCount="1" />
  <row Id="19785" PostTypeId="2" ParentId="19784" CreationDate="2011-02-01T01:03:18.740" Score="0" ViewCount="" Body="&lt;p&gt;Look at the table and subtract two values. In particular, it would be $112.5-50.10 = 62.4$&lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2011-02-01T01:03:18.740" CommentCount="1" />
  <row Id="19786" PostTypeId="2" ParentId="19782" CreationDate="2011-02-01T01:10:06.867" Score="6" ViewCount="" Body="&lt;p&gt;The elliptic integral of the second kind crops up since&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int\sqrt{\cos\;2\theta}\;\mathrm d\theta=\int\sqrt{1-2\sin^2\theta}\;\mathrm d\theta=E(\theta|2)+C$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $E(\phi|m)=\int_0^\phi \sqrt{1-m\sin^2\theta}\;\mathrm d\theta$ is the incomplete elliptic integral of the second kind.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All well and good, but traditionally one transforms elliptic integrals such that the parameter $m$ is within the interval $[0,1]$ (corresponding to the fact that ellipses have their eccentricity in the interval $(0,1)$). Using &lt;a href=&quot;http://dlmf.nist.gov/19.7#ii&quot; rel=&quot;nofollow&quot;&gt;certain transformations&lt;/a&gt;, one now has&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$E(\theta|2)=\sqrt{2}E\left(\arcsin(\sqrt{2}\sin\;\theta)|\frac12\right)-\frac1{\sqrt{2}}F\left(\arcsin(\sqrt{2}\sin\;\theta)|\frac12\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $F(\phi|m)=\int_0^\phi\frac{\mathrm d\theta}{\sqrt{1-m\sin^2\theta}}$ is the incomplete elliptic integral of the first kind, which is probably what you want.&lt;/p&gt;&#xA;" OwnerUserId="6458" LastActivityDate="2011-02-01T01:10:06.867" />
  <row Id="19788" PostTypeId="1" AcceptedAnswerId="19789" CreationDate="2011-02-01T01:19:00.117" Score="2" ViewCount="64" Body="&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/pP0cK.jpg&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My calculation was (5520*5845)*31% and I got value 10,001,964. Closest answer is 10,100,000. Is the answer right?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="6457" LastEditorUserId="5363" LastEditorDisplayName="" LastEditDate="2011-03-30T09:42:24.653" LastActivityDate="2011-03-30T09:42:24.653" Title="Percentage + abstract/substract question" Tags="&lt;arithmetic&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="19789" PostTypeId="2" ParentId="19788" CreationDate="2011-02-01T01:23:30.923" Score="1" ViewCount="" Body="&lt;p&gt;No, the 5520*5845 gives the number of good units produced.  This is 69% of the total.  The rejection is 31% of the total.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-02-01T01:23:30.923" CommentCount="1" />
  <row Id="19790" PostTypeId="1" AcceptedAnswerId="19791" CreationDate="2011-02-01T01:41:18.943" Score="2" ViewCount="233" Body="&lt;p&gt;&lt;em&gt;I'm not very good at this, so hopefully I'm not making a silly mistake here...&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming that $\infty$ is larger than any real number, we can then assume that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\dfrac{1}{\infty}$ is the smallest possible positive real number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It then stands to reason that anything &lt;em&gt;less&lt;/em&gt; than $\infty$ is a real number. Therefore, if we take the smallest possible quantity from $\infty$, we end up with:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\infty-\dfrac{1}{\infty}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does that expression represent the largest real number? If not, what am I doing wrong?&lt;/p&gt;&#xA;" OwnerUserId="4626" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-02T19:22:47.283" LastActivityDate="2011-02-02T19:22:47.283" Title="Does this expression represent the largest real number?" Tags="&lt;infinity&gt;&lt;real-analysis&gt;" AnswerCount="5" CommentCount="1" />
  <row Id="19791" PostTypeId="2" ParentId="19790" CreationDate="2011-02-01T01:53:05.547" Score="1" ViewCount="" Body="&lt;p&gt;You are looking at the &lt;a href=&quot;http://en.wikipedia.org/wiki/Extended_real_number_line&quot; rel=&quot;nofollow&quot;&gt;extended real number system&lt;/a&gt;. In particular, we adjoin $+ \infty$ and $- \infty$ to $\mathbb{R}$. So in the context of the extended real numbers, $+\infty- \frac{1}{+ \infty} = + \infty$.&lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2011-02-01T01:53:05.547" CommentCount="8" />
  <row Id="19792" PostTypeId="2" ParentId="19790" CreationDate="2011-02-01T01:54:12.317" Score="1" ViewCount="" Body="&lt;p&gt;Since $\infty$ is not a real number, you cannot assume that $\dfrac{1}{\infty}$ is a meaningful statement.  It is not a real number. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might want to investigate non-standard, hyperreal and surreal numbers and infinitesimals. &lt;/p&gt;&#xA;" OwnerUserId="6460" LastActivityDate="2011-02-01T01:54:12.317" />
  <row Id="19793" PostTypeId="2" ParentId="19790" CreationDate="2011-02-01T01:54:12.210" Score="1" ViewCount="" Body="&lt;p&gt;George, the symbol you have written $\infty$ is not a real number. It is a concept which we use (when we are dealing with numbers) to indicate something is unbounded.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The real numbers are unbounded so there is no largest or smallest real number.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2011-02-01T02:12:43.427" LastActivityDate="2011-02-01T02:12:43.427" />
  <row Id="19794" PostTypeId="1" CreationDate="2011-02-01T01:55:30.193" Score="1" ViewCount="46" Body="&lt;p&gt;I am trying to understand the proof of this fact. On page 183, Eisenbud defines a map from the formal power series ring $S[[x_1,x_2,...,x_n]]$ to $R/m^i$ where $R=S[x_1,x_2,...,x_n]$ sending $f$ to $f+m^i$. I have trouble understanding this map since, $f$ need not be an element of $R$ and so what does the coset $f+m^i$ mean?&#xA;Thanks for any help.&lt;/p&gt;&#xA;" OwnerUserId="6166" LastActivityDate="2011-02-01T01:55:30.193" Title="Completion of a polynomial ring w.r.t the homogeneous maximal ideal is the power series ring" Tags="&lt;commutative-algebra&gt;" CommentCount="2" />
  <row Id="19795" PostTypeId="2" ParentId="19589" CreationDate="2011-02-01T01:57:21.077" Score="2" ViewCount="" Body="&lt;p&gt;Picard iteration produces a sequence of approximations&#xA;that converges uniformly to a solution: &#xA;Let $X_0(z)=u$ and find $X_n(z)$ ($n\ge1$) to satisfy&#xA;$$ X_n(z) = u+ \int_{z_0}^z (A(w)X_{n-1}(w)+B(w))\,dw&#xD;&#xA;$$&#xA;integrating along the line segment path connecting $z_0$ to $z$.&#xA;Each $X_n$ is holomorphic on $\Omega$ since $\Omega$, being  convex,&#xA;is simply connected.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the given assumptions, &#xA;it is straightforward to prove by induction that&#xA;$$&#xD;&#xA;\|X_{n}(z)-X_{n-1}(z)\| \le \frac{C^n|z-z_0|^n}{n!} \qquad (n\ge1),&#xD;&#xA;$$&#xA;(in terms of a matrix norm) where&#xA;$C=\sup_{z\in\Omega}( \|A(z)\|(|u|+1)+\|B(z)\|)&amp;lt;\infty$. &#xA;Uniform convergence to a solution follows &#xA;since the telescoping series&#xA;$ \sum_{n=1}^\infty (X_n(z)-X_{n-1}(z) )$&#xA;is absolutely convergent.&lt;/p&gt;&#xA;" OwnerUserId="2947" LastActivityDate="2011-02-01T01:57:21.077" />
  <row Id="19796" PostTypeId="1" CreationDate="2011-02-01T02:00:35.517" Score="3" ViewCount="166" Body="&lt;p&gt;Again:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\int e^{\alpha x}\cos(\beta x) = \frac{e^{\alpha x} (\alpha \cos(\beta x)+\beta \sin(\beta x))}{\alpha^2+\beta^2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also the one for $\sin$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\int e^{\alpha x}\sin(\beta x) = \frac{e^{\alpha x} (\alpha \sin(\beta x)-\beta \cos(\beta x))}{\alpha^2+\beta^2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both help avoid integration by parts. I find these two proofs very useful and I'd like to know what they're names are.&lt;/p&gt;&#xA;" OwnerUserId="6394" LastEditorUserId="664" LastEditorDisplayName="" LastEditDate="2011-02-01T10:54:20.247" LastActivityDate="2011-02-01T10:54:20.247" Title="Name of this identity? $\int e^{\alpha x}\cos(\beta x) = \frac{e^{\alpha x} (\alpha \cos(\beta x)+\beta \sin(\beta x))}{\alpha^2+\beta^2}$" Tags="&lt;integral&gt;&lt;trigonometry&gt;" AnswerCount="1" CommentCount="7" />
  <row Id="19797" PostTypeId="1" CreationDate="2011-02-01T02:04:32.283" Score="2" ViewCount="148" Body="&lt;p&gt;Let P = (1, 1, -1), Q = (2, -2, -3), and R = (-1, 0, 4) be the vertices of a triangle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let L be a line passing through P and Q.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Find:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The point on the line x = y + 1 = z - 2 closest to the line L.&lt;/p&gt;&#xA;" OwnerUserId="6461" LastActivityDate="2011-02-01T02:41:53.243" Title="Find the closest point on one line to another line in 3D" Tags="&lt;calculus&gt;" AnswerCount="2" />
  <row Id="19798" PostTypeId="2" ParentId="19797" CreationDate="2011-02-01T02:20:06.567" Score="0" ViewCount="" Body="&lt;p&gt;The equation of the line $L$ through $P$ and $Q$ is $\frac{x-1}{2} = -\frac{y-1}{2} = -\frac{z+1}{3}$. &lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2011-02-01T02:20:06.567" />
  <row Id="19799" PostTypeId="2" ParentId="19796" CreationDate="2011-02-01T02:20:35.413" Score="5" ViewCount="" Body="&lt;p&gt;They are the real and imaginary parts of the following identity:&#xA;$$&#xD;&#xA;\int{e^{(\alpha+i\beta)x}dx} = \frac{e^{(\alpha+i\beta)x}}{\alpha+i\beta} = \frac{(\alpha - i\beta)e^{(\alpha+i\beta)x}}{\alpha^2 + \beta^2}.&#xD;&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="5546" LastActivityDate="2011-02-01T02:20:35.413" CommentCount="3" />
  <row Id="19800" PostTypeId="2" ParentId="19797" CreationDate="2011-02-01T02:41:53.243" Score="1" ViewCount="" Body="&lt;p&gt;Hint: A parameterized version of L is $(1+t,1-3t,-1-2t)$.  The other line is $(s, 1+s, s-2)$.  So you can just calculate the squared distance, take the derivative with respect to s and t, and get two equations in two unknowns. &lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-02-01T02:41:53.243" />
  <row Id="19801" PostTypeId="1" AcceptedAnswerId="19803" CreationDate="2011-02-01T02:53:06.803" Score="4" ViewCount="88" Body="&lt;p&gt;I remember there is some name for a    special topology on a set, such that    every subset is open i.e. every point    forms an open set, but I cannot    recall it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please tell me its name or say I am    wrong.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="1281" LastEditorUserId="1281" LastEditorDisplayName="" LastEditDate="2011-02-01T03:24:00.507" LastActivityDate="2011-02-01T03:24:00.507" Title="What is the name for the topology where every point forms an open set" Tags="&lt;topology&gt;&lt;terminology&gt;" AnswerCount="1" />
  <row Id="19802" PostTypeId="1" AcceptedAnswerId="19804" CreationDate="2011-02-01T02:55:54.760" Score="2" ViewCount="104" Body="&lt;p&gt;I have the following question: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/Tm25P.png&quot; alt=&quot;Verify that Stokes' theorem is true for...&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've managed to do the surface integral part of stokes theorem (and got an answer of 0) however I'm having a lot of trouble doing the line integral part in order to verify the theorem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As I understand it, I should be doing a line integral around the part of the plane $z = 2-2x-y$ that's in the $x,y,z &amp;gt; 0$ quadrant of the $x$ $y$ and $z$ axes but I don't know how to parametrize that particular part seeing as it's in 3D. All the examples I've done so far have been solely in the $xy$ plane. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Should I treat each side of the path I'm going to traverse as a separate line and then just do all four and add their answers up? Or am I missing something really simple?&lt;/p&gt;&#xA;" OwnerUserId="6462" LastEditorUserId="856" LastEditorDisplayName="" LastEditDate="2011-02-01T03:09:01.463" LastActivityDate="2011-02-01T03:17:24.040" Title="Verifying Stokes' theorem with a line integral" Tags="&lt;calculus&gt;&lt;vector-calculus&gt;" AnswerCount="1" CommentCount="0" />
  <row Id="19803" PostTypeId="2" ParentId="19801" CreationDate="2011-02-01T02:57:14.307" Score="5" ViewCount="" Body="&lt;p&gt;I believe it is called the &lt;a href=&quot;http://en.wikipedia.org/wiki/Discrete_topology&quot; rel=&quot;nofollow&quot;&gt;Discrete Topology&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2011-02-01T02:57:14.307" CommentCount="0" />
  <row Id="19804" PostTypeId="2" ParentId="19802" CreationDate="2011-02-01T03:17:24.040" Score="1" ViewCount="" Body="&lt;p&gt;Since your boundary is where that plane intersects the &lt;em&gt;coordinate planes&lt;/em&gt;, then you can just think in 2D by doing each part separately. You have the part in the xy-plane which is $2-2x-y=0$ by setting $z=0$. The part in the xz-plane by setting $y=0$ and getting the line $z=2-2x$. And lastly the part in the yz-plane by setting $x=0$ to get the line $z=2-y$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Parametrize those separately and do the integral. The last thing to be careful of is that when you add the integrals you &lt;em&gt;may&lt;/em&gt; get a sign wrong if you don't parametrize according to the orientation. I'd maybe sketch it, so that the orientation is clear, and then just make sure the parametrization starts and ends in the right place.&lt;/p&gt;&#xA;" OwnerUserId="2000" LastActivityDate="2011-02-01T03:17:24.040" CommentCount="1" />
  <row Id="19805" PostTypeId="2" ParentId="18445" CreationDate="2011-02-01T03:18:02.710" Score="0" ViewCount="" Body="&lt;p&gt;My webMathematica site hardmath is referring to is indeed down at the moment as I have problems getting webM work with the latest tomcat.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A few comments:&#xA;- the question of the o/p was about &quot;fastest way&quot;. That should be made a bit more precise. Usually people mean fast convergence rates with that. But &quot;fast&quot; in its original sense really means running time. And contractions of continued fractions are a good example of dramatically increasing the convergence rate, but at the expense of getting more and more complicated terms to compute for that. In other words, you need much fewer terms to attain higher precision, but every term is more complex than the simpler ones, of which you need more. Sometimes more cheap steps are faster, and sometimes few expensive terms are faster. It depends on how the computation is done. High-level language? A VM? Machine code? Assember?&#xA;- hardmath, regarding your third paragraph. Every power series, not just Taylor series, can be converted into an EQUIVALENT continued fraction expansion. The converse is not generally true, especially for delta fractions. The continued fraction expansion is (almost?) always more stable than a power series solution as the terms don't grow as fast. In a power series you are adding many terms of which numerator and denominator are growing at very fast rates, incurring a lot of numerical problems. The c/f expansions are (usually) much more stable, in fact, the parabola theorem states that Stieltjes, C-, and several other c/f expansions are even self-correcting if the backwards recursion is used. And many power series are outright trash if the function to be approximated has poles, which isn't the case here with exp(x), so I won't go into that (just consider that the p/s for tan(x) involves Bernoulli polynomials, converges slowly, converges only in a small interval between poles, whereas the c/f expansion converges rapidly with very simple terms in the whole complex plane EXCEPT at the poles -- which form a null set -- and is very stable with small terms in the convergents).&#xA;- there's also a few papers on the web that describe how sin, cos, exp, and log are computed by the chip manufacturers. The fastest way to compute these functions is to burn them on a chip and let the semiconductors do it for you, than writing software. The math they use for this is truly mind-boggling. A combination of numerical math, implementation on a chip, and sometimes GPU-based parallelization that is beyond fascinating.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Feel free to contact me through my website, www.lauschkeconsulting.com/contact.html if you need more information, I can provide you with my contraction formulas of the formulas hardmath is referring to (in the Cuyt et al book). The convergence rates are enormous, but the terms also get pretty complicated very fast, which makes numerical computations a bit slower again as more complicated terms have to be evaluated, than if you didn't contract the c/f.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Added on 20110201: My webMathematica site is live again, I solved the tomcat problem.&lt;/p&gt;&#xA;" OwnerUserId="6463" LastEditorUserId="6463" LastEditorDisplayName="" LastEditDate="2011-02-02T03:56:58.867" LastActivityDate="2011-02-02T03:56:58.867" />
  <row Id="19806" PostTypeId="2" ParentId="19672" CreationDate="2011-02-01T03:22:58.533" Score="0" ViewCount="" Body="&lt;p&gt;The mobius mu function is denoted by $\mu(x)$.&#xA;Another function that I got while studying Dirichlet eta function&#xA;$$\nu(x) = \rho\left(\frac{x}{2}\right)+\frac{1}{2} - \rho\left(\frac{x}{2}+ \frac{1}{2}\right)$$&#xA;where, $\rho(x)$ is the fractional part of $x$. Observe $\nu(x)$ only takes the values $0$ and $1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While working with a criterion (equivalent to the Nyman-Beurling approach) for Dirichlet eta function, I observed,&#xA;$$\sum_{k=1}^{n} \mu(k) \nu(n/k) = -1 $$ for all $n \geq 2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It has given me a recursive algorithm to calculate $\mu$ since calculating $\nu$ is much easier.&#xA;We start with&#xA;$$f_1(x) = \nu(x)$$&#xA;$$f_n(x) = f_{n-1}(x) + \mu(n)\nu(x/n)$$&#xA;Then,&#xA;$$\mu(n) = -1 - f_{n-1}(n)$$&lt;/p&gt;&#xA;" OwnerUserId="2320" LastActivityDate="2011-02-01T03:22:58.533" CommentCount="4" />
  <row Id="19807" PostTypeId="1" AcceptedAnswerId="19813" CreationDate="2011-02-01T03:26:51.363" Score="0" ViewCount="95" Body="&lt;p&gt;&lt;a href=&quot;http://i.imgur.com/6PLiG.jpg&quot; rel=&quot;nofollow&quot;&gt;http://i.imgur.com/6PLiG.jpg&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need help on this question. Thanks in advance!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;my calculation is like this:&#xA;Percentage difference: 31-27=4&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4% = £50,000&#xA;100% = £1.25mil&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Need to find total revenue:&#xA;Profit = Revenue - Cost&#xA;Not too sure about this part.&lt;/p&gt;&#xA;" OwnerUserId="6457" LastEditorUserId="6457" LastEditorDisplayName="" LastEditDate="2011-02-01T03:39:04.783" LastActivityDate="2011-02-01T04:19:47.310" Title="Calculate total revenue" Tags="&lt;arithmetic&gt;" AnswerCount="3" />
  <row Id="19808" PostTypeId="1" AcceptedAnswerId="19824" CreationDate="2011-02-01T03:31:57.480" Score="2" ViewCount="100" Body="&lt;p&gt;Im trying to write recursive formulas for sequences but it seems like there are different techniques depending on what type of sequence I'm dealing with. for example I want to the sequence:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} +.... \frac{1}{n}$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I see when I look at this is all the next terms are $\frac{1}{2}$ the previous term. So, I wrote the following recursive formula - which is incorrect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f(1) = 1, f(n) = f(n-1) + \frac{f(n-1)}{2}$ Why is this wrong? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is this correct? $f(1) = 1, f(n) = f(n-1) +\frac{1}{2}^{n-1}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Intuitively I thought that my original attempt looked correct. &lt;/p&gt;&#xA;" OwnerUserId="1040" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2011-06-03T13:50:23.217" LastActivityDate="2011-06-03T13:50:23.217" Title="Basic Recursion" Tags="&lt;recurrence-relations&gt;" AnswerCount="1" CommentCount="6" />
  <row Id="19809" PostTypeId="2" ParentId="19807" CreationDate="2011-02-01T03:33:47.923" Score="0" ViewCount="" Body="&lt;p&gt;Ahh, homework ;)&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Sanity check year 3: 20%+22%+31%+27% = 100%&lt;/li&gt;&#xA;&lt;li&gt;Salaries: 31%, Profit: 27%. Salaries - Profit = 4%.&lt;/li&gt;&#xA;&lt;li&gt;4% -&gt; 50k GBP&lt;/li&gt;&#xA;&lt;li&gt;Question reformulated: if 4% are 50k GBP, how much are 100%?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="5870" LastActivityDate="2011-02-01T03:33:47.923" CommentCount="4" />
  <row Id="19810" PostTypeId="2" ParentId="19807" CreationDate="2011-02-01T03:34:10.980" Score="0" ViewCount="" Body="&lt;p&gt;So what was the percentage difference between profit and salaries that year?&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-02-01T03:34:10.980" CommentCount="0" />
  <row Id="19811" PostTypeId="1" AcceptedAnswerId="19814" CreationDate="2011-02-01T03:50:41.560" Score="0" ViewCount="88" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I rewrite this equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$&#xD;&#xA;\hspace{2in}&#xD;&#xA;V_0=\frac{E_1+B_0-B_1}{(1+r)^1}&#xD;&#xA;    +&#xD;&#xA;    \frac{E_2+B_1-B_2}{(1+r)^2}&#xD;&#xA;    +&#xD;&#xA;   \frac{E_3+B_2-B_3}{1+r)^3}&#xD;&#xA;    +&#xD;&#xA;   \dots&#xD;&#xA;&#xD;&#xA;$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;into:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ &#xD;&#xA;\hspace{2in}&#xD;&#xA;V_0=B_0+&#xD;&#xA;\frac{E_1-rB_0}{(1+r)^1}&#xD;&#xA;+&#xD;&#xA;\frac{E_2-rB_1}{(1+r)^2}&#xD;&#xA;+&#xD;&#xA;\frac{E_3-rB_2}{(1+r)^3}&#xD;&#xA;+&#xD;&#xA;\dots&#xD;&#xA;$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My textbook simply states that this can be done without providing any hints.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would also appreciate if someone could provide some resources (preferably online) on how to do this sort of algebraic manipulation. I had referred to my textbooks and could not find any hints on how to do this sort of algebraic manipulation.&lt;/p&gt;&#xA;" OwnerUserId="2973" LastActivityDate="2011-02-01T04:15:40.027" Title="Rewriting Dividend Discount Model to derive residual income model" Tags="&lt;homework&gt;&lt;algebra&gt;&lt;finance&gt;" AnswerCount="1" />
  <row Id="19813" PostTypeId="2" ParentId="19807" CreationDate="2011-02-01T04:11:16.603" Score="0" ViewCount="" Body="&lt;p&gt;Given&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;(1) $revenue*0.27 = profit$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;(2) $revenue*0.31 = salaries$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;(3) $profit + 50000 = salaries$ (since salaries has higher allocated %)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;You can substitute the left side of (3) for $salaries$ in (2), then solve for $profit$:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;$revenue*0.31 = profit + 50000$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$profit = revenue*0.31 - 50000$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Now, set this and (1) equal to each other and solve for $revenue$&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$revenue*0.31 - 50000 = revenue*0.27$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.wolframalpha.com/input/?i=r%2a0.31+-+50000+%3D+r%2a0.27&quot; rel=&quot;nofollow&quot;&gt;WolframAlpha&lt;/a&gt; says revenue is about 1,250,000&lt;/p&gt;&#xA;" OwnerUserId="6394" LastEditorUserId="6394" LastEditorDisplayName="" LastEditDate="2011-02-01T04:19:47.310" LastActivityDate="2011-02-01T04:19:47.310" />
  <row Id="19814" PostTypeId="2" ParentId="19811" CreationDate="2011-02-01T04:15:40.027" Score="1" ViewCount="" Body="&lt;p&gt;Look at each E and B term on the right side.  For example, in the top equation you have $\frac{B_0}{1+r}$.  In the bottom equation you have $B_0-\frac{rB_0}{1+r}$.  Can you see that these are equal?  For $B_1$ you start with $\frac{-B_1}{1+r}+\frac{B_1}{(1+r)^2}$ and end with $\frac{-rB_1}{(1+r)^2}$.  So  $\frac{-B_1}{1+r}+\frac{B_1}{(1+r)^2}=\frac{-B_1(1+r)+B_1}{(1+r)^2}=\frac{-rB_1}{(1+r)^2}$.  The rest of the B's are similar.  The E's don't change at all.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-02-01T04:15:40.027" CommentCount="3" />
  <row Id="19815" PostTypeId="1" AcceptedAnswerId="19817" CreationDate="2011-02-01T04:16:08.843" Score="5" ViewCount="173" Body="&lt;p&gt;Hello,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Problem:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Prove that if gcd( a, b ) = 1, then gcd( a - b, a + b ) is either 1 or 2.   &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;From Bezout's Theorem, I see that am + bn = 1, and a, b are relative primes. However, I could not find a way to link this idea to a - b and a + b. I realized that in order to have gcd( a, b ) = 1, they must not be both even. I played around with some examples (13, 17), ...and I saw it's actually true :( ! Any idea?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks,&lt;br&gt;&#xA;Chan&lt;/p&gt;&#xA;" OwnerUserId="5711" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-01T04:29:45.467" LastActivityDate="2011-02-01T15:27:37.250" Title="What's special about the greatest common divisor of a + b and a - b?" Tags="&lt;elementary-number-theory&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="19816" PostTypeId="2" ParentId="19815" CreationDate="2011-02-01T04:20:31.253" Score="2" ViewCount="" Body="&lt;p&gt;Note that $d|(a-b)$ and $d|(a+b)$ where $d = \gcd(a-b, a+b)$. So $d$ divides the sum and difference (i.e. $2a$ and $2b$).&lt;/p&gt;&#xA;" OwnerUserId="2548" LastEditorUserId="2548" LastEditorDisplayName="" LastEditDate="2011-02-01T04:26:15.977" LastActivityDate="2011-02-01T04:26:15.977" CommentCount="1" />
  <row Id="19817" PostTypeId="2" ParentId="19815" CreationDate="2011-02-01T04:21:34.903" Score="9" ViewCount="" Body="&lt;p&gt;The gcd of $x$ and $y$ divides any linear combination of $x$ and $y$. And any number that divides $r$ and $s$ divides the gcd of $r$ and $s$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you add $a+b$ and $a-b$, you get &lt;code&gt;&amp;lt;blank&amp;gt;&lt;/code&gt;, so  $\mathrm{gcd}(a+b,a-b)$ divides &lt;code&gt;&amp;lt;blank&amp;gt;&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you subtract $a-b$ from $a+b$, you get &lt;code&gt;&amp;lt;blankity&amp;gt;&lt;/code&gt;, so $\mathrm{gcd}(a+b,a-b)$ divides &lt;code&gt;&amp;lt;blankity&amp;gt;&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So $\mathrm{gcd}(a+b,a-b)$ divides $\mathrm{gcd}($&lt;code&gt;&amp;lt;blank&amp;gt;,&amp;lt;blankity&amp;gt;&lt;/code&gt;$) = $&lt;code&gt;&amp;lt;blankety-blank&amp;gt;&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(For good measure, assuming the result is true you'll want to come up with examples where you get $1$ and examples where you get $2$, just to convince yourself that the statement you are trying to prove is the best you can do).&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-01T04:31:01.440" LastActivityDate="2011-02-01T04:31:01.440" CommentCount="6" />
  <row Id="19818" PostTypeId="1" AcceptedAnswerId="19819" CreationDate="2011-02-01T04:31:32.413" Score="1" ViewCount="86" Body="&lt;p&gt;I would like to solve the compound interest formula &#xA;$$V = Pz^n + c\left(\frac{z^{n+1}-z}{z-1}\right)$$&#xA;for $n$. Or, given what I would like the final value to be, the amount I can save a year, an interest rate, and how much money I have to invest initially, I would like to find how long it would take to reach my goal.&lt;/p&gt;&#xA;" OwnerUserId="6465" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-01T04:44:55.643" LastActivityDate="2011-02-01T04:44:55.643" Title="Solve for n in Compound Interest Formula" Tags="&lt;finance&gt;" AnswerCount="1" />
  <row Id="19819" PostTypeId="2" ParentId="19818" CreationDate="2011-02-01T04:43:22.293" Score="1" ViewCount="" Body="&lt;p&gt;We have:&#xA;\begin{align*}&#xA;V &amp;amp;= Pz^n + c\left(\frac{z^{n+1}-z}{z-1}\right)\\&#xA;V(z-1) &amp;amp;= Pz^n(z-1) + cz^{n+1} - cz\\&#xA;V(z-1) &amp;amp;= Pz^{n+1} - Pz^n + cz^{n+1} - cz\\&#xA;V(z-1) + cz &amp;amp;= z^n(Pz - P + cz)\\&#xA;\frac{V(z-1) + cz}{(P+c)z - P} &amp;amp;= z^n\\&#xA;\ln(V(z-1)+cz) - \ln((P+c)z - P) &amp;amp;= n\ln(z),&#xA;\end{align*}&#xA;so&#xA;$$n= \frac{\ln(V(z-1)+cz) - \ln((P+c)z-P)}{\ln z}.$$&#xA;You can use logarithms base 10 (or any base) if you prefer.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2011-02-01T04:43:22.293" />
  <row Id="19820" PostTypeId="1" AcceptedAnswerId="19823" CreationDate="2011-02-01T05:03:37.330" Score="4" ViewCount="370" Body="&lt;p&gt;Take the differentiable vector function $\vec{v}(t)$ (a velocity vector).  If its speed, $|\vec{v}(t)|=constant$, then prove that at any point which $\frac{d\vec{v}}{dt}$ is non-zero, $\frac{d\vec{v}}{dt}$ is perpendicular to $\vec{v}(t)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, if a velocity vector has constant speed, show that whenever its acceleration vector is non-zero, it is perpendicular to that velocity vector.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Intuitively this seems clear, since whenever the speed is remains constant, the acceleration is 0.  Why though are the velocity and acceleration vectors perpendicular though?  Because they share no intersection when the acceleration is non-zero since the speed is constant?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm assuming we should somehow show that $\vec{v} \cdot \vec{a} = 0$.  The thing is, I can't figure out what knowing that the speed is constant tells us about the velocity.  It could still be anything (like some trig functions squaring up and becoming 1 through an identity)?&lt;/p&gt;&#xA;" OwnerUserId="2533" LastEditorUserId="2533" LastEditorDisplayName="" LastEditDate="2011-02-03T13:24:27.580" LastActivityDate="2011-02-03T19:23:51.220" Title="Prove that a non-zero acceleration is perpendicular to a constant speed" Tags="&lt;homework&gt;&lt;vector-calculus&gt;" AnswerCount="4" CommentCount="5" />
  <row Id="19821" PostTypeId="2" ParentId="19820" CreationDate="2011-02-01T05:10:09.317" Score="4" ViewCount="" Body="&lt;p&gt;If $v(t) = \bigl(x_1(t),x_2(t),\ldots,x_n(t)\bigr)$ (I don't know what dimension you are working on), then the speed is given by&#xA;$$||v(t)|| = \sqrt{ x_1(t)^2  + \cdots + x_n(t)^2}.$$&#xA;Since this is constant, say equal to $k$, then you have&#xA;$$k^2 = x_1(t)^2 + \cdots + x_n(t)^2.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now take derivatives with respect to $t$. Compare that with $v\cdot a$ (remembering that $a(t) = v'(t)$). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Of course, if $\frac{dv}{dt}$ &lt;em&gt;is&lt;/em&gt; zero, then it is also perpendicular to $v(t)$, since $\mathbf{0}$ is perpendicular to every vector...)&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2011-02-01T05:10:09.317" />
  <row Id="19822" PostTypeId="2" ParentId="19820" CreationDate="2011-02-01T05:15:05.683" Score="0" ViewCount="" Body="&lt;p&gt;Lemma 1: Velocity vector is tangent to the velocity curve.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then consider the acceleration as a limit of velocity difference. Since the magnitude of the velocity vector is constant, the velocity vector traverse the circle of a constant radius. The difference of the two infinitely close velocity vectors is perpendicular to the radius, hence is perpendicular to velocity vector. Here the solution follows immediately.&lt;/p&gt;&#xA;" OwnerUserId="3806" LastActivityDate="2011-02-01T05:15:05.683" />
  <row Id="19823" PostTypeId="2" ParentId="19820" CreationDate="2011-02-01T05:20:00.437" Score="8" ViewCount="" Body="&lt;p&gt;You can apply the product rule when differentiating $\vec v\cdot\vec v =$constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The intuitive idea is that $\vec v(t)$ traces out a curve on a sphere centered at the origin (i.e., picture $\vec v(t)$ as a moving radius vector), while $\vec v&amp;#39;(t)=\vec a(t)$ is tangent to the sphere, and hence perpendicular to the radius at the point of tangency, namely $\vec v(t)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is also a geometric description of what this says in terms of the original curve, say $\vec x(t)$, of which $\vec v(t)$ gives the velocity.  Since $\vec v(t)$ gives a vector tangent to the trajectory of $\vec x(t)$, and in your case $\vec a(t)$ is perpendicular to $\vec v(t)$, $\vec a(t)$ is perpendicular to the trajectory of the original curve $\vec x(t)$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Acceleration can occur for &lt;a href=&quot;http://en.wikipedia.org/wiki/Acceleration#Tangential_and_centripetal_acceleration&quot; rel=&quot;nofollow&quot;&gt;2 reasons&lt;/a&gt;.  In general, the component of $\vec a(t)$ in the direction of the trajectory (or opposite this direction) tells you how the speed is changing, while the component of $\vec a(t)$ perpendicular to the trajectory tells you how the direction is changing.  So, again, in case the speed is constant, $\vec a(t)$ is perpendicular to the curve.  Its direction tells you which direction $\vec x(t)$ is turning, while its magnitude tells you how quickly $\vec x(t)$ is changing direction.  This last quantity is a constant multiple of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Curvature#Curvature_of_space_curves&quot; rel=&quot;nofollow&quot;&gt;curvature&lt;/a&gt; of $\vec x(t)$.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2011-02-03T19:23:51.220" LastActivityDate="2011-02-03T19:23:51.220" CommentCount="3" />
  <row Id="19824" PostTypeId="2" ParentId="19808" CreationDate="2011-02-01T05:20:51.523" Score="6" ViewCount="" Body="&lt;p&gt;As Leonidas suggests, your formula doesn't work (let's try it for $n=3$):&#xA;\begin{align*}&#xA;f(2) &amp;amp; = f(1) + \frac{f(1)}{2} = 1 + \frac{1}{2} = \frac{3}{2} \\&#xA;f(3) &amp;amp; = f(2) + \frac{f(2)}{2} = \frac{3}{2} + \frac{\frac{3}{2}}{2} = \frac{9}{4} \neq 1 + \frac{1}{2} + \frac{1}{4} = \frac{7}{4}.&#xA;\end{align*}&#xA;Your last summand in $f(n) = 1 + \frac{1}{2} + \frac{1}{4} + \cdots + \frac{1}{2^{n-1}}$ should be $\frac{1}{2^{n−1}}$ instead of $\frac{1}{n}$ (in the denominator there always is a power of $2$). If you write it this way then the &quot;recursion&quot; $f(n) = \left(1 + \frac{1}{2} + \cdots + \frac{1}{2^{n-2}}\right) + \frac{1}{2^{n-1}} = f(n-1) + \frac{1}{2^{n-1}}$ should be clear. But writing the parentheses in a different way yields&#xA;\begin{align*}&#xA;f(n) &amp;amp; = 1  + \left(\frac{1}{2} + \frac{1}{4} + \cdots + \frac{1}{2^{n-2}} + \frac{1}{2^{n-1}}\right) \\&#xA;&amp;amp; = 1 + \frac{1}{2}\left(1 + \frac{1}{2} + \cdots + \frac{1}{2^{n-3}} + \frac{1}{2^{n-2}}\right) \\&#xA;&amp;amp; = 1 + \frac{1}{2} f(n-1).&#xA;\end{align*}&#xA;This shows that you can define $f(1)=1$ and $f(n)=1+\frac{1}{2}f(n−1)$ recursively, and this seems to be nicer and closer to what you tried to do.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;One thing to take away from this: If you postulate a formula, you should check it for small values of $n$. If there's an obvious mistake, you'll catch it quickly and don't waste time trying to prove it. Playing with the formulas for small $n$ often gives you a feeling of what &lt;em&gt;could&lt;/em&gt; or &lt;em&gt;should&lt;/em&gt; be true.&lt;/p&gt;&#xA;" OwnerUserId="5363" LastEditorUserId="5363" LastEditorDisplayName="" LastEditDate="2011-02-01T05:27:07.773" LastActivityDate="2011-02-01T05:27:07.773" CommentCount="2" />
  <row Id="19825" PostTypeId="1" CreationDate="2011-02-01T05:22:29.407" Score="1" ViewCount="52" Body="&lt;p&gt;For a ray from the origin with a given angle in $R^n$, I am trying to find the radius at which that ray intersects the frontier of the unit &lt;em&gt;n&lt;/em&gt;-cube. In two dimensions, the picture is this:&#xA;&lt;img src=&quot;http://i.imgur.com/QLXkf.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given $\theta$, find the distance &lt;em&gt;r&lt;/em&gt; to the edge of the unit square. Or, in $n$ dimensions, given angles $\theta_1,...,\theta_{n-1}$, find the distance $r$ to the edge of the unit hypercube.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If somebody could provide a good source of information on doing trigonometry in arbitrary dimensions, that may be enough. As it is I don't know how to approach this problem.&lt;/p&gt;&#xA;" OwnerUserId="6467" LastEditorUserId="3971" LastEditorDisplayName="" LastEditDate="2011-02-01T08:20:09.303" LastActivityDate="2011-02-01T08:20:09.303" Title="Radius of a hypercube at a given angle" Tags="&lt;trigonometry&gt;&lt;euclidean-geometry&gt;&lt;convex-analysis&gt;" AnswerCount="2" />
  <row Id="19826" PostTypeId="2" ParentId="19687" CreationDate="2011-02-01T05:33:39.130" Score="2" ViewCount="" Body="&lt;p&gt;No factorization in rings without unique factorization is necessary.  Look at the proof of Theorem 2.2 in &lt;a href=&quot;http://www.math.uconn.edu/~kconrad/blurbs/gradnumthy/mordelleqn1.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.math.uconn.edu/~kconrad/blurbs/gradnumthy/mordelleqn1.pdf&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="619" LastActivityDate="2011-02-01T05:33:39.130" CommentCount="2" />
  <row Id="19827" PostTypeId="2" ParentId="19825" CreationDate="2011-02-01T05:35:38.567" Score="1" ViewCount="" Body="&lt;p&gt;In 2D it is min$(1/\cos\theta, 1/\sin\theta)$  Where are you measuring your angles?  In 3D, is $\theta_1$ the angle in the plane as you show and $\theta_2$ the angle out of the plane? &lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-02-01T05:35:38.567" />
  <row Id="19828" PostTypeId="2" ParentId="19790" CreationDate="2011-02-01T05:36:14.733" Score="5" ViewCount="" Body="&lt;p&gt;Another way to think of infinity is as follows. One way to construct the real numbers from the rationals is via Cauchy sequences, i.e. rational numbers &quot;converging&quot; (in effect) to the real number. For example, what is the meaning of $\pi$? Given a list of digits $3.14159\ldots$, we may think of $\pi$ as a sequence of rational converging to $\pi$, for example $$3, 3.1, 3.14, 3.141, 3.1415, 3.14159, \ldots$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can do the same with infinity! Let's an infinite number as a sequence diverging to infinity. One example is $$1,2,3,4,5,6,\ldots$$ We can then define all the usual arithmetic operations as usual, for example $$1,2,3,4,5,6,\ldots + 1,2,3,4,5,6,\ldots = 2,4,6,8,10,12,\ldots$$ There is now a very concrete meaning of &quot;the limit of a function $f$ at $1,2,3,4,5,6$&quot; - that is just the limit of the sequence $f(1),f(2),f(3),\ldots$. If a function converges to some value at $\infty$, then it will converge to the same value under all divergent sequence. Compare this to the definition of limit at a point $x$ - we should get the same limit whatever the sequence converging to $x$ is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Things become more complicated when we try to understand when $a &amp;lt; b$ for two infinite numbers. If $a,b$ are finite and represented by some sequences $a_i,b_i$ converging to them, then if $a&amp;lt;b$ we know that eventually $a_i &amp;lt; b_i$. However, that is not the case with infinite sequences - consider for example $1,2,3,4,5,6,\ldots$ vs. $0,4,0,8,0,12,\ldots$. This can be fixed using some &quot;decision rule&quot; implemented using an ultrafilter, which I'm not going to explain here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Having done this, we have a full fledged number system, and for each infinite &quot;number&quot; $\alpha$ we can form a corresponding infinitesimal $1/\alpha$ which is positive but smaller than any &quot;real&quot; number (i.e. smaller than any constant sequence $x,x,\ldots$ where $x$ is real).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you find this interesting, look up nonstandard analysis.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2011-02-01T05:36:14.733" />
  <row Id="19829" PostTypeId="2" ParentId="19790" CreationDate="2011-02-01T05:54:30.797" Score="1" ViewCount="" Body="&lt;p&gt;You could use the same reasoning to say that since $1$ is a real number then $\infty - 1$ has to be a real number. So you see that if you subtract anything from $\infty$ it does &lt;strong&gt;NOT&lt;/strong&gt; become a real number. Actually $\lim_{x \to \infty} x-1 = \infty$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To visualize this: Given that you have infinite money and spend 1 dollar. How much money have you left, yes its still infinite. This is a bit hard to understand when starting to deal with infinity, you can for example split your money up into two parts and both parts will remain infinite.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You see that infinity cannot be made finite by subtracting some finite number.&lt;/p&gt;&#xA;" OwnerUserId="3123" LastActivityDate="2011-02-01T05:54:30.797" />
  <row Id="19830" PostTypeId="1" AcceptedAnswerId="19831" CreationDate="2011-02-01T06:11:33.053" Score="3" ViewCount="233" Body="&lt;p&gt;Hello,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried to prove this theorem, but I'm really confused about its correctness.&lt;br&gt;&#xA;Let's say gcd( a, b ) = 1 -&gt; am + bn = 1 where m, n are integers. Then&lt;br&gt;&#xA;by multiply both side by c, we have: amc + bnc = c. &#xA;From here, is it sufficient to say c = gcd( ac, bc )?&lt;br&gt;&#xA;I feel like this proof needs some more explanation, but I can't think of one. Any idea?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks,&lt;br&gt;&#xA;Chan&lt;/p&gt;&#xA;" OwnerUserId="5711" LastActivityDate="2011-02-01T07:56:29.600" Title="If gcd( a, b ) = 1, then is it true to say gcd( ac, bc ) = c?" Tags="&lt;elementary-number-theory&gt;" AnswerCount="3" />
  <row Id="19831" PostTypeId="2" ParentId="19830" CreationDate="2011-02-01T06:16:27.070" Score="5" ViewCount="" Body="&lt;p&gt;What you have shown is that there is a linear combination of $ac$ and $bc$ that equals $c$. Now, for any $x$, $y$, the gcd of $x$ and $y$ divides any linear combination of $x$ and $y$. In this case, this means that the gcd of $ac$ and $bc$ &lt;em&gt;divides&lt;/em&gt; $c$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are almost done, but you also need to argue that $c$ divides the gcd of $ac$ and $bc$, to conclude the equality (if $z$ and $w$ are positive, and $z$ divides $w$, and $w$ divides $z$, then they are equal).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, any linear combination of $ac$ and $bc$ is (obviously) a multiple of $c$ since $c$ divides both $ac$ and $bc$. So we are in fact done.&lt;/p&gt;&#xA;" OwnerUserId="462" LastActivityDate="2011-02-01T06:16:27.070" CommentCount="1" />
  <row Id="19832" PostTypeId="2" ParentId="19830" CreationDate="2011-02-01T06:38:10.183" Score="3" ViewCount="" Body="&lt;p&gt;Generally, $a$ and $b$ can be presented as $a=a_{0}^{n_1} \cdot a_{1}^{n_2} \cdots a_{N}^{n_N}$ and $b=b_{0}^{m_1} \cdot b_{1}^{m_2} \cdots b_{M}^{m_M}$, where $a_0 \cdots a_N$ and $b_0 \cdots b_M$ are prime divisors of $a$ and $b$ respectively, and $n_1 \cdots n_N$ and $m_1 \cdots m_M$ are respective prime powers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $gcd(a,b)=1$ then $a$ and $b$ has no common prime divisors: $\forall i,j: a_i \ne b_j$. When you multiply both $a$ and $b$ by $c$, sets $a_0 \cdots a_N$ and $b_0 \cdots b_M$ are updated by prime divisors of $c$: $c_0 \cdots c_P$ if they didn't exist, or power sets are updated if some prime divisors already existed. Since $gcd(ac,bc)$ is composed of only common prime divisors of $ac$ and $bc$, and since the only common prime divisors they have are $c=c_{0}^{p_1} \cdot c_{1}^{p_2} \cdots c_{P}^{p_P}$, then $gcd(ac, bc)$ equals $c$.&lt;/p&gt;&#xA;" OwnerUserId="3806" LastActivityDate="2011-02-01T06:38:10.183" />
  <row Id="19833" PostTypeId="1" AcceptedAnswerId="19839" CreationDate="2011-02-01T07:01:45.967" Score="0" ViewCount="51" Body="&lt;p&gt;Consider a given inflation rates $I_1, I_2, \dots, I_N$ for years $1 \dots N$. What is the total (cumulative) inflation rate over the whole period of $N$ years?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was trying to chain-link the resulting &lt;code&gt;purchasing powers&lt;/code&gt;:&#xA;$P_1=P_0\frac{1}{1+I_1}$, $P_2=P_1\frac{1}{1+I_2}$, etc. yielding the $P_N=P_0\prod_{j=1}^{N}{\frac{1}{1+I_j}}$, but I cannot revert it back to &lt;code&gt;inflation rate&lt;/code&gt; $I_1^N$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT1: Okay, I got the &lt;code&gt;cumulative purchasing power rate&lt;/code&gt;: $P_1^N=\frac{P_N}{P_0}=\frac{1}{\prod_{j=1}^{N}{(1+I_j)}}$. I should express it in terms of &lt;code&gt;cumulative inflation rate&lt;/code&gt;: $P_1^N=\frac{1}{1+I_1^N}$. Thus, $\frac{1}{1+I_1^N}=\frac{1}{\prod_{j=1}^{N}{(1+I_j)}}$, that is, $1+I_1^N = \prod_{j=1}^{N}{(1+I_j)}$, which gives $I_1^N = \prod_{j=1}^{N}{(1+I_j)} - 1$. Looks like an answer.&lt;/p&gt;&#xA;" OwnerUserId="3806" LastEditorUserId="3806" LastEditorDisplayName="" LastEditDate="2011-02-01T07:41:57.960" LastActivityDate="2011-02-01T07:41:57.960" Title="What is the total inflation rate over a several year period given the each year inflation rate?" Tags="&lt;homework&gt;&lt;finance&gt;" AnswerCount="1" />
  <row Id="19834" PostTypeId="1" CreationDate="2011-02-01T07:05:27.770" Score="1" ViewCount="164" Body="&lt;p&gt;the question asks:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Find $f &amp;#39;(a)$, &#xA;$f(x) = \frac{x^2 + 1}{x - 5}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f&amp;#39;(a)=$ [answer box here]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand that $f&amp;#39;(a)$ is the derivative, but i don't understand what the derivative is. Or how to solve this. I am used to finding  a tangent slope with two points. This one doesn't have that. So i don't understand how i can use the definition to solve this. &#xA;Thanks,&lt;/p&gt;&#xA;" OwnerUserId="6470" LastEditorUserId="622" LastEditorDisplayName="" LastEditDate="2011-02-01T07:25:23.413" LastActivityDate="2011-02-01T13:52:19.890" Title="&quot;Find $f '(a)$&quot; explaining in english" Tags="&lt;calculus&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="1" />
  <row Id="19835" PostTypeId="2" ParentId="19834" CreationDate="2011-02-01T07:13:32.563" Score="1" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT:&lt;/strong&gt; Use the $\frac{f}{g}$ rule which states that $$\Bigl(\frac{f}{g}\Bigr)&amp;#39; = \frac{ g \cdot f&amp;#39; - f \cdot g&amp;#39;}{g^{2}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So here your $f(x)=x^{2}+1$ and $g(x) = x-5$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So your $$f&amp;#39;(x) = \frac{(x-5) \cdot \frac{d}{dx}\Bigl[x^{2}+1\Bigr] - (x^{2}+1) \cdot \frac{d}{dx}(x-5)}{(x-5)^{2}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After doing this plugin $x=a$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Okay if you are not comfortable with this we can do this by &lt;em&gt;Product rule&lt;/em&gt;. I hope you are familiar with the product rule which states: $$(fg)&amp;#39; = f \cdot g&amp;#39; + g \cdot f&amp;#39;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So your $f(x) = x^{2}+1$ and $g(x) = (x-5)^{-1}$. Then by Product rule you have $$(fg)&amp;#39;= (x^{2}+1) \Bigl[ (x-5)^{-2}\Bigr] + \frac{1}{x-5} \Bigl[\frac{d}{dx} (x^{2}+1)\Bigr]$$&lt;/p&gt;&#xA;" LastActivityDate="2011-02-01T07:30:35.513" CommentCount="2" />
  <row Id="19837" PostTypeId="1" CreationDate="2011-02-01T07:32:18.530" Score="0" ViewCount="55" Body="&lt;p&gt;How can i guarantee that a transportation of a vector, defined on the tangent space at a element in complete Riemannian manifold, by left-invariant translation is a parallel transport along geodesic path?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since parallel transport of a vector from a tangent space to one other depends on the curve we move along, how can i assure&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;it is along geodesic path? and&lt;/li&gt;&#xA;&lt;li&gt;parallel transport? &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I read in a text(boothby) that there exist a unique geodesic path for a complete Riemannian manifold, and also about levi-civita connection.&lt;/p&gt;&#xA;" OwnerUserId="6471" LastEditorUserId="367" LastEditorDisplayName="" LastEditDate="2011-02-01T22:43:52.887" LastActivityDate="2011-02-01T22:43:52.887" Title="Transportation of a tangent vector by left-invariant translation in complete Riemannian manifolds." Tags="&lt;lie-groups&gt;&lt;manifold&gt;&lt;riemannian-geometry&gt;" CommentCount="4" />
  <row Id="19838" PostTypeId="1" AcceptedAnswerId="19843" CreationDate="2011-02-01T07:37:11.307" Score="2" ViewCount="95" Body="&lt;p&gt;What is the idea behind defined a &lt;a href=&quot;http://en.wikipedia.org/wiki/Inverse_limit&quot; rel=&quot;nofollow&quot;&gt;Projective system&lt;/a&gt; of Groups/Rings. In our class an example for the Projective system was given by Taking the Ring $\mathbb{Z}/n\mathbb{Z}$ over $\mathbb{N}$. The order defined was: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$ m \leq n$ if $m \mid n$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Moreover then we calculated the &lt;em&gt;Projective Limit&lt;/em&gt; of this system and found it to be $\hat{\mathbb{Z}}$. I have not at all understood this concept fully so even while posing this question there might be errors. So people who can understand what i am talking about are welcome to pose answers. If one still has problems with the question please let me know i shall try to improve it or delete it.&lt;/p&gt;&#xA;" LastActivityDate="2011-02-01T13:26:29.353" Title="Idea behind defining a Projective System" Tags="&lt;abstract-algebra&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="19839" PostTypeId="2" ParentId="19833" CreationDate="2011-02-01T07:39:44.140" Score="0" ViewCount="" Body="&lt;p&gt;It's just $(1+I_1)\cdots(1+I_N) - 1$. The ratio $P_0 / P_N$ of the purchasing powers gives you 1 + the rate of inflation for the whole period.&lt;/p&gt;&#xA;" OwnerUserId="5191" LastActivityDate="2011-02-01T07:39:44.140" CommentCount="1" />
  <row Id="19840" PostTypeId="2" ParentId="19834" CreationDate="2011-02-01T07:40:37.090" Score="0" ViewCount="" Body="&lt;p&gt;In this case I would use the 'quotient rule': $\left( \frac{g}{h} \right)&amp;#39; = \frac{g&amp;#39;h - gh&amp;#39;}{h^2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means that you have to find the derivative of $g(x) = x^2 +1$ and $h(x) = x -5$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To find $g(x)$ you can use the power rule (a specific case of the chain rule), the sum rule, and the constant rule. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;First split $g(x)=x^2 + 1$ into $f(x) =x^2$ and $f(x)=1$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The power rule is if $f(x) = x^a$ where a is a constant, then $f&amp;#39;(x)=ax^{a-1}$. So looking at $x^2$ you can see that $a = 2$ so $f&amp;#39;(x) = 2x^{2-1} = 2x$, right?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then you take the constant rule and look at $f(x) = 1$. The constant rule is if $f(x) = a$ where $a$ is some constant, then $f&amp;#39;(x) = 0$. In this case $a=1$ and is constant so $f&amp;#39;(1) = 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now use the sum rule to put these together. The sum rule is: $(f + g)&amp;#39; = f&amp;#39; + g&amp;#39;$ basically saying that you can take the derivatives separately first and then add them together. So now we take the derivatives that we already calculated and add them: $2x + 0 = 2x = g&amp;#39;(x)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you now take another look at that original quotient rule, you have solved for $g&amp;#39;$... Since you already know $g$ and $h$ (without the $'$) all you need to do is find is $h&amp;#39;$ and $h^2$ and plug them in...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a good reference, it might be worth it to check out &lt;a href=&quot;http://en.wikipedia.org/wiki/Derivative&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Derivative&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="3711" LastEditorUserId="3711" LastEditorDisplayName="" LastEditDate="2011-02-01T07:57:48.503" LastActivityDate="2011-02-01T07:57:48.503" />
  <row Id="19841" PostTypeId="2" ParentId="19834" CreationDate="2011-02-01T07:48:02.883" Score="3" ViewCount="" Body="&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Derivative&quot; rel=&quot;nofollow&quot;&gt;derivative&lt;/a&gt; is a &lt;a href=&quot;http://en.wikipedia.org/wiki/Limit_of_a_function&quot; rel=&quot;nofollow&quot;&gt;limit&lt;/a&gt; of the slopes between two points that you're used to finding.  As the points get closer together, these slopes between two points approach the slope of the line tangent to the graph of the function at a point.  For an excellent explanation of the general ideas here, see &lt;a href=&quot;http://math.stackexchange.com/questions/12287/approaching-to-zero-but-not-equal-to-zero-then-why-do-the-points-get-overlapped/12290#12290&quot;&gt;this answer&lt;/a&gt; of Arturo Magidin.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you were finding the slope of the line connecting $(a,f(a))$ and a nearby point on the graph, say $(a+\Delta x, f(a+\Delta x))$, where $\Delta x$ is thought of as being small, then you would compute change in $y$ over change in $x$ as $\frac{f(a+\Delta x) - f(a)}{\Delta x}$.  The derivative $f&amp;#39;(a)$ is the limit of these quotients as $\Delta x$ goes to $0$.  One way to find $f&amp;#39;(a)$ in this case is to first do some algebra with the general expression $\frac{f(a+\Delta x) - f(a)}{\Delta x}$, until you get it in a form where it is clear what happens when $\Delta x$ goes to zero.  As indicated in other answers, there are general rules for calculating derivatives of algebraic expressions like this without appeal to limits, but if you haven't learned those yet then the direct method may be the way to go.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My approach to directly computing the limit would be to first break $f$ into simpler functions using &lt;a href=&quot;http://en.wikipedia.org/wiki/Polynomial_division&quot; rel=&quot;nofollow&quot;&gt;polynomial division&lt;/a&gt;.  This gives $f(x)=26\cdot\frac{1}{x-5}+x+5$.  The heaviest algebra will be simplying the combination of $\frac{1}{a+\Delta x - 5}$ and $\frac{1}{a-5}$ in the quotient, but you'll know you're on the right track when you can cancel the $\Delta x$ in the denominator of the whole thing.  At this point you should be able to &quot;set $\Delta x = 0$&quot; to compute the limit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In practice, after first becoming acquainted with the limit definition, everyone learns to use rules for finding derivatives of polynomials and quotients, as indicated in other answers.  Those methods are more efficient, but might seem mysterious if you haven't seen derivatives before.  An example of using the limit definition is given on the Wikipedia page &lt;a href=&quot;http://en.wikipedia.org/wiki/Derivative#Example&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2011-02-01T13:52:19.890" LastActivityDate="2011-02-01T13:52:19.890" />
  <row Id="19842" PostTypeId="2" ParentId="19830" CreationDate="2011-02-01T07:56:29.600" Score="3" ViewCount="" Body="&lt;p&gt;It's the special case $\rm\ (a,b) = 1\ $ of the following &lt;em&gt;distributive law&lt;/em&gt; for GCDs&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;LEMMA&lt;/strong&gt; $\rm\quad (a,b)\ =\ (ac,bc)/c$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof&lt;/strong&gt; $\rm\quad\ d\ |\ a,b\ \iff\ dc\ |\ ac,bc\ \iff\ dc\ |\ (ac,bc)\ \iff\ d|(ac,bc)/c$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The above holds more generally in any domain as long as $\rm\ (ac,bc)\ $ exists. See my &lt;a href=&quot;http://math.stackexchange.com/questions/16042/about-greatest-common-divisor/16043#16043&quot;&gt;post here&lt;/a&gt; for further discussion of this property and its relationship with Euclid's Lemma.&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2011-02-01T07:56:29.600" />
  <row Id="19843" PostTypeId="2" ParentId="19838" CreationDate="2011-02-01T08:12:05.897" Score="5" ViewCount="" Body="&lt;p&gt;A useful heuristic when thinking about projective systems is that you want to specify a congruence condition modulo lots of numbers (or ideals) in a compatible way. For example, you can specify any congruence condition modulo 3 and any congruence condition modulo 5. No matter, what you specify, the two can be satisfied at the same time, since 3 and 5 are co-prime: this is the Chinese remainder theorem. But if you specify a congruence condition modulo 5 and modulo 25, then the two could turn out to be incompatible. E.g. there is no number that is 0 modulo 5, but 1 modulo 25. Why? Because a number that is 0 modulo 5, is divisible by 5, so it can only by congruent to 0,5,10,15, or 20 modulo 25. Similarly, once you have chosen a congruence class modulo 3 and modulo 5, the congruence class modulo 15 is uniquely determined, while the congruence class modulo 30 is not unique but heavily restricted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The projective limit of $\mathbb{Z}/n\mathbb{Z}$ that you have described is the set of all legal simultaneous choices of congruence classes modulo all integers (with the obvious ring structure). The projective system $\mathbb{Z}/m\mathbb{Z}\rightarrow \mathbb{Z}/n\mathbb{Z}$ for $n|m$ expresses all the above rules, that I have spelled out for 3 and 5.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similraly, if you fix a prime $p$, then you can consider the projective system&#xA;$\mathbb{Z}/p^m\mathbb{Z}\rightarrow \mathbb{Z}/p^n\mathbb{Z}$ for $n\leq m$. The projective limit is denoted by $\mathbb{Z}_p$ and is the set of all possible legal simultaneous choices of congruence classes modulo higher and higher powers of $p$. Thus, the above example shows that $(0,5,\ldots)$ could represent an element of the projective limit, while $(0,1,\ldots)$ cannot.&lt;/p&gt;&#xA;" OwnerUserId="3212" LastActivityDate="2011-02-01T08:12:05.897" />
  <row Id="19844" PostTypeId="2" ParentId="19825" CreationDate="2011-02-01T08:14:36.333" Score="1" ViewCount="" Body="&lt;p&gt;In higher dimensions it is usually best to use a directional vector from the sphere $S^{n-1}$  instead of an angle $\theta$.  For convex bodies (or even star bodies) the radial function. $\rho$: $S^{n-1}$$\rightarrow R_{\ge0}$ gives the radius of the body in a particular direction.  In your case, you are working with the $l_\infty$ ball.  The radial function, when restricted to $S^{n-1}$ is simply the receprical of the minkowski functional:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Minkowski_functional&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Minkowski_functional&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In short, you need a &quot;nice&quot; formula for the radial function $\rho$ of the $l_\infty$ ball.  You may find such a formula in a convex geometry book.  In particular, Koldobsky has a book on Fourier analysis and convex geometry which contains extensive results on $l_p$ balls and radial functions.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll edit my answer if I think of more specific references.  &lt;/p&gt;&#xA;" OwnerUserId="3971" LastActivityDate="2011-02-01T08:14:36.333" />
  <row Id="19845" PostTypeId="2" ParentId="19815" CreationDate="2011-02-01T08:23:25.823" Score="1" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\rm\quad a-b\ +\ (a+b)\ i\ =\ (1+i)\ (a+b\ i)\ \ $ provides a slick proof using Gaussian integers. This reveals the arithmetical essence of the matter and, hence, suggests obvious generalizations.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2011-02-01T15:27:37.250" LastActivityDate="2011-02-01T15:27:37.250" />
  <row Id="19846" PostTypeId="2" ParentId="19749" CreationDate="2011-02-01T08:41:36.007" Score="1" ViewCount="" Body="&lt;p&gt;About the vector projection of $\vec{b}$ onto $\vec{a}$. I am taking its definition from here &lt;a href=&quot;http://en.wikipedia.org/wiki/Vector_projection&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Vector_projection&lt;/a&gt; where it is given as $(\vec{b} \cdot \vec{e}_a ) \cdot \vec{e}_a$ where $\vec{e}_a$ is the unit vector in $a$-direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First for some explanation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What does the scalar product &quot;$\cdot$&quot; do? If you look at this picture: &lt;a href=&quot;http://upload.wikimedia.org/wikipedia/commons/3/3e/Dot_Product.svg&quot; rel=&quot;nofollow&quot;&gt;http://upload.wikimedia.org/wikipedia/commons/3/3e/Dot_Product.svg&lt;/a&gt; you can see that it gives you the length of $A$ in $B$ direction. What does it mean? You can picture it like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the sun shines onto the vectors straight from above, the shadow of $A$ cast onto $B$ is exactly the length of $A$ in the direction of $B$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The scalar product is defined to be $\vec{A} \cdot \vec{B} = |\vec{A}|&amp;#160;|\vec{B}| \cos \Theta$ so you know how to calculate  this length: $|A| cos \Theta = \frac{\vec{A} \cdot \vec{B}}{|\vec{B}|}$. In your case $\vec{B} = \vec{e}_a$ is a unit vector so its length is one and therefore you get $\vec{b} \cdot \vec{e}_a = |\vec{b}| \cos \Theta$ which is the length of $\vec{b}$ in the direction of $\vec{a}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A vector always consists of a length and a direction so now you need to add a direction to the length you just computed. But the direction is the direction of $\vec{a}$. How do you get the direction only without its length? You make it a unit vector like this: $\vec{e}_a = \frac{1}{|\vec{a}|} \vec{a}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You see that $length$ $\cdot$ $direction$ is $(\vec{b} \cdot \vec{e}_a ) \cdot \vec{e}_a$ in this case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope this helps. As for the other two cases, scalar projection and orthogonal projection, I don't know what they are. Maybe you could post their definitions given in your lecture here?&lt;/p&gt;&#xA;" OwnerUserId="5798" LastActivityDate="2011-02-01T08:41:36.007" CommentCount="2" />
  <row Id="19847" PostTypeId="1" AcceptedAnswerId="19859" CreationDate="2011-02-01T08:49:15.053" Score="2" ViewCount="120" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to know how to find the number of (complex) roots of the polynomal $f(z) = z^4+3z^2+z+1$ inside the unit disk. The usual way to solve such a problem, via Rouché's theorem does not work, at least not in an &quot;obvious way&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any ideas?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;edit: here is a rough idea I had: For any $\epsilon &amp;gt;0$, let $f_{\epsilon}(z) = z^4+3z^2+z+1-\epsilon$. By Rouché's theorem, for each such $\epsilon$, $f_{\epsilon}$ has exactly 2 roots inside the unit disc. Hence, by continuity, it follows that $f$ has 2 roots on the closed unit disc, so it remains to determine what happens on the boundary. Is this reasoning correct? what can be said about the boundary?&lt;/p&gt;&#xA;" OwnerUserId="5491" LastEditorUserId="5491" LastEditorDisplayName="" LastEditDate="2011-02-01T09:07:51.473" LastActivityDate="2011-02-01T10:03:27.877" Title="Finding number of roots of a polynomial in the unit disk" Tags="&lt;complex-analysis&gt;&lt;polynomials&gt;" AnswerCount="1" CommentCount="0" />
  <row Id="19848" PostTypeId="1" AcceptedAnswerId="19857" CreationDate="2011-02-01T09:12:21.593" Score="1" ViewCount="216" Body="&lt;p&gt;&lt;strong&gt;{Xn}&lt;/strong&gt; is a sequence of independent random variables each with the same &#xA;&lt;strong&gt;Sample Space {0,1}&lt;/strong&gt; and &lt;strong&gt;Probability {1-1/$n^2$ ,1/$n^2$}&lt;/strong&gt;&#xA;    &lt;br&gt; &lt;em&gt;Does this sequence converge with probability one (Almost Sure) to the &lt;strong&gt;constant 0&lt;/strong&gt;?&lt;/em&gt;&#xA;&lt;br&gt;&lt;br&gt;Essentially the {Xn} will look like this (its random,but just to show that the frequency of ones drops)&lt;br&gt;&#xA;   010010000100000000100000000000000001......&lt;/p&gt;&#xA;" OwnerUserId="5982" LastEditorUserId="5982" LastEditorDisplayName="" LastEditDate="2011-02-28T05:26:37.160" LastActivityDate="2011-02-28T05:26:37.160" Title="Almost sure convergence" Tags="&lt;probability-theory&gt;&lt;probability&gt;" AnswerCount="4" />
  <row Id="19849" PostTypeId="1" CreationDate="2011-02-01T09:14:30.457" Score="6" ViewCount="93" Body="&lt;p&gt;I have a question concerning the formulation of an (3D) ellipsoid. The most common definition for an ellipsoid seems to be:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$E = \{ x=\left( x_1, \dots x_n \right)^T \in R^n: \sum_{i=1}^n \left( \frac{x_i}{r_i} \right)^2 = 1 \}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;based on the different radii $r_i$. However for an ellipse (2D) I also found the following definition:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$E = \{ x \in R^2: \| x - f_1 \| + \| x - f_2 \| = c \}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is based on the two focal points $f_1$ and $f_2$ of an ellipse. Now my question is, if there exist such kind of definitions also for ellipsoid in 3D or higher? How many focal points can an ellipsoid in 3D or higher actually have? I am especially interested in ellipsoids where all radii $r_1, \dots, r_n$ can be different (like a Scalene or triaxial ellipsoids).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many thanks in advance&lt;/p&gt;&#xA;" OwnerUserId="6197" LastActivityDate="2011-02-01T09:14:30.457" Title="Definition of an ellipsoid based on its focal points" Tags="&lt;conic-sections&gt;" CommentCount="3" FavoriteCount="1" />
  <row Id="19850" PostTypeId="1" AcceptedAnswerId="19855" CreationDate="2011-02-01T09:18:16.103" Score="3" ViewCount="188" Body="&lt;p&gt;I throw two coins (simultaneously). A student (very much a beginner in both math and probability theory) thought that the following 3 outcomes are equally likely: &quot;two heads&quot;, &quot;two tails&quot;, &quot;a head and a tail&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However much I tried, I couldn't find a clear and obvious explanation of why it's not the case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, I could say the coins are distinct, so we need to look at how each individual coin falls (leading to 4 equally likely outcomes: HH, HT, TH, TT). But I couldn't clearly explain the concept of &quot;distinct&quot; and why it's important for the probability calculation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyone can help with a simple, precise and very intuitive explanation?&lt;/p&gt;&#xA;" OwnerUserId="6474" LastActivityDate="2011-02-01T17:28:45.683" Title="how to explain that Prob[heads, tails] = 2 * Prob[heads, heads] to a student?" Tags="&lt;probability&gt;&lt;intuition&gt;&lt;education&gt;" AnswerCount="5" CommentCount="2" />
  <row Id="19851" PostTypeId="2" ParentId="19848" CreationDate="2011-02-01T09:21:09.443" Score="0" ViewCount="" Body="&lt;p&gt;If you have given $P(X_n = 0) = 1 - \frac{1}{n^2}$ and $P(X_n = 1) = \frac{1}{n^2}$ then $P(\lim_{n \rightarrow \infty} X_n = 0) = \lim_{n \rightarrow \infty} P(X_n = 0) = \lim_{n \rightarrow \infty} 1 - \frac{1}{n^2} = 1$ which per definition means that $X_n$ converges almost sure towards $X=0$.&lt;/p&gt;&#xA;" OwnerUserId="5798" LastActivityDate="2011-02-01T09:21:09.443" CommentCount="6" />
  <row Id="19852" PostTypeId="2" ParentId="16650" CreationDate="2011-02-01T09:26:50.120" Score="0" ViewCount="" Body="&lt;p&gt;Are you looking for the &lt;a href=&quot;http://en.wikipedia.org/wiki/Chebyshev%27s_inequality&quot; rel=&quot;nofollow&quot;&gt;Chebyshev's inequality&lt;/a&gt;? Obviously you need to know the variance of the distribution to apply it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Chebyshev inequality,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;P(|X−μ|≥kσ)≤ 1/$k^2$&lt;/strong&gt; &lt;br&gt;&lt;br&gt;&lt;em&gt;where μ = E(X) and σ = Standard Deviation&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;.&lt;/p&gt;&#xA;" OwnerUserId="5982" LastActivityDate="2011-02-01T09:26:50.120" />
  <row Id="19854" PostTypeId="2" ParentId="19850" CreationDate="2011-02-01T09:29:02.270" Score="1" ViewCount="" Body="&lt;p&gt;There are 4 outcomes, not 3: &quot;HH&quot;, &quot;TH&quot;, &quot;HT&quot; and &quot;TT&quot;. And they are all equally likely, $p = \frac{1}{4}$. So $P(HT) = \frac{1}{4}$. If you don't distinguish between &quot;HT&quot; and &quot;TH&quot;, then $P(&amp;quot;HT or TH&amp;quot;) = P(HT) + P(TH) = \frac{1}{2}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wouldn't write $P(HT) = 2 P(HH)$. This is just my opinion but expressing something in terms of something else that only works by coincidence in this particular case is very confusing, especially to a learner. When your coins are biased, it's broken.&lt;/p&gt;&#xA;" OwnerUserId="5798" LastEditorUserId="5798" LastEditorDisplayName="" LastEditDate="2011-02-01T17:28:45.683" LastActivityDate="2011-02-01T17:28:45.683" />
  <row Id="19855" PostTypeId="2" ParentId="19850" CreationDate="2011-02-01T09:30:03.847" Score="6" ViewCount="" Body="&lt;p&gt;Try painting one of the coins red.  Paint can't possibly affect the probabilities of the oucomes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternately, get ten coins and ask the student how many outcomes she thinks there are and whether ten heads is more or less likely than five heads and five tails.  &lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2011-02-01T10:32:32.823" LastActivityDate="2011-02-01T10:32:32.823" CommentCount="1" />
  <row Id="19856" PostTypeId="2" ParentId="19850" CreationDate="2011-02-01T09:34:29.257" Score="9" ViewCount="" Body="&lt;p&gt;Arrange with the student that you'll give them 1 dollar for each HH, and they'll give you 1 dollar for each HT. Then start tossing. Either they'll get the point quite fast, or you'll make a nice profit, so you win either way. &lt;/p&gt;&#xA;" OwnerUserId="5656" LastActivityDate="2011-02-01T09:34:29.257" CommentCount="1" />
  <row Id="19857" PostTypeId="2" ParentId="19848" CreationDate="2011-02-01T09:34:40.473" Score="1" ViewCount="" Body="&lt;p&gt;Yes, but how fast does the frequency drop ? The faster it does the more it is probable that (Xn) converges to 0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lemma : If (Xn) is a sequence of elements in {0,1}, Xn converges to 0 if and only if Xn has finitely many 1s :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If a sequence (Xn) converges to 0, then by definition of the limit, there exists some integer N such that for all n&gt;=N, |Xn| &amp;lt;= 1/2. Now since Xn is 0 or 1, |Xn| &amp;lt;= 1/2 implies that Xn = 0.&#xA;Thus, for all n &gt;= N, Xn = 0.&#xA;This means that the sequence (Xn) has finitely (in fact, less than N) many 1s.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conversely, if a sequence (Xn) has finitely many 1s, there exist an integer N (the index of the last 1 of the sequence) such that Xn = 0 for all n &gt; N.&#xA;Then the sequence (Xn) converges to 0 because for any $\varepsilon$ &gt; 0, we do have that for all n &gt; N, $|X_n - 0| = 0 &amp;lt; \varepsilon$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here, define $Y_n = \Sigma_{k=1}^n X_n$ and $Y_\infty = \lim_{n \rightarrow \infty} Y_n$ = the number of 1s in the sequence (Xn). The lemma tells us that (Xn) converges to 0 if and only if $Y_\infty &amp;lt; \infty$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You'll notice that $E[Y_\infty] = \lim E[Y_n] \lt \infty$ thanks to the fact that $\Sigma \frac{1}{n^2}$ is convergent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This implies that $Y_\infty &amp;lt; \infty$ almost surely, which in turn shows that (Xn) does converge with probability 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have the frequency drop too slowly, this won't work. For example if P(Xn = 1) = 1/log(n) for n&gt;=3, it will almost always diverge.&lt;/p&gt;&#xA;" OwnerUserId="6270" LastEditorUserId="6270" LastEditorDisplayName="" LastEditDate="2011-02-02T10:10:40.437" LastActivityDate="2011-02-02T10:10:40.437" CommentCount="7" />
  <row Id="19858" PostTypeId="2" ParentId="19850" CreationDate="2011-02-01T09:36:20.417" Score="1" ViewCount="" Body="&lt;p&gt;To distinguish &quot;T&quot; and &quot;H&quot; would be our task. Why not transfer the problem into other settings. For example. We call two people walking together as &quot;TT&quot;, while they riding two bycycles as &quot;HH&quot;. Then when A is riding one bycicle and B is sitting at the backseat, we denote it as &quot;TH&quot;. Finally, &quot;HT&quot; is also similiarly defined. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Such explanations can be various.&lt;/p&gt;&#xA;" OwnerUserId="5476" LastActivityDate="2011-02-01T09:36:20.417" />
  <row Id="19859" PostTypeId="2" ParentId="19847" CreationDate="2011-02-01T09:42:41.760" Score="2" ViewCount="" Body="&lt;p&gt;This one is slightly tricky, but you can apply Rouch&amp;eacute; directly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $g(z) = 3z^2 + 1$. Note that $|g(z)| \geq 2$ for $|z| = 1$ with equality only for $z = \pm i$ (because $g$ maps the unit circle onto the circle with radius $3$ centered at $1$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand for all $|z| = 1$ we have the estimate $h(z) = |f(z) - g(z)| = |z(z^3 + 1)| \leq 2$ and for $z = \pm i$ we have $h(\pm i) = \sqrt{2} &amp;lt; 2 \leq |g(\pm i)|$. Therefore $|f(z) - g(z)| &amp;lt; |g(z)|$ for all $|z| = 1$ and thus Rouch&amp;eacute; can be applied.&lt;/p&gt;&#xA;" OwnerUserId="5363" LastEditorUserId="5363" LastEditorDisplayName="" LastEditDate="2011-02-01T10:03:27.877" LastActivityDate="2011-02-01T10:03:27.877" />
  <row Id="19860" PostTypeId="2" ParentId="19850" CreationDate="2011-02-01T09:44:42.273" Score="8" ViewCount="" Body="&lt;p&gt;Try to draw the following diagram for him.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/rqCm3.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-Let him explain why every arrow represents the probability $\frac{1}{2}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;-Let him find all paths from Start to the result he wants (TH, or HT). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;-Let him calculate the total probability.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These steps should be elementary enough for him to get the correct answer.&lt;/p&gt;&#xA;" OwnerUserId="3123" LastActivityDate="2011-02-01T09:44:42.273" />
  <row Id="19861" PostTypeId="2" ParentId="19848" CreationDate="2011-02-01T10:38:50.877" Score="1" ViewCount="" Body="&lt;p&gt;Hint: The Borel-Cantelli lemma.&lt;/p&gt;&#xA;" OwnerUserId="822" LastActivityDate="2011-02-01T10:38:50.877" />
  <row Id="19862" PostTypeId="1" AcceptedAnswerId="19863" CreationDate="2011-02-01T10:42:43.860" Score="9" ViewCount="453" Body="&lt;p&gt;Are there unsolved problems in math that are large believed to be true, but for reasons other then statistical justification?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems that &lt;a href=&quot;http://mathoverflow.net/questions/31585/heuristic-justification-for-goldbachs-conjecture&quot; rel=&quot;nofollow&quot;&gt;Goldbach should be true&lt;/a&gt;, but this is based on heuristic justification.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am looking for conjectures that seem to be true, but where the 'why' is something other then a statistical justification, and I want know what exactly that 'why' is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: Can you please include the &lt;em&gt;reason&lt;/em&gt; it is widely believed to be true in your posting? That is the interesting part&lt;/p&gt;&#xA;" OwnerUserId="4929" LastEditorUserId="4929" LastEditorDisplayName="" LastEditDate="2011-02-01T18:49:24.140" LastActivityDate="2011-02-01T19:42:29.747" CommunityOwnedDate="2011-02-01T12:34:07.227" Title="Problems that are largely believed to be true, but are unresolved" Tags="&lt;soft-question&gt;&lt;intuition&gt;&lt;philosophy&gt;" AnswerCount="6" CommentCount="20" />
  <row Id="19863" PostTypeId="2" ParentId="19862" CreationDate="2011-02-01T10:54:18.780" Score="5" ViewCount="" Body="&lt;p&gt;What about $P \neq NP$? Scott Aaronson has made some excellent points on this:&#xA;&lt;a href=&quot;http://www.scottaaronson.com/blog/?p=122&quot; rel=&quot;nofollow&quot;&gt;http://www.scottaaronson.com/blog/?p=122&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2394" LastActivityDate="2011-02-01T10:54:18.780" CommunityOwnedDate="2011-02-01T12:34:07.227" CommentCount="4" />
  <row Id="19864" PostTypeId="1" CreationDate="2011-02-01T11:15:50.410" Score="0" ViewCount="68" Body="&lt;p&gt;Hello ! I have a random vector X = ( X1 , X2 ) that has a bidimensional normal repartition with mean 0 and covariation matrix :&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;               Sigma = | 1 q |&#xA;                       | q 1 |&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I need to calculate q , which is &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;q = Corr(X1,X2)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;How can i calculate Corr(X1,X2) ?&lt;/p&gt;&#xA;" OwnerUserId="2777" LastActivityDate="2011-02-01T11:15:50.410" Title="Calculate Correlation between 2 values" Tags="&lt;normal-distribution&gt;&lt;correlation&gt;" CommentCount="0" />
  <row Id="19865" PostTypeId="1" AcceptedAnswerId="19868" CreationDate="2011-02-01T11:31:48.153" Score="3" ViewCount="103" Body="&lt;p&gt;What does this set look like? $A = \{ (x, \frac{1}{x}) |&amp;#160;  x \in \mathbb{R} \backslash 0 \}$, where $A$ is a subset of $\mathbb{R}^2$ with the Euclidean topology.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I thought $A = (-\infty, 0] \cup [0, \infty)$ but I know that its projection $(x,y) \rightarrow x$ is not closed, so this must be wrong.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for your help.&lt;/p&gt;&#xA;" OwnerUserId="5798" LastEditorUserId="5798" LastEditorDisplayName="" LastEditDate="2011-02-01T11:40:08.750" LastActivityDate="2011-02-01T19:05:04.587" Title="Closure of $A = \{ (x, \frac{1}{x}) | x \in \mathbb{R} \backslash 0 \}$" Tags="&lt;topology&gt;&lt;general-topology&gt;" AnswerCount="2" />
  <row Id="19866" PostTypeId="1" AcceptedAnswerId="19872" CreationDate="2011-02-01T11:34:44.800" Score="0" ViewCount="120" Body="&lt;p&gt;In spatial euclidean vector spaces norm is an intuitive concept: It measures the distance from the null vector and from other vectors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The generalization to function spaces is quite a mental leap (at least for me).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My question:&lt;/strong&gt; &lt;br&gt;Is there some kind of intuition what &quot;norm&quot; or &quot;euclidean norm&quot; is even supposed to &quot;mean&quot; here? What is the null vector and how what does a &quot;distance&quot; between to functions reveal?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;: Please also see this follow up question: &lt;a href=&quot;http://math.stackexchange.com/questions/19902/visualization-of-2-dimensional-function-spaces&quot;&gt;Visualization of 2-dimensional function spaces&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="346" LastEditorUserId="346" LastEditorDisplayName="" LastEditDate="2011-02-01T15:24:39.390" LastActivityDate="2011-02-01T15:24:39.390" Title="What is the norm measuring in function spaces" Tags="&lt;real-analysis&gt;&lt;measure-theory&gt;&lt;intuition&gt;&lt;vector-space&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="1" />
  <row Id="19867" PostTypeId="1" AcceptedAnswerId="19929" CreationDate="2011-02-01T11:38:27.410" Score="3" ViewCount="162" Body="&lt;p&gt;What is the closure of $\{ (x_n) |&amp;#160;  x_n = 0 $ for almost all $n \}$ in $\mathbb{R}^\mathbb{N}$ with the product topology?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure how to think about it. A point $x$ is in the closure of a set $A$ if for every neighbourhood $N$ of $x$, $N \cap A \neq \emptyset$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I'm not sure what a neighbourhood of a point in $\mathbb{R}^\mathbb{N}$ looks like. In fact, I'm not even sure what an open ball looks like.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can someone tell me what open balls or neighbourhoods look like?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for your help.&lt;/p&gt;&#xA;" OwnerUserId="5798" LastActivityDate="2011-04-06T00:04:59.777" Title="Closure of $\{ (x_n) | x_n = 0 $ for almost all $n \}$" Tags="&lt;topology&gt;&lt;general-topology&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="19868" PostTypeId="2" ParentId="19865" CreationDate="2011-02-01T11:38:41.903" Score="4" ViewCount="" Body="&lt;p&gt;Interpreting your notation to mean that $A$ is a certain subset of $\mathbb{R}^2$...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint&lt;/strong&gt; $A$ can be treated as the graph of some function.&lt;/p&gt;&#xA;" OwnerUserId="1543" LastActivityDate="2011-02-01T11:38:41.903" CommentCount="3" />
  <row Id="19869" PostTypeId="2" ParentId="19776" CreationDate="2011-02-01T11:41:24.153" Score="4" ViewCount="" Body="&lt;p&gt;let $A$ be the invertible matrix whose eigenvalues are $x_1, . . . ,x_n$. We write &#xA;$e_k(A)$ for $e_k(x_1,. . ., x_n)$. The generating function of the elementary symmetric polynomials:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$det(1+sA) = \sum_{k=0}^ne_k(A)s^k$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sum_{k=0}^ne_k^2=(2 \pi i)^{-1}\oint \frac{1}{s}det(1+sA)det(1+s^{-1}A) ds $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where the integration is over a closed contour encircling the origin. (The contour integral just picks the &#xA;coeficient of $s^0$). Therefore: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sum_{k=0}^ne_k^2=(2 \pi i)^{-1}\oint \frac{1}{s} det(A) (s+s^{-1})^n det(1+ (s+s^{-1})^{-1}(A+A^{-1})) ds $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$=\sum_{k=0}(2 \pi i)^{-1}\oint \frac{1}{s} (s+s^{-1})^k e_{n-k}(A+A^{-1}) ds $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the polynomial $(s+s^{-1})^k$ has an $s^0$ only if k is even. Let k = 2j, then the binomial coefficient&#xA;of  $s^0$  is $2j \choose j $, which gives the required result.&lt;/p&gt;&#xA;" OwnerUserId="1430" LastEditorUserId="1430" LastEditorDisplayName="" LastEditDate="2011-02-01T11:51:31.260" LastActivityDate="2011-02-01T11:51:31.260" CommentCount="1" />
  <row Id="19870" PostTypeId="2" ParentId="16650" CreationDate="2011-02-01T11:54:50.550" Score="0" ViewCount="" Body="&lt;p&gt;Assuming I understand the question correctly, there is no such bound other than 1. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If S is well concentrated (say, a non zero constant), kS-E(S) can be made as large as you want by increasing E(S).&lt;/p&gt;&#xA;" OwnerUserId="6478" LastActivityDate="2011-02-01T11:54:50.550" CommentCount="1" />
  <row Id="19871" PostTypeId="1" AcceptedAnswerId="19882" CreationDate="2011-02-01T11:59:52.313" Score="2" ViewCount="82" Body="&lt;p&gt;To do some computations (such as finding copies of GL(2,3) in a larger of groups), I need presentation of GL(2,3). ( Because: Sylow-2 subgroups of GL(2,3) is semidihedral :$  $. It is easy to find copies of semidihedral groups in bigger groups. Therefore, to get a copy of GL(2,3) in larger group, it is sufficient to get element of order 3. If we have presentation of GL(2,3), the relations will provide some path to get elements of order 3, and will give a copy of GL(2,3). ) Can anyone help?&lt;/p&gt;&#xA;" OwnerUserId="8186" LastActivityDate="2011-02-01T13:06:28.217" Title="Presentation of GL(2,3)" Tags="&lt;group-theory&gt;" AnswerCount="1" FavoriteCount="2" />
  <row Id="19872" PostTypeId="2" ParentId="19866" CreationDate="2011-02-01T12:02:42.597" Score="4" ViewCount="" Body="&lt;p&gt;Since you used the word &quot;distance&quot;, I am going to focus more on metric spaces rather than norms. But once you see what I'm driving at, the basic idea is the same. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In usual, Euclidean space, there is the &lt;em&gt;intuitive&lt;/em&gt; sense of &quot;distance&quot; between two points. This allows us to intuitively define a metric function, which takes input two points in Euclidean space, and outputs the &quot;distance&quot; between them. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now there are a lot of things we know about the Euclidean distance (triangle inequality etc.) In an effort to find out the logical relationships between the various properties of the Euclidean distance, we try to distill out the bare necessities of the properties of the metric function. In particular, we try to see how much of the usual Euclidean intuition can be removed while still leaving a logically consistent whole. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fromr this we were eventually led to the definition of a &lt;a href=&quot;http://en.wikipedia.org/wiki/Metric_space&quot; rel=&quot;nofollow&quot;&gt;metric space&lt;/a&gt; as nothing more than a set $M$ and a distance function $d$, taking as input two points in $M$ and outputing a non-negative real number, with certain basic properties on $d$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can think of this &lt;em&gt;abstraction&lt;/em&gt; as a way to offer insights on &lt;em&gt;when can we draw an analogy between whatever it is we are looking at, and the distance function of Euclidean space&lt;/em&gt;. The abstraction says that &quot;for all intents and purposes, the really useful properties of the Euclidean distance is captured in such and such properties.&quot; So whenever we are faced with a situation where we have a set $M$ and a function $d: M\times M\to\mathbb{R}$ with those properties, we can &lt;em&gt;by analogy think of Euclidean space and its distance function&lt;/em&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I am trying to say is that perhaps looking for &quot;meaning&quot; of the norm-induced metric on a function space is going about the wrong way. It is not because we intuit certain qualities of functions which reminds us of the Euclidean distance that we consider the metrics. Rather, you should think of the norms and metrics being something we impose on the space of functions &lt;em&gt;so that&lt;/em&gt; we can appeal to our intuitions of the Euclidean distance function by analogy. &lt;/p&gt;&#xA;" OwnerUserId="1543" LastActivityDate="2011-02-01T12:02:42.597" CommentCount="1" />
  <row Id="19873" PostTypeId="2" ParentId="19866" CreationDate="2011-02-01T12:05:16.820" Score="2" ViewCount="" Body="&lt;p&gt;First understand what a norm means on 2 dimensional space (the plane). There are many norms: one for each convex symmetric body you can draw around the origin. Then the norm of a point is the ratio between the length of the line from the origin to the point, and the segment from the origin to its intersection with the edge of the convex symmetric body.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Same applies to R^n, and also to higher dimension vector spaces such as function spaces. So don't look for a &quot;one true meaning&quot; for all norms. Understand norms one at a time. I would start with the sup norm, L_1 norm and then L_2 norm.&lt;/p&gt;&#xA;" OwnerUserId="6478" LastActivityDate="2011-02-01T12:05:16.820" />
  <row Id="19874" PostTypeId="2" ParentId="19862" CreationDate="2011-02-01T12:11:41.020" Score="7" ViewCount="" Body="&lt;p&gt;The Riemann hypothesis is largely believed to be true, and further conjectures have been made based on its truth (e.g. statements about the distribution of prime numbers) but no one has ever proved it.&lt;/p&gt;&#xA;" OwnerUserId="3812" LastActivityDate="2011-02-01T12:11:41.020" CommunityOwnedDate="2011-02-01T12:34:07.227" CommentCount="1" />
  <row Id="19875" PostTypeId="1" AcceptedAnswerId="19883" CreationDate="2011-02-01T12:18:03.290" Score="4" ViewCount="235" Body="&lt;p&gt;I have to prove the following without any use of further mathematical theories except basic calculus and linear algebra:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $ \ell $ be a positive integer and $A$ a real, symmetric and positive definite $n \times n$-Matrix (square Matrix with $n$ columns and $n$ rows). Show that the following identity holds:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \int_{\mathbb{R}^n} x_{i_1} \cdots x_{i_{2 \ell}} e^{-\frac{1}{2} \left&amp;lt; \mathbf{ x}, A \&#xD;&#xA;&#xD;&#xA;    \mathbf{x} \right&amp;gt;} d^n x = \frac{(2 \pi)^{n/2}}{\ell! \sqrt{\det{A}}} \sum\limits_{&#xD;&#xA;    \begin{array}&#xD;&#xA;    \{ \{ k_1,k_1&amp;#39;\},\ldots,\{k_\ell,k_\ell&amp;#39;\}\in P \\&#xD;&#xA;    \cup_{j=1}^\ell \{ k_j, k_j&amp;#39; \} = \{ 1,\ldots,2 \ell \}&#xD;&#xA;    \end{array} }&#xD;&#xA;    ( A^{-1} )_{i_{k_1},i_{k_1&amp;#39;}} \cdots ( A^{-1} )_{i_{k_\ell},i_{k_\ell&amp;#39;}}&#xD;&#xA;    $$&#xA;    with $P = \left\{ \{k,k&amp;#39;\}, k \not= k&amp;#39; \in \{1,\ldots,2 \ell\} \right\}$ and $\left&amp;lt; \mathbf{x}, \mathbf{y} \right&amp;gt;$ the standard scalar product for $\mathbf{x},\mathbf{y} \in \mathbb{R}^n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Frankly speaking I have no idea what to do, where to begin. I'm also afraid that I cannot really explain the notation, because I don't get it 100% myself. In fact we are a group of approximately 15 students and we all don't know a smooth short way. This exercise could be part of a 2 hour exam with 5 other exercises. So we suppose there exists a more or less short solution.&lt;/p&gt;&#xA;" OwnerUserId="5450" LastEditorUserId="5450" LastEditorDisplayName="" LastEditDate="2011-02-01T17:09:02.783" LastActivityDate="2011-02-01T17:09:02.783" Title="WANTED: Short proof for challenging identity" Tags="&lt;homework&gt;&lt;mathematical-physics&gt;" AnswerCount="1" CommentCount="6" />
  <row Id="19876" PostTypeId="1" AcceptedAnswerId="19896" CreationDate="2011-02-01T12:30:51.717" Score="4" ViewCount="333" Body="&lt;p&gt;Very important in integrating things like $\int \cos^{2}(\theta) d\theta$ but it is hard for me to remember them. So how do you deduce this type of formulae? If I can remember right, there was some $e^{\theta i}=\cos(\theta)+i \sin(\theta)$ trick where you took $e^{2 i \theta}$ and $e^{-2 i \theta}$. While I am drafting, I want your ways to remember/deduce things (hopefully fast).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;[About replies]&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;About TPV's suggestion, how do you attack it geometrically??&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;$\cos^{2}(x) - \sin^{2}(x)=\cos(2x)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;plus $2\sin^{2}(x)$, then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\cos^{2}(x)+\sin^{2}(x)=\cos(2x)+2\sin^{2}(x)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and now solve homogenous equations such that &lt;code&gt;LHS=A&lt;/code&gt; and &lt;code&gt;RHS=B&lt;/code&gt;, where $A\in\mathbb{R}$ and $B\in\mathbb{R}$. What can we deduce from their solutions?&lt;/p&gt;&#xA;" OwnerUserId="5902" LastEditorUserId="5902" LastEditorDisplayName="" LastEditDate="2011-02-01T13:59:18.390" LastActivityDate="2011-05-18T18:56:54.597" Title="How to deduce trigonometric formulae like $2 \cos(\theta)^{2}=\cos(2\theta) +1$?" Tags="&lt;integral&gt;" AnswerCount="7" CommentCount="0" />
  <row Id="19877" PostTypeId="2" ParentId="19876" CreationDate="2011-02-01T12:36:39.083" Score="10" ViewCount="" Body="&lt;p&gt;By far, the best approach to trigonometric formulae is to use &lt;a href=&quot;http://en.wikipedia.org/wiki/Euler%27s_identity&quot; rel=&quot;nofollow&quot;&gt;Euler's identity&lt;/a&gt; and its corollary, &lt;a href=&quot;http://en.wikipedia.org/wiki/De_Moivre%27s_formula&quot; rel=&quot;nofollow&quot;&gt;de Moivre's formula&lt;/a&gt;, recalling that $\sin^2 + \cos^2 = 1$.&lt;/p&gt;&#xA;" OwnerUserId="589" LastActivityDate="2011-02-01T12:36:39.083" CommentCount="0" />
  <row Id="19878" PostTypeId="1" CreationDate="2011-02-01T12:41:12.717" Score="0" ViewCount="77" Body="&lt;p&gt;If I knew that by the year 2000 that there were 1 trillion humans, and that they started reproducing 2000 years before that, how would I calculate their birth rate?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume an initial population of 1 million, and a reproduction age of 15.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also take into account a standard death rate, where death occurs after 75 years, and reproduction age ends at 50.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update&lt;/strong&gt;&#xA;: Assume a uniform age distribution.&lt;/p&gt;&#xA;" OwnerUserId="6479" LastEditorUserId="6479" LastEditorDisplayName="" LastEditDate="2011-02-02T02:39:03.860" LastActivityDate="2011-02-02T02:39:03.860" Title="Calculate reproduction rate" Tags="&lt;algebra&gt;&lt;logarithms&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="19879" PostTypeId="2" ParentId="19876" CreationDate="2011-02-01T12:46:33.443" Score="0" ViewCount="" Body="&lt;p&gt;For this you need only two very basic trigonometric identities:&#xA;$$ \cos^2(x) + \sin^2(x) = 1 $$&#xA;and&#xA;$$ \cos^2(x) - \sin^2(x) = \cos(2x) $$&#xA;Add or subtract them, and you have the identities for $\cos^2(x)$ and $\sin^2(x)$.&lt;/p&gt;&#xA;" OwnerUserId="4216" LastActivityDate="2011-02-01T12:46:33.443" CommentCount="2" />
  <row Id="19880" PostTypeId="1" CreationDate="2011-02-01T12:52:08.787" Score="1" ViewCount="181" Body="&lt;p&gt;I want to write down $\ln(\cos(x))$  Maclaurian polynomial of degree 6. I'm having trouble understanding what I need to do, let alone explain why it's true rigorously.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The known expansions of $\ln(1+x)$ and $\cos(x)$ gives:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\forall x \gt -1,\ \ln(1+x)=\sum_{n=1}^{k} (-1)^{n-1}\frac{x^n}{n} + R_{k}(x)=x-\frac{x^2}{2}+\frac{x^3}{3}+R_{3}(x)$$&#xA;$$\cos(x)=\sum_{n=0}^{k} (-1)^{n}\frac{x^{2n}}{(2n)!} + T_{2k}(x)=1-\frac{x^2}{2}+\frac{x^4}{24}+T_{4}(x)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Writing $\ln(1+x)$ with $t=x-1$ gives:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\forall t \gt 0,\ \ln(t)=\sum_{n=1}^{k} (-1)^{n-1}\frac{(t+1)^n}{n} + R_{k}(t)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But now I'm clueless. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Do I just 'plug' $\cos(x)$ expansion in $\ln(t)$? Can I even do that?&lt;/li&gt;&#xA;&lt;li&gt;Isn't it a problem that $\ln(x)$ isn't defined for $x\leq 0$ but $|cos(x)| \leq 1$?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="6480" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2011-02-01T14:14:25.940" LastActivityDate="2011-02-01T17:42:21.917" Title="Maclaurin polynomial of $\ln(\cos(x))$" Tags="&lt;taylor-expansion&gt;" AnswerCount="4" />
  <row Id="19881" PostTypeId="1" AcceptedAnswerId="19886" CreationDate="2011-02-01T12:54:05.557" Score="0" ViewCount="37" Body="&lt;p&gt;So I have a large txt file with such contents (like 20mbs long)&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;20 30 40 550 60 70 80 91&#xA;20 30 40 50 60 70 80 92&#xA;20 30 40 50 60 70 80 93&#xA;20 30 40 50 64 70 80 90&#xA;20 30 40 50 60 70 80 90&#xA;20 30 40 50 60 70 80 90&#xA;20 30 40 40 60 70 80 90&#xA;40 30 40 50 60 70 80 90&#xA;&#xA;4 5 6 6&#xA;&#xA;20 30 40 50 60 70 80 91&#xA;20 30 40 50 60 70 80 92&#xA;20 30 40 50 60 70 80 93&#xA;2 30 40 50 64 70 80 90&#xA;20 30 20 50 60 70 80 90&#xA;20 30 40 50 60 70 80 90&#xA;20 30 40 40 60 70 80 90&#xA;40 30 40 50 60 70 80 90&#xA;&#xA;4 5 1 6&#xA;&#xA;20 30 40 50 60 70 80 91&#xA;20 30 40 50 60 70 80 92&#xA;20 30 40 50 60 70 80 93&#xA;20 30 40 50 64 70 80 90&#xA;20 30 40 50 60 70 80 90&#xA;20 30 40 50 60 70 80 90&#xA;20 1 40 40 60 70 80 90&#xA;40 30 40 50 60 70 80 90&#xA;&#xA;4 5 6 1&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I want to get out of that document an array of matrices 8x8 and an array of matrices 1*4 is such thing possible and how to do it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need such thing to get data to train NN.&lt;/p&gt;&#xA;" OwnerUserId="2065" LastActivityDate="2011-02-01T13:22:25.410" Title="Maple: How to parse such CSV (Comma Separated Values) document?" Tags="&lt;maple&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="19882" PostTypeId="2" ParentId="19871" CreationDate="2011-02-01T13:06:28.217" Score="4" ViewCount="" Body="&lt;p&gt;You can compute a presentation of any finite group of moderately small order on any generating set using GAP or Magma. But the resulting presentations will not always be illuminating.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;${\rm GL}(2,3)$ is of course a 2-generator group, so you might prefer a presentation on two generators.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But a presentation that exhibits its structure as a semidirect product of $Q_8$ and $D_6$ is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Generators: $a,b,c,d$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Relations: $a^2=b^2$, $b^{-1}ab=a^{-1}$, $c^3=1$, $d^2=1$, $(cd)^2=1$, $c^{-1}ac=b^{-1}$, $c^{-1}bc=a^{-1}b$, $d^{-1}ad=b$.&lt;/p&gt;&#xA;" OwnerUserId="2820" LastActivityDate="2011-02-01T13:06:28.217" />
  <row Id="19883" PostTypeId="2" ParentId="19875" CreationDate="2011-02-01T13:07:11.603" Score="6" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Hint&lt;/strong&gt; for the case $\ell = 1$. The rest can be done by induction. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $A$ is positive definite, can define the matrix $B = (A)^{-1/2}$. Consider the change of variables $x = By$, so $dx = |B| dy$ using the Jacobian determinant. Then your equation can be re-written as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \int_{\mathbb{R}^n} x_i x_j e^{-\frac12 xAx} dx = |B| \int_{\mathbb{R}^n} B_{ik}B_{jl}y_ky_l e^{-\frac12 y\cdot y} dy $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Notice that $|B| = |A|^{-1/2}$ as desired.). Now, using the linearity of the action by the matrix $B$, you can integrate by parts in $y_l$ using the derivative operator $\partial_l$. This will pick out a term of the form (using that $\partial_ly_k = \delta_{lk}$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ B_{il}B_{jl} \int_{\mathbb{R}^n} e^{-\frac12 y\cdot y} dy $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the integral is the Gaussian integral giving you the factor of $2\pi$. And $B_{il}B_{jl} = B^2 = A^{-1}_{ij}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The factorial factor comes from when you integrate by parts, when $\ell &amp;gt; 1$, by counting the terms you actually have. &lt;/p&gt;&#xA;" OwnerUserId="1543" LastEditorUserId="1543" LastEditorDisplayName="" LastEditDate="2011-02-01T13:57:33.080" LastActivityDate="2011-02-01T13:57:33.080" CommentCount="1" />
  <row Id="19884" PostTypeId="1" AcceptedAnswerId="19961" CreationDate="2011-02-01T13:09:49.653" Score="3" ViewCount="113" Body="&lt;p&gt;Let $k$ be an algebraically closed field. Let $X$ be the quasi-afine variety $X = A^2 - \{0,0\}$. I know from complex-analytic arguments that the ring of regular functions on $X$ is equal to $k[x,y]$. How can I prove this algebraically?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="5491" LastActivityDate="2011-02-01T23:27:32.183" Title="Regular functions on the plane minus a point" Tags="&lt;algebraic-geometry&gt;" AnswerCount="2" />
  <row Id="19885" PostTypeId="2" ParentId="19880" CreationDate="2011-02-01T13:16:49.303" Score="0" ViewCount="" Body="&lt;p&gt;Try starting from the definition of the MacLaurin series (e.g. as defined here: &lt;a href=&quot;http://mathworld.wolfram.com/MaclaurinSeries.html&quot; rel=&quot;nofollow&quot;&gt;http://mathworld.wolfram.com/MaclaurinSeries.html&lt;/a&gt;)?&lt;/p&gt;&#xA;" OwnerUserId="1470" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2011-02-01T14:15:29.077" LastActivityDate="2011-02-01T14:15:29.077" />
  <row Id="19886" PostTypeId="2" ParentId="19881" CreationDate="2011-02-01T13:22:25.410" Score="0" ViewCount="" Body="&lt;p&gt;Your problem is rather a technical one than a mathematical one. However this link should help you: &lt;a href=&quot;http://books.google.com/books?id=fTCpTakFIZQC&amp;amp;pg=PA103&amp;amp;lpg=PA103&amp;amp;dq=maple+ImportData&amp;amp;source=bl&amp;amp;ots=XmFhOksdP2&amp;amp;sig=82gVw7seRgp3xlzul0dLTP9i-cM&amp;amp;hl=en&amp;amp;ei=2gdITZnFPIGaOve_7cYE&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=8&amp;amp;ved=0CD8Q6AEwBw#v=onepage&amp;amp;q=maple%20ImportData&amp;amp;f=false&quot; rel=&quot;nofollow&quot; title=&quot;Google Books - Statistics with Maple&quot;&gt;Statistics with Maple&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And &lt;a href=&quot;http://www.maplesoft.com/support/help/Maple/view.aspx?path=ImportData&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; is the Maple Helpfile for the &quot;ImportData&quot; function. However, Maple is not as powerful as Mathematica or Matlab concerning external data and you might have to preprocess the data to a form which allows Maple to handle it correctly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also I agree that you should rather ask on a programming platform.&lt;/p&gt;&#xA;" OwnerUserId="3123" LastActivityDate="2011-02-01T13:22:25.410" />
  <row Id="19887" PostTypeId="2" ParentId="19838" CreationDate="2011-02-01T13:26:29.353" Score="2" ViewCount="" Body="&lt;p&gt;Basically, a projective system formalizes the idea of &quot;constructing elements by successive approximations&quot; in an algebraic setting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $G$ is a group (abelian, to fix ideas) and $G_n$ a sequence of smaller and smaller subgroups, the projective limit $\Gamma=\lim(G/G_n)$ is the set of elements that are defined by giving them up to an ambiguity that gets smaller and smaller.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When $G$ is also a topological space and the $G_n$ are a local basis around the identity element, an element of $\Gamma$ is morally an element given by assigning a Cauchy sequence converging to it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Incidentally, this is exactly the case for the $p$-adic integers ${\Bbb Z}_p$ because it is possible to define a metric in $\Bbb Z$ in such a way the ideal $p^n{\Bbb Z}$ are open spheres.&lt;/p&gt;&#xA;" OwnerUserId="688" LastActivityDate="2011-02-01T13:26:29.353" />
  <row Id="19888" PostTypeId="2" ParentId="19880" CreationDate="2011-02-01T13:30:36.213" Score="1" ViewCount="" Body="&lt;p&gt;Since you know the polynomial of $\ln(1+t)$ and you know that $\cos(x)= 1 + (-\frac {x^2}{2!}+\frac{x^4}{4!}+x^4\epsilon(x))$ then you can &quot;plug&quot; it in the polynomial of $\ln(1+t)$. You now have that $\ln(\cos(x)) = P(x) + R(x)$ where P is a polynomial and R is the error such that $\lim _{x\rightarrow 0} \frac{R(x)}{x^k} = 0$ and $k=deg(P)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Every function that has a Taylor polynomial actually has a unique polynomial (for a fixed degree).&#xA;If $f(x)=P_1 (x) +x^k \epsilon_1 (x) = P_2 (x)+x^k \epsilon_2 (x)$ are two representations, then taking the limit as x approaches 0 shows that $P_1 (0)=P_2 (0)=a_0$. subtract $a_0$ from both sides, divide by x, and again take the limit as x approaches 0 to see that the coefficients of x in $P_1$ and $P_2$ are the same, and continue this by induction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This shows that the polynomial that you found is the Taylor polynomial.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In you case, when x is near zero, then $\cos(x)$ is near 1, so $\ln(\cos(x))$ is well defined.&lt;/p&gt;&#xA;" OwnerUserId="2125" LastActivityDate="2011-02-01T13:30:36.213" />
  <row Id="19889" PostTypeId="2" ParentId="19878" CreationDate="2011-02-01T13:36:35.397" Score="1" ViewCount="" Body="&lt;p&gt;The simplest model would just say the population increases by a fixed fraction every year:  $P(t)=P_0g^t$ where $P_0$ is the initial population, $g$ the growth rate, and $t$ is time.  Given two points, you can solve this with logarithms.  Using the values you supply:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$10^{12}=10^6g^{2000}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$12=6+2000 \log_{10} g$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\log_{10} g = .003$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$g\approx 1.007$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to take into account the age distribution, you need to specify a starting one.  Then you will have a reproduction rate for those in a certain age bracket and you have to specify also what age people die at.  But if you define the inputs completely, you can write a set of equations to cover it-you might have one for every age, for example.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-02-01T13:36:35.397" CommentCount="4" />
  <row Id="19890" PostTypeId="1" AcceptedAnswerId="19895" CreationDate="2011-02-01T13:37:00.793" Score="-1" ViewCount="58" Body="&lt;p&gt;&lt;a href=&quot;http://i.imgur.com/xs2AR.jpg&quot; rel=&quot;nofollow&quot;&gt;http://i.imgur.com/xs2AR.jpg&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first method I used was find out Profits by&#xA;Sales Revenue - Operating Cost - Payroll Cost - Waste -Shrinkage&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have Geneva the lowest and the answer is wrong.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2nd method, I used sales revenue/number of employees&#xA;London was the lowest and it's wrong again.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could anyone help me out on this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you&lt;/p&gt;&#xA;" OwnerUserId="6457" LastActivityDate="2011-02-01T14:09:11.767" Title="Numerical Test Question" Tags="&lt;arithmetic&gt;" AnswerCount="2" />
  <row Id="19891" PostTypeId="2" ParentId="19880" CreationDate="2011-02-01T13:41:14.283" Score="0" ViewCount="" Body="&lt;p&gt;Another way to do the problem is to realize that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;\frac{d}{dx}\ln(\cos x)=-\tan x&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And so the series will just be $\ln(\cos 0)=0$ plus a slightly adjusted series for $-\tan x$.  This also might help you see that the series is valid, since you are doing it out the long way.&lt;/p&gt;&#xA;" OwnerUserId="2838" LastEditorUserId="2838" LastEditorDisplayName="" LastEditDate="2011-02-01T13:50:32.887" LastActivityDate="2011-02-01T13:50:32.887" />
  <row Id="19892" PostTypeId="2" ParentId="19890" CreationDate="2011-02-01T13:48:41.703" Score="0" ViewCount="" Body="&lt;p&gt;Maybe they want the least profitable (but not a loss?) site.  I think your first calculation is the correct approach.  Or they could want profit/employee, but it looks like Geneva is still the biggest loser/employee (I didn't run the numbers, just guessed)&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-02-01T13:48:41.703" CommentCount="1" />
  <row Id="19893" PostTypeId="1" AcceptedAnswerId="19894" CreationDate="2011-02-01T13:52:09.563" Score="0" ViewCount="66" Body="&lt;p&gt;hi&#xA;i'm looking for some historical informations about Gruenberger path.&#xA;it is a path based on conjecture that every prime could be written in the form $6k+1$ and $6k-1$ (but what is the name of this conjecture?)&#xA;i don't find a lot in the net.. the best one seems to be this &lt;a href=&quot;http://bit-player.org/2010/gruenbergers-prime-path&quot; rel=&quot;nofollow&quot;&gt;site&lt;/a&gt;.. &#xA;any other infos?&lt;/p&gt;&#xA;" OwnerUserId="5046" LastActivityDate="2011-02-01T13:59:17.343" Title="Gruenberger's prime path" Tags="&lt;prime-numbers&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="19894" PostTypeId="2" ParentId="19893" CreationDate="2011-02-01T13:59:17.343" Score="2" ViewCount="" Body="&lt;p&gt;It is not a conjecture that all primes (greater than 3) are of the form 6k±1, it is a fact. 6k+0,2,4 are all divisible by 2 and 6k+3 is divisible by 3. His path just uses that fact. An interesting article.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-02-01T13:59:17.343" />
  <row Id="19895" PostTypeId="2" ParentId="19890" CreationDate="2011-02-01T14:09:11.767" Score="0" ViewCount="" Body="&lt;p&gt;Your first method gives -57 for Geneva and -70 for Rome, so Rome would seem to be the answer. &lt;/p&gt;&#xA;" OwnerUserId="1301" LastActivityDate="2011-02-01T14:09:11.767" CommentCount="1" />
  <row Id="19896" PostTypeId="2" ParentId="19876" CreationDate="2011-02-01T14:17:39.620" Score="6" ViewCount="" Body="&lt;p&gt;In my experience, almost all trigonometric identities can be obtained by knowing a few values of $\sin x$ and $\cos x$, that $\sin x$ is odd and $\cos x$ is even, and the addition formulas:&#xA;\begin{align*}&#xA;\sin(\alpha+\beta) = \sin\alpha\cos\beta + \cos\alpha\sin\beta,\\&#xA;\cos(\alpha+\beta) = \cos\alpha\cos\beta - \sin\alpha\sin\beta.&#xA;\end{align*}&#xA;For example, to obtain the classic $\sin^2x + \cos^2x = 1$, simply set $\beta=-\alpha$ in the formula for the cosine, and use the facts that $\cos(0)=1$ and  that $\sin(-a)=-\sin(a)$ for all $a$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the one you have, we use the formula for the cosine with $\alpha=\beta=\theta$ to get&#xA;$$\cos(2\theta) = \cos^2\theta - \sin^2\theta.$$&#xA;Then using $\sin^2 \theta+\cos^2\theta=1$, so $-\sin^2\theta = \cos^2\theta-1$ gives&#xA;$$\cos(2\theta) = \cos^2\theta +\cos^2\theta - 1 = 2\cos^2\theta - 1.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you know the basic values (at $\theta=0$, $\frac{\pi}{6}$, $\frac{\pi}{4}$, $\frac{\pi}{3}$, $\frac{\pi}{2}$, $\pi$, $\frac{3\pi}{2}$), parity, and the addition formulas, you can get almost any of the formulas.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2011-02-01T14:17:39.620" CommentCount="6" />
  <row Id="19897" PostTypeId="1" CreationDate="2011-02-01T14:30:12.003" Score="1" ViewCount="45" Body="&lt;p&gt;I'm interested in what to call the length of the sequence of ones in the &lt;a href=&quot;http://en.wikipedia.org/wiki/Smith_normal_form&quot; rel=&quot;nofollow&quot;&gt;&lt;strong&gt;Smith normal form&lt;/strong&gt;&lt;/a&gt; of a matrix. Equivalently (or so it seems to me), this is the rank of the largest submodule of &lt;em&gt;M&lt;/em&gt;, on which &lt;em&gt;A&lt;/em&gt; acts injectively (by left-multiplication; there should be an equivalent formulation for right-multiplication). For a given matrix &lt;em&gt;A&lt;/em&gt;, let us denote this by &lt;strong&gt;R&lt;/strong&gt;(&lt;em&gt;A&lt;/em&gt;). Does this have a name?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a matrix &lt;em&gt;A&lt;/em&gt; over a vector space &lt;em&gt;M&lt;/em&gt;, this is simply the &lt;strong&gt;rank&lt;/strong&gt; of &lt;em&gt;A&lt;/em&gt;: its the largest number of linearly independent rows/columns in &lt;em&gt;A&lt;/em&gt;, the dimension of the image of &lt;em&gt;A&lt;/em&gt;, the number of non-zero rows in the RREF of &lt;em&gt;A&lt;/em&gt;... &lt;em&gt;etcetera&lt;/em&gt;. If we consider &lt;em&gt;A&lt;/em&gt; over an &lt;em&gt;R&lt;/em&gt;-module &lt;em&gt;M&lt;/em&gt; for an arbitrary Euclidean domain, things get a bit more complicated. Over a field, the Smith normal form has only 1s on a part of the diagonal, and 0s elsewhere; more generally, however, the coefficients on the diagonal form sequence where each diagonal coefficient evenly divides the one which follows (this being the definition of the Smith normal form).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the generalization &lt;strong&gt;R&lt;/strong&gt;(&lt;em&gt;A&lt;/em&gt;) of 'rank' to the number of 1s in the Smith normal form, we may easily form non-zero matrices &lt;em&gt;A&lt;/em&gt; for which &lt;strong&gt;R&lt;/strong&gt;(&lt;em&gt;A&lt;/em&gt;)&amp;nbsp;=&amp;nbsp;0: for instance, the matrix 2&lt;em&gt;I&lt;/em&gt; of any dimension over the integers, or over the integers modulo 2&lt;em&gt;N&lt;/em&gt;, for &lt;em&gt;N&lt;/em&gt;&amp;gt;1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is &lt;strong&gt;R&lt;/strong&gt;(&lt;em&gt;A&lt;/em&gt;) called?&lt;/p&gt;&#xA;" OwnerUserId="439" LastEditorUserId="439" LastEditorDisplayName="" LastEditDate="2011-02-02T14:38:07.467" LastActivityDate="2011-02-02T14:38:07.467" Title="Terminological question: generalization of rank to matrices over modules" Tags="&lt;terminology&gt;&lt;modules&gt;" CommentCount="2" />
  <row Id="19898" PostTypeId="2" ParentId="19754" CreationDate="2011-02-01T14:38:44.767" Score="3" ViewCount="" Body="&lt;p&gt;Other books on Convex Polytopes are: Arne Brondsted, An Introduction to Convex Polytopes, Branko Grünbaum, Convex Polytopes (there is a second edition that updates the 1967 version), A. D. Alexandrov, Convex Polyhedra (translation from Russian of a Russian book from 1950, but with update and notes). They all have their pros/cons. I, at least, think Ziegler's book is excellent.&lt;/p&gt;&#xA;" OwnerUserId="1369" LastActivityDate="2011-02-01T14:38:44.767" />
  <row Id="19899" PostTypeId="2" ParentId="19880" CreationDate="2011-02-01T14:41:47.890" Score="0" ViewCount="" Body="&lt;p&gt;The MacLaurin series of $f(x)$ is given by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f(x)= f(0)+f^{\prime}(0)x + \frac{ f^{\prime \prime }(0)}{2!}x^2 +&#xD;&#xA;\frac{ f^{\prime \prime \prime}(0)}{3!}x^3 + \cdots $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and so you need to calculate the values of $f(0),f^{\prime}(0),\ldots$ for your function&#xA;$f(x)=\log( \cos x).$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have&#xA;$$\begin{align}&#xD;&#xA;f(x) &amp;amp;= \log( \cos x ) \quad \textrm{ and so } f(0)=0, \\&#xD;&#xA;f^{\prime}(x) &amp;amp;= - \tan x \quad \textrm{ and so } f^{\prime}(0)=0, \\&#xD;&#xA;f^{\prime \prime }(x) &amp;amp;= - \sec^2 x \quad \textrm{ and so } f^{\prime \prime }(0)=-1.&#xD;&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just carry on differentiating and evaluating the derivatives at $x=0$ until you reach the term in $x^6 .$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's worth noting that since $\sec^2 = 1 + \tan^2 x$ we can make the evaluation easier by writing&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f^{ \prime \prime }(x)=-1-f^{\prime}(x)^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and so, using the chain rule,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f^{ \prime \prime \prime }(x)=-2f^{\prime}(x)f^{\prime \prime}(x),$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and so putting $x=0$ we have $f^{ \prime \prime \prime }(0)=0.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Differentiating again we get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f^{(4)}(x)=-2f^{ \prime \prime }(x)^2-2f^{\prime}(x)f^{\prime \prime \prime}(x)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and so $f^{(4)}(0)=-2,$ and so on...&lt;/p&gt;&#xA;" OwnerUserId="1301" LastEditorUserId="1301" LastEditorDisplayName="" LastEditDate="2011-02-01T17:42:21.917" LastActivityDate="2011-02-01T17:42:21.917" />
  <row Id="19900" PostTypeId="2" ParentId="18361" CreationDate="2011-02-01T14:43:05.423" Score="1" ViewCount="" Body="&lt;p&gt;Under the described conditions, I(i,j) will describe the order over systems induced by your empirical (average) measurement. This is good - the strategy is obvious, choose the top system. Analysis is more delicate. You need to know something about the dispersion of excepted values for the systems vs the concentration of a single measurement of a single system, and then use Hoeffding or Bernstein inequality to bound the probability that the some system is actually better than the chosen system, despite the empirical evidence. &#xA;Of course if all systems are almost equal and measurements are very noisy, only very high values of n will do. The effect of k on reliability depends on the distribution from which &lt;em&gt;systems&lt;/em&gt; are chosen, which you didn't mention. Highly dispersed systems lead to higher reliability with higher k. Bounded distribution for systems lead to decreasing reliability with higher k (many very systems are very similar to the top rated, one will be better).&lt;/p&gt;&#xA;" OwnerUserId="6478" LastActivityDate="2011-02-01T14:43:05.423" />
  <row Id="19901" PostTypeId="2" ParentId="19866" CreationDate="2011-02-01T15:05:52.183" Score="2" ViewCount="" Body="&lt;p&gt;Let's start with the $L_{2}[0,1]$ norm.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Start with a single function $f$.  The 2-norm of $f$ is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ \| f \|_{2}=\sqrt{\int_{0}^{1} f(x)^2 dx} $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, $\| f \|_2$ will be 0 only if the integral of $f(x)^2$ is 0.  This will certainly happen if $f(x)=0$, but it can also happen in cases where $f$ has some isolated points with discontinuities where the function values are non zero.  Thus the &quot;null vector&quot; in this space is the zero function and all of the functions that are zero &quot;almost everywhere&quot; are considered to be equivalent to $f(x)=0$ with respect to this norm. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're familiar with electrical engineering, where $f(x)$ might be a time varying voltage, then $\| f \|_2$ is essentially the energy associated with that signal.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, suppose that $f(x)$ and $g(x)$ are two functions defined on $[0,1]$, and consider the 2-norm of the difference:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ \| f-g \|_{2}=\sqrt{\int_{0}^{1} (f(x)-g(x))^2 dx} $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the functions $f$ and $g$ are very nearly identical on $[0,1]$, then this norm will be close to 0.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're familiar with electrical engineering terminology, you &#xA;might have heard of the &quot;root mean square (RMS)&quot; difference between two signals.  This norm is exactly the RMS difference between $f$ and $g$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $f(x)=g(x)$, except for perhaps some individual points where there are discontinuities, then the norm will actually be 0.  Thus the $L_{2}$ distance between two functions is 0 if the functions are equal &quot;almost everywhere.&quot;  &lt;/p&gt;&#xA;" OwnerUserId="6310" LastActivityDate="2011-02-01T15:05:52.183" />
  <row Id="19902" PostTypeId="1" CreationDate="2011-02-01T15:16:46.480" Score="0" ViewCount="53" Body="&lt;p&gt;As a follow-up question to &lt;a href=&quot;http://math.stackexchange.com/questions/19866/what-is-the-norm-measuring-in-function-spaces&quot;&gt;what is the norm measuring in function spaces&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I just had an idea: How about visualizing function spaces as normal planes. What I have in mind is to have an orthogonal coordinate system where the two axes represent two orthogonal functions (e.g. sin and cos). Different functions, that can be related to these, are then entered into the system according to some distance measure (most easily euclidean): sin would be 1 on the sin-axis, cos 1 on the cos-axis, other functions would be positioned according to the norm and their being composed of these building blocks. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course two dimensions are not very much but it would suffice for getting an intuition about these concepts and as a toy model (one could e.g. change the basis functions, the functions that are being dis-assembled, the norm, the boundaries etc.).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does this make sense? Has anyone ever done this? Are their perhaps even tools to play with?&lt;/p&gt;&#xA;" OwnerUserId="346" LastEditorUserId="346" LastEditorDisplayName="" LastEditDate="2011-02-01T15:25:41.713" LastActivityDate="2011-02-01T15:25:41.713" Title="Visualization of 2-dimensional function spaces" Tags="&lt;real-analysis&gt;&lt;measure-theory&gt;&lt;intuition&gt;&lt;vector-space&gt;&lt;visualization&gt;" CommentCount="2" FavoriteCount="1" />
  <row Id="19903" PostTypeId="1" AcceptedAnswerId="19923" CreationDate="2011-02-01T15:18:01.803" Score="0" ViewCount="77" Body="&lt;p&gt;Hi I am learning about Barker sequence. I have a problem, because I do not know, why for example 5, in Barker Code looks like this: +++-+. What is the base of this code. I am looking for some algorithm, because, I do not want to learn by hearth.&lt;/p&gt;&#xA;" OwnerUserId="6483" LastActivityDate="2011-02-02T15:00:34.507" Title="Barker sequence" Tags="&lt;sequences-and-series&gt;&lt;signal-processing&gt;" AnswerCount="2" />
  <row Id="19904" PostTypeId="2" ParentId="19876" CreationDate="2011-02-01T15:36:52.903" Score="3" ViewCount="" Body="&lt;p&gt;It might be useful to remember that $\cos^2 x$ oscillates twice as fast as $\cos x$. This is something that people who work with alternating current know very well; the effect (which is proportional to the &lt;em&gt;square&lt;/em&gt; of the current) has twice the frequency. For example, a light bulb flickers at 100 Hz if the AC frequency is 50 Hz. This means that $\cos^2 x$ should be &quot;something with $\cos 2x$&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next, since $\cos x$ oscillates between -1 and 1, $\cos^2 x$ will oscillate between 0 and 1. The average around which the curve oscillates will be 1/2, and the amplitude will also be 1/2 (so that you reach down to 0 and up to 1 from the central level 1/2). Thus $\cos^2 x$ should be &quot;1/2 + (1/2)*oscillating term&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Combining these two facts, it's not too hard to remember that&#xA;$$ \cos^2 x = \frac12 + \frac12 \cos 2x.$$&#xA;One has to be a little bit careful with the sign before the second term, but it must be plus if the formula is to hold when $x=0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you choose the minus sign, you get the related formula&#xA;$$ \sin^2 x = \frac12 - \frac12 \cos 2x.$$&#xA;(So $\sin^2 x$ also oscillates around 1/2 with amplitude 1/2 and twice the frequency. Note that when you add the two formulas up, the oscillations cancel, and you get $\cos^2 x + \sin^2 x = 1/2+1/2 = 1$.)&lt;/p&gt;&#xA;" OwnerUserId="1242" LastEditorUserId="1242" LastEditorDisplayName="" LastEditDate="2011-02-02T12:47:31.787" LastActivityDate="2011-02-02T12:47:31.787" CommentCount="1" />
  <row Id="19905" PostTypeId="1" AcceptedAnswerId="19907" CreationDate="2011-02-01T15:54:05.137" Score="2" ViewCount="96" Body="&lt;p&gt;What does it mean when, in (derivation of) (partial) differential equations, a term containing an operator like:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\delta t$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;gets replaced by:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$d t$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and what are the rules pertaining to this? What is the difference between the two? It seems black magic to me :).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="6484" LastActivityDate="2011-02-01T16:13:28.070" Title="Delta operator magic" Tags="&lt;differential-equations&gt;&lt;differential&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="19906" PostTypeId="2" ParentId="19884" CreationDate="2011-02-01T16:01:33.727" Score="3" ViewCount="" Body="&lt;p&gt;The ring of regular functions on $A^2 \setminus \{ x=0 \}$ is $k[x,y,x^{-1}]$ and the ring of regular functions on $A^2 \setminus \{ y=0 \}$ is $k[x,y,y^{-1}]$, and $A^2 \setminus \{0\}$ is the union of these open subsets, so its ring of regular functions is the intersection of the two, which is equal to $k[x,y]$.&lt;/p&gt;&#xA;" OwnerUserId="2660" LastActivityDate="2011-02-01T16:01:33.727" CommentCount="2" />
  <row Id="19907" PostTypeId="2" ParentId="19905" CreationDate="2011-02-01T16:13:28.070" Score="2" ViewCount="" Body="&lt;p&gt;Could you give a specific example? In some instances $\delta t$ is used to represent a variation or change in time, in this case. $dt$ just represents the &quot;differential&quot; of time in this case, i.e., the infinitesimal change. Think of it as the differential in calculus, which is obtained by passing to the limit in the expression&#xA;$$&#xD;&#xA;\frac{\Delta y}{\Delta x}&#xD;&#xA;$$&#xA;yielding&#xA;$$&#xD;&#xA;\frac{d y}{d x}.&#xD;&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="6209" LastActivityDate="2011-02-01T16:13:28.070" CommentCount="1" />
  <row Id="19908" PostTypeId="1" AcceptedAnswerId="19913" CreationDate="2011-02-01T16:27:48.767" Score="0" ViewCount="80" Body="&lt;p&gt;Two part question:&#xA;If I define a linear operator $L=\frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2}$ and then use that operator on a variable, say $L\alpha$ then is it accurate to say that the $L$ maps $\alpha$ onto itself like this: $L\alpha=\frac{\partial^2}{\partial x^2}\alpha + \frac{\partial^2}{\partial y^2}\alpha$. Is that how we talk about operators and why...so that we can get rid of the variable that's being acted on and just talk about the operation that's being performed?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's say I have an equation that has $p\alpha^2$. If I wanted to write the operator for this, how would I do it?&lt;/p&gt;&#xA;" OwnerUserId="6390" LastActivityDate="2011-02-01T17:36:53.113" Title="Operator for a Squared Term? - General Operator Question" Tags="&lt;differential-equations&gt;" AnswerCount="2" />
  <row Id="19909" PostTypeId="2" ParentId="19862" CreationDate="2011-02-01T16:39:04.910" Score="3" ViewCount="" Body="&lt;p&gt;I believe that the Jacobian Conjecture is widely considered true (is even given as a double-starred problem in Hartshorne's Algebraic Geometry book) but it has resisted attempts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the way, I haven't checked in a while, so what is the current status of the Jacobian Conjecture?&lt;/p&gt;&#xA;" OwnerUserId="688" LastActivityDate="2011-02-01T16:39:04.910" CommunityOwnedDate="2011-02-01T16:39:04.910" CommentCount="2" />
  <row Id="19910" PostTypeId="1" CreationDate="2011-02-01T16:48:31.663" Score="0" ViewCount="63" Body="&lt;p&gt;I have a system that is defined as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\dot{x}(t) = Ax(t) + Bu(t)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ t\geq0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;it has a well known solution:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x(t) = e^{tA}x_{0} + \int_0^t \! e^{(t-\tau)A}B \, \mathrm{d}\tau $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that one of the tasks during exam will be to draw the the graph of the solution given that A and B are numbers. If B is zero then this is no problem, but if it's not I don't really know how to calculate that integral. Could anyone please explain the steps that I need to perform to solve this? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS&gt; I've already found some examples but none of them showed the intermediate steps, so I can't figure out anything :/&lt;/p&gt;&#xA;" OwnerUserId="6485" LastActivityDate="2011-02-01T16:48:31.663" Title="First order dynamic system" Tags="&lt;dynamical-systems&gt;" CommentCount="5" />
  <row Id="19911" PostTypeId="2" ParentId="19848" CreationDate="2011-02-01T16:51:25.027" Score="1" ViewCount="" Body="&lt;p&gt;Let's elaborate on the answers already given, by considering the more general case where $X_n = 1$ with probability $a_n$ and $X_n = 0$ with probability $1-a_n$, where $a_n \downarrow 0$ (the typical cases being $a_n = 1/n$ and $a_n = 1/n^2$) and the $X_n$ are not necessarily independent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most natural approach is to use Borel-Cantelli lemma, as follows. Denote by $E_n$ the event $\lbrace X_n = 1 \rbrace$.&#xA;If $\sum\nolimits_{n = 1}^\infty  {{\rm P}(E_n )}  &amp;lt; \infty$, then the probability of $E_n$ occurring for infinitely many $n$ is $0$. Thus, if $\sum\nolimits_{n = 1}^\infty  {a_n}  &amp;lt; \infty$, then $X_n \to 0$ almost surely (that is, with probability $1$), since, almost surely, $X_n = 0$ for all $n$ sufficiently large. If, on the other hand, $\sum\nolimits_{n = 1}^\infty  {{\rm P}(E_n )}  = \infty$ and the events $E_n$ are independent, then the probability of $E_n$ occurring for infinitely many $n$ is $1$. Thus, if $\sum\nolimits_{n = 1}^\infty  {a_n}  = \infty$ and the $X_n$ are independent, then almost surely $X_n$ does not converge to $0$ (even though ${\rm P}(X_n = 0) \to 1$, which only gives that $X_n \to 0$ in probability).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for chandok's approach, define $Y_n = \sum\nolimits_{k = 1}^n {X_k }$, so that $0 \leq Y_1 \leq Y_2 \leq \cdots$. Since the sequence $Y_n$ is monotone, it converges, say to $Y$ (which might be infinite). By the Monotone Convergence Theorem, ${\rm E}(Y_n) \to {\rm E}(Y)$ (to put it otherwise, $\int_\Omega  {Y_n \,{\rm dP}}  \to \int_\Omega  {Y\,{\rm dP}}$), where the right-hand side might be infinite (and is obviously infinite if $Y$ is). Now, ${\rm E}(Y_n) = \sum\nolimits_{k = 1}^n {a_k }$. Hence, if $\sum\nolimits_{k = 1}^\infty {a_k }$ is finite, then so is ${\rm E}(Y)$, or $\int_\Omega  {Y\,{\rm dP}}$. From measure theory (since $Y$ is nonnegative), it follows that $Y$ is almost everywhere finite, or, in our setting, $Y$ is almost surely finite. But $Y$ is finite if and only if $X_n = 0$ for all $n$ sufficiently large. So, $\sum\nolimits_{k = 1}^\infty {a_k } &amp;lt; \infty$ implies $X_n \to 0$. However, if $\sum\nolimits_{k = 1}^\infty {a_k } = \infty$, then one cannot a priori conclude from ${\rm E}(Y) = \infty$ that $Y$ is almost surely infinite (simply since a random variable can have infinite expectation).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: With the above notation and assuming that the $X_n$ are independent, the characteristic function $\varphi _{n } (u)$ of $Y_n$ is given by&#xA;$$&#xD;&#xA;\varphi _{n } (u) = {\rm E}[e^{{\rm i}uY_n } ] = {\rm E}[e^{{\rm i}u(X_1  +  \cdots  + X_n )} ] = \prod\limits_{k = 1}^n {{\rm E}[e^{{\rm i}uX_k } ]}  = \prod\limits_{k = 1}^n {[(1 - a_k ) + a_k e^{{\rm i}u} ]}. &#xD;&#xA;$$&#xA;Now, $X_n \to 0$ if and only if $Y := \lim _{n \to \infty } Y_n $ is finite. So it is instructive to note here the following (general) result. If $\varphi _{n } (u)$ converges to a function $\varphi(u)$ for every $u \in \mathbb{R}$ and $\varphi(u)$ is continuous at $u=0$, then $\varphi(u)$ is the characteristic function of some distribution.&lt;/p&gt;&#xA;" OwnerUserId="2810" LastEditorUserId="2810" LastEditorDisplayName="" LastEditDate="2011-02-01T21:28:25.543" LastActivityDate="2011-02-01T21:28:25.543" />
  <row Id="19912" PostTypeId="2" ParentId="19862" CreationDate="2011-02-01T16:53:06.320" Score="3" ViewCount="" Body="&lt;p&gt;The rank conjecture for rational elliptic curves (for every $n\gt 0$ there is an elliptic cuve over $\mathbb{Q}$ whose group of rational points has rank at least $n$).&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2011-02-01T16:53:06.320" CommunityOwnedDate="2011-02-01T16:53:06.320" CommentCount="5" />
  <row Id="19913" PostTypeId="2" ParentId="19908" CreationDate="2011-02-01T16:53:12.577" Score="1" ViewCount="" Body="&lt;p&gt;For the first part of your question: indeed, if you define $L$ like that, the expression $L \alpha = \frac{\partial^2}{\partial x^2} \alpha + \frac{\partial^2}{\partial y^2} \alpha$ is correct. What is not so right is saying that &quot;$L$ maps $\alpha$ to itself - it doesn't, in the same sense the derivative of a function is a new function, usually different from the original function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which brings me to the next point: I'm not too sure of how you interpret the term &quot;variable&quot; here, but in this case $\alpha$ must be a &lt;em&gt;function&lt;/em&gt;, and not a regular variable. The idea behind working with operators like this is not so much talking about the operation being performed, but rather talking about &lt;em&gt;properties&lt;/em&gt; the operator as such might have. It is, in a way, a generalization of finite-dimensional linear algebra. There you have a linear equation, say $Tx = y$, and you are concerned with what the eigenvalues of $T$ look like and how they determine the kind of possible solutions to that equation. It's similar here, you can still find the eigenvalues of the Laplacian. But the function spaces we deal with here are infinite-dimensional.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the second part of your question: what is $p$? Is $\alpha$ still understood to be a function, as in the previous part?&lt;/p&gt;&#xA;" OwnerUserId="6209" LastActivityDate="2011-02-01T16:53:12.577" CommentCount="2" />
  <row Id="19914" PostTypeId="2" ParentId="19356" CreationDate="2011-02-01T16:55:06.550" Score="0" ViewCount="" Body="&lt;p&gt;Thank you for your hints, I'll check it. Actually I could not attach examples of my dataset, but imagine a set of trajectories in &quot;M&quot; shape. Going through all points of the dataset of trajectories I was intending to fit the data by some kind of regression.&#xA;But, probably you are right, by using splines perhaps I can do it. The main idea is: given a set of trajectories, to represent this dataset in a single trajectory, a kind of prototype to match in the future with other trajectories to verify if they are similar.&lt;/p&gt;&#xA;" OwnerUserId="6489" LastActivityDate="2011-02-01T16:55:06.550" />
  <row Id="19915" PostTypeId="1" AcceptedAnswerId="19970" CreationDate="2011-02-01T17:02:54.793" Score="1" ViewCount="95" Body="&lt;p&gt;For $x \in \mathbb{R}^n$ and $A,B \in \mathbb{R}^{m \times n}$,&#xA;$f(x) = ((Ax)^{2})^T((Bx)^2)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $^2$ denotes the power of 2, element-by-element of vector Ax or Bx.&#xA;(I wasn't sure how to notate this)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is $f(x)$ convex? How can it be shown?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the domain of $x$ is restricted to be $x$ nonnegative, then is it convex?&lt;/p&gt;&#xA;" OwnerUserId="5610" LastActivityDate="2011-02-02T00:49:09.033" Title="convexity of inner product of elementwise powers" Tags="&lt;vector-space&gt;&lt;convex-analysis&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="19916" PostTypeId="1" AcceptedAnswerId="19920" CreationDate="2011-02-01T17:27:08.913" Score="6" ViewCount="118" Body="&lt;p&gt;I have been working through the exercises in Tenenbaum's &quot;Introduction to analytic and probabilistic number theory&quot; book, and I am stumped here (Exercise I.1.6).  This is not a homework assignment, but just rather for my own edification.  I know Tenenbaum's exercises are usually considered quite hard, so I don't feel as embarrassed to ask this question!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For those not aware (I suspect most experts know this exercise and book inside and out), the homework problem reads as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Set $d_n = p_{n+1} - p_n$.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Show that $p_n \sim n \log n \qquad (n \to \infty)$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$\sum_{1 &amp;lt; n \leq x} d_n / \log x \sim x \qquad (x \to \infty)$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$\liminf_{n \to \infty} \frac{d_n}{\log n} \leq 1 \leq \limsup_{n \to \infty} \frac{d_n}{\log n}$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;For each $\alpha &amp;gt; 0$ there exists a sequence of integers $\{n_1, n_2, \dots \}$, increasing in the weak sense, such that $p_{n_j} \sim \alpha j \qquad (j \to \infty)$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The set of rational numbers of the form $p/p&amp;#39;$, where $p$ and $p&amp;#39;$ are prime is dense in $[0, \infty)$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;My questions are two-fold:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;How does part 5 of this exercise follow from part 4?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Are there different proofs (elementary or not) of question 5 that do not follows this route?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="6491" LastActivityDate="2011-02-01T17:46:42.693" Title="The set of rational numbers of the form p/p', where p and p' are prime, is dense in $[0, \infty)$" Tags="&lt;number-theory&gt;&lt;prime-numbers&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" />
  <row Id="19917" PostTypeId="1" CreationDate="2011-02-01T17:33:53.460" Score="2" ViewCount="168" Body="&lt;p&gt;how to show T(S^3) isomorphic to S^3 cross R^3? so can I say it for every odd dimension?I have shown it for n=1&lt;/p&gt;&#xA;" OwnerUserId="6493" LastActivityDate="2011-02-04T05:59:42.473" Title="Tangent Bundle on S^3" Tags="&lt;differential-geometry&gt;" AnswerCount="4" CommentCount="2" />
  <row Id="19918" PostTypeId="2" ParentId="19908" CreationDate="2011-02-01T17:36:53.113" Score="0" ViewCount="" Body="&lt;p&gt;You can think of an operator as being a map from some space of functions to itself or to another space of functions.  For instance, if you are considering only smooth real-valued functions on ${\mathbb{R}}^2$, you might look at operators $O: C^\infty(\mathbb{R}^2)\rightarrow C^\infty(\mathbb{R}^2)$.  If the domain and range are both vector spaces (over ${\mathbb{R}}$, say), then the &lt;em&gt;linear operators&lt;/em&gt; are just those maps that satisfy&#xA;$$&#xD;&#xA;a(O\cdot f) + b(O\cdot g) = O\cdot(af + bg)&#xD;&#xA;$$&#xA;for all $a,b\in\mathbb{R}$ and functions $f$ and $g$.  It is straightforward to verify that differential expressions like $\partial_x^{2} + \partial_y^{2}$ are linear operators on the appropriate function spaces.  Nonlinear operators can also be defined, like the one you mentioned, by specifying their effect on arbitrary functions: e.g., $(O \cdot f)(x) = p f(x)^2$.&lt;/p&gt;&#xA;" OwnerUserId="5546" LastActivityDate="2011-02-01T17:36:53.113" />
  <row Id="19919" PostTypeId="1" CreationDate="2011-02-01T17:43:06.520" Score="3" ViewCount="131" Body="&lt;p&gt;Let $\tau$ be the topology generated by half-open intervals of the form $[a,b)$ where $a$ is a rational number and $b$ is a real number. Let $C$ denote the space endowed with the previously described topology. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Prove/disprove: $C \times C$ is a Lindelof space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do you proceed: here $C$ is &lt;em&gt;not&lt;/em&gt; equal to the Sorgenfrey line (because of the rational endpoint). Do we have to use Jones lemma like when showing $\mathbb{R}_{l} \times \mathbb{R}_{l}$ is not Lindelof?&lt;/p&gt;&#xA;" OwnerUserId="6494" LastActivityDate="2011-04-27T03:42:45.553" Title="Topology generated by half open intervals with one endpoint rational" Tags="&lt;topology&gt;" AnswerCount="1" />
  <row Id="19920" PostTypeId="2" ParentId="19916" CreationDate="2011-02-01T17:46:42.693" Score="6" ViewCount="" Body="&lt;p&gt;From question 4 you can easily show that for any $\alpha &amp;gt; 0$, there exists a sequence $(q_n, q&amp;#39;_n)$ of prime numbers such that $q_n/q&amp;#39;_n$ converges to $\alpha$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You just have to take the sequence $\{n_1, n_2, \dots\}$ and use the subsequence of the terms with prime index. So you can build this sequence $\{(p_1, p_{n_{p_1}}), (p_2, p_{n_{p_2}}), \dots \}$.&#xA;Each pair in this new sequence in composed of prime numbers, and from the fact that $p_{n_j} / j \rightarrow \alpha$ you get that $p_{n_{p_i}}/p_i \rightarrow \alpha$ as $i \rightarrow \infty$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore $\alpha$ is in the closure of the rational numbers that are quotient of primes.&#xA;Since it is true for any $\alpha &amp;gt; 0$, this set is dense in $[0, \infty)$.&lt;/p&gt;&#xA;" OwnerUserId="6270" LastActivityDate="2011-02-01T17:46:42.693" CommentCount="2" />
  <row Id="19921" PostTypeId="2" ParentId="16948" CreationDate="2011-02-01T17:50:00.180" Score="1" ViewCount="" Body="&lt;p&gt;The way you get uncountably many loops is similar to the way you get uncountably many for the Hawaiian Earrings. Choose a sequence of loops $\alpha_i$ whose diameters approach $0$. Then consider the infinite concatenations $\alpha_1^{\epsilon_1}*\alpha_2^{\epsilon_2}\cdots*\alpha_n^{\epsilon_n}*\cdots$ where $\epsilon_n=\pm 1$. Since the diameters go to zero, this really represents a loop in the space. However, there are clearly uncountably many such loops. The tricky part is showing that no pair of them is homotopic.&lt;/p&gt;&#xA;" OwnerUserId="2941" LastActivityDate="2011-02-01T17:50:00.180" CommentCount="2" />
  <row Id="19922" PostTypeId="2" ParentId="19919" CreationDate="2011-02-01T17:50:54.680" Score="1" ViewCount="" Body="&lt;p&gt;Hint: You can show that $C$ is second countable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Update after the first 2 comments: Once you have that, what can you say about $C\times C$?  What can you say about the relationship between second countability and Lindelöfness?&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2011-02-01T18:05:50.250" LastActivityDate="2011-02-01T18:05:50.250" CommentCount="8" />
  <row Id="19923" PostTypeId="2" ParentId="19903" CreationDate="2011-02-01T18:07:38.440" Score="2" ViewCount="" Body="&lt;p&gt;You should look at &lt;a href=&quot;http://en.wikipedia.org/wiki/Barker_code&quot; rel=&quot;nofollow&quot;&gt;wikipedia&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There you will find the definition: &quot;A Barker code is a sequence of N values of +1 and −1, $a_j$ for $j= 1, 2, ..., N$ such that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$|\sum_{j=1}^{N-v} a_ja_{j+v}| \le 1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for all $1 \le v &amp;lt; N$.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Barker code for 3 is given by $+1$ $+1$ $-1$, i.e $a_1 = +1$, $a_2 = +1$, $a_3 = -1$. Here you can check that the definition works: if $v = 1$ you have that  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$|\sum_{j=1}^{3-1} a_ja_{j+1}| = |\sum_{j=1}^{2} a_ja_{j+1}| = | a_1a_2 + a_2a_3| = |1\cdot 1 + 1 \cdot (-1)| = 0 \le 1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And for $v = 2$ you have:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$|\sum_{j=1}^{3-2} a_ja_{j+2}| = |a_1a_3| = |1\cdot (-1)| \le 1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence $+1$ $+1$ $-1$ is a correct Barker code for $3$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in Barker code, 5 is given by the sequence $+1$ $+1$ $+1$ $−1$ $+1$. This is just a sequence of numbers that &lt;em&gt;works&lt;/em&gt; given your definition. You can check this by computing the 4 sums. &lt;/p&gt;&#xA;" OwnerUserId="5105" LastActivityDate="2011-02-01T18:07:38.440" />
  <row Id="19924" PostTypeId="2" ParentId="19917" CreationDate="2011-02-01T18:15:42.363" Score="3" ViewCount="" Body="&lt;p&gt;Identify the sphere with the set $S\subset\mathbb H$ of quaternions of norm $1$. Pick an ordered basis $(u_1,u_2,u_3)$ of the tangent space $T_1S$ to $S$ at the point $1$, and now consider the vector fields $$X_i:p\in S \mapsto pu_i\in T_pS,$$ where on the right hand side $pu_i$ is the quaternion product of the element $p$ with the quaternion $u_i$ (recall that one can identify the tangent space at $T_1S$ with a subspace of $\mathbb H$) You need to check that this makes sense: in particular, that $pu_i$ is, in fact, in the tangent space $T_pS$ to $S$ at $p$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This construction gives you three non-zero vector fields $X_1$, $X_2$ and $X_3$. Check that at each point they are a basis. Voilà. &lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2011-02-01T18:15:42.363" />
  <row Id="19925" PostTypeId="2" ParentId="16948" CreationDate="2011-02-01T18:23:02.077" Score="2" ViewCount="" Body="&lt;p&gt;In case anyone doesn't like the arguments from uncountability, here's a more concrete argument:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fix a base-point $(p,p)$ for an arbitrary $p\in\mathbb{R}-\mathbb{Q}$. For any $q\in\mathbb{R}-\mathbb{Q}$, $q\not=p$, consider the loop $L_q$ consisting of the following line segments: $(p,p)\rightarrow(p,q)\rightarrow(q,q)\rightarrow(q,p)\rightarrow(p,p)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We claim that for any $q_1&amp;lt;q_2$, the loops $L_{q_1}$ and $L_{q_2}$ are non-homotopic. To see this, pick a rational $r$ such that $q_1&amp;lt;r&amp;lt;q_2$. Embed $\mathbb{R}^2-\mathbb{Q}^2$ into $\mathbb{R}^2-(r,r)$ (aka, the punctured plane). Clearly, $L_{q_1}\equiv 0$ while $L_{q_2}\equiv 1$, so the loops are non-homotopic, as desired.&lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2011-02-01T18:23:02.077" />
  <row Id="19926" PostTypeId="2" ParentId="19867" CreationDate="2011-02-01T18:24:38.967" Score="1" ViewCount="" Body="&lt;p&gt;@Matt: Recall that in the product topology there are only finitely many restrictions (i.e finite number of open sets). The answer is that it is dense in the product topology. Can you see why? Pick a basic open set in the product topology and try to find a point which is in your set and in the basic open set. It helps to think about 0's.&lt;/p&gt;&#xA;" OwnerUserId="6495" LastActivityDate="2011-02-01T18:24:38.967" CommentCount="1" />
  <row Id="19928" PostTypeId="2" ParentId="19865" CreationDate="2011-02-01T19:05:04.587" Score="1" ViewCount="" Body="&lt;p&gt;@Matt: The set $A$ is closed so it is equal to its own closure. To show it is closed you can use the characterization of a closed set (in metric spaces) in terms of sequences.&lt;/p&gt;&#xA;" OwnerUserId="6495" LastActivityDate="2011-02-01T19:05:04.587" />
  <row Id="19929" PostTypeId="2" ParentId="19867" CreationDate="2011-02-01T19:11:42.903" Score="3" ViewCount="" Body="&lt;p&gt;A better way to think about $\mathbb{R^N}$ is as the set of all real-valued sequences, right?  You have $\mathbb N$ dimensions and each dimension corresponds to a real number, so $\mathbb{R^N}$ is just the set of all real-valued sequences (as a note $\mathbb{R^R}$ can be thought of as all the functions from $\mathbb R$ to $\mathbb R$). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So with this definition in mind, what does it mean for there only be a finite number of non-zero indexes in a sequence. Well it means that past a certain point the sequence is entirely zeros. Okay now keeping this in mind, think about what the product topology means. Well it means that any open set (We'll just work with basis sets) is a product of spaces $X_i$ where for only finitely many $i$ $X_i\neq \mathbb R$. Then if we want to show that our set is dense in $\mathbb{R^N}$, we need to show that it intersects every basis open set. Well we've just said that an open set has only finite many indices that aren't the whole space. So if we pick an $x \in X_i$ where each $X_i \neq \mathbb R$, then we'll get finitely many points that aren't zero and we can let everything else be zero. This will show that our set is dense in $\mathbb{R^N}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What happens if we put the box topology on $\mathbb{R^N}$?&lt;/p&gt;&#xA;" OwnerUserId="1930" LastEditorUserId="1930" LastEditorDisplayName="" LastEditDate="2011-04-06T00:04:59.777" LastActivityDate="2011-04-06T00:04:59.777" CommentCount="2" />
  <row Id="19930" PostTypeId="2" ParentId="19862" CreationDate="2011-02-01T19:14:07.990" Score="1" ViewCount="" Body="&lt;p&gt;The Kakeya conjecture.  It states that a set in $\mathbb{R}^d$ containing a line in every direction has Hausdorff dimension $d$.  It is solved for $d = 2$ and open for $d \geq 3$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The finite field analogue was recently solved by Dvir (Tao made a very nice post about it, &lt;a href=&quot;http://terrytao.wordpress.com/2008/03/24/dvirs-proof-of-the-finite-field-kakeya-conjecture/&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;).  Dvir's contributions have also led to a deeper understanding and an ultimate resolution by Guth and Katz of the Erdos distance problem in $\mathbb{R}^2$ (of which Tao also has a nice post, &lt;a href=&quot;http://terrytao.wordpress.com/2010/11/20/the-guth-katz-bound-on-the-erdos-distance-problem/&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When one constructs a Kakeya set of measure zero, it is typically visualized by Bourgain's &quot;Bush&quot; construction or Wolff's &quot;Hairbrush&quot; argument.  The iterations of these constructions has full Hausdorff dimension.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe a great number of people would be surprised if it were false.&lt;/p&gt;&#xA;" OwnerUserId="6491" LastActivityDate="2011-02-01T19:14:07.990" CommunityOwnedDate="2011-02-01T19:14:07.990" />
  <row Id="19932" PostTypeId="1" CreationDate="2011-02-01T19:20:08.837" Score="5" ViewCount="131" Body="&lt;p&gt;I encountered an interesting function which is called &quot;Eulerian&quot; by the Wolfram's MathWorld:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\phi(q)=\prod_{k=1}^{\infty} (1-q^{k})$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is interesting because it seems that roots of any polynomial can be expressed in this function and elementary functions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to know more about the properties of this function, where can I find the information?&lt;/p&gt;&#xA;" OwnerUserId="2513" LastEditorUserId="2513" LastEditorDisplayName="" LastEditDate="2011-02-01T20:02:55.037" LastActivityDate="2011-02-02T08:12:04.453" Title="What is Eulerian?" Tags="&lt;special-functions&gt;&lt;q-analogs&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="19933" PostTypeId="1" AcceptedAnswerId="19986" CreationDate="2011-02-01T19:33:57.300" Score="1" ViewCount="420" Body="&lt;p&gt;&lt;strong&gt;Second Update:&lt;/strong&gt;&#xA;I see that some answers that reference my image are more closely answering my question.&#xA;Here is a second image to clarify my point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take this image representing a checkerboard like configuration.  If I asked you to move the X negative 1 space it would be impossible.  You would have to make a guess as to which direction is negative. However, I can ask you to move the X 1 space/unit, since moving to any square from the center would satisfy moving 1 space/unit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But wait!  Once you complete moving the X 1 space/unit; I could now follow up and request you move negative 1 space, since you have now already travelled in one direction. Now you are able to negate the original request by 1 space/unit&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore negative doesn't exists until you have already begun moving in some direction/unit/space/time/apples.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/mmR9C.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;update&lt;/strong&gt;&#xA;&lt;img src=&quot;http://i.imgur.com/6nL2D.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe this helps illustrate my point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this image there is a line, going either direction will take you to infinity (A or B),  the number of spaces you move is &lt;strong&gt;always&lt;/strong&gt; a positive number. If I moved 20 widgets closer to infinity A. or I moved 20 widgets closer to infinity B.  So, negative numbers don't exist and are only positive numbers that are increasing in the opposite direction&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Negative indicates the direction of travel on a line.  the measure of movements will always be positive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does a negative number really exist?  I can't have -1 apples.  I can only imagine I had an apple when I really did not.  Thus, in reality I would have -1 apple as compared to my imagination.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;didn't know how to tag this&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="6497" LastEditorUserId="6497" LastEditorDisplayName="" LastEditDate="2011-02-02T14:10:03.800" LastActivityDate="2011-03-24T08:35:41.060" Title="Does a negative number really exist?" Tags="&lt;soft-question&gt;&lt;philosophy&gt;" AnswerCount="8" CommentCount="17" FavoriteCount="1" />
  <row Id="19934" PostTypeId="2" ParentId="19933" CreationDate="2011-02-01T19:34:56.067" Score="1" ViewCount="" Body="&lt;p&gt;$-1$ is an integer. It is not imaginary. It is a complex number with zero imaginary part (i.e. $-1 = -1 + 0i$).&lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2011-02-01T19:34:56.067" CommentCount="1" />
  <row Id="19935" PostTypeId="2" ParentId="19932" CreationDate="2011-02-01T19:36:49.277" Score="2" ViewCount="" Body="&lt;p&gt;You may want to look up &lt;a href=&quot;http://en.wikipedia.org/wiki/Weierstrass_factorization_theorem&quot; rel=&quot;nofollow&quot;&gt;Weierstrass factorization theorem&lt;/a&gt; which plays a crucial role in complex analysis for writing functions as infinite products. It is a simple but powerful idea. Euler's famous proof of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Basel_problem&quot; rel=&quot;nofollow&quot;&gt;Basel's problem&lt;/a&gt; exploited this infinite product for $\sin(x)$.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2011-02-01T19:42:32.777" LastActivityDate="2011-02-01T19:42:32.777" />
  <row Id="19936" PostTypeId="2" ParentId="19862" CreationDate="2011-02-01T19:42:29.747" Score="4" ViewCount="" Body="&lt;p&gt;It is widely believed that the fundamental axioms of set theory ZFC are consistent, but this has not been proved in ZFC and in fact provably cannot be proved in ZFC itself unless ZFC is inconsistent, by the second &lt;a href=&quot;http://en.wikipedia.org/wiki/G%C3%B6del%27s_incompleteness_theorems&quot; rel=&quot;nofollow&quot;&gt;G&amp;ouml;del Incompleteness theorem&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Indeed, whatever fundamental axioms you favor, whether PA or KP or Z or ZF or ZFC or ZFC+large cardinals, then it is natural to suppose also that since you believe that those axioms are true that you also believe that those axioms are consistent, but this is provably not provable from your axioms, unless they are inconsistent. &lt;/p&gt;&#xA;" OwnerUserId="413" LastActivityDate="2011-02-01T19:42:29.747" CommunityOwnedDate="2011-02-01T19:42:29.747" />
  <row Id="19937" PostTypeId="2" ParentId="19933" CreationDate="2011-02-01T19:44:43.593" Score="4" ViewCount="" Body="&lt;p&gt;I think this is rather a philosophical than a mathematical question. The question whether numbers exist is almost as old as mathematics itself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I think about this I always have to smile and think about something like the image from René Magritte:&#xA;&lt;img src=&quot;http://i.imgur.com/UpAiv.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It sais in french &quot;this is not a pipe&quot;. The question you ask is very simillar. Of course we encounter numbers in the nature, but are they already there or did we invent them?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe this link could also be interesting:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://queuea9.wordpress.com/category/philosophy/&quot; rel=&quot;nofollow&quot;&gt;http://queuea9.wordpress.com/category/philosophy/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3123" LastActivityDate="2011-02-01T19:44:43.593" />
  <row Id="19938" PostTypeId="2" ParentId="19933" CreationDate="2011-02-01T19:45:47.573" Score="3" ViewCount="" Body="&lt;p&gt;The term 'imaginary' in mathematics is usually reserved for a technical concept involving $\sqrt{-1}$. Because it is easy to confuse with the normal everday English usage of 'existing only in the imagination', mathematician's prefer to use 'complex number' instead.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are wondering whether a negative number 'exists only in the imagination', then I agree it is hard to be very literal and count a negative number of apples.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But numbers have many uses (and so follow many different rules). Think of money: suppose you have 5 dollars in your hand but you owe someone 7 dollars. Then in on sense/use of numbers you really have a usable amount of -2 dollars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The difficulty with negative numbers is that, well, the positive numbers are just so obviously represented by objects, but negative numbers are not. But really, both kinds, positive and negative, are in your imagination. You don't have 5 in your hands, you have 5 dollars in your hand and the 5 is really how your imagination will deal with the situation.&lt;/p&gt;&#xA;" OwnerUserId="1919" LastActivityDate="2011-02-01T19:45:47.573" CommentCount="1" />
  <row Id="19939" PostTypeId="2" ParentId="19932" CreationDate="2011-02-01T19:51:29.813" Score="6" ViewCount="" Body="&lt;p&gt;One notable property of this function is &lt;a href=&quot;http://en.wikipedia.org/wiki/Pentagonal_number_theorem&quot; rel=&quot;nofollow&quot;&gt;Euler's Pentagonal Number Theorem:&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\prod_{k=1}^\infty (1-x^k) = \sum_{k=-\infty}^\infty (-1)^k x^{k(3k-1)/2}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a very interesting &lt;a href=&quot;http://arxiv.org/PS_cache/math/pdf/0510/0510054v2.pdf&quot; rel=&quot;nofollow&quot;&gt;paper&lt;/a&gt; on the Pentagonal Number Theorem by Jordan Bell.&lt;/p&gt;&#xA;" OwnerUserId="1301" LastEditorUserId="1301" LastEditorDisplayName="" LastEditDate="2011-02-02T08:12:04.453" LastActivityDate="2011-02-02T08:12:04.453" />
  <row Id="19940" PostTypeId="2" ParentId="19917" CreationDate="2011-02-01T20:03:51.323" Score="5" ViewCount="" Body="&lt;p&gt;You want to show that the tangent bundle T(S^3) is a &lt;em&gt;trivial bundle&lt;/em&gt;. It is a (not so hard) theorem that a vector bundle being trivial is equivalent to the existence of a (global) basis of sections. A section of the tangent bundle is another word for a vector field. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For S^1, considered as the unit circle in the complex plane, you probably have considered the vector field $z\mapsto iz$, which forms a basis. For S^3, you can do something similar, just like Mariano has described: now you consider S^3 as the unit sphere in the quaternions, and look at the vector fields&#xA;$V_i:z\mapsto iz$&#xA;$V_j:z\mapsto jz$&#xA;$V_k:z\mapsto kz$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[Lack of reputation prevents me from being able to place this as a comment. I wanted to point out a similarity between the S^1 and S^3 case.]&lt;/p&gt;&#xA;" OwnerUserId="6490" LastActivityDate="2011-02-01T20:03:51.323" />
  <row Id="19941" PostTypeId="1" AcceptedAnswerId="19942" CreationDate="2011-02-01T20:41:19.737" Score="5" ViewCount="171" Body="&lt;p&gt;If A is a commutative integral domain that's not a field, and let $K$ be the quotient field of A. We know that $K$ is not finitely generated as an A-module. But can $K$ ever be finitely generated as an A-algebra?&lt;/p&gt;&#xA;" OwnerUserId="6369" LastActivityDate="2011-02-01T21:17:04.057" Title="Can a quotient field ever be finitely generated as an algebra?" Tags="&lt;abstract-algebra&gt;&lt;modules&gt;" AnswerCount="2" />
  <row Id="19942" PostTypeId="2" ParentId="19941" CreationDate="2011-02-01T20:48:50.353" Score="9" ViewCount="" Body="&lt;p&gt;Sure. Let $A$ be the integers localized at $(2)$; that is,&#xA;$$A = \left\{ \frac{a}{b}\in\mathbb{Q}\;\Bigm|\; a,b\in\mathbb{Z}, b\gt 0, \gcd(a,b)=\gcd(2,b)=1\right\}.$$&#xA;The field of quotients of $A$ is $\mathbb{Q}$, and is equal to $A[\frac{1}{2}]$, so it is generated as an $A$-algebra by $1$ and $\frac{1}{2}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;More generally, any UFD $R$ with only finitely many pairwise non-associated irreducibles will have a field of fractions that is finitely generated as an $R$-algebra: just take $m_1,\ldots,m_k$ to be a maximal list of pairwise non-associates irreducibles in $R$, and the field of fractions will be equal to the subalgebra $R[\frac{1}{m_1},\ldots,\frac{1}{m_k}]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added.&lt;/strong&gt; In fact:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Theorem.&lt;/strong&gt; &lt;em&gt;Let $R$ be a UFD. The following are equivalent:&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;em&gt;The field of fractions $K$ of $R$ is finitely generated as an $R$-algebra.&lt;/em&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;em&gt;$R$ has only finitely many pairwise non-associate irreducible elements.&lt;/em&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof.&lt;/strong&gt; If $R$ has only finitely many non-associate irreducible elements (possibly $0$), $m_1,\ldots,m_k$, then the subalgebra $R[\frac{1}{m_1},\ldots,\frac{1}{m_k}]$ in $K$ equals all of $K$: every element of $K$ can be written as $\frac{a}{b}$ with $a,b\in R$, $b\neq 0$, and $b$ can be factored into irreducibles $b=um_1^{\alpha_1}\cdots m_k^{\alpha^k}$, where $\alpha_1,\ldots,\alpha_k$ are nonnegative integers and $u$ is a unit of $R$. Then&#xA;$$\frac{a}{b} = au^{-1}\left(\frac{1}{m_1}\right)^{\alpha_1}\cdots\left(\frac{1}{m_k}\right)^{\alpha_k}\in R\left[\frac{1}{m_1},\ldots,\frac{1}{m_k}\right].$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conversely, suppose that $K$ is finitely generated as an $R$-algebra. We may assume that the set that generated $K$ is made up entirely of fractions of the form $\frac{1}{b}$, because any element $\frac{a}{b}$ can be replaced with $\frac{1}{b}$ and we still get the same $R$-subalgebra. Moreover, we can assume that $b$ is irreducible, because if $b=m_1^{\alpha_1}\cdots m_r^{\alpha_r}$, then we can replace $\frac{1}{b}$ with $\frac{1}{m_1},\ldots,\frac{1}{m_r}$. Thus, we may assume that $K$ is generated as an $R$-algebra by the multiplicative inverses of a finite set of pairwise non-associated irreducible elements of $R$, $m_1,\ldots,m_k$. Now let $m\in R$ be any irreducible. We can express $\frac{1}{m}$ as a sum of multiples of powers of the $m_i^{-1}$, so we have&#xA;$$\frac{1}{m} = \frac{a_1}{m_1^{\alpha_1}} + \cdots + \frac{a_k}{m_k^{\alpha_k}} = \frac{b_1a_1+\cdots+b_ka_k}{m_1^{\alpha_1}\cdots m_k^{\alpha_k}}$$&#xA;where &#xA;$$b_j = \frac{m_1^{\alpha_1}\cdots m_k^{\alpha_k}}{m_i^{\alpha_i}}.$$&#xA;Then we must have that $(b_1a_1+\cdots+b_km_k)m = m_1^{\alpha_1}\cdots m_k^{\alpha_k}$, so $m$ divides $m_1^{\alpha_1}\cdots m_k^{\alpha_k}$, and hence $m$ is an associate of one of $m_1,\ldots,m_k$. Thus, $R$ has only finitely many pairwise non-associate irreducibles, as claimed. $\Box$&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-01T21:17:04.057" LastActivityDate="2011-02-01T21:17:04.057" CommentCount="4" />
  <row Id="19943" PostTypeId="1" AcceptedAnswerId="19958" CreationDate="2011-02-01T20:53:19.417" Score="8" ViewCount="183" Body="&lt;p&gt;It is well known that one of the first books Ramanujan studied was &quot;Synopsis of Pure and Applied Mathematics&quot; and that it shaped the way Ramanujan thought and wrote about mathematics. Being interested in the works of Ramanujan, I was wondering if it would be of any help if I studied this &quot;Synopsis&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moreover, I would appreciate it if someone who has experience with learning about Ramanujan's work (such as his &quot;Notebooks&quot; and his &quot;Lost Notebook&quot;) could explain to me what is a good way to approach his works. What are the prerequisites? Would I be able to comprehend his works after having taken courses on, say, Analytic Number Theory and (Real/Complex) Analysis? Is it better to read his notebooks (supposedly after having taken these courses) right away, or start of with books such as &quot;Ramanujan: twelve lectures suggested by his life and work&quot; and &quot;Number theory in the spirit of Ramanujan&quot; (written by Hardy and Berndt respectively)? Should I learn about modular forms beforehand? Etcetera.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Max&lt;/p&gt;&#xA;" OwnerUserId="1055" LastEditorUserId="1055" LastEditorDisplayName="" LastEditDate="2011-02-01T22:23:38.460" LastActivityDate="2011-02-01T23:09:30.660" CommunityOwnedDate="2011-02-01T21:42:18.480" Title="Guidelines for learning about Ramanujan's work?" Tags="&lt;soft-question&gt;&lt;big-list&gt;&lt;education&gt;&lt;ramanujan&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" />
  <row Id="19944" PostTypeId="2" ParentId="19941" CreationDate="2011-02-01T21:03:09.463" Score="5" ViewCount="" Body="&lt;p&gt;In fact, the class of such $A$ is not only nonempty, but important enough to have a standard name: &lt;a href=&quot;http://en.wikipedia.org/wiki/Goldman_domain&quot; rel=&quot;nofollow&quot;&gt;Goldman domains&lt;/a&gt;.  See page 117 of Pete L. Clark's notes on commutative algebra &lt;a href=&quot;http://math.uga.edu/~pete/integral.pdf&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2011-02-01T21:03:09.463" CommentCount="1" />
  <row Id="19945" PostTypeId="1" AcceptedAnswerId="19947" CreationDate="2011-02-01T21:07:30.530" Score="0" ViewCount="104" Body="&lt;p&gt;I saw in my book that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;2x² – 2x + 2&lt;/code&gt; factored became &lt;code&gt;2(x² – x + 1)&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;why it does not became &lt;code&gt;2(x(x – 1) + 1)&lt;/code&gt;? Is it wrong or correct as well?&lt;/p&gt;&#xA;" OwnerUserId="4501" LastEditorUserId="4501" LastEditorDisplayName="" LastEditDate="2011-02-06T13:19:13.853" LastActivityDate="2011-02-06T13:19:13.853" Title="Doubt about factoring" Tags="&lt;algebra&gt;&lt;factorial&gt;" AnswerCount="1" CommentCount="0" />
  <row Id="19946" PostTypeId="2" ParentId="4146" CreationDate="2011-02-01T21:13:26.937" Score="1" ViewCount="" Body="&lt;p&gt;Here is a compilation of books taken from OR-Exchange.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://industrialengineertools.blogspot.com/2010/08/favorite-operations-research-books-from.html&quot; rel=&quot;nofollow&quot;&gt;http://industrialengineertools.blogspot.com/2010/08/favorite-operations-research-books-from.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="6501" LastActivityDate="2011-02-01T21:13:26.937" CommunityOwnedDate="2011-02-01T21:13:26.937" />
  <row Id="19947" PostTypeId="2" ParentId="19945" CreationDate="2011-02-01T21:15:03.093" Score="4" ViewCount="" Body="&lt;p&gt;Yes, it's true that $x(x-1)+1 = x^2 - x + 1$. However, when we think about &quot;factoring&quot;, we think about the &lt;em&gt;factors&lt;/em&gt; (thing being multiplied), so we don't think of $x(x-1)+1$ as different from $x^2-x+1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Just like if we say &quot;factor $77$&quot;, we can write $77 = 7\times 11$ or $77 = 7\times(10+1)$, but we don't really like to write $11$ as $10+1$ when thinking about factors because what we are interested in is the factor.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Writing $x^2-x+1$ as $x(x-1)+1$ is not that helpful &lt;em&gt;in this situation&lt;/em&gt;. &lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2011-02-01T21:15:03.093" CommentCount="3" />
  <row Id="19948" PostTypeId="1" CreationDate="2011-02-01T21:35:03.357" Score="2" ViewCount="121" Body="&lt;p&gt;Hey!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm trying to solve an homework question but I got stuck.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let A be a m x n matrix with a SVD decomposition $A = U \Sigma V^*$ and $A^+ = (A^* A)^{-1} A^*$ its pseudoinverse.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm trying to get $A^+ = V \Sigma^{-1} U^*$, but I'm missing something.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone help me with this please?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="6500" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2011-02-01T22:09:58.850" LastActivityDate="2011-02-01T22:31:16.747" Title="Pseudoinverse matrix and SVD decomposition" Tags="&lt;homework&gt;&lt;linear-algebra&gt;&lt;matrices&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="19949" PostTypeId="1" CreationDate="2011-02-01T21:45:22.723" Score="0" ViewCount="63" Body="&lt;p&gt;I'm looking at the &lt;strong&gt;one dimensional&lt;/strong&gt; wave equation with a perturbation term that has a small coefficient. Since the wave equation is time dependent, does that mean that I should be looking into time dependent perturbation theory? I'm trying to narrow down what it is exactly I should be looking into. The perturbation is a squared term, the same term that is acted on by the wave equation, times a very small constant.&lt;/p&gt;&#xA;" OwnerUserId="6503" LastEditorUserId="6503" LastEditorDisplayName="" LastEditDate="2011-02-01T23:39:34.683" LastActivityDate="2011-02-01T23:39:34.683" Title="Perturbation of the Wave Equation - Where to Start?" Tags="&lt;differential-equations&gt;" CommentCount="3" />
  <row Id="19950" PostTypeId="1" AcceptedAnswerId="19959" CreationDate="2011-02-01T21:50:26.540" Score="2" ViewCount="104" Body="&lt;p&gt;Is the image of the general Veronese embedding ever contained in a hyperplane of $P^{n}$? I'm guessing no, but I can't prove it.&lt;/p&gt;&#xA;" OwnerUserId="2298" LastActivityDate="2011-02-01T23:16:26.500" Title="Image of the Veronese Embedding" Tags="&lt;algebraic-geometry&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="19951" PostTypeId="2" ParentId="19933" CreationDate="2011-02-01T21:54:50.483" Score="7" ViewCount="" Body="&lt;p&gt;Asking whether a negative number &quot;really exists&quot; is not really a meaningful question, to my mind.  A number (of any kind) is an abstract mathematical idea, and it's not really clear how to talk about whether ideas &quot;exist&quot;.  Does justice really exist?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, what we can say is that there are situations in the real world for which the idea of a negative number gives a useful description or model: any situation where something is removed or decreased.&lt;/p&gt;&#xA;" OwnerUserId="822" LastActivityDate="2011-02-01T21:54:50.483" CommentCount="1" />
  <row Id="19953" PostTypeId="2" ParentId="19948" CreationDate="2011-02-01T22:08:20.050" Score="1" ViewCount="" Body="&lt;p&gt;Paulochf. It is SVD - Singular Value Decomposition and not SDV. I have edited the title. Also there is a homework tag which you need to apply when you are posting a homework question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To get your answer, first you need to assume that the matrix $A^*A$ is invertible. For which you need $n \leq m$ and rank($A$) is $n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So when $n \leq m$ and when rank($A$) is $n$, then the reduced SVD of $A$ is $A = U \Sigma V^*$ where $U \in \mathbb{R}^{m \times n}$, $\Sigma \in \mathbb{R}^{n \times n}$ and $V \in \mathbb{R}^{n \times n}$ such that $U^* U = I_{n \times n}$, $V^* V = I_{n \times n}$, $V V^* = I_{n \times n}$ and $\Sigma$ is a square diagonal matrix and has only (positive) real entries.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that $V^{-1}=V^*$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also note that $A^* = V \Sigma^* U^* = V \Sigma U^*$ since $\Sigma^* = \Sigma$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Further note that if $M_1,M_2 \text{and} M_3$ are invertible matrices then $(M_1 M_2 M_3)^{-1} = M_3^{-1} M_2^{-1} M_1^{-1}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Use these to get the final answer.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2011-02-01T22:22:13.190" LastActivityDate="2011-02-01T22:22:13.190" CommentCount="3" />
  <row Id="19954" PostTypeId="2" ParentId="19948" CreationDate="2011-02-01T22:31:16.747" Score="0" ViewCount="" Body="&lt;p&gt;$$&#xD;&#xA;\begin{align}&#xD;&#xA;A^+ &amp;amp;= (A^*A)^{-1}A^* \\&#xD;&#xA;    &amp;amp;=(V\Sigma U^*U\Sigma V^*)^{-1} V\Sigma U^* \\&#xD;&#xA;    &amp;amp;=(V\Sigma^2 V^*)^{-1} V\Sigma U^* \\&#xD;&#xA;    &amp;amp;=(V^*)^{-1} \Sigma^{-2} V^{-1} V\Sigma U^* \\&#xD;&#xA;    &amp;amp;= V \Sigma^{-2}\Sigma U^* \\&#xD;&#xA;    &amp;amp;= V\Sigma^{-1}U^*&#xD;&#xA;\end{align}&#xD;&#xA;$$&#xA;using the properties of the matrices $U,V,\Sigma$ in the &lt;a href=&quot;http://en.wikipedia.org/wiki/Singular_value_decomposition&quot; rel=&quot;nofollow&quot;&gt;Singular_value_decomposition&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="367" LastActivityDate="2011-02-01T22:31:16.747" CommentCount="1" />
  <row Id="19956" PostTypeId="1" CreationDate="2011-02-01T22:45:00.423" Score="0" ViewCount="127" Body="&lt;p&gt;I'm learning about lg (log base 2) for an algorithms class and theres a problem in the book that is confusing me.&lt;br&gt;&#xA;It asks:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Find a formula for lg(n!) using Stirling's approximation for n!, for large n.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Stirling's approximation for n! is sqrt(2pi(n))(n/e)^n&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anyone have guidance on how to go about creating this formula?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="3691" LastActivityDate="2011-02-01T22:45:00.423" Title="Logarithm base 2 and factorials" Tags="&lt;algorithms&gt;&lt;logarithms&gt;&lt;factorial&gt;" CommentCount="7" />
  <row Id="19957" PostTypeId="1" CreationDate="2011-02-01T22:55:02.830" Score="4" ViewCount="121" Body="&lt;p&gt;What do you think is the best way to rank universities participating in a competition if we know for each university the number of recognitions that it's students won. Each university can participate with 3 students, and approximately half of them get a recognition - which can be first (5%), second (15%) or third place (30%).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To be more specific, let's say 5 universities participated in a competition and the first university won 2 first place recognitions and 1 second place recognitions, the second university won 2 third place recognitions, etc... Probably a good way of measuring universities is by the total number of recognitions won, or the number of first place recognitions won, followed by number of second place recognitions won etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, my question is: How do we rank the quality of education in the universities if we know for each university how many students learn there? I'm asking this because it's probably a good idea to take into account the number of students? One possibility that I was thinking of is to divide the number of achievements with the number of students and use that to compare universities, but the problem here is that each school can participate with only 3 students (probably if the largest university participated with more than 3 students it would have won more recognitions). We can assume that the universities have a lot of students, and that the background of the students enrolled at different universities is the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you very much for your time!&lt;/p&gt;&#xA;" OwnerUserId="6506" LastActivityDate="2011-04-29T17:01:37.253" Title="Specific statistics problem: Ranking universities" Tags="&lt;statistics&gt;&lt;probability-distributions&gt;" AnswerCount="2" />
  <row Id="19958" PostTypeId="2" ParentId="19943" CreationDate="2011-02-01T23:09:30.660" Score="4" ViewCount="" Body="&lt;p&gt;I know one person who read (at least parts of) &lt;em&gt;Carr's Synopsis&lt;/em&gt; when they were young; they are now an exceptionally strong mathematician, but I don't think there is a causal relationship.  Rather, there is a correlation between their early interest in mathematics and their independence, and their interest in reading unusual original sources like &lt;em&gt;Carr's Synopsis&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regarding the question of reading original sources (especially old ones): I think this tends to be discouraged, and the reasons are related to Qiaochu's comment --- it is hard to both learn everything that is necessary to do modern mathematics &lt;em&gt;and&lt;/em&gt; spend time reading all the original sources.    On the other&#xA;hand there are advantages that can come from reading older sources that are often not mentioned: for example, one can find points of view or topics or techniques that are no longer emphasized, or are even forgotten, which can sometimes be helpful in a very specific way, and other times helpful in a more general inspirational sense.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To speak of myself for a moment, when I was learning mathematics I read many older and somewhat unusual sources, and I think that there are advantages to this, as well as the disadvantages mentioned above.   I think that I have an above average interest in the history of mathematics, and of the historical development of mathematical ideas, and so it was enjoyable and profitable for me to do this.   But if I had a different mathematical temperament this wouldn't have been a sensible choice; it was a reflection of my own mathematical personality.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you find it interesting to read older sources, or you think it might be interesting to try, then I would suggest that you just look at them and see what you get out of it.    Just bear in mind that you won't be able to learn modern mathematics in this way, so you wouldn't want such reading to be your only reading by any means; rather, you might want to regard it as a supplement to your regular study of modern expositions.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for learning specifically about Ramanujan's work, the best place to start is Hardy's book on Ramanujan.  This is a wonderful book.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, to approach Ramanujan's legacy in a modern context, the best thing to do is to learn the theory of modular forms.  For this, Serre's &lt;em&gt;Course in arithmetic&lt;/em&gt; is a very good source.  (Serre's book has three parts which are only loosely connected.  So you can read the section on modular forms without having read the other sections carefully beforehand.)&lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2011-02-01T23:09:30.660" CommunityOwnedDate="2011-02-01T23:09:30.660" CommentCount="8" />
  <row Id="19959" PostTypeId="2" ParentId="19950" CreationDate="2011-02-01T23:16:26.500" Score="3" ViewCount="" Body="&lt;p&gt;No.  To prove it, imagine what it would mean for the image to be contained in a hyperplane: this would mean that some non-zero linear combination of the degree $d$ monomials vanished identically, which is to say, that there is some non-zero degree $d$ homogeneous equation which vanishes identically on $\mathbb P^n$.  Hopefully you can convince yourself that this is not possible.&lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2011-02-01T23:16:26.500" />
  <row Id="19960" PostTypeId="1" AcceptedAnswerId="19964" CreationDate="2011-02-01T23:23:16.180" Score="3" ViewCount="117" Body="&lt;p&gt;In the chapter on completions in Atiyah Macdonald, they define a topological abelian group. Let $G$ be such a group. Denote $H$ to be the set of intersection of neighbourhoods of $0$. Then it is shown that $H$ is a subgroup. Then they say $G/H$ is Hausdorff because points are closed. I don't quite understand this. &lt;/p&gt;&#xA;" OwnerUserId="6166" LastActivityDate="2011-02-02T09:00:26.543" Title="Hausdorffness of a topological abelian group" Tags="&lt;topology&gt;&lt;commutative-algebra&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="19961" PostTypeId="2" ParentId="19884" CreationDate="2011-02-01T23:27:32.183" Score="2" ViewCount="" Body="&lt;p&gt;The answer of Plop is correct, but I'll try to explain it a little bit more. I know how confused you are. I'm assuming you are using Scheme Theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By definition, $\mathbb{A}^2_k=\text{Spec }k[x,y]$. The origin of the plane  is given by the maximal ideal $(x,y)$. Let $X=\text{Spec }k[x,y] \setminus (x,y)$. Then $X=D(x) \cup D(y)$. We got an affine cover for $X$. By the definition of the sheaf, a function on $\mathcal{O}_{\mathbb{A}^2_k}(X)$ can be represented as an ordered pair of a function in  $\mathcal{O}_{\mathbb{A}^2_k}(D(x))$ and the other in $\mathcal{O}_{\mathbb{A}^2_k}(D(y))$ wich coincides in the intersection $\mathcal{O}_{\mathbb{A}^2_k}(D(x)\cap D(y))$. The 'coincides' means that they are equal after applying the restrinction map.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, we know that $\mathcal{O}_{\mathbb{A}^2_k}(D(x))=k[x,y]_x=k[x,y,1/x]$ and $\mathcal{O}_{\mathbb{A}^2_k}(D(y))=k[x,y]_y=k[x,y,1/y]$. Since $D(x)\cap D(y)=D(xy)$, we get $\mathcal{O}_{\mathbb{A}^2_k}(D(x)\cap D(y))=k[x,y]_{xy}=k[x,y,1/(xy)]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you have to do now is to show that if you have a function $f\in k[x,y,1/x]$ and $g\in k[x,y,1/y]$ wich coincides in $k[x,y,1/(xy)]$, they must be equal and belong to $k[x,y]$. Remember that, in this case, the restriction map is a localisation map.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is an important example because it's one of the first (and maybe the simpler) examples of a non-affine scheme! Can you see why?&lt;/p&gt;&#xA;" OwnerUserId="1971" LastActivityDate="2011-02-01T23:27:32.183" CommentCount="1" />
  <row Id="19962" PostTypeId="1" CreationDate="2011-02-01T23:32:26.140" Score="0" ViewCount="71" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Covariance_matrix&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Covariance_matrix&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Cov(Xi,Xj) = E((Xi-Mi)(Xj-Mj))&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Is the above equivalent to:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;(Xi-Mi)(Xj-Mj)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I don't understand why the expectancy of (Xi-Mi)(Xj-Mj) would be different than just (Xi-Mi)(Xj-Mj)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Addendum:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's say I have two sets of data:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Set 1: 1,2,3 avg: 2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Set 2: 4,5 avg: 4.5&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is the following a covariance matrix?  &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;(1-2)*(4-4.5) , (2-2)*(4-4.5) , (3-2)*(4-4.5)&#xA;(1-2)*(5-4.5) , (2-2)*(5-4.5) , (3-2)*(5-5.5)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I'm reading online and it seems like a covariance matrix composed of two data sets should be a 2x2 matrix, but in this case, I have a 2x3 matrix.  How do I go from what I have to the correct 2x2 matrix?&lt;/p&gt;&#xA;" OwnerUserId="6333" LastEditorUserId="6333" LastEditorDisplayName="" LastEditDate="2011-02-01T23:53:49.117" LastActivityDate="2011-02-02T01:23:21.787" Title="Simple Question About Expectancy" Tags="&lt;probability&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="19963" PostTypeId="2" ParentId="19962" CreationDate="2011-02-01T23:41:45.680" Score="3" ViewCount="" Body="&lt;p&gt;It's like asking if $X$ is equivalent to $E(X)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$X$ is (asssumed to be) a random variable, it can take several values according to some probability law. In contrast $E(X)$ is the expected value of $X$ - so it's not a random variable which takes different values in different tries, it's a constant number.  So they are conceptually different entities (they coincide only in the degenerate-trivial case in which the random variable only takes one value).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same applies to the covariance matrix. In your formula $X_i$ is a random variable, so the product is also a random var (because it's a function of random variables!). The expectation operator &quot;takes out&quot; the randomness, and so the covariance matrix is a constant matrix.&lt;/p&gt;&#xA;" OwnerUserId="312" LastActivityDate="2011-02-01T23:41:45.680" CommentCount="2" />
  <row Id="19964" PostTypeId="2" ParentId="19960" CreationDate="2011-02-01T23:41:53.260" Score="4" ViewCount="" Body="&lt;p&gt;Suppose $G$ is a topological group such that $\{1\}$ is a closed set. Let $x$, $y\in G$ be two distinct elements. Since $x^{-1}y\neq1$ and $\{1\}$ is closed, there exists an open neighborhood $U$ of $1$ such that $x^{-1}y\not\in U$, so that $y\not\in xU$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, it is easy to prove using the axioms of topological groups that there exists an open neighborhood $V$ of $1$ such that $VV^{-1}\subseteq U$. And, finally, one can see easily that $xV\cap yV=\emptyset$, so that $xV$ and $yV$ are disjoint open neighborhoods of $x$ and $y$, respectively. $G$ is therefore Hausdorff.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, on a topological group even the fact that the topology be $T_0$ (a weaker condition that $T_1$,which is the one we used above) is enough to prove that it is Hausdorff. You can do that as an exercise :)&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2011-02-01T23:41:53.260" CommentCount="2" />
  <row Id="19965" PostTypeId="2" ParentId="19962" CreationDate="2011-02-01T23:45:52.777" Score="1" ViewCount="" Body="&lt;p&gt;There is a difference between a random variable and its expectation.  Take a standard fair six sided die.  It can take any of the values {1,2,3,4,5,6}, but its expectation is 3.5.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly $(X_i - \mu_i)(X_j - \mu_j)$ is a random variable with several possible values while $\mathrm{E}[(X_i - \mu_i)(X_j - \mu_j)]$ is a number depending on the joint distribution of $X_i$ and $X_j$.&lt;/p&gt;&#xA;" OwnerUserId="6460" LastActivityDate="2011-02-01T23:45:52.777" />
  <row Id="19966" PostTypeId="2" ParentId="14475" CreationDate="2011-02-02T00:09:45.290" Score="2" ViewCount="" Body="&lt;p&gt;Spivak and Munkres are both not bad, but I think Advanced Calculus by Loomis and Sternberg is even better. It rigorously covers linear algebra, calculus in severable variables, metric spaces, and multilinear algebra. Then it moves on to manifolds and integration, and ends with applications to classical mechanics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moreover, it's free! See &lt;a href=&quot;http://www.math.harvard.edu/~shlomo/&quot; rel=&quot;nofollow&quot;&gt;http://www.math.harvard.edu/~shlomo/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="6490" LastActivityDate="2011-02-02T00:09:45.290" />
  <row Id="19967" PostTypeId="1" CreationDate="2011-02-02T00:17:21.580" Score="1" ViewCount="315" Body="&lt;p&gt;Let $$\varphi(x) = e^{-x} + \frac{x}{1!} \cdot e^{-2x} + \frac{3x^{2}}{2!}\cdot e^{-3x} + \frac{4^{2} \cdot x^{3}}{3!} \cdot e^{-4x} + \cdots$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then what is the value of: $$ \displaystyle\lim_{t \to 0} \frac{\varphi(1+t) - \varphi(t)}{t}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not getting any idea as to how to proceed for this problem. I tried summing up the expression but no avail. I also tried differentiating $\varphi(x)$ since the &lt;em&gt;limit&lt;/em&gt; quantity which we require seems to involve derivative but again i couldn't find any pattern. Any ideas on how to solve this problem.&lt;/p&gt;&#xA;" LastActivityDate="2011-02-04T02:55:12.993" Title="Finding value of the limit: $ \lim_{t \to 0} \frac{\varphi(1+t) - \varphi(t)}{t}$" Tags="&lt;real-analysis&gt;&lt;analysis&gt;" AnswerCount="2" CommentCount="6" FavoriteCount="2" />
  <row Id="19968" PostTypeId="2" ParentId="18288" CreationDate="2011-02-02T00:23:30.193" Score="0" ViewCount="" Body="&lt;p&gt;It is odd that people here are hesitant to quantify things i.e. to break things up and to construct interesting structures. Did you get hooked from Arturo's reply? I did! For future readers so that the Fun will never end: &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;ProofWiki &lt;a href=&quot;http://www.proofwiki.org/wiki/De_Morgan%27s_Laws_%28Set_Theory%29#General_Proof&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; and more about quantification &lt;/li&gt;&#xA;&lt;li&gt;Tree Proof Generator (very good to correct mistakes in thinking) &lt;a href=&quot;http://www.umsu.de/logik/trees/&quot; rel=&quot;nofollow&quot;&gt;http://www.umsu.de/logik/trees/&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;or if you are lazy like me, you could use a math Study Search &lt;a href=&quot;http://www.google.com/coop/profile?user=007883453237583604479&amp;amp;preview_query=mathematics&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; (not to remember awkward urls)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;It is fun :)&lt;/p&gt;&#xA;" OwnerUserId="5902" LastEditorUserId="5902" LastEditorDisplayName="" LastEditDate="2011-02-02T00:48:05.607" LastActivityDate="2011-02-02T00:48:05.607" CommentCount="2" />
  <row Id="19969" PostTypeId="2" ParentId="19960" CreationDate="2011-02-02T00:31:39.577" Score="1" ViewCount="" Body="&lt;p&gt;G/H will be Hausdorff, essentially because quotienting by H identifies all the points that can't be discerned from 0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lemma : If $U \subset G$ is open, $U = U+H$ :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $x \in U$. $U-x$ is a neighboorhood of 0, so $H \subset U-x$. Thus $H+x \subset U-x+x = U$. Since we have this for all $x$, $U = U+H$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $x$ be an element of G/H.&#xA;$x$ is the projection of some $y$ from G, with $y \not\in H$.&#xA;Since $y \not\in H$, it means that there exists a neighboorhood $U$ of 0 that doesn't contain $y$.&#xA;Then, as Mariano said, since the application $x \rightarrow x-x$ is continuous, there exists an open $V$ containing 0 such that $V-V \subset U$. Finally, since we have $y \not\in V-V$, $y+V \cap V = \emptyset$, and $y+V+H \cap V+H = \emptyset$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This shows that, in the quotient G/H, 0 and x are separated.&lt;/p&gt;&#xA;" OwnerUserId="6270" LastEditorUserId="6270" LastEditorDisplayName="" LastEditDate="2011-02-02T08:33:02.430" LastActivityDate="2011-02-02T08:33:02.430" CommentCount="3" />
  <row Id="19970" PostTypeId="2" ParentId="19915" CreationDate="2011-02-02T00:49:09.033" Score="2" ViewCount="" Body="&lt;p&gt;No, it is not convex in general.  Taking $A = \left[\begin{smallmatrix}1 &amp;amp; 0\end{smallmatrix}\right]$ and $B = \left[\begin{smallmatrix}0 &amp;amp; 1\end{smallmatrix}\right]$ your function is $f(x,y) = x^2y^2$.  But $0.5 f(1,0) + 0.5 f(0,1) = 0 + 0 = 0 &amp;lt; 0.5^4 = f(0.5,0.5)$.&lt;/p&gt;&#xA;" OwnerUserId="884" LastActivityDate="2011-02-02T00:49:09.033" CommentCount="1" />
  <row Id="19971" PostTypeId="1" AcceptedAnswerId="19972" CreationDate="2011-02-02T00:56:19.450" Score="3" ViewCount="97" Body="&lt;p&gt;In this example problem in my textbook:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Find a power series representation for ln(1-x) and its radius of convergence.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They integrate both sides:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-ln(1-x) = integral (1/1-x)dx which comes out to be SUM x^n/n + C.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They solve for C, C=0. This is where I get stuck. They proceed to show what the series looks like when C=0, and show:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ln(1-x) = -x - x^2/2 - x^3/x - ... = -SUM x^n/n |x| &amp;lt; 1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How come these are all negative terms?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They then say: &quot;Notice what happens if we put x = 1/2 in the result of Example 6. Since ln(1/2) = -ln2 we see that: &quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ln2 = 1/2 + 1/8 + 1/24 + 1/64 + ... = SUM 1/(n2^n)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why was x = 1/2 put back into -ln(1-x)? Why not ln(1-x)? Now why are they all positive?&lt;/p&gt;&#xA;" OwnerUserId="3514" LastEditorUserId="3514" LastEditorDisplayName="" LastEditDate="2011-02-02T01:04:07.023" LastActivityDate="2011-02-02T01:04:07.023" Title="Integrating Power Series" Tags="&lt;homework&gt;&lt;power-series&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="19972" PostTypeId="2" ParentId="19971" CreationDate="2011-02-02T01:03:21.507" Score="3" ViewCount="" Body="&lt;p&gt;$\ln(1-x) \lt 0$ for $0 \lt x \lt 1$. So you expect negative terms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have a series $f(x) = \sum a_n x^n$ valid for $|x| \lt 1$, then we have that $-f(x) = g(x) = \sum (-a_n) x^n$ is also valid for $|x| \lt 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does that help?&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2011-02-02T01:03:21.507" CommunityOwnedDate="2011-02-02T01:03:21.507" CommentCount="3" />
  <row Id="19973" PostTypeId="2" ParentId="19962" CreationDate="2011-02-02T01:23:21.787" Score="1" ViewCount="" Body="&lt;p&gt;The covariance matrix is $Cov(X_i,X_j)=E[(X_i-\mu_i)(X_j-\mu_j)]$  Each $\mu_i=E[X_i]$. the expected value of variable $i$.  If your variables are independent, the off-axis covariance terms are zero because $E[(X_i-\mu_i)(X_j-\mu_j)]=E[(X_i-\mu_i)]E[(X_j-\mu_j)]$ and both terms on the right are zero.  The diagonal terms are just the variances of each variable, as you should be able to see from the definition.  You are right that the covariance matrix for your data should be 2x2, as the dimension comes from the number of variables, not from the number of data points.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-02-02T01:23:21.787" />
  <row Id="19974" PostTypeId="1" AcceptedAnswerId="19982" CreationDate="2011-02-02T02:07:45.143" Score="1" ViewCount="61" Body="&lt;p&gt;Suppose you have a deck of cards. Each card has n images on it. Any two cards will have exactly one matching image. What would be the formula for the maximum number of unique images and the maximum number of unique cards?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At first I thought the maximum number of images would be:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sum\limits_{i=1}^n i$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But that doesn't seem right as the pairs themselves don't have to be unique (ie, you can have three cards that share a single image).&lt;/p&gt;&#xA;" OwnerUserId="6510" LastActivityDate="2011-02-02T04:02:14.137" Title="Limit to Unique Combinations" Tags="&lt;combinatorics&gt;" AnswerCount="3" />
  <row Id="19975" PostTypeId="2" ParentId="19974" CreationDate="2011-02-02T02:30:30.720" Score="0" ViewCount="" Body="&lt;p&gt;I saw a game like this that had only two copies of any given image.  In that case, for $n$ images per card you can have a deck of $n+1$ cards and need $\frac{n(n+1)}{2}$ unique images, as you said.  I think the deck I saw had $n=8$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think for prime $p$ you can do it with $p^2+p$ images and $p^2$ cards, using $p+1$ images per card.  For $p=3$ you have $$\begin{array}{ccc} 1ADG &amp;amp; 1CEH &amp;amp; 1BFJ  \\  2BEG &amp;amp; 2AFH &amp;amp; 2CDJ \\  3CFG &amp;amp; 3BDH &amp;amp; 3AEJ \end{array}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the way to generalize this is to think of the cards as $(i,j)$ where $i$ and $j$ range from $0$ to $p-1$.  You have $p$ sets of images, noted as $k$ ranging from $0$ to $p-1$ and each card gets image $i+jk \pmod {p}$ of set $k$ plus image $j$ of set $p$ (this is the one I forgot before)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As I am always one to like a cheap solution, you can do it with only one image.  Think of the backs of a deck of cards.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2011-02-02T04:02:14.137" LastActivityDate="2011-02-02T04:02:14.137" CommentCount="2" />
  <row Id="19976" PostTypeId="1" CreationDate="2011-02-02T02:40:44.830" Score="4" ViewCount="112" Body="&lt;p&gt;Hi. Let $M$ and $N$ be graded $R$-modules (with $R$ a graded ring). Let $\varphi:M\rightarrow N$ be an homogeneous homomorphism of degree $i$ (that is $\varphi(M_n)\subset N_{n+i}$). Denote by $\mathrm{Hom}_i(M,N)$ the group of homogeneous homomorphisms of degree $i$. We define $^*\mathrm{Hom}_R(M,N)=\bigoplus_{i\in\mathbb{Z}}\mathrm{Hom}_i(M,N)$, it is a graded $R$-submodule of $\mathrm{Hom}_R(M,N)$. How can I prove that these two modules are equal if $M$ is finite? And do you know a counterexample if $M$ is not finite?&lt;/p&gt;&#xA;" OwnerUserId="6236" LastEditorUserId="6236" LastEditorDisplayName="" LastEditDate="2011-02-02T16:54:52.393" LastActivityDate="2011-02-02T16:54:52.393" Title="Homomorphisms of graded modules" Tags="&lt;abstract-algebra&gt;&lt;commutative-algebra&gt;&lt;modules&gt;&lt;graded-modules&gt;" CommentCount="8" />
  <row Id="19977" PostTypeId="2" ParentId="19876" CreationDate="2011-02-02T02:53:22.573" Score="0" ViewCount="" Body="&lt;p&gt;All of trig identities can be derived from $\exp{(\hat{i}\theta)}=\cos(\theta)+\hat{i}\sin\theta$. How?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Example&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\cos(2\,x) = \mathrm{Re}\left(\exp{(2\hat{i}\,x)}\right)$$ &#xA;$$ = \mathrm{Re}\left(\exp{(\hat{i}x)}\,\exp{(\hat{i}x)}\right) $$&#xA;$$ = \mathrm{Re}\left((\cos x+\hat{i}\sin x)\,(\cos x+\hat{i}\sin x)\right) $$&#xA;$$ = \mathrm{Re}\left( \cos^2 x - \sin^2 x +2\hat{i}\cos x \sin x \right) $$&#xA;$$ = \cos^2 x - \sin^2 x $$&lt;/p&gt;&#xA;" OwnerUserId="3301" LastActivityDate="2011-02-02T02:53:22.573" />
  <row Id="19978" PostTypeId="1" CreationDate="2011-02-02T02:59:34.483" Score="3" ViewCount="70" Body="&lt;p&gt;This is a homework question, so I'd appreciate hints (or perhaps explanations of concepts I've not properly digested)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyhow: This is exercise 1.3.6 in Weibel's book on homological algebra. Let $0 \to A \to B \to C \to 0$ be an exact sequence of double complexes of modules. Show that there is a short exact sequence of total complexes, and conclude that if Tot(C) is acyclic, then $Tot(A) \to Tot(B)$ is a quasi-isomorphism.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The last part of the exercise is clear. If Tot(C) is acyclic, then the long exact sequence is of the form&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ldots \to H_{i+1}C(=0) \to H_i(A) \to H_i(B) \to H_i(C)(=0) $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so the induced morphism on homology is an isomorphism.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first part of the question is unclear, however. The definition of an exact sequence of double complexes is not explicitly stated, but I assume it is such that $0 \to A_{ij} \to B_{ij} \to C_{ij} \to 0$ is exact for all i,j and everything commutes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $\alpha:A \to B$ be a morphism of double complexes. The induced morphism between the total complexes, $\alpha^*: Tot(A) \to  Tot(B)$, is then defined, I'd assume, as $\alpha^*=d_B^h \alpha + d_B^v \alpha$ (where $d_B^h$ and $d_B^v$ denotes the horizontal and vertical differentials, respectively). The problem is how I go about showing the induced sequence is exact.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;EDIT:&quot; After some thought, I guess a good first step would be to show that $\beta^* \circ \alpha^* = 0$, which shouldn't be too difficult. (where $\alpha^*,\beta^*$ denotes the induced morphisms of total complexes)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit2: I've clarified my notation a bit.&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="4284" LastEditorUserId="4284" LastEditorDisplayName="" LastEditDate="2011-02-02T04:30:08.933" LastActivityDate="2011-02-02T04:30:08.933" Title="Exact sequence of double complexes induces exact sequence on total complexes" Tags="&lt;homological-algebra&gt;" CommentCount="9" FavoriteCount="1" />
  <row Id="19979" PostTypeId="1" AcceptedAnswerId="19989" CreationDate="2011-02-02T03:24:04.823" Score="1" ViewCount="70" Body="&lt;p&gt;I think this is false, a counter example could be:&lt;br&gt;&#xA;c = 100,&lt;br&gt;&#xA;b = 10,&lt;br&gt;&#xA;a = 5  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the book answer is true :( ! Did I misunderstand the problem or the book's answer was wrong?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks,&lt;br&gt;&#xA;Chan&lt;/p&gt;&#xA;" OwnerUserId="5711" LastEditorUserId="5711" LastEditorDisplayName="" LastEditDate="2011-02-02T03:30:32.447" LastActivityDate="2011-02-02T13:12:21.180" Title="If b is the largest square divisor of a and a^2|c then a|b?" Tags="&lt;elementary-number-theory&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="19980" PostTypeId="2" ParentId="19979" CreationDate="2011-02-02T03:28:58.347" Score="0" ViewCount="" Body="&lt;p&gt;That is not a counterexample:  $10$ is not a divisor of $4$ and $4^2$ does not divide $100$.  You must have at least one typo.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps you are trying to ask this:  if $b^2 | c$ and $b$ is the largest number with this property (NOTE NOT $b^2 | a$), and also $a^2|c$ then must $a|b$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer to that would be yes, which you can prove by thinking about each prime that divides $a$ and showing it divides $b$ at least as many times.&lt;/p&gt;&#xA;" OwnerUserId="6071" LastEditorUserId="6071" LastEditorDisplayName="" LastEditDate="2011-02-02T13:12:21.180" LastActivityDate="2011-02-02T13:12:21.180" CommentCount="1" />
  <row Id="19981" PostTypeId="1" AcceptedAnswerId="20042" CreationDate="2011-02-02T03:37:18.847" Score="7" ViewCount="119" Body="&lt;p&gt;In reading &quot;Real Analysis: Modern Techniques and Their Applications&quot; (Folland), I've come across a few different notions of &quot;distribution&quot; or &quot;distribution functions.&quot;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;The &lt;strong&gt;distribution function&lt;/strong&gt; of a finite Borel measure $\mu$ on $\mathbb{R}$ is defined by $F(x) = \mu((-\infty, x])$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The &lt;strong&gt;distribution function&lt;/strong&gt; of a measurable function $f\colon X \to \mathbb{R}$ on a measure space $(X, M, \mu)$ is a function $\lambda_f\colon (0,\infty) \to [0,\infty]$ given by $\lambda_f(\alpha) = \mu(\{x\colon |f(x)| &amp;gt; \alpha\})$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;A &lt;strong&gt;distribution&lt;/strong&gt; on an open set $U \subset \mathbb{R}^n$ is a continuous linear functional on $C^\infty_c(U)$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Is there any sort of relationship between these concepts?&lt;/p&gt;&#xA;" OwnerUserId="640" LastActivityDate="2011-02-02T16:23:52.583" Title="How do different notions of &quot;distribution&quot; relate to one another?" Tags="&lt;real-analysis&gt;&lt;measure-theory&gt;&lt;distribution-theory&gt;" AnswerCount="2" />
  <row Id="19982" PostTypeId="2" ParentId="19974" CreationDate="2011-02-02T03:39:19.453" Score="2" ViewCount="" Body="&lt;p&gt;If $n$ is prime, one possibility is to take a projective plane over the field of $n$ elements for the underlying set of images, and then make one card for each line in the projective plane.  Since every two lines intersect in exactly one point, you will have what you want.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The projective plane has $1+n+n^2$ points, and the same number of lines.  So you could do it with this many images and cards.  I don't know if this is maximal, though, or what to do if $n$ isn't prime.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Have a look at&#xA;    &lt;a href=&quot;http://en.wikipedia.org/wiki/Projective_plane&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Projective_plane&lt;/a&gt; &#xA;if you don't know about projective planes.)&lt;/p&gt;&#xA;" OwnerUserId="6071" LastActivityDate="2011-02-02T03:39:19.453" />
  <row Id="19983" PostTypeId="2" ParentId="19974" CreationDate="2011-02-02T03:51:15.190" Score="2" ViewCount="" Body="&lt;p&gt;There are infinitely many cards with this property. Fix some picture which will appear on all cards, and make up $n-1$ new pictures for each card.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the total number of pictures is $m$, then the Ray-Chaudhury-Wilson theorem implies that the maximal number of cards is $m$, assuming each card contains &lt;i&gt;exactly&lt;/i&gt; $n$ pictures. Without that assumption, the Frankl-Wilson theorem states that the maximal number of cards is $m+1$.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2011-02-02T03:51:15.190" />
  <row Id="19984" PostTypeId="1" AcceptedAnswerId="19991" CreationDate="2011-02-02T03:58:26.213" Score="3" ViewCount="125" Body="&lt;p&gt;Suppose f(z) is a polynomial such that its derivatives are non-zero for all $|z| &amp;lt;1$.   Is the restriction of $f$ to $|z|&amp;lt;1$ 1 to 1?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that $f$ must be locally 1 to 1.   It is obvious that $f$ is 1-1 for polynomials of order 1.  The case of order 2 follows from Gauss-Lucas Thm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am stuck on how to prove the general case.&lt;/p&gt;&#xA;" OwnerUserId="832" LastEditorUserId="856" LastEditorDisplayName="" LastEditDate="2011-02-02T05:26:59.560" LastActivityDate="2011-02-02T06:56:55.577" Title="Roots of derivative of polynomial" Tags="&lt;analysis&gt;&lt;complex-analysis&gt;" AnswerCount="2" />
  <row Id="19985" PostTypeId="1" AcceptedAnswerId="19992" CreationDate="2011-02-02T04:03:55.940" Score="1" ViewCount="121" Body="&lt;p&gt;I've thought about trying to prove this by induction, and by contradiction, but I can't seem to phrase this in a formal way. Any help would be appreciated.&lt;/p&gt;&#xA;" OwnerUserId="6436" LastActivityDate="2011-02-02T07:38:18.723" Title="How can I prove that a number n is either of the form 3k+1 or 3k+2, or 3k" Tags="&lt;number-theory&gt;&lt;divisibility&gt;" AnswerCount="3" CommentCount="4" />
  <row Id="19986" PostTypeId="2" ParentId="19933" CreationDate="2011-02-02T04:04:08.883" Score="2" ViewCount="" Body="&lt;p&gt;Where the philosophy of negative numbers &quot;existing&quot; is concerned, I think Nate Eldredge answered the question very nicely.  However, let me post a separate answer (really an extended comment) to address the diagram.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think there's some confusion regarding the distinction between &lt;em&gt;position&lt;/em&gt; and &lt;em&gt;distance&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we want to talk about &lt;em&gt;position&lt;/em&gt; on the diagram that you've drawn, then one natural way of doing so is introducing some sort of coordinate system.  Let's start by labelling &quot;You are here&quot; as 0.  Going to the right, we can introduce 1, 2, 3,... (in whatever your favorite unit is).  Going to the left, we have a couple of options:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;We could perhaps label going to the left as 1, 2, 3,... -- but then we have to specify what side of &quot;here&quot; we're on: left or right.  In other words, one's position on the line would have to be specified as &quot;x units left&quot; or &quot;x units right.&quot;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;An &lt;em&gt;easier&lt;/em&gt; approach, then, might be to label the left side as -1, -2, -3....&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The point is, though, that these two approaches are equivalent: Here $-x$ means &quot;x units left.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if we want to talk about &lt;em&gt;distance&lt;/em&gt; -- that is, how &lt;em&gt;far&lt;/em&gt; we are from &quot;here,&quot; independent of our direction -- then, yes, our &lt;em&gt;distance&lt;/em&gt; will always be positive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mathematically, this can be represented by taking the absolute value of the position.  This makes sense since the absolute value of a number is often defined as &quot;distance from zero,&quot; thereby forgetting about the distinction between left and right.&lt;/p&gt;&#xA;" OwnerUserId="640" LastActivityDate="2011-02-02T04:04:08.883" CommentCount="5" />
  <row Id="19988" PostTypeId="1" AcceptedAnswerId="20084" CreationDate="2011-02-02T04:13:02.180" Score="4" ViewCount="81" Body="&lt;p&gt;Suppose $z,w \in \mathbb{C}$ s.t. $|z|,|w| &amp;lt; 1$. How would you show that $\frac{|z-w|}{|1-\bar{w}z|} \leq |z| + |w|$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I calculated that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align*}&#xA;&amp;amp;\frac{|z-w|}{|1-\bar{w}z|} \leq |z| + |w| \\&#xA;&amp;amp;\Rightarrow \frac{|z|^2-2Re(\bar{w}z)-|w|^2}{1-2Re(\bar{w}z) + |w|^2|z|^2} \leq (|z|+|w|)^2 \\&#xA;&amp;amp;\Rightarrow -2Re(\bar{w}z) \leq 2|z||w| + (|w|^2|z|^2 - 2 Re(\bar{w}z))(|z| +|w|)^2\&#xA;\end{align*}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $2Re(\bar{w}z) &amp;lt; 0$, then $2|z||w| \geq 2Re(\bar{w}z)$ and $(|w|^2|z|^2 - 2 Re(\bar{w}z)) \geq 0$ imply that the inequality is true.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I haven't been able to figure out what happens in the case that $2Re(\bar{w}z) &amp;gt;0$ though.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On a separate note, whenever I come across these types of problems, I always try to multiply everything out and see if I can get enough terms to cancel so that the inequality becomes obvious. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not quite sure if that's the smartest way to go about things, though and if there is some intuition that I should be using that I don't know about.&lt;/p&gt;&#xA;" OwnerUserId="1736" LastActivityDate="2011-02-02T16:34:11.140" Title="A bound on a mobius transform" Tags="&lt;complex-analysis&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="19989" PostTypeId="2" ParentId="19979" CreationDate="2011-02-02T04:19:25.087" Score="3" ViewCount="" Body="&lt;p&gt;With or without your edit, b does not divide a.  I suspect the question you want is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If b is the largest square divisor of c (not a) and a^2|c then a|b?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then the answer would be true.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-02-02T04:19:25.087" />
  <row Id="19990" PostTypeId="2" ParentId="19985" CreationDate="2011-02-02T04:29:46.780" Score="1" ViewCount="" Body="&lt;p&gt;Try the Division Algorithm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Division_algorithm&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Division_algorithm&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe this is overpowered for your purposes though.&lt;/p&gt;&#xA;" OwnerUserId="565" LastActivityDate="2011-02-02T04:29:46.780" />
  <row Id="19991" PostTypeId="2" ParentId="19984" CreationDate="2011-02-02T04:50:16.433" Score="5" ViewCount="" Body="&lt;p&gt;It's not true in general... consider $f(z) = (z-1)^n - 1$. Then its derivatives have zeroes only at $z = 1$, but it takes the value zero at any $z = 1 + e^{2\pi i / n}$. Since these are spaced evenly around the circle $|z - 1|= 1$, for large $n$ many of them will lie inside the unit disk.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want an example where the derivatives have zeroes only outside the ${\it closed}$ unit disk, then you can take $f(z) = (z-(1+\epsilon))^n - 1$ for small $\epsilon &amp;gt; 0$.&lt;/p&gt;&#xA;" OwnerUserId="3035" LastActivityDate="2011-02-02T04:50:16.433" />
  <row Id="19992" PostTypeId="2" ParentId="19985" CreationDate="2011-02-02T04:55:29.190" Score="7" ViewCount="" Body="&lt;p&gt;I wasn't going to say anything, but all these comments may be overly confusing. No need for Division Algorithm. If you're proving this fact it is probably leading up to the Div Alg and shouldn't be used to prove this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;No need for $\mathbb{Z}_3$. This probably hasn't been defined yet, or the eventual use of this fact you're proving will be related to that later.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;No need for contradiction (though there isn't anything wrong with how Bill Dubuque says to prove it, I'm just suggesting sticking with your gut instinct of induction).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Go back to the induction idea like Alon Amit says. If $n=3k+l$ where $l$ is 0, 1, or 2, then $n+1=3k+l+1$. You now have two cases. If $l=0$ or 1, then $(l+1)$ is 1 or 2, so $n+1=3k+(l+1)$ which is the right form. If $l=2$, then $n+1=3k+3=3(k+1)$, which is of the right form. So if you have the base case you have the positive integers by induction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $n$ be any negative number. Then $-n$ is positive, so $-n=3k+l$ where $l$ is 0,1, or 2. Thus $n=-(3k+l)=3(-k)-l=3(-k)-l+3-3=3(-k-1)+(3-l)$ and $3-l$ is 0,1, or 2.&lt;/p&gt;&#xA;" OwnerUserId="2000" LastEditorUserId="2000" LastEditorDisplayName="" LastEditDate="2011-02-02T05:30:02.580" LastActivityDate="2011-02-02T05:30:02.580" CommentCount="3" />
  <row Id="19993" PostTypeId="2" ParentId="19876" CreationDate="2011-02-02T05:11:36.737" Score="0" ViewCount="" Body="&lt;p&gt;Here's yet another method for your amusement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $y = \cos^2(t)$.  Then note that $y&amp;#39;(t) = 2\cos(t)\sin(t)$ and therefore &#xA;$$&#xD;&#xA;y&amp;#39;&amp;#39;(t) = 2\cos^2(t) - 2\sin^2(t) = 2\cos^2(t) - 2(1 - \cos^2(t)) = 4 y - 2.&#xD;&#xA;$$&#xA;The general solution to this nonhomogeneous ode is&#xA;$$&#xD;&#xA;y = C_1 \sin(2t) + C_2 \cos(2t) + 1/2&#xD;&#xA;$$&#xA;and since we know that $y$ satisfies the initial conditions $y(0) = 1$ and $y&amp;#39;(0) = 0$, we can solve for the coefficients and get $C_1 = 0$ and $C_2 = 1/2$.&lt;/p&gt;&#xA;" OwnerUserId="4229" LastActivityDate="2011-02-02T05:11:36.737" />
  <row Id="19994" PostTypeId="1" CreationDate="2011-02-02T05:19:47.340" Score="0" ViewCount="52" Body="&lt;p&gt;I have a kernel function generated by an inner product $k = x^T X$. Where $x$ is an 0th order dilated b-spline, and $X$ is a dictionary of 0th order b-splines. The kernel function plotted looks like:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/v5DRF.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have found an algorithm that generates this kernel function directly:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$h = 10;$&#xA;$z = q_i - Q_i;$&#xA;$k_i = max(h - z, 0)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $h$ is the dilation amount, $q$ is a scaled and quantized input value, and $Q$ is a dictionary of quantized inputs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I am trying to apply a form of regularisation of the form $k = x^T \times (I + \lambda D)^{-1} \times X$. Where I is the identity matrix, and $D$ is a diagonal matrix given by $D = D + diag(x)$, for each value of $x$ stored in the dictionary. For example, the following image shows how the kernel function deforms when $D = diag(x)$ where $x$ corresponds to the value 0.5.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/TDeky.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anyone have any ideas how I might be able to modify the algorithm that generates this kernel function directly to allow for this sort of deformation?&lt;/p&gt;&#xA;" OwnerUserId="1588" LastActivityDate="2011-02-02T05:19:47.340" Title="Kernel Function Regularization Geometry Help" Tags="&lt;geometry&gt;&lt;functions&gt;" CommentCount="2" />
  <row Id="19995" PostTypeId="2" ParentId="19984" CreationDate="2011-02-02T05:31:46.230" Score="2" ViewCount="" Body="&lt;p&gt;Zarrax has already given a great example.  Here's a way to see that there are polynomial examples from the fact that there are holomorphic examples.  The motivation for this is that the non-polynomial example $g(z)=e^{2\pi i z}$ came to mind when I read your question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $g$ be holomorphic in a neighborhood of the closed disk, not injective on the open disk, but with nonvanishing derivative on the closed disk.  Let $(p_n)_n$ be the sequence of partial sums of the Taylor series of $g$ centered at $0$, so $p_n\to g$ and $p_n&amp;#39;\to g&amp;#39;$ uniformly on a (smaller) neighborhood of the closed disk.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $g&amp;#39;$ is nonvanishing on the closed disk, it has a positive minimum modulus $m$ there, which by uniform convergence means that $p_n&amp;#39;$ eventually has modulus greater than $m/2$, and in particular is eventually nonvanishing on the closed disk.  Take $a\neq b$ in the open unit disk such that $g(a)=g(b)$.  Applying Hurwitz's theorem to the sequence of functions $p_n(z)-g(a)$ on small disjoint disks centered at $a$ and $b$ shows that $p_n$ eventually takes on the value $g(a)$ at more than one point in the open unit disk.  Thus, $p_n$ is eventually a counterexample.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, using WolframAlpha, it looks like $f(z)=\displaystyle{\sum_{k=0}^{25}\frac{(2\pi iz)^k}{k!}}$ gives an example.  It allegedly takes on the value $-1$ at points very close to $\pm\frac{1}{2}$, and all $24$ of the zeros of its derivative are outside the closed disk.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2011-02-02T06:56:55.577" LastActivityDate="2011-02-02T06:56:55.577" CommentCount="1" />
  <row Id="19996" PostTypeId="1" AcceptedAnswerId="19998" CreationDate="2011-02-02T05:43:11.933" Score="1" ViewCount="142" Body="&lt;p&gt;Hello, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In 1556, Tartaglia claimed that the sums&lt;br&gt;&#xA;1 + 2 + 4&lt;br&gt;&#xA;1 + 2 + 4 + 8&lt;br&gt;&#xA;1 + 2 + 4 + 8 + 16&lt;br&gt;&#xA;are alternative prime and composite. Show that his conjecture is false.     &lt;/p&gt;&#xA;&#xA;&lt;p&gt;With a simple counter example, 1 + 2 + 4 + 8 + 16 + 32 + 64 + 128 = 255, apparently it's false. However, I want to prove it in general case instead of using a specific counter example, but I got stuck :( !&#xA;I tried:&lt;br&gt;&#xA;The sum of 2^i where i = 0 -&gt; n is equal to 2^(n+1) - 1. I assumed that 2^(n+1) - 1 is prime, then we must show that 2^(n+1) - 1 + 2^(n+1) = 2^(n+2) - 1 is not composite. Or we assume 2^(n+1) is composite and we must show that 2^(n+2) - 1 is not prime. &#xA;But I have no clue how 2^(n+2) - 1 relates to its previous prime/composite. Any hint?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks,&lt;br&gt;&#xA;Chan&lt;/p&gt;&#xA;" OwnerUserId="5711" LastEditorUserId="5711" LastEditorDisplayName="" LastEditDate="2011-02-02T06:03:06.300" LastActivityDate="2011-02-02T06:07:41.707" Title="If 2^(n+1) - 1 is composite, is 2^(n+2) - 1 prime?" Tags="&lt;elementary-number-theory&gt;" AnswerCount="2" CommentCount="5" />
  <row Id="19997" PostTypeId="1" CreationDate="2011-02-02T06:01:09.267" Score="19" ViewCount="397" Body="&lt;p&gt;Here is a nice proof of the isoperimetric inequality (equality part ommited):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Isoperimetric Inequality&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $\gamma$ is any simple closed piecewise $C^1$ curve of length $l$, with it's interior having area $A$, then $4\pi A \le l^2$. Furthermore, if equality holds then $\gamma$ is a circle. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Proof.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take two parallel straight lines $L$ and $L&amp;#39;$ such that $\gamma$ is between them and move them together until they first touch the curve. See my nice picture below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/gctMU.png&quot; alt=&quot;Figure 1&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let C be a circle as in the picture. Take $x$ and $y$ axes as shown. Let $\gamma = (x,y)$ be a parametrization of $\gamma$. Pick points $\gamma (s_0)$ and $\gamma (s_1)$ on both $L$ and $L&amp;#39;$ wherever the lines touch the curve, respectively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $C$ be parametrized by $(x, \overline{y})$ where&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;\overline{y}(s) =&#xD;&#xA;\begin{cases}&#xD;&#xA; + \sqrt{r^2 - x^2 (s)}, &amp;amp; \text{if } s_0 \le s \le s_1 \\&#xD;&#xA; - \sqrt{r^2 - x^2 (s)}, &amp;amp; \text{if } s_1 \le s \le s_0 + l&#xD;&#xA;\end{cases}&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Denote the derivative of $f$ with respect to $s$ as $f_s$. Using Green's Theorem, we write:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$A + \pi r^2 = \int_{\gamma} x\,dy + \int_C -y\,dx = \int^l_0 x(s)y_s(s)\,ds - \int^l_0 \overline{y}(s)x_s(s)\,ds = $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ = \int^l_0 ( x(s)y_s(s) - \overline{y}(s)x_s(s)) \,ds \le \int^l_0 \sqrt{ (x(s)y_s(s) - \overline{y}(s)x_s(s))^2} \,ds \stackrel{*}{\le}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \stackrel{*}{\le} \int^l_0 \sqrt{ (x^2(s) + \overline{y}^2(s))} \,ds = lr$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where the starred inequality follows from the fact that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(x y_s - \overline{y} x_s)^2 = [(x, - \overline{y}) \cdot (y_s, x_s)]^2 \le (x^2 + \overline{y}^2) \cdot (y^2_s + x^2_s) = x^2 + \overline{y}^2 $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we have that $A + \pi r^2 \le lr$. Next we employ the Geometric-Arithmetic Mean Inequality to find that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sqrt{A \pi r^2} \le \dfrac{A + \pi r^2}{2} \le \dfrac{lr}{2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From which it directly follows that $4 \pi A \le l^2$, as needed. $\square$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have seen other proofs of this Theorem, for example the one using Wirtinger's Inequality. This proof was presented to me by my professor, who said this it was rather &lt;em&gt;mysterious&lt;/em&gt;, and I agree. I think this proof is rather beautiful and much simpler than the other proofs. Here are my questions: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;How?&lt;/em&gt; How does it work? I do not mean to ask how to we get from one step to another. I mean to ask what makes this work intrinsically. In particular, I am bothered by this constructed circle, and its radius. The next picture is what I have in mind:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/YO62b.png&quot; alt=&quot;Funny picture.&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For this curve $\gamma$, choosing two different pairs of lines $L$, $L&amp;#39;$ and $K$, $K&amp;#39;$ gives us two circles with different radii. Furthermore, note that it appears that the area of the smaller circle is less that the area traced out by $\gamma$, whereas it is not the case for the larger circle. I guess my question here can be rephrased as: why do the $r$'s &lt;em&gt;magically&lt;/em&gt; fall out of the equation in the last step? I am not looking for an anwser of the type &quot;because the math works out that way,&quot; rather some geometric insight/explanation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One observation I have thought about is that in case of equality $4 \pi A = l^2$, i.e. when $\gamma$ is a circle, any circle given by our &lt;em&gt;construction&lt;/em&gt; will always have equall radius, in fact the radius of the circle $\gamma$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From that I was led to this question: Suppose we take $n$ pairs of parallel lines (where two distinct pairs of pairs are not mutually parallel), and construct circles for each. Now, as $n \rightarrow \infty$, what will happen to the average radius of these circles? What will a circle with this radius represent? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt;&#xA;I have found an example where this last question turns out to be rather uninteresting. However, what will happen if we assume the curve to be &lt;strong&gt;convex&lt;/strong&gt; as well?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I do not know how to explore this last question with my knowledge whatsoever.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, I want to ask if anyone knows how this proof came to be; what is the hidden motivation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you.&lt;/p&gt;&#xA;" OwnerUserId="5105" LastEditorUserId="5105" LastEditorDisplayName="" LastEditDate="2011-02-05T21:37:17.513" LastActivityDate="2011-02-06T17:32:08.143" Title="A proof of the Isoperimetric Inequality - how does it work?" Tags="&lt;differential-geometry&gt;&lt;proof-strategy&gt;&lt;geometric-construction&gt;" AnswerCount="1" CommentCount="16" FavoriteCount="4" />
  <row Id="19998" PostTypeId="2" ParentId="19996" CreationDate="2011-02-02T06:04:08.773" Score="5" ViewCount="" Body="&lt;p&gt;It is known that $2^n-1$ can only be prime if $n$ is prime.  This is because if $jk=n$, $2^n-1=\sum_{i=0}^{n-1}2^i=\sum_{i=0}^{j-1} 2^i \sum_{i=0}^{k-1} 2^{ij}$ So they will only continue to alternate at twin primes.  In particular, $2^{6k+2}-1, 2^{6k+3}-1$ and $2^{6k+4}-1$ will all be composite&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-02-02T06:04:08.773" CommentCount="1" />
  <row Id="19999" PostTypeId="2" ParentId="19996" CreationDate="2011-02-02T06:07:41.707" Score="3" ViewCount="" Body="&lt;p&gt;There is simply no such correlation in general. If that were the case, Mersenne primes would not be as interesting to study as they are. You can not prove that $2^n-1$ composite $\Rightarrow 2^{n+1} - 1$ prime, because it isn't true. As Ross mentioned, $2^n-1$ can only be prime when $n$ is prime, so for any prime $p &amp;gt; 2$, if $2^p-1$ is prime, then $2^{p+1}-1$ must be composite, as $2 \mid (p+1)$.&lt;/p&gt;&#xA;" OwnerUserId="1016" LastActivityDate="2011-02-02T06:07:41.707" />
  <row Id="20000" PostTypeId="1" AcceptedAnswerId="20003" CreationDate="2011-02-02T06:15:04.787" Score="3" ViewCount="104" Body="&lt;p&gt;Prove:&#xA;$[ (A \cap B) \cup C ]\cap (A^c\cup C) =  C$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My attempt:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(A\cup C)\cap(B\cup C)\cap(A^c\cup C) = C$ I used the distributive property,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but this is where I'm lost at. I think that I need to get A and $A^c$ together. So I should use the commutative property.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(A\cup C)\cap(A^c\cup C)\cap(B\cup C) = C$  Should I distribute $(A\cup C)\cap(A^c\cup C)$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My original post of this question was incorrect. Sorry about that.&lt;/p&gt;&#xA;" OwnerUserId="1040" LastEditorUserId="1040" LastEditorDisplayName="" LastEditDate="2011-02-02T15:12:25.287" LastActivityDate="2011-02-02T15:37:54.570" Title="algebra of Sets" Tags="&lt;elementary-set-theory&gt;" AnswerCount="2" CommentCount="5" />
  <row Id="20001" PostTypeId="1" CreationDate="2011-02-02T06:17:04.800" Score="7" ViewCount="268" Body="&lt;p&gt;I wrote a program for myself in Mathematica to generate the answer for the first 300, which was really easy, but I can't find a pattern. The results are &lt;a href=&quot;http://dl.dropbox.com/u/1277378/fact.txt&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;. This is a problem in Underwood Dudley's Elementary Number Theory, but for the LIFE of me I can't figure it out! My initial thought was this: Let m be the smallest prime such that &lt;code&gt;k&amp;lt;m&lt;/code&gt;. Then the answer is &lt;code&gt;m-2&lt;/code&gt;. This is quite clearly not the answer based on the link I provided. My logic was that any number of the form &lt;code&gt;k!+n&lt;/code&gt; will be composite when &lt;code&gt;1&amp;lt;n&amp;lt;m&lt;/code&gt;. That is true. However, it is not the end of the story. Any ideas?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: For anyone curious, this is problem 2 part (b) in section 23.2 of the Second Edition of Underwood Dudley's &lt;em&gt;Elementary Number Theory&lt;/em&gt;. Part (a) asks what the smallest integer &lt;code&gt;n&lt;/code&gt; is such that &lt;code&gt;n+1, n+2, n+3,&lt;/code&gt; and &lt;code&gt;n+4&lt;/code&gt; are all composite.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second edit: Milcak's comment made me realize that, given some arbitrary &lt;code&gt;n&lt;/code&gt;, we can write &lt;code&gt;k!+n=k!+i+j&lt;/code&gt; so long as &lt;code&gt;i+j=n&lt;/code&gt;. I'm thinking about this now...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Third edit: Here is another problem with the concept for me. I feel like the primes less than or equal to k should help us predict the divisbility of &lt;code&gt;k!+i&lt;/code&gt;, and indeed they do, not when &lt;code&gt;i&lt;/code&gt; is a prime greater than &lt;code&gt;k&lt;/code&gt;. For example, &lt;code&gt;11!+13=199*200587&lt;/code&gt;. That sort of behavior seems unpredictable to me.&lt;/p&gt;&#xA;" OwnerUserId="6518" LastEditorUserId="6518" LastEditorDisplayName="" LastEditDate="2011-02-02T07:13:01.767" LastActivityDate="2011-02-02T14:53:37.677" Title="How many consecutive composite integers follow k!+1?" Tags="&lt;number-theory&gt;&lt;prime-numbers&gt;&lt;factorial&gt;" AnswerCount="4" CommentCount="1" />
  <row Id="20002" PostTypeId="1" CreationDate="2011-02-02T06:17:58.820" Score="0" ViewCount="121" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Possible Duplicate:&lt;/strong&gt;&lt;br&gt;&#xA;  &lt;a href=&quot;http://math.stackexchange.com/questions/1774/undergraduate-high-school-olympiad-level-introductory-number-theory-books-for-sel&quot;&gt;Undergraduate/High-School-Olympiad Level Introductory Number Theory Books For Self-Learning&lt;/a&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;p&gt;Hello everyone,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any recommended books for these two subjects? I'm looking for books at undergraduate level, not too advance, since I'm only using Math to solve Computer Science problems, so elementary books would suffice my need.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks,&lt;br&gt;&#xA;Chan &lt;/p&gt;&#xA;" OwnerUserId="5711" LastActivityDate="2011-02-02T06:17:58.820" ClosedDate="2011-02-02T11:38:08.500" Title="What are good books for Number Theory/Geometry at undergraduate level?" Tags="&lt;elementary-number-theory&gt;&lt;elementary-geometry&gt;" CommentCount="4" />
  <row Id="20003" PostTypeId="2" ParentId="20000" CreationDate="2011-02-02T06:23:55.447" Score="5" ViewCount="" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;What is $(A\cup C)\cap(A^c\cup C)$ ? Try finding the anwser this way: if $a \in (A\cup C)$ and $a \in (A^c\cup C)$, where must $a$ be?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;What must $C \cap (B \cup C)$ be?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Note that the operation of &lt;em&gt;intersection&lt;/em&gt; is associative, i.e. $(A \cap B) \cap C = A \cap (B \cap C)$.&lt;/p&gt;&#xA;" OwnerUserId="5105" LastActivityDate="2011-02-02T06:23:55.447" CommentCount="2" />
  <row Id="20005" PostTypeId="1" CreationDate="2011-02-02T06:45:28.197" Score="1" ViewCount="50" Body="&lt;p&gt;I recently learned about summation methods when dealing with divergent series to give them a finite value. An example of this isusing Cesàro summation on Grandi's series to get 1/2. However every method I know of is unable to sum the harmonic series. Are there any summation methods that work on the harmonic series or is it provably impossible to sum this series? &lt;/p&gt;&#xA;" OwnerUserId="6519" LastActivityDate="2011-02-02T08:05:33.790" Title="Is it possible to use regularization methods on the Harmonic Series?" Tags="&lt;sequences-and-series&gt;" AnswerCount="0" CommentCount="3" />
  <row Id="20007" PostTypeId="1" AcceptedAnswerId="20101" CreationDate="2011-02-02T07:00:07.150" Score="0" ViewCount="93" Body="&lt;p&gt;Implicit diff on $\frac{2x}{y} - \frac{3y}{x} = 8$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm doing this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ \frac{2x^2 - 3y^2}{xy} = 8  $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ 2x^2 - 3y^2 = 8xy  $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ 2x^2 - 3y^2 - 8xy = 0 $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Derive&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ 4x dx - 6y dy - 8( ydx + xdy ) = 0 $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ dx( 4x - 8y ) - dy ( 6y + 8x ) = 0 $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ \frac{dy}{dx} = \frac{4x - 8y}{ 8x + 6y } $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if you use the quotient rule from the beginning, you get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ \frac{dy}{dx} = \frac{y}{x} $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can someone point out to me where I've gone wrong?&lt;/p&gt;&#xA;" OwnerUserId="1069" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-02T19:12:12.300" LastActivityDate="2011-02-02T20:01:24.053" Title="Implicit diff on $\frac{2x}{y} - \frac{3y}{x} = 8$" Tags="&lt;calculus&gt;" AnswerCount="4" CommentCount="0" />
  <row Id="20008" PostTypeId="1" AcceptedAnswerId="20011" CreationDate="2011-02-02T07:06:03.147" Score="2" ViewCount="126" Body="&lt;p&gt;A laser is suspended 1 metre above an infinite plane. The laser swivels on its aperture, which remains at point A. The aperture is directly above point B. The laser shines a infinitely small dot onto point C. We'll call the angle BAC θ, and the distance BC d. Therefore, d = tan θ.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As θ approaches 90°, d approaches infinite. In other words, the dot starts from B, and keeps moving along this plane. Since it moves an infinite distance, it is moving forever away from B. Since it is moving forever on this plane, θ never reaches 90°, and the laser dot never comes off the plane (after all, why would the dot continue going along this plane and suddenly come off it?). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the above 'proof', the laser could never go past 90°. Yet reality contradicts this: anyone can grab a laser pointer and lift it past 90°; physically a laser wouldn't have a purely theoretical force stop it from moving past 90°.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, what is wrong with the above proof? If there is a fallacy, please make close reference to exactly &lt;em&gt;when&lt;/em&gt; and &lt;em&gt;how&lt;/em&gt; the dot parts with the plane and ceases to exist.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: Assume the light emitted from the laser instantaneously arrives at its endpoint on the plane.&lt;/p&gt;&#xA;" OwnerUserId="6520" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2011-02-02T07:29:20.430" LastActivityDate="2011-02-02T07:49:49.027" Title="Laser paradox - A dot restricted to moving across an infinite plane" Tags="&lt;physics&gt;&lt;paradox&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="20009" PostTypeId="2" ParentId="20007" CreationDate="2011-02-02T07:06:24.070" Score="3" ViewCount="" Body="&lt;p&gt;Did you try checking the possibility that $\displaystyle \frac{4x-8y}{8x+6y} = \frac{y}{x}$?&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2011-02-02T07:06:24.070" CommentCount="3" />
  <row Id="20010" PostTypeId="2" ParentId="18405" CreationDate="2011-02-02T07:14:58.417" Score="3" ViewCount="" Body="&lt;p&gt;There is a direct, and rather automatic, way to compute the probability to hit A=HTHTH first rather than B=HTHH. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both motives begin by HTH hence one can wait until HTH first appears. Then, either (1) the next letter is H, or (2) the two next letters are TH, or (3) the two next letters are TT. If (1) happens, B won. If (2) happens, A won. If (3) happens, one has to wait for more letters to know who won. The important fact in case (3) is that, since the last letters are TT, A or B &lt;em&gt;must be entirely produced again&lt;/em&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, $p_B=p_1+p_3p_B$ and $p_A=p_2+p_3p_A$, where $p_i$ for $i=$ 1, 2 and 3, is a shorthand for the conditional probability that ($i$) happens starting from the word HTH. Since $p_1=p$, $p_2=qp$ and $p_3=q^2$, one gets $p_B=p_1/(1-p_3)=p/(1-q^2)$, hence&#xA;$$&#xD;&#xA;p_B=1/(1+q),\quad p_A=q/(1+q).&#xD;&#xA;$$&#xA;Similarly, a standard, and rather automatic, way to compute the mean number of tosses before this happens is to consider a Markov chain on the state space made of the &lt;em&gt;prefixes&lt;/em&gt; of the words one wishes to complete. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here, the states are 0 (for the empty prefix), 1=H, 2=HT, 3=HTH, B=HTHH, 4=HTHT and A=HTHTH. The transitions are from 0 to 1 and 0, from 1 to 2 and 1, from 2 to 3 and 0, from 3 to B and 4 and from 4 to A and 0. The transitions from B and from A are irrelevant. The next step is to compute $n_s$ the number of tosses needed to produce A or B starting from any state $s$ amongst 0, 1, 2, 3 and 4, knowing that one is in fact only interested in $n_0$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The $n_s$ are solutions of a Cramér system which reflects the structure of the underlying Markov chain: &#xA;$$&#xD;&#xA;n_0=1+pn_1+qn_0,\quad n_1=1+pn_1+qn_2,\quad n_2=1+pn_3+qn_0,&#xD;&#xA;$$ &#xA;$$&#xD;&#xA;n_3=1+qn_4,\quad n_4=1+qn_0.&#xD;&#xA;$$&#xA;Solving this system of equations &lt;em&gt;backwards&lt;/em&gt;, that is, going from the last equation back to the first one, yields $n_3$ in terms of $n_0$, then $n_2$ in terms of $n_0$, then $n_1$ in terms of $n_0$, and finally an equation for $n_0$ alone, which yields Mike's formula for $n_0$, namely:&#xA;$$&#xD;&#xA;n_0=\frac{1+pq+p^2q+p^2q^2}{p^3q(1+q)}.&#xD;&#xA;$$&#xA;An accessible reference for these techniques (in the context of genomic sequence analysis) is the book &lt;em&gt;DNA, Words and Models&lt;/em&gt; by Robin, Rodolphe and Schbath, at Cambridge UP.&lt;/p&gt;&#xA;" OwnerUserId="6179" LastEditorUserId="6179" LastEditorDisplayName="" LastEditDate="2011-02-02T13:52:48.973" LastActivityDate="2011-02-02T13:52:48.973" CommentCount="6" />
  <row Id="20011" PostTypeId="2" ParentId="20008" CreationDate="2011-02-02T07:18:27.320" Score="7" ViewCount="" Body="&lt;p&gt;Of course, this is not a math problem because it has to do with physical objects, which math can only model, and only with varying degrees of appropriateness. In particular, there is obviously no such thing as an infinite plane in reality. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer is: The dot is off the plane when $\theta=90^\circ$. The dot is on the plane when $\theta&amp;lt;90^\circ$. One can choose real numbers that are arbitrarily close to $90^\circ$, without being equal; but this no more prevents the motion of the laser than the fact that there are real numbers arbitrarily close to 1 prevents the motion of a point moving with constant speed along the interval $[0,1]$ from reaching 1 in exactly the amount of time you would expect. This is just &lt;a href=&quot;http://en.wikipedia.org/wiki/Zeno%27s_paradoxes#The_dichotomy_paradox&quot; rel=&quot;nofollow&quot;&gt;Zeno's paradox&lt;/a&gt;, only involving an angle, instead of a distance.&lt;/p&gt;&#xA;" OwnerUserId="264" LastEditorUserId="264" LastEditorDisplayName="" LastEditDate="2011-02-02T07:49:49.027" LastActivityDate="2011-02-02T07:49:49.027" />
  <row Id="20012" PostTypeId="2" ParentId="19985" CreationDate="2011-02-02T07:38:18.723" Score="2" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\rm\ \ 3\:\mathbb N\cup\:(3\:\mathbb N+1)\cup (3\:\mathbb N+2)\ =\ \mathbb N\ $ since it contains $\:0\:$ and it is closed under $\rm\:n\to n+1\:$.&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2011-02-02T07:38:18.723" />
  <row Id="20013" PostTypeId="1" AcceptedAnswerId="20145" CreationDate="2011-02-02T07:50:36.467" Score="3" ViewCount="99" Body="&lt;p&gt;If we take $S^1$ acting on $S^2$ by rotation, then the height function $h: S^2\to R$ is an example of an invariant map (or an equivariant map where the action on $R$ is the trivial one).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm looking for an easy example of an equivariant map which is as simple as this one (and which can be visualized).&lt;/p&gt;&#xA;" OwnerUserId="1007" LastActivityDate="2011-02-04T17:40:59.943" Title="What's a very simple example of an equivariant map?" Tags="&lt;examples&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="20014" PostTypeId="1" CreationDate="2011-02-02T07:54:37.923" Score="3" ViewCount="75" Body="&lt;p&gt;We define the following subgroups of the automorphism group of $k[x_1, \ldots, x_n]$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;E(k,n) is the subgroup generated by elementary $k$-automorphisms of $k[x_1, \cdots, x_n]$ of the form $(x_1, x_2, \cdots, x_{i-1}, x_i + a, x_{i+1}, \cdots, x_n)$ where $a \in k[x_1,x_2, \cdots, x_{i-1},x_{i+1}, \cdots, x_n]$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Aff(k,n) is the subgroup of linear transformations&lt;/p&gt;&#xA;&#xA;&lt;p&gt;J(k,n) is the subgroup generated by elementary $k$-automorphisms of $k[x_1, \cdots, x_n]$ of the form $(x_1, x_2 + f_2, x_3 + f_3, \cdots, x_n+f_n)$ where $f_i \in k[x_1, \cdots, x_{i-1}]$ for $2 \leq i \leq n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've been having trouble figuring out a way to write an element of E(k,n) as a composition of linear transformations and elements of J(k,n). I know it involves some permutation of the $x_i$s.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any help would be appreciated. Thanks.&lt;/p&gt;&#xA;" OwnerUserId="6521" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-02T16:06:09.260" LastActivityDate="2011-03-04T16:39:26.130" Title="elementary k-automorphism of $k[x_1, \ldots, x_n]$" Tags="&lt;algebraic-geometry&gt;&lt;commutative-algebra&gt;" AnswerCount="1" />
  <row Id="20015" PostTypeId="1" AcceptedAnswerId="20017" CreationDate="2011-02-02T07:57:41.467" Score="6" ViewCount="144" Body="&lt;p&gt;I've stumbled upon the definition of exact sequence, particularly on &lt;a href=&quot;http://en.wikipedia.org/wiki/Exact_sequence&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;, and noted the use of $\hookrightarrow$ to denote a monomorphism and $\twoheadrightarrow$ for epimorphisms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was wondering whether this notation was widely used, or if it is common to define a morphism in the general form and indicate its characteristics explicitly (e.g. &quot;an epimorphism $f \colon X \to Y$&quot;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, if epimorphisms and monomorphisms have their own special arrows, are isomorphisms notated by a special symbol as well, maybe a juxtaposition of $\hookrightarrow$ and $\twoheadrightarrow$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, are there other kinds of morphisms (or more generally, relations) that are usually notated by different arrows depending on the type of morphism, particularly in the context of category theory?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="6427" LastActivityDate="2011-02-02T08:12:20.913" Title="Special arrows for notation of morphisms" Tags="&lt;notation&gt;&lt;morphism&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="20016" PostTypeId="1" AcceptedAnswerId="20048" CreationDate="2011-02-02T08:03:31.010" Score="12" ViewCount="251" Body="&lt;p&gt;I'm well aware of the standard proof based on cardinality arithmetic to show that these two sets have the same cardinality and the question of defining a bijection between the two sets came up. I worked on it a little while and was wondering if there were any gaps, or a simpler method of coming up with such a bijection. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The main idea I had was to take a family of sequences indexed by the real numbers. So that each one of these sequences corresponds to a unique real number. Then I can map each one of these sequences to another number, by applying the binary function from $\mathbb R$ to $\{0,1\}$ and mapping the infinite binary sequence to a real number, use one of the standard bijections.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First let $E$ be a family of sequences indexed by the real numbers. Such that $E_{xi}\neq E_{yj}$ for all $x,y \in \mathbb R$ and $i,j \in \mathbb N$ also with the property that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\mathbb R= \bigcup_{x \in \mathbb R, i \in \mathbb N} E_{xi}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What sort of theorems would I need to prove such a family existed, is it enough to note that the reals can be represented as a uncountable union of countably infinite sets?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also let $P: 2^{\mathbb N} \rightarrow \mathbb R$ be you favorite bijection.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now consider an element of $2^{\mathbb R}$ as a function $f: \mathbb R \rightarrow \{0,1\}$ and $g \in \mathbb{ R^R}$ in a similar manner. Then our bijection between the two sets will be the function $\Phi$ which maps $f$ to a function $g$, such that $g(x)=P(F(E_x))$ where $F$ applies $f$ to each element of the sequence $E_x$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm fairly sure this is an onto and one-to-one function. The one-to-one property should follow because the sequences are disjoint, cover $\mathbb R$ and $P$ is a bijection. The onto argument is a little hazier, but it should be possible to recreate the function $f$, by first inverting each element with $P$ and then defining $f$ based on these sequences. &lt;/p&gt;&#xA;" OwnerUserId="1930" LastActivityDate="2011-02-02T16:30:00.193" Title="Bijection between $2^{\mathbb R}$ and $\mathbb{ R ^ R}$" Tags="&lt;set-theory&gt;&lt;elementary-set-theory&gt;" AnswerCount="1" />
  <row Id="20017" PostTypeId="2" ParentId="20015" CreationDate="2011-02-02T08:05:32.293" Score="4" ViewCount="" Body="&lt;p&gt;Some people use those notations, some don't. Using $\hookrightarrow$ to mean that the map is a mono is not a great idea, in my opinion, and I much prefer $\rightarrowtail$ and use the former only to denote inclusions. Even when using that notation, I would say things like «consider the epimorphism $f:X\twoheadrightarrow Y$».&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In some contexts (for example, when dealing with exact categories) one uses $\rightarrowtail$ and $\twoheadrightarrow$ to denote that the map is not only a mono or an epi, but that it has certain special properties (for example, that it is a split mono, a cofibration, or what not)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Denoting isomorphisms by mixing $\twoheadrightarrow$ and $\rightarrowtail$ is something I don't recall seeing. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You will find that there are no rules on notation, and that everyone uses pretty much whatever they like---the only important thing is that when you use what &lt;em&gt;you&lt;/em&gt; like you make it clear to the reader.&lt;/p&gt;&#xA;" OwnerUserId="274" LastEditorUserId="274" LastEditorDisplayName="" LastEditDate="2011-02-02T08:12:20.913" LastActivityDate="2011-02-02T08:12:20.913" />
  <row Id="20020" PostTypeId="2" ParentId="20015" CreationDate="2011-02-02T08:12:06.540" Score="4" ViewCount="" Body="&lt;p&gt;I agree with everything Mariano says, especially about $\hookrightarrow$ and $\rightarrowtail$ only being for inclusions. I wanted to add that I usually denote an isomorphism by $\xrightarrow{\sim}$, which kind of evokes $\simeq$ and $\cong$. Another type of &quot;morphism&quot; that often gets a special arrow is a natural transformation, namely $\alpha:F\Rightarrow G$ for a natural transformation between the functors $F,G:\mathcal{A}\rightarrow\mathcal{B}$ (see &lt;a href=&quot;http://ncatlab.org/nlab/show/natural+transformation&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;).&lt;/p&gt;&#xA;" OwnerUserId="264" LastActivityDate="2011-02-02T08:12:06.540" />
  <row Id="20021" PostTypeId="2" ParentId="20013" CreationDate="2011-02-02T08:19:20.030" Score="3" ViewCount="" Body="&lt;p&gt;It is very easy to come up with those. A few examples:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;The function which maps each $k$-tuple of linearly independent vectors in $\mathbb C^n$ to the $k$-dimensional subspace they span is equivariant with respect to the obvious actions of $\mathrm{GL}(\mathbb C^n)$ on its domain and codomain;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The function mapping each triangle in $\mathbb R^2$ to its area, a real number, is equivariant with respect to the action of $\mathrm{GL}(\mathbb R^2)$ acting in the obvious way on the domain, and by multiplication by the determinant on the codomain;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;the function which takes a finite word in the letters $a$, $b$ and returns the word with the same letters but with all $a$s put before all the $b$s, is equivariant with respect to the action of $\mathbb Z_2$ on the domain by interchaning $a$s and $b$s in each word, and on the codomain, by interchanging $a$s and $b$s and reflecting the word backwards.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;the function which maps vertices to the diagonal to which it belongs is equivariant for the group of symmetries of a cube acting in the obvious way on the domain and codomain of the function.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="274" LastEditorUserId="274" LastEditorDisplayName="" LastEditDate="2011-02-02T22:01:54.223" LastActivityDate="2011-02-02T22:01:54.223" CommentCount="4" />
  <row Id="20022" PostTypeId="2" ParentId="19960" CreationDate="2011-02-02T09:00:26.543" Score="1" ViewCount="" Body="&lt;p&gt;Other ways to go about it:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Topological groups $G$ are homogeneous (for every $x$ and $y$ in $G$ we have a homeomorphism of $G$ that maps $x$ to $y$) and homogeneous $T_0$ spaces are $T_1$ (easy exercise).  Suppose there is a net $(x_i), i \in I$ that converges to 2 points $x, y$. Then the net $((x_i)^{-1} * x_i),  i \in I$ is constantly $e$ but also converges (using continuity of the group operations) to $x * y^{-1}$ which must equal $e$ as $\{e\}$ is closed in a $T_1$ space.&#xA;Hence $x = y$ and $G$ is Hausdorff (unique net limits).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact we can get Tychonoff (completely regular, $T_{3\frac{1}{2}}$) and the most general way to go about that is to use that the topology is induced by a uniformity. All uniform spaces that are $T_0$ are Tychonoff.&lt;/p&gt;&#xA;" OwnerUserId="4280" LastActivityDate="2011-02-02T09:00:26.543" CommentCount="2" />
  <row Id="20023" PostTypeId="1" AcceptedAnswerId="20076" CreationDate="2011-02-02T09:05:49.723" Score="3" ViewCount="145" Body="&lt;p&gt;Suppose we're given a Dihedral group $D_{30}$.&lt;br&gt;&#xA;a) Find a cyclic subgroup H of order 10 in $D_{30}$. List all generators of H.&lt;br&gt;&#xA;b) Let k and n be an integer such that k &gt;= 3 and k divides n. Prove that $D_n$ contains exactly one cyclic subgroup of order k.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My attempt:&lt;br&gt;&#xA;a) in $D_{30}$ we know that |r| = 30. So we can find a cyclic subgroup H generated by r such that it is of order 10. take &amp;lt; $r^{30/10}$ &gt; = &amp;lt; $r^3$ &gt;. Then &amp;lt; $r^3$ &gt; contains the identity element e and powers of $r^3$, to $r^{27}$. Thus the generator of H is then $r^3$. Would this be okay?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;b) Since k divides n, we can write n as n = kp  for some p. Then we see that gcd(n,k) = k and thus:&lt;br&gt;&#xA;|&amp;lt; r &gt;| =  |r| =  n.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then by Fundamental theorem of cyclic groups I can say that the group  has exactly one subgroup of order k, ie:  =  since n =  kp. And we are done.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is my first course in Group Theory so I'm rather shaky and insecure about my proofs. Your comments and help would be greatly appreciated.&lt;/p&gt;&#xA;" OwnerUserId="6523" LastActivityDate="2011-02-02T16:15:05.917" Title="Group Theory: Cyclic groups and Subgroup of Dihedrals, is my proof okay?" Tags="&lt;group-theory&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" />
  <row Id="20024" PostTypeId="2" ParentId="20014" CreationDate="2011-02-02T09:09:55.927" Score="1" ViewCount="" Body="&lt;p&gt;You have to use that E(k,n) contains the automorphisms that permute the variables x1...xn.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you pick a basic element $f_{i,a} = (x1,\dots,x_{i-1},x_i+a,x_{i+1},\dots,x_n) \in E(k,n)$,&#xA;then conjugate with the automorphism $g_i \in Aff(k,n)$ that permutes $x_i$ with $x_n$, you get $g_i f_{i,a} g_i = f_{n,ag_i} \in J(k,n)$ :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{array}{rcl}g_i f_{i,a} g_i (x_1 \dots x_i \dots x_n) &amp;amp; =&#xD;&#xA;  &amp;amp; g_i f_{i,a} (x_1 \dots x_n \dots x_i) = \\&#xD;&#xA; &amp;amp; = &amp;amp; g_i (x_1 \dots x_n + a(x_1 \dots \text{not} x_n \dots x_i) \dots x_i) \\&#xD;&#xA; &amp;amp; = &amp;amp; (x_1 \dots x_i \dots x_n + a(x_1 \dots \text{not} x_n \dots x_i)) \\&#xD;&#xA; &amp;amp; = &amp;amp; (x_1 \dots x_{n-1}, x_n + b(x_1 \dots x_{n-1}))\end{array}$$&lt;/p&gt;&#xA;" OwnerUserId="6270" LastActivityDate="2011-02-02T09:09:55.927" />
  <row Id="20025" PostTypeId="1" AcceptedAnswerId="20029" CreationDate="2011-02-02T09:13:24.190" Score="2" ViewCount="95" Body="&lt;p&gt;Let D be a p.i.d.,E a domain containing D as a subring.Is it true that if d is a g.c.d. of a and b in D,then d is also a g.c.d. of a and b in E?&lt;/p&gt;&#xA;" OwnerUserId="6526" LastActivityDate="2011-02-02T21:57:12.423" Title="A Question about Principal Ideal Domains and Euclidean Domains" Tags="&lt;abstract-algebra&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="20026" PostTypeId="2" ParentId="20001" CreationDate="2011-02-02T09:20:34.480" Score="6" ViewCount="" Body="&lt;p&gt;Let us write $N(k)$ for the number of consecutive composite numbers following $k! + 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $m$ is the smallest prime greater than $k$, then for all $n$ with $1 &amp;lt; n &amp;lt; m$, $n$ must be divisible by some prime $p$ with $p \leq k$ (for otherwise $n$ would be a nonempty product of primes each of which is at least $m$, so $n \geq m$).  Then $p$ &#xA;divides both $k!$ and $n$ so $k! + n$ is divisible by $p$ and has size at least $2p$, so is composite.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All this is to say that your argument is correct as far as it goes: what it establishes is the &lt;strong&gt;inequality&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$N(k) \geq m -2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(In particular, since $m \geq k+1$, it follows that $N(k) \geq k-1$, a weaker but more commonly made observation.)  Note that if $k = 2$, then the number of consecutive composites following $2! + 1 = 3$ is $1$: $4$ is composite, but $5$ is prime.  Here $m = 3$ and $m-2 = 1$.  Thus, your inequality is sharp when $k = 2$, i.e., it is an equality.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, as your computations show, it is usually the case that $N(k)$ is strictly larger than $m-2$, often much larger.  This is not surprising: the above argument &lt;em&gt;guarantees&lt;/em&gt; that the first $m-2$ numbers after $k! + 1$ will all be composite.  After that it doesn't tell us anything.  But most numbers are not prime: in a sense made precise by the Prime Number Theorem, if you choose a number randomly from the interval $[1,N]$, the chance it is prime is $\frac{1}{\log N}$, so if you choose a sufficiently large number at random, then the chance that it is prime is very close to zero, so we certainly expect several more values after $k! + (m-2)$ to be composite for most values of $k$.  However, we are far from being able to predict exactly where the primes and composites will fall.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;All this is to say that at least from an elementary perspective, I would be surprised if you were expected to say anything more than $N(k) \geq m -2$.  (You didn't directly quote the statement of the problem, but whatever it is, I am guessing that this inequality, or perhaps even $N(k) \geq k-1$, is the intended answer.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One could of course try to do better: for instance, surely there are infinitely many values of $k$ for which $N(k) &amp;gt; m-2$, and with sufficient knowledge and technique in analytic number theory one could probably prove that.  (I'd be interested to see people try!)  But I think it is exceedingly unlikely that there will be any kind of simple formula or expression for $N(k)$ in general, so you shouldn't be disappointed not to have found one!&lt;/p&gt;&#xA;" OwnerUserId="299" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2011-02-02T14:53:37.677" LastActivityDate="2011-02-02T14:53:37.677" CommentCount="2" />
  <row Id="20027" PostTypeId="1" AcceptedAnswerId="20032" CreationDate="2011-02-02T09:22:34.853" Score="2" ViewCount="73" Body="&lt;p&gt;(Inspired by &lt;a href=&quot;http://photo.stackexchange.com/q/8109/378&quot;&gt;this question&lt;/a&gt; on the photography site)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Say you have N copies of the same signal, each with a layer of noise on top. You average these copies together in an attempt to reduce the effect of noise. How much noise will the average have?&lt;/p&gt;&#xA;" OwnerUserId="6524" LastActivityDate="2011-02-02T15:29:08.403" Title="How much noise will the average of N noisy signals have?" Tags="&lt;statistics&gt;&lt;signal-processing&gt;&lt;noise&gt;" AnswerCount="4" FavoriteCount="1" />
  <row Id="20029" PostTypeId="2" ParentId="20025" CreationDate="2011-02-02T09:34:27.703" Score="1" ViewCount="" Body="&lt;p&gt;In any domain $D$, for $a,b \in D \setminus \{0\}$, if the ideal $\langle a,b \rangle_D = \{xa + yb \ | \ x,y \in D\}$ of $D$ is principal, then any generator $d$ of the ideal is a gcd of $a$ and $b$.  (Note that in general gcd's are unique precisely up to units, i.e., the corresponding principal ideal is unique.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if $D$ is a PID, then sure, there exists $d \in D$ such that $\langle a,b \rangle_D = dD$.  Now push forward these ideals to any extension domain $E$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\langle a,b \rangle_E = \langle a,b \rangle_D E = (d D) E = dE$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus the ideal of $E$ generated by $a$ and $b$ is still principal and still generated by the same element $d$ (now well-determined up to a unit &lt;em&gt;of $E$&lt;/em&gt;; note that the unit group of $E$ could be larger than the unit group of $D$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in summary: &lt;strong&gt;yes&lt;/strong&gt;.&lt;/p&gt;&#xA;" OwnerUserId="299" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2011-02-02T09:41:40.703" LastActivityDate="2011-02-02T09:41:40.703" CommentCount="1" />
  <row Id="20030" PostTypeId="2" ParentId="20001" CreationDate="2011-02-02T09:54:48.013" Score="0" ViewCount="" Body="&lt;p&gt;Let me just point out that also $k!-2$, $k!-3$, ..., $k!-k$ are all composite.&lt;/p&gt;&#xA;" OwnerUserId="688" LastActivityDate="2011-02-02T09:54:48.013" />
  <row Id="20031" PostTypeId="2" ParentId="20027" CreationDate="2011-02-02T10:03:23.917" Score="0" ViewCount="" Body="&lt;p&gt;Statistics states that if the noise is normally distributed, it is reduced by $\sqrt {N}$ times if summed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, noise patches being summed must be equally distributed (must have same distribution parameters) and independent.&lt;/p&gt;&#xA;" OwnerUserId="3806" LastEditorUserId="3806" LastEditorDisplayName="" LastEditDate="2011-02-02T10:11:12.617" LastActivityDate="2011-02-02T10:11:12.617" />
  <row Id="20032" PostTypeId="2" ParentId="20027" CreationDate="2011-02-02T10:03:37.217" Score="6" ViewCount="" Body="&lt;p&gt;Assuming the noise &quot;components&quot; of the $N$ copies are taken independently from the same normal distribution, then noise in the average will also be normally distributed with the same mean and with standard deviation equal to $1/\sqrt{N}$ times the standard deviation of the original noise distribution.  See &lt;a href=&quot;http://en.wikipedia.org/wiki/Sum_of_normally_distributed_random_variables&quot; rel=&quot;nofollow&quot;&gt; here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we assume the mean of the noise distribution is zero, then this amounts to saying that the averaged noise is reduced according to the factor $1/\sqrt{N}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This can be generalized in various ways, either by assuming different normal distributions of the noise in the copies or by assuming the noise in the copies is identically distributed and independent but not necessarily a normal distribution.  An illustration of the latter is the so-called &lt;a href=&quot;http://en.wikipedia.org/wiki/Random_walk&quot; rel=&quot;nofollow&quot;&gt; &quot;random walk&quot; phenomenon&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="3111" LastActivityDate="2011-02-02T10:03:37.217" />
  <row Id="20033" PostTypeId="2" ParentId="20001" CreationDate="2011-02-02T10:06:48.790" Score="0" ViewCount="" Body="&lt;p&gt;$k! + n$ is composite for all $n$ until $n$ is a prime larger than $k$. Otherwise, $n$ must be divisible by a prime smaller or equal to k.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is no general formula for the distance to a prime, so we can only give an inequality or look at the average maximum $n$.&lt;/p&gt;&#xA;" OwnerUserId="632" LastActivityDate="2011-02-02T10:06:48.790" CommentCount="0" />
  <row Id="20034" PostTypeId="2" ParentId="19988" CreationDate="2011-02-02T10:36:25.870" Score="3" ViewCount="" Body="&lt;p&gt;A hint: Think in terms of hyperbolic distance. If $d(z,w)$ is the hyperbolic distance between $z$ and $w\in D$ then $|z-w|/|1-\bar w z|= \tanh d(z,w)$, and in particular $|z|=\tanh d(0,z)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another hint: You may assume $w=\rho&amp;gt;0$ and $z=r e^{i\alpha}$. Now compute $|z-w|^2$ and $|z-{1\over \bar  w}|^2$ by means of the cosine theorem and show that the quotient of the two is largest when $t:=\cos\alpha=-1$.&lt;/p&gt;&#xA;" OwnerUserId="1303" LastEditorUserId="1303" LastEditorDisplayName="" LastEditDate="2011-02-02T12:08:51.117" LastActivityDate="2011-02-02T12:08:51.117" />
  <row Id="20035" PostTypeId="2" ParentId="20023" CreationDate="2011-02-02T10:38:02.587" Score="0" ViewCount="" Body="&lt;p&gt;For the first part, one solution is to use brute force as follow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let us consider $D_{30}$ acting on a set of cardinality $30$, generated by the two elements $(1, 2, 3, 4, 5,$ $6, 7, 8, 9, 10,$ $11, 12, 13, 14, 15,$ $16, 17, 18, 19, 20,$ $21, 22, 23, 24, 25,$ $26, 27, 28, 29, 30)$ and $(1, 30)(2, 29)(3,28)$ $(4, 27)(5, 26)(6, 25)$ $(7, 24)(8, 23)(9, 22)$ $(10, 21)(11, 20)(12, 19)$ $(13, 18)(14, 17)(15, 16)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then the cyclic subgroup of order 10 of $D_{30}$ is generated by $(1, 28, 25, 22, 19,$ $16, 13, 10, 7, 4)$ $(2, 29, 26, 23, 20,$ $17, 14, 11, 8, 5)$ $(3, 30, 27, 24, 21,$ $18, 15, 12, 9, 6)$ and $(1, 25, 19, 13, 7)$ $(2, 26, 20, 14, 8)$ $(3, 27, 21, 15, 9)$ $(4, 28, 22, 16, 10)$ $(5, 29, 23, 17, 11)$ $(6, 30, 24, 18, 12)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This can be checked easily with Magma or GAP.&#xA;However I agree it lacks subtelty. One can probably use part $b)$ to build a smarter answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the second part, I would suggest this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is well known that a dihedral group can be generated by two involutions as follows: $$D_{n}=\langle s,t \mid s^2=t^2=e, (st)^n=e \rangle$$&#xA;On the other hand, a cyclic group of order $k$ is generated by a unique element of order $k$: $$C_k= \langle r \mid r^k=e \rangle$$&#xA;Now take an integer $k\geq 3$ such that $k\mid n$. Hence there exists $p \in \mathbb{N}$ such that $pk=n$. Considering $s,t$ as in the presentation of $D_n$, one has $$(st)^n = e$$ and thus $$((st)^p)^k = e$$ that is to say, there exists an element $r := (st)^p$ of order $k$ in $D_n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore there exists $H &amp;lt; D_n$ such that $$H= \langle r \mid r^k=e \rangle$$ i.e. $H$ is a cyclic group of order $k$. Unicity is a consequence of the fact that $r$ is uniquely determined as soon as $k$ is not $2$. In this case, any involution generates a cyclic group of order $2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope I answered your question.&lt;/p&gt;&#xA;" OwnerUserId="5465" LastActivityDate="2011-02-02T10:38:02.587" CommentCount="1" />
  <row Id="20036" PostTypeId="2" ParentId="20007" CreationDate="2011-02-02T10:40:00.203" Score="0" ViewCount="" Body="&lt;p&gt;I seriously doubt the jump from equation before the word &lt;code&gt;Derive&lt;/code&gt; to equation after the word in your question. There are definitely some voluntaristic differentiations present.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Implicit differentiation of implicit function $F(x,y)=0$ is carried out in the following way:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{dy}{dx} = - \frac{\partial F/ \partial x}{\partial F/ \partial y}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We've got $F(x,y)= \frac{2x}{y} - \frac{3y}{x} - 8$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then $\frac{\partial F}{\partial x} = \frac{2}{y} + \frac{3y}{x^2} = \frac{2x^2+3y^2}{x^2y}$, and $\frac{\partial F}{\partial y} = -\frac{2x}{y^2} - \frac{3}{x} = -\frac{2x^2+3y^2}{xy^2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dividing finally gives $\frac{dy}{dx}= -\frac{\frac{2x^2+3y^2}{x^2y}}{-\frac{2x^2+3y^2}{xy^2}} = {\frac{2x^2+3y^2}{x^2y}} \cdot \frac{xy^2}{2x^2+3y^2} = \frac{y}{x}$.&lt;/p&gt;&#xA;" OwnerUserId="3806" LastActivityDate="2011-02-02T10:40:00.203" />
  <row Id="20037" PostTypeId="2" ParentId="19967" CreationDate="2011-02-02T10:41:55.687" Score="3" ViewCount="" Body="&lt;p&gt;Hmm, using Pari/GP, I just tried to find the coefficients of $ \varphi (t) $&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;polcoeffs(exp(-x)*sumalt(k=0,exp(-k*x)*x^k/k!*(k+1)^(k-1)*1.0))&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;(where the polcoeffs-function just retrieves the whole vector of coefficients up to the defined series-precision and I used&lt;/em&gt; x &lt;em&gt;instead of&lt;/em&gt; t &lt;em&gt;)&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The result of this is an (arbitrary near) approximation to &lt;strong&gt;&lt;em&gt;the formal powerseries&lt;/em&gt;&lt;/strong&gt; $ \varphi(t) = 1.0 $ So it may be useful to analyze the approximation of that powerseries to the constant in order to evaluate  your limit-formula.    &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;[update] If one expands the powerseries for $e^{- k x}$ and collect coefficients at like powers of &lt;em&gt;x&lt;/em&gt; then indeed all coefficients at a certain power of &lt;em&gt;x&lt;/em&gt; sum to zero, thus vanish &lt;em&gt;(at least heuristically up to $x^{64}$, I did not do final analysis)&lt;/em&gt; . Then all derivatives are also zero and even the L'Hospital-rule is of no use here. However, I don't know whether there could be some special reason that this expansion into powerseries might not be feasible/allowed because the result is an infinite sum of formal powerseries. [end update]    &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;[update 2]&lt;br&gt;&#xA;$ \varphi(x)=e^{-x}+xe^{-2x}\frac1{1!} &#xD;&#xA;+x^2e^{-3x}\frac3{2!} &#xD;&#xA;+x^3e^{-4x}\frac{4^2}{3!} + \ldots&#xD;&#xA;  $       &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\begin{array} {rrrr}&#xD;&#xA;\varphi(x)= ( &amp;amp;  1 &amp;amp; -x &amp;amp; + \frac{x^2}{2!}&amp;amp;+ \frac{x^3}{3!} &amp;amp; + \ldots )\\&#xD;&#xA;+ \frac{x}{1!}(&amp;amp; 1 &amp;amp; -2x &amp;amp; + \frac{2^2x^2}{2!}&amp;amp;+ \frac{2^3x^3}{3!} &amp;amp; + \ldots ) \\&#xD;&#xA;+ \frac{3x^2}{2!}(&amp;amp; 1 &amp;amp; -3x &amp;amp; + \frac{3^2x^2}{2!}&amp;amp;+ \frac{3^3x^3}{3!} &amp;amp; + \ldots ) \\&#xD;&#xA;+ \ldots \\&#xD;&#xA;\end{array} $      &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Collecting equal powers of &lt;em&gt;x&lt;/em&gt; means to add coefficients along the antidiagonal so&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\begin{array} {rrrr}&#xD;&#xA;\varphi(x)= 1 \\&#xD;&#xA;- x (&amp;amp;\frac{1}{1!}&amp;amp; - \frac{1}{1!} ) \\&#xD;&#xA;+ x^2 (&amp;amp;\frac{1}{2!}&amp;amp; - \frac{2}{1!}&amp;amp; + \frac{3^1}{2!}) \\&#xD;&#xA;- x^3 (&amp;amp;\frac{1}{3!}&amp;amp; - \frac{2^2}{2!}&amp;amp; + \frac{3^2}{2!}&amp;amp; - \frac{4^2}{3!}) \\&#xD;&#xA;+ \ldots \\&#xD;&#xA;\end{array} $      &lt;/p&gt;&#xA;&#xA;&lt;p&gt;All the parentheses seem to form binomially weighted sums of like powers with alternating signs which are known to evaluate to zero, so the coefficients at each power of &lt;em&gt;x&lt;/em&gt; should be zero and if this is so, the taylorseries is then $\varphi(x)=1 $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[end update 2]&lt;/p&gt;&#xA;" OwnerUserId="1714" LastEditorUserId="1714" LastEditorDisplayName="" LastEditDate="2011-02-02T17:24:47.320" LastActivityDate="2011-02-02T17:24:47.320" />
  <row Id="20038" PostTypeId="2" ParentId="19740" CreationDate="2011-02-02T10:57:44.433" Score="1" ViewCount="" Body="&lt;p&gt;Here's what I would do:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Initialise a large integer $N = 2^n - 1$. (This requires 1 or 2 kilobytes.)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For each prime $p \le n$, compute $2^n$ mod $p$ = $2^{n \mod (p-1)}$ mod $p$, using a square-and-multiply algorithm (see &lt;a href=&quot;http://en.wikipedia.org/wiki/Exponentiation_by_squaring&quot; rel=&quot;nofollow&quot;&gt;this Wikipedia article&lt;/a&gt; for details). If the result is $1$, then $N$ is divisible by $p$, and we want the highest power of $p$ that divides both $N$ and $n!$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The highest power of $p$ that divides $n!$ is just $a = \sum_{r=1}^K \lfloor \frac{n}{p^r}\rfloor$, where $K$ is any integer such that $p^K&amp;gt;n$ (see e.g. &lt;a href=&quot;http://myyn.org/m/article/prime-power-dividing-a-factorial/&quot; rel=&quot;nofollow&quot;&gt;this article&lt;/a&gt;). So try to divide $N$ by $p$ up to $a$ times, stopping if the remainder is non-zero. The number of times that this succeeds is the power that we want.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now just multiply together all these prime powers.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that $N$ gets smaller and smaller as the computation proceeds, limiting the amount of arithmetic required. It might be faster to process the primes in reverse order, to accelerate this trend; experimentation will tell.&lt;/p&gt;&#xA;" OwnerUserId="1508" LastActivityDate="2011-02-02T10:57:44.433" CommentCount="1" />
  <row Id="20039" PostTypeId="1" CreationDate="2011-02-02T11:01:58.127" Score="3" ViewCount="45" Body="&lt;p&gt;Let $\{Z_i\}$ a directed projective system of quasi-compact topological spaces with projective limit $Z$. Assume we are given open subsets $U_i \subseteq Z_i$ such that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) For every $i \leq j$, the preimage $(Z_j \to Z_i)^{-1}(U_i)$ is contained in $U_j$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) For every $i$, the preimage $(Z \to Z_i)^{-1}(U_i)$ equals $Z$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does it follow that $U_i = Z_i$ for some $i$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If not, assume that $\{Z_i\}$ is actually a system of affine schemes and scheme morphisms. Then it is true by (EGA IV, Corollaire 8.3.4). But I wonder if there is a direct proof which avoids all these nasty lemmas about ind-constructible sets ...&lt;/p&gt;&#xA;" OwnerUserId="1650" LastActivityDate="2011-02-02T11:01:58.127" Title="open subsets in a projective system" Tags="&lt;general-topology&gt;" />
  <row Id="20040" PostTypeId="2" ParentId="20001" CreationDate="2011-02-02T11:12:34.830" Score="2" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Bertrand%27s_postulate&quot; rel=&quot;nofollow&quot;&gt;Bertrand's postulate&lt;/a&gt; gives a crude upper bound on $N(k),$ the largest number of consecutive composite numbers following $k!+1.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $n&amp;gt;3$ there exists a prime $p$ such that $n&amp;lt;p&amp;lt;2n-2$ and so there is a prime&#xA;$$k!+1&amp;lt; p &amp;lt; 2 k!,$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but the first $(k-1)$ of $k!+2,k!+3, \ldots, 2k!-1$ are certainly composite and so&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$N(k) \le k! - k -1.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An additional observation is that if $k! \pm 1$ are not &lt;a href=&quot;http://en.wikipedia.org/wiki/Factorial_prime&quot; rel=&quot;nofollow&quot;&gt;factorial primes&lt;/a&gt; then Bertrand's postulate says that there is a prime&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ k!-k-1 &amp;lt; p &amp;lt; 2k! -2k -4 $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and so, taking into account Andrea's observation, the first such prime must be $\ge k!+k+1,$ but also $ \le 2k! -2k -5.$ Hence we have for $k&amp;gt;3$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$N(k) \le (2k!-2k-5)-(k!+2) = k! -2k-7,$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a slight improvement for the bound in the absence of factorial primes.&lt;/p&gt;&#xA;" OwnerUserId="1301" LastEditorUserId="1301" LastEditorDisplayName="" LastEditDate="2011-02-02T13:07:13.623" LastActivityDate="2011-02-02T13:07:13.623" />
  <row Id="20041" PostTypeId="2" ParentId="19981" CreationDate="2011-02-02T12:16:09.913" Score="2" ViewCount="" Body="&lt;p&gt;Let's consider the relationship between the first two notions. It is convenient and natural to consider this in a probability setting, so that the total measure is $1$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;First notion. Suppose that $X$ is a random variable on a probability space $(\Omega,\mathcal{F},P)$. Define $\mu$ as follows. For any $B \in \mathcal{B}(\mathbb{R})$, $\mu(B)=P(\lbrace\omega : X(\omega) \in B\rbrace)$. The right-hand side is written in short as $P(X \in B)$. Then $\mu$ is a probability measure on $\mathcal{B}(\mathbb{R})$, which we call the distribution of $X$. Now, the function $F:\mathbb{R} \to [0,1]$ defined by $F(x) = \mu((-\infty,x])$ is called the distribution function of $X$. Note that $F(x) = P(X \in (-\infty,x]) = P(X \leq x)$, as one would expect. Of course, $F$ satisfies all the usual properties from probability theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second notion. Let's first change the notation, in accordance with the previous notion, as follows. The distribution function of a random variable (that is, a measurable function) $X: \Omega \to \mathbb{R}$ on a probability space (that is, a measure space with total measure $1$) $(\Omega,\mathcal{F},P)$ is a function $\lambda_X:(0,\infty) \to [0,1]$ given by $\lambda _X (\alpha ) = P(\lbrace \omega :|X(\omega )| &amp;gt; \alpha \rbrace )$. As before, the right-hand side is written in short as $P(|X| &amp;gt; \alpha)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The relationship between the first two notions is thus established (in the setting of probability spaces). Specifically,&#xA;$$&#xD;&#xA;\lambda _X (\alpha ) = P(|X| &amp;gt; \alpha) = P(X &amp;gt; \alpha) + P(X &amp;lt; -\alpha) = [1 - F(\alpha)] + F(-\alpha^-),&#xD;&#xA;$$&#xA;where $F(-\alpha^-)=\lim _{s \uparrow -\alpha } F(s)$ (recall that $F$ is right-continuous with left limits).&lt;/p&gt;&#xA;" OwnerUserId="2810" LastActivityDate="2011-02-02T12:16:09.913" />
  <row Id="20042" PostTypeId="2" ParentId="19981" CreationDate="2011-02-02T12:18:04.367" Score="3" ViewCount="" Body="&lt;p&gt;The first two uses came from probability theory, and are somewhat related as terminlogy. They are however, as far as I know, not related at all with the third concept. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, given a measurable map $f:(X,M)\to(\mathbb{R},\mathcal{B})$, and a measure $\mu$ on $(X,M)$. $f$ (or I guess $|f|$ in your case) defines a push-forward measure $f_*\mu$ on $(\mathbb{R},\mathcal{B})$ by the definition that, for every $b\in\mathcal{B}$ the Borel sigma-algebra, $f_*\mu(b) = \mu(f^{-1}(b))$. Then the distribution function $\lambda_f$ is something like $1- F$ for the distribution function corresponding to the pushforward Borel measure $|f|_*\mu$. (The 1 should be replaced by the total mass of the measure $|f|_*\mu$ when it is not a probability measure.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See the website &lt;a href=&quot;http://jeff560.tripod.com/d.html&quot; rel=&quot;nofollow&quot;&gt;Earlist Known Uses of some of the Words of Mathematics&lt;/a&gt; for some references for what I write below. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the distribution in the sense of the continuous linear functional is introduced by Laurent Schwartz, in French. In French, however, the distribution function of your Borel measure (or of your measurable function) is called &lt;a href=&quot;http://fr.wikipedia.org/wiki/Fonction_de_r%C3%A9partition&quot; rel=&quot;nofollow&quot;&gt;&quot;la fonction de répartition&quot;&lt;/a&gt;, which strongly suggests that Schwartz's choice of terminology is completely independent of the probability/measure theory uses of the words. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In German, the language in which &quot;distribution functions&quot; were introduced, the probabilistic concept is &lt;a href=&quot;http://de.wikipedia.org/wiki/Verteilungsfunktion&quot; rel=&quot;nofollow&quot;&gt;Verteilungsfunktion&lt;/a&gt;, while the functional analytic concept is just taken straight from French/English as Distribution. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The above all strongly indicates that while senses 1 and 2 are related, they are disjoint for the 3rd use of the word distribution. In fact, English is one of the (perhaps few) unhappy languages in which they coincide. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Just to confuse you further, there is also a use of the word distribution in differential geometry, which also means something completely different and disjoint from the three senses you listed above.)&lt;/p&gt;&#xA;" OwnerUserId="1543" LastEditorUserId="1543" LastEditorDisplayName="" LastEditDate="2011-02-02T16:23:52.583" LastActivityDate="2011-02-02T16:23:52.583" CommentCount="1" />
  <row Id="20043" PostTypeId="1" AcceptedAnswerId="20044" CreationDate="2011-02-02T12:34:05.233" Score="3" ViewCount="205" Body="&lt;p&gt;I would like to apply Seifert van Kampen to a simple example taken from Wikipedia: I have $X = S^2$ and $A = S^2 - n$, where $n$ is the north pole and $B = S^2 - s$, where $s$ is the south pole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to my understanding, which might be wrong, Seifert van Kampen tells me that $\pi_1(X) = \pi_1(A) *_{\pi_1(A \cap B)} \pi_1(B)$, where the right hand side is the free product with amalgamation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$A \cap B$, the sphere minus the two points has a fundamental group isomorphic to $\mathbb{Z}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The free product with amalgamation of two groups $G, H$ is $ G * H / N$, where $N$ is the smallest normal subgroup in $G * H$ (according to the Wikipedia entry about free product with amalgamation).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Note&lt;/strong&gt; : I am aware that in the sphere example both $G$ and $H$ are trivial and so quotienting them with anything will be trivial again. This question is not about actually computing the fundamental group of $S^2$!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the sphere example, this means I have to find the smallest normal subgroup of $\mathbb{Z}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question 1:&lt;/strong&gt; Is my understanding of Seifert van Kampen correct?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question 2:&lt;/strong&gt; What is the smallest normal subgroup of $\mathbb{Z}$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for question 2, what do you think about $lim_{k \rightarrow \infty} k\mathbb{Z}$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for your help.&lt;/p&gt;&#xA;" OwnerUserId="5798" LastEditorUserId="5798" LastEditorDisplayName="" LastEditDate="2011-02-02T13:34:12.237" LastActivityDate="2011-05-23T10:53:46.883" Title="Seifert-van-Kampen and free product with amalgamation" Tags="&lt;group-theory&gt;&lt;algebraic-topology&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="1" />
  <row Id="20044" PostTypeId="2" ParentId="20043" CreationDate="2011-02-02T13:04:29.603" Score="3" ViewCount="" Body="&lt;p&gt;Your understanding of Seifert van Kampen seems correct. However, note that for any group, 'the smallest normal subgroup' is of course the trivial subgroup {0}. Indeed, you left out an important part of the wikipedia sentence:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;(...) the smallest normal subgroup N of G*H &lt;em&gt;containing all elements on the left-hand side of the above equation&lt;/em&gt;, which are tacitly being considered in G*H by means of the inclusions of G and H in their free product.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you said, A and B have trivial fundamental groups. So in this case you are computing $\{0\}\star_{\mathbb{Z}}\{0\}$. All maps involved (i1,i2,.. in the wikipedia diagram) are therefore trivial. So this is the trivial group, as you already knew.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a nice explanation of the amalgamated free product, with emphasis on the universal property, see &lt;a href=&quot;http://unapologetic.wordpress.com/2007/03/03/amalgamated-free-products/&quot; rel=&quot;nofollow&quot;&gt;The Unapologetic Mathematician&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not sure what you mean by $\lim_{k\to\infty} k\mathbb{Z}$.&lt;/p&gt;&#xA;" OwnerUserId="6490" LastEditorUserId="6490" LastEditorDisplayName="" LastEditDate="2011-02-02T13:13:19.120" LastActivityDate="2011-02-02T13:13:19.120" CommentCount="5" />
  <row Id="20045" PostTypeId="2" ParentId="19933" CreationDate="2011-02-02T13:15:06.817" Score="1" ViewCount="" Body="&lt;p&gt;In your image&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/u8oQh.jpg&quot; alt=&quot;number line&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;you have a line.  Suppose we want to label the point on this line for the purpose of describing things that we can measure (i.e. number of apples, temperature of sun or moon, charge on a particle, weight of a car, etc).  The line is not very useful in and of itself  for this purpose because it has no point of reference.  The ends will not do for such a purpose as they are at infinity.  We therefore define a reference point, a datum, or a landmark on the line (you call it &quot;you are here&quot;) called zero.  We then adopt conventions stating that a certain distance from that datum is one 'unit' which we call 1.  We use this point as a abstract way to refer to the number of a given set of objects.  It could be one apple, one car, one planet, one atom;  we use the same label to refer to one unit of anything.  We then define other labels (i.e. 2,3,4,...) as multiples of this unit to count.  Some things that we want to label are not discrete objects, but can be divided continuously such as distance (i.e. distance from home to work is 13.77 miles).  Here we are assuming that we are always on the same side of our datum '0'.  There are other quantities that we measure and wish to label that can be increased arbitrarily but also decreased arbitrarily such as electric charge, a force, a moment, a voltage, a DC current, etc.  In these cases, we not only want to signify how &lt;em&gt;far&lt;/em&gt; we are from our datum, but we also want to know which side of the datum we are on.  We use negative numbers to signify that our quantity is the same in magnitude but opposite in direction of what we decide to call positive (an arbitrary choice by the way).&lt;/p&gt;&#xA;" OwnerUserId="4726" LastActivityDate="2011-02-02T13:15:06.817" />
  <row Id="20046" PostTypeId="1" CreationDate="2011-02-02T13:20:10.247" Score="1" ViewCount="208" Body="&lt;p&gt;I want a convincing proof of this...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\nabla^2 (1/|r-r&amp;#39;|)$ = $-4\pi \delta^3 (r-r&amp;#39;)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(where r and r' are vectors and $\delta^3 (r-r&amp;#39;)$ represents the 3-D delta function...)&lt;/p&gt;&#xA;" OwnerUserId="6094" LastEditorUserId="6094" LastEditorDisplayName="" LastEditDate="2011-02-02T13:43:58.823" LastActivityDate="2011-02-02T13:43:58.823" Title="Proof of a vector identity..." Tags="&lt;vector-calculus&gt;" CommentCount="6" />
  <row Id="20047" PostTypeId="1" CreationDate="2011-02-02T13:34:10.490" Score="2" ViewCount="89" Body="&lt;p&gt;Is it possible to have a general expression for the binary sequence ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Eg.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for N= 3, and i=0, M_i = diag(0,0,0) i=1, M_i = diag(0,0,1) . . i=7, M_i = diag(1,1,1)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need a general matrix (M_i) or expression whose arguments are N,i.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S: I am not looking for a software code to generate this.&lt;/p&gt;&#xA;" OwnerUserId="5978" LastActivityDate="2011-06-02T17:09:14.690" Title="Matrix for Binary sequences" Tags="&lt;linear-algebra&gt;" AnswerCount="1" />
  <row Id="20048" PostTypeId="2" ParentId="20016" CreationDate="2011-02-02T13:42:10.580" Score="10" ViewCount="" Body="&lt;p&gt;Once you have the bijection&#xA;$P:2^\mathbb{N}\cong\mathbb{R}$, you can build your desired&#xA;bijection as follows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, note that $\mathbb{N}\times 2^\mathbb{N}\cong&#xD;&#xA;2^\mathbb{N}$, essentially  by the bijection that associates $(n,A)$ with&#xA;$\{n\}\cup (n+1+A)$, except that this will miss the empty set on the right, but we can fix this by composing with a map witnessing $2^\mathbb{N}-\{\varnothing\}\cong 2^\mathbb{N}$, such as shifting on a fixed countable subset.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now simply observe that&#xA;$$\mathbb{R}^\mathbb{R}\cong&#xD;&#xA;(2^\mathbb{N})^{(2^\mathbb{N})}\cong 2^{(\mathbb{N}\times&#xD;&#xA;2^\mathbb{N})} \cong 2^{(2^\mathbb{N})} \cong 2^\mathbb{R},&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which provides the desired bijection. The first map is conjugation by $P$, simply&#xA;composing with $P^{-1}$ and $P$ before and after; the second map is an easy exercise in parenthesis rearranging; the&#xA;third map applies the observation above; and the final map&#xA;applies $P$.&lt;/p&gt;&#xA;" OwnerUserId="413" LastEditorUserId="413" LastEditorDisplayName="" LastEditDate="2011-02-02T16:30:00.193" LastActivityDate="2011-02-02T16:30:00.193" CommentCount="1" />
  <row Id="20049" PostTypeId="2" ParentId="20043" CreationDate="2011-02-02T14:10:00.360" Score="2" ViewCount="" Body="&lt;p&gt;This is slightly too long for a comment, therefore I'm posting it here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In complete generality: Let $S$ be a subset of a group $G$. Then you're familiar with the subgroup $\langle S \rangle$ generated by $S$. This is the smallest subgroup of $G$ containing $S$. Similarly, there is $N = \langle \langle S \rangle \rangle$, the smallest &lt;em&gt;normal&lt;/em&gt; subgroup generated by $S$ (sometimes also called &lt;a href=&quot;http://en.wikipedia.org/wiki/Conjugate_closure&quot; rel=&quot;nofollow&quot;&gt;normal or conjugate closure&lt;/a&gt;). While $\langle S \rangle$ consists of precisely the words of the form $s_{i_1}^{\pm 1} \cdots s_{i_{n}}^{\pm 1}$ with $s_{i_{j}} \in S$, the smallest normal subgroup consists of the words of the form $g_{1}s_{i_1}^{\pm 1}g_{1}^{-1} \cdots g_{n}s_{i_{n}}^{\pm 1}g_{n}^{-1}$ with $g_{j} \in G$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, if you're writing $G = \langle S | R \rangle$, i.e. $G$ is given in terms of a &lt;a href=&quot;http://en.wikipedia.org/wiki/Presentation_of_a_group&quot; rel=&quot;nofollow&quot;&gt;presentation&lt;/a&gt; with generators $S$ and relations $R$, then you actually mean $G \cong F(S) / \langle\langle R \rangle \rangle$, or in words $G$ is the quotient of the free group on $S$ modulo the normal soubgroup generated by the relations. The normal subgroup generated by a set is tremendously hard to determine (that's why the &lt;a href=&quot;http://en.wikipedia.org/wiki/Word_problem_for_groups&quot; rel=&quot;nofollow&quot;&gt;Word problem&lt;/a&gt; is so difficult, i.e. not solvable in general).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now you should be able to understand what $G \ast_{A} H$ is: it's $G \ast H/ \langle\langle R \rangle \rangle$, with $R = \{\varphi(a)\psi(a)^{-1}\,:\,a \in A\}$, where $\varphi:A \to G$ and $\psi: A \to H$ are the given homomorphisms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your concrete situation, the situation is quite silly: As $G \ast H = \{1\}$, we must have that $N = \{1\}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to learn about free products with amalgamation in general, the standard reference is J.-P. Serre's &lt;em&gt;Arbres, amalgames et&lt;/em&gt; $SL_{2}$ (translated as &lt;em&gt;Trees&lt;/em&gt;). As for Seifert-van Kampen, I think Allen Hatcher has a pretty lucid and detailed explanation on pages 40ff of his algebraic topology book, available on his &lt;a href=&quot;http://www.math.cornell.edu/~hatcher&quot; rel=&quot;nofollow&quot;&gt;home page&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="5363" LastEditorUserId="5363" LastEditorDisplayName="" LastEditDate="2011-02-02T14:30:59.123" LastActivityDate="2011-02-02T14:30:59.123" CommentCount="6" />
  <row Id="20050" PostTypeId="1" CreationDate="2011-02-02T14:36:38.047" Score="2" ViewCount="97" Body="&lt;p&gt;I want to take the inverse of a dot product, where both vectors have complex components.  In other words, if $\textbf{A} \cdot \textbf{B} = d$, and I know $\textbf{A}$ and $d$, I want to find a $\textbf{B}$.  I know that I cannot do so uniquely, which is fine; I have a procedure for creating a set of vectors that will satisfy $\textbf{A} \cdot \textbf{B} = d$.  But, it relies on finding vectors that are orthogonal to $\textbf{A}$ and each other.  Normally this would not be a problem; just take $\textbf{A}$, zero all but two of its components, switch the last two, and negate one of them.  Make sure it's a different pair every time.  For more orthogonal vectors, you can take cross products.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The difficulties arise when I consider vectors with complex components.  I want to normalize each of the orthogonal vectors, which means divide by their magnitudes.  I have read that you need to divide by $\sqrt{|v_x|^2 + |v_y|^2 + |v_z|^2 ...}$.  What is the justification for this?  Since I am just normalizing vectors, do I absolutely have to take the magnitudes of the components?  Also, My technique of finding the inverse dot product relies on the identity $\textbf{A} \cdot \textbf{B} = d = |\textbf{A}||\textbf{B}|$.  What modifications might I need to make?  I can post more details if people want.  Also, if anyone has links to stuff to read about, especially with regards to the reasons why, I would be most appreciative.&lt;/p&gt;&#xA;" OwnerUserId="6534" LastActivityDate="2011-02-02T15:02:12.030" Title="What are the rules for complex-component vectors and why?" Tags="&lt;complex-analysis&gt;&lt;complex-numbers&gt;&lt;vector-space&gt;&lt;vector-analysis&gt;" AnswerCount="2" />
  <row Id="20051" PostTypeId="2" ParentId="19903" CreationDate="2011-02-02T14:37:15.143" Score="1" ViewCount="" Body="&lt;p&gt;As milcak said $1,1,1,-1,1$ satisfies the definition.   You can also verify that the reversal of the sequence also works and the sequence obtained by multiplying the entire sequence by $-1$ also works.&lt;/p&gt;&#xA;" OwnerUserId="832" LastEditorUserId="832" LastEditorDisplayName="" LastEditDate="2011-02-02T15:00:34.507" LastActivityDate="2011-02-02T15:00:34.507" />
  <row Id="20052" PostTypeId="1" AcceptedAnswerId="20069" CreationDate="2011-02-02T14:42:59.957" Score="4" ViewCount="67" Body="&lt;p&gt;Let's define a stair case polygon to be a polygon look like the following.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/vj9lQ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let the lower left most point to be the origin, all edges are axis-aligned.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One can see if we remove a stair case polygon from a rectangle, we get another stair case polygon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $S_k$ be some $k$ vertices stair case polygon.&#xA;Let $A(x)$ be the area of the polygon x.&#xA;Let $R_k$ be the largest axis-aligned rectangle that is contained in $S_k$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What can we say about the lower bound of $\frac{A(R_k)}{A(S_k)}$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Currently I know $\frac{A(R_k)}{A(S_k)} \geq \frac{1}{(k-4)/2 + 1}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Define the maximal rectangle as a rectangle that can't be stretched in any direction without going outside the polygon.&#xA;There are only $(k-4)/2 + 1$ maximal rectangles, the union covers the entire polygon. Then at least one have area $\geq \frac{1}{(k-4)/2 + 1}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wonder does there exist a better lower bound? Or can we prove that's the best possible?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Motivation, I originally had a conjecture, that for any bounded continuous monotonic decreasing convex function defined in $[0,a]$ as $f(x)$. The maximal $xf(x)$ over the area under $f(x)$ is a constant. It turns out to be false, counterexample $f(x) = 1/x$. So now I'm doing this for a discrete version, the maximal $xf_k(x)$ over the area under $f_k(x)$ is a function of $k$, where $f_k(x)$ is the average value of $f(x)$ in each of the $k$ partitions of $[0,a]$.&lt;/p&gt;&#xA;" OwnerUserId="96" LastEditorUserId="96" LastEditorDisplayName="" LastEditDate="2011-02-02T15:55:41.897" LastActivityDate="2011-02-02T16:06:07.480" Title="The largest rectangle in a stair case polygon with n vertices." Tags="&lt;geometry&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="20053" PostTypeId="2" ParentId="19933" CreationDate="2011-02-02T14:48:13.527" Score="0" ViewCount="" Body="&lt;p&gt;Think about it. Does a positive number really exist?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just like negative numbers, complex numbers, positive numbers have a status that is no different. And you can ask the same question for other mathematical objects, sets, groups, etc. (And depending on what you mean by &quot;exist&quot;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Think about complex numbers. At first glance, they look like non-sense. I mean how can $\sqrt{-1}$ exist? But when I read complex analysis, I realized what a beauty I was missing. My heart was more than ever, ready to accept the existence of complex numbers. &quot;It gave us a new way to model this world&quot;, and &quot;new way to derive meaning from complexity and confusion&quot;, and thats what most mathematical objects do in the end, even though they may take abstraction to the highest level.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you look at it more closely, the problem is not with &quot;numbers&quot;, the problem is with the word &quot;exist&quot;. My answer to your doubt will be &quot;Cogito ergo sum&quot; (I think therefore I am).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Peace :)&lt;/p&gt;&#xA;" OwnerUserId="2320" LastActivityDate="2011-02-02T14:48:13.527" />
  <row Id="20054" PostTypeId="2" ParentId="20027" CreationDate="2011-02-02T14:49:08.203" Score="0" ViewCount="" Body="&lt;p&gt;If each noise sample is independent,zero-mean and has identical variance, then the noise variance actually $increases $ by a factor of $N$.  However the noise variance relative to the squared magnitude of signal would decrease by $1/N$&lt;/p&gt;&#xA;" OwnerUserId="832" LastEditorUserId="832" LastEditorDisplayName="" LastEditDate="2011-02-02T14:58:21.217" LastActivityDate="2011-02-02T14:58:21.217" />
  <row Id="20056" PostTypeId="2" ParentId="20047" CreationDate="2011-02-02T14:52:29.887" Score="0" ViewCount="" Body="&lt;p&gt;It is not clear what you want your matrices to look like.  If you just put the binary expansion of $i$ down the diagonal, you can use the usual approach of extracting the binary bits.  So your example would have $i \pmod 2$ in the lower right, $i/2 \pmod 2$ in the center, $i/4 \pmod 2$ in the upper left, where all divisions are floor divisions.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-02-02T14:52:29.887" />
  <row Id="20057" PostTypeId="1" CreationDate="2011-02-02T15:01:17.267" Score="3" ViewCount="30" Body="&lt;p&gt;Hi all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am looking for an (almost everywhere) differentiable function $f(p_1,p_2,p_3,p_4)$ that given four points will give me a scale-agnostic measure for co-planarity. It is zero if the four points lie on the same plane and positive otherwise. Scale-agnostic means that, when I uniformly scale all points the planarity measure will return the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I came up with something that is quite complex and not easy to optimize. Define $u=p_2-p_1$, $v=p_3-p_1$, $w=p_4-p_1$. Then the planarity measure is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\left[(u \times v) \cdot w \right]^{2}}{\left(||u \times v||^{2} ||w||^{2} \right)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The numerator is simply (the square of) the volume of the tetrahedron defined by the four points, and the denominator is a normalizing factor that makes this measure become simply the cosine of an angle. Because angles do not changed under uniform scale, this function satisfies all my requirements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anybody know of something simpler?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alex.&lt;/p&gt;&#xA;" OwnerUserId="5073" LastEditorUserId="5073" LastEditorDisplayName="" LastEditDate="2011-02-02T15:18:33.487" LastActivityDate="2011-02-02T15:19:10.753" Title="Scale-agnostic, differentiable, co-planarity measure" Tags="&lt;vector-calculus&gt;&lt;3d&gt;" AnswerCount="0" />
  <row Id="20058" PostTypeId="2" ParentId="20050" CreationDate="2011-02-02T15:02:01.063" Score="1" ViewCount="" Body="&lt;p&gt;Yes, to take the length of a complex vector you need the squared magnitudes of the components.  So the vector $(1+i,1-i)$ has length $\sqrt{|1+i|^2+|1-i|^2}=\sqrt{2+2}=2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is still true that the dot product of orthogonal vectors is zero in a complex space.  So once you find a $\textbf{B}$ parallel to $\textbf{A}$ with $\textbf{A} \cdot \textbf{B} = d$ you can add any vector orthogonal to $\textbf{A}$ to it and the dot product will not change.  But when you take the dot product, remember to take the complex conjugate of the components of $\textbf{B}$&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-02-02T15:02:01.063" />
  <row Id="20059" PostTypeId="2" ParentId="20050" CreationDate="2011-02-02T15:02:12.030" Score="1" ViewCount="" Body="&lt;p&gt;I wouldn't call this the &quot;inverse of a dot product&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given complex vectors $v = (v_{1}, \ldots, v_{n})$ and $w = (w_{1}, \ldots, w_{n})$, their scalar (=dot) product is given by $v \cdot w = \sum_{j=1}^{n} v_{j} \overline{w}_{j} = v_{1} \overline{w}_{1} + \cdots + v_{n} \overline{w}_{n}$. Why? Well, you want to generalize the usual dot product on $\mathbb{R}^{n}$, but you also want that $v \cdot v \geq 0$ for all $v$, and the vector $(i,0,\ldots,0)$ shows that you can't do without the conjugation. You might ask why, I don't care if $v \cdot v \geq 0$ for all $v$. Most people do: the expression $\|v\| = \sqrt{v \cdot v}$ should define a norm and $\|v - w\|$ a metric on $\mathbb{C}^{n}$, and taking square-roots of non-positve numbers (or even complex numbers) simply isn't well-defined. Note that $\|v\| = \sqrt{v \cdot v} = \sqrt{\sum_{j=1}^{n} |v_{j}|^{2}}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Having settled this, let $v$ be a non-zero vector. Its orthogonal complement $U = v^{\perp} = \{u \in \mathbb{C}^{n}\,:u \cdot v = 0\}$ is the set of all vectors orthogonal to $v$. Since $U$ is determined by the single linear equation $u_{1}\overline{v}_{1} + \cdots + u_{n}\overline{v}_{n} = 0$, it is an $(n-1)$-dimensional subspace of $\mathbb{C}^{n}$. Finding solutions is easily achieved using &lt;a href=&quot;http://en.wikipedia.org/wiki/Gaussian_elimination&quot; rel=&quot;nofollow&quot;&gt;Gauss elimination&lt;/a&gt;, this will give you vectors $u_{1}, \cdots, u_{n-1}$, which you can make into an orthonormal basis of $U$ using&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Gram%2DSchmidt_process&quot; rel=&quot;nofollow&quot;&gt;Gram-Schmidt&lt;/a&gt; (note that the notation $\langle u, v \rangle = u \cdot v$ is to be understood. The fact that you're working with $\mathbb{C}$ and not with $\mathbb{R}$ is immaterial, just be careful to note that $v \cdot (\lambda w) = \overline{\lambda} (v \cdot w)$, i.e., the dot product is conjugate-linear in the second variable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, in order to solve the equation $w \cdot v = d$ simply take any $u \in U = v^{\perp}$ and put $w = u + \frac{d}{v \cdot v} v$, and note that $w \cdot v = (u + \frac{d}{v \cdot v}v) \cdot v = (u \cdot v) + \frac{d}{v \cdot v} v \cdot v = 0 + d = d$.&lt;/p&gt;&#xA;" OwnerUserId="5363" LastActivityDate="2011-02-02T15:02:12.030" />
  <row Id="20061" PostTypeId="2" ParentId="20027" CreationDate="2011-02-02T15:29:08.403" Score="1" ViewCount="" Body="&lt;p&gt;The answers you got are correct assuming (as they state) the noise runs are independent.  But if not, you can't guarantee that.  If you have a noise spike that repeats say 5 sec after the start of the signal, it will not average out.  This is the ongoing interplay between random and systematic error.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-02-02T15:29:08.403" />
  <row Id="20062" PostTypeId="1" CreationDate="2011-02-02T15:31:23.427" Score="3" ViewCount="114" Body="&lt;p&gt;I have come across a paper that has suggested a formula for &quot;NUMBER OF MULTINOMIAL COEFFICIENTS NOT DIVISIBLE BY A PRIME&quot;; but I don't understand the notation.Please help.&#xA;The formula is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ G(n,l,p)= \sum_{k=0}^m(l,p-1)^k\frac{a^k}{l}\prod_{i=k}^m(l-1,a_i) $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Link to the paper:http://www.jstor.org/pss/119039&lt;/p&gt;&#xA;" OwnerUserId="6536" LastEditorUserId="5966" LastEditorDisplayName="" LastEditDate="2011-02-02T17:34:50.877" LastActivityDate="2011-02-02T17:34:50.877" Title="Multinomial Coefficients !" Tags="&lt;combinatorics&gt;&lt;binomial-coefficients&gt;&lt;permutations&gt;" CommentCount="4" />
  <row Id="20063" PostTypeId="1" AcceptedAnswerId="20071" CreationDate="2011-02-02T15:31:44.823" Score="2" ViewCount="99" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Let $G$ be a finite group and $P$ is a p-Sylow subgroup of $G$.&lt;br&gt;&#xA;  Show that $N(N(P))=N(P)$ where $N(P)$ is the normalizer.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Not sure where to begin... Any hints will be appreciated.&lt;/p&gt;&#xA;" OwnerUserId="6535" LastActivityDate="2011-02-02T16:09:38.723" Title="p-Sylow subgroup" Tags="&lt;abstract-algebra&gt;&lt;group-theory&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="20064" PostTypeId="2" ParentId="20063" CreationDate="2011-02-02T15:35:28.943" Score="1" ViewCount="" Body="&lt;p&gt;Show that $P$ is normal in $N(N(P))$.  We know that $N(P) = \{g \in G: gPg^{-1} = P \}$. So $N(P)$ is the largest subgroup of $G$ for which $P$ is normal. Now $N(N(P)) = \{g \in G: gN(P)g^{-1} = N(P) \}$. &lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2011-02-02T15:35:28.943" />
  <row Id="20065" PostTypeId="2" ParentId="20000" CreationDate="2011-02-02T15:37:54.570" Score="1" ViewCount="" Body="&lt;p&gt;Another pattern to look for in this problem is to -un-distribute $C$, that is pull out the $C$ from the two main parts being intersected. That is,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$[(A\cap B)\cup C]∩(A^c\cup C) \rightarrow [((A\cap B) \cap A^c) \cup C]$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then you can notice something in that intersection about $A$ and $A^C$ that will eliminate that entire part not involving $C$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Two general strategies for solving equivalence problems by just playing with the syntax is to  &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;rewrite to simpler things, like using an identity like distribution to reduce the length of the 'phrases' (the shortest set of things operated on). This can be very automatic (apply any identity that fits but always in one direction), but sometimes you get stuck, where you don't think you can simplify any further.&lt;/li&gt;&#xA;&lt;li&gt;look for patterns to undo. Here you can ignore complexity in the syntax  and consider one complex part as replaced by some new variable. Then with this new variable the syntax you have might match one side of an identity you already know. This is where a little experience helps, knowing identities that might be useful.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Another strategy is to try to think of what the (convoluted) syntax -means- (for example by some visualization like a Venn diagram, or thinking of an intersection as somehow below' both the sets being intersected)&lt;/p&gt;&#xA;" OwnerUserId="1919" LastActivityDate="2011-02-02T15:37:54.570" CommentCount="3" />
  <row Id="20066" PostTypeId="1" AcceptedAnswerId="20068" CreationDate="2011-02-02T15:51:48.977" Score="3" ViewCount="164" Body="&lt;p&gt;Hi. I'm doing some exercises on group-theory but got stuck in the following question:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Let $G$ be a group such that for all $a \in G\setminus\{e\}$, $a^2 = e$. Prove that $|G|$ is even.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I tried using Lagrange's theorem but perhaps I still don't understand it fully for I was not capable of finishing it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could you shed some enlightening on the matter?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance.&lt;/p&gt;&#xA;" OwnerUserId="6538" LastEditorUserId="5966" LastEditorDisplayName="" LastEditDate="2011-02-02T16:10:54.010" LastActivityDate="2011-02-02T16:10:54.010" Title="Prove for G group that |G| is even." Tags="&lt;homework&gt;&lt;abstract-algebra&gt;&lt;group-theory&gt;" AnswerCount="2" CommentCount="5" />
  <row Id="20067" PostTypeId="2" ParentId="20066" CreationDate="2011-02-02T15:57:55.510" Score="5" ViewCount="" Body="&lt;p&gt;One must obviously assume $G$ is a nontrivial finite group.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case $G$ has a subgroup $H = \{e,a\}$ of order 2, for any $a\in G$ that is not the unit. Therefore by Lagranges theorem the order of H divides the order of G, in other words: $|G|$ is even.&lt;/p&gt;&#xA;" OwnerUserId="5189" LastActivityDate="2011-02-02T15:57:55.510" CommentCount="3" />
  <row Id="20068" PostTypeId="2" ParentId="20066" CreationDate="2011-02-02T15:59:02.480" Score="4" ViewCount="" Body="&lt;p&gt;HINT: Let $a\not=e$. If $a^2=e$, how many elements you got in the subgroup generated by $a$? Apply Lagrange......&lt;/p&gt;&#xA;&#xA;&lt;p&gt;NOTE: written at the same time as Myself's answer......&lt;/p&gt;&#xA;" OwnerUserId="688" LastActivityDate="2011-02-02T15:59:02.480" CommentCount="10" />
  <row Id="20069" PostTypeId="2" ParentId="20052" CreationDate="2011-02-02T16:06:07.480" Score="1" ViewCount="" Body="&lt;p&gt;I claim the estimate you have is sharp, but not attained. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That it is not attained is clear, since a decomposition by maximal polygons will always have some overlap. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To show that it is sharp, consider the following construction. Let $k$ be fixed. Let $1 &amp;gt; \epsilon &amp;gt; 0$ be a small positive number. Start with the unit square ($1\times 1$). Add to it the following maximal polygons, one at a time: $\epsilon^j \times \epsilon^{-j}$ for each $0 &amp;lt; j &amp;lt; (k-4)/2 + 1$. So each maximal polygon has area 1. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each additional maximal polygon adds an area of $\epsilon^j \times (\epsilon^{-j} - \epsilon^{-(j-1)}) = 1 - \epsilon$ (the amount it protrudes from what is already there). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the total area $A(S_k)$ for this figure is $\frac{k-4}{2}(1-\epsilon) + 1$. By choosing $\epsilon$ as small as you want you can make $A(R_k)/A(S_k) = \left((k-4)(1-\epsilon)/2 + 1\right)^{-1}$ as close to the bound you derived as possible. &lt;/p&gt;&#xA;" OwnerUserId="1543" LastActivityDate="2011-02-02T16:06:07.480" />
  <row Id="20070" PostTypeId="1" AcceptedAnswerId="20075" CreationDate="2011-02-02T16:06:19.417" Score="1" ViewCount="338" Body="&lt;p&gt;I am trying to measure complexity of the following code segment&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;int sum = 0;&#xA;for (int i = 1; i &amp;lt;= n/2; i++)&#xA;{&#xA;   sum++;&#xA;}&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;As far as I understand it can be represented by following sum $$\displaystyle\sum_{i=1}^{n/2}1$$ but I don't know how to evaluate it. &lt;/p&gt;&#xA;" OwnerUserId="6539" LastEditorUserId="1208" LastEditorDisplayName="" LastEditDate="2011-02-28T18:38:18.920" LastActivityDate="2011-03-01T09:19:00.293" Title="How to evaluate $\displaystyle\sum_{i=1}^{n/2}1$" Tags="&lt;algorithms&gt;&lt;sequences-and-series&gt;&lt;asymptotics&gt;" AnswerCount="4" />
  <row Id="20071" PostTypeId="2" ParentId="20063" CreationDate="2011-02-02T16:09:38.723" Score="1" ViewCount="" Body="&lt;p&gt;Since $P$ is normal in $N(P)$ it is the unique normal $p$- Sylow subgroup of $N(P)$. However if $x \in N(N(P))$ then $xN(P)x^{-1}=N(P)$ so $xPx^{-1} \subseteq xN(P)x^{-1}=N(P)$. This forces $xPx^{-1}=P$, so $x \in N(P)$. Hence $N(N(P)) \subseteq N(P)$. But $N(P) \subseteq N(N(P))$. Hence $N(N(P))=N(P)$.&lt;/p&gt;&#xA;" LastActivityDate="2011-02-02T16:09:38.723" CommentCount="1" />
  <row Id="20072" PostTypeId="2" ParentId="20070" CreationDate="2011-02-02T16:10:09.693" Score="1" ViewCount="" Body="&lt;p&gt;Since the summation is over a constant value.&#xA;The answer should be N/2 more precisely (upper limit - lower limit + 1) * constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here, constant = 1,upper limit = N/2,lower limit = 1.The +1 is because both limits are inclusive.&lt;/p&gt;&#xA;" OwnerUserId="6536" LastActivityDate="2011-02-02T16:10:09.693" />
  <row Id="20074" PostTypeId="1" AcceptedAnswerId="20078" CreationDate="2011-02-02T16:13:45.440" Score="1" ViewCount="199" Body="&lt;p&gt;How to solve $\int\tan^3(x)\,dx$ ? Do I use substitution?&lt;/p&gt;&#xA;" OwnerUserId="5889" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-02T16:18:36.147" LastActivityDate="2011-02-02T16:25:37.340" Title="How to solve $\int\tan^3(x)\,dx$ ?" Tags="&lt;integral&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="20075" PostTypeId="2" ParentId="20070" CreationDate="2011-02-02T16:14:13.880" Score="5" ViewCount="" Body="&lt;p&gt;In the sum I guess it's actually $\lfloor n/2 \rfloor$, the largest integer smaller than $n/2$ (basically, if $n$ is odd, throw away the .5 after division by 2), since $n$ isn't always even.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sum is quick to evaluate:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{i=1}^{\lfloor n/2 \rfloor} 1 = \underbrace{1 + 1 + \dots + 1}_{\lfloor n/2 \rfloor \textrm{ terms}} = \lfloor n/2 \rfloor$$&lt;/p&gt;&#xA;" OwnerUserId="5966" LastEditorUserId="5966" LastEditorDisplayName="" LastEditDate="2011-03-01T08:42:10.903" LastActivityDate="2011-03-01T08:42:10.903" CommentCount="3" />
  <row Id="20076" PostTypeId="2" ParentId="20023" CreationDate="2011-02-02T16:15:05.917" Score="2" ViewCount="" Body="&lt;p&gt;You are using $D_{30}$ to represent the dihedral group of order $60$, with $r$ corresponding to the &quot;rotation&quot;, and some other element, call it $s$, the &quot;reflection&quot;; that is,&#xA;$$D_{30}=\left\langle r,s\Bigm| r^{30} = s^2 = 1,\quad sr = r^{-1}s\right\rangle.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Part (a): Your first part begins well enough, but it seems like you did not read the question carefully. The last part of part (a) asks you to find &lt;strong&gt;all&lt;/strong&gt; generators of the subgroup $H$ you found.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You took $H=\langle r^3\rangle$. That's fine; this is a group of order $10$. But $r^3$ is not the &lt;em&gt;only&lt;/em&gt; generator of this group. Perhaps you know that, in general, the cyclic group of order $n$, $C_n$ has $\varphi(n)$ generators, where $\varphi$ is Euler's function that counts the number of positive integers less than $n$ and relatively prime to $n$ (if you don't, then try to prove it). So $H$, being cyclic of order $10$, should have $\varphi(10) = 4$ generators. You've found one, there are three more to go.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your part (b) also suffers a bit. You know that there is one and only one subgroup of $\langle r\rangle$ that has order $k$; that's fine. But you did not prove that &lt;em&gt;every&lt;/em&gt; subgroup of $D_n$ of order $k$ must be a subgroup of $\langle r\rangle$! You need to show that too if you want your argument to hold. So you would need to show that the &lt;em&gt;only&lt;/em&gt; elements of order $k$ all lie in $\langle r\rangle$; that's where you'll need the fact that $k\geq 3$. Perhaps you can show that every &lt;em&gt;other&lt;/em&gt; element has other ideas about what their order is?&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2011-02-02T16:15:05.917" />
  <row Id="20077" PostTypeId="2" ParentId="20074" CreationDate="2011-02-02T16:18:35.290" Score="1" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT:&lt;/strong&gt; Rewrite your integral in this form:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align*}&#xA;\int\tan^{3}{x} \ dx &amp;amp; = \int (\sec^{2}{x}-1) \cdot \tan{x} \ dx \\ &amp;amp;= \int \sec^{2}{x} \cdot \tan{x} \ dx - \int \tan{x} \ dx \\ &amp;amp;= \int \sec{x} \cdot (\sec{x}\tan{x}) \ dx - \int \tan{x} \ dx&#xA;\end{align*}&lt;/p&gt;&#xA;" LastActivityDate="2011-02-02T16:18:35.290" />
  <row Id="20078" PostTypeId="2" ParentId="20074" CreationDate="2011-02-02T16:19:10.283" Score="3" ViewCount="" Body="&lt;p&gt;One standard approach to integral involving powers of tangent is the following: Rewrite your power as something times $\tan^2(x)$. In this case: $\tan^3(x)=\tan(x)\cdot\tan^2(x)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, use that $\tan^2(x)=\sec^2(x)-1$, so&#xA;$$ \int\tan^3(x)dx=\int\tan(x)\sec^2(x)dx-\int\tan(x)dx.$$&#xA;The first integral on the right hand side can be solved using the substitution $u=\tan(x)$, so $du=\sec^2(x)dx$ and $$ \int\tan(x)\sec^2(x)dx=\int u du. $$ The second integral is $\ln|\sec x|$, and can be found writing $\tan(x)=\sin(x)/\cos(x)$ and using the substitution $v=\cos(x)$.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Note that the same idea allows you to solve $\displaystyle \int\tan^n(x) dx$ for any $n\ge 3$. For example:&#xA;$$\int\tan^5(x)dx=\int\tan^3(x)\tan^2(x)dx=\int\tan^3(x)(\sec^2(x)-1)dx,$$&#xA;as before we split this last integral into two:&#xA;$$\int\tan^3(x)\sec^2(x)dx-\int\tan^3(x)dx.$$&#xA;The first one is solved with the substitution $u=\tan(x)$. The second is the integral we solved before. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This, by the way, is an example of a &lt;em&gt;reduction method&lt;/em&gt; that transforms an integral that depends on an integer parameter $n$ (in this case, $\displaystyle \int\tan^n(x)dx$) into a similar integral, but that now depends on some $m&amp;lt;n$. This gives us an &lt;em&gt;algorithm&lt;/em&gt; to solve the integral by repeatedly applying the reduction procedure.&lt;/p&gt;&#xA;" OwnerUserId="462" LastEditorUserId="462" LastEditorDisplayName="" LastEditDate="2011-02-02T16:25:37.340" LastActivityDate="2011-02-02T16:25:37.340" CommentCount="2" />
  <row Id="20079" PostTypeId="2" ParentId="20074" CreationDate="2011-02-02T16:19:35.507" Score="2" ViewCount="" Body="&lt;p&gt;A simple way out is to write $\tan^3(x) = \frac{\sin^3(x)}{\cos^3(x)}$. Now let $\cos(x) = t$ and rewrite $\sin^2(x) = 1-t^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The integral now becomes $\int \frac{\sin^3(x)}{\cos^3(x)} dx$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\cos(x) = t \Rightarrow -\sin(x)dx = dt$ and $\sin^2(x) = 1-t^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note the numerator of the integral $\sin^3(x) dx$ can be written as $\sin^2(x) \times \sin(x) dx = (1-t^2) (-dt) = (t^2-1) dt$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the integral becomes $\int \frac{t^2-1}{t^3} dt = \int \frac{dt}{t} - \int \frac{dt}{t^3} = \log(t) + \frac{1}{2t^2} = \log(\cos(x)) + \frac{1}{2 \cos^2(x)} = \log(\cos(x)) + \frac{\sec^2(x)}{2}$&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2011-02-02T16:25:07.483" LastActivityDate="2011-02-02T16:25:07.483" />
  <row Id="20080" PostTypeId="2" ParentId="20070" CreationDate="2011-02-02T16:22:13.397" Score="1" ViewCount="" Body="&lt;p&gt;If $N$ is a non-negative integer, then&#xA;\begin{eqnarray}&#xA;\sum_{i = 1}^{N} 1 = N.&#xA;\end{eqnarray}&#xA;In your question, if you mean $\lfloor \frac{n}{2} \rfloor$, which is the greatest integer less than $\frac{n}{2}$, then the sum evaluates to $\lfloor \frac{n}{2} \rfloor$.&lt;/p&gt;&#xA;" OwnerUserId="2720" LastEditorUserId="2720" LastEditorDisplayName="" LastEditDate="2011-02-26T14:18:11.223" LastActivityDate="2011-02-26T14:18:11.223" CommentCount="2" />
  <row Id="20081" PostTypeId="1" AcceptedAnswerId="20089" CreationDate="2011-02-02T16:26:15.140" Score="2" ViewCount="91" Body="&lt;p&gt;What's statistically more likely to win you the UK lottery (6 numbers out of 1-49), &#xA;To play the lottery weekly or to save the money and buy multiple tickets &#xA;all in a oner?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a discussion in the office where a group of 6 people want to play weekly.&lt;/p&gt;&#xA;" OwnerUserId="6541" LastActivityDate="2011-02-02T17:13:17.573" Title="Lottery chances." Tags="&lt;probability&gt;&lt;statistics&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="20082" PostTypeId="1" AcceptedAnswerId="20085" CreationDate="2011-02-02T16:26:16.513" Score="2" ViewCount="64" Body="&lt;p&gt;My algebra books are in storage and Google is not being helpful...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In figure 6.5.5, how do I apply the angle operator to the matrix columns to determine a value for theta?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.w3.org/TR/SVG/implnote.html#ArcImplementationNotes&quot; rel=&quot;nofollow&quot;&gt;http://www.w3.org/TR/SVG/implnote.html#ArcImplementationNotes&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for any pointers,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Paul&lt;/p&gt;&#xA;" OwnerUserId="6542" LastEditorUserId="5966" LastEditorDisplayName="" LastEditDate="2011-02-02T16:58:54.743" LastActivityDate="2011-02-02T16:58:54.743" Title="How do I calculate theta, given the linked formula?" Tags="&lt;linear-algebra&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="20083" PostTypeId="1" AcceptedAnswerId="20088" CreationDate="2011-02-02T16:27:29.307" Score="1" ViewCount="89" Body="&lt;p&gt;This question came up on a pub quiz last night. The answer was parabola.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I disagreed, because I thought I remembered that angles could go off into infinity too. An angle is two rays whose endpoints are shared, so would that also count as a &quot;bend&quot; ? Or is there another name for this type of angle?&lt;/p&gt;&#xA;" OwnerUserId="6543" LastActivityDate="2011-02-02T17:09:38.740" Title="What do you call a shape with a single bend and lines going into infinity?" Tags="&lt;geometry&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="20084" PostTypeId="2" ParentId="19988" CreationDate="2011-02-02T16:28:57.120" Score="2" ViewCount="" Body="&lt;p&gt;As Christian Blatter mentioned, this has a geometric interpretation in terms of hyperbolic metrics, but you can also do it directly: First note that&#xA;$${|z - w|^2 \over |1 - \bar{w}z|^2} = {|z|^2 - 2Re(\bar{w}z) + |w|^2 \over 1 - 2Re(\bar{w}z) + |w|^2|z|^2}$$&#xA;(You made a small error in your calculation). Next, observe that $|z|^2 + |w|^2 \leq 1 + |w|^2|z|^2$, since this is equivalent to $(1 - |w|^2)(1 - |z|^2) \geq 0$. Thus the above expression is of the form ${\displaystyle {A + c \over B + c}}$, where $0 \leq A \leq B$ and both $A + c$ and $B + c$ are nonnegative. Such an expression increases as $c$ increases. Since $- 2Re(\bar{w}z)$ is at most $2|w||z|$, we thus have&#xA;$$ {|z|^2 - 2Re(\bar{w}z) + |w|^2 \over 1 - 2Re(\bar{w}z) + |w|^2|z|^2}&#xD;&#xA;\leq {|z|^2 + 2|w||z| + |w|^2 \over 1 + 2|w||z| + |w|^2|z|^2}$$&#xA;$$ = {(|z| + |w|)^2 \over (|w||z| + 1)^2}$$&#xA;Showing this is at most $(|z| + |w|)^2$ is equivalent to showing $(|w||z| + 1)^2 \geq 1$, which is clearly true. &lt;/p&gt;&#xA;" OwnerUserId="3035" LastEditorUserId="3035" LastEditorDisplayName="" LastEditDate="2011-02-02T16:34:11.140" LastActivityDate="2011-02-02T16:34:11.140" />
  <row Id="20085" PostTypeId="2" ParentId="20082" CreationDate="2011-02-02T16:34:00.837" Score="1" ViewCount="" Body="&lt;p&gt;Well, it's the angle between two vectors, in this case the vectors&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{pmatrix} 1 \\ 0 \end{pmatrix} ~~ \textrm{and} ~~ \begin{pmatrix} \frac{x_1 - c_x}{r_x} \\ \frac{y_1 - c_y}{r_y} \end{pmatrix}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, if you have two real vectors $v_1$ and $v_2$, and want to calculate the angle between them, use the scalar product, $\langle v_1, v_2 \rangle = v_2^T v_1$ and the formula&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\langle v_1, v_2 \rangle = \|v_1\| \|v_2\| \cos \theta$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\theta$ is the angle. This gives you:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\cos \theta = \frac{v_2^T v_1}{\|v_1\| \|v_2\|} = \frac{\sum_{i = 1}^n v^1_i v^2_i}{\|v_1\| \|v_\|}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $v^1_i$ and $v^2_i$ are the components at position $i$ of the vectors $v_1$ and $v_2$ respectively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your case you get:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\cos \theta = \frac{\frac{x_1-c_x}{r_x}}{\sqrt{\left( \frac{x_1 - c_x}{r_x} \right)^2 + \left(  \frac{y_1 - c_y}{r_y} \right)^2}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;use $\arccos$ to get a value for $\theta$.&lt;/p&gt;&#xA;" OwnerUserId="5966" LastEditorUserId="5966" LastEditorDisplayName="" LastEditDate="2011-02-02T16:39:19.687" LastActivityDate="2011-02-02T16:39:19.687" CommentCount="4" />
  <row Id="20086" PostTypeId="2" ParentId="20081" CreationDate="2011-02-02T16:48:51.000" Score="2" ViewCount="" Body="&lt;p&gt;If, in the second scenario, you are planning to buy tickets with distinct number combinations, then the odds are exactly the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you buy multiple tickets with the same numbers on them, your chances of winning go down.  However, your &quot;expected winnings&quot; (in the mathematical sense) don't change.&lt;/p&gt;&#xA;" OwnerUserId="6071" LastActivityDate="2011-02-02T16:48:51.000" />
  <row Id="20087" PostTypeId="1" AcceptedAnswerId="20109" CreationDate="2011-02-02T17:09:25.407" Score="3" ViewCount="143" Body="&lt;p&gt;Here is a question I've often wondered about, but have never figured out a satisfactory answer for. Here are the rules for the solitaire game &quot;clock patience.&quot; Deal out 12 piles of 4 cards each with an extra 4 card draw pile. (From a standard 52 card deck.) Turn over the first card in the draw pile, and place it under the pile corresponding to that card's number 1-12 interpreted as Ace through Queen. Whenever you get a king you place that on the side and draw another card from the pile. The game goes out if you turn over every card in the 12 piles, and the game ends if you get four kings before this happens. My question is what is the probability that this game goes out?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One thought I had is that the answer could be one in thirteen, the chances that the last card of a 52 card sequence is a king. Although this seems plausible, I doubt it's correct, mainly because I've played the game probably dozens of times since I was a kid, and have never gone out!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any light that people could shed on this problem would be appreciated!&lt;/p&gt;&#xA;" OwnerUserId="2941" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2011-02-02T20:55:17.517" LastActivityDate="2011-02-02T20:55:17.517" Title="Probability of &quot;clock patience&quot; going out" Tags="&lt;probability&gt;" AnswerCount="3" CommentCount="4" />
  <row Id="20088" PostTypeId="2" ParentId="20083" CreationDate="2011-02-02T17:09:38.740" Score="1" ViewCount="" Body="&lt;p&gt;This is the second worst pub quiz question that I have come across. There are so many &quot;curves with single bends that go to infinity&quot; that the question is meaningless.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;BTW, the worst question was this: &quot;Which continent is wider in the South than in the North?&quot; You might like to ponder that for a microsecond.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The official answer from the book was, if you can believe it, Antarctica. I was flabbergasted (and said so). Some years later, I stumbled upon the key to this idiocy, quite by accident, while browsing my Encyclopaedia Britannica (as you do). An article on the continents said, quite reasonably (I am paraphrasing here): &quot;With the exception of Antarctica, all continents are wider in the North than in the South.&quot; The question setter obviously read this exact same sentence and thought that a question could be forged out of it.&lt;/p&gt;&#xA;" OwnerUserId="1508" LastActivityDate="2011-02-02T17:09:38.740" CommentCount="3" />
  <row Id="20089" PostTypeId="2" ParentId="20081" CreationDate="2011-02-02T17:13:17.573" Score="2" ViewCount="" Body="&lt;p&gt;If you just want to say &quot;I've won the lottery&quot;, then all in a oner, provided you buy different numbers for the multiple tickets. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose there are $Q$ possible outcomes in the lottery. The chance of winning with 1 ticket is $1/Q$. Assuming (reasonably) that the individual drawings are independent, the chance you fail to win $k$ times in a row is $(1-1/Q)^k$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you buy $k$ tickets at one time, the probability that the outcomes is not one of the $k$ tickets you bought is $(Q-k)/Q = 1 - k/Q$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Observe that $(1-1/Q)^k &amp;gt; 1 - k/Q$ if $Q &amp;gt; 1$. So the chances of not having a winning ticket at all is greater if you play weekly. In other words, buying multiple tickets ever-so-slightly increase your chances of having a winning ticket. (The increase, however, is very, very small.)&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;But if you play for the pay-out: Assuming the pay-out is constant across the time-frame, let's call it $P$, buying $k$ tickets at once would give expected payout $kP/Q$: there's only one winning number, and the chance of you hitting it is $k/Q$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whereas if you play weekly, there is the chance that you win multiple times! The chances that you hit the jack pot $j$ times is  $\binom{k}{j}(1-1/Q)^{k-j}(1/Q)^j$. But the payout for that would be $jP$. Adding them all up (from $j = 0$ to $j=k$, with a bit of algebra and the binomial theorem, you see that the total expected payout for playing weekly ends up also exactly as $kP/Q$, for playing $k$ times. &lt;/p&gt;&#xA;" OwnerUserId="1543" LastActivityDate="2011-02-02T17:13:17.573" />
  <row Id="20090" PostTypeId="2" ParentId="20087" CreationDate="2011-02-02T17:19:06.703" Score="0" ViewCount="" Body="&lt;p&gt;If I understand you correctly, then the game is won if the $48$ cards on the regular piles contain fewer than $4$ kings. Let's try to calculate the probability that the game is lost instead. It is exactly $$\frac{\binom{48}{44}}{\binom{52}{48}} = \frac{38916}{54145} \approx 72\%.$$ There are $\binom{52}{48}$ possible choices of the regular piles, and out of these $\binom{48}{44}$ contain all four kings.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2011-02-02T17:19:06.703" CommentCount="1" />
  <row Id="20091" PostTypeId="1" AcceptedAnswerId="20094" CreationDate="2011-02-02T17:21:25.203" Score="3" ViewCount="132" Body="&lt;p&gt;This is targeted for elementary kids. One way to think is that add any number to itself zero times... Just wondering if there is a better way...&lt;/p&gt;&#xA;" OwnerUserId="6426" LastActivityDate="2011-02-02T21:57:10.147" Title="Is there any physical analogy to explain multiplication by zero ?" Tags="&lt;education&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="20092" PostTypeId="2" ParentId="77" CreationDate="2011-02-02T17:22:41.567" Score="0" ViewCount="" Body="&lt;p&gt;If you want me to name 2 concepts that are used in engineering calculations so frequently, they will be dot and cross products.There are several interpretations of the dot and cross product and can be applied in various scenarios: Angle between vectors, Projection of one vector in the direction of another as mentioned in the above posts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Going a few steps above, interesting facts to remember are that the cross product of two vectors can be written as the product of a matrix(skew symmetric) and a vector.The norm of a cross product can be expressed as a determinant.&#xA;The triple product of three vectors namely a.(b$\times$c) represents the area of a parallelopiped and this triple product can also be written as a determinant.This useful fact can be used to prove collinearity of 3 points (of course after representing/considering them as vectors)&lt;/p&gt;&#xA;" OwnerUserId="6050" LastActivityDate="2011-02-02T17:22:41.567" />
  <row Id="20093" PostTypeId="1" AcceptedAnswerId="20097" CreationDate="2011-02-02T17:28:08.270" Score="2" ViewCount="102" Body="&lt;p&gt;I tried to start by showing that $\frac{a}{\gcd(a,b)}$ is always an integer, let's call it $d$, because $a$ is always a multiple of $(a,b)$ based on the definition of a g.c.d. I then tried to show that if $a | b$ then $a|bc$ from the theorem that states that $a|b$ implies that $a|bc$ for any integer $c$, but I cant seem to prove that $a | b$ or that $a|bc$ only if $d \mid c$ (where $d = \frac{a}{\gcd(a,b)}$ ). Any help would be much appreciated.&lt;/p&gt;&#xA;" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-02T17:50:35.520" LastActivityDate="2011-02-02T19:42:34.690" Title="How can I prove that a|bc if and only if (a/(a,b))|c?" Tags="&lt;elementary-number-theory&gt;&lt;divisibility&gt;" AnswerCount="2" />
  <row Id="20094" PostTypeId="2" ParentId="20091" CreationDate="2011-02-02T17:31:06.593" Score="6" ViewCount="" Body="&lt;p&gt;Apples in bags. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Multiplying zero: you have a bunch of empty bags with no apples in them. You take 5 bags. How many apples do you have?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Multiplying by zero: you have many bags on the table, each with 5 apples in them. You don't take any bags, how many apples do you have?&lt;/p&gt;&#xA;" OwnerUserId="1543" LastActivityDate="2011-02-02T17:31:06.593" CommunityOwnedDate="2011-02-02T17:31:06.593" />
  <row Id="20095" PostTypeId="2" ParentId="20091" CreationDate="2011-02-02T17:31:45.377" Score="2" ViewCount="" Body="&lt;p&gt;The area of a &quot;rectangle&quot; with sides 0 and $x$ is zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I really don't this any such analogy will be of much help...&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2011-02-02T17:31:45.377" />
  <row Id="20096" PostTypeId="1" CreationDate="2011-02-02T17:34:13.270" Score="12" ViewCount="447" Body="&lt;p&gt;This is a tall order, but since this site is for everything mathematical, here goes:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are some nice games/puzzles/problems/problem-like activities that would be appropriate for a three year old?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I ask this because I'm not particularly interested in trying to accelerate &quot;number literacy&quot; or prepare for &quot;school mathematics&quot;. One may assume the three year old can count to 20, say, but the games don't necessarily need to involve number. I would like some activities that would introduce the surprise and delight of the problematic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An article by a Russian mathematician (I can't remember the reference...) pointed out that if you let a young child count three objects and then separate the objects, the child will think that the number of objects has increased. When prompted to actually count the separated objects, the child was surprised and amazed to see that the number was the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The goal is to bring a bit of the joy of mathematical &lt;em&gt;thought&lt;/em&gt; to a very young child. Any ideas?&lt;/p&gt;&#xA;" OwnerUserId="941" LastEditorUserId="941" LastEditorDisplayName="" LastEditDate="2011-02-02T18:28:26.570" LastActivityDate="2011-03-01T20:57:59.710" CommunityOwnedDate="2011-02-02T17:43:09.147" Title="Mathematics problems for three year olds?" Tags="&lt;education&gt;" AnswerCount="12" CommentCount="7" FavoriteCount="3" />
  <row Id="20097" PostTypeId="2" ParentId="20093" CreationDate="2011-02-02T17:40:15.780" Score="3" ViewCount="" Body="&lt;p&gt;Note that $x|y$ if and only if $(x,y) = x$; then use the properties of the gcd.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align*}&#xA;a|bc &amp;amp;\Longleftrightarrow (a,bc) = a\\&#xA;&amp;amp;\Longleftrightarrow \left(\frac{a}{(a,b)}, \frac{bc}{(a,b)}\right) = \frac{a}{(a,b)}\\&#xA;&amp;amp;\Longleftrightarrow \left(\frac{a}{(a,b)}, \frac{b}{(a,b)}c\right) = \frac{a}{(a,b)}\\&#xA;&amp;amp;\Longleftrightarrow \left(\frac{a}{(a,b)}, c\right) = \frac{a}{(a,b)}&amp;amp;&amp;amp;\mbox{(by Euclid's Lemma)}\\&#xA;&amp;amp;\Longleftrightarrow \frac{a}{(a,b)} \Bigm| c,&#xA;\end{align*}&#xA;using Euclid's Lemma (&quot;If $(x,y)=1$, then $x|yz\Leftrightarrow x|z$&quot;) and &#xA;$$1 = \frac{1}{(a,b)}(a,b) = \left(\frac{a}{(a,b)},\frac{b}{(a,b)}\right).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The manipulations are consequences of $x(y,z) = (xy,xz)$; see &lt;a href=&quot;http://math.stackexchange.com/questions/19830/if-gcd-a-b-1-then-is-it-true-to-say-gcd-ac-bc-c/19842#19842&quot;&gt;Bill Dubuque's proof here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-02T19:42:34.690" LastActivityDate="2011-02-02T19:42:34.690" CommentCount="2" />
  <row Id="20098" PostTypeId="2" ParentId="20096" CreationDate="2011-02-02T17:41:15.200" Score="1" ViewCount="" Body="&lt;p&gt;Try some simple graph theory. Show that them you can draw edges between vertices without drawing over the same edge twice, and explain the general result of Euler in a way they understand.&#xA;Edit:&#xA;Or just show them Bridges of königsberg, or what it's called, and challenge them and let them try to figure it out themselves. Then guide them etc.&lt;/p&gt;&#xA;" OwnerUserId="2394" LastActivityDate="2011-02-02T17:41:15.200" CommunityOwnedDate="2011-02-02T17:43:09.147" CommentCount="4" />
  <row Id="20099" PostTypeId="2" ParentId="20087" CreationDate="2011-02-02T17:50:55.740" Score="1" ViewCount="" Body="&lt;p&gt;The name clock patience (solitaire in the US) is appropriate, not just because of the shape of the layout but because it is about cycles in the permutation.  As you start with the King pile, a cycle ends when you find a King.  If four cycles (one for each King) include all 52 cards, you win.  You lose if the bottom card on any non-King pile matches its position, as that would be a one-card cycle in the permutation.  You also lose if the bottom card in the Ace pile is a Two and the bottom card in the Two pile is an Ace.  I'm trying to figure out the impact of the fact that suits are ignored.  Maybe you can give each card its particular destination (always put the spade ace on top, for example) and ask for a single cycle of the 52 cards.  In that case, it would be 1/52.  To make a single cycle, the first card cannot go to itself (51/52).  The card the first card goes to cannot go to the first (50/51).  Then the next card in the chain cannot go to the first (49/50) and so on.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-02-02T17:50:55.740" CommentCount="2" />
  <row Id="20100" PostTypeId="2" ParentId="20096" CreationDate="2011-02-02T17:53:03.990" Score="5" ViewCount="" Body="&lt;p&gt;My limited experience with three-year-olds who can count to 20 suggests that the counting is simply a recitation. Can your particular three-year-old reliably count the number of matches (10-20) in a matchbox? That would be a rare talent, and one worth encouraging.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edited to add:&lt;/strong&gt; An &lt;a href=&quot;http://www.newscientist.com/article/dn20095-without-language-numbers-make-no-sense.html&quot; rel=&quot;nofollow&quot;&gt;article in today's New Scientist&lt;/a&gt; agrees with this assessment:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Children learn this count list well before they actually understand that &quot;four&quot; refers to four objects rather than three or six, says Michael Frank at Stanford University in California.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1508" LastEditorUserId="1508" LastEditorDisplayName="" LastEditDate="2011-02-07T21:38:51.413" LastActivityDate="2011-02-07T21:38:51.413" CommunityOwnedDate="2011-02-02T17:53:03.990" CommentCount="2" />
  <row Id="20101" PostTypeId="2" ParentId="20007" CreationDate="2011-02-02T17:56:09.417" Score="3" ViewCount="" Body="&lt;p&gt;Note that you are introducing a lot of &quot;false positives&quot;. The function implicitly defined by&#xA;$$\frac{2x}{y} - \frac{3y}{x} = 8$$&#xA;is &lt;em&gt;not&lt;/em&gt; the same as the function implicitly defined by&#xA;$$2x^2 - 3y^2 = 8xy.$$&#xA;Why not? Well, the second function includes $y(0)=0$ (that is, the point $(0,0)$), but the first one is &lt;em&gt;not&lt;/em&gt; defined for $x=0$ or $y=0$!  Now, it &lt;em&gt;is&lt;/em&gt; true that any point on the &lt;em&gt;second&lt;/em&gt; curve in which $x\neq 0$ and $y\neq 0$ will be on your original curve, and that the only point in the second curve not on the orginal one is $(0,0)$, so you &lt;em&gt;can&lt;/em&gt; make some progress.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These are things you must be careful with when you perform that kind of easy manipulations with implicit functions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, what you seem to be forgetting is that the values of $x$ and of $y$ are connected by the equation $\frac{2x}{y}-\frac{3y}{x} = 8$. So it is entirely possible that the formula $\frac{y}{x}$ will give the exact same values as &#xA;$$\frac{4x-8y}{8x+6y} = \frac{2(x-y)}{4x+3y}$$&#xA;for any $(x,y)$ that satisfies $\frac{2x}{y}-\frac{3y}{x}=8$. Let's see if that's the case. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We must assume that $x\neq 0$ and that $y\neq 0$; in addition, your formula has the potential unpleasant side effect of &lt;em&gt;also&lt;/em&gt; prohibiting points on the line $4x+3y=0$. Are there any points on that line that are on the original curve? If $y = -\frac{4}{3}x$, then plugging into the left hand side of your original equation we have&#xA;$$-\frac{6x}{4x} + \frac{4x}{x} = -\frac{6}{4}+4 \neq 8$$&#xA;so, phew!, none of the points your formula excludes are on the original graph. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, some manipulation is in order. Since $x\neq 0$, $y\neq 0$, and $4x+3y\neq 0$, we have:&#xA;\begin{align*}&#xA;\frac{y}{x} = \frac{2x-4y}{4x+3y} &amp;amp;\Longleftrightarrow y(4x+3y) = (2x-4y)x\\&#xA;&amp;amp;\Longleftrightarrow 4xy + 3y^2 = 2x^2 - 4xy\\&#xA;&amp;amp;\Longleftrightarrow 8xy = 2x^2 - 3y^2\\&#xA;&amp;amp;\Longleftrightarrow 8 = \frac{2x}{y} - \frac{3y}{x}.&#xA;\end{align*}&#xA;So not only are the two expressions equal on the curve you are working on, they are equal &lt;em&gt;only&lt;/em&gt; on the curve you are working on. So the two answers are equivalent, though yours is by far more complicated.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2011-02-02T17:56:09.417" />
  <row Id="20102" PostTypeId="1" AcceptedAnswerId="22398" CreationDate="2011-02-02T15:46:13.190" Score="4" ViewCount="304" Body="&lt;p&gt;Hello&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to solve system of linear equations of size 300x300 (300 variables &amp;amp; 300 equations) &lt;strong&gt;exactly&lt;/strong&gt; (not with floating point, aka &lt;code&gt;dgesl&lt;/code&gt;, but with &lt;em&gt;fractions in answer&lt;/em&gt;). All coefficients in this system are integer (e.g. 32 bit), there is only 1 solution to it. There are non-zero constant terms in right column (b).&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;A*x = b&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;where &lt;code&gt;A&lt;/code&gt; is matrix of coefficients, &lt;code&gt;b&lt;/code&gt; is vector of constant terms.&lt;/li&gt;&#xA;&lt;li&gt;answer is &lt;code&gt;x&lt;/code&gt; vector, given in rational numbers (fractions of pairs of very long integers).&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The matrix A is dense (general case), but it can have up to 60 % of zero coefficients.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the best way (fastest algorithm for x86/x86_64) to solve such system?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS typical answer of such systems have integers in fraction up to 50-80 digits long, so, please don't suggest anything based only on float/doubles. They have no needed precision.&lt;/p&gt;&#xA;" OwnerUserId="6545" LastEditorUserId="6545" LastEditorDisplayName="" LastEditDate="2011-02-03T01:12:06.287" LastActivityDate="2011-02-16T20:58:45.613" Title="Best way to exactly solve a linear system (300x300) with integer coefficients" Tags="&lt;linear-algebra&gt;&lt;matrices&gt;&lt;fractions&gt;" AnswerCount="7" CommentCount="4" FavoriteCount="2" />
  <row Id="20103" PostTypeId="2" ParentId="20102" CreationDate="2011-02-02T15:49:29.103" Score="2" ViewCount="" Body="&lt;p&gt;Don't know if it's the best, but &lt;a href=&quot;http://docs.scipy.org/doc/scipy/reference/generated/scipy.linalg.solve.html&quot; rel=&quot;nofollow&quot;&gt;Scipy's linalg.solve&lt;/a&gt; works. You could look at the source code...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT for the cases that I tried &lt;code&gt;scipy.linalg.solve(A,b) * scipy.linalg.det(A)&lt;/code&gt; gives an exact result, e.g. gives a matrix corresponding to &lt;code&gt;(1/det(A)) * x&lt;/code&gt;, so the actual fractional result is each element divided by that common denominator.&lt;/p&gt;&#xA;" OwnerUserId="4833" LastActivityDate="2011-02-02T16:49:29.257" CommentCount="14" />
  <row Id="20104" PostTypeId="2" ParentId="20102" CreationDate="2011-02-02T17:58:46.400" Score="3" ViewCount="" Body="&lt;p&gt;For the algorithm, straightforward Gaussian elimination should be fine. This part may actually be simpler than when dealing with floating-point numbers, since you don't have to worry about the numerical stability.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Depending on your matrix, you might be able to do a bit better that Gaussian elimination, e.g. if your matrix is symmetric, you can use a Cholesky factorization. But 60% sparseness isn't that much in the scheme of things. If it were tridiagonal, banded, etc. then you could try some specialized methods.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You will need a good rational number library to handle the actual arithmetic operations. &lt;a href=&quot;http://gmplib.org/&quot; rel=&quot;nofollow&quot;&gt;GNU MP&lt;/a&gt; should fit the bill, and claims it can do rational numbers, but I've never used it.&lt;/p&gt;&#xA;" OwnerUserId="6547" LastEditorUserId="6547" LastEditorDisplayName="" LastEditDate="2011-02-02T20:13:58.527" LastActivityDate="2011-02-02T20:13:58.527" CommentCount="2" />
  <row Id="20105" PostTypeId="1" AcceptedAnswerId="20107" CreationDate="2011-02-02T18:16:47.673" Score="2" ViewCount="66" Body="&lt;p&gt;Hi, I would like to know, how do you simplify this:&#xA;$$\cos x\sin(x+y) + \sin x\cos(x+y)$$&#xA;to this:&#xA;$$\sin(2x+y).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wolfram alpha says so, but how does human being do so? :)&lt;/p&gt;&#xA;" OwnerUserId="4447" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-02T18:18:40.177" LastActivityDate="2011-02-02T18:26:45.177" Title="Trigonometric equality" Tags="&lt;trigonometry&gt;" AnswerCount="3" CommentCount="0" />
  <row Id="20106" PostTypeId="2" ParentId="20105" CreationDate="2011-02-02T18:18:54.747" Score="1" ViewCount="" Body="&lt;p&gt;We know that $\sin(x+y) = \sin x \cos y + \cos x \sin y$ and $\cos(x+y) = \cos x \cos y - \sin x \sin y$.&lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2011-02-02T18:18:54.747" CommentCount="1" />
  <row Id="20107" PostTypeId="2" ParentId="20105" CreationDate="2011-02-02T18:19:56.530" Score="3" ViewCount="" Body="&lt;p&gt;I hope you are aware of the $\sin(A+B)$ formula which is $$\sin(A+B)=\sin{A}\cos{B}+ \cos{A}\sin{B}$$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a complete list of Trigonometric identities please see:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/List_of_trigonometric_identities&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/List_of_trigonometric_identities&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" LastActivityDate="2011-02-02T18:26:45.177" />
  <row Id="20108" PostTypeId="2" ParentId="20105" CreationDate="2011-02-02T18:20:48.630" Score="2" ViewCount="" Body="&lt;p&gt;A human being uses the addition formula for the sine&#xA;$$\sin(\alpha+\beta) = \sin\alpha\cos\beta + \cos\alpha\sin\beta$$&#xA;And applies it with $\alpha=x+y$ and $\beta=x$.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2011-02-02T18:20:48.630" CommentCount="1" />
  <row Id="20109" PostTypeId="2" ParentId="20087" CreationDate="2011-02-02T18:35:18.377" Score="4" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://orion.math.iastate.edu/burkardt/puzzles/solitaire_solution.html&quot; rel=&quot;nofollow&quot;&gt;Here's an explanation&lt;/a&gt; of why it's 1/13.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We are essentially dealing out a randomly ordered deck into piles of a kind of four, and we require that the pile of kings be the last pile completed. And so the probability of winning is 1/13.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe this is more convincing. Imagine playing the game backwards with the deck facing you so that you can see the cards. Remove the top card, a seven, say, the next card facing you is a two, say, so you place the seven in the pile at the two o'clock position, the next card facing you is a five, so you place the two at the five o'clock position. You now take the five from the deck and see a king beneath it, so you place the five in the king pile, and so on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You continue until you've placed out all the cards and, if you are going to win this game (when it's played in the correct direction), the last card must be placed in the king pile because that's where you take your first card from when you run the game in the right direction. Now the only way that can happen is if you placed a king down as your first card. And that's a 1/13 chance with a random ordered deck.&lt;/p&gt;&#xA;" OwnerUserId="1301" LastEditorUserId="1301" LastEditorDisplayName="" LastEditDate="2011-02-02T20:19:36.507" LastActivityDate="2011-02-02T20:19:36.507" CommentCount="4" />
  <row Id="20110" PostTypeId="2" ParentId="20096" CreationDate="2011-02-02T18:39:52.250" Score="3" ViewCount="" Body="&lt;p&gt;You can let him count things, like a number of apples or bananas. It will be easier if the objects are something the kid knows well, nothing abstract.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also you can let him sort pens or something by their height, it will be interesting to see if he develops some technique to do this. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If he is very talented (probably not as he is just 3 years old) you could go on by letting him add small numbers or for example ask him how he would give 6 pieces of cake to 3 persons so everyone would get the same amount. Also pattern recognition is an important more advanced skill in mathematics.&lt;/p&gt;&#xA;" OwnerUserId="3123" LastActivityDate="2011-02-02T18:39:52.250" CommunityOwnedDate="2011-02-02T18:39:52.250" />
  <row Id="20111" PostTypeId="2" ParentId="20096" CreationDate="2011-02-02T18:45:35.053" Score="2" ViewCount="" Body="&lt;p&gt;I think any type of sorting or matching would be appropriate for a 3 year old.  Dominoes, or cards that match geometric distribution of numbers are a great way to building connections (this teaches about equivalence).  Algebra tiles would also work (e.g. make as many squares as you can with one large square, a few rectangles and a few small squares.) Note: algebra tiles come in the aforementioned denominations.  If this is done intentionally, you could draw connections later in life to algebra.  &lt;/p&gt;&#xA;" OwnerUserId="6243" LastActivityDate="2011-02-02T18:45:35.053" CommunityOwnedDate="2011-02-02T18:45:35.053" CommentCount="1" />
  <row Id="20112" PostTypeId="2" ParentId="20093" CreationDate="2011-02-02T18:52:26.373" Score="0" ViewCount="" Body="&lt;p&gt;It's easy: $\rm\ \ a\ |\ bc\ \iff\ a\ |\ ac,bc\ \iff\ a\ |\ (ac,bc) = (a,b)\:c\ \iff\ a/(a,b)\ |\ c\quad$   &lt;strong&gt;QED&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or, dually: $\rm\ a\ |\ bc\ \iff\ a,b\ |\ bc\ \iff\ [a,b]\ |\ bc\ \iff\ [a.b]/b\ |\ c\quad$ (here $\rm\ [m,n] = lcm(m,n)\:$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that combining the two yields the &lt;a href=&quot;http://math.stackexchange.com/questions/3118/if-a-m-and-a-1-m-prove-aa-1-m/3149#3149&quot;&gt;GCD * LCM law:&lt;/a&gt; $\rm\:\ [a,b]/b = a/(a,b)\:,\:$ i.e. $\rm\: (a,b)\ [a,b] = ab$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively, cancelling $\rm\ (a,b)\ $ reduces it to &lt;a href=&quot;http://math.stackexchange.com/questions/16042/about-greatest-common-divisor/16043#16043&quot;&gt;Euclid's Lemma&lt;/a&gt; $\rm\ (a,b)= 1,\ a\:|\:bc\ \Rightarrow\ a|c$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a proof of the $\ $ GCD $\ $ &lt;em&gt;distributive law&lt;/em&gt; $\rm\ \ (ac,bc) = (a,b)\:c\ \ $ see my &lt;a href=&quot;http://math.stackexchange.com/questions/19830/if-gcd-a-b-1-then-is-it-true-to-say-gcd-ac-bc-c/19842#19842&quot;&gt;post here.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More generally it is simply the special case $\rm\ a\:|\:bc\ $ of $\rm\ (a,bc) = (a,(a,b)\:c)\:,\:$ which holds true for both GCDs and ideals. As you can see these basic properties are all intimately connected, so it's useful to master these properties to become proficient with GCD arithmetic. For another example of applying these properties to obtain simple proofs see this proof of the &lt;a href=&quot;http://math.stackexchange.com/questions/10400/comaximal-ideals-in-a-commutative-ring/10416#10416&quot;&gt;Freshman's Dream&lt;/a&gt; $\rm\ (A,B)^n = (A^n,B^n)\ $ for GCDs and invertible ideals.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2011-02-02T19:32:20.160" LastActivityDate="2011-02-02T19:32:20.160" />
  <row Id="20113" PostTypeId="1" AcceptedAnswerId="20114" CreationDate="2011-02-02T18:56:35.870" Score="7" ViewCount="134" Body="&lt;p&gt;For some reason, I can't find a reference for $\pi_i GL(n,\mathbb C)$ nor can I figure what they are.  For most Lie groups, you can get a nice fibration and use the long exact sequence in homotopy to inductively compute the homotopy groups (e.g. the fibration $SO(n-1) \to SO(n) \to S^{n-1}$).  However, I can't think of a nice fibration; $GL(n)$ acts transitively on $\mathbb C^n$ but I don't know a nice description for the stabilizer subgroup.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is motivated by understanding the statement that $GL(n)/GL(k)$ is $k-1$ connected (for the real and complex cases), so if there's an easy explanation for that without appealing to $\pi_1 GL(n)$, then that would also be appreciated.&lt;/p&gt;&#xA;" OwnerUserId="9" LastActivityDate="2011-02-02T19:09:42.347" Title="What is $\pi_i(GL(n))$?" Tags="&lt;algebraic-topology&gt;&lt;lie-groups&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="20114" PostTypeId="2" ParentId="20113" CreationDate="2011-02-02T19:09:42.347" Score="10" ViewCount="" Body="&lt;p&gt;There's a fibration &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$GL(n, \mathbb C) \to GL(n+1, \mathbb C) \to \mathbb C^{n+1} \setminus \{0\}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By Gram-Schmidt, this fibration is fibre homotopy-equivalent to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$U_n \to U_{n+1} \to S^{2n+1}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;given by only remembering the 1st vector in the matrix, just as in your $SO(n)$ example. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The stable homotopy groups of the unitary groups are known.  Google &quot;Bott Periodicity&quot;. The unstable groups for $U_n$, just like for $SO_n$, are only known in a range. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe these fibrations are discussed in Bredon's book, as well as May, among others.  This is example 4.55 in Section 4.2 of Hatcher's book. &lt;/p&gt;&#xA;" OwnerUserId="642" LastActivityDate="2011-02-02T19:09:42.347" CommentCount="4" />
  <row Id="20115" PostTypeId="1" AcceptedAnswerId="20139" CreationDate="2011-02-02T19:37:00.267" Score="6" ViewCount="94" Body="&lt;p&gt;Let $f$ be a holomorphic function on an open, connected set $\Omega\subset \mathbb{C}$ with $z_0\in \Omega$ a fixed point, and $\{f^n\}_{n\in \mathbb{N}}$ the sequence of iterates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to prove that if $|f&amp;#39;(z_0)|&amp;lt;1$ then there is a neighborhood of $z_0$ such that $\{f^n\}$ is normal in it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't really know what to do, because I don't know how to handle the iterates. I think the fact that $|f(z)| &amp;lt; |z-z_0| + |z_0| + \epsilon |z-z_0|$ might be useful to apply Montel's theorem, but I didn't get very far.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is not homework, it's a problem I'm trying to solve as preparation for a complex analysis exam.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also: I'd be grateful to have a geometrical interpretation of $|f&amp;#39;(z_0)|&amp;lt;1$ or of $|f&amp;#39;|$ in general. I don't really understand what does it mean for $|f&amp;#39;(z_0)|&amp;lt;1$, for example in Schwarz's lemma.&lt;/p&gt;&#xA;" OwnerUserId="2614" LastEditorUserId="2614" LastEditorDisplayName="" LastEditDate="2011-02-02T21:32:43.177" LastActivityDate="2011-04-03T23:39:15.770" Title="Fixed, attracting points are Fatou points" Tags="&lt;complex-analysis&gt;&lt;dynamical-systems&gt;" AnswerCount="1" />
  <row Id="20116" PostTypeId="2" ParentId="20096" CreationDate="2011-02-02T19:51:16.797" Score="3" ViewCount="" Body="&lt;p&gt;This is complicated.. I find it really hard to make my 4 years old niece understand the concept of quantity. She can count, but she still thinks the numbers are like names for the objects. I remove the &quot;one&quot; object and ask her to recount and she goes &quot;two, three four&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So maybe the best way to introduce math elements is with visual things, like jigsaw puzzles for kids, with big wooden pieces. Matching shapes and colors.&lt;/p&gt;&#xA;" OwnerUserId="6153" LastActivityDate="2011-02-02T19:51:16.797" CommunityOwnedDate="2011-02-02T19:51:16.797" />
  <row Id="20117" PostTypeId="2" ParentId="20007" CreationDate="2011-02-02T20:01:24.053" Score="0" ViewCount="" Body="&lt;p&gt;You don't need derivation to solve this equation.&#xA;$x \neq 0$, so $3y^2 + 8xy - 2x^2 = 0$ is equivalent to $3\frac{y}{x}^2 + 8\frac{y}{x} - 2 = 0$, which is simply a quadratic equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then you get $y = k_ix$, where $k_1$, $k_2$ are the roots of the equation. Note that it does imply that $\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{y}{x} = k_i$ :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: if either of $k_i$ is zero, the corresponding function will obviously not be a solution to the original equation.&lt;/p&gt;&#xA;" OwnerUserId="3793" LastActivityDate="2011-02-02T20:01:24.053" CommentCount="2" />
  <row Id="20118" PostTypeId="1" AcceptedAnswerId="20201" CreationDate="2011-02-02T20:02:46.017" Score="5" ViewCount="120" Body="&lt;p&gt;The ideals $I=(X,Y)$ and $J=(X^2+Y^2)$ in $\mathbb R[X,Y]$ are such that $V(I)=V(J)$ and their radicals aren't the same contradicting the  Nullstellensatz (in case it was true for arbitrary fields). However, this shouldn't be a surprise, if we look at their varieties in $\mathbb C$ we find that they aren't the same as the second has a pair of lines that were hidden.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is if it is true the other way around: Suppose we have two ideals $I$ and $J$ of $\mathbb K[X_1,\ldots,X_n]$ where $\mathbb K$ is an arbitrary field. Such that $V(I)=V(J)$ and such that ${\bf V}(I)={\bf V}(J)=V(I)$; where ${\bf V}$ means the variety in the affine space of dimension $n$ over $ \mathbb{\bar K}$, the algebraic closure of $\mathbb K$. That is, there are no additional points hidden in the algebraic closure. &#xA;Is it true then that $\sqrt I=\sqrt J$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My motivation for the question is a problem of primary descomposition, where we had to find one for $J=(X + Y − X^2 + XY − Y^2,X(X + Y − 1))$ in $\mathbb K[X,Y]$. We had already proved that $V(J)=\{(0,0),(1,0),(0,1)\}$ over $\mathbb{ A}_{\mathbb K}^2$ for any field $\mathbb K$. It made some computations easier in the case of algebraically closed fields because we had automatically that $\sqrt J=I$ where $I$ was the ideal of the three points $I=(X,Y)\cap(X-1,Y)\cap (X,Y-1)$.&lt;/p&gt;&#xA;" OwnerUserId="6276" LastActivityDate="2011-02-03T05:49:45.427" Title="Is Nullstellensatz true for arbitrary fields if there aren't hidden points?" Tags="&lt;commutative-algebra&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="20119" PostTypeId="1" CreationDate="2011-02-02T20:18:09.347" Score="2" ViewCount="32" Body="&lt;p&gt;I am already able to prove that $g|s$ assuming $x+y=s$ and $(x,y)=g$, but I am having some trouble showing that assuming $g|s$, there exists an $x$ and $y$ such that $x+y=s$ and $(x,y)=g$.  So far, I've started with saying that $g|0$ necessarily.  Therefore we also know that $g|(0x+sy)$.  I'd like to be able to set the values of $x$ and $y$ to something to show that there exist an $x$ and a $y$ that satisfy $x+y=s$ and $(x,y)=g$, but I'm not really sure where to go from here.  Can anyone offer any help?&lt;/p&gt;&#xA;" OwnerUserId="6548" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-02T20:24:27.157" LastActivityDate="2011-02-02T21:08:59.453" Title="If s and g&gt;0 are integers, how can I prove that x and y exist satisfying x+y=s and (x,y)=g if and only if g|s?" Tags="&lt;elementary-number-theory&gt;&lt;divisibility&gt;" AnswerCount="2" />
  <row Id="20120" PostTypeId="2" ParentId="20119" CreationDate="2011-02-02T20:23:11.270" Score="1" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT.&lt;/strong&gt; $(gn,g) = g(n,1)=g$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The converse follows because $a|b$ and $a|c$ implies $a|b+c$. &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-02T20:31:22.120" LastActivityDate="2011-02-02T20:31:22.120" CommentCount="5" />
  <row Id="20121" PostTypeId="1" CreationDate="2011-02-02T20:23:11.443" Score="3" ViewCount="68" Body="&lt;p&gt;I have been trying to read a lot of literature concerning the above subject but I've not found anything useful to help my case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose you're given a linear diophantine in $a_1,a_2,\ldots,a_k$ where $k\leq 10$, and we are asked to tell if $a_1 x_1 + a_2 x_2 + \cdots+a_kx_k = N$ has a non negative solution or not?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We are given many such queries, so I think regular Euclid method wouldn't be sufficient.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also,since I anyway brought the topic, could we utilise the calculation of the Frobenius Number of the equation to answer the above query.&lt;/p&gt;&#xA;" OwnerUserId="6536" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-02T20:27:58.780" LastActivityDate="2011-02-02T20:36:27.733" Title="Non Negative Solutions To a Linear Diophantine?" Tags="&lt;linear-algebra&gt;&lt;diophantine-equations&gt;" AnswerCount="1" />
  <row Id="20122" PostTypeId="1" AcceptedAnswerId="21229" CreationDate="2011-02-02T20:23:50.897" Score="4" ViewCount="532" Body="&lt;p&gt;Let us assume that we have an infinite line of people, and each person can either move forwards or remain at the same place. They move only one step at a time. (They are jumping from one position to the next if that position is empty). All people either move or remain still simultaneously.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let us further assume that a person in spot $x$ will move forward to spot $x+1$ if spots $x+1$,$x+2$ and $x-1$ are empty. For all other people, they move with a probability $\alpha$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can it be shown that &quot;eventually&quot;, there will be only two regions, one where the density is high and one where the density is low? (the fast lane, and a jammed lane)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A way to think about this is to think about the people being on a finite circular loop. (obviously then there are not infinite people in the line)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the purposes of clarification since many people are worried about the initial configuration, if it is assumed that it is sufficiently dense. (more than 0.33). We can have any initial configuration, it should not matter as we are interested in long term behavior.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT 2: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The objection raised below in the comments is deterministic. However, the question is about stochastic, thus we can say that $\alpha \not = 0$ or $1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example: &#xA;Let 1 be a occupied spot and a 0 an unoccupied spot. &#xA;then we can start with &lt;/p&gt;&#xA;&#xA;&lt;p&gt;...1110001000...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and then next we have either&lt;/p&gt;&#xA;&#xA;&lt;p&gt;...1110000100...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or &lt;/p&gt;&#xA;&#xA;&lt;p&gt;...1101000100...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;depending on whether the person advanced who had the probability $\alpha$ of advance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: C++ PROGRAM&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Program that I wrote. &#xA;&lt;a href=&quot;http://www.2shared.com/file/HO_BeE1G/test2.html&quot; rel=&quot;nofollow&quot;&gt;File&lt;/a&gt;&#xA;You may have to add a .exe to the end if you want to run it in Windows OS.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Same program with colors fixed and much longer time frame. &#xA;&lt;a href=&quot;http://www.2shared.com/file/I5rJmw3I/test2.html&quot; rel=&quot;nofollow&quot;&gt;File2&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4728" LastEditorUserId="4728" LastEditorDisplayName="" LastEditDate="2011-02-09T16:55:11.127" LastActivityDate="2011-03-11T19:32:25.217" Title="Infinite line of people" Tags="&lt;stochastic-processes&gt;" AnswerCount="2" CommentCount="27" FavoriteCount="2" />
  <row Id="20123" PostTypeId="1" CreationDate="2011-02-02T20:24:59.297" Score="0" ViewCount="69" Body="&lt;p&gt;I need to graph the error between a set of observed&#xA;values and an expected values.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I calculate the relative error for each point as:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;(observed - expect) / expected&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This error however bounces up and down because the &#xA;behaviour is sinusoidal. I want to make this error&#xA;always a positive quantity. Is there an error measure&#xA;like this? if so, what's its name?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have looked for error measures, but all I find is&#xA;absolute and relative, and I am interested in a &#xA;positive relative error.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any help appreciated,&#xA;Ted.&lt;/p&gt;&#xA;" OwnerUserId="5434" LastActivityDate="2011-02-02T22:57:19.487" Title="Calculating a Positive Relative error." Tags="&lt;statistics&gt;" AnswerCount="1" CommentCount="6" />
  <row Id="20124" PostTypeId="2" ParentId="20121" CreationDate="2011-02-02T20:36:27.733" Score="1" ViewCount="" Body="&lt;p&gt;One useful approach is to combine integer linear programming with lattice reduction, e.g. see Lichtblau: &lt;a href=&quot;http://library.wolfram.com/infocenter/Conferences/7533/JMM2008_ILP_talk.pdf&quot; rel=&quot;nofollow&quot;&gt;Integer Linear Programming, Frobenius Instances, &#xA;and Frobenius Numbers&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2011-02-02T20:36:27.733" CommentCount="1" />
  <row Id="20125" PostTypeId="2" ParentId="20119" CreationDate="2011-02-02T20:45:21.643" Score="1" ViewCount="" Body="&lt;p&gt;$\rm\ \ g = (x,s-x) = (x,s)\ \iff\ 1 = (x/g,\:s/g)\:,\ $ so choose $\rm x/g\ $ coprime to $\rm s/g\:,\ $ e.g. $\rm\ s/g + 1$&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2011-02-02T21:08:59.453" LastActivityDate="2011-02-02T21:08:59.453" />
  <row Id="20126" PostTypeId="1" AcceptedAnswerId="20309" CreationDate="2011-02-02T20:59:49.993" Score="9" ViewCount="271" Body="&lt;p&gt;I was wondering how to prove that $$\lim_{n\to \infty}\int_{1}^{n}\frac{1}{(x^{2}+1)^{n}}dx\sim \frac{1}{n\cdot 2^{n}}?.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This appears to be asymptotic to $\frac{1}{n2^{n}}$, but how to prove it?.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I checked with larger and larger values of n, and it does get closer and closer to &#xA;$$\frac{1}{n2^{n}}.$$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;i.e &#xA;$$\int_{1}^{10}\frac{1}{(x^{2}+1)^{10}}dx\approx .00009843725636$$&#xA;and &#xA;$$\frac{1}{10\cdot 2^{10}}\approx .00009765625.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The larger $n$, the closer they get. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried using parts to no avail. I also thought $$\sum_{k=0}^{\infty}\binom{-n}{k}x^{2k}=\frac{1}{(1+x^{2})^{n}}$$ may be useful in some manner.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anyone have a good idea as to how to prove this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="6358" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-02T21:07:37.957" LastActivityDate="2011-02-04T00:03:52.697" Title="How to prove limit of integral is $\frac{1}{n2^n}$?" Tags="&lt;improper-integrals&gt;" AnswerCount="4" CommentCount="7" />
  <row Id="20127" PostTypeId="1" CreationDate="2011-02-02T21:11:14.840" Score="5" ViewCount="160" Body="&lt;p&gt;Please suggest me how to compute the GCD of theese really big numbers :&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;GCD of $2^{120547564397}-1$ and $2^{356946681940}-1$ &lt;/li&gt;&#xA;&lt;li&gt;GCD of $2^n-1$ and $n!$ where $n=3^{19}$&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Thanks to &lt;a href=&quot;http://math.stackexchange.com/users/242/bill-dubuque&quot;&gt;Bill Dubuque&lt;/a&gt;'s &lt;a href=&quot;http://math.stackexchange.com/questions/20127/how-to-compute-these-gcds/20130#20130&quot;&gt;answer&lt;/a&gt; I understood that the first problem could be solved by the property that $gcd(f(m), f(n)) = f(gcd(m,n))$ if $ f(n) \equiv f(n-m) (mod\ f(m)),\ \ \ f(0)\ =\ 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any hints for the second one?&lt;/p&gt;&#xA;" OwnerUserId="3744" LastEditorUserId="3744" LastEditorDisplayName="" LastEditDate="2011-02-03T06:05:01.913" LastActivityDate="2011-02-03T11:16:43.887" Title="How to compute these GCD's ?" Tags="&lt;number-theory&gt;&lt;arithmetic&gt;&lt;contest-math&gt;" AnswerCount="3" />
  <row Id="20128" PostTypeId="2" ParentId="20118" CreationDate="2011-02-02T21:24:54.293" Score="1" ViewCount="" Body="&lt;p&gt;Yes, take $f \in \sqrt{I}$, then $f^n \in I \subset \bar{K}I$ so $f \in \sqrt{\bar{K}I}=\sqrt{\bar{K}J}$ so $f^m \in \bar{K}J$ for some $m$.&#xA;Now we have to prove that $\bar{K}J \cap K[X_1,\ldots,X_n]=J$, which is obvious once you have taken a basis $(e_i)_i$ of $\bar{K}$ over $K$ with $e_{i_0}=1$: $\bar{K}J = \oplus_i e_i J$ and $K[X_1,\ldots,X_n]=e_{i_0}K[X_1,\ldots,X_n]$.&lt;/p&gt;&#xA;" OwnerUserId="2660" LastActivityDate="2011-02-02T21:24:54.293" />
  <row Id="20129" PostTypeId="1" AcceptedAnswerId="20134" CreationDate="2011-02-02T21:26:47.097" Score="0" ViewCount="116" Body="&lt;p&gt;I was given this on a practice exam:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Decide whether each of the following expressions is valid or invalid. Justify your answers (i.e.,if invalid, give an interpretation for which the expression is false;if valid explain why the expression is true for all interpretations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(a) $[(\forall x)P(x) \vee (\exists x)Q(x)] \implies (\forall x)[P(x) \vee Q(x) ]$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(b) $[(\exists x)Q(x) \implies (\forall x)P(x)] \implies (\exists x)[P(x) \vee Q(x)]$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answers are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a. False: Let P(x)= &quot;x != x&quot; and Q(x) = &quot;x=1&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;b. False: Let &quot;There exists x Q(x)&quot; be false and &quot;There exists x P(x)&quot; be false.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't understand what he's doing. What am I not seeing?&lt;/p&gt;&#xA;" OwnerUserId="1040" LastActivityDate="2011-02-02T21:48:10.237" Title="What is the concept here" Tags="&lt;logic&gt;" AnswerCount="2" />
  <row Id="20130" PostTypeId="2" ParentId="20127" CreationDate="2011-02-02T21:26:48.050" Score="6" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\rm\ (2^a-1,\:2^b-1)\ =\ 2^{(a,b)}-1\ $ where $\rm\ (a,b) := gcd(a,b)\:.\ $ For proofs see &lt;a href=&quot;http://math.stackexchange.com/questions/11567/gcdbx-1-by-1-b-z-1-b-gcdx-y-z-1/11636#11636&quot;&gt;here,&lt;/a&gt; or &lt;a href=&quot;http://math.stackexchange.com/questions/11567/gcdbx-1-by-1-b-z-1-b-gcdx-y-z-1/11636#11636&quot;&gt;here&lt;/a&gt; - which has a polynomial analog.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2011-02-02T21:35:56.123" LastActivityDate="2011-02-02T21:35:56.123" CommentCount="3" />
  <row Id="20131" PostTypeId="2" ParentId="20129" CreationDate="2011-02-02T21:34:45.610" Score="0" ViewCount="" Body="&lt;p&gt;For each statement, we have $\text{TRUE} \implies \text{FALSE}$ which makes both statements invalid. In other words, the premises are true but the conclusion is false (hence both statements are invalid).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An argument is valid if and only if whenever its premises are true its conclusion must also be true. Consider the following argument:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose I have a black cat Timmy. If all cats are black, then Timmy is black. Timmy is black. Therefore all cats are black. This is an instance of the following argument form:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If $p$ then $q$&lt;/li&gt;&#xA;&lt;li&gt;$q$&lt;/li&gt;&#xA;&lt;li&gt;Therefore $p$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This is invalid.&lt;/p&gt;&#xA;" OwnerUserId="2548" LastEditorUserId="2548" LastEditorDisplayName="" LastEditDate="2011-02-02T21:45:48.907" LastActivityDate="2011-02-02T21:45:48.907" CommentCount="1" />
  <row Id="20132" PostTypeId="1" AcceptedAnswerId="20140" CreationDate="2011-02-02T21:35:14.950" Score="4" ViewCount="131" Body="&lt;p&gt;During self-study, I ran across the question of how many ways six numbers can be chosen from the numbers 1 - 49 without replacement, stipulating that no two of the numbers be consecutive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can obtain a simple lower bound by saying that, in the worst-case scenario, when you choose a particular number, there are now three numbers that cannot be chosen next.  For example, if I first pick 17, then I can't choose 16, 17, or 18 for the remaining choices.  This gives me the lower bound&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{49*46*43*40*37*34}{6!} = 6,773,770 \frac{8}{9}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is about 48% of ${}_{49}C_6 =  13,983,816$.  The real answer must be bigger (and an integer).  I haven't found a way to calculate it, though.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The original problem asked to show that the probability of having non-consecutive integers when you choose six is greater than 50%, so if the problem is complicated to count exactly, better approximations that show the answer is above 50% would also be appreciated. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, I can use a computer to do the counting, but I'm interested in learning what methods I'm missing.&lt;/p&gt;&#xA;" OwnerUserId="3305" LastActivityDate="2011-02-02T22:38:23.853" Title="How many ways can $r$ nonconsecutive integers be chosen from the first $n$ integers?" Tags="&lt;homework&gt;&lt;counting&gt;" AnswerCount="2" CommentCount="6" FavoriteCount="1" />
  <row Id="20133" PostTypeId="2" ParentId="20126" CreationDate="2011-02-02T21:38:58.457" Score="0" ViewCount="" Body="&lt;p&gt;This one should be a comment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As PEV said you could use contour integration to show that $\int_{0}^{1} \frac{1}{(1+x^2)^n} dx$ is of the same order as $\int_{1}^{n} \frac{1}{(1+x^2)^n} dx$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To evaluate it directly you could try the following.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Plug in $x = \tan(\theta)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The integral becomes $\displaystyle \int_{\pi/4}^{\tan^{-1}n} \frac{\sec^2 (\theta)}{\sec^{2n} (\theta)} d \theta = \int_{\pi/4}^{\tan^{-1}n} \cos^{2n-2}(\theta) d \theta \approx \int_{\pi/4}^{\pi/2} \cos^{2n-2}(\theta) d \theta = \frac{\text{(Hypergeometric function)}}{2^{n-1}(2n-1)}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The approximation tends to better and better asymptotically since $\cos(\theta)$ is bounded.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2011-02-02T22:54:21.123" LastActivityDate="2011-02-02T22:54:21.123" CommentCount="2" />
  <row Id="20134" PostTypeId="2" ParentId="20129" CreationDate="2011-02-02T21:48:10.237" Score="1" ViewCount="" Body="&lt;p&gt;Okay, let's start with part a.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$[(\forall x)P(x) \vee (\exists x)Q(x)] \implies (\forall x)[P(x) \vee Q(x) ]$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That $P(x)$ is true for all $x$ or that there exists a $x$ such that $Q(x)$ is true implies that for all $x$, at least one of $P(x)$ or $Q(x)$ is true.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the examples given, $P(x)$ is never true (since $x = x$ always) and $Q(x)$ is true whenever $x = 1$. So we see that the part before the implication is true (since there exists a $x$ such that $Q(x)$ is true). The part after the implication is of course false, since for all $x \neq 1$, neither $P(x)$ nor $Q(x)$ is true. So the statement is false. For the statement to be true, the part to the right should always be true whenever the part to the left is true (if $A \Rightarrow B$, then $B$ should always be true when $A$ is true).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Part b.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$[(\exists x)Q(x) \implies (\forall x)P(x)] \implies (\exists x)[P(x) \vee Q(x)]$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it is true that $Q(x)$ is true for at least one $x$ implies that $P(x)$ is true for all $x$, then there exists a $x$ such that at least one of $P(x)$ and $Q(x)$ is true.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the example given, $Q(x)$ and $P(x)$ are never true, for any $x$. Thus the first implication can be taken to be true, but the last part is never true. The statement is false, by the same reasoning as in part a.&lt;/p&gt;&#xA;" OwnerUserId="5966" LastActivityDate="2011-02-02T21:48:10.237" />
  <row Id="20135" PostTypeId="2" ParentId="20091" CreationDate="2011-02-02T21:57:10.147" Score="0" ViewCount="" Body="&lt;p&gt;Suppose I have no apples, and I decide to look through all apples I have, and give you two for every one I have.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then the total number of apples I give you is 0*2. But looking it at another way, I just start looking through all apples, and immediately stop because I found none; I haven't given you anything yet, and now I'm done. &lt;/p&gt;&#xA;" OwnerUserId="6551" LastActivityDate="2011-02-02T21:57:10.147" />
  <row Id="20136" PostTypeId="2" ParentId="20025" CreationDate="2011-02-02T21:57:12.423" Score="0" ViewCount="" Body="&lt;p&gt;This is true simply because in a PID, the equations for $\rm\:d\:$ to be a gcd have existential form, so a solution is necessarily inherited by any extension ring. In more detail, in any PID $\rm\:D\:$ we have $\rm\ d = gcd(a,b)\ \iff\ (d) = (a,b)\:.\ $ Thus a common divisor $\rm\:d\:$ of $\rm\:a,b\:$ is a gcd iff it is a $\rm\:D$-linear combination of $\rm\:a,b\:.\ $ Clearly if this is true in $\rm\:D\:$ then it remains true in any extension ring $\rm\:E\:$, since it is equivalent to saying the equations $\rm\ a\: x = b\: y = d = a\: u + b\: v\ $ are solvable for $\rm\ x,y,u,v\in D\:,\ $ and such roots in $\rm\:D\:$ necessarily persist as roots in any extension ring $\rm\:E\:.$&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2011-02-02T21:57:12.423" />
  <row Id="20137" PostTypeId="2" ParentId="20132" CreationDate="2011-02-02T22:09:31.447" Score="3" ViewCount="" Body="&lt;p&gt;In analogy to the Pascal triangle, define D(n,r) as the number of ways to select r items from n with no two consecutive.  You can either use 1 or not.  If you do use it, you can't use 2 but only need to pick r-1.  If you don't use it you have n-1 items to pick r from.  So D(n,r)=D(n-2,r-1)+D(n-1,r).  The base case is D(2r-1,r)=1For values like this you can use a spreadsheet.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-02-02T22:09:31.447" CommentCount="1" />
  <row Id="20138" PostTypeId="2" ParentId="20102" CreationDate="2011-02-02T22:25:41.680" Score="4" ViewCount="" Body="&lt;p&gt;This kind of thing can be done by symbolic computation packages like Maple, Mathematica, Maxima, etc.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you need a subroutine to do this that you'll call from a larger program, then the answer will depend a lot on the programming language that you're using and what kinds of licensing you're willing to work with.  &lt;/p&gt;&#xA;" OwnerUserId="6310" LastActivityDate="2011-02-02T22:25:41.680" CommentCount="3" />
  <row Id="20139" PostTypeId="2" ParentId="20115" CreationDate="2011-02-02T22:34:26.837" Score="2" ViewCount="" Body="&lt;p&gt;I think I got it. Correct me if I'm wrong.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There exists a neighborhood $U$ of $z_0$ such that $|f(z)-z_0|\leq \rho |z-z_0|$, where $|f&amp;#39;(z_0)|&amp;lt;\rho&amp;lt;1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then for all $z\in U$, $|f^n(z_0)-z_0| = |f(f^{n-1}(z))-z_0| \leq \rho |f^{n-1}(z)-z_0| \leq \dots \leq \rho^n |z-z_0|$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means the sequence $\{f^n\}$ converges uniformly to $z_0$ in $U$, because $\rho^n\to 0$ as $n\to \infty$. In particular, it converges uniformly in compacts of $U$, whence $\{f^n\}$ is normal in $U$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is still the last part of the question to be answered, though.&lt;/p&gt;&#xA;" OwnerUserId="2614" LastActivityDate="2011-02-02T22:34:26.837" CommentCount="1" />
  <row Id="20140" PostTypeId="2" ParentId="20132" CreationDate="2011-02-02T22:38:23.853" Score="3" ViewCount="" Body="&lt;p&gt;Another good way of solving problems like this is to set up a correspondence with a problem you already know how to solve.  In this case, imagine that you've got a sorted set of six non-consecutive numbers $a_1, a_2, \dots a_6$ between 1 and 49.  What does it look like?  Well, the first number $a_1$ is essentially arbitrary; it can't be greater than a certain value (&lt;em&gt;i.e.&lt;/em&gt; 39) or else there isn't 'room' for five more non-consecutive numbers above it, but we can ignore that for now.  The second number $a_2$ has to be greater than $a_1+1$ &amp;mdash; that's what it means to be nonconsecutive, after all &amp;mdash; so we know that $a_2-1$ (call this $b_2$) is greater than $a_1$.  Similarly, $a_3$ has to be greater than $a_2+1$, so $a_3-1$ is greater than $a_2$, and $a_3-2$ is greater than $a_2-1 = b_2$; we can call this $b_3$.  And so on, and so forth, until we define $b_6 = a_6-5$.  But this correspondence works both ways &amp;mdash; given the $b_n$ it's easy to get the $a_n$ &amp;mdash; and so we have a one-to-one correspondence between our non-consecutive sets of numbers and an ordinary combination (but with a different upper bound - can you see what it should be?).  It takes a while to build this sort of instinct, but learning how to spot these correspondences is the most basic tool for proving most combinatorial identities.&lt;/p&gt;&#xA;" OwnerUserId="785" LastActivityDate="2011-02-02T22:38:23.853" CommentCount="3" />
  <row Id="20143" PostTypeId="2" ParentId="20123" CreationDate="2011-02-02T22:57:19.487" Score="1" ViewCount="" Body="&lt;p&gt;I think it is sensible to use the following error measure:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{(O - E)^2}{E}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is related to the test statistic for &lt;a href=&quot;http://en.wikipedia.org/wiki/Pearson%27s_chi-square_test#Calculating_the_test-statistic&quot; rel=&quot;nofollow&quot;&gt;Pearson's $\chi^2$ test&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="3567" LastActivityDate="2011-02-02T22:57:19.487" />
  <row Id="20144" PostTypeId="2" ParentId="20126" CreationDate="2011-02-02T22:59:27.423" Score="7" ViewCount="" Body="&lt;p&gt;Divide $\int_{1}^{n}\frac{2^n}{(x^{2}+1)^{n}}dx$ between $1$ and $1+n^{-2/3}$ and between $1+n^{-2/3}$ and $n$.&#xA;The second part is less than $n \left( \frac{2}{(1+n^{-2/3})^2+1}\right)^n$ which is equivalent to $n \exp (n(-n^{-2/3}+o(n^{-2/3}))$ which can be neglected considering the equivalent we are going to find for the other part.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the first part, we use $\frac{2}{(1+u)^2+1}=1-u+O(u^2)$ as $u \rightarrow 0$ to evaluate $\int_0^{n^{-2/3}}\left(\frac{2}{(1+u)^2+1}\right)^n du$.&#xA;This way we get that $e^{-nu-Cnu^2} \leq \left(\frac{2}{(1+u)^2+1}\right)^n \leq e^{-nu+Cnu^2}$ for some constant $C \gt 0$ and for all $u \in [0,1]$.&#xA;So the first part falls between $e^{-Cn^{-1/3}} \int_0^{n^{-2/3}} e^{-nu} du$ and $e^{Cn^{-1/3}} \int_0^{n^{-2/3}} e^{-nu} du$, so is equivalent to $\int_0^{n^{-2/3}} e^{-nu} du = \frac{1}{n}(1-e^{-n^{1/3}}) \sim 1/n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$n^{-2/3}$ could be replaced by any $\epsilon_n$ such that $n \epsilon_n \rightarrow + \infty$ and $n \epsilon_n^2 \rightarrow 0$&lt;/p&gt;&#xA;" OwnerUserId="2660" LastActivityDate="2011-02-02T22:59:27.423" CommentCount="2" />
  <row Id="20145" PostTypeId="2" ParentId="20013" CreationDate="2011-02-02T23:03:53.017" Score="1" ViewCount="" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;If you dig straight down into the Earth, where do you end up?  What if we take into account the Earth's rotation?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A silly example of a G-equivariant map f from X to Y takes X=Y={ (x,y) : x,y in &lt;strong&gt;R&lt;/strong&gt; s.t. x&lt;sup&gt;2&lt;/sup&gt;+y&lt;sup&gt;2&lt;/sup&gt; = 1 } to be the unit circle, and G to be the full group of rotations (and reflections, if you'd like) of X.  Taking f(x,y) = (−x,−y) to be the &quot;antipodal&quot; or opposite point of x turns out to be G-equivariant, since the &quot;opposite&quot; of a point is defined intrinsically.  This f is not invariant, but is equivariant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can do the same thing for any ball or sphere, and G can be any subgroup of the isometry group.  For instance take X=Y=S&lt;sup&gt;2&lt;/sup&gt; to be the sphere, and take G to be a group of rotations with a common axis of rotation.  If you want to know where you get after drilling straight down into the earth, you could ask before the Earth rotates or after, but it wouldn't really matter, since the opposite point rotates at the same rate.&lt;/p&gt;&#xA;" OwnerUserId="583" LastEditorUserId="583" LastEditorDisplayName="" LastEditDate="2011-02-04T17:40:59.943" LastActivityDate="2011-02-04T17:40:59.943" CommentCount="2" />
  <row Id="20146" PostTypeId="1" AcceptedAnswerId="20153" CreationDate="2011-02-02T23:05:56.197" Score="4" ViewCount="76" Body="&lt;p&gt;Consider a function $f$ over $\Re^n$ to $\Re$.  Suppose it is true that for every affine subspace with dimension strictly lesser than $n$ the function $f$ is concave.  Is the function $f$ concave over $\Re^n$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To start with, suppose $f: \Re^2 \to \Re$, such that for all $x \in \Re$, $f(x, .)$ is concave and for all $y \in \Re$, $f(., y)$ is concave.  Is it possible that $f$ is not concave?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT:&#xA;Not sure about etiquette here, so I'll leave my original post above.  I failed to state the question carefully.  Let me give it another try:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose we have real-valued function $f$ on $\Re^n$, and we know the following about the function:&#xA;Pick a dimension and call it dimension $k$.  Next, pick a real number $x$.  Now, suppose it is true that $f(., ..., x, ., ...)$ is strictly concave, for any $x$, and for any dimension $k$.  Is it true that $f$ is concave?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Theo's response below includes an example $f(x, y), = xy$ that I construe as a counterexample if the qualifier of &quot;strict&quot; were not included.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For what it's worth, this is not a homework question, but instead one born of curiosity.&lt;/p&gt;&#xA;" OwnerUserId="2404" LastEditorUserId="2404" LastEditorDisplayName="" LastEditDate="2011-02-03T00:39:43.253" LastActivityDate="2011-02-03T01:13:20.967" Title="Is concavity of a real-valued function on a Euclidean space implied by concavity of its restriction to every lower dimensional affine subspace?" Tags="&lt;real-analysis&gt;&lt;functions&gt;&lt;euclidean-geometry&gt;&lt;convex-analysis&gt;" AnswerCount="3" />
  <row Id="20147" PostTypeId="1" CreationDate="2011-02-02T23:08:14.677" Score="2" ViewCount="58" Body="&lt;p&gt;I'm reading a book (&lt;em&gt;Yangians and Classical Lie Algebras&lt;/em&gt; by Molev) which regularly uses (what appear to me to be) clever tricks with formal power series to encapsulate lots of relations.  For instance, if we let $S_n$ act on $(\mathbb{C}^N)^{\otimes n}$ by permuting tensor components, so that e.g. $P_{(1 2)} (a \otimes b \otimes c) = b \otimes a \otimes c$, then working in $({\rm End} \mathbb{C}^N)^{\otimes 3}(u, v)$ we have the identity&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\left(u - P_{(1 2)}\right)\left(u + v - P_{(1 3)}\right)\left(v - P_{(2 3)}\right) = \left(v - P_{(2 3)}\right)\left(u + v - P_{(1 3)}\right)\left(u - P_{(1 2)}\right)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is used to motivate the definition of an operator $R_{(j k)}(u) = 1 - P_{(j k)} u^{-1}$, the Yang R-matrix, which is then used to express an enormous family of relations on an algebra by multiplying by a matrix of formal power series.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course it's straightforward to verify that the above expression holds if we multiply out the terms.  That said, it seems considerably less straightforward to me how one would start from $S_3$ and end up at the equation above.  Is this just a marvelous ad-hoc construction, or does it belong to some class of examples?&lt;/p&gt;&#xA;" OwnerUserId="3296" LastActivityDate="2011-02-02T23:08:14.677" Title="Giving relations via formal power series" Tags="&lt;quantum-groups&gt;" CommentCount="3" FavoriteCount="1" />
  <row Id="20148" PostTypeId="1" AcceptedAnswerId="20150" CreationDate="2011-02-02T23:09:00.200" Score="2" ViewCount="118" Body="&lt;p&gt;Forgive me if this is elementary, but my analysis is quite rusty and I'm struggling to get back up to speed. Given a convergent series $\sum_{i=1}^\infty|x_i|^p$, does it follow that $(\sum_{i=1}^\infty|x_i|^p)^\frac{1}{p}$ also converges? If so, how do they relate, i.e. is one always greater than (or equal) to the other? &lt;/p&gt;&#xA;" OwnerUserId="4518" LastActivityDate="2011-02-03T03:02:26.607" Title="How can one determine if a convergent series raised to a power is also convergent?" Tags="&lt;real-analysis&gt;&lt;convergence&gt;" AnswerCount="4" />
  <row Id="20149" PostTypeId="2" ParentId="20096" CreationDate="2011-02-02T23:12:13.893" Score="3" ViewCount="" Body="&lt;p&gt;It is never too early to introduce count-by. Count by 1/2: 1/2, 1. Count by 1/4: 1/4, 1/2, 3/4, 1. Count by eggs: 1/12, 1/6, 1/4, 1/3, 5/12, 1/2, 7/12, 2/3, 3/4, 5/6, 11/12, one dozen. Count by 2s: 2,4,6,8,10, count by 3s: 3,6,9,12, etc. While pushing on the swing count by 1s to 10, 2s to 20, threes to 30, 4s to 40, 5s to 50, 6s to 60, 7s to 70, etc, until you get to 100. You do the counting, and teach the sequences. You want to get those hard wired like the zyxwvutsrqponmlkjihgfedcba sequence or its reverse. Always ask your child what is the derivative of a constant when you are putting him/her on your shoulders. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have square tiles in the kitchen teach about area. Play with blocks and build. Tie knots in rope and especially do math as a play activity.&lt;/p&gt;&#xA;" OwnerUserId="722" LastActivityDate="2011-02-02T23:12:13.893" CommunityOwnedDate="2011-02-02T23:12:13.893" />
  <row Id="20150" PostTypeId="2" ParentId="20148" CreationDate="2011-02-02T23:15:49.033" Score="3" ViewCount="" Body="&lt;p&gt;The sum is just a number (if it converges).  Call it S.  If S&gt;0 and $p \ne 0$ you can calculate $S^{(1/p)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe what you want to ask if if $\sum_{i=1}^\infty|x_i|$ converges does $\sum_{i=1}^\infty|x_i|^p$ converge?  It will if $p \ge 1$ as the $x_i$ are going to zero and raising them to a power greater than 1 will decrease them.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-02-02T23:15:49.033" CommentCount="1" />
  <row Id="20151" PostTypeId="2" ParentId="20148" CreationDate="2011-02-02T23:16:59.077" Score="1" ViewCount="" Body="&lt;p&gt;Denote $S=\sum_{i=1}^{\infty}|x_i|^p$. When does $S^{\frac{1}{p}}$ exist?&lt;/p&gt;&#xA;" OwnerUserId="752" LastActivityDate="2011-02-02T23:16:59.077" CommentCount="2" />
  <row Id="20152" PostTypeId="2" ParentId="20148" CreationDate="2011-02-02T23:18:14.663" Score="2" ViewCount="" Body="&lt;p&gt;You're just taking the $p$th root of the real number $s = \sum_{n=1}^{\infty} |x_{n}|^{p}$. So if $s \gt 1$ then $s \gt s^{\frac{1}{p}}$ and if $s \lt 1$ then $s \lt s^{\frac{1}{p}}$ finally, equality holds if $s = 1$.&lt;/p&gt;&#xA;" OwnerUserId="5363" LastActivityDate="2011-02-02T23:18:14.663" CommentCount="2" />
  <row Id="20153" PostTypeId="2" ParentId="20146" CreationDate="2011-02-02T23:33:03.590" Score="7" ViewCount="" Body="&lt;p&gt;The definition of concavity is $f((1-t)x + ty) \geq (1-t)f(x) + tf(y)$ for all $x, y$ and all $0 \leq t \leq 1$. This a condition on line segments (hence only involving points lying in $1$-dimensional affine subspaces), so your first question has the tautological answer &lt;strong&gt;yes&lt;/strong&gt; for $n \geq 2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second question has a negative answer. Consider $f(x,y) = xy$. Then $f(x,\cdot)$ and $f(\cdot,y)$ are linear, hence concave, but the function itself is not concave ($f(x,x) = x^{2}$ is convex and $f(x,-x) = -x^{2}$ is concave).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $C^{2}$-functions the following are equivalent:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$f$ is concave;&lt;/li&gt;&#xA;&lt;li&gt;the Hessian of $f$ is negative semi-definite at each point.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In contrast, the condition that $f(\cdot,y)$ and $f(x,\cdot)$ be concave just says that $f_{xx}(\cdot,y), f_{yy}(x,\cdot) \leq 0$, while concavity also involves the mixed derivatives $f_{xy} = f_{yx}$. In detail, concavity is equivalent to $f_{xx} + f_{yy} \leq 0$ and $f_{xx}f_{yy} - (f_{xy})^{2} \geq 0$ everywhere on $\mathbb{R}^{2}$ (a symmetric matrix is negative definite if and only if its eigenvalues are negative and for $2 \times 2$-matrices this is equivalent to negative trace and &lt;em&gt;positive&lt;/em&gt; determinant.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;. To make the example from before strictly concave in the $x$ and $y$-directions, just consider $f(x,y) - e^{x + y}$ or Jonas' example.&lt;/p&gt;&#xA;" OwnerUserId="5363" LastEditorUserId="5363" LastEditorDisplayName="" LastEditDate="2011-02-03T01:01:30.237" LastActivityDate="2011-02-03T01:01:30.237" CommentCount="2" />
  <row Id="20154" PostTypeId="2" ParentId="20146" CreationDate="2011-02-02T23:40:37.513" Score="0" ViewCount="" Body="&lt;p&gt;This is very likely to be a homework question.  Because of that I won't give you a solution to the homework problem but I will give you a suggestion.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You actually have two questions here, and it isn't clear that solving the simplified version (concave in x and y) will actually help you to solve the full version of the problem.  You'll want to be careful about that.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the simpler version of the problem, recall that a twice continuously differentiable function is concave if the Hessian matrix is negative definitede, concave in x if $f_{xx}$ is negative, and concave in $y$ if $f_{yy}$ is negative.  Can you construct a function (perhaps a quadratic function so that the Hessian is very simple) that has a $f_{xx}$ negative and $f_{yy}$ negative but does not have a negative definite Hessian?  Can you show that the function you've found is actually nonconvex?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the full version of the question, you might consider the definition of concavity.  How does this definition relate to &quot;every affine subspace&quot;?   &lt;/p&gt;&#xA;" OwnerUserId="6310" LastActivityDate="2011-02-02T23:40:37.513" CommentCount="1" />
  <row Id="20155" PostTypeId="1" AcceptedAnswerId="20215" CreationDate="2011-02-02T23:47:29.457" Score="1" ViewCount="105" Body="&lt;p&gt;After taking measurements on the deflection of stars during the 1919 eclipse in Borneo...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It was found that: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;n = 15&lt;br&gt;&#xA;Mean = 1.4824 angular units (AU)&lt;br&gt;&#xA;SD = 1.349237 AU&lt;br&gt;&#xA;SE = .348371 AU&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A) Test the hypothesis that Einstein was correct (that the offset was 1.70 angular units as predicted by his theory). If Isaac Newton was correct, the offset would be 0.85 angular units. State your conclusion and your observations about the data.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;B) Determine how many observations would be needed to be 90% sure that you would be able to reject each hypothesis if the other one was correct (using the above variability).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I solved A but I'm not quite sure what to do with B. Tried solving for t value given SE and Mean, SE and 1.7, and SE and 1.3(half way between two hypotheses) but none worked and I'm not sure if any of those are appropriate. Also think it might have something to do with power curves, but not really sure about that either.&lt;/p&gt;&#xA;" OwnerUserId="6555" LastActivityDate="2011-02-03T12:09:49.973" Title="Hypothesis Testing - Newton or Einstein?" Tags="&lt;statistics&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" />
  <row Id="20156" PostTypeId="1" AcceptedAnswerId="20162" CreationDate="2011-02-02T23:51:23.907" Score="3" ViewCount="96" Body="&lt;p&gt;I'm talking bout this phenomenon (The third one):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/n1tNv.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to know if with sound waves, happen the same as with (water) waves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, If two speakers were emitting the same sound, would there technically be a point where you wouldn't hear. (I guess this would only happen if you heard with only one ear).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this true for sound wave too?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does this happen in real life, say, in car?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance!&lt;/p&gt;&#xA;" OwnerUserId="4256" LastActivityDate="2011-02-03T00:43:38.723" Title="Is there a point of no hearing if two speakers are emitting the same sounds - Wave Superposition" Tags="&lt;physics&gt;&lt;mathematical-physics&gt;" AnswerCount="2" />
  <row Id="20157" PostTypeId="1" AcceptedAnswerId="20193" CreationDate="2011-02-03T00:02:37.950" Score="7" ViewCount="186" Body="&lt;p&gt;I'm a computer science student, I'm not a mathematician, I don't know anything about number or group theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm looking at RSA, and I want to understand it.&#xA;I know what Fermats's little theorem and Euler's totient function are, and this is what I've got:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$p, q$ primes, $n=pq$.&lt;/li&gt;&#xA;&lt;li&gt;$d$ relatively prime to of $\varphi(n)=(p-1)(q-1)$.&lt;/li&gt;&#xA;&lt;li&gt;$e$ so that $ed \pmod{\varphi(n)}= 1$. &lt;/li&gt;&#xA;&lt;li&gt;Hence $x^{ed} \pmod{n} = x$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;For encryption I have to use $c=w^e \pmod n$, and for decryption $w=c^d \pmod n$. But I don't understand what has happened beyond, and how those formulas are linked.&lt;/p&gt;&#xA;" OwnerUserId="5046" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2011-02-03T00:38:08.197" LastActivityDate="2011-02-03T14:08:29.700" Title="RSA in plain English" Tags="&lt;number-theory&gt;&lt;prime-numbers&gt;&lt;cryptography&gt;" AnswerCount="3" CommentCount="4" FavoriteCount="5" />
  <row Id="20158" PostTypeId="2" ParentId="8501" CreationDate="2011-02-03T00:05:06.963" Score="5" ViewCount="" Body="&lt;p&gt;New York City, in addition to a vibrant city life and diverse professional opportunities, has also a vibrant academic life, with &lt;a href=&quot;http://en.wikipedia.org/wiki/List_of_colleges_and_universities_in_New_York&quot; rel=&quot;nofollow&quot;&gt;an enormous number of colleges and universities&lt;/a&gt;, rivaling even Boston. The large number of math seminars and workshops at the &lt;a href=&quot;http://math.gc.cuny.edu&quot; rel=&quot;nofollow&quot;&gt;CUNY Graduate Center&lt;/a&gt;, for example, are usually quite welcoming of outsiders, and the typical pattern there has high-level specialized seminars in whatever subject populated by faculty and advanced graduate students from diverse regional institutions. I am certain that you would find something suitable, challenging and rewarding.&lt;/p&gt;&#xA;" OwnerUserId="413" LastActivityDate="2011-02-03T00:05:06.963" />
  <row Id="20159" PostTypeId="2" ParentId="20156" CreationDate="2011-02-03T00:14:03.630" Score="1" ViewCount="" Body="&lt;p&gt;Yes.  Take a look at this: &lt;a href=&quot;http://en.wikipedia.org/wiki/Noise-cancelling_headphones&quot; rel=&quot;nofollow&quot;&gt;Noise Cancelling Headphones&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4266" LastActivityDate="2011-02-03T00:14:03.630" CommentCount="7" />
  <row Id="20160" PostTypeId="2" ParentId="20157" CreationDate="2011-02-03T00:34:13.410" Score="3" ViewCount="" Body="&lt;p&gt;The RSA cryptosystem is composed of three steps:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Key generation: Each user $u$ generates two primes $p,q$, calculates $n = pq$ and $\varphi(n) = (p-1)(q-1)$, picks a random $e$ (which must be relatively prime to $\varphi(n)$) and calculates $d = e^{-1} \pmod{\varphi(n)}$. The user publishes the &lt;i&gt;public key&lt;/i&gt; $pub_u = (n,e)$ and keeps for herself the &lt;i&gt;private key&lt;/i&gt; $pri_u = d$.&lt;/li&gt;&#xA;&lt;li&gt;Encryption: If you want send $u$ a (short) message $m$ so that only she can understand it, you send $c=m^e \pmod{n}$; you can do that since $(n,e)$ are public.&lt;/li&gt;&#xA;&lt;li&gt;Decryption: Upon receipt of $c$, user $u$ can restore the original message $m$ by calculating $m=c^d \pmod{n}$ (this only works if $m &amp;lt; n$).&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;People hope that it's very difficult to break the cryptosystem, that is to decrypt messages unless you're $u$. However, that depends on how you use the system - you can use it wrongly. For example, if $m$ is small and an attacker knows that, he can recover $m$; other attacks are more sophisticated.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2011-02-03T00:34:13.410" />
  <row Id="20161" PostTypeId="2" ParentId="20157" CreationDate="2011-02-03T00:40:13.250" Score="3" ViewCount="" Body="&lt;p&gt;You are almost there.  The only math that you and Yuval didn't mention is that because $ed = 1 \pmod{\varphi(n)}$,   $(w^e)^d \pmod n=w^{ed} \pmod n = w$ so the encryption/decryption works and this is from Fermat's little theorem.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-02-03T00:40:13.250" />
  <row Id="20162" PostTypeId="2" ParentId="20156" CreationDate="2011-02-03T00:43:38.723" Score="4" ViewCount="" Body="&lt;p&gt;Since you've gotten one 'yes' answer, let me give you a 'no' answer. :-)  More precisely, there are two different questions here: Brian's 'yes' answer handles the question 'for any sound wave, is there some other wave that will cancel that wave out?'  The question you asked, on the other hand, was closer to 'for any sound wave, is there some shift of that wave that will cancel the original out' - in other words, does every wave have a set of peaks and troughs that can be shifted in time to cancel it out?  In general, the answer is no: there are wave functions for which no shift can possibly cancel out the function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What does it mean to be self-cancellable?  It means there's some shift of your wavefunction $f(t)$ - say, by $t_0$ - such that the sum of the two equals zero: $f(t)+f(t-t_0) = 0$.  But then, this means that $f(t) = -f(t-t_0)$, and $f(t-t_0) = -f(t-t0-t0) = -f(t-2t_0)$ - so $f(t) = -f(t-t_0) = -(-f(t-2t_0)) = f(t-2t_0)$ - in other words, $f(t)$ is periodic with period $2t_0$.  So since every self-cancellable wavefunction is periodic, no non-periodic function can be self-cancellable; as a simple example, any glissando (rising tone) won't be self-cancellable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, being periodic isn't sufficient to be self-cancellable, either.  The simplest example is probably the &lt;a href=&quot;http://en.wikipedia.org/wiki/Sawtooth_wave&quot; rel=&quot;nofollow&quot;&gt;Sawtooth wave&lt;/a&gt;: $f(t) = t - \lfloor t\rfloor$.  Note that this function has derivative (i.e. slope) 1 everywhere its derivative is defined, and so every shift of it (by, say, $t_0$) also has derivative 1 everywhere its derivative is defined.  But then $f(t) + f(t-t_0)$ has derivative 2 everywhere it's defined, whereas to cancel out they would need to add to the zero function - whose derivative is also zero.&lt;/p&gt;&#xA;" OwnerUserId="785" LastActivityDate="2011-02-03T00:43:38.723" CommentCount="6" />
  <row Id="20163" PostTypeId="2" ParentId="19776" CreationDate="2011-02-03T00:52:19.887" Score="2" ViewCount="" Body="&lt;p&gt;Consider the following experiment. There are $n$ boxes, with two identical coins each. We pick some $k \in \{0,\ldots,n\}$, pick $k$ coins from different boxes $A$, and then again pick $k$ coins from different boxes $B$. The left-hand side is exactly the generating function of all possible outcomes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is another way of performing the same experiment. First choose a number $j &amp;lt; n/2$; this is going to be the size of $|A \setminus B| = |B \setminus A|$. Instead of choosing the elements in $(A \setminus B) \cup (B \setminus A)$, we will choose which elements are in $A \cap B$ and which in $\overline{A \cup B}$; we have to decide on $n-2j$ elements. If we start with assuming that each element is picked once (this is the factor $x_1\cdots x_n$), then putting an element $i$ in $A \cap B$ is like multiplying by $x_i$, and putting it in $\overline{A \cup B}$ is like multiplying by $1/x_i$. We are left with $2j$ elements, $j$ of which are in $A \setminus B$ and the rest in $B \setminus A$. There are $\binom{2j}{j}$ choices for that. Writing everything out, we get the right-hand side.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2011-02-03T00:52:19.887" />
  <row Id="20164" PostTypeId="2" ParentId="20146" CreationDate="2011-02-03T00:59:14.370" Score="3" ViewCount="" Body="&lt;p&gt;The function $f(x,y)=-(x^2+1)(y^2+1)$ is a counterexample for the new question, being strictly concave in $x$ for each fixed $y$ and vice versa, but not being concave.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More generally, if $g$ is a twice differentiable, positive, strictly convex function on $\mathbb{R}$ that is not log convex, then $f(x,y)=-g(x)g(y)$ is a counterexample.  The log convexity criterion for $g(x)g(y)$ to be convex came up at &lt;a href=&quot;http://math.stackexchange.com/questions/15707/for-what-functions-fx-is-fxfy-convex&quot;&gt;this other question&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2011-02-03T01:13:20.967" LastActivityDate="2011-02-03T01:13:20.967" CommentCount="1" />
  <row Id="20165" PostTypeId="1" AcceptedAnswerId="20169" CreationDate="2011-02-03T01:09:35.727" Score="1" ViewCount="81" Body="&lt;p&gt;Let $k$ be a unital commutative ring, and A be a $k$-module.&#xA;Is there a homomorphism $f: A \otimes_k A \to \bigwedge^2 A$ such that $f(a \otimes a) \neq 0$ for some $a \in A$? I can take a hint :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Motivation&lt;/strong&gt;:&#xA;In his textbook on Lie algebra, Serre defines a Lie algebra as a k-module with a homomorphism $A \otimes_k A \to A$ that factors through $\Lambda^2 A$. He then goes on to explain that it means that [x,x]=0 for all $x \in A$.&lt;/p&gt;&#xA;" OwnerUserId="3793" LastEditorUserId="627" LastEditorDisplayName="" LastEditDate="2011-02-03T15:35:43.933" LastActivityDate="2011-02-03T15:35:43.933" Title="$A \otimes_k A \to \bigwedge^2 A$ homs" Tags="&lt;abstract-algebra&gt;&lt;lie-algebra&gt;" AnswerCount="2" CommentCount="5" />
  <row Id="20166" PostTypeId="2" ParentId="19776" CreationDate="2011-02-03T01:10:45.820" Score="1" ViewCount="" Body="&lt;p&gt;Let $L$ be a a multiset whose elements are elements of $\{1,\dots,n\}$. Then the coefficient of $x^L$ (which is the product of the $x_i$ with $i\in L$, taking into account multiplicities) in $\sum_ke_k^2$ is the cardinal of the set $$X_L=\{(I,J)\in\mathcal P(\mathbf n)\times\mathcal P(\mathbf n):|I|=|J|, I\cup J=L\}.$$ Here $\mathbf n$ is the set $\{1,\dots,n\}$, $\mathcal P(\mathbf n)$ is the set of subsets of $\mathbf n$ (not multisets), and $I\cup J$ is the union taking into account multiplicities, whose result is a multiset.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Count the elements of the set $X_L$ in another way to obtain your equality.&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2011-02-03T01:10:45.820" CommentCount="1" />
  <row Id="20167" PostTypeId="1" AcceptedAnswerId="20173" CreationDate="2011-02-03T01:12:31.390" Score="2" ViewCount="248" Body="&lt;p&gt;I'm having a hard time understanding what exact processes I have to take to represent a function as a power-series.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand that $\frac{1}{1-x} = 1 + x + x^2 + x^3 + ... = \sum_{0}^\infty x^n$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, in words, $\frac{1}{1-x}$ can be represented as the power series $\sum_{0}^\infty x^n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Jumping to the problem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Represent $f(x) = ln(5-x)$ as a power series.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ln(5-x) = -\int\frac{1}{5-x}dx = -\frac{1}{5}\int\frac{1}{1-\frac{x}{5}} = &#xD;&#xA;&#xD;&#xA;-\frac{1}{5}\int[\sum_{0}^\infty (\frac{x}{5})^n]dx = C - \frac{1}{5}\sum_{0}^\infty \frac{x^{n+1}}{5^n(n+1)} = C - \sum_{0}^\infty \frac{x^n}{n5^n}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this problem, it's difficult to get ln(5-x) in some sort of geometric form to turn into a power series...but we can differentiate it to get it into the convenient geometric form $\frac{1}{5-x}$. The constant is factored out and put into power series form with $r = (\frac{x}{5})^n$. Then it is integrated. But because we're integrating a derivative of $ln(5-x)$ in the form of a power series, we're actually getting ln(5-x) in the form of a power series...which is what we wanted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is this correct (btw, what happened to that 1/5 in that last step?)? This is one reason why we integrate/differentiate power series, right? To get some function in the form of power series where it can be difficult to do so?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another problem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This problem's solution confused me a bit (the original instruction says to use a power series to approximate the definite integral to six decimal places the limits of integration being 0 to 0.2, but I don't care about that right now):&#xA;$\int xarctan(3x) = \int x \sum_{0}^\infty (-1)^n\frac{(3x)^{2n+1}}{2n+1} dx = \int\sum_{0}^\infty(-1)^n\frac{(3)^{2n+1}x^{2n+2}}{2n+1}dx = C + \sum_{0}^\infty(-1)^n\frac{(3)^{2n+1}x^{2n+3}}{(2n+1)(2n+3)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case, we have a function $xarctan(3x)$ we want to turn into a power series (so we can approximate). (After skipping alot of steps) The arctan is put into geometric form, and then into a power series. The power series, arctan(3x) is then integrated...with the constant x factored out for the moment. Then it is factored in and then the x itself is integrated, giving us $C + ...power series$ so essentially, we now finally have xarctan(3x) in power series form. Is this correct?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just wanted a confirmation that I was seeing this correctly cause I had a hard time following in class what exactly the point of integrating/differentiating power series was.&lt;/p&gt;&#xA;" OwnerUserId="3514" LastEditorUserId="3971" LastEditorDisplayName="" LastEditDate="2011-02-03T08:04:52.313" LastActivityDate="2011-02-03T08:04:52.313" Title="Understanding and Integrating/Differentiating Power Series" Tags="&lt;calculus&gt;&lt;homework&gt;&lt;power-series&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="20168" PostTypeId="2" ParentId="20165" CreationDate="2011-02-03T01:18:04.267" Score="0" ViewCount="" Body="&lt;p&gt;&lt;b&gt;This answer is no loger relevant, since the OP now asked (in a comment...) what he had meant to askm which happens not to be what he asked :)&lt;/b&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There obviously exist examples of $k$s and $A$s such that such homomorphisms do exist (take, for example, $A$ free of rank at least $2$ over $k$...)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now it is not true that such homomorphisms exist for all choices of $k$ and $A$ (take, for example, $A$ free of rank $1$...)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not sure neither of these facts is what you intended to ask, though.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Later:&lt;/strong&gt; one thing you might have intended to ask if there is such a homomorphism depending &lt;em&gt;naturally&lt;/em&gt; on $A$. The answer to that is also no, because of my second example above.&lt;/p&gt;&#xA;" OwnerUserId="274" LastEditorUserId="274" LastEditorDisplayName="" LastEditDate="2011-02-03T02:24:10.160" LastActivityDate="2011-02-03T02:24:10.160" />
  <row Id="20169" PostTypeId="2" ParentId="20165" CreationDate="2011-02-03T01:29:21.633" Score="2" ViewCount="" Body="&lt;p&gt;What Serre means is the following.  You probably know that the tangent square $A \otimes_k A$ is universal for $k$-bilinear maps out of $A \times A$.  The exterior square $\Lambda^2 A$ is universal for &lt;em&gt;alternating&lt;/em&gt; $k$-bilinear maps out of $A \times A$.  Any such map is first of all a $k$-bilinear map, so it factors through $A \otimes_k A$.  But the alternating property causes it to further factor through the &lt;em&gt;natural&lt;/em&gt; quotient $A \otimes_k A \to \Lambda^2 A$.  Serre has in mind a specific map here.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2011-02-03T01:29:21.633" />
  <row Id="20170" PostTypeId="1" AcceptedAnswerId="20171" CreationDate="2011-02-03T01:45:43.360" Score="5" ViewCount="187" Body="&lt;p&gt;is there a measurable set A such that m(A ∩ B)=m(B)/2 for every open set B?&lt;/p&gt;&#xA;" OwnerUserId="6556" LastActivityDate="2011-02-03T02:35:55.637" Title="is there a measurable set A such that m(A ∩ B)=m(B)/2 for every open set B?" Tags="&lt;measure-theory&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" />
  <row Id="20171" PostTypeId="2" ParentId="20170" CreationDate="2011-02-03T01:54:25.457" Score="3" ViewCount="" Body="&lt;p&gt;Hint: Lebesgue density theorem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively, approximate $A\cap[0,1]$ with a finite union of intervals.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;On second thought, those hints are overly complicated.  You can use the definition of Lebesgue measure to find an open set $B$ containing $A\cap[0,1]$ with measure close to that of $A\cap[0,1]$.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2011-02-03T02:35:55.637" LastActivityDate="2011-02-03T02:35:55.637" CommentCount="3" />
  <row Id="20172" PostTypeId="1" AcceptedAnswerId="20176" CreationDate="2011-02-03T02:00:38.037" Score="4" ViewCount="102" Body="&lt;p&gt;Find an equation of the plane that passes through the line of intersection of the planes $x-y=1$ and $y+2z=3$ and is perpendicular to the plane $x+y-2z=1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm having a hard time with this. I'm very confused by this section in the book, as the definitions and equations are all over the place. Can anyone give me any hints or tips on how to start this one and things I should be looking for?&lt;/p&gt;&#xA;" OwnerUserId="2985" LastEditorUserId="2985" LastEditorDisplayName="" LastEditDate="2011-02-03T02:13:07.773" LastActivityDate="2011-02-03T02:38:41.587" Title="Finding an equation of a plane" Tags="&lt;calculus&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="20173" PostTypeId="2" ParentId="20167" CreationDate="2011-02-03T02:05:36.563" Score="1" ViewCount="" Body="&lt;p&gt;Are these the notes you took in class?  ie.. do you just want some explanation on some of the details?  Well for your first example, in the last step the $$1\over 5$$ is multiplied into the summation giving the term $$x^{n+1}\over {5^{n+1}(n+1)}$$.  Then the power series is shifted by replacing n with n-1 (which explains the term you have written in the last step).  However, in doing this, the series now starts at $n=1$ not $n=0$.  Was that a typo?  Clearly the term to the right of the last summation is undefined at $n=0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the second example (and I guess the first one too) the term $C$ is an arbitrary constant of integration.  So to evaluate the definite integral in your second question, you can choose $C$ to be zero as any antiderivative can be used in the fundamental theorem of calculus.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, power series are nice because polynomials are by far the easiest functions to work with, and a power series allows you approximate a much harder function as a polynomial.  There are only a few well known power series in elementary calculus (namely, $sin(x)$, $cos(x)$, $e^x$, and $1\over {x-1}$ ) but by differentiating and integrating you can produce power series for many, many functions.  That's the whole point of this. &lt;/p&gt;&#xA;" OwnerUserId="3971" LastEditorUserId="3971" LastEditorDisplayName="" LastEditDate="2011-02-03T08:01:36.950" LastActivityDate="2011-02-03T08:01:36.950" CommentCount="1" />
  <row Id="20174" PostTypeId="1" AcceptedAnswerId="20177" CreationDate="2011-02-03T02:06:33.163" Score="5" ViewCount="240" Body="&lt;p&gt;Either works when trying to find the eigenvalues, but only the former works when trying to find corresponding eigenvectors. I can understand how it makes a difference, but what I don't understand how one is supposed to &quot;know&quot; the former is the &quot;correct&quot; form, since it starts from here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;AX = λX&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And from there, you can end up with either:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;AX - λX = 0&#xA;or &#xA;λX - AX = 0&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And finally:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(A-λI)X = 0&#xA;or&#xA;(λI-A)X = 0&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And so should they not both be correct? Furthermore, once it's in that form, can you not multiply both sides by -1 to flip them?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I guess I'm overlooking some sort of rule of algebra when dealing with matrices. Let me know. Thanks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is an example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;{{5,3},{6,-2}} has an eigenvalue -4&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we do (A-λI)X = 0:&#xA;{{9,3},{6,2}}*X = 0&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This solves to X = {-1,3} (and multiples of it)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we do (λI-A)X = 0:&#xA;{{-9,3},{6,-2}}*X = 0&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This solves to X = {1,3} (and multiples of it)&lt;/p&gt;&#xA;" OwnerUserId="6558" LastEditorUserId="6558" LastEditorDisplayName="" LastEditDate="2011-02-03T02:17:52.307" LastActivityDate="2011-02-03T12:49:45.270" Title="(A-λI)X=0 works, but (λI-A)X=0 does not when finding eigenvectors. Why?" Tags="&lt;matrices&gt;&lt;eigenvectors&gt;" AnswerCount="2" CommentCount="6" />
  <row Id="20175" PostTypeId="1" CreationDate="2011-02-03T02:09:09.213" Score="3" ViewCount="96" Body="&lt;p&gt;Hi guys,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am creating a spreadsheet to calculate Adword formulas and I am stuck in how to calculate the &lt;strong&gt;Monthly Net Income&lt;/strong&gt; for each keyword. I have created a formula which calculates it but can't figure out how to limit the &lt;strong&gt;Monthly Budget&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The formula I've created is this one:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Monthly Net Income = ( DailyClicks x ConversionRate x SaleProfit) - ( CPC x DailyClicks )&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There is an example of the formula in the file which is a Google Spreadsheet publicly available here: &lt;a href=&quot;https://spreadsheets.google.com/ccc?key=0AnQMyM9XJ0EidDB6TUF0OTdaZ2dFb2ZGNmhQeE5lb0E&amp;amp;hl=en_GB#gid=2&quot; rel=&quot;nofollow&quot;&gt;https://spreadsheets.google.com/ccc?key=0AnQMyM9XJ0EidDB6TUF0OTdaZ2dFb2ZGNmhQeE5lb0E&amp;amp;hl=en_GB#gid=2&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;(you can create your own copy going to File &gt; Make a copy...)&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am releasing this set of tools as Public Domain so feel free to use it :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any help is much appreciated!&lt;/p&gt;&#xA;" OwnerUserId="6559" LastEditorUserId="6565" LastEditorDisplayName="" LastEditDate="2011-02-03T20:49:53.587" LastActivityDate="2011-02-03T20:49:53.587" Title="Help with a formula for Google Adwords" Tags="&lt;algebra&gt;" AnswerCount="1" />
  <row Id="20176" PostTypeId="2" ParentId="20172" CreationDate="2011-02-03T02:10:55.667" Score="4" ViewCount="" Body="&lt;p&gt;If the equation of a plane is given by $ax+by+cz+d=0$, then the normal vector for that plane is $(a,b,c)$, if $a$, $b$, $c$ are not all $0$. Since every line contained in a plane is perpendicular to the normal vector, you can cross the two normal vectors of $x-y=1$ and $y+2z=3$ to find the vector in the direction of the line of intersection, which is contained in your desired plane. The reason for crossing the normals is that the line of intersection is contained in both planes, and thus perpendicular to both normal vectors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now two planes are perpendicular if their normal vectors are perpendicular, so to find the normal vector of your desired plane, you can cross the vector in the direction of the line of intersection, and the normal vector of $x+y-2z=1$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lastly, a plane is determined by a point in the plane and a normal vector, so all that remains is to find a point in the plane. You should be able to find one in the line of intersection.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: Your situation looks something like this at first. Maybe it'll help in visualizing the space in which you're working.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/mtIcE.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1583" LastEditorUserId="1583" LastEditorDisplayName="" LastEditDate="2011-02-03T02:38:41.587" LastActivityDate="2011-02-03T02:38:41.587" CommentCount="7" />
  <row Id="20177" PostTypeId="2" ParentId="20174" CreationDate="2011-02-03T02:14:41.327" Score="0" ViewCount="" Body="&lt;p&gt;See &lt;a href=&quot;http://tutorial.math.lamar.edu/Classes/LinAlg/EVals_Evects.aspx&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt;(explanation of eigenvalues and eigenvectors). Both equations should give you the corresponding eigenvectors. Perhaps your book just mentioned the first equation and you assumed that it couldn't be of the second form as well.&lt;/p&gt;&#xA;" OwnerUserId="2548" LastEditorUserId="2548" LastEditorDisplayName="" LastEditDate="2011-02-03T02:20:57.343" LastActivityDate="2011-02-03T02:20:57.343" />
  <row Id="20178" PostTypeId="1" CreationDate="2011-02-03T02:46:37.820" Score="4" ViewCount="217" Body="&lt;p&gt;If $\varphi:R^{m}\to R^{n}$ is an epimorphism of free modules over a commutative ring, does it follow that $m \geq n$?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is obviously true for vector spaces over a field, but how would one show this over just a commutative ring?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-----Edit&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any way to use the following?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $\varphi : M \to M&amp;#39;$ is an epimorphism of left $S$-modules and $N$ is any right $S$-module then $id_N \otimes \varphi $ is an epimorphism.&lt;/p&gt;&#xA;" OwnerUserId="6560" LastEditorUserId="6560" LastEditorDisplayName="" LastEditDate="2011-02-03T05:48:32.347" LastActivityDate="2011-02-03T05:48:32.347" Title="Given a commutative ring $R$ and an epimorphism $R^m \to R^n$ is then $m \geq n$?" Tags="&lt;abstract-algebra&gt;&lt;commutative-algebra&gt;&lt;ring-theory&gt;&lt;modules&gt;&lt;tensors&gt;" CommentCount="16" FavoriteCount="2" />
  <row Id="20179" PostTypeId="2" ParentId="12423" CreationDate="2011-02-03T02:53:52.027" Score="2" ViewCount="" Body="&lt;p&gt;Let ${\mathbf{x}}(s)$ be a curve in ${\mathbb{R}}^3$ with natural parameter $s$.  We will need the following lemma, the proof of which is given at the end of this answer.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Lemma&lt;/strong&gt;: Choose a fixed point ${\mathbf{y}}$.  Then the curvature $\kappa$ satisfies&#xA;  $$&#xD;&#xA;\kappa \ge \left|\dot{\theta} + \frac{1}{r}\sin{\theta}\right|,&#xD;&#xA;$$&#xA;  where $r = \lVert\mathbf{x} - \mathbf{y}\rVert$ and $\theta$ is the angle between the&#xA;  velocity $\dot{\mathbf{x}}$ and $\mathbf{x} - \mathbf{y}$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Take $\mathbf{y} = \mathbf{x}(0)$; then $r(0)=0$ and $\theta(0)=0$.  The function $r(s)$ is monotonically increasing as long as $0 \le \theta &amp;lt; \pi/2$ (since $\dot{r}(s) = \cos{\theta}$), so $\theta$ and $\kappa$ can be considered as single-valued functions of $r$ until that point.  The above lemma then gives&#xA;$$&#xD;&#xA;\kappa(r) \ge \left|{\theta}&amp;#39;(r)\dot{r} + \frac{1}{r}\sin{\theta(r)}\right| = \left|{\theta}&amp;#39;(r)\cos{\theta(r)} + \frac{1}{r}\sin{\theta(r)}\right| = \left|\frac{1}{r}(r \sin\theta(r) )&amp;#39;\right|.&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Until the first turning point of the motion (where the velocity becomes perpendicular to the radius), we have&#xA;$$&#xD;&#xA;R\sin\theta(R) \le \int_{0}^{R} r \kappa(r) dr.&#xD;&#xA;$$&#xA;If the curvature is strictly below a fixed value (say, $\kappa &amp;lt; K$), then the integral is less than $\frac{1}{2}KR^2$ for $R&amp;gt;0$, and we have the result that&#xA;$$&#xD;&#xA;\sin{\theta(r)} &amp;lt; \frac{1}{2}Kr&#xD;&#xA;$$&#xA;for $r&amp;gt;0$.  A turning point is reached when $\theta=\pi/2$; this equation shows that the first such turning point must be at a radius greater than $2/K$, and hence the curve cannot be confined within a ball of diameter $2/K$.  Finally, the arclength before reaching a given radius $R$ is bounded by&#xA;$$&#xD;&#xA;\begin{eqnarray}&#xD;&#xA;s(R) &amp;amp;=&amp;amp; \int_{0}^{R} \frac{ds}{dr}dr \\ &amp;amp;=&amp;amp; \int_{0}^{R} \frac{dr}{\cos\theta(r)} \\ &amp;amp;&amp;lt;&amp;amp; \int_{0}^{R} \frac{dr}{\sqrt{1 - \frac{1}{4}K^2 r^2}} \\ &amp;amp;=&amp;amp; \frac{2}{K}\sin^{-1}\left(\frac{1}{2}KR\right)&#xD;&#xA;\end{eqnarray}&#xD;&#xA;$$&#xA;for $R \le 2/K$.  We conclude that any curve contained in the open unit ball with curvature $\kappa &amp;lt; 1$ must have length $s(2) &amp;lt; 2\sin^{-1}(1) = \pi$.  Moreover, this bound is tight, since a circular arc joining the points at $\pm (1-\epsilon^2)\hat{\mathbf{z}}$ and the point at $(1-\epsilon)\hat{\mathbf{x}}$ has length approaching $\pi$ as $\epsilon \rightarrow 0$.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof of Lemma&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We will work in spherical coordinates centered at $\mathbf{y}$; then&#xA;$$&#xD;&#xA;\begin{eqnarray}&#xD;&#xA;{\mathbf{x}}&#xD;&#xA;  &amp;amp;=&amp;amp; r\hat{\mathbf{r}}, \\&#xD;&#xA;{\dot{\mathbf{x}}}&#xD;&#xA;  &amp;amp;=&amp;amp; \dot{r}{\hat{\mathbf{r}}} + r\dot{\hat{\mathbf{r}}} \\&#xD;&#xA;  &amp;amp;=&amp;amp; \dot{r}{\hat{\mathbf{r}}} + r v_{\perp} \hat{\mathbf{v}}_{\perp}.&#xD;&#xA;\end{eqnarray}&#xD;&#xA;$$&#xA;Here $\hat{\mathbf{r}}$ is the unit vector from the origin to ${\mathbf{x}}$, and $\dot{\hat{\mathbf{r}}} = v_{\perp} \hat{\mathbf{v}}_{\perp}$ is its rate of change.  Because $\hat{\mathbf{r}}$ has constant length, we have $\hat{\mathbf{v}}_{\perp}\cdot \hat{\mathbf{r}} = 0$.  Taking the time derivative of this gives&#xA;$$&#xD;&#xA;0 = \dot{\hat{\mathbf{v}}}_{\perp}\cdot \hat{\mathbf{r}} + \hat{\mathbf{v}}_{\perp}\cdot \dot{\hat{\mathbf{r}}} = v_{\perp} + \dot{\hat{\mathbf{v}}}_{\perp}\cdot \hat{\mathbf{r}},&#xD;&#xA;$$&#xA;which we will use later.  Now, because $s$ is a natural parameter, $$\lVert\dot{\mathbf{x}}\rVert^2 = \left(\dot{r}\right)^2 + \left(rv_{\perp}\right)^2 = 1;$$&#xA;so we can define $\theta \in [0,\pi]$ such that $\dot{r} = \cos{\theta}$ and $rv_{\perp} = \sin{\theta}$.  The velocity and acceleration become&#xA;$$&#xD;&#xA;\begin{eqnarray}&#xD;&#xA;{\dot{\mathbf{x}}}&#xD;&#xA;  &amp;amp;=&amp;amp; \left(\cos{\theta}\right){\hat{\mathbf{r}}} + \left(\sin{\theta}\right)\hat{\mathbf{v}}_{\perp}, \\&#xD;&#xA;{\ddot{\mathbf{x}}}&#xD;&#xA;  &amp;amp;=&amp;amp; -\left(\sin{\theta}\dot{\theta}\right){\hat{\mathbf{r}}} + \left(\cos{\theta}\right)\dot{\hat{\mathbf{r}}} + \left(\cos{\theta}\dot{\theta}\right)\hat{\mathbf{v}}_{\perp} + \left(\sin{\theta}\right)\dot{\hat{\mathbf{v}}}_{\perp} \\&#xD;&#xA;&amp;amp;=&amp;amp; -\left(\sin{\theta}\dot{\theta}\right){\hat{\mathbf{r}}} +  \left(\cos{\theta}\right)\left(\dot{\theta} + \frac{1}{r}\sin{\theta}\right)\hat{\mathbf{v}}_{\perp} + \left(\sin{\theta}\right)\dot{\hat{\mathbf{v}}}_{\perp},&#xD;&#xA;\end{eqnarray}&#xD;&#xA;$$&#xA;and the acceleration has (two of its three) components&#xA;$$&#xD;&#xA;\begin{eqnarray}&#xD;&#xA;{\ddot{\mathbf{x}}}\cdot\hat{\mathbf{r}} &amp;amp;=&amp;amp; -\left(\sin{\theta}\dot{\theta}\right) + \left(\sin{\theta}\right)\left(\dot{\hat{\mathbf{v}}}_{\perp} \cdot \hat{\mathbf{r}}\right) \\&#xD;&#xA;&amp;amp;=&amp;amp; -\left(\sin{\theta}\right)\left(\dot{\theta} + \frac{1}{r}\sin{\theta}\right), \\&#xD;&#xA;{\ddot{\mathbf{x}}}\cdot\hat{\mathbf{v}}_{\perp} &amp;amp;=&amp;amp; +\left(\cos{\theta}\right)\left(\dot{\theta} + \frac{1}{r}\sin{\theta}\right).&#xD;&#xA;\end{eqnarray}&#xD;&#xA;$$&#xA;This brings us to the result that the squared curvature&#xA;$$&#xD;&#xA;\kappa^2 = \lVert\ddot{\mathbf{x}}\rVert^2 \ge \left({\ddot{\mathbf{x}}}\cdot\hat{\mathbf{r}}\right)^{2} +&#xD;&#xA;\left({\ddot{\mathbf{x}}}\cdot\hat{\mathbf{v}}_{\perp}\right)^{2} = \left(\dot{\theta} + \frac{1}{r}\sin{\theta}\right)^{2},&#xD;&#xA;$$&#xA;where $\theta$ is the angle between the velocity and the outward radial vector.  The lemma follows by taking the square root of both sides.&lt;/p&gt;&#xA;" OwnerUserId="5546" LastEditorUserId="5546" LastEditorDisplayName="" LastEditDate="2011-02-05T20:17:46.387" LastActivityDate="2011-02-05T20:17:46.387" CommentCount="5" />
  <row Id="20180" PostTypeId="2" ParentId="20148" CreationDate="2011-02-03T03:02:26.607" Score="0" ViewCount="" Body="&lt;p&gt;For your second question, recall the elementary inequality $(a_1+...+a_n)^r\leq a_1^r+...+a_n^r$ valid for any natural number $n$, postive numbers $a_1,...,a_n$ and $0&amp;lt;r&amp;lt;1$. Now just apply this to the partial sums: If $p\geq 1$ then $\left( \sum_{k=1}^{m} |a_k|^p \right) ^{\frac{1}{p}} \leq \sum_{k=1}^m |a_k|$ and then take limits. If $0&amp;lt;p&amp;lt;1$ then the inequality is reversed.  &lt;/p&gt;&#xA;" OwnerUserId="4829" LastActivityDate="2011-02-03T03:02:26.607" />
  <row Id="20181" PostTypeId="1" AcceptedAnswerId="20182" CreationDate="2011-02-03T03:21:24.280" Score="1" ViewCount="53" Body="&lt;p&gt;Find the value of n so that the error obtained by approximating sin(x) by the nth degree Maclaurin polynomial Tn(x) on the interval [-0.5,0.5] will be less than 5 x 10^-6.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm really stuck and don't even know where to start.&#xA;Any help would be appreciated.&lt;/p&gt;&#xA;" OwnerUserId="6252" LastActivityDate="2011-02-03T05:41:47.873" Title="Help with Maclaurin Polynomial involving Error" Tags="&lt;power-series&gt;" AnswerCount="2" />
  <row Id="20182" PostTypeId="2" ParentId="20181" CreationDate="2011-02-03T03:25:05.213" Score="1" ViewCount="" Body="&lt;p&gt;The error is given by the following: $$f(x) - T_{n}(x) = \frac{1}{(n+1)!} x^{n+1} f^{(n+1)} (c_x)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $c_x \in [0,x]$.&lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2011-02-03T03:25:05.213" CommentCount="3" />
  <row Id="20183" PostTypeId="1" CreationDate="2011-02-03T03:27:41.587" Score="2" ViewCount="94" Body="&lt;p&gt;OK, so I am choosing 4 letters/numbers out of 10 possible letters. These letters are X,Y,LT,LB,RT,RB,U,D,L,R. The first three letters can be any one of these, but must contain at least one X or Y. The last letter can be an U,D,L, or right. Can someone please figure out the possible combinations?&lt;/p&gt;&#xA;" OwnerUserId="6561" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2011-02-03T05:54:01.007" LastActivityDate="2011-02-03T05:54:01.007" Title="Figuring out possible combinations with restrictions" Tags="&lt;combinatorics&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="20184" PostTypeId="1" CreationDate="2011-02-03T03:34:18.887" Score="1" ViewCount="74" Body="&lt;p&gt;I have a question concerning the proof of theorem 3.5 in Milnor's Morse Theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This theorem states that if $f$ is a differentiable function on a Manifold M with no critical points, and if each $M^a = \{x\in M | f(x)\leq a \} $ is compact then $M$ has the homotopy type of a CW-complex with one cell of dimension $\lambda$ for each critical point of index $\lambda$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically the proof shows via induction and appealing to previous theorems that $M^a$ has the homotopy type of a CW complex satisfying the conclusion above. But then Milnor says &quot;If $M$ is compact this completes the proof&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't understand how this just follows directly? Is there some theorem involving Compact manifolds or CW complexes that I am missing?  Does $M^a$ have to be homotopic to $M$ for some $a$ if $M$ is compact?&lt;/p&gt;&#xA;" OwnerUserId="6562" LastActivityDate="2011-02-03T03:45:52.543" Title="Compact manifold/Morse theory" Tags="&lt;differential-geometry&gt;&lt;differential-topology&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="20186" PostTypeId="2" ParentId="20184" CreationDate="2011-02-03T03:37:54.827" Score="3" ViewCount="" Body="&lt;p&gt;If $M$ is compact, one of the $M^a$ is &lt;em&gt;equal&lt;/em&gt; to $M$!&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2011-02-03T03:37:54.827" CommentCount="2" />
  <row Id="20187" PostTypeId="2" ParentId="20183" CreationDate="2011-02-03T03:38:18.213" Score="2" ViewCount="" Body="&lt;p&gt;There are $10^3$ combinations without a restriction in the first three letters. Now there are $8^3$ possible combinations for the first three letters that do not contain a $Y$ nor $X$, so there are $10^3 - 8^3 = 488$ possibilities with your restrictions in the first three letters. There are 4 choices for the last letter, hence there are $488 \cdot 4 = 1952$ choices.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Oh, in case you don't allow repetition (it says &lt;em&gt;&quot;at least one $X$ or $Y$,&quot;&lt;/em&gt; that's a bit misleading):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fix some letter from $U, D, L, R$ in the last position. Then you have in general $9 \cdot 8 \cdot 7 = 504$ choices for the first letter, without restrictions. $7 \cdot 6 \cdot 5 = 210$ of these choices do not have an $X$ or $Y$. So total you have $(504 - 210) \cdot 4 = 1176$ choices without repetition. &lt;/p&gt;&#xA;" OwnerUserId="5105" LastEditorUserId="5105" LastEditorDisplayName="" LastEditDate="2011-02-03T03:51:20.583" LastActivityDate="2011-02-03T03:51:20.583" />
  <row Id="20188" PostTypeId="1" CreationDate="2011-02-03T03:42:35.743" Score="1" ViewCount="116" Body="&lt;p&gt;Suppose $R$ is a commutative ring, $f\colon F_1\to F_2$ is a homomorphism of free modules, and $M$ is an $R$-module.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $f$ is a surjective homomorphism, then $f\otimes_R \mathrm{id}_M$ is surjective.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it true that if $f$ is a monomorphism, then $f\otimes_R \mathrm{id}_M$ is a monomorphism? I can't think of any counterexamples.&lt;/p&gt;&#xA;" OwnerUserId="6560" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-03T04:14:30.200" LastActivityDate="2011-02-03T04:14:30.200" Title="Tensoring a monomorphism of free modules with an identity map" Tags="&lt;commutative-algebra&gt;&lt;ring-theory&gt;&lt;modules&gt;&lt;tensors&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="20190" PostTypeId="2" ParentId="20188" CreationDate="2011-02-03T03:55:20.267" Score="3" ViewCount="" Body="&lt;p&gt;Take $f:\mathbb Z\to\mathbb Z$ be multiplication by $2$, and $M=\mathbb Z/2\mathbb Z$. This example captures &lt;em&gt;precisely&lt;/em&gt; why it does not work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;N.B.&lt;/strong&gt; If you have an injective map $f:P\to Q$ with $Q$ is free, you can complete it to a short exact sequence $$0\longrightarrow P\xrightarrow{\;f\;}Q\longrightarrow Q/f(P)\longrightarrow 0$$ If $M$ is another module, then there is a short exact sequence $$0\longrightarrow\mathrm{Tor}_1^R(Q/f(P),M)\longrightarrow P\otimes_RM\xrightarrow{\;f\otimes\mathrm{id}_M\;} Q\otimes_RM\longrightarrow Q/f(P)\otimes_RM\longrightarrow 0$$ where $\mathrm{Tor}_1^R(Q/f(P),M)$ is a certain $R$-module one learns to construct when one learns about &lt;em&gt;homological algebra&lt;/em&gt;; it measures how far $f\otimes\mathrm{id}_M$ is from being a monomorphism. In the example above we have $$\mathrm{Tor}_1^R(Q/f(P),M)\cong \mathbb Z/2\mathbb Z,$$ as one can easily check as soon as one knows how this is done, and in fact this is how I found the example: I looked for a $\mathrm{Tor}$ I knew was not zero.&lt;/p&gt;&#xA;" OwnerUserId="274" LastEditorUserId="274" LastEditorDisplayName="" LastEditDate="2011-02-03T04:06:46.880" LastActivityDate="2011-02-03T04:06:46.880" CommentCount="2" />
  <row Id="20191" PostTypeId="1" AcceptedAnswerId="20194" CreationDate="2011-02-03T03:55:24.500" Score="4" ViewCount="174" Body="&lt;p&gt;Define $D^+f(x) = \limsup\limits_{h\to 0^+}{\left(\dfrac{f(x+h)-f(x)}{h}\right)}$. Given the function $f(x)=\begin{cases}x\,\sin{\left(\frac{1}{x}\right)} &amp;amp; x\neq 0\\ 0 &amp;amp; x=0\end{cases}\,,$ find $D^+f(x)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are also three other approximate derivatives to work out (lim inf from the right, lim sup and lim inf from the left), but I figure if I can get one I can get the others. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is that, while I understand lim sup's for sequences, I still haven't grokked lim sup's for functions... The definition seems totally unclear. (This is also the first time I've ever had to calculate the lim sup of a function, somehow.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Edited title for accuracy. Original title was &quot;Calculating an 'approximate' derivative&quot;&lt;/p&gt;&#xA;" OwnerUserId="2313" LastEditorUserId="2313" LastEditorDisplayName="" LastEditDate="2011-02-03T18:25:55.507" LastActivityDate="2011-02-03T19:39:00.857" Title="Calculating Dini derivatives" Tags="&lt;real-analysis&gt;&lt;homework&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="20192" PostTypeId="2" ParentId="20191" CreationDate="2011-02-03T04:11:28.233" Score="3" ViewCount="" Body="&lt;p&gt;Note that if $x\neq 0$, the function is differentiable at $x$, the limsups and liminfs equal the limit, and you know how to find this from calculus.  If $x=0$, the expression becomes quite simple, and unwinding the definitions and using what you know about values of $\sin(1/h)$ for arbitrarily small values of $h$ will give you the limsups and liminfs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Roughly speaking, the limsup of $g$ at $x$ is the largest value that $g(y)$ gets arbitrarily close to for $y$ arbitrarily close to $x$.  Similarly for liminfs, but with &quot;largest&quot; replaced with &quot;smallest&quot;.  In your case, the limsups and liminfs at $0$ are actually attained as values of the expression for $h$ arbitrarily small.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2011-02-03T04:11:28.233" CommentCount="1" />
  <row Id="20193" PostTypeId="2" ParentId="20157" CreationDate="2011-02-03T04:55:37.107" Score="11" ViewCount="" Body="&lt;p&gt;Since you ask for &quot;plain English&quot;, let me discuss a bit of what the issue is and how RSA hopes to address it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As I mentioned in &lt;a href=&quot;http://math.stackexchange.com/questions/7377/why-are-very-large-prime-numbers-important-in-cryptography/7381#7381&quot;&gt;an earlier answer&lt;/a&gt; the underlying problem is this: you want to communicate securely with someone. There are two standard techniques: one is &lt;em&gt;steganography&lt;/em&gt;, which consists of trying to hide the very existence of the communication; that's not usually very mathematical, and it is difficult to do. The other is &lt;em&gt;cryptography&lt;/em&gt;, which consists of obscuring the message so that even if the message is overheard, the people overhearing it will not understand it. A method for doing this is called a &quot;cryptosystem.&quot; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cryptosystems usually rely on a &lt;em&gt;key&lt;/em&gt;, a way for the speaker to encrypt his message, and a way for the listener to decrypt it (figure out what the message says). Historically, most cryptosystems used what are called &quot;symmetric keys&quot;: the piece of information that was needed to encrypt was the same as the piece of information that was needed to decrypt (or at least, knowing one was &quot;equivalent&quot; to knowing the other). Both sides needed to know the key. This leads to a difficulty: how do we agree on a key? If we cannot communicate without being overheard, then we cannot agree on a key without everyone finding out what the key is, which makes the system useless. And if we &lt;em&gt;can&lt;/em&gt; communicate without being overheard, then there's no point in the whole song and dance, we can just talk using that communication.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There were a number of ways of solving this issues: traditionally, you had a secure way of communication with limited bandwidth (only small amounts of information could be exchanged through it, and only at certain times), so that they key could be exchanged but communication was too difficult (diplomatic bags, spies, etc). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But there is another type of cryptosystems that were discovered in the 70s (it's known that the UK's GCHQ knew about them before, and  there are always rumors that the NSA knew about them before they were &quot;publically discovered&quot; as well), which are not symmetric, in the following sense: knowing the encryption key does not give you enough information to find the decryption key. They are called &quot;public key cryptosystems&quot;, because you can tell the entire world how to encrypt messages for you, but knowing this information does not give them a way to decrypt messages sent to you. That requires the &quot;decryption key&quot;, which you only know.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[&lt;em&gt;Sorry, we'll have to throw some math in now, but you had to expect it.&lt;/em&gt;] These systems rely on what are called &lt;em&gt;trapdoor functions&lt;/em&gt;. The idea of a trapdoor function is that you have a function $f$ that encrypts, but for which it is difficult to find the inverse, $f^{-1}$ (which decrypts); so that even if you know exactly what $f$ is, and you know that the output of $f$ was, say, $D$, it is still very hard to find $f^{-1}(D)$ (which would be the message being sent). However, if you happen to know an &lt;em&gt;extra piece of information&lt;/em&gt;, $k$, then from $f$, $D$, and $k$ you can easily find $f^{-1}(D)$ (think of $k$ as the lever that &quot;opens the trapdoor&quot; and lets the inverse fall through). So the idea of these public key systems is to tell everyone what $f$ is, but keep $k$ secret. Nobody needs to know $k$ in order to compute $f(M)$, which they can do in secret. Then they can yell $f(M)$ at you; it doesn't matter if people overhear $f(M)$, because it's very hard to find $M=f^{-1}(f(M))$ (very hard to decrypt). But since &lt;em&gt;you&lt;/em&gt; know the secret piece of information $k$, when you hear $f(M)$, you can use $k$ to find $M$, and voilá!, you know the secret message but nobody else does.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;RSA is one of these systems. In RSA, the messages are numbers (it is straightforward to convert text into a number). I want to communicate a number to you secretly, even though everyone can hear me talking.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The way RSA work is by using modular exponentiation. You select a big number $n$, and two numbers, $e$ and $d$, such that $ed\equiv 1\pmod{\varphi(n)}$ (Euler's $\varphi$ function). You tell everyone what $n$ is, and you tell everyone what $e$ is; these are the &lt;em&gt;public keys&lt;/em&gt;, but you keep $d$ secret. If I want to send you a message $M$, I first make sure that $\gcd(M,n) = 1$ and $1\lt M\lt n$ (if not, I modify the message a bit so that it is, or I break it up into pieces if it is too big). Then I compute $R=M^e \bmod n$, which is easy to do with computers, and I tell everyone that I'm sending you the message $R$. When you receive $R$, you can figure out what the original message $M$ was, simply by computing $R^d \bmod n$ (which again is easy to do computationally). The reason this will tell you what $M$ was is due to Euler's Theorem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Euler's Theorem.&lt;/strong&gt; If $n$ is a positive integer, and $\gcd(a,n)=1$, then $a^{\varphi(n)} \equiv 1 \pmod{n}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a generalization of Fermat's Little Theorem (in Fermat's Little Theorem, you have $n=p$ a prime, so $\varphi(n)=\varphi(p)=n-1$, and the equation tells you that $a^{p-1}\equiv 1\pmod{p}$ if $\gcd(a,p)=1$). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $ed\equiv 1 \pmod{\varphi(n)}$, we can write $ed = k\varphi(n)+1$ for some integer $k$. Then we have:&#xA;$$ R^d \equiv (M^e)^d \equiv M^{ed} \equiv M^{k\varphi(n)+1} = (M^{\varphi(n)})^kM \equiv 1^kM = M \pmod{n}$$&#xA;(by using the rules of exponentiation, and because $M^{\varphi(n)}\equiv 1\pmod{n}$). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, this means that &lt;em&gt;you&lt;/em&gt; can figure out $M$, because you know $d$. Can other people figure out $M$ &lt;em&gt;without&lt;/em&gt; knowing $d$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you can figure out $\varphi(n)$, then it turns out to be very easy to figure out $d$ from knowing $e$: because you know that $\gcd(e,\varphi(n))=1$, then using the Euclidean Algorithm you can find integers $x$ and $y$ such that $ex+\varphi(n)y = 1$; and then $d\equiv x\pmod{\varphi(n)}$, and all of this can be done easily. So you want to make sure that computing $\varphi(n)$ is not easy, even if you know $n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you can &lt;em&gt;factor&lt;/em&gt; $n$ into primes, though, then $\varphi(n)$ is very easy to find. So you want $n$ to &lt;em&gt;not&lt;/em&gt; be a prime, but to have very few prime factors (because every time you find a prime factor, you divide it out, and you have a smaller problem left). So ideally, we want $n$ to just be a product of two &lt;em&gt;really big&lt;/em&gt; primes, $n=pq$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the way this works is: you first find two really big primes secretly (there are ways of doing this), $p$ and $q$. Then you compute $n=pq$. Since you know $p$ and $q$, you also know that $\varphi(n) = (p-1)(q-1)$, so $\varphi(n)$ is easy &lt;em&gt;for you&lt;/em&gt;. Then you pick an $e$, and since you know $\varphi(n)$, you can find $d$ easily (there are &quot;bad choices&quot; of $e$ that will make finding $d$ not too hard, but they are known; you avoid them; there are also bad choices of $p$ and $q$, so you avoid them too, don't worry too much about this).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, now you know $p$, $q$, $d$ and $e$. You tell everyone $pq$ and $e$, but you keep $d$, $p$, and $q$ secret.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I want to send you message $M$, $1\lt M \lt pq$, I don't tell anyone what $M$ is, I compute $M^e\bmod n$, and I tell everyone $M^e$. You can figure out $M$ because you know $d$, so as above, $(M^e)^d = M^{ed} \equiv M\pmod{n}$ by Euler's Theorem. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The hope is that just from knowing $n$, $d$, and $M^d \bmod{n}$, it is hard to figure out $M$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;strong&gt;RSA Problem&lt;/strong&gt; is precisely that problem: if you know $n$ (and you know that $n$ is the product of two primes, but you don't know which primes ahead of time), you know $d$, and you know $M^d\bmod{n}$, can you figure out $M$? The security of the RSA cryptosystem depends on how hard the RSA Problem is to solve.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I can factor $n$, then I can solve the RSA Problem: I can find $\varphi(n)$, then I can use $\varphi(n)$ and $d$ to find $e$, and once I have $e$ and $M^d$, I just figure out $(M^d)^e \equiv M \pmod{n}$. So the RSA Problem is no harder than the problem of factoring a product of two primes. Luckily, in so far as we know the problem of factoring a product of two large primes is pretty hard.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is not known, however, if there might be a different way of solving the RSA problem that may not rely on factoring $n$. Maybe, maybe not. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope this helps. &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-03T14:08:29.700" LastActivityDate="2011-02-03T14:08:29.700" CommentCount="4" />
  <row Id="20194" PostTypeId="2" ParentId="20191" CreationDate="2011-02-03T04:58:47.550" Score="5" ViewCount="" Body="&lt;p&gt;The four &quot;approximate derivatives&quot; are usually called &lt;a href=&quot;http://en.wikipedia.org/wiki/Dini_derivative&quot; rel=&quot;nofollow&quot;&gt;Dini derivatives&lt;/a&gt;. They encapsulate best approximations to the usual derivative (and are always defined even if the latter fails to exist) in four ways: they are upper and lower approximations to the derivative from the left and the right.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you know what the graph of $x\sin{\frac{1}{x}}$ looks like then it should be rather easy to imagine what they are supposed to be:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/aAhdI.png&quot; alt=&quot;&quot; title=&quot;graph&quot; /&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the graph of $x \sin{\frac{1}{x}}$ touches each of the four rays issuing from the origin infinitely often. Every ray with smaller slope will intersect infinitely often, while every ray with larger slope will be too far away&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Concerning the definition of $\limsup$, remember the reason for the notation. If $(x_n)_{n \in \mathbb{N}}$ is a sequence of real numbers, then $\limsup_{n\to \infty} x_{n} = \lim_{n \to \infty} \sup_{k \geq n} x_{k}$. If the sequence is bounded then this limit exists because $s_{n} = \sup_{k \geq n} x_{k}$ is a &lt;em&gt;decreasing&lt;/em&gt; sequence and because the real numbers are complete. You can also write $$\limsup_{n \to \infty} x_{n} = \inf_{n \geq 0} \sup_{k \geq n} x_{k}.$$ As for the generalization to functions remember that a sequence is a function $x(n)$ from the natural numbers to the real numbers. Now if $f: (0,\varepsilon) \to \mathbb{R}$ is a function you can generalize the last expression for the $\limsup$ by writing&#xA;$$\limsup_{h \searrow 0} f(h) = \inf_{h \geq 0}{\sup_{x \in (0, h)}} f(x).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you understand the definition of infimum and supremum properly, you should have no problem in solving the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Exercise:&lt;/strong&gt; If $a = \limsup_{h \searrow 0} f(h)$ then there exists a sequence $x_{n} \searrow 0$ such that $a = \limsup_{n \to \infty} f(x_{n})$ and for every other sequence $\limsup_{n &#xD;&#xA;to \infty} y_{n} \leq a$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, in order to calculate $\limsup_{h \searrow 0} f(h)$, find the maximum possible $\limsup_{n \to \infty} f(x_{n})$ among all sequences $x_{n} \searrow 0$.&lt;/p&gt;&#xA;" OwnerUserId="5363" LastEditorUserId="5363" LastEditorDisplayName="" LastEditDate="2011-02-03T14:24:30.133" LastActivityDate="2011-02-03T14:24:30.133" CommentCount="1" />
  <row Id="20195" PostTypeId="1" CreationDate="2011-02-03T05:26:25.363" Score="2" ViewCount="118" Body="&lt;p&gt;I am working on a project that needs to be able to calculate the probability of rolling a given value $k$ given a set of dice, &lt;em&gt;not&lt;/em&gt; necessarily all the same size.  So for instance, what is the distribution of rolls of a D2 and a D6?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An equivalent question, if this is any easier, is how can you take the mass function of one dice and combine it with the mass function of a second dice to calculate the mass function for the sum of their rolls?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Up to this point I have been using the combinatorics function at the bottom of the probability section of Wikipedia's article on &lt;a href=&quot;http://en.wikipedia.org/wiki/Dice#Probability&quot; rel=&quot;nofollow&quot;&gt;dice&lt;/a&gt;, however I cannot see how to generalize this to different sized dice.&lt;/p&gt;&#xA;" OwnerUserId="1887" LastActivityDate="2011-02-03T14:08:20.417" Title="Probability of Sum of Different Sized Dice" Tags="&lt;probability&gt;&lt;combinatorics&gt;&lt;dice&gt;" AnswerCount="3" CommentCount="3" />
  <row Id="20196" PostTypeId="2" ParentId="20195" CreationDate="2011-02-03T05:36:07.343" Score="0" ViewCount="" Body="&lt;p&gt;For D2+D6, you can only get 2 or 8 in one case each, probability 1/12.  Other totals 3-7 you can attain two ways (for example 6=5+1 or 4+2), probability 2/12=1/6.  A good check is the total over all the values is 1. The  check is 2*(1/12)+(7-3+1)*(1/6)=1&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-02-03T05:36:07.343" CommentCount="2" />
  <row Id="20197" PostTypeId="2" ParentId="20195" CreationDate="2011-02-03T05:41:05.673" Score="5" ViewCount="" Body="&lt;p&gt;You can use generating functions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I presume D2 means dice with numbers 1 and 2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In which case the probability generating function is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(x/2 + x^2/2)(x/6 + x^2/6 + x^3/6 + x^4/6 + x^5/6 + x^6/6) = \frac{x^2(x^2-1)(x^6 - 1)}{12(x-1)^2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You need to find the coefficient of $x^k$ in this to get the probability that the sum is $k$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can use binomial theorem to expand out $\frac{1}{(x-1)^2}$ in the form $\sum_{n=0}^{\infty} a_n x^n$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can generalized it to any number of dice with varying sides.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I will leave the formula to you.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2011-02-03T05:41:05.673" CommentCount="7" />
  <row Id="20198" PostTypeId="2" ParentId="20181" CreationDate="2011-02-03T05:41:47.873" Score="0" ViewCount="" Body="&lt;p&gt;There are many ways to find the power series that  corresponds to a function.  As is often the case in mathematics, you try to reduce the problem at hand to a problem you already know how to solve.  So if you know the geometric power series you try to reduce new problems to that, which the examples do.  An example would be if you want to do $\sum_{n=0}^{\infty}  n x^n$.  Since you know $\sum_0^{\infty}x^n=\frac{1}{1-x}, \sum_{n=0}^{\infty}  nx^n=\sum_0^{\infty}x\frac{d}{dx}x^n$  Now you need to justify taking the derivative outside the sum.   There are beautiful theorems that tell you when this is justified.  We physicists ignore them, as all of our functions satisfy them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another example would be the series for $\sin(x)$ near 0 (the Maclaurin series).  Taylor's theorem says $$\sin(x)=\sum_{n=0}^{\infty}\frac{\sin^{(n)}(0)}{n!}x^n$$ where $\sin^{(n)}(0)$ is the $n^{th}$ derivative of $\sin(x)$ evaluated at 0.  If you know that the $n^{th}$ derivative of $\sin (x)$ at 0 is 0 if $n$ is even and $-1^{(n-1)/2}$ if $n$ is odd this becomes the familiar $$\sin(x)=\sum_{n=1,n \text{ odd}}^{\infty}\frac{-1^{\left(\frac{n-1}{2}\right)}x^n}{n!}$$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you have this (and the corresponding ones for other functions like e^x, cos(x), etc) you can reduce problems to these as well.  Then, as Arjang commented, if you try to operate with differential and integral operators, you need to worry about whether this is justified.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-02-03T05:41:47.873" />
  <row Id="20199" PostTypeId="2" ParentId="20175" CreationDate="2011-02-03T05:42:46.250" Score="2" ViewCount="" Body="&lt;p&gt;the formula as presented is not calculating the monthly net income, it is calculating the daily net income&lt;/p&gt;&#xA;&#xA;&lt;p&gt;assuming 30 days/month, divide the monthly budget by 30 to get a daily budget&lt;/p&gt;&#xA;&#xA;&lt;p&gt;then limit the DailyClicks to be no more than DailyBudget / CPC&lt;/p&gt;&#xA;" OwnerUserId="6565" LastActivityDate="2011-02-03T05:42:46.250" />
  <row Id="20200" PostTypeId="1" AcceptedAnswerId="20219" CreationDate="2011-02-03T05:48:40.670" Score="3" ViewCount="68" Body="&lt;p&gt;I have a matrix-vector inner product multiplication $G = X D x$ where $D$ is a diagonal matrix. Now let's say I already know $E = Xx$. Is there a method that I can use to change $E$ into $G$ using $D$ without having to calculate $G$ in full?&lt;/p&gt;&#xA;" OwnerUserId="1588" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2011-02-03T05:55:51.330" LastActivityDate="2011-02-03T10:00:08.927" Title="Matrix Diagonal Multiplication" Tags="&lt;linear-algebra&gt;&lt;matrices&gt;" AnswerCount="2" />
  <row Id="20201" PostTypeId="2" ParentId="20118" CreationDate="2011-02-03T05:49:45.427" Score="2" ViewCount="" Body="&lt;p&gt;There is a form of the Nullstellensatz (see &lt;a href=&quot;http://en.wikipedia.org/wiki/Hilbert%27s_Nullstellensatz&quot; rel=&quot;nofollow&quot;&gt;the wikipedia entry&lt;/a&gt;)&#xA;which is valid for arbitrary fields (and even for more general rings --- so-called &lt;a href=&quot;http://en.wikipedia.org/wiki/Jacobson_ring&quot; rel=&quot;nofollow&quot;&gt;Jacobson rings&lt;/a&gt;).  One way to formulate it (in the case of an arbitrary field) is as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $k$ is a field, then:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;For any ideal $I$ in $k[x_1,\ldots,x_n]$, &#xA;the radical $\sqrt{I}$ is the intersection of all maximal ideals $\mathfrak m$&#xA;in $k[x_1,\ldots,x_n]$ containing $I$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If $\mathfrak m$ is a maximal ideal of $k[x_1,\ldots,x_n]$, then there is a homomorphism of $k$-algebras&#xA;$k[x_1,\ldots,x_n] \to \overline{k}$ whose kernel is precisely $\mathfrak m$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;These two facts taken together say that you can recover $\sqrt{I}$ by knowing all of the points of what you call ${\mathbf V}(I)$.  (With a little work one&#xA;can deduce these statements from the Nullstellensatz for $\overl&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[Note also that many (I would guess most) people, when they write $V(I)$,&#xA;would mean what you call ${\mathbf V}(I)$; i.e. even when the ideal consists of polynomials in a non-algebraically closed field $k$, they would take the variety $V(I)$ attached to $I$ to consist of all the common zeroes of the polynomials wiht coordinates lying in $\overline{k}$ (not just those with coordinates lying in $k$).   Of course this is just a matter of convention; but it is a common convention which it might be helpful to be aware of.]&lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2011-02-03T05:49:45.427" />
  <row Id="20202" PostTypeId="2" ParentId="20174" CreationDate="2011-02-03T05:55:46.137" Score="0" ViewCount="" Body="&lt;p&gt;You usally use $(\lambda I-A)X$ to calculate the Eigenvalues because the polynom you will get is normalized in this case ( its highest degree coefficient is 1).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For Eigenvektors it totally doesn't matter which way you do it as you have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(\lambda I-\vec A)\vec v=\vec0 \Leftrightarrow -(\lambda I-A)\vec v=\vec 0 \Leftrightarrow (A-\lambda I)\vec v=\vec 0$&lt;/p&gt;&#xA;" OwnerUserId="3123" LastEditorUserId="3123" LastEditorDisplayName="" LastEditDate="2011-02-03T12:49:45.270" LastActivityDate="2011-02-03T12:49:45.270" CommentCount="4" />
  <row Id="20203" PostTypeId="2" ParentId="20200" CreationDate="2011-02-03T05:57:22.227" Score="0" ViewCount="" Body="&lt;p&gt;I love cheap answers because they help in asking good questions.  So you have $G=XDX^{-1}E$  How do you intend to use $G$ that this does not satisfy?  I don't understand where diagonal comes into this.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-02-03T05:57:22.227" CommentCount="1" />
  <row Id="20204" PostTypeId="1" AcceptedAnswerId="20210" CreationDate="2011-02-03T06:12:20.477" Score="5" ViewCount="167" Body="&lt;p&gt;I have shown that two of the three elementary operations will not change the image of the row space of the matrix: given a row vector $\vec{v}$, $k\vec{v}$ will span the same (scalar multiplication), and switching the order of the rows won't change anything because the span of the row vectors will be the same (vectors will be in a different order). However, how would one show that adding/subtracting vectors does not change the row space?&lt;/p&gt;&#xA;" OwnerUserId="5157" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2011-02-03T10:49:56.927" LastActivityDate="2011-02-03T11:48:45.187" Title="Why do elementary matrix operations not affect the row space of a given matrix?" Tags="&lt;linear-algebra&gt;" AnswerCount="5" />
  <row Id="20206" PostTypeId="2" ParentId="20204" CreationDate="2011-02-03T06:58:47.573" Score="3" ViewCount="" Body="&lt;p&gt;Suppose the rows of the matrix are $A=(v_1 ,v_2 ,...,v_n)$, then switching two rows, is just switching the order of the vectors, and as you said, it doesn't change the span of this set. The same is true for multiplying by a scalar ($\neq 0$) one of the rows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose you want to change $v_2$ to the vector $v_2 + \alpha v_1$ which is the last elementary operation. The span of $B=(v_1 ,v_2 + \alpha v_1 ,...,v_n)$ will contain $(v_2 + \alpha v_1) - \alpha v_1 = v_2$ so $A \subseteq span(B) \Rightarrow span(A) \subseteq span(B)$. In the same way $v_2 + \alpha v_1 \in span(A)$ so $B\subseteq span(A) \Rightarrow span(B) \subseteq span (A)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;now you have that $span(B)=span(A)$ so adding one row (times a scalar) to another doesn't change the row space.&lt;/p&gt;&#xA;" OwnerUserId="2125" LastActivityDate="2011-02-03T06:58:47.573" />
  <row Id="20207" PostTypeId="2" ParentId="20204" CreationDate="2011-02-03T07:00:47.690" Score="1" ViewCount="" Body="&lt;p&gt;Consider two vectors, &lt;strong&gt;X&lt;/strong&gt; and &lt;strong&gt;Y&lt;/strong&gt;, in the plane, beginning at the origin and enclosing a nonzero angle. How many times you add &lt;strong&gt;Y&lt;/strong&gt; to &lt;strong&gt;X&lt;/strong&gt;, geometrically continue the vector &lt;strong&gt;X&lt;/strong&gt; into the direction of &lt;strong&gt;Y&lt;/strong&gt; , meaning: &lt;strong&gt;Z = X + k*Y&lt;/strong&gt;  - the angle between &lt;strong&gt;Y&lt;/strong&gt; and &lt;strong&gt;Z&lt;/strong&gt; diminuishes, but never vanishes.&lt;/p&gt;&#xA;" OwnerUserId="1714" LastEditorUserId="1714" LastEditorDisplayName="" LastEditDate="2011-02-03T11:44:22.097" LastActivityDate="2011-02-03T11:44:22.097" />
  <row Id="20208" PostTypeId="2" ParentId="20127" CreationDate="2011-02-03T07:27:01.747" Score="1" ViewCount="" Body="&lt;p&gt;I feel that the second one is only going to be possible to find computationally (although feel free to prove me wrong).  The primes dividing Mersenne numbers are difficult to guess -- so we can't exclude some prime $p&amp;lt;n$ from dividing $2^n-1$ without testing (although, of course we can exclude every $p&gt;n$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, for each prime $p&amp;lt;n$ we can find the largest $x$ such that $p^x$ divides both $n!$ and $2^n-1$, then combine the results using the Chinese Remainder Theorem.  Some tips:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The largest $a$ such that $p^a$ divides $n!$ is $a=\sum_{k \geq 1} \lfloor n/p^k \rfloor$.&lt;/li&gt;&#xA;&lt;li&gt;Then we can use &lt;a href=&quot;http://en.wikipedia.org/wiki/Euler%27s_theorem&quot; rel=&quot;nofollow&quot;&gt;Euler's Theorem&lt;/a&gt; to reduce the amount of computation required to find $2^n-1 \pmod {p^a}$ (or if you use some computer algebra system it'll do this for you).&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;There's approximately 60 million primes to process, so it will take a while, but it's not out-of-the-question.&lt;/p&gt;&#xA;" OwnerUserId="139" LastEditorUserId="139" LastEditorDisplayName="" LastEditDate="2011-02-03T07:37:41.970" LastActivityDate="2011-02-03T07:37:41.970" />
  <row Id="20209" PostTypeId="1" AcceptedAnswerId="20211" CreationDate="2011-02-03T07:28:43.743" Score="5" ViewCount="279" Body="&lt;p&gt;Hi-&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was just browsing through Ravi Vakil's website when i found a nice article written on what he demands from his students. Here is the webpage. (For those interested!)&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://math.stanford.edu/~vakil/potentialstudents.html&quot; rel=&quot;nofollow&quot;&gt;http://math.stanford.edu/~vakil/potentialstudents.html&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I actually found a line which caught my attention. He says:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;em&gt;&quot;Mathematics isn't just about answering questions; even more so, it is about asking the right questions, and that skill is a difficult one to master.&quot;&lt;/em&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I would like somebody to explain the meaning of this statement. What does asking the right questions mean here?&lt;/p&gt;&#xA;" LastActivityDate="2011-02-03T10:07:08.003" CommunityOwnedDate="2011-02-03T10:29:01.743" Title="Explanation of a Phrase from Prof. Ravi Vakil's Website" Tags="&lt;soft-question&gt;" AnswerCount="3" CommentCount="3" />
  <row Id="20210" PostTypeId="2" ParentId="20204" CreationDate="2011-02-03T07:33:09.307" Score="2" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\ \ $ If $\rm\ v\in V\ $ then $\rm\ w\in V\ \iff\ w+c\:v\in V\:.\ $ Thus $\rm\ span(v,w)\ =\ span(v,w+c\:v)\:.$&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2011-02-03T07:33:09.307" />
  <row Id="20211" PostTypeId="2" ParentId="20209" CreationDate="2011-02-03T07:57:47.223" Score="7" ViewCount="" Body="&lt;p&gt;Here's my take; I'm sure there are other, better answers.  (community wiki?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the one hand, you could ask a question that isn't very meaningful (e.g. why is 2 the only even prime), while on the other hand you could ask a question that in all likelihood you'll never be able to answer (e.g. describe explicitly the group of diffeomorphisms of a given manifold).  The trick is to shoot somewhere in between, to find fruit hanging high enough that it's interesting and deep but low enough that you actually have a fighting chance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another (related) interpretation is that there are certain questions that have motivated a whole lot of fascinating research and discovery (why isn't there a quintic equation, what are the maps between spheres up to homotopy, are there or aren't there an infinite number of twin primes, etc.), and it takes a good understanding of the meta-structure of the field of mathematics and the way it progresses (or perhaps occasionally a blatant disregard for that meta-structure!) to hit upon such questions.  Anyone can plug away through book after book, but it's as least as important to try to come to grips with the bigger picture.&lt;/p&gt;&#xA;" OwnerUserId="2159" LastActivityDate="2011-02-03T07:57:47.223" CommunityOwnedDate="2011-02-03T10:29:01.743" />
  <row Id="20214" PostTypeId="2" ParentId="20204" CreationDate="2011-02-03T09:11:44.107" Score="1" ViewCount="" Body="&lt;p&gt;Yet another attempt to convey the answer succinctly:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By definition, the rowspan is the collection of vectors you can get when you take (the sums of) all possible combinations of the row-vectors (together with scalar multiples).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you add one vector to another, you may think of this as &quot;doing some of the combining ahead of time&quot;. You're not going to change the collection of &quot;all possible combinations of the rows&quot; by doing some of the combining ahead of time.&lt;/p&gt;&#xA;" OwnerUserId="439" LastActivityDate="2011-02-03T09:11:44.107" />
  <row Id="20215" PostTypeId="2" ParentId="20155" CreationDate="2011-02-03T09:38:03.317" Score="2" ViewCount="" Body="&lt;p&gt;Hint:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You basically have two hypotheses, each has a distribution for the deflection. This distribution depends on the number of observations as well. So, you compute the probability of rejection of one hypothesis under assumption of the other and you impose it to be 90% . This leads to a t-value which you equate to your test statistic to determine the size of the sample. You do this for both cases and take the maximum of the results.&lt;/p&gt;&#xA;" OwnerUserId="3567" LastEditorUserId="3567" LastEditorDisplayName="" LastEditDate="2011-02-03T12:09:49.973" LastActivityDate="2011-02-03T12:09:49.973" />
  <row Id="20216" PostTypeId="2" ParentId="20209" CreationDate="2011-02-03T09:41:26.540" Score="4" ViewCount="" Body="&lt;p&gt;Once you move on to original research, there will no longer be exercise questions on some sheet of papert. There might be some open problems in your area of interest. But in general, it will have to be you yourself who finds an interesting problem, formulates a (new) question about this problem and answers it. A set of such question-answer pairs might then lead to a solution of the problem.&lt;/p&gt;&#xA;" OwnerUserId="3330" LastActivityDate="2011-02-03T09:41:26.540" CommunityOwnedDate="2011-02-03T10:29:01.743" />
  <row Id="20217" PostTypeId="1" CreationDate="2011-02-03T09:51:21.500" Score="4" ViewCount="384" Body="&lt;p&gt;I was wondering why an integral domain is required to be commutative and have a multiplicative neutral element.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example the quaternions would be a non-commutative integral domain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The interesting property is that it has no zero divisors, so why require it to be commutative and have a $1$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words: what consequences follow from having a one vs. not having one when investigating rings without zero divisors?&lt;/p&gt;&#xA;" OwnerUserId="5798" LastEditorUserId="5798" LastEditorDisplayName="" LastEditDate="2011-02-03T13:22:34.307" LastActivityDate="2011-02-05T09:08:32.617" Title="Why is an integral domain a commutative ring with unity?" Tags="&lt;abstract-algebra&gt;&lt;soft-question&gt;&lt;ring-theory&gt;" AnswerCount="4" CommentCount="9" FavoriteCount="1" />
  <row Id="20218" PostTypeId="1" CreationDate="2011-02-03T09:59:32.393" Score="0" ViewCount="99" Body="&lt;p&gt;Hi, everyone:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Say X is an m-manifold (m&gt;2) , and Sg is the genus-g surface. If I obtain a space X' by embedding Sg into X. How does that change the homology and homotopy of X? More specifically, what is the homology H* and (if H* not Abelian) the homotopy of X'? I know of formulas for&#xA;the homology of product spaces, a bit for quotients; but I cannot think of how to work with embeddings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks For any Ideas, Comments, Refs.&lt;/p&gt;&#xA;" OwnerUserId="6600" LastActivityDate="2011-02-04T04:41:45.630" Title="Embedded Surfaces and Homology; Fundamental Group" Tags="&lt;algebraic-topology&gt;&lt;general-topology&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="20219" PostTypeId="2" ParentId="20200" CreationDate="2011-02-03T10:00:08.927" Score="0" ViewCount="" Body="&lt;p&gt;Merely knowing the vector $E$ and the matrix $D$ doesn't seem to be enough to determine $G$, since $E$ is some vector in the column space of $X$, and $G$ can be more or less any other vector in that column space. I don't see how you can tell what $G$ is without using what the individual columns in $X$ are and how they were combined to obtain $E$.&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2011-02-03T10:00:08.927" />
  <row Id="20220" PostTypeId="2" ParentId="20209" CreationDate="2011-02-03T10:07:08.003" Score="3" ViewCount="" Body="&lt;p&gt;I would say that most of the advancements in Mathematics were made as a result of trying to resolve conjectures.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Good conjectures themselves arise from asking questions, picking the right ones and failing to answer them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even during problem solving, many times, one finds that you need to make assumptions (which are really questions in disguise) which help solve the problem, and then verify/prove that the assumptions are true. It is probably easier to verify/prove the assumptions than coming up with them, and without making the right assumptions the problem would probably be unsolved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Without new questions, there would be very little advancement.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2011-02-03T10:07:08.003" CommunityOwnedDate="2011-02-03T10:07:08.003" />
  <row Id="20221" PostTypeId="2" ParentId="19917" CreationDate="2011-02-03T10:09:43.913" Score="0" ViewCount="" Body="&lt;p&gt;Actually, doesn't every S^(2n-1) admit the nowhere-zero tangent vector field:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(-x2,x1,-x4,x3,..,-x2n )? (using that S^(2n-1)={ x in R^2n , ||x||=1} )&lt;/p&gt;&#xA;" OwnerUserId="6600" LastActivityDate="2011-02-03T10:09:43.913" CommentCount="3" />
  <row Id="20222" PostTypeId="2" ParentId="19820" CreationDate="2011-02-03T10:38:46.180" Score="4" ViewCount="" Body="&lt;p&gt;The shortest way is to notice that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$v^2=\vec{v}\cdot\vec{v}$$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;After differentiating: $$2v\frac{dv}{dt}=2\vec{v}\cdot\frac{d\vec{v}}{dt}$$ Done.&lt;/p&gt;&#xA;" OwnerUserId="2323" LastActivityDate="2011-02-03T10:38:46.180" CommentCount="2" />
  <row Id="20223" PostTypeId="1" AcceptedAnswerId="20227" CreationDate="2011-02-03T10:52:53.070" Score="1" ViewCount="132" Body="&lt;p&gt;Hi, how do I get modulo equations to satisfy a given X in CRT.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example say I have &lt;code&gt;X = 1234&lt;/code&gt;. I choose &lt;code&gt;m&lt;sub&gt;i&lt;/sub&gt;&lt;/code&gt; as &lt;code&gt;5, 7, 11, 13&lt;/code&gt;. This satisfies the simple requirements of &lt;a href=&quot;http://en.wikipedia.org/wiki/Secret_sharing_using_the_Chinese_remainder_theorem#Mignotte.27s_threshold_secret_sharing_scheme&quot; rel=&quot;nofollow&quot;&gt;Mignotte's threshold secret sharing scheme&lt;/a&gt;. More precisely given in my example &lt;code&gt;k = n = 4&lt;/code&gt;, and the product of any &lt;code&gt;k - 1&lt;/code&gt; is smaller then &lt;code&gt;X&lt;/code&gt; how come simply computing the remainder of each won't give equations that solve to &lt;code&gt;X = 1234&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of the example,&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;x = 4 mod 5&#xA;x = 2 mod 7&#xA;x = 2 mod 11&#xA;x = 12 mod 13&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Which resolves to &lt;a href=&quot;http://www.mathcelebrity.com/chinese.php?matrix1=x+%3D+4+mod+5%0D%0Ax+%3D+2+mod+7%0D%0Ax+%3D+2+mod+11%0D%0Ax+%3D+12+mod+13&amp;amp;pl=Solve+using+Chinese+Remainder+Theorem&quot; rel=&quot;nofollow&quot;&gt;31264&lt;/a&gt; (won't CRT produce the smallest?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any hints?&lt;/p&gt;&#xA;" OwnerUserId="6570" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2011-02-03T10:56:46.383" LastActivityDate="2011-02-04T01:42:03.523" Title="Getting an X for Chinese Remainder Theorem (CRT)" Tags="&lt;calculus&gt;&lt;modular-arithmetic&gt;" AnswerCount="3" />
  <row Id="20224" PostTypeId="1" CreationDate="2011-02-03T10:54:22.417" Score="2" ViewCount="80" Body="&lt;p&gt;How do I prove that dot product of two free vectors ( inner product in 2D and 3D vectors) does not depend on the choice of frames in which their coordinates are defined?&lt;/p&gt;&#xA;" OwnerUserId="6572" LastActivityDate="2011-02-03T14:57:52.310" Title="Inner product and the choice of coordinate frames" Tags="&lt;linear-algebra&gt;&lt;inner-product-space&gt;" AnswerCount="2" CommentCount="6" />
  <row Id="20225" PostTypeId="1" AcceptedAnswerId="20231" CreationDate="2011-02-03T10:56:04.897" Score="1" ViewCount="58" Body="&lt;p&gt;Let $R$ be a commutative ring, and let $M,N,P$ be $R$-modules.  In what generality can we say that $Hom_R(M,N\otimes_R P)\cong Hom_R(M,N)\otimes_R Hom_R(M,P)$.  This is true in a cartesian monoidal category, and from that, it seems like we might be able to force it if all of the modules are free (look at basis-to-basis mappings and distribute the Hom first), but do we need any finiteness conditions?  Do we need freeness at all?&lt;/p&gt;&#xA;" OwnerUserId="6573" LastActivityDate="2011-02-03T15:06:11.590" Title="In what generality does the second argument of Hom distribute over tensor?" Tags="&lt;commutative-algebra&gt;" AnswerCount="1" />
  <row Id="20226" PostTypeId="1" CreationDate="2011-02-03T11:07:38.630" Score="6" ViewCount="74" Body="&lt;p&gt;Let $T$ be the operator from $L^2(\mathbb R^n)$ to $L^2(\mathbb R^n)$&lt;br&gt;&#xA;defined as composition of convolution and multiplication, &#xA;$Tf := (af) * g$ where $g$ is in $L^2$ and $a$ is a bounded function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can we find the spectrum of $T$? &#xA;For $a$ identically equal 1, the spectrum is the essential range of the &#xA;Fourier transform of $g$. &#xA;I am interested in the more general case.&#xA;If both $a$ and the Fourier transform of $g$ are positive functions,&#xA;I assume that the spectrum of $T$ will also be positive but don't have a proof. &lt;/p&gt;&#xA;" OwnerUserId="6574" LastActivityDate="2011-02-03T11:07:38.630" Title="Spectrum of composition of convolution and multiplication" Tags="&lt;analysis&gt;&lt;functional-analysis&gt;" CommentCount="4" FavoriteCount="1" />
  <row Id="20227" PostTypeId="2" ParentId="20223" CreationDate="2011-02-03T11:10:44.220" Score="3" ViewCount="" Body="&lt;p&gt;The final result of the CRT calculation must be reduced modulo 5 x 7 x 11 x 13 = 5005. This gives the correct answer.&lt;/p&gt;&#xA;" OwnerUserId="1508" LastActivityDate="2011-02-03T11:10:44.220" CommentCount="1" />
  <row Id="20228" PostTypeId="2" ParentId="20127" CreationDate="2011-02-03T11:16:43.887" Score="0" ViewCount="" Body="&lt;p&gt;Funny, your second question &lt;a href=&quot;http://math.stackexchange.com/questions/19740/efficient-method-to-compute-mathrmgcd2n-1-n&quot;&gt;came up here&lt;/a&gt; a couple of days ago. I answered it &lt;a href=&quot;http://math.stackexchange.com/questions/19740/efficient-method-to-compute-mathrmgcd2n-1-n/20038#20038&quot;&gt;here&lt;/a&gt;. Perhaps $n=3^{19}$ is too large to make this method practical, but Douglas explains how to do without using $N$.&lt;/p&gt;&#xA;" OwnerUserId="1508" LastActivityDate="2011-02-03T11:16:43.887" />
  <row Id="20230" PostTypeId="1" CreationDate="2011-02-03T11:27:33.670" Score="1" ViewCount="203" Body="&lt;p&gt;I know this is a very standard question widely popular in the Internet and the Mathworld.&#xA;I myself have solved the above problem is N^2 Log N avoiding floating arithmetic.However, can anyone give me a good resource/detailed explanation of how it can be solved in O(N^2) using point line duality concepts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sorry to create such a confusion regarding the statement.&#xA;The text would be:&#xA;&quot;N different points with integer coordinates are given in a plane. You are to write a program that finds the maximum number of collinear points (they all belong to the same line).&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You may also refer to &lt;a href=&quot;http://www.spoj.pl/problems/CHASE/&quot; rel=&quot;nofollow&quot;&gt;this site&lt;/a&gt; which is the problem I have solved using a N^2Log N approach.&lt;/p&gt;&#xA;" OwnerUserId="6536" LastEditorUserId="1543" LastEditorDisplayName="" LastEditDate="2011-02-03T16:45:28.093" LastActivityDate="2011-02-03T17:09:26.413" Title="maximum number of collinear points?" Tags="&lt;algorithms&gt;&lt;euclidean-geometry&gt;&lt;online-resources&gt;&lt;intersection&gt;" AnswerCount="2" CommentCount="5" FavoriteCount="1" />
  <row Id="20231" PostTypeId="2" ParentId="20225" CreationDate="2011-02-03T11:29:12.940" Score="1" ViewCount="" Body="&lt;p&gt;This is not true even if all the modules are free; let $R$ be a field, let $M, N, P$ be finite-dimensional, and compare dimensions. (It &lt;em&gt;is&lt;/em&gt; true if $\otimes$ is replaced with $\oplus$; this is because $\oplus$ is both a finite product and a finite coproduct.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The tensor product is not a Cartesian product, and in this sense it does not even behave like a Cartesian product. The sense in which it does behave like a Cartesian product is the adjunction&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\text{Hom}_R(M \otimes_R N, P) \simeq \text{Hom}_R(M, \text{Hom}_R(N, P))$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\text{Hom}_R(N, P)$ is the $R$-module of all $R$-module homomorphisms. This is analogous to the adjunction&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\text{Hom}(M \times N, P) \simeq \text{Hom}(M, P^N)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;in $\text{Set}$, where $P^N$ is the &lt;em&gt;set&lt;/em&gt; of all functions $N \to P$. This and the hom above are examples of &quot;internal homs,&quot; and if internal hom has a left adjoint it defines a notion of tensor product which does not necessarily have to behave like a Cartesian product; when it does, you might be in a &lt;a href=&quot;http://en.wikipedia.org/wiki/Topos#Elementary_toposes_.28toposes_in_logic.29&quot; rel=&quot;nofollow&quot;&gt;topos&lt;/a&gt;, and the category of $R$-modules is not a topos but an abelian category.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2011-02-03T11:29:12.940" />
  <row Id="20232" PostTypeId="1" AcceptedAnswerId="20236" CreationDate="2011-02-03T11:37:39.510" Score="2" ViewCount="52" Body="&lt;p&gt;Let $R$ be a CRing and let $M,N$ be $R$-modules.  Let $M^*:=Hom_R(M,R)$.  I have seen the following isomorphism asserted in the case where $R$ is a field and $M$ and $N$ are f.g. vector spaces:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$M^*\otimes_R N\cong Hom_R(M,N)$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can give a proof of this using a basis, but in what generality can we say this kind of thing?  Does it have a nice arrow-theoretic proof?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For some reason, I can't accept answers, so I apologize for not doing it if I am having trouble.  Maybe the mods can force my account to accept the answers? (To answer the gentleman's question, I am having javascript problems, so I fear that it won't help).&lt;/p&gt;&#xA;" OwnerUserId="6573" LastEditorUserId="6573" LastEditorDisplayName="" LastEditDate="2011-02-03T11:43:56.693" LastActivityDate="2011-02-03T12:06:40.443" Title="In what generality does the following isomorphism involving tensors and homs hold?" Tags="&lt;commutative-algebra&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="20233" PostTypeId="2" ParentId="20224" CreationDate="2011-02-03T11:39:17.143" Score="2" ViewCount="" Body="&lt;p&gt;Since I guess it's homework, I'll just give a hint. A change of basis (or 'frame') is a linear mapping, i.e. applying an certain matrix $A$, sending each vector $v \mapsto A.v.$ What properties does $A$ have?&lt;/p&gt;&#xA;" OwnerUserId="6004" LastActivityDate="2011-02-03T11:39:17.143" CommentCount="1" />
  <row Id="20234" PostTypeId="2" ParentId="20204" CreationDate="2011-02-03T11:48:45.187" Score="0" ViewCount="" Body="&lt;p&gt;The main point is that elementary operations are &lt;em&gt;invertible&lt;/em&gt;, so you can go back to the original set of vectors.&lt;/p&gt;&#xA;" OwnerUserId="589" LastActivityDate="2011-02-03T11:48:45.187" CommentCount="3" />
  <row Id="20236" PostTypeId="2" ParentId="20232" CreationDate="2011-02-03T12:06:40.443" Score="2" ViewCount="" Body="&lt;p&gt;Again ol' Bourbaki gives you an answer: when $M$ or $N$ are finitely projective modules, the canonical morphism $M^* \otimes N \longrightarrow \mathrm{Hom}(M,N)$ is an isomorphism (Algebra I, chapter II, 4.2, proposition 2).&lt;/p&gt;&#xA;" OwnerUserId="664" LastActivityDate="2011-02-03T12:06:40.443" />
  <row Id="20237" PostTypeId="2" ParentId="20126" CreationDate="2011-02-03T12:10:59.820" Score="6" ViewCount="" Body="&lt;p&gt;Since $\int_2^n {\frac{1}{{(x^2  + 1)^n }}\,{\rm d}x}  \le \frac{{n - 2}}{{5^n }}$, it suffices to show that&#xA;$$&#xD;&#xA;\frac{1}{{(n + 1)2^n }} \le \int_1^2 {\frac{1}{{(x^2  + 1)^n}}\,{\rm d}x}  \le \frac{1}{{(n - 1)2^n }}.&#xD;&#xA;$$&#xA;This follows straight from&#xA;$$&#xD;&#xA;\frac{{2 - x}}{2} \le \frac{1}{{x^2  + 1}} \le \frac{1}{{2x}}.&#xD;&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="2810" LastActivityDate="2011-02-03T12:10:59.820" CommentCount="1" />
  <row Id="20238" PostTypeId="1" AcceptedAnswerId="20240" CreationDate="2011-02-03T13:03:16.353" Score="3" ViewCount="244" Body="&lt;p&gt;How many 6-letter permutations can be formed using only the letters of the word, &lt;strong&gt;MISSISSIPPI&lt;/strong&gt;? I understand the trivial case where there are no repeating letters in the word (for arranging smaller words) but I stumble when this isn't satisfied. I'm wondering if there's a simpler way to computing all the possible permutations, as an alternative to manually calculating all the possible 6-letter combinations and summing the corresponding permutations of each. In addition, is there a method to generalize the result based on any &lt;strong&gt;P&lt;/strong&gt; number of letters in a set with repeating letters, to calculate the number of &lt;em&gt;n&lt;/em&gt;-letter permutations?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sorry if this question (or similar) has been asked before, but if it has, please link me to the relevant page. I also apologize if the explanation is unclear, or if I've incorrectly used any terminology (I'm only a high-school student.) If so, please comment. :-) &lt;/p&gt;&#xA;" OwnerUserId="6577" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-03T14:39:28.343" LastActivityDate="2011-02-03T17:34:52.020" Title="6-letter permutations in MISSISSIPPI" Tags="&lt;combinatorics&gt;&lt;permutations&gt;&lt;counting&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="20239" PostTypeId="2" ParentId="3444" CreationDate="2011-02-03T13:19:04.937" Score="3" ViewCount="" Body="&lt;p&gt;This is another answer in terms of Poisson processes and the Gamma distribution, and it still uses a bit of calculus which you might call a trick, but I think at least it does add another bit of intuition:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the &lt;a href=&quot;http://en.wikipedia.org/wiki/Poisson_process&quot; rel=&quot;nofollow&quot;&gt;homogeneous Poisson process&lt;/a&gt; with rate parameter 1; this means we are counting the number of occurrences of an event that happens with rate 1. Let's calculate the probabilities $p_k(t)$ that we are in state $k$ at time $t$, i.e., that the event occurs $k$ times in the interval $[0,t]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the event happens with rate 1, probability mass flows from $p_k(t)$ to $p_{k+1}(t)$ with rate 1. This means that ${p&amp;#39;}_0(t) = -p_0(t)$ and ${p&amp;#39;}_{k+1}(t) = p_k(t) - p_{k+1}(t)$. Also, $p_0(0) = 1$ and $p_{k+1}(0) = 0$. Here comes the bit of calculus: these equations have the solution $p_k(t) = \frac{t^k}{k!} e^{-t}$. (A bit fuzzily, we can read this in two parts: $\frac{t^k}{k!}$ is 1 integrated $k$ times, and $e^{-t}$ represents probability mass being lost at rate 1 to states further down the line. See below for yet another fuzzy explanation.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now consider the waiting time $T_k$ until the $k$'th occurrence. Clearly, $T_k = t$ means that the transition from state $k-1$ to state $k$ happens at time $t$, so the probability of $T_k \le t$ is the probability that the transition happens before time $t$, and the density is the derivative of this, i.e., the rate at which probability mass flows from state $k-1$ to $k$. This equals the occurrence rate (i.e., 1) times $p_{k-1}(t)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the probability density of the random variable $T_k$ is $1 \cdot p_{k-1}(t) = \frac{t^{k-1}}{(k-1)!} e^{-t}$ (for $t \ge 0$). Since the probability that there is no occurrence ever is obviously zero, $\int_{0}^{\infty} \frac{t^{k-1}}{(k-1)!} e^{-t} dt = 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Incidentally, this is related to a way of thinking about why $\sum_{k=0}^\infty \frac{t^k}{k!} = \lim_{n\to\infty} (1 + \frac{t}{n})^n$. Suppose you start with one unit of money in an account and get 100% interest, continuously compounded. However, the interest from the original account (number 0) is paid not the original account, but to account #1; interest from account #1 is paid to account #2, and so on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then the money $m_k(t)$ in account $k$ equals $1$ integrated $k$ times, and the total money $m(t)$ is $\sum_{k=0}^\infty m_k(t) = \sum_{k=0}^\infty \frac{t^k}{k!}$. But on the other hand, $m(t)$ is continuously compounded at 100% interest, so $m(t) = \lim_{n\to\infty} (1 + \frac{t}{n})^n$ by the usual reasoning.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This gives another fuzzy argument why we should have $p_k(t) = \frac{t^k}{k!} e^{-t}$. The change in the $p_k(t)$ over time consists of two parts: on the other hand, each $p_{k+1}(t)$ increases at rate $p_k(t)$; on the other hand, each $p_k(t)$ decreases at rate $p_k(t)$. If we view the $p_k(t)$ as accounts, since we are taking money out of every account at the constant rate 1, the effect is to decrease the total amount at the constant rate 1, i.e. by a factor of $e^{-t}$, which cancels out the increase of $e^t$ due to the accruing interest.  It makes some intuitive sense that we can model this effect by simply rescaling the amount of money in each of the accounts by $e^{-t}$.&lt;/p&gt;&#xA;" OwnerUserId="6578" LastEditorUserId="6578" LastEditorDisplayName="" LastEditDate="2011-02-03T13:41:30.760" LastActivityDate="2011-02-03T13:41:30.760" />
  <row Id="20240" PostTypeId="2" ParentId="20238" CreationDate="2011-02-03T13:54:56.220" Score="3" ViewCount="" Body="&lt;p&gt;I can think of a generating function type of approach. You have 1 M, 2 P's, 4 I's and 4 S's in the word MISSISSIPPI. Suppose you picked the two P's and four I's, the number of permutations would be $\frac{6!}{4! 2!}$. However, we need to sum over all possible selections of 6 letters from this group.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer will be the coefficient of $x^6$ in&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;6!\left(1 + \frac{x}{1!}\right)\left(1 + \frac{x}{1!} + \frac{x^2}{2!}\right)\left(1 + \frac{x}{1!} + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!}\right)^2&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each polynomial term corresponds to the ways in which you could make a selection of a given letter. So you have $1 + x$ for the letter M and $1 + x + x^2/2$ for the 2 letters P and so on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The coefficient of $x^6$ comes out to 1610 in this case. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;: (I'm elaborating a bit in response to George's comment). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a pretty standard approach to such counting problems. The value of $x$ is not relevant to the problem at all. The benefit of using such polynomials is that it gives you a nice tool to &quot;mechanically&quot; solve the problem. The idea is that by looking at the coefficient of a particular term in the expanded polynomial, you get the answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I wrote a term (1+x) corresponding to the letter M, it captures two possibilities&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) I could either leave out M (which corresponds to the coefficient of x^0 which is 1)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) I could include M, which corresponds to the coefficient of x^1 which is one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose you select 1M, 2I's 2P's and 1 S. This is encoded by the term $x^1\cdot x^2 \cdot x^2 \cdot x^1$. The first $x^1$ term corresponds to selecting the single M. The first $x^2$ term corresponds to selecting 2 I's (which are identical). Using similar logic, the next $x^2$ is for 2P's and the last $x^1$ is for 1S. Since you are interested in permutations with repetition, the number of permutations for this case will be $\frac{6!}{2!2!}$, which should be the coefficient of this term.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How would you encode 0 M, 3I's, 2P's and 1S? The term would then be $x^0 \cdot x^3 \cdot x^2 \cdot x^1$. However, this term would have to be multiplied by $\frac{6!}{3!2!}$ to keep the count correct. The $6!$ in the numerator will be common to all such terms. The denominator depends on your selection of letters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You need to add all such possibilities. Instead of listing them all out, which will be laborious, you represent the possibility of choosing each letter by a polynomial. As an example, for 4 S's. you have $1 + \frac{x}{1!} + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!}$. You divide $x^n$ by $n!$ to keep the count correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You then multiply the polynomials and look at the appropriate coefficient.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;6!\left(1 + \frac{x}{1!}\right)\left(1 + \frac{x}{1!} + \frac{x^2}{2!}\right)\left(1 + \frac{x}{1!} + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!}\right)^2&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You expand out this polynomial and look at the coefficient of $x^6$ which gives you the answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For more powerful uses of this kind of approach, please read the book at &lt;a href=&quot;http://www.math.upenn.edu/~wilf/gfologyLinked2.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.math.upenn.edu/~wilf/gfologyLinked2.pdf&lt;/a&gt; (Warning -  it a big PDF file).&lt;/p&gt;&#xA;" OwnerUserId="2641" LastEditorUserId="2641" LastEditorDisplayName="" LastEditDate="2011-02-03T17:34:52.020" LastActivityDate="2011-02-03T17:34:52.020" CommentCount="3" />
  <row Id="20241" PostTypeId="2" ParentId="15330" CreationDate="2011-02-03T13:55:08.510" Score="1" ViewCount="" Body="&lt;p&gt;You can certainly find the Taylor series of exp(-x) around 0.25.  &lt;a href=&quot;http://www.wolframalpha.com/input/?i=expand+exp%28-x%29+around+.25&quot; rel=&quot;nofollow&quot;&gt;Wolfram Alpha&lt;/a&gt; gives an answer.  Then you can plug large numbers into it if you want.  It just won't be at all accurate.  But I don't think I am understanding what you mean by your boldface question.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-02-03T13:55:08.510" CommentCount="3" />
  <row Id="20242" PostTypeId="2" ParentId="20217" CreationDate="2011-02-03T14:03:49.517" Score="0" ViewCount="" Body="&lt;p&gt;There are two things that come to my mind.  One thing is that the polynomial ring isn't as nice.  For example over a field, the number of roots of a polynomial is bounded by the degree.  This isn't the case over a skew-field (a ring satisfying all the axioms of a field except for commutativity).  For example, in the quaternions, the innocent looking polynomial $x^2 + 1$ has infinitely many roots (any purely imaginary quaternion with length 1)!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another thing that gets tricky is linear algebra over skew-fields.  Here you have to distinguish between left actions and right actions.  Also nice and useful formulas don't hold, e.g. it is not the case that $tr(AB) = tr(BA)$ or that $\det(AB) = \det A \det B$. &lt;/p&gt;&#xA;" OwnerUserId="9" LastActivityDate="2011-02-03T14:03:49.517" CommentCount="1" />
  <row Id="20243" PostTypeId="2" ParentId="20195" CreationDate="2011-02-03T14:08:20.417" Score="0" ViewCount="" Body="&lt;p&gt;Suppose we have dice $Da$ and $Db$, with $a \le b$. Then there are three cases:  &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;If $2 \le n \le a$, the probability of throwing $n$ is $\frac{n-1}{ab}$.&lt;/li&gt;&#xA;&lt;li&gt;If $a+1 \le n \le b$, the probability of throwing $n$ is $\frac{1}{b}$.&lt;/li&gt;&#xA;&lt;li&gt;If $b+1 \le n \le a+b$, the probability of throwing $n$ is $\frac{a+b+1-n}{ab}$.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="1508" LastActivityDate="2011-02-03T14:08:20.417" />
  <row Id="20244" PostTypeId="2" ParentId="20217" CreationDate="2011-02-03T14:17:17.960" Score="1" ViewCount="" Body="&lt;p&gt;From one general algebraic perspective, the &quot;point&quot; of defining integral domains is really to define their &lt;a href=&quot;http://en.wikipedia.org/wiki/Field_of_fractions&quot; rel=&quot;nofollow&quot;&gt;fraction fields&lt;/a&gt;. Integral domains are precisely the (unital) subrings of (commutative) fields, so in that sense the study of integral domains is a natural extension of the study of fields. This is why the basic objects in, say, algebraic number theory are certain integral domains which are subrings of number fields.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the perspective of number theory, the &quot;point&quot; of defining integral domains is to impose an obvious requirement on any ring with a nice theory of factorization: if $0$ can have a nontrivial factorization, then there are going to be problems. Again, this is why in algebraic number theory we study integral domains.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the perspective of &lt;a href=&quot;http://en.wikipedia.org/wiki/Algebraic_geometry&quot; rel=&quot;nofollow&quot;&gt;algebraic geometry&lt;/a&gt;, the point of defining integral domains is as follows. Associated to any nice integral domain $R$ is a certain topological space, its &lt;a href=&quot;http://en.wikipedia.org/wiki/Spectrum_of_a_ring&quot; rel=&quot;nofollow&quot;&gt;spectrum&lt;/a&gt;, and one should roughly speaking think of elements of $R$ as functions on this space $\text{Spec } R$. Then the requirement that $R$ is an integral domain translates into requiring a very strong connectivity hypothesis on this space. If $a, b \in R$ such that $ab = 0$, then the set of points in $\text{Spec } R$ such that $a = 0$ and the set of points in $\text{Spec } R$ such that $b = 0$ are closed subsets of $\text{Spec } R$ whose union is all of $\text{Spec } R$, and under certain hypotheses on $R$ these subsets are proper. Requiring that $R$ has no zero divisors is (again, under certain hypotheses) equivalent to requiring that it is not possible to partition $\text{Spec } R$ in this way. This is a very strong form of connectivity called &lt;a href=&quot;http://en.wikipedia.org/wiki/Irreducible_component&quot; rel=&quot;nofollow&quot;&gt;irreducibility&lt;/a&gt;. It turns out that any (nice) &lt;a href=&quot;http://en.wikipedia.org/wiki/Algebraic_variety&quot; rel=&quot;nofollow&quot;&gt;variety&lt;/a&gt; is a unique disjoint union of a finite number of irreducible components, so it's natural to study irreducible varieties.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for requiring units, all of the natural examples I can think of have units, and all of the natural morphisms I can think of preserve them. The category of rings without units is messy and, while it has its place, I don't think it's appropriate for any of the applications I've described above. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Geometrically, consider the following basic example: if $X$ is a topological space, then there is a ring $C(X)$ of continuous functions $X \to \mathbb{R}$. This ring is commutative, and it has a unit, although it is not usually an integral domain. In any case, the idea here is that we can study topological properties of $X$ by studying algebraic properties of $C(X)$, and in particular any continuous map $X \to Y$ gives an $\mathbb{R}$-algebra homomorphism $C(Y) \to C(X)$ in the other direction by precomposition. This ring homomorphism automatically preserves units, so it's not natural to ignore the units here. If $X$ and $Y$ are compact Hausdorff, something amazing happens: &lt;em&gt;all&lt;/em&gt; $\mathbb{R}$-algebra homomorphisms $C(Y) \to C(X)$ preserving units come from continuous maps $X \to Y$, so one really can get all the topological properties of compact Hausdorff spaces using commutative &lt;em&gt;unital&lt;/em&gt; rings.&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2011-02-03T14:22:52.273" LastActivityDate="2011-02-03T14:22:52.273" />
  <row Id="20245" PostTypeId="2" ParentId="20217" CreationDate="2011-02-03T14:32:56.143" Score="6" ViewCount="" Body="&lt;p&gt;Yes, people do study rings without unity (see for example &lt;a href=&quot;http://math.stackexchange.com/questions/16168/applications-of-rings-without-identity/16171#16171&quot;&gt;this previous answer&lt;/a&gt; for some comments on rings without unity). So people do study commutative-rings-without-unit-that-have-no-zero-divisors. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Likewise, people do study non-commutative rings (with or without identity), and study such rings that have no zero divisors (though from what I can tell from my colleagues, zero divisors don't worry them quite as much as nilpotent elements do, so &quot;reduced rings&quot; seem to play a bigger role). So certainly, people study things that are &quot;almost domains except for one of the properties&quot; (nobody drops the $1\neq 0$ condition, though, because that makes it too easy). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A ring with 1 which is not the zero ring and has no zero divisors is called a &lt;strong&gt;domain&lt;/strong&gt; (at least in Lam's &lt;strong&gt;First Course in Noncommutative Rings&lt;/strong&gt;). They are defined on page 3 of the book, so pretty early. They can get pretty hairy: if you have a domain, it may not be embeddable in a division ring (in contrast to the integral domain case, where you &lt;em&gt;always&lt;/em&gt; have the field of fractions); this was proven by Mal'cev, and is presented in Lam's &lt;strong&gt;Lectures on Rings and Modules&lt;/strong&gt;, section 9B.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So really your question is more along the lines of &lt;em&gt;&quot;Why do we distinguish precisely integral domains, and not some of these other objects?&quot;&lt;/em&gt; Maybe we should study rings without zero divisors, give them a special name (say, &quot;wuzeds&quot;, for &quot;without zero divisors&quot;, and then call integral domains &quot;commutative wuzeds&quot;?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Partly, history. Ring theory was born from ideas of Dedekind, and later Noether and Artin, which arose from number theory considerations; they all took place in the commutative setting, and most of what Dedekind worked with were in fact (a special family of examples of) what we now call &quot;integral domains&quot;. Integral domains were simply studied more. Also, a lot of ring theory originally arose as a way to try to abstract these ideas, generalize the results; a lot of the results were based on polynomials, on matrices, and other objects constructed &lt;em&gt;from&lt;/em&gt; the rings, which people were familiar with in the case of the integers and other prototypes. Most of these objects become &lt;em&gt;very&lt;/em&gt; hard when you switch to rings without identity (for example, whereas every ring with identity can be embedded in both its polynomial and its matrix rings in canonical ways, rings without identity are far more difficult to embed), or when you switch to noncommutative rings. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So people were studying special cases of &lt;em&gt;commutative rings&lt;/em&gt; before they were studying noncommutative rings. That's also why we call them &lt;em&gt;fields&lt;/em&gt; and &lt;em&gt;skew-fields&lt;/em&gt; (or &lt;em&gt;division rings&lt;/em&gt;), and not &quot;commutative division rings&quot; and &quot;division rings&quot;. (Bourbaki attempted to change this, by defining &quot;field&quot; to be a ring with identity, $1\neq 0$, in which every nonzero element has an inverse, i.e. what we usually call a division ring, and then talking about fields and commutative fields; it did not stick outside of France). &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-03T16:02:48.047" LastActivityDate="2011-02-03T16:02:48.047" />
  <row Id="20246" PostTypeId="1" CreationDate="2011-02-03T14:49:21.513" Score="4" ViewCount="202" Body="&lt;p&gt;I have the following problem:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In a triangle $ABC$ the line joining incentre and circumcentre is parallel to side $BC$. Prove that $\cos B + \cos C=1$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Could someone help me solve it?&lt;/p&gt;&#xA;" OwnerUserId="3786" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-03T16:09:45.013" LastActivityDate="2011-02-06T19:21:24.917" Title="Geometry problem on the incentre and circumcenter of a triangle" Tags="&lt;geometry&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="2" />
  <row Id="20247" PostTypeId="2" ParentId="20224" CreationDate="2011-02-03T14:57:52.310" Score="1" ViewCount="" Body="&lt;p&gt;The dot-product is in the first place a bilinear function of two vector variables with the additional property that ${\bf x}\bullet{\bf x}&amp;gt;0$ for ${\bf x}\ne{\bf 0}$. For any basis $({\bf e}_i)_{1\leq i\leq n}$ of your space the dot-product is encoded in a certain symmetric matrix $G:=(g_{ik})_{1\leq i\leq n,1\leq k\leq n}$. This matrix changes with the basis in a characteristic way. When a dot product is given then there are certain distinguished bases, namely the orthonormal ones. For these the matrix $G$ is just the identity matrix, and in the corresponding coordinates the dot-product computes as $(*)\ {\bf x}\bullet{\bf y} =\sum_{i=1}^n x_i y_i$. In particular, if one starts with the &quot;standard space&quot; ${\mathbb R}^n$ then the standard basis is considered orthonormal, whence one has the standard formula $(*)$ there. &lt;/p&gt;&#xA;" OwnerUserId="1303" LastActivityDate="2011-02-03T14:57:52.310" />
  <row Id="20248" PostTypeId="1" AcceptedAnswerId="20250" CreationDate="2011-02-03T15:43:04.867" Score="1" ViewCount="85" Body="&lt;p&gt;I'm working with the universal and the existential quantifiers and I was wondering is there an order of operations that applies to them? What is the difference between &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(\forall x)P(x)(\exists x)Q(x)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(\exists x)Q(x)(\forall x)P(x)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lets say we have this relationship&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(a) $[(\forall x)P(x) \vee (\exists x)Q(x)] \implies (\forall x)[P(x) \vee Q(x) ]$&lt;/p&gt;&#xA;" OwnerUserId="1040" LastEditorUserId="1040" LastEditorDisplayName="" LastEditDate="2011-02-03T15:54:47.473" LastActivityDate="2011-02-03T15:54:47.473" Title="Does the order of quantification matter?" Tags="&lt;elementary-set-theory&gt;" AnswerCount="1" CommentCount="7" />
  <row Id="20249" PostTypeId="2" ParentId="20230" CreationDate="2011-02-03T15:46:00.733" Score="-1" ViewCount="" Body="&lt;p&gt;The maximum number of collinear points among N points in a plane is N (i.e., if they happen to be collinear.)&lt;/p&gt;&#xA;" OwnerUserId="1782" LastActivityDate="2011-02-03T15:46:00.733" CommentCount="2" />
  <row Id="20250" PostTypeId="2" ParentId="20248" CreationDate="2011-02-03T15:54:40.373" Score="3" ViewCount="" Body="&lt;p&gt;Here is the question you probably meant to ask, and if it isn't, then here is the question you should have asked. Is there a difference between the statement $(\forall x)(\exists y) P(x, y)$ and the statement $(\exists y)(\forall x) P(x, y)$?  The answer is yes. Consider the following simple example:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;For all men, there exists a woman who loves them.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;is not the same as&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;There exists a woman who loves all men.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The point is that when we say $(\forall x)(\exists y)$, the $y$ that is meant is a &lt;em&gt;function of $x$&lt;/em&gt;, and this is not true in the other order. This is unfortunately hidden by the notation.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2011-02-03T15:54:40.373" CommentCount="4" />
  <row Id="20251" PostTypeId="1" AcceptedAnswerId="20253" CreationDate="2011-02-03T16:01:19.197" Score="6" ViewCount="130" Body="&lt;p&gt;Firstly, I'm not a mathematician as will become evident in a quick moment. I was pondering some maths the other day and had an interesting thought: If you encased an integrable function over some range in a primitive with an easily computable area, the probability that a random point within said primitive also exists below that function's curve, scaled by the area of the primitive, is the indefinite integral of the function over that domain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So let's say I want to &quot;solve&quot; for $\pi$. Exploiting a circle's symmetry, I can define $\pi$ as:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$4 \int_{0}^{1}\sqrt{1-x^2} \,dx$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which I can &quot;encase&quot; in the unit square. Since the area of the unit square is 1, $\pi$ is just 4 * the probability that a point chosen at random within the unit square is below the quarter-circle's arc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm sure this is well known, and so my questions are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;What is this called?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Is there anything significant about&#xA;this--for instance, is the&#xA;relationship between the integral and&#xA;the encasing object of interest--or&#xA;is it just another way of phrasing&#xA;indefinite integrals?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Sorry if this is painfully elementary!&lt;/p&gt;&#xA;" OwnerUserId="6580" LastActivityDate="2011-02-03T16:29:20.803" Title="Integrals as Probabilities" Tags="&lt;calculus&gt;&lt;probability&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="20252" PostTypeId="2" ParentId="20251" CreationDate="2011-02-03T16:08:25.657" Score="2" ViewCount="" Body="&lt;p&gt;This is known as &lt;a href=&quot;http://en.wikipedia.org/wiki/Geometric_probability&quot; rel=&quot;nofollow&quot;&gt;geometric probability&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2011-02-03T16:08:25.657" CommentCount="1" />
  <row Id="20253" PostTypeId="2" ParentId="20251" CreationDate="2011-02-03T16:23:18.300" Score="6" ViewCount="" Body="&lt;p&gt;No, this is a very good observation! It is the basis of the modern definition of probability, where &lt;em&gt;all&lt;/em&gt; probabilities are essentially defined as integrals. Your particular observation about $\pi$ being closely related to the probability that a point lands in a circle is also very good, and actually leads to a probabilistic algorithm to &lt;em&gt;compute&lt;/em&gt; $\pi$ (an example of a &lt;a href=&quot;http://en.wikipedia.org/wiki/Monte_Carlo_method&quot; rel=&quot;nofollow&quot;&gt;Monte Carlo method&lt;/a&gt;). The subject in which probabilities are studied as integrals is, broadly speaking, called &lt;a href=&quot;http://en.wikipedia.org/wiki/Measure_(mathematics)&quot; rel=&quot;nofollow&quot;&gt;measure theory&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Monte Carlo methods are also used to numerically compute other integrals; this is called &lt;a href=&quot;http://en.wikipedia.org/wiki/Monte_Carlo_integration&quot; rel=&quot;nofollow&quot;&gt;Monte Carlo integration&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now that you have discovered this wonderful fact, here are some interesting exercises. I recommend that you try to draw the relevant regions when $n = 2, 3$ before tackling the general case.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Choose $n$ numbers randomly in the interval $[0, 1]$. What is the probability that the first number you chose is the biggest one? &lt;/li&gt;&#xA;&lt;li&gt;Choose $n$ numbers randomly in the interval $[0, 1]$. What is the probability that they are in decreasing order?&lt;/li&gt;&#xA;&lt;li&gt;Choose $n$ numbers randomly in the interval $[0, 1]$. What is the probability that their sum is less than $1$?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2011-02-03T16:29:20.803" LastActivityDate="2011-02-03T16:29:20.803" CommentCount="3" />
  <row Id="20254" PostTypeId="1" AcceptedAnswerId="20263" CreationDate="2011-02-03T16:39:13.347" Score="3" ViewCount="132" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Let $f(x)$ be a decreasing, continuous function in $[0, \infty)$.&lt;br&gt;&#xA;  If $\int_{0}^{\infty}f(x)dx$ converges then $\int_{0}^{\infty}\sin(f(x))dx$ converges.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The only improper point is $\infty$ since $f(x)$ is continuous in $[0, \infty)$.&lt;/li&gt;&#xA;&lt;li&gt;$f(x)$ has a bounded anti-derivate: if $\int_{0}^{\infty}f(x)dx$ converges, then $F(x)=\int_{0}^{x}f(t)dt$ is bounded.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;If I set $g(x)=\frac{\sin(f(x))}{f(x)}$ and look at $\int_{0}^{\infty}g(x)f(x)dx$ then Dirichlets test seems like a possible candidate but I'm not sure if $g(x)\searrow 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I feel close but seems like I'm missing something obvious here. Hints are appreciated.&lt;/p&gt;&#xA;" OwnerUserId="3237" LastActivityDate="2011-02-03T19:37:45.943" Title="Show $\int_{0}^{\infty}\sin(f(x))dx$ converges if $\int_{0}^{\infty}f(x)dx$ converges and $f(x)$ is a decreasing, continuous function in $[0, \infty)$" Tags="&lt;calculus&gt;&lt;improper-integrals&gt;" AnswerCount="1" CommentCount="10" />
  <row Id="20255" PostTypeId="1" AcceptedAnswerId="20256" CreationDate="2011-02-03T16:47:09.437" Score="0" ViewCount="186" Body="&lt;p&gt;Heya everyone,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm currently starting a course about Graph Theory, and I've been explained about the Havel &amp;amp; Hakimi degree sequence. However, I'm not too sure about it yet. First of all, I want to know if I understand it correctly, so I'm going to try to explain what I think the theory proves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, I've shamelessly stolen the following statement from the presentation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/0t2u4.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Okay, so what I want to ask is the following.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;What does &quot;is graphic&quot; mean exactly. Does that mean it's a graph? I know stupid question, but I cannot think of a degree sequence that's not a graph...&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;What's the practical use of this. What problems could you solve with this. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="3605" LastActivityDate="2011-02-03T18:04:28.613" Title="Havel &amp; Hakimi degree sequence theory" Tags="&lt;graph-theory&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="20256" PostTypeId="2" ParentId="20255" CreationDate="2011-02-03T16:53:28.020" Score="4" ViewCount="" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;I think the term is usually &quot;graphical.&quot; It just means that $s^{*}$ is a sequence of degrees of the vertices of some graph $G$. So if you can create a graph with degrees from $s^{*}$ then you can create a graph with degrees from $s$ and vice versa. Note it doesn't mean that you create the same graph from $s$ and $s^{*}$. We just need to be able to create graphs from the sets of degrees (e.g. if $s$ is graphical, then we can create a graph $G$ from $s$ and a graph $G&amp;#39;$ from $s^{*}$).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;It can be used to figure out whether two graphs are isomorphic. If the degree sequences of two graphs are different, then they can't be &lt;a href=&quot;http://mathworld.wolfram.com/IsomorphicGraphs.html&quot; rel=&quot;nofollow&quot;&gt;isomorphic&lt;/a&gt;. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="2548" LastEditorUserId="2548" LastEditorDisplayName="" LastEditDate="2011-02-03T17:04:12.877" LastActivityDate="2011-02-03T17:04:12.877" CommentCount="1" />
  <row Id="20258" PostTypeId="1" AcceptedAnswerId="20261" CreationDate="2011-02-03T17:05:37.057" Score="0" ViewCount="97" Body="&lt;p&gt;Hi! I'm returning from an exam on group-theory and there were 2 questions I couldn't solve (and still can't), so I'm asking here for any hint you could possibly give.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Let G be a group and H and K subgroups such that $|H| = n$, $|K| = m$ and $gcd(n, m) = 1$. Show that $H \cap K = \{e\}$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I wish I could show you some of my attempts before hand, but they're all rubbish that didn't get me anywhere. Essentially, the only (and last) thing I remembered and thought it could be useful was to see if if H and K are partitions of G. I think I've read something similar somewhere but can't recall where, so, am uncertain about it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other question I couldn't solve is, I think, related to this, so I shall try it once I understand this one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for taking the time to read! Any tip is appreciated.&lt;/p&gt;&#xA;" OwnerUserId="6538" LastActivityDate="2011-02-03T17:10:39.370" Title="What happens when the order of two finite subgroups are relatively prime?" Tags="&lt;homework&gt;&lt;abstract-algebra&gt;&lt;group-theory&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="20259" PostTypeId="2" ParentId="20223" CreationDate="2011-02-03T17:06:20.997" Score="2" ViewCount="" Body="&lt;p&gt;Here is a much, much simpler way to immediately obtain the sought answer. Contrast the solution below to the much longer solution in your link, which involves calculations with much larger numbers and performs 4 inversions vs. the single simple inversion below. Always look for hidden innate symmetries in a problem before diving head-first into brute-force mechanical calculations! Here the key insight is that the equations can be split into pairs that have obvious solutions, viz.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\quad\quad\quad\quad x\equiv \ \ \ 2\ \ \:(mod\ 7),\ \ x\equiv \ \ \ 2\ \ \:(mod\ 11)\ \iff\ x\equiv \ \ \ 2\ \ (mod\ 77)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\quad\quad\quad\quad x\equiv -1\ \ (mod\ 5),\ \ x\equiv -1\ \ (mod\ 13)\ \iff\ x\equiv -1\ \ (mod\ 65)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus we've reduced the four original LHS equations to the two RHS equations. $\ $ Now we can apply a simple version of the Chinese Remainder Theorem to solve the RHS equations. $\ $ By &lt;strong&gt;CRT&lt;/strong&gt; below&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\quad\quad\quad\quad x\equiv\ 2 + 77\ \bigg[\displaystyle\frac{-3}{77}\ mod\ 65\bigg]\ \ (mod\ 77\cdot65)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now $\rm\displaystyle\ \ \ \ \  mod\ \ 65:\ \ \ \ \ \frac{-3}{77}\ \equiv\ \frac{-3}{12}\ \equiv\ \frac{-1}4\ \equiv\ \frac{64}4\ \equiv\ 16$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence $\rm\ \ x\ \equiv\ 2 + 77\cdot 16\ \equiv\ 1234\ \ (mod\ 77\cdot 65)\quad\quad\ $ &lt;strong&gt;QED&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;THEOREM (CRT)&lt;/strong&gt; $\rm\ \ $ If $\rm\ m,\:n\:$ are coprime integers then $\rm\ m^{-1}\ $ exists $\rm\ (mod\ n)\ \ $ and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\displaystyle\quad\quad\quad\quad\quad x\equiv a\ (mod\ m),\ \ x\equiv b\ (mod\ n)\ \ \iff\ \ x\ \equiv\ a + m\ \bigg[\frac{b-a}{m}\ mod\ n\:\bigg]\ \ (mod\ m\:n)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof&lt;/strong&gt; $\rm\ (\Leftarrow)\ \ \ mod\ m\::\ x\equiv a + m\ (\cdots)\equiv a\:,\ $ and $\rm\ mod\ n\::\ x\equiv a + (b-a)\ m/m \equiv b\:.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\ (\Rightarrow)\ \ $ The solution is unique $\rm\ (mod\ mn)\ $ since if $\rm\ x&amp;#39;,\:x\ $ are solutions then $\rm\ x&amp;#39;\equiv x\ $ mod $\rm\:m,n\:$ therefore $\rm\ m,\:n\ |\ x&amp;#39;-x\ \Rightarrow\ m\:n\ |\ x&amp;#39;-x\ \ $ since $\rm\ \:m,\:n\:$ coprime $\rm\:\Rightarrow\ lcm(m,n) = m\:n\:.\quad\quad$ &lt;strong&gt;QED&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Below is the solution you linked to on&quot;Math Celebrity&quot; (cached to avoid link rot).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/nnRe6.jpg&quot; alt=&quot;enter image description here&quot;&gt;&#xA;&lt;img src=&quot;http://i.imgur.com/x3Jg7.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2011-02-04T01:42:03.523" LastActivityDate="2011-02-04T01:42:03.523" CommentCount="6" />
  <row Id="20260" PostTypeId="2" ParentId="20230" CreationDate="2011-02-03T17:09:26.413" Score="1" ViewCount="" Body="&lt;p&gt;By point-line duality, this is equivalent to the question 'given a configuration of $n$ lines in the plane, what is the maximum number of them which intersect at one point?'; since there are $O(n^2)$ pairwise intersection checks, it becomes a question of whether there's some bucketing/perfect hashing scheme that allows for finding a point in a dynamically-built list in $O(1)$ time.  While I don't know of any specific bucketing schemes applicable to this problem, AFAIK it's very common to have $O(1)$ or at least $O(\alpha(n))$ approaches for this sort of thing ($\alpha(n)$ being the inverse-Ackermann function).&lt;/p&gt;&#xA;" OwnerUserId="785" LastActivityDate="2011-02-03T17:09:26.413" CommentCount="4" />
  <row Id="20261" PostTypeId="2" ParentId="20258" CreationDate="2011-02-03T17:10:39.370" Score="0" ViewCount="" Body="&lt;p&gt;Let $g \in H\cap K$. By Lagrange's Theorem, the order of g divides m and n, but the greatest common divisor of m and n is 1, so g=e.&lt;/p&gt;&#xA;" OwnerUserId="2934" LastActivityDate="2011-02-03T17:10:39.370" CommentCount="1" />
  <row Id="20262" PostTypeId="1" AcceptedAnswerId="20274" CreationDate="2011-02-03T17:14:34.870" Score="0" ViewCount="44" Body="&lt;p&gt;What is the relation between homogenous spaces and principle bundles. I've been reading the two definitions and am left confused as to whether one is a subset of the other or whether no such relation exists.&lt;/p&gt;&#xA;" OwnerUserId="2099" LastActivityDate="2011-02-03T18:56:46.383" Title="Relation between homogenous spaces and principle Bundles" Tags="&lt;differential-geometry&gt;" AnswerCount="2" />
  <row Id="20263" PostTypeId="2" ParentId="20254" CreationDate="2011-02-03T17:14:42.477" Score="3" ViewCount="" Body="&lt;p&gt;As the commenters above pointed out, the only way for $\int_0^{\infty} f(x)$ to converge is for $\lim_{x \rightarrow \infty} f(x)$ to be zero: Since $f(x)$ is decreasing it converges to a lower bound $L = \inf_x f(x)$ which may be $-\infty$. If $L$ were negative or $-\infty$, then there'd be some $N$ and some $\epsilon &amp;gt; 0$ such that for $x &amp;gt; N$, we would have $f(x) &amp;lt; -{\epsilon}$. In this case $\int_{N}^{\infty} f(x)$ would be $-\infty$, which can't happen. Similarly, if $L$ were positive, there would be some $N$ and some $\epsilon &amp;gt; 0$ such that for $x &amp;gt; N$, we would have $f(x) &amp;gt; {\epsilon}$. In this case we'd similarly have  $\int_{N}^{\infty} f(x) = \infty$, impossible. We conclude that $L = 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So $f(x)$ decreases to zero. Thus $f(x) = |f(x)|$ and we have that $\int_0^{\infty} |f(x)|$ converges. Since $|\sin(f(x))| = |{\sin(f(x)) \over f(x)}| |f(x)|$ and since $|{\sin(y) \over y}| &amp;lt; M$ for all $y$ for some $M$, $\int_0^{\infty}|\sin f(x)|$ must also be finite. Hence $\int_0^{\infty} \sin f(x)$ converges.&lt;/p&gt;&#xA;" OwnerUserId="3035" LastEditorUserId="3035" LastEditorDisplayName="" LastEditDate="2011-02-03T19:37:45.943" LastActivityDate="2011-02-03T19:37:45.943" CommentCount="3" />
  <row Id="20264" PostTypeId="1" AcceptedAnswerId="20267" CreationDate="2011-02-03T17:20:50.937" Score="5" ViewCount="221" Body="&lt;p&gt;In his book Lectures on the theory of algebraic numbers, Hecke says that the content of the quadratic reciprocity theorem, formulated and proved entirely in terms of rationals (integers) points beyond the domain of rational numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;He is talking about the algebraic numbers but how is it seen that it points elsewhere?&lt;/p&gt;&#xA;" OwnerUserId="4290" LastEditorUserId="4290" LastEditorDisplayName="" LastEditDate="2011-02-03T19:47:34.317" LastActivityDate="2011-02-03T21:03:19.373" Title="Where does Quadratic Reciprocity point?" Tags="&lt;number-theory&gt;&lt;quadratic-reciprocity&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="1" />
  <row Id="20265" PostTypeId="1" AcceptedAnswerId="20268" CreationDate="2011-02-03T17:31:30.703" Score="2" ViewCount="74" Body="&lt;p&gt;Hi, the question is exactly &quot;why a reduced ring (commutative with 1) can be embedded into a sum of integral rings?&quot; Is this simply because in the normalization process we can have many irreducible component?&lt;/p&gt;&#xA;" OwnerUserId="6583" LastActivityDate="2011-02-03T18:34:33.173" Title="why a reduced ring (commutative with 1) can be embedded into a sum of integral rings?" Tags="&lt;commutative-algebra&gt;" AnswerCount="2" />
  <row Id="20266" PostTypeId="2" ParentId="20255" CreationDate="2011-02-03T18:04:28.613" Score="1" ViewCount="" Body="&lt;p&gt;There is a nice treatment of the Havel-Hakimi theorem on pages 44-46 of Doug West's book, Introduction to Graph Theory (Second edition). As far as uses of this kind of result one might be using graph theory to model some physical molecule that one would want to construct. One first step might be to see if there is a graph with specified valences (degrees). Often one wants a graph with the specified valences and additional properties such as that the graph with these valences is a tree or can be embedded in the plane where the edges meet only at vertices. Many additional specialized results about &quot;realizing&quot; graphs with specific properties and specified valences have been found.&lt;/p&gt;&#xA;" OwnerUserId="1369" LastActivityDate="2011-02-03T18:04:28.613" />
  <row Id="20267" PostTypeId="2" ParentId="20264" CreationDate="2011-02-03T18:05:51.630" Score="6" ViewCount="" Body="&lt;p&gt;Quadratic reciprocity suggests that there ought to be a cubic or quartic reciprocity law. There are, but even to &lt;em&gt;state&lt;/em&gt; them cleanly (let alone prove them) it is best to work in $\mathbb{Z}[\omega]$ or $\mathbb{Z}[i]$ respectively instead of the integers, roughly speaking because of &lt;a href=&quot;http://en.wikipedia.org/wiki/Kummer_theory&quot; rel=&quot;nofollow&quot;&gt;Kummer theory&lt;/a&gt;. See, for example, the &lt;a href=&quot;http://en.wikipedia.org/wiki/Cubic_reciprocity#Eisenstein_integers&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt; &lt;a href=&quot;http://en.wikipedia.org/wiki/Quartic_reciprocity&quot; rel=&quot;nofollow&quot;&gt;articles&lt;/a&gt;, and in particular the following quotes from Gauss on quartic reciprocity:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The theorems on biquadratic residues gleam with the greatest simplicity and genuine beauty only when the field of arithmetic is extended to imaginary numbers, so that without restriction, the numbers of the form a + bi constitute the object of study ... we call such numbers integral complex numbers.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and cubic reciprocity: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The theory of cubic residues must be based in a similar way on a consideration of numbers of the form a + bh where h is an imaginary root of the equation h3 = 1 ... and similarly the theory of residues of higher powers leads to the introduction of other imaginary quantities.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The proof of quadratic reciprocity by Gauss sums also naturally leads to an elegant proof using the Galois theory of number fields which generalizes to the &lt;a href=&quot;http://en.wikipedia.org/wiki/Kronecker%E2%80%93Weber_theorem&quot; rel=&quot;nofollow&quot;&gt;Kronecker-Weber theorem&lt;/a&gt; and suggests the beginnings of class field theory.&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2011-02-03T20:45:50.190" LastActivityDate="2011-02-03T20:45:50.190" CommentCount="10" />
  <row Id="20268" PostTypeId="2" ParentId="20265" CreationDate="2011-02-03T18:09:15.863" Score="6" ViewCount="" Body="&lt;p&gt;Consider the obvious map from your ring $R$ to the direct product $\prod_{\mathfrak p}R/\mathfrak p$ of all the quotients of $R$ by all its prime ideals $\mathfrak p$. The kernel is zero precisely because the ring is reduced.&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2011-02-03T18:09:15.863" />
  <row Id="20269" PostTypeId="1" AcceptedAnswerId="20271" CreationDate="2011-02-03T18:15:27.747" Score="2" ViewCount="59" Body="&lt;p&gt;Hello, How could one prove the following?&#xA;Find a finite set of 3-connected graphs from which all can be constructed iteratively by the following operation, or show that no such set can exist. The operation consists of adding a new vertex to the graph H constructed so far and joinig it by t least three edges to some subdivision of H.&#xA;I think it might be K_4 and K_{3,3} but I'm not sure.&lt;/p&gt;&#xA;" OwnerUserId="6584" LastActivityDate="2011-02-03T18:42:28.597" Title="Constructing 3-connected graphs" Tags="&lt;graph-theory&gt;" AnswerCount="1" CommentCount="0" />
  <row Id="20270" PostTypeId="2" ParentId="20265" CreationDate="2011-02-03T18:34:33.173" Score="3" ViewCount="" Body="&lt;p&gt;Just a comment to supplement Mariano's answer (left as an answer for ease of formatting):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a (very) basic commutative algebra fact which is being taken as known here, namely that in any commutative ring, the intersection of all prime ideals -- called the &lt;strong&gt;nil radical&lt;/strong&gt; of $R$ -- is also equal to the set of all nilpotent elements of $R$.  Among infinitely many other places, see Proposition 53 in Section 4 of &lt;a href=&quot;http://math.uga.edu/~pete/integral.pdf&quot; rel=&quot;nofollow&quot;&gt;these notes&lt;/a&gt; for a proof.&lt;/p&gt;&#xA;" OwnerUserId="299" LastActivityDate="2011-02-03T18:34:33.173" />
  <row Id="20271" PostTypeId="2" ParentId="20269" CreationDate="2011-02-03T18:42:28.597" Score="3" ViewCount="" Body="&lt;p&gt;Take a look at this web page which addresses the much more general issue of inductive constructions for families of graphs. &lt;a href=&quot;http://vlado.fmf.uni-lj.si/vlado/projects/indcla.htm&quot; rel=&quot;nofollow&quot;&gt;http://vlado.fmf.uni-lj.si/vlado/projects/indcla.htm&lt;/a&gt; For planar 3-connected graphs Steinitz's Theorem, in one form, states that all such graph can be constructed from the graph of the tetrahedron (K4) by three kinds of &quot;face splits.&quot; Planar 3-connected graphs are the edge-vertex graphs of convex 3-dimensional polyhedra.&lt;/p&gt;&#xA;" OwnerUserId="1369" LastActivityDate="2011-02-03T18:42:28.597" />
  <row Id="20272" PostTypeId="1" AcceptedAnswerId="20286" CreationDate="2011-02-03T18:43:03.633" Score="2" ViewCount="53" Body="&lt;p&gt;If it true that for every $b$-normal number (where the base $b,b^2,b^3,\ldots$ digits are asymptotically equiprobable) that the first differences of the base-$b$ digits, interpreted as a (signed) base-$b$ expansion, form a $b$-normal number?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems 'obviously' true but I wanted to check.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;This is actually a practical question in disguise!  I want to detect when a (short) string of digits is 'probably' a normal number or something else.  Of course in pure theory there's no reason the first billion digits of a normal number wouldn't be 7, but in this application I'm just looking for a best guess.  I found that there are many times that the digits themselves have roughly even distributions but the first differences are decidedly not well-distributed -- the numbers are in 'clumps', perhaps, with small first differences.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Generally speaking my thought is to apply some sort of chi-square test to the numbers and first differences, but I wanted to make sure the theory was sound first.  (If so, the next step will be testing it on the decimal expansions of random constants to see how well it works.)&lt;/p&gt;&#xA;" OwnerUserId="1778" LastEditorUserId="1778" LastEditorDisplayName="" LastEditDate="2011-02-03T19:05:48.310" LastActivityDate="2011-04-05T20:33:50.740" Title="Do the first differences of a b-normal number form a b-normal number? [Now with application!]" Tags="&lt;number-theory&gt;&lt;transcendence-theory&gt;" AnswerCount="1" />
  <row Id="20273" PostTypeId="2" ParentId="20262" CreationDate="2011-02-03T18:44:43.733" Score="1" ViewCount="" Body="&lt;p&gt;An homogeneous space is the base space of principal bundle (which has the acting group as total space and the isotropy group as fiber)&#xA;In general, though, a principal bundle does not arise from this situation: for example, the principal $S^1$-bundle $S^{2n-1}\to \mathbb CP^{n-1}$ is not of that form for most $n$, because for most $n$ the $n$-sphere is not a group.&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2011-02-03T18:44:43.733" CommentCount="2" />
  <row Id="20274" PostTypeId="2" ParentId="20262" CreationDate="2011-02-03T18:56:46.383" Score="2" ViewCount="" Body="&lt;p&gt;Both notions involve actions of groups, but there is a huge difference: In a homogeneous space, the group acts (transitively) on a (topological) space $X$. In a principal bundle $E\to X$, there is a base space $X$ and there are fibers $E_x$, one for each $x\in X$. The group acts transitively and freely on each of these fibers.&#xA;Of course the actions on fibers give an action on the total space $E$, but this action is not transitive (unless $X$ has only one point) and fairly restricted by the property that points in $E$ are only mapped to other points in the same fiber.&lt;/p&gt;&#xA;" OwnerUserId="1609" LastActivityDate="2011-02-03T18:56:46.383" />
  <row Id="20276" PostTypeId="1" AcceptedAnswerId="20290" CreationDate="2011-02-03T19:02:56.280" Score="0" ViewCount="94" Body="&lt;p&gt;Let $\mathbb{H}$ be a Hilbert space on which $\Phi$ is a bounded linear functional.  Define $f : \mathbb{H} \rightarrow \mathbb{R} : x \mapsto ||x||^2 + \Phi (x)$.  Prove that each (nonempty) closed convex $K\subset \mathbb{H}$ contains a unique point at which the restriction $f|K$ is minimized.&lt;/p&gt;&#xA;" OwnerUserId="2505" LastActivityDate="2011-02-03T19:52:29.770" Title="Convexity in Hilbert Spaces" Tags="&lt;analysis&gt;&lt;hilbert-spaces&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="20277" PostTypeId="1" AcceptedAnswerId="20278" CreationDate="2011-02-03T19:05:20.993" Score="1" ViewCount="156" Body="&lt;p&gt;I've been thinking about this particular problem and I'm stumped.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For all n &gt;= 5, show that there exists a graph, G = (V,E) such that all vertices of V have degree of 4.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've tried using Havel-Hakami with a degree sequence of n-fours (ie: [4,4,4,4,4,4....]), but there were too many cases. I can't think of anything else. Any suggestions or ideas?&lt;/p&gt;&#xA;" OwnerUserId="5166" LastActivityDate="2011-02-03T19:24:23.530" Title="Graph Theory: Proving That A Degree Sequence is Graphical (Havel-Hakami)" Tags="&lt;graph-theory&gt;" AnswerCount="2" />
  <row Id="20278" PostTypeId="2" ParentId="20277" CreationDate="2011-02-03T19:18:04.353" Score="2" ViewCount="" Body="&lt;p&gt;Try drawing a physical diagram of such a graph (all vertices of degree (valence) 4) on a piece of paper. Now see if you can subdivide some edges with new vertices and adjust all these new verrtices to have degree 4. How many new vertices did you need to add? So now, what would you have to do to guarantee that you could find a collection of graphs with all longer sequences of only 4's? This approach should avoid too many cases.&lt;/p&gt;&#xA;" OwnerUserId="1369" LastActivityDate="2011-02-03T19:18:04.353" CommentCount="4" />
  <row Id="20279" PostTypeId="2" ParentId="20102" CreationDate="2011-02-03T19:23:20.963" Score="4" ViewCount="" Body="&lt;p&gt;This was meant to be comment -- in Mathematica it takes me 0.5 seconds for 300x300 matrix with around 60% zeros and 40% positive integers uniformly sampled from 1-10.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To integrate Mathematica library with C, you need to use &lt;a href=&quot;http://reference.wolfram.com/mathematica/tutorial/MathLinkAndExternalProgramCommunicationOverview.html&quot; rel=&quot;nofollow&quot;&gt;MathLink&lt;/a&gt; interface. I'm not sure if you can statically link it, it may require launching Mathematica's kernel along with your program in order to use the interface.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The relevant function is &lt;code&gt;LinearSolve&lt;/code&gt;. Here's the complete code I used to run timing comparison&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;mat1 = RandomVariate[BernoulliDistribution[.4], {300, 300}];&#xA;mat2 = RandomInteger[{1, 10}, {300, 300}];&#xA;mat3 = mat1*mat2;&#xA;answer = RandomInteger[{1, 10}, 300];&#xA;result = LinearSolve[mat3, answer]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I don't know the algorithm they used. They say version 8 is about &lt;a href=&quot;http://www.wolfram.com/mathematica/new-in-8/new-and-improved-core-algorithms/fast-solving-of-systems-of-linear-equations-with-i.html&quot; rel=&quot;nofollow&quot;&gt;1000 times faster&lt;/a&gt; that previous version, so it's probably something specialized and proprietary&lt;/p&gt;&#xA;" OwnerUserId="998" LastEditorUserId="998" LastEditorDisplayName="" LastEditDate="2011-02-03T19:29:49.537" LastActivityDate="2011-02-03T19:29:49.537" CommentCount="7" />
  <row Id="20280" PostTypeId="2" ParentId="20277" CreationDate="2011-02-03T19:24:23.530" Score="1" ViewCount="" Body="&lt;p&gt;A degree sequence $\{d_1, \dots, d_n \}$ is graphical iff the sum of the vertex degrees is even and $$ \sum_{i=1}^{r} d_i \leq r(r-1) + \sum_{i=r+1}^{n} \min(r, d_i)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for each integer $r \leq n-1$. So $4r \leq r(r-1)+4(n-r)$. So there exists a graph of order $n \geq 5$ which has all vertices of degree $4$.&lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2011-02-03T19:24:23.530" CommentCount="1" />
  <row Id="20281" PostTypeId="1" AcceptedAnswerId="20285" CreationDate="2011-02-03T19:26:19.777" Score="2" ViewCount="139" Body="&lt;p&gt;I gave my trigonometry students the following example: Solve $\triangle ABC\ $ , where AC=0.923, AB=.387, and $\measuredangle A\  = 43.33^\circ\ $.  First I found BC using the law of cosines, then I found $\measuredangle B\ $ using the law of sines, and $\measuredangle C\ $ using the sum of the interior angles of the triangle.&lt;br&gt;&#xA;Some of my students found BC first, then found &lt;em&gt;$\measuredangle C\ $&lt;/em&gt; and got a different value then if they had solved it my way.  Why is this?  I was under the impression that the ambiguous case existed when you're in the situation of SSA.  Note: the textbook says that the convention is to solve for the &lt;strong&gt;largest&lt;/strong&gt; angle first.  Why is this? &lt;/p&gt;&#xA;" OwnerUserId="6243" LastActivityDate="2011-02-03T19:41:50.713" Title="Why are there two possible triangles when given SAS?" Tags="&lt;trigonometry&gt;&lt;convention&gt;" AnswerCount="3" CommentCount="0" />
  <row Id="20282" PostTypeId="1" CreationDate="2011-02-03T19:29:11.410" Score="2" ViewCount="224" Body="&lt;p&gt;Consider the quotient space obtained by identifying the north and south pole of S^2. I think the fundamental group should be infinite cyclic, but I do not know how to prove this. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it is infinite cyclic, would this and the S^1 be an example of two spaces which have isomorphic fundamental groups but are not of the same homotopy type?&lt;/p&gt;&#xA;" OwnerUserId="6587" LastActivityDate="2011-02-19T23:22:19.200" Title="Fundamental group of S^2 with north and south pole identified" Tags="&lt;algebraic-topology&gt;" AnswerCount="2" />
  <row Id="20283" PostTypeId="2" ParentId="20281" CreationDate="2011-02-03T19:33:45.103" Score="0" ViewCount="" Body="&lt;p&gt;Does the value they found for $\angle C$ &lt;em&gt;actually&lt;/em&gt; solve the problem, or did they find one that is only partially consistent with the information given? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, if they only take into account $AC$, $BC$, and the angle $\angle A$, then they may obtain two possible values for the angle $\angle C$, but only one of them will be consistent with the known value of $AB$. If you ignore the value of $AB$, you get in trouble. So you should check if the &quot;other value&quot; they found for $C$ is consistent with the length of $AB$, and I'll bet you it is not. &lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2011-02-03T19:33:45.103" CommentCount="1" />
  <row Id="20284" PostTypeId="2" ParentId="20281" CreationDate="2011-02-03T19:36:38.627" Score="0" ViewCount="" Body="&lt;p&gt;Could it be because when taking the inverse sine somebody forgot that there are two angles with the same sine, one less than 90 degrees and one greate?.  In this case I get $\measuredangle B=119.6^\circ$ but you could think it is $61.4^\circ$&lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2011-02-03T19:41:50.713" LastActivityDate="2011-02-03T19:41:50.713" CommentCount="1" />
  <row Id="20285" PostTypeId="2" ParentId="20281" CreationDate="2011-02-03T19:36:44.430" Score="0" ViewCount="" Body="&lt;p&gt;I believe the error that you or your students made is in assuming that $\arcsin (\sin(\theta))=\theta$.  This is false for an obtuse angle.  In this case, since $AB &amp;lt; AC$, you know that $\measuredangle C$ is acute, so it sounds like your students' method is correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When using the law of cosines to find an angle, there is no problem, because $\arccos (\cos(\theta))=\theta$ is correct for the angles in a triangle.  The law of sines is easier to use however, so it is nice to reduce to the acute case.  That is why if you have SSS, it is a good idea to find the largest angle (opposite the largest side) first, because then you can safely use $\arcsin$ to find the other angles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, if you &lt;em&gt;know&lt;/em&gt; in advance that the angle $\theta$ you're trying to find is obtuse, you can use $\theta=\pi - \arcsin(\sin(\theta))$.  The ambiguity from the $\sin$ function is that supplementary angles have the same $\sin$.  Given SAS, you can first find the other side using law of cosines, and next use law of sines to find the angle opposite one of the shorter sides, because you know it will be acute.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2011-02-03T19:36:44.430" CommentCount="1" />
  <row Id="20286" PostTypeId="2" ParentId="20272" CreationDate="2011-02-03T19:37:34.380" Score="3" ViewCount="" Body="&lt;p&gt;The first differences of a normal sequence will be asymptotically equiprobable if they are computed modulo $b$.  This follows from the fact that each pair of digits is equiprobable and there are exactly $b$ pairs $(x, (x+y)\text{ mod }b)$ that give rise to each first difference $y$.  Similarly, the sequence of first differences modulo $b$ of a normal sequence must contain every $n$-digit string with asymptotic density $b^{-n}$, because the normal sequence contained every $(n+1)$-digit string with density $b^{-(n+1)}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We conclude that, for any sequence $S$ of digits in base $b$, the sequence $\Delta S\text{ mod }b$ is normal if $S$ is normal; but we cannot conclude that $\Delta S\text{ mod }b$ is simply normal just because $S$ is simply normal.&lt;/p&gt;&#xA;" OwnerUserId="5546" LastEditorUserId="1778" LastEditorDisplayName="" LastEditDate="2011-04-05T20:33:50.740" LastActivityDate="2011-04-05T20:33:50.740" CommentCount="5" />
  <row Id="20287" PostTypeId="2" ParentId="20191" CreationDate="2011-02-03T19:39:00.857" Score="1" ViewCount="" Body="&lt;p&gt;Here is an elementary characterization of what the superior limit means for functions, without hiding behind other definitions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $f:(0,a)\to \mathbb{R}$ be a bounded function. Then $\limsup_{x\to 0^+} f(x)$ is the unique number $\mu\in \mathbb{R}$ such that the following two statements hold:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;For each $V&amp;gt;\mu$ there exists $\delta&amp;gt;0$ such that for each $x\in (0,\delta)$ we have $f(x)&amp;lt;V$.&lt;/li&gt;&#xA;&lt;li&gt;For each $v&amp;lt;\mu$ and each $\delta&amp;gt;0$ there exists $x\in (0,\delta)$ such that $f(x)&amp;gt;v$.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I think the the most straightfoward way to prove that the superior limit is some number is to show these two statements.&lt;/p&gt;&#xA;" OwnerUserId="1609" LastActivityDate="2011-02-03T19:39:00.857" CommentCount="1" />
  <row Id="20288" PostTypeId="2" ParentId="20282" CreationDate="2011-02-03T19:40:18.213" Score="7" ViewCount="" Body="&lt;p&gt;A sphere with two points identified is homotopy equivalent to the wedge of a sphere with a circle (this is proved in &lt;a href=&quot;http://www.math.cornell.edu/~hatcher/AT/ATchapters.html&quot; rel=&quot;nofollow&quot;&gt;Hatcher's book&lt;/a&gt;, on page 11 of chapter 0). Thus, Van Kampen's theorem implies that the fundamental group is infinite cyclic. However, the second homology group is also infinite cyclic, so it's not homotopy equivalent to the circle.&lt;/p&gt;&#xA;" OwnerUserId="5203" LastActivityDate="2011-02-03T19:40:18.213" />
  <row Id="20289" PostTypeId="2" ParentId="20223" CreationDate="2011-02-03T19:42:28.633" Score="1" ViewCount="" Body="&lt;p&gt;Hello,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's Don from MathCelebrity.  I saw this thread come up on my webstat reports.  Any shortcut solution that you propose, I would be happy to program it.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you list more detail on a shortcut?  Or e-mail me on the contact page, I would like to simplify it more if it can be done.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I originally did this method because some students who were studying it at the time learned it from the book this way.&lt;/p&gt;&#xA;" OwnerUserId="6588" LastActivityDate="2011-02-03T19:42:28.633" />
  <row Id="20290" PostTypeId="2" ParentId="20276" CreationDate="2011-02-03T19:52:29.770" Score="3" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Hint&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;By Riesz representation, you can complete the square and it suffices to consider $f(x) = \|x\|^2$. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Let $x,y$ attain the minimum. Denote $z = (x+y)/2$. By convexity $z$ is also in $K$. Show that $f(z) &amp;lt; f(x)$. This shows uniqueness.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;For existence, you should recall that closed, bounded, convex sets in a Hilbert space is weakly compact (Hanh-Banach + Banach-Alaoglu). Then it suffices to show that $f$ is sequentially lower semi-continuous with respect to the weak topology. To show this last step note that if $x_n \to x$ weakly, then $\|x\|^2 = \lim \langle x_n,x\rangle$. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="1543" LastActivityDate="2011-02-03T19:52:29.770" />
  <row Id="20291" PostTypeId="1" CreationDate="2011-02-03T20:27:08.340" Score="3" ViewCount="228" Body="&lt;p&gt;1.Let $A \in GL_n(\mathbb{C})$. Show that $\det(I+A)=1+\operatorname{tr}(A)+ \epsilon(A)$ where Modulas of epsilon(A) by norm of A=0 as A tends to 0,for any matrix norm.&#xA;If I define J(A)= det(A) for A is in GLn(C),then J is differentiable at all such A and that,if H is in Mn(C),then J'(A)(H)=det(A)tr(A*H). where A* is inverse of A.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can any one say how to write latex here?&lt;/p&gt;&#xA;" OwnerUserId="6990" LastEditorUserId="4171" LastEditorDisplayName="" LastEditDate="2011-02-03T23:29:24.450" LastActivityDate="2011-04-03T22:39:15.423" Title="Matrix Analysis in $\mathbb{C}$" Tags="&lt;matrices&gt;&lt;functional-analysis&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" />
  <row Id="20293" PostTypeId="1" CreationDate="2011-02-03T20:47:10.290" Score="0" ViewCount="74" Body="&lt;p&gt;Let $V$ be a normed linear space and let $W$ be a subspace of $V$, let $X$ be a finite dimensional space and let $T\colon W\to X$ be a continuous linear transformation. Show that there exist $T^*\colon V\to X$ which extends T.&lt;/p&gt;&#xA;" OwnerUserId="6990" LastEditorUserId="622" LastEditorDisplayName="" LastEditDate="2011-02-03T20:55:10.353" LastActivityDate="2011-02-03T20:55:10.353" Title="extension of continous linear transformation" Tags="&lt;functional-analysis&gt;" CommentCount="3" />
  <row Id="20294" PostTypeId="2" ParentId="20264" CreationDate="2011-02-03T21:03:19.373" Score="5" ViewCount="" Body="&lt;p&gt;Just to elaborate on Qiaochu's answer: one way to prove quadratic reciprocity is to observe that, for an odd prime $p$, the quadratic field $\mathbb Q(\sqrt{\pm p})$ (the sign being chosen so that $\pm p \equiv 1 \bmod 4$) is contained in $\mathbb Q(\zeta_p)$ (the field obtained by adjoining a primitive $p$th root of unity&#xA;to $\mathbb Q$), as can be seen by using Gauss sums, and combining this with&#xA;the irreducibility of the $p$th cyclotomic polynomial (which shows that&#xA;$Gal(\mathbb Q(\zeta_p)/\mathbb Q) = (\mathbb Z/p)^{\times}$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All these concepts go back to Gauss's &lt;em&gt;Disquitiones Arithmeticae&lt;/em&gt;, which served to ispire and guide all the subsequent developments in number theory in the 19th century.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gauss himself introduced his Gaussian integers (i.e. the ring $\mathbb Z[i]$) as part of his investigations of biquadratic (i.e. fourth power) reciprocity.  His student Eisenstein investigated cubic reciprocity (and introduced the ring&#xA;$\mathbb Z[\zeta_3]$ as a tool to this end).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Later in the 19th century Kummer investigated higher prime power reciprocity laws, and was led to the invention of the main concepts of algebraic number theory (ideals, unique factorization into prime ideas, the class group and class number, all in the context of the fields $\mathbb Q(\zeta_p)$) as part of his investigation.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The investigation of higher reciprocity laws continued.  When the class group&#xA;of $\mathbb Q(\zeta_p)$ is non-trivial, especially when it has order divisible by $p$, new phenomena emerged, which led Hilbert to the concept of Hilbert class field.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Out of all this the general conception of class field theory emerged, and was finally established by Takagi in the early 20th century.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hecke was aware of all this tradition, and it to this tradition and these developments that he is referring in his remark.&lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2011-02-03T21:03:19.373" />
  <row Id="20295" PostTypeId="1" AcceptedAnswerId="20296" CreationDate="2011-02-03T21:15:34.283" Score="1" ViewCount="57" Body="&lt;p&gt;Again, this is a homework question, so (for the sake of learning) I'd be happy I get subtle hints only.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyhow, the setup is as follows: &quot;$K=\mathbb{Q}(\alpha)$ is of degree n over $\mathbb{Q}$, where $\alpha$ is in the ring of integers $O_K$ of $K$, let the index of $\alpha$ be the index of $[O_K:\mathbb{Z}[\alpha]]$. Suppose that the minimal polynomial of $\alpha$ is Eistenstein at $p$. Then $p$ does not divide the index of $\alpha$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proceed as follows: If p divides the index, then there exists $\beta \in O_K, \beta \notin \mathbb{Z}[\alpha]$ such that $p\beta \in \mathbb{Z}[\alpha]$. [...]&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rest of the exercise is clear. It is only the last sentence I don't follow. &lt;strong&gt;That is, why should such a $\beta$ exist?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(We say that a monic polynomial $x^n+a_{n-1}+x^{n-1}+\ldots+a_0$ is Eistenstein at p if p divides all the $a_i$ but $p^2$ does not divide $a_0$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(my creativity failed finding a good title for this question)&lt;/p&gt;&#xA;" OwnerUserId="4284" LastEditorUserId="4284" LastEditorDisplayName="" LastEditDate="2011-02-03T21:58:06.210" LastActivityDate="2011-02-03T21:58:06.210" Title="If p divides the index, then there exists $\beta \in O_K, \beta \notin \mathbb{Z}[\alpha]$ such that $p\beta \in \mathbb{Z}[\alpha]$" Tags="&lt;homework&gt;&lt;algebraic-number-theory&gt;&lt;field-theory&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="20296" PostTypeId="2" ParentId="20295" CreationDate="2011-02-03T21:22:29.607" Score="2" ViewCount="" Body="&lt;p&gt;Consider the quotient group (under addition) $\mathcal O_K/\mathbb Z[\alpha].$&#xA;You are assuming that $p$ divides its order. Therefore ... [apply a basic theorem in group theory here!].&lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2011-02-03T21:22:29.607" CommentCount="3" />
  <row Id="20297" PostTypeId="1" AcceptedAnswerId="20310" CreationDate="2011-02-03T21:27:03.390" Score="7" ViewCount="153" Body="&lt;p&gt;I am studying valuation theory on the way to local class field theory, and the texts I have looked at immediately focus on discrete valuations in developing the theory of nonarchimedean valuations.  Why? Are there nondiscrete nonarchimedean valuations? If so, why do we ignore them? (it is true that if a field is locally compact with respect to a nonarchimedean valuation, then that valuation must be discrete, and local compactness is very important, but I wonder if there isn't more to be said here).&lt;/p&gt;&#xA;" OwnerUserId="5265" LastActivityDate="2011-02-04T00:12:11.420" Title="Are all nonarchimedean valuations discrete?" Tags="&lt;number-theory&gt;&lt;p-adic-number-theory&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="1" />
  <row Id="20298" PostTypeId="2" ParentId="20297" CreationDate="2011-02-03T22:17:25.993" Score="8" ViewCount="" Body="&lt;p&gt;No, there is a ridiculous number (i.e., a big proper class) of nondiscrete non-Archimedean valuations.  To see some of them, you need only consult a text or section of a text which treats general valuations, e.g. Chapter 17 of &lt;a href=&quot;http://math.uga.edu/~pete/integral.pdf&quot; rel=&quot;nofollow&quot;&gt;these notes&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In them I include a proof of the following fact: for any torsionfree commutative group $G$, there exists a total ordering $\leq$ on $G$ and a valuation ring $R$ with value group isomorphic to $(G,\leq)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know what to make of the question &quot;Why do we ignore them?&quot;  &lt;strong&gt;We&lt;/strong&gt; don't.  In some branches of mathematics -- like the theory of local fields -- discrete valuations are more important than non-discrete valuations, and in other branches of mathematics -- e.g. commutative algebra, certain parts of algebraic geometry -- one definitely needs to consider more general valuation rings.&lt;/p&gt;&#xA;" OwnerUserId="299" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2011-02-03T22:42:59.137" LastActivityDate="2011-02-03T22:42:59.137" CommentCount="7" />
  <row Id="20299" PostTypeId="1" CreationDate="2011-02-03T22:22:33.050" Score="3" ViewCount="110" Body="&lt;p&gt;Hi math.se,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I asked myself a question during my first year and never find any interesting clue about it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of the first thing you learned in integral calculus is that $\int fg \neq \int f \int g$ &lt;strong&gt;in general&lt;/strong&gt;. One can ask the following question : When do we have $\int_I fg = \int_I f \int_I g$ ? Can I find a maximal subset of $CM(K)^2$ or $L^1(\mathbb{R})^2$ for example ? A full &quot;algebric&quot; characterization of such couples ? Sure there are some trivial couples, but the more the time passes, the more I have the feeling that the answer is : this set exists, and there is nothing more interesting to say about it. But it is not really a big satisfaction ... Any ideas/answers about these questions ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS : It's my first post on math.se, and it seems that my questions are appropriate after reading the FAQ. I hope so, but please redirect me if you think that I'm not on the right place.&lt;/p&gt;&#xA;" OwnerUserId="6582" LastEditorUserId="6582" LastEditorDisplayName="" LastEditDate="2011-03-06T23:49:58.317" LastActivityDate="2011-03-06T23:49:58.317" Title="Biggest subset of $L^1$ for which products and integrals commute" Tags="&lt;calculus&gt;&lt;analysis&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="0" />
  <row Id="20300" PostTypeId="1" AcceptedAnswerId="20307" CreationDate="2011-02-03T22:53:18.740" Score="3" ViewCount="159" Body="&lt;p&gt;Let $f: X \times Y \rightarrow \mathbb{R}$ be a continuous map. Show that if $Y$ is compact then the function $g: X \rightarrow \mathbb{R}$ defined by $g(x) = \inf \{f(x,y): y \in Y\}$ is also continuous.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;No clue here. Can you please help?&lt;/p&gt;&#xA;" OwnerUserId="6495" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-03T23:07:34.833" LastActivityDate="2011-02-03T23:49:39.587" Title="Infimum is a continuous function, compact set" Tags="&lt;topology&gt;" AnswerCount="1" CommentCount="7" />
  <row Id="20301" PostTypeId="1" AcceptedAnswerId="20305" CreationDate="2011-02-03T23:09:34.473" Score="3" ViewCount="254" Body="&lt;p&gt;Is there a way of go from $a^3+b^3$ to $(a+b)(a^2-ab+b^2)$ other than know the property by heart?&lt;/p&gt;&#xA;" OwnerUserId="4501" LastEditorUserId="4501" LastEditorDisplayName="" LastEditDate="2011-02-06T13:17:02.950" LastActivityDate="2011-02-06T13:17:02.950" Title="How to know that $a^3+b^3 = (a+b)(a^2-ab+b^2)$" Tags="&lt;algebra&gt;&lt;polynomials&gt;" AnswerCount="4" CommentCount="2" FavoriteCount="1" />
  <row Id="20302" PostTypeId="1" AcceptedAnswerId="20315" CreationDate="2011-02-03T23:14:20.553" Score="2" ViewCount="125" Body="&lt;p&gt;Hello all!&#xA;My situation is as follows: I have a symmetric positive semi-definite matrix L (the Laplacian matrix of a graph) and a diagonal matrix S with positive entries $s_i$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's plenty of literature on the spectrum of $L$, and I'm most interested in bounds on the second-lowest eigenvalue, $\lambda_2$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the thing is that I'm not using the Laplacian $L$ itself, but rather the 'generalized' Laplacian $L S^{-1}$. I still need results on its second lowest eigenvalue $\lambda_2$ (note that the lowest eigenvalue of the Laplacian, both the normal and the generalized, is 0).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is: Are there some readily available theorems/lemmata that allow me to relate the spectra of $L$ and $L S^{-1}$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Of course, $LS^{-1}$ is not a symmetric matrix any more, so I'm talking about its right-eigenvectors. The eigenvalues of $LS^{-1}$ are the same as those of &#xA;$S^{-1/2} L S^{-1/2}$ which again is a symmetric positive semi-definite matrix, so I know an eigenbasis actually exists.&lt;/p&gt;&#xA;" OwnerUserId="4171" LastActivityDate="2011-02-04T00:42:10.813" Title="Eigenvalues of product of a matrix and a diagonal matrix" Tags="&lt;linear-algebra&gt;&lt;matrices&gt;&lt;eigenvalues&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" />
  <row Id="20303" PostTypeId="2" ParentId="20301" CreationDate="2011-02-03T23:17:26.393" Score="3" ViewCount="" Body="&lt;p&gt;So I guess the answer to this question is to expand the right-hand side of the equation, although it might not be too clear on how to do this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It might be clearer if you make the substitution $u=a+b$.  Then the right hand side is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align*}&#xA;(a+b)(a^2-ab+b^2) &amp;amp; = u(a^2-ab+b^2) \\&#xA;&amp;amp; = ua^2-uab+ub^2 \\&#xA;&amp;amp; = (a+b)a^2-(a+b)ab+(a+b)b^2 \\&#xA;&amp;amp; = a^3+ba^2-a^2b-ab^2+ab^2+b^3 \\&#xA;&amp;amp; = a^3+b^3&#xA;\end{align*}&#xA;using the distributive law (and commutivity).&lt;/p&gt;&#xA;" OwnerUserId="139" LastActivityDate="2011-02-03T23:17:26.393" CommentCount="4" />
  <row Id="20304" PostTypeId="2" ParentId="20301" CreationDate="2011-02-03T23:19:01.437" Score="6" ViewCount="" Body="&lt;p&gt;Keeping $a$ fixed and treating $a^3 + b^3$ as a polynomial in $b$, you should immediately notice that $-a$ will be a root of that polynomial. This tells you that you can divide it by $a + b$. Then you apply long division as mentioned in one of the comments to get the other factor.&lt;/p&gt;&#xA;" OwnerUserId="4171" LastActivityDate="2011-02-03T23:19:01.437" CommentCount="3" />
  <row Id="20305" PostTypeId="2" ParentId="20301" CreationDate="2011-02-03T23:27:24.007" Score="10" ViewCount="" Body="&lt;p&gt;If you want to know if $a^n + b^n$ is divisible by $a+b$ (or by $a-b$, perhaps), you can always try &lt;a href=&quot;http://en.wikipedia.org/wiki/Polynomial_long_division&quot; rel=&quot;nofollow&quot;&gt;long division&lt;/a&gt;, whether explicitly or in your head. I can't figure out a way to do the LaTeX or ASCII art for it here to do it explicitly, so I'll show you how one would do it &quot;in one's head&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, for $a^3+b^3$, to divide $a^3+b^3$ by $a+b$, start by writing $a^3+b^3= (a+b)(\cdots)$. Then: we need to multiply $a$ by $a^2$ to get the $a^3$, so we will get $a^3+b^3=(a+b)(a^2+\cdots)$. The $a^2$ makes an unwanted $a^2b$ when multiplied by $b$, so we need ot get rid of it; how? We multiply $a$ by $-ab$. So now we have&#xA;$$a^3+b^3 = (a+b)(a^2-ab+\cdots).$$&#xA;But now you have an unwanted $-ab^2$ you get when you multiply $b$ by $-ab$; to get rid of that $-ab^2$, you have to &quot;create&quot; an $ab^2$. How? We multiply $a$ by $b^2$. So now we have:&#xA;$$a^3 + b^3 = (a+b)(a^2-ab+b^2+\cdots)$$&#xA;and then we notice that it comes out exactly, since we &lt;em&gt;do&lt;/em&gt; want the $b^3$ that wee get when we multiply $b^2$ by $b$; so&#xA;$$a^3 + b^3 = (a+b)(a^2-ab+b^2).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the expression you want is not divisible by what you are trying, you'll run into problems which require a &quot;remainder&quot;. For instance, if you tried to do the same thing with $a^4+b^4$, you would start with $a^4+b^4 = (a+b)(a^3+\cdots)$; then to get rid of the extra $a^3b$, we subtract $a^2b$: $a^4+b^4 = (a+b)(a^3 - a^2b+\cdots)$. Now, to get rid of the unwanted $-a^2b^2$, we add $ab^2$, to get $a^4+b^4 = (a+b)(a^3-a^2b+ab^2+\cdots)$. Now, to get rid of the unwanted $ab^3$, we subtract $b^3$, to get&#xA;$$a^4+b^4 = (a+b)(a^3-a^2b+ab^2 - b^3+\cdots).$$&#xA;At this point we notice that we get an unwanted $-b^4$ (unwanted because we want $+b^4$, not $-b^4$). There is no way to get rid of it with the $a$, so this will be a &quot;remainder&quot;. We need to cancel it out (by adding $b^4$) and then add what is still missing (another $b^4$), so&#xA;$$a^4 + b^4 = (a+b)(a^3-a^2b+ab^2 -b^3) + 2b^4.$$&#xA;(Which, by the way, tells you that $a^4-b^4 = (a+b)(a^3-a^2b+ab^2-b^3)$, by moving the $2b^4$ to the left hand side).&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-04T00:11:16.540" LastActivityDate="2011-02-04T00:11:16.540" CommentCount="1" />
  <row Id="20307" PostTypeId="2" ParentId="20300" CreationDate="2011-02-03T23:49:39.587" Score="1" ViewCount="" Body="&lt;p&gt;Following the suggestion of Joe Johnson (I was going using intervals $(a,b)$, but this is simpler) (do you see why Joe's suggestion is enough? The sets $(a,\infty)$ and $(-\infty,a)$ form a subbasis for the topology of $\mathbb{R}$, and for a function to be continuous, it suffices to show that the inverse image of every set in a given subbasis is open).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $a\in\mathbb{R}$. What is $g^{-1}(-\infty,a)$? It consists of all $x\in X$ such that $g(x)\lt a$. If $g(x)\lt a$, there exists $y_x$ such that $f(x,y_x)\lt a$. Since $f$ is continuous, $f^{-1}(-\infty,a)$ is open, and $(x,y_x)\in f^{-1}(-\infty,a)$, so there exist open sets $U_x$ and $V_y$ of $X$ and $Y$, respectively, such that $U_x\times V_y\subseteq f^{-1}(-\infty,a)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, suppose $x&amp;#39;\in U_x$. Then for all $y\in V_y$ (in particular, for $y_x$) you have $f(x&amp;#39;,y_x)\in (-\infty,a)$. What does that tell you about $g(x&amp;#39;)$? What does that tell you about $g^{-1}(-\infty,a)$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now try to do something along those lines with $g^{-1}(a,\infty)$. &lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2011-02-03T23:49:39.587" CommentCount="3" />
  <row Id="20308" PostTypeId="1" AcceptedAnswerId="20342" CreationDate="2011-02-03T23:58:21.060" Score="3" ViewCount="103" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the only proof I know of that $\mathbb{R}^n$ cannot be homeomorphic to $\mathbb{R}^{n+1}$ requires quite a lot of mathematical apparatus, namely singular homology theory with its exact sequences and pretty abstract setting, as well as some previous (though admittedly not too difficult) calculations of the homology groups of spheres to be used after &quot;excising&quot; a point in $\mathbb{R}^n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is anybody aware of an alternate route? Original research papers are not a problem, as long as they don't involve even more elaborate frameworks or some arcane theory, since eluding those is precisely the point. (To the algebraic-theorists around: it's not that I have any particular bad feelings about homology groups, just curiosity ;-)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks,&lt;/p&gt;&#xA;" OwnerUserId="6280" LastActivityDate="2011-02-04T05:21:46.283" Title="Another (non-homological) proof of the invariance of dimension" Tags="&lt;topology&gt;&lt;reference-request&gt;&lt;algebraic-topology&gt;" AnswerCount="2" />
  <row Id="20309" PostTypeId="2" ParentId="20126" CreationDate="2011-02-04T00:03:52.697" Score="5" ViewCount="" Body="&lt;p&gt;Motivated by the simplicity of integrating from $0$, rather than $1$, substituting $x = 1+u$ gives&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$2^n\int_1^n{\frac{dx}{(1+x^2)^n}} = \int_0^{n-1}{\frac{du}{(1+u+u^2/2)^n}}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We seek estimates--an upper bound and a lower bound for the integrand--that will be simple to integrate.  Dropping the &quot;complicated&quot; $u^2/2$ term gives one obviously integrable estimate; continuing the pattern $1+u+u^2/2 + \cdots + u^k/k! + \cdots = e^u$ leads to another.  The resulting inequalities $1 + u &amp;lt; 1 + u + u^2/2 &amp;lt; e^u$ yield (for $n \ge 1$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\eqalign{&#xD;&#xA;\frac{1}{n}\left(1 - e^{-n(n-1)}\right) &amp;amp;= \int_0^{n-1}{e^{-n u}du} \cr&#xD;&#xA;&amp;amp;\lt \int_0^{n-1}{\frac{du}{(1+u+u^2/2)^n}} \cr&#xD;&#xA;&amp;amp;\lt \int_0^{n-1}{\frac{du}{(1+u)^n}} = \frac{1}{n-1}\left(1 - \frac{1}{n^{n-1}}\right) \cr&#xD;&#xA;&amp;amp;\lt \frac{1}{n-1}.&#xD;&#xA;}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both estimates are asymptotically $1/n$, QED.&lt;/p&gt;&#xA;" OwnerUserId="1489" LastActivityDate="2011-02-04T00:03:52.697" CommentCount="1" />
  <row Id="20310" PostTypeId="2" ParentId="20297" CreationDate="2011-02-04T00:12:11.420" Score="3" ViewCount="" Body="&lt;p&gt;As Pete says, many of us do not ignore non-discrete valuations. However, I can explain why a text on class field theory might. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $K$ is a finite extension of $\mathbb{Q}$, then all nonarchimedean valuations on $K$ are discrete. If your text expects to spend most of its time focused on such fields, that would explain its focus.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof:&lt;/strong&gt; Any valuation on $K$ gives rise to a valuation on $\mathbb{Q}$. By the classification of valuations on $\mathbb{Q}$, it must be the $p$-adic valuation for some $p$. Normalize $v(p)$ to $1$. If you read your textbooks description of extending valuations from $\mathbb{Q}$ to $K$, you should see that the image lands in $(1/e) \mathbb{Z}$, where $e$ is the ramification degree, and is bounded by $[K:\mathbb{Q}]$. QED&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For an example of a non-discrete valuation of interest in number theory, let $K$ be the extension of $\mathbb{Q}$ obtained by adjoining every $p^k$ root of unity, for every $k$. If $\zeta_{p^k}$ is a $p^k$-th root of $1$, then $v_p(\zeta_{p^k} -1 ) = 1/((p-1)p^{(k-1)})$. In particular, the extension of $v_p$ to $K$ is not discrete.&lt;/p&gt;&#xA;" OwnerUserId="448" LastActivityDate="2011-02-04T00:12:11.420" CommentCount="1" />
  <row Id="20311" PostTypeId="1" AcceptedAnswerId="20312" CreationDate="2011-02-04T00:14:36.687" Score="1" ViewCount="69" Body="&lt;p&gt;I'm trying to simplify:&#xA;$$\frac{x^2}{x^2-4}-\frac{x+1}{x+2}$$&#xA;but I can't get to the answer:&#xA;$$\frac{1}{x-2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How to do it?&lt;/p&gt;&#xA;" OwnerUserId="4501" LastEditorUserId="4501" LastEditorDisplayName="" LastEditDate="2011-02-06T13:18:08.097" LastActivityDate="2011-02-06T13:18:08.097" Title="How to simplify this to the given answer?" Tags="&lt;algebra&gt;&lt;polynomials&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="20312" PostTypeId="2" ParentId="20311" CreationDate="2011-02-04T00:20:56.133" Score="3" ViewCount="" Body="&lt;p&gt;First, do the operation by finding the least common denominator. Since $x^2-4 = (x-2)(x+2)$, the least common denominator is already $x^2-4$. Then do some simple algebra:&#xA;\begin{align*}&#xA;\frac{x^2}{x^2-4} - \frac{x+1}{x+2} &amp;amp;= \frac{x^2 - (x+1)(x-2)}{x^2-4} = \frac{x^2-(x^2-x-2)}{x^2-4}\\&#xA; &amp;amp;= \frac{x+2}{x^2-4} = \frac{x+2}{(x-2)(x+2)} = \frac{1}{x-2}.&#xA;\end{align*}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you didn't realize that $x+2$ already divides $x^2-4$, you probably would get&#xA;\begin{align*}&#xA;\frac{x^2}{x^2-4} - \frac{x+1}{x+2} &amp;amp;= \frac{x^2(x+2) - (x+1)(x^2-4)}{(x^2-4)(x+2)}\\&#xA;&amp;amp;= \frac{x^3 + 2x^2 - (x^3 +x^2 - 4x - 4)}{(x^2-4)(x+2)}\\&#xA;&amp;amp;= \frac{x^2 +4x + 4}{(x^2-4)(x+2)} = \frac{(x+2)^2}{(x^2-4)(x+2)}\\&#xA;&amp;amp;= \frac{x+2}{x^2-4} = \frac{1}{x-2},&#xA;\end{align*}&#xA;with the extra work for not noticing.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2011-02-04T00:20:56.133" CommentCount="3" />
  <row Id="20313" PostTypeId="2" ParentId="20302" CreationDate="2011-02-04T00:26:07.753" Score="0" ViewCount="" Body="&lt;p&gt;If you are interested in the second smallest eigenvalue $\lambda_2$ of $L$, then this is just the &lt;a href=&quot;http://en.wikipedia.org/wiki/Algebraic_connectivity&quot; rel=&quot;nofollow&quot;&gt;algebraic connectivity&lt;/a&gt; of $G$ (i.e. $\mu = \mu(G)$). Also &lt;a href=&quot;http://portal.acm.org/citation.cfm?id=1456002&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; paper and &lt;a href=&quot;http://mathoverflow.net/questions/47711/spectrum-of-the-laplacian-on-gn-p-and-gn-m&quot; rel=&quot;nofollow&quot;&gt;MO question&lt;/a&gt; might be relevant.&lt;/p&gt;&#xA;" OwnerUserId="2548" LastEditorUserId="2548" LastEditorDisplayName="" LastEditDate="2011-02-04T00:42:10.813" LastActivityDate="2011-02-04T00:42:10.813" CommentCount="1" />
  <row Id="20314" PostTypeId="1" CreationDate="2011-02-04T00:34:47.420" Score="3" ViewCount="91" Body="&lt;p&gt;In our matrices class we were given a problem that I'm having trouble with.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $E$ and $H$ be subspaces of $\mathbb{C}^n$ such that $\mathbb{C}^n = E \oplus H$. Construct a projection $\mathbf P$ such that $R(\mathbf P) = E$ and $R(\mathbf I − \mathbf P) = H$. Hint: set $F = H ^\perp$ , take a basis $\mathbf v_1, \dots, \mathbf v_d$ of $E$, and use theorems 3.4 and 3.3.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are the related theorems:&lt;br&gt;&#xA;&lt;img src=&quot;http://i.imgur.com/XQrjG.png&quot; alt=&quot;Theorem 3.4 (and the related lemma)&quot;&gt;&#xA;&lt;img src=&quot;http://i.imgur.com/A6IOb.png&quot; alt=&quot;Theorem 3.3&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, apparently I need such $\mathbf P$ that $\mathbf P^2 = \mathbf P$. Also, it needs to have $R(\mathbf P) = E$ and $R(\mathbf I − \mathbf P) = H$. But I have no idea how to utilize the result we get from using theorem 3.4...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any help would be appreciated.&lt;/p&gt;&#xA;" OwnerUserId="2076" LastEditorUserId="856" LastEditorDisplayName="" LastEditDate="2011-02-04T01:15:10.683" LastActivityDate="2011-03-06T01:39:34.303" Title="Constructing a projection" Tags="&lt;linear-algebra&gt;&lt;matrices&gt;" AnswerCount="1" />
  <row Id="20315" PostTypeId="2" ParentId="20302" CreationDate="2011-02-04T00:41:50.847" Score="3" ViewCount="" Body="&lt;p&gt;Let $\mu_i$ be the eigenvalues of $L S^{-1}$. Then $(\lambda_i, \mu_i, s_i)$ obey the multiplicative version of &lt;a href=&quot;http://www.google.com/url?sa=t&amp;amp;source=web&amp;amp;cd=1&amp;amp;ved=0CBcQFjAA&amp;amp;url=http%3A%2F%2Fmath.ucr.edu%2F~jdolan%2Fschubert2.pdf&amp;amp;rct=j&amp;amp;q=quantum%20horn%20monthly&amp;amp;ei=8kpLTcCsGsycgQeDmY3WDw&amp;amp;usg=AFQjCNG8nRr6mtbyVvaPfT2LKMQcwyMpKw&amp;amp;cad=rja&quot; rel=&quot;nofollow&quot;&gt;Horn's inequalities&lt;/a&gt;. The most basic of these, if $\lambda_1 \geq \lambda_2 \geq \cdots \geq \lambda_n$ and $s_1^{-1} \geq \cdots \geq s_n^{-1}$ and $\mu_1 \geq \mu_2 \geq \cdots \geq \mu_n$ is that &#xA;$$\mu_{i+j-1} \leq \lambda_i s_j^{-1} \ \mbox{and}\ \mu_{i+j-n} \geq \lambda_i s_j^{-1}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proof: Let $X=\sqrt{L}$ and $T=\sqrt{S^{-1}}$. So the &lt;a href=&quot;http://en.wikipedia.org/wiki/Singular_values&quot; rel=&quot;nofollow&quot;&gt;singular values&lt;/a&gt; of $X$ and $T$ are $\sqrt{\lambda_i}$ and $\sqrt{s_i^{-1}}$. Then $\sqrt{\mu_i}$ are the singular values of $XT$. By a result of &lt;a href=&quot;http://www.math.neu.edu/~suciu/GAS/klyachko/kl.pdf&quot; rel=&quot;nofollow&quot;&gt;Klyachko&lt;/a&gt;, the singular values of a product obey the exponentiated version of Horn's inequalities.&lt;/p&gt;&#xA;" OwnerUserId="448" LastActivityDate="2011-02-04T00:41:50.847" CommentCount="1" />
  <row Id="20316" PostTypeId="1" AcceptedAnswerId="20317" CreationDate="2011-02-04T00:47:35.757" Score="1" ViewCount="63" Body="&lt;p&gt;The fraction:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{y^2-x^2}{x-y}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;should simplify to the answer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$-(x+y)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but the best I could do was expand it to:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{(y+x)(y-x)}{x-y}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What's next step?&lt;/p&gt;&#xA;" OwnerUserId="4501" LastEditorUserId="4501" LastEditorDisplayName="" LastEditDate="2011-02-06T13:18:42.713" LastActivityDate="2011-02-06T13:18:42.713" Title="What's next step to simplify this?" Tags="&lt;algebra&gt;&lt;polynomials&gt;" AnswerCount="1" />
  <row Id="20317" PostTypeId="2" ParentId="20316" CreationDate="2011-02-04T00:49:02.367" Score="3" ViewCount="" Body="&lt;p&gt;For $x \neq y$, $$\frac{(y+x)(y-x)}{x-y} = \frac{(y+x)(y-x)}{-(y-x)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ = -(y+x)$$&#xA;$$ = -(x+y)$$&lt;/p&gt;&#xA;" OwnerUserId="2548" LastEditorUserId="2548" LastEditorDisplayName="" LastEditDate="2011-02-04T00:56:41.753" LastActivityDate="2011-02-04T00:56:41.753" CommentCount="2" />
  <row Id="20318" PostTypeId="2" ParentId="19594" CreationDate="2011-02-04T00:50:21.167" Score="0" ViewCount="" Body="&lt;p&gt;@ elgeorges:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I mean by a single problem point is that we only need to do a single&#xA;   vertical cut on the bundle to be able to &quot;flatten it&quot;, i.e., make it be&#xA;   the union of a copy of two disjoint product spaces. I  don't know if my statement&#xA;   is too informal, or maybe even wrong, but the fact that we only need two charts&#xA;   for the bundle would seem to reinforce this claim that there is one main problem&#xA;   point. I am also referring to the fact that --please correct me if I am wrong--&#xA;   that all the fibers except for one are of R with one orientation, and one fiber&#xA;   only has orientation opposite that of all the others; it is only at the attaching&#xA;   parts of the fundamental domain that the bundle becomes non-trivial. If the map&#xA;   y--&gt;1-y  (after assignining coordinates (0,0), (0,1),(1,0),(1,1) to the sides of&#xA;   the square were homotopic to the map y-&gt;y , the bundle would be trivial. Is this&#xA;   correct?&lt;/p&gt;&#xA;" OwnerUserId="6595" LastActivityDate="2011-02-04T00:50:21.167" />
  <row Id="20319" PostTypeId="1" AcceptedAnswerId="20325" CreationDate="2011-02-04T00:53:37.143" Score="3" ViewCount="195" Body="&lt;p&gt;How does one find all integer solutions to this equation: $$ \biggl\lfloor{\frac{x}{1!}\biggr\rfloor} + \biggl\lfloor{\frac{x}{2!}\biggr\rfloor} + \cdots +\biggl\lfloor{\frac{x}{10!}\biggr\rfloor}=1001$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can't think about methods for solving this. Though i did something i am not sure of it.&lt;/p&gt;&#xA;" LastActivityDate="2011-04-19T04:19:53.133" Title="$ \biggl\lfloor{\frac{x}{1!}\biggr\rfloor} + \biggl\lfloor{\frac{x}{2!}\biggr\rfloor} + \cdots \biggl\lfloor{\frac{x}{10!}\biggr\rfloor}=1001$" Tags="&lt;number-theory&gt;&lt;elementary-number-theory&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="20321" PostTypeId="1" CreationDate="2011-02-04T01:03:02.657" Score="2" ViewCount="150" Body="&lt;p&gt;I'm working a little program that converges on vector-based approximations of raster images, inspired by &lt;a href=&quot;http://rogeralsing.com/2008/12/07/genetic-programming-evolution-of-mona-lisa/&quot; rel=&quot;nofollow&quot;&gt;Roger Alsing's genetic Mona Lisa&lt;/a&gt;. (I started on this after his first blog post two years ago, played a bit, and it's been on the back burner. But I'd like to un-back-burner it because I have some interesting further ideas.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Roger Alsing's program uses randomly-generated sometimes-self-overlapping n-gons. Instead, I'm using bézier splines linked into what, for lack of a math education, I am calling &quot;bézier petal&quot; shapes. You can see this in action &lt;a href=&quot;http://www.youtube.com/watch?v=abmd920csu0&quot; rel=&quot;nofollow&quot;&gt;in this little youtube video&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The shapes I'm currently using are constructed by generating six random points. These are then sorted radially around their center of gravity, and two opposite points become the end-points of two quadratic Bézier curves, with the other four points used as control points where the fit into the order.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because I, as I mentioned, have no math background, I can't prove it, but empirically this guarantees a non-degenerate petal, with no overlaps or twists. But, it's frustratingly limiting, since it will never produce crescents or S shapes. (The results are not always convex, but concavities can only happen at the end points.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Here's &lt;a href=&quot;http://www.youtube.com/watch?v=sO2ZmfZ3uPk&quot; rel=&quot;nofollow&quot;&gt;a video of what happens if I allow self-intersecting shapes&lt;/a&gt; — not graceful.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd love suggestions for a process which generates all possible non-intersecting shapes. It's clearly better for the application if the process can be done with minimal computation, although I'm willing to pay a bit of time in exchange for nicer results.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A further constraint is that small changes in the &quot;source&quot; data — in the current case, the six random points — needs to produce relatively small changes in the output, or else my stochastic hill-climbing approach can't get anywhere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, it's fine if some input values produce &lt;em&gt;no&lt;/em&gt; output — those will just get discarded.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="4270" LastEditorUserId="4270" LastEditorDisplayName="" LastEditDate="2011-02-04T18:04:15.427" LastActivityDate="2011-02-08T15:57:09.713" Title="How can I generate non-self-intersecting &quot;bézier petals&quot;?" Tags="&lt;geometry&gt;&lt;bezier-curve&gt;&lt;intersection&gt;" CommentCount="5" />
  <row Id="20322" PostTypeId="2" ParentId="20314" CreationDate="2011-02-04T01:03:48.300" Score="1" ViewCount="" Body="&lt;p&gt;If the basis is orthonormal, then consider $\sum_{j=1}^d |e_j\rangle \langle e_j|$ where $|e_j\rangle$ represents $e_j$ as a column vector and $\langle e_j|$ is its transpose. &lt;/p&gt;&#xA;" OwnerUserId="722" LastActivityDate="2011-02-04T01:03:48.300" />
  <row Id="20323" PostTypeId="1" AcceptedAnswerId="20324" CreationDate="2011-02-04T01:19:47.620" Score="2" ViewCount="49" Body="&lt;p&gt;I have the expression (for some $k$ and $r$ natural numbers):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sum_{l=0}^r {l \choose k}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a way to bound this expression using a polynomial of degree which is linear in $k$ (or polynomial in $k$)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am pretty sure that each $l \choose k$ is a polynomial of the form $O(l^k)$.&lt;/p&gt;&#xA;" OwnerUserId="7076" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2011-02-04T09:22:31.373" LastActivityDate="2011-02-04T09:22:31.373" Title="what is some polynomial bound of the following expression?" Tags="&lt;combinatorics&gt;&lt;binomial-coefficients&gt;" AnswerCount="2" />
  <row Id="20324" PostTypeId="2" ParentId="20323" CreationDate="2011-02-04T01:25:00.313" Score="2" ViewCount="" Body="&lt;p&gt;You sum is same as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \sum_{l = k}^{r} \binom{l}{k}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Note, from $k$ to $r$, assuming $\binom{a}{b} = 0$ if $a \lt b$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is actually equal to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\binom{r+1}{k+1}$$ by using&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\binom{l+1}{k+1} - \binom{l}{k+1} = \binom{l}{k}$$&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2011-02-04T01:25:00.313" />
  <row Id="20325" PostTypeId="2" ParentId="20319" CreationDate="2011-02-04T01:25:06.567" Score="10" ViewCount="" Body="&lt;p&gt;I noticed that $x=6!$ is too big. So the solution solves:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\biggl\lfloor{\frac{x}{1!}\biggr\rfloor} + \biggl\lfloor{\frac{x}{2!}\biggr\rfloor} + \cdots \biggl\lfloor{\frac{x}{5!}\biggr\rfloor}=1001.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then I divided $\frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} +\frac{1}{4!} + \frac{1}{5!}$ into $1001$ and got $583.107$. Trying $584$ in the equation works. Any integer larger or smaller will not solve the equation, but $584 \leq x &amp;lt; 585$.&lt;/p&gt;&#xA;" OwnerUserId="6272" LastEditorUserId="6272" LastEditorDisplayName="" LastEditDate="2011-02-04T01:30:07.967" LastActivityDate="2011-02-04T01:30:07.967" />
  <row Id="20326" PostTypeId="2" ParentId="20323" CreationDate="2011-02-04T01:25:48.997" Score="0" ViewCount="" Body="&lt;p&gt;According to &lt;a href=&quot;http://www.wolframalpha.com/input/?i=sum+from+l+%3D+0+to+l+%3D+r++of+Binomial%28l,k%29&quot; rel=&quot;nofollow&quot;&gt;Wolfram Alpha&lt;/a&gt;, this is equivalent to $$ \frac{(-k+r+1) \binom{r+1}{k}+k \binom{0}{k}}{k+1}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So perhaps it will be easier to bound that with a polynomial of degree $k$.&lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2011-02-04T01:25:48.997" CommentCount="1" />
  <row Id="20327" PostTypeId="2" ParentId="20311" CreationDate="2011-02-04T01:28:43.177" Score="1" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\ \ $ Taking fractional parts (by subtracting $\:1\:$ from both terms) it reduces to the following&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\rm \frac{4}{x^2-4}\ +\ \frac{1}{x+2}\ =\ \frac{1}{x-2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which you'll probably find easier to derive.&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2011-02-04T01:28:43.177" />
  <row Id="20328" PostTypeId="1" CreationDate="2011-02-04T01:34:52.640" Score="4" ViewCount="152" Body="&lt;p&gt;Hi, Everyone:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is from a post in sci.math that did not get a full answer; I may repost it for the OP there:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am  interested on the issue I read in another site of when an embedding from a closed set extends into a homeomorphism, i.e., if $C$ is closed in $X$, and $f:C \rightarrow Y$ is an embedding, when can we extend $f$ to $F$ so that $F:X \rightarrow Y$ is a homeomorphism, and $F|_C=f$ (i.e., $f=F$ in $C$ )?  Of course there are trivial cases&#xA;like when $f$ is the identity on $C$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know of, e.g., Tietze Extension, and I think there are results about extending&#xA;maps from a space into its compactification; I think the map must be regular (inverse&#xA;image of a compact set is compact). But I don't know of any general result.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I will learn Latex as soon as I can; my apologies for using ASCII&lt;/p&gt;&#xA;" OwnerUserId="6595" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2011-02-04T03:10:52.387" LastActivityDate="2011-05-11T04:56:28.680" Title="When does an Embedding extend into a Homeomorphism?" Tags="&lt;general-topology&gt;" AnswerCount="3" />
  <row Id="20329" PostTypeId="1" AcceptedAnswerId="20331" CreationDate="2011-02-04T01:36:18.157" Score="4" ViewCount="98" Body="&lt;p&gt;Hello, please bear with me as I'm not a mathematician and this is difficult to word properly. :]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need the equation for a &lt;a href=&quot;http://en.wikipedia.org/wiki/Logarithmic_spiral&quot; rel=&quot;nofollow&quot;&gt;logarithmic spiral&lt;/a&gt; (let's call it $S(\theta)$) that meets certain constraints for a music visualizer I'm working on. Let's call the arc-length of the spiral $A(\theta)$.  I'm looking for a spiral that meets the following requirements:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$A(0)=27.5$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$A(2\pi)=55.0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$A(4\pi)=110.0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$A(6\pi)=220.0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$A(8\pi)=440.0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm basically wanting the arc-length of this spiral to correspond to a frequency of a musical note.  The requirements I gave basically plot the octaves of &quot;A&quot; notes. (&lt;a href=&quot;http://www.phy.mtu.edu/~suits/notefreqs.html&quot; rel=&quot;nofollow&quot;&gt;http://www.phy.mtu.edu/~suits/notefreqs.html&lt;/a&gt;). This way I can create a &quot;directional&quot; spectrogram. All &quot;A&quot; notes will point toward $\theta=0,2\pi,4\pi,...$, all &quot;E&quot; notes will point roughly in the direction of $\theta=\pi,3\pi,...$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I spent some time grinding out rough approximations of $a$ and $b$ values that give me proper values between arclengths of 27.5 and 14080.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The closes constant values I've landed on are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a=1.5145$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$b=0.0551625$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm looking for a way to generate values of expressions for $a$ and $b$ that will produce very accurate arclength results for the extent of the average human hearing range (20Hz-20KHz).  If someone could explain the process I would need to go through to get these values I would be very grateful.  Let me know if something doesn't make sense. My mind is a horribly tangled place.&lt;/p&gt;&#xA;" OwnerUserId="6593" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2011-05-13T23:25:31.213" LastActivityDate="2011-05-13T23:33:09.307" Title="What constants do I need to create this specific logarithmic spiral?" Tags="&lt;logarithms&gt;&lt;plane-curves&gt;&lt;music-theory&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" />
  <row Id="20330" PostTypeId="2" ParentId="20319" CreationDate="2011-02-04T01:50:55.340" Score="9" ViewCount="" Body="&lt;p&gt;The left hand side is strictly less than $(e-1)x$ and so you can start testing at $x=\biggl\lceil{\dfrac{1001}{e-1}}\biggr\rceil = 583$ and continue until you get a sum larger than 1001, something which will happened very quickly; in this particular case $x=585$ produces a sum which is too large. &lt;/p&gt;&#xA;" OwnerUserId="6460" LastActivityDate="2011-02-04T01:50:55.340" CommentCount="3" />
  <row Id="20331" PostTypeId="2" ParentId="20329" CreationDate="2011-02-04T01:55:31.407" Score="5" ViewCount="" Body="&lt;p&gt;The arc-length of a logarithmic spiral is given &lt;a href=&quot;http://mathworld.wolfram.com/LogarithmicSpiral.html&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;. You seem to want $A(\theta + 2\pi) = 2 A(\theta)$. This gives you $b = \ln 2/(2\pi)$. Use the value you want for $A(0)$ to find $a$.&lt;/p&gt;&#xA;" OwnerUserId="589" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2011-05-13T23:33:09.307" LastActivityDate="2011-05-13T23:33:09.307" CommentCount="2" />
  <row Id="20332" PostTypeId="2" ParentId="20301" CreationDate="2011-02-04T01:57:31.387" Score="4" ViewCount="" Body="&lt;p&gt;It's a &lt;em&gt;homogenization&lt;/em&gt; of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Cyclotomic_polynomial&quot; rel=&quot;nofollow&quot;&gt;cyclotomic factorization&lt;/a&gt; $\rm\ x^3 + 1 = (x+1)\ (x^2 - x + 1)\:.\ $ Recall that the homogenization of a degree $\rm\:n\:$ polynomial $\rm\ f(x)\ $ is the polynomial $\rm\ y^n\ f(x/y)\:.\ $ This maps $\rm\ x^{k}\ \to\ x^{k}\ y^{n-k}\ $ so the result is a &lt;a href=&quot;http://en.wikipedia.org/wiki/Homogeneous_polynomial&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;homogeneous&lt;/em&gt; polynomial&lt;/a&gt; of degree $\rm\:n\:.\ $ While this cyclotomic factorization is rather trivial, other cyclotomic homogenizations can be far less trivial.&#xA; For example, &lt;a href=&quot;http://en.wikipedia.org/wiki/Cyclotomic_polynomial&quot; rel=&quot;nofollow&quot;&gt;Aurifeuille, Le Lasseur and Lucas&lt;/a&gt; discovered so-called &lt;a href=&quot;http://en.wikipedia.org/wiki/Homogeneous_polynomial&quot; rel=&quot;nofollow&quot;&gt;&lt;strong&gt;Aurifeuillian factorizations&lt;/strong&gt;&lt;/a&gt; of cyclotomic polynomials $\rm\;\Phi_n(x) = C_n(x)^2 - n\ x\  D_n(x)^2\;$. These play a role in factoring numbers of the form $\rm\; b^n \pm 1\:$, cf. the &lt;a href=&quot;http://homes.cerias.purdue.edu/~ssw/cun/&quot; rel=&quot;nofollow&quot;&gt;Cunningham Project&lt;/a&gt;. Below are some simple examples of such factorizations:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{array}{rl}&#xD;&#xA;x^4 + 2^2 \quad=&amp;amp;  (x^2 + 2x + 2)\;(x^2 - 2x + 2) \\\\&#xD;&#xA;\frac{x^6 + 3^2}{x^2 + 3} \quad=&amp;amp;  (x^2 + 3x + 3)\;(x^2 - 3x + 3) \\\\&#xD;&#xA;\frac{x^{10} - 5^5}{x^2 - 5} \quad=&amp;amp;  (x^4 + 5x^3 + 15x^2 + 25x + 25)\;(x^4 - 5x^3 + 15x^2 - 25x + 25) \\\\&#xD;&#xA;\frac{x^{12} + 6^6}{x^4 + 36} \quad=&amp;amp;  (x^4 + 6x^3 + 18x^2 + 36x + 36)\;(x^4 - 6x^3 + 18x^2 - 36x + 36) \\\\&#xD;&#xA;\end{array}$$&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2011-02-04T06:19:14.360" LastActivityDate="2011-02-04T06:19:14.360" />
  <row Id="20333" PostTypeId="1" AcceptedAnswerId="20343" CreationDate="2011-02-04T02:05:50.390" Score="1" ViewCount="96" Body="&lt;p&gt;The notation of my teacher confuses me in the title. If &lt;code&gt;x&lt;/code&gt; is the same on the both sides, it seems trivial. So I suspect the &lt;code&gt;x&lt;/code&gt; is different so it becomes:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x_{1}\leq x_{2}$ for every $x_{1} \in S$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so what it would mean is that every $x_{1} \in S$ has a maximum element $x_{2}$ that does not need be in $S$. Attacking the problem with an example, why not trivial with the explicit interpretation in the following. I can imagine a partial ordering that would satisfy the condition like cardinality with a $n$-cube where $n$ is the number of nodes and the number of nodes is $2^{n}$. To prove the condition in the case would require to show that the subsets form a chain with each nodes like the case with 3-cube $card(\emptyset) \leq card(\{1\}) \leq card(\{2\}) ... \leq card(\{1,2\}) \leq card(\{1,2,3\})$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what does the reflexive -term mean? Am I interpreting it right about maximum element or am I leaping to conclusion?&lt;/p&gt;&#xA;" OwnerUserId="5902" LastActivityDate="2011-02-04T02:53:59.593" Title="Does the reflexivity mean maximum element in partial ordering $x \leq x$ for every $x \in S$" Tags="&lt;set-theory&gt;" AnswerCount="4" />
  <row Id="20334" PostTypeId="2" ParentId="20333" CreationDate="2011-02-04T02:13:58.703" Score="2" ViewCount="" Body="&lt;p&gt;No, it should in fact be $x \le x$, for every $x \in S$. I myself find it best to think of power sets, partially ordered by inclusion $\subset$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So let $\mathcal{P}(S)$ be the power set of some set $S$, and let $A \in \mathcal{P}(S)$. Here it's easy to see what $A \subset A$ means: every set is a subset of itself. You can check that transitivity and antisymmetry hold here as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And no, it doesn't mean that there is a maximal element. For example, $\mathbb{R}$ is partially ordered by $\le$.&lt;/p&gt;&#xA;" OwnerUserId="5105" LastEditorUserId="5105" LastEditorDisplayName="" LastEditDate="2011-02-04T02:42:16.083" LastActivityDate="2011-02-04T02:42:16.083" CommentCount="2" />
  <row Id="20335" PostTypeId="2" ParentId="20333" CreationDate="2011-02-04T02:14:42.387" Score="3" ViewCount="" Body="&lt;p&gt;It seems trivial just because you're using a familiar symbol $\le$ for a relation that you really do not know. Try a different symbol such as $\preceq$ or $\triangleleft$.&lt;/p&gt;&#xA;" OwnerUserId="589" LastActivityDate="2011-02-04T02:14:42.387" CommentCount="2" />
  <row Id="20336" PostTypeId="2" ParentId="20333" CreationDate="2011-02-04T02:14:44.087" Score="3" ViewCount="" Body="&lt;p&gt;For an order, reflexive means just what you said in the title.  For all $x \in S$ you have $x \le x$  This is a useful requirement to place on orders.  Without it, you could have an &quot;order&quot; that was simply $\forall x,y \in S \neg (x\le y)$  That doesn't seem like an order and would ruin lots of what we want to say about orders.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-02-04T02:14:44.087" CommentCount="1" />
  <row Id="20337" PostTypeId="1" AcceptedAnswerId="20339" CreationDate="2011-02-04T02:15:33.147" Score="0" ViewCount="187" Body="&lt;p&gt;Hello,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Has anyone experienced using TI-89 BASIC programming to solve math problems? I want to create an array using TI-89 but I don't know how? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update&lt;/strong&gt;&lt;br&gt;&#xA;I tried both&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;{0, 0, 0, 0}-&amp;gt;A&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;[0, 0, 0, 0]-&amp;gt;A&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Basically, I want to create a boolean array to check for duplicate digits of a number. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update 1&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;dup(n)&#xA;Prgm&#xA;    {0,0,0,0,0,0,0,0,0,0}-&amp;gt;a&#xA;    1-&amp;gt;d // digit&#xA;    While n &amp;gt; 0&#xA;        mod(10,n)-&amp;gt;d&#xA;        If a[d] = 0 Then 1-&amp;gt;a[d]&#xA;        Else Return 1&#xA;        EndIf&#xA;        floor(n/10)-&amp;gt;n&#xA;    EndWhile    &#xA;    Return 0&#xA;EndPrgm&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Thanks,&lt;br&gt;&#xA;Chan&lt;/p&gt;&#xA;" OwnerUserId="5711" LastEditorUserId="5711" LastEditorDisplayName="" LastEditDate="2011-02-06T23:34:18.903" LastActivityDate="2011-02-06T23:34:18.903" Title="How to create an array in TI-89 Titanium calculator?" Tags="&lt;calculator&gt;" AnswerCount="2" />
  <row Id="20338" PostTypeId="1" AcceptedAnswerId="20341" CreationDate="2011-02-04T02:35:15.613" Score="1" ViewCount="62" Body="&lt;p&gt;I'm not Even sure I'm using the correct terminology here, but I'm helping out my high school daughter with her algebra and was presented with the following rule:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sqrt3/\sqrt2 = \sqrt{3/2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Accepting that this is true (which a calculator did demonstrate), I should be able to step through a proof using the process for simplifying radicals in the denominator.  So, given:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sqrt3/\sqrt2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I then multiply by the value of the denominator divided by itself:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sqrt3/\sqrt2 * \sqrt2/\sqrt2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;to get:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(\sqrt3 * \sqrt2)/2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which equals:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sqrt6/2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And that's where I get stuck.  What can I do to get to:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sqrt{3/2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other than using a calculator of course.  :)&lt;/p&gt;&#xA;" OwnerUserId="6596" LastActivityDate="2011-02-04T03:49:11.370" Title="Proof of simplification of radicals in fraction" Tags="&lt;algebra&gt;&lt;roots&gt;" AnswerCount="2" />
  <row Id="20339" PostTypeId="2" ParentId="20337" CreationDate="2011-02-04T02:36:50.667" Score="1" ViewCount="" Body="&lt;p&gt;I don't have an 89 at hand to check, but I suspect that you can use lists (contained in &lt;code&gt;{}&lt;/code&gt;, comma-separated) in much the same way you'd use an array, including subscript-based accessing.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;As it turns out, I do have a TI Voyage 200, which is functionally identical to a TI-89 Titanium (runs the same OS, but more memory, better screen, QWERTY keyboard).  I am able to store a list into a variable:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;{2,4,6,8}-&amp;gt;a&lt;/code&gt; (where &lt;code&gt;-&amp;gt;&lt;/code&gt; is the &quot;sto&gt;&quot; key)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and access items in that list:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;a[3]&lt;/code&gt; is reformatted as $a_3$ and returns &lt;code&gt;6&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In what way was &lt;code&gt;{0, 0, 0, 0}-&amp;gt;A&lt;/code&gt; not working for you?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Okay, here are some relatively messy thoughts, but hopefully some of it will be helpful.  You can create a list of the digits in a number &lt;code&gt;n&lt;/code&gt; with&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;seq(mod(floor(n/10^k),10),k,0,floor(log(n)))-&amp;gt;digitList&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;(the ones digit will be the first item in the list, ...)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, having a list of the digits, you can sort the list using&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;SortA digitList&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Construct a list of the differences between successive terms of &lt;code&gt;digitList&lt;/code&gt; (where there will be 0s for duplicated digits):&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;∆List(digitList)-&amp;gt;deltaList&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Now, you can do something like looping over &lt;code&gt;deltaList&lt;/code&gt; and whenever you find a zero, check the corresponding location in &lt;code&gt;digitList&lt;/code&gt; to see which digit is repeated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternately, you could step through the sorted &lt;code&gt;digitList&lt;/code&gt;, tracking the previous entry, and find duplicates that way.&lt;/p&gt;&#xA;" OwnerUserId="72" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2011-02-04T06:24:57.977" LastActivityDate="2011-02-04T06:24:57.977" CommentCount="8" />
  <row Id="20340" PostTypeId="2" ParentId="20328" CreationDate="2011-02-04T02:39:43.843" Score="0" ViewCount="" Body="&lt;p&gt;This will never happen unless $C=X$. Consider the identity embedding $C \to C$; by your hypothesis, this extends to a homeomorphism $X \to C$ extending $C \to C$. This is an inverse to the inclusion $C \to X$. Thus $C=X$.&lt;/p&gt;&#xA;&#xA;&lt;del&gt;&#xA;In general, this is unlikely. Consider any $Y$ between $X$ and $C$. Then there is an embedding $C \to Y$ which by your condition extends to a homeomorphism $X \simeq Y$. So $X$ is homeomorphic to all its closed subspaces that contain $C$ (with a homeomorphism that extends $1: C \to C$). This generally will not happen in natural examples (though it will if $X$ is a [Toronto space][1] and $C$ has the same cardinality as $X$): for instance, it means that $X/C$ is homeomorphic to any of its closed subspaces.&#xA;&lt;/del&gt;&#xA;" OwnerUserId="536" LastEditorUserId="536" LastEditorDisplayName="" LastEditDate="2011-02-04T03:19:17.070" LastActivityDate="2011-02-04T03:19:17.070" CommentCount="4" />
  <row Id="20341" PostTypeId="2" ParentId="20338" CreationDate="2011-02-04T02:39:47.000" Score="1" ViewCount="" Body="&lt;p&gt;From my perspective (this is probably not the only way to arrive at the property you're asking about):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sqrt{\frac{a}{b}}=\frac{\sqrt{a}}{\sqrt{b}}$$&#xA;(with $a,b\ge 0$) is a special case of&#xA;$$\left(\frac{a}{b}\right)^x=\frac{a^x}{b^x}$$&#xA;(or, equivalently, $(ab)^x=a^xb^x$), the Distributive Property of Exponentiation over Division (or Multiplication).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Distributive Property of Exponentiation over Multiplication, for integer exponents, comes from associativity and commutativity of multiplication:&#xA;$$(ab)^n=\underset{n\text{ factors of }ab}{\underbrace{(ab)\cdots(ab)}}=\left(\underset{n\text{ factors of }a}{\underbrace{a\cdots a}}\right)\left(\underset{n\text{ factors of }b}{\underbrace{b\cdots b}}\right)=a^nb^n$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The property extends naturally to non-integer exponents.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;original answer:&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{\sqrt{6}}{2}=\frac{\sqrt{6}}{\sqrt{4}}=\sqrt{\frac{6}{4}}=\sqrt{\frac{3}{2}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is that what you were looking for?&lt;/p&gt;&#xA;" OwnerUserId="72" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2011-02-04T02:54:59.403" LastActivityDate="2011-02-04T02:54:59.403" CommentCount="6" />
  <row Id="20342" PostTypeId="2" ParentId="20308" CreationDate="2011-02-04T02:42:36.960" Score="3" ViewCount="" Body="&lt;p&gt;You could use homotopy groups instead :).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More seriously, there is a notion of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Inductive_dimension&quot; rel=&quot;nofollow&quot;&gt;topological dimension&lt;/a&gt; of a space&#xA;(to be distinguished from the combinatorial dimension one studies in algebraic geometry). One can show that n-space has dimension n (cf. &lt;a href=&quot;http://books.google.com/books?id=eMseAQAAIAAJ&amp;amp;q=dimension+theory&amp;amp;dq=dimension+theory&amp;amp;hl=en&amp;amp;ei=F2dLTaeKLMmr8AalrPypDg&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=2&amp;amp;ved=0CC0Q6AEwAQ&quot; rel=&quot;nofollow&quot;&gt;the book by Hurewicz and Wallman&lt;/a&gt;), which proves invariance of dimension. This can be done in an elementary manner, free of any reference to singular homology or homotopy theory.&lt;/p&gt;&#xA;" OwnerUserId="536" LastActivityDate="2011-02-04T02:42:36.960" CommentCount="1" />
  <row Id="20343" PostTypeId="2" ParentId="20333" CreationDate="2011-02-04T02:53:59.593" Score="3" ViewCount="" Body="&lt;p&gt;A relation $R$ on a set $S$ is just a subset of $S\times S$, that is, a set of pairs of elements from $S$. If $(a,b) \in R$ then $aRb$. As an example, here is the $&amp;lt;$ relation on $\{1,2,3\}$: $$&amp;lt; = \{(1,2),(1,3),(2,3)\}.$$ So $1&amp;lt;3$ because $(1,3) \in &amp;lt;$. On the other hand, $1 \nless 1$ since $(1,1) \notin &amp;lt;$. So $&amp;lt;$ is an example of a non-reflexive (in fact, anti-reflexive) relation. The corresponding reflexive relation $\leq$ is $$\leq = \{(1,1),(1,2),(1,3),(2,2),(2,3),(3,3)\}.$$ Both relations are \emph{transitive}, that is $aRb$ and $bRc$ together imply $aRc$. That need not be true in general, for example it fails in $$R = \{(1,2),(2,3)\}$$ since $1R2$ and $2R3$ but $1\not{R}3$. In this case the relation $R$ represents the successor function, but in general a relation can be arbitrary. A third useful property is antisymmetry. It is the property that guarantees that if $x \leq y$ and $y \leq x$ then $x = y$. It usually fails, for example, in the case of an equivalence relation. A fourth property is that for every $x,y \in S$, either $x \leq y$ or $y \leq x$; along with reflexivity and transitivity these are the defining properties of a total order. This property fails, for example, for the set inclusion relation $\subseteq$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As Ross mentioned, usually we're not interested in general relations but in some &quot;well-behaved&quot; ones. A good example is a function, when viewed as a relation (try to think yourself how to do that). Other common examples are partial orders (reflexive and transitive), total orders and equivalence relations (reflexive, symmetric and transitive).&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2011-02-04T02:53:59.593" CommentCount="3" />
  <row Id="20344" PostTypeId="2" ParentId="19967" CreationDate="2011-02-04T02:55:12.993" Score="5" ViewCount="" Body="&lt;p&gt;The function $\varphi(x)$ can be written as&#xA;$$&#xD;&#xA;\begin{eqnarray}&#xD;&#xA;\varphi(x)&#xD;&#xA;  &amp;amp;=&amp;amp; \frac{1}{x} \sum_{n=1}^{\infty} \frac{n^{n-1}}{n!}\left(xe^{-x}\right)^{n} \\&#xD;&#xA;  &amp;amp;=&amp;amp; -\frac{1}{x} \sum_{n=1}^{\infty} \frac{(-n)^{n-1}}{n!}\left(-xe^{-x}\right)^{n} \\&#xD;&#xA;  &amp;amp;=&amp;amp; -\frac{1}{x} W_0\left(-xe^{-x}\right),&#xD;&#xA;\end{eqnarray}&#xD;&#xA;$$&#xA;where $W_0$ is the Lambert W-function's main branch.  For sufficiently small $y$, the definition of the W-function ensures that $W_0(ye^y)=y$, so for small $t$,&#xA;$$&#xD;&#xA;\begin{eqnarray}&#xD;&#xA;\varphi(t)&#xD;&#xA;  &amp;amp;=&amp;amp; -\frac{1}{t}W_0\left(-te^{-t}\right) \\&#xD;&#xA;  &amp;amp;=&amp;amp; -\frac{1}{t}(-t) \\&#xD;&#xA;  &amp;amp;=&amp;amp; 1.&#xD;&#xA;\end{eqnarray}&#xD;&#xA;$$&#xA;In fact this equality holds for all $0 \le t &amp;lt; 1$.  For values slightly larger than $1$, we use&#xA;$$&#xD;&#xA;-(1+t)e^{-(1+t)} = \frac{-(1+t)}{1+t+\frac{1}{2}t^2+O(t^3)} = -1 + \frac{1}{2}t^2+O(t^3) = -(1-t)e^{-(1-t)},&#xD;&#xA;$$&#xA;so&#xA;$$&#xD;&#xA;\begin{eqnarray}&#xD;&#xA;\varphi(1+t) &amp;amp;=&amp;amp; -\frac{1}{1+t}W_0(-(1+t)e^{-(1+t)}) \\ &amp;amp;\approx&amp;amp; -\frac{1}{1+t}W_0(-(1-t)e^{-(1-t)}) \\ &amp;amp;=&amp;amp; \frac{1-t}{1+t} \\ &amp;amp;=&amp;amp; 1 - 2t + O(t^2).&#xD;&#xA;\end{eqnarray}&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The desired limit, then, is&#xA;$$&#xD;&#xA;\lim_{t\rightarrow 0}\frac{\varphi(1+t) - \varphi(t)}{t} = \lim_{t\rightarrow 0}\frac{(1-2t) - 1}{t} = -2.&#xD;&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="5546" LastActivityDate="2011-02-04T02:55:12.993" />
  <row Id="20345" PostTypeId="1" CreationDate="2011-02-04T02:55:42.887" Score="3" ViewCount="77" Body="&lt;p&gt;I am trying to prove the following assertion:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If an algebraic number field $K$ contains a complex root of unity, then the norm of every nonzero element of $K$ is positive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think this is supposed to be a simple application of the theory of geometry of numbers, but I can't see how to prove it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can someone please help me?&lt;/p&gt;&#xA;" OwnerUserId="6597" LastActivityDate="2011-02-04T02:55:42.887" Title="How to show that if K contains a complex root of unity, then every nonzero element of K has positive norm?" Tags="&lt;number-theory&gt;&lt;algebraic-number-theory&gt;" CommentCount="2" />
  <row Id="20346" PostTypeId="2" ParentId="20338" CreationDate="2011-02-04T03:49:11.370" Score="1" ViewCount="" Body="&lt;p&gt;While it's a simple consequence of the distributive law for exponents, one can view it from a more general perspective. Namely, both $\rm\ \sqrt{3/2}\ $ and $\rm\ \sqrt{3}/\sqrt{2}\ $ are roots of $\rm\ x^2 - 3/2  $. But this polynomial has a &lt;em&gt;unique&lt;/em&gt; positive root since it is increasing from $\rm\: -3/2\ $ to $\: +\infty\ $ on $\ [0,\infty)\:.\ $ Therefore this &lt;em&gt;uniqueness theorem&lt;/em&gt; implies that the two roots are equal. One can frequently apply analogous techniques to more much complicated expressions involving messy nested radicals - inferences far removed from the laws of exponents. Thus I point out once again what I have emphasized in &lt;a href=&quot;http://math.stackexchange.com/search?tab=newest&amp;amp;q=user:242%20uniqueness&quot;&gt;many varied posts here:&lt;/a&gt; uniqueness theorems provide powerful tools for proving equalities.&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2011-02-04T03:49:11.370" CommentCount="2" />
  <row Id="20347" PostTypeId="1" AcceptedAnswerId="20348" CreationDate="2011-02-04T03:51:58.573" Score="1" ViewCount="79" Body="&lt;p&gt;Hello,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there  a way to perform these operations in a particular base b but 10? I can convert it back to base 10 and perform the operation, however, I think it's kinda odd by doing this way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks,&lt;br&gt;&#xA;Chan&lt;/p&gt;&#xA;" OwnerUserId="5711" LastActivityDate="2011-02-04T04:15:19.193" Title="How to do +, -, *, / with number in a base b?" Tags="&lt;elementary-number-theory&gt;" AnswerCount="1" CommentCount="0" />
  <row Id="20348" PostTypeId="2" ParentId="20347" CreationDate="2011-02-04T04:05:43.997" Score="3" ViewCount="" Body="&lt;p&gt;If what you are interested in are paper-and-pencil methods, then the same algorithms that one uses in base 10 work in base $b$, provided you remember to perform &lt;em&gt;all&lt;/em&gt; operations in base $b$. (Of course, it may be that the proliferation of calculators has atrophied some people's ability to do it even in base $10$...)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, if you want to add $573641$ and $561373$ in base $9$ (I just made them up), then we add can add them right to left, with carries; note that in base $9$, $7+4 = 12$, $6+3=10$, etc. So we get:&#xA;$$\begin{array}{r}&#xD;&#xA;573641\\\&#xD;&#xA;\underline{+\ 561373}\\\&#xD;&#xA;1245124&#xD;&#xA;\end{array}$$&#xA;Similar with subtraction, multiplication, and long division.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-04T04:15:19.193" LastActivityDate="2011-02-04T04:15:19.193" CommentCount="5" />
  <row Id="20349" PostTypeId="1" CreationDate="2011-02-04T04:11:01.000" Score="2" ViewCount="171" Body="&lt;p&gt;Hi, first time here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My professor gave us a puzzle problem that we discussed in class that I could elaborate on if requested. But I interpreted the puzzle and formed a recursive function to model it which is as follows:&#xA;$$f(n) = \frac{n f(n-1)}{n - 1} + .01 \hspace{1cm} \textrm{where } f(1) = .01\text{  and } n\in\mathbb{N}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question that is asked is when (if ever!) does $f(x) = 1000x$. About half concluded that it eventually will equal (they didn't have the formula I made) and that x would be damn near infinity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My personal question is, can the function be reduced so that it isn't recursive? And so that it doesn't need to be solved by brute force computer algorithm (which would be about 3 lines of code).&lt;/p&gt;&#xA;" OwnerUserId="6599" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2011-06-03T13:50:05.900" LastActivityDate="2011-06-03T13:50:05.900" Title="Converting recursive function to closed form" Tags="&lt;algorithms&gt;&lt;puzzle&gt;&lt;recurrence-relations&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="0" />
  <row Id="20350" PostTypeId="2" ParentId="20337" CreationDate="2011-02-04T04:18:20.147" Score="1" ViewCount="" Body="&lt;p&gt;I don't know about TI-89 but as for TI-83+,84, they have Lists stored in L1, L2, etc that can be written to by {1,2,3,4}-&gt;L1.  You can find L1 as 2nd 1, 2nd 2, 2nd 3, etc on the older models.  I can't say exactly for the 89.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can also do things like seq(Y1,X,0,4,.2)-&gt;L1 to quickly evaluate sequences and store in Lists.&lt;/p&gt;&#xA;" OwnerUserId="6599" LastActivityDate="2011-02-04T04:18:20.147" CommentCount="2" />
  <row Id="20351" PostTypeId="2" ParentId="20328" CreationDate="2011-02-04T04:29:41.100" Score="0" ViewCount="" Body="&lt;p&gt;Thanks both for your comments.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am sorry if I did not make this clear; I don't know if you were assuming this,&#xA;   but the embedding is &lt;em&gt;into&lt;/em&gt; Y, and not onto it, i.e., we have an embedding&#xA;   e:C--&gt;e(C) , and not e:C--&gt;Y, which,  as you said, cannot be 1-1. I can see&#xA;   how my using the identity on C was misleading, sorry; what I meant was if, e.g.,&#xA;   we had a map from (0,1) as a subspace of R embedded by identity into another&#xA;   copy of (0,1) in R. Then we would define f^ (R-(0,1))-&gt;R-(0,1) by f(x)=x (though&#xA;   there are many other options, using, e.g., bump functions and partitions of unity.)&#xA;   But R, R^n are too nice in too many ways, and it would be great to have more general&#xA;   results.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And, re your second question, Akhil: I would like , if workable, to fix Y, and&#xA;   not look for a result for all possible spaces Y, which may be too ambitious for&#xA;   me at this point. so maybe we could look for properties that Y may have to be&#xA;   able to have this extension.&lt;/p&gt;&#xA;" OwnerUserId="6600" LastActivityDate="2011-02-04T04:29:41.100" />
  <row Id="20352" PostTypeId="2" ParentId="20218" CreationDate="2011-02-04T04:41:45.630" Score="-1" ViewCount="" Body="&lt;p&gt;@Mariano:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;K' is the space obtained by embedding a copy of Sg , the genus-g surface into&#xA;   the space K. Maybe a good idea to start by assuming K is a manifold. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we start with a manifold K, then we embed an Sg in K, to end up with a&#xA;   new space. Just curious of how embedding a copy of an Sg--a reasonably nice space&#xA;   which is not &quot;too nice&quot; (meaning that, unlike, e.g., R^n, we have H*(Sg) ==/0 ) into X.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;@Jim: Good idea, I will give that a try.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;@Sean: good catch!. I was thinking of the case of H_1 and Pi_1 , re Hurewicz&lt;/p&gt;&#xA;" OwnerUserId="6600" LastActivityDate="2011-02-04T04:41:45.630" CommentCount="7" />
  <row Id="20353" PostTypeId="2" ParentId="20349" CreationDate="2011-02-04T04:53:37.490" Score="5" ViewCount="" Body="&lt;p&gt;If you define $g(n) = \frac{f(n)}{n}$ then your expression is $g(n) = g(n-1) + \frac{0.01}{n}$ and $g(0) = 0$.  So $g(n) = 0.01\sum_{i=1}^n \frac{1}{n} = 0.01 h(n)$ where $h(n)$ is the $n^{\text{th}}$ harmonic number.  Therefore $f(n) = 0.01n \cdot h(n)$.  I think this is about as closed of a form as you're going to get.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the sequence $h(n)$ diverges, you will eventually get $h(n)\geq 100000$ so $g(n) \geq 1000$, which means $f(n)\geq 1000n$.&lt;/p&gt;&#xA;" OwnerUserId="884" LastActivityDate="2011-02-04T04:53:37.490" CommentCount="1" />
  <row Id="20354" PostTypeId="2" ParentId="20349" CreationDate="2011-02-04T04:53:58.663" Score="3" ViewCount="" Body="&lt;p&gt;$$\frac{f(n)}{n}=\frac{f(n-1)}{n-1}+\frac{.01}{n}$$  Define $g(n)=100\frac{f(n)}{n}$ with $g(1)=1$ Then $g(n)=g(n-1)+\frac{1}{n}=\sum_{i=1}^{n}\frac{1}{i}=H_{n}$ where $H_n$ are the harmonic numbers.  So we want $H_{n}\gt 1000$ so $n=\exp(1000-\gamma) \pm 1$ (cuz I took the limit of $H_n$ to be true at this point. It's probably right without the $\pm 1$)&lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2011-02-04T17:42:15.070" LastActivityDate="2011-02-04T17:42:15.070" CommentCount="3" />
  <row Id="20355" PostTypeId="2" ParentId="20349" CreationDate="2011-02-04T04:54:19.880" Score="1" ViewCount="" Body="&lt;p&gt;Well, it can be proved by induction that a &quot;closer&quot; form for that sequence is $f(n) = \frac{H_n}{100}$, where $H_n$ is the nth &lt;a href=&quot;http://en.wikipedia.org/wiki/Harmonic_number&quot; rel=&quot;nofollow&quot;&gt;harmonic number&lt;/a&gt;. Since the harmonic series diverges, so does your function, which will grow logarithmically.&lt;/p&gt;&#xA;" OwnerUserId="6427" LastActivityDate="2011-02-04T04:54:19.880" CommentCount="2" />
  <row Id="20356" PostTypeId="2" ParentId="20308" CreationDate="2011-02-04T05:21:46.283" Score="0" ViewCount="" Body="&lt;p&gt;I don't know if this is what you are looking for, but there is a proof using invariance of domain: if U is  an open  subset of R^n and f:U--&gt;R^n is a continuous injection,  then V=f(U) is open in R^n, and f:U--&gt;V is an embedding. IOW:if U is open in R^n and f: R^n --&gt; R^m is injective and continuous, then f: U --&gt; f(U) is a homeomorphism.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Use the (assumed) existence of this homeomorphism , together with the inclusion map i:R^n--&gt;R^m : (x1,...,xn)--&gt;(x1,..,xn,0,..,0) and then i(R^n) is open in R^m, which cannot happen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sorry, I am in a hurry at the moment to give more detail; will come back later.&lt;/p&gt;&#xA;" OwnerUserId="6600" LastActivityDate="2011-02-04T05:21:46.283" CommentCount="1" />
  <row Id="20357" PostTypeId="1" AcceptedAnswerId="20585" CreationDate="2011-02-04T05:46:52.237" Score="1" ViewCount="159" Body="&lt;p&gt;Suppose you are given two oriented manifolds with boundary $M$ say $B, B&amp;#39;$ and $\partial B = M = \partial B&amp;#39;$.  Identify the boundaries and form $C = B \sqcup_{Id: M \to M} B&amp;#39;$. I want to see why, with the orientation induced by being submanifolds of $C$, $B$ and $B&amp;#39;$ induce opposite orientations to their boundary $M$. I'm particularly interested in the way the fundamental classes of $C,B,B&amp;#39;$ and $M$ behave.&#xA;Thanks a lot!&lt;/p&gt;&#xA;" OwnerUserId="6603" LastEditorUserId="642" LastEditorDisplayName="" LastEditDate="2011-02-04T06:14:32.150" LastActivityDate="2011-02-05T20:59:53.137" Title="Orientation induced on submanifolds" Tags="&lt;algebraic-topology&gt;&lt;differential-topology&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="20358" PostTypeId="1" AcceptedAnswerId="20396" CreationDate="2011-02-04T05:49:04.670" Score="4" ViewCount="100" Body="&lt;p&gt;Among other things, the Fourier transform maps functions from $L^2(\mathbb{R}^n) \to L^2(\mathbb{R}^n)$, $L^1(\mathbb{R}^n) \to C_0(\mathbb{R}^n)$ (continuous functions vanishing at infinity), and $\mathcal{S}(\mathbb{R}^n)\to\mathcal{S}(\mathbb{R}^n)$ (Schwartz space, or the space of rapidly decreasing functions). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm interested in looking more closely at the codomain of the second mapping. Since $C_0 \subset L^\infty$, every $L^1$ function is mapped by the Fourier transform into an $L^\infty$ function. However, I was wondering if it is easy to find specific examples of functions that are only in $L^1$, but are mapped into $L^1$ or $L^2$ (or possibly any other $L^p$ for $1 \leq p &amp;lt; \infty$). &lt;/p&gt;&#xA;" OwnerUserId="1736" LastEditorUserId="1736" LastEditorDisplayName="" LastEditDate="2011-02-04T05:59:17.087" LastActivityDate="2011-02-04T15:40:59.933" Title="Codomain of the Fourier transform" Tags="&lt;fourier-analysis&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="2" />
  <row Id="20359" PostTypeId="1" CreationDate="2011-02-04T05:55:13.743" Score="3" ViewCount="160" Body="&lt;p&gt;Hi, my first time here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am trying to show the completeness of R, using the LUB property.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Problem is that I don't know, given a Cauchy sequence,where the limit would come from; I can check if a sequence {an} converges to a specific value, but I don't know how to come up with the limit value that the sequence would converge to.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I imagine as the intervals containing the terms am (m&gt;N) become smaller, as |am-ak|N, maybe an converges to the limiting point of intervals (am,ak).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I used limsup and iminf to show that an  converges , but this does not tell me&#xA; what an should converge to.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any Suggestions?&lt;/p&gt;&#xA;" OwnerUserId="6600" LastActivityDate="2011-02-04T06:08:14.890" Title="Cauchy Sequences in R Converge: to What?" Tags="&lt;real-analysis&gt;" AnswerCount="1" />
  <row Id="20360" PostTypeId="2" ParentId="19917" CreationDate="2011-02-04T05:59:42.473" Score="1" ViewCount="" Body="&lt;p&gt;It does not work for any odd dimension, if I recall correctly it will only work for 1,3 and 7 which correspond to real-division algebras...&lt;/p&gt;&#xA;" OwnerUserId="6603" LastActivityDate="2011-02-04T05:59:42.473" />
  <row Id="20361" PostTypeId="2" ParentId="1064" CreationDate="2011-02-04T06:01:21.833" Score="0" ViewCount="" Body="&lt;p&gt;Let $C$ represent the Cantor set. Consider the set $C+\alpha$, where $\alpha=\sum_{n=1}^{\infty}10^{2^n}$. &lt;/p&gt;&#xA;" OwnerUserId="5959" LastActivityDate="2011-02-04T06:01:21.833" CommentCount="1" />
  <row Id="20362" PostTypeId="1" CreationDate="2011-02-04T06:02:28.790" Score="1" ViewCount="102" Body="&lt;p&gt;I have been trying to derive &lt;a href=&quot;http://i.imgur.com/192GI.jpg&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; approximation but have been unsuccessful in doing so. Any help would be greatly appreciated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/azgCf.jpg&quot; alt=&quot;The image at the above mentioned link.&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="6604" LastEditorUserId="2404" LastEditorDisplayName="" LastEditDate="2011-02-04T07:13:39.807" LastActivityDate="2011-02-04T17:47:14.897" Title="Approximation For Difference Of Two Sides Of A Triangle" Tags="&lt;geometry&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="20363" PostTypeId="2" ParentId="20359" CreationDate="2011-02-04T06:08:14.890" Score="4" ViewCount="" Body="&lt;p&gt;Sure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Step 1: Show that any Cauchy sequence is bounded.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(This is true in any metric space and has nothing to do with completeness.)  It follows that the limsup and liminf of (the underlying set of terms of) your sequence are finite.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Step 2: Show that any sequence with limsup $L &amp;lt; \infty$ has a subsequence converging to $L$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Step 3: Now you have something specific to try to show the Cauchy sequence converges to: namely, $L$.  Show in fact that if a subsequence of a Cauchy sequence converges to some $L$, then the sequence itself converges to $L$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Step 4 (optional): Think about what would happen if in Steps 2 and 3 you chose to use the liminf instead of the limsup.&lt;/p&gt;&#xA;" OwnerUserId="299" LastActivityDate="2011-02-04T06:08:14.890" CommentCount="6" />
  <row Id="20364" PostTypeId="1" AcceptedAnswerId="29241" CreationDate="2011-02-04T06:16:48.947" Score="5" ViewCount="187" Body="&lt;p&gt;The Stanford Encyclopedia of Philosophy's &lt;a href=&quot;http://plato.stanford.edu/entries/category-theory/&quot; rel=&quot;nofollow&quot;&gt;article on category theory&lt;/a&gt; claims that adjoint functors can be thought of as &quot;conceptual inverses&quot; of each other. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, the forgetful functor &quot;ought to be&quot; the &quot;conceptual inverse&quot; of the free-group-making functor. Similarly, in multigrid the restriction operator &quot;ought to be&quot; the conceptual inverse of it's adjoint prolongation operator. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think there is some deep and important intuition here, but so far I can only grasp it in specific cases and not in the abstract sense. Can anyone help shed light on what is meant by this statement about adjoint functors being conceptual inverses? &lt;/p&gt;&#xA;" OwnerUserId="3060" LastActivityDate="2011-03-27T14:16:19.497" Title="Adjoint functors as &quot;conceptual inverses&quot;" Tags="&lt;category-theory&gt;&lt;adjoint&gt;&lt;inverse&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" />
  <row Id="20365" PostTypeId="1" AcceptedAnswerId="20375" CreationDate="2011-02-04T06:25:33.360" Score="3" ViewCount="120" Body="&lt;p&gt;Let's consider this simple dice game: A coin is faked so it has &lt;code&gt;p&lt;/code&gt; chance to land on heads, and &lt;code&gt;1-p&lt;/code&gt; chance to land on tails. Every round costs &lt;code&gt;$1&lt;/code&gt;, and gives you &lt;code&gt;$2&lt;/code&gt; if you win (for a total of &lt;code&gt;+$1&lt;/code&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume you're starting with &lt;code&gt;$n&lt;/code&gt;. What are your odds to &quot;go infinite&quot; - be able to play the game forever? This sounds like Markov Chains 101, it's just been ages since I read anything about Markov Chains.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Also&lt;/strong&gt; - given any constant &lt;code&gt;m&lt;/code&gt;, what are the odds of ever reaching &lt;code&gt;$m&lt;/code&gt; in this game?&lt;/p&gt;&#xA;" OwnerUserId="947" LastActivityDate="2011-02-04T07:52:33.127" Title="What are the odds of never losing in a loaded coin game?" Tags="&lt;probability&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="20366" PostTypeId="1" AcceptedAnswerId="20367" CreationDate="2011-02-04T06:30:53.417" Score="2" ViewCount="76" Body="&lt;p&gt;Hello,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given x is a real number and c is a natural number. Prove that:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;floor( floor(x) / c ) = floor( x / c )&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;When x &gt; 0, I can prove it by let x = n + z where 0 &amp;lt; z &amp;lt; 1.&lt;br&gt;&#xA;Then rhs = floor( n/c + z/c ) which equals to floor( floor(x) ) since z/c -&gt; 0.&lt;br&gt;&#xA;However, when x &amp;lt; 0, it's wrong when letting x = -( n + z ). Because it always give me a smaller negative number. Is my logic wrong here?  Any hint?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks,&lt;br&gt;&#xA;Chan &lt;/p&gt;&#xA;" OwnerUserId="5711" LastActivityDate="2011-02-04T07:16:43.530" Title="Question regarding floor() function" Tags="&lt;elementary-number-theory&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="20367" PostTypeId="2" ParentId="20366" CreationDate="2011-02-04T06:33:53.053" Score="1" ViewCount="" Body="&lt;p&gt;You can see it here:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://chandrumath.wordpress.com/2010/10/07/more-on-the-floor-function/&quot; rel=&quot;nofollow&quot;&gt;http://chandrumath.wordpress.com/2010/10/07/more-on-the-floor-function/&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" LastActivityDate="2011-02-04T06:33:53.053" CommentCount="2" />
  <row Id="20368" PostTypeId="2" ParentId="20362" CreationDate="2011-02-04T07:01:22.570" Score="0" ViewCount="" Body="&lt;p&gt;I have only the first term in your approximation.&lt;br&gt;&#xA;Here's how:&#xA;First, the angles opposite a side are denoted by the capital letter.  Next, notice that the angle $B$ is equal to $\frac{\pi}{2} - (\theta + \gamma)$.&#xA;Now, the Law of Cosines states&#xA;$ b^2 = a^2 + c^2 - 2ac \cos B $&#xA;and&#xA;$ a^2 = b^2 + c^2 - 2bc \cos A $&#xA;so if you take the difference of the two and simplify you will get&#xA;$b^2 - a^2 = -c(a \cos B - b \cos A)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So far we have made no use of the fact that $c &amp;lt;&amp;lt; a$.  Let's now use this to note that $C \approx 0$ and so $A \approx \pi - B$ which implies that $\cos A \approx \cos (\pi - B) = - \cos B$.  Substitute this into the difference of the two cosine laws we derived earlier.  Thus,&#xA;$b^2 - a^2 \approx -c(a \cos B - b (- \cos B)) = -c \cos B (a + b)$.  Now divide both sides by $(a + b)$ to get&#xA;$b - a \approx -c \cos B$.  Finally, since $B = \frac{\pi}{2} - (\theta + \gamma)$, $\cos B = sin (\theta + \gamma)$, and so we've shown that&#xA;$b - a \approx -c \sin (\theta + \gamma)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I have some time later, I'll return to this.&lt;/p&gt;&#xA;" OwnerUserId="2404" LastActivityDate="2011-02-04T07:01:22.570" />
  <row Id="20369" PostTypeId="1" CreationDate="2011-02-04T07:08:49.767" Score="1" ViewCount="57" Body="&lt;p&gt;I'm trying create a function that returns a matrix containing a variable &quot;l&quot; which is an independent variable to be swept for a plot later on. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would calculate &quot;phi&quot; based on user inputs which include &quot;n&quot; and &quot;d&quot;, then I would use &quot;n&quot;, &quot;d&quot;, and &quot;phi&quot; to find &quot;a&quot;, &quot;b&quot;, &quot;c&quot;, and &quot;d&quot; to create a matrix &quot;m&quot; with. This matrix &quot;m&quot; will be a function of &quot;l&quot;.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;phi = 2*pi*n*d/l;&#xA;a = cos(phi);&#xA;b = 1i*sin(phi)/n;&#xA;c = 1i*n*sin(phi);&#xA;d = cos(phi);&#xA;m = [a b;c d];&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I'm really not enjoying Matlab's coding style as compared to C++ and Python... How would you guys implement this functionality?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Summary: I want a function that returns a matrix which contains an independent variable to be swept for a plot later.&lt;/p&gt;&#xA;" OwnerUserId="6605" LastActivityDate="2011-02-04T07:08:49.767" Title="Matlab - Matrix Function with an Independent Variable?" Tags="&lt;matrices&gt;&lt;matlab&gt;" CommentCount="2" />
  <row Id="20370" PostTypeId="1" AcceptedAnswerId="20390" CreationDate="2011-02-04T07:14:24.500" Score="1" ViewCount="106" Body="&lt;p&gt;Hello,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a little question inspired from Hartshorne's Geometry, which I've been juggling around for a while. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Suppose that $\Pi$ is the Cartesian plane $F^2$ for some field $F$, with the set of ordered pairs of elements of $F$ being the points and lines those subsets defined by linear equations. Let $\Pi&amp;#39;$ be the associated projective plane. Here $\Pi&amp;#39;$ is just $\Pi$ together with the points at infinity, which are the pencils of sets of parallel lines. A line in $\Pi&amp;#39;$ will be the subset consisting of a line of $\Pi$ plus its unique point at infinity.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Also, let $V=F^3$ be a three-dimensional vector space over $F$. Let $\Sigma$ be the set of $1$-dimensional subspaces of $V$, and call them points. If $W\subseteq V$ is a $2$-dimensional subspace of $V$, then the set of all 'points' contained in $W$ will be called a 'line,' and $\Sigma$ forms a projective plane.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I would like to see how $\Pi&amp;#39;$ and $\Sigma$ are isomorphic. I figure I should construct a bijection from the points of $\Pi&amp;#39;$ to the points of $\Sigma$, however, I don't see a natural mapping. At first I figured if $(a,b)\in\Pi&amp;#39;$, then maybe $(a,b)\mapsto\text{span}(a,b,0)$, but this isn't injective or surjective. Not only that, but I figure if $A$ and $B$ are points in $\Pi&amp;#39;$ on a line, then the images of all points in that line in $\Pi&amp;#39;$ should be in the same $2$-dimensional subspace in $\Sigma$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't quite see why they should be isomorphic. Suppose to lines $l$ and $m$ intersect in $\Pi&amp;#39;$, shouldn't their intersection point map to the intersection of the lines in $\Sigma$? But what if this intersection doesn't even pass through the origin, and hence isn't even a $1$-dimensional subspace? Thank you for any insight on how to solve this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: For my last question above, I just realized that all the $2$-dimensional subspaces pass through the origin (whoops!), and thus the intersection will contain the origin as well, so that's not an issue.&lt;/p&gt;&#xA;" OwnerUserId="1583" LastEditorUserId="1583" LastEditorDisplayName="" LastEditDate="2011-02-04T07:32:56.357" LastActivityDate="2011-02-04T10:29:43.710" Title="Is the projective plane of $F^2$ isomorphic to the set of $1$-dimensional subspaces of $F^3$?" Tags="&lt;projective-geometry&gt;" AnswerCount="2" />
  <row Id="20371" PostTypeId="2" ParentId="20366" CreationDate="2011-02-04T07:16:43.530" Score="1" ViewCount="" Body="&lt;p&gt;This is the special case $\rm\ m = 1\ $ in a proof I &lt;a href=&quot;http://math.stackexchange.com/questions/6748/adjoint-functors/6793#6793&quot;&gt;presented here&lt;/a&gt;. See that thread for more on the &lt;em&gt;universal&lt;/em&gt; viewpoint that explains the simplicity of this proof. For convenience, here is the proof.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;LEMMA&lt;/strong&gt; $\rm\: \  \lfloor x/(mn)\rfloor\ =\ \lfloor{\lfloor x/m\rfloor}/n\rfloor\ \  $&#xA;for $\rm\ \ n &amp;gt; 0$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof&lt;/strong&gt; $\rm\quad\quad\quad\quad\quad\ \ \ k\ \le \lfloor{\lfloor x/m\rfloor}/n\rfloor$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\quad\quad\quad\quad\iff\quad\ \ k\ \le\ \:{\lfloor x/m\rfloor}/n$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\quad\quad\quad\quad\iff\ \ nk\ \le\ \ \lfloor x/m\rfloor$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\quad\quad\quad\quad\iff\ \ nk\ \le\:\ \ \ x/m$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\quad\quad\quad\quad\iff\ \ \ \ k\ \le\:\ \ \ x/(mn)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\quad\quad\quad\quad\iff\ \ \ \  k\ \le\ \ \lfloor x/(mn)\rfloor $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Compare the above trivial proof to more traditional proofs, e.g. the &lt;a href=&quot;http://chandrumath.wordpress.com/2010/10/07/more-on-the-floor-function/&quot; rel=&quot;nofollow&quot;&gt;special case $\rm\ m = 1\ $ here.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2011-02-04T07:16:43.530" CommentCount="1" />
  <row Id="20372" PostTypeId="1" AcceptedAnswerId="20376" CreationDate="2011-02-04T07:42:49.977" Score="3" ViewCount="88" Body="&lt;p&gt;I have attempted as follows:&#xA;$|f(x) - f(y)| = |x + 1/x - y - 1/y|$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\leq |x - y| + |1/x - 1/y|$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Struck here.&#xA;Any help.&lt;/p&gt;&#xA;" OwnerUserId="5049" LastActivityDate="2011-02-04T08:06:52.530" Title="How to prove $|f(x) - f(y)| &lt; |x - y|$ if $f(x) = x + 1/x$ where $x &gt; 1$" Tags="&lt;functions&gt;&lt;inequality&gt;" AnswerCount="2" />
  <row Id="20373" PostTypeId="2" ParentId="20372" CreationDate="2011-02-04T07:48:07.403" Score="3" ViewCount="" Body="&lt;p&gt;Use the mean value theorem. See Example 3.4 in &lt;a href=&quot;http://www.math.uconn.edu/~kconrad/blurbs/analysis/contractionshort.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.math.uconn.edu/~kconrad/blurbs/analysis/contractionshort.pdf&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="619" LastActivityDate="2011-02-04T07:48:07.403" />
  <row Id="20374" PostTypeId="1" CreationDate="2011-02-04T07:51:43.893" Score="0" ViewCount="70" Body="&lt;p&gt;There are 5 questions in a quiz designated by:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a, b, c, d, e&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each of the questions can only have  one of 4 possible answers:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Possible Answer #1: 5&#xA;Possible Answer #2: -5&#xA;Possible Answer #3: 7&#xA;Possible Answer #4: -7&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Am I correct to say that there are 625 possible variations of answers that we can receive? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;-&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can any of the variations equal to zero when we would add all the answers up? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example on variation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(5) + (-5) + (7) + (-7) + (5) = 5&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the core question is can the sum of all the answers to a particular variation ever be zero.&lt;/p&gt;&#xA;" OwnerUserId="6606" LastEditorUserId="6606" LastEditorDisplayName="" LastEditDate="2011-02-04T08:10:04.633" LastActivityDate="2011-02-04T08:39:38.983" Title="(a)+(b)+(c)+(d)+(e) odd numbers" Tags="&lt;algebra&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="20375" PostTypeId="2" ParentId="20365" CreationDate="2011-02-04T07:52:33.127" Score="4" ViewCount="" Body="&lt;p&gt;Define $f(n)$ as the probability of playing forever when starting out with n coins. Also, assume that the probability p of winning a round is bigger than $\frac{1}{2}$ (otherwise, the probability of playing forever is 0). Then, we get the recurrence relation&#xA;$$&#xD;&#xA;f(n) = p f(n + 1) + (1-p) f(n-1)&#xD;&#xA;$$&#xA;with the boundary conditions $f(0) = 0$ and $\lim_{n \to \infty} f(n) = 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The general solution of the recurrence relation is&#xA;$$&#xD;&#xA;f(n) = a + b \left( \frac{1-p}{p} \right)^n&#xD;&#xA;$$&#xA;and from the boundary conditions we get $a = -b = 1$. So, the probability to play infinitely is&#xA;$$&#xD;&#xA;f(n) = 1 - \left( \frac{1-p}{p} \right)^n .&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second question can be answered just as the first one, but with different boundary conditions: $f(0) = 0$ and $f(m) = 1$. This leads to the probability of reaching m as&#xA;$$&#xD;&#xA;\frac{1 - \left( \frac{1-p}{p} \right)^n}{1 - \left( \frac{1-p}{p} \right)^m} .&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the way, the keyword for this kind of problems is 'Random Walk'.&lt;/p&gt;&#xA;" OwnerUserId="960" LastActivityDate="2011-02-04T07:52:33.127" />
  <row Id="20376" PostTypeId="2" ParentId="20372" CreationDate="2011-02-04T08:06:52.530" Score="5" ViewCount="" Body="&lt;p&gt;Suppose $x,y&amp;gt;1$. If $x=y$ then both sides are 0 in your inequality, so I assume $x\ne y$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Say that $x&amp;gt;y$. Then $f(x)-f(y)=(x-y)+(1/x-1/y)=\displaystyle (x-y)(1-\frac1{xy})$. Now, the fraction $1/(xy)$ is positive, but strictly less than 1, as $1&amp;lt;x,y$. Then the product is clearly positive but strictly less than $x-y$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, if $y&amp;gt;x$, we get $0&amp;lt;f(y)-f(x)=\displaystyle(y-x)(1-\frac1{xy})&amp;lt;y-x$.&lt;/p&gt;&#xA;" OwnerUserId="462" LastActivityDate="2011-02-04T08:06:52.530" />
  <row Id="20377" PostTypeId="2" ParentId="20374" CreationDate="2011-02-04T08:08:16.950" Score="1" ViewCount="" Body="&lt;p&gt;The answer is no. The sum of a single odd number is odd (since it is just the original number), the sum of two odd numbers is even (&lt;a href=&quot;http://en.wikipedia.org/wiki/Parity_%28mathematics%29&quot; rel=&quot;nofollow&quot;&gt;see here&lt;/a&gt;), and in general, the sum of $n$ odd numbers is odd if $n$ is odd, and even if $n$ is even.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your problem, you are adding 5 odd numbers, and because 5 is odd, we must have that the sum is odd, regardless of which of the values $5,-5,7,-7$ the numbers $a,b,c,d,e$ take on. However, 0 is an even number. Thus, the sum $a+b+c+d+e$ can never be 0.&lt;/p&gt;&#xA;" OwnerUserId="264" LastActivityDate="2011-02-04T08:08:16.950" />
  <row Id="20378" PostTypeId="1" AcceptedAnswerId="20381" CreationDate="2011-02-04T08:19:11.183" Score="5" ViewCount="182" Body="&lt;p&gt;a_n converges more slowly then b_n if there exist an x such that for all m&gt;x, a_m&gt;b_m, and both sum a_n and sum b_n converges for n=1 to n=inf.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ignoring constant factors, which type of function converges the most slowly?&lt;/p&gt;&#xA;" OwnerUserId="5904" LastActivityDate="2011-02-04T08:34:26.147" Title="Which series converges the most slowly ?" Tags="&lt;sequences-and-series&gt;&lt;convergence&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="20379" PostTypeId="1" CreationDate="2011-02-04T08:24:51.110" Score="5" ViewCount="145" Body="&lt;p&gt;In the Eculidean Space, one can automatically define a norm if a specific scalar product is given. On the contrary, it's not always true. A p-norm is a scalar product if and only if p=2. My question is what condition do we need in order to move back?&lt;/p&gt;&#xA;" OwnerUserId="2177" LastActivityDate="2011-05-29T08:09:58.407" Title="From norm to scalar product" Tags="&lt;norm&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="20380" PostTypeId="1" AcceptedAnswerId="20388" CreationDate="2011-02-04T08:33:55.757" Score="2" ViewCount="62" Body="&lt;p&gt;Given a metric space, we can check what the largest number of equidistant points are (ie, such that the distance between any two of these points is the same). Of course, this might not be finite, as in the case of the discrete metric on an infinite space.&#xA;This is not an invariant of the induced metric, however, as we can see by looking at the usual euclidian distance in $\mathbb{R}^2$ and the metric where one adds the vertical and horizontal distances. These gave the same topology, but the first has at most 3 equidistant points whereas the other has 4.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is whether this quantity has been studied and if it has a name. Also, can two metrics have an arbitrarily large difference in this number even though they induce the same topology? Can one have a finite number where the other has an infinite?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am asking purely out of curiosity, by the way.&lt;/p&gt;&#xA;" OwnerUserId="2538" LastActivityDate="2011-02-04T09:21:49.277" Title="Largest number of equidistant points" Tags="&lt;topology&gt;&lt;metric-spaces&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="20381" PostTypeId="2" ParentId="20378" CreationDate="2011-02-04T08:34:26.147" Score="8" ViewCount="" Body="&lt;p&gt;There is not such a function! Walter Rudin in his &quot;Principles of Mathematical Analysis&quot; has a series of exercises trying to indicate that there is no function at the &quot;boundary&quot; between convergence and divergence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There was an interesting series of answers about this very issue at MathOverflow (MO): &lt;a href=&quot;http://mathoverflow.net/questions/49415/nonexistence-of-boundary-between-convergent-and-divergent-series&quot; rel=&quot;nofollow&quot;&gt;http://mathoverflow.net/questions/49415/nonexistence-of-boundary-between-convergent-and-divergent-series&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A quick way of seeing that there is no &quot;slowest&quot; convergent series is Rudin's exercise 12.b, mentioned at the link above: If $\sum a_n$ converges, and the $a_n$ are positive, then $\sum a_n/\sqrt {r_n}$ converges, where $r_n$ is the tail $\sum_{i\ge n}a_i$. Note $r_n\to 0$, so $a_n/\sqrt{r_n}&amp;gt;a_n$ for $n$ large enough.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I recommend that you take a look at the answers at MO and at the references they suggest, for more subtle examples.&lt;/p&gt;&#xA;" OwnerUserId="462" LastActivityDate="2011-02-04T08:34:26.147" CommentCount="3" />
  <row Id="20382" PostTypeId="2" ParentId="20379" CreationDate="2011-02-04T08:36:54.303" Score="10" ViewCount="" Body="&lt;p&gt;I'm not sure if this is what you're asking:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A norm on a vector space is induced from a scalar product if and only if the &lt;em&gt;parallelogram law&lt;/em&gt; $\| x - y\|^{2} + \|x + y\|^{2} = 2\|x\|^{2} + 2\|y\|^{2}$ is satisfied.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One direction is obtained by expanding the scalar products. If the parallelogram law holds, then one can verify that the expression given by &lt;em&gt;polarization&lt;/em&gt; is a scalar product inducing the norm:&#xA;\[&#xA;\langle x, y\rangle = \frac{1}{4}( \|x + y\|^{2} - \|x - y\|^{2})&#xA;\]&#xA;in the real case and&#xA;\[&#xA;\langle x, y\rangle = \frac{1}{4} \sum_{k = 1}^{4} i^{k} \|x + i^{k} y\|^{2}&#xA;\]&#xA;in the complex case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added much later:&lt;/strong&gt; For a good outline of the somewhat painful proof of the non-trivial direction, see &lt;a href=&quot;http://math.stackexchange.com/questions/21792/norms-induced-by-inner-products&quot;&gt;Arturo's answer&lt;/a&gt; here.&lt;/p&gt;&#xA;" OwnerUserId="5363" LastEditorUserId="5363" LastEditorDisplayName="" LastEditDate="2011-05-29T08:09:58.407" LastActivityDate="2011-05-29T08:09:58.407" />
  <row Id="20383" PostTypeId="2" ParentId="20374" CreationDate="2011-02-04T08:39:38.983" Score="1" ViewCount="" Body="&lt;p&gt;As to the first question (&quot;Am I correct to say that there are 625 possible variations of answers that we can receive?&quot;), if I understand correctly the setting, you have four possible answers for the first question, four for the second and so on. So the number of possible sets of five answers is 4 × 4 × 4 × 4 × 4, i.e. 4^5, i.e 1024.&lt;/p&gt;&#xA;" OwnerUserId="6507" LastActivityDate="2011-02-04T08:39:38.983" />
  <row Id="20384" PostTypeId="2" ParentId="20362" CreationDate="2011-02-04T08:40:18.847" Score="0" ViewCount="" Body="&lt;p&gt;From the law of cosines we know that $b^2 - a^2 = c^2 - 2ac \cos B$, so&#xA;$ b - a = { {-2ac \cos B} \over {b+a} } + { {c^2} \over {b+a} }$. Asuuming that $ c &amp;lt;&amp;lt; a \Rightarrow b \approx a $, gets you close but the second term does not match. &lt;/p&gt;&#xA;" OwnerUserId="6607" LastActivityDate="2011-02-04T08:40:18.847" />
  <row Id="20385" PostTypeId="1" AcceptedAnswerId="20387" CreationDate="2011-02-04T08:48:53.627" Score="9" ViewCount="239" Body="&lt;p&gt;Suppose we don't believe the continuum hypothesis. Using Von Neumann cardinal  assignment (so I guess we believe well-ordering?), is there any &quot;familiar&quot; ordinal number $\alpha$ such that, for non-tautological reasons, $\aleph_\alpha$ is provably larger than the cardinality of the continuum? I would hope not since it would seem pretty silly if something like $\alpha = \omega_0$ worked and we could say &quot;well gee we can't prove that $c = \aleph_1$, but it's definitely one of $\aleph_1, \aleph_2, \ldots , \aleph_{73}, \ldots$&quot;. I (obviously) don't know jack squat about set theory, so this is really just idle curiosity. If a more precise question is desired I guess I would have to make it&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;For any countable ordinal $\alpha$ is the statement: $c &amp;lt; \aleph_\alpha$ independent of ZFC in the same sense as the continuum hypothesis?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;assuming that even makes sense. Thanks!&lt;/p&gt;&#xA;" OwnerUserId="6608" LastEditorUserId="622" LastEditorDisplayName="" LastEditDate="2011-02-04T09:55:10.433" LastActivityDate="2011-02-04T09:55:10.433" Title="bound on the cardinality of the continuum? I hope not..." Tags="&lt;set-theory&gt;&lt;cardinals&gt;" AnswerCount="1" />
  <row Id="20386" PostTypeId="2" ParentId="20379" CreationDate="2011-02-04T09:00:17.627" Score="0" ViewCount="" Body="&lt;p&gt;Maybe more generally, while every norm gives us a metric, a metric is generated by&#xA;  a norm, I think, if the norm is translation-invariant.&lt;/p&gt;&#xA;" OwnerUserId="6600" LastActivityDate="2011-02-04T09:00:17.627" CommentCount="4" />
  <row Id="20387" PostTypeId="2" ParentId="20385" CreationDate="2011-02-04T09:01:19.017" Score="18" ViewCount="" Body="&lt;p&gt;Mike: If you fix an ordinal $\alpha$, then it is consistent that ${\mathfrak c}&amp;gt;\aleph_\alpha$. More precisely, there is a (forcing) extension of the universe of sets &lt;em&gt;with the same cardinals&lt;/em&gt; where the inequality holds. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you begin with a model of GCH, then you can go to an extension where ${\mathfrak c}=\aleph_\alpha$ and no cardinals are changed, as long as $\alpha$ is not a limit ordinal of countable cofinality. For example, $\aleph_{\aleph_\omega}$ is &lt;em&gt;not&lt;/em&gt; a valid size for the continuum. But it can be larger.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here, the cofinality of the limit ordinal $\alpha$ is the smallest $\beta$ such that there is an unbounded function $f:\beta\to\alpha$. There is a result of König that says that $\kappa^\lambda&amp;gt;\kappa$ if $\lambda$ is the cofinality of $\kappa$. If $\kappa={\mathfrak c}$, this says that $\lambda&amp;gt;\omega=\aleph_0$, since ${\mathfrak c}=2^{\aleph_0}$ and $(2^{\aleph_0})^{\aleph_0}=2^{\aleph_0}$. Since $\aleph_{\aleph_\omega}$ has cofinality $\omega$, it cannot be ${\mathfrak c}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But this is the only restriction! The technique to prove this (forcing) was invented by Paul Cohen and literally transformed the field.&lt;/p&gt;&#xA;" OwnerUserId="462" LastActivityDate="2011-02-04T09:01:19.017" CommentCount="13" />
  <row Id="20388" PostTypeId="2" ParentId="20380" CreationDate="2011-02-04T09:21:49.277" Score="3" ViewCount="" Body="&lt;p&gt;These types of problems are studied for subsets of the Euclidean plane, but not in topology, since as you say the question is not topologically invariant; rather they are studied in combinatorial geometry. One classic problem along these lines is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Hadwiger%E2%80%93Nelson_problem&quot; rel=&quot;nofollow&quot;&gt;Hadwiger-Nelson problem&lt;/a&gt;. Another question along these lines is the &lt;a href=&quot;http://www.cs.umd.edu/~gasarch/erdos_dist/erdos_dist.html&quot; rel=&quot;nofollow&quot;&gt;Erdős distance problem&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer to your second question is trivially yes. Consider the metric space $M_1$ on, say, countably many points where all distinct points are at distance $1$ from each other and the metric space $M_2$ on countably many points where all distinct points are at different distances from each other in, say, the interval $[0.5, 1]$. Both induce the discrete topology.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(These types of problems are also unlikely to be studied by anyone studying metric spaces, since it is only very rarely that anyone cares about metric spaces up to &lt;em&gt;isometry&lt;/em&gt;. Most people only care about metric spaces up to homeomorphism, a few up to uniform isomorphism or quasi-isometry...)&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2011-02-04T09:21:49.277" />
  <row Id="20389" PostTypeId="1" AcceptedAnswerId="20401" CreationDate="2011-02-04T10:21:36.887" Score="1" ViewCount="112" Body="&lt;p&gt;A triangle has rational side lengths and rational angles measured as degrees. Is such a triangle necessary equilateral?&lt;/p&gt;&#xA;" OwnerUserId="6610" LastActivityDate="2011-02-04T14:41:44.837" Title="Triangle and rational numbers" Tags="&lt;euclidean-geometry&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="20390" PostTypeId="2" ParentId="20370" CreationDate="2011-02-04T10:29:33.820" Score="3" ViewCount="" Body="&lt;p&gt;Maybe picking $(a,b) \mapsto \text{span}(a,b,1)$ would be a better choice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This map seem awfully arbitrary and unmotivated, and I find it easier to explain how there can be an isomorphism if I go the other way : start from $\Sigma$ and go to $\Pi&amp;#39;$ :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to see the points of $\Sigma$ as points on a normal plane, pick a 2-dimensional affine subspace $P$ of $V$ that doesn't contain the origin and put a coordinate system on it so that we can call it $\Pi$ (for example the plane z=1 if you want to compare with the map I gave).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can construct an isomorphism as follows :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pick $W$ a line in $\Sigma$ (a 2-dimensional linear subspace of V). If $W$ intersects $P$ on a line $w$ of $P$, we want to take $f(W)$ = the line of $\Pi&amp;#39;$ containing $w$.&#xA;So in $\Pi&amp;#39;$, $f(W) = w$ plus the pencil of parallel lines that contains $w$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pick $A$ a point if $\Sigma$ (a 1-dimensional linear subspace of V). If $A$ intersects $P$ on a point $a$ of $P$, we define $f(A) = a$. It is consistent with our choice of images of lines, because if a $W$ contains $A$ then $W \cap P$ contains $A \cap P$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $A$ is parallel to $P$, consider a line $W$ containing $A$ and intersecting $P$.&#xA;$f(W) = $ the intersection of $P$ and $W$ = a line of $P$ &quot;parallel&quot; to $A$. Thus our only choice is to pick $f(A) =$ the point lying at the intersection of those $f(W)$ = the pencil of parallel lines that are the lines &quot;parallel&quot; to $A$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, if $W$ is the 2-dimensional subspace &quot;parallel&quot; to $P$, we have to chose $f(W) =$ the reunion for the points $A$ in $W$ of $f(A)$ = the line of $\Pi&amp;#39;$ whose points are the pencils of parallel lines, that is, the &quot;line at infinity&quot;.&lt;/p&gt;&#xA;" OwnerUserId="6270" LastActivityDate="2011-02-04T10:29:33.820" CommentCount="2" />
  <row Id="20391" PostTypeId="2" ParentId="20370" CreationDate="2011-02-04T10:29:43.710" Score="2" ViewCount="" Body="&lt;p&gt;This is too long for a comment [I hope that this time I'm not using anything I'm not allowed to use ;-)]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would start by identifying $\Pi$ with the affine plane $(x,y,1) \subset F^{3}$. Obviously, every $1$-dimensional linear subspace of $F^{3}$ that doesn't lie in the $(x,y)$-plane intersects this affine plane in a unique point and every $2$-dimensional linear subspace that isn't the $(x,y)$-plane intersects this affine plane in a $1$-dimensional affine subspace. Conversely, every point in $\Pi$ determines a $1$-dimensional linear subspace of $F^{3}$ and every line in $\Pi$ determines a $2$-dimensional linear subspace of $F^{3}$. Now it should be obvious what happens with the points at infinity: they determine (and are determined by) a $1$-dimensional linear subspace $\ell$ of the $(x,y)$-plane and they themselves lie on the line at infinity in $\Pi&amp;#39;$ (which you've forgotten in your description of lines of $\Pi'$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This construction extends of course to higher dimensions. Hope that helps.&lt;/p&gt;&#xA;" OwnerUserId="5363" LastActivityDate="2011-02-04T10:29:43.710" CommentCount="3" />
  <row Id="20392" PostTypeId="1" CreationDate="2011-02-04T10:50:01.167" Score="0" ViewCount="111" Body="&lt;p&gt;I want to simulate a heat source (eg a cpu) connected to a heatsink without any cooling fans. The simulation will run indefinitely using small time steps. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What i want to measure at each time step is the temperature of the heat source and the temperature of the heatsink at each tick (of time).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The heat source will be generating a known amount of power in watts to be converted to heat, which may vary during the simulation. Think of the cpu reducing its clockspeed thereby lowering its power consumption and heat emitted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe what i need to perform each step is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[1] calculate heat generated by source&#xA;[2] emit some of that heat in the form of radiation&#xA;[3] determine some kind of cooling&#xA;[4] subtract cooling from heat generated to get a temperature for heatsource, and for the heatsink.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have been messing around with heat equations but am struggling to join them together to model this system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any help would be very much appreciated.&lt;/p&gt;&#xA;" OwnerUserId="6613" LastActivityDate="2011-06-04T20:11:37.693" Title="passive heatsink simulation" Tags="&lt;simulation&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="20393" PostTypeId="1" CreationDate="2011-02-04T11:49:03.793" Score="0" ViewCount="126" Body="&lt;p&gt;What is closed form equation for the tail probability of multivariate normal distribution ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;y&lt;/strong&gt; ~ $\mathcal{N}$($0$, $\sigma^2I_n$ ) ; &lt;strong&gt;y&lt;/strong&gt; $\in\mathbb{R}^n$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pr(&lt;strong&gt;y&lt;/strong&gt; &gt; $\gamma$) = $\int_\gamma^\infty$ pr(&lt;strong&gt;y&lt;/strong&gt;) d &lt;strong&gt;y&lt;/strong&gt; ; $\gamma\in\mathbb{R}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-Sun&lt;/p&gt;&#xA;" OwnerUserId="5978" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2011-03-06T14:30:35.983" LastActivityDate="2011-03-06T17:10:53.387" Title="tail probability of multivariate normal distribution" Tags="&lt;probability&gt;" AnswerCount="2" CommentCount="5" />
  <row Id="20394" PostTypeId="2" ParentId="20393" CreationDate="2011-02-04T12:52:54.237" Score="1" ViewCount="" Body="&lt;p&gt;There are two ways to make sense of this problem: rectangular and radial.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rectangular approach is to make $\gamma$ a vector and interpret $y &amp;gt; \gamma$ to mean that every component of $y$ is greater than the corresponding component of $\gamma$. This is a common problem, but it cannot be solved in closed form. See &lt;a href=&quot;http://www.math.wsu.edu/faculty/genz/papers/mvn/mvn.html&quot; rel=&quot;nofollow&quot;&gt;this article&lt;/a&gt; for numerical algorithms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The radial approach is to ask for the probability that $|y| &amp;gt; \gamma$, i.e. the probability that $y$ is further than a distance $\gamma$ from the origin. That's an easier problem.&lt;/p&gt;&#xA;" OwnerUserId="247" LastActivityDate="2011-02-04T12:52:54.237" />
  <row Id="20395" PostTypeId="2" ParentId="20392" CreationDate="2011-02-04T13:18:58.010" Score="0" ViewCount="" Body="&lt;p&gt;1) is just your input.  You make it up.  2) the &lt;a href=&quot;http://en.wikipedia.org/wiki/Thermal_radiation&quot; rel=&quot;nofollow&quot;&gt;radiation law&lt;/a&gt; is the power out (energy per time) $P=\sigma A T^4$, where $\sigma$ is the Stefan-Boltzman constant, A the area and T the absolute temperature.  The net radiation (because you receive from the environment) is then $P=\sigma A (T_{heat sink}^4-T_{environment}^4)$  At these temperatures, unless your sink is polished metal, you can consider it black.  Conduction and convection are just $P=k(T_{heat sink}-T_{environment})$ where the hard part is figuring our reasonable constant k's (they are different).  As you say, you have a thermal balance, and the temperature changes as $\delta T=\frac{(P_{in}-P_{out})\delta t}{C}$ where $C$ is the heat capacity.  You can look that up for the material of your heat sink.l  Aluminum is about $0.9 \frac{J}{gK}$ for example.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-02-04T13:18:58.010" />
  <row Id="20396" PostTypeId="2" ParentId="20358" CreationDate="2011-02-04T13:40:18.010" Score="3" ViewCount="" Body="&lt;p&gt;By only in $L^1$ I understand that the function $f$ is in $L^1$ but not in $L^p$ for any $p&amp;gt;1$. Such a function is for instance $f(x)=\min(|x|^{-a},|x|^{-b})$, where $0&amp;lt;a&amp;lt;1&amp;lt;b$ (and $f(0)=0$). Then $\hat f\in L^\infty$, but $\hat f\not\in L^p$ for $1\le p\le2$. This a consequence of the inversion formula and the Haussdorf Young inequality: if $g\in L^p$, $1\le p\le2$, then $\|\hat g\|_q\le C\|g\|_p$ for a constant $C$ independent of $g$ and $p^{-1}+q^{-1}=1$. However, it could still happen $\hat f\in L^p$ for some $p\in(2,\infty)$.&lt;/p&gt;&#xA;" OwnerUserId="4791" LastEditorUserId="4791" LastEditorDisplayName="" LastEditDate="2011-02-04T15:40:59.933" LastActivityDate="2011-02-04T15:40:59.933" CommentCount="2" />
  <row Id="20397" PostTypeId="1" CreationDate="2011-02-04T14:04:04.247" Score="19" ViewCount="920" Body="&lt;p&gt;What are your favorite applications of integration by parts?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(The answers can be as lowbrow or highbrow as you wish. I'd just like to get a bunch of these in one place!)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for your contributions, in advance! &lt;/p&gt;&#xA;" OwnerUserId="941" LastEditorUserId="3793" LastEditorDisplayName="" LastEditDate="2011-02-04T20:25:10.780" LastActivityDate="2011-06-01T21:55:57.530" CommunityOwnedDate="2011-02-04T14:20:24.457" Title="Striking applications of integration by parts" Tags="&lt;calculus&gt;&lt;big-list&gt;" AnswerCount="15" CommentCount="2" FavoriteCount="16" />
  <row Id="20398" PostTypeId="1" CreationDate="2011-02-04T14:28:00.620" Score="2" ViewCount="129" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm totally ignorant about the Universal Coefficient Spectral Sequence (I used to work only with principal ideal domains, where the Universal Coefficient Theorem only amounts to a short exact sequence) but I need now to understand it. The book I was redirected to is the infamous &lt;em&gt;User's Guide to SS&lt;/em&gt; where I have only been able to find very general versions (one about spectra and the other with functors on general abelian categories). I guess that an exposition in the simple case of chain complexes of A-modules exists somewhere, but I don't know where to look.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do you have any suggestion?&lt;/p&gt;&#xA;" OwnerUserId="6617" LastActivityDate="2011-02-07T19:31:42.917" Title="Reference for the Universal Coefficient Spectral Sequence" Tags="&lt;reference-request&gt;&lt;algebraic-topology&gt;&lt;homological-algebra&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="1" />
  <row Id="20400" PostTypeId="2" ParentId="20397" CreationDate="2011-02-04T14:40:12.420" Score="4" ViewCount="" Body="&lt;p&gt;A lowbrow favorite of mine:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int \frac{1}{x} dx = \frac{1}{x} \cdot x - \int x \cdot\left(-\frac{1}{x^2}\right) dx = 1 + \int \frac{1}{x} dx$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, $1=0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A bit more highbrow, I like the use of partial integration to establish recursive formulas for integrals.&lt;/p&gt;&#xA;" OwnerUserId="3567" LastActivityDate="2011-02-04T14:40:12.420" CommunityOwnedDate="2011-02-04T14:40:12.420" CommentCount="4" />
  <row Id="20401" PostTypeId="2" ParentId="20389" CreationDate="2011-02-04T14:41:44.837" Score="4" ViewCount="" Body="&lt;p&gt;As Sebastian pointed out, if the angle $\theta$ is rational measured in degrees, so if $\theta / \pi$ is rational, and if $\cos \theta$ is also rational, then $\theta \in \left\{ 0, \pi \pm \pi/2, \pm \pi/3, \pm/ 2\pi/3 \right\}$.&#xA;The &quot;modern&quot; proof of this fact is that if $\theta / \pi$ is rational, $z=e^{i\theta}$ is an algebraic integer (it is a root of $1$), so $2 \cos \theta = z + \bar{z}$ also, and the only rationals which are algebraic integers are the integers, so $2 \cos \theta \in \mathbb{Z}$, and the result follows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Back to your question, the law of cosines implies that the cosines of the three angles are rational, and from the previous lemma, the (undirected) angles can be $\pi/3$, $\pi/2$ or $2 \pi/3$ (I assume the triangle is not flat).&#xA;The only possible case is if the three angles are $\pi/3$ (the sum has to be $\pi$), so yes, the triangle is equilateral.&lt;/p&gt;&#xA;" OwnerUserId="2660" LastActivityDate="2011-02-04T14:41:44.837" CommentCount="0" />
  <row Id="20402" PostTypeId="2" ParentId="392" CreationDate="2011-02-04T14:54:34.200" Score="0" ViewCount="" Body="&lt;p&gt;Let's talk about the first one,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{d}{dx}sin(x) = cos(x)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take a look at the plot:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/v7yb1.png&quot; /&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sin$ is red, $\cos$ is blue.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;strong&gt;rate of change&lt;/strong&gt; of the &lt;em&gt;red curve&lt;/em&gt; ($\sin$) is exactly the current y-value of the &lt;strong&gt;blue curve&lt;/strong&gt; ($\cos$) &lt;strong&gt;at every point&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pointing out some salient points:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;@ x=$\frac{\pi}{2}$, $\sin(\frac{\pi}{2})=1$ and $\cos(\frac{\pi}{2})=0$.  This means the rate of the sin curve @ x=$\frac{\pi}{2}$is NOTHING, which you can see clearly in the graph - a local maximum.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;@ x=0, $\sin(0)=0$ and $\cos(0)=1$, which means sin(x) should appear to travel along the straight line y=x at the origin, which it does.  In fact, near x=0 we have the approximation sin(x)=x.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;@ x=$\frac{\pi}{2}^+$, you can see $\sin(\frac{\pi}{2}^+)$ starts to go downward.  At this point, $\cos(\frac{\pi}{2}^+)$ ALSO dips below the x-axis, i.e. for the first time the rate of change of sin(x) becomes negative.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1069" LastEditorUserId="1069" LastEditorDisplayName="" LastEditDate="2011-02-04T15:01:22.633" LastActivityDate="2011-02-04T15:01:22.633" />
  <row Id="20403" PostTypeId="1" CreationDate="2011-02-04T14:59:07.607" Score="4" ViewCount="259" Body="&lt;p&gt;Let $f,g:\mathbb C \to \mathbb C$ be two analytic functions such that $f(z)(g(z)+z^2)=0$ for all $z$ .Then prove that either $f(z)=0$ or $g(z)=-z^2$.&lt;/p&gt;&#xA;" OwnerUserId="6618" LastEditorUserId="5966" LastEditorDisplayName="" LastEditDate="2011-02-04T16:36:41.267" LastActivityDate="2011-04-05T23:14:53.473" Title="A question on complex analysis" Tags="&lt;complex-analysis&gt;" AnswerCount="2" CommentCount="7" />
  <row Id="20405" PostTypeId="1" CreationDate="2011-02-04T15:23:05.740" Score="1" ViewCount="81" Body="&lt;p&gt;Hello,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We all know Dijkstra and so on.&#xA;But is there a way to extend this to get the N best shortest paths between 2 nodes? ...and still with &quot;acceptable&quot; complexity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks,&#xA;Arnaud&lt;/p&gt;&#xA;" OwnerUserId="6619" LastActivityDate="2011-02-04T15:41:28.660" Title="n-best shortest path?" Tags="&lt;graph-theory&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="20406" PostTypeId="2" ParentId="5396" CreationDate="2011-02-04T15:26:37.073" Score="3" ViewCount="" Body="&lt;p&gt;I am just going to answer the first question.  For a two-tensor we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$|T|^2 = \left&amp;lt; T,T \right&amp;gt; = g^{ik}g^{jl}T_{ij}T_{kl},$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $g$ is the metric, and the summation convention is understood.  Note that this is the standard inner product structure induced by $g$, as mentioned by Jason DeVito.  You might like to try and prove that this really does define a norm in the Riemannian case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More generally, for a $(k,l)$ ($k$ times contravariant, $l$ times covariant) tensor field $T$ (on the manifold) we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$|T|^2 = \left&amp;lt; T,T \right&amp;gt; = g^{j_1q_1}g^{j_2q_2}\cdots g^{j_lq_l}g_{i_1p_1}g_{i_2p_2}\cdots g_{i_kp_k}T^{i_1i_2\cdots i_k}_{j_1j_2\cdots j_l}T^{p_1p_2\cdots p_k}_{q_1q_2\cdots q_l}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For your other questions, you could do worse than look in 'Einstein Manifolds' by Besse.&lt;/p&gt;&#xA;" OwnerUserId="4322" LastActivityDate="2011-02-04T15:26:37.073" />
  <row Id="20407" PostTypeId="1" AcceptedAnswerId="20408" CreationDate="2011-02-04T15:32:19.373" Score="0" ViewCount="22" Body="&lt;p&gt;Hey guys,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is kind of a programming question, but really it's math in disguise!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm working on a little project that will look at the size of a file and make sure it's in within an acceptable range based on previous file sizes. I know how to get the average file size of previous files and all that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm wondering if there is an equation or a method that I can dynamically create an acceptable range depending on avg file size, number of files, or some other statistical value. I can provide more details if needed, but I really don't know where to start.&lt;/p&gt;&#xA;" OwnerUserId="6620" LastActivityDate="2011-02-04T15:36:47.223" Title="File Size Stats" Tags="&lt;analysis&gt;&lt;statistics&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="20408" PostTypeId="2" ParentId="20407" CreationDate="2011-02-04T15:36:47.223" Score="1" ViewCount="" Body="&lt;p&gt;You may want to try testing whether the size of the file in question is within say 1 standard deviation of the average size.&lt;/p&gt;&#xA;" OwnerUserId="589" LastActivityDate="2011-02-04T15:36:47.223" CommentCount="1" />
  <row Id="20409" PostTypeId="2" ParentId="20405" CreationDate="2011-02-04T15:41:28.660" Score="1" ViewCount="" Body="&lt;p&gt;From &lt;a href=&quot;http://en.wikipedia.org/wiki/Dijkstra%27s_algorithm&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The functionality of Dijkstra's original algorithm can be extended with a variety of modifications. For example, sometimes it is desirable to present solutions which are less than mathematically optimal. To obtain a ranked list of less-than-optimal solutions, the optimal solution is first calculated. A single edge appearing in the optimal solution is removed from the graph, and the optimum solution to this new graph is calculated. Each edge of the original solution is suppressed in turn and a new shortest-path calculated. The secondary solutions are then ranked and presented after the first optimal solution.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;However, this probably doesn't cover &quot;acceptable&quot; complexity. Also, it doesn't necessarily find all near-optimal paths. (E.g. consider the case where there exists a very small loop from one of the cities on the optimal path to itself; the optimal path plus this loop might be the second-most-optimal path, but wouldn't be picked up by the above.)&lt;/p&gt;&#xA;" OwnerUserId="5001" LastActivityDate="2011-02-04T15:41:28.660" />
  <row Id="20410" PostTypeId="1" AcceptedAnswerId="20411" CreationDate="2011-02-04T15:46:46.683" Score="1" ViewCount="111" Body="&lt;p&gt;While I was solving a problem something interesting came up if we know the dot product of 2 vectors, and one of the vectors is known. Can we find the other? The interesting bit about this is that even though the dot product behaves pretty much like normal algebraic multiplication &lt;strong&gt;1&lt;/strong&gt;, but that does not extend to properties like division for example if $a * b = c$ then $a = \frac{b}{c}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is this so (other than the fact that we're talking about vectors)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1 As Qiaochu Yuan points out that statement is simply inaccurate. Sorry.&lt;/p&gt;&#xA;" OwnerUserId="5576" LastEditorUserId="5576" LastEditorDisplayName="" LastEditDate="2011-02-04T19:27:00.850" LastActivityDate="2011-02-04T19:27:00.850" Title="Differences Between Algebraic Multiplication &amp; The Dot Product" Tags="&lt;vector-analysis&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="20411" PostTypeId="2" ParentId="20410" CreationDate="2011-02-04T16:01:37.907" Score="1" ViewCount="" Body="&lt;p&gt;The reason why the dot product of two vectors has no inverse is that it is not an injective function. In fact, if you fix a vector $v$ and a number $r$, then the set of all vectors $w$ that satisfy $v * w = r$ is a hyperplane of your vector space, and those usually contain more than one element.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In contrast to this, multiplication by a non-zero real number is an injective function (in fact, a bijective function), and so it has an inverse function, namely division by that number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You should also note that multiplication of real numbers and dot products are fundamentally different in the sense that multiplication of two real numbers gives you back a real number, whereas the dot product of two vectors in general does not give you back a vector of the same space, but a real number (or an element of the field you're working over).&lt;/p&gt;&#xA;" OwnerUserId="1766" LastActivityDate="2011-02-04T16:01:37.907" />
  <row Id="20412" PostTypeId="1" CreationDate="2011-02-04T16:14:25.920" Score="1" ViewCount="222" Body="&lt;p&gt;Is there a notation for element-wise (or pointwise) operations?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, take the element-wise product of two vectors x and y (in Matlab, x .* y, in numpy x*y), producing a new vector of same length z, where $z_i = x_i * y_i$ . &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In mathematical notation, there doesn't seem to be a standard for this, am I wrong?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is $x \cdot y$, the dot product. There is $x*y$, which is usually considered the cross product. I need to find a notation for element-wise multiplication. I was aiming at maybe using the . as is done in Matlab, but it looks a little off :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$z = x .* y$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What do you guys think? &lt;/p&gt;&#xA;" OwnerUserId="3158" LastEditorUserId="3158" LastEditorDisplayName="" LastEditDate="2011-02-04T16:27:56.693" LastActivityDate="2011-03-01T17:13:51.920" Title="Element-wise (or pointwise) operations notation?" Tags="&lt;notation&gt;" AnswerCount="2" FavoriteCount="0" />
  <row Id="20413" PostTypeId="2" ParentId="20412" CreationDate="2011-02-04T16:21:54.987" Score="6" ViewCount="" Body="&lt;p&gt;If you need to do this, generally a good idea is to write your &quot;vectors&quot; as functions $I \to \mathbb{R}$ (or whatever) where $I$ is your index set, and then say that $fg(i) = f(i) g(i)$ is the &lt;em&gt;pointwise&lt;/em&gt; product. Mathematicians don't have a special notation for this because&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;nobody takes the pointwise product of &lt;em&gt;vectors&lt;/em&gt; (in the geometric sense) because it's not invariant under change of coordinates, and&lt;/li&gt;&#xA;&lt;li&gt;if you are taking the pointwise product of &lt;em&gt;functions&lt;/em&gt; then it is generally clear that you are doing this from context (e.g. if $f, g$ are continuous functions $X \to \mathbb{R}$ where $X$ is a topological space).&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;If you are working in a context where both dot products and pointwise products make sense, the answer is to change your notation for dot products to something like $\langle f, g \rangle$.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2011-02-04T16:21:54.987" CommentCount="2" />
  <row Id="20414" PostTypeId="2" ParentId="18776" CreationDate="2011-02-04T16:22:36.917" Score="0" ViewCount="" Body="&lt;p&gt;Since I could not find a formula for an expectation value of a generalized chi square distribution, I decided to ignore the dependency between the components of $\hat{X}$ and continue with a diagonal covariance matrix. This then gives me:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$E_{\tilde{X}} \left[ \sqrt{ \tilde{X}^T \tilde{X} } \right] = E_X \left[ \sqrt{ \sum_{i=1}^n \tilde{x}^2_i } \right]$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However each $\tilde{x}_i$ still has its own variance $\sigma_i^2$. Therefore I guess the expectation value of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Noncentral_chi_distribution&quot; rel=&quot;nofollow&quot;&gt;non-central chi distribution&lt;/a&gt; won't help here since I can not factor out one common variance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My approach now is the following: Since I need this expectation value as part of an upper bound, I apply the  triangle inequality in order to approximate the square root of a sum:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$E_{\tilde{X}} \left[ \sqrt{ \sum_{i=1}^n \tilde{x}^2_i } \right] \le E_{\tilde{X}} \left[ \sum_{i=1}^n \sqrt{ \tilde{x}_i^2 } \right] =  \sum_{i=1}^n E_{\tilde{x_i}} \left[  \| \tilde{x}_i \| \right]$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think this is equivalent to exchanging the euclidean norm with the manhattan norm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What do you think about this idea? Is there maybe another approximation which might be more close to the real expectation value?&lt;/p&gt;&#xA;" OwnerUserId="6197" LastEditorUserId="6197" LastEditorDisplayName="" LastEditDate="2011-02-04T16:29:51.623" LastActivityDate="2011-02-04T16:29:51.623" />
  <row Id="20415" PostTypeId="2" ParentId="20397" CreationDate="2011-02-04T16:32:46.873" Score="27" ViewCount="" Body="&lt;p&gt;I always liked the derivation of Taylor's formula with error term:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{array}{rl}&#xD;&#xA;f(x) &amp;amp;= f(0) + \int_0^x f&amp;#39;(x-t) \,dt\\&#xD;&#xA;&amp;amp;= f(0) + xf&amp;#39;(0) + \int_0^x tf&amp;#39;&amp;#39;(x-t)\,dt\\&#xD;&#xA;&amp;amp;= f(0) + xf&amp;#39;(0) + \frac{x^2}2f&amp;#39;&amp;#39;(0) + \int_0^x \frac{t^2}2 f&amp;#39;&amp;#39;&amp;#39;(x-t)\,dt&#xD;&#xA;\end{array}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and so on. Using the mean value theorem on the final term readily gives the &lt;em&gt;Cauchy form&lt;/em&gt; for the remainder.&lt;/p&gt;&#xA;" OwnerUserId="434" LastActivityDate="2011-02-04T16:32:46.873" CommunityOwnedDate="2011-02-04T16:32:46.873" />
  <row Id="20417" PostTypeId="2" ParentId="20397" CreationDate="2011-02-04T16:39:42.457" Score="4" ViewCount="" Body="&lt;p&gt;Lowbrow: $\int e^x\sin x\ dx$ and its ilk.&lt;/p&gt;&#xA;" OwnerUserId="4992" LastActivityDate="2011-02-04T16:39:42.457" CommunityOwnedDate="2011-02-04T16:39:42.457" />
  <row Id="20418" PostTypeId="1" CreationDate="2011-02-04T16:46:43.933" Score="2" ViewCount="275" Body="&lt;p&gt;How can I find the sum of the following series?&#xA;$$&#xD;&#xA;\sum_{n=0}^{+\infty}\frac{n^2}{2^n}&#xD;&#xA;$$&#xA;I know that it converges, and &lt;a href=&quot;http://www.wolframalpha.com/input/?i=%5Csum_%7Bn%3D0%7D%5E%7B%2B%5Cinfty%7D%28n%5E2/%282%5E%28n%29%29%29&quot; rel=&quot;nofollow&quot;&gt;Wolfram Alpha&lt;/a&gt; tells me that its sum is 6 .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which technique should I use to prove that the sum is 6?&lt;/p&gt;&#xA;" OwnerUserId="354" LastEditorUserId="10557" LastEditorDisplayName="" LastEditDate="2011-05-14T21:48:45.863" LastActivityDate="2011-05-15T06:54:04.323" Title="How to find the sum of the following series" Tags="&lt;sequences-and-series&gt;&lt;convergence&gt;" AnswerCount="3" CommentCount="3" />
  <row Id="20420" PostTypeId="2" ParentId="20418" CreationDate="2011-02-04T16:59:14.047" Score="8" ViewCount="" Body="&lt;p&gt;It is equal to $f(x)=\sum_{n \geq 0} n^2 x^n$ evaluated at $x=1/2$.&#xA;To compute this function of $x$, write $n^2 = (n+1)(n+2)-3(n+1)+1$, so that $f(x)=a(x)+b(x)+c(x)$ with:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a(x)= \sum_{n \geq 0} (n+1)(n+2) x^n = \frac{d^2}{dx^2} \left( \sum_{n \geq 0} x^n\right) = \frac{2}{(1-x)^3}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$b(x)=\sum_{n \geq 0} 3 (n+1) x^n = 3\frac{d}{dx} \left( \sum_{n \geq 0} x^n \right) = \frac{3}{(1-x)^2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$c(x)= \sum_{n \geq 0} x^n = \frac{1}{1-x}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So $f(1/2)=\frac{2}{(1/2)^3}-\frac{3}{(1/2)^2} + \frac{1}{1/2} = 16-12+2=6$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &quot;technique&quot; is to add a parameter in the series, to make the multiplication by $n+1$ appear as differentiation.&lt;/p&gt;&#xA;" OwnerUserId="2660" LastActivityDate="2011-02-04T16:59:14.047" CommentCount="1" />
  <row Id="20421" PostTypeId="1" AcceptedAnswerId="20433" CreationDate="2011-02-04T17:15:45.810" Score="1" ViewCount="198" Body="&lt;p&gt;I'm trying to find a set which is Lebesgue measurable but not Borel measurable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I was thinking of taking a Lebesgue set of measure zero and intersecting it with something so that the result is not Borel measurable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this a good approach? Can someone give a hint what set I would take (so please no full answers, I want to find it myself in the end ;-))&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, I seem to remember that to construct a non-Lebesgue measurable set one needs to use the axiom of choice. Is this also the case for non-Borel measurable sets?&lt;/p&gt;&#xA;" OwnerUserId="1120" LastActivityDate="2011-02-04T17:53:38.340" Title="Lebesgue measurable but not Borel measurable." Tags="&lt;measure-theory&gt;" AnswerCount="2" CommentCount="9" />
  <row Id="20422" PostTypeId="2" ParentId="20392" CreationDate="2011-02-04T17:16:02.537" Score="0" ViewCount="" Body="&lt;p&gt;I think a good starting place would be to look at Laplace's Equation which can be very helpful in heat conduction.  I assume you might only care about two dimensions which would help considerably in continuous calculations.&lt;/p&gt;&#xA;" OwnerUserId="6599" LastActivityDate="2011-02-04T17:16:02.537" />
  <row Id="20423" PostTypeId="2" ParentId="20418" CreationDate="2011-02-04T17:17:37.987" Score="3" ViewCount="" Body="&lt;p&gt;You can also apply the formula for the sum of a geometric series three times in a row :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\begin{array}{rcl}&#xD;&#xA;\Sigma_{n=1}^\infty \frac{n^2}{2^n} &amp;amp; = &amp;amp; \Sigma_{n=1}^\infty (2n-1) \Sigma_{k=n}^\infty \frac{1}{2^k}  = \Sigma_{n=1}^\infty \frac{2n-1}{2^{n-1}} \\&#xD;&#xA;&amp;amp;=&amp;amp; \Sigma_{k=0}^\infty \frac{1}{2^k} + \Sigma_{n=1}^\infty 2 \Sigma_{k=n}^\infty \frac{1}{2^k} = 2 + \Sigma_{n=1}^\infty \frac{2}{2^{n-1}} \\&#xD;&#xA;&amp;amp;= &amp;amp; 2 + 2*2 = 6 \end{array}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, if cutting up those geometric series is too hard, there is an easier way to do the same thing :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\begin{array}{rcl}&#xD;&#xA;(1-\frac{1}{2})^3 * \Sigma_{n=1}^\infty \frac{n^2}{2^n} &amp;amp; = &amp;amp;&#xD;&#xA;(1-\frac{1}{2})^2 * \Sigma_{n=1}^\infty \frac{n^2-(n-1)^2}{2^n} =&#xD;&#xA;(1-\frac{1}{2})^2 * \Sigma_{n=1}^\infty \frac{2n-1}{2^n} \\&#xD;&#xA;&amp;amp;=&amp;amp; (1-\frac{1}{2}) * (\frac{1}{2} + \Sigma_{n=2}^\infty \frac{(2n-1)-(2n-3)}{2^n}) \\&#xD;&#xA;&amp;amp;=&amp;amp; (1-\frac{1}{2}) * (\frac{1}{2} + \Sigma_{n=2}^\infty \frac{2}{2^n}) \\&#xD;&#xA;&amp;amp;=&amp;amp; \frac{1}{4} + \frac{2}{4} = \frac{3}{4} = (1-\frac{1}{2})^3 * 6&#xD;&#xA;\end{array}$&lt;/p&gt;&#xA;" OwnerUserId="6270" LastEditorUserId="6270" LastEditorDisplayName="" LastEditDate="2011-02-04T19:45:50.547" LastActivityDate="2011-02-04T19:45:50.547" />
  <row Id="20424" PostTypeId="2" ParentId="20397" CreationDate="2011-02-04T17:22:32.933" Score="16" ViewCount="" Body="&lt;p&gt;Repeated integration by parts gives&#xA;$$\int_0^\infty x^n e^{-x} dx=n!$$&lt;/p&gt;&#xA;" OwnerUserId="940" LastActivityDate="2011-02-04T17:22:32.933" CommunityOwnedDate="2011-02-04T17:22:32.933" CommentCount="3" />
  <row Id="20425" PostTypeId="2" ParentId="20421" CreationDate="2011-02-04T17:24:42.400" Score="3" ViewCount="" Body="&lt;p&gt;There is some good stuff here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://mathoverflow.net/questions/32720/non-borel-sets-without-axiom-of-choice&quot; rel=&quot;nofollow&quot;&gt;http://mathoverflow.net/questions/32720/non-borel-sets-without-axiom-of-choice&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="940" LastActivityDate="2011-02-04T17:24:42.400" CommentCount="1" />
  <row Id="20426" PostTypeId="1" AcceptedAnswerId="20428" CreationDate="2011-02-04T17:24:54.900" Score="0" ViewCount="181" Body="&lt;p&gt;&lt;strong&gt;Puzzle&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a country in which people only want boys, every family continues to have children until they have a boy. If they have a girl, they have another child. If they have a boy, they stop. What is the proportion of boys to girls in the country?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My solution (not finished)&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we assume that the probability of having a girl is 50%, the set of possible cases are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Boy (50%)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Girl, Boy (25%)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Girl, Girl, Boy (12.5%)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, if we call G the number of girls that a family had and B the number of boys that a family had, we have:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$B = 1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$P(G = x) = (1/2)^{x+1}*x$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$G = \Sigma (1/2)^{x+1}*x$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I feel like the sum of this infinite serie is 1 and that the proportion of girls/boys in this country will be 50%, but I don't know how to prove it!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="6623" LastActivityDate="2011-02-04T17:39:00.453" Title="Famous puzzle: Girl/Boy proportion problem (Sum of infinite serie)" Tags="&lt;puzzle&gt;&lt;sequences-and-series&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" />
  <row Id="20427" PostTypeId="2" ParentId="20403" CreationDate="2011-02-04T17:26:46.973" Score="0" ViewCount="" Body="&lt;p&gt;If $f=0$ you have done, so can assume this is not the case. Can assume there is a point where both $f$ and $g-z^2$ are both different from zero otherwise you have done. Up to traslation can assume this point is the origin. So there exists integers $h,k$ such that $f=z^hf&amp;#39;$ and $g=z^kg&amp;#39;$ with $f&amp;#39;,g&amp;#39;$ different from zero at the origin. then you have $f&amp;#39;(z^{h+k}g&amp;#39;-z^{2+k})=0$. Since $f&amp;#39;$ not zero at the origin this means that locally $z^{h+k}g&amp;#39;-z^{2+k}=0$ from which the thesis follows.&lt;/p&gt;&#xA;" OwnerUserId="6583" LastActivityDate="2011-02-04T17:26:46.973" />
  <row Id="20428" PostTypeId="2" ParentId="20426" CreationDate="2011-02-04T17:39:00.453" Score="1" ViewCount="" Body="&lt;p&gt;Mike Scott is correct that you don't need to sum the series, but suppose you want to.  Each family has 1 boy-that is easy.  Each family has 50% chance of no girls, 25% chance of 1, etc.  So the average number of girls is $$\sum_{i=0}^\infty \frac{i}{2^{i+1}}$$  The way to sum this is to remember that $$\sum_{i=0}^\infty  a^{-i} = \frac{1}{1-1/a}$$  Now if you take the derivative with respect to a and evaluate it at a=2 So $$\frac{d}{da}\sum_{i=0}^\infty  a^{-i} =\frac{d}{da} \frac{1}{1-1/a}=\frac{1}{(a-1)^2}=1$$ for $a=2$.  So there is an average of one girl per family as well&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-02-04T17:39:00.453" />
  <row Id="20429" PostTypeId="2" ParentId="20362" CreationDate="2011-02-04T17:47:14.897" Score="3" ViewCount="" Body="&lt;p&gt;The vectors ${\mathbf{a}}$ and ${\mathbf{c}}$, in terms of their magnitudes and the angles $\theta$ and $\gamma$, are&#xA;$$&#xD;&#xA;{\mathbf{a}} = (a \sin \theta, a \cos \theta)&#xD;&#xA;$$&#xA;and&#xA;$$&#xD;&#xA;{\mathbf{c}} = (c \cos \gamma, c \sin \gamma).&#xD;&#xA;$$&#xA;Assuming $c \ll a$, we can expand $b-a$ in powers of the small parameter $\varepsilon = c/a$ as follows:&#xA;$$&#xD;&#xA;\begin{eqnarray}&#xD;&#xA;b-a &amp;amp;=&amp;amp; -a + \lVert{\mathbf{b}}\rVert \\&#xD;&#xA;  &amp;amp;=&amp;amp; -a + \lVert{\mathbf{a}} - {\mathbf{c}}\rVert \\&#xD;&#xA;  &amp;amp;=&amp;amp; -a + \sqrt{(\mathbf{a} - \mathbf{c})\cdot(\mathbf{a} - \mathbf{c})} \\&#xD;&#xA;  &amp;amp;=&amp;amp; -a + \sqrt{a^2 - 2{\mathbf{a}}\cdot{\mathbf{c}} + c^2} \\&#xD;&#xA;  &amp;amp;=&amp;amp; -a + a \sqrt{1 - 2 \varepsilon\left(\sin\theta \cos\gamma + \cos\theta \sin\gamma\right) + \varepsilon^2} \\&#xD;&#xA;  &amp;amp;=&amp;amp; -a + a \sqrt{1 - 2 \varepsilon\sin(\theta+\gamma) + \varepsilon^2} \\&#xD;&#xA;  &amp;amp;=&amp;amp; -a + a \left(1 + \frac{1}{2}\left(- 2 \varepsilon\sin(\theta+\gamma) + \varepsilon^2\right) - \frac{1}{8}\left(- 2 \varepsilon\sin(\theta+\gamma) + \varepsilon^2\right)^2 + O(\varepsilon^3)\right) \\&#xD;&#xA;  &amp;amp;=&amp;amp; a \left(- \varepsilon\sin(\theta+\gamma) + \frac{1}{2}\varepsilon^2 - \frac{1}{2}\varepsilon^2\sin^2(\theta+\gamma) + O(\varepsilon^3)\right) \\&#xD;&#xA;  &amp;amp;=&amp;amp; - c\sin(\theta+\gamma) + \frac{c^2}{2a}\cos^2(\theta+\gamma) + O\left(\frac{c^3}{a^2}\right),&#xD;&#xA;\end{eqnarray}&#xD;&#xA;$$&#xA;which is the desired result.  The series expansion $\sqrt{1+x} = 1 + (1/2)x - (1/8)x^2 + O(x^3)$ was used to approximate the square root.&lt;/p&gt;&#xA;" OwnerUserId="5546" LastActivityDate="2011-02-04T17:47:14.897" />
  <row Id="20430" PostTypeId="1" CreationDate="2011-02-04T17:49:22.260" Score="1" ViewCount="145" Body="&lt;p&gt;What is the number of positive integers that, when 2004 is divided by them, leave a remainder of 24?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A.  036&#xA;B.  020&#xA;C.  022&#xA;D.  021&#xA;E.  014&lt;/p&gt;&#xA;" OwnerUserId="6625" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-04T20:16:20.293" LastActivityDate="2011-02-05T14:04:00.797" Title="The number of positive integers which divide 2004 to leave a remainder of 24" Tags="&lt;arithmetic&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="20431" PostTypeId="1" CreationDate="2011-02-04T17:49:44.927" Score="3" ViewCount="38" Body="&lt;p&gt;On a normal variety (possibly singular) one can make sense of the canonical class $K_X$ by computing it on the smooth locus and then extending over the singular part.  I'm wondering how to actually do this in practice: for example, on the singular cone $C$ in $\mathbb P^3$ given by $y^2-xz=0$, what is the canonical class?  What's a good example of a variety where $K_X$ is not Cartier (but preferably is $\mathbb Q$-Cartier) and I can still compute it easily?&lt;/p&gt;&#xA;" OwnerUserId="6626" LastActivityDate="2011-02-04T17:49:44.927" Title="Canonical class of a singular variety" Tags="&lt;algebraic-geometry&gt;" />
  <row Id="20432" PostTypeId="2" ParentId="20397" CreationDate="2011-02-04T17:51:09.110" Score="12" ViewCount="" Body="&lt;p&gt;As with Taylor's Theorem, the &lt;a href=&quot;http://en.wikipedia.org/wiki/Euler%E2%80%93Maclaurin_formula#Derivation_by_mathematical_induction&quot; rel=&quot;nofollow&quot;&gt;Euler-Maclaurin summation formula&lt;/a&gt; (with remainder) can be derived using repeated application of integration by parts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Tom Apostol's paper &quot;&lt;a href=&quot;http://www.myoops.org/twocw/nctu/upload/fourier/supplement/euler%20summation.pdf&quot; rel=&quot;nofollow&quot;&gt;An Elementary View of Euler's Summation Formula&lt;/a&gt;&quot; (&lt;em&gt;American Mathematical Monthly&lt;/em&gt; 106 (5): 409–418, 1999) has a more in-depth discussion of this.  See also Vito Lampret's &quot;&lt;a href=&quot;http://carma.newcastle.edu.au/~jb616/Preprints/Papers/Published-InPress/Euler-Boole/Related/euler-taylor.pdf&quot; rel=&quot;nofollow&quot;&gt;The Euler-Maclaurin and Taylor Formulas: Twin, Elementary Derivations&lt;/a&gt;&quot; (&lt;em&gt;Mathematics Magazine&lt;/em&gt; 74 (2): 109-122, 2001).&lt;/p&gt;&#xA;" OwnerUserId="2370" LastActivityDate="2011-02-04T17:51:09.110" CommunityOwnedDate="2011-02-04T17:51:09.110" />
  <row Id="20433" PostTypeId="2" ParentId="20421" CreationDate="2011-02-04T17:53:38.340" Score="1" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Bit of a spoiler:&lt;/strong&gt; Your approach seems on the way to what I've seen done, but instead of trying to intersect your set, you might want to map a non measurable one into it using a measurable map and remember how preimages of borel sets behave.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Spoiler:&lt;/strong&gt; your map could be one from the unit interval onto that very famous set by that very famous guy born in 1845 who suffered from depression and the dislike of many of his contemporaries... ;-)&lt;/p&gt;&#xA;" OwnerUserId="6280" LastActivityDate="2011-02-04T17:53:38.340" CommentCount="1" />
  <row Id="20434" PostTypeId="2" ParentId="20397" CreationDate="2011-02-04T17:58:10.633" Score="3" ViewCount="" Body="&lt;p&gt;Lowbrow: $\int\sin(x)\cos(x)dx=\sin^2x-\int\sin(x)\cos(x)dx+C$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finding the unknown integral again after integrating by parts is an interesting case. &#xA;Solving the resulting equation immediately gives the result&#xA;$\int\sin(x)\cos(x)dx=\dfrac12\sin^2x$&lt;/p&gt;&#xA;" OwnerUserId="6537" LastEditorUserId="6537" LastEditorDisplayName="" LastEditDate="2011-04-16T05:45:49.143" LastActivityDate="2011-04-16T05:45:49.143" CommunityOwnedDate="2011-02-04T17:58:10.633" CommentCount="4" />
  <row Id="20435" PostTypeId="2" ParentId="20299" CreationDate="2011-02-04T18:02:08.210" Score="1" ViewCount="" Body="&lt;p&gt;I lied when I said that this condition is not interesting. If $f, g$ are random variables and $I$ a probability space, then this is condition is satisfied if $f, g$ are independent. So that's an interesting source of examples.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2011-02-04T18:02:08.210" CommentCount="2" />
  <row Id="20436" PostTypeId="1" CreationDate="2011-02-04T18:07:09.150" Score="0" ViewCount="56" Body="&lt;p&gt;How can we make good estimates of the energies of arbitrary linear combinations from a collection of discrete signals, using only some compact measurements from the collection?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given a matrix $A$ made up of a set of real column vectors, $a_1, a_2, ..., a_n \in \mathbb{R}^m$, or a matrix $B$ of complex column vectors, $b_1, b_2, ..., b_n \in \mathbb{C}^m$, what are good ways to predict $\|Ax\|_2^2$ or $\|Bx\|_2^2$ for arbitrary $x \in \mathbb{R}^n$ storing only, say $k \ll m$ numbers derived from each vector.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;A simple approximation for $k=1$:&lt;br&gt;&#xA;Store the energies of the individual vectors $e_i = \|a_i\|_2^2, i=1...n$. Use $(x^2)^Te$ as your estimate, where $x^2$ is the x squared elementwise.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;One nice aspect of $1$, the estimate is a linear function of $x^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure why this would work, I guess it takes a middle ground in between the cases where components of the sum are completely opposite/out-of-phase (zero energy) and when components are in-phase and the same (4x energy).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a way we can improve this estimate by storing a few more numbers? For an estimate that is (or isn't) some kind of linear function?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What difference does real or complex make?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does it make a difference if $x$ are nonnegative or not?&lt;/p&gt;&#xA;" OwnerUserId="5610" LastEditorUserId="5610" LastEditorDisplayName="" LastEditDate="2011-02-07T17:40:00.270" LastActivityDate="2011-02-07T17:40:00.270" Title="estimating energy of vector sums from reduced information" Tags="&lt;statistics&gt;&lt;signal-processing&gt;&lt;nonlinear-optimization&gt;&lt;machine-learning&gt;" />
  <row Id="20437" PostTypeId="1" CreationDate="2011-02-04T18:10:18.913" Score="1" ViewCount="67" Body="&lt;p&gt;We showed in class that every strongly, exactly consistent s.c.f is strongly firm (I don't know if this is the right translation - we defined strong firmness as the equivalence of $*,\alpha,\beta$ effectiveness, which simply means that $\beta$ effectiveness $\Rightarrow$ * effectiveness).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was wondering - is the converse statement true? That is: is every strongly firm s.c.f also strongly and exactly consistent? I thought about this for a bit and it seems the answer should be negative, however I haven't been able to come up with a suitable example.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="5573" LastEditorUserId="766" LastEditorDisplayName="" LastEditDate="2011-02-23T18:46:15.590" LastActivityDate="2011-02-23T18:46:15.590" Title="Question on social choice functions" Tags="&lt;game-theory&gt;&lt;social-choice-theory&gt;" CommentCount="1" />
  <row Id="20438" PostTypeId="1" CreationDate="2011-02-04T18:13:45.740" Score="7" ViewCount="130" Body="&lt;p&gt;I am trying to prove the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $l$ be a prime and let $\zeta$ be a $l$th root of unity.  Show that,  in the order $\{ 1, \zeta, \ldots, \zeta^{l-2}  \}$ of the field $\mathbb{Q}(\zeta)$, if a product $\alpha \beta$ is divisible by $1-\zeta$, then $\alpha$ or $\beta$ must be divisible by $1-\zeta$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that $1-\zeta$ is irreducible in the maximal order $\mathbb{Z}[ \zeta ]$, and I am trying to mimic the proof of that statement, but I'm stuck.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can someone please help?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I see that all the products I am dealing with are of the form $\zeta^k$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also does the order contain $\zeta^{l-1}$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a feeling that &quot;the order ${ 1, \zeta, \ldots, \zeta^{l-2}  }$&quot; is actually just $\mathbb{Z}[ \zeta ]$.  Is this true?&lt;/p&gt;&#xA;" OwnerUserId="6597" LastEditorUserId="6597" LastEditorDisplayName="" LastEditDate="2011-02-04T20:23:26.137" LastActivityDate="2011-02-05T02:50:43.977" Title="How to show that $1-\zeta$ is prime in the order $\{ 1, \zeta, \ldots, \zeta^{l-2} \}$?" Tags="&lt;number-theory&gt;&lt;algebraic-number-theory&gt;" AnswerCount="1" CommentCount="6" FavoriteCount="0" />
  <row Id="20440" PostTypeId="2" ParentId="20357" CreationDate="2011-02-04T18:20:01.147" Score="0" ViewCount="" Body="&lt;p&gt;I don't know if this answers what you meant by &quot;why&quot;, but to my mind this is why the orientations are opposite: Given a point $P$ on the boundary, consider a neighbourhood $U$ of $P$ in $C$ which is homeomorphic to an open ball in $\mathbb{R}^n$, with the boundary mapped to a hyperplane bisecting the ball. An orientation of $C$ determines an orientation of this ball. For the ball, it is clear that its two halves induce opposite orientations on the hyperplane bisecting it. Then this has to be true also of the intersections of $M$, $B$ and $B&amp;#39;$ with $U$, and hence, since $P$ was arbitrary, of $M$, $B$ and $B&amp;#39;$ in general.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If this wasn't the sort of answer you were looking for, please elaborate.&lt;/p&gt;&#xA;" OwnerUserId="6622" LastActivityDate="2011-02-04T18:20:01.147" CommentCount="3" />
  <row Id="20441" PostTypeId="1" AcceptedAnswerId="20460" CreationDate="2011-02-04T18:22:15.903" Score="11" ViewCount="270" Body="&lt;p&gt;There are two identities that have a seemingly dual correspondence:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$e^x = \sum_{n\ge0} {x^n\over n!}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$n! = \int_0^{\infty} {x^n\over e^x}\ dx.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there anything to this comparison? (I vaguely remember a generating function/integration correspondence)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there similar sum/integration pairs for other well-known (or not-so-well-known) functions?&lt;/p&gt;&#xA;" OwnerUserId="1919" LastEditorUserId="1543" LastEditorDisplayName="" LastEditDate="2011-05-03T09:49:33.147" LastActivityDate="2011-05-03T09:49:33.147" Title="Factorial and exponential dual identities" Tags="&lt;analysis&gt;&lt;power-series&gt;&lt;generating-functions&gt;" AnswerCount="2" CommentCount="5" FavoriteCount="4" />
  <row Id="20442" PostTypeId="1" AcceptedAnswerId="20446" CreationDate="2011-02-04T18:27:49.560" Score="4" ViewCount="120" Body="&lt;p&gt;This is a question I like to bring up in a multivariable calculus class from time to time:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A sphere of radius 4 is dropped into a bowl-shaped paraboloid given by $z = x^2+y^2$.  How close will the sphere come to the vertex of the paraboloid?  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The problem can be solved using standard optimization techniques discussed in multivariable calculus, such as Lagrange multipliers or finding the zeros of the derivative(s) of the appropriate objective function.  It can also be easily reduced to a problem in 2D by exploiting symmetry or introducing polar coordinates, and even to a problem in one variable, but the obvious solution approaches still seem to require Lagrange multipliers or finding the zeros of the derivative(s) of the objective function.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I've wondered, though, if there is a way (particularly a clever one) to solve this problem that does not use calculus.  Has anyone seen such a solution, or can anyone think of one?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(I'll refrain from posting how close the sphere gets to the bottom of the bowl for those who would enjoy working on the problem for its own sake.)&lt;/p&gt;&#xA;" OwnerUserId="2370" LastActivityDate="2011-02-04T19:17:25.680" Title="Drop a ball into a bowl in the shape of a paraboloid. How far from the bottom of the bowl will the ball come to rest?" Tags="&lt;geometry&gt;&lt;analytic-geometry&gt;" AnswerCount="1" />
  <row Id="20443" PostTypeId="2" ParentId="20398" CreationDate="2011-02-04T18:29:52.053" Score="0" ViewCount="" Body="&lt;p&gt;Switzer sends the interested reader to Adams' &lt;em&gt;Lectures on generalised cohomology&lt;/em&gt; and &lt;em&gt;Stable homotopy and generalised homology&lt;/em&gt;.  I couldn't promise that these don't deal with spectra, but at least they'll be geared towards topology...  Why are you trying to avoid spectra, by the way?&lt;/p&gt;&#xA;" OwnerUserId="2159" LastActivityDate="2011-02-04T18:29:52.053" />
  <row Id="20444" PostTypeId="1" AcceptedAnswerId="20651" CreationDate="2011-02-04T18:45:34.440" Score="2" ViewCount="142" Body="&lt;p&gt;If I'm given a curve as an implicit function, is there some algorithm for obtaining a parametrization of the curve?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, how would I parameterize&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{2x}{y} - \frac{3y}{x} = 8$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know I might use $\cosh$ and $\sinh$, but I'm wondering if there's a surefire way to parameterize a curve, when you have a formula for it like this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can't isolate y, clearly, but you can re-arrange it to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ 2x^2 - 3y^2 - 8xy = 0 $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm hoping there's a way to find a trig rep for an arbitrary equation that I don't know about.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where do you go from here?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it just trial and error / recognizing the form and choosing trig functions?&lt;/p&gt;&#xA;" OwnerUserId="1069" LastEditorUserId="1069" LastEditorDisplayName="" LastEditDate="2011-02-05T03:38:14.063" LastActivityDate="2011-02-06T04:30:59.503" Title="Parameterizing a function where you cannot isolate y" Tags="&lt;algebra&gt;&lt;parameteric&gt;" AnswerCount="3" CommentCount="8" />
  <row Id="20445" PostTypeId="2" ParentId="20444" CreationDate="2011-02-04T18:52:10.863" Score="0" ViewCount="" Body="&lt;p&gt;Solve a quadratic equation for $y$.    &lt;/p&gt;&#xA;" OwnerUserId="589" LastActivityDate="2011-02-04T18:52:10.863" CommentCount="6" />
  <row Id="20446" PostTypeId="2" ParentId="20442" CreationDate="2011-02-04T19:01:11.937" Score="4" ViewCount="" Body="&lt;p&gt;As you say, this is really a 2-dimensional problem. We seek a point $(x,z)$ on the parabola $z=x^2$ such that the length of the normal from ($x,z)$ to the $z$-axis is $4$. The slope of the parabola is $2x$, so the slope of the normal is $-1/(2x)$. So the normal meets the $z$-axis at $(0,z+\frac{1}{2})$. The squared length of the line segment from $(x,z)$ to $(0,z+\frac{1}{2})$ is $x^2+\frac{1}{4}$, which is $16$ by hypothesis. Thus $x = \frac{3}{2}\sqrt 7$, and the height of the centre of the sphere is $z+\frac{1}{2} = x^2+\frac{1}{2} = \frac{65}{4}$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you don't need Lagrange multipliers. You just need the slope of a parabola.&lt;/p&gt;&#xA;" OwnerUserId="1508" LastEditorUserId="1508" LastEditorDisplayName="" LastEditDate="2011-02-04T19:17:25.680" LastActivityDate="2011-02-04T19:17:25.680" CommentCount="3" />
  <row Id="20447" PostTypeId="2" ParentId="213" CreationDate="2011-02-04T19:04:12.253" Score="0" ViewCount="" Body="&lt;p&gt;Integrals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My high school didn't cover them, so in 1Y uni, it was all new to me.&lt;/p&gt;&#xA;" OwnerUserId="1069" LastActivityDate="2011-02-04T19:04:12.253" CommunityOwnedDate="2011-02-04T19:04:12.253" />
  <row Id="20450" PostTypeId="1" CreationDate="2011-02-04T19:09:32.753" Score="2" ViewCount="108" Body="&lt;p&gt;I have a metric space with a probability measure (implying that the measure is regular). What conditions shouldIi ask of the measure or of the metric to get that&#xA;a function in $\mathcal{L}^2(d\mu)$ can be approximated in $\mathcal{L}^2$ by a uniformly continuous function?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="6630" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-04T21:55:45.370" LastActivityDate="2011-02-06T17:05:54.047" Title="approximation by continuous functions in a metric space" Tags="&lt;topology&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" />
  <row Id="20451" PostTypeId="1" AcceptedAnswerId="20466" CreationDate="2011-02-04T19:10:16.090" Score="2" ViewCount="72" Body="&lt;p&gt;Let $f(x)$ defined on $[0,a]$ with the following property.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$f(0) = b$, $f(x)\geq 0$&lt;/li&gt;&#xA;&lt;li&gt;$f(x)$ is a continuous and non-increasing function&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Let $g(x) = xf(x) + k(ab - \int_0^a f(x) dx)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does there exist a $m$, such that $\max(g(x)) \geq mab$ for every $f$? If so, how does $m$ related to $k$? I'm assuming $m = h(k)$ for some function $h$ independent of $a$ and $b$, but I might be wrong.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Graphically this means, given a function $f(x)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/WGcu2.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The area of $x f(x)$ and the area between $h(x)=b$ and $f(x)$ times $k$ is always larger than the area of the rectangle times $m$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know what area of mathematics deals with this kind of problem. Please retag if this is not analysis related. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$k=0$ then $m=0$ is the only thing I know. I have problem even proving the case when $k = 1$. I conjecture it's $m=3/4$ when $k=1$.  &lt;/p&gt;&#xA;" OwnerUserId="96" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-04T20:28:19.080" LastActivityDate="2011-02-04T21:32:03.410" Title="The set of maximum value of a set of function is larger than some number" Tags="&lt;analysis&gt;" AnswerCount="1" />
  <row Id="20452" PostTypeId="1" AcceptedAnswerId="20455" CreationDate="2011-02-04T19:15:01.047" Score="2" ViewCount="246" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Possible Duplicate:&lt;/strong&gt;&lt;br&gt;&#xA;  &lt;a href=&quot;http://math.stackexchange.com/questions/15524/riddle-simple-arithmetic-problem-illusion&quot;&gt;Riddle (simple arithmetic problem/illusion)&lt;/a&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;p&gt;&#xA;Three men go to a shop and take some food which is cost $25. Waiter came and gave them bill. &lt;pre&gt;Each of them gave $10.&lt;/pre&gt; So, waiter take total &lt;pre&gt;$10+$10+$10=$30.&lt;/pre&gt; Waiter came back with &lt;pre&gt;$5&lt;/pre&gt;. Three men took &lt;pre&gt;$1+$1+$1&lt;/pre&gt; and they decide to gave rest $2 to waiter. Now if we calculate the above thing then it should look like this:&#xA;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&#xA;Man A total spend $10 - $1  = $9&#xA;Man B total spend $10 - $1  = $9&#xA;Man C total spend $10 - $1  = $9&#xA;Three men Total = $27&#xA;and waiter take =    $2&#xA;So , three men total + waiter = $27 + $2 = $29 !!!! &#xA;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Where is another Dollar gone! Can anybody please explain it for me please?&lt;/p&gt;&#xA;" OwnerUserId="6629" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-04T20:03:26.340" LastActivityDate="2011-02-04T20:03:26.340" ClosedDate="2011-02-05T08:30:12.227" Title="Where is $1 gone!" Tags="&lt;puzzle&gt;" AnswerCount="2" CommentCount="6" />
  <row Id="20453" PostTypeId="1" CreationDate="2011-02-04T19:21:27.703" Score="1" ViewCount="54" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Possible Duplicate:&lt;/strong&gt;&lt;br&gt;&#xA;  &lt;a href=&quot;http://math.stackexchange.com/questions/20362/approximation-for-difference-of-two-sides-of-a-triangle&quot;&gt;Approximation For Difference Of Two Sides Of A Triangle&lt;/a&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;p&gt;I cannot figure out how this approximation was arrived at. Any help in figuring it out would be greatly appreciated. A drawing of the problem is &lt;a href=&quot;http://i.imgur.com/C5RFV.jpg&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; -- it would not let me post the image directly since I am not logged in. Essentially, it is just as an approximation for the law of cosines when one of the sides is much greater than another, viz., &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ b^2 - a^2 = c^2 - 2ac \cos(B) \Rightarrow b-a = { {-2ac \cos(B)} \over {b+a} } + { {c^2} \over {b+a} }$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$c &amp;lt;&amp;lt; a \Rightarrow b-a \approx -c \cos(B) + { {c^2} \over {2a} } \sin^2(B) $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If someone could show me how this approximation is derived, I would be most grateful.&lt;/p&gt;&#xA;" OwnerUserId="6631" LastActivityDate="2011-02-04T19:21:27.703" ClosedDate="2011-02-05T08:27:12.277" Title="Law of Cosines Approximation" Tags="&lt;geometry&gt;" CommentCount="0" />
  <row Id="20454" PostTypeId="2" ParentId="20452" CreationDate="2011-02-04T19:23:56.327" Score="1" ViewCount="" Body="&lt;p&gt;The $\$27$ includes the $\$25$ they paid for food, plus the $\$2$ to the waiter.  They pocketed the other three dollars.&lt;/p&gt;&#xA;" OwnerUserId="6491" LastActivityDate="2011-02-04T19:23:56.327" />
  <row Id="20455" PostTypeId="2" ParentId="20452" CreationDate="2011-02-04T19:24:56.453" Score="2" ViewCount="" Body="&lt;p&gt;The wrong numbers are added and subtracted. It should be:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The men pay 30&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The waiter gives 3 back, so the men paid 27 and the waiter got 27&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The food was 25, so the waiter can keep 2 dollars.&lt;/p&gt;&#xA;" OwnerUserId="6632" LastActivityDate="2011-02-04T19:24:56.453" CommentCount="3" />
  <row Id="20456" PostTypeId="2" ParentId="20430" CreationDate="2011-02-04T19:30:40.677" Score="5" ViewCount="" Body="&lt;p&gt;Let $a$ be the positive integer you're looking for. We know that &#xA;$$ 2004 \equiv 24 \mod{a}$$&#xA;So we know, that there is an integer, let's say $k$ for which&#xA;$$ k \cdot a + 24 = 2004$$&#xA;Subtracting 24 from both sides yields&lt;br&gt;&#xA;$$ 1980 \equiv 0 \mod{a}$$&#xA;From&#xA;$$1980 = 2^2 \cdot 3^2 \cdot 5 \cdot 11$$&#xA;we can tell that $1980$ has &#xA;$$3 \cdot 3 \cdot 2 \cdot 2  = 36 $$ &#xA;divisors, but we only need the ones that are greater than 24, otherwise there couldn't remain 24 when we divide 2004 with $a$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The divisors lesser than 25 are: 1, 2, 3, 4, 5, 6, 9, 10, 11, 12, 15, 18, 20,  and 22;  14 divisors total. &#xA;So the answer is $36-14 = 22$, that's option C.     &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(And here's the full list of solutions (I used brute force and ruby):&#xA;30,&#xA;33,&#xA;36,&#xA;44,&#xA;45,&#xA;55,&#xA;60,&#xA;66,&#xA;90,&#xA;99,&#xA;110,&#xA;132,&#xA;165,&#xA;180,&#xA;198,&#xA;220,&#xA;330,&#xA;396,&#xA;495,&#xA;660,&#xA;990,&#xA;1980.)&lt;/p&gt;&#xA;" OwnerUserId="4216" LastEditorUserId="4216" LastEditorDisplayName="" LastEditDate="2011-02-04T21:03:41.843" LastActivityDate="2011-02-04T21:03:41.843" CommentCount="2" />
  <row Id="20457" PostTypeId="1" AcceptedAnswerId="20459" CreationDate="2011-02-04T19:50:14.293" Score="2" ViewCount="119" Body="&lt;p&gt;Hi everyone,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A professor in my grad program gave me the impression that random graphs are an interesting and as of yet largely unexplored area of mathematics (feel free to correct this impression if I'm wrong). I've found the Erdos and Renyi papers from the late 50s/early 60s and nabbed a copy of &lt;a href=&quot;http://books.google.com/books?id=o9WecWgilzYC&amp;amp;dq=bollobas+random+graphs&amp;amp;source=gbs_navlinks_s&quot; rel=&quot;nofollow&quot;&gt;Random Graphs&lt;/a&gt; by Bollobas. I was wondering if anyone could recommend references/reading materials regarding the current state of research in this area, something perhaps more recent than these.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Update: I appreciate the responses and plan to read the information linked. To close out the question,I'm going to (semi-arbitrarily) pick the response with more references in it. Thanks everybody!&lt;/p&gt;&#xA;" OwnerUserId="4689" LastEditorUserId="4689" LastEditorDisplayName="" LastEditDate="2011-02-05T17:54:05.297" LastActivityDate="2011-02-05T17:54:05.297" Title="Reference request: random graphs" Tags="&lt;probability&gt;&lt;reference-request&gt;&lt;graph-theory&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="20458" PostTypeId="2" ParentId="20457" CreationDate="2011-02-04T19:56:20.880" Score="1" ViewCount="" Body="&lt;p&gt;Something like this?&#xA;&lt;a href=&quot;http://arxiv.org/PS_cache/math/pdf/0503/0503745v1.pdf&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/PS_cache/math/pdf/0503/0503745v1.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="6632" LastActivityDate="2011-02-04T19:56:20.880" CommentCount="1" />
  <row Id="20459" PostTypeId="2" ParentId="20457" CreationDate="2011-02-04T20:01:17.463" Score="2" ViewCount="" Body="&lt;p&gt;I haven't read either of these thoroughly, but they are recent &#xA;and look interesting. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.math.duke.edu/~rtd/RGD/RGD.html&quot; rel=&quot;nofollow&quot;&gt;Random Graph Dynamics&lt;/a&gt; by Rick Durrett&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.statslab.cam.ac.uk/~grg/books/pgs.html&quot; rel=&quot;nofollow&quot;&gt;Probability on Graphs&lt;/a&gt; by Geoffrey Grimmett&lt;/p&gt;&#xA;" OwnerUserId="940" LastEditorUserId="940" LastEditorDisplayName="" LastEditDate="2011-02-04T20:08:38.330" LastActivityDate="2011-02-04T20:08:38.330" CommunityOwnedDate="2011-02-04T20:01:17.463" CommentCount="1" />
  <row Id="20460" PostTypeId="2" ParentId="20441" CreationDate="2011-02-04T20:37:09.187" Score="6" ViewCount="" Body="&lt;p&gt;There is a close relationship between the two identities, but I don't know if the exact formal similarity is anything other than a neat coincidence along the lines of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Sophomore%27s_dream&quot; rel=&quot;nofollow&quot;&gt;Sophomore's dream&lt;/a&gt; (although I could of course be wrong about this). First note that the second identity can be written as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$1 = \int_{0}^{\infty} e^{-x} \frac{x^n}{n!} \, dx$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and therefore it is equivalent to the identity&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{1}{1 - t} = \sum_{n=0}^{\infty} t^n \int_0^{\infty} e^{-x} \frac{x^n}{n!} \, dx = \int_0^{\infty} e^{-x} e^{tx} \, dx.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is an application of the first identity. (This new identity is easy to prove, since the integrand is just $e^{(t-1)x}$ so it has antiderivative $\frac{1}{t-1} e^{(t-1)x}$ and the identity follows from here.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know of interesting explanations of the two identities separately which are somewhat related, but not another direct connection like the one above: for the first see &lt;a href=&quot;http://math.stackexchange.com/questions/3507/purely-combinatorial-proof-that-d-dxex-ex&quot;&gt;this math.SE question&lt;/a&gt; and for the second see &lt;a href=&quot;http://math.stackexchange.com/questions/3444/intuition-for-the-definition-of-the-gamma-function&quot;&gt;this math.SE question&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2011-02-04T21:21:30.843" LastActivityDate="2011-02-04T21:21:30.843" CommentCount="1" />
  <row Id="20461" PostTypeId="2" ParentId="20430" CreationDate="2011-02-04T20:49:06.230" Score="2" ViewCount="" Body="&lt;p&gt;Tyler, it has 36 divisors because a number of the form $n = p_1^{\alpha_1} \dots p_k^{\alpha_k}$ has&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;\prod_{i=1}^k (\alpha_i + 1)&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;divisors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(If I had the reputation I would leave this as a comment!)&lt;/p&gt;&#xA;" OwnerUserId="6491" LastActivityDate="2011-02-04T20:49:06.230" CommentCount="1" />
  <row Id="20462" PostTypeId="1" CreationDate="2011-02-04T20:53:14.120" Score="0" ViewCount="115" Body="&lt;p&gt;I realize most people work in &quot;convenient categories&quot; where this is not an issue.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In most topology books there is a proof of the fact that there is a natural homeomorphism of function spaces (with the compact-open topology): $$F(X\times Y,Z)\cong F(X,F(Y,Z))$$ when $X$ is Hausdorff and $Y$ is locally compact Hausdorff. There is also supposed to be a homeomorphism in the based case with the same conditions on $X$ and $Y$:&#xA;$$F_{\ast}(X\wedge Y,Z)\cong F_{\ast}(X,F_{\ast}(Y,Z))$$involving spaces of based maps and the smash product $\wedge$. This, for instance, is asserted on &lt;a href=&quot;http://ncatlab.org/nlab/show/exponential+law+for+spaces&quot; rel=&quot;nofollow&quot;&gt;n-lab&lt;/a&gt;. I checked the references listed on this page and many other texts but have not found a proof of this &quot;well-known fact.&quot; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems pretty clear if $X$ and $Y$ are compact Hausdorff (EDIT: in fact this is Theorem 6.2.38 of Maunder's &lt;em&gt;Algebraic Topology&lt;/em&gt;) but can this really be proven in this generality? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone provide a reference for a proof?&lt;/p&gt;&#xA;" OwnerUserId="2423" LastEditorUserId="2423" LastEditorDisplayName="" LastEditDate="2011-02-10T05:44:04.133" LastActivityDate="2011-05-11T07:56:32.703" Title="Exponential Law for based spaces" Tags="&lt;reference-request&gt;&lt;general-topology&gt;&lt;adjoint&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="20463" PostTypeId="1" AcceptedAnswerId="21238" CreationDate="2011-02-04T21:04:27.783" Score="0" ViewCount="94" Body="&lt;p&gt;This is a really basic question:&#xA;Say this is a question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A painted line is increased by length $x$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can $x$ be a negative number? just by the wording of the question. What would be a better way of rephrasing the question to make it clear that $x$ can be negative.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(The reason I'm asking is that I'm programming and a method name needs to indicate that the parameter can be negative or positive, in order to decrease or increase a size)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically I'm looking for a generic term for decrease and increase.&lt;/p&gt;&#xA;" OwnerUserId="2075" LastEditorUserId="2075" LastEditorDisplayName="" LastEditDate="2011-02-04T21:19:17.583" LastActivityDate="2011-02-09T19:58:16.373" Title="Terminology:Increase with negative number" Tags="&lt;terminology&gt;" AnswerCount="1" CommentCount="7" />
  <row Id="20466" PostTypeId="2" ParentId="20451" CreationDate="2011-02-04T21:30:01.937" Score="3" ViewCount="" Body="&lt;p&gt;First, let's work out the possible pairs $\phi(f) = ( \sup_x \{xf(x)\}, \int_a^b f(t)dt)$ when $f$ is such a function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose $xf(x)$ takes some value $z \leq ab$. Then the smallest possible value for the integral of $f$ is $z$ :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pick $x \leq a$ and $y \leq b$ such that $z = xy$. For any $\varepsilon &amp;gt; 0$, we can build a piecewise affine function $f$ by choosing $f(0) = b, f(\varepsilon) = f(x) = y, f(x+\varepsilon) = f(b) = 0$. Then, $\phi(f) = (z,z+b\varepsilon /2)$.&#xA;Thus we can build $f$ with $\phi(f)$ as close to $(z,z)$ as we want.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conversely, if $z = \sup_x xf(x)$, the highest possible value for the integral of $f$ is for $f(x) = \inf (b, z/x)$.&#xA;We can compute the integral, and find $z(1 - \ln(z/ab))$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, we know that the image of $\phi$ is $\{(z,t), 0 \leq z \leq t \leq  z(1 - \ln(z/ab)), z \leq ab\}$.&#xA;$$ \begin{array}{rcl} m &amp;amp; = &amp;amp; \frac{1}{ab} \inf_f (\sup_x g(x)))&#xD;&#xA; = k + \frac{1}{ab} \inf_f (\sup_x (xf(x)) - k\int_0^a f(t) dt) \\&#xD;&#xA;&amp;amp;=&amp;amp; k + \frac{1}{ab} \inf_{(z,t) \in \text{Im}\phi} (z-kt)\end{array}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $k\leq 0$, we get $m = k + \frac{1}{ab} \inf_z z(1-k) = k$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $k\geq 0$, we get $m = k + \inf_z \frac{z}{ab} (1-k(1- \ln(z/ab)))&#xD;&#xA; = k + \inf_{u \in [0,1]} u(1-k(1- \ln(u)))$.&#xA;The minimum occurs for $z/ab = u = e^{-1/k}$, which leads to $m = k(1-e^{-1/k})$.&lt;/p&gt;&#xA;" OwnerUserId="6270" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-04T21:32:03.410" LastActivityDate="2011-02-04T21:32:03.410" />
  <row Id="20467" PostTypeId="2" ParentId="213" CreationDate="2011-02-04T21:36:15.533" Score="6" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0521784514&quot; rel=&quot;nofollow&quot;&gt;Lattices and order theory&lt;/a&gt;. While these concepts are so ubiquitous, they seem to be banned from mathematics courses. Also, if you know something about order theory, many concepts from category theory turn out to be quite familiar. (E.g. view a poset as a category, then product resp. coproduct become infimum resp. supremum, the slice and coslice category are up and down set, etc.)&lt;/p&gt;&#xA;" OwnerUserId="6490" LastActivityDate="2011-02-04T21:36:15.533" CommunityOwnedDate="2011-02-04T21:36:15.533" />
  <row Id="20468" PostTypeId="1" AcceptedAnswerId="20515" CreationDate="2011-02-04T21:41:22.670" Score="3" ViewCount="324" Body="&lt;p&gt;I keep tossing a fair coin until I see HTHT appears. What is the probability that I have already observed the sequence TTH before I stop?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Okay. This problem is the same as before. I am trying to think of the following. Could anyone please tell me if this is also the same as before, and if not, how to solve it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Two players A and B toss two different fair coins. A tosses his coin until he sees HTHT appearing for the first time, and records the number of tosses, $X$. B tosses his coin until he sees TTH appearing for the first time, and records the number of tosses made, $Y$. What is the probability that $Y&amp;lt;X$?&lt;/p&gt;&#xA;" OwnerUserId="6016" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-05T03:31:04.983" LastActivityDate="2011-02-05T21:55:59.383" Title="a coin toss probability question" Tags="&lt;probability&gt;" AnswerCount="2" CommentCount="13" />
  <row Id="20469" PostTypeId="2" ParentId="20397" CreationDate="2011-02-04T21:44:40.917" Score="10" ViewCount="" Body="&lt;p&gt;Perhaps not really an application, but the definition of the derivative of a distribution is based on partial integration:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;if $u\in C^1(X)$ and $\phi\in C^\infty_c(X)$ is a test function, then &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\left&amp;lt;\partial_i u,\phi\right&amp;gt;=\int\phi\partial_i u=-\int u\partial_i\phi=-\left&amp;lt;u,\partial_i\phi\right&amp;gt;$ by partial integration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Extending this, for a distribution $u$ we then define its derivative $\partial_i u$ by this formula.&lt;/p&gt;&#xA;" OwnerUserId="6490" LastActivityDate="2011-02-04T21:44:40.917" CommunityOwnedDate="2011-02-04T21:44:40.917" CommentCount="1" />
  <row Id="20470" PostTypeId="2" ParentId="20468" CreationDate="2011-02-04T22:25:21.133" Score="1" ViewCount="" Body="&lt;p&gt;&lt;em&gt;This answer addresses the question as originally posted; I made it a community wiki because it is really an extended comment that would not have fit as a comment.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The previous question asked, &lt;em&gt;inter alia&lt;/em&gt;, the following: given a two-sided coin with probability $p$ of showing heads, and the coin is tossed until either HTHTH or HTHH appears, how to compute the probability that the pattern we got was HTHTH.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://math.stackexchange.com/questions/18405/a-question-related-to-two-competing-patterns-in-coin-tossing/18503#18503&quot;&gt;Mike Spivey's answer&lt;/a&gt; cast the problem as a two-player game, in which one player wins if the first pattern to show up is HTHTH, and the second player wins if the pattern that shows up first is HTHH, and showed how to obtain the probability that either the first player wins, or the second player wins.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The current problem has a fair coin, and asks that the coin be tossed until a particular pattern, HTHT, shows up, and then asks what is the probability that another pattern, TTH, actually occurs before HTHT shows up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Think of this problem as a two-player game again: the coin is tossed, until either HTHT or TTH show up. The first player wins if HTHT shows up first, the second if TTH shows up first. The exact same approach as used in the previous question works here to obtain the probability that the second player wins, i.e., the probability that TTH will show up first.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, you state in comments that you don't see &lt;em&gt;this&lt;/em&gt; problem as the same as that game, because in this problem you keep tossing until HTHT appears, instead of stopping if TTH shows up. Well, take the game described above, and then imagine that the two players are utterly bored and so, if player two wins, they keep tossing the coin until HTHT shows up. Will that change the probability that the second player wins? No. The second player still wins if TTH shows up first, regardless of whether you keep tossing the coin or not after he wins. So the probability that TTH showed up at least once before you hit HTHT is the same, whether you keep tossing after TTH shows up or not, because the number of sequences in which TTH shows up but HTHT never does is negligible, so it will not affect the probability. (The probability that a particular finite pattern &lt;em&gt;never&lt;/em&gt; shows up is $0$).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in the end, the answer to your question is &lt;em&gt;still&lt;/em&gt; &quot;the probability that TTH showed up before HTHT shows up is equal to the probability that Player Two wins the game&quot;. &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-05T04:30:28.980" LastActivityDate="2011-02-05T04:30:28.980" CommunityOwnedDate="2011-02-04T22:25:21.133" CommentCount="2" />
  <row Id="20471" PostTypeId="1" CreationDate="2011-02-04T22:33:16.983" Score="9" ViewCount="234" Body="&lt;p&gt;Part of the reason I think algebraic topology has acquired something of a fearsome reputation is that the terrible properties of the topological category (e.g. the existence of space-filling curves) force us to work very hard to prove the main theorems setting up all of the big machinery to get the payoff we want (e.g. invariance of domain, fixed point theorems). But why should I care about these arbitrary and terrible spaces and functions in the first place when, as far as I can tell, any manifold which occurs in applications is at least piecewise-differentiable and any morphism which occurs in applications is at least homotopic to a piecewise-differentiable one? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, do topological manifolds really naturally occur in the rest of mathematics (without some extra structure)?&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2011-02-04T23:22:41.723" Title="What's the point of studying topological (as opposed to smooth, PL, or PDiff) manifolds?" Tags="&lt;topology&gt;&lt;manifold&gt;" AnswerCount="3" CommentCount="4" FavoriteCount="1" />
  <row Id="20472" PostTypeId="2" ParentId="20471" CreationDate="2011-02-04T23:02:33.393" Score="7" ViewCount="" Body="&lt;p&gt;As far as I know, historically smooth manifolds were the first manifolds studied by people like Poincare, Riemann, up to Whitney.   There were a few major events that caused people to take things like topological and PL manifolds seriously, but originally people were not motivated to study these kinds of objects.  Here are some of the big events/ideas that come to mind:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Poincare's original proof of Poincare duality was a proof for triangulated manifolds.  That smooth manifolds had triangulations (and whether or not they were essentially unique) was a problem that took some time to solve.  So the study of triangulations and PL manifolds picked up. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Smale's proof of the h-cobordism theorem, although written up for the smooth category when you look at it carefully there's a lot of &quot;smoothing the corners&quot; going on.  You can think carefully about it and determine all the smoothing of the corners does not kill the proof but I know many strong mathematicians that were hesitant to accept Smale's proof, insisting that it was only a PL-category proof.  FYI, the smoothing of the corners issue has been settled, there's a very nice write-up in Kosinski's manifolds text.  But this was another issue that kept people thinking about the PL category. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) If anything, topological manifolds play a role simply for comparison sake -- after all the forgetful functor from the smooth to the topological category is an interesting functor.  Perhaps for different people in different ways.  I've yet to be interested by a topological manifold that admits no smooth structure but I do find multiple smooth structures on the same topological manifold interesting.  Is this purely psychological?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;4) Topological and PL-manifold theory is where some &quot;nasty&quot; constructions work, like the Alexander trick.  There are different versions of it, one being that the restriction map $Aut(D^n) \to Aut(S^{n-1})$ admits a section in the topological or PL categories.  It does not in the smooth category.  If anything, I find these kinds of facts informative on the smooth category.  The smooth category is interesting largely because of facts like these.  There's a similar Alexander trick for knots, for example, the space of topological or PL embeddings $\mathbb R^j \to \mathbb R^n$ which restrict to the standard inclusion $x \longmapsto (x,0)$ outside of the unit ball, this space is contractible, by &quot;pulling the knot tight&quot;.  But in the smooth category, this spage isn't contractible. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think one of the major events in the development of this subject is simply pragmatic.  To get smooth manifold theory off the ground you need Sard's theorem and transversality.  This requires analysis to the level of measure theory, and a solid multi-variable calculus background, which in many undergraduate educations is skimped on (especially since it comes early, and many curriculums are too service-based to teach calculus &quot;well&quot;).  PL manifolds are inherently more combinatorial and so the learning curve for people with weak analysis backgrounds is easier to deal with.  I think also some people really appreciate the combinatorial nature of the subject.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyhow, those are some thoughts off the top of my head. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Getting to your conversation with Mariano:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PL manifold theory by-and-large isn't terribly different from smooth manifold theory.  So I think once you learn one, adapting to the other isn't so hard. But topological manifolds are really quite different.  This may be my ignorance speaking to some extent, but I'm still working my way through Kirby-Siebenmann.  I'm told various people are working on re-writing the main theorems of that text, to make it easier reading for people that are not named Larry Siebenmann.  But we're probably still several years from that.  I suspect in the next 10 years there should be several different accounts of most of that material.  But I'm still some ways from understanding smoothing theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Manifolds when they come up &quot;in nature&quot; like in physics or engineering applications tend to always be smooth, and usually with plenty of extra structure.  Sometimes the objects that come up aren't manifolds, but algebraic varieties, or even more degenerate (but smooth) stratified spaces.   &lt;/p&gt;&#xA;" OwnerUserId="642" LastEditorUserId="642" LastEditorDisplayName="" LastEditDate="2011-02-04T23:20:20.370" LastActivityDate="2011-02-04T23:20:20.370" CommentCount="4" />
  <row Id="20473" PostTypeId="2" ParentId="7376" CreationDate="2011-02-04T23:12:14.703" Score="1" ViewCount="" Body="&lt;p&gt;This is from a set of class notes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You need a lemma (interior and closure operators are $\mathrm{int},\mathrm{cl}$, resp.):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Lemma&lt;/strong&gt;. If $C$ a convex subset of a topological space, and if $x\in \mathrm{int} C$ and $y\in \mathrm{cl} C$, then $[x,y)\subset \mathrm{int} C$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The half-open segment $[x,y)$ is a right-open convex combination. If $\mathrm{int}C$ is empty, then the above lemma is true. Try to prove why this is true when the interior is not empty.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proposition&lt;/strong&gt;. $\mathrm{int} C = \mathrm{int}\;\mathrm{cl}\;C$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $y\in \mathrm{int\; cl\;} C$; position it in an open ball, noting that the in the interior of the closure of $C$ is open, and hence $\exists r&amp;gt;0$ such that $B_r(y)\subset \mathrm{cl\;} C$. Pick a $y&amp;#39;\in \mathrm{int}\;C$. Then there exists an $\epsilon&amp;gt;0$, such that $y&amp;#39;+(1+\epsilon)(y-y&amp;#39;) = y+\epsilon(y-y&amp;#39;)\in B_r(y)\subset \mathrm{cl}\;C$. At the same time, $y$ belongs to the segment $[y&amp;#39;,y+\epsilon(y-y&amp;#39;))$, so that, by the previous lemma, $y$ belongs to $\mathrm{int\;}C$. The reverse inclusion is straightforward, since $C\subseteq \mathrm{cl\;}C$, and thus obtaining the result.&lt;/p&gt;&#xA;" OwnerUserId="6637" LastActivityDate="2011-02-04T23:12:14.703" />
  <row Id="20474" PostTypeId="2" ParentId="20471" CreationDate="2011-02-04T23:15:48.120" Score="2" ViewCount="" Body="&lt;p&gt;Topological manifolds arise naturally as the background of other phenomena.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, nature throws at you examples of spaces which admit several smooth structures and when you try to describe that phenomenon you need to say something like «there are many different $Y$s one can put on an $X$». In the situation of exotic smooth structures, a natural class one can use for the role of $X$s is that of topological manifolds. &lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2011-02-04T23:15:48.120" CommentCount="7" />
  <row Id="20475" PostTypeId="1" CreationDate="2011-02-04T23:19:23.417" Score="2" ViewCount="262" Body="&lt;p&gt;I'm trying to prove that ${n \choose r}$ is equal to ${{n-1} \choose {r-1}}+{{n-1} \choose r}$ when $1\leq r\leq n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suppose I could use the counting rules in probability, perhaps combination= ${{n} \choose {r}}=\frac{n!}{r!(n-r!)}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to see an actual proof behind this equation.  Does anyone have any ideas?&lt;/p&gt;&#xA;" OwnerUserId="6639" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2011-02-05T05:54:09.543" LastActivityDate="2011-02-05T15:37:39.577" Title="Proving ${{n} \choose {r}}={{n-1} \choose {r-1}}+{{n-1} \choose r}$ when $1\leq r\leq n$" Tags="&lt;probability&gt;&lt;combinatorics&gt;&lt;binomial-coefficients&gt;" AnswerCount="5" FavoriteCount="1" />
  <row Id="20476" PostTypeId="2" ParentId="20471" CreationDate="2011-02-04T23:22:41.723" Score="4" ViewCount="" Body="&lt;p&gt;The topological category is inherently beautiful. Some very pretty and subtle phenomena happen. My favorite example of this is that some knots bound locally flat (that is they admit a locally trivial normal bundle) topological disks into the 4-ball, but they don't bound smooth disks. That is, there are topologically slice knots which are not smoothly slice. The topological disks themselves are practically impossible to visualize, coming from a limiting construction due to very deep work of Mike Freedman, and are a little bit fractal in nature. The world would be a poorer place if we never knew about this amazing structure.&lt;/p&gt;&#xA;" OwnerUserId="2941" LastActivityDate="2011-02-04T23:22:41.723" CommentCount="8" />
  <row Id="20477" PostTypeId="2" ParentId="20475" CreationDate="2011-02-04T23:25:34.887" Score="2" ViewCount="" Body="&lt;p&gt;See this Wikipedia page:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Binomial_coefficient&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Binomial_coefficient&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Under the subsection &lt;strong&gt;Recursion formula&lt;/strong&gt;, &lt;em&gt;HINT&lt;/em&gt; for proving this formula is given. Hope you can do it from there.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The formula follows either from tracing the contributions to $X^{k}$ in $(1 + X)^{n−1}(1 + X)$, or by counting k-combinations of {1, 2, ..., n} that contain n and that do not contain n separately&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" LastActivityDate="2011-02-04T23:25:34.887" />
  <row Id="20478" PostTypeId="2" ParentId="20475" CreationDate="2011-02-04T23:27:40.640" Score="5" ViewCount="" Body="&lt;p&gt;Consider $n$ balls in a basket. Let there be $1$ red ball and $n-1$ blue balls. Now look at the number of ways of choosing $r$ balls in two different ways&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One way is to choose $r$ balls out of the $n$ balls. So the number of ways is $C(n,r)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other way is to look at the cases when out of the $r$ balls chosen if we have a red ball or not. We have only two options namely out of the $r$ balls we could have one red ball or no red balls&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The number of ways of having $1$ red ball is to choose the one red ball which can be done in $C(1,1)$ ways and choose the remaining $(r-1)$ balls from the $(n-1)$ blue balls which can be done in $C(n-1,r-1)$ ways&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, the number of ways of having no red balls is to choose all the balls as blue balls which can be done in $C(n-1,r)$ ways&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These are the only two cases and these are mutually exclusive and hence the total number of ways is $C(n-1,r-1)+C(n-1,r)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, we get $$C(n,r) = C(n-1,r-1) + C(n-1,r)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same idea could be extended to prove a generalization of the above $$C(m+n,r) = \displaystyle \sum_{k=\max(0,r-n)}^{\min(r,m)} C(m,k) C(n,r-k)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider a basket with $m$ red balls and $n$ blue balls and we want to count the number of ways in which $r$ balls can be drawn. Argue by two different ways to count (same as above) to prove this.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2011-02-05T00:06:50.693" LastActivityDate="2011-02-05T00:06:50.693" CommentCount="2" />
  <row Id="20479" PostTypeId="2" ParentId="20397" CreationDate="2011-02-04T23:29:53.677" Score="8" ViewCount="" Body="&lt;p&gt;My favorite example is getting an &lt;em&gt;asymptotic&lt;/em&gt; expansion: for example, suppose we want to compute $\int_x^\infty e^{-t^2}\cos(\beta t)dt$ for large values of $x$.  Integrating by parts multiple times we end up with&#xA;$$ \int_x^\infty e^{-t^2}\cos(\beta t)dt&#xD;&#xA;   \sim e^{-x^2}\sum_{k=1}^\infty(-1)^n\frac{H_{k-1}(x)}{\beta^k}&#xD;&#xA;   \begin{cases}&#xD;&#xA;     \cos(\beta x) &amp;amp; k=2n \\&#xD;&#xA;     \sin(\beta x) &amp;amp; k=2n+1&#xD;&#xA;   \end{cases}$$&#xA;where the Hermite polynomials are given by&#xA;$H_n(x) = (-1)^ne^{x^2}\frac{d^n}{dx^n}e^{-x^2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This expansion follows mechanically applying IBP multiple times and gives a nice asymptotic expansion (which is divergent as a power series).&lt;/p&gt;&#xA;" OwnerUserId="3852" LastEditorUserId="434" LastEditorDisplayName="" LastEditDate="2011-02-05T15:11:26.360" LastActivityDate="2011-02-05T15:11:26.360" CommunityOwnedDate="2011-02-04T23:29:53.677" />
  <row Id="20480" PostTypeId="2" ParentId="20475" CreationDate="2011-02-04T23:45:29.067" Score="2" ViewCount="" Body="&lt;p&gt;As Sivaram and Chandru1 suggested, a &lt;strong&gt;combinatorial argument&lt;/strong&gt; is often a very good way to understand/prove that kind of identities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other way would be, as you said, to use the explicit formula for the Binomial coefficient:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$${{n-1} \choose {r-1}}+{{n-1} \choose r}=\frac{(n-1)!}{(n-k)!(k-1)!}+\frac{(n-1)!}{(n-k-1)!k!}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which reduces to $\frac{n!}{(n-k)!k!}={n\choose k}$.&lt;/p&gt;&#xA;" OwnerUserId="524" LastEditorUserId="524" LastEditorDisplayName="" LastEditDate="2011-02-05T00:00:04.640" LastActivityDate="2011-02-05T00:00:04.640" />
  <row Id="20481" PostTypeId="2" ParentId="20397" CreationDate="2011-02-05T00:03:14.300" Score="10" ViewCount="" Body="&lt;p&gt;My favorite example of integration by parts (there are other nice tricks as well in this example but integration by parts starts it off) is this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $I_n = \displaystyle \int_{0}^{\frac{\pi}{2}} \sin^n(x) dx$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$I_n = \displaystyle \int_{0}^{\frac{\pi}{2}} \sin^{n-1}(x) d(-\cos(x)) = -\sin^{n-1}(x) \cos(x) |_{0}^{\frac{\pi}{2}} + \int_{0}^{\frac{\pi}{2}} (n-1) \sin^{n-2}(x) \cos^2(x) dx$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first expression on the right hand side is zero since $\sin(0) = 0$ and $\cos(\frac{\pi}{2}) = 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now rewrite $\cos^2(x) = 1 - \sin^2(x)$ to get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$I_n = (n-1) (\displaystyle \int_{0}^{\frac{\pi}{2}} \sin^{n-2}(x) dx - \int_{0}^{\frac{\pi}{2}} \sin^{n}(x) dx) = (n-1) I_{n-2} - (n-1) I_n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rearranging we get  $n I_n = (n-1) I_{n-2}$, $I_n = \frac{n-1}{n}I_{n-2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using this recurrence we get&#xA;$$I_{2k+1} = \frac{2k}{2k+1}\frac{2k-2}{2k-1} \cdots \frac{2}{3} I_1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$I_{2k} = \frac{2k-1}{2k}\frac{2k-3}{2k-2} \cdots \frac{1}{2} I_0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$I_1$ and $I_0$ can be directly evaluated to be $1$ and $\frac{\pi}{2}$ respectively and hence,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$I_{2k+1} = \frac{2k}{2k+1}\frac{2k-2}{2k-1} \cdots \frac{2}{3}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$I_{2k} = \frac{2k-1}{2k}\frac{2k-3}{2k-2} \cdots \frac{1}{2} \frac{\pi}{2}$$&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2011-06-01T21:55:57.530" LastActivityDate="2011-06-01T21:55:57.530" CommunityOwnedDate="2011-02-05T00:03:14.300" CommentCount="2" />
  <row Id="20482" PostTypeId="2" ParentId="213" CreationDate="2011-02-05T00:09:46.027" Score="0" ViewCount="" Body="&lt;p&gt;Non-linear Dynamics and Chaos!&lt;/p&gt;&#xA;" OwnerUserId="3302" LastActivityDate="2011-02-05T00:09:46.027" CommunityOwnedDate="2011-02-05T00:09:46.027" />
  <row Id="20483" PostTypeId="1" AcceptedAnswerId="20486" CreationDate="2011-02-05T00:15:51.870" Score="2" ViewCount="75" Body="&lt;p&gt;It's possible that this question is trivial and I've overlooked something.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let us impose the following constraint on the well-known &lt;a href=&quot;http://en.wikipedia.org/wiki/Partition_problem&quot; rel=&quot;nofollow&quot;&gt;PARTITION&lt;/a&gt; problem:&#xA;in all inputs of a new problem the number of elements is fixed and equals $k$?&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Does such constraint make PARTITION polynomially solvable?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;P.S. There is pseudopolynomial algorithm for it with running time $\mathcal{O}(nA)$, where $A$ is the sum of the numbers in the given input. However, fixing $n$&#xA;doesn't seem to make it polynomial in $\log A$. Or am I missing something?&lt;/p&gt;&#xA;" OwnerUserId="1856" LastActivityDate="2011-02-05T00:59:17.480" Title="Does fixing the number of elements in PARTITION send it in P?" Tags="&lt;computer-science&gt;" AnswerCount="1" />
  <row Id="20484" PostTypeId="2" ParentId="20450" CreationDate="2011-02-05T00:25:42.400" Score="1" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Hint:&lt;/strong&gt; Suppose $f$ is orthogonal to all bounded, uniformly continuous functions. Show that $\int_F f(x)\mu(dx)=0$ for all closed sets $F$. Deduce that $\int_B f(x)\mu(dx)=0$ for all Borel sets $B$. &lt;/p&gt;&#xA;" OwnerUserId="940" LastActivityDate="2011-02-05T00:25:42.400" />
  <row Id="20485" PostTypeId="2" ParentId="20444" CreationDate="2011-02-05T00:34:55.313" Score="0" ViewCount="" Body="&lt;p&gt;You can define $z=\frac{y}{x}$ and get $2z^{-1}-3z=8$, $3z^2+8z-2=0$, $z=\frac{-8 \pm \sqrt{88}}{6}$.  Plugging them in, both solutions check.  So you can have $x=t, y=\frac{-8 \pm \sqrt{88}}{6}t$ for $t \ne 0$.  Why did you expect trigonometry?&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-02-05T00:34:55.313" />
  <row Id="20486" PostTypeId="2" ParentId="20483" CreationDate="2011-02-05T00:59:17.480" Score="2" ViewCount="" Body="&lt;p&gt;Yes. The problem becomes in $P$. The reason is that there would be a fixed number of partitions bounded by $O(2^k)$ which is a constant. You do not need the $O(nA)$ dynamic programming algorithm.&lt;/p&gt;&#xA;" OwnerUserId="2459" LastActivityDate="2011-02-05T00:59:17.480" CommentCount="1" />
  <row Id="20487" PostTypeId="1" AcceptedAnswerId="20490" CreationDate="2011-02-05T01:31:07.473" Score="8" ViewCount="272" Body="&lt;p&gt;Here is a problem which i have not been able to do for quite sometime.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;For $n \geq 2$, show that $n \nmid 2^{n}-1$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I have thought of proving this in two ways: One by using induction which didn't actually work. Next by Fermat's little theorem we have $2^{p-1} \equiv 1 \ (\text{mod} \ p)$, which actually says that for $p \mid 2^{p}-2$. But i couldn't proceed more than this. Any ideas by which i can actually solve the problem?&lt;/p&gt;&#xA;" OwnerUserId="9413" LastEditorUserId="9413" LastEditorDisplayName="" LastActivityDate="2011-02-13T15:45:48.983" Title="For $n \geq 2$, show that $n \nmid 2^{n}-1$." Tags="&lt;number-theory&gt;&lt;elementary-number-theory&gt;" AnswerCount="2" />
  <row Id="20488" PostTypeId="1" CreationDate="2011-02-05T01:49:57.100" Score="2" ViewCount="135" Body="&lt;p&gt;I just read a very short paper on the De Bruijn-Erdos theorem. It essentially states that if we have $n$ points and $m$ lines, where any two points lie on a line, and each line has at least two points, then there are at least as many lines as points.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The proof is in this &lt;a href=&quot;http://www.renyi.hu/~p_erdos/1948-01.pdf&quot; rel=&quot;nofollow&quot;&gt;short paper&lt;/a&gt;. It is stated at the very top, and the proof I'm curious about begins in the middle of page 2 and ends at the top of page 3. I don't quite follow the inequalities &#xA;$$&#xD;&#xA;s_j\leq k_n\ \text{for}\ j&amp;gt;\nu&#xD;&#xA;$$&#xA;that are found at the very top of page 3. Why would that fact that an index $j$ being greater than the minimum number of lines passing through a given point require that the number of points on that indexed line is at most the number of lines passing through that point? Apart from that I follow the rest. Does it violate one of the assumptions about the space? Thanks.&lt;/p&gt;&#xA;" OwnerUserId="6644" LastEditorUserId="1583" LastEditorDisplayName="" LastEditDate="2011-02-05T02:49:50.047" LastActivityDate="2011-02-05T02:54:17.440" Title="A question on the proof of the De Bruijn-Erdos theorem" Tags="&lt;geometry&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="20489" PostTypeId="2" ParentId="20487" CreationDate="2011-02-05T01:59:14.510" Score="5" ViewCount="" Body="&lt;p&gt;You can look at this modulo the smallest prime factor $p$ of $n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Write $n=p^rm$ where $m$ is a product of primes greater than $p$. Then $m$ will be coprime to $p-1$, so $ma=1$ (mod $p-1$) for some positive integer $a$. Now use Fermat's little theorem&#xA;$$&#xD;&#xA;(2^n)^a=2^{p^rma}=2^{p^r}=2\not=1\qquad{\rm(mod\ }p{)}.&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[Edit: Even better - $n$ will be coprime to $p-1$, as Bill notes in his answer, so there is no need to extract out the $p^r$ factor.]&lt;/p&gt;&#xA;" OwnerUserId="1321" LastEditorUserId="1321" LastEditorDisplayName="" LastEditDate="2011-02-05T02:25:28.540" LastActivityDate="2011-02-05T02:25:28.540" />
  <row Id="20490" PostTypeId="2" ParentId="20487" CreationDate="2011-02-05T02:20:04.580" Score="6" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\ \ $ Prime $\rm\ P|N\ \Rightarrow\ 2^{P-1}\equiv 1\equiv 2^N\ \Rightarrow\ 2^{GCD(P-1,N)} \equiv 1\ \ (mod\ P)\ $   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if $\rm\:P\:$ is the &lt;em&gt;least&lt;/em&gt; prime factor of $\rm\:N\:$ then $\rm\ GCD(P-1,N) = 1\:.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I.e. $\rm\ mod\ P\ $ the order of $\rm\:2\:$ is $\:1\:$ since it divides the coprime integers $\rm\:P-1\:$ and $\rm\:N\:.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;NOTE&lt;/strong&gt; $\ $ This problem was posted to &lt;a href=&quot;http://groups.google.com/group/sci.math/browse_frm/thread/494db4354fe1ae19/cdb16b082acb9559&quot; rel=&quot;nofollow&quot;&gt;sci.math on 2009\11\03&lt;/a&gt;. $\ $ There I remarked that the proof shows that if $\rm\ A^N = 1\ $ then the order of $\rm\:A\:$ is $\ge$ the least prime factor $\rm\:P\:$ of $\rm\:N\:$. In particular this implies that $\rm\ A^{P-1}\ne 1\:,\ $ which settles the problem at hand.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2011-02-05T06:56:21.083" LastActivityDate="2011-02-05T06:56:21.083" CommentCount="4" />
  <row Id="20491" PostTypeId="2" ParentId="20462" CreationDate="2011-02-05T02:23:50.867" Score="0" ViewCount="" Body="&lt;p&gt;I'm pretty sure that this result is proved in the book of &lt;a href=&quot;http://books.google.es/books?id=YkyizIcJdK0C&amp;amp;printsec=frontcover&amp;amp;dq=inauthor%3a%22C.+R.+F.+Maunder%22&amp;amp;hl=es&amp;amp;ei=HLJMTfHYFYmZOsK63AQ&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=1&amp;amp;ved=0CCkQ6AEwAA#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;Maunder&lt;/a&gt; and it must be somewhere between proposition 6.2.37 and theorem 6.3.4, pages 221-222, that unfortunately Google doesn't show right now.&lt;/p&gt;&#xA;" OwnerUserId="664" LastActivityDate="2011-02-05T02:23:50.867" CommentCount="1" />
  <row Id="20492" PostTypeId="1" CreationDate="2011-02-05T02:40:02.947" Score="8" ViewCount="356" Body="&lt;p&gt;So today I was challenged by my Uncle because he says that we are going to run out of drinkable water on earth before we run out of Oil, I told him that he is just saying that to scare me but he honestly thinks that this is true. So I wanted to do the math for him to prove that we have nothing to worry about. But I got a lot of it finished but I have hit a brick wall, I do not think my math is correct so I would like you guys to help me prove that my math is correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My variables:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Volume of Earth in Cubic meters:&lt;/strong&gt; 260,000,000,000 cubic miles&lt;br&gt;&#xA;&lt;strong&gt;Volume of (any) water on Earth: &lt;/strong&gt;75% of earth's volume, 260,000,000,000 * 0.75 = 195,000,000,000 cubic miles&lt;br&gt;&#xA;&lt;strong&gt;Volume of drinkable water on earth:&lt;/strong&gt; 1% of Water, 195,000,000,000 * 0.01 =  1,950,000,000 cubic miles&lt;br&gt;&#xA;&lt;strong&gt;Conversion of Cubic Miles to Gallons:&lt;/strong&gt; 1 Cubic Mile = 1.101117147e+12&lt;br&gt;&#xA;&lt;strong&gt;Gallons of Drinkable water on Earth:&lt;/strong&gt; 2.147178437e+21&lt;br&gt;&#xA;&lt;strong&gt;Conversion of Gallons to Ounces:&lt;/strong&gt; 1 Gallon = 128 Ounces, Total of 5.359357378752e+32 Ounces&lt;br&gt;&#xA;&lt;strong&gt;Average person consumes:&lt;/strong&gt; 64 Ounces of Drinking Water per day&lt;br&gt;&#xA;&lt;strong&gt;People on earth:&lt;/strong&gt; 6.5 Billion&lt;br&gt;&#xA;&lt;strong&gt;Average amount of ounces each person drinks of water per day:&lt;/strong&gt; 64 * 6,500,000,000 = 416,000,000,000 Ounces of Drinking Water per day for every person on earth&lt;br&gt;&#xA;&lt;strong&gt;How many days earth drinking water would last with this amount:&lt;/strong&gt; 5.359357378752e+32 / 416,000,000,000 = 1288307062200000000000 &lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I just want some math people to check my work. I am only 15 trying to debate with a 40yo haha, I do not want to screw up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for any help and it is well appreciated!&lt;/p&gt;&#xA;" OwnerUserId="6646" LastActivityDate="2011-02-05T12:35:06.950" ClosedDate="2011-02-05T15:23:33.663" Title="How many gallons of water is on earth to this day?" Tags="&lt;calculus&gt;&lt;algebra-precalculus&gt;" AnswerCount="3" CommentCount="6" FavoriteCount="1" />
  <row Id="20493" PostTypeId="2" ParentId="20438" CreationDate="2011-02-05T02:50:43.977" Score="2" ViewCount="" Body="&lt;p&gt;First, the ring you are talking about is just the ring of integers $\mathbb{Z}[\zeta]$, as Arturo has remarked in the comments. Actually, to see that $\mathbb{Z}[\zeta]$ is precisely the ring of integers is not entirely trivial, but I will assume this. One reference that spontaneously comes to my mind is Washington's book on Cyclotomic Fields, but I'm sure there are more elementary ones.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is one easy way to see that $1-\zeta$ is prime, avoiding questions like &quot;if norms of elements divide each other, do the elements divide each other?&quot; :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Step 1.&lt;/strong&gt; Show that an element $x$ of an integral domain is prime if and only if the ideal $(x)$ is prime.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Step 2.&lt;/strong&gt; Show that if an ideal has prime norm, then it is prime. Recall that the norm of an ideal is its index in the ring. (Hint: any maximal ideal is prime. Prove it!)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Step 3.&lt;/strong&gt; Show that in the case of rings of integers of number fields, the norm of the ideal $(x)$ is equal to the absolute norm of $x$. Now, what is $\text{Norm}_{\mathbb{Q}(\zeta)/\mathbb{Q}}(1-\zeta)$? (Hint: the norm of an element of a number field is the constant coefficient of its minimal polynomial. Why?)&lt;/p&gt;&#xA;" OwnerUserId="3212" LastActivityDate="2011-02-05T02:50:43.977" />
  <row Id="20494" PostTypeId="2" ParentId="20488" CreationDate="2011-02-05T02:54:17.440" Score="2" ViewCount="" Body="&lt;p&gt;All the inequalities in (3) are derived from (2). The first $\nu$ inequalities, up to the semicolon, follow because $a_i$ is not on $A_j$ for $i\neq j$, $i \leq \nu$, $j \leq \nu$. The remaining inequalities, $s_k \leq k_n$ for $j &amp;gt; \nu$, follow because $A_1$, $A_2$, ..., $A_\nu$ are the lines through $a_n$, and hence $a_n$ is not on $A_j$ for $j &amp;gt; \nu$.&lt;/p&gt;&#xA;" OwnerUserId="6622" LastActivityDate="2011-02-05T02:54:17.440" />
  <row Id="20495" PostTypeId="1" CreationDate="2011-02-05T02:58:57.407" Score="2" ViewCount="85" Body="&lt;p&gt;I've been looking at Janusz book Algebraic Number Fields. I can't seem to figure out the following correspondence:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $\mathfrak{p}$ be a finite prime of a field K and L/K a finite Galois extension. Then we get a valuation ring R from $\mathfrak{p}$ by picking one valuation. Janusz then writes (p. 123) &quot;Identify $\mathfrak{p}$ as the maximal ideal of R&quot;. The problem is that given the maximal ideal in the valuation ring R, I can't see how we get back a valuation on K unless we assume that the ring R is a DVR?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I must be missing something here? Finally if we then let R' be the integral closure of R in L and $\mathfrak{P}$ some prime containing the maximal ideal of R should this somehow define a prime on L?&lt;/p&gt;&#xA;" OwnerUserId="6647" LastActivityDate="2011-02-05T02:58:57.407" Title="Correspondence between valuations and prime ideals" Tags="&lt;algebraic-number-theory&gt;" CommentCount="4" FavoriteCount="1" />
  <row Id="20496" PostTypeId="2" ParentId="20492" CreationDate="2011-02-05T03:32:03.210" Score="1" ViewCount="" Body="&lt;p&gt;Your volume of the earth line confuses meters and miles.  The figure is right in miles $4\pi/3*4000^3=4^4*1000^3=256*10^9$.  The volume of water is much smaller, though.  The deepest ocean is 7 miles out of those 4000.  99% is salt or ice, and most of the rest is inaccessible.  While you may only drink 2 quarts a day (and I abhor the fashion that says you must) &lt;a href=&quot;http://en.wikipedia.org/wiki/Water_supply_and_sanitation_in_the_United_States&quot; rel=&quot;nofollow&quot;&gt;Wikipedia says &lt;/a&gt;the American city average is more like 100 gallons/person/day consumption (think washing,  irrigation and swimming pools).  I have seen figures like that elsewhere.  So I think you shouldn't be arguing on one pass, as Moron commented.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That said, I think doing calculations like this is an extremely valuable activity.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-02-05T03:32:03.210" />
  <row Id="20497" PostTypeId="2" ParentId="20492" CreationDate="2011-02-05T03:33:00.690" Score="8" ViewCount="" Body="&lt;p&gt;It is not the case that 75% of the earth's volume is water.  You are thinking of the statistic that about 75% (closer to 70% really) of the earth's &lt;em&gt;surface&lt;/em&gt; is covered by water; there isn't any water on the inside.  I'd estimate the volume of water on earth as follows:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Circumference of the earth: $c = 4\times 10^7 m$ (almost by definition of the meter)&#xA;&lt;li&gt;Radius of the earth $r = c/2\pi \approx 7 \times 10^6 m$&#xA;&lt;li&gt;Surface area of the earth: $A = \frac{4}{3} \pi r^2 \approx 2 \times 10^{14} m^2$ &#xA;&lt;li&gt;Area covered by water: $A_W = 0.7 A \approx 1.4\times 10^{14} m^2$&#xA;&lt;li&gt;Average depth of the ocean: $D = 2\times 10^3 m$ (wild guess)&#xA;&lt;li&gt;Total volume of water: $A_W D \approx 3 \times 10^{17} m^3$&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This is about  $7 \times 10^7 mi^3$ or $8 \times 10^{19}$ gallons, which is about 5 orders of magnitude smaller than your value.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is also worth mentioning that humans consume water for many more things than just drinking (watering crops and livestock, industry, etc), and the rest of life on the planet needs water too.  64 ounces per human per day would not keep the human race alive.  On the other hand, it isn't necessarily gone after being used once.&#xA;&lt;/p&gt;&#xA;" OwnerUserId="822" LastActivityDate="2011-02-05T03:33:00.690" CommentCount="2" />
  <row Id="20498" PostTypeId="2" ParentId="20475" CreationDate="2011-02-05T05:51:55.850" Score="2" ViewCount="" Body="&lt;p&gt;Here's a take on this formula from a different direction.  Suppose you were given the recurrence relation $R(n,k) = R(n-1,k) + R(n-1,k-1)$, for $0 \leq k \leq n$, and boundary condition $R(0,k) = [k=0]$.  How would you derive the solution $R(n,k)$ if you didn't already know what it was?  You could use generating functions.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(The following is borrowed from Wilf's &lt;em&gt;&lt;a href=&quot;http://www.math.upenn.edu/~wilf/gfologyLinked2.pdf&quot; rel=&quot;nofollow&quot;&gt;Generatingfunctionology&lt;/a&gt;&lt;/em&gt;, 2nd edition, p. 14).  Let $$G_n(x) = \sum_{k=0}^{\infty} R(n,k) x^k.$$  Multiplying the recurrence relation above by $x^k$ and summing $k$ from $1$ to $\infty$ yields $$G_n(x) -1 = G_{n-1}(x) - 1 + x G_{n-1}(x).$$&#xA;Thus $G_n(x) = (x+1)G_{n-1}(x)$, with $G_0(x) = 1$.  Thus $G_n(x) = (x+1)^n$.  Since $R(n,k)$ is the coefficient of $x^k$ in $(x+1)^n$, applying the binomial theorem tells us that $R(n,k) = \binom{n}{k}$.&lt;/p&gt;&#xA;" OwnerUserId="2370" LastActivityDate="2011-02-05T05:51:55.850" />
  <row Id="20500" PostTypeId="1" CreationDate="2011-02-05T06:27:11.973" Score="-3" ViewCount="208" Body="&lt;p&gt;From &lt;a href=&quot;http://www.boardofstudies.nsw.edu.au/hsc_exams/hsc2005exams/pdf_doc/maths_ext2_05.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.boardofstudies.nsw.edu.au/hsc_exams/hsc2005exams/pdf_doc/maths_ext2_05.pdf&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/IvRdE.png&quot; alt=&quot;question 8(a), (i)-(iv)&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="6649" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2011-02-07T01:49:19.663" LastActivityDate="2011-02-07T01:49:19.663" Title="Polynomial problem" Tags="&lt;algebra&gt;&lt;polynomials&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="20501" PostTypeId="1" CreationDate="2011-02-05T06:59:22.110" Score="0" ViewCount="162" Body="&lt;p&gt;Let's say I have many measures of a random variable - x.&#xA;Moreover, I have many measures of y and z.&#xA;Meaning I can approximate their PDF by the histogram of the measures.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I estimate the joint PDF of (x, y, z)?&#xA;I don't know whether they are uncorrelated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S.&#xA;References to a practical demos and info on the subject are welcome.&lt;/p&gt;&#xA;" OwnerUserId="33" LastEditorUserId="33" LastEditorDisplayName="" LastEditDate="2011-02-05T07:05:05.507" LastActivityDate="2011-03-03T01:52:42.403" Title="Estimating Joint Probability Density Function" Tags="&lt;probability&gt;&lt;probability-theory&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="20502" PostTypeId="1" CreationDate="2011-02-05T07:29:28.843" Score="0" ViewCount="113" Body="&lt;p&gt;a= initial point &lt;/p&gt;&#xA;&#xA;&lt;p&gt;b=final point &lt;/p&gt;&#xA;&#xA;&lt;p&gt;n= numbers of rectangles &lt;/p&gt;&#xA;&#xA;&lt;p&gt;how to find endpoints ?&lt;/p&gt;&#xA;" OwnerUserId="6650" LastActivityDate="2011-02-05T16:23:09.643" Title="formula to find right endpoints of n rectangles in curve ?" Tags="&lt;calculus&gt;&lt;algebra&gt;&lt;differential-geometry&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="20503" PostTypeId="1" AcceptedAnswerId="20505" CreationDate="2011-02-05T07:31:22.630" Score="0" ViewCount="128" Body="&lt;p&gt;I'm reading the beginning of chapter 3 of Ahlfors where he starts to introduce analytic functions in regions, and I had a question on a part of Theorem 11:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $f=u+iv$ be an analytic function in a region $\Omega$. If the modulus of $f$ is constant, then $f$ must reduce to a constant. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;His proof starts out as follows: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $|f|=u^2+v^2$ is constant, then $ u\frac{\partial u}{\partial x} + v \frac{\partial v}{\partial x} = 0$, and $ u \frac{\partial u}{\partial y} + v \frac{\partial v}{\partial y} = -u \frac{\partial v}{\partial x} + v \frac{\partial u}{\partial x} = 0$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is why does this imply that either $ \frac{\partial u}{\partial x} = \frac{\partial v}{\partial x} = 0$ or that the determinant $u^2+v^2$ vanishes?&lt;/p&gt;&#xA;" OwnerUserId="1736" LastActivityDate="2011-02-05T07:36:55.207" Title="Ahlfors's proof of analytic functions reducing to a constant" Tags="&lt;complex-analysis&gt;" AnswerCount="1" />
  <row Id="20504" PostTypeId="2" ParentId="20217" CreationDate="2011-02-05T07:32:39.953" Score="4" ViewCount="" Body="&lt;p&gt;Why do you say that the interesting property of an integral domain is that it have zero divisors?  This is giving short shrift to the other properties, i.e. having 1, and being commutative.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question of rings without 1, and their relevance within mathematics, has been discussed in various places here and on MO: see e.g. &lt;a href=&quot;http://math.stackexchange.com/questions/16168/applications-of-rings-without-identity&quot;&gt;here&lt;/a&gt;, &lt;a href=&quot;http://mathoverflow.net/questions/22579/what-are-the-reasons-for-considering-rings-without-identity&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;,&#xA;and &lt;a href=&quot;http://mathoverflow.net/questions/34332/consequences-of-not-requiring-ring-homomorphisms-to-be-unital&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The differences between commutative and non-commutative rings are immense.  Essentially, in commutative rings one can localize, and in non-commutative rings one cannot.  See e.g. &lt;a href=&quot;http://mathoverflow.net/questions/7917/non-commutative-algebraic-geometry/15196#15196&quot; rel=&quot;nofollow&quot;&gt;this answer&lt;/a&gt; for more discussion of this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The theory of integral domains is based on all three properties (commutativity, existence of 1, no zero-divisors).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To give a concrete example, the enveloping algebra of a  Lie algebra over a field has no zero divisors.   If the Lie algebra is of finite dimension, it is Noetherian.  (One sees this because the enveloping algebra admits a filtration,&#xA;the filtration by degree, whose associated graded is a polynomial ring --- so in particular a domain --- in as many variables as the dimension of the Lie algebra --- and so Noetherian for a finite dimensional Lie algebra.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now lets suppose our Lie algebra $\mathfrak g$ is finite dimensional and semi-simple over a field of characteristic zero, say $\mathbb C$ just to fix ideas.  The enveoping algebra $U(\mathfrak g)$ contains an augmentation ideal (the kernel of the action on the trivial representation, if you like), call it $I$;&#xA;so there is a short exact sequence&#xA;$$0 \to I \to U(\mathfrak g) \to \mathbb C \to 0.$$&#xA;Now because $\mathfrak g$ is semi-simple, any extension of the trivial representation $\mathbb C$ by itself splits, and so is again trivial.  This&#xA;implies that $I^2 = I$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, in a Noetherian integral domain (commutative!), one has&#xA;the result that $I^2 = I$ implies $I = 0$. (See e.g. &lt;a href=&quot;http://math.stackexchange.com/questions/8937/why-is-the-artin-rees-lemma-used-here/19524#19524&quot;&gt;this answer&lt;/a&gt;.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Geometrically, the idea is that if $A$ is a commutative ring with one and $I$ is an ideal such that $I = I^2$, then Spec $A/I$ is a closed subscheme of Spec $A$&#xA;which has no non-trivial normal directions into Spec $A$.  If one pictures this, you will see that the only way this seems possible, intuitively, is if Spec $A/I$ is open as well as closed in Spec $A$ (otherwise there would be some directions pointing out of Spec $A/I$ into the rest of Spec $A$).  One can then actually prove that if $A$ is furthermore Noetherian, and if $I = I^2$, then $A$ factors as a product $A/I \times B$, so Spec $A$ is the disjoint union of Spec $A/I$ and Spec $B$.  In particular, if $A$ is a domain (so that Spec $A$ is irreducible, and so connected), one finds that either $I = A$ or $I = 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The example of the enveloping algebra shows that the non-commutative situation is completely different: we have the ring $U(\mathfrak g)$ without zero divisors, and Noetherian, but with a non-trivial idempotent ideal.  This is (at least for me) quite hard to interpret geometrically, and certainly suggests that it is reasonable to separate the study of (commutative) integral domains from the more general study of (possibly non-commutative) rings without zero divisors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Final remark: I have no objection to using the term &lt;em&gt;integral domain&lt;/em&gt;&#xA;or &lt;em&gt;domain&lt;/em&gt; in the non-commutative context.  (I should note, that like many&#xA;commutative algebraists, I frequently say and write &lt;em&gt;domain&lt;/em&gt; to mean commutative integral domain.)  My answer here is not intended to advocate any position on terminology, but just to explain why the commutative context is quite different from the more general non-commutative one, and hence why it makes sense to make a special study of the commutative case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Incidentally, my answer could reasonably be taken in a more general sense as (at least partly) explaining why we make a special study of commutative rings (the subject of commutative algebra) rather than just studying all rings at once.&lt;/p&gt;&#xA;" OwnerUserId="221" LastEditorUserId="221" LastEditorDisplayName="" LastEditDate="2011-02-05T09:08:32.617" LastActivityDate="2011-02-05T09:08:32.617" />
  <row Id="20505" PostTypeId="2" ParentId="20503" CreationDate="2011-02-05T07:36:55.207" Score="3" ViewCount="" Body="&lt;p&gt;notice that you now have a solution&#xA;$$\left(\begin{array}{cc}&#xD;&#xA;u &amp;amp; v\\&#xD;&#xA;v &amp;amp; -u\end{array}\right)\left(\begin{array}{c}&#xD;&#xA;\frac{\partial u}{\partial x}\\&#xD;&#xA;\frac{\partial v}{\partial x}\end{array}\right)=0$$&#xA;If the modulus (which is the determinant) is not zero then the matrix is invertible, so the solution to the system above must be zero.&lt;/p&gt;&#xA;" OwnerUserId="2125" LastActivityDate="2011-02-05T07:36:55.207" CommentCount="1" />
  <row Id="20506" PostTypeId="1" CreationDate="2011-02-05T07:44:10.933" Score="0" ViewCount="127" Body="&lt;p&gt;This is probably not a very ethical question to ask but I need to have a fast introduction to a range of concepts about the representation theory of the $SO(n)$ and I would be happy to see some online references which will help me do this journey. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am listing below the specific concepts about it that I need to know the most. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;I want to know about the concept of &quot;lowest weights&quot; and &quot;highest weights&quot; and how a string of $ [ \frac{n}{2}]$ numbers (integers?) say $(h_i, i = 1,2,..,[ \frac{n}{2}])$ label a representation of $SO(n)$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I want to know what is a &quot;quadratic Casimir&quot; of such a representation (vaguely I understand that to be the eigenvalue of an operator which commutes with all the basis of the group's Lie algebra). For such a representation as above the quadratic Casimir is &#xA;$$c_2(\{h_i\}) = \sum _{i=1} ^{ i= [ \frac{n}{2}]} (h_i ^2 + (n-2i)h_i)$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;which will help see that the quadratic Casimir for a &quot;scalar representation&quot; is $0$, for a &quot;vector representation&quot; it is $n-1$ and for a &quot;spinor representation&quot; it is $\frac{n(n-1)}{8}$ &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I hope to know how to convert the above in quotes standard physics terminology into the language of weights. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If $\{ H_i\}$ form a set of Cartan generators of the group $SO(2n+1)$ then in the &quot;vector representation&quot; the character of the element labelled by the real numbers say $\{t_i\}$ is $1+\sum _{i=1} ^n 2Cosh(t_i)$ and for the &quot;spinor representation&quot; it is given by $1+\prod _{i=1} ^n 2Cosh(t_i)$ and in general it is given as, &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;$$\chi (h_i,t_i) = \frac{det (sinh [ t_i(h_j +(n-j) +\frac{1}{2}  ]) }{det (sinh [t_i((n-j) +\frac{1}{2}) } $$&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;The above equation leads to a Clebsch-Gordan (which I am familiar for $SO(3)$) like thinking that $\{h_i\} \times \{vector\} = \{h_i\} +$ All representations obtained from $\{h_i\}$ by adding or subtracting unity from a single $h_i$ such that in the resulting set $h_n \geq 0$ and $h_i \geq h_{i+1}$ for other $is$. And similarly for $\{h_i\} \times \{spinor\} = \{h_i\} +$ All representations obtained from $\{h_i\}$ by adding or subtracting half from every $h_i$ such that in the resulting set $ h_n  \geq 0$ and $h_i \geq h_{i+1}$ for other $is$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Similarly for $SO(2n)$ the corresponding character formula looks like, &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;$$\chi (h_i,t_i) = \frac{det (sinh [ t_i(h_j + n-j)]) + det (cosh [ t_i(h_j + n-j)]) }{det (sinh [t_i(n-j)] } $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And similar interpretations lead to the thinking that $\{h_i\} \times \{vector\} = \{h_i\} +$ All representations obtained from $\{h_i\}$ by adding or subtracting unity from a single $h_i$ such that in the resulting set $\vert h_n \vert  \geq 0$ and $h_i \geq h_{i+1}$ for other $is$. And similarly for $\{h_i\} \times \{\pm chirality spinor\} = \{h_i\} +$ All representations obtained from $\{h_i\}$ by adding or subtracting half from every $h_i$ with number of subtractions being even/odd, such that in the resulting set $\vert h_n \vert  \geq 0$ and $h_i \geq h_{i+1}$ for other $is$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not sure I am reading everything right but I hope to get corrected and I would be happy to see references hopefully online which will explain to me the above things.&lt;/p&gt;&#xA;" OwnerUserId="965" LastActivityDate="2011-02-05T13:40:52.773" Title="Representation theory of $SO(n)$" Tags="&lt;reference-request&gt;&lt;lie-groups&gt;&lt;representation-theory&gt;&lt;lie-algebra&gt;" AnswerCount="1" />
  <row Id="20507" PostTypeId="1" CreationDate="2011-02-05T07:51:17.357" Score="2" ViewCount="94" Body="&lt;p&gt;Why is a smooth connected scheme (say over a field) necessarily irreducible? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Intuitively it makes sense because we might very well expect points in the intersection of two irreducible components to be singular points.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But what is a proof? Feel free to add any extra hypotheses if needed (e.g., separated if that is required).&lt;/p&gt;&#xA;" OwnerUserId="6652" LastActivityDate="2011-02-05T08:32:14.753" Title="Why is a smooth connected scheme irreducible?" Tags="&lt;algebraic-geometry&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="20508" PostTypeId="2" ParentId="20507" CreationDate="2011-02-05T08:32:14.753" Score="6" ViewCount="" Body="&lt;p&gt;The local rings of a smooth scheme over a field are regular, and a regular local ring is a domain.  Thus a smooth scheme over a field has all local rings being domains.  Thus the intersection of any two components must be empty (a point lying on the intersection would not have its local ring being a domain).&lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2011-02-05T08:32:14.753" CommentCount="1" />
  <row Id="20509" PostTypeId="1" CreationDate="2011-02-05T08:39:21.057" Score="1" ViewCount="96" Body="&lt;p&gt;1.if (a,m)=(b,m)=1 then (ab,m)=1&#xA; 2. if (a,b)=(a,c) then (a^2\b^2)=&lt;/p&gt;&#xA;" OwnerUserId="6654" LastActivityDate="2011-02-05T08:51:48.367" Title="1.if (a,m)=(b,m)=1 then (ab,m)=1 2. if (a,b)=(a,c) then (a^2\b^2)=" Tags="&lt;number-theory&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="20511" PostTypeId="1" AcceptedAnswerId="20513" CreationDate="2011-02-05T08:50:05.137" Score="1" ViewCount="48" Body="&lt;p&gt;Hi all,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have several numbers, say 3, 1, 4, 8, 5, and I wanted them to be mapped into a fixed domain [0.5, 3]. In this case, 1 should be mapped as 0.5 and 8 is 3. Then the rest numbers should be scaled down to their correspondences.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, my question is what should I do to deal with this case? what's the name of this processing?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks,&#xA;Mike&lt;/p&gt;&#xA;" OwnerUserId="6653" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2011-02-05T10:56:44.027" LastActivityDate="2011-02-05T10:56:44.027" Title="How to map discrete numbers into a fixed domain?" Tags="&lt;number-theory&gt;" AnswerCount="1" />
  <row Id="20512" PostTypeId="2" ParentId="20509" CreationDate="2011-02-05T08:51:48.367" Score="2" ViewCount="" Body="&lt;p&gt;HINT: For question one, start by assuming $p$ is a prime divisor of $(ab,m).$ You will quickly arrive at a contradiction.&lt;/p&gt;&#xA;" OwnerUserId="1301" LastActivityDate="2011-02-05T08:51:48.367" CommentCount="2" />
  <row Id="20513" PostTypeId="2" ParentId="20511" CreationDate="2011-02-05T09:02:51.013" Score="1" ViewCount="" Body="&lt;p&gt;Suppose you have $I=[a,b]$ and you want to map it to $J=[c,d]$ such that there is a constant C and for every $x,y \in I$ and their images $f(x),f(y)\in J$ you have $(x-y)C = (f(x)-f(y))$ (I hope this is what you mean by &quot;scaled down...&quot;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You also want that $a\mapsto c$ and $b\mapsto d$ (in your case $1\mapsto 0.5,\; 8 \mapsto 3$), then you see that the constant is $\frac{c-d}{a-b}$. Now, for every other $x\in I$ you have (with y=a)&#xA;$$ f(x) = C(x-a) + f(a) = Cx +(f(a)-Ca) $$&lt;/p&gt;&#xA;" OwnerUserId="2125" LastActivityDate="2011-02-05T09:02:51.013" CommentCount="1" />
  <row Id="20514" PostTypeId="1" AcceptedAnswerId="20516" CreationDate="2011-02-05T09:06:48.147" Score="5" ViewCount="125" Body="&lt;p&gt;I'm trying to wrap my head around this concept of extracting sets from a given set to show that a subset is indeed a subset.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose I want to show that the set of all sets of two elements does not exist. I approach it like this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose such a set $V$ exists of all elements containing exactly the sets of two elements. Then for any nonempty set $X$, the set $\{X,\emptyset\}$ is a set, by the pairing axiom, and since $X\neq\emptyset$, $\{X,\emptyset\}$ has two elements, so $\{X,\emptyset\}\in V$ for all sets $X$. Then by the power set axiom, $\bigcup V$ is the set of all sets, since $\emptyset\in\bigcup V$, and any nonempty set $X\in\bigcup V$ as well. But the set of all sets does not exist, a contradiction, so such a set $V$ does not exist. Does this approach work?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if I'm given a nonempty set $A$, is the following a proper formulation that shows that the set of all sets of the form $\{a,b\}$ for $a,b\in A$ is a set?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $a,b\in A$, then $\{a,b\}\subset A$, and $\{a,b\}\in\mathscr{P}(A)$. Assuming that $a$ and $b$ need not be distinct, consider&#xA;$$&#xD;&#xA;\{t\in\mathscr{P}(A)\ |\ \exists x\exists y(x\in A\land y\in A\land t=\{x,y\}\}.&#xD;&#xA;$$&#xA;This is a set by the subset schema. Have I correctly specified the type of sets I want, or do I need to include something about if $z\in t\implies z=x\vee z=y$?, or is that unnecessary? Thanks.&lt;/p&gt;&#xA;" OwnerUserId="6655" LastActivityDate="2011-02-05T09:55:32.897" Title="The set of all sets of two elements does not exist, but the set of sets of two elements of a given set does?" Tags="&lt;elementary-set-theory&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="20515" PostTypeId="2" ParentId="20468" CreationDate="2011-02-05T09:53:54.053" Score="2" ViewCount="" Body="&lt;p&gt;By the standard techniques explained &lt;a href=&quot;http://math.stackexchange.com/questions/18405/a-question-related-to-two-competing-patterns-in-coin-tossing/20010#20010&quot;&gt;here&lt;/a&gt; in a similar context, one can show that the first time $T_A$ when the word A=HTHT is produced has generating function&#xA;$$&#xD;&#xA;E(s^{T_A})=\frac{s^4}{a(s)},\qquad a(s)=(2-s)(8-4s-s^3)-2s^3,&#xD;&#xA;$$&#xA;and that the first time $T_B$ when the word B=TTH is produced has generating function&#xA;$$&#xD;&#xA;E(s^{T_B})=\frac{s^3}{b(s)},\qquad b(s)=(2-s)(4-2s-s^2),&#xD;&#xA;$$&#xA;The next step would be to develop these as series of powers of $s$, getting&#xA;$$&#xD;&#xA;E(s^{T_A})=\sum_{n\ge4}p_A(n)s^n,\quad E(s^{T_B})=\sum_{k\ge3}p_B(k)s^k,&#xD;&#xA;$$&#xA;and finally, to compute the sum&#xA;$$&#xD;&#xA;P(T_B&amp;lt;T_A)=\sum_{k\ge3}\sum_{n\ge k+1}p_B(k)p_A(n).&#xD;&#xA;$$&#xA;An alternative solution is to consider the Markov chain on the space of the couples of prefixes of the words A and B and to solve directly the associated first hitting time problem for this Markov chain, as explained, once again, &lt;a href=&quot;http://math.stackexchange.com/questions/18405/a-question-related-to-two-competing-patterns-in-coin-tossing/20010#20010&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(+1 to Arturo's and Mike's comments.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added later&lt;/strong&gt; Introduce the decomposition into simple elements&#xA;$$&#xD;&#xA;\frac1{(1-s)b(s)}=\sum_{i=1}^4\frac{c_i}{1-s\gamma_i}.&#xD;&#xA;$$&#xA;Then, one can show that&#xA;$$&#xD;&#xA;P(T_B&amp;lt;T_A)=\sum_{i=1}^4\frac{c_i}{a(\gamma_i)},&#xD;&#xA;$$&#xA;by decomposing the rational fractions $1/b(s)$ and $1/a(s)$ into simple elements, by relying on the fact that all their poles are simple, and by using some elementary algebraic manipulations. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;At this point, I feel it is my duty to warn the OP that to ask &lt;a href=&quot;http://math.stackexchange.com/questions/18093&quot;&gt;question&lt;/a&gt; after &lt;a href=&quot;http://math.stackexchange.com/questions/18405&quot;&gt;question&lt;/a&gt; on math.SE or &lt;a href=&quot;http://mathoverflow.net/questions/53011&quot; rel=&quot;nofollow&quot;&gt;elsewhere&lt;/a&gt; along the lines of this post is somewhat pointless. Much more useful would be to learn once and for all the basics of the theory of finite Markov chains, for instance by delving into the quite congenial &lt;a href=&quot;http://www.statslab.cam.ac.uk/~james/Markov/&quot; rel=&quot;nofollow&quot;&gt;Markov chains&lt;/a&gt; by James Norris, and, to learn to manipulate power series, into the magnificent &lt;a href=&quot;http://www.math.upenn.edu/~wilf/DownldGF.html&quot; rel=&quot;nofollow&quot;&gt;generatingfunctionology&lt;/a&gt; by Herbert Wilf.&lt;/p&gt;&#xA;" OwnerUserId="6179" LastEditorUserId="6179" LastEditorDisplayName="" LastEditDate="2011-02-05T21:55:59.383" LastActivityDate="2011-02-05T21:55:59.383" CommentCount="10" />
  <row Id="20516" PostTypeId="2" ParentId="20514" CreationDate="2011-02-05T09:55:32.897" Score="4" ViewCount="" Body="&lt;p&gt;The problem is that $V$ is not a set, it is in fact a proper class. A good rule of thumb when it comes to these sort of problems is that if you can map each set into your collection - it's a proper class.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If, on the other hand, you started with a set $A$ (which you already know that it is a set) then you can have the collection of all pairs from the set. The formula which you wrote above is indeed the collection of all singletons and pairs from the set $A$, if you want to exclude the singletons you have to add $x\not= y$ as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a result from the Subset Axiom/Replacement Axiom - both are schemes and not actual axiom - which state that if you have a formula in one free variable, then the collection of elements which satisfies the formula coming from a set $A$ is a set itself, namely - if it's a subcollection of a set then it is a set.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The replacement axiom is slightly more complicated in that matter, but it says that the range of a function whose domain is a set is also a set, from this the subset axiom can be deduced easily.&lt;/p&gt;&#xA;" OwnerUserId="622" LastActivityDate="2011-02-05T09:55:32.897" CommentCount="2" />
  <row Id="20517" PostTypeId="1" CreationDate="2011-02-05T10:16:04.180" Score="1" ViewCount="144" Body="&lt;p&gt;How to solve this trigonometric equation $x = \frac 1 2 \cos\left(\frac 2 3 \sin\left(\frac 3 4 x\right)\right) + 1$ ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The iterative solution seems to be 1.417.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anybody suggest an algebraic solution ?&#xA;Does it really exist ?&lt;/p&gt;&#xA;" OwnerUserId="5049" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2011-02-05T12:24:39.963" LastActivityDate="2011-02-05T12:24:39.963" Title="How to solve algebraically the equation $x = \frac{1}{2}\cos\left(\frac 2 3 \sin\left(\frac 3 4 x\right)\right) + 1$" Tags="&lt;trigonometry&gt;&lt;functional-equations&gt;" CommentCount="3" />
  <row Id="20518" PostTypeId="1" AcceptedAnswerId="21424" CreationDate="2011-02-05T12:04:31.467" Score="1" ViewCount="148" Body="&lt;p&gt;In a previous question &lt;a href=&quot;http://math.stackexchange.com/questions/9836/summing-something-with-random-and-non-random-parts&quot;&gt;Summing something with random and non-random parts.&lt;/a&gt; I asked how one could estimate (specifically I am interested in an upper bound).&#xA;$$\sum_{k=1}^n e^{-k B_k^2} $$&#xA;where $B_k$ is a uniformly distributed sequence on [0,1]. The response was to take the expected value of the summand. But what can I do if my $B_k^2$ are deterministic (and not random variables?). To clarify, by uniformly distributed I mean $\{B_k\}$ satisfies&#xA;$$ \lim_{n \rightarrow \infty} \frac{|[a,b] \cap \{B_1,B_2,\ldots B_n\}|}{n} = b-a$$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: Suppose $B_n = \{ \alpha n \}$ is the fractional part of $\alpha n$  for some irrational number $\alpha$. By Weyl's Criterion I know that $\{B_n\}$ is uniformly distributed. From some numerical calculations it seems that there exists some $K &amp;gt; 0$ that only depends on $\alpha$ such that&#xA;$$\sum_{k=1}^n e^{-k B_k^2} &amp;lt; K \sqrt{n}$$&#xA;for all sufficiently large $n$. How would I go about demonstrating this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit 2&#xA;Here is an idea I had.&#xA;$$&#xD;&#xA;\sum_{k=1}^n e^{-k B_k^2} = \sum_{k \leq n \: : e^{-k B_k^2} \leq n^{-1/2} } e^{-k B_k^2} + \sum_{k \leq n \: : e^{-k B_k^2} &amp;gt; n^{-1/2}} e^{-k B_k^2}&#xD;&#xA;$$&#xA;which is less than&#xA;$$&#xD;&#xA;\sum_{k = 1}^n n^{-1/2} + \sum_{k \leq n \: : e^{-k B_k^2} &amp;gt; n^{-1/2}} 1&#xD;&#xA;$$&#xA;Now the left sum is just $n^{1/2}$. Now $ \#\{e^{-k B_k^2} &amp;gt; n^{-1/2} : k \leq n\} = \#\{B_k &amp;lt; \sqrt{\log (n)/(2k)}  : k \leq n\}$.&#xA;I believe that&#xA;$$\#\{B_k &amp;lt; \sqrt{\log (n)/(2k)}  : k \leq n\} \approx \int_1^n \sqrt{\frac{\log (n)}{2t}} \, dt$$&#xA;Or maybe that integral should be a sum, I don't know. But that implies there exists $\epsilon &amp;gt; 0$ such that for sufficiently large $n$,&#xA;$$\#\{B_k &amp;lt; \sqrt{\log (n)/(2k)}  : k \leq n\}  &amp;lt; n^{1/2+\epsilon}$$&#xA;So finally we have&#xA;$$ \sum_{k=1}^n e^{-k B_k^2} &amp;lt; n^{1/2} + n^{1/2+\epsilon} &amp;lt; 2 n^{1/2+\epsilon}$$&#xA;Clearly the above argument has quite a few gaps in it. I am hoping that someone can give me some feedback as to whether or not the gaps can be filled.&lt;/p&gt;&#xA;" OwnerUserId="899" LastEditorUserId="899" LastEditorDisplayName="" LastEditDate="2011-02-09T11:22:38.220" LastActivityDate="2011-02-10T19:52:28.313" Title="Estimating a sum containing a uniformly distributed but deterministic term." Tags="&lt;probability&gt;&lt;analysis&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="20519" PostTypeId="1" AcceptedAnswerId="20535" CreationDate="2011-02-05T12:18:07.453" Score="4" ViewCount="222" Body="&lt;p&gt;The Weierstrass M-test tells that given a function sequence $(u_{n}(x))$ where $x \in I$, if there exists a convergent series $\sum a_{n}$ such that $|u_{n}(x)|\leq a_{n}$ for all $n$ and $x\in I$, then $\sum u_{n}(x)$ converges uniformly in $I$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What about the opposite of it? &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If $\sum u_{n}(x)$ converges uniformly in $I$, then there exists a convergent series $\sum g_{n}$ such that $|u_{n}(x)|\leq g_{n}$ for all $n$ and $x \in I$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Since the theorem isn't in the form of 'if and only if', I'm trying to think of an example to counter the above:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My attempt was to define a function sequence such as this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/Yjzgp.png&quot; alt=&quot;example&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then taking $u_{1}(x)=f_{1}(x),\ u_{n}(x)=f_{n}(x)-f_{n-1}(x)$. It can be shown that $f_{n}(x)$ converges uniformly in $[1, \infty)$, therefore $u_{n}(x)$ does aswell.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But $|u_{n}(1)|=|f_{n}(1)|$ which is the constant sequence: $1, 1, 1, ...$. If we assume by contradiction that there exists such a sequence $1\leq g_{n}$ that $\sum g_{n}$ converges then by the comparison test $\sum 1$ converges which is obviously not true.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, I'd like to know if the above is true. It took me quite a while to come up with something and I'm not even sure it's true. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, It's very difficult for me to visualize these complicated functions (like the one above) where both $x$ and $n$ play a role. Is there an easier way to deal with these questions?&lt;/p&gt;&#xA;" OwnerUserId="3237" LastActivityDate="2011-02-05T19:46:38.100" Title="Question regarding Weierstrass M-test" Tags="&lt;real-analysis&gt;&lt;sequences-and-series&gt;" AnswerCount="3" CommentCount="9" />
  <row Id="20520" PostTypeId="2" ParentId="20492" CreationDate="2011-02-05T12:26:28.780" Score="1" ViewCount="" Body="&lt;p&gt;As Alex notes, the meat of this question is not mathematical, rather it has to do with economics. Keep in mind that even if you were able to calculate precisely the amount of water on earth today, it would still be unclear whether we would run out of drinkable water before oil. When it comes to consumption of a good such as oil, the price and quantity will be determined by the intersection of the supply and demand curves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Supply_and_demand&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Supply_and_demand&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Try holding the demand for oil constant (unrealistic assumption of course!). Now if you also assume that oil on earth is all known reserves with no difference in quality, you can think of the supply as constantly being reduced (slowly shifting the supply curve to the left). &quot;Cetaris paribus&quot; (all other things equal) we can see that this will have an effect of increasing the price while decreasing the quantity demanded (as the intersection moves along the fixed demand curve). In this oversimplified version, the closer we got to the &quot;last reserves,&quot; the less people are demanding oil at the given price, so of course, the supply is depleted at a lower rate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now try holding the supply for oil constant (also unrealistic!). There are many factors influencing the demand curve for oil: taxes/regulations, subsititute goods/alternative energies, increased/decreased wealth to name a few. Think about whether these factors will increase demand (shifting the curve to the right) or decrease it (shifting the curve to the left).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is of course a very complex interplay between the supply and the demand which extends to many other goods...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Overall, the point is that both water and oil will most likely not simply be used up at a constant rate until they are gone. In both cases, the invisible hand of the market has ways of applying pressure in (hopefully) the right spots to ensure that we run out of neither!&lt;/p&gt;&#xA;" OwnerUserId="3711" LastEditorUserId="3711" LastEditorDisplayName="" LastEditDate="2011-02-05T12:35:06.950" LastActivityDate="2011-02-05T12:35:06.950" CommentCount="3" />
  <row Id="20521" PostTypeId="2" ParentId="20519" CreationDate="2011-02-05T12:34:17.980" Score="3" ViewCount="" Body="&lt;p&gt;$u_n = f_n - f_{n-1}$ and $u_n(1) = 0$ for n&gt;1.&#xA;However, if $a_n = \sup_{x\in I} |u_n(x)|$, you have $a_n \geq u_n(n) = 1/n$ is still a divergent sequence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your idea to &quot;spread&quot; where the functions $u_n$ are adding things along the x axis is good :&#xA;You never add too much at a specific x so that the sum converges, but you are still able to add enough &quot;big&quot; things at each n so that $\Sigma a_n$ doesn't converge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can use this idea to make a more clear counter example :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pick $B(x)$ a bump function on $\mathbb{R}$ with support in [0,1], for example a piecewise affine function with $B(0)=B(1) = 0$ and $B(1/2) = 1$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then define $u_n(x) = B(x-n)/n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Sigma_{n\geq 1} u_n$ converges uniformly because at every x, $\Sigma_{k\geq n} u_k(x) \leq 1/n$.&#xA;However, $\Sigma_{n\geq 1} \sup_x |u_n(x)| = \Sigma_{n\geq 1} 1/n$ will diverge&lt;/p&gt;&#xA;" OwnerUserId="6270" LastActivityDate="2011-02-05T12:34:17.980" CommentCount="1" />
  <row Id="20522" PostTypeId="1" AcceptedAnswerId="22923" CreationDate="2011-02-05T12:41:19.387" Score="7" ViewCount="260" Body="&lt;p&gt;Given a property P, is there some rules that are sufficient or necessary to determine if there exists a set of all sets with property P?&lt;/p&gt;&#xA;" OwnerUserId="4929" LastActivityDate="2011-02-20T15:05:46.347" Title="For which P can one claim the existence of the set of all sets that satisfy P ?" Tags="&lt;logic&gt;&lt;set-theory&gt;&lt;elementary-set-theory&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="1" />
  <row Id="20523" PostTypeId="2" ParentId="20518" CreationDate="2011-02-05T13:05:06.447" Score="0" ViewCount="" Body="&lt;p&gt;I would have thought the lower bound was 0.5819767068693 which can be approached by taking $B_k$ very close to $1$ for small $k$.  Plouffe's inverse symbolic calculator tells me this is the solution to &#xA;$$\log_e(1+x) = 1+\log_e(x)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, there is no upper bound, since taking $B_k$ very close to $0$ for small $k$ will give an arbitrary large sum.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Forget that, but I still think your $K \sqrt{n}$ may not be an upper bound.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take for example &#xA;$$B_k = \dfrac{2k-1}{2n},$$ &#xA;i.e. $\frac{1}{2n}, \frac{3}{2n}, \frac{5}{2n}, \ldots, \frac{2n-1}{2n},$ which might be described as being uniformly distributed on $[0,1]$.  Then I think you will find that for any $n$ you care to check&#xA;$$0.77 n^{2/3} &amp;lt; \sum_{k=1}^n e^{-k B_k^2} &amp;lt; 0.9 n^{2/3} $$&#xA;so increasing faster than $K \sqrt{n}$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This may not be what you describe as uniformly distributed (in particular the $B_k$ depend on $n$).  Nor is it necessarily an upper bound, as I have not checked whether there may be another way of ordering the $B_k$ to give a higher sum.&lt;/p&gt;&#xA;" OwnerUserId="6460" LastEditorUserId="6460" LastEditorDisplayName="" LastEditDate="2011-02-06T21:43:40.950" LastActivityDate="2011-02-06T21:43:40.950" CommentCount="1" />
  <row Id="20524" PostTypeId="2" ParentId="20518" CreationDate="2011-02-05T13:25:38.427" Score="0" ViewCount="" Body="&lt;p&gt;Upper bound: $$\sum_{k=1}^n e^{-0}=n $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rest of this question makes no sense, what does 'what can I do' mean ?&#xA;Either you know B_k and you calculate the sum, or you dont and you estimate it.&lt;/p&gt;&#xA;" OwnerUserId="4929" LastEditorUserId="4929" LastEditorDisplayName="" LastEditDate="2011-02-05T13:38:28.293" LastActivityDate="2011-02-05T13:38:28.293" />
  <row Id="20525" PostTypeId="2" ParentId="20506" CreationDate="2011-02-05T13:40:52.773" Score="6" ViewCount="" Body="&lt;p&gt;Read the book by Fulton and Harris on &lt;em&gt;Representation Theory&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2011-02-05T13:40:52.773" CommentCount="5" />
  <row Id="20526" PostTypeId="2" ParentId="15831" CreationDate="2011-02-05T14:04:07.320" Score="2" ViewCount="" Body="&lt;p&gt;$$\pi(x)=\sum _{n=1}^{x}u(n)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The formula for f_2(n) in mathematica does not return the sequence above, but returns fractions etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Seeing as derivation is missing, can anyone confirm that they have gotten any of these &quot;new&quot; formulas to &lt;em&gt;work&lt;/em&gt; or to &lt;em&gt;not work&lt;/em&gt;?&lt;/p&gt;&#xA;" OwnerUserId="4929" LastEditorUserId="4929" LastEditorDisplayName="" LastEditDate="2011-02-06T09:18:58.087" LastActivityDate="2011-02-06T09:18:58.087" CommentCount="4" />
  <row Id="20527" PostTypeId="1" AcceptedAnswerId="20528" CreationDate="2011-02-05T14:18:20.830" Score="0" ViewCount="132" Body="&lt;p&gt;Working on a pre-lab, some of the directions for drawing graphs is that the &quot;x-axis&quot; is for the non-variables, and the &quot;y-axis&quot; is for the variables (what you have measured).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the first thing I have to do is to make a standard curve for &quot;Glucose concentration&quot; and &quot;Absorbtion&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The next step is to measure the &quot;Enzyme activity&quot; and use the standard curve done to deternmine the amount of glucose since I will be having the &quot;Absorbance&quot; values in the &quot;Enzyme activity&quot; process.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, we are asked to plot a graph that shows the &lt;strong&gt;&lt;em&gt;glucose production against time&lt;/em&gt;&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Based on that, what (glucose production, time) goes where (x-axis, y-axis)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="6659" LastActivityDate="2011-02-05T14:31:04.533" Title="Plotting glucose production against time , x-axis &amp; y-axis issue" Tags="&lt;graph&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="20528" PostTypeId="2" ParentId="20527" CreationDate="2011-02-05T14:31:04.533" Score="1" ViewCount="" Body="&lt;p&gt;time is x axis almost always, glucose production y axis&lt;/p&gt;&#xA;" OwnerUserId="4929" LastActivityDate="2011-02-05T14:31:04.533" CommentCount="2" />
  <row Id="20529" PostTypeId="1" AcceptedAnswerId="20565" CreationDate="2011-02-05T14:32:14.613" Score="5" ViewCount="178" Body="&lt;p&gt;I am trying to compute Nth &lt;a href=&quot;http://oeis.org/A005117&quot; rel=&quot;nofollow&quot;&gt;Squarefree numbers&lt;/a&gt; using Mathematica. What I am trying to utilize is the &lt;code&gt;SquareFreeQ[i]&lt;/code&gt; function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is my solution :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;NSqFree[N_] := &#xA; For[n = 1; i = 1, n &amp;lt;=  N + 1, &#xA;  If[SquareFreeQ[i], If[n == N, Print[i] ] n++] i++]&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I am &lt;a href=&quot;http://mathalon.in/?page=show_problem.php&amp;amp;pid=111&quot; rel=&quot;nofollow&quot;&gt;supposed to compute&lt;/a&gt; &lt;code&gt;NSqFree[1000000000]&lt;/code&gt; but seems like my approach is taking for ever. Any faster method ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;ADDED&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here an &lt;a href=&quot;http://www.topcoder.com/stat?c=problem_statement&amp;amp;pm=2342&amp;amp;rd=4770&quot; rel=&quot;nofollow&quot;&gt;exactly identical topcoder question&lt;/a&gt; and the &lt;a href=&quot;http://www.topcoder.com/tc?module=Static&amp;amp;d1=match_editorials&amp;amp;d2=srm190&quot; rel=&quot;nofollow&quot;&gt;corresponding editorial&lt;/a&gt; for the same. &lt;/p&gt;&#xA;" OwnerUserId="2109" LastEditorUserId="2109" LastEditorDisplayName="" LastEditDate="2011-02-05T18:55:15.760" LastActivityDate="2011-02-05T18:55:15.760" Title="Fast method for Nth Squarefree number (using mathematica)" Tags="&lt;number-theory&gt;&lt;contest-math&gt;&lt;mathematica&gt;" AnswerCount="4" />
  <row Id="20530" PostTypeId="1" CreationDate="2011-02-05T14:52:08.510" Score="2" ViewCount="101" Body="&lt;p&gt;I am looking for software/algorithm for the smallest context free grammar describing a finite &lt;strong&gt;set&lt;/strong&gt; of words (and no other words).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a single word I found &lt;a href=&quot;http://sequitur.info/&quot; rel=&quot;nofollow&quot;&gt;sequitur&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Related to this seems: given a CFG what is the shortest functionally equivalent CFG?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would prefer the size to be measured by the CNF (for a precise estimate of the size).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To clarify I know the set can be described by a regular expression, this question is about the &lt;strong&gt;smallest CFG&lt;/strong&gt;.&lt;/p&gt;&#xA;" OwnerUserId="4750" LastActivityDate="2011-02-05T21:14:05.190" Title="Software/algorithm for the smallest context free grammar describing a set of words?" Tags="&lt;computer-science&gt;&lt;math-software&gt;&lt;formal-systems&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="20531" PostTypeId="1" AcceptedAnswerId="20534" CreationDate="2011-02-05T15:01:11.400" Score="2" ViewCount="137" Body="&lt;p&gt;Tried the squeeze theorem but it didn't get me anywhere since:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$0 \leftarrow \frac{1}{n^{1-1/n}}=\frac{n^{1/n}}{n}\leq\frac{(1^n+2^n+...+n^n)^{1/n}}{n}\leq \ n^{1/n}\to 1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;None of the other tools we studied seem to work. Any tips? Thanks.&lt;/p&gt;&#xA;" OwnerUserId="6661" LastActivityDate="2011-02-05T15:22:47.580" Title="Find $\lim_{n\to \infty}\frac{(1^n+2^n+...+n^n)^{1/n}}{n}$" Tags="&lt;calculus&gt;&lt;limit&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="20533" PostTypeId="1" CreationDate="2011-02-05T15:13:08.820" Score="1" ViewCount="66" Body="&lt;p&gt;If we want to define &lt;code&gt;circa&lt;/code&gt;, do we just say it means &lt;em&gt;around&lt;/em&gt; or there is a better definition to that?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="6659" LastActivityDate="2011-02-05T15:13:08.820" ClosedDate="2011-02-05T15:21:59.107" Title="Defining circa (Ca.)" Tags="&lt;definition&gt;" CommentCount="1" />
  <row Id="20534" PostTypeId="2" ParentId="20531" CreationDate="2011-02-05T15:15:04.637" Score="4" ViewCount="" Body="&lt;p&gt;$$\frac{\left(1^n+2^n+...+n^n\right)^{1/n}}{n} = \left(\left(\frac{1}{n}\right)^n+\left(\frac{2}{n}\right)^n+...+1^n\right)^{1/n} \geq (1)^{1/n} = 1 \rightarrow 1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rest is as you suggest.&lt;/p&gt;&#xA;" OwnerUserId="6662" LastEditorUserId="6662" LastEditorDisplayName="" LastEditDate="2011-02-05T15:22:47.580" LastActivityDate="2011-02-05T15:22:47.580" CommentCount="2" />
  <row Id="20535" PostTypeId="2" ParentId="20519" CreationDate="2011-02-05T15:27:45.937" Score="6" ViewCount="" Body="&lt;p&gt;The first thing I would think of is to consider a constant series, that is one that does not depend on any variable. This will certainly converge uniformly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the next step is to think about your second condition, for this to fail it would work if the series does not converge absolutely.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, this leads to the most simple example I can think of:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{n \geq 1} \frac{(-1)^n}{n}$$&lt;/p&gt;&#xA;" OwnerUserId="1120" LastActivityDate="2011-02-05T15:27:45.937" />
  <row Id="20536" PostTypeId="2" ParentId="20475" CreationDate="2011-02-05T15:29:21.260" Score="1" ViewCount="" Body="&lt;p&gt;Since you mentioned that you were having a hard time visualising this and I always seem to find myself visualising it whenever I have the need to write it down here is what goes through my mind as the pen is moving across the paper:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We are placing $r$ identical balls in $n$ boxes (at most one in each) that are in a straight line, so&#xA;${ n \choose r}$ ways to do this, now either the last box is empty, that's ${n-1 \choose r}$ ways, or the last box is full, that's ${ n-1 \choose r-1}$ ways. QED&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is equivalent to Sivaram's answer but does away with the colours, which for the purposes of visualising is probably slightly easier.&lt;/p&gt;&#xA;" OwnerUserId="1301" LastEditorUserId="1301" LastEditorDisplayName="" LastEditDate="2011-02-05T15:37:39.577" LastActivityDate="2011-02-05T15:37:39.577" />
  <row Id="20537" PostTypeId="1" CreationDate="2011-02-05T15:32:50.320" Score="-4" ViewCount="190" Body="&lt;p&gt;What is the best known attack on RSA given just the public key $(n,e)$?&lt;/p&gt;&#xA;" OwnerUserId="6663" LastEditorUserId="7162" LastEditorDisplayName="" LastEditDate="2011-05-23T03:27:29.130" LastActivityDate="2011-05-23T03:27:29.130" Title="hacking rsa with n" Tags="&lt;cryptography&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="20538" PostTypeId="2" ParentId="20529" CreationDate="2011-02-05T15:42:27.390" Score="1" ViewCount="" Body="&lt;p&gt;You won't do a billion (if I counted the zeros right) loops any time soon, so you need a better approach.  &lt;a href=&quot;http://en.wikipedia.org/wiki/Square-free_integer&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt; has an approximation under the section Distribution of square-free numbers.  So you could start with $10^9 \pi^2/6$ and calculate how many below that are square-free, then correct from there.  You will need to use inclusion/exclusion, but there are only about 40,000 to consider.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-02-05T15:42:27.390" />
  <row Id="20539" PostTypeId="1" AcceptedAnswerId="20547" CreationDate="2011-02-05T15:59:42.120" Score="1" ViewCount="132" Body="&lt;p&gt;I'm learning about transformation in mathematics in general. There are many of them like for example Jacobian (which I'm trying to understand now). The &quot;how ?&quot; part is very easy. There are formulas and we use them but the &quot;why ?&quot; part is what I'm looking for. I fail to understand the intuition behind it. Why are we using transformations ? Any links, suggestions, are all welcome. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My apologies for ambiguity. Here is an example question. &#xA;Let X1, X2, . . ., Xn be independent identically distributed U (0, θ) random variables. What is the distribution of (X1-X2) ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can solved this question directly but it is very simple to use Jacobian to solve this. I've the solution so I know how to use it. But why are we using Jacobian and how do I know when to use it ?&lt;/p&gt;&#xA;" OwnerUserId="6335" LastEditorUserId="6335" LastEditorDisplayName="" LastEditDate="2011-02-05T18:29:21.673" LastActivityDate="2011-02-05T18:29:21.673" Title="What are transformations in mathematics?" Tags="&lt;transformation&gt;" AnswerCount="2" CommentCount="5" />
  <row Id="20540" PostTypeId="2" ParentId="20529" CreationDate="2011-02-05T16:00:07.327" Score="0" ViewCount="" Body="&lt;p&gt;I think that your method can be improved by sieving: if $n$ is not squarefree you don't need to check any of its multiples. Start with a reasonable estimate (like the one Ross Millikan gave) and sieve like you would do in a prime sieve.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this viable? I don't know! I can't program in Mathematica, but later today I'll try to implement this in Python.&lt;/p&gt;&#xA;" OwnerUserId="4471" LastActivityDate="2011-02-05T16:00:07.327" />
  <row Id="20541" PostTypeId="2" ParentId="20529" CreationDate="2011-02-05T16:04:28.480" Score="1" ViewCount="" Body="&lt;p&gt;I'm not an expert of number-theoretic algorithms, but it seems to me that you can employing the Chinese Remainder Theorem to obtain a decent first stab at a &quot;sieving&quot; algorithm.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Use several registers r[p], each storing a residue modulo p&lt;sup&gt;2&lt;/sup&gt; for some prime p. Define such a register for various small primes p&amp;nbsp;&amp;isin;&amp;nbsp;{2,&amp;thinsp;3,&amp;thinsp;5,&amp;thinsp;...&amp;thinsp;,&amp;thinsp;P} for some suitably large prime P. You will use these to represent an integer R, such that R&amp;nbsp;&amp;equiv;&amp;nbsp;r[p] (mod p&lt;sup&gt;2&lt;/sup&gt;). You shouldn't need too many such registers to faithfully represent even reasonably large non-negative numbers (the registers can uniquely determine any integer from 1 to 2&lt;sup&gt;2&lt;/sup&gt;&amp;nbsp;&amp;times;&amp;nbsp;3&lt;sup&gt;2&lt;/sup&gt;&amp;nbsp;&amp;times;&amp;nbsp;5&lt;sup&gt;2&lt;/sup&gt;&amp;nbsp;&amp;times;&amp;nbsp;...&amp;nbsp;&amp;times;&amp;nbsp;P&lt;sup&gt;2&lt;/sup&gt;).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Whenever you wish to increment R, increment each of the registers r[p] as well. If R is square-free, none of these registers will be zero modulo the square of the appropriate prime. For sufficiently small integers N, you can even characterize N as being square-free if none of these residues are zero. Put another way, the more registers r[p] you maintain, the larger the range of square-free numbers you can automatically detect using these registers.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Suppose that you want to test for square-freeness up to some upper limit N. What do you need to test square-freeness using nothing but registers such as I've described? What you need is for any composite number less than N to have a prime factor from the list of registers that you maintain; that is, you need registers for each of the primes up to &amp;radic;N.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're going to iterate through all integers anyway, you can discover the list of primes that you need to characterize square-freeness at the same time by the Sieve of Erasthotenes, and construct the list of residues modulo squares-of-primes as you go. Any time you find a new prime P, so long as P&lt;sup&gt;2&lt;/sup&gt;&amp;nbsp;&amp;lt;&amp;nbsp;N, add P to the list of primes for which you maintain registers and initialize a register for residues modulo P&lt;sup&gt;2&lt;/sup&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As Ross suggests in his answer, you can use results on the distribution of square-free numbers to obtain an upper bound for the N&lt;sup&gt;th&lt;/sup&gt; square-free number (it should grow more slowly than N&amp;nbsp;ln(N) or so, in any case). This gives you an upper bound on the number of registers that you have to maintain as you search for square-free integers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From this, you should actually be able to show a polynomial-time asymptotic upper bound on the runtime of your procedure.&lt;/p&gt;&#xA;" OwnerUserId="439" LastActivityDate="2011-02-05T16:04:28.480" />
  <row Id="20542" PostTypeId="2" ParentId="20398" CreationDate="2011-02-05T16:05:45.950" Score="-1" ViewCount="" Body="&lt;p&gt;As mentioned in the responding question, learn spectral sequences and use the one resulting from the derived tensor product.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Do you already know spectral sequences?  If not, Weibel's introduction to homological algebra is a reasonable place to start. (Your goal would be to understand Section 5.6.) Some people like the start of Bott and Tu's Differential Forms in Algebraic Topology.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) You know spectral sequences? Are you sure? I think you should rephrase your question &quot;how do I learn about spectral sequences&quot;.&lt;/p&gt;&#xA;" OwnerUserId="6665" LastEditorUserId="6665" LastEditorDisplayName="" LastEditDate="2011-02-05T17:23:23.073" LastActivityDate="2011-02-05T17:23:23.073" />
  <row Id="20543" PostTypeId="2" ParentId="20537" CreationDate="2011-02-05T16:11:54.233" Score="0" ViewCount="" Body="&lt;p&gt;It's a long time since I dabbled with algebra, but my guess is your main problem is finding k to satisfy your condition not knowing m. This is a big if.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And anyway, if I recall correctly, factoring e gives you the answer, so if you first need to find k to satisfy your condition and then find the gcd of k and e, have you not just factored e? If my math is correct then you've just spent more time trying to crack e in a more complicated way. But I'm rusty with algebra, so I may be wrong.&lt;/p&gt;&#xA;" OwnerUserId="6258" LastEditorUserId="6258" LastEditorDisplayName="" LastEditDate="2011-02-05T17:12:00.017" LastActivityDate="2011-02-05T17:12:00.017" CommentCount="1" />
  <row Id="20544" PostTypeId="2" ParentId="20502" CreationDate="2011-02-05T16:23:09.643" Score="1" ViewCount="" Body="&lt;p&gt;Let's suppose you are asking how to divide an interval from $a$ to $b$ with $a&amp;lt;b$ into $n$ equal subintervals.  Then each subinterval has length $\frac{b-a}{n}$ and the endpoints of the intervals are at $$a+j\frac{b-a}{n} = \dfrac{a(n-j)+bj}{n}$$ with $j=0,1,2,...n$. &#xA;When $j=0$ this gives $a$, and when $j=n$ it gives $b$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are you asking something more complicated?&lt;/p&gt;&#xA;" OwnerUserId="6460" LastActivityDate="2011-02-05T16:23:09.643" />
  <row Id="20545" PostTypeId="2" ParentId="20397" CreationDate="2011-02-05T16:25:40.707" Score="8" ViewCount="" Body="&lt;p&gt;High brow: Let $f(\theta)$ be a smooth function from the circle to $\mathbb{R}$. The Fourier coefficients of $f$ are given by $a_n = 1/(2 \pi) \int f(\theta)  e^{-i n \theta} d \theta$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Integrating by parts:&#xA;$$a_n = \frac{1}{n} \frac{i}{2 \pi} \int f&amp;#39;(\theta) e^{- i n \theta} d \theta = \frac{1}{n^2} \frac{-1}{2 \pi} \int f&amp;#39;&amp;#39;(\theta) e^{- i n \theta} d \theta = \cdots$$&#xA;$$\cdots = \frac{1}{n^k} \frac{i^k}{2 \pi} \int f^{(k)}(\theta) e^{- i n \theta} d \theta = O(1/n^k)$$&#xA;for any $k$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, if $f$ is smooth, it Fourier coefficients die off faster than $1/n^k$ for any $k$. More generally, if $f$ has $k$ continuous derivatives, then $a_n = O(1/n^k)$. &lt;/p&gt;&#xA;" OwnerUserId="448" LastActivityDate="2011-02-05T16:25:40.707" CommunityOwnedDate="2011-02-05T16:25:40.707" CommentCount="1" />
  <row Id="20546" PostTypeId="2" ParentId="20539" CreationDate="2011-02-05T16:30:28.580" Score="2" ViewCount="" Body="&lt;p&gt;As PEV hinted, changing coordinates can simplify a problem.  In $\mathbb{R}^2$, if all the action is along a couple axes, it helps to have those aligned with $x$ and $y$.  If you want to find the area of a square by integration, if it is $[0,1]\times [0,1]$ it is much easier than if it is $[(-\sqrt{2}/2,-\sqrt{2}/2) \text{to} (-\sqrt{2}/2,\sqrt{2}/2)]\times[(-\sqrt{2}/2,-\sqrt{2}/2) \text{to} (\sqrt{2}/2,-\sqrt{2}/2)]$  Point charge electrostatics is much easier in spherical coordinates.  There are many more examples like this.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-02-05T16:30:28.580" />
  <row Id="20547" PostTypeId="2" ParentId="20539" CreationDate="2011-02-05T16:32:15.787" Score="4" ViewCount="" Body="&lt;p&gt;Here is an extremely general answer to go with your extremely general question. If you have trouble seeing something because it's hidden behind a tree, generally it's a good idea to move to a different position so that the tree isn't blocking your view. All you've done is apply an affine change of coordinates to your view of the universe. Mathematics is no different: when we want to see something clearly in a problem, we do it in a view that makes it as easy to see as possible.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2011-02-05T16:32:15.787" CommentCount="1" />
  <row Id="20548" PostTypeId="1" AcceptedAnswerId="20549" CreationDate="2011-02-05T16:43:27.760" Score="5" ViewCount="296" Body="&lt;p&gt;After looking at the list of trigonometric identities, I can't seem to find a way to solve this. Is it solvable?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\cos(\theta) + \sin(\theta) = x.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What if I added another equation to the problem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$-\sin(\theta) + \cos(\theta) = y,$$&#xA;where $\theta$ is the same and $y$ is also known?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;OK, so using the linear combinations I was able to whip out:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$a \sin(\theta) + b \cos(\theta) = x = \sqrt{a^2 + b^2} \sin(\theta + \phi),$$&#xA;where $\phi = \arcsin \left( \frac{b}{\sqrt{a^2 + b^2}} \right) = \frac{\pi}{4}$ (as long as $a\geq 0$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Giving me:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x = \sin(\theta + \frac{\pi}{4}) \text{ and } \arcsin(x) - \frac{\pi}{4} = \theta.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All set! Thanks!&lt;/p&gt;&#xA;" OwnerUserId="3158" LastEditorUserId="1269" LastEditorDisplayName="" LastEditDate="2011-02-05T17:19:32.983" LastActivityDate="2011-02-06T00:14:20.527" Title="Solve $\cos(\theta) + \sin(\theta) = x$ for known $x$, unknown $\theta$?" Tags="&lt;trigonometry&gt;" AnswerCount="3" CommentCount="3" />
  <row Id="20549" PostTypeId="2" ParentId="20548" CreationDate="2011-02-05T16:45:13.070" Score="5" ViewCount="" Body="&lt;p&gt;Yeah you can write $\frac{1}{\sqrt{2}}\Bigl[\cos{\theta} + \sin{\theta}\Bigl]$ as $\sin\Bigl(\frac{\pi}{4}+\theta\Bigr)$ and solve for $x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Multiply both sides by $\frac{1}{\sqrt{2}}$ and then try something.&lt;/p&gt;&#xA;" OwnerUserId="9413" LastEditorUserId="9413" LastEditorDisplayName="anonymous" LastEditDate="2011-02-05T16:50:27.883" LastActivityDate="2011-02-05T16:50:27.883" CommentCount="4" />
  <row Id="20550" PostTypeId="1" AcceptedAnswerId="20554" CreationDate="2011-02-05T16:55:14.223" Score="1" ViewCount="77" Body="&lt;p&gt;Hi.&#xA;i'm studing probabilistic algorithms: the ones that - with a great gain in efficency - sometimes could return a false response.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They return the true response with a probability of $\frac{3}{4}$.&#xA;The the way to use them to lower the error probability is: runs $t$ times, compare results and gets the more frequent answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Those algos (that are for decision problems) can respond True or False, so two discrete random variables, and the binomial probability is used. Chernoff's inequality is important to get the number of $t$ for gain the desired fixed error.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, in my book there is this foruma:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$prob \, error &amp;lt;= e^{-2t(x-p)^2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$p = \frac{3}{4}$ because is the probability of true response&lt;/li&gt;&#xA;&lt;li&gt;$x = \frac{1}{2}$ because i want that the response is True for more of the half of $t$ indipendent executions&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;i use&#xA;$prob \, error &amp;lt;= e^{-\frac{t}{8}}$&#xA;to get $t$ for a fixed $prob error$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;now my question is.. what exactly are $x$ and $p$ in more formal way?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;sorry for my bad english, i hope it's clear enough to understand&lt;/p&gt;&#xA;" OwnerUserId="5046" LastActivityDate="2011-02-07T11:51:41.917" Title="clarification on Chernoff's inequality" Tags="&lt;probability&gt;&lt;algorithms&gt;" AnswerCount="1" />
  <row Id="20551" PostTypeId="1" CreationDate="2011-02-05T17:01:47.177" Score="2" ViewCount="140" Body="&lt;p&gt;I am teaching a class in Number Theory and using a new textbook by James Pommersheim. &#xA;There is a proof in the first chapter that states:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let m and n be integers. If both m and n are odd, then mn is odd.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They then prove this by saying m=2q+1 and n=2r+1. There's a simple foil, and then you can factor and find that mn=2(something)+1 which means it's odd, thus proving the statement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then under the indirect proof section they make another statement:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let m and n be integers. Prove that if mn is odd, then both m and n are odd.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They then prove this by contradiction, saying m=2k. Then multiply both sides of the equation by n, giving mn=2kn, which states mn is even, which contradicts the first statement. So, therefore, without loss of generality, both m and n must be odd.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My students had issue with the fact that they couldn't just use the proof for the first statement to prove the second statement. My understanding is that the two statements are converse of one another, they are both true, but need to be proven in different ways. In the first one, you start with m and n, in the second one you start with mn. I'd like to give a more formal explanation for why we have a different method for proving the converse statement (using contradiction indirectly), but I could only just say &quot;because it is the converse you have to go about it indirectly. But I'm not exactly sure &lt;/p&gt;&#xA;" OwnerUserId="6668" LastActivityDate="2011-02-05T17:48:54.347" Title="Can this be proved both directly and indirectly?" Tags="&lt;proof-strategy&gt;" AnswerCount="3" />
  <row Id="20552" PostTypeId="2" ParentId="20551" CreationDate="2011-02-05T17:11:06.067" Score="6" ViewCount="" Body="&lt;p&gt;Statements are simply not equivalent to their converses. To formally explain this, all you need to do is provide a counterexample. For example, the statement&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If I break my leg, then I won't go to work tomorrow&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;is not equivalent to the statement&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If I'm not at work tomorrow, then I broke my leg.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There are plenty of other reasons not to go to work tomorrow (e.g. you missed your bus, you're sick, you really didn't want to).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case, the two statements you're relating happen to be equivalent, but you have to &lt;em&gt;prove&lt;/em&gt; this. Here is a very similar-looking situation where we do not have equivalence: it's true that&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If m and n are even then mn is even&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;but it is &lt;em&gt;not&lt;/em&gt; true that&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If mn is even then m and n are even.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I just reread your question and noticed that you are really asking &lt;em&gt;two&lt;/em&gt; questions, and possibly I have misread one of them. The other one is &quot;why do we need contradiction to prove the second statement?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We don't. As Divakaran says, we can take the contrapositive instead, and proving a statement is equivalent to proving its contrapositive. The difference between a proof by contradiction and a proof by contrapositive is somewhat subtle. When you want to prove that $P \Rightarrow Q$ by contradiction, you assume $P$ and $\neg Q$ and try to derive a contradiction. When you prove it by contrapositive, you assume $\neg Q$ and try to derive $\neg P$. Note that you cannot assume $P$ if you do this.&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2011-02-05T17:48:54.347" LastActivityDate="2011-02-05T17:48:54.347" />
  <row Id="20553" PostTypeId="2" ParentId="20551" CreationDate="2011-02-05T17:12:36.203" Score="3" ViewCount="" Body="&lt;p&gt;Consider the analog when you replace &quot;odd&quot; with &quot;even&quot;: if $m,n$ are even then $mn$ is even, but it's not true that if $mn$ is even then $m,n$ are even.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first statement is a simple one: given two numbers with given structure, I can tell you what you get if you multiply them. That's not very surprising. What the second statement is saying is more complicated: the only way to get to an odd number by multiplication is if you multiply two &lt;i&gt;odd&lt;/i&gt; numbers. This is much more surprising, and a deeper fact. To prove that fact, you &quot;go over all possibilities&quot; of the product of two numbers, and show that only &quot;odd times odd&quot; gives you an odd number.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2011-02-05T17:12:36.203" />
  <row Id="20554" PostTypeId="2" ParentId="20550" CreationDate="2011-02-05T17:25:14.197" Score="3" ViewCount="" Body="&lt;p&gt;Chernoff's bound is a &lt;i&gt;large deviation bound&lt;/i&gt;. Consider some binomial experiment $X \sim \mathrm{Bin}{n,p}$; thus $X$ is the sum of $n$ independent coins, each $1$ with probability $p$ and $0$ otherwise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $n$ is large then $X$ will be concentrated around its expectation $np$. Close to the expectation, $X$ behaves very similar to a normal variable with the same expectation and variance, i.e. $\mathcal{N}{np,np(1-p)}$. However, far from the center the probability decays much faster than a normal distribution, and one resorts to large deviation bounds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For any constant $\Delta$, as $n$ grows bigger it becomes exponentially more unlikely that $|X-np| \geq n\Delta$; the quantitative version of this statement is given by Chernoff's bound. Put differently, we have a bound on $$\Pr[X \geq n(p+\Delta)] + \Pr[X \leq n(p-\Delta)];$$ the connection between the parameters $p,\Delta,x$ is that $x = p \pm \Delta$; $x$ is the threshold beyond which it is highly improbable that the average finds itself.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Original answer&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Chernoff's bound is a bound on a probability that $\mathrm{Bin}(n,p)$ trials cross the $nx$ threshold. In your case, the algorithm is successful if the majority of answers are right, which means that the threshold in $n/2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $p &amp;gt; x$, then Chernoff bounds the probability that $\mathrm{Bin}(n,p)$ is smaller than $nx$. (If $p &amp;lt; x$ then replace &quot;smaller than&quot; by &quot;larger than&quot;.)&lt;/p&gt;&#xA;" OwnerUserId="1277" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2011-02-07T11:51:41.917" LastActivityDate="2011-02-07T11:51:41.917" CommentCount="1" />
  <row Id="20555" PostTypeId="1" AcceptedAnswerId="20609" CreationDate="2011-02-05T17:30:26.173" Score="10" ViewCount="2081" Body="&lt;p&gt;Are there any infinite series about which we don't know whether it converges or not? Or are the convergence tests exhaustive, so that in the hands of a competent mathematician any series will eventually be shown to converge or diverge?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: People were kind enough to point out that without imposing restrictions on the terms, it's trivial to find such &quot;open problem&quot; sequences. So, to clarify, what I had in mind were sequences whose terms are composed of &quot;simple&quot; functions, the kind you would find in an introductory calculus text: exponential, factorial, etc.&lt;/p&gt;&#xA;" OwnerUserId="6131" LastEditorUserId="6131" LastEditorDisplayName="" LastEditDate="2011-02-05T21:54:42.263" LastActivityDate="2011-02-05T22:51:39.347" Title="Are there any series whose convergence is unknown?" Tags="&lt;real-analysis&gt;&lt;sequences-and-series&gt;&lt;soft-question&gt;&lt;infinity&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="3" />
  <row Id="20558" PostTypeId="2" ParentId="20551" CreationDate="2011-02-05T17:41:08.283" Score="2" ViewCount="" Body="&lt;p&gt;I would like to add to what Qiaochu Yuan said.  It is easy to see that &quot;If I break my leg, then I won't go to work tomorrow&quot; is equivalent to &quot;If i am at work tomorrow then i didnt break my leg&quot;.  This is what you did in the proof: called proving the contrapositive.  May be this link will explain better: &lt;a href=&quot;http://en.wikipedia.org/wiki/Contraposition&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Contraposition&lt;/a&gt;   &lt;/p&gt;&#xA;" OwnerUserId="2893" LastActivityDate="2011-02-05T17:41:08.283" />
  <row Id="20559" PostTypeId="2" ParentId="20397" CreationDate="2011-02-05T17:46:29.700" Score="18" ViewCount="" Body="&lt;p&gt;My favorite this week, since I learned it just yesterday: $n$ integrations by parts produces $$ \int_0^1 \frac{(-x\log x)^n}{n!}dx = (n+1)^{-(n+1)}.$$ Then summing on $n$ yields $$\int_0^1 x^{-x}\,dx = \sum_{n=1}^\infty n^{-n}.$$ &lt;/p&gt;&#xA;" OwnerUserId="2947" LastActivityDate="2011-02-05T17:46:29.700" CommunityOwnedDate="2011-02-05T17:46:29.700" CommentCount="1" />
  <row Id="20560" PostTypeId="2" ParentId="20548" CreationDate="2011-02-05T18:06:07.437" Score="3" ViewCount="" Body="&lt;p&gt;Linear equations in $\sin \theta $ and $\cos \theta $ can be solved by a&#xA;resolvent quadratic equation, using the two identities (also &lt;a href=&quot;http://math.stackexchange.com/questions/9138/solve-trigonometric-equation-1-m-textcos-alpha-textsin-alpha/9185#9185&quot;&gt;here&lt;/a&gt;):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\cos \theta =\frac{1-\tan ^{2}\frac{\theta }{2}}{1+\tan ^{2}\frac{\theta }{2%&#xD;&#xA;}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sin \theta =\frac{2\tan \frac{\theta }{2}}{1+\tan ^{2}\frac{\theta }{2}}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case, we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\cos \theta +\sin \theta =x\Leftrightarrow \left( 1-\tan ^{2}\dfrac{\theta }{2}\right) +2\tan \dfrac{%&#xD;&#xA;\theta }{2}=x\left( 1+\tan ^{2}\dfrac{\theta }{2}\right) $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Leftrightarrow (x+1)\tan ^{2}\dfrac{\theta }{2}-2\tan \dfrac{\theta }{2}%&#xD;&#xA;+x-1=0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Leftrightarrow \tan \dfrac{\theta }{2}=\dfrac{2\pm \sqrt{4-4(x+1)(x-1)}}{%&#xD;&#xA;2(x+1)}=\dfrac{1\pm \sqrt{2-x^{2}}}{x+1}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Leftrightarrow \theta =2\arctan \dfrac{1\pm \sqrt{2-x^{2}}}{x+1}$.&lt;/p&gt;&#xA;" OwnerUserId="752" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2011-02-06T00:14:20.527" LastActivityDate="2011-02-06T00:14:20.527" />
  <row Id="20561" PostTypeId="1" CreationDate="2011-02-05T18:07:18.090" Score="1" ViewCount="78" Body="&lt;p&gt;Hello, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a vecotr like a = [123;456;789] and I want to get the value of the second item as scalar. If I try a[2] I get [456] but I'd like to get the scalar value without the matrix brackets. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for your help!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;best regards&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Simon&lt;/p&gt;&#xA;" OwnerUserId="6669" LastActivityDate="2011-02-05T18:07:18.090" ClosedDate="2011-02-05T18:55:02.840" Title="TI89 access matrix item as scalar" Tags="&lt;matrices&gt;&lt;calculator&gt;" CommentCount="3" />
  <row Id="20562" PostTypeId="2" ParentId="20537" CreationDate="2011-02-05T18:20:47.113" Score="1" ViewCount="" Body="&lt;p&gt;As mentioned previously, finding such a $k$ may be rather tough. Suppose we found one, though, and furthermore suppose $k \le \varphi(n)$ (which makes sense since $n$ is large). Note that $m^k \equiv 1 \equiv m^{\varphi(n)} \mod n$; this means $k$ divides $\varphi(n)$. Since $k \mid \varphi(n)$ and $\gcd(e, \varphi(n)) = 1$, this means $gcd(e, k)$ would have to be one as well because $k$ is a divisor of $\varphi(n)$.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Update, after @Nudan changed question&lt;/em&gt;: note first that $n$ is always given, as is $e$; hence you'd actually have both at your disposal, not just $n$. If you have absolutely no access to the decryption process, the best bet may be to factor $n$ into $p$ and $q$, which should (hopefully) take a long time (provided the person implementing the cryptosystem involved has made smart choices). There are a number of attacks that rely on more information/access than this scenario, though, like chosen plaintext or chosen ciphertext attacks. See, e.g., the &quot;Attacks agains plain RSA&quot; portion of &lt;a href=&quot;http://en.wikipedia.org/wiki/RSA&quot; rel=&quot;nofollow&quot;&gt;Wikipedia's entry&lt;/a&gt; on RSA, or the &quot;Security and Practical Concerns&quot; section. For a more thorough/mathematical treatment, see section 8.2 (specifically 8.2.2, &quot;Security of RSA&quot;) of &lt;a href=&quot;http://www.amazon.com/exec/obidos/ISBN=0849385237/7181-7381933-595174&quot; rel=&quot;nofollow&quot;&gt;Handbook of Applied Cryptography&lt;/a&gt;. Sample chapters are available from the authors &lt;a href=&quot;http://www.cacr.math.uwaterloo.ca/hac/&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="4689" LastEditorUserId="4689" LastEditorDisplayName="" LastEditDate="2011-02-05T20:54:18.120" LastActivityDate="2011-02-05T20:54:18.120" CommentCount="0" />
  <row Id="20563" PostTypeId="1" AcceptedAnswerId="20568" CreationDate="2011-02-05T18:21:20.713" Score="5" ViewCount="68" Body="&lt;p&gt;I think the title pretty much says it all. I'm getting confused in the subtle parts of a proof, and would appreciate some help.&lt;/p&gt;&#xA;" OwnerUserId="6670" LastActivityDate="2011-02-05T18:58:25.910" Title="Does $\mathbb{Q}_p \cap \overline{\mathbb{Q}}$ depend on the embedding of $\overline{\mathbb{Q}}$ into $\overline{\mathbb{Q}_p}$?" Tags="&lt;number-theory&gt;&lt;p-adic-number-theory&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="20564" PostTypeId="1" CreationDate="2011-02-05T18:34:00.133" Score="13" ViewCount="421" Body="&lt;p&gt;As one can notice every integer greater than $1$ is a sum of two squarefree numbers.(numbers that are not divided by some prime square  power) .Can we prove that?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit :Can we have bounds for the length of these numbers?( meaning the number of the primes that divide it)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Chenn's theorem asserts that for large enough even  numbers the length (2,1) is enough Goldbach's conjecture says that (1,1) would be enough too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and one cojecture:every odd number can be written as a sum of two squarefree numbers of length at most (2,1) (meaning as a sum of a prime and a double of a prime or a sum of a prime plus 2 or as a sum of 1 plus a double of a prime) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Questions&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;do i really need the prime plus 2 or the 1 plus the double of a prime in oredr to have all the odd numbers?I think i do not need them but can we prove that??&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;What is the relation of this conjecture to Goldbach's conjecture? does the one implies the other?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;EDIT Searching wikipedia i realised that this is a well known conjecture ,for more details see &lt;a href=&quot;http://en.wikipedia.org/wiki/Lemoine%27s_conjecture&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Lemoine%27s_conjecture&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5010" LastEditorUserId="5010" LastEditorDisplayName="" LastEditDate="2011-02-08T12:18:03.223" LastActivityDate="2011-02-08T12:18:03.223" Title="sums of square free numbers , is this conjecture equivalent to goldbach's conjecture?" Tags="&lt;number-theory&gt;" AnswerCount="2" />
  <row Id="20565" PostTypeId="2" ParentId="20529" CreationDate="2011-02-05T18:35:02.263" Score="4" ViewCount="" Body="&lt;p&gt;You have to us the Inclusion-Exclusion principle: suppose you want to find the number of square free numbers up to $N$, then from $N$ you have to substract the number of integers divisible by the square of a prime, but then you have to add any multiple of the square of the product of two discinct primes and so on, in formulas the number looked for is&#xA;$$ N - \sum_{p^2 \le N} \left\lfloor\frac{N}{p^2}\right\rfloor + \sum_{p^2q^2 \le N} \left\lfloor\frac{N}{p^2q^2}\right\rfloor - \sum_{p^2q^2r^2 \le N} \left\lfloor\frac{N}{p^2q^2r^2}\right\rfloor + \cdots -\cdots $$&#xA;using the moebius $\mu$ function you can write this last formula as&#xA;$$ \sum_{n \le \sqrt{N}} \mu(n) \left\lfloor\frac{N}{n^2}\right\rfloor $$&#xA;I don't know how to write this in mathematica but it should take a negligible fraction of the time it takes your current method. &lt;/p&gt;&#xA;" OwnerUserId="3274" LastActivityDate="2011-02-05T18:35:02.263" CommentCount="4" />
  <row Id="20566" PostTypeId="1" AcceptedAnswerId="20600" CreationDate="2011-02-05T18:46:31.117" Score="4" ViewCount="387" Body="&lt;p&gt;Assume that a proof that pi is normal existed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it then possible that starting at some finite position x in pi, from there on every p(n)'th digit is 0, where p(n) is the n'th prime?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know probability arguments say no, but do they also &lt;em&gt;prove&lt;/em&gt; that it is impossible?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a way to disprove the statement?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, the digits of Pi can tell us certain mathematical truths, for instance that the circumference of a circle is less then any other shape with same area.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is, is there a limit to the information one can extract from the digits of Pi?&#xA;Is it possible to reconstruct all theorems from the digits of Pi?&#xA;What sort of truths can be extracted from the digits of Pi?&#xA;Is entire math and all truths somehow encoded within the digits of Pi ?&lt;/p&gt;&#xA;" OwnerUserId="4929" LastEditorUserId="4929" LastEditorDisplayName="" LastEditDate="2011-02-05T20:42:13.950" LastActivityDate="2011-02-24T15:30:01.067" Title="Prove there are no hidden messages in Pi" Tags="&lt;probability&gt;&lt;logic&gt;&lt;decimal-expansion&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="20567" PostTypeId="1" AcceptedAnswerId="20650" CreationDate="2011-02-05T18:55:01.533" Score="3" ViewCount="199" Body="&lt;p&gt;Per now, I have basically come upon proofs of the irrationality of $\sqrt{2}$ (and so on) and the proof of the irrationality of $e$. However, both proofs were by contradiction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When thinking about it, it seems like the defintion of irrationality itself demands proofs by contradiction. An irrational number is a number that is &lt;strong&gt;not&lt;/strong&gt; a rational number. It seems then that if we were to find direct irrationality proofs, this would rely on some equivalent definition of irrational numbers, not involving rational numbers themselves. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there any irrationality proofs not using contradiction?&lt;/p&gt;&#xA;" OwnerUserId="4284" LastActivityDate="2011-02-06T05:36:26.830" Title="Irrationality proofs not by contradiction" Tags="&lt;irrational-numbers&gt;" AnswerCount="3" CommentCount="7" />
  <row Id="20568" PostTypeId="2" ParentId="20563" CreationDate="2011-02-05T18:58:25.910" Score="5" ViewCount="" Body="&lt;p&gt;Once you fix an algebraic closure $\overline{\mathbb{Q}_p}$, the subfield $\overline{\mathbb{Q}}$ is uniquely determined as the smallest algebraically closed subfield containing the prime field, so all embeddings have the same image.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2011-02-05T18:58:25.910" />
  <row Id="20569" PostTypeId="1" AcceptedAnswerId="20571" CreationDate="2011-02-05T19:14:25.507" Score="1" ViewCount="104" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm trying to factorise a 20-term polynomial of the 4th degree with four variables. Ideally, I'd like to do it by hand, but I get the idea that this is pretty improbable. Can anyone point me to some helpful resources on this subject?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="6672" LastEditorUserId="3731" LastEditorDisplayName="" LastEditDate="2011-02-05T21:01:53.387" LastActivityDate="2011-02-05T22:10:15.620" Title="Factoring a complicated 20-term polynomial" Tags="&lt;polynomials&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="20570" PostTypeId="1" CreationDate="2011-02-05T19:14:34.997" Score="2" ViewCount="80" Body="&lt;p&gt;Hello,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have three points on a map: origin, position, and destination. For each point, I have a latitude and longitude. The position point includes the lat/long plus a heading. The heading is in degrees between $0^{\circ}$ and $360^{\circ}$. $0^{\circ}$ and $360^{\circ}$ both represent absolute North.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to draw an arrow that points from towards the destination from the current position, considering the heading. How do I determine the degree to draw the line from and two. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Essentially, I'm trying to figure out how to build a compass where the destination is always north.&lt;/p&gt;&#xA;" OwnerUserId="6673" LastEditorUserId="2736" LastEditorDisplayName="" LastEditDate="2011-02-05T19:27:18.297" LastActivityDate="2011-03-07T23:39:38.543" Title="How do I determine a direction?" Tags="&lt;coordinate-systems&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="20571" PostTypeId="2" ParentId="20569" CreationDate="2011-02-05T19:44:06.307" Score="1" ViewCount="" Body="&lt;p&gt;Try &lt;a href=&quot;http://www.wolframalpha.com/&quot; rel=&quot;nofollow&quot;&gt;Wolfram Alpha&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="589" LastActivityDate="2011-02-05T19:44:06.307" CommentCount="1" />
  <row Id="20572" PostTypeId="2" ParentId="20519" CreationDate="2011-02-05T19:46:38.100" Score="3" ViewCount="" Body="&lt;p&gt;Almost a hundred years ago G.H. Hardy showed that there are power series over $\mathbb{C}$ that converge uniformly on the closed unit disk but not absolutely on the boundary, which is the same as saying that the hypothesis of the Weierstrass M-test doesn't hold (A theorem concerning Taylor's series, Quart. J. Pure Appl. Math. 44 (1913), 147-160).  I posted some references including links to examples &lt;a href=&quot;http://mathoverflow.net/questions/12968/mode-of-convergence-of-a-power-series&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;. &lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2011-02-05T19:46:38.100" CommentCount="1" />
  <row Id="20573" PostTypeId="1" CreationDate="2011-02-05T19:47:53.513" Score="3" ViewCount="86" Body="&lt;p&gt;Hi, I'm searching for a good book or web page that has a good amount of problems and their solutions, at undergraduate level, of divisibility, inequalities, induction, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance&lt;/p&gt;&#xA;" OwnerUserId="6675" LastEditorUserId="3731" LastEditorDisplayName="" LastEditDate="2011-02-05T20:50:51.110" LastActivityDate="2011-02-05T20:50:51.110" Title="Good resource of maths problems with solutions" Tags="&lt;reference-request&gt;&lt;inequality&gt;&lt;induction&gt;&lt;divisibility&gt;&lt;online-resources&gt;" AnswerCount="1" />
  <row Id="20574" PostTypeId="1" AcceptedAnswerId="20576" CreationDate="2011-02-05T20:05:46.870" Score="12" ViewCount="145" Body="&lt;p&gt;I begin losing intuition when I start dealing with infinitely generated fields over $\mathbb{Q}$...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The naive guess is that it is noetherian of krull dimension 1. Is this correct?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A related question is what sort of morphism is $Spec($the integral closure of $\mathbb{Z}$ in $\overline{\mathbb{Q}}) \rightarrow Spec(O_K)$ for some number field $K$? For example, is it a flat morphism?&lt;/p&gt;&#xA;" OwnerUserId="6670" LastActivityDate="2011-03-19T06:01:00.750" Title="What sort of ring is the integral closure of $\mathbb{Z}$ in $\overline{\mathbb{Q}}$?" Tags="&lt;number-theory&gt;&lt;algebraic-number-theory&gt;" AnswerCount="3" FavoriteCount="1" />
  <row Id="20575" PostTypeId="2" ParentId="20574" CreationDate="2011-02-05T20:10:28.167" Score="6" ViewCount="" Body="&lt;p&gt;No, it is not noetherian. The ideals generated by $2^{1/2^n}, n = 1, 2, \dots$ form a strictly ascending chain that does not stabilize. It is of Krull dimension one (every nonzero prime ideal is maximal) by lying over and going up.&lt;/p&gt;&#xA;" OwnerUserId="536" LastActivityDate="2011-02-05T20:10:28.167" />
  <row Id="20576" PostTypeId="2" ParentId="20574" CreationDate="2011-02-05T20:13:46.037" Score="11" ViewCount="" Body="&lt;p&gt;This is just the ring of algebraic integers, which is a &lt;a href=&quot;http://en.wikipedia.org/wiki/B%C3%A9zout_domain&quot; rel=&quot;nofollow&quot;&gt;Bézout domain&lt;/a&gt;. That means that every finitely generated ideal is principal. As there do exist non-finitely generated ideals which are not principal, it is not a Noetherian domain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, consider the ideal generated by $\{2^{1/n}\colon n\ge1\}$. As no generating set is contained in any finite extension of $\mathbb{Q}$, it is not finitely generated. On the other hand, any quotient by a nonzero prime ideal $\mathfrak{p}$ gives an algebraic extension of $\mathbb{Z}/\mathfrak{p}\cap\mathbb{Z}$, so is a field, and $\mathfrak{p}$ is maximal. So it does have Krull dimension 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other common example of a Bézout domian which is not Noetherian is the ring of entire functions on the complex plane.&lt;/p&gt;&#xA;" OwnerUserId="1321" LastEditorUserId="1321" LastEditorDisplayName="" LastEditDate="2011-02-05T20:30:29.193" LastActivityDate="2011-02-05T20:30:29.193" />
  <row Id="20577" PostTypeId="2" ParentId="20573" CreationDate="2011-02-05T20:22:59.570" Score="2" ViewCount="" Body="&lt;p&gt;&lt;em&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0387982191&quot; rel=&quot;nofollow&quot;&gt;Problem Solving Strategies&lt;/a&gt;&lt;/em&gt; by Arthur Engel. &lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2011-02-05T20:22:59.570" CommentCount="1" />
  <row Id="20578" PostTypeId="1" AcceptedAnswerId="20636" CreationDate="2011-02-05T20:38:55.593" Score="14" ViewCount="627" Body="&lt;p&gt;If a function is a combination of other functions whose derivatives are known via composition, addition, etc., the derivative can be calculated using the chain rule and the like. But even the product of integrals can't be expressed in general in terms of the integral of the products, and forget about composition! Why is this?&lt;/p&gt;&#xA;" OwnerUserId="6676" LastActivityDate="2011-04-20T11:00:40.927" Title="Why is integration so much harder than differentiation?" Tags="&lt;calculus&gt;" AnswerCount="5" CommentCount="10" FavoriteCount="4" />
  <row Id="20579" PostTypeId="1" AcceptedAnswerId="20594" CreationDate="2011-02-05T20:49:11.613" Score="4" ViewCount="168" Body="&lt;p&gt;If W is a subspace, is it also a linear space? If V is a linear space, is it also a subspace? I am having trouble wrapping my head around the difference between the two, as it seems that the way the book defines them is the following: both have to have a zero (neutral) element, both are closed under addition and scalar multiplication. Thanks!&lt;/p&gt;&#xA;" OwnerUserId="5157" LastActivityDate="2011-02-05T21:19:17.633" Title="What is the difference between linear space and a subspace?" Tags="&lt;linear-algebra&gt;" AnswerCount="3" />
  <row Id="20580" PostTypeId="2" ParentId="20578" CreationDate="2011-02-05T20:53:10.863" Score="3" ViewCount="" Body="&lt;p&gt;Think about what the two represent. If you have some function (it need not be &lt;em&gt;too&lt;/em&gt; complicated) such as $f(x) = x^2 \sin(x)$, what does $\frac{df}{dx}(3)$ and $\int_0^3 f(s) ds$ mean?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose you don't know any theory, and you are given the graph of the function, which will be easier to find?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then it's not surprising that you can calculate the derivative of complicated functions whereas the integral might not even have a closed form.&lt;/p&gt;&#xA;" OwnerUserId="5105" LastActivityDate="2011-02-05T20:53:10.863" />
  <row Id="20581" PostTypeId="2" ParentId="20564" CreationDate="2011-02-05T20:55:27.713" Score="5" ViewCount="" Body="&lt;p&gt;A majority of numbers are squarefree.  More precisely, you can show that there is a constant $c&amp;lt;\frac{1}{2}$ such that for each positive integer $n$, less than $cn$ of the numbers $1$ through $n$ are nonsquarefree.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One way to get a rough bound for $c$ is to observe that at most $n/4$ of the numbers from $1$ to $n$ are divisible by $4$, at most $n/9$ divisible by $9$, at most $n/25$ divisible by $25$, etc.  If we add up all these inequalities, there are less than $(1/4+1/9+1/25+\cdots)n\lt .46n$ nonsquarefree numbers from $1$ to $n$.  The sum $\displaystyle{\sum_{p\text{ prime}}\frac{1}{p^2}=.4522474200410654985065...}$ is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Prime_zeta_function&quot; rel=&quot;nofollow&quot;&gt;prime zeta function&lt;/a&gt; &lt;a href=&quot;http://oeis.org/A085548&quot; rel=&quot;nofollow&quot;&gt;evaluated&lt;/a&gt; at $2$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you can take $c=.46$.  For $n\geq 13$, $\lfloor n/2\rfloor \gt .46n$, so not all of the $\lfloor n/2\rfloor$ pairwise disjoint sets $\{1,n-1\},\{2,n-2\},\ldots,\{\lfloor n/2\rfloor,n-\lfloor n/2\rfloor\}$ can contain a nonsquarefree number.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2011-02-06T04:05:20.617" LastActivityDate="2011-02-06T04:05:20.617" CommentCount="7" />
  <row Id="20582" PostTypeId="2" ParentId="20579" CreationDate="2011-02-05T20:56:43.183" Score="0" ViewCount="" Body="&lt;p&gt;Yes, a subspace is a space in itself. The conditions required for a subspace $+$ the conditions that the subspace &lt;em&gt;inherits&lt;/em&gt; from the space make it a space in it's own right. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You should check &lt;a href=&quot;http://en.wikipedia.org/wiki/Linear_subspace&quot; rel=&quot;nofollow&quot;&gt;here.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why are they important? A subspace &lt;em&gt;lives&lt;/em&gt; inside a space. The fact that it has in itself all the nice properties of a space make it interesting. For example, if you have a space $V$ and your subspace $W$ is generated by linearly independent vectors $v_1, v_2, ... , v_k$, then any vector that is a linear combination of these $k$ vectors also belongs to the same subspace. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Sorry about those comments, pressed enter by accident)&lt;/p&gt;&#xA;" OwnerUserId="5105" LastEditorUserId="5105" LastEditorDisplayName="" LastEditDate="2011-02-05T21:14:33.460" LastActivityDate="2011-02-05T21:14:33.460" CommentCount="4" />
  <row Id="20583" PostTypeId="1" AcceptedAnswerId="20587" CreationDate="2011-02-05T20:57:54.297" Score="4" ViewCount="236" Body="&lt;p&gt;The following expression:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\sqrt{4+h}-2}{h}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;should be simplified to:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{1}{\sqrt{4+h}+2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(even if I don't agree that this second is more simple than the first).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is that I have no idea of the first step to simplify that.. any help?&lt;/p&gt;&#xA;" OwnerUserId="4501" LastEditorUserId="4501" LastEditorDisplayName="" LastEditDate="2011-02-06T13:17:40.047" LastActivityDate="2011-02-06T13:17:40.047" Title="How to simplify $\frac{\sqrt{4+h}-2}{h}$" Tags="&lt;algebra&gt;&lt;polynomials&gt;" AnswerCount="4" CommentCount="2" />
  <row Id="20584" PostTypeId="2" ParentId="20530" CreationDate="2011-02-05T20:58:37.587" Score="1" ViewCount="" Body="&lt;p&gt;Any finite set of words is recognized by a finite state automaton (i.e., a regular expression) and therefore does not need the expressive power of a CFG. You can build a deterministic or non-deterministic automaton and &lt;a href=&quot;http://en.wikipedia.org/wiki/Finite-state_machine&quot; rel=&quot;nofollow&quot;&gt;minimize it&lt;/a&gt;. To do this programmatically, use a Lex tool from the Lex/Yacc tool family. The size of the automaton (typically the number of states is used) is a measure for the complexity of the set.&lt;/p&gt;&#xA;" OwnerUserId="6678" LastEditorUserId="6678" LastEditorDisplayName="" LastEditDate="2011-02-05T21:14:05.190" LastActivityDate="2011-02-05T21:14:05.190" CommentCount="1" />
  <row Id="20585" PostTypeId="2" ParentId="20357" CreationDate="2011-02-05T20:59:53.137" Score="0" ViewCount="" Body="&lt;p&gt;I'm assuming you're JuanOS from MO, and this question corresponds to the MO question: &lt;a href=&quot;http://mathoverflow.net/questions/54278/orientation-of-a-glued-manifold&quot; rel=&quot;nofollow&quot;&gt;http://mathoverflow.net/questions/54278/orientation-of-a-glued-manifold&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's how I interpret your question.  You have an oriented manifold $C$ which is compact without boundary, and you've decomposed it into the union of two submanifolds $B$ and $B&amp;#39;$ with $B \cap B&amp;#39; = M$, $M$ a compact manifold. Let $n=dim(C)$, so $n-1=dim(M)$.  The global orientation class for $C$ is a generator $\mu_C \in H_n C$.  There are restriction maps $H_n C \to H_n(B,M)$ and $H_n C \to H_n(B&amp;#39;,M)$ which give the corresponding global orientations $\mu_B$ for $B$ and $\mu_{B&amp;#39;}$ for $B&amp;#39;$ respectively.   Then there is the pairs $M \to B \to (B,M)$ and $M \to B&amp;#39; \to (B&amp;#39;,M)$ and you want to know how the two generators of $H_n(B,M)$ and $H_n(B&amp;#39;,M)$ compare when mapped to elements of $H_{n-1}M$ via the two connecting maps for the above pairs, specifically you want to show that $\partial \mu_{B} + \partial \mu_{B&amp;#39;} = 0$.  Moreover, you want an argument that's fairly generic, in particular not specific to triangulated smooth manifolds or anything like that. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The above &quot;restriction maps&quot; are formally induced by inclusion $C \to (C, C \setminus int(B&amp;#39;) ) \leftarrow (B,M)$, one being an excision inclusion, the other just an inclusion. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't believe this is as complicated as Kuperberg makes out -- the complication comes when attempting to bridge the gap between the smooth or simplicial views with the strictly homological view, especially say in the singular homology setting.  But the above formulation side-steps those complications as your question is phrased entirely in terms of Mayer-Vietoris type constructions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Okay, so here's a cheap way to check that it's true.  Since $M$ is a submanifold of $C$, given a point $p \in M$ you can find an orientation-preserving degree 1 map $f : C \to S^n$ such that $f(M) \subset S^{n-1} \times \{0\} \subset S^n$. Moreover, you can ensure $f$ when restricted to $M$ is an isomorphism on the top homology groups of $M$ and $S^{n-1}$ respectively, and that $f$ sends $B$ to the top hemi-sphere, and $B&amp;#39;$ to the bottom hemi-sphere.  So by naturality, you've reduced your problem to the case $D^n \sqcup_{S^{n-1}} D^n = S^n$, i.e. $C= S^n$, and $B$ and $B&amp;#39;$ both discs, which one way or another boils down to a cellular homology computation (we are using singular homology but what I'm saying is this is singular homology of a CW-complex to effectively cellular homology).  This is the equivalent step to using outward pointing normals and determinants for smooth manifolds. &lt;/p&gt;&#xA;" OwnerUserId="642" LastActivityDate="2011-02-05T20:59:53.137" CommentCount="2" />
  <row Id="20586" PostTypeId="2" ParentId="20583" CreationDate="2011-02-05T21:00:03.233" Score="-2" ViewCount="" Body="&lt;p&gt;So its not about simplification. You just want to show they are equal. What you do is put an equal sign between them, and cancel everything you can, if you get 1=1 or similar you are done.&lt;/p&gt;&#xA;" OwnerUserId="4929" LastActivityDate="2011-02-05T21:00:03.233" CommentCount="4" />
  <row Id="20587" PostTypeId="2" ParentId="20583" CreationDate="2011-02-05T21:03:14.677" Score="10" ViewCount="" Body="&lt;p&gt;If you multiply both the top and the bottom by $\sqrt{4+h}+2$, you get $\frac{(\sqrt{4+h}-2)(\sqrt{4+h}+2)}{h(\sqrt{4+h}+2)}$, which simplifies to $\frac{h}{h(\sqrt{4+h}+2)}$. Then, divide both by $h$ (assuming $h\neq 0$), and you get $\frac{1}{\sqrt{4+h}+2}$.&lt;/p&gt;&#xA;" OwnerUserId="3731" LastEditorUserId="3731" LastEditorDisplayName="" LastEditDate="2011-02-05T21:08:23.220" LastActivityDate="2011-02-05T21:08:23.220" CommentCount="1" />
  <row Id="20588" PostTypeId="2" ParentId="20583" CreationDate="2011-02-05T21:07:01.193" Score="3" ViewCount="" Body="&lt;p&gt;It is really simple. Let us just do what is most intuitive, multiply numerator and denominator with what you want to have in denominator. You get: $$ \frac{(\sqrt{4+h} - 2)(\sqrt{4+h} + 2)}{h(\sqrt{4+h}+2)} $$&#xA;Then observe the numerator has a difference of squares. Multiply the numerator easily using that and then your left with&#xA;$$\frac{h}{h(\sqrt{4+h}+2)}$$&#xA;Just assume $ h \neq 0 $ and get &quot;rid&quot; of it.&lt;/p&gt;&#xA;" OwnerUserId="5501" LastActivityDate="2011-02-05T21:07:01.193" CommentCount="1" />
  <row Id="20589" PostTypeId="2" ParentId="20583" CreationDate="2011-02-05T21:08:00.410" Score="3" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\rm\displaystyle\quad\quad g^2 = 4+h\ \ \Rightarrow\ \ \frac{g-2}h\ =\ \frac{g-2}{g^2-4}\ =\ \frac{1}{g+2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Usually the &quot;simplification&quot; is the opposite inference - known as &lt;a href=&quot;http://www.google.com/search?q=rationalize+denominator&quot; rel=&quot;nofollow&quot;&gt;rationalizing the denominator.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2011-02-05T21:55:27.317" LastActivityDate="2011-02-05T21:55:27.317" CommentCount="2" />
  <row Id="20590" PostTypeId="2" ParentId="20566" CreationDate="2011-02-05T21:08:14.633" Score="1" ViewCount="" Body="&lt;p&gt;(this is a too-long comment)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You're describing a set $S$ of numbers, those for which there is a $k$ with the $(k+p)$-th digit equal to 0 when p is a prime, and asking whether $\pi\in S$.  (The problem doesn't essentially change whether you fix a base b or allow any base.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We don't know much about set $S$.  Finite methods are useless, since by assumption $\pi$ already contains arbitrarily long finite segments that match this description (as well as the Declaration of Independence and this post).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, $S$ is a tricky character.  It is dense in the reals so we can't separate $\pi$ from its members by an $\varepsilon$.  It is uncountable so we can't use something like Roth's theorem to separate it from $\pi$.  We don't really have any methods for working on these sorts of problems.&lt;/p&gt;&#xA;" OwnerUserId="1778" LastEditorUserId="1778" LastEditorDisplayName="" LastEditDate="2011-02-24T15:30:01.067" LastActivityDate="2011-02-24T15:30:01.067" CommentCount="0" />
  <row Id="20591" PostTypeId="1" AcceptedAnswerId="20607" CreationDate="2011-02-05T21:08:54.293" Score="1" ViewCount="73" Body="&lt;p&gt;I was given this question by a friend:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Let $f(x)$ be 2 times differentiable in $R$ and $x_{0}$ a local extremum.  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Show that there are $a, b, c \in R$ such that a function $g(x)=f(x+a)+b$ exists. And $g(x)=cx^2+h(x)$ where $\lim_{x\to 0}\frac{h(x)}{x^2}=0$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I began by writing a first degree Taylor expansion of $f(x)$ around $x_{0}=a$ using Lagrange's remainder:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f(x)=f(a)+\frac{f&amp;#39;&amp;#39;(\xi)}{2}(x-a)^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the above we see that $f(x+a)=f(a)+\frac{f&amp;#39;&amp;#39;(\xi)}{2}x^2$. Now comes the part I'm not sure of: Can I say that $g(x)=\frac{f&amp;#39;&amp;#39;(\xi)}{2}x^2 \implies g(x)=f(x+a)-f(a)$? Is $g(x)$ defined properly? The reason I'm not sure is because the point $\xi$ is different for every $x$.&lt;/p&gt;&#xA;" OwnerUserId="6679" LastEditorUserId="6679" LastEditorDisplayName="" LastEditDate="2011-02-05T21:29:24.917" LastActivityDate="2011-02-05T22:52:12.603" Title="Stuck on Taylor expansion question" Tags="&lt;calculus&gt;&lt;taylor-expansion&gt;" AnswerCount="1" CommentCount="12" />
  <row Id="20592" PostTypeId="2" ParentId="20579" CreationDate="2011-02-05T21:09:57.990" Score="0" ViewCount="" Body="&lt;p&gt;I think some of your confusion my arise from being a bit imprecise in what you are saying. One generally says W is a subspace of V. To mean W is a subset of V containing 0 which is closed under the operations of addition and scalar multiplication which were defined for V.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your second question (If V is a linear space is it also a subspace?) is a bit difficult to answer because you did not specify what V is a subspace of? Trivially, V is a subspace of itself, but I don't think this is what you meant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's consider a geometric example. Consider the vector spaces $\mathbb{R}$ and $\mathbb{R}^{2}$ defined in the usual way. Intuitively, many people want to say $\mathbb{R}$ is a subspace of  $\mathbb{R}^{2}$ but this is technically incorrect. we say instead that $\mathbb{R}$ is isomorphic to any 1-dimensional subspace of $\mathbb{R}^{2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope this helps clarify the two concepts. As to why we bother with the concept of 2 different definitions, we can consider a set of homogeneous linear equations (things like 2x+3y+4z=0, the 0 on the RHS is what makes them homogeneous) in, say $\mathbb{R}^3$ and note that their solution set would form a subspace of $\mathbb{R}^3$&lt;/p&gt;&#xA;" OwnerUserId="249" LastActivityDate="2011-02-05T21:09:57.990" />
  <row Id="20593" PostTypeId="1" AcceptedAnswerId="20596" CreationDate="2011-02-05T21:18:36.527" Score="1" ViewCount="98" Body="&lt;p&gt;I'd like to calculate a standard deviation for a very large (but known) number of sample values, with the highest accuracy possible.  The number of samples is larger than can be efficiently stored in memory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The basic variance formula is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sigma^2 = \frac{1}{N}\sum (x - \mu)^2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;... but this formulation depends on knowing the value of $\mu$ already.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\mu$ can be calculated cumulatively -- that is, you can calculate the mean without storing every sample value.  You just have to store their sum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But to calculate the variance, is it necessary to store every sample value?  Given a stream of samples, can I accumulate a calculation of the variance, without a need for memory of each sample?  Put another way, is there a formulation of the variance which doesn't depend on foreknowledge of the exact value of $\mu$ before the whole sample set has been seen?&lt;/p&gt;&#xA;" OwnerUserId="6677" LastActivityDate="2011-02-05T22:31:42.793" Title="Calculate variance from a stream of sample values" Tags="&lt;statistics&gt;&lt;algorithms&gt;" AnswerCount="1" />
  <row Id="20594" PostTypeId="2" ParentId="20579" CreationDate="2011-02-05T21:19:17.633" Score="3" ViewCount="" Body="&lt;p&gt;The main difference between refering to a vector spaces as a &lt;strong&gt;linear space&lt;/strong&gt; or as a &lt;strong&gt;subspace&lt;/strong&gt; is, unsurprisingly, &lt;em&gt;context.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When one talks about a &quot;subspace&quot;, one is thinking of it as being &quot;inside&quot; &lt;em&gt;another&lt;/em&gt; vector space, with the inherited operations. So, if we refer to the set of all polynomials of degree at most $2$ with coefficients in $\mathbb{R}$ as a &quot;subspace of the vector space of all polynomials&quot; (note that &lt;em&gt;subspace&lt;/em&gt; is relation term; something is a subspace &lt;strong&gt;of something else&lt;/strong&gt;, thought the &quot;something else&quot; may be left implied or understood from context), then we are not merely thinking of this set as a vector space, we are thinking of it as a vector space sitting inside a larger vector space. On the other hand, if we refer to the set as &quot;the vector space of all polynomials of degree at most $2$ with coefficients in $\mathbb{R}$&quot;, then we are not really interested in the fact that it is sitting inside a larger vector space, but we want to consider it as a vector space in and of itself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We do that in set theory all the time, since a set may be a set or a subset of something else. The  natural numbers form a &lt;em&gt;set&lt;/em&gt;, and we refer to &quot;the set of natural numbers&quot; when we are focusing on the natural numbers exclusively; on the other hand, sometimes we want to think about the natural numbers and their relationship with the larger set of rationals or real numbers, so we can talk about &quot;the subset of the real numbers consisting of the natural numbers.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, &lt;em&gt;every&lt;/em&gt; vector space is a subspace of itself; every subspace of a vector space is itself a vector space (with the inherited operations). So whether we call them &quot;space&quot; or &quot;subspace of &lt;code&gt;&amp;lt;something&amp;gt;&lt;/code&gt;&quot; is entirely a question of whether we want to focus on the object itself, or consider it in context with something else.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2011-02-05T21:19:17.633" />
  <row Id="20595" PostTypeId="2" ParentId="20564" CreationDate="2011-02-05T21:33:13.827" Score="4" ViewCount="" Body="&lt;p&gt;It is known that the &lt;a href=&quot;http://en.wikipedia.org/wiki/Schnirelmann_density&quot; rel=&quot;nofollow&quot;&gt;Schnirelmann Density&lt;/a&gt; of square free numbers is $\displaystyle \frac{53}{88}$ for instance, as mentioned here: &lt;a href=&quot;http://www.jstor.org/pss/2040089&quot; rel=&quot;nofollow&quot;&gt;http://www.jstor.org/pss/2040089&lt;/a&gt;. Note this is different from the natural density which is known to be $\displaystyle \frac{6}{\pi^2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This implies that for any $n$, the number of square free numbers $\displaystyle \le n$ is at least $\displaystyle \frac{53n}{88}$. Since $\displaystyle \frac{53}{88} \gt \frac{1}{2}$, this implies that $n+1$ can be written as the sum of &lt;em&gt;exactly two&lt;/em&gt; squarefree numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;btw, your title does not match the question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Additive Basis means that you also include $\displaystyle 0$ in the set which you seem to exclude by saying two instead of at most two.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2011-02-06T00:03:14.493" LastActivityDate="2011-02-06T00:03:14.493" CommentCount="7" />
  <row Id="20596" PostTypeId="2" ParentId="20593" CreationDate="2011-02-05T21:35:04.900" Score="3" ViewCount="" Body="&lt;p&gt;You can keep two running counters - one for $\sum_i x_i$ and another for $\sum_i x_i^2$. Since variance can be written as&#xA;$$ \sigma^2 = \frac{1}{N} \left[ \sum_i x_i^2 - \frac{(\sum_i x_i)^2}{N} \right] $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;you can compute the variance of the data that you have seen thus far with just these two counters. Note that the $N$ here is not the total length of all your samples but only the number of samples you have observed in the past.&lt;/p&gt;&#xA;" OwnerUserId="618" LastEditorUserId="618" LastEditorDisplayName="" LastEditDate="2011-02-05T22:31:42.793" LastActivityDate="2011-02-05T22:31:42.793" CommentCount="3" />
  <row Id="20597" PostTypeId="1" AcceptedAnswerId="20599" CreationDate="2011-02-05T21:46:16.537" Score="2" ViewCount="183" Body="&lt;p&gt;The exercise asks to &quot;Re-write completing the square&quot;: $$x^2+x+1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer is: $$(x+\frac{1}{2})^2+\frac{3}{4}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't even understand what it means with &quot;Re-write completing the square&quot;..&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What's the steps to solve this?&lt;/p&gt;&#xA;" OwnerUserId="4501" LastEditorUserId="4501" LastEditorDisplayName="" LastEditDate="2011-02-06T13:16:11.297" LastActivityDate="2011-02-06T16:06:02.883" Title="How to &quot;Re-write completing the square&quot;: $x^2+x+1$" Tags="&lt;algebra&gt;&lt;polynomials&gt;" AnswerCount="5" CommentCount="1" />
  <row Id="20598" PostTypeId="2" ParentId="20597" CreationDate="2011-02-05T21:58:47.470" Score="3" ViewCount="" Body="&lt;p&gt;&quot;Completing the square&quot; is a standard step in solving a quadratic equation. To see how it helps, consider the following: The general formulation of a quadratic equation is $ax^2+bx+c = 0$ with $a \neq 0$. Let us say we are tasked with solving this equation, i.e., finding values of $x$ that satisfy this equation. To start with, note that this equation is easy to solve if $b=0$. Then the equation looks like $ax^2 + c = 0$ which would simplify to $x = \pm \sqrt{\frac{-c}{a}}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Completing the square&quot; is a step that takes a general quadratic and reduces it to the form of the simple quadratic above. It does this with a substitution $y = x + \frac{b}{2a}$. Then, $ay^2 = ax^2 + bx + \frac{b^2}{4a}$ which means the general quadratic can be written as $ay^2-\frac{b^2}{4a} + c = 0$ or equivalently as&#xA;$$ a(x+\frac{b}{2a})^2 = \frac{b^2}{4a} - c $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This equation is easy to solve and yields the two roots of the general quadratic equation. Note that the above equation is equivalent to the one we started with ($ax^2+bx+c = 0$) for the purposes of finding the roots. This process of rewriting is called completing the square. This is the point behind rewriting $x^2+x+1$ as $(x+\frac{1}{2})^2 + \frac{3}{4}$.&lt;/p&gt;&#xA;" OwnerUserId="618" LastEditorUserId="618" LastEditorDisplayName="" LastEditDate="2011-02-05T22:22:04.330" LastActivityDate="2011-02-05T22:22:04.330" CommentCount="3" />
  <row Id="20599" PostTypeId="2" ParentId="20597" CreationDate="2011-02-05T21:59:35.740" Score="7" ViewCount="" Body="&lt;p&gt;Remember the formula for the square of a binomial:&#xA;$$(a+b)^2 = a^2 + 2ab + b^2.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, when you see $x^2+x+1$, you want to think of $x^2+x$ as the first two terms you get in expanding the binomial $(x+c)^2$ for some $c$; that is,&#xA;$$x^2 + x + \cdots = (x+c)^2.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the middle term should be $2cx$, and you have $x$, that means that you want $2c=1$, or $c=\frac{1}{2}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if you have $(x+\frac{1}{2})^2$, you get $x^2 + x + \frac{1}{4}$. Since all you have is $x^2+x$, you &lt;em&gt;complete the square&lt;/em&gt; by adding the missing &quot;$\frac{1}{4}$&quot;. Since you are not allowed to just add constants willy-nilly, you must also cancel it out by subtracting $\frac{1}{4}$. So:&#xA;\begin{align*}&#xA;x^2 + x + 2 &amp;amp;= (x^2 + x + \cdots) + 1\\&#xA;&amp;amp;=\left( x^2 + 2\left(\frac{1}{2}\right)x + \cdots \right) + 1 &amp;amp;&amp;amp;\mbox{figuring out what $c$ is}\\&#xA;&amp;amp;= \left(x^2 + 2\left(\frac{1}{2}\right)x + \left(\frac{1}{2}\right)^2\right) -\frac{1}{4} + 1 &amp;amp;&amp;amp;\mbox{completing the square}\\&#xA;&amp;amp;= \left(x+\frac{1}{2}\right)^2 + \frac{3}{4}.&#xA;\end{align*}&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-05T22:10:39.427" LastActivityDate="2011-02-05T22:10:39.427" CommentCount="2" />
  <row Id="20600" PostTypeId="2" ParentId="20566" CreationDate="2011-02-05T22:09:39.747" Score="19" ViewCount="" Body="&lt;p&gt;If $\pi$ is normal, then in fact not only does $\pi$ have&#xA;secret messages, but it will contain in its digits every&#xA;possible finite message infinitely often! This is simply&#xA;because it is part of the meaning of &lt;a href=&quot;http://en.wikipedia.org/wiki/Normal_number&quot; rel=&quot;nofollow&quot;&gt;normal&#xA;number&lt;/a&gt; that&#xA;every finite sequence appears with the expected density.&#xA;And so in particular, every normal number will contain long&#xA;blocks of digits that exactly express the collected works&#xA;of William Shakespeare, and also versions with the plays&#xA;translated (and mis-translated) into every other language,&#xA;in all possible ways, and so on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Indeed, any normal number will contain within it all the theorems of mathematics, with fully correct proofs (as well as with false proofs, in all possible ways). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But of course, this is why you asked about coding on an&#xA;infinite set, asking for a kind of infinitary regularity&#xA;property. Leaving aside the particular question about&#xA;primes, let me consider your latter question: &lt;em&gt;is there a&#xA;limit to the amount of information that we can extract from&#xA;$\pi$?&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For this, the answer is yes, there is a limit, if we make the problem precise with a particular meaning of what it should mean to extract information. The reason is that&#xA;$\pi$ is a computable number; there is a computable&#xA;procedure that will tell us the $n$-th digit. Thus, any&#xA;procedure that extracts information from $\pi$ by means of&#xA;a computable procedure inspecting the digits will necessarily be altogether&#xA;computable. So for example, there can be no computable&#xA;procedure that extracts information from $\pi$ so as to&#xA;answer yes-or-no the question of whether a given Turing&#xA;machine program halts, since the halting problem is not&#xA;decidable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More generally, there are only countably many sets that are&#xA;computable from any given real, no matter the complexity of&#xA;that real (since there are only countably many programs),&#xA;and so there is a robust sense in which most sets of&#xA;natural numbers are not reducible to $\pi$ or any other&#xA;fixed real in this way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lastly, apart from $\pi$, it seems that there are normal&#xA;numbers $z$ that have the property that the $p(n)$-th digit&#xA;of $z$ is $0$, where $p(n)$ is the $n$-th prime. Since the&#xA;primes have asymptotic density $0$, then unless I am&#xA;mistaken, it appears that we could simply place $0$'s in&#xA;the prime digits, and choose the rest of the digits&#xA;randomly, to arrive at a normal number exhibiting your&#xA;property.&lt;/p&gt;&#xA;" OwnerUserId="413" LastEditorUserId="413" LastEditorDisplayName="" LastEditDate="2011-02-05T22:23:56.193" LastActivityDate="2011-02-05T22:23:56.193" CommentCount="1" />
  <row Id="20601" PostTypeId="2" ParentId="20569" CreationDate="2011-02-05T22:10:15.620" Score="4" ViewCount="" Body="&lt;p&gt;My vote goes to &lt;a href=&quot;http://www.sagemath.org/&quot; rel=&quot;nofollow&quot;&gt;Sage&lt;/a&gt;, a free (in two senses) incredible piece of software. Also available to try online without installation.&lt;/p&gt;&#xA;" OwnerUserId="4689" LastActivityDate="2011-02-05T22:10:15.620" CommentCount="2" />
  <row Id="20602" PostTypeId="2" ParentId="20548" CreationDate="2011-02-05T22:14:16.060" Score="4" ViewCount="" Body="&lt;p&gt;Another method goes by noting that $\cos^2\theta+\sin^2\theta=1$. We have $\cos\theta+\sin\theta=x$, so $$\cos^2\theta+2\cos\theta\sin\theta+\sin^2\theta=x^2,$$ or $2\cos\theta\sin\theta=x^2-1$. But $2\cos\theta\sin\theta=\sin(2\theta)$, so $2\theta=\sin^{-1}(x^2-1)$, or $$ \theta=\frac12\sin^{-1}(x^2-1).$$&lt;/p&gt;&#xA;" OwnerUserId="462" LastActivityDate="2011-02-05T22:14:16.060" />
  <row Id="20603" PostTypeId="2" ParentId="20597" CreationDate="2011-02-05T22:14:56.590" Score="0" ViewCount="" Body="&lt;p&gt;As Arturo points out what you have to observe is the coefficient.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align*}&#xA;x^{2}+x+1 &amp;amp; = x^{2} + x  + \frac{1+3}{4} \\ &amp;amp; = x^{2} + x + \frac{1}{4} + \frac{3}{4} \\ &amp;amp; = \Bigl(x+\frac{1}{2}\Bigr)^{2} + \frac{3}{4}&#xA;\end{align*}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am sure once you get used to such type of things you shall not have trouble in doing such problems. Solve more problems based on this type. Suppose you have the coefficient of $x$ as $a$ note that $a^{2}/4$ should be added and subtracted from the constant term. What i mean by this is: Suppose you have something of this type $x^{2}+ax + b^{2}$ then you can write this as $(x+a/2)^{2} + b^{2}-a^{2}/{4}$.&lt;/p&gt;&#xA;" LastActivityDate="2011-02-05T22:14:56.590" />
  <row Id="20604" PostTypeId="2" ParentId="20597" CreationDate="2011-02-05T22:27:41.813" Score="1" ViewCount="" Body="&lt;p&gt;One can rewrite a degree $\rm\:n &amp;gt; 1\:$ polynomial $\rm\ f(x)\ =\ x^n + b\ x^{n-1} +\ \cdots\ $ into a form such that its two highest degree terms are &quot;absorbed&quot; into a &lt;em&gt;perfect&lt;/em&gt; $\rm\: n$'th power of a linear polynomial, namely&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\quad f(x)\ =\ (x + b/n)^n\ -\ g(x)\ \ $ where $\rm\ \ g(x)\ =\  (x+b/n)^n - f(x)\ $ has degree $\rm\:\le\: n-2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When $\rm\ n = 2\ $ this is called &lt;em&gt;completing the square&lt;/em&gt; - esp. when used to solve a quadratic equation.$\ \ $ If $\rm\ \ g(x)\ = g\ $ is constant (as is always true when $\rm\ n = 2\:$) then this yields a closed form for the roots of $\rm\:\ f(x)\:,\ $ namely $\rm\ x\ =\ \sqrt[n]{g}-b/2\:.$&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2011-02-06T00:46:10.743" LastActivityDate="2011-02-06T00:46:10.743" />
  <row Id="20605" PostTypeId="1" CreationDate="2011-02-05T22:27:57.420" Score="1" ViewCount="77" Body="&lt;p&gt;Given a symmetric positive definite matrix $A$ and a mostly-zeros non-negative diagonal matrix $D$, is there a way to cheaply update the eigenvalues and/or eigenvectors of $A$ to that of $A+D$? Ideally I'm looking for something akin to the Woodbury matrix identity.&lt;/p&gt;&#xA;" OwnerUserId="6681" LastActivityDate="2011-05-29T14:05:14.660" Title="updating eigendecomposition of symmetric matrix A to eigendecomposition of A+D where D is low-rank diagonal" Tags="&lt;linear-algebra&gt;&lt;matrices&gt;&lt;eigenvalues&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="20606" PostTypeId="1" AcceptedAnswerId="20611" CreationDate="2011-02-05T22:37:15.143" Score="4" ViewCount="178" Body="&lt;p&gt;This is a problem from an olympiad I took today. I tried but couldn't solve it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $A$ and $B$ rectangular matrices with real entries, of dimensions $k\times n$ and $m\times n$ respectively. Prove that if for every $n\times l$ matrix $X$ ($l\in \mathbb{Z}^+$) the equality $AX=0$ implies $BX=0$, then there exists a matrix $C$ such that $B=CA$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried to define a commutative diagram, but failed. Anyhow, I'd prefer a solution that works with the matrices explicitly. But I'll welcome and appreciate any solutions.&lt;/p&gt;&#xA;" OwnerUserId="4786" LastActivityDate="2011-02-05T22:58:46.027" Title="Olympiad linear algebra problem" Tags="&lt;matrices&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" />
  <row Id="20607" PostTypeId="2" ParentId="20591" CreationDate="2011-02-05T22:44:09.543" Score="4" ViewCount="" Body="&lt;p&gt;The key here is to look at the condition&#xA;$$g(x) = cx^2 + h(x)\qquad\mbox{with $\lim\limits_{x\to 0}\frac{h(x)}{x^2} = 0$.}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you think about a potential Taylor expansion of $g(x)$ of degree $2$, this tells you that $h(x)$ has to consists of the degree $3$ and above terms only; that means that the Taylor expansion of $g(x)$ around $0$ cannot have a linear or constant terms, so $g(0)=0$ and $g&amp;#39;(0)=0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, since we will want $g(x) = f(x+a)+b$, then $g&amp;#39;(x) = f&amp;#39;(x+a)$, so in order to get $g&amp;#39;(0)=0$, we need $f&amp;#39;(a)=0$. The only point where we know $f&amp;#39;(a)=0$ is at $a=x_0$, so we'll pick $a=x_0$. Thus, $g(x) = f(x+x_0) + b$. In order to also get $g(0) = 0$, we need $f(x_0) + b = 0$, so $b=-f(x_0)$. Thus, the function we expect will work is&#xA;$$g(x) = f(x+x_0) - f(x_0).$$&#xA;Now, to see this works, we proceed as you did: taking the Taylor expansion for $f$ around $x_0$ up to degree $2$, we get&#xA;$$f(x) = f(x_0) + \frac{f&amp;#39;&amp;#39;(x_0)}{2}(x-x_0)^2 + R_3(x).$$&#xA;So &#xA;$$g(x) = f(x+x_0) - f(x_0) = \frac{f&amp;#39;&amp;#39;(x_0)}{2}x^2 + R_3(x+x_0).$$&#xA;Thus, $c=\frac{f&amp;#39;&amp;#39;(x_0)}{2}$, and $h(x) = R_3(x+x_0)$. We just need to show that $h(x)/x^2 \to 0$ as $x\to 0$, but this is equivalent to $R_3(x+x_0)/x^2\to 0$ as $x\to 0$, which in turn is equivalent to &#xA;$$\lim_{x\to x_0}\frac{R_3(x)}{(x-x_0)^2} = 0.$$&#xA;This hold precisely because $R_3$ is the degree $3$ remainder at $x_0$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, in summary, we set $a=x_0$, $b=-f(x_0)$, $c=\frac{f&amp;#39;&amp;#39;(x_0)}{2}$, $g(x) = f(x+x_0) - f(x_0)$, and $h(x) = R_3(f,x_0,x)$ (the Remainder of degree $3$ in the Taylor expansion of $f$ around $x_0$, evaluated at $x$). &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-05T22:52:12.603" LastActivityDate="2011-02-05T22:52:12.603" CommentCount="1" />
  <row Id="20608" PostTypeId="2" ParentId="20606" CreationDate="2011-02-05T22:50:26.960" Score="2" ViewCount="" Body="&lt;p&gt;Think of $A$ as a linear transformation from $\mathbb{R}^n$ to $\mathbb{R}^k$, and $B$ as a linear transformation from $\mathbb{R}^n$ to $\mathbb{R}^m$. White $L_M$ for the linear transformation defined by multiplication by the matrix $M$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the matrix $C$ to exist, you must have a linear transformation $L$ from $\mathbb{R}^k$ to $\mathbb{R}^m$ such that $L\circ L_A = L_B$. Then you can define $C$ to be the standard matrix representation of $L$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A necessary condition for this to be possible is that $\mathbf{N}(L_A) \subseteq \mathbf{N}(L_B)$ (if there is $\mathbf{v}\in\mathbf{N}(L_A)$ that is not in $\mathbf{N}(L_B)$, then $L\circ L_A(\mathbf{v})=\mathbf{0}\neq L_B(\mathbf{v})$ and you're sunk). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, this is &lt;em&gt;sufficient&lt;/em&gt;: find a basis $\mathbf{v}_1,\ldots,\mathbf{v}_s$ for the nullspace of $A$, and then extend it to a basis for $\mathbb{R}^n$, $\mathbf{v}_{s+1},\dots,\mathbf{v}_n$. Note that the image of $\mathbf{v}_{s+1},\ldots,\mathbf{v}_n$ in $\mathbb{R}^k$ under $L_A$ is linearly independent (this is essentially the Rank-Nullity Theorem), so you can complete $A\mathbf{v}_{s+1},\ldots,A\mathbf{v}_n$ to a basis $\mathbf{w}_1,\ldots,\mathbf{w}_{(k-n+s)}$ of $\mathbb{R}^k$. Define $L\colon\mathbb{R}^k\to\mathbb{R}^n$ by letting $L(A\mathbf{v}_j) = B\mathbf{v}_j$ for $j=s+1,\ldots,n$, and arbitrarily for $\mathbf{w}_i$. Then let $C$ be the standard matrix for this linear transformation.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-05T22:56:01.023" LastActivityDate="2011-02-05T22:56:01.023" CommentCount="2" />
  <row Id="20609" PostTypeId="2" ParentId="20555" CreationDate="2011-02-05T22:51:39.347" Score="19" ViewCount="" Body="&lt;p&gt;It is unknown whether&#xA;$$&#xD;&#xA;\sum_{n=1}^\infty\frac{1}{n^3\sin^2n}&#xD;&#xA;$$&#xA;converges or not. The difficulty here is that convergence depends on the term $n\sin n$ not being too small, which in turn depends on how well $\pi$ can be approximated by rational numbers. It is possible that, if $\pi$ can be approximated `too well' enough by rationals, then this will diverge. See &lt;a href=&quot;http://mathoverflow.net/questions/24579/convergence-of-a-series&quot; rel=&quot;nofollow&quot;&gt;this MathOverflow question&lt;/a&gt; for a discussion of this particular series.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another even simpler example of a sequence (no summation) for which it is not known whether it converges or not is&#xA;$$&#xD;&#xA;x_n=\frac{1}{n^2\sin n}.&#xD;&#xA;$$&#xA;We would expect this to tend to zero, but the proof is beyond what is currently known. Suppose that there were only finitely many rational numbers $p/q$ with $\vert p/q-\pi\vert\le q^{-3+\epsilon}$ (for any $\epsilon &gt; 0$), then $x_n$ would tend to zero at rate $O(n^{-\epsilon})$. If, on the other hand, there were infinitely many rationals satisfying $\vert p/q-\pi\vert\le q^{-3-\epsilon}$, then infinitely many $x_n$ would be of order at least $n^\epsilon$, so it diverges. This can be expressed in terms of the &lt;a href=&quot;http://mathworld.wolfram.com/IrrationalityMeasure.html&quot; rel=&quot;nofollow&quot;&gt;irrationality measure&lt;/a&gt; of $\pi$. The sequence $x_n$ converges to zero if the irrationality measure of $\pi$ is less than 3, and diverges if it is geater than 3. Currently, the best known bound for the irrationality measure is that it is no more than about $7.63$ (see the link to the mathworld page above). It is expected that the irrationality measure of $\pi$ is 2 (it is known that all but a zero-measure set of real numbers have irrationality measure 2). Therefore, it is expected that $x_n$ tends to zero, but there is currently no proof of this.&lt;/p&gt;&#xA;" OwnerUserId="1321" LastActivityDate="2011-02-05T22:51:39.347" />
  <row Id="20610" PostTypeId="2" ParentId="20570" CreationDate="2011-02-05T22:53:42.977" Score="1" ViewCount="" Body="&lt;p&gt;(Too long for a comment) How do you want to use the heading information?  Rahul has given two reasonable choices for a path from the current position to the destination, but neither needs heading information.  Think of the corresponding problem in the plane-you just put a ruler between the current position and the destination and the direction is whatever it is, not some prespecified heading.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The last sentence seems unrelated to the question above.  To know north you don't need a destination, but it seems you are  thinking of using the north pole.  It is true that if the destination is the north pole the direction to take is north, but that doesn't tell you how to find north looking at the ground.  You need some outside information, such as the magnetic field, position of the sun, or GPS signals to do that.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-02-05T22:53:42.977" />
  <row Id="20611" PostTypeId="2" ParentId="20606" CreationDate="2011-02-05T22:58:46.027" Score="1" ViewCount="" Body="&lt;p&gt;$AX=0$ means that the rows of $A$ are all orthogonal to the columns of $X$, and likewise for $BX=0$. $B=CA$ means that the rows of $B$ are linear combinations of the rows of $A$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume that there is a row of $B$ that is not a linear combination of the rows of $A$. Orthogonalize that row to the rows of $A$, and choose $X$ as the column vector corresponding to this orthogonalized row. Then $AX=0$, but $BX\neq0$. Hence the rows of $B$ are linear combinations of the rows of $A$.&lt;/p&gt;&#xA;" OwnerUserId="6622" LastActivityDate="2011-02-05T22:58:46.027" CommentCount="7" />
  <row Id="20612" PostTypeId="1" AcceptedAnswerId="20639" CreationDate="2011-02-05T23:02:48.757" Score="3" ViewCount="281" Body="&lt;p&gt;This is the other problem I couldn't solve in the olympiad test I took today.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $c_1, ... , c_n$ be complex numbers with unitary norm, and $S_k=\sum_{i=1}^n c_i^k$, $k\in \mathbb{N}$. Suppose $S_1, S_2, ...$ converges. Prove that $c_i=1$ for every $i=1, ..., n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By using De Moivre's formula, the problem is equivalent to showing that $\cos(k\theta_1) + ... +\cos(k\theta_n)$ has no limit as $k\to\infty$ unless all $\theta_i$'s are $0$ (mod $2\pi$) (and an analogous claim for the sum of sines).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know how to prove that.&lt;/p&gt;&#xA;" OwnerUserId="4786" LastActivityDate="2011-02-08T12:09:32.177" Title="Olympiad - sequence of sum of complex numbers" Tags="&lt;contest-math&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="2" />
  <row Id="20613" PostTypeId="1" AcceptedAnswerId="20616" CreationDate="2011-02-05T23:07:26.463" Score="3" ViewCount="112" Body="&lt;p&gt;The first step is clear -- the zero vector is in both, so it will be in the intersection, too. Scalar multiplication is pretty obvious too -- if the vector is in the intersection, it can be represented as some linear combination of vectors in $V$ = some linear combination of vectors in $W$. Multiplying both by $k$ preserves the equality. How can we prove that it is closed under addition, too? I can think of a 'cheat' way, saying that the intersection of $V$ and $W$ is the kernal of the matrix with spanning vectors of $V$ and $W$, and by properties of kernal, etc. but I was looking for a more simply-formulated solution. Thanks!&lt;/p&gt;&#xA;" OwnerUserId="5157" LastEditorUserId="5157" LastEditorDisplayName="" LastEditDate="2011-02-06T01:23:52.107" LastActivityDate="2011-02-06T01:23:52.107" Title="If $\mathbf{V}$ and $\mathbf{W}$ are subspaces of $\mathbb{R}^{n}$, prove that so is $\mathbf{V} \cap \mathbf{W}$" Tags="&lt;linear-algebra&gt;&lt;homework&gt;" AnswerCount="2" CommentCount="0" />
  <row Id="20614" PostTypeId="2" ParentId="20613" CreationDate="2011-02-05T23:10:25.647" Score="1" ViewCount="" Body="&lt;p&gt;Let $a,b \in V \cap W$. Then $a+b \in V$ and $a+b \in W$. Hence $a+b \in V \cap W$. &lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2011-02-05T23:10:25.647" CommentCount="1" />
  <row Id="20615" PostTypeId="1" CreationDate="2011-02-05T23:12:34.193" Score="3" ViewCount="186" Body="&lt;p&gt;How would I evaluate these limits?&#xA;$$ \lim_{n \to \infty} \int_0^\infty \frac{n}{1+(nx)^2} \ dx$$&#xA;and&#xA;$$ \lim_{n \to \infty} \int_0^\infty \frac{(1+(nx)^2)}{(1+nx^2)^n} \ dx$$&lt;/p&gt;&#xA;" OwnerUserId="6683" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-05T23:18:15.557" LastActivityDate="2011-02-06T14:35:18.653" Title="Limits and Measure Theory" Tags="&lt;measure-theory&gt;" AnswerCount="3" CommentCount="8" />
  <row Id="20616" PostTypeId="2" ParentId="20613" CreationDate="2011-02-05T23:14:04.863" Score="7" ViewCount="" Body="&lt;p&gt;You are overcomplicating things. There is absolutely no need to think about linear combinations. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $\mathbf{x}\in\mathbf{V}\cap\mathbf{W}$, then $\mathbf{x}\in\mathbf{V}$ and $\mathbf{x}\in\mathbf{W}$. If $k$ is a scalar, then $k\mathbf{x}\in\mathbf{V}$ (because $\mathbf{V}$ is a subspace, and $\mathbf{x}$ is in $\mathbf{V}$), and likewise $k\mathbf{x}\in\mathbf{W}$ because $\mathbf{x}\in\mathbf{W}$ and $\mathbf{W}$ is a subspace. Since $k\mathbf{x}$ is in &lt;em&gt;both&lt;/em&gt; $\mathbf{V}$ and $\mathbf{W}$, then it is in $\mathbf{V}\cap\mathbf{W}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For sums, do the same thing: take $\mathbf{x}$ and $\mathbf{y}$ which are both in $\mathbf{V}\cap\mathbf{W}$. Then $\mathbf{x}$ and $\mathbf{y}$ are both in &lt;em&gt;each&lt;/em&gt; of $\mathbf{V}$ and in $\mathbf{W}$. Since $\mathbf{V}$ is a subspace, and $\mathbf{x}$ and $\mathbf{y}$ are both in $\mathbf{V}$, then so is $\mathbf{x}+\mathbf{y}$. Now show that $\mathbf{x}+\mathbf{y}$ is &lt;em&gt;also&lt;/em&gt; in $\mathbf{W}$ to conclude that it must be in $\mathbf{V}\cap\mathbf{W}$, which is what you want.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2011-02-05T23:14:04.863" CommentCount="1" />
  <row Id="20617" PostTypeId="1" AcceptedAnswerId="20901" CreationDate="2011-02-05T23:14:29.057" Score="6" ViewCount="261" Body="&lt;p&gt;Can someone explain in complete detail with the appropriate convergence arguments of the Gibbs Phenomenon for Fourier Series? I know that the overshoot near a jump does not die out as the frequency increases, but approaches a finite limit. We derived the overshoot in an engineering class and there was a lot of hand-waving. I understand it but I would like to see a complete picture explained here for a general function $f(x)$ on a bounded domain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Further, the Gibbs Phenomena doesn't occur for some other eigen function expansions like Chebyshev expansion etc. The explanation given in the class was again very crude. Again for these cases could someone explain in detail the connection to the Sturm-Liouville equation and which of these eigenfunction expansions lead to overshoot and which ones do not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="1543" LastEditorDisplayName="" LastEditDate="2011-02-06T00:49:06.283" LastActivityDate="2011-02-07T23:22:29.233" Title="Gibbs Phenomenon" Tags="&lt;fourier-series&gt;" AnswerCount="2" CommentCount="6" FavoriteCount="2" />
  <row Id="20618" PostTypeId="1" CreationDate="2011-02-05T23:14:29.707" Score="2" ViewCount="53" Body="&lt;p&gt;This has been annoying me for the past two days.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose we have an algebra $A$ over some field and a subalgebra $B \subseteq A$. Suppose we also have that $A$ is a finitely-generated free module over $B$, so as a $B$-module $A =Bx_1 \oplus ... \oplus Bx_n$ for some $x_1,...,x_n \in A$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can I assume $x_1=1$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or is it possible there is some sort of example where although $A$ is just $n$ copies of $B$ as a &lt;em&gt;module&lt;/em&gt;, the only way it contains $B$ as a &lt;em&gt;subalgebra&lt;/em&gt; is a different one where the structure is more complicated and it doesn't decompose into that direct sum?&lt;/p&gt;&#xA;" OwnerUserId="6682" LastActivityDate="2011-02-06T00:21:46.903" Title="Algebras that are free modules over a subalgebra" Tags="&lt;abstract-algebra&gt;&lt;ring-theory&gt;" AnswerCount="1" />
  <row Id="20619" PostTypeId="1" CreationDate="2011-02-05T23:53:04.610" Score="3" ViewCount="109" Body="&lt;p&gt;I'm reading lectures for the first time starting next week ^_^ The subject is calculus on manifolds I was told that the students I'll be lecturing are not motivated (at all), so I need to kick off the series with an impressive demonstration of what an cool subject it will be that I'll be able to periodically return to later on when I'll need to motivate specific concepts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So far I came up with:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Projective plane, I'll demonstrate how conics morph into each other on different models, should be cool enough and motivate general-topological manifolds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Plücker coordinates and their applications to line geometry and computer graphics (not sure if it will be easy to demonstrate), should motivate forms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) Expanding Universe, motion in relativity and perception of time, should motivate tangent vectors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure they will work, and I could always use more ideas. I would greatly appreciate examples and general advice too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S.: Please, wiki-hammer this question.&lt;/p&gt;&#xA;" OwnerUserId="3793" LastActivityDate="2011-02-06T00:46:13.323" CommunityOwnedDate="2011-02-05T23:57:32.547" Title="Motivating differentiable manifolds" Tags="&lt;soft-question&gt;&lt;differential-geometry&gt;&lt;big-list&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="20620" PostTypeId="2" ParentId="20619" CreationDate="2011-02-05T23:59:45.680" Score="0" ViewCount="" Body="&lt;p&gt;See &lt;a href=&quot;http://books.google.com/books?id=sWXcAyR37acC&amp;amp;pg=PR16&amp;amp;lpg=PR16&amp;amp;dq=motivating+differential+manifolds&amp;amp;source=bl&amp;amp;ots=pC5btOBOVu&amp;amp;sig=VQbmVOXrtb-6snbqeWKK5ndB_70&amp;amp;hl=en&amp;amp;ei=G-RNTe6AFcKC8gbguLj4Dg&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=1&amp;amp;ved=0CBgQ6AEwAA#v=onepage&amp;amp;q=motivating%20differential%20manifolds&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; book by W.D Curtis and F.W Miller. Essentially they say to view the &quot;evolution of a physical system as a curve in the space of states.&quot; Vector fields on state spaces describe the behavior an evolving physical system. &lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2011-02-05T23:59:45.680" CommunityOwnedDate="2011-02-05T23:59:45.680" CommentCount="1" />
  <row Id="20621" PostTypeId="2" ParentId="20619" CreationDate="2011-02-06T00:02:11.217" Score="3" ViewCount="" Body="&lt;p&gt;The universe is a manifold! And we don't know which. I think that is already plenty of motivation, but you might be interested in reading Weeks' &lt;a href=&quot;http://books.google.com/books?id=A8WBiUWy3SgC&amp;amp;printsec=frontcover&amp;amp;dq=weeks+shape+of+space&quot; rel=&quot;nofollow&quot;&gt;The Shape of Space&lt;/a&gt; for some more motivation along these lines, as well as some interesting visual exercises and good exposition overall (although some of the physical possibilities he suggests later are, I think, out of date). I would also suggest that you draw a lot of pictures, e.g. of 2-manifolds.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2011-02-06T00:02:11.217" CommunityOwnedDate="2011-02-06T00:02:11.217" CommentCount="2" />
  <row Id="20622" PostTypeId="1" AcceptedAnswerId="20623" CreationDate="2011-02-06T00:10:49.797" Score="3" ViewCount="112" Body="&lt;p&gt;In the last two days, I got a bit obsessed with the following game, partly because I beat others most of the time I play it. The gamers are expected to have a sound logical reasoning coupled with the ability to analyze data.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is the rule of the game:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) The game is for two players.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) Each player holds a 4-digit positive number in his mind. (Numerals 0 to 9 are allowed. For instance you can hold 0176 or5678 or 2384, etc.)  The numbers are to be kept secret.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(3) Each then guesses the other's 4-digit number. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(4) Each person gives a mark for the other person's guess. The mark consists of two parts. The first part counts the number of digits the other person rightly guessed. The second  part counts the number of digits that were guessed in the right place. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(5) The fist person who comes to the right 4-digit the other person held is the winner.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a sample where the players are Mr.A and Mr.B. Assume A has held 3476 and that B has 7609. Let A begin. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\begin{vmatrix}&#xD;&#xA; \hline&#xD;&#xA;   Player(A)   &amp;amp; Correct digits &amp;amp; Correct places &amp;amp; Player(B)  &amp;amp; Correct digits &amp;amp; Correct places \\ guesses &amp;amp; (B-  marks) &amp;amp; (B-marks)&#xD;&#xA;&amp;amp;  guesses &amp;amp; ( A- marks) &amp;amp; ( A-  marks)\\&#xD;&#xA;   4521             &amp;amp; 0                       &amp;amp; 0                       &amp;amp; 5735             &amp;amp; 2                        &amp;amp; 0                        \\ &#xD;&#xA;   8309             &amp;amp; 2                       &amp;amp; 2                       &amp;amp; 8762             &amp;amp; 2                        &amp;amp; 0                        \\ &#xD;&#xA;   ...              &amp;amp; ...                     &amp;amp; ...                     &amp;amp; ...              &amp;amp; ...                      &amp;amp; ...                      \\ &#xD;&#xA;   7609             &amp;amp; 4                       &amp;amp; 4                       &amp;amp; 3467             &amp;amp; 4                        &amp;amp; 2                        \\&#xD;&#xA;  \hline&#xD;&#xA;\end{vmatrix}&#xD;&#xA;&#xD;&#xA;$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So A is the winner. There is a lot of mathematical elimination strategy going in there  which makes the play a very interesting pastime at least for me. If this is a known game, could you please point me to a reference? Thank you.&lt;/p&gt;&#xA;" OwnerUserId="3992" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2011-02-06T00:39:27.113" LastActivityDate="2011-02-06T00:39:27.113" Title="Is this a well-known game?" Tags="&lt;reference-request&gt;" AnswerCount="1" CommentCount="7" />
  <row Id="20623" PostTypeId="2" ParentId="20622" CreationDate="2011-02-06T00:14:18.757" Score="7" ViewCount="" Body="&lt;p&gt;Seems like a variant of &lt;a href=&quot;http://en.wikipedia.org/wiki/Mastermind_%28board_game%29&quot; rel=&quot;nofollow&quot;&gt;Mastermind&lt;/a&gt;. I believe it is called &lt;a href=&quot;http://en.wikipedia.org/wiki/Bulls_and_cows&quot; rel=&quot;nofollow&quot;&gt;Cows and Bulls&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2011-02-06T00:21:25.287" LastActivityDate="2011-02-06T00:21:25.287" CommentCount="1" />
  <row Id="20625" PostTypeId="2" ParentId="2949" CreationDate="2011-02-06T00:20:04.973" Score="3" ViewCount="" Body="&lt;p&gt;Bounded holomorphic function is constant; integration of a meromorphic function.&lt;/p&gt;&#xA;" OwnerUserId="3793" LastActivityDate="2011-02-06T00:20:04.973" CommunityOwnedDate="2011-02-06T00:20:04.973" />
  <row Id="20626" PostTypeId="2" ParentId="20618" CreationDate="2011-02-06T00:21:46.903" Score="1" ViewCount="" Body="&lt;p&gt;&lt;em&gt;I guess the following works:&lt;/em&gt; (I have the roles of $A$ and of $B$ reversed...)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $A$ is either local notherian or non-negatively graded and connected, and if the map $A\to B$ is injective (it has to be, I think...) then $B/A$ is free as an $A$-module. It follows that the s.e.s. $0\to A\to B\to B/A\to0$ splits and that $A$ is a summand in $B$: that means you can suppose $x_1=1_B$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To prove that $B/A$ is free, it is enough to see that $\mathrm{Tor}_1^A(K,B/A)=0$ for $K$ the residue field in the first hypothesis, and the base field in the second one. Looking at the long exact sequence for $\mathrm{Tor}_1^A(K,\mathord-)$ strongly enough does it.&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2011-02-06T00:21:46.903" CommentCount="2" />
  <row Id="20627" PostTypeId="2" ParentId="20612" CreationDate="2011-02-06T00:26:52.373" Score="2" ViewCount="" Body="&lt;p&gt;Let $c_i = \exp(2\pi i \alpha_i)$. Choose a basis $\beta_j$ (over the rationals) for all $\alpha_i$, such that in the resulting representation, the coefficient of every $\beta_j$ is integral, and the coefficients of $\beta_j$ don't have a common multiple.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there is any rational $\beta_j$, let $M$ be the maximal coefficient of $\beta_j$, and replace $c_i$ with $c_i^M$; now the &quot;rational&quot; part of the sum is constant; adjust the basis accordingly so that the other properties still hold. We can remove the rational basis member (if any) from the basis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now construct a sequence of $k$s $S_1$ such that $\{k \beta_i\} \approx 0$, and another sequence of $k$s $S_2$ such that $\{k \beta_1\} \approx 1/2$ and $\{k \beta_i\} \approx 0$ for $i \geq 2$. The limit for $S_1$ and $S_2$ is different (since the coefficients of $\beta_1$ are all integers and not all even); this contradiction shows that all coefficients must be rational.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now go back to the original $c_i$, and write $\alpha_i = p_i/q_i$. If not all $\alpha_i$ are $0$, choose some prime $Q$ dividing one of the $q_i$. Let $M$ be the least common multiple of all $q_i$. Form a new set of $c_i$ by replacing $c_i$ with $c_i^{M/Q}$. We now have either $c_1 = 1$ or $c_i = p_i/Q$. Now for multiples of $Q$ all $c_i^k = 1$ and so the sum is some constant $C$, whereas for anything else, this won't be the case, and so in particular the real part of the sum will be smaller than $C$. This contradiction shows that, in fact, all $\alpha_i$ must be $0$, i.e. all $c_i$ are $1$.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2011-02-06T00:26:52.373" CommentCount="3" />
  <row Id="20628" PostTypeId="2" ParentId="20537" CreationDate="2011-02-06T00:28:21.000" Score="1" ViewCount="" Body="&lt;p&gt;The most efficient (publicly) known way to factor $n$, assuming it is &quot;big&quot;, is the &lt;a href=&quot;http://en.wikipedia.org/wiki/General_number_field_sieve&quot; rel=&quot;nofollow&quot;&gt;Number-Field Sieve&lt;/a&gt; (NFS). This is assuming that $n$ is a product of two large primes of roughly the same magnitude.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2011-02-06T01:23:16.117" LastActivityDate="2011-02-06T01:23:16.117" CommentCount="0" />
  <row Id="20629" PostTypeId="2" ParentId="20612" CreationDate="2011-02-06T00:37:09.110" Score="6" ViewCount="" Body="&lt;p&gt;I've not done olympiads, so I'm not sure if this is the intended method.. but anyhow... we may assume without loss of generality that no $c_i = 1$, and that we are working by induction and know the result for $n-1$ and are seeking it for $n$ ($n=1$ is trivial.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One fact which is often useful is that if a sequence $a_1,a_2,...$ converges then the means $b_m = {1 \over m}( a_1 + ... + a_m) $ will converge to the same limit. Applying it to the sequence $S_1, S_2,...$, the averages $T_m =  {1 \over m}\sum_{k=1}^m (\sum_{i=1}^n c_i^k) =  \sum_{i=1}^n {1 \over m}(\sum_{k=1}^m  c_i^k)$ will also converge to $L$. But the sum ${1 \over m} (\sum_{k=1}^m c_i^k) $ can be summed directly as a geometric series to &#xA;${1 \over m}{c_i^{m+1} - c_i \over c_i - 1}$. As $m$ goes to infinity this goes to zero. So the limit $L$ must be zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $S_1, S_2,..$ converges to zero and $|c_1| = 1$, $T_m = (c_1)^{-m}S_m$ also goes to zero. But $T_m$ is exactly &#xA;$1 + \sum_{i=2}^m (c_i/c_1)^m$. So the sequence $T_m - 1 = \sum_{i=2}^m (c_i/c_1)^m$ converges to $-1$, contradicting the induction hypothesis. &lt;/p&gt;&#xA;" OwnerUserId="3035" LastEditorUserId="3035" LastEditorDisplayName="" LastEditDate="2011-02-06T00:50:04.747" LastActivityDate="2011-02-06T00:50:04.747" CommentCount="2" />
  <row Id="20630" PostTypeId="2" ParentId="20619" CreationDate="2011-02-06T00:46:13.323" Score="0" ViewCount="" Body="&lt;p&gt;Personally I always loved to link maths and physics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, why not introduce hysteresis? Some hysteresis phenomenons can be modelled using manifolds I believe.&lt;/p&gt;&#xA;" OwnerUserId="6258" LastActivityDate="2011-02-06T00:46:13.323" CommunityOwnedDate="2011-02-06T00:46:13.323" />
  <row Id="20631" PostTypeId="1" AcceptedAnswerId="20637" CreationDate="2011-02-06T00:48:47.890" Score="4" ViewCount="104" Body="&lt;p&gt;This is question from Spivak's Calculus.&#xA;Question statement (paraphrased): For any even function $f$ there is infinite number of functions $g$ such that $f(x) = g(|x|)$&lt;br&gt;&#xA;I have made attempt at proof, here is my work. My main concern is whether my proof that there are infinite such functions is correct.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Let $g$ be a function such that for any positive $x$, $f(x) = g(x)$ and let $f$ be some even function. We will show that above equality then holds even when $x$ is negative: $f(-x) = g(|-x|) \Leftrightarrow f(x) = g(x)$, which is true by definition of these functions. It now remains to show that there is infinite number of those functions. I understand that is so because $g$ can be defined as one whishes for negative numbers and it won't change equality. But I cannot think of formal proof. I'm thinking about assigning values to functions $g$ in a way that I define $g_{1}(-1) = k$ and then making infinite number of them by induction: for any function $g_{n}(x) = k$, if x is negative, we define $g_{n+1}(x-1) = k$. Then by the fact that there is infinite natural numbers, and for every natural number there is going to be unique function, we can conclude there is infinite number of functions.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Is the concept of induction applicable here or is it not?&lt;/p&gt;&#xA;" OwnerUserId="5501" LastActivityDate="2011-02-06T01:49:59.273" Title="For any even function $f$ there is infinite number of functions $g$ such that $f(x) = g(|x|)$" Tags="&lt;calculus&gt;&lt;functions&gt;" AnswerCount="3" CommentCount="20" FavoriteCount="1" />
  <row Id="20632" PostTypeId="2" ParentId="20631" CreationDate="2011-02-06T00:55:48.400" Score="3" ViewCount="" Body="&lt;p&gt;It is enough to observe that $g$ can have any value whatsoever for $x &amp;lt; 0$. Since there is an infinite number of values for each $f(x)$ for $x &amp;lt; 0$, we're done.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(you really don't need to do it any more difficult than that.)&lt;/p&gt;&#xA;" OwnerUserId="4284" LastEditorUserId="4284" LastEditorDisplayName="" LastEditDate="2011-02-06T01:03:08.380" LastActivityDate="2011-02-06T01:03:08.380" CommentCount="4" />
  <row Id="20633" PostTypeId="2" ParentId="20631" CreationDate="2011-02-06T01:00:02.647" Score="1" ViewCount="" Body="&lt;p&gt;You could use induction but it's really not necessary. Just define $g_k(x) = k$ when $x$  is negative. Then you have a infinite family of functions that are different.&lt;/p&gt;&#xA;" OwnerUserId="414" LastActivityDate="2011-02-06T01:00:02.647" CommentCount="1" />
  <row Id="20634" PostTypeId="1" AcceptedAnswerId="21432" CreationDate="2011-02-06T01:05:30.500" Score="5" ViewCount="265" Body="&lt;p&gt;I'm having trouble solving the following exercise in Hatcher's Algebraic Topology(1.3 #5):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Let $X$ be the subspace of $R^2$ consisting of the four sides of the square $[0,1] \times [0,1]$ together with the segments of the vertical lines $x=\frac{1}{2},\frac{1}{3},\frac{1}{4},\ldots$ inside the square. Show that for every covering space $\tilde{X}\to X$ there is some neighbourhood of the left edge of $X$ that lifts homeomorphically to $\tilde{X}$. Deduce that $X$ has no simply-connected cover.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried piecing together open sets in $\tilde{X}$ which were homeomorphic to open square shaped $\epsilon$-neighbourhoods of points in the left edge of $X$, but I couldn't glue them together coherently.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My second idea was to use the path lifting property to lift the left edge, but I couldn't extend it to a lift of an open neighbourhood of the left edge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;: Some more detail on my first attempt:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose I have two open squares $U_1$ and $U_2$ in $X$ (containing the left edge) with nonempty intersection, and that have preimages $\coprod_i U_{1,i}$ and $\coprod_j U_{2,j}$ with each piece of that mapping homeomorphically onto $U_1$ and $U_2$ respectively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose I've chosen a specific $U_{1,i}$, and I want to choose a specific $U_{2,j}$ so that their union maps homeomorphically to $U_1$ union $U_2$. Such a choice can be specified by a point in the chosen $U_{1,i}$ that maps into $U_2$, but for each such point, there could be a different choice of $j$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My intution tells me that to make this choice, one needs to appeal to the bad local nature of the left edge of $X$, (so that you're certain to get some of the vertical lines inside of the desired open set) but I'm not sure how to proceed with that.&lt;/p&gt;&#xA;" OwnerUserId="5290" LastEditorUserId="5290" LastEditorDisplayName="" LastEditDate="2011-02-06T02:46:11.527" LastActivityDate="2011-02-10T21:31:39.860" Title="Comb space has no simply connected cover" Tags="&lt;topology&gt;&lt;algebraic-topology&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="20635" PostTypeId="2" ParentId="20578" CreationDate="2011-02-06T01:12:51.797" Score="11" ViewCount="" Body="&lt;p&gt;The family of functions you generally consider (e.g., &lt;a href=&quot;http://en.wikipedia.org/wiki/Elementary_function&quot; rel=&quot;nofollow&quot;&gt;elementary functions&lt;/a&gt;) is closed under differentiation, that is, the derivative of such function is still in the family. However, the family is not in general closed under integration. For instance, even the family of rational functions is not closed under integration because you $\int 1/x = \log$.&lt;/p&gt;&#xA;" OwnerUserId="589" LastActivityDate="2011-02-06T01:12:51.797" />
  <row Id="20636" PostTypeId="2" ParentId="20578" CreationDate="2011-02-06T01:28:09.723" Score="22" ViewCount="" Body="&lt;p&gt;Here is an extremely generic answer. Differentiation is a &quot;local&quot; operation: to compute the derivative of a function at a point you only have to know how it behaves in a neighborhood of that point. But integration is a &quot;global&quot; operation: to compute the definite integral of a function in an interval you have to know how it behaves on the &lt;em&gt;entire&lt;/em&gt; interval (and to compute the indefinite integral you have to know how it behaves on &lt;em&gt;all&lt;/em&gt; intervals). That is a lot of information to summarize. Generally, local things are generally much easier than global things.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, if you can do the global things, they tend to be &lt;em&gt;useful&lt;/em&gt; because of how much information goes into them. That's why theorems like the fundamental theorem of calculus, the full form of Stokes' theorem, and the main theorems of complex analysis are so powerful: they let us calculate global things in terms of slightly less global things.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2011-02-06T01:28:09.723" CommentCount="1" />
  <row Id="20637" PostTypeId="2" ParentId="20631" CreationDate="2011-02-06T01:31:25.607" Score="3" ViewCount="" Body="&lt;p&gt;Okay, from the comments I was able to guess at what you were trying to do.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are correct that the key is that as long as $g(x)=f(x)$ for $x\geq 0$, everything will work out, so you only need to worry about defining $g$ on the negative numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You attempted to do so by letting $g_1$ be a function that is defined only on the nonnegative numbers and $-1$, and setting $g_1(-1)=k$. Then $g_2$ would be an &lt;em&gt;extension&lt;/em&gt; of $g_1$, which is also defined as $k$ at $-2$; and so on. In general, $g_{n}$ would be define on $\{-n,-n+1,\ldots,-1\}\cup[0,\infty)$, by $g(x) = f(x)$ if $x\geq 0$ and $g(x)=k$ if $x\lt n$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As I noted in the comments, your formulas didn't really say that; instead, they only specified $g_1$ at $-1$, and then said, for example, that $g_2(x-1)=k$. That is at best confusing. What is $g_2(-0.5)$? According to this, I have to think of $-0.5$ as $0.5 - 1$, and then it's $k$, and ... Well, a bit of a confusing issue arises...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In any case, whether this works as an answer or not depends on whether you are assuming that your functions need to be defined on the same set or not. Normally, we would be looking for functions $g$ such that $f(x)=g(|x|)$ and we want both $f$ and $g$ to have &lt;em&gt;the same domain&lt;/em&gt;. Remember that two functions are equal if and only if they have the same &lt;em&gt;domain&lt;/em&gt;, the same codomain, and the same value at every element of the domain.  So even if you could have set up the induction properly to get the functions you wanted (or if you wanted to define $g_1$ on $[-1,0)$, then $g_2$ extended to all of $[-2,0)$, and so on sothat $g_n$ was defined on $[-n,\infty)$) it still would not give a good answer to the problem because of the restrictive domains of your $g$s. (You are really composing the absolute value with $g$; I think Spivak wants you to play with functions that are defined everywhere here, rather than on artificial domains; I could be wrong, though).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you do not require your functions $g$ to have the same domain as $f$, then your &lt;em&gt;intended&lt;/em&gt; answer would also work; the functions $g_n$ are different because they have different domains, even though they all have the same values where their domains agree. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think, though, that the intended answer relies instead in defining $g_n(x)$ for $x\lt 0$ as different things for different $n$. For example, you could set&#xA;$$g_n(x) = \left\{\begin{array}{ll}&#xA;f(x) &amp;amp;\mbox{if $x\geq 0$,}\\&#xA;n &amp;amp;\mbox{if $x\lt 0$;}&#xA;\end{array}\right.$$&#xA;and this would work. There is no need to state it as induction, because the values depend only on the labels, and you simply have infinitely many distinct labels to choose from.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That said: yes, you can use induction to define a (countably infinite) series of functions. If you wanted to do that here, you could express it &lt;em&gt;explicitly&lt;/em&gt; stating what $g_1$ is (giving its domain clearly; if all you say is $g(-1)=k$, then you are telling us the value at $-1$, but not saying anything about values elsewhere, not even that it is not defined there). Then saying that assuming you have defined $g_n$ with a domain of, say, $[-n,\infty)$, you define $g_{n+1}$ on $[-n-1,\infty)$ by&#xA;$$g_{n+1}(x) = \left\{\begin{array}{ll}&#xA;g_n(x) &amp;amp;\mbox{if $x\in[-n,\infty)$.}\\&#xA;\mbox{whatever} &amp;amp;\mbox{if $x\in [-n-1,-n)$;}&#xA;\end{array}\right.$$&#xA;This would indeed give an inductive definition for your $g_n$, defined on $[-n,\infty)$, each extending the previous one.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-06T01:49:59.273" LastActivityDate="2011-02-06T01:49:59.273" CommentCount="3" />
  <row Id="20638" PostTypeId="1" AcceptedAnswerId="20648" CreationDate="2011-02-06T01:39:38.037" Score="5" ViewCount="98" Body="&lt;p&gt;Given a nowhere vanishing vector field, say $E_1$, on a manifold $M^n$, it should be possible to extend this locally to a basis of vector fields $E_1,\ldots , E_n$ so that $E_i = \frac{\partial}{\partial x_i}$ for some coordinates $x_i$. (For example, locally we can choose a chart where $E_1$ looks like $\frac{\partial}{\partial x_1}$ and then just define $E_i = \frac{\partial}{\partial x_i}$). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For such a frame there we may define a co-frame on $T^*M$, $\varphi^i$ so that $\varphi^i(E_j) = \delta_{ij}$. By the intrinsic definition of the exterior derivative,&#xA;\begin{equation} d\varphi^1(E_i,E_j) = E_i(\varphi^1(E_i)) - E_j(\varphi^1(E_j) - \varphi^1([E_i,E_j]) . \end{equation}&#xA;The first two terms will be zero as they are the derivative of either the constant function 0 or 1. Similarly, $[E_i,E_j] = [\frac{\partial}{\partial x_i},\frac{\partial}{\partial x_j}] \equiv 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, since $d\varphi^1$ is a two form, we see by linearity that $d\varphi^1 \equiv 0$ .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, any nonvanishing 1-form $\omega$ can (locally) be expressed as the dual of a nonvanishing vector in some coordinate framing. This would seem to imply that any nowhere zero 1-form has zero exterior derivative.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since I do not believe this is true, there must be some flaw to my reasoning, but I cannot see it.&lt;/p&gt;&#xA;" OwnerUserId="6687" LastEditorUserId="6687" LastEditorDisplayName="" LastEditDate="2011-02-06T01:45:40.160" LastActivityDate="2011-02-07T06:47:40.600" Title="Is the dual covector field of a nonzero vector field closed?" Tags="&lt;geometry&gt;&lt;differential-geometry&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="20639" PostTypeId="2" ParentId="20612" CreationDate="2011-02-06T01:49:01.520" Score="4" ViewCount="" Body="&lt;p&gt;I've edited the answer twice without deleting the previous versions, to keep the history of the answer and the comments transparent. That makes it much longer than it originally was, but if you're just interested in the end result, the first and last paragraphs make up a concise and complete proof.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $S_k$ converges, then the differences $\Delta S_k := S_{k+1} - S_k$ must converge to zero. Hence the differences $\Delta^2 S_k := \Delta S_{k+1} - \Delta S_k$ must also converge to zero, and so on for $\Delta^n S_k$ for all $n$. But $\Delta^n S_k = \sum_{i=1}^{n} (c_i - 1)^n c_i^k$. There can be either one or two values of $c_i$ for which the absolute value of $c_i - 1$ is maximal. If there is only one, the corresponding term will dominate the sum for sufficiently large $n$, in the sense that its absolute value becomes greater than the absolute value of the sum of all the other contributions. Since this term only converges to zero if $c_i - 1 = 0$, it follows that $c_i = 1$ for all $i$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there are two (conjugate) maximal values of $c_i - 1$, there is a linear combination of $S_k$ and its complex conjugate in which only one of these maximal values occurs; if $S_k$ had a limit, so would its complex conjugate and this linear combination.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit in response to kevincuadros's question about the linear combination part:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In trying to clarify this, I can see now why you didn't &quot;fully get&quot; it -- because I hadn't thought it through properly :-)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I had in mind was this: If there are two different values $c_i$ with maximal absolute value of $c_i - 1$, they are conjugate. Both may occur more than once; let $\mu_1$ and $\mu_2$ be their multiplicities. Then $\mu_1 S_k - \mu_2 S_k^\mathrm{*}$ will contain one of them with multiplicity $\mu_1^2 - \mu_2^2$ and the other with multiplicity $\mu_1\mu_2 - \mu_2\mu_1=0$. I was thinking that we could then reason that since this linear combination contains only one of the two, we can apply the above proof to it, and then argue that if $S_k$ had a limit, then so would $S_k^\mathrm{*}$ and any linear combination of them. But I see now that that doesn't work out because we could have $\mu_1=\mu_2$, and hence $\mu_1^2-\mu_2^2$, and in that case neither of the two would occur in that linear combination, so we still have to deal with that case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in that case, the contributions from these conjugates cancel in the imaginary part and add up in the real part. If $S_k$ is to converge, its real part must converge. But $(c_i - 1)^n c_i^k$ comes arbitrarily close to being real infinitely often; thus its real part comes arbitrarily close to $|c_i - 1|^n$ infinitely often, and hence the argument for the general case carries through.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To make the full proof more concise, we can forget about the whole linear combination thing and just argue as follows: If there are two distinct values of $c_i$ with equal maximal absolute value of $c_i - 1$, they are conjugate. Then consider the sum of the real parts  of their contributions, which comes arbitrarily close to the sum of their absolute values infinitely often, and hence the argument for the general case carries through. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Further edit to salvage the style of the proof:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't really like the above fix, since part of the point of the original proof was to avoid saying something like &quot;x gets arbitrarily close to y an infinite number of times&quot;. So here's a slightly nicer fix:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there are two distinct values of $c_i$ with equal maximal absolute value of $c_i - 1$, they are conjugate. Then the real parts of their contributions add up to $|c_i-1|^n$ times the cosine of an angle that changes with $k$. Since the ratio of $|c_i-1|^n$ to the absolute value of the sum of all other terms goes to infinity with increasing $n$, increasing $n$ will yield arbitrarily small upper bounds on the cosine. But since $c_i \neq \pm 1$, the cosine necessarily violates these bounds. (Note that this doesn't use the property that the cosine gets arbitrarily close to 1, only that it doesn't remain arbitrarily close to 0, which is much less and doesn't require an argument using rational and irrational numbers.)&lt;/p&gt;&#xA;" OwnerUserId="6622" LastEditorUserId="6622" LastEditorDisplayName="" LastEditDate="2011-02-08T12:09:32.177" LastActivityDate="2011-02-08T12:09:32.177" CommentCount="3" />
  <row Id="20640" PostTypeId="2" ParentId="20615" CreationDate="2011-02-06T02:28:50.153" Score="0" ViewCount="" Body="&lt;p&gt;From $(1+a)^n\geq1+na\;\;\;\forall a\geq0$ with $a=nx^2$, you have:&#xA;$\frac{1+n^2x^2}{(1+nx^2)^n}\leq1$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore, you have:$\frac{1+n^2x^2}{(1+nx^2)^n}&amp;lt;\frac{1+n^2x^2}{(nx^2)^n}=\frac{1}{n^nx^{2n}}+\frac{1}{n^{n-2}x^{2(n-1)}}&amp;lt;\frac{2}{x^2}$ if $n\geq2$ and $x&amp;gt;0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, by the Lebesgue dominated convergence theorem, you need only consider the integral of the limit function to evaluate the limit &#xA;(let $g(x)=\begin{cases}&#xD;&#xA;1 &amp;amp; \text{ if } 0\leq x\leq1 \\ &#xD;&#xA; \frac{2}{x^2}&amp;amp; \text{ if } x&amp;gt;1 &#xD;&#xA;\end{cases}$ ).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now $0\leq\frac{1+n^2x^2}{(1+nx^2)^n}\leq\frac{n^2+n^2x^2}{(1+x^2)^n}=\frac{n^2}{(1+x^2)^{n-1}}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence $lim_{n\rightarrow\infty}\frac{1+n^2x^2}{(1+nx^2)^n}=0$ by the squeeze theorem (we can neglect $x=0$ since it constitutes a set of measure zero).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence the desired integral is equal to $0$.&lt;/p&gt;&#xA;" OwnerUserId="5959" LastEditorUserId="5959" LastEditorDisplayName="" LastEditDate="2011-02-06T03:49:02.427" LastActivityDate="2011-02-06T03:49:02.427" CommentCount="3" />
  <row Id="20641" PostTypeId="2" ParentId="20578" CreationDate="2011-02-06T02:32:38.903" Score="1" ViewCount="" Body="&lt;p&gt;This is more of a philosophical argument. In general, the inverse problems are usually harder than the direct problems. A matrix vector product is $\mathcal{O}(N^2)$ whereas solving a linear system is $\mathcal{O}(N^3)$. So the question is more of why is differentiation &quot;direct&quot; and why not integration? To which I think Qiaochu answer fits in.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastActivityDate="2011-02-06T02:32:38.903" CommentCount="2" />
  <row Id="20643" PostTypeId="1" CreationDate="2011-02-06T02:41:05.137" Score="2" ViewCount="188" Body="&lt;p&gt;Hello everyone,&#xA;Do you know of a good book on linear programming? To be more specific, i am taking linear optimization class and my textbook sucks. Teacher is not too involved in this class so can't get too much help from him either,&#xA;Any help will be appreciated.&#xA;Thank you&lt;/p&gt;&#xA;" OwnerUserId="6212" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2011-02-06T03:18:05.383" LastActivityDate="2011-02-06T17:25:33.843" Title="Linear Programming Books" Tags="&lt;optimization&gt;&lt;books&gt;&lt;linear-programming&gt;" AnswerCount="4" FavoriteCount="2" />
  <row Id="20644" PostTypeId="2" ParentId="20643" CreationDate="2011-02-06T02:46:55.210" Score="1" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://www.google.com/products/catalog?hl=en&amp;amp;biw=1366&amp;amp;bih=667&amp;amp;q=operations+research+winston&amp;amp;um=1&amp;amp;ie=UTF-8&amp;amp;cid=3682745626224534069&amp;amp;ei=DQtOTafeGIOC8gbc8tHEDg&amp;amp;sa=X&amp;amp;oi=product_catalog_result&amp;amp;ct=result&amp;amp;resnum=3&amp;amp;ved=0CDAQ8wIwAg#&quot; rel=&quot;nofollow&quot;&gt;This&lt;/a&gt;&lt;em&gt;(Operations Research: Applications and Algorithms&lt;/em&gt; by Winston) is a good book. I learned linear optimization techniques from this book(e.g. the simplex method, assignment problems, etc..).&lt;/p&gt;&#xA;" OwnerUserId="2548" LastEditorUserId="2548" LastEditorDisplayName="" LastEditDate="2011-02-06T02:51:57.437" LastActivityDate="2011-02-06T02:51:57.437" />
  <row Id="20646" PostTypeId="2" ParentId="20643" CreationDate="2011-02-06T03:07:55.387" Score="0" ViewCount="" Body="&lt;p&gt;The other classics besides Winston are Hillier and Lieberman's &lt;em&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0077298349&quot; rel=&quot;nofollow&quot;&gt;Introduction to Operations Research&lt;/a&gt;&lt;/em&gt; and Chvátal's &lt;em&gt;&lt;a href=&quot;http://books.google.com/books?id=DN20_tW_BV0C&amp;amp;printsec=frontcover&amp;amp;dq=chvatal+linear+programming&amp;amp;source=bl&amp;amp;ots=4cNHLAaXjJ&amp;amp;sig=Wmmw_-v_mT9xTp5ZzKmf3AvxIIM&amp;amp;hl=en&amp;amp;ei=MBFOTYWQFY3GsAOg2ID3Cg&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=3&amp;amp;ved=0CDIQ6AEwAg#v=onepage&amp;amp;q=chvatal%20linear%20programming&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;Linear Programmming&lt;/a&gt;&lt;/em&gt;.  I learned linear programming out of Bob Vanderbei's &lt;em&gt;&lt;a href=&quot;http://www.princeton.edu/~rvdb/LPbook/&quot; rel=&quot;nofollow&quot;&gt;Linear Programming: Foundations and Extensions&lt;/a&gt;,&lt;/em&gt; which is also a fine book.  The last time I taught linear programming I used Dave Rader's new book, &lt;a href=&quot;http://books.google.com/books?id=VqjhIZhsITsC&amp;amp;printsec=frontcover&amp;amp;dq=dave+rader+deterministic&amp;amp;source=bl&amp;amp;ots=mbLk0q5SAL&amp;amp;sig=d4Q-ECU2nn43ZOAnvFr5_Ci8Ca8&amp;amp;hl=en&amp;amp;ei=ZhFOTai-NIuasAOThrGRCg&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=2&amp;amp;ved=0CBwQ6AEwAQ#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;Deterministic Operations Research&lt;/em&gt;&lt;/a&gt;, and was happy with it.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for a comparison, Winston focuses on how the different methods work and gives lots of examples but doesn't spend much time on theory.  Hillier and Lieberman is at a slightly higher level than Winston, with a more leisurely pace and a little more theory but with fewer examples.  Chvátal and Vanderbei have a more noticeable focus on the theory.  Rader takes a different approach in that the simplex method does not appear until about halfway through the book.  Instead, he spends a lot of time early on algorithm design and on what an algorithm for solving linear programs might look like so that when you finally do see the simplex method the reaction is closer to &quot;Of course&quot; than to &quot;Where in the world did that come from?&quot;  He doesn't do the tableau form of the simplex method, which, while a plus in my opinion, may make it hard to understand his version of the simplex method if you're used to the tableau.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many LP books spend little time on how to construct linear programming models (i.e, how to come up with variables, an objective function, and constraints that describe the problem you're trying to solve).  Of these five, Winston and Rader discuss construction of LP models the most.&lt;/p&gt;&#xA;" OwnerUserId="2370" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2011-02-06T06:09:44.887" LastActivityDate="2011-02-06T06:09:44.887" CommentCount="1" />
  <row Id="20647" PostTypeId="2" ParentId="20643" CreationDate="2011-02-06T03:41:30.980" Score="1" ViewCount="" Body="&lt;p&gt;I'm taking a class that uses &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0898716438&quot; rel=&quot;nofollow&quot;&gt;Linear Programming with Matlab&lt;/a&gt;, but it's really hard to read.  I got &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/048643284X&quot; rel=&quot;nofollow&quot;&gt;Linear Programming: Methods and Applications&lt;/a&gt; to supplement it and it's much more readable.  It's targeted at undergraduates, I guess.  It has a lot of proofs, but most of the text is explanatory.&lt;/p&gt;&#xA;" OwnerUserId="6059" LastActivityDate="2011-02-06T03:41:30.980" />
  <row Id="20648" PostTypeId="2" ParentId="20638" CreationDate="2011-02-06T03:45:52.350" Score="0" ViewCount="" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;In fact, any nonvanishing 1-form ω can (locally) be expressed as the dual of a nonvanishing vector in some coordinate framing. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is the part I question.  How do you show this?  If you run your argument in reverse, you need that any nonvanishing form $\omega$ is $dx^1$ for some coordinate $x^1$, which now implies that $\omega$ is closed.  So perhaps while every non-zero vector field can be taken to be a coordinate field, the corresponding statement for one-forms is not true.&lt;/p&gt;&#xA;" OwnerUserId="9" LastActivityDate="2011-02-06T03:45:52.350" CommentCount="2" />
  <row Id="20649" PostTypeId="2" ParentId="20567" CreationDate="2011-02-06T03:50:49.403" Score="4" ViewCount="" Body="&lt;p&gt;Cantor's diagonal&#xA;argument &#xA;shows that for any countable list of numbers, one can&#xA;construct a number not on that list. Cantor used this&#xA;argument to show, for example, that there are&#xA;transcendental numbers, since one may describe a way to&#xA;list all the polynomials with integer coefficients and&#xA;their roots and hence to list all the algebraic numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One sometimes hears it asserted that Cantor's proof of the&#xA;existence of transcendental numbers is a &quot;pure-existence&quot;&#xA;proof, showing merely that transcendental numbers exist,&#xA;but not that any particular number is transcendental. But&#xA;that view is not correct, for the argument is completely&#xA;constructive: one may explicitly describe an enumeration of&#xA;the algebraic numbers and the diagonal procedure produces a&#xA;definite real number that is not algebraic. (I once saw an&#xA;article, I think in one of the MAA volumes, with the title&#xA;something like &lt;em&gt;&quot;0.24543... is transcendental&quot;&lt;/em&gt;, where they&#xA;implemented Cantor's actual algorithm.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The relevance to this question is that Cantor&#xA;diagonalization also can be used to prove that specific&#xA;real numbers are not rational, by producing real numbers&#xA;that are explicitly different from every rational number.&#xA;Specifically, we may enumerate the rational numbers as&#xA;$q_k$ in any of the usual effective manners, and define a&#xA;real number $z$ so that the $k$-th digit of $z$ is $4$,&#xA;say, if $r_k$ does not have $k$-th digit $4$, and otherwise&#xA;the $k$-th digit of $z$ is $5$. It now follows by&#xA;construction that $z\neq r_k$ for each $k$, which is what it means for $z$ to be irrational.&lt;/p&gt;&#xA;" OwnerUserId="413" LastActivityDate="2011-02-06T03:50:49.403" CommentCount="2" />
  <row Id="20650" PostTypeId="2" ParentId="20567" CreationDate="2011-02-06T04:05:22.343" Score="4" ViewCount="" Body="&lt;p&gt;An irrational number can be defined as having infinite continued fraction&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The continued fraction of sqrt(2) is [1, 2, 2, 2, ...] so it's irrational.&lt;/p&gt;&#xA;" OwnerUserId="4290" LastActivityDate="2011-02-06T04:05:22.343" CommentCount="2" />
  <row Id="20651" PostTypeId="2" ParentId="20444" CreationDate="2011-02-06T04:30:59.503" Score="-1" ViewCount="" Body="&lt;p&gt;Quoting Arturo Magidin:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;There is no general algorithm that will always yield a good parametrization of any curve. Relatively simple trigonometric parametrizations cannot work in general, because they always yield closed curves (being periodic). Parametrizing curves generally takes ingenuity, not unthinking application of general rules, though practice and familiarity with a variety of tricks will help (much as with integration)&#xA;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1069" LastActivityDate="2011-02-06T04:30:59.503" CommunityOwnedDate="2011-02-06T04:30:59.503" CommentCount="2" />
  <row Id="20652" PostTypeId="1" AcceptedAnswerId="20655" CreationDate="2011-02-06T04:56:04.270" Score="4" ViewCount="231" Body="&lt;p&gt;This is from an exercise (1.2.2) in introduction to algorithms that I'm working on privately.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To find at what point a $n \lg n$ function will run faster than a $n^2$ function I need to figure out for what value $n$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$8n^2 &amp;gt; 64n \lg n$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(with lg here being the binary log)&#xA;after some elementary simplification we get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$n &amp;gt; 8\lg n$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Playing around with properties of log I can further get this to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$n^8 &amp;lt; 2^n$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$n &amp;lt; 2^{n/8}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While I'm sure it's something very elementary I've lost somewhere over the years, after checking out a few logarithm tutorials I'm just not finding how to get any further on this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any help with solving for $n$ would be appreciated.&lt;/p&gt;&#xA;" OwnerUserId="4801" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2011-02-06T22:13:22.777" LastActivityDate="2011-02-06T22:13:22.777" Title="How to solve $n &lt; 2^{n/8}$ for $n$?" Tags="&lt;inequality&gt;&lt;logarithms&gt;&lt;asymptotics&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="20655" PostTypeId="2" ParentId="20652" CreationDate="2011-02-06T05:22:01.377" Score="3" ViewCount="" Body="&lt;p&gt;It's not hard to show that $f(x) = x - 8 \lg x$ only has one critical point. Since $f(2) \lt 0$ it suffices to find $x \gt 2$ such that $f(x) &amp;gt; 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take $n = 2^{k}$ then you get $2^{k} &amp;gt; 8k$ from your second inequality and plugging in $k$ from $1$ on shows that $n = 2^6 = 64$ will do. You can check by hand that $n = 44$ is optimal, as $f(44) \gt 0 \gt f(43)$. &lt;/p&gt;&#xA;" OwnerUserId="5363" LastActivityDate="2011-02-06T05:22:01.377" CommentCount="3" />
  <row Id="20656" PostTypeId="2" ParentId="20652" CreationDate="2011-02-06T05:27:19.300" Score="6" ViewCount="" Body="&lt;p&gt;Since it doesn't take long for $2^{n/8}$ to get larger than $n$, it would be reasonable to experiment to find roughly where this happens, using multiples of $8$ to make it easy.  E.g., $40\gt 2^5$, but $48&amp;lt;2^6$.  The smallest solution (not counting $n=1$) is thus somewhere between $41$ and $48$.  It is geometrically clear from the shape of the curves $y=x$ and $y=2^{x/8}$ that all greater $n$ will work, but to prove it you could use mathematical induction.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Claim: If $n\geq 44$, then $n\lt 2^{n/8}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proof: The base case is $n=44$, for which I'll omit the verification.  Suppose for some $n\geq 44$ that $n\lt 2^{n/8}$.  Then $2^{(n+1)/8}=2^{1/8}2^{n/8}\gt 2^{1/8}n$.  Since $n\geq 44$, $\frac{n+1}{n}=1+\frac{1}{n}\leq 1+\frac{1}{44}\lt 2^{1/8}$.  Thus $2^{1/8}n\gt n+1$, and combining with the previous inequality completes the inductive step.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll also omit the verification that $43\gt 2^{43/8}$.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2011-02-06T05:27:19.300" />
  <row Id="20657" PostTypeId="2" ParentId="20567" CreationDate="2011-02-06T05:36:26.830" Score="0" ViewCount="" Body="&lt;p&gt;A very simple proof follows from Gauss' lemma.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Gauss%27s_lemma_(polynomial&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Gauss%27s_lemma_(polynomial&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $p$ is prime, then clearly, $\sqrt{p}$ is a root of $f(x)=x^2-p$.  Gauss's lemma shows that $f(x)=x^2-p$ has no rational roots (since it clearly has no integer roots).  Thus $\sqrt{p}$ is irrational.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also note the proof of Gauss' lemma is not a proof by contradiction so this entire result can be proven directly. &lt;/p&gt;&#xA;" OwnerUserId="3971" LastActivityDate="2011-02-06T05:36:26.830" CommentCount="1" />
  <row Id="20658" PostTypeId="1" AcceptedAnswerId="21273" CreationDate="2011-02-06T05:56:01.447" Score="5" ViewCount="363" Body="&lt;p&gt;Initially the deck is randomly ordered. The aim is to sort the deck in order. Now in each turn the deck is shuffled randomly. If any of the initial or last cards are in sorted order then they are kept aside, and the process is followed with rest of the cards. The question now is what is the expected number of shuffles for the entire deck to get sorted in this manner.&lt;br&gt;&#xA;e.g. let's say initially the deck is (3,5,6,2,7,1,4). After the first shuffle it becomes (1,2,6,4,5,3,7), then the next shuffle will be done with (6,4,5,3) as (1,2) and (7) are already in proper order. Check that even if 5 in also at right place, but not in continuous order hence considered for shuffle.&lt;br&gt;&#xA;What I know is that the expected number of shuffles to sort the deck in one attempt is $n!$. (Something known as bogo-sort). But can't think any further. Any help will be appreciated. &lt;/p&gt;&#xA;" OwnerUserId="5403" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2011-02-06T06:15:37.343" LastActivityDate="2011-02-10T13:02:00.787" Title="expected number of shuffles to sort the cards" Tags="&lt;permutations&gt;&lt;probability&gt;&lt;sorting&gt;" AnswerCount="3" FavoriteCount="1" />
  <row Id="20659" PostTypeId="2" ParentId="20658" CreationDate="2011-02-06T06:27:51.790" Score="2" ViewCount="" Body="&lt;p&gt;Some thoughts:  there will be a recurrence.  The idea is to find the expected time to reduce the number of cards in the deck, then to find the probability that if you reduce it, what is the chance that you reduce it by 1, 2, ... First the probability that at least one of the ends is in proper position is 2/n, as I claim the fact that the top card is not right doesn't change the chance  that the bottom card is right-there is 1/n chance that the top card should be on the bottom, then of the 1-1/n=(n-1)/n chance it isn't there is 1/(n-1) chance the right one is on the bottom.  So the expected time to reduction is n/2.  Then sort out $P_k(n)$, the probability that if there is a reduction it is exactly k.  By the previous argument, $P_1(n)=\frac{n-3}{n-1}$  Then, as in the &lt;a href=&quot;http://en.wikipedia.org/wiki/Coupon_collector%27s_problem&quot; rel=&quot;nofollow&quot;&gt;coupon collector problem&lt;/a&gt; you have N(n) (the number of shuffles of n cards)=n/2(probability of reduction of 1 given that there is a reduction)*N(n-1)+probability of reduction of 2 given that there is a reduction)*N(n-2)+...)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Added: 2 is special because if one card is right, so is the other.  This chance is 1/2 and you are done.  So the expected number of shuffles is 2.  For 3, you have 1/3 chance of nothing correct, 1/6 chance of all correct, and 1/2 chance of 1 correct.  So N(3)=(1/3)(N(3)+1)+(1/2)(N(2)+1)+(1/6)(N(0)+1)  As you know N(2)=2 and N(0)=0 you can solve for N(3)&lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2011-02-06T16:04:48.730" LastActivityDate="2011-02-06T16:04:48.730" CommentCount="2" />
  <row Id="20660" PostTypeId="1" CreationDate="2011-02-06T06:45:13.197" Score="5" ViewCount="189" Body="&lt;p&gt;What possible choices do fresh Phds have if they can't find a job,especially for pure math,provided current job market is tight? &lt;/p&gt;&#xA;" OwnerUserId="1944" LastActivityDate="2011-02-06T20:56:04.340" CommunityOwnedDate="2011-02-06T20:11:20.733" Title="about career choice" Tags="&lt;soft-question&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="20661" PostTypeId="1" AcceptedAnswerId="20714" CreationDate="2011-02-06T06:56:59.647" Score="8" ViewCount="267" Body="&lt;p&gt;Claim:If $(x_\alpha)_{\alpha\in A}$ is a collection of real numbers $x_\alpha\in [0,\infty]$&#xA;such that $\sum_{\alpha\in A}x_\alpha&amp;lt;\infty$, then $x_\alpha=0$ for all but at most countably many $\alpha\in A$ (A need not be countable).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proof: Let $\sum_{\alpha\in A}x_\alpha=M&amp;lt;\infty$. Consider $S_n=\{\alpha\in A|x_\alpha&amp;gt;1/n\}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then $M\geq\sum_{\alpha\in S_n}x_\alpha&amp;gt;\sum_{\alpha\in S_n}1/n=\frac{N}{n}$, where $N\in\mathbb{N}\cup\{\infty\}$ is the number of elements in $S_n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus $S_n$ has at most $Mn$ elements. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence $\{\alpha\in A|x_\alpha&amp;gt;0\}=\bigcup_{n\in\mathbb{N}}S_n$ is countable as the countable union of finite sets. $\square$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, is my proof correct? Second, are there more concise/elegant proofs?&lt;/p&gt;&#xA;" OwnerUserId="5959" LastEditorUserId="5959" LastEditorDisplayName="" LastEditDate="2011-02-06T07:02:47.753" LastActivityDate="2011-02-06T18:13:00.033" Title="The sum of an uncountable number of positive numbers..." Tags="&lt;measure-theory&gt;" AnswerCount="1" CommentCount="8" />
  <row Id="20662" PostTypeId="2" ParentId="20652" CreationDate="2011-02-06T07:33:03.160" Score="5" ViewCount="" Body="&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Lambert_W_function#Applications&quot; rel=&quot;nofollow&quot;&gt;Lambert $W$ function&lt;/a&gt; can be used to solve equations of the form $p^{ax+b} = cx+d.$  In fact, the Wikipedia page gives the solution to $n = 2^{n/8}$ to be&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$n = \frac{-8}{\ln 2} W\left(\frac{-\ln 2}{8}\right).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Via Mathematica, which implements the $W$ function as &lt;code&gt;ProductLog&lt;/code&gt;, this comes out to be 43.5592 (although you have to take the lower branch of the $W$ function), in agreement with the Newton iteration solution Theo Buehler mentions.&lt;/p&gt;&#xA;" OwnerUserId="2370" LastActivityDate="2011-02-06T07:33:03.160" />
  <row Id="20663" PostTypeId="1" AcceptedAnswerId="20689" CreationDate="2011-02-06T07:40:54.927" Score="4" ViewCount="128" Body="&lt;p&gt;&lt;strong&gt;Motivation:&lt;/strong&gt;  In Spanier's Algebraic Topology we are introduced to direct and indirect limits in the first few pages.  I partially understand the idea behind it but since I am an example-driven learner I'd actually like to compute some direct limits.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; Where can I find explicit examples of a direct (or even indirect) limit being calculated?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What I've done so far:&lt;/strong&gt; So far, I've computed the direct limit for some finite directed sets.  On the &lt;a href=&quot;http://en.wikipedia.org/wiki/Direct_limit&quot; rel=&quot;nofollow&quot;&gt;wikipedia&lt;/a&gt; page, there are a number of examples which I've been trying to work out.  In particular, the third example (which is the infinite chain of ${\mathbb Z}/p^{n}{\mathbb Z}\rightarrow {\mathbb Z}/p^{n+1}{\mathbb Z}$ with multiplication by $p$ is the map) I'm unsure of how they come up with the roots of unity for some power of $p$.  Any help on this one would also be greatly appreciated.&lt;/p&gt;&#xA;" OwnerUserId="2959" LastActivityDate="2011-02-06T15:31:14.747" Title="Computing a Direct Limit." Tags="&lt;algebra&gt;&lt;examples&gt;" AnswerCount="1" />
  <row Id="20664" PostTypeId="1" CreationDate="2011-02-06T07:44:11.513" Score="7" ViewCount="130" Body="&lt;p&gt;Let $A$, $B$, $C$, and $D$ be finitely generated modules over a PID $R$ such that $A\oplus $ $B$ $\cong$ $C\oplus $ $D$ and $A\oplus $ $D$ $\cong$ $C\oplus $ $B$ . Prove that $A$ $\cong$ $C$ and $B$ $\cong $$D$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only tool I have is the theorem about finitely generated modules, but I don't quite see the connection. Please Help. Thanks. &lt;/p&gt;&#xA;" OwnerUserId="6697" LastEditorUserId="6697" LastEditorDisplayName="" LastEditDate="2011-02-06T18:42:30.097" LastActivityDate="2011-02-06T18:42:30.097" Title="Finitely generated modules over PID" Tags="&lt;abstract-algebra&gt;&lt;modules&gt;&lt;principal-ideal-domains&gt;" AnswerCount="1" FavoriteCount="2" />
  <row Id="20665" PostTypeId="1" AcceptedAnswerId="20666" CreationDate="2011-02-06T08:19:13.957" Score="2" ViewCount="141" Body="&lt;p&gt;Hello guys!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have the following series: &#xA;$\sum_{n=1}^{\infty} \frac{n}{\sqrt{n(n + 1)(n + 2)}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I must investigate if it is divergent or convergent. I have tried different approaches like the root test, but it didn't help me:(. Could someone point me in the right direction?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Best regards, Petar&lt;/p&gt;&#xA;" OwnerUserId="6698" LastActivityDate="2011-02-06T08:23:00.327" Title="Convergent or divergent series" Tags="&lt;sequences-and-series&gt;&lt;convergence&gt;" AnswerCount="1" />
  <row Id="20666" PostTypeId="2" ParentId="20665" CreationDate="2011-02-06T08:23:00.327" Score="8" ViewCount="" Body="&lt;p&gt;For large $n$, the $n$th term of your series is roughly $\frac{1}{\sqrt{n}}$. Can you think of another series whose convergence/divergence behavior is known and whose terms can be compared with $\frac{1}{\sqrt{n}}$? Try using the &lt;a href=&quot;http://en.wikipedia.org/wiki/Comparison_test&quot; rel=&quot;nofollow&quot;&gt;comparison test&lt;/a&gt; between your original series and the new series.&lt;/p&gt;&#xA;" OwnerUserId="618" LastActivityDate="2011-02-06T08:23:00.327" CommentCount="3" />
  <row Id="20667" PostTypeId="1" AcceptedAnswerId="20669" CreationDate="2011-02-06T08:52:46.493" Score="1" ViewCount="54" Body="&lt;p&gt;Hello,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was wondering if it's possible to prove the following statement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lets say we have a value K. I want to find the value where&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A+B = K (both A,B,K are integers)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where A*B is the highest possible value.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've found out that it's:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a = K/2; b = K/2; when K is even &lt;/p&gt;&#xA;&#xA;&lt;p&gt;a = (K-1)/2; b = ((K-1)/2)+1; when K is odd.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But is there a way to prove this.&lt;/p&gt;&#xA;" OwnerUserId="3605" LastEditorUserId="3605" LastEditorDisplayName="" LastEditDate="2011-02-06T08:57:56.257" LastActivityDate="2011-02-06T09:15:11.360" Title="Prove product maximum" Tags="&lt;proof-strategy&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="20668" PostTypeId="2" ParentId="20667" CreationDate="2011-02-06T08:55:15.613" Score="1" ViewCount="" Body="&lt;p&gt;We know that $B = K-A$. So you want to maximize $AB = A(K-A)$. So we have:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$AB = A(K-A)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ = AK-A^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Find the critical points and use some derivative tests.&lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2011-02-06T08:55:15.613" />
  <row Id="20669" PostTypeId="2" ParentId="20667" CreationDate="2011-02-06T09:15:11.360" Score="3" ViewCount="" Body="&lt;p&gt;In the even case, assuming $A\leq B$, the numbers will be of the form $A=\frac{K}{2}-m$ and $B=\frac{K}{2}+m$ in order to get $A+B=K$, with $m=0,1,\ldots,\frac{K}{2}$.  Then multiplying gives $AB=\left(\frac{K}{2}\right)^2 - m^2 \leq \left(\frac{K}{2}\right)^2$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the odd case, you'll have $A=\frac{K}{2}-m-\frac{1}{2}$ and $B=\frac{K}{2}+m+\frac{1}{2}$, with $m=0,1,\ldots,\frac{K-1}{2}$.  Multiplying gives &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\begin{align*}&#xD;&#xA;AB &amp;amp;= \left(\frac{K}{2}\right)^2 - \left(m+\frac{1}{2}\right)^2 = \frac{K^2}{4} - m^2 - m - \frac{1}{4} \\&#xD;&#xA;   &amp;amp; = \frac{K-1}{2}\left(\frac{K-1}{2}+1\right) - m^2 - m \leq \frac{K-1}{2}\left(\frac{K-1}{2}+1\right).&#xD;&#xA;\end{align*}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively, you can think of optimizing a quadratic function by &lt;a href=&quot;http://math.stackexchange.com/questions/709/derivation-of-the-formula-for-the-vertex-of-a-parabola&quot;&gt;finding the vertex&lt;/a&gt;.  If you write $AB$ as a quadratic function of $A$ or $B$ alone using the constraint $A+B=K$ (as indicated in PEV's answer), then the closest integer to the vertex will solve your problem because the parabola opens down.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2011-02-06T09:15:11.360" />
  <row Id="20670" PostTypeId="1" CreationDate="2011-02-06T09:16:44.273" Score="11" ViewCount="359" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The more general version of this theorem in Munkres' 'Topology' (p. 290 - 2nd edition) states that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given a locally compact Hausdorff space $X$ and a metric space $(Y,d)$; a family $\mathcal F$ of continuous functions has compact closure in $\mathcal C (X,Y)$ (topology of compact convergence) if and only if it is equicontinuous under $d$ and the sets&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \mathcal F _a = \{f(a) | f \in \mathcal F\} \qquad a \in X$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;have compact closure in $Y$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I do not see why the Hausdorff condition on $X$ should be necessary? Why include it then? Am I maybe even missing something here (and there are counterexamples)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;btw if you are looking up the proof: Hausdorffness is needed for the evaluation map $e: X \times \mathcal C(X,Y) \to Y, \, e(x,f) = f(x)$ to be continuous. But the only thing really used in the proof is the continuity of $e_a: \mathcal C(X,Y) \to Y, \, e_a(f) = f(a)$ for fixed $a \in X$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cheers, S.L.&lt;/p&gt;&#xA;" OwnerUserId="3208" LastActivityDate="2011-02-06T09:16:44.273" Title="Theorem of Arzelà-Ascoli" Tags="&lt;topology&gt;" CommentCount="5" FavoriteCount="4" />
  <row Id="20671" PostTypeId="2" ParentId="20615" CreationDate="2011-02-06T09:51:40.863" Score="1" ViewCount="" Body="&lt;p&gt;As already noted, the integrals considered in the first case do not depend on $n$ hence their limit is not very mysterious.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $I_n$ denote the integral considered in the second case. By the change of variable $u=nx$, &#xA;$$&#xD;&#xA;nI_n=\displaystyle\int_0^\infty(1+u^2)f_n(u)\mathrm{d}u,\qquad\mbox{with}\ f_n(u)=(1+u^2/n)^{-n}.&#xD;&#xA;$$&#xA;The sequence of positive functions $(f_n)_n$ is nonincreasing (*) and $f_n\to f$ pointwise when $n\to\infty$, with $f(u)=\mathrm{e}^{-u^2}$. Since $f_2\ge f_n$ for very $n\ge2$ and $I_2$ is finite, by Lebesgue dominated convergence theorem, &#xA;$$&#xD;&#xA;nI_n\to J=\displaystyle\int_0^\infty(1+u^2)f(u)\mathrm{d}u=1+(3\pi/2).&#xD;&#xA;$$ &#xA;In particular, $I_n\to0$ when $n\to\infty$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(*) This is implied by the fact that, for every $c\ge0$, the function $z\mapsto(1+c/z)^z$ is nondecreasing on $z\ge0$. To show this last fact, differentiate twice the logarithm of this function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added later on&lt;/strong&gt; Following Shai's suggestion, I mention that the proof above shows that the sequence $(nI_n)_n$ is nonincreasing. For instance $nI_n\le2I_2$ for every $n\ge2$ (note that $I_1=+\infty$ but that $I_2$ is finite).&lt;/p&gt;&#xA;" OwnerUserId="6179" LastEditorUserId="6179" LastEditorDisplayName="" LastEditDate="2011-02-06T14:35:18.653" LastActivityDate="2011-02-06T14:35:18.653" />
  <row Id="20672" PostTypeId="1" CreationDate="2011-02-06T10:00:48.440" Score="1" ViewCount="92" Body="&lt;p&gt;Wikipedia tells me about Max Noether's &lt;a href=&quot;http://en.wikipedia.org/wiki/AF%2BBG_theorem&quot; rel=&quot;nofollow&quot;&gt;$AF + BG$ theorem&lt;/a&gt; but only gives one reference and one external link. I've had a look at the MathWorld link but it seems to be an entirely geometric formulation which I'm not even sure is the same theorem. Does anyone know where I can find an algebraic statement of the theorem, together with a proof? I've tried searching online but the name of the theorem is somewhat unfortunate...&lt;/p&gt;&#xA;" OwnerUserId="5191" LastActivityDate="2011-02-06T10:11:57.690" Title="Max Noether's $AF + BG$ theorem" Tags="&lt;algebraic-geometry&gt;&lt;reference-request&gt;&lt;commutative-algebra&gt;" AnswerCount="1" />
  <row Id="20673" PostTypeId="2" ParentId="20672" CreationDate="2011-02-06T10:11:57.690" Score="3" ViewCount="" Body="&lt;p&gt;You can find it in Fulton's wonderful book - algebraic curves, available online here:&#xA;&lt;a href=&quot;http://people.reed.edu/~davidp/332/CurveBook.pdf&quot; rel=&quot;nofollow&quot;&gt;http://people.reed.edu/~davidp/332/CurveBook.pdf&lt;/a&gt;&#xA;page 61 (page 69 in the above file).&lt;/p&gt;&#xA;" OwnerUserId="5491" LastActivityDate="2011-02-06T10:11:57.690" CommentCount="1" />
  <row Id="20674" PostTypeId="2" ParentId="20617" CreationDate="2011-02-06T10:36:37.453" Score="0" ViewCount="" Body="&lt;p&gt;The periodic jump function  defined by $j(t):=(\pi -t)/2$ on $]0,2\pi[$ has $\sum_{k=1}^\infty \sin(kt)/k$ as its Fourier series. We now compute the value of the partial sum $s_N$ at the point $t_N:=\pi/N$:&#xA;$$s_N(t_N)=\sum_{k=1}^N{\sin(k\pi/N)\over k}= \sum_{k=1}^N{\pi\over N}{\sin(k\pi/N)\over k\pi/N} \doteq \int_0^\pi {\rm sinc}(t)dt\doteq 1.852$$&#xA;(the last sum can be interpreted as a Riemann sum for the sinc-integral). On the other hand $\lim_{t\to 0+} j(t)=\pi/2\doteq 1.571$ so that we have a guaranteed overshoot of about 17%.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The choice of the point $t_N$ is nearly &quot;optimal&quot;, but I won't go into this here.&lt;/p&gt;&#xA;" OwnerUserId="1303" LastActivityDate="2011-02-06T10:36:37.453" />
  <row Id="20675" PostTypeId="2" ParentId="20615" CreationDate="2011-02-06T11:32:38.407" Score="0" ViewCount="" Body="&lt;p&gt;Actually, the second integral, like the first, could have been a &quot;calc 1 problem&quot; (but it is useful to practice the dominated convergence theorem). &#xA;Indeed, the inequality $\frac{{1 + n^2 x^2 }}{{(1 + nx^2 )^n }} \le 1$ (also used by bobobinks) gives&#xA;$$&#xD;&#xA;\int_0^\varepsilon  {\frac{{1 + n^2 x^2 }}{{(1 + nx^2 )^n }}\,{\rm d}x}  \le \varepsilon, &#xD;&#xA;$$&#xA;for any $\varepsilon &amp;gt; 0$ (arbitrarily small, but fixed). On the other hand, for any $n \geq 4$, &#xA;$$&#xD;&#xA;\int_\varepsilon ^\infty  {\frac{{1 + n^2 x^2 }}{{(1 + nx^2 )^n }}\,{\rm d}x}  \le \int_\varepsilon ^\infty  {\frac{{\varepsilon ^{ - 2} x^2  + n^2 x^2 }}{{(2\sqrt n x)^n }}\,{\rm d}x}  \le \frac{{\varepsilon ^{ - 2}  + n^2 }}{{(2\sqrt n )^n }}\int_\varepsilon ^\infty  {x^{2 - n} \,{\rm d}x} = \frac{{\varepsilon ^{ - 2}  + n^2 }}{{(2\sqrt n )^n }}\frac{{\varepsilon ^{3 - n} }}{{n - 3}}. &#xD;&#xA;$$&#xA;Since, for $\varepsilon &amp;gt; 0$ fixed, the expression on the right-hand side tends to $0$ as $n \to \infty$, it follows that $\int_0^\infty  {\frac{{1 + n^2 x^2 }}{{(1 + nx^2 )^n }}\,{\rm d}x} \to 0$ too.&lt;/p&gt;&#xA;" OwnerUserId="2810" LastActivityDate="2011-02-06T11:32:38.407" CommentCount="5" />
  <row Id="20677" PostTypeId="1" AcceptedAnswerId="20679" CreationDate="2011-02-06T12:55:03.470" Score="5" ViewCount="250" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm just working through Conway's book on complex analysis and I stumbled across this lovely exercise (quote):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Use Cauchy's Integral Formula to prove the Cayley-Hamilton Theorem: If $A$ is an $n \times n$ matrix over $\mathbb C$ and $f(z)  = \det(z-A)$ is the characteristic polynomial of $A$ then $f(A) = 0$. (This exercise was taken from a paper by C. A. McCarthy, Amer. Math. Monthly, 82 (1975), 390-391)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, I was not able to find said paper. I'm completely lost with this exercise. I can't even start to imagine how one could possibly make use of Cauchy here...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for any hints.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regards, S.L.&lt;/p&gt;&#xA;" OwnerUserId="3208" LastActivityDate="2011-02-06T15:38:44.383" Title="Cauchy's Integral Formula for Cayley-Hamilton Theorem" Tags="&lt;linear-algebra&gt;&lt;complex-analysis&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" />
  <row Id="20678" PostTypeId="1" AcceptedAnswerId="20684" CreationDate="2011-02-06T13:44:12.810" Score="1" ViewCount="91" Body="&lt;p&gt;Let $\hat{f}(t) := \int_\mathbb{R} f(x) e^{-2 \pi i xt} d\lambda(x)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I quote from my lecture notes:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Equation (3.4)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ sup_{t \in \mathbb{R}} | \hat{f} (t)|  \leq  ||&amp;#160;f ||_1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remark:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the language of normed vector spaces the inequality (3.4) means that the Fourier transform map $L^1 (\mathbb{R}) \rightarrow C_b (\mathbb{R})$, $f \rightarrow \hat{f}$ is continuous and has norm $\leq 1$, where $C_b (\mathbb{R})$ is the space of bounded continuous functions on $\mathbb{R}$, which is a normed vector space with the norm $|| f&amp;#160;||_\infty := sup_{x \in \mathbb{R}}|f(x)|$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question: what is the norm on the vector space of functions from $L^1 (\mathbb{R})$ to $C_b (\mathbb{R})$? Is it something like $ ||&amp;#160;\hat{} || := sup_{f \in L^1} \frac{sup_{t \in \mathbb{R}} | \hat{f} (t)|}{||&amp;#160;f ||_1} $?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many thanks for your help!&lt;/p&gt;&#xA;" OwnerUserId="5798" LastActivityDate="2011-02-06T14:55:14.523" Title="Continuity of Fourier transform map $\hat{}$" Tags="&lt;analysis&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" />
  <row Id="20679" PostTypeId="2" ParentId="20677" CreationDate="2011-02-06T13:52:19.800" Score="7" ViewCount="" Body="&lt;p&gt;The idea is to use &lt;a href=&quot;http://en.wikipedia.org/wiki/Holomorphic_functional_calculus&quot; rel=&quot;nofollow&quot;&gt;holomorphic functional calculus&lt;/a&gt; and to show that for a matrix $A$ and a polynomial $p(z)$ you have for $r \gt \|A\|$ &#xA;\begin{equation}\tag{$\ast$}&#xA;p(A) = \frac{1}{2\pi i} \int_{|z| = r} p(z) \cdot (z - A)^{-1}\,dz&#xA;\end{equation}&#xA;in complete analogy with the Cauchy formula for complex numbers. The integral of a matrix of holomorphic functions is defined by integrating each entry separately.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By &lt;a href=&quot;http://en.wikipedia.org/wiki/Cramer%27s_rule&quot; rel=&quot;nofollow&quot;&gt;Cramer's rule&lt;/a&gt; the $(k,l)$-entry of $(z-A)^{-1}$ is $((z-A)^{-1})_{k,l} = \frac{1}{\det(z-A)} c_{k,l}(z)$ where $c_{k,l}(z)$ is some polynomial in $z$. Let $p(z) = \det(z-A)$ be the characteristic polynomial of $A$, use $(\ast)$ and apply &lt;a href=&quot;http://en.wikipedia.org/wiki/Cauchy_integral_theorem&quot; rel=&quot;nofollow&quot;&gt;Cauchy's integral theorem&lt;/a&gt; to $c_{k,l}$.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;To see that the identity $(\ast)$ holds, you can proceed as follows (this is a slight variant of McCarthy's argument):&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The usual &lt;a href=&quot;http://en.wikipedia.org/wiki/Matrix_norm&quot; rel=&quot;nofollow&quot;&gt;matrix norm&lt;/a&gt; induced by the Euclidean norm on $\mathbb{C}^{n}$ satisfies $\|A^{n}\| \leq \|A\|^{n}$.&lt;/li&gt;&#xA;&lt;li&gt;Use this to show that $(z - A)^{-1} = \sum_{n = 0}^{\infty} \frac{A^{n}}{z^{n+1}}$, where the right hand side converges uniformly for all $|z| \gt \|A\| + \varepsilon$.&lt;/li&gt;&#xA;&lt;li&gt;From this it follows that you can interchange integration and summation and use this to conclude that $A^{k} = \int_{|z| = r} z^{k} (z - A)^{-1}\,dz$, from which $(\ast)$ follows immediately.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Here's a link to &lt;a href=&quot;http://www.jstor.org/stable/2318415&quot; rel=&quot;nofollow&quot;&gt;McCarthy's article&lt;/a&gt; (you need a university subscription to download it, but the first page is almost the entire article).&lt;/p&gt;&#xA;" OwnerUserId="5363" LastEditorUserId="5363" LastEditorDisplayName="" LastEditDate="2011-02-06T15:38:44.383" LastActivityDate="2011-02-06T15:38:44.383" CommentCount="3" />
  <row Id="20680" PostTypeId="1" CreationDate="2011-02-06T13:57:51.523" Score="8" ViewCount="424" Body="&lt;p&gt;How can I find the minimum of $x^2+\frac{a}{x}$ $\forall x&amp;gt;0$ WITHOUT Calculus?&lt;/p&gt;&#xA;" OwnerUserId="6703" LastActivityDate="2011-02-07T17:39:20.583" Title="Minimum of $x^2+\frac{a}{x}$ without Calculus" Tags="&lt;analysis&gt;" AnswerCount="5" CommentCount="8" />
  <row Id="20681" PostTypeId="2" ParentId="20680" CreationDate="2011-02-06T14:06:43.080" Score="-1" ViewCount="" Body="&lt;p&gt;You can plot it for a few values of $a$ and guess the answer.  Then plug it in and prove that small changes of $x$ increase the value.  You may argue that this approach just hides the calculus and there is some truth to that.  But you can do this without knowing how to take a derivative.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-02-06T14:06:43.080" />
  <row Id="20682" PostTypeId="2" ParentId="20680" CreationDate="2011-02-06T14:12:19.270" Score="17" ViewCount="" Body="&lt;p&gt;We have: $x^2+\frac{a}{x}=x^2+\frac{a}{2x}+\frac{a}{2x}\geq3\sqrt[3]{(x^2)(\frac{a}{2x})(\frac{a}{2x})}=3\sqrt[3]{\frac{a^2}{4}}$, where AM-GM was used. Equality occurs if $x^2=\frac{a}{2x}$.&lt;/p&gt;&#xA;" OwnerUserId="5959" LastActivityDate="2011-02-06T14:12:19.270" />
  <row Id="20683" PostTypeId="2" ParentId="20680" CreationDate="2011-02-06T14:23:34.103" Score="0" ViewCount="" Body="&lt;p&gt;By the change of variable $x=az/2$, the answer is $(a/2)^{2/3}$ times the minimum $m$ of $u(z)=z^2+2/z$ over $z\ge0$. Since $u(1)=3$, $m\le3$. Hence $m=3$ iff $u(z)\ge3$ for every $z\ge0$ iff $p(z)\ge0$ for every $z\ge0$, where $p$ is the polynomial defined as $p(z)=z(u(z)-3)=z^3-3z+2$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $p(1)=0$, $p(z)=(z-1)q(z)$ where $q$ is a polynomial of degree $2$. One finds $q(z)=z^2+z-2$. Since $q(1)=0$, $q(z)=(z-1)r(z)$ where $r$ is a poynomial of degree $1$. One finds $r(z)=z+2$. Hence $p(z)=(z-1)^2(z+2)$ and $p(z)\ge0$ for every $z\ge0$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I do not know whether you consider this proof as being without calculus.&lt;/p&gt;&#xA;" OwnerUserId="6179" LastActivityDate="2011-02-06T14:23:34.103" />
  <row Id="20684" PostTypeId="2" ParentId="20678" CreationDate="2011-02-06T14:34:51.700" Score="2" ViewCount="" Body="&lt;p&gt;To expand on the Wikipedia-Link I gave as a comment: Let $(V,\Vert\cdot\Vert_{V})$ and $(W,\Vert\cdot\Vert_{W})$ be normed linear spaces. Then a standard basic result of functional analysis asserts that for linear operator $T: V \to W$ the following are equivalent:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$T$ is continuous at $0$&lt;/li&gt;&#xA;&lt;li&gt;$T$ is continuous&lt;/li&gt;&#xA;&lt;li&gt;$T$ is uniformly continuous&lt;/li&gt;&#xA;&lt;li&gt;The &lt;strong&gt;norm of&lt;/strong&gt; $T$ defined by &#xA;\[&#xA;\|T\| = \sup_{0 \neq v \in V} \frac{\Vert Tv\Vert_{W}}{\Vert v \|_{V}} \lt \infty&#xA;\]&#xA;is finite.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;To see this, note that $\|T\| \lt \infty$ implies that $\Vert Tv_{1} - Tv_{2} \Vert_{W} \leq \|T\|\, \Vert v_{1} - v_{2}\Vert_{V}$ so that $T$ is in fact Lipschitz continuous with Lipschitz constant $\|T\|$. Hence we have the obvious chain of implications $(4.) \Rightarrow (3.) \Rightarrow (2.) \Rightarrow (1.)$. It remains to prove that $(1.) \Rightarrow (4.)$. Suppose $(4.)$ fails. Then there exists for each $n$ an $v_{n}$ with $\Vert T v_{n}\Vert_{W} \geq n \Vert v_{n} \Vert_{V}$ by the definition of $\|T\|$. Put $x_{n} = \frac{v_{n}}{n\cdot\Vert v_{n}\Vert_{V}}$ and note that $\Vert x_{n} \Vert_{V} = \frac{1}{n} \to 0$ so that $x_{n} \to 0$. On the other hand, $\|Tx_{n}\|_{W} \geq 1$ and hence $T$ can't be continuous at $0$.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Applying this to the Fourier transform $\mathcal{F}: L^{1} \to C_{b}$ you get $\|\mathcal{F}\| = \sup_{0 \neq f \in L^{1}} \frac{\Vert \mathcal{F}(f) \Vert_{C_{b}}}{\Vert f\Vert_{L^{1}}}$, which is precisely your guess (up to the minor detail $f \neq 0$). Recall that the norm $\|g\|_{C_{b}} = \sup_{t \in \mathbb{R}} |g(t)|$. Now the lemma you quote clearly says that $\|\mathcal{F}\| \leq 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me finish by mentioning that in fact the image of $L^{1}$ under the Fourier transform is contained in $C_{0} \subset C_{b}$, the space of functions vanishing at infinity. This is the content of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Riemann%2DLebesgue_lemma&quot; rel=&quot;nofollow&quot;&gt;Riemann-Lebesgue lemma&lt;/a&gt;, see also the first paragraph of &lt;a href=&quot;http://math.stackexchange.com/questions/19526/limits%2Dand%2Dconvolution/19543%2319543&quot;&gt;this answer&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added&lt;/strong&gt;: I forgot to say why $\|\mathcal{F}\| = 1$, but Akhil has explained this in his comment to your question.&lt;/p&gt;&#xA;" OwnerUserId="5363" LastEditorUserId="5363" LastEditorDisplayName="" LastEditDate="2011-02-06T14:55:14.523" LastActivityDate="2011-02-06T14:55:14.523" CommentCount="1" />
  <row Id="20685" PostTypeId="1" CreationDate="2011-02-06T14:41:41.690" Score="4" ViewCount="249" Body="&lt;p&gt;In differential calculus, We know that dy/dx is the ratio of between the rate of change in y and the rate of change in x. In other words, the rate of change in y with respect to x.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, why is dy/dx written as d/dx ( y)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is, why and how d/dx is considered as an operator?&lt;/p&gt;&#xA;" OwnerUserId="6704" LastActivityDate="2011-02-06T15:05:47.820" Title="In differential calculus, why is dy/dx written as d/dx ( y)?" Tags="&lt;calculus&gt;&lt;differential&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="3" />
  <row Id="20686" PostTypeId="1" CreationDate="2011-02-06T14:52:12.590" Score="1" ViewCount="57" Body="&lt;p&gt;I have some conditional questions.  First, in the category of topological pairs (as used in cohomology theories), are these only pairs of the form $(X,A)$ with $A \subset X$?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the answer to this question is yes, then my next question is: is a morphism $(X,A) \to (Y,B)$ a map $f:X \to Y$ with $f(A) \subset B$ or pairs of maps $f_1: X \to Y$ and $f_2: A \to B$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="9" LastActivityDate="2011-02-06T21:04:26.917" Title="Questions on the category of pairs" Tags="&lt;algebraic-topology&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="20687" PostTypeId="2" ParentId="20685" CreationDate="2011-02-06T14:56:48.700" Score="3" ViewCount="" Body="&lt;p&gt;The way you're phrasing it, $x$ and $y$ play similar roles, and the question naturally arises why they should be treated differently, as in $\mbox{d}/\mbox{d}x (y)$. However, in calculus, one usually considers functions of variables, such as $f (x)$ or $y (x)$ -- here the symbols for the independent variable and the function play quite different roles, and in order to be able to think of differentiation more abstractly as an operation applied to functions (and yielding new functions), it is helpful to &quot;factorize&quot; the notation so that the function stands alone at the right and &quot;what is being done to it&quot;, the operator, is separate and applied from the left -- hence this notation.&lt;/p&gt;&#xA;" OwnerUserId="6622" LastActivityDate="2011-02-06T14:56:48.700" />
  <row Id="20688" PostTypeId="2" ParentId="20685" CreationDate="2011-02-06T15:00:32.563" Score="6" ViewCount="" Body="&lt;p&gt;It is productive to regard $D = \frac{d}{dx}$ as a linear operator, say from the space of smooth functions on $\mathbb{R}$ to itself, for several reasons. The simplest reason I can think of is that it makes the theory of linear homogeneous differential equations very simple. For a linear homogeneous differential equation is nothing more than an attempt to find a nullspace of the operator $p(D)$ where $p$ is some polynomial. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To do this we need to find the spectrum of $D$. It's not hard to see that there is a unique eigenvector with eigenvalue $\lambda$ given by $e^{\lambda x}$, and from here it follows that the nullspace of $p(D)$ at least contains (and, if $p$ has distinct roots, is entirely made of) the functions $e^{\lambda x}$ where $p(\lambda) = 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Said another way, if $p(x) = \prod_{i=1}^n (x - \lambda_i)$ then we can factor the &lt;em&gt;operator&lt;/em&gt; $p(D)$ as $\prod_{i=1}^n (D - \lambda_i)$, and it's not hard to see that $f$ is in the nullspace of this operator whenever $(D - \lambda_i) f = 0$, or $f(x) = e^{\lambda_i x}$ (up to initial conditions). In fact, we get a solution $f$ whenever $(D - \lambda_i)^{e_i} f = 0$ where $e_i$ is the multiplicity of $\lambda_i$, and studying this condition readily leads to the complete set of solutions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, thinking of $D$ as an operator in its own right essentially reduces the study of linear homogeneous differential equations to linear algebra (modulo some existence and uniqueness arguments), specifically the study of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Jordan_decomposition&quot; rel=&quot;nofollow&quot;&gt;Jordan decomposition&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course one can go much, much &lt;a href=&quot;http://en.wikipedia.org/wiki/Functional_analysis&quot; rel=&quot;nofollow&quot;&gt;further&lt;/a&gt; with this idea: for example we can factor differential operators in more than one variable in the same way. The Laplacian $D_x^2 + D_y^2$ where $D_x$ is the derivative with respect to $x$ and $D_y$ the derivative with respect to $y$ factors as $\left( D_x + D_y i \right) \left( D_x - D_y i \right)$ and this immediately gives the connection between harmonic functions and holomorphic functions via the &lt;a href=&quot;http://en.wikipedia.org/wiki/Cauchy%E2%80%93Riemann_equations&quot; rel=&quot;nofollow&quot;&gt;Cauchy-Riemann equations&lt;/a&gt;. And the &lt;a href=&quot;http://en.wikipedia.org/wiki/Dirac_equation&quot; rel=&quot;nofollow&quot;&gt;Dirac equation&lt;/a&gt; in quantum mechanics was discovered through a similar factorization process, but with &lt;em&gt;matrix&lt;/em&gt; rather than merely complex coefficients.&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2011-02-06T15:05:47.820" LastActivityDate="2011-02-06T15:05:47.820" />
  <row Id="20689" PostTypeId="2" ParentId="20663" CreationDate="2011-02-06T15:31:14.747" Score="2" ViewCount="" Body="&lt;p&gt;A standard example of a direct limit is the construction of the ring of germs of functions at a point (on a topological space), which generalizes to the stalk of a &lt;a href=&quot;http://en.wikipedia.org/wiki/Sheaf_%28mathematics%29&quot; rel=&quot;nofollow&quot;&gt;sheaf&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the germs of holomorphic functions at $0 \in \mathbb{C}$. By definition, an element of this ring is a pair $(U, f)$ where $U$ is an open subset containing the origin and $f: U \to \mathbb{C}$ is a holomorphic function. Two pairs $(U, f), (V, g)$ are equivalent if there is a neighborhood $W \subset U \cap V$ containing $0$ such that $f = g$ on $W$. (We can take $W=U \cap V$ if we make our neighborhoods connected by the rigidity of holomorphic functions.) This is an example of a direct limit: it is the limit of $Hol(U)$ as the neighborhood $U$ shrinks to zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is a good exercise to see that this direct limit is the ring of convergent power series in $\mathbb{C}[[z]]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is another exercise you might try. Let $A$ be a ring, $S$ a multiplicatively closed subset. Then one can define a preorder (it's not really an order) on the elements of $S$ by $s&amp;#39; \leq s$ if $s&amp;#39;$ divides $s$. Then the localization $A_S$ is the direct limit of the $A_s$ as $s \in S$.&lt;/p&gt;&#xA;" OwnerUserId="536" LastActivityDate="2011-02-06T15:31:14.747" CommentCount="1" />
  <row Id="20692" PostTypeId="2" ParentId="20597" CreationDate="2011-02-06T16:06:02.883" Score="1" ViewCount="" Body="&lt;p&gt;See &lt;a href=&quot;http://www.youtube.com/watch?v=_GBtlR4m67g&quot; rel=&quot;nofollow&quot;&gt; this video &lt;/a&gt; and&#xA;&lt;a href=&quot;http://www.youtube.com/watch?v=YW1MVXyhJZE&quot; rel=&quot;nofollow&quot;&gt; its sequel &lt;/a&gt; to see the process worked in real time.&lt;/p&gt;&#xA;" OwnerUserId="722" LastActivityDate="2011-02-06T16:06:02.883" />
  <row Id="20693" PostTypeId="2" ParentId="20680" CreationDate="2011-02-06T16:13:14.813" Score="3" ViewCount="" Body="&lt;p&gt;In the overkill-but-doesn't-use-calculus department: We assume $a &amp;gt; 0$, and the goal is to determine the minimal $K$ such that the equation $x^2 + {a \over x} = K$ has a positive root. This is equivalent to finding how many positive roots $p_K(x) = x^3 - Kx +a$ has. The total number of real roots is determined by the discriminant of $p_K(x)$, which is given by $D_K = 4k^3 - 27a^2$ here. If $D_K &amp;gt; 0$, it has three distinct real roots. If $D_K &amp;lt; 0$ is has one real and two complex conjugate roots. If $D_K = 0$ it has three real roots including at least one repeated root. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next, note from the graphs of $x^2$ and ${a \over x}$ that the function $x^2 + {a \over x}$ decreases from $\infty$ to $-\infty$ as $x$ goes from $-\infty$ to zero, so for any $K$ the equation $x^2 + {a \over x} = K$ has exactly one negative root. $p_K(0) \neq 0$ as long as $a &amp;gt; 0$. Hence by the above considerations $p_K(x)$ has a real positive root if and only if $D_K \geq 0$, which translates into $K^3 \geq 27{a^2 \over 4}$, or $K \geq 3({a^2 \over 4})^{2 \over 3}$. Hence the minimal value of $x^2 + {a \over x}$ for $x &amp;gt; 0$ is given by $3({a^2 \over 4})^{2 \over 3}$.&lt;/p&gt;&#xA;" OwnerUserId="3035" LastEditorUserId="3035" LastEditorDisplayName="" LastEditDate="2011-02-06T16:30:57.330" LastActivityDate="2011-02-06T16:30:57.330" />
  <row Id="20694" PostTypeId="1" AcceptedAnswerId="20712" CreationDate="2011-02-06T16:21:15.650" Score="0" ViewCount="260" Body="&lt;p&gt;Given a matrix $A$ and column vector $x$, what is the derivative of $Ax$ with respect to $x^T$ i.e. $\frac{d(Ax)}{d(x^T)}$, where $x^T$ is the transpose of $x$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Side note - my goal is to get the known derivative formula $\frac{d(x^TAx)}{dx} = x^T(A^T + A)$ from the above rule and the chain rule.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks,&lt;br&gt;&#xA;Asaf&lt;/p&gt;&#xA;" OwnerUserId="6706" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2011-02-06T19:15:16.897" LastActivityDate="2011-02-06T21:57:51.033" Title="Vector derivative w.r.t its transpose $\frac{d(Ax)}{d(x^T)}$" Tags="&lt;calculus&gt;&lt;matrices&gt;&lt;vector-calculus&gt;" AnswerCount="3" />
  <row Id="20695" PostTypeId="1" AcceptedAnswerId="20700" CreationDate="2011-02-06T16:26:05.607" Score="0" ViewCount="74" Body="&lt;p&gt;These four inequalities define a hollow (zero volume) cube.  Is it possible to describe a hollow cube using a smaller system of (real scalar generalized-polynomial) inequalities?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{eqnarray}&#xA;x^2 &amp;amp; \leq &amp;amp; 1 \\&#xA;y^2 &amp;amp; \leq &amp;amp; 1 \\&#xA;z^2 &amp;amp; \leq &amp;amp; 1 \\&#xA;(x^2 - 1)^2(y^2 - 1)^2(z^2 - 1)^2 &amp;amp; \leq &amp;amp; 0&#xA;\end{eqnarray}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strike&gt;EDIT: After Rahul made me think in terms of norms, I came up with the following very simple form based on the 1-norm:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left ( \sqrt{x^2} + \sqrt{y^2} + \sqrt{z^2} - 1 \right ) ^2 \leq 0$$&#xA;&lt;/strike&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rahul is right, the 1-isonorm and $\infty$-isonorm are similar in $\Re^2$ but not in $\Re^3$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left ( \left ( \sqrt{x^{2p}} + \sqrt{y^{2p}} + \sqrt{z^{2p}} \right )^{1/p} - 1 \right ) ^2 \leq 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;approaches a cube in the limit as $p \rightarrow \infty$&lt;/p&gt;&#xA;" OwnerUserId="4923" LastEditorUserId="4923" LastEditorDisplayName="" LastEditDate="2011-02-06T19:29:57.243" LastActivityDate="2011-02-06T19:29:57.243" Title="Fewest inequalities to define a hollow cube" Tags="&lt;geometry&gt;&lt;inequality&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="20696" PostTypeId="2" ParentId="19997" CreationDate="2011-02-06T16:32:09.303" Score="1" ViewCount="" Body="&lt;p&gt;My attempt at an answer has two parts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, I think the geometric mean thing slightly obscures what's going on and contributes to your bafflement why &quot;the $r$'s magically fall out of the equation in the last step&quot;. A more geometric view of this step is: We know that the curve has &quot;width&quot; $w=2r$ (along the chosen direction). Then $A = w\overline{h} = 2r\overline{h}$, where $\overline{h}$ is the &quot;average height&quot; of the curve (along the chosen direction). Thus we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\[A + \pi r^2 \le lr \;\Leftrightarrow\; 2r\overline{h} + \pi r^2 \le lr \;\Leftrightarrow\; 2\overline{h} + \pi r \le l \;\Leftrightarrow\; 2\overline{h} + \frac{\pi}{2} w \le l\;.\]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the cancellation of the $r$'s seems more natural, and the inequality gives an upper bound for the &quot;average height&quot; we can achieve for a given &quot;width&quot; $w$ and length $l$. This bound implies the isomperimetric inequality:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\[l^2 \ge (2\overline{h} + \frac{\pi}{2} w)^2=(2\overline{h} - \frac{\pi}{2} w)^2 + 4\pi w\overline{h} \ge 4\pi w\overline{h}=4\pi A\;.\]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(There may be a connection here to the inequality by Bonnesen cited in Christian Blatter's comment.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second, regarding the proof as a whole, it seems useful to think of it as a way of transforming the difficult global optimization problem implied by the isoperimetric inequality (how to enclose the greatest possible area within a given circumference) into a trivial local optimization problem through some clever bookkeeping. In a sense, what makes the problem difficult is that how much area you can enclose with a curve element (say, with respect to the origin) depends both on where you are and in which direction you move, but in which direction you move in turn determines where you will end up, and hence how much area you will be able to enclose later.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The proof decouples this by adding to the area element a suitable penalty which has two crucial properties: It exactly cancels the &quot;where you are&quot; part of how much area you can enclose, and because it is itself the area element of a circle, it automatically adds up to a constant. There are no longer any variable lengths in the integrand, only the angle between the tangent vector at the curve and the tangent vector at the corresponding point of the circle, and it is then obvious that the integral is maximized by always choosing the tangent vector of the curve parallel to the tangent vector of the circle -- which necessarily results in a circle.&lt;/p&gt;&#xA;" OwnerUserId="6622" LastEditorUserId="6622" LastEditorDisplayName="" LastEditDate="2011-02-06T16:42:30.747" LastActivityDate="2011-02-06T16:42:30.747" />
  <row Id="20697" PostTypeId="1" AcceptedAnswerId="20741" CreationDate="2011-02-06T16:35:23.137" Score="5" ViewCount="413" Body="&lt;p&gt;I have some exam questions that were left unanswered for me:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Suppose that for every $\alpha&amp;lt;\kappa$ there is a subset $A_\alpha$ of $\kappa$ of cardinality $\kappa$. Show that there is a subset X of $\kappa$ so that for every $\alpha&amp;lt;\kappa$ : |$A_\alpha\cap$ X|=|$A_\alpha$ \ X|=$\kappa$. &lt;/li&gt;&#xA;&lt;li&gt;{$A_\alpha | \alpha&amp;lt;\aleph_1$} a collection of non-stationary disjoint sets. Suppose that their union is a stationary set. Prove that the set {$minA_\alpha | \alpha&amp;lt;\aleph_1$} is stationary.&lt;/li&gt;&#xA;&lt;li&gt;For every couple of sets A,B $\in P(\omega)$ we will denote A $\subseteq^*$ B if $|A\setminus B|&amp;lt;\aleph_0$. Prove that there exist a sequence of length $\aleph_1$ in relation to $\subseteq^*$ .&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I'll appreciate any help regarding those questions.&#xA;Thanks in advance,&#xA;Pavel&lt;/p&gt;&#xA;" OwnerUserId="6707" LastEditorUserId="6707" LastEditorDisplayName="" LastEditDate="2011-02-06T18:45:57.083" LastActivityDate="2011-03-13T20:56:56.387" Title="Some questions in Set Theory" Tags="&lt;set-theory&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="20698" PostTypeId="2" ParentId="20695" CreationDate="2011-02-06T16:38:05.570" Score="1" ViewCount="" Body="&lt;p&gt;What sort of functions do you allow? If you allow square roots then you can encode multiple inequalities in a single one:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$A \geq 0 \wedge B \geq 0 \Leftrightarrow (\sqrt{A^2}-A)^2 + (\sqrt{B^2}-B)^2 \leq 0$&lt;/p&gt;&#xA;" OwnerUserId="3747" LastActivityDate="2011-02-06T16:38:05.570" />
  <row Id="20699" PostTypeId="1" AcceptedAnswerId="20701" CreationDate="2011-02-06T16:40:10.373" Score="1" ViewCount="65" Body="&lt;p&gt;To calculate an average in reverse polish notation (RPN) is the following correct (assuming a-j are variables representing numbers):&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;a,b,+,c,+,d,+,e,+,f,+,g,+,h,+,i,+,j,+,10,/&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;If it is correct, are there other ways that this could be expressed in RPN?&lt;/p&gt;&#xA;" OwnerUserId="2713" LastActivityDate="2011-02-06T17:05:55.353" Title="Calculate an Average in RPN (Reverse Polish Notation)" Tags="&lt;polish-notation&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="20700" PostTypeId="2" ParentId="20695" CreationDate="2011-02-06T16:40:58.153" Score="1" ViewCount="" Body="&lt;p&gt;&lt;em&gt;Note: When I first posted this answer, the question did not contain the condition that the equations should involve polynomials.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Certainly there exist infinitely many possibilities of a single scalar equation that will do the job, using any function that is zero on the boundary of the unit cube and nonzero elsewhere. Here is one explicit example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\max(|x|,|y|,|z|) - 1 = 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you prefer inequalities, you can just square the left-hand side and set it $\le 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But is this just a mathematical version of code golf, or is there some other reason for this question?&lt;/p&gt;&#xA;" OwnerUserId="856" LastEditorUserId="856" LastEditorDisplayName="" LastEditDate="2011-02-06T17:58:46.763" LastActivityDate="2011-02-06T17:58:46.763" CommentCount="1" />
  <row Id="20701" PostTypeId="2" ParentId="20699" CreationDate="2011-02-06T16:55:27.760" Score="3" ViewCount="" Body="&lt;p&gt;This is correct. There are more ways to compute this but I would call your solution fairly canonical. For example, you could have started like this:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;a,b,c,+,+,...&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;or like this:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;10,1/x,a,b,+,...,*&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;However, these are fairly obvious variants and you are probably looking for a completely different way. &lt;/p&gt;&#xA;" OwnerUserId="6678" LastEditorUserId="6678" LastEditorDisplayName="" LastEditDate="2011-02-06T17:05:55.353" LastActivityDate="2011-02-06T17:05:55.353" />
  <row Id="20702" PostTypeId="2" ParentId="20694" CreationDate="2011-02-06T17:01:39.370" Score="1" ViewCount="" Body="&lt;p&gt;I think there is no such thing. $\mbox{d}(x^\mbox{T}Ax)/\mbox{d}x$ is something that, when multiplied by the change $\mbox{d}x$ in $x$, yields the change $\mbox{d}(x^\mbox{T}Ax)$ in $x^\mbox{T}Ax$. Such a thing exists and is given by the formula you quote. $\mbox{d}(Ax)/\mbox{d}(x^\mbox{T})$ would have to be something that, when multiplied by the change $\mbox{d}x^\mbox{T}$ in $x^\mbox{T}$, yields the change $\mbox{d}Ax$ in $Ax$. No such thing exists, since $x^\mbox{T}$ is a $1 \times n$ row vector and $Ax$ is an $n \times 1$ column vector.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If your main goal is to derive the derivative formula, here's a derivation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(x^\mbox{T} + \mbox{d}x^\mbox{T})A(x + \mbox{d}x) = x^\mbox{T}Ax + \mbox{d}x^\mbox{T}Ax + x^\mbox{T}A\mbox{d}x + \mbox{d}x^\mbox{T}A\mbox{d}x =$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$=x^\mbox{T}Ax + x^\mbox{T}A^\mbox{T}\mbox{d}x + x^\mbox{T}A\mbox{d}x + O (\lVert \mbox{d}x \rVert^2) = x^\mbox{T}Ax + x^\mbox{T}(A^\mbox{T} + A)\mbox{d}x + O (\lVert \mbox{d}x \rVert^2)$&lt;/p&gt;&#xA;" OwnerUserId="6622" LastEditorUserId="6622" LastEditorDisplayName="" LastEditDate="2011-02-06T18:59:14.357" LastActivityDate="2011-02-06T18:59:14.357" CommentCount="12" />
  <row Id="20703" PostTypeId="2" ParentId="20450" CreationDate="2011-02-06T17:05:54.047" Score="0" ViewCount="" Body="&lt;p&gt;If I understood it well (did I?), Byron Schmuland's solution needs no additional conditions on your space $X$, since it relies on the theorem of characterization of maximal orthonormal systems (of arbitrary cardinal), although you'd have to exhibit one such system inside the bounded and uniformly continuous functions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An alternate solution could use Lusin's theorem, for which you need to throw in local compactness of $X$. For the relevant setup, you can follow Rudin's &quot;&lt;em&gt;Real and complex analysis&lt;/em&gt;&quot;, theorems &lt;strong&gt;2.24&lt;/strong&gt; and &lt;strong&gt;3.14&lt;/strong&gt; in the third edition. The advantage here is that you get your result cheaply (local compactness is not that weird for a base space ;-) and for any $1 \leq p &amp;lt;  \infty$, not just for the Hilbert case.&lt;/p&gt;&#xA;" OwnerUserId="6280" LastActivityDate="2011-02-06T17:05:54.047" />
  <row Id="20704" PostTypeId="2" ParentId="20664" CreationDate="2011-02-06T17:13:06.037" Score="4" ViewCount="" Body="&lt;p&gt;According to the theorem about finitely generated modules, any finitely generated module over a PID $R$ can be expressed as&#xA;$$&#xD;&#xA;R^r \oplus R/(p_1^{k_1}) \oplus R/(p_2^{k_2}) \oplus \cdots \oplus R/(p_n^{k_n})&#xD;&#xA;$$&#xA;where $p_1^{k_1},\ldots,p_n^{k_n}$ are prime powers in $R$.  The $R^r$ is called the &lt;em&gt;free part&lt;/em&gt;, and the other summands are &lt;em&gt;elementary divisors&lt;/em&gt;.  According to the theorem, these summands are unique up to permutation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $p^k$ be a prime power in $R$.  Let $a$, $b$, $c$, and $d$ denote the numbers of times that $R/(p^k)$ appears as an elementary divisor of $A$, $B$, $C$, and $D$, respectively.  Since $A\oplus B \cong C \oplus D$, the uniqueness part of the theorem tells us that $a+b=c+d$.  Similarly, since $A\oplus D = C\oplus B$, we know that $a+d=c+b$.  Since $a$, $b$, $c$, and $d$ are nonnegative, it follows that $a=c$ and $b=d$.  This holds for all $p^k$, and a similar argument works for the rank of the free parts, which proves that $A\cong C$ and $B\cong D$&lt;/p&gt;&#xA;" OwnerUserId="1726" LastEditorUserId="1726" LastEditorDisplayName="" LastEditDate="2011-02-06T18:24:48.157" LastActivityDate="2011-02-06T18:24:48.157" />
  <row Id="20705" PostTypeId="1" AcceptedAnswerId="20713" CreationDate="2011-02-06T17:14:43.810" Score="1" ViewCount="183" Body="&lt;p&gt;I am having trouble in solving the following problem.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Let $p_{n}$ denote the $n$-th prime. Then prove that $$\pi(\sqrt{p_{1}p_{2}\cdots p_{n}})&amp;gt;2n$$ for $n \geq 6$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;No idea how to start.&lt;/p&gt;&#xA;" LastActivityDate="2011-02-06T19:31:13.683" Title="Proving $\pi(\sqrt{p_{1}p_{2}\cdots p_{n}})&gt;2n$ for $n \geq 6$" Tags="&lt;number-theory&gt;&lt;analysis&gt;&lt;analytic-number-theory&gt;" AnswerCount="1" CommentCount="6" />
  <row Id="20706" PostTypeId="2" ParentId="20660" CreationDate="2011-02-06T17:21:04.440" Score="1" ViewCount="" Body="&lt;p&gt;Presuming you mean when an -academic- job market is tight (research, or separately teaching at the university level), there's:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;finance - supposedly they're always looking for people who can come up with mathematical models&lt;/li&gt;&#xA;&lt;li&gt;software development - not necessarily to -do- programming, but to help with analysis of problems. depends on the application itself of course, and helps to have some computer skills&lt;/li&gt;&#xA;&lt;li&gt;secondary school - yes, this sounds like a downgrade of expectations, and there's the extra overhead of extra credentialing for the public school system, but it -is- academic.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;What I'm saying is that for pure math, well, applied math is where the jobs are outside of academia. MS Research, whatever the AT&amp;amp;T version that still exists, and the NSA might be a places where one can do -some- pure math as long as there is some ostensible, tenuous link to an application.&lt;/p&gt;&#xA;" OwnerUserId="1919" LastActivityDate="2011-02-06T17:21:04.440" CommunityOwnedDate="2011-02-06T20:11:20.733" />
  <row Id="20707" PostTypeId="2" ParentId="20643" CreationDate="2011-02-06T17:25:33.843" Score="1" ViewCount="" Body="&lt;p&gt;I feel like I have to mention Schrijver's &quot;Theory of Linear and Integer Programming&quot; as it is a classic, but I don't feel like it is the best text for an introduction or for application. &lt;/p&gt;&#xA;" OwnerUserId="1919" LastActivityDate="2011-02-06T17:25:33.843" CommentCount="1" />
  <row Id="20712" PostTypeId="2" ParentId="20694" CreationDate="2011-02-06T18:07:24.523" Score="1" ViewCount="" Body="&lt;p&gt;Let $f(x) = x^TAx$ and you want to evaluate $\frac{df(x)}{dx}$. This is nothing but the gradient of $f(x)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are two ways to represent the gradient one as a row vector or as a column vector. From what you have written, your representation of the gradient is as a row vector.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First make sure to get the dimensions of all the vectors and matrices in place. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here $x \in \mathbb{R}^{n \times 1}$, $A \in \mathbb{R}^{n \times n}$ and $f(x) \in \mathbb{R}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This will help you to make sure that your arithmetic operations are performed on vectors of appropriate dimensions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now lets move on to the differentiation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;All you need to know are the following rules for vector differentiation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{d(x^Ta)}{dx} = \frac{d(a^Tx)}{dx} = a^T$$ where $x,a \in \mathbb{R}^{n \times 1}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that $x^Ta = a^Tx$ since it is a scalar and the equation above can be derived easily.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Some people follow a different convention i.e. treating the derivative as a column vector instead of a row vector. Make sure to stick to your convention and you will end up with the same conclusion in the end)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Make use of the above results to get,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{d(x^TAx)}{dx} = x^T A^T + x^T A$$&#xA;Use chain rule to get the above result i.e. first take $Ax$ as constant and then take $x^T A$ as constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, $$\frac{df(x)}{dx} = x^T(A^T + A)$$&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2011-02-06T19:04:04.113" LastActivityDate="2011-02-06T19:04:04.113" CommentCount="3" />
  <row Id="20713" PostTypeId="2" ParentId="20705" CreationDate="2011-02-06T18:07:49.043" Score="2" ViewCount="" Body="&lt;p&gt;The case $n=6$ can be verified by hand (you only need to find $13$ primes less than $\sqrt{2\cdot 3\cdot 5\cdot 7\cdot 11\cdot 13}$), and once you have it this follows by induction and Bertrand's postulate, because for $n\gt 6$, $p_n\gt 16\Rightarrow \sqrt{p_n}\gt 2^2$.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2011-02-06T19:31:13.683" LastActivityDate="2011-02-06T19:31:13.683" CommentCount="2" />
  <row Id="20714" PostTypeId="2" ParentId="20661" CreationDate="2011-02-06T18:13:00.033" Score="4" ViewCount="" Body="&lt;p&gt;Just so the question gets an answer: &lt;strong&gt;yes&lt;/strong&gt;, your proof is correct and is one of several phrasings of the shortest proof that I know.  &lt;/p&gt;&#xA;" OwnerUserId="299" LastActivityDate="2011-02-06T18:13:00.033" CommunityOwnedDate="2011-02-06T18:13:00.033" />
  <row Id="20715" PostTypeId="1" CreationDate="2011-02-06T18:48:08.640" Score="1" ViewCount="121" Body="&lt;p&gt;Let $\{r_n\}_{n\in\mathbb{N}}$ be a enumeration over the rationals&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let &#xA;$$g(x)=\sum_1^\infty \frac{1}{2^n} \frac{1}{\sqrt{x-r_n}} \chi_{(0,1]}$$&#xA;where &#xA;$$\chi_{(0,1]} = \left\{\begin{array}{ll}&#xA;1&amp;amp;\mbox{if $x-r_n \in (0,1]$,}\\&#xA;0&amp;amp;\mbox{otherwise.}&#xA;\end{array}\right.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Show that $g$ is not continuous except possibly on a set of measure $0$.&lt;/p&gt;&#xA;" OwnerUserId="6556" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-06T19:57:04.303" LastActivityDate="2011-02-06T19:57:04.303" Title="Show that this function is not continuous except on a set of measure zero" Tags="&lt;measure-theory&gt;" CommentCount="4" />
  <row Id="20717" PostTypeId="1" AcceptedAnswerId="20727" CreationDate="2011-02-06T19:08:49.023" Score="2" ViewCount="184" Body="&lt;p&gt;Hello,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to find a set of integer solutions of Diophantine equation: $ax + by = c$, and apparently $gcd(a,b)|c$. Then by what formula can I use to find $x$ and $y$ ? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried to play around with it:&lt;br&gt;&#xA;$x = (c - by)/a$, hence $a|(c - by)$.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a$, $c$ and $b$ are known. So to obtain integer solution for $a$, then $c - by = ak$, and I lost from here, because $y = (c - ak)/b$. I kept repeating this routine and could not find a way to get rid of it? Any hint?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks,&lt;br&gt;&#xA;Chan&lt;/p&gt;&#xA;" OwnerUserId="5711" LastEditorUserId="5711" LastEditorDisplayName="" LastEditDate="2011-02-06T21:37:48.157" LastActivityDate="2011-02-28T01:06:01.610" Title="How to find solutions of linear Diophantine ax + by = c ?" Tags="&lt;elementary-number-theory&gt;" AnswerCount="3" CommentCount="3" />
  <row Id="20719" PostTypeId="2" ParentId="20246" CreationDate="2011-02-06T19:21:24.917" Score="1" ViewCount="" Body="&lt;p&gt;This one has been untouched for a few days now, so here's a solution. It does the job but it feels like I beat it into submission with a stick.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would be interested to know if anyone has a more elegant approach, as it was tougher than I was expecting. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/E2lw5.jpg&quot; alt=&quot;triangle&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the above diagram $I$ is the incentre and $K$ is the circumcentre of $\Delta$ ABC.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$PS \parallel BC $ therefore&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \frac{CP}{BS}=\frac{AC}{AB}= \frac{ \sin B}{\sin C}, \quad (1)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;by the sine rule.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore, $\Delta$ CIP and $\Delta$ BIS are isosceles and so&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \frac{PS}{BC}=\frac{CP+BS}{BC}= 1 - \frac{BS}{AB},$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the last equality is implied by $PS \parallel BC .$ And so&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ CP = \left( 1 - \frac{BS}{AB} \right) BC - BS = \frac{\sin B}{\sin C} \cdot BS,$$&#xA;using $(1).$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ BC = BS \left( 1+ \frac{\sin B}{\sin C} + \frac{BC}{AB} \right)&#xD;&#xA;= BS \left( 1+ \frac{\sin B}{\sin C} + \frac{\sin (B+C)}{\sin C} \right), \quad (2)$$&#xA;by the sine rule.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $ \angle CBK=\theta$ then, again by the sine rule and noting that $BK=CK,$&#xA;we have&#xA;$$\frac{\sin B}{\sin (B+C)} = \frac{AC}{BC} = \frac{\cos (C-\theta)}{\cos \theta} =&#xD;&#xA;\cos C + \sin C \tan \theta.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \tan \theta = \frac{ \frac{\sin B}{\sin (B+C)} - \cos C}{\sin C}. $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$BC = 2 RK \cot \theta = 2 BS \sin B \cot \theta&#xD;&#xA;= \frac{2BS \sin B \sin C}{ \frac{\sin B}{\sin(B+C)} - \cos C}, \quad (3)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;using $RK = BS \sin B$ and the expression for $\tan \theta.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Combining $(2)$ and $(3)$ we obtain&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ 1+ \frac{\sin B}{\sin C} + \frac{\sin (B+C)}{\sin C}&#xD;&#xA;=\frac{2\sin B \sin C}{ \frac{\sin B}{\sin(B+C)} - \cos C}. \quad (4)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we need to show that $(4) \Rightarrow \cos B + \cos C = 1.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ABC is a non-degenerate triangle and so $C \ne 0$ and $\cos B + \cos C \ne -1$ and so&#xA;$\cos B + \cos C = 1 \Longleftrightarrow$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$2\sin B \sin^2 C = 2 \sin B (1- \cos^2 C) + \sin C&#xD;&#xA;\lbrace 1 - (\cos B + \cos C)^2 \rbrace $$&#xA;$$= 2\sin B - 2\sin B \cos^2 C + \sin C (-2\cos B \cos C + \sin^2 B - \cos^2 C)$$&#xA;$$=\lbrace 2 + \cos(B-C) \rbrace \lbrace \sin B - \cos C \sin(B+C) \rbrace, \quad (5)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;noting that $\sin(B+C)\cos(B-C) = \sin B \cos B + \sin C \cos C.$ When we expand the RHS of $(5)$ the two terms $\sin B \cos B \cos C$ and $-\sin B \cos B \cos C$ cancel, leaving precisely the RHS of the second to last line of $(5).$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence we have $\cos B + \cos C = 1 \Longleftrightarrow$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \frac{2+ \cos(B-C)}{\sin C} = \frac{2 \sin B \sin C}{\sin B - \cos C \sin (B+C)}. \quad (6)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sin (B+C) \lbrace 1+ \cos(B-C) \rbrace = \sin(B+C) + \sin(B+C) \cos(B-C)$$&#xA;$$=\sin B \cos C + \cos B\sin C + \sin B \cos B + \sin C \cos C $$&#xA;$$=(\sin B + \sin C)(\cos B + \cos C) = \sin B + \sin C \quad (7)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Longleftrightarrow \cos B + \cos C = 1,$ since $\sin B + \sin C \ne 0.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And so the LHS of $(4)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ 1+ \frac{\sin B}{\sin C} + \frac{\sin (B+C)}{\sin C} =&#xD;&#xA;\frac{\sin B + \sin C + \sin(B+C)}{\sin C}$$&#xA;and using $(7)$ to substitute for $\sin B + \sin C$&#xA;$$=\frac{2\sin(B+C)+ \sin(B+C)\cos(B-C)}{\sin C}&#xD;&#xA;=\sin(B+C)\frac{2+\cos(B-C)}{\sin C}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and using $(6)$ to substitute for the fraction&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$=\frac{2\sin B \sin C \sin(B+C)}{\sin B - \cos C \sin(B+C)}&#xD;&#xA;=\frac{2\sin B \sin C}{ \frac{\sin B}{\sin(B+C)} - \cos C}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And so we have shown that when $B$ and $C$ are angles of a triangle then&#xA;$(4) \Rightarrow \cos B + \cos C = 1$ and the result is proved.&lt;/p&gt;&#xA;" OwnerUserId="1301" LastActivityDate="2011-02-06T19:21:24.917" CommentCount="2" />
  <row Id="20720" PostTypeId="2" ParentId="20717" CreationDate="2011-02-06T19:33:20.767" Score="1" ViewCount="" Body="&lt;p&gt;Do you mean $\gcd(a,b)$ divides $c$? If so, you can divide both sides of the equation to get&#xA;$$&#xD;&#xA;\frac{a}{g}x+\frac{b}{g}y=\frac{c}{g}&#xD;&#xA;$$&#xA;where $g=\gcd(a,b)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But since $\gcd(a/g,b/g)=1$, you can use the extended Euclidean algorithm to find a solution $(x_0,y_0)$ to the equation&#xA;$$&#xD;&#xA;\frac{a}{g}x+\frac{b}{g}y=1.&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you have that, the solution $(X,Y)=(\frac{c}{g}\cdot x_0,\frac{c}{g}\cdot y_0)$ is a solution to your original equation. Furthermore, the values&#xA;$$&#xD;&#xA;x=X + \frac{b}{g} t\quad y=Y - \frac{a}{g} t&#xD;&#xA;$$&#xA;give all solutions when $t$ ranges over $\mathbb{Z}$, I believe.&lt;/p&gt;&#xA;" OwnerUserId="1583" LastActivityDate="2011-02-06T19:33:20.767" CommentCount="1" />
  <row Id="20721" PostTypeId="1" CreationDate="2011-02-06T19:47:04.170" Score="3" ViewCount="113" Body="&lt;p&gt;How does one show that the prime spectrum of a domain is irreducible in the Zariski topology?&lt;/p&gt;&#xA;" OwnerUserId="6166" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2011-02-06T21:50:41.083" LastActivityDate="2011-02-07T02:40:27.237" Title="Why is the prime spectrum of a domain irreducible in the Zariski topology" Tags="&lt;algebraic-geometry&gt;&lt;commutative-algebra&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="20722" PostTypeId="1" AcceptedAnswerId="20735" CreationDate="2011-02-06T19:50:46.120" Score="1" ViewCount="75" Body="&lt;p&gt;I am not sure where to begin on this:&#xA;Determine the images of all conic sections with a focus at the origin under the principal branch of the complex square root.&#xA;I probably have to use the formula u(x, y) + i v(x, y) = sqrt((x, y)) and view square roots via the polar form. &#xA;Any and every suggestion is appreciated. Thanks!&lt;/p&gt;&#xA;" OwnerUserId="6399" LastActivityDate="2011-02-06T21:38:36.470" Title="Complex Square Root" Tags="&lt;homework&gt;" AnswerCount="1" />
  <row Id="20723" PostTypeId="2" ParentId="20694" CreationDate="2011-02-06T19:51:39.067" Score="0" ViewCount="" Body="&lt;p&gt;As Sivaram points out, you must define your convention about rows/colums derivatives and just be consistent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, you could define the derivative of a column vector with respect to a row vector as (assuming the letters represent column vectors) as matrix:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle \frac{d(y)}{dx^T} = D$ with $d_{i,j} = \frac{d(y_i)}{dx^j}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And that will work (it will be consistent). For example, you get $\displaystyle  \frac{d(Ax)}{dx^T} = A$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But it's not so simple to apply this -and the product rule of derivation- to deduce your identity, because you get to different derivatives: a row with respect to a row and a column respect to row, and you can't (at least without further justification) mix them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, if the matrix is simmetric all is simpler.&lt;/p&gt;&#xA;" OwnerUserId="312" LastEditorUserId="312" LastEditorDisplayName="" LastEditDate="2011-02-06T21:57:51.033" LastActivityDate="2011-02-06T21:57:51.033" CommentCount="1" />
  <row Id="20724" PostTypeId="2" ParentId="20721" CreationDate="2011-02-06T20:15:40.917" Score="7" ViewCount="" Body="&lt;p&gt;A topological space is irreducible if and only if it is not empty and any two nonempty open subsets have nontrivial intersection. Equivalently, if and only if the space is not empty, and cannot be written as a union of two proper closed subsets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(To see the equivalence: suppose any two nonempty open subsets have nontrival intersection, and let $C_1$ and $C_2$ be closed subsets such that $C_1\cup C_2 = X$. Let $O_1=X-C_1$, $O_2=X-C_2$; since $O_1\cap O_2 = (X-C_1)\cap(X-C_2) = X-(C_1\cup C_2) = \emptyset$, either $O_1$ is empty or $O_2$ is empty, hence either $C_1 =X$ or $C_2=X$. Conversely, suppose that $X$ is not the union of two proper closed subsets, and let $O_1$ and $O_2$ be nonempty open sets. Then $C_1=X-O_1$ and $C_2=X-O_2$ are proper closed subsets, so $\emptyset\neq X-(C_1\cup C_2) = (X-C_1)\cap(X-C_2) = O_1\cap O_2$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a ring $R$, $\mathrm{Spec}(R)$ as a set consists of all prime ideals of $R$, and the closed sets are the sets of the form $V(E)$, where $E\subseteq R$ is a subset, and $V(E) = \{\mathfrak{p}\in\mathrm{Spec}(R) \mid E\subseteq \mathfrak{p}\}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, suppose $R$ is a domain. Is $\mathrm{Spec}(R)$ empty? &lt;strong&gt;Hint:&lt;/strong&gt; Remember that if $S$ is a commutative ring with identity, then $\mathfrak{I}$ is a prime ideal if and only if $S/\mathfrak{I}$ is an integral domain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now suppose that $E$ and $F$ are subsets of $R$ such that $\mathrm{Spec}(R) = V(E) \cup V(F)$. That means: &lt;em&gt;every&lt;/em&gt; prime ideal of $R$ contains either $E$ or $F$ (or both). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, there is a very special prime ideal that must contain either $E$ or $F$, which tells you that $E$ or $F$ is actually equal to &lt;code&gt;blank.&lt;/code&gt; And therefore, either $V(E)$ or $V(F)$ is &lt;code&gt;blah&lt;/code&gt;. Which means that $\mathrm{Spec}(R)$ is &lt;code&gt;blankety-blank.&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For an encore, try Exercise I.19 of Atiyah-MacDonald, which says that in general, $\mathrm{Spec}(A)$ is irreducible if and only if the nilradical of $A$ is a prime ideal (and notice that your proposition is just one implication of a special case of this).&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-07T02:40:27.237" LastActivityDate="2011-02-07T02:40:27.237" CommentCount="2" />
  <row Id="20725" PostTypeId="1" CreationDate="2011-02-06T20:25:50.347" Score="1" ViewCount="91" Body="&lt;p&gt;I was doing some thought experiments for a game project, and while considering something related to pathfinding, this problem came into my head.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Say we have an infinite plane that is covered with an even random distribution of connected points, such that the average distance of a point to any of its connected neighbours is some constant $k$. Given an arbitrary subset of these points of size $n$, is it possible to say anything about the average distance $d$ to the nearest of the infinite number of congruent subsets elsewhere in the plane?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://dl.dropbox.com/u/1612665/stackexchange/math-2075-1.png&quot; alt=&quot;Graph Example&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's basically a graph embedded in the plane with non-intersecting edges. For the purpose of the problem, I can assume that it always subdivides the plane into triangles. The average distance between $v$ and its 6 neighbours is approximately $k$. The subset $S$ of size $n=3$ is congruent to the subset $S&amp;#39;$, and the subset $T$ of size $n=2$ is congruent to the subset $T&amp;#39;$. It's not shown here, but a subset need not be fully connected: you could have a subset $R=S \cup T$, for instance, to which $S&amp;#39; \cup T&amp;#39;$ is &lt;em&gt;not&lt;/em&gt; congruent. Because the plane is infinite, for any given subset there are infinitely many congruent subsets. Since subsets of size 1 are an average of $k$ units away from the nearest congruent subset, is the average distance between congruent subsets proportional to $n$, and if so, how so?&lt;/p&gt;&#xA;" OwnerUserId="6717" LastEditorUserId="6717" LastEditorDisplayName="" LastEditDate="2011-02-07T00:19:54.147" LastActivityDate="2011-02-07T00:19:54.147" Title="Average distance between features in plane of connected points with known average distance" Tags="&lt;geometry&gt;&lt;graph-theory&gt;" CommentCount="12" />
  <row Id="20726" PostTypeId="2" ParentId="20680" CreationDate="2011-02-06T20:32:51.127" Score="3" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Revised (neater) answer&lt;/strong&gt;: Fix $a&amp;gt;0$. For any $x&amp;gt;0$,&#xA;$$&#xD;&#xA;\Big(x^2  + \frac{a}{x}\Big) - \Big(m^2  + \frac{a}{m}\Big) = \frac{{(x - m)(x^2 m + m^2 x - a)}}{{mx}} \geq 0,&#xD;&#xA;$$&#xA;if $2m^3 = a$. We are done. (The above also shows that the minimum is strict.)&lt;/p&gt;&#xA;" OwnerUserId="2810" LastEditorUserId="2810" LastEditorDisplayName="" LastEditDate="2011-02-07T17:39:20.583" LastActivityDate="2011-02-07T17:39:20.583" CommentCount="2" />
  <row Id="20727" PostTypeId="2" ParentId="20717" CreationDate="2011-02-06T20:46:40.440" Score="5" ViewCount="" Body="&lt;p&gt;The diophantine equation $ax+by = c$ has solutions if and only if $\gcd(a,b)|c$. If so, it has infinitely many solutions, and any one solution can be used to generate all the other ones.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To see this, note that the greatest common divisor of $a$ and $b$ divides both $ax$ and $by$, hence divides $c$ if there is a solution. This gives the necessity of the condition &lt;strike&gt;(which you have backwards).&lt;/strike&gt; &lt;em&gt;(fixed in edits)&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The converse is actually a &lt;em&gt;constructive&lt;/em&gt; proof, that you can find in pretty much every elementary number theory course or book, and which is essentially the same as yunone's answer above (but without dividing through first).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the &lt;a href=&quot;http://en.wikipedia.org/wiki/Extended_Euclidean_algorithm&quot; rel=&quot;nofollow&quot;&gt;Extended Euclidean Algorithm&lt;/a&gt;, given any integers $a$ and $b$ you can find integers $s$ and $t$ such that $as+bt = \gcd(a,b)$; the numbers $s$ and $t$ are not unique, but you only need one pair. Once you find $s$ and $t$, since we are assuming that $\gcd(a,b)$ divides $c$, there exists an integer $k$ such that $\gcd(a,b)k = c$. Multiplying $as+bt=\gcd(a,b)$ through by $k$ you get&#xA;$$a(sk) + b(tk) = \gcd(a,b)k = c.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So this gives one solution, with $x=sk$ and $y=tk$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now suppose that $ax_1 + by_1 = c$ is a solution, and $ax+by=c$ is some other solution. Taking the difference between the two, we get&#xA;$$a(x_1-x) + b(y_1-y) = 0.$$&#xA;Therefore, $a(x_1-x) = b(y-y_1)$. That means that $a$ divides $b(y-y_1)$, and therefore $\frac{a}{\gcd(a,b)}$ divides $y-y_1$. Therefore, $y = y_1 + r\frac{a}{\gcd(a,b)}$ for some integer $r$. Substituting into the equation $a(x_1-x) = b(y-y_1)$ gives&#xA;$$a(x_1 - x) = rb\left(\frac{a}{\gcd(a,b)}\right)$$&#xA;which yields&#xA;$$\gcd(a,b)a(x_1-x) = rba$$&#xA;or $x = x_1 - r\frac{b}{\gcd(a,b)}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, if $ax_1+by_1 = c$ is any solution, then all solutions are of the form&#xA;$$x = x_1 - r\frac{b}{\gcd(a,b)},\qquad y = y_1 + r\frac{a}{\gcd(a,b)}t$$&#xA;exactly as yunone said.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;hr/&gt;&#xA;To give you an example of this in action, suppose we want to find all integer solutions to $$258x + 147y = 369.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, we use the Euclidean Algorithm to find $\gcd(147,258)$; the parenthetical equation on the far right is how we will use this equality after we are done with the computation.&#xA;\begin{align*}&#xA;258 &amp;amp;= 147(1) + 111 &amp;amp;\quad&amp;amp;\mbox{(equivalently, $111=258 - 147$)}\\&#xA;147 &amp;amp;= 111(1) + 36&amp;amp;&amp;amp;\mbox{(equivalently, $36 = 147 - 111$)}\\&#xA;111 &amp;amp;= 36(3) + 3&amp;amp;&amp;amp;\mbox{(equivalently, $3 = 111-3(36)$)}\\&#xA;36 &amp;amp;= 3(12).&#xA;\end{align*}&#xA;So $\gcd(147,258)=3$. Since $3|369$, the equation has integral solutions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then we find a way of writing $3$ as a linear combination of $147$ and $258$, using the Euclidean algorithm computation above, and the equalities on the far right. We have:&#xA;\begin{align*}&#xA;3 &amp;amp;= 111 - 3(36)\\&#xA;&amp;amp;= 111 - 3(147 - 111) = 4(111) - 3(147)\\&#xA;&amp;amp;= 4(258 - 147) - 3(147)\\&#xA;&amp;amp;= 4(258) -7(147).&#xA;\end{align*}&#xA;Then, we take $258(4) + 147(-7)=3$, and multiply through by $123$; why $123$? Because $3\times 123 = 369$. We get:&#xA;$$258(492) + 147(-861) = 369.$$&#xA;So one solution is $x=492$ and $y=-861$. All other solutions will have the form&#xA;\begin{align*}&#xA;x &amp;amp;= 492 - \frac{147r}{3} = 492 - 49r,\\&#xA;y &amp;amp;= -861 + \frac{258r}{3} =86r - 861, &amp;amp;\qquad&amp;amp;r\in\mathbb{Z}.&#xA;\end{align*}&#xA;You can reduce those constants by making a simple change of variable. For example, if we let $r=t+10$, then &#xA;\begin{align*}&#xA;x &amp;amp;= 492 - 49(t+10) = 2 - 49t,\\&#xA;y &amp;amp;= 86(t+10) - 861 = 86t - 1,&amp;amp;\qquad&amp;amp;t\in\mathbb{Z}.&#xA;\end{align*}&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-28T01:06:01.610" LastActivityDate="2011-02-28T01:06:01.610" CommentCount="4" />
  <row Id="20728" PostTypeId="2" ParentId="20660" CreationDate="2011-02-06T20:56:04.340" Score="0" ViewCount="" Body="&lt;p&gt;In addition to the above, you can start taking the actuarial exams and try to become an actuary. But you'll need a couple of exams passed to be competitive for jobs in that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most of the pure math PhD's I know who left academia ended out in jobs involving a lot of programming, either in finance or for the US government, the latter normally at the NSA, IDA, or defense contractors. If you're foreign I'm not sure what the analogues are. &lt;/p&gt;&#xA;" OwnerUserId="3035" LastActivityDate="2011-02-06T20:56:04.340" CommunityOwnedDate="2011-02-06T20:56:04.340" />
  <row Id="20729" PostTypeId="2" ParentId="20686" CreationDate="2011-02-06T21:04:26.917" Score="1" ViewCount="" Body="&lt;p&gt;As Akhil points out it is the first. But you could also think of morphisms in the category of pairs as commutative squares. This is just the typical definition in the category where objects are morphisms and morphisms are squares.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is response to your comment:&#xA;They are in fact homotopic in the category of pairs. The disk bundle is homotopic to the whole of the total space, and $E-B$ really means $E$ with out the image of the zero section which is equivalent to $B$. Think about what happens when you remove $0$ from each fiber. You get something homotopic to the sphere in that vector space.&lt;/p&gt;&#xA;" OwnerUserId="627" LastActivityDate="2011-02-06T21:04:26.917" CommentCount="1" />
  <row Id="20730" PostTypeId="1" AcceptedAnswerId="20733" CreationDate="2011-02-06T21:05:46.040" Score="5" ViewCount="147" Body="&lt;p&gt;Let $M$ denote the additive group of $ \mathbb{C} $. Why is $ \mathbb{C}$, as a ring, not isomorphic to $\mathrm{End}(M)$, where addition is defined pointwise, and multiplication as endomorphism composition?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="766" LastEditorUserId="6725" LastEditorDisplayName="" LastEditDate="2011-02-06T23:36:25.207" LastActivityDate="2011-02-06T23:36:25.207" Title="$ \mathbb{C} $ is not isomorphic to the endomorphism ring of its additive group" Tags="&lt;abstract-algebra&gt;&lt;ring-theory&gt;" AnswerCount="2" CommentCount="5" />
  <row Id="20731" PostTypeId="1" CreationDate="2011-02-06T21:09:52.150" Score="0" ViewCount="69" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Possible Duplicate:&lt;/strong&gt;&lt;br&gt;&#xA;  &lt;a href=&quot;http://math.stackexchange.com/questions/16179/function-is-smooth-iff-the-composition-with-any-smooth-curve-is-again-smooth&quot;&gt;function is smooth iff the composition with any smooth curve is again smooth&lt;/a&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;p&gt;Suppose $f:\mathbb{R}^n \rightarrow \mathbb{R}$ has the property that for all smooth $\gamma:\mathbb{R} \rightarrow \mathbb{R}^n$, $f\gamma:\mathbb{R} \rightarrow \mathbb{R}$ ($f$ following $\gamma$) is smooth. Does $f$ have to be smooth?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: smooth means $C^\infty$, i.e. partial derivatives of arbitray orders.&lt;/p&gt;&#xA;" OwnerUserId="6720" LastActivityDate="2011-02-06T21:47:50.317" ClosedDate="2011-02-06T22:24:46.013" Title="Detecting smoothness by looking at paths" Tags="&lt;calculus&gt;&lt;real-analysis&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="20732" PostTypeId="1" CreationDate="2011-02-06T21:09:02.837" Score="3" ViewCount="405" Body="&lt;p&gt;This could be mathematics or computer science, but also statistical physics, so I hope it qualifies for interest :)&#xA;I am aware that there were reservations about the proof but no fatal flaws. I have followed Terence Tao's blog and Tim Gowers, both of whom have reservations, but Deolalikar is sticking with his assertions and was supposedly preparing an updated response to the critics.&#xA;I haven't seen anything much since posts in August. Is anyone aware of any new updates more recent than Sept?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/P_versus_NP_problem&quot; rel=&quot;nofollow&quot;&gt;P=NP problem statement&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://michaelnielsen.org/polymath1/index.php?title=Deolalikar%27s_P!%3DNP_paper&quot; rel=&quot;nofollow&quot;&gt;status of paper&lt;/a&gt;&lt;/p&gt;&#xA;" LastEditorUserId="856" LastEditorDisplayName="" LastEditDate="2011-02-06T21:19:58.273" LastActivityDate="2011-02-07T04:01:37.660" Title="What is the current status of Vinay Deolalikar's proof that P is not equal to NP" Tags="&lt;computer-science&gt;" AnswerCount="2" />
  <row Id="20733" PostTypeId="2" ParentId="20730" CreationDate="2011-02-06T21:18:18.780" Score="7" ViewCount="" Body="&lt;p&gt;Method 1: $\operatorname{End}(M)$ is not an integral domain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Method 2: Count idempotents.  There are only 2 idempotent elements of $\mathbb{C}$, but lots in $\operatorname{End}(M)$, including for example real and imaginary projections along with the identity and zero maps.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Method 3: Cardinality.  Assuming a Hamel basis for $\mathbb{C}$ as a $\mathbb{Q}$ vector space, there are $2^\mathfrak{c}$ $\mathbb{Q}$ vector vector space endomorphisms of $\mathbb{C}$, but only $\mathfrak{c}$ elements of $\mathbb{C}$.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2011-02-06T21:18:18.780" />
  <row Id="20734" PostTypeId="2" ParentId="20730" CreationDate="2011-02-06T21:19:09.350" Score="5" ViewCount="" Body="&lt;p&gt;The endomorphism ring of $M$ is its endomorphism group as a vector space &lt;strong&gt;over $\mathbb{Q}$&lt;/strong&gt; (since any additive function on $\mathbb{C}$ is in fact $\mathbb{Q}$-linear, and any $\mathbb{Q}$-linear map is necessarily additive). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How many endomorphisms does a vector space over $\mathbb{Q}$ of dimension $\kappa\geq\aleph_0$ have? Turns out it has $2^{\kappa}$ endomorphisms. Any endomorphism is completely determined by the image of a basis. Fix a basis $\beta$ for $\mathbf{V}$ over $\mathbb{Q}$, which is of cardinality $\kappa$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each endomorphism corresponds to a function $\beta\to\mathop{\oplus}\limits_{b\in\beta}\mathbb{Q} = \mathbb{Q}^{(\beta)}$. So the cardinality of the endomorphism ring is the cardinality of $(\mathbb{Q}^{(\beta)})^{\beta}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The cardinality of $\mathbb{Q}^{(\beta)}$ is $|\beta|=\kappa$, so the entire thing has cardinality &#xA;$$|\mathrm{end}(M)| = \left|\mathbb{Q}^{(\beta)}\right|^{|\beta|} = \kappa^{\kappa} = 2^{kappa}.$$&#xA;(For a proof that if $\kappa$ is infinite and $2\leq\lambda\leq 2^{\kappa}$, then $\lambda^{\kappa} = 2^{\kappa}$, see &lt;a href=&quot;http://math.stackexchange.com/questions/17914/cardinality-of-the-set-of-all-real-functions-of-real-variable/17915#17915&quot;&gt;this previous answer&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of $M$ and $\mathbb{C}$, since $M$ is of dimension $\mathfrak{c}=2^{\aleph_0}$ as a $\mathbb{Q}$-vector space, you have that $|\mathrm{End}(M)| = 2^{\mathfrak{c}}\gt \mathfrak{c}=|\mathbb{C}|$. So there aren't enough elements in $\mathbb{C}$ to give you all the endomorphisms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(The elements of $\mathbb{C}$ correspond to the $\mathbb{C}$-linear maps, of course, by simply multiplication; but here you want a lot more maps). &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-06T21:29:50.620" LastActivityDate="2011-02-06T21:29:50.620" />
  <row Id="20735" PostTypeId="2" ParentId="20722" CreationDate="2011-02-06T21:38:36.470" Score="0" ViewCount="" Body="&lt;p&gt;Polar form sounds like a good idea. Then you need to know what the conic sections look like in polar coordinates: &lt;a href=&quot;http://en.wikipedia.org/wiki/Conic_section#Polar_coordinates&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Conic_section#Polar_coordinates&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2011-02-06T21:38:36.470" CommentCount="1" />
  <row Id="20736" PostTypeId="2" ParentId="20731" CreationDate="2011-02-06T21:39:14.960" Score="0" ViewCount="" Body="&lt;p&gt;Sure, take $\gamma_i(t)=a+te_i$ ($e_i$ the i.th unit vector). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;edit: wrong. if $f\gamma_i$ is smooth, then $\frac{\partial f}{\partial x_i}$ is smooth?&lt;/p&gt;&#xA;" OwnerUserId="6721" LastEditorUserId="6721" LastEditorDisplayName="" LastEditDate="2011-02-06T21:47:50.317" LastActivityDate="2011-02-06T21:47:50.317" CommentCount="0" />
  <row Id="20737" PostTypeId="1" AcceptedAnswerId="20743" CreationDate="2011-02-06T21:40:06.810" Score="0" ViewCount="163" Body="&lt;p&gt;If we have some prime $p$ and a natural number $k$, is there a formula for the largest natural number $n_k$ such that $p^{n_k} | k!$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This came up while doing an unrelated homework problem, but it is not itself homework. I haven't had any good ideas yet worth putting here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The motivation came from trying to figure out what power of a prime you can factor out of a binomial coefficient.  Like $\binom{p^m}{k}$.&lt;/p&gt;&#xA;" OwnerUserId="565" LastActivityDate="2011-02-06T22:24:30.073" Title="Prime powers that divide a factorial." Tags="&lt;prime-numbers&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="20738" PostTypeId="2" ParentId="20717" CreationDate="2011-02-06T21:43:33.667" Score="3" ViewCount="" Body="&lt;p&gt;As others have mentioned one may employ the extended Euclidean algorithm. It deserves to be better known that this is most easily performed via row-reduction on an augmented matrix - analogous to methods used in linear algebra. See this excerpt from one of my old sci.math posts:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;For example, to solve  mx + ny = gcd(x,y) one begins with&#xA;two rows  [m   1    0], [n   0    1], representing the two&#xA;equations  m = 1m + 0n,  n = 0m + 1n. Then one executes&#xA;the Euclidean algorithm on the numbers in the first column,&#xA;doing the same operations in parallel on the other columns,&#xA;&#xA;Here is an example:  d =  x(80) + y(62)  proceeds as:&#xA;&#xA;                      in equation form   | in row form&#xA;                    ---------------------+------------&#xA;                    80 =   1(80) + 0(62) | 80   1   0&#xA;                    62 =   0(80) + 1(62) | 62   0   1&#xA; row1 -   row2  -&amp;gt;  18 =   1(80) - 1(62) | 18   1  -1&#xA; row2 - 3 row3  -&amp;gt;   8 =  -3(80) + 4(62) |  8  -3   4&#xA; row3 - 2 row4  -&amp;gt;   2 =   7(80) - 9(62) |  2   7  -9&#xA; row4 - 4 row5  -&amp;gt;   0 = -31(80) -40(62) |  0 -31  40&#xA;&#xA;Above the row operations are those resulting from applying&#xA;the Euclidean algorithm to the numbers in the first column,&#xA;&#xA;        row1 row2 row3 row4 row5&#xA;namely:  80,  62,  18,   8,   2  = Euclidean remainder sequence&#xA;               |    |&#xA;for example   62-3(18) = 8, the 2nd step in Euclidean algorithm&#xA;&#xA;becomes:   row2 -3 row3 = row4  on the identity-augmented matrix.&#xA;&#xA;In effect we have row-reduced the first two rows to the last two.&#xA;The matrix effecting the reduction is in the bottom right corner.&#xA;It starts as the identity, and is multiplied by each elementary&#xA;row operation matrix, hence it accumulates the product of all&#xA;the row operations, namely:&#xA;&#xA;       [  7 -9] [ 80  1  0]  =  [2   7  -9]&#xA;       [-31 40] [ 62  0  1]     [0 -31  40]&#xA;&#xA;The 1st row is the particular  solution: 2 =   7(80) -  9(62)&#xA;The 2nd row is the homogeneous solution: 0 = -31(80) + 40(62),&#xA;so the general solution is any linear combination of the two:&#xA;&#xA;       n row1 + m row2  -&amp;gt;  2n = (7n-31m) 80 + (40m-9n) 62&#xA;&#xA;The same row/column reduction techniques tackle arbitrary&#xA;systems of linear Diophantine equations. Such techniques&#xA;generalize easily to similar coefficient rings possessing a&#xA;Euclidean algorithm, e.g. polynomial rings F[x] over a field, &#xA;Gaussian integers Z[i]. There are many analogous interesting&#xA;methods, e.g. search on keywords: Hermite / Smith normal form, &#xA;invariant factors, lattice basis reduction, continued fractions,&#xA;Farey fractions / mediants, Stern-Brocot tree / diatomic sequence.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2011-02-06T22:00:22.903" LastActivityDate="2011-02-06T22:00:22.903" CommentCount="4" />
  <row Id="20739" PostTypeId="2" ParentId="20721" CreationDate="2011-02-06T21:49:22.210" Score="1" ViewCount="" Body="&lt;p&gt;A basic fact about the topological space $\operatorname{Spec} R$ is that the irreducible components of $\operatorname{Spec} R$ correspond naturally to the minimal prime ideals of $R$.  For a proof of that see Section 13.4 of &lt;a href=&quot;http://math.uga.edu/~pete/integral.pdf&quot; rel=&quot;nofollow&quot;&gt;my commutative algebra notes&lt;/a&gt;.  (I don't know why I haven't stated this in the form &quot;Theorem: ...&quot; It just appears in the text.  I'll fix that eventually...)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you admit this fact, then your question becomes &quot;What are the minimal primes in a domain?&quot; which is very easy to answer.&lt;/p&gt;&#xA;" OwnerUserId="299" LastActivityDate="2011-02-06T21:49:22.210" />
  <row Id="20740" PostTypeId="2" ParentId="20737" CreationDate="2011-02-06T21:49:23.817" Score="2" ViewCount="" Body="&lt;p&gt;It should be fairly obvious that the answer is &#xA; $$n_k = \biggl\lfloor{\frac{k}{p}\biggr\rfloor} + \biggl\lfloor{\frac{k}{p^2}\biggr\rfloor}  +\biggl\lfloor{\frac{k}{p^3}\biggr\rfloor} + \cdots$$ &lt;/p&gt;&#xA;" OwnerUserId="6460" LastActivityDate="2011-02-06T21:49:23.817" />
  <row Id="20741" PostTypeId="2" ParentId="20697" CreationDate="2011-02-06T21:51:09.257" Score="4" ViewCount="" Body="&lt;p&gt;Question 3 admits an easy answer, once you check two facts:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$A\subseteq^* B\subseteq^* C$ implies $A\subseteq^* C$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is completely straightforward: If $A\setminus B\subseteq m$ and $B\setminus C\subseteq n$, then $A\setminus C\subseteq \max\{m,n\}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Given a (strictly) $\subseteq^*$-increasing sequence $(A_n\mid n\in\omega)$ (so $A_n\setminus A_{n+1}$ is finite but $A_{n+1}\setminus A_n$ is infinite), there is a $B$ with $A_n\subseteq^* B$ for all $n$ (so, in particular, $B\setminus A_n$ is infinite), and such that $\omega\setminus B$ is infinite as well.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For this: Define two strictly increasing sequences of numbers $m_0&amp;lt;m_1&amp;lt;\dots$ and $$t_{00}&amp;lt;t_{01}&amp;lt;t_{10}&amp;lt;t_{11}&amp;lt;t_{20}&amp;lt;t_{21}&amp;lt;\dots$$ so that  $A_n\setminus A_{n+1}\subseteq m_n$ for all $n$, $[t_{n0},t_{n1})\cap (\omega\setminus A_{n+1})\ne\emptyset$, and $m_n&amp;lt;t_{n0}&amp;lt;t_{n1}&amp;lt;m_{n+1}$ for all $n$. (Check that this is possible.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $$ B=\bigcup_n [m_n,m_{n+1})\cap A_{n+1}. $$ Check that $A_n\setminus B\subseteq m_n$ for any $n$: If $s\in A_n\setminus B$ and $s\ge m_n$ then $s\in A_{n+1}$ (by definition of $m_n$), so $s\ge m_{n+1}$ (or else $s\in B$); inductively, this gives us that $s\ge m_k$ for all $k$, which of course is absurd.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moreover, by construction, $\omega\setminus B$ contains at least one element of each interval $[t_{n0},t_{n1})$, so it is infinite.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using these two facts, a straightforward transfinite construction of length $\omega_1$ gives us the desired $\subseteq^*$ increasing sequence $(A_\alpha\mid\alpha&amp;lt;\omega_1)$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Start with any $A_0\subseteq\omega$ infinite and coinfinite. Given $A_\alpha$ infinite and coinfinite, let $A_{\alpha+1}$ consist of $A_\alpha$ and an infinite-coinfinite subset of $\omega\setminus A_\alpha$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At limit stages $\alpha&amp;lt;\omega_1$, given $(A_\beta\mid \beta&amp;lt;\alpha)$, fix $(\beta_n\mid n&amp;lt;\omega)$ increasing and cofinal in $\alpha$ and use the second fact to find $A_\alpha$ with $A_{\beta_n}\subseteq^* A_\alpha$ for all $n$ and $\omega\setminus A_\alpha$ infinite. That $A_\beta\subseteq A_\alpha$ for all $\beta&amp;lt;\alpha$ now follows from the first fact.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;For question 2, use Fodor's lemma: Suppose, towards a contradiction that the $A_\alpha$ are pairwise disjoint non-stationary subsets of $\omega_1$ such that $\bigcup_\alpha A_\alpha$ is stationary, and yet $$B=\{{\rm min}(A_\alpha)\mid\alpha&amp;lt;\omega_1\}$$ is non-stationary. Let $A=\bigcup_\alpha A_\alpha\setminus B$. Then $A$ is stationary. Define a regressive $f:A\to\omega_1$ as follows: Given $\beta\in A$, there is a unique $\alpha$ such that $\beta\in A_\alpha$. Then set $f(\beta)=\min(A_\alpha)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By Fodor, $f$ is constant on a stationary set. This is, by construction, a subset of a fixed $A_\alpha$ (since the $A_\alpha$ are pairwise disjoint), contradicting that no $A_\alpha$ is stationary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a side remark, note that you need something like the assumption that the $A_\alpha$ are disjoint. An easy counterexample otherwise is obtained by setting $A_\alpha=\alpha+1$. Then $\bigcup_\alpha A_\alpha=\omega_1$, yet $B=\{0\}$.&lt;/p&gt;&#xA;" OwnerUserId="462" LastEditorUserId="462" LastEditorDisplayName="" LastEditDate="2011-02-07T07:05:26.077" LastActivityDate="2011-02-07T07:05:26.077" CommentCount="2" />
  <row Id="20742" PostTypeId="1" AcceptedAnswerId="20747" CreationDate="2011-02-06T21:56:05.630" Score="5" ViewCount="153" Body="&lt;p&gt;I need some hints for this problem from Dummit and Foote. (edit: added the full question verbatim for context)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $p$ be an odd prime and let $n$ be a positive integer.  Use the binomial theorem to show that $(1+p)^{p^{n-1}} \equiv 1 \pmod{p^n}$ but $(1+p)^{p^{n-2}} \not\equiv 1\pmod{p^n}$. Deduce that $1+p$ is an element of order $p^{n-1}$ in the multiplicative group $(\mathbb{Z}/p^n\mathbb{Z})^{\times}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the first part I just need to show that $$p^n \;\left|\; \sum\limits_{k=1}^{p^{n-1}}\binom{p^{n-1}}{k}p^k\right.$$ and it seems obvious enough that it should, since I think I can factor a $p^n$ out of each term, but Im having trouble actually proving this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the second part I am unsure what to do.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For reference it is problem 21. on page 60 of the third edition. (section 2.3).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt; Ok I have managed to show the first part.  From the lemma in the post Bill Dubuque linked to, which I could prove easily, if $z\equiv 1$ mod $p^n$ and $z = p+1$ then $z^p \equiv 1$ mod $p^{n+1}$. Since $z^p-1 = (z-1)(1+z+...+z^{p-1})$ and $(1+z+...+z^{p-1})\equiv 1+1+...+1 = p \equiv 0$ mod $p$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So it follows by induction that $(p+1)^{p^{n-1}} \equiv 1$ mod $p^n$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My problem here is that what Ive proven really only shows that $ord_p(z^p-1) \geq ord_p(z-1)+1$ as far as I can tell, so Im still not sure how I can show that $(1+p)^{p^{n-2}} \not\equiv 1\pmod{p^n}$. I think Im missing something here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit II&lt;/strong&gt; Ok, I snuck a peek at Pete's notes and I see what I was missing. Really the crucial step was writing $z = 1 + xp$, then I could see how to do it.&lt;/p&gt;&#xA;" OwnerUserId="565" LastEditorUserId="565" LastEditorDisplayName="" LastEditDate="2011-02-07T17:56:09.063" LastActivityDate="2011-02-07T17:56:09.063" Title="Modular arithmetic problem" Tags="&lt;modular-arithmetic&gt;" AnswerCount="1" CommentCount="6" FavoriteCount="1" />
  <row Id="20743" PostTypeId="2" ParentId="20737" CreationDate="2011-02-06T21:58:51.657" Score="1" ViewCount="" Body="&lt;p&gt;This follows from a famous result of Kummer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Theorem.&lt;/strong&gt; (Kummer, 1854) &lt;em&gt;The highest power of $p$ that divides the binomial coefficient $\binom{m+n}{n}$ is equal to the number of &quot;carries&quot; when adding $m$ and $n$ in base $p$.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Equivalently, the highest power of $p$ that divides $\binom{m}{n}$, with $0\leq n\leq m$ is the number of carries when you add $m-n$ and $n$ in base $p$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a corollary, you get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Corollary.&lt;/strong&gt; For a positivie integer $r$ and a prime $p$, let $[r]_p$ denote the exact $p$-divisor of $r$; that is, we say $[r]_p=a$ if $p^a|r$ but $p^{a+1}$ does not divide $r$. If $0\lt k\leq p^n$, then&#xA;$$\left[\binom{p^n}{k}\right]_p = n - [k]_p.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Proof.&lt;/em&gt; To get a proof, assuming Kummer's Theorem: when you add $p^n - k$ and $k$ in base $p$, you get a $1$ followed by $n$ zeros. You start getting a carry as soon as you don't have both numbers in the column equal to $0$, which is as soon as you hit the first nonzero digit of $k$ in base $p$ (counting from right to left). So you really just need to figure out what is the first nonzero digit of $k$ in base $p$, from right to left. This is exactly the $([k]_p+1)$th digit. So the number of carries will be $(n+1)-([k]_p+1) = n-[k]_p$, hence this is the highest power of $p$ that divides $\binom{p^n}{k}$, as claimed. $\Box$&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-06T22:24:30.073" LastActivityDate="2011-02-06T22:24:30.073" CommentCount="1" />
  <row Id="20744" PostTypeId="2" ParentId="20732" CreationDate="2011-02-06T21:59:18.397" Score="0" ViewCount="" Body="&lt;p&gt;Even though Deolalikar's paper was the most seriously considered recently, and a lot of work was put into it by important people in the few weeks after it came out, but very little public judgement since then. There were enough reservations that people probably won't bother looking at it again until the update comes out.   &lt;/p&gt;&#xA;" OwnerUserId="1919" LastActivityDate="2011-02-06T21:59:18.397" />
  <row Id="20746" PostTypeId="1" AcceptedAnswerId="20760" CreationDate="2011-02-06T22:05:12.620" Score="1" ViewCount="124" Body="&lt;p&gt;I can't find an example of a countable injective module over a non-Noetherian ring.&lt;/p&gt;&#xA;" OwnerUserId="6236" LastActivityDate="2011-02-07T00:29:32.743" Title="Example of an injective module" Tags="&lt;abstract-algebra&gt;&lt;commutative-algebra&gt;&lt;ring-theory&gt;&lt;modules&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="1" />
  <row Id="20747" PostTypeId="2" ParentId="20742" CreationDate="2011-02-06T22:05:18.347" Score="3" ViewCount="" Body="&lt;p&gt;Hint: use the binomial formula to establish the following fact.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $p$ be an odd prime number and $z$ an integer which is $1$ mod $p$.  Then &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\operatorname{ord}_p(z^p − 1) = \operatorname{ord}_p(z − 1) + 1.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here, for a nonzero integer $N$, $\operatorname{ord}_p(N)$ is the largest power of $p$ which divides $N$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(This is actually taken from a result in some notes of mine from an elementary number theory course, but since the OP asked for a hint only, I wish to obey by not posting the link.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: the hint should serve to solve both parts of the question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Added&lt;/b&gt;: Okay, for more help see Lemma 3 &lt;a href=&quot;http://math.uga.edu/~pete/4400primitiveroots.pdf&quot; rel=&quot;nofollow&quot;&gt;of these notes&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="299" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2011-02-07T01:29:55.080" LastActivityDate="2011-02-07T01:29:55.080" CommentCount="4" />
  <row Id="20748" PostTypeId="2" ParentId="20746" CreationDate="2011-02-06T22:11:11.423" Score="2" ViewCount="" Body="&lt;p&gt;Let $R$ be a countable non-Noetherian integral domain, e.g. a polynomial ring in countably infinitely many indeterminates over $\mathbb{Z}$ (or, more appealingly to a number theorist, the ring $\overline{\mathbb{Z}}$ of all algebraic integers).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then the fraction field $K$ of $R$ is a countable, injective $R$-module.  In fact it is the &lt;strong&gt;injective hull&lt;/strong&gt; of the $R$-module $R$: see &lt;a href=&quot;http://en.wikipedia.org/wiki/Injective_hull&quot; rel=&quot;nofollow&quot;&gt;this wikipedia article on injective hulls&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, I confess that I have not thought as much as I should on injective hulls -- I haven't even gone through the proof that they always exist! -- but I would have to think that if $M$ is any infinite $R$-module, then its injective hull has the same cardinality as $M$.  This would give many examples.&lt;/p&gt;&#xA;" OwnerUserId="299" LastActivityDate="2011-02-06T22:11:11.423" CommentCount="2" />
  <row Id="20749" PostTypeId="1" CreationDate="2011-02-06T22:26:27.353" Score="10" ViewCount="236" Body="&lt;p&gt;Any hexagon in Pascal's triangle, whose vertices are 6 binomial coefficients surrounding any entry, has the property that:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;the product of non-adjacent vertices is constant.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;the greatest common divisor of non-adjacent vertices is constant.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Below is one such hexagon. As an example, here we have that $4 \cdot 10  \cdot 15 = 6 \cdot 20  \cdot 5$, as well as $\gcd(4, 10, 15) = \gcd(6,20,5)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/pChFw.png&quot; alt=&quot;triangle&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a quick proof &lt;a href=&quot;http://www.fq.math.ca/Scanned/12-1/gupta.pdf&quot; rel=&quot;nofollow&quot;&gt;here (pdf).&lt;/a&gt; The original proof should be in &lt;em&gt;V. E. Hoggatt, Jr., &amp;amp; W. Hansell. &quot;The Hidden Hexagon Squares.&quot; The Fibonacci Quarterly 9(1971):120, 133.&lt;/em&gt; but I cannot access it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am, however, intereseted in a purely &lt;em&gt;combinatorial&lt;/em&gt; proof. I do not know how to approach this at all: I cannot see what the non-adjacent vertices represent and/or I do not know how to remodel their meaning. Can anyone help?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt; To specify my question more closely, what I am looking for is some natural bijection between the two sets of triads that create the hexagon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="5105" LastEditorUserId="5105" LastEditorDisplayName="" LastEditDate="2011-02-21T06:12:14.837" LastActivityDate="2011-02-21T06:12:14.837" Title="The Hexagonal Property of Pascal's Triangle" Tags="&lt;combinatorics&gt;&lt;binomial-coefficients&gt;&lt;alternative-proof&gt;" AnswerCount="1" />
  <row Id="20751" PostTypeId="1" AcceptedAnswerId="20794" CreationDate="2011-02-06T23:08:26.463" Score="11" ViewCount="339" Body="&lt;p&gt;I got the following question by mail from someone I don't know from Adam. (Quoted in part.)&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;if $f(t)$ continuously diff. on $[0,1]$ and&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;a) $\int_0^1f(t)\ dt=0$&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;b) $m\le f\,&amp;#39;\le M$ on $[0,1]$&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Prove&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$\frac m{12}\le\int_0^1t\cdot f(t)\ dt\le\frac M{12}$&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;I suspect it might be an error&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I assumed immediately that it's an error, but my first two thoughts as counterexamples were $f(t)=\frac12-t$ and $f(t)=\sin(2\pi t)$, both of which satisfy the result. Anyone with a proof or counterexample?&lt;/p&gt;&#xA;" OwnerUserId="4992" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2011-02-13T18:15:27.797" LastActivityDate="2011-04-02T16:55:48.747" Title="Proof of bound on $\int t\,f(t)\ dt$ given well-behaved $f$" Tags="&lt;integral&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="3" />
  <row Id="20752" PostTypeId="1" CreationDate="2011-02-06T23:11:45.913" Score="3" ViewCount="224" Body="&lt;p&gt;How would I start in solving this recursive sequence?  Thanks!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sequence: 3, 5, -2, 7, -9, 16, -25, 41, ...&lt;/p&gt;&#xA;" OwnerUserId="6726" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2011-06-03T13:49:48.163" LastActivityDate="2011-06-03T13:49:48.163" Title="Recursive Sequence &quot;3, 5, -2, 7, -9, 16, -25, 41&quot;" Tags="&lt;discrete-mathematics&gt;" AnswerCount="2" CommentCount="6" />
  <row Id="20753" PostTypeId="2" ParentId="20752" CreationDate="2011-02-06T23:17:17.783" Score="1" ViewCount="" Body="&lt;p&gt;How does:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$-2$ relate to $3$ and $5$?&lt;/li&gt;&#xA;&lt;li&gt;$7$ relate to $5$ and $-2$?&lt;/li&gt;&#xA;&lt;li&gt;$-9$ relate to $-2$ and $7$?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;etc. If we write $f_i$ to be the $i$-th term in your sequence, these questions are just asking you: &lt;em&gt;How can you express $f_n$ in terms of $f_{n-1}$ and $f_{n-2}$?&lt;/em&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that you may always define the first few numbers in a recursion by hand.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Of course this is not some general strategy, but if you have such a problem, the desired formula won't be something &lt;em&gt;too&lt;/em&gt; complicated - here I just &lt;em&gt;guessed&lt;/em&gt; it would involve the two previous terms somehow, but again in general, that's probably the first thing you should try.)&lt;/p&gt;&#xA;" OwnerUserId="5105" LastEditorUserId="5105" LastEditorDisplayName="" LastEditDate="2011-02-06T23:25:12.213" LastActivityDate="2011-02-06T23:25:12.213" />
  <row Id="20755" PostTypeId="1" AcceptedAnswerId="20757" CreationDate="2011-02-06T23:28:07.437" Score="0" ViewCount="102" Body="&lt;p&gt;I am referring to a paper which is over here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;www.scipub.org/fulltext/jcs/jcs410879-889.pdf&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The paper is not hard to read, but I think notation is rather sloppy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am specifically referring page 880, &quot;Materials and Methods&quot; in the beginning.&#xA;Elementary sets are defined as the equivalence classes of some equivalence relation $A=(Ob,R)$ where $Ob$ is the universe and $R$ is the relation (an approximation space it is called?), and then in turn we define $com(A)$ to be the set of all finite unions of elementary sets (the equivalence relations).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This $com(A)$ is supposed to be a topology, according to the paper, but I don't see why. What about infinite unions? They should also be there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another problem... In proposition 2, they mention $\tau_2 \subseteq \tau_1$. Does that mean that the open sets of the second topology should be a subset of the open sets of the first topology?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Like I said, I think notation is rather sloppy there, but I may be misunderstanding something. Any help appreciated.&lt;/p&gt;&#xA;" OwnerUserId="7076" LastActivityDate="2011-02-06T23:51:15.503" Title="could I get help with a paper about topology?" Tags="&lt;topology&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="20756" PostTypeId="1" CreationDate="2011-02-06T23:37:20.957" Score="1" ViewCount="82" Body="&lt;p&gt;Hello.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to obtain a bound from below for the function&lt;br&gt;&#xA;$$f(p) = \frac{pe - (1-p)^{d+1}v}{p^{v-1}}$$ subject to $0 &amp;lt; p \leq 1$ and $e,v,d &amp;gt; 0.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The usual method is to check the boundary cases and the value at which the derivative is zero. Solving the equation involving the derivative is rather messy in this case, so I am looking at an elegant way to first bound $f(p)$ by some other function $g(p)$ and then find a bound for g  that is as tight as possible. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any hints on that? For $0 &amp;lt; p \leq 0.5$ one could use  the fact that $(1-p)^{d+1} \geq exp(-2p(d+1))$ which still yields a function with an &quot;ugly&quot; derivative.&lt;/p&gt;&#xA;" OwnerUserId="4677" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-07T00:25:36.150" LastActivityDate="2011-02-07T00:25:36.150" Title="Bounding function from below" Tags="&lt;inequality&gt;" CommentCount="1" />
  <row Id="20757" PostTypeId="2" ParentId="20755" CreationDate="2011-02-06T23:51:15.503" Score="1" ViewCount="" Body="&lt;p&gt;Since $\mathrm{Ob}$ is assumed to be finite (first page, 2nd column), you do not need to worry about infinite unions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are correct about the meaning of proposition 2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(By the way, it seems to me to be a very trivial example of a topology, and I'm a bit skeptical that the paper will be able to make anything interesting out of it.)&lt;/p&gt;&#xA;" OwnerUserId="1726" LastActivityDate="2011-02-06T23:51:15.503" />
  <row Id="20758" PostTypeId="1" CreationDate="2011-02-06T23:58:21.937" Score="5" ViewCount="97" Body="&lt;p&gt;Can anyone explain to me why the minimal prime ideals of a commutative ring (with the subspace topology inherited from the Zariski topology) form a totally disconnected space, or give a reference? I remember that this is true but can't seem to prove it myself or find the proof anywhere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would be especially happy if there is some proof that does not use the (easy) fact that this space is Hausdorff. This is because I am trying to prove that the primitive spectrum of a certain noncommutative ring, which I know is not Hausdorff, is totally disconnected. Hopefully the proof works in my situation when phrased in the appropriate way.&lt;/p&gt;&#xA;" OwnerUserId="4842" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2011-02-07T01:43:29.497" LastActivityDate="2011-02-07T01:43:29.497" Title="Minimal spectrum of a commutative ring" Tags="&lt;commutative-algebra&gt;&lt;ring-theory&gt;&lt;noncommutative-geometry&gt;" CommentCount="1" />
  <row Id="20759" PostTypeId="1" CreationDate="2011-02-07T00:23:21.460" Score="1" ViewCount="85" Body="&lt;p&gt;&lt;strong&gt;Background:&lt;/strong&gt;&#xA;In the context of divergent summation I'm analyzing the matrix of eulerian numbers for a regular matrix-summation method. Beginning indexes at zero (r for &quot;row&quot;, c for &quot;column&quot;) the entries of M are &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$m_{r,c}=\frac{eul(r,c)}{r!}  $$ and $ eul(r,c)=\displaystyle\sum_{k=0}^{c}(-1)^k \binom{r+1}{k} (c+1-k)^r $. &lt;em&gt;(see for instance &lt;a href=&quot;http://en.wikipedia.org/wiki/Eulerian_numbers&quot; rel=&quot;nofollow&quot;&gt;wikipedia&lt;/a&gt;)&lt;/em&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The top-left segment of this (infinite, triangular) matrix is&lt;br&gt;&#xA;$$ \small{ &#xD;&#xA;\begin{array} {rrrrr}&#xD;&#xA; 1 &amp;amp; . &amp;amp; . &amp;amp; . &amp;amp; . &amp;amp; . \\&#xD;&#xA; 1 &amp;amp; 0 &amp;amp; . &amp;amp; . &amp;amp; . &amp;amp; . \\&#xD;&#xA; 1/2! &amp;amp; 1/2! &amp;amp; 0 &amp;amp; . &amp;amp; . &amp;amp; . \\&#xD;&#xA; 1/3! &amp;amp; 4/3! &amp;amp; 1/3! &amp;amp; 0 &amp;amp; . &amp;amp; . \\&#xD;&#xA; 1/4! &amp;amp; 11/4! &amp;amp; 11/4! &amp;amp; 1/4! &amp;amp; 0 &amp;amp; . \\&#xD;&#xA; 1/5! &amp;amp; 26/5! &amp;amp; 66/5! &amp;amp; 26/5! &amp;amp; 1/5! &amp;amp; 0&#xD;&#xA; \end{array} } $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The idea is, to sum a sequence $ \{a_k\}_{k=0..\infty} $ using the double sum&lt;br&gt;&#xA;  $$ s = \sum_{r=0}^{\infty} a_r = \sum_{r=0}^{\infty} ( a_r \sum_{c=0}^r m_{r,c} ) &#xD;&#xA;= \sum_{c=0}^{\infty} ( \sum_{r=0}^{\infty} a_r m_{r,c} )= \sum_{c=0}^{\infty} b_c $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was studying that summation with various sequences $ {a_k}$ but I wanted to optimize the computation. For instance, if $ \{a_k\}_{k=0..\infty} = q^k $ and thus define a geometric series with quotient &lt;em&gt;q&lt;/em&gt; the $ b_c $ are finite compositions of exponentialseries of $q $ and of powers of $q$:&#xA;$$b_c = e^{q(1+c)}+\sum_{k=1}^c (-1)^k \frac{z^k+kz^{k-1}}{k!}e^z $$&#xA;where for sum-term $ z=(c+1-k)q $ is simply a shortcut.     &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Writing the formula for the partial sums $$ ps(q,c)= \sum_{k=0}^c b_k $$ this boils down to the following series-transformation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ ps(x,c) = e^{x(1+c)} \sum_{k=0}^c \frac{(c+1-k)^k}{k!}(-x e^{-x})^k $$&#xA;and we have in the limit $$ \lim_{c\to \infty} ps(x,c) = \frac{1}{1-x} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After arriving at the term $-x e^{-x} $ I've a vague impression I should have seen this transformation; but even if: I cannot remember. On the other hand - this summation-procedure is powerful, so this transformation is possibly interesting in more general use.     &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; Does someone know this transformation and/or can provide a source where I can read more about it?&lt;/p&gt;&#xA;" OwnerUserId="1714" LastEditorUserId="1714" LastEditorDisplayName="" LastEditDate="2011-02-07T01:08:59.947" LastActivityDate="2011-02-07T01:08:59.947" Title="Is that series-transformation known in the context of divergent summation" Tags="&lt;number-theory&gt;&lt;matrices&gt;&lt;reference-request&gt;&lt;sequences-and-series&gt;" />
  <row Id="20760" PostTypeId="2" ParentId="20746" CreationDate="2011-02-07T00:29:32.743" Score="0" ViewCount="" Body="&lt;p&gt;Another type of example: Let A be any non-Noetherian ring, let B be the field of 2 elements, and let C be the field of rational numbers.  Then R = A×B×C is a non-Noetherian ring, its module 0×B×0 is injective and finite, and its module 0×0×C is injective and countably infinite.  Injective modules are at least somewhat local ideas, so even if the ring is bad somewhere, it doesn't mean all the injectives are bad.&lt;/p&gt;&#xA;" OwnerUserId="583" LastActivityDate="2011-02-07T00:29:32.743" />
  <row Id="20761" PostTypeId="2" ParentId="20752" CreationDate="2011-02-07T01:03:51.763" Score="3" ViewCount="" Body="&lt;p&gt;I would try to denote the problem using a matrix-formulation, with a guessed linear recursion involving for instance three elements:&lt;br&gt;&#xA;$$ \begin{array} {rrrrrr}&#xD;&#xA;&amp;amp; &amp;amp; &amp;amp; &amp;amp; x_1 &amp;amp; \\&#xD;&#xA;&amp;amp; &amp;amp; &amp;amp; &amp;amp; x_2 &amp;amp; \\&#xD;&#xA;&amp;amp; &amp;amp; &amp;amp; &amp;amp; x_3 &amp;amp; \\&#xD;&#xA;--&amp;amp;--&amp;amp;--&amp;amp;-&amp;amp;-- \\&#xD;&#xA;3&amp;amp;5&amp;amp;-2&amp;amp; &amp;amp;7\\&#xD;&#xA;5&amp;amp;-2&amp;amp;7&amp;amp;=&amp;amp;-9 \\&#xD;&#xA;-2&amp;amp;7&amp;amp;-9&amp;amp; &amp;amp;16\\&#xD;&#xA;&#xD;&#xA;\end{array} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where the left matrix is &lt;strong&gt;P&lt;/strong&gt;, the vector &lt;strong&gt;X&lt;/strong&gt; contains the unknowns $x_1,x_2,x_3$ and the right vector is &lt;strong&gt;Q&lt;/strong&gt;, with the obvious insertions. We ask: is the following matrix-equation solvable:&#xA;$$ P * X = Q $$ $$ X = P^{-1} * Q $$&#xA;thus first we check, whether &lt;strong&gt;P&lt;/strong&gt; is invertible. Maybe we need only a &lt;em&gt;2x2&lt;/em&gt;-matrix &lt;strong&gt;P&lt;/strong&gt; or possibly we need higher dimension, if the the result is not correct for the following members of the given sequence. &#xA;In our example we come along with a &lt;em&gt;2x2&lt;/em&gt; matrix &lt;strong&gt;P&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \begin{array}{} P=\begin{pmatrix} 3&amp;amp;5 \\\ 5&amp;amp;-2 \end{pmatrix} &amp;amp; &amp;amp; &#xD;&#xA;P^{-1}= \begin{pmatrix}  2/31 &amp;amp; 5/31 \\\ 5/31 &amp;amp; -3/31  \end{pmatrix} \end{array}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and get $X = (1,-1)$ so that $$a(k+2)=x_1 a(k)+x_2 a(k+1) = 1 a(k) - 1 a(k+1)$$ &lt;/p&gt;&#xA;" OwnerUserId="1714" LastActivityDate="2011-02-07T01:03:51.763" CommentCount="2" />
  <row Id="20762" PostTypeId="1" AcceptedAnswerId="20763" CreationDate="2011-02-07T01:32:53.890" Score="0" ViewCount="83" Body="&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/s95E2.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the solution to the problem. However, I dont understand how they got to the part that says &quot;and the parallel vector is i+4j..&quot;. What does this mean and how did they derive that?&lt;/p&gt;&#xA;" OwnerUserId="2985" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-07T01:52:04.690" LastActivityDate="2011-02-07T01:52:04.690" Title="Finding the unit vectors parallel to a tangent line" Tags="&lt;vector-calculus&gt;" AnswerCount="1" />
  <row Id="20763" PostTypeId="2" ParentId="20762" CreationDate="2011-02-07T01:41:09.157" Score="1" ViewCount="" Body="&lt;p&gt;For any real number $m$, the vector $(1,m)$ determines a line of slope $m$ through the origin: simply note that the line through $(0,0)$ and $(1,m)$ has rise $m$ and run $1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case, your $m=4$, so the vector $(1,4) = 1i + 4j = i+4j$ is a vector of the appropriate slope, hence parallel to the tangent. &lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2011-02-07T01:41:09.157" CommentCount="4" />
  <row Id="20764" PostTypeId="2" ParentId="20500" CreationDate="2011-02-07T01:47:02.037" Score="2" ViewCount="" Body="&lt;p&gt;(i) Minima of a function occur at critical points—where the derivative of the function is undefined or zero.  Have you found the critical points of the function?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(ii) Are you stuck on the &quot;Show that...&quot; part, or the &quot;... deduce that...&quot; part?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(iii) If the three positive real roots are $a$, $b$, and $c$, then $x^3-px^2+qx-r=(x-a)(x-b)(x-c)$—what are $p$ and $r$ in terms of $a$, $b$, and $c$?  How do these relate to (ii)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(iv) For a polynomial with real coefficients, any nonreal complex roots occur in conjugate pairs.  Consequently, for a cubic polynomial, there must be exactly __ or __ real roots (fill in the blanks).&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2011-02-07T01:47:02.037" />
  <row Id="20765" PostTypeId="1" CreationDate="2011-02-07T01:53:27.623" Score="2" ViewCount="138" Body="&lt;p&gt;Take&#xA;$$   \frac{x-a}{b-a}(b) + \frac{b-x}{b-a}(a) = x.$$&lt;br&gt;&#xA;I need an interpretation of this, using concrete example.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;well i dont have difficulty doing it and is not HW. IM ust drawing a blank right now trying to interpret this. example would be on a number like 2-9 i want 6 then applying this (4/7)(9)+(3/7)2 =6 IM TRYING to see whats going on in , not making sense to me but it works&lt;/p&gt;&#xA;" OwnerUserId="6731" LastEditorUserId="1543" LastEditorDisplayName="" LastEditDate="2011-05-03T10:04:24.743" LastActivityDate="2011-05-03T10:04:24.743" Title="I need an interpretation of this formula using concrete example" Tags="&lt;algebra-precalculus&gt;" AnswerCount="1" CommentCount="9" />
  <row Id="20766" PostTypeId="2" ParentId="20749" CreationDate="2011-02-07T02:27:43.250" Score="6" ViewCount="" Body="&lt;p&gt;In symbols, the identity is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left({n-1\atop m-1}\right)\left({n\atop m+1}\right)\left({n+1\atop m}\right) = &#xD;&#xA;\left({n\atop m-1}\right)\left({n-1\atop m}\right)\left({n+1\atop m+1}\right).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The usual combinatorial interpretation of a binomial coefficient $\left({n-1\atop m-1}\right)$ is that it counts subsets of size $m$ from a set of size $n$. Multiplication is usually interpreted as mutually exclusive choice ($f(n)g(n)$ counts the process of picking $f(n)$ configurations, then picking (independently) $g(n)$ items.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Putting this together, the LHS counts subsets of size $m-1$ from a set of size $n-1$, then subsets of size $m$ from an (independent) set of size $n+1$, then (again independently) subsets of size $m+1$ from a set of size $n$. This corresponds one-to-one with the RHS because the things counted by the LHS can be counted in a different way by the RHS: For the RHS distinguish an element of the $n$ set and one of the $n+1$ set. What's left over for those two sets can be chosen by $\left({n-1\atop (m+1)-1}\right)$ and $\left({(n+1)-1\atop m-1}\right)$ respectively, and then the two distinguished elements can be included to be (possibly) chosen in the $n-1$ set to account for $\left({(n-1) +2 \atop (m-1)+2}\right)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To be clearer about the combinatorial interpretation, there are three sets, of size $n-1$, $n$, and $n+1$, from which you choose subsets of size $m-1$, $m+1$, and $m$, respectively. Another way to count this situation is to, take 1 item each out of the $n$ and $n+1$ sets, and add them to the $n-1$ set. So now you're counting out of sets of size $n+1$, $n-1$, and $n$, from which you choose subsets of size $m+1$, $m$, and $m-1$, respectively.&lt;/p&gt;&#xA;" OwnerUserId="1919" LastEditorUserId="1919" LastEditorDisplayName="" LastEditDate="2011-02-07T18:07:14.987" LastActivityDate="2011-02-07T18:07:14.987" CommentCount="7" />
  <row Id="20767" PostTypeId="2" ParentId="20765" CreationDate="2011-02-07T02:56:08.467" Score="4" ViewCount="" Body="&lt;p&gt;Building on Shai's interpretation, consider the following more general question: suppose $l \leq x \leq r$. Since $x$ is inside the interval $[l,r]$, it can be represented as a weighted average of $l$ and $r$. Let $a = x-l$ be the distance from $x$ to $l$, and $b = r-x$ be the distance from $x$ to $r$. Then clearly $$x = \frac{a}{a+b} l + \frac{b}{a+b} r.$$ The left endpoint $l$ can be recovered by starting with $x$ and going $a$ backwards, so $l = x-a$. Similarly, the right endpoint $r$ can be recovered by starting with $x$ and going $b$ forwards, so $r = x + b$. So $$x = \frac{a}{a+b} (x-a) + \frac{b}{a+b} (x+b).$$ Your formula replaces $b$ with $-b$ for some reason, but this formula is actually more intuitive.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2011-02-07T02:56:08.467" CommentCount="1" />
  <row Id="20768" PostTypeId="2" ParentId="20751" CreationDate="2011-02-07T03:20:58.570" Score="1" ViewCount="" Body="&lt;p&gt;I presume that the answer comes from integrating $t\cdot f(t)$ by parts; $\int t\cdot f(t)dt = (t\cdot \int_0^t f(x) dx)|_0^1 - \int_0^1(\int_0^t f(x)dx)dt$, and since the first term evaluates to $0$ (the value at $0$ is $0$ because of the factor of $t$, and the value at $1$ is $0$ because $\int_0^1 f(x) dx = 0$) it simplifies to $-\int_0^1(\int_0^t f(x)dx)dt$ ; now, this is $0$ at both ends of the $t$ interval, and so there should be relatively straightforward ways of bounding it.&lt;/p&gt;&#xA;" OwnerUserId="785" LastActivityDate="2011-02-07T03:20:58.570" />
  <row Id="20769" PostTypeId="1" CreationDate="2011-02-07T03:44:03.917" Score="3" ViewCount="117" Body="&lt;p&gt;I have an optimization problem that at first sounds quite textbook.  I have a convex objective function in $D$-dimensional space that is twice differentiable everywhere and has no local optima.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ordinarily it would be a perfect candidate for numerical Newton-Raphson methods.  However, Newton-Raphson requires solving a system of linear equations of size $D$.  This takes $O(D^3)$ computations, at least with any reasonably implementable algorithm I'm aware of.  In my case $D$ is on the order of several thousand.  Can anyone suggest an optimization algorithm that is typically more efficient than Newton-Raphson for $D$ this large?  I tried gradient descent, but empirically it seemed absurdly slow to converge.&lt;/p&gt;&#xA;" OwnerUserId="6088" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2011-02-07T07:19:27.497" LastActivityDate="2011-02-07T09:53:51.270" Title="High Dimensional Optimization Algorithm?" Tags="&lt;numerical-methods&gt;&lt;optimization&gt;&lt;numerical-optimization&gt;" AnswerCount="3" FavoriteCount="1" />
  <row Id="20770" PostTypeId="2" ParentId="20769" CreationDate="2011-02-07T03:53:59.940" Score="1" ViewCount="" Body="&lt;p&gt;Have you tried conjugate gradients? (&lt;a href=&quot;http://en.wikipedia.org/wiki/Conjugate_gradient_method&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Conjugate_gradient_method&lt;/a&gt;) I did a lot of work with that method some time ago; if you have questions about it feel free to ask.&lt;/p&gt;&#xA;" OwnerUserId="6622" LastActivityDate="2011-02-07T03:53:59.940" CommentCount="2" />
  <row Id="20771" PostTypeId="1" AcceptedAnswerId="20779" CreationDate="2011-02-07T03:59:45.343" Score="0" ViewCount="55" Body="&lt;p&gt;I'm reading a book called 'Rotations, Quaternions, and Double Groups' and I'm having a great deal of difficulty trying to understand the definition of bilateral rotations,  bilateral binary rotation, and semiaxes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is the snippet I'm confused about,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Binary rotations and their corresponding axes, or binary axes, will be found to be most important in the study of the rotation group. One particularly significant property which they have is this: if a rotation $C_m$ has a binary axis perpendicular to it, then the two semiaxes of $C_m$ are interchanged by the binary rotation&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) What are the semiaxes of $C_m$? My guess is if we take the +Z to be the binary rotation axis and $C_m$'s rotation axis to be +X then I'm guessing when he says semiaxes he must mean +X and +Y? I can't find any good google results for semiaxes. If so when Altmann says interchanged he means to say that +Y and +X will now be pointing in the opposite directions, correct?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;He continues on with,&#xA;&quot;Rotations $C_m$ thus related to a binary rotation are called for this reason bilateral rotations.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;s&gt;2) What does mean when he says 'thus related to a binary rotation'.&lt;/s&gt; &lt;i&gt;I understand what he means after a few more readings&lt;/i&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To finish the paragraph he goes on to say,&#xA;&quot;If the rotation $C_m$ here is itself binary, $C_2$, then it is called a bilateral binary rotation. It is clear in this case that the second $C_2$ is also bilateral binary; that is, that bilateral binary rotations must always appear as pairs of mutually perpendicular binary axes.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you, I've been having a hard time finding other sources of information on this so I'm quite confused on his definitions.&lt;/p&gt;&#xA;" OwnerUserId="6732" LastEditorUserId="6732" LastEditorDisplayName="" LastEditDate="2011-02-07T04:13:48.067" LastActivityDate="2011-04-08T06:39:44.340" Title="Confusion about bilateral binary rotations" Tags="&lt;group-theory&gt;" AnswerCount="1" />
  <row Id="20772" PostTypeId="2" ParentId="20732" CreationDate="2011-02-07T04:01:37.660" Score="5" ViewCount="" Body="&lt;p&gt;It was my understanding that Terrence Tao felt that there was no hope of recovery:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;To give a (somewhat artificial) analogy: as I see it now, the paper is like a lengthy blueprint for a revolutionary new car, that somehow combines a high-tech engine with an advanced fuel injection system to get 200 miles to the gallon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The FO(LFP) objections are like a discovery of serious wiring faults in the engine, but the inventor then claims that one can easily fix this by replacing the engine with another, slightly less sophisticated engine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The XORSAT and solution space objections are like a discovery that, according to the blueprints, the car would run just as well if the gasoline fuel was replaced with ordinary tap water. As far as I can tell, the response to this seems to be equivalent to “That objection is invalid – everyone knows that cars can’t run on water.”&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The pp/ppp objection is like a discovery that the fuel is in fact being sent to a completely different component of the car than the engine.&#xA;&quot;&#xA;&lt;a href=&quot;http://rjlipton.wordpress.com/2010/08/15/the-p%E2%89%A0np-proof-is-one-week-old/#comment-5647&quot; rel=&quot;nofollow&quot;&gt;Terrence Tao at P=NP and Godel's Lost Letter&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1930" LastActivityDate="2011-02-07T04:01:37.660" />
  <row Id="20773" PostTypeId="1" AcceptedAnswerId="20774" CreationDate="2011-02-07T04:06:28.360" Score="0" ViewCount="130" Body="&lt;p&gt;Let's say I got stuck with a problem in integration, and I decided to peek at an answer, and I differentiate the answer to get the question, and then I flip from question to answer, and I have my solution ready.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, I have many times seen that this approach does not match with the way we find answer with question, it somehow looks illogical, because I feel like I had never seen the answer, then I would have never guessed this could be in my entire life.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this approach acceptable, or standard way? I would like to have your opinion!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pardon me if I put the wrong type of question. &#xA;And I couldn't find any tag for my question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regards,&lt;/p&gt;&#xA;" OwnerUserId="2199" LastActivityDate="2011-02-07T07:18:13.587" ClosedDate="2011-02-07T12:59:23.853" Title="Is this way of solving integration problem correct?" Tags="&lt;calculus&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="20774" PostTypeId="2" ParentId="20773" CreationDate="2011-02-07T04:17:08.570" Score="3" ViewCount="" Body="&lt;p&gt;Well, if you guess the correct answer, it's not wrong. In fact, it's important to have good guesses. But you can't guess a more complicated integral, so it's also important to know the integration methods.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Integration is an art. You got to have a bag of some known primitives, some tricks and some methods. What you have to do is combine them to get a solution.&lt;/p&gt;&#xA;" OwnerUserId="1971" LastActivityDate="2011-02-07T04:17:08.570" />
  <row Id="20776" PostTypeId="1" CreationDate="2011-02-07T04:25:59.177" Score="4" ViewCount="74" Body="&lt;p&gt;what exactly is the meaning of &quot;obtaining a structural result&quot; in mathematics, for example in graph theory?&lt;/p&gt;&#xA;" LastActivityDate="2011-02-07T17:29:41.950" Title="structural result" Tags="&lt;graph&gt;&lt;graph-theory&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="20777" PostTypeId="1" CreationDate="2011-02-07T04:55:57.153" Score="0" ViewCount="70" Body="&lt;p&gt;Imagine a sphere of liquid in $\mathbb{R}^3$.The sphere is rotating around a vector with some coordinates (1,0,0) and reached a hydrostatic equilibrium so to become ellipsoid. What would happen if it was rotating around a vector with coordinates  (i,0,0)?&lt;/p&gt;&#xA;" OwnerUserId="2513" LastActivityDate="2011-02-07T04:55:57.153" Title="Rotation around a vector with complex coordinates" Tags="&lt;euclidean-geometry&gt;" CommentCount="3" />
  <row Id="20778" PostTypeId="1" CreationDate="2011-02-07T05:02:30.350" Score="2" ViewCount="118" Body="&lt;p&gt;Give the parameterized solution for the dependent system represented by the matrix:&#xA;$$\left(\begin{array}{ccc|r}&#xD;&#xA;1 &amp;amp; 0 &amp;amp; 3 &amp;amp; 6\\ &#xD;&#xA;0 &amp;amp; 1 &amp;amp; 2 &amp;amp; -4\\ &#xD;&#xA;0 &amp;amp; 0 &amp;amp; 0 &amp;amp; 0&#xD;&#xA;\end{array}\right).$$&lt;/p&gt;&#xA;" OwnerUserId="6735" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-07T17:19:10.103" LastActivityDate="2011-02-07T17:19:10.103" Title="How to parameterize a solution of a system of equations?" Tags="&lt;matrices&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="20779" PostTypeId="2" ParentId="20771" CreationDate="2011-02-07T05:14:58.667" Score="1" ViewCount="" Body="&lt;p&gt;I haven't come across the term &quot;semiaxis&quot; before, either, but this book doesn't seem to use precise terminology, anyway -- there's no such thing as an axis perpendicular to a rotation -- they must mean the axis of rotation there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From what you quote, &quot;if a rotation $C_m$ has a binary axis perpendicular to it, then the two semiaxes of $C_m$ are interchanged by the binary rotation&quot;, it seems that by a semiaxis they mean the two rays making up an axis -- so in your example, if the binary rotation axis is the $Z$ axis and $C_m$ is a rotation about the $X$ axis, then the semiaxes would be the $+X$ ray and the $-X$ ray.&lt;/p&gt;&#xA;" OwnerUserId="6622" LastActivityDate="2011-02-07T05:14:58.667" CommentCount="2" />
  <row Id="20780" PostTypeId="2" ParentId="20776" CreationDate="2011-02-07T05:26:05.093" Score="6" ViewCount="" Body="&lt;p&gt;There is no exact meaning of the term &quot;structural result&quot; across mathematics (or even across graph theory).  In fact, I fear it is one of these impressive sounding terms which means very little apart from what the writer wants it to mean in any particular case (cue Humpty Dumpty and the obligatory reminder that Lewis Carroll's day job was mathematics professor).   Or at least there is relatively little that it can be &lt;em&gt;counted on&lt;/em&gt; to mean.  For many mathematicians (including me) it does call &lt;em&gt;something&lt;/em&gt; to mind, and indeed this and similar terminology appears in some of my papers and notes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me to try to give an example.  Consider the problem of determining whether a finite graph is planar.  Perhaps the graph is given to me as an &lt;strong&gt;immersed&lt;/strong&gt; planar graph, i.e., with edges crossing each other.  In this case, one could ask about algorithms for making the graph planar possibly by changing the original diagram in a minimal way (in some sense).  [Note that there is a game which is based upon solving this problem: &lt;a href=&quot;http://www.planarity.net/&quot; rel=&quot;nofollow&quot;&gt;planarity&lt;/a&gt;.] One could imagine very nice -- and certainly, very useful -- results of this kind, but somehow they are not &quot;structural&quot; in the sense that they are paying too much attention to the particular representation of the graph one starts with rather than its &lt;em&gt;abstract structure&lt;/em&gt; as a graph.  On the other hand, the &lt;a href=&quot;http://en.wikipedia.org/wiki/Planar_graph#Kuratowski.27s_and_Wagner.27s_theorems&quot; rel=&quot;nofollow&quot;&gt;theorem of Kuratowski-Wagner&lt;/a&gt; that a finite graph is planar iff it does not have $K_5$ or $K_{3,3}$ as a graph minor is more structural: it manifestly depends only on the isomorphism class of the graph.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This example is not fantastic.  If I wanted to criticize it, I would start by pointing out that an immersed planar graph is a graph with &lt;em&gt;additional&lt;/em&gt; structure of a certain kind.  But I guess the idea of the nonstructuralist approach is that one does not make that additional structure explicit and try to prove theorems which mention the structure explicitly.  A structuralist would say exactly what kind of metric / differential geometric object an immersed graph is, try to construct moduli spaces of such things and then study their properties, and so forth...&lt;/p&gt;&#xA;" OwnerUserId="299" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2011-02-07T05:50:01.013" LastActivityDate="2011-02-07T05:50:01.013" CommentCount="2" />
  <row Id="20781" PostTypeId="1" AcceptedAnswerId="20788" CreationDate="2011-02-07T05:31:09.427" Score="1" ViewCount="143" Body="&lt;p&gt;Consider a random sample from a distribution with the p.d.f. (for some $c &gt; 0$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation} f(x) = cx^2 (1 − x), \text{if } 0 &amp;lt; x &amp;lt; 1; \end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation} 0, \text{ elsewhere}. \end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is an approximate distribution of $\bar{X}_n$ ? For a sample of size 45, find approximate mean and variance for $\bar{X}_n$ ? Find an approximate distribution of $\bar{X}^3_n$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How should I go about solving this problem ? More importantly, I'd like to have an intuition of this question. I'm new to statistics so more details would help me more. This may be naive/dumb but we know that we are taking a sample from uniform distribution and why would we want to find its distribution again and how is it leading to an another distribution ? I'm very confused.&lt;/p&gt;&#xA;" OwnerUserId="6335" LastEditorUserId="6335" LastEditorDisplayName="" LastEditDate="2011-02-07T16:14:57.490" LastActivityDate="2011-02-07T16:14:57.490" Title="Probability and Distribution" Tags="&lt;homework&gt;&lt;probability&gt;&lt;statistics&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="20782" PostTypeId="1" AcceptedAnswerId="20785" CreationDate="2011-02-07T05:35:51.660" Score="4" ViewCount="125" Body="&lt;p&gt;Hello,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Problem statement&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Five sailors were cast away in an&#xA;  island, inhabited only by monkeys. To&#xA;  provide food, they collected all the&#xA;  coconuts that they could find. During&#xA;  the night, one of the sailors woke up&#xA;  and decided to take his share of the&#xA;  coconuts. He divided them into 5 equal&#xA;  piles and noticed that one coconut was&#xA;  left over, so he threw it to the&#xA;  monkeys; he then hid his pile and went&#xA;  to sleep. A little later a second&#xA;  sailor awoke and had the same idea as&#xA;  the first one. He divided the&#xA;  remainder of the coconuts into 5 equal&#xA;  piles, discovered also that one&#xA;  coconut was left over, and threw it to&#xA;  the monkeys. He then hid his share and&#xA;  went back to sleep. One by one the&#xA;  other three sailors did the same&#xA;  thing, each throwing one coconuts to&#xA;  the money. The next morning, all&#xA;  sailors looking sharp and ready for&#xA;  breakfast, divided the remaining pile&#xA;  of coconuts into five equal parts, and&#xA;  no coconuts was left over this time.&#xA;  Find the smallest number of coconuts&#xA;  in the original pile.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My solution was, each time a sailer take his share, I recalculate the number of coconut:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;n = 5*q0 + 1&lt;br&gt;&#xA;-&gt; # left = (4/5)&lt;em&gt;q0 = (4/5)&lt;/em&gt;(n - 1)/5  &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;(4/5)*q0 = 5*q1 + 1&lt;br&gt;&#xA;-&gt; # left = (4/5)*q1 ....&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Continuing this process, I ended up with a very strange fraction:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(256*n - 464981)/1953125  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then I set this fraction to 5*k, since the last time # coconuts is divisible by 5, to solve for n.&lt;br&gt;&#xA;Am I in the right direction? Any hint would greatly appreciated.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks,&lt;br&gt;&#xA;Chan&lt;/p&gt;&#xA;" OwnerUserId="5711" LastActivityDate="2011-02-07T06:52:08.133" Title="Division problem" Tags="&lt;elementary-number-theory&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" />
  <row Id="20783" PostTypeId="2" ParentId="20778" CreationDate="2011-02-07T05:36:20.843" Score="2" ViewCount="" Body="&lt;p&gt;I LaTeX'ed your original post so that the matrix you are trying to convey is obvious. Seems like the matrix you are talking about is the following (got it from the 'edit' text):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\begin{bmatrix}&#xD;&#xA;1 &amp;amp; 0 &amp;amp; 3 &amp;amp; | &amp;amp; 6\\ &#xD;&#xA;0 &amp;amp; 1 &amp;amp; 2 &amp;amp; | &amp;amp; -4\\ &#xD;&#xA;0 &amp;amp; 0 &amp;amp; 0 &amp;amp; | &amp;amp; 0&#xD;&#xA;\end{bmatrix}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The way to go here is from the bottom to up. You see that the third coordinate can be anything (it's a free variable), since $x_{3}$ times $0$ equals $0$ in all cases (I am looking at the third coefficient of the third equation in your system). So let $x_{3}=t$. Then the second equation is of form $0x_{1}+1x_{2}+2t=-4$. So, $x_{2}=-4-2t$. And, finally, the first equation can be represented as $1x_{1}+0x_{2}+3t=6$, which amounts to $x_{1}=6-3t$. So your solution vector is $\begin{bmatrix}&#xD;&#xA;6-3t\\ &#xD;&#xA;-4-2t\\ &#xD;&#xA;t&#xD;&#xA;\end{bmatrix}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S. You can use &lt;a href=&quot;http://www.codecogs.com/latex/eqneditor.php&quot; rel=&quot;nofollow&quot;&gt;http://www.codecogs.com/latex/eqneditor.php&lt;/a&gt; to convert your math to LaTeX.&lt;/p&gt;&#xA;" OwnerUserId="3731" LastActivityDate="2011-02-07T05:36:20.843" />
  <row Id="20784" PostTypeId="1" CreationDate="2011-02-07T05:39:33.620" Score="2" ViewCount="69" Body="&lt;p&gt;The underlying set of the product $X \times Y$ of two schemes is by no means the set-theoretic product of the underlying sets of $X$ and $Y$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Although I am happy with the abstract definition of fiber products of schemes, I'm not confident with some very basic questions one might ask. One I am specifically thinking about has to do with the connected components of a fiber product.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Say $X$ and $Y$ are schemes over $S$ and let's suppose that $Y$ is connected and that $X = \coprod_{i \in I} X_i$ is the decomposition of $X$ into connected components. Is the connected component decomposition of $X \times_S Y$ simply $\coprod_{i \in I} X_i \times_S Y$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If so, how can we see this? What about if we replace &quot;connected&quot; with &quot;irreducible&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you for your consideration.&lt;/p&gt;&#xA;" OwnerUserId="6652" LastActivityDate="2011-02-07T14:04:37.277" Title="Connected components of a fiber product of schemes" Tags="&lt;algebraic-geometry&gt;&lt;schemes&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="20785" PostTypeId="2" ParentId="20782" CreationDate="2011-02-07T05:50:22.297" Score="3" ViewCount="" Body="&lt;p&gt;You wrote that $n = 5q_0 + 1$, so then the first sailor threw away a coconut, and kept $q_0$ coconuts. &lt;strong&gt;The remaining number of coconuts is $4q_0$!&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So then you write:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$n = 5q_0 + 1$$&#xA;$$4 q_0 = 5q_1 + 1$$&#xA;$$4 q_1 = 5q_2 + 1$$&#xA;$$4 q_2 = 5q_3 + 1$$&#xA;$$4 q_3 = 5q_4 + 1$$&#xA;$$4 q_4 = 5q_5$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now you know that $q_5 &amp;gt; 0$, so $n$ will be smallest when $q_5 = 1$. (Of course if they had nothing for breakfast, do the following with $q_0 = 0$). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You must also consider that when the fifth sailor was taking his share, after throwing away a coconut, the remaining number of coconuts was divisible by $5$. So &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ q_4 = \frac{5q_5}{4}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;must be an integer. Let's write $q_5 = 4k_0$. Then writing the equations backwards, and expressing $q_3$ with $q_{4}$ yields:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$4 q_4 = 5q_5 = 20k_0 \text{  so  } q_4 = 5k_0$$&#xA;$$4 q_3 = 5q_4 + 1 = 5^2k_0 +1 $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again you must make sure that $\frac{5^2k_0 +1}{4}$ is an integer. Continue on until you reach $n$. You will be reexpressing the $k_0$ from here on (that's why I added the subscript). &lt;/p&gt;&#xA;" OwnerUserId="5105" LastEditorUserId="5105" LastEditorDisplayName="" LastEditDate="2011-02-07T06:52:08.133" LastActivityDate="2011-02-07T06:52:08.133" CommentCount="9" />
  <row Id="20786" PostTypeId="2" ParentId="20782" CreationDate="2011-02-07T06:02:07.390" Score="1" ViewCount="" Body="&lt;p&gt;You asked for a hint, so I won't give the full answer: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Step (1): Realize that if K is an answer, so is K + n*5^(5+1) for any n.&#xA;Step (2): The hint for this step will probably give the whole thing away so it is withheld&#xA;Step (3): &lt;/p&gt;&#xA;" OwnerUserId="6737" LastActivityDate="2011-02-07T06:02:07.390" CommentCount="1" />
  <row Id="20787" PostTypeId="2" ParentId="19773" CreationDate="2011-02-07T06:09:21.917" Score="1" ViewCount="" Body="&lt;p&gt;This question was answered at MathOverflow:&#xA;&lt;a href=&quot;http://mathoverflow.net/questions/54051/random-permutations-of-z-n&quot; rel=&quot;nofollow&quot;&gt;http://mathoverflow.net/questions/54051/random-permutations-of-z-n&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2011-02-07T06:09:21.917" />
  <row Id="20788" PostTypeId="2" ParentId="20781" CreationDate="2011-02-07T06:17:43.863" Score="4" ViewCount="" Body="&lt;p&gt;First of all, the p.d.f $f$ corresponds to a beta distribution with parameters $\alpha=3$ and $\beta=2$ (hence, not a uniform distribution). Now, by the central limit theorem, $\frac{{\sum\nolimits_{i = 1}^n {X_i }  - n \mu }}{{\sigma \sqrt n }}$ converges in distribution to the ${\rm N}(0,1)$ distribution, where $\mu = {\rm E}(X_1)$ and $\sigma^2 = {\rm Var}(X_1)$. This means that, for $n$ sufficiently large (for example, $n=45$), we may write $&#xD;&#xA;\sum\nolimits_{i = 1}^n {X_i }  \approx n \mu + \sigma \sqrt n Z$, where $Z \sim {\rm N}(0,1)$. Dividing by $n$ thus gives $\bar X_n \approx \mu + (\sigma/{\sqrt{n}})Z$. Thus, $\bar X_n$ is approximated by the normal distribution with mean $\mu$ and small variance $\sigma^2/n$ (this is exactly the distribution of the right-hand side, $\mu + (\sigma/{\sqrt{n}})Z$). As for the mean and variance of $\bar X_n$, this is a simple exercise, and you should be able to show that the mean is $\mu$ and the variance is $\sigma^2/n$. Finally, to find approximate distribution of $\bar X_n^3$, you may use the approximation $\bar X_n \approx \mu + (\sigma/{\sqrt{n}})Z$. &lt;/p&gt;&#xA;" OwnerUserId="2810" LastActivityDate="2011-02-07T06:17:43.863" CommentCount="2" />
  <row Id="20789" PostTypeId="1" AcceptedAnswerId="20803" CreationDate="2011-02-07T06:25:57.757" Score="2" ViewCount="118" Body="&lt;p&gt;If $\varphi: R^m \times R^n \to R$ is a non-degenerate bilinear map and $R$ is an integral domain then we must have $m=n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;edit:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By &quot;non-degenerate&quot; bilinear map I mean that for every nonzero $m \in R^m$ there is an $n \in R^n$ such that $f(m,n)\neq 0$. The reverse also holds: for all non-zero $n \in R^n$ there is an $m \in R^m$ with $f(m,n)\neq 0$.&lt;/p&gt;&#xA;" OwnerUserId="6560" LastEditorUserId="6560" LastEditorDisplayName="" LastEditDate="2011-02-07T07:04:01.297" LastActivityDate="2011-02-07T07:26:18.827" Title="Non-degenerate bilinear map on modules over an integral domain." Tags="&lt;abstract-algebra&gt;&lt;ring-theory&gt;&lt;modules&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="20790" PostTypeId="1" CreationDate="2011-02-07T06:31:54.497" Score="1" ViewCount="83" Body="&lt;p&gt;A company produces three combinations of mixed vegetables that sell in 1Kg package. Italian style combines 0.3kg of zucchini, 0.3kg of broccoli and 0.4kg of carrots. French style combines 0.6kg of broccoli and 0.4kg of carrots. Oriental style combines 0.2kg of zucchini, 0.5kg of broccoli and 0.3kg of carrots. The company has a stock of 16,200 kg of zucchini, 41400 kg of broccoli and 29,400 kg of carrots. How many packages of each style should it prepare with the existing supplies?&lt;/p&gt;&#xA;" OwnerUserId="6740" LastActivityDate="2011-03-09T07:39:49.903" Title="How to solve this problem using matrix" Tags="&lt;matrices&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="20791" PostTypeId="1" AcceptedAnswerId="20793" CreationDate="2011-02-07T06:37:09.533" Score="1" ViewCount="42" Body="&lt;p&gt;Let's say $X$ is a topological space and let's consider the categories of sheaves and presheaves of abelian groups on $X$. Suppose we have a presheaf that is a direct summand (in the category of presheaves) of a sheaf. Can we conclude that the presheaf is also a sheaf? What if we assume further that the complementary direct summand is a sheaf too--does that help us?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that in general colimits in the category of sheaves differs than those in the category of presheaves, and in general the quotient of two sheaves (in the category of presheaves) need not be a sheaf (I think). But maybe the fact that things are split helps us?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What about in the general context of sheaves and presheaves on a site?&lt;/p&gt;&#xA;" OwnerUserId="6652" LastActivityDate="2011-02-07T06:40:39.517" Title="Is a presheaf which is a direct summand of a sheaf necessarily a sheaf?" Tags="&lt;algebraic-geometry&gt;&lt;sheaf-theory&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" />
  <row Id="20792" PostTypeId="2" ParentId="20790" CreationDate="2011-02-07T06:40:16.797" Score="2" ViewCount="" Body="&lt;p&gt;If $x$ is the number of italian style packages, $y$ the number of french style, and $z$ the number of oriental style, then the amount of zucchini is&#xA;$0.3x + 0y + 0.2z$.&#xA;The amount of broccoli is $0.3x + 0.6y + 0.5z$; and the amount of carrots is $0.4x + 0.4y + 0.3z$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You want the quantity of zucchini to equal 16,200, broccoli to equal 41,400, and carrots to equal 29,400, so your system of equations is:&#xA;\begin{align*}&#xA;0.3x &amp;amp; + &amp;amp;&amp;amp;  0y &amp;amp; + &amp;amp;&amp;amp; 0.2x &amp;amp; = &amp;amp;&amp;amp; 16200\\&#xA;0.3x &amp;amp; + &amp;amp;&amp;amp; 0.6y &amp;amp; + &amp;amp;&amp;amp; 0.5z &amp;amp; = &amp;amp;&amp;amp; 41400\\&#xA;0.4x &amp;amp; + &amp;amp;&amp;amp; 0.4y &amp;amp; + &amp;amp;&amp;amp; 0.3z &amp;amp; = &amp;amp;&amp;amp; 29400.&#xA;\end{align*}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I expect you know how to solve a system of linear equations using matrices.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2011-02-07T06:40:16.797" />
  <row Id="20793" PostTypeId="2" ParentId="20791" CreationDate="2011-02-07T06:40:39.517" Score="5" ViewCount="" Body="&lt;p&gt;If the presheaf $\mathcal G$ is a summand of the sheaf $\mathcal F$ seen as a presheaf, then there is a morphism of presheaves $\mathcal F\to\mathcal F$ which has $\mathcal G$ as its kernel. Now a map of presheaves between sheaves is a map of sheaves, and the kernel of a map of sheaves is a sheaf.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Introducing the word &lt;em&gt;site&lt;/em&gt; into the picture does not change anything, as far as I can see...)&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2011-02-07T06:40:39.517" CommentCount="1" />
  <row Id="20794" PostTypeId="2" ParentId="20751" CreationDate="2011-02-07T06:43:57.617" Score="13" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; Incorporated simplification and shortening of the argument suggested by Didier. Thanks!&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;By a) we know that $\frac{1}{2} \int_{0}^{1} f(x)\,dx = 0$. Therefore &#xA;\begin{align*}&#xA;\int_{0}^{1} xf(x)\,dx &amp;amp; = \int_{0}^{1} \left(x - \frac{1}{2}\right)f(x)\,dx \\&#xA;&amp;amp; =&#xA;\left. \frac{1}{2}(x^{2} - x) f(x)\right\vert_{0}^{1} -&#xA;\int_{0}^{1}  \frac{1}{2}(x^{2} - x)f'(x)\,dx \\ &#xA;&amp;amp; = \frac{1}{2} \int_{0}^{1}  (x - x^{2})f'(x)\,dx&#xA;\end{align*}&#xA;using integration by parts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now note that $x - x^{2} \geq 0$ on $[0,1]$ and by b) we have $m \leq f&amp;#39;(x) \leq M$, hence &#xA;$$&#xD;&#xA;C m \leq \int_{0}^{1} x f(x)\,dx \leq C M&#xD;&#xA;$$&#xA;with&#xA;$$&#xD;&#xA;C = \frac{1}{2} \int_{0}^{1} (x - x^{2})\,dx = \left.\frac{1}{2}\left(\frac{1}{2}x^{2} - \frac{1}{3}x^{3}\right)\right\vert_{0}^{1} = \frac{1}{12}&#xD;&#xA;$$&#xA;as we wanted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me add that these estimates arise in the Euler summation method and are often used in proofs of the Stirling formula.&lt;/p&gt;&#xA;" OwnerUserId="5363" LastEditorUserId="6179" LastEditorDisplayName="" LastEditDate="2011-04-02T16:55:48.747" LastActivityDate="2011-04-02T16:55:48.747" CommentCount="5" />
  <row Id="20795" PostTypeId="2" ParentId="20638" CreationDate="2011-02-07T06:47:40.600" Score="1" ViewCount="" Body="&lt;p&gt;Just realized I can add my answer as an answer and not a comment:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What bothered me originally was the question in the title of the post. However, I think the answer to my question is this: When I say the dual of a covector field, I implicitly assume some framing on the tangent bundle (not necessarily one from a coordinate frame.) However, in the proof I choose a different (coordinate) framing, whose dual coframe will be different from the one I am trying to prove is zero. The conclusion only holds for the new framing, and I have proved nothing about the one I started with.&lt;/p&gt;&#xA;" OwnerUserId="6687" LastActivityDate="2011-02-07T06:47:40.600" />
  <row Id="20796" PostTypeId="1" AcceptedAnswerId="20805" CreationDate="2011-02-07T06:49:47.540" Score="2" ViewCount="194" Body="&lt;p&gt;1) $(P \vee Q) \Rightarrow R$&lt;br&gt;&#xA;2) $(P \Rightarrow R) \wedge (Q \Rightarrow R)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Explain why 1 and 2 are equivalent statements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Attempt at solution:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I converted 1 and 2 to these statements which are equivalent:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) $(P \cup Q) \subseteq R$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) $(P \subseteq R) \cap (Q \subseteq R)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This now seems logical but I dont know how to prove it? Any help is appreciated, thanks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: I should have said truth tables are not necessary.&lt;/p&gt;&#xA;" OwnerUserId="5505" LastEditorUserId="5505" LastEditorDisplayName="" LastEditDate="2011-02-07T16:42:31.193" LastActivityDate="2011-02-07T17:26:32.187" Title="Proof of Statements involving Sets" Tags="&lt;logic&gt;&lt;elementary-set-theory&gt;" AnswerCount="4" CommentCount="5" />
  <row Id="20797" PostTypeId="1" AcceptedAnswerId="20802" CreationDate="2011-02-07T06:56:11.343" Score="0" ViewCount="60" Body="&lt;p&gt;Hello,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While solving the monkey problem, I encountered a problem that require solving this equation:&lt;br&gt;&#xA;$$ax - by = c$$&lt;br&gt;&#xA;And I was confused when $b &amp;lt; 0$. Does it hold for EEA or this require another special case?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks,&lt;br&gt;&#xA;Chan&lt;/p&gt;&#xA;" OwnerUserId="5711" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-07T14:01:45.157" LastActivityDate="2011-02-07T14:01:45.157" Title="Does the Extended Euclidean Algorithm hold for the case ax - by = c?" Tags="&lt;elementary-number-theory&gt;" AnswerCount="1" />
  <row Id="20798" PostTypeId="2" ParentId="8312" CreationDate="2011-02-07T06:56:37.207" Score="1" ViewCount="" Body="&lt;p&gt;In response to Qiaochu,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$Disc(f)/Disc(\mathcal{O}_K)$ is the square of the index of $\mathbb{Z}[ \alpha _1, \ldots , \alpha _n ]$ in $\mathcal{O}_K$. The index itself is the determinant of the change of basis matrix from $(\alpha _1, \ldots , \alpha _n )$ to an integral basis for $\mathcal{O}_K$. This matrix is squared when taking the discriminant.&lt;/p&gt;&#xA;" OwnerUserId="6741" LastActivityDate="2011-02-07T06:56:37.207" />
  <row Id="20799" PostTypeId="2" ParentId="20781" CreationDate="2011-02-07T07:11:20.757" Score="5" ViewCount="" Body="&lt;p&gt;Shai Covo's answer shows you how to find the solutions to your homework problems; the key idea is to use the central limit theorem.  My answer is going to address your big picture question about intuition behind why one would care about the central limit theorem and about the distribution of the sample mean.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The central limit theorem says that the mean of a sufficiently large random sample from any distribution with finite variance has approximately a normal distribution.  This is an extremely surprising and powerful result.  It is probably the most important theoretical result in statistics.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's surprising because many simple transformations of random variables don't have &quot;nice&quot; (i.e., named) distributions.  Even adding two independent and identically distributed random variables from the same distribution rarely yields a nice distribution, and when it does, it yields something different depending on the original distribution (e.g., the sum of two geometrics is negative binomial, the sum of two exponentials is gamma, the sum of two bernoullis is binomial).  But the central limit theorem says that if you add up enough independent random variables from the same distribution, regardless of what that distribution is, the resulting sum has approximately a normal distribution.  (Again, the original distribution must have finite variance.)  There is no &lt;em&gt;a priori&lt;/em&gt; reason that one would expect a result like this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The central limit theorem is powerful because knowing the approximate distribution of the sample mean makes it much easier to use the sample to draw inferences about the original population.  To take an example, suppose the water quality in a stream is good if its mean oxygen content is about 5 mg per liter.  Suppose your random sample of 45 observations has a mean of 4.62 mg per liter.  Assuming no flaws in your sampling design, the difference between what your sample produces and what you were hoping it would produce is either due to chance or due to the stream actually having poorer water quality.  Knowing that the sample mean has approximately a particular normal distribution allows you to say what the probability is that the difference is due to chance.  If, say, $\sigma = 0.92$ mg per liter, then you can determine that the probability that the difference is due to chance is about 0.28%.  Thus it's extremely likely that the difference is not due to chance and that the stream's water quality actually is lower than it should be.  (What I've described here is what's called a &lt;em&gt;test of significance&lt;/em&gt;.  There are other inference procedures that the sample mean can be used for, too.  You should know that in practice, the picture with statistical inference is generally more complicated than what I've described here.  For instance, you normally don't know $\sigma$, which causes more problems.  But this is the basic idea.) &lt;/p&gt;&#xA;" OwnerUserId="2370" LastActivityDate="2011-02-07T07:11:20.757" CommentCount="2" />
  <row Id="20800" PostTypeId="2" ParentId="20796" CreationDate="2011-02-07T07:18:08.407" Score="4" ViewCount="" Body="&lt;p&gt;You should either use &lt;a href=&quot;http://en.wikipedia.org/wiki/Truth_table&quot; rel=&quot;nofollow&quot;&gt;truth tables&lt;/a&gt;, or you can argue by contrapositive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To prove $(P \vee Q) \Rightarrow R$ then $(P \Rightarrow R) \wedge (Q \Rightarrow R)$, it is equivalent to show that when $(P \Rightarrow R) \wedge (Q \Rightarrow R)$ is false, then $(P \vee Q) \Rightarrow R$ is false as well. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When is $(P \Rightarrow R) \wedge (Q \Rightarrow R)$ false? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then to prove the other direction reverse the formulas above, and again; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When is$(P \vee Q) \Rightarrow R$ false?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In both of these cases you can find sufficient conditions on $P,Q,R$, then establish the other statement.&lt;/p&gt;&#xA;" OwnerUserId="5105" LastActivityDate="2011-02-07T07:18:08.407" />
  <row Id="20801" PostTypeId="2" ParentId="20773" CreationDate="2011-02-07T07:18:13.587" Score="3" ViewCount="" Body="&lt;p&gt;The idea is a reasonable one, and can be used even when there is no ``back of the book'' available.  For example, suppose that we want to find $\int xe^x dx$.  It is not unreasonable to guess that $xe^x$ might be an antiderivative.  If we differentiate, we get $xe^x +e^x$: not quite right.  But it is now easy to see that $xe^x -e^x$ will do the job.&lt;/p&gt;&#xA;" OwnerUserId="6312" LastActivityDate="2011-02-07T07:18:13.587" />
  <row Id="20802" PostTypeId="2" ParentId="20797" CreationDate="2011-02-07T07:25:58.557" Score="2" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Extended_Euclidean_algorithm&quot; rel=&quot;nofollow&quot;&gt;YES! Read about it here.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5105" LastActivityDate="2011-02-07T07:25:58.557" CommunityOwnedDate="2011-02-07T07:25:58.557" CommentCount="1" />
  <row Id="20803" PostTypeId="2" ParentId="20789" CreationDate="2011-02-07T07:26:18.827" Score="2" ViewCount="" Body="&lt;p&gt;Suppose that $n&amp;gt;m$ and let $e_i,f_j$ be a basis for $R^m, R^n$ respectively. Define the matrix $A\in R^{m\times n}$ by $A_{i,j}=\varphi(e_i,f_j)$ then the columns of the matrix are linear dependent in $K=Frac(R)$, so there is a vector $0 \neq v\in K^n$ such that $Av=0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;since K is the fraction field, then you can find $r\in R$ such that all the entries of $0\neq r\cdot v$ are in $R$ (for example, let r by the multiplication of the denumerators). You now have $A(rv)=0$ so &#xA;$$ \varphi (e_i , \sum_j (rv)_j f_j) = \sum_j (rv)_j \varphi(e_i,f_j) = (A(rv))_i = 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sum_j (rv)_i f_j \neq 0$ since $rv\neq 0$ and the $f_j$ are a basis, so $\varphi$ is degenerate.&lt;/p&gt;&#xA;" OwnerUserId="2125" LastActivityDate="2011-02-07T07:26:18.827" CommentCount="1" />
  <row Id="20804" PostTypeId="2" ParentId="20769" CreationDate="2011-02-07T07:41:28.783" Score="0" ViewCount="" Body="&lt;p&gt;How about the &lt;a href=&quot;http://en.wikipedia.org/wiki/BFGS_method&quot; rel=&quot;nofollow&quot;&gt;BFGS-method&lt;/a&gt;? I behaves asymptotically like Newton's method but builds the Hessian implicitely without solving any equations.&lt;/p&gt;&#xA;" OwnerUserId="3148" LastActivityDate="2011-02-07T07:41:28.783" />
  <row Id="20805" PostTypeId="2" ParentId="20796" CreationDate="2011-02-07T07:49:25.110" Score="6" ViewCount="" Body="&lt;p&gt;If you are unwilling to use truth tables then the following are equivalent&#xA;$$(P \lor Q) \rightarrow R$$&#xA;$$\lnot(P \lor Q) \lor R$$&#xA;$$(\lnot P \land \lnot Q) \lor R$$&#xA;$$(\lnot P \lor R)  \land (\lnot Q \lor R)$$&#xA;$$(P \rightarrow R)  \land (Q \rightarrow R)$$&lt;/p&gt;&#xA;" OwnerUserId="6460" LastActivityDate="2011-02-07T07:49:25.110" CommentCount="3" />
  <row Id="20806" PostTypeId="1" AcceptedAnswerId="20807" CreationDate="2011-02-07T07:55:23.837" Score="4" ViewCount="67" Body="&lt;p&gt;I was reading about the &lt;a href=&quot;http://en.wikipedia.org/wiki/Catalan_number#Applications_in_combinatorics&quot; rel=&quot;nofollow&quot;&gt;Catalan numbers&lt;/a&gt; on wikipedia, because it seems like they show up quite a bit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm reading some of the examples in the applications to combinatorics section. Some of them make sense fairly easily for me. For example, I see that the number of ways to parenthesize a product of $n+1$ terms $x_0,x_1,\dots, x_n$ is $C_n$, since the final multiplication $\cdot$ based on the parentheses must show up between two factors $x_k$ and $x_{k+1}$, and the way to parenthesize those terms $x_0,\dots,x_k$ and $x_{k+1},\dots,x_n$ are $C_k$ and $C_{n-k-1}$, respectively . Since this final $\cdot$ could show up anywhere, I get the recurrence&#xA;$$&#xD;&#xA;C_n=C_0C_{n-1}+C_1C_{n-2}+\cdots +C_{n-1}C_0&#xD;&#xA;$$&#xA;which is the same as the Catalan numbers, given that the initial conditions are the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the same way, I see that the number of binary rooted trees with $n+1$ leaves is also given by $C_n$, since a binary tree essentially determines a parenthesization of a product of $n+1$ leaves if two leaves are enclosed in parentheses if they have the same parent node, and you follow the tree from the leaves up. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also understand the way it counts monotonic paths not passing the diagonal, as the first time a path touches the diagonal is like the position of the final multiplication on a product of $n$ terms, and splits the product into two smaller, but essentially similar subcases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But how does triangulating an $n+2$-gon give the same counting situation? I figured if you choose a vertex and draw a triangle with it as one of the vertices, it should give two other polygons that should suggest a recurrence relation like the one above, just like the Catalan numbers. Could someone point it out, or possibly show a way to relate it to one of the other identical counting situations which I understood? I don't quite see how this situation is the same. Thanks.&lt;/p&gt;&#xA;" OwnerUserId="6742" LastActivityDate="2011-02-07T08:09:25.327" Title="Explanation/Intuition behind why $C_n$ counts the number of triangulations of a convex $n+2$-gon?" Tags="&lt;combinatorics&gt;&lt;catalan-numbers&gt;" AnswerCount="1" />
  <row Id="20807" PostTypeId="2" ParentId="20806" CreationDate="2011-02-07T08:09:25.327" Score="3" ViewCount="" Body="&lt;p&gt;We can find a 1-1 correspondence between a triangulation of a polygon and a binary tree, by considering a triangle to be a vertex and connecting two vertices by an edge if they share a side. You fix one side of the polygon as the &quot;root&quot; side, and the triangle on that side becomes the root vertex of the tree.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance see this: &lt;a href=&quot;http://www.cs.sunysb.edu/~jgao/CSE548-fall07/David-mount-DP.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.cs.sunysb.edu/~jgao/CSE548-fall07/David-mount-DP.pdf&lt;/a&gt;, see pages 20 and 21.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another page: &lt;a href=&quot;http://www.ics.uci.edu/~eppstein/260/011023/&quot; rel=&quot;nofollow&quot;&gt;http://www.ics.uci.edu/~eppstein/260/011023/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2011-02-07T08:09:25.327" CommentCount="1" />
  <row Id="20808" PostTypeId="2" ParentId="20784" CreationDate="2011-02-07T08:48:58.413" Score="2" ViewCount="" Body="&lt;p&gt;&quot;No&quot; in both cases:&#xA;let $f\in K[X]$ be a separable irreducible polynomial of degree $d&amp;gt;1$ over some field $K$.&#xA;Let $L$ be the splitting field of $f$ over $K$.&#xA;Let $X:=\mathrm{Spec} (K[X]/fK[X])$, $Y:=\mathrm{Spec}(L)$ and $S:=\mathrm{Spec}(K)$.&#xA;Then $X$ is irreducible and $X\times_K Y=\mathrm{Spec}(L[X]/fL[X])$ consists of $d$ points,&#xA;which are the irreducible components of  $X\times_K Y$. Moreover these points are also&#xA;the connected components of $X\times_K Y$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;H&lt;/p&gt;&#xA;" OwnerUserId="2479" LastEditorUserId="2479" LastEditorDisplayName="" LastEditDate="2011-02-07T14:04:37.277" LastActivityDate="2011-02-07T14:04:37.277" />
  <row Id="20809" PostTypeId="1" CreationDate="2011-02-07T09:42:50.260" Score="1" ViewCount="64" Body="&lt;p&gt;I just started studying a small &lt;a href=&quot;http://www.cse.yorku.ca/~kosta/CompVis_Notes/harris_detector.pdf&quot; rel=&quot;nofollow&quot;&gt;paper&lt;/a&gt; about the Harris Corner Detection. The problem is I don't understand how step 7 is derived from step 6. In step 7 the expression is expanded in a way that we get a structure tensor $C$ for $x$ and $y$. If one would multiply the three matrices again, I see that we would end up with 6 again (and that it's correct). However I do not see given step 6 how one can derive step 7.&lt;/p&gt;&#xA;" OwnerUserId="6441" LastEditorUserId="825" LastEditorDisplayName="" LastEditDate="2011-02-07T17:16:06.190" LastActivityDate="2011-02-07T17:16:06.190" Title="Trouble deriving the Harris Corner Detection" Tags="&lt;matrices&gt;&lt;tensors&gt;&lt;image-processing&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="20810" PostTypeId="1" AcceptedAnswerId="20822" CreationDate="2011-02-07T09:51:43.030" Score="7" ViewCount="144" Body="&lt;p&gt;I've seen Van-Kampen's theorem presented algebraically many times; and although it provides a useful method of calculation; I don't have a very clear picture for &quot;why&quot; it should be true.  Does anyone know of a more visual argument; or even an example that makes the it easier to see?&lt;/p&gt;&#xA;" OwnerUserId="4772" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-07T14:21:29.133" LastActivityDate="2011-02-09T03:42:58.817" Title="geometric argument for van-kampen?" Tags="&lt;algebraic-topology&gt;&lt;intuition&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="3" />
  <row Id="20811" PostTypeId="2" ParentId="20769" CreationDate="2011-02-07T09:53:51.270" Score="1" ViewCount="" Body="&lt;p&gt;Even if your matrix is of order several thousand, a linear solve should still be pretty fast, and Newton methods should not require many iterations. If your matrix is sparse or has special structure, it will be even faster. My Matlab solves a rank 2000 system in 3 seconds. If you need even more speed, you can try to solve the system only approximately using an iterative method with some fixed number of steps (so then it becomes $O(D^2)$, and see if the Newton iteration still converges.&lt;/p&gt;&#xA;" OwnerUserId="398" LastActivityDate="2011-02-07T09:53:51.270" />
  <row Id="20812" PostTypeId="2" ParentId="20809" CreationDate="2011-02-07T09:57:00.720" Score="1" ViewCount="" Body="&lt;p&gt;$$\big(a^Tb\big)^2 = \big(a^Tb\big)\big(a^Tb\big) = \big(b^Ta\big)\big(a^Tb\big) = b^T\big(aa^T\big)b.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One sees this often enough in linear algebra (or at least, certain applications thereof) that the author presumably saw fit to elide this step.&lt;/p&gt;&#xA;" OwnerUserId="856" LastActivityDate="2011-02-07T09:57:00.720" CommentCount="2" />
  <row Id="20813" PostTypeId="1" AcceptedAnswerId="20815" CreationDate="2011-02-07T10:20:59.823" Score="3" ViewCount="162" Body="&lt;p&gt;I know next to nothing about intuitionism, so my question is probably silly :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As I understand from Wikipedia, intuitionism (at least finitism) doesn't 'trust' in the existence of irrational numbers, because they cannot be constructed (at least in finite number of steps). How does it deal with circles then? Or, even better, how does it deal with bilateral right triangles? Do such things exist in this framework? Do metric concepts exist at all?&lt;/p&gt;&#xA;" OwnerUserId="3793" LastEditorUserId="3793" LastEditorDisplayName="" LastEditDate="2011-02-07T10:28:39.467" LastActivityDate="2011-02-21T12:39:48.363" Title="Intuitionism and circles" Tags="&lt;soft-question&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="20814" PostTypeId="1" CreationDate="2011-02-07T11:06:21.737" Score="1" ViewCount="137" Body="&lt;p&gt;Suppose I have a neural network, with input variables $a,b,c,d,f,g$ and output variables $m,n,o,p,q$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given different input values, the neural network will output corresponding $m,n,o,p,q$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I want find out the best input values which can maxmize m,n, while minimize $o,p,q$ with different weights as well. So how can I find the best $a,b,c,d,f,g$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Currently I use a simple way, which calculate $x=w_1m+w_2n+\frac{w_3}{o}+\frac{w_4}{p}+\frac{w_5}{q}$, then simply search all the possible inputs to find the input to get maxmization of x. However this simple method is not efficient and it also assume m,n,o,p,q are independent, which is not the case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Meanwhile, normally Neural network will not give you a formula how the output related to the input, all the optimization approach I known all need a specific function how the output related to input.   Even for genetic algorithm,  I need a fitness function like $x=w_1m+w_2n+\frac{w_3}{o}+\frac{w_4}{p}+\frac{w_5}{q}$, but in this case, x has to related to inputs $a,b,c,d,f,g$ instead of outputs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any suggestions? Many thanks&lt;/p&gt;&#xA;" OwnerUserId="6743" LastEditorUserId="5610" LastEditorDisplayName="" LastEditDate="2011-02-07T20:30:43.910" LastActivityDate="2011-03-09T23:46:03.270" Title="how to find the input for this optimization problem?" Tags="&lt;optimization&gt;&lt;machine-learning&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="0" />
  <row Id="20815" PostTypeId="2" ParentId="20813" CreationDate="2011-02-07T11:52:29.573" Score="3" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Constructivism_%28mathematics%29&quot; rel=&quot;nofollow&quot;&gt;Constructive mathematics&lt;/a&gt; allows irrational numbers and much more, including of course circles. See the books by &lt;a href=&quot;http://en.wikipedia.org/wiki/Errett_Bishop&quot; rel=&quot;nofollow&quot;&gt;Bishop&lt;/a&gt; and his followers on &lt;a href=&quot;http://en.wikipedia.org/wiki/Constructive_analysis&quot; rel=&quot;nofollow&quot;&gt;constructive analysis&lt;/a&gt;. Constructivism is often identified with intuitionism but it's not the same thing. See also this &lt;a href=&quot;http://mathdl.maa.org/mathDL/19/?pa=reviews&amp;amp;sa=viewBook&amp;amp;bookId=67400&quot; rel=&quot;nofollow&quot;&gt;review&lt;/a&gt; of a book by Bridger.&lt;/p&gt;&#xA;" OwnerUserId="589" LastActivityDate="2011-02-07T11:52:29.573" CommentCount="1" />
  <row Id="20816" PostTypeId="1" AcceptedAnswerId="20819" CreationDate="2011-02-07T12:31:54.160" Score="0" ViewCount="82" Body="&lt;p&gt;I need a function that will produce a graph similar to the one below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://img155.imageshack.us/img155/1812/graphn.png&quot; alt=&quot;This is the graph (png)&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This function is odd, symmetrical relatively to origin in III quarter. A is an asymptote (the top part is similar to hyper-tangent); B is the length of &quot;dead zone&quot; (straight line), where the function value is 0. By changing A and B params the user can control the look of the function, so they have to be included in equation somewhere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also it's interesting to know, is there some unified procedure for restoring the function by graph? Maybe some super-computer can do that? I can imagine the approximation by using NURBS methods or similar, but hardly imagine this using analytical way.&lt;/p&gt;&#xA;" OwnerUserId="6744" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2011-04-30T04:15:59.170" LastActivityDate="2011-04-30T04:15:59.170" Title="Restoring the function by its graph" Tags="&lt;functions&gt;&lt;approximation&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="20817" PostTypeId="1" CreationDate="2011-02-07T12:34:44.407" Score="0" ViewCount="46" Body="&lt;p&gt;A rectangular body of water undergoes oscillation at a frequency of $\sqrt{\frac{12hg}{L^2}}$ where h is the depth and L is the width of the larger side of the rectangular body of water. How can you derive this frequency by showing a parabolic trajectory of $y = \frac{6h}{L^2}x^2$ where $x$ and $y$ are the coordinates measured from the equilibrium position?&lt;/p&gt;&#xA;" OwnerUserId="6745" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2011-02-07T15:13:00.943" LastActivityDate="2011-02-07T15:13:00.943" Title="Correlating frequency of oscillation to shape of wave" Tags="&lt;geometry&gt;&lt;physics&gt;" CommentCount="3" />
  <row Id="20818" PostTypeId="2" ParentId="17320" CreationDate="2011-02-07T13:42:01.413" Score="-1" ViewCount="" Body="&lt;p&gt;I know how to answer your question. I developed one formulates exact&lt;br&gt;&#xA;that  answers your question and  gives more information. If you are&lt;br&gt;&#xA;interested , write for :&lt;br&gt;&#xA;paulo.santarita@gmail.com  &lt;/p&gt;&#xA;" OwnerUserId="6748" LastActivityDate="2011-02-07T13:42:01.413" CommentCount="1" />
  <row Id="20819" PostTypeId="2" ParentId="20816" CreationDate="2011-02-07T13:54:46.083" Score="2" ViewCount="" Body="&lt;p&gt;For the more general question, there are basically two approaches.  One is what Joshua Shane Lieberman did, to think about the characteristics you want in the function and create a function that does that.  There are a number of &lt;a href=&quot;http://en.wikipedia.org/wiki/Sigmoid_function&quot; rel=&quot;nofollow&quot;&gt;sigmoid functions&lt;/a&gt; beside the arctangent (the logistic function is another.) and working from a graph (as opposed to measured points) it is hard to tell one from another.  The other is to have some mathematical model with adjustable parameters and do a numeric fit to determine them.  Any numerical analysis text will describe this, for  example see chapters 12 and 15 of &lt;a href=&quot;http://www.nr.com/&quot; rel=&quot;nofollow&quot;&gt;Numerical Recipes&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-02-07T13:54:46.083" CommentCount="1" />
  <row Id="20820" PostTypeId="1" CreationDate="2011-02-07T13:57:57.720" Score="2" ViewCount="71" Body="&lt;p&gt;Let &lt;em&gt;C&lt;/em&gt; be a smooth, genus &lt;em&gt;g&lt;/em&gt; curve defined over a number field &lt;em&gt;K'&lt;/em&gt;.  Suppose we have a &lt;em&gt;K'&lt;/em&gt;-valued point &lt;em&gt;P&lt;/em&gt; on &lt;em&gt;C&lt;/em&gt;.  We can view &lt;em&gt;C&lt;/em&gt; as a Riemann surface; then the space of holomorphic differential forms has dimension &lt;em&gt;g&lt;/em&gt; over &lt;strong&gt;C&lt;/strong&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems to me that we can choose a uniformizer &lt;em&gt;x&lt;/em&gt; at &lt;em&gt;P&lt;/em&gt; and a basis of differential forms $\{ \omega_i \}$ such that the expansion of each $\omega_i$ in terms of &lt;em&gt;x&lt;/em&gt; has algebraically integral coefficients; that is, there is some finite extension &lt;em&gt;K&lt;/em&gt; of &lt;em&gt;K'&lt;/em&gt;, with ring of integers &lt;em&gt;A&lt;/em&gt;, such that&#xA;$\omega_i = \sum_{n=0}^\infty a_{i,n} x^n dx$ with all $a_{i,n} \in A$.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I would really appreciate critiques, help, or references for my reasoning.  We can start with &lt;em&gt;dx&lt;/em&gt;, which is a meromorphic differential form on &lt;em&gt;X&lt;/em&gt;.  We just need to multiply &lt;em&gt;dx&lt;/em&gt; by rational functions with zeros in the right places to get holomorphic differential forms, but we want to choose rational functions with integral expansions in &lt;em&gt;x&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, we make an initial choice of uniformizer &lt;em&gt;x&lt;/em&gt;.  If &lt;em&gt;P&lt;/em&gt;=(C&lt;sub&gt;0&lt;/sub&gt;:C&lt;sub&gt;1&lt;/sub&gt;: ... :C&lt;sub&gt;n&lt;/sub&gt;) then we know each monomial C&lt;sub&gt;i&lt;/sub&gt;X&lt;sub&gt;j&lt;/sub&gt;-C&lt;sub&gt;j&lt;/sub&gt;X&lt;sub&gt;i&lt;/sub&gt; has a zero of some order at &lt;em&gt;P&lt;/em&gt;; we can just choose some convenient quotient of powers of these monomials to get a zero of order one and call that expression &lt;em&gt;x&lt;/em&gt; (for now).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The poles of &lt;em&gt;dx&lt;/em&gt; will be at various points of &lt;em&gt;C&lt;/em&gt; with coordinates in some finite extension of &lt;em&gt;K'&lt;/em&gt; (as will the zeros of &lt;em&gt;dx&lt;/em&gt;).  It seems to me that we can build enough rational functions from monomials involving these coordinates to get what we need.  (Similar to how we chose the uniformizer &lt;em&gt;x&lt;/em&gt;.)  We can then expand these rational functions in terms of &lt;em&gt;x&lt;/em&gt;.  We may get some denominators (say powers of &lt;em&gt;N&lt;/em&gt;) in the expansions of our functions, but then we can replace &lt;em&gt;x&lt;/em&gt; with &lt;em&gt;x/N&lt;/em&gt; to absorb them.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What do you think?  Many thanks for the assistance!&lt;/p&gt;&#xA;" OwnerUserId="6749" LastEditorUserId="6749" LastEditorDisplayName="" LastEditDate="2011-02-07T16:53:16.193" LastActivityDate="2011-02-07T16:53:16.193" Title="Algebraically integral expansions of differential forms?" Tags="&lt;algebraic-geometry&gt;" CommentCount="5" />
  <row Id="20821" PostTypeId="2" ParentId="20796" CreationDate="2011-02-07T14:11:57.237" Score="1" ViewCount="" Body="&lt;p&gt;It is usually the other way around: the basic properties of propositionsal calculus and logical connectives are established from first principles, and then the corresponding properties for sets and set-theoretic operations are concluded from them. So the properties of $\cap$ are deduced from the properties of $\wedge$; the properties of $\subseteq$ are deduced from the properties of $\Rightarrow$, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You seem to be trying to go the other way. Unless this is how it was done in your class or book, chances are this would not be acceptable. Moreover, the second set-theoretic statement that you write doesn't really make a lot of sense as a set-theoretic statement: &#xA;$$(P\subseteq R)\cap(Q\subseteq R)$$&#xA;is not a well-formed formula of set theory: $\cap$ is an operator between sets, but $\subseteq$ is a relation; $P\subseteq R$ is not a set, but a statement about sets, and $\cap$ cannot operate on &lt;em&gt;statements&lt;/em&gt; about sets, only on &lt;em&gt;sets&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-07T17:15:24.817" LastActivityDate="2011-02-07T17:15:24.817" CommunityOwnedDate="2011-02-07T14:11:57.237" />
  <row Id="20822" PostTypeId="2" ParentId="20810" CreationDate="2011-02-07T14:38:04.780" Score="6" ViewCount="" Body="&lt;p&gt;I'm only going to try to give some intuition why the Seifert-van Kampen Theorem is geometrically reasonable, rather than attempt a geometric proof (which would be well beyond my abilities, I fear).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Seifert-van Kampen Theorem (&quot;van Kampen&quot; was the last name, so it should not be hyphenated) says that if $X=U\cup V$, with $U$ and $V$ arcwise connected open subsets, and $U\cap V$ is nonempty and arcwise connected, then choosing a basepoint $x_0\in U\cap V$, then&#xA;$$\pi(X) \cong \pi(U)*_{\pi(U\cap V)}\pi(V);$$&#xA;that is, the fundamental group of $X$ is the free amalgamated product of the fundamental groups of $U$ and $V$, amalgamated over the subgroup corresponding to the fundamental group of $U\cap V$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Intuitively: it should be clear that $\pi(X)$ contains subgroups isomorphic to $\pi(U)$ and $\pi(V)$, and since $X=U\cup V$, with both open, it should be reasonable that $\pi(X)$ is generated by these two subgroups: if you have an arbitrary loop in $X$ that meanders between $U$ and $V$, you can try decomposing it into an equivalent loop that alternates being entirely in $U$ and entirely in $V$. Also, $\pi(V)$ and $\pi(U)$ will each have subgroups corresponding to $\pi(U\cap V)$. So the real question is &lt;em&gt;why&lt;/em&gt; this should be the only interaction between $\pi(U)$ and $\pi(V)$ (that is, why you get the &lt;em&gt;free&lt;/em&gt; product, and not just some quotient of it). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A relation would correspond to some loop in $X$ which goes through $U$, then $V$, then $U$, then $V$, etc., and is homotopy equivalent to the trivial loop. Imagine that you find such a loop, and pick one with the smallest possible number of &quot;crossings&quot; from one set to the other which satisfies this. You can imagine setting up your deformation so that it first &quot;shrinks&quot; the &quot;last&quot; part, until it is completely contained in $U\cap V$ (since this is open and that is where the basepoint is). But that would give you a new loop which is also equivalent to the identity, but with a fewer number of &quot;crossings&quot; (since you can call the portion in $U\cap V$ part of $U$ or part of $V$, as convenient). This tells you that the smallest possible number of crossings has to be $0$; that is, the entire thing was contained inside $U\cap V$ in the first place. So you should not expect any relations between the $\pi(U)$ and $\pi(V)$, except those that come from identifying their common subgroup $\pi(U\cap V)$. This is precisely the free product with amalgamation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A good place to try to get some intuition is the case where the intersection is simply connected, like Qiaochu suggests in the comments; I find bouquets of circles particularly fertile ground, but then I'm a group theorist, so this is almost the only kind of fundamental group I play with on a semi-regular basis.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-07T17:11:53.883" LastActivityDate="2011-02-07T17:11:53.883" />
  <row Id="20823" PostTypeId="1" CreationDate="2011-02-07T14:47:03.463" Score="4" ViewCount="328" Body="&lt;p&gt;If one gets a phd in math, is it able to be admitted to another phd program in math? It is not for me, for someone else.&lt;/p&gt;&#xA;" OwnerUserId="1944" LastActivityDate="2011-02-07T16:39:28.500" Title="two phds in math" Tags="&lt;soft-question&gt;" AnswerCount="1" CommentCount="11" FavoriteCount="1" />
  <row Id="20824" PostTypeId="1" AcceptedAnswerId="20826" CreationDate="2011-02-07T14:56:59.103" Score="0" ViewCount="69" Body="&lt;p&gt;How can we prove $\bigtriangledown = \bigtriangleup E^{-1}$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where,$\bigtriangleup \rightarrow \text{ Forward difference operator }$&#xA;$\bigtriangledown \rightarrow \text{ Backward difference operator }$&#xA;$E \rightarrow \text{ Shift operator }$ I tried to utilize the method of seperation of symbol but not quite there... &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Give some hints for this proof.&lt;/p&gt;&#xA;" OwnerUserId="2109" LastActivityDate="2011-02-07T15:22:50.593" Title="How to prove $\bigtriangledown = \bigtriangleup E^{-1}$?" Tags="&lt;numerical-methods&gt;" AnswerCount="1" />
  <row Id="20825" PostTypeId="2" ParentId="20814" CreationDate="2011-02-07T15:01:19.817" Score="1" ViewCount="" Body="&lt;p&gt;@zhang: Your goal is optimizing some objective function that has some unknown, black box relationship with the inputs. Either you have some samples of that black box or are able to sample the function yourself. You've chosen to use some very general regression method (artificial neural networks) to model the black box function--but with this approach your objective function will be smooth but non-convex, meaning you might not be to use convex techniques for the optimization. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;@Yuval: He wants to generalize his data to a smooth function with the neural network, and then design or optimize the input for some objective function. The standard case only covers the generalization/prediction part.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Some of your options:&lt;br&gt;&#xA;1. Use your neural network and plug it in to the objective function of some general nonlinear programming technique such as Matlab's &lt;a href=&quot;http://www.mathworks.com/help/toolbox/optim/ug/fmincon.html&quot; rel=&quot;nofollow&quot;&gt;fmincon&lt;/a&gt;. To do this, you won't even need to necessarily know the form of the relation.&lt;br&gt;&#xA;2. Another option is using a more restricted regression (such as linear or convex quadratic) for either the relation or the objective function itself. In this case, once you have the regression, you can solve a convex problem to get the optimum.&lt;br&gt;&#xA;3. Use a trust region method (which #1 may be using anyway) directly on your samples of the relation/objective. This will build local convex approximations of the relation/objective, which obviates the need for training a global approximation (the neural net) as in #1.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;About the structure of the relation from the neural net: you'll need to know exactly what kind of net you are using and any parameters. Below is an explanation of the structure of a common neural net variety.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, if your network is a &lt;a href=&quot;http://en.wikipedia.org/wiki/Multilayer_perceptron&quot; rel=&quot;nofollow&quot;&gt;multilayer perceptron&lt;/a&gt; (one of the most common types), the output for a &lt;em&gt;single output or hidden node&lt;/em&gt; will follow this form: $\phi(x) = f( w^Tx+b )$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x$ is a vector of inputs &lt;em&gt;for that node&lt;/em&gt; (in the input layer, these will be your inputs $a,b,c,d,f,g$)&lt;br&gt;&#xA;$w$ is the learned weights for &lt;em&gt;for that node&lt;/em&gt;&lt;br&gt;&#xA;$b$ is a bias term for the node, like another weight&lt;br&gt;&#xA;$f$ is some nonlinear function, which for a multilayer perceptron can be $\tanh$, or $(1 + e^{-x})^{-1}$ (there are other choices)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, if you know the structure of the network, and the weights at each node (you should be able to get this from your neural network package), you can compose the functions to get expressions for the output nodes. Then you can plug in the output expressions into your formula for x, your objective function.&lt;/p&gt;&#xA;" OwnerUserId="5610" LastEditorUserId="5610" LastEditorDisplayName="" LastEditDate="2011-03-09T23:46:03.270" LastActivityDate="2011-03-09T23:46:03.270" />
  <row Id="20826" PostTypeId="2" ParentId="20824" CreationDate="2011-02-07T15:22:50.593" Score="4" ViewCount="" Body="&lt;p&gt;How about writing $\bigtriangleup(f(x))=f(x+1)-f(x)$, $E^{-1}(f(x))=\ldots$, so $\bigtriangleup E^{-1}(f(x))=$?&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-02-07T15:22:50.593" CommentCount="4" />
  <row Id="20827" PostTypeId="1" AcceptedAnswerId="20830" CreationDate="2011-02-07T15:38:55.557" Score="0" ViewCount="65" Body="&lt;p&gt;let's consider regression problem. Given set of training data $\{(x_i,y_i)\}_{i=1}^N$, $x_i \in \mathbb{R}^n$ and $y_i \in \mathbb{R}$, find prediction function $y = f(x)$, e.g. in RBF regression case $y = \sum_{i=0}^m a_i g(x_i,c_i)$, where $g(x_i,c_i)$ is radial function with center $c_i$.&#xA;It's well known that solution can be find by least-squares method from training data as $a = (G^\mathrm{T} G)^{-1} G^\mathrm{T} y$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is another question, if we don't have training sample, but we know the joint distribution density $p(x,y)$, e.g. $p(x,y) = \sum_{i=1}^r b_i k_i(x,y)$. How to estimate regression coefficients $a_i$ from distribution $p(x,y)$ (coefficients $b_i$)?&lt;/p&gt;&#xA;" OwnerUserId="6273" LastActivityDate="2011-02-07T23:33:09.027" Title="dependency between regression coefficients and probability distribution" Tags="&lt;statistics&gt;&lt;regression&gt;&lt;machine-learning&gt;" AnswerCount="1" />
  <row Id="20828" PostTypeId="1" CreationDate="2011-02-07T15:51:08.800" Score="2" ViewCount="143" Body="&lt;p&gt;It is easy to produce a random variable with Dirichlet distribution using Gamma variables with the same scale parameter. If:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ X_i \sim \text{Gamma}(\alpha_i, \beta) $ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ \left(\frac{X_1}{\sum_j X_j},\; \ldots\; , \frac{X_n}{\sum_j X_j}\right) \sim \text{Dirichlet}(\alpha_1,\;\ldots\;,\alpha_n) $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Problem&lt;/em&gt;&#xA;What happens if the scale parameters are not equal?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ X_i \sim \text{Gamma}(\alpha_i, \beta_i) $ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then what is the distribution this variable?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ \left(\frac{X_1}{\sum_j X_j},\; \ldots\; , \frac{X_n}{\sum_j X_j}\right) \sim \; ? $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For me it would be sufficient to know the expected value of this distribution.&lt;br&gt;&#xA;I need a approximate closed algebraic formula that can be evaluated very very quickly by a computer.&lt;br&gt;&#xA;Let's say approximation with accurancy of 0.01 is sufficient.&lt;br&gt;&#xA;You can assume that:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ \alpha_i, \beta_i \in \mathbb{N} $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Note&lt;/em&gt; In short, the task is to find an approximation of this integral: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ f(\vec{\alpha}, \vec{\beta}) = \int_{\mathbb{R}^n_+} \;\frac{x_1}{\sum_j x_j} \cdot \prod_j \frac{\beta_j^{\alpha_j}}{\Gamma(\alpha_j)} x_j^{\alpha_j - 1} e^{-\beta_j x_j} \;\;  dx_1\ldots dx_n$&lt;/p&gt;&#xA;" OwnerUserId="5552" LastEditorUserId="5552" LastEditorDisplayName="" LastEditDate="2011-02-07T17:13:36.873" LastActivityDate="2011-02-07T17:13:36.873" Title="What is the expected value of modified Dirichlet distribution? (integration problem)" Tags="&lt;probability&gt;&lt;integral&gt;&lt;probability-distributions&gt;" CommentCount="7" FavoriteCount="1" />
  <row Id="20829" PostTypeId="1" AcceptedAnswerId="20834" CreationDate="2011-02-07T16:01:37.223" Score="1" ViewCount="74" Body="&lt;p&gt;Hi,&#xA;let $A$ a finite dimensional simple algebra over a field $k$. Denote with $K$ the center of the algebra. Why the dimension of $A$ over $K$ is a square?&lt;/p&gt;&#xA;" OwnerUserId="6583" LastActivityDate="2011-02-07T20:07:48.950" Title="degree of an algebra" Tags="&lt;algebra&gt;" AnswerCount="1" />
  <row Id="20830" PostTypeId="2" ParentId="20827" CreationDate="2011-02-07T16:01:43.680" Score="2" ViewCount="" Body="&lt;p&gt;You can consider the probability distribution as a continuum of sample points and use the same formula, $a=(G^\mathrm{T}G)^{-1}G^\mathrm{T}y$, except that matrix multiplication now means integrating over all $x$ and $y$, weighted by the probability distribution, instead of summing over a finite number of points. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit in response to TheBug's comment:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sorry, I was too vague and general in my original answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know what &quot;radial function with center $c_i$&quot; is in your problem statement; I'll write it down as I understand it and I think you'll be able to adapt it to your notation/case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We want to find $a_i$ to make $y=\sum_i a_i g_i (x)$ a best fit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Discrete case with sample points:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sum_i \left(y_i - \sum_j a_j g_j(x_i)\right)^2 \rightarrow \mathrm{min}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sum_i \left(y_i - \sum_j a_j g_j(x_i)\right)g_k(x_i) = 0$ for all $k$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In matrix form: $G^\mathrm{T}y - G^\mathrm{T}Ga=0$ with $G_{ik} = g_k(x_i)$ and $(G^\mathrm{T}G)_{kj} = \sum_i G_{ik}G_{ij}$, and thus $a=(G^\mathrm{T}G)^{-1}G^\mathrm{T}y$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Continuous case with probability distribution:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\[\int \mathrm{d}^nx \;\mathrm{d}y\; p (x,y) \left(y - \sum_j a_j g_j(x))\right)^2 \rightarrow \mathrm{min}\]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\[\int \mathrm{d}^nx \;\mathrm{d}y\; p (x,y) \left(y - \sum_j a_j g_j(x))\right)g_k(x)= 0\]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the equivalent of $G$ is an object with one discrete index (j/k) and one continuous index (x and y). The equivalent of the matrix multiplication $G^\mathrm{T}G$ is the integration&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\[\int \mathrm{d}^nx \;\mathrm{d}y\; p (x,y) g_j(x) g_k(x) =: A_{kj}\;,\]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which integrates over the continuous index (corresponding to the sum over $i$ in the discrete case) and leaves a &quot;normal&quot; matrix $A$ with the two discrete indices $j$ and $k$ which plays the role of $G^\mathrm{T}G$. The equivalent of the matrix multiplication $G^\mathrm{T}y$ is the integration&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\[\int \mathrm{d}^nx \;\mathrm{d}y\; p (x,y) y g_k(x) =: b_k\;,\]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which integrates over the continuous index and leaves a &quot;normal&quot; vector $b$ with the discrete index $k$. The result is again a &quot;normal&quot; matrix equation with discrete indices: $Aa=b$, and thus $a=A^{-1}b$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope that was more helpful; sorry for being a bit terse the first time around; feel free to ask more questions if I haven't made it clear enough.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit in response to the additional question about convergence with increasing training set size:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think phrasing the question as we both did (i.e. the probability in every neighbourhood of the &quot;correct&quot; values tending to 1), we can prove it using the continuity of the inverse.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we were trying to calculate the expected values of the fitted parameters, we'd have to deal with the non-linearity and potential singularity of the matrix inversion. The average of $G^\mathrm{T}G$ over the probability distribution is already the matrix $A$, independent of the size of the training set, but the same is not true for the inverses of these matrices. In fact the inverse of $G^\mathrm{T}G$ doesn't even exist until the training set contains at least as many points as we're trying to fit, and even then, there's always a subspace (of measure 0) of training sets (of measure 0) that lead to a singular matrix, and in integrating over the inverse we'd be integrating over all these singularties.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if we're only interested in the probability in neighbourhoods tending to 1, I think we don't have to worry about all that. The expected values of $G^\mathrm{T}G$ and of $G^\mathrm{T}y$ are the &quot;correct&quot; values, and the variances of their entries decrease as $n^{-1}$ with the size $n$ of the training set. Thus, the probability of $G^\mathrm{T}G$ and $G^\mathrm{T}y$ being within a given neighbourhood of their &quot;correct&quot; values tends to 1. If we choose the neighbourhood small enough, it will not contain any singular matrices. By the continuity of matrix inversion (restricted to such a well-behaved neighbourhood), for every given neighbourhood of the &quot;correct&quot; parameters we can choose a sufficiently small neighbourhood of the &quot;correct&quot; values of $G^\mathrm{T}G$ and $G^\mathrm{T}y$ such that $a=(G^\mathrm{T}G)^{-1}G^\mathrm{T}y$ will lie in the given neighbourhood, and hence the probability of the fitted parameter values lying within the given neighbourhood will also tend to 1.&lt;/p&gt;&#xA;" OwnerUserId="6622" LastEditorUserId="6622" LastEditorDisplayName="" LastEditDate="2011-02-07T23:33:09.027" LastActivityDate="2011-02-07T23:33:09.027" CommentCount="5" />
  <row Id="20831" PostTypeId="2" ParentId="20823" CreationDate="2011-02-07T16:08:56.840" Score="6" ViewCount="" Body="&lt;p&gt;Here is the &lt;a href=&quot;http://grad.berkeley.edu/admissions/admis_require.shtml&quot; rel=&quot;nofollow&quot;&gt;statement from the Graduate Division at Berkeley&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Duplication of Degree.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Students who already hold a doctoral level degree are not admitted and duplication of degree or admission to a lesser degree is not permitted. However, in extraordinary circumstances, the faculty of the department may request an exception from the Dean of the Graduate Division. The department must demonstrate that the second degree field of study and program are distinctly different from that of the original degree, and that there is a professional or scholarly purpose that requires this second degree.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Similar restrictions apply throughout the UC system. &lt;a href=&quot;http://www.gsas.harvard.edu/prospective_students/admissions_overview.php&quot; rel=&quot;nofollow&quot;&gt;Harvard&lt;/a&gt; likewise states:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Persons holding a PhD or its equivalent, or who have completed most of the work required to earn the PhD elsewhere, may apply to a PhD program in the Graduate School only if it is an unrelated field of study.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I know that at least until the late 1990's MIT's Computer Science Department had a rule that nobody with a Ph.D. (in any field) would be admitted to their Ph.D. program; the rule was changed in the early 2000s so that a Ph.D. in a field other than Computer Science was no longer a bar for admission, but it still placed applicants at a disadvantage. I would expect similar rules to exist in other Departments at MIT, but I did not spot any explicit rule in the mathematics admission page. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similar policies are explicit in many US universities; even in those where it is not explicit, space limitations will usually lead admission committees to consider such applications very skeptically, prefering to give spots to new students, and if they are very keen on the applicant, will likely suggest a post-doc instead. &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-07T16:39:28.500" LastActivityDate="2011-02-07T16:39:28.500" />
  <row Id="20832" PostTypeId="1" CreationDate="2011-02-07T16:25:46.077" Score="0" ViewCount="54" Body="&lt;p&gt;An electronics company produces three models of stereo speakers, models A, B and C, and can deliver them by station wagon, truck, or van. A station wagon holds 3 boxes of model A, 3 boxes of model B and 2 boxes of model C. A truck holds 7 boxes of model A, 6 boxes of model B and 4 boxes of model C. A van holds 4 boxes of model A, 3 boxes of model B and 2 boxes of model C. If 23 boxes of model A, 21 boxes of model B and 14 boxes of model C are to be delivered and the owner wants each vehicle to operate at full capacity, the parameterized solution to the resulting system of equations is: (z+3,-z+2,Z). for (station wagons, trucks, vans). Use the parameterized solution to find the possible configurations of vehicles.&lt;/p&gt;&#xA;" OwnerUserId="6740" LastActivityDate="2011-02-07T16:36:54.197" Title="How to use parameterized solution to find solutions?" Tags="&lt;matrices&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="20833" PostTypeId="2" ParentId="20832" CreationDate="2011-02-07T16:36:54.197" Score="3" ViewCount="" Body="&lt;p&gt;What the parametrized solution tells you is that&#xA;$$\left(\begin{array}{c}&#xD;&#xA;\text{Station wagons}\\&#xD;&#xA;\text{Trucks}\\&#xD;&#xA;\text{Vans}&#xD;&#xA;\end{array}\right) = \left(\begin{array}{c}&#xD;&#xA;3\\&#xD;&#xA;2\\&#xD;&#xA;0&#xD;&#xA;\end{array}\right) + \left(\begin{array}{r}&#xD;&#xA;1\\&#xD;&#xA;-1\\&#xD;&#xA;1&#xD;&#xA;\end{array}\right)z.$$&#xA;So each value of $z$ will give you a configuration. That is, the number of vans will determine how many station wagons, how many trucks, and how many vans you need. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, since you want the answer to give you the number of vehicles, that places some restrictions on $z$: it cannot be, for instance, $\pi$, because that would mean you have $\pi$ vans, and that's going to be a little difficult to arrange (not to mention, to get insurance coverage for). And you cannot have $z=-500$, because that would require the number of trucks to be $-498$, which again seems somewhat difficult to accomplish. So there are only so many values of $z$ for which the mathematical answers actually make sense for the problem as it is to be applied. Figure out what those values are, and then use them to figure out what each of them means in terms of the number of vans, trucks, and station wagon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(And please try to phrase your questions as &lt;em&gt;questions&lt;/em&gt; in the future, instead of as orders)&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2011-02-07T16:36:54.197" />
  <row Id="20834" PostTypeId="2" ParentId="20829" CreationDate="2011-02-07T17:16:30.353" Score="5" ViewCount="" Body="&lt;p&gt;This follows from the Artin-Wedderburn classification of central simple algebras, together with a sneaky little argument involving tensor products.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, the big theorem: let $A$ be a finite dimensional simple (associative, unital) $K$-algebra with precise center $K$.  Then $A$ is isomorphic to $M_n(D)$, where $D$ is a &#xA;$K$-central division algebra.  Here, for any associative algebra $B$, $M_n(B)$ is the $n \times n$ matrix algebra with entries in $B$.  Clearly $\operatorname{dim}_K M_n(B) = n^2 \operatorname{dim}_K B$, so it is enough to show that the dimension of a $K$-central division algebra $D$ is a square.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For this, tensor from $K$ to the algebraic closure $\overline{K}$.  On the one hand, &#xA;$\operatorname{dim}_{\overline{K}} D \otimes \overline{K} = \operatorname{dim}_K D$.  But on the other hand, an algebraically closed field admits no nontrivial finite-dimensional division algebras -- easy exercise -- so $D \otimes \overline{K} \cong M_N(\overline{K})$ and thus $\operatorname{dim}_{\overline{K}} M_N(\overline{K}) = N^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To prove the &quot;easy exercise&quot;, let $D$ be a finite-dimensional division algebra over an algebraically closed field $K$, let $x \in D$ and consider the minimal polynomial of &#xA;multiplication by $x$, viewed as a $K$-linear endomorphism of $D$...&lt;/p&gt;&#xA;" OwnerUserId="299" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2011-02-07T20:07:48.950" LastActivityDate="2011-02-07T20:07:48.950" />
  <row Id="20835" PostTypeId="1" AcceptedAnswerId="20841" CreationDate="2011-02-07T17:23:10.957" Score="2" ViewCount="130" Body="&lt;p&gt;How to calculate the probability that one random sample say X1 will be greater than the other random sample say X2 if they are from a uniform distribution with [0,1] ? This is not a homework. I'm trying to solve some exercise problem and this is a part of the problem.&lt;/p&gt;&#xA;" OwnerUserId="6335" LastActivityDate="2011-02-07T18:20:57.953" Title="Probability and uniform distribution" Tags="&lt;probability&gt;" AnswerCount="5" CommentCount="2" />
  <row Id="20836" PostTypeId="2" ParentId="20796" CreationDate="2011-02-07T17:26:32.187" Score="0" ViewCount="" Body="&lt;p&gt;Instead of working with truth-tables or Venndiagrams you could just reason with these statements:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume (1) holds. Then R is true, whenever P or Q is true. So assume P holds, then &quot;P or Q&quot; holds and therefore R holds. Likewise Q implies R. We have that (2) holds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now assume (2) holds. Let &quot;P or Q&quot; be true, one can assume without loss that P is true. But from (2) we know that P implies R, therefore R holds. The means that (1) holds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now (1) implies (2) and (2) implies (1), this means both are indeed equivalent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is certainly less 'formal' than what you might be expecting (although it can be made formal). But it is this kind of reasoning one will usually make in a mathematical proof and because of that it's useful to be able to think about it this way.&lt;/p&gt;&#xA;" OwnerUserId="5189" LastActivityDate="2011-02-07T17:26:32.187" />
  <row Id="20837" PostTypeId="2" ParentId="20776" CreationDate="2011-02-07T17:29:41.950" Score="1" ViewCount="" Body="&lt;p&gt;Here is a nice structural result for general graphs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Every graph is the disjoint union of its connected components.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Every connected graph is a tree of biconnected components. A &lt;i&gt;biconnected component&lt;/i&gt; is one in which the removal of any one edge leaves the graph connected. An edge disconnecting the graph is known as a &lt;i&gt;bridge&lt;/i&gt;. Removing all bridges, we get a collection of biconnected components. It's easy to see that if we shrink all biconnected components to a dot, then the original graph becomes a tree.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Every biconnected component is obtained from triconnected components (those requiring the removal of at least three edges to disconnect the graph) using two specific constructions, &quot;series&quot; and &quot;parallel&quot;; see &lt;a href=&quot;http://en.wikipedia.org/wiki/SPQR_tree&quot; rel=&quot;nofollow&quot;&gt;SPQR tree&lt;/a&gt; for more details.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Recent results in the same vein concern the structure of claw-free graphs (Seymour and Chudnovsky) and bull-free graphs (Chudnovsky); the same method (and a lot of hard work) was used to prove the &lt;a href=&quot;http://en.wikipedia.org/wiki/Perfect_graph#Characterizations_and_the_perfect_graph_theorems&quot; rel=&quot;nofollow&quot;&gt;Strong Perfect Graph Theorem&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And let's not forget the &lt;a href=&quot;http://en.wikipedia.org/wiki/Robertson-Seymour_theorem&quot; rel=&quot;nofollow&quot;&gt;Graph Minor Theorem&lt;/a&gt;, which is (in some sense) a vast generalization of Kuratowski's theorem.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2011-02-07T17:29:41.950" />
  <row Id="20838" PostTypeId="2" ParentId="20835" CreationDate="2011-02-07T17:31:43.680" Score="2" ViewCount="" Body="&lt;p&gt;Just plot the posible pair ocurrences  (in the X1-X2 plane: it lies inside the unit square), identify the &quot;success&quot; region (in which X1&gt;X2 : a triangle). The probability is given (assuming that the variables are not only uniform but also independent) by the area of the triangle. That is, 1/2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this particular problem, it's easy to guess the result from its symmetry. &lt;/p&gt;&#xA;" OwnerUserId="312" LastActivityDate="2011-02-07T17:31:43.680" CommentCount="5" />
  <row Id="20839" PostTypeId="2" ParentId="20835" CreationDate="2011-02-07T17:31:47.860" Score="2" ViewCount="" Body="&lt;p&gt;Consider the following experiment. Pick $X_1,X_2$ independently, uniformly at random from $[0,1]$; now, with probability $1/2$, switch $X_1$ and $X_2$. What does this tell you about the probability you're looking for?&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2011-02-07T17:31:47.860" />
  <row Id="20840" PostTypeId="2" ParentId="20835" CreationDate="2011-02-07T17:31:53.367" Score="4" ViewCount="" Body="&lt;p&gt;There's no calculation to perform -- the answer must be $1/2$ by symmetry.&lt;/p&gt;&#xA;" OwnerUserId="6622" LastActivityDate="2011-02-07T17:31:53.367" />
  <row Id="20841" PostTypeId="2" ParentId="20835" CreationDate="2011-02-07T17:32:03.707" Score="4" ViewCount="" Body="&lt;p&gt;You could do it by integration or argue out directly. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To argue out directly, all you need to recognize is that since the two samples are independent, $P(X_1&amp;gt;X_2) = P(X_1&amp;lt;X_2)$. And we have $P(X_1&amp;gt;X_2)+P(X_1=X_2)+P(X_1&amp;lt;X_2) = 1$ since all three are mutually exclusive. $P(X_1=X_2) = 0$ and hence we get $P(X_1&amp;gt;X_2) = P(X_1&amp;lt;X_2) = \frac{1}{2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To do it by integration, first find the $P(X_1&amp;gt;X_2 | X_2=x)$. Since the distribution is uniform, $P(X_1&amp;gt;X_2 | X_2=x) = 1-x$. Now $P(X_1&amp;gt;X_2) = \displaystyle \int_{0}^1 P(X_1&amp;gt;X_2 | X_2=x) f_{X_2}(x) dx = \int_{0}^1 (1-x) \times 1 dx = \frac{1}{2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As Yuval points out this is true irrespective of the distribution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The direct argument holds good irrespective of the distribution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, leonbloy's argument  based on areas still work out fine irrespective of the distribution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the argument based on integration,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$P(X_1&amp;gt;X_2 | X_2=x) = 1-F_{X}(x)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$P(X_1&amp;gt;X_2) = \displaystyle \int_{ll}^{ul} P(X_1&amp;gt;X_2 | X_2=x) dF_{X_2}(x) = \int_{ll}^{ul} (1-F_X(x)) dF_X(x)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$P(X_1&amp;gt;X_2) = \displaystyle \int_{ll}^{ul} (1-F_X(x)) dF_X(x) = \int_{ll}^{ul} dF_X(x) -\int_{ll}^{ul} d(\frac{F_X^2(x)}{2})$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$P(X_1&amp;gt;X_2) = F_X(ul) - F_X(ll) - \frac{F_X^2(ul) - F_X^2(ll)}{2} = 1 - 0 - \frac{1-0}{2} = \frac{1}{2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All these seemingly different arguments are fundamentally the same way of expressing the same idea but I thought it would be good to write it out explicitly.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2011-02-07T17:49:15.580" LastActivityDate="2011-02-07T17:49:15.580" CommentCount="6" />
  <row Id="20842" PostTypeId="2" ParentId="20810" CreationDate="2011-02-07T17:38:20.763" Score="1" ViewCount="" Body="&lt;p&gt;I will say it less acurately than Arturo and sure enough too much roughly. Consider it as a first approxiation to the theorem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The way I see Seifert-Van Kampen theorem is like a sophisticated version of the dimension of the sum of vector subspaces:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;\mathrm{dim} (F + G) = \mathrm{dim} F + \mathrm{dim} G - \mathrm{dim} (F\cap G) \ .&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because, forgetting torsion, the fundamental group counts the number of &quot;holes&quot; (non contractible loops, generators of $\pi_1$) in your space $X$. Right? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, if you have $X = U \cup V$, then the number of holes in $X$ must be equal to the number of holes in $U$ plus the number of holes in $V$..., minus the number of holes in $U\cap V$, because these ones you have already counted them twice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, in order to count all the holes in $U$ plus those in $V$, you take the free product $\pi_1 (U) * \pi_1(V)$, but you must &quot;identify&quot; the holes &quot;shared&quot; by $U$ and $V$. This is the reason for the amalgamated product $\pi_1(U) *_{\pi_1(U\cap V)} \pi_1(V)$.&lt;/p&gt;&#xA;" OwnerUserId="664" LastEditorUserId="664" LastEditorDisplayName="" LastEditDate="2011-02-09T03:42:58.817" LastActivityDate="2011-02-09T03:42:58.817" CommentCount="0" />
  <row Id="20843" PostTypeId="1" CreationDate="2011-02-07T17:51:01.723" Score="4" ViewCount="124" Body="&lt;p&gt;Let $A$ be an invertible $n\times n$ matrix.  Define the &lt;strong&gt;matrix of minors&lt;/strong&gt; $\Delta(A)$ of $A$ to be the matrix whose $(i,j)$ entry is the determinant of the minor of $A$ with the $i$th row and $j$ column omitted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Certainly, the most important property of $\Delta(A)$ is that it is instrumental in computing the inverse of $A$:&#xA;$$ A^{-1}= \frac{1}{det(A)}\sigma\cdot \Delta(A)^T\cdot \sigma$$&#xA;where here, $\sigma$ is the diagonal matrix whose diagonal entries alternate $1$ and $-1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From this fact and elementary properties of the inverse, it is easy to prove the following&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Taking the matrix of minors is an involution up to scaling; that is, $\Delta(\Delta(A))=det(A)^{2-n}\cdot A$.&lt;/li&gt;&#xA;&lt;li&gt;Taking the matrix of minors is an group homomorphism; that is, $\Delta(AB)=\Delta(A)\Delta(B)$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;If you actually write out either of these identities in terms of minors, you get a series of non-trivial-looking identities on the minors of an invertible matrix.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this the easiest way to obtain these identities on minors?  It would bother me if it was, for a couple of reasons.  First, it seems extremely likely that the second identity above holds for non-invertible matrices.  Second, if $A$ is totally positive or non-negative (that is, every minor of every size is positive or non-negative), then so is $\Delta(A)$.  However, the inverse of a totally positive matrix will almost never be totally positive, and so the above proof of the identities will stray off the totally positive path.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Can the identities on minors implied by the above relations on $\Delta$ be proven in a more direct way?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Ideally, this proof would be 'subtraction-free', to make it more natural in the totally-positive setting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; I should be a little bit more clear about what sort of thing I am looking for.  I am interested in spaces of totally-positive matrices and the corresponding algebras generated by minors on them (specifically, they are cluster algebras).  An important involution of these spaces and algebras is the matrix of minors defined above.  Note that taking inverses or cofactors does not preserve totally positivity, and so it doesn't act on this space.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mainly, I asked this question because I was annoyed at using a map whose properties couldn't be proven without passing to a different, bigger space (not that this doesn't happen all the time in math).  I was curious if there was a proof along algebraic lines, following only from the simplest relations on minors (the 3-term Plucker relations).&lt;/p&gt;&#xA;" OwnerUserId="1062" LastEditorUserId="1062" LastEditorDisplayName="" LastEditDate="2011-02-07T18:31:02.940" LastActivityDate="2011-02-07T20:08:09.450" Title="A direct proof of the properties of the matrix of minors" Tags="&lt;linear-algebra&gt;&lt;minors&gt;&lt;positivity&gt;" AnswerCount="1" CommentCount="6" />
  <row Id="20845" PostTypeId="1" AcceptedAnswerId="20846" CreationDate="2011-02-07T18:02:42.190" Score="3" ViewCount="53" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;If $|a_{n}-a_{m}|\leq \frac{1}{\min{(m, n)}}$ then $\sum_{n=1}^{\infty}(a_{n}-a_{n+2})$ converges&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Using the fact that $|a_{n}-a_{m}|\leq \frac{1}{\min{(m, n)}}$ I showed that $a_{n}$ is Cauchy hence $a_{n}\to L$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we let $b_{n}=a_{n}-a_{n+2}$ then by looking at the partial sums: $$S_{2n}=\sum_{k=1}^{2n}b_{k}=a_{2}-a_{2n}\to a_{2}-L$$ and $$S_{2n-1}=\sum_{k=1}^{2n-1}b_{k}=a_{1}-a_{2n-1}\to a_{1}-L$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I seem to missing something because if $a_{1}\neq a_{2}$ the above partial sums don't converge to the same limit and the series diverges. Where is the error?&lt;/p&gt;&#xA;" OwnerUserId="6751" LastActivityDate="2011-02-07T19:13:30.467" Title="If $|a_{n}-a_{m}|\leq \frac{1}{\min{(m, n)}}$ then $\sum_{n=1}^{\infty}(a_{n}-a_{n+2})$ converges" Tags="&lt;sequences-and-series&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="20846" PostTypeId="2" ParentId="20845" CreationDate="2011-02-07T18:10:38.203" Score="2" ViewCount="" Body="&lt;p&gt;Your formulas for $S_{2n}$ and for $S_{2n-1}$ are incorrect. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, take $n=2$. You claim that $S_4 = a_2 - a_4$, but in fact,&#xA;\begin{align*}&#xA;S_4 &amp;amp;=\mathop{\sum}\limits_{k=1}^4b_k = (a_1-a_3) + (a_2-a_4) + (a_3-a_5) + (a_4-a_6)\\&#xA;&amp;amp;= a_1+a_2 - (a_5+a_6).&#xA;\end{align*}&#xA;And you claim $S_3=a_1 - a_3$, when in fact&#xA;\begin{align*}&#xA;S_3 &amp;amp;=\mathop{\sum}\limits_{k=1}^3b_k = (a_1-a_3) + (a_2-a_4) + (a_3-a_5)\\&#xA;&amp;amp;= a_1+a_2 - (a_4+a_5).&#xA;\end{align*}&#xA;Get the right formulas for $S_{2n}$ and $S_{2n-1}$, and things should work out fine. &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-07T19:13:30.467" LastActivityDate="2011-02-07T19:13:30.467" CommentCount="1" />
  <row Id="20847" PostTypeId="2" ParentId="13657" CreationDate="2011-02-07T18:11:26.017" Score="1" ViewCount="" Body="&lt;p&gt;Geodesic incompleteness occurs when a space has &quot;holes&quot; in it, or if it is possible to &quot;fall off the edge&quot; of the space.  Here are some examples:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The punctured plane $\mathbb{R}^2 - \{(0,0)\}$ is geodesically incomplete in the standard metric.  For example, a horizontal geodesic starting at $(-1,0)$ and traveling in the positive $x$-direction can only be extended for one unit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that this space is homeomorphic to an infinite cylinder $S^1\times\mathbb{R}$, which is complete under the standard metric.  The difference is that it isn't possible to get to infinity in a finite distance on the cylinder, while it is on the punctured plane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Covers of a geodesically incomplete space are also geodesically incomplete.  Thus all of the covers of the punctured plane are geodesically incomplete, including the universal cover.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The open unit disc $\{(x,y)\in\mathbb{R}^2 \mid x^2+y^2 &amp;lt; 1\}$ is geodesically incomplete in the standard metric.  No geodesic can be extended for more than $\pi$ units.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The open upper half plane $\{(x,y)\in\mathbb{R}^2\mid y &amp;gt;0\}$ is geodesically incomplete in the standard metric.  As with the disc, the idea is that geodesics can fall off the edge of the space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the upper half plane is complete under the hyperbolic metric.  The difference is that it takes infinite time to reach the $x$-axis under the hyperbolic metric, while it can  be reached in finite time under the Euclidean metric.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Most of the examples so far are proper open subsets of complete Riemannian manifolds, but not every geodesically incomplete space has this form.  For example, consider the infinite cone $C=\{(x,y,z)\in\mathbb{R}^3 \mid x^2+y^2=z^2\text{ and }z&amp;gt;0\}$ under the standard metric, and note that the vertex $(0,0,0)$ does not lie in $C$. This space is incomplete because geodesics can approach the point $(0,0,0)$, but it doesn't work to add this point because the closed cone $C\cup\{(0,0,0)\}$ isn't a Riemannian manifold.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The covers of the punctured plane mentioned in example #1 also aren't open subsets of complete Reimannian manifolds.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The most basic way to prove that a Riemannian manifold $M$ is incomplete is to identify a geodesic $\gamma\colon[0,a)\to M$ that cannot be extended any further.  The way to prove that $\gamma$ cannot be extended is to show that $\lim_{t\to a} \gamma(t)$ does not exist in $M$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the Hopf-Rinow theorem, you can also prove that $M$ is geodesically incomplete by proving that $M$ is incomplete as a metric space.  Thus, you can also show that $M$ is geodesically incomplete by demonstrating a Cauchy sequence that doesn't converge, or by demonstrating that some closed and bounded subset of $M$ in not compact.&lt;/p&gt;&#xA;" OwnerUserId="1726" LastActivityDate="2011-02-07T18:11:26.017" />
  <row Id="20848" PostTypeId="1" AcceptedAnswerId="20874" CreationDate="2011-02-07T18:11:31.663" Score="2" ViewCount="157" Body="&lt;p&gt;This is Problem 3.19.3 of Dieudonné's &lt;em&gt;Foundations of Modern Analysis&lt;/em&gt; (in my words). For $x$ a rational number, let $E_x=\{x\}\times\left[-1,0\right[$, and for $x$ an irrational number, let $E_x=\{x\}\times[0,1]$. Let $E=\bigcup_{x\in\mathbf{R}}E_x$ with the subspace topology. Show that $E$ is connected.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is the following hint: &quot;Use (3.19.1) and (3.19.6) to study the structure of a subset of $E$ which is both open and closed.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(3.19.1) is the fact that the connected subspaces of $\mathbf{R}$ are intervals and that intervals are connected.&#xA;(3.19.6) is the fact that any open set of $\mathbf{R}$ is a countable disjoint union of open intervals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My thoughts: Let $A$ be a clopen subset of $E$. It is fairly obvious that, for any $x\in\mathbf{R}$, $A$ contains either all elements or no element of $E_x$. So we define a subset $B$ of the real line by $B=\{x\in\mathbf{R}:E_x\subset A\}$. Now I thought that $B$ has to be clopen as well (in $\mathbf{R}$). But I can't prove it. I have no problems showing that any irrational point of $B$ is an interior point, and that any irrational point of $\overline{B}$ is in $B$, but nothing about rational points. I was pretty sure that this is the way to go, since the hint is exclusively about subsets of $\mathbf{R}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any (further) hint or comment is much appreciated. &lt;/p&gt;&#xA;" OwnerUserId="1878" LastActivityDate="2011-02-07T23:37:06.620" Title="A connected space which is neither locally connected nor path connected." Tags="&lt;topology&gt;&lt;general-topology&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="20849" PostTypeId="2" ParentId="20845" CreationDate="2011-02-07T18:11:38.060" Score="2" ViewCount="" Body="&lt;p&gt;There's an error in your calculation of the partial sums, but otherwise you're on the right track. Notice that &#xA;$$\sum_{n=1}^{N} a_n -a_{n+2} = \sum_{n=1}^{N} a_n - \sum_{m=3}^{N+2} a_m.$$&lt;/p&gt;&#xA;" OwnerUserId="5253" LastActivityDate="2011-02-07T18:11:38.060" />
  <row Id="20850" PostTypeId="2" ParentId="20835" CreationDate="2011-02-07T18:20:57.953" Score="2" ViewCount="" Body="&lt;p&gt;Let's consider the more general case of the probability ${\rm P}(X_1 \geq a X_2)$, where $a \geq 1$ is constant.&#xA;You can find this probability as follows. Use the law of total probability, conditioning on $X_2$, to obtain&#xA;$$&#xD;&#xA;{\rm P}(X_1 \geq a X_2) = \int_0^1 {{\rm P}(X_1  \ge aX_2 |X_2  = u)\,{\rm d}u} = \int_0^1 {{\rm P}(X_1  \ge au)\,{\rm d}u}, &#xD;&#xA;$$&#xA;where the last equality follows from independence. Hence, since $a \geq 1$,&#xA;$$&#xD;&#xA;{\rm P}(X_1 \geq a X_2) = \int_0^{1/a} {{\rm P}(X_1  \ge au)\,{\rm d}u}  + \int_{1/a}^1 {{\rm P}(X_1  \ge au)\,{\rm d}u}.&#xD;&#xA;$$&#xA;Finally, we obtain&#xA;$$&#xD;&#xA;{\rm P}(X_1 \geq a X_2) = \int_0^{1/a} {(1 - au)\,{\rm d}u}  + 0 = \frac{1}{a} - a\frac{1}{{2a^2 }} = \frac{1}{{2a}}.&#xD;&#xA;$$&#xA;In particular, &#xA;$$&#xD;&#xA;{\rm P}\bigg(X_1  \ge \frac{{3X_2 }}{2}\bigg) = \frac{1}{{2(3/2)}} = \frac{1}{3}.&#xD;&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="2810" LastActivityDate="2011-02-07T18:20:57.953" CommentCount="1" />
  <row Id="20851" PostTypeId="1" CreationDate="2011-02-07T18:31:18.247" Score="2" ViewCount="50" Body="&lt;p&gt;Suppose $\mathcal{C}$ is a category such that for every $c \in \mathrm{Ob}(\mathcal{C})$, the slice category $\mathcal{C}/c$ is equivalent to a small category. I need to show that the category of presheaves $[\mathcal{C}^\mathrm{op}, \mathbf{Set}]$ is an elementary topos.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand the standard argument used when $\mathcal{C}$ is a small category — for example, to construct the exponential $G^F$ of two presheaves, we apply the Yoneda lemma and see that we are forced to set $G^F (c) = \mathrm{Hom}(\mathbf{y}c \times F, G)$, where $\mathbf{y}c = \mathrm{Hom}(-, c) : \mathcal{C}^\mathrm{op} \to \mathbf{Set}$ is the contravariant hom-functor. The main obstruction, then, to using this argument is showing that $\mathrm{Hom}(\mathbf{y}c \times F, G)$ is indeed a set under these weaker assumptions. Well, actually, I'd first have to show that $\mathbf{y}c$ is actually a set-valued functor... but isn't this the same as showing that $\mathcal{C}$ is locally small? It's intuitively plausible that $\mathcal{C}/c$ being equivalent to a small category implies $\mathcal{C}$ itself is locally small, but I imagine, from the phrasing of the problem, that it's not the case.&lt;/p&gt;&#xA;" OwnerUserId="5191" LastActivityDate="2011-02-07T18:31:18.247" Title="The category of presheaves on a possibly-large category" Tags="&lt;homework&gt;&lt;set-theory&gt;&lt;category-theory&gt;&lt;topos-theory&gt;" />
  <row Id="20852" PostTypeId="1" AcceptedAnswerId="20857" CreationDate="2011-02-07T19:00:54.097" Score="0" ViewCount="78" Body="&lt;p&gt;Suppose we have a simply connected domain in the complex plane containing 0. I would like to show that there is no holomorphic n'th root function defined on the domain. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using some student logic, I feel that I have to use the log function and its properties. Indeed, log is not defined at 0 so it seems I could come up with a contradiction if I could relate a hypothetical n'th root function on the domain to log. &lt;/p&gt;&#xA;" OwnerUserId="2298" LastActivityDate="2011-02-07T19:15:50.247" Title="Holomorphic n-th roots" Tags="&lt;complex-analysis&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="20853" PostTypeId="1" AcceptedAnswerId="20855" CreationDate="2011-02-07T19:06:42.830" Score="0" ViewCount="68" Body="&lt;p&gt;In general it is not true that a vector bundle $E$ is isomorphic to its dual bundle $E^*$. But it is true when the vector bundle is the tangnet space of a manifold (at least I think it's true). How does one prove this?&lt;/p&gt;&#xA;" OwnerUserId="2099" LastActivityDate="2011-02-07T19:14:55.537" Title="Tangent Bundle and its (Isomorphic?) Dual Bundle" Tags="&lt;differential-geometry&gt;" AnswerCount="2" />
  <row Id="20854" PostTypeId="2" ParentId="20852" CreationDate="2011-02-07T19:12:45.007" Score="2" ViewCount="" Body="&lt;p&gt;Here's a simple proof by contradiction: Cauchy's theorem tells us that the integral of a holomorphic function over a closed curve in a simply-connected domain vanishes. Show that this is not the case for $$\oint_C z^{\frac{1}{n}} \, dz$$ where C is the unit circle.&lt;/p&gt;&#xA;" OwnerUserId="5191" LastActivityDate="2011-02-07T19:12:45.007" CommentCount="1" />
  <row Id="20855" PostTypeId="2" ParentId="20853" CreationDate="2011-02-07T19:14:46.287" Score="7" ViewCount="" Body="&lt;p&gt;An isomorphism between a vector bundle and its dual, fibre-by-fibre is just an isomorphism between a fibre and its dual.  An isomorphism between a vector space and its dual is given by a non-degenerate bilinear function -- so for example, an inner product suffices.   So if you had an inner product on your vector bundle, you would have an isomorphism between $E$ and $E^*$ simply by the operation $v \longmapsto \langle v, \cdot \rangle$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Generally speaking, vector bundles have inner products.  For example, if the base space is paracompact and the fibres are finite-dimensional.   &lt;/p&gt;&#xA;" OwnerUserId="642" LastActivityDate="2011-02-07T19:14:46.287" />
  <row Id="20856" PostTypeId="2" ParentId="20853" CreationDate="2011-02-07T19:14:55.537" Score="4" ViewCount="" Body="&lt;p&gt;Pick a metric on $M$ and use it to identify each tangent vector space to its dual. This gives a smooth isomorphism $TM\cong T^*M$.&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2011-02-07T19:14:55.537" />
  <row Id="20857" PostTypeId="2" ParentId="20852" CreationDate="2011-02-07T19:15:50.247" Score="4" ViewCount="" Body="&lt;p&gt;Pick a circle around the origin that lies within the domain and move around it once. If you have a holomorphic $n$-th root function, you get the same function value again when you get back to where you started. But that can't be, since the argument (angle in polar coordinates) of your point and the argument of the function value must vary continuously with each other, with the argument of the function value increasing at one n-th of the rate of the argument of your point. Since the argument of your point has increased by $2\pi$, the argument of the function value must have increased by $2\pi/n$, so you can't get the same function value again. The contradiction shows that there can be no such function.&lt;/p&gt;&#xA;" OwnerUserId="6622" LastActivityDate="2011-02-07T19:15:50.247" />
  <row Id="20858" PostTypeId="1" AcceptedAnswerId="20860" CreationDate="2011-02-07T19:28:31.233" Score="2" ViewCount="142" Body="&lt;p&gt;Okay, so in my math homework I'm supposed to draw a graph of the following function:&#xA;$$f(x)=\sin x \cos x.$$&#xA;I have the solution in the textbook, but I just can't figure out how they got to that. So, if someone could please post (a slightly more detailed) explanation for this, it would be really appreciated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have to turn the homework this Wednesday, but since I already have the solution the answer doesn't have to be really swift. Bonus points if it is, tohugh&lt;/p&gt;&#xA;" OwnerUserId="6753" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-07T19:31:06.647" LastActivityDate="2011-02-07T19:41:45.070" Title="Graph for $f(x)=\sin x\cos x$" Tags="&lt;homework&gt;&lt;trigonometry&gt;&lt;graph&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="20859" PostTypeId="2" ParentId="20398" CreationDate="2011-02-07T19:31:42.917" Score="1" ViewCount="" Body="&lt;p&gt;Well, I'm not sure that my answer is what you want.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I learned the universal coefficient spectral sequence from the following two sources.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;J. Levine, &quot;&lt;a href=&quot;http://www.ams.org/mathscinet/search/publdoc.html?arg3=&amp;amp;co4=AND&amp;amp;co5=AND&amp;amp;co6=AND&amp;amp;co7=AND&amp;amp;dr=all&amp;amp;pg4=AUCN&amp;amp;pg5=TI&amp;amp;pg6=PC&amp;amp;pg7=ALLF&amp;amp;pg8=ET&amp;amp;review_format=html&amp;amp;s4=&amp;amp;s5=Knot%20module&amp;amp;s6=&amp;amp;s7=&amp;amp;s8=All&amp;amp;vfpref=html&amp;amp;yearRangeFirst=&amp;amp;yearRangeSecond=&amp;amp;yrop=eq&amp;amp;r=11&amp;amp;mx-pid=461518&quot; rel=&quot;nofollow&quot;&gt;Knot modules I&lt;/a&gt;&quot; (1977, Tran. AMS)&#xA;Cochran-Orr-Teichner &quot;&lt;a href=&quot;http://www.ams.org/mathscinet/search/publdoc.html?amp=&amp;amp;loc=refcit&amp;amp;refcit=461518&amp;amp;vfpref=html&amp;amp;r=28&amp;amp;mx-pid=1973052&quot; rel=&quot;nofollow&quot;&gt;Knot Concordance, Whitney Towers and L2-Signatures&lt;/a&gt;&quot; (2003, Ann. Math.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In Levine's paper (see Thm. 2.3 in page 5), Levine gives a quick and readable construction of UCSS. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In Cochran-Orr-Teichner's one (see Remark. 2.8, proof of Thm. 2.13), they use UCSS to control the dimension of homology groups with coefficient in (noncommutative) quotient field of Ore domain. Also, they gives some examples (Reamrk 2.8) about collapsing conditions (you might want) of UCSS. &lt;/p&gt;&#xA;" OwnerUserId="3100" LastActivityDate="2011-02-07T19:31:42.917" />
  <row Id="20860" PostTypeId="2" ParentId="20858" CreationDate="2011-02-07T19:40:13.317" Score="4" ViewCount="" Body="&lt;p&gt;Here is a hint:  Can you draw the graph of $\sin x$?  What about $a\sin{(bx)}$?  Then recall the identity $$\sin{x} \cos{x} = \frac{1}{2} \sin {2x}.$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe that helps.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit:  This is just one way, there are many.  As asked in the above comments, how are you supposed to solve it?  What tools do you have your disposal?  What type of things have you being taught so far?&lt;/p&gt;&#xA;" OwnerUserId="6075" LastEditorUserId="6075" LastEditorDisplayName="" LastEditDate="2011-02-07T19:41:45.070" LastActivityDate="2011-02-07T19:41:45.070" CommentCount="2" />
  <row Id="20861" PostTypeId="1" AcceptedAnswerId="20869" CreationDate="2011-02-07T19:46:20.470" Score="1" ViewCount="115" Body="&lt;p&gt;I have been reading about finite fields because they are used in cryptography quite a bit. However I am having some trouble conceptualizing how they work exactly. Using AES as the example. When you multiply two numbers you then repeatedly xor with 283 until some point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ex, my book states using Hex to binary and GF(2^8)&#xA;87*02 is&lt;br&gt;&#xA;10000111&lt;br&gt;&#xA;00000010&lt;br&gt;&#xA;-is 100001110. Then xored with 283?&lt;br&gt;&#xA;xor100011011&lt;br&gt;&#xA;==000010101&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now how do I know when to stop xoring? Why am I xoring? Just until there are 8 or less digits so I get the correct modulo?&lt;/p&gt;&#xA;" OwnerUserId="6754" LastActivityDate="2011-02-07T20:34:57.760" Title="Help with finite fields" Tags="&lt;finite-fields&gt;" AnswerCount="1" CommentCount="6" />
  <row Id="20862" PostTypeId="1" CreationDate="2011-02-07T20:02:02.117" Score="2" ViewCount="58" Body="&lt;p&gt;I barely remember that I could identify them by attempts. I would get the coefficient, see what numbers can divide it and put them into the Ruffini's rule.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I don't even know what such method is called.&lt;/p&gt;&#xA;" OwnerUserId="6756" LastActivityDate="2011-02-07T20:50:04.273" Title="Finding polynomial roots by attempts" Tags="&lt;algebra&gt;&lt;polynomials&gt;&lt;roots&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="20863" PostTypeId="1" CreationDate="2011-02-07T20:04:53.110" Score="0" ViewCount="146" Body="&lt;p&gt;equation 1:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$-i\omega\underline \xi - 2i(\underline k \cdot \underline \Omega )\underline u = 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\underline \xi = i \underline k \times \underline u$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\underline k \cdot \underline u = 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;taking:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$i\underline k \times \text{(1)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;derive:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\omega = \pm 2\frac{(\underline k \cdot \underline \Omega )}{|\underline k|}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can get pretty close and can post my working if required - but I don't do anything complicated - any help = many thanks&lt;/p&gt;&#xA;" OwnerUserId="6757" LastEditorUserId="536" LastEditorDisplayName="" LastEditDate="2011-02-07T20:26:22.110" LastActivityDate="2011-02-08T00:12:33.970" Title="Vector Analysis Question" Tags="&lt;analysis&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="20864" PostTypeId="2" ParentId="20843" CreationDate="2011-02-07T20:08:09.450" Score="1" ViewCount="" Body="&lt;h3&gt;Combinatorial interpretation&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Let's start with a combinatorial interpretation of $\Delta(A)_{i,j}$. It is a weighted sum of terms, each of which corresponds to a directed $1$-regular graph. Each term is the product of the corresponding entries of $A$, only we don't count the $(j,i)$ edge. In other words, the graph is not really $1$-regular, since $i$ has no in-edge and $j$ has no out-edge; so there is an $i$ to $j$ path, and the rest are circles. Each term is weighted according to the sum of the permutation.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Multiplicativity&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Now $\Delta(AB)_{i,j}$ is the same thing, only each $(x,y)$ edge really corresponds to a pair of edges $(x,z),(z,y)$, with $z$ arbitrary. Suppose a vertex $z$ appears twice, in both $(x_1,z),(z,y_1)$ and $(x_2,z),(z,y_2)$. Replace this wiring to the wiring $(x_1,z),(z,y_2)$ and $(x_2,z),(z,y_1)$. This corresponds to a transposition, and so the sign is opposite and the two cancel. We are left with only the terms where each vertex appears at most once as an intermediate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, in $(\Delta(A)\Delta(B))_{i,j}$ we have two graphs superimposed, with an $i$ to $j$ path composed of an $A$ part followed by a $B$ part. We can now construct a $\Delta(AB)_{i,j}$ path in the obvious way; this is a reversible process (the missing $z$ vertex is the one in the middle of the $i$ to $j$ path). Since sign is multiplicative, the signs agree.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt; Involution up to scaling &lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Consider now $(\Delta(\Delta(A))_{i,j}$. On the first level, we have an almost $1$-regular graph, with some $i$ to $j$ path. On the second level, each edge $x,y$ on the first level corresponds to some $1$-regular graph with some $x$ to $y$ path.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's count the in- and out-degrees of all the vertices. We start with $n-1$ each. For each edge $(x,y)$, remove $1$ from the in-degree of $x$, and $1$ from the out-degree of $y$. Every vertex other than $i$ has an incoming edge, and every vertex other than $j$ has an outgoing edge. Let's assume that $i \neq j$ (the other case is simpler). Thus the out-degree of $i$ is $n-1$, and of every other vertex it is $n-2$; the in-degree of $j$ is $n-1$, and of every other vertex it is $n-2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose the path from $i$ to $j$ is not direct, for example $i$ goes to $k$ goes to $l$ (which might equal $j$). It is possible that if you're cunning enough, you will be able to permute $k$ and $j$ in such a way that the result will have the same value but opposite sign. Given that, we are left only with terms in which the $i$ to $j$ path is direct. The rest of the multigraph is just an arbitrary $(n-2)$-regular graph, which explains the factor $(\det A)^{n-2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It remains for someone in the audience to find the correct cut-and-paste argument...&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2011-02-07T20:08:09.450" />
  <row Id="20865" PostTypeId="1" AcceptedAnswerId="20868" CreationDate="2011-02-07T20:10:20.187" Score="2" ViewCount="86" Body="&lt;p&gt;Is a symmetric real matrix with diagonal entries strictly greater than 1 and off-diagonal entries positive but strictly less than 1 necessarily positive-semidefinite?&lt;/p&gt;&#xA;" OwnerUserId="2404" LastActivityDate="2011-02-07T23:27:48.413" Title="positive symmetric matrices and positive-definiteness" Tags="&lt;linear-algebra&gt;&lt;matrices&gt;&lt;optimization&gt;&lt;quadratic-forms&gt;" AnswerCount="3" FavoriteCount="1" />
  <row Id="20866" PostTypeId="1" AcceptedAnswerId="20871" CreationDate="2011-02-07T20:22:28.357" Score="0" ViewCount="95" Body="&lt;p&gt;x is an integer, and i can write it with $\log_2 x$ bit, and, viceversa, with $n$ bit i can write a number till $2^n$..&#xA;but..&#xA;how many bits to write $\sqrt x$ ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: the integer part!&lt;/p&gt;&#xA;" OwnerUserId="5046" LastEditorUserId="5046" LastEditorDisplayName="" LastEditDate="2011-02-07T21:06:52.993" LastActivityDate="2011-02-07T21:38:01.447" Title="how many bits to write $\sqrt x $ ?" Tags="&lt;homework&gt;" AnswerCount="2" CommentCount="7" />
  <row Id="20867" PostTypeId="1" AcceptedAnswerId="20879" CreationDate="2011-02-07T20:22:53.020" Score="1" ViewCount="80" Body="&lt;p&gt;I'm sure &quot;jaggedness&quot; isn't the right term to use here, so please correct me. I'm trying to quantify how jagged a distribution is. For example, this is moderately jagged:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://imgur.com/kPcwt&quot; rel=&quot;nofollow&quot;&gt;distribution #1&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is really jagged:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://imgur.com/mSLU0&quot; rel=&quot;nofollow&quot;&gt;distribution #2&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Essentially, I want to quantify the number of peaks and the disparity between peaks and valleys in a meaningful way, normalized for size of data. So in the above example, distribution #2 has a larger amount of data than distribution #1, but I don't want that to affect the quantification. How would I do this?&lt;/p&gt;&#xA;" OwnerUserId="6634" LastActivityDate="2011-02-09T10:53:37.833" Title="Calculating the &quot;jaggedness&quot; of a distribution" Tags="&lt;statistics&gt;&lt;normal-distribution&gt;&lt;standard-deviation&gt;" AnswerCount="2" />
  <row Id="20868" PostTypeId="2" ParentId="20865" CreationDate="2011-02-07T20:23:46.917" Score="3" ViewCount="" Body="&lt;p&gt;Nope.  Just playing around with my computer, I found the matrix&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[$\frac{11}{10}$ $\frac{1}{100}$ $\frac{99}{100}$]  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;[$\frac{1}{100}$  $\frac{11}{10}$ $\frac{99}{100}$]  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;[$\frac{99}{100}$ $\frac{99}{100}$  $\frac{11}{10}$]  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;with determinant $\frac{-25179}{31250}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this perhaps a misremembering of the definition of a &lt;a href=&quot;http://en.wikipedia.org/wiki/Diagonally_dominant_matrix&quot; rel=&quot;nofollow&quot;&gt;diagonally dominant matrix&lt;/a&gt;?&lt;/p&gt;&#xA;" OwnerUserId="299" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2011-02-07T20:30:14.383" LastActivityDate="2011-02-07T20:30:14.383" CommentCount="2" />
  <row Id="20869" PostTypeId="2" ParentId="20861" CreationDate="2011-02-07T20:24:54.217" Score="3" ViewCount="" Body="&lt;p&gt;The finite field $GF(2^8)$ is better thought of a collection of $7$th degree polynomials modulo $2$ and some (irreducible) $8$th degree polynomial $P$. Let's see what all of this means.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, each member of $GF(2^8)$ is of the form $$\sum_{i=0}^7 a_i x^i$$, where the $a_i$ are coefficients &quot;modulo $2$&quot;, i.e. bits. When you add two of the $a_i$s, you calculate the answer modulo $2$, so it's the same as XORing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Adding two polynomials is easy: $$\sum_{i=0}^7 a_i x^i + \sum_{i=0}^7 b_i x^i = \sum_{i=0}^7 (a_i+b_i) x^i.$$ So addition corresponds to XOR.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Multiplication is more involved. For AES the polynomial in question is $$P(x) = x^8 + x^4 + x^3 + x + 1.$$ Written in bits, it is $(100011011)_2 = (283)_{10}$. The point is that $x^8 = x^4 + x^3 + x + 1$ (since everything's mod $2$), so in order to multiply a polynomial by $x$, you do the following:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Remember the MSB (that's the power of $x^7$).&lt;/li&gt;&#xA;&lt;li&gt;Shift the number left once (replacing $x^i$ with $x^{i+1}$).&lt;/li&gt;&#xA;&lt;li&gt;If the old MSB was $1$, XOR $(11011)_2$ (since $x^8 = x^4 + x^3 + x + 1$).&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Using this primitive, you can multiply two polynomials $A = \sum_{i=0}^7 a_i x^i$ and $B = \sum_{i=0}^7 b_i x^i$ as follows:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Initialize the result $C = 0$.&lt;/li&gt;&#xA;&lt;li&gt;Add to $C$ the value of $a_0 B$, that is, if the LSB of $A$ is $1$, XOR $B$ to $C$.&lt;/li&gt;&#xA;&lt;li&gt;Add to $C$ the value of $a_1 x B$, that is, if the second least bit of $A$ is $1$, XOR $xB$ to $C$; to calculate $xB$, use the method above.&lt;/li&gt;&#xA;&lt;li&gt;And so on, until you get to $a_7$.&lt;/li&gt;&#xA;&lt;li&gt;Return $C$.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;In practice, you implement it as a loop:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Initialize $C = 0$.&lt;/li&gt;&#xA;&lt;li&gt;If $LSB(A)=1$, $C = C + B$.&lt;/li&gt;&#xA;&lt;li&gt;Set $B = xB$, $C=C/2$ (i.e. shift $C$ right once).&lt;/li&gt;&#xA;&lt;li&gt;Repeat the previous two steps $7$ more times.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;In a real implementation, this multiplication table is stored in some condensed form. At the very least, you store the table of multiplication by $x$. The other extreme is storing all $2^{16}$ possible products. In between, you can store the product of any $A$ by any $2$-bit $B$ (size $2^{10}$ bytes), any $3$-bit $B$ (size $2^{11}$ bytes) or any $4$-bit $B$ (size $2^{12}$ bytes). It all depends on how much memory you can spare, and on the trade-off between memory access (i.e. cache sizes) and ALU performance.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2011-02-07T20:34:57.760" LastActivityDate="2011-02-07T20:34:57.760" CommentCount="6" />
  <row Id="20870" PostTypeId="2" ParentId="7966" CreationDate="2011-02-07T20:25:28.227" Score="1" ViewCount="" Body="&lt;p&gt;I really like the book J.Davis and P.Kirk, &quot;&lt;a href=&quot;http://www.ams.org/bookstore/pspdf/gsm-35-prev.pdf&quot; rel=&quot;nofollow&quot;&gt;Lecture notes in Algebraic topology&lt;/a&gt;&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In their book (in Chapter 9,10), they give a friendly and thorough introduction to Spectral Sequence with &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;full-detailed calculations&lt;/li&gt;&#xA;&lt;li&gt;elegant interpretations&lt;/li&gt;&#xA;&lt;li&gt;vivid examples.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Here are some nice interpretations and jokes which I like it. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Computing with spectral&#xA;  sequences is somewhat like computing integrals in calculus  - in page 237 (introduction part of spectral sequence). &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;As a service to the reader, we will explicitly unravel the statement of&#xA;  the above theorem...&#xA;  - in page 242, giving detailed and helpful explanation.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;We will use the following lemma, which shows why Serre classes are&#xA;  tailor-made to be used with spectral sequences. - in page 271. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Unfortuantely, there are so many errata in their book. &lt;a href=&quot;http://www.indiana.edu/~lniat/typos.pdf&quot; rel=&quot;nofollow&quot;&gt;The collection of errata&lt;/a&gt; is available from J. Davis's page. Actually, I read it from cover to cover. But, I remember that there are some errata not in J.Davis's collection. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, I would like to emphasize &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;J.Davis and P.Kirk book is a perfect&#xA;  introduction to Spectral Sequence.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="3100" LastEditorUserId="3100" LastEditorDisplayName="" LastEditDate="2011-02-08T00:58:56.677" LastActivityDate="2011-02-08T00:58:56.677" />
  <row Id="20871" PostTypeId="2" ParentId="20866" CreationDate="2011-02-07T20:26:19.063" Score="1" ViewCount="" Body="&lt;p&gt;If $\sqrt{x}$ is not an integer, it will take a &lt;strong&gt;lot&lt;/strong&gt; of bits.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint:  If it is an integer (or you are just writing the integer part) you should have a rule of logarithms that will help.  Do you know another way to express $\sqrt{x}$?&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-02-07T20:26:19.063" />
  <row Id="20873" PostTypeId="1" AcceptedAnswerId="20876" CreationDate="2011-02-07T20:31:29.213" Score="1" ViewCount="134" Body="&lt;p&gt;So I know that you can use the Strassen Algorithm to multiply two matrices in seven operations, but what about multiplying two matrices that are exactly the same. Is there a faster way to go about doing that (ie. by reducing the number of multiplications per iteration to something less than 7) ? &lt;/p&gt;&#xA;" OwnerUserId="6758" LastActivityDate="2011-02-07T20:39:28.860" Title="What's the fastest way to take powers of a square matrix?" Tags="&lt;linear-algebra&gt;&lt;algorithms&gt;&lt;matrices&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="20874" PostTypeId="2" ParentId="20848" CreationDate="2011-02-07T20:32:28.480" Score="3" ViewCount="" Body="&lt;p&gt;Building on what you have already proved for irrational points:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume that a rational point $b$ of $\overline{B}$ is not in $B$. Then $E_b$ is not in $A$. Since $b$ is in $\overline{B}$, there is a sequence in $B$ that converges to $b$. Each irrational point $x$ of this sequence is an interior point of $B$, and hence has a neighbourhood contained in $B$. This neighborhood contains a rational number that is closer to $b$ than $x$. Replace all irrational points in the sequence with the rational points thus chosen. This yields a sequence of rational numbers in $B$ converging to $b$, and hence corresponding sequences in $A$ converging to points in $E_b$, contradicting the fact that $A$ is clopen and $E_b$ is not in $A$. Hence all rational points of $\overline{B}$ lie in $B$, and $B$ is closed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Analogously, assume that a rational point $b$ of $B$ is not an interior point. Then there is a sequence of points not in $B$ that converges to $b$. Each irrational point $x$ of this sequence has a neighbourhood that lies entirely outside $B$ (since otherwise it would be in $\overline{B}$, and hence in $B$). This neighbourhood contains a rational point $y$ that is closer to $b$ than $x$. Again, replace all irrational points in the sequence by the rational points thus chosen. This yields a sequence of rational numbers not in $B$ converging to $b$, and hence corresponding sequences not in $A$ converging to points in $E_b$, contradicting the fact that $A$ is clopen and $E_b$ is in $A$. Hence all rational points of $B$ are interior points, and $B$ is open.&lt;/p&gt;&#xA;" OwnerUserId="6622" LastEditorUserId="6622" LastEditorDisplayName="" LastEditDate="2011-02-07T23:37:06.620" LastActivityDate="2011-02-07T23:37:06.620" />
  <row Id="20875" PostTypeId="2" ParentId="20865" CreationDate="2011-02-07T20:32:38.453" Score="1" ViewCount="" Body="&lt;p&gt;If the sum of each row is positive, then no eigenvalue can be non-negative: take any eigenvector, and pick the largest (in magnitude) coordinate. After applying the matrix, even if the signs are all perfectly against us, still $0$ will not be crossed (or reached to).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the correct formulation is $\geq 1$ on the diagonal, and the off-diagonals &lt;i&gt;in each row&lt;/i&gt; sum to less than $1$.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2011-02-07T20:32:38.453" />
  <row Id="20876" PostTypeId="2" ParentId="20873" CreationDate="2011-02-07T20:38:47.293" Score="3" ViewCount="" Body="&lt;p&gt;Consider $$M = \begin{pmatrix} 0 &amp;amp; A &amp;amp; 0 \\ 0 &amp;amp; 0 &amp;amp; B \\ 0 &amp;amp; 0 &amp;amp; 0 \end{pmatrix}.$$ We have $$M^2 = \begin{pmatrix} 0 &amp;amp; 0 &amp;amp; AB \\ 0 &amp;amp; 0 &amp;amp; 0 \\ 0 &amp;amp; 0 &amp;amp; 0 \end{pmatrix}.$$&#xA;So matrix squaring is not asymptotically better than regular product. Of course, in practice the constant factor difference is very significant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Taking a power is best achieved by repeated squaring. The other method is diagonalizing, but I think usually eigenvalues are found via the power method rather than vice versa.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2011-02-07T20:38:47.293" CommentCount="5" />
  <row Id="20877" PostTypeId="2" ParentId="20873" CreationDate="2011-02-07T20:39:28.860" Score="3" ViewCount="" Body="&lt;p&gt;It really depends on the matrix. For instance if you have a diagonalizable matrix then you can decrease the amount of multiplications by diagonalizing so that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$A^k=P^{-1}B^k P,$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and since $B$ is a diagonal matrix it only takes $2$ multiplications to multiply the matrices. Somewhat similar things can be done with jordan normal form matrices. A general method consists of using binary exponentiation to reduce the number of matrix multiplications that need to be done from N to $\log(N)$. Technically speaking matrix multiplication can be done &quot;faster&quot; than Strassen as well, but this will only be the case for very large matrices, due to the large constant coefficient hidden in the Coppersmith–Winograd algorithm.&lt;/p&gt;&#xA;" OwnerUserId="1930" LastActivityDate="2011-02-07T20:39:28.860" />
  <row Id="20878" PostTypeId="2" ParentId="20848" CreationDate="2011-02-07T20:46:14.287" Score="2" ViewCount="" Body="&lt;p&gt;Consider the set $B$ as you have defined it. I claim that $B$ is an open subset of $\mathbb{R}$. If $x\in B$ and $x$ is irrational, then $(x,0)\in A$ and so $A$ contains a ball around $(x,0)$, and so $A$ contains points from $E_y$ for all $y$ sufficiently close to $x$. So $x$ is in the interior of $B$, as desired. If $x$ is rational, in contrast, then $A$ contains points from $E_y$ for all rational $y$ sufficiently close to $x$, and so $E_y\subset A$ for all rational $y$ sufficiently close to $x$. In this case, since the complement of $A$ is open, it means that the closure of such $E_y$ must be contained in $A$, but the closure of such $E_y$ includes all points of the form $(z,0)$ for irrational $z$ sufficiently close to $x$.  Thus again, $x$ is in the interior of $B$, as desired.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It now follows by considering the complement of $A$ that $B$ is also clopen, which would be a contradiction unless $A$ and hence $B$ was either empty or the whole space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I see now that Joriki has answered while I was writing this...)&lt;/p&gt;&#xA;" OwnerUserId="413" LastActivityDate="2011-02-07T20:46:14.287" CommentCount="1" />
  <row Id="20879" PostTypeId="2" ParentId="20867" CreationDate="2011-02-07T20:47:27.177" Score="1" ViewCount="" Body="&lt;p&gt;Try calculating the $L_2$ norm. It's going to be much bigger for distribution #2. If the histogram is $h_i$, normalized so that the $L_1$ norm is $1$, i.e. $\sum_i h_i = 1$, then the $L_2$ norm (squared) is $$L_2^2 = \sum_i h_i^2.$$ One problem you're going to encounter is that the noise is proportional to $\sqrt{n}$, so you should normalize (divide) the $L_2^2$ norm by $n$ (I think). To make sure you have the correct normalization, sample your input and compare the normalized values; they should be about the same if you take all points or, say, only half of them (try even taking only, say, a tenth).&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2011-02-07T20:47:27.177" CommentCount="2" />
  <row Id="20880" PostTypeId="2" ParentId="20862" CreationDate="2011-02-07T20:50:04.273" Score="2" ViewCount="" Body="&lt;p&gt;The OP was after the &lt;a href=&quot;http://en.wikipedia.org/wiki/Rational_root_theorem&quot; rel=&quot;nofollow&quot;&gt;Rational root theorem&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Added so the question could be officially answered.)&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2011-02-07T20:50:04.273" />
  <row Id="20881" PostTypeId="1" CreationDate="2011-02-07T20:50:20.280" Score="2" ViewCount="81" Body="&lt;p&gt;Helly's Theorem states the following: Suppose that $X_1,X_2,...,X_n$ are convex sets in $\mathbb{R}^d$, such that for any $|I|\leq d+1$, $\cap_{i\in I}X_i \neq \emptyset$.  Then $\cap_{i=1}^{n}X_i \neq \emptyset$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;I'm looking for interesting problems, the solutions of which use this theorem.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is one example:&#xA;Let $K_1,K_2,...,K_n$ be closed intervals parallel to the $y$ axis. Assume that for any $|I|\leq d+2$ there exists a polynomial of degree at most $d$, the graph of which intersects all $K_i$ where $i\in I$. Show that there exists a polynomial of degree at most $d$, the graph of which intersects all the intervals $K_1,K_2,...,K_n$.&lt;/p&gt;&#xA;" OwnerUserId="766" LastActivityDate="2011-02-07T20:50:20.280" Title="Applications of Helly's theorem to problem solving" Tags="&lt;combinatorics&gt;&lt;big-list&gt;&lt;convex-sets&gt;" CommentCount="2" FavoriteCount="1" />
  <row Id="20882" PostTypeId="2" ParentId="20866" CreationDate="2011-02-07T21:38:01.447" Score="2" ViewCount="" Body="&lt;p&gt;$log_2 \sqrt n = log_2 n^{\frac{1}{2}} = \frac{1}{2} log_2 x$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;thanks to Yuval Filmus and Ross Millikan for comments&lt;/p&gt;&#xA;" OwnerUserId="5046" LastActivityDate="2011-02-07T21:38:01.447" CommentCount="0" />
  <row Id="20883" PostTypeId="2" ParentId="20697" CreationDate="2011-02-07T21:41:36.360" Score="4" ViewCount="" Body="&lt;p&gt;Here is a solution for the first problem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We already know that for every cardinal $\kappa$ we have $\kappa=\kappa\cdot\kappa$. So separate every $A_\alpha$ into $\kappa$ partitions of size $\kappa$ and let's call them $A_{\alpha,\beta}$. Now let $\{B_\alpha : \alpha&amp;lt;\kappa\}$ be an enumeration of all the $A_{\alpha,\beta}$. Observe now that we need to find sets $X,Y$ such that for every $\alpha&amp;lt;\kappa$ we have $B_\alpha\cap X\neq\varnothing\neq B_\alpha\cap Y$ and $X\cap Y=\varnothing$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each of these sets will satisfy the property we want since in every $A_{\alpha,\beta}$ there will exist an element in $X$ and one in $Y$ and therefore $\kappa$ elements of $A_\alpha$ in $X$ and $\kappa$ in $Y$ and furthermore $X\cap Y=\varnothing$ will give $Y\subset\kappa\setminus X$ and $X\subset\kappa\setminus Y$. I will construct two such sets $\mathcal{C}$, $\mathcal{D}$ inductively:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $\gamma_0$ be the least element of $B_0$ and make $\mathcal{C}_0=\{\gamma_0\}$. Now let $\delta_0$ be the least element of $B_0\setminus\mathcal{C}_0$ and let $\mathcal{D}_0=\{\delta_0\}$. Assume now that for every $\xi&amp;lt;\alpha$ we have defined $\mathcal{C}_\xi$ and $\mathcal{D}_\xi$ such that each of them have at most $|\xi+1|$ elements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $(\bigcup_{\xi&amp;lt;\alpha}\mathcal{C}_\xi)\cap B_\alpha$ is non-empty let $\mathcal{C}_\alpha=\bigcup_{\xi&amp;lt;\alpha}\mathcal{C}_\xi$  Otherwise let $\gamma_\alpha$ be the least element of $B_\alpha\setminus(\bigcup_{\xi&amp;lt;\alpha}\mathcal{D}_\xi)$ (this element has to exist since the union has at most $|\alpha|$ elements while $B_\alpha$ has $\kappa$) and let $\mathcal{C}_\alpha=(\bigcup_{\xi&amp;lt;\alpha}\mathcal{C}_\xi)\cup\{\gamma_\alpha\}$ (observe that this has at most $|\alpha+1|$ elements). Now if $(\bigcup_{\xi&amp;lt;\alpha}\mathcal{D}_\xi)\cap B_\alpha$ is non-empty let $\mathcal{D}_\alpha=\bigcup_{\xi&amp;lt;\alpha}\mathcal{D}_\xi$. Otherwise if $\delta_\alpha$ is the least element of $B_\alpha\setminus\mathcal{C}_\alpha$ (again this exists) let $\mathcal{D}_\alpha=(\bigcup_{\xi&amp;lt;\alpha}\mathcal{D}_\xi)\cup\{\delta_\alpha\}$. Finally let $\mathcal{C}=\bigcup_{\alpha&amp;lt;\kappa}\mathcal{C}_\alpha$ and $\mathcal{D}=\bigcup_{\alpha&amp;lt;\kappa}\mathcal{D}_\alpha$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, notice that $\mathcal{C}\cap\mathcal{D}=\varnothing$ because of the above construction and for every $\alpha&amp;lt;\kappa$ we have $B_\alpha\cap\mathcal{C}\neq\varnothing$ and $B_\alpha\cap\mathcal{D}\neq\varnothing$.&lt;/p&gt;&#xA;" OwnerUserId="1772" LastEditorUserId="1772" LastEditorDisplayName="" LastEditDate="2011-03-13T20:56:56.387" LastActivityDate="2011-03-13T20:56:56.387" CommentCount="1" />
  <row Id="20884" PostTypeId="1" AcceptedAnswerId="20885" CreationDate="2011-02-07T21:51:13.220" Score="6" ViewCount="139" Body="&lt;p&gt;Using the standard notation $(x,y)$ for cartesian coordinates, and $(r, \theta)$ for polar coordinates, it is true that $$ x = r \cos \theta$$ and so we can infer that &#xA;$$ \frac{\partial x}{\partial r} = \cos \theta.$$ This means that if we perturb $r$ to $r+\Delta r$, $x$ gets perturbed to approximately $x+ \cos(\theta) \Delta r$. This is equivalent to saying that if we perturb $x$ to $x+\Delta x$, then we perturb $r$ to approximately $r+ (1/\cos \theta) \Delta x$. Correspondingly, we expect that $$ \frac{\partial r}{\partial x} = \frac{1}{\cos \theta}.$$ In fact, &#xA;$$ \frac{ \partial r}{\partial x} = \frac{\partial}{\partial x} \sqrt{x^2+y^2}= \frac{2x}{2 \sqrt{x^2+y^2}} = \frac{x}{r} = \cos \theta.$$ What gives?&lt;/p&gt;&#xA;" OwnerUserId="6760" LastActivityDate="2011-02-07T22:26:45.747" Title="polar coordinates and derivatives" Tags="&lt;calculus&gt;&lt;polar-coordinates&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" />
  <row Id="20885" PostTypeId="2" ParentId="20884" CreationDate="2011-02-07T22:02:24.490" Score="7" ViewCount="" Body="&lt;p&gt;When you take partial derivatives, what you are doing is to hold the other variable fixed. So for instance if $f$ is a function of $n$ variables, $f(x_1,x_2,\ldots,x_n)$, $\frac{\partial f}{\partial x_i}$ means you hold all the other $x_j$'s constant except $x_i$ and vary $x_i$ by $\delta x_i$ and find out what happens to $\delta f$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x= r \cos(\theta)$,$y= r \sin(\theta)$, $r^2= x^2 + y^2$ and $\tan(\theta) = \frac{y}{x}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We are transforming from the $(x,y)$ space to $(r,\theta)$ space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the $(x,y)$ space, $x$ and $y$ are independent variables and In the $(r,\theta)$ space, $r$ and $\theta$ are independent variables.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{\partial x}{\partial r}$ means you are fixing $\theta$ and finding out how changing $r$ affects $x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So $\frac{\partial x}{\partial r} = \cos(\theta)$ since $\theta$ is fixed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{\partial r}{\partial x}$ means you are fixing $y$ and finding out how changing $x$ affects $r$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So $\frac{\partial r}{\partial x} = \frac{x}{\sqrt{x^2+y^2}}$ since $y$ is fixed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you do $r = \frac{x}{\cos(\theta)}$ and argue that $\frac{\partial r}{\partial x} = \frac{1}{\cos (\theta)}$ you are &lt;strong&gt;not&lt;/strong&gt; holding $y$ constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you were to hold $y$ constant, then $\theta$ would change as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am adding this in the hope that this will make it a bit more clear. If you want to use $r = \frac{x}{\cos(\theta)}$ and still derive it, you need to do as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$r = \frac{x}{\cos(\theta)}$, $\delta r = \frac{\delta x}{\cos(\theta)} + \frac{-x}{\cos^2(\theta)} (-\sin(\theta)) \delta \theta$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\tan(\theta) = \frac{y}{x} \Rightarrow x \tan(\theta) = y$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\delta x \tan(\theta) + x \sec^2(\theta) \delta \theta = 0$ (Since $y$ is held constant)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x \delta \theta = - \frac{\delta x \tan(\theta)}{\sec^2{\theta}} = - \sin(\theta) \cos(\theta) \delta x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Plugging the above in the previous expression, we get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\delta r = \frac{\delta x}{\cos(\theta)} + \frac{\sin(\theta)}{\cos^2(\theta)} (-\sin(\theta) \cos(\theta) \delta x) = \frac{1-\sin^2(\theta)}{\cos(\theta)} \delta x = \cos(\theta) \delta x$ and hence we get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{\partial r}{\partial x} = \cos(\theta) = \frac{x}{\sqrt{x^2+y^2}}$&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2011-02-07T22:26:45.747" LastActivityDate="2011-02-07T22:26:45.747" CommentCount="3" />
  <row Id="20886" PostTypeId="1" CreationDate="2011-02-07T22:02:55.390" Score="1" ViewCount="143" Body="&lt;p&gt;Find a 2-connected planar graph whose drawings are all topologically isomorphic, but whose planar embeddings are not all equivalent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think $K_{2,3}$ might be an example, but I'm not sure how to show this at all. Anything would be welcome.&lt;/p&gt;&#xA;" OwnerUserId="6584" LastEditorUserId="279" LastEditorDisplayName="" LastEditDate="2011-04-10T02:38:23.523" LastActivityDate="2011-05-10T04:56:15.073" Title="Finding a 2-connected planar graph" Tags="&lt;graph-theory&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="20887" PostTypeId="1" AcceptedAnswerId="20890" CreationDate="2011-02-07T22:23:32.830" Score="1" ViewCount="41" Body="&lt;p&gt;Hello,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $0 \leq r \leq n$, is it possible to use this equation:&#xA;$$C( n, r ) = C( n - 1, r ) + C( n-1, r - 1 ).$$&#xA;The problem is I want to prove that $C( n, r )$ is integers by induction. In other words, I have to prove $C( n + 1, r )$ is integer deducted from $C( n, r )$.  I thought of the equation above but I found the condition for r is slightly different.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So is there any other equality that I can use for this problem? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks,&lt;br&gt;&#xA;Chan&lt;/p&gt;&#xA;" OwnerUserId="5711" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-07T22:26:02.547" LastActivityDate="2011-02-07T22:30:05.617" Title="Question regarding combination" Tags="&lt;elementary-number-theory&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="20888" PostTypeId="1" CreationDate="2011-02-07T22:23:51.487" Score="3" ViewCount="73" Body="&lt;p&gt;Struggling with some equations here. I've got a particle I need to track in an absolute reference frame, but each step I move/rotate it relative to its own reference frame. I need to track it's absolute position while moving it relatively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm trying to figure out the proper set of rotation matrices. It propagates along the x-axis and I can describe its current direction with [ux, uy, uz], now I need to rotate it about it's own reference frame by the deflection $\theta$ (deflection from incident direction) and azimuthal rotation $\phi$ (rotation about incident direction), and then find its new position, [ux', uy', uz']. Should I multiply by the rotation matrix, ROT[y,$\theta$], then ROT[x,$\phi$] or the opposite order?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm just getting myself confused. By way of reference, I'm trying to derive the set of equations on (pdf) page 20 of &lt;a href=&quot;http://www.dtic.mil/cgi-bin/GetTRDoc?Location=U2&amp;amp;doc=GetTRDoc.pdf&amp;amp;AD=ADA426624&quot; rel=&quot;nofollow&quot;&gt;this document&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's a) not derived very well and b) derived for propagation along the z-axis, not the x-axis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;TL;DR&lt;/strong&gt; - I have a vector in space pointing in an arbitrary direction, described by its projections on the world x,y,z-axis. I need to rotate the vector, in relation to its own reference frame, and find the world axis projections. Essentially this is a ray-tracing problem. Incoming ray needs to point in a new direction - what is that direction?&lt;/p&gt;&#xA;" OwnerUserId="6761" LastEditorUserId="6761" LastEditorDisplayName="" LastEditDate="2011-02-08T14:13:42.517" LastActivityDate="2011-02-08T14:13:42.517" Title="rotation matrix for particle track" Tags="&lt;matrices&gt;" CommentCount="2" />
  <row Id="20889" PostTypeId="1" AcceptedAnswerId="20893" CreationDate="2011-02-07T22:27:51.097" Score="6" ViewCount="715" Body="&lt;p&gt;Hello,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know it might be too easy for you guys here. I'm practicing some problems in the textbook, but this one really drove me crazy.&lt;br&gt;&#xA;From $\gcd( a, b ) = 1$, I have $ax + by = 1$, where should I go from here? The extra $ac$ is so annoying. Any hint?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks,&lt;br&gt;&#xA;Chan&lt;/p&gt;&#xA;" OwnerUserId="5711" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-07T22:35:07.127" LastActivityDate="2011-02-08T18:47:12.617" Title="Prove that if gcd( a, b ) = 1 then gcd( ac, b ) = gcd( c, b )" Tags="&lt;elementary-number-theory&gt;" AnswerCount="5" CommentCount="2" />
  <row Id="20890" PostTypeId="2" ParentId="20887" CreationDate="2011-02-07T22:30:05.617" Score="1" ViewCount="" Body="&lt;p&gt;Yes, you can use the equation, if interpreted correctly: since $C(m,k)$ is the number of ways of selecting a $k$-element subset from a set with $m$ elements, then we get $C(m,k) = 0$ if $k\gt m$ or if $k\lt 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With the corrected equation, it should follow easily by induction on $n$, with the induction hypothesis being that $C(k,r)$ is an integer for every $r$ with $0\leq r\leq k$ (you'll have to deal with $r=0$ separately when trying to use the formula). Alternatively, if you feel uneasy about assuming the result for &lt;strong&gt;all&lt;/strong&gt; $r$ corresponding to $k$, you can do induction on $n+r$ (which will also require you to deal with $r=0$ separately).&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2011-02-07T22:30:05.617" CommentCount="2" />
  <row Id="20891" PostTypeId="2" ParentId="20867" CreationDate="2011-02-07T22:31:04.633" Score="1" ViewCount="" Body="&lt;p&gt;You can even extend Yuval's proposal and compute:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$R(\alpha) = \frac{1}{1-\alpha}\ln \sum_i h_i^{\alpha}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where the case $\alpha = 1$ is to be understood as a limit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These are called Rényi indices or entropies and are a standard tool of multifractal analysis which is the study of &quot;jagged&quot; geometrical (or other) objects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's hard to find good references, I'd need to spend more time searching, and anyway that's why I started my own &lt;a href=&quot;http://math.stackexchange.com/questions/16618/need-good-material-on-multifractal-analysis&quot;&gt;question here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found &lt;a href=&quot;http://chaosbook.dk/chapters/thermodyn.pdf&quot; rel=&quot;nofollow&quot;&gt;this paper&lt;/a&gt; which explains Rényi entropies and their connection to fractal dimensions. There's some talk about dynamical systems as well, but I think you can skip over most of it.&lt;/p&gt;&#xA;" OwnerUserId="3567" LastEditorUserId="3567" LastEditorDisplayName="" LastEditDate="2011-02-09T10:53:37.833" LastActivityDate="2011-02-09T10:53:37.833" />
  <row Id="20892" PostTypeId="2" ParentId="20889" CreationDate="2011-02-07T22:38:05.627" Score="5" ViewCount="" Body="&lt;p&gt;My attempt at a Bill Dubuquesque argument:&#xA;\begin{align*}&#xA;r|ac,\;b &amp;amp;\Longleftrightarrow r|ac,\;bc,\;b\\&#xA;&amp;amp;\Longleftrightarrow r|\gcd(ac,bc),\;b\\&#xA;&amp;amp;\Longleftrightarrow r|\gcd(a,b)c,\;b &amp;amp;\quad&amp;amp;\mbox{(since $\gcd(ac,bc)=\gcd(a,b)c$)}\\&#xA;&amp;amp;\Longleftrightarrow r|c,\;b&#xA;\end{align*}&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-07T22:45:27.430" LastActivityDate="2011-02-07T22:45:27.430" CommentCount="4" />
  <row Id="20893" PostTypeId="2" ParentId="20889" CreationDate="2011-02-07T22:40:58.483" Score="5" ViewCount="" Body="&lt;p&gt;Let $d_1 = \gcd(c,b)$ and $d_2 = \gcd(ac,b)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we have $cx_1 + by_1 = d_1$ and $acx_2 + by_2 = d_2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Further we have $ax + by = 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now multiply $ax+by = 1$ by $d_1$, write $d_1 = cx_1 + by_1$ and rearrange to show that $d_2|d_1$ i.e.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$d_1 \times (ax+by = 1) \Rightarrow ax(cx_1 + by_1) + bd_1y = d_1 \Rightarrow a c (x x_1) + b(a x y_1 + d_1 y) = d_1$. $d_2 = \gcd(ac,b)$ divides any integer linear combination of $ac$ and $b$ and hence $d_2 | d_1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And by a similar argument multiply $ax+by = 1$ by $d_2$, write $d_2 = acx_2 + by_2$ and rearrange to show that $d_1|d_2$ i.e.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$d_2 \times (ax+by = 1) \Rightarrow ax(acx_2 + by_2) + bd_2y = d_2 \Rightarrow c (a^2 x x_2) + b(a x y_2 + d_2 y) = d_2$. $d_1 = \gcd(c,b)$ divides any integer linear combination of $c$ and $b$ and hence $d_1 | d_2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we have $d_1 | d_2$ and $d_2 | d_1$ and $d_1$,$d_2$ are non-negative since they are the gcd of two integers and hence $d_1 = d_2$&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2011-02-07T22:52:03.263" LastActivityDate="2011-02-07T22:52:03.263" CommentCount="7" />
  <row Id="20894" PostTypeId="1" AcceptedAnswerId="20895" CreationDate="2011-02-07T22:43:27.383" Score="1" ViewCount="90" Body="&lt;p&gt;I'd like to say logarithmically decreasing but it does not have to decrease to zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An example of an increasing function which flattens out at around 4.5%.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/ulKwz.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="825" LastEditorUserId="825" LastEditorDisplayName="" LastEditDate="2011-02-07T23:14:11.477" LastActivityDate="2011-02-07T23:14:11.477" Title="What is the correct term for an increasing or decreasing function which &quot;flattens out&quot;?" Tags="&lt;terminology&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="20895" PostTypeId="2" ParentId="20894" CreationDate="2011-02-07T22:45:51.703" Score="4" ViewCount="" Body="&lt;p&gt;I don't know what &quot;flattens out&quot; means (and your image doesn't load), but maybe you're thinking of functions of a variable $x$ which tend monotonically to a limit as $x$ tends to infinity. Having a &quot;horizontal asymptote&quot; is another common related term. &lt;/p&gt;&#xA;" OwnerUserId="308" LastActivityDate="2011-02-07T22:45:51.703" CommentCount="2" />
  <row Id="20896" PostTypeId="2" ParentId="20889" CreationDate="2011-02-07T22:46:15.843" Score="3" ViewCount="" Body="&lt;p&gt;Not sure if I can prove it algebraically but can try to explain logically.  There is no common denominator between a and b (this is what gcd(a,b) = 1 means).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you multiply a * c you introduce c's prime factors to the new number &quot;ac&quot;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, if any new common factors (between ac and b) are introduced, we know that they came solely from c.  We know this because prior to multiplying by c there were no common denominators.  So, the only commonality that could originate here is from c and b.  &lt;/p&gt;&#xA;" OwnerUserId="6764" LastActivityDate="2011-02-07T22:46:15.843" />
  <row Id="20898" PostTypeId="1" CreationDate="2011-02-07T23:01:25.783" Score="3" ViewCount="114" Body="&lt;p&gt;I was wondering if someone could recommend good books on stochastic processes &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;with measure theory treatment&lt;/li&gt;&#xA;&lt;li&gt;with not much or no measure theory&#xA;treatment&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;for each, it would be nice to have some books for introductory level, mid-level to comprehensive level respectively. They can be either classical or recently published, either continuous, discrete or both.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance!&lt;/p&gt;&#xA;" OwnerUserId="1281" LastEditorUserId="1281" LastEditorDisplayName="" LastEditDate="2011-02-07T23:15:26.987" LastActivityDate="2011-02-07T23:15:26.987" CommunityOwnedDate="2011-02-07T23:14:46.873" Title="Recommendation on stochastic process books" Tags="&lt;reference-request&gt;&lt;stochastic-processes&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="20900" PostTypeId="2" ParentId="20898" CreationDate="2011-02-07T23:08:49.250" Score="2" ViewCount="" Body="&lt;p&gt;Some good books include:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;em&gt;Stochastic Processes&lt;/em&gt; by Sheldon Ross (no measure theory)&lt;/li&gt;&#xA;&lt;li&gt;&lt;em&gt;Introduction to Probability Models&lt;/em&gt; by Sheldon Ross (no measure theory)&lt;/li&gt;&#xA;&lt;li&gt;&lt;em&gt;Stochastic Processes&lt;/em&gt; by J.L.Doob (measure theory)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2011-02-07T23:08:49.250" CommunityOwnedDate="2011-02-07T23:14:46.873" />
  <row Id="20901" PostTypeId="2" ParentId="20617" CreationDate="2011-02-07T23:22:29.233" Score="2" ViewCount="" Body="&lt;p&gt;Chapter 3 of the book &lt;a href=&quot;http://books.google.com/books?id=y2c00LAMnFUC&quot; rel=&quot;nofollow&quot;&gt;The Gibbs phenomenon in Fourier analysis, splines, and wavelet approximations&lt;/a&gt;&#xA;by Abdul J. Jerri seems to cover the case of a general orthogonal expansion along eigenfunctions of a Sturm-Liouville problem.&lt;/p&gt;&#xA;" OwnerUserId="6179" LastActivityDate="2011-02-07T23:22:29.233" />
  <row Id="20902" PostTypeId="2" ParentId="20865" CreationDate="2011-02-07T23:27:48.413" Score="0" ViewCount="" Body="&lt;p&gt;By &lt;a href=&quot;http://en.wikipedia.org/wiki/Gershgorin_circle_theorem&quot; rel=&quot;nofollow&quot;&gt;Gershgorin's Theorem&lt;/a&gt; we can conclude that the absolute sum of all non-diagonal entries in each row must be less than 1.  Then the matrix will have positive eigenvalues for diagonal entries &gt; 1.&lt;/p&gt;&#xA;" OwnerUserId="4726" LastActivityDate="2011-02-07T23:27:48.413" />
  <row Id="20903" PostTypeId="2" ParentId="20889" CreationDate="2011-02-07T23:43:32.200" Score="1" ViewCount="" Body="&lt;p&gt;We could prove the contrapositive: $\gcd(ac,b) \neq \gcd(c,b) \Longrightarrow \gcd(a,b) \neq 1$. Let $d = \gcd(ac,b)$ and $d&amp;#39; = \gcd(c,b)$. Now $d&amp;#39; &amp;lt; d$. So $d$ does not divide $c$ and $b$ and $d$ divides $ac$ and $b$. So $\gcd(a,b) \neq 1$.&lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2011-02-07T23:43:32.200" CommentCount="1" />
  <row Id="20904" PostTypeId="2" ParentId="20889" CreationDate="2011-02-07T23:43:52.857" Score="3" ViewCount="" Body="&lt;p&gt;This follows easily from basic laws of GCD arithmetic. First I will present the&#xA;proof using the common notation $\rm\ (a,b) = gcd(a,b)\:,\ $ then I will show how to present the proof more intuitively by employing a more suggestive arithmetical notation. First let's consider the less-intuitive proof. The proof employs only basic GCD laws, namely the associative and commutative laws combined with the key $\:$ &lt;a href=&quot;http://math.stackexchange.com/questions/19830/if-gcd-a-b-1-then-is-it-true-to-say-gcd-ac-bc-c/19842#19842&quot;&gt;distributive law&lt;/a&gt; $\rm\: \ (ac,bc)\ =\ (a,b)\:c\ $ and a peculiar law $\rm\: (c,1) = 1\:,\: $ i.e. $\rm\: gcd(c,1)= 1\:.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\ \ \ (ac,b)\ =\ (ac,b\:(c,1))\ =\ (ac,bc,b)\ =\ ((a,b)\:c,b)\ \ $ [$\rm\: =\ (c,b)\ $ if  $\rm\ (a,b) = 1\:]\quad\quad$ &lt;strong&gt;QED&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because the arithmetic of GCDs shares many of the same basic laws as the arithmetic of integers, we can make the above proof more intuitive by choosing a notation that highlights this common arithmetical structure. Reconsider the above proof using the notation $\rm\:A+B\ $ for the gcd $\rm\:(A,B)\ $ and employing the law $\rm\  C + 1 = 1\:,\: $ i.e. $\rm\ gcd(C,1) = 1\:,\ $ the only law which is peculiar to GCDs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\ \:  AC + B\ =\ AC + B\:(C + 1)\ =\ AC+BC+B\ =\ (A+B)\:C + B\ \ $ [$\rm =\ C + B\ \: $ if $\rm\: \ A+B = 1\:]$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice how this suggestive notation has allowed us to exploit our well-honed arithmetical intuition regarding the associative, commutative and distributive laws. $\ $ For a less trivial example see my analogous proof of the &lt;a href=&quot;http://math.stackexchange.com/questions/10400/comaximal-ideals-in-a-commutative-ring/10416#10416&quot;&gt;Freshman's Dream&lt;/a&gt; $\rm\ (A,B)^n\ =\ (A^n,B^n)\ $ for GCDs and ideals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The motivation behind this powerful abstract axiomatic approach will become much clearer should you go on to study ideal theory and/or divisor theory (and you should, for there lies much beauty).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For further discussion and generalizations see my &lt;a href=&quot;http://math.stackexchange.com/search?q=user:242+gcd&quot;&gt;menagerie of GCDs.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2011-02-08T18:47:12.617" LastActivityDate="2011-02-08T18:47:12.617" CommentCount="1" />
  <row Id="20905" PostTypeId="2" ParentId="20863" CreationDate="2011-02-08T00:06:12.260" Score="2" ViewCount="" Body="&lt;p&gt;You have&#xA;$$\underline 0=i\underline k\times(-i\omega\underline \xi - 2i(\underline k \cdot \underline \Omega )\underline u))=-i^2\omega(\underline k\times \underline \xi)-2i^2(\underline k \cdot \underline \Omega)(\underline k\times\underline u)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But by the &lt;a href=&quot;http://en.wikipedia.org/wiki/Cross_product#Triple_product_expansion&quot; rel=&quot;nofollow&quot;&gt;triple cross product expansion formula&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\underline k\times \underline \xi=\underline k\times (i\underline k\times \underline u)=i|k|^2\underline u$$ since $\underline u\cdot\underline k=0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus the first equation can be restated as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$-i^3\omega |k|^2\underline u=2i^2(\underline k \cdot \underline \Omega)(\underline k\times\underline u)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Taking norms, you will get what you want.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remember that $|\underline k\times\underline u|=|\underline k||\underline u|$ because $\underline u\cdot\underline k=0$ (see &lt;a href=&quot;http://en.wikipedia.org/wiki/Cross_product#Alternative_formulation&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt;).&lt;/p&gt;&#xA;" OwnerUserId="524" LastEditorUserId="524" LastEditorDisplayName="" LastEditDate="2011-02-08T00:12:33.970" LastActivityDate="2011-02-08T00:12:33.970" />
  <row Id="20906" PostTypeId="1" AcceptedAnswerId="20944" CreationDate="2011-02-08T00:09:13.040" Score="2" ViewCount="143" Body="&lt;p&gt;Hello,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know how to solve &lt;code&gt;ax + by = c&lt;/code&gt; using Extended Algorithm. But with more than variables, I'm lost :(. To verify if it has an integer solution is easy, since we only need to check for gcd(a,b,c)|d. Other than that, how can we find an integer solution for this equation?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks,&lt;br&gt;&#xA;Chan &lt;/p&gt;&#xA;" OwnerUserId="5711" LastEditorUserId="1778" LastEditorDisplayName="" LastEditDate="2011-02-24T15:48:20.670" LastActivityDate="2011-02-24T15:48:20.670" Title="How to find an integer solution for general Diophantine equation ax + by + cz + dt... = N" Tags="&lt;elementary-number-theory&gt;&lt;diophantine-equations&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="20907" PostTypeId="1" AcceptedAnswerId="21066" CreationDate="2011-02-08T00:17:49.330" Score="6" ViewCount="124" Body="&lt;p&gt;That   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) Ext$^n_\mathbb{Z}(M, \mathbb{Q})=0$ for every module $M$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;follows easily from the fact that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) $\mathbb{Q}$ is injective.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the only proof I have seen of the injectivity of $\mathbb{Q}$ relies on Baer's Criterion.  While the proof of Baer's Criterion is not difficult, it seems stronger than the injectivity of $\mathbb{Q}$ (for example, the proof uses Zorn's Lemma).  Is there a different (and preferably simpler) proof of (1) or (2)?&lt;/p&gt;&#xA;" OwnerUserId="5265" LastActivityDate="2011-02-08T22:33:42.067" Title="Proof that Ext$^n_\mathbb{Z}(M, \mathbb{Q})=0$ and Baer's Criterion" Tags="&lt;algebra&gt;&lt;homological-algebra&gt;" AnswerCount="4" CommentCount="1" />
  <row Id="20908" PostTypeId="1" AcceptedAnswerId="20936" CreationDate="2011-02-08T00:18:36.950" Score="10" ViewCount="151" Body="&lt;p&gt;If one computes the average width of a triangle, then one gets $(s_1+s_2+s_3)/\pi$, where $s_1$, $s_2$, $s_3$ are the side lengths.  I did this by brute force, using an integral which went through an interval of length $\pi$.  For each angle $\theta$, the &quot;width&quot; $w_\theta$ of the triangle is the smallest distance one may place two parallel lines at that angle without penetrating the triangle.  (Imagine using calipers for each angle.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My questions:  Is there an easier way (without using an integral) to see that the ratio of the perimeter $s_1+s_2+s_3$ to this average width is $\pi$?  Does this generalize to any convex polygon (or polytope)?&lt;/p&gt;&#xA;" OwnerUserId="6766" LastActivityDate="2011-02-08T03:17:28.210" Title="What's the average width of a convex polygon?" Tags="&lt;euclidean-geometry&gt;" AnswerCount="1" FavoriteCount="2" />
  <row Id="20909" PostTypeId="1" CreationDate="2011-02-08T00:19:00.237" Score="3" ViewCount="65" Body="&lt;p&gt;Let $X=C(S)$ where $S$ is compact. Suppose $T\subset S$ is a closed&#xA;subset such that for every $g\in C(T),$ there is an $f\in C(S)$&#xA;such that: $f\mid_{T}=g$. Show that there exists a constant $C&amp;gt;0$&#xA;such that every $g\in C(T)$ can be continuously extended to $f\in C(S)$&#xA;such that: $\sup_{x\in S}\left|f(x)\right|\leq C\sup_{y\in T}\left|g(y)\right|$&lt;/p&gt;&#xA;" OwnerUserId="6767" LastActivityDate="2011-02-08T01:49:08.027" Title="Uniform Boundedness/Hahn-Banach Question" Tags="&lt;functional-analysis&gt;" AnswerCount="2" />
  <row Id="20911" PostTypeId="1" AcceptedAnswerId="20954" CreationDate="2011-02-08T00:20:43.407" Score="2" ViewCount="158" Body="&lt;p&gt;I need to sketch a diagram which shows what happens when a 2x2 matrix is applied to a lattice. What would be the best software to perform such a thing?&lt;/p&gt;&#xA;" OwnerUserId="4728" LastActivityDate="2011-02-11T09:10:26.003" Title="Sketching vector diagram" Tags="&lt;math-software&gt;&lt;transformation&gt;" AnswerCount="2" />
  <row Id="20912" PostTypeId="1" AcceptedAnswerId="20917" CreationDate="2011-02-08T00:25:52.367" Score="2" ViewCount="130" Body="&lt;p&gt;I don't know where to begin to calculate the expectation value of the random variable $1/V$, where $V$ is a random variable with chi-square distribution $\chi^2(\nu)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could somebody help me?&lt;/p&gt;&#xA;" OwnerUserId="4786" LastActivityDate="2011-02-08T00:55:05.557" Title="Calculation of inverse of chi-square's expectation." Tags="&lt;probability-distributions&gt;" AnswerCount="1" CommentCount="0" />
  <row Id="20913" PostTypeId="1" AcceptedAnswerId="20923" CreationDate="2011-02-08T00:29:25.707" Score="10" ViewCount="280" Body="&lt;p&gt;I hope this doesn't turn out to be a silly question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are lots of nice examples of continuous bijections $X\to Y$ between topological spaces that are not homeomorphisms.  But in the examples I know, either $X$ and $Y$ are not homeomorphic to one another, or they are (homeomorphic) disconnected spaces.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My Question: Is there a connected topological space $X$ and a continuous bijection $X\to X$ that is not a homeomorphism?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the record, my example of a continuous bijection $X\to X$ that is not a homeomorphism is the following.  Roughly, the idea is to find an ordered family of topologies $\tau_i$ (&#xA;$i\in \mathbb Z$) on a set $S$ and use the shift map to create a continuous bijection from $\coprod_{i\in \mathbb Z} (S, \tau_i)$ to itself.  Let $S = \mathbb{Z} \coprod \mathbb Z$.  The topology $\tau_i$ is as follows: if $i&amp;lt;0$, then the left-hand copy of $\mathbb Z$ is topologized as the disjoint union of the discrete topology on $[-n, n]$ and the indiscrete topology on its complement, while the right-hand copy of $\mathbb Z$ is indiscrete.  The space $(S, \tau_0)$ is then indiscrete.  For $i&amp;gt;0$, the left-hand copy of $\mathbb Z$ is indiscrete, while the right-hand copy is the disjoint union of the indiscrete topology on $[-n, n]$ with the discrete topology on its complement.  Now the map $\coprod_{i\in \mathbb Z} (S, \tau_i)\to \coprod_{i\in \mathbb Z} (S, \tau_i)$ sending $(S, \tau_i) \to (S, \tau_{i+1})$ by the identity map of $S$ is a continuous bijection, but not a homeomorphism.&lt;/p&gt;&#xA;" OwnerUserId="1392" LastActivityDate="2011-03-31T10:14:05.377" Title="Are continuous self-bijections of connected spaces homeomorphisms?" Tags="&lt;general-topology&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="3" />
  <row Id="20914" PostTypeId="1" AcceptedAnswerId="20927" CreationDate="2011-02-08T00:31:21.370" Score="4" ViewCount="201" Body="&lt;p&gt;OK, so I know that the Lebesgue measure of the ternary Cantor set is $0$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, in class the prof briefly mentioned that if we build a Lebesgue-Stieltjes measure $\mu$ out of the Cantor function, then $\mu(C) = 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't understand why this is true, can someone please explain?&lt;/p&gt;&#xA;" OwnerUserId="6556" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2011-05-15T17:24:24.583" LastActivityDate="2011-05-15T22:30:12.673" Title="Confused about Cantor function and measure of Cantor set" Tags="&lt;measure-theory&gt;" AnswerCount="2" />
  <row Id="20915" PostTypeId="1" CreationDate="2011-02-08T00:34:57.537" Score="1" ViewCount="74" Body="&lt;p&gt;I am having some trouble understand essential prime implicants. So if a minterm is not covered by another overlapping rectangle, then that is an EPI. However, if we make a K-map for $f(x,y,z)=xy+xz&amp;#39;+y&amp;#39;z$, we have minterms m4, m6, and m7 not covered by overlapping rectangles. If what I did is correct, then there should be a total of only 2 rectangles in the table-one horizontal one that completely encapsulated the 2nd row, and a vertical rectangle for the 2nd column. And so it is a rule that EPI must appear in the minimal sum of products, in which case I'd have a lot of terms, but in fact I only see that it simplifies to $y&amp;#39;z+x$. What am I missing here?&lt;/p&gt;&#xA;" OwnerUserId="2985" LastActivityDate="2011-02-08T00:34:57.537" Title="Question on essential prime implicants" Tags="&lt;boolean-algebra&gt;" CommentCount="0" />
  <row Id="20917" PostTypeId="2" ParentId="20912" CreationDate="2011-02-08T00:42:57.873" Score="2" ViewCount="" Body="&lt;p&gt;The pdf of a chi-square distribution is $$\frac{1}{2^{\nu/2} \Gamma(\nu/2)} x^{\nu/2-1} e^{-x/2}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you want to calculate &#xA;$$\int_0^{\infty} \frac{1}{x} \frac{1}{2^{\nu/2} \Gamma(\nu/2)} x^{\nu/2-1} e^{-x/2} dx = \int_0^{\infty} \frac{1}{2^{\nu/2} \Gamma(\nu/2)} x^{\nu/2-2} e^{-x/2} dx.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rewrite the integrand so that it is the pdf of a $\chi^2(\nu-2)$ random variable, which will then integrate to 1.  The leftover constant factor will be the expected value you're looking for.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want a more detailed hint, just ask.  &lt;/p&gt;&#xA;" OwnerUserId="2370" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2011-02-08T00:55:05.557" LastActivityDate="2011-02-08T00:55:05.557" />
  <row Id="20919" PostTypeId="2" ParentId="20913" CreationDate="2011-02-08T01:02:10.883" Score="4" ViewCount="" Body="&lt;p&gt;Yes (to your body question, not your title question; it is confusing when people do this). Take $X = \mathbb{Z}$ with the topology generated by an open set containing $n$ for every positive integer $n$. (This space is connected because the smallest open set containing a non-positive integer is the entire space.) Consider the continuous bijection given by sending $x$ to $x - 1$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is what might be a Hausdorff example: take $X = \mathbb{R}$ with the topology generated by the usual topology together with the open set $(0, \infty) \cap \mathbb{Q}$, and again consider the continuous bijection $x \to x-1$. Unfortunately I am not sure if $X$ is connected.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most general situation I know where a continuous bijection $X \to Y$ is automatically a homeomorphism is if $X$ is compact and $Y$ is Hausdorff. This is a nice exercise and extremely useful.&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2011-02-08T01:33:53.867" LastActivityDate="2011-02-08T01:33:53.867" CommentCount="6" />
  <row Id="20920" PostTypeId="1" AcceptedAnswerId="20938" CreationDate="2011-02-08T01:06:32.150" Score="7" ViewCount="127" Body="&lt;p&gt;I was reading about the &lt;a href=&quot;http://en.wikipedia.org/wiki/Fano_plane&quot; rel=&quot;nofollow&quot;&gt;Fano plane&lt;/a&gt;, the smallest possible projective plane. After playing around with it, it seems that any projective plane of 7 points will be isomorphic to the Fano plane. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I've always been troubled with showing isomorphisms between sets of points and lines, because it seems like any function between two sets is really just an assignment of points to other points that preserves lines, but there is no nice fixed function that maps elements to others based on some hard and fast rule, like say $x\mapsto x^2$ or $(x,y)\mapsto x-y$, etc. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there an elegant way to show that any projective plane of 7 points is necessarily isomorphic to the Fano plane? I could only think of exhausting all permutations of points and lines and saying &quot;Look, if these points $A$, $B$, $C$ are on a line $l$ here, then $f(A), f(B), f(C)$ are on a line $f(l)$ here.&quot; But this seems like it's brute forcing the matter, and not very efficient at all, considering that there are many ways to connect the points. What is the best way to go about something like this? Thanks.&lt;/p&gt;&#xA;" OwnerUserId="6742" LastActivityDate="2011-02-08T17:27:26.180" Title="Is there a slick way to show that finite projective planes of $7$ points are unique up to isomorphism?" Tags="&lt;graph-theory&gt;&lt;projective-geometry&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="20921" PostTypeId="2" ParentId="20909" CreationDate="2011-02-08T01:09:37.810" Score="1" ViewCount="" Body="&lt;p&gt;$C(S) \to C(T)$ is a surjective bounded linear map of Banach spaces (with sup norms), so there is a closed linear subspace $M \subset C(S)$ such that $C(S)/M \to T$ is bijective and bounded with the quotient norm. Inverse mapping theorem says that the inverse is a bounded linear map. The statement then follows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the way, how is this related to Banach-Steinhaus/Hahn-Banach?&lt;/p&gt;&#xA;" OwnerUserId="325" LastActivityDate="2011-02-08T01:09:37.810" />
  <row Id="20922" PostTypeId="2" ParentId="20907" CreationDate="2011-02-08T01:15:52.220" Score="2" ViewCount="" Body="&lt;p&gt;The main use of Baer's criterion in this case is that injective modules over $\mathbb{Z}$ are the same as divisible groups. Since $Ext$ is closely related to injectivity, there isn't much reason to avoid a clean description of injective modules (via Baer's criterion).&lt;/p&gt;&#xA;" OwnerUserId="325" LastActivityDate="2011-02-08T01:15:52.220" CommentCount="4" />
  <row Id="20923" PostTypeId="2" ParentId="20913" CreationDate="2011-02-08T01:31:59.440" Score="12" ViewCount="" Body="&lt;p&gt;Here's a nice geometric example.  Let $X\subset\mathbb{R}^2$ be the union of the $x$-axis, the line segments $\{n\}\times[0,2\pi)$ for $n\in \{\ldots,-3,-2,-1,0\}$, and circles in the upper half plane of radius $1/3$ tangent to the $x$-axis at the points $(1,0),(2,0),\ldots$.  Note that $X$ is connected.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Define a map $f\colon X\to X$ by&#xA;$$&#xD;&#xA;f(x,y) \;=\; \begin{cases}(x+1,y) &amp;amp; \text{if }x\ne 0 \\ \left(1+\frac{\sin y}{3},\frac{1-\cos y}{3}\right) &amp;amp; \text{if }x=0\end{cases}.&#xD;&#xA;$$&#xA;That is, $f$ translates most points to the right by $1$, and maps the line segment $\{0\}\times[0,2\pi)$ onto the circle that's tangent to the $x$-axis at the point $(1,0)$.  Then $f$ is continuous and bijective, but is not a homeomorphism.&lt;/p&gt;&#xA;" OwnerUserId="1726" LastActivityDate="2011-02-08T01:31:59.440" CommentCount="2" />
  <row Id="20924" PostTypeId="2" ParentId="20909" CreationDate="2011-02-08T01:32:39.830" Score="0" ViewCount="" Body="&lt;p&gt;So $A:C(S)/M \rightarrow C(T)$, then $A^{-1}:C(T)\rightarrow C(S)/M$,&#xA;apply the inverse mapping theorem, we obtain $\left\Vert A^{-1}g\right\Vert \leq\left\Vert A^{-1}\right\Vert \left\Vert g\right\Vert $,&#xA;then $\inf_{f&amp;#39;\in M}\left\Vert f+f^{&amp;#39;}\right\Vert \leq C\left\Vert g\right\Vert $.&#xA;We must show that there exists a function in $f+M$ whose restriction&#xA;is $g$ and it's norm is less than C$\left\Vert g\right\Vert $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But How to show the existence of such function?&lt;/p&gt;&#xA;" OwnerUserId="6769" LastEditorUserId="6769" LastEditorDisplayName="" LastEditDate="2011-02-08T01:49:08.027" LastActivityDate="2011-02-08T01:49:08.027" CommentCount="1" />
  <row Id="20925" PostTypeId="1" AcceptedAnswerId="26120" CreationDate="2011-02-08T01:41:59.910" Score="1" ViewCount="151" Body="&lt;p&gt;If a Cat has 4 kittens, and in 6 months those kittens have 4 kittens, and so on for 7 years, such that 14 batches of kittens are produced, from the one kitten. I produce the summation:&#xA;$$\sum_{x=1}^{14} 4^x$$&#xA;Where added up looks like $4 + 16 + 64 + \cdots +4^{14}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there some kind of trick in maths for working this huge number out without a calculator?&lt;/p&gt;&#xA;" OwnerUserId="6770" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-10T03:39:46.217" LastActivityDate="2011-03-10T03:39:46.217" Title="Exponential Summation" Tags="&lt;sequences-and-series&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="20926" PostTypeId="2" ParentId="20925" CreationDate="2011-02-08T01:51:33.733" Score="2" ViewCount="" Body="&lt;p&gt;Look up &lt;a href=&quot;http://en.wikipedia.org/wiki/Geometric_progression&quot; rel=&quot;nofollow&quot;&gt;Geometric progression&lt;/a&gt;. The answer to your problem is $4 \times \frac{4^{14}-1}{3} = 357913940$&lt;/p&gt;&#xA;" OwnerUserId="3302" LastActivityDate="2011-02-08T01:51:33.733" />
  <row Id="20927" PostTypeId="2" ParentId="20914" CreationDate="2011-02-08T02:00:20.660" Score="6" ViewCount="" Body="&lt;p&gt;The Cantor function is constant on each of the open intervals that makes up the complement of the Cantor set. The distribution puts zero mass on the union of these open intervals. This gives $u([0,1]\setminus C)=0$, so that $u(C)=1$. &lt;/p&gt;&#xA;" OwnerUserId="940" LastActivityDate="2011-02-08T02:00:20.660" CommentCount="2" />
  <row Id="20928" PostTypeId="1" CreationDate="2011-02-08T02:08:39.540" Score="-5" ViewCount="108" Body="&lt;p&gt;Hey, I have a couple of questions, I was hoping someone could help me with?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;\displaystyle\sum_{n=0}^{\infty} (- \frac{6}{5})^n&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;\displaystyle\sum_{n=0}^{\infty} (- \frac{3}{10})^n&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;\displaystyle\sum_{n=0}^{\infty} 20(\frac{2}{7})^n&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;using the formula&lt;/p&gt;&#xA;&#xA;&lt;p&gt;s= 1 - r^n / 1 - r&lt;/p&gt;&#xA;&#xA;&lt;p&gt;answers&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1 - 6/5 = -0.2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1 - -0.2 = 1.2 or 6/5&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Answer: 6/5 or infinity&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1 - -3/10 = 1.3&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1 - 1.3 = -0.3 = -3/10&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Answer: -3/10&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1 - 2/7 = 0.714285714&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1 - 0.714285714 = 0.285714286 or 5 2/7&lt;/p&gt;&#xA;&#xA;&lt;p&gt;5 2/7 or infinity&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the a/b denotes a fraction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you in advance.&lt;/p&gt;&#xA;" OwnerUserId="6771" LastEditorUserId="6771" LastEditorDisplayName="" LastEditDate="2011-02-08T15:04:27.457" LastActivityDate="2011-02-08T15:04:27.457" Title="sum of a geometric series" Tags="&lt;homework&gt;&lt;sequences-and-series&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="20929" PostTypeId="2" ParentId="20907" CreationDate="2011-02-08T02:10:42.470" Score="1" ViewCount="" Body="&lt;p&gt;Well, I'm not sure that this is what you really want.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The injective hull of an integral domain is its quotient field. See Lam &quot;Lectures on Modules and Rings&quot;, Example 3.35&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, $\mathbb{Q}$ is injective because it is injective hull of an integral domain $\mathbb{Z}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, isn't Baer's crieterion is equivalent to the injectivity?&lt;/p&gt;&#xA;" OwnerUserId="3100" LastActivityDate="2011-02-08T02:10:42.470" CommentCount="1" />
  <row Id="20930" PostTypeId="2" ParentId="20928" CreationDate="2011-02-08T02:15:51.570" Score="1" ViewCount="" Body="&lt;p&gt;This would appear to be homework-please tag as such.  Do you know the usual formula for the sum of a &lt;a href=&quot;http://en.wikipedia.org/wiki/Geometric_series&quot; rel=&quot;nofollow&quot;&gt;geometric series&lt;/a&gt;?  And when it applies?  These all fit in nicely.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-02-08T02:15:51.570" CommentCount="3" />
  <row Id="20931" PostTypeId="1" AcceptedAnswerId="20935" CreationDate="2011-02-08T02:26:04.837" Score="3" ViewCount="177" Body="&lt;p&gt;I'm trying to prove (or disprove) that in the set $C[0,1]$ of continuous (bounded) functions on the real interval [0,1] with the integral norm $\|f(x)\|_1 = \int_0^1|f(x)|dx$ that a sequence of functions $f_n$ is convergent to $f$ if and only if $f_n$ is pointwise-convergent to $f$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My intuition tells me that this is true, but I'm having trouble formulating an attack (particularly for the convergence =&gt; pointwise convergence case).&#xA;Given $\forall \epsilon &amp;gt; 0 \; \exists N \in \mathbb{N} \;$&#xA;$\forall n \geq N \; \int_0^1 |f_n(x) - f(x)|dx &amp;lt; \epsilon $,&#xA;I'm trying to prove that for an arbitrary $x \in [0,1]$, $\forall \epsilon_0 &amp;gt; 0$ there is an $N_0$ such that for $n_0 \geq N_0 |f_{n_0}(x) - f(x)| &amp;lt; \epsilon_0$. I think I want to make $N_0$ somehow depend on the that integral (or a 'slice' of it), but I don't see it yet. Any pointers/advice/hints are welcome.&lt;/p&gt;&#xA;" OwnerUserId="4518" LastEditorUserId="4518" LastEditorDisplayName="" LastEditDate="2011-02-08T04:01:20.303" LastActivityDate="2011-02-09T04:15:38.930" Title="How to prove that convergence is equivalent to pointwise convergence in $C[0,1]$ with the integral norm?" Tags="&lt;homework&gt;&lt;analysis&gt;&lt;convergence&gt;&lt;metric-spaces&gt;" AnswerCount="3" CommentCount="5" />
  <row Id="20932" PostTypeId="2" ParentId="20913" CreationDate="2011-02-08T02:30:12.150" Score="4" ViewCount="" Body="&lt;p&gt;Zipping up halfway gives a continuous bijection from your pants with the fly down to your pants with the fly at half mast and this is not a homeomorphism. However, the two spaces are homeomorphic no?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/cZs9n.png&quot; alt=&quot;pants&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One can well-imagine this phenomena persists for various other &quot;manifolds with tears&quot; - even in higher dimensions.&lt;/p&gt;&#xA;" OwnerUserId="6608" LastEditorUserId="6608" LastEditorDisplayName="" LastEditDate="2011-03-31T10:14:05.377" LastActivityDate="2011-03-31T10:14:05.377" CommentCount="1" />
  <row Id="20934" PostTypeId="2" ParentId="20931" CreationDate="2011-02-08T02:58:56.230" Score="1" ViewCount="" Body="&lt;p&gt;Hint: Consider $f_n(x) = 2xn^2$ if $0 \leq x \leq 1/(2n)$, $f_n(x) = 2n - 2xn^2$ if $1/(2n) \leq x \leq 1/n$, and ...&lt;/p&gt;&#xA;" OwnerUserId="2810" LastActivityDate="2011-02-08T02:58:56.230" />
  <row Id="20935" PostTypeId="2" ParentId="20931" CreationDate="2011-02-08T03:00:10.720" Score="2" ViewCount="" Body="&lt;p&gt;It is possible for a sequence of continuous functions to converge in the $1$-norm to a continuous function, while converging pointwise nowhere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is also possible for a sequence of continuous functions to converge pointwise everywhere to a continuous function without converging in $1$-norm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The last claim is the easier of the 2 to verify.  Consider functions whose graphs are spikes of rapidly increasing height with base on say $[0,1/n]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the first claim, you can consider a more complicated sequence of spikes, again getting narrower, but this time not increasing in height, and this time have them move all over the place (but in an orderly enough fashion to ensure the lack of convergence at any point).&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2011-02-08T03:00:10.720" />
  <row Id="20936" PostTypeId="2" ParentId="20908" CreationDate="2011-02-08T03:17:28.210" Score="11" ViewCount="" Body="&lt;p&gt;The most basic statement along these lines is that the average width of any line segment is equal to $2L/\pi$, where $L$ is the length of the segment.  This is highly related to the &lt;a href=&quot;http://en.wikipedia.org/wiki/Buffon%27s_needle&quot; rel=&quot;nofollow&quot;&gt;Buffon's needle problem&lt;/a&gt;, where the length of the needle is the same as the width of the strips.  It is also related to the fact that the average value of $|x|$ on the unit circle is equal to $2/\pi$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now consider a convex polygon in the plane. No matter how you turn the polygon, its horizontal width is always equal to twice the sum of the horizontal widths of the edges.  Specifically, the sum of the widths of the edges along the bottom is equal to the width of the polygon, and the same is true of the widths of the edges along the top.  Thus, the average width of any convex polygon is equal to half the sum of the average widths of the side lengths:&#xA;$$&#xD;&#xA;\text{avg. width} \;=\; \frac{1}{2}\left(\frac{2s_1}{\pi} + \cdots + \frac{2s_n}{\pi}\right) \;=\; \frac{s_1+\cdots+s_n}{\pi}.&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a similar formula in three dimensions involving area.  Specifically, suppose we take a polyhedron and rotate it in all possible ways.  For each rotation, we measure the area of the projection of the polyhedron onto the $xy$-plane.  Then&#xA;$$&#xA;\text{avg. $xy$-area} \;=\; \frac{A_1+\cdots+A_n}{4}&#xA;$$&#xA;where $A_1,\ldots,A_n$ are the areas of the faces.  Thus, if you take a unit cube outside on a sunny day and rotate it at random, the average area of its shadow will be $(1+1+1+1+1+1)/4 = 3/2$.  Again, the factor of $4$ comes from the fact that a polyhedron has top faces and bottom faces, and the average $xy$-area of a flat object is $A/2$.  The latter follows from the fact that the average value of $|z|$ on the unit sphere is $1/2$.&lt;/p&gt;&#xA;" OwnerUserId="1726" LastActivityDate="2011-02-08T03:17:28.210" CommentCount="1" />
  <row Id="20937" PostTypeId="2" ParentId="20931" CreationDate="2011-02-08T03:18:27.180" Score="0" ViewCount="" Body="&lt;p&gt;Consider $\displaystyle f_n(x) = \begin{cases} nx &amp;amp; \text{ for } 0 \leq x \leq \tfrac{1}{2n} \\ 1-nx &amp;amp; \text{ for } \tfrac{1}{2n} \lt x \lt \tfrac{1}{n} \\ 0 &amp;amp; \text{ otherwise}\end{cases}.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clearly $\lVert f_n\rVert_1$ is constant (the integral is just the area of a triangle), but $f_n \to 0$ pointwise and $(f_n)$ isn't convergent w.r.t. $\lVert \cdot \rVert_\infty$.&lt;/p&gt;&#xA;" OwnerUserId="1269" LastActivityDate="2011-02-08T03:18:27.180" CommentCount="1" />
  <row Id="20938" PostTypeId="2" ParentId="20920" CreationDate="2011-02-08T03:25:48.123" Score="3" ViewCount="" Body="&lt;p&gt;It seems you can use the enormous symmetry of the Fano plane for this.  You can pick a line ABC and a point off the line D and pick f(A) and f(B) arbitrarily, f(C) is the third point on the line AB, and pick f(D) arbitrarily different from the first three.  Then if the lines through D are ADE, BDF, and CDG you can define f(E), f(F) and f(G) in the obvious way.  Now you just need to prove that f(CEF), f(AGF), and f(BEG) are lines.  But there has to be a meet between the lines f(CE) and f(AG) and it can't be f(B) or f(D), so must be f(F).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Added:  support for this comes from that fact that any isomorphism should be able to be composed with all the automorphisms of the destination plane to make another.  This isomorphism has 7*6*4 possibilities as the first, second, and fourth points can be chosen at random and 7*6*4=168 is the number of automorphisms of the Fano plane.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2011-02-08T13:55:29.047" LastActivityDate="2011-02-08T13:55:29.047" CommentCount="2" />
  <row Id="20939" PostTypeId="1" CreationDate="2011-02-08T03:37:38.593" Score="3" ViewCount="71" Body="&lt;p&gt;Let $C_*$ be a chain complex of abelian groups.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it true that $H_i(C_*\otimes \mathbb{Z}/p)=0$ for all $i$ if and only if $H_i(C_*\otimes \mathbb{Z}_p)=0$ for all $i$, where $\mathbb{Z}_p$ is localization of $\mathbb{Z}$ away from $p$? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And I want to know that the precise definition of localization of $\mathbb{Z}$ away from $p$. Is it equal to $\mathbb{Z}[\frac{1}{p}]$ or $(\mathbb{Z}-p\mathbb{Z})^{-1}\mathbb{Z}$?&lt;/p&gt;&#xA;" OwnerUserId="6774" LastActivityDate="2011-02-08T04:16:42.740" Title="Application of universal coefficient theorem." Tags="&lt;algebraic-topology&gt;&lt;homological-algebra&gt;" AnswerCount="2" />
  <row Id="20940" PostTypeId="2" ParentId="20939" CreationDate="2011-02-08T03:55:56.330" Score="1" ViewCount="" Body="&lt;p&gt;No. Consider for instance the chain complex $0 \to \mathbb{Z}_p \stackrel{p}{\to} \mathbb{Z}_p$. This is acyclic but its reduction mod $p$ isn't. The problem is that $\mathbb{Z}/p$ is not a &lt;a href=&quot;http://en.wikipedia.org/wiki/Flat_module&quot; rel=&quot;nofollow&quot;&gt;flat module&lt;/a&gt; over the localization $\mathbb{Z}_p$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for your second question, it's the latter one. You are localizing &lt;em&gt;away&lt;/em&gt; from $p$, so you invert the elements not in $(p)$.&lt;/p&gt;&#xA;" OwnerUserId="536" LastActivityDate="2011-02-08T03:55:56.330" />
  <row Id="20941" PostTypeId="2" ParentId="20939" CreationDate="2011-02-08T04:05:04.747" Score="0" ViewCount="" Body="&lt;p&gt;Dear Akhil, I'm the one who asked the question. Is it also false in the case that&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$C_*$ is chain complex of free abelian groups? (In here, each $C_i$ is not necessarily finitely generated.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here, I think that if the additional condition that each $C_i$ is free abelian group is given, then we can use the Universal Coefficient theorem to get the desired result. Is it right?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More explicitly, $H_i(C_*\otimes \mathbb{Z}_p)=H_i(C_*)\otimes \mathbb{Z}_p$, because the localization is an exact functor. Hence, $H_i(C_*\otimes \mathbb{Z}_p)=0$ for every $i$ if and only if $H_i(C_*)$ is torsion group and every element has order relative prime to $p$ for every $i$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And $H_i(C_*\otimes \mathbb{Z}/p)=H_i(C_*)\otimes \mathbb{Z}/p\oplus \operatorname{Tor}(H_{i-1}(C_*),\mathbb{Z}/p)=H_i(C_*)\otimes {}_{p}H_{i-1}(C_*)$, where ${}_{p}M=\lbrace x\in M | px=0\rbrace$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, $H_i(C_*\otimes \mathbb{Z}/p)=0$ for every $i$ if and only if $H_i(C_*)$ is torsion group and order of every element is relatively prime to $p$ for every $i$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think that your nice example shows that the condition that $C_*$ is free is needed, indeed. I hope that this is right.&lt;/p&gt;&#xA;" OwnerUserId="6775" LastEditorUserId="6775" LastEditorDisplayName="" LastEditDate="2011-02-08T04:16:42.740" LastActivityDate="2011-02-08T04:16:42.740" CommentCount="4" />
  <row Id="20942" PostTypeId="1" AcceptedAnswerId="20943" CreationDate="2011-02-08T04:11:02.677" Score="3" ViewCount="171" Body="&lt;p&gt;If I had a recursive function (&lt;code&gt;f(n) = f(n-1) + 2*f(n-2)&lt;/code&gt; for example), how would I derive a formula to solve this? For example, with the Fibonacci sequence, Binet's Formula can be used to find the nth term.&lt;/p&gt;&#xA;" OwnerUserId="1110" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2011-06-03T13:49:34.887" LastActivityDate="2011-06-03T13:49:34.887" Title="Deriving formulas for recursive functions" Tags="&lt;functions&gt;&lt;recurrence-relations&gt;&lt;fibonacci-number&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="20943" PostTypeId="2" ParentId="20942" CreationDate="2011-02-08T04:18:12.770" Score="7" ViewCount="" Body="&lt;p&gt;If the recurrence relation is linear, homogeneous and has constant coefficients, here is the way to solve it. First obtain the characteristic equation. To do this, assume $f(n) = m^n$. Plug it in to get a quadratic in $m$. Solve for $m$. Get the two roots say $m_1$ and $m_2$. Now the general solution is given by the linear combination namely $f(n) = a_1 m_1^n + a_2 m_2^n$. Solve for $a_1$ and $a_2$ using the initial conditions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For your recurrence, the corresponding equation becomes, $m^2-m-2 = 0 \Rightarrow (m-2)(m+1) = 0$. So $f(n) = a 2^n + b (-1)^n$. You need to specify two initial conditions to get $a$ and $b$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This methodology is analogous to plugging in $y=e^{mx}$ when you want to solve a linear, homogeneous ODE with constant coefficients. Here you have a difference equation instead of a differential equation.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2011-02-08T05:05:32.810" LastActivityDate="2011-02-08T05:05:32.810" />
  <row Id="20944" PostTypeId="2" ParentId="20906" CreationDate="2011-02-08T04:26:34.663" Score="3" ViewCount="" Body="&lt;p&gt;Suppose you need to solve &#xA;$$a_1x_1 + a_2x_2 + a_3x_3 = c\qquad (1)$$&#xA;in integers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I claim this is equivalent to solving &#xA;$$\gcd(a_1,a_2)y + a_3x_3 = c\qquad (2)$$&#xA;in integers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To see this, note that any solution to (1) produces a solution to (2): letting $g=\gcd(a_1,a_2)$, we can write $a_1 = gk_1$, $a_2=gk_2$, so then we have:&#xA;$$c = a_1x_1 + x_2x_2 + a_3x_3 = g(k_1x_1) + g(k_2x_2) + a_3x_3 = g(k_1x_1+k_2x_2) + a_3x_3,$$&#xA;solving (2). Conversely, suppose you have a solution to (2). Since we can find $r$ and $s$ such that $g=ra_1+sa_2$, we have&#xA;$$c = gy+a_3x_3 = (ra_1+sa_2)y +a_3x_3 = a_1(ry) + a_2(sy) + a_3x_3,$$&#xA;yielding a solution to (1). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This should tell you how to solve the general case&#xA;$$a_1x_1+\cdots+a_nx_n = c$$&#xA;in terms of $\gcd(a_1,\ldots,a_n)$, which can in turn be computed recursively.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2011-02-08T04:26:34.663" CommentCount="1" />
  <row Id="20945" PostTypeId="1" AcceptedAnswerId="20947" CreationDate="2011-02-08T05:25:06.597" Score="7" ViewCount="266" Body="&lt;p&gt;Let $F$ be an extension field of $K$. let $L$ and $M$ be intermediate fields,  with both finite algebraic extensions of $K$. &#xA; Suppose {$a_1,...,a_n$} is a basis for $L$ over $K$ and {$b_1, ...,b_m$} is a basis for $M$ over $K$. Show that {$a_ib_j$} is a spanning set for the field $LM$ ($LM$ is the smallest field between $K$ and $F$ containing both $L$ and $M$) as a vector space over $K$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I have done so far is this:&#xA;Let $x$ $\in$ $L$. Then $x$ can be written as a linear combination of $a_i$. Also $y$ $\in$ $M$ implies that $y$ can be written as a linear combination of $b_j$. This is where I'm stuck.&lt;/p&gt;&#xA;" OwnerUserId="4426" LastEditorUserId="4426" LastEditorDisplayName="" LastEditDate="2011-02-08T05:58:52.200" LastActivityDate="2011-02-15T03:55:38.017" Title="Field Extension" Tags="&lt;abstract-algebra&gt;" AnswerCount="2" CommentCount="5" FavoriteCount="2" />
  <row Id="20946" PostTypeId="1" CreationDate="2011-02-08T05:33:07.917" Score="11" ViewCount="505" Body="&lt;p&gt;Bonjour, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;J'ai rencontré le problème suivant dans le livre &quot;Real and Functional Analysis&quot; de Lang, au chapitre 3. J'explique d'abord le contexte, puis j'en viendrai à la question précise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Il faut démontrer que les fonctions de la forme $e^{-x}p(x)$, où $p$ est un polynôme, sont denses dans l'ensemble des fonctions continues sur $[0, +\infty[$ qui tendent vers zéro en $+\infty$, muni de la norme sup.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pour cela, on applique Stone-Weierstrass au complété $[0,+\infty]$, en considérant l'algèbre des fonctions de la forme $\sum_{n=1}^{N}{e^{-nx}p_n(x)}$ définies sur le complété (auxquelles on doit rajouter aussi des constantes). Ensuite, il faut montrer qu'on peut approcher uniformément les fonctions $e^{-nx}p(x)$ par des fonctions de la forme $e^{-x}q(x)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lang suggère d'approcher d'abord $e^{-2x}$ par des fonctions $e^{-x}q(x)$ à l'aide de la formule de Taylor avec reste, puis $e^{-nx}p(x)$ en général. Je crois avoir réussi la première partie, et la deuxième partie en découle assez facilement pour $n \geq 3$. Par contre, je n'arrive pas à traiter le cas $n = 2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alors ma question est la suivante: &lt;strong&gt;Pour $m \in \mathbf{N}$, peut-on approcher uniformément $x^m e^{-2x}$ sur $[0,+\infty[$ par des fonctions de la forme $e^{-x}q(x)$, où $q$ est un polynôme?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;J'apprécierais beaucoup une référence ou une démonstration (ou les deux). (Vous pouvez répondre dans une autre langue.)&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;[Mod: attempt at translation below]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hello, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've encountered the following problem in the book &quot;Real and Functional Analysis&quot; by Lang, in chapter 3. I'll first explain the context, after which I'll pose my precise question. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We want to prove that the functions of the form $e^{-x}p(x)$, where $p$ is a polynomial, are dense in the set of continuous functions defined on $[0,\infty)$ which tends to 0 at $+\infty$, equipped with the sup norm. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For this, we apply Stone-Weierstrass on the completion $[0,+\infty]$, and consider the algebra generated from the functions of the form $\sum_{n=1}^N e^{-nx}p_n(x)$ (to which we add also the constant functions). Next, we need to show that we can uniformly approximate the functions $e^{-nx}p(x)$ by functions of the form $e^{-x}q(x)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lang suggests to first approximate $e^{-2x}$ by functions of the form $e^{-x}q(x)$ using Taylor's theorem with remainders. Then consider $e^{-nx}p(x)$ in general. I think I know how to do the first step. For the second step, the $n\geq 3$ cases are easy. On the other hand, I don't know how to treat the case $n=2$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So here's my question: &lt;strong&gt;can we approximate a function of the form $x^me^{-2x}$, where $m\in\mathbb{N}$, uniformly over $[0,\infty)$, by functions of the form $e^{-x}q(x)$, where $q$ is a polynomial?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would appreciate a reference or a proof (or both). (You can post your answer in another language.)&lt;/p&gt;&#xA;" OwnerUserId="6778" LastEditorUserId="1543" LastEditorDisplayName="" LastEditDate="2011-02-08T20:42:19.760" LastActivityDate="2011-04-10T12:39:58.440" Title="Application de Stone-Weierstrass" Tags="&lt;analysis&gt;&lt;approximation&gt;" AnswerCount="2" CommentCount="10" />
  <row Id="20947" PostTypeId="2" ParentId="20945" CreationDate="2011-02-08T05:34:34.430" Score="5" ViewCount="" Body="&lt;p&gt;Consider the field $L(b_1,\ldots,b_m)=L[b_1,\ldots,b_m]$ (the equality because we are dealing with finite algebraic extensions). Suppose you can prove that it is &lt;em&gt;equal&lt;/em&gt; to $LM$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then every element of $LM$ can be written as an $L$-linear combination of $b_1,\ldots,b_m$ (there may be a bit of work to be done here if it is not clear; certainly, you can write it as an $L$-linear combination of products of powers of the $b_i$, but since each of those lies in $M$, you can write &lt;em&gt;them&lt;/em&gt; as $K$-linear combinations of the $b_i$; take it from there).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And every coefficient in that linear combination can be written as a $K$-linear combination of $a_1,\ldots,a_n$. See where that leads you (assuming you can prove $L[b_1,\ldots,b_m]=LM$, of course). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Corrected.&lt;/strong&gt; In comments you ask about showing that if $[LM:K]=[L:K][M:K]$, then $L\cap M=K$. I messed up my first attempt in a rather silly way (feel free to look at the edit history to see the screw-up!). Sorry about that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again, let $E=L\cap M$ for simplicity. Then:&#xA;\begin{align*}&#xA;[L:K][M:K] = [LM:K] &amp;amp;= [LM:E][E:K]\\&#xA;&amp;amp;\leq [L:E][M:E][E:K]\\&#xA;&amp;amp;= [L:E][M:K].&#xA;\end{align*}&#xA;Can you take it from here?&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-15T03:55:38.017" LastActivityDate="2011-02-15T03:55:38.017" CommentCount="14" />
  <row Id="20948" PostTypeId="1" AcceptedAnswerId="21016" CreationDate="2011-02-08T05:42:28.207" Score="5" ViewCount="192" Body="&lt;p&gt;Hello,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider this Fibonacci equation:&#xA;$$f_{n+1}^2 - f_nf_{n+2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem asked to write a program with given &lt;code&gt;n&lt;/code&gt;, output the the result of this equation. I could use the formula &#xA;$$f_n = \frac{(1+\sqrt{5})^n - ( 1 - \sqrt{5} )^n}{2^n\sqrt{5}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, from mathworld, I found this formula &lt;code&gt;Cassini's identity&lt;/code&gt; &#xA;$$f_{n-1}f_{n+1} - f_{n}^2 = (-1)^n$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, I decided to play around with the equation above, and I have:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \text{Let } x = n + 1 $$&#xA;$$ \text{then the equation above becomes } f_x^2 - f_{x-1}f_{x+1} $$&#xA;$$ \Rightarrow -( f_{x-1}f_{x+1}  - f_x^2 ) = -1(-1)^x = (-1)^{x+1} = (-1)^{n+1+1} = (-1)^{n+2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So this equation either is 1 or -1. Am I in the right track?   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks,&lt;br&gt;&#xA;Chan&lt;/p&gt;&#xA;" OwnerUserId="5711" LastEditorUserId="5711" LastEditorDisplayName="" LastEditDate="2011-02-08T05:51:22.403" LastActivityDate="2011-02-28T04:06:39.193" Title="Fibonacci identity" Tags="&lt;elementary-number-theory&gt;&lt;fibonacci-number&gt;" AnswerCount="2" CommentCount="6" />
  <row Id="20949" PostTypeId="1" AcceptedAnswerId="20959" CreationDate="2011-02-08T05:56:38.067" Score="4" ViewCount="97" Body="&lt;p&gt;Hello Everyone,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Recently, I was going over a curious recreational math paper titled &quot;On the Diagonal Queens Domination Problem&quot;. The main result of the paper is establishing the minimum value $diag(n)$ of number of queens needed to be kept on the diagonal of a $n \times n$ chessboard so that they attack all the squares on the board.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The authors make use of some clever lemma and make an observation that $diag(n) = n + 1 - r_3(\lceil \frac{n}{2} \rceil)$. Here, $r_3(.)$ denotes the minimum value $k$ such that any $k$-element subset of $[n]$ contains a $k$ term arithmetic progression (a la Roth's theorem statement). In order to prove this statement, authors make use of a number theoretic statement which they do not prove. I also cannot prove i by myself because of my limited number theory background (and maybe because I am dense). The statement says the following&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A collection of $k$ even numbers or $k$ odd numbers (the numbers in the collection have same parity, this will keep their sum even) is midpoint free even sum subset of $[n]$ $\mathit{if\ and\ only\ if}$ all the half of the numbers in the collection (rounded down if not integer) is a midpoint-free subset of $\lceil \frac{n}{2} \rceil$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A few definitions to clarify the question. A subset $X$ of $[n]$ is midpoint free if for any two elements $i,j \in X$ $(i+j)/2 \not\in X$. A subset is even sum if $i+j$ is even for any $2$ elements $i,j \in X$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope the question is clear. In case, its not, I apologize and would request you to kindly let me know whats not clear.&lt;/p&gt;&#xA;" OwnerUserId="5391" LastActivityDate="2011-02-08T09:38:26.380" Title="A number theoretic function to characterize midpoint free subsets" Tags="&lt;number-theory&gt;&lt;combinatorics&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="20950" PostTypeId="2" ParentId="20945" CreationDate="2011-02-08T06:01:14.867" Score="2" ViewCount="" Body="&lt;p&gt;Comme l'a signalé Arturo Magidin, $LM$ est en fait &lt;em&gt;l'anneau&lt;/em&gt; engendré par $L$ et $b_1, \ldots, b_m$. Pour montrer que tout élément de cet anneau est combinaison linéaire des $b_i$, il suffit de voir que $A = \sum Lb_i$ est un anneau. On se ramène donc à prouver que chaque produit $b_i b_j$ (et 1) est dans $A$. Mais en fait, $b_i b_j \in \sum Kb_i$ (et 1 aussi) par l'hypothèse que les $b_i$ forment une base de $M$ sur $K$. &lt;/p&gt;&#xA;" OwnerUserId="6778" LastActivityDate="2011-02-08T06:01:14.867" CommentCount="1" />
  <row Id="20951" PostTypeId="1" CreationDate="2011-02-08T06:04:07.610" Score="3" ViewCount="124" Body="&lt;p&gt;I haven't been able to find a reference that tells what word (if a word) the L is short for. &lt;/p&gt;&#xA;" OwnerUserId="890" LastActivityDate="2011-02-08T06:04:07.610" Title="what does the &quot;L&quot; in &quot;L-function&quot; stand for?" Tags="&lt;terminology&gt;&lt;special-functions&gt;&lt;l-functions&gt;" CommentCount="2" />
  <row Id="20952" PostTypeId="1" AcceptedAnswerId="20953" CreationDate="2011-02-08T06:11:04.797" Score="7" ViewCount="114" Body="&lt;p&gt;Ok I need to evaluate the following function and then prove by taking the derivative of my answer to check:&#xA;$$\int e^{ax}\cos(bx)\,dx$$&#xA;where $a$ is any real number and $b$ is any positive real number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that you set $u=\cos(bx)$  and $dv=e^{ax} dx$,&#xA;and the second time you need to integrate again you set $u=\sin(bx)$ and $dv=e^{ax}dx$ again.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It eventually simplifies down to&#xA;$$\int e^{ax}\cos(bx)dx = \frac{1}{a}e^{ax}\cos(bx) + \frac{b}{a}\left(\frac{1}{a}e^{ax}\sin(bx) - \frac{b}{a}\int e^{ax}\cos(bx)\,dx\right).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I know to move the integral on the left side to the right side so that I can just divide by the constant to solve.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is my problem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know I need to solve the right side to be: &#xA;$$\frac{e^{ax}\left(a\cos(bx) + b\sin(bx)\right)}{a^2+b^2} + C.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To divide by the constant, I multiplied everything on the right side by &#xA;$$\frac{a^2}{b^2+1}.$$&#xA;but this leads me to get $b^2 + 1$ on the bottom instead of $a^2 + b^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I will show what I am doing in detail:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After setting the initial integral equal to:&#xA;$$\frac{1}{a}e^{ax}\cos(bx) + \frac{b}{a}\left(\frac{1}{a}e^{ax}\sin(bx) - \frac{b}{a}\int e^{ax}\cos(bx)\,dx\right) + C$$&#xA;I simplify:&#xA;$$\int e^{ax}\cos(bx)\,dx = \frac{a^2}{b^2+1}\left(\frac{1}{a}e^{ax}\cos(bx) + \frac{b}{a}\left(\frac{1}{a}e^{ax}\sin(bx)\right)\right) + C$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If this is already wrong, can someone point be in the right direction?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I have not gone wrong yet, I can edit to show the rest of my work. &lt;/p&gt;&#xA;" OwnerUserId="6423" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-08T06:18:13.610" LastActivityDate="2011-02-08T06:40:21.700" Title="Integration by parts." Tags="&lt;calculus&gt;&lt;homework&gt;&lt;integral&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="20953" PostTypeId="2" ParentId="20952" CreationDate="2011-02-08T06:25:43.813" Score="5" ViewCount="" Body="&lt;p&gt;So you have&#xA;$$\int e^{ax}\cos(bx)dx = \frac{1}{a}e^{ax}\cos(bx) + \frac{b}{a}\left(\frac{1}{a}e^{ax}\sin(bx) - \frac{b}{a}\int e^{ax}\cos(bx)\,dx\right).$$&#xA;Multiplying out you get&#xA;$$\int e^{ax}\cos(bx)\,dx = \frac{1}{a}e^{ax}\cos(bx) + \frac{b}{a^2}e^{ax}\sin(bx) - \frac{b^2}{a^2}\int e^{ax}\cos(bx)\,dx.$$&#xA;At this point, you should move that last integral on the right hand side to the left hand side and add in the constant of integration on the right.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moving the last integral to the left hand side, you get&#xA;$$\left(1 + \frac{b^2}{a^2}\right)\int e^{ax}\cos(bx)\,dx = \frac{1}{a}e^{ax}\cos(bx) + \frac{b}{a^2}e^{ax}\sin(bx) + C,$$&#xA;and I think this is where you made your mistake.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You tried to clear that $1 + \frac{b^2}{a^2}$ by  multiplying through by $\frac{a^2}{1+b^2}$. But this is incorrect:&#xA;$$1 + \frac{b^2}{a^2} = \frac{a^2+b^2}{a^2} \neq \frac{1+b^2}{a^2}$$&#xA;so that what you multiplied through did &lt;strong&gt;not&lt;/strong&gt; clear that factor. You need to multiply by $\frac{a^2}{a^2+b^2}$ (or, more horribly, by&#xA;$$\frac{1}{1+\frac{b^2}{a^2}}$$&#xA;which is too horrible for words) for things to cancel out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you do that, from&#xA;$$\frac{a^2+b^2}{a^2}\int e^{ax}\cos(bx)\,dx = \frac{1}{a}e^{ax}\cos(bx) + \frac{b}{a^2}e^{ax}\sin(bx) + C,$$&#xA;multiplying both sides by $\frac{a^2}{a^2+b^2}$, we get:&#xA;$$\int e^{ax}\cos(bx)\,dx = \frac{a}{a^2+b^2}e^{ax}\cos(bx) + \frac{b}{a^2+b^2}e^{ax}\sin(bx) + C$$&#xA;the intended answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the way: you don't need to have the &quot;$+C$&quot; on the right hand side until there are no more indefinite integrals there; the constant of integration is implicit in the indefinite integral, so for example, in your penultimate displayed equation, the &quot;$+C$&quot; is superfluous.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-08T06:40:21.700" LastActivityDate="2011-02-08T06:40:21.700" CommentCount="4" />
  <row Id="20954" PostTypeId="2" ParentId="20911" CreationDate="2011-02-08T06:33:37.890" Score="3" ViewCount="" Body="&lt;p&gt;I've used MATLAB to plot such transformations.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I am assuming you have a lattice of 2D points $x$ and your matrix $A: \mathbb{R}^2 \rightarrow \mathbb{R}^2$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you'll get another set of 2D points $x&amp;#39;$ where $x&amp;#39; = Ax$. I use MATLAB to plot both sets of points.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;E.g. $A$ is the 2x2 rotation matrix defined below with $\theta = \pi/4$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ A = &#xD;&#xA;\begin{bmatrix}&#xD;&#xA;\cos(\theta) &amp;amp; -\sin(\theta) \\&#xD;&#xA;\sin(\theta) &amp;amp; \cos(\theta)&#xD;&#xA;\end{bmatrix}&#xD;&#xA;$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The MATLAB code:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;theta = pi/4;&#xA;A = [cos(theta) -sin(theta); sin(theta) cos(theta)];&#xA;x = [1 1 1 0 0 0 -1 -1 -1;-1 0 1 -1 0 1 -1 0 1];&#xA;xp = A*x;&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The new display code:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;v = xp - x;&#xA;axis([-2 2 -2 2]);&#xA;hold on;&#xA;plot(x(1,:),x(2,:),'*k');&#xA;plot(xp(1,:),xp(2,:),'*r');&#xA;quiver(x(1,:),x(2,:),v(1,:),v(2,:),0)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;And the diagram:&#xA;&lt;img src=&quot;http://i.imgur.com/ECUzr.png&quot; alt=&quot;Diagram&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="825" LastEditorUserId="825" LastEditorDisplayName="" LastEditDate="2011-02-09T17:46:45.607" LastActivityDate="2011-02-09T17:46:45.607" CommentCount="4" />
  <row Id="20956" PostTypeId="2" ParentId="20949" CreationDate="2011-02-08T07:40:22.300" Score="-1" ViewCount="" Body="&lt;p&gt;Je vais répondre en français, et j'espère que ça ne posera pas de problème.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Supposons que $X$ est un ensemble de $k$ nombres pairs, et $X&amp;#39; = \{\frac{n}{2} \mid n \in X \}$. Alors étant donnés $m, n \in X$, posons $p = \frac{1}{2}(m+n)$.&#xA;Alors, si $m&amp;#39; = \frac{m}{2}$, $n&amp;#39; = \frac{n}{2}$, et $p&amp;#39; = \frac{p}{2}$, on a $p&amp;#39; = \frac{1}{2}(m&amp;#39; + n&amp;#39;)$. Or $p \in X$ si et seulement si $p&amp;#39; \in X&amp;#39;$. Il s'ensuit que $X$ est &quot;midpoint-free&quot; si et seulement si $X&amp;#39;$ l'est.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Si $X$ est un ensemble de nombres impairs, alors on fait la même chose, mais avec $n&amp;#39; = \frac{1}{2}(n-1)$.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Voici une autre formulation que j'espère plus claire.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dans le cas où tous les éléments de l'ensemble $X$ sont pairs (resp. impairs), posons $f(x) = x/2$ (resp. $f(x) = (x-1)/2$). Alors $f$ est une bijection de $\mathbf{R}$ sur $\mathbf{R}$, avec la propriété que $p$ est le milieu de $m$ et $n$ si et seulement si $f(p)$ est le milieu de $f(m)$ et $f(n)$. En d'autres termes, la propriété &quot;est le milieu de&quot; est préservée par $f$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;L' &quot;ensemble des moitiés&quot; de $X$ est $X&amp;#39; = f(X)$. Il découle de ce qui précède que $X&amp;#39;$ est midpoint-free si et seulement si $X$ l'est. ($X$ est dit &lt;em&gt;midpoint-free&lt;/em&gt; s'il ne contient pas trois éléments distincts $m$, $n$, $p$ tels que $p$ soit le milieu de $m$ et $n$.)&lt;/p&gt;&#xA;" OwnerUserId="6780" LastEditorUserId="6780" LastEditorDisplayName="" LastEditDate="2011-02-08T09:38:26.380" LastActivityDate="2011-02-08T09:38:26.380" CommentCount="7" />
  <row Id="20957" PostTypeId="1" AcceptedAnswerId="20958" CreationDate="2011-02-08T07:49:04.757" Score="2" ViewCount="86" Body="&lt;p&gt;I've got:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{e^{ax}(b^2\cos(bx)+a^2\cos(bx))}{a^2+b^2}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could someone show me how it simplifies to:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$e^{ax}cos(bx)$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It looks like the denominator is canceled by the terms that are being added, but then how do I get rid of one of the cosines?&lt;/p&gt;&#xA;" OwnerUserId="6423" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2011-02-08T09:40:59.097" LastActivityDate="2011-02-08T09:40:59.097" Title="Showing $\frac{e^{ax}(b^2\cos(bx)+a^2\cos(bx))}{a^2+b^2}=e^{ax}\cos(bx).$" Tags="&lt;homework&gt;&lt;algebra&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="20958" PostTypeId="2" ParentId="20957" CreationDate="2011-02-08T07:54:30.633" Score="5" ViewCount="" Body="&lt;p&gt;You use the distributive law, which says that $(X+Y)\cdot Z=(X\cdot Z)+(Y\cdot Z)$ for any $X$, $Y$, and $Z$. In your case, we have $X=b^2$, $Y=a^2$, and $Z=\cos(bx)$, and so&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{e^{ax}(b^2\cos(bx)+a^2\cos(bx))}{a^2+b^2}=\frac{e^{ax}((b^2+a^2)\cos(bx))}{(a^2+b^2)}=\frac{e^{ax}((a^2+b^2)\cos(bx))}{(a^2+b^2)}=e^{ax}\cos(bx).$$&lt;/p&gt;&#xA;" OwnerUserId="264" LastEditorUserId="264" LastEditorDisplayName="" LastEditDate="2011-02-08T08:13:15.440" LastActivityDate="2011-02-08T08:13:15.440" />
  <row Id="20959" PostTypeId="2" ParentId="20949" CreationDate="2011-02-08T08:02:06.337" Score="2" ViewCount="" Body="&lt;p&gt;I don't know if understand the problem correctly, since some of the information seems extraneous and I think there is one mistake, so I will first reformulate the question before I answer it. Please let me know if I misunderstood.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instead of &quot;midpoint-free subset of $\lceil \frac{n}{2} \rceil$&quot;, I think you meant &quot;midpoint-free subset of $[\lfloor \frac{n}{2} \rfloor]$&quot;. Firstly, the brackets for turning the number into a set were missing, but more importantly, we need the floor and not the ceiling, since otherwise the theorem would fail simply because the set of halves could contain $\lceil \frac{n}{2} \rceil$ and this would prevent the original set from being a subset of $[n]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With this out of the way, it seems that neither $k$ nor $n$ nor $[n]$ are actually relevant to the problem, so I will drop all references to them. Also, the property &quot;even-sum&quot; is guaranteed by the collection containing either only even numbers or only odd numbers, so it's a bit confusing to state it as part of the equivalence to be proved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With these clarifications, I would restate the theorem as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A set $S$ containing either only even numbers or only odd numbers is midpoint-free &lt;em&gt;iff&lt;/em&gt; the set $H$ of all halves of its elements (rounded down if not integer) is midpoint-free.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This seems to be a purely &quot;local&quot; property that can be proved by just looking at every pair of numbers individually:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, let the numbers in $S$ be even, and take two numbers $2m$ and $2n$. Then the halves of these numbers are $m$ and $n$, respectively, and $(2m + 2n) / 2 = m + n$, so the theorem is proved if we can show that $m + n \notin S \Leftrightarrow (m + n) / 2 \notin H$. Now if $m + n$ is odd, this is vacuously true since the odd number $m + n$ cannot be in the set $S$ of even numbers and the non-integer $(m + n) / 2$ cannot be in the set $H$ of integers. If $m + n$ is even, the &quot;$\Leftarrow$&quot; direction is obvious, and the &quot;$\Rightarrow$&quot; direction follows because &quot;S&quot; contains only even integers, so the other number of which $(m + n)/2$ is  a (rounded) half, $m + n + 1$, is not in $S$ either. This proves the theorem for $S$ containing only even numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let the numbers in $S$ be odd, and  take two numbers $2m+1$ and $2n+1$. Then the halves of these numbers, rounded down, are $m$ and $n$, respectively, and $(2m+1 + 2n+1) / 2 = m + n + 1$, so the theorem is proved if we can show that $m + n + 1\notin S \Leftrightarrow (m + n) / 2 \notin H$. Again, if $m + n$ is odd, this is vacuously true since the even number $m + n + 1$ cannot be in the set $S$ of odd numbers and the non-integer $(m + n) / 2$ cannot be in the set $H$ of integers. If $m + n$ is even, the &quot;$\Leftarrow$&quot; direction is again obvious, and the &quot;$\Rightarrow$&quot; direction follows because &quot;S&quot; contains only odd integers, so the other number of which $(m + n)/2$ is  a (rounded) half, $m + n$, is not in $S$ either. This proves the theorem for $S$ containing only odd numbers.&lt;/p&gt;&#xA;" OwnerUserId="6622" LastActivityDate="2011-02-08T08:02:06.337" CommentCount="1" />
  <row Id="20960" PostTypeId="1" AcceptedAnswerId="20997" CreationDate="2011-02-08T08:40:22.873" Score="1" ViewCount="81" Body="&lt;p&gt;I have to prove that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x \sec x - \ln |\sec x + \tan x| + C$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is the indefinite integral of:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x \sec x \tan x $$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;by taking the derivative.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've got far enough to get:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x\sec x\tan x + \sec x -\dfrac{|\sec x+\tan x|(\sec^2 x + \sec x \tan x)}{|\sec x + \tan x|}.$$   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Kind of stuck here.  Am I able to cancel out the $|\sec x + \tan x|$ on top and bottom and then set $-\sec x$ equal to $\sec^2 x + \sec x \tan x$?  I'm guessing that's not right though.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sorry for the crummy way I have it setup, feel free to edit it.&lt;/p&gt;&#xA;" OwnerUserId="6423" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2011-02-08T14:07:34.010" LastActivityDate="2011-02-08T14:07:34.010" Title="Simplifying a Trigonometric Expression" Tags="&lt;calculus&gt;&lt;homework&gt;&lt;algebra&gt;&lt;trigonometry&gt;" AnswerCount="2" />
  <row Id="20961" PostTypeId="1" CreationDate="2011-02-08T08:41:46.080" Score="4" ViewCount="138" Body="&lt;p&gt;Hello!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the following integral, p(x) and q(x) are probability distributions. Can you help me to determine in what situation this integral is equal to infinity. For example, I think that such a situations is when only p(x) has an infinite peak.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_{-\infty}^{\infty}\{\log\frac{p(x)}{q(x)}\}p(x)dx$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you very much!&lt;/p&gt;&#xA;" OwnerUserId="6781" LastEditorUserId="622" LastEditorDisplayName="" LastEditDate="2011-02-08T09:00:04.727" LastActivityDate="2011-02-09T08:08:41.027" Title="In what situations is the integral equal to infinity ?" Tags="&lt;integral&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="20962" PostTypeId="1" AcceptedAnswerId="20968" CreationDate="2011-02-08T08:54:39.843" Score="1" ViewCount="41" Body="&lt;p&gt;The following equation arises in the water-filling problem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sum_{i=1}^{n} max \left\{ 0, \frac{1}{\nu} - \alpha_i \right\} = 1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming that all $\alpha_i$s are known, how does one solve for $\nu$?&lt;/p&gt;&#xA;" OwnerUserId="1058" LastActivityDate="2011-02-08T09:58:53.070" Title="Solve for variable in argument of multiple max(a,b) functions" Tags="&lt;calculus&gt;" AnswerCount="1" CommentCount="0" />
  <row Id="20963" PostTypeId="1" AcceptedAnswerId="21112" CreationDate="2011-02-08T09:05:03.850" Score="1" ViewCount="156" Body="&lt;p&gt;Could someone take me through the steps of dividing&#xA;$2x^2 + 11x$ by $x^2+11x+30$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that the denominator is a perfect square of $(x + 5)(x + 6)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer should end up being:&#xA;$$-\dfrac{6}{x+6}-\dfrac{5}{x+5}+2$$ &lt;/p&gt;&#xA;" OwnerUserId="6423" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-09T02:44:14.270" LastActivityDate="2011-02-09T04:08:51.027" Title="Partial fraction decomposition of a quotient of Polynomials" Tags="&lt;homework&gt;&lt;algebra&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="20964" PostTypeId="2" ParentId="20960" CreationDate="2011-02-08T09:13:51.577" Score="3" ViewCount="" Body="&lt;p&gt;Your derivative for $x\sec\;x$ is correct; for the second term, note that $\frac{\mathrm d}{\mathrm dx}\ln(f(x))=\frac{f^{\prime}(x)}{f(x)}$ . Apply the formula accordingly.&lt;/p&gt;&#xA;" OwnerUserId="6783" LastActivityDate="2011-02-08T09:13:51.577" CommentCount="2" />
  <row Id="20965" PostTypeId="2" ParentId="20963" CreationDate="2011-02-08T09:15:04.480" Score="2" ViewCount="" Body="&lt;p&gt;I can't really TeX out polynomial long division in the normal sense, but essentially you want to solve &#xA;$$&#xD;&#xA;2x^2+11x=g(x)\cdot (x^2+11x+30)+r(x)&#xD;&#xA;$$&#xA;by finding polynomials $g(x)$ and $r(x)$ such that $\deg(r)=0$ or $\deg(r)&amp;lt;\deg(x^2+11x+30)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You want to get the leading terms to match on both sides, right? So the leading term of $g(x)$ should be $2$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the constant term of the polynomial, so you've reached the end of the line in a sense. &#xA;So &#xA;$$&#xD;&#xA;2x^2+11x=2\cdot (x^2+11x+30)+r(x)&#xD;&#xA;$$&#xA;which implies&#xA;$$&#xD;&#xA;r(x)=2x^2+11x-2\cdot (x^2+11x+30)=-11x-60.&#xD;&#xA;$$&#xA;It follows that &#xA;$$&#xD;&#xA;\frac{2x^2+11x}{x^2+11x+30}=2+\frac{-11x-60}{x^2+11x+30}.&#xD;&#xA;$$&#xA;You can check that &#xA;$$&#xD;&#xA;\frac{-11x-60}{x^2+11x+30}=-\frac{6}{x+6}-\frac{5}{x+5}.&#xD;&#xA;$$&#xA;To actually get the remainder in that desired form, you could use the method of partial fractions to split the your remainder with quadratic denominator into a sum of terms with linear denominators. I can expand more on that if you're not familiar with it. It also may be quite helpful to follow the &lt;a href=&quot;http://en.wikipedia.org/wiki/Polynomial_long_division&quot; rel=&quot;nofollow&quot;&gt;nice example&lt;/a&gt; given on wikipedia.&lt;/p&gt;&#xA;" OwnerUserId="1583" LastEditorUserId="1583" LastEditorDisplayName="" LastEditDate="2011-02-08T22:27:15.453" LastActivityDate="2011-02-08T22:27:15.453" CommentCount="3" />
  <row Id="20966" PostTypeId="2" ParentId="20961" CreationDate="2011-02-08T09:33:27.520" Score="3" ViewCount="" Body="&lt;p&gt;The integral that you have is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Kullback%E2%80%93Leibler_divergence&quot; rel=&quot;nofollow&quot;&gt;Kullback-Leibler divergence&lt;/a&gt; between distributions P and Q, $D_{KL}(P \parallel Q)$. This divergence is roughly a kind of a &quot;distance&quot; between the two distributions. The reason &quot;distance&quot; is in quotes is because this divergence is not symmetric and is hence not a metric. However, a useful way to think about the divergence $D_{KL}(P \parallel Q)$ is that it is the penalty paid for mistaking distribution $P$ as distribution $Q$. This statement can be made precise using information theory. If $D_{KL}(P \parallel Q)$ is infinity, it is because the two distributions are quite unlike each other that you incur an infinite penalty for mistaking $P$ as $Q$. This can happen for instance when distribution $P$ can produce values that distribution $Q$ can never do - in this case, mistaking $P$ as $Q$ is indeed a grievous error. I will leave it to you to interpret this in terms of the integral above to derive the condition under which the divergence is infinite.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update&lt;/strong&gt;: Adding more information in response to Marco's comment below. It got too unwieldy to be left as a comment: Given any $M&amp;gt;0$ and any distribution $P$, we can find $Q$ such that $D_{KL}(P \parallel Q) &amp;gt; M$. But note that as $M$ grows, we need to adaptively change $Q$ to make sure the divergence grows larger than $M$. This is different from saying $D_{KL}(P \parallel Q) = \infty$ for a given $P$, $Q$ which is what the question is stating. I think this can happen only if the support of $P$ includes points not in the support of $Q$.&lt;/p&gt;&#xA;" OwnerUserId="618" LastEditorUserId="618" LastEditorDisplayName="" LastEditDate="2011-02-09T08:08:41.027" LastActivityDate="2011-02-09T08:08:41.027" CommentCount="4" />
  <row Id="20967" PostTypeId="1" AcceptedAnswerId="20973" CreationDate="2011-02-08T09:53:24.433" Score="0" ViewCount="137" Body="&lt;p&gt;Given a a polynomial with coefficients in $GF(p)$ and degree $d$, will that polynomial always have $d$ roots in $GF(p^d)$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The text I am reading seems to be implying that this is true but I can't see why.&lt;/p&gt;&#xA;" OwnerUserId="4764" LastActivityDate="2011-02-08T23:32:24.523" Title="Which field will contain all the roots of a polynomial over $GF(p)$" Tags="&lt;algebra&gt;&lt;polynomials&gt;&lt;roots&gt;&lt;finite-fields&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" />
  <row Id="20968" PostTypeId="2" ParentId="20962" CreationDate="2011-02-08T09:58:53.070" Score="1" ViewCount="" Body="&lt;p&gt;One possible method: arrange all $\alpha_i$'s into an increasing sequence:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\alpha_{(1)} \leq \alpha_{(2)} \leq \alpha_{(3)} \leq \ldots \leq \alpha_{(n)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For now, introduce the variable $x=\frac{1}{\nu}$. Then, the problem can be restated as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{i=1}^n \max\{0,x-\alpha_{(i)}\} = 1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We are looking for an $m$ such that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$m x - \sum_{i=1}^m \alpha_{(i)} = 1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ x = \frac{1+\sum_{i=1}^m \alpha_{(i)}}{m} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and we must also have that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \frac{1+\sum_{i=1}^m \alpha_{(i)}}{m} \geq \alpha_{(j)} \; , \; \forall j \in \{1,\ldots,m\} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \frac{1+\sum_{i=1}^m \alpha_{(i)}}{m} &amp;lt; \alpha_{(j)} \; , \; \forall j \in \{m+1,\ldots,n\} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This gives a method for finding the solution.&lt;/p&gt;&#xA;" OwnerUserId="3567" LastActivityDate="2011-02-08T09:58:53.070" CommentCount="1" />
  <row Id="20969" PostTypeId="1" CreationDate="2011-02-08T10:02:44.160" Score="2" ViewCount="319" Body="&lt;p&gt;prove  from first principles that 0! is equal to 1&lt;/p&gt;&#xA;" OwnerUserId="6785" LastEditorUserId="5001" LastEditorDisplayName="" LastEditDate="2011-02-08T10:38:26.907" LastActivityDate="2011-03-08T22:14:48.677" Title="Prove $0! = 1$ from first principles" Tags="&lt;algebra-precalculus&gt;" AnswerCount="3" CommentCount="3" />
  <row Id="20970" PostTypeId="2" ParentId="20967" CreationDate="2011-02-08T10:10:06.883" Score="1" ViewCount="" Body="&lt;p&gt;C'est faux tel qu'énoncé. Il faut supposer le polynôme &lt;em&gt;irréductible&lt;/em&gt;, auquel cas c'est vrai.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Si $P$ est irréductible sur $F_p$ et $P(\alpha) = 0$, alors $[F_p(\alpha):F_p] = \deg P = d$. Par conséquent, $F_p(\alpha) = F_{p^d}$, donc $\alpha \in F_{p^d}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Par contre, sur $F_3$, posons, $P(x) = (x^2 + 1)(x + 1)$. Si $\alpha^2 + 1 = 0$, on a $F_p(\alpha) = F_{9}$, qui n'est pas un sous-corps de $F_{27}$.&lt;/p&gt;&#xA;" OwnerUserId="6780" LastActivityDate="2011-02-08T10:10:06.883" CommentCount="5" />
  <row Id="20971" PostTypeId="2" ParentId="20969" CreationDate="2011-02-08T10:32:11.353" Score="6" ViewCount="" Body="&lt;p&gt;I'm not sure that there is anything to prove. I think it follows directly from the definition of factorial: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ n! := \Pi_{k = 1}^n k$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if $n=0$ the right hand side is the empty product which is $1$ by convention.&lt;/p&gt;&#xA;" OwnerUserId="5798" LastActivityDate="2011-02-08T10:32:11.353" CommentCount="3" />
  <row Id="20972" PostTypeId="1" CreationDate="2011-02-08T10:33:07.453" Score="1" ViewCount="145" Body="&lt;p&gt;Let's take a number $x$. &#xA;The square of $x$ is $x^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Find the value of $x$ such that there will be no prime number between $x$ and $x^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any number exists?&lt;/p&gt;&#xA;" OwnerUserId="5179" LastEditorUserId="2468" LastEditorDisplayName="" LastEditDate="2011-02-08T22:24:25.867" LastActivityDate="2011-02-08T22:24:25.867" Title="Finding a number such that there will be no prime number between number and square of number." Tags="&lt;prime-numbers&gt;&lt;number&gt;" AnswerCount="1" CommentCount="6" />
  <row Id="20973" PostTypeId="2" ParentId="20967" CreationDate="2011-02-08T10:37:13.520" Score="6" ViewCount="" Body="&lt;p&gt;In my opinion the theory of finite fields is much clearer if one works in the algebraic closure $\overline{\mathbb{F}_p}$ from the get-go. The Frobenius map $x \mapsto x^p$ acts on the algebraic closure and its fixed points are precisely the roots of $x^p - x$, or the elements of $\mathbb{F}_p$. It follows that the orbits of the roots of a polynomial over $\mathbb{F}_p$ under the action of the Frobenius map are its irreducible factors. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence an element of $\overline{\mathbb{F}_p}$ is a root of an irreducible polynomial of degree $d$ if and only if it has order exactly $d$ under the Frobenius map, if and only if it divides $x^{p^d} - x$ and does not divide $x^{p^k} - x$ for $k &amp;lt; d$, if and only if it is contained in the fixed field of $x \mapsto x^{p^d}$ but not in the fixed field of $x \mapsto x^{p^k}$ for $k &amp;lt; d$. The fixed field of $x \mapsto x^{p^d}$ is a canonical copy of $\mathbb{F}_{p^d}$ inside the algebraic closure, and we have proven both (the correct version of) your statement and uniqueness of finite fields in one go.&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2011-02-08T23:32:24.523" LastActivityDate="2011-02-08T23:32:24.523" CommentCount="5" />
  <row Id="20974" PostTypeId="1" AcceptedAnswerId="20976" CreationDate="2011-02-08T10:48:32.873" Score="2" ViewCount="66" Body="&lt;p&gt;I'm trying to learn about the &quot;functor of points&quot; approach to algebraic geometry, and I get everything &lt;a href=&quot;http://planetmath.org/encyclopedia/ExampleOfFunctorOfPointsOfAScheme.html&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; except for why a $k[\epsilon]/(\epsilon^2)$-point of a scheme corresponds to choosing a point and a tangent vector. Am I correct that the $\kappa_i$ must satisfy the $f_j$'s because we can always mod out by $\epsilon$ and get back a $k$-point? Why do the $\epsilon$'s produce information about the derivative? Also, how exactly do we put together all of the $k[\epsilon]/(\epsilon^2)$-points to make a &quot;tangent bundle&quot;? Such a thing seems like it should have the structure of a scheme (by analogy of the case of manifolds), but as far as I can tell, the collection of all $k[\epsilon]/(\epsilon^2)$-valued points, namely $\text{Hom}(\text{Spec}(k[\epsilon]/(\epsilon^2)),X)$, is just a set (this being the whole idea behind the &quot;functor of points&quot; being from $Schemes$ to $Sets$ in the first place).&lt;/p&gt;&#xA;" OwnerUserId="264" LastActivityDate="2011-02-08T12:44:16.000" Title="Correspondence between $k[\epsilon]/(\epsilon^2)$-points of a scheme, and points with tangent vectors" Tags="&lt;algebraic-geometry&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="20975" PostTypeId="2" ParentId="20969" CreationDate="2011-02-08T11:08:29.220" Score="5" ViewCount="" Body="&lt;p&gt;We need $0!$ to be defined as $1$ so that many mathematical formulae work. For example we would like&#xA;$$n! = n \times (n-1)!$$&#xA;to work when $n=1,$ ie $1! = 1 \times 0!.$&#xA;Also we require that the formula for the number of ways of choosing $k$ objects from $n$ is valid for $k=n.$ ie&#xA;$${n \choose k} = \frac{n!}{k!(n-k)!}$$&#xA;is valid when $k=n.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Things need to work when we extend our definition of the factorial via the &lt;a href=&quot;http://en.wikipedia.org/wiki/Gamma_function&quot; rel=&quot;nofollow&quot;&gt;gamma function.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Gamma(z) = \int_0^\infty t^{z-1} e^{-t} \textrm{ d}z.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The above gives $\Gamma(n)=(n-1)!$ and so we require $0!=1,$ since $\Gamma(1)=1.$&lt;/p&gt;&#xA;" OwnerUserId="1301" LastActivityDate="2011-02-08T11:08:29.220" />
  <row Id="20976" PostTypeId="2" ParentId="20974" CreationDate="2011-02-08T11:10:50.793" Score="5" ViewCount="" Body="&lt;p&gt;Let $k[V] = k[x_1, ... x_n]/(f_1, ... f_r)$ be the ring of functions on the variety $V = \{ f_1 = ... = f_r = 0 \}$. Then a homomorphism $\phi : k[x_1, ... x_n]/(f_1, ... f_r) \to k[\epsilon]/(\epsilon^2)$ is precisely determined by the images $\phi(x_i) = p_i + \epsilon q_i \in k[\epsilon]/(\epsilon^2)$ subject to the condition that $f_j(p + \epsilon q) = 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The key point here is that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f_j(p + \epsilon q) = f_j(p) + \epsilon \sum_i q_i \frac{\partial f_j}{\partial x_i}(p).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is just truncated Taylor expansion, and the corresponding statement is true for $k[\epsilon]/(\epsilon^n)$ for every finite $n$. Hence this condition holds if and only if the $p_i$ define a point of $V$ and the $q_i$ define a vector orthogonal to the gradients of each of the $f_j$; this is precisely the condition that they define a tangent vector over $\mathbb{R}$ so we adopt it as our definition of tangent vector in general. It's not hard to see that we can add tangent vectors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The equations above allow us to give the following definition of the tangent bundle: define the polynomials $g_j = \sum_i y_i \frac{\partial f_j}{\partial x_i}(x) \in k[x_1, ... x_n, y_1, ... y_n]$, and then the tangent bundle ought to be $\text{Spec } k[x_1, ... x_n, y_1, ... y_n]/(f_1, ... f_r, g_1, ... g_r)$. There is probably a coordinate-independent way to state this definition.&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2011-02-08T12:44:16.000" LastActivityDate="2011-02-08T12:44:16.000" />
  <row Id="20977" PostTypeId="1" AcceptedAnswerId="20979" CreationDate="2011-02-08T11:17:34.573" Score="0" ViewCount="33" Body="&lt;p&gt;In a book there is a derivation for y' that comes from&lt;/p&gt;&#xA;&#xA;&lt;p&gt;y' / [1+(y')^2]^1/2 = c&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where c is a constant. The result they had was&lt;/p&gt;&#xA;&#xA;&lt;p&gt;y' = sqrt[c^2 / (1-c^2)]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How did they get this? I tried expanding the square, and other tricks, I cant seem to get their result. &lt;/p&gt;&#xA;" OwnerUserId="6786" LastActivityDate="2011-02-08T11:23:50.363" Title="Equation for y' from y' / [1+(y')^2]^1/2 = c" Tags="&lt;algebra&gt;" AnswerCount="1" />
  <row Id="20978" PostTypeId="1" CreationDate="2011-02-08T11:19:52.467" Score="2" ViewCount="123" Body="&lt;p&gt;Is the AND logic function considered to be a bent function. If so, how would you make a hyper-bent function using logic gates? Thanks!&lt;/p&gt;&#xA;" OwnerUserId="6787" LastActivityDate="2011-03-14T05:35:27.477" Title="bent and hyper-bent boolean functions" Tags="&lt;boolean-algebra&gt;" AnswerCount="0" CommentCount="2" />
  <row Id="20979" PostTypeId="2" ParentId="20977" CreationDate="2011-02-08T11:23:50.363" Score="2" ViewCount="" Body="&lt;p&gt;Square both sides, solve for ${y^{\prime}}^2$ and then take the square root.&lt;/p&gt;&#xA;" OwnerUserId="1301" LastActivityDate="2011-02-08T11:23:50.363" CommentCount="1" />
  <row Id="20980" PostTypeId="1" CreationDate="2011-02-08T11:29:56.220" Score="1" ViewCount="67" Body="&lt;p&gt;in advance excuse my use of non technical language.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This function $f(\alpha)$ and the set $S$ have the following property ($\lambda \in (0,1)$):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\alpha_1,\alpha_2 \in S \nRightarrow \lambda\alpha_1+(1-\lambda) \alpha_2 \in S$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\forall \alpha_1, \alpha_2 \in S$ we have that $f(\lambda\alpha_1+(1-\lambda) \alpha_2) \leq \lambda f(\alpha_1)+(1-\lambda) f(\alpha_2)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm asking whether such a function $f$ has a technical name.  &lt;/p&gt;&#xA;" OwnerUserId="1963" LastEditorUserId="1963" LastEditorDisplayName="" LastEditDate="2011-02-08T12:29:22.957" LastActivityDate="2011-02-08T12:29:22.957" Title="Is there a precise name for a function with the following property:" Tags="&lt;convex-analysis&gt;" CommentCount="6" />
  <row Id="20981" PostTypeId="2" ParentId="20942" CreationDate="2011-02-08T11:31:46.937" Score="4" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Formal_power_series&quot; rel=&quot;nofollow&quot;&gt;Formal power series&lt;/a&gt; can be also be used to solve recurrence relations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $a_n=f(n)$ and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{eqnarray}&#xD;&#xA;S=\sum a_nx^n&amp;amp;=&amp;amp;a_0+a_1x+\sum_2 a_{n-1}x^n+2\sum_2 a_{n-2}x^n\\&#xD;&#xA;&amp;amp;=&amp;amp;a_0+a_1x+\sum_1 a_n x^{n+1}+2\sum_0 a_n x^{n+2}\\&#xD;&#xA;&amp;amp;=&amp;amp;a_0+a_1x+x(S-a_0)+2x^2S&#xD;&#xA;\end{eqnarray}&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the example, suppose that $a_0=0, \ a_1=1$. Then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{eqnarray}&#xD;&#xA;S&amp;amp;=&amp;amp;\frac{-x}{2x^2+x-1}\\&#xD;&#xA;&amp;amp;=&amp;amp;\frac{1}{3}\left(\frac{1}{x+1}-\frac{1}{2x-1}\right)\\&#xD;&#xA;&amp;amp;=&amp;amp;\frac{1}{3}\sum_0 (-(-1)^n+2^n)x^n&#xD;&#xA;\end{eqnarray}&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus $a_n=\frac{1}{3}\left(-(-1)^n+2^{n}\right)$.&lt;/p&gt;&#xA;" OwnerUserId="524" LastEditorUserId="524" LastEditorDisplayName="" LastEditDate="2011-02-08T12:18:36.100" LastActivityDate="2011-02-08T12:18:36.100" />
  <row Id="20982" PostTypeId="1" CreationDate="2011-02-08T12:08:46.313" Score="1" ViewCount="203" Body="&lt;p&gt;how to draw a graph with 5 vertices which is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a)Eulerian cycle but not Hamiltonnian cycle.&#xA;b)Hamiltonnian cycle but not Eulerian cycle.&#xA;c)Eulerian cycle and Hamiltonnian cycle.&lt;/p&gt;&#xA;" OwnerUserId="6788" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2011-02-10T19:09:18.437" LastActivityDate="2011-02-10T19:09:18.437" Title="Eulerean and Hamiltonian cycles on 5 vertices" Tags="&lt;graph-theory&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="20983" PostTypeId="2" ParentId="7966" CreationDate="2011-02-08T12:10:25.140" Score="2" ViewCount="" Body="&lt;p&gt;A nice place to start is Timothy Chow's &quot;&lt;a href=&quot;http://www.ams.org/notices/200601/fea-chow.pdf&quot; rel=&quot;nofollow&quot;&gt;You Could Have Invented Spectral Sequences&lt;/a&gt;&quot;. &lt;/p&gt;&#xA;" OwnerUserId="448" LastActivityDate="2011-02-08T12:10:25.140" />
  <row Id="20984" PostTypeId="1" AcceptedAnswerId="20991" CreationDate="2011-02-08T12:13:27.883" Score="9" ViewCount="187" Body="&lt;p&gt;It it an easy exercise to show that if $X$ is first-countable then for every point $x$ and every subset $A$ we have $x \in \text{cl}A$ iff there exists a sequence $(x_n)_n$ that converges to $x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, this uses the axiom of choice to create the sequence (I think). What would happen if we don't have that? (I know that in topology it is much better to have AC but I want to figure out what happens).&lt;/p&gt;&#xA;" OwnerUserId="1120" LastActivityDate="2011-02-08T16:34:13.987" Title="Topology and axiom of choice." Tags="&lt;topology&gt;" AnswerCount="3" FavoriteCount="2" />
  <row Id="20985" PostTypeId="2" ParentId="20634" CreationDate="2011-02-08T12:18:21.697" Score="0" ViewCount="" Body="&lt;p&gt;The trick in gluing here is, as you correctly pointed out, to fix a point in $U_{1,i}$.  Fix a point in $U_{1,i}$ such that its image under the covering map goes to $U_1\cap U_2$.  Thus it specifies a point which belongs to $U_{2,j}$ for some $j$.  And hence we can get a common neighbourhood.&lt;br&gt;&#xA;The bad local nature is used in the second step.  As a neighbourhood of the left edge can be lifted homeomorphically, we can clearly see that there are non trivial loops in the $\tilde{X}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope the explanation is clear enough. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: This is my answer to the comment made by joriki.  I couldn't write the whole thing as a comment hence I am writing it here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't think one will be able to prove that the circle lifts homeomorphically to an open interval using the idea I suggested.  The issue that will occur is the following: consider a cover of circle with two open sets $U_1$,$U_2$ such that both $U_1$ and $U_2$ can be lifted homeomorphically and choose a $U_{1,i}$ and $U_{2,j}$ as suggested.  What happens in this case is that the projection map will not be injective on the union of $U_{1,i}$ and $U_{2,j}$, as $U_{1,i}\cap U_{2,j}$ will have two components.  As soon as we fix a pre-image for a point in one intersection the preimage of $U_2$ is determined. Now $\pi(U_{2,j}-U_{1,i})\cap U_1$ will not be empty(Here $\pi$ is the projection map).  In the general case when we take $n$ open sets what happens is that $\pi(U_{n,i_n}-\cup_{1}^{n-1}U_{k,i_k})\cap U_1$ will be non empty and hence the projection map will not be injective.  The thing is that, any proper subset of the circle can be lifted homeomorphically can be proved by the arguement i have suggested.  The issue occurs precisely because the circle closes on itself.&lt;/p&gt;&#xA;" OwnerUserId="2893" LastEditorUserId="2893" LastEditorDisplayName="" LastEditDate="2011-02-09T19:33:32.110" LastActivityDate="2011-02-09T19:33:32.110" CommentCount="1" />
  <row Id="20986" PostTypeId="1" AcceptedAnswerId="20992" CreationDate="2011-02-08T12:27:12.133" Score="5" ViewCount="112" Body="&lt;p&gt;I know that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) A function $f:\mathbb{R}^2\to \mathbb{R}^2$, when differentiable at a point, has a $2\times 2$ matrix as a derivative, which is a linear transformation from $\mathbb{R}^2\to \mathbb{R}^2$ best approximating the function linearly in some neighbourhood.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) There is a ring homomorphism $\mathbb{C} \to Mat_{2x2}(\mathbb{R})$ as $a+ib \longmapsto \left[\begin{array}{11}a &amp;amp; -b\\b &amp;amp; a \end{array}\right]$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) For a function $f:\mathbb{C} \to \mathbb{C}$, I can define complex differentiabilty as the best $\mathbb{C}$-linear approximation of the function locally at a point, i.e., $f&amp;#39;(z_0):h \mapsto f&amp;#39;(z_0)h$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, I want to combine these three observations, so that the Cauchy-Riemann equation falls out by considering a complex differentiable function as a function from $\mathbb{R}^2\to \mathbb{R}^2$ and connect the jacobian with the $\mathbb{C}$-linear transformation via the homomorphism. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I am having trouble even formulating a proposition that I can prove. Do I define something called 'Complexfying an $\mathbb{R}^2$-operator'? Any help will be appreciated.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The upshot will be that I can then 'shift' the proofs of some of the basic results of holomorphic functions (such as the fact that if the partial derivatives of the co-ordinate functions exist and are continuous then the function will be holomorphic, etc) to that of multivariable calculus. &lt;/p&gt;&#xA;" OwnerUserId="4806" LastActivityDate="2011-02-08T13:19:02.703" Title="Complex differentiabilty" Tags="&lt;complex-analysis&gt;" AnswerCount="2" FavoriteCount="3" />
  <row Id="20987" PostTypeId="2" ParentId="20986" CreationDate="2011-02-08T12:31:47.993" Score="1" ViewCount="" Body="&lt;p&gt;Nice question. The definition of a differentiable function $f : \mathbb{C} \to \mathbb{C}$ carries over verbatim from multivariable calculus: it is a function such that we can write $f(x + h) = f(x) + df_x(h) + o(|h|)$ for any $x$ and sufficiently small $h$, where $x, h \in \mathbb{C}$ and $df_x(h)$ is a linear operator $\mathbb{C} \to \mathbb{C}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we identify $\mathbb{C}$ with $\mathbb{R}^2$. Then the &quot;realification&quot; of $df_x(h)$ is the Jacobian of $f$ as a differentiable function $\mathbb{R}^2 \to \mathbb{R}^2$, and it is given by multiplication by a complex number, so your observations make sense.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2011-02-08T12:31:47.993" />
  <row Id="20989" PostTypeId="1" CreationDate="2011-02-08T12:43:21.707" Score="4" ViewCount="167" Body="&lt;p&gt;Find four weights such that given four weights and weighing pan (balance scale) you can measure all weights between $1$ to $80$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found this one &lt;a href=&quot;http://mathalon.in/?page=show_problem.php&amp;amp;pid=1&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.Any idea how to solve?&lt;/p&gt;&#xA;" OwnerUserId="3744" LastEditorUserId="3744" LastEditorDisplayName="" LastEditDate="2011-02-08T12:56:21.837" LastActivityDate="2011-02-08T19:58:32.483" Title="Finding four numbers" Tags="&lt;number-theory&gt;&lt;arithmetic&gt;" AnswerCount="2" CommentCount="6" FavoriteCount="1" />
  <row Id="20990" PostTypeId="1" CreationDate="2011-02-08T12:51:27.527" Score="12" ViewCount="5564" Body="&lt;p&gt;A girl you like tells you that there's a 1% chance she'll go out on a date with you. What is the expected number of times you have to ask her out before you get a date?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Not a homework problem - I figure it's like flipping a coin with 99 % heads, 1 % tails, stop flipping when you get a tail.)&lt;/p&gt;&#xA;" OwnerUserId="6790" LastActivityDate="2011-02-12T08:01:14.480" Title="Chance of getting a date" Tags="&lt;puzzle&gt;" AnswerCount="4" CommentCount="12" FavoriteCount="3" />
  <row Id="20991" PostTypeId="2" ParentId="20984" CreationDate="2011-02-08T12:53:30.093" Score="8" ViewCount="" Body="&lt;p&gt;It is consistent with the ZF axioms that there is a dense set of reals $D\subset\mathbb{R}$ having no countable subset. Such a set is infinite, but &lt;a href=&quot;http://en.wikipedia.org/wiki/Dedekind_finite&quot; rel=&quot;nofollow&quot;&gt;Dedekind finite&lt;/a&gt;. It follows that any point in $\mathbb{R}-D$ is in the closure of $D$, but not a limit of any sequence from $D$, since any such sequence would give rise to a countable subset of $D$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Meanwhile, your argument does not require full AC, but only countable AC, since you are making countably many choices of points closer and closer to $x$. &lt;/p&gt;&#xA;" OwnerUserId="413" LastActivityDate="2011-02-08T12:53:30.093" CommentCount="1" />
  <row Id="20992" PostTypeId="2" ParentId="20986" CreationDate="2011-02-08T12:55:40.370" Score="2" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Lemma.&lt;/strong&gt; Let $T:\mathbb{C}\to\mathbb{C}$ be $\mathbb{R}$-linear (considering $\mathbb{C}$ as an $\mathbb{R}$-vector space just by restricting the scalar multiplication). Then the following are equivalent:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;(i) T is $\mathbb{C}$-linear&lt;/li&gt;&#xA;&lt;li&gt;(ii) There exists $\lambda\in\mathbb{C}$ such that $T(z)=\lambda z$ for all $z\in\mathbb{C}$ (i.e. T is multiplication by $\lambda$).&lt;/li&gt;&#xA;&lt;li&gt;(iii) The matrix of $T$ w.r.t. the standard $\mathbb{R}^2$-basis is $\left[\begin{array}{11}a &amp;amp; -b\\b &amp;amp; a \end{array}\right]$, where $a+bi=\lambda$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Now suppose $f:\mathbb{C}\to\mathbb{C}$ is complex differentiable at $z_0$, i.e. $T:\mathbb{C}\to\mathbb{C}$, $z\mapsto f&amp;#39;(z_0)z$ is the $\mathbb{C}$-linear derivative. Then $T$ is also the best $\mathbb{R}$-linear approximation of $f:\mathbb{R}^2\to \mathbb{R}^2$:&#xA;$\frac{|f(x)-f(z_0)-T(z-z_0)|}{|z-z_0|}=\left|\frac{f(z)-f(z_0)}{z-z_0}-f&amp;#39;(z_0)\right|\to 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence the matrix of $T$ is the jacobian of $f$. Recalling that the matrix entries of the jacobian are the partial derivatives, from the Lemma we get the Cauchy-Riemann equations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conversely, suppose $f$ is totally differentiable as map $f:\mathbb{R}^2\to \mathbb{R}^2$ and the Cauchy-Riemann equations hold. Then there is a unique $\mathbb{R}$-linear map $T:\mathbb{C}\to\mathbb{C}$ which is the best approximation, i.e.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{|f(x)-f(z_0)-T(z-z_0)|}{|z-z_0|}\to 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know its matrix entries are the partial derivatives. Since the Cauch-Riemann equations hold, the Lemma says that $T$ is $\mathbb{C}$-linear, say $T(z)=\lambda z$. Note that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\left|\frac{f(z)-f(z_0)}{z-z_0}-\lambda\right|=\frac{\left|f(z)-f(z_0)-c(z-z_0)\right|}{|z-z_0|}=\frac{|f(x)-f(z_0)-T(z-z_0)|}{|z-z_0|}\to 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence $T$ is also the complex derivative.&lt;/p&gt;&#xA;" OwnerUserId="6490" LastEditorUserId="6490" LastEditorDisplayName="" LastEditDate="2011-02-08T13:19:02.703" LastActivityDate="2011-02-08T13:19:02.703" CommentCount="1" />
  <row Id="20993" PostTypeId="2" ParentId="20989" CreationDate="2011-02-08T13:02:54.483" Score="5" ViewCount="" Body="&lt;p&gt;Try using base 3 to express weights in that range.&lt;/p&gt;&#xA;" OwnerUserId="589" LastActivityDate="2011-02-08T13:02:54.483" CommentCount="6" />
  <row Id="20994" PostTypeId="1" CreationDate="2011-02-08T13:09:01.113" Score="1" ViewCount="105" Body="&lt;p&gt;I have a binary vector &lt;strong&gt;b&lt;/strong&gt; $\in \{0,1\}^n$ , and &lt;strong&gt;C&lt;/strong&gt; = &lt;strong&gt;diag&lt;/strong&gt;(b) is a $n \times n$ diagonal matrix.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any  decomposition or transformation like &lt;strong&gt;bx&lt;/strong&gt; = &lt;strong&gt;C&lt;/strong&gt; , so that we can generalize &lt;strong&gt;x&lt;/strong&gt; ?&lt;/p&gt;&#xA;" OwnerUserId="5978" LastActivityDate="2011-05-09T15:55:56.757" Title="vector to diagonal matrix operations/decomposition" Tags="&lt;linear-algebra&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="20995" PostTypeId="2" ParentId="20994" CreationDate="2011-02-08T14:01:46.527" Score="0" ViewCount="" Body="&lt;p&gt;The way you have defined it, for all $i$ where $b_i=1$, $x$ must have a 1 on the main diagonal and zeros elsewhere.  Is this what you mean by &quot;generalize x&quot;?&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-02-08T14:01:46.527" />
  <row Id="20996" PostTypeId="1" CreationDate="2011-02-08T14:02:58.253" Score="4" ViewCount="209" Body="&lt;p&gt;Dear All&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was just having a look at Gödels incompletness theorem as found in:&#xA;&lt;a href=&quot;http://www.research.ibm.com/people/h/hirzel/papers/canon00-goedel.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.research.ibm.com/people/h/hirzel/papers/canon00-goedel.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I noticed that Gödel used a higher order logic. At least such a logic&#xA;was introduced via the Axioms IV and V in the above paper:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Axiom IV (looks to me like comprehension, but is called reducibility axiom):&#xA;$$\exists u \forall v(u(v) \leftrightarrow a)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Axiom V (looks to me like extensionality, but is called type-lift/set axiom):&#xA;$$\forall x_1(x_2(x_1) \leftrightarrow y_2(x_1)) \rightarrow x_2 = y_2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It looks to me higher order, since it uses the formulation $p(x)$ for a kind of member-ship. What alternative formulations are there around of Gödels incompletness theorem that don't make use of higher order logic. Will they be free of variables&#xA;ranging over sets?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Best Regards&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S.: I am looking for proofs that preserve the original content of Gödels proof. So I am not looking for later discovered proofs eventually not based on the same paradox.&lt;/p&gt;&#xA;" OwnerUserId="4414" LastActivityDate="2011-03-13T13:00:49.513" Title="Non Higher-Order Formulation of Gödels Incompletness Theorem" Tags="&lt;logic&gt;&lt;arithmetic&gt;&lt;proof-strategy&gt;" AnswerCount="2" CommentCount="9" FavoriteCount="1" />
  <row Id="20997" PostTypeId="2" ParentId="20960" CreationDate="2011-02-08T14:06:30.940" Score="1" ViewCount="" Body="&lt;p&gt;Two issues—first, as suggested in &lt;a href=&quot;http://math.stackexchange.com/questions/20960/simplifying-a-trigonometric-expression/20964#20964&quot;&gt;Jerry's answer&lt;/a&gt;, you have a factor of $|\sec x+\tan x|$ in the numerator of the last term of your derivative that does not belong there.  Second, the derivative of $\ln|x|$ is $\frac{1}{x}$ (no absolute value), so $\frac{d}{dx}\ln|f(x)|=\frac{1}{f(x)}\cdot f&amp;#39;(x)$ with the chain rule.&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2011-02-08T14:06:30.940" CommentCount="5" />
  <row Id="20998" PostTypeId="2" ParentId="20989" CreationDate="2011-02-08T14:12:13.250" Score="3" ViewCount="" Body="&lt;p&gt;OK, I'll write an answer since I have a bit more to say than will fit into comments. I won't give away the answer though; if you want it given away, please indicate that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please see my comments above regarding the problem formulation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we have two different problems, one where &quot;measure&quot; means &quot;balance&quot; and we want to be able to balance the weights from 1 to 40, and one where &quot;measure&quot; means &quot;infer&quot;, and we want to be able to infer the weights from 1 to 81. (80 in the original question, but 81 makes more sense). Let's first think why we can't do better than 1 to 40 when &quot;measure&quot; means &quot;balance&quot;. We have 81 different combinations of weights available, as we can choose $+$, $-$ or $0$ for each weight, corresponding to putting it on the one scale, the other scale or neither scale. There is a symmetry in that reversing the signs of a combination reverses the sign of the weight we can balance, which gives us nothing new since it just means we have to put the weight to be balanced on the other scale. So we don't actually have 81 different values, but one zero value and $2\times 40$ paired positive/negative values, i.e. 40 useful values. That's why we can't do better than 1 to 40 if &quot;measure&quot; means &quot;balance&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now if &quot;measure&quot; means &quot;infer&quot;, we don't have to be able to balance each weight. In fact, what we need is to be able to balance precisely every &lt;em&gt;other&lt;/em&gt; weight, since two consecutive weights that we can't balance can't be distinguished. And 81 is the greatest number such that we can balance 40 numbers and not leave any consecutive weights unbalanced, namely by balancing all even numbers from 2 to 80. Thats' why we can't do better than 1 to 81 if &quot;measure&quot; means &quot;infer&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That actually already tells you a lot about how to find the answer, and in particular how to map between answers for the &quot;measure&quot;=&quot;balance&quot; 1 to 40 problem and answers for the &quot;measure&quot;=&quot;infer&quot; 1 to 81 problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One more thing about brute force searching: It's not a priori clear that we can restrict the search to integers (there was no such requirement in the question), but the above considerations imply that we can, for if we need to hit 40 integers with 40 available combinations, that only works if all the weights are integers.&lt;/p&gt;&#xA;" OwnerUserId="6622" LastEditorUserId="6622" LastEditorDisplayName="" LastEditDate="2011-02-08T19:58:32.483" LastActivityDate="2011-02-08T19:58:32.483" CommentCount="5" />
  <row Id="20999" PostTypeId="1" CreationDate="2011-02-08T14:13:36.693" Score="0" ViewCount="100" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Possible Duplicate:&lt;/strong&gt;&lt;br&gt;&#xA;  &lt;a href=&quot;http://math.stackexchange.com/questions/20928/sum-of-a-geometric-series&quot;&gt;sum of a geometric series&lt;/a&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;p&gt;I have the following questions, answers are below, can anyone tell me if the following is correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/O4za9.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;using the formula &lt;/p&gt;&#xA;&#xA;&lt;p&gt;s= 1 - r^n / 1 - r &lt;/p&gt;&#xA;&#xA;&lt;p&gt;answers&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1 - 6/5 = -0.2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1 - -0.2 = 1.2 or 6/5 &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Answer: 6/5 or infinity&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1 - -3/10 = 1.3&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1 - 1.3 = -0.3 = -3/10&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Answer: -3/10&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1 - 2/7 = 0.714285714&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1 - 0.714285714 = 0.285714286 or 5 2/7&lt;/p&gt;&#xA;&#xA;&lt;p&gt;5 2/7 or infinity&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the a/b denotes a fraction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you in advance.&lt;/p&gt;&#xA;" OwnerUserId="6771" LastEditorUserId="6771" LastEditorDisplayName="" LastEditDate="2011-02-08T14:49:28.917" LastActivityDate="2011-02-08T15:11:23.613" ClosedDate="2011-02-08T15:39:33.980" Title="infinity geometric series" Tags="&lt;homework&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="21000" PostTypeId="2" ParentId="20990" CreationDate="2011-02-08T14:15:49.280" Score="3" ViewCount="" Body="&lt;p&gt;First of all, if a girl tells you that, I figure it's nothing like flipping a coin with 99% heads, 1% tails, and stopping when you get tails.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if we accept that model, it's a well-defined problem with a clear answer.  The expected number of times you have to ask is just the average number of times you would have to ask if you repeated this process many times.  (It's possible you would be really lucky, and she'd agree the first time; it's possible you would be really unlucky, and she'd turn you down 200 times before accepting.)  We can apply the same kind of reasoning as in &lt;a href=&quot;http://math.stackexchange.com/questions/20426/famous-puzzle-girl-boy-proportion-problem-sum-of-infinite-serie&quot;&gt;this question&lt;/a&gt;. (If people keep having children until they get a boy, how many children will the average family end up with?)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose 1 million guys ask this girl out again and again until she says yes, and her responses really are random, with just a 1% chance of acceptance each time.  The 1% chance of acceptance means that about 1 out of every 100 responses was a yes.  So once all 1 million guys have gotten their dates, we have 1 million yes responses and about 99 million no responses.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So on average, each guy had to ask 100 times.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See the answer to the linked question for the infinite sum to obtain the same answer.&lt;/p&gt;&#xA;" OwnerUserId="1461" LastActivityDate="2011-02-08T14:15:49.280" />
  <row Id="21001" PostTypeId="2" ParentId="20990" CreationDate="2011-02-08T14:21:54.607" Score="15" ViewCount="" Body="&lt;p&gt;Under your assumption that it's like flipping a coin—that is, that each time asking is independent of every other time—the probability that it takes asking exactly $n$ times (first $n-1$ no's, then 1 yes) is $(1-p)^{n-1}p$, where $p=1\%$ is the probability of a yes, so the expected value of the number of times to ask to get to the first yes is&#xA;$$\begin{align}\sum_{n=1}^{\infty}n(1-p)^{n-1}p&#xD;&#xA;&amp;amp;=p+2p(1-p)+3p(1-p)^2+4p(1-p)^3+\cdots&#xD;&#xA;\\&#xD;&#xA;&amp;amp;\begin{align}=p+p(1-p)+p(1-p)^2+p(1-p)^3+\cdots&amp;amp;&#xD;&#xA;\\+p(1-p)+p(1-p)^2+p(1-p)^3+\cdots&amp;amp;&#xD;&#xA;\\+p(1-p)^2+p(1-p)^3+\cdots&amp;amp;&#xD;&#xA;\\+p(1-p)^3+\cdots&amp;amp;&#xD;&#xA;\\+\cdots&amp;amp;\end{align}&#xD;&#xA;\\&#xD;&#xA;&amp;amp;\begin{align}=&amp;amp;\sum_{n=1}^{\infty}p(1-p)^{n-1}+\sum_{n=1}^{\infty}p(1-p)(1-p)^{n-1}\\&amp;amp;+\sum_{n=1}^{\infty}p(1-p)^2(1-p)^{n-1}+\sum_{n=1}^{\infty}p(1-p)^3(1-p)^{n-1}+\cdots\end{align}&#xD;&#xA;\\&#xD;&#xA;&amp;amp;=\frac{p}{1-(1-p)}+\frac{p(1-p)}{1-(1-p)}+\frac{p(1-p)^2}{1-(1-p)}+\frac{p(1-p)^3}{1-(1-p)}+\cdots&#xD;&#xA;\\&#xD;&#xA;&amp;amp;=\sum_{n=1}^{\infty}(1-p)^{n-1}&#xD;&#xA;\\&#xD;&#xA;&amp;amp;=\frac{1}{1-(1-p)}=\frac{1}{p}&#xD;&#xA;\end{align}$$&#xA;so for $p=1\%$, the expected number of times is $100$.&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2011-02-08T14:21:54.607" CommentCount="6" />
  <row Id="21002" PostTypeId="2" ParentId="20972" CreationDate="2011-02-08T14:38:13.207" Score="1" ViewCount="" Body="&lt;p&gt;Despite the comments about Bertrand's postulate, there is still the range $-\sqrt{2} \le x \le \sqrt{2}$ If you want x a natural there is 1 and maybe 0.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-02-08T14:38:13.207" />
  <row Id="21003" PostTypeId="1" CreationDate="2011-02-08T14:55:27.973" Score="3" ViewCount="149" Body="&lt;p&gt;With a few colleagues, we're trying to design an (intermediate) algebra course (US terminology) where we stress the interplay between algebra and geometry. The algebraic topics we would like to cover are (1) linear equation in two variables, (2) quadratic equations in two variables, (3) polynomials in one variable, (4) rational functions in one variable (though we're not sure we want to introduce functions), (5) radicals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For (1) and (2) there are obvious geometric counterparts: lines and conic sections.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question&lt;/strong&gt;: Are there natural geometric counterparts for (3), (4) and (5)?  Are there elementary geometric constructions that naturally lead to these algebraic objects? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Side question&lt;/strong&gt;: Are there (affordable) textbooks or lecture notes out there which have this kind of approach?&lt;/p&gt;&#xA;" OwnerUserId="6791" LastEditorUserId="6791" LastEditorDisplayName="" LastEditDate="2011-02-09T02:52:51.953" LastActivityDate="2011-05-10T12:56:18.690" Title="The elementary coordinate geometry of polynomials? Of rational expressions? Of radicals?" Tags="&lt;algebra&gt;&lt;education&gt;&lt;teaching&gt;&lt;elementary-geometry&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="1" />
  <row Id="21004" PostTypeId="1" CreationDate="2011-02-08T14:59:14.440" Score="1" ViewCount="58" Body="&lt;p&gt;If I have two lognormal processes (X and Y) with mean and volatiilty for each and also correlation between the two, what is the characteristic formula of X + Y (i.e. what is the new mean and new volatility of X + Y)?  &lt;/p&gt;&#xA;" OwnerUserId="6764" LastActivityDate="2011-02-08T16:21:17.473" Title="What is the characteristic formula for the addition of two lognormal distributions?" Tags="&lt;statistics&gt;&lt;probability-distributions&gt;&lt;correlation&gt;" AnswerCount="1" />
  <row Id="21005" PostTypeId="2" ParentId="20999" CreationDate="2011-02-08T15:11:23.613" Score="0" ViewCount="" Body="&lt;p&gt;Your formula s= 1 - r^n / 1 - r is incorrect.  It should be $s= (1 - r^n )/( 1 - r)$ and that is when the upper limit is $n$, not $\infty$  &lt;strong&gt;Parentheses matter&lt;/strong&gt;.  You can never have two answers to a question like this, so 6/5 or infinity cannot be correct.  Have another go.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-02-08T15:11:23.613" CommentCount="4" />
  <row Id="21006" PostTypeId="2" ParentId="20990" CreationDate="2011-02-08T15:13:50.470" Score="9" ViewCount="" Body="&lt;p&gt;The number of times you'll have to ask her, given you've not asked her yet, is the sum of the following:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;1 (because you're about to ask her)&lt;/li&gt;&#xA;&lt;li&gt;the probability she says no, multiplied by the number of times &lt;em&gt;more&lt;/em&gt; you'll expect to ask her if she says no.&lt;/li&gt;&#xA;&lt;li&gt;the probability she says yes, multiplied by the number of times &lt;em&gt;more&lt;/em&gt; you'll expect to ask her if she says no.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The last bullet is zero, because if she says yes, you won't have to ask again.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then note that the number of times &lt;em&gt;more&lt;/em&gt; you'll expect to ask her if she says no is the same as the number of times you'd expect to ask &lt;em&gt;in the first place&lt;/em&gt;, because - as with flipping a coin - they're all independent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus we have $E = 1 + pE$ where $E$ is the number of times you can expect to have to ask and $p$ is the probability she says no.&lt;/p&gt;&#xA;" OwnerUserId="5001" LastActivityDate="2011-02-08T15:13:50.470" CommentCount="3" />
  <row Id="21007" PostTypeId="1" AcceptedAnswerId="21032" CreationDate="2011-02-08T15:30:50.850" Score="4" ViewCount="133" Body="&lt;p&gt;e.g. given two numbers 5 and 6 the maximum number is 19, as after 19 each number can be formed using equation 5*n+6*m by putting different (non-negative) values for n and m.&lt;br&gt;&#xA;Such number is not possible for 4 &amp;amp; 6.&lt;br&gt;&#xA;I am not sure how to go-forward. The only thing I have till now is that the number is possible only if the given numbers are pseudo-prime. Any help is appreciated. &lt;/p&gt;&#xA;" OwnerUserId="5403" LastActivityDate="2011-02-08T19:43:57.700" Title="Given 2 numbers a and b, what is the maximum number which can not be formed using n*a + m*b" Tags="&lt;number-theory&gt;&lt;elementary-number-theory&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="21008" PostTypeId="2" ParentId="20984" CreationDate="2011-02-08T15:47:26.700" Score="8" ViewCount="" Body="&lt;p&gt;Some papers:&#xA;&lt;a href=&quot;http://www.emis.de/journals/CMUC/pdf/cmuc0201/Tachtsis.pdf&quot; rel=&quot;nofollow&quot;&gt;Disasters in metric topology without choice&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.sciencedirect.com/science?_ob=ArticleURL&amp;amp;_udi=B6V1K-4C6JJCS-7&amp;amp;_user=10&amp;amp;_coverDate=04%2F21%2F1995&amp;amp;_rdoc=1&amp;amp;_fmt=high&amp;amp;_orig=search&amp;amp;_origin=search&amp;amp;_sort=d&amp;amp;_docanchor=&amp;amp;view=c&amp;amp;_searchStrId=1634422408&amp;amp;_rerunOrigin=google&amp;amp;_acct=C000050221&amp;amp;_version=1&amp;amp;_urlVersion=0&amp;amp;_userid=10&amp;amp;md5=49429ba007325c4849a7edcbad409a5e&amp;amp;searchtype=a&quot; rel=&quot;nofollow&quot;&gt;Continuing horrors of topology without choice&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and references therein.&lt;/p&gt;&#xA;" OwnerUserId="4280" LastActivityDate="2011-02-08T15:47:26.700" CommentCount="1" />
  <row Id="21009" PostTypeId="2" ParentId="21007" CreationDate="2011-02-08T15:50:22.733" Score="1" ViewCount="" Body="&lt;p&gt;The general formula for coprime $a$ and $b$ is that you can make all numbers from $(a-1)(b-1)$ onwards but there are numbers before it that cannot be made as such positive integer multiples. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Start by observing that for coprime $a$ and $b$ we can write $1 = na + mb$ for some $n,m \in \mathbb{Z}$. So for an integer $x$ we have $x = nxa + mxb$. Now substract or add $0 = a*b - b*a$ to both sides of the equation to make coefficients positive &lt;/p&gt;&#xA;" OwnerUserId="4280" LastEditorUserId="4280" LastEditorDisplayName="" LastEditDate="2011-02-08T15:56:00.213" LastActivityDate="2011-02-08T15:56:00.213" CommentCount="1" />
  <row Id="21010" PostTypeId="1" CreationDate="2011-02-08T15:55:30.043" Score="3" ViewCount="64" Body="&lt;p&gt;Hey,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Need guidance on this problem. Given the following ordinals, determine the order relation (and equalities) - ($\Omega = 2^{\aleph_0}$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\omega_1$, $\Omega \cdot  3$, $\omega \cdot \omega_1$, $3^{\Omega}$, $\Omega^{\omega_1}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My solution (partial):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Omega\cdot 3 \gt \Omega\cdot 2 \gt \Omega$, $3^{\Omega} = \lim_{n \lt \Omega}3^{n} = \Omega$ ($\Omega$ is a limit ordinal)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not certain about how to deal with $\omega \cdot \omega_1$ and $\Omega^{\omega_1}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The inequalities:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Omega\cdot 3 \gt 3^{\Omega} \gt \omega_1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="3637" LastActivityDate="2011-02-08T16:22:55.223" Title="Order relation between Ordinals" Tags="&lt;elementary-set-theory&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="21011" PostTypeId="2" ParentId="21004" CreationDate="2011-02-08T16:21:17.473" Score="1" ViewCount="" Body="&lt;p&gt;$$E(X+Y) = E(X) + E(Y)$$&#xA;and&#xA;$$\sigma_{X+Y}^2 = \sigma_X^2 + \sigma_Y^2 + 2 \rho \sigma_X \sigma_Y$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\rho$ is the correlation coefficient.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastActivityDate="2011-02-08T16:21:17.473" />
  <row Id="21012" PostTypeId="2" ParentId="21010" CreationDate="2011-02-08T16:22:55.223" Score="2" ViewCount="" Body="&lt;p&gt;A hint for $\omega\cdot\omega_1$: What does it look like?  Can you simplify it?  (For a simpler example, compare $2\cdot\omega$ and $\omega\cdot2$...)&lt;/p&gt;&#xA;" OwnerUserId="3852" LastActivityDate="2011-02-08T16:22:55.223" CommentCount="2" />
  <row Id="21013" PostTypeId="2" ParentId="20984" CreationDate="2011-02-08T16:34:13.987" Score="3" ViewCount="" Body="&lt;p&gt;Note that depending on the way you've defined the topology and how you want to use it, you may need choice to get the countable base at each point in your space.  Thus even if you have dependent (countable) choice, there may be subtleties.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, suppose we work in ZF+DC+AD.  Then $\omega_1$ with the usual topology is first-countable and we can even exhibit a countable local base at each point $\aleph\in\omega_1$, namely the collection of half-open intervals&#xA;$\{(\beta,\alpha] : \beta&amp;lt;\alpha\}$ --- we can even order this in order-type $\omega$.  However, we cannot &lt;em&gt;uniformly&lt;/em&gt; order all the bases in order-type $\omega$.  That is, there is no function $f:\omega_1\times\omega\to P(\omega_1)$ such that $\{f(\alpha,n) : n\in\omega\}$ is a local base at $\alpha$.  (Recall that AD implies that there is no sequence $\{C_\lambda\subseteq\lambda\}_{\lambda\in\omega_1}$ such that $C_\lambda$ is a cofinal subset of $\lambda$ with order-type $\omega$.&lt;/p&gt;&#xA;" OwnerUserId="3852" LastActivityDate="2011-02-08T16:34:13.987" CommentCount="1" />
  <row Id="21014" PostTypeId="1" AcceptedAnswerId="21022" CreationDate="2011-02-08T16:43:44.860" Score="8" ViewCount="114" Body="&lt;p&gt;Imagine a rectangle (&lt;em&gt;x&lt;sub&gt;1&lt;/sub&gt;&lt;/em&gt; by &lt;em&gt;y&lt;sub&gt;1&lt;/sub&gt;&lt;/em&gt;) always has to be drawn with horizontal and vertical lines (so it can't have lines at 45 degrees). If the rectangle is rotated by angle &lt;em&gt;θ&lt;/em&gt;, it needs to have a rectangle drawn inside it so that this new rectangle still follows the rules. In the diagram, the black rectangle is the original, the red one is the rotated rectangle, and the green ones are three possible options (the two extremes, and one which maintains the aspect ratio).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/d2Ufw.png&quot; alt=&quot;Rotated rectangles diagram&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I work out (in terms of &lt;em&gt;x&lt;sub&gt;1&lt;/sub&gt;&lt;/em&gt;, &lt;em&gt;y&lt;sub&gt;1&lt;/sub&gt;&lt;/em&gt; and &lt;em&gt;θ&lt;/em&gt;):&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;the dimensions (&lt;em&gt;x&lt;sub&gt;2&lt;/sub&gt;&lt;/em&gt; and &lt;em&gt;y&lt;sub&gt;2&lt;/sub&gt;&lt;/em&gt;) of the rectangle which maintains the aspect ratio of the original?&lt;/li&gt;&#xA;&lt;li&gt;the dimensions (&lt;em&gt;x&lt;sub&gt;2&lt;/sub&gt;&lt;/em&gt; and &lt;em&gt;y&lt;sub&gt;2&lt;/sub&gt;&lt;/em&gt;) of the rectangle which has the largest possible area within the given boundaries?&lt;/li&gt;&#xA;&lt;li&gt;&lt;em&gt;x&lt;sub&gt;2&lt;/sub&gt;&lt;/em&gt; for a given &lt;em&gt;y&lt;sub&gt;2&lt;/sub&gt;&lt;/em&gt; (and vice-versa) within a valid range?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Answers to only one part are quite welcome. I've tried to work this out myself by looking at what I know about the resulting triangles, and can't seem to get anywhere (it's been a while since I studied Maths), so an answer to one of these might point me in the right direction.&lt;/p&gt;&#xA;" OwnerUserId="6795" LastEditorUserId="6795" LastEditorDisplayName="" LastEditDate="2011-02-08T17:13:14.507" LastActivityDate="2011-02-08T18:10:08.653" Title="New size of a rotated-then-cropped rectangle" Tags="&lt;geometry&gt;&lt;trigonometry&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" />
  <row Id="21016" PostTypeId="2" ParentId="20948" CreationDate="2011-02-08T17:05:49.473" Score="3" ViewCount="" Body="&lt;p&gt;Hint:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\begin{pmatrix}&#xD;&#xA;1 &amp;amp; 1 \\&#xD;&#xA;1 &amp;amp; 0 \\&#xD;&#xA;\end{pmatrix}^n =&#xD;&#xA;\begin{pmatrix}&#xD;&#xA;f_{n+1} &amp;amp; f_n \\&#xD;&#xA;f_n &amp;amp; f_{n-1} \\&#xD;&#xA;\end{pmatrix}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the determinant of the matrix on the right side?&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2011-02-08T17:11:06.317" LastActivityDate="2011-02-08T17:11:06.317" CommentCount="1" />
  <row Id="21017" PostTypeId="1" AcceptedAnswerId="21037" CreationDate="2011-02-08T17:13:20.677" Score="7" ViewCount="117" Body="&lt;p&gt;Let $R$ be a ring, and let $I_1,\ldots,I_n$ be ideals in $R$ (or submodules of some $R$-module). Consider the sequence&#xA;$$&#xD;&#xA;\bigoplus_{1\leq j &amp;lt; k\leq n} I_j\cap I_k\quad\xrightarrow{f}\quad\bigoplus_{l=1}^n I_l\quad\xrightarrow{g}\quad\sum_{k=1}^n I_k,&#xD;&#xA;$$&#xA;where $g$ is given by addition, and $f$ maps $x\in I_j\cap I_k$ to $x\in I_j$ and to $-x\in I_k$ (and to zero in all other components).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clearly, $g$ is surjective and the composition $g\circ f$ vanishes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; Is the above sequence exact in the middle?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(This seems to be easy for $n=2$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Concerning the title: I am aware of the fact that $f$ won't be injective in general.)&lt;/p&gt;&#xA;" OwnerUserId="367" LastEditorUserId="367" LastEditorDisplayName="" LastEditDate="2011-02-09T12:44:47.190" LastActivityDate="2011-02-09T12:44:47.190" Title="What is the kernel of the summation map from the direct sum to the sum?" Tags="&lt;algebra&gt;&lt;homological-algebra&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="2" />
  <row Id="21018" PostTypeId="1" AcceptedAnswerId="21020" CreationDate="2011-02-08T17:14:42.597" Score="1" ViewCount="86" Body="&lt;p&gt;Hi.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So it is clear than the nondeterministic versions of computational models such as the Turing Machine is equivalent in &quot;power&quot; to the deterministic model.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other than showing this fact, what would be the &quot;purpose&quot; in learning nondeterminism?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My instictive answer to this is that the nondeterministic version of a TM is easier to describe than the deterministic version.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would there be other reasons that you can think of?&lt;/p&gt;&#xA;" OwnerUserId="6299" LastActivityDate="2011-02-08T22:43:51.167" Title="Nondeterminism and computational models" Tags="&lt;computer-science&gt;&lt;computability&gt;" AnswerCount="2" />
  <row Id="21019" PostTypeId="2" ParentId="20920" CreationDate="2011-02-08T17:16:59.247" Score="0" ViewCount="" Body="&lt;p&gt;Let $P$ be a projective plane on $7$ points. Fix an ordered $4$-tuple of distinct points $(p_1,p_2,p_3,p_4)$ in $P$ such that no line is incident with more than two of them. If $\pi=\{\{u,v\},\{w,t\}\}$ is a partition of $\{1,2,3,4\}$ in two parts of size $2$, let $q_\pi$ be the point of intersection of the line joining $p_u$ and $p_v$ with the line joining $p_w$ and $p_t$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This provides a labeling $p_1$, $p_2$, $p_3$, $p_4$, $q_{\{\{1,2\},\{3,4\}\}}$, $q_{\{\{1,3\},\{2,4\}\}}$, $q_{\{\{1,4\},\{2,3\}\}}$ of the $7$ points.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now show from the labeling alone you can reconstruct the lines.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This proves uniqueness.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;NB:&lt;/strong&gt; Notice that there are the $4$-element &lt;em&gt;set&lt;/em&gt; $\{p_1,p_2,p_3,p_4\}$ is determined  by its complement, which is a line. It follows that there are $7$ possible sets, one for each line, and each such set can be turned into an ordered $4$-tuple in $4!$ ways. The above reasoning shows then that the automorphism group of $P$ has order $4!\cdot 7=168$.&lt;/p&gt;&#xA;" OwnerUserId="274" LastEditorUserId="274" LastEditorDisplayName="" LastEditDate="2011-02-08T17:27:26.180" LastActivityDate="2011-02-08T17:27:26.180" />
  <row Id="21020" PostTypeId="2" ParentId="21018" CreationDate="2011-02-08T17:48:08.683" Score="2" ViewCount="" Body="&lt;p&gt;Non-determinism underlies the all important P=NP question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Non-deterministic computation has a different interpretation. Consider for example primality testing. It is not clear how to test for primality (although [provably] efficient tests are now known), but it is clear how to get convinced that a number is &lt;i&gt;composite&lt;/i&gt;: all you need to do is find some non-trivial factorization. This is a property which is easy to &lt;i&gt;verify&lt;/i&gt; given a &lt;i&gt;witness&lt;/i&gt;, in this case the non-trivial factorization. We can therefore build a very simple non-deterministic verifier for compositeness:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;int verify_composite(int n, int a, int b) {&#xA;  return (a &amp;gt; 1) &amp;amp;&amp;amp; (b &amp;gt; 1) &amp;amp;&amp;amp; (n == a* b);&#xA;}&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This algorithm is much simpler and faster than the known algorithms deciding primality. Now in principle, a TM could go over all possible factorizations; however, that is pretty slow, and people conjecture that you can't do it efficiently (even though you can do better than exhaustive search). That illustrates the power of non-determinism.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are lots of combinatorial problems which are easy to check given a witness, in a one-sided fashion. One example is the Traveling Salesman Problem: what is the most efficient route a traveling salesman should take if she wishes to visits all major cities in New York? It is easy to give a witness showing that there's a route taking X days: just present the route. However, it is not so clear whether the reverse is possible; in fact, people strongly believe it is impossible to give an efficiently verifiable witness that there is &lt;i&gt;no&lt;/i&gt; route taking X days.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Problems which are efficiently solvable are captured by the complexity class P. Decision problems which are efficiently verifiable are captured by NP. The problem of deciding whether there's a route for the salesman taking at most X days is thus in NP, but the opposite problem is not believed to line in NP (and in particular, not in P).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Karp found in the 70s that a lot of combinatorial problems in NP reduce to each other, in a sense which implies that if one of them is in P, then all of them are in P. Cook was able to show that one of these problems captures all of NP, in other words if that particular problem is in P, that all of NP is in P; such problems are called NP-complete. One of the problems studied by Karp was the one used by Cook, so all of the problems he studies are NP-complete. (Karp was in fact preceded by Cook.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;People strongly believe that NP-complete problems are not in P; this is the well-known P=NP problem (people believe the two classes are different). Thus if your problem is NP-complete, you know that it is hard to solve &lt;i&gt;in the worst case&lt;/i&gt;. Though many such problems tend to be solvable (or at least approximately solvable) in practice (i.e. in the average case).&lt;/p&gt;&#xA;" OwnerUserId="1277" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2011-02-08T22:43:51.167" LastActivityDate="2011-02-08T22:43:51.167" CommentCount="4" />
  <row Id="21022" PostTypeId="2" ParentId="21014" CreationDate="2011-02-08T17:57:41.987" Score="4" ViewCount="" Body="&lt;p&gt;Let's say that the origin of a coordinate system is at the center of the original black rectangle.  This is the center of the rotation and also the center of each of the green rectangles.  Consider only the first quadrant (from the origin up and to the right).  Exactly 1/4 of the black rectangle and each green rectangle is in the first quadrant.  Let's further assume that, as in your picture, the angle of rotation is relatively small (say, less than 45°) and in the clockwise direction, and that $y_1&amp;lt;x_1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let the vertex of the black rectangle that is in the first quadrant be $(a,b)=(\frac{1}{2}x_1,\frac{1}{2}y_1)$.  Call the magnitude of the angle of rotation $\alpha$ ($0°&amp;lt;\alpha&amp;lt;45°$).  The coordinates of the vertex of the red rectangle in the first quadrant are $(a\cos\alpha+b\sin\alpha,-a\sin\alpha+b\cos\alpha)$.  The slope of the upper side of the red rectangle is $-\tan\alpha$.  An equation for the upper side of the red rectangle in the first quadrant is $y-(-a\sin\alpha+b\cos\alpha)=-\tan\alpha(x-(a\cos\alpha+b\sin\alpha))$ for $0\le x\le a\cos\alpha+b\sin\alpha$.  An equivalent simplified equation is $y=\sec\alpha(b-x\sin\alpha)$.  This red upper side intersects the original black upper side at $x=b\cot\alpha(\sec\alpha-1)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;for $b\cot\alpha(\sec\alpha-1)\le x\le a$, the vertex of the green rectangle in the first quadrant will be on the red side, at $(x,\sec\alpha(b-x\sin\alpha))$, and it will have area $4x\sec\alpha(b-x\sin\alpha)$.&lt;/li&gt;&#xA;&lt;li&gt;for $0\le x&amp;lt;b\cot\alpha(\sec\alpha-1)$, the vertex of the green rectangle in the first quadrant will be on the black side (here, I'm assuming from your picture that the green rectangle must be contained entirely within the black rectangle), at $(x,b)$, and it will have area $4xb$.&lt;/li&gt;&#xA;&lt;li&gt;assuming that the green rectangle must be contained entirely within the black rectangle, $x$ cannot be greater than $a$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Assuming the first case:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;the maximum area is $b^2\csc\alpha\sec\alpha$ when $x=\frac{1}{2}b\csc\alpha$.&lt;/li&gt;&#xA;&lt;li&gt;the green rectangle will have the same aspect ratio as the black rectangle when $x=\frac{ab}{b\cos\alpha+a\sin\alpha}$ (and the area is $\frac{4ab^3}{(b\cos\alpha+a\sin\alpha)^2}$).&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;edit&lt;/em&gt;: I originally forgot the factors of 4 on the area, moving from my picture of just one-fourth of things back to the whole rectangles.  Similarly, note that my $x$- and $y$-values are half of your $x_2$ and $y_2$, respectively.  The maximum possible area for the second case (where the vertex is at $(x,b)$) is $4b^2\cot\alpha(\sec\alpha-1)$, which Mathematica seems to be telling me cannot be greater than the maximum area for the first case under my assumptions about the problem.  &lt;/p&gt;&#xA;" OwnerUserId="72" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2011-02-08T18:10:08.653" LastActivityDate="2011-02-08T18:10:08.653" CommentCount="2" />
  <row Id="21023" PostTypeId="2" ParentId="21017" CreationDate="2011-02-08T18:05:32.853" Score="3" ViewCount="" Body="&lt;p&gt;If you take $I_1, \ldots I_n$ submodules of an $R$-module $M$, you can easily find a counterexample when their intersection is zero :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pick $R = \mathbb{Z}, M = \mathbb{Z}^3, x_1 = (1,-1,0), x_2 = (0,1,-1), x_3 = (-1,0,1)$.&#xA;Then you have that $x_1+x_2+x_3 = 0$, $x_i$ is in its submodule $I_i = \mathbb{Z}x_i$, and for $i \not = j, I_i \cap I_j = \{0\}$.&lt;/p&gt;&#xA;" OwnerUserId="6270" LastActivityDate="2011-02-08T18:05:32.853" />
  <row Id="21024" PostTypeId="2" ParentId="21018" CreationDate="2011-02-08T18:07:51.183" Score="2" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Reason A:&lt;/strong&gt; For some other useful computation models, nondeterminism makes a difference, for example pushdown automata.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Reason B:&lt;/strong&gt; Nondeterminism changes runtime results. In theory, consider $P$ vs $NP$; in practice, consider (non)deterministic Quicksort.&lt;/p&gt;&#xA;" OwnerUserId="3330" LastActivityDate="2011-02-08T18:07:51.183" />
  <row Id="21025" PostTypeId="2" ParentId="20886" CreationDate="2011-02-08T18:13:01.820" Score="0" ViewCount="" Body="&lt;p&gt;I think this might be K_{2,4} and I'm more or less sure. If anyone could confirm or say whether this is correct or not, would be nice. Just see that there's an automorphism that doesn't translate to a topological isomorphism.&lt;/p&gt;&#xA;" OwnerUserId="6584" LastActivityDate="2011-02-08T18:13:01.820" />
  <row Id="21026" PostTypeId="2" ParentId="11551" CreationDate="2011-02-08T18:40:33.610" Score="8" ViewCount="" Body="&lt;p&gt;The statement is consistently false.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In vaughn's article &quot;Countably compact and sequentially compact spaces&quot; in the Handbook of Set-Theoretic Topology, he gives the following equivalences for an infinite cardinal $\kappa$:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\{0,1\}^\kappa$ is sequentially compact&lt;/li&gt;&#xA;&lt;li&gt;$\kappa &amp;lt; \mathfrak{s}$ (where $\mathfrak{s}$ is the &quot;splitting number&quot;)&lt;/li&gt;&#xA;&lt;li&gt;every compact space of weight $\leq\kappa$ is sequentially compact&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Thus if we are in a universe where $\omega&amp;lt;\omega_1=\kappa&amp;lt;\mathfrak{s}\leq\mathfrak{c}&amp;lt;2^\kappa$ then the space&#xA;$\{0,1\}^{\omega_1}$ gives a separable space which is sequentially compact but has cardinality greater than $\mathfrak{c}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Added: including the forcing construction to get such a model (from van Douwen's article):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Start with ground model&#xA;$M\models(\mathfrak{c}=\omega_2 \wedge 2^{\omega_1}=\omega_3)$&#xA;then obtain an iterated ccc extension $\langle M_\eta : \eta\in\omega_2\rangle$ by adding $X_\eta\in[\omega]^\omega$ at stage $\eta\in\omega_2$ s.t. $\forall Y\in M\cap[\omega]^\omega$ either $X_\eta\subseteq^* Y$ or $X_\eta\subseteq^* (\omega\setminus Y)$.  Then in $M_{\omega_2}$ we have $\mathfrak{c}=\omega_2&amp;lt;2^{\omega_1}$, $\mathfrak{t}=\omega_1$, $\mathfrak{s}=\omega_2$, exactly as desired.  (Details are left as an exercise).&lt;/p&gt;&#xA;" OwnerUserId="3852" LastEditorUserId="3852" LastEditorDisplayName="" LastEditDate="2011-02-08T19:34:55.300" LastActivityDate="2011-02-08T19:34:55.300" CommentCount="6" />
  <row Id="21027" PostTypeId="1" CreationDate="2011-02-08T18:45:15.503" Score="0" ViewCount="49" Body="&lt;p&gt;I'm trying to show that complements of hyperplanes in $P^{n}$ are affine varieties on my way to a bigger result but I've gotten stuck. Am I missing something- or will the complement be a copy of affine space? I'm not seeing it.&lt;/p&gt;&#xA;" OwnerUserId="6800" LastActivityDate="2011-02-08T21:04:02.467" Title="Complements of hyperplanes in projective space are affine." Tags="&lt;algebraic-geometry&gt;" AnswerCount="2" />
  <row Id="21028" PostTypeId="1" AcceptedAnswerId="21030" CreationDate="2011-02-08T18:59:47.510" Score="1" ViewCount="72" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;If $f(x)\nearrow\infty$ in $[a, \infty)$ and $f&amp;#39;(x)$ is continuous in $[a, \infty)$ then $\int_{a}^{\infty}\frac{f&amp;#39;(x)}{f(x)}\sin(f(x))$ converges&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm not exactly sure what to try here. I thought of Dirichlets test but I can't seem to identify the right functions to make it work.&lt;/p&gt;&#xA;" OwnerUserId="6801" LastActivityDate="2011-02-08T19:21:34.670" Title="If $f(x)\nearrow\infty$ in $[a, \infty)$ and $f'(x)$ is continuous in $[a, \infty)$ then $\int_{a}^{\infty}\frac{f'(x)}{f(x)}\sin(f(x))$ converges" Tags="&lt;calculus&gt;&lt;improper-integrals&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="21029" PostTypeId="1" AcceptedAnswerId="21031" CreationDate="2011-02-08T19:02:27.120" Score="1" ViewCount="80" Body="&lt;p&gt;There are five real numbers $a,b,c,d,e$ such that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$a + b + c + d + e = 7$$$$a^2+b^2+c^2+d^2+e^2 = 10$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can we find the maximum and minimum possible values of any one of the numbers ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://mathalon.in/?page=show_problem.php&amp;amp;pid=137&quot; rel=&quot;nofollow&quot;&gt;Source&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="6802" LastEditorUserId="6802" LastEditorDisplayName="" LastEditDate="2011-02-08T19:05:08.367" LastActivityDate="2011-02-08T19:40:40.313" Title="Minimum and maximum" Tags="&lt;contest-math&gt;" AnswerCount="3" />
  <row Id="21030" PostTypeId="2" ParentId="21028" CreationDate="2011-02-08T19:08:59.493" Score="2" ViewCount="" Body="&lt;p&gt;Try a change of variables to reduce it to an integral of a sinc function.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2011-02-08T19:08:59.493" CommunityOwnedDate="2011-02-08T19:08:59.493" CommentCount="2" />
  <row Id="21031" PostTypeId="2" ParentId="21029" CreationDate="2011-02-08T19:12:41.513" Score="2" ViewCount="" Body="&lt;p&gt;Maximum value that can be taken by one of the variables is $\frac{9}{5}$ and the minimum value is $1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This can be argued based on symmetry as follows (and I hope I have not overlooked anything in my symmetry argument).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Say we want to find the extreme values of $a$. By symmetry when $a$ reaches the extremum the remaining variables have to be equal. So when $a$ takes the extremum $b=c=d=e$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, $a + 4b = 7$ and $a^2 + 4b^2 = 10$ i.e. $a^2 + 4 (\frac{7-a}{4})^2 = 10$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;i.e. $a^2 + \frac{(a-7)^2}{4} = 10$ i.e. $5a^2 -14a + 49 = 40 $ i.e. $5a^2 - 14a + 9 =0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, $(5a-9)(a-1)=0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, minimum is $a=1$ and maximum is $a = \frac{9}{5}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The round about way would be to use Lagrange multiplier technique and set up 7 equations in 7 unknowns and solve it.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2011-02-08T19:18:17.170" LastActivityDate="2011-02-08T19:18:17.170" CommentCount="2" />
  <row Id="21032" PostTypeId="2" ParentId="21007" CreationDate="2011-02-08T19:14:39.003" Score="2" ViewCount="" Body="&lt;p&gt;Here's an excerpt from one of my old sci.math posts which explains a geometric viewpoint.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice that we may &lt;em&gt;normalize&lt;/em&gt; any representation $\rm\ N\ =\ P\ X + Q\ Y\ $  so  that $\rm\ 0 \le X &amp;lt; Q\ $ by adding a certain integral multiple of $\rm\ (-Q,P)\ $ to $\rm\:(X,Y)\:.\ $ From this observation follows this&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;LEMMA&lt;/strong&gt; $\rm\ \  N = P\ X + Q\ Y\ $ for some integers  $\rm\ X,Y \ge 0\ $&#xA;iff its normalization has $\rm\: Y \ge 0\:$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof&lt;/strong&gt; $\rm\ \ \  X,Y \ge 0\ $ implies that normalization requires addition of $\rm\:(-Q,P)\:$ zero or more times,$\ $ and this preserves the condition $\rm\: Y \ge 0\:.\ $ Conversely if the normalization has $\rm\: Y &amp;lt; 0\:,\ $ then $\rm\:N\:$ has no &#xA;representation with $\rm\ X, Y \ge 0\:,\ $ because to shift $\rm\: Y &amp;gt; 0\: $ &#xA;requires adding $\rm\ (-Q,P)\ $ at least once, which shifts $\rm\: X &amp;lt; 0\:.\ $ &#xA;Finally, since $\rm\ X\ P + Y\ Q\ $  is increasing in both $\rm\: X,Y\:,\ $&#xA;it is clear that the largest non-representable number $\rm\: N\:$ has normalization&#xA;$\rm\: (X,Y)\ =\ (Q-1,-1)\:,\: $ therefore $\rm\ N\ =\ PQ - P - Q\:.\quad\quad $ &lt;strong&gt;QED&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice that the proof has a vivid geometric picture: &#xA;representations of $\rm\:N\:$ correspond to lattice points $\rm\:(X,Y)\:$ &#xA;on the line  $\rm\ N = P\ X + Q\ Y\ $  with negative slope $\rm = -P/Q\:.\ $ &#xA;Normalization is achieved by shifting forward/backward &#xA;along the line by integral multiples of vector $\rm\:(-Q,P)\:$ &#xA;until you land in the normal strip where $\rm\ 0 \le X &amp;lt; Q-1\:.\ $ &#xA;From this viewpoint, the proof becomes crystal clear. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In case you're interested, the class I'm teaching &#xA;  is linear algebra, but as you can see I like to &#xA;  give puzzles to the class which are not necessarily &#xA;  related to the material (in an obvious way). &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here the underlying linear structure is a $\mathbb Z$-module, &#xA;a generalization of a vector space; i.e. here the &#xA;scalars are the integers so have only the structure &#xA;of a ring, not a field. Unless you've already taught &#xA;some module theory, it might be tricky to precisely &#xA;explain the relationship to vector spaces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally it should be mentioned that there has been much &#xA;written on this classical problem. To locate such work &#xA;you should ensure that you search on the many aliases, &#xA;e.g. postage stamp problem, Sylvester/Frobenius problem, &#xA;Diophantine problem of Frobenius, Frobenius conductor, &#xA;money changing, coin changing, change making problems, &#xA;h-basis and asymptotic bases in additive number theory, &#xA;integer programming algorithms and Gomory cuts, &#xA;knapsack problems and greedy algorithms, etc. &lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2011-02-08T19:43:57.700" LastActivityDate="2011-02-08T19:43:57.700" />
  <row Id="21033" PostTypeId="1" AcceptedAnswerId="21035" CreationDate="2011-02-08T19:20:42.087" Score="0" ViewCount="56" Body="&lt;p&gt;For a polynomial of order n with unknown coefficients, what are the ways to find the coefficients from n+1 points on its plot? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I remember one way is to construct a fractional for each point, and the polynomial is the sum of the fractionals, s.t. for each fractional, its value is the desired value for its corresponding point and zero for other points. Is there a name for the method?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="1281" LastActivityDate="2011-02-08T19:28:45.920" Title="Find coefficients of a polynomial given several points on its plot" Tags="&lt;polynomials&gt;" AnswerCount="1" />
  <row Id="21034" PostTypeId="2" ParentId="21028" CreationDate="2011-02-08T19:21:34.670" Score="0" ViewCount="" Body="&lt;p&gt;Integration by parts gives&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_a^b \frac{f&amp;#39;(x)}{f(x)}\sin(f(x)) \, dx = \frac{-\cos(f(x))}{f(x)}\Bigg \vert_a^b - \int_a^b \frac{f&amp;#39;(x)}{f(x)^2}\cos(f(x)) \, dx $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But now &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left| \int_a^b \frac{f&amp;#39;(x)}{f(x)^2}\cos(f(x)) \, dx \right| \le \int_a^b  \frac{f&amp;#39;(x)}{f(x)^2}  \, dx = \frac{1}{f(x)}\Bigg \vert_b^a$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: Looking at it again, there might be some trouble with f being zero somewhere...&lt;/p&gt;&#xA;" OwnerUserId="3208" LastActivityDate="2011-02-08T19:21:34.670" CommentCount="3" />
  <row Id="21035" PostTypeId="2" ParentId="21033" CreationDate="2011-02-08T19:23:27.083" Score="2" ViewCount="" Body="&lt;p&gt;The method you are talking about is called &lt;a href=&quot;http://en.wikipedia.org/wiki/Lagrange_polynomial&quot; rel=&quot;nofollow&quot;&gt;Lagrange Interpolation&lt;/a&gt;. If you want to find out the individual coefficients in the original &quot;basis&quot;, then you would need to solve a $(n+1) \times (n+1)$ Linear system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What we are essentially doing in Lagrange Interpolation is to shift to a new &quot;basis&quot; which depends on the points at which we observe the value, so that the system we need to solve is just the Identity matrix and hence we do not need to invert it as such to get the coefficients.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2011-02-08T19:28:45.920" LastActivityDate="2011-02-08T19:28:45.920" />
  <row Id="21036" PostTypeId="1" AcceptedAnswerId="21044" CreationDate="2011-02-08T19:28:21.340" Score="5" ViewCount="63" Body="&lt;p&gt;The question contains 2 stages: &lt;br&gt;&#xA;1. Prove that ${\aleph_n} ^ {\aleph_1} = {2}^{\aleph_1}\cdot\aleph_n$ &lt;br&gt;&#xA;This one is pretty clear by induction and by applying Hausdorf's formula. &lt;br&gt; &lt;br&gt;&#xA;2. Prove  ${\aleph_\omega} ^ {\aleph_1} = {2}^{\aleph_1}\cdot{\aleph_\omega}^{\aleph_0}$ &lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There exist an easy proof for the second one?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance.&lt;/p&gt;&#xA;" OwnerUserId="6707" LastActivityDate="2011-02-08T19:45:14.970" Title="Cardinal equality question" Tags="&lt;set-theory&gt;&lt;cardinals&gt;" AnswerCount="1" />
  <row Id="21037" PostTypeId="2" ParentId="21017" CreationDate="2011-02-08T19:34:21.167" Score="3" ViewCount="" Body="&lt;p&gt;You can convert chandok's example to ideals in a commutative ring as well.  Take R=&lt;b&gt;Z&lt;/b&gt;[x,y,z]/(3,xx,xy,xz,yy,yz,zz), a 4-dimensional algebra over Z/3Z.  It has ideals of (vector space) dimension 1 generated by x−y, y−z, and z−x.  There is clearly a non-zero kernel of &quot;g&quot; containing the triple ( x−y, y−z, z−x ).  However, the domain of &quot;f&quot; is 0, since all pairwise intersections of the ideals are 0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The obvious translation of chandok's example to one over the ring &lt;b&gt;Z&lt;/b&gt;&lt;sup&gt;&lt;sup&gt;3&lt;/sup&gt;&lt;/sup&gt; fails, since the ideals generated by the x&lt;sub&gt;i&lt;/sub&gt; get bigger.  This ring fixes that, since it has a bunch of pointless ring elements that leave most abelian subgroups as ideals.&lt;/p&gt;&#xA;" OwnerUserId="583" LastActivityDate="2011-02-08T19:34:21.167" CommentCount="2" />
  <row Id="21038" PostTypeId="1" AcceptedAnswerId="21039" CreationDate="2011-02-08T19:36:02.113" Score="4" ViewCount="253" Body="&lt;p&gt;Most math texts claim that $\pi$ is an irrational number. However, I'm having a little bit of trouble understanding that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since nobody has calculated all of the digits of $\pi$, how can we know that either:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;one of the digits repeats (as in $\frac{10}{3}$)&lt;/li&gt;&#xA;&lt;li&gt;the number eventually terminates&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Note: Please be very descriptive in your answers... I don't have anything beyond high school math.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="4626" LastActivityDate="2011-02-08T20:37:24.830" Title="Is there a proof that $\pi$ is an irrational number?" Tags="&lt;irrational-numbers&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="21039" PostTypeId="2" ParentId="21038" CreationDate="2011-02-08T19:39:36.213" Score="5" ViewCount="" Body="&lt;p&gt;There are quite a number of &lt;a href=&quot;http://en.wikipedia.org/wiki/Proof_that_%CF%80_is_irrational&quot; rel=&quot;nofollow&quot;&gt;proofs on Wikipedia&lt;/a&gt;. Not sure how accessible you will find them, though -- feel free to ask about one of them that you might be able to understand.&lt;/p&gt;&#xA;" OwnerUserId="6622" LastEditorUserId="1269" LastEditorDisplayName="" LastEditDate="2011-02-08T20:37:24.830" LastActivityDate="2011-02-08T20:37:24.830" CommentCount="1" />
  <row Id="21040" PostTypeId="2" ParentId="21029" CreationDate="2011-02-08T19:39:48.630" Score="1" ViewCount="" Body="&lt;p&gt;I just wanted to point out that if you don't want to appeal to symmetry there is a Lagrange multiplier argument which isn't too bad: Plugging in the first equation into the second, the last 4 variables satisfy the equation&#xA;$$(7 - b - c - d - e)^2 + b^2 + c^2 + d^2 + e^2 = 10$$&#xA;Suppose you want to maximize $b$ for example. Then the gradient of $(7 - b - c - d - e)^2 + b^2 + c^2 + d^2 + e^2$ has to be a multiple of the gradient of $b$, which means there is some $\lambda$ such that&#xA;$$-2(7 - b - c - d - e) + 2b = \lambda$$&#xA;$$-2(7 - b - c - d - e) + 2c = 0$$&#xA;$$-2(7 - b - c - d - e) + 2d = 0$$&#xA;$$-2(7 - b - c - d - e) + 2e = 0$$&#xA;But the last three equations are the same as $-2a + 2c = 0$, $-2a + 2d = 0$, and&#xA;$-2a + 2e = 0$, which lead to $a = c = d = e$. Thus the $4$ variables other than $b$ are the same and one can now proceed as in Sivaram's argument.&lt;/p&gt;&#xA;" OwnerUserId="3035" LastActivityDate="2011-02-08T19:39:48.630" />
  <row Id="21041" PostTypeId="1" AcceptedAnswerId="21073" CreationDate="2011-02-08T19:40:40.380" Score="2" ViewCount="118" Body="&lt;p&gt;I've two questions&#xA;(1) How do I determine the distribution of the first order and the highest order statistic for sample of random size N is taken from the continuous uniform(0, $\theta$) and  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation} \rm{P} (N = n) = \frac{1}{n! (e − 1)} \text{for n = 1, 2, 3, . . . .} \end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) In general without any distribution of any random variables given, the distribution of highest order statistic is $F_{X(n)}(a) = [F_{X}(a)]^n$ and the distribution of the lowest order statistic is $F_{X(1)}(a) = 1 - [1 - F_X(a)]^n$. I understand this derivation. But what does these values say about the distribution ?&lt;/p&gt;&#xA;" OwnerUserId="6335" LastActivityDate="2011-02-08T22:52:51.310" Title="Order statistic" Tags="&lt;statistics&gt;&lt;probability-distributions&gt;" AnswerCount="2" />
  <row Id="21042" PostTypeId="2" ParentId="21029" CreationDate="2011-02-08T19:40:40.313" Score="1" ViewCount="" Body="&lt;p&gt;So you want $x\in \mathbb{R}^5$ such that $\| x\|^2 = 10$  and $(x,1)=7$.  (Here 1 refers to the all 1's vector)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Geometrically, we are looking at a circle around the all 1's vector in $\mathbb{R}^5$, and we just need to find the extremes of the circle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the above discussion it should be clear that the maximum and minimum will be close to $\sqrt {2}$, but the question is how much can it deviate?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now lets actually solve the problem:  Consider $a=b=c=d$ so our equations are in two variables, $$4a+e=7$$&lt;br&gt;&#xA;$$4a^2+e^2=10$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solving $a=\frac{7-e}{4}$ and plugging into the second equation we get the quadratic equation:  $$5e^2-14e+9=0$$  Which factors nicely as $$(5e-9)(e-1)=0$$  So either $e=1$ or $e=1.8$.  I now claim that these are indeed the minimum and maximum values, but I will leave that for you to prove.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint:  The vector $x$ in $\mathbb {R}^n$ with $\|x\|\leq r$ that maximizes $(x,1)$ will be in the direction of $1$ (the all 1's vector).  In other words, maximizing $a+b+c+d$ given $a^2+b^2+c^2+d^2&amp;lt;10-1.8^2$ occurs when $a=b=c=d$.  Similarly for minimizing.&lt;/p&gt;&#xA;" OwnerUserId="6075" LastActivityDate="2011-02-08T19:40:40.313" />
  <row Id="21043" PostTypeId="2" ParentId="11551" CreationDate="2011-02-08T19:43:33.070" Score="2" ViewCount="" Body="&lt;p&gt;Here is some progress in the positive direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Theorem.&lt;/b&gt; In any Hausdorff space, the closure of a&#xA;countable set has size at most $2^{\mathfrak{c}}$, where $\mathfrak{c}$ is the&#xA;continuum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proof. Suppose that $X$ is a Hausdorff topological space&#xA;with a countable set $D$. For any point $a$ in the closure&#xA;$\bar D$, we may consider the collection of open sets&#xA;containing $a$, and their trace on $D$. That is, consider&#xA;$F_a=\{U\cap D\mid a\in U\text{ open }\}$. Since there are&#xA;only continuum many subsets of $D$, there are therefore at&#xA;most $2^{\mathfrak{c}}$ many possible such families.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the closure $\bar D$ had size larger than $2^{\mathfrak{c}}$, then&#xA;there would be at least two (in fact many) distinct points&#xA;$a$ and $b$ in $\bar D$ for which $F_a=F_b$. Let $U$ and&#xA;$V$ be disjoint neighborhoods of $a$ and $b$. Let $U_1$ be&#xA;another neighborhood of $a$ such that $U_1\cap D=V\cap D$,&#xA;which must exist since $F_a=F_b$. Thus, $U\cap U_1$ is a&#xA;neighborhood of $a$ that is disjoint from $D$,&#xA;contradicting $a\in\bar D$. QED&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The bound is sharp, even for compact Hausdorff spaces, in&#xA;the sense that the &lt;a href=&quot;http://en.wikipedia.org/wiki/Stone%E2%80%93%C4%8Cech_compactification&quot; rel=&quot;nofollow&quot;&gt;Stone–Čech &#xA;compactification&lt;/a&gt;&#xA;$\beta\mathbb{N}$ is a Hausdorff topological space of size&#xA;$2^c$ with a countable dense set. But $\beta\mathbb{N}$ is&#xA;not sequentially compact, so this is not actually a&#xA;counterexample to your question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What the argument actually shows is that if $a$ and $b$ are&#xA;in the closure of the countable set $D$, then $F_b$ is not a&#xA;subset of $F_a$. If it were, we could find disjoint&#xA;neighborhoods $U$ and $V$ of $a$ and $b$, respectively, and&#xA;then $V\cap D\in F_b$, and so there is a neighborhood $U_1$&#xA;of $a$ with $U_1\cap D=V\cap D$, making $U\cap U_1$ a&#xA;neighborhood of $a$ having no points from $D$, a&#xA;contradiction. By symmetry, we conclude $F_a$ and $F_b$ are&#xA;incomparable with respect to $\subset$ for all $a,b\in\bar&#xD;&#xA;D$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suspect that such a line of reasoning could be improved&#xA;when there is sequential compactness, perhaps by using a&#xA;cardinal characteristic, such as the splitting number.&lt;/p&gt;&#xA;" OwnerUserId="413" LastEditorUserId="413" LastEditorDisplayName="" LastEditDate="2011-02-08T19:48:36.590" LastActivityDate="2011-02-08T19:48:36.590" CommentCount="1" />
  <row Id="21044" PostTypeId="2" ParentId="21036" CreationDate="2011-02-08T19:45:14.970" Score="5" ViewCount="" Body="&lt;p&gt;As you mention, the first equation is a consequence of Hausdorff's formula and induction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the second: Clearly the right hand side is at most the left hand side. Now: Either $2^{\aleph_1}\ge\aleph_\omega$, in which case in fact $2^{\aleph_1}\ge{\aleph_\omega}^{\aleph_1}$, and we are done, or $2^{\aleph_1}&amp;lt;\aleph_\omega$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I claim that in this case we have ${\aleph_\omega}^{\aleph_1}={\aleph_\omega}^{\aleph_0}$. Once we prove this, we are done.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that $\aleph_\omega={\rm sup}_n\aleph_n\le\prod_n\aleph_n$, so&#xA;$$ {\aleph_\omega}^{\aleph_1}\le\left(\prod_n\aleph_n\right)^{\aleph_1}=\prod_n{\aleph_n}^{\aleph_1}. $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now use part 1, to conclude that ${\aleph_n}^{\aleph_1}&amp;lt;\aleph_\omega$ for all $n$, since we are assuming that $2^{\aleph_1}&amp;lt;\aleph_\omega$. This shows that the last product is at most $\prod_n \aleph_\omega={\aleph_\omega}^{\aleph_0}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This shows that ${\aleph_\omega}^{\aleph_1}\le {\aleph_\omega}^{\aleph_0}$. But the other inequality is obvious, and we are done. &lt;/p&gt;&#xA;" OwnerUserId="462" LastActivityDate="2011-02-08T19:45:14.970" />
  <row Id="21045" PostTypeId="2" ParentId="20122" CreationDate="2011-02-08T19:56:31.557" Score="1" ViewCount="" Body="&lt;p&gt;I suspect that the restriction &quot;if spots x+1,x+2 and x−1 are empty&quot; turns the problem more complex without adding anything conceptually relevant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't claim to have an answer, but want to point that the model has some resemblance to some physical models. In particular, consider the Tonk gas (or &quot;&lt;a href=&quot;http://www.sklogwiki.org/SklogWiki/index.php/1-dimensional_hard_rods&quot; rel=&quot;nofollow&quot;&gt;hard rods&lt;/a&gt;&quot; model). Here, we have one dimensional particles of a fixed width that move in a line, and only interact through hard elastic colitions. One way to simulate this statistical models (not is dynamics, but its average configurations) is through &lt;a href=&quot;http://en.wikipedia.org/wiki/Monte_Carlo_molecular_modeling&quot; rel=&quot;nofollow&quot;&gt;Montecarlo&lt;/a&gt; simulations. In this procedure, we produce random configurations by some rule that produces a representative sampling of the ensemble - in the hard-rods model a possible rule would be to displace the particles (one at a time or all simultaneously) a random distance and &quot;accept&quot; the new configuration iff no particles overlap. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the model presented here has (to me) some resemblance  to such a Montecarlo simulation of a system of hard rods with a &quot;drift&quot; added. I'm not sure if that similarity can be exactly worked out, but I suspect so - and also I suspect that there is no &quot;phase separation&quot; as you seem to expect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I made some little simulation here (only modern browsers with canvas support: Chrome, Firefox - perhaps IE 9)&#xA;&lt;a href=&quot;http://hjg.com.ar/varios/mat/queuesimul.html&quot; rel=&quot;nofollow&quot;&gt;http://hjg.com.ar/varios/mat/queuesimul.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;UPdate: sorry, I had misread the model. My thoughts (and simulation) correspond to another model (people move if probability 'a' if forward neighbours are empty )&lt;/p&gt;&#xA;" OwnerUserId="312" LastEditorUserId="312" LastEditorDisplayName="" LastEditDate="2011-02-09T18:52:00.280" LastActivityDate="2011-02-09T18:52:00.280" CommentCount="8" />
  <row Id="21046" PostTypeId="1" AcceptedAnswerId="21054" CreationDate="2011-02-08T20:04:06.607" Score="1" ViewCount="135" Body="&lt;p&gt;Hello guys!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am trying to factorize $-6x^5+15x^4-30x^2+30x-13$ for hours:( Could someone help me? I tried making a system of equations from $(Ax^3 + Bx^2 + Cx + D) (Ex^2 + Fx + G)$ but it is a nightmare:(&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In case you are interested, the system is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$AE = -6$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$AF + BE = 15$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$AG + BF + CE = 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$BG + CF + DE = -30$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$CG + DF = 30$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$DG = -13$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: The original task is to draw the following function: $\ln \dfrac{x^2 - 3x + 2}{x^2 + 1}$. The polynomial above is the numerator of the second derivative of the function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Best regards,&#xA;Petar&lt;/p&gt;&#xA;" OwnerUserId="6698" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2011-02-08T22:02:13.857" LastActivityDate="2011-02-08T23:04:20.057" Title="Factorize polynomial" Tags="&lt;polynomials&gt;" AnswerCount="2" CommentCount="7" />
  <row Id="21047" PostTypeId="2" ParentId="21027" CreationDate="2011-02-08T20:04:35.003" Score="1" ViewCount="" Body="&lt;p&gt;Well, take $H=\{(x_0:\cdots:x_n):x_0\neq0\}$ and show that $P_n\setminus H\cong\mathbb A^n$ by explicitly writing down an isomorphism.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(You did not tell us anything about what you know, so it is difficult to be more explicit here... For example, do you think of varieties as schemes, as algebraic subsets of $\mathbb A^n$$?)&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2011-02-08T20:04:35.003" />
  <row Id="21048" PostTypeId="2" ParentId="21041" CreationDate="2011-02-08T20:11:46.643" Score="4" ViewCount="" Body="&lt;p&gt;The idea is to consider simultaneously the minimum $Y$ and the maximum $Z$ of the sample. For $y\le z$, the event $[y\le Y,Z\le z]$ is equivalent to the whole sample being between $y$ and $z$, hence, for a sample of size $n$, $P(y\le Y,Z\le z)$ would be $P(y\le X\le z)^n$. Here one considers a sample of random size $N$,&#xA;hence&#xA;$$&#xD;&#xA;P(y\le Y,Z\le z)=\sum_{n\ge1}P(N=n)P(y\le X\le z)^n=c(\mathrm{e}^{P(y\le X\le z)}-1),&#xD;&#xA;$$&#xA;with&#xA;$$&#xD;&#xA;c=1/(\mathrm{e}-1).&#xD;&#xA;$$&#xA;To get the joint distribution of $(Y,Z)$, one should differentiate $P(y\le Y,Z\le z)$ twice, with respect to $y$ and $z$, yielding&#xA;$$&#xD;&#xA;P(Y\in\mathrm{d}y,Z\in\mathrm{d}z)=c\mathrm{e}^{P(y\le X\le z)}P(X\in\mathrm{d}y)P(X\in\mathrm{d}z).&#xD;&#xA;$$&#xA;To get the distribution of $Y$ alone is even simpler, one differentiates $P(y\le Y,Z\le z)$ once with respect to $y$ and one lets $z\to+\infty$, hence&#xA;$$&#xD;&#xA;P(Y\in\mathrm{d}y)=c\mathrm{e}^{P(X\ge y)}P(X\in\mathrm{d}y).&#xD;&#xA;$$&#xA;Likewise, to get the distribution of $Z$ alone, one differentiates $P(y\le Y,Z\le z)$ once with respect to $z$ and one lets $y\to-\infty$, hence&#xA;$$&#xD;&#xA;P(Z\in\mathrm{d}z)=c\mathrm{e}^{P(X\le z)}P(X\in\mathrm{d}z).&#xD;&#xA;$$&#xA;In the special case where the sample is uniform on $(0,\theta)$, for $0\le y\le z\le\theta$, the density of the distribution of $(Y,Z)$ at $(y,z)$ is&#xA;$$&#xD;&#xA;f_{Y,Z}(y,z)=(c/\theta^{2})\mathrm{e}^{(z-y)/\theta}.&#xD;&#xA;$$&#xA;Finally, the densities of the distributions of $Y$ and $Z$ on $(0,\theta)$ are&#xA;$$&#xD;&#xA;f_Y(y)=(c\mathrm{e}/\theta)\mathrm{e}^{-y/\theta},&#xD;&#xA;\quad&#xD;&#xA;f_Z(z)=(c/\theta)\mathrm{e}^{z/\theta}.&#xD;&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="6179" LastEditorUserId="6179" LastEditorDisplayName="" LastEditDate="2011-02-08T22:07:55.113" LastActivityDate="2011-02-08T22:07:55.113" CommentCount="8" />
  <row Id="21049" PostTypeId="2" ParentId="21027" CreationDate="2011-02-08T20:16:15.690" Score="1" ViewCount="" Body="&lt;p&gt;Start by showing that you may assume the hyperplane is of form $H\colon x_n=0$ (where the points in $\mathbb{P}^n$ are $(x_0\colon\cdots\colon x_n)$. That will simplify things.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then you want to define a bijection between $\mathbb{A}^n$ and $\mathbb{P}^n-H$. There is a standard embedding of affine space into projective space, so you can start there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, the trick is to show that this bijection is in fact a homeomorphism in the Zariski topology. This involves making a correspondence between zeros of homogeneous polynomials in, say, $z_0,\ldots,z_n$, and zeros of polynomials in $z_0,\ldots,z_{n-1}$ (to make closed sets in $\mathbb{P}^n$ corresponds to closed sets in $\mathbb{A}^n$). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do you know any way to take a polynomial in $k$ variables and use it to define a set of zeroes in projective $k$-space? (&lt;em&gt;Hint:&lt;/em&gt; If your milk doesn't have clumps of cream in it, then it is &lt;code&gt;&amp;lt;blanked&amp;gt;&lt;/code&gt;.) And how would take a homogeneous polynomial in $k+1$ variables and use it to define a set of zeros in affine $k$-space? &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-08T21:04:02.467" LastActivityDate="2011-02-08T21:04:02.467" CommentCount="2" />
  <row Id="21050" PostTypeId="2" ParentId="21038" CreationDate="2011-02-08T20:24:44.657" Score="4" ViewCount="" Body="&lt;p&gt;If you know a bit of calculus and have come across induction, then here's an outline of a standard exercise (see Burkill - A first Course in Analysis) to prove $\pi$ irrational.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$I_n(\alpha)=\int_{-1}^1 (1-x^2)^n \cos \alpha x \textrm{ d}x$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;then integrate by parts to show that for $n \ge 2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\alpha^2 I_n = 2n(2n-1)I_{n-1}-4n(n-1)I_{n-2}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Use induction to show that for positive integer $n$ we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\alpha^{2n+1}I_n(\alpha)=n!(P(\alpha) \sin \alpha + Q(\alpha) \cos \alpha),$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $P(\alpha)$ and $Q(\alpha)$ are polynomials of degree less than $2n+1$ in $\alpha$ with integral coefficients.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Show that if $\pi/2 = b/a,$ where $a$ and $b$ are integers, then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{b^{2n+1}I_n(\pi/2)}{n!} \quad (1)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;would be an integer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$I_n(\pi/2) &amp;lt; \int_{-1}^1 (1-x^2)^n \textrm{ d}x &amp;lt; 2 \textrm{ and } &#xD;&#xA;\frac{b^{2n+1}}{n!} \rightarrow 0 \textrm{ as } n \rightarrow \infty$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which results in contradiction since $(1)$ is supposed to be an integer but we can show that it is as small as one desires.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This was the first proof of the irrationality of $\pi$ that I came across, and think it is very accessible for those willing to give it a go. &lt;/p&gt;&#xA;" OwnerUserId="1301" LastActivityDate="2011-02-08T20:24:44.657" CommentCount="1" />
  <row Id="21051" PostTypeId="2" ParentId="21046" CreationDate="2011-02-08T20:27:01.720" Score="2" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\ $ If you let $\rm\ z = x-1\ $ then the 2nd derivative is $\rm - 6 \ z^5 - 15\ z^4 - 4\ $ so the roots satisfy $\rm 6\ z + 15 = -4/z^4\ $ which one easily sees has a &lt;em&gt;unique&lt;/em&gt; real root roughly $\rm\ z\ \approx\: -2.5$&#xA;&lt;img src=&quot;http://i.imgur.com/VM10D.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2011-02-08T21:17:35.310" LastActivityDate="2011-02-08T21:17:35.310" CommentCount="5" />
  <row Id="21052" PostTypeId="1" AcceptedAnswerId="21058" CreationDate="2011-02-08T20:38:44.073" Score="3" ViewCount="71" Body="&lt;p&gt;Hope I'm not spamming too much by asking questions on separate threads. &lt;br&gt; &lt;br&gt;&#xA;I have 2 more questions, not connected one to another, in any way: &lt;br&gt;&#xA;1. Show that every stationary set in $\aleph_1$, contains, for every $\alpha&amp;lt;\omega_1$, a closed set of otp $\alpha$. &lt;br&gt;&#xA;2. Show without using AC, that there exist a surjective function from $\mathbb{R}$ on $\aleph_1$. &lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second one seems pretty obvious, so I'm not sure what I need to prove.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="6707" LastEditorUserId="6707" LastEditorDisplayName="" LastEditDate="2011-02-08T21:06:03.060" LastActivityDate="2011-02-08T22:12:18.477" Title="Closed subset of stationary set and AC questions" Tags="&lt;set-theory&gt;&lt;axiom-of-choice&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="21054" PostTypeId="2" ParentId="21046" CreationDate="2011-02-08T21:35:57.680" Score="3" ViewCount="" Body="&lt;p&gt;First note that the domain over which the function makes sense in real variables is when $x^2 - 3x + 2 &amp;gt; 0$ i.e. when $(x-1)(x-2) &amp;gt; 0$ i.e. when $x &amp;gt; 2$ or $x &amp;lt; 1$. Now the way out is to rewrite &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\log \frac{x^2 - 3x + 2}{x^2 + 1}$ as $\log (x-1) + \log (x-2) - \log (x^2+1)$ if $x &amp;gt; 2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\log \frac{x^2 - 3x + 2}{x^2 + 1}$ as $\log (1-x) + \log (2-x) - \log (x^2+1)$ if $x &amp;lt; 1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The individual plots for the three terms can be drawn trivially. All you need to do now is to superpose these three together.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could do further analysis which will help you with your plotting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, when $\frac{x^2-3x+2}{x^2+1} \geq 1$, the function will be non-negative and when $\frac{x^2-3x+2}{x^2+1} &amp;lt; 1$, the function will be negative.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the function will be non-negative when $\frac{x^2-3x+2}{x^2+1} \geq 1 \Rightarrow -3x+2 \geq 1 \Rightarrow x \leq \frac{1}{3}$ and will be negative when $x &amp;gt; \frac{1}{3}$. The zero crossing is at $x= \frac{1}{3}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As mentioned previously the function is not-defined for $1 \leq x \leq 2$. And the function tends to $-\infty$ as $x \rightarrow 2^+$ or as $x \rightarrow 1^{-}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Further as $x \rightarrow \pm \infty$, the function tends to $0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The derivative when $x&amp;gt;2$ is $-\frac{2x}{x^2+1} + \frac{1}{x-2} + \frac{1}{x-1} &amp;gt; - \frac{2x}{x^2} + \frac{1}{x} + \frac{1}{x} = 0$ when $x&amp;gt;2$. So in the domain $(2,\infty)$ we have the function to be increasing and $f(2^+) = - \infty$ and $\displaystyle \lim_{x \rightarrow \infty}f(x) = 0$. Hence, $f(x) &amp;lt; 0$, $\forall x \in (2, \infty)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we have $f(x)$ negative and it increases from $-\infty$ to $0$ in the domain $(2,\infty)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the domain $(-\infty,1)$, we know that $\displaystyle \lim_{x \rightarrow -\infty}f(x) = 0$ and $f(1^-) = -\infty$ and we know that there is only one zero crossing, which means there has to be at least one maximum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Setting the derivative to zero, we get a quadratic in $x$ which gives $x = \frac{1 \pm \sqrt{10}}{3}$. The positive root falls in $[1,2]$ and hence can be ruled out. The negative root is where the maximum occurs. And there is only one maximum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Further, $x=0$ gives $f(0) = \log (2) &amp;gt; 0$ as expected since $f(\frac{1}{3}) = 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the summary is,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The function increases from $0$ to $f(\frac{1 - \sqrt{10}}{3})$ in the domain $(-\infty,\frac{1 - \sqrt{10}}{3})$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The function decreases from $f(\frac{1 - \sqrt{10}}{3})$ to $-\infty$ in the domain $[f(\frac{1 - \sqrt{10}}{3}),1)$ with zero crossing at $x = \frac{1}{3}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The function is not defined in the domain $[1,2]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The function $f(x)$ negative and it increases from $-\infty$ to $0$ in the domain $(2,\infty)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This information should be enough to help you make a sketch of the plot.&#xA;&lt;img src=&quot;http://i.imgur.com/rXNEP.png&quot; alt=&quot;Plot of the function using grapher&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2011-02-08T23:04:20.057" LastActivityDate="2011-02-08T23:04:20.057" CommentCount="4" />
  <row Id="21055" PostTypeId="1" AcceptedAnswerId="21057" CreationDate="2011-02-08T21:36:27.147" Score="8" ViewCount="183" Body="&lt;p&gt;The sum $$ I + A + A^2 + A^3 + \cdots $$ equals $$(I-A)^{-1}$$ under the assumption $\rho(A)&amp;lt;1$, which is necessary to make the sum converge. My question: what does the sum &#xA;$$ I + A^T A + (A^2)^T A^2 + (A^3)^T A^3 + \cdots$$ equal under the same assumption? Is there a similarly neat expression?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is true that this last sum converges under the assumption $\rho(A)&amp;lt;1$. The ``obvious'' guess $(I-A^T A)^{-1}$ for the sum is not true because $ (A^2)^T A^2 \neq (A^T A)^2$; in fact, this guess does not even make sense because $\rho(A)&amp;lt;1$ does not rule out $\rho(A^T A)=1$. &lt;/p&gt;&#xA;" OwnerUserId="6760" LastEditorUserId="6760" LastEditorDisplayName="" LastEditDate="2011-02-08T22:15:46.643" LastActivityDate="2011-02-09T14:13:20.997" Title="A ``geometric'' infinite sum of matrices" Tags="&lt;linear-algebra&gt;" AnswerCount="3" FavoriteCount="1" />
  <row Id="21057" PostTypeId="2" ParentId="21055" CreationDate="2011-02-08T21:42:39.803" Score="8" ViewCount="" Body="&lt;p&gt;FWIW: Call $S$ the sum, provided it makes sense. Then $A^TSA=S-I$, so the series converges to a root of this equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $n=2$, then the determinant of the endomorphism $S\mapsto A^TSA-S$ of $M_n(\mathbb C)$ is $(\det A-1)^2\det(A^2-I)$, so for most $A$s, the solution is unique. For larger $n$, I have no idea.&lt;/p&gt;&#xA;" OwnerUserId="274" LastEditorUserId="274" LastEditorDisplayName="" LastEditDate="2011-02-08T21:56:30.530" LastActivityDate="2011-02-08T21:56:30.530" CommentCount="2" />
  <row Id="21058" PostTypeId="2" ParentId="21052" CreationDate="2011-02-08T21:46:01.500" Score="2" ViewCount="" Body="&lt;p&gt;For question 2, one way is as follows: We can identify the set ${\mathbb R}\setminus{\mathbb Q}$ of irrationals with the set ${}^\omega\omega$ of functions from $\omega$ to $\omega$, for example, by looking at continued fraction representations. (There are other ways, just playing with the decimal expansions, for example.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, each natural can be seen as coding two, say $a$ codes $(b,c)$ iff $a+1=2^b(2c+1)$. (Again, there are other ways.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The point is that this gives us a way to associate to each irrational $r$ a binary relation on $\omega$, namely we have associated $r$ with the set $\{(b_0,c_0),(b_1,c_1),\dots\}$. Here, first $r$ was associated with a sequence $(a_0,a_1,\dots)$, and then each $a_i$ was identified with a pair $(b_i,c_i)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ok. &lt;em&gt;If&lt;/em&gt; the relation associated to $r$ is a well-ordering of $\omega$ of type $\omega+\alpha$, &lt;em&gt;then&lt;/em&gt; map $r$ to $\alpha$. &lt;em&gt;Else&lt;/em&gt;, map $r$ to 0. This is a surjection of ${\mathbb R}$ onto $\aleph_1$. We did not use choice. That it is surjective comes from noting that for any $\alpha&amp;lt;\omega_1$ there is a well-ordering of $\omega$ in type $\omega+\alpha$, and the codings described above are reversible, so any such well-ordering is the image of some $r$ under the coding we described.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The set of $r$ that code well-orderings is usually called WO and plays an important role in descriptive set theory; the construction I described is &quot;canonical&quot; in some sense. On the other hand, without choice we cannot prove the dual result that there is an injection from $\omega_1$ into ${\mathbb R}$.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;For question 1, one argues by induction that a stationary set contains closed copies of $\alpha+1$ for any $\alpha&amp;lt;\omega_1$. The following is from a set of notes I wrote for a course I taught a while ago:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $ S\subseteq\omega_1$ be a given stationary set, and argue by induction on $ \alpha&amp;lt;\omega_1$ that $ S$ contains closed copies $ t$ of $ \alpha+1$ with $ \min(t)$ arbitrarily large. (Of course, if the result holds, this must be the case: Given any $ \gamma&amp;lt;\omega+1$, notice that $ S\setminus(\gamma+1)$ is stationary, so it must contain a closed copy $ t$ of $ \alpha+1$, and $ \min(t)&amp;lt;\gamma$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This strengthened version holds trivially for $ \alpha$ finite or successor, by induction. So it suffices to show it for $ \alpha$ limit, assuming it holds for all smaller ordinals. Define a club $ C\subseteq\omega_1$ with increasing enumeration $ \{\gamma_\beta:\beta&amp;lt;\omega_1\}$ as follows: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $ (\alpha_n:n&amp;lt;\omega)$ be strictly increasing and cofinal in $ \alpha$. Since $ S$ contains closed copies $ A_n$ of $ \alpha_n+1$ for all $ n$, with their minima arbitrarily large, by choosing such copies $ A_n$ with $ \min(A_{n+1})&amp;gt;\max(A_n)$ and taking their union, we see that $ S$ must contain copies of $ \alpha$, &lt;em&gt;closed in their supremum&lt;/em&gt;, with arbitrarily large minimum element. (I am not claiming that $ A=\bigcup_n A_n$ built this way has order type $ \alpha$. For example, if $ \alpha=\omega+\omega$ and $ \alpha_n=\omega+n$, then $ A$ would have order type $ \omega^2$; but for sure $ A\subset S$ is closed in its supremum and has order type &lt;em&gt;at least&lt;/em&gt; $ \alpha$. So a suitable initial segment of $ A$ is as wanted.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $ \gamma_0$ be the supremum of such a copy of $ \alpha$. At limit ordinals $ \beta$, let $ \gamma_\beta=\sup_{\delta&amp;lt;\beta}\gamma_\delta$. Once $ \gamma_\beta$ is defined, find such a copy of $ \alpha$ inside $ S$ with minimum larger than $ \gamma_\beta$, and let $ \gamma_{\beta+1}$ be its supremum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The set $ C$ so constructed is club, so it meets $ S$. If they meet in $ \gamma_0$ or in a $ \gamma_{\beta+1}$, this immediately gives us a closed copy of $ \alpha+1$ inside $ S$. If they meet in a $ \gamma_\beta$ with $ \beta$ limit, let $ (\beta_n:n&amp;lt;\omega)$ be strictly increasing and cofinal in $ \beta$, and consider an appropriate initial segment of $ A=(\bigcup_n A_n)\cup\{\gamma_\beta\}$, where $ A_n$ is a closed copy of $ \alpha_n+1$ in $ S\cap[\gamma_{\beta_n},\gamma_{\beta_{n+1}})$.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Let me add that if one is familiar with the method of forcing, there is a nice argument for question 1. (This is perhaps overkill, but a nice one in any case.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Namely, given any stationary subset $S$ of $\omega_1$, there is a forcing poset that adds a club subset of $S$ while not adding any new countable sequences of countable ordinals. This means that the proper initial segments of the club are subsets of $S$ that are in $V$, and of course are closed in their supremum, so this immediately gives the result.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The poset is the natural one: Conditions are closed initial segments of $S$, and the order is end-extension. It takes a bit of work to see that this poset indeed does the work; this is usually presented in the context of proper forcing, since this is an example of an $S$-proper poset. This is an argument that goes back to the mid-seventies and is due to Baumgartner-Harrington-Kleinberg. It is specific to $\omega_1$, there are no natural generalizations for stationary subsets of larger cardinals.&lt;/p&gt;&#xA;" OwnerUserId="462" LastEditorUserId="462" LastEditorDisplayName="" LastEditDate="2011-02-08T22:12:18.477" LastActivityDate="2011-02-08T22:12:18.477" />
  <row Id="21059" PostTypeId="2" ParentId="21055" CreationDate="2011-02-08T21:46:21.933" Score="3" ViewCount="" Body="&lt;p&gt;If $F:M_n\to M_n$ is defined by $F(X)=A^\mathrm{T}XA$ and $\mathrm{id}:M_n\to M_n$ is the identity map, then your sum is $(\mathrm{id} - F)^{-1}(I)$.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2011-02-08T21:46:21.933" CommentCount="2" />
  <row Id="21060" PostTypeId="1" AcceptedAnswerId="21065" CreationDate="2011-02-08T22:16:33.793" Score="0" ViewCount="77" Body="&lt;p&gt;I am reading a cryptography paper, and the authors introduce a function $f(x) = x^{-\omega(1)}$ and call it a negligible function in $x$. What is the possible meaning of this?&lt;/p&gt;&#xA;" OwnerUserId="2468" LastActivityDate="2011-02-08T22:30:57.137" Title="Notation: What does $f(x) = x^{-\omega(1)}$ mean?" Tags="&lt;notation&gt;&lt;asymptotics&gt;" AnswerCount="1" />
  <row Id="21061" PostTypeId="2" ParentId="20907" CreationDate="2011-02-08T22:17:34.317" Score="1" ViewCount="" Body="&lt;p&gt;I got an answer from a professor in person today, so I will post it.  Please let me know if something is wrong with this argument, or if you don't think it is a more direct proof of (2) than using Baer's criterion. Since $\mathbb{Z}$ is a P.I.D., we may take a two stage projective (and free) resolution of any module $M$:&#xA;$$0 \rightarrow P_1 \rightarrow P_0 \rightarrow M \rightarrow 0$$&#xA;Suppose we have a homomorphism $\phi: P_1 \rightarrow \mathbb{Q}$.  We will tensor everything with $\mathbb{Q}$.  We require the fact that $\mathbb{Q}$ is flat, but it seems to me that the proof of this fact is more basic than the proof of Baer's criterion.  After tensoring, we have the following exact sequence&#xA;$$0 \rightarrow P_1 \otimes_\mathbb{Z} \mathbb{Q} \rightarrow P_0 \otimes_\mathbb{Z} \mathbb{Q} \rightarrow ...$$&#xA;With a homomorphism $P_1 \otimes_\mathbb{Z} \mathbb{Q} \rightarrow \mathbb{Q} \otimes_\mathbb{Z} \mathbb{Q} \cong \mathbb{Q}$.  Now treating $P_1 \otimes_\mathbb{Z} \mathbb{Q}$ as a $\mathbb{Q}$-subspace of $P_0 \otimes_\mathbb{Z} \mathbb{Q}$, we have a homomorphism on the subspace, which may then be extended on the whole space (just take a basis for the subspace, and extend it to the whole space, sending all of the nonsubspace basis vectors to 0).  We have thus found the map $P_0 \rightarrow P_0 \otimes_\mathbb{Z} \mathbb{Q}\rightarrow \mathbb{Q}$ that we were looking for.  &lt;/p&gt;&#xA;" OwnerUserId="5265" LastActivityDate="2011-02-08T22:17:34.317" />
  <row Id="21062" PostTypeId="1" CreationDate="2011-02-08T22:22:01.360" Score="3" ViewCount="144" Body="&lt;p&gt;Does anyone know if there exists any small-scale game-theoretical models of the game of contract bridge, similar to the many simplified versions of poker that have been analysed?&lt;/p&gt;&#xA;" OwnerUserId="4623" LastActivityDate="2011-02-09T07:46:17.837" Title="game theory models of bridge" Tags="&lt;game-theory&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="21063" PostTypeId="1" AcceptedAnswerId="21064" CreationDate="2011-02-08T22:25:48.953" Score="2" ViewCount="248" Body="&lt;p&gt;Hi!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming I have a always positive random variable $X$, $X \in \mathbb{R}$, $X &amp;gt; 0$. Then  I am now interested in the difference between the following two expectation values:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$E \left[ \ln X \right]$&lt;/li&gt;&#xA;&lt;li&gt;$\ln E \left[ X \right]$&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Is one maybe always a lower/upper bound of the other?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many thanks in advance...&lt;/p&gt;&#xA;" OwnerUserId="6197" LastActivityDate="2011-02-13T20:23:55.467" Title="Difference between logarithm of an expectation value and expectation value of a logarithm" Tags="&lt;probability&gt;&lt;random&gt;&lt;estimation-theory&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="21064" PostTypeId="2" ParentId="21063" CreationDate="2011-02-08T22:30:32.143" Score="10" ViewCount="" Body="&lt;p&gt;Since the function logarithm is concave, &lt;a href=&quot;http://en.wikipedia.org/wiki/Jensen%27s_inequality&quot; rel=&quot;nofollow&quot;&gt;Jensen's inequality&lt;/a&gt; shows that $E(\ln(X))\le \ln E(X)$ and that equality occurs iff $X$ is almost surely constant.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt; (This is to expand on a remark made by Shai.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Shai's answer explains how to prove $E(\ln(X))\le \ln E(X)$ using only AM-GM inequality and the strong law of large numbers. These very tools yield the following refinement (adapted from J. M. Aldaz, &lt;a href=&quot;http://jmi.ele-math.com/03-21&quot; rel=&quot;nofollow&quot;&gt;Self-improvement of the inequality between arithmetic and geometric means&lt;/a&gt;). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Apply AM-GM inequality to the &lt;em&gt;square roots&lt;/em&gt; of an i.i.d. sequence of positive random variables $(X_i)$, that is,&#xA;$$&#xD;&#xA;\sqrt[n]{\sqrt{X_1}\cdots\sqrt{X_n}}\le\frac1n(\sqrt{X_1}+\cdots+\sqrt{X_n}).&#xD;&#xA;$$&#xA;In the limit $n\to\infty$, the strong law of large numbers yields&#xA;$$&#xD;&#xA;\exp(E(\ln\sqrt{X}))\le E(\sqrt{X}),&#xD;&#xA;$$&#xA;that is,&#xA;$$&#xD;&#xA;E(\ln X)\le 2\ln E(\sqrt{X})=\ln (E(X)-\mbox{var}(\sqrt{X})).&#xD;&#xA;$$&#xA;Finally:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt; &#xA;For every positive integrable $X$, &#xA;$$&#xD;&#xA;E(\ln X)\le [\ln E(X)]-\delta(X)\quad\mbox{where}\ \delta(X)=\ln[E(X)/E(\sqrt{X})^2].&#xD;&#xA;$$&#xA;&lt;/blockquote&gt; &#xA;&#xA;&lt;p&gt;The correction term $\delta(X)$ is nonnegative for every $X$, and $\delta(X)=0$ iff $X$ is almost surely constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Naturally, this also obtains directly through Jensen's inequality applied to $\sqrt{X}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And this result is just a special case of the fact that, for every $s$ in $(0,1)$,&#xA;$$&#xD;&#xA;E(\ln X)\le [\ln E(X)]-\delta_s(X)\quad\mbox{where}\ \delta_s(X)=\ln[E(X)/E(X^s)^{1/s}].&#xD;&#xA;$$&#xA;The quantity $\delta_s(X)$ is a nonincreasing function of $s$ hence the upper bound $[\ln E(X)]-\delta_s(X)$ is better and better when $s$ decreases to $0$. For every $X$, $\delta_1(X)=0$, $\delta_{1/2}(X)=\delta(X)$ and $\delta_0(X)=[\ln E(X)]-E(\ln X)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The interesting point in all this is that one has &lt;em&gt;quantified&lt;/em&gt; the discrepancy between $E(\ln X)$ and $\ln E(X)$ and, simultaneously, recovered the fact that $E(\ln X)=\ln E(X)$ iff $X$ is almost surely constant.&lt;/p&gt;&#xA;" OwnerUserId="6179" LastEditorUserId="6179" LastEditorDisplayName="" LastEditDate="2011-02-13T20:23:55.467" LastActivityDate="2011-02-13T20:23:55.467" CommentCount="2" />
  <row Id="21065" PostTypeId="2" ParentId="21060" CreationDate="2011-02-08T22:30:57.137" Score="2" ViewCount="" Body="&lt;p&gt;This is a relative of &lt;a href=&quot;http://en.wikipedia.org/wiki/Big_O_notation#Family_of_Bachmann.E2.80.93Landau_notations&quot; rel=&quot;nofollow&quot;&gt;big-O notation&lt;/a&gt;. We say that $f(x) = \omega(g(x))$ if $f(x) \ge k g(x)$ for all positive constants $k$, at least asymptotically. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the notation should really be read $- \log f(x) = \omega(\log x)$, or $- \log f(x) \ge k \log x$ for every positive constant $k$ asymptotically. In other words, $f(x) \le x^{-k}$ for every positive constant $k$ asymptotically, hence $f$ vanishes faster than polynomially as $x \to \infty$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Keep in mind that this does not define a &lt;em&gt;function&lt;/em&gt; in the usual sense; rather, it specifies the asymptotic behavior of that function in a way that is good enough for whatever the paper is proving.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2011-02-08T22:30:57.137" />
  <row Id="21066" PostTypeId="2" ParentId="20907" CreationDate="2011-02-08T22:33:42.067" Score="2" ViewCount="" Body="&lt;p&gt;Let $M$ be an abelian group and consider an extension $$0\to\mathbb Q\to E\to M\to0$$ of abelian groups. Since $\mathbb Q$ is flat, there is an induced exact sequence $$0\to\mathbb Q\otimes_{\mathbb Z}\mathbb Q\to \mathbb Q\otimes_{\mathbb Z} E\to Q\otimes_{\mathbb Z} M\to 0$$ The latter, being a short exact sequence of rational vector spaces, splits and there is a map $\mathbb Q\otimes_{\mathbb Z}E\to\mathbb Q\otimes_{\mathbb Z}\mathbb Q$ such that the composition $$\mathbb Q\otimes_{\mathbb Z}\mathbb Q\to\mathbb Q\otimes_{\mathbb Z}E\to\mathbb Q\otimes_{\mathbb Z}\mathbb Q$$ is the identity. Now the composition $$E\to \mathbb Q\otimes_{\mathbb Z}E\to\mathbb Q\otimes_{\mathbb Z}\mathbb Q\to\mathbb Q$$ with the first and the last maps being the obvious morphisms, splits the original extension. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It follows that $\mathrm{Ext}_{\mathbb Z}^1(M,\mathbb Q)=0$.&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2011-02-08T22:33:42.067" CommentCount="3" />
  <row Id="21067" PostTypeId="1" AcceptedAnswerId="21069" CreationDate="2011-02-08T22:35:47.313" Score="2" ViewCount="89" Body="&lt;p&gt;What does the notation $x\equiv y$ mod $2\mathbb{Z}[\zeta_3]$ mean?&lt;/p&gt;&#xA;" OwnerUserId="2212" LastEditorUserId="6335" LastEditorDisplayName="" LastEditDate="2011-02-08T22:46:57.823" LastActivityDate="2011-02-08T23:07:37.417" Title="What does the notation $x\equiv y$ mod $2\mathbb{Z}[\zeta_3]$ mean?" Tags="&lt;number-theory&gt;" AnswerCount="2" />
  <row Id="21068" PostTypeId="1" AcceptedAnswerId="21135" CreationDate="2011-02-08T22:46:43.910" Score="3" ViewCount="175" Body="&lt;p&gt;Everyone likes to prove that Dirichlet convolution is a group operation on the multiplicative arithmetic functions, but what consequence does this have?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;does any important theorem use this fact?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;can general group theory lead to results about these functions (or even better, about numbers) from the this theorem?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Furthermore there are two ring structures on this set, the usual pointwise ring as well as the ring with convolution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to extend the same question for these.&lt;/p&gt;&#xA;" OwnerUserId="4290" LastEditorUserId="4290" LastEditorDisplayName="" LastEditDate="2011-02-09T00:49:50.650" LastActivityDate="2011-02-19T19:27:41.813" Title="Why prove that muliplicative functions are a group with Dirichlet convolution?" Tags="&lt;number-theory&gt;&lt;group-theory&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="0" />
  <row Id="21069" PostTypeId="2" ParentId="21067" CreationDate="2011-02-08T22:46:52.583" Score="4" ViewCount="" Body="&lt;p&gt;It means that there is an element $z$ of the ring $\mathbb{Z}[\zeta_3] = \{a+b\left(\frac{1+\sqrt{-3}}{2}\right) \ |  \ a,b \in \mathbb{Z} \}$ such that $x-y = 2z$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;More generally, if $R$ is a &lt;a href=&quot;http://en.wikipedia.org/wiki/Ring_%28mathematics%29&quot; rel=&quot;nofollow&quot;&gt;ring&lt;/a&gt; and $I$ is an &lt;a href=&quot;http://en.wikipedia.org/wiki/Ideal_%28ring_theory%29&quot; rel=&quot;nofollow&quot;&gt;ideal&lt;/a&gt; of $R$, then $x \equiv y \pmod I$ means that $x-y \in I$.  &lt;/p&gt;&#xA;" OwnerUserId="299" LastActivityDate="2011-02-08T22:46:52.583" CommentCount="1" />
  <row Id="21070" PostTypeId="1" AcceptedAnswerId="21087" CreationDate="2011-02-08T22:47:26.543" Score="2" ViewCount="57" Body="&lt;p&gt;Let $f:X \rightarrow Y$ be a continuous map of topological spaces and let $\cal{F}$ be a sheaf on $X$. Is there an obvious map $H^\ast(Y,f_\ast \mathcal{F} ) \rightarrow H^\ast (X,\cal{F})$?&lt;/p&gt;&#xA;" OwnerUserId="6652" LastActivityDate="2011-02-09T05:56:40.197" Title="Pushing forward sheaves and the result on sheaf cohomology." Tags="&lt;cohomology&gt;&lt;sheaf-theory&gt;" AnswerCount="1" />
  <row Id="21071" PostTypeId="2" ParentId="21067" CreationDate="2011-02-08T22:47:26.920" Score="2" ViewCount="" Body="&lt;p&gt;It means $\rm\ x - y\ =\ 2\ z\ $ for some $\rm\ z \in \mathbb Z[\zeta_3]\:,\ $ i.e. $\rm\ z = m + n\ \zeta_3,\ $ where $\rm\ \zeta_3^2 + \zeta_3 + 1 = 0\:.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a generalization of the notion of congruence in the ring $\rm\ \mathbb Z/n\ $ of integers modulo $\rm\:n\:$. Recall that when solving integer problems it often proves convenient to attempt to decompose the problem into simpler problems in the &lt;em&gt;finite&lt;/em&gt; rings/fields $\rm\ \mathbb Z/n\:.\ $ The same technique proves useful for general rings, where one consider &quot;simpler&quot; images $\rm\ R/I \ $ where $\rm\:I\:$ is the congruence class of $0\:,\ $ the set of all elements that are mapped to $0$ in the quotient ring. This has a fundamental structure known as an &lt;a href=&quot;http://en.wikipedia.org/wiki/Ideal_%28ring_theory%29&quot; rel=&quot;nofollow&quot;&gt;ideal,&lt;/a&gt; of which the above &lt;em&gt;principal&lt;/em&gt; ideal $\rm\ n\ R\ =\ \{ n\:r\ :\ r\in R\}\ $ is a prototypical example.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2011-02-08T23:07:37.417" LastActivityDate="2011-02-08T23:07:37.417" CommentCount="2" />
  <row Id="21072" PostTypeId="1" AcceptedAnswerId="21074" CreationDate="2011-02-08T22:52:09.540" Score="0" ViewCount="233" Body="&lt;p&gt;I have two points. I need to draw an arc (&amp;lt;180°) between them, and I know how long it should be, but nothing else about it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Knowing either the radius length or the coordinates of the center point of the circle should be enough to draw it directly. I could also calculate the radius if I knew the subtended angle θ. But I don't know how to calculate either of these. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that somebody answered a &lt;a href=&quot;http://math.stackexchange.com/questions/8487/calculating-the-height-of-a-circular-segment-at-all-points-provided-only-chord-an&quot;&gt;similar question&lt;/a&gt; a while ago. But for some reason, the formula they give for the answer includes θ. Maybe there is an easy way to calculate it and my knowledge is just too rusty, but I can't solve my problem with this information. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried deriving an answer by myself, but arrived at an equation containing both an angle's cosine and the angle squared, and I don't know how to solve such an equation. Besides, there is probably an easier way than what I used. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is OK if the answer is derived by using trigonometry, analytical geometry or anything else, as long as I can use a pocket calculator to get my radius/coordinates. But if you don't want to lose me along the way, it would be nice if you could use calculations at undergraduate level. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a drawing of what I mean: &#xA;&lt;img src=&quot;http://i.imgur.com/Xxpd0.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="6346" LastActivityDate="2011-02-12T13:56:38.757" Title="Finding out an arc's radius by arc length and endpoints" Tags="&lt;geometry&gt;" AnswerCount="4" />
  <row Id="21073" PostTypeId="2" ParentId="21041" CreationDate="2011-02-08T22:52:51.310" Score="2" ViewCount="" Body="&lt;p&gt;Didier gave much more than the OP requested. My solution below addresses the original question (and is thus somewhat simpler, in my opinion).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $Z$ denotes the maximum, then&#xA;$$&#xD;&#xA;{\rm P}(Z \le z) = \sum\limits_{n = 1}^\infty  {{\rm P}(Z \le z|N = n){\rm P}(N = n)}  = \sum\limits_{n = 1}^\infty  {F_{X_{(n)} } (z)\frac{1}{{n!({\rm e} - 1)}}}. &#xD;&#xA;$$&#xA;Since $F_{X_{(n)} } (z) = [F_X (z)]^n = (\frac{z}{\theta })^n$, $0 \leq z \leq \theta$, we get &#xA;$$&#xD;&#xA;{\rm P}(Z \le z) = \frac{1}{{{\rm e} - 1}}\sum\limits_{n = 1}^\infty  {\frac{{( z/\theta)^n }}{{n!}}}  = \frac{1}{{{\rm e} - 1}}({\rm e}^{z/\theta}  - 1).&#xD;&#xA;$$&#xA;Hence $Z$ has density $f_Z$ given by&#xA;$$&#xD;&#xA;f_Z (z) = \frac{1}{{({\rm e} - 1)\theta }}{\rm e}^{ z/\theta}, \;\; 0 &amp;lt; z &amp;lt; \theta. &#xD;&#xA;$$&#xA;Similarly, if $Y$ denotes the minimum, then&#xA;$$&#xD;&#xA;{\rm P}(Y \le y) = \sum\limits_{n = 1}^\infty  {{\rm P}(Y \le y|N = n){\rm P}(N = n)}  = \sum\limits_{n = 1}^\infty  {F_{X_{(1)} } (y)\frac{1}{{n!({\rm e} - 1)}}}. &#xD;&#xA;$$&#xA;Since $F_{X_{(1)} } (y) = 1 - [1-F_X (y)]^n = 1 - [1 - \frac{y}{\theta }]^n$, $ 0 \leq y \leq \theta$, we get&#xA;$$&#xD;&#xA;{\rm P}(Y \le y) = \sum\limits_{n = 1}^\infty  {\frac{{1 - [1 - y/\theta ]^n }}{{n!({\rm e}-1)}}}  = 1 -  \sum\limits_{n = 1}^\infty  {\frac{{(1 - y/\theta )^n }}{{n!({\rm e}-1)}}}  = 1 - \frac{{{\rm e}^{1 - y/\theta }  - 1}}{{{\rm e} - 1}}.&#xD;&#xA;$$&#xA;Hence $Y$ has density $f_Y$ given by&#xA;$$&#xD;&#xA;f_Y (y) = \frac{1}{{({\rm e} - 1)\theta }}{\rm e}^{ 1 - y/\theta}, \;\; 0 &amp;lt; y &amp;lt; \theta. &#xD;&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="2810" LastActivityDate="2011-02-08T22:52:51.310" CommentCount="8" />
  <row Id="21074" PostTypeId="2" ParentId="21072" CreationDate="2011-02-08T23:01:05.533" Score="2" ViewCount="" Body="&lt;p&gt;As you said you had, I arrive at an equation that cannot be solved algebraically, $\cos\frac\theta 2=\frac{c\theta}{2l}$, so I suspect that there is no algebraic solution, only a numerical approximation.&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2011-02-08T23:01:05.533" />
  <row Id="21075" PostTypeId="1" CreationDate="2011-02-08T23:11:37.990" Score="2" ViewCount="108" Body="&lt;p&gt;Hi!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming I have a Gamma distributed random Variable $x \sim Gamma( \alpha, \beta )$. Now I like to have the following two expectation values (integrals):&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$E \left[ x \ln x \right]$&lt;/li&gt;&#xA;&lt;li&gt;$E \left[ \ln \Gamma \left( x \right) \right]$&#xA;with $\Gamma \left( x \right)$ being the &lt;a href=&quot;http://en.wikipedia.org/wiki/Gamma_function&quot; rel=&quot;nofollow&quot;&gt;Gamma function&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Many thanks in advance&lt;/p&gt;&#xA;" OwnerUserId="6197" LastActivityDate="2011-02-09T23:35:43.363" Title="Some expectation values for a Gamma distribution" Tags="&lt;probability&gt;&lt;gamma-function&gt;&lt;probability-distributions&gt;&lt;estimation-theory&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="21076" PostTypeId="1" CreationDate="2011-02-08T23:13:32.517" Score="6" ViewCount="153" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Possible Duplicate:&lt;/strong&gt;&lt;br&gt;&#xA;  &lt;a href=&quot;http://math.stackexchange.com/questions/14970/rational-numbers&quot;&gt;Rational Numbers&lt;/a&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;p&gt;Baby Rudin has a very nice construction showing that, given a positive rational number whose square is less than (greater than) two, one can always find a larger (smaller) rational whose square is also less than (greater than) two.  Basically, if $p$ is a rational whose square is less (greater than) than 2, then let $q$ be $$ q=p-\frac{p^2-2}{p+2}=\frac{2p+2}{p+2}$$ so that $$q^2 - 2 = \frac{2(p^2-2)}{(p+2)^2} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's easy to see that $q$ satisfies the requirements.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question: how would one come up with this on one's own? What approach could I have taken to derive this gem by myself? What magic happened behind the scenes to get here?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tackled it from scratch before going back to Rudin, and came up with my own solution (applying Newton's method to find an approximation for the positive root of $(x^2-2)^2$, giving me $q=\frac{3p}{4}+\frac{1}{2p}$), but the resulting proof is uglier and requires restricting p to certain intervals and handling the rest as a trivial special case.&lt;/p&gt;&#xA;" OwnerUserId="6449" LastActivityDate="2011-02-08T23:13:32.517" ClosedDate="2011-02-08T23:19:21.150" Title="Constructing a larger rational whose square is also less than two" Tags="&lt;analysis&gt;&lt;proof-strategy&gt;" CommentCount="10" />
  <row Id="21077" PostTypeId="2" ParentId="20096" CreationDate="2011-02-08T23:20:49.073" Score="0" ViewCount="" Body="&lt;p&gt;Three years old seems very young to do more than instill a sense of wonder about things that happen in the world which it will turn out in the future are related to mathematics. If you trust a three year old with a scissors than try having the child cut down the middle a long thin rectangle after its thin edges are taped, once with a half-turn before taping and once without the half-turn before taping. There are lots of nice paper cutting and paper folding examples that are fun for kids. I have done this, including the Moebius band work, with kindergarten children: &lt;a href=&quot;http://www.york.cuny.edu/~malk/kindergarten/kindergarten.html&quot; rel=&quot;nofollow&quot;&gt;http://www.york.cuny.edu/~malk/kindergarten/kindergarten.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1369" LastActivityDate="2011-02-08T23:20:49.073" CommunityOwnedDate="2011-02-08T23:20:49.073" />
  <row Id="21078" PostTypeId="2" ParentId="18161" CreationDate="2011-02-08T23:25:54.567" Score="0" ViewCount="" Body="&lt;p&gt;There are also &lt;a href=&quot;http://en.wikipedia.org/wiki/Hardware_random_number_generator&quot; rel=&quot;nofollow&quot;&gt;hardware circuits&lt;/a&gt; that can produce random numbers.&lt;/p&gt;&#xA;" OwnerUserId="5864" LastActivityDate="2011-02-08T23:25:54.567" />
  <row Id="21079" PostTypeId="2" ParentId="20096" CreationDate="2011-02-08T23:31:44.560" Score="1" ViewCount="" Body="&lt;p&gt;I used to do simple mental addition and subtraction with my kids when they were that age.  Questions like:  There are five cookies in this jar, if I eat two of them how many do you get.&lt;/p&gt;&#xA;" OwnerUserId="5864" LastActivityDate="2011-02-08T23:31:44.560" CommunityOwnedDate="2011-02-08T23:31:44.560" />
  <row Id="21080" PostTypeId="2" ParentId="16133" CreationDate="2011-02-08T23:36:52.877" Score="3" ViewCount="" Body="&lt;p&gt;Glib answer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you feel like it, you can assume that&#xA;$|\mathbb{R}|=\aleph_{7}$, &#xA;$|P(\mathbb{R})|=\aleph_{13}$, &#xA;and&#xA;$|P(P(\mathbb{R}))|=\aleph_{\aleph_{\omega_3+4}}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Less glib:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;as other posters have pointed out, the descriptions as power-sets is generally the simplest description.  One of the reasons is that the standard axioms of set-theory (ZFC) do not &quot;pin down&quot; the cardinalities of these sets in terms of the alephs ($\aleph$).  A technique called &quot;forcing&quot; (in particular &quot;Easton forcing&quot;) shows that it is consistent that the cardinalities of power-sets can be just about anything.  There are a few constraints coming from cardinal arithmetic ($\alpha&amp;lt;\beta$ implies $2^\alpha\leq2^\beta$, $\text{cof}(2^\alpha)&gt;\text{cof}(\alpha)$), but they are very weak.  Thus it is consistent that we have:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\aleph_0 &amp;lt; \aleph_1 &amp;lt; \aleph_2=2^{\aleph_0}=|\mathbb{R}| &amp;lt; \aleph_3 $&#xA;${} &amp;lt; \cdots &amp;lt; \aleph_{\omega+1} = 2^{\aleph_1} = 2^{\aleph_2} = |P(\mathbb{R})| $&#xA;${} &amp;lt; \aleph_{\omega+2}  = 2^{\aleph_3} = 2^{\aleph_4}&#xD;&#xA;= \cdots = 2^{\aleph_\omega} = 2^{\aleph_{\omega+1}} = |P(P(\mathbb{R}))|$&#xA;etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(But we could &lt;em&gt;not&lt;/em&gt; have $2^{\aleph_0}=\aleph_\omega$ due to the cofinality restriction.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, GCH would say that $2^{\aleph_{\alpha}}=\aleph_{\alpha+1}$ for any ordinal $\alpha$ (in particular, $|\mathbb{R}|=\aleph_1$, $|P(\mathbb{R})|=\aleph_2$, and $|P(P(\mathbb{R}))|=\aleph_3$ as pointed out above)&#xA;and GCH is also consistent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that I am not claiming that these wild cardinal arithmetics are natural or reasonable, simply that they are consistent and one needs further assumptions to pin down the cardinalities of power-sets in terms of alephs; and thus it's hard to give the kind of answers we might want for &quot;what is the cardinality of $P(\mathbb{R})$?&quot; for example...&lt;/p&gt;&#xA;" OwnerUserId="3852" LastEditorUserId="3852" LastEditorDisplayName="" LastEditDate="2011-02-08T23:49:22.450" LastActivityDate="2011-02-08T23:49:22.450" />
  <row Id="21081" PostTypeId="2" ParentId="21062" CreationDate="2011-02-08T23:50:52.073" Score="2" ViewCount="" Body="&lt;p&gt;There are many applications of game theory to bridge. Perhaps the simplest is the Principle of Restricted Choice:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your trump suit is AJ1098 in dummy and four small in hand. You play small to the Jack, which loses to the Queen. On regaining the lead, you play small towards dummy, and left-hand opponent plays small. Do you finesse again, or play the Ace hoping to drop the King?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a kind of pons asinorum for bridge players. The probability is nearly 2/3 that the finesse will succeed, but some players never grasp it. The reason is that if right-hand opponent held both honours, they might have won with the King on the first round. Conditional probability does the rest. The game-theoretical aspect of this is that right-hand opponent, holding both the King and the Queen, must decide at random which one to play on the first round, otherwise declarer can improve the odds by remembering how this particular defender played on previous occasions.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are more complicated situations, where in principle both declarer and defender must play randomly. But in practice they are so rare that gathering statistics on a player's habits is impossible, so the best play in a particular situation is more likely to depend on your estimation of your opponents' technical skill. I once (thirty years ago) wrote a letter to an internationally popular bridge magazine, explaining how game theory could be used to resolve a supposedly intractable problem discussed in their pages. I was gently mocked for my pains.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, surprisingly, there are situations in the bidding that (in theory) call for randomised choices. Such situations are easiest to construct in the context of a head-to-head teams-of-four match.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If any of this interests anybody, let me know and I will try to construct some concrete examples.&lt;/p&gt;&#xA;" OwnerUserId="1508" LastEditorUserId="1508" LastEditorDisplayName="" LastEditDate="2011-02-08T23:56:48.927" LastActivityDate="2011-02-08T23:56:48.927" CommentCount="5" />
  <row Id="21083" PostTypeId="1" AcceptedAnswerId="21095" CreationDate="2011-02-09T00:01:21.603" Score="2" ViewCount="155" Body="&lt;p&gt;I'm having a hard time understanding how to find the don't-care values in a Kmap. What does it even mean? If I have a boolean function, say $f(a,b,c,d)=a&amp;#39;bc+abc&amp;#39;+bc&amp;#39;d+a&amp;#39;bc&amp;#39;d$, how would I determine don't-care values? What would I be looking for?&lt;/p&gt;&#xA;" OwnerUserId="2985" LastEditorUserId="2985" LastEditorDisplayName="" LastEditDate="2011-02-09T01:48:09.057" LastActivityDate="2011-02-09T01:48:09.057" Title="Determining dont care values in a Karnaugh Map" Tags="&lt;boolean-algebra&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="21084" PostTypeId="1" AcceptedAnswerId="21085" CreationDate="2011-02-09T00:03:50.457" Score="0" ViewCount="33" Body="&lt;p&gt;I'm looking at some class examples on basic postulates and I can't figure out how the 2nd part is reduced (see below). Could someone explain it to me?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;eq:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;X='a'c+ac+'ac+a'c&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;this is the part I do not understand, and could use some explanation using the formulas below.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;= 'a('c+c) + a('c+c)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The example uses theses formulas:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;[p6a] x*y = y*x&#xA;[p8a] x(y+z) = x*y + x*z&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&lt;br&gt;&#xA;and to finish it all off: (which I understand)&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;= 'a*1 + a*1&#xA;= 'a+a&#xA;= 1&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="6328" LastActivityDate="2011-02-09T00:30:18.687" Title="Can someone explain how this question is reduced using basic postulates" Tags="&lt;boolean-algebra&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="21085" PostTypeId="2" ParentId="21084" CreationDate="2011-02-09T00:23:21.740" Score="1" ViewCount="" Body="&lt;p&gt;To be completely explicit: we begin with the formula $$X = a&amp;#39; c&amp;#39; + ac + a&amp;#39;c+ac&amp;#39;$$ (where I have written 'c as $c&amp;#39;$ in order to make the latex easier). We then switch the second and third terms:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$X = a&amp;#39; c&amp;#39; + a&amp;#39; c + ac + ac&amp;#39;.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can now apply p8a (in &quot;reverse&quot;, as it were):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$X = a&amp;#39; (c&amp;#39; + c) + a (c + c&amp;#39;).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, we reorder the terms inside the second set of parentheses:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$X = a&amp;#39; (c&amp;#39; + c) + a (c&amp;#39; + c),$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and you can take it from here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that this derivation doesn't use p6a (the commutativity of multiplication), but it &lt;em&gt;does&lt;/em&gt; use the commutativity of addition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; As Henry pointed out in the comments above, your class might be calling the commutativity of addition p6b.&lt;/p&gt;&#xA;" OwnerUserId="4743" LastEditorUserId="4743" LastEditorDisplayName="" LastEditDate="2011-02-09T00:30:18.687" LastActivityDate="2011-02-09T00:30:18.687" CommentCount="1" />
  <row Id="21086" PostTypeId="1" CreationDate="2011-02-09T00:28:58.860" Score="0" ViewCount="76" Body="&lt;p&gt;Say I have a flat rectangular bar of aluminum with a known volume. Sitting directly on top of that piece of aluminum is a rectangular circuit board made of FR4 (standard circuit board material).  I can raise and lower the temperature of the aluminum very rapidly and I can monitor the temperature of the aluminum instantaneously but I cannot monitor the temperature of the circuit board directly. We can assume that the entire aluminum piece heats and cools evenly as does the circuit board.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I know that I want to raise the temperature of the circuit board from 30°C to 200°C linearly in 90 seconds what calculations would I need to do to determine how fast i need to raise the temperature of the aluminum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;( I am tagging this incorrectly because there isn't a good tag for it and I can't add tags yet!)&lt;/p&gt;&#xA;" OwnerUserId="6745" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2011-02-09T04:24:32.183" LastActivityDate="2011-02-09T04:33:56.590" Title="Thermal transfer calculation" Tags="&lt;physics&gt;" AnswerCount="2" />
  <row Id="21087" PostTypeId="2" ParentId="21070" CreationDate="2011-02-09T00:35:09.437" Score="4" ViewCount="" Body="&lt;p&gt;There is always a natural map $$\Gamma(Y, \mathcal{G}) \to \Gamma(X, f^{-1}(\mathcal{G}))$$ for any sheaf $\mathcal{G}$ on $Y$. However, these are both left-exact functors in the sheaf $\mathcal{G}$ (since $f^{-1}$ is an exact functor). By general (Tohoku) nonsense, there is an induced natural transformation of $\delta$-functors $H^*(Y, \mathcal{G}) \to H^*(X, f^{-1}(\mathcal{G}))$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take $\mathcal{G} = f_*(\mathcal{F})$ now. We get a map&#xA;$$H^*(Y, f_*(\mathcal{F})) \to H^*(X, f^{-1} f_*(\mathcal{F})).$$ However, the adjunction between push-forwards and inverse images produces a map $f^{-1} f_*(\mathcal{F})) \to \mathcal{F}$. This induces a morphism in cohomology $H^*(X, f^{-1} f_*(\mathcal{F})) \to H^*(X, \mathcal{F})$. Composing the two maps just described gives the map that you ask for.&lt;/p&gt;&#xA;" OwnerUserId="536" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2011-02-09T05:56:40.197" LastActivityDate="2011-02-09T05:56:40.197" CommentCount="1" />
  <row Id="21088" PostTypeId="2" ParentId="21075" CreationDate="2011-02-09T00:40:22.220" Score="0" ViewCount="" Body="&lt;p&gt;1.&#xA;E(XlnX)=(formula of expectation, group x) =E(lnY) with Y~&lt;a href=&quot;http://en.wikipedia.org/wiki/Gamma_distribution&quot; rel=&quot;nofollow&quot;&gt;gamma&lt;/a&gt;(alpha+1,beta)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;E(lnY)=ψ(alpha+1)-ln(beta) where ψ(k) is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Digamma_function&quot; rel=&quot;nofollow&quot;&gt;digamma&lt;/a&gt; function.&lt;/p&gt;&#xA;" OwnerUserId="6812" LastActivityDate="2011-02-09T00:40:22.220" CommentCount="3" />
  <row Id="21089" PostTypeId="2" ParentId="21072" CreationDate="2011-02-09T00:48:21.997" Score="0" ViewCount="" Body="&lt;p&gt;If you have 2 points and arc length, it seems like you should be able to find the diameter (length of the segment connecting 2 points) by using C=πd or C/2= πr. Since you know the value of C/2 (where C= circumference and C/2 is your arc length) and you can plug in π on your calculator, you can get r, which will tell you the center of the circle. Or did I miss something?&lt;/p&gt;&#xA;" OwnerUserId="6813" LastActivityDate="2011-02-09T00:48:21.997" CommentCount="2" />
  <row Id="21090" PostTypeId="2" ParentId="21072" CreationDate="2011-02-09T00:49:51.010" Score="2" ViewCount="" Body="&lt;p&gt;If $r$ is the radius then you have presumably found, as Isaac did, that $l=r\theta$ and $\frac{c}{2}=r \sin(\theta/2).$  You could eliminate $\theta$ to get $$2r \sin\left(\frac{l}{2r}\right) = c.$$  That you can solve for $r$ using numerical methods.&lt;/p&gt;&#xA;" OwnerUserId="6460" LastEditorUserId="6460" LastEditorDisplayName="" LastEditDate="2011-02-12T13:56:38.757" LastActivityDate="2011-02-12T13:56:38.757" CommentCount="4" />
  <row Id="21091" PostTypeId="2" ParentId="20946" CreationDate="2011-02-09T00:50:36.500" Score="1" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; I only achieved the part that C.R. took for granted in his argument. Since it's already written, I leave it here for the sake of completeness.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Let $p_{n}(x) = \sum_{k=0}^{n} \frac{(-1)^{k}}{k!} x^{k}$ be the $n$th Taylor polynomial of $e^{-x}$. Then by the Lagrange remainder theorem we have $e^{-x} = p_{n}(x) + \frac{(-1)^{n}}{(n+1)!} e^{-\xi} x^{n+1}$ for some $\xi \in [0,x]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore $|e^{-2x} - e^{-x} p_{n}(x)| \leq \frac{e^{-\xi}}{(n+1)!} e^{-x} x^{n+1}$ and we want to show that the right hand side converges to zero uniformly in $x$. As $e^{-\xi} \leq 1$, it suffices to find the maximum of $e^{-x}x^{n+1}$. Differentiation yields that the maximum is located at $x = n+1$ and its value is $\frac{(n+1)^{n+1}}{e^{n+1}}$. On the other hand, Stirling's formula tells us that $n! = \sqrt{2\pi n} \left(\frac{n}{e}\right)^{n} e^{r_{n}}$ with $|\frac{1}{12n} - r_{n}| \leq \frac{1}{120n^2}$. Putting these things together we get&#xA;\[&#xA;|e^{-2x} - e^{-x} p_{n}(x)| \leq \frac{1}{\sqrt{2\pi (n+1)}} e^{-r_{n+1}} \xrightarrow{n\to\infty} 0&#xA;\]&#xA;as we wanted.&lt;/p&gt;&#xA;" OwnerUserId="5363" LastEditorUserId="5363" LastEditorDisplayName="" LastEditDate="2011-02-09T12:10:20.310" LastActivityDate="2011-02-09T12:10:20.310" />
  <row Id="21092" PostTypeId="2" ParentId="21075" CreationDate="2011-02-09T01:02:48.540" Score="2" ViewCount="" Body="&lt;p&gt;First integral. $X$ has density function &#xA;$$&#xD;&#xA;f(x;\alpha ,\beta ) = \frac{{\beta ^\alpha  }}{{\Gamma (\alpha )}}x^{\alpha  - 1} e^{ - \beta x},\;\; x &amp;gt; 0, &#xD;&#xA;$$&#xA;with $\alpha,\beta &amp;gt; 0$ fixed.&#xA;It follows readily from&#xA;$$&#xD;&#xA;\Gamma &amp;#39;(\alpha ) = \int_0^\infty  {x^{\alpha  - 1} e^{ - x} \ln x\,{\rm d}x} &#xD;&#xA;$$&#xA;and&#xA;$$&#xD;&#xA;\Gamma &amp;#39;(\alpha ) = \Gamma (\alpha )\psi _0 (\alpha ),&#xD;&#xA;$$&#xA;where $\psi_0$ is the  digamma function,&#xA;that &#xA;$$&#xD;&#xA;{\rm E}[X\ln X] = \frac{{\beta ^\alpha  }}{{\Gamma (\alpha )}}\int_0^\infty  {(x\ln x)x^{\alpha  - 1} e^{ - \beta x} dx} = \frac{\alpha }{\beta }[\psi_0 (\alpha  + 1) - \ln \beta ].&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;: Elaborating (in response to the OP's request). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, you can find &lt;a href=&quot;http://en.wikipedia.org/wiki/Gamma_function&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; the above formulas for $\Gamma&amp;#39;(\alpha)$. Now, using a change of variable, we have&#xA;$$&#xD;&#xA;\Gamma&amp;#39;(\alpha+1) = \int_0^\infty  {\beta ^{\alpha+1}  x^{\alpha} e^{ - \beta x} \ln (\beta x)\,{\rm d}x} =  \ln \beta \int_0^\infty  {\beta ^{\alpha+1} x^{\alpha} e^{ - \beta x} \,{\rm d}x}  +  \int_0^\infty  { \beta ^{\alpha+1} x^{\alpha} e^{ - \beta x} \ln x\,{\rm d}x}. &#xD;&#xA;$$&#xA;The first integral on the right-hand side is equal to $\Gamma(\alpha+1)\ln \beta$, and the second integral to $\Gamma(\alpha)\beta {\rm E}[X\ln X]$. It thus follows that&#xA;$$&#xD;&#xA;\Gamma (\alpha  + 1)\psi _0 (\alpha  + 1) = \Gamma (\alpha  + 1)\ln \beta  + \Gamma (\alpha )\beta {\rm E}[X\ln X].&#xD;&#xA;$$&#xA;Finally, by $\Gamma(\alpha+1)=\alpha \Gamma(\alpha)$, we get&#xA;$$&#xD;&#xA;{\rm E}[X\ln X] = \frac{\alpha }{\beta }[\psi_0 (\alpha  + 1) - \ln \beta ].&#xD;&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="2810" LastEditorUserId="2810" LastEditorDisplayName="" LastEditDate="2011-02-09T23:35:43.363" LastActivityDate="2011-02-09T23:35:43.363" CommentCount="1" />
  <row Id="21093" PostTypeId="2" ParentId="21086" CreationDate="2011-02-09T01:02:58.023" Score="1" ViewCount="" Body="&lt;p&gt;I don't think there's enough information to solve the problem. You write that you &quot;cannot monitor the temperature of the circuit board directly&quot;, but you also don't describe any way of monitoring it indirectly, so it seems you can't monitor it at all. There's also no information about the efficiency of heat transport between the aluminum and the circuit board. If you can really assume that the entire aluminum and the entire circuit board heat and cool evenly (which seems like a bad approximation to me if they're in contact), and if you can also assume that heat is transferred only through heat transport (and not e.g. through radiation reflecting off a case), then the problem is characterized by a single number quantifying the efficiency of heat transport between the aluminum and the circuit board -- this is the constant of proportionality between the rate of change of the temperature of the circuit board and the temperature difference. (This is assuming, as you seem to be applying, that you fully control the temperature of the aluminum, so the heat loss of the aluminum to the circuit board doesn't have to be taken into account. Otherwise we'd need a second quantity, like the ratio of the heat capacities of the two bodies.) You're going to have to know that constant of proportionality to be able to calculate anything; if you do know it, the rest should be a relatively simple exercise in ordinary differential equations.&lt;/p&gt;&#xA;" OwnerUserId="6622" LastActivityDate="2011-02-09T01:02:58.023" CommentCount="7" />
  <row Id="21094" PostTypeId="1" AcceptedAnswerId="21096" CreationDate="2011-02-09T01:09:42.977" Score="5" ViewCount="140" Body="&lt;p&gt;Does there exist $K \subseteq \mathbb{R} \backslash \mathbb{Q}$ such that $K$ is compact, and has Lebesgue measure greater than $0$? As I have been trying to think of examples, I suspect that any subset of $\mathbb{R} \backslash \mathbb{Q}$ that is closed can be at most countable, since the closure of an uncountable subset of irrationals should contain some rationals. And, the Lebesgue measure of a countable set is $0$. If there are any examples of such a set, I would be very interested to know how it is constructed.&lt;/p&gt;&#xA;" OwnerUserId="6814" LastActivityDate="2011-02-09T01:17:21.917" Title="Is there a compact subset of the irrationals with positive Lebesgue measure?" Tags="&lt;real-analysis&gt;&lt;measure-theory&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="0" />
  <row Id="21095" PostTypeId="2" ParentId="21083" CreationDate="2011-02-09T01:11:10.460" Score="3" ViewCount="" Body="&lt;p&gt;You are not looking for anything. Don't-care values in a Karnaugh map relate to output that is generally not reachable under normal circumstances and allow you to simplify the logic more than you would normally be able to. Because of this, don't-care values must be explicitly given, or understood from some statement akin to &quot;the values $a&amp;#39;bcd$ and $ab&amp;#39;c&amp;#39;$ will never be used as input&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For an example, consider the input logic to a seven-segment display (the display you see on digital alarm clocks, microwaves, etc.) The input to a seven-segment display is often a number between $0$ and $9$, represented with 4 bits. Since the numbers $10$ through $15$ are also valid combinations of those 4 bits, the logic would theoretically work and determine output values if that was the input received. But because that input is never expected, it doesn't matter what would happen on the display if the number $12$ were received. The display could show a $4$ or an $8$ or any combination of the segments. Thus, when determining the logic for such things, you can put an X in the Karnaugh map under the values that correspond to $10, \dots, 15$ and they will often simplify the logic. This ultimately results in fewer logic gates necessary in hardware, which is always a good thing.&lt;/p&gt;&#xA;" OwnerUserId="1016" LastActivityDate="2011-02-09T01:11:10.460" CommentCount="5" />
  <row Id="21096" PostTypeId="2" ParentId="21094" CreationDate="2011-02-09T01:17:21.917" Score="6" ViewCount="" Body="&lt;p&gt;The answer is yes.  Count the rationals in $[0,1]$ as $r_1,r_2,\ldots$, let $I_k$ be an open interval containing $r_k$ of length $3^{-k}$, and let $K=[0,1]\setminus\cup_k I_k$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This question is somewhat related to the question &lt;a href=&quot;http://math.stackexchange.com/questions/1064/perfect-set-without-rationals&quot;&gt;Perfect set without rationals&lt;/a&gt;, but there measure did not come up.  For example, the Cantor set-like construction given there by JDH could be made to have positive measure.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2011-02-09T01:17:21.917" CommentCount="7" />
  <row Id="21097" PostTypeId="1" AcceptedAnswerId="21098" CreationDate="2011-02-09T01:22:32.323" Score="6" ViewCount="102" Body="&lt;p&gt;Prompted by &lt;a href=&quot;http://math.stackexchange.com/questions/16133/cardinality-of-p-mathbbr-and-pp-mathbbr&quot;&gt;this question&lt;/a&gt;: of course Godel's constructibility axiom implies that $P(S)$ is minimal for any set $S$ and so handily answers the question of the size of the power set of the continuum in $L$.  Martin's Maximum is the only other principle I know of that has implications for the size of the continuum (caveat: most of my knowledge comes from Kanamori's &lt;i&gt;The Higher Infinite&lt;/i&gt;, a couple of other set theory books and the occasional dip into Wikipedia), and it's not clear to me what its implications are for power set operations on higher cardinals, in particular $2^{\mathbb{R}} (=2^{\aleph_2})$ - my impression is that it implies that $2^{\aleph_1} = \aleph_2$ but I'm not 100% sure on that front either.  Can anyone point to good information on the implications of MM for other cardinalities than just $\mathfrak{c}$?&lt;/p&gt;&#xA;" OwnerUserId="785" LastActivityDate="2011-02-09T21:44:10.123" Title="What does Martin's Maximum imply for $P(\mathbb{R})$?" Tags="&lt;set-theory&gt;&lt;cardinals&gt;&lt;forcing&gt;" AnswerCount="2" />
  <row Id="21098" PostTypeId="2" ParentId="21097" CreationDate="2011-02-09T01:40:59.647" Score="5" ViewCount="" Body="&lt;p&gt;Martin's Maximum is known to be consistent relative to a supercompact cardinal $\kappa$ by a revised countable support forcing iteration that results in $\mathfrak{c}=\aleph_2=\kappa$. This forcing does not affect the values of the continuum function $\gamma\mapsto 2^\gamma$ for values of $\gamma$ at $\kappa$ or above.  But it is known by the Laver preparation that if $\kappa$ is supercompact, then there is a forcing extension in which the supercompactness of $\kappa$ is preserved by all ${\lt}\kappa$-directed closed forcing. Such forcing can modify the values of the continuum function at $\kappa$ or above to be in accordance with any patter that conforms with Easton's theorem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So by combining the two facts, if it is consistent with ZFC that there is a supercompact cardinal, then it is consistent with ZFC that MM holds, and the continuum function has any desired pattern at $\aleph_2$ and above, in acordance with Easton's theorem. In particular, MM does not determine the value of $2^\mathbb{R}$. &lt;/p&gt;&#xA;" OwnerUserId="413" LastActivityDate="2011-02-09T01:40:59.647" CommentCount="6" />
  <row Id="21099" PostTypeId="1" AcceptedAnswerId="21109" CreationDate="2011-02-09T01:46:20.040" Score="4" ViewCount="148" Body="&lt;p&gt;I can see how this works out with various examples of the sum of three or more consecutive integers being prime, but I cant seem to prove it for all odd integers n&gt;1. Any help would be great.&lt;/p&gt;&#xA;" LastActivityDate="2011-02-09T02:37:48.417" Title="Prove that an odd integer n&gt;1 is prime if and only if it is not expressible as a sum of three or more consecutive integers." Tags="&lt;number-theory&gt;&lt;prime-numbers&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="21100" PostTypeId="1" AcceptedAnswerId="21107" CreationDate="2011-02-09T01:49:17.933" Score="4" ViewCount="269" Body="&lt;p&gt;What are the importance of rank of a matrix. I know that rank of a matrix is the number of linearly independent rows/columns (whichever is smaller). Why is it a problem if a matrix is rank deficient?. Also why is the smaller value between row and column is the rank? A intuitive or a descriptive answer (also in terms of geometry) would help a lot. Thanks.&lt;/p&gt;&#xA;" OwnerUserId="6335" LastActivityDate="2011-02-09T03:33:28.080" Title="Importance of rank of a matrix." Tags="&lt;linear-algebra&gt;&lt;matrices&gt;" AnswerCount="4" CommentCount="3" FavoriteCount="2" />
  <row Id="21101" PostTypeId="2" ParentId="20096" CreationDate="2011-02-09T01:52:26.667" Score="0" ViewCount="" Body="&lt;p&gt;Well, I have a three year old granddaughter, and she can (more or less) count up to 100.  She pretty much did it for me, with only a few glitches.  I asked her if 100 was the biggest number, and she thought the answer was &quot;yes&quot;.  I just accepted that.  It'll be interesting to see when the answer changes to &quot;no&quot;.&lt;/p&gt;&#xA;" OwnerUserId="2579" LastActivityDate="2011-02-09T01:52:26.667" CommunityOwnedDate="2011-02-09T01:52:26.667" />
  <row Id="21102" PostTypeId="2" ParentId="21100" CreationDate="2011-02-09T01:54:22.440" Score="2" ViewCount="" Body="&lt;p&gt;The rank of the matrix $A$ is equal to the dimension of the Image of $A$ (which is spanned by columns of $A$), if that's a sufficient enough geometrical explanation. You can read about vector spaces at &lt;a href=&quot;http://en.wikipedia.org/wiki/Vector_space&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Vector_space&lt;/a&gt; and about image of a matrix at &lt;a href=&quot;http://goo.gl/hKyvN&quot; rel=&quot;nofollow&quot;&gt;http://goo.gl/hKyvN&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="3731" LastEditorUserId="3731" LastEditorDisplayName="" LastEditDate="2011-02-09T02:03:30.640" LastActivityDate="2011-02-09T02:03:30.640" CommentCount="2" />
  <row Id="21103" PostTypeId="2" ParentId="21099" CreationDate="2011-02-09T01:59:02.770" Score="2" ViewCount="" Body="&lt;p&gt;Hint: The number of terms in the sum and the middle term are especially relevant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To elaborate a little, $m\cdot n = n + n + \cdots +n = \cdots + (n-1) + n + (n+1) + \cdots$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conversely, $(n-k) + \cdots + (n+k) = \cdots$.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2011-02-09T02:37:30.240" LastActivityDate="2011-02-09T02:37:30.240" CommunityOwnedDate="2011-02-09T02:09:21.707" />
  <row Id="21104" PostTypeId="1" AcceptedAnswerId="21110" CreationDate="2011-02-09T02:05:28.387" Score="2" ViewCount="164" Body="&lt;p&gt;Find&#xA;$$\lim\limits_{x \to \infty}\frac{2^x}{3^{x^2}}$$&#xA;I can only reason with this intuitively. since $3^{x^2}$ grows much faster than $2^x$ the limit as $x \to \infty$ of $f$ must be 0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a more rigorous way to show this?&lt;/p&gt;&#xA;" OwnerUserId="1069" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-09T02:42:15.273" LastActivityDate="2011-02-09T06:24:37.060" Title="$\lim\limits_{x \to \infty}\frac{2^x}{3^{x^2}}$" Tags="&lt;limit&gt;" AnswerCount="5" />
  <row Id="21105" PostTypeId="2" ParentId="21104" CreationDate="2011-02-09T02:07:50.010" Score="4" ViewCount="" Body="&lt;p&gt;How about comparison to $\displaystyle{\frac{2^x}{3^x}=\left(\frac{2}{3}\right)^x}$?&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2011-02-09T02:07:50.010" CommunityOwnedDate="2011-02-09T02:07:50.010" CommentCount="3" />
  <row Id="21106" PostTypeId="2" ParentId="21104" CreationDate="2011-02-09T02:08:13.163" Score="5" ViewCount="" Body="&lt;p&gt;Hint: $2^x \lt 3^x$ for sufficiently large $x$.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2011-02-09T02:08:13.163" CommunityOwnedDate="2011-02-09T02:08:43.550" CommentCount="4" />
  <row Id="21107" PostTypeId="2" ParentId="21100" CreationDate="2011-02-09T02:17:56.810" Score="9" ViewCount="" Body="&lt;p&gt;A rank of the matrix is probably the most important concept you learn in Matrix Algebra. There are two ways to look at the rank of a matrix. One from a theoretical setting and the other from a applied setting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From a theoretical setting, if we say that a linear operator has a rank $p$, it means that the range of the linear operator is a $p$ dimensional space.&#xA;From a matrix algebra point of view, column rank denotes the number of independent columns of a matrix while row rank denotes the number of independent rows of a matrix. An interesting, and I think a non-obvious (though the proof is not hard), fact is the row rank is same as column rank. When we say a matrix $A \in \mathbb{R}^{n \times n}$ has rank $p$, what it means is that if we take all vectors $x \in \mathbb{R}^{n \times 1}$, then $Ax$ spans a $p$ dimensional sub-space. Let us see this in a 2D setting. For instance, if &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$A = \left( \begin{array}{cc}&#xD;&#xA;1 &amp;amp; 2 \\&#xD;&#xA;2 &amp;amp; 4 \end{array} \right) \in \mathbb{R}^{2 \times 2}$ and let $x = \left( \begin{array}{c}&#xD;&#xA;x_1 \\&#xD;&#xA;x_2 \end{array} \right) \in \mathbb{R}^{2 \times 1}$, then $\left( \begin{array}{c}&#xD;&#xA;y_1 \\&#xD;&#xA;y_2 \end{array} \right) = y = Ax = \left( \begin{array}{c}&#xD;&#xA;x_1 + 2x_2 \\&#xD;&#xA;2x_1 + 4x_2 \end{array} \right)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rank of matrix $A$ is $1$ and we find that $y_2 = 2y_1$ which is nothing but a line passing through the origin in the plane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What has happened is the points $(x_1,x_2)$ on the $x_1 - x_2$ plane have all been mapped on to a line $y_2 = 2y_1$. Looking closely, the points in the $x_1 - x_2$ plane  along the line $x_1 + 2x_2 = c = \text{const}$, have all been mapped onto a single point $(c,2c)$ in the $y_1 - y_2$ plane. So the single point $(c,2c)$  on the $y_1 - y_2$ plane represents a straight line $x_1 + 2x_2 = c$ in the $x_1 - x_2$ plane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the reason why you cannot solve a linear system when it is rank deficient. The rank deficient matrix $A$ maps $x$ to $y$ and this transformation is neither onto (points in the $y_1 - y_2$ plane not on the line $y_2 = 2y_1$ e.g. $(2,3)$ are not mapped onto, which results in no solutions) nor one-to-one (every point $(c,2c)$ on the line $y_2 = 2y_1$ corresponds to the line $x_1 + 2x_2 =c$ in the $x_1 - x_2$ plane, which results in infinite solutions).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An observation you can make here is that the product of the slopes of the line $x_1 + 2x_2 = c$ and $y_2 = 2y_1$ is $-1$. This is true in general for higher dimensions as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From an applied setting, rank of a matrix denotes the &lt;strong&gt;information content&lt;/strong&gt; of the matrix. The lower the rank, the lower is the &quot;information content&quot;. For instance, when we say a rank $1$ matrix, the matrix can be written as a product of a column vector times a row vector i.e. if $u$ and $v$ are column vectors, the matrix $uv^T$ is a rank one matrix. So all we need to represent the matrix is $2n-1$ elements. In general, if we know that a matrix $A \in \mathbb{R}^{m \times n}$ is of rank $p$, then we can write $A$ as $U V^T$ where $U \in \mathbb{R}^{m \times p}$ and is of rank $p$ and $V \in \mathbb{R}^{n \times p}$ and is of rank $p$. So if we know that a matrix $A$ is of rank $p$ all we need is only $2np-p^2$ of its entries. So if we know that a matrix is of low rank, then we can compress and store the matrix and can do efficient matrix operations using it. The above ideas can be extended for any linear operator and these in fact form the basis for &lt;a href=&quot;http://en.wikipedia.org/wiki/Compressed_sensing&quot; rel=&quot;nofollow&quot;&gt;Compressed Sensing&lt;/a&gt;. You might also want to look up &lt;a href=&quot;http://en.wikipedia.org/wiki/Singular_value_decomposition&quot; rel=&quot;nofollow&quot;&gt;Singular Value Decomposition&lt;/a&gt; which gives us a nice (though expensive) way to make low rank approximations of a matrix which allows for compression.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From solving a linear system point of view, when the square matrix is rank deficient, it means that we do not have complete information about the system ergo we cannot solve the system.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2011-02-09T03:16:29.693" LastActivityDate="2011-02-09T03:16:29.693" CommentCount="1" />
  <row Id="21108" PostTypeId="2" ParentId="21104" CreationDate="2011-02-09T02:26:23.673" Score="4" ViewCount="" Body="&lt;p&gt;How about rewriting $\frac{2^x}{3^{x^2}}$ as $\frac{2^x}{3^x3^{x^2-x}}=(\frac{2}{3})^x\cdot\frac{1}{3^{x^2-x}}$?&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2011-02-09T02:26:23.673" />
  <row Id="21109" PostTypeId="2" ParentId="21099" CreationDate="2011-02-09T02:32:46.927" Score="5" ViewCount="" Body="&lt;p&gt;First of all, you can assume you're adding only positive numbers; otherwise the question isn't correct as written.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the sum of the numbers from $1$ to $n$ is ${\displaystyle {n^2 + n \over 2}}$. So the sum of the numbers from $m+1$ to $n$ is ${\displaystyle {n^2 + n \over 2} -  {m^2 + m \over 2}&#xD;&#xA;= {n^2 - m^2 + n - m \over 2} = {(n - m)(n + m + 1) \over 2}}$. You want to know which odd numbers $k$ can be written in this form for $n - m \geq 3$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $k$ were a prime $p$ that could be expressed this way, then you'd have $(n- m )(n+m+1) = 2p$. But $n - m \geq 3$, and $n + m + 1$ would only be bigger than that. Since $2p$ has only the factors $2$ and $p$, that can't happen. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So suppose $k$ is an odd non-prime, which you can write as $k_1k_2$ where $k_1 \geq k_2$ are odd numbers that are at least $3$. You now want to solve $(n-m)(n+m+1) = 2k_1k_2$. It's natural to set $n - m  = k_2$ (the smaller factor), and $2k_1 = n + m + 1$, the larger factor. Solving for $n$ and $m$ one gets ${\displaystyle n = {2k_1 + k_2 - 1 \over 2}}$ and ${\displaystyle m = {2k_1 - k_2 - 1 \over 2}}$. Since $k_1$ and $k_2$ are odd these are both integers. And since $k_1 \geq k_2$, the numbers $m$ and $n$ are nonnegative.&lt;/p&gt;&#xA;" OwnerUserId="3035" LastEditorUserId="3035" LastEditorDisplayName="" LastEditDate="2011-02-09T02:37:48.417" LastActivityDate="2011-02-09T02:37:48.417" />
  <row Id="21110" PostTypeId="2" ParentId="21104" CreationDate="2011-02-09T02:33:30.030" Score="2" ViewCount="" Body="&lt;p&gt;Rewriting the fraction as $$\frac{e^{x \ln 2}}{e^{x^2 \ln 3}} = e^{x \ln 2 - x^2 \ln 3}$$ may also help you (although perhaps no more than the other answers given already).&lt;/p&gt;&#xA;" OwnerUserId="5226" LastActivityDate="2011-02-09T02:33:30.030" CommentCount="4" />
  <row Id="21111" PostTypeId="2" ParentId="21100" CreationDate="2011-02-09T02:38:32.110" Score="3" ViewCount="" Body="&lt;p&gt;The rank of a matrix is of &lt;em&gt;major&lt;/em&gt; importance. It is closely connected to the nullity of the matrix (which is the dimension of the solution space of the equation $A\mathbf{x}=\mathbf{0}$), via the Dimension Theorem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Dimension Theorem.&lt;/strong&gt; Let $A$ be an $m\times n$ matrix. Then $\mathrm{rank}(A)+\mathrm{nullity}(A) = n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even if all you know about matrices is that they can be used to solve systems of linear equations, this tells you that the rank is &lt;em&gt;very&lt;/em&gt; important, because it tells you whether $A\mathbf{x}=\mathbf{0}$ has a single solution or multiple solutions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you think of matrices as being linear transformations (there is a correspondence between $m\times n$ matrices with coefficients in a field $\mathbf{F}$, and the linear transformations between a given vector space over $\mathbf{F}$ of dimension $n$ with a given basis, and a vector space of dimension $m$ with a given basis), then the rank of the matrix is the dimension of the image of that linear transformation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The simplest way of computing the Jordan Canonical Form of a matrix (an important way of representing a matrix) is to use the ranks of certain matrices associated to $A$; the same is true for the Rational Canonical Form.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Really, the rank just shows all over the place, it is usually relatively easy to compute, and has a lot of applications and important properties. They will likely not be completely apparent until you start seeing the myriad applications of matrices to things like vector calculus, linear algebra, and the like, but trust me, they're there.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2011-02-09T02:38:32.110" />
  <row Id="21112" PostTypeId="2" ParentId="20963" CreationDate="2011-02-09T02:50:04.603" Score="6" ViewCount="" Body="&lt;p&gt;What you are trying to do is a partial fraction decomposition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Idea.&lt;/strong&gt; Imagine your calculator is broken and a bunch of hoodlums stop you on the street, and demand at knifepoint that you compute $\frac{191}{105}$ as a decimal (part of their initiation into the Mathies Gang, you see), or else they will slit your throat. Unfortunately, since your calculator is broken, you can really only do divisions if the divisor is a single digit number (so you can use your fingers to do the operations; you're &lt;em&gt;very&lt;/em&gt; good with those, because you know the  multiplication tables of single digit numbers...). Luckily, you do notice that $105 = 3\times 5 \times 7$. Is there some way you can save your neck using this observation?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, you do know that $191 = 105 + 86$, so at least you know that $\frac{191}{105} = 1 + \frac{86}{105}$, so that takes care of the integer part of the fraction. What about $\frac{86}{105}$? Aha! Here's a clever idea: &lt;em&gt;maybe&lt;/em&gt; $\frac{86}{105}$ is really the &lt;em&gt;result&lt;/em&gt; of a sum of fractions! If you had a sum of fractions of the form&#xA;$$\frac{A}{3} + \frac{B}{5} + \frac{C}{7}$$&#xA;then to write it as a single fraction you would find the common denominator, $105$, and then do a bunch of operations, and end up with a fraction $\frac{\mathrm{something}}{105}$. If you can find an $A$, $B$, and $C$ so that the &lt;code&gt;something&lt;/code&gt; is $86$, then instead of computing $\frac{86}{105}$ you can do $\frac{A}{3}$, $\frac{B}{5}$, and $\frac{C}{7}$ (which you can do, since the denominators are single digit numbers), and then &lt;em&gt;add&lt;/em&gt; those decimals to get the answer. Can we? We do a bit of algebra:&#xA;$$\frac{A}{3} + \frac{B}{5} + \frac{C}{7} = \frac{35A + 21B + 15C}{105}$$&#xA;so you want $35A + 21B + 15C = 86$. As luck would have it, $A=B=1$ and $C=2$ works, so&#xA;$$\frac{86}{105} = \frac{1}{3} + \frac{1}{5} + \frac{2}{7}.$$&#xA;And now, all is well:&#xA;\begin{align*}&#xA;\frac{191}{105} &amp;amp;= 1 + \frac{86}{105}\\&#xA;&amp;amp;= 1 + \frac{1}{3} + \frac{1}{5} + \frac{2}{7}\\&#xA;&amp;amp;= 1 + (0.3333\ldots) + (0.2) + (0.285714285714\overline{285714}\ldots)&#xA;\end{align*}&#xA;and you can give those dangerous hoodlums their answer, and live to derive another day.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Your problem.&lt;/strong&gt; You want to do something similar with the polynomial quotient, with denominators that &quot;easy&quot;; in this case, degree $1$. The first task is to make the fraction &quot;less than $1$&quot;, by making sure the denominator has degree less than the numerator. You do this with &lt;a href=&quot;http://en.wikipedia.org/wiki/Polynomial_long_division&quot; rel=&quot;nofollow&quot;&gt;polynomial long division&lt;/a&gt; (see also this &lt;a href=&quot;http://math.stackexchange.com/questions/20301/how-to-know-that-a3b3-aba2-abb2/20305#20305&quot;&gt;recent answer&lt;/a&gt;). Doing the long division mentally, we have: to get $2x^2$ from $x^2+11x+30$ we multiply by $2$:&#xA;$$2x^2 + 11x = (x^2+11x+30)(2+\cdots)$$&#xA;that produces unwanted $11x + 60$, (well, you get $22x + 60$, but you &lt;em&gt;do&lt;/em&gt; want $11x$ of those, so you only have a leftover of $11x+60$); nothing to do about them, except cancel them out after the product is done. So you have&#xA;$$2x^2 + 11x = (x^2+11x+30)(2) - (11x+60).$$&#xA;So you can write&#xA;$$\frac{2x^2+11x}{x^2+11x+30} = 2 + \frac{-(11x+60)}{x^2+11x+30}.$$&#xA;Now we got the &quot;integer part&quot;, and we work on the &quot;fraction part&quot;. The denominator factors as $(x+5)(x+6)$, so we want to think of that fraction as the end result of doing a sum of the form&#xA;$$\frac{A}{x+5} + \frac{B}{x+6}.$$&#xA;Because the sum is &quot;smaller than $1$&quot; (numerator of degree smaller than the denominator) each of these fractions should also be &quot;smaller than one&quot;. So both the $A$ and $B$ will be constants.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we have:&#xA;\begin{align*}&#xA;\frac{-11x - 60}{x^2+11x+30} &amp;amp;= \frac{A}{x+5} + \frac{B}{x+6} \\&#xA;&amp;amp;= \frac{A(x+6)}{(x+5)(x+6)} + \frac{B(x+5)}{(x+6)(x+5)}\\&#xA;&amp;amp;= \frac{A(x+6) + B(x+5)}{(x+5)(x+6)}\\&#xA;&amp;amp;= \frac{Ax + 6A + Bx + 5B}{x^2+11x+30} = \frac{(A+B)x + (6A+5B)}{x^2+11x+30}.&#xA;\end{align*}&#xA;For this to work out, you need $(A+B)x + (6A+5B) = -11x-60$. That means we need $A+B=-11$ (so the $x$s agree) and $6A+5B = -60$ (so the constant terms agree).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That means $A=-11-B$ (from the first equation). Plugging into the second equation, we get&#xA;$$-60 = 6A+5B = 6(-11-B)+5B = -66 -6B + 5B = -66-B.$$&#xA;So that means that $B=60-66 = -6$. And since $A+B=-11$, then $A=-5$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is,&#xA;$$\frac{-11x-60}{x^2+11x+30} = \frac{-5}{x+5} + \frac{-6}{x+6}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Putting it all together, we have:&#xA;$$\frac{2x^2+11x}{x^2+11x+30} = 2 + \frac{-11x-60}{x^2+11x+30} = 2 - \frac{5}{x+5} - \frac{6}{x+6}.$$&#xA;And &lt;code&gt;&amp;lt;triumphant trumpet cue&amp;gt;&lt;/code&gt;&lt;em&gt;ta da!&lt;/em&gt;&lt;code&gt;&amp;lt;/triumphant trumpet cue&amp;gt;&lt;/code&gt; Your task is done. You've decomposed the quotient into partial fractions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Caveat.&lt;/strong&gt; You need to be careful if the denominator has repeated factors, or has factors that are of degree $2$ and cannot be further factored (e.g., $x^2+1$); but we'll leave that to another day/homework.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, let's hope the Mathies Gang doesn't ask for a proof of Goldbach's Conjecture from its next batch of pledges...&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-09T04:08:51.027" LastActivityDate="2011-02-09T04:08:51.027" CommentCount="1" />
  <row Id="21113" PostTypeId="2" ParentId="21003" CreationDate="2011-02-09T03:08:39.910" Score="-1" ViewCount="" Body="&lt;p&gt;For 5, for square roots, it seems almost too obvious to use the hypotenuse of right triangles. For higher roots, diagonals on cubes of higher dimensions?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For 3, a quadratic in one variable is also a conic section. For higher degrees...so this is for high school right? ...yeah this one isn't obvious. &lt;/p&gt;&#xA;" OwnerUserId="1919" LastActivityDate="2011-02-09T03:08:39.910" />
  <row Id="21114" PostTypeId="2" ParentId="21097" CreationDate="2011-02-09T03:11:52.810" Score="6" ViewCount="" Body="&lt;p&gt;About the size of $2^{\mathfrak c}$: MM is preserved by $\omega_2$-directed closed forcing, so we can change the size of $2^{\aleph_2}$ by forcing directly over a model of MM and preserving MM. This is a result of Paul Larson. (To contrast with Joel's answer: He shows how we can ensure models of MM with different sizes for $2^{\aleph_2}$ by manipulating the exponential function above a supercompact and then forcing MM. Paul's argument allows us to manipulate the relevant cardinal regardless of how the model of MM is originally obtained.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for implications beyond the continuum: MM implies PFA, and PFA implies SCH, the singular cardinal hypothesis (a very nice recent result of Matteo Viale), so it has a direct influence on (singular) cardinals larger than $\aleph_2$. For example, this implies that $2^\kappa=\kappa^+$ holds in models of MM for a proper class of cardinals $\kappa$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(The same is true in ZFC above a strongly compact.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PFA also implies the failure of $\square_\kappa$ at all $\kappa&amp;gt;\omega$. This was first shown by Todorcevic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Again, this also holds above strongly compact cardinals. This is all indirect evidence that the strength of PFA ought to be in the neighborhood of strong compactness, and that the method we know for establishing its consistency is essentially the only possible method. There are recent results of Matteo Viale and Christoph Weiss strengthening this connection, showing that the &lt;em&gt;standard&lt;/em&gt; forcing argument requires supercompactness. On the other hand, there are also recent &lt;a href=&quot;http://www.math.ucla.edu/~ineeman/fwsc.pdf&quot; rel=&quot;nofollow&quot;&gt;results&lt;/a&gt; of Itay Neeman exhibiting a different method of forcing PFA, although also from a supercompact and using properness.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a result that follows from MM and not from PFA: Magidor showed that MM implies that the principle &quot;very weak square&quot; fails at singulars of cofinality $\omega$. This shows that the singular cardinal combinatorics in models of MM is very interesting (and not yet well understood.)&lt;/p&gt;&#xA;" OwnerUserId="462" LastEditorUserId="462" LastEditorDisplayName="" LastEditDate="2011-02-09T21:44:10.123" LastActivityDate="2011-02-09T21:44:10.123" CommentCount="2" />
  <row Id="21115" PostTypeId="2" ParentId="21072" CreationDate="2011-02-09T03:21:44.733" Score="1" ViewCount="" Body="&lt;p&gt;The only easier thing comes when the arc length is not much greater than the length of the straight line segment from A to B, which means the angle of the circle is small.  Then you can use the small angle approximation $\cos\frac{\theta}{2}=1-\frac{\theta^2}{8}=\frac{c\theta}{2l}$, which can be solved with the quadratic formula: $\theta^2+\frac{4c\theta}{l}-88=0$  $\theta=\frac{2c}{l}+\sqrt{\frac{4c^2}{l^2}+8}$  Of course, when the difference is very small, you can just draw the straight line.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-02-09T03:21:44.733" />
  <row Id="21116" PostTypeId="2" ParentId="21100" CreationDate="2011-02-09T03:33:28.080" Score="3" ViewCount="" Body="&lt;p&gt;If you are interested in learning the why's of linear algebra I highly recommend viewing Gilbert Strang's &lt;a href=&quot;http://ocw.mit.edu/courses/mathematics/18-06-linear-algebra-spring-2010/&quot; rel=&quot;nofollow&quot;&gt;Linear Algebra Course&lt;/a&gt; and purchase his &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0980232716&quot; rel=&quot;nofollow&quot;&gt;book&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="4726" LastActivityDate="2011-02-09T03:33:28.080" CommentCount="1" />
  <row Id="21118" PostTypeId="1" CreationDate="2011-02-09T04:23:17.050" Score="-1" ViewCount="137" Body="&lt;p&gt;I was requesting somebody to help me discuss how the prime number rule helps to solve easily the sudoku game in just a few minutes and an example showing the relevant steps on how the rule is used&lt;/p&gt;&#xA;" OwnerUserId="6785" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-09T04:30:42.897" LastActivityDate="2011-02-09T06:15:41.447" Title="Prime number rule" Tags="&lt;prime-numbers&gt;&lt;puzzle&gt;" AnswerCount="1" CommentCount="6" />
  <row Id="21119" PostTypeId="2" ParentId="21086" CreationDate="2011-02-09T04:33:56.590" Score="1" ViewCount="" Body="&lt;p&gt;Given your assumption that the circuit board is isothermal and so is the aluminum, the heat transfer across the interface will be proportional to the temperature difference between them.  This means if you want to raise the circuit board temperature linearly with time (and no heat leaks out of the board) you need to maintain a constant temperature difference between the board and the heat source.  If you insulate the sides or have the board and aluminum large compared to the thickness you have a one-dimensional problem.  You have $\frac{dT_{board}}{dt}=\frac{CdQ}{dt}=Ck(T_{sink}-T_{board})$ where C is the heat capacity of the board and k is the thermal transfer constant&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-02-09T04:33:56.590" />
  <row Id="21120" PostTypeId="1" AcceptedAnswerId="21293" CreationDate="2011-02-09T05:15:23.623" Score="3" ViewCount="233" Body="&lt;p&gt;consider the ternary cantor set C, and the asscoiated cantor function f, and the associated Lebesgue-Stieltjes measure u.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;what is the integral of f over all of R with respect to u?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;my attempt:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i know that under the Lebesgue measure, the integral of the cantor function is 1/2 using a symmetry argument. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;but under this measure, u, is the integral still the same?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i was thinking about breaking up the integral into 3 parts: (-infty, 0), (0, 1), and (1, infty).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;since the cantor function is constant on (-infty, 0) and (1, infty), it follows that the integral is 0 on those intervals...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;now im kind of stuck here because im not sure how to handle (0, 1)...help would be appreciated&lt;/p&gt;&#xA;" OwnerUserId="6556" LastActivityDate="2011-02-27T16:18:37.067" Title="integral of the cantor function" Tags="&lt;measure-theory&gt;" AnswerCount="1" CommentCount="6" FavoriteCount="1" />
  <row Id="21121" PostTypeId="1" AcceptedAnswerId="21123" CreationDate="2011-02-09T06:05:54.253" Score="3" ViewCount="99" Body="&lt;p&gt;I'm trying to follow/understand a research paper that I have, and well, it's been a while since I've done this kind of math. At this point I have an nxn matrix &lt;strong&gt;H&lt;/strong&gt; and from that construct an (n-1)xn matrix &lt;strong&gt;H'&lt;/strong&gt; = $[\textbf{h}_1-\textbf{h}_n$, ..., $\textbf{h}_{1-n}-\textbf{h}_n]^T$. Now &quot;using orthogonal decomposition&quot; I'm to obtain &lt;strong&gt;H'&lt;/strong&gt;=$\textbf{QU}$, where $\textbf{Q}$ is an (n-1)x(n-1) orthogonal matrix and $\textbf{U}$, which is an (n-1)xn upper diagonal matrix.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I guess I'm hoping someone can better explain orthogonal decomposition and help me write the elements of $\textbf{Q}$ and $\textbf{U}$ in terms of the elements of &lt;strong&gt;H'&lt;/strong&gt;.&lt;/p&gt;&#xA;" OwnerUserId="6816" LastActivityDate="2011-02-09T06:30:03.963" Title="Orthogonal Decomposition of A Matrix" Tags="&lt;linear-algebra&gt;&lt;matrices&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="21122" PostTypeId="1" CreationDate="2011-02-09T06:09:19.740" Score="1" ViewCount="93" Body="&lt;p&gt;If $f$ is a real-valued function defined in a convex open set $E$ of $\mathbb{R}^n$, such that $\frac{\partial f}{\partial x_1}$ is $0$ for every $x\in E$, prove that $f(x)$ does not depend on the coordinate $x_1$. Why do we need convexity of $E$ in order to show this?&lt;/p&gt;&#xA;" OwnerUserId="6817" LastEditorUserId="1543" LastEditorDisplayName="" LastEditDate="2011-02-09T11:30:37.490" LastActivityDate="2011-02-09T17:07:23.390" Title="Real functions of n variables" Tags="&lt;real-analysis&gt;" AnswerCount="2" />
  <row Id="21123" PostTypeId="2" ParentId="21121" CreationDate="2011-02-09T06:12:14.337" Score="4" ViewCount="" Body="&lt;p&gt;Look up &lt;a href=&quot;http://en.wikipedia.org/wiki/QR_decomposition&quot; rel=&quot;nofollow&quot;&gt;QR decomposition&lt;/a&gt;. QR decomposition decomposes the matrix as a product of an orthonormal matrix and an upper triangular matrix. i.e. $QQ^T = Q^TQ = I$ and $R$ is upper triangular. QR decomposition is nothing but &lt;a href=&quot;http://en.wikipedia.org/wiki/Gram%E2%80%93Schmidt_process&quot; rel=&quot;nofollow&quot;&gt;Gram-Schmidt&lt;/a&gt; orthogonalization process. It is relatively easy to code up QR based on Gram Schmidt orthogonalization process. Wikipedia does a nearly complete job of explaining both QR and Gram-Schmidt. The main idea behind QR/Gram Schmidt is to construct an orthonormal basis for the column space/range of $A$ using the columns of the matrix $A$. The idea is relatively simple. &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Take the first column of $A$ and make it into a unit vector by dividing out by the norm&lt;/li&gt;&#xA;&lt;li&gt;Take the second column of $A$ and remove the component along the first column of $A$ and make it into a unit vector by dividing out by the norm&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;In general, take the $k^{th}$ column of $A$ and remove the components from the previous $k-1$ columns of $A$ (i.e. remove the subspace spanned by the first $k-1$ columns of $A$) and make it into a unit vector by dividing out by the norm&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This orthonormal set of vectors give you the $Q$ matrix and the (rotation) operations you perform on the vectors of the $Q$ matrix go into the $R$ (rotation) matrix.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;However, the QR algorithm using Gram-Schmidt orthogonalization process can become unstable. Hence couple of other algorithms based on &lt;a href=&quot;http://en.wikipedia.org/wiki/Givens_rotation&quot; rel=&quot;nofollow&quot;&gt;Givens Rotation&lt;/a&gt;, &lt;a href=&quot;http://en.wikipedia.org/wiki/Householder_reflection&quot; rel=&quot;nofollow&quot;&gt;Householder reflection&lt;/a&gt; are used in practise since they tend to be more stable than Gram-Schmidt algorithm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a one to one correspondence between the algorithm based on Givens Rotation and the algorithm based on Householder reflection. (Just like the correspondence between solving a system by Gauss Elimination and LU). So technically the stability of the algorithm based on Givens rotation is same as the stability of the algorithm based on Householder reflection.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2011-02-09T06:30:03.963" LastActivityDate="2011-02-09T06:30:03.963" CommentCount="2" />
  <row Id="21124" PostTypeId="2" ParentId="21118" CreationDate="2011-02-09T06:15:41.447" Score="1" ViewCount="" Body="&lt;p&gt;Perhaps this is what you mean by &quot;the prime number rule:&quot;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The problem with deducing missing numbers does not arise if we use a form of Gödel numbers. Here we don't just multiply the Sudoku numbers together we use the corresponding prime number. So for a '1' we use the first prime number '2', for '2' use the second prime '3'; for '3' use '5'; for '4' use '7' and so on...&#xA;  We can use this trick for finding any number of missing numbers in a Sudoku group (of any size) just by a little multiplication and division. Why does this work? Because prime numbers are well, prime, we can't make a prime number by multiplying two other numbers together.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;From the Gödel numbers section on &lt;a href=&quot;http://www.sudokudragon.com/sudokutheory.htm&quot; rel=&quot;nofollow&quot;&gt;Sodoku Dragon&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="6816" LastActivityDate="2011-02-09T06:15:41.447" CommentCount="1" />
  <row Id="21125" PostTypeId="2" ParentId="21104" CreationDate="2011-02-09T06:24:37.060" Score="0" ViewCount="" Body="&lt;p&gt;After so many equivalent examples one more should not be missed because of its form. Here we get the numerator constant and keep the notation in powers of &lt;em&gt;2&lt;/em&gt; and &lt;em&gt;3&lt;/em&gt;. Denote $ &amp;#223;=\frac{\ln(2)}{\ln(3)} $ Then &#xA;$$ \frac{2^x}{3^{x^2}}=3^{&amp;#223;x-x^2}=3^{(&amp;#223;/2)&amp;#178;-(x-\frac&amp;#223;2)^2} = \frac{2^{&amp;#223;/4}}{3^{(x-&amp;#223;/2)^2}} $$ where the numerator is constant. We see also, that the denominator is minimal if $x=&amp;#223;/2$ and the whole expression is maximal by the exponent $&amp;#223;x-x^2$ of the second term, its first derivative $&amp;#223;-2x$ (which is zero at $x=ß/2$) and its second derivative $-2$ which is negative and shows, that the whole expression has a maximum there.&lt;/p&gt;&#xA;" OwnerUserId="1714" LastActivityDate="2011-02-09T06:24:37.060" />
  <row Id="21126" PostTypeId="2" ParentId="21122" CreationDate="2011-02-09T06:32:33.180" Score="0" ViewCount="" Body="&lt;p&gt;Imagine a line that is parallel to the x1 axis. If E were not convex, it might happen that its intersection with the line yields multiple segments. Then, f could be constant in every segment (i.e. the partial derivatives are zero where defined), but &quot;tacitly&quot; change its value in between segments.&lt;/p&gt;&#xA;" OwnerUserId="6818" LastActivityDate="2011-02-09T06:32:33.180" CommentCount="1" />
  <row Id="21127" PostTypeId="1" CreationDate="2011-02-09T06:40:20.957" Score="2" ViewCount="107" Body="&lt;p&gt;Every month, a girl gets An allowance. Assume last year she had no money, and kept all the money she has earned up to now. Then she spends $\frac{1}{2}$ of her money on clothes, then $\frac{1}{3}$ of the remaining money on games, and then $\frac{1}{4}$ of the remaining money on toys. After she bought all of that, she had $7777$ left. Assuming she only gets money by allowance, how much money does she earn every month? &lt;/p&gt;&#xA;" OwnerUserId="6785" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2011-02-09T20:22:34.217" LastActivityDate="2011-02-10T23:19:38.810" Title="Interests and Amounts" Tags="&lt;algebra&gt;&lt;arithmetic&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="21128" PostTypeId="1" AcceptedAnswerId="21139" CreationDate="2011-02-09T06:42:52.333" Score="14" ViewCount="606" Body="&lt;p&gt;Hi, I'm a undergraduate who wishes to learn category theory but I only have basic knowledge of linear algebra and set theory, I've also had a short course on number theory which used some basic concepts about groups and modular arithmetic. Is it too early to start learning category theory? should I wait to take a course on abstract algebra?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it very important to use category theory facts in a first course in group theory, ring theory, fields and Galois theory, modules and tensor products (each of those is a one semester course), would that make it a 'better' course?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was unsure to learn category theory early but this post &lt;a href=&quot;http://math.stackexchange.com/questions/213/mathematical-subjects-you-wish-you-learned-earlier&quot;&gt;Mathematical subjects you wish you learned earlier&lt;/a&gt; inspired me to ask you given my background.&lt;/p&gt;&#xA;" OwnerUserId="1938" LastEditorUserId="1938" LastEditorDisplayName="" LastEditDate="2011-02-10T03:45:26.657" LastActivityDate="2011-02-10T03:45:26.657" Title="When to learn category theory?" Tags="&lt;abstract-algebra&gt;&lt;category-theory&gt;" AnswerCount="7" CommentCount="7" FavoriteCount="5" />
  <row Id="21129" PostTypeId="1" CreationDate="2011-02-09T07:02:17.837" Score="2" ViewCount="99" Body="&lt;p&gt;Let $f(x):\mathbb R^n \to \mathbb R$ be a smooth function, and let $f(0)=0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I alway see that someone rewrite the function in the form $f(x)=x_1b_1(x_1)+x_2b_2(\overline {x_2})+...+x_nb_n(\overline x)$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where the $\overline {x_i}=[x_1\ \ x_2\ \ ...\ \ x_i]^\mathrm{T}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can someone provide a proof?&lt;/p&gt;&#xA;" OwnerUserId="6820" LastEditorUserId="6820" LastEditorDisplayName="" LastEditDate="2011-02-10T02:11:58.033" LastActivityDate="2011-02-10T02:11:58.033" Title="Proof : Given that function $f(x)$ vanishes at 0, can we rewrite $f(x) = x_1b_1(x_1)+x_2 b_2(\overline { x_2})+\cdots+x_nb_n( x)$?" Tags="&lt;functions&gt;" CommentCount="5" />
  <row Id="21130" PostTypeId="2" ParentId="21122" CreationDate="2011-02-09T07:04:04.873" Score="1" ViewCount="" Body="&lt;p&gt;Here is an instructive example why:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider two disjoint open balls $B_{\frac{1}{2}}((-1,0)) \subset \mathbb{R}^2$ and $B_{\frac{1}{2}}((1,0)) \subset \mathbb{R}^2$. Let your set $E$ be their union - clearly not a convex set. Let $f:E \rightarrow \mathbb{R}$ be given by:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ f((x,y)) =\left\{\begin{matrix}&#xD;&#xA; y&amp;amp; \text{ if   } x&amp;gt;0\\ &#xD;&#xA; -y&amp;amp; \text{ if   } x&amp;lt;0&#xD;&#xA;\end{matrix}\right.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clearly, $f$ is differentiable and $\dfrac{\partial f}{\partial x} = 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here, the non-convexity allows you to define the cases, while keeping the function continuous.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Above $E$ was disconnected. So let's suppose $E$ is an open, non-convex, &lt;strong&gt;connected&lt;/strong&gt; set. Let $g:E \rightarrow \mathbb{R}^m$ be any function, such that $\frac{\partial g}{\partial x_1} = 0$. We can write any $v \in E$ as $v=(x_1,y)$, where $y$ is an $n-1$ dimensional vector. Fix some vector $y_0$, let $A = \{ (x_1,y) \in E: y = y_0 \}$. We can think of $A$ as a subset of $\mathbb{R}$ - it is just some affine line. It is clear that if $A \subset \mathbb{R}$ is connected, then $g(A) = \{ z \}$ for some $z \in \mathbb{R}^m$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now suppose that $A \subset \mathbb{R}$ is not connected. Suppose that in this case there are two points $(x_i,y_0), (x_j, y_0)$ such that $g(x_i,y_0) \neq g(x_j,y_0)$. Then since we assumed that $E$ is connected and open, it follows that $E$ is in fact path-connected. So there is some smooth $\gamma: [0,1] \rightarrow E$ such that $\gamma(0) = (x_i,y_0)$ and $\gamma(1) = (x_j,y_0)$. Note here that $\gamma ([0,1]) \subset E$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $\gamma(c) = (x,y)$. Then by the chain rule we have that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$[D(g \circ \gamma)(c)] = [Dg(\gamma(c))]\cdot [D\gamma(c)]$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here $  D(g \circ \gamma)(x,y) $ is an $m \mathbb{x} 1$ matrix. But we have that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$[D(g \circ \gamma)(c)]_{i1} = \sum_{k=1}^n \left ( [Dg(\gamma(c))]_{ik} \cdot [D\gamma(c)]_{k1} \right )$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now notice that even though $[Dg(\gamma(c))]_{i1} \cdot [D\gamma(c)]_{11}$ will always be zero, the rest of the sum doesn't have to be. This captures the fact that $E$ is not a convex set; i.e. you can change the values along the $x_1$ axis without the function having any dependency on $x_1$ whatsoever. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So $g(x_i,y_0) \neq g(x_j,y_0)$ is possible, even though $\frac{\partial g}{\partial x_1} = 0$.&lt;/p&gt;&#xA;" OwnerUserId="5105" LastEditorUserId="5105" LastEditorDisplayName="" LastEditDate="2011-02-09T17:07:23.390" LastActivityDate="2011-02-09T17:07:23.390" />
  <row Id="21131" PostTypeId="1" AcceptedAnswerId="21132" CreationDate="2011-02-09T07:16:28.537" Score="7" ViewCount="295" Body="&lt;p&gt;What is an &lt;em&gt;intuitive&lt;/em&gt; meaning of the null space of a matrix? Why is it useful?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not looking for textbook definitions... my textbook gives me the definition, but I just don't &quot;get&quot; it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;E.g.: I think of the &lt;em&gt;rank&lt;/em&gt; &lt;code&gt;R&lt;/code&gt; of a matrix as the minimum number of dimensions that a linear combination of its columns would have; it tells me that, if I combined the vectors in its columns in some order, I'd get a set of coordinates for an &lt;code&gt;R&lt;/code&gt;-dimensional space, where &lt;code&gt;R&lt;/code&gt; is minimum (please correct me if I'm wrong). So that means I can relate &lt;em&gt;rank&lt;/em&gt; (and also dimension) to actual coordinate systems, and so it makes sense to me. But I can't think of any physical meaning for a null space... could someone explain what its meaning would be, for example, in a coordinate system?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="4890" LastEditorUserId="5105" LastEditorDisplayName="" LastEditDate="2011-02-09T14:23:44.280" LastActivityDate="2011-02-09T18:17:13.410" Title="Physical Meaning of Null Space of a Matrix" Tags="&lt;linear-algebra&gt;&lt;matrices&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="3" />
  <row Id="21132" PostTypeId="2" ParentId="21131" CreationDate="2011-02-09T07:33:32.100" Score="5" ViewCount="" Body="&lt;p&gt;If $A$ is your matrix, the null-space is simply put, the set of all vectors $v$ such that $A \cdot v = 0$. It's good to think of the matrix as a linear transformation; if you let $h(v) = A \cdot v$, then the null-space is again the set of all vectors that are sent to the zero vector by $h$. Think of this as the set of vectors that &lt;em&gt;lose their identity&lt;/em&gt; as $h$ is applied to them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the null-space is equivalently the set of solutions to the homogeneous equation $A \cdot v = 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nullity is the complement to the rank of a matrix. They are both really important; here is a &lt;a href=&quot;http://math.stackexchange.com/questions/21100/importance-of-rank-of-a-matrix&quot;&gt;similar question&lt;/a&gt; on the rank of a matrix, you can find some nice answers why there.&lt;/p&gt;&#xA;" OwnerUserId="5105" LastEditorUserId="5105" LastEditorDisplayName="" LastEditDate="2011-02-09T12:03:07.333" LastActivityDate="2011-02-09T12:03:07.333" CommentCount="1" />
  <row Id="21133" PostTypeId="2" ParentId="21131" CreationDate="2011-02-09T07:41:17.407" Score="3" ViewCount="" Body="&lt;p&gt;The rank $r$ of a matrix $A \in \mathbb{R}^{m \times n}$, as you have said is the dimension of the column space ($r$ is also the dimension of the row space as well) i.e. the dimension of the space spanned by vectors which are obtained by a linear combination of the columns of $A$, equivalently the range of $A$. (The use of the word &quot;minimum&quot; in the question is unnecessary). However each column vector has $m$ components and the vectors in the range of $A$ has $m$ components as such but span only a $r (\leq m)$ dimensional subspace instead of a $m$ dimensional space. So we are missing out spanning the remaining $m-r$ dimensional subspace of the $m$ dimensional space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The left null-space now plays the roll of spanning the remaining $m-r$ dimensional subspace. This is why the left null-space is orthogonal to the column space. So the left null-space along with the column space now spans the entire $m$ dimensional space i.e. if $C = \{y \in \mathbb{R}^{m \times 1}: y = Ax\text{ for some }x \in \mathbb{R}^{n \times 1} \}$ and $Z_L = \{z \in \mathbb{R}^{m \times 1}:z^T A = 0 \}$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;then $Z_L \cup C = \mathbb{R}^{m}$ and $Z_L \perp C$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The right null-space plays the analogous roll for the rows. The rows span only a $r$ dimensional subspace of the $n$ dimensional space. The right null-space now plays the roll of spanning the remaining $n-r$ dimensional subspace. This is why the right null-space is orthogonal to the row space. So the right null-space along with the row space now spans the entire $n$ dimensional space i.e. if $R = \{y \in \mathbb{R}^{n \times 1}: y = A^Tx\text{ for some }x \in \mathbb{R}^{m \times 1} \}$ and $Z_R = \{z \in \mathbb{R}^{n \times 1}: Az = 0 \}$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;then $Z_R \cup R = \mathbb{R}^{n}$ and $Z_R \perp R$&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2011-02-09T18:17:13.410" LastActivityDate="2011-02-09T18:17:13.410" CommentCount="1" />
  <row Id="21134" PostTypeId="2" ParentId="21128" CreationDate="2011-02-09T07:41:44.270" Score="3" ViewCount="" Body="&lt;p&gt;I tried reading Mac Lane's classic &lt;em&gt;Categories for the working mathematician&lt;/em&gt; the summer after I finished first year. That didn't go very well, and I didn't learn much. I suspect the reason was that the examples were too inaccessible at that stage. On the other hand, I did understand Lawvere and Rosebrugh's &lt;em&gt;Sets for Mathematics&lt;/em&gt; at that time, but again the lack of examples meant that I didn't appreciate the significance and elegance of categorification.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would recommend not learning category theory until you've seen enough concrete examples to be able motivate its study properly — at the very least one course in group theory, one in linear algebra, and one in general point-set topology. Generalised abstract nonsense is better appreciated when you realise, for example, that the Cartesian product of sets, the direct product of groups, the direct product of vector spaces, and the product topology all satisfy the same universal property: Given two objects $A$ and $B$, their product is an object $A \times B$ together with a pair of arrows (structure-preserving maps) $p_1 : A \times B \to A$, $p_2 : A \times B \to B$ which are &lt;em&gt;universal&lt;/em&gt;, in the sense that for any pair of arrows $f: X \to A$, $g: X \to B$, there is a unique arrow $(f, g): X \to A \times B$ such that $p_1 \circ (f, g) = f$ and $p_2 \circ (f, g) = g$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, learning category theory can also lead to some insights: for example, it is a remarkable fact that in the category of vector spaces, the direct sum of finitely many vector spaces is the same as the direct product of them. This is a mysterious coincidence, since in the other categories you are familiar with, the constructions corresponding to direct sum and direct product are distinct and in general not the same. This, together with some other facts, should convince you that something special and very nice is going on in linear algebra.&lt;/p&gt;&#xA;" OwnerUserId="5191" LastActivityDate="2011-02-09T07:41:44.270" CommentCount="5" />
  <row Id="21135" PostTypeId="2" ParentId="21068" CreationDate="2011-02-09T07:41:46.797" Score="3" ViewCount="" Body="&lt;p&gt;James Delaney's paper &quot;&lt;a href=&quot;http://www.math.hmc.edu/~benjamin/papers/SuryLetter.pdf&quot; rel=&quot;nofollow&quot;&gt;Groups of Arithmetical Functions&lt;/a&gt;&quot; (&lt;em&gt;Mathematics Magazine&lt;/em&gt; 78 (2), 83-97, 2005) may be worth a look.  Here are some of his main results.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The group $U$ of units of the ring of arithmetic functions (pointwise addition and Dirichlet convolution ($\ast$) as multiplication) can be expressed as the direct sum $U = C \oplus U_M \oplus U_A$, where $C$ is the subgroup of scalar functions, $U_M$ is the subgroup of multiplicative functions, and $U_A$ is the subgroup of what he calls &quot;anti-multiplicative&quot; functions defined by $f(1) = 1$ and $f(p^k) = 0$ when $p^k$ is a prime power with $k &amp;gt; 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;He then shows, if $U_1 = U_M \oplus U_A$, then $(U_1, \ast)$ is a divisible torsion-free group and thus can be viewed as a vector space over the rationals.  Therefore $U_M$ and $U_A$ are divisible subgroups and are thus complementary subspaces when $U_1$ is regarded as a vector space.  This also implies that the $n$th root of a multiplicative function is multiplicative.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, he proves that the functions $\{\epsilon_{\alpha} \lambda_{\beta} | \alpha, \beta \in \mathbb{C}, \beta \neq 0\}$ are linearly independent, where $\epsilon_{\alpha}(n) = n^{\alpha}$ and $\lambda_{\beta}(p_1^{k_1} \cdots p_r^{k_r}) = \beta^{k_1 + \cdots + k_r}$ (a generalization of Liouville's $\lambda$).    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have pretty much paraphrased straight from Delaney's paper.  My algebra background is too deficient for me to be able to comment on how interesting (or uninteresting) these results actually are or how well they satisfy what the OP is looking for.  (Perhaps someone else could help out with that?)&lt;/p&gt;&#xA;" OwnerUserId="2370" LastActivityDate="2011-02-09T07:41:46.797" CommentCount="3" />
  <row Id="21136" PostTypeId="2" ParentId="21062" CreationDate="2011-02-09T07:46:17.837" Score="1" ViewCount="" Body="&lt;p&gt;If you are looking for &quot;solving completely&quot; the game of bridge (like some variants of poker), I would say you definitely won't find any current literature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are just looking for applications of game theory, I believe Bridge World did have a couple of articles by Jeff Rubens published, which talked about game theory.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Mixing It Up, published in December 2005.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Mixing It Up, II, published in January 2006.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Hope that helps.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2011-02-09T07:46:17.837" CommentCount="2" />
  <row Id="21137" PostTypeId="1" AcceptedAnswerId="21278" CreationDate="2011-02-09T08:09:49.590" Score="0" ViewCount="114" Body="&lt;p&gt;Hello,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The title is self-explanatory . :-)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;thanks,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Actually my description is not comprehensive : I want to draw the graphical solution of an operation research problem &lt;/p&gt;&#xA;&#xA;&lt;p&gt;maximize $z = 7x_1 + 9x_2$. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;under $x_1 + x_2 \leq 8$.&lt;/li&gt;&#xA;&lt;li&gt;$x_2\leq 4$. &lt;/li&gt;&#xA;&lt;li&gt;$2x_1 + 3x_2 \leq 19$.&lt;/li&gt;&#xA;&lt;li&gt;$x_1\geq 0$.&lt;/li&gt;&#xA;&lt;li&gt;$x_2\geq 0$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="4475" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-09T17:20:51.527" LastActivityDate="2011-02-09T23:28:07.327" Title="How to draw the equation of the form ax + by = 0 please ?" Tags="&lt;geometry&gt;" AnswerCount="2" CommentCount="5" />
  <row Id="21138" PostTypeId="2" ParentId="21137" CreationDate="2011-02-09T08:14:12.087" Score="1" ViewCount="" Body="&lt;p&gt;To draw any straight line, all you need is two points. In the case of $ax+by = 0$, the line clearly passes through $(0,0)$. Further, the line passes through couple of other points which are easy to recognize namely $(b,-a)$ and $(-b,a)$. So just mark these points on the sheet and draw it. If $a=0$, then the line is nothing but the $X$ axis while if $b=0$, then the line is nothing but the $Y$ axis.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastActivityDate="2011-02-09T08:14:12.087" CommentCount="1" />
  <row Id="21139" PostTypeId="2" ParentId="21128" CreationDate="2011-02-09T08:17:22.843" Score="12" ViewCount="" Body="&lt;p&gt;I very often find some knowledge of category theory useful to understand things conceptually.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A book that one could read before studying mathematics at the university is Lawvere's and Schanuel's &lt;a href=&quot;http://books.google.com/books?id=o1tHw4W5MZQC&amp;amp;printsec=frontcover&amp;amp;dq=Schanuel+introduction+categor&amp;amp;source=bl&amp;amp;ots=4u4sAtE60d&amp;amp;sig=KlEyA-zEXxMtAKrx-SLO8pQRr64&amp;amp;hl=de&amp;amp;ei=9UtSTfe6GMiaOvGZ0ZQI&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=3&amp;amp;sqi=2&amp;amp;ved=0CCsQ6AEwAg#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;Conceptual mathematics&lt;/a&gt;. This is an introduction to category theoretic ideas on a most elementary level.&lt;/p&gt;&#xA;" OwnerUserId="367" LastActivityDate="2011-02-09T08:17:22.843" CommentCount="2" />
  <row Id="21140" PostTypeId="1" AcceptedAnswerId="21143" CreationDate="2011-02-09T08:30:02.613" Score="2" ViewCount="48" Body="&lt;p&gt;I have a &lt;a href=&quot;http://en.wikipedia.org/wiki/Polyhedron&quot; rel=&quot;nofollow&quot;&gt;polyhedron&lt;/a&gt; $P$, in 3D, which consists of $f$ faces, each face consists of $V$ vertexes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is, how to choose a consistent normal orientation for all the faces? Consistent here means that if I unfold all the faces of that polyhedron onto a common plane all the normals will agree. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the sake of argument, the polyhedron can be either convex or concave. However, it can be unfolded onto a plane without overlap. &lt;/p&gt;&#xA;" OwnerUserId="550" LastEditorUserId="550" LastEditorDisplayName="" LastEditDate="2011-02-09T09:03:51.027" LastActivityDate="2011-02-09T09:06:45.880" Title="Algorithm to Choose Consistent Normals for All Faces on a Polyhedron" Tags="&lt;geometry&gt;&lt;computational-geometry&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="21141" PostTypeId="1" AcceptedAnswerId="21169" CreationDate="2011-02-09T08:38:52.863" Score="2" ViewCount="74" Body="&lt;p&gt;In a group of permutations of $n$ elements, there are two permutations $P_1$ and $P_2$ such that $P_2=P_1^e$. $P_1$ and $P_2$ have the same order $o$: $P_1^o = P_2^o$. How can I find $e$? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$n$, $P_1, P_2$ and $o$ are known, $e$ is not known. Problem is to find $e$. &lt;/p&gt;&#xA;" OwnerUserId="6823" LastEditorUserId="6823" LastEditorDisplayName="" LastEditDate="2011-02-09T11:48:12.357" LastActivityDate="2011-02-09T13:09:12.053" Title="Equation on permutations" Tags="&lt;permutations&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="21142" PostTypeId="1" AcceptedAnswerId="21145" CreationDate="2011-02-09T08:49:43.837" Score="4" ViewCount="68" Body="&lt;p&gt;I got this from Thomson et al.'s freely available &quot;&lt;a href=&quot;http://classicalrealanalysis.com/download.aspx&quot; rel=&quot;nofollow&quot;&gt;Elementary Real Analysis&lt;/a&gt;&quot; p.356.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They introduce Baire's category theorem through a game where, given two players (A) and (B)&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Player (A) is given a subset $A$ of $\mathbb{R}$, and player (B) is given the complementary set $B = \mathbb{R} \backslash A$. Player (A) first selects a closed interval $I_1 \subset \mathbb{R}$; then player (B) chooses a closed interval $I_2 \subset I_1$. The players alternate moves, a move consisting of selecting a closed interval inside the previously chosen interval.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The play of the game thus determines a descending sequence of closed intervals&#xA;  \begin{align} &#xA;  I_1 \supset I_2 \supset \ldots I_n \supset &#xA;  \ldots\end{align}&#xA;  where player (A) chooses those with odd index and player (B) those with even index. If&#xA;  \begin{align}&#xA;  A \: \bigcap_n^{\infty} I_n \neq \emptyset&#xA;  \end{align}&#xA;  then player (A) wins; otherwise player (B) wins.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Then they argue that if player (A) is dealt the irrational set and (B) is dealt the rational set, (A) always has a strategy to win.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm confused in several ways by this argument. One confusion is about the term &quot;closed interval&quot;. Does, for example, the closed interval [1,1] count as an interval? Because if that's the case, can't whoever has the first turn just end the game then and there without regard to whether he has the rationals or irrationals? Say, if (A) received the rationals, he can just pick [0.5,0.5]. Game over. (But I'm guessing probably not, because the game wouldn't happen)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If that is not the case, and an interval has to be defined $[a,b]$ s.t. $a &amp;lt; b$, isn't it always true that for any $I_{2n+1} = [a_{2n+1}, b_{2n+1}]$ for the odd numbered turns of (A), and where (A) has the irrationals, there exists some $q \in [a_{2n+1},b_{2n+1}]$ s.t. $q$ is rational? Because $\mathbb{Q}$ is also dense on the real line. Or is this argument relying on the fact that a countable intersection of closed sets is always closed? And that it can converge to a single point. But if it converges to a single point, it might be a closed set, but is it still a closed interval? And does that still count as winning? And can't (A) still play this same game even if he were dealt the rationals if all that is needed is that his choice of intervals converge to a single point?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've been thinking about this for a few days now and none of the ways I've approached it convince me that their discussion of the game is true (though I do trust that it is true, since the authors are mathematicians and I'm not), so any help would be appreciated.&lt;/p&gt;&#xA;" OwnerUserId="6690" LastActivityDate="2011-02-09T10:04:00.933" Title="Help understanding game version of Baire category theorem" Tags="&lt;real-analysis&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="21143" PostTypeId="2" ParentId="21140" CreationDate="2011-02-09T09:06:45.880" Score="2" ViewCount="" Body="&lt;p&gt;You didn't specify any data structures you have for your polyhedron (e.g. adjacency lists/matrices between vertices, edges, faces), and this is a math site, so I'll only address the mathematical aspect of the question and not provide an efficient implementation with data structures.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any two adjacent faces share an edge. A normal defines an orientation of a face, and thus of its edges. If you don't now the orientation of the edges with respect to their faces yet, either, you can determine the orientation a face normal induces on the edges as follows (you need at least something like an ordered list of the vertices to define the face): Traverse the edges (i.e. the vectors between consecutive pairs of vertices) of the face and add up the cross products of all pairs of consecutive edges; then the order in which you traversed the edges is or isn't the one corresponding to the normal according as the dot product of the sum of the cross products with the normal is positive or negative.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So that gives you the relationship between the choice of normals and the orientation of the edges these choices induce. Then you choose a normal on an arbitrary one of the faces and for a pair of adjacent faces where you've chosen a normal on one and not on the other, choose the other normal so that the induced orientations of the edges are opposite.&lt;/p&gt;&#xA;" OwnerUserId="6622" LastActivityDate="2011-02-09T09:06:45.880" CommentCount="1" />
  <row Id="21144" PostTypeId="1" CreationDate="2011-02-09T09:10:25.970" Score="4" ViewCount="248" Body="&lt;p&gt;Can someone explain intuitively what the Fundamental Theorem of Linear Algebra states? and why specifically it is called the above? Specifically, what makes it 'Fundamental' in the broad scope of the theory.&lt;/p&gt;&#xA;" OwnerUserId="6824" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-09T17:18:02.560" LastActivityDate="2011-05-17T15:48:55.960" Title="Intuitive explanation of the Fundamental Theorem of Linear Algebra" Tags="&lt;linear-algebra&gt;&lt;intuition&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="21145" PostTypeId="2" ParentId="21142" CreationDate="2011-02-09T09:53:12.657" Score="2" ViewCount="" Body="&lt;p&gt;A closed interval is of the form $[a,b]$, $[a, \infty)$ or $(-\infty,b]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $A$ deals first, he can start with a bounded interval, so let's assume that all intervals $I_{n}$ are bounded (I leave it to you to describe a winning strategy for $B$ if $A$ is silly enough to always deal an unbounded interval). The only really interesting case is if the length of the interval converges to zero, for if not, the intersection will be a non-trivial interval, which will then contain an irrational number. So, if $|I_{n}| \to 0$ then there will be a unique real number $x \in \bigcap_{n} I_{n}$ by the priniple of &lt;a href=&quot;http://en.wikipedia.org/wiki/Nested_intervals&quot; rel=&quot;nofollow&quot;&gt;nested intervals&lt;/a&gt; and the question becomes: can $B$ force this number to be rational so that $A$ loses? The answer is &lt;strong&gt;no&lt;/strong&gt;. The reason is that the decimal expansion of a rational number is eventually periodic, so $A$ only has to take care to break any possible period of the decimal expansion of $x$ at each step and this is easily achieved by choosing an interval of sufficiently small length.&lt;/p&gt;&#xA;" OwnerUserId="5363" LastEditorUserId="5363" LastEditorDisplayName="" LastEditDate="2011-02-09T10:04:00.933" LastActivityDate="2011-02-09T10:04:00.933" CommentCount="2" />
  <row Id="21146" PostTypeId="1" AcceptedAnswerId="21147" CreationDate="2011-02-09T09:53:42.313" Score="4" ViewCount="242" Body="&lt;p&gt;Hello,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I prove that $$n^4 + 4$$ is composite for all n &gt; 5?&lt;br&gt;&#xA;This problem looked very simple, but I took 6 hours and ended up with nothing :(. I broke it into cases base on quotient remainder theorem, but it did not give any useful information.&lt;br&gt;&#xA;Plus, I try to factor it out:&#xA;$$n^4 - 16 + 20 = ( n^2 - 4 )( n^2 + 4 ) - 5.4$$&#xA;If a composite add a number that is a multiple of 5, is there anything special? A hint would be suffice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks,&lt;br&gt;&#xA;Chan &lt;/p&gt;&#xA;" OwnerUserId="5711" LastEditorUserId="5711" LastEditorDisplayName="" LastEditDate="2011-02-09T10:02:11.223" LastActivityDate="2011-02-12T17:34:57.790" Title="Prove a number is composite" Tags="&lt;prime-numbers&gt;&lt;elementary-number-theory&gt;" AnswerCount="4" CommentCount="1" />
  <row Id="21147" PostTypeId="2" ParentId="21146" CreationDate="2011-02-09T10:00:26.020" Score="6" ViewCount="" Body="&lt;p&gt;Try factoring as the product of two quadratic expressions: $n^4+4=(n^2+an+b)(n^2+cn+d)$.&lt;/p&gt;&#xA;" OwnerUserId="589" LastActivityDate="2011-02-09T10:00:26.020" CommentCount="5" />
  <row Id="21148" PostTypeId="1" AcceptedAnswerId="21153" CreationDate="2011-02-09T10:01:14.327" Score="3" ViewCount="162" Body="&lt;p&gt;Hello,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Prove that $$(2^k + 1)|( 2^{ k( 2n + 1 ) } + 1 )$$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I approach this problem? Using algebra manipulation with the number $$2^{ k( 2n + 1 ) } + 1 $$ (this is what I tried, but failed :( ), or there are some other properties are more useful. Any hint?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks,&lt;br&gt;&#xA;Chan&lt;/p&gt;&#xA;" OwnerUserId="5711" LastActivityDate="2011-02-11T10:15:07.123" Title="What special properties does the number of the form $2^k + 1$ have?" Tags="&lt;elementary-number-theory&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="21149" PostTypeId="2" ParentId="21146" CreationDate="2011-02-09T10:06:49.833" Score="1" ViewCount="" Body="&lt;p&gt;In fact, it is true for $n&amp;gt;1$. You just need a clever way of factoring the expression:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$n^4 + 4 = n^4 + 4 + 4n^2 - 4n^2 = \left ( \cdots + \cdots - \cdots \right ) \cdot \left ( \cdots + \cdots + \cdots \right )$$&lt;/p&gt;&#xA;" OwnerUserId="5105" LastEditorUserId="5105" LastEditorDisplayName="" LastEditDate="2011-02-09T10:47:27.890" LastActivityDate="2011-02-09T10:47:27.890" />
  <row Id="21150" PostTypeId="1" CreationDate="2011-02-09T10:10:56.480" Score="-2" ViewCount="187" Body="&lt;p&gt;If $x^2 + 2x - 35 = 0$,&#xA;then $x = $? &lt;/p&gt;&#xA;" OwnerUserId="6785" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2011-04-30T04:14:31.620" LastActivityDate="2011-04-30T15:02:50.237" Title="Solving $x^2 + 2x - 35 = 0$" Tags="&lt;algebra-precalculus&gt;" AnswerCount="4" CommentCount="2" />
  <row Id="21151" PostTypeId="2" ParentId="21150" CreationDate="2011-02-09T10:14:40.313" Score="5" ViewCount="" Body="&lt;p&gt;This is a quadratic equation. Try completing the square.&lt;/p&gt;&#xA;" OwnerUserId="589" LastActivityDate="2011-02-09T10:14:40.313" />
  <row Id="21152" PostTypeId="2" ParentId="21150" CreationDate="2011-02-09T10:20:58.633" Score="5" ViewCount="" Body="&lt;p&gt;Well you can do this: Observe that $7 \times 5 =35$ and the difference between $7$ and $5$ is $2$, so you can write your equation as $$x^{2}+7x - 5x -35=0$$ which then can be written as $(x+7)\cdot (x-5)$.&lt;/p&gt;&#xA;" LastActivityDate="2011-02-09T10:20:58.633" />
  <row Id="21153" PostTypeId="2" ParentId="21148" CreationDate="2011-02-09T10:44:22.833" Score="3" ViewCount="" Body="&lt;p&gt;Hints:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;$2n+1$ is odd.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$2^{k(2n+1)} + 1 = 2^{k(2n+1)} + 1 + 2^{2kn} - 2^{2kn}$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;try to show that $2^k + 1 | (2^{k(2n+1)} + 1)$ if and only if $2^k + 1 | (2^{k(2n-1)} + 1)$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$\cdots$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$2^k + 1 | 2^k + 1 $  $\square$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint $2$ to Hint $3$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$2^{k(2n+1)} + 1 = 2^{k(2n+1)} + 1 + 2^{2kn} - 2^{2kn} = 2^{2kn+k} + 1 + 2^{2kn} - 2^{2kn} =$$ &#xA;$$=2^{2kn}2^k + 1 + 2^{2kn} - 2^{2kn} = 2^{2kn}(2^k + 1) + 1 - 2^{2kn} =$$&#xA;Let $a = 2^{2kn}(2^k + 1)$, now the equality is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$= a + 1 - 2^{2kn} = a + 1 - 2^{k(2n-1)+k} = a + 1 - 2^{k(2n-1)}2^k =$$&#xA;$$= a + 1 - 2^{k(2n-1)}2^k +2^{k(2n-1)} - 2^{k(2n-1)}= a + 1 - 2^{k(2n-1)}(2^k +1) +2^{k(2n-1)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let $b= 2^{k(2n-1)}(2^k +1)$, and you are left with:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$2^{k(2n+1)} + 1 = (a - b) +( 2^{k(2n-1)} + 1)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that $a,b$ are divisible by $2^k +1$ to reach Hint $3$ (I read in some other post that you haven't learned modular arithmetic yet - once you do this becomes so so so so much easier!).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now it smells like &lt;em&gt;induction.&lt;/em&gt; Apply Hint $1$ and draw a square.  &lt;/p&gt;&#xA;" OwnerUserId="5105" LastEditorUserId="5105" LastEditorDisplayName="" LastEditDate="2011-02-11T10:15:07.123" LastActivityDate="2011-02-11T10:15:07.123" CommentCount="10" />
  <row Id="21154" PostTypeId="2" ParentId="21128" CreationDate="2011-02-09T10:49:55.357" Score="4" ViewCount="" Body="&lt;p&gt;The professor of the short algebraic topology course I took during my second year refused to give us the definition of functor, saying that it is better to start using them and building up interesting examples before looking at the abstract definitions.&#xA;I remember that at the time I tough that this was quite stupid, but now I actually agree with him. I was happy to learn the basics of category theory the following year having more examples in mind.&#xA;Most of the concepts of category theory are extremely natural but to realize that you need a good background. Otherwise you will most likely struggle against the abstractness with few chances to understand what's truly going on.&#xA;Moreover, I don't think that you will profit much from knowing some category theory this early, before an advanced course in abstract algebra. I learned it while studying module theory, quite a long time after a first course in abstract algebra and a course in Galois theory, and I felt that that was perfect.&lt;/p&gt;&#xA;" OwnerUserId="6314" LastActivityDate="2011-02-09T10:49:55.357" CommentCount="4" />
  <row Id="21155" PostTypeId="2" ParentId="16618" CreationDate="2011-02-09T10:56:05.037" Score="0" ViewCount="" Body="&lt;p&gt;For now, I'll have to answer my own question. I noticed that on the page of Benoît Mandelbrot, the man himself, there is a nice list of &lt;a href=&quot;http://www.math.yale.edu/mandelbrot/&quot; rel=&quot;nofollow&quot;&gt;books on fractals&lt;/a&gt;. I've seen some titles that look like what I need.&lt;/p&gt;&#xA;" OwnerUserId="3567" LastActivityDate="2011-02-09T10:56:05.037" />
  <row Id="21156" PostTypeId="1" AcceptedAnswerId="21162" CreationDate="2011-02-09T11:00:29.567" Score="0" ViewCount="100" Body="&lt;p&gt;Suppose I have a deterministic sequence $\{t_n\}$ that is uniformly distributed on $[0,1]$ (for example $t_n = \{ \pi n \}$, i.e the fractional part of $\pi n$) and a decreasing function $f : \mathbb{R} \rightarrow [0,1]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think it is reasonable to expect that&#xA;$\#\{t_k &amp;lt; f(k) : k \leq n \} \approx \sum_{k = 1}^n f(k)$&#xA;But I'm not sure how I could demonstrate something like that, nor what the correct error would be.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that $\#\{t_i &amp;lt; f(k) : i \leq n \} \approx f(k)n$ but I am unsure how I can translate that into a (rigorous) statement about $\#\{t_k &amp;lt; f(k) : k \leq n \} = \sum [t_k \leq f(k) \text{ and } k \leq n]$ where $[x]$ is Iverson's Bracket which is equal to $1$ if $x$ is true and $0$ if $x$ is false.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: As chandok has pointed out, it is possible to pick a function and sequence where the above does not hold. If we require that $\#\{t_k &amp;lt; f(n) : k \leq n\}$ be unbounded, can we avoid &quot;bad&quot; functions?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit 2: The motivation for this question comes from &lt;a href=&quot;http://math.stackexchange.com/questions/20518/estimating-a-sum-containing-a-uniformly-distributed-but-deterministic-term&quot;&gt;Estimating a sum containing a uniformly distributed but deterministic  term.&lt;/a&gt; .&lt;/p&gt;&#xA;" OwnerUserId="899" LastEditorUserId="899" LastEditorDisplayName="" LastEditDate="2011-02-09T13:01:16.483" LastActivityDate="2011-02-09T13:01:16.483" Title="The &quot;expected value&quot; of a deterministic sequence." Tags="&lt;analysis&gt;" AnswerCount="1" />
  <row Id="21157" PostTypeId="1" CreationDate="2011-02-09T11:06:22.613" Score="0" ViewCount="29" Body="&lt;p&gt;Suppose X is normed space,Prove that X* is rotund iff X/M smooth whenever M is a  closed subspace of X such that X/M is two-dimensional.&#xA;I know that &quot;a norm space is smooth iff each of its two-dimensional subspaces is smooth.&quot;,maybe that will be used.&lt;/p&gt;&#xA;" OwnerUserId="4706" LastActivityDate="2011-02-09T11:06:22.613" Title="rotund and smooth space" Tags="&lt;functional-analysis&gt;&lt;banach-spaces&gt;" />
  <row Id="21158" PostTypeId="2" ParentId="20946" CreationDate="2011-02-09T11:21:20.693" Score="0" ViewCount="" Body="&lt;p&gt;Je crois avoir trouvé la solution. C'est une petite astuce à laquelle je n'avais pas pensé. J'espère ne pas faire d'erreur, mais dites-le-moi si j'en fais une. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Tenons pour acquis que $e^{-2x}$ est la limite uniforme d'une suite de fonctions $e^{-x}p_n(x)$, ce qui se démontre par un argument avec la formule de Taylor, lequel j'omettrai si personne ne demande à le voir.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Soit $c &amp;gt; 0$. Par un changement de variable, on voit que $e^{-2cx}$ est la limite uniforme de la suite $e^{-cx}p_n(cx)$. Soit maintenant $r$ un polynôme. Puisque la fonction $e^{-x}r(x)$ est bornée, on a que $e^{-cx}p_n(cx)e^{-x}r(x)$ tend uniformément vers $e^{-2cx}e^{-x}r(x)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Autrement dit, les fonctions de la forme $e^{-(2c+1)x}p(x)$ sont dans la clôture uniforme de celles de la forme $e^{-(c+1)x}p(x)$. En posant $\alpha = \frac{2c+1}{c+1}$ (qui peut prendre n'importe quelle valeur $1 &amp;lt; \alpha &amp;lt; 2$), on a par un nouveau changement de variable que pour tout $d &amp;gt; 0$, les fonctions $e^{-\alpha dx}p(x)$ sont dans la clôture uniforme des fonctions $e^{-dx}p(x)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Il suffit maintenant de choisir $c$ pour que $\alpha = \sqrt{2}$, et d'appliquer deux fois ce qui précède.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remarques:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Souvent quand je rencontre un argument dans un livre de Lang qui semble marcher presque, mais pas tout à fait, je suis tenté de croire que c'est une erreur. Dans ce cas-ci, je suis allé un peu vite en besogne.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Merci aux traducteurs. Traduire peut aider plus de gens à comprendre. Par contre, je ne voudrais pas que les gens se sentent &lt;em&gt;obligés&lt;/em&gt; de traduire s'ils n'y voient pas d'intérêt. J'ai vu sur une autre page quelqu'un parler d'une &quot; 'translate this and do my work for me' mentality.&quot; Je n'ai pas posé la question pour obliger quelqu'un d'autre à faire ce travail. Personnellement, j'aurais été parfaitement satisfait que la question soit laissée telle quelle, même si cela voulait dire que certaines personnes ne pourraient pas la comprendre.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Je suis passablement choqué d'avoir fait couler autant d' &quot;encre&quot; (si je puis dire). J'aurais préféré n'attirer l'attention que relativement à la question de maths.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Un grand merci à ceux qui ont réfléchi au contenu mathématique de la question. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="6828" LastEditorUserId="6828" LastEditorDisplayName="" LastEditDate="2011-02-09T11:28:28.847" LastActivityDate="2011-02-09T11:28:28.847" CommentCount="3" />
  <row Id="21159" PostTypeId="1" AcceptedAnswerId="21163" CreationDate="2011-02-09T11:27:09.400" Score="0" ViewCount="37" Body="&lt;p&gt;In the Bishop Book on Machine Learning, when discussinf Variational Inference, he gives an example of the univariate Gaussian.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;He defines in 10.1.3 the mean prior as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$p(\mu|\tau)=N(\mu|\mu_0,(\lambda_0*\tau)^{-1})$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the meaning of the conditional mean over an initial $\mu_0$ and what would be the analytical form for that kind of expression.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank You&lt;/p&gt;&#xA;" OwnerUserId="2780" LastActivityDate="2011-02-09T12:44:28.983" Title="Gaussian Distribution with conditional mean" Tags="&lt;normal-distribution&gt;" AnswerCount="1" />
  <row Id="21161" PostTypeId="2" ParentId="21146" CreationDate="2011-02-09T12:09:53.613" Score="5" ViewCount="" Body="&lt;p&gt;You can factor out $n^{4}+4$ algebraically by finding the four roots of $n^{4}+4=0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $n^{4}+4=0\Leftrightarrow n^{4}=4e^{i\pi }$, we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{eqnarray*}&#xD;&#xA;n &amp;amp;=&amp;amp;4^{1/4}e^{i\left( \dfrac{\pi +2k\pi }{4}\right) }\quad k=0,1,2,3 \\&#xD;&#xA;&amp;amp;&amp;amp; \\&#xD;&#xA;n &amp;amp;=&amp;amp;\sqrt{2}e^{i\left( \dfrac{\pi }{4}\right) }=1+i\quad \left( k=0\right) &#xD;&#xA;\\&#xD;&#xA;n &amp;amp;=&amp;amp;\sqrt{2}e^{i\left( \dfrac{3\pi }{4}\right) }=-1+i\quad \left( k=1\right) &#xD;&#xA;\\&#xD;&#xA;n &amp;amp;=&amp;amp;\sqrt{2}e^{i\left( \dfrac{5\pi }{4}\right) }=-1-i\quad \left( k=2\right) &#xD;&#xA;\\&#xD;&#xA;n &amp;amp;=&amp;amp;\sqrt{2}e^{i\left( \dfrac{7\pi }{4}\right) }=1-i\quad \left( k=3\right). &#xD;&#xA;\end{eqnarray*}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And so&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{eqnarray*}&#xD;&#xA;n^{4}+4 &amp;amp;=&amp;amp;\left( n-1-i\right) \left( n+1-i\right) \left( n+1+i\right)&#xD;&#xA;\left( n-1+i\right)  \\&#xD;&#xA;&amp;amp;=&amp;amp;\left( \left( n+1-i\right) \left( n+1+i\right) \right) \left( \left(&#xD;&#xA;n-1-i\right) \left( n-1+i\right) \right)  \\&#xD;&#xA;&amp;amp;=&amp;amp;\left( n^{2}+2n+2\right) \left( n^{2}-2n+2\right), &#xD;&#xA;\end{eqnarray*}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;combining the complex conjugates factors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: for $n&amp;gt;1$, $n^2+2n+2&amp;gt;5$ and $n^2-2n+2&amp;gt;1$.&lt;/p&gt;&#xA;" OwnerUserId="752" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2011-02-09T14:05:17.337" LastActivityDate="2011-02-09T14:05:17.337" />
  <row Id="21162" PostTypeId="2" ParentId="21156" CreationDate="2011-02-09T12:28:26.113" Score="1" ViewCount="" Body="&lt;p&gt;The problem with deterministic &quot;random&quot; sequences is that you can easily use them to provide an object (here a decreasing function) against which they won't be &quot;random&quot; at all :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you pick $f(k) = \inf \{t_m, 1 \leq m \leq k\} /2$, you have that $\#\{t_k &amp;lt; f(k) : k \leq n \} = \#\{t_k = 0 : k \leq n \}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can easily pick &quot;uniformly distributed&quot; sequences where this is $0$, while $\Sigma f(k)$ gets as large as you want (in fact I think it will diverge almost surely if you pick a random uniformly distributed sequence).&lt;/p&gt;&#xA;" OwnerUserId="6270" LastActivityDate="2011-02-09T12:28:26.113" CommentCount="2" />
  <row Id="21163" PostTypeId="2" ParentId="21159" CreationDate="2011-02-09T12:44:28.983" Score="1" ViewCount="" Body="&lt;p&gt;The expression $N(x \mid \mu_0, \sigma_0)$ is the notation of Bishop to denote the Gaussian distribution under the condition that the mean and standard deviation are given by $\mu_0$ and $\sigma_0$ respectively. So, your expression means: $\mu$ is normally distributed with mean $\mu_0$ and standard deviation $( \lambda_0 \tau )^{-1}$, which translates to the pdf&#xA;$$&#xD;&#xA;\frac{\lambda_0 \tau}{\sqrt{ 2 \pi }} e^{\frac{(\lambda_0 \tau)^2}{2} (\mu - \mu_0)^2} .&#xD;&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="960" LastActivityDate="2011-02-09T12:44:28.983" />
  <row Id="21164" PostTypeId="1" CreationDate="2011-02-09T12:45:47.623" Score="6" ViewCount="80" Body="&lt;p&gt;I have just started learining Reimann Surfaces and I am using the book by Rick Miranda: Algebraic curves and Reimannn Surfaces. #F in section 1.3 asks to determine the genus of the curve in $\mathbb{P}^3$ defined by the two equations $x_0x_3=2x_1x_2$ and $x_0^2 + x_1^2 +x_2^2 +x_3^2 = 0$.  #G also has a similar question in which he asks to determine the genus of the twisted cubic.  Please explain how to approch this type of question.  &lt;/p&gt;&#xA;" OwnerUserId="2893" LastActivityDate="2011-02-09T12:45:47.623" Title="Problem in Rick Miranda: finding genus of a Projective curve" Tags="&lt;algebraic-geometry&gt;" CommentCount="1" FavoriteCount="0" />
  <row Id="21165" PostTypeId="1" CreationDate="2011-02-09T12:59:14.743" Score="1" ViewCount="75" Body="&lt;p&gt;Taken from &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Cambridge University Press&#xA;0521830842 - Foundations of Cryptography: Basic Applications, Volume 2&#xA;CHAPTER FIVE - Encryption Schemes&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which can be downloaded &lt;a href=&quot;http://bilder.buecher.de/zusatz/21/21269/21269809_lese_1.pdf&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Notation. In the rest of this text, we&#xA;  write $E_e (\alpha)$ instead of $E(e, \alpha)$ and&#xA;  $D_d (\beta)$ instead of $D(d, \beta)$. Sometimes,&#xA;  when there is little risk of&#xA;  confusion, we drop these subscripts.&#xA;  Also, we let $G_1 (1^n)$ (resp., $G_2 (1^n&#xA;  )$) denote the first (resp., second)&#xA;  element in the pair G(1n ). That is,&#xA;  $G(1n ) = (G_1 (1^n ), G_2 (1^n))$.&#xA;  Without loss of generality, we may&#xA;  assume that $|G_1 (1^n )|$ and $|G_2 (1^n&#xD;&#xA;&amp;gt; )|$ are &lt;strong&gt;polynomially related to n&lt;/strong&gt;,&#xA;  and that each of these integers can be&#xA;  efficiently computed from the other.&#xA;  (In fact, we may even assume that $|G_1&#xD;&#xA;&amp;gt; (1^n )| = |G_2 (1^n)| = n$; see Exercise&#xA;  6.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="6830" LastEditorUserId="6830" LastEditorDisplayName="" LastEditDate="2011-02-09T13:09:57.637" LastActivityDate="2011-04-10T13:39:58.320" Title="What does it mean &quot;polynomial related&quot;?" Tags="&lt;cryptography&gt;" AnswerCount="1" />
  <row Id="21166" PostTypeId="1" AcceptedAnswerId="21167" CreationDate="2011-02-09T13:03:32.587" Score="0" ViewCount="47" Body="&lt;p&gt;I have a polygon and I only know the length of its sides, how would I go about figuring out what it's points are?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume that the first side starts at the origin (0,0) and extends along the x axis to (length of side a,0).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I realise that this is trivial for, say a triangle, but is there a rule that can be used for &lt;em&gt;all&lt;/em&gt; polygons?&lt;/p&gt;&#xA;" OwnerUserId="5709" LastActivityDate="2011-02-09T13:07:10.500" Title="General rule/algorithm for finding the points of a polygon, given only the lengths of the sides" Tags="&lt;geometry&gt;" AnswerCount="1" />
  <row Id="21167" PostTypeId="2" ParentId="21166" CreationDate="2011-02-09T13:07:10.500" Score="1" ViewCount="" Body="&lt;p&gt;No, you can never do this for for any polygon with more than three sides (unless it's degenerate, so that the length of one side is equal to the sum of the lengths of all the other sides -- but then you just get a straight line). For instance, if the polygon has four sides, all the same length, then it can be a square, but it can also be a &lt;a href=&quot;http://en.wikipedia.org/wiki/Rhombus&quot; rel=&quot;nofollow&quot;&gt;rhombus&lt;/a&gt; with any interior angle.&lt;/p&gt;&#xA;" OwnerUserId="1508" LastActivityDate="2011-02-09T13:07:10.500" CommentCount="1" />
  <row Id="21168" PostTypeId="2" ParentId="21055" CreationDate="2011-02-09T13:07:26.013" Score="0" ViewCount="" Body="&lt;p&gt;I've possibly another useful reformulation, don't know whether this fits the scope of the question. I assume that A is diagonalizable and do not consider a range of convergence - it's just a sketch.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For ease of notation I introduce $ \rm\ B = transpose(A) $, small letters denote the inverse of the related capital-letters denoted matrix. $ \rm\ I $ is the identity matrix.    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have from the statement of the problem $ \rm\ S = I + B S A $       &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we introduce the diagonalization of &lt;strong&gt;A&lt;/strong&gt; we have $ \rm\ A = W * D * w $ where &lt;strong&gt;D&lt;/strong&gt; is diagonal. Use &lt;strong&gt;M&lt;/strong&gt; and &lt;strong&gt;m&lt;/strong&gt; for the transpose of &lt;strong&gt;W&lt;/strong&gt; and &lt;strong&gt;w&lt;/strong&gt;, then the diagonalization of &lt;strong&gt;B&lt;/strong&gt; is $ \rm\ B = m * D * M $     &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we have $$ S = I + BSA = I + mDM*S*WDw $$ and thus $$  MSW = MW + D MSW D $$     &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's denote $ \rm\ MSW = U$ and $ \rm\ MW = t$ &#xA;Then this also means: $$  U = t + D U D = t + D t D + D^2tD^2 + D^3tD^3+...  $$&#xA;Note that &lt;strong&gt;t&lt;/strong&gt; is symmetric and thus &lt;strong&gt;U&lt;/strong&gt; is symmetric and $ \rm\ S = mUw $ is then symmetric as well.    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because &lt;strong&gt;D&lt;/strong&gt; is diagonal this can now be described more easily for each matrix-element individually. We have, using &lt;em&gt;r&lt;/em&gt; and &lt;em&gt;c&lt;/em&gt; for row and column-inbdex respectively&#xA;$$ U_{r,c} = t_{r,c}*(1 + D_rD_c + (D_rD_c)^2 + ...) = t_{r,c}/(1-D_rD_c) $$&#xA;Having the matrix &lt;strong&gt;U&lt;/strong&gt; by this, &lt;strong&gt;S&lt;/strong&gt; is $$S = m * U * w $$&#xA;[update note: I edited the first version because I'd taken &lt;strong&gt;A&lt;/strong&gt; and &lt;strong&gt;B&lt;/strong&gt; in the wrong order]&lt;br&gt;&#xA;[edit 2]: the denominators in the description of elements of &lt;strong&gt;U&lt;/strong&gt; seem to describe the range of applicability. Since geometric series with quotient &lt;em&gt;q&lt;/em&gt; is defined also for the divergent case (except for &lt;em&gt;q=1&lt;/em&gt;) we might discard the condition $ \varrho(A) \lt 1 $ and replace it with something like &lt;em&gt;: &lt;strong&gt;A&lt;/strong&gt; cannot have at the same time a value x and its reciprocal as eigenvalues&lt;/em&gt;&lt;br&gt;&#xA;[edit 3]: removed the &quot;surprise&quot; when writing about symmetry - actually obvious ;-) &lt;/p&gt;&#xA;" OwnerUserId="1714" LastEditorUserId="1714" LastEditorDisplayName="" LastEditDate="2011-02-09T14:13:20.997" LastActivityDate="2011-02-09T14:13:20.997" />
  <row Id="21169" PostTypeId="2" ParentId="21141" CreationDate="2011-02-09T13:09:12.053" Score="3" ViewCount="" Body="&lt;p&gt;Knowing the cycle decomposition of $P_1$ is useful :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a $k$-cycle $(a_0, a_2, \ldots a_{k-1})$ of $P_1$, $P_2(a_0)$ will be an element of that cycle, $a_p$ for some $p \in \{0 \ldots k-1\}$. Since $P_1^e(a_0) = a_{e \mod k}$, this information is equivalent to $e \mod k = p$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gather those congruences from enough cycles of $P_1$ (so that the least common multiple of the cycle lengths is $o$), and the system you obtain will give you the unique solution of $e \mod o$.&lt;/p&gt;&#xA;" OwnerUserId="6270" LastActivityDate="2011-02-09T13:09:12.053" CommentCount="3" />
  <row Id="21170" PostTypeId="1" AcceptedAnswerId="21452" CreationDate="2011-02-09T13:16:06.927" Score="4" ViewCount="173" Body="&lt;p&gt;As a follow-up to &lt;a href=&quot;http://math.stackexchange.com/questions/21017/&quot;&gt;this question&lt;/a&gt;, I'd like to ask:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are examples of rings $R$ with the property that for all finite sets of ideals $I_1,\ldots,I_n$ in $R$ the sequence&#xA;$$&#xD;&#xA;\bigoplus_{1\leq j &amp;lt; k\leq n}^n I_j\cap I_k\quad\xrightarrow{f}\quad\bigoplus_{l=1}^n I_l\quad\xrightarrow{g}\quad\sum_{k=1}^n I_k&#xD;&#xA;$$&#xA;is &lt;strong&gt;exact in the middle&lt;/strong&gt;? Here $g$ is given by addition, and $f$ maps $x\in I_j\cap I_k$ to $x\in I_j$ and to $-x\in I_k$ (and to zero in all other components).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A complete description of this class of rings would be even better, of course.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obvious examples seem to be rings with at most two proper ideals. Those I would consider pathological in this context.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd be happy to restrict to complex algebras if that is useful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, is there a name for rings with this property?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; &lt;em&gt;In order to make the question more answerable, let's just consider the case $R=\mathbb Z$. Can we find ideals in $\mathbb Z$ such that the above sequence is not exact, or can we prove that this is impossible?&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="367" LastEditorUserId="367" LastEditorDisplayName="" LastEditDate="2011-02-10T21:37:23.263" LastActivityDate="2011-02-11T23:21:48.527" Title="Does the ring of integers have the following property?" Tags="&lt;number-theory&gt;&lt;algebra&gt;&lt;homological-algebra&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="21171" PostTypeId="2" ParentId="21165" CreationDate="2011-02-09T13:20:10.653" Score="1" ViewCount="" Body="&lt;p&gt;It just means that the functions mentioned in your question can be written as a polynomial in $n$ (the independent variable is $n$). For example, $n^2+3n+4$ or something like that. It cannot be an exponential function or super-polynomial in $n$ or sublinear (i.e. $o(n)$).&lt;/p&gt;&#xA;" OwnerUserId="1330" LastActivityDate="2011-02-09T13:20:10.653" CommentCount="2" />
  <row Id="21172" PostTypeId="2" ParentId="21063" CreationDate="2011-02-09T13:34:21.437" Score="4" ViewCount="" Body="&lt;p&gt;To add on Didier's answer, it is instructive to note that the inequality ${\rm E}(\ln X) \le \ln {\rm E}(X)$&#xA;can be seen as a consequence of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Inequality_of_arithmetic_and_geometric_means&quot; rel=&quot;nofollow&quot;&gt;AM-GM inequality&lt;/a&gt; combined with the &lt;a href=&quot;http://en.wikipedia.org/wiki/Law_of_large_numbers&quot; rel=&quot;nofollow&quot;&gt;strong law of large numbers&lt;/a&gt;, upon writing &#xA;the AM-GM inequality&#xA;$$&#xD;&#xA;\sqrt[n]{{X_1  \cdots X_n }} \le \frac{{X_1  +  \cdots  + X_n }}{n}&#xD;&#xA;$$&#xA;as&#xA;$$&#xD;&#xA;\exp \bigg(\frac{{\ln X_1  +  \cdots  + \ln X_n }}{n}\bigg) \le \frac{{X_1  +  \cdots  + X_n }}{n},&#xD;&#xA;$$&#xA;and letting $n \to \infty$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: For completeness, let me note that ${\rm E}[\ln X]$ might be equal to $-\infty$. For example, if $X$ has density function &#xA;$$&#xD;&#xA;f(x) = \frac{{\ln a}}{{x\ln ^2 x}},\;\;0 &amp;lt; x &amp;lt; \frac{1}{a},&#xD;&#xA;$$&#xA;where $a&amp;gt;1$ (note that $\int f  = 1$), then &#xA;$$&#xD;&#xA;{\rm E}[\ln X] = &#xD;&#xA;\int_0^{1/a} {\frac{{\ln a}}{{x\ln x}}} \,{\rm d}x = -\infty.&#xD;&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="2810" LastEditorUserId="2810" LastEditorDisplayName="" LastEditDate="2011-02-13T17:28:55.793" LastActivityDate="2011-02-13T17:28:55.793" CommentCount="3" />
  <row Id="21173" PostTypeId="1" AcceptedAnswerId="21325" CreationDate="2011-02-09T14:10:04.727" Score="2" ViewCount="85" Body="&lt;p&gt;I have figured out that:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;given N = 10^K, D = K + 1&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;What is the formula for a natural number D = ???(N)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How to solve this in Mathematica/WolframAlpha ?&lt;/p&gt;&#xA;" OwnerUserId="5537" LastEditorUserId="5537" LastEditorDisplayName="" LastEditDate="2011-02-09T16:27:28.183" LastActivityDate="2011-02-10T07:02:51.220" Title="How to calculate the number of digits(D) of a natural number(N) in base 10?" Tags="&lt;number-theory&gt;" AnswerCount="2" />
  <row Id="21175" PostTypeId="1" AcceptedAnswerId="21181" CreationDate="2011-02-09T14:12:20.577" Score="12" ViewCount="397" Body="&lt;p&gt;$\sum_{n=2}^{\infty} \frac{ (-1)^n} { \ln(n) +\cos(n)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;any ideas? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;thanks!&lt;/p&gt;&#xA;" OwnerUserId="5491" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2011-02-09T14:17:58.670" LastActivityDate="2011-05-15T01:48:05.987" Title="Does the following series converge?" Tags="&lt;sequences-and-series&gt;&lt;classical-analysis&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="1" />
  <row Id="21176" PostTypeId="2" ParentId="21173" CreationDate="2011-02-09T14:14:21.107" Score="1" ViewCount="" Body="&lt;p&gt;You have it (presuming that you see 000 as having 2 leading zeros and make a special case for it-the log won't work well).  Just round Z up to the next $10^k-1$, or use the ceiling function:  $\lceil log_{10}(N+1)\rceil$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With the change to the question, see the Hristo's comment below&lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2011-02-09T14:32:49.220" LastActivityDate="2011-02-09T14:32:49.220" CommentCount="3" />
  <row Id="21177" PostTypeId="1" CreationDate="2011-02-09T14:14:30.687" Score="1" ViewCount="124" Body="&lt;p&gt;In my programming project I need to evenly distribute various sized circles in ellipse shaped area. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/MNZw5.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The ellipse area is defined by major and minor axis (4 points).&lt;/li&gt;&#xA;&lt;li&gt;There is finite number of various sized circles that will all fit into the ellipse.&lt;/li&gt;&#xA;&lt;li&gt;The order of the circles does not matter.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;What is the criterion that will make them look evenly distributed? I don't think it's same distance between them, this can not even be achieved, can it? What algorithm will make them look as homogeneously distributed as possible? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for any help.&lt;/p&gt;&#xA;" OwnerUserId="6811" LastEditorUserId="6811" LastEditorDisplayName="" LastEditDate="2011-02-09T19:47:42.030" LastActivityDate="2011-02-10T21:54:33.473" Title="How to evenly distribute various sized circles in ellipse shaped area?" Tags="&lt;geometry&gt;&lt;packing-problem&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="21180" PostTypeId="2" ParentId="21177" CreationDate="2011-02-09T14:30:03.590" Score="0" ViewCount="" Body="&lt;p&gt;&quot;Evenly distributed&quot; almost sounds like an aesthetical criterion. But I guess this could be achieved by picking points randomly. Say, you pick a point within the ellipse and attribute it as a center for a circle (take the biggest circle such that it still is within the ellipse). Then repeat the procedure, but now, taking into account that the new circle must not overlap with circles that are already present. Whenever no circle satisfies the criteria and you are not done yet with distributing all the circles, pick a new random point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I guess the method is not very efficient if you're stuck with a big circle at the end of the day and all other circles have been distributed. One way to overcome that is to start with the biggest circle, determine the ellipse of points within which the center of that bigger circle can be placed and choose a random point from that. Then, pick the next biggest circle and go over the same procedure: determining an ellipse of points that can be possible centers and pick a random point again, but now you have to see that there is no overlap with the already placed circle. Etc...&lt;/p&gt;&#xA;" OwnerUserId="3567" LastActivityDate="2011-02-09T14:30:03.590" CommentCount="2" />
  <row Id="21181" PostTypeId="2" ParentId="21175" CreationDate="2011-02-09T14:44:05.823" Score="13" ViewCount="" Body="&lt;p&gt;This replaces my original answer, which was faulty but led in the right direction (I think).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll let the series start at some $n_0\ge2$; the reason for this will become apparent later.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Expand in $\cos(n)/\ln(n)$ (with $\lvert\cos(n)/\ln(n)\rvert &amp;lt; 1$ for $n \ge 2$):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\[\sum_{n=n_0}^\infty \sum_{k=0}^\infty \frac{(-1)^n}{\ln(n)} \frac{\cos^k(n)}{\ln^k(n)}\]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's interchange the summations and worry later whether that was OK:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\[\sum_{k=0}^\infty \sum_{n=n_0}^\infty \frac{(-1)^n}{\ln(n)} \frac{\cos^k(n)}{\ln^k(n)}\]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the inner sum converges according to &lt;a href=&quot;http://en.wikipedia.org/wiki/Dirichlet%27s_test&quot; rel=&quot;nofollow&quot;&gt;Dirichlet's test&lt;/a&gt;. In fact we can proceed as in the proof of Dirichlet's test to bound its limit (suppressing indices $k$ while we work at fixed $k$):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\[a_n=\ln^{-k-1}(n)\]&#xA;\[b_n=(-1)^n\cos^k(n)=\sum_{j=0}^k\alpha_{jk}\cos((j + \pi)n)\;,\]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with $\lvert\alpha_{jk}\rvert \le 1$, using &lt;a href=&quot;http://en.wikipedia.org/wiki/Trigonometric_identities#Power-reduction_formula&quot; rel=&quot;nofollow&quot;&gt;the power-reduction formula for the cosine&lt;/a&gt; and absorbing the factor $(-1)^n$ into the cosines by adding $\pi$ to their frequency.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\[B_m=\sum_{n=n_0}^m \; b_n\]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\[\lvert B_m\rvert = \left\lvert\sum_{n=n_0}^m \sum_{j=0}^k&#xA;\alpha_{jk}\cos((j + \pi)n)\right\rvert&#xA;\le \sum_{j=0}^k\left\lvert\sum_{n=n_0}^m\cos((j + \pi)n)\right\rvert \le \sum_{j=0}^k\frac{2}{1 - \cos(j + \pi)}=:M_k\]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\[\left\lvert\sum_{n=n_0}^m a_n b_n\right\rvert = \left\lvert B_m a_{m+1}+\sum_{n=n_0}^{m}B_n(a_n-a_{n+1})\right\rvert\le\]&#xA;\[\le M_k a_{m+1} + \sum_{n=n_0}^{m}M_k(a_n-a_{n+1})=&#xA;M_k a_{m+1} + M_k(a_{n_0}-a_{m+1}) = M_ka_{n_0}\]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, we can apply the &lt;a href=&quot;http://en.wikipedia.org/wiki/Comparison_test&quot; rel=&quot;nofollow&quot;&gt;comparison test&lt;/a&gt; with the following series:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\[\sum_{k=0}^\infty M_k\ln^{-k-1}(n_0)\]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This shows why we need to start at some $n_0$ instead of $2$, which wouldn't work since $\ln 2 &amp;lt; 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now everything depends on the behaviour of $M_k$. The cosine comes arbitrarily close to $1$; when $j$ comes within $\epsilon$ of an odd multiple of $\pi$, then $1/(1-\cos (j + \pi)) \approx 1/(1-(1 - \epsilon^2/2))= 2/\epsilon^2$. Heuristically speaking, we might expect this to happen every $1/\epsilon$ integers, so on average these spikes would not destroy the long-term decay with $\ln^{-k-1}(n_0)$ (which can be made quantitively, though not qualitatively, stronger by increasing $n_0$). However, I don't know how to make this argument rigorous, or whether that is even possible given the &quot;unpredictability&quot; of $\pi$. It's interesting that this proof might depend on the details of $\pi$ in this way. [Update: Thanks to George Lowther for the comments pointing to &lt;a href=&quot;http://mathworld.wolfram.com/IrrationalityMeasure.html&quot; rel=&quot;nofollow&quot;&gt;this MathWorld page&lt;/a&gt;, which gives an upper bound on the irrationality measure of $\pi$ (never heard of that before :-). That this is finite implies that $M_k$ can be bounded by a power of $k$, and hence the outer sum converges (uniformly) -- I believe that completes the proof.] &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming that the spike problem can be solved, we still have the interchangeability of the sums to come back to. &lt;a href=&quot;http://math.asu.edu/~jss/courses/fall06/mat472/limit_interchange.pdf&quot; rel=&quot;nofollow&quot;&gt;This theorem&lt;/a&gt; shows that we may interchange the sums if the convergence of&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\[\sum_{k=0}^\infty \sum_{n=n_0}^m \frac{(-1)^n}{\ln(n)} \frac{\cos^k(n)}{\ln^k(n)}\]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is uniform in $m$ -- which it is, since $m$ dropped out in the derivation of the bound $M_ka_{n_0}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To summarize, this derivation shows that the series can be shown to converge if the irregular spikes in $1/(1-\cos (j + \pi))$ that occur because $j$ gets arbitrarily close to odd multiples of $\pi$ can be shown not to destroy the convergence of the outer sum over $k$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for an interesting question :-).&lt;/p&gt;&#xA;" OwnerUserId="6622" LastEditorUserId="6622" LastEditorDisplayName="" LastEditDate="2011-02-10T17:50:40.623" LastActivityDate="2011-02-10T17:50:40.623" CommentCount="8" />
  <row Id="21182" PostTypeId="1" AcceptedAnswerId="21184" CreationDate="2011-02-09T14:57:39.383" Score="2" ViewCount="78" Body="&lt;p&gt;What is the simplification of $\sin^2 x/(1+ \sin^2 x +\sin^4 x +\sin^6 x + \cdots)$?&lt;/p&gt;&#xA;" OwnerUserId="6834" LastEditorUserId="5363" LastEditorDisplayName="" LastEditDate="2011-02-09T15:08:55.367" LastActivityDate="2011-02-09T15:08:55.367" Title="What is the simplification of $\sin^2 x/(1+ \sin^2 x +\sin^4 x +\sin^6 x + \cdots)$?" Tags="&lt;sequences-and-series&gt;&lt;trigonometry&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="21183" PostTypeId="2" ParentId="21182" CreationDate="2011-02-09T15:05:39.377" Score="3" ViewCount="" Body="&lt;p&gt;What does $1 + \sin^2 x + \sin^4 x + \sin^6 x + ....$ simplify to? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or better, what does $1 + x^2 + x^4 + x^6 + ....$ simplify to?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or better, what does $1 + x + x^2 + x^3 + ....$ simplify to?&lt;/p&gt;&#xA;" OwnerUserId="1919" LastActivityDate="2011-02-09T15:05:39.377" />
  <row Id="21184" PostTypeId="2" ParentId="21182" CreationDate="2011-02-09T15:07:29.593" Score="2" ViewCount="" Body="&lt;p&gt;Assuming that $x \notin \frac{\pi}{2} + \pi \mathbb{Z}$ you can write $q = \sin^{2}{x}$ with $|q| &amp;lt; 1$ and use the geometric series $\sum_{n=0}^{\infty} q^{n} = \frac{1}{1-q}$.&lt;/p&gt;&#xA;" OwnerUserId="5363" LastActivityDate="2011-02-09T15:07:29.593" CommentCount="4" />
  <row Id="21185" PostTypeId="2" ParentId="21128" CreationDate="2011-02-09T15:11:03.660" Score="4" ViewCount="" Body="&lt;p&gt;I agree a lot with user1728's answer. IMO category theory is a beautiful subject, but one that does not make a whole lot of sense without examples. I think it would be pretty hard to learn category theory while also learning all of the standard examples. I consider having an abstract algebra and/or a topology course an absolute prerequisite to understanding category theory. For me, the best part about category theory is that it is a framework that makes life easier.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are some subjects that it is absolutely necessary to know category theory before hand, but a pass familiarity with the definitions should suffice for a first algebra or galois theory course.&lt;/p&gt;&#xA;" OwnerUserId="627" LastActivityDate="2011-02-09T15:11:03.660" />
  <row Id="21186" PostTypeId="1" AcceptedAnswerId="21213" CreationDate="2011-02-09T15:17:01.190" Score="3" ViewCount="185" Body="&lt;p&gt;Why is it that no one seems to factor quadratics into just one bracket Eg:&#xA;$$2x^2+8x+6$$&#xA;into&#xA;$$2x\left(x+4+3\cdot\frac{1}{x}\right)\quad\text{or}\quad 2x\left(x+4+\frac{3}{x}\right)\quad ?$$&lt;/p&gt;&#xA;" OwnerUserId="2075" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-09T18:02:03.347" LastActivityDate="2011-02-10T02:50:09.317" Title="Why are quadratics factored into 2 brackets?" Tags="&lt;quadratics&gt;" AnswerCount="5" CommentCount="1" />
  <row Id="21188" PostTypeId="2" ParentId="21186" CreationDate="2011-02-09T15:31:33.923" Score="4" ViewCount="" Body="&lt;p&gt;Here's a few reasons:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;No one likes division. It is difficult.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;This factoring doesn't tell you what the roots might be.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Either way you have $4$ &quot;terms.&quot;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;It is not as neat.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Note that: $2x(x+8+3\cdot \frac{1}{x}) = (2x)(x+8+3\cdot \frac{1}{x})$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="5105" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-09T17:15:30.237" LastActivityDate="2011-02-09T17:15:30.237" />
  <row Id="21189" PostTypeId="2" ParentId="21186" CreationDate="2011-02-09T15:32:13.397" Score="2" ViewCount="" Body="&lt;p&gt;Factoring a constant out of quadratics in order to make it monic is fairly common. However, factoring it in the method that you have introduces a removable singularity at $x=0$.&#xA;Using your example $p(x) = 2x^2+8x+6$, $p(0)=6$. However if we call $g(x) = 2x(x+8+\frac{3}{x})$, $g(0)$ is not defined. Clearly $p(x) = g(x) \forall x \neq 0$, but we lose continuity and differentiability at $x = 0$ for $g$.&lt;/p&gt;&#xA;" OwnerUserId="1016" LastActivityDate="2011-02-09T15:32:13.397" />
  <row Id="21190" PostTypeId="1" AcceptedAnswerId="21202" CreationDate="2011-02-09T15:40:39.050" Score="0" ViewCount="49" Body="&lt;p&gt;As I understand the &lt;a href=&quot;http://en.wikipedia.org/wiki/Kullback%E2%80%93Leibler_divergence&quot; rel=&quot;nofollow&quot;&gt;KL Divergence&lt;/a&gt;, it measures how different two probability distributions $P$ and $Q$ are.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, say the two distributions are:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;P = [0.2 0.2 0.2 0.4];&#xA;Q = [0 0.5 0.1 0.4];&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Then the KL-divergence is infinity. What is the justification for these distributions being infinitely different?&lt;/p&gt;&#xA;" OwnerUserId="4965" LastActivityDate="2011-02-09T16:43:07.067" Title="Justification for infinite KL Divergence" Tags="&lt;infinity&gt;&lt;probability-distributions&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="21191" PostTypeId="1" AcceptedAnswerId="21194" CreationDate="2011-02-09T15:41:59.187" Score="1" ViewCount="158" Body="&lt;p&gt;I was just reading about the e&lt;a href=&quot;http://en.wikipedia.org/wiki/Exponential_function#Complex_plane&quot; rel=&quot;nofollow&quot;&gt;-function in the complex plane&lt;/a&gt; and was trying to understand the differences between the real and the complex case. Part of the problem is that the mapping of a 2-D plane to another 2-D plane is hard to visualize.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My question&lt;/strong&gt;&lt;br&gt;&#xA;What are the properties of the complex exponential function in terms of being injective, surjective and/or bijective - and how is this different from the real case?&#xA;How can you proof these attributes in the real and in the complex case?&lt;/p&gt;&#xA;" OwnerUserId="346" LastEditorUserId="1016" LastEditorDisplayName="" LastEditDate="2011-02-09T15:44:53.940" LastActivityDate="2011-02-09T16:05:06.860" Title="Is the complex exponential function injective, surjective and/or bijective - and why?" Tags="&lt;complex-analysis&gt;" AnswerCount="3" CommentCount="4" FavoriteCount="1" />
  <row Id="21193" PostTypeId="2" ParentId="21191" CreationDate="2011-02-09T15:50:57.183" Score="1" ViewCount="" Body="&lt;p&gt;Richard Palais has a wonderful program for visualizing some complex maps called 3D-XplorMath which can be downloaded for free at &lt;a href=&quot;http://3d-xplormath.org/&quot; rel=&quot;nofollow&quot;&gt;http://3d-xplormath.org/&lt;/a&gt;.  It also has some nice visual for 3d surfaces and fractals, among other things.  This should at least be able to help you visualize exponentiation in $\mathbb{C}$.&lt;/p&gt;&#xA;" OwnerUserId="6491" LastActivityDate="2011-02-09T15:50:57.183" />
  <row Id="21194" PostTypeId="2" ParentId="21191" CreationDate="2011-02-09T15:57:52.077" Score="1" ViewCount="" Body="&lt;p&gt;The complex exponential is never zero, so it's not surjective; this is like the real case. It is also periodic, so it's not injective either; this is unlike the real case.&lt;/p&gt;&#xA;" OwnerUserId="589" LastActivityDate="2011-02-09T15:57:52.077" />
  <row Id="21195" PostTypeId="2" ParentId="21191" CreationDate="2011-02-09T15:59:48.363" Score="3" ViewCount="" Body="&lt;p&gt;The complex exponential $e^z : \mathbb{C} \to \mathbb{C}$ is neither injective nor surjective. The former is because $e^{z + 2 \pi i k} = e^z \forall k \in \mathbb{Z}$ and the latter is because $e^z \neq 0$. In fact, if $e^z = e^w$ then $e^{z-w} = 1 \Rightarrow z - w \in 2 \pi i \mathbb{Z}$, so this is the only source of non-injectivity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Geometrically, $e^{x + iy} = e^x (\cos y + i \sin y)$ is just the point $(e^x, y)$ in polar coordinates. One can think of the complex exponential as specifying a conformal isomorphism between the following two Riemann surfaces:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Its domain mod $2 \pi i \mathbb{Z}$. One should think of this as the complex plane cut up into horizontal strips $S_k = \{ x + iy : 2 \pi k \le y &amp;lt; 2 \pi (k+1) \}$, all of which are then identified, and all of which have their top and bottom borders identified. Roughly speaking this is an infinitely long tube open at both ends.&lt;/li&gt;&#xA;&lt;li&gt;Its range $\mathbb{C} \setminus \{ 0 \}$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Both of these Riemann surfaces can be identified with the Riemann sphere minus two points. In the first picture one first &quot;puffs out&quot; the tube and then shrinks the points. In the second picture one first identifies $\mathbb{C}$ with the Riemann sphere minus a point via &lt;a href=&quot;http://en.wikipedia.org/wiki/Stereographic_projection&quot; rel=&quot;nofollow&quot;&gt;stereographic projection&lt;/a&gt; and then removes an additional point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(This picture works fairly well, I think. For example vertical lines in the domain are mapped to circles in the range in a natural way, and shifting the lines left or right corresponds to varying the radius of the circle, or varying its relative position between the poles on the Riemann sphere.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a nice discussion about how to visualize holomorphic functions in Needham's &lt;a href=&quot;http://usf.usfca.edu/vca/&quot; rel=&quot;nofollow&quot;&gt;Visual Complex Analysis&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2011-02-09T16:05:06.860" LastActivityDate="2011-02-09T16:05:06.860" />
  <row Id="21196" PostTypeId="1" AcceptedAnswerId="21198" CreationDate="2011-02-09T16:08:46.670" Score="1" ViewCount="196" Body="&lt;p&gt;If $\bar{X}_n$ if a mean of a random sample size ${n}$ from exponential distribution with $\lambda$ &gt; 0, How does exp$ \left(-\frac{1}{\bar{X}_n} \right)$ $\xrightarrow{\rm{P}}$ exp (-$\lambda$) ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More clearly what does the $\rm{P}$ on top of the arrow mean ? I understand it's probability but what is the difference between some expression converges to some value in terms of distribution and probability ?  &lt;/p&gt;&#xA;" OwnerUserId="6335" LastEditorUserId="6335" LastEditorDisplayName="" LastEditDate="2011-02-09T16:25:13.973" LastActivityDate="2011-02-14T14:57:16.550" Title="Question on exponential distribution ?" Tags="&lt;probability&gt;&lt;probability-distributions&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="21197" PostTypeId="1" AcceptedAnswerId="21204" CreationDate="2011-02-09T16:14:19.363" Score="4" ViewCount="139" Body="&lt;p&gt;While working with elliptic curves for cryptography reasons, I found the notion of a primitive point, but no definition.&lt;br&gt;&#xA;For example, $P(0,6)$ is a primitive point on the elliptic curve $y^2\equiv x^3+2x+2 \mod 17$.&#xA;What does that mean? How can I tell if a point is primitive or not?&lt;/p&gt;&#xA;" OwnerUserId="894" LastActivityDate="2011-04-11T22:36:37.373" Title="What is a primitive point on an elliptic curve?" Tags="&lt;modular-arithmetic&gt;&lt;elliptic-curves&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="21198" PostTypeId="2" ParentId="21196" CreationDate="2011-02-09T16:22:26.877" Score="3" ViewCount="" Body="&lt;p&gt;We say that $X_n\to X$ &lt;em&gt;in distribution&lt;/em&gt; if $F_n(x)\to F(x)$ for all continuity points $x$ of $F$. Here, $F_n$ is the distribution function of $X_n$ and $F$ the distribution function of $X$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, $X_n\to X$ &lt;em&gt;in probability&lt;/em&gt; means that &#xA;for every $\varepsilon &amp;gt;0$, we have $$P(|X_n-X|&amp;gt;\varepsilon)\to 0\mbox{ as } n\to\infty.$$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The two concepts are similar, but not quite the same. In fact, convergence in probability is stronger, in the sense that if $X_n\to X$ in probability, then  $X_n\to X$ in distribution. It doesn't work the other way around though; convergence in distribution does not guarantee convergence in probability.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As to your question &quot;How?&quot;, the answer is that for any continuous function $g$, if &#xA;$X_n\to X$ then $g(X_n)\to g(X)$. This holds for both convergence in distribution and for convergence in probability.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $\bar X_n\to1/\lambda$ in probability by the weak law of large numbers, it follows that $\exp(-1/\bar X_n)\to \exp(-\lambda)$ in probability as $n\to \infty$. &lt;/p&gt;&#xA;" OwnerUserId="940" LastActivityDate="2011-02-09T16:22:26.877" CommentCount="3" />
  <row Id="21199" PostTypeId="1" AcceptedAnswerId="21209" CreationDate="2011-02-09T16:23:43.880" Score="30" ViewCount="1205" Body="&lt;p&gt;In the book Thomas's Calculus (11th edition) it is mentioned (Section 3.8 pg 225) that the derivative dy/dx is not a ratio. Couldn't it be interpreted as a ratio, because according to the formula dy = f'(x)dx we are able to plug in values for dx and calculate a dy (differential). Then if we rearrange we get dy/dx which could be seen as a ratio. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wonder if the author say this because dx is an independent variable, and dy is a dependent variable, for dy/dx to be a ratio both variables need to be independent.. maybe?  &lt;/p&gt;&#xA;" OwnerUserId="6786" LastActivityDate="2011-02-10T09:44:15.487" Title="dy/dx is not a ratio?" Tags="&lt;calculus&gt;" AnswerCount="5" CommentCount="3" FavoriteCount="15" />
  <row Id="21200" PostTypeId="2" ParentId="20658" CreationDate="2011-02-09T16:30:47.557" Score="3" ViewCount="" Body="&lt;p&gt;One gets the recurrence as follows. Let $N(n)$ denote the mean number of shuffles needed to sort a deck of $n$ cards and $p_n(k)$ the probability to fix exactly $k$ cards by shuffling once a deck of $n$ cards. Then $N(1)=0$ (the deck is already ordered) and, for every $n\ge2$,&#xA;$$&#xD;&#xA;N(n)=1+\sum_{k=0}^{n-2}p_n(k)N(n-k).&#xD;&#xA;$$&#xA;Let $T_n$ and $B_n$ denote the random numbers of cards fixed at the top and at the bottom of the deck by one shuffle, thus &#xA;$$&#xD;&#xA;p_n(k)=P(T_n+B_n=k).&#xD;&#xA;$$ &#xA;The event $[T_n\ge i,B_n\ge j]$ corresponds to the fact that $i+j$ cards have their position imposed and the remaining $n-i-j$ cards are shuffled freely. Hence &#xA;$$&#xD;&#xA;P(T_n\ge i,B_n\ge j)=(n-k)!/n!=r_n(n-k)&#xD;&#xA;$$ &#xA;for every $(i,j)$ such that $i+j=k$. Now $[T_n=i,B_n=j]$&#xA;is&#xA;$$&#xD;&#xA;[T_n\ge i,B_n\ge j]-[T_n\ge i+1,B_n\ge j]-[T_n\ge i,B_n\ge j+1]+[T_n\ge i+1,B_n\ge j+1],&#xD;&#xA;$$&#xA;hence&#xA;$$&#xD;&#xA;P(T_n=i,B_n=j)=r_n(n-k)-2r_n(n-k-1)+r_n(n-k-2).&#xD;&#xA;$$ &#xA;Since $p_n(k)$ is $(k+1)$ times this, one gets&#xA;$$&#xD;&#xA;N(n)=1+\sum_{k=0}^{n-2}(k+1)(r_n(n-k)-2r_n(n-k-1)+r_n(n-k-2))N(n-k),&#xD;&#xA;$$&#xA;that is,&#xA;$$&#xD;&#xA;n!N(n)=n!+\sum_{k=2}^{n}(n-k+1)\left(k!-2(k-1)!+(k-2)!\right)N(k).&#xD;&#xA;$$&#xA;The rest should be a story of generating functions techniques, as explained &lt;a href=&quot;http://www.math.upenn.edu/~wilf/DownldGF.html&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="6179" LastActivityDate="2011-02-09T16:30:47.557" />
  <row Id="21201" PostTypeId="1" AcceptedAnswerId="21228" CreationDate="2011-02-09T16:42:07.660" Score="1" ViewCount="147" Body="&lt;p&gt;How does one show this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Show that the set of all continuous functions $f:[0,1] \rightarrow R$ is a &lt;strong&gt;closed subset&lt;/strong&gt; of set of all functions $g:[0,1] \rightarrow R$, when endowed with the box topology.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What does it mean if a set is endowed with a box topology? What function will fit this? What happens if I change the box topology to product topology?&lt;/p&gt;&#xA;" OwnerUserId="6838" LastActivityDate="2011-02-09T18:52:49.467" Title="A set of continuous functions is closed subsets set of functions" Tags="&lt;topology&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="21202" PostTypeId="2" ParentId="21190" CreationDate="2011-02-09T16:43:07.067" Score="3" ViewCount="" Body="&lt;p&gt;One interpretation of the Kullback-Leibler divergence $D_{KL}(P \Vert Q)$ is that it measures the extra message length per character needed to send a message drawn from distribution $P$ using an encoding scheme optimized for distribution $Q$.  If one character is assigned a very small probability by $Q$, then the encoding scheme will use a very large number of bits to encode that character, since it is expected to occur so rarely.  (By &lt;a href=&quot;http://en.wikipedia.org/wiki/Kraft%27s_inequality&quot; rel=&quot;nofollow&quot;&gt;Kraft's inequality&lt;/a&gt;, the number of bits should be $-\log_2(q_i)$, where $q_i$ is the probability of the unlikely character.)  If in fact it occurs frequently in distribution $P$, then a message drawn from $P$ will take a large number of bits to encode (specifically, $-p_i \log_2(q_i)$ per character on average).  Therefore, if the $i$-th component of $Q$ goes to zero while the corresponding component of $P$ remains fixed and positive, the KL divergence $D_{KL}(P \Vert Q)$ should (and does) diverge logarithmically as $-p_i \log_2(q_i)$.&lt;/p&gt;&#xA;" OwnerUserId="5546" LastActivityDate="2011-02-09T16:43:07.067" CommentCount="2" />
  <row Id="21203" PostTypeId="1" AcceptedAnswerId="21205" CreationDate="2011-02-09T16:46:09.440" Score="4" ViewCount="190" Body="&lt;p&gt;For the past couple of days I have been encountering the word &quot;Torus&quot; quite often. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;em&gt;I would like to know what special properties does the Torus possess that it is studied very much in Mathematics.&lt;/em&gt; &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;A recent article on Toral Automorphisms was given out to students by one of our Professors, which i have posted here.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://chandrumath.wordpress.com/2011/01/31/toral-automorphisms/&quot; rel=&quot;nofollow&quot;&gt;http://chandrumath.wordpress.com/2011/01/31/toral-automorphisms/&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This is one example which illustrates what properties &quot;Torus&quot; possesses. I am looking for more exciting properties of the Torus which makes it ubiquitous in Mathematics. Another question :&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Are Torus and Doughnuts both same? Or is there any topological difference between them.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" LastActivityDate="2011-02-09T23:59:42.733" Title="What makes Torus Special" Tags="&lt;geometry&gt;&lt;topology&gt;" AnswerCount="4" CommentCount="2" />
  <row Id="21204" PostTypeId="2" ParentId="21197" CreationDate="2011-02-09T16:50:50.010" Score="5" ViewCount="" Body="&lt;p&gt;The points on an elliptic curve (plus a 'point at infinity') form a group under a certain addition law, explained in &lt;a href=&quot;http://en.wikipedia.org/wiki/Elliptic_curve#The_group_law&quot; rel=&quot;nofollow&quot;&gt;this Wikipedia article&lt;/a&gt;. (You probably know this already.) A primitive point $P$ is simply a generator of this group: all elements of the group can be expressed as $P+P+...+P$ ($k$ times) for some $k$. If the elliptic curve has a prime number of points, then all its points (except the point at infinity) are primitive; but in general, the elliptic curve may or may not have a primitive point.&lt;/p&gt;&#xA;" OwnerUserId="1508" LastActivityDate="2011-02-09T16:50:50.010" CommentCount="2" />
  <row Id="21205" PostTypeId="2" ParentId="21203" CreationDate="2011-02-09T16:53:00.553" Score="5" ViewCount="" Body="&lt;p&gt;It is a basic fact that any complex, connected, compact Lie group is a torus of some dimension. So, topologically and analytically, abelian varieties (in particular, elliptic curves) are tori. The fact that an elliptic curve over $\mathbb{C}$ is a torus enables one to determine easily what the group of torsion points looks like. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The torus is an easy example of a space with a nontrivial cohomology ring. The reason is that it is a tensor product (by the Kunneth theorem) of the cohomology ring of $S^1$ with itself, and consequently becomes an exterior algebra (this works for the torus in several dimensions). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The torus in two dimensions is also a simple interesting example of an orientable surface of positive genus and, indeed, any orientable surface of positive genus can be obtained from tori.&lt;/p&gt;&#xA;" OwnerUserId="536" LastActivityDate="2011-02-09T16:53:00.553" CommentCount="4" />
  <row Id="21206" PostTypeId="1" AcceptedAnswerId="21220" CreationDate="2011-02-09T16:54:12.110" Score="0" ViewCount="163" Body="&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/G4Nhf.png&quot; alt=&quot;The question is this&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please tell me if what I did is correct or if there's any faster alternatives.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I set $x$ and $y$ axes on the center of the circle with radius $r$, therefore this can be seen as an area described by $x^2+y^2=r$ revolving around $x=-R-r$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$dV$ can be written as&#xA;$$dV = 2\pi(R+x)\cdot 2y \cdot dx =2\pi(R+x)\cdot 2 \sqrt{r-x^2} \cdot dx$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$V$ is then&lt;br&gt;&#xA;$$ \int_R^{R+2r} 4\pi(R+x)\sqrt{r-x^2} dx $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this correct?&lt;/p&gt;&#xA;" OwnerUserId="5889" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-09T17:54:23.573" LastActivityDate="2011-02-09T17:54:23.573" Title="How to integrate the volume of a solid torus (donut-shaped solid)?" Tags="&lt;integral&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="21207" PostTypeId="1" AcceptedAnswerId="21245" CreationDate="2011-02-09T16:58:58.670" Score="1" ViewCount="177" Body="&lt;p&gt;I've split the integral into:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_0^1\frac{1-x^\alpha}{1-x}\mathrm dx=\int_0^{1/2}\frac{1-x^\alpha}{1-x}\mathrm dx+\int_{1/2}^1\frac{1-x^\alpha}{1-x}\mathrm dx$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm trying to find a suitable function in the form of $g(x)=\frac1{(x-1)^\text{?}}$ so I can use the LCT when $x\to 1$ but I can't figure out what needs to be the exponent. I need to somehow extract $(1-x)$ from $1-x^\alpha$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can this be done?&lt;/p&gt;&#xA;" OwnerUserId="6840" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2011-04-20T16:21:36.803" LastActivityDate="2011-04-20T16:21:36.803" Title="For which values of $\alpha$ does $\int_0^1\frac{1-x^\alpha}{1-x}\mathrm dx$ converge?" Tags="&lt;calculus&gt;&lt;improper-integrals&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="21208" PostTypeId="2" ParentId="21199" CreationDate="2011-02-09T17:00:58.450" Score="1" ViewCount="" Body="&lt;p&gt;The notation $dy/dx$ - in elementary calculus - is simply that: notation to denote the derivative of, in this case, $y$ w.r.t. $x$. (In this case $f&amp;#39;(x)$ is another &lt;em&gt;notation&lt;/em&gt; to express essentially the same thing, i.e. $df(x)/dx$ where $f(x)$ signifies the function $f$ w.r.t. the dependent variable $x$. According to what you've written above, $f(x)$ is the function which takes values in the &lt;em&gt;target space&lt;/em&gt;  $y$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore, by definition, $dy/dx$ &lt;em&gt;at a &lt;strong&gt;specific&lt;/strong&gt; point $x_0$ within the domain $x$&lt;/em&gt; is the real number $L$, if it exists. Otherwise, if no such number exists, then the function $f(x)$ does not have a derivative at the point in question, (i.e. in our case $x_0$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For further information you can read the Wikipedia article: &lt;a href=&quot;http://en.wikipedia.org/wiki/Derivative&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Derivative&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="6824" LastActivityDate="2011-02-09T17:00:58.450" />
  <row Id="21209" PostTypeId="2" ParentId="21199" CreationDate="2011-02-09T17:05:55.190" Score="83" ViewCount="" Body="&lt;p&gt;Historically, when Leibnitz conceived of the notation, $\frac{dy}{dx}$ &lt;em&gt;was&lt;/em&gt; supposed to be a quotient: it was the quotient of the &quot;infinitesimal change in $y$ produced by the change in $x$&quot; divided by the &quot;infinitesimal change in $x$&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the formulation of calculus with infinitesimals in the usual setting of the real numbers leads to a lot of problems. For one thing, infinitesimals can't exist in the usual setting of real numbers! Because the real numbers satisfy an important property, called the Archimedean Property: given &lt;em&gt;any&lt;/em&gt; positive integer $\epsilon\gt 0$, no matter how small, and given any positive real number $M\gt 0$, no matter how big, there exists a natural number $n$ such that $n\epsilon\gt M$. But an &quot;infinitesimal&quot; $\xi$ is supposed to be so small that no matter how many times you add it to itself, it never gets to $1$, contradicting the Archimedean Property. Other problems: Leibnitz defined the tangent to the graph of $y=f(x)$ at $x=a$ by saying &quot;Take the point $(a,f(a))$; then add an infinitesimal amount to $a$, $a+dx$, and take the point $(a+dx,f(a+dx))$, and draw the line through those two points.&quot; But if they are two different points on the graph, then it's not a tangent, and if it's just one point, then you can't define the line because you just have one point. That's just two of the problems with infinitesimals. (See below where it says &quot;&lt;strong&gt;However...&lt;/strong&gt;&quot;, though).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So Calculus was essentially rewritten from the ground up in the following 200 years to avoid these problems, and you are seeing the results of that rewriting (that's where limits came from, for instance). Because of that rewriting, the derivative is no longer a &lt;em&gt;quotient&lt;/em&gt;, now it's a &lt;strong&gt;limit&lt;/strong&gt;:&#xA;$$\lim_{h\to0 }\frac{f(x+h)-f(x)}{h}.$$&#xA;And because we cannot express this limit-of-a-quotient as a-quotient-of-the-limits (both numerator and denominator go to zero), then the derivative is &lt;em&gt;not&lt;/em&gt; a quotient.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, Leibnitz's notation is &lt;em&gt;very&lt;/em&gt; suggestive and very useful; even though derivatives are not really quotients, in many ways they behave &lt;em&gt;as if&lt;/em&gt; they were quotients. So we have the Chain Rule:&#xA;$$\frac{dy}{dx} = \frac{dy}{du}\;\frac{du}{dx}$$&#xA;which looks very natural if you think of the derivatives as &quot;fractions&quot;. You have the Inverse Function theorem, which tells you that&#xA;$$\frac{dx}{dy} = \frac{1}{\quad\frac{dy}{dx}\quad},$$&#xA;again, almost &quot;obvious&quot; if you think of the derivatives as fractions. So, because the notation is so nice and so suggestive, we keep the notation even though the notation no longer represents an &lt;em&gt;actual&lt;/em&gt; quotient, it now represents a single limit.  In fact, Leibnitz's notation is so good, so superior to the prime notation and to Newton's notation, that England fell behind all of Europe for &lt;em&gt;centuries&lt;/em&gt; in mathematics and science because, due to the fight between Newton's and Leibnitz's camp over who had invented Calculus and who stole it from whom (consensus is that they each discovered it independently), England's scientific establishment decided to ignore what was being done in Europe with Leibnitz notation and stuck to Newton's... and got stuck in the mud in large part because of it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Differentials are part of this same issue: originally, $dy$ and $dx$ really &lt;em&gt;did&lt;/em&gt; mean the same thing as those symbols do in $\frac{dy}{dx}$, but that leads to all sorts of logical problems, so they no longer mean the same thing, even though they behave as if they did).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, even though we write $\frac{dy}{dx}$ as if it were a fraction, and many computations look like we are working with it &lt;em&gt;like&lt;/em&gt; a fraction, it isn't &lt;em&gt;really&lt;/em&gt; a fraction (it just plays one on television). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;However...&lt;/strong&gt; There is a way of getting around the logical difficulties with infinitesimals; this is called &lt;em&gt;nonstandard analysis&lt;/em&gt;. It's pretty difficult to explain how one sets it up, but you can think of it as creating two classes of real numbers: the ones you are familiar with, that satisfy things like the Archimedean Property, the Supremum Property, and so on, and then you add another, separate class of reals numbers that includes infinitesimals and a bunch of other things. If you do that, then you &lt;em&gt;can&lt;/em&gt;, if you are careful, define derivatives exactly like Leibnitz, in terms of infinitesimals and actual quotients; if you do that, then all the rules of Calculus that make use of $\frac{dy}{dx}$ &lt;em&gt;as if&lt;/em&gt; it were a fraction are justified because, &lt;em&gt;in that setting&lt;/em&gt;, it &lt;strong&gt;is&lt;/strong&gt; a fraction. Still, one has to be careful because you have to keep infinitesimals and regular real numbers separate and not let them get confused, or you can run into some serious problems. &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-09T21:24:36.747" LastActivityDate="2011-02-09T21:24:36.747" CommentCount="4" />
  <row Id="21210" PostTypeId="2" ParentId="21199" CreationDate="2011-02-09T17:06:07.750" Score="0" ViewCount="" Body="&lt;p&gt;It is not a ratio, just as $dx$ is not a product.&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2011-02-09T17:06:07.750" CommentCount="2" />
  <row Id="21211" PostTypeId="1" AcceptedAnswerId="21221" CreationDate="2011-02-09T17:08:16.267" Score="0" ViewCount="110" Body="&lt;p&gt;Hello, I've found the following derivative in my Calculus book and I can't get my my head around the algebra involved. Can anybody help me?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/jd7gQ.png&quot; alt=&quot;problem &amp;amp; solution&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="6839" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2011-02-09T17:47:26.513" LastActivityDate="2011-02-09T21:12:55.197" Title="Derivative of a rational function" Tags="&lt;calculus&gt;" AnswerCount="3" CommentCount="3" />
  <row Id="21212" PostTypeId="2" ParentId="21177" CreationDate="2011-02-09T17:09:57.367" Score="1" ViewCount="" Body="&lt;p&gt;I would resize the circles by a factor $k$ (in the figure $k=2$) so that the maximum number of circles is tangent to the ellipse and among them. To determine their centers I would try to solve this complex packing problem by trial and error, starting with the biggest circle (corresponding to the light blue one in the figure) and proceeding with the smaller ones (corresponding to the dark blue and the two equal sized black ones). The original circles would have the same centers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/hGCfO.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="752" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2011-02-10T21:54:33.473" LastActivityDate="2011-02-10T21:54:33.473" CommentCount="2" />
  <row Id="21213" PostTypeId="2" ParentId="21186" CreationDate="2011-02-09T17:14:18.443" Score="11" ViewCount="" Body="&lt;p&gt;The natural domain of the function $f(x) = 2x^2 +8x+6$ is all real numbers (remember, the &quot;natural domain&quot; is the collection of all real numbers for which the formula 'makes sense', or yields a real number).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The natural domain of $g(x) = 2x\left(x + 4 + \frac{3}{x}\right)$ is all real numbers &lt;strong&gt;except $x=0$&lt;/strong&gt;, because you cannot plug in $x=0$ into the formula (division by zero makes the universe explode, after all). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So one &lt;em&gt;major&lt;/em&gt; problem with writing $2x^2 + 8x + 6 = 2x\left(x + 4 + \frac{3}{x}\right)$ is that &lt;strong&gt;it is not true&lt;/strong&gt;. They are not equal! The left hand side makes sense at $x=0$, but the right hand side does not. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another major problem is that polynomials are nice and easy, while rational functions are less nice and less easy (just wait until you get to integrals: if you have to integrate a rational function, you'll groan and settle yourself in for some strenuous work; if you have to integrate a polynomial, you'll smile at how easy your life is going to be). So you would much rather deal with polynomials than rational functions. Here you start with a polynomial, $2x^2+8x+6$, and end up with a product of a polynomial and a rational function, so you've made your life that much harder.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another issue is that one is very interested in knowing when a function is equal to $0$. If you factor as usual, $2x^2+8x+6 = 2(x^2+4x+3) = 2(x+1)(x+3)$, then it is very easy to figure out where $2x^2+8x+6$ is zero, because a product is zero if and only if at least one factor is zero, so you would need either $2=0$ (impossible), $x+1=0$ (which means $x=-1$) or $x+3=0$ (which means $x=-3$). With a bit of practice you can basically just &quot;read off&quot; where the product is zero. If we factor like you suggest, $2x^2+8x+6 = 2x\left(x + 4 + \frac{3}{x}\right)$, first you might be misled into thinking that $x=0$ will make it zero, since you have the factor $2x$ (which would be zero at $x=0$; but unfortunately, you cannot plug in $x=0$ into the second factor, so that causes problems). And second, to figure out when $x+4+\frac{3}{x}$ is $0$, you would end up having to figure out where $x^2+4x+3=0$, that is, your original problem. So the factorization you propose doesn't simplify the problem, and introduces the possibility of error.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's a couple of reasons, at any rate.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-09T18:58:55.343" LastActivityDate="2011-02-09T18:58:55.343" />
  <row Id="21214" PostTypeId="1" AcceptedAnswerId="21217" CreationDate="2011-02-09T17:17:22.447" Score="1" ViewCount="64" Body="&lt;p&gt;As I was a student, I got the following problem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $f$ be a mapping from $[a,b]$ to $\mathbb{R}$ satisfying&#xA;$$(1)\quad f((1-\lambda)x+\lambda y)\leq (1-\lambda )f(x)+\lambda f(y)$$&#xA;for all $x,y\in [a,b]$ and $\lambda\in [0,1]$. Suppose $f$ is differentiable on $]a,b[$. Show that if $f$ is continuous and $f&amp;#39;$ increasing then $(1)$ holds and vice versa.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a part of the solution, I fixed $y$, assumed $x&amp;lt;y$, and denoted $z:=(1-\lambda)x+\lambda y$ where $\lambda\ne 0,1$. Then we have $x\leq z$. I also proved that&#xA;$$\frac{f(z)-f(x)}{z-x}\leq \frac{f(y)-f(z)}{y-z}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I thought that &#xA;$$\lim_{z\to x}\frac{f(z)-f(x)}{z-x}\leq \lim_{z\to x}\frac{f(y)-f(z)}{y-z}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My advisor said it is wrong and it should be&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\lim_{z\to x+}\frac{f(z)-f(x)}{z-x}\leq \lim_{z\to x+}\frac{f(y)-f(z)}{y-z}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I never got a proper reasoning why the formula I wrote is wrong. I assumed that $x\leq z$ so I thought $\lim_{z\to x}$ means $\lim_{z\to x+}$ as $z$ can't approach $x$ from left. Could anyone explain me why limit does not mean one sided limit if I can't compute the two-sided limit?&lt;/p&gt;&#xA;" OwnerUserId="5462" LastActivityDate="2011-02-09T17:34:26.180" Title="One sided limit" Tags="&lt;limit&gt;" AnswerCount="1" />
  <row Id="21215" PostTypeId="2" ParentId="21211" CreationDate="2011-02-09T17:20:34.750" Score="0" ViewCount="" Body="&lt;p&gt;Just try using the quotient rule for derivatives and see where you get stuck on the algebra. I'm guessing the book just skipped over some intermediate steps and that's why it's not immediately clear how it got to one step to the other. I suggest trying it out on your own first with pencil and paper, then ask again if you get stuck.&lt;/p&gt;&#xA;" OwnerUserId="5369" LastActivityDate="2011-02-09T17:20:34.750" CommentCount="1" />
  <row Id="21216" PostTypeId="2" ParentId="21203" CreationDate="2011-02-09T17:34:23.200" Score="4" ViewCount="" Body="&lt;p&gt;(Below I only consider compact orientable surfaces, &lt;em&gt;i.e.&lt;/em&gt;, those than embed in $3$-space)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The torus is the only surface which can be endowed with a metric of vanishing curvature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is the only parallelizable surface.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is the only surface which can be turned into a topological group.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Is it the only surface where a differential equation can have all its orbits dense? one orbit dense?)&lt;/p&gt;&#xA;" OwnerUserId="274" LastEditorUserId="274" LastEditorDisplayName="" LastEditDate="2011-02-09T20:08:29.517" LastActivityDate="2011-02-09T20:08:29.517" CommentCount="2" />
  <row Id="21217" PostTypeId="2" ParentId="21214" CreationDate="2011-02-09T17:34:26.180" Score="0" ViewCount="" Body="&lt;p&gt;As you said $z\geq x$, so $z$ cannot approach $x$ from the left.  The notation $\lim_{z\to x}$ implies that $z$ could be approaching from either side.&lt;/p&gt;&#xA;" OwnerUserId="2838" LastActivityDate="2011-02-09T17:34:26.180" />
  <row Id="21218" PostTypeId="2" ParentId="21128" CreationDate="2011-02-09T17:47:38.497" Score="5" ViewCount="" Body="&lt;p&gt;Luckily these days there is a beautiful text that teaches algebra and category theory at the same time: &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0821847813&quot; rel=&quot;nofollow&quot;&gt;Aluffi - Chapter 0&lt;/a&gt;. It deserves to be more well-known. Besides the fact that it uses (basic) category language from the outset, it is very well-written. If I would ever teach an algebra course, this would probably the text I would use.&lt;/p&gt;&#xA;" OwnerUserId="6490" LastActivityDate="2011-02-09T17:47:38.497" CommentCount="1" />
  <row Id="21219" PostTypeId="2" ParentId="21207" CreationDate="2011-02-09T17:48:22.420" Score="5" ViewCount="" Body="&lt;p&gt;If $\alpha$ is a natural number, try factoring the numerator into something manageable. Notice that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$1-x^{\alpha} = (1-x)(1+x+...+x^{\alpha-2}+x^{\alpha-1})$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now you can integrate each term.&lt;/p&gt;&#xA;" OwnerUserId="1919" LastActivityDate="2011-02-09T17:48:22.420" />
  <row Id="21220" PostTypeId="2" ParentId="21206" CreationDate="2011-02-09T17:51:33.420" Score="1" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;hint:&lt;/strong&gt; in addition to what Raskolnikov said, it may be simpler to consider the washer method instead of the shell method for setting up your integral. (That is, integrate in $y$ and consider the area of annuli of the constant $y$ slices.)&lt;/p&gt;&#xA;" OwnerUserId="1543" LastActivityDate="2011-02-09T17:51:33.420" CommentCount="2" />
  <row Id="21221" PostTypeId="2" ParentId="21211" CreationDate="2011-02-09T17:52:41.640" Score="5" ViewCount="" Body="&lt;p&gt;\begin{align*}&#xA;\frac{d}{dx}\left(-\frac{x^2+1}{(x^2-1)^2}\right) &amp;amp;= -\frac{d}{dx}\left(\frac{x^2+1}{(x^2-1)^2} \right)&amp;amp;\quad&amp;amp;\mbox{(1)}\\&#xA;&amp;amp;= -\left(\frac{(x^2-1)^2(x^2+1)' - (x^2+1)\left((x^2-1)^2\right)'}{\left((x^2-1)^2\right)^2}\right)&amp;amp;&amp;amp;\mbox{(2)}\\&#xA;&amp;amp;= - \frac{(x^2-1)^2(2x) - (x^2+1)\left(2(x^2-1)(x^2-1)'\right)}{(x^2-1)^4}&amp;amp;&amp;amp;\mbox{(3)}\\&#xA;&amp;amp;= -\frac{2x(x^2-1)^2 - (x^2+1)(2(x^2-1)2x)}{(x^2-1)^4}&amp;amp;&amp;amp;\mbox{(4)}\\&#xA;&amp;amp;= -\frac{2x(x^2-1)^2 - 4x(x^2+1)(x^2-1)}{(x^2-1)^4}&amp;amp;&amp;amp;\mbox{(5)}\\&#xA;&amp;amp;= - \frac{2x(x^2-1)\left((x^2-1) - 2(x^2+1)\right)}{(x^2-1)^4}&amp;amp;&amp;amp;\mbox{(6)}\\&#xA;&amp;amp;= - \frac{2x(x^2-1)\left(x^2-1-2x^2-2\right)}{(x^2-1)^4}&amp;amp;&amp;amp;\mbox{(7)}\\&#xA;&amp;amp;= - \frac{2x(x^2-1)(-x^2-3)}{(x^2-1)^4}&amp;amp;&amp;amp;\mbox{(8)}\\&#xA;&amp;amp;= - \frac{2x(-x^2-3)}{(x^2-1)^3}&amp;amp;&amp;amp;\mbox{(9)}\\&#xA;&amp;amp;= -\frac{-2x(x^2+3)}{(x^2-1)^3}&amp;amp;&amp;amp;\mbox{(10)}\\&#xA;&amp;amp;= -(-2)\frac{x(x^2+3)}{(x^2-1)^3}&amp;amp;&amp;amp;\mbox{(11)}\\&#xA;&amp;amp;= 2\frac{x(x^2+3)}{(x^2-1)^3}.&amp;amp;&amp;amp;\mbox{(12)}&#xA;\end{align*}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Notes.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Pull out the minus sign fromt he derivative.&lt;/li&gt;&#xA;&lt;li&gt;Use the Quotient Rule.&lt;/li&gt;&#xA;&lt;li&gt;Do the derivatives in the numerator, using the Chain Rule for $(x^2-1)^2$.&lt;/li&gt;&#xA;&lt;li&gt;Finish the derivative.&lt;/li&gt;&#xA;&lt;li&gt;Do some of the algebra in the numerator. Notice that both summands in the numerator have a factor of $2x(x^2-1)$.&lt;/li&gt;&#xA;&lt;li&gt;Factor out $2x(x^2-1)$ from both summands in the numerator.&lt;/li&gt;&#xA;&lt;li&gt;Do the operations in the other factor.&lt;/li&gt;&#xA;&lt;li&gt;Do the algebra in the numerator.&lt;/li&gt;&#xA;&lt;li&gt;Cancel the $x^2-1$ in the numerator with one in the denominator.&lt;/li&gt;&#xA;&lt;li&gt;Pull out the minus sign from $(-x^2-3)$.&lt;/li&gt;&#xA;&lt;li&gt;Pull out the $-2$ from the fraction.&lt;/li&gt;&#xA;&lt;li&gt;Simplify $-(-2)$ to $2$, and rejoice for your answer matches the one in the book.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-09T18:57:23.473" LastActivityDate="2011-02-09T18:57:23.473" CommentCount="2" />
  <row Id="21222" PostTypeId="1" CreationDate="2011-02-09T17:53:17.427" Score="3" ViewCount="52" Body="&lt;p&gt;Hi,&#xA;I am looking for complete references on places, valuations and so on. In particular I need to understand the meaning of &quot;a finite place does not divide a prime number $\ell$&quot; and what is the Frobenius automorphism at a place.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you&lt;/p&gt;&#xA;" OwnerUserId="6583" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-09T17:55:13.243" LastActivityDate="2011-02-09T17:55:13.243" Title="places valuations and so on" Tags="&lt;algebraic-number-theory&gt;&lt;arithmetic-geometry&gt;" CommentCount="3" FavoriteCount="1" />
  <row Id="21223" PostTypeId="2" ParentId="21128" CreationDate="2011-02-09T17:57:11.290" Score="2" ViewCount="" Body="&lt;p&gt;I think there are good reasons to learn category theory early on for the unifying perspective it gives and also good reasons to wait until you have a healthy amount of examples on hand. One thing I want to emphasize, since no one here has mentioned it thus far: elementary category theory is largely a matter of learning a new language, with many definitions and few results. In fact, I would go so far as to say there are &lt;em&gt;no&lt;/em&gt; results in elementary category theory. The Yoneda lemma might count, depending on whether it is actually elementary and whether its proof is actually more than a tautology. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, I think this helps the case of those who would encourage you to learn the subject earlier, since reading or talking about category theory is pretty smooth sailing in the sense that there are no intricate proofs to follow. Note, however, that the definitions of basic categorical concepts may themselves be somewhat intricate, but hopefully after a while they will seem natural and not cumbersome.&lt;/p&gt;&#xA;" OwnerUserId="4842" LastActivityDate="2011-02-09T17:57:11.290" CommentCount="3" />
  <row Id="21224" PostTypeId="2" ParentId="21203" CreationDate="2011-02-09T18:26:26.137" Score="2" ViewCount="" Body="&lt;p&gt;Tori (plural of torus) are mathematical objects which are very rich of structure because their algebraic and geometric features interact deeply and nicely:&#xA;1. Geometrically, they can be described has a product of copies of $S^1$, making evident their nature of compact variety, whose homology and cohomology are non-trivial,well-understood and &quot;easy&quot; to manipulate.&#xA;2. Algebraically, they're a group (a Lie group, in fact) and there are more or less obvious links with the multiplicative algebraic group and the theory of lattices.&#xA;3. Analitically, tori are the natural &quot;home&quot; of periodic functions, elliptic functions and modular forms.&#xA;4. Finally, under some restrictions, tori are also endowed with an arithmetic structure which is deeply intertwined with all the previous ones.&lt;/p&gt;&#xA;" OwnerUserId="688" LastActivityDate="2011-02-09T18:26:26.137" />
  <row Id="21225" PostTypeId="2" ParentId="21186" CreationDate="2011-02-09T18:31:34.867" Score="0" ViewCount="" Body="&lt;p&gt;Why writing something in a complicated way, while you can write it in a simpler way?&lt;/p&gt;&#xA;" OwnerUserId="688" LastActivityDate="2011-02-09T18:31:34.867" CommentCount="1" />
  <row Id="21226" PostTypeId="2" ParentId="21207" CreationDate="2011-02-09T18:36:02.053" Score="4" ViewCount="" Body="&lt;p&gt;This is just an expansion on Shai Covo's comment above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to Gradshteyn &amp;amp; Ryzhik's &lt;a href=&quot;http://books.google.com/books?id=aBgFYxKHUjsC&amp;amp;printsec=frontcover&amp;amp;dq=table+of+integrals+series+and+products&amp;amp;source=bl&amp;amp;ots=bJmzPYG5KA&amp;amp;sig=DcY0WN_wayVB-HGTukp97nu2M5o&amp;amp;hl=en&amp;amp;ei=ZtxSTb2uEJG_gQfOj_nLCA&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=3&amp;amp;ved=0CC8Q6AEwAg#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;Table of Integrals, Series, and Products&lt;/a&gt;,&#xA;$$\begin{align}\int_{0}^{1}\frac{1-x^\alpha}{1-x}dx &amp;amp;= \psi(\alpha + 1) - \psi(1)\\&#xD;&#xA;                                       &amp;amp;= \psi(\alpha + 1) + \gamma \end{align}$$&#xA;and converges for $\text{Re} (\alpha) &amp;gt; 0$.&lt;/p&gt;&#xA;" OwnerUserId="1016" LastActivityDate="2011-02-09T18:36:02.053" CommunityOwnedDate="2011-02-09T18:36:02.053" CommentCount="3" />
  <row Id="21227" PostTypeId="2" ParentId="21175" CreationDate="2011-02-09T18:40:05.027" Score="1" ViewCount="" Body="&lt;p&gt;Inserting &lt;code&gt;\sum_{n=2}^{infty} (-1)^n/(ln(n)+cos(n))&lt;/code&gt; into the search box of &lt;a href=&quot;http://www.wolframalpha.com/input/?i=%5Csum_%7Bn=2%7D%5E%7Binfty%7D%20%28-1%29%5En/%28ln%28n%29%20+cos%28n%29%29&quot; rel=&quot;nofollow&quot;&gt;Wolfram Alpha&lt;/a&gt; gives an approximate value of –4.552162944044021. However unbelievable the actual sum may be, you can be a bit sure about its convergence.&lt;/p&gt;&#xA;" OwnerUserId="5595" LastEditorUserId="10557" LastEditorDisplayName="" LastEditDate="2011-05-15T01:48:05.987" LastActivityDate="2011-05-15T01:48:05.987" CommentCount="1" />
  <row Id="21228" PostTypeId="2" ParentId="21201" CreationDate="2011-02-09T18:46:51.813" Score="2" ViewCount="" Body="&lt;p&gt;One way to think about the box topology on the space of all functions $f:[0,1]\to\mathbb{R}$ is that the basic open neighborhoods of such an $f$ consist of all functions that lay inside a tiny tube around $f$, whose radius can vary (even discontinuously). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the point is that if every such tube around $f$ contains a continuous function, then $f$ itself must be continuous, by a simple $\epsilon/3$ argument. That is, for any $x\in[0,1]$ and any $\epsilon\gt 0$, pick a tube of radius $\epsilon/3$ and a continuous function $g$ within that tube. Since $g$ is continuous at $x$, there is some $\delta\gt 0$ such that any $y$ within $\delta$ of $x$ has $g(y)$ within $\epsilon/3$ of $g(x)$, and also $f(y)$ within $\epsilon/3$ of $g(y)$ and $f(x)$ within $\epsilon/3$ of $g(x)$. So altogether, $f(y)$ is within $\epsilon$ of $f(x)$, as desired.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the product topology, in contrast, the basic open sets can be thought of as specifying tiny windows at finitely many points in the domain, such that any function that goes through those windows is counted as in the open set. (In the box topology, in contrast, we in effect get to specify a window at &lt;em&gt;every&lt;/em&gt; point in the domain.) Consider the functions $f_n(x)=x^n$. The pointwise limit of these functions is not continuous, since it has value $0$ on $[0,1)$ and value $1$ at $x=1$. But any basic open neighborhood of that limit function will contain some $f_n$, and so the limit function is in the closure of the continuous functions, but not itself continuous.&lt;/p&gt;&#xA;" OwnerUserId="413" LastEditorUserId="413" LastEditorDisplayName="" LastEditDate="2011-02-09T18:52:49.467" LastActivityDate="2011-02-09T18:52:49.467" CommentCount="2" />
  <row Id="21229" PostTypeId="2" ParentId="20122" CreationDate="2011-02-09T19:04:02.773" Score="0" ViewCount="" Body="&lt;p&gt;Having seen the simulations, here are some definitions that may help clarify the problem. (Though I don't have a solution.)&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;First, to define the free region versus the jammed region, the expected density of either region should be dependent on $\alpha$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;i. Say you start with a single solid block. For $\alpha = 0$ there will be no motion, so the &quot;free region&quot; would be empty. But for $\alpha = 1$, the people would peel off from the front of the queue, so the expected free region should have about 50% density. In general you probably will expect the free region to have $\alpha / (\alpha + 1)$ density.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ii. On the other hand, say you start with initially alternating configuration. For $\alpha = 0$, this leads to no motion, and should be considered as jammed. While for $\alpha = 1$, this leads to continuous flow, and should be considered as non-jammed. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we can probably define free and non-free based on the expected proportion of moving guys. Say a stretch of $K$ spots has density $\delta$. Then you should expect $(1-\delta)^3K$ number individuals that can flow freely, $[(1-\delta)\delta^2 + 2\delta(1-\delta)^2]K$ individuals to flow with probability $\alpha$, and $\delta K$ individuals to be necessarily trapped. So the expected proportion of moving individuals is $$ P(\delta,\alpha) = (1-\delta)^3 + 2\alpha (1-\delta)^2\delta + (1-\delta)\delta^2 \alpha$$ so maybe we can arbitrarily define a free-flowing region to be one with density $\delta$ such that $P(\delta,\alpha) &amp;gt; 2/3$ and a jammed region to be one with density such that $P(\delta,\alpha) &amp;lt; 1/3$ (the numbers 1/3 and 2/3 I just pulled out of a hat). In any case, since $\alpha$ and $\delta$ will be between 0 and 1, you can check that $P(\delta,\alpha)$ is a decreasing function of $\delta$. So you can define $\delta_{max}$ the lower bound for the jammed region and $\delta_{min}$ the upper bound for the free region.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;However, to compute $\delta$, you will need a window of size $K$ over which to compute. So perhaps what we need to do is to set the initial number of people really large, say $N &amp;gt; 10000$. Then at every point we compute its neighboring density $\delta_K(x)$ as the density of people in the region $[x-50,x+50]$ (again, here I picked $K = 101$ rather arbitrarily, just much smaller than $N$). Then you can properly define the density near a point $x$. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;So you can rephrase your question as asking for the following. Given $K$ much bigger than 1 and $N$ much bigger than $K$. Let your process run with some initial density $\delta_{min} &amp;lt; \Delta &amp;lt; \delta_{max}$ (so you don't start with something that is already completely free or completely jammed) under probability $\alpha$. At every point $x$ compute $\delta_K(x)$ as above. Your question then is one about the expected number of of components of $\{x| \delta_K(x) &amp;gt; \delta_{max}\}$ and the expected number of components of $\{x|\delta_K(x) &amp;lt; \delta_{min}\}$, the former being the number of jammed regions and the latter being the number of free regions. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Or, perhaps an even better formulation (instead of steps 2 and 3 above) of the question can be taken as the following. Start with a large number $N$ and probability $\alpha$, with initial configuration with density $\delta_{min} &amp;lt; \Delta &amp;lt; \delta_{max}$. Prove or disprove that there is a constant $k$ depending on $\delta_{max}$, $\delta_{min}$ and $\alpha$, such that there must exist (with probability 1), after a long time evolution, a contiguous set of size at least $N/k$ on which the density of people is larger than $\delta_{max}$ and a contiguous set of size at least $N/k$ on which the density of people is less than $\delta_{min}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This way, I think, you will have a problem that is well-formulated and tractable mathematically. Good luck!&lt;/p&gt;&#xA;" OwnerUserId="1543" LastActivityDate="2011-02-09T19:04:02.773" CommentCount="12" />
  <row Id="21230" PostTypeId="1" CreationDate="2011-02-09T19:12:13.570" Score="0" ViewCount="47" Body="&lt;p&gt;Hi, I have a question about the relationship between parameters of weakly regular graphs (if it exists)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider two weakly regular graphs with parameters (n,k,lambda,beta) where n is number of vertices, k is the degree, lambda is the number of common neighbors of adjacent vertices, and beta is the number of common neighbors of nonadjacent vertices. (At least one of the latter two parameters (lambda,beta) will have two values for a weakly regular graph).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose (n,k) is the same for both graphs. And suppose one graph has lambda=0 for all vertices (zero clustering, triangle-free) and the other one has lambda=1 for all.. Can we say/show that when lambda is 1 (or in general higher), beta's of the remaining vertices in that graph is lower than the other graph? In other words, the number of common neighbours of nonadjacent vertices is less when the number of common neighbours of adjacent vertices are higher in a graph.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;thank you very much in advance..&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gizem&lt;/p&gt;&#xA;" OwnerUserId="6842" LastActivityDate="2011-02-09T19:12:13.570" Title="weakly regular graphs" Tags="&lt;graph-theory&gt;" />
  <row Id="21231" PostTypeId="1" AcceptedAnswerId="27747" CreationDate="2011-02-09T19:21:04.803" Score="3" ViewCount="89" Body="&lt;p&gt;A function $f$ is called fully time-constructible if there exists a Turing machine $M$ which, given a string $1^n$ consisting of $n$ ones, stops after exactly $f(n)$ steps.&#xA;Analogously, we can call a function $f$ fully space-constructible if there exists a Turing machine $M$ which, given a string $1^n$ consisting of $n$ ones, halts after using exactly $f(n)$ cells.&#xA;Consider this function:&#xA;\begin{equation*}&#xA;  f(n) = &#xA;  \begin{cases}&#xA;      n^3, &amp;amp; \text{if}\ n = 3k \ for \ k \in \mathbb{N} \ and&#xA;      \ 2^n, &amp;amp; \text{otherwise}&#xA;  \end{cases} &#xA; \end{equation*}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is $f(n)$ time-constructible and/or space constructible? Also, what is the intuition behind considering time/space constructible functions in complexity theory? Will things be any different if we include non-constructible time/space functions?&lt;/p&gt;&#xA;" OwnerUserId="1441" LastActivityDate="2011-03-18T06:06:16.173" Title="Importance of Constructible functions" Tags="&lt;algorithms&gt;&lt;computer-science&gt;" AnswerCount="1" FavoriteCount="2" />
  <row Id="21232" PostTypeId="1" AcceptedAnswerId="21236" CreationDate="2011-02-09T19:27:45.647" Score="2" ViewCount="75" Body="&lt;p&gt;The problem i am supposed to solve for $x$ by completing the square: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$3x^2+9x+5 = 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;step 1. move constant to right: $3x^2+9x\quad\quad+5 = 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;step 2. divide by $3$: $x^2+3\quad\quad+\frac{5}{3}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;step 3. $(\frac{1}{2}b)^2$:  $(\frac{1}{2}\cdot 3)^2 = \frac{9}{4}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;step 4. add and subtract result from step 3 into equation: $(x^2+3+\frac{9}{4}) -\frac{9}{4} + \frac{5}{3} = 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;step 5. simplify: $(x+\frac{3}{2})^2 -\frac{7}{12} = 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;step 6. subtract constant on both sides: $(x+\frac{3}{2})^2 = \frac{7}{12}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;step 7. solve for $x$: $x = -\frac{3}{2} + \sqrt{\frac{7}{12}}$ or $-\frac{3}{2} - \sqrt{\frac{7}{12}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wolfram seems to be telling me my answer is wrong.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the correct answer listed is: $x = \frac{-9-\sqrt{21}}{6}$ or $x = \frac{\sqrt{21}-9}{6}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If anyone could let me know where my error is I would greatly appreciate it!&lt;/p&gt;&#xA;" OwnerUserId="5032" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2011-02-09T19:52:57.730" LastActivityDate="2011-02-09T19:52:57.730" Title="Completing the square problem" Tags="&lt;algebra&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="21233" PostTypeId="1" CreationDate="2011-02-09T19:31:27.547" Score="3" ViewCount="176" Body="&lt;p&gt;Let $G=&amp;lt;a,b  | a^2b=ba^2, b^2a=ab^2, a^2=b^4&amp;gt;$.  Consider a homomorphism $\phi$ from $G$ to $&amp;lt;c| &amp;gt;$ where $\phi(a)=c^2$ and $\phi(b)=c$.  Can someone help me decide what group the kernel of this map is?  Thanks!&lt;/p&gt;&#xA;" OwnerUserId="6077" LastActivityDate="2011-02-10T21:30:31.717" Title="help finding the kernel of a group homomorphism" Tags="&lt;group-theory&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="21234" PostTypeId="1" AcceptedAnswerId="21251" CreationDate="2011-02-09T19:36:17.133" Score="0" ViewCount="104" Body="&lt;p&gt;In &lt;a href=&quot;http://users.jyu.fi/~laurikah/REP/REPtext2010_2.pdf&quot; rel=&quot;nofollow&quot;&gt;these Karen Smith's notes on representation of finite groups&lt;/a&gt;, on page 50 the irreducible representations of $S_3$ are found.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $\sigma=(1 2 3)$ and $V$ is a complex representation of $S_3$, I don't understand what &quot;the action of $\sigma$ on $V$&quot; means. Is it this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\langle\sigma\rangle \times V \to V$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sigma \cdot (v_1,\dots,v_n)=(v_{\sigma(1)},v_{\sigma(2)},v_{\sigma(3)},v_4,\dots,v_n)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\langle\sigma\rangle$ means the subgroup of $S_3$ generated by $\sigma$?&lt;/p&gt;&#xA;" OwnerUserId="2614" LastActivityDate="2011-02-09T21:13:10.757" Title="Representations of $S_3$" Tags="&lt;representation-theory&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="21235" PostTypeId="1" AcceptedAnswerId="21237" CreationDate="2011-02-09T19:39:28.943" Score="2" ViewCount="190" Body="&lt;p&gt;$$\int{96\cos^4(16x)} \ dx$$&lt;br&gt;&#xA;Setting $u=16x$, $du=16dx$,&#xA;$${\frac{96}{16}}\int{\cos^4(u)} \ du$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Kinda stuck here, I checked Wolfram Alpha but it suggests using some reduction formula that we haven't learned in my class yet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;@Myself:  Using what you posted, I end up getting&#xA;$${\frac{3}{2}}\left(\frac{\sin4u}{4}+\frac{3u}{4}\right)$$&#xA;Sub back in my $u$ and I've got:&#xA;$$\frac{3\sin(64x)}{8} + \frac{9x}{8}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't think I did it right...&lt;/p&gt;&#xA;" OwnerUserId="6423" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-10T04:39:56.217" LastActivityDate="2011-04-23T19:20:10.477" Title="Evaluating ${\frac{96}{6}}\int{\cos^4(16x)} \ dx$" Tags="&lt;calculus&gt;&lt;homework&gt;&lt;integral&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="21236" PostTypeId="2" ParentId="21232" CreationDate="2011-02-09T19:45:38.547" Score="2" ViewCount="" Body="&lt;p&gt;As Theo Buehler said in a comment, your answer and technique are correct, just yielding a different form of the same answer (rationalize the denominator in your answer, then combine the fractions by finding a common denominator).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an aside, while I'm sure that you're applying the technique as you were taught (those steps are fairly common in first-year algebra courses), I prefer a slightly different process for completing the square.  The process, described below, is a bit more compatible with uses of completing the square that show up in later courses.  Using a quadratic expression (not the whole equation):&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$3x^2+9x+5$&lt;/li&gt;&#xA;&lt;li&gt;Factor out the leading coefficient from the variable terms: $3(x^2+3x)+5$&lt;/li&gt;&#xA;&lt;li&gt;Determine the desired perfect-square constant term: $(\frac{1}{2}\cdot 3)^2=\frac{9}{4}$&lt;/li&gt;&#xA;&lt;li&gt;Add and subtract the desired constant term inside the parentheses: $3(x^2+3x+\frac{9}{4}-\frac{9}{4})+5$&lt;/li&gt;&#xA;&lt;li&gt;Factor the perfect square and distribute the leading coefficient: $3(x+\frac{3}{2})^2-3\cdot\frac{9}{4}+5$&lt;/li&gt;&#xA;&lt;li&gt;Combine the constant terms: $3(x+\frac{3}{2})^2-\frac{7}{4}$&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="72" LastActivityDate="2011-02-09T19:45:38.547" CommentCount="1" />
  <row Id="21237" PostTypeId="2" ParentId="21235" CreationDate="2011-02-09T19:48:11.540" Score="4" ViewCount="" Body="&lt;p&gt;[note: this is wrong! see below]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $\cos 2u = 2\cos^2 u -1$ we have that&#xA;$$ \cos^4 u = \frac{\frac{\cos 4u +1}{2}+1}{2} = \frac{\cos 4u}{4} + \frac{3}{4}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This should get you started.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[edit]&#xA;Woops there's a big mistake in here! :-)&#xA;Let's try again and start from&#xA;$$\cos 2v = 2\cos^2 v - 1$$&#xA;therefore&#xA;$$ \cos^2 v = \frac{1 + \cos 2v}{2}$$&#xA;Now squaring on both sides&#xA;$$ \cos^4 v = \frac{ 1 + 2\cos 2v + \cos^2 2v }{4} $$&#xA;Now use $\cos^2 2v = (\cos 4v +1)/2$:&#xA;$$ \cos^4 v = \frac{ 1 + 2\cos 2v + \frac{1 + \cos 4v}{2}}{4}$$&#xA;So the correct formula is:&#xA;$$ \cos^4 v = \frac{3}{8} + \frac{1}{2}\cos(2u) + \frac{1}{8}\cos(4u)$$&#xA;Now all terms can be easily integrated.&lt;/p&gt;&#xA;" OwnerUserId="5189" LastEditorUserId="5189" LastEditorDisplayName="" LastEditDate="2011-02-10T12:04:03.707" LastActivityDate="2011-02-10T12:04:03.707" CommentCount="4" />
  <row Id="21238" PostTypeId="2" ParentId="20463" CreationDate="2011-02-09T19:58:16.373" Score="2" ViewCount="" Body="&lt;p&gt;As a &quot;neutral&quot; verb regarding increase or decrease, I propose &quot;adjust&quot; so that the method would be &lt;code&gt;adjustLengthBy&lt;/code&gt;.  This is concise but conforms to the MS Code Complete dictum to use strong verbs in function names.&lt;/p&gt;&#xA;" OwnerUserId="3111" LastActivityDate="2011-02-09T19:58:16.373" />
  <row Id="21239" PostTypeId="2" ParentId="21233" CreationDate="2011-02-09T20:00:03.693" Score="2" ViewCount="" Body="&lt;p&gt;This may help to understand this group:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Every element can be written as $a^{i_1} b^{j_1} a^{i_2} b^{j_2}\cdots$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now since $a^2$ and $b^2$ both commute with both $a$ and $b$ we can collect even powers on one side, for instance $a^7b^3ab = abab\cdot a^6b^2$. Now substituting $a^2$ by $b^4$ on the right, on obtains an expression of the form $ b^\epsilon (ab)^n a^{\epsilon^\prime} b^{2\ell}$, where $\epsilon,\epsilon^\prime \in \{0,1\}$ and $n \in \mathbb N$ and $\ell \in \mathbb Z$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now it should be easy to see which of these elements are in the kernel of the map.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[edit] I've given it some more thought, this might help too. I haven't double checked the details yet though so there may be a mistake somewhere. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The kernel consists of elements of this form&#xA;$$ (ab)^{2x} b^{-6x} , (ab)^{2x} a b^{-6x-2}, (ba)^{2x} b^{-6x}, (ba)^{2x+1}b^{-6x-3}$$&#xA;Note for instance that $(ab)^{2x} a b^{-6x-2} = (ab)^{2x+1} b^{-6x-3}$ so one can collect these into&#xA;$$ (ab)^n b^{-3n} , (ba)^n b^{-3n}$$&#xA;These are still related because&#xA;$$ (ba)^{-1} = a^{-1} b^{-1} = ab b^{-2}a^{-2} = ab b^{-2}b^{-4} = ab b^{-6},$$&#xA;therefore with $m = -n$ we have that&#xA;$$ (ba)^n b^{-3n} = (ba)^{-m} b^{3m} = (ab b^{-6})^m b^{3m} = (ab)^m b^{3m-6m} = (ab)^{-n} b^{3n}$$&#xA;therefore the kernel is the set of all &#xA;$$ (ab)^n b^{-3n},$$&#xA;where $n \in \mathbb Z$. Denote $(ab)^n b^{-3n}$ by $g_n$, then we can attempt to determine the group law as function of $n$ and we find that&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$g_u g_v = g_{u+v}$ if $u$ is even&lt;/li&gt;&#xA;&lt;li&gt;$g_u g_v = g_{u-v}$ if $u$ is odd&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This would imply the kernel is isomorphic to a group $(\mathbb Z,\circ)$ where $a\circ b = a + (-1)^a b$. Note that $2\mathbb Z$ is a normal subgroup and note that all odd numbers are involutions that act  nontrivial on $2\mathbb Z$ by sending $a$ to $-a$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think that would make the exact structure $C_\infty \rtimes C_2$, where $C_\infty \cong (\mathbb Z,+)$ is the subgroup $2\mathbb Z$ and $C_2$ is (for instance) the subgroup ${0,1}\cong C_2$ acting nontrivially on $C_\infty$ by sending every element to its inverse.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[edit 2] And now I notice the comment underneath the question that points to a solution making this a bit pointless...&lt;/p&gt;&#xA;" OwnerUserId="5189" LastEditorUserId="5189" LastEditorDisplayName="" LastEditDate="2011-02-10T21:30:31.717" LastActivityDate="2011-02-10T21:30:31.717" CommentCount="3" />
  <row Id="21240" PostTypeId="2" ParentId="21127" CreationDate="2011-02-09T20:06:30.673" Score="1" ViewCount="" Body="&lt;p&gt;Let $x$ be the amount of money the girl started with.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then we can derive this equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x-\frac{1}{2}x-(\frac{1}{2}x(\frac{1}{3}))-(\frac{1}{2}x(\frac{2}{3}(\frac{1}{4})))=7777$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Multiply out the fractions and factor out x, you get:&lt;br&gt;&#xA;$x(1-\frac{1}{2}-\frac{1}{6}-\frac{1}{12})=7777$    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So $x(\frac{12}{12}-\frac{6}{12}-\frac{2}{12}-\frac{1}{12})=7777$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus :&lt;br&gt;&#xA;$\frac{1}{4}x=7777$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Multiply both sides by $4$ to get:&lt;br&gt;&#xA;$x=31108$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Divide this by 12, for each month,&lt;br&gt;&#xA;$31108/12\approx2592.33$    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the girl gets an allowance of $2592.33$$/month  &lt;/p&gt;&#xA;" OwnerUserId="2533" LastEditorUserId="2533" LastEditorDisplayName="" LastEditDate="2011-02-09T22:04:58.780" LastActivityDate="2011-02-09T22:04:58.780" CommentCount="3" />
  <row Id="21241" PostTypeId="2" ParentId="5772" CreationDate="2011-02-09T20:13:28.650" Score="5" ViewCount="" Body="&lt;p&gt;Hi&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$A \rightarrow \lnot \lnot A$ is intuitionistically valid. Therefore you should be able to proof it already from MP plus Axiom 1 and Axiom 2 plus Ex Falso Quodlibet (An axiom that amounts to $\bot \rightarrow A$). No need to use Axiom 3.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But there is a twist, you need to represent $\lnot A$ as $A \rightarrow \bot$. Here you see a Hilbert style proof of $A \rightarrow ((A \rightarrow \bot) \rightarrow \bot)$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We need first a little lemma, namely that $A \rightarrow A$ is derivable:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1: $(A \rightarrow ((B \rightarrow A) \rightarrow A)) \rightarrow ((A \rightarrow (B \rightarrow A)) \rightarrow (A \rightarrow A))$ (Axiom 2)&lt;br&gt;&#xA;2: $A \rightarrow ((B \rightarrow A) \rightarrow A)$ (Axiom 1)&lt;br&gt;&#xA;3: $(A \rightarrow (B \rightarrow A)) \rightarrow (A \rightarrow A)$ (MP 1, 2)&lt;br&gt;&#xA;4: $A \rightarrow (B \rightarrow A)$ (Axiom 1)&lt;br&gt;&#xA;5: $A \rightarrow A$ (MP 3, 4)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we can proof what we desire:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1: $(A \rightarrow (((A \rightarrow \bot) \rightarrow A) \rightarrow ((A \rightarrow \bot) \rightarrow \bot))) \rightarrow ((A \rightarrow ((A \rightarrow \bot) \rightarrow A)) \rightarrow (A \rightarrow ((A \rightarrow \bot) \rightarrow \bot)))$ (Axiom 2)&lt;br&gt;&#xA;2: $(((A \rightarrow \bot) \rightarrow A) \rightarrow ((A \rightarrow \bot) \rightarrow \bot)) \rightarrow (A \rightarrow (((A \rightarrow \bot) \rightarrow A) \rightarrow ((A \rightarrow \bot) \rightarrow \bot)))$ (Axiom 1)&lt;br&gt;&#xA;3: $((A \rightarrow \bot) \rightarrow (A \rightarrow \bot)) \rightarrow (((A \rightarrow \bot) \rightarrow A) \rightarrow ((A \rightarrow \bot) \rightarrow \bot))$ (Axiom 2)&lt;br&gt;&#xA;4: $(A \rightarrow \bot) \rightarrow (A \rightarrow \bot)$ (Lemma)&lt;br&gt;&#xA;5: $((A \rightarrow \bot) \rightarrow A) \rightarrow ((A \rightarrow \bot) \rightarrow \bot)$ (MP 3, 4)&lt;br&gt;&#xA;6: $A \rightarrow (((A \rightarrow \bot) \rightarrow A) \rightarrow ((A \rightarrow \bot) \rightarrow \bot))$ (MP 2, 5)&lt;br&gt;&#xA;7: $(A \rightarrow ((A \rightarrow \bot) \rightarrow A)) \rightarrow (A \rightarrow ((A \rightarrow \bot) \rightarrow \bot))$ (MP 1, 6)&lt;br&gt;&#xA;8: $A \rightarrow ((A \rightarrow \bot) \rightarrow A)$ (Axiom 1)&lt;br&gt;&#xA;9: $A \rightarrow ((A \rightarrow \bot) \rightarrow \bot)$ (MP 7, 8)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The given proof shows something even stronger, $A \rightarrow \lnot \lnot A$ is not only intuitionistically valid, it is already valid in minimal logic, since we did not make use of Ex Falso Quodlibet. We were able to derive it from MP plus Axiom 1 and Axiom 2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Best Regards&lt;/p&gt;&#xA;" OwnerUserId="4414" LastEditorUserId="4414" LastEditorDisplayName="" LastEditDate="2011-02-09T20:40:41.433" LastActivityDate="2011-02-09T20:40:41.433" CommentCount="2" />
  <row Id="21242" PostTypeId="2" ParentId="21186" CreationDate="2011-02-09T20:14:59.787" Score="3" ViewCount="" Body="&lt;p&gt;In fact in some contexts it is fruitful to employ such &lt;em&gt;rational&lt;/em&gt; factorizations. For example,  see &lt;a href=&quot;http://math.stackexchange.com/questions/8058/how-to-solve-this/8072#8072&quot;&gt;this question&lt;/a&gt; whose solution involves rewriting symmetric polynomials as polynomials in $\rm\ x + 1/x\ $ e.g.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\rm a\ x^4 + b\ x^3+ 2\:a\ x^2 + b\ x + a\ \ =\ \ \bigg(a\ \bigg(x+\frac{1}x\bigg)^2 + b\ \bigg(x+\frac{1}x\bigg)+c\bigg)\ x^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Generally contexts enjoying some sort of innate rational structure or symmetry may similarly profit from factorizations or compositions involving rational function. You will discover some pretty examples of such if you study Galois theory.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2011-02-10T02:50:09.317" LastActivityDate="2011-02-10T02:50:09.317" />
  <row Id="21243" PostTypeId="1" AcceptedAnswerId="21246" CreationDate="2011-02-09T20:19:29.403" Score="3" ViewCount="95" Body="&lt;p&gt;Hello everyone.&#xA;I know that this question is some sort of bridge between Informatics and Mathematics, not knowing the best place where to post this question, I opted for this place because of the type of answer I want (which concerns Math more than anything).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider to have a graph represented by a collection of nodes and connections. The best way to describe the graph is through the adjacency matrix (AM): a matrix that has 0 or 1 if one node is connected to another (we consider non-directed graphs, so we have all bidirectional connections).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anyone know whether the eigenvalues of the matrix implies something for the graph??? properties, topology....&#xA;I ask this question because I've almost finished studying Markov Chains. In a chain, the matrix of transition probabilities P (for discrete time markov chains), or the matrix of transition frequencies Q (for cont-time markov chains), can be evaluated (their eigenvalues) in order to inspect whether the chain is ergodic or not (with a parallelism to Control Theory: eigenvalues in the unitary circle or in the negative half-plane).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am trying to find something similar for graphs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you &lt;/p&gt;&#xA;" OwnerUserId="4175" LastActivityDate="2011-02-22T05:45:14.780" Title="Knowledge about Graph Spectral Theory and correlation between a graph Weighted Adjacency matrix and its eigenvalues" Tags="&lt;matrices&gt;&lt;graph-theory&gt;&lt;eigenvalues&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" />
  <row Id="21244" PostTypeId="2" ParentId="21148" CreationDate="2011-02-09T20:21:45.130" Score="3" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\rm\ \ \ 2^K\ \equiv\: -1\ \ \Rightarrow\ \ (2^K)^{\:2N+1}\ \equiv\: -1\ \ \ (mod\ \ 2^K+1)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice how converting the problem from &lt;em&gt;relational&lt;/em&gt; form (divisibility) to &lt;em&gt;functional&lt;/em&gt; form (equality or congruence) has the effect of greatly simplifying the problem, reducing it to the obvious fact that $-1$ to an odd power is $-1\:$. This is one of the great powers of congruence arithmetic - converting obfuscated divisibility problems into intuitive arithmetical problems. It is essential to master this reduction in order to succeed in elementary number theory.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2011-02-09T20:36:27.863" LastActivityDate="2011-02-09T20:36:27.863" CommentCount="3" />
  <row Id="21245" PostTypeId="2" ParentId="21207" CreationDate="2011-02-09T20:40:23.850" Score="4" ViewCount="" Body="&lt;p&gt;Evidently, $\alpha=0$ results in a convergent integral.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose $\alpha&amp;gt;0$.&#xA;Clearly, $\frac{1-x^{\alpha}}{1-x}$ is continuous on [0,1).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, $\frac{1-x^{\alpha}}{1-x}$ is bounded on $[0,1)$ since $\displaystyle \lim_{x\to 1}\frac{1-x^{\alpha}}{1-x}=\lim_{x\to 1}\;\alpha x^{\alpha-1}=\alpha$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It follows that the integral converges $\forall\alpha\geq0$.&lt;/p&gt;&#xA;" OwnerUserId="5959" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2011-04-20T12:45:58.197" LastActivityDate="2011-04-20T12:45:58.197" CommentCount="3" />
  <row Id="21246" PostTypeId="2" ParentId="21243" CreationDate="2011-02-09T20:44:11.830" Score="3" ViewCount="" Body="&lt;p&gt;There are a &lt;strong&gt;lot&lt;/strong&gt; of properties on eigenvalues of adjacency matrix.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the diameter of $G$ is d, then $A$ has at least $d+1$ distinct eigenvalues.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, if a graph is $d$-regular, then its largest eigenvalue is bound by d. He is connected if and only the multiplicity of d is 1. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A complete bipartite graph $K_{m,n}$ has three eigenvalues : 0, $\lambda$, $-\lambda$ where $\lambda = \sqrt{mn}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For more, you should check a book on algebraic graph theory like Algebraic Graph Theory, Godsile and Royle, Springer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note : The AM is not the &quot;best&quot; way to describe a graph. It's one way, very interesting in case you're looking to compute paths.&lt;/p&gt;&#xA;" OwnerUserId="6582" LastActivityDate="2011-02-09T20:44:11.830" CommentCount="1" />
  <row Id="21247" PostTypeId="2" ParentId="11012" CreationDate="2011-02-09T20:45:52.683" Score="1" ViewCount="" Body="&lt;p&gt;The conditions &quot;locally free&quot;, &quot;finitely generated projective&quot; and &quot;dualizable&quot; are equivalent, and the latter one can be formulated in terms of tensor products (namely of the &quot;unit&quot; $A \to M \otimes M^*$ and the &quot;counit&quot; $M^* \otimes M \to A$ satisfying the two triangular identities).&lt;/p&gt;&#xA;" OwnerUserId="1650" LastActivityDate="2011-02-09T20:45:52.683" />
  <row Id="21248" PostTypeId="2" ParentId="21128" CreationDate="2011-02-09T21:02:58.917" Score="4" ViewCount="" Body="&lt;p&gt;Echoing a bit of user1728 and Sean's comments, Category Theory is a wonderful unifying language that ties together a lot of ideas and makes certain things much easier, but it is pretty rough going to learn in the abstract. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was lucky enough that the professor that taught Abstract Algebra at the National University in Mexico when I took it did a lot of the proofs as if they were category theory, but without actually &lt;em&gt;saying&lt;/em&gt; &quot;Category Theory&quot;. So he proved that the product of groups has the universal property of a product, and the uniqueness up to unique isomorphism, and so on, with diagrams; did the same thing with rings. Etc. By the end of the course, he was mentioning that all of these ideas were special cases of a general theory called &quot;Category Theory&quot;. And so on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the time I got to an actual course &lt;em&gt;in&lt;/em&gt; Category Theory, I had a whole library of mental examples to draw upon when looking at all the different concepts, amplified with some of the less algebraic-flavored examples (such as considering a partially ordered set as a category, etc) that the professor for that course gave. With that in hand, the first couple of chapters of Mac Lane's book became easier to digest and understand, and use elsewhere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, this may slant my view; I tend to view Category Theory more as a useful unifying language than as a particular subject (in which I am at least somewhat wrong, if not more). But I suspect you'll be able to get into, and get a lot more out of, Category Theory if you have the library of examples on hand. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, as I said, I was lucky: I was primed for Category Theory with examples that were essentially Category Theory without saying so. You may not benefit from that. Still, I think that waiting until you study some abstract algebra and see some of these constructions in action might be a good idea.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2011-02-09T21:02:58.917" />
  <row Id="21249" PostTypeId="2" ParentId="21211" CreationDate="2011-02-09T21:06:19.897" Score="0" ViewCount="" Body="&lt;p&gt;The only nontrivial simplification employed in the derivation you give is factoring $\rm\ 2x\:(x^2+1)\ $ out of both terms in the numerators of the second term. You can avoid this by doing the derivative generically first, i.e. instead of using the formula for the derivative of $\rm\ f/g\ $ instead work out the formula for $\rm\ f/g^n\ $, and do the simplification in this general form, before specializing $\rm\:f,g\:.$ Namely&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\rm\displaystyle \bigg(\frac{f}{g^n}\bigg)&amp;#39;\ =\ \frac{f\ &amp;#39;}{g^n} - \ \frac{n\ f\ g&amp;#39;}{g^{n+1}}\ =\ \frac{f\:&amp;#39;g-n\ f\ g&amp;#39;}{g^{n+1}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now if you specialize $\rm\ n = 2,\ f = -1-x^2,\ g = -1 + x^2\ $ you'll find the arithmetic is a bit simpler. While this &quot;generic preprocessing&quot; seems a bit trivial here, it can provide immense simplifications in other contexts,$\ $ e.g. $\ $ see my &lt;a href=&quot;http://math.stackexchange.com/questions/17831/sylvesters-determinant-identity/17837#17837&quot;&gt;proof here&lt;/a&gt; of $\ $ &lt;strong&gt;Sylvester's identity&lt;/strong&gt; $\rm\ \ det(1+AB) = det(1+BA)\ $ that proceeds by taking $\rm\ det\ $ of $\rm\ (1+A\ B)\ A\ =\ A\ (1+B\ A)\ $ then generically cancelling $\rm\ det(A)\:\ \ $ - a cancellation that is much less trivial then the cancellation of $\rm g^n $ that occurs when you do the derivative non-generically above. Indeed, non-generic proofs of Sylvester's identity must resort to less elementary non-algebraic methods to effect the cancellation, e.g. density arguments. Moral of the story: a little bit of generic algebra can go a long way towards avoiding dense arguments.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2011-02-09T21:12:55.197" LastActivityDate="2011-02-09T21:12:55.197" CommentCount="4" />
  <row Id="21250" PostTypeId="1" CreationDate="2011-02-09T21:12:41.097" Score="5" ViewCount="110" Body="&lt;p&gt;Hi, Everyone:&#xA;  Please forgive that I do not yet know LaTex, bro, and my English ( I am from UCV in Venezuela).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think I understand  concept of bundles almost well, and that, once a vector bundle with a fiber is known/given, that we can define a new fiber pointwise, in manipulating each of the fibers, e.g., we may change the fiber from being R (over itself) to being R(+)R, or from R to R(x)R ; RxR*, (dual)etc.&#xA;    What I  not too clear on, is on how one put together all the new fibers coherently into a bundle, i.e., how one construct new trivializations and transition functions to turn the space with altered fiber into a new bundle. I am particularly interest in the quotient bundle, if someone  knows. I imagine we use initial charts, trivialization to construct the altered bundle, but I don't see fully how, other than I pretty sure we use multilinear algebra and functoriality somehow.   Would be great if someone knew about how to do this for general fiber bundles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks You from Caracas.&lt;/p&gt;&#xA;" OwnerUserId="6845" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2011-02-10T01:59:17.913" LastActivityDate="2011-02-10T03:08:39.660" Title="Pasting Together Fibers of a Vector Bundle" Tags="&lt;linear-algebra&gt;&lt;topology&gt;&lt;vector-bundles&gt;" AnswerCount="1" FavoriteCount="2" />
  <row Id="21251" PostTypeId="2" ParentId="21234" CreationDate="2011-02-09T21:13:10.757" Score="2" ViewCount="" Body="&lt;p&gt;She is considering &lt;em&gt;a&lt;/em&gt; complex representation &lt;em&gt;V&lt;/em&gt; which comes with an action of the group elements on &lt;em&gt;V&lt;/em&gt;. &lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2011-02-09T21:13:10.757" />
  <row Id="21252" PostTypeId="2" ParentId="12833" CreationDate="2011-02-09T21:24:43.430" Score="4" ViewCount="" Body="&lt;p&gt;There can be locally-compact metrizable spaces with non-first-countable Alexandroff compactifications but with other first-countable compactifications.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This follows from a theorem of Banakh and Leiderman in &quot;Uniform Eberlein compactifications of metrizable spaces&quot; which states that a metrizable space $X$ has a first-countable uniform Eberlein compactification if and only if $|X|\leq\mathfrak{c}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This implies that the discrete space on $\omega_1$ (which is trivially locally-compact and metrizable) has a first-countable compactification.  But the Alexandroff compactification is not first-countable, since the open subsets of the &quot;point at infinity&quot; consist of all co-finite subsets of $\omega_1$ and any (sub-)basis must be uncountable.&lt;/p&gt;&#xA;" OwnerUserId="3852" LastActivityDate="2011-02-09T21:24:43.430" />
  <row Id="21253" PostTypeId="1" AcceptedAnswerId="21255" CreationDate="2011-02-09T21:33:34.400" Score="1" ViewCount="46" Body="&lt;p&gt;If $X$ has the density function &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ f_\vartheta (x) = \Big \{&amp;#160;\begin{array}{cc}  &#xD;&#xA;         (\vartheta - 1)x^{-\vartheta} &amp;amp; x \geq 1\\&#xD;&#xA;         0 &amp;amp; otherwise&#xD;&#xA;         \end{array}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I see that $\log X \sim Exp(\vartheta - 1)$? &#xA;I had the idea to look at $f(\log x)$ but I think that's not right.&#xA;Many thanks for your help!&lt;/p&gt;&#xA;" OwnerUserId="5798" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2011-02-09T21:50:14.057" LastActivityDate="2011-02-09T21:50:14.057" Title="Distribution of $\log X$" Tags="&lt;probability&gt;" AnswerCount="1" FavoriteCount="0" />
  <row Id="21254" PostTypeId="2" ParentId="21250" CreationDate="2011-02-09T21:47:50.337" Score="6" ViewCount="" Body="&lt;p&gt;Suppose $T$ is a functor from the category of finite dimensional (real, say) vector spaces to itself. We can say that $T$ is &lt;em&gt;continuous&lt;/em&gt; if for every finite dimensional vector space $V$ the associated map $T:\hom(V,V)\to\hom(TV,TV)$ is continuous, when we view its domain and codomain as real vector spaces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, the functor $T=\Lambda^2(\mathord-)$ which computes exterior squares is continuous, as many others.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now suppose $E$ is a locally trivial vector bundle of dimension $n$ over a space $B$, and suppose $\mathcal U$ is an open covering of $B$ over whose open sets $E$ is trivial. Then for each pair $U$, $V\in\mathcal U$ such that $U\cap V\neq\emptyset$ we have a corresponding transition function $g_{U,V}:U\cap V\to\mathrm{GL}(\mathbb R,n)$, as explained for example in the Wikipedia page for vector bundles. Moreover, we can reconstruct $E$ up to isomorphism from the knowledge of $\mathcal U$ and the family $\{g_{U,V}:U,V\in\mathcal U,U\cap V\neq\emptyset\}$ alone.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, suppose $T(\mathbb R^n)$ has dimension $m$. Then the vector bundle $T(E)$ can be defined to be the vector bundle which one can construct from the covering $\mathcal U$ and the family $\{\tilde g_{U,V}:U,V\in\mathcal U,U\cap V\neq\emptyset\}$ of transition functions, where for each $U$, $V\in\mathcal U$ such that $U\cap V\neq\emptyset$, the map $$\tilde g:p\in U\cap V\mapsto T(g_{U,V}(p))\in GL(m).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This construction gives a meaning to $T(E)$ for all continuous functors $T$, and it can be generalized for continuous functors of many variables, both covariant and contravariant. If I recall correctly, this is discussed in Milnor and Stasheff's book on &lt;em&gt;Characteristic Classes&lt;/em&gt;, for example.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;NB:&lt;/strong&gt; there is another way of doing this...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $E$ is a vector bundle of dimension $n$ over a space $B$, then one can attach to it a &lt;em&gt;principal $\mathrm{GL(n)}$-bundle&lt;/em&gt; $P_E$, called the &lt;a href=&quot;http://en.wikipedia.org/wiki/Frame_bundle&quot; rel=&quot;nofollow&quot;&gt;frame bundle&lt;/a&gt;. The fiber of $P_E$ over a point $b\in B$ is the set of all ordered bases of the fiber of $E$ over $b$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, given any $\mathrm{GL}(n)$-representation $V$, we can form the bundle $P_E\times_{\mathrm{GL}(n)}V$. If we take $V=V_{\mathrm{taut}}=\mathbb R^n$ with the tautological action of $\mathrm{GL}(n)$, then  $P_E\times_{\mathrm{GL}(n)}V$ is isomorphic to $E$; if we take $V=\Lambda^3(V_{\mathrm{taut}})$, then &#xA;$P_E\times_{\mathrm{GL}(n)}V$ is the bundle $\Lambda^3(E)$, and so on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&amp;nbsp;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both constructions do the same thing: they pick some gadget which precisely captures the way the fibers of the vector bundle you start with are put together (the set of transition functions, the associated principal bundle) and then pick some other fiber and glue copies together using the same prescription. The two approaches are equivalent, of course. The second one is somewhat more «geometric» because the gadget used to record the way the fibers of $E$ are put together is in fact a geometric object and not some vaporous family of transition functions, also known, if you want to scare the kids, as a $1$-cocycle.&lt;/p&gt;&#xA;" OwnerUserId="274" LastEditorUserId="274" LastEditorDisplayName="" LastEditDate="2011-02-10T03:08:39.660" LastActivityDate="2011-02-10T03:08:39.660" CommentCount="2" />
  <row Id="21255" PostTypeId="2" ParentId="21253" CreationDate="2011-02-09T21:47:56.837" Score="3" ViewCount="" Body="&lt;p&gt;One way is to use the cdf technique.  Let $Y = \log X$.  Then &#xA;$$P(Y \leq y) = P(\log X \leq y) = P(X \leq e^y) = \int_1^{e^y} (\vartheta - 1)x^{-\vartheta} dx = \left.-x^{-\vartheta+1}\right|_1^{e^y} = 1-e^{-(\vartheta-1)y}.$$&#xA;Differentiating with respect to $y$ yields $(\vartheta-1) e^{-(\vartheta-1)y}$ as the pdf of $Y$, which is also the pdf of an $Exp(\vartheta - 1)$ random variable.&lt;/p&gt;&#xA;" OwnerUserId="2370" LastActivityDate="2011-02-09T21:47:56.837" CommentCount="3" />
  <row Id="21256" PostTypeId="1" AcceptedAnswerId="21261" CreationDate="2011-02-09T21:49:03.597" Score="2" ViewCount="120" Body="&lt;p&gt;Suppose A is the set composed of all ordered pairs of positive integers. Let R be the relation defined on A where (a,b) R (c,d) means that a + d = b + c.&#xA;(a) Prove that R is an equivalence relation. &#xA;Here is what I have so far. Is this correct?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reflexive:&#xA;a ~ a &#xA;=&gt; a+b=a+b; (a,b) R (c,d)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Symmetric: &#xA;if a ~ b then b ~ a&#xA;=&gt; if a+d=b+c then c+b=d+a &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Transitive: &#xA;if a ~ b and b ~ c then a ~ c&#xA;=&gt; if a+d=b+c and c+f=d+e then a+d=d+e&lt;/p&gt;&#xA;&#xA;&lt;p&gt;b) Find [(1,1)] &#xA;I'm not sure how to approach this. &lt;/p&gt;&#xA;" OwnerUserId="6331" LastActivityDate="2011-02-09T22:59:41.953" Title="Equivalence Relations" Tags="&lt;discrete-mathematics&gt;" AnswerCount="3" CommentCount="4" />
  <row Id="21257" PostTypeId="1" AcceptedAnswerId="21356" CreationDate="2011-02-09T21:58:19.620" Score="4" ViewCount="218" Body="&lt;p&gt;The function space $H^{\alpha} (\Omega)$ for $0 &amp;lt; \alpha \le 1$,  is the set of functions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\{ f \in C^0(\Omega) : \sup_{x \neq y} \dfrac{|f(x) - f(y)|}{|x-y|^{\alpha}} &amp;lt; \infty \}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with the metric $d_{H^{\alpha}} = || f - g ||_{H^{\alpha}}$, where &#xA;$$||f||_{H^{\alpha}} = ||f||_{sup} + [f]_{H^{\alpha}} \text{ ,        } [f]_{H^{\alpha}} = \sup_{x \neq y}  \dfrac{|f(x) - f(y)|}{|x-y|^{\alpha}} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if $0 &amp;lt; \alpha &amp;lt; \beta \le 1$, then &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$[f]_{H^{\alpha}} \le 2 ||f||_{sup}^{1-\frac{\alpha}{\beta}}  [f]_{H^{\beta}}^{\frac{\alpha}{\beta}} \space \forall f \in H^{\beta}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And also, there is some constant $M$ so that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$||f||_{H^{\alpha}} \le M ||f||_{sup}^{1-\frac{\alpha}{\beta}}  ||f||_{H^{\beta}}^{\frac{\alpha}{\beta}} \space \forall f \in H^{\beta}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These were some questions on a problem set: I have checked that $d_{H^{\alpha}}$ is a metric, and proved the two properties (in the second I found that $M = 2$ is sufficient). However, rather blindly. It's easy to show from the first that if $0 &amp;lt; \alpha &amp;lt; \beta \le 1$, then $H^{\beta} \subset H^{\alpha}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What else do these formulas mean? Are they just some useful inequalities, or do they establish some connection between $H^{\beta}$ and $H^{\alpha}$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="5105" LastEditorUserId="5105" LastEditorDisplayName="" LastEditDate="2011-02-10T02:44:55.633" LastActivityDate="2011-02-10T12:56:42.113" Title="On Some Properties of Hölder Continuous Functions" Tags="&lt;real-analysis&gt;&lt;functions&gt;&lt;functional-analysis&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="21258" PostTypeId="2" ParentId="21256" CreationDate="2011-02-09T22:00:20.367" Score="0" ViewCount="" Body="&lt;p&gt;HINT: The equivalence class of $(1,1)$ is made up of all pairs $(x,y)\sim(1,1)$. Write explicitely what the latter means and get a relation that need to be satisfied by $x$ and $y$.&lt;/p&gt;&#xA;" OwnerUserId="688" LastActivityDate="2011-02-09T22:00:20.367" />
  <row Id="21259" PostTypeId="2" ParentId="21146" CreationDate="2011-02-09T22:00:42.160" Score="7" ViewCount="" Body="&lt;p&gt;This is a special case of a class of cyclotomic factorizations due to &lt;a href=&quot;http://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.114.548&quot; rel=&quot;nofollow&quot;&gt;Aurifeuille, Le Lasseur and Lucas,&lt;/a&gt; the so-called &lt;a href=&quot;http://www.ams.org/journals/mcom/2006-75-253/S0025-5718-05-01766-7/S0025-5718-05-01766-7.pdf&quot; rel=&quot;nofollow&quot;&gt;&lt;strong&gt;Aurifeuillian factorizations&lt;/strong&gt;&lt;/a&gt; of cyclotomic polynomials $\rm\;\Phi_n(x) = C_n(x)^2 - n\ x\  D_n(x)^2\;$. These play a role in factoring numbers of the form $\rm\; b^n \pm 1\:$, cf. the &lt;a href=&quot;http://homes.cerias.purdue.edu/~ssw/cun/&quot; rel=&quot;nofollow&quot;&gt;Cunningham Project&lt;/a&gt;. Below are some simple examples of such factorizations:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{array}{rl}&#xD;&#xA;x^4 + 2^2 \quad=&amp;amp;  (x^2 + 2x + 2)\;(x^2 - 2x + 2) \\\\&#xD;&#xA;\frac{x^6 + 3^3}{x^2 + 3} \quad=&amp;amp;  (x^2 + 3x + 3)\;(x^2 - 3x + 3) \\\\&#xD;&#xA;\frac{x^{10} - 5^5}{x^2 - 5} \quad=&amp;amp;  (x^4 + 5x^3 + 15x^2 + 25x + 25)\;(x^4 - 5x^3 + 15x^2 - 25x + 25) \\\\&#xD;&#xA;\frac{x^{12} + 6^6}{x^4 + 36} \quad=&amp;amp;  (x^4 + 6x^3 + 18x^2 + 36x + 36)\;(x^4 - 6x^3 + 18x^2 - 36x + 36) \\\\&#xD;&#xA;\end{array}$$&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2011-02-12T17:34:57.790" LastActivityDate="2011-02-12T17:34:57.790" CommentCount="2" />
  <row Id="21260" PostTypeId="1" AcceptedAnswerId="21267" CreationDate="2011-02-09T22:00:47.980" Score="0" ViewCount="117" Body="&lt;p&gt;I'm trying to solve&#xA;$$\lim\limits_{x \to \infty}\frac{\ln^{1000} x}{x^5}$$&#xA;Here's what I get:&#xA;$$e^{\lim\limits_{x \to \infty}\ln{\frac{\ln^{1000}x}{x^5} }}$$&#xA;Dropping the $e$ for ease,&#xA;$$\lim\limits_{x \to \infty} 1000\ln{(\ln{(x)})} - 5 \ln{x} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I have $\infty - \infty$.. I know there must be a next step, but I don't know what it would be.&lt;/p&gt;&#xA;" OwnerUserId="1069" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-09T22:30:37.843" LastActivityDate="2011-02-09T23:10:38.607" Title="$\lim\limits_{x \to \infty}\frac{\ln^{1000} x}{x^5}$" Tags="&lt;limit&gt;" AnswerCount="3" CommentCount="4" />
  <row Id="21261" PostTypeId="2" ParentId="21256" CreationDate="2011-02-09T22:01:17.090" Score="1" ViewCount="" Body="&lt;p&gt;Your proof regarding the fact that $R$ is in fact an equivalence relation is indeed correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now think what the equivalence class of $(1,1)$ could be, since $(1,1) R (1,1)$ you know that everyone in the equivalence class has the property that $(a,b)$ holds $a+1=b+1$, therefore $a=b$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means that the equivalence class of $(1,1)$ is $(a,a)$ for $a\in\mathbb{N}$.&lt;/p&gt;&#xA;" OwnerUserId="622" LastActivityDate="2011-02-09T22:01:17.090" />
  <row Id="21262" PostTypeId="1" AcceptedAnswerId="21265" CreationDate="2011-02-09T22:01:34.427" Score="1" ViewCount="78" Body="&lt;p&gt;I would like so find the function $n(z)$ that solves the following equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;n(z) = \frac{.2 + .24 z \int_1^{\infty} \frac{n(z)}{z^2}dz}{.5 + .24 z \int_1^{\infty} \frac{n(z)}{z^2}dz}&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is, $n(z)$ is defined in terms of it's own integral.  Even if there's no nice analytic solution to this, Numerical approximations of some sort are good enough.  In fact this is just a parametrized instance of a more general equation I am interested in.  What methods should I be using to find such a solution?  Does Mathematica or Matlab has nice built in routines for this kind of thing?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For your further information I am interested only in $n(z)$ on $[1,\infty]$, and it is required to be between 0 and 1, so those integrals lie between 0 and 1, so $n(z)$ should be monotonically increasing.  It seems like it should be well behaved, but on the face of it Mathematica doesn't like it.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Background: it is the steady state of a dynamic system in a network; think epidemic diffusion.  $n(z)$ is the infection rate among agents with z links, and in the steady state it is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;n(z)  = (g + (1-g)t v z )/(g + r + (1-g) t v z ),&#xD;&#xA;$$&#xA;where $g$ is the new infection rate, $r$ is the cure rate,and $v$ is the new infections from your peers in a networks, and $f(z)$ is the distribution of that networks; the proportion of peers withg $z$ links.    One such distribution often of interest is power-law, which is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;f(z) = 2 z^{-3}&#xD;&#xA;$$&#xA;and t is the infection rate among a random peer, given by&#xA;$$&#xD;&#xA;t = \int_1^{\infty} z n(z) f(z)dz,&#xD;&#xA;$$&#xA;which, along with a paramtetrization of the above rates, gave the problem above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other distributions of interest are geometric, where $f(z) = Log(4) 2^{-z}$.  So we are solving for the steady state level of infection among guys with z peers. &lt;/p&gt;&#xA;" OwnerUserId="2464" LastEditorUserId="2464" LastEditorDisplayName="" LastEditDate="2011-02-09T23:43:56.057" LastActivityDate="2011-02-09T23:43:56.057" Title="Trying to solve for an unknown function, numerically or otherwise" Tags="&lt;calculus&gt;&lt;numerical-methods&gt;" AnswerCount="1" />
  <row Id="21263" PostTypeId="1" AcceptedAnswerId="21271" CreationDate="2011-02-09T22:06:06.897" Score="2" ViewCount="141" Body="&lt;p&gt;Given a sequence $a_n$, I know that if I can find a divergent subsequence of $a_n$, or two subsequences of $a_n$ that converge to different values, $a_n$ diverges, since, if I have understood correctly, a sequence $a_n$ converges to a limit $L$ if and only if every subsequence of $a_n$ converges to that value $L$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've been wondering if this last condition was equivalent to showing that some subsequences converge to $L$, picking the subsequences such that every element of the original sequence is in at least one of the subsequences. Is it? I would guess that the terms &quot;partition&quot; or &quot;covering&quot; fit this description.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="6427" LastEditorUserId="6427" LastEditorDisplayName="" LastEditDate="2011-02-10T02:05:21.820" LastActivityDate="2011-02-10T02:05:21.820" Title="Sufficient conditions on subsequences for convergence of a sequence" Tags="&lt;calculus&gt;&lt;homework&gt;&lt;sequences-and-series&gt;" AnswerCount="2" CommentCount="5" />
  <row Id="21265" PostTypeId="2" ParentId="21262" CreationDate="2011-02-09T22:18:26.083" Score="6" ViewCount="" Body="&lt;p&gt;Well you know how $n(z)$ looks like. Note that $\displaystyle \int_{1}^{\infty} \frac{n(z)}{z^2} dz$ is just a number. Call it $a$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So now we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$n(z) = \frac{0.2+0.24 a z}{0.5+0.24 a z} = 1 - \frac{0.3}{0.5+0.24 a z}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $a = \displaystyle \int_{1}^{\infty} \frac{n(z)}{z^2} dz$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Use the above equation to get an equation solely in $a$ which you can solve for numerically.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have done the solving part below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a = \displaystyle \int_{1}^{\infty} \frac{n(z)}{z^2} dz = \int_{1}^{\infty} \frac{1}{z^2} dz - \int_{1}^{\infty} \frac{0.3}{0.5 z^2 + 0.24 a z^3} dz = 1 - \int_{1}^{\infty} \frac{15}{25 z^2 + 12 a z^3} dz$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$a = \frac{2}{5} + \frac{36}{125}a \log(\frac{25}{12a}+1)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can now try to solve this numerically or plug it in mathematica as I did, which gives me $a \approx 0.673338$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you expected, $a \in (0,1)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the function now is $$n(z) \approx 1 - \frac{15}{25+8.080056z} = \frac{10+8.080056z}{25+8.080056z}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and the function $n(z) \in (0,1)$, $\forall z \in \mathbb{R}^+$. In fact when $z \in [1,\infty)$, $n(z) \in (\frac{6}{11},1)$.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2011-02-09T22:58:07.843" LastActivityDate="2011-02-09T22:58:07.843" CommentCount="4" />
  <row Id="21266" PostTypeId="2" ParentId="21263" CreationDate="2011-02-09T22:21:50.530" Score="1" ViewCount="" Body="&lt;p&gt;Consider the sequence $(x_n)_n$ such that $x_n=0$ if $n$ is even and $x_n=1$ if $n$ is odd.  Let $m$ be a positive integer, and consider the subsequence $(x_{n_k})_k$ such that $x_{n_k}=x_k$ if $k\leq m$, and $x_{n_k}=x_{2k}$ if $k\gt m$.  Now consider the collection of such subsequences as $m$ varies over the positive integers.  They all are eventually constantly $0$, and they &quot;cover&quot; the original sequence.  An analogous covering construction could be given for any sequence with a convergent subsequence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Related to your remarks before the question, it is often useful to use the criterion that a sequence converges to $L$ if (and only if) every subsequence has a further subsequence that converges to $L$.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2011-02-09T22:21:50.530" />
  <row Id="21267" PostTypeId="2" ParentId="21260" CreationDate="2011-02-09T22:33:35.860" Score="2" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\ $ Changing variables $\rm\ Z = ln\ X\ $ yields $\displaystyle\rm\ \lim_{\ Z\ \to\ \infty}\ \frac{Z^{1000}}{e^{5\:Z}}\ $ which is easily handled either by power series, L'Hopital or related techniques.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2011-02-09T23:10:38.607" LastActivityDate="2011-02-09T23:10:38.607" CommentCount="3" />
  <row Id="21269" PostTypeId="2" ParentId="21260" CreationDate="2011-02-09T22:37:58.150" Score="3" ViewCount="" Body="&lt;p&gt;Taking $\ln^{1000}x=(\ln x)^{1000}$ you can apply L'Hopital's rule 999 times to reduce to &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;\lim_{x\to\infty}\frac{1000!\ln x}{5^{1000}x^5}.&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then one more application gives&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;\lim_{x\to \infty}\frac{1000!}{5^{1001}x^5}=0.&#xD;&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="2838" LastActivityDate="2011-02-09T22:37:58.150" CommentCount="2" />
  <row Id="21271" PostTypeId="2" ParentId="21263" CreationDate="2011-02-09T22:40:37.897" Score="2" ViewCount="" Body="&lt;p&gt;If you can partition a sequence into finitely many subsequences, each of which converges to $L$, then the original sequence must converge to $L$ as well.  This is clear from the following definition of the limit: $a_n \rightarrow L$ iff for all $\epsilon&amp;gt;0$ there exists $N_\epsilon$ such that $|a_n - L| &amp;lt; \epsilon$ whenever $n \ge N_\epsilon$.  But then for any $\epsilon &amp;gt; 0$, because the $i$-th subsequence converges to $L$, it is within $\epsilon$ of the limit for $n \ge N^{(i)}_\epsilon$; so the sequence itself is within $\epsilon$ of the limit for $n \ge N_\epsilon = \max_{i}N^{(i)}_\epsilon$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the result does not hold for a partition into infinitely many subsequences.  For instance, consider the case where $a_n=1$ when $n$ is prime and $a_n=0$ otherwise.  This can be partitioned into infinitely many subsequences, where $a_n$ is in the $i$-th subsequence if the smallest prime factor of $n$ is the $i$-th prime.  (Just put $a_1$ anywhere.)  Each subsequence converges (immediately) to zero, but the original sequence does not converge, because it has sporadic $1$'s as far out as you care to look. &lt;/p&gt;&#xA;" OwnerUserId="5546" LastActivityDate="2011-02-09T22:40:37.897" CommentCount="1" />
  <row Id="21272" PostTypeId="2" ParentId="21260" CreationDate="2011-02-09T22:40:57.847" Score="3" ViewCount="" Body="&lt;p&gt;Be careful using limit operation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, let show that $\lim_{x \rightarrow +\infty} \dfrac{\ln x}{x} = 0$. For $t \geq 1$, we have $t \geq \sqrt{t}$ which imply for $x \geq 1$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ 0 \leq \ln x = \int_1^x \dfrac{dt}{t} \leq \int_1^x \dfrac{dt}{\sqrt{t}} = 2 \sqrt{x} - 2 \leq 2 \sqrt{x} $$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, for any $a,b &amp;gt; 0$ and $x &amp;gt; 1$, we have&#xA;$$ \dfrac{\ln^b x}{x^a} = \left( \dfrac{\ln x}{x^{\frac{a}{b}}} \right)^b = \left( \dfrac{b}{a} \right)^b \left( \dfrac{\ln (x^\frac{a}{b})}{x^{\frac{a}{b}}} \right)^b $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which answer your question.&lt;/p&gt;&#xA;" OwnerUserId="6582" LastActivityDate="2011-02-09T22:40:57.847" CommentCount="1" />
  <row Id="21273" PostTypeId="2" ParentId="20658" CreationDate="2011-02-09T22:43:25.757" Score="3" ViewCount="" Body="&lt;p&gt;Let  $r$ denote the number of cards that, after shuffling $n$ cards, remain unsorted (the middle portion of the deck, that is to be suffled again). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $t_n$ be the needed number or shuffles starting from $n$ cards till all cards are sorted. We want $E(t_n)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, lets consider that expectation conditioned on the value of $r_n$, i.e. suppose we know exactly how many cards remain unsorted &lt;strong&gt;after the first try&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then we can write: $E[t_n | r_n=r] = 1 + E[t_r]$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(note that this includes the extreme cases $r_n=0$ and $r_n = n$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Applying the property $E[E[x|y]] = E[x]$ we have:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle E[t_n] = \sum_{r=0}^n (1 + E[t_r]) \; p(n,r) $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $p(n,r) = Prob(r_n = r)$ (we'll compute it later)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We must take out the term $r=n$ for the sum, and by trivial manipulation we get the desired recursion:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle&#xD;&#xA;E[t_n] = \frac{1}{1-p(n,n)} \left( p(n,n) + \sum_{r=0}^{n-1} (1 + E[t_r]) \; p(n,r) \right) $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As initial values we have  $E[t_0] = E[t_1] = 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To compute $p(n,r)$ (probability that after one shuffling of $n$ cards there remains -with the rules of our game- exactly $r$ cards to shuffle again) it's more simple (but more tedious), just combinatorial counting - i'll explain if you ask to.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My result is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle  p(n,r) = \frac{(n + 1 -r) \; (r^2 - 3r + 3) \; (r-2)!}{n!}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;if $r&amp;gt;1$. Elsewhere, $p(n,0) = 1/n!$ and $p(n,1)=0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are some values (in javascript) &lt;a href=&quot;http://hjg.com.ar/varios/mat/cards.html&quot; rel=&quot;nofollow&quot;&gt;http://hjg.com.ar/varios/mat/cards.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;BTW: This problem can be seen as in between two problems, one simpler and one more complex/general:&#xA;As compared to the Coupon Collector problem, this is more complicated, because we can't divide the road into &quot;states&quot; that will be visited sequentially - and hence we can simply express the desired expectation as a sum of expectations of those pieces.&#xA;It can be consisdered as a Markov Model with one absorbing state; but the markovian formulation, though apt, is too general, and the desired result (expected numbers of iterations to reach the absorbing state) is less straigforward that the solution proposed here. That's because our problem has some special restrictions (in terms of a Markov chain, we'd say that its transition matrix is triangular)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ADDED: To compute $p(n,r)$ , for $r \ge 2$ :&#xA;After  shuffling $n$ cards, we get $r$ unsorted cards, say $(n1;r;n2)$ Lets count how many ways (with $n1$ $n2$ fixed) there are. They correspond to choose a permutation of the $r$ middle cards, with the condition that both extremes are &quot;wrong&quot;. This is given by:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$r! - 2(r - 1)! + (r-2)! = (r^2 - 3 r + 3) (r-2)!$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(This is exclusion-inclusion counting: the first term counts all possibilities, the second count the cases where the first OR the last is right, the third count the cases where the first AND the last are right).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, for a given $r$ there are several choices of $(n1;n2)$, they are only required to sum up $n-r$. Actually, there are $n - r +1$ alternatives. Then we get the total number of permutations that give $r$ unsorted cards, and divide over all the permutations:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle  p(n,r) = \frac{(n + 1 -r) \; (r^2 - 3r + 3) \; (r-2)!}{n!}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;UPDATED: An asymptotic can be guessed (theoretically and empirically): &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle E(t_n) \approx \frac{n(n-1)}{4} \approx \frac{n^2}{4}$&lt;/p&gt;&#xA;" OwnerUserId="312" LastEditorUserId="312" LastEditorDisplayName="" LastEditDate="2011-02-10T13:02:00.787" LastActivityDate="2011-02-10T13:02:00.787" CommentCount="0" />
  <row Id="21274" PostTypeId="1" CreationDate="2011-02-09T22:54:42.723" Score="4" ViewCount="122" Body="&lt;p&gt;Suppose a statement form &quot;phi&quot; always has value T or U. Show &quot;phi&quot; is a classical tautology.&lt;/p&gt;&#xA;" OwnerUserId="6848" LastActivityDate="2011-02-10T19:01:40.983" Title="Lukasiewicz Logic Tautology" Tags="&lt;fuzzy-logic&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="21275" PostTypeId="2" ParentId="21256" CreationDate="2011-02-09T22:59:41.953" Score="0" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\ $ This is geometrically  obvious once you note that the equivalence class containing $\rm\ (a,b)\ $ is simply the slope $1$ line through it (with positive integer coordinates). In effect you are discovering the negative integers as those lines having no positive $x$-intercept (= normal-form), i.e. whose intercept is &quot;negative&quot;. This &quot;group of differences&quot; construction is the additive analog of the field of fractions construction.&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2011-02-09T22:59:41.953" />
  <row Id="21276" PostTypeId="1" AcceptedAnswerId="21327" CreationDate="2011-02-09T23:19:56.170" Score="1" ViewCount="160" Body="&lt;p&gt;Total differential for a function of two variables is known as &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$df = f_x(x_0, y_0)dx + f_y(x_0,y_0)dy$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's not clear how to derive this. Thomas' Calculus 11th Ed pg 1021 says to use the definition of linearization L(x,y). The definition of linearization for a function f(x) at page 223 is only for one variable&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f(x) \approx L(x) = f(x_0) + f&amp;#39;(x_0)(x-a)$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;or&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$L(x) = f(x_0) + f&amp;#39;(x_0)∆x$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I had the idea to rewrite the formula $L(x)$ for $L(x,y)$, so I had&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$L(x,y) = f(x_0,y) + f_x(x_0,y)∆x$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then I thought I could maybe linearize $f(x_0,y)$ to get &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$L(x,y) = L(x_0,y_0) + f_y(x_0,y_0)∆y + f_x(x_0,y_0)∆x$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$∆L = L(x,y) - L(x_0,y_0) = f_y(x_0,y_0)∆y + f_x(x_0,y_0)∆x$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would this be right? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also found this page from Google&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.physicsforums.com/showthread.php?p=3109168&quot; rel=&quot;nofollow&quot;&gt;http://www.physicsforums.com/showthread.php?p=3109168&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="6786" LastActivityDate="2011-02-10T07:12:07.873" Title="Total differential for a function with two variables" Tags="&lt;calculus&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="21277" PostTypeId="2" ParentId="18776" CreationDate="2011-02-09T23:26:18.880" Score="0" ViewCount="" Body="&lt;p&gt;Try a change of variable to get the density of a standard multivariate normal multiplied by something else. It should make the problem clearer.&lt;/p&gt;&#xA;" OwnerUserId="6849" LastActivityDate="2011-02-09T23:26:18.880" />
  <row Id="21278" PostTypeId="2" ParentId="21137" CreationDate="2011-02-09T23:28:07.327" Score="0" ViewCount="" Body="&lt;p&gt;Sivaram has answered your title problem: you simply plot the points $(0,0)$, $(b,-a)$ and $(-b,a)$ or any two of them.  Then you draw the straight line through these points, extending it if necessary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's extend it in two ways to your optimisation problem. First, how to draw $ax+by=c$ where $c$ is not $0$.  Plot the points $(\frac{c}{a},0)$ and $(0,\frac{c}{b})$; this also works if one of $a$ or $b$ is $0$.  Then you draw the straight line through these points, extending it if necessary.  That, together with Sivaram's advice, is enough to draw your five constraints. This will give you a feasible region which is a simple convex polygon, in this case with five vertices.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second extension is to draw a line parallel to $ax+by=0$ which passes through the point $(d,e)$.  The answer is to plot the points $(d,e)$, $(d+b,e-a)$ and $(d-b,e+a)$ or any two of them.  Then you draw the straight line through these points, extending it if necessary. You can do this with your function to be maximised and each of the five points.  One or more of the five resulting lines will be further to the top-right than the others.  The point(s) in your feasible region this passes though will give where $z$ is maximised.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If after all this you do not find that the maximum possible value of $z$ is $62$ then either you or I have made a mistake. &lt;/p&gt;&#xA;" OwnerUserId="6460" LastActivityDate="2011-02-09T23:28:07.327" CommentCount="1" />
  <row Id="21279" PostTypeId="1" AcceptedAnswerId="21281" CreationDate="2011-02-09T23:30:40.030" Score="0" ViewCount="77" Body="&lt;p&gt;&lt;strong&gt;Definition of closure from &lt;a href=&quot;http://www.proofwiki.org/wiki/Definition%3aBoundary&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Boundary (Topology): The points in the&#xA;  closure of a set which are not in the&#xA;  interior of that set.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Definition of Exterior from &lt;a href=&quot;http://www.proofwiki.org/wiki/Definition%3aExterior_%28Topology%29&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Let $T$ be a topological space. Let $H \subseteq T$. The exterior of $H$ is the complement of the closure of $H$ in $T$. Alternatively, the exterior of $H$ is the interior of the complement of $H$ in $T$. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5902" LastActivityDate="2011-02-09T23:44:13.860" Title="When are the topological Exterior and Boundary the same thing?" Tags="&lt;topology&gt;&lt;exterior&gt;&lt;closure&gt;" AnswerCount="1" />
  <row Id="21280" PostTypeId="2" ParentId="21199" CreationDate="2011-02-09T23:42:32.197" Score="2" ViewCount="" Body="&lt;p&gt;It is best to think of &quot;d/dx&quot; as an &lt;strong&gt;operator&lt;/strong&gt; which takes the derivative, with respect to x, of whatever expression follows.&lt;/p&gt;&#xA;" OwnerUserId="2191" LastActivityDate="2011-02-09T23:42:32.197" />
  <row Id="21281" PostTypeId="2" ParentId="21279" CreationDate="2011-02-09T23:44:13.860" Score="0" ViewCount="" Body="&lt;p&gt;In general, if $T$ is a topological space, then any set $H$ in $T$ partitions $T$ into three disjoint sets: the interior of $H$, the boundary of $H$, and the exterior of $H$. So the exterior only equals the boundary when both are empty, that is when all of $T$ is in the interior of $H$, i.e. when $H$=$T$.&lt;/p&gt;&#xA;" OwnerUserId="5203" LastActivityDate="2011-02-09T23:44:13.860" CommentCount="7" />
  <row Id="21282" PostTypeId="1" CreationDate="2011-02-09T23:52:09.707" Score="1" ViewCount="150" Body="&lt;p&gt;I need to show that every positive integer $n$ can be written uniquely in the  form $n = ab$, where $a$ is square-free and $b$ is a square.  Then I need to show that $b$ is then the largest square dividing $n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem I have here is that I can't even see how this is true.  How can $1$ be represented this way?  Is $1$ a square number?  If not, then I cannot see how this is possible.  If it is, then I cannot see how $1$, $2$, or $3$ can be represented this way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please, any help you can offer giving me a push in the right direction here is greatly appreciated.&lt;/p&gt;&#xA;" OwnerUserId="6548" LastEditorUserId="825" LastEditorDisplayName="" LastEditDate="2011-02-10T01:11:35.443" LastActivityDate="2011-02-10T01:11:35.443" Title="Show that every positive integer n can be written uniquely in the form n = ab, where a is square-free and b is a square." Tags="&lt;number-theory&gt;&lt;prime-numbers&gt;&lt;algebraic-number-theory&gt;&lt;number&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="21283" PostTypeId="2" ParentId="21282" CreationDate="2011-02-09T23:57:18.207" Score="1" ViewCount="" Body="&lt;p&gt;$1=1^2$, so $1$ is square. There are two natural ways to prove this result. One is using the prime factorization, i.e., the fundamental theorem of arithmetic. The other is by induction: if $n$ does not have a square factor, then $n$ is squarefree and $n$=$n\cdot 1$. Otherwise, divide $n$ by a square factor and use the induction hypothesis on the quotient.&lt;/p&gt;&#xA;" OwnerUserId="589" LastActivityDate="2011-02-09T23:57:18.207" />
  <row Id="21284" PostTypeId="1" AcceptedAnswerId="21286" CreationDate="2011-02-09T23:57:44.453" Score="0" ViewCount="95" Body="&lt;p&gt;The wikipedia example of &lt;a href=&quot;http://en.wikipedia.org/wiki/Polynomial_long_division&quot; rel=&quot;nofollow&quot;&gt;Polynomial Long Division&lt;/a&gt; starts with:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Divide the first term of the numerator&#xA;  by the highest term of the denominator&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;if the denominator is $x-3$, and we don't know $x$, how to know which is highest? $x$ or $3$?&lt;/p&gt;&#xA;" OwnerUserId="4501" LastActivityDate="2011-04-10T07:22:25.333" Title="Polynomial Long Division explanation" Tags="&lt;algebra&gt;&lt;polynomials&gt;&lt;division-algebra&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="21285" PostTypeId="2" ParentId="21203" CreationDate="2011-02-09T23:59:42.733" Score="2" ViewCount="" Body="&lt;p&gt;One of your questions was &quot;Is a torus a doughnut?&quot; The 2-dimensional &lt;em&gt;surface&lt;/em&gt; that we know as a torus is the 2-dimensional surface of a standard doughnut or bagel. It is also the 2-dimensional surface of a coffee cup up to homeomorphism (read topological equivalence). This is why we say that a topologist cannot tell the difference between a doughnut and a coffee cup.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If a piece of rope (or better extension cord) is tied loosely into a knot and the ends joined together (or plugged into one another, respectively) then the boundary of the resulting figure is represented as torus. Three dimensional compact spaces that have no incompressible tori are important because of the geometrization conjecture. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To paraphrase Homer (Simpson, not the poet), &quot;MMM. Tori...&quot;&lt;/p&gt;&#xA;" OwnerUserId="722" LastActivityDate="2011-02-09T23:59:42.733" />
  <row Id="21286" PostTypeId="2" ParentId="21284" CreationDate="2011-02-09T23:59:51.907" Score="3" ViewCount="" Body="&lt;p&gt;&quot;highest&quot; here means not &quot;greatest&quot;, but the one with the highest power of $x$ (which in this case is $x$).&lt;/p&gt;&#xA;" OwnerUserId="6622" LastActivityDate="2011-02-09T23:59:51.907" CommentCount="1" />
  <row Id="21287" PostTypeId="2" ParentId="21282" CreationDate="2011-02-10T00:02:00.370" Score="1" ViewCount="" Body="&lt;p&gt;As others have mentioned, the number 1 is both squarefree and a square. When $n=1$, we have $a=b=1$. When $n$ is a prime, we have $a=n$ and $b=1$. This is perfectly valid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me expand on the first method mentioned by lhf. In general, let &#xA;$$N=\prod_{\text{primes }p_i}p_i^{e_i}$$&#xA;be the prime decomposition of a number $N$. Note that $N$ is squarefree if and only if each $e_i$ is either 0 or 1, and $N$ is a square if and only if every $e_i$ is even. So if we want to write $n=ab$ where $a$ is squarefree and $b$ is a square, then if their prime factorizations are as follows,&#xA;$$n=\prod_{\text{primes }p_i}p_i^{x_i}\hskip0.3in a=\prod_{\text{primes }p_i}p_i^{y_i}\hskip0.3in b=\prod_{\text{primes }p_i}p_i^{z_i}$$&#xA;then we want to solve&#xA;$$n=\prod_{\text{primes }p_i}p_i^{x_i}=\prod_{\text{primes }p_i}p_i^{y_i+z_i}=ab$$&#xA;such that each $y_i$ is either 0 or 1, and each $z_i$ is even. Can you see why this uniquely specifies all the $y_i$'s and $z_i$'s, and hence uniquely specifies $a$ and $b$?&lt;/p&gt;&#xA;" OwnerUserId="264" LastActivityDate="2011-02-10T00:02:00.370" CommentCount="3" />
  <row Id="21288" PostTypeId="2" ParentId="21282" CreationDate="2011-02-10T00:15:28.563" Score="2" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\ $ The problem is &lt;em&gt;multiplicative&lt;/em&gt; so it suffices to show that it's true for a prime power $\rm\ P^N\:.\ $ But that's trivial: $\rm\ P^{2N} =\ (P^N)^2,\ \ P^{\:2N+1} =\ P\ (P^N)^2\:,\ $ uniquely.&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2011-02-10T00:15:28.563" CommentCount="6" />
  <row Id="21289" PostTypeId="1" AcceptedAnswerId="21290" CreationDate="2011-02-10T00:15:39.833" Score="3" ViewCount="112" Body="&lt;p&gt;&lt;a href=&quot;http://www.flickr.com/photos/31602616@N05/5432388226/&quot; rel=&quot;nofollow&quot; title=&quot;wedge by stephanoc, on Flickr&quot;&gt;&lt;img src=&quot;http://farm6.static.flickr.com/5095/5432388226_286765f4df.jpg&quot; width=&quot;463&quot; height=&quot;229&quot; alt=&quot;wedge&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I've just got my cheese from the cheesemonger and he's cut it along the axis $x$,$y$, and $z$ so I can do my math homework with it.  The lengths of my cheese are:&lt;br&gt;&#xA;$x$=6&lt;br&gt;&#xA;$y$=4&lt;br&gt;&#xA;$z$=2  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm asked to find the volume using calculus, because using geometry would be cheating, evidently.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The three ways I am asked to &quot;cut up&quot; the cheese are &lt;strong&gt;perpendicular&lt;/strong&gt; to the $y$-axis (triangles), the $x$-axis (rectangles), and $z$-axis (also rectangles).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is easy, and it turns out the volume is 24 units cubed (all three times).  E.g:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4$\int_0^6 \frac{x}{3} \mathrm{d}x = 24$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, I ask myself what if I want to take the &lt;strong&gt;other&lt;/strong&gt; rectangle, and use that as my slice.  I know that width is 4 ($y$-axis), and the length is z$\sqrt{10}$ (pythagoras with $\sqrt{z^2 + (3z)^2}$).  But how do I find my limits of integration?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$4\sqrt{10}\int_a^b z \mathrm{d}z.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would need to know the distance from the $y$-axis to the top (in this photo) of the wedge.  However, I've exhausted my math knowledge here. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I asked my calc teacher, and he told me that I find this number by &quot;taking more calculus classes&quot; (a good joke actually).  Since I don't have time to go bug him in his office hours, can someone explain some of the different ways to find this distance?&lt;/p&gt;&#xA;" OwnerUserId="5248" LastEditorUserId="5248" LastEditorDisplayName="" LastEditDate="2011-02-10T01:06:35.000" LastActivityDate="2011-02-10T02:32:35.457" Title="Calc question regarding the volume of my cheese." Tags="&lt;calculus&gt;" AnswerCount="1" CommentCount="6" />
  <row Id="21290" PostTypeId="2" ParentId="21289" CreationDate="2011-02-10T00:23:32.947" Score="2" ViewCount="" Body="&lt;p&gt;If we were to look at the cheese with the $Y$ axis going into the board, we will be seeing a right triangle in the $X-Z$ plane right angled at the origin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sides of the right triangle are given by $x=0$, $z=0$ and $\frac{x}{6} + \frac{z}{2} = 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A cut parallel to the &quot;top&quot; side of the cheese will then be the equation of the form $\frac{x}{6} + \frac{z}{2} = c$ in the $X-Z$ plane, where $c \in (0,1)$ which is parallel to the line $\frac{x}{6} + \frac{z}{2} = 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/D6B9f.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Staying in the $X-Z$ plane, the perpendicular distance, say $l$, of the line $\frac{x}{6} + \frac{z}{2} = c$ from the origin can be found as follows. We want a line passing through the origin and perpendicular to $\frac{x}{6} + \frac{z}{2} = c$. So we just need to take the negative inverse of the slope of the above line and we get $z = 3x$. The intersection of these two lines is the point where the perpendicular from the origin meets the line. The point of intersection is $(\frac{3c}{5},\frac{9c}{5})$. So $l = \sqrt{(\frac{3c}{5})^2 + (\frac{9c}{5})^2} = \frac{3c \sqrt{10}}{5}$. There are also other ways to find $l$ like equating the area of the triangle calculated by two different ways. i.e. $(6c) \times (2c) = l \times (2 \sqrt{10} c) \Rightarrow l = \frac{3c \sqrt{10}}{5}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the volume element becomes $dV = 4 \times \text{Length of the line segment} \{\frac{x}{6} + \frac{z}{2} = c\} \text{bounded by the axes} \times dl$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that $4$ comes from the span along the $y$ direction which is constant for all slices.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$dV = 4 \times 2 \sqrt{10} c \times \text{ }dl$ and $dl = \frac{3 \sqrt{10}}{5} dc$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$dV = 8 \sqrt{10} \frac{3 \sqrt{10}}{5} c dc = 48 c dc$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, the volume is given by $V = \displaystyle \int_{0}^{1} 48 c dc = 24$.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2011-02-10T02:32:35.457" LastActivityDate="2011-02-10T02:32:35.457" CommentCount="8" />
  <row Id="21291" PostTypeId="2" ParentId="21257" CreationDate="2011-02-10T00:32:39.087" Score="0" ViewCount="" Body="&lt;p&gt;I wish I could comment...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, I don't think the second inequality holds at all, for instance all constant functions violate it, and even taking these out will not solve the problem; for example: Assume the inequality holds, take $f\in H^\beta$, $f\geq 0$ and $c&amp;gt;0$, then since the seminorm doesn't see constants we get $(\| f\| _{sup} +c)^{\alpha /\beta} = \| f+c\| _{sup}^{\alpha / \beta} \leq M[f]_{\beta } ^{\alpha / \beta }$, and this is for all $c&amp;gt;0$.  &lt;/p&gt;&#xA;" OwnerUserId="4829" LastActivityDate="2011-02-10T00:32:39.087" CommentCount="4" />
  <row Id="21292" PostTypeId="1" AcceptedAnswerId="21347" CreationDate="2011-02-10T00:43:30.237" Score="1" ViewCount="73" Body="&lt;p&gt;I am trying to prove that morphisms of finite type are stable under base change, but I am having some trouble moving from the case where everything is affine to the general case.  Suppose $f:X \rightarrow Y$ is a morphism of finite type and $Y&amp;#39;$ is a $Y$-scheme.  I want to show that the morphism $g: X \times_Y Y&amp;#39; \rightarrow Y&amp;#39;$ is of finite type.  In the case that $X$, $Y$, and $Y&amp;#39;$ are affine, I understand why this is true.  For the general case, by a lemma in Liu's book, it is enough to show that there is an affine open cover $\{V_i\}_i$ of $Y&amp;#39;$ such that for each $i$, $g^{-1}(V_i)$ is a finite union of affine open subsets $U_{ij}$ such that for each $i$ and $j$, $O_X(U_{ij})$ is a finitely generated algebra over $O_Y(V_i)$. Here is my attempt at proving this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Choose an affine open cover $\{V_i\}_i$ of $Y&amp;#39;$.  Is it true that $g^{-1}(V_i)=X \times_Y V_i$?  I think this should follow from how we constructed the fibered product by gluing.  Since $f:X \rightarrow Y$ is of finite type, we may choose an affine open cover $\{Y_j\}$ of $Y$ such that $f^{-1}(Y_j)$ is covered by a finite number of affine opens $W_{jk}$.  Now, $W_{jk} \times_Y V_i$ are open subschemes that cover $X \times_Y V_i$, but since $Y$ is not necessarily affine, these schemes are not necessarily affine, right?  Furthermore, if we are using the $W_{jk}$ to cover all of $X$, there could be infinitely many of them.  To make the $W_{jk} \times_Y V_i$ affine, we could further cover them with $W_{jk} \times_{Y_k} V_i$, but we are not guaranteed finitely many $Y_k$ either, so while these schemes will be affine, there will not necessarily be finitely many.  I have been having some trouble with these sorts of arguments where one can immediately reduce to the affine case, and some help here would be greatly appreciated.&lt;/p&gt;&#xA;" OwnerUserId="5265" LastActivityDate="2011-02-10T11:01:03.413" Title="Morphisms of finite type are stable under base change" Tags="&lt;algebraic-geometry&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" />
  <row Id="21293" PostTypeId="2" ParentId="21120" CreationDate="2011-02-10T00:58:29.393" Score="1" ViewCount="" Body="&lt;p&gt;You should get that the answer is $1/2$, immediately from integration by parts. See &lt;a href=&quot;http://en.wikipedia.org/wiki/Lebesgue%E2%80%93Stieltjes_integration&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt;, in particular the sections &quot;Integration by parts&quot; and &quot;Related concepts&quot;. Further, see &lt;a href=&quot;http://www.math.mcgill.ca/labute/courses/255w03/L3.pdf&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; for a proof of the integration by parts formula.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Explicitly, since $f$ is continuous and non-decreasing, and is constant on $\mathbb{R}-[0,1]$, $\int_{\mathbb R} {f(x)du(x)}  = \int_0^1 {f(x)df(x)} $, and it holds&#xA;$$&#xD;&#xA;\int_0^1 {f(x)df(x)}  = f(1)f(1) - f(0)f(0) - \int_0^1 {f(x)df(x)} &#xD;&#xA;$$&#xA;(integration by parts). Since $f(0)=0$ and $f(1)=1$, it thus follows that $\int_{\mathbb R} {f(x)du(x)} = 1/2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;: More generally, if $F$ is any continuous distribution function (the Cantor function $f$ is a particular example), then $\int_{\mathbb R} {F(x)dF(x)}  = 1/2$. As before, this can be proved using integration by parts, which is allowed since $F$ is continuous and non-decreasing. (Under certain conditions, this also follows from a change of variable.) Indeed, for any $a &amp;lt; b$,&#xA;$$&#xD;&#xA;\int_a^b {F(x)dF(x)}  = F(b)F(b) - F(a)F(a) - \int_a^b {F(x)dF(x)}.&#xD;&#xA;$$&#xA;Hence the result follows by letting $a \to -\infty$ and $b \to \infty$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another way to obtain the general result is as follows. Let $X$ be an arbitrary random variable with continuous distribution function $F$. Then, $\int_{\mathbb R} {F(x)dF(x)}$ expresses the expectation of the random variable $F(X)$. As is well known, in this case $F(X) \sim {\rm uniform}(0,1)$. Hence,  $\int_{\mathbb R} {F(x)dF(x)}  = 1/2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remark. With $f$ as above, integration by parts gives&#xA;$$&#xD;&#xA;\int_0^1 {xdf(x)}  = xf(x)\big|_0^1  - \int_0^1 {f(x)dx} = 1 - 1/2 = 1/2.&#xD;&#xA;$$&#xA;The left-hand side, $\int_0^1 {xdf(x)}$, expresses the expectation of the Cantor distribution. &lt;/p&gt;&#xA;" OwnerUserId="2810" LastEditorUserId="2810" LastEditorDisplayName="" LastEditDate="2011-02-27T16:18:37.067" LastActivityDate="2011-02-27T16:18:37.067" CommentCount="0" />
  <row Id="21296" PostTypeId="1" CreationDate="2011-02-10T01:17:03.447" Score="1" ViewCount="51" Body="&lt;p&gt;I'm working on a homework assignment in which part of the question statement says that each of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Classical_group&quot; rel=&quot;nofollow&quot;&gt;classical Lie algebras&lt;/a&gt; can be described as the set of all matrices $X \in gl(n,\mathbb{C})$ satisfying&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$XM + MX^T = 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for some particular choice of $M$. Specifically, it says that if $M$ is the $n\times n$ unit matrix $\mathbb{I}_n$, then $X \in A_{(n-1)} = su(n)$. But that doesn't seem to make sense. For $M = \mathbb{I}_n$, if my math is correct, the equation above indicates that $X$ should be limited to $n\times n$ antisymmetric matrices, which is not the same set as the $su(n)$ Lie algebra. So it seems that there is a mistake in the question statement. Is that correct?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a secondary question, it'd be nice to get confirmation for the given definitions of the other classical algebras:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Setting $M = \begin{pmatrix}1 &amp;amp; 0 &amp;amp; 0 \\ 0 &amp;amp; 0 &amp;amp; \mathbb{I}_n \\ 0 &amp;amp; \mathbb{I}_n &amp;amp; 0\end{pmatrix}$ implies $X \in B_n = so(2n+1)$&lt;/li&gt;&#xA;&lt;li&gt;Setting $M = \begin{pmatrix}0 &amp;amp; -\mathbb{I}_n \\ \mathbb{I}_n &amp;amp; 0\end{pmatrix}$ implies $X \in C_n = sp(n)$&lt;/li&gt;&#xA;&lt;li&gt;Setting $M = \begin{pmatrix}\mathbb{I}_n &amp;amp; 0 \\ 0 &amp;amp; \mathbb{I}_n\end{pmatrix}$ implies $X \in D_n = so(2n)$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I didn't tag this &lt;code&gt;[homework]&lt;/code&gt; because the question I'm asking is not the question I'm supposed to be solving in the assignment. I'm just looking for clarification on the instructions. (Also, I'm asking here because the instructor isn't available to contact right now)&lt;/p&gt;&#xA;" OwnerUserId="1190" LastActivityDate="2011-02-10T01:17:03.447" Title="Do the classical Lie algebras all satisfy $XM + MX^T = 0$?" Tags="&lt;group-theory&gt;&lt;lie-algebra&gt;" CommentCount="3" />
  <row Id="21297" PostTypeId="2" ParentId="4414" CreationDate="2011-02-10T01:29:28.293" Score="1" ViewCount="" Body="&lt;p&gt;Hi&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I assume we are talking here about arbitrary deductive systems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem here is that non-primitive logical connectors cannot necessarely be introduced by a definition statement of the following form. Let $\psi$ be the non-primitive logical connector, then one cannot necesseraley find a formula $\phi$, such that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\psi(x_1,..,x_n) \leftrightarrow \phi(x_1,..,x_n)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem begins with the question whether we have $\leftrightarrow$ available or not.&#xA;But what can be done is the following. Assume we have at least the consequence relation $\vdash$ available. Then in the general case a new connective $\psi$ will be a set $S$ of axioms AND/OR inference rules. Let $\vdash_S$ denote your deductive system extended by this set $S$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we can ask two important questions. 1) Does the new connective exist, i.e. is $\vdash_S$ consistent. 2) Is the new connective uniquely determined, i.e. do we have the following where S' denotes a copy of the new axioms and/or inference rules:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\psi(x_1,..,x_n) \vdash_{S+S&amp;#39;} \psi&amp;#39;(x_1,..,x_n)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\psi&amp;#39;(x_1,..,x_n) \vdash_{S+S&amp;#39;} \psi(x_1,..,x_n)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lets show an example. Assume we have a deductive system with only the following few rules. The square brakets indicate that a premisse is optional:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\quad \over A \vdash A$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Gamma, [A] \vdash B \over \Gamma \vdash A \rightarrow B$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Gamma \vdash A \quad \quad \quad \Delta \vdash A \rightarrow B &#xD;&#xA;\over \Gamma, \Delta \vdash B$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Via cut elimination the above deductive system can be shown logically consistent. Now I introduce a new connective $\circ$ via the following inference rules.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Gamma, [A], [B] \vdash C \over \Gamma, A \circ B \vdash C$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Gamma \vdash A \quad \quad \quad \Delta \vdash B &#xD;&#xA;\over \Gamma, \Delta \vdash A \circ B$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The new deductive system is still logically consistent, the cut elimination can also be carried out for the new connective $\circ$. And the new connective is unique, since we can show:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$${\over A \vdash A} \quad {\over B \vdash B}&#xD;&#xA;\over {A, B \vdash A \circ B&#xD;&#xA;\over A \circ&amp;#39; B \vdash A \circ B}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I give everybody a bounty who can bring the inference rules for $\circ$ into the canonical form of a definition. What is possible is to replace the two inference rules by two axioms. Namely the following two axioms will do as well instead of the two inference rules:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$A \rightarrow (B \rightarrow A \circ B)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(A \rightarrow (B \rightarrow C)) \rightarrow (A \circ B \rightarrow C)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Still I don't believe this can be brought into the canonical form of a definition, since the definition would need $A \leftrightarrow B$, which is usually $(A \rightarrow B) \circ (B \rightarrow A)$. But we can exactly not do this, since we are in the process of defining $\circ$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bottom line is: Yes axioms and/or inference rules do, but the definition schema not in general. And inference rules can be most likely replaced by axioms, provided for example your basic deductive system has at least modus ponens and the deduction rule.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Best Regards&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S.: That the definition via axioms and/or inference rules is proper you have to do by yourself, i.e. show existence and uniqueness. When you would use the definition schema this would come for free most often.  &lt;/p&gt;&#xA;" OwnerUserId="4414" LastEditorUserId="4414" LastEditorDisplayName="" LastEditDate="2011-02-10T01:43:20.323" LastActivityDate="2011-02-10T01:43:20.323" />
  <row Id="21298" PostTypeId="2" ParentId="21276" CreationDate="2011-02-10T02:06:48.697" Score="1" ViewCount="" Body="&lt;p&gt;Sorry, the &quot;comments&quot; section is disabled, so I will post in here. My apologies&#xA;   if this is against the protocol.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yes; the total differential is what you just described: the equation of the plane&#xA;   ( as a 2-d linear object) tangent to your function at a point;&#xA;   this tangent plane is the linearization of f, or, in a precise del-eps. sense,&#xA;   the best linear approximation to f, in a small 'hood (neighborhood) of a point.&lt;/p&gt;&#xA;" OwnerUserId="6850" LastActivityDate="2011-02-10T02:06:48.697" />
  <row Id="21299" PostTypeId="1" AcceptedAnswerId="21300" CreationDate="2011-02-10T02:31:19.850" Score="1" ViewCount="59" Body="&lt;p&gt;Say our question requires us to use traces to sketch the quadric surface with equation $x^2+y^2/9+z^2/4=1$. This is a worked example in our book, but I'm confused by their steps. So first we set $z=k$ to get $x^2+y^2/9=1-k^2/4$. First of all, is $k$ just a constant? And what is the difference between setting $z=k$ rather than $z=0$? Also, if $k$ is a constant (which I'm not sure if it is or not, or what it even represents for that matter), then how come when graphing using Wolfram, if I put in the equation with $k$, I get an ellipsoid, but if I just substitute say $0.5$ for $k^2/4$, I get a circle? And so then we repeat the same steps above with $x$ and $y$, and the same questions apply. &lt;/p&gt;&#xA;" OwnerUserId="2985" LastActivityDate="2011-02-10T02:51:17.223" Title="Question on Quadric Surfaces" Tags="&lt;calculus&gt;" AnswerCount="1" />
  <row Id="21300" PostTypeId="2" ParentId="21299" CreationDate="2011-02-10T02:51:17.223" Score="2" ViewCount="" Body="&lt;p&gt;Yes, $k$ is a constant. The way you should think of this is that $z=k$ is the plane parallel to the $xy$-coordinate plane at &quot;height&quot; $k$. So $k=0$ corresponds to exactly the $xy$-plane, and as $k$ increases you are getting different planes higher above that one and $k&amp;lt;0$ gives you planes &quot;below&quot; the $xy$-coordinate plane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Substituting in different numbers gives you a &quot;slice&quot; of the 3D picture, so it is just the curve that you get at that height. This is why when you substitute in a value for $k$ you get just a curve. When you typed it into Wolfram with the $k$, it interpreted $k$ as a variable instead of a constant so it gave you the full 3D ellipsoid. When you just look at a plane section of the ellipsoid you get ellipses (or circles).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The purpose of this method is that if you know what a bunch of the 2D slices of the 3D thing look like (which is much easier since you just keep getting curves that should be familiar), then you can piece them together to sketch the 3D thing.&lt;/p&gt;&#xA;" OwnerUserId="2000" LastActivityDate="2011-02-10T02:51:17.223" CommentCount="2" />
  <row Id="21301" PostTypeId="1" AcceptedAnswerId="21302" CreationDate="2011-02-10T02:55:01.897" Score="1" ViewCount="121" Body="&lt;p&gt;I have two random variables $X$ and $Y$. Both are distributed according to $N(0,1)$. If their covariance is 0, are they independent?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that this is not true for other distributions, say the Wikipedia example: $X$ chosen uniformly in $[-1,1]$ and $Y=X^2$.&lt;/p&gt;&#xA;" OwnerUserId="915" LastActivityDate="2011-02-10T02:57:35.480" Title="Does zero covariance imply independence of random variables here?" Tags="&lt;probability-theory&gt;" AnswerCount="1" />
  <row Id="21302" PostTypeId="2" ParentId="21301" CreationDate="2011-02-10T02:57:35.480" Score="4" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Normally_distributed_and_uncorrelated_does_not_imply_independent&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Normally_distributed_and_uncorrelated_does_not_imply_independent&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="822" LastActivityDate="2011-02-10T02:57:35.480" CommentCount="1" />
  <row Id="21304" PostTypeId="1" AcceptedAnswerId="21346" CreationDate="2011-02-10T03:16:35.003" Score="7" ViewCount="165" Body="&lt;p&gt;Suppose $M$ is a n-dimensional (compact) manifold and $f$ is a differentiable function with exactly three (non-degenerate) critical points. Then one can show, using Morse theory, that $M$ is homeomorphic to a $\frac{n}{2}$ sphere with an $n$ cell attached.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand why we will have critical points with index $0$ and $n$ (since $M$ is compact and it achieves its max and min). My question is exactly why there one of index $\frac{n}{2}$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By Poincare duality, we have that $H^{k}(M) \cong H_{n-k}(M)$.  So if $k$ is not $\frac{n}{2}$, then $n-k \neq k$ so we will have two additional nonzero (co)homology classes.  Does this somehow correspond to having two additional critical points, contradicting that there are only 3?&lt;/p&gt;&#xA;" OwnerUserId="6851" LastActivityDate="2011-02-20T05:32:19.350" Title="Manifold with 3 nondegenerate critical points" Tags="&lt;algebraic-topology&gt;&lt;differential-topology&gt;&lt;cohomology&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" />
  <row Id="21305" PostTypeId="1" AcceptedAnswerId="21310" CreationDate="2011-02-10T03:27:12.510" Score="6" ViewCount="83" Body="&lt;p&gt;If you take a look at &lt;a href=&quot;http://www.wolframalpha.com&quot; rel=&quot;nofollow&quot;&gt;WolframAlpha&lt;/a&gt;, or other computer algebraic system, you will find that it is able to do symbolic manipulation like real humans.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, if you type in an integral, it can show you step by step on how to solve the integration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are the algorithms behind all this?&lt;/p&gt;&#xA;" OwnerUserId="550" LastActivityDate="2011-02-10T04:36:10.533" Title="Algorithms for Symbolic Manipulation" Tags="&lt;symbolic-computation&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="21306" PostTypeId="2" ParentId="21304" CreationDate="2011-02-10T03:38:29.297" Score="1" ViewCount="" Body="&lt;p&gt;Warning: I don't really know what I'm talking about.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That said: It seems like $-f$ will also be a morse function with 3 critical points. Clearly the index 0 critical point from $f$ will correspond to an index $n$ one for $-f$ and similarly for the index $n$ critical point from $f$. We are left with our critical point of unknown index... But I think it would make sense that if we put two morse functions on a manifold, $f$ and $g$, and they have the same number of critical points and the indices all match except for one unknown index for each function, then these remaining two must match. So it seems like the unknown critical point must have the same index in $f$ and $-f$, which means it should have index $n/2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact... the statement that if $f$ is a Morse function so is $-f$ is kind of a baby version of Poincare duality (it gives the correspondence between cells that you want, but it doesn't immediately give the map with the cap product). So maybe this argument is pretty much what you were saying...&lt;/p&gt;&#xA;" OwnerUserId="423" LastActivityDate="2011-02-10T03:38:29.297" />
  <row Id="21307" PostTypeId="1" AcceptedAnswerId="21323" CreationDate="2011-02-10T03:46:17.220" Score="5" ViewCount="132" Body="&lt;p&gt;Let $A$ be a symmetric stochastic matrix, such that the sum over the columns, for each row, is 1, and all elements are positive. $A$ dimensions are $n \times n$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's say that $B$ is a matrix which is not necessarily of the same column and row dimension, $B$'s dimensions are $n \times m$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We also have the property for $B$, like $A$, that all elements are positive, and the sum over columns is 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What can we say about the product $A \times B$?&#xA;What if we continuously apply $B_{t+1} \leftarrow A \times B_t$ where $B_1 = B$?&#xA;Is there anything I can read about?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any clues appreciated.&#xA;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="7076" LastEditorUserId="2404" LastEditorDisplayName="" LastEditDate="2011-02-10T20:40:07.233" LastActivityDate="2011-02-10T20:40:07.233" Title="stochastic matrices?" Tags="&lt;linear-algebra&gt;&lt;stochastic-processes&gt;" AnswerCount="2" FavoriteCount="3" />
  <row Id="21308" PostTypeId="1" AcceptedAnswerId="21309" CreationDate="2011-02-10T04:09:27.993" Score="2" ViewCount="67" Body="&lt;p&gt;I know asin(sqrt(2)/2) is equal to pi/4. However I don't understand why, I've done some searching on google about arc trig functions and I haven't found any webpages that explain it very well. Can any of you help?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="6852" LastActivityDate="2011-02-10T16:51:28.023" Title="Trig Question - asin(sqrt(2)/2) and arc trig functions in general" Tags="&lt;trigonometry&gt;" AnswerCount="2" />
  <row Id="21309" PostTypeId="2" ParentId="21308" CreationDate="2011-02-10T04:16:16.473" Score="0" ViewCount="" Body="&lt;p&gt;You have to understand how the inverse function is defined. If suppose $x = \sin\Bigl(\frac{\pi}{4}\Bigr)=\frac{1}{\sqrt{2}}$, then $\arcsin\Bigl(\frac{1}{\sqrt{2}}\Bigr) = \frac{\pi}{4}$. You can see more of this on&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Inverse_trigonometric_functions&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Inverse_trigonometric_functions&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" LastActivityDate="2011-02-10T04:16:16.473" />
  <row Id="21310" PostTypeId="2" ParentId="21305" CreationDate="2011-02-10T04:25:31.980" Score="7" ViewCount="" Body="&lt;p&gt;The algorithms behind symbolic integration (due to Liouville, Ritt, Risch, Bronstein et al.) are discussed in prior questions here, e.g. the &lt;a href=&quot;http://math.stackexchange.com/questions/155/how-can-you-prove-that-a-function-has-no-closed-form-integral/5723#5723&quot;&gt;transcendental case&lt;/a&gt; and &lt;a href=&quot;http://math.stackexchange.com/questions/3543/evaluating-indefinite-integrals-of-the-form-fracfxpx-sqrtqx/3566#3566&quot;&gt;algebraic case&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For general references on symbolic computation see any of the standard texbooks, e.g. Geddes et al. &lt;a href=&quot;http://books.google.com/books?id=TXfY891RiiEC&amp;amp;pg=PP1&quot; rel=&quot;nofollow&quot;&gt;Algorithms for computer algebra&lt;/a&gt;, Grabmeier et al: &lt;a href=&quot;http://books.google.com/books?id=Pnlxei_XfFQC&amp;amp;pg=PP1&quot; rel=&quot;nofollow&quot;&gt;Computer algebra handbook&lt;/a&gt;,von zur Gathen: &lt;a href=&quot;http://books.google.com/books?id=NuEHj0wPwgIC&amp;amp;pg=PP1&quot; rel=&quot;nofollow&quot;&gt;Modern computer algebra&lt;/a&gt;, and Zippel: &lt;a href=&quot;http://books.google.com/books?id=--Ih9bv0m_AC&amp;amp;pg=PP1&quot; rel=&quot;nofollow&quot;&gt;Effective polynomial computation&lt;/a&gt;, and many other books. See also the &lt;a href=&quot;http://www.elsevier.com/wps/find/journaldescription.cws_home/622902/description#description&quot; rel=&quot;nofollow&quot;&gt;Journal of Symbolic Computation&lt;/a&gt; and various conferences: &lt;a href=&quot;http://www.sigsam.org/index.phtml&quot; rel=&quot;nofollow&quot;&gt;SIGSAM ISSAC&lt;/a&gt;, EUROCAL, etc.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2011-02-10T04:36:10.533" LastActivityDate="2011-02-10T04:36:10.533" CommentCount="2" />
  <row Id="21311" PostTypeId="1" CreationDate="2011-02-10T04:30:25.997" Score="3" ViewCount="269" Body="&lt;p&gt;Hey everyone,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So our calc teacher is being tricky and sending us off to fend for ourselves in the world of mathematics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Here's the question:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We need to set up an integral to find the volume of the solid formed by the area bounded by the functions y=ln(x), x=3, and the x-axis revolved around the line x=3.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The problem?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most of us put something similar to $\pi \int_0^{\ln 3} (e^y - 3)^2\ dy$, which is believed to be correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;He claims it can also be written as $\pi \int_0^{e^3} (3 - e^y)^2\ dy$, but hasn't given us any idea as to why.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Many thanks to anyone who can help us out.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Multiple choice options given:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\pi \int_0^{e^3} (3 - e^y)^2\ dy$&lt;/li&gt;&#xA;&lt;li&gt;$\pi \int_0^{e^3} (9 - e^{2y})\ dy$&lt;/li&gt;&#xA;&lt;li&gt;$\pi \int_1^{3} (\ln_e(x))^2\ dy$&lt;/li&gt;&#xA;&lt;li&gt;None of These &amp;lt;- Not this answer!&lt;/li&gt;&#xA;&lt;li&gt;$\pi \int_1^{3} (9 - (\ln_e(x))^2)\ dy$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="6853" LastEditorUserId="6853" LastEditorDisplayName="" LastEditDate="2011-02-10T05:42:11.600" LastActivityDate="2011-02-10T05:42:11.600" Title="Calculus Teacher being Tricky (Integrals)" Tags="&lt;calculus&gt;&lt;integral&gt;&lt;graph&gt;" AnswerCount="2" CommentCount="8" />
  <row Id="21312" PostTypeId="2" ParentId="21307" CreationDate="2011-02-10T04:33:42.997" Score="0" ViewCount="" Body="&lt;p&gt;I assume that when you say that for $B$ the &quot;sum over the columns equals $1$,&quot; that means that the sum of the elements in each column of $B$ is $1$. Is that right?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In that case, the answer is that the iteration $B(t+1) = A B(t)$ will converge, in the limit as $t \rightarrow \infty$, to a matrix whose every entry is $1/n$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To prove this, it suffices to show that for any vector $v(0)$ the iteration&#xA;$$ v(t+1) = A v(t)$$ converges to a vector whose every entry is the average of the vector $v(0)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that a &lt;em&gt;positive&lt;/em&gt; stochastic matrix will have a single eigenvalue at $1$, and every other eigenvalue will have magnitude strictly less than $1$. Moreover, the eigenvalue $1$ will be associated with the all-ones eigenvector, which I'll denote by ${\bf 1}$. Thus the limit $\lim_t A^t v(0)$ will be a multiple of ${\bf 1}$, i.e. a vector $c {\bf 1}$ for some number $c$ ( to see this, note that because $A$ is symmetric, it has a set of orthogonal eigenvectors; expand $v(0)$ in the basis of these eigenvectors, and look at what multiplication by $A$ does). Next, we must figure out what $c$ is; observe that by symmetry of the matrix $A$, multiplication by the matrix $A$ preserves the sum of the entries of a vector $$ \sum_{i=1}^n (Ax)_i = {\bf 1}^T A x = (A {\bf 1})^T x = {\bf 1}^T x = \sum_{i=1}^m x_i,$$ which tells you that the constant $c$ must be the average of the entries of $x(0)$. &lt;/p&gt;&#xA;" OwnerUserId="317" LastEditorUserId="3253" LastEditorDisplayName="" LastEditDate="2011-02-10T05:54:16.053" LastActivityDate="2011-02-10T05:54:16.053" />
  <row Id="21313" PostTypeId="1" CreationDate="2011-02-10T05:17:57.967" Score="4" ViewCount="278" Body="&lt;p&gt;What is the difference between a topological and a metric space?&lt;/p&gt;&#xA;" OwnerUserId="6854" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-10T05:19:45.733" LastActivityDate="2011-02-10T18:42:28.380" Title="What is the difference between topological and metric spaces?" Tags="&lt;topology&gt;&lt;metric-spaces&gt;" AnswerCount="4" CommentCount="5" />
  <row Id="21314" PostTypeId="2" ParentId="21311" CreationDate="2011-02-10T05:32:04.993" Score="5" ViewCount="" Body="&lt;p&gt;From my calculations, it seems that your integral is correct.  The other one is not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The volume is indeed given by $$V=\pi \int_0^{\ln 3} (e^y-3)^2 dy=5.929$$ (We go from 0 to $\ln 3$, and the radius of the circle at height $y$ is $3-e^y$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively, integrating the $x$-coordinate we find the volume is $$V=2\pi \int_{1}^{3} (3-x)\ln x dx=5.929$$ (Using the so called &quot;shells&quot; method)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope that helps, sometimes people just make typos.&lt;/p&gt;&#xA;" OwnerUserId="6075" LastActivityDate="2011-02-10T05:32:04.993" CommentCount="2" />
  <row Id="21315" PostTypeId="2" ParentId="21311" CreationDate="2011-02-10T05:34:26.553" Score="3" ViewCount="" Body="&lt;p&gt;Let $f$ be some function. I'll demonstrate how to calculate the volume $V$ of the solid obtained by revolving the region under the graph $f \ge 0$ on some interval $I$ around the &lt;em&gt;horizontal axis.&lt;/em&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $P = \{ t_0, \cdots, t_n \}$ be any partition of $I$. Let $m_i = \inf \{ f(x) : t_{i-1} \le x \le t_i \}$ and $M_i = \sup \{ f(x) : t_{i-1} \le x \le t_i \}$, be as always. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's easy to see, that $\pi m_i^2(t_i - t_{i-1}) $ is the volume of some cylinder that lies inside $V$ on $[t_{i-1}, t_i]$. Similarly, $\pi M_i^2(t_i - t_{i-1}) $ is some cylinder that contains a part of $V$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know that as we sum over the partition of $I$, the lower sum will be a lower bound for the volume of $V$ and the upper sum an upper bound. So we have:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{i=1}^n \pi m_i^2 (t_i - t_{i-1}) \le volume(V) \le \sum_{i=1}^n \pi M_i^2 (t_i - t_{i-1})$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These sums are the lower and upper sums for $f^2$ on $I$, so it follows that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$volume(V) = \pi \int_a^b f(x)^2 dx$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, we need to adapt your region to this enviroment. Here we need to rotate the region about a line that is parallel to the $y$-axis. So we can shift back the region $3$ units towards the $y$-axis. Now our region is just the area under $\ln(x+3)$ on the interval $[-2,0]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As we are rotating about the $y$-axis, we must express the function in terms of $y$, i.e. $x=e^y - 3$, on the interval from $0$ to $\ln(0+3) = \ln 3$, so indeed, plugging this into our formula, yields:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$volume (V) = \pi \int_0^{\ln 3} (e^y - 3)^2\ dy$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so yes, your first integral is correct.&lt;/p&gt;&#xA;" OwnerUserId="5105" LastActivityDate="2011-02-10T05:34:26.553" CommentCount="1" />
  <row Id="21317" PostTypeId="1" CreationDate="2011-02-10T05:47:22.410" Score="2" ViewCount="119" Body="&lt;p&gt;What is the difference between limit point of a sequence and limit of a sequence. Can it be unique?&lt;/p&gt;&#xA;" OwnerUserId="6854" LastActivityDate="2011-04-11T06:40:04.047" Title="Difference between limit point and limit" Tags="&lt;metric-spaces&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="21318" PostTypeId="2" ParentId="21313" CreationDate="2011-02-10T05:50:09.037" Score="3" ViewCount="" Body="&lt;p&gt;A topological space is a set $X$ along with another set usually denoted by $\tau$ which is a collection of subsets of $X$, satisfying the following properties:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$\emptyset, X \in \tau$&lt;/li&gt;&#xA;&lt;li&gt;Countable or Uncountable union of sets in $\tau$ is again in $\tau$&lt;/li&gt;&#xA;&lt;li&gt;Finite intersection of sets in $\tau$ is again in $\tau$&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The space $(X,\tau)$ is called the topological space and the set $\tau$ is called a topology on $X$. The elements of $\tau$ are called open sets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A metric space is a set $X$ and a function $d:X \times X \rightarrow \mathbb{R}^+ \cup \{0\}$ called the &quot;metric&quot; which takes in two elements from the set and pops out a non-negative real number. This metric has to satisfy certain properties:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$d(x,y) \geq 0$, $\forall x,y \in X$&lt;/li&gt;&#xA;&lt;li&gt;$d(x,y) = 0$, iff $x=y$&lt;/li&gt;&#xA;&lt;li&gt;$d(x,y) = d(y,x)$, $\forall x,y \in X$&lt;/li&gt;&#xA;&lt;li&gt;$d(x,y) \leq d(x,z) + d(z,y)$, $\forall x,y,z \in X$&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The space $(X,d)$ is called the metric space and $d$ is the metric i.e. a function such that $d:X \times X \rightarrow \mathbb{R}^+ \cup \{0\}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using this metric, we can define &quot;certain&quot; sets. The set of these sets call it $\tau$, along with the original set $X$ can now shown to be a topological space. So with every metric space $(X,d)$, we can associate a topological space $(X,\tau)$. The elements of the set $\tau$ are open sets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, topological spaces need not arise out of a metric space. There are non-metrizable topological spaces.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2011-02-10T06:01:51.273" LastActivityDate="2011-02-10T06:01:51.273" CommentCount="9" />
  <row Id="21319" PostTypeId="2" ParentId="21317" CreationDate="2011-02-10T05:52:17.410" Score="1" ViewCount="" Body="&lt;p&gt;Gaurav Please look at the definitions and tell what trouble you are facing in understanding the definitions.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Limit_point&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Limit_point&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Limit_of_a_sequence&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Limit_of_a_sequence&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" LastActivityDate="2011-02-10T05:52:17.410" CommunityOwnedDate="2011-02-10T05:52:17.410" CommentCount="3" />
  <row Id="21320" PostTypeId="1" AcceptedAnswerId="21321" CreationDate="2011-02-10T05:56:28.480" Score="3" ViewCount="84" Body="&lt;p&gt;By identity representation I mean the representation sending each element of $SL(2,k)$ to itself. Is there a simple way to see this isomorphism? I feel like I am missing something incredibly basic here.&lt;/p&gt;&#xA;" OwnerUserId="251" LastActivityDate="2011-02-10T06:37:42.083" Title="Why is the identity representation of $SL(2,k)$ isomorphic to its dual representation?" Tags="&lt;linear-algebra&gt;&lt;representation-theory&gt;" AnswerCount="1" />
  <row Id="21321" PostTypeId="2" ParentId="21320" CreationDate="2011-02-10T06:08:57.227" Score="3" ViewCount="" Body="&lt;p&gt;Call $V$ that representation. The wedge product in the exterior algebra gives you a map $\mu:V\otimes V\to\Lambda^2V$, which is a morphism of $\mathrm{SL}(2,k)$-modules. Now $\Lambda^2V$ is one-dimensional, and the induced action of $\mathrm{SL}(2,k)$ on it is trivial (precisely because the elements of the group have determinant one!), so we can identify it with the trivial module $k$. Partially transposing $\mu$ (that is, considering its image under the natural isomorphism $\hom_k(V\otimes V,k)\to\hom_k(V,V^*)$), we obtain a map $\mu^t:V\to V^*$ which is also of $\mathrm{SL}(2,k)$-modules. Since $\mu$ is non-degenerate, $\mu^t$ is an isomorphism. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively, the module $V$ is simple and, for reasonable characteristics (I guess this means in this case different from $2$...) and reasonable notions of what a representation is (see Matt's comment below), the group $\mathrm{SL}(2,k)$ has exactly one simple module of dimension $2$. Since $V^*$ is also simple and of dimension $2$, so we must necessarily have $V\cong V^*$.&lt;/p&gt;&#xA;" OwnerUserId="274" LastEditorUserId="274" LastEditorDisplayName="" LastEditDate="2011-02-10T06:37:42.083" LastActivityDate="2011-02-10T06:37:42.083" CommentCount="6" />
  <row Id="21322" PostTypeId="2" ParentId="21313" CreationDate="2011-02-10T06:16:48.623" Score="4" ViewCount="" Body="&lt;p&gt;Just in terms of ideas: a metric space has a notion of distance, while a topological space only has a notion of closeness. If we have a notion of distance then we can say when things are close to each other. However, distance is not necessary to determine when things are close to each other.&lt;/p&gt;&#xA;" OwnerUserId="627" LastActivityDate="2011-02-10T06:16:48.623" />
  <row Id="21323" PostTypeId="2" ParentId="21307" CreationDate="2011-02-10T06:52:08.123" Score="1" ViewCount="" Body="&lt;p&gt;Firstly, note that if $A$ is row-stochastic (i.e. the sum across columns is equal to 1 for every row and the entries are non-negative) and symmetric then it must also be column-stochastic, hence doubly stochastic.  Also, since the entries are positive, the Perron-Frobenius Theorem would be very useful here.  Basically, the matrix $A$ represents a discrete time Markov chain with a finite state space, and understanding the effect of repeatedly multiplying $B$ by $A$ requires an understanding of whether the following limit exists: $\lim_{t \to \infty} A^t$.  This is a well-studied problem in the theory of Markov Processes (coming from results in linear algebra), and the answer is that if the matrix is irreducible and aperiodic (equivalently, if there exists some $k$ such that for all $m &amp;gt; k$, $A^k$ has all positive entries) then this limit exists.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given that in this case $A$ has all positive entries, the infinite product of the matrix will have a limit, which is a matrix with identical rows $s$ given by the equation: $sA = s$.  In other words, the limiting matrix $\lim_{t \to \infty} A^t$ is the $n \times n$ matrix with every row equaling the left eigenvector associated with the eigenvalue 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, as you multiply $B$ repeatedly, you will find that $B_t$ will converge to a matrix with identical rows, where the entry in column $i$ is a weighted average of the column $i$ of $B$, with weights given by $s$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A great reference for you would be Carl Meyer's book &lt;a href=&quot;http://www.matrixanalysis.com/&quot; rel=&quot;nofollow&quot;&gt;Matrix Analysis and Applied Linear Algebra&lt;/a&gt;.  Alternatively, you'll find the essence of what you need in the following journal article's appendix: &lt;a href=&quot;http://www.stanford.edu/~bgolub/papers/naivelearning.pdf&quot; rel=&quot;nofollow&quot;&gt;Jackson, Golub - Naive  Learning in Social Networks and the Wisdom of Crowds&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2404" LastEditorUserId="2404" LastEditorDisplayName="" LastEditDate="2011-02-10T06:59:53.790" LastActivityDate="2011-02-10T06:59:53.790" />
  <row Id="21324" PostTypeId="1" CreationDate="2011-02-10T06:56:28.773" Score="3" ViewCount="218" Body="&lt;p&gt;Inspired by this &lt;a href=&quot;http://mathoverflow.net/questions/2144/a-single-paper-everyone-should-read&quot; rel=&quot;nofollow&quot;&gt;question&lt;/a&gt;, at &lt;strong&gt;MO&lt;/strong&gt; i am asking this question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone list some elementary articles at ArXiv which can be understood by High-School/Undergrad Students. I am asking this because, i would like to see some interesting papers and articles to learn something new. If the paper is very much advanced then its of no use to me since i haven't learnt enough Math. I have been searching around for past two hours and i couldn't find a single paper which i could comprehend. Article which i would like to see:-&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Interesting proofs of some Elementary number Theory results.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Interesting identity involving Infinite series.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Interesting articles in Basic Abstract Algebra ( concerning Groups and Rings.)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Your favorite article. ( Make sure its elementary enough!)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The best example of what type of articles i am looking for can be found in the answer given by Bill here:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://math.stackexchange.com/questions/5763/characterizing-continuous-functions-based-on-the-graph-of-the-function&quot;&gt;Characterizing continuous functions based on the graph of the function.&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This is an interesting article. I really enjoyed reading it since i could understand it.&lt;/p&gt;&#xA;" LastActivityDate="2011-02-10T11:37:49.213" CommunityOwnedDate="2011-02-10T07:16:49.757" Title="Elementary Papers at ArXiv" Tags="&lt;soft-question&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="1" />
  <row Id="21325" PostTypeId="2" ParentId="21173" CreationDate="2011-02-10T07:02:51.220" Score="0" ViewCount="" Body="&lt;p&gt;I found out that the answer is&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;D = Floor[Log10[N]] + 1&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;http://www.wolframalpha.com/input/?i=Floor[Log10[2011]]+%2B+1&lt;/p&gt;&#xA;" OwnerUserId="5537" LastActivityDate="2011-02-10T07:02:51.220" />
  <row Id="21326" PostTypeId="1" AcceptedAnswerId="21328" CreationDate="2011-02-10T07:06:07.473" Score="2" ViewCount="48" Body="&lt;p&gt;I was able to prove an extension of the Ham Sandwich Theorem (namely; that given any 3 integrable functions on $\mathbb{R}^3$, there exists a plane which simultaneously divides their total integrals in half), and am attempting to show that this does not carry over to 4 functions.  I have reduced the proof to one fact which I think is true; but don't know how to prove.&#xA;Given two sets of points in $\mathbb{R}^4$, $S$ and $T$, is there a continuous function $f:\mathbb{R}^4\to\mathbb{R}$ which takes the value $a$ on all points in $S$ and $b$ on all points in $T$?&lt;/p&gt;&#xA;" OwnerUserId="4772" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-10T07:09:27.083" LastActivityDate="2011-02-10T07:21:32.833" Title="Extension to a continuous function of a set of specified values" Tags="&lt;topology&gt;" AnswerCount="1" />
  <row Id="21327" PostTypeId="2" ParentId="21276" CreationDate="2011-02-10T07:12:07.873" Score="0" ViewCount="" Body="&lt;p&gt;It seems the derivation is correct, thanks Jim Conant.&lt;/p&gt;&#xA;" OwnerUserId="6786" LastActivityDate="2011-02-10T07:12:07.873" />
  <row Id="21328" PostTypeId="2" ParentId="21326" CreationDate="2011-02-10T07:15:23.110" Score="2" ViewCount="" Body="&lt;p&gt;Depends on the sets. For example, if you could find a sequence of points in $S$ that converges to a point in $T$, and $a\neq b$, then you would not be able to do so, since the value of $f$ at the limit (that is, $b$), would have to equal the limit of the values (that is, $a$). Or worse, if $S\cap T\neq\emptyset$ (though I suspect you meant the two sets to be disjoint, you failed to mention this).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if both $S$ and $T$ are &lt;em&gt;closed&lt;/em&gt; and &lt;em&gt;disjoint&lt;/em&gt;, then the answer is yes: any two disjoint closed sets in $\mathbb{R}^4$ (in fact, in $\mathbb{R}^n$) can be separated by a continuous function $f\colon\mathbb{R}^4\to \mathbb{R}$ such that $S$ is precisely the set of points that map to $0$, and $T$ is precisely the set of points that map to $1$. From there to your function is an easy step. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(This follows because $\mathbb{R}^4$ is a metric space, and hence perfectly normal. See &lt;a href=&quot;http://en.wikipedia.org/wiki/Normal_space&quot; rel=&quot;nofollow&quot;&gt;Wikipedia's page on  normal spaces&lt;/a&gt;.) &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-10T07:21:32.833" LastActivityDate="2011-02-10T07:21:32.833" CommentCount="2" />
  <row Id="21329" PostTypeId="2" ParentId="21324" CreationDate="2011-02-10T07:25:08.627" Score="0" ViewCount="" Body="&lt;p&gt;That's a bit of a stretch, but the beautiful paper &lt;a href=&quot;http://www.math.lsa.umich.edu/~ablass/7trees.pdf&quot; rel=&quot;nofollow&quot;&gt;Seven Trees in One&lt;/a&gt; by Andreas Blass (which has a faulty ArXiv version) is perhaps possible to understand up to section 4 (which is exactly where my understanding stops). The first three sections are not really difficult, even though they mention some abstract non-sense (which you can skip), some enumerative combinatorics (which you can also skip), and some really basic definitions in abstract algebra, which are unfortunately too &quot;basic&quot; to include in a normal course; but all you need to know about them is how to calculate within them. You also need to know some very basic set theory (mainly $\aleph_0^2 = \aleph_0$).&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2011-02-10T07:25:08.627" CommunityOwnedDate="2011-02-10T07:25:08.627" CommentCount="2" />
  <row Id="21330" PostTypeId="1" CreationDate="2011-02-10T07:37:53.297" Score="14" ViewCount="574" Body="&lt;p&gt;Earlier today, I was talking with my friend about some &quot;cool&quot; infinite series and the value they converge to like the Basel problem, Madhava-Leibniz formula for $\displaystyle \pi/4, \log 2$ and similar alternating series etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One series that popped into our discussion was $\displaystyle \sum_{n=1}^{\infty} \frac{1}{n^n}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proving the convergence of this series is trivial but finding the value to which converges has defied me so far. Mathematica says this series converges to $\displaystyle \sim 1.29129$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried googling about this series and found very little information about this series (which is actually surprising since the series looks cool enough to arise in some context).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We were joking that it should have something to do with $\pi$,$e$,$\phi$, $\gamma$  or at the least it must be a transcendental number :-).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My questions are:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;What does this series converge to?&lt;/li&gt;&#xA;&lt;li&gt;Does this series arise in any context and are there interesting trivias to be known about this series?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I am actually slightly puzzled that I have not been able to find much about this series on the internet. (At least my google search did not yield any interesting results).&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2011-05-07T21:24:58.460" LastActivityDate="2011-05-07T21:24:58.460" Title="&quot;Closed&quot; form for $\sum \frac{1}{n^n}$" Tags="&lt;sequences-and-series&gt;" AnswerCount="4" CommentCount="7" FavoriteCount="2" />
  <row Id="21332" PostTypeId="2" ParentId="21330" CreationDate="2011-02-10T07:59:20.767" Score="11" ViewCount="" Body="&lt;p&gt;Certainly you need to check out &lt;a href=&quot;http://en.wikipedia.org/wiki/Sophomore%27s_dream&quot; rel=&quot;nofollow&quot;&gt;Sophomore's Dream&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2937" LastActivityDate="2011-02-10T07:59:20.767" CommentCount="3" />
  <row Id="21333" PostTypeId="1" AcceptedAnswerId="21337" CreationDate="2011-02-10T07:59:45.410" Score="1" ViewCount="174" Body="&lt;p&gt;Hello,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the proof from the book:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Theorem.&lt;/strong&gt; There are infinitely many primes of the form $4n+3$. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Lemma.&lt;/strong&gt; If $a$ and $b$ are integers, both of the form $4n + 1$, then the product $ab$ is also in this form.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof of Theorem:&lt;/strong&gt;&#xA;Let assume that there are only a finite number of primes of the form $4n + 3$, say &#xA;$$p_0, p_1, p_2, \ldots, p_r.$$&lt;br&gt;&#xA;Let $$Q = 4p_1p_2p_3\cdots p_r + 3.$$&lt;br&gt;&#xA;Then there is at least one prime in the factorization of $Q$ of the form $4n + 3$. Otherwise, all of these primes would be of the form $4n + 1$, and by the Lemma above, this would imply that $Q$ would also be of this form, which is a contradiction. However, none of the prime $p_0, p_1,\ldots, p_n$ divides $Q$. The prime $3$ does not divide $Q$, for if $3|Q$  then $$3|(Q-3) = 4p_1p_2p_3\cdots p_r,$$ which is a contradiction. Likewise, none of the primes $p_j$ can divides $Q$, because $p_j | Q$ implies $p_j | ( Q - 4p_1p_2\cdots p_r ) = 3$, which is absurd. Hence, there are infinitely many primes of the form $4n +3$.   &lt;strong&gt;END&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From &quot;however, none of the prime ....&quot; to the end, I totally lost!  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My questions:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Is the author assuming $Q$ is prime or is not?&lt;/li&gt;&#xA;&lt;li&gt;Why none of the primes $p_0, p_1,\ldots, p_r$ divide $Q$? Based on what argument?  &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Can anyone share me a better proof?   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="5711" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-03T19:09:39.860" LastActivityDate="2011-03-03T19:09:39.860" Title="Help understand the proof of infinitely many primes of the form $4n+3$" Tags="&lt;elementary-number-theory&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="21334" PostTypeId="2" ParentId="21330" CreationDate="2011-02-10T08:04:14.287" Score="5" ViewCount="" Body="&lt;p&gt;This series turns out to be equal to the integral of $1/x^x$ from 0 to 1. Sorry I am bad with the notation needed for math symbols here; but if you rewrite $x^{-x}$ as the exponential of $x\log x$, you can expand it as a taylor series and integrate term by term; to find it equals your sum.&#xA;This doesn't give an explicit value; but I think it's a pretty cool identity.&lt;/p&gt;&#xA;" OwnerUserId="4772" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2011-02-10T15:00:44.517" LastActivityDate="2011-02-10T15:00:44.517" CommentCount="2" />
  <row Id="21335" PostTypeId="2" ParentId="21313" CreationDate="2011-02-10T08:11:58.430" Score="0" ViewCount="" Body="&lt;p&gt;An important difference in terms of technique is that in a metric space there are distinguished neighborhoods, namely the open balls of radius $r$ around a point $x$:&#xA;$$&#xD;&#xA;B_r(x) = \{y\ :\ d(x,y)&amp;lt;r\}.&#xD;&#xA;$$&#xA;These open balls form a local base for the topology and hence, carry all information about the topology of the metric space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While in topological spaces the notion of a neighborhood is just an abstract concept which reflects somehow the properties a &quot;neighborhood&quot; should have, a metric space really have some notion of &quot;nearness&quot; and hence, the term neighborhood somehow reflects the intuition a bit more.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moreover, in a metric space it is more convenient to work with sequences than in a topological space. For example it makes total sense, to memorize convergence of a sequence $(x_n)$ in a metric space to a point $x$ as &quot;from some point on all $x_n$ are arbitrarily close to $x$&quot;. A statment which is quite useless in a topological space.&lt;/p&gt;&#xA;" OwnerUserId="3148" LastActivityDate="2011-02-10T08:11:58.430" />
  <row Id="21336" PostTypeId="2" ParentId="21330" CreationDate="2011-02-10T08:13:00.770" Score="0" ViewCount="" Body="&lt;p&gt;This has been asked here as well:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;http://www.artofproblemsolving.com/Forum/resources.php?c=2&amp;amp;cid=59&amp;amp;year=2005&quot; rel=&quot;nofollow&quot;&gt;http://www.artofproblemsolving.com/Forum/resources.php?c=2&amp;amp;cid=59&amp;amp;year=2005&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;http://math.stackexchange.com/questions/2536/finding-f-such-that-int-f-sum-f&quot;&gt;Finding $f$ such that $ \int f = \sum f$&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" LastActivityDate="2011-02-10T08:13:00.770" />
  <row Id="21337" PostTypeId="2" ParentId="21333" CreationDate="2011-02-10T08:19:47.410" Score="4" ViewCount="" Body="&lt;p&gt;This is an adaption of Euclid's classical proof of the infinitude of prime. Suppose that $p_1,...,p_t$ are all the primes and consider the number $N=p_1\cdots p_t+1$. The number $N$ must be divisible by some prime (possibly itself, but this is irrelevant for the argument) but since noone of the $p_i$ divides $N$, this gives a contradiction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The proof you report is similar in concept but is adapted to show that this &quot;extra prime&quot; obtained by looking at divisors of a suitably constructed auxiliary number (the $Q$ in the proof) is actually of the form $4k+3$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe that a slight correction in the proof is in order: namely, take $p_0=3$. The important technical point is that you DON'T include $p_0=3$ in the product defining $Q$. Thus, you can show that none of the $p_i$ (INCLUDING $p_0$) divides $Q$ and you're done by the Lemma.&lt;/p&gt;&#xA;" OwnerUserId="688" LastEditorUserId="688" LastEditorDisplayName="" LastEditDate="2011-02-10T09:36:15.580" LastActivityDate="2011-02-10T09:36:15.580" CommentCount="1" />
  <row Id="21338" PostTypeId="2" ParentId="21333" CreationDate="2011-02-10T08:24:51.930" Score="5" ViewCount="" Body="&lt;p&gt;That part of the proof is simply a variant of Euclid's classical method for producing a new prime. Instead of $\rm\ 1+ p\cdot p_1\cdots p_n\ $ it employs $\rm\ p+ p_0\cdots p_n\:,\: $ where $\rm\ (p,\ p_i) = 1\:.\: $ (above $\rm\: p=3,\ p_0 = 4\:$)&lt;br&gt;&#xA;It is easy to verify that this newly constructed integer is coprime to all the prior $\rm\: p $'s, namely: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\quad\quad\quad\quad\quad\quad (p,\ \ \: p+p_0\cdots p_n)\ =\ (p,\  p_0\cdots p_n)\ =\ 1\ \ $ via $\rm\ \ (p,\ p_i)\ =\ 1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\quad\quad\quad\quad\quad\quad (p_i,\ p+ p_0\cdots p_n)\ =\ (p_i,\ p)\ =\ 1\ $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Essentially this proof relies on the fact that $\rm\ (p\:q,\ p+q) = 1\ \iff\ (p,\ q) = 1\:.\  $ Hence to produce a number coprime to $\rm\ n\ $ we can simply sum the factors $\rm\ p,\:q\ $ from any coprime splitting $\rm\ n = p\:q\:.\ $ Euclid's classic proof uses the trivial splitting where $\rm\ q = 1\ $ (and $\rm\:p\:$ is a product of given primes). &#xA;Ribenboim credits this splitting-form generalization of Euclid's proof to Stieltjes (1890).&#xA;For a handful of proofs of said gcd property see &lt;a href=&quot;http://math.stackexchange.com/questions/21545/gcda-b-gcda-b-mathrmlcma-b/21626#21626&quot;&gt;my post here.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2011-03-03T18:32:57.143" LastActivityDate="2011-03-03T18:32:57.143" CommentCount="1" />
  <row Id="21339" PostTypeId="1" AcceptedAnswerId="21350" CreationDate="2011-02-10T08:56:07.860" Score="1" ViewCount="42" Body="&lt;p&gt;I have $\log X \sim Exp(\vartheta - 1)$ and I would like to show &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ P \Big [ \Big |\frac{1}{\frac{1}{n} \sum_{i=1}^n \log X_i}&amp;#160;- (\vartheta - 1) \Big | &amp;gt; \varepsilon \Big ] \rightarrow 0  \hspace{5 mm} \forall \vartheta (n \rightarrow \infty)$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the answer to this question it states that all moments of the exponential distribution that are necessary for the strong law of large numbers exist. Therefore $\frac{1}{\frac{1}{n} \sum_{i=1}^n \log X_i}$ converges almost surely towards $\frac{1}{\vartheta - 1}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can someone explain to me why the moments need to exist for the law of large numbers? And how this proof works? Many thanks for your help!&lt;/p&gt;&#xA;" OwnerUserId="5798" LastActivityDate="2011-02-10T11:49:45.810" Title="Showing consistency of an estimator" Tags="&lt;homework&gt;&lt;probability&gt;&lt;statistics&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="0" />
  <row Id="21340" PostTypeId="2" ParentId="21199" CreationDate="2011-02-10T09:25:27.780" Score="8" ViewCount="" Body="&lt;p&gt;Just to add some variety to the list of answers, I'm going to go against the grain here and say that you can, in an albeit silly way, interpret dy/dx as a ratio of real numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For every (differentiable) function $f$, we can define a function $df(x; dx)$ of two real variables $x$ and $dx$ via $$df(x; dx) = f&amp;#39;(x)\,dx.$$&#xA;Here, $dx$ is just a real number, and no more.  (In particular, it is not a differential 1-form, nor an infinitesimal.)  So, when $dx \neq 0$, we can write:&#xA;$$\frac{df(x;dx)}{dx} = f&amp;#39;(x).$$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;All of this, however, should come with a few remarks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is clear that these notations above do not constitute a definition of the derivative of f.  Indeed, we needed to know what the derivative $f&amp;#39;$ meant before defining the function $df$.  So in some sense, it's just a clever choice of notation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if it's just a trick of notation, why do I mention it at all?  The reason is that in higher dimensions, the function $df(x;dx)$ actually becomes the focus of study, in part because it contains information about all the partial derivatives.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To be more concrete, for multivariable functions $f\colon R^n \to R$, we can define a function $df(x;dx)$ of two n-dimensional variables $x, dx \in R^n$ via&#xA;$$df(x;dx) = df(x_1,\ldots,x_n; dx_1, \ldots, dx_n) = \frac{\partial f}{\partial x_1}dx_1 + \ldots + \frac{\partial f}{\partial x_n}dx_n.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice that this map $df$ is &lt;em&gt;linear&lt;/em&gt; in the variable $dx$.  That is, we can write:&#xA;$$df(x;dx) = (\frac{\partial f}{\partial x_1}, \ldots, \frac{\partial f}{\partial x_n})&#xD;&#xA;\begin{pmatrix}&#xD;&#xA; dx_1 \\&#xD;&#xA; \vdots \\&#xD;&#xA; dx_n \\&#xD;&#xA;\end{pmatrix}&#xD;&#xA;= A(dx),$$&#xA;where $A$ is the 1xn row matrix of partial derivatives.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, the function $df(x;dx)$ can be thought of as a linear function of $dx$, whose matrix has variable coefficients (depending on $x$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So for the 1-dimensional case, what is really going on is a trick of &lt;em&gt;dimension&lt;/em&gt;.  That is, we have the variable 1x1 matrix (f'(x)) acting on the vector $dx \in R^1$ -- and it just so happens that vectors in $R^1$ can be identified with scalars, and so can be divided.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, I should mention that, as long as we are thinking of $dx$ as a real number, mathematicians multiply and divide by $dx$ all the time -- it's just that they'll usually use another notation.  The letter &quot;h&quot; is often used in this context, so we usually write $$f&amp;#39;(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h},$$&#xA;rather than, say,&#xA;$$f&amp;#39;(x) = \lim_{dx \to 0} \frac{f(x+dx) - f(x)}{dx}.$$&#xA;My guess is that the main aversion to writing $dx$ is that it conflicts with our notation for differential 1-forms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Just to be even more technical, and at the risk of being confusing to some, we really shouldn't even be regarding $dx$ as an element of $R^n$, but rather as an element of the tangent space $T_xR^n$.  Again, it just so happens that we have a canonical identification between $T_xR^n$ and $R^n$ which makes all of the above okay, but I like distinction between tangent space and euclidean space because it highlights the different roles played by $x \in R^n$ and $dx \in T_xR^n$.&lt;/p&gt;&#xA;" OwnerUserId="640" LastEditorUserId="640" LastEditorDisplayName="" LastEditDate="2011-02-10T09:44:15.487" LastActivityDate="2011-02-10T09:44:15.487" CommentCount="9" />
  <row Id="21341" PostTypeId="1" AcceptedAnswerId="21342" CreationDate="2011-02-10T09:27:55.967" Score="6" ViewCount="112" Body="&lt;p&gt;Hi,&#xA;what does it means that a place divide a prime on an algebraic number field?&lt;/p&gt;&#xA;" OwnerUserId="6583" LastEditorUserId="1508" LastEditorDisplayName="" LastEditDate="2011-02-10T12:24:37.500" LastActivityDate="2011-02-10T20:02:04.263" Title="places and primes" Tags="&lt;number-theory&gt;&lt;algebraic-geometry&gt;&lt;algebraic-number-theory&gt;&lt;arithmetic-geometry&gt;" AnswerCount="1" />
  <row Id="21342" PostTypeId="2" ParentId="21341" CreationDate="2011-02-10T09:48:21.530" Score="3" ViewCount="" Body="&lt;p&gt;A place is a valuation (or an equivalence class of) $v:K^\times\rightarrow{\Bbb Z}$. The place $v$ divides the rational prime $p$ when $v(p)&amp;gt;0$, or equivalently when $v$ extends the $p$-adic valuation on $\Bbb Q$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since (non-archimedean) places in a number field $K$ correspond to prime non-zero ideals in the ring of integers ${\cal O}_K$, another formulation is that the valuation $v$ corresponding to the prime ideal ${\cal P}_v\subset{\cal O}_K$ divides $p$ when ${\cal P}_v$ appears in the primary decomposition of the ideal $p{\cal O}_K$.&lt;/p&gt;&#xA;" OwnerUserId="688" LastEditorUserId="688" LastEditorDisplayName="" LastEditDate="2011-02-10T20:02:04.263" LastActivityDate="2011-02-10T20:02:04.263" CommentCount="2" />
  <row Id="21343" PostTypeId="1" AcceptedAnswerId="21364" CreationDate="2011-02-10T09:59:54.173" Score="1" ViewCount="70" Body="&lt;p&gt;Hello!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What does $N.\hat{N}_{fa}(S)$ mean, when $\hat{N}_{fa}(S)$ is a cumulative distribution function? $\hat{N}_{fa}(S) = \sum\limits_{s=0}^{S}\hat{p}_{fa}(s)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The formula is from this paper (section 4.2): &lt;a href=&quot;http://kusu.comp.nus.edu.sg/proceedings/mm09/mm/p581.pdf&quot; rel=&quot;nofollow&quot;&gt;Logo Retrieval with A Contrario Visual Query Expansion&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="6857" LastEditorUserId="6857" LastEditorDisplayName="" LastEditDate="2011-02-10T13:29:04.877" LastActivityDate="2011-02-10T14:07:20.217" Title="What does this Notation mean: N.&quot;some cumulative distribution function&quot;?" Tags="&lt;notation&gt;&lt;probability-distributions&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="21344" PostTypeId="1" CreationDate="2011-02-10T10:01:52.923" Score="11" ViewCount="550" Body="&lt;p&gt;Now I'm not a graduate student, but I was hoping to compile a list of what may be considered definitive texts in various branches of mathematics. I'm curious to what books are considered good and useful to people who have a decent level of mathematical maturity (like say a grad student), in hopes that I may understand some of these such books within the next 5-10 years or so. Feel free to add topics that I have forgotten. I'm open to multiple books per subject, too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've gone through all questions tagged as &quot;books&quot; and found a few:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Algebra: (1) Algebra by S. Lang; &#xA;         (2) Algebra by T. W. Hungerford&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Category Theory: Categories for the Working mathematician by S. Mac Lane&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Commutative Algebra:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Homological Algebra:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Representation Theory: Representation Theory: A First Course by W. Fulton / J. Harris&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Linear Algebra:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Real Analysis: (1) Principles of Mathematical Analysis, 3rd. by W. Rudin; &#xA;               (2) Real and Complex Analysis by W. Rudin&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Complex Analysis: Complex Analysis by L. Ahlfors&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Functional Analysis: Functional Analysis by W. Rudin&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Measure Theory:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;General Topology: Topology, 2nd. by J. Munkres&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Differential Geometry: [Reference] Foundations of Differential Geometry by S. Kobayashi / &#xA;                                   K. Nomizu (2 vols.);&#xA;                       Fundamentals of Differential Geometry by S. Lang&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Algebraic Geometry: Algebraic Geometry by R. Hartshorne&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Algebraic Topology: Differential Forms in Algebraic Topology by R. Bott / L. Tu&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Geometric Topology:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Knot Theory:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Combinatorics: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Graph Theory:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Logic:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Set Theory: Naive Set Theory by P. Halmos&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Number Theory: A Classical Introduction to Modern Number Theory by K. Ireland / M. Rosen&lt;/p&gt;&#xA;&#xA;&lt;p&gt;General / Companion(s): All the Mathematics You Missed: But Need to Know for Graduate &#xA;                            School by T. Garrity / L. Pedersen&#xA;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="6742" LastEditorUserId="6824" LastEditorDisplayName="" LastEditDate="2011-02-10T11:58:31.677" LastActivityDate="2011-02-10T17:00:55.117" CommunityOwnedDate="2011-02-10T11:59:10.753" Title="What should be in every grad student's library?" Tags="&lt;big-list&gt;&lt;books&gt;" AnswerCount="9" CommentCount="5" FavoriteCount="10" />
  <row Id="21345" PostTypeId="2" ParentId="21330" CreationDate="2011-02-10T10:42:50.920" Score="1" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Liouville_number&quot; rel=&quot;nofollow&quot;&gt;Liouville's theorem&lt;/a&gt; implies that if your constant is irrational then it is in fact transcendental. Unfortunately, the &lt;a href=&quot;http://en.wikipedia.org/wiki/Proof_that_e_is_irrational&quot; rel=&quot;nofollow&quot;&gt;trivial way&lt;/a&gt; of proving irrationality doesn't work. On second thought, the same problem also prevents us from applying Liouville's theorem!&lt;/p&gt;&#xA;" OwnerUserId="1277" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2011-02-11T04:44:41.303" LastActivityDate="2011-02-11T04:44:41.303" CommentCount="3" />
  <row Id="21346" PostTypeId="2" ParentId="21304" CreationDate="2011-02-10T10:51:38.603" Score="4" ViewCount="" Body="&lt;p&gt;You can always do $\mathbb{Z}/2$ Morse homology, because then your manifold is guaranteed to be orientable with respect to your coefficient system, and so it's got to be that $H_0(M;\mathbb{Z}/2)=\mathbb{Z}/2$ and $H_n(M;\mathbb{Z}/2)=\mathbb{Z}/2$.  So not only do we know we have critical points of index $0$ and $n$, but we know that they actually have to descend to generators in homology.  This means that all differentials in our Morse complex will be zero, so generators in the complex are the same as generators for homology.  This proves that the last guy has to be in dimension $n/2$.*  As you point out, this means that $M$ is homeomorphic to $S^{n/2}$ with an $n$-cell attached.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;*&lt;/em&gt; Just to put what I said in the comments right here in the answer: This last critical point has some index $k\in [0,n]$, and it becomes a generator of $H_k(X)$ (with coefficients, if you'd like).  But it's trivial to see that when $f$ is a Morse function then $-f$ is a Morse function too, and that this critical point of index $k$ will become a critical point of index $n-k$ for $-f$.  And $-f$ equally well satisfies what I just said, i.e. all its critical points become homology generators too.  So if you believe that (Morse) homology is a topological invariant, then it must be true that $n-k=k$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; As Jason explains in answer to &lt;a href=&quot;http://math.stackexchange.com/questions/21488/when-can-we-recover-a-manifold-when-we-attach-a-2n-cell-to-sn&quot;&gt;my related question&lt;/a&gt;, the only manifolds this question can even apply to are homotopy equivalent to $\mathbb{R}P^2$, $\mathbb{C}P^2$, $\mathbb{H}P^2$, and $\mathbb{O}P^2$!  Crazy.&lt;/p&gt;&#xA;" OwnerUserId="2159" LastEditorUserId="2159" LastEditorDisplayName="" LastEditDate="2011-02-20T05:32:19.350" LastActivityDate="2011-02-20T05:32:19.350" CommentCount="12" />
  <row Id="21347" PostTypeId="2" ParentId="21292" CreationDate="2011-02-10T11:01:03.413" Score="1" ViewCount="" Body="&lt;p&gt;You can do a slight change to make your argument work and give you an affine cover of $X\times_Y Y&amp;#39;$ as follows.&#xA;Call $h$ the base change morphism $Y&amp;#39;\to Y$. Consider a cover of $Y$ by open affines $V_i=Spec(A_i)$ and cover any $h^{-1}(V_i)$ by open affines $V_{ij}=Spec(A_{ij})$ of $Y&amp;#39;$. The preimage of $V_{ij}$ by $g$ is as you said $X\times_Y V_{ij}$ which by the properties of the fibered product coincides with $X_i\times_{V_i}V_{ij}$, where $X_i=f^{-1}(V_i)$ (this is crucial to end up with affine schemes as you will see in a moment). You can cover $X_i$ with open affines $X_{ik}=Spec(B_{ik})$ with all $B_{ik}$ finite type $A_i$-algebras. So $g^{-1}(V_{ij})$ is covered by the affines $X_{ik}\times_{V_i}V_{ij}=Spec(B_{ik}\otimes_{A_i}A_{ij})$, which are $A_{ij}$-algebras of finite type.&#xA;This proves that the base change of $f$ is locally of finite type (actually we didn't use that $f$ is quasi compact, so we proved that locally of finite type morphisms are stable under base change).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If now $f$ is quasi compact, you just need a finite number of $X_{ik}$ to cover $X_{i}$ and so a finite number of $X_{ik}\times_{V_i}V_{ij}$ is enough to cover $g^{-1}(V_{ij})$, proving the quasi-compactness of $g$.&lt;/p&gt;&#xA;" OwnerUserId="6314" LastActivityDate="2011-02-10T11:01:03.413" CommentCount="2" />
  <row Id="21348" PostTypeId="1" AcceptedAnswerId="21507" CreationDate="2011-02-10T11:21:49.977" Score="1" ViewCount="185" Body="&lt;p&gt;Consider an alchemist that has many ($N$) sorts of ingredients in his possession. Initial amounts of each of the ingredients is expressed by vector $C^0=(C_1, C_2, \dots, C_N)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alchemist knows several ($M$) recipes of ingredient transmutation, expressed as a set of recipes: $R=\{ R^1, R^2, \dots, R^M\}$. Each recipe $R^i$ is a vector that describes the reagents and the products: $R^i=(R^i_1, R^i_2, \dots, R^i_N)$ such that if $R^i_j, i \in [1 \dots M], j \in [1 \dots N]$ is zero, that means $j&amp;#39;th$ ingredient is not used in mutation, if it is positive, than the ingredient appears in that quantity as a product of a mutation, and if it is negative, than it's quantity is consumed in mutation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, a single mutation can be expressed as a vector sum: $C^1=C^0+R^i$, where $C^0$ are ingredients before mutation, $C^1$ - after mutation, $R^i$ - mutation recipe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider that we have a market where ingredients are traded. Market prices are fixed and are described by a value vector $v=(v_1, v_2, \dots, v_N)$. Thus, a value of the alchemist's supplies on the $k$-th step can be expressed as a dot product: $V^k=(C^k \cdot v)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Question: having the initial supply of ingredients $C^0$, book of recipes $R$ and market prices $v$, how can the alchemist derive such sequence of $L$ mutations $S=(S^0, S^1, S^2, \dots, S^L), \forall t : S^t \in R$ that the price $V^L=(C^L \cdot v)$ of the final set of products $C^L=C^0+S^1+S^2+\dots + S^L$  would be maximal?&lt;/p&gt;&#xA;" OwnerUserId="3806" LastEditorUserId="6622" LastEditorDisplayName="" LastEditDate="2011-02-10T13:49:09.350" LastActivityDate="2011-03-19T20:00:55.347" Title="Alchemist's problem" Tags="&lt;discrete-mathematics&gt;&lt;optimization&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="21349" PostTypeId="2" ParentId="21324" CreationDate="2011-02-10T11:37:49.213" Score="2" ViewCount="" Body="&lt;p&gt;A similar paper to Yuval's answer, which is also quite elementary, is Doyle (and Conway)'s &lt;a href=&quot;http://www.math.dartmouth.edu/~doyle/docs/three/three.pdf&quot; rel=&quot;nofollow&quot;&gt;Division by Three&lt;/a&gt;. I don't think it requires any advanced knowledge, and is rather interesting.  &lt;/p&gt;&#xA;" OwnerUserId="2941" LastActivityDate="2011-02-10T11:37:49.213" CommunityOwnedDate="2011-02-10T11:37:49.213" />
  <row Id="21350" PostTypeId="2" ParentId="21339" CreationDate="2011-02-10T11:44:33.737" Score="2" ViewCount="" Body="&lt;p&gt;First, the convergence of $\frac{1}{{\frac{1}{n}\sum\nolimits_{i = 1}^n {\log X_i } }}$ is towards $\vartheta  - 1$, not $\frac{1}{{\vartheta  - 1}}$, of course. Second, as Yuval indicated, it suffices that the first moment (expectation) exists. This guarantees that the average converges to the expectation almost surely, hence also in probability. (An assumption of finite variance is often used for convenience, but is not necessary.) Finally, it is relevant to note that continuous functions are limit-preserving even if their arguments are sequences of random variables, see &lt;a href=&quot;http://en.wikipedia.org/wiki/Continuous_mapping_theorem&quot; rel=&quot;nofollow&quot;&gt;continuous mapping theorem&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2810" LastEditorUserId="2810" LastEditorDisplayName="" LastEditDate="2011-02-10T11:49:45.810" LastActivityDate="2011-02-10T11:49:45.810" />
  <row Id="21351" PostTypeId="1" AcceptedAnswerId="21352" CreationDate="2011-02-10T12:07:35.897" Score="2" ViewCount="103" Body="&lt;p&gt;E.g. 1/8=0.125 has three decimals when written out in base 10, but what is a good example of a simple fraction where the decimal sequence terminates but is very large?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there some sort of rule which determines how many decimals the terminating exact decimal expansion can have based on the amount of digits in the numerator and denominator?&lt;/p&gt;&#xA;" OwnerUserId="6861" LastEditorUserId="6861" LastEditorDisplayName="" LastEditDate="2011-02-11T12:56:40.280" LastActivityDate="2011-02-11T12:56:40.280" Title="Upper bound/exact length of decimal expansion of simple fraction" Tags="&lt;fractions&gt;&lt;decimal-expansion&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="21352" PostTypeId="2" ParentId="21351" CreationDate="2011-02-10T12:15:00.247" Score="8" ViewCount="" Body="&lt;p&gt;Given a fraction $p/q$, first get it into its lowest terms (so that $p$ and $q$ have no common factor). Then, if $q$ is of the form $2^a5^b$ for integers $a,b$, its decimal expansion has max$(a,b)$ digits after the decimal point. If it's not of this form, its decimal expansion is non-terminating (but repeating).&lt;/p&gt;&#xA;" OwnerUserId="1508" LastActivityDate="2011-02-10T12:15:00.247" CommentCount="4" />
  <row Id="21353" PostTypeId="2" ParentId="18505" CreationDate="2011-02-10T12:27:18.297" Score="3" ViewCount="" Body="&lt;p&gt;As described below, any $n$-gon is a &quot;sum&quot; of regular $\left\lbrace\frac{n}{k}\right\rbrace$-gons, with $k = 0, 1, 2, \cdots, n-1$. This can give rise to a vector of complex numbers that serves to characterize the shape of the polygon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given polygon $P$, we start by choosing a &quot;feature point&quot; in the form of a distinguished starting vertex, $v_0$, as well as a preferred tracing direction --in the universe of convex polygons, we can unambiguously take that direction as &quot;always counter-clockwise&quot;-- to get a list of successive vertices $v_0, v_1, \dots, v_{n-1}$. Write $[P]$ for the vector whose $j$-th element is the point of the Complex Plane at which the $j$-th vertex of $P$ lies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Define the &quot;standard regular $\left\lbrace\frac{n}{k}\right\rbrace$-gon&quot;, $P_k$, as the polygon whose $j$-th vertex coincides with the complex number  $\exp\frac{2\pi i j k}{n}$. (As shapes, $P_k$ and $P_{n-k}$ (for $k \ne 0$) are identical, but they are traced in opposing directions.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, any $n$-gon is the &quot;sum&quot; of rotated-and-scaled images of the $P_k$s, in the sense that we can write&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$[P] = r_0 [P_0] + r_1 [P_1] + \cdots + r_{n-1} [P_{n-1}]$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with each complex $r_j$ effecting the corresponding rotation-and-scale. (Determine the $r_j$s by reading the above as $n$ component-wise equations. The solution is, clearly, unique.) Therefore, the vector $R(P) := (r_0, r_1, \dots, r_{n-1} )$ exactly encodes the polygon as a figure in the plane. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that, for $k &amp;gt; 0$, polygon $P_k$ is centered at the origin, while all the vertices of polygon $P_0$ coincide at the complex number $1$. Consequently, the $P_0$ component of the decomposition amounts to a translational element, identifying the centroid (average of vertex-points) of the figure. As we are concerned about &lt;em&gt;shape&lt;/em&gt; without regard for &lt;em&gt;position&lt;/em&gt;, we can suppress (or just ignore) the $r_0$ component of $R(P)$. Since a polygon's shape is independent of the figure's rotational orientation, we choose to normalize $R(P)$ by rotating the elements through an angle that would align $v_0$ with the positive-real axis, arriving at our $C(P)$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$C(P) := \frac{1}{\exp(i\arg{v_0})} R(P) = \frac{|v_0|}{v_0} (r_1,r_2,\dots,r_{n-1})$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If polygons $P$ and $Q$ are congruent (with compatible distinguished vertices and tracing directions), then we have $C(P) = C(Q)$. When $P$ and $Q$ are nearly-congruent, $|C(P)-C(Q)|$ will be small, and vice-versa.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: When $P$ and $Q$ are similar  (with compatible distinguished vertices and tracing directions), we have $\frac{C(P)}{|C(P)|} = \frac{C(Q)}{|C(Q)|}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Edit&lt;/b&gt;&#xA;As noted in comments, this $C(P)$ isn't invariant under cyclic permutations of the vertices. It's worth investigating exactly what effect a cyclic permutation has.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the triangle $P$ with $[P] = ( v_0, v_1, v_2 )$. The corresponding regular $P_k$ figures are given by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$[P_0] := ( 1, 1, 1 )$$&#xA;$$[P_1] := ( 1, w, w^2 )$$&#xA;$$[P_2] := ( 1, w^2, w )$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $w = \exp\frac{2\pi i}{3}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can easily solve the decomposition equation to get &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$R(P) = (r_0, r_1, r_2) = \frac{1}{3} \left( v_0+v_1+v_2 \;,\; v_0 + v_1 w^2 + v_2 w \;,\; v_0 + v_1 w + v_2 w^2 \right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $P&amp;#39;$ is identical to $P$, but with cyclically re-ordered vertices, $[P&amp;#39;] = ( v_1, v_2, v_0 )$, then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$R(P&amp;#39;) =  \frac{1}{3} \left( v_1+v_2+v_0 \;,\; v_1 + v_2 w^2 + v_0 w \;,\; v_1 + v_2 w + v_0 w^2 \right) = ( r_0 \;,\; w r_1 \;,\; w^2 r_2 )$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Observe that $w r_1 [P_1] = r_1 ( w, w^2, 1 )$ yields the &lt;em&gt;same polygon&lt;/em&gt; as $r_1 [P_1] = r_1 ( 1, w, w^2 )$, except that &lt;em&gt;its&lt;/em&gt; vertices have been cyclically re-ordered. Likewise for $w^2 r_2 [P_2]$ (and $w^0 r_0 [P_0]$, for that matter). The same holds for arbitrary $n$-gons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, as a &lt;em&gt;family of polygonal shapes&lt;/em&gt;, the decomposition into regular components is independent of cyclic permutation, as is the correspondence between the vertices of the components and the vertices of the polygon. That is, in our triangles $P$ and $P&amp;#39;$, we have $v_0 = r_0 + r_1 + r_2$, and $v_1 = r_0 + w r_1 + w^2 r_2$, and $v_2 = r_0 + w^2 r_1 + w r_2$, regardless of where each $v_k$ appears in $[P]$ or $[P&amp;#39;]$. Unfortunately, the $R(\cdot)$ vector doesn't suffice to capture this invariance; and $C(\cdot)$'s dependence on the distinguished vertex doesn't help matters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$R(\cdot)$ and $C(\cdot)$ aren't entirely useless, however. The moduli, $|r_k|$, which yield the radii of the regular components, are invariants for the polygons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit 2.&lt;/strong&gt; Perhaps my $C(\cdot)$ provides a workable, &lt;em&gt;comparable&lt;/em&gt;, characterization, after all ... with the caveat that we don't require &lt;em&gt;equality&lt;/em&gt; between $C(P)$ and $C(Q)$ for congruent $P$ and $Q$, but, rather, an appropriate notion of &lt;em&gt;equivalence&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To incorporate the feature point, we'll assume that our polygons are positioned with feature point at origin; the translational component, $P_0$, then becomes significant, so we won't suppress the corresponding element from $C(\cdot)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $r = C(P) = \frac{|u_0|}{u_0}(r_0, r_1, r_2, \dots, r_{n-1})$ and $s = C(Q) = &#xD;&#xA;\frac{|v_0|}{v_0} (s_0, s_1, s_2, \dots, s_{n-1})$ be two $C$-vectors with respect to starting vertices $u_0$ and $v_0$ in polygons $P$ and $Q$, respectively. Define &quot;$r \equiv s$&quot; iff, for all $k$ and some fixed integer $m$, we have $\frac{|v_0|}{v_0} s_k = \frac{|u_0|}{u_0} r_k w^{km}$, where $w = \exp \frac{2 \pi i}{n}$. That is, $|s_k| = |r_k|$, and $\arg(r_k) - \arg(s_k) + 2 \pi k m/n \equiv \arg(u_0) - \arg(v_0) \mod 2 \pi$. (I suspect there's a cleaner way to express this.) Then $P \cong Q$, with compatible feature points, if and only if $C(P) \equiv C(Q)$. (If we don't need feature points, we can position our polygons with their average-of-vertices centroids at the origin and suppress the $0$-th components of $C(\cdot)$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With this, we just need to determine the best way to measure the degree of non-equivalence for incongruent figures.&lt;/p&gt;&#xA;" OwnerUserId="409" LastEditorUserId="409" LastEditorDisplayName="" LastEditDate="2011-02-12T21:09:32.570" LastActivityDate="2011-02-12T21:09:32.570" CommentCount="9" />
  <row Id="21354" PostTypeId="2" ParentId="21344" CreationDate="2011-02-10T12:43:28.743" Score="2" ViewCount="" Body="&lt;p&gt;basic commutative algebra: Atiyah &amp;amp; MacDonald book - Introduction To Commutative Algebra&lt;/p&gt;&#xA;" OwnerUserId="5491" LastActivityDate="2011-02-10T12:43:28.743" CommunityOwnedDate="2011-02-10T12:43:28.743" />
  <row Id="21355" PostTypeId="1" AcceptedAnswerId="21373" CreationDate="2011-02-10T12:49:30.197" Score="1" ViewCount="67" Body="&lt;p&gt;I am trying to work out the formula used for this graph, I don't really know where to start, does anyone have any suggestions? Its a bit unclear, its showing FAT, Mach No, CIT, FWD B/P Position.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.sr-71.org/blackbird/manual/1/1-36.php&quot; rel=&quot;nofollow&quot;&gt;Link to graph&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="6862" LastEditorUserId="312" LastEditorDisplayName="" LastEditDate="2011-02-10T14:56:38.980" LastActivityDate="2011-02-10T15:11:49.107" Title="Calculating the formula for a graph" Tags="&lt;functions&gt;&lt;graph&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="21356" PostTypeId="2" ParentId="21257" CreationDate="2011-02-10T12:56:42.113" Score="4" ViewCount="" Body="&lt;p&gt;Those two final inequalities are known as &quot;interpolation inequalities&quot;. The point being the following: you can &quot;extend&quot; the Holder norms to $\alpha = 0$ with the formal expression &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ [ f ]_{H^0} = \sup_{x\neq y} \frac{|f(x) - f(y)|}{|x-y|^0} = \sup_{x\neq y} \frac{|f(x) - f(y)|}{1} \leq 2 [f]_{sup} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or, in other words, you identify $H^0$ with $C^0$ equipped with the sup norm. As you observed, it gives you that $H^\alpha \subset H^\beta$ if $\alpha &amp;gt; \beta$. What's more, however, is that now, using the sup-norm factor in the interpolation inequality, you can use Arzela-Ascoli to show that the inclusion of $H^\alpha\subset H^\beta$ is pre-compact! That is, any bounded sequence in $H^\alpha$ would have a converging subsequence in $H^\beta$, for $\beta &amp;lt; \alpha$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think you understand how, whenever something allows you to extract a converging subsequence, it is very useful in analysis indeed. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lastly, the expression illustrates a phenomenon that &lt;a href=&quot;http://terrytao.wordpress.com/2009/03/30/245c-notes-1-interpolation-of-lp-spaces/&quot; rel=&quot;nofollow&quot;&gt;happens with regularity in classical analysis&lt;/a&gt;, which is that &lt;em&gt;good&lt;/em&gt; &quot;scales&quot; of function spaces are often log-convex in the exponent. Your family of Holder space norms $H^\alpha$, parametrized by $\alpha$, by your two inequalities, is log-convex.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There presumably are very nice applications of the log convexity in interpolation theory etc for the Holder spaces, but unfortunately none comes to mind immediately at the moment. &lt;/p&gt;&#xA;" OwnerUserId="1543" LastActivityDate="2011-02-10T12:56:42.113" />
  <row Id="21357" PostTypeId="2" ParentId="21344" CreationDate="2011-02-10T13:10:36.640" Score="0" ViewCount="" Body="&lt;p&gt;linear algebra: Greub - Linear Algebra&lt;/p&gt;&#xA;" OwnerUserId="370" LastActivityDate="2011-02-10T13:10:36.640" CommunityOwnedDate="2011-02-10T13:10:36.640" />
  <row Id="21358" PostTypeId="2" ParentId="21344" CreationDate="2011-02-10T13:26:38.267" Score="4" ViewCount="" Body="&lt;p&gt;Jack Lee, Introduction to Smooth Manifolds.&lt;/p&gt;&#xA;" OwnerUserId="9" LastActivityDate="2011-02-10T13:26:38.267" CommunityOwnedDate="2011-02-10T13:26:38.267" CommentCount="2" />
  <row Id="21359" PostTypeId="1" CreationDate="2011-02-10T13:39:42.203" Score="-1" ViewCount="33" Body="&lt;p&gt;I want to verify that 1000 pieces of data are valid within a certain probability.  The way I understand to do this is by doing the following:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Take a random piece of unverified data&lt;/li&gt;&#xA;&lt;li&gt;if it isn't valid make it valid&lt;/li&gt;&#xA;&lt;li&gt;mark it verified&lt;/li&gt;&#xA;&lt;li&gt;Repeat&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;In the first repetition i verify that 1/1000.  The next repetition I have verified that 1/999. and so on. If i keep adding these probabilities then i should have the probability that my data is valid. &#xA;$f(n) = \sum_{1}^{k}1/(1000-n).$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But this can't be right because after 633rd test it becomes larger than one.  Any help would be greatly appreciated.  Thank you&lt;/p&gt;&#xA;" OwnerUserId="6866" LastActivityDate="2011-02-10T13:56:03.057" Title="Probability that sampled data is valid" Tags="&lt;probability&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="21360" PostTypeId="2" ParentId="21348" CreationDate="2011-02-10T13:52:11.923" Score="0" ViewCount="" Body="&lt;p&gt;This is precisely the standard &lt;a href=&quot;http://en.wikipedia.org/wiki/Linear_programming&quot; rel=&quot;nofollow&quot;&gt;linear programming&lt;/a&gt; problem.  Any text will discuss it in detail.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-02-10T13:52:11.923" CommentCount="6" />
  <row Id="21361" PostTypeId="1" CreationDate="2011-02-10T13:54:38.367" Score="11" ViewCount="265" Body="&lt;p&gt;I found this explanation in a journal paper but I could not understand it. Can someone give me an explanation or possibly a proof that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If&#xA;$$\frac{\mathrm{d}V(t)}{\mathrm{d}t}=\sqrt{2}\sum_{h=1}^{H}h\omega V_{h}\cos\left(h\omega t+\frac{\pi }{2}\right),$$&#xA;then why integration over whole period is:&#xA;$$\frac{1}{T}\int_{0}^{T}  \left( \frac{\mathrm{d} V(t)}{\mathrm{d} t}  \right)^{2}dt=\omega \sum_{h=1}^{H}h^{2}V_{h}^{2}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have problem with the power of $\omega$; my solution returns $\omega^2$, while the power of $\omega$ in answer is one. Here is my solution:&#xA;$$\frac{1}{T}\int_{0}^{T}\ \left( \frac{dV}{dt} \right)^{2}dt=\frac{2\omega ^{2}}{T}\int_{0}^{T}\sum_{h=1}^{H}h^{2}V_{h}^{2}\sin^{2}(h\omega t)dt$$&#xA;and over whole period:&#xA;$$\frac{1}{T}\int_{0}^{T}\sin^{2}(h\omega t)dt=\frac{1}{2}$$&#xA;then we will have &#xA;$$\omega ^{2}\sum h^{2}V_{h}^{2} $$&#xA;not&lt;br&gt;&#xA;$$\omega \sum h^{2}V_{h}^{2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why?&lt;/p&gt;&#xA;" OwnerUserId="6856" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2011-02-11T12:16:41.127" LastActivityDate="2011-02-11T12:16:41.127" Title="integration of a function" Tags="&lt;calculus&gt;&lt;integral&gt;" AnswerCount="2" CommentCount="13" FavoriteCount="2" />
  <row Id="21362" PostTypeId="2" ParentId="21359" CreationDate="2011-02-10T13:56:03.057" Score="1" ViewCount="" Body="&lt;p&gt;How do you verify the data is valid?  Then what do you mean &quot;verify that 1/1000&quot;?  If you are keeping track how much of the data is verified, when you do the second one you have now verified 2/1000, not 1/1000 + 1/999, as the first point is still in the universe.  This will not exceed 1.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-02-10T13:56:03.057" />
  <row Id="21363" PostTypeId="1" AcceptedAnswerId="21446" CreationDate="2011-02-10T13:57:32.700" Score="2" ViewCount="97" Body="&lt;p&gt;I have two urns, urn $1$ contains $k$ white balls and $l$ red balls, urn $2$ contains $n - k$ white balls and $n -l$ red balls. I have the total probability &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ P($ &quot;white ball is drawn&quot; $) = \frac{k}{k + l} \frac{1}{2}&amp;#160;+ \frac{n - k}{2n - (k + l)}&amp;#160;\frac{1}{2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and I would like to find $k$ and $l$ such that this probability is maximal. How can I do that? I tried to set the derivative of $P$ with respect to $k$ zero but this gets awfully complicated. Is there a better way to do this? Many thanks for your help!&lt;/p&gt;&#xA;" OwnerUserId="5798" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2011-02-10T14:10:09.957" LastActivityDate="2011-02-10T22:15:36.613" Title="Maximise conditional probability of urns" Tags="&lt;probability&gt;&lt;puzzle&gt;" AnswerCount="3" />
  <row Id="21364" PostTypeId="2" ParentId="21343" CreationDate="2011-02-10T14:07:20.217" Score="1" ViewCount="" Body="&lt;p&gt;I'm pretty sure $N$ in the context means the size of the set $\Omega$, the total number of images. See beginning of section 3.1 of the linked paper. And in $N.\hat{N}_{fa}(S)$, the $.$ should just be multiplication. So the entire term should be interpreted as a distribution of the number of false alarms (total number times the cumulative distribution). &lt;/p&gt;&#xA;" OwnerUserId="1543" LastActivityDate="2011-02-10T14:07:20.217" CommunityOwnedDate="2011-02-10T14:07:20.217" CommentCount="2" />
  <row Id="21365" PostTypeId="2" ParentId="21363" CreationDate="2011-02-10T14:09:53.873" Score="1" ViewCount="" Body="&lt;p&gt;Hint: First try with the same number of black and white balls, as it will lead you to the solution in the more general case.  Note that if you have the same number of balls in each urn, the probability is 1/2 regardless of the distribution.  So to get the probability away from 1/2 it needs to be unbalanced.  Also if each urn is equally split, the probability will be 1/2 regardless of how many balls are in each urn.  So imagine transferring a black ball from one urn to the other and see what it does to the probability.  There is a solution that is easy to convince yourself is optimal once you see it.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-02-10T14:09:53.873" CommentCount="1" />
  <row Id="21366" PostTypeId="1" AcceptedAnswerId="22768" CreationDate="2011-02-10T14:15:29.267" Score="7" ViewCount="140" Body="&lt;p&gt;I am reading Max Karoubi's &quot;K-Theory&quot; and I think I'm overlooking some trivial fact.  We have a vector bundle $E\rightarrow X$ and a morphism $p:E\rightarrow E$ with $p^2=p$.  He is showing that $\ker p$ is locally trivial.  First he assumes that $E=X\times V$ for a vector space $V$.  Here's where I'm stuck:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;He defines $f:X\longrightarrow \operatorname{End}(V)$ by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;f(x)=1-p_{x_0}-p_x+2p_xp_{x_0},&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $p_x$ is the restriction of $p$ to the fiber over $x$ and $x_0$ is a basepoint.  The claim is that $p_{x_0}\circ f(x)=f(x)\circ p_x$.  When I compute both sides I get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;2p_{x_0}p_xp_{x_0}-p_{x_0}p_x=2p_{x}p_{x_0}p_{x}-p_{x_0}p_x&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which says&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;2p_{x_0}p_xp_{x_0}=2p_{x}p_{x_0}p_{x}.&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is that true?  Thanks.&lt;/p&gt;&#xA;" OwnerUserId="2838" LastEditorUserId="1543" LastEditorDisplayName="" LastEditDate="2011-04-23T11:45:20.480" LastActivityDate="2011-04-23T11:45:20.480" Title="Why Are These Two Morphisms the Same?" Tags="&lt;algebraic-topology&gt;&lt;topological-k-theory&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="21367" PostTypeId="1" CreationDate="2011-02-10T14:20:15.707" Score="2" ViewCount="83" Body="&lt;p&gt;What do 1, 2, 3 represent in u(1)+su(2)+su(3)?&#xA;if they are dimensions, how they can be added? or plus has another meaning?&lt;/p&gt;&#xA;" OwnerUserId="6867" LastActivityDate="2011-02-10T22:18:06.440" Title="What do 1, 2, 3 represent in u(1)+su(2)+su(3)?" Tags="&lt;lie-groups&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="21368" PostTypeId="2" ParentId="21363" CreationDate="2011-02-10T14:28:31.750" Score="1" ViewCount="" Body="&lt;p&gt;Hint: change variables. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Call $b_1 = l_1 + k_1 $ the balls in urn 1, $b_2 = l_2 + k_2 $ the balls in urn 2.&#xA;We may assume  $l_1 , l_2$ are fixed, and that $b_1,b_2$ are our variables, subject to the restriction $b_1 + b_2 = 2n$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You want then to maximize&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle \frac{b_1-l_1}{b_1}\frac{1}{2} +\frac{b_2-l_2}{b_2}\frac{1}{2}$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is the same that minimize  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle g(b_1,b_2) = \frac{l_1}{b_1} + \frac{l_2}{b_2}$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(with the restriction $b_1 + b_2 = 2n$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From this, you can continue (Lagrange multipliers or whatever).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;FYI, I get (assuming real values) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle \frac{b_1^2}{l_1} = \frac{b_2^2}{l_2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which leds to &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle k_1 = \frac{2n}{1+\sqrt{l_2/l_1}}-l_1$&lt;/p&gt;&#xA;" OwnerUserId="312" LastEditorUserId="312" LastEditorDisplayName="" LastEditDate="2011-02-10T15:55:33.267" LastActivityDate="2011-02-10T15:55:33.267" CommentCount="2" />
  <row Id="21369" PostTypeId="2" ParentId="21344" CreationDate="2011-02-10T14:42:51.770" Score="1" ViewCount="" Body="&lt;p&gt;graph theory - Doug West, Introduction to Graph Theory&lt;/p&gt;&#xA;" OwnerUserId="1919" LastActivityDate="2011-02-10T14:42:51.770" CommunityOwnedDate="2011-02-10T14:42:51.770" />
  <row Id="21370" PostTypeId="2" ParentId="21344" CreationDate="2011-02-10T14:44:11.797" Score="6" ViewCount="" Body="&lt;p&gt;There is no such thing as a book which should be in every grad student's library. I know several excellent Ph.D. students who only own a handful of books related to their own field and do just fine. As Gunnar Magnusson pointed out, all the books you listed will anyway be available at the library of any mathematics department.&lt;/p&gt;&#xA;" OwnerUserId="677" LastActivityDate="2011-02-10T14:44:11.797" CommunityOwnedDate="2011-02-10T14:44:11.797" />
  <row Id="21371" PostTypeId="2" ParentId="21367" CreationDate="2011-02-10T14:44:42.453" Score="5" ViewCount="" Body="&lt;p&gt;They indicate the dimension of the complex vector space on which the group acts. The &lt;a href=&quot;http://en.wikipedia.org/wiki/Unitary_group&quot; rel=&quot;nofollow&quot;&gt;unitary group&lt;/a&gt; $\text{U}(n)$ and the &lt;a href=&quot;http://en.wikipedia.org/wiki/Special_unitary_group&quot; rel=&quot;nofollow&quot;&gt;special unitary group&lt;/a&gt; $\text{SU}(n)$ both act on $\mathbb{C}^n$. The &quot;plus&quot; actually indicates the &lt;a href=&quot;http://en.wikipedia.org/wiki/Direct_product&quot; rel=&quot;nofollow&quot;&gt;direct product&lt;/a&gt; $G \times H$ of groups $G, H$, which is the set of pairs $(g, h), g \in G, h \in H$ under pointwise multiplication. This is sometimes conflated with the &lt;a href=&quot;http://en.wikipedia.org/wiki/Direct_sum&quot; rel=&quot;nofollow&quot;&gt;direct sum&lt;/a&gt;, but in my opinion it is better to avoid the sum notation for non-abelian groups.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; Ah, you might be talking about the &lt;a href=&quot;http://en.wikipedia.org/wiki/Lie_algebra&quot; rel=&quot;nofollow&quot;&gt;Lie algebras&lt;/a&gt; $\mathfrak{u}(n), \mathfrak{su}(n)$, in which case the answer is still that $n$ is the dimension of the complex vector space on which the Lie algebra acts, except that the notion of an action is different. &lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2011-02-10T22:18:06.440" LastActivityDate="2011-02-10T22:18:06.440" CommentCount="1" />
  <row Id="21372" PostTypeId="1" CreationDate="2011-02-10T14:50:36.587" Score="3" ViewCount="71" Body="&lt;p&gt;Let $ y =$ min {$(x + 6), (4 – x)$}, then find $y$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How to solve this problem?&lt;/p&gt;&#xA;" OwnerUserId="6868" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2011-02-10T14:56:28.870" LastActivityDate="2011-02-10T17:04:46.733" Title="Let $ y =$ min {$(x + 6), (4 – x)$}, then $y = ? $" Tags="&lt;algebra-precalculus&gt;&lt;algebra&gt;" AnswerCount="3" />
  <row Id="21373" PostTypeId="2" ParentId="21355" CreationDate="2011-02-10T14:53:46.753" Score="0" ViewCount="" Body="&lt;p&gt;First you need to figure out what are the inputs and what are the outputs.  For the top graph, I would guess Mach No and FAT (is that some air temperature?) are the inputs and B/P positions and CIT are the outputs-you find a point in the center grid and read off horizontally and vertically.  This looks like a reference for flight-you look up the flight conditions and it tells you how to set the aircraft controls.  Collect a bunch of data by reading off points-for example at the intersections of the grid.  Each point should be (Mach, FAT, B/P position, CIT).  Then look at the curves and guess a functional form.  You might even want to replot the data as two graphs with Mach and FAT on the axes and each of the other parameters as z and look at what it looks like.  Finally feed your form to a multidimensional fitting routine to find the parameters and see how good the fit is.  If you're not happy, try another form until you are.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If this sounds non-rigorous, that's because it is.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-02-10T14:53:46.753" />
  <row Id="21374" PostTypeId="2" ParentId="21372" CreationDate="2011-02-10T14:55:52.133" Score="2" ViewCount="" Body="&lt;p&gt;Hint:  if $x$ is quite large and positive, which of the two choices will be smaller?  If $x$ is large and negative, which will be smaller?  Can you find the crossover point?&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-02-10T14:55:52.133" />
  <row Id="21375" PostTypeId="1" CreationDate="2011-02-10T15:04:51.283" Score="2" ViewCount="233" Body="&lt;p&gt;Which sum shows up most often when you roll 10 dice?&lt;/p&gt;&#xA;" OwnerUserId="6785" LastEditorUserId="367" LastEditorDisplayName="" LastEditDate="2011-02-10T16:12:04.330" LastActivityDate="2011-03-26T17:37:03.967" Title="Which sum shows up most often when you roll 10 dice?" Tags="&lt;probability&gt;" AnswerCount="3" CommentCount="3" />
  <row Id="21376" PostTypeId="2" ParentId="21355" CreationDate="2011-02-10T15:11:49.107" Score="0" ViewCount="" Body="&lt;p&gt;You could try plotting each of the paths as a separate graph and work out the formulas for each one, then try working out a formula that links the separate graphs together.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or read all the data off and break it down into a spreadsheet so the data can be analyzed more easily. Then you can re-plot the data in different ways and see how that works out.&lt;/p&gt;&#xA;" OwnerUserId="6864" LastActivityDate="2011-02-10T15:11:49.107" />
  <row Id="21377" PostTypeId="1" AcceptedAnswerId="21422" CreationDate="2011-02-10T15:19:53.330" Score="3" ViewCount="201" Body="&lt;p&gt;Hey guys,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I arrived at something during my maths ponderings which is really exciting for me. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is clearly stated in the book on Riemann Hypothesis by Borwein that the convergence of $\sum_{n=1}^{\infty} \frac{\mu(n)}{n^s}$ for $\Re(s) &amp;gt; 1/2$ is necessary and sufficient for RH. This is ofcourse valid since, $\sum \mu(n)/n^s = 1/\zeta(s)$ for $\Re(s) &amp;gt; 1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Having said that, I have reached a point where I got, for $\Re(s) &amp;gt; 1/2$&#xA;$$ \left| \frac{\eta(s)}{s} \sum_{n=1}^{\infty} \frac{\mu(n)}{n^s} - \frac{1-2^{1-s}}{s} \right| &amp;lt; \infty $$&#xA;$\eta(s)$ is the Dirichlet eta function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What do I interpret out of this formula. Does this result imply RH, or falls short of it?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I believe the second option might be more correct, because this result does not say anything about the zeros of the eta function. But at least it is clear that if $\sum \mu(n)/n^s$ blows up then $\eta(s)$ also must have a zero to bring it down.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any elaborated answer will be highly appreciated, because it is a current work in progress. :)&lt;/p&gt;&#xA;" OwnerUserId="2320" LastEditorUserId="2320" LastEditorDisplayName="" LastEditDate="2011-02-10T15:25:52.513" LastActivityDate="2011-02-10T20:39:22.040" Title="On the convergence of $\sum \mu(n)/n^s$" Tags="&lt;analytic-number-theory&gt;" AnswerCount="1" />
  <row Id="21378" PostTypeId="1" CreationDate="2011-02-10T15:31:46.283" Score="2" ViewCount="29" Body="&lt;p&gt;First I am a new user of this site.&#xA;Second my math background is very limited, although I do have a lot of experience in applied statistics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Component or piece part failures on high value parts($1000 to $1,000,000 each). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many of the theoretical statistical forecasting functions I use are based on the Poisson, exponential (single, double, and triple), moving average,weibull,and others.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many actual failure distributions fit these theoretical distributions fairly well. Others don't. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some have multiple series of zero failures per period, others don't.  Many times the mean variance ratio is large i.e. the standard deviation is many multiples of the theoretical distribution mean.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am looking for a practical tool or set of tools that can be used to derive functions from actual failure distributions if that is possible.  Practical means fairly easy to program, preferably 1 or two parameter.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any help is appreciated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gary&lt;/p&gt;&#xA;" OwnerUserId="6870" LastActivityDate="2011-02-10T15:31:46.283" Title="Deriving functions for empiracal distributions -very applied mathatics" Tags="&lt;probability&gt;&lt;functions&gt;&lt;special-functions&gt;&lt;mathematical-modeling&gt;" CommentCount="1" />
  <row Id="21379" PostTypeId="1" AcceptedAnswerId="21391" CreationDate="2011-02-10T15:32:53.463" Score="3" ViewCount="188" Body="&lt;p&gt;The symbols $\forall$ and $\exists$ denote &quot;for all&quot; and &quot;there exists&quot; quantifiers. In some papers, I saw the (not so common) quantifiers $Я$ and $\exists^+$, denoting &quot;for a randomly chosen element of&quot; and &quot;for most elements in&quot;, respectively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there other symbols for quantifiers?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm specially interested in quantifiers for:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;for all but finitely many elements of...&lt;/li&gt;&#xA;&lt;li&gt;for infinitely many elements of...&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; After seeing some of the comments, I found the &lt;a href=&quot;http://en.wikipedia.org/wiki/List_of_logic_symbols&quot; rel=&quot;nofollow&quot;&gt;list of logic symbols&lt;/a&gt; and the &lt;a href=&quot;http://en.wikipedia.org/wiki/Table_of_mathematical_symbols&quot; rel=&quot;nofollow&quot;&gt;table of mathematical symbols&lt;/a&gt;, which I could be useful for others.&lt;/p&gt;&#xA;" OwnerUserId="2420" LastEditorUserId="2420" LastEditorDisplayName="" LastEditDate="2011-02-10T18:46:05.303" LastActivityDate="2011-02-10T19:22:38.703" Title="Symbols for Quantifiers Other Than $\forall$ and $\exists$" Tags="&lt;quantifiers&gt;" AnswerCount="2" CommentCount="6" />
  <row Id="21380" PostTypeId="1" CreationDate="2011-02-10T15:44:43.637" Score="3" ViewCount="115" Body="&lt;p&gt;Hey guys and gals,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've been suggested to this site by some nice people at mathoverflow.net&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Before I get started, let me tell you a little about myself. I’m a fourth year Mechanical Engineering student at the University of Michigan, Dearborn Campus. This said, I am not a mathematician, however I do know a bit about math, and I also enjoy doing math.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please excuse any redundancies I may make and my “lack of rigor”.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem I’m working on now is an interesting one, and it goes as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose we have some unknown function of some real variable x, namely f(x), which is “well-behaved” on some known, finite interval, {a,b}. What is meant by “well-behaved” is that it is at least once differentiable, once integrate-able, smooth, and continuous on {a,b}.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose further that we are given some finite area-under-the-curve A, and some finite arc-length S. Suppose finally that we are given f(x=a) and f(x=b).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given the above restraints on f(x), prove that there always exists some unique function f(x) that “fits the bill”.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Say f(x) = Sin(x), a = pi/2, and b = pi.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;f(a) = 1, f(b) = 0, A = 1, S = Sqrt[2]*EllipticE[1/2] ≈ 1.9101.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem in this case would be to prove that f(x) = Sin(x) is the only function that has the known area A, and known arc-length S.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Side Note: f(x) in this case could also be Cos(x - pi/2), however, this would not constitute two unique functions, f(x).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A corollary question could be posed if a proof to the first problem was given, and is as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given the same criterion for f(x) as before, devise a method of determining what f(x) would have to be to “fit the bill”. That is, like in our previous example, given: A = 1, S = 1.9101; devise a method of determining that f(x) must equal Sin(x).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another Side Note: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have realized (with help from mathoverflow.net) that f(x)=(x-1)^2(x+1) and f(-x) have the same Area and arc-length on [-1,1]. Making a potentially erroneous proposition, let us say that for all polynomials of nth order P(x), P(-x) will have the same Area and arc-length on any [-c,c]. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This would prove difficult to prove analytically due to the &quot;ugly&quot; expression for arc-length. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;While in general f(x) and f(-x) are unique functions, there must be some way of thinking about them as similar enough to &quot;lump&quot; them into one unique function... Just a thought &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you all. I look forward to any insight you can shed on the problem.&lt;/p&gt;&#xA;" OwnerUserId="6871" LastEditorUserId="640" LastEditorDisplayName="" LastEditDate="2011-02-10T19:09:33.780" LastActivityDate="2011-02-10T19:09:33.780" Title="Proving that there exists a unique f(x) given Area and arc-length of f(x) on a given interval" Tags="&lt;calculus&gt;&lt;integral&gt;" AnswerCount="3" />
  <row Id="21381" PostTypeId="1" AcceptedAnswerId="21386" CreationDate="2011-02-10T15:54:47.387" Score="5" ViewCount="217" Body="&lt;p&gt;Hi there!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wish to calculate a power like 2.14 ^ 2.14.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I ask my calculator to do it, I just get an answer, but I want to see the calculation.&lt;br&gt;&#xA;So my question is, how to calculate this with a pen, paper and a bunch of brains.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance,&lt;br&gt;&#xA;Mixxiphoid&lt;/p&gt;&#xA;" OwnerUserId="6872" LastEditorUserId="6872" LastEditorDisplayName="" LastEditDate="2011-02-10T16:15:23.563" LastActivityDate="2011-02-10T16:29:01.303" Title="How to calculate a decimal power of a number." Tags="&lt;powers&gt;" AnswerCount="3" FavoriteCount="2" />
  <row Id="21382" PostTypeId="2" ParentId="12181" CreationDate="2011-02-10T16:03:27.240" Score="0" ViewCount="" Body="&lt;p&gt;Follow-up&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Much to my chagrin at the time (happy ending though!), I discovered that restricting Universal Instantiation alone was not enough to avoid problems with dependent variables. By indirect methods you could still use UI to introduce new free variables as follows:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Ax P(x) &amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp; (Premise)&lt;/li&gt;&#xA;&lt;li&gt;~P(y) &amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;  (Premise)&lt;/li&gt;&#xA;&lt;li&gt;P(y) &amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;(UI, 1)&lt;/li&gt;&#xA;&lt;li&gt;P(y) &amp;amp; ~P(y) (Join 2,3)&lt;/li&gt;&#xA;&lt;li&gt;~~P(y) &amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp; (Conclusion, 2)&lt;/li&gt;&#xA;&lt;li&gt;P(y) &amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp; (Remove ~~)  &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I managed to close off this possibility, however, by formalizing what mathematicians habitually seem to do when writing proofs: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you want to prove something is true in general, you introduce an assumption with one or more new free variables in it, say P(x). Then you prove, say Q(x). Then, if Q(x) is not a contradiction, you can then generalize Ax (P(x) =&gt; Q(x)). In this case, you must ensure that no free variables introduced in the interim occur free in Q(x). If Q(x) is a contradiction, then Ax ~P(x).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By implementing this rule in my program, the conclusion on line 5 becomes:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ax ~~P(x) (Conclusion, 2)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My new conclusion rule will automatically generalize on variables introduced in the corresponding premise (line 2 in this case). There is now no other way introduce universal quantifiers. It just doesn't seem to be necessary. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In addition to automatically doing universal generalizations, the new conclusion rule will also automatically do existential generalizations if necessary so that no new free variables show up in the conclusion. See the example at: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.dcproof.com/EA-AE.html&quot; rel=&quot;nofollow&quot;&gt;http://www.dcproof.com/EA-AE.html&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This HTML document was generated using my program. Comments are in a blue font.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My DC Proof program may not serve everyone's need, but if students want to write mathematical proofs in, say, number theory, algebra or analysis, it can serve as an excellent introduction to the rules that mathematicians actually use when writing proofs. In my opinion, there is simply no need to burden such students with esoterica about dependent variables that never seem to come up outside of rarefied philosophical discussions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can download Version 2.0 of my program at my website &lt;a href=&quot;http://www.dcproof.com&quot; rel=&quot;nofollow&quot;&gt;http://www.dcproof.com&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dan&lt;/p&gt;&#xA;" OwnerUserId="3515" LastEditorUserId="3515" LastEditorDisplayName="" LastEditDate="2011-02-10T16:21:11.857" LastActivityDate="2011-02-10T16:21:11.857" />
  <row Id="21383" PostTypeId="2" ParentId="21375" CreationDate="2011-02-10T16:10:18.280" Score="5" ViewCount="" Body="&lt;p&gt;The &lt;em&gt;sum&lt;/em&gt; that shows up most often when you roll 10 dice is &lt;strong&gt;35&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are you also interested to see a reason why this is so?&lt;/p&gt;&#xA;" OwnerUserId="367" LastActivityDate="2011-02-10T16:10:18.280" />
  <row Id="21384" PostTypeId="2" ParentId="21344" CreationDate="2011-02-10T16:10:33.223" Score="8" ViewCount="" Body="&lt;p&gt;Overview of Mathematics: The Princeton Companion to Mathematics, editor Timothy Gowers, Princeton U. Press.&lt;/p&gt;&#xA;" OwnerUserId="1369" LastActivityDate="2011-02-10T16:10:33.223" CommunityOwnedDate="2011-02-10T16:10:33.223" />
  <row Id="21385" PostTypeId="2" ParentId="21344" CreationDate="2011-02-10T16:14:01.540" Score="1" ViewCount="" Body="&lt;p&gt;Two wonderful texts on measure theory are :&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Measure theory by Paul Halmos&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Measure theory by G.Folland&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" LastActivityDate="2011-02-10T16:14:01.540" CommunityOwnedDate="2011-02-10T16:14:01.540" />
  <row Id="21386" PostTypeId="2" ParentId="21381" CreationDate="2011-02-10T16:15:36.883" Score="6" ViewCount="" Body="&lt;p&gt;For positive bases $a$, you have the general rule&#xA;$$a^b = \exp(b\ln(a)) = e^{b\ln a}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This follows from the fact that exponentials and logarithms are inverses of each other, and that the logarithm has the property that&#xA;$$\ln(x^r) = r\ln(x).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you have, for example,&#xA;\begin{align*}&#xA;(2.14)^{2.14} &amp;amp;= e^{\ln\left((2.14)^{2.14}\right)} &amp;amp;\quad&amp;amp;\mbox{(because $e^{\ln x}=x$)}\\&#xA;&amp;amp;= e^{(2.14)\ln(2.14)} &amp;amp;&amp;amp;\mbox{(because $\ln(x^r) = r\ln x$)}&#xA;\end{align*}&#xA;Or more generally,&#xA;$$a^b = e^{\ln(a^b)} = e^{b\ln a}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, this is formula can be taken as the &lt;em&gt;definition&lt;/em&gt; of $a^b$ for $a\gt 0$ and arbitrary exponent $b$ (that is, not an integer, not a rational).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As to computing $e^{2.14\ln(2.14)}$, there are reasonably good methods for approximating numbers like $\ln(2.14)$, and numbers like $e^r$ (e.g., Taylor polynomials or other methods).&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2011-02-10T16:25:12.413" LastActivityDate="2011-02-10T16:25:12.413" CommentCount="2" />
  <row Id="21387" PostTypeId="2" ParentId="21381" CreationDate="2011-02-10T16:15:39.507" Score="4" ViewCount="" Body="&lt;p&gt;You use $\exp(2.14 \ln 2.14)$ or any base for logarithms you choose.  But if you want pen and paper, you can help with the properties of exponents.  $2.14^{2.14}=2.14^2\cdot2^{.14}=2.14^2\exp(.14(\ln 2 + \ln1.07))$ will converge more quickly, especially if you are willing to look up $\ln 2$. &lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-02-10T16:15:39.507" />
  <row Id="21388" PostTypeId="1" AcceptedAnswerId="21392" CreationDate="2011-02-10T16:24:10.230" Score="2" ViewCount="149" Body="&lt;p&gt;You have six triangles. Two are red, two are blue, and two are green. How many truly different hexagons can you make by combining these triangles?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have two possible approachtes to solving this question:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;In general, you can arrange $n$ objects, of which $a$ are of type one, $b$ are of type two, and $c$ are of type three, in $\frac{n!}{a! \cdot b! \cdot c!}$ ways. In this case, $n = 6$, and $a = b = c = 2$. There are 90 possible ways to arrange the six triangles. However, the triangles are in a circle, which means that six different arrangements are really one truly different arrangement. Division by six results in 15 possible hexagons.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;It is possible to enumerate every different arrangement, and count how many truly different arrangements you can make. There are six different ways to arrange the triangles with the two red triangles next to each other. There are six different ways to arrange the triangles with one triangle between two red triangles. There are four different ways to arrange the triangles with two triangles between two red triangles. This results in 16 possible hexagons. I also wrote a simple computer program that tries every possible combination and counts how many are different, and it confirms the answer 16.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;It turns out that the second approach is the right one, and 16 is the right answer. I can enumerate 16 different hexagons that are all different. Now my question is, what is wrong with the first approach? Where is the error?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remarks:&#xA;When you arrange the 16 different hexagons in lines, you can create six different arrangements for each hexagon, but this results in doubles. There are less than 96 different arrangements in a line. This does not contradict the first approach, in which there are no doubles.&lt;/p&gt;&#xA;" OwnerUserId="6873" LastActivityDate="2011-02-15T20:22:57.433" Title="Arrangement of six triangles in a hexagon" Tags="&lt;permutations&gt;&lt;factorial&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="21389" PostTypeId="2" ParentId="21344" CreationDate="2011-02-10T16:24:20.777" Score="2" ViewCount="" Body="&lt;p&gt;When looking at mathematics textbooks, I always find myself &lt;a href=&quot;http://www.ocf.berkeley.edu/~abhishek/chicmath.htm&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;. I often just read it for the poignant comments. The more difficult textbooks can be found at the bottom. It's nice to see comments about the book from people who have worked through it; amazon.com is notorious for its poor reviews.&lt;/p&gt;&#xA;" OwnerUserId="2465" LastActivityDate="2011-02-10T16:24:20.777" CommunityOwnedDate="2011-02-10T16:24:20.777" CommentCount="2" />
  <row Id="21390" PostTypeId="2" ParentId="21381" CreationDate="2011-02-10T16:29:01.303" Score="1" ViewCount="" Body="&lt;p&gt;A decimal power can be seen as a fraction:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x^{\frac{a}{b}} = \sqrt[b]{x^a}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course you cannot write every number as a fraction, but you can at least approximate every number by a fraction.&lt;/p&gt;&#xA;" OwnerUserId="6873" LastActivityDate="2011-02-10T16:29:01.303" CommentCount="1" />
  <row Id="21391" PostTypeId="2" ParentId="21379" CreationDate="2011-02-10T16:34:37.303" Score="5" ViewCount="" Body="&lt;p&gt;The most commonly used symbols to express &quot;for all but finitely many&quot; and &quot;there are infinitely many&quot; are $\forall^\infty$ and $\exists^\infty$, respectively.&lt;/p&gt;&#xA;" OwnerUserId="25" LastActivityDate="2011-02-10T16:34:37.303" CommentCount="3" />
  <row Id="21392" PostTypeId="2" ParentId="21388" CreationDate="2011-02-10T16:36:11.360" Score="2" ViewCount="" Body="&lt;p&gt;The problem in approach 1 is your assumption that every arrangement has 6 distinguishable rotations of that arrangement.  This is not true, because some arrangements are preserved by a one-half rotation of 180 degrees.  (We should also check for any arrangements preserved by a one-third rotation of 120 degrees, but there aren't any of those.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What permutations are preserved by a half-turn?  RGBRGB (identified with GBRGBR and BRGBRG) and RBGRBG (identified with BGRBGR and GRBGRB).  So, among the 90 total permutations, we have 84 that come from 14 possibilities times the 6 rotations and 6 that come from 2 possibilities times the 3 rotations.&lt;/p&gt;&#xA;" OwnerUserId="1461" LastActivityDate="2011-02-10T16:36:11.360" CommentCount="1" />
  <row Id="21393" PostTypeId="1" CreationDate="2011-02-10T16:36:59.193" Score="0" ViewCount="108" Body="&lt;p&gt;If a set has n elements then what are maximum no of equivalence classes and equivalence relations possible on it ??&lt;/p&gt;&#xA;" OwnerUserId="6874" LastActivityDate="2011-02-11T19:42:24.983" Title="Equivalence Relation &amp; Equivalence Classes" Tags="&lt;algebra&gt;" AnswerCount="2" />
  <row Id="21394" PostTypeId="2" ParentId="21372" CreationDate="2011-02-10T16:40:37.423" Score="1" ViewCount="" Body="&lt;p&gt;Try graphing $y=x+6$ and $y=4-x$ together on one graph, then highlight or otherwise mark the parts of those graphs that make up $y=\min{x+6,4-x}$.  The resulting shape should be a familiar type of basic function, perhaps translated and/or reflected.&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2011-02-10T16:40:37.423" />
  <row Id="21395" PostTypeId="2" ParentId="21393" CreationDate="2011-02-10T16:50:23.797" Score="4" ViewCount="" Body="&lt;p&gt;The maximum number of equivalence classes is $n$ -the identity relation $\{ (x,x) \ | \ x \in X \}$ is an equivalence relation.  The number of equivalence relations is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Bell_numbers&quot; rel=&quot;nofollow&quot;&gt;Bell number&lt;/a&gt;.  The series is in &lt;a href=&quot;http://oeis.org/A000110&quot; rel=&quot;nofollow&quot;&gt;A000110 of OEIS&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2011-02-10T17:20:05.723" LastActivityDate="2011-02-10T17:20:05.723" CommentCount="4" />
  <row Id="21396" PostTypeId="2" ParentId="21393" CreationDate="2011-02-10T16:51:19.883" Score="2" ViewCount="" Body="&lt;p&gt;The Stirling numbers of the second kind $S(n,k)$ are by definition the number of ways you can partition some set of $n$ elements into $k$ non-empty and disjoint subsets. But since an equivalence relation may partition your set into any amount of subsets, you need to sum over all possibilities:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{k=1}^n S(n,k)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;gives you the anwser.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can read about them &lt;a href=&quot;http://en.wikipedia.org/wiki/Stirling_numbers_of_the_second_kind&quot; rel=&quot;nofollow&quot;&gt;here.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5105" LastActivityDate="2011-02-10T16:51:19.883" />
  <row Id="21397" PostTypeId="2" ParentId="21308" CreationDate="2011-02-10T16:51:28.023" Score="0" ViewCount="" Body="&lt;p&gt;The arc functions are sort of inverse functions for the trig functions.  &quot;Sort of&quot; because the trig functions are periodic, so their inverse relations are not functions, but if we consider only a smaller part of the trig function, the inverse relation is a function.  This is why the arc functions produce angles in a specific range.  For sine, cosine, secant, and cosecant, the range of the corresponding arc function is the first quadrant angles ($0$ to $\frac\pi 2$) where the original function is positive, plus either the second or the fourth quadrant angles, whichever quadrant the original function is negative.  For tangent and cotangent, the range is the first quadrant plus whichever of the second or fourth quadrants gives a contiguous range (because tangent and cotangent are discontinuous).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, $\arcsin(\frac{\sqrt{2}}{2})$ is asking &quot;For what angle $\theta$ (with $-\frac\pi 2\le\theta\le\frac\pi 2$) is $\sin\theta=\frac{\sqrt{2}}{2}$?&quot;&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2011-02-10T16:51:28.023" />
  <row Id="21398" PostTypeId="2" ParentId="21388" CreationDate="2011-02-10T16:54:46.217" Score="3" ViewCount="" Body="&lt;p&gt;You want to count each collection of hexagons interrelated by rotations as a single &quot;truly different&quot; hexagon.  If $X$ is the set of hexagons, and $G$ is the group of rotations, then you want to find $|X/G|$: the number of orbits of elements of $X$ under $G$.  &lt;a href=&quot;http://en.wikipedia.org/wiki/Burnside%27s_lemma&quot; rel=&quot;nofollow&quot;&gt;Burnside's lemma&lt;/a&gt; gives the equation for this:&#xA;$$&#xD;&#xA;|X/G| = \frac{1}{|G|}\sum_{g \in G}|X^{g}| = \frac{|X|}{|G|} + \frac{1}{|G|}\sum_{g \in G | g \neq e}|X^g|,&#xD;&#xA;$$&#xA;where $X^g$ is the set of elements fixed by $g$, and the second sum excludes the identity element of $G$.  In this case, you correctly counted $|G|=6$ and $|X|=90$, and so if no hexagons were fixed by a non-zero rotation, then $90/6=15$ would be the right answer.  However, there are $6$ hexagons fixed by the $180^{\circ}$ rotation, giving an additional contribution of $6/6=1$, for a total of $15+1=16$.&lt;/p&gt;&#xA;" OwnerUserId="5546" LastActivityDate="2011-02-10T16:54:46.217" />
  <row Id="21399" PostTypeId="2" ParentId="21344" CreationDate="2011-02-10T17:00:55.117" Score="0" ViewCount="" Body="&lt;p&gt;I agree with Gunnar Magnusson's comment, repeated in Dan Petersen's answer, that you don't need all these books. But since you asked:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Knots: Rolfsen. Also a black book by someone with a Japanese name, which I forget.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basic algebraic topology: Hatcher (free online).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also: Ask the math librarian of any university that has one: he'll be able to tell you which books are popular.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also: If (for example) you're in algebra and want a good book on topology, ask any algebraist.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wikipedia serves as a tolerably good resource when you need to look something up which is outside your field but is basic in the field it's in (but doesn't do as well with more specialized things in my experience). Wolfram's MathWorld is also tolerable.&lt;/p&gt;&#xA;" OwnerUserId="4992" LastActivityDate="2011-02-10T17:00:55.117" CommunityOwnedDate="2011-02-10T17:00:55.117" />
  <row Id="21400" PostTypeId="1" CreationDate="2011-02-10T17:00:58.033" Score="2" ViewCount="81" Body="&lt;p&gt;Hey,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Problem: What is the cardinality of all lines $l$ on $\mathbb R^{2}$ which do not contain a point $(x,y)\in l$ where $x, y \in \mathbb Q$ (call it $A$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My solution: I was thinking of using CB theorem for this problem. It's easy to show that the cardinality of all lines in $\mathbb R^{2}$ is $2^{\aleph_0}$, so it's obvious that  $|A|\le 2^{\aleph_0}$, but I'm having trouble of showing that the opposite direction ($|A|\ge 2^{\aleph_0}$). I thought about this injective function ($f:\mathbb R \rightarrow A$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\forall r \in \mathbb R$ &#xA;$f(r)=\left\{\begin{matrix}&#xD;&#xA;(r,0), r \in \mathbb R-\mathbb Q &amp;amp; \\&#xD;&#xA;(g(r),0), r \in \mathbb Q&amp;amp; &#xD;&#xA;\end{matrix}\right.&#xD;&#xA;$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $g(r) = min{(x\in\mathbb R-\mathbb Q, x \lt r)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is that injective correct? Thanks!&lt;/p&gt;&#xA;" OwnerUserId="3637" LastActivityDate="2011-02-10T17:20:11.510" Title="Cardinality of all lines on $\mathbb R^{2}$ which do not contain point $(x,y)\in l$ where $x, y \in \mathbb Q$" Tags="&lt;elementary-set-theory&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="21401" PostTypeId="1" CreationDate="2011-02-10T17:02:31.180" Score="4" ViewCount="61" Body="&lt;p&gt;Hi,&#xA;consider $f(x)=x^4+5x^3+1$. Let $\alpha$ be a root of this polynomial and consider $K=\mathbb{Q}(\alpha)$. Which are the absolute values on $K$ extending the usual absolute values on $\mathbb{Q}$? Are these finitely many? In general, if $K$ is a number field how to find them?&#xA;Thanks&lt;/p&gt;&#xA;" OwnerUserId="6583" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2011-02-10T17:18:34.943" LastActivityDate="2011-02-10T17:18:34.943" Title="find valuations" Tags="&lt;number-theory&gt;&lt;algebraic-geometry&gt;&lt;arithmetic-geometry&gt;" CommentCount="3" FavoriteCount="2" />
  <row Id="21402" PostTypeId="2" ParentId="21372" CreationDate="2011-02-10T17:04:46.733" Score="0" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\ $ For any continuous functions $\rm\:f,g,\:$ the intermediate value theorem implies that $\rm\ f-g\ $ will have constant sign between it's roots. So you need only partition $\mathbb R$ by these  roots and then evaluate the function at any test point of each interval to determine the sign on the whole interval.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact this decomposition technique generalizes to higher dimensions. It is known as the &lt;a href=&quot;http://math.stackexchange.com/questions/13654/problems-with-inequalities/13681#13681&quot;&gt;cyclindrical decomposition algorithm&lt;/a&gt; - an effective interpretation of Tarksi's decision procedure for the first-order theory of the reals.&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2011-02-10T17:04:46.733" />
  <row Id="21403" PostTypeId="2" ParentId="12637" CreationDate="2011-02-10T17:06:32.033" Score="1" ViewCount="" Body="&lt;p&gt;If $X$ is locally compact, then it is easy, since $X$ is open in $\beta X$, hence the basis of $x$ in $X$ is also a basis in $\beta X$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's an argument for normal spaces (using my favorite of the many characterizations of $\beta X$ (for normal spaces)):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We're going to define $\beta X$ as the space of ultrafilters in the algebra of closed subsets of $X$.  That is, an element of $\beta X$ is a maximal set $\mathfrak{a}$ of closed sets that is closed under finite intersections and such that if $f\in\mathfrak{a}$ and $g\supseteq f$ is closed, then $g\in\mathfrak{a}$.&#xA;A base for the closed sets consists of sets of the form&#xA;$F_f=\{\mathfrak{a}\in\beta X : f\in\mathfrak{a}\}$ for closed sets $f\subseteq X$ (meaning the closed sets are intersections of such $F_f$s.)  Finally, define the embedding $X\to\beta X$ by letting $\hat{x}=\{f\subseteq X:x\in f\}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, suppose that $\{U_n\}_{n\in\omega}$ is a local base at $x$ in $X$.  Then&#xA;$\{Z_n\}_{n\in\omega}$ where $Z_n=X\setminus U_n$ is a local closed base at $x$, meaning that $x\notin Z_n$ for all $n$ and if $f\subseteq X$ is any closed set with $x\notin f$ then there is an $n$ with $f\subseteq Z_n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I claim that $\{F_{Z_n}\}_{n\in\omega}$ is a closed base at $\hat{x}$ in $\beta X$.  For, let $A\subseteq\beta X$ be closed with $\hat{x}\notin A$.  Thus there is some closed $f\subseteq X$ with $\hat{x}\notin A_f$ and $A\subseteq A_f$. This implies that $x\notin f$ hence $f\subseteq Z_n$ for some $n$.  But $x\notin Z_n$ so $\hat x\notin F_{Z_n}$ and $F_{Z_n}\supseteq F_f\supseteq A$.&lt;/p&gt;&#xA;" OwnerUserId="3852" LastActivityDate="2011-02-10T17:06:32.033" CommentCount="2" />
  <row Id="21404" PostTypeId="2" ParentId="21400" CreationDate="2011-02-10T17:10:12.417" Score="5" ViewCount="" Body="&lt;p&gt;Consider how many horizontal lines there are with the $y$ coordinate irrational, such lines certainly can't intersect $\mathbb{Q}\times\mathbb{Q}$.&lt;/p&gt;&#xA;" OwnerUserId="3852" LastActivityDate="2011-02-10T17:10:12.417" CommentCount="4" />
  <row Id="21407" PostTypeId="2" ParentId="21380" CreationDate="2011-02-10T17:23:02.410" Score="4" ViewCount="" Body="&lt;p&gt;Consider the infinite family of functions $\sin(nx)$ on $[0,2\pi]$.  Each will have an integral of $0$, and will have the same endpoints, namely $0$ at each end.  Can we choose a constant so the arclength's are the same?  Yes we can.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take the slightly different family of functions $\frac{1}{n}\sin(nx)  $.  Again, the endpoints are $0$ and the integrals are the same.  But this time the arclength is the same for any $n$.  We have $$\text{arclength}(\frac{1}{n}\sin(nx))=\int_0^{2\pi} \sqrt{1+(\cos nx)^2}dx$$ making the substitution $x=\frac{u}{n}$ this becomes  $$\frac{1}{n}\int_0^{2n\pi}\sqrt{1+\cos^2 x}dx=\int_0^{2\pi}\sqrt{1+\cos^2 x}dx$$ which is independent of $n$. (and where the last equality follows from the periodicity of $\cos x$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus the answer to your question is no.  The example above is an infinite family of analytic functions with the same arc-length, integral and endpoints.&lt;/p&gt;&#xA;" OwnerUserId="6075" LastEditorUserId="6075" LastEditorDisplayName="" LastEditDate="2011-02-10T17:51:28.480" LastActivityDate="2011-02-10T17:51:28.480" />
  <row Id="21408" PostTypeId="2" ParentId="21380" CreationDate="2011-02-10T17:48:04.647" Score="3" ViewCount="" Body="&lt;p&gt;Perhaps I don't understand the problem, but is this proposition true at all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose let a=0, b=1, and f(x) equal 1 everywhere except on some short interval (x0, x1) where there is a smooth bump. The bump could be anywhere on the interval. &lt;/p&gt;&#xA;" OwnerUserId="1312" LastActivityDate="2011-02-10T17:48:04.647" CommentCount="2" />
  <row Id="21409" PostTypeId="1" AcceptedAnswerId="21560" CreationDate="2011-02-10T17:50:50.243" Score="3" ViewCount="188" Body="&lt;p&gt;I'm trying to show that given $(X_i)$ i.i.d., $E[X_i^2] &amp;lt; \infty$, $\mu := E[X_i]$ then $P\Big [  \lim_{n \rightarrow \infty} \frac{S_n}{n} = \mu \Big ] = 1$ where $ S_n := \sum_{k=1}^n X_k$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So far, I have rewritten  $P\Big [  \lim_{n \rightarrow \infty} \frac{S_n}{n} = \mu \Big ]$ as &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \lim_{k \rightarrow \infty} P \Big [ \cap_{n \geq k} \{ \omega \Big | |\frac{S_n}{n} - \mu| &amp;lt; \varepsilon \}  \Big ]$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I'm not sure how to proceed from here. I have&#xA;$$ \sum_{n} P\Big [ |X_n - X|&amp;#160; &amp;gt; \varepsilon \Big] &amp;lt; \infty  \Rightarrow P\Big [ \lim_n X_n = X \Big ] = 1$$ which I think I should apply but I don't see how. Can anyone help me with this? Also, I don't see where $E[X_i^2] &amp;lt; \infty$ comes in. Many thanks for your help!&lt;/p&gt;&#xA;" OwnerUserId="5798" LastActivityDate="2011-02-11T17:26:01.880" Title="Proof that $\frac{S_n}{n}$ converges almost surely to $\mu$" Tags="&lt;probability&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" />
  <row Id="21410" PostTypeId="2" ParentId="21380" CreationDate="2011-02-10T18:12:55.913" Score="4" ViewCount="" Body="&lt;p&gt;As explained by Eric and phv, such a function is not unique in general.  However, there are some cases in which it is unique.  For example, if the arclength is $\sqrt{(b-a)^2 + (f(b)-f(a))^2}$, then the only continuous function from &lt;em&gt;(a,f(a))&lt;/em&gt; to &lt;em&gt;(b,f(b))&lt;/em&gt; is the straight line.  Then there is only one possibility for the are under the curve, &lt;em&gt;(b-a)(f(a)+f(b))&lt;/em&gt;/2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For any bigger choice of arclength from &lt;em&gt;(a,f(a))&lt;/em&gt; to &lt;em&gt;(b,f(b))&lt;/em&gt;, there is a maximum possible area under the curve and a minimum possible area under the curve.  (These are symmetric problems-- if you solve one, you'll solve the other.)  For any area between these maximum and minimum values, there will be infinitely many solutions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is an interesting question to identify these curves with fixed arclength that give maximum and minimum area under the curve.  Can you think of any solutions, or perhaps the general solution?  The straight line is the most trivial example.&lt;/p&gt;&#xA;" OwnerUserId="1461" LastActivityDate="2011-02-10T18:12:55.913" CommentCount="8" />
  <row Id="21411" PostTypeId="1" AcceptedAnswerId="21417" CreationDate="2011-02-10T18:18:19.777" Score="2" ViewCount="144" Body="&lt;p&gt;Let $A$, $B$, and $C$ be nonempty sets. I need to find conditions to guarantee that there is an injective function from $B^A$, the set of all functions from $A$ to $B$, to $\displaystyle (C^B)^{C^A}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Afterwards, I need to show this kind of a function and to prove that it is injective one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried to do that with a power equation.&#xA;Any suggestions? Thanks&#xA;(and if you may try to write it in simple as you can due to my math-English barriers :) )&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-Nir&lt;/p&gt;&#xA;" OwnerUserId="6163" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-16T04:05:20.927" LastActivityDate="2011-02-16T04:27:05.690" Title="Injective maps from $B^A$ to $(C^B)^{C^A}$" Tags="&lt;functions&gt;&lt;elementary-set-theory&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="0" />
  <row Id="21413" PostTypeId="1" AcceptedAnswerId="21430" CreationDate="2011-02-10T18:24:56.350" Score="1" ViewCount="50" Body="&lt;p&gt;Hi,&#xA;let $G$ be a group (can assume finite) acting on a vector space $V$, and assume that the action is semisimple. Let $H$ a normal subgroup of $G$. Why also the action of $H$ on $V$ is semisimple too?&lt;/p&gt;&#xA;" OwnerUserId="6583" LastActivityDate="2011-02-10T22:28:14.180" Title="semisimplicity for subgroup" Tags="&lt;group-theory&gt;&lt;representation-theory&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="21415" PostTypeId="1" AcceptedAnswerId="21420" CreationDate="2011-02-10T18:31:20.403" Score="0" ViewCount="56" Body="&lt;p&gt;Say I define a probability distribution on $P$ dimensional symmetric matrices such that the diagonals are strictly positive but the off diagonals are unrestricted (except for the symmetry constraint). Does this induce a proper probability distribution on the space of symmetric positive definite matrices?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A simple example like what I'm thinking about: $\sigma_{ii}\sim TN(0, 1)$ ($TN$ is the normal distribution truncated to be positive) and $\sigma_{ij}\sim N(0, 1)$ independently with $\sigma_{ji}=\sigma_{ij}$ so that $\pi(\Sigma) \propto [\prod_{p=1}^P N(0, 1) 1(\sigma_{ii}&amp;gt;0)][ \prod_{i&amp;lt;j} N(0, 1)]$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So my question is if $A$ is the set of SPD matrices, is $\pi_0(\Sigma) \propto \pi(\Sigma)1(\Sigma \in A)$ a probability distribution? Does this hold generally, or does it depend on the distributions for $\sigma_{ii}, \sigma_{ij}$ (assuming that they are all continuous), or on the size of $P$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Leaving aside for the moment if this is actually useful or a good idea, does it work? To me it seems like...maybe. Intuitively it seems that it should, but I've learned not to trust my intuition. And I'm not sure about a good way to demonstrate that it does or doesn't, either. &lt;/p&gt;&#xA;" OwnerUserId="6389" LastEditorUserId="6389" LastEditorDisplayName="" LastEditDate="2011-02-10T18:57:26.213" LastActivityDate="2011-02-10T18:57:26.213" Title="Does this induce a proper probability distribution on the space of covariance matrices?" Tags="&lt;probability&gt;&lt;statistics&gt;&lt;probability-theory&gt;&lt;measure-theory&gt;" AnswerCount="1" />
  <row Id="21417" PostTypeId="2" ParentId="21411" CreationDate="2011-02-10T18:38:07.077" Score="5" ViewCount="" Body="&lt;p&gt;First, define your mapping from $B^A$ to $(C^B)^{C^A}$: we're essentially forced in our definition.  Namely, given $f:A\to B$ and $g:B\to C$ we need to produce a function $A\to C$ but this clearly must be given by $g \circ f$.  (That's the only way to get domain and co-domain to match up correctly.)  In other words, we define $\eta:B^A\to (C^B)^{C^A}$ by&#xA;$\eta(f)(g)(x)=g(f(x))$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next we need to show that this map is injective.  Suppose that $f_1,f_2:A\to B$ with $f_1\neq f_2$, then we need to show that $\eta(f_1)\neq\eta(f_2)$.  Consider some point $a\in A$ with $f_1(a)\neq f_2(a)$ and go from there (basically just push through the definitions).  (Remaining details left as an exercise as well as arguing why there is such an $a$.)&lt;/p&gt;&#xA;" OwnerUserId="3852" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-16T03:56:01.547" LastActivityDate="2011-02-16T03:56:01.547" CommentCount="1" />
  <row Id="21418" PostTypeId="2" ParentId="21313" CreationDate="2011-02-10T18:42:28.380" Score="1" ViewCount="" Body="&lt;p&gt;A metric space gives rise to  a topological space on the same set (generated by the open balls in the metric). Different metrics can give the same topology. A topology that arises in this way is a metrizable topology. Using the topology we can define notions that are purely topological, like convergence, compactness, continuity, connectedness, dimension etc. Using the metric we can talk about other things that are more specific to metric spaces, like uniform continuity, uniform convergence and stuff like Hausdorff dimension, completeness etc, and other notions that do depend on the metric we choose. Different metrics that yield the same topology on a set can induce different notions of Cauchy sequences, e.g., so that the space is complete in one metric, but not in the other. In analysis e.g. one often is interested in both of these types of notions, while in topology only the purely topological notions are studied. In topology we can in fact characterize those topologies that are induced from metrics. Such topologies are quite special in the realm of all topological spaces. So in short: all metric spaces are also topological spaces, but certainly not vice versa.&lt;/p&gt;&#xA;" OwnerUserId="4280" LastActivityDate="2011-02-10T18:42:28.380" />
  <row Id="21419" PostTypeId="1" CreationDate="2011-02-10T18:42:59.707" Score="1" ViewCount="69" Body="&lt;p&gt;This Image:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://img821.imageshack.us/i/nonconopt.png/&quot; rel=&quot;nofollow&quot;&gt;http://img821.imageshack.us/i/nonconopt.png/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Shows the non continous target function I'm talking about. That's just to illustrate the problem. I'm looking for an algorithm that searches for the minimum/minima of this function or similar functions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you can see, it consists of rectangular areas with constant function value (represented by color). That's a problem, as all the algorithms I know expect the objective function to be continous. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Algorithms like PSO or GA do actually some kind of optimization if I apply them, but I guess that's not really the most effective way to do this, they converge way to early. No surprise there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any suggestions on what algorithms could be applicable? &#xA;Maybe even some key words for searching would be of help.&lt;/p&gt;&#xA;" OwnerUserId="6877" LastActivityDate="2011-03-17T12:33:35.427" Title="Algorithms to solve a non continous global optimization problem" Tags="&lt;algorithms&gt;&lt;optimization&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="21420" PostTypeId="2" ParentId="21415" CreationDate="2011-02-10T18:44:24.807" Score="1" ViewCount="" Body="&lt;p&gt;There is no guarantee that the matrix you suggest will be positive symmetric with full probability. Already in dimension $2$, your matrix is $\left(\begin{matrix}|X| &amp;amp; Z \\ Z&amp;amp; |Y|\end{matrix}\right)$ with $X$, $Y$ and $Z$ independent standard Gaussian, hence its determinant is $|XY|-Z^2$, which may very well be negative. The argument works as soon as the coefficients are independent and the distribution of an off diagonal coefficient is unbounded.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Natural distributions on the space of symmetric random matrices are the so-called generalized inverse gaussian (GIG) measures, see &lt;a href=&quot;http://math.uc.edu/~brycw/preprint/rmq.pdf&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="6179" LastEditorUserId="6179" LastEditorDisplayName="" LastEditDate="2011-02-10T18:53:38.380" LastActivityDate="2011-02-10T18:53:38.380" CommentCount="5" />
  <row Id="21421" PostTypeId="2" ParentId="20982" CreationDate="2011-02-10T18:47:01.060" Score="2" ViewCount="" Body="&lt;p&gt;Two days later, this is probably worth an answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If your points are A, B, C, D, E then the edges AB, BC, CD, DE, EA clearly provide one of the smallest graphs which has both a &lt;a href=&quot;http://en.wikipedia.org/wiki/Eulerian_cycle&quot; rel=&quot;nofollow&quot;&gt;Eulerian cycle&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/Hamiltonian_cycle&quot; rel=&quot;nofollow&quot;&gt;Hamiltonian cycle&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Add another edge, for example AC, and you have a graph which has the same Hamiltonian cycle but not an Eulerian cycle, since A and C have an odd number of edges.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively, the edges AB, BC, CA, AD, DE, EA, provide a Eulerian cycle but not a Hamiltonian cycle, since you can only get from B to D via A, and you cannot get back without visiting A again.    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Incidentally, the neat Flash game at &lt;a href=&quot;http://neamar.fr/Res/Icosien/&quot; rel=&quot;nofollow&quot;&gt;http://neamar.fr/Res/Icosien/&lt;/a&gt; asks you to find 20 Eulerian cycle and Hamiltonian cycles for given graphs. &lt;/p&gt;&#xA;" OwnerUserId="6460" LastActivityDate="2011-02-10T18:47:01.060" />
  <row Id="21422" PostTypeId="2" ParentId="21377" CreationDate="2011-02-10T19:00:39.227" Score="3" ViewCount="" Body="&lt;p&gt;Let me make sure I understand what you mean by the formula &#xA;$$ \left| \frac{\eta(s)}{s} \sum_{n=1}^{\infty} \frac{\mu(n)}{n^s} - \frac{1-2^{1-s}}{s} \right| &amp;lt; \infty. $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do you actually mean that you have proven that this sum is convergent for $s$ in the critical strip and that, for such $s$, this inequality holds? Because, if you have shown that this sum converges in that range, you have already shown RH. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can think of other interpretations of your statement, but I will wait for clarification before elaborating on them. Here is a basic point to remember: The statement that $\sum_{n=1}^{\infty} a_n$ converges is a statement about the partial sums $\sum_{n=1}^N a_n$. (Namely, that they form a Cauchy sequence.) If you never say anything about these partial sums, it is unlikely you have proved convergence.&lt;/p&gt;&#xA;" OwnerUserId="448" LastActivityDate="2011-02-10T19:00:39.227" CommentCount="5" />
  <row Id="21423" PostTypeId="2" ParentId="21274" CreationDate="2011-02-10T19:01:40.983" Score="1" ViewCount="" Body="&lt;p&gt;Suppose that a statement $\varphi$ of propositional logic (using only $\wedge$, $\vee$ and $\to$) is not a propositional tautology in classical logic. It follows that some row of the truth table for $\varphi$ has value $F$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, consider the truth table of $\varphi$ in &lt;a href=&quot;http://en.wikipedia.org/wiki/Three-valued_logic&quot; rel=&quot;nofollow&quot;&gt;Łukasiewicz three-valued logic&lt;/a&gt;, which is called Kleene logic on the Wikipedia page. The key observation is that Łukasiewicz three-valued logic agrees with classical logic on classical logic input. In other words, if all the propositional variables of $\varphi$ are given classical T/F values, then the truth value of $\varphi$ will agree with the classical logic value. Thus, since $\varphi$ had a classical row with value $F$, the very same row will have value F in the Łukasiewicz truth table of $\varphi$, contrary to the assumption that $\varphi$ had only values T or U. &lt;/p&gt;&#xA;" OwnerUserId="413" LastActivityDate="2011-02-10T19:01:40.983" />
  <row Id="21424" PostTypeId="2" ParentId="20518" CreationDate="2011-02-10T19:18:01.017" Score="3" ViewCount="" Body="&lt;p&gt;There are some uniformly distributed sequences $B_k$ for which $\Sigma_k=0^n e^{-kB_k^2}$ grows faster than $\sqrt n$ :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Simply pick $B_k = 0$ about $n^{2/3}$ times out of $n$ and pad the rest of the values with any uniformly distributed sequence. You will still get a uniformly distributed sequence because the density $n^{2/3} / n$ is still convergent to $0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But with such a sequence, you have $\Sigma_{k=0}^n e^{-kB_k^2} &amp;gt; n^{2/3}$. (and we could pick any function $g(n)$ provided g(n)/n converges to $0$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Truly random sequences have a lot of better properties than &quot;uniformly distributed&quot;.&#xA;Here, a property that would work and that random sequences have, is that for any interval $I$, &#xA;$$ \lim_{n \rightarrow \infty} \frac{\#\{k \in \{0\ldots n\}, \sqrt{k}B_k \in I\}}{\sqrt{n} \text{length}(I)} = 2$$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then you can show that the sum $\Sigma_{k=0}^n e^{-kB_k^2}$, behaves the sames as random sequences, like: &#xA;$$\begin{array}{rcl}&#xD;&#xA;\Sigma_{k=0}^n \int_{t=0}^1 e^{-kt^2} dt &amp;amp;&amp;lt;&amp;amp; \Sigma_{k=0}^n \int_{t=0}^\infty e^{-kt^2} dt&#xD;&#xA; = \Sigma_{k=0}^n \int_{t=0}^\infty \frac{e^{-t^2}}{\sqrt{k}} dt \\&#xD;&#xA;&amp;amp;=&amp;amp; \left(\int_{t=0}^\infty e^{-t^2} dt \right) \left(\Sigma_{k=0}^n \frac{1}{\sqrt{k}} \right)&#xD;&#xA;\sim \sqrt{\pi n} \end{array}&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The property described above implies that in an interval $I = [x;x+dx]$, there are about $2\sqrt{n} dx$ integers $k$ such that $\sqrt{k}B_k$ is in $I$. By dividing $\mathbb{R}^+$ in such intervals and regrouping the terms according to which interval they land into, you obtain bounds for $\Sigma_{k=0}^n e^{-k B_k^2}$. The finer you divide, the closer you get to the value of the integral, until you arrive at $\sqrt{\pi n}$&lt;/p&gt;&#xA;" OwnerUserId="6270" LastEditorUserId="6270" LastEditorDisplayName="" LastEditDate="2011-02-10T19:52:28.313" LastActivityDate="2011-02-10T19:52:28.313" />
  <row Id="21425" PostTypeId="2" ParentId="21379" CreationDate="2011-02-10T19:22:38.703" Score="2" ViewCount="" Body="&lt;p&gt;Not quite a symbol per se, but of course there is &quot;a.e.&quot; (&quot;almost everywhere&quot;):  &quot;for all but a set of measure zero&quot;.  Probabilists call it &quot;a.s.&quot; (&quot;almost surely&quot;).  They are also in the habit of writing &quot;a.a.&quot; (&quot;almost all&quot;): &quot;for all but finitely many&quot; and &quot;i.o.&quot; (&quot;infinitely often&quot;): &quot;for infinitely many&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In potential theory, one also sees &quot;q.e.&quot; (&quot;quasi-everywhere&quot;): &quot;for all but a set of capacity zero&quot;.&lt;/p&gt;&#xA;" OwnerUserId="822" LastActivityDate="2011-02-10T19:22:38.703" CommunityOwnedDate="2011-02-10T19:22:56.363" />
  <row Id="21426" PostTypeId="1" CreationDate="2011-02-10T19:30:57.700" Score="0" ViewCount="144" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Possible Duplicate:&lt;/strong&gt;&lt;br&gt;&#xA;  &lt;a href=&quot;http://math.stackexchange.com/questions/17026/what-can-we-say-about-f-if-int-01-fxpxdx-0-for-all-polynomials-p&quot;&gt;What can we say about $f$ if $\int_0^1 f(x)p(x)dx=0$ for all polynomials $p$?&lt;/a&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;p&gt;suppose that $f \in L^2[0,1]$ such that $\int_0^1 x^n f(x) dx=0$ for all $n=0,1,2,\ldots. $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;show that it follows that  $f(x)=0$ almost everywhere&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i saw some related questions posted, but according to the posted answers, this statement is not true, so im really confused because according to my assignment i have to show that it is, in fact, true&lt;/p&gt;&#xA;" OwnerUserId="6689" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2011-02-10T19:39:16.603" LastActivityDate="2011-02-10T19:39:16.603" ClosedDate="2011-02-10T19:38:10.117" Title="show that $\int_0^1 f(x)x^n dx = 0$ for $n=0,1,2,\ldots$ imply $f=0$ a.e." Tags="&lt;measure-theory&gt;" CommentCount="3" />
  <row Id="21427" PostTypeId="2" ParentId="21361" CreationDate="2011-02-10T19:41:39.020" Score="6" ViewCount="" Body="&lt;p&gt;Your solution is right. It should be a typo in the paper. Here is my evaluation confirming yours. Since&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{eqnarray*}&#xD;&#xA;\frac{dV(t)}{dt} &amp;amp;=&amp;amp;\sqrt{2}\sum_{h=1}^{H}h\omega V_{h}\cos \left( h\omega t+%&#xD;&#xA;\frac{\pi }{2}\right)  \\&#xD;&#xA;&amp;amp;=&amp;amp;-\sqrt{2}\sum_{h=1}^{H}h\omega V_{h}\sin h\omega t,&#xD;&#xA;\end{eqnarray*}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and assuming $\omega$ is the angular frequency given by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\omega =\frac{2\pi }{T},$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left( \frac{dV(t)}{dt}\right) ^{2}=2\omega ^{2}\left(&#xD;&#xA;\sum_{h=1}^{H}hV_{h}\sin \left( h\omega t\right) \right) ^{2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{eqnarray*}&#xD;&#xA;\frac{1}{T}\int_{0}^{T}\left( \frac{dV(t)}{dt}\right) ^{2}dt &amp;amp;=&amp;amp;\frac{\omega &#xD;&#xA;}{2\pi }\int_{0}^{2\pi /\omega }\left( \frac{dV(t)}{dt}\right) ^{2}dt \\&#xD;&#xA;&amp;amp;=&amp;amp;\frac{\omega }{2\pi }\int_{0}^{2\pi /\omega }2\omega ^{2}\left(&#xD;&#xA;\sum_{h=1}^{H}hV_{h}\sin \left( h\omega t\right) \right) ^{2}dt \\&#xD;&#xA;&amp;amp;=&amp;amp;\frac{\omega ^{3}}{\pi }\int_{0}^{2\pi /\omega }\left(&#xD;&#xA;\sum_{h=1}^{H}hV_{h}\sin \left( h\omega t\right) \right) ^{2}dt.&#xD;&#xA;\end{eqnarray*}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The integrand $\left( \sum_{h=1}^{H}hV_{h}\sin \left( h\omega t\right)&#xD;&#xA;\right) ^{2}$ is a sum of terms of two different types:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i) $h^{2}V_{h}^{2}\sin ^{2}\left( h\omega t\right) $ and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ii) $k\left( pV_{p}\sin \left( p\omega t\right) \cdot qV_{q}\sin \left(&#xD;&#xA;q\omega t\right) \right) \,$, with $p\neq q$ and $p,q,k\in\mathbb{N}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second type terms do not contribute to the last integral, because the $\sin nx$ ($n\in\mathbb{N}$) functions form an orthogonal system over $[0,2\pi ]$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_{0}^{2\pi /\omega }k\left( pV_{p}\sin \left( p\omega t\right) \cdot&#xD;&#xA;qV_{q}\sin \left( q\omega t\right) \right) dt=0\quad p\neq q$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sum of the first type ones is $\sum_{h=1}^{H}h^{2}V_{h}^{2}\sin ^{2}\left(&#xD;&#xA;h\omega t\right) $. Thus  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{eqnarray*}&#xD;&#xA;\frac{1}{T}\int_{0}^{T}\left( \frac{dV(t)}{dt}\right) ^{2}dt &amp;amp;=&amp;amp;\frac{\omega&#xD;&#xA;^{3}}{\pi }\int_{0}^{2\pi /\omega }\sum_{h=1}^{H}h^{2}V_{h}^{2}\sin&#xD;&#xA;^{2}\left( h\omega t\right) dt \\&#xD;&#xA;&amp;amp;=&amp;amp;\frac{\omega ^{3}}{\pi }\sum_{h=1}^{H}h^{2}V_{h}^{2}\int_{0}^{2\pi&#xD;&#xA;/\omega }\sin ^{2}\left( h\omega t\right) dt \\&#xD;&#xA;&amp;amp;=&amp;amp;\frac{\omega ^{3}}{\pi }\sum_{h=1}^{H}h^{2}V_{h}^{2}\cdot \frac{\pi }{%&#xD;&#xA;\omega } \\&#xD;&#xA;&amp;amp;=&amp;amp;\omega ^{2}\sum_{h=1}^{H}h^{2}V_{h}^{2},&#xD;&#xA;\end{eqnarray*}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;because&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{eqnarray*}&#xD;&#xA;\int \sin ^{2}\left( h\omega t\right) dt &amp;amp;=&amp;amp;\frac{1}{h\omega }\left( -\frac{1%&#xD;&#xA;}{2}\cos h\omega t\sin h\omega t+\frac{1}{2}h\omega t\right)  \\&#xD;&#xA;\int_{0}^{2\pi /\omega }\sin ^{2}\left( h\omega t\right) dt &amp;amp;=&amp;amp;\frac{\pi }{%&#xD;&#xA;\omega }.&#xD;&#xA;\end{eqnarray*}$$&lt;/p&gt;&#xA;" OwnerUserId="752" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2011-02-11T11:29:49.307" LastActivityDate="2011-02-11T11:29:49.307" CommentCount="6" />
  <row Id="21428" PostTypeId="2" ParentId="21361" CreationDate="2011-02-10T19:44:18.357" Score="2" ViewCount="" Body="&lt;p&gt;I agree, it should be $\omega^2$. The whole thing is in fact just the Pythagorean theorem: the functions $\sqrt{2} \cos(\dots)$ are orthonormal in the space $L^2([0,T])$, and the integral is the square of the $L^2$ norm of $dV/dt$, hence the sum of the squares of the coefficients: $\sum (h\omega V_h)^2$.&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2011-02-10T19:44:18.357" CommentCount="1" />
  <row Id="21429" PostTypeId="1" AcceptedAnswerId="21466" CreationDate="2011-02-10T20:36:22.323" Score="5" ViewCount="175" Body="&lt;p&gt;I am studying Linear Algebra II, and I came across several questions in which, for a certain linear transformation ($T\colon\mathbf{V}\to\mathbf{V}$) I was told that:&#xA;$$||T(a)|| \leq ||a||.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not completely certain how to use this information. For instance, consider the following question (please forgive my translation, it's the first time I write math in English):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;For a linear transformation $T\colon\mathbf{V}\to\mathbf{V}$in a unitary space [i.e., complex inner product space], such that&lt;/p&gt;&#xA;  &#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;$|c|=1$ for every eigenvalue $c$ of $T$;&lt;/li&gt;&#xA;  &lt;li&gt;$||T(a)|| \leq ||a||$ for every vector $a$ in $\mathbf{V}$;&#xA;  prove that T is a unitary operator.&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;How does the fact that $||T(a)|| \leq ||a||$ help me?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="6879" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-11T01:06:43.843" LastActivityDate="2011-02-11T01:19:29.580" Title="Linear transformations and norm" Tags="&lt;homework&gt;&lt;linear-algebra&gt;" AnswerCount="1" CommentCount="15" FavoriteCount="1" />
  <row Id="21430" PostTypeId="2" ParentId="21413" CreationDate="2011-02-10T20:38:47.650" Score="3" ViewCount="" Body="&lt;p&gt;This is called &lt;a href=&quot;http://en.wikipedia.org/wiki/Clifford_theory&quot; rel=&quot;nofollow&quot;&gt;Clifford's theorem&lt;/a&gt;.  Roughly:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Theorem.&lt;/strong&gt; Take a simple G-submodule W of V.  Then W is a direct sum of simple H-submodules U1⊕…⊕Un.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Let U be any simple H-submodule of W.  Then the G-conjugates of U are also H-submodules, since H is normal.  They intersect as proper H-submodules of U, and so intersect trivially.  They span W since their span is a G-submodule of W.  Since every H-submodule of V is a direct sum of simple G-submodules (G acts semisimply), and each simple G-submodule is a direct sum of simple H-submodules (as we just showed), V is a direct sum of simple H-submodules, and so is semi-simple.&lt;/p&gt;&#xA;" OwnerUserId="583" LastEditorUserId="274" LastEditorDisplayName="" LastEditDate="2011-02-10T22:28:14.180" LastActivityDate="2011-02-10T22:28:14.180" CommentCount="2" />
  <row Id="21432" PostTypeId="2" ParentId="20634" CreationDate="2011-02-10T20:39:34.537" Score="0" ViewCount="" Body="&lt;p&gt;I believe that I found an answer to my problem. The earlier approach of playing with open covers didn't pan out because I couldn't extend local injectivity of $p$ on a bunch of open sets to injectivity on their union.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I found the following theorem, which solves the problem quite neatly. I found it in [1], along with a sketch proof. (But I'll prove it fully):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Theorem: Let $E$ be any space and $F$ be a metric space. Let $K\subset E$ be compact, $f:E\to F$ be continuous, locally injective near every point $p\in K$, and (globally) injective on $K$. Then $f$ is also injective on some open neighbourhood $N$ of $K$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proof: Let $g:E\times E\to R$ be the function defined by $g(x,x&amp;#39;)=d(f(x),f(x&amp;#39;))$. Around each point of the form $(k,k)\in K\times K$, there is an open product neighbourhood $V_k \times V_k$ such that $(V_k\times V_k)\cap g^{-1}(0)^c \subset \{ (x,x)\in E\times E \}$ (by local injectivity). Let $V=\cup_k V_k$. Then the compact set $K\times K$ is contained in&#xA;the open set $g^{-1}(0)^c \cup V$. By the generalized tube lemma, there are open sets $W,W&amp;#39;$ of $E$ so that $K\times K\subset W\times W&amp;#39;\subset g^{-1}(0)^c \cup V$. Then let&#xA;$N=W\cap W&amp;#39;$. $\square$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Applying this theorem to $p:\tilde{X} \to X$, with $K$ being a lift of the left edge of&#xA;$X$, gives an open neighbourhood of that lift which maps homeomorphically to an open&#xA;neighbourhood of the left edge of $X$. Then we use the bad local nature of $X$ to find a&#xA;loop in $X$ which lifts to a loop in $\tilde{X}$ which is not contractible. (Compactness is used again to find an open set which contains the entirety of a vertical line)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[1] Wagner,M., &quot;Existence of schlicht integral manifolds for ordinary differential equations&quot;, Archiv der Mathematik, Volume 61, Number 6, 529-542 (1993)&lt;/p&gt;&#xA;" OwnerUserId="5290" LastEditorUserId="5290" LastEditorDisplayName="" LastEditDate="2011-02-10T21:31:39.860" LastActivityDate="2011-02-10T21:31:39.860" />
  <row Id="21433" PostTypeId="1" AcceptedAnswerId="21436" CreationDate="2011-02-10T21:19:31.803" Score="2" ViewCount="102" Body="&lt;p&gt;I am doing some background research on sensitivity and elasticity analysis, and I came across the following &lt;a href=&quot;http://www.ag.unr.edu/sedinger/Courses/ERS488-688/lectures/ers485lecture10.pdf&quot; rel=&quot;nofollow&quot;&gt;definitions of elasticity&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$e_{ij}=\frac{a_{ij}\partial \lambda}{\lambda \partial a_{ij}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$e_{ij} = \frac{\partial \log \lambda}{\partial \log a_{ij}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why are these equivalent?&lt;/p&gt;&#xA;" OwnerUserId="3733" LastEditorUserId="1543" LastEditorDisplayName="" LastEditDate="2011-04-17T20:31:40.383" LastActivityDate="2011-04-17T20:31:40.383" Title="why does $\frac{b\partial a}{a \partial b} = \frac{\partial \log a}{\partial \log b}$?" Tags="&lt;calculus&gt;&lt;mathematical-modeling&gt;&lt;mathematical-physics&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="21434" PostTypeId="2" ParentId="21433" CreationDate="2011-02-10T21:21:23.067" Score="3" ViewCount="" Body="&lt;p&gt;Because $\partial \log x=\frac{\partial x}{x}$, whatever it is that $\partial$ may mean.&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2011-02-10T21:21:23.067" CommentCount="3" />
  <row Id="21435" PostTypeId="1" AcceptedAnswerId="21439" CreationDate="2011-02-10T21:26:08.077" Score="4" ViewCount="184" Body="&lt;p&gt;I'm trying to prove that an Archimedian field is a subfield of the real numbers, my plan is to use the fact that the rationals are dense within the field and their Dedekind completion is the real numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I seem to run into some trouble writing a neat proof for the fact that the two properties of Archimedian fields are equivalent:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $F$ be an ordered field, the following are equivalent:&lt;br&gt;&#xA; 1. For every $x\in F$ there exists a natural number $n$ such that $x &amp;lt; n$&lt;br&gt;&#xA; 2. The rationals are dense in $F$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(In fact, I only need $(1)\implies (2)$ to be proved)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have the essence of the proof, assume by contradiction some $a&amp;lt;b$ are two numbers that have no rational numbers in between, wlog we can assume $a,b\in (0,1)$, take some $n$ which is large enough and show some $\frac{k}{n}$ has to be between $a$ and $b$ otherwise there is some $x$ which is an upper bound of $\mathbb{N}$ in the field. However it seems that I can't write it elegantly.&lt;/p&gt;&#xA;" OwnerUserId="622" LastActivityDate="2011-02-11T03:36:01.407" Title="Equivalency of Archimedian Fields Properties" Tags="&lt;field-theory&gt;" AnswerCount="3" />
  <row Id="21436" PostTypeId="2" ParentId="21433" CreationDate="2011-02-10T21:29:40.290" Score="5" ViewCount="" Body="&lt;p&gt;From the Chain Rule:&#xA;$$\frac{\partial\log\lambda}{\partial \log a_{ij}}\;\frac{\partial\log a_{ij}}{\partial t} = \frac{\partial\log\lambda}{\partial t} = \frac{1}{\lambda}\frac{\partial\lambda}{\partial t}.$$&#xA;Since $$\frac{\partial\log a_{ij}}{\partial t} = \frac{1}{a_{ij}}\frac{\partial a_{ij}}{\partial t}$$&#xA;&quot;solving&quot; for $\frac{\partial\log\lambda}{\partial \log a_{ij}}$ gives&#xA;$$\frac{\partial\log\lambda}{\partial\log a_{ij}} = \frac{\quad\frac{1}{\lambda}\frac{\partial \lambda}{\partial t}\quad}{\frac{1}{a_{ij}}\frac{\partial a_{ij}}{\partial t}} = \frac{a_{ij}\partial \lambda}{\lambda\partial a_{ij}}.$$&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2011-02-10T21:29:40.290" />
  <row Id="21437" PostTypeId="1" AcceptedAnswerId="21513" CreationDate="2011-02-10T21:34:01.453" Score="2" ViewCount="108" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd appreciate some help with the following problem form Conway's book on functions of one complex variable:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $f$ be analytic in $\overline B (0;R)$ with $|f(z)|\le M$ for $|z|\le R$ and $|f(0)|=a&amp;gt;0$. Show that the number of zeros of f in $B(0;R/3)$ is less than or equal to $$\frac{1}{\log(2)}\log\left(\frac M a\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that the number of zeros is given by &#xA;$$n = \frac 1 {2\pi i}\int_{|z|=R/3} \frac{f&amp;#39;}{f} \, dz&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And there is a hint to look at $g(z) = f(z) \prod_{k=1}^n (1-z/z_k)^{-1}$, where the $z_k$ are the zeros of $f$. I have given it some time now, but don't seem to get anywhere. In particular I don't see how the logarithm, $M, a$ come into play.&#xA;The problem is in the chapter on the maximum modulus theorem, if that's of any help.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Might someone maybe give me a hint?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cheers, S.L.&lt;/p&gt;&#xA;" OwnerUserId="3208" LastEditorUserId="3208" LastEditorDisplayName="" LastEditDate="2011-02-10T22:15:31.997" LastActivityDate="2011-02-11T13:21:40.520" Title="Upper bound for zeros of holomorphic function" Tags="&lt;complex-analysis&gt;" AnswerCount="1" />
  <row Id="21438" PostTypeId="1" AcceptedAnswerId="21467" CreationDate="2011-02-10T21:38:28.020" Score="19" ViewCount="771" Body="&lt;p&gt;Better yet, what I'm asking is how do you actually &lt;em&gt;write&lt;/em&gt; your mathematics?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think I need to give brief background: Through most of my childhood, I'd considered myself pretty good at math, up through the high school level. I easily followed mathematical concepts introduced in my classes and even did a few competitions; I definitely wouldn't say I was a star of the caliber one meets when one ventures out into the bigger ponds, but I thought I was decent and convinced myself I would major in math when I entered college.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That changed after a couple of years when I hit my first Real Analysis class that used Rudin's book; that was the first class, I think, I took that really required more than &quot;expand-a-definition&quot; type proofs and my struggle to find intuition and understanding there impacted my mathematical self-confidence.   I eventually switched majors, with a bit of regret.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One thing that got me, I think, was the veritable explosion of superscripts and subscripts that one encounters for the first time in Real Analysis.  I'd often find myself struggling to set up the machinery of what I was trying to prove, lost in the notation.  &lt;strong&gt;How do good mathematicians format their work on paper so as not to get lost in the $i$s, $j$s, and $k$s and keep track of what they're investigating?&lt;/strong&gt;  I remember dealing with subsequences of sequences to show that limits did or did not exist got especially hairy in this way...writing things like $s_{n_{k_{\epsilon}}}$ and remembering what my goal at each &quot;level&quot; was difficult.  I'd be interested in knowing if aspiring mathematicians and/or professional mathematicians scribble marginalia or have a system to overcome such problems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another thing that got me were what I personally called &quot;consider...&quot; statements.  Many times, on this site, the most talented commenters will say &quot;Consider $f(n)$&quot; or &quot;Consider transformation $T:$ $U \rightarrow V$&quot; that in the first case gives a summation that wonderfully telescopes/has an obvious bound, or in the second case transforms the problem into a trivial application of the rank-nullity theorem, or something like that.  Mathematics is a subject replete with geniuses , I understand that, but how do mere mortals investigate such functions and &quot;massage&quot; them into doing what they want?  &lt;strong&gt;When good mathematicians get intuitionistic ideas, what (explicit) steps do they take to formalize them, especially when it is likely that first idea is murky or wrong&lt;/strong&gt;? (Aside: I've been given &quot;use numerical examples&quot; as advice before, but sometimes I think to myself, &quot;I've been dealing with $\mathbb{Z}$ since I was 6 years old, and not so much with Dedekind's definition of the real numbers...&quot;) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's lots more I could ask, but I want to keep this question tractable, so I guess I might summarize by asking: &lt;strong&gt;How do you [professional and aspiring mathematicians] organize your math &quot;notebook&quot;, and what perhaps idiosyncratic methods do you employ to be original and clever within it?&lt;/strong&gt; I know there will be no strict formulas anyone can give; mathematicians are scientists of the abstract; I understand that the subject is acclaimed partly because it's so intellectually and individually demanding.   But I think even acclaimed scientists draw on Springer's Protocols and &lt;em&gt;Nature Methods&lt;/em&gt;...There seems to me a bit of a jump between the dryly algorithmic way one is taught to do math in high school and the more abstruse methods at the undergraduate level.  I'd be interested if anyone here could help me bridge that gap, if only for my personal fulfillment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Apologies in advance if the question is ill-posed or too subjective in its current form to meet the requirements of the FAQ; I'd certainly appreciate any suggestions for its modification if need be.)&lt;/p&gt;&#xA;" OwnerUserId="6296" LastEditorUserId="6296" LastEditorDisplayName="" LastEditDate="2011-03-05T00:24:08.717" LastActivityDate="2011-03-05T00:24:08.717" Title="How Do You Actually Do Your Mathematics?" Tags="&lt;soft-question&gt;&lt;notation&gt;&lt;education&gt;" AnswerCount="6" CommentCount="8" FavoriteCount="6" />
  <row Id="21439" PostTypeId="2" ParentId="21435" CreationDate="2011-02-10T21:40:41.200" Score="3" ViewCount="" Body="&lt;p&gt;There is a positive integer $n$ greater than $(b-a)^{-1}$, and there is a least positive integer $k$ greater than $n\cdot a$.  Thus $a\lt\frac{k}{n}$, and because $\frac{k-1}{n}\leq a$, $\frac{k}{n}=\frac{k-1}{n}+\frac{1}{n}\lt a +(b-a) = b$.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2011-02-10T21:40:41.200" CommentCount="11" />
  <row Id="21440" PostTypeId="1" AcceptedAnswerId="21442" CreationDate="2011-02-10T21:41:38.913" Score="3" ViewCount="164" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Where does the &quot;naive definition&quot; of the product of two ideals $I$ and $J$, $IJ = \{ ij \mid i \in I, j \in J \}$ fall apart?&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;(The product of two ideals $I$ and $J$ is defined to be $IJ := \sum_i a_ib_i$, where each $a_i$ is in $I$ and each $b_i$ is in $J$.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Note: This is a question from a problem set I'm working on. I am not expecting a full solution for the answers, but I'd like some help knowing where to look. It seems to work fine in $\mathbb{Z}$ and $\mathbb{C}[x]$.&lt;/p&gt;&#xA;" OwnerUserId="541" LastEditorUserId="541" LastEditorDisplayName="" LastEditDate="2011-02-10T22:16:01.007" LastActivityDate="2011-02-10T22:37:46.017" Title="Defining the Product of Ideals" Tags="&lt;homework&gt;&lt;abstract-algebra&gt;&lt;algebra&gt;&lt;ring-theory&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="21442" PostTypeId="2" ParentId="21440" CreationDate="2011-02-10T21:50:29.313" Score="4" ViewCount="" Body="&lt;p&gt;Suppose $I=J=(x,y)\subseteq \mathbb R[x,y]$. Then $x^2$ and $y^2$ are in your &lt;em&gt;naive&lt;/em&gt; definition, but their sum is not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is an enlightening exercise to try to see &lt;em&gt;what&lt;/em&gt; exactly $\mathbb Z$ and $\mathbb C[x]$ have that make the naive definition work...&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2011-02-10T21:50:29.313" CommentCount="5" />
  <row Id="21443" PostTypeId="2" ParentId="21438" CreationDate="2011-02-10T21:54:41.723" Score="3" ViewCount="" Body="&lt;p&gt;I think these might be actually three different questions. I'll give a brief answer that touches all of them, however.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A good thing to do is: when you have the feeling that you have found a solution to a problem, write it down formally (preferably with latex). This sounds self-evident but it is not a trivial step. Often you find out that your solution is incorrect or incomplete only this way. Also it is obviously a good way to keep your notes organised.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See also: &lt;a href=&quot;http://mathoverflow.net/questions/1785/how-do-you-keep-your-research-notes-organized&quot; rel=&quot;nofollow&quot;&gt;http://mathoverflow.net/questions/1785/how-do-you-keep-your-research-notes-organized&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="367" LastActivityDate="2011-02-10T21:54:41.723" CommentCount="2" />
  <row Id="21444" PostTypeId="2" ParentId="21435" CreationDate="2011-02-10T22:02:20.527" Score="3" ViewCount="" Body="&lt;p&gt;For my take on this, see the proof of Proposition 108 in &lt;a href=&quot;http://math.uga.edu/~pete/FieldTheory.pdf&quot; rel=&quot;nofollow&quot;&gt;these notes&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="299" LastActivityDate="2011-02-10T22:02:20.527" />
  <row Id="21445" PostTypeId="2" ParentId="21438" CreationDate="2011-02-10T22:09:51.993" Score="6" ViewCount="" Body="&lt;p&gt;Though I left the comment about Gauss saying that &quot;Notions are more important then Notations&quot;, I find that when I do mathematics (in my case more of computational and applied mathematics) having good notations/symbols helps me out. Good notations/symbols in some sense makes you &quot;lazy&quot; and if your notations/symbols is good enough, all you might need to do is some &quot;blind&quot; algebra without actually thinking deep. It is like having good variable names when you code up a computer program. You want the variable names to be suggestive and at the same time not too long or messy. I sometimes get too finicky with notations and symbols. If I find that I should have used some other notation or symbol at the beginning, then I actually take pains and the efforts to go back to the beginning and change them throughout. Having good notations/symbols I believe lets your mathematical thought process flow smoothly, especially when you are explaining to others. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another habit which I believe is helpful is to always write out your thoughts and ideas. I always face the problem that if I don't write them out, all the thoughts and ideas in my head gets disorganized. If you want to learn a language, you need to speak in the language to learn it. Similarly, if you want to learn Mathematics, you need to think and write mathematics to learn it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another thing which I usually do (which sometimes lead others to frown upon me) is to write out the details and pay attention to details, even though it might seem trivial. I understand that this is an infinite process i.e. writing out all the details. Somewhere we need to draw a line, between paying attention to details and keeping track of the overall problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also once every quarter, I take a look at all I have done over the last quarter and &quot;try&quot; to get them in order and organize them. (though I have often failed on this front of organizing them in some nice order).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another thing I find, especially when I study pure mathematics on my own, is that I am totally lost to find out why certain things are even done or discussed about in the first place. I still remember the day when I did my first rigorous real analysis course when they defined, a closed set as &quot;A set which contains all its limit points&quot;. I was like &quot;what?&quot;. I was asking myself &quot;What has an innocent looking sequence and its limit got to do with defining a closed set&quot;. Where and how did the limits come into picture?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think this is where a good teacher can help by providing the right motivation for why things are done in a specific way. In fact this is one of the main reasons why I visit this website daily. You have a lot of wonderful Professors and some great students here actively participating and answering questions raised by students like me. Reading some of the questions and answers are really enlightening even though you might have already looked at the problem. Each answer gives you a different perspective of the problem which essentially helps to understand the problem and its intricacies better and helps to understand why things are done in a certain way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another important thing to do when learning or doing mathematics is to look for counter examples. I really believe counterexamples play a crucial role in understanding a problem better. Sometimes counter examples help to understand a problem/proof better than the actual proof itself.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2011-02-10T23:08:14.467" LastActivityDate="2011-02-10T23:08:14.467" />
  <row Id="21446" PostTypeId="2" ParentId="21363" CreationDate="2011-02-10T22:15:36.613" Score="1" ViewCount="" Body="&lt;p&gt;If you set $m = k+l$, then the probability of selecting a white ball can be expressed as&#xA;$$&#xD;&#xA;P_w = \frac{k}{2m} + \frac{n-k}{2(2n-m)} = \frac{k(2n-m) + (n-k)m}{2m(2n-m)} = \frac{2k(n-m) + nm}{2m(2n-m)}.&#xD;&#xA;$$&#xA;This is invariant under $(k,m)\rightarrow(n-k,2n-m)$ (which just exchanges the urns), so we can assume $m \le n$ with no loss of generality.  If $m=n$, we have $P_w = 1/2$.  For $m&amp;lt;n$, the probability is clearly maximized by taking $k$ as large as possible, which is $k=m$ (so $l=0$).  In that case,&#xA;$$&#xD;&#xA;P_w = \frac{2m(n-m) + nm}{2m(2n-m)} = \frac{3n-2m}{4n-2m} = \frac{3}{4} - \frac{m}{4(2n-m)}.&#xD;&#xA;$$&#xA;This is a decreasing function of $m$, so $m$ should be taken as small as possible, which is $m=1$, in which case&#xA;$$&#xD;&#xA;P_w = \frac{3}{4} - \frac{1}{4(2n-1)}.&#xD;&#xA;$$&#xA;This is at least $1/2$ for all $n\ge 1$, so the answer $(k,l)=(1,0)$ always gives the best odds of drawing a white ball... probability $P=1$ if the left urn is picked, and just under $P=1/2$ if the right urn is picked, for an overall probability of just under $3/4$.&lt;/p&gt;&#xA;" OwnerUserId="5546" LastActivityDate="2011-02-10T22:15:36.613" />
  <row Id="21447" PostTypeId="2" ParentId="21440" CreationDate="2011-02-10T22:37:46.017" Score="0" ViewCount="" Body="&lt;p&gt;I think you'll find of interest my old sci.math post below.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;From: Bill Dubuque &amp;lt;w...@nestle.ai.mit.edu&amp;gt;&#xA;Date: 30 Jul 2003 23:54:07 -0400&#xA;Message-ID: &amp;lt;y8zhe538i74.fsf@nestle.ai.mit.edu&amp;gt;&#xA;&#xA;Bill Dubuque &amp;lt;w...@nestle.ai.mit.edu&amp;gt; wrote:&#xA;&amp;gt;Rasmus Villemoes &amp;lt;burner+use...@imf.au.dk&amp;gt; wrote:&#xA;&amp;gt;&amp;gt;&#xA;&amp;gt;&amp;gt; In my algebra textbook, the product of two ideals I,J is defined as&#xA;&amp;gt;&amp;gt;&#xA;&amp;gt;&amp;gt; { sum_{i=1..n} a_i b_i | n &amp;gt;= 1 , a_i in I and b_i in J }&#xA;&amp;gt;&amp;gt;&#xA;&amp;gt;&amp;gt; Now it is rather easy to prove that IJ is an ideal in R. The last&#xA;&amp;gt;&amp;gt; question of the exercise is:&#xA;&amp;gt;&amp;gt;&#xA;&amp;gt;&amp;gt; Is A = { ab | a in I, b in J } an ideal of R.&#xA;&amp;gt;&amp;gt;&#xA;&amp;gt;&amp;gt; Now the preceding questions strongly suggest that the answer in&#xA;&amp;gt;&amp;gt; general is no, but I can't find a counterexample. Clearly, (since it&#xA;&amp;gt;&amp;gt; is understood that R is commutative), if one of I or J is a principal&#xA;&amp;gt;&amp;gt; ideal, the set A is an ideal, so a counterexample has to consist of a&#xA;&amp;gt;&amp;gt; non-PID and two ideals generated by at least two elements each  [...]&#xA;&amp;gt; &#xA;&amp;gt; HINT: Find proper ideals whose product contains an irreducible element,&#xA;&amp;gt; &#xA;&amp;gt; e.g.  p in (p,a)(p,b)  if  (a,b) = (1)&#xA;&amp;gt;&#xA;&amp;gt; Examples abound.&#xA;&#xA;Domains where ideals multiply simply as  IJ = { ij : i in I, j in J },&#xA;are called condensed domains. Below are reviews of related papers.&#xA;------------------------------------------------------------------------------&#xA;84a:13019  13F99 &#xA;Anderson, David F.; Dobbs, David E.&#xA;On the product of ideals. &#xA;Canad. Math. Bull. 26 (1983), no. 1, 106-114.&#xA;------------------------------------------------------------------------------&#xA;In this paper the authors define an integral domain R to be a condensed&#xA;domain provided  IJ = {ij: i in I, j in J}  for all ideals I and J of R.&#xA;Bezout domains are condensed domains. The main results of the paper&#xA;characterize condensed domains within some large class of domains. For&#xA;example, it is shown that a GCD-domain is condensed if and only if it is &#xA;a Bezout domain, and a Krull domain is condensed if and only if it is a&#xA;principal ideal domain. For a Noetherian domain  R  to be condensed it &#xA;is necessary that  dim R &amp;lt;= 1  and that the integral closure of  R  be &#xA;a principal ideal.&#xA;    Reviewed by J. T. Arnold&#xA;------------------------------------------------------------------------------&#xA;86h:13017  13F05 (13B20 13G05)&#xA;Anderson, David F.(1-TN); Arnold, Jimmy T.(1-VAPI); Dobbs, David E.(1-TN)&#xA;Integrally closed condensed domains are Bezout. &#xA;Canad. Math. Bull. 28 (1985), no. 1, 98-102.&#xA;------------------------------------------------------------------------------&#xA;An integral domain  R  is termed quasicondensed if  I^n = {i_1 i_2...i_n :&#xA;i_j in I  for  1 &amp;lt;= j &amp;lt;= n}  for each positive integer  n  and each &#xA;two-generated ideal  I = (a,b)  of  R.  R  is said to be condensed if  &#xA;IJ = {ij: i in I, j in J} for all ideals  I  and  J  of  R. The main theorem&#xA;shows that an integral domain is a Bezout domain if and only if it is &#xA;integrally closed and condensed. An example (a  D+M  construction) is given&#xA;of an integrally closed quasicondensed domain which is not a Bezout domain.&#xA;    Reviewed by Anne Grams&#xA;------------------------------------------------------------------------------&#xA;90e:13019  13F30 (13B20 13G05)&#xA;Gottlieb, Christian (S-STOC)&#xA;On condensed Noetherian domains whose integral closures are discrete&#xA;valuation rings. &#xA;Canad. Math. Bull. 32 (1989), no. 2, 166-168.&#xA;------------------------------------------------------------------------------&#xA;Following D. F. Anderson and the reviewer [same journal 26 (1983), no. 1, &#xA;106-114; MR 84a:13019] an integral domain  R  is said to be condensed in case&#xA; IJ = {ij : i in I, j in J}  for all ideals  I,J  of  R. The author defines an&#xA;integral domain  R  to be strongly condensed if for every pair  I,J  of ideals&#xA;of R, either IJ = aJ for some  a in I  or IJ = Ib for some  b in J. Suppose&#xA;henceforth that  R  is a Noetherian integral domain whose integral closure  R'&#xA;is a discrete valuation ring. It is proved that if  R  is condensed, then  R&#xA;contains an element of value 2 (in the associated discrete rank 1 valuation).&#xA;It is not known whether the converse holds, nor whether all condensed domains&#xA;are strongly condensed. As a partial converse, it is proved that R is strongly&#xA;condensed under the following conditions: (R',M')  is a finitely generated&#xA;R-module,  R'/M' is isomorphic to  R/M  and R contains an element of value 2.&#xA;    Reviewed by David E. Dobbs&#xA;------------------------------------------------------------------------------&#xA;1 955 608  13A15 (13Bxx)&#xA;Anderson, D. D.; Dumitrescu, Tiberiu&#xA;Condensed domains. &#xA;Canad. Math. Bull. 46 (2003), no. 1, 3-13. &#xA;http://journals.cms.math.ca/cgi-bin/vault/view/anderson8107&#xA;------------------------------------------------------------------------------&#xA;Abstract: &#xA;An integral domain D with identity is condensed (resp., strongly condensed) if&#xA;for each pair of ideals I,J of D, IJ = {ij : i in I, j in J} (resp., IJ = iJ&#xA;for some i in I or IJ = Ij for some j in J). We show that for a Noetherian&#xA;domain D, D is condensed if and only if Pic(D) = 0 and D is locally condensed,&#xA;while a local domain is strongly condensed if and only if it has the &#xA;two-generator property. An integrally closed domain D is strongly condensed &#xA;if and only if D is a Bezout generalized Dedekind domain with at most one&#xA;maximal ideal of height greater than one. We give a number of equivalencies&#xA;for a local domain with finite integral closure to be strongly condensed.&#xA;Finally, we show that for a field extension k &amp;lt; K, the domain D = k + XK[[X]]&#xA;is condensed if and only if [K:k] &amp;lt;= 2 or [K:k] = 3 and each degree-two&#xA;polynomial in  k[X] splits over  k, while D is strongly condensed if and &#xA;only if [K:k] &amp;lt;= 2.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="242" LastActivityDate="2011-02-10T22:37:46.017" />
  <row Id="21448" PostTypeId="2" ParentId="21127" CreationDate="2011-02-10T22:44:46.397" Score="1" ViewCount="" Body="&lt;p&gt;Solution without equations, only fractions. If we represent by $1$ the&#xA;total money earned by the girl before spending on cloths, games and toys, we&#xA;can split this unit according to the fractions she spent:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$1=\overset{3/4}{\overbrace{\underset{2/3}{\underbrace{\frac{1}{2}+\frac{1}{3}&#xD;&#xA;\cdot \frac{1}{2}}}+\frac{1}{4}\cdot \frac{1}{3}}}+\frac{1}{4}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fraction $\frac{1}{4}$ represents the money left, which we know is $7777&#xD;&#xA;$. So the money she earned in a year is $4\cdot 7777$. An in a month $4\cdot&#xD;&#xA;7777\cdot \frac{1}{12}=\frac{7777}{3}$.&lt;/p&gt;&#xA;" OwnerUserId="752" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2011-02-10T23:19:38.810" LastActivityDate="2011-02-10T23:19:38.810" />
  <row Id="21449" PostTypeId="1" CreationDate="2011-02-10T22:49:01.367" Score="3" ViewCount="152" Body="&lt;p&gt;title says everything. How do I evaluate the limit given ?&lt;/p&gt;&#xA;" OwnerUserId="559" LastActivityDate="2011-02-10T23:23:10.317" Title="How to evaluate the following limit $\lim_{x\rightarrow \infty} \frac{x^x - (x-1)^x}{x^x}$?" Tags="&lt;limit&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="21450" PostTypeId="2" ParentId="21449" CreationDate="2011-02-10T22:53:31.463" Score="7" ViewCount="" Body="&lt;p&gt;Perhaps try dividing? Then&#xA;$$&#xD;&#xA;\lim_{x\to\infty}\frac{x^x-(x-1)^x}{x^x}=\lim_{x\to\infty}\left[1-\left(\frac{x-1}{x}\right)^x\right]=\lim_{x\to\infty}\left[1-\left(1-\frac{1}{x}\right)^x\right].&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice &#xA;$$&#xD;&#xA;\lim_{x\to\infty}\left(1-\frac{1}{x}\right)^x=e^{\lim_{x\to\infty}x\ln(1-\frac{1}{x})}.(*)&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Try making a substitution like $u=1/x$ to get a situation in which l'Hôpital's rule applies to find this limit. Remember this will also change the value the limit approaches. It should look something like this:&#xA;$$&#xD;&#xA;e^{\lim_{u\to 0}\frac{\ln(1-u)}{u}}=e^{\lim_{u\to 0}\frac{1}{u-1}}.&#xD;&#xA;$$&#xA;Apologies for the poor legibility, I hope it at least gets you started.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;*Edit: As Sivaram kindly pointed out, you could use the fact that $\lim_{x\to\infty}(1-\frac{1}{x})^x=e^{-1}$ to get the result right off the bat.&lt;/p&gt;&#xA;" OwnerUserId="1583" LastEditorUserId="1583" LastEditorDisplayName="" LastEditDate="2011-02-10T23:23:10.317" LastActivityDate="2011-02-10T23:23:10.317" CommentCount="7" />
  <row Id="21451" PostTypeId="2" ParentId="21409" CreationDate="2011-02-10T22:55:59.857" Score="2" ViewCount="" Body="&lt;p&gt;This is the strong law of large numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want a measure theoretic proof, check out these lecture notes: &lt;a href=&quot;http://staff.science.uva.nl/~spreij/onderwijs/master/mtp.pdf&quot; rel=&quot;nofollow&quot;&gt;http://staff.science.uva.nl/~spreij/onderwijs/master/mtp.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is in paragraph 10.4.&lt;/p&gt;&#xA;" OwnerUserId="1120" LastActivityDate="2011-02-10T22:55:59.857" />
  <row Id="21452" PostTypeId="2" ParentId="21170" CreationDate="2011-02-10T23:18:35.267" Score="4" ViewCount="" Body="&lt;p&gt;I can see that your property does hold in $\mathbb{Z}$. In fact, it holds in any principal ideal domain, and even in every &lt;a href=&quot;http://en.wikipedia.org/wiki/Dedekind_domain&quot; rel=&quot;nofollow&quot;&gt;Dedekind domain&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[&lt;strong&gt;Edit:&lt;/strong&gt; As pointed out by Rasmus below, for the commutative case, the property is equivalent to the ring being &lt;a href=&quot;http://en.wikipedia.org/wiki/Arithmetic_ring&quot; rel=&quot;nofollow&quot;&gt;arithmetical&lt;/a&gt;. This is because arithmetical rings are defined by the distributive lattice property I state in (2) below. So, for integral domains, the property is equivalent to the ring being &lt;a href=&quot;http://en.wikipedia.org/wiki/Pr%C3%BCfer_domain&quot; rel=&quot;nofollow&quot;&gt;Prüfer&lt;/a&gt;. This certainly includes all Dedekind domains and, in the case of Noetherian domains, it is actually equivalent to the ring being Dedekind.]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think it's just as easy to start by considering an arbitrary ring $R$. Choosing any $x\in\bigoplus_kI_k$ such that $g(x)=0$, then we have $x_1=-\sum_{k\ge2}x_k\in I_1\cap\sum_{k\ge2}I_k$. Conversely, if $x_1\in I_1\cap\sum_{k\ge2}I_k$ then this can be extended to an $x\in\bigoplus_kI_k$ with $g(x)=0$.&#xA;We want this to mean that there exists a $y\in\bigoplus_{j &amp;lt; k}I_j\cap I_k$ with $f(y)=x$. This implies that $x_1=\sum_{k\ge2}y_{1k}\in\sum_{k\ge2}I_1\cap I_k$. So, a necessary condition for the given sequence to be exact in the middle is that&#xA;$$&#xD;&#xA;I_1\cap\sum_{k=2}^nI_k\subseteq\sum_{k=2}^nI_1\cap I_k.\qquad\qquad{\rm(1)}&#xD;&#xA;$$&#xA;The reverse inclusion is immediate for any ring. In particular, just considering $n=3$ gives the necessary condition&#xA;$$&#xD;&#xA;I\cap(J+K)=I\cap J+ I\cap K,\qquad\qquad{\rm(2)}&#xD;&#xA;$$&#xA;for all ideals $I,J,K\subseteq R$. In fact, this is both a necessary and sufficient condition. It can be seen that your original condition and (2) are both just statements about a collection of subgroups of an abelian group (here, the group is $R$ under addition and the subgroups are the ideals). In fact, you are asking when the first homology group of a specific chain complex vanishes (much like in &lt;a href=&quot;http://en.wikipedia.org/wiki/%C4%8Cech_cohomology&quot; rel=&quot;nofollow&quot;&gt;Cech cohomology&lt;/a&gt;). Also, (2) is the same as saying that the ideals form a &lt;a href=&quot;http://en.wikipedia.org/wiki/Distributive_lattice&quot; rel=&quot;nofollow&quot;&gt;distributive lattice&lt;/a&gt; $({\rm Id}_R,+,\cap)$. However, I'm no expert in this, and don't know what the name of rings satisfying your particular condition is (Edit: In the commutative case, these are &lt;em&gt;arithmetical&lt;/em&gt; rings as mentioned above and in a comment below).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To see that (2) implies (1), use induction for $n\ge2$,&#xA;$$&#xD;&#xA;I_1\cap\sum_{k=2}^nI_k=I_1\cap\sum_{k=2}^{n-1}I_k+I_1\cap I_n=\sum_{k=2}^nI_1\cap I_k.&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To see that (1) implies exactness, consider $x\in\bigoplus_kI_k$ with $g(x)=0$. Then, $x_n\in I_n\cap\sum_{k &amp;lt; n}I_k=\sum_{k &amp;lt; n}I_k\cap I_n$. So, there will be a $y\in\bigoplus_{j &amp;lt; k}I_j\cap I_k$ with $y_{jk}=0$ for $j &amp;lt; k &amp;lt; n$ and $f(y)_n=x_n$. Replacing $x$ by $\tilde x=x-f(y)$ reduces to the case with $\tilde x_n=0$, so exactness follows by induction on $n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, (2) holds in any Dedekind domain. It is clear if any of $I,J,K$ are zero, so suppose that they are nonzero. For any ideal $I$ and nonzero prime ideal $\mathfrak{p}$, write $v_{\mathfrak{p}}(I)$ for the index of $\mathfrak{p}$ in the factorization of $I$ (the $\mathfrak{p}$-adic valuation). Two nonzero ideals $I,J$ are equal if and only if $v_{\mathfrak{p}}(I)=v_{\mathfrak{p}}(J)$ for all primes $\mathfrak{p}$. Then, using $v_{\mathfrak{p}}(I+J)=\min(v_{\mathfrak{p}}(I),v_{\mathfrak{p}}(J))$ and $v_{\mathfrak{p}}(I\cap J)=\max(v_{\mathfrak{p}}(I),v_{\mathfrak{p}}(J))$ implies that (2) holds.&lt;/p&gt;&#xA;" OwnerUserId="1321" LastEditorUserId="1321" LastEditorDisplayName="" LastEditDate="2011-02-11T23:21:48.527" LastActivityDate="2011-02-11T23:21:48.527" CommentCount="3" />
  <row Id="21453" PostTypeId="1" AcceptedAnswerId="21455" CreationDate="2011-02-10T23:20:36.853" Score="1" ViewCount="172" Body="&lt;p&gt;Taken from elsewhere on the web : -&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;The derivative of a sine function is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;d SIN(ω*t) / dt = ω * COS(ω &lt;em&gt;t)&#xA;where ω = angular frequency = 2&lt;/em&gt; π *frequency&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and this derivative leads the original function by 90&#xA;degrees (a quarter period) because a cosine wave leads a sine wave by&#xA;90 degrees. This derivative is also different in amplitude from&#xA;the original sine wave because the cosine wave is multiplied by the&#xA;angular frequency. If we take the simple difference of equally spaced successive points of the sine wave (i.e. 20 points in a 20 period sine wave, 21 points in a 21 period sine wave etc.) we can cause that difference to have the same amplitude as the sine wave if we “normalize” the difference. Normalization is done by multiplying the difference by (1/ω). In the case the sine wave the angular frequency is 2*Pi divided by the period of the sine wave therefore the amplitude normalization factor is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Normalizer = Period / (2*π)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the normalizer, this difference function of sine wave now has the same amplitude as the sine wave and leads it by 90 degrees in phase (a quarter cycle). The amount of phase lead can be reduced with a little vector arithmetic. Vector addition of the sine wave and difference function results in a vector that leads the sine wave by only 45 degrees. Since this vector forms the hypotenuse of a right angled triangle, it is also larger than the sine wave by a factor of 1.414. We simply divide the sum by 1.414 to achieve the&#xA;correct amplitude for our leading signal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In summary, we create a leading signal by taking the&#xA;difference of equally spaced successive samples of a sine wave; multiply that difference by Period/(2*π); add that product to the sine wave; and divide the sum by 1.414.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to use this approach to create a leading signal for a sine wave which always has a set constant lead e.g. the above approach gives a 2.5 point lead for a 20 period sine wave but will give a 5 point lead for a 40 period sine wave. Since I will always know the period and the difference between successive points of the sine wave, how can I adjust the vector arithmetic so that the lead will always be a constant 2.5 points (or constant 3.5 points, whatever?) irrespective of the sine wave period?&lt;/p&gt;&#xA;" OwnerUserId="6882" LastEditorUserId="6882" LastEditorDisplayName="" LastEditDate="2011-02-10T23:31:47.210" LastActivityDate="2011-02-10T23:47:09.577" Title="Mathematics of a leading signal for a sine wave" Tags="&lt;signal-processing&gt;" AnswerCount="1" />
  <row Id="21454" PostTypeId="1" AcceptedAnswerId="21456" CreationDate="2011-02-10T23:40:30.987" Score="3" ViewCount="227" Body="&lt;p&gt;Let $A$ be a block upper triangular matrix:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$A = \left( \begin{matrix} A_{1,1}&amp;amp;A_{1,2}\\ 0&amp;amp;A_{2,2} \end{matrix} \right)$ where $A_{1,1} ∈ C^{p&amp;#215;p}$, $A_{2,2} ∈ C^{(n-p)&amp;#215;(n-p)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Show that the eigenvalues of $A$ are the combined eigenvalues of $A_{1,1}$ and $A_{2,2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've been pretty much stuck looking at this for a good hour and a half, so any help would be much appreciated. Thanks.&lt;/p&gt;&#xA;" OwnerUserId="2076" LastActivityDate="2011-02-11T02:14:34.653" Title="Prove that the eigenvalues of a block matrix are the combined eigenvalues of its blocks" Tags="&lt;matrices&gt;&lt;eigenvalues&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="21455" PostTypeId="2" ParentId="21453" CreationDate="2011-02-10T23:47:09.577" Score="1" ViewCount="" Body="&lt;p&gt;As $\sin(\omega t + \phi)=\sin(\omega t) \cos(\phi)+\sin(\phi)\cos(\omega t)$.  $\phi$ is the lead angle.  As you say, $\cos(\omega t)=\frac{1}{\omega}\frac{d\sin(\omega t)}{dt}$.  So $\sin(\omega t + \phi)=\sin(\omega t) \cos(\phi)+\frac{\sin(\phi)}{\omega}\frac{d\sin(\omega t)}{dt}$.  Is this what you were looking for?&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-02-10T23:47:09.577" CommentCount="1" />
  <row Id="21456" PostTypeId="2" ParentId="21454" CreationDate="2011-02-10T23:50:30.993" Score="4" ViewCount="" Body="&lt;p&gt;Let $A$ be the original matrix of size $n \times n$. One way out is to use the identity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\det \left( \begin{matrix} B_{1,1}&amp;amp;B_{1,2}\\ B_{2,1 }&amp;amp;B_{2,2} \end{matrix} \right) = \det(B_{1,1}) \times \det(B_{22} - B_{21}B_{11}^{-1}B_{12})$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know that $\lambda$ is a number such that $Ax = \lambda x$. From which we get $\det(A-\lambda I_n) = 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your case, the matrix $A_{21}$ is a zero matrix and hence, we get $\det(A-\lambda I_n) = \det \left( \left( \begin{matrix} A_{1,1}&amp;amp;A_{1,2}\\ 0&amp;amp;A_{2,2} \end{matrix}\right) - \lambda I_n \right) = \det \left( \begin{matrix} A_{1,1} - \lambda I_{k}&amp;amp;A_{1,2}\\ 0&amp;amp;A_{2,2} - \lambda I_{n-k} \end{matrix}\right)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence $\det(A-\lambda I_n) = \det(A_{1,1} - \lambda I_{k}) \times \det(A_{22} - \lambda I_{n-k})$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we get that if $\lambda$ is an eigen value of $A_{11}$ or $A_{22}$, then either $\det(A_{11}-\lambda I_k) = 0$ or $\det(A_{11}-\lambda I_{n-k}) = 0$ and hence $\det(A-\lambda I_n) = 0$ and hence $\lambda$ is an eigenvalue of $A$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, if $\lambda$ is an eigenvalue of $A$, then $\det(A-\lambda I_n) = 0$, then either $\det(A_{11}-\lambda I_k) = 0$ or $\det(A_{11}-\lambda I_{n-k}) = 0$ and hence $\lambda$ is an eigen value of $A_{11}$ or $A_{22}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is actually a small error in the above argument.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might wonder that if $\lambda$ is an eigenvalue of $A_{11}$, then $A_{11} - \lambda I_k$ is not invertible and hence the identity $\det \left( \begin{matrix} B_{1,1}&amp;amp;B_{1,2}\\ B_{2,1 }&amp;amp;B_{2,2} \end{matrix} \right) = \det(B_{1,1}) \times \det(B_{22} - B_{21}B_{11}^{-1}B_{12})$ is false since $B_{11}$ is not invertible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, there is an another identity $\det \left( \begin{matrix} B_{1,1}&amp;amp;B_{1,2}\\ 0&amp;amp;B_{2,2} \end{matrix} \right) = \det(B_{1,1}) \times \det(B_{22})$ which is always true. (Prove both the identites as an exercise).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can make use of this identity to get&#xA; $\det(A-\lambda I_n) = \det(A_{1,1} - \lambda I_{k}) \times \det(A_{22} - \lambda I_{n-k})$.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2011-02-11T00:25:03.773" LastActivityDate="2011-02-11T00:25:03.773" />
  <row Id="21457" PostTypeId="2" ParentId="21435" CreationDate="2011-02-10T23:50:32.943" Score="2" ViewCount="" Body="&lt;p&gt;This is quite trivial if viewed correctly. In a nutshell it boils down to the observation that we can employ rational-based rulers, i.e. we can change our unit of measurement on the number line from $1$ to any rational $\rm\:s\:$. More precisely, suppose we wish to find a rational in the real interval $\rm\ (a,b),\ a,b\in \mathbb R\:$. Simply choose some rational $\rm\:r &amp;lt; a\ $ and a positive rational step size $\rm\ s &amp;lt; b-a\:.\ $ Now, starting from $\rm\:r\:,\ $ keep taking steps of size $\rm\:s\:.\ $ By the Archimedean Property (AP) eventually you'll surpass $\rm\:b\:$, and taking a step back from the first such point necessarily lands you at a rational point in $\rm\ (a,b)\:,\ $ since the step size is smaller than the interval. Since the proof uses only the Archimedean property it works for any Archimedean field. Essentially it's simply employing the (Euclidean) division algorithm for rationals. Equivalently, instead of choosing the unit $1$ for our number line we instead choose the unit to be the rational number $\rm\:s\:$ that is our step size. In fact these ideas go all the way back to Euclid, who used it to (effectively) compute the gcd of rationals (greatest common measure of line segments).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It proves illuminating to examine the proof in JM's answer from this geometric viewpoint: by AP we can scale the interval $\rm\ I = (a,b)\ $ by an integer factor $\rm\:n\:$ to obtain $\rm\:n\:I = (na,nb)$ of length $&amp;gt; 1\:.\ $ Thus by AP there is an integer $\rm\:k\in n\:I\:$ hence a rational $\rm\:k/n\in I\:.\ $ &lt;strong&gt;QED&lt;/strong&gt; $\ $ Notice how much more intuitive the proof is when presented in this geometric form. The key to its success is that the problem is &lt;strong&gt;invariant under rational scaling symmetries&lt;/strong&gt;. Thus  we can scale it to reduce it to what amounts to a trivial floor or ceiling computation (these functions exist by AP and induction).&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2011-02-11T03:36:01.407" LastActivityDate="2011-02-11T03:36:01.407" />
  <row Id="21458" PostTypeId="2" ParentId="21438" CreationDate="2011-02-10T23:58:10.477" Score="1" ViewCount="" Body="&lt;p&gt;I'm struggling with this myself right now! And one of the problems I often have is having a good notation. Normally, I use one notation in my notebooks, but when I make a clean version in Latex I fix them. Once I finish with the latex file, I re-read my proofs, and look for ways to improve the exposition and the notation. I find this a good way to find mistakes, and to make it easier to read. There are always ways to find better notation when you write your proofs in latex and read from the nice version. I find this process of writing your proofs immediately in latex a good practice. I also write dates in my notebooks to know when I did it, and the ideas on that day. Also, I put the name of the latex file corresponding to that part.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I developed this approach for myself without nobody telling me, and it was based on trial and error. Before that, I gave my advisor a lot of headaches. &lt;/p&gt;&#xA;" OwnerUserId="1330" LastActivityDate="2011-02-10T23:58:10.477" />
  <row Id="21459" PostTypeId="1" AcceptedAnswerId="21469" CreationDate="2011-02-11T00:23:31.730" Score="6" ViewCount="173" Body="&lt;p&gt;This is Exercise I.11.4 of Bourbaki's &lt;em&gt;General Topology&lt;/em&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $X$ be a connected space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a) Let $A$ be a connected subset of $X$, $B$ a subset of $\complement A$ which is both open and closed in $\complement A$. Show that $A\cup B$ is connected.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;b) Let $A$ be a connected subset of $X$ and $B$ a component of the set $\complement A$. Show that $\complement B$ is connected (use a)). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have managed to show a). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My attempts to show b): Let $C$ be a nonempty clopen subset of $\complement B$. By a), $B\cup C$ is connected. Since $B$ is a component of $\complement A$, $B\cup C$ cannot be a subset of $\complement A$. So $C$ has to contain an element of $A$. Since $A$ is connected, $A\subset C$. But I'm stuck here. I can't see how this implies $C=\complement B$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can someone point me in the right direction?  &lt;/p&gt;&#xA;" OwnerUserId="1878" LastActivityDate="2011-02-11T10:36:31.163" Title="Bourbaki exercise on connected sets" Tags="&lt;topology&gt;&lt;general-topology&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" />
  <row Id="21460" PostTypeId="1" CreationDate="2011-02-11T00:29:31.007" Score="3" ViewCount="214" Body="&lt;p&gt;Suppose $1&amp;lt;p_1&amp;lt;p_2&amp;lt;\infty$, then show $L^{p_1}[a,b] \supset L^{p_2}[a,b]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was able to show $||f||_{p_1} \le ||f||_{p_2} (b-a)^{1/p_1 - 1/p_2}$ but I'm not sure how to proceed from here.&lt;/p&gt;&#xA;" OwnerUserId="6689" LastEditorUserId="6296" LastEditorDisplayName="" LastEditDate="2011-03-24T13:10:05.903" LastActivityDate="2011-03-24T19:07:50.280" Title="How to show that $L^p$ spaces are nested?" Tags="&lt;measure-theory&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="21461" PostTypeId="2" ParentId="21454" CreationDate="2011-02-11T00:32:26.223" Score="4" ViewCount="" Body="&lt;p&gt;A simpler way is from the definition. Is is easy to show that if $\lambda_1$ is an eigenvalue of the upper diagonal block $A_{1,1}$, with eigenvector $p_1$,  (size $n_1$) then it's also an eigenvalue of the full matrix, with the same eigenvector augmented with zeros.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$A_{1,1} \; p_1 =  \lambda_1 p_1$ with  $p_1 \ne 0 $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ \left( \begin{matrix} A_{1,1}&amp;amp;A_{1,2} \\ 0 &amp;amp;A_{2,2} \end{matrix} \right) &#xD;&#xA;\left( \begin{matrix} p_1 \\ 0 \end{matrix} \right) = &#xD;&#xA;\left( \begin{matrix} A_{1,1} \; p_1 \\ 0 \end{matrix} \right) = &#xD;&#xA;\left( \begin{matrix} \lambda_1 p_1 \\ 0 \end{matrix} \right) = &#xD;&#xA;\lambda_1 \left( \begin{matrix} p_1 \\ 0 \end{matrix} \right) $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are $n_1$ (counting multiplicity) such eigenvalues. &lt;s&gt;The same applies to the lower diagonal block $A_{2,2}$. So we have found the $n_1$ + $n_2 = n$ eigenvalues of the full matrix. &lt;/s&gt; (Wrong! This only applied to block diagonal matrix - Fixed below)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suposse now that $\lambda_2$ is eigenvalue of $A_{2,2}$ with eigenvector $p_2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\left( \begin{matrix} A_{1,1}&amp;amp;A_{1,2} \\ 0 &amp;amp;A_{2,2} \end{matrix} \right) &#xD;&#xA;\left( \begin{matrix} x \\ p_2 \end{matrix} \right) =&#xD;&#xA;\left( \begin{matrix} A_{1,1} x + A_{1,2} p_2 \\ \lambda_2 p_2 \end{matrix} \right) &#xD;&#xA;$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $\lambda_2$  is not an eigenvalue of $A_{1,1}$ , then we &#xA;can make $ A_{1,1} x + A_{1,2} p_2 = \lambda x$ by choosing $x = - (A_{1,1} - \lambda_2 I)^{-1} A_{1,2} \; p_2$ &#xA;and then we found an eigenvector for $A$ with $\lambda_2$ eigenvalue.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It this way, we showed that if $\lambda$ is eigenvalue of $A_{1,1}$ or $A_{2,2}$, then it's an eigenvalue of $A$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To complete the proof, one should show the other way round: that if $\lambda$  is eigenvalue of $A$ then it's eigenvalue of $A_{1,1}$ or $A_{2,2}$. But that's easy:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\left( \begin{matrix} A_{1,1}&amp;amp;A_{1,2} \\ 0 &amp;amp;A_{2,2} \end{matrix} \right) &#xD;&#xA;\left( \begin{matrix} x_1 \\ x_2 \end{matrix} \right) =&#xD;&#xA;\left( \begin{matrix} A_{1,1} \;  x_1 + A_{1,2} \; x_2 \\ A_{2,2} \; x_2 \end{matrix} \right) &#xD;&#xA;= \left( \begin{matrix} \lambda \; x_1 \\ \lambda \; x_2 \end{matrix} \right) &#xD;&#xA;$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, either $x_2 = 0$ or not.  If not, then $\lambda$ is eigenvalue of $A_{2,2}$. If yes,&#xA;it's eigenvalue of $A_{1,1}$.  &lt;/p&gt;&#xA;" OwnerUserId="312" LastEditorUserId="312" LastEditorDisplayName="" LastEditDate="2011-02-11T02:14:34.653" LastActivityDate="2011-02-11T02:14:34.653" CommentCount="4" />
  <row Id="21462" PostTypeId="2" ParentId="1747" CreationDate="2011-02-11T00:42:42.493" Score="0" ViewCount="" Body="&lt;p&gt;I think mathematics/statistics will give us a lot of nice continuous analyses of the data, including quantitative measures of confidence, but the discrete decision to presume that the coin is loaded is arbitrary, even if based on those measures of confidence.&lt;/p&gt;&#xA;" OwnerUserId="7681" LastActivityDate="2011-02-11T00:42:42.493" />
  <row Id="21463" PostTypeId="1" CreationDate="2011-02-11T01:11:32.610" Score="3" ViewCount="118" Body="&lt;p&gt;I'm having a really hard time understanding how to figure out if a mapping is linear or not. Here is my homework question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Determine which of the following mappings F are linear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(a) $F: \mathbb{R}^3 \to \mathbb{R}^2$ defined by $F(x,y,z) = (x, z)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(b) $F: \mathbb{R}^4 \to \mathbb{R}^4$ defined by $F(X) = -X$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(c) $F: \mathbb{R}^3 \to \mathbb{R}^3$ defined by $F(X) = X + (0, -1, 0)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sorry about my formatting. I'm not sure how to write exponents and the arrow showing that the mapping is from R^n to R^m. Any help is greatly appreciated!!&lt;/p&gt;&#xA;" OwnerUserId="6886" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2011-02-11T01:41:00.087" LastActivityDate="2011-02-11T22:24:07.097" Title="Determine which of the following mappings F are linear." Tags="&lt;linear-algebra&gt;&lt;homework&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="21464" PostTypeId="2" ParentId="21438" CreationDate="2011-02-11T01:14:31.857" Score="1" ViewCount="" Body="&lt;p&gt;There are already several good answers here. I'd like to add a point that hasn't yet been made.  When you are studying mathematics (or almost any subject) in college, you typically follow a path of progression starting with a survey of broad mathematical concepts and methods and, assuming you go to grad school; you end up with a more directed focus on at least a branch of mathematics, if not much more narrowly focused than that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you study different areas, you are introduced to different notational methods and symbols. You remember some, but probably don't hold many in memory that you don't use in whatever course you are studying at the time.  It's easy to get overwhelmed. But as you narrow your focus, the notation also narrows and it becomes more important to you because you have invested more time and energy in learning it. And hopefully, because it excites you.  As you become more focused on a particular body of work, the notation becomes second nature and fades into the background while you focus on whatever it is that you are using the notation for.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In my case, my daily work is far from the number theory I studied and I have lost some familiarity with the notation used there.  But I have built my vocabulary in the more applied mathematics I use regularly in my day-job.  In fact, when I read this question, my first thought was that it did not apply to me because the notation I use is so trivial it's obvious to anyone. Then I realized much of it is far out of the mainstream of &quot;general&quot; mathematics and is only obvious because I use it frequently.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To summarize, right now you are wrestling with the notation. But as you get deeper into the mathematics behind it, the notation will sink into the background and you will not see it as a significant challenge.  At least that has been my experience. &lt;/p&gt;&#xA;" OwnerUserId="6713" LastActivityDate="2011-02-11T01:14:31.857" />
  <row Id="21466" PostTypeId="2" ParentId="21429" CreationDate="2011-02-11T01:19:29.580" Score="3" ViewCount="" Body="&lt;p&gt;Every linear transformation on a finite dimensional complex inner product space is unitarily triangularizable.  See for example Hogben's &lt;em&gt;&lt;a href=&quot;http://books.google.com/books?id=n2g-x1OIbvYC&amp;amp;lpg=PP1&amp;amp;pg=SA7-PA3#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;Handbook of linear algebra&lt;/a&gt;&lt;/em&gt;.  This means you can find an orthonormal basis $e_1,\ldots e_n$ such that $Te_k=\sum_{i\leq k}a_{ik}e_i$.  The eigenvalues of a triangular matrix are the diagonal entries, so $|a_{kk}|=1$ for each $k$.  Thus $1\geq\|Te_k\|^2=\sum_{i\leq k}|a_{ik}|^2\geq|a_{kk}|^2=1$, forcing $a_{ik}=0$ for $i\lt k$.  So the basis actually diagonalizes $T$, and it is now straightforward to show that $T$ is unitary regardless of which definition you use. &lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2011-02-11T01:19:29.580" CommentCount="8" />
  <row Id="21467" PostTypeId="2" ParentId="21438" CreationDate="2011-02-11T01:19:48.703" Score="12" ViewCount="" Body="&lt;p&gt;In many ways, I am atypical in the way that I approach a problem, but it works for me. Specifically, I try to understand an example in as much detail as I possibly can. If the example, is too complicated, then I make a simpler example. As much of the intricate detail that I can bring to bear on the example is brought. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, instead of trying to understand Lie groups and Lie algebras in general, start with the circle and the line that is tangent at the point (1,0). What is the exponential map? Oh, OK. Now how about $SU(2)$ and $su(2)$? Can you understand that the Lie group is the $3$-dimensional sphere? Can you understand the coordinates? Can you understand the equators? How do $i,j$ and $k$ really work?&#xA;What is the difference between the multiplication rule $i\times i =0$ and $i^2=-1$? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I spend time pondering. And often my notebooks will contain tangential problems or specific computations. I will keep doing the computation until I get it right! If necessary, I will write a program to complete the computation. When I understand the example completely, it is usually easy to abstract. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then I follow up, usually writing in a notebook or several notebooks before I begin writing on the computer. I have an advantage in that I have long-distance collaborators, so it becomes necessary to explain the idea to the collaborator(s). That is the first writing stage: write for someone who knows your short-hand and your metaphors. the second stage is to write for someone who does not. Then I write with a set of colleagues in mind, but I assume the colleagues do not remember anything from the previous work. I also try to explicate the notation writing for example &quot;the function $f$, the knot $k$, or the tubular neighborhood $N$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A complex analytical colleague only uses $z$ for a complex number, $x$ for a real variable, and $n$ for an integer. These variable choices are culturally determined, and so one keeps with the culture of the discipline unless there is good reason to deviate. As a final example of this, the variable $A$ in the bracket polynomial is known to everyone in the field. The variables $q$, $t$, $X$ etc. are less known and involve different normalizations. So it is the burden of the author to relate these to the more well known choices.&lt;/p&gt;&#xA;" OwnerUserId="722" LastActivityDate="2011-02-11T01:19:48.703" CommentCount="1" />
  <row Id="21468" PostTypeId="2" ParentId="21463" CreationDate="2011-02-11T01:21:43.630" Score="2" ViewCount="" Body="&lt;p&gt;To check if a mapping is linear in general, all you need is verify the two properties.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$f(x+y) = f(x) + f(y)$&lt;/li&gt;&#xA;&lt;li&gt;$f(\alpha x) = \alpha f(x)$&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The above two can be combined into one property: $f(\alpha x + \beta y) = \alpha f(x) + \beta f(y)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, if we want to show say $F(x) = f(x_1,x_2,x_3) = x_1 - 4x_2 + x_3$ is linear, where $x = (x_1,x_2,x_3)$, then all you need to do is as follows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align*}&#xA;F(\alpha x + \beta y) &amp;amp; = f(\alpha x_1 + \beta y_1,\alpha x_2 + \beta y_2,\alpha x_3 + \beta y_3) \\&#xA; &amp;amp; = (\alpha x_1 + \beta y_1) - 4(\alpha x_2 + \beta y_2) + (\alpha x_3 + \beta y_3)\\&#xA; &amp;amp; = (\alpha x_1 - 4 \alpha x_2 + \alpha x_3) + (\beta y_1 - 4 \beta y_2 + \beta y_3)\\&#xA; &amp;amp; = \alpha (x_1 - 4 x_2 + x_3) + \beta (y_1 - 4 y_2 + y_3)\\&#xA; &amp;amp; = \alpha f(x_1,x_2,x_3) + \beta f(y_1,y_2,y_3)\\&#xA; &amp;amp; = \alpha F(x) + \beta F(y)\\&#xA;\end{align*}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence the above function is linear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As Arturo points out problem $c$ is not a linear map because of the constant hanging around. Such maps are called affine maps. Affine maps are those for which $f(x) - f(0)$ is a linear map.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2011-02-11T22:24:07.097" LastActivityDate="2011-02-11T22:24:07.097" CommentCount="6" />
  <row Id="21469" PostTypeId="2" ParentId="21459" CreationDate="2011-02-11T02:26:32.310" Score="3" ViewCount="" Body="&lt;p&gt;Suppose $B^{c} = U \cup V$ with $U,V$ disjoint and open. Then $U$ and $V$ are also closed and if they are non-empty, they both must contain $A$ by your argument, so they can't be disjoint.&lt;/p&gt;&#xA;" OwnerUserId="5363" LastEditorUserId="5363" LastEditorDisplayName="" LastEditDate="2011-02-11T10:36:31.163" LastActivityDate="2011-02-11T10:36:31.163" CommentCount="2" />
  <row Id="21470" PostTypeId="1" AcceptedAnswerId="21495" CreationDate="2011-02-11T02:28:37.593" Score="2" ViewCount="107" Body="&lt;p&gt;I am wondering the validity and the procedure to condense/group several states into one in markov chain, namely, if it is possible, how to transform the state vector and the transition matrix? Many thanks.&lt;/p&gt;&#xA;" OwnerUserId="6016" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2011-02-11T03:38:06.653" LastActivityDate="2011-02-11T21:11:55.197" Title="What's the procedure to condense several states into one in Markov chain?" Tags="&lt;probability&gt;&lt;stochastic-processes&gt;" AnswerCount="2" />
  <row Id="21471" PostTypeId="1" AcceptedAnswerId="21476" CreationDate="2011-02-11T02:34:02.833" Score="4" ViewCount="51" Body="&lt;p&gt;Notation: $|A|$ is the Lebesgue measure of $A \subset \mathbb{R}^d$, and $A_\delta = \{ x : \text{dist}(x,A) \leq \delta \} $ is the $\delta$-neighborhood of $A$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to show that there is a constant $C$ such that $|E_{2\delta}| \leq C |E_{\delta}|$ for all compact sets $E \subseteq \mathbb{R}^d$ and all $\delta &amp;gt; 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have been told I need to apply the Hardy-Littlewood maximal theorem to $f = \mathbb{1}_{E_{\delta}}$.  For $p &amp;gt; 1$, this gives the estimate $||Mf||_p^p \leq C ||f||_p^p = C|E_{\delta}|$.  For $p = 1$, it is the estimate $\lambda \cdot |\{x : Mf(x) &amp;gt; \lambda    \}| \leq C ||f||_1 = C |E_{\delta}|$.  Here&#xA;$$&#xD;&#xA;Mf(x) = \sup_{r &amp;gt; 0} \frac{1}{B(x,r)} \int_{B(x,r)} |f(y)| dy.&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think I should try to show that $|E_{2 \delta}| \leq C ||Mf||_p^p$.  I see that &#xA;$$Mf(x) = \sup_{r &amp;gt; 0} \frac{|E_{\delta} \cap B(x,r)|  }{|B(x,r)|}.$$  But I am stuck after that.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know I need to use the compactness of $E$ somehow.  My thought is that I should look at a finite covering by balls $B(x,r)$ or balls $B(x,\delta)$ or something like that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can someone please help?&lt;/p&gt;&#xA;" OwnerUserId="6887" LastEditorUserId="6887" LastEditorDisplayName="" LastEditDate="2011-02-11T21:15:17.403" LastActivityDate="2011-02-11T21:15:17.403" Title="Bound on the relative measure of $\delta$-neighbourhoods of compact sets using the Hardy-Littlewood maximal theorem" Tags="&lt;real-analysis&gt;&lt;harmonic-analysis&gt;" AnswerCount="1" />
  <row Id="21472" PostTypeId="1" CreationDate="2011-02-11T02:35:15.157" Score="9" ViewCount="150" Body="&lt;p&gt;Hi, I'm hoping the userbase here doesn't mind if I do a little crowd-sourcing. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm curious to find out about popular general-interest mathematics or statistics classes that are offered universities that you know about.  By popular I mean &lt;em&gt;crowd pleasing&lt;/em&gt;, in that these courses are suitable to large audiences, the course is aimed at the student that wants to know a little bit about the subject -- but they don't want to take a foundational course, or a course on proofs. This course isn't intended for majors in any particular subject, more for students that just want to dabble in some ideas but not &quot;dive in&quot; -- so there would likely be little homework of much intensity (by a math major's standards, at least). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Physics or astronomy departments sometimes have courses with titles like &quot;astrophysics for poets&quot;.  I'm curious about math offerings you might know of that are of that flavour. I'm particularly interested in course-catalogue listings and course webpages, if possible.   I'm not interested in one-off type courses -- I want courses that have been repeated with success, and preferrably having more than one instructor that has been able to offer it with success (sorry if this is getting demanding).  I'm not very interested in upper-level courses (like a fractals or dynamics course that assume students know a fair bit of calculus).  I wonder if many people have been able to turn topics like gambing or knot theory into accessible &quot;no background required&quot; offerings or not. &lt;/p&gt;&#xA;" OwnerUserId="642" LastActivityDate="2011-02-11T03:52:41.060" CommunityOwnedDate="2011-02-11T03:03:02.903" Title="Popular general-interest math courses" Tags="&lt;education&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="21473" PostTypeId="2" ParentId="21438" CreationDate="2011-02-11T02:36:30.897" Score="3" ViewCount="" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;How do good mathematicians format their work on paper so as not to get lost in the is, js, and ks and keep track of what they're investigating?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To be honest, part of this is just getting used to juggling several things in your head at once. Fortunately, as with many other skills, this is trainable: try, for example, doing Sudoku puzzles with a pen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are, of course, other ways to keep your work organized. The first thing to do when solving a problem is to write down all of the data you're given and write down the goal of the problem. The second thing is to unravel all of the definitions, working through all the quantifiers. This will get you surprisingly far, at least in real analysis. And then the third thing is to actually think about it. If you can't hold all the quantifiers in your head at once, practice reading the statements aloud slowly until you can (and see the first paragraph).&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;When good mathematicians get intuitionistic ideas, what (explicit) steps do they take to formalize them, especially when it is likely that first idea is murky or wrong?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Write down an argument that follows the lines of the intuition, at least for a simpler case or version of the problem. If it doesn't work on the simple version, it probably won't work on the hard version. If it does, you can try to figure out whether it extends to the original question and, if not, what the barrier is. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is easier said than done. The bottom line is you need to practice the skill of turning your intuitions into proofs. This comes in a few steps: first you need to learn how to prove things, then you need to learn how to train your intuition to help you prove things more easily. Like anything else, this takes hard work and practice and there isn't a magical shortcut. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Terence Tao has written some &lt;a href=&quot;http://terrytao.wordpress.com/career-advice/&quot; rel=&quot;nofollow&quot;&gt;very clear stuff&lt;/a&gt; on this and related subjects.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;How do you [professional and aspiring mathematicians] organize your math &quot;notebook&quot;, and what perhaps idiosyncratic methods do you employ to be original and clever within it?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm not completely sure what this means.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;There seems to me a bit of a jump between the dryly algorithmic way one is taught to do math in high school and the more abstruse methods at the undergraduate level. I'd be interested if anyone here could help me bridge that gap, if only for my personal fulfillment.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Many people I know bridged the gap through competitions such as the &lt;a href=&quot;http://amc.maa.org/&quot; rel=&quot;nofollow&quot;&gt;AMC&lt;/a&gt;. Training for competitions is not everybody's style, but it is a chance to be exposed to interesting topics not covered in the high school curriculum and also a chance to hone problem-solving and proof-writing skills (at the Olympiad level). A book I benefited from enormously while doing this is Engel's &lt;a href=&quot;http://books.google.com/books?id=wj3_FxnRz5kC&amp;amp;printsec=frontcover&amp;amp;dq=engel+problem-solving+strategies&quot; rel=&quot;nofollow&quot;&gt;Problem-Solving Strategies&lt;/a&gt;, which is geared fairly specifically to Olympiad preparation but is a great source of problems and elementary techniques for solving them. For more general advice, Polya's &lt;a href=&quot;http://en.wikipedia.org/wiki/How_to_Solve_It&quot; rel=&quot;nofollow&quot;&gt;How to Solve It&lt;/a&gt; comes highly recommended (although I have not read it myself).&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2011-02-11T02:36:30.897" CommentCount="2" />
  <row Id="21474" PostTypeId="2" ParentId="21470" CreationDate="2011-02-11T02:46:41.123" Score="2" ViewCount="" Body="&lt;p&gt;If all the transition probabilities out of a set of states ${\cal S} = \{S_1, S_2, ... S_n\}$ are equal (that is, $P(S_i\rightarrow T)=P(S_j\rightarrow T)$ for all $i$ and $j$ and for all states $T \notin {\cal S}$), then those states may be consolidated into a single new state $S^{\prime}$ without affecting the dynamics.  The state vector will no longer record the individual probabilities of the states in ${\cal S}$, but at any time $P(S^{\prime})$ will be equal to $\sum_i P(S_i)$ in the original model.  The transition probabilities into the new state should be $P(T \rightarrow S^{\prime}) = \sum_i P(T \rightarrow S_i)$ for $T \notin {\cal S}$; the transition probabilities out of the new state should be $P(S^{\prime} \rightarrow T) = P(S_i \rightarrow T)$ for $T \notin {\cal S}$ (this is independent of the choice of $S_i$); and the probability for remaining in the new state should be $P(S^{\prime} \rightarrow S^{\prime}) = \sum_j P(S_i \rightarrow S_j)$ (again independent of the choice of $S_i$).&lt;/p&gt;&#xA;" OwnerUserId="5546" LastActivityDate="2011-02-11T02:46:41.123" />
  <row Id="21475" PostTypeId="1" AcceptedAnswerId="21492" CreationDate="2011-02-11T03:23:59.530" Score="8" ViewCount="182" Body="&lt;p&gt;After reading about Russell's paradox, I see that the set of all sets does not exist, so instead it is called a class.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What other commonly known classes exist that are not sets? I know the class of all singleton sets is not a set, because you can unite that class to get the class of all sets. This seems like the class of all sets of a fixed finite size is not a set either, correct? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are some examples of other 'things' that can't be gathered up and put into a set?&lt;/p&gt;&#xA;" OwnerUserId="6891" LastActivityDate="2011-02-11T06:53:41.327" Title="What are some examples of classes that are not sets?" Tags="&lt;elementary-set-theory&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="21476" PostTypeId="2" ParentId="21471" CreationDate="2011-02-11T03:27:30.820" Score="2" ViewCount="" Body="&lt;p&gt;You can use the Vitali covering lemma to show this. Let $E_\delta = \{x \in R^d: dist(x,A) &amp;lt; \delta\}$; it's a little easier to work with the $E_\delta$ and the result is entirely equivalent since $E_{\delta} \subset A_{\delta} \subset A_{2\delta} \subset E_{4\delta}$ and you can just apply the $E_{\delta}$ result twice. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;By Vitali, there is a countable collection of disjoint balls $B(x_n,2\delta)$ with $x_n \in A$ for all $n$ such that &#xA;$$E_{2\delta} = \cup_{x \in A}B(x,2\delta) \subset \cup_n B(x_n,10\delta)$$&#xA; So we have&#xA;$$|E_{2\delta}| \leq \sum_n |B(x_n,10\delta)| $$&#xA;$$= 10^d \sum_n |B(x_n,\delta)|$$&#xA;Since these balls are disjoint subsets of $E_{\delta}$ this is at most&#xA;$$ \leq 10^d |E_{\delta}|$$&#xA;This gives the desired result. &lt;/p&gt;&#xA;" OwnerUserId="3035" LastActivityDate="2011-02-11T03:27:30.820" CommentCount="3" />
  <row Id="21477" PostTypeId="1" CreationDate="2011-02-11T03:36:52.610" Score="1" ViewCount="61" Body="&lt;p&gt;I am trying to estimate, in either direction, the following integral:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\int_{B} a^{i,j} X_j D_ig dx$ . We are integrating over a unit ball, $B$, and $[a^{i,j}]$ is positive. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have tried to use ellipticity, but it didn't work. I was wondering if some one can show me:&#xA;a. How to use ellipticity to estimate the integral, or&#xA;b. Is there any other inequality that can be useful to estimating the integral.&#xA;Thanks in advance.&lt;/p&gt;&#xA;" OwnerUserId="6892" LastEditorUserId="5253" LastEditorDisplayName="" LastEditDate="2011-02-11T05:47:43.337" LastActivityDate="2011-02-13T05:57:41.157" Title="Estimating using ellipticity" Tags="&lt;differential-equations&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="21478" PostTypeId="2" ParentId="21472" CreationDate="2011-02-11T03:45:39.870" Score="3" ViewCount="" Body="&lt;p&gt;You might check out &lt;em&gt;&lt;a href=&quot;http://www.cambridge.org/gb/knowledge/isbn/item1161841/?site_locale=en_GB&quot; rel=&quot;nofollow&quot;&gt;When Topology meets Chemistry&lt;/a&gt;&lt;/em&gt; by Erica Flapan or &lt;em&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/071672393X&quot; rel=&quot;nofollow&quot;&gt;The Knot Book&lt;/a&gt;&lt;/em&gt; by Colin Adams.&lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2011-02-11T03:45:39.870" CommunityOwnedDate="2011-02-11T03:45:39.870" CommentCount="1" />
  <row Id="21479" PostTypeId="2" ParentId="21472" CreationDate="2011-02-11T03:52:41.060" Score="2" ViewCount="" Body="&lt;p&gt;My university offers a 2nd year course on Discrete math which is very popular.  It requires no prerequisite knowledge except highschool algebra.  This is basically a university level course on problem solving.  It includes puzzles, games, number theory, riddles, as well as basic coding thoery, crytopgraphy and graph theory.  The course follows the text &quot;The puzzling Adventures of Dr Echo&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://books.google.ca/books?id=UvYGqvXfgbcC&amp;amp;printsec=frontcover&amp;amp;dq=%5CThe+Puzzling+Adventures+of+Dr.+Ecco&amp;amp;source=bl&amp;amp;ots=8LlBwI0kE9&amp;amp;sig=ZbW7f9d45RamYiL7G62Nb6UNq8U&amp;amp;hl=en&amp;amp;ei=6LFUTZSCEYa0lQe7pcmTBw&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=2&amp;amp;ved=0CB8Q6AEwAQ#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;http://books.google.ca/books?id=UvYGqvXfgbcC&amp;amp;printsec=frontcover&amp;amp;dq=%5CThe+Puzzling+Adventures+of+Dr.+Ecco&amp;amp;source=bl&amp;amp;ots=8LlBwI0kE9&amp;amp;sig=ZbW7f9d45RamYiL7G62Nb6UNq8U&amp;amp;hl=en&amp;amp;ei=6LFUTZSCEYa0lQe7pcmTBw&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=2&amp;amp;ved=0CB8Q6AEwAQ#v=onepage&amp;amp;q&amp;amp;f=false&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a required course for all students wishing to become math teachers as it explains much of the theory behind the games and puzzles used in elementary school.  There is no course website but here are some assignments so you can see what I'm talking about.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.math.ualberta.ca/~bpowell/Math%20222/Exam/Exam%20Review.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.math.ualberta.ca/~bpowell/Math%20222/Exam/Exam%20Review.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.math.ualberta.ca/~bpowell/Math%20222/Assignment%203/Ast%203.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.math.ualberta.ca/~bpowell/Math%20222/Assignment%203/Ast%203.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3971" LastActivityDate="2011-02-11T03:52:41.060" CommunityOwnedDate="2011-02-11T03:52:41.060" CommentCount="4" />
  <row Id="21480" PostTypeId="2" ParentId="21475" CreationDate="2011-02-11T03:53:16.137" Score="4" ViewCount="" Body="&lt;p&gt;First, any description which doesn't uniquely specify the elements of a set tends to give a proper class.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, a group is an set together with operations satisfying a certain collection of axioms.  Since these axioms don't pin down the elements of the group, one might expect the collection of all groups to not be a set.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This holds in a similar fasion for topological spaces, rings, vector spaces, fields, manifolds, metric spaces, and many of the other commonly studied objects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This also applies to your &quot;collection of all singletons&quot; example, and similarly to any sort of &quot;collection of all sets of a fixed size&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But just because a description DOES pin down the elements doesn't mean you have a set.  For example, the collection of all ordinal numbers is not a set, despite the fact that if there is an order preserving bijection between two ordinals $\alpha$ and $\beta$, then one must have $\alpha = \beta$ as sets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, the collection of all cardinals doesn't form a set.&lt;/p&gt;&#xA;" OwnerUserId="331" LastActivityDate="2011-02-11T03:53:16.137" CommentCount="2" />
  <row Id="21481" PostTypeId="1" AcceptedAnswerId="21482" CreationDate="2011-02-11T03:56:34.370" Score="1" ViewCount="90" Body="&lt;p&gt;Given that $y&amp;#39;(x)+p(x)y(x)\geq 0$ and $y(x_0)\geq 0$, how does one go about showing that $y(x)\geq0$ for all $x\in [x_0,\infty)$&lt;/p&gt;&#xA;" OwnerUserId="4728" LastActivityDate="2011-03-22T23:04:45.080" Title="Differential equation with inequalities" Tags="&lt;homework&gt;&lt;differential-equations&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="21482" PostTypeId="2" ParentId="21481" CreationDate="2011-02-11T04:03:58.343" Score="2" ViewCount="" Body="&lt;p&gt;Define $b=y&amp;#39;+py$, so that you have, well, $y&amp;#39;+py=b$ and $b\geq0$. Solve the inhomogeneous equation for $y$, and use now that $b$ is non-negative.&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2011-02-11T04:03:58.343" CommentCount="1" />
  <row Id="21483" PostTypeId="1" AcceptedAnswerId="21485" CreationDate="2011-02-11T04:16:26.910" Score="2" ViewCount="114" Body="&lt;p&gt;The book says that $$\lim_{x \rightarrow 0^{-}} \left( \frac{1}{x} - \frac{1}{|x|} \right) \mbox{does not exist}$$&lt;br&gt;&#xA;But, given any $M \lt 0$ of large magnitude, if I choose $\delta = \frac{-x^{2}M}{2}$ then any value of x where $|x-0|&amp;lt; \delta$ and $x &amp;lt;0$ (as we are coming from the left) will lead to $\left( \frac{1}{x} - \frac{1}{|x|} \right) &amp;lt; M$. To me, that says that my text book is incorrect in saying that this limit &quot;d.n.e.&quot; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm a little bothered that my $\delta$ depends on $x$, but I tried a few numerical examples and it worked fine. Perhaps the function is not uniformly continuous when $x \lt 0$? I have not done enough work to answer that question yet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe the book meant to say&lt;br&gt;&#xA;$$\lim_{x \rightarrow 0} \left( \frac{1}{x} - \frac{1}{|x|} \right) \mbox{does not exist?}$$&#xA;Or maybe I have missed something elementary. &lt;/p&gt;&#xA;" OwnerUserId="2563" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-11T14:34:46.397" LastActivityDate="2011-02-11T14:34:46.397" Title="Is the book wrong about this left-hand limit with absolute value? (But, my delta depends on x.)" Tags="&lt;calculus&gt;&lt;limit&gt;&lt;absolute-value&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="1" />
  <row Id="21484" PostTypeId="1" CreationDate="2011-02-11T04:28:43.173" Score="1" ViewCount="74" Body="&lt;p&gt;The goal of this problem is to solve the initial value problem&lt;/p&gt;&#xA;&#xA;&lt;p&gt;y' = -f(x)y;  y(0) = 1;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where&#xA;f(x) =&#xA;(1 if x&amp;lt;=2,&#xA;3/2 if x&gt;2):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since f is discontinuous, it is necessary to solve the above ODE separately in each of&#xA;the intervals where f is continuous.&#xA;(a) Determine the intervals where f is continuous.&#xA;(b) Solve the equation in each of these intervals. Note that each of the solutions&#xA;obtained will have a dierent constant of integration.&#xA;(c) Match the solutions at the points where f is discontinuous, in order to make the&#xA;solution y continuous on R. Note that in this case it is impossible to make y'&#xA;continuous at the points where f is discontinuous.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I just dont understand what part (c) is asking, if anyone could help me with the concepts I would really appreciate it. &lt;/p&gt;&#xA;" OwnerUserId="6895" LastActivityDate="2011-02-11T05:10:13.833" Title="solving an initial value ODE with a twist" Tags="&lt;homework&gt;&lt;differential-equations&gt;" AnswerCount="1" />
  <row Id="21485" PostTypeId="2" ParentId="21483" CreationDate="2011-02-11T04:34:37.877" Score="4" ViewCount="" Body="&lt;p&gt;One problem with your $\delta$ is that $|x|\lt \frac{-x^2M}{2}$ with $x\neq0$ implies $|x|\gt-\frac{2}{M}$.  Having this positive lower bound on $|x|$ means that you are not actually approaching $0$.  As Jason DeVito indicates, taking advantage of the fact that $x\lt0$ to write $|x|=-x$ makes it easier to see why the limit doesn't exist.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a comment you just mentioned that the limit is $-\infty$.  That is true, but then there is a divide in terminology as to what having a limit means.  Infinite limits can be dealt with in terms of convergence in the extended reals, $[-\infty,+\infty]$, but often $\pm\infty$ are treated separately with respect to limits, and the existence of a limit (without further qualification) often only refers to existence of a limit in the real number system, $(-\infty,+\infty)$. &lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2011-02-11T04:53:59.937" LastActivityDate="2011-02-11T04:53:59.937" CommunityOwnedDate="2011-02-11T04:34:37.877" CommentCount="3" />
  <row Id="21486" PostTypeId="2" ParentId="21484" CreationDate="2011-02-11T04:51:02.350" Score="2" ViewCount="" Body="&lt;p&gt;On $x\le 2$, the solution is $y=\exp(-x)$, on x&gt;2 it is $y=C\exp(-3x/2)$.  The reason there is no constant in the first is you are given $y(0)$, and can find the constant is 1.  If you take the limit of $f(x)$ as $x \to 2^-$ and  as $x \to 2^+$ the second will depend on $C$ and the first will not, so you can choose $C$ to make $f(2)$ consistent across $x=2$.The note says that if you take the derivative as a limit as $x \to 2^-$ you get one answer and if you take the limit as $x \to 2^+$ you get a different one.  If you plot it there will be a kink in the curve.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2011-02-11T05:10:13.833" LastActivityDate="2011-02-11T05:10:13.833" CommentCount="1" />
  <row Id="21488" PostTypeId="1" AcceptedAnswerId="21512" CreationDate="2011-02-11T05:21:41.487" Score="8" ViewCount="82" Body="&lt;p&gt;I have a question related to &lt;a href=&quot;http://math.stackexchange.com/questions/21304/manifold-with-3-nondegenerate-critical-points/21346#21346&quot;&gt;this one&lt;/a&gt;.  In my answer I was going to try and say something about the possible manifolds that might arise in this way, i.e. as mapping cones of elements of $\pi_{2n-1}(S^n)$.  Certainly not all of them will be manifolds.  At first I felt like there should be a reason why we'd need to be using a torsion-free homotopy generator if we want a manifold (in which case we could appeal to Serre's theorem that the only non-torsion part of the homotopy groups of spheres is $\pi_n(S^n)=\mathbb{Z}$ and $\pi_{4n-1}(S^{2n})=\mathbb{Z}$), but then I realized I couldn't think of any reason why that should be true.  Also I was hoping the Hopf invariant would enter into the picture too, since that's a pretty obvious tool at our disposal, but I couldn't get anywhere with that either...and I think at this point I'm more or less all out of tricks.  Does anyone have any ideas?&lt;/p&gt;&#xA;" OwnerUserId="2159" LastActivityDate="2011-02-11T21:30:04.900" Title="When can we recover a manifold when we attach a $2n$-cell to $S^n$?" Tags="&lt;algebraic-topology&gt;&lt;manifold&gt;" AnswerCount="1" />
  <row Id="21489" PostTypeId="1" AcceptedAnswerId="21509" CreationDate="2011-02-11T05:42:28.087" Score="2" ViewCount="317" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikibooks.org/wiki/Haskell&quot; rel=&quot;nofollow&quot;&gt;Haskell&lt;/a&gt; is a programming language which uses some concepts from category theory like functor, monad, etc. My question is: Learning intuitive concepts about category from Haskell will ruin my intuition &lt;a href=&quot;http://math.stackexchange.com/questions/21128/when-to-learn-category-theory&quot;&gt;when I learn category theory&lt;/a&gt; as a mathematician or it could help developing it?&lt;/p&gt;&#xA;" OwnerUserId="1938" LastActivityDate="2011-02-13T04:05:56.627" Title="Is learning haskell a bad thing for a beginner mathematician?" Tags="&lt;computer-science&gt;&lt;category-theory&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="1" />
  <row Id="21490" PostTypeId="1" AcceptedAnswerId="21556" CreationDate="2011-02-11T06:34:40.427" Score="5" ViewCount="98" Body="&lt;p&gt;Consider an $n$-sided &lt;em&gt;convex&lt;/em&gt; polygon $P$ that contains the origin in the complex plane. Let the $j$-th vertex be denoted $z_j = r_j e^{i\theta_j}$ ($0 \leq \theta_j &amp;lt; 2 \pi$) for $j= 1 \dots n$. I'm interested in non-zero values of &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ a_k(P)= \sum_{j=1}^{n} \frac{z_{j}^{k}}{|z_{j}|^{k-1}}=\sum_{j=1}^{n} r_j e^{ik\theta_j} \textrm{ for } k \geq 1.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Lemma:&lt;/em&gt; Given a integer $m \geq 2$, if, for every $k \geq 1$ where $m$ does not divide $k$, $a_k(P)=0$, then the polygon $P$ is $m$-fold rotationally symmetric, that is, a rotation of $e^{i\frac{2 \pi}{m}}$ rotates the polygon into itself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Pseudo-Proof:&lt;/em&gt; Re-imagine the $n$-sided polygon $P$ as a $2 \pi$-periodic function $f(\theta)$ of the angle $\theta$ where each vertex $z_j$ is represented as a Dirac delta function at $\theta_j$ with integral $r_j$, that is,&#xA;$$f(\theta)=\sum_{j=1}^{n} r_j \delta (\theta - \theta_j).$$ The calculation $a_k(P)$ is then just $2 \pi$ times the $k$-th Fourier coefficient for $f(\theta)$. If, for all $k$ where $m$ does not divide $k$, $a_k(P)=0$, then the corresponding Fourier coefficients of $f(\theta)$ are all zero, implying that $f(\theta)$ is $\frac{2 \pi}{m}$-periodic. Hence the polygon will be $m$-fold rotationally symmetric. $\square$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Firstly, is there a good way to prove this lemma without resorting to non-converging Fourier series?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, in the same vein, the lemma implies that, if $P$ is &lt;em&gt;not&lt;/em&gt; rotationally symmetric, then for every $m$, there are values of $k$, that are not multiples of $m$ for which $a_k(P) \neq 0$. But I believe much more is true, namely, that for 'almost' all $k$, $a_k(P) \neq 0$. In particular, if $k$ is the smallest so that $a_k(P) \neq 0$, I'd like to show that there is a $k&amp;#39;$ relatively prime to $k$ so that $a_{k&amp;#39;}(P) \neq 0$, but I'm not sure how to approach the issue. Any thoughts?&lt;/p&gt;&#xA;" OwnerUserId="6272" LastEditorUserId="6272" LastEditorDisplayName="" LastEditDate="2011-02-11T16:21:30.007" LastActivityDate="2011-02-11T17:04:48.360" Title="For a polygon on complex plane, when are the vertex 'Fourier coefficients' non-zero." Tags="&lt;geometry&gt;&lt;complex-numbers&gt;&lt;fourier-series&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="21491" PostTypeId="1" AcceptedAnswerId="21494" CreationDate="2011-02-11T06:47:07.063" Score="4" ViewCount="312" Body="&lt;p&gt;The question is prompted by change of basis problems -- the book keeps multiplying the bases by matrix $S$ from the left in order to keep subscripts nice and obviously matching, but in examples bases are multiplied by $S$ (the change of basis matrix) from whatever side. So is matrix multiplication commutative if at least one matrix is invertible?&lt;/p&gt;&#xA;" OwnerUserId="5157" LastActivityDate="2011-02-11T18:15:26.340" Title="If a matrix is invertible, is its multiplication commutative?" Tags="&lt;linear-algebra&gt;" AnswerCount="3" CommentCount="6" />
  <row Id="21492" PostTypeId="2" ParentId="21475" CreationDate="2011-02-11T06:53:41.327" Score="4" ViewCount="" Body="&lt;p&gt;To add on Jason's good answer, operations which are defined on all the sets can be seen as functions, which are usually treated as a collection of $n$-tuples (which are also sets)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So when the domain of a function is a class then the function itself is a proper class.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example $\{\langle x,P(x)\rangle | x\in V\}$ is the Power Set operation, and $\{\langle x,y,x\cap y\rangle | x,y\in V\}$ is the function which takes two sets and returns their intersection.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both of these functions are proper classes.&lt;/p&gt;&#xA;" OwnerUserId="622" LastActivityDate="2011-02-11T06:53:41.327" CommentCount="5" />
  <row Id="21493" PostTypeId="2" ParentId="21491" CreationDate="2011-02-11T07:01:26.410" Score="6" ViewCount="" Body="&lt;p&gt;In general, two matrices (invertible or not) do not commute. For example&#xA;$$\left(\begin{array}{cc}&#xD;&#xA;1 &amp;amp; 1\\&#xD;&#xA;0 &amp;amp; 1\end{array}\right)\left(\begin{array}{cc}&#xD;&#xA;1 &amp;amp; 0\\&#xD;&#xA;1 &amp;amp; 1\end{array}\right)  =  \left(\begin{array}{cc}&#xD;&#xA;2 &amp;amp; 1\\&#xD;&#xA;1 &amp;amp; 1\end{array}\right)&#xD;&#xA;$$&#xA;$$&#xD;&#xA;\left(\begin{array}{cc}&#xD;&#xA;1 &amp;amp; 0\\&#xD;&#xA;1 &amp;amp; 1\end{array}\right)\left(\begin{array}{cc}&#xD;&#xA;1 &amp;amp; 1\\&#xD;&#xA;0 &amp;amp; 1\end{array}\right)  =  \left(\begin{array}{cc}&#xD;&#xA;1 &amp;amp; 1\\&#xD;&#xA;1 &amp;amp; 2\end{array}\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, to change a basis you usually need to conjugate and not just multiply from the left (or just right).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you do know is that a matrix A commutes with $A^n$ for all $n$ (negative too if it is invertible, and $A^0 = I$), so for every polynomial P (or Laurent polynomial if A is invertible) you have that A commutes with $P(A)$.&lt;/p&gt;&#xA;" OwnerUserId="2125" LastActivityDate="2011-02-11T07:01:26.410" CommentCount="1" />
  <row Id="21494" PostTypeId="2" ParentId="21491" CreationDate="2011-02-11T07:03:28.940" Score="7" ViewCount="" Body="&lt;p&gt;Definitely not.  Yuan's comment is also not correct, diagonal matrices do not necessarily commute with non-diagonal matrices.  Consider $$\left[\begin{array}{cc}&#xD;&#xA;1 &amp;amp; 1\\&#xD;&#xA;0 &amp;amp; 1\end{array}\right]\left[\begin{array}{cc}&#xD;&#xA;a &amp;amp; 0\\&#xD;&#xA;0 &amp;amp; b\end{array}\right]=\left[\begin{array}{cc}&#xD;&#xA;a &amp;amp; b\\&#xD;&#xA;0 &amp;amp; b\end{array}\right]&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Changing the order I get &#xA;$$&#xD;&#xA;\left[\begin{array}{cc}&#xD;&#xA;a &amp;amp; 0\\&#xD;&#xA;0 &amp;amp; b\end{array}\right]\left[\begin{array}{cc}&#xD;&#xA;1 &amp;amp; 1\\&#xD;&#xA;0 &amp;amp; 1\end{array}\right]=\left[\begin{array}{cc}&#xD;&#xA;a &amp;amp; a\\&#xD;&#xA;0 &amp;amp; b\end{array}\right]&#xD;&#xA;$$&#xA;Which is different for $a\neq b$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope that helps.  (Sometimes change of basis matrices can go on different sides for different reasons, but without seeing the exact text you are talking about I can't comment)&lt;/p&gt;&#xA;" OwnerUserId="6075" LastActivityDate="2011-02-11T07:03:28.940" CommentCount="3" />
  <row Id="21495" PostTypeId="2" ParentId="21470" CreationDate="2011-02-11T07:51:33.653" Score="3" ViewCount="" Body="&lt;p&gt;To generalize slightly the result mentioned by mjqxxxx, assume that one groups the original states $S_i$ into classes $C_a$. Hence the collection of classes $(C_a)$ is a partition of the state space and each class can be reduced to one state, or not. Then a sufficient condition for the resulting process to still be Markov is that the function $\varphi$ defined, for every state $S$ and class $C$, by the formula&#xA;$$&#xD;&#xA;\varphi(S,C)=\sum_{T\in C}P(S\to T)&#xD;&#xA;$$&#xA;depends on the state $S$ through its class only. In other words, one asks that, for every classes $C$ and $D$ and every states $S$ and $S&amp;#39;$ in $C$, &#xA;$$&#xD;&#xA;\varphi(S,D)=\varphi(S&amp;#39;,D).&#xD;&#xA;$$&#xA;Note that the condition is always true if every class is reduced to one state, and that it is also true if there is only one class.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Caveat: if this condition is not met, it can nevertheless happen that the resulting process is Markov for some, but not all, initial conditions.&lt;/p&gt;&#xA;" OwnerUserId="6179" LastEditorUserId="6179" LastEditorDisplayName="" LastEditDate="2011-02-11T21:11:55.197" LastActivityDate="2011-02-11T21:11:55.197" />
  <row Id="21496" PostTypeId="1" CreationDate="2011-02-11T08:01:26.723" Score="2" ViewCount="151" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Show that if a $3\times 3$ matrix $A$ represents the reflection about  a plane, then  $A$ is similar to the matrix $\begin{bmatrix}&#xD;&#xA;1 &amp;amp; 0 &amp;amp; 0\\ &#xD;&#xA;0 &amp;amp; 1 &amp;amp; 0\\ &#xD;&#xA;0 &amp;amp; 0 &amp;amp; -1&#xD;&#xA;\end{bmatrix}$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is not homework, just me preparing for a test -- but how can this be solved? I know the definition of similarity, but can't find a way to approach the problem.&lt;/p&gt;&#xA;" OwnerUserId="6900" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-11T14:31:52.717" LastActivityDate="2011-02-11T14:31:52.717" Title="Representation of a reflection about a plane in $\mathbb{R}^3$" Tags="&lt;linear-algebra&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="1" />
  <row Id="21497" PostTypeId="2" ParentId="21496" CreationDate="2011-02-11T08:12:08.353" Score="5" ViewCount="" Body="&lt;p&gt;Suppose that $A$ is a reflection about a plane spanned by $u,v$ and let $w$ be orthogonal to this plane, then $Au=u$ and $Av=v$ since the reflection through the plane doesn't change them and $Aw = -w$ (this is the reflection). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now switch the basis to $u,v,w$ and you will get the result.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;If you have any two diagonal matrices, that have the same entries on their diagonal, but maybe in a different order, then they are similar. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can see this by defining the matrix $E_\sigma$ for a permutation $\sigma \in S_n$ on n elements as $(E_\sigma)_{i,\sigma(i)}=1$ and all the other entries are zero. If $D$ is diagonal then $E_\sigma D E_\sigma ^{-1}$ is just the permutation of the entries on the diagonal.&lt;/p&gt;&#xA;" OwnerUserId="2125" LastEditorUserId="2125" LastEditorDisplayName="" LastEditDate="2011-02-11T09:43:29.103" LastActivityDate="2011-02-11T09:43:29.103" />
  <row Id="21498" PostTypeId="2" ParentId="15332" CreationDate="2011-02-11T08:14:38.733" Score="2" ViewCount="" Body="&lt;p&gt;Depending on what you mean by representation theory the answer is: yes!&lt;br&gt;&#xA;Certainly, you can prove it using &quot;analysis&quot;. The theorem is: the space of automorphic forms (on a reductive group) of a certain level with a specific $\mathcal Z$-type and $K$-type is finite dimensional. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The proof is based on the fact that a closed subspace of $L^2(X,\mu)$, with $\mu(X)$ finite, made of essentially bounded functions is finite dimensional.&lt;br&gt;&#xA;This is Theorem 5.2 in Rudin's &quot;Functional Analysis&quot;, where it is attributed to Grothendieck. It is also Lemma 8.3 in Borel's &quot;Automorphic forms&quot;, where it is attributed to Godement. The statements are slightly different, so there is not necessarily an attribution error.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me sketch the proof for the cusp form case. The general case follows from the cuspidal case and an induction argument (on the rank of the group).  For simplicity, $X$ will be $\Gamma\backslash SL_2({\mathbb R})$, where $\Gamma$ is an arithmetic subgroup.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Before you begin, you need:&lt;br&gt;&#xA;1) Cusp forms are rapidly decreasing (I should add &quot;in Siegel sets&quot;). This is standard, if somewhat technical.&lt;br&gt;&#xA;2) Distributions on $X$ with prescribed $\mathcal Z$- and $K$-types are actually smooth functions. This is essentially an application of elliptic regularity.&lt;br&gt;&#xA;3) $L^2_{\rm cusp}(X)$ is closed in $L^2(X)$. This is either obvious or straightforward, depending on what you already know.&lt;br&gt;&#xA;(Note that while cusp forms are rapidly decreasing, $L^2_{\rm cusp}(X)$ is not made up of essentially bounded functions, because we have stronger criteria for a function to be a cusp form than just being in $L^2_{\rm cusp}(X)$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now that we have these facts at hand, the proof is pretty straightforward.&lt;br&gt;&#xA;Cusp forms of specific type are rapidly decreasing (1), so essentially bounded.  A convergent (in $L^2$) sequence of cusp forms of a specific type converges weakly to a distribution of the same type (it is necessary to use distributions to talk about the $\mathcal Z$-type of a non-smooth function), hence it was actually a smooth function (2), so it converges to a cusp form (3). So we can deduce that the space of cusp forms of a given type is finite dimensional.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is not much more than an abbreviated transcription of the argument in Borel's &quot;Automorphic forms on $SL_2({\mathbb R})$&quot;. The theorem is originally from Harish-Chandra, in &quot;Automorphic forms on semisimple Lie groups&quot;.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The theorem I think Matt E refers to is the much stronger assertion that &quot;irreducible unitary representations of reductive groups over local fields are admissible&quot;. This implies that irreducible unitary representations of adele groups are admissible. Which implies that subrepresentations of the space of cusp forms are admissible, which implies automorphic representations are admissible, since they are built up from cuspidal representations in ways that preserve admissibility (parabolic induction and subquotients). This theorem is provable in representation-theoretic terms (though it is surprisingly algebraic: the main tricks are about algebras). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is important because it lets us factor automorphic representations over primes. It is not hard to show that irreducible admissible representations of adele groups factor over primes, but it wasn't proven that automorphic representations were admissible until it was proven that irreducible unitary representations were.&lt;/p&gt;&#xA;" OwnerUserId="5773" LastEditorUserId="5773" LastEditorDisplayName="" LastEditDate="2011-02-12T20:39:49.600" LastActivityDate="2011-02-12T20:39:49.600" CommentCount="2" />
  <row Id="21502" PostTypeId="2" ParentId="20911" CreationDate="2011-02-11T09:10:26.003" Score="0" ViewCount="" Body="&lt;p&gt;There are many diagramming software available, however, I use &lt;a href=&quot;http://www.visiotoolbox.com/2010&quot; rel=&quot;nofollow&quot;&gt;Visio&lt;/a&gt; 2010 which is one of the best diagramming software. I will recommend you to try Visio and see whether it works for you or not.&lt;/p&gt;&#xA;" OwnerUserId="6903" LastActivityDate="2011-02-11T09:10:26.003" />
  <row Id="21503" PostTypeId="2" ParentId="21489" CreationDate="2011-02-11T09:22:46.233" Score="2" ViewCount="" Body="&lt;p&gt;No, on the contrary, I think it will help your understanding.&lt;/p&gt;&#xA;" OwnerUserId="5882" LastActivityDate="2011-02-11T09:22:46.233" />
  <row Id="21504" PostTypeId="1" AcceptedAnswerId="21541" CreationDate="2011-02-11T09:46:24.090" Score="4" ViewCount="118" Body="&lt;p&gt;There are $m$ integer points on a line of length $m-1$ (including the end point of the line). We randomly mark $n$ points of the $m$ points ($n\leq m$), dividing the line into several segments, so each segment have an integer length. What is the probability the max segment's length is equal to $k$?&lt;/p&gt;&#xA;" OwnerUserId="6904" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-11T14:36:17.220" LastActivityDate="2011-02-11T15:22:52.817" Title="What's the probability of the max segment is equal to k?" Tags="&lt;probability&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="21506" PostTypeId="2" ParentId="21348" CreationDate="2011-02-11T10:31:53.507" Score="0" ViewCount="" Body="&lt;p&gt;Since the market prices are fixed and $V^L = C^0 \cdot v + S^0 \cdot v + S^1 \cdot v + \ldots + S^L \cdot v$, you can also assign a value $R^i \cdot v$ to each recipe and use Dijkstras algorithm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The exact implementation works as follows:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;A vertex in the graph is represented by some choice of $n \leq N$ recipes. (You can identify to vertices if the same recipes were chosen in a possibly different order.)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Two vertices are connected by a directed edge if it is possible to go from one vertex to the next by using one recipe. At this point, one has to take into account that $C^i_k \geq 0$, because if we run out of an ingredient, it is not possible to use it.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Each edge gets a weight of $O - S \cdot v$ where $O$ is an offset such that all weights are positive (otherwise Dijkstra won't work here) and $S$ is the recipe applied along the edge.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Identify all vertices that are L steps away from $C^0$ and make this vertex the target vertex.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Let Dijkstra's algorithm find the shortest way. It has length $L \cdot O - V^L$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="5416" LastEditorUserId="5416" LastEditorDisplayName="" LastEditDate="2011-02-12T12:33:49.463" LastActivityDate="2011-02-12T12:33:49.463" CommentCount="5" />
  <row Id="21507" PostTypeId="2" ParentId="21348" CreationDate="2011-02-11T10:43:26.913" Score="1" ViewCount="" Body="&lt;p&gt;If you want to solve it via linear programming, you have to make it an integer linear problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could for example define the (boolean) vectors $s^t_i \in \{0, 1\}$ that are 1 when recipe $i$ is applied at time $t$ and 0 if it is not applied.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You then have to add the constraint that $\sum_i s^t_i = 1 \forall t$, in words: Use exactly one recipe at a time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The next constraint to be added is that $C_j + \sum_i \sum_{t&amp;#39; = 1}^{t} s^{t&amp;#39;}_i R^i_j \geq 0 \forall t, j$, in words: Ingredients can never exist in negative amounts. This is a linear constraint since $R^i_j$ are constants.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then optimise for $V^L$. You can use e.g. the Gnu Linear Programming Kit to implement this problem.&lt;/p&gt;&#xA;" OwnerUserId="5416" LastEditorUserId="5416" LastEditorDisplayName="" LastEditDate="2011-03-19T20:00:55.347" LastActivityDate="2011-03-19T20:00:55.347" />
  <row Id="21508" PostTypeId="1" AcceptedAnswerId="21565" CreationDate="2011-02-11T10:46:51.420" Score="2" ViewCount="118" Body="&lt;p&gt;Consider subset $K\subset \mathbb C^2$ constuted with pairs $(z,\bar z)$ such &#xA;$|z|=1$ . Is there easy way see that continous functions on $K$ can be uniformly approximed by polinomials in two variables ? Problem is not so trivial because if you take $L\subset \mathbb C^2$,, set of pairs  $(z,0)$ such &#xA;$|z|=1$ then  $K$ is homeomorphic $L$ but it is clear that it is impossible approxime continous $\frac{1}{z}$ on $L$ by polynomials because Maximum Principle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you.&lt;/p&gt;&#xA;" OwnerUserId="2872" LastEditorUserId="2872" LastEditorDisplayName="" LastEditDate="2011-02-11T12:19:10.387" LastActivityDate="2011-02-11T17:56:08.340" Title="Uniform approximation of continous functions by polynomials in two variables" Tags="&lt;complex-analysis&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="21509" PostTypeId="2" ParentId="21489" CreationDate="2011-02-11T12:09:37.880" Score="4" ViewCount="" Body="&lt;p&gt;Just to echo what Mariano said: you may also be interested in reading the various responses to &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://mathoverflow.net/questions/3721/programming-languages-based-on-category-theory&quot; rel=&quot;nofollow&quot;&gt;Programming Languages Based on Category Theory&lt;/a&gt; over at MO.&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://mathoverflow.net/questions/903/resources-for-learning-practical-category-theory&quot; rel=&quot;nofollow&quot;&gt;Resources to learning Practical Category Theory&lt;/a&gt;, ibid&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://mathoverflow.net/questions/11084/what-programming-languages-do-mathematicians-use&quot; rel=&quot;nofollow&quot;&gt;What programming languages do Mathematicans use&lt;/a&gt;, ibid&lt;/li&gt;&#xA;&lt;li&gt;and &lt;a href=&quot;http://blog.sigfpe.com/2006/01/eleven-reasons-to-use-haskell-as.html&quot; rel=&quot;nofollow&quot;&gt;Sigfpe's 11 reasons to learn Haskell as a mathematician&lt;/a&gt;. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1543" LastActivityDate="2011-02-11T12:09:37.880" CommunityOwnedDate="2011-02-11T12:09:37.880" />
  <row Id="21510" PostTypeId="2" ParentId="21477" CreationDate="2011-02-11T12:40:03.073" Score="3" ViewCount="" Body="&lt;p&gt;Estimating above is just a matter of Cauchy-Schwarz (if you also assume $a^{ij}$ is symmetric). Since $a^{ij}$ is positive, you can define the matrix $[b] = [a]^{1/2}$, and rewrite &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \int_B a^{ij}X_jD_ig dx = \int_B \sum_k b^{ik}b^{kj}X_jD_ig dx $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;apply Cauchy-Schwarz to the terms $V^k = b^{ik}D_ig$ and $W^k = b^{kj}X_j$ you get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \left\lvert\int_B a^{ij}X_jD_ig dx\right\rvert \leq \left(\int_B V^kV^k dx\right)^{1/2}\left(\int_B W^kW^k dx\right)^{1/2} = \left(\int_B a^{ij}X_jX_i dx\right)^{1/2} \left(\int_B a^{ij}D_ig D_jg dx\right)^{1/2} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general: Cauchy-Schwarz can be used for any positive semi-definite symmetric bilinear form.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One direction of the ellipticity definition can then let you control the RHS by a constant factor times&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \left(\int_B |X|^2 dx\right)^{1/2} \left(\int_B  |Dg|^2 dx\right)^{1/2} $$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Another thing you can do, if you know the behaviour of $u$ on the boundary of $B$, is to integrate by parts:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \int_{B} a^{ij}X_j D_i g dx = \int_{\partial B} a^{ij}X_j n_i g d\sigma - \int_{B} g D_i( a^{ij}X_j ) dx $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which can be useful if you know something about the derivatives of the vector $a^{ij}X_j$. $n_i$ in the first term on the right hand side is just the unit out-ward normal on $\partial B$. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;In general you cannot estimate from below. Ellipticity does not rule out the possibility that $X_j$ and $D_i g$ are pointwise orthogonal relative to the bilinear form $a^{ij}$. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;If you say more precisely what kinds of estimates you are looking for, or what possible additional properties about $X_j$ and $D_i g$ that you have, I may be able to give more precise answers. &lt;/p&gt;&#xA;" OwnerUserId="1543" LastEditorUserId="5363" LastEditorDisplayName="" LastEditDate="2011-02-11T12:51:29.737" LastActivityDate="2011-02-11T12:51:29.737" CommentCount="3" />
  <row Id="21512" PostTypeId="2" ParentId="21488" CreationDate="2011-02-11T13:13:47.910" Score="3" ViewCount="" Body="&lt;p&gt;I believe this is related to the Hopf invariant one problem.  Let $X = S^n \cup_f B^{2n}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The attaching map from $B^{2n}$ is a map $f$ from $S^{2n-1}$ to $S^n$.  I think one way of thinking about the Hopf invariant of this map is as follows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $x\in H^n(X)$.  Then $x^2 = k y$ where $y$ generates $H^{2n}(X)$.  The number $k$ is the Hopf invariant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order for $X$ to be a manifold, it must satisfy Poincare duality (so long as $n&amp;gt;1$ - if $n=1$ different stuff can happen), which implies that $k = \pm 1$.  But Adams has shown you can only have Hopf invariant 1 when $n = 2, 4, 8$, getting you something homotopy equivalent to $\mathbb{C}P^2$, $\mathbb{H} P^2$ or $\mathbb{O}P^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The $n=1$ case, you're attaching a disc to a circle and getting a manifold.  Here, since $\pi_1(X)$ may be nontrivial, one cannot neccesarily use Poincare duality.  However, by a simple analysis, the only manifold this can give rise to is $\mathbb{R}P^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Edit&lt;/em&gt;  If one wants to avoid cases ($n = 1$ vs $n &gt; 1$) then one can work with cohomology with $\mathbb{Z}/2\mathbb{Z}$ coefficients instead.&lt;/p&gt;&#xA;" OwnerUserId="331" LastEditorUserId="331" LastEditorDisplayName="" LastEditDate="2011-02-11T21:30:04.900" LastActivityDate="2011-02-11T21:30:04.900" CommentCount="6" />
  <row Id="21513" PostTypeId="2" ParentId="21437" CreationDate="2011-02-11T13:21:40.520" Score="2" ViewCount="" Body="&lt;p&gt;The function $g$ is holomorphic in $B(0,R)$ and continuous on $\bar{B}(0,R)$. For $|z|=R$ we have $\left|\frac{z}{z_k}\right|\geq 3$ and one obtains $|g(z)|\leq 2^{-n}M$. From the maximum principle one can infer that this also holds for $|z|&amp;lt;R$, in particular for $z=0$. Now solve for $n$.&lt;/p&gt;&#xA;" OwnerUserId="1403" LastActivityDate="2011-02-11T13:21:40.520" CommentCount="1" />
  <row Id="21514" PostTypeId="1" AcceptedAnswerId="21554" CreationDate="2011-02-11T13:25:39.597" Score="6" ViewCount="234" Body="&lt;p&gt;Is there a good source of undergraduates linear algebra problems available on line?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do you have any better idea of how to resharpen my linear algebra skills?&lt;/p&gt;&#xA;" OwnerUserId="52" LastActivityDate="2011-02-12T10:47:01.157" Title="Linear Algebra exercises available online?" Tags="&lt;linear-algebra&gt;&lt;online-resources&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="2" />
  <row Id="21515" PostTypeId="1" AcceptedAnswerId="21517" CreationDate="2011-02-11T13:38:32.723" Score="6" ViewCount="189" Body="&lt;p&gt;First of all, my background in Homology and Algebraic Topology is very limited and doesn't include much except for the basic concepts; I'm trying to understand some Homology techniques as applications relevant to a course in combinatorics I'm taking.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, in class we've seen the example of computing the homology groups of $\mathbb{P}^2$ - The projective plane over $\mathbb{R}$. It is known that $\beta_0(\mathbb{P}^2 ; \mathbb{F})=1$ since the projective plane is connected. Now, it turns out that $H_2(\mathbb{P}^2 ; \mathbb{Q})=0$ while $H_2(\mathbb{P}^2 ; \mathbb{F}_2) \cong \mathbb{F}_2$, but the lecturer hasn't really explained how it is computed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any explanation on the computing method would be greatly appreciated, as will be any intuition on this whole concept, or references to further reading (I know Hatcher's book on Algebraic Topology, but didn't find it very helpful).&lt;/p&gt;&#xA;" OwnerUserId="766" LastActivityDate="2011-02-14T05:25:44.253" Title="Computing homology groups over different fields" Tags="&lt;algebraic-topology&gt;&lt;homology&gt;" AnswerCount="3" />
  <row Id="21516" PostTypeId="1" AcceptedAnswerId="21821" CreationDate="2011-02-11T13:40:38.617" Score="3" ViewCount="150" Body="&lt;p&gt;Wikipedia gives this solution:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/NF1lU.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I have no idea how I should exactly understand the partial part: $\left(\frac{\partial}{\partial c}\right)\frac{e^{cx}}{c}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for your responses so far. I should add that n is not necessarily an integer. Can be for example $n = 1.2$.&#xA;I'll see how far I get on learning about fractional derivatives.&lt;/p&gt;&#xA;" OwnerUserId="6909" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2011-02-13T17:12:52.473" LastActivityDate="2011-02-13T17:12:52.473" Title="Is there a closed form solution for $\int x^n e^{cx}$?" Tags="&lt;integral&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="21517" PostTypeId="2" ParentId="21515" CreationDate="2011-02-11T13:52:17.823" Score="4" ViewCount="" Body="&lt;p&gt;It turns out that the homology groups $H_{\bullet}(X,\mathbb{Z})$ of a space $X$ with $\mathbb{Z}$-coefficients determine the homology groups $H_{\bullet}(X,A)$ with coefficients in any abelian group $A$.  The key result here is the (very well named!) &lt;a href=&quot;http://en.wikipedia.org/wiki/Universal_coefficient_theorem&quot; rel=&quot;nofollow&quot;&gt;Universal Coefficient Theorem&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The basic idea is that our first guess at $H_i(X,A)$ is simply $H_i(X,\mathbb{Z}) \otimes A$.  This is a good first guess in that in all cases there is a natural injective map $H_i(X,\mathbb{Z}) \otimes A \rightarrow H_i(X,A)$.  As you have seen, this map need not be an isomorphism.  The universal coefficient theorem tells you that its cokernel is $\operatorname{Tor}(H_{i-1}(X,\mathbb{Z}),A)$ and also that the sequence is (non-canonically) split, i.e.,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$H_i(X,A) \cong (H_i(X,\mathbb{Z}) \otimes A) \oplus \operatorname{Tor}(H_{i-1}(X,\mathbb{Z}),A).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here $\operatorname{Tor}( \ , \ )$ is the first &quot;Tor group&quot; of homological algebra.  It may well be that you don't know what this gadget is.  (I didn't when I first learned algebraic topology.)  So I found it helpful to write down a &quot;cheatsheet&quot; for $\operatorname{Tor}(X,Y)$ when $X$ and $Y$ are both finitely generated abelian groups.  Indeed, since $\operatorname{Tor}$ is bi-additive and symmetric, it is enough to know that for all $m,n \in \mathbb{Z}^+$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\operatorname{Tor}(\mathbb{Z},\mathbb{Z}/n\mathbb{Z}) = 0$ and&lt;br&gt;&#xA;$\operatorname{Tor}(\mathbb{Z}/m\mathbb{Z},\mathbb{Z}/n\mathbb{Z}) \cong \mathbb{Z}/\operatorname{gcd}(m,n) \mathbb{Z}$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a first exercise, try to use all this information to confirm that the homology groups of $\mathbb{R} \mathbb{P}^2$ with $\mathbb{Z}/2\mathbb{Z}$-coefficients are as you said.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(There is also a Universal Coefficient Theorem for cohomology in which the correction term involves $\operatorname{Ext} = \operatorname{Ext}^1$ instead of $\operatorname{Tor}$...)&lt;/p&gt;&#xA;" OwnerUserId="299" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2011-02-11T15:41:51.023" LastActivityDate="2011-02-11T15:41:51.023" />
  <row Id="21518" PostTypeId="2" ParentId="21515" CreationDate="2011-02-11T13:53:30.197" Score="1" ViewCount="" Body="&lt;p&gt;My favorite way to compute the (co)homology is using the Serre spectral sequence, but I think that is a bit of over kill for this problem. I think the best way to do it would be to come up with a model for $\mathbb{R}P^2$ as a cell complex and then compute the cellular homology. To do this look at the model for $S^2$ that has a 2 cells in every dimension (less than or equal to 2). Now $\mathbb{R}P^2$ is a quotient of $S^2$ by a free $\mathbb{Z}/2$ action so we identify use as a model for $\mathbb{R}P^2$ the quotient of the model for $S^2$ under this action. Then we compute the cellular Homology. Tensor the complex with whatever coefficients you like and then take homology.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I will give you some time to think about the model for $S^2$ and come back later with more.&lt;/p&gt;&#xA;" OwnerUserId="627" LastActivityDate="2011-02-11T13:53:30.197" CommentCount="8" />
  <row Id="21519" PostTypeId="1" AcceptedAnswerId="21520" CreationDate="2011-02-11T13:54:41.813" Score="2" ViewCount="92" Body="&lt;p&gt;I need to figure out what is the cardinality of the set of functions from $\mathbb{R}$ to $\mathbb{R}$ that for each $x$ that is not in $\mathbb{Q}$, $f(x)\in\{x,x+1\}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can make a mapping for each function to be $\mathbb{Q}\to\mathbb{R}\times\left(\frac{R}{\mathbb{Q}}\right)\to\{x,x+1\}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does it ok?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="6163" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-11T17:44:33.797" LastActivityDate="2011-02-11T17:46:56.790" Title="Cardinality of a set." Tags="&lt;elementary-set-theory&gt;" AnswerCount="2" />
  <row Id="21520" PostTypeId="2" ParentId="21519" CreationDate="2011-02-11T14:05:25.173" Score="1" ViewCount="" Body="&lt;p&gt;Hint: Do you know the cardinality(power) of the set(group) of functions from $\mathbb{R}$ to $\mathbb{R}$ with no restriction?  Can you see how to argue that this must be the same?  Alternately, you should be able to follow through the argument of the power of the unrestricted functions and make it work the same.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2011-02-11T17:46:56.790" LastActivityDate="2011-02-11T17:46:56.790" CommentCount="2" />
  <row Id="21521" PostTypeId="2" ParentId="21519" CreationDate="2011-02-11T14:06:11.170" Score="2" ViewCount="" Body="&lt;p&gt;For each $x$ in $\mathbb{R}-\mathbb{Q}$, you have two possible values, so the cardinality is at least $2^{|\mathbb{R}-\mathbb{Q}|}$, which is the same as $2^{|\mathbb{R}|}$, also known as $\beth_2$, since $|\mathbb{R}-\mathbb{Q}|=|\mathbb{R}|$. On the other hand, the cardinality is also at most $\beth_2$, since that is the cardinality of the set of all functions $\mathbb{R}\to\mathbb{R}$.&lt;/p&gt;&#xA;" OwnerUserId="413" LastActivityDate="2011-02-11T14:06:11.170" CommentCount="1" />
  <row Id="21522" PostTypeId="2" ParentId="21516" CreationDate="2011-02-11T14:06:27.683" Score="1" ViewCount="" Body="&lt;p&gt;It means you differentiate with respect to c, n times&lt;/p&gt;&#xA;" OwnerUserId="4929" LastActivityDate="2011-02-11T14:06:27.683" />
  <row Id="21523" PostTypeId="2" ParentId="21516" CreationDate="2011-02-11T14:10:25.170" Score="1" ViewCount="" Body="&lt;p&gt;It is the (n-fold because of the exponent) derivative of $\frac{e^{cx}}{c}$ with respect to $c$, considering $x$ to be fixed.  So for $n=1$ it is $\frac{c^2e^{cx}-e^{cx}}{c^2}$&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-02-11T14:10:25.170" CommentCount="1" />
  <row Id="21524" PostTypeId="1" AcceptedAnswerId="21527" CreationDate="2011-02-11T14:18:56.857" Score="1" ViewCount="117" Body="&lt;p&gt;Given integers a,b,c and d; is the absolute value of the determinant of [a c;b d] less than or equal to the absolute value of [a e;b f] + absolute value of [e c;f d] for all a,b,c,d?&#xA;In other words, is |ad-bc| less than or equal to |af-be| + |de-cf| for all integers a,b,c and ; with the additional constraint that the ordered pairs (a,b), (c,d), and (e,f) cannot equal each other and none of the ordered pairs can be (0,0), and if any of the ordered pairs is of the form (x,0) then the other two can't be of the form (y,0) and if one is (0,x) the other two cannot be (o,y).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for your time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried to upload a picture but I wasn't allowed to.I want to know whether the sum of the area of the two paralellograms with sides {(a,b),(e,f)} and {(e,f),(c,d)} will always be equal to or greater than associated with {(a,b),(c,d)}. It's just the counter-examples you gave me were of case where the areas become degenrate.&lt;/p&gt;&#xA;" OwnerUserId="6910" LastEditorUserId="6910" LastEditorDisplayName="" LastEditDate="2011-02-11T14:56:25.423" LastActivityDate="2011-02-11T19:28:19.653" Title="determinant inequality." Tags="&lt;inequality&gt;" AnswerCount="2" />
  <row Id="21525" PostTypeId="1" AcceptedAnswerId="21540" CreationDate="2011-02-11T14:23:58.160" Score="8" ViewCount="324" Body="&lt;p&gt;Hello,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;what is the general definition for some partial differential equation being called elliptic, parabolic or hyperbolic - in particular, if the PDE is nonlinear and above second-order.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So far, I have not found any precise definition in literature.&lt;/p&gt;&#xA;" OwnerUserId="3557" LastActivityDate="2011-02-11T20:08:28.263" Title="Mathematical precise definition of a PDE being elliptic, parabolic or hyperbolic" Tags="&lt;differential-equations&gt;&lt;pde&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" />
  <row Id="21526" PostTypeId="2" ParentId="21524" CreationDate="2011-02-11T14:27:51.523" Score="1" ViewCount="" Body="&lt;p&gt;No. Take $(a,b)=(4,1)$, $(c,d)=(1,1)$, $(e,f)=(1,0)$.&lt;/p&gt;&#xA;" OwnerUserId="6314" LastEditorUserId="6314" LastEditorDisplayName="" LastEditDate="2011-02-11T14:44:11.063" LastActivityDate="2011-02-11T14:44:11.063" CommentCount="5" />
  <row Id="21527" PostTypeId="2" ParentId="21524" CreationDate="2011-02-11T14:27:56.370" Score="1" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Edited.&lt;/strong&gt; Here is what is going on. If the first determinant is zero, then the inequality follows immediately, But if the first determinant is not zero, then it need not hold. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is exactly what happens, and how to get as many counterexamples as you please. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take any matrix whose determinant is not zero:&#xA;$$\left(\begin{array}{cc}&#xD;&#xA;a&amp;amp;c\\&#xD;&#xA;b&amp;amp;d&#xD;&#xA;\end{array}\right),$$&#xA;and let&#xA;$$\left|\det\left(\begin{array}{cc}&#xD;&#xA;a &amp;amp; c\\&#xD;&#xA;b &amp;amp; d\end{array}\right)\right| = D.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $D\neq 0$, that means that the columns, $(a,b)$ and $(c,d)$, form a basis for $\mathbb{R}^2$. Hence, &lt;em&gt;every&lt;/em&gt; vector $(e,f)$ can be written uniquely as $(e,f)=\alpha(a,b) + \beta(c,d)$ for some choice of scalars $\alpha$ and $\beta$. Thus, by the properties of the determinant, we have:&#xA;\begin{align*}&#xA;\left|\det\left(\begin{array}{cc}&#xA;a &amp;amp; e\\&#xA;b &amp;amp; f&#xA;\end{array}\right)\right| &amp;amp;= \left|\det\left(\begin{array}{cc}&#xA;a &amp;amp; \alpha a+\beta c\\&#xA;b &amp;amp; \alpha b + \beta d&#xA;\end{array}\right)\right| \\&#xA;&amp;amp;= \left|\det\left(\begin{array}{cc}&#xA;a&amp;amp;\alpha a\\&#xA;b &amp;amp;\alpha b&#xA;\end{array}\right) + \det\left(\begin{array}{cc}&#xA;a &amp;amp; \beta c\\&#xA;b &amp;amp; \beta d&#xA;\end{array}\right)\right|\\&#xA;&amp;amp;= \left| 0 + \beta\det\left(\begin{array}{cc}&#xA;a &amp;amp; c\\&#xA;b &amp;amp; d&#xA;\end{array}\right)\right|\\&#xA;&amp;amp;=|\beta|D.&#xA;\end{align*}&#xA;Symmetrically, you will have that&#xA;$$\left|\det\left(\begin{array}{cc}&#xD;&#xA;e &amp;amp; c\\\&#xD;&#xA;f &amp;amp; d&#xD;&#xA;\end{array}\right)\right| = |\alpha|D.$$&#xA;So you will have&#xA;$$&#xD;&#xA;\left|\det\left(\begin{array}{cc}&#xD;&#xA;a &amp;amp; e\\\&#xD;&#xA;b &amp;amp; f&#xD;&#xA;\end{array}\right)\right| + \left|\det\left(\begin{array}{cc}&#xD;&#xA;e &amp;amp; c\\\&#xD;&#xA;f &amp;amp; d&#xD;&#xA;\end{array}\right)\right| = |\alpha|D + |\beta|D = (|\alpha|+|\beta|)D,$$&#xA;where $\alpha$ and $\beta$ are the unique scalars such that $(e,f) = \alpha(a,b) + \beta(c,d)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the inequality you want is equivalent to&#xA;$$D\leq (|\alpha|+|\beta|)D,$$&#xA;which holds if and only if $|\alpha|+|\beta|\geq 1$, but there are plenty of choices that will make $|\alpha|+|\beta| \lt 1$, in which case your inequality cannot hold. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My first example, now deleted, took $(a,b)=(1,0)$, $(c,d) = (0,1)$, $\alpha=e$ and $\beta=0$. The second example (also now deleted in favor of this exposition) took $\alpha=\beta=e$. As long as neither of $\alpha$ and $\beta$ are zero, you'll get nondegenerate parallelograms, but picking them small in absolute value will ensure the inequality cannot hold. &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-11T19:28:19.653" LastActivityDate="2011-02-11T19:28:19.653" />
  <row Id="21528" PostTypeId="1" AcceptedAnswerId="21564" CreationDate="2011-02-11T14:39:34.683" Score="2" ViewCount="86" Body="&lt;p&gt;How can I calculate the cardinality of&#xA;$$\left(\{1,2,3\}^{\mathbb{N}} - \{1,2\}^{\mathbb{N}}\right)\cap\mathcal{P}(\mathbb{N}\times\mathbb{N}).$$&#xA;where $A^B$ is the set of all functions $f\colon B\to A$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;thanks&lt;/p&gt;&#xA;" OwnerUserId="6163" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-11T17:39:08.780" LastActivityDate="2011-02-11T17:53:08.517" Title="one more question with cardinality" Tags="&lt;elementary-set-theory&gt;&lt;cardinals&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="21529" PostTypeId="1" AcceptedAnswerId="21535" CreationDate="2011-02-11T14:53:56.010" Score="2" ViewCount="71" Body="&lt;p&gt;given p&gt;1, whats an example of  f where $\int_{-\infty}^{\infty} |f| &amp;lt; \infty$ but $\int_{-\infty}^{\infty} |f|^p = \infty$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;what about vica-versa? that is whats an example of g where $\int_{-\infty}^{\infty} |g| = \infty$ but $\int_{-\infty}^{\infty} |g|^p &amp;lt; \infty$&lt;/p&gt;&#xA;" OwnerUserId="6556" LastActivityDate="2011-02-11T15:14:02.320" Title="given p&gt;1, whats an example of f where $\int_{-\infty}^{\infty} |f| &lt; \infty$ but $\int_{-\infty}^{\infty} |f|^p = \infty$" Tags="&lt;integral&gt;" AnswerCount="1" />
  <row Id="21530" PostTypeId="1" AcceptedAnswerId="21637" CreationDate="2011-02-11T14:55:38.200" Score="1" ViewCount="188" Body="&lt;p&gt;Show that if $a$, $b$, and $c$ are positive integers with $\gcd(a, b) = 1$ and $ab = c^n$, then there are positive integers $d$, and $e$ such that $a = d^n$ and $b = e^n$.&lt;/p&gt;&#xA;" OwnerUserId="6912" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-11T19:45:54.147" LastActivityDate="2011-02-12T06:00:16.473" Title="If a product of relatively prime integers is an $n$th power, then each is an $n$th power" Tags="&lt;elementary-number-theory&gt;" AnswerCount="3" CommentCount="6" />
  <row Id="21531" PostTypeId="2" ParentId="21528" CreationDate="2011-02-11T15:00:12.740" Score="1" ViewCount="" Body="&lt;p&gt;So you have all the functions from $\mathbb{N} \to \{1,2,3\}$ that are not in $\mathbb{N} \to \{1,2\}$?  Each of these functions is a subset of P($\mathbb{N} \times \mathbb{N}$), so that part is not important.  Hint: you just need one 3 in the image to take it out of $\mathbb{N} \to \{1,2\}$&lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2011-02-11T15:28:11.670" LastActivityDate="2011-02-11T15:28:11.670" CommentCount="4" />
  <row Id="21532" PostTypeId="1" CreationDate="2011-02-11T15:02:09.623" Score="0" ViewCount="248" Body="&lt;p&gt;How to prove that&#xA;In an isoceles triangle circumcentre,centroid,orthocentre,incentre are collinear.&lt;/p&gt;&#xA;" OwnerUserId="6913" LastActivityDate="2011-02-13T04:13:47.873" Title="Centroid,orthocentre,incentre,circumcentre problem" Tags="&lt;geometry&gt;&lt;homework&gt;&lt;euclidean-geometry&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="21533" PostTypeId="1" AcceptedAnswerId="21537" CreationDate="2011-02-11T15:08:04.120" Score="1" ViewCount="295" Body="&lt;p&gt;I need tricks or shortcuts to finding inverse of  2x2 and 3x3 matrix. I have to give a time based exam, in which I have to find inverse of matrices and square of a matrix.&lt;/p&gt;&#xA;" OwnerUserId="6911" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2011-04-17T13:21:19.560" LastActivityDate="2011-04-17T13:21:19.560" Title="shortcut for finding a inverse  of matrix" Tags="&lt;linear-algebra&gt;&lt;matrices&gt;" AnswerCount="3" CommentCount="10" />
  <row Id="21534" PostTypeId="2" ParentId="21533" CreationDate="2011-02-11T15:13:44.573" Score="1" ViewCount="" Body="&lt;p&gt;Your best bet is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Gauss-Jordan_elimination&quot; rel=&quot;nofollow&quot;&gt;Gauss-Jordan method&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="589" LastActivityDate="2011-02-11T15:13:44.573" CommentCount="2" />
  <row Id="21535" PostTypeId="2" ParentId="21529" CreationDate="2011-02-11T15:14:02.320" Score="7" ViewCount="" Body="&lt;p&gt;The former is caused by functions that blow up too rapidly near a point.  Think about functions of the form $f(x) = x^{-a} 1_{[0,1]}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The latter is caused by functions that decay too slowly near $\infty$.  Think about functions of the form $f(x) = x^{-a} 1_{[1,\infty)}$.&lt;/p&gt;&#xA;" OwnerUserId="822" LastActivityDate="2011-02-11T15:14:02.320" CommentCount="1" />
  <row Id="21536" PostTypeId="1" AcceptedAnswerId="21546" CreationDate="2011-02-11T15:14:36.853" Score="0" ViewCount="36" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need to get the axis angle from a matrix of the form:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;\begin{matrix}&#xD;&#xA;\cos \theta &amp;amp; -\sin \theta &amp;amp; tx\\&#xD;&#xA;\sin \theta &amp;amp; \cos \theta &amp;amp; ty\\&#xD;&#xA;0 &amp;amp; 0 &amp;amp; 1\\&#xD;&#xA;\end{matrix}&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Into an axis angle. I've tried the standard formula:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\theta$ = $\arccos$ ( (m[0][0] + m[1][1] + m[2][2] - 1) / 2 )&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However this formula inverts when the angle reaches PI. For example, given a rotational matrix for an angle of 225 degrees, we get a returned value of about 135 degrees, which in itself is perfectly correct, but I really need the full value.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any way of handling these cases?&lt;/p&gt;&#xA;" OwnerUserId="6914" LastActivityDate="2011-02-11T15:38:51.833" Title="Calculating axis angle from matrix without wrapping at PI" Tags="&lt;matrices&gt;" AnswerCount="1" />
  <row Id="21537" PostTypeId="2" ParentId="21533" CreationDate="2011-02-11T15:14:57.630" Score="0" ViewCount="" Body="&lt;p&gt;For a 2x2 matrix, the inverse is:&#xA;$$&#xD;&#xA;\left(\begin{array}{cc}&#xD;&#xA;a&amp;amp;b\\&#xD;&#xA;c&amp;amp;d&#xD;&#xA;\end{array}\right)^{-1}&#xD;&#xA;=&#xD;&#xA;{1 \over a d - b c}&#xD;&#xA;\left(\begin{array}{rr}&#xD;&#xA;d&amp;amp;-b\\&#xD;&#xA;-c&amp;amp;a&#xD;&#xA;\end{array}\right)&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;just swap the 'a' and 'd', negate the 'b' and 'c', then divide all by the determinant $a d - b c$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's really the most straightforward 'trick', just memorize that pattern.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For 3x3, it's lot more complicated but there is a pattern. As usual compute the determinant first (kind of a pain; but surely you already know the pattern to compute that quickly).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;\left(\begin{array}{ccc}&#xD;&#xA;a&amp;amp;b&amp;amp;c\\&#xD;&#xA;d&amp;amp;e&amp;amp;f\\&#xD;&#xA;g&amp;amp;h&amp;amp;i&#xD;&#xA;\end{array}\right)^{-1}&#xD;&#xA;=&#xD;&#xA;{1 \over {\rm{det}}}&#xD;&#xA;\left(\begin{array}{rrr}&#xD;&#xA;e i - f h&amp;amp;-(b i - c h)&amp;amp;b f - c e\\&#xD;&#xA;-(f g - d i)&amp;amp;a i - c g&amp;amp;-(a f -c d)\\&#xD;&#xA;d h - e g&amp;amp;-(a h - b g)&amp;amp;a e - b d&#xD;&#xA;\end{array}\right)&#xD;&#xA;$$.&#xA;The pattern is that each entry is&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;the determinant of the submatrix gotten by removing that row and column. I.e. for row 2 column 3 (the $f$ position, the determinant is $a h - b g$:&#xA;$$&#xD;&#xA;\det\left(\begin{array}{cc}&#xD;&#xA;a&amp;amp;b\\&#xD;&#xA;g&amp;amp;h&#xD;&#xA;\end{array}\right)&#xD;&#xA;= a h - b g&#xD;&#xA;$$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;then multiply in the checkerboard pattern. (i.e.1x1 is positive, 1x2 is negative... mathematically it's multiply by $(-1)^r(-1)^c$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;Then transpose.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;See? There's a pattern, but I feel it's about the same symbolic complexity as just doing it brute force Gaussian-elimination style.&lt;/p&gt;&#xA;" OwnerUserId="1919" LastEditorUserId="1919" LastEditorDisplayName="" LastEditDate="2011-02-11T15:45:14.860" LastActivityDate="2011-02-11T15:45:14.860" CommentCount="10" />
  <row Id="21538" PostTypeId="1" AcceptedAnswerId="21566" CreationDate="2011-02-11T15:17:31.827" Score="3" ViewCount="128" Body="&lt;p&gt;I have the 7 following reccurence relations:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$A_n = B_{n-1} + C_{n-1}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$B_n = A_n + C_{n-1}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$C_n = B_n + C_{n-1}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$D_n = E_{n-1} + G_{n-1}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$E_n = D_n + F_{n-1}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$F_n = G_n + C_n$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$G_n = E_n + F_{n-1}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which I would like to solve, with the goal of eventually finding an explicit form of $E_n$. I started out by looking at only $A_n$, $B_n$ and $C_n$, and found a formula for $A_n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$A_n = 1/3 \sqrt{3} (2+ \sqrt{3})^n - 1/3 \sqrt{3} (2 - \sqrt{3})^n$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but I cant seem to find the right trick this time.&lt;/p&gt;&#xA;" OwnerUserId="3317" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2011-06-03T13:48:36.913" LastActivityDate="2011-06-03T13:48:36.913" Title="Solving a set of recurrence relations" Tags="&lt;recurrence-relations&gt;" AnswerCount="3" />
  <row Id="21539" PostTypeId="2" ParentId="21538" CreationDate="2011-02-11T15:20:57.797" Score="3" ViewCount="" Body="&lt;p&gt;Write the system in matrix form. Then try to diagonalize the matrix.&lt;/p&gt;&#xA;" OwnerUserId="589" LastActivityDate="2011-02-11T15:20:57.797" />
  <row Id="21540" PostTypeId="2" ParentId="21525" CreationDate="2011-02-11T15:22:01.117" Score="11" ViewCount="" Body="&lt;p&gt;The short answer is: there is none. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The slightly less short answer is: there are some nice descriptions of ellipticity for higher order equations and systems. There are some nice descriptions of hyperbolicity for higher order equations and systems. There is generally no description for parabolicity. &lt;strong&gt;And&lt;/strong&gt; elliptic, parabolic, and hyperbolic do not come close to exhausting the possible PDEs that can be written down beyond second order scalar equations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The long answer is, well, long. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;To scratch at the surface of this problem, we need to journey back to 1926, when the definitions of elliptic, parabolic, and hyperbolic PDEs are given by Jacques Hadamard. In his study of scalar linear partial differential equations of second order (the work has since been compiled and published as &lt;em&gt;Lectures on Cauchy's problem in linear partial differential equations&lt;/em&gt; by Dover publications in 1953), Hadamard made the following definitions. (As an aside, it is also in those lectures that Hadamard made the first modern definition of well-posedness of the Cauchy problem of a PDE.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given a linear partial differential operator of second order with real coefficients, its principal part can be represented by a (symmetric) matrix of coefficients $a^{ij}\partial^2_{ij}$. The operator is then said to be &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;Elliptic&lt;/strong&gt; if $a^{ij}$ is positive definite or negative definite&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Parabolic&lt;/strong&gt; if $a^{ij}$ is positive or negative semi-definite, and admits precisely one 0 eigenvalue. &lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Hyperbolic&lt;/strong&gt; if $a^{ij}$ is indefinite, but is non-degenerate; that is, for every vector $v_j$, there exists some vector $w_i$ such that $w_ia^{ij}v_j \neq 0$. In other words, it has no zero eigenvalue. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Already you see that the classification is incomplete: any operator with principle part having nullity $&amp;gt;1$, or having nullity $= 1$ but with indefinite sign is not classified. Furthermore, the definition of hyperbolic is different from the modern one. Indeed, Hadamard made the additional definition that the operator is &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;Normal hyperbolic&lt;/strong&gt; if $a^{ij}$ is hyperbolic, and furthermore all but one of its eigenvalues have the same sign. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The wave operator is a normal hyperbolic operator. Whereas what now-a-days we call the &lt;a href=&quot;http://en.wikipedia.org/wiki/Ultrahyperbolic_wave_equation&quot; rel=&quot;nofollow&quot;&gt;ultrahyperbolic operators&lt;/a&gt; (in order to distinguish them from hyperbolic ones) were nominally considered to be hyperbolic by Hadamard's standards.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hadamard was able to show that linear, normal, hyperbolic PDEs admit well-posed initial value (Cauchy) problems for data prescribed on hypersurfaces over which tangent space the restriction of $a^{ij}$ is elliptic. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&lt;strong&gt;Hyperbolicity redefined&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let us first talk about hyperbolic equations (because I'm more familiar with its history). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The next advance came around the mid 20th century, with contributions by people like Jean Leray and Lars Gårding. Rather than using Hadamard's geometric definition of hyperbolicity, in order to deal with higher order equations and systems of equations, hyperbolicity becomes more-or-less synonymous with &quot;having a well-posed initial value problem&quot;. This recognition does not mean the definition is separate from the geometry of the coefficients. Here I give a theorem (which you can find in Lars Hörmander's &lt;em&gt;Analysis of Linear Partial Differential Operators&lt;/em&gt;):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Theorem&lt;/strong&gt; A linear partial differential operator of order $m$ with smooth coefficients admit a well-posed Cauchy problem for arbitrary smooth data if and only if its symbol is a &lt;a href=&quot;http://eom.springer.de/h/h048300.htm&quot; rel=&quot;nofollow&quot;&gt;hyperbolic polynomial&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically, a linear partial differential operator $P = \sum_{|\alpha| \leq m} (i)^{|\alpha|}a^{\alpha}\partial_\alpha$ (where $\alpha$ is a &lt;a href=&quot;http://en.wikipedia.org/wiki/Multi-index_notation&quot; rel=&quot;nofollow&quot;&gt;multi index&lt;/a&gt;) has corresponding symbol $p(\xi) = \sum_{|\alpha| \leq m} a^\alpha \xi_\alpha$. The symbol $p$ is said to by a hyperbolic polynomial in $\xi\in\mathbb{R}^n$ if there exists a vector $\tau\in\mathbb{R}^n$ such that the $\sum_{|\alpha| = m} a^\alpha \tau_\alpha \neq 0$, and that there exists some number $s_0 &amp;gt; 0$ such that for all $s &amp;lt; s_0$, the polynomial&#xA;$$ p(\xi + i s \tau) \neq 0 \qquad \forall \xi\in\mathbb{R}^n $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To apply to a system of equations, We consider the system given by &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ P_{ij}u^j = 0 $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and $P_{ij} = \sum (i)^{|\alpha|} a^{\alpha}_{ij} \partial_\alpha$ is a matrix of scalar partial differential operators, the condition is analogous, with the symbol defined instead by &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ p(\xi) = \det( \sum_{|\alpha| \leq m} a^{\alpha}_{ij} \xi_\alpha ) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;via the determinant. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem with this definition is that it is not-so-easy to check, and does not generalize easily to non-linear equations. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As it turns out, to consider quasilinear equations, one realizes that what is really important for the well-posedness of the initial value problem is only the principal part of the operator (that is, the part with the highest number derivatives). There have been many attempts to provide sufficient conditions on the principal part of the operator that guarantees the well-posedness of the Cauchy problem, and all of these conditions are named as &quot;(blank) hyperbolicity&quot;. Notice that none of them are necessary conditions: they are all strictly stronger assumptions than the conditions given above in the theorem. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I will not attempt to give all the definitions here. I'll refer you to some keywords for searching in textbooks or literature. First there is the notion of strict hyperbolicity. This is the direct analogue of the above characterisation using hyperbolic polynomials in the context of principal symbols. In fact, we have that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Theorem&lt;/strong&gt; If the principal part of an operator is strictly hyperbolic, than regardless of what the lower order terms are, the operator is hyperbolic. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The definition of strict hyperbolicity is as follows: construct the principal part of the symbol as above, but just using the highest order terms $p_m(\xi) = \sum_{|\alpha| = m} a^\alpha\xi_\alpha$, so it is a homogeneous polynomial. It is strictly hyperbolic if there against exists some vector $\tau$ such that for any $\xi\neq 0$, the $m$-th degree polynomial in $s$ given by $p(\xi + s\tau)$ has $m$ distinct real roots. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A rather different definition of hyperbolicity is called &lt;a href=&quot;https://ccrma.stanford.edu/~bilbao/master/node46.html&quot; rel=&quot;nofollow&quot;&gt;symmetric hyperbolicity&lt;/a&gt;. This relies on re-writing your higher-order equation/system as a first order system of conservation laws. It requires being able to represent/convert the system into a special algebraic form, with certain symmetry and positivity conditions on its principal part. There are many, many good books written about this subject. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One interesting facet of symmetric hyperbolic systems is that sometimes the usual/simple way of reducing a system does not suffice to exhibit the symmetric hyperbolic structure, and one may be forced to &quot;augment&quot; the system with dummy dependent variables. Much of this is discussed in C. Dafermos's book &lt;em&gt;Hyperbolic conservation laws in continuum physics&lt;/em&gt;, and has been used by, e.g. &lt;a href=&quot;http://www.umpa.ens-lyon.fr/~serre/DPF/Conv.pdf&quot; rel=&quot;nofollow&quot;&gt;Denis Serre to show that complicated systems such as the Maxwell-Born-Infeld&lt;/a&gt; model of non-linear electrodynamics is hyperbolic. (Though the same can be shown, perhaps slightly more easily, using the next method; a proof can be found in &lt;a href=&quot;http://arxiv.org/abs/1008.5018&quot; rel=&quot;nofollow&quot;&gt;Jared Speck's recent paper&lt;/a&gt;.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A third method of defining hyperbolicity is the notion of &lt;em&gt;regular hyperbolicity&lt;/em&gt; due to Demetrios Christodoulou in his book &lt;em&gt;Action Principle and Partial Differential Equations&lt;/em&gt; (though the germ of the idea was already considerd by Hughes, Kato, and Marsden in the 1970s). This is a method specially adapted for studying second order quasilinear systems of differential equations. I gave a &lt;a href=&quot;http://arxiv.org/abs/1011.3029v1&quot; rel=&quot;nofollow&quot;&gt;quick discussion in a recent paper of mine&lt;/a&gt;. In particular, it is noted in that paper that there is a large gap between regular hyperbolicity, which is perhaps the least restrictive in the context of second order systems of equations of the three above, and ellipticity. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are also other more special cases where the &quot;time direction&quot; is distinguished in the system of equations. In particular there is a large class of equations of the form &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \partial_t^{2m}u = F(u,\partial u, \ldots, \partial^{2m-1} u) + A u $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which can easily to be seen as hyperbolic as long as $A$ is a quasilinear elliptic operator of order $2m$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general a principal feature of hyperbolic equations is that their solution tends to rely on the existence of $L^2$ energy estimates. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&lt;strong&gt;Ellipticity&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Compared to hyperbolicity, ellipticity is better understood (but unfortuantely, not by me. So I'll point you to references).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For scalar nonlinear equations, there are very well developed theory available in various textbooks. See, for example, Gilbarg and Trudinger, &lt;em&gt;Elliptic partial differential equations of second order&lt;/em&gt; (pay attention to the last few chapters), or perhaps Caffarelli and Cabre, &lt;em&gt;Fully nonlinear elliptic equations&lt;/em&gt; for development in the second order case. For higher order operators, say $P$ is of order $2m$, a genearally used criterion is that its principal symbol satisfies the elliptic inequality $p_{2m}(\xi) \geq \lambda |\xi|^{2m}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For systems, on the other hand, ellipticity is also a tricky business. In so far as second order systems are concerned, a very useful condition is the so-called Legendre-Hadamard condition, otherwise known as  rank-1 convexity condition. This condition goes back to (if I am not mistaken) Jesse Douglas and C.B. Morrey in their works on the Plateau problem in Riemannian geometry. The effectiveness of this condition is still a much studied subject related to nonlinear elasticity and calculus of variations. You can learn quite a bit about it by looking at some of &lt;a href=&quot;http://people.maths.ox.ac.uk/ball/&quot; rel=&quot;nofollow&quot;&gt;John Ball's&lt;/a&gt; survey articles. &lt;/p&gt;&#xA;" OwnerUserId="1543" LastEditorUserId="1543" LastEditorDisplayName="" LastEditDate="2011-02-11T20:08:28.263" LastActivityDate="2011-02-11T20:08:28.263" CommentCount="5" />
  <row Id="21541" PostTypeId="2" ParentId="21504" CreationDate="2011-02-11T15:22:52.817" Score="2" ViewCount="" Body="&lt;p&gt;I adjusted your question again for further clarity.  As far as I can tell, you are trying to decompose an integer $m$ into $n$ positive integer parts.  These are called compositions and there are ${m-1}\choose{n-1}$ equally likely ways of doing it as you have to choose $n-1$ points from $m-1$ inside the line.  So for $m=10$ and $n=3$ there are 36 ways of doing this.  This is the denominator in your probability.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finding the number of compositions where the largest is a particular value is slightly harder.  Fortunately I have a Java &lt;a href=&quot;http://www.btinternet.com/~se16/js/partitions.htm&quot; rel=&quot;nofollow&quot;&gt;partition and composition calculator&lt;/a&gt;.  So to find how many compositions of 10 into 3 parts where the largest is 4, you choose &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&quot;Compositions of:&quot; 10, &lt;/li&gt;&#xA;&lt;li&gt;&quot;Exact number of terms:&quot; 3, &lt;/li&gt;&#xA;&lt;li&gt;&quot;Largest term exactly:&quot; 4. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Click the button and the answer is 6. [To check: 10 = 4+4+2 = 4+3+3 = 4+2+4 = 3+4+3 = 3+3+4 = 2+4+4.] So the probability is 6/36 or 1/6.  Similarly with the largest part 5 it is 12/36 = 1/3, with largest part 6 it is 9/36 = 1/4, with largest part 7 it is 6/36 = 1/6, and with largest part 7 it is 3/36 = 1/12.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The calculator uses a recurrence based on removing the last part, and adding up these earlier numbers, so the number of compositions of 10 into 3 parts where the largest is 4 is equal to the sum of: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;the number of compositions of 6 into 2 parts where the largest is 1 (0)&lt;/li&gt;&#xA;&lt;li&gt;the number of compositions of 6 into 2 parts where the largest is 2 (0)&lt;/li&gt;&#xA;&lt;li&gt;the number of compositions of 6 into 2 parts where the largest is 3 (1)&lt;/li&gt;&#xA;&lt;li&gt;the number of compositions of 6 into 2 parts where the largest is 4 (2)&lt;/li&gt;&#xA;&lt;li&gt;the number of compositions of 7 into 2 parts where the largest is 4 (2)&lt;/li&gt;&#xA;&lt;li&gt;the number of compositions of 8 into 2 parts where the largest is 4 (1)&lt;/li&gt;&#xA;&lt;li&gt;the number of compositions of 9 into 2 parts where the largest is 4 (0)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;and 0+0+1+2+2+1+0 = 6, as indicated by the calculator.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I doubt there is a closed form for the probabilities.   &lt;/p&gt;&#xA;" OwnerUserId="6460" LastActivityDate="2011-02-11T15:22:52.817" CommentCount="1" />
  <row Id="21542" PostTypeId="1" AcceptedAnswerId="21574" CreationDate="2011-02-11T15:27:58.550" Score="1" ViewCount="120" Body="&lt;p&gt;We know that &lt;a href=&quot;http://en.wikipedia.org/wiki/Stirling_numbers_of_the_second_kind&quot; rel=&quot;nofollow&quot;&gt;Stirling numbers of the second kind&lt;/a&gt; is the number of ways to partition a set of $n$ elements into $m$ nonempty sets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is what's the number ways if the max cardinality of all the partitioned sets is $k$.&lt;/p&gt;&#xA;" OwnerUserId="6904" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2011-05-31T17:17:35.423" LastActivityDate="2011-05-31T17:17:35.423" Title="Stirling numbers of the second kind with max cardinality" Tags="&lt;combinatorics&gt;&lt;stirling-numbers&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="21543" PostTypeId="2" ParentId="21538" CreationDate="2011-02-11T15:30:01.777" Score="4" ViewCount="" Body="&lt;p&gt;Here's what I would do: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;let $A(z) = \sum_{n=0}^\infty A_n z^n$ be the ``generating function'' of $A_n$. Define $B(z), C(z)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then multiply the recurrence $A_n = B_{n-1} + C_{n-1}$ by $z^n$ to get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ A_n z^n = B_{n-1} z^n + C_{n-1} z^n $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and sum over $z^n$ to get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \sum_{n=1}^\infty A_n z^n = \sum_{n=1}^\infty B_{n-1} z^n + \sum_{n=1}^\infty C_{n-1} z^n. $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The left-hand side is $A(z) - A_0$ and the right-hand size is $z B(z) + z C(z)$; so you get $A(z) - A_0 = zB(z) + zC(z)$. Notice that I had to write $A(z) - A_0$ because you have not provided the initial conditions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can do the same with the second and third equations and solve the resulting three-by-three system, which shouldn't be too hard. This will give you $A(z)$.  Now you just need to find the coefficients in $A(z)$; you can do this by writing $A(z)$ in terms of partial fractions. For a written-out example of this, see Section 1.3 of &lt;a href=&quot;http://www.math.upenn.edu/~wilf/gfology2.pdf&quot; rel=&quot;nofollow&quot;&gt;generatingfunctionology&lt;/a&gt; by Wilf. (Link goes to the full text, available online.)&lt;/p&gt;&#xA;" OwnerUserId="173" LastActivityDate="2011-02-11T15:30:01.777" />
  <row Id="21544" PostTypeId="1" AcceptedAnswerId="21600" CreationDate="2011-02-11T15:31:25.847" Score="2" ViewCount="61" Body="&lt;p&gt;I was trying to answer this question - whether a subsequence of a regular sequence is regular in a Noetherian ring which is not local. In the local case, regular sequences can be permuted and so a subsequence can be considered to be the initial subsequence of a regular sequence and hence regular. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's consider the following case. Let $R$ be a Noetherian ring which is not local. Let $x_1,x_2,x_3$ be a regular sequence. Is $x_1,x_3$ regular?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, $x_1$ is a nonzerodivisor on $R$. So it's really a question of whether $x_3$ is a nonzerodivisor on $R/(x_1)$. Suppose not, then there is an element $y\in R\setminus (x_1)$ s.t. $x_3 y\in (x_1)R$. But since, $x_3$ is a nonzerodivisor on $R/(x_1,x_2)$, we must have $y\in (x_1,x_2)$. So we may assume, $y\in (x_2)$. Write, $y=rx_2$. Then, $x_3 rx_2 \in (x_1)$. But $x_2$ is a nonzerodivisor on $R/(x_1)$. So, $rx_3\in (x_1)$. I am stuck here. Can someone help with the proof or know a counterexample? Thanks.&lt;/p&gt;&#xA;" OwnerUserId="6166" LastActivityDate="2011-02-11T22:45:36.550" Title="Subsequences of regular sequence" Tags="&lt;commutative-algebra&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="21545" PostTypeId="1" AcceptedAnswerId="21626" CreationDate="2011-02-11T15:38:22.557" Score="2" ViewCount="192" Body="&lt;p&gt;Show that if $a$, $b$ are positive integers, then we have:&#xA;$\gcd(a,b) = \gcd(a + b, \mathrm{lcm}[a,b])$.&lt;/p&gt;&#xA;" OwnerUserId="6912" LastEditorUserId="6296" LastEditorDisplayName="" LastEditDate="2011-03-03T20:45:49.893" LastActivityDate="2011-03-03T20:45:49.893" Title="$\gcd(a,b) = \gcd(a + b, \mathrm{lcm}[a,b])$" Tags="&lt;homework&gt;&lt;elementary-number-theory&gt;" AnswerCount="4" CommentCount="7" />
  <row Id="21546" PostTypeId="2" ParentId="21536" CreationDate="2011-02-11T15:38:51.833" Score="0" ViewCount="" Body="&lt;p&gt;You may have an &lt;a href=&quot;http://en.wikipedia.org/wiki/Atan2&quot; rel=&quot;nofollow&quot;&gt;Atan2&lt;/a&gt; function available, which sorts out the quadrants.  If so, you want Atan2$(\cos\theta,\sin\theta)$  Otherwise, you can do $\arccos\theta$ but then you have to sort out the quadrants afterward using the value for $\sin\theta$&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-02-11T15:38:51.833" CommentCount="2" />
  <row Id="21547" PostTypeId="2" ParentId="21530" CreationDate="2011-02-11T15:45:09.860" Score="1" ViewCount="" Body="&lt;p&gt;Been a while since I did any maths, so this is probably the wrong way of going about it. I'd take logs with base $a$ of $ab=c^n$, this will give:&#xA;$$1 + \log_a(b) = n\log_a(c)  \rightarrow  1 = n\log_a(c) - \log_a(b).$$&#xA;For the right hand side to equal 1, we need $b = e^n$ (this wouldn't necessarily be the case if $\gcd(a,b) \neq 1$):&#xA;$$1 = n(\log_a(c) - \log_a(e)).$$&#xA;I'd take the inverse log from here, giving $a = \left(\frac{c}{e}\right)^n$. Simple to explain why $\frac{c}{e}$ must be a whole number, $d$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bet that's the worst possible solution to this problem&lt;/p&gt;&#xA;" OwnerUserId="6915" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-12T06:00:16.473" LastActivityDate="2011-02-12T06:00:16.473" />
  <row Id="21548" PostTypeId="1" AcceptedAnswerId="21553" CreationDate="2011-02-11T16:00:36.297" Score="4" ViewCount="210" Body="&lt;p&gt;Using congruences, show that the following is always an integer for every integer&#xA;value of $n$:&#xA;$$\frac{1}{5}n^5 + \frac{1}{3}n^3 + \frac{7}{15}n.$$&lt;/p&gt;&#xA;" OwnerUserId="6912" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-11T16:18:47.367" LastActivityDate="2011-02-18T00:21:31.667" Title="Using congruences, show $\frac{1}{5}n^5 + \frac{1}{3}n^3 + \frac{7}{15}n$ is integer for every $n$" Tags="&lt;elementary-number-theory&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="21549" PostTypeId="2" ParentId="21548" CreationDate="2011-02-11T16:04:44.417" Score="3" ViewCount="" Body="&lt;p&gt;Taking the lcm we have $\displaystyle \frac{1}{15} \cdot \Bigl[ 3n^{5}+5n^{3} + 7n\Bigr]$ Now show that the quantity $3n^{5} + 5n^{3}+7n$ is always divisible by $15$. Induction may be useful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clearly for $n=1$, $3+5+7=15$ is divisible by $15$. Assume that it is true for $n=k$. That is assume that $3k^{5}+5k^{3}+7k$ is divisible by $15$. Use this to show that the quantity &lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align*}&#xA;3(k+1)^{5}+5(k+1)^{3}+7(k+1) &amp;amp;= 3 \Bigl[k^{5} + {5 \choose 1}k^{4} + \cdots +1\Bigr] + 5(k+1)^{3}+ 7(k+1)&#xA;\end{align*}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is divisible by $15$.&lt;/p&gt;&#xA;" LastActivityDate="2011-02-11T16:41:35.403" CommentCount="6" />
  <row Id="21550" PostTypeId="2" ParentId="21542" CreationDate="2011-02-11T16:42:02.657" Score="3" ViewCount="" Body="&lt;p&gt;I Googled &lt;code&gt;restricted partitions stirling&lt;/code&gt; and the third hit was &lt;a href=&quot;http://dlmf.nist.gov/26.9&quot; rel=&quot;nofollow&quot;&gt;http://dlmf.nist.gov/26.9&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Looks pretty thorough.&lt;/p&gt;&#xA;" OwnerUserId="5676" LastActivityDate="2011-02-11T16:42:02.657" CommentCount="1" />
  <row Id="21551" PostTypeId="2" ParentId="21545" CreationDate="2011-02-11T16:42:30.817" Score="0" ViewCount="" Body="&lt;p&gt;Start by writing $a=d a&amp;#39;$, $b=d b&amp;#39;$, where $d=(a,b)$.&lt;/p&gt;&#xA;" OwnerUserId="589" LastActivityDate="2011-02-11T16:42:30.817" CommentCount="6" />
  <row Id="21552" PostTypeId="1" CreationDate="2011-02-11T16:44:10.860" Score="1" ViewCount="97" Body="&lt;p&gt;Hi,&#xA;I'm working on to prove the following:&#xA;Show that the convergence in the space w (space of all sequences with respect to the Frechet metric ) is the coordinate convergence.&#xA;Any hint is appreciated, thanx.&lt;/p&gt;&#xA;" OwnerUserId="6919" LastActivityDate="2011-02-12T03:05:14.980" Title="w - space of all sequences with Frechet metric" Tags="&lt;functional-analysis&gt;" AnswerCount="1" CommentCount="9" />
  <row Id="21553" PostTypeId="2" ParentId="21548" CreationDate="2011-02-11T16:48:05.360" Score="15" ViewCount="" Body="&lt;p&gt;Show $P(n)=3n^5+5n^3+7n$ is divisible by 15 for every $n$.  Recall $x^p\equiv x \pmod{p}$ (Fermat's Little Theorem).  Looking modulo 5 we see &#xA;$$P(n)\equiv 3n^5+7n\equiv 3n+7n=10n\equiv 0.$$&#xA;Now looking modulo 3 we see $$P(n)\equiv 5n^3+7\equiv 5n+7n=12n\equiv 0.$$ Thus by application of the chinese remainder theorem, $P(n)$ is divisible by 15 for every $n$.&lt;/p&gt;&#xA;" OwnerUserId="6075" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-11T21:06:45.713" LastActivityDate="2011-02-11T21:06:45.713" CommentCount="4" />
  <row Id="21554" PostTypeId="2" ParentId="21514" CreationDate="2011-02-11T16:54:46.193" Score="2" ViewCount="" Body="&lt;p&gt;I am not an expert in guiding you but do problems to make your concepts more clearer and clearer. As Klaus points out there are very good texts in Linear Algebra. Just borrow them and work out the problems. Let me lists some good texts:- &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Linear Algebra by Hoffman Kunze&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Linear Algebra by S.Friedberg, A.Insel and Lawrence Spencer.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Linear Algebra Problem Book,by Paul Halmos.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Finite Dimensional Vector Spaces, by Paul Halmos.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;PROBLEMS AND THEOREMS IN LINEAR ALGEBRA V. Prasolov link &lt;a href=&quot;http://mathsouls.110mb.com/Ebook/Math/En/Linear%20Algebra/Problems%20and%20Theorems%20in%20Linear%20Algebra.pdf&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Just go through all of them ans see which one suits you the best. I liked the Friedberg book on linear Algebra. Michael Artin's Algebra book also has good amount of Linear Algebra in it. You may like to see that as well. These books will all be available at &lt;strong&gt;library.nu&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moreover, if you are unable to solve a problem or understand a concept, please visit this site and the experts will be happy to help you out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Happy Reading!!!&lt;/p&gt;&#xA;" LastActivityDate="2011-02-11T17:02:59.237" CommentCount="1" />
  <row Id="21555" PostTypeId="1" AcceptedAnswerId="21836" CreationDate="2011-02-11T17:02:44.340" Score="5" ViewCount="48" Body="&lt;p&gt;Hi, I have a basic question :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a) Suppose $\gamma $ and $ \gamma&amp;#39; $ be conjugate Fuchsian groups acting freely and properly discontinuously on the upper half-plane H to produce two Riemann surfaces $ S $ and $S&amp;#39;$. How can we prove that $S$ and $S$ are isometric ? A detailed answer would be appreciated .Do I need to have $S$ and $S&amp;#39;$ closed for this question ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was trying to send the equivalence class  $ [\gamma(z) ] $ in $S$ to $[\delta\gamma\delta^-1] $ . Would that work ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;b) Also, how can we prove that two hyperbolic surfaces ( again, do we need closed surfaces ) are isometric by $f$ implies the lift $\tilde{f}$ of $f$ is an isometry ( it is obviously a local isometry, but why is it bijective,i.e. have an inverse ) of the upper half plane $H$ ? Is the result true for arbitrary quotients of differential manifolds and diffeomorphism of quotients rather than isometry ? Thanks !&lt;/p&gt;&#xA;" OwnerUserId="6920" LastActivityDate="2011-02-13T12:50:03.360" Title="Two hyperbolic surfaces corresponding to conjugate Fuchsian groups are isometric" Tags="&lt;riemann-surface&gt;&lt;hyperbolic-geometry&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="21556" PostTypeId="2" ParentId="21490" CreationDate="2011-02-11T17:04:48.360" Score="3" ViewCount="" Body="&lt;p&gt;Here is a proof of the lemma, where I now  assume $0\leq \theta_1&amp;lt;\ldots&amp;lt;\theta_n&amp;lt;2\pi$ and all $r_j&amp;gt;0$; furthermore I replace $e^{i k\theta_j}$ by $e^{-i k\theta_j}$ in the definition of $a_k(P)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $b(t)$ be the $2\pi$-periodic box function which is $=1$ for $0\leq t&amp;lt;{2\pi\over m}$ and $=0$ for ${2\pi\over m}\leq t&amp;lt;2\pi$, and consider the function&#xA;$f(t):=\sum_{j=1}^n r_j b(t-\theta_j)$. For $k\ne0$ the Fourier coefficients of $f$ compute to&#xA;$$\hat f(k)={i\over 2\pi k}(e^{-2\pi i k/m}-1)\&amp;gt; a_k(P)=0,$$&#xA;so $f$ is a constant. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider a summand $r_j b(t-\theta_j)$ of $f$. As this summand jumps down to zero at the point $t=\theta_j+{2\pi\over m}$ there has to be another summand $r_l b(t-\theta_l)$ to compensate for this jump; in fact one necessarily has $\theta_l= \theta_j+{2\pi\over m}$ and $r_l=r_j$. It follows that $P$ is invariant under a rotation by ${2\pi\over m}$.&lt;/p&gt;&#xA;" OwnerUserId="1303" LastActivityDate="2011-02-11T17:04:48.360" CommentCount="1" />
  <row Id="21557" PostTypeId="1" AcceptedAnswerId="21563" CreationDate="2011-02-11T17:15:42.520" Score="4" ViewCount="89" Body="&lt;p&gt;Let's say we have a basis $B$ of $\mathbb{R}^n$ consisting of vectors $\vec{v}_1$ through $\vec{v}_n$, and some other basis $C$ of $\mathbb{R}^n$. Then, would $[\vec{v}_1]_C$ through $[\vec{v}_n]_C$ be a basis for $\mathbb{R}^n$ as well? It seems obvious, but I am not sure how to go about this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am trying to prove this by creating a matrix $S$ that is a change of basis matrix, so it has $[\vec{v}_1]_C$ through $[\vec{v}_n]_C$ as its columns. So I get $C$$S$=$B$. I know that kernels of both $C$ and $B$ are equal to zero, since they are square matrices with independent columns as they are bases, but I am not sure what that tells me about $S$. Maybe there is a way to show that a matrix with independent columns ($C$) times a matrix with not-independent columns ($S$) can't (always/sometimes?) yield another matrix with independent columns ($B$).&lt;/p&gt;&#xA;" OwnerUserId="5157" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-11T17:54:08.050" LastActivityDate="2011-02-11T17:55:58.267" Title="Prove that if $S$ is a change of basis matrix, its columns are a basis for $\mathbb{R}^n$" Tags="&lt;linear-algebra&gt;" AnswerCount="2" CommentCount="0" />
  <row Id="21558" PostTypeId="2" ParentId="21532" CreationDate="2011-02-11T17:16:35.930" Score="3" ViewCount="" Body="&lt;p&gt;For each one, go back to the definition and prove it lies on the axis of symmetry.  If you work with coordinates, putting the triangle at (-1,0), (1,0), (0,a) can make it much easier.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-02-11T17:16:35.930" />
  <row Id="21559" PostTypeId="1" AcceptedAnswerId="21594" CreationDate="2011-02-11T17:25:35.810" Score="1" ViewCount="64" Body="&lt;p&gt;For functions $f: \mathbb{R} \rightarrow \mathbb{C}$, define&#xA;$$&#xD;&#xA;M f(x) =  \sup_{t &amp;gt;0} \frac{1}{2}| f(x+t) + f(x-t) |.&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given $p \geq 1$, I want to construct a sequence of smooth functions $f_n$ such that $||f_n||_p \leq 1$, $|| M f_n   ||_p &amp;lt; \infty$, but $|| M f_n   ||_p \rightarrow \infty$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it even possible?  What if the smoothness condition is relaxed?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In an earlier version of the question, I forgot the $|| M f_n   ||_p &amp;lt; \infty$ condition.&lt;/p&gt;&#xA;" OwnerUserId="6887" LastEditorUserId="6887" LastEditorDisplayName="" LastEditDate="2011-02-11T21:08:16.770" LastActivityDate="2011-02-11T22:06:55.230" Title="Sequence of smooth functions whose image under a maximal operator diverges in $L_p$ norm" Tags="&lt;real-analysis&gt;&lt;harmonic-analysis&gt;" AnswerCount="2" />
  <row Id="21560" PostTypeId="2" ParentId="21409" CreationDate="2011-02-11T17:26:01.880" Score="5" ViewCount="" Body="&lt;p&gt;A proof of the strong law of large numbers can be more&#xA;or less complicated depending on your hypotheses.&#xA;In your case, since you assume that $E[X_i^2]&amp;lt;\infty$ there is &#xA;a straightforward proof. I am taking this from section 7.4 of the third&#xA;edition of &lt;em&gt;Probability and Stochastic Processes&lt;/em&gt; by Grimmett and Stirzaker. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, by splitting into positive and negative parts we can assume (without &#xA;loss of generality) that $X_i\geq 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second, using the positivity, it suffices to &#xA;prove that  $S_{n^2}/n^2\to\mu$ almost surely; that is, we only need &#xA;convergence along that subsequence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next, Chebyshev's inequality gives&lt;br&gt;&#xA;$$P(|S_{n^2}/n^2-\mu|&amp;gt;\varepsilon_n)\leq{E[X_i^2]\over n^2\varepsilon_n^2}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Choosing $\varepsilon_n\downarrow 0$ so slowly that the right hand side above&#xA; is summable, Borel-Cantelli finishes the job since then &#xA;$$P(|S_{n^2}/n^2-\mu| \leq \varepsilon_n \mbox{ for all but finitely many }n) = 1.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, the strong law of large numbers holds under the weaker hypothesis&#xA;that $E[|X_i|]&amp;lt;\infty$. There are various proofs in the literature, but every &#xA;student of probability ought to be familiar with Etemadi's &lt;a href=&quot;http://www.springerlink.com/content/q53k44k209804w95/&quot; rel=&quot;nofollow&quot;&gt;tour de force elementary&#xA;proof&lt;/a&gt;. Etemadi uses a clever truncation argument and similar tools to those  above, and only needs pairwise independence of the $X_i$'s, not full independence. &#xA;Some good textbooks like Grimmett and Stirzaker (section 7.5), Billingsley's &lt;em&gt;Probability and Measure (2nd edition)&lt;/em&gt;, or Durrett's &lt;em&gt;Probability: Theory and Examples (2nd edition)&lt;/em&gt; include Etemadi's treatment.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;N. Etemadi, &lt;em&gt;An elementary proof of the strong law of large numbers&lt;/em&gt;, &#xA; Z. Wahrscheinlichkeitstheorie verw. Gebeite &lt;strong&gt;55&lt;/strong&gt;, 119-122 (1981)&lt;/p&gt;&#xA;" OwnerUserId="940" LastActivityDate="2011-02-11T17:26:01.880" CommentCount="2" />
  <row Id="21562" PostTypeId="2" ParentId="21557" CreationDate="2011-02-11T17:31:37.647" Score="0" ViewCount="" Body="&lt;p&gt;Yes. If $SB = C$ then $S = CB^{-1}$, which is also an invertible matrix since $B$ and $C$ are invertible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $B$ and $C$ are basis, there is a basis change matrix $S$ (or rather, a linear application $f$ associated to $S$) that changes $B$ into $C$.&#xA;But $f$ is not only the application that turns one basis into another, it is an application that turns any basis into another basis.&#xA;If $(e_1,\ldots,e_n)$ is the canonical basis of $\mathbb{R}^n$, $f$ turns this basis into a new basis, whose vectors are the column vectors of its matrix representation $S$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Those applications $f$ who turn basis into basis are precisely the bijective linear applications from $\mathbb{R}^n$ to $\mathbb{R}^n$.&#xA;It is easy to check that if $f$ is a change of basis, then it is bijective ; and if $f$ is bijective, it turns any basis into another basis.&lt;/p&gt;&#xA;" OwnerUserId="6270" LastActivityDate="2011-02-11T17:31:37.647" />
  <row Id="21563" PostTypeId="2" ParentId="21557" CreationDate="2011-02-11T17:32:54.667" Score="3" ViewCount="" Body="&lt;p&gt;Yes, the coordinate vectors of $\mathbf{v}_1,\ldots,\mathbf{v}_n$ with respect to the matrix $C$ will be a basis for $\mathbb{R}^n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that you are abusing notation somewhat: if $B$ and $C$ are bases, then they aren't matrices; what you really want is to have a basis $\beta$, and $B$ is the matrix whose columns are the vectors in $\beta$; and another basis $\gamma$, and $C$ is the matrix whose columns are the vectors in $\gamma$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From what you have: since $CS = B$, and the kernel of $B$ is zero, then so is the kernel of $S$: if $\mathbf{x}$ lies in the kernel of $S$, then &#xA;$$\mathbf{0} = C\mathbf{0} = C(S\mathbf{x}) = (CS)\mathbf{x} = B\mathbf{x}.$$&#xA;But since the kernel of $B$ is zero, that means that $\mathbf{x}=\mathbf{0}$. So the kernel of $S$ is zero, which means that its columns are linearly independent, hence a basis. But the columns of $S$ are the coordinate vectors of the elements of $\beta$ written in terms of $\gamma$, which gives your result.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For your final question, yes, you are correct. Here's a general statement you can prove (using the same technique as above):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If $A$ is $n\times p$, $B$ is $p\times m$, and $C$ is $n\times m$ with $AB=C$, then $\mathrm{ker}(B)\subseteq\mathrm{ker}(C)$. That is, if $\mathbf{x}\in\mathbb{R}^m$ is in the kernel of $B$, then it is also in the kernel of $C$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Another way of attacking the problem is to prove that $[\mathbf{v}_1]_{\gamma},\ldots,[\mathbf{v}_n]_{\gamma}$ are linearly independent directly. But this is easy: suppose that&#xA;$$\alpha_1[\mathbf{v}_1]_{\gamma} + \cdots + \alpha_n[\mathbf{v}_n]_{\gamma} = \mathbf{0}.$$ Then we have:&#xA;$$\mathbf{0} = \alpha_1[\mathbf{v}_1]_{\gamma} + \cdots + \alpha_n[\mathbf{v}_n]_{\gamma}&#xD;&#xA;= [\alpha_1\mathbf{v}_1+\cdots+\alpha_n\mathbf{v}_n]_{\gamma}.$$&#xA;But this means that $\alpha_1\mathbf{v}_1+\cdots+\alpha_n\mathbf{v}_n=\mathbf{0}$, and since $\mathbf{v}_1,\ldots,\mathbf{v}_n$ are linearly independent, that means $\alpha_1=\cdots=\alpha_n=0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In essence, it's the same argument as you have above, but playing directly with the vectors instead of the associated change-of-basis matrices. &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-11T17:55:58.267" LastActivityDate="2011-02-11T17:55:58.267" />
  <row Id="21564" PostTypeId="2" ParentId="21528" CreationDate="2011-02-11T17:41:19.823" Score="2" ViewCount="" Body="&lt;p&gt;Since the set of all functions from $\mathbb{N}$ to $\{1,2,3\}$ has cardinality $2^{\aleph_0}$, you certainly know that you have at most $2^{\aleph_0}$ elements in your difference.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So to show that $\{1,2,3\}^{\mathbb{N}} - \{1,2\}^{\mathbb{N}}$ has cardinality exactly $2^{\aleph_0}$, it is enough to produce $2^{\aleph_0}$ functions from $\mathbb{N}$ to $\{1,2,3\}$ whose image is not contained in $\{1,2\}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, hopefully you agree that there are just as many functions from $\mathbb{N}$ to $\{1,2\}$ as there are from $\mathbb{N}-\{1\}$ to $\{1,2\}$? Now, take a function from $\mathbb{N}-\{1\}$ to $\{1,2\}$, and extend it to all of $\mathbb{N}$ in such a way that you &lt;em&gt;guarantee&lt;/em&gt; the image includes $3$. That will produce at least $2^{\aleph_0}$ functions in $\{1,2,3\}^{\mathbb{N}}$ that are not in $\{1,2\}^{\mathbb{N}}$. &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-11T17:53:08.517" LastActivityDate="2011-02-11T17:53:08.517" CommentCount="2" />
  <row Id="21565" PostTypeId="2" ParentId="21508" CreationDate="2011-02-11T17:56:08.340" Score="3" ViewCount="" Body="&lt;p&gt;If $f$ is a continuous function on $K$ then $g(t):=f(e^{i t}, e^{-it})$ is a continuous $2\pi$-periodic function of the real variable $t$. It follows that $g$ can be approximated on ${\mathbb R}$ by trigonometric polynomials $\sum_{k=-N}^N a_k e^{i k t}$. Therefore one has $f(z,\bar z)\doteq \sum_{k=1}^N a_{-k}\bar z^k + \sum_{k=0}^N a_{k} z^k$ on $K$.&lt;/p&gt;&#xA;" OwnerUserId="1303" LastActivityDate="2011-02-11T17:56:08.340" />
  <row Id="21566" PostTypeId="2" ParentId="21538" CreationDate="2011-02-11T18:10:47.127" Score="2" ViewCount="" Body="&lt;p&gt;Substitute and eliminate. You already know how to solve a recurrence on one function; eliminate functions from the system until you get to only one, by substituting one function for another. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, substitute $B_{n-1}$ into the definition of $C_n$ to get $$C_n = A_{n-1} + C_{n-2} + C_{n-1}.$$ $B$ has been eliminated, and we're one step closer to having an equation involving just $C$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By inspection, $A$, $B$, and $C$ are mutually defining, and separately $D$, $E$, $F$, $G$ are mutually defining except for $C$. So solve for $C$ first, using $A$ and $B$, then continue on with just $D$, $E$, $F$, $G$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is very similar to Gaussian elimination, but you have the extra subscripts to deal with in $C_{n-1}$, $C_{n-2}$, etc.&lt;/p&gt;&#xA;" OwnerUserId="1919" LastActivityDate="2011-02-11T18:10:47.127" CommentCount="2" />
  <row Id="21567" PostTypeId="2" ParentId="21491" CreationDate="2011-02-11T18:15:26.340" Score="-1" ViewCount="" Body="&lt;p&gt;If we ignore zero matrices, for commmutivity to hold it must also be part of the same automorphism group as the other matrix. Consider the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$AB=BA$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;=&gt; $B^{-1}AB=B^{-1}BA$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;=&gt; $A=B^{-1}AB$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus invertibility is required, but it is not sufficient.&lt;/p&gt;&#xA;" OwnerUserId="166" LastActivityDate="2011-02-11T18:15:26.340" />
  <row Id="21568" PostTypeId="1" CreationDate="2011-02-11T18:42:32.470" Score="1" ViewCount="112" Body="&lt;p&gt;I want a very quick &lt;code&gt;non recursive method&lt;/code&gt; for finding the HCF (Highest Common Factor) of 2 &lt;code&gt;very large numbers&lt;/code&gt; &lt;/p&gt;&#xA;" OwnerUserId="6094" LastActivityDate="2011-02-11T18:52:20.120" Title="Fastest way to compute HCF of 2 numbers" Tags="&lt;algorithms&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="21569" PostTypeId="2" ParentId="21568" CreationDate="2011-02-11T18:52:20.120" Score="1" ViewCount="" Body="&lt;p&gt;For example, &lt;a href=&quot;http://www.lysator.liu.se/~nisse/archive/sgcd.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.lysator.liu.se/~nisse/archive/sgcd.pdf&lt;/a&gt; gives a discussion of sub-quadratic methods for finding the GCD (in particular Schoenhage's algorithm.)&lt;/p&gt;&#xA;" OwnerUserId="3852" LastActivityDate="2011-02-11T18:52:20.120" />
  <row Id="21570" PostTypeId="1" CreationDate="2011-02-11T18:52:37.327" Score="5" ViewCount="148" Body="&lt;p&gt;This is exercise 38 from Chapter 3. Modules and Vector Spaces in Algebra by Adkins and Weintraub (GTM). How do you solve this problem? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let &#xA;\begin{equation*}&#xA;R = \lbrace f : [0, 1] \to \Re : f \;\text{ is continuous and} \; f (0) = f (1) \rbrace&lt;br&gt;&#xA;\end{equation*}&#xA;and let&#xA;\begin{equation*}&#xA;M = \lbrace f : [0, 1]\to \Re : f \;\text{is continuous and} \; f (0) = - f (1) \rbrace.&#xA;\end{equation*}&#xA;Then $R$ is a ring under addition and multiplication of functions, and $M$ is an $R$-module. Show that $M$ is a projective $R$-module that is not free. &lt;/p&gt;&#xA;" OwnerUserId="6924" LastActivityDate="2011-02-16T00:20:33.080" Title="Projective but not free (exercise from Adkin - Weintraub)" Tags="&lt;abstract-algebra&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="21571" PostTypeId="2" ParentId="21530" CreationDate="2011-02-11T19:00:33.693" Score="-1" ViewCount="" Body="&lt;p&gt;obvious assuming the fundamental theorem of arithmetic&lt;/p&gt;&#xA;" OwnerUserId="6925" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-11T19:47:11.900" LastActivityDate="2011-02-11T19:47:11.900" />
  <row Id="21572" PostTypeId="1" CreationDate="2011-02-11T19:06:39.343" Score="1" ViewCount="130" Body="&lt;p&gt;Tom prefers A over B and B over C. Also, Tom is indifferent between a lottery in which he gets C with probability p and A with probability 1-p and a lottery in which he gets B with a probability p and C with probability 1-p. The value of p in both lotteries is the same. For what values of p would Tom's preference be rational in the sense of von Neumann-Morgenstern's expected utility theory?&lt;/p&gt;&#xA;" OwnerUserId="6926" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-11T19:17:08.867" LastActivityDate="2011-05-12T22:57:01.137" Title="game theory hw. von Neumann-Morgenstern's expected utility theory" Tags="&lt;homework&gt;&lt;game-theory&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="21573" PostTypeId="2" ParentId="21570" CreationDate="2011-02-11T19:24:32.340" Score="2" ViewCount="" Body="&lt;p&gt;I made a similar comment on MO where this question was first posted.  Here is an elaboration:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the circle $S^1$ can be thought of as the unit interval $[0,1]$ with the two endpoints identified, $R$ may be viewed as the ring of all real-valued continuous functions on $S^1$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My hint is to view $M$ as the module of global sections of the Mobius band.  For this, think about building the Mobius band as an identifcation space of $[0,1] \times \mathbb{R}$: you glue the two ends together with a half-twist.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is only fair to mention that I have implicitly in mind the celebrated theorem of Richard Swan which gives an equivalence between vector bundles over a compact base and modules over the ring of continuous functions on the base: see e.g. Chapter 6 &lt;a href=&quot;http://math.uga.edu/~pete/integral.pdf&quot; rel=&quot;nofollow&quot;&gt;of these notes&lt;/a&gt;.  Perhaps it is possible to give a more elementary solution of this problem: I would be happy to see one myself.  &lt;/p&gt;&#xA;" OwnerUserId="299" LastActivityDate="2011-02-11T19:24:32.340" CommentCount="4" />
  <row Id="21574" PostTypeId="2" ParentId="21542" CreationDate="2011-02-11T19:26:27.453" Score="2" ViewCount="" Body="&lt;p&gt;I you want for an exact closed formula, I suspect that it would be fairly complex, if it can be found - perhaps by generating functions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are interested in approximate/asymptotic formulas for large values of $n, m$, (and $n/m \gtrsim 3$) you can try the following trick. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are in total $m^n$ ways of placing distinct objects in $m$ distinct cells  But we want the subset that consist of those configurations that have from 1 to $k$ objects in each cell. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To estimate the relative size of this subset, we notice that the experiment of placing randomly $n$ distinct objects (equiprobably) in $m$ distinct cells is asymptotically equivalent to throwing $m$ independent Poisson variables with mean $\lambda = n/m$. &#xA;The probability that this configuration fits our restricted subset is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ \displaystyle \left( \sum_{j=1}^k e^{-\lambda} \frac{\lambda^j}{j!} \right)^m $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From this we can estimate our desider number, multiplying this expression by $m^n$ ,and dividing by $m!$ if we want to consider the cells as undistinguishable, as Stirling numbers do. So&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle  S_k(n,m) \approx \frac{m^n}{m!} \left( e^{-\lambda} \sum_{j=1}^k  \frac{\lambda^j}{j!} \right)^m &#xD;&#xA; $ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;with $\lambda = n/m$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;BTW setting $k = \infty$ we have an approximation of the standard Stirling numbers of the second kind (further simplified-approximated by the Stirling approximation of factorials).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle  S(n,m) \approx \left( \frac{m}{e} \right)^{n-m} \; \frac{ \left( e^{\lambda} -1 \right)^m} {\sqrt{2 \pi \; m}}&#xD;&#xA; $ &lt;/p&gt;&#xA;" OwnerUserId="312" LastEditorUserId="312" LastEditorDisplayName="" LastEditDate="2011-02-11T19:52:06.060" LastActivityDate="2011-02-11T19:52:06.060" />
  <row Id="21575" PostTypeId="2" ParentId="21514" CreationDate="2011-02-11T19:32:08.963" Score="0" ViewCount="" Body="&lt;p&gt;So here are three links to online resources on linear algebra:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://mathsouls.110mb.com/Ebook/Math/En/Linear%20Algebra/Problems%20and%20Theorems%20in%20Linear%20Algebra.pdf&quot; rel=&quot;nofollow&quot;&gt;http://mathsouls.110mb.com/Ebook/Math/En/Linear%20Algebra/Problems%20and%20Theorems%20in%20Linear%20Algebra.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://ocw.mit.edu/courses/mathematics/18-06-linear-algebra-spring-2010/&quot; rel=&quot;nofollow&quot;&gt;http://ocw.mit.edu/courses/mathematics/18-06-linear-algebra-spring-2010/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.math.brown.edu/~treil/papers/LADW/LADW.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.math.brown.edu/~treil/papers/LADW/LADW.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5736" LastEditorUserId="5736" LastEditorDisplayName="" LastEditDate="2011-02-12T10:47:01.157" LastActivityDate="2011-02-12T10:47:01.157" />
  <row Id="21577" PostTypeId="2" ParentId="21559" CreationDate="2011-02-11T19:43:27.663" Score="2" ViewCount="" Body="&lt;p&gt;hmm... $M$ doesn't even map $L^p$ to $L^p$, never mind continuously. Take $f$ to be the characteristic function of $[-1/2,1/2]$. For any $p$, its $L^p$ norm is 1. $Mf(x) = 1$ if $x\in (-1/2,1/2)$ and equals $1/2$ other wise. It it not in $L^p$ for any $p$ other than infinity. By definition $\|Mf(x)\|_\infty &amp;lt; \|f\|_\infty$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, for any function of compact support, $Mf$ is not in any $L^p$ except possibly $L^\infty$. Are you sure you have the definition correct? &lt;/p&gt;&#xA;" OwnerUserId="1543" LastActivityDate="2011-02-11T19:43:27.663" />
  <row Id="21578" PostTypeId="2" ParentId="21570" CreationDate="2011-02-11T20:05:16.513" Score="6" ViewCount="" Body="&lt;p&gt;I'll consider the interval $[0,2\pi]$ for notational simplicity. Consider the matrix &#xA;$$&#xD;&#xA;A = \left(&#xD;&#xA;\begin{array}{cc}&#xD;&#xA; \sin ^2\tfrac{\theta }{2} &amp;amp; - \sin&#xD;&#xA;   \tfrac{\theta }{2} \cos&#xD;&#xA;   \tfrac{\theta }{2} \\&#xD;&#xA; -\sin \tfrac{\theta }{2}\cos&#xD;&#xA;   \tfrac{\theta }{2} &amp;amp; \cos&#xD;&#xA;   ^2\tfrac{\theta }{2}&#xD;&#xA;\end{array}&#xD;&#xA;\right),&#xD;&#xA;$$ &#xA;which defines an $R$-linear map $p:R^2\to R^2$. Computing $A^2$ we see that $p^2=p$, so $p$ is idempotent, and its kernel is a projective $R$-module $P$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now consider the map $$&#xD;&#xA;\phi : f\in M \mapsto(f(\theta)\cos \tfrac{\theta }{2},f(\theta)\sin&#xD;&#xA;   \tfrac{\theta }{2}) \in R^2.&#xD;&#xA;$$&#xA;It is clearly an $R$-linear injective map, whose image is precisely the kernel $P$ of $p$. It follows that $M\cong P$, and this shows projectivity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Non-freeness is more subtle...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a morphism of rings $\varepsilon:R\to\mathbb R$ given by evaluation at $0$. One can see that $P\otimes_R\mathbb R$ is of dimension $1$ over $\mathbb R$, so that if $P$ is free, then it is free of rank $1$. In that case, $M$ would be free of rank $1$: suppose so, and let $h\in M$ be a generator. It is immediate then that every element of $M$ has to vanish where $h$ vanishes. But one can easily find an element of $M$ whose only zero is not a zero of $h$.&lt;/p&gt;&#xA;" OwnerUserId="274" LastEditorUserId="274" LastEditorDisplayName="" LastEditDate="2011-02-16T00:20:33.080" LastActivityDate="2011-02-16T00:20:33.080" CommentCount="4" />
  <row Id="21579" PostTypeId="1" AcceptedAnswerId="21583" CreationDate="2011-02-11T20:19:47.343" Score="1" ViewCount="80" Body="&lt;p&gt;Given a graph with an adjacency matrix $\bf A$, powers of this matrix give the number of walks from vertices. That is, $({\bf A}^k)_{ij}$ gives the number of walks from nodes $i$ to $j$ in $k$ steps. Is there any nice physical iterpertation for the powers of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Laplacian_matrix&quot; rel=&quot;nofollow&quot;&gt;Laplacian&lt;/a&gt; matrix?&lt;/p&gt;&#xA;" OwnerUserId="196" LastActivityDate="2011-02-11T20:33:02.203" Title="Powers of the Laplacian matrix" Tags="&lt;linear-algebra&gt;&lt;graph-theory&gt;" AnswerCount="1" />
  <row Id="21580" PostTypeId="2" ParentId="21555" CreationDate="2011-02-11T20:22:26.313" Score="2" ViewCount="" Body="&lt;p&gt;Here are pointers to get you started. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) If the two groups $\Gamma$ and $\Gamma&amp;#39;$ are conjugate by an isometry of the upper half plane, then you can use that given isometry to define an isometry of the quotients.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) On the other hand, using some covering space theory, an isometry of the quotients lifts to a conjugacy of the deck groups.   For example, you can use the lifting property for covering maps (see this &lt;a href=&quot;http://en.wikipedia.org/wiki/Covering_space&quot; rel=&quot;nofollow&quot;&gt;page&lt;/a&gt; on covering spaces). &lt;/p&gt;&#xA;" OwnerUserId="1307" LastActivityDate="2011-02-11T20:22:26.313" />
  <row Id="21581" PostTypeId="1" AcceptedAnswerId="21585" CreationDate="2011-02-11T20:24:03.257" Score="7" ViewCount="225" Body="&lt;p&gt;I've always heard this reasoning, and it makes obvious sense, but how do you actually show it for some arbitrary product? Would it be something like this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(a(b(cd)))e=((ab)(cd))e=(((ab)c)d)e=abcde?$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do you just say that the grouping of the parentheses now corresponds to just multiplying straight through? Thanks.&lt;/p&gt;&#xA;" OwnerUserId="6927" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-11T21:04:35.920" LastActivityDate="2011-02-12T18:52:04.160" Title="How does one actually show from associativity that one can drop parentheses?" Tags="&lt;algebra&gt;" AnswerCount="4" CommentCount="1" FavoriteCount="1" />
  <row Id="21582" PostTypeId="2" ParentId="21581" CreationDate="2011-02-11T20:27:11.423" Score="4" ViewCount="" Body="&lt;p&gt;That «one can drop the parentheses» really means that «no matter how you put the parentheses in, the result will be the same».&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To prove such a statement, what one usually does is pick &lt;em&gt;one&lt;/em&gt; specific way of putting in parentheses, and shows that any other way gives the same result as the one we picked. This is done by induction in the number of factors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some enterprising soul might post a complete proof...&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2011-02-11T20:27:11.423" CommentCount="1" />
  <row Id="21583" PostTypeId="2" ParentId="21579" CreationDate="2011-02-11T20:33:02.203" Score="1" ViewCount="" Body="&lt;p&gt;In my opinion, it is not natural to take powers of the Laplacian in isolation. That is, the Laplacian should not be thought of as a linear operator that one is interested in iterating. There are two points of view that I know of about what the Laplacian &quot;is&quot;:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;It's the matrix representation of the quadratic form $\sum_{(v, w) \in E} (f(v) - f(w))^2$ where $f$ is a function on the vertices. Matrices of quadratic forms are not meant to be composed.&lt;/li&gt;&#xA;&lt;li&gt;It generates a one-dimensional Lie algebra acting on the space of functions on the vertices. This perspective is relevant to the answer I gave to &lt;a href=&quot;http://math.stackexchange.com/questions/19292/connection-between-graphs-and-the-eigenvectors-of-their-matrix-representation/19298#19298&quot;&gt;this math.SE&lt;/a&gt; question, where by using the Laplacian to define certain differential equations it becomes natural to &lt;em&gt;exponentiate&lt;/em&gt;, rather than take powers of, the Laplacian. The powers appear in the exponential, but indirectly.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="232" LastActivityDate="2011-02-11T20:33:02.203" CommentCount="2" />
  <row Id="21584" PostTypeId="2" ParentId="21581" CreationDate="2011-02-11T20:37:07.763" Score="0" ViewCount="" Body="&lt;p&gt;A more &lt;del&gt;formal&lt;/del&gt; pedantic version of &lt;del&gt;part&lt;/del&gt; a very small part of Mariano Suárez-Alvarez' answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Associativity is a property of multiplication, which is a binary operation. More precisely, multiplication is a map $m:R \times R \to R, (a,b) \mapsto m(a,b)=ab$.&lt;br&gt;&#xA;The associative law says that (ab)c = a(bc). Using m, that is the same as m(m(a,b),c) = m(a,m(b,c)). (Note that m takes exactly two elements of $R$ as &quot;input&quot;). A consequence of the associative law is that there is no ambiguity in writing abc, since «no matter how you put the parentheses in, the result will be the same». &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Edit:&lt;/b&gt;As Arturo writes in a comment below: &quot;associativity is not a property of multiplication per se ([...] addition [is] associative [as is] composition of functions) but rather, a property that a binary operation may or may not have.&quot; Of course, Arturo's proof in his answer works for any associative binary operation.&lt;/p&gt;&#xA;" OwnerUserId="6791" LastEditorUserId="6791" LastEditorDisplayName="" LastEditDate="2011-02-11T21:58:31.827" LastActivityDate="2011-02-11T21:58:31.827" CommentCount="8" />
  <row Id="21585" PostTypeId="2" ParentId="21581" CreationDate="2011-02-11T20:42:36.577" Score="13" ViewCount="" Body="&lt;p&gt;You can prove by induction on the number of factors that all such products are equal. We show that if you have $a_1,\ldots,a_n$, then any way to place the parenthesis will yield the same result as $a_1(a_2(\cdots(a_{n-1}a_n)\cdots))$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the number of factors is $1$ or $2$, then the result is immediate. If the number of factors is $3$, then it follows precisely by associativity: $(a_1a_2)a_3 = a_1(a_2a_3)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume the result holds for any $k$, $1\leq k\lt n$, and you have some product of $n$ elements $a_1,\ldots,a_n$, in order, with parentheses placed in some arbitrary (but valid) manner. Then there exists $k$, $1\leq k\lt n$, such that the product is of the form &#xA;$$w_1(a_1,\ldots,a_k)\cdot w_2(a_{k+1},\ldots,a_n)$$&#xA;where $w_1(a_1,\ldots,a_k)$ is just an expression involving the product of $a_1,\ldots,a_k$ with parentheses placed in some fashion, and similarly with $w_2$. (We are just looking at the &quot;last&quot; product to be done; e.g., if you have $((a_1a_2)a_3)(a_4a_5)$, we would have $k=3$, $w_1(a_1,a_2,a_3) = (a_1a_2)a_3$, and $w_2(a_4,a_5) = a_4a_5$; if you have $a_1((a_2a_3)(a_4a_5))$, then $k=1$, $w_1(a_1) = a_1$, and $w_2(a_2,a_3,a_4,a_5) = (a_2a_3)(a_4a_5)$; etc).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By induction we can write&#xA;$$w_1(a_1,\ldots,a_k) = a_1(a_2(\cdots(a_{k-1}a_k)\cdots))$$&#xA;so&#xA;\begin{align*}&#xA;w_1(a_1,\ldots,a_k)w_2(a_{k+1},\ldots,a_n) &amp;amp;= \Bigl(a_1\bigl(a_2(\cdots(a_{k-1}a_k)\cdots)\Bigr)\cdot w_2(a_{k+1},\ldots,a_n)\\&#xA;&amp;amp;= a_1\cdot\Bigl(a_2(\cdots(a_{k-1}a_k)\cdots)w_2(a_{k+1},\ldots,a_n)\Bigr) \\&#xA;&amp;amp;= a_1\cdot\Bigl( a_2(a_3(\cdots(a_{n-1}a_n)\cdots)\Bigr)&#xA;\end{align*}&#xA;where the last equality follows from the induction hypothesis applied to a product with $n-1$ factors, and the immediately preceding equality by simple associativity of three factors, applied to $a_1$, $a_2(\cdots(a_{k-1}a_k)\cdots)$, and $w_2(a_{k+1},\ldots,a_n)$.  (&lt;em&gt;Note:&lt;/em&gt; If $k=1$, then you can skip directly from the first line to the last line without having to invoke associativity; the argument still holds.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, any expression of a product of $a_1,\ldots,a_n$, in that order, with parentheses place in an arbitrary but valid manner, is equal to&#xA;$$a_1(a_2(\cdots(a_{n-1}a_n)\cdots)),$$&#xA;This completes the induction and the proof.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since all ways of associating a product of $n$ factors, $a_1,\ldots,a_n$ yield results that are equal to one another, we may freely write any such product simply as $a_1\cdots a_n$ to represent their common value. &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-11T21:36:35.347" LastActivityDate="2011-02-11T21:36:35.347" CommentCount="4" />
  <row Id="21586" PostTypeId="2" ParentId="21572" CreationDate="2011-02-11T21:24:23.280" Score="0" ViewCount="" Body="&lt;p&gt;Use the fact that the von Neumann-Morgenstern utility is only determined up to a scale factor and an additive constant. Thus, you can arbitrarily choose the utilities of two of the three outcomes as values that are easy to calculate with -- say, 0 and 1. Then only the utility of the third outcome remains as a variable. Then you can use the equation from the given indifference to determine that variable as a function of $p$, and find the values of $p$ which make the order of the three outcomes come out right.&lt;/p&gt;&#xA;" OwnerUserId="6622" LastEditorUserId="6622" LastEditorDisplayName="" LastEditDate="2011-02-11T21:43:09.240" LastActivityDate="2011-02-11T21:43:09.240" />
  <row Id="21587" PostTypeId="2" ParentId="21545" CreationDate="2011-02-11T21:24:50.380" Score="3" ViewCount="" Body="&lt;p&gt;Another Dubuquesque attempt; for legibility, write $d=\gcd(a,b)$:&#xA;\begin{align*}&#xA;\gcd\Bigl(d(a+b), ab\Bigr) &amp;amp;= \gcd\Bigl(d(a+b), ab, ab\Bigr)\\&#xA;&amp;amp;=\gcd\Bigl(d(a+b),\ ab-a(a+b),\ ab-b(a+b)\Bigr)\\&#xA;&amp;amp;=\gcd\Bigl(d(a+b),\ a^2,\ b^2\Bigr)\\&#xA;&amp;amp;=\gcd\Bigl(d(a+b),\ \gcd(a^2,b^2)\Bigr)\\&#xA;&amp;amp;=\gcd\Bigl(d(a+b),\ \gcd(a,b)^2\Bigr)\\&#xA;&amp;amp;=\gcd\Bigl(d(a+b),\ d^2\Bigr)\\&#xA;&amp;amp;= d\gcd\Bigl(a+b,d\Bigr)\\&#xA;&amp;amp;= d\gcd\Bigl(a+b,\gcd(a,b)\Bigr)\\&#xA;&amp;amp;= d\gcd(a,b)\\&#xA;&amp;amp;= \gcd(a,b)\gcd(a,b).&#xA;\end{align*}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Second line uses the fact that $a(a+b)$ and $b(a+b)$ are both multiples of $d(a+b)$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now divide through by $\gcd(a,b)$ to get the desired result.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-11T22:24:09.830" LastActivityDate="2011-02-11T22:24:09.830" CommentCount="3" />
  <row Id="21588" PostTypeId="1" AcceptedAnswerId="21590" CreationDate="2011-02-11T21:26:37.357" Score="2" ViewCount="83" Body="&lt;p&gt;The question is this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In how many partitions of $\{1,2,\ldots,n\}$ into three disjoint sets, $A$, $B$, and $C$,  are there no two consecutive numbers in the set $A$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;thanks!&lt;/p&gt;&#xA;" OwnerUserId="6163" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-11T21:28:40.407" LastActivityDate="2011-02-11T21:43:37.437" Title="combinatorics - partitions of sets" Tags="&lt;combinatorics&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="21589" PostTypeId="1" AcceptedAnswerId="21591" CreationDate="2011-02-11T21:35:07.403" Score="5" ViewCount="152" Body="&lt;p&gt;Is this correct? I thought it would be but when I entered it into wolfram alpha, I got a different answer. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int (\cos^3x)(\sin^2x)dx  = \int(\cos x)(\cos^2x)(\sin^2x)dx&#xD;&#xA;&#xD;&#xA;= \int (\cos x)(1-\sin^2x)(\sin^2x)dx.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;let $u = \sin x$, $du = \cos xdx$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int(1-u^2)u^2du = \int(u^2-u^4)du&#xD;&#xA;&#xD;&#xA;= \frac{u^3}{3} - \frac{u^5}{5} +C$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Plugging in back $u$, we get $\displaystyle\frac{\sin^3 x}{3} - \frac{\sin^5 x}{5}$ + C&lt;/p&gt;&#xA;" OwnerUserId="6331" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-11T22:01:30.067" LastActivityDate="2011-02-11T22:01:30.067" Title="Evaluate $\int \cos^3 x\;\sin^2 xdx$" Tags="&lt;calculus&gt;&lt;integral&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="21590" PostTypeId="2" ParentId="21588" CreationDate="2011-02-11T21:43:37.437" Score="2" ViewCount="" Body="&lt;p&gt;Number of ways of selecting $k$ numbers from $1,2,\dots,n$ such that no two are consecutive is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle \binom{n-k+1}{k}$, for instance see my answer here: &lt;a href=&quot;http://math.stackexchange.com/questions/18268/consecutive-birthdays-probability/18363#18363&quot;&gt;Consecutive birthdays probability&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you do that, count the number of partitions of the remaining into $B$ and $C$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which is given by $2S(n-k,2) + 2$ where $S(a,b)$ are the Stirling numbers of the second kind (for and explanation see the answer linked above).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I guess your answer is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{k=0}^{n} \binom{n-k+1}{k} (2S(n-k,2) + 2)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Note: I haven't tried to confirm it, I leave that to you)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope that helps.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2011-02-11T21:43:37.437" CommentCount="7" />
  <row Id="21591" PostTypeId="2" ParentId="21589" CreationDate="2011-02-11T21:46:14.103" Score="4" ViewCount="" Body="&lt;p&gt;\begin{align*}&#xA;\frac{\sin^3 (x)}{3} - \frac{\sin^5 (x)}{5} &amp;amp;= \sin^3 (x) (\frac{1}{3} - \frac{\sin^2 (x)}{5});\\&#xA;\cos(2x) &amp;amp;= 1 - 2 \sin^2(x)\\&#xA;\sin^2(x) &amp;amp;= \frac{1- \cos(2x)}{2}.&#xA;\end{align*}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, we get,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align*}&#xA;\sin^3 (x) \left(\frac{1}{3} - \frac{\sin^2 (x)}{5}\right) &amp;amp;= \sin^3 (x) \left(\frac{1}{3} - \frac{1 - \cos(2x)}{10}\right)\\&#xA;&amp;amp; = \sin^3(x) \left(\frac{10 - 3 + 3 \cos(2x)}{30}\right)\\&#xA;&amp;amp; = \frac{\sin^3(x)}{30} (3 \cos(2x)+7)&#xA;\end{align*}&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-11T21:48:32.293" LastActivityDate="2011-02-11T21:48:32.293" CommentCount="3" />
  <row Id="21593" PostTypeId="1" CreationDate="2011-02-11T21:50:17.970" Score="1" ViewCount="98" Body="&lt;p&gt;If ${\bf x}(s)$ is a curve in $\mathbb{R}^3$ on a surface parameterized by its arc length $s$, and ${\bf N}$ is the surface normal at ${\bf x}$, consider the following equality (with &quot;$\cdot$&quot; being the dot product):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align*}&#xA;\frac{d{\bf x}}{ds} \cdot \frac{d{\bf N}}{ds} = \frac{d{\bf x} \cdot d{\bf N}}{d{\bf x} \cdot d{\bf x}}&#xA;\end{align*}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This comes from a book on differential geometry. Now, the LHS is a perfectly fine dot product between two vectors. But to me, the RHS does not make any sense at all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I do understand that this result can easily be arrived at if you treat $d{\bf x}$ and $d{\bf N}$ as real vectors (e.g. the part where $d{\bf x} \cdot d{\bf x}=ds^2$). But this kind of manipulation seems rather sloppy...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would really like to understand this from first principles. A derivative like $\frac{d{\bf x}}{ds}$ is defined in terms of limits,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align*}&#xA;\frac{d{\bf x}}{ds}(s) = \lim_{\Delta s \rightarrow \infty} \frac{{\bf x}(s + \Delta s) - {\bf x}(s)}{\Delta s}&#xA;\end{align*}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The extension from scalar $x$ to a vector-valued $\bf x$ is obvious. But how does this definition apply to the RHS of the equation given above? Does a dot product of differentials like $d{\bf x} \cdot d{\bf x}$ even mean anything? How do you set up the limits so that the expression can be evaluated (after all, the RHS should evaluate to a real number)?&lt;/p&gt;&#xA;" OwnerUserId="6932" LastActivityDate="2011-02-11T21:50:17.970" Title="Strange use of differentials - is $d{\bf x} \cdot d{\bf x}$ a dot product?" Tags="&lt;differential&gt;" CommentCount="2" />
  <row Id="21594" PostTypeId="2" ParentId="21559" CreationDate="2011-02-11T21:55:53.237" Score="2" ViewCount="" Body="&lt;p&gt;This is not possible (updated question).  If $0\lt p\lt \infty$, $f$ is in $L^p$, and $Mf$ is in $L^p$, then $f$ is zero a.e..  In other words, if $0\lt\int_\mathbb{R}|f|^p\lt\infty$, then $\int_\mathbb{R}|Mf|^p=\infty$ (and in fact, |Mf| has a positive lower bound outside of some bounded interval).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose that $0\lt\|f\|_p\lt\infty$.  There is some bounded interval $[a,b]$ with $\int_a^b|f|^p\gt0$.  WLOG (rescale and shift) assume $[a,b]=[-1,0]$.  Thus the essential supremum of $|f|$ on $[-1,0]$ is positive, so there is a positive number $c$ and a set $E\subseteq[-1,0]$ of positive measure $m$ such that $|f(x)|\gt c$ for all $x\in E$.  There is a $K\gt 0$ such that the measure of the set $\{x&amp;gt;K:|f(x)|\geq \frac{c}{2}\}$ is less than $\frac{m}{2}$.  For $x\gt K/2$, as $t$ ranges over the interval $[x,x+1]$, $x-t$ ranges over $[-1,0]$ while $x+t$ is contained in $(K,\infty)$.  Thus the set of such $t$ with $|f(x-t)|&amp;gt;c$ and $|f(x+t)|&amp;lt;\frac{c}{2}$ has measure greater than $\frac{m}{2}\gt 0$ (and in particular is nonempty, but this would also work if you changed the sup to an essential sup).  Therefore $|Mf(x)|\gt \frac{c}{4}$ for all $x\gt K/2$.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2011-02-11T22:06:55.230" LastActivityDate="2011-02-11T22:06:55.230" CommentCount="1" />
  <row Id="21596" PostTypeId="1" AcceptedAnswerId="21623" CreationDate="2011-02-11T22:15:32.500" Score="2" ViewCount="74" Body="&lt;p&gt;Let $\{f_j\}$ be a family of self-homeomorphisms of $\overline{\mathbb{R}}^n$ and $x,y\in\overline{\mathbb{R}}^n$, where $\overline{\mathbb{R}}^n$ is the one-point compactification of $\mathbb{R}^n$.  What does it mean for $\{f_j\}$ to converge c-uniformly to y in $\overline{\mathbb{R}}^n\setminus \{x\}$? That is, what does &quot;$\lim_{j\rightarrow\infty} f_j =y$, c-uniformly in $\overline{\mathbb{R}}^n\setminus {x}$&quot; mean?  I'm only confused about what function the point $y$ represents.  I would have assumed it meant the function that sends everything to the point $y$, but this is not a bijection on $\overline{\mathbb{R}}^n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The source for this question is the paper &quot;Discrete Quasiconformal Groups I&quot; by Gehring and Martin, Proc. London Math. Soc. (3) 55, 1987.  Definition 3.3 and Theorem 3.7.  My interest here is in understanding convergence groups, in particular their use in defining relatively hyperbolic groups.&lt;/p&gt;&#xA;" OwnerUserId="6931" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-11T22:17:52.563" LastActivityDate="2011-02-12T00:59:59.850" Title="What does it mean for a sequence of self-homeomorphism of $\mathbb{R}^n$ to converge to a point?" Tags="&lt;analysis&gt;&lt;group-theory&gt;&lt;analytic-geometry&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="21597" PostTypeId="1" CreationDate="2011-02-11T22:19:02.997" Score="0" ViewCount="51" Body="&lt;p&gt;Hw do i estimate which pattern is the most likely in the distribution of the number of cards in each suit in a hand among 5-3-3-2; 4-4-3-2 and 4-3-3-3. This is a Bridge Game related problem&lt;/p&gt;&#xA;" OwnerUserId="6935" LastActivityDate="2011-02-11T22:58:30.840" Title="Probabilities of cards distribution" Tags="&lt;probability&gt;&lt;statistics&gt;" AnswerCount="1" />
  <row Id="21599" PostTypeId="2" ParentId="21597" CreationDate="2011-02-11T22:44:13.377" Score="1" ViewCount="" Body="&lt;p&gt;Try to count out the number of possible hand types.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;5-3-3-2 can be chosen in $\displaystyle 12 \ \binom{13}{5} \ \binom{8}{3} \ \binom{5}{3} = 8648640$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(the 12 factor comes from 5=3=3=2, 3=5=3=2, etc).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4-4-3-2 : $\displaystyle 12 \ \binom{13}{4} \ \binom{9}{4} \ \binom{5}{3} = 10810800 $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4-3-3-3 : $\displaystyle 4 \ \binom{13}{4} \ \binom{9}{3} \ \binom{6}{3} = 4804800$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So 4-4-3-2 is most likely.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, there might be errors and I am counting a particular suit split, which I suppose, should give the same proportion of probabilities as the hand pattern you seek.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope that helps.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2011-02-11T22:58:30.840" LastActivityDate="2011-02-11T22:58:30.840" />
  <row Id="21600" PostTypeId="2" ParentId="21544" CreationDate="2011-02-11T22:45:36.550" Score="2" ViewCount="" Body="&lt;p&gt;I cannot answer in a comment (not enough rep), but what is wrong with Akhil's reply? $(x-1)y$ is clearly a subsequence of $x,(x-1)y$.&#xA;You could just as well take $R=k[x,y,z,w]/(x-1)z$, $x_1=w$, $x_2=x$, $x_3=(x-1)y$. Then, $x_1,x_2,x_3$ is regular, while $x_1,x_3$ isn't - which is just what you were trying to prove.&lt;/p&gt;&#xA;" OwnerUserId="586" LastActivityDate="2011-02-11T22:45:36.550" CommentCount="1" />
  <row Id="21601" PostTypeId="1" AcceptedAnswerId="21603" CreationDate="2011-02-11T22:59:17.220" Score="1" ViewCount="103" Body="&lt;p&gt;Given $a^{b^c}\mod p = x$, where $a,b,c$ are real positive integers and $p$ is a prime, what is the most efficient algorithm to compute $x$ (ideally in polynomial time)?&lt;/p&gt;&#xA;" OwnerUserId="6936" LastActivityDate="2011-02-11T23:41:10.053" Title="What is an efficient algorithm to compute modular exponentiation of stacked exponents?" Tags="&lt;number-theory&gt;&lt;algorithms&gt;" AnswerCount="2" FavoriteCount="0" />
  <row Id="21602" PostTypeId="1" AcceptedAnswerId="21604" CreationDate="2011-02-11T23:04:22.517" Score="0" ViewCount="218" Body="&lt;p&gt;How does complex numbers form a group under multiplication (closure, association, identity) ?&lt;/p&gt;&#xA;" OwnerUserId="6938" LastActivityDate="2011-02-12T00:20:32.003" Title="How does complex numbers form a group under multiplication ?" Tags="&lt;complex-numbers&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="21603" PostTypeId="2" ParentId="21601" CreationDate="2011-02-11T23:11:23.097" Score="2" ViewCount="" Body="&lt;p&gt;The first step I would do would to reduce the exponent $b^c$ modulo $p-1$ to get $d$; if you can factor $p-1$ you could use CRT, otherwise just compute.  Then compute $a^d$ modulo $p$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For computing these powers the binary method is an easy and efficient method --- it involves just squaring and multiplying.&lt;/p&gt;&#xA;" OwnerUserId="3852" LastActivityDate="2011-02-11T23:11:23.097" CommentCount="2" />
  <row Id="21604" PostTypeId="2" ParentId="21602" CreationDate="2011-02-11T23:14:04.263" Score="2" ViewCount="" Body="&lt;p&gt;A group will satisfy the four conditions: closure, identity, existence of inverses, and associativity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\mathbb{C}$ under multiplication is not a group. Can you see which axiom fails?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, $\mathbb{C}-\{0\}$ will be a group, the only axiom that is non-trivial is the existence of a multiplicative inverse. Here the formula $z\overline{z} = |z|^2$ is useful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you can prove this, then a nice next exercise is to show that the set $S = \{z\in &#xD;&#xA;\mathbb{C} |\,|z| = 1 \}$ is also a group under multiplication. Then show $z\mapsto z^2$ is an automorphism of this group.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: I see you have changed your question.  To show that the $n$th roots of unity are a group under multiplication, think about the identity $e^{a+b} = e^ae^b$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An easy extension of this problem is to prove that the union of all the groups of $n$th roots of unity is a group, and that it is exactly the set $T = \{e^{ik\pi} | k\in \mathbb{Q}\}$. &lt;/p&gt;&#xA;" OwnerUserId="565" LastEditorUserId="565" LastEditorDisplayName="" LastEditDate="2011-02-12T00:20:32.003" LastActivityDate="2011-02-12T00:20:32.003" CommentCount="2" />
  <row Id="21605" PostTypeId="1" AcceptedAnswerId="21609" CreationDate="2011-02-11T23:24:32.407" Score="1" ViewCount="81" Body="&lt;p&gt;Let $x$ and $b$ represent cardinals. Assume that  $b\geq x &amp;gt; 1$ and $b^2=b$.&#xA;Prove that $x^b=2^b$. Thanks!&lt;/p&gt;&#xA;" OwnerUserId="6163" LastEditorUserId="1321" LastEditorDisplayName="" LastEditDate="2011-02-11T23:56:00.020" LastActivityDate="2011-02-11T23:56:00.020" Title="Cardinal arithmetic" Tags="&lt;cardinals&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="21609" PostTypeId="2" ParentId="21605" CreationDate="2011-02-11T23:33:01.970" Score="5" ViewCount="" Body="&lt;p&gt;Since $b^2=b$ we know that $b$ is an infinite cardinal.  Then we have&#xA;$2^b \leq x^b \leq b^b$ since $2\leq x$ and $x\leq b$.  But $2^b=b^b$ for infinite cardinals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It was pointed out that we should show the claimed $2^b=b^b$ in the above proof.  So let's give a proof of that.  First, we note that as $b\geq2$ then we have $2^b\leq b^b$, so we just need to show $b^b\leq 2^b$.  We know (left as exercise) that $2^b=|P(b)|$ and $b^b=|\{f:b\to b\}|$.  So it will suffice to construct an injection from $\{f:b\to b\}$ (the set of all functions from $b$ to $b$) into $P(b)$ (the powerset of $b$).  Now, we know that $b=|b\times b|$ (another exercise), so it is good enough to inject $\{f:b\to b\}$ into $P(b\times b)$.  But any function $f:b\to b$ corresponds to the set&#xA;$\{\langle\alpha,f(\alpha)\rangle\ :\ \alpha\in b\}\in P(b\times b)$. Note that distinct functions will map to distinct elements of $P(b\times b)$.  Thus&#xA;$b^b = |\{f:b\to b\}| \leq |P(b\times b)|=2^b$.&lt;/p&gt;&#xA;" OwnerUserId="3852" LastEditorUserId="3852" LastEditorDisplayName="" LastEditDate="2011-02-11T23:51:15.103" LastActivityDate="2011-02-11T23:51:15.103" CommentCount="5" />
  <row Id="21610" PostTypeId="2" ParentId="21601" CreationDate="2011-02-11T23:41:10.053" Score="1" ViewCount="" Body="&lt;p&gt;As stated in Apollo's answer, binary method is an efficient method. More precisely, the number of multiplication is a $\Theta(\log n)$ which show that you cannot beat binary method only by a constant factor. For more detail, check about addition-chain multiplication.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Check &lt;a href=&quot;http://en.wikipedia.org/wiki/Modular_exponentiation&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Modular_exponentiation&lt;/a&gt; for precise definition of binary method.&lt;/p&gt;&#xA;" OwnerUserId="6582" LastActivityDate="2011-02-11T23:41:10.053" />
  <row Id="21611" PostTypeId="1" AcceptedAnswerId="21612" CreationDate="2011-02-11T23:46:14.123" Score="1" ViewCount="79" Body="&lt;p&gt;Suppose $m(E)=1$ and $f,g \geq 0$ with $fg \geq 1$. Show that $\displaystyle \int_E f dm \int_E g dm \geq 1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can someone please point me to the right direction?&lt;/p&gt;&#xA;" OwnerUserId="6556" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2011-02-12T08:21:08.287" LastActivityDate="2011-02-12T08:21:08.287" Title="Show Integral Inequality" Tags="&lt;integral&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="21612" PostTypeId="2" ParentId="21611" CreationDate="2011-02-11T23:51:54.200" Score="3" ViewCount="" Body="&lt;p&gt;Hint : Use Cauchy-Scharwz inequality.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint 2 : If $f,g \geq 0, fg \geq 1$ then $\sqrt{f}\sqrt{g} \geq 1$ also.&lt;/p&gt;&#xA;" OwnerUserId="6582" LastActivityDate="2011-02-11T23:51:54.200" CommentCount="2" />
  <row Id="21613" PostTypeId="2" ParentId="21611" CreationDate="2011-02-11T23:53:52.747" Score="1" ViewCount="" Body="&lt;p&gt;I assume that $f$ and $g$ are integrable. Outside a null-set we have $g \geq \frac{1}{f} \gt 0$, in particular $\frac{1}{f}$ is integrable. Now apply&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Jensen%27s_inequality&quot; rel=&quot;nofollow&quot;&gt;Jensen's inequality&lt;/a&gt; $\varphi\left(\int f\right) \leq \int \varphi \circ f$ for the convex function $\varphi(x) = \frac{1}{x}$ (here we use $f \geq 0$ again), so $\frac{1}{\int f} \leq \int \frac{1}{f} \leq  \int g$.&lt;/p&gt;&#xA;" OwnerUserId="5363" LastEditorUserId="5363" LastEditorDisplayName="" LastEditDate="2011-02-12T00:25:25.943" LastActivityDate="2011-02-12T00:25:25.943" CommentCount="0" />
  <row Id="21614" PostTypeId="1" AcceptedAnswerId="21617" CreationDate="2011-02-11T23:54:50.400" Score="15" ViewCount="388" Body="&lt;p&gt;In the &lt;strong&gt;few&lt;/strong&gt; linear algebra texts I have read, the determinant is introduced in the following manner;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Here is a formula for what we call $detA$. Here are some other formulas. And finally, here are some nice properties of the determinant.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, in very elementary textbooks it is introduced by giving the co-factor expansion formula.  In Axler's &quot;Linear Algebra Done Right&quot; it is defined, for $T\in L(V)$ to be $(-1)^{dimV}$ times the constant term in the characteristic polynomial of $T$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However I find this somewhat unsatisfactory. Its like the real definition of the determinant is hidden. Ideally, wouldn't the determinant be defined in the following manner:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Given a matrix $A$, let $detA$ be an element of $\mathbb{F}$ such that x, y and z.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then one would proceed to prove that this element is unique, and derive the familiar formulae.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So my question is: Does a definition of the latter type exist, is there some minimal set of properties sufficient to define what a determinant is? If not, can you explain why?&lt;/p&gt;&#xA;" OwnerUserId="565" LastActivityDate="2011-02-18T19:09:48.373" Title="Is there a definition of determinants that does not rely on how they are calculated?" Tags="&lt;linear-algebra&gt;" AnswerCount="5" CommentCount="22" FavoriteCount="5" />
  <row Id="21615" PostTypeId="2" ParentId="21614" CreationDate="2011-02-12T00:00:08.180" Score="10" ViewCount="" Body="&lt;p&gt;Let $B$ a basis of a vector space $E$ of dimension $n$ over $\Bbbk$. Then $det_B$ is the only $n$-alternating multilinear form with $det_B(B) = 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A $n$-multilinear form is a map of $E^n$ in $\Bbbk$ which is linear for each variable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A $n$- alternated multilinear form is a multilinear form which verify for all $i,j$&#xA;$$ f(x_1,x_2,\dots,x_i,\dots, x_j, \dots, x_n) = -f(x_1,x_2,\dots,x_j,\dots, x_i, \dots, x_n) $$&#xA;In plain english, the sign of the application change when you switch two argument. You understand why you use the big sum over permutations to define the determinant with a closed formula. &lt;/p&gt;&#xA;" OwnerUserId="6582" LastEditorUserId="6582" LastEditorDisplayName="" LastEditDate="2011-02-12T00:27:40.120" LastActivityDate="2011-02-12T00:27:40.120" CommentCount="4" />
  <row Id="21616" PostTypeId="2" ParentId="21614" CreationDate="2011-02-12T00:02:42.343" Score="1" ViewCount="" Body="&lt;p&gt;This paper: &lt;a href=&quot;http://www.axler.net/DwD.html&quot; rel=&quot;nofollow&quot;&gt;http://www.axler.net/DwD.html&lt;/a&gt; &quot;Down with determinants&quot; by Sheldon Axler, is probably what you are looking for. He proves many of the familiar properties of finite-dimensional vector spaces without the use of determinants - and in the end he gives an easy definition of the determinant as the product of the eigenvalues of the matrix, and finally proves the usual defining formula for the determinant.&lt;/p&gt;&#xA;" OwnerUserId="4284" LastActivityDate="2011-02-12T00:02:42.343" CommentCount="2" />
  <row Id="21617" PostTypeId="2" ParentId="21614" CreationDate="2011-02-12T00:12:51.697" Score="18" ViewCount="" Body="&lt;p&gt;Let $V$ be a vector space of dimension $n$. For any $p$, the construction of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Exterior_algebra&quot; rel=&quot;nofollow&quot;&gt;exterior power&lt;/a&gt; $\Lambda^p(V)$ is &lt;a href=&quot;http://en.wikipedia.org/wiki/Functor&quot; rel=&quot;nofollow&quot;&gt;functorial&lt;/a&gt; in $V$: it is the universal object for alternating multilinear functions out of $V^p$, that is, functions&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\phi : V^p \to W$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $W$ is any other vector space satisfying $\phi(v_1, ... v_i + v, ... v_p) = \phi(v_1, ... v_i, ... v_p) + \phi(v_1, ... v_{i-1}, v, v_{i+1}, ... v_p)$ and $\phi(v_1, ... v_i, ... v_j, ... v_p) = - \phi(v_1, ... v_j, ... v_i, ... v_p)$. What this means is that there is a map $\psi : V^p \to \Lambda^p(V)$ (the exterior product) which is alternating and multilinear which is universal with respect to this property; that is, given any other map $\phi$ as above with the same properties, $\phi$ factors uniquely as $\phi = f \circ \psi$ where $f : \Lambda^p(V) \to W$ is linear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Intuitively, the universal map $\psi : V^p \to \Lambda^p(V)$ is the universal way to measure the oriented $p$-dimensional volumes of &lt;a href=&quot;http://en.wikipedia.org/wiki/Parallelepiped#Parallelotope&quot; rel=&quot;nofollow&quot;&gt;paralleletopes&lt;/a&gt; defined by $p$-tuples of vectors in $V$, the point being that for geometric reasons oriented $p$-dimensional volume is alternating and multilinear. (It is instructive to work out how this works when $n = 2, 3$ by explicitly drawing some diagrams.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Functoriality means the following: if $T : V \to W$ is any map between two vector spaces, then there is a natural map $\Lambda^p T : \Lambda^p V \to \Lambda^p W$ between their $p^{th}$ exterior powers satisfying certain natural conditions. This natural map comes in turn from the natural action $T(v_1, ... v_p) = (Tv_1, ... Tv_p)$ defining a map $T : V^p \to W^p$ which is compatible with the passing to the exterior powers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The top exterior power $\Lambda^n(V)$ turns out to be one-dimensional. We then define the &lt;strong&gt;determinant&lt;/strong&gt; $T : V \to V$ to be the scalar $\Lambda^n T : \Lambda^n(V) \to \Lambda^n(V)$ by which $T$ acts on the top exterior power. This is equivalent to the intuitive definition that $\det T$ is the constant by which $T$ multiplies oriented $n$-dimensional volumes. But it requires &lt;em&gt;no arbitrary choices&lt;/em&gt;, and the standard properties of the determinant (for example that it is multiplicative, that it is equal to the product of the eigenvalues) are extremely easy to verify.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this definition of the determinant, all the work that would normally go into showing that the determinant is the unique function with such-and-such properties goes into showing that $\Lambda^n(V)$ is one-dimensional. If $e_1, ... e_n$ is a basis, then $\Lambda^n(V)$ is in fact spanned by $e_1 \wedge e_2 \wedge ... \wedge e_n$. This is not so hard to prove; it is essentially an exercise in row reduction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that this definition does not even require a definition of oriented $n$-dimensional volume as a number. Abstractly such a notion of volume is given by a choice of isomorphism $\Lambda^n(V) \to k$ where $k$ is the underlying field, but since $\Lambda^n(V)$ is one-dimensional its space of endomorphisms is already &lt;em&gt;canonically&lt;/em&gt; isomorphic to $k$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note also that just as the determinant describes the action of $T$ on the top exterior power $\Lambda^n(V)$, the $p \times p$ minors of $T$ describe the action of $T$ on the $p^{th}$ exterior power $\Lambda^p(V)$. In particular, the $(n-1) \times (n-1)$ minors (which form the matrix of cofactors) describe the action of $T$ on the second-to-top exterior power $\Lambda^{n-1}(V)$. This exterior power has the same dimension as $V$, and with the right extra data can be identified with $V$, and this leads to a quick and natural proof of the explicit formula for the inverse of a matrix.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;As an advance warning, the determinant is sometimes defined as an alternating multilinear function on $n$-tuples of vectors $v_1, ... v_n$ satisfying certain properties; this properly defines a linear transformation $\Lambda^n(V) \to k$, not a determinant of a linear transformation $T : V \to V$. If we fix a basis $e_1, ... e_n$, then this function can be thought of as the determinant of the linear transformation sending $e_i$ to $v_i$, but this definition is basis-dependent. &lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2011-02-12T00:39:31.720" LastActivityDate="2011-02-12T00:39:31.720" CommentCount="9" />
  <row Id="21618" PostTypeId="1" CreationDate="2011-02-12T00:23:07.303" Score="4" ViewCount="145" Body="&lt;p&gt;I'm currently attending a somewhat disorganized seminar on combinatorics that follows no textbook. So far we have covered the orbit-stabilizer theorem, some recursion, and we're heading into the Mobius Inversion formula. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone suggest a text that approaches combinatorics at this level for a 2nd-3rd year undergrad who already knows some algebra and the more basic combinatorics like combinations, permutations, stars-and-bars, generating functions? Most introductory combinatorics books I've found are more suited to a discrete math class and cover stuff which I already know. I'm looking for something to supplement this lecture. Thank you.&lt;/p&gt;&#xA;" OwnerUserId="6950" LastActivityDate="2011-04-05T19:29:55.337" Title="Intermediate Text in Combinatorics?" Tags="&lt;combinatorics&gt;&lt;books&gt;" AnswerCount="5" CommentCount="1" FavoriteCount="3" />
  <row Id="21620" PostTypeId="2" ParentId="21618" CreationDate="2011-02-12T00:33:46.927" Score="1" ViewCount="" Body="&lt;p&gt;If you read French, you &lt;strong&gt;must&lt;/strong&gt; read &lt;em&gt;Analyse Combinatoire&lt;/em&gt; of L. Comtet at PUF edition. It's two short books full of fascinating materials.&lt;/p&gt;&#xA;" OwnerUserId="6582" LastActivityDate="2011-02-12T00:33:46.927" CommentCount="1" />
  <row Id="21621" PostTypeId="2" ParentId="21618" CreationDate="2011-02-12T00:39:30.613" Score="0" ViewCount="" Body="&lt;p&gt;How about &lt;a href=&quot;http://en.wikipedia.org/wiki/Concrete_Mathematics&quot; rel=&quot;nofollow&quot;&gt;Concrete Mathematics&lt;/a&gt;?&lt;/p&gt;&#xA;" OwnerUserId="589" LastActivityDate="2011-02-12T00:39:30.613" CommentCount="2" />
  <row Id="21623" PostTypeId="2" ParentId="21596" CreationDate="2011-02-12T00:59:59.850" Score="1" ViewCount="" Body="&lt;p&gt;I'm pretty sure it is meant that for every compact subset $K \subset \overline{\mathbb{R}}^{n} \smallsetminus \{x\}$ and every neighborhood $U$ of $y$ there is $N$ such that $f_{j}(K) \subset U$ for all $j \geq N$. Think of a hyperbolic automorphism of the (closed) Poincar&amp;eacute; disk. It has an attractive point $y$ and a repelling point $x$ and precisely this behaviour.&lt;/p&gt;&#xA;" OwnerUserId="5363" LastActivityDate="2011-02-12T00:59:59.850" CommunityOwnedDate="2011-02-12T01:14:32.057" CommentCount="2" />
  <row Id="21624" PostTypeId="2" ParentId="21618" CreationDate="2011-02-12T01:03:25.817" Score="5" ViewCount="" Body="&lt;p&gt;I can't say I can predict what direction your seminar will head in, but an amazing one-stop-shop for combinatorics is Richard Stanley's &lt;a href=&quot;http://www-math.mit.edu/~rstan/ec/&quot; rel=&quot;nofollow&quot;&gt;Enumerative Combinatorics&lt;/a&gt; (both volumes). You will learn a tremendous amount if you get through both of these (and the revised version of volume 1 is currently available online!). In particular Volume I has a thorough discussion of Möbius inversion.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2011-02-12T01:03:25.817" />
  <row Id="21625" PostTypeId="2" ParentId="21545" CreationDate="2011-02-12T01:08:12.957" Score="0" ViewCount="" Body="&lt;p&gt;Perhaps overkill, but if you accept the 'distribution law' $(x,[y,z])=[(x,y),(x,z)]$ stated at &lt;a href=&quot;http://en.wikipedia.org/wiki/Greatest_common_divisor#Properties&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;, then it is easy:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(a+b,[a,b])=[(a+b,a),(a+b,b)]=[(b,a),(a,b)]=(a,b)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where in the second equality I used the easy fact $(x,y)=(x,y \mod x)$.&lt;/p&gt;&#xA;" OwnerUserId="6490" LastEditorUserId="6490" LastEditorDisplayName="" LastEditDate="2011-02-12T01:14:58.013" LastActivityDate="2011-02-12T01:14:58.013" />
  <row Id="21626" PostTypeId="2" ParentId="21545" CreationDate="2011-02-12T01:13:47.680" Score="4" ViewCount="" Body="&lt;p&gt;Below are a few proofs. First, here's a couple from my &lt;a href=&quot;http://groups.google.com/group/alt.math.undergrad/msg/a687e26e7db0396d&quot; rel=&quot;nofollow&quot;&gt;sci.math post on 2001/11/10&lt;/a&gt;&#xA;&lt;img src=&quot;http://i.imgur.com/e1ALN.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;br&gt;&#xA;For variety here is yet another using $\rm\ (a,b)\ \ [a,b]\ =\ ab\ \ $ and basic gcd laws:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad (a,b)\ (a+b, [a,b])\ =\ (aa+ab,\ ab+bb,\ ab)\ =\ (aa,\:bb,\:ab)\ =\ (a,b)^2\quad $ &lt;strong&gt;QED&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the way, recall that the key identity in the second proof arose the other day in our discussion of &lt;a href=&quot;http://math.stackexchange.com/questions/21333/help-understand-the-proof-of-infinitely-many-prime-of-the-form-ax-b/21338#21338&quot;&gt;Stieltjes $\rm\ 4\:n+3\ $ generalization&lt;/a&gt; of Euclid's proof of infinitely many primes. Here's a slicker proof:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;LEMMA&lt;/strong&gt; $\rm\ \ (a+b,\:ab) = 1\ \iff\ (a,\:b) = 1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof&lt;/strong&gt; $\rm\ \ \ (a,\:b)^2 \subset (a+b,\:ab) \subset (a,\:b)\ \ $ since e.g. $\rm\ \ a^2 = a\:(a+b)-ab\in (a+b,\:ab)$    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus $\rm\ 1\in (a+b,\ ab)\ \Rightarrow\ 1\in (a,\:b)\:.\:$ Conversely $\rm\ 1 \in (a,\:b)\ \Rightarrow\ 1 \in (a,\:b)^2 \subset (a+b,\:ab)\quad$ &lt;strong&gt;QED&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2011-03-03T20:42:21.707" LastActivityDate="2011-03-03T20:42:21.707" />
  <row Id="21627" PostTypeId="1" AcceptedAnswerId="21636" CreationDate="2011-02-12T01:19:56.913" Score="3" ViewCount="107" Body="&lt;p&gt;If $X\le Y$, that is, $Y-X$ is positive semi-definite, does it imply that $Tr\{XZ\}\le Tr\{YZ\}$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it's true, how should one go about proving it? Is it a result from some other theorem?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;*edit: $Z$ is also positive, all matrices are Hermitian.&lt;/p&gt;&#xA;" OwnerUserId="4683" LastEditorUserId="4683" LastEditorDisplayName="" LastEditDate="2011-02-13T01:29:35.163" LastActivityDate="2011-02-13T01:29:35.163" Title="If $X$, $Y$, $Z$ positive, does $Tr\{XZ\}\le Tr\{YZ\}$ hold?" Tags="&lt;linear-algebra&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="21628" PostTypeId="1" AcceptedAnswerId="21629" CreationDate="2011-02-12T01:33:34.537" Score="1" ViewCount="203" Body="&lt;p&gt;Hello,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Problem:&#xA;$$\text{ If } 2^{n} - 1 \text{ is prime then n is prime}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof 1:&lt;/strong&gt;&#xA;$$\text{If } n = kl \text{ with } 2 \leq k, l &amp;lt; n \text{ then } (2^{k} - 1)|(2^{n} - 1). \text{ Hence if } 2^{n} - 1 \text{ is prime, n must be prime }$$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof 2&lt;/strong&gt;: Using the identity:&#xA;$$2^{n} - 1 = ( 2 - 1 )( 2^{k-1} + 2^{k-2} + ... + 1 )$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\text{ Let }n = rs, \text{ then } 2^{n} - 1 = ( 2^{r} - 1 )( 2^{r(s-1)} + 2^{r(s-2)} + ... + 1 )$$ &#xA;$$\text{ However each factor on the right exceeds 1 contradicting with the fact that } 2^{n} - 1 \text{ is prime }$$ &#xA;Hence, n is prime.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proof 1: I have no idea what the author's purpose.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proof 2: I understood a little bit. Can anyone help me explain why n must be prime? I am totally lost with the above proofs. If we let $ n = rs $, then what can we say about $n$ with $2^{n} - 1$ is prime?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks,&#xA;Chan&lt;/p&gt;&#xA;" OwnerUserId="5711" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2011-02-12T02:26:15.393" LastActivityDate="2011-02-12T06:21:29.960" Title="Help in understanding the proof of Mersenne Prime" Tags="&lt;elementary-number-theory&gt;" AnswerCount="2" />
  <row Id="21629" PostTypeId="2" ParentId="21628" CreationDate="2011-02-12T01:58:47.193" Score="6" ViewCount="" Body="&lt;p&gt;Both the proofs are the same. This is an example of a proof by transposition or proof by contrapositive where you prove the contrapositive. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The statement $A \Rightarrow B$ is equivalent to the statement $\neg B \Rightarrow \neg A$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, proving $A \Rightarrow B$ is equivalent to proving $\neg B \Rightarrow \neg A$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The statement $A$ in your problem is &quot;$2^n - 1$ is a prime&quot; and the statement $B$ in your problem is &quot;$n$ is a prime&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this context, we have &quot;If $2^n - 1$ is a prime, then $n$ is a prime&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The contrapositive is &quot;If $n$ is not a prime then $2^n - 1$ is not a prime&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The author now proves the contrapositive statement. Hence, the author takes $n$ to be not a prime i.e. he takes $n$ to be a composite number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $n$ is a composite number, $n$ can be written as $n = kl$, where $2 \leq k,l &amp;lt; n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So $2^n - 1 = 2^{kl} - 1 = (2^k)^l-1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now observe that $(a-1)|(a^l - 1)$, $\forall l \in \mathbb{N}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This can be seen from the remainder theorem or by factorizing $(a^l - 1) = (a-1) (a^{l-1} + a^{l-2} + \cdots + a + 1)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that $2 &amp;lt; 2^k-1 &amp;lt; 2^n-1$ since $2 \leq k&amp;lt;n$ and plug in $a=2^k$, we get $((2^k)^l - 1) = ((2^k)-1) ((2^k)^{l-1} + (2^k)^{l-2} + \cdots + (2^k) + 1)$. Hence, we have $(2^k - 1)|(2^n-1)$, which implies $2^n-1$ is a composite number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, we have &quot;If $n$ is not a prime then $2^n - 1$ is not a prime&quot; which implies the contrapositive i.e. &quot;If $2^n - 1$ is a prime, then $n$ is a prime&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Note&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Say, we denote the original statement by $A$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $A$ be the statement &quot;If $2^n - 1$ is a prime, then $n$ is a prime&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $B$ be the statement &quot;If $n$ is not a prime, then $2^n-1$ is not a prime&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $C$ be the statement &quot;If $2^n-1$ is not a prime, then $n$ is not a prime&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $D$ be the statement &quot;If $n$ is a prime, then $2^n - 1$ is a prime&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Statements $A$ and $B$ are contrapositives of each other. So if $A$ is true then so is $B$ and if $B$ is true then so is $A$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, Statements $C$ and $D$ are contrapositives of each other.  So if $C$ is true then so is $D$ and if $D$ is true then so is $C$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Statements $C$(equivalently statement $D$) and $A$(equivalently statement $B$) are converses of each other. In general, we cannot conclude that if a statement is true then its converse is true and we cannot conclude that if a converse is true then the statement is true.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this example, statement $A$(equivalently statement $B$) is true.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, statements $C$(equivalently statement $D$) which is the converse of statement $A$(equivalently statement $B$) is false.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An example is $2^{11} - 1 = 2047 = 23 &amp;#215; 89$ is not a prime but $11$ is a prime. So $C$(equivalently statement $D$) is false.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2011-02-12T06:21:29.960" LastActivityDate="2011-02-12T06:21:29.960" CommentCount="3" />
  <row Id="21630" PostTypeId="1" AcceptedAnswerId="21632" CreationDate="2011-02-12T02:07:09.863" Score="6" ViewCount="259" Body="&lt;p&gt;I am truly baffled as to go on to prove this by double induction:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://i.imgur.com/Zvrzt.png&quot; rel=&quot;nofollow&quot;&gt;http://i.imgur.com/Zvrzt.png&lt;/a&gt; (snap shot of question)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This question seems rather trivial on first glimpse, however trying to prove it via double induction is another thing for me. &lt;strong&gt;I'm having trouble deciding what the variables that I will be applying induction on will be&lt;/strong&gt;. Obviously one of the two variables will be the string input, but what about the other? I've thought of using length but I don't see what that can do to help. What do you guys think?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm sorry for the long question, I'd appreciate any input &amp;amp; help &amp;amp; insight regarding this question or in double induction in general as my Google-fu seems to be coming short on information of the latter :(.&lt;/p&gt;&#xA;" OwnerUserId="6523" LastEditorUserId="6523" LastEditorDisplayName="" LastEditDate="2011-02-14T11:40:42.463" LastActivityDate="2011-02-14T11:40:42.463" Title="Proof by double induction on strings (SOLVED)" Tags="&lt;homework&gt;&lt;logic&gt;&lt;computer-science&gt;&lt;induction&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="21631" PostTypeId="1" AcceptedAnswerId="21824" CreationDate="2011-02-12T02:16:14.700" Score="0" ViewCount="97" Body="&lt;p&gt;For each given $p$, let $X$ have a binomial distribution with parameters $p$ and $N$. Suppose $p$ is distributed according to a beta distribution with parameters r and s. Find the resulting distribution of $X$. When is this distribution uniform on $x=0,1,2,...N$?&lt;/p&gt;&#xA;" OwnerUserId="6912" LastActivityDate="2011-02-13T10:58:38.547" Title="Binomial distribution with parameters p and N, p is distributed to a beta distribution with parameters r and s." Tags="&lt;homework&gt;&lt;stochastic-processes&gt;" AnswerCount="2" />
  <row Id="21632" PostTypeId="2" ParentId="21630" CreationDate="2011-02-12T02:27:08.040" Score="3" ViewCount="" Body="&lt;p&gt;Let $a$ be the first digit of the string, and let $b$ be&#xA;the value of the remaining digits. Let's prove that the&#xA;operation is reducing with respect to the lexical order of&#xA;$\langle a,b\rangle$, which is one version of interpreting&#xA;your phrase &quot;double induction.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, note that the operation of inserting the $0$ to the&#xA;right of the leading digit affects neither the value of $a$&#xA;nor $b$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next, note that if $b$ is all zero but $a$ is not $0$, then&#xA;the operation will end up with $a$ being reduced, because&#xA;when you subtract one, you will have to borrow from $a$,&#xA;thereby reducing. Thus, the operation will result with a&#xA;string having lower $\langle a,b\rangle$ in the lexical&#xA;order.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $b$ is not all zero, however, then the operation will&#xA;end up with the $b$ value being value $1$ less (despite the&#xA;extra $0$), and so will reduce $\langle a,b\rangle$ in the&#xA;lexical order.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the lexical order on $\mathbb{N}\times\mathbb{N}$ is a well-order, it follows that&#xA;the operation must eventually hit all $0$s.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The argument shows that you can insert any number of $0$s, rather than just one $0$ at each step, although if you insert more, then it will take longer to get to the all $0$ string, since when you borrow, you will get more $9$s. &lt;/p&gt;&#xA;" OwnerUserId="413" LastEditorUserId="413" LastEditorDisplayName="" LastEditDate="2011-02-12T02:38:27.543" LastActivityDate="2011-02-12T02:38:27.543" CommentCount="8" />
  <row Id="21633" PostTypeId="2" ParentId="21552" CreationDate="2011-02-12T02:28:23.373" Score="3" ViewCount="" Body="&lt;p&gt;In the following $\mathbb{K}$ always denotes a &lt;a href=&quot;http://en.wikipedia.org/wiki/Topological_field&quot; rel=&quot;nofollow&quot;&gt;topological field&lt;/a&gt;. In most applications this will be the real or complex numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Definition.&lt;/b&gt; A vector space with a &lt;a href=&quot;http://en.wikipedia.org/wiki/Hausdorff_space&quot; rel=&quot;nofollow&quot;&gt;Hausdorff topology&lt;/a&gt; over $\mathbb{K}$ such that the vector space operations are continuous is said to be a &lt;em&gt;topological vector space&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note in particular that any normed vector space or metric linear space (i.e. a metric space $(E,d)$ where $E$ is a vector space over the real or complex numbers such that $d$ is translation invariant and the vector space operations are continuous) is a TVS.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Definition.&lt;/b&gt; A family of &lt;a href=&quot;http://en.wikipedia.org/wiki/Seminorm&quot; rel=&quot;nofollow&quot;&gt;seminorms&lt;/a&gt; $(\lVert \cdot \rVert_\alpha)_{\alpha \in A}$ on an $\mathbb{K}$-vector space $E$ is said to be a &lt;em&gt;fundamental system of seminorms&lt;/em&gt; (or a &lt;em&gt;sufficient family of seminorms&lt;/em&gt;) if $x \neq 0 \implies \exists \alpha \in A : \lVert x \rVert_\alpha \neq 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Proposition.&lt;/b&gt; If $(\lVert \cdot \rVert_\alpha)$ is a fundamental system of seminorms on a $\mathbb{K}$-vector space $E$, then the topology generated by the open balls of the seminorms gives $E$ the structure of a TVS.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is fairly easy to prove. Sufficiency is (of course) needed to prove that the topology is Hausdorff. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Proposition.&lt;/b&gt; If $(\lVert \cdot \rVert_n)$ is a countable fundamental system of seminorms on a real or complex vector space $E$, then &#xA;$$d(x,y) = \sum_{j=1}^\infty \frac{\lVert x-y \rVert_j}{1 + \lVert x-y \rVert_j} 2^{-j} $$&#xA;defines a metric on $E$ that gives it the structure of a metric linear space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Proposition.&lt;/b&gt; If $(\lVert \cdot \rVert_n)$ is a countable fundamental system of seminorms on a real or complex vector space $E$ such that for all sequences $(x_k)$ that are Cauchy w.r.t. each seminorm $\lVert\cdot\rVert_j$ there exists an $x \in E$ such that $\lVert x-x_k\rVert_j \to 0$ as $k \to \infty$ for all $j$, then the metric defined in the proposition above is complete, i.e. it gives $E$ the structure of a Fréchet space. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course checking that $\lVert (x_n) \rVert_j := \lvert x_j \rvert$ defines a countable fundamental system of seminorms on $\mathbb{C}^\mathbb{N}$ that satisfies the conditions of the above proposition is easy to do. Now proving that proposition on the other hand is probably going to be rather tedious but not terribly difficult since you have a lot to work with.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sources: Wikipedia's article on &lt;a href=&quot;http://en.wikipedia.org/wiki/Fr%C3%A9chet_space&quot; rel=&quot;nofollow&quot;&gt;Fréchet spaces&lt;/a&gt; &amp;amp; other (linked) articles and Meise &amp;amp; Vogt - &lt;em&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0198514859&quot; rel=&quot;nofollow&quot;&gt;Introduction to Functional Analysis&lt;/a&gt;&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1269" LastEditorUserId="1269" LastEditorDisplayName="" LastEditDate="2011-02-12T03:05:14.980" LastActivityDate="2011-02-12T03:05:14.980" />
  <row Id="21634" PostTypeId="2" ParentId="21627" CreationDate="2011-02-12T02:44:22.787" Score="2" ViewCount="" Body="&lt;p&gt;If the matrices are not hermitian, here is an example to prove that the statement is false.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The statement is equivalent to is $\text{Tr}((Y-X)Z) \geq 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, take &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$Y-X = \left(\begin{array}{c c c c}&#xD;&#xA;1 &amp;amp; -1 &amp;amp; -1 &amp;amp; -1 \\&#xD;&#xA;0 &amp;amp; 1 &amp;amp; -1 &amp;amp; -1 \\&#xD;&#xA;0 &amp;amp; 0 &amp;amp; 1 &amp;amp; -1 \\&#xD;&#xA;0 &amp;amp; 0 &amp;amp; 0 &amp;amp; 1&#xD;&#xA;\end{array} \right)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$Z = \left(\begin{array}{c c c c}&#xD;&#xA;1 &amp;amp; 0 &amp;amp; 0 &amp;amp; 0 \\&#xD;&#xA;1 &amp;amp; 1 &amp;amp; 0 &amp;amp; 0 \\&#xD;&#xA;1 &amp;amp; 1 &amp;amp; 1 &amp;amp; 0 \\&#xD;&#xA;1 &amp;amp; 1 &amp;amp; 1 &amp;amp; 1 \\&#xD;&#xA;\end{array} \right)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clearly, $Y&amp;gt;X$ and $Z&amp;gt;0$, however the product is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(Y-X)Z = \left(\begin{array}{c c c c}&#xD;&#xA;-2 &amp;amp; -1 &amp;amp; 0 &amp;amp; 1 \\&#xD;&#xA;-3 &amp;amp; -1 &amp;amp; 0 &amp;amp; 1 \\&#xD;&#xA;-2 &amp;amp; -2 &amp;amp; 0 &amp;amp; 1 \\&#xD;&#xA;-1 &amp;amp; -1 &amp;amp; -1 &amp;amp; 1 \\&#xD;&#xA;\end{array} \right)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and the trace of the above is negative and hence $\text{Tr}(XZ) \leq \text{Tr}(YZ)$ is false.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the fact that the matrices are Hermitian might turn out to be helpful in proving that the statement is true. My hunch is that the statement is true but I do not have a proof now.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastActivityDate="2011-02-12T02:44:22.787" />
  <row Id="21635" PostTypeId="2" ParentId="21628" CreationDate="2011-02-12T03:04:05.637" Score="3" ViewCount="" Body="&lt;p&gt;The essence of the matter is the following: $\rm\ f\:(gcd(n,m))\ =\ gcd(f(n),f(m))\ $ for $\rm\ f(n)\ =\ 2^n-1\:$. Because the map $\rm\:f\:$ preserves gcds it must preserve many multiplicative properties (e.g. primes). You can find  further discussion and proofs in my &lt;a href=&quot;http://math.stackexchange.com/questions/7473/prove-that-gcdan-1-am-1-a-gcdn-m-1/7561#7561&quot;&gt;posts here&lt;/a&gt;. If you go on to study &lt;em&gt;divisor theory&lt;/em&gt; then these properties will become crystal clear.&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2011-02-12T03:04:05.637" CommentCount="2" />
  <row Id="21636" PostTypeId="2" ParentId="21627" CreationDate="2011-02-12T03:11:21.630" Score="1" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Yes.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hints: &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;You can write $Y - X = S^2$, $Z = T^2$ with $S,T \geq 0$.&lt;/li&gt;&#xA;&lt;li&gt;Recall $\text{Tr}(AB) = \text{Tr}(BA)$ and $\text{Tr}(C^{\ast}C) \geq 0$.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="5363" LastActivityDate="2011-02-12T03:11:21.630" />
  <row Id="21637" PostTypeId="2" ParentId="21530" CreationDate="2011-02-12T03:30:56.407" Score="1" ViewCount="" Body="&lt;p&gt;Of course it's trivial using unique factorization. Here's a more general proof using gcd's (or ideals) that has the benefit of giving an explicit closed form:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;LEMMA&lt;/strong&gt; $\rm\ \ c|ab,\ (a,b,c)=1\ \ \Rightarrow\ \ (a,c)^n\ (b,c)^n\ =\ (c)^n$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof&lt;/strong&gt; $\rm\quad (a,c)^n\ (b,c)^n\ =\ ((a,c)\:(b,c))^n\ =\ (ab,c(a,b,c))^n\ =\ (ab,c)^n\ =\ (c)^n$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This may be considered as the essence of Fermat's method of &lt;em&gt;infinite descent&lt;/em&gt;. It generalizes to rings of algebraic integers but depends upon much deeper results in this more general context, viz. the finiteness of the class number and Dirichlet's unit theorem. For further discussion see my &lt;a href=&quot;http://math.stackexchange.com/questions/18154/diophantine-equations-solved-using-algebraic-numbers/18187#18187&quot;&gt;post here&lt;/a&gt;, esp. the quote by Weil.&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2011-02-12T03:30:56.407" CommentCount="1" />
  <row Id="21638" PostTypeId="1" CreationDate="2011-02-12T03:38:45.940" Score="6" ViewCount="128" Body="&lt;p&gt;It seems that most introductory books on elliptic curves simply state the definition of the j-invariant of an elliptic curve without giving any background on how that definition was conceived.  Of course, for moduli reasons, it is clear why one might want such an invariant, but the actual formula has always seemed quite mysterious to me.  Does anyone know of a nice self-contained source that explains the definition of the j-invariant?&lt;/p&gt;&#xA;" OwnerUserId="6960" LastActivityDate="2011-02-12T11:20:43.273" Title="Definition of the j-invariant of an elliptic curve" Tags="&lt;elliptic-curves&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="21639" PostTypeId="1" CreationDate="2011-02-12T03:41:13.483" Score="1" ViewCount="82" Body="&lt;p&gt;If $X$ is a compact Riemann surface, $Y$ is a universal cover of $X$, and $f\colon X\rightarrow X$ is a biholomorphic map, then, can $f$ be lifted to a $biholomorphic$ map on $Y$? (I mean, topologically, it can be lifted to a homeomorphism from $Y$ to $Y$, just by using &quot;lifting criteria&quot;. But can the lift be &quot;biholomorphic&quot;?)&lt;/p&gt;&#xA;" OwnerUserId="8186" LastActivityDate="2011-05-26T10:02:11.277" Title="lifting of automorphisms of Riemann surface" Tags="&lt;riemann-surface&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="2" />
  <row Id="21640" PostTypeId="1" AcceptedAnswerId="21794" CreationDate="2011-02-12T03:44:22.763" Score="2" ViewCount="118" Body="&lt;p&gt;The rule (v1,w1)⋅(v2,w2)=(v1+v2,w1+w2+(v1∧v2)) defines a group structure on the vector space V⊕(V∧V) whenever V is itself a vector space over some field F.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is a more common name for this group?&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The group is nilpotent of class at most 2, with commutator [(v1,w1),(v2,w2)] = (0,2(v1∧v2)).&lt;/li&gt;&#xA;&lt;li&gt;The group has exponent char(F) since (v1,w1)&lt;sup&gt;n&lt;/sup&gt; = (v1&lt;sup&gt;n&lt;/sup&gt;,w1&lt;sup&gt;n&lt;/sup&gt;).&lt;/li&gt;&#xA;&lt;li&gt;When char(F) = 2, this is just an elementary abelian 2-group.&lt;/li&gt;&#xA;&lt;li&gt;When dim(V) = 1, this is just the abelian group V, so has a faithful 2-dimensional F-module.&lt;/li&gt;&#xA;&lt;li&gt;When dim(V) = 2 and char(F)≠2, this is a maximal unipotent subgroup of GL(3,F), so has a faithful 3-dimensional F-module.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I don't recognize it when dim(V) = 3.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;When dim(V) = 3 and char(F)≠2, does the group have a faithful F-module of dimension independent of F?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Is this a maximal unipotent subgroup of some classical group or does it otherwise have a standard description as a matrix group?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am primarily concerned with the case that F is a finite prime field (and am hoping to do better than a permutation module of dimension |F|&lt;sup&gt;5&lt;/sup&gt;), but I assume many sorts of answers should be mostly independent of the field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't have much hope for large dim(V), since the nilpotency class of the group remains at 2, while maximal unipotent subgroups should increase their nilpotency class with their rank.  Maybe dim(V)=3 is small enough though.&lt;/p&gt;&#xA;" OwnerUserId="583" LastActivityDate="2011-02-13T04:48:50.433" Title="What is a more formal name for the wedgey group?" Tags="&lt;group-theory&gt;&lt;exterior-algebra&gt;" AnswerCount="1" CommentCount="7" />
  <row Id="21641" PostTypeId="1" AcceptedAnswerId="21652" CreationDate="2011-02-12T03:59:46.227" Score="2" ViewCount="142" Body="&lt;p&gt;For a non-negative integer $\ell$, define the arithmetic functions $\sigma_{\ell}(k) = \sum_{d \mid k} d^{\ell}$ and $\bar{\sigma}_{\ell}(k) = \sum_{k = dd^{\prime}} (-1)^{(d^{\prime} - 1)/2} d^{\ell}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The function $\sigma(k) = \sigma_{1}(k)$ is the divisor-sum function. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; Let $k$ be an odd integer. Is the following congruence known? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align*}&#xA;3\sigma(k) &amp;amp;\equiv \sigma_{3}(k) + 2 \bar{\sigma}_{0}(k) &amp;amp;\pmod{16}&#xA;\end{align*}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Update&lt;/em&gt;: What about this one? &#xA;\begin{align*}&#xA;\bar{\sigma}_{2}(k) &amp;amp;\equiv \bar{\sigma}_{0}(k) &amp;amp;\pmod{4}&#xA;\end{align*}&lt;/p&gt;&#xA;" OwnerUserId="2720" LastEditorUserId="2720" LastEditorDisplayName="" LastEditDate="2011-02-13T18:38:44.913" LastActivityDate="2011-02-13T18:38:44.913" Title="Congruences for Divisors" Tags="&lt;number-theory&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" />
  <row Id="21642" PostTypeId="2" ParentId="21638" CreationDate="2011-02-12T04:22:38.380" Score="5" ViewCount="" Body="&lt;p&gt;Actually, Ravi Vakil's notes give a great reason (modulo the strange constant out front). This explanation is given somewhere in the Foundations of Algebraic Geometry notes here &lt;a href=&quot;http://math.stanford.edu/~vakil/preprints.html#coursenotes&quot; rel=&quot;nofollow&quot;&gt;http://math.stanford.edu/~vakil/preprints.html#coursenotes&lt;/a&gt;. I'm just going off memory, so don't attribute any errors I make to him.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice that once you prove that every elliptic curve has an affine model given by $y^2=x(x-1)(x-\lambda)$, then you know the $j$-invariant has to be independent of the different $\lambda$ you get by permuting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can just explicitly work it out that the six choices of lambda then are $\lambda, \frac{1}{\lambda}, 1-\lambda, \frac{1}{1-\lambda}, \frac{\lambda}{\lambda -1}, \frac{\lambda -1}{\lambda}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some obvious first choices for an invariant with respect to all of these is to multiply them all together. You get $1$, oops, that isn't a good invariant since it not only doesn't depend on the choice of $\lambda$, but also is independent of curve. So you could also try adding them all, oops again, they come in pairs each adding to $1$, so they add to $3$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So let's try the next best thing which is to sum the squares. If you check, this is exactly the $j$ invariant (but without the constant which is stuck in for characteristic $2$ reasons).&lt;/p&gt;&#xA;" OwnerUserId="2000" LastActivityDate="2011-02-12T04:22:38.380" CommentCount="1" />
  <row Id="21643" PostTypeId="2" ParentId="21631" CreationDate="2011-02-12T04:44:14.260" Score="1" ViewCount="" Body="&lt;p&gt;The key concept you need to use is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Law_of_total_probability&quot; rel=&quot;nofollow&quot;&gt;law of total probability&lt;/a&gt;. Let $P$ be the random variable that denotes the success probability of the Binomial distribution. We know the distribution of $P$. Suppose we also know that $P=p$ for some $0 &amp;lt; p &amp;lt; 1$. Given this information, you should be able to write down the probability of the event $\{X=k\}$. Mathematically speaking, we know how to write down $P(X=k \mid P=p)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $P$ were a discrete distribution, the law of total probability will read as $$P(X=k) = \sum P(X=k \mid P = p) Pr(P = p)$$ where the sum is over all the values taken by the discrete random variable $P$. But in your example, $P$ is continuous which means this summation is replaced by integration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The final step in finding $P(X=k)$ is to substitute the expressions for $P(X=k \mid P = p)$ and $Pr(P=p)$ into the integral and evaluate the integral. You might find &lt;a href=&quot;http://en.wikipedia.org/wiki/Beta_function&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; link useful for evaluating the integral. If you are still getting stuck, show us your work and the difficulties you are facing. I am sure someone will chip in with more hints.&lt;/p&gt;&#xA;" OwnerUserId="618" LastActivityDate="2011-02-12T04:44:14.260" />
  <row Id="21644" PostTypeId="2" ParentId="21514" CreationDate="2011-02-12T06:05:31.070" Score="0" ViewCount="" Body="&lt;p&gt;Those above are probably right about just getting a book, in particular as chandru says, Hoffman and Kunze. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However you might be happy with this book: &lt;a href=&quot;http://www.math.brown.edu/~treil/papers/LADW/LADW.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.math.brown.edu/~treil/papers/LADW/LADW.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is actually a text book, but It has some good problems.  &lt;/p&gt;&#xA;" OwnerUserId="16" LastActivityDate="2011-02-12T06:05:31.070" />
  <row Id="21645" PostTypeId="1" AcceptedAnswerId="21779" CreationDate="2011-02-12T06:47:58.173" Score="1" ViewCount="60" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Composition_%28number_theory%29&quot; rel=&quot;nofollow&quot;&gt;Number Composition&lt;/a&gt; studies the number ways of compositing a number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wanna know the number of compositions of $m$ with $n$ parts with the size of the max part equal to or less than $k$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a closed form for this problem?&lt;/p&gt;&#xA;" OwnerUserId="6904" LastActivityDate="2011-02-13T02:53:05.200" Title="Restricted Compositions" Tags="&lt;combinatorics&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="21646" PostTypeId="1" AcceptedAnswerId="21647" CreationDate="2011-02-12T06:56:07.580" Score="4" ViewCount="81" Body="&lt;p&gt;How to prove that, if a is an ordinal and b is in a, then b is an ordinal?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are the definitions I'm using.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A set is an ordinal number if it is transitive and well-ordered by ∈.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A set T is transitive if every element of T is a subset of T.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My difficulty is mainly that I can't prove b is transitive. What's the magic?&lt;/p&gt;&#xA;" OwnerUserId="5079" LastActivityDate="2011-02-12T07:05:39.553" Title="Every member of an ordinal is an ordinal" Tags="&lt;set-theory&gt;" AnswerCount="1" />
  <row Id="21647" PostTypeId="2" ParentId="21646" CreationDate="2011-02-12T07:05:39.553" Score="5" ViewCount="" Body="&lt;p&gt;Hi Zirui,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose $\beta\in\alpha$ and $\alpha$ is an ordinal. Let $\gamma\in\beta$. We need to prove that $\gamma\subseteq\beta$. Accordingly, let $\delta\in\gamma$. We need to show that $\delta\in\beta$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $\alpha$ is an ordinal, and $\beta\in\alpha$, then $\beta\subseteq\alpha$. Therefore, $\gamma\in\alpha$. Again, this gives us that $\gamma\subseteq\alpha$. Thus $\delta\in\alpha$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now consider the set $\{\beta,\delta\}$. Since $\alpha$ is well-ordered by $\in$, we have that one of the following holds: $\beta=\delta$, $\beta\in\delta$, $\delta\in\beta$. We need to rule out the first two options.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $\beta=\delta$, consider the set $\{\beta,\gamma\}$ and note that it contradicts well-foundedness: $\gamma\in\beta$ and $\beta=\delta\in\gamma$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $\beta\in\delta$, consider the set $\{\beta,\gamma,\delta\}$ and note that $\beta\in\delta\in\gamma\in\beta$, again contradicting well-foundedness.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only option we have left is that $\delta\in\beta$, as we wanted.&lt;/p&gt;&#xA;" OwnerUserId="462" LastActivityDate="2011-02-12T07:05:39.553" CommentCount="1" />
  <row Id="21648" PostTypeId="1" CreationDate="2011-02-12T07:43:10.043" Score="0" ViewCount="59" Body="&lt;p&gt;A recent assignment asked us to compute the push-forward of $\phi _*Z$ to $\mathbb{R}^2$ under $\phi$ of a vector field Z restricted to U.  The previous question asked us to compute the integral curve of Z through a point.  The lecturer's solutions declared that we could use the integral curve of Z to compute the push-forward, but I'm unsure how, like so:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $(u,v)\in \mathbb{R}^2$ and $\gamma$ the integral curve of Z through $\phi^{-1}(u,v)$.  Then by the definition of the push-forward and the integral,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;${\phi _*Z|_{(u,v)}}$ == ${{d\phi }|_{\phi ^{-1}(u,v)}(Z|_{\phi ^{-1}(u,v)}}$ == ${\frac{d}{\text{dt}}(\phi  o \gamma (t))|_{t=0}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How did he get from the second expression to the third?&lt;/p&gt;&#xA;" OwnerUserId="6968" LastActivityDate="2011-02-12T07:43:10.043" Title="Compute push-forward with integral curves" Tags="&lt;manifold&gt;" CommentCount="1" />
  <row Id="21649" PostTypeId="2" ParentId="20990" CreationDate="2011-02-12T08:01:14.480" Score="1" ViewCount="" Body="&lt;pre&gt;&lt;code&gt;1% = 1 in 100  &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;So you ask her 100 times for one date.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[The second date may take 0, 1, 100 or ∞ times, depending on your first date :-) ]&lt;/p&gt;&#xA;" OwnerUserId="5703" LastActivityDate="2011-02-12T08:01:14.480" />
  <row Id="21650" PostTypeId="2" ParentId="21645" CreationDate="2011-02-12T08:30:18.350" Score="1" ViewCount="" Body="&lt;p&gt;I found the solution in the following paper:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.fq.math.ca/Scanned/14-5/abramson.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.fq.math.ca/Scanned/14-5/abramson.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P441 Formula(E) is the just the solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I have another question about how to calculate the formula, when n and k is big, say n=2^24, k= 10,000, is there some approximation to the formula, the preciser, the better. &lt;/p&gt;&#xA;" OwnerUserId="6904" LastActivityDate="2011-02-12T08:30:18.350" />
  <row Id="21651" PostTypeId="1" AcceptedAnswerId="21653" CreationDate="2011-02-12T08:45:36.687" Score="2" ViewCount="61" Body="&lt;p&gt;I was looking at some source code that was described as 'maintaining a moving average over the previous 10 minutes', and noticed it uses the exponential function. Now its been a while since my university study, and i didnt take as many numerical analysis type math topics as I wish I had (mostly vector calc and diff equations), but I cannot undertand how this algorithm works and I really would like to:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Geodesic g = new Geodesic(lat1, lon1, lat2, lon2);&#xA;double t = (cal.getTimeInMillis() - ts.getTimeInMillis()) / 1000.0D;&#xA;if (t &amp;gt; 0.0D)&#xA;{&#xA;    double s = 3600.0D * g.distance() / t;&#xA;    if ((0.0D != sog) &amp;amp;&amp;amp; (0.0D != s))&#xA;    {&#xA;        double d = 1.0D - Math.exp(-t / 600.0D);&#xA;        sog = (d * s + (1.0D - d) * sog);&#xA;    }&#xA;    else&#xA;    {&#xA;        sog = s;&#xA;    }&#xA;}&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Basically position is updated every minute or so, with some uncertainty in the exact time values between the actual updates, hence some averaging over time is needed, I just can't figure out how the 2 lines within the if() statement a acheive that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only link I can think of that may have relevance is that the 1st derivative of ln(x) is 1/x.&lt;/p&gt;&#xA;" OwnerUserId="6969" LastEditorUserId="6969" LastEditorDisplayName="" LastEditDate="2011-02-12T09:03:48.253" LastActivityDate="2011-02-12T09:34:44.573" Title="Exp function in an averaging algorithm." Tags="&lt;analysis&gt;&lt;algorithms&gt;" AnswerCount="1" />
  <row Id="21652" PostTypeId="2" ParentId="21641" CreationDate="2011-02-12T08:57:05.750" Score="4" ViewCount="" Body="&lt;p&gt;This is not actually a statement about divisor sums; it's just a sum of the corresponding statements for the individual divisors:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$3d \equiv d^3 + 2 (-1)^{(d-1)/2}\pmod{16}\; \mbox{for odd } d\;.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you substitute $4n+d$ for $d$ in this equation, $n$ drops out, so you only have to check the two cases $d=1$ and $d=3$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Update in response to the update in the question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again, that result is just simpler results in a complicated &quot;disguise&quot;. In fact $\bar{\sigma}_{2}(k) \equiv \bar{\sigma}_{0}(k) \pmod{4}$, since $d^2 \equiv d^0\pmod{4}$ for the individual odd divisors. That you can stick an additional sign factor in is due to the fact that it doesn't make a difference if the number of divisors is even, only squares have an odd number of divisors and the sign factor is $1$ for $k$ an odd square.&lt;/p&gt;&#xA;" OwnerUserId="6622" LastEditorUserId="6622" LastEditorDisplayName="" LastEditDate="2011-02-12T16:41:58.423" LastActivityDate="2011-02-12T16:41:58.423" CommentCount="11" />
  <row Id="21653" PostTypeId="2" ParentId="21651" CreationDate="2011-02-12T09:34:44.573" Score="3" ViewCount="" Body="&lt;p&gt;The code can be abstracted as follows: we have a time series $S_t$ whose noisy values are observed at discrete samples of time. The averaging estimate used is $$\bar{S}_{t+1} = d_t S_t + (1-d_t) \bar{S}_t $$ This is a type of averaging called the &lt;a href=&quot;http://en.wikipedia.org/wiki/Exponential_smoothing&quot; rel=&quot;nofollow&quot;&gt;exponential moving average&lt;/a&gt;. It is exponential because we can expand out the above expression recursively to obtain $$\bar{S}_{t+1} = d_t S_t + (1-d_t)d_{t-1} S_{t-1} + (1-d_t)(1-d_{t-1})d_{t-2} S_{t-2} + \dots $$ If we for a moment imagine that $d_t$ is a constant, we can see that past samples get weighed down by a factor that decreases exponentially. This is a useful feature to have in an averaging procedure where you want to give more weighting to your recent samples than to your older samples. The exponential weighting gradually forgets the past at a rate determined by the magnitude of $d_t$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is one more slight wiggle in your setup which is that the spacing between the samples is not necessarily uniform. Consider again the equation $\bar{S}_{t+1} = d_t S_t + (1-d_t) \bar{S}_t $. If the time interval $\Delta t$ between the last sample $S_{t-1}$ and the current sample $S_t$ is too large, we want to make sure that the past (i.e., $\bar{S}_t$) gets less weight, i.e., $d_t$ should be close to $1$ (since the series could have changed its behavior in the elapsed time $\Delta t$). $d_t$ could have been chosen using any arbitrary function that approaches $1$ as $t$ grows. In this case, the function chosen is $d_t = 1-e^{-\Delta t/t_0}$.&lt;/p&gt;&#xA;" OwnerUserId="618" LastActivityDate="2011-02-12T09:34:44.573" CommentCount="5" />
  <row Id="21654" PostTypeId="2" ParentId="21638" CreationDate="2011-02-12T11:20:43.273" Score="2" ViewCount="" Body="&lt;p&gt;The $j$-invariant has the following classical interpretation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider a model $E\subset{\Bbb P}^2$ of the elliptic curve (one knows that $E$ is a cubic). Let $P\in E$. There are 4 lines through $P$ that are tangent to $E$ and one can show that the set of cross-ratios $c$ of these 4 lines are independent of $P$. Then&#xA;$$&#xD;&#xA;j=\frac{(c^2-c+1)^3}{c^2(c-1)^2}&#xD;&#xA;$$&#xA;is invariant under the 24 permutations of the 4 tangents (which give up to 6 different values of the cross-ratio) and is the $j$-invariant of the elliptic curve, up to normalization.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the curve $E$ is given in Legendre form $y^2=x(x-1)(x-\lambda)$ the formula for the $j$-invariant is obtained taking $P$ the point at infinity and the 4 tangents the line at infinity and the lines $x=0$, $x=1$ and $x=\lambda$.&lt;/p&gt;&#xA;" OwnerUserId="688" LastActivityDate="2011-02-12T11:20:43.273" />
  <row Id="21655" PostTypeId="1" AcceptedAnswerId="21676" CreationDate="2011-02-12T11:38:46.250" Score="3" ViewCount="153" Body="&lt;p&gt;I made this observation and it seems reasonable to me to ask :if $n$ is a natural number then the number of the primes less than or equal to $n$ is denoted by $π(n)$ . is that true that in any interval of length $n$ there are at most $π(n)+1$ primes?(the $+1$ is needed for the trivial occasion where $n=p-1$ and the interval of length $n$ is $[2,p]$)&#xA;Alternative we can say that in any interval of length $n-1$ there are at most $π(n)$ primes.&lt;/p&gt;&#xA;" OwnerUserId="5010" LastEditorUserId="6075" LastEditorDisplayName="" LastEditDate="2011-03-24T18:41:23.923" LastActivityDate="2011-03-24T18:41:23.923" Title="bound of the number of the primes on an interval of length n" Tags="&lt;number-theory&gt;&lt;prime-numbers&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="1" />
  <row Id="21656" PostTypeId="1" CreationDate="2011-02-12T12:48:21.060" Score="1" ViewCount="68" Body="&lt;p&gt;Suppose we choose a random number n from 0..100.Y and X are discrete random independent variables that have uniform distribution with 0...n.If Y+X equals to 15 what is expectancy of n?&lt;/p&gt;&#xA;" OwnerUserId="5145" LastEditorUserId="5145" LastEditorDisplayName="" LastEditDate="2011-02-12T13:01:14.030" LastActivityDate="2011-02-12T14:47:20.157" Title="random variables with uniform probability and expectancy." Tags="&lt;homework&gt;&lt;probability&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="21657" PostTypeId="1" AcceptedAnswerId="21729" CreationDate="2011-02-12T12:57:09.493" Score="2" ViewCount="131" Body="&lt;p&gt;I recently came across two properties that imply that a group is abelian:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If the order of each element except the identity element of a group is two then the group is abelian.&lt;/li&gt;&#xA;&lt;li&gt;If every element of the group is its inverse then the group is abelian.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Obviously this comes from my module with no proof, I also looked into my book but none like this there too, so I am interested in knowing how can we prove them?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="2109" LastEditorUserId="312" LastEditorDisplayName="" LastEditDate="2011-02-13T02:49:09.700" LastActivityDate="2011-02-13T02:49:09.700" Title="How to prove these properties ?" Tags="&lt;abstract-algebra&gt;&lt;algebra&gt;" AnswerCount="4" CommentCount="7" />
  <row Id="21658" PostTypeId="2" ParentId="21657" CreationDate="2011-02-12T13:15:25.423" Score="3" ViewCount="" Body="&lt;p&gt;By converse I assume you mean&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$G$ abelian $\Rightarrow$ the order of $g \in G$ is two&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Counterexample: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $G$ be $\mathbb{Z} / 3 \mathbb{Z}$ with addition mod 3, i.e. $G = \{ 0, 1, 2\}$. This group is clearly abelian but $1 + 1 \neq 0$, so the order of $1$ is not $2$, so the converse doesn't hold.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I assume you are aware that both statements you gave are equivalent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a proof of the claim:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;claim:&lt;/strong&gt; $\forall g \in G: \hspace{3mm} gg = e \hspace{3mm}$ then $G$ is abelian&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;proof:&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $a$,$b$ $\in G$. Then $a = a ^{-1}$ and $b = b^{-1}$. Let $e$ be the neutral element of $G$. We want to show &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ ab = ba$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because $G$ is a group we know if $a,b \in G$ then $ ab \in G$ and by assumption $abab = e$. But&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ e = abab = ab a^{-1}b^{-1} = ab (ba)^{-1}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Multiplying both sides with $ba$ yields&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ ba = ab$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that I have used $(ab)^{-1} = b^{-1}a^{-1}$ which holds for any group and you have to prove as well.&lt;/p&gt;&#xA;" OwnerUserId="5798" LastEditorUserId="5798" LastEditorDisplayName="" LastEditDate="2011-02-12T13:31:29.000" LastActivityDate="2011-02-12T13:31:29.000" />
  <row Id="21659" PostTypeId="2" ParentId="21657" CreationDate="2011-02-12T13:15:58.617" Score="3" ViewCount="" Body="&lt;p&gt;Suppose $x^2=1$ for all $x$ in the group. This is equivalent to $x=x^{-1}$ for all $x$. Now consider $x=ab$.&lt;/p&gt;&#xA;" OwnerUserId="589" LastActivityDate="2011-02-12T13:15:58.617" />
  <row Id="21660" PostTypeId="1" AcceptedAnswerId="21662" CreationDate="2011-02-12T13:17:34.530" Score="4" ViewCount="141" Body="&lt;p&gt;Hello.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm trying to show why &lt;strong&gt;it isn't possible&lt;/strong&gt; to define an order of magnitude on ℤn (modular artitmethic) that satisfies the ordering properties of ℤ. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;By letting addition to be ⊕ and multiplication ⊗, I know the following about ℤn:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Closed&lt;/strong&gt; under ⊕ and ⊗&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;⊕ and ⊗ are &lt;strong&gt;commutative&lt;/strong&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;⊕ and ⊗ are &lt;strong&gt;associative&lt;/strong&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;0&lt;/strong&gt; is ⊕ identity and &lt;strong&gt;1&lt;/strong&gt; is ⊗ identity&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Cancellation&lt;/strong&gt; [(-a) ⊕ a = 0]&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;a ⊗ (b ⊕ c) = a ⊗ b ⊕ a ⊗ c &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(b ⊕ c) ⊗ a = b ⊗ a ⊕ c ⊗ a&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;How would I prove that at least one of the ordering properties of ℤ does not hold for ℤn?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I will be definitely using the &lt;em&gt;ordering properties of ℤ&lt;/em&gt;:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;exactly one of a &amp;lt; b or a = b or a &gt; b holds&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;if a &amp;lt; b and b &amp;lt; c then a &amp;lt; c&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;if a &amp;lt; b then a + c &amp;lt; b + c&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;if a &amp;lt; b and c &gt; 0 then a * c &amp;lt; b * c&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;In a way I can see that one of these properties would collapse in ℤn, but I cannot prove it. By &lt;em&gt;contradiction&lt;/em&gt; perhaps? (start with a &amp;lt; b and find end up with a &gt; b ?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any help much appreciated.&lt;/p&gt;&#xA;" OwnerUserId="6974" LastActivityDate="2011-02-12T20:42:52.537" Title="How can I prove that an order (&quot;&lt;&quot; say) on ℤn cannot be defined?" Tags="&lt;modular-arithmetic&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="21661" PostTypeId="1" AcceptedAnswerId="21707" CreationDate="2011-02-12T13:17:49.553" Score="1" ViewCount="180" Body="&lt;p&gt;Ok so i am stuck at this:&#xA;I need to calculate distance between 2 points...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example:&#xA;I have 30x30 square and point1 is at X4,Y5 and point2 is at X30,Y23 now i need to get the shortest distance from point1 to point2. By wich way is the shortest North,East,West,South...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know i have to do that by &quot;pythagorean theorem&quot; but problem is if point1 is (X4,Y5) and point2 is (X28,Y6) for example... now the shortest way would be to go East and then you come to the right side out and just go one squeare to South. And the shortest distance would be (5)squares.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I dont know exactly to say what i need, but you will probably see on image2 on the link!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is an example of 30x30 and &lt;a href=&quot;http://oi54.tinypic.com/2rhphmo.jpg&quot; rel=&quot;nofollow&quot;&gt;here is a full image of what i am talking about&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;ADDED MORE EXAMPLES:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://i54.tinypic.com/2j1rlt2.gif&quot; rel=&quot;nofollow&quot;&gt;Here would the shortest be (6).&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://i54.tinypic.com/2wq9yr9.gif&quot; rel=&quot;nofollow&quot;&gt;Here would the shortest be (3).&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://i51.tinypic.com/nwde9u.gif&quot; rel=&quot;nofollow&quot;&gt;Here would the shortest be (21).&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://i53.tinypic.com/zm06k6.gif&quot; rel=&quot;nofollow&quot;&gt;Here would the shortest be (5, something).&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://i56.tinypic.com/2mx493r.gif&quot; rel=&quot;nofollow&quot;&gt;Here would the shortest be (4).&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/Hy2ah.gif&quot; alt=&quot;Example&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you for any help people! :)&lt;/p&gt;&#xA;" OwnerUserId="6975" LastEditorUserId="6975" LastEditorDisplayName="" LastEditDate="2011-02-12T19:36:13.143" LastActivityDate="2011-02-12T19:36:13.143" Title="Calculate distance between two points N,W,E,S" Tags="&lt;mathematica&gt;&lt;famous-theorems&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="21662" PostTypeId="2" ParentId="21660" CreationDate="2011-02-12T13:19:59.363" Score="6" ViewCount="" Body="&lt;p&gt;Prove that $1&amp;gt;0$. Conclude that $n-1 &amp;gt;0$. Now add $1$ to both sides.&lt;/p&gt;&#xA;" OwnerUserId="589" LastActivityDate="2011-02-12T13:19:59.363" CommentCount="6" />
  <row Id="21663" PostTypeId="2" ParentId="21657" CreationDate="2011-02-12T13:23:46.673" Score="0" ViewCount="" Body="&lt;p&gt;Look at this if it helps:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Since all your elements are of order $2$ you have $x^{2}=e$ for all $x \in G$ which implies $x=x^{-1}$. If you have two elements $a,b \in G$, then by above you have $a \cdot b = a^{-1} \cdot b^{-1}=(ba)^{-1}=ba$. Hence your group is abelian.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;You have $x =x^{-1}$ for all $x \in G$. Continue doing as I proceeded for the first one. You should be able to get it.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" LastActivityDate="2011-02-12T13:23:46.673" />
  <row Id="21664" PostTypeId="1" AcceptedAnswerId="21665" CreationDate="2011-02-12T13:42:06.737" Score="1" ViewCount="78" Body="&lt;p&gt;I have a problem solving an integration. This is my approach:&#xA;\begin{align*}&#xA;u &amp;amp;= \frac{1}{T}\cdot\intop_{0}^{T}(U_{0}+\hat{u}\cdot \sin(\omega t))dt\qquad\text{with }\omega=\frac{2\pi}{T}\\&#xA;u &amp;amp;= \frac{1}{T}\cdot\intop_{0}^{T}(U_{0}+\hat{u}\cdot \sin(\frac{2\pi}{T}\cdot t))dt\\&#xA;u &amp;amp;= \frac{1}{T}\cdot\left[U_{0}t-\hat{u}\cdot\frac{T}{2\pi}\cdot\cos\frac{2\pi}{T}\cdot t\right]_{0}^{T}\\&#xA;u &amp;amp;= \frac{U_{0}T}{T}-\hat{u}\cdot\frac{T}{2\pi\cdot T}\cdot\cos\frac{2\pi\cdot T}{T}\\&#xA;u &amp;amp;= U_{0}-\hat{u}\cdot\frac{1}{2\pi}\cdot\cos2\pi\\&#xA;u &amp;amp;= U_{0}-\frac{\hat{u}}{2\pi}.&#xA;\end{align*}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the solution should be ${u = U_{0}}$. Where is my mistake?&lt;/p&gt;&#xA;" OwnerUserId="2280" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-12T20:29:10.530" LastActivityDate="2011-02-12T20:29:10.530" Title="Problem with integrating sine function" Tags="&lt;integral&gt;" AnswerCount="1" CommentCount="0" />
  <row Id="21665" PostTypeId="2" ParentId="21664" CreationDate="2011-02-12T13:47:02.687" Score="5" ViewCount="" Body="&lt;p&gt;It's $\cos 0=1$, not $\cos 0=0$.&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2011-02-12T13:47:02.687" CommentCount="3" />
  <row Id="21666" PostTypeId="2" ParentId="21661" CreationDate="2011-02-12T13:50:14.267" Score="2" ViewCount="" Body="&lt;p&gt;If I understand correctly, you are interested in the distances computed by only moving along horizontal and vertical directions, and not diagonally. Is that right?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next, you are assuming your square is actually the surface of a &lt;a href=&quot;http://en.wikipedia.org/wiki/Torus&quot; rel=&quot;nofollow&quot;&gt;torus&lt;/a&gt;, so that by &quot;going out&quot; of the right side you re-enter from the left one and so on. Is that right too?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If both are true, then the answer to your question is pretty simple. First of all, consider the case when the two points are on a same row, one of them in column X and the other in column X'. Then their distance is the minimum between |X-X'| (where || denote the modulus, or absolute value) and 30-|X-X'| (of course in the case when the width of the square is different from 30, the new value has to be substituted). If the shortest way is &quot;moving within&quot; the square, the minimum is |X-X'|, while if it is shortest to take a shortcut through the sides, then 30-|X-X'| is the distance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the points are on different rows, you have just to add this &quot;horizontal&quot; distance to the &quot;vertical&quot; distance, computed in the same way, but using the Y-coordinates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope this helps!&lt;/p&gt;&#xA;" OwnerUserId="6507" LastActivityDate="2011-02-12T13:50:14.267" CommentCount="6" />
  <row Id="21667" PostTypeId="1" CreationDate="2011-02-12T14:23:13.067" Score="0" ViewCount="61" Body="&lt;p&gt;Given some element A of SL(2,Z) which is associated with a particular orientation preserving homeomorphism up to isotopy (by the isomorphism phi), what orientation preserving homeomorphisms up to isotopy will the negative multiple of A in SL(2,Z), that is (-1)* A correspond to with respect to A? For instance, when considering the generators of the fundamental group of the torus, do both A and (-1)*A &quot;act&quot; on these two generators similarly?&#xA;Thank you.&lt;/p&gt;&#xA;" OwnerUserId="6977" LastEditorUserId="6977" LastEditorDisplayName="" LastEditDate="2011-02-12T23:27:41.357" LastActivityDate="2011-02-12T23:27:41.357" Title="elements of SL(2,Z) and their negative multiples acting with the mapping class group of the two-torus T^2" Tags="&lt;algebraic-topology&gt;" CommentCount="2" />
  <row Id="21668" PostTypeId="1" AcceptedAnswerId="21686" CreationDate="2011-02-12T14:26:06.470" Score="5" ViewCount="167" Body="&lt;p&gt;Seems like I &lt;a href=&quot;http://math.stackexchange.com/questions/21429/linear-transformations-and-norm&quot;&gt;still don't get it&lt;/a&gt;, I think I am missing something important.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Let $V$ be an $n$ dimensional inner product space ($n \geq 1$), and $T\colon\mathbf{V}\to\mathbf{V}$ be a linear transformation such that:&lt;/p&gt;&#xA;  &#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;$T^2 = T$&lt;/li&gt;&#xA;  &lt;li&gt;$||T(a)|| \leq ||a||$ for every vector $a$ in $\mathbf{V}$;&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;  &#xA;  &lt;p&gt;Prove that a subspace $U \subseteq V$ exist, such that $T$ is the orthogonal projection on $U$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now, I know these things:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The fact hat $T^2 = T$ guarantees that $T$ is indeed a projection, so I need to prove that T is an &lt;strong&gt;orthogonal&lt;/strong&gt; projection (I guess this is where $||T(a)|| \leq ||a||$ kicks in).&lt;/li&gt;&#xA;&lt;li&gt;To do this I can prove that:&#xA;&lt;ul&gt;&#xA;&lt;li&gt;For every $v$ in $ImT^{\perp}$, $T(v) = 0$&lt;/li&gt;&#xA;&lt;li&gt;Alternatively, I can prove that for every $v$ in $ImT$ and $u$ in $KerT$, $(v,u)=0$.&lt;/li&gt;&#xA;&lt;li&gt;$T$ is &lt;a href=&quot;http://en.wikipedia.org/wiki/Projection_%28linear_algebra%29#Orthogonal_projections&quot; rel=&quot;nofollow&quot;&gt;self-adjoint&lt;/a&gt; (according to Wikipedia)&lt;/li&gt;&#xA;&lt;li&gt;The matrix $A = [T]_{E}$ when $E$ is an orthonormal basis, is hermitian (this is equivalent to the previous point).&lt;/li&gt;&#xA;&lt;li&gt;What else?&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I've been thinking about it for quite some time now, and I'm pretty sure there is something big I'm missing, &lt;strong&gt;again&lt;/strong&gt;. I just don't know how to use the data to prove any of these things.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="6879" LastActivityDate="2011-02-12T16:02:39.770" Title="Orthogonal Projection" Tags="&lt;homework&gt;&lt;linear-algebra&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="21669" PostTypeId="1" AcceptedAnswerId="21673" CreationDate="2011-02-12T14:28:09.733" Score="5" ViewCount="97" Body="&lt;ul&gt;&#xA;&lt;li&gt;Z is a 1D lattice&lt;/li&gt;&#xA;&lt;li&gt;Gaussian and Eisenstein integers are 2D lattices&lt;/li&gt;&#xA;&lt;li&gt;But the golden integers (for example) are dense on the real line.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Are there rings of integers which have 3D, 4D, ... lattices?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Here is a plot of $(a + \tfrac{1}{2}(1+\sqrt{5})b,a + \tfrac{1}{2}(1-\sqrt{5})b)$ for $-10\le a,b \le 10$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/7eVzp.png&quot; alt=&quot;lattice&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is the lattice corresponding to the golden integers, if I understand correctly. The green points represent rational integers and the blue points represent multiples of $\varphi$.&lt;/p&gt;&#xA;" OwnerUserId="4290" LastEditorUserId="4290" LastEditorDisplayName="" LastEditDate="2011-03-22T18:23:28.410" LastActivityDate="2011-03-22T18:23:28.410" Title="Lattice of Gauss and Eisenstein Integers" Tags="&lt;number-theory&gt;&lt;algebraic-number-theory&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="21671" PostTypeId="2" ParentId="21655" CreationDate="2011-02-12T14:37:39.937" Score="2" ViewCount="" Body="&lt;p&gt;A combination of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Brun%E2%80%93Titchmarsh_theorem&quot; rel=&quot;nofollow&quot;&gt;Brun-Titchmarsh inequality&lt;/a&gt; and the &lt;a href=&quot;http://en.wikipedia.org/wiki/Prime_number_theorem&quot; rel=&quot;nofollow&quot;&gt;Prime Number Theorem&lt;/a&gt; will yield the following:  For every $\epsilon&amp;gt;0$ there exists $N$ such that for $y&amp;gt;N$ and for every $x&amp;gt;0$ we have $$\pi(x+y)-\pi(x)&amp;lt;(2+\epsilon)\pi(y).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, this is not quite as good as what you are asking, since you want for every $M,N$ $$\pi(M+N)-\pi(N)\leq \pi(M)$$&#xA;Is this true or not?  According to my analytic number theory text (Montgomery and Vaughn):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;It was once conjectured that $$\pi (M+N)\leq \pi (M)+\pi (N)$$ for $M&amp;gt;1$, $N&amp;gt;1$, but there is now serious doubt as to the validity of this inequality.  Indeed, it seems likely that $\rho(y)&amp;gt;\pi (y)$ for all large $y$.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here, $\rho(y)$ is defined as $$\limsup_{x\rightarrow\infty}(\pi (x+y)-\pi (x)).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope that helps.&lt;/p&gt;&#xA;" OwnerUserId="6075" LastEditorUserId="6075" LastEditorDisplayName="" LastEditDate="2011-03-24T18:40:24.157" LastActivityDate="2011-03-24T18:40:24.157" CommentCount="3" />
  <row Id="21672" PostTypeId="2" ParentId="21656" CreationDate="2011-02-12T14:47:20.157" Score="2" ViewCount="" Body="&lt;p&gt;It is probably worth starting by working out the answer to: If $Y$ and $X$ are discrete random independent variables that have uniform distribution on the integers $0,\ldots...n$, then what is the probability that $Y+X=15$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could then use these probabilities as likelihoods in your original question.&lt;/p&gt;&#xA;" OwnerUserId="6460" LastActivityDate="2011-02-12T14:47:20.157" CommentCount="2" />
  <row Id="21673" PostTypeId="2" ParentId="21669" CreationDate="2011-02-12T15:02:02.113" Score="6" ViewCount="" Body="&lt;p&gt;The example of the golden integers (and more generally rings of integers in quadratic number fields with positive discriminant) shows that the single embedding into $\mathbb{R}$ is inadequate. Instead, if $K = \mathbb{Q}(\sqrt{d}), d &amp;gt; 0$ then the appropriate way to embed $K$ as a lattice in the plane is to look at &lt;em&gt;both&lt;/em&gt; embeddings $\sigma_1, \sigma_2 : K \to \mathbb{R}$. The first one sends $\sqrt{d}$ to $\sqrt{d}$ and the second one sends $\sqrt{d}$ to $-\sqrt{d}$. Together they give an embedding $(\sigma_1, \sigma_2)$ of $K$ into $\mathbb{R}^2$, and relative to this embedding the ring of integers $\mathcal{O}_K$ really is a lattice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More generally, if $K$ is a number field of degree $n$, then there are $n = r + 2s$ embeddings $\sigma_i : K \to \mathbb{C}$, $r$ of which have image in $\mathbb{R}$ and $2s$ of which have image outside of $\mathbb{R}$, which come in complex conjugate pairs. Here the appropriate generalization of the above embedding is to use all of the real embeddings $\sigma_1, ... \sigma_r$ and one representative of each complex conjugate pair of complex embeddings $\sigma_{r+1}, ... \sigma_{r+s}$. This gives an embedding $K \to \mathbb{R}^r \times \mathbb{C}^s$, and embedding $\mathbb{C}$ into $\mathbb{R}^2$ gives an embedding $K \to \mathbb{R}^n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Relative to this embedding, it's a standard exercise that $\mathcal{O}_K$ is a lattice in $\mathbb{R}^n$ of rank $n$. This is the standard construction used to prove the finiteness of the class group and Dirichlet's unit theorem, and details can be found in any book on algebraic number theory.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2011-02-12T15:02:02.113" CommentCount="1" />
  <row Id="21675" PostTypeId="2" ParentId="21655" CreationDate="2011-02-12T15:05:08.587" Score="1" ViewCount="" Body="&lt;p&gt;This is indeed a known open problem, &lt;a href=&quot;http://www.opertech.com/primes/k-tuples.html&quot; rel=&quot;nofollow&quot;&gt; the Hardy-Littlewood conjecture&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\pi(x+y) - \pi(x) \le \pi(y)$$&lt;/p&gt;&#xA;" OwnerUserId="3111" LastActivityDate="2011-02-12T15:05:08.587" />
  <row Id="21676" PostTypeId="2" ParentId="21655" CreationDate="2011-02-12T15:11:21.113" Score="5" ViewCount="" Body="&lt;p&gt;This is a well-known conjecture. It even has a name: the &lt;a href=&quot;http://en.wikipedia.org/wiki/Second_Hardy%E2%80%93Littlewood_conjecture&quot; rel=&quot;nofollow&quot;&gt;Second Hardy-Littlewood Conjecture&lt;/a&gt;, in the form: $\pi(x+y) \le \pi(x)+\pi(y)$ for $x, y \ge 2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a long time, this was generally thought to be true. Then in 1974, &lt;a href=&quot;http://projecteuclid.org/DPubS?service=UI&amp;amp;version=1.0&amp;amp;verb=Display&amp;amp;handle=euclid.bams/1183535510&quot; rel=&quot;nofollow&quot;&gt;Ian Richards showed&lt;/a&gt; that it was incompatible with the &lt;a href=&quot;http://www.wordiq.com/definition/First_Hardy-Littlewood_conjecture&quot; rel=&quot;nofollow&quot;&gt;First Hardy-Littlewood Conjecture&lt;/a&gt;! He did this by constructing explicitly an admissible prime constellation of length $x$ and size larger than $\pi(x)$. Computers were involved. See &lt;a href=&quot;http://www.ieeta.pt/~tos/apc.html&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; for more details.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The First H-L Conjecture is considered a sure thing, which has led most mathematicions to abandon the Second H-L Conjecture (although any counterexamples are likely to be extremely large).&lt;/p&gt;&#xA;" OwnerUserId="1508" LastActivityDate="2011-02-12T15:11:21.113" CommentCount="2" />
  <row Id="21677" PostTypeId="1" AcceptedAnswerId="23356" CreationDate="2011-02-12T15:15:13.430" Score="3" ViewCount="134" Body="&lt;p&gt;If I have three lines $a,$ $b$ and $c$ in the euclidean 3D space, which are pairwise non-coplanar, is there always a fourth line $x$, that intersects theses three lines?&lt;/p&gt;&#xA;" OwnerUserId="5282" LastActivityDate="2011-02-23T15:27:39.880" Title="Three non-coplanar lines in the 3D-space always have a fourth one that intersect them all?" Tags="&lt;euclidean-geometry&gt;&lt;3d&gt;&lt;intersection&gt;" AnswerCount="3" />
  <row Id="21680" PostTypeId="2" ParentId="21677" CreationDate="2011-02-12T15:38:50.833" Score="1" ViewCount="" Body="&lt;p&gt;I think so, but have not finished the proof.  I would argue that if you pick two of the lines you can find an affine transformation to make the them (x,0,0) and (0,y,1) where x and y parameterize the lines.  The general line through these points is (x-xt,yt,t) where t parameterizes this line.  The third line is (u+u's, v+v's, w+w's).  So to get an intersection, you have three equations in four unknowns: x,y,t,s.  There ought to be a one-parameter family of solutions unless something goes wrong.  You have to show that non-coplanar is enough to make sure nothing goes wrong.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Added:  the equations are $\begin {align} x-xt&amp;amp;=u+u&amp;#39;s \\ yt&amp;amp;=v+v&amp;#39;s \\ t&amp;amp;=w+w&amp;#39;s \end{align}$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $u,u&amp;#39;,v,v&amp;#39;,w,w&amp;#39;$ are the fixed parameters of the third line.  So you can pick $s$ at will, calculate $t$ from $w, w&amp;#39;$, calculate $y$ from $v, v&amp;#39;$ as long as $t \neq 0$ and calculate $x$ from $u, u&amp;#39;$ as long as $t \neq 1$.  This gives only two values of $s$ that are not acceptable.  So we have the promised one dimensional family of solutions.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2011-02-23T15:27:39.880" LastActivityDate="2011-02-23T15:27:39.880" CommentCount="1" />
  <row Id="21681" PostTypeId="1" CreationDate="2011-02-12T15:40:00.427" Score="2" ViewCount="112" Body="&lt;p&gt;What goes wrong with this argument?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose I have a curve $\gamma: [a, b] \rightarrow R^n$. We express this in polar coordinates as $\gamma(t) = (r(t), \theta(t))$. The derivative of $\gamma$ is just taken componentwise, so $\gamma&amp;#39;(t) = (r&amp;#39;(t), \theta&amp;#39;(t))$. Now the length of $\gamma$ is given by $\int_a^b \| \gamma&amp;#39;(t) \| \; dt = \int \|(r&amp;#39;(t), \theta&amp;#39;(t))\| \; dt$. In polar coordinates, the norm of $(r&amp;#39;(t), \theta&amp;#39;(t))$ is simply $r&amp;#39;(t)$. It follows that the length is just $r(b) - r(a)$, but this is clearly wrong. Where are the mistakes?&lt;/p&gt;&#xA;" OwnerUserId="6981" LastActivityDate="2011-02-14T01:16:19.533" Title="Length of a curve" Tags="&lt;calculus&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="21682" PostTypeId="1" CreationDate="2011-02-12T15:40:15.000" Score="4" ViewCount="99" Body="&lt;p&gt;Apparently this should be a straightforward / standard homework problem, but I'm having trouble figuring it out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $D$ be a square-free integer not divisible by 3. Let $\theta = \sqrt[3]{D}$, $K = \mathbb{Q}(\theta)$. Let $\mathcal{O}_K$ be the ring of algebraic integers inside $K$. I need to find explicitly elements generating $\mathcal{O}_K$ as a $\mathbb{Z}$-module. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is reasonably clear that $\theta$ is itself an algebraic integer and that $\mathbb{Z}[\theta] \le \mathcal{O}_K$, but I strongly suspect it isn't the whole ring. I'm not sure where the hypotheses on $D$ come in at all... any hints would be much appreciated.&lt;/p&gt;&#xA;" OwnerUserId="5191" LastActivityDate="2011-02-12T15:47:18.577" Title="Algebraic integers of a cubic extension" Tags="&lt;homework&gt;&lt;algebraic-number-theory&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" />
  <row Id="21683" PostTypeId="2" ParentId="21681" CreationDate="2011-02-12T15:46:12.043" Score="6" ViewCount="" Body="&lt;p&gt;The differential of arc length is $\sqrt{(dr)^2+(r d\theta)^2}$ in polar coordinates.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-02-12T15:46:12.043" />
  <row Id="21684" PostTypeId="2" ParentId="21682" CreationDate="2011-02-12T15:47:18.577" Score="4" ViewCount="" Body="&lt;p&gt;I thought this was a fun problem. Let $z = a + b \theta + c \theta^2$ be an algebraic integer, $a, b, c \in \mathbb{Q}$. Then the coefficients of the minimal polynomial of $z$ must be integers. This tells you, for example, that $\text{tr}(z) = 3a$ is an integer. The other two coefficients are slightly harder to work with, but here's a start: since $\theta z$ and $\theta^2 z$ are algebraic integers, their traces are also integers, so...? And then you work with the other coefficients of the minimal polynomial.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The hypotheses on $D$ come from some divisibility arguments you will need to make later; without them the problem is harder. Note that the discriminant of $\mathbb{Z}[\theta]$ has absolute value $27D^2$, which means that the index of $\mathbb{Z}[\theta]$ in $\mathcal{O}_K$ divides $3D$. (Actually it will turn out to divide $3$.)&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2011-02-12T15:47:18.577" CommentCount="2" />
  <row Id="21685" PostTypeId="1" AcceptedAnswerId="21687" CreationDate="2011-02-12T15:54:12.380" Score="2" ViewCount="49" Body="&lt;p&gt;I am working on a texturing algorithm for 3d graphics and I am trying to rotate triangles that are attached at a point so that they are connected (share a single edge) I am using &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\arctan( \frac{m1 - m2}{1 + m1*m2} )$&#xA;where $m1 \gt m2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is working well in some test cases but not in all... I have two lines one with slope ~1 and ~-.3 which gets me &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\arctan(\frac{1.05555556+0.368567038}{1 + 1.05555556*-0.368567038)})$ = 1.16553753 radians = ~66 degrees&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is correct if the line with slope -.3 is moving down to the right while the line with slope 1 is moving up to the right however the direction of the -.3 slope is moving up and to the left. Is there a way to parameterize this function so that I can provide the direction the lines moving in? Thanks.&lt;/p&gt;&#xA;" OwnerUserId="6983" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2011-02-12T15:58:47.417" LastActivityDate="2011-02-12T16:03:53.667" Title="Finding the appropriate slope between two lines (slopes)" Tags="&lt;geometry&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="21686" PostTypeId="2" ParentId="21668" CreationDate="2011-02-12T16:02:39.770" Score="3" ViewCount="" Body="&lt;p&gt;One approach to this is by using Pythagorean Theorem. You fill in the details.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Indeed, suppose that $T \psi \in Im T$. Note that $V = Ker T \oplus Ker T^\bot$. Now write $T \psi = k + k&amp;#39;$ where $k \in Ker T$ and $k&amp;#39; \in Ker T^\bot$. We get that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\|k&amp;#39;\|^2 \ge \|T k&amp;#39;\|^2 = \|T k + T k&amp;#39;\|^2 = \|T(T \psi)\|^2 = \|T \psi\|^2 = \|k\|^2 + \|k&amp;#39;^2\|,$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;from which we gather that $k = 0$. Therefore $T \psi \in Ker T^\bot$.&#xA;Thus $Im T \subset Ker T^\bot$. On the other hand if $\phi \in Im T^\bot \cap Ker T^\bot$, then one can show that $\phi \in Im T^* \cap Ker T^*$. (Because $Im T^\bot = Ker T^*$.) One can also show that $T^*$ is a projection. Let $\phi = T^* \psi$. Then $T^* \psi = T^* T^* \psi = T^* \phi = 0$. Hence $\phi = 0$. Since $V = Im T \oplus Im T^\bot$, we get that $Ker T^\bot \subset Im T$.&lt;/p&gt;&#xA;" OwnerUserId="3776" LastActivityDate="2011-02-12T16:02:39.770" CommentCount="3" />
  <row Id="21687" PostTypeId="2" ParentId="21685" CreationDate="2011-02-12T16:03:53.667" Score="1" ViewCount="" Body="&lt;p&gt;There are two angles to choose from.  One will be 180 degrees minus the other.  If you take a vector along each direction and take the dot product, it will be negative when you want the angle greater than 90.  So in your example, if both lines are moving to the right $(1,1)\cdot(1,-0.3)=0.7\gt0$ so you want the 66 degree solution.  In the second case $(1,1)\cdot(-1,0.3)=-0.7\lt0$ so you want the 180-66=114 degree solution.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-02-12T16:03:53.667" CommentCount="5" />
  <row Id="21688" PostTypeId="1" CreationDate="2011-02-12T16:18:51.130" Score="3" ViewCount="130" Body="&lt;p&gt;Grandma has 8 grandchildren, and 4 different types of popsicles:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;6 Vanilla popsicles&lt;/li&gt;&#xA;&lt;li&gt;6 Strawberry popsicles&lt;/li&gt;&#xA;&lt;li&gt;5 Banana popsicles&lt;/li&gt;&#xA;&lt;li&gt;3 Chokolate popsicles&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This morning, all of her grandchildren came together and asked for one popsicle each (every grandchild asked for a particular flavor). What is the total number of different sets of requests that Grandma can fulfill?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think this is related to the Inclusion-Exclusion principle because it was taught in the same class. Can you help me solve it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I did reach the following sum, but I imagine the question's author had something simpler in mind...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ E(0) = W(0)-W(1) = 3^8 - 4\cdot C(8,8)-4\cdot 3\cdot C(7, 8)-2\cdot C(6, 8)\cdot 3^2-C(5, 8)\cdot 3^3 - C(4, 8)\cdot 3^4$&lt;/p&gt;&#xA;" OwnerUserId="947" LastActivityDate="2011-02-12T19:41:10.937" Title="A question about the Inclusion-Exclusion principle" Tags="&lt;homework&gt;&lt;combinatorics&gt;" AnswerCount="1" />
  <row Id="21689" PostTypeId="1" AcceptedAnswerId="21698" CreationDate="2011-02-12T16:20:20.287" Score="2" ViewCount="76" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the problem of a point &lt;strong&gt;p&lt;/strong&gt; rotating around a given axis.&#xA;The axis of rotation is ω , |ω| = 1 , and &lt;strong&gt;q&lt;/strong&gt; is a point on the axis.&#xA;Assuming that the point rotates around the axis with unit angular velocity, then the velocity of the point p(t) is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;p'(t) = ω x ( p(t) - q )&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm looking for a more detailed explanation for this equation, as it is not the same with the classic equation of angular velocity. I'm confused with the term p(t) - q in the right hand side.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="6984" LastEditorUserId="6984" LastEditorDisplayName="" LastEditDate="2011-02-12T16:26:33.900" LastActivityDate="2011-02-12T19:25:24.960" Title="Angular Velocity around a given axis" Tags="&lt;vector-calculus&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="21690" PostTypeId="1" AcceptedAnswerId="21720" CreationDate="2011-02-12T16:23:07.787" Score="5" ViewCount="190" Body="&lt;p&gt;While doing some self-study, a friend posed this problem to me:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Let $a$ be a sequence, defined as following:&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$$a_0 = 0,\quad a_1= 1, \qquad a_{n+2}=\frac{a_n+a_{n+1}}2$$&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Figure out, whether $a$ converges, and if yes to which value. Find a closed form for $a_n$. If you want to, you may also have a look at the common case with $a_0= \alpha$ and $a_1 = \beta$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It's easy to see, that this converges to $\frac2 3$, and the other parts are also easy to solve with quessing a result and afterwards proving it correct, but I dislike this approach, as it is not productive if the solution isn't obvious.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a systematically approach for such problems? I would be really happy if somebody can explain me, how to systematically solve problems like this.&lt;/p&gt;&#xA;" OwnerUserId="5282" LastEditorUserId="4335" LastEditorDisplayName="" LastEditDate="2011-02-12T17:18:11.593" LastActivityDate="2011-02-12T21:42:56.157" Title="How can I systematically find a solution to this problem?" Tags="&lt;sequences-and-series&gt;&lt;proof-strategy&gt;" AnswerCount="4" CommentCount="1" FavoriteCount="2" />
  <row Id="21691" PostTypeId="2" ParentId="21690" CreationDate="2011-02-12T16:38:06.357" Score="4" ViewCount="" Body="&lt;p&gt;There are several.  My favorite, and one of the most generalizable, is the use of &lt;a href=&quot;http://en.wikipedia.org/wiki/Generating_function&quot; rel=&quot;nofollow&quot;&gt;Generating functions&lt;/a&gt;.  These offer a systematic way of solving many recurrence relations of any order.  (Yours is a 2nd order equation)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An excellent place to start learning about them is &lt;a href=&quot;http://www.math.upenn.edu/~wilf/gfologyLinked2.pdf&quot; rel=&quot;nofollow&quot;&gt;this online book&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope that helps.&lt;/p&gt;&#xA;" OwnerUserId="6075" LastActivityDate="2011-02-12T16:38:06.357" />
  <row Id="21692" PostTypeId="1" AcceptedAnswerId="21709" CreationDate="2011-02-12T16:43:18.110" Score="5" ViewCount="113" Body="&lt;p&gt;I have to prove that $f \colon x \mapsto e^{4x} + x^5 + 2$ ($f\colon \mathbb{R}\to\mathbb{R}$) is bijective. The argument given in the solution is that since the first two summands of the image is a bijective function of $x$, then so is $f$. Nonetheless, this &quot;proof&quot; doesn't seem at all rigorous to me, since there are many counterexamples to this argument.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I proved that $f$ is strictly increasing by looking at its derivative, thus injective, and that every $c \in \mathbb{R}$ has at least one preimage, applying Bolzano's theorem to $f(x) - c$ and evaluating its limit at $-\infty$ and $+\infty$, and so, demonstrating $f$ to be surjective. In consequence, $f$ is bijective.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am much more pleased about this proof than the one given in the solutions, but I want to know if I missed something, or if my hypothesis are insufficient. I gave an example to illustrate my argument, but the question I want to ask in the general form is in the title. Also, I would like to know whether the converse holds as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Update: I've been thinking about this, and realized that only monotonicity and continuity (together with unboundedness, as Mariano pointed out) are necessary for bijectiveness, and derivability only helps to prove monotonicity. Is this correct?&lt;/p&gt;&#xA;" OwnerUserId="6427" LastEditorUserId="6427" LastEditorDisplayName="" LastEditDate="2011-02-12T18:21:05.017" LastActivityDate="2011-02-12T18:50:32.830" Title="Does monotonicity and derivability of a function $f\colon \mathbb{R}\to\mathbb{R}$ imply bijectiveness?" Tags="&lt;calculus&gt;&lt;homework&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="1" />
  <row Id="21693" PostTypeId="1" AcceptedAnswerId="23865" CreationDate="2011-02-12T17:02:50.603" Score="0" ViewCount="172" Body="&lt;p&gt;I've edited this question to make it more understandable:&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;newly worded&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See the following pairs of descriptions of sets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;'n mod 2 = 0', '2 * int(n)'&lt;/p&gt;&#xA;&#xA;&lt;p&gt;'sqrt(x^2+y^2) = 1', 'x = sin(n) y = cos(n)'&lt;/p&gt;&#xA;&#xA;&lt;p&gt;'?', 'collatz algorithm starting from 27, returning the nth element'&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The left of each pair can be queried for if a given number is part of the set, from the first example '4 mod 2 = 0' and the answer in this case is yes. This is usefull but it may take forever to find even one number from a set with only a few elements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The right part represents the same set but can be queried for a number from the set in no particular order, so for example to get 5 numbers from the set is easy and fast but one can't ask if a given number is part of the set easily if at all for a set of infinite size.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can I convert between thease two forms, if so then how?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;old form of the question&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;'n mod 2 = 0' can be plotted to show a row of equally spaced peeks (where peak is a word I use to mean a value of true sourounded by values of false), you can ask if a given position is a peek by replacing n with the value you want like this '2 mod 2 = 0', in this case yes, there is a peak at position 2. The equation can presumably be rearranged to get the position of the n'th peek, for example 2n would work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thease two different approaches can apparently be refered to as push and pull methods of getting the information.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a way to convert between thease two ideas?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;'n mod 2 = 0' &amp;lt;-&gt; '2n'&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A less trivial example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;x = sin(n) y = cos(n) &lt;a href=&quot;http://goo.gl/5tqEV&quot; rel=&quot;nofollow&quot;&gt;Wolram Alpha&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;sqrt(x^2+y^2) = 1 &lt;a href=&quot;http://goo.gl/G9rkQ&quot; rel=&quot;nofollow&quot;&gt;Wolram Alpha&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where the first verstion gives every part of the outside of the circle when run for every possible n and the second version gives if a position is on the outside of the circle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;goo.gl/wZ9Ox,&#xA;goo.gl/2W6uU&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Returning to the first example. Completing all peaks using '2n' to find out if any lay on a given position would take forever, however you may only have an algorithm that works by repeatedly placing a peak and moving forward two units and you may not know about the mod function or that 'n mod 2 = 0' is the reverse of your algorithm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A more complicated but perhaps meaningfull example is a en.wikipedia.org/wiki/Collatz_conjecture starting from a given number might go on forever, or for a verry long time, described algorythmically it is like 'n2' in the first example but if we want to know if a given number is part of the seqence then we need the 'n mod 2 = 0' verstion of the equation.&lt;/p&gt;&#xA;" OwnerUserId="6982" LastEditorUserId="6982" LastEditorDisplayName="" LastEditDate="2011-02-14T12:28:17.030" LastActivityDate="2011-02-26T16:38:21.657" Title="'n mod 2 = 0' vs '2n'" Tags="&lt;number&gt;" AnswerCount="2" CommentCount="7" FavoriteCount="1" />
  <row Id="21694" PostTypeId="1" AcceptedAnswerId="21701" CreationDate="2011-02-12T17:11:21.817" Score="4" ViewCount="108" Body="&lt;p&gt;(Please rephrase if you think of a better title)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In combinatorics, there are several different of the form &quot;How many integer solutions does the equation $\Sigma_{1\le i\le k}X_i=n$ have?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some variations include:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$X_i\in\{0, 1\}$&lt;/li&gt;&#xA;&lt;li&gt;$0\le X_i$&lt;/li&gt;&#xA;&lt;li&gt;$1\le X_i$&lt;/li&gt;&#xA;&lt;li&gt;$0\le X_i\le m$&lt;/li&gt;&#xA;&lt;li&gt;$m_1\le X_i \le m_2$&lt;/li&gt;&#xA;&lt;li&gt;...&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Is there a complete guide to solving these kinds of questions?&lt;/p&gt;&#xA;" OwnerUserId="947" LastEditorUserId="1508" LastEditorDisplayName="" LastEditDate="2011-02-12T20:15:41.647" LastActivityDate="2011-02-12T20:15:41.647" Title="A complete guide to solving questions of the form &quot;how many integer solutions does this sum have&quot; ?" Tags="&lt;combinatorics&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="1" />
  <row Id="21695" PostTypeId="2" ParentId="21690" CreationDate="2011-02-12T17:16:17.440" Score="2" ViewCount="" Body="&lt;p&gt;Less often effective than generating functions, but in my opinion less work, is to guess that the solution to a linear homogeneous recurrence is of the form ar^n.  The a is like a constant of integration  Plug it in and in your case you get $ar^2=(a+ar)/2$ or $2r^2-r-1=0$.  There are two roots, $r_1$ and $r_2$ so the solution is $ar_1^n+br_2^n$  Now use your initial values to find $a$ and $b$.  If you have a repeated root you will have a term in $anr^n$ as well.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2011-02-12T20:54:23.870" LastActivityDate="2011-02-12T20:54:23.870" CommentCount="2" />
  <row Id="21696" PostTypeId="1" AcceptedAnswerId="21699" CreationDate="2011-02-12T17:20:44.773" Score="2" ViewCount="75" Body="&lt;p&gt;Let $A$ be a (commutative) noetherian ring, and let $I \subseteq A$ be an ideal. It is not hard to show that if $I$ is generated by a length $n$ regular sequence, then the $A/I$-module $I/I^2$ is a free module of rank $n$. Is the converse true? if we know that $I/I^2$ is free, does it follows that $I$ is generated by a regular sequence?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="5491" LastActivityDate="2011-02-17T04:36:43.367" Title="Local complete intersections and the cotangent module" Tags="&lt;algebraic-geometry&gt;&lt;commutative-algebra&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="21697" PostTypeId="2" ParentId="21694" CreationDate="2011-02-12T17:22:34.720" Score="0" ViewCount="" Body="&lt;p&gt;The general answers come in &lt;a href=&quot;http://en.wikipedia.org/wiki/Partition_%28number_theory%29&quot; rel=&quot;nofollow&quot;&gt;Partition theory&lt;/a&gt;.  For the first you just have $\binom{k}{n}$.  For the rest there are various ways.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-02-12T17:22:34.720" CommentCount="2" />
  <row Id="21698" PostTypeId="2" ParentId="21689" CreationDate="2011-02-12T17:28:36.150" Score="1" ViewCount="" Body="&lt;p&gt;Normally, $\omega$ is a vector, pointing along the axis with magnitude the rotation rate.  The x you have is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Cross_product&quot; rel=&quot;nofollow&quot;&gt;cross product&lt;/a&gt; of two vectors, not a simple multiply.  It only depends upon perpendicular components of the vectors.  So the projection of $\vec{p(t)}-\vec{q}$ perpendicular to $\vec{\omega}$ is just the radius and your expression is the usual $v=\omega r$ of circular motion&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-02-12T17:28:36.150" />
  <row Id="21699" PostTypeId="2" ParentId="21696" CreationDate="2011-02-12T17:45:35.667" Score="3" ViewCount="" Body="&lt;p&gt;No. Here are two counterexamples:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is possible to have $I=I^2$, in which case $I$ is generated by an idempotent, which is not a regular element.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $A$ is local, and $I$ the maximal ideal, then $I/I^2$ is always free over the residue field $A/I$, but in general $I$ is not generated by a regular sequence. (If this is true, then $A$ is called a regular local ring.)&lt;/p&gt;&#xA;" OwnerUserId="536" LastActivityDate="2011-02-12T17:45:35.667" CommentCount="1" />
  <row Id="21700" PostTypeId="2" ParentId="20996" CreationDate="2011-02-12T17:47:20.320" Score="6" ViewCount="" Body="&lt;p&gt;There is nothing inherently higher-order in Gödel's original proof. The object theory in the proof is &quot;$P$&quot;, which is based on the system of type theory from &lt;em&gt;Principia Mathematica&lt;/em&gt;, but in principle there is no difficulty applying exactly the same techniques to first-order systems such as Peano arithmetic. The metatheory in Gödel's paper is entirely finitistic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are two reasons that the paper was written this way. First, there is an obvious motivation when presenting a new technique to pick some &quot;well known&quot; system to empirically demonstrate the applicability of the results. Because the system of &lt;em&gt;Principia&lt;/em&gt; was well known at the time, using that system made it clear that the results could be applied to systems of real interest. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second, and more important, there was no general definition of a &quot;formal system&quot; at the time that Gödel published the paper in 1931. The notion of computability would not be discovered for a few more years. When he wrote the paper on the incompleteness theorems, Gödel originally planned to write a second part that would contain generalizations of his theorems to other systems. The final paragraph in Gödel's paper states (from &quot;From Frege to Gödel&quot;):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In the present paper, we have on the whole restricted ourselves to the system  $P$, and we have only indicated the applications to other systems. The results will be stated and proved in full generality in a sequel to be published soon. In that paper, also, the proof of Theorem XI, only sketched here, will be given in detail&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;After the paper was published, the results were accepted more readily than Gödel had hoped. Once the notion of computability was developed a few years later by Church and Turing, everyone immediately recognized that this was the key property needed to give a general definition of &quot;formal system&quot;, and that the same techniques that Gödel used for $P$ would apply to any effective formal system. Therefore, the second part of the paper was never written, because it was so clear how to apply the same techniques of the 1931 paper to any effective formal system. &lt;/p&gt;&#xA;" OwnerUserId="630" LastActivityDate="2011-02-12T17:47:20.320" CommentCount="4" />
  <row Id="21701" PostTypeId="2" ParentId="21694" CreationDate="2011-02-12T17:52:33.440" Score="4" ViewCount="" Body="&lt;p&gt;All of the problems you describe can be quickly solved using generating functions. The standard free reference for this is Wilf's &lt;a href=&quot;http://www.math.upenn.edu/~wilf/DownldGF.html&quot; rel=&quot;nofollow&quot;&gt;generatingfunctionology&lt;/a&gt;. You can even add coefficients in front of the $X_i$.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2011-02-12T17:52:33.440" />
  <row Id="21702" PostTypeId="2" ParentId="21693" CreationDate="2011-02-12T17:54:44.390" Score="0" ViewCount="" Body="&lt;p&gt;You seem to be talking about implicit and explicit descriptions of sets. Implicit ones are descriptions of sets like $A = \{ n \in \mathbb{N} | n \equiv 0 (\mod 2) \}$ or $B = \{ (x,y) \in \mathbb{R}^2 | x^2 + y^2 = 1 \},$ while explicit (think: closed-form) ones are `parametrisations' $n \mapsto 2n$ or $z \mapsto (\cos z,\sin z).$ What I think you're trying to say with the 'reverse' algorithm, is that in general you don't have a way to generate a surjective parametrisation, i.e. one which covers the entire set (for example: if you just take natural numbers and take $n \mapsto (\cos n, \sin n)$, you'll almost get the circle (technically: a dense subset), but not the entire thing. With the mod 2-thing, you &lt;em&gt;will&lt;/em&gt; get the entire set A, since the equation n = 0 (mod 2) just means that n is even, and every even integer can be written as n = 2k for some k. Hope this helps! &lt;/p&gt;&#xA;" OwnerUserId="6004" LastActivityDate="2011-02-12T17:54:44.390" CommentCount="2" />
  <row Id="21703" PostTypeId="2" ParentId="21660" CreationDate="2011-02-12T17:57:52.920" Score="1" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\ \: $ ordered $\Rightarrow$ torsion-free:  $\rm\ x&amp;gt;0\ \Rightarrow\ n\cdot x &amp;gt; 0\ \Rightarrow\ n\cdot x \ne 0\ \ $ for integral $\rm\ n &amp;gt; 0$&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2011-02-12T17:57:52.920" />
  <row Id="21704" PostTypeId="1" CreationDate="2011-02-12T17:59:29.367" Score="3" ViewCount="165" Body="&lt;p&gt;If $a$ is a positive constant,then show that $\displaystyle \lim_{n \rightarrow \infty} \prod_{k=1}^{n} (1-e^{-ka})$ exists and is strictly positive.&lt;/p&gt;&#xA;" OwnerUserId="6986" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2011-02-13T18:07:31.833" LastActivityDate="2011-02-13T18:07:31.833" Title="A limit problem" Tags="&lt;calculus&gt;&lt;limit&gt;" AnswerCount="2" FavoriteCount="0" />
  <row Id="21705" PostTypeId="1" AcceptedAnswerId="21714" CreationDate="2011-02-12T18:04:02.117" Score="3" ViewCount="171" Body="&lt;p&gt;Can contractible subspace be ignored/collapsed when computing $\pi_n$ or $H_n$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Motivation: I took this for granted for a long time, as I thought collapsing the contractible subspace does not change the homotopy type. Now it seems that this is only true for a CW pair... &lt;/p&gt;&#xA;" OwnerUserId="325" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-12T21:54:30.717" LastActivityDate="2011-02-12T21:54:30.717" Title="Can contractible subspace be ignored/collapsed when computing $\pi_n$ or $H_n$?" Tags="&lt;algebraic-topology&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="2" />
  <row Id="21706" PostTypeId="2" ParentId="21704" CreationDate="2011-02-12T18:09:21.067" Score="4" ViewCount="" Body="&lt;p&gt;I assume your question is a product of the form $\displaystyle \lim_{n \rightarrow \infty}\prod_{k=1}^n (1 - e^{-ka})$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any product of the form $\displaystyle \lim_{n \rightarrow \infty}\prod_{k=1}^n (1 - a_k)$ converges iff $\displaystyle \lim_{n \rightarrow \infty}\sum_{k=1}^n a_k$ converges.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your cases, $a_k = e^{-ka}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle \lim_{n \rightarrow \infty}\sum_{k=1}^n  a_k = \lim_{n \rightarrow \infty}\sum_{k=1}^n e^{-ka}$ converges to $\frac{1}{e^a-1}$ $(\text{geometric series with }e^{-a}&amp;lt;1\text{ as }a&amp;gt;0)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, the infinite product $\displaystyle \lim_{n \rightarrow \infty}\prod_{k=1}^n (1 - e^{-ka})$ converges.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The equivalence of the convergence of $\displaystyle \prod_{n=1}^{\infty} (1 + a_n)$ and $\displaystyle \sum_{n=1}^{\infty} a_n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume that $\displaystyle \sum_{n=1}^{\infty} a_n$ converges. This implies $\displaystyle \lim_{n \rightarrow \infty} a_n = 0$. Hence, $\displaystyle \lim_{n \rightarrow \infty} \frac{\log(1+a_n)}{a_n} = 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now consider $b_n = \log(1+a_n)$. By limit comparison test, since $0&amp;lt;\frac{b_n}{a_n}&amp;lt;\infty$, we have that $\displaystyle \sum_{n=1}^{\infty} b_n$ converges. (Intuitively, what the limit comparison test means is that the tail sums differ only by a factor and hence the convergence of one implies the other.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, $\displaystyle \sum_{n=1}^{\infty} \log(1+a_n)$ converges which essentially means $\displaystyle \prod_{n=1}^{\infty} (1+a_n)$ converges.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, you can argue out that if $\displaystyle \prod_{n=1}^{\infty} (1+a_n)$ converges, then $\displaystyle \sum_{n=1}^{\infty} a_n$ converges.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2011-02-13T17:21:39.450" LastActivityDate="2011-02-13T17:21:39.450" CommentCount="21" />
  <row Id="21707" PostTypeId="2" ParentId="21661" CreationDate="2011-02-12T18:10:10.987" Score="2" ViewCount="" Body="&lt;p&gt;What you're looking for is &lt;strong&gt;not the Euclidean distance&lt;/strong&gt;, calculated using the Pythagorean theorem (which is good for an infinitely extending plane with a continuum of points (and a few other restrictions)), but instead, as mentioned, the &lt;strong&gt;taxicab&lt;/strong&gt; or &lt;strong&gt;Manhattan&lt;/strong&gt; distance, with the additional restriction that it is on a finite set of points that 'wrap around' (are on a torus).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Manhattan distance between two points is just&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$d( p_1, p_2 ) = |x_1 - x_2| + |y_1 - y_2|. $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(if $p_1 = (x_1,y_1)$ and $p_2 = (x_2, y_2)$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the sum of the differences in each of the coordinates - the absolute value is just to make sure it doesn't matter which point you start with.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The above works if you have an infinite plane. For the additional constraint that you really have a finite area (it is on a torus) to account for the fact that it may be shorter to go South rather than North or East instead of West, take the min of all possible directions you could go, East/West, and North/South both mod the width/height of the bounded plane. So the better distance function uses that above and tries the minimum in two directions for each coordinate:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{eqnarray}&#xD;&#xA;d_t(p_1, p_2) = \min(&amp;amp;&amp;amp;d(p_1, p_2) ,\\&#xD;&#xA;&amp;amp;&amp;amp; d((x_1+30, y_1),p_2),\\&#xD;&#xA;&amp;amp;&amp;amp; d((x_1, y_1+30),p_2),\\&#xD;&#xA;&amp;amp;&amp;amp; d((x_1+30, y_1+30),p_2)).&#xD;&#xA;\end{eqnarray}$$     &lt;/p&gt;&#xA;&#xA;&lt;p&gt;These four possibilities are displayed in your image and the above is just taking the smallest one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that this latter calculation will work on a torus also for the Euclidean or other distance measures.&lt;/p&gt;&#xA;" OwnerUserId="1919" LastActivityDate="2011-02-12T18:10:10.987" CommentCount="5" />
  <row Id="21708" PostTypeId="2" ParentId="21660" CreationDate="2011-02-12T18:22:26.093" Score="4" ViewCount="" Body="&lt;p&gt;It is a 1943 theorem of F.W. Levi that for a commutative group $(G,+)$ the following are &#xA;equivalent:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(i) There exists a total ordering relation $\leq$ on $G$ which is compatible with the group law in the sense that for all $x_1,x_2,y_1,y_2 \in G$, $x_1 \leq y_1, \ x_2 \leq y_2 \implies x_1 + x_2 \leq y_1 + y_2$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(ii) $G$ is torsionfree: for all $x \in G$ and $n \in \mathbb{Z}^+$, $nx = 0 \implies x = 0$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A proof (and a citation to Levi's paper) is given in $\S 17.2$ &lt;a href=&quot;http://math.uga.edu/~pete/integral.pdf&quot; rel=&quot;nofollow&quot;&gt;of these notes&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the implication (i) $\implies$ (ii) -- which is what was being asked about here -- is by far the easier one, and the argument I give for this is no different than those in the other answers.  (So this answer is more for those with a more general interest in ordered commutative groups.)&lt;/p&gt;&#xA;" OwnerUserId="299" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2011-02-12T20:42:52.537" LastActivityDate="2011-02-12T20:42:52.537" CommentCount="3" />
  <row Id="21709" PostTypeId="2" ParentId="21692" CreationDate="2011-02-12T18:37:36.940" Score="6" ViewCount="" Body="&lt;p&gt;A function $f: \mathbb{R} \rightarrow \mathbb{R}$ which is &lt;em&gt;strictly&lt;/em&gt; monotonic -- i.e., either (for all $x_1,x_2 \in \mathbb{R}$, $x_1 &amp;lt; x_2 \implies f(x_1) &amp;lt; f(x_2)$) or (for all $x_1,x_2 \in \mathbb{R}$, $x_1 &amp;lt; x_2 \implies f(x_1) &gt; f(x_2)$) -- is necessarily injective.  This is immediate from the definition, and no continuity or differentiability is necessary for this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But what about bijectivity?  Let me assume WLOG that $f$ is strictly increasing.  Then a necessary condition for bijectivity is that $\lim_{x \rightarrow \pm \infty} f(x) = \pm \infty$.  (This is not absolutely immediate, but I hope and believe it will become clear after only a little thought.)  For future use, let us call this the &lt;em&gt;infinite limits property&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This observation can be used to see that the answer to your title question is &lt;strong&gt;no&lt;/strong&gt;: as Mariano says, consider the arctangent function.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moreover, being strictly increasing and having the infinite limits property is not enough to guarantee surjectivity: consider for instance the function defined piecewise as $x$ for $x \leq 0$ and as $x+1$ for $x &amp;gt; 0$.  This example seems to indicate that we are missing the condition of &lt;em&gt;continuity&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proposition&lt;/strong&gt;: Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be strictly increasing and &#xA;have the infinite limits property.  The following are equivalent:&lt;br&gt;&#xA;(i) $f$ is continuous.&lt;br&gt;&#xA;(ii) $f$ is surjective.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the word &quot;homework&quot; is being thrown around, I leave the proof to the OP.  (Anyway, it is a fun and not so difficult exercise.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the OP says, differentiability is not part of the fundamental picture here, but comes up in practice because having strictly positive derivative on an interval ensures that a function is strictly increasing.&lt;/p&gt;&#xA;" OwnerUserId="299" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2011-02-12T18:50:32.830" LastActivityDate="2011-02-12T18:50:32.830" />
  <row Id="21710" PostTypeId="1" CreationDate="2011-02-12T18:42:42.117" Score="-3" ViewCount="132" Body="&lt;p&gt;Let $f$ be a measurable function on $\mathbb{R}$ satisfying $\displaystyle \int\mid f(x)\mid dm(x)&amp;lt; \infty$,  where $m$ is the Lebesgue measure on the Borel subsets of  $\mathbb{R}$. Show that the function $\displaystyle g(t)=\int_\mathbb{R}e^{itx}f(x)dm(x)$ is uniformly continuous.&lt;/p&gt;&#xA;" OwnerUserId="6990" LastEditorUserId="5191" LastEditorDisplayName="" LastEditDate="2011-02-12T18:56:28.700" LastActivityDate="2011-02-12T19:01:10.533" Title="Uniform continuity question" Tags="&lt;measure-theory&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="21711" PostTypeId="2" ParentId="21581" CreationDate="2011-02-12T18:52:04.160" Score="2" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\: $ It's actually very easy: keep pushing ')'s rightward using the rewrite rule $\rm\ (xy)\:z\ \to\ x(yz)\:.$ This process terminates with the &lt;em&gt;right-associated normal form&lt;/em&gt; where all ')'s are at the right end,&#xA;namely $\rm\ a(b(c(d(\cdots)))\:,\ $ written $\rm\ a\ b\ c\ \cdots $ (without ambiguity - all possible bracketings agree).   &lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2011-02-12T18:52:04.160" />
  <row Id="21712" PostTypeId="2" ParentId="21694" CreationDate="2011-02-12T18:56:35.157" Score="3" ViewCount="" Body="&lt;p&gt;Because the $X_i$ are ordered, these are integer compositions with restrictions. But by suitable conversion (bijection) one can usually count these using combinations (and binomial coefficients). For your particular examples,&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$X_i \in \{0,1\}$, these are the number of combinations of size $n$ from $k$ items or ${k \choose n}$.&lt;/li&gt;&#xA;&lt;li&gt;$X_i \ge 0$, this is the 'stars and bars' problem, where you're counting the number of configurations of $n$ stars and $k-1$ bars in sequence (that is $n$ items separated into $k$ ordered parts by the $k-1$ bars). E.g. for $n = 2$ and $k = 3$, the ways of doing it are: &quot;##||&quot; ,&quot;#|#|&quot;,&quot;#||#&quot;,&quot;|##|&quot;,&quot;|#|#&quot;,&quot;||##&quot;. This is counted by ${n+k-1 \choose n}$ (the positions of $n$ stars in a string of length $n+k-1$ or equivalently ${n+k-1 \choose k-1}$ the position of the $k-1$ bars).&lt;/li&gt;&#xA;&lt;li&gt;$X_i \ge 1$, just subtract $k$ from $n$ to get the same problem (that is, if you -have- to have at least one item in each bin, just count the ways to do it as if you have 1 less item in each bin). This is ${n-1 \choose k-1}$.&lt;/li&gt;&#xA;&lt;li&gt;for more general restrictions on the number of items in each bin, you may need to do some inclusion-exclusion. For lower bounds $m_1$, just subtract off like in the previous problem. For upper bounds, count the number of ways when the lower bound is $m_2+1$ (again using the previous problem) and subtract from the solution when it is unbounded.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So these are strategies geared towards the specific problem given. You might see much more complex restrictions than you've given, and there you'll probably want to use a more general method like generating functions as mentioned in other answers.&lt;/p&gt;&#xA;" OwnerUserId="1919" LastActivityDate="2011-02-12T18:56:35.157" />
  <row Id="21713" PostTypeId="2" ParentId="21710" CreationDate="2011-02-12T19:01:10.533" Score="3" ViewCount="" Body="&lt;p&gt;Hint: Continuity follows from dominated convergence.  Vanishing at infinity follows from the Riemann-Lebesgue Lemma.  The combination implies uniform continuity.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2011-02-12T19:01:10.533" />
  <row Id="21714" PostTypeId="2" ParentId="21705" CreationDate="2011-02-12T19:07:24.753" Score="5" ViewCount="" Body="&lt;p&gt;Yes, you need a hypothesis like being a CW pair. A more general condition you could impose is that the inclusion of the contractible subspace is a cofibration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For an example of why this is necessary, consider $X = S^1$ and $A = X\setminus\{\ast\}$. Then collapsing &lt;em&gt;A&lt;/em&gt; gives you a two-point space which is contractible. There are several ways to see the contractibility of this space, one is that this space is the non-Hausdorff cone over a point.&lt;/p&gt;&#xA;" OwnerUserId="677" LastActivityDate="2011-02-12T19:07:24.753" CommentCount="4" />
  <row Id="21715" PostTypeId="1" CreationDate="2011-02-12T19:09:11.383" Score="-6" ViewCount="234" Body="&lt;p&gt;Let $L\colon \mathbb{R}^n\to\mathbb{R}^n$&#xA;be an invertible linear transformation and let $V\subset \mathbb{R}^n$ be a subspace with $L(V)\subseteq V$. Show that $L(V)=V$.&lt;/p&gt;&#xA;" OwnerUserId="6990" LastEditorUserId="367" LastEditorDisplayName="" LastEditDate="2011-02-12T19:14:52.247" LastActivityDate="2011-02-12T19:18:33.467" Title="Invertible linear operator" Tags="&lt;linear-algebra&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="21716" PostTypeId="1" AcceptedAnswerId="21719" CreationDate="2011-02-12T19:11:49.917" Score="4" ViewCount="111" Body="&lt;p&gt;Let $I(f)$ be defined as $\int_{0}^{1}f(x)dx$ from $C$[0, 1] to $\mathbb{R}$, where $C$[0, 1] is a linear space of all continuous functions from closed interval [0, 1] to $\mathbb{R}$. Is there a systematic way to find the kernel of $I$? Clearly, the zero function is there, as integral of $0$ from 1 to 0 will be equal to zero. Some trig functions might be in the kernel too (not sure if 1 in the integral is being interpreted as radian or degrees), but what would be an exhaustive way to find the basis of the kernel of $I$? Thanks. &lt;/p&gt;&#xA;" OwnerUserId="3731" LastActivityDate="2011-02-12T19:47:38.020" Title="What is the kernel of $I$ from $C$[0, 1] to $\mathbb{R}$?" Tags="&lt;linear-algebra&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="3" />
  <row Id="21717" PostTypeId="2" ParentId="21715" CreationDate="2011-02-12T19:18:33.467" Score="0" ViewCount="" Body="&lt;p&gt;If $L(V)\subsetneq V$, then $L|_V\colon V\to V$ is not surjective and hence not injective (because $V$ is a finite dimensional vector space). Therefore $L$ is not injective, which contradicts the assumtion that $L$ is invertible.&lt;/p&gt;&#xA;" OwnerUserId="367" LastActivityDate="2011-02-12T19:18:33.467" CommentCount="1" />
  <row Id="21718" PostTypeId="2" ParentId="21689" CreationDate="2011-02-12T19:25:24.960" Score="1" ViewCount="" Body="&lt;p&gt;The point ${\bf q}$ on the axis is the natural origin pertaining to your problem, it does not have to be the origin of the coordinate system. Choosing another point $\bar{\bf q}$ on the axis does not change the value of the vector product since ${\bf q}$ and $\bar{\bf q}$ differ by a vector parallel to ${\vec \omega}$. The equation expresses the fact that ${\bf p}&amp;#39;(t)$ is perpendicular to  ${\vec \omega}$ and to ${\bf p}(t)-{\bf q}$, and has the right absolute value $r \thinspace |{\vec \omega}|$ where $r$ denotes the distance of the point ${\bf p}(t)$ from the axis.&lt;/p&gt;&#xA;" OwnerUserId="1303" LastActivityDate="2011-02-12T19:25:24.960" />
  <row Id="21719" PostTypeId="2" ParentId="21716" CreationDate="2011-02-12T19:25:54.257" Score="4" ViewCount="" Body="&lt;p&gt;Since you are looking at a surjective linear map from an infinite dimensional vector space to a one-dimensional vector space, the kernel is a subset in $C[0,1]$ of co-dimension one. Co-dimension one means that you would need to add one more vector to a basis of the kernel to get a basis of $C[0,1]$. This means that the kernel is very large.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For every function $f\in C[0,1]$ you can find a number $c\in\mathbb R$, such that $f-c\cdot 1$ is in the kernel (here $1$ denotes the constance function with value 1).&lt;/p&gt;&#xA;" OwnerUserId="367" LastEditorUserId="367" LastEditorDisplayName="" LastEditDate="2011-02-12T19:47:38.020" LastActivityDate="2011-02-12T19:47:38.020" CommentCount="3" />
  <row Id="21720" PostTypeId="2" ParentId="21690" CreationDate="2011-02-12T19:28:05.483" Score="2" ViewCount="" Body="&lt;p&gt;If the recurrence relation is linear, homogeneous and has constant coefficients, here is a systematic way to solve it. Similar method will work if you have linear, inhomogeneous equation with constant coefficients. (For instance if the inhomogeneity is a constant, the inhomogeneity can be removed if possible by defining $b_n = a_n + c$ and figuring out $c$ such that the equation in $b_n$ is homogeneous. If it is not possible, which happens when the sum of the coefficients of $a_n$ add up to zero, then there is a linear growth term added to $a_n$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First obtain the characteristic equation. To do this, assume $a_n = m^n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Plug it in to get a quadratic in $m$ in this case. Solve for $m$. (If you have $a_n$ depending on $a_{n-1},a_{n-2},\ldots,a_{n-p}$, you will get a $p^{th}$ order polynomial.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Get the two roots say $m_1, m_2$. (In general, the $p$ roots).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the general solution is given by the linear combination of the roots namely $a_n = c_1 m_1^n + c_2 m_2^n$. (In general, $a_n = \displaystyle \sum_{k=1}^p c_k m_k^n$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solve for $c_1$ and $c_2$ (In general, $c_1,c_2,\ldots,c_p$) using the initial conditions&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For your recurrence, the corresponding equation becomes, $2m^2-m-1 = 0 \Rightarrow (2m+1)(m-1) = 0 \Rightarrow m = -\frac{1}{2},1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, $a_n = c (-\frac{1}{2})^n + d$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $a_0 =0, a_1 =1$, we get $c + d = 0$ and $d - \frac{c}{2} = 1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$d = \frac{2}{3}$ and $c = -\frac{2}{3}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, $a_n = \frac{2}{3} - \frac{2}{3} (-\frac{1}{2})^{n}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, $\displaystyle \lim_{n \rightarrow \infty} a_n = \frac{2}{3}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $a_0 =\alpha, a_1 =\beta$, we get $c + d = \alpha$ and $d - \frac{c}{2} = \beta$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$c = \frac{2}{3}(\alpha-\beta)$, $d = \frac{\alpha + 2 \beta}{3}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, $a_n = \frac{\alpha + 2 \beta}{3} + \frac{2}{3}(\alpha-\beta) (\frac{-1}{2})^n$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, $\displaystyle \lim_{n \rightarrow \infty} a_n = \frac{\alpha + 2 \beta}{3}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This methodology is analogous to plugging in $y=e^{mx}$ when you want to solve a linear, homogeneous ODE with constant coefficients. You get the constants for the ODE using the boundary conditions. Here you have a difference equation instead of a differential equation.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2011-02-12T19:52:59.187" LastActivityDate="2011-02-12T19:52:59.187" CommentCount="3" />
  <row Id="21721" PostTypeId="2" ParentId="21705" CreationDate="2011-02-12T19:28:19.443" Score="5" ViewCount="" Body="&lt;p&gt;You are right. An interesting example of this kind of behavior consists of taking two copies of the Hawaiian Earring space and connecting their basepoints by a line segment. Contracting the middle segment gives you the standard Hawaiian Earrings. However this contraction is not a homotopy equivalence! The fundamental groups are different!&lt;/p&gt;&#xA;" OwnerUserId="2941" LastActivityDate="2011-02-12T19:28:19.443" CommentCount="8" />
  <row Id="21722" PostTypeId="1" CreationDate="2011-02-12T19:36:09.167" Score="-2" ViewCount="100" Body="&lt;p&gt;is it true that a finite, hausdorff  topological space is metrizable?i.e (X,T) be the topological space wher T can be obtained from a metr&lt;/p&gt;&#xA;" OwnerUserId="6990" LastActivityDate="2011-02-12T19:38:20.630" Title="metrizable problem" Tags="&lt;topology&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" />
  <row Id="21723" PostTypeId="2" ParentId="21722" CreationDate="2011-02-12T19:38:20.630" Score="7" ViewCount="" Body="&lt;p&gt;A finite Hausdorff space is discrete, so metrizable by the discrete metric.&lt;/p&gt;&#xA;" OwnerUserId="5203" LastActivityDate="2011-02-12T19:38:20.630" />
  <row Id="21724" PostTypeId="2" ParentId="21688" CreationDate="2011-02-12T19:41:10.937" Score="3" ViewCount="" Body="&lt;p&gt;I don't quite understand how you arrived at your result, so I'll try to sketch my solution. As you tagged the question as homework, there are no computations - just the idea.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are at first $4$ possibilities to consider, depending how many chocolate popsicles the kids want: $0,1,2$ or $3$. Say $A(i)$ is the number of requests that have $i$ chocolate popsicles chosen. Let $T$ be the desired total. Then clearly:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$T = A(0) + A(1) + A(2) + A(3)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;as the possibilities are mutually exclusive and exhaustive.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;$A(3)$ is easy: $3$ kids picked chocolate, and the remaining $5$ are choosing amongst the other three flavours. Note that they can all choose the same flavour.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$A(2)$ is just as easy: you have $6$ kids left to consider. Assume first that &#xA;there are $6$ banana popsicles as well. Count all possible choices, then discount the one case when all six kids pick the banana popsicle.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$A(1)$ and $A(0)$: this would be hard to compute as directly as the previous two. Try it this way: counting $A(1)$ is equivalent to the problem: &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Grandma has 7 grandchildren, and 3 different types of popsicles:&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;- 6 Vanilla popsicles&#xA;- 6 Strawberry popsicles&#xA;- 5 Banana popsicles&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;This morning, all of her grandchildren came together and asked for one popsicle each (every grandchild asked for a particular flavor). What is the total number of different sets of requests that Grandma can fulfill?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Which is just the original problem minus one child and the chocolate popsicles. You approach it as above. Let $B(i)$ be the number of requests where $i$ banana popsicles are picked.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then you have:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$A(1) = B(0) + B(1) + B(2) + B(3) + B(4) + B(5)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All these $B(i)$ are now easy to compute. A similar calculation works for $A(0)$.&lt;/p&gt;&#xA;" OwnerUserId="5105" LastActivityDate="2011-02-12T19:41:10.937" />
  <row Id="21725" PostTypeId="1" CreationDate="2011-02-12T19:57:01.560" Score="-1" ViewCount="131" Body="&lt;p&gt;I am not getting a fact whether if $X\,\times\,Y$ is homeomorphic to $Y\,\times\,Y$ then is it true that $X$ is homeomorphic to $Y$? Actually I want to know the result for say $X=[0,1]$ and $Y=[0,1)$.&lt;/p&gt;&#xA;" OwnerUserId="6990" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-12T20:18:51.140" LastActivityDate="2011-02-12T21:51:30.767" Title="a problem on homeomorphism" Tags="&lt;general-topology&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="21726" PostTypeId="1" CreationDate="2011-02-12T20:06:37.770" Score="1" ViewCount="103" Body="&lt;p&gt;I got a little stuck on a simple proof of the following probability identity. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\mathbb{P}(A^c \cap B^c)=1-\mathbb{P}(A)-\mathbb{P}(B)+\mathbb{P}(A\cap B)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;how to prove for any set $X$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\mathbb{P}(X \cap A^c \cap B^c)=\mathbb{P}(X)-\mathbb{P}(X\cap A)-\mathbb{P}(X\cap B)+\mathbb{P}(X\cap A\cap B)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Looks very intuitive; just replace the whole space by $X$. But how to prove it simply and rigorously? Thanks.&lt;/p&gt;&#xA;" OwnerUserId="6016" LastActivityDate="2011-02-12T23:02:21.807" Title="simple probability identity" Tags="&lt;probability&gt;" AnswerCount="2" CommentCount="6" />
  <row Id="21728" PostTypeId="2" ParentId="21705" CreationDate="2011-02-12T20:17:33.373" Score="4" ViewCount="" Body="&lt;p&gt;Let me note a general fact: if the inclusion $A \hookrightarrow X$ (for $A$ a closed subspace) is a cofibration, and $A$ is contractible, then the map $X \to X/A$ is a homotopy equivalence. See Corollary 5.13 in chapter 1 of Whitehead's &quot;Elements of homotopy theory.&quot;&lt;/p&gt;&#xA;" OwnerUserId="536" LastActivityDate="2011-02-12T20:17:33.373" />
  <row Id="21729" PostTypeId="2" ParentId="21657" CreationDate="2011-02-12T20:26:55.007" Score="2" ViewCount="" Body="&lt;p&gt;In fact, the following is a stronger condition:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Prop.&lt;/strong&gt; Let $G$ be a group, and let $x$ and $y$ be elements of $G$. If&#xA;$$(xy)^2 = x^2y^2$$&#xA;then $x$ commutes with $y$; that is, $xy=yx$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If every element of the group has order dividing $2$, then both sides of the equation above are equal to $1$, so the equality holds for any two elements of $G$; if every element is its own inverse, then every element has order $1$ or $2$, then again both sides equal the identity, so the result follows for every pair of elements. If any pair of elements commute, then the group is abelian. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a stronger result, though, because it also applies in other cases, when neither side is the identity but they are nonetheless equal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Proof.&lt;/em&gt; Write the equation explicitly as&#xA;$$xyxy = xxyy.$$&#xA;Multiplying on the left by $x^{-1}$ and on the right by $y^{-1}$, We get&#xA;$$x^{-1}xyxyy^{-1} = x^{-1}xxyyy^{-1}$$&#xA;which is the same as&#xA;$$yx = xy.$$&#xA;So $x$ and $y$ commute. $\Box$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, try proving the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proposition.&lt;/strong&gt; Let $G$ be a group, and let $x$ and $y$ be two elements of $G$. The following are equivalent (that is, if one of them holds, then they all hold; if any on of them is false, then they are all false):&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$xy=yx$.&lt;/li&gt;&#xA;&lt;li&gt;$(xy)^2 = x^2y^2$.&lt;/li&gt;&#xA;&lt;li&gt;$(xy)^{-1} = x^{-1}y^{-1}$.&lt;/li&gt;&#xA;&lt;li&gt;There exists an integer $n$ such that&#xA;$$(xy)^{n+i} = x^{n+i}y^{n+i}$$&#xA;holds for $i=0,1,2$. &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="742" LastActivityDate="2011-02-12T20:26:55.007" />
  <row Id="21730" PostTypeId="1" AcceptedAnswerId="21737" CreationDate="2011-02-12T20:30:11.220" Score="9" ViewCount="99" Body="&lt;p&gt;My birthday this year (2011) is on a Friday. In most years, one's birthday the following year is on the subsequent day of the week, and in that pattern, my birthday next year (2012) it is on a Saturday.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, due to 2012 being a leap year, my birthday in 2013 will be on Monday - I miss out on having a Sunday birthday.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I quite like having weekend birthdays, so that disappoints me a bit. However it leads me to a question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Over a reasonable-length life-time, does any given person's number of weekend birthdays even out to an average, or are some people significantly more blessed than others?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If some people do have an advantage in this respect, is there any way mathematical way to work out whether being born on any given day of any given year will confer an advantage? And further, what are the probabilities of being 'lucky' in your number of weekend birthdays.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My guess is that for any given year, on each side of Feb 29th, birthdays on any given day of the week throughout the year will have the same score.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To make it a bit easier, lets say for a 'reasonable length life-time', we mean a fixed length of 80 years (though I for one intend to beat that!). That said, I would also be interested in how the maths change as we vary the life-span.&lt;/p&gt;&#xA;" OwnerUserId="6992" LastEditorUserId="1543" LastEditorDisplayName="" LastEditDate="2011-04-16T18:33:15.570" LastActivityDate="2011-04-16T18:33:15.570" Title="Weekend birthdays" Tags="&lt;combinatorics&gt;&lt;calendar-computations&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="21731" PostTypeId="2" ParentId="21725" CreationDate="2011-02-12T20:32:05.647" Score="2" ViewCount="" Body="&lt;p&gt;I could be wrong, but apparently no, that is not true. For example, $\mathbb{R}^{\omega} \times \mathbb{R}^{\omega}$ seems to be homeomorphic to $\mathbb{R} \times \mathbb{R}^{\omega}$, both homeomorphic to $\mathbb{R}^{\omega}$ but clearly, $\mathbb{R}$ is not homeomorphic to $\mathbb{R}^{\omega}$. (Here $\mathbb{R}^{\omega}$ is the countable product of $\mathbb{R}$ in the product topology.)&lt;/p&gt;&#xA;" OwnerUserId="6884" LastActivityDate="2011-02-12T20:32:05.647" CommentCount="1" />
  <row Id="21732" PostTypeId="2" ParentId="21681" CreationDate="2011-02-12T20:38:48.337" Score="5" ViewCount="" Body="&lt;p&gt;When you have $\gamma(t) = (x(t),y(t))$, it is then right to say $\gamma&amp;#39;(t) = (x&amp;#39;(t),y&amp;#39;(t))$. However, when you are in other coordinate systems, you need to be careful. For instance, if you move to the $r-\theta$ coordinate system, and parameterize $\gamma(t)$ by $(r(t),\theta(t))$, it is incorrect to say that the curve $\gamma&amp;#39;(t)$ is $(r&amp;#39;(t),\theta&amp;#39;(t))$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To make it clear, let $\gamma_1(t) = (x(t),y(t))$ and $\gamma_2(t) = (r(t),\theta(t))$ be the parameterizations of the same curve in different coordinate systems. $\gamma_1$ is in the $X-Y$ coordinate system, while $\gamma_2$ is the parameterization of the same curve in $r-\theta$ coordinate system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that $x(t) = r(t) \cos(\theta(t))$, $y(t) = r(t) \sin(\theta(t))$, $r(t) = \sqrt{x(t)^2 + y(t)^2}$, $\tan(\theta(t)) = \frac{y(t)}{x(t)}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\gamma_1&amp;#39;(t) = (x&amp;#39;(t),y&amp;#39;(t)) = (r&amp;#39; \cos(\theta) - r \sin(\theta) \theta&amp;#39;,r&amp;#39; \sin(\theta) + r \cos(\theta) \theta&amp;#39;)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you express this in $r-\theta$ coordinate system, $\gamma_2&amp;#39;(t) = (\sqrt{r&amp;#39;^2 + (r \theta&amp;#39;)^2},\tan^{-1}(\frac{r&amp;#39; \sin(\theta) + r \cos(\theta) \theta&amp;#39;}{r&amp;#39; \cos(\theta) - r \sin(\theta) \theta&amp;#39;})) = (r_1(t),\theta_1(t))$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the length of the curve is $\int {||\gamma_1&amp;#39;(t)|| dt}=\int {\sqrt{x&amp;#39;(t)^2 + y&amp;#39;(t)^2} dt}=\int {\sqrt{r&amp;#39;^2 + (r \theta&amp;#39;)^2} dt}=\int {r_1(t) dt}=\int {||\gamma_2&amp;#39;(t)|| dt}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Note&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$r(t) = ||\gamma_1(t)||$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that $r&amp;#39;(t) = ||\gamma_1(t)||&amp;#39; \neq ||\gamma_1&amp;#39;(t)||$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You need to apply chain rule to get the relation between $r&amp;#39;(t)$ and $\gamma_1&amp;#39;(t)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$r&amp;#39;(t) = \frac{\partial r}{\partial x} x&amp;#39;(t) + \frac{\partial r}{\partial y} y&amp;#39;(t) = \frac{x}{\sqrt{x^2+y^2}} x&amp;#39;(t) + \frac{y}{\sqrt{x^2+y^2}} y&amp;#39;(t)$&#xA;$r&amp;#39;(t) = \cos(\theta(t)) x&amp;#39;(t) + \sin(\theta(t)) y&amp;#39;(t) = (\cos(\theta),\sin(\theta)) \cdot \gamma_1&amp;#39;(t)$&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2011-02-14T01:16:19.533" LastActivityDate="2011-02-14T01:16:19.533" />
  <row Id="21733" PostTypeId="1" AcceptedAnswerId="21734" CreationDate="2011-02-12T20:50:55.293" Score="1" ViewCount="83" Body="&lt;p&gt;Hi&#xA;i need to figure out how can i add given number of squares in a rectangle&#xA;i want to create this.&#xA; &lt;a href=&quot;http://i.imgur.com/MHoda.jpg&quot; rel=&quot;nofollow&quot;&gt;http://i.imgur.com/MHoda.jpg&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The whole scenario is to create a function that take a number and then create a layout of it, for a quick reference i did this with predefined values for rows, columns, x&amp;amp;y scale etc.&#xA;The size of the rectangle is fixed and it will always be something predefined, everything else need to be calculated with these values.&#xA;Can anybody help me please??&lt;/p&gt;&#xA;" OwnerUserId="6991" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2011-02-12T21:18:03.267" LastActivityDate="2011-02-12T22:22:38.363" Title="How to evenly distribute squares in a rectangular area?" Tags="&lt;geometry&gt;&lt;soft-question&gt;&lt;algorithms&gt;" AnswerCount="1" />
  <row Id="21734" PostTypeId="2" ParentId="21733" CreationDate="2011-02-12T21:05:43.077" Score="0" ViewCount="" Body="&lt;p&gt;The first step is to figure out how many rows $R$ and columns $C$ you want.  If you want N squares, you need $R*C\ge N$.  If you are given the dimensions of the rectangle, say w is the ratio of the width $W$ to the height $H$(presumed $\ge 1$).  Then you probably want $C/R$ to be about this, so $R^2w \ge N$  Now you can round up to get $R$ and divide and round up to get $C$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To get the coordinates of each square of side $s$, there is some space $t$ between the squares.  So $Cs+(C-1)t\le W$ and $Rs+(R-1)t\le H$ and you lay them out starting at (0,0) if that is top left in increments of $s+t$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Added:  for 34 tiles in 600 wide by 200 high, you can either have a 4x10 or 3x12 layout (with some blanks like you showed in the original question).  If we take the 4x10 option, then you can have, for example, the tiles 45 high, which uses up 180 of the 200.  This leaves 20 units to divide by 3, so the gaps would be 6. But going the other way, 10 tiles would use 450, leaving 50 to divide by 9, so the gaps could only be 5.  Then the top left tile is at (0,0), the next one down is at (50,0), the next at (100,0) and the last at (150,0) with its bottom at 195.  In width they go 0, 50, 100, 150,...450 with the right edge at 495.  Is that what you were looking for?&lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2011-02-12T22:22:38.363" LastActivityDate="2011-02-12T22:22:38.363" CommentCount="3" />
  <row Id="21735" PostTypeId="2" ParentId="21690" CreationDate="2011-02-12T21:21:57.197" Score="2" ViewCount="" Body="&lt;p&gt;I realize you're asking for a systematic answer to the problem, and others have supplied such, but there is a neat correspondence with the binary representation of 1/3. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Every iteration of the recurrence, you're dividing everything by two each time, you're going up and down each time by a negative power of 2, $1 - 1/2 + 1/4 - 1/8 + ...$ or $1/2 + 1/8 + 1/32 + ...$ essentially creating a binary encoding .1010101... which, as a geometric series comes out to $${1\over 2}\cdot{1\over 1- {1\over 4}} = {2\over 3}.$$&lt;/p&gt;&#xA;" OwnerUserId="1919" LastEditorUserId="1919" LastEditorDisplayName="" LastEditDate="2011-02-12T21:42:56.157" LastActivityDate="2011-02-12T21:42:56.157" CommentCount="1" />
  <row Id="21736" PostTypeId="1" AcceptedAnswerId="21739" CreationDate="2011-02-12T21:22:39.640" Score="1" ViewCount="88" Body="&lt;p&gt;&quot;Given $\vec{u}_1,\ldots ,\vec{u}_n$ mutually orthogonal non-zero vectors, explain why for $\vec{v}=c_1\vec{u}_1+\ldots +c_n\vec{u}_n$ $c_k=\frac{\vec{v} \cdot \vec{u}_k}{\vec{u}_k \cdot \vec{u}_k}$&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This I explained by dotting both sides with $\vec{u}_k$ and simplifying everything. However, the question I now have is, how using this achieved result, can I show that $\vec{u}_1,\ldots ,\vec{u}_n$ are linearly independent? I was thinking of saying that in accordance to $c_k=\frac{\vec{v} \cdot \vec{u}_k}{\vec{u}_k \cdot \vec{u}_k}$, every coefficient can be of only one fixed value, so there is no room for changing one at the expense of another (as one could with coefficients of linearly dependent vectors), but I am not sure if this is right, and whether I am phrasing this correctly. Thanks for your help!&lt;/p&gt;&#xA;" OwnerUserId="5157" LastEditorUserId="5157" LastEditorDisplayName="" LastEditDate="2011-02-12T21:33:01.617" LastActivityDate="2011-02-12T21:36:59.820" Title="Explain why a set of mutually orthogonal non-zero vectors is linearly independent given a clause" Tags="&lt;linear-algebra&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="21737" PostTypeId="2" ParentId="21730" CreationDate="2011-02-12T21:34:08.777" Score="10" ViewCount="" Body="&lt;p&gt;The calculations are slightly different depending on whether it is the Julian or Gregorian calendar, and whether the date in question occurs between January 1st and February 28, or March 1st and later (we'll ignore February 29, I think). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider first the situation in the Julian calendar, where there is a leap year every $4$ years without fail.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose you have January 1st on a Monday, in the year immediately after a leap year. When will January 1st again fall on a Monday in the year immediately after a leap year? The pattern is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;M, T, W, R, Sa, Su, M, T, R, F, Sa, Su, T, W, R, F, Su, M, T, W, F, Sa, Su, M, W, R, F, Sa,&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and then repeats; that is, a 28 year cycle. So, ignoring the leap year rule for Gregorian calendar, you could check for a period of 84 years (three full cycles).  The situation is similar if your birthday occurs March 1st or later, except that the leap year affects you in position 4 instead of 5; you would have:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;M, T, W, F, Sa, Su, M, W, R, F, Sa, M, T, W, R, Sa, Su, M, T, R, F, Sa, Su, T, W, R, F, Su,&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and the pattern repeats. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In our putative example, which began on a Monday, we got each day of the week exactly $4$ times; that is, they all occur equally likely. If you chop it off at 80 years, then you'd be one short on four days. Changing the day of the week, or the time of the most recent leap year, would not affect the outcome: it cycles every 28 years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the Gregorian calendar has a slightly different rule for leap years: in the Gregorian calendar, a year is leap year if and only if it is a multiple of $4$ that is not a multiple of $100$, &lt;em&gt;except&lt;/em&gt; for the multiples of $400$ (that's why $1900$ was not a leap year, $2000$ &lt;em&gt;was&lt;/em&gt;, but $2100$ will not). For someone born before $2016$, assuming a period of $84$ years, this will not matter because we will have a cycle just as above, with leap years every four years. But if you extend the lifetime so that it includes the year $2100$, then you start running into trouble because you skip a leap year.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can think of the Julian calendar as repeating the pattern of leap years every four years; the Gregorian calendar, on the other hand, has a cycle of 400 years for repeating leap years. &lt;em&gt;Added:&lt;/em&gt; And as Ross pointed out, while the Julian calendar has a cycle of 28 years, the Gregorian calendar has a cycle of 400. So you want to check what happens over a period of 400 years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Over a cycle of 400 years, the number of times a given date (other than February 29) falls on a particular day &lt;em&gt;does&lt;/em&gt; change (it's not uniform, like in the Julian calendar). If you start counting on a Monday, over the next 400 years, that date will fall on Monday, Wednesday, and Saturday 58 times each, Thursday and Friday 57 times each; and Tuesday and Sunday 56 times each. It's a pretty small difference over a period of 400 years, but there you go.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-12T21:53:21.203" LastActivityDate="2011-02-12T21:53:21.203" CommentCount="1" />
  <row Id="21739" PostTypeId="2" ParentId="21736" CreationDate="2011-02-12T21:36:59.820" Score="5" ViewCount="" Body="&lt;p&gt;To show they are linearly independent, you want to show that any linear combination equal to zero is the trivial linear combination. So, suppose you have a linear combination&#xA;$$\alpha_1\mathbf{u}_1 + \cdots + \alpha_n\mathbf{u}_n = \mathbf{0}.$$&#xA;Now, using the formula you got, taking $\mathbf{v}=\mathbf{0}$, will give you the value of each $\alpha_i$, namely:&#xA;$$\alpha_i = \frac{\mathbf{0}\cdot\mathbf{u}_i}{\mathbf{u}_i\cdot\mathbf{u}_i}.$$&#xA;What does that tell you about $\alpha_1,\ldots,\alpha_n$?&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2011-02-12T21:36:59.820" CommentCount="2" />
  <row Id="21740" PostTypeId="2" ParentId="21725" CreationDate="2011-02-12T21:51:30.767" Score="1" ViewCount="" Body="&lt;p&gt;A more advanced example was given by Bing (&lt;a href=&quot;http://www.jstor.org/pss/1970322&quot; rel=&quot;nofollow&quot;&gt;reference&lt;/a&gt;) and shows that the problem can be far more intrincate than a cardinality argument (whether of the spaces or the indexing set).&lt;/p&gt;&#xA;" OwnerUserId="4829" LastActivityDate="2011-02-12T21:51:30.767" />
  <row Id="21741" PostTypeId="2" ParentId="21726" CreationDate="2011-02-12T22:01:17.057" Score="4" ViewCount="" Body="&lt;p&gt;Write $X$ as a disjoint union:&#xA;$$&#xD;&#xA;X=(X\cap A\cap B)\sqcup (X\cap A^c\cap B)\sqcup (X\cap A\cap B^c)\sqcup (X\cap A^c\cap B^c).&#xD;&#xA;$$&#xA;This gives:&#xA;$$&#xD;&#xA;P(X)=P(X\cap A\cap B)+P(X\cap A^c\cap B)+P(X\cap A\cap B^c)+P(X\cap A^c\cap B^c).&#xD;&#xA;$$&#xA;Hence&#xA;$$&#xD;&#xA;P(X\cap A^c\cap B^c)=P(X)-P(X\cap A\cap B)-P(X\cap A^c\cap B)-P(X\cap A\cap B^c).&#xD;&#xA;$$&#xA;Now the result follows from $P(X\cap A\cap B)+P(X\cap A^c\cap B)=P(X\cap B)$ and $P(X\cap A\cap B)+P(X\cap A\cap B^c)=P(X\cap A)$.&lt;/p&gt;&#xA;" OwnerUserId="367" LastActivityDate="2011-02-12T22:01:17.057" CommentCount="2" />
  <row Id="21742" PostTypeId="2" ParentId="21726" CreationDate="2011-02-12T22:05:07.097" Score="-1" ViewCount="" Body="&lt;p&gt;so.. i am using the latter to derive the former..&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;write P(X)  as $\mathbb{P}(X\cap (A U A^c))$ , equal to $P(X \cap A)+P(X \cap A^c)$, cancel some terms&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;then expand $P(X \cap A^c)$ to $P(X \cap A^c \cap (B U B^c))$, which is equal to $P(X \cap A^c \cap B)+ P(X\cap A^c \cap B^c)$, cancel the terms to get the answer.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="6995" LastActivityDate="2011-02-12T22:05:07.097" CommentCount="3" />
  <row Id="21743" PostTypeId="1" CreationDate="2011-02-12T22:10:15.447" Score="2" ViewCount="79" Body="&lt;p&gt;Let $u$ and $v$ be two $L^1(\mathbb{R})$ functions such that &#xA;$\|u\|_{L^1} \le \|v\|_{L^1}$ and $f$ is non-negative $L^1(\mathbb{R})$&#xA;with non-negative inverse Fourier transform.&#xA;Is it true that for the convolution&#xA;$\|u*f\| \le \|v*f\|$?&#xA;If not, maybe someone know additional condition that&#xA;will give the last inequality.&lt;/p&gt;&#xA;" OwnerUserId="6997" LastEditorUserId="367" LastEditorDisplayName="" LastEditDate="2011-02-12T22:16:57.083" LastActivityDate="2011-02-12T22:50:00.070" Title="Convolution inequality" Tags="&lt;analysis&gt;&lt;fourier-analysis&gt;&lt;harmonic-analysis&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" />
  <row Id="21744" PostTypeId="1" AcceptedAnswerId="21756" CreationDate="2011-02-12T22:18:46.237" Score="7" ViewCount="182" Body="&lt;p&gt;I am starting to see a pattern in my math studies, and I'm interested in your thoughts on the subject.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems like there are certain things that are much easier for me to learn than others. For instance, I am a pretty quick learner when I can &quot;visualize&quot; what I'm learning, and logical thinking and rigid formalism seem pretty natural to me. I am also good with abstractions. On the other hand, I am painfully bad at calculations. I can make 5 mistakes when solving the simplest of equations (often not seeing my mistake even when I look for it, and then making 3 others when I try to fix it), and many mathematical tasks seem like meaningless symbol manipulations to me (calculating integrals, identities of all sorts, everything that involves unpleasant functions manipulations, complex numbers can reduce me to tears).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, naturally I am a bit worried about my studies. Here are my questions:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;I think there may be something I am missing. For the things I &lt;strong&gt;am&lt;/strong&gt; good at, I often have a strategy: an abstraction or a certain way of thinking that helps me deal with them. Do you have such a strategy for dealing with the problems I mentioned? How do you start thinking about them?&lt;/li&gt;&#xA;&lt;li&gt;How can I get better at these things? Trying &lt;strong&gt;really hard&lt;/strong&gt; to do it &lt;strong&gt;a lot&lt;/strong&gt; never seems to help. I just give up after making the same mistakes and reaching dead ends for two hours, look at the solution and don't understand how I could have come up with it on my own.&lt;/li&gt;&#xA;&lt;li&gt;Is all lost? What can I do with the skills I have?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="6879" LastActivityDate="2011-02-13T00:40:34.037" Title="General question about mathematical thinking" Tags="&lt;soft-question&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="21745" PostTypeId="2" ParentId="1747" CreationDate="2011-02-12T22:32:31.080" Score="2" ViewCount="" Body="&lt;p&gt;I am surprised that no one has mentioned &lt;a href=&quot;http://en.wikipedia.org/wiki/Statistical_hypothesis_testing&quot; rel=&quot;nofollow&quot;&gt;Hypothesis Testing&lt;/a&gt; so far. Hypothesis testing lets you to decide, with a certain level of significance, whether you have sufficient evidence to reject the underlying (Null) hypothesis or you have do not sufficient evidence against the Null Hypothesis and hence you accept the Null Hypothesis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am explaining the Hypothesis testing below assuming that you want to determine if a coin comes up heads more often than tails. If you want to determine, if the coin is biased or unbiased, the same procedure holds good. Just that you need to do a two-sided hypothesis testing as opposed to one-sided hypothesis testing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this question, your Null hypothesis is $p \leq 0.5$ while your Alternate hypothesis is $p &amp;gt; 0.5$, where $p$ is the probability that the coin shows up a head. Say now you want to perform your hypothesis testing at $10\%$ level of significance. What you do now is to do as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $n_H$ be the number of heads observed out of a total of $n$ tosses of the coin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take $p=0.5$ (the extreme case of the Null Hypothesis). Let $x \sim B(n,0.5)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Compute $n_H^c$ as follows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$P(x \geq n_H^c) = 0.1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$n_H^c$ gives you the critical value beyond which you have sufficient evidence to reject the Null Hypothesis at $10\%$ level of significance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i.e. if you find $n_H \geq n_H^c$, then you have sufficient evidence to reject the Null Hypothesis at $10\%$ level of significance and conclude that the coin comes up heads more often than tails.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to determine if the coin is unbiased, you need to do a two-sided hypothesis testing as follows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your Null hypothesis is $p = 0.5$ while your Alternate hypothesis is $p \neq 0.5$, where $p$ is the probability that the coin shows up a head. Say now you want to perform your hypothesis testing at $10\%$ level of significance. What you do now is to do as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $n_H$ be the number of heads observed out of a total of $n$ tosses of the coin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $x \sim B(n,0.5)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Compute $n_H^{c_1}$ and $n_H^{c_2}$ as follows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$P(x \leq n_H^{c_1}) + P(x \geq n_H^{c_2}) = 0.1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;($n_H^{c_1}$ and $n_H^{c_2}$ are symmetric about $\frac{n}{2}$ i.e. $n_H^{c_1}$+$n_H^{c_2} = n$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$n_H^{c_1}$ gives you the left critical value and $n_H^{c_2}$ gives you the right critical value.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you find $n_H \in (n_H^{c_1},n_H^{c_2})$, then you have do not have sufficient evidence against Null Hypothesis and hence you accept the Null Hypothesis at $10\%$ level of significance. Hence, you accept that the coin is fair at $10\%$ level of significance.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2011-02-12T22:45:09.153" LastActivityDate="2011-02-12T22:45:09.153" CommentCount="4" />
  <row Id="21746" PostTypeId="2" ParentId="21743" CreationDate="2011-02-12T22:50:00.070" Score="1" ViewCount="" Body="&lt;p&gt;If you additionally assume that the functions are in $L^2$ you could use Plancherel to obtain $\|u \ast v\|_2 = \|\hat{u} \hat{f}\|_2$. So now $\hat{f} \geq 0$ so the multiplication operator is order preserving.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I assume you mean the $L^2$ norm in your desired inequality (natural for convolutions).&lt;/p&gt;&#xA;" OwnerUserId="1120" LastActivityDate="2011-02-12T22:50:00.070" CommentCount="1" />
  <row Id="21748" PostTypeId="1" AcceptedAnswerId="21749" CreationDate="2011-02-12T23:05:36.943" Score="4" ViewCount="475" Body="&lt;p&gt;Hello,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I did some experiments, using C++, investigating the values of $\sqrt{1+24n}$. &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; n: 1 -&amp;gt; 5&#xA; n: 2 -&amp;gt; 7&#xA; n: 5 -&amp;gt; 11&#xA; n: 7 -&amp;gt; 13&#xA; n: 12 -&amp;gt; 17&#xA; n: 15 -&amp;gt; 19&#xA; n: 22 -&amp;gt; 23&#xA; n: 35 -&amp;gt; 29&#xA; n: 40 -&amp;gt; 31&#xA; n: 57 -&amp;gt; 37&#xA; n: 70 -&amp;gt; 41&#xA; n: 77 -&amp;gt; 43&#xA; n: 92 -&amp;gt; 47&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I wonder,  if &#xA;$$\sqrt{1+24n}$$ &#xA;is an integer, will it also be a prime?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any interesting theory about this formula?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks,&lt;br&gt;&#xA;Chan&lt;/p&gt;&#xA;" OwnerUserId="5711" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-12T23:11:37.127" LastActivityDate="2011-03-05T01:39:06.597" Title="Does the formula $\sqrt{ 1 + 24n }$ always yield prime ?" Tags="&lt;prime-numbers&gt;&lt;elementary-number-theory&gt;" AnswerCount="5" CommentCount="6" />
  <row Id="21749" PostTypeId="2" ParentId="21748" CreationDate="2011-02-12T23:10:20.350" Score="16" ViewCount="" Body="&lt;p&gt;How about $n=26$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, take a composite number of the form $12k+1$ and take $n = k + 6k^2$ to arrive at a contradiction for your statement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$k=2 \Rightarrow n=26 \Rightarrow \sqrt{1+24n} = 25$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$k=4 \Rightarrow n=100 \Rightarrow \sqrt{1+24n} = 49$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$k=7 \Rightarrow n=301 \Rightarrow \sqrt{1+24n} = 85$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and so on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are infinite composite numbers of the form $(12k+1)$ which gives infinite counterexamples to your claim.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your observation though is a nice one, since $24 | (p^2-1)$, $\forall \text{ primes } p &amp;gt; 3$. So you will find that all the primes $&amp;gt;3$ can be written as $\sqrt{1+24n}$.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2011-02-13T05:03:32.957" LastActivityDate="2011-02-13T05:03:32.957" CommentCount="10" />
  <row Id="21750" PostTypeId="2" ParentId="21748" CreationDate="2011-02-12T23:12:50.150" Score="2" ViewCount="" Body="&lt;p&gt;$\sqrt{1+24\cdot 26} = \sqrt{625} = 25$!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sqrt{1+24\cdot n} = x$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$${1+24\cdot n} = x^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ n = \dfrac{x^2 -1}{24}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if $x=25$, $\dfrac{x^2 -1}{24}$ is an integer.&lt;/p&gt;&#xA;" OwnerUserId="5105" LastEditorUserId="5105" LastEditorDisplayName="" LastEditDate="2011-02-12T23:14:26.310" LastActivityDate="2011-02-12T23:14:26.310" CommentCount="1" />
  <row Id="21751" PostTypeId="2" ParentId="21748" CreationDate="2011-02-12T23:15:10.193" Score="2" ViewCount="" Body="&lt;p&gt;Nope. $\sqrt{1+24*381276} = 3025 = 605 * 5$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are many such formulars which seem to yield only primes, but most of them aren't.&lt;/p&gt;&#xA;" OwnerUserId="5282" LastEditorUserId="5282" LastEditorDisplayName="" LastEditDate="2011-02-12T23:20:12.533" LastActivityDate="2011-02-12T23:20:12.533" CommentCount="4" />
  <row Id="21752" PostTypeId="1" AcceptedAnswerId="21753" CreationDate="2011-02-12T23:49:23.963" Score="2" ViewCount="216" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;Let $P$ be the change-of-basis matrix&#xA;  from a basis $S$ to a basis $S&amp;#39;$ in a&#xA;  vector space $V$. Then, for any vector&#xA;  $v \in V$, we have $$P[v]_{S&amp;#39;}=[v]_{S}&#xD;&#xA; \text{ and hence, }  P^{-1}[v]_{S} =&#xD;&#xA; [v]_{S&amp;#39;}$$&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Namely, if we multiply the coordinates&#xA;  of $v$ in the original basis $S$ by&#xA;  $P^{-1}$, we get the coordinates of&#xA;  $v$ in the new basis $S&amp;#39;$.&quot; - Schaum's&#xA;  Outlines: Linear Algebra. 4th Ed.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am having a lot of difficulty keeping these matrices straight. Could someone please help me understand the reasoning behind (what appears to me as) the counter-intuitive naming of $P$ as the change of basis matrix from $S$ to $S&amp;#39;$? It seems like $P^{-1}$ is the matrix which actually changes a coordinate vector in terms of the 'old' basis $S$ to a coordinate vector in terms of the 'new' basis $S&amp;#39;$...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Added:&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;Consider a basis $S =&#xD;&#xA; \{u_1,u_2,...,u_n\}$ of a vector space&#xA;  $V$ over a field $K$. For any vector&#xA;  $v\in V$, suppose $v = a_1u_1&#xD;&#xA; +a_2u_2+...+a_nu_n$&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Then the coordinate vector of $v$&#xA;  relative to the basis $S$, which we&#xA;  assume to be a column vector (unless&#xA;  otherwise stated or implied), is&#xA;  denoted and defined by $[v]_S =&#xD;&#xA; [a_1,a_2,...,a_n]^{T}$. &quot;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&quot;Let $S = \{ u_1,u_2,...,u_n\}$ be a&#xA;  basis of a vector space $V$, and let&#xA;  $S&amp;#39;=\{v_1,v_2,...,v_n\}$ be another&#xA;  basis. (For reference, we will call&#xA;  $S$ the 'old' basis and $S&amp;#39;$ the 'new'&#xA;  basis.) Because $S$ is a basis, each&#xA;  vector in the 'new' basis $S&amp;#39;$ can be&#xA;  written uniquely as a linear&#xA;  combination of the vectors in S; say,&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$\begin{array}{c} v_1 = a_{11}u_1 +&#xD;&#xA; a_{12}u_2 + \cdots +a_{1n}u_n \\ v_2 =&#xD;&#xA; a_{21}u_1 + a_{22}u_2 + \cdots&#xD;&#xA; +a_{2n}u_n \\ \cdots \cdots \cdots \\ v_n = a_{n1}u_1 + a_{n2}u_2 + \cdots&#xD;&#xA; +a_{nn}u_n \end{array}$&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Let $P$ be the transpose of the above&#xA;  matrix of coefficients; that is, let&#xA;  $P = [p_{ij}]$, where $p_{ij} =&#xD;&#xA; a_{ij}$. Then $P$ is called the&#xA;  \textit{change-of-basis matrix} from&#xA;  the 'old' basis $S$ to the 'new' basis&#xA;  $S&amp;#39;$.&quot; - Schaum's Outline: Linear Algebra 4th Ed.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am trying to understand the above definitions with this example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basis vectors of $\mathbb{R}^{2}: S= \{u_1,u_2\}=\{(1,-2),(3,-4)\}$ and $S&amp;#39; = \{v_1,v_2\}= \{(1,3), (3,8)\}$ the change of basis matrix from $S$ to $S&amp;#39;$ is $P = \left( \begin{array}{cc} -\frac{13}{2} &amp;amp; -18 \\ \frac{5}{2} &amp;amp; 7 \end{array} \right)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My current understanding is the following: normally vectors such as $u_1, u_2$ are written under the assumption of the usual basis that is $u_1 = (1,-2) = e_1 - 2e_2 = [u_1]_E$. So actually $[u_1]_S = (1,0)$ and I guess this would be true in general... But I am not really understanding what effect if any $P$ is supposed to have on the basis vectors themselves (I think I understand the effect on the coordinates relative to a basis). I guess I could calculate a matrix $P&amp;#39;$ which has the effect  $P&amp;#39;u_1, P&amp;#39;u_2,...,P&amp;#39;u_n = v_1, v_2,..., v_n$ but would this be anything?&lt;/p&gt;&#xA;" OwnerUserId="3711" LastEditorUserId="3711" LastEditorDisplayName="" LastEditDate="2011-02-13T21:37:11.067" LastActivityDate="2011-02-13T21:37:11.067" Title="Why is the 'change-of-basis matrix' called such?" Tags="&lt;linear-algebra&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="4" />
  <row Id="21753" PostTypeId="2" ParentId="21752" CreationDate="2011-02-13T00:04:03.777" Score="4" ViewCount="" Body="&lt;p&gt;The situation here is closely related to the following situation: say you have some real function $f(x)$ and you want to shift its graph to the right by a positive constant $a$. Then the correct thing to do to the function is to shift $x$ over to the &lt;em&gt;left&lt;/em&gt;; that is, the new function is $f(x - a)$. In essence you have shifted the graph to the right by shifting the coordinate axes to the left. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this situation, if you have a vector $v$ expressed in some basis $e_1, ... e_n$, and you want to express it in a new basis $Pe_1, .... Pe_n$ (this is why $P$ is called the change of basis matrix), then you multiply the numerical vector $v$ by $P^{-1}$ in order to do this. You should carefully work through some numerical examples to convince yourself that this is correct. Consider, for example, the simple case that $P$ is multiplication by a scalar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The lesson here is that one must carefully distinguish between vectors and the components used to express a vector in a particular basis. &lt;a href=&quot;http://en.wikipedia.org/wiki/Covariance_and_contravariance_of_vectors&quot; rel=&quot;nofollow&quot;&gt;Vectors transform covariantly, but their components transform contravariantly&lt;/a&gt;. &lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2011-02-13T00:04:03.777" CommentCount="6" />
  <row Id="21754" PostTypeId="2" ParentId="21748" CreationDate="2011-02-13T00:08:04.433" Score="6" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\rm\: \mod\ 24\::\ \ x^2 \equiv 1\ \Rightarrow\ (5x)^2 \equiv 1\:,\ $ but $\rm\:5\:x\:$ is prime iff $\rm\: x= \pm1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that this yields a general &lt;em&gt;structural&lt;/em&gt; reason explaining why such integers can't all be primes. Namely, the integers you describe are simply those integers that, when reduced modulo $24\:,$ yield square roots of $1\:.\:$ But such roots are closed under multiplication: $\rm\ x^2\equiv 1,\ y^2\equiv 1\ \Rightarrow\ (xy)^2\equiv 1\:.\:$ But primes are &lt;em&gt;not&lt;/em&gt; closed under multiplication. For example, one can take any of your prime solutions and multiply them to obtain a composite solution, e.g. $\rm\ 5^2 = 25,\: \ 5\cdot 7 = 35\:,\:$ etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice that there are precisely $8\:$ square-roots of $\rm 1\ (mod\ 24)\ $ viz. $\rm \pm 1,\:\pm 5,\:\pm 7,\: \pm 11\:,\:$ corresponding (by $\rm CRT$) to the product of the two roots $\rm\ \pm 1\ (mod\ 3)\ $ times the four roots $\rm\ \pm 1,\: \pm 3\ (mod\ 8)\:.\:$ Note that these are precisely the congruence classes of all the integers coprime to $\:3\:$ and $\rm\:2\:,\:$ which includes &lt;em&gt;all&lt;/em&gt; primes $&amp;gt; 3$. This explains your empirical observations above. The key observation, that $\rm\ x^2\equiv 1\ (mod\ 24)\ \iff\ x\:$ is coprime to $\:6\:,\:$ is nothing but a very special case computation of Carmichael's generalization of Euler's phi-function - see my &lt;a href=&quot;http://math.stackexchange.com/questions/855/for-any-prime-p-3-why-is-p2-1-always-divisible-by-24/910#910&quot;&gt;post here&lt;/a&gt; for details.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2011-02-13T20:13:44.023" LastActivityDate="2011-02-13T20:13:44.023" />
  <row Id="21755" PostTypeId="1" AcceptedAnswerId="21761" CreationDate="2011-02-13T00:29:22.263" Score="5" ViewCount="75" Body="&lt;p&gt;Let $\mathbb{R}(t)$ be the field of rational functions over $\mathbb{R}$ (the fraction field of $\mathbb{R}[x]$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am looking for elements in the Brauer group of the field, and the current idea I have to follow on is to find infinitely many cyclic field extensions, and use those to create cyclic division algebras.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My Galois theory experience is not very rich with transcendental extensions of $\mathbb{R}$ and I'm a bit lost. Am I even on the right path towards the Brauer group? Any ideas on how to prove there are many cyclic extensions?&lt;/p&gt;&#xA;" OwnerUserId="622" LastActivityDate="2011-02-13T01:01:52.320" Title="Cyclic Extensions of $\mathbb{R}(t)$" Tags="&lt;abstract-algebra&gt;&lt;field-theory&gt;&lt;galois-theory&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" />
  <row Id="21756" PostTypeId="2" ParentId="21744" CreationDate="2011-02-13T00:40:34.037" Score="5" ViewCount="" Body="&lt;p&gt;Mistakes in calculations plague mathematicians at all levels. Here's what Vladimir Arnol'd had to say about it:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Every working mathematician knows that if one does not control oneself (best of all by examples), then after some ten pages half of all the signs in formulae will be wrong and twos will find their way from denominators into numerators.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The best way to catch mistakes in calculations is using 'sanity checks'. The most basic versions of these go back to elementary school (did you get a negative number for the volume?) and at the more advanced level you might plug in different values of x in your formula, let x go to infinity, etc. If you proved something about a group or a manifold, see what it gives you in a specific example. If you can think of another way to do the same calculation, that's best of all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you say &quot;I just give up after making the same mistakes and reaching dead ends for two hours, look at the solution and don't understand how I could have come up with it on my own&quot;, it makes me think you're working on something too difficult. Find easier problems to work on first, build up your skills, and then come back to the harder problems later.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, keep poking around different subjects until you find where your strengths are. I've found in my own life that abstract algebra has always been and remains confusing and difficult. Calculus, differential equations, asymptotic analysis came much more easily. Complex analysis used to be torture, but now I've gotten pretty good at it and consider it one of my strengths. If something is really making you suffer, drop it for a while (maybe months or even years!) and then if necessary come back to it later when hopefully your increased mathematical maturity will make it easier. &lt;/p&gt;&#xA;" OwnerUserId="7000" LastActivityDate="2011-02-13T00:40:34.037" CommentCount="1" />
  <row Id="21757" PostTypeId="1" CreationDate="2011-02-13T00:44:07.050" Score="2" ViewCount="215" Body="&lt;p&gt;Hi, I am solving these problems in preparation to a midterm. Here's the problem I have solved before tackling the one I am asking help with:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was given 1 basis vector for $\mathbf{W}$ - $\begin{bmatrix}&#xD;&#xA;1\\ &#xD;&#xA;2\\ &#xD;&#xA;3\\ &#xD;&#xA;4\\&#xD;&#xA;5&#xD;&#xA;\end{bmatrix}$, and I had to find a basis for $\mathbf{W}^{\perp}$. I set up an equation $\begin{bmatrix}&#xD;&#xA;1\\ &#xD;&#xA;2\\ &#xD;&#xA;3\\ &#xD;&#xA;4\\&#xD;&#xA;5&#xD;&#xA;\end{bmatrix}$ dotted with $\begin{bmatrix}&#xD;&#xA;a &amp;amp; b &amp;amp; c &amp;amp; d &amp;amp; e &#xD;&#xA;\end{bmatrix}$ = 0 and found the kernel (it was &#xA;$\begin{bmatrix}&#xD;&#xA;-2b-3c-4d-5e\\ &#xD;&#xA;b\\ &#xD;&#xA;c\\ &#xD;&#xA;d\\ &#xD;&#xA;e&#xD;&#xA;\end{bmatrix}$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, I know that Span of $\mathbb{W}$ = $\begin{bmatrix}&#xD;&#xA;1\\ &#xD;&#xA;2\\ &#xD;&#xA;3\\ &#xD;&#xA;0&#xD;&#xA;\end{bmatrix}$, $\begin{bmatrix}&#xD;&#xA;1\\ &#xD;&#xA;2\\ &#xD;&#xA;3\\ &#xD;&#xA;4&#xD;&#xA;\end{bmatrix}$. I need to find a basis for $\mathbf{W}^{\perp}$, or the sub-space that contains vectors such that if two vectors from $\mathbf{W}$ and $\mathbf{W}^{\perp}$ were dotted, the result would be 0. &#xA;But I am not sure how to go about this similar problem when I am given two vectors. Thanks!&lt;/p&gt;&#xA;" OwnerUserId="7001" LastActivityDate="2011-02-13T01:01:29.387" Title="Find a basis for $\mathbf{W}^{\perp}$ given spanning vectors of $\mathbf{W}$." Tags="&lt;linear-algebra&gt;" AnswerCount="2" />
  <row Id="21758" PostTypeId="2" ParentId="21757" CreationDate="2011-02-13T00:52:57.917" Score="2" ViewCount="" Body="&lt;p&gt;Your same ideas as before will work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take a vector&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\begin{bmatrix}&#xD;&#xA;a\\ &#xD;&#xA;b\\ &#xD;&#xA;c\\ &#xD;&#xA;d&#xD;&#xA;\end{bmatrix}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You want this vector to be perpendicular to $\begin{bmatrix}&#xD;&#xA;1\\ &#xD;&#xA;2\\ &#xD;&#xA;3\\ &#xD;&#xA;0&#xD;&#xA;\end{bmatrix}$ and to $\begin{bmatrix}&#xD;&#xA;1\\ &#xD;&#xA;2\\ &#xD;&#xA;3\\ &#xD;&#xA;4&#xD;&#xA;\end{bmatrix}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You get $a + 2b + 3c = 0$ and $a+ 2b + 3c + 4d = 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From which you get $d = 0$ and $a+2b+3c = 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, $W^{\perp}$ is a $2D$ subspace, with $d=0$ and $a+2b+3c = 0$ i.e. $\begin{bmatrix}&#xD;&#xA;-2b-3c\\ &#xD;&#xA;b\\ &#xD;&#xA;c\\ &#xD;&#xA;0&#xD;&#xA;\end{bmatrix}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Select two pairs $(a,b,c)$ such that the two are linearly independent to get a pair of basis for $W^{\perp}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An example for the two basis are $[3,0,-1,0]$ and $[2,-1,0,0]$&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2011-02-13T00:59:20.610" LastActivityDate="2011-02-13T00:59:20.610" />
  <row Id="21759" PostTypeId="1" AcceptedAnswerId="21764" CreationDate="2011-02-13T00:52:52.097" Score="0" ViewCount="134" Body="&lt;p&gt;This must be a very basic question but I am a finance student just learning some basics about Fourier Transformation to apply to time series analysis. I did a fourier transform on a function in time domain to get the following function in frequency domain:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$Y_1[\omega] = \frac{1}{1-\phi_1 e^{-jw}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$Y_2[\omega] = \frac{1}{1-(\phi_1 + \phi_2)e^{-jw} +\phi_1\phi_2e^{-2jw}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do I find the spectrum of this function for given $\phi_1$ and $\phi_2$ coefficients and in the discretization interval $w = [-\pi:.1*\pi: \pi]$? Then, how do I find the 'magnitude' of spectrum and 'phase' of spectrum?&lt;/p&gt;&#xA;" OwnerUserId="957" LastEditorUserId="957" LastEditorDisplayName="" LastEditDate="2011-02-13T01:34:05.313" LastActivityDate="2011-02-13T01:39:50.787" Title="spectrum and phase of function in frequency domain" Tags="&lt;fourier-analysis&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="21760" PostTypeId="2" ParentId="21757" CreationDate="2011-02-13T01:01:29.387" Score="2" ViewCount="" Body="&lt;p&gt;Okay, first of all you can simplify your basis vectors a bit. You can write&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$W = \operatorname{span} \left\{ \begin{pmatrix} 1 \\ 2 \\ 3 \\ 0 \end{pmatrix}, \begin{pmatrix} 0 \\ 0 \\ 0 \\ 1 \end{pmatrix} \right\}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general you can apply Gram-Schmidt before to get an ON-basis for the subspace.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Call the vectors $v_1$ and $v_2$. Now, if $u \in W^\perp$, $\langle u, v_1 \rangle = \langle u, v_2 \rangle = 0$. Let $u = (a,b,c,d)^T$. Immidiately you get:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\langle u, v_2 \rangle = 0 \Leftrightarrow d = 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\langle u, v_1 \rangle = 0 \Leftrightarrow a + 2b + 3c = 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which gives a parametric solution (set $c = t$ and $b = s$, solve fo $a$):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$u = s \begin{pmatrix} 3 \\ 0 \\ -1 \\ 0 \end{pmatrix} + t \begin{pmatrix} 2 \\ -1 \\ 0 \\ 0 \end{pmatrix}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$W^\perp = \operatorname{span} \left\{ \begin{pmatrix} 3 \\ 0 \\ -1 \\ 0 \end{pmatrix}, \begin{pmatrix} 2 \\ -1 \\ 0 \\ 0 \end{pmatrix} \right\}$$&lt;/p&gt;&#xA;" OwnerUserId="5966" LastActivityDate="2011-02-13T01:01:29.387" />
  <row Id="21761" PostTypeId="2" ParentId="21755" CreationDate="2011-02-13T01:01:52.320" Score="5" ViewCount="" Body="&lt;p&gt;By a famous theorem of Tsen and Lang, the Brauer group of $\mathbb{C}(t)$ is zero.  Thus if you take any class in the Brauer group of $\mathbb{R}(t)$ and restrict it to the quadratic extension $\mathbb{C}(t)$, it becomes zero. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From this it follows that every element of $\operatorname{Br}(\mathbb{R}(t))$ can be represented by a quaternion algebra, so you're on the right track by considering cyclic (quadratic) extensions, of which there are infinitely many.  You need to think about when $\mathbb{R}(t)(\sqrt{f(t)}) = \mathbb{R}(t)(\sqrt{g(t)})$ when $f$ and $g$ are rational functions.  Hint: it is enough to consider the case of squarefree polynomials, and then the above equality implies that $f$ and $g$ have the same roots (in $\mathbb{C}$).  It is very tempting for me to speak geometrically in terms of &lt;em&gt;ramification&lt;/em&gt; but I don't know if you would be comfortable with that.&lt;/p&gt;&#xA;" OwnerUserId="299" LastActivityDate="2011-02-13T01:01:52.320" CommentCount="2" />
  <row Id="21762" PostTypeId="1" AcceptedAnswerId="21765" CreationDate="2011-02-13T01:04:18.157" Score="1" ViewCount="60" Body="&lt;p&gt;Given an $n$-sided polygon, how many ways can you color the vertices using $k$ colors so that no two adjacent vertices have the same color?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Inspired by 2011 AMC 12 A #16 – I'm able to do this for small cases but not in general.)&lt;/p&gt;&#xA;" OwnerUserId="45" LastActivityDate="2011-02-13T01:16:17.020" Title="Coloring points on an n-gon" Tags="&lt;counting&gt;&lt;coloring&gt;" AnswerCount="2" />
  <row Id="21763" PostTypeId="2" ParentId="21748" CreationDate="2011-02-13T01:04:20.353" Score="0" ViewCount="" Body="&lt;p&gt;take n=$24k^2$+$2k$  , $\Rightarrow \sqrt{1+24n}=24k+1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;24k+1, is composite infinitely, to give one such case..&#xA;if $k=24^{2r}$ r=0,1,2,3... then 25 divides $24k+1$ always&lt;/p&gt;&#xA;&#xA;&lt;p&gt;it follows, for $n$=$24^{4r+1}$+$(2.24^{2r})$ , r=0,1,2... the value $\sqrt{1+24n}$ is divisible by 25, and hence definitely not prime.&lt;/p&gt;&#xA;" OwnerUserId="6995" LastEditorUserId="6995" LastEditorDisplayName="" LastEditDate="2011-02-13T02:58:53.800" LastActivityDate="2011-02-13T02:58:53.800" />
  <row Id="21764" PostTypeId="2" ParentId="21759" CreationDate="2011-02-13T01:08:09.557" Score="0" ViewCount="" Body="&lt;p&gt;Let&#xA;\begin{align*}&#xA;Z(\omega) &amp;amp; = 1-\phi e^{-jw} = 1-\phi (\cos(\omega) - j \sin(\omega))\\&#xA; &amp;amp; = 1-\phi \cos(\omega) + j \phi \sin(\omega)\\&#xA;\end{align*}&#xA;Hence, we get that &#xA;\begin{align*}&#xA;|Z(\omega)| = &amp;amp; \sqrt{(1-\phi (\cos(\omega))^2 + (\phi \sin(\omega))^2} = \sqrt{1 + \phi^2 - 2 \phi \cos(\omega)}\\&#xA;\angle{Z(\omega)} = &amp;amp; \tan^{-1}(\frac{\phi \sin(\omega)}{1-\phi \cos(\omega)})&#xA;\end{align*}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$Y(\omega) = \frac{1}{Z(\omega)}$$&#xA;Hence, we get that &#xA;\begin{align*}&#xA;|Y(\omega)| = &amp;amp; \frac{1}{\sqrt{1 + \phi^2 - 2 \phi \cos(\omega)}}\\&#xA;\angle{Y(\omega)} = &amp;amp; -\tan^{-1}(\frac{\phi \sin(\omega)}{1-\phi \cos(\omega)})&#xA;\end{align*}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finding the magnitude and phase for $|Y_2(\omega)|$ is a trivial process once you have done the previous one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$Y_2(\omega) = \frac{1}{1-\phi_1 e^{-jw}} \times \frac{1}{1-\phi_2 e^{-jw}}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence,&#xA;\begin{align*}&#xA;|Y_2(\omega)| = &amp;amp; \frac{1}{\sqrt{1 + \phi_1^2 - 2 \phi_1 \cos(\omega)}} \times \frac{1}{\sqrt{1 + \phi_2^2 - 2 \phi_2 \cos(\omega)}}\\&#xA;\angle{Y_2(\omega)} = &amp;amp; -\tan^{-1}(\frac{\phi_1 \sin(\omega)}{1-\phi_1\cos(\omega)}) -\tan^{-1}(\frac{\phi_2 \sin(\omega)}{1-\phi_2\cos(\omega)})&#xA;\end{align*}&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2011-02-13T01:39:50.787" LastActivityDate="2011-02-13T01:39:50.787" CommentCount="6" />
  <row Id="21765" PostTypeId="2" ParentId="21762" CreationDate="2011-02-13T01:09:48.337" Score="3" ViewCount="" Body="&lt;p&gt;Here are two ways to do the problem, both of which use graph theory. The problem is equivalent to counting the number of closed walks of length $n$ on the complete graph $K_k$ on $k$ vertices. The number of closed walks of length $n$ on a graph $G$ with &lt;a href=&quot;http://en.wikipedia.org/wiki/Adjacency_matrix&quot; rel=&quot;nofollow&quot;&gt;adjacency matrix&lt;/a&gt; $A$ is given by $\text{tr } A^n = \sum_{i=1}^k \lambda_i^n$ where $\lambda_i$ are the eigenvalues of $A$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So it suffices to compute the eigenvalues of the adjacency matrix of $K_k$. This is a nice exercise, but for the sake of completeness I will tell you that they are $k-1, -1, -1, ...$. So the final answer is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(k-1)^n + (k-1)(-1)^n.$$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The second method is to observe that the problem is equivalent to computing the &lt;a href=&quot;http://en.wikipedia.org/wiki/Chromatic_polynomial&quot; rel=&quot;nofollow&quot;&gt;chromatic polynomial&lt;/a&gt; of the cyclic graph $C_n$. This can be done by induction and the &lt;a href=&quot;http://en.wikipedia.org/wiki/Chromatic_polynomial#Deletion.E2.80.93contraction&quot; rel=&quot;nofollow&quot;&gt;deletion-contraction recurrence&lt;/a&gt;, and you get the same answer.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Note that these two solutions generalize in different directions. The first solution suggests a generalization of the problem where you replace $K_k$ with a more complicated graph, whereas the second solution suggests a generalization of the problem where you replace $C_n$ with a more complicated graph. Both problems are in turn special cases of the following problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Definition:&lt;/strong&gt; Let $G, H$ be two (undirected, simple) graphs. A &lt;em&gt;morphism&lt;/em&gt; $f : G \to H$ is a function $f : V_G \to V_H$ sending vertices of $G$ to vertices of $H$ such that, if $u, v \in G$ are connected by an edge, then $f(u), f(v)$ are also connected by an edge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Problem:&lt;/strong&gt; How many morphisms are there from a given graph $G$ to a given graph $H$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The walk problem corresponds to letting $G = C_n$ and letting $H$ be arbitrary, while the coloring problem corresponds to letting $H = K_k$ and letting $G$ be arbitrary. Both of these are very special cases. The general problem is interesting; I have no idea what's known about it.&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2011-02-13T01:16:17.020" LastActivityDate="2011-02-13T01:16:17.020" CommentCount="1" />
  <row Id="21766" PostTypeId="2" ParentId="21762" CreationDate="2011-02-13T01:10:05.293" Score="1" ViewCount="" Body="&lt;p&gt;Do you know about the chromatic polynomial? That has a simple skein formula which can be easily calculated for n-gons. It is a polynomial in k, the number of possible colorings.&lt;/p&gt;&#xA;" OwnerUserId="2941" LastActivityDate="2011-02-13T01:10:05.293" CommentCount="2" />
  <row Id="21767" PostTypeId="1" AcceptedAnswerId="21768" CreationDate="2011-02-13T01:39:38.490" Score="1" ViewCount="104" Body="&lt;p&gt;My book introduces $\mathbf{C}^{\infty}$ as subspace of $F(\mathbb{R},\mathbb{R})$ that consists of 'smooth' functions, that is, functions that are differentiable infinitely many times. It then asks me to tell whether or not $\mathbf{C}^{0}$=$(f\in(\mathbb{R},\mathbb{R})$ such that $f$  is continuous$)$ is a subspace. Is $\mathbf{C}^{0}$ a collection of undiffirentiable (because of 0 as opposed to $\infty$) functions that are continuous? Does it have any elements then? Thanks a lot for clarifying the confusion!&lt;/p&gt;&#xA;" OwnerUserId="5157" LastEditorUserId="5157" LastEditorDisplayName="" LastEditDate="2011-02-13T01:49:33.937" LastActivityDate="2011-02-13T01:59:47.150" Title="Meaning of $\mathbf{C}^{0}$?" Tags="&lt;linear-algebra&gt;&lt;notation&gt;" AnswerCount="1" />
  <row Id="21768" PostTypeId="2" ParentId="21767" CreationDate="2011-02-13T01:42:35.043" Score="7" ViewCount="" Body="&lt;p&gt;Typically, $C^{0}(\Omega)$ denotes the space of functions which are continuous over $\Omega$. The higher derivatives may or may not exist.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$C^{\infty}(\Omega) \subset C^{0}(\Omega)$ since if the function is infinitely &quot;smooth&quot; it has to continuous.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Typically, people use the notation $C^{(n)}(\Omega)$ where $n \in \mathbb{N}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f(x) \in C^{(n)}(\Omega)$, means that $f(x)$ has $n$ derivatives in the entire domain ($\Omega$ denotes the domain of the function) and the $n^{th}$ derivative of $f(x)$ is continuous i.e. $f^{n}(x)$ is continuous.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By convention, $f(x) \in C^{(0)}(\Omega)$ denotes the space of continuous functions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f(x) \in C^{(\infty)}(\Omega)$ if the function is differentiable any number of times. For instance, $e^{x} \in C^{(\infty)}(\mathbb{R})$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An example to illustrate is to consider the following function $f: \mathbb{R} \rightarrow \mathbb{R}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f(x) = 0, x \leq 0$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f(x) = x^2, x&amp;gt;0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This function is in $C^{(1)}(\mathbb{R})$ but not in $C^{(2)}(\mathbb{R})$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, when the domain of the function is the largest set over which the function definition makes sense, then we omit $\Omega$ and write that $f \in C^{(n)}$ the domain being understood as the largest set over which the function definition makes sense.&#xA;Also, note the obvious embedding $C^{(n)} \subseteq C^{(m)}$ whenever $n&amp;gt;m$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For &quot;most&quot; functions, if a function is differentiable $n$ times it is $C^{(n)}$. However, there are functions for which the derivative might exist but the derivative is not continuous. Some people might argue that the ramp function has derivative but the derivative is not continuous. It is incorrect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f(x) = 0$, when $x&amp;lt;0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f(x) = x$ when $x \geq 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the derivative doesn't even exist at $x=0$. So the ramp function is not even differentiable in the first place.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let us take a look at the function $f(x) = x^2 \sin(\frac{1}{x})$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first question is &quot;Is the function even in $C^{(0)}$&quot;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer is not yet since the function is ill-defined at the origin. However if we define $f(0) = 0$, then yes the function is in $C^0$. This can be seen from the fact that $\sin(\frac{1}{x})$ is bounded and hence the function is bounded above by $x^2$ and below by $-x^2$. So as we go towards $0$, the function is bounded by functions which themselves tend to $0$. And the limit is $0$ and thereby the function is continuous.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the next question &quot;Is the function differentiable everywhere?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is obvious that the function is differentiable everywhere except at $0$. At $0$, we need to pay little attention. If we were to blindly differentiate $f(x)$ using the conventional formulas, we get $g(x) = f&amp;#39;(x) = 2x \sin(\frac{1}{x}) + x^2 \times \frac{-1}{x^2} \cos(\frac{1}{x})$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now $g(x)$ is ill-defined for $x=0$. Further $\displaystyle \lim_{x \rightarrow 0} g(x)$ doesn't exist. This is what we get if we use the formula. So can we say that $f(x)$ is not differentiable at the origin. Well no! All we can say is $g(x)$ is discontinuous at $x=0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what about the derivative at $x=0$? Well as I always prefer to do, get back to the definition of $f&amp;#39;(0)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f&amp;#39;(0) = \displaystyle \lim_{\epsilon \rightarrow 0} \frac{f(\epsilon) - f(0)}{\epsilon} = \displaystyle \lim_{\epsilon \rightarrow 0} \frac{\epsilon^2 \sin(\frac{1}{\epsilon})}{\epsilon} = \displaystyle \lim_{\epsilon \rightarrow 0} \epsilon \sin(\frac{1}{\epsilon}) = 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Since $|\sin(\frac{1}{\epsilon})| \leq 1$ so it is bounded).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we find that the function $f(x)$ has a derivative at the origin whereas the function $g(x) = f&amp;#39;(x)$, $\forall x \neq 0$ is not continuous or even well-defined at the origin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we have this function whose derivative exists everywhere but then $f(x) \notin C^{(1)}$ since the derivative is not continuous at the origin.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2011-02-13T01:59:47.150" LastActivityDate="2011-02-13T01:59:47.150" />
  <row Id="21769" PostTypeId="1" CreationDate="2011-02-13T01:46:22.007" Score="7" ViewCount="127" Body="&lt;p&gt;Is there any efficient way , to find for a particular n, the cardinality of set consisting of all numbers coprimes to n, but bigger than m(assuming i know the prime factorisation of n and m)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am looking for the implementation which is simple+fast (like the euler totient/phi function, which given the factorisation of n will just need O(logn) steps). &lt;/p&gt;&#xA;" OwnerUserId="6995" LastActivityDate="2011-03-10T04:24:07.663" Title="Variation on euler totient/phi function" Tags="&lt;number-theory&gt;" AnswerCount="1" FavoriteCount="0" />
  <row Id="21770" PostTypeId="2" ParentId="21704" CreationDate="2011-02-13T01:57:13.430" Score="4" ViewCount="" Body="&lt;p&gt;Here's a slightly different argument than @Sivaram's answer. I'm not sure his makes explicit (or even proves) that the limit is &lt;em&gt;strictly positive&lt;/em&gt;, as requested.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First note that if $a &amp;gt; 0$, then $0 &amp;lt; 1 - e^{-an} &amp;lt; 1$ for all $n$ and so the &quot;partial product&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;P_n = \prod_{k=1}^n (1 - e^{-a k})&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is decreasing and bounded below. The fact that $\lim_n\, P_n$ &lt;strong&gt;&lt;em&gt;exists&lt;/em&gt;&lt;/strong&gt; follows immediately.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To show that the limit is strictly positive, note that for all $n \geq \frac{1}{a} \log 2$, we have that $e^{-a n} \leq \frac{1}{2}$. Let $N(a) \equiv N = \lceil \frac{1}{a} \log 2\rceil$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since for $0 \leq x \leq 1/2$, $\log(1-x) \geq - x - x^2$, we get that&#xA;$$&#xD;&#xA;\prod_{n=N}^\infty (1 - e^{-a n}) = \exp\big( \sum_{n=N}^\infty \log(1-e^{-an}) \big) \geq \exp( -\sum_{n=N}^\infty (e^{-an}+e^{-2an}) ) \geq \exp( - 2\sum_{n=N}^\infty e^{-an} ) .&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, $\sum_{n=N}^\infty \,e^{-a n} \leq \frac{1}{1-e^{-a}}$ and so&#xA;$$&#xD;&#xA;\lim_n \, P_n = P_{N-1} \prod_{m=N}^\infty (1 - e^{-a m}) \geq e^{-2/(1-e^{-a})} P_{N-1} &amp;gt; 0 .&#xD;&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="7003" LastActivityDate="2011-02-13T01:57:13.430" CommentCount="2" />
  <row Id="21771" PostTypeId="2" ParentId="21752" CreationDate="2011-02-13T02:03:13.313" Score="1" ViewCount="" Body="&lt;p&gt;Everybody studying the change of basis affair should work out some simple examples like the following. Consider this basis in $\mathbb{R}^2$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;v_1 = (1,1) \qquad \text{and} \qquad v_2 = (1,-1) \ .&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or, since we are going to stress the bases and coordinates thing, we could write it this way&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;v_1 = (1,1)_e \qquad \text{and} \qquad v_2 = (1,-1)_e \ ,&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;since these are coordinates in the standard basis&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;e_1 = (1,0) \qquad \text{and} \qquad e_2 = (0,1) \ .&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The change of basis matrix from $v$ to $e$ is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;P = &#xD;&#xA;\begin{pmatrix}&#xD;&#xA;1 &amp;amp; 1 \\\&#xD;&#xA;1 &amp;amp; -1&#xD;&#xA;\end{pmatrix} \ .&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, take the vector &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;u = 2v_1 - 3v_2 \ .&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Its coordinates in the $v$ basis are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;u = (2,-3)_v \ .&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to obtain its coordinates in the $e$ (standard) basis, you can do it by hand:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;u = 2v_1 - 3v_2 = 2(1,1)_e -3(1,-1)_e = (2-3, 2+3)_e = (-1, 5)_e \ .&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, you realise that these are &lt;em&gt;exactly the same&lt;/em&gt; operations that you do when performing this matrix multiplication:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;P &#xD;&#xA;\begin{pmatrix}&#xD;&#xA;2 \\\&#xD;&#xA;-3&#xD;&#xA;\end{pmatrix}&#xD;&#xA;=&#xD;&#xA;\begin{pmatrix}&#xD;&#xA;1 &amp;amp; 1   \\\&#xD;&#xA;1 &amp;amp; -1&#xD;&#xA;\end{pmatrix}&#xD;&#xA;\begin{pmatrix}&#xD;&#xA;2 \\\&#xD;&#xA;-3&#xD;&#xA;\end{pmatrix}&#xD;&#xA;=&#xD;&#xA;\begin{pmatrix}&#xD;&#xA;2 - 3 \\\&#xD;&#xA;2 + 3&#xD;&#xA;\end{pmatrix}&#xD;&#xA;=&#xD;&#xA;\begin{pmatrix}&#xD;&#xA;-1 \\\&#xD;&#xA;5&#xD;&#xA;\end{pmatrix}  \ .&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Exercise.&lt;/strong&gt; Maybe now you could redo yourself the proof of the change of basis theorem: take two arbitrary bases $v$ and $e$ in no matter which vector space, related by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;v_i = a^1_i e_1 + \cdots + a^n_i e_n \ , \qquad i = 1, \dots , n \ .&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Write down the change of basis matrix from $v$ to $e$ (that is, put the coordinates of the $v$ vectors as columns, like in the previous example):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;P = &#xD;&#xA;\begin{pmatrix}&#xD;&#xA;a^1_1    &amp;amp; \dots      &amp;amp; a^1_n     \\\&#xD;&#xA;\vdots   &amp;amp; \ddots     &amp;amp; \vdots    \\\&#xD;&#xA;a^n_1    &amp;amp; \dots      &amp;amp; a^n_n&#xD;&#xA;\end{pmatrix}   \ ,&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;take any vector&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;u = b^1v_1 + \cdots + b^nv_n \ ,&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and write down its coordinates in the $v$ basis. Finally, find out its coordinates in the $e$ basis (by hand and with the help of the matrix $P$).&lt;/p&gt;&#xA;" OwnerUserId="664" LastEditorUserId="664" LastEditorDisplayName="" LastEditDate="2011-02-13T02:14:00.417" LastActivityDate="2011-02-13T02:14:00.417" />
  <row Id="21772" PostTypeId="1" AcceptedAnswerId="21773" CreationDate="2011-02-13T02:28:58.550" Score="1" ViewCount="120" Body="&lt;p&gt;Ok this is probably the most absurd question you'll ever read, but it came to my mind, and I cant shake it off. Eulers Identity states that:&#xA;$e^{i\pi}+1=0$. So my ridiculous question is why was it stated this way? Why couldnt it have been $e^{i\pi}=-1$? Are there any reasons for this, or it could have been either of the two, but this one was chosen?&lt;/p&gt;&#xA;" OwnerUserId="2985" LastActivityDate="2011-02-13T02:39:11.993" Title="Why is Euler's Identity stated the way it is?" Tags="&lt;calculus&gt;" AnswerCount="1" CommentCount="6" FavoriteCount="1" />
  <row Id="21773" PostTypeId="2" ParentId="21772" CreationDate="2011-02-13T02:30:56.293" Score="9" ViewCount="" Body="&lt;p&gt;The reason is to get just the $5$ &quot;fundamental&quot; numbers $\pi,e,i,0,1$ into one equation.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2011-02-13T02:39:11.993" LastActivityDate="2011-02-13T02:39:11.993" />
  <row Id="21774" PostTypeId="1" AcceptedAnswerId="21777" CreationDate="2011-02-13T02:33:09.997" Score="1" ViewCount="121" Body="&lt;p&gt;Hello, if I have a function&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;f(1) = 2&#xA;f(2) = 2^2&#xA;f(3) = 2^(2^2)&#xA;f(4) = 2^(2^(2^2))&#xA;etc&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;how can I find a closed form solution to this? Thanks.&lt;/p&gt;&#xA;" OwnerUserId="7005" LastEditorUserId="1543" LastEditorDisplayName="" LastEditDate="2011-05-03T09:48:57.720" LastActivityDate="2011-05-03T09:48:57.720" Title="Finding 2^2^2^2^2 ... n times" Tags="&lt;notation&gt;&lt;functional-equations&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="21775" PostTypeId="2" ParentId="21774" CreationDate="2011-02-13T02:40:51.167" Score="5" ViewCount="" Body="&lt;p&gt;This operation is known as &lt;a href=&quot;http://en.wikipedia.org/wiki/Tetration&quot; rel=&quot;nofollow&quot;&gt;tetration&lt;/a&gt;. There is no closed form solution in terms of the usual arithmetic operations other than the one you wrote.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2011-02-13T02:40:51.167" />
  <row Id="21776" PostTypeId="1" AcceptedAnswerId="21787" CreationDate="2011-02-13T02:43:35.683" Score="6" ViewCount="84" Body="&lt;p&gt;Number theory-question again.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $K$ be the biquadratic field $K=\mathbb{Q}[\sqrt{m},\sqrt{n}]$ where $m,n$ are distinct squarefree integers. Let $\mathcal{O}_K$ denote the ring of integers of $K$. If $\alpha \in K$, prove that $\alpha \in \mathcal{O}_K$ if and only if the trace and the norm of $\alpha$ over $\mathbb{Q}(\sqrt{m})$ are algebraic integers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've been stuck on this one for quite some time now. The left-right direction is easy, but I'm stuck on the other direction. My &quot;intuition&quot; tells me that I should somehow use that an element of a quadratic extension is integral if and only if its norm and its trace are.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd like subtle hints in the right direction. Thanks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: Some minor corrections for clarity.&lt;/p&gt;&#xA;" OwnerUserId="4284" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-13T06:04:29.700" LastActivityDate="2011-02-13T06:04:29.700" Title="If $\alpha \in K$, prove that $\alpha \in \mathcal{O}_K$ if and only if the trace and the norm are integral." Tags="&lt;homework&gt;&lt;algebraic-number-theory&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="21777" PostTypeId="2" ParentId="21774" CreationDate="2011-02-13T02:44:11.593" Score="7" ViewCount="" Body="&lt;p&gt;Well the first thing that might be natural to do is define your function recursively, such as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f(0)=1$$&#xA;$$f(n)=2^{f(n-1)}$$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You'll get stuck pretty quickly solving this though if you aren't familiar with &lt;a href=&quot;http://en.wikipedia.org/wiki/Tetration&quot; rel=&quot;nofollow&quot;&gt;tetration&lt;/a&gt;. Defining a closed form would be quite tricky, especially in terms of basic operations because tetration is actually a basic operation. If you consider the hierarchy of arithmetic functions (also known as &lt;a href=&quot;http://en.wikipedia.org/wiki/Hyper_operator&quot; rel=&quot;nofollow&quot;&gt;hyperoperators&lt;/a&gt; you have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$A_0(a,b)=b+1, \mbox{ this is just the successor function.} $$&#xA;$$A_1(a,b)=a+b, \mbox{ addition, right?}$$&#xA;$$A_2(a,b)=a\times b $$&#xA;$$A_3(a,b)=a^b, $$&#xA;$$A_4(a,b)=a\uparrow \uparrow b $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and so forth. If you look at the wikipedia article there's actually an infinite hierarchy of functions that can be defined recursively. But back to your original question, well you could define it in several ways but they're all going to need to use tetration (or some very funky notation). So we might write:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f(n)=A_4(2,n)$$&#xA;$$f(n)=2\uparrow \uparrow n$$&#xA;$$f(n)=\underset{\mbox{$n$ times}}{\underbrace{2^{2^{\dots^{2^2}}}}}$$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope this helps.&lt;/p&gt;&#xA;" OwnerUserId="1930" LastActivityDate="2011-02-13T02:44:11.593" CommentCount="1" />
  <row Id="21778" PostTypeId="1" AcceptedAnswerId="21789" CreationDate="2011-02-13T02:47:54.913" Score="3" ViewCount="121" Body="&lt;p&gt;For an assignment, I was asked to find the vertical asymptote of the function $$g(x)= \frac{\frac{1}{2}x^3-4x^2+6x}{7x^2-56x+84}.$$ &#xA;According to my text, a reliable method of finding the asymptote is to factor the numerator and denominator, and what left in the denominator that was not cancelled out is the asymptote.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I factored the numerator to $\frac{1}{2}(x^2-6x)(x-2)$, and the denominator factored to $7(x-2)(x-6)$, therefore $(x-2)$ cancelled out, leaving $(x-6)$ in the denominator.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, 6 was not accepted as the answer, and I would like to know why.&lt;/p&gt;&#xA;" OwnerUserId="7006" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-13T04:22:10.507" LastActivityDate="2011-02-13T17:58:49.770" Title="Find the vertical asymptote of a function" Tags="&lt;calculus&gt;&lt;asymptotics&gt;" AnswerCount="3" CommentCount="7" />
  <row Id="21779" PostTypeId="2" ParentId="21645" CreationDate="2011-02-13T02:53:05.200" Score="2" ViewCount="" Body="&lt;p&gt;You want the coefficient of $x^m$ in&#xA;$$ \left(\frac{x^{k+1}-x}{x-1}\right)^n = x^n\left(\frac{x^k-1}{x-1}\right)^n. $$&#xA;The asymptotics really depend on $m,n,k$: you can start with the formula for \emp{all} &quot;compositions&quot;, and multiply it by the probability that a composition has all its numbers bounded. In the regime where the maximum is usually less than $k$, you can approximate the probability by $1$; otherwise it gets more complicated... the formula you quote can be used here by estimating the relative magnitude of the terms, but doing it in general is probably a bit laborious. If you have some $m,n,k$ in mind, try punching the numbers and look which of the summands is significant: probably the order of magnitude is only determined by a few of them.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2011-02-13T02:53:05.200" />
  <row Id="21780" PostTypeId="1" CreationDate="2011-02-13T03:01:56.620" Score="7" ViewCount="141" Body="&lt;p&gt;Let $T(f(t))$=&#xA;$\begin{bmatrix}&#xD;&#xA;f(0) &amp;amp; f(1)\\ &#xD;&#xA;f(2) &amp;amp; f(3)&#xD;&#xA;\end{bmatrix}$ from $P_2$ to $\mathbb{R}^{2\times 2}$. To show that it is not an isomorphism, I need to show that either kernel of the transformation is not equal to the zero element only, or that the image is not the whole target space. I am struggling in showing that either of these is false, dealing with polynomials in transformations is very counter-intuitive. Thanks for help!&lt;/p&gt;&#xA;" OwnerUserId="7008" LastEditorUserId="664" LastEditorDisplayName="" LastEditDate="2011-02-13T07:43:52.620" LastActivityDate="2011-02-13T07:43:52.620" Title="Why is this not an isomorphism?" Tags="&lt;linear-algebra&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" />
  <row Id="21781" PostTypeId="2" ParentId="21780" CreationDate="2011-02-13T03:09:34.227" Score="9" ViewCount="" Body="&lt;p&gt;A polynomial of degree at most 2 is determined by its values at three points. So if $f(0)=f(1)=f(2)=0$, then $f(3)=0$ also. Thus $T$ is not surjective.&lt;/p&gt;&#xA;" OwnerUserId="5203" LastEditorUserId="5203" LastEditorDisplayName="" LastEditDate="2011-02-13T03:14:43.243" LastActivityDate="2011-02-13T03:14:43.243" CommentCount="2" />
  <row Id="21782" PostTypeId="1" CreationDate="2011-02-13T03:13:07.750" Score="1" ViewCount="143" Body="&lt;p&gt;What does twice/thrice as much mean? Is it the same as two/three times as much?&lt;/p&gt;&#xA;" OwnerUserId="7010" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2011-02-13T18:13:53.197" LastActivityDate="2011-02-13T18:13:53.197" Title="Twice as much vs Two times as much vs Double" Tags="&lt;terminology&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="21783" PostTypeId="2" ParentId="21782" CreationDate="2011-02-13T03:16:48.120" Score="5" ViewCount="" Body="&lt;p&gt;This question is probably more a question about the English language than mathematics, but the answer is nonetheless yes. (possibly depending on the context)&lt;/p&gt;&#xA;" OwnerUserId="4284" LastActivityDate="2011-02-13T03:16:48.120" CommentCount="1" />
  <row Id="21784" PostTypeId="1" AcceptedAnswerId="21786" CreationDate="2011-02-13T03:37:48.103" Score="0" ViewCount="34" Body="&lt;p&gt;I'm still confused about order of operations on exponents:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is $x^{a^b}$ equal to $x^{(a^b)}$ or $(x^a)^b$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;*Hmm... $x^{a^b}$ has the source of &lt;code&gt;x^{a^b}&lt;/code&gt;. Whenever I try &lt;code&gt;x^a^b&lt;/code&gt;, it requires me to &quot;clarify by using braces&quot;.&lt;/p&gt;&#xA;" OwnerUserId="2736" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-13T04:00:15.697" LastActivityDate="2011-02-13T04:00:15.697" Title="Exponents Order of Operation" Tags="&lt;arithmetic&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="21785" PostTypeId="1" AcceptedAnswerId="21814" CreationDate="2011-02-13T03:39:46.533" Score="4" ViewCount="199" Body="&lt;p&gt;If I have $m$ distinct boxes, and $n$ distinct balls. I put all of these balls into the boxes with one box possibly containing more than one balls. What is the expected maximum number of balls in one box?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Appreciate your thoughts on solving this problem.&lt;/p&gt;&#xA;" OwnerUserId="6016" LastActivityDate="2011-02-14T05:27:11.477" Title="what is the expected maximum number of balls in the same box" Tags="&lt;probability&gt;" AnswerCount="1" CommentCount="23" />
  <row Id="21786" PostTypeId="2" ParentId="21784" CreationDate="2011-02-13T03:49:07.687" Score="3" ViewCount="" Body="&lt;p&gt;Since $(a^b)^c = a^{bc}$, it makes much more sense to use the convention $$a^{b^c} = a^{\left(b^c\right)}.$$ This is knows as &quot;right associativity&quot;.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2011-02-13T03:49:07.687" />
  <row Id="21787" PostTypeId="2" ParentId="21776" CreationDate="2011-02-13T04:04:33.820" Score="1" ViewCount="" Body="&lt;p&gt;Two hints:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;If $R\subseteq S\subseteq T$ are domains, and $S$ is integral over $R$, and $T$ is integral over $S$, then $T$ is integral over $R$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;What is $[\mathbb{Q}(\sqrt{m})(\alpha):\mathbb{Q}(\sqrt{m})]$ ?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="742" LastActivityDate="2011-02-13T04:04:33.820" CommentCount="3" />
  <row Id="21788" PostTypeId="2" ParentId="21489" CreationDate="2011-02-13T04:05:56.627" Score="0" ViewCount="" Body="&lt;p&gt;One problem is that &quot;class Functor&quot;, &quot;class Monad&quot; are special cases of categorical concepts, namely the strong ones. With Haskell, you are working in the specific category. That may hinder learning category theory in full generality. Of course, this is just my POV, someone can perceive that obstacle as a no-brainer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A concrete example. Try to define &quot;instance Monad&quot;, i.e. a monad, on the category of rings which sends a ring R to the ring of polynomials with coefficients in R.&lt;/p&gt;&#xA;" OwnerUserId="7011" LastActivityDate="2011-02-13T04:05:56.627" CommentCount="7" />
  <row Id="21789" PostTypeId="2" ParentId="21778" CreationDate="2011-02-13T04:07:59.697" Score="3" ViewCount="" Body="&lt;p&gt;You did not finish factoring the numerator. That's where the problem is.&#xA;You started out all right, but then you had to keep factoring that $x^2-6x$: it is not degree $1$, and it is &lt;em&gt;not&lt;/em&gt; irreducible quadratic, so it can still be factored. In fact, it has a factor of $x$. So you really have:&#xA;$$\frac{1}{2}x^3 - 4x^2+6x = \frac{1}{2}\left(x^3 - 8x^2 + 12x\right) = \frac{1}{2}x\left(x^2-8x+12\right) = \frac{1}{2}x(x-6)(x-2).$$&#xA;In fact, &lt;em&gt;both&lt;/em&gt; $x-2$ and $x-6$ cancel out in $g(x)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a function, $g(x)$ is equal to &#xA;$$g(x) = \left\{\begin{array}{ll}&#xA;\frac{1}{14}x &amp;amp;\mbox{if $x\neq 2$ and $x\neq 6$,}\\&#xA;\mbox{undefined} &amp;amp;\mbox{if $x=2$ or $x=6$.}&#xA;\end{array}\right.$$&#xA;So it does not have any vertical asymptotes. &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-13T05:57:34.153" LastActivityDate="2011-02-13T05:57:34.153" CommentCount="1" />
  <row Id="21790" PostTypeId="2" ParentId="21532" CreationDate="2011-02-13T04:13:47.873" Score="0" ViewCount="" Body="&lt;p&gt;As suggested in comments, consider the types of lines that determine each point (perpendicular bisectors of sides, medians, altitudes, and angle bisectors) and how these are affected by the triangle being isosceles.  In particular, consider the symmetry that comes from the triangle being isosceles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In addition, it's worth noting that the circumcenter, centroid, and orthocenter of a triangle are always collinear.  As long as the triangle is not equilateral, they determine a line, which is called the &lt;em&gt;Euler line&lt;/em&gt; of the triangle.&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2011-02-13T04:13:47.873" />
  <row Id="21791" PostTypeId="2" ParentId="21778" CreationDate="2011-02-13T04:23:04.413" Score="3" ViewCount="" Body="&lt;p&gt;Note that $\displaystyle f(x) = \frac{\frac{1}{2}x^3-4x^2+6x}{7x^2-56x+84} = \frac{1}{2}\frac{x^3-8x^2+12x}{7x^2-56x+84} = \frac{1}{2}\frac{1}{7}\frac{x^3-8x^2+12x}{x^2-8x+12} = \frac{x}{14}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the function is &quot;almost&quot; a straight line passing through origin with a slope $\frac{1}{14}$ except at $x=2$ and $x=6$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The function is not defined at $x=2$ and $x=6$. But $\displaystyle \lim_{x \rightarrow 2} = \frac{1}{7}$, $\displaystyle \lim_{x \rightarrow 6} = \frac{3}{7}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are no asymptotes for the problem.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2011-02-13T04:27:26.063" LastActivityDate="2011-02-13T04:27:26.063" CommentCount="3" />
  <row Id="21792" PostTypeId="1" AcceptedAnswerId="21795" CreationDate="2011-02-13T04:37:10.230" Score="13" ViewCount="207" Body="&lt;p&gt;Let $ V $ be a normed vector space (over $\mathbb{R}$, say, for simplicity) with norm $ || \cdot || $.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's not hard to show that if $|| \cdot || = \sqrt{\langle \cdot, \cdot \rangle}$ for some (real) inner product $\langle \cdot, \cdot \rangle$, then the parallelogram equality&#xA;$$ 2||u||^2 + 2||v||^2 = ||u + v||^2 + ||u - v||^2 $$&#xA;holds for all pairs $u, v \in V$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm having difficulty with the converse.  Assuming the parallelogram identity, I'm able to convince myself that the inner product should be&#xA;$$ \langle u, v \rangle = \frac{||u||^2 + ||v||^2 - ||u - v||^2}{2} = \frac{||u + v||^2 - ||u||^2 - ||v||^2}{2} = \frac{||u + v||^2 - ||u - v||^2}{4} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I cannot seem to get that $\langle \lambda u,v \rangle = \lambda \langle u,v \rangle$ for $\lambda \in \mathbb{R}$.  How would one go about proving this?&lt;/p&gt;&#xA;" OwnerUserId="3537" LastActivityDate="2011-02-13T05:11:39.437" Title="Norms Induced by Inner Products" Tags="&lt;functional-analysis&gt;&lt;norm&gt;&lt;inner-product-space&gt;" AnswerCount="1" FavoriteCount="3" />
  <row Id="21793" PostTypeId="2" ParentId="370" CreationDate="2011-02-13T04:39:26.037" Score="0" ViewCount="" Body="&lt;p&gt;Lawvere, Rosebrugh. Sets for Mathematics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pierce B. C. Basic category theory for computer scientists.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;José L. Fiadeiro. Categories for Software Engineering.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Martini. Elements of Basic Category Theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Burstall, Rydeheard. Computational category theory. Requires &lt;a href=&quot;http://en.wikipedia.org/wiki/ML_(programming_language)&quot; rel=&quot;nofollow&quot;&gt;ML&lt;/a&gt; background.&lt;/p&gt;&#xA;" OwnerUserId="7011" LastActivityDate="2011-02-13T04:39:26.037" CommunityOwnedDate="2011-02-13T04:39:26.037" />
  <row Id="21794" PostTypeId="2" ParentId="21640" CreationDate="2011-02-13T04:48:50.433" Score="1" ViewCount="" Body="&lt;p&gt;I &lt;em&gt;think&lt;/em&gt; you have a nil-2 product of $n$ copies of $F$ (rather than two copies of $V$, as I originally wrote in comments). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $A_1,\ldots,A_n$ are abelian groups, then their 2-nilpotent product is their free product modulo the second term of the lower central series of the free product (in general, you take the free product and mod out by the intersection of the second term of the lower central series with the &lt;strong&gt;cartesian&lt;/strong&gt;, the kernel of the map from the free to the direct product).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the abelian case, the elements of the 2-nil product can be written uniquely as&#xA;$$a_1\cdots a_n\gamma$$&#xA;where $a_i\in A_i$, and $\gamma$ is in the commutator subgroup. The commutator subgroup is isomorphic to&#xA;$$\sum_{1\leq i\lt j\leq n} A_j\otimes A_i.$$&#xA;(In the general case, you take the abelianizations). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The multiplication rule is&#xA;$$(a_1\cdots a_n\gamma)(b_1\cdots b_n\delta) = (a_1b_1\cdots a_nb_n)\left(\gamma+\delta+\sum_{1\leq i\lt j\leq n}b_i\otimes a_j\right).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we take each $A_i$ isomorphic to $F$, then the commutator subgroup will be isomorphic to an $F$-vector space of dimension $\binom{n}{2}$, with basis given by $v_{ji}$, where $1\leq i\lt j\leq n$ (identifying $1_j\otimes 1_i$ with $v_{ji}$). If you make the &quot;convention&quot; that $v_{ij} = -v_{ji}$ and $v_{ii}=0$, then you get exactly the wedge $F^n\wedge F^n$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am a little worried about that factor of $2$ in the commutator; there would be no problem with defining the $2$-nil product of groups of exponent $2$ and getting a 2-nil group (of exponent 4), which you cannot do with your set up. Indeed, if I think of this as the $2$-nilpotent product of $n$-copies of $F$, with elements expressed as sums of multiples of $\mathbf{1}_1,\ldots,\mathbf{1}_n$, I would write:&#xA;\begin{align*}&#xA;[\alpha_1\mathbf{1}_1+\cdots+\alpha_n\mathbf{1}_n,\beta_1\mathbf{1}_1+\cdots+\beta_n\mathbf{1}_n] &amp;amp;= \prod_{i,j=1}^n [\alpha_i\mathbf{1}_i,\beta_j\mathbf{1}_j]\\&#xA;&amp;amp;= \prod_{1\leq i\lt j\leq n} [\mathbf{1}_j,\mathbf{1}_i]^{\alpha_j\beta_i-\alpha_i\beta_j}.&#xA;\end{align*}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, you would have&#xA;\begin{align*}&#xA;2(\alpha_1\mathbf{v}_1+\cdots+\alpha_n\mathbf{v}_n)\wedge(\beta_1\mathbf{v}_1 + \cdots + \beta_n\mathbf{v}_n) &amp;amp;=&#xA;2\sum_{i,j=1}^n \alpha_i\mathbf{v}_i\wedge\beta_j\mathbf{v}_j\\&#xA;&amp;amp;= 2\sum_{i,j=1}^n (\alpha_i\beta_j)\mathbf{v}_i\wedge\mathbf{v}_j\\&#xA;&amp;amp;= 2\sum_{1\leq i\lt j\leq n} (\alpha_j\beta_i - \beta_j\alpha_i)\mathbf{v}_i\wedge\mathbf{v}_j,&#xA;\end{align*}&#xA;that is, twice as much as I do. &lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2011-02-13T04:48:50.433" CommentCount="3" />
  <row Id="21795" PostTypeId="2" ParentId="21792" CreationDate="2011-02-13T05:03:22.850" Score="16" ViewCount="" Body="&lt;p&gt;It's not immediate or trivial, so I wouldn't feel too bad for having trouble. This is an exercise in Friedberg, Insel, and Spence's &lt;strong&gt;Linear Algebra&lt;/strong&gt;, 4th Edition, which has an extensive 8 part &quot;Hint.&quot; Here's an edited sequence of hints, following theirs:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;First, prove that the result holds for $\lambda = 2$, that is, $\langle 2u,v\rangle = 2\langle u,v\rangle$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Then, prove that the inner product is linear in the first component: $\langle x+u,v\rangle = \langle x,v\rangle + \langle u,v\rangle$. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Then, prove the result holds for $\lambda$ any positive integer. Then for the reciprocal $\frac{1}{m}$ of any positive integer. Then for any rational number.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Then prove that $|\langle u,v\rangle|\leq ||u||\,||v||$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Then prove that for every $\lambda\in\mathbb{R}$, every $r\in\mathbb{Q}$, you have&#xA;$$|\lambda\langle u,v\rangle - \langle \lambda  u,v\rangle | = |(\lambda-r)\langle u,v\rangle - \langle(\lambda-r)u,v\rangle|\leq 2|\lambda-r|\,||u||\,||v||.$$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Finally, use that to prove homogeneity: for every $\lambda\in\mathbb{R}$, $\langle\lambda u,v\rangle = \lambda\langle u,v\rangle$. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-13T05:11:39.437" LastActivityDate="2011-02-13T05:11:39.437" CommentCount="1" />
  <row Id="21796" PostTypeId="1" CreationDate="2011-02-13T05:27:40.207" Score="9" ViewCount="358" Body="&lt;p&gt;I'm interested in why $$\int_0^{\pi/2} \sin x\,dx = 1.$$ I know how to do the integral the conventional way but am more interested in what makes radians special for this problem. If we instead compute $$\int_{0}^{90} \sin x^\circ\,dx,$$ we won't get $1$ as the answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What about the definition of radians makes this integral evaluate to $1$? I'm looking for an intuitive (presumably geometric) explanation.&lt;/p&gt;&#xA;" OwnerUserId="45" LastEditorUserId="45" LastEditorDisplayName="" LastEditDate="2011-02-13T19:59:48.453" LastActivityDate="2011-02-14T00:00:32.917" Title="Finding $\int_0^{\pi/2} \sin x\,dx$" Tags="&lt;calculus&gt;&lt;integral&gt;&lt;trigonometry&gt;" AnswerCount="3" CommentCount="6" FavoriteCount="1" />
  <row Id="21797" PostTypeId="2" ParentId="21796" CreationDate="2011-02-13T05:41:59.213" Score="6" ViewCount="" Body="&lt;p&gt;Think of the definite integral as giving displacement, and use the fact that the key feature of radians as an angle measure is that the arc length of the unit circle subtended by an angle of radian measure $x$ is $x$ units.    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;More specifically, imagine you are traveling along the unit circle at a speed of 1 rad/s starting at (-1,0) and moving clockwise.  Since you are in radians, this means that you are traveling 1 m/s (to pick a distance unit) on the circle.  When you have traveled $x$ units, your position vector is $(-\cos x, \sin x)$.  The velocity vector must be tangent to the circle and thus is perpendicular to the position vector.  Thus the velocity vector is in the direction of $(\sin x, \cos x)$.  Because the speed is 1 m/s, the velocity vector actually is $(\sin x, \cos x)$ and not some other scalar multiple of it.  Thus $\int_0^{\pi/2} \sin x dx$ when $x$ is in radians is just the horizontal displacement on the unit circle when traveling $\pi/2$ radians along the circle's perimeter from (-1,0) to (0,1).  Since the horizontal displacement is 1, $\int_0^{\pi/2} \sin x dx = 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note the effect of radians in making the speed 1 m/s.  If you were traveling at 1 degree/s, the speed along the circle would be $\frac{\pi}{180}$ m/s, and you would have $\int_0^{90} \sin x \frac{\pi}{180} dx = 1$ as in Sivaram's answer. &lt;/p&gt;&#xA;" OwnerUserId="2370" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2011-02-13T10:23:15.387" LastActivityDate="2011-02-13T10:23:15.387" CommentCount="1" />
  <row Id="21799" PostTypeId="1" AcceptedAnswerId="21801" CreationDate="2011-02-13T05:58:54.103" Score="2" ViewCount="92" Body="&lt;p&gt;If you know the eigenvectors of a specific matrix, is that enough information to get the original matrix back? So for example, for some 2x2 matrix, it is known that its two distinct eigenvectors are [1, u] and [1, v] after normalization, can the original matrix be recovered?&lt;/p&gt;&#xA;" OwnerUserId="7014" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2011-02-13T06:22:42.157" LastActivityDate="2011-02-13T06:22:42.157" Title="Recover matrix from eigenvectors" Tags="&lt;linear-algebra&gt;" AnswerCount="2" />
  <row Id="21800" PostTypeId="2" ParentId="21796" CreationDate="2011-02-13T06:01:16.070" Score="19" ViewCount="" Body="&lt;p&gt;I have had a similar question from my friend. He was surprised to find out $\displaystyle \lim_{x^{\circ} \rightarrow 0} \frac{\sin(x^{\circ})}{x^{\circ}} = \frac{\pi}{180}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason is that we are actually overloading the function name $\sin$. I think the confusion would be subsided if you were to look at $\sin_r(x)$ and $\sin_d(x)$ as two different functions where in the first function you input $x$ in radians and in the second function you input $x$ in degrees and these two are related by $\sin_r(x^r) = \sin_d(y^{\circ})$, where $y^{\circ} =\frac{180}{\pi}x^r$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know that $\displaystyle \int_{0}^{\pi/2} \sin_r(x^r) dx^r = 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, $\displaystyle \int_{0}^{\pi/2} \sin_d(y^{\circ}) dx^r = 1$, since $\sin_r(x^r) = \sin_d(y^{\circ})$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have $(\frac{180}{\pi}x^r) = y^{\circ}$. Hence $(\frac{180}{\pi}dx^r) = dy^{\circ}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As $x^r$ goes from $0$ to $\pi/2$, $y^{\circ}$ goes from $0$ to $90$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, we now have $\displaystyle \int_{0}^{90} \sin_d(y^{\circ}) \frac{\pi}{180} dy^{\circ} = 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, $\displaystyle \int_{0}^{90} \sin_d(y^{\circ}) dy^{\circ} = \frac{180}{\pi}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&quot;Why is $\int_{0}^{\pi/2} \sin(\theta) d\theta = 1$ when $\theta$ is in radians?&quot;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What follows is my attempt for a purely geometric argument for this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The geometric argument ultimately hinges (as one would expect) on&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&quot;The length of an arc of a circle subtending an angle $\theta$ at the center is $r \theta$ where $\theta$ is in radians&quot;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;This essentially comes from the way a radian is defined.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The main crux of the problem is the question&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&quot;Why is the average value of $\sin$ over quarter its period is $\frac{2}{\pi}$?&quot;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Note when we talk of average value of $\sin$ over quarter its period, there is no difference between radians or degrees).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is easy to see that the average value of $\sin$ over quarter its period lies between $0$ and $1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The average value of of $\sin$ over quarter its period is nothing but the average value, with respect to $\theta$, of the height of the line segment as the line segment moves from the right end point of the circle towards the origin such that it is always perpendicular to the $X$ axis, with one end point of the segment on the $X$ axis and the other end point of the segment on the circumference of the circle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/lSRyM.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now here comes the claim. The average value, with respect to $\theta$, of the height of the line segment times the circumference of the quarter of the circle is the area of a unit square.&#xA;&lt;img src=&quot;http://i.imgur.com/jbeBK.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $h$ denote the average value, with respect to $\theta$, of the height of the line segment i.e. $$\displaystyle h = \frac{\int_{0}^{\pi/2} y d \theta}{\int_{0}^{\pi/2} d \theta}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The claim is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle h \times \frac{\pi}{2} = 1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First note that the area of the square with vertices at $(0,0),(0,1),(1,0),(1,1)$ is obviously $1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle \text{Area of square} = 1 = \int_{0}^{1} 1 \times dx = \int_{0}^{1} dx$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For points on the circle, $\displaystyle x^2 + y^2 = 1 \Rightarrow xdx + ydy = 0 \Rightarrow dy = -\frac{x}{y} dx$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Further, for the small triangle, we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle (dx)^2 + (dy)^2 = (1 \times d \theta)^2$$&#xA;$$\displaystyle (dx)^2 + \frac{x^2}{y^2}(dx)^2 = (d \theta)^2$$&#xA;$$\displaystyle \frac{x^2 + y^2}{y^2}(dx)^2 = (d \theta)^2$$&#xA;$$\displaystyle (dx)^2 = y^2 (d \theta)^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that as $\theta$ increases from $0$ to $\frac{\pi}{2}$, $x$ decreases from $1$ to $0$ and hence&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle dx = -y d \theta$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, $$\displaystyle \text{Area of square} = 1 = \int_{0}^{1} 1 \times dx = \int_{0}^{1} dx = \int_{\pi/2}^0 (-y) d \theta = \int_0^{\pi/2} y d \theta$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, $\displaystyle 1 = \int_0^{\pi/2} y d \theta = \frac{\int_0^{\pi/2} y d \theta}{\int_0^{\pi/2} d \theta} \times \int_0^{\pi/2} d \theta$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that $\displaystyle \int_0^{\pi/2} d \theta = \frac{\pi}{2}$ which is the circumference of the quadrant and $\displaystyle h = \frac{\int_0^{\pi/2} y d \theta}{\int_0^{\pi/2} d \theta}$ is the average value, with respect to $\theta$, of the height of the line segment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, we get $\displaystyle h \times \frac{\pi}{2} = 1$ i.e. average value, with respect to $\theta$, of the height of the line segment if $\frac{2}{\pi}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, we get that $$\displaystyle \int_{0}^{\pi/2} \sin(\theta) d \theta = 1$$ when $\theta$ is in radians&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2011-02-14T00:00:32.917" LastActivityDate="2011-02-14T00:00:32.917" CommentCount="3" />
  <row Id="21801" PostTypeId="2" ParentId="21799" CreationDate="2011-02-13T06:03:41.587" Score="2" ViewCount="" Body="&lt;p&gt;No, it's not enough information. For instance, both&#xA;$$\left(\begin{array}{cc}&#xD;&#xA;1 &amp;amp; 0\\&#xD;&#xA;0 &amp;amp; 2&#xD;&#xA;\end{array}\right)\quad\text{and}\quad\left(\begin{array}{cc}&#xD;&#xA;7 &amp;amp; 0\\&#xD;&#xA;0 &amp;amp; -1&#xD;&#xA;\end{array}\right)$$&#xA;have $[1,0]$ and $[0,1]$ as eigenvectors, but they are not even similar matrices (let alone equal).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if you have &lt;em&gt;enough&lt;/em&gt; eigenvectors ($n$ linearly independent eigenvectors in the case of an $n\times n$ matrix) &lt;strong&gt;and&lt;/strong&gt; you know their corresponding eigenvalues, then you can recover the original matrix. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $\mathbf{v}_1,\ldots,\mathbf{v}_n$ are linearly independent eigenvectors with corresponding eigenvalues $\lambda_1,\ldots,\lambda_n$, then letting $P$ be the  matrix that has $\mathbf{v}_1,\ldots,\mathbf{v}_n$ as its columns, in that order, then the &quot;original&quot; matrix will be&#xA;$$A = P\left(\begin{array}{cccc}&#xD;&#xA;\lambda_1 &amp;amp; 0 &amp;amp; \cdots &amp;amp; 0\\&#xD;&#xA;0 &amp;amp; \lambda_2 &amp;amp; \cdots &amp;amp; 0\\&#xD;&#xA;\vdots &amp;amp; \vdots &amp;amp; \ddots &amp;amp; \vdots\\&#xD;&#xA;0 &amp;amp; 0 &amp;amp; 0\cdots &amp;amp; \lambda_n&#xD;&#xA;\end{array}\right)P^{-1}.$$&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2011-02-13T06:03:41.587" />
  <row Id="21802" PostTypeId="2" ParentId="21196" CreationDate="2011-02-13T06:06:46.457" Score="2" ViewCount="" Body="&lt;p&gt;In response to the OP's request, I give two elaborated examples in order to clarify the difference between convergence in probability (denoted by $\stackrel{{\rm P}}{\to}$) and convergence in distribution (denoted by $\stackrel{{\rm D}}{\to}$). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example 1. Suppose that $X_i$, $i=1,2,\ldots$, are non-constant random variables taking values in $[0,M]$, and let $(a_n)$ be a sequence of positive numbers such that $\sum\nolimits_{i = 1}^\infty  {a_i }  =  c &amp;lt; \infty $. Define $S_n = \sum\nolimits_{i = 1}^n {a_i X_i }$. Then, since the sequence $S_n$ is monotone increasing, $S_n$ converges pointwise (that is, for all $\omega \in \Omega$) to a random variable $S$ taking values in $[0,Mc]$. Here, we have the strongest type of convergence (sure convergence), which implies all the other kinds of convergence. In particular, as one would expect, $S_n \stackrel{{\rm P}}{\to} S$. Indeed, this can be shown directly as follows. Fix $\varepsilon &amp;gt; 0$. Then, for all sufficiently large $n$,&#xA;$$&#xD;&#xA;{\rm P}(|S_n  - S| &amp;gt; \varepsilon) = {\rm P}\bigg(\sum\limits_{i = n + 1}^\infty  {a_i X_i }  &amp;gt; \varepsilon \bigg) \leq {\rm P}\bigg(M \sum\limits_{i = n + 1}^\infty  {a_i }  &amp;gt; \varepsilon \bigg) = 0.&#xD;&#xA;$$&#xA;Now, since convergence in probability implies convergence in distribution, $S_n \stackrel{{\rm D}}{\to} S$ as well. However, the limit random variable $S$ plays no special role with regard to the convergence in distribution. Indeed, take, for example, an independent copy $S&amp;#39;$ of $S$. Then, trivially, $S_n \stackrel{{\rm D}}{\to} S&amp;#39;$ (simply because $S$ and $S&amp;#39;$ have the same distribution function). On the other hand, the limit $S$ plays an essential role with regard to the convergence in probability. In fact, it is easy to prove the following general statement: the limit of convergence in probability is unique in the sense that if $Z_n \stackrel{{\rm P}}{\to} X$ and $Z_n \stackrel{{\rm P}}{\to} Y$, then $X = Y$ almost surely, that is ${\rm P}(X \neq Y) = 0$.&#xA;Finally, it is worth noting that if $Z_n \stackrel{{\rm D}}{\to} Z$, and $Z$ is distributed according to a distribution $F$, then we can write $Z_n \stackrel{{\rm D}}{\to} F$. For example, if $Z_n \stackrel{{\rm D}}{\to} Z$ where $Z \sim {\rm exponential}(\lambda)$, then we can write $Z_n \stackrel{{\rm D}}{\to} {\rm exponential}(\lambda)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To further clarify the difference between convergence in probability and convergence in distribution, let's consider the fundamental case of the central limit theorem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example 2. Suppose that $X_1,X_2,\ldots$ is a sequence of i.i.d. random variables with expectation $\mu$ and (finite) variance $\sigma^2 &amp;gt; 0$. Define $S_n = X_1 + \cdots + X_n$ and $Z_n  = \frac{{S_n  - n\mu }}{{\sigma \sqrt n }}$. The central limit theorem states that $Z_n$ converges in distribution to the standard normal distribution ${\rm N}(0,1)$, that is $Z_n \stackrel{{\rm D}}{\to} {\rm N}(0,1)$. So, given any random variable $Z \sim {\rm N}(0,1)$ (which, in particular, may be defined on a different probability space), we can write $Z_n \stackrel{{\rm D}}{\to} Z$. On the other hand, there is no random variable $Z$ such that $Z_n \stackrel{{\rm P}}{\to} Z$. Indeed, suppose for a contradiction that $Z_n \stackrel{{\rm P}}{\to} Z$. It is an easy exercise to show that&#xA;$$&#xD;&#xA;{\rm P}(|Z_n  - Z_m | &amp;gt; 2 \varepsilon ) \le {\rm P}(|Z_n  - Z| &amp;gt; \varepsilon ) + {\rm P}(|Z_m  - Z| &amp;gt; \varepsilon ).&#xD;&#xA;$$&#xA;Now, in order to reach a contradiction, it suffices to realize that $Z_n$ and $Z_m$ become asymptotically independent  as $n,m \to \infty$ with $n/m \to 0$; indeed,&#xA;$$&#xD;&#xA;Z_m  = \sqrt {\frac{n}{m}} Z_n + \sqrt {\frac{{m - n}}{m}} \frac{{\sum\nolimits_{i = n + 1}^m {X_i }  - (m - n)\mu }}{{\sigma \sqrt {m - n} }},&#xD;&#xA;$$&#xA;from which it is also seen that&#xA;$$&#xD;&#xA;{\rm Cov}(Z_n,Z_m) = \sqrt {\frac{n}{m}}. &#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, especially in view of the first example, it is worth noting that convergence in probability, though quite strong relative to convergence in distribution, does not imply almost sure convergence. A short but sophisticated example is given in my answer to &lt;a href=&quot;http://math.stackexchange.com/questions/19848/almost-sure-convergence/19911#19911&quot;&gt;this question&lt;/a&gt;, at the end of the second paragraph.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;: As I commented below, I intentionally gave the non-trivial example of the central limit theorem. Here are two trivial examples. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;First (elaborating on Didier's example), if $X_1,X_2,\ldots$ are i.i.d. from a distribution $F$, then, trivially, $X_n \stackrel{{\rm D}}{\to} F$ (since $X_i \sim F$ for each $i$). But, unless the $X_i$ are deterministic, the sequence never converges in probability. Indeed, suppose that $X_n \stackrel{{\rm P}}{\to} X$. Let $\varepsilon &amp;gt;0 $ be arbitrary but fixed. By the triangle inequality, the event $\lbrace |X_{n+1} - X_n| &amp;gt; 2 \varepsilon \rbrace$ is contained in the event $\lbrace |X_{n+1} - X| &amp;gt; \varepsilon \rbrace \cup \lbrace |X_{n} - X| &amp;gt; \varepsilon \rbrace$. Hence,&#xA;$$&#xD;&#xA;{\rm P}(|X_{n+1}-X_n| &amp;gt; 2 \varepsilon) \leq {\rm P}(|X_{n + 1}  - X| &amp;gt; \varepsilon ) + {\rm P}(|X_n  - X| &amp;gt; \varepsilon ).&#xD;&#xA;$$&#xA;Since, by our assumption, the right-hand side tends to $0$ as $n \to \infty$, and since $|X_{n+1}-X_n|$ is equal in distribution to $Y:=|X_1 - X_2|$, we get ${\rm P}(Y &amp;gt; 2 \varepsilon) = 0$. Since $Y$ is nonnegative, this implies that $Y = 0$ almost surely (exercise), that is, $X_1 = X_2$ almost surely. Hence, the $X_i$ are deterministic (since they are independent).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As another example, suppose that ${\rm P}(X_1 = 1) = {\rm P}(X_1 = -1) = 1/2$, and let $X_{n+1}=-X_n$. Then, trivially, $X_n \stackrel{{\rm D}}{\to} X_1$, but either $(X_n) = (1,-1,1,-1,\ldots)$ or $(X_n) = (-1,1,-1,1,\ldots)$.&lt;/p&gt;&#xA;" OwnerUserId="2810" LastEditorUserId="2810" LastEditorDisplayName="" LastEditDate="2011-02-14T14:57:16.550" LastActivityDate="2011-02-14T14:57:16.550" CommentCount="4" />
  <row Id="21803" PostTypeId="1" AcceptedAnswerId="21805" CreationDate="2011-02-13T06:14:59.463" Score="5" ViewCount="154" Body="&lt;p&gt;I am given a subspace of all polynomials $f(t)$ in $\mathbf{P}_2$ such that $f(1)=0$. I know that a basis for this space is $1-t$, $1-t^{2}$, and when I look at it, it makes perfect sense as to why. I was just wondering what is a systematic way of finding it, without eyeballing. Thanks!&lt;/p&gt;&#xA;" OwnerUserId="7016" LastEditorUserId="7016" LastEditorDisplayName="" LastEditDate="2011-02-13T06:30:54.053" LastActivityDate="2011-02-13T06:36:07.877" Title="How to find a basis for this sub-space?" Tags="&lt;linear-algebra&gt;" AnswerCount="1" />
  <row Id="21804" PostTypeId="2" ParentId="21799" CreationDate="2011-02-13T06:20:59.990" Score="2" ViewCount="" Body="&lt;p&gt;It is not enough, as Arturo points out. Lets stick with $2-$D for a geometric explanation. The geometric reason for this is as follows. The eigenvectors give the direction of maximum and minimum scaling. However, without the eigenvalues we do not know what the scaling is going to be. So we could have two different matrices whose directions of maximum and minimum scaling are the same but the scale factors along these directions are different. So one matrix could deform a circle into an ellipse of high eccentricity while another matrix could deform into an ellipse of low eccentricity though the axes for both the ellipse are the same.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastActivityDate="2011-02-13T06:20:59.990" />
  <row Id="21805" PostTypeId="2" ParentId="21803" CreationDate="2011-02-13T06:24:32.903" Score="10" ViewCount="" Body="&lt;p&gt;Note that there is no such thing as &lt;strong&gt;the&lt;/strong&gt; basis for this subspace: there are only two vector spaces in the entire universe that have just one basis: the zero vector space (okay, this may count as more than one vector space: the zero vector space over &lt;em&gt;any&lt;/em&gt; field), and the one dimensional vector space over a field of $2$ elements. This is neither, so there is no such thing as &quot;the&quot; basis of the subspace: the subspace has &lt;em&gt;lots&lt;/em&gt; of different bases, this is just one of them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, as to this particular problem. &lt;em&gt;Added:&lt;/em&gt; You don't actually need the following paragraph, but it gives you a way to know ahead of time how &quot;big&quot; a basis you are looking for, which may be useful in other situations. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The subspace you want is the nullspace of the linear tranformation $\mathbf{P}_2\to\mathbb{R}$ given by &quot;evaluation at $1$.&quot; By the Rank-Nullity Theorem, you know the nullspace has dimension $2$, so you are looking for two linearly independent polynomials that are $0$ at $1$. &lt;em&gt;Added.&lt;/em&gt; To see how the Rank-Nullity Theorem comes into play, remember that when you interpret the Rank-Nullity Theorem in terms of linear transformations it tells you that if $T\colon\mathbf{V}\to\mathbf{W}$ is a linear transformation, then &#xA;$$\dim(\mathbf{V}) = \mathrm{rank}(T) + \mathrm{nullity}(T) = \dim(\mathrm{Im}(T)) + \dim(\mathrm{ker}(T)).$$&#xA;Here, $\mathbf{V}=\mathbf{P}_2$, $\mathbf{W}=\mathbb{R}$, and $T$ is &quot;evaluation at $1$&quot;. Since $\dim(\mathbb{R}) = 1$, the only two possibilities for the rank of $T$ are $0$ and $1$; but the rank of $T$ is not $0$, because $T$ is not constant $0$ (it maps the polynomial $x$ to $1\neq 0$). So the rank is $1$, and since $\dim(\mathbf{P}_2) = 3$, that means the nullity is $2$. So the subspace you are looking for has dimension $2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A way to think about the Rank-Nullity Theorem is like the &quot;law of conservation of matter&quot;: dimensions are not created nor destroyed, they are just transformed. If you have $\mathbf{V}$ of dimension $n$, then if you add the dimensions you &quot;transform&quot; to $\mathbf{0}$ and the dimensions you get in the image, you should get &lt;em&gt;all&lt;/em&gt; the dimensions (they add up to $n$). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And now, finally, how to get a basis:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A polynomial is $0$ at $1$ if and only if it is divisible by $t-1$. So you are looking for a basis for the polynomials that are multiples of $t-1$, that is, of the form $(t-1)q(t)$, where $q(t)$ is a polynomial of degree $0$ or $1$ (cannot be degree two, since you are in $\mathbf{P}_2$). So if you pick any basis for the subspace of polynomials of degree at most $1$, say $1$ and $t+1$, then you get a basis for your subspace by considering the products $(t-1)1$ and $(t-1)(t+1)$: just take your arbitrary $(t-1)q(t)$, express $q(t)$ in terms of the vectors you took, $1$ and $t+1$, and this gives you how to express $(t-1)q(t)$ in terms of $(t-1)1$ and $(t-1)(t+1)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could also have gotten a basis by taking $1$ and $t$, to get $t-1$ and $t^2-t$; or &lt;em&gt;any&lt;/em&gt; basis for the subspace of polynomials of degree at most $1$ will do.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-13T06:36:07.877" LastActivityDate="2011-02-13T06:36:07.877" CommentCount="3" />
  <row Id="21806" PostTypeId="1" CreationDate="2011-02-13T06:48:54.257" Score="2" ViewCount="186" Body="&lt;p&gt;Given a non-parabolic transformation which is also an orientation preserving isometry in the hyperbolic upper half plane union the boundary, if I know the two fixed points and they are two different irreducible fractions on the boundary, how can I find the corresponding Mobius transformation?&lt;/p&gt;&#xA;" OwnerUserId="7017" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2011-02-13T07:12:18.130" LastActivityDate="2011-02-13T20:11:00.223" Title="Finding mobius transformation from fixed points" Tags="&lt;hyperbolic-geometry&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="21807" PostTypeId="1" AcceptedAnswerId="21813" CreationDate="2011-02-13T07:12:02.143" Score="2" ViewCount="223" Body="&lt;p&gt;Given that:&#xA;$$AB= \left[ {\matrix{&#xD;&#xA;   3 &amp;amp; 1 \cr &#xD;&#xA;   2 &amp;amp; 1 \cr &#xD;&#xA;&#xD;&#xA; } } \right]$$&#xA;and&#xA;$$BA= \left[ {\matrix{&#xD;&#xA;   5 &amp;amp; 3 \cr &#xD;&#xA;   -2 &amp;amp; -1 \cr &#xD;&#xA;&#xD;&#xA; } } \right]$$&#xA;find $A$ and $B$.&lt;/p&gt;&#xA;" OwnerUserId="7018" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-13T19:58:45.447" LastActivityDate="2011-02-15T00:06:54.857" Title="Find matrices $A$ and $B$ given $AB$ and $BA$" Tags="&lt;matrices&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="21808" PostTypeId="1" CreationDate="2011-02-13T07:42:31.477" Score="1" ViewCount="179" Body="&lt;p&gt;I am working on my homework for a class, and I am really stuck&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the question I was given was:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;prove that S(a ⊗ b) = (Sa) ⊗ b&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anyone have any tips on how to solve this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For further explanation this is problem 6a from this &lt;a href=&quot;http://books.google.com/books?id=W-0b09yIgq4C&amp;amp;lpg=PP1&amp;amp;dq=an%20introduction%20to%20continuum%20mechanics&amp;amp;pg=PA9#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;book&lt;/a&gt; on the page the link goes to&lt;/p&gt;&#xA;" OwnerUserId="7019" LastEditorUserId="16" LastEditorDisplayName="" LastEditDate="2011-02-13T15:36:53.567" LastActivityDate="2011-03-15T17:11:05.063" Title="Tensor Algebra help" Tags="&lt;linear-algebra&gt;&lt;homework&gt;" AnswerCount="2" CommentCount="5" FavoriteCount="2" />
  <row Id="21809" PostTypeId="1" AcceptedAnswerId="21982" CreationDate="2011-02-13T07:49:48.367" Score="0" ViewCount="145" Body="&lt;p&gt;First of all, the polynomial $p(x)=x^{3}-x-4$ must have one real root. Let $\theta$ be one of its roots, L=Q($\theta$), $O$ the integral closure of Z in L.&lt;br&gt;&#xA;Then Kummer's theorem tells us then and there that for a prime q not to be nonsplit it is necessary and sufficient that $p(x) \bmod q$ is reducible.&lt;br&gt;&#xA;Nevertheless, it in general is pretty hard to determine whether or not it is true, e.g. hard enough for the case q=the biggest prime ever known until now.&lt;br&gt;&#xA;So, my question is that is there an efficient way determining the solvability of this kind of polynomials, or if I mistakenly thought of something pretty easy as very difficult, please inform me, thanks.&lt;br&gt;&#xA;Edit:An efficient way means it might work for cases such as q=109 since for small enough primes, it suffices to check by hands, such as for 29, (29)=$P_1$$P_2$$P_3$ where $P_1$=29$O$+($\theta-5)O$,$P_2$=29$O$+($\theta-19)O$, $P_3$=29$O$+($\theta+15)O$, and some similar result for (5) while (5) is not totally split in $O$.&lt;/p&gt;&#xA;" OwnerUserId="6792" LastEditorUserId="6792" LastEditorDisplayName="" LastEditDate="2011-02-15T09:02:38.927" LastActivityDate="2011-02-15T09:02:38.927" Title="Decomposition of prime ideals" Tags="&lt;algebraic-number-theory&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="2" />
  <row Id="21811" PostTypeId="2" ParentId="21807" CreationDate="2011-02-13T07:53:47.733" Score="1" ViewCount="" Body="&lt;p&gt;Since $AB$ and $BA$ are both defined, we know that $A$ and $B$ are both $2x2$ matrices.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let:&lt;br&gt;&#xA;$A= \left[ {\matrix{&#xD;&#xA;   x_1 &amp;amp; x_2 \cr &#xD;&#xA;   x_3 &amp;amp; x_4 \cr &#xD;&#xA;&#xD;&#xA; } } \right]$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$B= \left[ {\matrix{&#xD;&#xA;   y_1 &amp;amp; y_2 \cr &#xD;&#xA;   y_3 &amp;amp; y_4 \cr &#xD;&#xA;&#xD;&#xA; } } \right]$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then we get the following equations :  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From $AB$:&lt;br&gt;&#xA;$x_1y_1+x_2y_3=3$&lt;br&gt;&#xA;$x_1y_2+x_2y_4=1$&lt;br&gt;&#xA;$x_3y_1+x_4y_3=2$&lt;br&gt;&#xA;$x_3y_3+x_4y_4=1$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From $BA$:&lt;br&gt;&#xA;$y_1x_1+y_2x_3=5$&lt;br&gt;&#xA;$y_1x_2+y_2x_4=3$&lt;br&gt;&#xA;$y_3x_1+y_4x_3=-2$&lt;br&gt;&#xA;$y_3x_2+y_3x_4=-1$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then we have eight equations in 8 unknowns, and it should be possible to find A and B.  &lt;/p&gt;&#xA;" OwnerUserId="2533" LastEditorUserId="2533" LastEditorDisplayName="" LastEditDate="2011-02-13T08:00:41.203" LastActivityDate="2011-02-13T08:00:41.203" CommentCount="2" />
  <row Id="21812" PostTypeId="1" CreationDate="2011-02-13T07:54:36.960" Score="6" ViewCount="127" Body="&lt;p&gt;I think it is probable because we can consider &#xA;$$ \displaystyle y = &#xD;&#xA;\left\{&#xD;&#xA;\begin{array}{ll}&#xD;&#xA;\sin \left( \frac{1}{x} \right) &amp;amp; \mbox{if }x \neq 0\\&#xD;&#xA;0 &amp;amp; \mbox{if }x=0&#xD;&#xA;\end{array}&#xD;&#xA;\right.$$&#xA;This fuction has intermediate value property but is discontinuous on $x=0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Inspired by this example, let $r_n$ denote the rational number,and define &#xA;$$ \displaystyle y = &#xD;&#xA;\left\{&#xD;&#xA;\begin{array}{ll}&#xD;&#xA;\sum_{n=1}^{\infty} \frac{1}{2^n} \left |\sin \left( \frac{1}{x-r_n} \right) \right| &amp;amp; \mbox{if }x \notin \mathbb{Q}\\&#xD;&#xA;0 &amp;amp; \mbox{if }x \in \mathbb{Q}\\&#xD;&#xA;\end{array}&#xD;&#xA;\right.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is easy to see this funtion is discontinuons if $x$ is not a rational number. But I can't verify its intermediate value property.&lt;/p&gt;&#xA;" OwnerUserId="7020" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2011-02-13T09:06:10.047" LastActivityDate="2011-02-13T09:06:10.047" Title="Can we construct a function $f:\mathbb{R} \rightarrow \mathbb{R}$ such that it has intermediate value property and discontinuous everywhere?" Tags="&lt;analysis&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="2" />
  <row Id="21813" PostTypeId="2" ParentId="21807" CreationDate="2011-02-13T07:55:28.347" Score="11" ViewCount="" Body="&lt;p&gt;First note that both the products are of full rank and hence $A$ and $B$ are also of full rank. So we would expect one set of solutions with one degree of freedom since if $A$ and $B$ satisfy our equations, so will $kA$ and $\frac{1}{k}B$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $$A = \left[ {\matrix{&#xD;&#xA;   a_1 &amp;amp; a_2 \cr &#xD;&#xA;   a_3 &amp;amp; a_4 \cr &#xD;&#xA; } } \right],B = \left[ {\matrix{&#xD;&#xA;   b_1 &amp;amp; b_2 \cr &#xD;&#xA;   b_3 &amp;amp; b_4 \cr &#xD;&#xA; } } \right]$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$AB= \left[ {\matrix{&#xD;&#xA;   3 &amp;amp; 1 \cr &#xD;&#xA;   2 &amp;amp; 1 \cr&#xD;&#xA; } } \right], BA= \left[ {\matrix{&#xD;&#xA;   5 &amp;amp; 3 \cr &#xD;&#xA;   -2 &amp;amp; -1 \cr &#xD;&#xA; } } \right]$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, $$B \left[ {\matrix{&#xD;&#xA;   3 &amp;amp; 1 \cr &#xD;&#xA;   2 &amp;amp; 1 \cr &#xD;&#xA; } } \right] = B(AB) = (BA)B= \left[ {\matrix{&#xD;&#xA;   5 &amp;amp; 3 \cr &#xD;&#xA;   -2 &amp;amp; -1 \cr &#xD;&#xA; } } \right] B$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align*}&#xA;3b_1 + 2b_2 &amp;amp; = 5b_1 + 3b_3\\&#xA;b_1 + b_2 &amp;amp; = 5b_2 + 3b_4\\&#xA;3b_3 + 2b_4 &amp;amp; = -2b_1 - b_3\\&#xA;b_3 + b_4 &amp;amp; = -2b_2 - b_4&#xA;\end{align*}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rearranging, we get,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align*}&#xA;2b_2 &amp;amp; = 2b_1 + 3b_3\\&#xA;b_1 &amp;amp; = 4b_2 + 3b_4\\&#xA;2b_1 + 4b_3 + 2b_4 &amp;amp; = 0\\&#xA;2b_2 + b_3 + 2b_4 &amp;amp; = 0&#xA;\end{align*}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Set $b_2 = k$, and $b_4 = m$, we get $b_1 = 4k+3m$ and $b_3 = -2(k+m)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, $B = \left[ {\matrix{&#xD;&#xA;   4k+3m &amp;amp; k \cr &#xD;&#xA;   -2(k+m) &amp;amp; m \cr &#xD;&#xA; } } \right]$ where $k,m \neq 0$, $B^{-1} = \frac{1}{3m^2+6mk+2k^2} \left[ {\matrix{&#xD;&#xA;   m &amp;amp; -k \cr &#xD;&#xA;   2(k+m) &amp;amp; 4k+3m \cr &#xD;&#xA; } } \right]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, $A = B^{-1} \times \left[ {\matrix{&#xD;&#xA;   5 &amp;amp; 3 \cr &#xD;&#xA;   -2 &amp;amp; -1 \cr &#xD;&#xA; } } \right]$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$A = \frac{1}{3m^2+6mk+2k^2} \left[ {\matrix{&#xD;&#xA;   5m+2k &amp;amp; 3m+k \cr &#xD;&#xA;   2k+4m &amp;amp; 2k+3m \cr &#xD;&#xA; } } \right]$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are two degrees of freedom.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks to Qiang Li for pointing out the other degree of freedom&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2011-02-15T00:06:54.857" LastActivityDate="2011-02-15T00:06:54.857" CommentCount="3" />
  <row Id="21814" PostTypeId="2" ParentId="21785" CreationDate="2011-02-13T08:14:46.317" Score="4" ViewCount="" Body="&lt;p&gt;This is answered in a paper by &lt;a href=&quot;http://www.google.ca/url?sa=t&amp;amp;source=web&amp;amp;cd=5&amp;amp;ved=0CDsQFjAE&amp;amp;url=http%3A%2F%2Fwww14.informatik.tu-muenchen.de%2Fpersonen%2Fsteger%2Fpub%2Fballs.ps.gz&amp;amp;rct=j&amp;amp;q=balls%20into%20bins%20a%20simple%20and%20tight%20analysis&amp;amp;ei=RotXTcqEJM6TtwfynMTPDA&amp;amp;usg=AFQjCNH2RhSSUTerFto4t64Z0y2cq0thzg&amp;amp;cad=rja&quot; rel=&quot;nofollow&quot;&gt;Raab and Steger&lt;/a&gt; (they switch your $n$ and $m$). The case $n=m$ is simpler and had been known before (see their introduction).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Intuitively, in order to find the &quot;correct&quot; answer, you calculate the expected number of bins $X_t$ which get at least $t$ balls; the value of $t$ such that $X_t \approx 1$ should be &quot;close&quot; to the expectation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to make this rigorous, follow these steps:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Find a critical value $t_0$ such that if $t \ll t_0$ then the probability that a given box gets at least $t$ balls is very close to $1$, and if $t \gg t_0$ then it is very close to $0$.&lt;/li&gt;&#xA;&lt;li&gt;When $t \gg t_0$, a union bound shows that with high probability no box gets at least $t$ balls.&lt;/li&gt;&#xA;&lt;li&gt;When $t \ll t_0$, the expected number of boxes with at least $t$ balls is very close to $m$, and so the probability that no box gets at least $t$ balls is very small.&lt;/li&gt;&#xA;&lt;li&gt;Deduce that most of the probability mass of the expectation is concentrated around $t_0$, and so the expectation itself is close to $t_0$.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;When doing the calculations, we hit a problem in step 3: the number of expected boxes with at least $t$ balls is not $m$ but somewhat smaller. Raab and Steger show that the variable $X_t$ is concentrated around its expectation (using the &quot;second moment method&quot;, i.e. Chebyshev's inequality), and so $\Pr[X_t = 0]$ is indeed small.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most of the work is estimating the binomial distribution of the number of balls in each box, finding the correct $t_0$ for each case; the method fails when the number of balls is significantly smaller than the number of bins.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Here is some Sage code:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;def analyze_partition(partition, bins):&#xA;    last = partition[0]&#xA;    counts = [1]&#xA;    for part in partition[1:]:&#xA;        if part != last:&#xA;            last = part&#xA;            counts += [1]&#xA;        else:&#xA;            counts[-1] += 1&#xA;    counts.append(bins - sum(counts))&#xA;    return multinomial(*partition) * multinomial(*counts)&#xA;&#xA;def expected_max(balls, bins):&#xA;    return sum([max(partition) * analyze_partition(partition, bins)&#xA;                for partition in partitions(balls)&#xA;                if len(partition) &amp;lt;= bins])/bins^balls&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;When plugging $10$ balls and $5$ bins, I get $$1467026/390625 \approx 3.76.$$&lt;/p&gt;&#xA;" OwnerUserId="1277" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2011-02-14T05:27:11.477" LastActivityDate="2011-02-14T05:27:11.477" CommentCount="17" />
  <row Id="21816" PostTypeId="1" AcceptedAnswerId="21820" CreationDate="2011-02-13T08:26:57.337" Score="11" ViewCount="351" Body="&lt;p&gt;Is there a geometric way of looking at the relationship between the positive real numbers $a$, $b$ and $c$ if $\frac{1}{a}+\frac{1}{b}=\frac{1}{c}$?&lt;/p&gt;&#xA;" OwnerUserId="1004" LastActivityDate="2011-02-13T14:32:02.743" Title="A geometric look at $\frac{1}{a}+\frac{1}{b}=\frac{1}{c}$?" Tags="&lt;geometry&gt;&lt;algebra&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="1" />
  <row Id="21817" PostTypeId="1" CreationDate="2011-02-13T08:27:11.843" Score="-1" ViewCount="227" Body="&lt;p&gt;I know that if $F$ is a field, and $x,y$ are roots of an irreducible polynomial over $F$ lying in an algebraic extension of $F$, then $F(x) \cong F(y)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But it seems it is not true (in general) that if $x,y,z$ are roots of an irreducible polynomial in an extension of $F$ then $F(x,y) \cong F(x,z)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What might a counterexample look like? I assume I should not be looking at algebraic extensions? Or is there such an example of an algebraic extension?&lt;/p&gt;&#xA;" OwnerUserId="6560" LastEditorUserId="16" LastEditorDisplayName="" LastEditDate="2011-02-13T15:40:58.253" LastActivityDate="2011-02-13T18:59:36.957" Title="Conjoining elements to fields which are the roots of irreducible polynomials." Tags="&lt;abstract-algebra&gt;&lt;algebra&gt;&lt;field-theory&gt;&lt;galois-theory&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="1" />
  <row Id="21818" PostTypeId="2" ParentId="21812" CreationDate="2011-02-13T08:38:54.933" Score="14" ViewCount="" Body="&lt;p&gt;Sure.  The class of functions satisfying the conclusion of the Intermediate Value Theorem is actually vast and well-studied: such functions are called &lt;a href=&quot;http://en.wikipedia.org/wiki/Darboux_function#Darboux_function&quot; rel=&quot;nofollow&quot;&gt;Darboux functions&lt;/a&gt; in honor of &lt;a href=&quot;http://en.wikipedia.org/wiki/Jean_Gaston_Darboux&quot; rel=&quot;nofollow&quot;&gt;Jean Gaston Darboux&lt;/a&gt;, who showed that any derivative is such a function (the point being that not every derivative is continuous).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A standard example of an everywhere discontinuous Darboux function is &lt;a href=&quot;http://en.wikipedia.org/wiki/Conway_base_13_function&quot; rel=&quot;nofollow&quot;&gt;Conway's base 13 function&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Perhaps it is worth noting that the existence of such functions is not due to Conway: his is just a particularly nice, elementary example.  I believe such functions were already well known to &lt;a href=&quot;http://en.wikipedia.org/wiki/Rene_Baire&quot; rel=&quot;nofollow&quot;&gt;Rene Baire&lt;/a&gt;, and indeed possibly to Darboux himself.)&lt;/p&gt;&#xA;" OwnerUserId="299" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2011-02-13T08:45:17.640" LastActivityDate="2011-02-13T08:45:17.640" CommentCount="2" />
  <row Id="21819" PostTypeId="2" ParentId="21812" CreationDate="2011-02-13T09:03:30.113" Score="3" ViewCount="" Body="&lt;p&gt;Any function that give all the reals numbers on any open interval is an example. You can find other examples of this in the answers to &lt;a href=&quot;http://mathoverflow.net/questions/32126/function-with-range-equal-to-whole-reals-on-every-open-set/49137#49137&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; question on MathOverflow.&lt;/p&gt;&#xA;" OwnerUserId="3078" LastActivityDate="2011-02-13T09:03:30.113" />
  <row Id="21820" PostTypeId="2" ParentId="21816" CreationDate="2011-02-13T09:25:27.433" Score="16" ViewCount="" Body="&lt;p&gt;If you have two poles of length $a$ and $b$, and string wire from the top of one pole to the bottom of the other, and vice versa, the height $c$ at which the wires  intersect will be given by half the harmonic mean of the height of the two poles, $\displaystyle \dfrac{1}{c} = \dfrac{1}{a} + \dfrac{1}{b}$, no matter how far apart they are.  This can be easily shown using similar triangles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can also think of one more place where the harmonic mean is commonly found in a geometric context:  Given a right triangle with legs $a$, $b$, the height of the altitude to the hypotenuse is given by half the harmonic mean of the squared lengths. &lt;a href=&quot;http://en.wikipedia.org/wiki/Right_triangle&quot; rel=&quot;nofollow&quot;&gt;Wikipedia provides&lt;/a&gt; a handy diagram so I'll just quickly write it out:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://upload.wikimedia.org/wikipedia/commons/6/68/Teorema.png&quot; alt=&quot;Similar triangles using the altitude&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Here $A$, $B$, $C$ are the angles opposite the lowercase sides, as is usual, and $P$ is the point where the altitude intersects the hypotenuse.) Then, since $\bigtriangleup ABC \; \sim \; \; \bigtriangleup CBP \;$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\dfrac{a}{f} = \dfrac{c}{b}  \Rightarrow \dfrac{a^2}{f^2} = \dfrac{c^2}{b^2} = \dfrac{a^2 + b^2}{b^2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, by cross-multiplying we get,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\dfrac{1}{f^2} = \dfrac{a^2 + b^2}{a^2b^2} = \dfrac{1}{a^2} + \dfrac{1}{b^2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;half the harmonic mean of the squared lengths.&lt;/p&gt;&#xA;" OwnerUserId="6296" LastEditorUserId="6296" LastEditorDisplayName="" LastEditDate="2011-02-13T11:52:18.873" LastActivityDate="2011-02-13T11:52:18.873" CommentCount="0" />
  <row Id="21821" PostTypeId="2" ParentId="21516" CreationDate="2011-02-13T10:02:25.587" Score="2" ViewCount="" Body="&lt;p&gt;Use &lt;a href=&quot;http://integrals.wolfram.com/index.jsp&quot; rel=&quot;nofollow&quot;&gt;Wolfram Online Integrator&lt;/a&gt;, for example. The general answer is given in terms of the &lt;a href=&quot;http://mathworld.wolfram.com/IncompleteGammaFunction.html&quot; rel=&quot;nofollow&quot;&gt;Incomplete Gamma Function&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2810" LastActivityDate="2011-02-13T10:02:25.587" CommentCount="4" />
  <row Id="21822" PostTypeId="2" ParentId="21806" CreationDate="2011-02-13T10:03:21.247" Score="1" ViewCount="" Body="&lt;p&gt;It is an important and beautiful accident that the orientation preserving isometry group of the upper half plane $\bf{H}$ is $PSL(2,\bf{R})$.  The action of the matrix $A = [[a,b],[c,d]]$ is given by $A\cdot z = (az + b)/(cz + d)$.  In addition, we allow the point $\infty = 1/0$: this has image $a/c$.  Likewise the real number $-d/c$ is sent to infinity.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now to your question: Suppose that $p, q$ are distinct real numbers, thus two points in $\partial \bf{H}$. Then there is a hyperbolic geodesic connecting them.  A calculation using the above reveals that there is a one-parameter family of hyperbolic isometries fixing $p, q$.   To get you started here is the first line of the computation supposing that you have explicit, simple, values of $p, q \in \bf{R}$ in mind. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fix $A$.  Then $A \cdot p = p$ implies that $ap + b = cp^2 + dp$.  Solve this, do the same for $q$ and remember that $kA = A$ for non-zero $k \in \bf{R}$. (That is what the $P$ stands for in $PSL$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a final note - as the comment suggests it is also possible to find a matrix $B$ sending $p, q$ to $0, \infty$, solve the problem for those very special values, and conjugate back.  But perhaps it is important to understand the problem both ways. &lt;/p&gt;&#xA;" OwnerUserId="1307" LastActivityDate="2011-02-13T10:03:21.247" CommentCount="3" />
  <row Id="21824" PostTypeId="2" ParentId="21631" CreationDate="2011-02-13T10:58:38.547" Score="3" ViewCount="" Body="&lt;p&gt;What you want (and much more), is given at &lt;a href=&quot;http://en.wikipedia.org/wiki/Beta-binomial_distribution&quot; rel=&quot;nofollow&quot;&gt;Beta-binomial distribution&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2810" LastActivityDate="2011-02-13T10:58:38.547" CommentCount="1" />
  <row Id="21825" PostTypeId="1" AcceptedAnswerId="21831" CreationDate="2011-02-13T11:00:31.750" Score="3" ViewCount="93" Body="&lt;p&gt;Suppose $\{\alpha n\}$ is the fractional part of $\alpha n$. Put&#xA;$$A_{\alpha}(n) = \#\{\{\alpha k\} &amp;lt; 1/\sqrt{k} : k \leq n\}.$$&#xA;If $\alpha$ is irrational, can I find some constant $K$ such that $A_{\alpha}(n) &amp;lt; K \sqrt{n}$ for all $n$? Does the order of $A_{\alpha}(n)$ depend on $\alpha$? Suppose $1/\sqrt{k}$ is replaced by some function $f(k)$. What can I say about the number of $\{\alpha n\}$ less than $f(n)$ as $n$ tends to infinity?&lt;/p&gt;&#xA;" OwnerUserId="899" LastEditorUserId="434" LastEditorDisplayName="" LastEditDate="2011-02-13T11:30:36.263" LastActivityDate="2011-02-13T11:55:04.967" Title="Estimating $\#\{\{\alpha k\} &lt; 1/\sqrt{k} : k \leq n\}$ for irrational $\alpha$." Tags="&lt;analysis&gt;&lt;irrational-numbers&gt;&lt;diophantine-approximation&gt;&lt;ergodic-theory&gt;" AnswerCount="1" />
  <row Id="21826" PostTypeId="1" AcceptedAnswerId="21846" CreationDate="2011-02-13T11:09:17.270" Score="0" ViewCount="99" Body="&lt;p&gt;I would like to show that if $K$ is a field with $p^n$ elements then its characteristic has to be $p$, $p$ prime. I'm not sure where to start. It's clear to me that I can construct a field of order $p^n$ if I have a ring of characteristic $p$ by taking the ring of polynomials and quotienting it by an irreducible polynomial of degree $n$. But this doesn't help here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many thanks for your help!&lt;/p&gt;&#xA;" OwnerUserId="5798" LastActivityDate="2011-02-13T14:19:37.253" Title="Proving $|K| = p^n \Rightarrow char(K) = p$" Tags="&lt;abstract-algebra&gt;&lt;field-theory&gt;&lt;finite-fields&gt;" AnswerCount="4" CommentCount="1" FavoriteCount="1" />
  <row Id="21827" PostTypeId="2" ParentId="21826" CreationDate="2011-02-13T11:20:46.203" Score="2" ViewCount="" Body="&lt;p&gt;Since the additive group has order $p^n$ it is obivious that the (additive) order of every element is a power of $p$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now assume $py \neq 0$ for some $y$ (with $p = 1 + 1 + ... + 1$, p times) then $p\neq 0$. Now $px = 0$ implies that $x = 0$, so there are no elements of order $p$; a contradiction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore $py = 0$ and every element has additive order p.&lt;/p&gt;&#xA;" OwnerUserId="5189" LastActivityDate="2011-02-13T11:20:46.203" />
  <row Id="21828" PostTypeId="2" ParentId="21826" CreationDate="2011-02-13T11:35:58.520" Score="4" ViewCount="" Body="&lt;p&gt;There is a unique map of rings with unity ${\Bbb Z}\rightarrow K$. Since the target is finite, the kernel must be a non trivial ideal $I$ of ${\Bbb Z}$. Moreover, $I$ must be a prime ideal, since the target is a domain (a field, in fact). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand the map maps $p^n=|K|$ to $0$, so that $p^n\in I$ and &#xA;$p^n{\Bbb Z}\subset I$. But the only prime ideal containing $p^n{\Bbb Z}$ is $p{\Bbb Z}$. Therefore $I=p{\Bbb Z}$ and &#xA;${\Bbb F}_p={\Bbb Z}/p{\Bbb Z}\hookrightarrow K$ so that $K$ has characteristic $p$.&lt;/p&gt;&#xA;" OwnerUserId="688" LastActivityDate="2011-02-13T11:35:58.520" />
  <row Id="21829" PostTypeId="1" CreationDate="2011-02-13T11:41:22.070" Score="1" ViewCount="95" Body="&lt;p&gt;It is known that if K is a ﬁnite Galois extension of Q with Galois group G, then G is generated by the inertia groups of ramiﬁed primes in the extension.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does the statement hold for infinite Galois extensions?&lt;/p&gt;&#xA;" OwnerUserId="7024" LastActivityDate="2011-03-17T02:33:32.423" Title="Inertia groups generating Galois groups" Tags="&lt;algebraic-number-theory&gt;" AnswerCount="0" CommentCount="1" FavoriteCount="1" />
  <row Id="21830" PostTypeId="2" ParentId="21816" CreationDate="2011-02-13T11:42:11.787" Score="0" ViewCount="" Body="&lt;p&gt;(should be a comment to Billare's answer) In your pole-and-wire example, the crossing height is correctly given in the formula, but $c$ is &lt;em&gt;half&lt;/em&gt; the harmonic mean of $a$ and $b$. See &lt;a href=&quot;http://en.wikipedia.org/wiki/Harmonic_mean&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Harmonic_mean&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="875" LastEditorUserId="875" LastEditorDisplayName="" LastEditDate="2011-02-13T14:32:02.743" LastActivityDate="2011-02-13T14:32:02.743" CommentCount="1" />
  <row Id="21831" PostTypeId="2" ParentId="21825" CreationDate="2011-02-13T11:55:04.967" Score="4" ViewCount="" Body="&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Equidistribution_theorem&quot; rel=&quot;nofollow&quot;&gt;Weyl equidistribution theorem&lt;/a&gt; says that for irrational $\alpha$ and sufficiently many $k$, the fractional parts $\{\alpha k\}$ will be equidistributed in the interval $[0,1]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To apply this to your particular problem, consider a large interval $k\in[N,2N]$. Because of equidistribution, the numbers $k$ will &quot;hit&quot; the condition $\{\alpha k\} &amp;lt; 1/\sqrt{N}$ roughly $1/\sqrt{N}$ of the time, i.e. there will be roughly $N&amp;#183;1/\sqrt{N} = \sqrt{N}$ numbers $k$ from the interval that fulfill the condition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, we are actually interested in the condition $\{\alpha k\} &amp;lt; 1/\sqrt{k}$, so we have overestimated things a bit, but it will still work out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, piecing intervals together by choosing $N=2^M$ as a sequence of powers of two, we obtain an estimate along the lines of&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$A_\alpha(n=2^M) \lesssim \sqrt{1} + \sqrt{2^1} + \sqrt{2^2} + .. + \sqrt{2^{M-1}} \leq K \sqrt{2}^M = K\sqrt{n} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;as desired. You might have to fill in some epsilons and stuff to make the proof precise, but this is the core argument.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the general case, a similar argument will yield a good bound as long as the function $f(k)$ doesn't vanish too fast. If it does vanish very fast, then it can only get better, though a better bound might be harder to prove.&lt;/p&gt;&#xA;" OwnerUserId="434" LastActivityDate="2011-02-13T11:55:04.967" CommentCount="3" />
  <row Id="21832" PostTypeId="1" CreationDate="2011-02-13T12:20:04.720" Score="0" ViewCount="42" Body="&lt;p&gt;I know this is rather simple, but despite my best efforts and quite a bit of searching through google I can't seem to find a satisfactory answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An example of what I'm trying to do:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given a value N= 6.8e-4 and upper and lower errors for that value of Nuperr = 9.1e-4 and Nlowerr = 3.9e-4 &lt;/p&gt;&#xA;&#xA;&lt;p&gt;With an equation such as 1.6e-9*(N)*x&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How would I go about plotting this point at x=1 on a log-log scale, with error-bars given by those I've specified?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whenever I try to plot this I get error bars which are far too large, since from what I've read online simply adding N+Nuperr doesn't work for log log plots.&lt;/p&gt;&#xA;" OwnerUserId="7023" LastActivityDate="2011-02-13T12:33:19.133" Title="Log-log plotting question" Tags="&lt;logarithms&gt;" AnswerCount="1" />
  <row Id="21833" PostTypeId="2" ParentId="21816" CreationDate="2011-02-13T12:21:08.360" Score="6" ViewCount="" Body="&lt;p&gt;Not necessarily geometric, but if $a$ and $b$ are electrical resistances placed in parallel, then their combined resistance is $c$.&lt;/p&gt;&#xA;" OwnerUserId="4726" LastActivityDate="2011-02-13T12:21:08.360" />
  <row Id="21834" PostTypeId="2" ParentId="21832" CreationDate="2011-02-13T12:33:19.133" Score="1" ViewCount="" Body="&lt;p&gt;I'll only note that when performing an arbitrary transformation $f(u)$ on an uncertain number of the form $x\pm\epsilon$, the formula you'll need goes like&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f(x\pm\epsilon)=f(x)\pm f^{\prime}(x)\epsilon$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(hint: Taylor)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now apply this to the transformations needed for doing log-log plots...&lt;/p&gt;&#xA;" OwnerUserId="7025" LastActivityDate="2011-02-13T12:33:19.133" CommentCount="1" />
  <row Id="21836" PostTypeId="2" ParentId="21555" CreationDate="2011-02-13T12:50:03.360" Score="1" ViewCount="" Body="&lt;p&gt;As $\gamma$ is generally used for curves, i will be using $\Gamma$ and $\Gamma&amp;#39;$ for the fuschian groups and $p$ and $p&amp;#39;$ the corresponding covering maps. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;a) Let the two groups be conjugate by a function $h$ i.e $h\Gamma h^{-1}= \Gamma&amp;#39;$. Define $f(x)=h(y)$ for some $y\in p^{-1}(x)$.  We will check that this is well defined.  That is we need to check $p(y)=p(y&amp;#39;)$ then $p&amp;#39;(h(y))=p&amp;#39;(h(y&amp;#39;))$.  $p(y)=p(y&amp;#39;)$ implies that $y&amp;#39;=g(y)$ for some $g\in \Gamma$.  $h\circ g\circ h^{-1} = g&amp;#39;\in \Gamma&amp;#39;$ which implies that $h\circ g = g^{-1}\circ h$. $h(y&amp;#39;)=h(g(y))=g&amp;#39;(h(y))$ thus $p&amp;#39;(h(y))=p&amp;#39;(h(y&amp;#39;))$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This map is clearly a local isometry. An assumption that $f$ is nor injective(surjective) will imply a similar statement for $h$, a contradiction.  Thus $f$ is bijective.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;To prove part b) i will use the following standard results from covering space theory(c.f Hatcher: Algebraic topology)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) For a fixed point $z_0\in p^{-1}(x_0)$ ($p$ is the covering map) the set $\Gamma$ corresponds one-one to $p^{-1}(x_0)$ via $g\in\Gamma\leftrightarrow g(z_0)\in p^{-1}(x_0)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) For a fixed point $z_0\in p^{-1}(x_0)$ the group of deck transformations $\Gamma$ and $\pi_1(\mathbb{H}/\Gamma,x_0)$ are isomorphic.  The isomorphism is: an element $\gamma\in \pi_1(\mathbb{H}/\Gamma,x_0)$ is sent to the deck transformation such that $z_0\mapsto \tilde{\gamma}(1)$.($\tilde{\gamma}$ denotes the lift of $\gamma$).&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;b)If $S$ and $S&amp;#39;$ are isometric, the isometry being $f$.  Fix points $x_0\in S$, $y_0\in p^{-1}(x_0)$, $x_1\in S&amp;#39;$ and $y_1\in p&amp;#39;^{-1}(x_1)$.  This induces a map $\tilde{f}$ from $\mathbb{H}$ to itself as follows: Consider any point $y\in \mathbb{H}$ and take a curve $\gamma\in \pi_1(S,x_0)$ such that the lift starting at $y_0$ ends at $y$.  Define $\tilde{f}(y):= \tilde{\alpha}(1))$ where $\alpha = f\circ \gamma$.  This clearly a local isometry as you said.  I will prove that it is bijective also.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First i will prove that it is injective.  Suppose $\tilde{f}(z)=\tilde{f}(z&amp;#39;)=y$.  Let $p(z)=x$ and $\tilde{\gamma}$ be any curve from $z$ to $z&amp;#39;$ and $\gamma:= \tilde{\gamma}$.  Now, $f(\gamma)\in \pi_1(S&amp;#39;,f(x))$ and $\tilde{f(\gamma)}$ starts and ends at $y$ hence $f{(\gamma)}$ is the identity in $\pi_1(S&amp;#39;,f(x))$.  But a homeomorphism between two spaces induces an isomorphism between their fundamental groups.  Hence $\gamma$ has to be identity, which implies that $x=x&amp;#39;$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Surjectivity:Consider a $y\in \mathbb{H}$.  If $x=p&amp;#39;(y)$ then there exists atleast one $y&amp;#39;\in p^{-1}(x)$ such that $y&amp;#39;=\tilde{f}(z)$ for some $z$, as otherwise $x$ will not be in the image of $f$.  Fix this $y&amp;#39;$.  Take an element $\gamma&amp;#39;\in\pi_1(S&amp;#39;,x)$ such that $\tilde{\gamma&amp;#39;}(1)=y$. If $\gamma$ is defined as $f^{-1}(\gamma&amp;#39;)$ then $\tilde{f}(\tilde{\gamma}(1))=y$. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Pardon me if i have made some silly error.&lt;/p&gt;&#xA;" OwnerUserId="2893" LastActivityDate="2011-02-13T12:50:03.360" CommentCount="1" />
  <row Id="21837" PostTypeId="1" AcceptedAnswerId="21848" CreationDate="2011-02-13T13:19:45.363" Score="2" ViewCount="192" Body="&lt;p&gt;Given a Lie group, how are you meant to find its Lie algebra?  The lie algebra of a lie group is the set of all the left invariant vector fields, but how would you determine them?  My group is the set of all affine maps (x -&gt; A.x+v) from Rn to Rn under function composition.&lt;/p&gt;&#xA;" OwnerUserId="6968" LastActivityDate="2011-02-13T16:37:01.520" Title="How do you find the Lie algebra of a Lie group (in practice)?" Tags="&lt;lie-groups&gt;&lt;lie-algebra&gt;" AnswerCount="3" CommentCount="4" />
  <row Id="21838" PostTypeId="1" AcceptedAnswerId="21841" CreationDate="2011-02-13T13:20:47.263" Score="2" ViewCount="82" Body="&lt;p&gt;A group $G$ is called an FC-group if $|x^G|$ is finite for each $x \in G$ (here $x^G$ is the conjugacy class of $x$ in $G$).  Equivalently, $|G:C_G(x)|$ is finite for each $x \in G$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think it should be completely obvious that a homomorphic image of an FC-group is an FC-group &amp;amp; maybe I'm just tired, but I've been trying to think of a proof of this for a few minutes now and am coming up blank.  It's one of these things that is left unproved whenever FC-groups are mentioned, presumably because the proof is straightforward.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could somebody throw a proof my direction to ease my troubled mind?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you.&lt;/p&gt;&#xA;" OwnerUserId="2209" LastActivityDate="2011-02-13T14:00:03.407" Title="Homomorphic images of FC groups" Tags="&lt;group-theory&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="21839" PostTypeId="1" AcceptedAnswerId="21842" CreationDate="2011-02-13T13:43:23.930" Score="1" ViewCount="225" Body="&lt;p&gt;I've never seen this notation before, and I'm having trouble finding a reference through search.  Could someone explain what these notations mean for me?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In context, the statement they're in is the following:&#xA;a bounded $f$ is Riemann integrable iff&#xA;$$\varliminf_{||C||\to 0}\mathcal{L}(f; C)\ge\varlimsup_{||C||\to 0}\mathcal{U}(f;C)$$&#xA;where $C$ is a non-overlapping, finite, exact cover of a rectangular region $J$ in $\mathbb{R}^N$, $||C||$ denotes mesh size, and $\mathcal{L}, \mathcal{U}$ represent the lower and upper Riemann sums, respectively.&lt;/p&gt;&#xA;" OwnerUserId="1769" LastEditorUserId="5363" LastEditorDisplayName="" LastEditDate="2011-02-13T14:31:50.293" LastActivityDate="2011-02-14T02:59:16.300" Title="What does limit notation with an underline or an overline mean?" Tags="&lt;notation&gt;" AnswerCount="3" CommentCount="9" FavoriteCount="1" />
  <row Id="21840" PostTypeId="2" ParentId="21826" CreationDate="2011-02-13T13:50:22.840" Score="0" ViewCount="" Body="&lt;p&gt;Consider the prime field of $K$, $P$ ,and then $K$ is a vector space over $P$ so that $x^{d}=p^{n}$ for some natural integer d in which x is the order of $P$.&lt;br&gt;&#xA;Moreover, since the order of the prime field must be a prime, it is equal to p in fact, hence we obtain the result.&lt;/p&gt;&#xA;" OwnerUserId="6792" LastActivityDate="2011-02-13T13:50:22.840" />
  <row Id="21841" PostTypeId="2" ParentId="21838" CreationDate="2011-02-13T14:00:03.407" Score="4" ViewCount="" Body="&lt;p&gt;Applying the &lt;a href=&quot;http://en.wikipedia.org/wiki/Isomorphism_theorem#First_isomorphism_theorem&quot; rel=&quot;nofollow&quot;&gt;First isomorphism theorem&lt;/a&gt;, this boils down to proving that the quotient of an FC group by a normal subgroup is again FC.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is true, because if two classes in the quotient are conjugate, you can find representatives in the group that were already conjugate. The conjugacy class can therefore only get smaller after dividing out.&lt;/p&gt;&#xA;" OwnerUserId="367" LastActivityDate="2011-02-13T14:00:03.407" CommentCount="2" />
  <row Id="21842" PostTypeId="2" ParentId="21839" CreationDate="2011-02-13T14:00:24.843" Score="9" ViewCount="" Body="&lt;p&gt;It's definitely liminf and limsup. Maybe this notation is more common in Europe than in America? For example, the &lt;a href=&quot;http://de.wikipedia.org/wiki/Limes_superior_und_Limes_inferior&quot; rel=&quot;nofollow&quot;&gt;German Wikipedia page&lt;/a&gt; mentions it as an alternative. A well-known book that uses this notation is Hörmander's &lt;em&gt;The Analysis of Linear Partial Differential Operators&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2011-02-13T14:00:24.843" CommentCount="2" />
  <row Id="21843" PostTypeId="2" ParentId="21839" CreationDate="2011-02-13T14:01:30.860" Score="1" ViewCount="" Body="&lt;p&gt;I, however, have seen this notation, called upper and lower limits defined as follows.&lt;br&gt;&#xA;Given a sequence of numbers, ${a_n}$, we may consider the lower bound: inf{$a_n$|n&gt;m} as $b_m$. And then take the upper bound of it, $sup_m$$b_m$, called as the greatest lower bound denoted by the limit notation with an underline.&lt;br&gt;&#xA;If, nevertheless, you have been acquainted with this notation and found that it's not what you want, then it must be because of my limited sight; if it is the case, please inform me, thanks.        &lt;/p&gt;&#xA;" OwnerUserId="6792" LastEditorUserId="6792" LastEditorDisplayName="" LastEditDate="2011-02-14T02:38:24.477" LastActivityDate="2011-02-14T02:38:24.477" CommentCount="2" />
  <row Id="21844" PostTypeId="2" ParentId="21837" CreationDate="2011-02-13T14:07:00.313" Score="2" ViewCount="" Body="&lt;p&gt;Like Qiaochu says, thinking of this as a matrix group is a good first step. Then you need to find one-parameter smooth paths $A(t)$ through the unit element $I$ corresponding to every tangent direction, where $A(0)=I$. Taking the derivative of these smooth paths at $0$ yields a basis for the Lie algebra.&lt;/p&gt;&#xA;" OwnerUserId="2941" LastActivityDate="2011-02-13T14:07:00.313" CommentCount="2" />
  <row Id="21845" PostTypeId="2" ParentId="21837" CreationDate="2011-02-13T14:08:13.617" Score="2" ViewCount="" Body="&lt;p&gt;To expand Qiaochu's comment: Your $G$ can be seen as the (closed) subgroup of $\operatorname{GL}_{n+1}(\mathbb{R})$ given by the matrices of the form&#xA;$$\begin{pmatrix} A &amp;amp; v \\\ 0 &amp;amp; 1 \end{pmatrix}$$ with the usual group multiplication and acting on the affine subspace of vectors of the form $\begin{pmatrix} x \\\ 1 \end{pmatrix}$. If you understand why $\mathfrak{gl}_{n}(\mathbb{R}) = \operatorname{Mat}_{n}(\mathbb{R})$ with the commutator bracket $[X,Y] = XY - YX$ then you should have no difficulty in proving that $\mathfrak{g} = \operatorname{Lie} G$ is given by the matrices of the form $\begin{pmatrix} X &amp;amp; y \\\ 0 &amp;amp; 0 \end{pmatrix}$ with the commutator bracket.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For me, the easiest way to determine the Lie algebra of a matrix group is to think of it as the tangent space at the identity, constructed by tangent vectors of paths through the identity.&lt;/p&gt;&#xA;" OwnerUserId="5363" LastActivityDate="2011-02-13T14:08:13.617" CommentCount="4" />
  <row Id="21846" PostTypeId="2" ParentId="21826" CreationDate="2011-02-13T14:19:37.253" Score="1" ViewCount="" Body="&lt;p&gt;First of all, thanks for your help. I used your helpful posts to come up with this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;claim:&lt;/strong&gt; $|K|=p^n  \Rightarrow char(K)=p$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;proof:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the additive group of $K$. Then $\forall g \in K: \hspace{2mm} |g| \Big | p^n$. In particular, $|1| \Big | p^n$. Now let $k$ be the smallest integer such that  $$ 0 = 1^{p^k} = (1 + \dots + 1)+ \dots +(1 + \dots +1) = \bar{p} + \dots + \bar{p}$$ $k$ times. By definition, this is the characteristic of $K$. But if $$ 0 = \bar{p}^{k} = \bar{p}\bar{p}^{k-1}$$ then there is a contradiction because either $\bar{p} = 0$ or  $\bar{p}^{k-1} = 0$ and both $1$ and $k-1$ are smaller than $k$. Therefore the characteristic has to be $p$.&lt;/p&gt;&#xA;" OwnerUserId="5798" LastActivityDate="2011-02-13T14:19:37.253" />
  <row Id="21847" PostTypeId="1" CreationDate="2011-02-13T14:25:59.300" Score="0" ViewCount="112" Body="&lt;p&gt;This is a dice game where the host throws the dice and the players bet on 3 numbers - numbers below 7, 7 and numbers above 7. On numbers below and above 7 the players get double their stake and on 7 the players get 5 times their stake. For example if i bet 100 bucks on &quot;below&quot; and 7 comes, I will lose, but if 6 comes, then i win 200 bucks. Similarly if someone has betted 100 bucks on 7 and if 7 comes, he wins 500 bucks. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the best way to bet in this game?&lt;/p&gt;&#xA;" OwnerUserId="7030" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2011-02-13T18:11:07.290" LastActivityDate="2011-02-13T18:11:07.290" Title="Strategy for 7 up 7 down game" Tags="&lt;probability-theory&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="21848" PostTypeId="2" ParentId="21837" CreationDate="2011-02-13T14:28:45.680" Score="5" ViewCount="" Body="&lt;p&gt;Here a worked out example:&#xA;What is the Lie algebra of the group of rotations in 3-dimensional space, $SO(3)$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Matrices $A\in SO(3)$ are defined by the property that they are invertible and that the scalar product is invariant $\langle A\vec x,A\vec y\rangle = \langle \vec x,\vec y\rangle$ for all vectors $\vec x,\vec y\in \mathbb R^3$. Expanding the latter condition into coordinates, you can see it is equivalent to $A^TA = I$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To find the Lie algebra, take a smooth path $A(t)$ with $A(0) = I$. In first order, it can be written as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$A(t) = I + t&amp;#183;H + \mathcal O(t^2) .$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Plugging this into the condition $A^TA=I$, we get in first order&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$I = (I+t&amp;#183;H)^T(I+t&amp;#183;H) = I + t&amp;#183;(H^T + H) .$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, the condition on the tanget vector $H$ is that it is antisymmetric&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ H^T = -H .$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, the Lie algebra of the rotation group $SO(3)$ consists of antisymmetric 3x3 matrices, which must have the form&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ H = \begin{pmatrix}0 &amp;amp; -\omega_3 &amp;amp; \omega_2 \\ \omega_3 &amp;amp; 0 &amp;amp; -\omega_1 \\ -\omega_2 &amp;amp; \omega_1 &amp;amp; 0\end{pmatrix} .$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is not difficult to show that exponentiating any such matrix will yield an element that preserves the scalar product. Hence, this is the whole Lie algebra.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the way, the elements of the Lie algebra $so(3)$ are usually represented by the angular velocity $\vec\omega=(\omega_1,\omega_2,\omega_3)$ such that multiplication becomes the cross product&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ H&amp;#183;\vec x = \vec\omega \times \vec x .$$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;A similar method applies to your original problem. For instance, you can embed affine transformations of $\mathbb R^n$ into $GL_{n+1}(\mathbb R)$ by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ (x \mapsto Ax + v) \iff \begin{pmatrix} A &amp;amp; v \\ 0 &amp;amp; 1 \end{pmatrix} \in GL_{n+1}(\mathbb R).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can express this as an algebraic condition: a matrix $B\in GL_{n+1}(\mathbb R)$ represents an affine mapping of $\mathbb R^n$ if and only if&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ (0,0,…,1)&amp;#183;B = (0,0,…,1) .$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This will induce an equation for the first derivative of a path $B(t)$ and as above, you will obtain the Lie algebra as a set of matrices.&lt;/p&gt;&#xA;" OwnerUserId="434" LastEditorUserId="434" LastEditorDisplayName="" LastEditDate="2011-02-13T16:37:01.520" LastActivityDate="2011-02-13T16:37:01.520" CommentCount="3" />
  <row Id="21849" PostTypeId="2" ParentId="21847" CreationDate="2011-02-13T14:33:37.707" Score="7" ViewCount="" Body="&lt;p&gt;As they say, &quot;the only winning move is not to play.&quot; I assume there are two dice. In that case, the expected value of all three bets is the same, and it's negative: on average, you lose $\frac{1}{6}$ of what you bet. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some details: when rolling $2$ dice the probabilities of getting a sum of $2, 3, 4, ... 12$ are $\frac{1}{36}, \frac{2}{36}, \frac{3}{36}$, and so on, increasing until $7$ and then decreasing. This is a nice exercise if you've never done it; you just need to count the number of ways a pair of numbers between $1$ and $6$ can add up to a given number. It follows that the probability of getting a sum of $7$ is $\frac{1}{6}$, and by symmetry the probability of getting above $7$ or below $7$ are each $\frac{5}{12}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It follows that if you bet $7$ then on average you multiply your stake by $\frac{5}{6}$, whereas if you bet either below or above then on average you also multiply your stake by $\frac{5}{6}$. So all three options are equally bad, in the long run.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some meta-lessons to learn from this computation:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Generally the expected value of a game like this is negative, or the host would never bother to offer it. For example, the expected value of every game at a casino and the lottery has to be negative, or else the casino (resp. the institution running the lottery) would not make money. This is as much a lesson about incentives as it is about probability.&lt;/li&gt;&#xA;&lt;li&gt;If one of the options was better than the others, at some point people would be aware of this and play only that option.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2011-02-13T14:38:45.717" LastActivityDate="2011-02-13T14:38:45.717" />
  <row Id="21850" PostTypeId="2" ParentId="21817" CreationDate="2011-02-13T14:34:17.063" Score="2" ViewCount="" Body="&lt;p&gt;The problem is, although p(x) is irreducible over F, it might not be so over F(x), and hence F(x,y) and F(x,z) might not be isomorphic.&lt;br&gt;&#xA;I hope you will accept this intuitive explanation as a supplement of the comment above by@KCd.&lt;/p&gt;&#xA;" OwnerUserId="6792" LastActivityDate="2011-02-13T14:34:17.063" />
  <row Id="21851" PostTypeId="1" AcceptedAnswerId="21930" CreationDate="2011-02-13T14:48:22.493" Score="18" ViewCount="266" Body="&lt;p&gt;I stumbled across the problem of connecting the points on a $n, n$ grid with a minimal amount of straight lines without lifting the pen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $n=1, n=2$ it is trivial. For $n=3$ you can find the solution with a bit trial and error (I will leave this to the reader as it is a fun to do, you can do it with 4 lines). I found one possible solution for a $4,4$ grid and animated it, it uses 6 lines and is probably optimal (will hopefully help you to understand the problem better, the path doesn't have to be closed like in the animation, open ends are allowed!):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/4D4a1.gif&quot; alt=&quot;Problem&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now my question is, for higher $n$, is there a way to get the amount of minimal lines to use and does an algorithm exist to find a actual solution? I think its quite hard to model the &quot;straight lines&quot; with graph theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit:&#xA;Reading Erics excellent answer I found the following website: &lt;a href=&quot;http://www.mathpuzzle.com/dots.html&quot; rel=&quot;nofollow&quot;&gt;http://www.mathpuzzle.com/dots.html&lt;/a&gt; that also gives an algorithm to connect the points in $2n-2$ steps, solutions up to $10,10$ and mentions:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Toshi Kato conjectures: On&#xA;  $(2N+1)x(2N+1)$ grid, $N \geq 2$, Using $4N$&#xA;  continuous lines, and not lifting your&#xA;  pencil from the paper, can go through&#xA;  all the dots of a $(2N+1)x(2N+1)$ grid,&#xA;  ending at the same place started. But&#xA;  must visit at least one dot twice in&#xA;  the route.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;On $(2N)x(2N)$ grid, $N \geq 2$, Using $4N-2$&#xA;  continuous lines, and not lifting your&#xA;  pencil from the paper, can go through&#xA;  all the dots of a $(2N)x(2N)$ grid,&#xA;  ending at the same place started. And&#xA;  can visit each dots just once.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It seems to be an open problem to show that $2n-2$ is optimal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also I found the following page with a proof that in the $3,3$ grid there cannot be $2$ parallel lines: &lt;a href=&quot;http://fahim-patel.blogspot.com/2011/01/proof.html&quot; rel=&quot;nofollow&quot;&gt;http://fahim-patel.blogspot.com/2011/01/proof.html&lt;/a&gt; I think it might be interesting for coming up with a proof that $2n-2$ is optimal (however maybe there is no such proof, as we only saw solutions for very small $n$, for bigger $n$ there might be some developments we don't know about).&lt;/p&gt;&#xA;" OwnerUserId="3123" LastEditorUserId="3123" LastEditorDisplayName="" LastEditDate="2011-02-14T09:10:39.717" LastActivityDate="2011-02-14T22:26:48.727" Title="Connecting a $n, n$ point grid" Tags="&lt;geometry&gt;&lt;graph-theory&gt;&lt;puzzle&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="3" />
  <row Id="21852" PostTypeId="1" AcceptedAnswerId="21900" CreationDate="2011-02-13T15:01:18.927" Score="8" ViewCount="208" Body="&lt;p&gt;In asymptotic analysis and analytic number theory, one often has to deal with complex integrals over infinite contours in the complex plane, and the required techniques to do so often go beyond the standard courses of complex analysis in one variable. In particular, I am interested in the following type of argument which I don´t fully understand and which I will try to illustrate by an example (taken from Paris and Kaminski, &quot;Asymptotics and Mellin-Barnes Integrals&quot;):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the so called Cahen-Mellin integral&#xA;$$e^{-z}=\frac{1}{2\pi i}\int_{(c)}\Gamma(s)z^{-s}ds,\ \arg(z)&amp;lt;\frac{\pi}{2}, z\neq 0,$$&#xA;which is an integral representation of the exponential function by taking the inverse Mellin transform of the Gamma function, where the integration contour $(c)$ stands for the vertical line $\{\Re(s)=c\}$ with some $c&amp;gt;0$. It can be shown that the integrand has &quot;the controlling behavior&quot; $|z|^{-\sigma}O(|t|^{\sigma-{1\over 2}}e^{t\arg(z)-{1\over 2}\pi|t|})$ as $|t|\to\infty$, where $s=\sigma + it$. Now, aside from the obvious way to show the validity of the above integral representation, Paris and Kaminski argue that because of the aforementioned exponential decay of the integrand we are allowed to move the contour of integration over the poles of the Gamma function and use the residues of the latter to obtain the exponential series. This is precisely the argument I want to understand, so I will try to break it down into few smaller questions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;(Q1)&lt;/strong&gt; How does the exponential decay of the integrand allow us to displace the contour of intagration over singularities? Is there a more general setup where the asymptotic behavior of the integrand allows for moving the contour of integration through and over singularities?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;(Q2)&lt;/strong&gt; After the displacement of the integration contour (still a vertical line), what kind of a residue theorem allows for considering all of the infinitely many singularities of the gamma function? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The version of the residue theorem I know uses bounded interior of a (simply) closed contour in the complex plane and only finitely many residues contained in there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Remark 1:&lt;/strong&gt; In order to compute the above integral in a classical way, I would take a finite line segments of the vertical line, symmetric with respect to the positive real axis, i.e. $\{Re(s)=c, -r_n\leq\Im(s)\leq r_n\}$, construct circle segments with radii $r_n$ encompassing each of the poles, with $r_n\uparrow \infty$ suitably chosen such that no poles lie on the segment contours, and then show that the integrals over the half-circles tend to zero as $n\to\infty$, thus obtaining on the one hand the integral over the infinite vertical line and on the other hand the infinite sum of the residues. However, I haven´t really checked whether the exponential decay of the integrand would suffice for the half-circle integrals to vanish in the limit. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Remark 2:&lt;/strong&gt; I could imagine that there might be a version of the residue theorem suitably formulated for the Riemann sphere resp. $\bar{\mathbb{C}}:=\mathbb{C}\cup\{\infty\}$, where basically infinite contours from the complex plane correspond to closed ones on the sphere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;(Q3)&lt;/strong&gt; Where could one find a more systematic treatment of integrals over infinite contours in the complex plane, including contour shifting over poles, other types of contour modifications, usage of infinitely many residues, as well as other techniques for the exact computation of such integrals? I understand that such techniques are often to be applied &quot;individually&quot;, thus such literature would ideally contain a few good examples. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance for your attention and sorry if I appear to sound too confused :-), I am only trying to fill in certain &quot;gaps&quot; in my knowledge of complex analysis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS: I am &lt;em&gt;not&lt;/em&gt; interested in numerical computations or &lt;em&gt;general&lt;/em&gt; asymptotic expansions for contour integrals (even though in the above example the residues &quot;expansion&quot; appears as a special case thereof).  &lt;/p&gt;&#xA;" OwnerUserId="5994" LastEditorUserId="5994" LastEditorDisplayName="" LastEditDate="2011-02-13T21:10:05.747" LastActivityDate="2011-02-13T22:29:24.760" Title="Techniques to compute complex integrals over infinite contours" Tags="&lt;complex-analysis&gt;&lt;reference-request&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="3" />
  <row Id="21853" PostTypeId="1" CreationDate="2011-02-13T15:04:45.527" Score="-2" ViewCount="182" Body="&lt;p&gt;Let $(\Omega,A)$ and  $(\Omega&amp;#39;,A&amp;#39;)$ be two measurable space and $\mu$ be a $\sigma$-finite measure on $(\Omega,A)$. Let $T$ be a measurable function from $\Omega$ to $\Omega&amp;#39;$. Define the measure $\mu^*$ on $(\Omega&amp;#39;,A&amp;#39;)$ by $ \mu^*(B&amp;#39;)=\mu(T^{-1}(B))$. Show that $\mu^*$ is a measure. Is it necessarily $\sigma$- finite? Prove or give an counter example.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose that both $\mu$  and $\mu^*$ are $\sigma$-finite,then for any $\mu^*$ -integrable function $f$, Show that $\int f(\omega&amp;#39;)d\mu^*(\omega&amp;#39;)=\int f\circ T(w)d\mu(\omega)$.&lt;/p&gt;&#xA;" OwnerUserId="6990" LastEditorUserId="367" LastEditorDisplayName="" LastEditDate="2011-02-13T15:11:47.687" LastActivityDate="2011-02-13T15:11:47.687" Title="Change of measure" Tags="&lt;homework&gt;&lt;measure-theory&gt;" CommentCount="4" />
  <row Id="21854" PostTypeId="2" ParentId="21808" CreationDate="2011-02-13T15:08:44.507" Score="0" ViewCount="" Body="&lt;p&gt;I think this is an easy problem using the universal property of tensor products provided that I can understand the question, it seems that the precise definition in the linked book is blocked and it is truly weird o find S operating on both tensors , $a$ tensor product $b$ and vectors, $a$.&lt;br&gt;&#xA;As far as I am concerned, it is either a typing problem or an unexplained notation arising error.&lt;/p&gt;&#xA;" OwnerUserId="6792" LastActivityDate="2011-02-13T15:08:44.507" />
  <row Id="21855" PostTypeId="2" ParentId="20751" CreationDate="2011-02-13T15:09:21.987" Score="3" ViewCount="" Body="&lt;p&gt;This is a posed problem in the College Mathematics Journal. January 2011. Is there really an Adam?&lt;/p&gt;&#xA;" OwnerUserId="7032" LastActivityDate="2011-02-13T15:09:21.987" CommentCount="1" />
  <row Id="21856" PostTypeId="1" CreationDate="2011-02-13T15:12:14.577" Score="0" ViewCount="40" Body="&lt;p&gt;I have a dataset with yearly levels of corruption in a number of countries.&#xA;year, corruption&#xA;2001, 5&#xA;2002, 7&#xA;etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to test whether corruption is affected by a change in power (defined as the election of a new president who isn't part of the same political party as the previous one).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My take at this is to look at the slope in the corruption measure (∂corruption / ∂year) two years before and after the new president takes office. So any change change in slope after the new president (good or bad) is treated as a change. It's likely that some presidents will increase corruption, others reduce it and some will have no impact.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any thoughts on a good statistical method to test this?&lt;/p&gt;&#xA;" OwnerUserId="7033" LastActivityDate="2011-02-13T15:12:14.577" Title="Measure change in two slopes for a set of data" Tags="&lt;statistics&gt;" CommentCount="2" />
  <row Id="21857" PostTypeId="2" ParentId="21817" CreationDate="2011-02-13T15:17:54.357" Score="4" ViewCount="" Body="&lt;p&gt;Let $P$ be an irreducible polynomial in $F[X]$ and pick $x,y,z$ three roots of $P$ in an extension.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You do have $F(x) \cong F[X]/P \cong F(y)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to go further, you have to look at the factorization of $P$ in $F(x)[X]$ : $P(X) = (X-x)Q(x)$ where $Q$ has coefficients in $F(x)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $Q$ is irreducible, then $F(x,y) = F(x)(y) \cong F(x)[X]/Q \cong F(x)(z) = F(x,z)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $Q$ factorizes further in $F(x)[X]$, the minimal polynomials of $y$ and $z$ on the field $F(x)$ can be different, and then $F(x,y) \not \cong F(x,z)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In KCd's example, $P(X) = X^4-2 = (X-x)(X^3+xX^2+x^2X+x^3) = (X-x)(X+x)(X^2+x^2)$.&#xA;You can see that if you know one root $x$, then the others are $-x, ix, -ix$, and $\mathbb{Q}(x,-x) \not \cong \mathbb{Q}(x,ix)$.&lt;/p&gt;&#xA;" OwnerUserId="6270" LastActivityDate="2011-02-13T15:17:54.357" />
  <row Id="21859" PostTypeId="1" AcceptedAnswerId="21860" CreationDate="2011-02-13T16:08:45.143" Score="2" ViewCount="68" Body="&lt;p&gt;I'm taking basic linear algebra now. During a lecture on orthogonal complements, a professor said that there's a thing that is more 'general' than orthogonal complements and it's called the annihilator. He didn't provide further details, so I hopped on Google and found some explanations of this concept. Now I think I understand what the annihilator is, but I can't seem to prove a basic fact about it..&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Theorem 3.14, part (1) in Advanced Linear Algebra by Steven Roman (&lt;a href=&quot;http://books.google.co.uk/books?id=bSyQr-wUys8C&amp;amp;pg=PA102&quot; rel=&quot;nofollow&quot;&gt;Google Books&lt;/a&gt;) says that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$M \subseteq N \Rightarrow N^0 \subseteq M^0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know how to prove this (the author leaves proof of it for the reader). Intuitively I understand why $M^0$ can't be smaller than $N^0$, for if we extend the subset $M$ by adding some $n \in N$ to it, then, in case (a), some functionals from the 'original' $M^0$ won't be able to send the new element of $M$ to zero and $M^0$ will get smaller. In case (b), every functional from the 'original' $M^0$ will send $n$ to zero, and $M^0$ will stay the same. But $M^0$ can't get bigger, because the functionals that annihilate $n$ AND all the vectors in $M$ must be in $M^0$ already.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So can you please help me convert my vision of the problem into a formal proof? &lt;/p&gt;&#xA;" OwnerUserId="7026" LastActivityDate="2011-02-13T16:21:49.823" Title="Linear algebra: showing that $M \subseteq N \Rightarrow N^0 \subseteq M^0$" Tags="&lt;linear-algebra&gt;" AnswerCount="1" />
  <row Id="21860" PostTypeId="2" ParentId="21859" CreationDate="2011-02-13T16:21:49.823" Score="3" ViewCount="" Body="&lt;p&gt;Well, as you mention, the definition of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Dual_space#Quotient_spaces_and_annihilators&quot; rel=&quot;nofollow&quot;&gt;annihilator&lt;/a&gt; of a subset $S\subset V$ of your vector space is the set $S^0$ of all elements $f\in V^*$ in the dual of $V$ which satisfy $f(s)=0$ for all $s\in S$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, $f\in N^0$ by definition means $f(n)=0$ for all $n\in N$. Since $M\subset N$, we also have $f(n)=0$ for all $n\in M$. By definition, this means $f\in M^0$. This shows $N^0\subset M^0$.&lt;/p&gt;&#xA;" OwnerUserId="367" LastActivityDate="2011-02-13T16:21:49.823" CommentCount="2" />
  <row Id="21861" PostTypeId="2" ParentId="309" CreationDate="2011-02-13T16:33:38.740" Score="0" ViewCount="" Body="&lt;p&gt;I suspect you did not read carefully that exercise. A subobjects is not an object with some special property. A subobject is an object and a monomorphism, like $(A_0,m_0)$. So if you define $A_0\le A_1$ iff $\exists f:A_0\to A_1. mono(f)$, this does not hire $m_0, m_1$, and this is intuitively wrong. And there is a standard solution: define $(A_0,m_0)\le (A_1,m_1)$ iff $m_0$ factors through $m_1$ ($\exists f:A_0\to A_1. m_0 = m_1\circ f$). $(A_0,m_0)\le (A_1,m_1) \land (A_1,m_1)\le (A_0,m_0)$ and factoring gives you morphisms to construct an isomorphism between $A_0,A_1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the context where subobjects are used, see “Robert Goldblatt. Topoi. The Categorial Analysis of Logic.” or any other textbook on categorical logic or Wikipedia.&lt;/p&gt;&#xA;" OwnerUserId="7011" LastActivityDate="2011-02-13T16:33:38.740" CommentCount="0" />
  <row Id="21863" PostTypeId="1" AcceptedAnswerId="21865" CreationDate="2011-02-13T17:12:28.197" Score="1" ViewCount="44" Body="&lt;p&gt;Of the six questions regarding finding vertical asymptotes of graphs, I've had problems with two.  The second is using the graph of $$g(x)= \frac{3+x}{x^{2}(3-x)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, looking at the function, it seems that both 0 and 3 will be the vertical asymptotes, since $$g(0)= \frac{3+0}{0^{2}(3-0)}= \frac{3}{0(3)}= \frac{3}{0}= undefined$$ and&#xA;$$g(3)= \frac{3+3}{3^{2}(3-3)}= \frac{6}{9(0)}= \frac{6}{0}= undefined$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, those answers were not accepted, why? &lt;/p&gt;&#xA;" OwnerUserId="7006" LastActivityDate="2011-02-13T17:51:57.267" Title="More help with vertical asymptotes" Tags="&lt;calculus&gt;&lt;asymptotics&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="21864" PostTypeId="1" AcceptedAnswerId="21870" CreationDate="2011-02-13T17:22:35.347" Score="2" ViewCount="123" Body="&lt;p&gt;Hello!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the relation $M=E-\epsilon\cdot\sin(E)$, I need to find the value of E, &lt;em&gt;knowing the two other parameters&lt;/em&gt;. How should I go about this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is part of a computation which will be done quite a number of times per second. I hope there's a quick way to get E out of this equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you very much,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;MJ&lt;/p&gt;&#xA;" OwnerUserId="4556" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2011-02-14T09:01:07.447" LastActivityDate="2011-02-14T09:01:07.447" Title="Solving short trigo equation with sine - need some help!" Tags="&lt;trigonometry&gt;&lt;numerical-methods&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="21865" PostTypeId="2" ParentId="21863" CreationDate="2011-02-13T17:46:30.533" Score="3" ViewCount="" Body="&lt;p&gt;The usual definition for a vertical asymptote at $x=a$ is that at least one of the following conditions must be satisfied.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$\displaystyle \lim_{x \rightarrow a^+} f(x) = \pm \infty$&lt;/li&gt;&#xA;&lt;li&gt;$\displaystyle \lim_{x \rightarrow a^-} f(x) = \pm \infty$&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/w2lUn.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is clear from the equations and from the graph of the function that at least one of these conditions is satisfied at $x=0$ and $x=3$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, $x=0$ and $x=3$ are both vertical asymptotes to the function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I have seen some people defining vertical asymptote as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle \lim_{x \rightarrow a^-} f(x) = \displaystyle \lim_{x \rightarrow a^+} f(x) = \pm \infty$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i.e. the left limit and the right limit must both tend to $+\infty$ or the left limit and the right limit must both tend to $-\infty$. In that case, $x=3$ won't be consider a vertical asymptote.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, I prefer to work with the first definition.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2011-02-13T17:51:57.267" LastActivityDate="2011-02-13T17:51:57.267" CommentCount="1" />
  <row Id="21866" PostTypeId="2" ParentId="21851" CreationDate="2011-02-13T17:55:09.200" Score="2" ViewCount="" Body="&lt;p&gt;There is some discussion of this problem here:  &lt;a href=&quot;http://www.tek-tips.com/viewthread.cfm?qid=1370890&amp;amp;page=2&quot; rel=&quot;nofollow&quot;&gt;http://www.tek-tips.com/viewthread.cfm?qid=1370890&amp;amp;page=2&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1369" LastActivityDate="2011-02-13T17:55:09.200" CommentCount="1" />
  <row Id="21867" PostTypeId="2" ParentId="21778" CreationDate="2011-02-13T17:58:49.770" Score="3" ViewCount="" Body="&lt;p&gt;No factorization is needed. First, separate the rational function $\rm\:f/g\:$ into its integral and fractional parts by using the division algorithm to divide the numerator by the denominator, namely&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\rm f\ =\ q\ g + r\ \ \Rightarrow\ \ \frac{f}g\ =\ q + \frac{r}g,\quad\quad deg\ r\ &amp;lt;\ deg\ g$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the integral part $\rm\:q\:$ is a polynomial it has no vertical asymptotes. Rather, such asymptotes arise only from the &quot;proper&quot; fractional part $\rm\:r/g\:,\: $ namely at roots of the denominator $\rm\:g\:,\: $ which are not also roots of the numerator $\rm\:r\:.\: $ But in your example, performing the simple division shows that remainder $\rm\:r = 0\:,\: $ so there are no such asymptotes. Generally this will be a much simpler approach than immediate factorization since generally division (and gcd) algorithms are much simpler than factorization algorithms.&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2011-02-13T17:58:49.770" CommentCount="3" />
  <row Id="21868" PostTypeId="1" CreationDate="2011-02-13T18:27:12.870" Score="7" ViewCount="93" Body="&lt;p&gt;Jacobi first noticed the connection between the functions that bear his name and counting the representations of &lt;a href=&quot;http://mathworld.wolfram.com/SumofSquaresFunction.html&quot; rel=&quot;nofollow&quot;&gt;sums-of-squares&lt;/a&gt;, &#xA;\begin{eqnarray}&#xA;\theta_{3}^{n}(q) = \left( \sum_{k \in \mathbb{Z}} q^{k^{2}} \right)^{n} = \sum_{k \geq 0} r_{2,n}(k) \ q^{k},&#xA;\end{eqnarray}&#xA;where $q = e^{\pi i \tau}$ (or $e^{2 \pi i \tau}$ depending on which text you consult) and $r_{2,n}(k)$ is the number of ways to represent $k$ as a sum of $n$ squares, &lt;em&gt;i.e.&lt;/em&gt;, the integral solutions of the equation $k = x_{1}^{2} + \dots + x_{n}^{2}$. The &lt;a href=&quot;http://en.wikipedia.org/wiki/Theta_function&quot; rel=&quot;nofollow&quot;&gt;Jacobi theta functions&lt;/a&gt;, including $\theta_{3}$, are known to satisfy a myriad of symmetries and identities. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; Are these identities related in any way to the family of curves $\{ X^{2} + Y^{2} = k \}_{k \in \mathbb{N}}$? If so, how?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose, more generally, I'd like to count $m$-powers instead of squares. The corresponding $q$-series is then&#xA;\begin{eqnarray}&#xA;\left( \sum_{k \in \mathbb{Z}} q^{k^{m}} \right)^{n} =  \sum_{k \geq 0} r_{m,n}(k) \ q^{k},&#xA;\end{eqnarray}&#xA;If the answer to the first question is yes, then this function (known?) and its symmetries should then be intimately connected to $\{ X_{1}^{m} + X_{2}^{m} + \dots + X_{n}^{m} = k \}_{k \in \mathbb{N}}$, a family of deformations of the homogeneous hypersurface $F = \sum_{i = 1}^{n} X_{i}^{m} = 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Generalizing still further, let $F = F(X_1, \dots, X_n)$ be a $\mathbb{Z}$-polynomial in $n$ indeterminates. Let $a(k)$ count the number of ways that an integer $k$ can be represented by $F$ over the integers. What general characteristics are known about the function $\sum_{k \geq 0} a(k) \ q^{k}$. Can one determine its symmetries by the symmetries of $\{ F(X_1, \dots, X_n) = k \}_{k \in \mathbb{N}}$ as a family of deformations of the hypersurface $F = 0$? Can one define a related $L$-function and presuppose its functional identities similarly? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any clarification is certainly appreciated!&lt;/p&gt;&#xA;" OwnerUserId="2720" LastEditorUserId="2720" LastEditorDisplayName="" LastEditDate="2011-02-14T17:04:35.457" LastActivityDate="2011-02-14T17:04:35.457" Title="Curves and Sums-of-Powers Representations" Tags="&lt;number-theory&gt;&lt;algebraic-geometry&gt;" FavoriteCount="2" />
  <row Id="21869" PostTypeId="1" AcceptedAnswerId="21876" CreationDate="2011-02-13T18:42:06.277" Score="5" ViewCount="494" Body="&lt;p&gt;I have $y$ is a function of $t$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have reached a situation here where I need to evaluate&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle \int_0^b{\frac{dy}{dt}dt}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now clearly $y$ has dependence on $t$, otherwise $\displaystyle \frac{dy}{dt}$ should be 0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So now I write&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle \int_{y(0)}^{y(b)}{dy} = y \rvert_{y(0)}^{y(b)} = y(b) - y(0)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that &lt;a href=&quot;http://math.stackexchange.com/questions/21199/dy-dx-is-not-a-ratio&quot;&gt;dt's don't &lt;em&gt;cancel&lt;/em&gt;&lt;/a&gt;, but what do you call it, then?  Just a &quot;change of variables&quot;?  How do we justify where $dt$ went?&lt;/p&gt;&#xA;" OwnerUserId="1069" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2011-02-13T18:54:56.220" LastActivityDate="2011-03-18T19:47:08.270" Title="If $\frac{dy}{dt}dt$ doesn't cancel, then what do you call it?" Tags="&lt;calculus&gt;&lt;notation&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="5" />
  <row Id="21870" PostTypeId="2" ParentId="21864" CreationDate="2011-02-13T18:54:07.993" Score="4" ViewCount="" Body="&lt;p&gt;I assume $\epsilon$ is a small quantity and propose one of the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(a) Write your equation in the form $E=M+\epsilon \sin(E)=: f(E)$ and consider this as a fixed point problem for the function $f$. Starting with $E_0:=M$ compute numerically successive iterates $E_{n+1}:=f(E_n)$; these will converge to the desired solution of the given equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(b) $E$ depends in an analytic way on the parameter $\epsilon$. Make the &quot;Ansatz&quot; $E:=M +\sum_{k=1}^\infty a_k \epsilon^k$ and determine the coefficients $a_k$ recursively. You will find $a_1=\sin(M)$, $a_2=\cos(M)\sin(M)$ and so on.&lt;/p&gt;&#xA;" OwnerUserId="1303" LastActivityDate="2011-02-13T18:54:07.993" CommentCount="2" />
  <row Id="21871" PostTypeId="2" ParentId="21869" CreationDate="2011-02-13T18:54:40.043" Score="10" ViewCount="" Body="&lt;p&gt;This is where the &lt;a href=&quot;http://en.wikipedia.org/wiki/Fundamental_theorem_of_calculus&quot; rel=&quot;nofollow&quot;&gt;Fundamental theorem of calculus&lt;/a&gt; comes in handy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It comes in two flavors:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The first flavor (Riemann Setting)&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $f(x)$ be a continuous function in $x$ on an interval $[a,b]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle F(x) = \int_{a_0}^{x} f(y) dy$, where $a_0 \in [a,b]$ is called a primitive of $f(x)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(The existence of a primitive can be proved since $f(x)$ is continuous.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then the Fundamental theorem of calculus states that&#xA;\begin{align*}&#xA;\displaystyle F(x) \in C^1{(a,b)}\\&#xA;\displaystyle \frac{dF(x)}{dx} = f(x)&#xA;\end{align*}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The second flavor (Riemann Setting)&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $f(x)$ be a continuous function on an interval $[a,b]$ and let its derivative $f&amp;#39;(x)$ exist on $(a,b)$ and let $f&amp;#39;(x)$ be integrable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then the Fundamental theorem of calculus states that&#xA;\begin{align*}&#xA;\displaystyle \int_{a_0}^{b_0} \frac{df(y)}{dy} dy = f(b_0) - f(a_0)&#xA;\end{align*}&#xA;where $a_0,b_0 \in [a,b]$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second flavor is the one you use in your argument above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might be able to relax some conditions that I have stated to obtain the same conclusion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again, this is a place where a good notation helps. $\displaystyle \frac{dy}{dt} dt$ can be &quot;treated&quot; like usual ratio and the $dt$'s can be &quot;canceled&quot; out though the actual reasoning for the &quot;canceling&quot; out is different.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As I always believe counterexamples are a great way to study a theorem or a property. Below are examples where you cannot &quot;cancel out the dt's&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In case of Riemann Integration, if you take $y(t)$ as the Volterra function (Thanks to Theo Buehler for pointing that one), then $y(t)$ is differentiable i.e. $\frac{dy}{dt}$ exists for all $t$. However, $\frac{dy}{dt}$ is &lt;strong&gt;not&lt;/strong&gt; &lt;em&gt;Riemann&lt;/em&gt; integrable i.e.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle \int_{a}^{b} y&amp;#39;(t) dt$ doesn't exists where the integral is interpreted in Riemann sense.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is however, &lt;em&gt;Lebesgue&lt;/em&gt; integrable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In case of Lebesgue Integration, there exists continuous functions, $f(x)$, which are differentiable almost everywhere i.e. $f&amp;#39;(x)$ exists almost everywhere, and its derivative, $f&amp;#39;(x)$, is Lebesgue integrable but the Lebesgue integral of $f&amp;#39;(x)$ is not equal to the change in $f(x)$ i.e.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle \int_{a}^{b} f&amp;#39;(x) dx \neq f(b) - f(a)$ where the integral is interpreted in Lebesgue sense.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The famous function here is the Cantor function $C(x)$. It is continuous everywhere and has zero derivative almost everywhere. Hence, the Lebesgue integral $\displaystyle \int_{0}^{1} C&amp;#39;(x) dx$ in the interval $[0,1]$ is $0$. However, $C(1)-C(0) = 1$&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2011-02-14T17:08:24.873" LastActivityDate="2011-02-14T17:08:24.873" CommentCount="7" />
  <row Id="21872" PostTypeId="1" CreationDate="2011-02-13T18:57:05.527" Score="1" ViewCount="222" Body="&lt;p&gt;I know you may find definitions out there for &lt;em&gt;standard curve&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, how can we define it in a way that makes it more understandable and clear of what it does?&lt;/p&gt;&#xA;" OwnerUserId="6659" LastEditorUserId="1208" LastEditorDisplayName="" LastEditDate="2011-03-27T21:41:34.587" LastActivityDate="2011-03-27T21:41:34.587" Title="What is a standard curve?" Tags="&lt;definition&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="21873" PostTypeId="2" ParentId="21817" CreationDate="2011-02-13T18:59:36.957" Score="8" ViewCount="" Body="&lt;p&gt;Please note that every question USER6560 has asked has come from the HW assigned in MATH 620 at university at buffalo. The link is &lt;a href=&quot;http://www.math.buffalo.edu/~badzioch/MTH620/Homework.html&quot; rel=&quot;nofollow&quot;&gt;http://www.math.buffalo.edu/~badzioch/MTH620/Homework.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;USER6560: stop asking questions from the HW.&lt;/p&gt;&#xA;" OwnerUserId="7042" LastActivityDate="2011-02-13T18:59:36.957" CommentCount="7" />
  <row Id="21875" PostTypeId="1" CreationDate="2011-02-13T19:37:53.823" Score="1" ViewCount="81" Body="&lt;p&gt;Let $i:N\to M$ be a smooth embedding, $\pi:E\to N$ a vector bundle and $s_0:N\to E$ is its zero section.&lt;br&gt;&#xA;I have an open neighborhood $U$ of $s_0(N)$ in $E$, and $f:U\to M$ is a smooth map such that $f\circ s_0=i$ and $T_xf$ is bijective for any $x\in s_0(N)$.&lt;br&gt;&#xA;How could I prove that there exists an open neighborhood $V$ of $s_0(N)$ in $U$ such that $f|_V$ is a diffeomorphism onto its image?&lt;/p&gt;&#xA;" OwnerUserId="6752" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2011-02-14T03:43:49.863" LastActivityDate="2011-02-14T03:43:49.863" Title="costructing a diffeomorphism" Tags="&lt;differential-geometry&gt;" CommentCount="4" />
  <row Id="21876" PostTypeId="2" ParentId="21869" CreationDate="2011-02-13T19:46:58.707" Score="17" ViewCount="" Body="&lt;p&gt;This is closely related to the &lt;a href=&quot;http://math.stackexchange.com/questions/21199/dy-dx-is-not-a-ratio/21209#21209&quot;&gt;recent question&lt;/a&gt; asking whether $\frac{dy}{dt}$ was a fraction or not, as you note.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As in that question, when Leibnitz first came up with the notation for integrals (the $\int$ sign was really a capital $S$, standing for &quot;summa&quot;, summation), and he wrote $S_a^b f(x)\,dx$, he was &lt;em&gt;really&lt;/em&gt; thinking of this as a sum of products, with $dx$ representing an &quot;infinitesimal change in $x$.&quot; He was thinking of dividing the interval $[a,b]$ into an infinite number of &quot;infinitesimally thin&quot; rectangles, each of height $f(x_i)$ (for whatever $x_i$ you happened to be in), and then adding up all the areas of these infinitesimally thin rectangles. Such a rectangle would have area $f(x_i)dx$ (height times base), and adding them together would yield the area.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you take this point of view (ignoring for a moment the fact that infintesimals can't exist in the usual real numbers), then the First Fundamental Theorem of Calculus follows by &quot;simple&quot; arithmetic with a telescoping sum (assuming $\frac{dy}{dt}$ is continuous, say). You have that the integral &#xA;$$\int_a^b\frac{dy}{dt}\,dt$$&#xA;is really the sum of the quotient of infinitesimal changes in $y$, divided by infinitesimal changes in $t$, multiplied by the infinitesimal change in $t$. If you think of $[a,b]$ as divided into infinitesimally thin subintervals&#xA;$$a=t_0\lt t_0+dt\lt t_0+2dt\lt\cdots \lt b$$&#xA;then $dy = y(t_0+(k+1)dt) - y(t_0+kdt)$, so that the integral becomes the telescoping sum&#xA;$$\sum \Bigl(y(t_0+(k+1)dt) - y(t_0+kdt)\Bigr) = y(b)-y(a)$$&#xA;because all the &quot;middle terms&quot; cancel out. So here, $dt$ really &lt;em&gt;does&lt;/em&gt; cancel out with the $dt$ in $\frac{dy}{dt}$, the sum is telescoping, and the final answer comes out exactly as desired.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, as with the derivative, there is a legion of logical problems with this way of thinking, not the least of which is that infinitesimals can't really exist in the usual setting of real numbers. So calculus had to be rewritten. There were proposals by Cauchy on how to define integrals, and eventually we had Riemann's way of defining integrals as limits. So that when we write $\int_a^b f(t)\,dt$ we no longer mean a sum of products of the form $f(t)\,dt$, but rather we mean a limit of certain Riemann sums. In this view, the &quot;$dt$&quot; is rather more like the symbol to balance the $\int_a^b$. Think of the integral sign on the left as a &quot;left parenthesis&quot;, and the $dt$ on the right as the &quot;right parenthesis&quot; that closes out the expression.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, just like $\frac{df}{dt}$ no longer literally means &quot;the quotient of an infinitesimal change in $f$ by an infinitesimal change in $t$&quot; but rather means &quot;the limit as $h$ goes to $0$ of $(f(t+h)-f(t))/h$, so $\displaystyle \int_a^b f(t)\,dt$ no longer literally means &quot;the sum of infinitesimally thin rectangles of height $f(t)$ from $a$ to $b$&quot;, but instead means &quot;the limit of Riemann sums of $f(t)$ over partitions of $[a,b]$ as the mesh goes to $0$.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But one of the great advantages of Leibnitz notation is that it is very suggestive. So you get the First Fundamental Theorem of Calculus, which looks very natural in Leibnitz notation:&#xA;$$\int _a^b \frac{df}{dt}\;dt = f(b) - f(a),$$&#xA;giving the suggestion that you are &quot;cancelling $dt$&quot;, even though you're not really doing that. But since infinitesimals don't &lt;em&gt;really&lt;/em&gt; exist, this is not &lt;em&gt;literally&lt;/em&gt; true. But good notation is not something to be cast aside when it comes along, and Leibnitz notation, being suggestive, is &lt;em&gt;very good&lt;/em&gt; notation, so we keep it because it helps with calculations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where did the &quot;$dt$&quot; go? Well, one might ask where the &quot;$)$&quot; goes in the following calculation:&#xA;$$2\times(3+5) = 16.$$&#xA;So... where did the &quot;$)$&quot; go (or where did &quot;$\times$&quot;, &quot;$+$,&quot; and &quot;$($&quot; all go) ? Same place as the &quot;$dt$&quot; went: since it is part of the notation, it &quot;goes away&quot; when we are done with the evaluation.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Note.&lt;/em&gt; As with derivatives, with Nonstandard Analysis one can write calculus so that the $dt$ in the integral really represents a quantity you are multiplying by and then adding, so that in nonstandard analysis the First Fundamental Theorem of Calculus really is just the observation that if you divide by $dx$ and multiply by $dx$, then the two cancel out.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-18T16:18:15.353" LastActivityDate="2011-03-18T16:18:15.353" CommentCount="2" />
  <row Id="21877" PostTypeId="1" AcceptedAnswerId="21892" CreationDate="2011-02-13T19:50:52.073" Score="1" ViewCount="152" Body="&lt;p&gt;Can someone tell me if this is right:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to find the minimal polynomial of &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(i) $\sqrt[4]{2}i$ over $\mathbb{Q}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(ii) $\sqrt[4]{2}i$ over $\mathbb{R}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(i):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sqrt[4]{2}i$ is a root of $f(x) = x^4-2$. This is already monic, so to show that this is a minimal polynomial I only need to show that it is irreducible. &lt;strong&gt;Edit:&lt;/strong&gt; To do that, I can use the Eisenstein: $p=2$ does not divide $a_1 = 1$ and $p^2$ does not divide $a_0 = -2$, therefore it is irreducible over $\mathbb{Q}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(ii):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This time, $\sqrt[4]{2}i$ is a root of $f(x) = x^2+\sqrt{2}$. For polynomials of degree two it's enough to check if they have a root. The only roots this one has are complex therefore it is irreducible over $\mathbb{R}$ and therefore the minimal polynomial.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So my more general question is: is the way to find a minimal polynomial of an element $a$ in general to find a polynomial $f$ such that $f(a) = 0$ and then to norm $f$ and then to show that $f$ is irreducible?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many thanks for your help!&lt;/p&gt;&#xA;" OwnerUserId="5798" LastEditorUserId="5798" LastEditorDisplayName="" LastEditDate="2011-02-13T20:14:33.760" LastActivityDate="2011-02-13T20:57:50.833" Title="Minimal polynomials" Tags="&lt;abstract-algebra&gt;&lt;field-theory&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="1" />
  <row Id="21878" PostTypeId="1" AcceptedAnswerId="21881" CreationDate="2011-02-13T19:53:19.133" Score="4" ViewCount="218" Body="&lt;p&gt;To better train my intuition, what are some illustrative examples of function sequences in C[0,1] that are Cauchy but do not converge under the integral norm?&lt;/p&gt;&#xA;" OwnerUserId="4518" LastActivityDate="2011-02-13T20:32:13.673" Title="Examples of function sequences in C[0,1] that are Cauchy but not convergent" Tags="&lt;analysis&gt;&lt;convergence&gt;&lt;metric-spaces&gt;" AnswerCount="3" />
  <row Id="21879" PostTypeId="2" ParentId="21878" CreationDate="2011-02-13T20:00:29.300" Score="1" ViewCount="" Body="&lt;p&gt;How about the sequence of functions given by $f_n(x)=x^n$? This sequence is Cauchy in $(C([0,1]),\|\cdot\|_{L^1(0,1)})$ but does not converge in $C([0,1])$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: This example doesn't work. Thanks for pointing that out. An example that &lt;em&gt;does&lt;/em&gt; work is given by the sequence of continuous functions $f_n$ with&#xA;$f_n(x)=\frac{2}{n}(x-\frac{1}{2}+\frac{1}{n})$ for $|x-\frac{1}{2}|\leq \frac{1}{n}$, $f_n(x)=0$ for $x&amp;lt;\frac{1}{2}-\frac{1}{n}$ and $f_n(x)=1$ for $x&amp;gt;\frac{1}{2}+\frac{1}{n}$. That's basically a continuous approximation of the characteristic function of the interval $[\frac{1}{2},1]$, which is not continuous.&lt;/p&gt;&#xA;" OwnerUserId="1403" LastEditorUserId="1403" LastEditorDisplayName="" LastEditDate="2011-02-13T20:20:51.780" LastActivityDate="2011-02-13T20:20:51.780" CommentCount="5" />
  <row Id="21880" PostTypeId="1" AcceptedAnswerId="21891" CreationDate="2011-02-13T20:03:40.420" Score="2" ViewCount="132" Body="&lt;p&gt;Can anyone help me with this problem?&#xA;The random variable $X_n$ takes the values $\frac{k}{n}$, $k=1,2,\ldots,n$, each with probability $\frac{1}{n}$. Find its characteristic function and the limit as $n\to\infty$. Identify the random variable of the limit characteristic function. &#xA;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="6912" LastEditorUserId="5363" LastEditorDisplayName="" LastEditDate="2011-02-13T20:18:07.963" LastActivityDate="2011-03-25T08:41:56.007" Title="Finding Characteristic Function" Tags="&lt;homework&gt;&lt;stochastic-processes&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="21881" PostTypeId="2" ParentId="21878" CreationDate="2011-02-13T20:04:57.043" Score="8" ViewCount="" Body="&lt;p&gt;You can get examples by considering elements of $L^1[0,1]$ that are not equal almost everywhere to any continuous function, and considering sequences of continuous functions converging in the $L^1$ norm to these discontinuous functions.  Because convergent sequences are Cauchy and $L^1$ limits are unique up to equality almost everywhere, such sequences will be Cauchy and nonconvergent in $C[0,1]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;E.g., let $f$ be $1$ on $[0,\frac{1}{2}]$ and $0$ elsewhere.  Let $f_n$ be the continuous function that is $1$ on $[0,\frac{1}{2}]$, $0$ on $[\frac{1}{2}+\frac{1}{n},1]$, and linear on $[\frac{1}{2},\frac{1}{2}+\frac{1}{n}]$.  Then because $f_n\to f$ in $L^1$, $(f_n)$ is Cauchy.  (The Cauchy-ness is also easy to verify directly.)  If there were a limit function $g\in C[0,1]$, you would have $g=f$ a.e..  But this is impossible, because the left-hand and right-hand limits at $\frac{1}{2}$ would not agree.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;More generally, a Cauchy sequence in a metric space $X$ with completion $\overline{X}$ that does not converge in $X$ is basically the same as a sequence in $X$ that converges to an element of $\overline{X}\setminus X$.  In a case like this where $X=C[0,1]$ with $L^1$ norm and $\overline{X}=L^1[0,1]$ have explicit descriptions, you can find examples by starting with an element of $\overline{X}\setminus X$, and find a sequence in $X$ converging to that element.  The same idea applies to demonstrating nonconvergent Cauchy sequences in $\mathbb{Q}$, where you can take any irrational number and consider the sequence of truncated decimal expansions.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2011-02-13T20:28:29.090" LastActivityDate="2011-02-13T20:28:29.090" />
  <row Id="21882" PostTypeId="2" ParentId="21878" CreationDate="2011-02-13T20:06:22.990" Score="4" ViewCount="" Body="&lt;p&gt;By measure theory you can always pass to a subsequence that converges pointwise a.e. so it's usually quite easy to determine the only possible limit. Then you should try to build the examples in such a way that the pointwise (a.e.) limit has no chance of being continuous.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's an example of what I have in mind:&#xA;$f_{n}(x) = \begin{cases} 0 &amp;amp; 0 \leq x \leq \frac{1}{2} \\\ nx - \frac{n}{2} &amp;amp; \frac{1}{2} \leq x \leq \frac{1}{2} + \frac{1}{n} \\\ 1 &amp;amp; \frac{1}{2} + \frac{1}{n} \leq x \leq 1 \end{cases}$ defined for $n \geq 2$. It's easy to see that it converges in the $L^{1}$-norm to the characteristic function of $[\frac{1}{2},1]$.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added:&lt;/strong&gt; Since an $L^{1}$-Cauchy sequence will always converge ($L^{1}$ is complete by the theorem of Riesz-Fischer (or Fr&amp;eacute;chet-Riesz)), the only thing you have to do is to pick a function in $L^{1}$ that doesn't have a continuous representative. The easiest example is a step function (that's probably the reason why all three answerers came up with essentially the same example). This suffices, because every function in $L^{1}$ is the $L^1$-limit of a sequence of continuous (even smooth) functions, as can be seen by convolving with &lt;a href=&quot;http://en.wikipedia.org/wiki/Mollifiers&quot; rel=&quot;nofollow&quot;&gt;mollifiers&lt;/a&gt;, for example. In other words, $C[0,1]$ (or even $C^{\infty}[0,1]$) are dense in $L^{1}$.&lt;/p&gt;&#xA;" OwnerUserId="5363" LastEditorUserId="5363" LastEditorDisplayName="" LastEditDate="2011-02-13T20:32:13.673" LastActivityDate="2011-02-13T20:32:13.673" />
  <row Id="21883" PostTypeId="1" CreationDate="2011-02-13T20:09:16.263" Score="23" ViewCount="497" Body="&lt;p&gt;The first irregular prime is 37. Does FLT(37)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x^{37} + y^{37} = z^{37}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;have any solutions in the ring of integers of $\mathbb Q(\zeta_{37})$, where $\zeta_{37}$ is a primitive 37th root of unity?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe it's not true, but how could I go about finding a counter-example? (for any cyclotomic ring, not necessarily 37)&lt;/p&gt;&#xA;" OwnerUserId="4290" LastEditorUserId="6296" LastEditorDisplayName="" LastEditDate="2011-02-14T01:03:05.033" LastActivityDate="2011-02-14T01:03:05.033" Title="Does Fermat's Last Theorem hold for cyclotomic integers in $\mathbb{Q(\zeta_{37})}$?" Tags="&lt;number-theory&gt;" CommentCount="11" FavoriteCount="3" />
  <row Id="21884" PostTypeId="2" ParentId="21806" CreationDate="2011-02-13T20:11:00.223" Score="1" ViewCount="" Body="&lt;p&gt;Let $T:z\to z&amp;#39;$ be such a Moebius transform and let $p, q\in{\mathbb R}$, $\  p&amp;lt;q$, be its two fixed points. As $z&amp;#39;=p$ iff $z=p$ and $z&amp;#39;=q$ iff $z=q$, the two variables $z$ and $z&amp;#39;$ have to be related by a formula of the form&#xA;$${z&amp;#39;-q \over z&amp;#39;-p}=\lambda {z-q \over z- p}$$&#xA;for some complex constant $\lambda$. This constant has to fulfill conditions to guarantee that (a) $T$ maps the real axis to itself and (b) $T$ maps, e.g., the point $i$ to a point in the upper half plane.&lt;/p&gt;&#xA;" OwnerUserId="1303" LastActivityDate="2011-02-13T20:11:00.223" />
  <row Id="21885" PostTypeId="2" ParentId="21796" CreationDate="2011-02-13T20:21:13.653" Score="4" ViewCount="" Body="&lt;p&gt;Based on your comments in response to answers, you seem to be asking rather why radians are &quot;special&quot; in giving the answer $1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This really comes down to why is it that when we use radians, we have&#xA;$$\lim_{x\to 0}\frac{\sin x}{x} = 1$$&#xA;because it comes down to why $(\cos x)&amp;#39; = -\sin x$ and not some other constant multiple of $\sin x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Think about the unit circle, parametrized by&#xA;\begin{align*}&#xA;x&amp;amp;=\cos t,\\ &#xA;y&amp;amp;=\sin t&#xA;\end{align*}&#xA;(in whatever units you want to pick for $t$: radians, gradians, degrees, anything, doesn't matter here). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whenever you parametrize a curve, you can change the parameter so that you traverse the curve slower, or faster, or in a different direction, or in a different way; here, if you replace $t$ with $-t$, you reparametrize the unit circle so that it is traversed clockwise, instead of counterclockwise. If you change $t$ to $2t$, you traverse it twice as fast but in the same direction; if you replace $t$ with $t+17$, then you start somewhere else; etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is, however, what one might call a &quot;normalized&quot; parametrization, which normalizes the &lt;em&gt;speed&lt;/em&gt; at which we traverse the curve. It is called the &quot;arc-length parametrization&quot;, and the objective is to choose the parameter $t$ so that, if $t$ varies from $a$ to $b$, then the length of curve that is traversed is exactly $b-a$; that is, it takes exactly one unit of &quot;time&quot; to traverse one unit of &quot;distance&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The arc-length parametrization has a number of advantages. For example, if you compute the tangent vector by taking $(x&amp;#39;(t),y&amp;#39;(t))$, in the arc length parametrization of a curve the tangent vector always has unit length. It also allows you to find the curvature of the curve much easier than with a run-of-the-mill parametrization.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, what is the arc-length parametrization of the unit circle? It is exactly the one in which it takes from $t=0$ to $t=2\pi$ to traverse the entire circle once; it is precisely the one in which an angle measured by $t$ cuts a circle arc of length $t$. It is, in short, measuring $t$ in radians. Radians give the arclength parametrization of the curve, which is why there are no pesky constants of proportionality when we take derivatives of the coordinate functions $\sin t$ and $\cos t$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, no matter how you measure your angles, there will be &lt;strong&gt;some&lt;/strong&gt; value of $b$ such that&#xA;$$\int_0^b \sin_x t\,dt = 1.$$&#xA;(where by $\sin_x t$ I mean &quot;sine, measuring the argument in $x$s&quot;). The only question is what is the value of $b$; it exists for degrees, for radians, for gradians, for half-degrees, anything, really. Why does it turn out to be exactly $\pi/2$ for radians? Because when we do radians, which give the arc length parametrization, the lack of proportionality constants yield that the integral of the sine is the negative of the total change in the $x$ coordinate (under other parametrizations, it would be some constant multiple of this). If you start at $t=0$, the point $(1,0)$, when has the $x$-coordinate changed by exactly one unit? When we get to $(0,1)$, that is, when $t=\pi/2$.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2011-02-13T20:21:13.653" />
  <row Id="21886" PostTypeId="1" AcceptedAnswerId="21889" CreationDate="2011-02-13T20:31:00.583" Score="1" ViewCount="117" Body="&lt;p&gt;Can anyone help me with it:&#xA;Using the central limit theorem for suitable Poisson random variables, prove that $$ \lim_{n\to\infty} e^{-n} \sum_{k=0}^{n} \frac{n^k}{k!}=1/2$$&#xA;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="6912" LastEditorUserId="524" LastEditorDisplayName="" LastEditDate="2011-02-13T21:09:16.043" LastActivityDate="2011-02-13T21:09:16.043" Title="Using Central Limit Theorem" Tags="&lt;homework&gt;&lt;stochastic-processes&gt;" AnswerCount="1" />
  <row Id="21887" PostTypeId="2" ParentId="21880" CreationDate="2011-02-13T20:39:14.113" Score="3" ViewCount="" Body="&lt;p&gt;Hint 1: You should know how to calculate ${\rm E}[f(X)]$, hence, in particular, the characteristic function of $X$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint 2: How can $\int_0^1 {f(x)\,{\rm d}x}$ be approximated by Riemann sum?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint 3: ${\rm d}x$ corresponds to which distribution?&lt;/p&gt;&#xA;" OwnerUserId="2810" LastActivityDate="2011-02-13T20:39:14.113" />
  <row Id="21888" PostTypeId="2" ParentId="21877" CreationDate="2011-02-13T20:40:52.803" Score="5" ViewCount="" Body="&lt;p&gt;In general, you can find the minimum polynomial for an algebraic number $\alpha$ by determining the smallest power of $\alpha$ for which $\{1,\alpha,\alpha^2,\ldots,\alpha^n\}$ is linearly dependent over $\mathbb{Q}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, let $\alpha=\sqrt{2} + \sqrt{3}$.  Then&#xA;\begin{align*}&#xA;\alpha^2 &amp;amp;= 5+2\sqrt{6} \\&#xA;\alpha^3 &amp;amp;= 11\sqrt{2}+9\sqrt{3} \\&#xA;\alpha^4 &amp;amp;= 49+20\sqrt{6}&#xA;\end{align*}&#xA;Since $\{1,\sqrt{2},\sqrt{3},\sqrt{6}\}$ are linearly independent over $\mathbb{Q}$, we can think of the powers of $\alpha$ as vectors:&#xA;$$&#xD;&#xA;1 = \begin{bmatrix}1\\\ 0\\\ 0\\\ 0\end{bmatrix},&#xD;&#xA;\qquad&#xD;&#xA;\alpha = \begin{bmatrix}0\\\ 1\\\ 1\\\ 0\end{bmatrix},&#xD;&#xA;\qquad&#xD;&#xA;\alpha^2 = \begin{bmatrix}5\\\ 0\\\ 0\\\ 6\end{bmatrix},&#xD;&#xA;\qquad&#xD;&#xA;\alpha^3 = \begin{bmatrix}0\\\ 11\\\ 9\\\ 0\end{bmatrix},&#xD;&#xA;\qquad&#xD;&#xA;\alpha^4 = \begin{bmatrix}49\\\ 0\\\ 0\\\ 20\end{bmatrix}&#xD;&#xA;$$&#xA;As you can see, $\{1,\alpha,\alpha^2,\alpha^3\}$ is linearly independent, so $\alpha$ is not the root of any cubic polynomial.  However, $\{1,\alpha,\alpha^2,\alpha^3,\alpha^4\}$ is linearly dependent, with&#xA;$$&#xD;&#xA;\alpha^4 - 10\alpha^2 + 1 \;=\; 0&#xD;&#xA;$$&#xA;It follows that $x^4-10x^2+1$ is the minimum polynomial for $\alpha$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This technique depends on being able to recognize a useful set of linearly independent elements like $\{1,\sqrt{2},\sqrt{3},\sqrt{6}\}$.  Depending on how much Galois theory you know, it may be hard to prove that elements like this are linearly independent over $\mathbb{Q}$.  In this case, you can use this technique to guess the minimum polynomial, and then prove that you are correct by proving that the polynomial you found is irreducible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, if $\alpha = i\sqrt[4]{2}$, then&#xA;$$&#xD;&#xA;\alpha^2 = -\sqrt{2},\qquad \alpha^3 = -i\sqrt[4]{8},\qquad \alpha^4 = 2.&#xD;&#xA;$$&#xA;It seems clear that $\{1,\alpha,\alpha^2,\alpha^3\}$ is linearly independent, while $\{1,\alpha,\alpha^2,\alpha^3,\alpha^4\}$ satisfies $\alpha^4-2 = 0$.  Thus $x^4-2$ should indeed be the minimum polynomial over $\mathbb{Q}$, though the easiest way to prove it is to show that $x^4 -2$ is irreducible.  This can be done using &lt;a href=&quot;http://en.wikipedia.org/wiki/Eisenstein%27s_criterion&quot; rel=&quot;nofollow&quot;&gt;Eisenstein's criterion&lt;/a&gt;, or by checking that $x^4-2$ has no roots and does not factor into quadratics modulo $4$.&lt;/p&gt;&#xA;" OwnerUserId="1726" LastActivityDate="2011-02-13T20:40:52.803" CommentCount="2" />
  <row Id="21889" PostTypeId="2" ParentId="21886" CreationDate="2011-02-13T20:44:36.483" Score="5" ViewCount="" Body="&lt;p&gt;Hint: A Poisson$(n)$ random variable can be represented as the sum of $n$ i.i.d. Poisson$(1)$ rv's. &lt;/p&gt;&#xA;" OwnerUserId="2810" LastActivityDate="2011-02-13T20:44:36.483" CommentCount="7" />
  <row Id="21890" PostTypeId="1" AcceptedAnswerId="21904" CreationDate="2011-02-13T20:47:56.413" Score="0" ViewCount="32" Body="&lt;p&gt;Let $G$ represent an undirected graph, let $a$ and $b$ represent vertices and let $k$ represent a non-negative integer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have two languages:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$L_1 = \left \{   \left \langle G, a, b, k \right \rangle  \mid G \text{ contains a path of at most } k \text{ from } a \text{ to } b \right \}$, and&lt;br/&gt;&#xA;$L_2 = \left \{   \left \langle G, a, b, k \right \rangle  \mid G \text{ contains a path of at least } k \text{ from } a \text{ to } b \right \}.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to show that $L_1$ is in &lt;strong&gt;P&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have thought that I could show it in $\text{Satisfiability}$, but I am unsure of what I am doing: $$(a \vee \lnot b )\wedge (b \vee \lnot a)\wedge(k \vee \lnot a) \wedge (k \vee \lnot b)\wedge (a \vee \lnot k) \wedge (b \vee \lnot k)$$ &lt;br/&gt;&#xA;If someone could walk me through this, that would be great, because I don't think I can use $\text{2-COL}$ to $\text{2-SAT reducibility}$ on this instance of $G$.&lt;br/&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Secondly, I want to show that $L_2$ is &lt;strong&gt;NP&lt;/strong&gt;-complete.&lt;br/&gt;Now I understand that I can use the fact that $\text{Hamiltonian-Path}$ is &lt;strong&gt;NP&lt;/strong&gt;-complete, but any further guidance would be gratefully appreciated.&lt;br/&gt;&lt;br/&gt;&lt;/p&gt;&#xA;" OwnerUserId="6796" LastActivityDate="2011-02-13T22:54:45.080" Title="Language complexity from instances of graph" Tags="&lt;homework&gt;&lt;computer-science&gt;" AnswerCount="1" />
  <row Id="21891" PostTypeId="2" ParentId="21880" CreationDate="2011-02-13T20:48:59.720" Score="2" ViewCount="" Body="&lt;p&gt;$$E(e^{itX})=\frac{1}{n}\sum_{k=1}^n {e^{is\frac{k}{n}}}=&#xD;&#xA;\frac{1}{n}\frac{e^{\frac{is}{n}}(1-e^{is})}{1-e^{\frac{is}{n}}}\to \frac{e^{is}-1}{is}$$ as $n \to \infty$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is cf of the uniform distribution on [0,1]&lt;/p&gt;&#xA;" OwnerUserId="6016" LastEditorUserId="6016" LastEditorDisplayName="" LastEditDate="2011-02-13T22:26:15.247" LastActivityDate="2011-02-13T22:26:15.247" />
  <row Id="21892" PostTypeId="2" ParentId="21877" CreationDate="2011-02-13T20:49:46.053" Score="3" ViewCount="" Body="&lt;p&gt;Part (i) as given is incorrect: lack of roots does not imply irreducibility when the polynomial is of degree greater than 3, because the polynomial could split into irreducible polynomials of degrees greater than 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have shown that $x^4-2$ has no linear factors modulo $3$, but you cannot conclude from this that it is irreducible modulo $3$: it could have two irreducible quadratic factors. And indeed, as chandok writes,&#xA;\begin{align*}&#xA;(x^2+x+2)(x^2+2x+2) &amp;amp;= x^4 + 2x^3 + 2x^2 + x^3 + 2x^2 + 2x + 2x^2 + 4x + 4\\&#xA;&amp;amp;= x^4 + 3x^3 + 6x^2 + 6x + 4 = x^4 + 1 = x^4 - 2,&#xA;\end{align*}&#xA;so the polynomial is not irreducible over $\mathbb{F}_3$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instead, I would suggest using Eisenstein's Criterion to show $x^4-2$ is irreducible over $\mathbb{Q}$, as it is pretty easy to apply there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Part (ii) is correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As to your final question: there is a top-down approach and a bottoms-up approach. In the top down approach, if you can find &lt;em&gt;any&lt;/em&gt; polynomial $f(x)$ that has $a$ as a root, then you know that the minimal polynomial will divide $f(x)$; in fact, it will be an irreducible factor of $f$. So you can try to find which irreducible factor of $f$ has $a$ as a root. This may be $f$ itself, or some factor. For example, to find the minimal polynomial of a $7$th root of unity, you could use the fact that it satisfies $x^7-1$. Then factor $x^7-1 = (x-1)(x^6+x^5+x^4+x^3+x^2+x+1)$, and you know that $a$ is a root of either $x-1$ or the second factor. Then you would look at the second factor (since presumably your $a$ is not $1$). And so on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The bottoms-up approach is described by Jim Belk in his answer, where you &quot;build up&quot; the polynomial by considering powers of $a$ (or alternatively, images under the action of some Galois group). &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-13T20:57:50.833" LastActivityDate="2011-02-13T20:57:50.833" />
  <row Id="21893" PostTypeId="1" AcceptedAnswerId="21980" CreationDate="2011-02-13T20:52:41.710" Score="3" ViewCount="141" Body="&lt;p&gt;Hi! I am trying to prove that the sequence $I_n(p)=\int_1^n \frac{dx}{x^p}$ does or does not converge uniformly to the function $I(p)=\int_1^\infty \frac{dx}{x^p}$ for $p&amp;gt;1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems to me that $\sup |I_n(p)-I(p)|= \lim_{p \to 1} (I(p)-I_n(p))$. Is it enough, then, to show that for all $\epsilon&amp;gt;0$ there exists a natural number $N$ such that for $p\to 1^+$ and all $n\geq N$, $I(p)-I_n(p) &amp;lt; \epsilon$? Can I set $p = 1$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit:&lt;br&gt;&#xA;$I(p)-I_n(p) = \int_1^\infty \frac{dx}{x^p} - \int_1^n \frac{dx}{x^p} = \int_n^\infty \frac{dx}{x^p}$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\lim_{n\to\infty} \lim_{p\to 1} \int_n^\infty \frac{dx}{x^p} = \lim_{n\to\infty} \int_n^\infty \frac{dx}{x} = 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this correct?&lt;/p&gt;&#xA;" OwnerUserId="7044" LastEditorUserId="7044" LastEditorDisplayName="" LastEditDate="2011-02-14T08:53:37.067" LastActivityDate="2011-02-14T08:56:53.117" Title="Proving uniform convergence between $I_n(p)=\int_1^n \frac{dx}{x^p}$ and $I(p)=\int_1^\infty \frac{dx}{x^p}$" Tags="&lt;real-analysis&gt;&lt;homework&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="21894" PostTypeId="1" AcceptedAnswerId="21895" CreationDate="2011-02-13T21:12:31.173" Score="6" ViewCount="154" Body="&lt;p&gt;...and if it's important, do those ideas have any generalization to more &quot;exotic&quot; number systems?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The motivation for my question comes from reading some of the excellent answers posted to other questions, such as &lt;a href=&quot;http://math.stackexchange.com/questions/21748/does-the-formula-sqrt-1-24n-always-yield-prime&quot;&gt;a recent one&lt;/a&gt; asking whether $\sqrt{1 + 24n}$ always yields primes. In particular, I've been struck by commenter Bill Dubuque's repeated use of what he terms &quot;modular reduction&quot;, casting a problem in $\mathbb{Z}_n$ to make it much easier to solve in the general case for $\mathbb{Z}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I don't quite grok is why this works; why can we do this? Is there a deep reason?  At first glance, to me, there doesn't seem anything inherent in the axioms of a given $\mathbb{Z}_k$ that necessarily ties it intimately with  $\mathbb{Z}$; all we care is that is has $k$ elements and it is closed under the binary operator of addition.  It doesn't seem to encode information about $\mathbb{Z}$'s other elements.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, in saying that, I'm not sure I'm on the right foot here at all, so I'll analogize to something I know a little better: One sees, in some textbook definitions, an identification of $\mathbb{C}$ with an ordered pair $(a,b)$, $(c,d)$ with rules for addition and multiplication that go to $(a + c, b + d)$ and $(ac - bd, ad + bc)$, respectively, with no immediate hint about the importance of $\mathbb{C}$ in that is the algebraic closure of $\mathbb{R}$, which is a highly nontrivial theorem that needs to be proved through the FTA.  Is there a similar relationship between $\mathbb{Z}_n$ and $\mathbb{Z}$, and is that extensible to other systems?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd appreciate any answers and references tailored to someone who's taken up to the middling undergraduate math level. (e.g. Linear Algebra, elementary abstract Algebra, undergraduate Complex Analysis...)&lt;/p&gt;&#xA;" OwnerUserId="6296" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-13T21:18:37.873" LastActivityDate="2011-02-13T22:04:51.677" Title="What is the term for the mathematical relationship between $\mathbb{Z}_n$ and $\mathbb{Z}$?" Tags="&lt;elementary-number-theory&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="2" />
  <row Id="21895" PostTypeId="2" ParentId="21894" CreationDate="2011-02-13T21:18:46.257" Score="13" ViewCount="" Body="&lt;p&gt;Well, the first thing to say is that the ring $\mathbb{Z}/n\mathbb{Z}$ is the &lt;strong&gt;quotient&lt;/strong&gt; of the ring $\mathbb{Z}$ by the &lt;strong&gt;ideal&lt;/strong&gt; $n \mathbb{Z}$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this already familiar to you?  If not, you should take a course and/or read a book on basic abstract algebra -- and, in the meantime, check out &lt;a href=&quot;http://en.wikipedia.org/wiki/Quotient_ring&quot; rel=&quot;nofollow&quot;&gt;this wikipedia article&lt;/a&gt;.  If so, could you clarify your question: what more than this do you want to know?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Added&lt;/b&gt;: well, I certainly understand and appreciate that you're looking for insight.  In this case I'm not sure exactly how to provide it.  But I'll try...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a number theorist's trick: there is in fact a unique homomorphism $f$ from $\mathbb{Z}$ to any ring $R$.  From this it follows that if $P = P(x_1,\ldots,x_n) = 0$ is any polynomial equation with $\mathbb{Z}$-coefficients, then any solution $(a_1,\ldots,a_n) \in \mathbb{Z}^n$ induces a solution $(f(a_1),\ldots,f(a_n)) \in R^n$.  Thinking contrapositively, if you can find a ring -- any ring -- for which the &#xA;equation $P = 0$ has no solutions in $R^n$, then you know right away that it has no solutions in $\mathbb{Z}$.  The most classical -- and also, not coincidentally, effective -- choices of rings $R$ to choose are $\mathbb{R}$ (the real numbers) and $\mathbb{Z}/n\mathbb{Z}$.  For instance:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $P(x,y) = x^2 + y^2 + 3 = 0$ has no solutions over $\mathbb{R}$, it has no solutions over $\mathbb{Z}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $P(x,y) = x^2 + y^2 - 3 = 0$ has no solutions over $\mathbb{Z}/4\mathbb{Z}$, it has no solutions over $\mathbb{Z}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for your more foundational question as to whether the finite rings $\mathbb{Z}/n\mathbb{Z}$ are defined in reference to $\mathbb{Z}$: by coincidence you have hit upon one of my standard pedagogical rants.  The answer is a resounding &lt;em&gt;yes&lt;/em&gt;.  If you don't think of the elements of $\mathbb{Z}/n\mathbb{Z}$ as equivalence classes of integers, how do you know that the addition and multiplication operations are commutative and associative, and that multiplication distributes over addition?  To show this, you do not have to &lt;em&gt;say&lt;/em&gt; &quot;quotient map&quot; if you really don't want to, but you have to use it, i.e., that the operations $+,\cdot$ in $\mathbb{Z}/n\mathbb{Z}$ are defined in terms of those in $\mathbb{Z}$.&lt;/p&gt;&#xA;" OwnerUserId="299" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2011-02-13T22:04:51.677" LastActivityDate="2011-02-13T22:04:51.677" CommentCount="7" />
  <row Id="21896" PostTypeId="1" AcceptedAnswerId="21897" CreationDate="2011-02-13T21:54:20.093" Score="2" ViewCount="100" Body="&lt;p&gt;If I understand correctly, the definition of the degree of a field extension $L/K$ is the dimension of $L$ over $K$ interpreted as a vector space. Now if the degree is $n &amp;lt; \infty$, the basis looks like $\{ 1, a_1, \dots , a_n\}$ where $a_i \in L - K$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it right to write $\{ 1, a, a^2, \dots , a^n\}$ for some $a \in L - K$ because of the primitive element theorem? If not, why exactly is the basis usually written $\{ 1, a, a^2, \dots , a^n\}$ instead of $\{ 1, a_1, \dots a_n\}$? &lt;/p&gt;&#xA;" OwnerUserId="5798" LastActivityDate="2011-02-13T22:04:51.497" Title="A question regarding finite field extensions" Tags="&lt;abstract-algebra&gt;&lt;field-theory&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="21897" PostTypeId="2" ParentId="21896" CreationDate="2011-02-13T22:02:39.113" Score="4" ViewCount="" Body="&lt;p&gt;You are correct - there is a basis of the form $\{1,a,a^2,\ldots,a^{n-1}\}$ for some $a\in L-K$ precisely when the primitive element theorem does in fact hold, which is when $L/K$ is a &lt;strong&gt;finite&lt;/strong&gt;, &lt;strong&gt;separable&lt;/strong&gt; extension. Note that the exponents only go up to $n-1$, since the dimension, and hence the number of elements in a basis, is $n$.&lt;/p&gt;&#xA;" OwnerUserId="264" LastActivityDate="2011-02-13T22:02:39.113" CommentCount="3" />
  <row Id="21898" PostTypeId="2" ParentId="21896" CreationDate="2011-02-13T22:04:51.497" Score="3" ViewCount="" Body="&lt;p&gt;$1$ doesn't necessarily have to be in the basis; we are ignoring the multiplication and just thinking about the vector space structure, so it should just look like $\{ a_1, ... a_n \}$ for some linearly independent $a_i \in L$ (and it doesn't matter whether they're in $K$ or not). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When people say that there exists $a$ such that $\{ 1, a, a^2, ... a^{n-1} \}$ is a basis, they are appealing to the primitive element theorem.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2011-02-13T22:04:51.497" CommentCount="2" />
  <row Id="21900" PostTypeId="2" ParentId="21852" CreationDate="2011-02-13T22:24:22.300" Score="2" ViewCount="" Body="&lt;p&gt;The passage in Paris and Kaminski's &quot;Asymptotics and Mellin-Barnes Integrals&quot; is available &lt;a href=&quot;http://books.google.de/books?id=LtOfHnClUSEC&amp;amp;lpg=PP1&amp;amp;ots=s2gZUP5D9n&amp;amp;dq=Asymptotics%20and%20Mellin-Barnes%20Integrals&amp;amp;pg=PA89#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;. It seems to me that the formulation &quot;because of this exponential decay, we are free to displace the contour of integration [...] over the poles [...] to produce the exponential series&quot; is not intended to mean that by displacing the contour over the poles we don't change the value of the integral, but rather that because of the exponential decay, we can displace the contour of integration and the value of the integral will change only due to the poles, and we can produce the exponential series by adding up the residues since the value of the remaining integral can be shown to go to zero as we move it to (negative) infinity. So perhaps there is nothing as mysterious here as you suspected? (To justify displacing the contour of integration, we have to integrate around infinite rectangular strips between two parallels of the imaginary axis. Because of the exponential decay, the pieces at infinity don't contribute, so the difference between the integrals along the two parallels of the imaginary axis must be given by the residues of the poles that lie within the strip.)&lt;/p&gt;&#xA;" OwnerUserId="6622" LastEditorUserId="6622" LastEditorDisplayName="" LastEditDate="2011-02-13T22:29:24.760" LastActivityDate="2011-02-13T22:29:24.760" CommentCount="1" />
  <row Id="21901" PostTypeId="1" AcceptedAnswerId="21909" CreationDate="2011-02-13T22:43:47.557" Score="5" ViewCount="347" Body="&lt;p&gt;I am in a biological field (medicine) but I have genuine passion for mathematics. I want to learn it on my own , in my spare time.  Mathematics , as I gather,  is learned best when you have grasped the prerequisite concepts for the area you are currently interested in.  Kindly suggest a sequence of study for the following areas ...  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Algebra&lt;br&gt;&#xA;2) Calculus&lt;br&gt;&#xA;3) Discrete math&lt;br&gt;&#xA;4) Geometry&lt;br&gt;&#xA;5) Probability and statistics&lt;br&gt;&#xA;6) Mathematics Software Packages (which one do you suggest for primarily educational, nonprofessional use)&lt;br&gt;&#xA;6) Any other areas of fundamental importance that I may have missed  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT: does first order logic and set theory belong up there ?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="2556" LastEditorUserId="2556" LastEditorDisplayName="" LastEditDate="2011-02-13T22:53:07.500" LastActivityDate="2011-02-14T03:33:31.777" Title="In what order should the following areas of mathematics be learned ?" Tags="&lt;self-learning&gt;" AnswerCount="5" CommentCount="4" FavoriteCount="3" />
  <row Id="21902" PostTypeId="2" ParentId="21901" CreationDate="2011-02-13T22:46:49.010" Score="4" ViewCount="" Body="&lt;p&gt;This is the order I would suggest:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Linear Algebra&lt;/li&gt;&#xA;&lt;li&gt;Calculus&lt;/li&gt;&#xA;&lt;li&gt;Discrete Math&lt;/li&gt;&#xA;&lt;li&gt;Probability and Statistics&lt;/li&gt;&#xA;&lt;li&gt;Geometry&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The first two are interchangeable. According to some &lt;a href=&quot;http://www.artofproblemsolving.com/Resources/Papers/DiscreteMath.pdf&quot; rel=&quot;nofollow&quot;&gt;people&lt;/a&gt;(article about teaching math), discrete math should really be taught &lt;a href=&quot;http://www.artofproblemsolving.com/Resources/articles.php?page=calculustrap&quot; rel=&quot;nofollow&quot;&gt;first&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2011-02-13T22:46:49.010" CommentCount="2" />
  <row Id="21903" PostTypeId="1" AcceptedAnswerId="21925" CreationDate="2011-02-13T22:50:27.970" Score="2" ViewCount="57" Body="&lt;p&gt;What's the usual method for proving the criticality of a graph?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've been trying out different methods and theorems but I can't find a decent method that's really convincing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks a lot in advance!&lt;/p&gt;&#xA;" OwnerUserId="5166" LastActivityDate="2011-02-14T01:07:28.780" Title="Criticality of a graph" Tags="&lt;graph-theory&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="21904" PostTypeId="2" ParentId="21890" CreationDate="2011-02-13T22:54:45.080" Score="0" ViewCount="" Body="&lt;p&gt;In order to show that a problem is in $\mathbf{P}$, you need to provide a polynomial-time algorithm for it. Check out the &lt;a href=&quot;http://en.wikipedia.org/wiki/Shortest_path_problem&quot; rel=&quot;nofollow&quot;&gt;shortest path problem&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's not clear what you mean by &quot;a path of at least $k$&quot; - it seems that $(G,a,b,k) \in L_2$ if $a,b$ are in the same connected component of $G$ and $(G,a,b,k-1) \notin L_1$, so that $L_2$ is also in $\mathbf{P}$.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2011-02-13T22:54:45.080" CommentCount="1" />
  <row Id="21905" PostTypeId="1" CreationDate="2011-02-14T00:01:41.920" Score="2" ViewCount="41" Body="&lt;p&gt;Is there a way to construct a linear feedback shift register (LFSR) which outputs no more than &lt;code&gt;k&lt;/code&gt; consecutive 1s or 0s? (It would have to be not a maximal-length LFSR, of course, if k &amp;lt; the degree of the characteristic polynomial)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;edit: I am also looking for one where the period is at least 1/16 the maximum period of the LFSR.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(clarification: exclude the trivial case of all zeros. Also, I can initialize the LFSR to an arbitrary vector.)&lt;/p&gt;&#xA;" OwnerUserId="120" LastEditorUserId="120" LastEditorDisplayName="" LastEditDate="2011-02-14T02:49:32.080" LastActivityDate="2011-02-14T02:49:32.080" Title="LFSR with limited numbers of runs?" Tags="&lt;discrete-mathematics&gt;&lt;lfsr&gt;" CommentCount="1" />
  <row Id="21906" PostTypeId="1" AcceptedAnswerId="21910" CreationDate="2011-02-14T00:07:08.257" Score="2" ViewCount="80" Body="&lt;p&gt;We want to reparametrize the curve&lt;br&gt;&#xA;$$\displaystyle \vec{r}(t)=&amp;lt;t^3+1, t^2-1, \frac{\sqrt{5}}{2}t^2&amp;gt;$$&#xA;in terms of the arc length measured from the point t=0 in the direction of increasing t.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is what I tried, but I've hit a snag:&lt;br&gt;&#xA;$$\displaystyle \vec{v}(t)=&amp;lt;3t^2, 2t, \sqrt{5}t&amp;gt;$$&#xA;$$\displaystyle |\vec{v}(t)|=3t\sqrt{t^2+1}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle s=\int_{0}^{t}3t\sqrt{t^2+1}d\tau=3t^2\sqrt{t^2+1}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think I'm missing something here. but assuming everything is correct, we need to solve:&lt;br&gt;&#xA;$$\frac{s^2}{9}=t^4(t^2+1)$$ for $t$, and then we are nearly done. I can't seem to solve for $t$ however, brain fart?  Assuming we did, we just plug $t$ in for the expression in terms of $s$ in the original equation and we are done?&#xA;P.S. this is exam review, not homework!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for reading!&lt;/p&gt;&#xA;" OwnerUserId="2533" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2011-02-14T00:12:16.217" LastActivityDate="2011-02-14T00:21:53.483" Title="Reparametrizing a curve in terms of the arc length" Tags="&lt;calculus&gt;&lt;parametric&gt;" AnswerCount="2" />
  <row Id="21908" PostTypeId="2" ParentId="21901" CreationDate="2011-02-14T00:10:25.990" Score="1" ViewCount="" Body="&lt;p&gt;In my opinion you want to learn the topics in this order :  &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Algebra &lt;/li&gt;&#xA;&lt;li&gt;Geometry &lt;/li&gt;&#xA;&lt;li&gt;Calculus/linear algebra, (statistics if you want, not necesarry)  &lt;/li&gt;&#xA;&lt;li&gt;real analysis&lt;/li&gt;&#xA;&lt;li&gt;complex analysis  &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Any duplicates on a line means that you can learn them simultaneously!  I would assume that you would cover some calculus rigorously in analysis and also touch upon some set theory, and of course, learn the fundamentals of &quot;proofs.&quot;&lt;/p&gt;&#xA;" OwnerUserId="2533" LastEditorUserId="2533" LastEditorDisplayName="" LastEditDate="2011-02-14T00:18:44.453" LastActivityDate="2011-02-14T00:18:44.453" CommentCount="0" />
  <row Id="21909" PostTypeId="2" ParentId="21901" CreationDate="2011-02-14T00:14:15.940" Score="6" ViewCount="" Body="&lt;p&gt;I'm unclear what you mean by &quot;Algebra&quot;; if you mean stuff like working with polynomials, basic equations, symbolic manipulation, etc., then that goes first. If you mean &quot;abstract algebra&quot;, then you can wait.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Added.&lt;/em&gt; Likewise: if by &quot;geometry&quot; you mean classical geometry, or even projective geometry, then the following applies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Calculus, Discrete Mathematics, and Geometry, are independent enough that their order doesn't matter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Added.&lt;/em&gt; However, if by &quot;geometry&quot; you mean analytic geometry, then it should definitely &lt;em&gt;precede&lt;/em&gt; calculus, and the same is true if it means trigonometry.  I think it unlikely that you meant &quot;differential geometry&quot; or &quot;algebraic geometry&quot;, but if you did those are very advanced topics that should wait until well after calculus, abstract algebra, and real/complex analysis. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For introductory probability and statistics you'll find Discrete Mathematics &lt;em&gt;very&lt;/em&gt; useful; for more advanced probability and statistics, Calculus is a must.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An &quot;introduction to proofs&quot;, which would include some set theory, some basic logic, etc., can be done at the same time as Discrete Mathematics, or immediately after. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;After all this, then you can hit linear algebra, abstract algebra, real or complex analysis, in pretty much any order (though complex analysis should follow real). Abstract algebra is a bit easier if you've taken linear algebra, but this is not strictly necessary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you happen to find probability and statistics very interesting, then you should do some measure theory after the real analysis.  &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-14T00:25:49.107" LastActivityDate="2011-02-14T00:25:49.107" CommentCount="2" />
  <row Id="21910" PostTypeId="2" ParentId="21906" CreationDate="2011-02-14T00:14:35.160" Score="2" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle s(t)=\int_{0}^{t} \left|\vec{v}(\tau) \right| d\tau=\int_{0}^{t}3\tau \sqrt{\tau^2+1}d\tau$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $\tau^2+1 = x^2$, then $\tau d \tau = x dx$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, the integral becomes $s(t) = \displaystyle \int_{1}^{\sqrt{1+t^2}}3x^2dx = x^3 \rvert_{1}^{\sqrt{1+t^2}} = (1+t^2)^{3/2} - 1$.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2011-02-14T00:21:53.483" LastActivityDate="2011-02-14T00:21:53.483" CommentCount="0" />
  <row Id="21911" PostTypeId="2" ParentId="21906" CreationDate="2011-02-14T00:16:09.183" Score="1" ViewCount="" Body="&lt;p&gt;You evaluated the integral incorrectly.  You have (as Sivaram also posted before I finished)&#xA;$$s=\int_0^t 3x\sqrt{x^2+1}dx,$$ which would best be handled by a change of variables. With the correct evaluation, $t$ will only appear once in the equation, making it easier to solve for $t$ in terms of $s$. &lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2011-02-14T00:16:09.183" CommentCount="4" />
  <row Id="21912" PostTypeId="2" ParentId="21901" CreationDate="2011-02-14T00:19:33.420" Score="1" ViewCount="" Body="&lt;p&gt;I'm doing the same thing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I find its best to interleave your study of these topics.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&#xA;1   algebra / geometry / calculus          | math&#xA;2   discrete math                          | software pkgs&#xA;3   probability+stats                      | (matlab/octave/maple)&#xA;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;You'll find a math software package takes a while to get used to (esp if you don't have a programming background).  So starting with one right away is a good idea.  &lt;a href=&quot;http://www.gnu.org/software/octave/&quot; rel=&quot;nofollow&quot;&gt;Octave&lt;/a&gt; is a free version of &lt;a href=&quot;http://www.mathworks.com/products/matlab/&quot; rel=&quot;nofollow&quot;&gt;MATLAB&lt;/a&gt;.  Be warned:  Octave can be painful to use at times.  MATLAB is the Cadillac.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also if you're doing proof-based study, you'll find a lot of calculus and algebra (at least as it appears to me at the moment) is mostly orthogonal.  That is, the Fundamental Theorem of Calculus won't really help you to prove the Triangle inequality.  But &lt;a href=&quot;http://math.stackexchange.com/questions/7743/getting-better-at-proofs&quot;&gt;how to prove&lt;/a&gt; and your way of thinking should be exercised by proving in either topic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Algebra is closely related to geometry.  Linear transformations, geometric intersection - to me geometry seems to boil down to simply applied algebra.&lt;/p&gt;&#xA;" OwnerUserId="1069" LastEditorUserId="1069" LastEditorDisplayName="" LastEditDate="2011-02-14T00:25:31.603" LastActivityDate="2011-02-14T00:25:31.603" CommunityOwnedDate="2011-02-14T00:19:33.420" />
  <row Id="21913" PostTypeId="1" AcceptedAnswerId="21963" CreationDate="2011-02-14T00:29:37.107" Score="7" ViewCount="211" Body="&lt;p&gt;From Hatcher, &quot;Algebraic Topology,&quot; &lt;a href=&quot;http://www.math.cornell.edu/~hatcher/AT/ATch2.pdf&quot; rel=&quot;nofollow&quot;&gt;Chapter 2, &quot;Singular Homology&quot; section&lt;/a&gt; (p. 108-109 in my copy):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Cycles in singular homology are defined algebraically, but they can be given a somewhat more geometric interpretation in terms of maps from finite $\Delta$-complexes.  To see this,  note that a singular $n$-chain $\xi$ can always be written in the form $\sum_i \varepsilon_i \sigma_i$ with $\varepsilon_i = \pm 1$, allowing repetitions of the singular $n$-simplices $\sigma_i$.  Given such an $n$-chain $\xi = \sum_i \varepsilon_i \sigma_i$, when we compute $\partial \xi$ as a sum of singular $(n-1)$-simplices with signs $\pm 1$, there may be some &lt;em&gt;canceling pairs&lt;/em&gt; consisting of two identical singular $(n-1)$-simplices with opposite signs. Choosing a maximal collection of canceling pairs, construct an $n$-dimensional $\Delta$ complex $K_\xi$ from a disjoint union of $n$-simplices, one for each $\sigma_i$, by identifying the pairs of $(n-1)$-dimensional faces corresponding to the chosen canceling pairs.  The $\sigma_i$'s then induce a map $K_\xi \rightarrow X$. If $\xi$ is a cycle, all the $(n-1)$ simplices of $K_\xi$ come from canceling pairs, hence are faces of exactly two $n$-simplices of $K_\xi$.  Thus $K_\xi$ is a manifold, locally homeomorphic to $\mathbb{R}^n$, except at a subcomplex of dimension at most $n - 2$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It's this last bit that has me confused, perhaps because I'm unable to visualize a nontrivial 3-manifold.  We're building a $\Delta$-complex by identifying the $(n-1)$-faces of simplices; there are only finitely many such simplices; we're not required to identify the faces in any unusual way; and so forth.  How, then, can we end up with something which is &lt;em&gt;not&lt;/em&gt; a manifold?  I'd really appreciate an explicit example.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've been told that this is a general example of a &quot;&lt;a href=&quot;http://en.wikipedia.org/wiki/Pseudomanifold&quot; rel=&quot;nofollow&quot;&gt;pseudomanifold&lt;/a&gt;,&quot; but all the examples of pseudomanifolds that I've been able to locate and follow wind up making a non-manifold by essentially identifying vertices to get pinched spaces.  This can't happen under the present construction because we're always identifying the largest proper faces.  So I'm quite confused by the situation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Looks like there are some more questions raised in the comments as to what the construction actually means.  If there's a name for this construction, or any other source that discusses it, I'd appreciate a reference.  And of course if anyone can shed any additional light on the topic that would be quite welcome too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: I may have inadvertently overwritten someone else's edit just then, judging by a message that popped up.  Apologies if so.  (How do I tell / fix this?)&lt;/p&gt;&#xA;" OwnerUserId="3296" LastEditorUserId="3296" LastEditorDisplayName="" LastEditDate="2011-02-14T04:55:05.947" LastActivityDate="2011-02-14T06:17:14.593" Title="Question about Singular Homology section in Hatcher" Tags="&lt;algebraic-topology&gt;&lt;manifold&gt;&lt;homology&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="2" />
  <row Id="21915" PostTypeId="1" AcceptedAnswerId="21924" CreationDate="2011-02-14T00:43:35.030" Score="5" ViewCount="161" Body="&lt;p&gt;So I have been studying the multi-variable chain rule.  Most importantly, and this is what I must have overlooked, is it's not always clear to me how to see which variables are functions of other variables, so that you know when to use the chain rule.  For example, if you have:&lt;br&gt;&#xA;$$\displaystyle x^2+y^2-z^2+2xy=1$$&#xA;$$\displaystyle x^3+y^3-5y=8$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, say we want to find $\frac{dz}{dt}$ but z is a function of x, then we get:&lt;br&gt;&#xA;$$\displaystyle \frac{dz}{dt}=\frac{dz}{dx}\frac{dx}{dt}$$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And if z is a funtion of both y and t, we get:&lt;br&gt;&#xA;$$\displaystyle \frac{dz}{dt}=\frac{dz}{dx}\frac{dx}{dt}+\frac{dz}{dy}\frac{dy}{dt}$$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case, we have 2 equations.  One involving all 3 variables $x,y,z$ and one involving just $x,y$.  Say we want to find $\frac{dz}{dx}$. What does this mean for this case?  How should we interpret this rule in general?&lt;/p&gt;&#xA;" OwnerUserId="2533" LastEditorUserId="2533" LastEditorDisplayName="" LastEditDate="2011-02-14T03:54:51.733" LastActivityDate="2011-03-24T18:42:36.947" Title="Chain rule for multi-variable functions" Tags="&lt;calculus&gt;&lt;multivariable-calculus&gt;" AnswerCount="3" />
  <row Id="21916" PostTypeId="1" AcceptedAnswerId="21920" CreationDate="2011-02-14T00:47:06.137" Score="8" ViewCount="102" Body="&lt;p&gt;Howdy all, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For some reason I cannot figure out how the book is finding the solution to this problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;find the inverse function of $f(x)=3x\sqrt{x}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My steps seem to lead to a dead end:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;step 1. switch $f(x)$ with $y$: $y = 3x\sqrt{x}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;step 2. swap $x$ and $y$: $x = 3y\sqrt{y}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;step 3. solve for $y$: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;step 3.1: $\displaystyle \frac{x}{3y} = \sqrt{y}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;step 3.2: $\displaystyle \left(\frac{x}{3y}\right)^2 = \left(\sqrt{y}\right)^2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;step 3.3: $\displaystyle \frac{x^2}{9y^2} = y$ ... uhhh?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The book answer is $\displaystyle y = \left(\frac{x}{3}\right)^{2/3}$.&lt;/p&gt;&#xA;" OwnerUserId="5032" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2011-02-14T01:04:37.580" LastActivityDate="2011-02-14T01:04:37.580" Title="Finding an Inverse function with multiple occurences of $y$" Tags="&lt;homework&gt;&lt;algebra-precalculus&gt;" AnswerCount="5" CommentCount="2" />
  <row Id="21917" PostTypeId="2" ParentId="21913" CreationDate="2011-02-14T00:47:33.843" Score="1" ViewCount="" Body="&lt;p&gt;Edit: The example below doesn't give a cycle, so it's not really what we're looking for here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Original post: The simplest example of what you're looking for is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Dunce_hat_(topology)&quot; rel=&quot;nofollow&quot;&gt;dunce cap.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's a compact 2-dimensional complex, and it's contractible, so it's definitely not a manifold w/o boundary. &lt;/p&gt;&#xA;" OwnerUserId="1392" LastEditorUserId="1392" LastEditorDisplayName="" LastEditDate="2011-02-14T04:33:45.157" LastActivityDate="2011-02-14T04:33:45.157" CommentCount="11" />
  <row Id="21918" PostTypeId="2" ParentId="21916" CreationDate="2011-02-14T00:50:56.397" Score="2" ViewCount="" Body="&lt;p&gt;$x=3y\sqrt y$ implies $x^2=9y^3$.&lt;/p&gt;&#xA;" OwnerUserId="589" LastActivityDate="2011-02-14T00:50:56.397" />
  <row Id="21919" PostTypeId="2" ParentId="21916" CreationDate="2011-02-14T00:52:52.380" Score="2" ViewCount="" Body="&lt;p&gt;The algorithm you are applying is correct, and will yield the inverse function.  In step 3 you should square both sides.  Then we get $$x^2=9y^2\cdot y=9y^3$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you solve the problem from here?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint:  Divide by $9$ and then take cube roots.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope that helps.&lt;/p&gt;&#xA;" OwnerUserId="6075" LastActivityDate="2011-02-14T00:52:52.380" />
  <row Id="21920" PostTypeId="2" ParentId="21916" CreationDate="2011-02-14T00:53:20.137" Score="3" ViewCount="" Body="&lt;p&gt;Nothing wrong with what you've done so far, you just haven't finished!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From $\displaystyle \frac{x^2}{9y^2} = y$, move all the $y$'s to the same side:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;step 3.4: $x^2 = 9y^3$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now isolate $y$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;step 3.5: $\displaystyle\frac{x^2}{9} = y^3$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now solve for $y$ by taking cubic roots; you might also notice that the left hand side is a perfect square...&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2011-02-14T00:53:20.137" />
  <row Id="21921" PostTypeId="2" ParentId="21916" CreationDate="2011-02-14T00:54:47.657" Score="2" ViewCount="" Body="&lt;p&gt;Two hints:  $\sqrt{x}=x^{\underline{?}}$ and $x^a\cdot x^b=x^{\underline{?}}$&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2011-02-14T00:54:47.657" />
  <row Id="21922" PostTypeId="2" ParentId="21916" CreationDate="2011-02-14T01:00:32.673" Score="3" ViewCount="" Body="&lt;p&gt;Note that $\frac{x^2}{9}=\left(\frac{x}{3}\right)^2$.  Also note that you can multiply both sides of your equation by $y^2$ to get just one $y$ term, and that the cube root is the same as the $\frac{1}{3}$ power.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could, as lhf indicates, avoid ever breaking up the $y$ terms in the first place.  Technically this would be preferred, because when you divided by $y$ you made the assumption that $y\neq 0$, which is not always true.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Beware of squaring when solving equations in general, because it can lead to &lt;a href=&quot;http://hotmath.com/hotmath_help/topics/extraneous-solutions.html&quot; rel=&quot;nofollow&quot;&gt;extraneous solutions&lt;/a&gt;. In this case it is a valid step, but for a silly example consider trying to solve the equation $\sqrt{y}=-1$ by squaring both sides.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A final remark.  The original function has an implicit domain, $x\geq 0$.  Your inverse function has a formula definition that does not appear to require any domain restriction, but it is misleading to say that $f^{-1}(x)=\left(\frac{x}{3}\right)^{2/3}$ without mentioning that this is only valid when $x\geq 0$.  This domain restriction arises from the fact that the range of $f$ is $[0,\infty)$.  The function $h(x)=\left(\frac{x}{3}\right)^{2/3}$ on the whole real line is not invertible (it fails the horizontal line test), whereas if $g(x)=-f(x)=-3x\sqrt{x}$, then the inverse function would be $g^{-1}(x)=\left(\frac{x}{3}\right)^{2/3}$ on $(-\infty,0]$.  Looking at the graphs of these functions and keeping in mind how inverses are reflected across the line $y=x$ (resulting from switching $x$ and $y$) should make it clearer what is going on.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2011-02-14T01:00:32.673" CommunityOwnedDate="2011-02-14T01:00:32.673" />
  <row Id="21923" PostTypeId="2" ParentId="21915" CreationDate="2011-02-14T01:01:25.853" Score="3" ViewCount="" Body="&lt;p&gt;If we have an explicit function $z = f(x_1,x_2,\ldots,x_n)$, then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle \frac{dz}{dt} = \frac{\partial z}{\partial x_1} \frac{dx_1}{dt} + \frac{\partial z}{\partial x_2} \frac{dx_2}{dt} + \cdots +\frac{\partial z}{\partial x_n} \frac{dx_n}{dt}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we have an implicit function $f(z,x_1,x_2,\ldots,x_n) = 0$, then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle \frac{\partial f}{\partial z} \frac{dz}{dt} + \frac{\partial f}{\partial x_1} \frac{dx_1}{dt} + \frac{\partial f}{\partial x_2} \frac{dx_2}{dt} + \cdots +\frac{\partial f}{\partial x_n} \frac{dx_n}{dt} = 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle \frac{dz}{dt} = - \frac{ \frac{\partial f}{\partial x_1} \frac{dx_1}{dt} + \frac{\partial f}{\partial x_2} \frac{dx_2}{dt} + \cdots +\frac{\partial f}{\partial x_n} \frac{dx_n}{dt}}{\frac{\partial f}{\partial z} }$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the first example,&#xA;\begin{align*}&#xA;\displaystyle x^2 + y^2 - z^2 + 2y &amp;amp; = 1\\&#xA;\displaystyle 2x\frac{dx}{dt} + 2y\frac{dy}{dt} - 2z\frac{dz}{dt} + 2\frac{dy}{dt} &amp;amp; = 0\\&#xA;\displaystyle x\frac{dx}{dt} + y\frac{dy}{dt} + \frac{dy}{dt} &amp;amp; = z\frac{dz}{dt}\\&#xA;\displaystyle \frac{dz}{dt} &amp;amp; = \frac{x\frac{dx}{dt} + y\frac{dy}{dt} + \frac{dy}{dt}}{z}&#xA;\end{align*}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the second example,&#xA;\begin{align*}&#xA;\displaystyle x^3 + y^3 - 5y &amp;amp; = 8\\&#xA;\displaystyle 3x^2 \frac{dx}{dt} + 3y^2 \frac{dy}{dt} - 5 \frac{dy}{dt} = 0\\&#xA;\displaystyle \frac{dy}{dt} &amp;amp; = \frac{3x^2}{5-3y^2} \frac{dx}{dt}&#xA;\end{align*}&lt;/p&gt;&#xA;" OwnerUserId="3302" LastActivityDate="2011-02-14T01:01:25.853" />
  <row Id="21924" PostTypeId="2" ParentId="21915" CreationDate="2011-02-14T01:07:05.520" Score="2" ViewCount="" Body="&lt;p&gt;You have &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle x^2+y^2-z^2+2y=1$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle x^3+y^3-5y=8$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can take the derivative of each with respect to $x$ to get &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle 2x+2y\frac{dy}{dx}-2z\frac{dz}{dx}+2\frac{dy}{dx}=0$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle 3x^2+3y^2\frac{dy}{dx}-5\frac{dy}{dx}=0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second gives us  $\frac{dy}{dx}=\frac{3x^2}{3y^2-5}$ which you can substitute into the first to find $\frac{dz}{dx}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Response to edit of the question:&#xA;The new first equation:  $\displaystyle x^2+y^2-z^2+2xy=1$  when we take the derivative by x gives $\displaystyle 2x+2y\frac{dy}{dx}-2z\frac{dz}{dx}+2x\frac{dy}{dx}+2y=0$, which can still be solved for dz/dx given the dy/dx of the second equation.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2011-02-14T04:11:50.000" LastActivityDate="2011-02-14T04:11:50.000" CommentCount="2" />
  <row Id="21925" PostTypeId="2" ParentId="21903" CreationDate="2011-02-14T01:07:28.780" Score="1" ViewCount="" Body="&lt;p&gt;If $G$ is $n$-critical (i.e. $\chi(G) = n$), then $\delta(G) \geq n-1$. So if $\delta(G) &amp;lt; n-1$ then $G$ is not $n$-critical.&lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2011-02-14T01:07:28.780" CommentCount="2" />
  <row Id="21926" PostTypeId="2" ParentId="21915" CreationDate="2011-02-14T01:11:16.820" Score="0" ViewCount="" Body="&lt;p&gt;What you want to look up is the chain rule for &lt;a href=&quot;http://en.wikipedia.org/wiki/Chain_rule#The_chain_rule_in_higher_dimensions&quot; rel=&quot;nofollow&quot;&gt;multivalued functions&lt;/a&gt;.  What follows deals with the second example, where $z$ is a function of $x$ and $y$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose $z:\mathbb{R}^n\rightarrow \mathbb{R}$ and $\boldsymbol{T} : \mathbb{R} \rightarrow \mathbb{R}^n$ are two differentiable functions.  Then $z\circ \boldsymbol{T} : \mathbb{R}\rightarrow \mathbb{R}$ and is differentiable, so we can ask what is $\frac {d(z(\boldsymbol{T}(t)))}{dt}$?  It turns out that at $t=t_0$ we have $$\frac {d(z(\boldsymbol{T}(t)))}{dt} = \nabla z(\boldsymbol{T}(t_0))\cdot \boldsymbol{T}^&amp;#39;(t_0) $$  Where $\nabla$ is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Gradient&quot; rel=&quot;nofollow&quot;&gt;gradient&lt;/a&gt; and the $\cdot$ refers to the &lt;a href=&quot;http://en.wikipedia.org/wiki/Dot_product&quot; rel=&quot;nofollow&quot;&gt;dot product&lt;/a&gt; between vectors.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope that helps,&lt;/p&gt;&#xA;" OwnerUserId="6075" LastEditorUserId="6075" LastEditorDisplayName="" LastEditDate="2011-03-24T18:42:36.947" LastActivityDate="2011-03-24T18:42:36.947" />
  <row Id="21927" PostTypeId="2" ParentId="2949" CreationDate="2011-02-14T01:34:28.523" Score="5" ViewCount="" Body="&lt;p&gt;A beautiful result by Erdős:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In any sequence of distinct $n^2 + 1$ integers, there is always some increasing or decreasing subsequence of length $n + 1$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pigeonhole!&lt;/p&gt;&#xA;" OwnerUserId="5105" LastActivityDate="2011-02-14T01:34:28.523" CommunityOwnedDate="2011-02-14T01:34:28.523" />
  <row Id="21928" PostTypeId="1" AcceptedAnswerId="21929" CreationDate="2011-02-14T01:41:15.430" Score="5" ViewCount="180" Body="&lt;p&gt;How would you compute for the definite integral of &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_0^{\infty}\frac{dx}{(1+x^2)^4}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that integral of $\displaystyle \frac1{(1+x^2)}$ equals $\tan^{-1}x$. I tried using integration by parts without much luck.  My teacher pointed me to special functions by which I found out about the hypergeometric distribution. Although I don't know how to apply it to this problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anybody know how to use special functions or how to go about this problem?&lt;/p&gt;&#xA;" OwnerUserId="7049" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2011-04-20T02:23:36.903" LastActivityDate="2011-04-20T02:23:36.903" Title="Tricky integration?" Tags="&lt;calculus&gt;&lt;integral&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="21929" PostTypeId="2" ParentId="21928" CreationDate="2011-02-14T01:49:40.817" Score="8" ViewCount="" Body="&lt;p&gt;The method given below works in general for any integral of the form $\displaystyle \int_{0}^{\infty} \frac{dx}{(1+x^2)^n}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Plug in $x = \tan(\theta)$. The integral becomes&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle \int_{0}^{\infty} \frac{dx}{(1+x^2)^4} = \int_{0}^{\pi/2} \frac{\sec^2(\theta) d \theta}{(1+\tan^2(\theta))^4} = \int_{0}^{\pi/2} \cos^{6}(\theta) d\theta = \frac{5}{6}\frac{3}{4}\frac{1}{2}\frac{\pi}{2} = \frac{5}{32}\pi$$&#xA;where the last integral has been done in a previous post &lt;a href=&quot;http://math.stackexchange.com/questions/20397/striking-applications-of-integration-by-parts/20481#20481&quot;&gt;here&lt;/a&gt;. The integration over there has been done for $\sin^n(\theta)$ but the same method works for $\cos^n(\theta)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are familiar with complex analysis, you could solve it using complex analysis by extending the function into the complex domain, choose a semicircular contour with the diameter along the real axis. Look for the poles inside the contour, (there are four poles at $z=+i$) and compute the residues. Let the radius of the semicircle tend to infinity and evaluate the integral using &lt;a href=&quot;http://en.wikipedia.org/wiki/Residue_theorem&quot; rel=&quot;nofollow&quot;&gt;Cauchy residue theorem&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2011-02-14T01:55:08.797" LastActivityDate="2011-02-14T01:55:08.797" CommentCount="3" />
  <row Id="21930" PostTypeId="2" ParentId="21851" CreationDate="2011-02-14T01:52:33.303" Score="4" ViewCount="" Body="&lt;p&gt;Interesting question.  In what follows consider the $n\times n$ square grid.  Notice that the trivial solution obtained by following a square spiral towards the center starting from an outside corner yields a solution with $2n-1$ lines.&lt;br&gt;&#xA;To see why, notice that 2 lines reduces the grid to a $(n-1)\times (n-1)$ grid, and since the $1\times 1$ grid requires only 1 line, induction yields $2n-1$ lines.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can we do better?  Based on the posts in the forum and my own attempts, I believe the answer is that $2n-2$ lines is optimal.  Showing this is possible is again easy.  Start at a corner, and spiral towards the center until there is only a $3\times 3$ grid remaining.  Recall from above that 2 lines in the spiral will reduce the grid by a dimension, so thus far we will have used $2\cdot (n-3)=2n-6$ lines.  On the last line, end it so that we are in a position to go through the diagonal of the $3\times 3 $ grid.  Since the $3\times 3$ grid has a solution with $4$ lines starting diagonally from a corner we have found a solution to the $n\times n$ grid using only $2n-2$ lines.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the question remains, is $2n-2$ optimal?  The more I think about it, the more I believe it, but a proof does not leap into mind.  I will think more.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Of course $n=1,2$ are exceptions, and required $2n-1$ lines.  Interestingly, the method I presented can be modified slightly to produce a closed path.  All that must be changed is the way the final $3\times 3$ grid is traversed, and perhaps moving the starting position of the first line to a spot slightly outside of the original $n\times n$ grid.  In other words the conjecture Toshi Kato is true.  &lt;/p&gt;&#xA;" OwnerUserId="6075" LastEditorUserId="6075" LastEditorDisplayName="" LastEditDate="2011-02-14T22:26:48.727" LastActivityDate="2011-02-14T22:26:48.727" CommentCount="1" />
  <row Id="21931" PostTypeId="1" AcceptedAnswerId="22415" CreationDate="2011-02-14T02:24:14.637" Score="6" ViewCount="127" Body="&lt;p&gt;Let $x_i$ be iid nonnegative discrete random variables $E[x_i]=N/M$ for some integers $N, M$, variance $\sigma^2$ and higher moments known (finite). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, the sum $\displaystyle S = \sum_{i=1}^M x_i$ will have $E[S]=N$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm interested in the probability that &#xA;$S$ takes that precise value: $A=P\left(S=E[S]\right)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Applying the central limit theorem, I can write&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle A \approx \frac{1}{\sqrt{ 2 \pi M \sigma^2}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is: can this approximation be refined?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ADDED: To add some example-context-motivation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lets consider $X$ as a sum of $N$ Bernoullis (0/1) with prob=$p$, such that $E(X)=E(N p)$ is an integer. We can compute exactly the probability that $X$ attains its expected value, it's a Binomial:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle P = P(X= N p) = {N \choose N p} p^{N p} q^{N q} \hspace{2cm}$  [1a]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We might also get an approximate value of that probability using the CTL (Central Limit Theorem)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle P \approx \frac{1}{\sqrt{2 \pi N p q}} \hspace{2cm} $  [2a]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we take [1a] and use the Stirling approximation, with $K \approx (K/e)^K \sqrt{2 \pi K}$, we get the same value. Fine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, we may try to refine the approximation, both from [1a] and [2a].&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Plugging the next orden Stirling approximation in [1a], we get (I am not mistaken)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle P \approx \frac{1}{\sqrt{2 \pi N p q}} \left(1 - \frac{1- p q}{12 N p q} \right) \hspace{2cm} $  [1b]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To refine the CTL, one can think of &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;use some &quot;continuity correction&quot; to evaluate more precisely the (hipothetical) gaussian integral&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;add some terms from the Edgeworth expansions&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;do nothing of the above - because the CLT does not justify those procedures in this scenario (just one value of a discrete variable)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I'm not sure which is the correct way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But let's try the first one:  the next order approximation of the integral gives me (again, if I'm not mistaken)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle P \approx \frac{1}{\sqrt{2 \pi N p q}} \left(1 - \frac{1}{24 N p q} \right) \hspace{2cm} $  [2b]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is not the same as [1b], but it's close.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this just casual? Was it a reasonable thing to do? Should I look (also/instead) after the Edgeworth expansions?&lt;/p&gt;&#xA;" OwnerUserId="312" LastEditorUserId="312" LastEditorDisplayName="" LastEditDate="2011-02-15T21:13:33.377" LastActivityDate="2011-02-16T22:47:37.667" Title="Refining the central limit theorem on discrete random vars" Tags="&lt;probability&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="1" />
  <row Id="21932" PostTypeId="1" AcceptedAnswerId="21935" CreationDate="2011-02-14T02:27:14.267" Score="5" ViewCount="107" Body="&lt;p&gt;Consider the following diagram:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://upload.wikimedia.org/wikipedia/commons/thumb/8/84/First_isomorphism_theorem_%28plain%29.svg/200px-First_isomorphism_theorem_%28plain%29.svg.png&quot; rel=&quot;nofollow&quot;&gt;commutative diagram&lt;/a&gt; (Sorry, It wouldn't let me directly post the image.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What does it mean precisely to say &quot;$f$ factors through $G/\text{ker}(f)$&quot;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does it mean $f = \tilde{f} \circ \pi$, for some $\tilde{f}$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've seen texts use the phrase, but never a definition of this notion.&lt;/p&gt;&#xA;" OwnerUserId="7029" LastActivityDate="2011-02-14T02:32:27.343" Title="What does it mean to say a map &quot;factors through&quot; a set?" Tags="&lt;abstract-algebra&gt;&lt;set-theory&gt;" AnswerCount="1" FavoriteCount="2" />
  <row Id="21933" PostTypeId="1" AcceptedAnswerId="21936" CreationDate="2011-02-14T02:30:04.740" Score="6" ViewCount="193" Body="&lt;p&gt;I recently found out that there is no algorithm which, given an arbitrary group presentation, will determine in finite time if it represents the trivial group*.  Additionally, in a lecture I recently attended, it was portrayed that given any desired property P of a group, there is no algorithm that will determine if an arbitrary group (presentation) has this property.  (For additional info, see the second answer to &lt;a href=&quot;http://mathoverflow.net/questions/31355/what-can-be-said-about-a-group-from-its-presentation&quot; rel=&quot;nofollow&quot;&gt;this question&lt;/a&gt; on MathOverflow).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is about the scope of this notion of &quot;algorithm.&quot;  With regard to group presentations, my intuition tells me that there must be some way to classify or otherwise look at these arbitrary group presentations that would enable us to determine if they are the trivial group (or have some property P).  However, I am not very well versed on what it means for something to be solved via an algorithm vs. other mathematical methods.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, if something has been shown to be unsolvable via algorithm, does it mean that it is entirely unsolvable (and thus futile to pursue in research)?  Or, are there cases where we have shown that despite being unable to use an algorithm to determine if a property exists, we have used other mathematical methods with success?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;*Clarification by Arturo.&lt;/p&gt;&#xA;" OwnerUserId="7051" LastEditorUserId="7051" LastEditorDisplayName="" LastEditDate="2011-02-14T02:48:48.057" LastActivityDate="2011-02-20T21:51:25.647" Title="Does algorithmic unsolvability imply unsolvability in general?" Tags="&lt;group-theory&gt;&lt;logic&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="1" />
  <row Id="21934" PostTypeId="1" AcceptedAnswerId="22005" CreationDate="2011-02-14T02:30:10.123" Score="1" ViewCount="148" Body="&lt;p&gt;It is well-known that any set $E \subseteq \mathbb{R}$ with positive outer measure contains a nonmeasurable subset $V$. I know that $0 &amp;lt; m^*(V) \le m^*(E)$. Nevertheless, my question is the following: given $r \in \mathbb{R}$ such that $r&amp;gt;0$, is there a nonmeasurable subset of $\mathbb{R}$ whose outer measure is exactly $r$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you in advance.&lt;/p&gt;&#xA;" OwnerUserId="5956" LastEditorUserId="5956" LastEditorDisplayName="" LastEditDate="2011-02-14T06:16:33.633" LastActivityDate="2011-02-14T14:42:48.910" Title="Nonmeasurable set with positive outer measure" Tags="&lt;measure-theory&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="21935" PostTypeId="2" ParentId="21932" CreationDate="2011-02-14T02:32:27.343" Score="6" ViewCount="" Body="&lt;p&gt;It means exactly what you write: that you can express $f$ as  &quot;product&quot; (composition) of two functions, with the first function going through $G/\mathrm{ker}(f)$; by implication, that map will be the &quot;natural&quot; map into the quotient, i.e., $\pi$. Under more general circumstances, you would also indicate the map in question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason for the term &quot;factors&quot; is that if you write composition of functions by juxtaposition, which is fairly common, then the equation looks exactly as if you &quot;factored&quot; $f$: $f=\tilde{f}\pi$. &lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2011-02-14T02:32:27.343" CommentCount="3" />
  <row Id="21936" PostTypeId="2" ParentId="21933" CreationDate="2011-02-14T02:52:11.967" Score="15" ViewCount="" Body="&lt;p&gt;First: just exactly what &lt;em&gt;is&lt;/em&gt; and what is not an algorithm is in fact a fairly deep philosophical question. Intuitively, we are talking about a &quot;recipe&quot; that, given the input, will yield an answer in an automatic manner, following certain pre-set rules and procedures.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There have been several independent attempts at formalizing the notion: Turing machines/Turing computability, Church's $\lambda$-calculus, Goedel's (amplified by Kleene and Rosser) notion of general recursive functions. It turned out that all of these notions are &lt;em&gt;equivalent&lt;/em&gt;, in that if some process satisfies one of the definitions, then it satisfies all of them. This led Church to state his famous &lt;a href=&quot;http://en.wikipedia.org/wiki/Church%27s_thesis&quot; rel=&quot;nofollow&quot;&gt;thesis&lt;/a&gt;, that in fact these notions &lt;em&gt;do&lt;/em&gt; capture the idea of what an &quot;algorithm&quot; is; that is, that anything we can reasonably call an &quot;algorithm&quot; will in fact satisfy these definitions. (Everyone agrees anything that is Turing computable/$\lambda$-definable/given by general recursive functions is an algorithm; the question is whether everything that &quot;is&quot; an algorithm satisfies the definition). It is not 100% acceptable, but most people seem to accept it. It's not something that can be proven formally; it can only be settled in the negative (by producing something that everyone agrees should be called an &quot;algorithm&quot;, but which can be shown to not be Turing computable/$\lambda$-definable/etc). We cannot prove it formally, because that would require a formal definition of what an &quot;algorithm&quot; is, and this is precisely the crux of the matter!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming you take Church's thesis as true, then you can essentially think of &quot;algorithm&quot; as something you can program your computer to do, and which will, without any further action, &lt;em&gt;eventually&lt;/em&gt; produce an answer, positive or negative, even if the &quot;eventually&quot; is so far off that the Sun would burn out before you got the answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, second: what does it mean to say that there is no algorithm which, given an arbitrary group presentation, will always be able to decide if the group is trivial? It means that there is no way to program a computer to figure it out. If you try writing a program to do it, either it will not always give you the correct answer, or it will sometimes &lt;em&gt;choke&lt;/em&gt;, meaning, it will be caught in an endless loop and never give you an answer (even theoretically). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a strong result: it doesn't just say that nobody has found a way to program a computer to do it, it says that &lt;em&gt;nobody ever will&lt;/em&gt;; there is simply no set of instructions that will do it. This is much like saying that one cannot trisect the general angle using only straightedge and compass: it doesn't mean nobody has been able to figure out how, it means that nobody ever &lt;em&gt;will&lt;/em&gt;, that no such procedure can exist.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That &lt;strong&gt;does not&lt;/strong&gt; mean that given a group presentation &lt;em&gt;you&lt;/em&gt; cannot figure out a way of determining if it is the trivial group, or whether the group it defines has property $P$. That's because &lt;em&gt;you&lt;/em&gt; are not bound by a set of rigid instructions; you can make guesses, inspired at times, you can use tricks, you can come up with clever maps into nontrivial groups, etc. Again, for an analogy, think about writing down an elementary function (addition, product, division, powers, radicals, exponentials, trigonometric functions, logarithms, etc). You can program a computer to write down its derivative, because there is an &lt;em&gt;algorithmic&lt;/em&gt; procedure for computing the formal derivative, using the product rule, sum rule, quotient rule, chain rule, etc. However, in order to &lt;em&gt;integrate&lt;/em&gt; the function, we don't have an &lt;em&gt;algorithm&lt;/em&gt;: we don't have a recipe that will always work. Instead, we have a bunch of heuristics of things to try, common tricks, and so on. That's why Mathematica can always give you the derivative, but sometimes chokes when you ask it to compute an indefinite integral.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Added.&lt;/em&gt; It also does not mean that there aren't &lt;em&gt;subclasses&lt;/em&gt; of groups for which you can solve the problem. For example, if you are given a finite presentation for an &lt;em&gt;abelian&lt;/em&gt; group, then there is a simple algorithm that will decide if that abelian group is trivial or not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, saying that the word problem (determining if a given finite group presentation yields the trivial group) does not have an algorithmic solution does not make it &quot;entirely unsolvable&quot;; it means there is no &quot;recipe&quot; that will &lt;em&gt;always&lt;/em&gt; work. Now there are &lt;em&gt;other&lt;/em&gt; interesting questions one can ask: is there a large and natural class of presentations where we &lt;em&gt;do&lt;/em&gt; have an algorithm? If so, what are they, and what is the algorithm? Can we trace where the boundary is between &quot;easy&quot; groups (those for which we have a general algorithm) and the &quot;hard&quot; ones (those for which we cannot)? Can we classify the kinds of presentations for which there is &lt;em&gt;no&lt;/em&gt; algorithm? Can we come up with &lt;em&gt;heuristics&lt;/em&gt; to help try to solve the problem? Given a class of groups, can we find an algorithm that works for &lt;em&gt;that&lt;/em&gt; class (e.g., nilpotent groups, solvable groups, etc). And so on. Particular instances of the problem (that is, particular presentations) may be dealt with in an ad hoc manner (a method specifically tailored to that particular presentation), so that determining if a particular presentation is the trivial group is not &quot;unsolvable&quot;. It's just that we cannot do it mindlessly.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-20T21:51:25.647" LastActivityDate="2011-02-20T21:51:25.647" CommentCount="3" />
  <row Id="21937" PostTypeId="2" ParentId="21839" CreationDate="2011-02-14T02:59:16.300" Score="1" ViewCount="" Body="&lt;p&gt;I agree : it is definitely Lim Sup, Lim Inf.; I have seen it used many times. If you do not see the top or bottom, you still have a Lim, but ---No Sup For You!&lt;/p&gt;&#xA;" OwnerUserId="7052" LastActivityDate="2011-02-14T02:59:16.300" />
  <row Id="21938" PostTypeId="2" ParentId="10375" CreationDate="2011-02-14T03:12:18.930" Score="2" ViewCount="" Body="&lt;p&gt;A topos with an NNO has intuitionistic internal logic in general (if you assume Choice then Dionescu's theorem tells you that the internal logic is classical), so any proof that relies on proof by contradiction will not work (not to say the results won't, but the proof needs to be fixed, or your concepts altered).&lt;/p&gt;&#xA;" OwnerUserId="3835" LastActivityDate="2011-02-14T03:12:18.930" />
  <row Id="21940" PostTypeId="2" ParentId="2949" CreationDate="2011-02-14T03:22:37.670" Score="1" ViewCount="" Body="&lt;p&gt;If $G$ is a (Hausdorff, locally compact) totally disconnected abelian group which is a filtered union of its compact open subgroups (e.g. the additive group of $\mathbb{Q}_p$), then the category of smooth complex representations of $G$ (smooth meaning the action map is continuous when the vector space has the discrete topology) is canonically equivalent to the category of sheaves of complex vector spaces on the Pontryagin dual of $G$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a beautiful example of an algebro-geometric duality in representation theory and quite shocking to me. The situation is sort of analogous to the equivalence, for a commutative ring $R$, of $R$-modules with quasi-coherent sheaves on $\text{Spec} \ R$.&lt;/p&gt;&#xA;" OwnerUserId="4842" LastEditorUserId="4842" LastEditorDisplayName="" LastEditDate="2011-02-14T04:59:50.933" LastActivityDate="2011-02-14T04:59:50.933" CommunityOwnedDate="2011-02-14T03:22:37.670" />
  <row Id="21941" PostTypeId="2" ParentId="21864" CreationDate="2011-02-14T03:23:58.557" Score="3" ViewCount="" Body="&lt;p&gt;Know that the solution to Kepler's equation can in fact be expressed as a series of Bessel functions of the first kind:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$E=M+2\sum_{k=1}^\infty \frac{\sin(k M)J_k(k\epsilon)}{k}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but since the Bessel function is a bit more complicated to use, one is better off using Newton-Raphson or Halley to solve Kepler's equation (particularly convenient due to the properties of the derivatives of the sine and cosine, so you can recycle common subexpressions in numerical evaluation). That leaves you with the problem of starting up the iteration; I personally prefer &lt;a href=&quot;http://adsabs.harvard.edu/full/1979CeMec..19..163S&quot; rel=&quot;nofollow&quot;&gt;the starting approximation due to G.R. Smith&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$E\approx M+\frac{\epsilon\;\sin\;M}{1+\sin\;M-\sin(\epsilon+M)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is easily polished off subsequently with Halley, Newton-Raphson, or the simple fixed-point iteration mentioned by Christian. the paper shows how one can arrive at this via linear interpolation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;FWIW, the first thing you should have done before asking was to search the archives of the &lt;em&gt;Celestial Mechanics and Dynamical Astronomy&lt;/em&gt; journal at Springer's web page or The SAO/NASA Astrophysics Data System; there are quite a number of survey articles on the efficient routes for evaluating the solutions to Kepler's (elliptic) equation, as well as methods for the parabolic ($\epsilon=1$) and hyperbolic ($\epsilon &gt; 1$) cases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lastly, as a meta-note since I can't comment, this should be tagged &lt;code&gt;[astronomy]&lt;/code&gt; and/or &lt;code&gt;[celestial-mechanics]&lt;/code&gt;&lt;/p&gt;&#xA;" OwnerUserId="7054" LastActivityDate="2011-02-14T03:23:58.557" CommentCount="1" />
  <row Id="21942" PostTypeId="1" AcceptedAnswerId="21944" CreationDate="2011-02-14T03:27:07.633" Score="3" ViewCount="60" Body="&lt;p&gt;Hi all,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need to simplify the following equation, by completing the square:&#xA;$$\frac{\left(x-z\right)^{2}}{2(u-s)}+\frac{(z-y)^{2}}{2(t-u)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As $\displaystyle\frac{\left(x-y\right)^{2}}{2(t-s)}+C$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I do this? I  can't seem to be able to deal with the fractions effectively when completing this square...&lt;/p&gt;&#xA;" OwnerUserId="7053" LastEditorUserId="3731" LastEditorDisplayName="" LastEditDate="2011-02-14T03:38:43.800" LastActivityDate="2011-02-14T03:38:43.800" Title="How do I complete this square?" Tags="&lt;algebra-precalculus&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="21943" PostTypeId="2" ParentId="21901" CreationDate="2011-02-14T03:33:31.777" Score="0" ViewCount="" Body="&lt;p&gt;I think I can recommend some books on algebra, such as the book &lt;em&gt;An introduction to field theory&lt;/em&gt; by &lt;em&gt;Iain.T.Adamson&lt;/em&gt; or the book &lt;em&gt;Galois Theory&lt;/em&gt; by &lt;em&gt;Joseph Rotman&lt;/em&gt; of which I think as pretty suitable for beginners.&lt;br&gt;&#xA;As for further study, I am wondering if you are interested in algebraic number theory or other ones(As &lt;em&gt;Gauss&lt;/em&gt; once said, Mathematics is the king of science and Number Theory is the queen of Mathematics). As &lt;em&gt;Jurgen Neukirch&lt;/em&gt; said, Number theory &lt;strong&gt;is&lt;/strong&gt; Geometry, you might firstly be familiar with analysis or geometry to study Numbers. Of course you might not be interested in numbers, nonetheless, if you do, get the book by &lt;em&gt;Hilbert&lt;/em&gt; whenever you can fully understand it.&lt;br&gt;&#xA;As for calculus, the books by &lt;em&gt;Richard Courant&lt;/em&gt; is absolutely good and worth studying.  &lt;/p&gt;&#xA;" OwnerUserId="6792" LastActivityDate="2011-02-14T03:33:31.777" />
  <row Id="21944" PostTypeId="2" ParentId="21942" CreationDate="2011-02-14T03:34:10.907" Score="4" ViewCount="" Body="&lt;p&gt;The only way to have $$\frac{\left(x-z\right)^{2}}{2(u-s)}+\frac{(z-y)^{2}}{2(t-u)}$$ equal to $$\frac{\left(x-y\right)^{2}}{2(t-s)}+C$$ is to let $$C=\frac{(-s y+s z+t x-t z-u x+u y)^2}{2 (s-t)&#xD;&#xA;   (s-u) (t-u)}.$$&#xA;I'm not sure this is &lt;em&gt;completing a square&lt;/em&gt;, though.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For extra fun, the numerator in $C$ is exactly $$\left|\begin{array}{ccc}1&amp;amp;1&amp;amp;1\\s&amp;amp;t&amp;amp;u\\x&amp;amp;y&amp;amp;z\end{array}\right|^2,$$ while the denominator is $$-2\left|\begin{array}{ccc}1&amp;amp;1&amp;amp;1\\s&amp;amp;t&amp;amp;u\\s^2&amp;amp;t^2&amp;amp;u^2\end{array}\right|.$$&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2011-02-14T03:34:10.907" CommentCount="1" />
  <row Id="21945" PostTypeId="1" AcceptedAnswerId="21947" CreationDate="2011-02-14T03:39:18.747" Score="8" ViewCount="183" Body="&lt;ol&gt;&#xA;&lt;li&gt;Suppose there is a family (can be&#xA;infinite) of measurable spaces. What&#xA;are the usual ways to define a sigma&#xA;algebra on their Cartesian product?&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;There is one way in the context of&#xA;defining product measure on&#xA;&lt;a href=&quot;http://planetmath.org/encyclopedia/InfiniteProductMeasure.html&quot; rel=&quot;nofollow&quot;&gt;planetmath&lt;/a&gt;. Let $(E_i, B_i)$ be&#xA;measurable spaces, where $i \in I$&#xA;is an index set, possibly infinite.&#xA;We define their product as follows:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;let $E= \prod_{i \in I} E_i$  , the    Cartesian product of $E_i$,&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;let $B=\sigma((B_i)i \in I)$ , the    smallest sigma algebra&#xA;containing    subsets of E of the&#xA;form $\prod_{i    \in I}B_i$  where&#xA;$B_i=E_i$ for all    but a finite&#xA;number of $i \in I$ .&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I was wondering why it is required&#xA;that &quot;$B_i=E_i$ for all but a finite&#xA;number of $i \in I$&quot;?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Thanks and regards!&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;ADDED:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was wondering if the product sigma algebra defined in 2 is the smallest sigma algebra such that any tuple composed of one measurable set from each individual sigma algebra is measurable? &lt;/p&gt;&#xA;" OwnerUserId="1281" LastEditorUserId="1281" LastEditorDisplayName="" LastEditDate="2011-02-14T13:52:25.110" LastActivityDate="2011-02-14T13:52:25.110" Title="Infinite product of measurable spaces" Tags="&lt;elementary-set-theory&gt;&lt;measure-theory&gt;&lt;sigma-algebra&gt;" AnswerCount="1" CommentCount="7" FavoriteCount="2" />
  <row Id="21946" PostTypeId="1" CreationDate="2011-02-14T03:45:52.107" Score="5" ViewCount="97" Body="&lt;p&gt;If for each $t\in I=[0,1]$ I have a measure space $(X_t,\Sigma_t)$, is there a standard notion which will give a measure space deserving to be called the integral $\int_I X_t\,\mathrm d t$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Motivated by this &lt;a href=&quot;http://math.stackexchange.com/q/21945/274&quot;&gt;question&lt;/a&gt; and curiosity...&lt;/p&gt;&#xA;" OwnerUserId="274" LastEditorUserId="274" LastEditorDisplayName="" LastEditDate="2011-02-14T03:58:34.917" LastActivityDate="2011-02-14T03:58:34.917" Title="Integral of measure spaces" Tags="&lt;measure-theory&gt;" CommentCount="4" FavoriteCount="1" />
  <row Id="21947" PostTypeId="2" ParentId="21945" CreationDate="2011-02-14T04:18:42.373" Score="8" ViewCount="" Body="&lt;p&gt;This is a good question.  I haven't yet worked out a complete answer myself, but Mariano's comment above is definitely part of it: the given product $\sigma$-algebra has the property which is analogous to that of the product topology (which it resembles and is surely modelled on): it is the smallest $\sigma$-algebra which makes all of the projections measurable.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because of this, I believe that if you make a category out of all measurable spaces and measurable functions in the obvious way, the product $\sigma$-algebra you have defined turns out to be the categorical product: i.e., it satisfies the requisite &lt;a href=&quot;http://en.wikipedia.org/wiki/Universal_property#Products&quot; rel=&quot;nofollow&quot;&gt;universal mapping property&lt;/a&gt;.  Again, this is the situation for the product topology.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I think the rest of the explanation has to do with the fact that this $\sigma$-algebra gives you the theorems you want, just as the product topology -- and not the &quot;box topology&quot; for instance -- has nice properties, especially &lt;a href=&quot;http://en.wikipedia.org/wiki/Tychonoff%27s_theorem&quot; rel=&quot;nofollow&quot;&gt;Tychonoff's theorem&lt;/a&gt;.  (The product topology was introduced for the first time in Tychonoff's paper, and his theorem played a large role in convincing mathematicians that it was the &quot;right&quot; topology on an infinite Cartesian product.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure exactly what the analogous result to Tychonoff's theorem is here, but I do know that this &quot;coarsest&quot; product $\sigma$-algebra enables one to define arbitrary products of probability spaces: see this &lt;a href=&quot;http://www.math.uga.edu/~pete/saeki.pdf&quot; rel=&quot;nofollow&quot;&gt;lovely paper of S. Saeki&lt;/a&gt; for an incredibly short proof of that.  I hope it is at least clear where the &quot;coarseness&quot; of the chosen product $\sigma$-algebra comes in handy: if (say in the case of a countably infinite index set, to fix ideas) we allowed arbitrary products $Y = \prod_{i} Y_i$ of &#xA;measurable subsets $Y_i \subset X_i$ to be measurable, then what should the measure of &#xA;$Y$ be?  If it's $\prod_{i = 1}^{\infty} \mu_i (Y_i)$, that's problematic, as we have an infinite product which is not guaranteed to converge.  (And a moment's thought suggests that the situation for an uncountably infinite product is even worse!)  By requiring $Y_i = X_i$ for almost every $i$, we get that $\mu_i(Y_i) = 1$ for almost every $i$ and the infinite &#xA;product is really a finite product.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there more to the story than this?  Is the above construction of the product probability measure the &quot;right&quot; analogue of Tychonoff's theorem in this context (is there even a &quot;right&quot; analogue of Tychonoff's theorem in this context?)?  I'm not sure, and I would be interested to hear more from others.&lt;/p&gt;&#xA;" OwnerUserId="299" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2011-02-14T13:10:33.797" LastActivityDate="2011-02-14T13:10:33.797" CommentCount="5" />
  <row Id="21948" PostTypeId="1" AcceptedAnswerId="21952" CreationDate="2011-02-14T04:23:00.767" Score="2" ViewCount="67" Body="&lt;p&gt;Let $C_1, C_2$ be two concentric circles in the extended complex plane. Is it true that if another &quot;circle&quot; $C$ is orthogonal to both $C_1$ and $C_2$, then $C$ must be a line?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think that this should be true, because at the point of intersection between $C_1$ and $C$, the tangent of $C$ must intersect the center of $C_1$. Similarly, this must also hold for $C_2$. But since the centers of concentric circles are the same, I think that this might constrain $C$ to taking the form of a line. I'm not sure if this is actually true though, or how one would show the statement.&lt;/p&gt;&#xA;" OwnerUserId="1736" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2011-02-14T05:02:09.947" LastActivityDate="2011-02-14T05:02:09.947" Title="Concentric circles and orthogonality" Tags="&lt;geometry&gt;&lt;euclidean-geometry&gt;" AnswerCount="2" />
  <row Id="21949" PostTypeId="1" CreationDate="2011-02-14T04:26:56.897" Score="7" ViewCount="236" Body="&lt;p&gt;I don't know the proper terms to type into Google, so please pardon me for asking here first.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While jingling around a few coins, I realized that one nice puzzle might be to figure out which $n$ or so coins of a currency (let us, say, use the coins of the American dollar as an example) can be used to total to a given amount. For instance, one can have five coins totaling to forty cents: three dimes and two nickels.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) How does one find other $n$ combinations of coins that can total to a set amount?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To use my example, are there other five coins that can total to forty cents? How might one algorithmically prove that those are the only solutions?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Given an amount not equal to a denomination, what is the minimum number of coins needed to be equivalent to the given amount?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, one can have forty cents in three coins: a quarter, a dime, and a nickel. How does one algorithmically show that the &quot;magic number&quot; for forty cents is three (i.e., one cannot find two coins whose amounts total to forty cents)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As mentioned already, this isn't homework; just idle curiosity. Any pointers to algorithms would be appreciated!&lt;/p&gt;&#xA;" OwnerUserId="7055" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2011-03-07T03:52:43.430" LastActivityDate="2011-03-07T03:52:43.430" Title="On problems of coins totaling to a given amount" Tags="&lt;combinatorics&gt;&lt;algorithms&gt;&lt;optimization&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="2" />
  <row Id="21950" PostTypeId="2" ParentId="21948" CreationDate="2011-02-14T04:35:27.747" Score="1" ViewCount="" Body="&lt;p&gt;(This probably should be a comment)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's one way to go about it: construct two concentric circles of arbitrary radius: $x^2+y^2=r_1^2$ and $x^2+y^2=r_2^2$; and a third circle of arbitrary radius that is orthogonal to one of the circles. Now, try to prove that the tangents at the intersection points of the other concentric circle and the circle orthogonal to the first circle cannot be perpendicular...&lt;/p&gt;&#xA;" OwnerUserId="7055" LastActivityDate="2011-02-14T04:35:27.747" CommentCount="1" />
  <row Id="21951" PostTypeId="1" AcceptedAnswerId="21958" CreationDate="2011-02-14T04:42:05.067" Score="6" ViewCount="130" Body="&lt;p&gt;I've been trying to wrap my head around the basic concepts of category theory, and I thought I would attempt to illustrate what I understand with the category of sets, probably the easiest example. Particularly, I've been trying to prove that $id_A$ (the identity morphism on $A$, for all $A \in Obj(\mathbf{Set})$) is $1_A \colon A \rightarrow A, x \mapsto x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a very intuitive and reasonable statement, and it's trivial to prove that $1_A$ is indeed an identity morphism on $A$, and I suppose uniqueness of $id_A$ can be demonstrated analogously to uniqueness of the identity element in a monoid (considering the subcategory which has $A$ as its only object, and endofunctions on $A$ as its only morphisms).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this manner, it is not hard to prove that the proposition in the title is true, but this demonstration requires to make an assumption or guess as to what could $id_A$ be. Specifically, the scheme of the proof is: assume $id_A$ = $1_A$, see that it works with the definition of an identity morphism, show that the identity morphism is unique, and, in conclusion, $id_A$ can only be $1_A$. What I'm looking for, nonetheless, is a somehow more direct proof, that doesn't assume $id_A$ = $1_A$ at the start. I want to place myself in a state of little or no knowledge about sets and functions, and under this assumption, why would I assume $id_A$ = $1_A$ at first? Why not try with $id_\mathbb{Z}$ = $f \colon \mathbb{Z} \to \mathbb{Z}, x \mapsto x^2 + 1$, for example? It wouldn't work, but I don't have any reason to think that $1_\mathbb{Z}$ is a better guess for $id_\mathbb{Z}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suppose that the proof for which I'm asking would work for categories of sets with additional structure, and probably for posets as well, although I'm not clear as to what modifications it would require to work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="6427" LastEditorUserId="6427" LastEditorDisplayName="" LastEditDate="2011-02-14T05:48:27.843" LastActivityDate="2011-02-20T14:30:24.497" Title="The identity morphism in $\mathbf{Set}$ is the identity function" Tags="&lt;category-theory&gt;" AnswerCount="2" CommentCount="11" FavoriteCount="1" />
  <row Id="21952" PostTypeId="2" ParentId="21948" CreationDate="2011-02-14T04:48:53.903" Score="1" ViewCount="" Body="&lt;p&gt;Your remark that the tangent of $C$ at the intersection with $C_1$ goes through the center of $C_1$ is the key.  Similarly, the tangent of $C_1$ at the intersection with $C$ goes through the center of $C$.  If the radius of $C$ is $r$ and the radius of $C_1$ is $r_1$, the distance between the centers is $\sqrt{r^2+r_1^2}$.  By the same token, if the radius of $C_2$ is $r_2$ the distance between the centers of $C_2$ and $C$ is $\sqrt{r^2+r_2^2}$  These disagree unless $r$ is infinite.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-02-14T04:48:53.903" />
  <row Id="21954" PostTypeId="2" ParentId="2949" CreationDate="2011-02-14T04:54:46.513" Score="3" ViewCount="" Body="&lt;p&gt;The primitive element theorem is quite surprising.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Theorem: Let $E \supseteq F$ be a finite degree separable extension. Then $E=F[\alpha]$ for some $\alpha \in E$.&lt;/p&gt;&#xA;" OwnerUserId="4284" LastActivityDate="2011-02-14T04:54:46.513" CommunityOwnedDate="2011-02-14T04:54:46.513" />
  <row Id="21956" PostTypeId="2" ParentId="21515" CreationDate="2011-02-14T05:25:44.253" Score="3" ViewCount="" Body="&lt;p&gt;I think that we can get along just fine without spectral sequences or the universal coefficient theorem :) (even though both of those are probably more useful in general- it's nice to actually do a real computation at least once in your life!).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the case of $\mathbb{R}P^2$ we happen to have a nice description of this as a $\Delta$-complex (I'm assuming you saw this in Hatcher). So you can compute the homology groups just as Hatcher described for a $\Delta$-complex in general, except that when he uses $\mathbb{Z}$ (i.e. looking at the &quot;free abelian group generated by&quot;) just replace it with $\mathbb{Z}/2$ or $\mathbb{Q}$ and look at the vector space. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;More specifically: Given a $\Delta$-complex $X$, let $\Delta_n(X, F)$ be the vector space with basis the open $n$-simplices of $X$. Now look at Hatcher's Example 2.4 and see what happens with these new coefficients!&lt;/p&gt;&#xA;" OwnerUserId="423" LastActivityDate="2011-02-14T05:25:44.253" />
  <row Id="21957" PostTypeId="1" CreationDate="2011-02-14T05:36:16.447" Score="4" ViewCount="168" Body="&lt;p&gt;Let $x$ and $y$ be continuously differentiable real-valued functions defined on $\mathbf{R}$ and suppose that $x&amp;#39;(t) = f\,(x(t),y(t))$ and $y&amp;#39;(t) = g(x(t),y(t))$ where $f$ and $g$ are continuously differentiable on $\mathbf{R}^2$. According to the ODE book I am reading, if $f$ is never 0, then &#xA;\[&#xA;\frac{dy}{dx} = \frac{g(x,y)}{f(x,y)}&#xA;\]&#xA;\[&#xA;y(x_0) = y_0 &#xA;\]&#xA;where $x_0,y_0$ are constants, is an initial value problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now what does $dy/dx$ mean here? Is it just the function $y&amp;#39;(x)$? Or is the author being sloppy and $x$ is actually a dummy variable? If the former is true, how is this an initial value problem? Normally, an initial value problem is given as $y&amp;#39;(t) = h(y(t),t)$, $y(t_0) = y_0$ where $t$ is variable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS. Why is it that ODE books are sloppy with notation and unshamefully nonrigorous?&lt;/p&gt;&#xA;" OwnerUserId="1427" LastActivityDate="2011-02-15T04:41:24.177" Title="dy/dx when x and y are functions" Tags="&lt;differential-equations&gt;" AnswerCount="4" CommentCount="1" />
  <row Id="21958" PostTypeId="2" ParentId="21951" CreationDate="2011-02-14T05:38:50.587" Score="3" ViewCount="" Body="&lt;p&gt;Okay, based on the extensive discussion in comments, it seems that you want to consider the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose that we have a well-defined collection of sets, which we want to make into a category by letting them be the objects, taking the collection of morphisms to be all set-theoretic functions between the two sets, and using regular composition and domain/codomain identifications. Can we prove that under these circumstances, for us to have a category then the categorical identity arrow &lt;em&gt;must&lt;/em&gt; be the identity function for the set?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The key is that you have enough functions to &quot;separate points&quot;. Given any $a,b\in A$, $a\neq b$, there exists a function $g\colon A\to A$ such that $g(a)\neq g(b)$. For example, define $g$ to be the function that maps $b$ to $a$, and maps everything else to $b$.  (Compare this with the example I gave in the comments, where this does not hold).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, fix a set $A$, and suppose that $f\colon A\to A$ is the arrow that satisfies the identity conditions (for all objects $B$ and $C$, and all arrows $g\colon A\to B$ and $h\colon C\to A$, $gf = g$ and $fh = h$). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pick any $a$ and $b$ in $A$, $a\neq b$. Let $g$ be a function with $g(a)\neq g(b)$; then $gf(a) = g(a)$, so it follows that $f(a)\neq b$. This holds for &lt;em&gt;every&lt;/em&gt; $b\in A-\{a\}$, so the only possibility is that $f(a)=a$. This holds for all $a\in A$, so $f$ must be the identity map.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can generalize this to any set-based category in which you can either separate points, or &quot;hit&quot; any point: if for every object $A$ and every elements $a,b\in A$ with $a\neq b$, there either exists an object $B$ and a morphism $h\in\mathcal{C}(A,B)$ such that $h(a)\neq h(b)$; or else there exists an object $C$, and a morphism $g\in\mathcal{C}(C,A)$ for which there exists an element $c\in C$ such that $g(c)=a$; then the identity morphism of $A$ must be the identity map of $A$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Indeed, suppose that $f$ is the identity morphism, and let $a\in A$. For each $b\in A$, $b\neq a$, either we have $B$ and $h$ as above, and $hf(a) = h(a)$ implies that $f(a)\neq b$; or else there exists $C$, $c$ and $g$ as above with $g(c) = a$. Then $fg = g$ gives that $f(a)\neq b$ (since $f(g(c))=b$ and $g(c)=a$ implies $a=b$). Either way, you get that for all $b\in A$ with $b\neq a$, $f(a)\neq b$. So the only possibility left is that $f(a)=a$. This holds for all $a\in A$, so $f=1_A$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Note.&lt;/em&gt; In a sense, the condition is both necessary and sufficient, though for silly reasons: if the condition is not met by $A$ and $a$, then the identity map of $A$ cannot be the identity morphism, simply because the identity map of $A$ satisfies the given conditions: for all $b\neq a$ you have $1_A(a)\neq 1_A(b)$, and $1_A(a)=a$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Added.&lt;/em&gt; This argument applies to categories such as topological spaces (because you always have the map from the $1$-element topological space to your toplogical space mapping the unique point to $a$); pointed topological spaces (the discrete 2-element pointed topological space maps the non-distinguished point to your favorite point); groups (you have maps from the cyclic group to any group, mapping the generator to your element $a$); and others. It's hard to make it work with posets as categories, because posets as categories are not really set-based categories (the objects are not usually sets and arrows set-theoretic functions between them); you can model them as set-based categories, but then the result need not hold: the example I gave in the comments can be thought of as the totally ordered set with two elements, for example, and here you don't have $id_A = 1_A$.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-14T05:54:53.753" LastActivityDate="2011-02-14T05:54:53.753" CommentCount="1" />
  <row Id="21959" PostTypeId="1" CreationDate="2011-02-14T05:49:33.060" Score="3" ViewCount="166" Body="&lt;p&gt;I have a quite small data set (on the order of 8-20) from an essentially unknown system and would like to predict a value that will be higher than the next number generated by the same system 90% of the time. Both underestimation and overestimation are problematic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the mathematically &quot;correct&quot; way to do this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I could also generate a level-of-confidence estimate, it would wow my manager.&#xA;Also, let me say I'm not a math major, so thanks for any help, however remedial it may be :)&lt;/p&gt;&#xA;" OwnerUserId="7056" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2011-02-14T18:12:33.127" LastActivityDate="2011-02-14T21:25:55.650" Title="Statistics: Predict 90th percentile with small sample set" Tags="&lt;statistics&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="1" />
  <row Id="21960" PostTypeId="1" CreationDate="2011-02-14T05:56:15.703" Score="2" ViewCount="58" Body="&lt;p&gt;A flat two-torus, T^2 that is the torus with Euclidean metric needs to at least be embedded in R^4. If we puncture the torus and leave the Euclidean metric on it as inherited (ignoring the issues of completeness), what ambient space could you embed the punctured torus with Euclidean metric?&lt;/p&gt;&#xA;" OwnerUserId="7057" LastActivityDate="2011-02-14T05:56:15.703" Title="embedding of punctured torus with euclidean metric" Tags="&lt;geometry&gt;" CommentCount="2" />
  <row Id="21961" PostTypeId="2" ParentId="21957" CreationDate="2011-02-14T06:00:05.427" Score="6" ViewCount="" Body="&lt;p&gt;If $x$ and $y$ are functions of $t$, $\frac{\mathrm dy}{\mathrm dx}$ means, in all likelihood, the following: where possible by the inverse function theorem, write $t$ in terms of $x$; then $y=y(t)=y(t(x))$ is a function of $x$: differentiate &lt;em&gt;that&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you think about it a bit, you'll see that $\frac{\mathrm dy}{\mathrm dx}$ &lt;em&gt;is&lt;/em&gt; in fact the ratio of change of $y$ with respect to changes in $x$, as it should be!&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2011-02-14T06:00:05.427" CommentCount="2" />
  <row Id="21962" PostTypeId="1" AcceptedAnswerId="21989" CreationDate="2011-02-14T06:06:27.813" Score="2" ViewCount="104" Body="&lt;p&gt;I'm learning about sufficient statistics and I understand basic stuff that it is the minimum information we need to represent a statistic (very vague definition, I know. I'm learning). I'm trying to solve problems in that. The problem is suppose a sample of size $n$ is taken from the geometric distribution with parameter $\pi$. How can I find a real valued sufficient statistic for this ?&lt;/p&gt;&#xA;" OwnerUserId="6335" LastActivityDate="2011-02-14T11:42:54.687" Title="Sufficient statistics" Tags="&lt;statistics&gt;" AnswerCount="1" FavoriteCount="0" />
  <row Id="21963" PostTypeId="2" ParentId="21913" CreationDate="2011-02-14T06:11:05.160" Score="6" ViewCount="" Body="&lt;p&gt;First, a remark that I think is relevant:&#xA;understanding the construction $K_{\xi}$ is related to the question of whether a homology class can be represented by a submanifold.   This is not always possible;&#xA;see e.g. &lt;a href=&quot;http://mathoverflow.net/questions/21171/when-is-a-homology-class-represented-by-a-submani&quot; rel=&quot;nofollow&quot;&gt;this MO answer&lt;/a&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, it is always possible for $n$-cycles for small values of $n$ (if we are considering the homology of a manifold),&#xA;which may be one reason why it is hard to visualize non-manifold examples of $K_{\xi}$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;E.g. in the case $n = 2$, we are gluing triangles along their edges, with the rule that exactly two triangles meet along a common edge.  This always gives a closed $2$-manifold.  In general, unless I am confused, the fact that for $n = 2$ we always get a manifold implies in general that the non-manifold points are in codimension at least $n - 3$ (not just $n - 2$).  So to find a counterexample we should take $n = 3$.  I will give such a counterexample in a moment, but first let me describe a general approach to thinking about this situation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When gluing simplices like this, the way to investigate whether the resulting object is a manifold or not is to consider the link of each vertex. Namely, if a bunch of $n$-simplices meet at a vertex $v$, take a transverse slice to each simplex just below the vertex (I am imagining that simplex is sitting on the face opposite to $v$, so that $v$ is at the top of the simplex),&#xA;to get an $n-1$-simplex.  Now these glue together to make a closed simplicial $n-1$-complex that surrounds $v$; this is the link of $v$.  Note that these $n-1$-simplices meet along $n-2$-dimensional faces, and so the link is a $K_{\xi}$-type object, but of one dimension less.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case that $n = 2$, we get a bunch of segments being joined at their vertices, and hence a circle.  No drama there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But now suppose that $n = 3$.  Then the link is formed by a bunch of triangles gluing together, and we have agreed that this gives a $2$-manifold.  But which one?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $K_{\xi}$ is locally Euclidean at $v$, then this surface would have to be a&#xA;$2$-sphere.    But &lt;em&gt;a priori&lt;/em&gt; it doesn't have to be, and so in this case we can&#xA;get a non-manifold example!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In practice, to make an example we should take a cone on a surface that is not&#xA;a $2$-sphere, e.g. a cone on a $2$-torus.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Precisely: triangulate a $2$-torus, e.g. by choosing two triangles &#xA;$\Delta_1$ and $\Delta_2$, and identifying the three edges of the first with the three edges of the second in the appropriate manner.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now to form the cone on this, replace $\Delta_i$ by a $3$-simplex&#xA;$\tilde{\Delta_i}$.  Regard $\Delta_i$ as one of the faces of $\tilde{\Delta}_i$,&#xA;and label the other three faces according to the edge along which they&#xA;meet $\Delta_i$.  Now glue the three faces of $\tilde{\Delta}_1$ other than&#xA;$\Delta_1$ with the three faces of $\tilde{\Delta}_2$ other than $\Delta_2$&#xA;according to the same gluing scheme we used previously to construct the $2$-torus.  What we end up with is a three dimensional simplicial complex whose boundary is equal to $\Delta_1 + \Delta_2$, i.e. a $2$-torus.  But it is not a $3$-manifold; rather it is a cone on the $2$-torus.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we let $\xi$ be the $3$-chain $\tilde{\Delta}_1 + \tilde{\Delta}_2$,&#xA;then $K_{\xi}$ is the cone on the $2$-torus, and so is not a $3$-manifold.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, this $\xi$ is not a cycle (so the boundary of $K_{\xi}$ is non-empty), but we could take two such cones and glue&#xA;them along their common $2$-torus boundary to get a three-dimensional simplicial complex without boundary, and then take $\xi$ to be the sum of the four $3$-simplices involved; then $\xi$ would be a cycle, but $K_{\xi}$ would not be a manifold; it has two vertices whose links are $2$-tori rather than $2$-spheres.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Concluding remark: In dimension $2$, the only way to get pseudo-manifolds that are not manifolds is to explicitly glue together vertices, as you observed.  But in dimension $3$ or higher, there are other examples, e.g. in dimension $3$ we can take cones on positive genus surfaces, as in the preceding construction.&lt;/p&gt;&#xA;" OwnerUserId="221" LastEditorUserId="221" LastEditorDisplayName="" LastEditDate="2011-02-14T06:17:14.593" LastActivityDate="2011-02-14T06:17:14.593" CommentCount="9" />
  <row Id="21964" PostTypeId="2" ParentId="21957" CreationDate="2011-02-14T06:24:46.520" Score="3" ViewCount="" Body="&lt;p&gt;The condition that f is never 0 is important .by the continuity of f,f must be always positive or negative ,which means x(t) is monotonous,so it has inverse function t(x),which is also smooth.so y(t)=y(t(x)).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;by&lt;/p&gt;&#xA;" OwnerUserId="7020" LastActivityDate="2011-02-14T06:24:46.520" />
  <row Id="21966" PostTypeId="1" AcceptedAnswerId="21967" CreationDate="2011-02-14T06:42:37.120" Score="5" ViewCount="93" Body="&lt;p&gt;I want to evaluate the following:&#xA;$$\lim_{r \rightarrow 0}\frac{-r^2}{2 \left(\sqrt{1-\frac{r^2}{4}}-1 \right)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I look at the graph and see that it seems to be going to zero. This makes sense to me because if I replace r with zero this function is defined and continuous near zero and the value of the function is zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, I think:&#xA;$$\lim_{r \rightarrow 0}\frac{-r^2}{2 \left(\sqrt{1-\frac{r^2}{4}}-1 \right)}=0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But next I try something else I rationalize the denominator since this is a good technique for solving limits.&#xA;$$\lim_{r \rightarrow 0}\frac{-r^2 \left(\sqrt{1-\frac{r^2}{4}}+1 \right)}{2 \left(1-\frac{r^2}{4}-1 \right)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then,&#xA;$$\lim_{r \rightarrow 0}\frac{-r^2 \left(\sqrt{1-\frac{r^2}{4}}+1 \right)}{\frac{-r^2}{2}}$$&#xA;so...&#xA;$$\lim_{r \rightarrow 0} 2 \left(\sqrt{1-\frac{r^2}{2}}+1 \right)=4$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What have I done? How can rationalizing change the graph? My guess is that the denominator $2 \left(\sqrt{1-\frac{r^2}{2}}-1 \right)$ is &quot;divisible by $r^2$ in some not obvious way? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know my original reasoning was sloppy, (not a proof) but the fact that the &lt;strong&gt;graph&lt;/strong&gt; shows a limit of zero has me very confused. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do I avoid this error? Just always rationalize everything? Why would I do that?&lt;/p&gt;&#xA;" OwnerUserId="4681" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-14T14:21:13.570" LastActivityDate="2011-02-14T14:21:13.570" Title="The limit (and function) changes after rationalizing?" Tags="&lt;calculus&gt;&lt;limit&gt;" AnswerCount="1" />
  <row Id="21967" PostTypeId="2" ParentId="21966" CreationDate="2011-02-14T06:47:24.773" Score="6" ViewCount="" Body="&lt;p&gt;It looks pretty clear to me that the limit is 4 from the graph:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/24Q9x.png&quot; alt=&quot;graph&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps you mis-entered it on your graphing tool?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Beyond that, when $r=0$, the numerator of the expression is $-r^2=-0^2=0$ and the denominator is $2 \left(\sqrt{1-\frac{r^2}{4}}-1 \right)=2 \left(\sqrt{1-\frac{0^2}{4}}-1 \right)=2 \left(\sqrt{1}-1 \right)=0$—that is, the expression is of the form $\frac{0}{0}$, which is an indeterminate form, so it needs further investigation.&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2011-02-14T06:47:24.773" CommentCount="1" />
  <row Id="21968" PostTypeId="2" ParentId="21928" CreationDate="2011-02-14T06:54:15.053" Score="2" ViewCount="" Body="&lt;p&gt;Integration by parts does in fact help.&#xA;Let $I_n = \int (1+x^2)^{-n}dx$.&#xA;Then multiply by 1 and integrate by parts:&#xA;$$I_n = x (1+x^2)^{-n} - \int x \cdot (-n)2x (1+x^2)^{-n-1}dx = \dots = x (1+x^2)^{-n} + 2n(I_n - I_{n+1}).$$&#xA;(In the &quot;...&quot; step, write $x^2=(x^2+1)-1$ in the numerator and then divide.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From this you can solve for $I_{n+1}$ in terms of $I_n$, and since you know $I_1 = \arctan x + C$, you can recursively compute $I_2$, $I_3$, $I_4$, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(This is pretty much the same method that Sivaram pointed you to for finding $\int \cos^6 \theta \, d\theta$.)&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2011-02-14T06:54:15.053" />
  <row Id="21969" PostTypeId="1" CreationDate="2011-02-14T06:55:12.470" Score="26" ViewCount="667" Body="&lt;p&gt;I'm primarily a programmer, so forgive me if I don't know the proper nomenclature or notation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Last night, an old teacher of mine told me about a question that had caused some noodle-scratching for him:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For any two sequences of consecutive integers, can the sums of their reciprocals be equal?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, I gather that these sums are called &quot;harmonic numbers&quot; if the consecutive sequence begins with 1.  But what if it doesn't begin with 1?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How might we go about proving that for any two such sequences, their sums are unequal?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have written a quick Python script that returns, in the form of a reduced fraction, the sum of any $\sigma(m, k)$ where $m$ is the first number to consider and $k$ is the length of the sequence.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, I guess I have two questions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Where can I find out about the current state of research on this question?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) What are the most likely approaches, or &quot;hooks&quot; that I might grasp onto to arrive at a proof?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Via the comments, I'll add the following for clarity:)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sigma(m,k)$ is $\frac{1}{m} + \frac{1}{m+1}+\dots +\frac{1}{m+k−1}.$ My question is: Can there exist distinct pairs $(m_1,k_1)$ and $(m_2,k_2)$ such that the corresponding sums are equal?&lt;/p&gt;&#xA;" OwnerUserId="7059" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2011-02-14T22:33:53.920" LastActivityDate="2011-02-25T18:09:33.147" Title="Can the sums of two sequences of reciprocals of consecutive integers be equal?" Tags="&lt;number-theory&gt;" AnswerCount="3" CommentCount="5" FavoriteCount="7" />
  <row Id="21970" PostTypeId="2" ParentId="21969" CreationDate="2011-02-14T07:09:09.130" Score="3" ViewCount="" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;How might we go about proving that for any two such sequences, their sums are unequal?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You can't. For example, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{1}{25}+\frac{1}{757}+\frac{1}{763309}+\frac{1}{873960180913}+\frac{1}{1527612795642093418846225}$$&#xA;is the same as $$\frac{1}{33}+\frac{1}{121}+\frac{1}{363}$$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One keyword is «egyptian fraction»; google should link to lots of information.&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2011-02-14T07:09:09.130" CommentCount="10" />
  <row Id="21972" PostTypeId="1" CreationDate="2011-02-14T07:39:16.630" Score="4" ViewCount="152" Body="&lt;p&gt;Is there any reason that this generalization of the mean value theorem would fail?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let A be a subset of Rn that is differentiably connected, and let f : A --&gt; R be continuously differentiable at every point in A. If x and y are two points in A then there exists a point c in A such that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;f(y)-f(x) = (grad(f(c)))*(y-x)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems like this would work to me, however every generalization I've run into specifies that A must be open, so I assume if it failed it would be on account of that, however I can't figure out why.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="7060" LastActivityDate="2011-02-14T19:15:50.490" Title="Mean Value Theorem for Multiple Variables" Tags="&lt;real-analysis&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" />
  <row Id="21973" PostTypeId="1" AcceptedAnswerId="21974" CreationDate="2011-02-14T07:40:11.683" Score="3" ViewCount="74" Body="&lt;p&gt;This question was inspired by the question on &lt;a href=&quot;http://math.stackexchange.com/questions/21475/what-are-some-examples-of-classes-that-are-not-sets&quot;&gt;examples of classes that are not sets&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the discussion in the comments there, it seems there does not exist a set of all sets of a given cardinality.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To me, it seems easy to see that this is true for any nonzero finite cardinality $n$. Given any set $x$, you can always make a set of size $n$ by adding $n-1$ other sets to $x$ to make a set of size $n$, which exists by repeated use of the pairing axiom. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But how would you do this for infinite cardinalities? For suppose $\kappa$ is some given infinite cardinality. To show that the set of all sets of cardinality $\kappa$ does not exist, it seems you would have to show that for any set $x$, there exists a set of that size with $x$ as an element, and then you could take the union of that set to find the set of all sets, and thus a contradiction. But it doesn't seem reasonable to simply say, for a given set of size $\kappa$ if $x$ is in the set, we have no problem. If not, just take an element out of the set and put $x$ in. Something about that seems like it would not be allowed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So how would you do this for infinite cardinalities? &lt;/p&gt;&#xA;" OwnerUserId="6742" LastActivityDate="2011-02-14T08:00:43.740" Title="Class of sets of a given infinite cardinality." Tags="&lt;elementary-set-theory&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="21974" PostTypeId="2" ParentId="21973" CreationDate="2011-02-14T07:50:34.013" Score="5" ViewCount="" Body="&lt;p&gt;Suppose there is a set $x$ whose elements are precisely the sets of size $\kappa$. We will define a function $f$ with domain $x$ whose range is the universe $V$ of all sets. Since the latter is known not to be a set (and functions map sets to sets---this is the replacement axiom), it follows that $x$ itself cannot be a set either.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fix a set $Y$ of size $\kappa$. Given a set $a\in x$, let $f(a)=0$, unless there is a set $z$ such that $a=\{(b,z)\mid b\in Y\}$, in which case we set $f(a)=z$. This $f$ works. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Let me add: The universe $V$ of sets can be seen as &quot;constructed by stages.&quot; More precisely, $$V=\bigcup_{\alpha\in ORD}V_\alpha,$$ where the set $V_\alpha$ is obtained by iterating the power set operation $\alpha$ times, starting with the empty set, and $ORD$ is the (proper) class of all ordinals. The details do not matter much right now. The point is that something is a set if it appears at some stage $V_\alpha$ of the construction, and a set cannot appear until all its elements do (i.e., the stage $\alpha$ at which a set $t$ appears is larger than any stage $\beta$ at which an element of $t$ appears). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But something like &quot;the collection of all sets of size $\kappa$&quot; cannot possibly appear at any stage $V_\alpha$, since we can always build a set of size $\kappa$ with $V_\alpha$ as one of its elements, forcing the set to appear at a stage larger than $\alpha$. In general, one can verify whether a collection is a set or a proper class by checking whether it contains sets that appear at arbitrarily large stages (proper class) or, instead, all its elements appear by some stage $\alpha$ (set).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This suggests the heuristic that a proper class ought to be as large as the universe, or at least, as large as the class ORD of all ordinals. This, however, is independent of the usual axioms of set theory. (An additional problem appears here, since the standard presentation of set theory does not treat proper classes as objects, so one cannot even make the statement &quot;all proper classes have the same size&quot;. But there are ways around this obstacle.)&lt;/p&gt;&#xA;" OwnerUserId="462" LastEditorUserId="462" LastEditorDisplayName="" LastEditDate="2011-02-14T08:00:43.740" LastActivityDate="2011-02-14T08:00:43.740" CommentCount="3" />
  <row Id="21975" PostTypeId="2" ParentId="21197" CreationDate="2011-02-14T08:00:00.067" Score="2" ViewCount="" Body="&lt;p&gt;Yes, primitive point means the group is cyclic and that P is a generator.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let E be an elliptic curve over a finite field F_q.  The group E( F_q ) is cyclic if gcd( #E( F_q ), q-1 ) = 1.  To test whether a point P generates E( F_q ) it is necessary to factorise #E( F_q ).  The case when #E( F_q ) is prime is an easy special case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If M = gcd( #E( F_q ), q-1 ) &gt; 1 then one can determine whether E( F_q ) is cyclic or not (both cases can arise) in expected (randomised) polynomial time using the Weil pairing.  This algorithm is due to Victor Miller and is explained in his paper in the Journal of Cryptology, volume 17 (2004).&lt;/p&gt;&#xA;" OwnerUserId="7061" LastActivityDate="2011-02-14T08:00:00.067" CommentCount="1" />
  <row Id="21976" PostTypeId="2" ParentId="21972" CreationDate="2011-02-14T08:04:27.307" Score="4" ViewCount="" Body="&lt;p&gt;A hint: Consider the auxiliary function $\phi(t):=f(x+t(y-x))$ for $0\leq t\leq 1$. In using this function you make an essential assumption about the two points $x,y\in A$. This assumption is the missing link; it has nothing to do with $A$ being open or what.&lt;/p&gt;&#xA;" OwnerUserId="1303" LastActivityDate="2011-02-14T08:04:27.307" />
  <row Id="21977" PostTypeId="2" ParentId="21951" CreationDate="2011-02-14T08:37:09.833" Score="2" ViewCount="" Body="&lt;p&gt;I think the most you can say in general is that $\text{id}_A$ acts as the identity function on the hom-sets $\text{Hom}(B, A)$ for all $B$ in the category. That uniquely specifies it by the Yoneda lemma, so there's no ambiguity here, and it is the appropriate generalization of &quot;acts as the identity on points of $A$&quot; to arbitrary categories: you should think of it as &quot;acts as the identity on generalized points of $A$.&quot; &lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2011-02-14T08:37:09.833" />
  <row Id="21978" PostTypeId="2" ParentId="21957" CreationDate="2011-02-14T08:42:25.927" Score="2" ViewCount="" Body="&lt;p&gt;Just wanted to comment ( but I don't have enough points), that , most likely,&#xA;  when you see a reference to &quot;f never being 0&quot; in problems of this sort; other&#xA;  than the obvious not wanting to divide by 0, this is a reference to the implicit/inverse&#xA;  function theorem, which guarantees a(n) (at least) local differentiable inverse.&#xA;    It then often ends up being a case of playing around with the chain rule until you get the&#xA; right expressions.&lt;/p&gt;&#xA;" OwnerUserId="7052" LastActivityDate="2011-02-14T08:42:25.927" />
  <row Id="21979" PostTypeId="1" AcceptedAnswerId="21981" CreationDate="2011-02-14T08:47:39.690" Score="4" ViewCount="37" Body="&lt;p&gt;Let $\mathbb{Q}^{alg}$ be the algebraic closure of the rationals. Given a point $P\in \mathbb{A}^n(\mathbb{Q}^{alg})$, $P = (a_1,\dots,a_n)$, we define the degree of $P$ to be the degree of the minimal field extension of $\mathbb{Q}$ over which $P$ is defined: $\text{deg}(P) = [\mathbb{Q}(a_1,\dots,a_n):\mathbb{Q}]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If a variety $X$ in $\mathbb{A}^n(\mathbb{Q}^{alg})$ has infinitely many points, must it have infinitely many points of bounded degree? That is, is there a positive integer $d$ such that $\{P\in X:\text{deg}(P)\leq d\}$ is infinite?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems like this should be an easy consequence of some more general theorem, but my knowledge is limited.&lt;/p&gt;&#xA;" OwnerUserId="7062" LastEditorUserId="7062" LastEditorDisplayName="" LastEditDate="2011-02-14T08:55:53.287" LastActivityDate="2011-02-14T09:09:16.337" Title="Points of bounded degree on varieties" Tags="&lt;algebraic-geometry&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="21980" PostTypeId="2" ParentId="21893" CreationDate="2011-02-14T08:56:53.117" Score="1" ViewCount="" Body="&lt;p&gt;The sequence doesn't converge uniformly -- $\lim_{p \to 1} (I(p)-I_n(p))$ doesn't exist, since &lt;/p&gt;&#xA;&#xA;&lt;p&gt;\[I(p)-I_n(p)=\frac{1}{(p-1)n^{p-1}}\;,\]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which diverges for $p\to 1^+$. That is, for given $n$ the difference is unbounded in $p$, and thus you can't find an $n_0$ such that the difference is bounded by $\epsilon$ for all $n&amp;gt;n_0$ and all $p$. This is because the integral diverges for $p=1$, and no matter how large you choose $n$ you're missing larger and larger contributions from the logarithmic tail as you take $p$ to $1$.&lt;/p&gt;&#xA;" OwnerUserId="6622" LastActivityDate="2011-02-14T08:56:53.117" />
  <row Id="21981" PostTypeId="2" ParentId="21979" CreationDate="2011-02-14T09:09:16.337" Score="2" ViewCount="" Body="&lt;p&gt;Yes, this is true.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint: we may assume that $X$ is irreducible (take any one irreducible component of positive dimension).  Apply Noether Normalization.&lt;/p&gt;&#xA;" OwnerUserId="299" LastActivityDate="2011-02-14T09:09:16.337" CommentCount="4" />
  <row Id="21982" PostTypeId="2" ParentId="21809" CreationDate="2011-02-14T09:34:50.060" Score="2" ViewCount="" Body="&lt;p&gt;For a particular prime, you can use &lt;a href=&quot;http://en.wikipedia.org/wiki/Cantor%E2%80%93Zassenhaus_algorithm&quot; rel=&quot;nofollow&quot;&gt;Cantor-Zassenhaus&lt;/a&gt; or another factoring algorithm to factor the polynomial modulo the prime. &lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2011-02-14T09:34:50.060" CommentCount="4" />
  <row Id="21983" PostTypeId="1" AcceptedAnswerId="21984" CreationDate="2011-02-14T09:40:27.187" Score="3" ViewCount="151" Body="&lt;p&gt;I'm having trouble keeping the definitions straight, so to help me remember the definitions: is it correct to say that a function is:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Injective (one-to-one) iff it is not many-to-one.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Surjective iff its codomain and range are equal.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;?&lt;/p&gt;&#xA;" OwnerUserId="4890" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-14T14:45:38.127" LastActivityDate="2011-02-14T21:26:08.597" Title="Clarification on Surjection and Injection" Tags="&lt;elementary-set-theory&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="21984" PostTypeId="2" ParentId="21983" CreationDate="2011-02-14T10:08:18.357" Score="2" ViewCount="" Body="&lt;p&gt;The first statement is true because if the function $f$ injective every set of two different elements $a,b$ is sent to different images $f(a),f(b)$ and if two different elements $a,b$ are sent to the same image $f(a)=f(b)$ the function is not injective.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second statement is also true. That is exactly the definition of surjective as the image is identical with the set that is mapped onto, so each element is hit at least once. Note that the term &quot;image&quot; is better than range because range can also stand for codomain.&lt;/p&gt;&#xA;" OwnerUserId="3123" LastActivityDate="2011-02-14T10:08:18.357" CommentCount="1" />
  <row Id="21985" PostTypeId="2" ParentId="21983" CreationDate="2011-02-14T10:13:59.163" Score="0" ViewCount="" Body="&lt;p&gt;First of all, I cannot understand why it is tagged linear algebra instead of set-theory.&lt;br&gt;&#xA;Then, when I was learning set theory, I had the same problem as @Mehrdad has, and in the end I found the most intuitive and useful definition is that of Bourbaki:&lt;br&gt;&#xA;A function is injective if it has a retraction;&lt;br&gt;&#xA;A function is surjective if it has a section,&lt;br&gt;&#xA;whose definition is in &lt;a href=&quot;http://en.wikipedia.org/wiki/Section_%28category_theory%29&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;br&gt;&#xA;This definition is easy to imagine and useful in application both, hope you will be satisfied.&#xA;As for the two definitions mentioned, they are as @user3123 stated.   &lt;/p&gt;&#xA;" OwnerUserId="6792" LastActivityDate="2011-02-14T10:13:59.163" CommentCount="2" />
  <row Id="21986" PostTypeId="2" ParentId="2949" CreationDate="2011-02-14T10:26:28.467" Score="1" ViewCount="" Body="&lt;p&gt;When I was a freshman at the University of Rome I took a course in Geometry in which some projective geometry was covered. I was really amazed by two facts:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;a Moebius strip sits inside the real projective plane.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;the group of rotations of the real plane fixes two (complex) lines and all the circumferences pass through the points at infinity of these two lines (the socalled cyclic points). A few years later I realized that I was the only grad student in David Eisenbud's class at Brandeis to know this (actually simple) fact!&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="688" LastActivityDate="2011-02-14T10:26:28.467" CommunityOwnedDate="2011-02-14T10:26:28.467" />
  <row Id="21987" PostTypeId="1" AcceptedAnswerId="21991" CreationDate="2011-02-14T10:50:49.243" Score="3" ViewCount="150" Body="&lt;p&gt;I saw definitions and theorem about power series are in the form of $\sum_{k=0}^n a_k (x-x_0)^k$. And it definitely doesn't include negative or noninteger powers. Nevertheless, I saw the theorems like&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;If the two series converge to the same value on some interval, then the corresponding coefficients are the same.&lt;/li&gt;&#xA;  &lt;li&gt;The series can be differentiated or integrated termwisely.&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;be used without justification. I couldn't find relevant theorems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is: are the series containing negative and noninteger power terms still called power series, and thus the theorems could apply? If not, do the theorems I mentioned as well as other common theorems like termwise multiplication (Cauchy product) hold?&lt;/p&gt;&#xA;" OwnerUserId="3066" LastEditorUserId="3066" LastEditorDisplayName="" LastEditDate="2011-02-14T10:57:11.810" LastActivityDate="2011-02-14T22:59:01.340" Title="Are negative or noninteger powers still power series?" Tags="&lt;definition&gt;&lt;power-series&gt;" AnswerCount="2" CommentCount="5" />
  <row Id="21988" PostTypeId="1" CreationDate="2011-02-14T11:32:04.263" Score="0" ViewCount="120" Body="&lt;p&gt;The bound formula is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$H(M,N,n,k) \leq (( \frac{p}{p+t} )^{p+t} (\frac{1-p}{1-p-t})^{1-p-t})^n$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So how to proof it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some paper discuss it:&#xA;&lt;a href=&quot;http://www.math.hmc.edu/~kluh/thesis/kluh-2011-prop.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.math.hmc.edu/~kluh/thesis/kluh-2011-prop.pdf&lt;/a&gt;&#xA;http://ansuz.sooke.bc.ca/professional/hypergeometric.pdf.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They say the proof is in the following paper:&#xA;The tail of the hypergeometric distribution. Discrete Mathematics&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I can not find the online version.&lt;/p&gt;&#xA;" OwnerUserId="6904" LastActivityDate="2011-02-14T11:32:04.263" Title="How to derive the bound formula of hypergeometric distribution?" Tags="&lt;probability&gt;&lt;combinatorics&gt;" CommentCount="9" />
  <row Id="21989" PostTypeId="2" ParentId="21962" CreationDate="2011-02-14T11:34:42.403" Score="3" ViewCount="" Body="&lt;p&gt;The joint probability density function (pdf) of the sample is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\prod_{i=1}^n (1-\pi)^{x_i} \pi = \pi^n (1-\pi)^{\sum_{i=1}^n x_i}.$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $T(x) = \sum_{i=1}^n x_i$.  We see that the pdf can be expressed as a function that depends on the sample only through $T(x)$.  Thus $T(x)$ is a sufficient statistic for $\pi$, by the &lt;a href=&quot;http://en.wikipedia.org/wiki/Sufficient_statistic#Fisher.E2.80.93Neyman_factorization_theorem&quot; rel=&quot;nofollow&quot;&gt;Neyman-Fisher factorization theorem&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Neyman-Fisher factorization theorem is a very useful tool for finding sufficient statistics.  The &lt;a href=&quot;http://en.wikipedia.org/wiki/Sufficient_statistic#Examples&quot; rel=&quot;nofollow&quot;&gt;Wikipedia page&lt;/a&gt; has several examples.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(This answer assumes one form of the geometric distribution.  The form of $T(x)$ would be the same for the other.)&lt;/p&gt;&#xA;" OwnerUserId="2370" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2011-02-14T11:42:54.687" LastActivityDate="2011-02-14T11:42:54.687" />
  <row Id="21990" PostTypeId="1" AcceptedAnswerId="21994" CreationDate="2011-02-14T11:44:18.980" Score="1" ViewCount="123" Body="&lt;p&gt;Hello. Could someone help me with the proof that a square matrix has maximal rank if and only if it is invertible?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks to everybody&lt;/p&gt;&#xA;" OwnerUserId="6049" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-14T14:17:42.590" LastActivityDate="2011-02-14T14:17:42.590" Title="Proof - Square Matrix has maximal rank if and only if it is invertible" Tags="&lt;matrices&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="21991" PostTypeId="2" ParentId="21987" CreationDate="2011-02-14T12:00:34.590" Score="5" ViewCount="" Body="&lt;p&gt;&quot;Power series&quot; does not have a fixed meaning. Generally, but not always, it means a series with non-negative integer exponents, and most theorems you have seen proved about power series are about this case. Many of these theorems extend to &lt;a href=&quot;http://en.wikipedia.org/wiki/Laurent_series&quot; rel=&quot;nofollow&quot;&gt;Laurent series&lt;/a&gt; with a fixed negative lower bound away from $x_0$, and many of them can be proven just by multiplying by the appropriate power of $x - x_0$ and using the corresponding theorem for ordinary power series. If you want more assurance than this, Laurent series are usually treated in books on complex analysis.&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2011-02-14T22:21:12.070" LastActivityDate="2011-02-14T22:21:12.070" CommentCount="4" />
  <row Id="21992" PostTypeId="1" CreationDate="2011-02-14T12:15:45.040" Score="-6" ViewCount="146" Body="&lt;p&gt;What number is the most essential or fundamental to mathematics? This can be numbers such a Pi.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am asking as I would like to know if numbers such a 0 or infinity have more real value to mathematical calculations, for solving complex mathematical problems than real numbers such as 1,2,3, 1,000,000. Infinity seems to have many uses in physics, Cosmology and computing even though it is actually an inconceivable number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps it may be to much of a theoretical question? &lt;/p&gt;&#xA;" OwnerUserId="6864" LastEditorUserId="6864" LastEditorDisplayName="" LastEditDate="2011-02-14T12:48:16.293" LastActivityDate="2011-02-14T12:48:16.293" ClosedDate="2011-02-14T17:27:23.693" Title="What is the most important number in maths" Tags="&lt;number&gt;" AnswerCount="1" CommentCount="7" />
  <row Id="21993" PostTypeId="2" ParentId="21992" CreationDate="2011-02-14T12:18:25.333" Score="4" ViewCount="" Body="&lt;p&gt;The number 1. At least if you agree with &lt;a href=&quot;http://en.wikipedia.org/wiki/Leopold_Kronecker&quot; rel=&quot;nofollow&quot;&gt;Kronecker&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="589" LastActivityDate="2011-02-14T12:18:25.333" />
  <row Id="21994" PostTypeId="2" ParentId="21990" CreationDate="2011-02-14T12:32:02.423" Score="5" ViewCount="" Body="&lt;p&gt;Suppose $A\in F^{n \times n}$. If A is invertible then there is a matrix B  such that $AB=I$ so the standard basis $e_i$ (the columns of I) is in the image of A (these vectors are just the image Av where v are the columns of B) - this shows that $\dim (Im(A)) = n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, if $\dim (Im (A))=n$ then for every i there is $v_i$ such that $A v_i = e_i$. Let B be the matrix with columns $v_i$ then $AB=I$ and A is invertible.&lt;/p&gt;&#xA;" OwnerUserId="2125" LastActivityDate="2011-02-14T12:32:02.423" CommentCount="1" />
  <row Id="21996" PostTypeId="1" AcceptedAnswerId="21997" CreationDate="2011-02-14T13:19:55.600" Score="2" ViewCount="77" Body="&lt;p&gt;I have a matrix of the following form:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$&#xD;&#xA;\begin{matrix}&#xD;&#xA;a_1 &amp;amp; 0 &amp;amp; \ldots &amp;amp; 0 \\&#xD;&#xA;a_2 &amp;amp; a_1 &amp;amp; \ldots &amp;amp; 0 \\&#xD;&#xA;\vdots &amp;amp; \vdots &amp;amp; \ddots &amp;amp; \vdots \\&#xD;&#xA;a_n &amp;amp; a_{n-1} &amp;amp; \ldots &amp;amp; a_1&#xD;&#xA;\end{matrix}&#xD;&#xA;$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd like to know if this matrix has a special name (or even if it's a sum/product of special matrices).&#xA;The thing is, I need to build this matrix in matlab, and I could use some for loops, but I'd rather use some clean builtin functions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: I just had a thought, maybe it can be a Vandermonde matrix times some other (unknown) matrix to nullify the terms above the diagonal...&lt;/p&gt;&#xA;" OwnerUserId="7070" LastActivityDate="2011-02-14T13:28:18.500" Title="Name the matrix (not the game show)" Tags="&lt;matrices&gt;" AnswerCount="1" />
  <row Id="21997" PostTypeId="2" ParentId="21996" CreationDate="2011-02-14T13:24:38.353" Score="6" ViewCount="" Body="&lt;p&gt;It's a lower triangular Toeplitz matrix.&lt;/p&gt;&#xA;" OwnerUserId="589" LastActivityDate="2011-02-14T13:24:38.353" CommentCount="1" />
  <row Id="21999" PostTypeId="2" ParentId="21969" CreationDate="2011-02-14T13:38:37.380" Score="6" ViewCount="" Body="&lt;p&gt;Three points which may or may not help.  The first two are obvious.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, if there is an example where the two sums are equal and the sequences overlap, then removing the overlap from both will provide another example of equal sums.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second, the consecutive sequence with larger integers and so smaller reciprocals must have more terms than the other sequence if the sums of the reciprocals are equal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Third, &lt;a href=&quot;http://mathworld.wolfram.com/HarmonicNumber.html&quot; rel=&quot;nofollow&quot;&gt;Mathworld&lt;/a&gt; says that for the Harmonic number $H_n$, &quot;the denominator is always divisible by the largest power of 2 less than or equal to $n$, and also by any prime $p$  with $n/2 &amp;lt; p \le n$&quot;.  A similar statement about primes therefore applies to those appearing in your sequences (at least providing that the primes are over half the end point), and that severely restricts the possibilities for achieving equality.  Something similar may be true with powers of two, and there may be statements to be made about other small primes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My guess is that it may be possible to show that the denominators of the reduced forms of the sums cannot be equal in any case where the sums are of similar size and the second point above is met.  &lt;/p&gt;&#xA;" OwnerUserId="6460" LastActivityDate="2011-02-14T13:38:37.380" CommentCount="3" />
  <row Id="22000" PostTypeId="1" AcceptedAnswerId="22001" CreationDate="2011-02-14T13:43:38.780" Score="2" ViewCount="82" Body="&lt;p&gt;I was following an example in my text, and there was one step I got stuck on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given $\frac{2x(\Delta x) + (\Delta x)^{2}}{\Delta x}$, how does the denominator cancel out to produce $2c+\Delta x$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My thinking is that the formula expands to $\frac{2x(\Delta x) + (\Delta x * \Delta x)}{\Delta x}$, but how does the denominator cancel out on both sides of the + sign if the numerator is one formula?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;*Edited the formula to be multiplication instead of addition&lt;/p&gt;&#xA;" OwnerUserId="7006" LastEditorUserId="7006" LastEditorDisplayName="" LastEditDate="2011-02-14T15:17:48.940" LastActivityDate="2011-02-14T15:17:48.940" Title="How does this denominator cancel out to create the next step?" Tags="&lt;fractions&gt;" AnswerCount="2" />
  <row Id="22001" PostTypeId="2" ParentId="22000" CreationDate="2011-02-14T13:47:42.283" Score="4" ViewCount="" Body="&lt;p&gt;$\frac{2x(\Delta x) + (\Delta x)^{2}}{\Delta x}=\frac{2x(\Delta x)}{\Delta x} + \frac{(\Delta x)^{2}}{\Delta x}= 2x +\Delta x$.&lt;/p&gt;&#xA;" OwnerUserId="6064" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2011-02-14T13:50:51.940" LastActivityDate="2011-02-14T13:50:51.940" CommentCount="2" />
  <row Id="22002" PostTypeId="1" AcceptedAnswerId="22181" CreationDate="2011-02-14T13:49:13.093" Score="15" ViewCount="591" Body="&lt;p&gt;Let $X$ be the following topological space (with the subspace topology): Connect the rational points of $([0,1]\cap \mathbb{Q})\times \{0\}$ with the point $(0,1)$ and connect the points of $([-1,0]\cap \mathbb{Q})\times \{1\}$ with $(0,0)$, as shown in the figure. Is $X$ contractible?   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/9vgna.jpg&quot; alt=&quot;Well, is it?&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="6064" LastEditorUserId="1307" LastEditorDisplayName="" LastEditDate="2011-02-28T21:45:37.743" LastActivityDate="2011-03-01T09:31:55.497" Title="Is this space contractible?" Tags="&lt;topology&gt;&lt;algebraic-topology&gt;&lt;general-topology&gt;" AnswerCount="1" CommentCount="7" FavoriteCount="2" />
  <row Id="22003" PostTypeId="2" ParentId="22000" CreationDate="2011-02-14T14:03:56.977" Score="0" ViewCount="" Body="&lt;p&gt;Jason, you have mistakenly rewritten $(\Delta x)^2$ as $\Delta x+\Delta x$, rather than as $\Delta x \cdot \Delta x$. Is this the problem?&lt;/p&gt;&#xA;" OwnerUserId="6507" LastActivityDate="2011-02-14T14:03:56.977" />
  <row Id="22004" PostTypeId="1" CreationDate="2011-02-14T14:12:58.353" Score="3" ViewCount="116" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;On &lt;a href=&quot;http://planetmath.org/encyclopedia/InfiniteProductMeasure.html&quot; rel=&quot;nofollow&quot;&gt;Planetmath&lt;/a&gt;, product measure&#xA;is roughly defined as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $(E_i, \mathbb{B}_i, u_i)$ be&#xA;measure spaces, where $i\in I$ an&#xA;index set, possibly infinite. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When each $u_i$ is totally finite,&#xA;there is a unique measure on the&#xA;product measurable space $(E,&#xD;&#xA;    \mathbb{B})$ of $(E_i, \mathbb{B}_i,&#xD;&#xA;    u_i)$. such that &quot;taking measure&quot;&#xA;and &quot;taking product&quot; can be&#xA;&quot;exchanged&quot; for any $B=\prod_{i \in&#xD;&#xA;    I} B_i$ with $B_i \in \mathbb{B_i}$&#xA;and $B_i=E_i$ for all $i \in I$&#xA;except on a finite subset $J$ of&#xA;$I$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was wondering if there is also a&#xA;unique measure on the product&#xA;measurable space, such that &quot;taking&#xA;measure&quot; and &quot;taking product&quot; can be&#xA;&quot;exchanged&quot; for any $B=\prod_{i \in&#xD;&#xA;    I} B_i$ with $B_i \in \mathbb{B_i}$,&#xA;without requiring &quot;$B_i=E_i$ for all&#xA;$i \in I$ except on a finite subset&#xA;$J$ of $I$&quot;? This is not used in the&#xA;definition of product measure, and is&#xA;it only because the product might be for&#xA;infinite number of terms?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;blockquote&gt;&#xA;  &lt;p&gt;If $I$ is infinite, one sees that the&#xA;  total finiteness of $u_i$ can not be&#xA;  dropped. For example, if $I$ is the&#xA;  set of positive integers, assume&#xA;  $u_1(E_1) &amp;lt; \infty$ and&#xA;  $u_2(E_2)=\infty$ . Then $u(B)$ for&#xA;  $$B:=B_1 \times \prod_{i&amp;gt;1} E_i=B_1\times E_2 \times \prod_{i&amp;gt;2} E_i,$$&#xA;  where&#xA;  $B_1 \in \mathbb{B}_1$  would not be&#xA;  well-defined (on the one hand, it is&#xA;  $u_1(B_1)&amp;lt;\infty$ , but on the other&#xA;  it is $u_1(B_1)u_2(E_2)=\infty$ ).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I don't understand the example.&#xA;Specifically how does the last&#xA;sentence in parenthesis show that&#xA;the measure $u$ is not well-defined&#xA;on $B$?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Thanks and regards!&lt;/p&gt;&#xA;" OwnerUserId="1281" LastEditorUserId="1281" LastEditorDisplayName="" LastEditDate="2011-02-14T23:08:58.963" LastActivityDate="2011-02-14T23:08:58.963" Title="Questions about product measure" Tags="&lt;measure-theory&gt;" CommentCount="4" FavoriteCount="1" />
  <row Id="22005" PostTypeId="2" ParentId="21934" CreationDate="2011-02-14T14:42:48.910" Score="1" ViewCount="" Body="&lt;p&gt;On can take a Vitali nonmeasurable subset of $[0,1]$, which has positive and finite outer measure, and just scale it appropriately.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As Jonas points out, this is closely related to &lt;a href=&quot;http://math.stackexchange.com/questions/14591/vitali-type-set-with-given-outer-measure&quot;&gt;this previous question&lt;/a&gt;, but much easier.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2011-02-14T14:42:48.910" CommunityOwnedDate="2011-02-14T14:42:48.910" />
  <row Id="22006" PostTypeId="1" AcceptedAnswerId="22009" CreationDate="2011-02-14T14:56:29.827" Score="0" ViewCount="59" Body="&lt;p&gt;I'm trying to find the root space decomposition of a lie algebra wrt a toral subalgebra h.  Both a matrix lie algebras.  I'm confused about how do I find the linear forms $\lambda \in \mathfrak{h}^*$ on $\mathfrak{h}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\mathfrak{g}_{\lambda } {\text{:=} \{ X\in \mathfrak{g}|\text{ad}_H(X).=\lambda (H)X \forall H\in \mathfrak{h} \} } $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the above formula for the root space is simple enough, but how do I find all the $\lambda$ such that $\mathfrak{g}_{\lambda }$ is nonzero?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In my example, $\mathfrak{h}= \{ \left(&#xD;&#xA;\begin{array}{cccc}&#xD;&#xA; 0 &amp;amp; u &amp;amp; 0 &amp;amp; 0 \\&#xD;&#xA; -u &amp;amp; 0 &amp;amp; 0 &amp;amp; 0 \\&#xD;&#xA; 0 &amp;amp; 0 &amp;amp; 0 &amp;amp; v \\&#xD;&#xA; 0 &amp;amp; 0 &amp;amp; -v &amp;amp; 0&#xD;&#xA;\end{array}&#xD;&#xA;\right) | u,v \in \mathbb{C} \}$ and $\mathfrak{g} = \mathfrak{so_4}\mathbb{C}$ but I'm looking for more general help on how to find all the elements of $\mathfrak{h}^*$ that give nonzero root spaces.&lt;/p&gt;&#xA;" OwnerUserId="6968" LastEditorUserId="6968" LastEditorDisplayName="" LastEditDate="2011-02-14T15:13:30.120" LastActivityDate="2011-02-14T23:00:24.033" Title="Finding roots for a Lie algebra g, wrt toral subalgebra h" Tags="&lt;homework&gt;&lt;matrices&gt;&lt;lie-algebra&gt;&lt;roots&gt;" AnswerCount="1" />
  <row Id="22007" PostTypeId="1" AcceptedAnswerId="22042" CreationDate="2011-02-14T15:02:35.703" Score="4" ViewCount="100" Body="&lt;ol&gt;&#xA;&lt;li&gt;Given a random vector, what are the&#xA;domain, range and sigma algebras on&#xA;them for each of its components to&#xA;be a random variable i.e. measurable&#xA;mapping? Specifically: &#xA;&lt;ul&gt;&#xA;&lt;li&gt;is the domain of each component    random variable same as the domain&#xA;of    the random vector, and are the&#xA;sigma    algebras on their domains&#xA;also the    same?&lt;/li&gt;&#xA;&lt;li&gt;Is the range of each component random    variable the component&#xA;space in the    cartesian product&#xA;for the range of    the random&#xA;vector? What is the sigma    algebra&#xA;on the range of each    component&#xA;random variable and how is    it&#xA;induced from the sigma algebra on&lt;br&gt;&#xA;the range of the random vector?&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;Please correct me if I am wrong. If&#xA;I understand correctly, given a&#xA;random vector, the probability&#xA;measure induced on the range (which&#xA;is a Cartesian product space) by the&#xA;random vector, is called the joint&#xA;probability measure of the random&#xA;vector. The probability measure&#xA;induced on the component space of&#xA;the range by each component of&#xA;the random vector, is called the&#xA;marginal probability measure of the&#xA;component random variable of the&#xA;random vector.&lt;/li&gt;&#xA;&lt;li&gt;Consider the concept of the component random variables of&#xA;a random vector being&#xA;independent. I read from &lt;a href=&quot;http://www.math.ku.dk/~richard/download/courses/Sand1MI_2008/week40_comments.html&quot; rel=&quot;nofollow&quot;&gt;a webpage&lt;/a&gt; that it is said so when the joint probability measure is the product measure of the individual marginal probability measures. I was wondering if the sigma algebra for the joint probability must be the same as the product sigma algebra for the individual probability measures,  or the former can just contain the latter?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Thanks and regards!&lt;/p&gt;&#xA;" OwnerUserId="1281" LastEditorUserId="1281" LastEditorDisplayName="" LastEditDate="2011-02-14T17:46:42.800" LastActivityDate="2011-02-14T19:54:22.940" Title="How to understand joint distribution of a random vector?" Tags="&lt;probability-theory&gt;&lt;measure-theory&gt;" AnswerCount="1" />
  <row Id="22008" PostTypeId="1" CreationDate="2011-02-14T15:06:42.537" Score="1" ViewCount="44" Body="&lt;p&gt;Hi,&#xA;why a smooth surjective morphism of schemes admits a section etale-locally?&lt;/p&gt;&#xA;" OwnerUserId="6583" LastActivityDate="2011-02-14T15:06:42.537" Title="etale sections...." Tags="&lt;algebraic-geometry&gt;&lt;arithmetic-geometry&gt;" />
  <row Id="22009" PostTypeId="2" ParentId="22006" CreationDate="2011-02-14T15:09:11.293" Score="0" ViewCount="" Body="&lt;p&gt;I'm not sure if there's a better way, but one way is to write a basis of $\mathfrak{g}$ and a basis ${ h_1, h_2}$for $\mathfrak{h}$ and then create the matrix for each $Ad_{h_i}$ in the basis of $\mathfrak{g}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, one solves the usual eigenvalue problem for both of the $Ad_{h_i}$.  Since $Ad_{h_1}$ and $Ad_{h_2}$ commute, you will end up finding they have a common eigenspaces.  Then, reading off the eigenvalues for the common eigenspaces and using linearity gives the $\lambda$ functionals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't have my copy of Fulton and Harris's Representation Theory book handy, but they might even work this out as an example.  At the very least, they'd work out &lt;em&gt;some&lt;/em&gt; example, which may give you better tools to work with than the one I've described.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Edit&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a start on working it out.  Let $h_1 = E_{12}$, the matrix with a 1 in the $1,2$ slot an a zero else.  Let $h_2 = E_{34}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the matrix $X = \begin{bmatrix} 0 &amp;amp; 0 &amp;amp; a &amp;amp; b \\ 0 &amp;amp; 0 &amp;amp; c &amp;amp; d \\ -a &amp;amp; -b &amp;amp; 0 &amp;amp; 0 \\ -b &amp;amp; -d &amp;amp; 0 &amp;amp; 0 \end{bmatrix}$ in $\mathfrak{g}$.  (I've set the entries corresponding to $\mathfrak{h}$ equal to 0 because we know what $Ad_{h_i}$ does to those).  To simplify notation, I'll only describe the top right 2 x 2 block at all times.  The top left and bottom right 2 x 2 blocks are always 0s, while the bottom left 2 x 2 block is always whatever is necessary to guarantee I'm looking at an element of $\mathfrak{g}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Computing, one finds $$ Ad_{h_1}X = \begin{bmatrix}  c &amp;amp; d \\ -a &amp;amp; -b\end{bmatrix}$$ (meaning, by the above convention, that the bottom left is what it needs to be to make $Ad_{h_1}X$ antisymmetric and the top left and bottom right are both 0s).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order for $X$ to be an eigenvector of $Ad_{h_i}$ with eigenvalue $\eta$, we must have $$\begin{align}&#xD;&#xA; c &amp;amp;=&amp;amp; \eta a \\ d &amp;amp;=&amp;amp; \eta b\\ -a &amp;amp;=&amp;amp; \eta c\\ -b &amp;amp;=&amp;amp; \eta d\end{align}$$ simultaneously holding.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This implies that $\eta = \pm i$.  When $\eta = i$, the eigenvectors are $\begin{bmatrix} 1 &amp;amp; 0 \\ i &amp;amp; 0\end{bmatrix}$ and $\begin{bmatrix} 0 &amp;amp; 1 \\ 0 &amp;amp; i \end{bmatrix}$.  Let $V = E_i$ be the eigenspace to $Ad_{h_1}$ corresponding to the eigenvalue $i$ (so $V$ is just the span of these two $X$s).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I claim that $Ad_{h_2}:V\rightarrow V$.  In fact, this is general:  if $A$ and $B$ commute, then $B$ maps any eigenvector of $A$ to a new eigenvector of $A$ with the same eigenvalue. For, if $Ax = \lambda x$, then $ABx = BAx = B\lambda x = \lambda B x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We now wish to find eigenvalues of $Ad_{h_2}|_V$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $Y = \begin{bmatrix} a &amp;amp; b \\ ai &amp;amp; bi\end{bmatrix}$ be an arbitrary element of $V$.  Then $Ad_{h_2}Y = \begin{bmatrix} b &amp;amp; -a\\ bi &amp;amp; - ai\end{bmatrix}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We see that $Y$ is an eigenvector of $Ad_{h_2}$ with eigenvalue $\mu$ iff $$\begin{align} b &amp;amp;=&amp;amp; \mu a \\ -a &amp;amp;=&amp;amp; \mu b\\ bi &amp;amp;=&amp;amp; \mu a i \\ -ai &amp;amp;=&amp;amp; \mu bi\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again, this implies that $\mu = \pm i$.  Choosing $\mu = i$, we see that $Y = \begin{bmatrix} 1 &amp;amp; i \\ i &amp;amp; -1\end{bmatrix}$ is both in $V$ and is an eigenvector for $Ad_{h_2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This implies that $span\{Y\}$ is an eigenspace for both $Ad_{h_1}$ and $Ad_{h_2}$, both with eigenvalue $i$.  Thus, if $\lambda(h_1) = \lambda(h_2) = i$ and we extend $\lambda$ to all of $\mathfrak{h}$ by linearity, then $Y\in \mathfrak{g}_\lambda$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From here, you find the other 3 eigenspaces by considering the cases $\eta = \pm i$ and $\mu = \pm i$ separately.  (I've just done the case where we choose plus signs for both).&lt;/p&gt;&#xA;" OwnerUserId="331" LastEditorUserId="331" LastEditorDisplayName="" LastEditDate="2011-02-14T23:00:24.033" LastActivityDate="2011-02-14T23:00:24.033" CommentCount="8" />
  <row Id="22010" PostTypeId="1" AcceptedAnswerId="22090" CreationDate="2011-02-14T15:17:01.857" Score="10" ViewCount="164" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;For $A=\mathbb{Z}[x]/(f)$ with quotient field $K$ and ring of integers $B$, does $U(B)/U(A)$ have a name?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For instance $u = \tfrac{1+\sqrt{5}}{2}$ is a unit in $\mathbb{Q}[\sqrt{5}]$, but neither $u$ nor $u^2$ has integer coefficients in the basis $\{ 1, \sqrt{5} \}$.  Of course $u^3$ has integer coefficients (spooky if you haven't tried it!) and in fact $u^n$ has integer coefficients iff $0 \equiv n \mod 3$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For quadratic fields with basis $\{ 1, \sqrt{n} \}$ for $n$ square-free, one almost always has $U(A) = U(B)$.  If not, then $[ U(B) : U(A) ] = 3$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's crazy, and it should have a name.  For instance, I'd like to find out if the following is true, but I don't even know what to look for:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Is $U(B)/U(A)$ always finite? [ where $B$ is the ring of integers of an order $A$ in a number ring ]&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="583" LastEditorUserId="583" LastEditorDisplayName="" LastEditDate="2011-02-14T23:09:27.317" LastActivityDate="2011-02-15T03:19:42.947" Title="Cubing 1/2 + √5/2 gives 2 + √5, but squaring is just 3/2 + √5/2. What is the name of the group that measures the power that makes units look nice?" Tags="&lt;terminology&gt;&lt;algebraic-number-theory&gt;" AnswerCount="1" FavoriteCount="4" />
  <row Id="22011" PostTypeId="1" AcceptedAnswerId="22017" CreationDate="2011-02-14T15:54:29.000" Score="0" ViewCount="43" Body="&lt;p&gt;I'm stuck on this question from a past paper.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A parachutist jumps out of a hot air balloon. The resistance is proportional to the speed of the parachutist. The parachutist has a terminal velocity of 4 ms&lt;sup&gt;-1&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How far did he fall after 3 seconds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm really not sure how to go about this question, any help much appreciated. &lt;/p&gt;&#xA;" OwnerUserId="2874" LastActivityDate="2011-02-14T16:33:48.397" Title="Variable Resistance" Tags="&lt;physics&gt;" AnswerCount="1" />
  <row Id="22012" PostTypeId="2" ParentId="2949" CreationDate="2011-02-14T15:54:46.387" Score="2" ViewCount="" Body="&lt;p&gt;The consequences of busy beaver problem are really suprising. For every conjecture/hypothesis about countable number of cases if we can write a program that can verify whether this conjecture holds for some case then we need to check only &lt;strong&gt;FINITE&lt;/strong&gt; number of cases to prove that this conjecture is true for every case. &lt;/p&gt;&#xA;" OwnerUserId="4811" LastActivityDate="2011-02-14T15:54:46.387" CommunityOwnedDate="2011-02-14T15:54:46.387" />
  <row Id="22013" PostTypeId="1" CreationDate="2011-02-14T16:00:04.023" Score="4" ViewCount="89" Body="&lt;p&gt;Markov's inequality states that &lt;code&gt;Pr(X &amp;gt;= t*E(X)) &amp;lt;= 1/t&lt;/code&gt;.  This is great for asking &quot;What is the probability that we get more than t times our expected value&quot;.  However, if rather than more, we want to ask what is the probability that we get LESS than our expected value, how is this handled?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Say our expected value is n/4.  If we wanted to ask what the probability is that we'd get less than n/6, intuitively I'd say &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;It is 1 - the probability that we get&#xA;  greater than or equal to n/6&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;, but that doesn't seem to be working out for me.  How is this accomplished with Markov's inequality?&lt;/p&gt;&#xA;" OwnerUserId="7074" LastActivityDate="2011-02-14T16:38:19.403" Title="Interpretation of Markov's Inequality" Tags="&lt;probability&gt;&lt;statistics&gt;&lt;inequality&gt;" AnswerCount="1" />
  <row Id="22014" PostTypeId="1" CreationDate="2011-02-14T16:04:13.393" Score="0" ViewCount="46" Body="&lt;p&gt;Hi,&#xA;why nodal curves over fields are Gorenstein? why a torsion-free sheaf on such curves without associated primes of height one has injective dimension=1?&lt;/p&gt;&#xA;" OwnerUserId="6583" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2011-02-14T18:10:46.093" LastActivityDate="2011-02-14T18:10:46.093" Title="Gorenstein curves" Tags="&lt;algebraic-geometry&gt;&lt;arithmetic-geometry&gt;" />
  <row Id="22015" PostTypeId="2" ParentId="2949" CreationDate="2011-02-14T16:08:26.960" Score="6" ViewCount="" Body="&lt;p&gt;The function $f(x)=\begin{cases}e^{-1/x^2} &amp;amp; \text{ if } x\neq 0\\ 0 &amp;amp; \text{ if } x=0   \end{cases}$&#xA;has Maclaurin series equal to $0$.&lt;/p&gt;&#xA;" OwnerUserId="5253" LastActivityDate="2011-02-14T16:08:26.960" CommunityOwnedDate="2011-02-14T16:08:26.960" />
  <row Id="22016" PostTypeId="1" AcceptedAnswerId="22128" CreationDate="2011-02-14T16:33:06.330" Score="3" ViewCount="230" Body="&lt;p&gt;I have $m$ continues integer points on a line, randomly uniform select $n$ points from the $m$ point without replacement. Order the points ascendingly. &#xA;Let the random variable $A_i$ is the position (coordination on the line) of the $i$th point. So, $$P(A_i=k)=\frac{{k-1\choose i-1} {m-k \choose n-i}}{{m \choose n}} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How to derive the tail inequality for this probability.&#xA;The tail probability look something like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$P(|A_i - E(A_i)| &amp;gt; t) &amp;lt; \sigma$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want the bound ($\sigma$) to be as tight as possible. The Chebyshev inequality is too loose. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Updated:&lt;/strong&gt;&#xA;Some supplement about the question: &lt;a href=&quot;http://www.math.uah.edu/stat/urn/OrderStatistics.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.math.uah.edu/stat/urn/OrderStatistics.pdf&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="6904" LastEditorUserId="6904" LastEditorDisplayName="" LastEditDate="2011-02-15T08:26:26.927" LastActivityDate="2011-02-15T16:31:56.523" Title="What this kind probability should be called?" Tags="&lt;probability&gt;&lt;combinatorics&gt;" AnswerCount="1" />
  <row Id="22017" PostTypeId="2" ParentId="22011" CreationDate="2011-02-14T16:33:48.397" Score="1" ViewCount="" Body="&lt;p&gt;The downward acceleration  in the absence of resistance is g, and you are told there is an upward force proportional to speed.  So (letting primes be d/dt) z&quot;=-g+kz' where k is the constant of proportionality between speed and resistance.  The terminal velocity lets you figure k, then you integrate the equation with respect to t.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-02-14T16:33:48.397" CommentCount="1" />
  <row Id="22018" PostTypeId="2" ParentId="21983" CreationDate="2011-02-14T16:34:28.677" Score="4" ViewCount="" Body="&lt;p&gt;As I mentioned in a comment, I would suggest shying away from &quot;many-to-one.&quot; This often has a very specific interpretation which &lt;em&gt;implies&lt;/em&gt; non-injectivity, but is not equivalent to non-injectivity. For instance, one says a function is &quot;$n$-to-1&quot; to mean that for every point $a$ in the image, there exist &lt;em&gt;exactly&lt;/em&gt; $n$ distinct points $x_1,\ldots,x_n$ such that $f(x_1)=\cdots=f(x_n)=a$ (hence &quot;1-to-1&quot; means each element in the image has &lt;em&gt;exactly&lt;/em&gt; one pre-image, i.e., injectivity). But this means that &quot;many-to-one&quot; is often used to mean that &lt;em&gt;every&lt;/em&gt; point in the image has multiple pre-images.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, any (nonempty) function that satisfies the condition that every point in the image has multiple pre-images is not injective, but the converse does not hold: the function $f\colon\mathbb{R}\to\mathbb{R}$ given by $f(x)=x^2$ is not injective (since $f(1)=f(-1)$ and $1\neq -1$), but there is one and only one point of the domain that maps to $0$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would instead suggest&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Injective (aka one-to-one) if and only if it sends different elements to different elements;&lt;/li&gt;&#xA;&lt;li&gt;Surjective (aka &lt;em&gt;onto&lt;/em&gt;) if and only if every element of the codomain is the image of someone. (Alternatively, as suggested by user3123, if and only if the &lt;em&gt;image&lt;/em&gt; equals the codomain/range).&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;As far as remembering that &quot;injective&quot; means &quot;one-to-one&quot;, you can think of a one-to-one function as a function that &lt;em&gt;injects&lt;/em&gt; the domain into the codomain, so that you end up with a &quot;copy&quot; of the domain inside the codomain (much like an injection puts everything in the syringe into your body). &quot;Surjective&quot; comes from the French &lt;em&gt;sur&lt;/em&gt;, meaning &quot;onto&quot;; or you can think of the Latin &lt;em&gt;super&lt;/em&gt; (above, over), imagining that the domain is complete above and covering the codomain. &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-14T21:26:08.597" LastActivityDate="2011-02-14T21:26:08.597" CommentCount="3" />
  <row Id="22019" PostTypeId="2" ParentId="22013" CreationDate="2011-02-14T16:38:04.033" Score="2" ViewCount="" Body="&lt;p&gt;The probability you want cannot be bounded with the given data. On the one hand it could be $0$, if for example the random variable is zero, and on the other hand it could be close to $1$, if for example the random variable is $0$ w.p. $1-\epsilon$ and $n/4\epsilon$ w.p. $\epsilon$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want an upper bound on such a probability then you need to know that your original random variable $X$ is bounded by some upper bound $B$. Then you consider the non-negative random variable $B-X$ and use the usual Markov's inequality.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2011-02-14T16:38:04.033" />
  <row Id="22021" PostTypeId="2" ParentId="19764" CreationDate="2011-02-14T16:44:26.013" Score="10" ViewCount="" Body="&lt;p&gt;The answer is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\[33817088 = 2^9 \cdot 257^2 = 2^9 \cdot (2^8 + 1)^2\]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\[\phi(33817088) = 16842752 = 2^{16} \cdot (2^8 + 1) = 2^{16} \cdot 257\;.\]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's the Java code to find it (it takes a couple of seconds to run on a MacBook Pro):&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;import java.util.Arrays;&#xA;&#xA;public class Totient {&#xA;    final static int mod4 = 0;     // remainder mod 4 to test                                                                                                                                                                                                                      &#xA;    final static int n = 40000000; // highest number to test                                                                                                                                                                                                                       &#xA;&#xA;    static boolean [] prime = new boolean [n / 2]; // prime [i] : is 2i + 1 prime?                                                                                                                                                                                                 &#xA;    static boolean [] seen = new boolean [n];      // seen [i] : we've seen phi (n) = i                                                                                                                                                                                            &#xA;    static int [] phi = new int [n];               // phi [i] : phi (i)                                                                                                                                                                                                            &#xA;&#xA;    public static void main (String [] args) {&#xA;        Arrays.fill (prime,true);&#xA;        Arrays.fill (seen,false);&#xA;        Arrays.fill (phi,1);&#xA;&#xA;        // calculate the primes we need                                                                                                                                                                                                                                            &#xA;        int limit = (int) Math.sqrt (n); // highest factor to test                                                                                                                                                                                                                 &#xA;        for (int p = 3;p &amp;lt;= limit;p += 2) // loop over odd integers                                                                                                                                                                                                                &#xA;            if (prime [p &amp;gt;&amp;gt; 1])            // only test primes p                                                                                                                                                                                                                   &#xA;                for (int k = 3*p;k &amp;lt; n;k += 2*p) // loop over odd multiples of p                                                                                                                                                                                                   &#xA;                    prime [k &amp;gt;&amp;gt; 1] = false;      // sieve them out                                                                                                                                                                                                                 &#xA;&#xA;        // fill phi by looping over all primes                                                                                                                                                                                                                                     &#xA;        fill (2);&#xA;        for (int p = 3;p &amp;lt; n;p += 2)&#xA;            if (prime [p &amp;gt;&amp;gt; 1])&#xA;                fill (p);&#xA;&#xA;        // now go through phi, remembering which values we've already seen                                                                                                                                                                                                         &#xA;        for (int i = 1;i &amp;lt; n;i++) {&#xA;            if ((i &amp;amp; 3) == mod4 &amp;amp;&amp;amp; !seen [phi [i]]) {&#xA;                System.out.println (&quot;found &quot; + i + &quot; with phi (&quot; + i + &quot;) = &quot; + phi [i]);&#xA;                return;&#xA;            }&#xA;            seen [phi [i]] = true;&#xA;        }&#xA;    }&#xA;&#xA;    // multiply all phi values by their factors of prime p                                                                                                                                                                                                                         &#xA;    static void fill (int p) {&#xA;        // once for the first factor of p                                                                                                                                                                                                                                          &#xA;        for (int i = p;i &amp;lt; n;i += p)&#xA;            phi [i] *= p - 1;&#xA;&#xA;        // and then for the remaining factors                                                                                                                                                                                                                                      &#xA;        long pow = p * (long) p; // long to avoid 32-bit overflow                                                                                                                                                                                                                  &#xA;        while (pow &amp;lt; n) {&#xA;            for (int i = (int) pow;i &amp;lt; n;i += pow)&#xA;                phi [i] *= p;&#xA;            pow *= p;&#xA;        }&#xA;    }&#xA;}&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="6622" LastActivityDate="2011-02-14T16:44:26.013" CommentCount="1" />
  <row Id="22022" PostTypeId="1" CreationDate="2011-02-14T16:53:47.670" Score="0" ViewCount="50" Body="&lt;p&gt;Hi,&#xA;how it is described the normalization of $Spec(k[[x,y]]/(xy))$? In particular call this $N$. Why $N$ can be written as the inersection of 2 local rings and the same holds for the maximal ideal (with 2 maximal ideals)?&lt;/p&gt;&#xA;" OwnerUserId="6583" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2011-02-14T18:10:19.463" LastActivityDate="2011-02-14T18:10:19.463" Title="Normalization of a node" Tags="&lt;algebraic-geometry&gt;&lt;arithmetic-geometry&gt;" CommentCount="4" />
  <row Id="22024" PostTypeId="1" AcceptedAnswerId="22025" CreationDate="2011-02-14T17:21:53.293" Score="0" ViewCount="22" Body="&lt;p&gt;Hello,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm trying to solve coordinate related problem. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a large set of X,Y coordinates (logged by the GPS device during circuit racing, the driver makes 10-20 laps during one racing session) and a vector (a line that marks start/finish of the circuit/lap). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need to divide large set of coordinates into sub-sets using intersections with the start/finish vector. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I solve this problem?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank You&lt;/p&gt;&#xA;" OwnerUserId="3906" LastActivityDate="2011-02-14T17:34:40.817" Title="Divide set of coordinates into sub-sets using intersections with the vector" Tags="&lt;vector-space&gt;&lt;coordinate-systems&gt;" AnswerCount="1" />
  <row Id="22025" PostTypeId="2" ParentId="22024" CreationDate="2011-02-14T17:34:40.817" Score="1" ViewCount="" Body="&lt;p&gt;Are the subsets supposed to be the those of each lap?  Note that your coordinates are really (X,Y,t).   If you list them in time order, you can just look for pairs where the segment from $(X_i,Y_i)$ to $(X_{i+1},Y_{i+1})$ crosses start/finish.  Is that what you are after?&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-02-14T17:34:40.817" CommentCount="1" />
  <row Id="22026" PostTypeId="2" ParentId="10086" CreationDate="2011-02-14T17:40:01.423" Score="2" ViewCount="" Body="&lt;p&gt;I don’t know of any set theories using symmetry to avoid the paradoxes, but there are a number of counterexamples to your claim that “the set of all sets must be excluded.”  The most famous is Quine’s New Foundations; my preferred theory is Church’s (first) Set Theory with a Universal Set (presented 1971, published 1974).  For more details, see the &lt;a href=&quot;http://en.wikipedia.org/w/index.php?title=Universal_set#Set_theories_with_a_universal_set&quot; rel=&quot;nofollow&quot; title=&quot;Universal set&quot;&gt;Wikipedia article on Universal Set&lt;/a&gt; (Disclaimer: I started the article.  I don’t know of any published accounts of Church’s later theory(s), though I discuss them in my forthcoming rewritten doctoral thesis.)&lt;/p&gt;&#xA;" OwnerUserId="7075" LastActivityDate="2011-02-14T17:40:01.423" CommentCount="1" />
  <row Id="22027" PostTypeId="1" CreationDate="2011-02-14T17:56:02.733" Score="1" ViewCount="57" Body="&lt;p&gt;I have two real vectors $v = (v_1,\ldots,v_n)$ and $u = (u_1,\ldots,u_n)$.&#xA;I know that the dot product of $u$ and $v$ is larger than $\delta &amp;gt; 0$: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\langle u,v \rangle \ge \delta$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What would be an interesting condition on $u$, $v$ or both such that I have $u_i v_i \ge f(\delta)$ for each coordinate $i$ with some real function $f()$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, one condition that I thought about is that for any $j \le n$ we have:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sum_{i \neq j} u_i v_i \le \delta/2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and then we can get that $u_i v_i \ge \delta/2$ for every $i$ using triangle inequality.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can assume that $||u|| = 1$ and that $||v||$ is bounded by some $M$ (L-2 norms here).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any help appreciated.&lt;/p&gt;&#xA;" OwnerUserId="7076" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2011-02-14T18:09:46.157" LastActivityDate="2011-02-14T18:09:46.157" Title="Is there a condition for the following consequence?" Tags="&lt;linear-algebra&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="22028" PostTypeId="2" ParentId="22027" CreationDate="2011-02-14T18:03:47.173" Score="2" ViewCount="" Body="&lt;p&gt;If I read you correctly, what you are trying to do is not possible in general. But it may be doable if you give more precisely what $u$'s and $v$'s are allowed, as well as what the desired function $f$ is. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Case in point: let $u$ be the unit vector in the $x_1$ direction. Then $u_iv_i = 0$ for any $i\neq 1$. So for any $f(\delta)$ such that $f$ sends positive numbers to strictly positive numbers, the inequality you are looking for is impossible. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, it is not generally very doable to try to estimate &lt;em&gt;coordinate dependent&lt;/em&gt; quantities (individual $u_iv_i$) by manifestly &lt;em&gt;coordinate independent&lt;/em&gt; ones (the dot product). At best you can expect only something trivial in the general case. &lt;/p&gt;&#xA;" OwnerUserId="1543" LastActivityDate="2011-02-14T18:03:47.173" />
  <row Id="22029" PostTypeId="2" ParentId="21959" CreationDate="2011-02-14T18:16:24.603" Score="3" ViewCount="" Body="&lt;p&gt;This is where the technique of &quot;&lt;a href=&quot;http://en.wikipedia.org/wiki/Bootstrapping_%28statistics%29&quot; rel=&quot;nofollow&quot;&gt;Bootstrap&lt;/a&gt;&quot; comes in extremely handy. You do not need to know anything about the underlying distribution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your question fits in perfectly for a good example of &quot;Bootstrap&quot; technique. The bootstrap technique would also let you determine the confidence intervals. Bootstrap is very elementary to implement on computer and can be done really quick. The typical number of bootstrap samples you take is around $100-200$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Go through the wiki page and let me know if you need more information on &quot;Bootstrap&quot; technique and I am willing to help you out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://books.google.com/books?id=gLlpIUxRntoC&amp;amp;printsec=frontcover&amp;amp;dq=bootstrap+efron&amp;amp;hl=en&amp;amp;ei=sXFZTdilKJH4sAPu-eibDA&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=1&amp;amp;ved=0CCcQ6AEwAA#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;book&lt;/a&gt; by Bradley Efron covers this technique from an application point of view in great detail. The bootstrap algorithm for estimating standard errors is explained on Page $47$, Algorithm $6.1$. You can use this algorithm to construct confidence intervals and finding the quantiles.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastActivityDate="2011-02-14T18:16:24.603" CommentCount="3" />
  <row Id="22030" PostTypeId="2" ParentId="21959" CreationDate="2011-02-14T18:34:40.220" Score="0" ViewCount="" Body="&lt;p&gt;Chances are that the best answer is in some book titled &quot;Non-parametric statistics.&quot; I don't know the texts in the field so I can't suggest one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, it's probably not that hard to get close to what you want using a heuristic method, guided perhaps by some experimentation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the sample was very large, you could take the 90 percentile point. I would adapt this idea to small sample sizes with the added rule that the prediction would be equal to one of the values in the set. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For set S with N observations sorted in ascending order indexed by i, the rule would be something like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For N &amp;lt;=9, prediction = x[9]&#xA;For N &gt; 9 and N &amp;lt; 20, prediction = x[N-1]&#xA;For N &gt; 20 and N &amp;lt; 40, prediction = x[N-2]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that I'm only suggesting the form of the answer, not the actual rule, and in fact, I'm really only suggesting that you base your prediction on the upper tail of the sample.&lt;/p&gt;&#xA;" OwnerUserId="1312" LastEditorUserId="1312" LastEditorDisplayName="" LastEditDate="2011-02-14T21:03:00.560" LastActivityDate="2011-02-14T21:03:00.560" />
  <row Id="22031" PostTypeId="1" CreationDate="2011-02-14T18:47:17.620" Score="19" ViewCount="596" Body="&lt;p&gt;This sort of question is probably in bad taste for math.stackexchange, but is probably in high demand. (I tried to start a site on Area 51 to house questions like this, but my request was closed due to the existence of math.stackexchange.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am advising talented students about graduate school. I believe that the most important thing one must be sure to do is pick the right PhD thesis advisor.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Question: What is your best advice on picking a good thesis advisor. (For the sake of levity, feel free to answer the dual question regarding how to pick a bad thesis advisor. Just be sure you are clear whether you are indicating how to pick a good or bad advisor!)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="941" LastEditorUserId="941" LastEditorDisplayName="" LastEditDate="2011-02-15T11:22:36.540" LastActivityDate="2011-02-16T02:04:22.830" CommunityOwnedDate="2011-02-14T19:01:29.090" Title="How to pick a thesis advisor?" Tags="&lt;soft-question&gt;&lt;education&gt;" AnswerCount="5" CommentCount="6" FavoriteCount="11" />
  <row Id="22032" PostTypeId="1" AcceptedAnswerId="22036" CreationDate="2011-02-14T18:52:40.267" Score="5" ViewCount="98" Body="&lt;p&gt;Let $X$ be a topological space, $Y$ and $Z$ subspaces of $X$. Let $C$ be a connected subset of $Y\cap Z$ such that $C$ is a component of $Y$ and a component of $Z$. Does it follow that $C$ is a component of $Y\cup Z$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Intuitively, I would say yes, but I don't know how to prove it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In case further assumptions are necessary, you can go as far as: $X$ is a compact metric space, $Y$ is open, $Z$ (and therefore $C$) is closed, and $C=Y\cap Z$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any help is much appreciated.&lt;/p&gt;&#xA;" OwnerUserId="1878" LastActivityDate="2011-02-14T19:06:58.780" Title="If $C$ is a component of $Y$ and a component of $Z$, is it a component of $Y\cup Z$?" Tags="&lt;topology&gt;&lt;general-topology&gt;" AnswerCount="1" FavoriteCount="2" />
  <row Id="22033" PostTypeId="2" ParentId="4170" CreationDate="2011-02-14T18:53:34.697" Score="0" ViewCount="" Body="&lt;p&gt;A book that should be read by everyone in mathematics regardless of level is Wolfe's &lt;em&gt;A Tour Through Mathematical Logic&lt;/em&gt;. It's simply a compulsory read,I couldn't put it down. It gives a broad overview of mathematical logic and set theory along with it's history and is absolutely beautifully written. That's the best place for anyone to begin. &lt;/p&gt;&#xA;" OwnerUserId="7012" LastActivityDate="2011-02-14T18:53:34.697" CommunityOwnedDate="2011-02-14T18:53:34.697" />
  <row Id="22034" PostTypeId="2" ParentId="22031" CreationDate="2011-02-14T19:00:32.913" Score="7" ViewCount="" Body="&lt;p&gt;I'm quite fond of the advice given in the article &lt;a href=&quot;http://www.cs.indiana.edu/how.2b/how.2b.html&quot; rel=&quot;nofollow&quot;&gt;How to be good graduate student&lt;/a&gt; although it's actually aimed towards students of computer science. There's also a section on how to pick one's advisor. I think a link to this article was posted on mathoverflow some time ago.&lt;/p&gt;&#xA;" OwnerUserId="1403" LastActivityDate="2011-02-14T19:00:32.913" CommunityOwnedDate="2011-02-14T19:01:29.090" CommentCount="2" />
  <row Id="22035" PostTypeId="2" ParentId="4170" CreationDate="2011-02-14T19:01:34.110" Score="0" ViewCount="" Body="&lt;p&gt;Cori, Lascar, Pelletier, &lt;em&gt;Mathematical Logic : A course with exercises -- Part I and Part II&lt;/em&gt;. Especially the second one.&#xA;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0198500483&quot; rel=&quot;nofollow&quot;&gt;Amazon link&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="6582" LastActivityDate="2011-02-14T19:01:34.110" CommunityOwnedDate="2011-02-14T19:01:34.110" />
  <row Id="22036" PostTypeId="2" ParentId="22032" CreationDate="2011-02-14T19:06:58.780" Score="5" ViewCount="" Body="&lt;p&gt;In general, no.  Let $X=[0,1]$, $Y=\mathbb{Q}\cap[0,1]$, $Z=([0,1]\setminus\mathbb{Q})\cup\{0\}$, and $C=\{0\}$.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2011-02-14T19:06:58.780" CommentCount="4" />
  <row Id="22037" PostTypeId="2" ParentId="22031" CreationDate="2011-02-14T19:07:54.637" Score="7" ViewCount="" Body="&lt;p&gt;This is an interesting take: &lt;a href=&quot;http://mathgradblog.williams.edu/?p=1111&quot; rel=&quot;nofollow&quot;&gt;http://mathgradblog.williams.edu/?p=1111&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="6491" LastActivityDate="2011-02-14T19:07:54.637" CommunityOwnedDate="2011-02-14T19:07:54.637" CommentCount="1" />
  <row Id="22038" PostTypeId="1" AcceptedAnswerId="22044" CreationDate="2011-02-14T19:10:12.843" Score="1" ViewCount="101" Body="&lt;p&gt;What exactly does this notation mean when using it to define a function?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f(x) = f(\text{ sum of the digits of }x) \{ &#xD;&#xA;\mathrm{if}( x \ge 10) \} =  x \{ \mathrm{if}( x \lt 10) \} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="7079" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-14T20:15:16.110" LastActivityDate="2011-02-14T20:42:37.810" Title="What does $f(x) = f(\text{ sum of the digits of }x ) \{ \mathrm{if}( x \ge 10) \}$ $= x \{ \mathrm{if}( x \lt 10) \} $ mean?" Tags="&lt;notation&gt;" AnswerCount="2" CommentCount="8" />
  <row Id="22039" PostTypeId="2" ParentId="22038" CreationDate="2011-02-14T19:14:11.103" Score="5" ViewCount="" Body="&lt;p&gt;If $x\lt 10$, $f(x)=x$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $x \ge 10$, to get $f(x)$ add up the digits of x, presumably in base 10, then repeat until you get below 10.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So for example, $f(6)=6, f(18)=9, f(987)=f(24)=6, f(1234)=f(10)=1$&lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2011-02-14T20:42:37.810" LastActivityDate="2011-02-14T20:42:37.810" CommentCount="4" />
  <row Id="22040" PostTypeId="2" ParentId="21972" CreationDate="2011-02-14T19:15:50.490" Score="0" ViewCount="" Body="&lt;p&gt;The important thing is that you have a $C^1$ curve between your points $x,y \in A$, where the curve lies wholly inside the domain where your function $f$ is differentiable. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now if you imagine a &quot;dumbbell&quot; like set (say, the unit circles centered on (1,0) and (-1,0) in the plane, it is not clear what you mean when you say that $f$ is differentiable inside this domain; specifically, the nature of $\nabla f$ at the origin becomes an interesting question. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can finesse the definition there, but the standard definition of the derivative assumes that there is some neighborhood about the point where the function is defined. &lt;/p&gt;&#xA;" OwnerUserId="1560" LastActivityDate="2011-02-14T19:15:50.490" />
  <row Id="22041" PostTypeId="1" CreationDate="2011-02-14T19:23:04.360" Score="0" ViewCount="128" Body="&lt;p&gt;My apologies if this comes out as stupid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Problem (Domain : stock exchanges):&#xA;       I have been given the sum of orders placed per instrument(security or company) for about 25000+ instruments. So my dataset looks something like this:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;[Instrument] [Order count]   [Average Order Size(AOS)]  [Orders above 10*AOS]&#xA;&#xA;   AAA            20                 10000.0                 ?&#xA;   BBB            5000               24334.5                 ?&#xA;   ...................&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I know the average order size placed for an instrument as shown above. I want to calculate an estimate for the number of orders placed which had a order size above &lt;code&gt;10*[average order size]&lt;/code&gt; per instrument. I don't have an idea about the distribution. Assuming it's a normal distribution, I don't have an idea about the standard deviation. Is there any way to get a practical answer (I don't care if it is a value in a range.) for my problem? I hope this question is relevant here. Thanks.&lt;/p&gt;&#xA;" OwnerUserId="7078" LastActivityDate="2011-02-14T22:13:15.480" Title="How to figure out this problem" Tags="&lt;normal-distribution&gt;" AnswerCount="4" CommentCount="3" />
  <row Id="22042" PostTypeId="2" ParentId="22007" CreationDate="2011-02-14T19:54:22.940" Score="3" ViewCount="" Body="&lt;p&gt;I'll take a stab at answering these:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;(Part a) Yes, the domain probability space of a random vector is the same as the domain probability space of its components.  Think of a random vector as a vector-valued random variable.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I'm having trouble parsing these questions, because I can't tell whether you're using the word &quot;range&quot; to mean &quot;image&quot; or &quot;codomain&quot;.  I'll assume you mean &quot;codomain&quot;.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;(Part b) Given a probability space $\Omega$ and measurable spaces $E_1,\ldots,E_n$, a random vector is a random variable $\Omega \to E_1\times\cdots\times E_n$.  Typically, each $E_i$ starts with a $\sigma$-algebra on it, and then the $\sigma$-algebra on $E = E_1\times\cdots\times E_n$ is defined as the product algebra.  (That is, the $\sigma$-algebra on $E$ is generated by all products of the form $A_1\times\cdots\times A_n$, where $A_i$ is measurable in $E_i$ for each $i$.)  Sometimes it is helpful to expand the $\sigma$-algebra on the product slightly, e.g. if you want some probability measure on the product to be complete.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;That seems right.  &quot;Marginal probability measure&quot; would also refer to the measure obtained on a product like $\prod_{i\in S} E_i$, where $S$ is some subset of $\{1,\ldots,n\}$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I suppose it's fine for the $\sigma$-algebra on the product to be slightly larger than the product $\sigma$-algebra, e.g. if we want the measure on the product to be complete.  However, the measure on the product should have the property that it is the unique extension of the product measure to the $\sigma$-algebra on the product.  That is, the measure on the product should either be the product measure, or the completion of the product measure, or something in between.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="1726" LastActivityDate="2011-02-14T19:54:22.940" CommentCount="1" />
  <row Id="22043" PostTypeId="2" ParentId="16547" CreationDate="2011-02-14T19:58:59.653" Score="0" ViewCount="" Body="&lt;p&gt;Assume that $h_n \to h$ uniformly on the domain $D$ of $\{h_n\}$, where&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$h_n(x) = &#xD;&#xA;\begin{cases} &#xD;&#xA; \frac{x}{n}  \biggl(1 + \frac{1}{n} \biggr), &amp;amp; \text{if } x \text{ is   irrational} \\&#xD;&#xA; a + \frac{a}{n} + \frac{a}{b} \biggl(1 + \frac{1}{n} \biggr) \biggl(\frac{1}{n} \biggr), &amp;amp; \text{if } x \text{ is rational, say } x = a/b, b &amp;gt; 0&#xD;&#xA;\end{cases}$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;and $$h(x) = &#xD;&#xA;\begin{cases}&#xD;&#xA; 0, &amp;amp; \text{if }x\text{ is irrational} \\&#xD;&#xA; a, &amp;amp; \text{if }x\text { is rational, say }x = a/b, b &amp;gt; 0.&#xD;&#xA;\end{cases}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there is a positive rational in $D$ we can find an interval $A = [s,t]$ contained in $D$ with $s$ rational because $D$ is a bounded interval.  Then given $\epsilon = 1$ there exists a positive integer $M$ such that $n \ge M$ implies that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA; \biggl|h_n\biggl(\frac{a}{b}\biggr) - h\biggl(\frac{a}{b}\biggr) \biggr|        &amp;amp;=   \&#xA; \biggl| \frac{a}{n} + \frac{a}{b} \biggl( \frac{1}{n} + \frac{1}{n^2} \biggr) \biggr| &amp;amp;&amp;lt; 1 &#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;whenever $n \ge$ M and $a/b \in A$ where $a$ and $b$ are positive integers.&#xA;But then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA; \frac{a}{M} &amp;amp;&amp;lt; \frac{a}{M} + \frac{a}{b} \biggl( \frac{1}{M} + \frac{1}{M^2} \biggr) \&#xA;  &amp;amp;= \biggl| \frac{a}{M} + \frac{a}{b} \biggl( \frac{1}{M} + \frac{1}{M^2} \biggr) \biggr| \&#xA;  &amp;amp;&amp;lt; 1. &#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Writing $s = c/d$ where $s$ is the left endpoint of $A$ and $c$ and $d$ are positive integers, we can choose a positive integer $k$ large enough so that both $(ck + 1)/d \in A$ and $ck + 1 &amp;gt; M$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This contradicts $(ck + 1)/M &amp;lt; 1$.  Therefore the assumption that $h_n \to h$ uniformly is false.  A similar argument holds if no positive rationals are in $D$&lt;/p&gt;&#xA;" OwnerUserId="6809" LastActivityDate="2011-02-14T19:58:59.653" />
  <row Id="22044" PostTypeId="2" ParentId="22038" CreationDate="2011-02-14T20:02:57.563" Score="4" ViewCount="" Body="&lt;p&gt;It means that which formula applies depends on which condition holds. Explicilty, it is equivalent to&#xA;$$f(x) = \left\{\begin{array}{ll}&#xA;f(\mbox{sum of the digits of $x$}) &amp;amp;\mbox{if $x\geq 10$}\\&#xA;x &amp;amp;\mbox{if $x\lt 10$.}&#xA;\end{array}\right.$$&#xA;which recursively defines the function for every positive integer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, if $x=481754$, then $x\geq 10$, so&#xA;\begin{align*}&#xA;f(x) &amp;amp;= f(481754) = f(4+8+1+7+5+4)\\&#xA;&amp;amp;= f(29) = f(2+9)\\&#xA;&amp;amp;= f(11) = f(1+1)\\&#xA;&amp;amp;= f(2) = 2.&#xA;\end{align*}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the function that reduces an integer to its residue class modulo $9$ (&quot;casting out nines&quot;), using $9$ as the representative for the multiples of $9$ (instead of $0$).  &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-14T20:09:11.360" LastActivityDate="2011-02-14T20:09:11.360" CommunityOwnedDate="2011-02-14T20:02:57.563" CommentCount="1" />
  <row Id="22045" PostTypeId="1" CreationDate="2011-02-14T20:12:21.680" Score="1" ViewCount="40" Body="&lt;p&gt;I have a question more about terminology than anything else:  If $f:X\rightarrow Y$ is an isomorphism of schemes, then what does it mean to say &quot;the invertible sheaf $\mathscr{L}$ on $X$ corresponds to the invertible sheaf $\mathscr{M}$ on $Y$?&quot;  Naturally, it makes sense to assume that this statement means $f_{\ast}\mathscr{L}=\mathscr{M}$ or $\mathscr{L}=f^{\ast}\mathscr{M}$, but I wasn't sure...&lt;/p&gt;&#xA;" OwnerUserId="7082" LastActivityDate="2011-02-14T20:12:21.680" Title="Isomorphism of schemes and invertible sheaves" Tags="&lt;soft-question&gt;&lt;schemes&gt;&lt;sheaf-theory&gt;" CommentCount="2" />
  <row Id="22047" PostTypeId="1" CreationDate="2011-02-14T21:00:28.427" Score="4" ViewCount="101" Body="&lt;p&gt;How do you prove that $\sin(n)$ with $n=0,1,2...$ is not uniformly distributed mod 1? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(This is an exercise in Uniform Distribution of Sequences by Kuipers and Niederreiter.)&lt;/p&gt;&#xA;" OwnerUserId="7084" LastActivityDate="2011-02-14T21:07:44.260" Title="$\sin(n)$ is not u.d. mod 1" Tags="&lt;sequences-and-series&gt;" AnswerCount="1" />
  <row Id="22048" PostTypeId="2" ParentId="22047" CreationDate="2011-02-14T21:07:44.260" Score="5" ViewCount="" Body="&lt;p&gt;You &lt;a href=&quot;http://en.wikipedia.org/wiki/Equidistribution_theorem&quot; rel=&quot;nofollow&quot;&gt;know&lt;/a&gt; that the sequence of powers of $e^{i}$ are equidistributed in the unit circle of $\mathbb C$. The sequence you have is obtained from this one by projecting to the imaginary axis. Can you see what happens?&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2011-02-14T21:07:44.260" />
  <row Id="22049" PostTypeId="1" AcceptedAnswerId="22052" CreationDate="2011-02-14T21:10:28.150" Score="3" ViewCount="78" Body="&lt;p&gt;Hi!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a module M over some commutative ring R. A square matrix $\Omega$ of size $m$ with coefficients in $R$ defines a $R$-linear map from $M^m$ to itself.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Proposition. If $x = \left( \begin{matrix} x_1 \\ \vdots \\ x_m \end{matrix} \right) \in M^m$ is in the kernel of $\Omega$, then $\det \Omega$ annihilates all the $x_i$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I think I have a proof of this fact, roughly going as follows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's call &quot;pseudomatrix&quot; a matrix whose coefficients are all in R except for a column, where the coefficients are in M. The remark is that the formulas for the determinant of a matrix and the product of two matrices still make sense in that context: the determinant of a pseudomatrix is an element of M and the product of a matrix and a pseudomatrix is a pseudomatrix. One even has that if A is a matrix and $B$ a pseudomatrix, then $\det(AB) = \det(A)\det(B)$. (I think that no proof is needed here: this formula is true in the classical setting, so it can be seen as a formal identity and can be specified in our setting). I can then easily construct a pseudomatrix whose first column is $x$ and whose determinant is $x_i$, and the proposition follows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm convinced that this proof should be easily expressed in an &quot;algebraic nonsensical&quot; language, using words such as &quot;$\otimes_R M$&quot; and &quot;$\bigwedge$&quot; instead of my ugly hybrid &quot;pseudomatrices&quot; but I've been unable to do the translation. Can you help me?&lt;/p&gt;&#xA;" OwnerUserId="7085" LastEditorUserId="7085" LastEditorDisplayName="" LastEditDate="2011-02-15T00:24:35.283" LastActivityDate="2011-02-15T00:24:35.283" Title="Matrices: injectivity and determinant" Tags="&lt;linear-algebra&gt;&lt;ring-theory&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="22050" PostTypeId="2" ParentId="21959" CreationDate="2011-02-14T21:25:55.650" Score="0" ViewCount="" Body="&lt;p&gt;You should make some assumptions about your distribution, if you are looking into confidence intervals. With 20 data points that is possible. You should first plot your distribution, and then use a chi-square test to see if it fits a hypothesized one (normal, exponential, poisson etc.)  Only after you have specified the distribution can you bootstrap.  Non-parametrics do not deal with confidence intervals per se, since they are based upon the median and ranked values. However you can use ranked measures such as quartiles or the 90Th decile to describe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-Ralph Winters&lt;/p&gt;&#xA;" OwnerUserId="7086" LastActivityDate="2011-02-14T21:25:55.650" CommentCount="1" />
  <row Id="22052" PostTypeId="2" ParentId="22049" CreationDate="2011-02-14T21:29:51.987" Score="2" ViewCount="" Body="&lt;p&gt;The matrix $\Omega$ and its &lt;a href=&quot;http://en.wikipedia.org/wiki/Adjugate_matrix&quot; rel=&quot;nofollow&quot;&gt;adjugate&lt;/a&gt; $\mho$ determine maps $\Omega:R^m\to R^m$ and $\mho:R^m\to R^m$ such that $\mho\circ\Omega=\det\Omega\cdot\mathrm{id}_{R^m}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now suppose $m\in M^m=R^m\otimes_R M$ is such that what you write $\Omega m$ and I will write $(\Omega\otimes\mathrm{id}_M)(m)$ is zero. Then &#xA;\begin{align}&#xA;0 &amp;amp;= (\mho\otimes\mathrm{id}_M)\bigl((\Omega\otimes\mathrm{id}_M)(m)\bigr) \\&#xA;  &amp;amp;= ((\mho\circ\Omega)\otimes\mathrm{id}_M)(m) \\&#xA;  &amp;amp;= \det\Omega\cdot m&#xA;\end{align}&lt;/p&gt;&#xA;" OwnerUserId="274" LastEditorUserId="274" LastEditorDisplayName="" LastEditDate="2011-02-14T21:43:37.067" LastActivityDate="2011-02-14T21:43:37.067" CommentCount="1" />
  <row Id="22053" PostTypeId="2" ParentId="22041" CreationDate="2011-02-14T21:38:56.817" Score="0" ViewCount="" Body="&lt;p&gt;Assuming a normal distribution, and given that you only known the average value, there is no way of estimating the standard deviation and thus the number of outlier values (values with size greater than ten times the average).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is not the case, however, for other families of distributions. There are, for instance, families of distribution depending on only one parameter which therefore determines both the expectation value and the variance.&lt;/p&gt;&#xA;" OwnerUserId="367" LastEditorUserId="367" LastEditorDisplayName="" LastEditDate="2011-02-14T21:44:03.223" LastActivityDate="2011-02-14T21:44:03.223" CommentCount="1" />
  <row Id="22054" PostTypeId="2" ParentId="21987" CreationDate="2011-02-14T21:51:54.907" Score="0" ViewCount="" Body="&lt;ul&gt;&#xA;&lt;li&gt;A &lt;a href=&quot;http://en.wikipedia.org/wiki/Formal_power_series&quot; rel=&quot;nofollow&quot;&gt;&quot;formal power series&quot;&lt;/a&gt; has only non-negative exponents&lt;/li&gt;&#xA;&lt;li&gt;A &lt;a href=&quot;http://en.wikipedia.org/wiki/Laurent_series&quot; rel=&quot;nofollow&quot;&gt;&quot;Laurent series&quot;&lt;/a&gt; allows negative exponents too&lt;/li&gt;&#xA;&lt;li&gt;A &lt;a href=&quot;http://en.wikipedia.org/wiki/Puiseux_series&quot; rel=&quot;nofollow&quot;&gt;&quot;Puiseux series&quot;&lt;/a&gt; allows fractional exponents too (with restriction)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1919" LastEditorUserId="1919" LastEditorDisplayName="" LastEditDate="2011-02-14T22:59:01.340" LastActivityDate="2011-02-14T22:59:01.340" />
  <row Id="22055" PostTypeId="2" ParentId="22041" CreationDate="2011-02-14T21:52:44.990" Score="2" ViewCount="" Body="&lt;p&gt;It is quite unlikely to be a normal distribution.  There is the heuristic 80/20 rule-80% of your orders come from 20% of the customers.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-02-14T21:52:44.990" />
  <row Id="22056" PostTypeId="1" CreationDate="2011-02-14T21:53:57.790" Score="0" ViewCount="73" Body="&lt;p&gt;Hi, i am developing an application for iPhone, and i need some help to solve this difficult problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am working with the device compass, from where i get the angle where the iPhone is oriented.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need to check what points are on a angle view(oriented with the compass) from the current location.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/Da873.png&quot; alt=&quot;diagram&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have no idea how to make it, i have been thinking all the afternoon. Any consideration is apreciated. Thanks!!!!&lt;/p&gt;&#xA;" OwnerUserId="7081" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2011-02-14T22:25:41.877" LastActivityDate="2011-02-15T15:43:16.830" Title="Angle points calculation" Tags="&lt;geometry&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" />
  <row Id="22057" PostTypeId="2" ParentId="22056" CreationDate="2011-02-14T22:00:04.343" Score="2" ViewCount="" Body="&lt;p&gt;So from the compass you have bearings that are the limits of visibility.  In your figure they appear to be $k+20^{\circ}$ and $k-20^{\circ}$  Then take the difference in position between the phone and each point.  You can feed this to &lt;a href=&quot;http://en.wikipedia.org/wiki/Atan2&quot; rel=&quot;nofollow&quot;&gt;the Atan2 function&lt;/a&gt; to get the bearing (remember to convert between degrees and radians).  If it is in range, you are good.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-02-14T22:00:04.343" CommentCount="3" />
  <row Id="22058" PostTypeId="2" ParentId="22041" CreationDate="2011-02-14T22:02:47.207" Score="4" ViewCount="" Body="&lt;p&gt;Given that the order size is positive, You could use Chebyshev's inequality or &lt;a href=&quot;http://en.wikipedia.org/wiki/Markov%27s_inequality&quot; rel=&quot;nofollow&quot;&gt;Markov inequality&lt;/a&gt; as people call it. This does not have any assumption on the underlying distribution. If you know the underlying distribution, then you could come up with a tighter bound.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your case, $P(\text{Order size} &amp;gt; 10 \times \text{AOS}) \leq \frac{1}{10} $&lt;/p&gt;&#xA;" OwnerUserId="3302" LastActivityDate="2011-02-14T22:02:47.207" CommentCount="1" />
  <row Id="22059" PostTypeId="2" ParentId="22041" CreationDate="2011-02-14T22:06:23.757" Score="0" ViewCount="" Body="&lt;p&gt;OK, I already suggested using &lt;a href=&quot;http://en.wikipedia.org/wiki/Markov%27s_inequality&quot; rel=&quot;nofollow&quot;&gt;Markov's inequality&lt;/a&gt;, but this only gives a very rough estimate for an upper bound on the probability of having an order of large size. Also see Sivaram's explanation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But maybe what you really ought to do is model the distribution of the order sizes. I think it's not unreasonable to assume some exponential distribution which depends only on one parameter, the average order size. Or maybe a power law distribution, &lt;a href=&quot;http://en.wikipedia.org/wiki/Pareto_distribution&quot; rel=&quot;nofollow&quot;&gt;Pareto type&lt;/a&gt;, it would give rise to 80-20 type rules. If you possess historical data, you can try to check how well your data fit this model.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe you should provide us the context of the question. Is it homework? If so, for what discipline? &lt;/p&gt;&#xA;" OwnerUserId="3567" LastEditorUserId="3567" LastEditorDisplayName="" LastEditDate="2011-02-14T22:13:15.480" LastActivityDate="2011-02-14T22:13:15.480" />
  <row Id="22060" PostTypeId="1" AcceptedAnswerId="22061" CreationDate="2011-02-14T22:13:28.757" Score="2" ViewCount="77" Body="&lt;p&gt;On the Wikipedia article for &lt;a href=&quot;http://en.wikipedia.org/wiki/Codomain&quot; rel=&quot;nofollow&quot;&gt;codomain&lt;/a&gt;, in the third paragraph, it roughly says:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;When the domain of a function is a&#xA;  proper class X, in which case there is&#xA;  formally no such thing as a triple (X,&#xA;  Y, F). (?)  With such a definition&#xA;  functions do not have a codomain.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As a proper class is a class that cannot be a member of some class, i.e. cannot be a set, I was wondering why a function with its domain being a proper class does not have a codomain?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks and regards!&lt;/p&gt;&#xA;" OwnerUserId="1281" LastActivityDate="2011-02-15T00:46:23.630" Title="Why a function whose domain is a proper class does not have a codomain?" Tags="&lt;elementary-set-theory&gt;" AnswerCount="1" CommentCount="0" />
  <row Id="22061" PostTypeId="2" ParentId="22060" CreationDate="2011-02-14T22:23:59.413" Score="4" ViewCount="" Body="&lt;p&gt;My guess is: proper classes cannot belong to sets, and triples are sets, so the triple in question does not make sense. &lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2011-02-14T22:23:59.413" CommentCount="5" />
  <row Id="22062" PostTypeId="2" ParentId="22031" CreationDate="2011-02-14T22:41:32.153" Score="13" ViewCount="" Body="&lt;p&gt;My lecturer in homological algebra, Hal Schenck, has some advice on this &lt;a href=&quot;http://www.math.uiuc.edu/~schenck/grad.html&quot; rel=&quot;nofollow&quot;&gt;http://www.math.uiuc.edu/~schenck/grad.html&lt;/a&gt; (item II in the list). I quote the relevant paragraph:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;&lt;strong&gt;1) Compatibility.&lt;/strong&gt;&#xA;It is really important that you and your advisor get along well; grad school is difficult enough without having a chilly (or worse, hostile) relationship with your advisor. The best time to figure this out is BEFORE you ask someone if they'll work with you. One way to determine this is by taking a course with the person, and seeing how you interact when you go to office hours to ask questions. It goes without saying that you should make sure you do a good job in the class, because if you do poorly, it is likely that the potential advisor will not be interested in taking you on as a student.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;2) Track record and reputation.&lt;/strong&gt;&#xA;It is important that your advisor be respected as a scholar. You will quickly find that one of the first questions mathematicians ask a graduate student is: &quot;and who is your advisor?&quot;. You may also want to find out how many theses the professor has directed, and how many students have dropped out (this should be done tactfully, of course, by talking to former students or the director of graduate studies). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;3) Good thesis topic.&lt;/strong&gt;&#xA;Most professors supervising dissertations have a set of problems that they think might make an appropriate thesis; you should ask about this, 'cause getting a bad problem can cause you to waste years of your life.&quot;&lt;/p&gt;&#xA;" OwnerUserId="4284" LastActivityDate="2011-02-14T22:41:32.153" CommunityOwnedDate="2011-02-14T22:41:32.153" />
  <row Id="22063" PostTypeId="1" AcceptedAnswerId="22222" CreationDate="2011-02-14T23:16:19.470" Score="2" ViewCount="88" Body="&lt;p&gt;I'm working in Einsiedler and Ward's book on Ergodic Theory and in Exercise 2.5.4 they want to prove the following&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\lim_{N - M \to \infty} \frac{1}{N - M} \sum_{n = M}^{N - 1} U_T^n f \to P_T f.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I'm wondering what this limit really says, it must be stronger than pointwise convergence since we can pick $M = 0$. It we let $N = 2n$ and $M = n$, then we seem to get more and more terms but adding them up from the &quot;tail&quot;. To me it seems that this is a very strong convergence condition or am I wrong? What does the limit intuitively say?&lt;/p&gt;&#xA;" OwnerUserId="1120" LastEditorUserId="5363" LastEditorDisplayName="" LastEditDate="2011-02-16T11:54:31.117" LastActivityDate="2011-02-16T11:54:31.117" Title="Uniform mean ergodic theorem." Tags="&lt;functional-analysis&gt;&lt;ergodic-theory&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="22064" PostTypeId="1" AcceptedAnswerId="22067" CreationDate="2011-02-14T23:23:35.713" Score="2" ViewCount="111" Body="&lt;p&gt;Hi, I need to find a point (A on this diagram) given the center point of the ellipse as well as an angle. I've been melting my brain all day (as well as searching through questions here) testing out different equations. What's the best way to do this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/YytNu.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I intend to grab point A at 225 degrees as well as another point at approximately 250 degrees using the same math. These need to be fetched regardless of elliptic width and height.&lt;/p&gt;&#xA;" OwnerUserId="7089" LastActivityDate="2011-02-14T23:32:30.840" Title="Calculating a Point that lies on an Ellipse given an Angle" Tags="&lt;geometry&gt;&lt;conic-sections&gt;" AnswerCount="2" />
  <row Id="22065" PostTypeId="1" CreationDate="2011-02-14T23:27:34.960" Score="15" ViewCount="434" Body="&lt;p&gt;How can set theory and category theory both be plausible theories for the foundations of mathematics? If these two theories are not mathematically equivalent, does it not mean that the rest of mathematics, when taken exclusively with either of these two theories as foundation, will be distinct; that is, unless these two theories are, in some way, equivalent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="7485" LastActivityDate="2011-04-09T12:04:41.397" Title="How can there be alternatives for the foundations of mathematics?" Tags="&lt;set-theory&gt;&lt;category-theory&gt;" AnswerCount="2" CommentCount="6" FavoriteCount="3" />
  <row Id="22066" PostTypeId="1" AcceptedAnswerId="22091" CreationDate="2011-02-14T23:28:42.603" Score="9" ViewCount="184" Body="&lt;p&gt;I am still studying the topics in forcing and did not yet study much about forcing with a class of conditions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know from Jech's &lt;em&gt;Set Theory&lt;/em&gt; that you can force that the class of ordinals in the world will be countable in the generic extension, which means that you can take a proper class and turn it into a set.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was wondering if there is a nice characterization of classes that can be forced into sets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clearly it is sufficient to assume that the class is well-orderable, and if you assume AC then if you collapsed a class to a set, in the generic extension it will be well-orderable, but does that mean that you can construct from the function you've added one which shows that you actually collapsed $Ord^M$ to some $\dot\alpha$ in the generic extension? (i.e. is the sufficient condition is also necessary?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I went to see my advisor today, and we talked about this question a little bit. He wasn't able to give me a full answer but he gave me a nice direction. We discussed generic extensions with global choice, namely you only add a class function which is a well-ordering of the world, without introducing new sets into the world. Jech wrote about it right between Easton Forcing and Levy Collapse in chapter 15 of his book.&lt;br&gt;&#xA;I searched further and found some mailing list in which Solovay said (quite recently too!) that he and Jensen did something like that already way back. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In light of everything, I came up with some new questions which I will ponder on my own for a while; one of the question is worth mentioning here as it related very much to my original question - especially in light of Joel's answer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it possible to collapse the entire universe to some inaccessible cardinal in a larger universe? (If so, doesn't that imply that $Con(ZFC)\rightarrow Con(ZFC+In)$? (where $In$ is the existence of an inaccessible cardinal)) If not in $ZFC$ how about $NBG$?&lt;/p&gt;&#xA;" OwnerUserId="622" LastEditorUserId="622" LastEditorDisplayName="" LastEditDate="2011-02-15T23:09:39.473" LastActivityDate="2011-02-16T00:26:59.593" Title="Forcing Classes Into Sets" Tags="&lt;set-theory&gt;&lt;large-cardinals&gt;&lt;forcing&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" />
  <row Id="22067" PostTypeId="2" ParentId="22064" CreationDate="2011-02-14T23:31:13.790" Score="3" ViewCount="" Body="&lt;p&gt;If the ellipse is centered at the origin, the equation of the ellipse is $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$.  The equation of the line is $y=\tan \theta x$  So you have $\frac{x^2}{a^2}+\frac{(\tan \theta x)^2}{b^2}=1$ or $x=\pm \frac{ab}{\sqrt{b^2+a^2(\tan \theta)^2}}$ where the sign is + if $ -\pi/2 \lt x \lt \pi/2$&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-02-14T23:31:13.790" CommentCount="1" />
  <row Id="22068" PostTypeId="2" ParentId="22064" CreationDate="2011-02-14T23:32:30.840" Score="2" ViewCount="" Body="&lt;p&gt;If the ellipse is centered at $(0,0)$, $2a$ wide in the $x$-direction, $2b$ tall in the $y$-direction, and the angle you want is $\theta$ from the positive $x$-axis, the coordinates of the point of intersection are&#xA;$$\left(\frac{a b}{\sqrt{b^2+a^2\tan^2(\theta)}},\frac{a b \tan(\theta)}{\sqrt{b^2 + a^2\tan^2(\theta)}}\right) \text{ if }0\le\theta&amp;lt; 90&amp;#176;\text{ or }270&amp;#176;&amp;lt;\theta\le360&amp;#176;$$&#xA;or&#xA;$$\left(-\frac{a b}{\sqrt{b^2+a^2\tan^2(\theta)}},-\frac{a b \tan(\theta)}{\sqrt{b^2 + a^2\tan^2(\theta)}}\right) \text{ if }90&amp;#176;&amp;lt;\theta&amp;lt; 270&amp;#176;.$$&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2011-02-14T23:32:30.840" CommentCount="3" />
  <row Id="22069" PostTypeId="1" AcceptedAnswerId="22073" CreationDate="2011-02-14T23:58:39.807" Score="5" ViewCount="328" Body="&lt;p&gt;I was wondering if there is a name for such kind of function &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f:\mathbb{R} \rightarrow \mathbb{R}$ and $f(x+y)=f(x) \times f(y)$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks and regards!&lt;/p&gt;&#xA;" OwnerUserId="1281" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-15T03:19:21.057" LastActivityDate="2011-02-15T03:19:21.057" Title="Is there a name for such kind of function?" Tags="&lt;calculus&gt;&lt;functions&gt;" AnswerCount="3" FavoriteCount="1" />
  <row Id="22070" PostTypeId="2" ParentId="22069" CreationDate="2011-02-15T00:11:09.470" Score="6" ViewCount="" Body="&lt;p&gt;If you enforce continuity on $f(x)$ (continuity even at a single point should do), then the only function which satisfies this is an exponential function of the form $f(x) = e^{ax}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You need to enforce continuity to extend this from rationals to reals (as you would expect).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have $f: \mathbb{Q} \rightarrow \mathbb{R}$, then $f(x) = e^{ax}$ is the only function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(The proof for this is analogous to proving &lt;a href=&quot;http://en.wikipedia.org/wiki/Cauchy%27s_functional_equation&quot; rel=&quot;nofollow&quot;&gt;$f(x) = cx$ when $f(x+y) = f(x) + f(y)$&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact one of the definitions of $f(x) = a^x$ is as follows. If there exists a function $f(x)$ satisfying the following properties,&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;$f(x) : \mathbb{R} \rightarrow \mathbb{R}$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$f(x+y) = f(x) \times f(y)$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$f(x) \text{ is continuous at-least at one point in } \mathbb{R}$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$f(1) = a$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Then, $f(x) = a^x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Note&lt;/strong&gt;: $e^x$ could be defined like above with $f(1) = e$, where $\displaystyle e = \lim_{n \rightarrow \infty} \left( 1+\frac{1}{n} \right)^n$. This is one of the many definitions for $e^x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Without continuity you could define a function as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know that along all rationals $f(x) = e^{ax}$ where $x \in \mathbb{Q}$. Now lets define $f(x)$ over $x \in \mathbb{R} \backslash \{ \mathbb{Q} \}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For this, consider $\mathbb{R}$ as a vector space over $\mathbb{Q}$. This is an infinite-dimensional vector space and the axiom of choice guarantees the existence of a basis for this space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now consider a basis for this space. Let the set of basis be $\{h_{\alpha}\}$ i.e. every real number can now be written as $r = \displaystyle \sum_{i=1}^{N(r)} q_i h_{\alpha_i}$, where $q_i \in \mathbb{Q}$, $N(r) \in \mathbb{N}$, and $h_{\alpha_i} \in \{h_{\alpha}\}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now you have complete freedom to decide what the value of $f$ should take on each element in the set $\{ h_{\alpha} \}$.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2011-02-15T02:22:16.183" LastActivityDate="2011-02-15T02:22:16.183" CommentCount="13" />
  <row Id="22071" PostTypeId="1" AcceptedAnswerId="22077" CreationDate="2011-02-15T00:12:48.660" Score="5" ViewCount="126" Body="&lt;p&gt;Came across this conundrum while going over the proof that&#xA;$$A \cdot \sin(bx) + B \cdot \cos(bx) = C \cdot \sin(bx + k)$$&#xA;for some numbers $C$ and $k$. ($A$, $B$ and $b$ are known.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The usual method is to expand the RHS using the compound angle identity&#xA;\begin{align}&#xA;C \cdot \sin(bx + k) &amp;amp;= C \cdot \bigl( \sin(bx)\cos(k) + \cos(bx)\sin(k) \bigl) \\&#xA;&amp;amp;= C\cos(k) \cdot \sin(bx) + C\sin(k) \cdot \cos(bx)&#xA;\end{align}&#xA;and thus set&#xA;\begin{align}&#xA;C\cos(k) &amp;amp;= A \\&#xA;C\sin(k) &amp;amp;= B&#xA;\end{align}&#xA;My trouble comes with what happens at this point - we then proceed to divide the second equation by the first, obtaining&#xA;$$ \tan(k) = \frac{B}{A} $$&#xA;which we then solve to obtain $k$, etc. etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is: how do we know that this is &quot;legal&quot;? We have reduced the original two-equation system to a single equation. How do we know that the values of $k$ that satisfy the single equation are equal to the solution set of the original system?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While thinking about this I drew up this other problem:&#xA;\begin{align}&#xA;\text{Find all }x\text{ such that} \\&#xA;\sin(x) &amp;amp;= 1 \\&#xA;\cos(x) &amp;amp;= 1&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously this system has no solutions ($\sin x$ and $\cos x$ are never equal to $1$ simultaneously). But if we apply the same method we did for the earlier example, we can say that since $\sin(x) = 1$ and $\cos(x) = 1$, let's divide $1$ by $1$ and get&#xA;$$ \tan(x) = 1 $$&#xA;which does have solutions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So how do we know when it's safe to divide simultaneous equations by each other? (If ever?)&lt;/p&gt;&#xA;" OwnerUserId="7090" LastEditorUserId="7090" LastEditorDisplayName="" LastEditDate="2011-02-15T00:26:36.813" LastActivityDate="2011-02-15T01:04:10.833" Title="Simultaneous equations, trig functions and the existence of solutions" Tags="&lt;algebra&gt;&lt;trigonometry&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="22072" PostTypeId="2" ParentId="22069" CreationDate="2011-02-15T00:13:20.867" Score="4" ViewCount="" Body="&lt;p&gt;$\log f$ satisfies &lt;a href=&quot;http://en.wikipedia.org/wiki/Cauchy%27s_functional_equation&quot; rel=&quot;nofollow&quot;&gt;Cauchy's functional equation&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="589" LastActivityDate="2011-02-15T00:13:20.867" CommentCount="1" />
  <row Id="22073" PostTypeId="2" ParentId="22069" CreationDate="2011-02-15T00:14:41.983" Score="17" ViewCount="" Body="&lt;p&gt;If $f(x_0)= 0$ for some $x_0\in\mathbb{R}$, then for all $x\in\mathbb{R}$, $f(x)=f(x_0+(x-x_0))=f(x_0)\cdot f(x-x_0)=0$.  Therefore, either $f$ is identically $0$ or never $0$.  If $f$ is not $0$, then it is a homomorphism from the group $\mathbb{R}$ with addition to the group $\mathbb{R}\setminus\{0\}$ with multiplication.  If $f(x)&amp;lt;0$ for some $x$, then $f(\frac{x}{2})^2\lt 0$, which is impossible, so $f$ is actually a homomorphism into the positive real numbers with multiplication.  By composing with the isomorphism $\log:(0,\infty)\to\mathbb{R}$, such $f$ can be analyzed by first analyzing all additive maps on $\mathbb{R}$.  Assuming continuity, these all have the form $x\mapsto cx$ for some $c\in \mathbb{R}$, and hence $f(x)=\exp(cx)$.  Assuming the axiom of choice, there are discontinuous additive functions on $\mathbb{R}$ that can be constructed using a Hamel basis for $\mathbb{R}$ over $\mathbb{Q}$, and thus there are also discontinuous homomorphisms from $\mathbb{R}$ to $(0,\infty)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So for an actual answer to the question: Yes, they are called (the zero map or) homomorphisms from the additive group of real numbers to the multiplicative group of positive real numbers.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2011-02-15T00:36:28.690" LastActivityDate="2011-02-15T00:36:28.690" CommentCount="1" />
  <row Id="22074" PostTypeId="1" CreationDate="2011-02-15T00:16:00.360" Score="1" ViewCount="83" Body="&lt;p&gt;In the link &lt;a href=&quot;http://en.wikipedia.org/wiki/Moore%E2%80%93Penrose_pseudoinverse&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Moore%E2%80%93Penrose_pseudoinverse&lt;/a&gt;, it talks about solving $Ax=b$ by $x = A^+b + [I − A^+A]w$ for any vector $w$. Let's say $A$ is $m\times n$, and $b$ and $x\in \mathbb{R}^n$. My question is: since $w$'s $n$ components are in general more than needed given the kernel of $A$ might be of dimension $n-r\gt 0$, how to use Moore–Penrose pseudoinverse $A^+$ to give an explicit solution with the least number of free parameters?&lt;/p&gt;&#xA;" OwnerUserId="6016" LastEditorUserId="6016" LastEditorDisplayName="" LastEditDate="2011-02-15T17:40:01.707" LastActivityDate="2011-02-15T17:40:01.707" Title="a question regarding the use of Moore–Penrose pseudoinverse" Tags="&lt;linear-algebra&gt;&lt;matrices&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="22075" PostTypeId="2" ParentId="8501" CreationDate="2011-02-15T00:31:41.567" Score="2" ViewCount="" Body="&lt;p&gt;I would think it depends largely on your personality type. Do you thrive in slow, peaceful&#xA;  environments, or do you prefer a faster environment with more variety?. I am in NYC&#xA;  and love it, but my friend prefers U of Florida in Gainesville, which is not that slow&#xA;  of a place, but he has the option of going to the country--real country, on the weekends&#xA;  ( if you want real country in NYC , you need to travel a few hours). Also, there are&#xA;  school environments that are more casual than others. I took  a class at Columbia,&#xA;  and many of the people there were a bunch of a-holes, who pretended not to know you when they saw you outside of class, even after you had taken classes with them for a whole semester, even a whole year. I think there are books out there that rate graduate programs, using many different parameters. If you are interested in getting a hold of these books and can't find thme, let me know and I will get you the actual title(s). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Good Luck!&lt;/p&gt;&#xA;" OwnerUserId="7094" LastActivityDate="2011-02-15T00:31:41.567" CommentCount="1" />
  <row Id="22076" PostTypeId="2" ParentId="22071" CreationDate="2011-02-15T00:41:35.380" Score="0" ViewCount="" Body="&lt;p&gt;Good question!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is similar to the following question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;If we have two linear polynomials say, $ax + b$ and $cx+d$, and we know that they are equal at two distinct points then can we conclude that the two linear polynomials are the same?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Getting back to your problem, First note that what we have is not a single equation but &lt;strong&gt;an infinite set of equations&lt;/strong&gt; since we want the &lt;strong&gt;equation to hold true for all $x \in \mathbb{R}$&lt;/strong&gt;. This is the key here and this is why we are allowed to &quot;equate the corresponding coefficients&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A somewhat detailed explanation is below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in fact, we should be surprised that a solution actually exists for these $2$ variables and satisfies this infinite set of equations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, if a solution were to exist, then it better satisfy &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$A \sin(bx) + B \cos(bx) = C \sin(bx+k)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;at $x=0$ and $x = \frac{\pi}{2b}$ (assume $b$ is non-zero).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Plugging in $x=0$ and $x = \frac{\pi}{2b}$ gives us&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$B = C \sin(k)$ and $A = C \sin(\frac{\pi}{2} + k) = C \cos(k)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you get to this stage you can solve for $C$ and $k$ by the usual algebra.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After solving this, you now go back and feed it into the original equation to check if the original equation is consistent. And you will find that it is consistent and does satisfy for all $x$ since we have the identity&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sin(bx+k) = \cos(k) \sin(bx) + \sin(k) \cos(bx)$$ which holds good $\forall$  $b,x,k \in \mathbb{R}$ and hence the only possible solution to this system is $$C = \sqrt{A^2+B^2}, \tan(k) = \frac{B}{A}$$&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2011-02-15T01:03:12.633" LastActivityDate="2011-02-15T01:03:12.633" CommentCount="1" />
  <row Id="22077" PostTypeId="2" ParentId="22071" CreationDate="2011-02-15T00:42:26.750" Score="2" ViewCount="" Body="&lt;p&gt;First, the question seems to be:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Show that there are some $C,k$ such that for every $x$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$A \sin bx + B \cos bx = C \sin(bx + k)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice that the step from &#xA;$$A \sin bx + B \cos bx = C \sin(bx) \cos k + C \cos bx \sin k$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$C\cos k = A, \ \ C\sin k = B$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is actually the implication&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ C\cos k = A, \ \ C\sin k = B \Longrightarrow \ A \sin bx + B \cos bx = C \sin(bx) \cos k + C \cos bx \sin k$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, if you manage to find $C,k$ such that $C\cos k = A, \ \ C\sin k = B$, you have proved the statement. Notice that this does not help you find &lt;em&gt;all&lt;/em&gt; such $C,k$. Just one such pair. (The reverse implication is true too, but needs a little more justification).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now if $C\cos k = A, \ \ C\sin k = B$, then we &lt;em&gt;must&lt;/em&gt; have for non-zero $A$ that $\tan k = \frac{B}{A}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is just the implication in one direction:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ C\cos k = A, \ \ C\sin k = B \Longrightarrow \tan k = \frac{B}{A}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It only means that any solution of &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$C\cos k = A, \ \ C\sin k = B$ satisfies $\tan k = \frac{B}{A}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i.e. the set of $k$ which satisfies $C\cos k = A, \ \ C\sin k = B$ is a &lt;em&gt;subset&lt;/em&gt; of the set of solutions of $\tan k = \frac{B}{A}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It does not mean that any solution of $\tan k = \frac{B}{A}$ satisfies $C\cos k = A, \ \ C\sin k = B$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance consider the equation $x = 1$. Squaring gives $x^2 = 1$, which has $-1$ as a solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a more extreme example: $x^2 = -1$ implies $0 \times x^2 = 0$, which is true for every $x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the step is &quot;legal&quot; in the sense that it is a valid implication, but you cannot just read off the solutions of the original equation from the implied ones, as you yourself noticed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If statement X implies statement Y it is not necessary that statement Y should imply statement X.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When solving equations, some manipulations are equivalent, i.e. the resulting equation you get also implies the original (like adding 1), while some are not, like squaring.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2011-02-15T00:57:38.763" LastActivityDate="2011-02-15T00:57:38.763" />
  <row Id="22079" PostTypeId="2" ParentId="8501" CreationDate="2011-02-15T00:58:20.700" Score="2" ViewCount="" Body="&lt;p&gt;An obvious answer here is Paris. Both historically and currently, Paris is a bit of a Mecca for mathematicians. With the 13 universities of Paris and ihes so close there are more seminars than could be attended per week. Additionally the level of the mathematicians is top tier if anything. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Culturally, and generally, Paris is a real city. Lots going on, lots to see, and many people find the atmosphere very inviting. I have been told by many mathematicians, and experienced it myself, that doing math in Paris seems somehow more relaxing. Additionally, public transportation, food, and general facilities are very accommodating. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Paris may be great, but it is no nirvana, it is known for it's insane housing issues, and busy-ness. Moreeover some actually dislike this French culture. &lt;/p&gt;&#xA;" OwnerUserId="16" LastActivityDate="2011-02-15T00:58:20.700" CommentCount="1" />
  <row Id="22080" PostTypeId="2" ParentId="22071" CreationDate="2011-02-15T01:04:10.833" Score="1" ViewCount="" Body="&lt;p&gt;In general, you don't. You know that all of the solutions of the pair of equations you started with are solutions of the single equation you ended up with (barring division-by-zero issues), but you generally don't know the converse. In this case, the reason you can get away with the converse is that you can choose $C$. Knowing $\tan k$ is the same as knowing $(\cos k, \sin k)$ up to a multiplicative constant; draw a unit circle if you don't believe this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, the only way you know whether it is &quot;legal&quot; to do anything is to prove or disprove that you can do it.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2011-02-15T01:04:10.833" CommentCount="2" />
  <row Id="22082" PostTypeId="1" CreationDate="2011-02-15T02:15:48.820" Score="5" ViewCount="218" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Let $F\subseteq\mathbb{C}$ be a Galois extension of $\mathbb{Q}$ such that $[F:\mathbb{Q}]=2^n$; then all elements in $F$ are constructible.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;hr/&gt;&#xA;&lt;em&gt;Added.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is what I have so far.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since there exist the finite normal extension $F\subseteq\mathbb{C}$ over $\mathbb{Q}$ which contains an element say, s, then s has a degree of power of 2. From group theory we have 2 propositons we could use. 'A group $G$ is solvable iff it has a normal series whose factors are abelian', and also 'If $G$ is solvable and $N\triangleleft G$, then $G/N$ is solvable'. Then we can say that the galois group F has a normal series $\{e\}=G_0\triangleleft G_1\triangleleft\cdots\triangleleft G_n=\mathrm{Gal}(F/\mathbb{Q})$, whose factors have order 2. Thus there must exist intermediate fields $\mathbb{Q}=F_0 \subseteq F_1 \subseteq F_2 \subseteq\cdots\subseteq F_n=F$, such that $[F_{j+1}:F_j]=2$ for all $j$. Now this is where I got stuck. Next I need to somehow show that $F_{j+1}=F_j(s_j)$, where $s^2_j$ is an element in $F_j$. Then I think I can conclude that if $s_j$ is constructible than every element in $F_{j+1}$ is constructible... then with $F_n$ is constructible...&lt;/p&gt;&#xA;" OwnerUserId="7097" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-15T14:42:07.260" LastActivityDate="2011-02-15T20:46:03.707" Title="Let $F$ be a Galois extension over $\mathbb{Q}$ with $[F:\mathbb{Q}]=2^n$, then all elements in $F$ are constructible." Tags="&lt;homework&gt;&lt;galois-theory&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="2" />
  <row Id="22084" PostTypeId="1" AcceptedAnswerId="22086" CreationDate="2011-02-15T02:32:52.570" Score="0" ViewCount="33" Body="&lt;p&gt;So we have an equation $x^2 + By^2=z^2$, and we're supposed to find for what values of $B$ this graph would show two intersecting planes. That value is $0$. Now I need to find the equation of these two planes. I'm not really sure how to start this one. I'm very shakey on this stuff. Any hints?&lt;/p&gt;&#xA;" OwnerUserId="2985" LastActivityDate="2011-02-15T02:37:39.537" Title="Finding equation of a plane" Tags="&lt;calculus&gt;" AnswerCount="1" />
  <row Id="22085" PostTypeId="1" CreationDate="2011-02-15T02:34:04.563" Score="0" ViewCount="39" Body="&lt;p&gt;We consider differential forms that are defined on all $R^{n}$, without singularity. Is Hodge decomposition always possible in this situation?&lt;/p&gt;&#xA;" OwnerUserId="3871" LastActivityDate="2011-02-15T02:34:04.563" Title="Hodge decomposition" Tags="&lt;differential-geometry&gt;&lt;differential-forms&gt;" FavoriteCount="0" />
  <row Id="22086" PostTypeId="2" ParentId="22084" CreationDate="2011-02-15T02:37:39.537" Score="3" ViewCount="" Body="&lt;p&gt;You write:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;x^2=z^2&#xD;&#xA;$$&#xA;$$&#xD;&#xA;x^2-z^2=0&#xD;&#xA;$$&#xA;$$&#xD;&#xA;(x+z)(x-z)=0&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;that is satisfed by $z+x=0$ and $z-x=0$ that are you planes.&lt;/p&gt;&#xA;" OwnerUserId="3871" LastActivityDate="2011-02-15T02:37:39.537" />
  <row Id="22087" PostTypeId="1" AcceptedAnswerId="22100" CreationDate="2011-02-15T02:56:59.540" Score="0" ViewCount="73" Body="&lt;p&gt;For a homework assignment, I was asked to find the derivative of the function $$f(x)= \frac{x^{3}-6x^{2}+6}{x^{2}}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the product rule, I worked through the following:&#xA;$$f&amp;#39;(x)=\frac{\frac{d}{dy}(x^{3}-6x^{2}+6)}{x^{2}}+(x^{3}-6x^{2}+6)\left[\frac{d}{dy}\frac{1}{x^{2}}\right]$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the derivative of $\frac{1}{x{^2}}$ is $-\frac{2}{x^{3}}$, the left side of the formula resulted in &#xA;$$\frac{-2(x^{3}-6x^{2}+6}{x^{3}}= -\frac{2x^{3}+12x^{2}-12}{x^{3}}.$$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The right side of the equation split up by the product rule into $$\frac{\frac{d}{dy}(x^{3})-6(\frac{d}{dy}(x^{2}))+\frac{d}{dy}(6)}{x^{2}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The derivatives of $x^{3}$ is $3x^{2}$; $x^{2}$ is $2x$, and 6 is 0, thus making the formula \&#xA;$$\frac{3x^{2}+12x}{x^{2}}-\frac{2x^{3}-12x{^2}+12}{x^{3}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's where I got stuck.  Now, when I plugged the formula into WolframAlpha, the derivative calculated was $1-\frac{12}{x^{3}}$  How can I go from the formula above to the calculated result?&lt;/p&gt;&#xA;" OwnerUserId="7006" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-15T03:11:13.297" LastActivityDate="2011-02-15T04:21:03.960" Title="Help finding the derivative of the function" Tags="&lt;calculus&gt;" AnswerCount="3" CommentCount="3" />
  <row Id="22088" PostTypeId="2" ParentId="22087" CreationDate="2011-02-15T03:00:31.873" Score="1" ViewCount="" Body="&lt;p&gt;Are you specifically told to use the product rule with no simplication? If not, you should probably try dividing through by $x^2$ first to simplify what you're doing. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice&#xA;$$&#xD;&#xA;f(x)=\frac{x^3-6x^2+6}{x^2}=x-6+6x^{-2}.&#xD;&#xA;$$&#xA;This should be much easier to work with and gives you the derivative given by WolframAlpha as well.&lt;/p&gt;&#xA;" OwnerUserId="1583" LastEditorUserId="1583" LastEditorDisplayName="" LastEditDate="2011-02-15T03:06:18.233" LastActivityDate="2011-02-15T03:06:18.233" />
  <row Id="22089" PostTypeId="2" ParentId="22087" CreationDate="2011-02-15T03:03:32.713" Score="1" ViewCount="" Body="&lt;p&gt;OK. I am confused so i am working the solution out again completely. What i feel is that somewhere you have a made a mistake of $\pm$ sign. That's all not a major error.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have $$\Bigl[x^{3}-6x^{2}+6\Bigr] \cdot -\frac{2}{x^{3}} \quad \ (1)$$ and &#xA;$$\Bigl[3x^{2}-12x\Bigr] \cdot \frac{1}{x^{2}}\quad \ (2)$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you have to add these two. Eq (1) is : $ -2 + \frac{12}{x} -\frac{12}{x^{3}}$ and Eq (2) is: $3 - \frac{12}{x}$. Adding these two gives $1- \frac{12}{x^{3}}$ which is what you wanted.&lt;/p&gt;&#xA;" LastActivityDate="2011-02-15T03:28:21.973" CommentCount="1" />
  <row Id="22090" PostTypeId="2" ParentId="22010" CreationDate="2011-02-15T03:14:35.760" Score="4" ViewCount="" Body="&lt;p&gt;In general, if $A \subset B$ is an extension of rings, I would call $B^{\times}/A^{\times}$ the &lt;strong&gt;relative unit group&lt;/strong&gt;.  (I am probably not the only one, but I couldn't say how widespread this is.)  I feel reasonably confident that there is no specialized terminology in the case of nonmaximal orders.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To answer your non-terminological question: &lt;strong&gt;yes&lt;/strong&gt;, if $A \subset B$ are orders in the same number field, the group $B^{\times}/A^{\times}$ is finite.  This follows from the fact that one can prove the Dirichlet Unit Theorem equally well for a nonmaximal order $A$ in a number field $K$: $A^{\times}$ is still finitely generated with torsion subgroup equal to the roots of unity in $A$ and free rank equal to $r+s-1$, where $r$ is the number of real places and $s$ is the number of complex places of $K$.  Thus we have finitely generated abelian groups $A^{\times} \subset B^{\times}$ with the same free rank, so $B^{\times}/A^{\times}$ is finite.  &lt;/p&gt;&#xA;" OwnerUserId="299" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2011-02-15T03:19:42.947" LastActivityDate="2011-02-15T03:19:42.947" CommentCount="4" />
  <row Id="22091" PostTypeId="2" ParentId="22066" CreationDate="2011-02-15T03:16:00.983" Score="7" ViewCount="" Body="&lt;p&gt;For any set $X$, we may consider the partial order $\mathbb{P}$ consisting of all finite partial functions from $\omega$ to $X$, ordered by extension. If $G\subset\mathbb{P}$ is $V$-generic for this forcing notion, then $f=\cup G$ is a function from $\omega$ onto $X$. It is a function, since conditions in $G$ are compatible; it is total, since it is dense to add any given point to the domain; and it is surjective, since for any $x\in X$ the set of conditions $p$ with $x\in\mathop{ran}(p)$ is dense. Thus, in the forcing extension $V[G]$, the set $X$ becomes countable. This is how any set can be made countable by forcing. Since the forcing notion $\mathbb{P}$ is a set, it also follows that $V[G]\models ZFC$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case that $X$ is an arbitrary proper class, one can still form the partial order $\mathbb{P}$ consisting of all finite partial functions from $\omega$ to $X$, and it will still be true that the union of the generic filter will be a function from $\omega$ to $X$, thereby making $X$ countable in $V[G]$. But what will no longer be true in this class forcing case is that $V[G]\models ZFC$. Indeed, if $X$ is a proper class, then the corresponding forcing extension will definitely not satisfy ZFC, since $X$ will contain members of arbitrarily large ordinal rank (appearing unboundedly high in the $V_\alpha$ hierarchy), the forcing extension will have a function from $\omega$ unbounded in the ordinals, in violation of the Replacement axiom. So the price you pay for making a proper class $X$ countable is that you must give up ZFC in the forcing extension. Indeed, if $X$ is a proper class in $V$, then $X$ contains elements of unboundedly high rank, and this will remain true in any extension of $V$ to a model $W$ with the same ordinals; thus, $X$ will not be a set in any such $W$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, to answer your question: every set can be countable in a forcing extension by set forcing (and thus while preserving ZFC), but no proper class can become a set in any extension of the universe to a model of ZF with the same ordinals. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Edit.&lt;/b&gt; I see now that you want to consider upward&#xA;extensions of the models, not just forcing extensions.&#xA;There are a few observations to make about this. I have&#xA;been interested in this topic for some time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is the extent to which a model $M$ of set&#xA;theory can become the $(V_\theta)^N$, for some taller of a&#xA;taller model of set theory.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;First, I claim that every model $M$ of set theory can be&#xA;elementarily embedded into the $V_\theta$ of another model&#xA;of set theory $N$, and you can even arrange that $(V_\theta)^N\prec N$.&#xA;This is a sense in which $M$ is&#xA;continued as you desire to a taller model. You can prove this by a&#xA;simple compactness argument using the elementary diagram&#xA;of $M$, augmenting the theory with a new constant symbol&#xA;$\dot\theta$ and the assertions that the right theory&#xA;holds. This theory is finitely consistent by an&#xA;application of the Reflection theorem.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Second, some models $M$ of set theory cannot be realized&#xA;directly as the $V_\theta$ of a taller model. For&#xA;example, this is true in any pointwise definable model (a&#xA;model in which every element is definable without&#xA;parameters), since the larger model would then see that&#xA;every element of $V_\theta$ is definable in $V_\theta$,&#xA;which would contradict that the larger model thinks it is&#xA;uncountable.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Third, every countable computably saturated model $M$ of set&#xA;theory is realized as the $V_\theta$ of a taller model.&#xA;Indeed, every countable computably saturated model of ZFC&#xA;is isomorphic to one of its own $V_\theta$'s, and by&#xA;looking at things from the perspective of that copy, the&#xA;result obtains. Harvey Friedman was the first to prove&#xA;results of this kind, concentrating at first on&#xA;nonstandard models of PA. You can find an account of the&#xA;argument in my recent paper with Victoria Gitman in &lt;a href=&quot;http://projecteuclid.org/DPubS?verb=Display&amp;amp;version=1.0&amp;amp;service=UI&amp;amp;handle=euclid.ndjfl/1285765800&amp;amp;page=record&quot; rel=&quot;nofollow&quot;&gt;&quot;A natural model of the multiverse axioms,&quot;&#xA;Notre Dame Journal of Formal logic, vol 51,&#xA;2010&lt;/a&gt;.&#xA;One of the things we prove is that if $M$ is a countable&#xA;computably saturated model of ZFC, then $M$ is also&#xA;isomorphic to an element of $M$ that thinks it is an&#xA;$\omega$-nonstandard model of set theory. Thus, every&#xA;countable computably saturated model of ZFC exists as a&#xA;nonstandard model inside another better model.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Fourth, it is possible to show that every countable transitive model of set theory can be end-extended to a (possibly nonstandard) model of V=L, which is well-founded to any desired countable height. The reason is that this assertion is true in $L$ itself, and the statement has complexity $\Pi^1_2$ in a code for the structure, so it is true in $V$ by Shoenfield absolutenss.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="413" LastEditorUserId="413" LastEditorDisplayName="" LastEditDate="2011-02-16T00:26:59.593" LastActivityDate="2011-02-16T00:26:59.593" CommentCount="6" />
  <row Id="22092" PostTypeId="2" ParentId="22082" CreationDate="2011-02-15T03:40:19.157" Score="7" ViewCount="" Body="&lt;p&gt;The idea is to try to deduce the result for $F$ by induction on $n$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, to get your feet wet, you want to show that if $[F:\mathbb{Q}]=2$, then all elements of $F$ are constructible. Feel comfortable with that one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, how would the general proof by induction work?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You want to argue that an $F$ that is Galois over $\mathbb{Q}$ with $[F:\mathbb{Q}]=2^{n}$ has a subextension $K$, $\mathbb{Q}\subseteq K\subseteq F$, with $K$ Galois over $\mathbb{Q}$, $[F:K]=2$, and $[K:\mathbb{Q}]=2^{n-1}$. Then you would use induction to show that everything in $K$ is constructible. And then you'd use an argument similar to the one you used for the case $n=1$ in order to show that since everything in $K$ is constructible, and $[F:K]=2$, then everything in $F$ is constructible. So your induction hypothesis would look something like&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If $L$ is Galois over $\mathbb{Q}$, and has $[L:\mathbb{Q}]=2^k$ for some $k\lt n$, then every element of $L$ is constructible.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;To do it that way you would want to use a subgroup of $\mathrm{Aut}(F/\mathbb{Q})$ which is of &lt;em&gt;index&lt;/em&gt; $2^{n-1}$ (rather than of &lt;em&gt;order&lt;/em&gt; $2^{n-1}$), and which is normal (so that the corresponding extension $K$ is normal over $\mathbb{Q}$). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can approach it going &quot;the other way&quot;, with a subgroup of order $2^{n-1}$ (which is &lt;em&gt;necessarily&lt;/em&gt; normal, being of index $2$), as youo ask. How would an inductive argument look in that case? The subgroup $H$ of order $2^{n-1}$ gives you an intermediate field $K$, $\mathbb{Q}\subseteq K\subseteq F$, with $[F:K] = 2^{n-1}$, $[K:\mathbb{Q}]=2$. So you would know, from the case $n=1$, that everything in $K$ is constructible, and you would want to argue inductively that everything in $F$ is constructible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So here, your induction hypothesis should be somewhat different: the induction hypothesis I quote above would &lt;em&gt;not&lt;/em&gt; let you conclude that everything in $F$ is constructible from the fact that everything in $K$ is constructible, because $K$ is not $\mathbb{Q}$ so the induction hypothesis would not apply.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You need a different induction hypothesis: one that doesn't &quot;care&quot; what the base field is, as long as everything in it is constructible. So your induction hypothesis should look like:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If $K$ is an extension of $\mathbb{Q}$ in which all elements are constructible, and $L$ is a field extension of $K$ with $[L:K]=2$, then every element of $L$ is constructible.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Then&lt;/strong&gt; you could use that induction hypothesis to conclude everything in $F$ is constructible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But this introduces a new problem: in order to prove the $n=1$ of the induction hypothesis I proposed first, you would need to prove that if $K$ is a field with $[K:\mathbb{Q}]=2$, then everything in $K$ is constructible; this is not too hard (I hope). But in order to use the &lt;em&gt;second&lt;/em&gt; induction hypothesis, the case $n=1$ needs to be more general: now you need to show that if $K$ is &lt;em&gt;any&lt;/em&gt; extension of $\mathbb{Q}$ in which every element is constructible, and $L$ is a field extension of $K$ with $[L:K]=2$, then every element of $L$ is constructible. Because the second induction hypothesis assumes more than the first, you need to prove more in the base case if you want to use it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is exactly what Chris pointed out: his &quot;first&quot; is the proof of the $n=1$ case of the second proposed induction hypothesis above; his &quot;then&quot; is the inductive step. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Of course, you are just trading when and where you will prove that if everything in $K$ is constructible and $[L:K]=2$, then everything in $L$ is constructible; in the first argument, you need to do that to finish off the inductive step; in the second argument, you need it to get started. The one advantage of the method you propose is that you only need to know that a group of order $2^n$ has a subgroup of order $2^{n-1}$; with the first method, you need to know it has a normal subgroup of order $2$. This is not very hard, though, one just shows the center is nontrivial.) &lt;/p&gt;&#xA;&#xA;&lt;hr/&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added after edit to the question.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think you are complicating your life again by trying to attack the entire problem in one fell swoop somehow, rather than just taking it one chunk at a time. Your original idea of trying to use induction somehow has a &lt;em&gt;lot&lt;/em&gt; of merit, and means that you don't really need to invoke theorems on solvability and normal series, just the fact that a group of order $2^{n+1}$ must have a normal/central subgroup of order $2$ (or if you want to go by your original attempt, just using that a group of order $2^{n+1}$ must have a subgroup of order $2^n$, see the final comments after the next horizontal rule). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The key is indeed showing that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Result we hope to prove.&lt;/strong&gt; &lt;em&gt;If&lt;/em&gt; $[F:K]=2$ (and $F/K$ is a Galois extension; but this is immediate, because &lt;em&gt;any&lt;/em&gt; extension of degree $2$ in a field of characteristic different from $2$ is always a Galois extension), &lt;em&gt;and all elements of&lt;/em&gt; $K$ &lt;em&gt;are constructible, then all elements of&lt;/em&gt; $F$ &lt;em&gt;are constructible.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume for a moment that you have already managed to prove this. How will the induction argument go? If $[F:\mathbb{Q}]=2^1$ is a Galois extension (that is, $n=1$) then the result will follow because every element of $\mathbb{Q}$ is certainly constructible, so by the result we are assuming you have proven, every element in $F$ is constructible. Done.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, assume inductively that for any finite Galois extension $K$ of the rationals, if $[K:\mathbb{Q}]=2^k$, then every element of $K$ is constructible. We want to show that if $F$ is an extension of $\mathbb{Q}$ with $[F:\mathbb{Q}]=2^{k+1}$, then every element of $F$ is constructible. If we can prove this, then this will prove the result you want by induction on $n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, say $F$ is a Galois extension of $\mathbb{Q}$ with $[F:\mathbb{Q}]=2^{n+1}$. Then we know that $\mathrm{Gal}(F/\mathbb{Q})$ is a group with $2^{n+1}$ elements, and thus has nontrivial center; the center is abelian of order $2^r$ for some $r\geq 1$, so there is a central subgroup of order $2$, call it $N$. Then $N\triangleleft G$, $[G:N]=2^k$, so the fixed field of $N$, call it $K$, satisfies $\mathbb{Q}\subseteq K\subseteq F$, $[K:\mathbb{Q}]=[G:N]=2^k$, $[F:K]=2$, and $K$ is Galois over $\mathbb{Q}$ because $N$ is normal in $G$. By the induction hypothesis, every element of $K$ is constructible. Now, we have $[F:K]=2$, and every element of $K$ is constructible, so by the Result-we-hope-to-prove, we conclude that every element of $F$ is constructible, and we are done. &lt;strong&gt;QED&lt;/strong&gt;, RIP, $\Box$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, how about that &quot;Result-we-hope-to-prove&quot;? Well, suppose that, as we've suggested, you manage to prove that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Lemma-we-hope-to-prove.&lt;/strong&gt; If $K\subseteq F\subseteq\mathbb{C}$ are field extensions, and $[F:K]=2$, then there exists $\xi\in F$ with $\xi^2\in K$ such that $F=K(\xi)=K[\xi]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then: assuming every element of $K$ is constructible, then notice that every element of $F$ can be written (uniquely) as $a+b\xi$ with $a,b\in K$. Now, $a$ and $b$ are both constructible, so if $\xi$ is constructible, then $a+b\xi$ is constructible (product of constructible numbers is constructible, sums of constructible numbers are constructible), so every element of $F$ is constructible. So, assuming the Lemma, it will all fall to showing $\xi$ is constructible. Since $\xi^2 = k\in K$, and $k$ is constructible, then &lt;code&gt;&amp;lt;insert valid argument here&amp;gt;&lt;/code&gt;. This proves the Result-we-hope-to-prove, modulo proving the Lemma-we-hope-to-prove.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So now we are down to proving the Lemma-we-hope-to-prove.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $[F:K]=2$, if you take $\alpha\in F$, $\alpha\notin K$ (it must exist), then $\{1,\alpha\}$ is a basis for $F$ over $K$ (linearly independent because $\alpha\notin K$, and of the correct cardinality to be a basis). Since $\{1,\alpha,\alpha^2\}$ is linearly dependent (too many vectors), but $\{1,\alpha\}$ is linearly independent, then $\alpha^2$ is a linear combination of $1$ and $\alpha$. So we can find $b,c\in K$ such that $\alpha^2+b\alpha+c = 0$. That is, the minimal polynomial of $\alpha$ over $K$ is $f(x) = x^2+bx+c$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But we know &lt;em&gt;exactly&lt;/em&gt; what the roots of $f(x)$ are: they are&#xA;$$\frac{-b+\sqrt{b^2-4c}}{2}\quad\text{and}\quad\frac{-b-\sqrt{b^2-4c}}{2}.$$&#xA;So $\alpha$ must be one of them. Letting $r$ be the first root, note that the second root is $-r-b$, so replacing $\alpha$ by $-\alpha-b$ if necessary (remember that $b$ is in $K$), we may assume that $\alpha = \frac{-b+\sqrt{b^2-4c}}{2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now... what about that $\sqrt{b^2-4c}$ ? Is it in $F$? Is it in $K$? What is $K\left(\sqrt{b^2-4c}\right)$, anyway?&lt;/p&gt;&#xA;&#xA;&lt;hr/&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Note.&lt;/strong&gt; If you change what you are trying to prove to the apparently more general:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Let $F\subseteq\mathbb{C}$ be a Galois extension of $K$ such that $[F:K]=2^n$. If all elements of $K$ are constructible then all elements of $F$ are constructible.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Then you get a result which implies the one you want, by taking $K=\mathbb{Q}$ (since certainly every element of $\mathbb{Q}$ is constructible). In this approach, the base of the induction would be the Result-we-hope-to-prove, and the induction hypothesis would be:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;em&gt;Induction hypothesis.&lt;/em&gt; If $M\subseteq\mathbb{C}$ is a Galois extension of $L$, $[M:L]=2^k$, and every element of $L$ is constructible, then every element of $M$ is constructible.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Then instead of using a central subgroup $N$ of $\mathrm{Aut}(F/K)$ of order $2$, you can take the subgroup $H$ of order $2^k$ that you know exists inside $\mathrm{Gal}(F/K)$. Letting $L$ be the fixed field of $H$, you have that $[L:K]=2$, $[F:L]=2^k$, and $F$ is Galois over $L$. By the Result-we-hope-to-prove you know every element of $L$ is constructible (since $H\triangleleft G$, so $L$ is Galois over $K$), and then by the induction hypothesis applied to $F/L$ you know that every element of $F$ is constructible. It all still comes down, though, to the Result-we-hope-to-prove. &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-15T20:46:03.707" LastActivityDate="2011-02-15T20:46:03.707" CommentCount="12" />
  <row Id="22093" PostTypeId="1" AcceptedAnswerId="22966" CreationDate="2011-02-15T03:59:54.860" Score="4" ViewCount="216" Body="&lt;p&gt;How to solve this limit?&#xA;$$\lim_{x \to \infty}{\sqrt[5]{x^5 - 3x^4 + 17} - x}$$&lt;/p&gt;&#xA;" OwnerUserId="880" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2011-02-20T19:08:42.043" LastActivityDate="2011-02-20T19:11:21.227" Title="Limit of algebraic function $\ \lim_{x\to\infty} \sqrt[5]{x^5 - 3x^4 + 17} - x$" Tags="&lt;limit&gt;" AnswerCount="4" CommentCount="5" />
  <row Id="22094" PostTypeId="1" AcceptedAnswerId="22099" CreationDate="2011-02-15T04:01:03.807" Score="0" ViewCount="51" Body="&lt;p&gt;Consider the plane defined by $2x+3y+5z=30$. Is the line parametrized by $l(s):=(-1,0,2)+s(-4,-12,18)$ for all s orthogonal to the plane?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I know this is easy, but I'm missing one key piece. If the normal vector of the plane cross the line is equal to zero, then this confirms the line is indeed orthogonal. But now how would I cross the normal vector of the plane with this line?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And the question also asks to find a parametrization of the plane. How would I go about doing this? There are no similar examples in the textbook..&lt;/p&gt;&#xA;" OwnerUserId="2985" LastEditorUserId="2985" LastEditorDisplayName="" LastEditDate="2011-02-15T04:06:38.777" LastActivityDate="2011-02-15T04:13:13.190" Title="Determining if a line is orthogonal to a plane" Tags="&lt;calculus&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="22095" PostTypeId="1" AcceptedAnswerId="22344" CreationDate="2011-02-15T04:03:33.450" Score="3" ViewCount="77" Body="&lt;p&gt;I'm looking for a module $E$ of a commutative ring $A$ with two submodules $E_1$ and $E_2$ such that the associated primes of $E_1 + E_2$ strictly contain the union of the associated primes of $E_1$ and $E_2$. There's a short exact sequence&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;0 \to E_1 \to E_1 + E_2 \to (E_1 + E_2)/E_1 \to 0&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where the last term is isomorphic to $E_2/(E_1 \cap E_2)$. So if they have trivial intersection the sequence splits and we have equality $\textrm{Ass}(E_1) \cup \textrm{Ass}(E_2) = \textrm{Ass}(E_1 + E_2)$. But I suspect that if the intersection is nontrivial there's a counterexample to this, I just can't think of what it would be. It won't work for something simple like $\mathbb{Z}$-modules, and I don't think it will work in general for integral domains. Or you might even need a non-Noetherian ring. I'm not sure.&lt;/p&gt;&#xA;" OwnerUserId="6723" LastActivityDate="2011-02-16T12:33:48.747" Title="Find two submodules $E_1$ and $E_2$ such that $\textrm{Ass}(E_1) \cup \textrm{Ass}(E_2) \subsetneq \textrm{Ass}(E_1 + E_2)$" Tags="&lt;commutative-algebra&gt;&lt;modules&gt;" AnswerCount="1" FavoriteCount="0" />
  <row Id="22096" PostTypeId="1" AcceptedAnswerId="22097" CreationDate="2011-02-15T04:05:23.020" Score="5" ViewCount="98" Body="&lt;p&gt;For some reason I can't seem to see why&lt;br&gt;&#xA;$$\frac{az+b}{cz+d} = \frac{bc-ad}{c^2\left(z+\frac{d}{c}\right)} + \frac{a}{c}.$$&#xA;Its been so many years since I've taken Algebra I, so when this was stated in class, without even the motions of manipulation, I felt lost. Sorry if this question is too dumb to belong on the site. &lt;/p&gt;&#xA;" OwnerUserId="6758" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-15T04:07:50.933" LastActivityDate="2011-02-15T04:52:13.950" Title="Simple Algebraic Manipulation I am Stuck On?" Tags="&lt;algebra-precalculus&gt;" AnswerCount="2" />
  <row Id="22097" PostTypeId="2" ParentId="22096" CreationDate="2011-02-15T04:10:35.173" Score="9" ViewCount="" Body="&lt;p&gt;Working on the right hand side, first multiply through by one $c$ to get rid of the compound fraction; then get the common denominator $c(zc+d)$, and do the addition:&#xA;\begin{align*}&#xA;\frac{bc-ad}{c^2\left(z+\frac{d}{c}\right)}+\frac{a}{c} &amp;amp;= \frac{bc-ad}{c(zc+d)} + \frac{a}{c}\\&#xA;&amp;amp;= \frac{bc-ad}{c(zc+d)} + \frac{a(zc+d)}{c(zc+d)}\\&#xA;&amp;amp;= \frac{bc-ad+a(zc+d)}{c(zc+d)}\\&#xA;&amp;amp;= \frac{bc-ad + azc + ad}{c(zc+d)}\\&#xA;&amp;amp;= \frac{bc+azc}{c(zc+d)}\\&#xA;&amp;amp;= \frac{c(b+az)}{c(zc+d)}\\&#xA;&amp;amp;= \frac{b+az}{zc+d} = \frac{az+b}{cz+d}.&#xA;\end{align*}&#xA;Of course, this doesn't really address how one can go from one to the other &quot;mentally&quot;, or how one can &quot;spot&quot; the equality. You are probably thinking, &quot;And how on Earth would anybody ever come up with &lt;em&gt;that&lt;/em&gt; on the fly?!&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a piece of information you may not have that is being used &quot;in the background&quot;: you are dealing with a &lt;a href=&quot;http://en.wikipedia.org/wiki/Moebius_transformation&quot; rel=&quot;nofollow&quot;&gt;Moebius transformation&lt;/a&gt;, a function from $\mathbb{C}$ to $\mathbb{C}$ that is given by&#xA;$$z\longmapsto \frac{az+b}{cz+d}\qquad\mbox{with $ad-bc\neq 0$.}$$&#xA;Any Moebius transformation can be decomposed as a composition of a translation, an inversion, a dilation, and a translation in a standard way (see &lt;a href=&quot;http://en.wikipedia.org/wiki/Moebius_transformation#Decomposition_and_elementary_properties&quot; rel=&quot;nofollow&quot;&gt;the Decomposition section in the Wikipedia page&lt;/a&gt;). First you take $z$ and you translate it by $\frac{d}{c}$, which gives you the expression $z+\frac{d}{c}$. Then you invert, giving you $$\frac{1}{z+\frac{d}{c}}.$$ Then you dilate by multiplying by $\frac{bc-ad}{c^2}$, to get&#xA;$$\frac{bc-ad}{c^2\left(z+\frac{d}{c}\right)}$$&#xA;and finally you translate by $\frac{a}{c}$ by adding it to the running total; this is exactly the expression you have. If you are accustomed to playing with Moebius transformations, this is a very common decomposition, so it is an identity that you would have &quot;at your fingertips&quot;, as it were. The reason for this particular decomposition is that the pieces (translations, inversions, and dilations) are very easy to understand, so they make particularly good pieces to break the Moebius transformation into.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-15T04:31:25.460" LastActivityDate="2011-02-15T04:31:25.460" CommentCount="2" />
  <row Id="22099" PostTypeId="2" ParentId="22094" CreationDate="2011-02-15T04:13:13.190" Score="4" ViewCount="" Body="&lt;p&gt;As you said in your comment, the normal vector of the plane is $\langle 2, 3, 5\rangle$; the line contains the point $(-1,0,2)$ and is in the direction $\langle -4,-12,18\rangle$.  You want to see if the cross-product of the normal vector to the plane and the direction vector of the line is zero.&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2011-02-15T04:13:13.190" CommentCount="4" />
  <row Id="22100" PostTypeId="2" ParentId="22087" CreationDate="2011-02-15T04:21:03.960" Score="1" ViewCount="" Body="&lt;p&gt;As noted in other answers, you have a +/- transcription error in the numerator of the first term of your answer—it should be:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xD;&#xA;\frac{3x^{2}-12x}{x^{2}}-\frac{2x^{3}-12x{^2}+12}{x^{3}}&#xD;&#xA;&amp;amp;=\frac{(3x^{2}-12x)\cdot x}{(x^{2})\cdot x}-\frac{2x^{3}-12x{^2}+12}{x^{3}}&#xD;&#xA;\\&#xD;&#xA;&amp;amp;=\frac{3x^{3}-12x^2}{x^{3}}-\frac{2x^{3}-12x{^2}+12}{x^{3}}&#xD;&#xA;\\&#xD;&#xA;&amp;amp;=\frac{3x^{3}-12x^2-(2x^{3}-12x{^2}+12)}{x^{3}}&#xD;&#xA;\\&#xD;&#xA;&amp;amp;=\frac{3x^{3}-12x^2-2x^{3}+12x{^2}-12}{x^{3}}&#xD;&#xA;\\&#xD;&#xA;&amp;amp;=\frac{x^{3}-12}{x^{3}}&#xD;&#xA;\\&#xD;&#xA;&amp;amp;=\frac{x^{3}}{x^3}-\frac{12}{x^{3}}&#xD;&#xA;\\&#xD;&#xA;&amp;amp;=1-\frac{12}{x^{3}}&#xD;&#xA;\end{align}$$&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2011-02-15T04:21:03.960" CommentCount="1" />
  <row Id="22101" PostTypeId="1" AcceptedAnswerId="22156" CreationDate="2011-02-15T04:33:41.477" Score="2" ViewCount="70" Body="&lt;p&gt;The following has come up during an undergraduate research project. It's rather elementary, but we are having trouble finding anywhere that mentions it. It might have something to do with rigidity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider two tetrahedrons that share a base.&lt;img src=&quot;http://i.imgur.com/eZ01H.png&quot; alt=&quot;enter image description here&quot;&gt; What is the relationship between the lengths of its sides and the diagonal? We suspect there should be some polynomial relationship between the lengths (that's what happens in the plane case). By introducing Cartesian coordinates and assuming the triangle is ABC is in the xy-plane, A is the origin, and AB is on the x-axis, I was able to compute the coordinates of C using sines and cosines. I'm assuming I know the lengths of all sides except for the orange diagonal in the picture. I can then look at the intersection of the three spheres centered at A,B and C with radii |AD|, |BD|, and |CD| to compute |D|. I can do the same thing for E, and then I can use the distance formula to solve for |DE|. I can replace my sine in compute C with $\sqrt{1-\cos^2(\theta)}$, and then use the law of cosines to write cosines in terms of the lengths of the triangle. In this way I get a relationship just between the lengths of the tetrahedron. I'm not very interested in computing |DE| from the lengths of the other sides, we're really just after a relationship between the sides that is an invariant for all &quot;double tetrahedrons&quot; (if anyone knows what this shape is called, please let me know). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Computing the relationship in the manner I did above worked, with the help of Maple, the the expression was had lots of square roots in it. The problem is that the y coordinate of C contained a square root ($\sqrt{1-cos^2(\theta)}$) where $\theta$ is the ABC angle and $\cos(\theta)$ turns out to be $\frac{AB^2+BC^2-AC^2}{2AB\cdot BC}$. My professor suggested to take my original relation, solve to the y coordinate of C, and square it and hope this leaves up with a polynomial. Maple, however, was unable to solve for this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anybody have any idea on how to approach this problem, or any reference to similar problems? In the plane case - with polygons and diagonals - we were able to work out the relationships between the lengths for pentagons and were able to give results for $n$-gons. The relationships turn out to be cubic polynomials with the square of the lengths as variables). This stuff is very elementary, but we cannot find anyplace where it has been done. I've ran out of time right now, but I will return later tonight to try and make the problem more clear or go through more details. Thanks in advance for any insights or comments. &lt;/p&gt;&#xA;" OwnerUserId="416" LastActivityDate="2011-02-15T13:31:54.960" Title="What is the relationship between the lengths of segments connecting 5 points in space?" Tags="&lt;geometry&gt;&lt;reference-request&gt;" AnswerCount="1" />
  <row Id="22102" PostTypeId="1" CreationDate="2011-02-15T04:38:55.047" Score="1" ViewCount="56" Body="&lt;p&gt;Given a subset of the hyperbolic upper half plane, say an ideal triangle (so with vertices on the boundary), what is the cardinality of all points contained in the interior?&lt;/p&gt;&#xA;" OwnerUserId="7485" LastActivityDate="2011-02-15T05:04:19.473" Title="What is the cardinality of a subset of the hyperbolic upper half plane?" Tags="&lt;set-theory&gt;&lt;hyperbolic-geometry&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="22103" PostTypeId="2" ParentId="21957" CreationDate="2011-02-15T04:41:24.177" Score="0" ViewCount="" Body="&lt;p&gt;I asked my ODE professor and he said that $dy/dx$ is the function whose value at $t$ gives the slope of the graph of $(x,y)$ at the point $(x(t), y(t))$. This means that $dy/dx$ is the function defined by&#xA;\[&#xA;\frac{dy}{dx}(t) = \lim_{s \to t} \frac{y(t+s) - y(t)}{x(t+s) - x(t)}&#xA;\]&#xA;and this is why $(dy/dx)(t) = y&amp;#39;(t)/x&amp;#39;(t) = g(x(t),y(t))/f(x(t),y(t))$.&lt;/p&gt;&#xA;" OwnerUserId="1427" LastActivityDate="2011-02-15T04:41:24.177" />
  <row Id="22104" PostTypeId="2" ParentId="22093" CreationDate="2011-02-15T04:43:59.057" Score="4" ViewCount="" Body="&lt;p&gt;Note that &#xA;$$&#xD;&#xA;x^5  - 3x^4  + 17 = \bigg(x - \frac{3}{5}\bigg)^5  + O(x^3 ),&#xD;&#xA;$$&#xA;and consider the mean value theorem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Specifically, show that&#xA;$$&#xD;&#xA;\sqrt[5]{{\bigg(x - \frac{3}{5}\bigg)^5  + O(x^3 )}} - \sqrt[5]{{\bigg(x - \frac{3}{5}\bigg)^5 }} \to 0.&#xD;&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="2810" LastEditorUserId="2810" LastEditorDisplayName="" LastEditDate="2011-02-15T05:14:24.040" LastActivityDate="2011-02-15T05:14:24.040" />
  <row Id="22105" PostTypeId="2" ParentId="22093" CreationDate="2011-02-15T04:49:46.380" Score="5" ViewCount="" Body="&lt;p&gt;I suggest writing $(x^5 - 3x^4 + 17)^{1 \over 5}$ as $x(1 - {3 \over x} + {17 \over x^5})^{1 \over 5}$ and then using that $(1 - \epsilon)^{1 \over n} =  1 - {\epsilon \over n} + O(\epsilon^2)$.&lt;/p&gt;&#xA;" OwnerUserId="3035" LastActivityDate="2011-02-15T04:49:46.380" />
  <row Id="22106" PostTypeId="1" CreationDate="2011-02-15T04:51:51.190" Score="4" ViewCount="181" Body="&lt;p&gt;Is there a central wiki or something where open questions (and relevant research on them) takes place?&lt;/p&gt;&#xA;" OwnerUserId="7059" LastEditorUserId="7059" LastEditorDisplayName="" LastEditDate="2011-02-15T04:57:35.380" LastActivityDate="2011-02-15T08:44:33.193" Title="Best place to find open questions / latest research" Tags="&lt;math-history&gt;&lt;open-problem&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="2" />
  <row Id="22107" PostTypeId="2" ParentId="22096" CreationDate="2011-02-15T04:52:13.950" Score="1" ViewCount="" Body="&lt;p&gt;Multiplying through by the common denominator $\rm\ c\ (c\:z + d)\ $ reduces the &quot;fractional&quot; identity to the equivalent &quot;integral&quot; identity $\rm\ c\ (a\:z+b)\ =\ bc-ad + a\ (c\:z + d)\:,\ $ which is easily verified.&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2011-02-15T04:52:13.950" />
  <row Id="22108" PostTypeId="1" AcceptedAnswerId="22127" CreationDate="2011-02-15T04:56:54.023" Score="1" ViewCount="62" Body="&lt;p&gt;Suppose you have a set of twelve distinct objects, and you were interested in ways of arranging them into a 3 by 4 rectangle. You wonder: does it matter if I arrange them in 3 bunches of 4 or 4 bunches of 3?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Label them by primes. (I could, and probably should write $a_{2}$ in the rest of this post, but I don't feel like messing around with tables that much.) The numbers after the equal signs are row and column products.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&#xA; 2   3   5   7&#xA;11  13  17  19&#xA;23  29  31  37&#xA;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Let's switch the 17 and the 19, again:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&#xA; 2             3         5       7         = 210&#xA;11            13        17      19         = 46189&#xA;23            29        31      37         = 1028859&#xA;&#xA;=              =         =       =&#xA;506         1131      2635    4921&#xA;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;If we switch the 17 and the 19:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&#xA; 2             3         5      7            = 210&#xA;11            13        19     17            = 46189&#xA;23            29        31     37            = 1028859&#xA;&#xA;=              =         =       =&#xA;506          1131     2945    4403&#xA;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Now let's swap the 5 and 19:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&#xA; 2             3        19        7          = 798&#xA;11            13         5       17          = 12155&#xA;23            29        31       37          = 1028859&#xA;&#xA;=              =         =        =&#xA;506         1131      2945     4403&#xA;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;What's interesting about all of these is that each of these&#xA;arrangements (factorizations if we're talking about numbers) has different uniquenesses about the rows and the columns&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Look at the last two tables:&#xA;In particular, the columns are pairwise the same, but the rows are pairwise different,&#xA;so in this case it does make a difference whether we're talking about four bunches of&#xA;three or three bunches of four, because there's a way of arranging it such that the four bunches&#xA;of three are the same, but the three bunches of four are different. Is there an explanation of this &quot;noncommutativity&quot; in terms of subgroups of the symmetric group?&lt;/p&gt;&#xA;" OwnerUserId="890" LastEditorUserId="890" LastEditorDisplayName="" LastEditDate="2011-02-15T05:53:20.130" LastActivityDate="2011-02-15T06:43:25.163" Title="&quot;noncommutativity&quot; of arranging 12 distinct objects in (4 bunches of 3) or (3 bunches of 4)" Tags="&lt;combinatorics&gt;&lt;group-theory&gt;" AnswerCount="1" />
  <row Id="22109" PostTypeId="2" ParentId="22106" CreationDate="2011-02-15T04:59:15.857" Score="-2" ViewCount="" Body="&lt;p&gt;No.  Math doesn't work that way.&lt;/p&gt;&#xA;" OwnerUserId="5067" LastActivityDate="2011-02-15T04:59:15.857" CommentCount="10" />
  <row Id="22110" PostTypeId="2" ParentId="22102" CreationDate="2011-02-15T05:04:19.473" Score="1" ViewCount="" Body="&lt;p&gt;Uncountably infinite.  For example, considered as a subset of the plane, it contains the set $\{(x+t,y) | 0 \leq t \leq \epsilon\}$ for some $x,y \in \mathbb{R}$ and $\epsilon&amp;gt;0$.  Such a set can be easily put in bijection with $\mathbb{R}$.&lt;/p&gt;&#xA;" OwnerUserId="5067" LastActivityDate="2011-02-15T05:04:19.473" CommentCount="4" />
  <row Id="22111" PostTypeId="2" ParentId="22031" CreationDate="2011-02-15T05:08:37.860" Score="-1" ViewCount="" Body="&lt;p&gt;You should consider two things and two things only.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;How successful have the advisor's previous students been?  What sorts of postdocs have they gone onto and have they managed to become independent researchers?  This is by far the most important thing.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;How well do you get along with them?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Notice that neither of these has anything to do with the subject they work in.  Except for the prodigies (who don't need and won't take advice), no beginning grad students have any idea what different areas of math are like.  They think they do, but they are wrong.  If you choose someone you enjoy talking to who is regularly cranking out good students, they you will have no problem learning to love whatever they work on.&lt;/p&gt;&#xA;" OwnerUserId="5067" LastActivityDate="2011-02-15T05:08:37.860" CommunityOwnedDate="2011-02-15T05:08:37.860" CommentCount="5" />
  <row Id="22112" PostTypeId="1" AcceptedAnswerId="22180" CreationDate="2011-02-15T05:13:19.000" Score="2" ViewCount="182" Body="&lt;p&gt;What is the definition of an envelope of a function ? For example if we multiply a sinusoid function of certain frequency ($1/f$ &amp;lt; support of bump) with a bump function we get a function whose envelope looks like bump function. The points where the envelope (bump here) and the given function (the product of sine and bump) meet, their derivatives are equal. Is there any formal definition for such an envelope of any given smooth function (say with compact support) ?&lt;/p&gt;&#xA;" OwnerUserId="2987" LastActivityDate="2011-03-17T15:33:35.567" Title="What is an envelope of a function ?" Tags="&lt;real-analysis&gt;&lt;functions&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="22113" PostTypeId="2" ParentId="262" CreationDate="2011-02-15T05:15:58.593" Score="0" ViewCount="" Body="&lt;p&gt;Title: The Mathematical Experience&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Authors: Davis and Hirsh&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Short Description: A really accessible and funny introduction to the philosophy of mathematics. I think the description of the &quot;ideal mathematician&quot; is particularly hilarious.&lt;/p&gt;&#xA;" OwnerUserId="7104" LastActivityDate="2011-02-15T05:15:58.593" CommunityOwnedDate="2011-02-15T05:15:58.593" />
  <row Id="22114" PostTypeId="1" CreationDate="2011-02-15T05:22:49.397" Score="0" ViewCount="38" Body="&lt;p&gt;Hi. I wonder, what are the profinite groups $G$ that have surjective homomorphisms from $G\rightarrow\mathbb{Z}/n\mathbb{Z}$ onto the group of integers mod $n$ for any $n$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, what are the profinite groups $G$ that have surjective homomorphisms from $G\rightarrow\mathbb{Z}_p$ onto the group of $p$-adic integers for any prime $p$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you. &lt;/p&gt;&#xA;" OwnerUserId="7098" LastActivityDate="2011-02-15T05:22:49.397" Title="Existence of homomorphisms from a profinite group to Z/nZ" Tags="&lt;algebra&gt;" CommentCount="1" />
  <row Id="22115" PostTypeId="1" AcceptedAnswerId="22209" CreationDate="2011-02-15T05:29:03.823" Score="4" ViewCount="87" Body="&lt;p&gt;Suppose I show that:&#xA;$$x^{f(z)/g(z)} = y \pmod{4}$$&#xA;is impossible for some given positive integers $x$ and $y$, where,&#xA;\begin{align*}&#xA;f(z) &amp;amp;= \phi(4)  k_1(z) + 1 \\&#xA;&amp;amp;= 2 k_1(z) + 1\\&#xA;g(z) &amp;amp;= \phi(4) k_2(z) + 1 \\&#xA;&amp;amp;= 2  k_2(z) + 1&#xA;\end{align*}&#xA;and $k_1(z)$ and $k_2(z)$ are integer functions, that approach infinity, such that $f(z)/g(z)$ approaches some irrational number. Can I then say, the equation:&#xA;$$x^{f(z)/g(z)} = y$$&#xA;has no solutions integer solutions, with the same $x$ and $y$, as $z$ goes to infinity as well?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is, if I let,&#xA;$$d = \lim_{z-&amp;gt;\infty}\frac{f(z)}{g(z)}$$&#xA;be the irrational number in the limit, then would it be true that,&#xA;$$x^d \neq y$$&#xA;for the same $x$ and $y$ ?&lt;/p&gt;&#xA;" OwnerUserId="7105" LastEditorUserId="7105" LastEditorDisplayName="" LastEditDate="2011-02-15T07:31:18.350" LastActivityDate="2011-02-15T19:02:19.250" Title="Are limits on exponents in moduli possible?" Tags="&lt;number-theory&gt;&lt;limit&gt;&lt;irrational-numbers&gt;&lt;powers&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="1" />
  <row Id="22116" PostTypeId="1" AcceptedAnswerId="22118" CreationDate="2011-02-15T05:38:08.823" Score="6" ViewCount="120" Body="&lt;p&gt;This is a very basic question, and its funny that I'm able to solve more advanced problems like this, but I was presented with a basic one and got stumped. I have the equation &#xA;$$y=-\frac{3}{4}x+6.$$ &#xA;In $\mathbb{R}^2$, this is a line. I want to find a parametrization of this line. I guess the problem here is I never really understood the concept of parametrization. I'm just a robot-I follow the steps the book tells me, but I really dont understand the intuition behind it. What is the point of parametrization (in laymen's terms, which is hard to find anywhere), and how would I do it for this equation?&lt;/p&gt;&#xA;" OwnerUserId="2985" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-15T05:44:00.830" LastActivityDate="2011-02-15T06:20:08.117" Title="Parametrization of a line" Tags="&lt;calculus&gt;&lt;parametric&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="1" />
  <row Id="22117" PostTypeId="1" AcceptedAnswerId="22166" CreationDate="2011-02-15T05:38:49.167" Score="1" ViewCount="56" Body="&lt;p&gt;I've been reading about Hilbert's Axioms of geometry, and I have two models on $\mathbb{R}^2$ which satisfy the usual axioms of incidence, betweenness, and congruence. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first model has a distance function given by &#xA;$$&#xD;&#xA;d(A,B)=|a_1-b_1|+|a_2-b_2|&#xD;&#xA;$$&#xA;where $A=(a_1,a_2)$, $B=(b_1,b_2)$, and two segments $AB\cong CD$ if $d(A,B)=d(C,D)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second model has a distance function given by&#xA;$$&#xD;&#xA;d(A,B)=\sup\{|a_1-b_1|,|a_2-b_2|\},&#xD;&#xA;$$&#xA;and again $AB\cong CD$ if $d(A,B)=d(C,D)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The author claims that these two models are isomorphic in that there is a bijection between the sets of points in $\mathbb{R}^2$ that preserves lines, betweenness and congruence. I've been toying with it for a while, but it seems to me that the notions of congruence in particular are too different. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, in the first model, if $A=(0,0)$ and $B=(2,1)$, and $C=(2,0)$ and $D=(5/2,5/2)$, then $AB\cong CD$ since both segments have length $3$. To preserve congruence, I suppose two segments are congruent if the maximum of the displacement along the $x$-axis or $y$-axis by traveling along one segment is equal to the maximum of the displacement along either axis when traveling along the other segment. In this sense, how could you shape the segment $CD$ so that the maximum of the displacements along either axis is the same as the maximum of the displacement for whatever the image of $AB$ is?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the first model, $AB$ is congruent to any segment which is the hypotenuse of triangle with legs that sum to $3$, and this sum could split into any two terms. I just don't see a function that could handle all these possibilities at once. Does anyone know of a particular isomorphism between these two? Thanks.&lt;/p&gt;&#xA;" OwnerUserId="1583" LastActivityDate="2011-02-15T13:02:32.220" Title="Isomorphism between two models of Hilbert's Axioms with different notions of congruence" Tags="&lt;geometry&gt;" AnswerCount="1" />
  <row Id="22118" PostTypeId="2" ParentId="22116" CreationDate="2011-02-15T05:49:08.077" Score="6" ViewCount="" Body="&lt;p&gt;Think of a parametrization as describing the &quot;trace&quot; of the curve, with $t$ representing time. You want to write equations&#xA;\begin{align*}&#xA;x &amp;amp;= f(t),\\&#xA;y &amp;amp;= g(t)&#xA;\end{align*}&#xA;that describe someone tracing the line as $t$ varies. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How would you trace the graph of a function $y=g(x)$? Well, the points on the graph are all of the form $(x,g(x))$, so the simplest is to use a parametrization like&#xA;\begin{align*}&#xA;x &amp;amp;= t,\\&#xA;y &amp;amp;= g(t).&#xA;\end{align*}&#xA;If you think about what the points $(x(t),y(t))$ look like as $t$ ranges from $a$ to $b$, you'll see that you are giving the graph of $y=g(x)$ from $x=a$ to $x=b$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That gives you one way to give the line.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another is to pick a point on the line, and just say &quot;go along this direction, backwards or forwards&quot;. A direction is given by a vector $(u,v)$. So if your line goes through the point $(a,b)$, and has direction $(u,v)$, then one possible parametrization just says: &quot;start at $x=a$ and $y=b$, and then move $x$ and $y$ in the direction $(u,v)$: if you move in the $x$ direction by $ku$, then you need to move in the $y$ direction by $kv$.&quot; That is,&#xA;\begin{align*}&#xA;x &amp;amp;= a + tu\\&#xA;y &amp;amp;= b + tv.&#xA;\end{align*}&#xA;You can do that with the graph you have above by determining a point and a direction. For the direction: the slope of the line with direction $(u,v)$ is $\frac{v}{u}$, because $v$ is the rise and $u$ is the run.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2011-02-15T05:49:08.077" CommentCount="2" />
  <row Id="22119" PostTypeId="2" ParentId="22116" CreationDate="2011-02-15T05:49:54.330" Score="1" ViewCount="" Body="&lt;p&gt;There are an infinite amount of ways to parametrize this line.  A good way here would be to let say&lt;br&gt;&#xA;$$\displaystyle x=t$$&lt;br&gt;&#xA;replacing x with t in the first equation, you know that&lt;br&gt;&#xA;$$\displaystyle y=-\frac{3}{4}t+6$$&lt;br&gt;&#xA;And there you are!&lt;br&gt;&#xA;When you want to parametrize an equation, start by letting one of your variables equal to t, and then solve for the other variable(s), replacing t in the equation.&lt;br&gt;&#xA;Note the great explanation from Arturo Magidin, you want to think of the position of a particle at some time $t$.&lt;/p&gt;&#xA;" OwnerUserId="2533" LastEditorUserId="2533" LastEditorDisplayName="" LastEditDate="2011-02-15T05:57:54.120" LastActivityDate="2011-02-15T05:57:54.120" />
  <row Id="22120" PostTypeId="2" ParentId="22116" CreationDate="2011-02-15T05:58:00.147" Score="3" ViewCount="" Body="&lt;p&gt;I may be misunderstanding your question, but you have a parametrization given by your equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ x\mapsto (x,-\frac 34 x+6) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or if you want something with integer coefficients:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ t\mapsto (4t, 6-3t).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The point of paramterization is that on one hand you reduce the number of variables you;re working with (in this case from two: $x,y$ to one $t$), but more importantly you make an &lt;em&gt;implicit&lt;/em&gt; situation, that is, one defined by equations into an &lt;em&gt;explicit&lt;/em&gt; one, that is, a way to generate the solutions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case the point is perhaps not obvious because the equation is so simple, but imagine that if you have a very difficult equation then how much easier it is to work with the solutions when you are given a way to generate them (that is, by a parametrization) than when they are just given as the solutions of the equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps a better example to work with is the solutions to the equation &#xA;$$ x^2+y^2=1.$$&#xA;A possible parametrization is $$t\mapsto (\cos t, \sin t).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sad truth about parametrizations is that they actually rarely exist. I guess you could interpret this as saying that &lt;em&gt;if&lt;/em&gt; you can find a parametrization, then you should be happy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Addendum&lt;/strong&gt;&#xA;To answer the question in the comments: Imagine that you have to solve a system of equations in two variables. In other words you have two curves given by their equation on the plane and you have to find their intersection points. If you can parametrize one of them, then you can plug the parametrization into the other and solve an equation of a single variable. Here is an example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Say you need to find the solution to the equation system&#xA;$$ x^2+y^2=0 \qquad 2x^3y-2xy^3=1.$$&#xA;(this is a rather random choice)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you just go head-to-the-wall approach, you might solve the first as a quadratic equation and plug that into the second an end up with a degree six equation. Good luck with that. (Actually it might be solvable in this case, I have no idea, but in general a degree six equation cannot be solved with a formula).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, if you use the parametrization of the first equation given above, you end up with the equation&#xA;$$ 2\cos^3t\sin t-2\cos t\sin^3 t=1.$$&#xA;This can easily be reduced to&#xA;$$\sin 4t =\frac 12$$&#xA;which can be easily solved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The point is that parametrization helped you solve a system of equations that would normally be very difficult or impossible to solve explicitly.&lt;/p&gt;&#xA;" OwnerUserId="4654" LastEditorUserId="4654" LastEditorDisplayName="" LastEditDate="2011-02-15T06:20:08.117" LastActivityDate="2011-02-15T06:20:08.117" CommentCount="3" />
  <row Id="22121" PostTypeId="1" AcceptedAnswerId="22124" CreationDate="2011-02-15T06:09:50.687" Score="4" ViewCount="384" Body="&lt;p&gt;I can see that this works for any integer n, but I cant figure out why this works, or why the number 42 has this property&lt;/p&gt;&#xA;" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-15T06:13:44.380" LastActivityDate="2011-02-15T16:35:29.287" Title="How can I prove that $n^7 - n$ is divisible by 42 for any integer n?" Tags="&lt;elementary-number-theory&gt;" AnswerCount="5" CommentCount="7" FavoriteCount="2" />
  <row Id="22122" PostTypeId="1" AcceptedAnswerId="22126" CreationDate="2011-02-15T06:23:14.510" Score="2" ViewCount="36" Body="&lt;p&gt;Suppose I have a discrete sequence (or would it be a set?) of numbers labeled by two integer indices, $\{a_{ik}|i,k\in 1,\ldots,n\}$. Also, I know that these numbers satisfy the properties&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$a_{ij} + a_{jk} = a_{ik}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$a_{ik} = -a_{ki}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are these conditions sufficient for me to conclude that there exists a sequence $\{\lambda_b|b\in 1,\ldots,n\}$ which satisfies&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$a_{ik} = \lambda_i - \lambda_k$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for all $i,k$? If so, are both conditions necessary? If not, what other conditions would I require?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I really have no idea what tags to put on this, so I invite someone to retag it appropriately and remove this sentence.&lt;/p&gt;&#xA;" OwnerUserId="1190" LastActivityDate="2011-02-15T06:37:03.563" Title="Necessary/sufficient conditions to express numbers as differences of other numbers" Tags="&lt;sequences-and-series&gt;" AnswerCount="1" />
  <row Id="22123" PostTypeId="2" ParentId="22121" CreationDate="2011-02-15T06:29:50.443" Score="4" ViewCount="" Body="&lt;p&gt;There are many equivalent ways of proving it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First observe that $42$ divides a number iff $2,3$ and $7$ divides the number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Since $42 = 2 \times 3 \times 7$ and $\gcd(2,3) = \gcd(3,7) = \gcd(2,7) = 1$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Divisibility by $2$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clearly, $2|(n^7-n)$ since $n^7$ and $n$ are of the same parity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Equivalently you could argue out that $2|(n^2-n)$ directly from Fermat's little Theorem. (This is an overkill of Fermat's Little Theorem.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Divisibility by $3$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$n^7-n = n(n^6-1) = n(n^2-1)(n^4+n^2+1)=n(n+1)(n-1)(n^4+n^2+1)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$3|n$ or $3|(n-1)$ or $3|(n+1)$ and hence $3|(n^7-n)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Equivalently you could argue out that $3|(n^3-n)$ directly from Fermat's little Theorem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Divisibility by $7$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that $n$ can be either $7k$ or $7k \pm 1$ or $7k \pm 2$ or $7k \pm 3$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $n=7k$ or $n=7k \pm 1$, we are again done since then $7|n$ or $7|(n+1)$ or $7|(n-1)$ and hence $7|(n^7-n)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $n=7k \pm 2$, then $n^2 = (7k \pm 2)^2 = 7m + 4$ and $n^4 = (7m+4)^2 = 7l+2$.&#xA;Hence $n^4 + n^2 + 1 = 7l+2 + 7m + 4 + 1 = 7(l+m+1)$ and hence $7|(n^4 + n^2 + 1) \Rightarrow 7|(n^7-n)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $n=7k \pm 3$, then $n^2 = (7k \pm 3)^2 = 7m + 2$ and $n^4 = (7m+2)^2 = 7l+4$.&#xA;Hence $n^4 + n^2 + 1 = 7l+4 + 7m + 2 + 1 = 7(l+m+1)$ and hence $7|(n^4 + n^2 + 1) \Rightarrow 7|(n^7-n)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, $7|(n^7-n)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Equivalently you could argue out that $7|(n^7-n)$ directly from Fermat's little Theorem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, we have that $2|(n^7-n)$ and $3|(n^7-n)$ and $7|(n^7-n)$, $\forall n \in \mathbb{N}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, $42|(n^7-n)$, $\forall n \in \mathbb{N}$.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2011-02-15T06:35:43.463" LastActivityDate="2011-02-15T06:35:43.463" />
  <row Id="22124" PostTypeId="2" ParentId="22121" CreationDate="2011-02-15T06:32:02.583" Score="15" ViewCount="" Body="&lt;p&gt;Problems like this appear frequently here. There are a couple of standard approaches. One is to use &lt;a href=&quot;http://en.wikipedia.org/wiki/Fermat%27s_little_theorem&quot; rel=&quot;nofollow&quot;&gt;Fermat's little theorem&lt;/a&gt;, which says that if $p$ is a prime number, then $n^p-n$ is divisible by $p$ for all $n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $42=2\times 3\times 7$, what we need to do is to check that 2, 3, and 7 divide $n^7-n$, no matter what $n$ is. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That 7 does is direct from Fermat's little theorem. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The theorem also ensures that 3 divides $n^3-n$. Now: $$n^7-n=n(n^6-1)=n((n^2)^3-1)=n(n^2-1)(n^4+n^2+1)=(n^3-n)(n^4+n^2+1),$$ so 3 indeed divides $n^7-n$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, $n$ and $n^7$ always have the same parity, so $n^7-n$ is even.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;We can actually argue without Fermat's little theorem in this case. An approach that only requires patience is as follows: The idea is to factor the polynomial $x^7-x$ and then analyze the result when $x=n$ is an integer. (This is a trick that Bill Dubuque suggests sometimes in his solutions.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have: $x^7-x=x(x^6-1)=x(x^3+1)(x^3-1)=x(x+1)(x^2-x+1)(x-1)(x^2+x+1)$. When $x=n$, we have&#xA;$$ n^7-n=(n-1)n(n+1)(n^2-n+1)(n^2+n+1). $$&#xA;Now we analyze this prime by prime, as before. Note that one of $n$ and $n-1$ is always even, so the product is even. Also, of 3 consecutive numbers, such as $n-1,n,n+1$, one is always divisible by 3, so it only remains to verify divisibility by 7.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We may assume that $n=7k+b$ where $b=\pm2$ or $\pm3$, since otherwise $(n-1)n(n+1)$ is a multiple of 7. In that case, $n^2\equiv 4$ or $2\pmod 7$, and one of $n^2+n$, $n^2-n$ is $\equiv 6\pmod 7$, so $(n^2-n+1)(n^2+n+1)$ is a multiple of 7. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The disadvantage of this approach over the previous one, of course, is the need to analyze different cases. Fermat's little theorem allows us to analyze all cases simultaneously, which typically (as here) results in a much faster approach.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;If you are comfortable with the method of induction, this gives us a way of verifying divisibility by 7 which is not without some elegance (divisibility by 2 and 3 is probably best approached as before). Note that $(-n)^7-(-n)=-(n^7-n)$, so we may as well assume that $n\ge 0$. If $n=0$ it is obvious that 7 divides $n^7-n$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose then that $7|n^7-n$, and argue that $7|(n+1)^7-(n+1)$. For this, actually expand $(n+1)^7$ using the binomial theorem: $$ (n+1)^7=n^7+7n^6+{7\choose 2}n^5+{7\choose 3}n^4+\dots+1.$$ The point is that $${7\choose k}=\frac{7!}{k!(7-k)!}$$ is obviously divisible by 7 as long as $k\ne0,7$, so (modulo 7) we have that $(n+1)^7-(n+1)\equiv (n^7+1)-(n+1)=(n^7-n)$. Now we invoke the induction hypothesis, that precisely says that the latter is divisible by 7, and we are done. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, exactly the same inductive argument gives us a proof of Fermat's little theorem: $p|n^p-n$ for any $p$ prime and any integer $n$.&lt;/p&gt;&#xA;" OwnerUserId="462" LastEditorUserId="462" LastEditorDisplayName="" LastEditDate="2011-02-15T06:41:49.377" LastActivityDate="2011-02-15T06:41:49.377" />
  <row Id="22125" PostTypeId="2" ParentId="22121" CreationDate="2011-02-15T06:34:04.703" Score="1" ViewCount="" Body="&lt;p&gt;F$\ell$T$^*$ (if $p\not\vert n$, then $p\vert n^{p-1}-1$) $\Rightarrow$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;if $2\not\vert n$, then $2\not\vert n^6$, and then $2\vert (n^6)^1-1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;if $3\not\vert n$, then $3\not\vert n^3$, and then $3\vert (n^3)^2-1=n^6-1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;if $7\not\vert n$, then $7\vert n^6-1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so $2,3,7\vert n^7-n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$^*$:  $\ell$=little.&lt;/p&gt;&#xA;" OwnerUserId="4654" LastEditorUserId="4654" LastEditorDisplayName="" LastEditDate="2011-02-15T06:41:03.963" LastActivityDate="2011-02-15T06:41:03.963" />
  <row Id="22126" PostTypeId="2" ParentId="22122" CreationDate="2011-02-15T06:37:03.563" Score="1" ViewCount="" Body="&lt;p&gt;Let $\lambda_1$ be arbitrary and define $\lambda_i = \lambda_1 + a_{1i}$. Now you have $a_{1i} = \lambda_i - \lambda_1$, and so $$a_{ij} = a_{1j} - a_{1i} = \lambda_j - \lambda_i.$$ Conversely, if there are numbers $\lambda_i$ such that $a_{ij} = \lambda_i - \lambda_j$ then clearly everything you wrote is true. So your conditions are both necessary and sufficient.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2011-02-15T06:37:03.563" CommentCount="1" />
  <row Id="22127" PostTypeId="2" ParentId="22108" CreationDate="2011-02-15T06:43:25.163" Score="2" ViewCount="" Body="&lt;p&gt;The general case is &lt;a href=&quot;http://en.wikipedia.org/wiki/Young_tableau&quot; rel=&quot;nofollow&quot;&gt;Young tableaux&lt;/a&gt;. They generalize your rectangular arrangements. They're useful in the &lt;a href=&quot;http://en.wikipedia.org/wiki/Representation_theory_of_the_symmetric_group&quot; rel=&quot;nofollow&quot;&gt;representation theory of $S_n$&lt;/a&gt;, where one usually considers them with &quot;unordered rows&quot;, and sometimes also with &quot;unordered columns&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The operation of taking a $3\times 4$ table and turning it into a $4\times 3$ table corresponds to moving to the &quot;transposed&quot; or &quot;conjugate&quot; tableau (or rather &lt;a href=&quot;http://en.wikipedia.org/wiki/Partition_%28number_theory%29&quot; rel=&quot;nofollow&quot;&gt;Ferrers diagram&lt;/a&gt;, which is the tableau without the numbers), which has a simple interpretation in terms of the resulting representation (it is multiplied by the sign representation).&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2011-02-15T06:43:25.163" CommentCount="1" />
  <row Id="22128" PostTypeId="2" ParentId="22016" CreationDate="2011-02-15T06:46:11.057" Score="4" ViewCount="" Body="&lt;p&gt;&lt;b&gt;Edit:&lt;/b&gt; See Didier's comment below. The binomial coefficients are &quot;upside down&quot; and so what's written below is meaningless. It is worthwhile, however, to see which tools are used to obtain tail estimates on the hypergeometric distribution, to get some ideas. Perhaps all they do is use Stirling's approximation and integrate it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your distribution is very close to a &lt;a href=&quot;http://en.wikipedia.org/wiki/Hypergeometric_distribution&quot; rel=&quot;nofollow&quot;&gt;hypergeomtric distribution&lt;/a&gt; (as noted in an earlier version of the question). In fact, it is related to it via a factor of $i/k$. So tail estimates for it should transfer to tail estimates for your distribution. &lt;/p&gt;&#xA;" OwnerUserId="1277" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2011-02-15T16:31:56.523" LastActivityDate="2011-02-15T16:31:56.523" CommentCount="6" />
  <row Id="22129" PostTypeId="1" AcceptedAnswerId="22137" CreationDate="2011-02-15T06:52:29.580" Score="0" ViewCount="90" Body="&lt;p&gt;Hello!&#xA;I'm confused with these definitions. The metric of certain space and the second fundamental form seem to be the same object.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know what else to say, this is a pretty straight forward question.&lt;/p&gt;&#xA;" OwnerUserId="6153" LastActivityDate="2011-02-15T07:35:01.843" Title="Relation between metric tensor and second fundamental form" Tags="&lt;differential-geometry&gt;&lt;tensors&gt;" AnswerCount="1" />
  <row Id="22131" PostTypeId="2" ParentId="22074" CreationDate="2011-02-15T07:09:43.390" Score="1" ViewCount="" Body="&lt;p&gt;The vector $w \in \mathbb{R}^{n \times 1}$ is an arbitrary vector i.e. all the indices are completely free. The reasoning is as follows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are solving $Ax = b$ where $A \in \mathbb{R}^{m \times n}$, $x \in \mathbb{R}^{n \times 1}$ and $b \in \mathbb{R}^{m \times 1}$ with $m \leq n$. Assume that $A$ is full rank i.e. rank($A$) = $m$. This guarantees that $AA^{*}$ is invertible and hence the existence of Moore-Penrose pseudo-right-inverse is guaranteed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we define $A^{+} = A^{*}(AA^{*})^{-1}$, then this the Moore-Penrose pseudo-right-inverse.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The general solution to $Ax = b$ is then given by $x = A^{+}b + z$ where $z \in \text{Null}(A)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Claim now is that $(I-A^{+}A)w$, where $w \in \mathbb{R}^{n \times 1}$ spans the null space of $A$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First observe that $(I-A^{+}A)w \in \text{Null}(A)$, $\forall w \in \mathbb{R}^{n \times 1}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is because $A \times (I-A^{+}A)w = (A-AA^{+}A)w = (A-A)w = 0$. Further note that the rank of $(I-A^{+}A)$ is $n-m$ and hence $(I-A^{+}A)w$ spans the null-space of $A$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What happens is, though you choose $w$ with all $n$ components independent, $(I-A^{+}A)w$ which is nothing but a projection of $w$ onto the null-space of $A$ ensures that there are only $(n-m)$ &quot;free&quot; components in the vector $(I-A^{+}A)w$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the summary is, you are free to choose all the $n$ components of $w$.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastActivityDate="2011-02-15T07:09:43.390" CommentCount="1" />
  <row Id="22132" PostTypeId="2" ParentId="7896" CreationDate="2011-02-15T07:12:48.673" Score="2" ViewCount="" Body="&lt;p&gt;A theorem of Higman, Neumann, and Neumann says that every countable group (no matter what horrible properties it might have) can be embedded as a subgroup of a group generated by $2$ elements.  Thus subgroups of finitely generated groups can be pretty much anything.&lt;/p&gt;&#xA;" OwnerUserId="5067" LastActivityDate="2011-02-15T07:12:48.673" />
  <row Id="22133" PostTypeId="1" AcceptedAnswerId="22146" CreationDate="2011-02-15T07:23:10.557" Score="2" ViewCount="168" Body="&lt;p&gt;I have been looking around for this question, but all results I found only describe the definition and not the answer I seek.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is &quot;kernel&quot; basically a synonym of &quot;function&quot;? When should be the time we should use the word &quot;kernel&quot; instead of &quot;function&quot;?&lt;/p&gt;&#xA;" OwnerUserId="5373" LastActivityDate="2011-02-15T10:02:01.527" Title="Difference between kernel and function?" Tags="&lt;functions&gt;&lt;definition&gt;" AnswerCount="6" />
  <row Id="22135" PostTypeId="2" ParentId="22133" CreationDate="2011-02-15T07:32:09.120" Score="0" ViewCount="" Body="&lt;p&gt;They aren't synonymous. A kernel is a property of a function. Most generically, if you have a function $f: X \to Y$, it is defined as the equivalence relation on X which identifies $x_1$ and $x_2$ if and only if $f(x_1) = f(x_2)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are specialisations of this depending on the category $f$ lives in: for example, if $f$ is a group homomorphism, the homogeneity of the group structure means that we can represent this equivalence relation as the normal subgroup $\{ x \in X : f(x) = e \}$. Similarly for $f$ a linear map of vector spaces, or modules, or rings. &lt;/p&gt;&#xA;" OwnerUserId="5191" LastActivityDate="2011-02-15T07:32:09.120" CommentCount="2" />
  <row Id="22136" PostTypeId="2" ParentId="22121" CreationDate="2011-02-15T07:34:37.180" Score="9" ViewCount="" Body="&lt;p&gt;This is a special case of the following global-form of Fermat's little theorem.$\: $  For naturals $\rm\: a,k,n$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ $ if $\rm\ a,k &amp;gt; 1\ $ then $\rm\ a\ |\ n^k - n\ $ for all $\rm\:n\:$ $\rm \iff\ a\:$ is squarefree, and $\rm\ p-1\ |\ k-1\ $ for all primes $\rm\:p\:|\:a$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence for $\rm\: a = 42\: = 2\cdot 3\cdot 7\ $ we deduce:  $\rm\ \ 42\ |\ n^k-n\ $ for all $\rm\:n\ \iff\ 6\ |\ k-1$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the simple proof and further discussion see my &lt;a href=&quot;http://groups.google.com/groups?selm=y8zk55tqcie.fsf%40nestle.csail.mit.edu&quot; rel=&quot;nofollow&quot;&gt;2009/04/10 sci.math post&lt;/a&gt; - which also presents the analogous generalization of Euler's $\phi$ function, and Korselt's criterion for Carmichael numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: to fix rotted Google Groups links in the cited sci.math post it may be necessary to change $\ $ &lt;a href=&quot;http://google.com/&quot; rel=&quot;nofollow&quot;&gt;http://google.com/&lt;/a&gt;... $\ $ to$\ $ &lt;a href=&quot;http://groups.google.com/&quot; rel=&quot;nofollow&quot;&gt;http://groups.google.com/&lt;/a&gt;... i.e. insert &quot;groups.&quot; before &quot;google.com&quot;.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2011-02-15T16:35:29.287" LastActivityDate="2011-02-15T16:35:29.287" />
  <row Id="22137" PostTypeId="2" ParentId="22129" CreationDate="2011-02-15T07:35:01.843" Score="3" ViewCount="" Body="&lt;p&gt;The &lt;em&gt;first&lt;/em&gt; fundamental form and the metric tensor are indeed one and the same. The second fundamental form is not. It's a sort of derivative of the metric tensor.&lt;/p&gt;&#xA;" OwnerUserId="5191" LastActivityDate="2011-02-15T07:35:01.843" CommentCount="1" />
  <row Id="22138" PostTypeId="1" AcceptedAnswerId="22152" CreationDate="2011-02-15T07:40:28.153" Score="1" ViewCount="136" Body="&lt;p&gt;In the book I'm reading there is a question,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Prove that given a rotation $R(\phi \hat{n})$ and a reflection $\sigma$, they commute iff $\sigma \perp \hat{n}$ and $\phi=\pi$. (Binary rotation normal to the reflection plane.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The answer in the back of the book is,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Take $R(\phi z)$. For $\phi=\pi$, the eigenvectors, with eigenvalues in brackets, are: $\vec{z}(+1), \vec{x}(-1), \vec{y}(-1)$. If $\sigma$ is the $\vec{x}\vec{y}$ plane its eigenvectors are $\vec{x}(+1), \vec{y}(+1), \vec{z}(-1)$ and coincide with those of $R(\pi\vec{z})$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I thought an eigenvector of a symmetry operation is one that after the symmetry operation $g\vec{r}$ leaves r changed only by the eigenvalues +1 or -1. In my mind, this means that the eigenvectors of a reflection operator $\sigma$ would be all vectors on the reflection plane and normal to the plane. Not just the two particular vectors that span the plane.&lt;/p&gt;&#xA;" OwnerUserId="6732" LastEditorUserId="6732" LastEditorDisplayName="" LastEditDate="2011-02-15T07:56:15.407" LastActivityDate="2011-04-16T11:40:30.457" Title="The eigenvectors of a reflection and rotation symmetry operation." Tags="&lt;group-theory&gt;&lt;symmetric-groups&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="22139" PostTypeId="2" ParentId="22106" CreationDate="2011-02-15T08:05:28.817" Score="2" ViewCount="" Body="&lt;p&gt;Going by the title of your question if you are trying to find out the &quot;&lt;strong&gt;Best place to find... latest Research&lt;/strong&gt;&quot; I think I would go by &lt;a href=&quot;http://arxiv.org&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org&lt;/a&gt;. Lots of papers out there. Many papers are junk (including some of mine... hehe) But you can also find papers on open problems by serious mathematicians.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For me, reading papers from arxiv and trying to understand them has helped me a lot to learn new stuff.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a wiki like site, you might check, &lt;a href=&quot;http://garden.irmacs.sfu.ca/&quot; rel=&quot;nofollow&quot;&gt;http://garden.irmacs.sfu.ca/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For collaborative mathematics:&#xA;You might also do google search on the polymath project which started from this blog post: &lt;a href=&quot;http://gowers.wordpress.com/2009/01/27/is-massively-collaborative-mathematics-possible/&quot; rel=&quot;nofollow&quot;&gt;http://gowers.wordpress.com/2009/01/27/is-massively-collaborative-mathematics-possible/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2320" LastEditorUserId="2320" LastEditorDisplayName="" LastEditDate="2011-02-15T08:44:33.193" LastActivityDate="2011-02-15T08:44:33.193" />
  <row Id="22140" PostTypeId="2" ParentId="22133" CreationDate="2011-02-15T08:15:39.290" Score="0" ViewCount="" Body="&lt;p&gt;They are not the same. A kernel is a set of elements in the domain whereas a function is a mapping from the domain to the codomain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Compare &lt;a href=&quot;http://en.wikipedia.org/wiki/Kernel_%28mathematics%29&quot; rel=&quot;nofollow&quot;&gt;kernel&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/Function_%28mathematics%29&quot; rel=&quot;nofollow&quot;&gt;function.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1301" LastActivityDate="2011-02-15T08:15:39.290" />
  <row Id="22141" PostTypeId="2" ParentId="22133" CreationDate="2011-02-15T08:16:40.080" Score="0" ViewCount="" Body="&lt;p&gt;Hello Karl.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The word &quot;kernel&quot; in mathematics has several different&#xA;meanings. Zhen Lin addressed one of them - and maybe&#xA;not the one you are referring too.&#xA;So maybe you should specify which area of mathematics&#xA;you are considering in your question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;H&lt;/p&gt;&#xA;" OwnerUserId="2479" LastActivityDate="2011-02-15T08:16:40.080" CommentCount="1" />
  <row Id="22142" PostTypeId="2" ParentId="22133" CreationDate="2011-02-15T08:18:39.960" Score="1" ViewCount="" Body="&lt;p&gt;A kernel is not a synonym for a function. As Zhen Lin said, it is a property of of a function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I will try to explain the kernel of a map in linear algebra. Suppose we have a linear map $L:V\rightarrow W$. Then the kernel of $L$ consists of all vectors in $V$ which are mapped to zero. $0$ is always mapped to zero, thus the kernel is never empty. It is easy to prove that the kernel of $L$ is always a linear subspace of $V$. If $v,w\in V$ are vectors in the kernel of $L$, that is $Lv=Lw=0$, then $L(\lambda v+\mu w)=\lambda Lv+\mu L w=\lambda 0+\mu 0=0$. Thus any linear combination of vectors in the kernel is an element of the kernel. Hence it is a linear subspace of $V$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This idea generalizes to a lot of other mathematical constructions (e.g. groups), this was what Zhen Lin was talking about. &lt;/p&gt;&#xA;" OwnerUserId="5882" LastActivityDate="2011-02-15T08:18:39.960" />
  <row Id="22143" PostTypeId="1" CreationDate="2011-02-15T08:37:09.553" Score="1" ViewCount="40" Body="&lt;p&gt;I'm using Macdonald's &quot;Symmetric Functions and Hall Polynomials&quot; as a reference and did not find what I was looking for -- apologies if I only missed it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an example, let us consider the partition $\lambda=(3,2,2,1)$ of $8$. Is the notation $2\lambda$ clear / in use / ... for describing the partition $(3,3,2,2,2,2,1,1)$ of $16$? (This actually corresponds to what Macdonald would write as $\lambda\cup\lambda$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If yes, do you have a reference? If no, what should I use?&lt;/p&gt;&#xA;" OwnerUserId="689" LastActivityDate="2011-05-16T09:58:04.770" Title="Notation for &quot;duplicating&quot; partitions" Tags="&lt;reference-request&gt;&lt;notation&gt;&lt;partitions&gt;" AnswerCount="1" CommentCount="0" />
  <row Id="22144" PostTypeId="5" CreationDate="2011-02-15T08:40:01.280" Score="0" ViewCount="" Body="&lt;p&gt;Questions related to the different ways of expressing an integer as a sum of integers; or, questions related to the subdivision of the edge or vertex set of a (hyper)graph.&lt;/p&gt;&#xA;" OwnerUserId="-1" LastEditorUserId="689" LastEditorDisplayName="" LastEditDate="2011-02-15T09:04:21.270" LastActivityDate="2011-02-15T09:04:21.270" />
  <row Id="22145" PostTypeId="4" CreationDate="2011-02-15T08:40:01.280" Score="0" ViewCount="" Body="Questions related to the different ways of expressing an integer as a sum of integers; or, questions related to the subdivision of the edge or vertex set of a (hyper)graph." OwnerUserId="-1" LastEditorUserId="689" LastEditorDisplayName="" LastEditDate="2011-02-15T09:04:23.250" LastActivityDate="2011-02-15T09:04:23.250" />
  <row Id="22146" PostTypeId="2" ParentId="22133" CreationDate="2011-02-15T09:00:40.870" Score="10" ViewCount="" Body="&lt;p&gt;&quot;Kernel&quot; is an old-fashioned term for the function you use to define certain &lt;a href=&quot;http://en.wikipedia.org/wiki/Integral_transform&quot; rel=&quot;nofollow&quot;&gt;integral operators&lt;/a&gt;. (I assume this is the sense you mean, not the more common modern sense, which is &lt;a href=&quot;http://en.wikipedia.org/wiki/Kernel_(algebra)&quot; rel=&quot;nofollow&quot;&gt;completely different&lt;/a&gt;.) Like many other words in mathematics (although people generally never tell you this), it has less to do with denotation than connotation: when you use the word &quot;kernel&quot; you are thinking of your function in terms of integral operators. &lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2011-02-15T09:00:40.870" CommentCount="2" />
  <row Id="22147" PostTypeId="1" CreationDate="2011-02-15T09:16:07.340" Score="0" ViewCount="44" Body="&lt;p&gt;In the book &quot;Markov Random Field Modeling In Image Analysis&quot; by Stan. Z Li, 3d Edition. Page 15.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An observation d of a field f (with m nodes) with i.i.d. gaussian noise:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$d_i = f_i + e_i$, there $e_i$ ~ $N(\mu, \sigma^2)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$p(d|f) = \frac{1}{\prod_1^m\sqrt{2\pi\sigma^2}} e^{-U(d|f)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$U(d|f) = \sum_{i=1}^{m} \frac{(f_i - d_i)^2}{2\sigma^2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would have thought that $f_i - d_i$ should be $d_i - f_i - \mu$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why does $\mu$ not impact $p(d|f)$ ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks   &lt;/p&gt;&#xA;" OwnerUserId="5464" LastActivityDate="2011-02-15T09:16:07.340" Title="Likelihood with gaussian noise" Tags="&lt;statistics&gt;&lt;probability-distributions&gt;" CommentCount="2" />
  <row Id="22148" PostTypeId="1" AcceptedAnswerId="22165" CreationDate="2011-02-15T09:29:23.240" Score="2" ViewCount="160" Body="&lt;p&gt;Let $f$ be a holomorphic function on the open unitary disk $\mathbb{D}$ and continous on $\mathbb{\overline{D}}$. If $f(\frac{z}{2})= \frac{1}{2}f(z)$ for all $z\in \mathbb{\overline{D}}$ and $f(1)=1$, then $f(z)=z$ for all $z\in \mathbb{\overline{D}}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Got this as homework. Any hints would be highly appreciated.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="5209" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2011-02-15T17:19:59.590" LastActivityDate="2011-02-15T20:24:06.077" Title="For holomorphic $f$, $f(\frac{z}{2})= \frac{1}{2}f(z) \Longrightarrow f(z) = z$" Tags="&lt;homework&gt;&lt;complex-analysis&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="1" />
  <row Id="22149" PostTypeId="2" ParentId="22143" CreationDate="2011-02-15T09:30:07.173" Score="1" ViewCount="" Body="&lt;p&gt;I don't think the notation $2\lambda$ is clear as it could be taken to mean the partition&#xA;$\lbrace 6,4,4,2 \rbrace$ of $16$. I would write $\lambda \cup \lambda, $ as Macdonald does.&#xA;Sorry, I don't have any recommendation for a more succinct and unambiguous notation. &lt;/p&gt;&#xA;" OwnerUserId="1301" LastActivityDate="2011-02-15T09:30:07.173" />
  <row Id="22150" PostTypeId="1" AcceptedAnswerId="22154" CreationDate="2011-02-15T09:46:27.750" Score="0" ViewCount="59" Body="&lt;p&gt;I have this deinition&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A field $F$ is &lt;strong&gt;perfect&lt;/strong&gt; if it has characteristic $0$ or if it has characteristic $p$ and $F^p = F$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From Wikipedia, I have this fact about separable polynomials:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Irreducible polynomials over perfect fields are separable. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can show that if $p(x)$ is irreducible and $char(F) = 0$, then $p$ cannot have multiple zeros, i.e. it's separable. I can also show that, if $p(x)$ has multiple zeros, it has to be of the form $p(x) = q(x^p)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now my &lt;strong&gt;question&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the comment on Wikipedia, how does the second part follow? That is, that if $char(F)=p$ for $p$ prime and $f(x)$ is irreducible over $F$ then it cannot have multiple zeros?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many thanks for your help.&lt;/p&gt;&#xA;" OwnerUserId="5798" LastActivityDate="2011-02-15T10:49:47.260" Title="Question regarding separability of polynomials and perfect fields" Tags="&lt;abstract-algebra&gt;&lt;field-theory&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="22151" PostTypeId="2" ParentId="22133" CreationDate="2011-02-15T10:02:01.527" Score="-1" ViewCount="" Body="&lt;p&gt;I think it's a slip of a tongue, morphisms are generally not functions.&lt;/p&gt;&#xA;" OwnerUserId="3793" LastActivityDate="2011-02-15T10:02:01.527" CommentCount="1" />
  <row Id="22152" PostTypeId="2" ParentId="22138" CreationDate="2011-02-15T10:25:08.360" Score="1" ViewCount="" Body="&lt;p&gt;Eigenvectors are not unique in general and can be with respect to any eigenvalue. Generally what one does instead is pick a basis for a given &lt;em&gt;eigenspace&lt;/em&gt; (the space of eigenvectors with respect to a particular eigenvalue) in the cases where such a basis exists. In this case the eigenspace associated to eigenvalue $1$ is the entire reflection plane, and any basis of this works.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, in the generic case eigenspaces have dimension $1$ and this issue doesn't occur.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2011-02-15T10:25:08.360" CommentCount="7" />
  <row Id="22153" PostTypeId="2" ParentId="21419" CreationDate="2011-02-15T10:28:47.093" Score="1" ViewCount="" Body="&lt;p&gt;I don't pretend this is an answer but I could not put it as a comment. Perhaps this may help.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your objective function is non-convex, it has local minimas. You want to find the global minima, there exist no algorithm that can find it with efficiency in both time and space.  Your regions could be represented by points, so you have a distribution of points and you want to find the maximum by searching in this distribution using the objective function. However, you know almost nothing about these points? (can the difference between neighbors be bounded?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you know almost nothing about these points, stochastic optimization algorithms (such as PSO or GA) should be useful, perhaps try with other parameters? if they get stuck on a local minima , perhaps try simulated annealing and start with a high temperature?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps you could also try regularization, for example by approximating each region with a center point and assume continuity between center-points, this would allow you to use gradient-descent techniques.&lt;/p&gt;&#xA;" OwnerUserId="5464" LastEditorUserId="5464" LastEditorDisplayName="" LastEditDate="2011-02-15T10:45:44.980" LastActivityDate="2011-02-15T10:45:44.980" CommentCount="1" />
  <row Id="22154" PostTypeId="2" ParentId="22150" CreationDate="2011-02-15T10:49:47.260" Score="1" ViewCount="" Body="&lt;p&gt;Ok, I found a proof in another book. The fact that I didn't see was that if $F$ has characteristic $p$ then $(x + y)^p = x^p + y^p$. Then if $f(x) = g(x^p)$ with coefficients in $F$ and $F^p = F$ then $f(x) = a_n^px^{pn} + \dots + a_0^p = (a_nx^n \dots + a_0)^p = g(x)^p$ because all $a_i \in F$ can be written as powers of $p$ by assumption. But then by assumption $f$ is irreducible over $F$ therefore contradiction. Therefore $f$ cannot have multiple zeros, therefore $f$ is separable.&lt;/p&gt;&#xA;" OwnerUserId="5798" LastActivityDate="2011-02-15T10:49:47.260" />
  <row Id="22155" PostTypeId="1" CreationDate="2011-02-15T11:02:11.103" Score="3" ViewCount="113" Body="&lt;p&gt;Hi.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have the following setting: Let (M,g) be a Riemannian manifold and $\iota: M \to R^N$ some isometric embedding. This means especially that the connection $\nabla^{TM}$ of M is given by the ordinary Euklidean connection $\nabla$ on R^N composed with the projection on the subbundle TM of TR^N. I also have the normal bundle NM of M and can equip it with the via $\iota$ pulled back metric and connection.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I want to bound the Christoffel symbols (and its derivatives) of $\nabla^{NM}$ by the Christoffel symbols of $\nabla^{TM}$ and also bound the entries of the projection matrices $p_{TM}: R^N \to TM$ and $p_{NM}: R^N \to NM$. And vice versa.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the computation of $\Gamma^{TM}$ I use Riemannian normal coordinates $\partial_{x_i}$. For the computation of $\Gamma^{NM}$ I take an orthonormal basis ${n_i}$ of NM at a point p and then parallel translate it along the radial geodesics of the Riemannian normal coordinate system. The entries of the projections I want to compute with respect to the standard coordinates $e_i$ of R^N.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have the following: Let $g_{ij}$ and $h_{ij}$ be the local expressions for the metrics g and h (w.r.t. the frame $\partial_{x_i}$, resp. $n_i$) and let $g^{ij}$ and $h^{ij}$ be the inverse matrices. Then the projections are given by $p_{TM}(v) = g^{kl} \langle v, \partial_{x_l} \rangle \partial_{x_k}$ and analogously for $p_{NM}$. Then the Christoffel symbols are given by $\Gamma_{ij}^{k, TM} = g^{kl} \langle \nabla_{\partial_{x_i}} \partial_{x_j}, \partial_{x_l}\rangle$ and analogously for $\Gamma_{ij}^{k, NM}$ and the entries of the projection metrices are $(p_{TM})_{ij} = g^{kl} \langle e_j, \partial_{x_l}\rangle \langle e_i, \partial_{x_k}\rangle$ and analogously for $(p_{NM})_{ij}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now if I suppose that all the $\Gamma_{ij}^{k, TM}$ are bounded as are all their derivatives, I do not see how to show the boundedness of the $(p_{TM})_{ij}$ and their derivatives. And even if somehow could show this, I don't see how to show then the bundedness of the $\Gamma_{ij}^{k, NM}$ and their derivatives.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Motivation for the whole question: A Riemannian manifold has bounded geometry, if the metric is complete, the injectivity radius positive and the curvature tensor and its covariant derivatives are uniformly bounded. One can show that this is equivalent to the statement, that the Christoffel symbols and its derivates are all uniformly bounded when computed in Riemannian normal coordinates (where the radii of all the coordinate balls are fixed). A bundle over a Riemannian manifold equipped with a connection has bounded geometry, if its curvature tensor and all its covariant derivatives are bounded. Here one can also show that this is equivalent to the statement, that the Christoffel symbols and its derivatives are uniformly bounded, when computed in such a frame as above (parallel translation of an orthonormal basis at a point along radial geodesics). Now I want to show, that if a Riemannian manifold has bounded geometry, its normal bundle (w.r.t. some isometric embedding into some R^N) has also bounded geometry. And I am also interested in a formulation of this via the entries of the projection matrices.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks,&#xA;Alexander Engel&lt;/p&gt;&#xA;" OwnerUserId="7110" LastActivityDate="2011-02-15T11:02:11.103" ClosedDate="2011-03-04T16:36:48.700" Title="Boundedness of the Christoffel symbols of a connection on the normal bundle." Tags="&lt;differential-geometry&gt;" CommentCount="6" FavoriteCount="1" />
  <row Id="22156" PostTypeId="2" ParentId="22101" CreationDate="2011-02-15T11:07:58.057" Score="1" ViewCount="" Body="&lt;p&gt;This is an exercise in lots of substitution. There are ways to get Mathematica to handle it somewhat mechanically --my first run-through used one of my favorite tools, &lt;code&gt;Resultant&lt;/code&gt;-- but you can also do pretty well &quot;manually&quot; (using Mathematica to perform the straightforward, but messy, algebraic manipulations).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll relabel your figure so that the &quot;central&quot; vertex is $O$, and the &quot;outer&quot; tetrahedron has vertices $A$, $B$, $C$, and $D$. We may assume that $O$ lies at the origin, $A$ lies on the $x$-axis, and $B$ lies in the $xy$-plane. So&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;\begin{align}&#xD;&#xA;O &amp;amp;= (0,0,0) \\&#xD;&#xA;A &amp;amp;= (x_a,0,0) \\&#xD;&#xA;B &amp;amp;= (x_b,y_b,0) \\&#xD;&#xA;C &amp;amp;= (x_c,y_c,z_c) \\&#xD;&#xA;D &amp;amp;= (x_d,y_d,z_d)&#xD;&#xA;\end{align}&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;\begin{align}&#xD;&#xA;OA\;\; &amp;amp;= x_a \\&#xD;&#xA;OB^2 &amp;amp;= x_b^2 + y_b^2 \\&#xD;&#xA;OC^2 &amp;amp;= x_c^2 + y_c^2 + z_c^2 \\&#xD;&#xA;OD^2 &amp;amp;= x_d^2 + y_d^2 + z_d^2 \\&#xD;&#xA;AB^2 &amp;amp;= ( x_a - x_b )^2 + y_b^2 = OA^2 + OB^2 - 2 \; OA \; x_b \\&#xD;&#xA;AC^2 &amp;amp;= ( x_a - x_c )^2 + y_c^2 + z_c^2 = OA^2 + OC^2 - 2 \; OA \; x_c \\&#xD;&#xA;AD^2 &amp;amp;= ( x_a - x_d )^2 + y_d^2 + z_d^2 = OA^2 + OD^2 - 2 \; OA \; x_d \\&#xD;&#xA;BC^2 &amp;amp;= ( x_b - x_c )^2 + ( y_b - y_c )^2 + z_c^2 = OB^2 + OC^2 - 2 x_b x_c - 2 y_b y_c \\&#xD;&#xA;BD^2 &amp;amp;= ( x_b - x_d )^2 + ( y_b - y_d )^2 + z_d^2 = OB^2 + OD^2 - 2 x_b x_d - 2 y_b y_d \\&#xD;&#xA;CD^2 &amp;amp;= ( x_c - x_d )^2 + ( y_c - y_d )^2 + ( z_c - z_d )^2 = OC^2 + OD^2 - 2 x_c x_d - 2 y_c y_d - 2 z_c z_d&#xD;&#xA;\end{align}&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The goal is to combine these equations in such a way that the $x$s, $y$s, and $z$s disappear, leaving only the lengths of edges. To start, we can easily solve for $x_b$, $x_c$, and $x_d$ ...&#xA;$$\begin{align}&#xD;&#xA;x_b &amp;amp;= \frac{1}{2 \; OA} \left( OA^2 + OB^2 - AB^2 \right) \\&#xD;&#xA;x_c &amp;amp;= \frac{1}{2 \; OA} \left( OA^2 + OC^2 - AC^2 \right) \\&#xD;&#xA;x_d &amp;amp;= \frac{1}{2 \; OA} \left( OA^2 + OD^2 - AD^2 \right) \\&#xD;&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can then isolate the $y_b$, $y_c$, and $y_d$ elements in terms of the &lt;em&gt;known&lt;/em&gt; $x$ values (which I'll mark with hats, rather than substituting-in) and still-unknown $z$ values. To save space later, I introduce the symbols $[OBC]$, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;\begin{align}&#xD;&#xA;y_b^2 &amp;amp;= OB^2 - \hat{x}_b^2 \\&#xD;&#xA;y_c^2 &amp;amp;= OC^2 - \hat{x}_c^2 - z_c^2 \\&#xD;&#xA;y_d^2 &amp;amp;= OD^2 - \hat{x}_d^2 - z_d^2 \\&#xD;&#xA;2 y_b y_c &amp;amp;= OB^2 + OC^2 - BC^2 - 2 \hat{x}_b \hat{x}_c =: [OBC] - 2 \hat{x}_b \hat{x}_c \\&#xD;&#xA;2 y_b y_d &amp;amp;= OB^2 + OD^2 - BD^2 - 2 \hat{x}_b \hat{x}_d =: [OBD] - 2 \hat{x}_b \hat{x}_d \\&#xD;&#xA;2 y_c y_d &amp;amp;= OC^2 + OD^2 - CD^2 - 2 \hat{x}_c \hat{x}_d - 2 z_c z_d =: [OCD] - 2 \hat{x}_c \hat{x}_d - 2 z_c z_d&#xD;&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore ...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xD;&#xA;4 \left( OB^2 - \hat{x}_b^2 \right)\left( OC^2 - \hat{x}_c^2 - z_c^2 \right) &amp;amp;= \left( [OBC] - 2 \hat{x}_b \hat{x}_c \right)^2 \\&#xD;&#xA;4 \left( OB^2 - \hat{x}_b^2 \right) \left( OD^2 - \hat{x}_d^2 - z_d^2 \right) &amp;amp;= \left( [OBD] - 2 \hat{x}_b \hat{x}_d \right)^2 \\&#xD;&#xA;2 y_c y_d = \frac{2 (2 y_b y_c) ( 2 y_b y_d )}{4 y_b^2} &amp;amp;= \\&#xD;&#xA;\frac{\left( [OBC] - 2 \hat{x}_b \hat{x}_c\right)\left( [OBD] - 2 \hat{x}_b \hat{x}_d \right)}{2\left( OB^2 - \hat{x}_b^2\right)^2 } &amp;amp;= [OCD] - 2 \hat{x}_c \hat{x}_d - 2 z_c z_d &#xD;&#xA;\end{align}&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[&lt;strong&gt;EDIT&lt;/strong&gt; My third equation had a stray square. Closing out the solution is a bit easier than I had indicated.]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From here, the first two equations give you $z_c^2$ and $z_d^2$, while the third can be solved for $z_c z_d$. Substitute the expressions into the identity $z_c^2 z_d^2 = ( z_c z_d )^2$ and you'll have obtain a polynomial in the edge lengths. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason I only outlined each step is because that final polynomial has about 65,000 terms. That's where Mathematica comes in handy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As I mentioned, you can use Mathematica's &lt;code&gt;Resultant&lt;/code&gt; to churn through this exercise as well. The resultant (or &quot;Sylvester's Eliminant&quot;) of two polynomials is a polynomial with one variable eliminated. With a larger system like the one above, proceed with a kind of enhanced Gaussian Elimination strategy: Pick your favorite equation, then pair it with each remaining equation, in each case computing a resultant that eliminates (say) $x_b$. Then choose a favorite among those resultants, pair it with the others to eliminate (say) $x_c$. Then $x_d$, $y_b$, $y_c$, $y_d$, $z_c$, and $z_d$. You end up with a (high-degree) polynomial in any remaining variables ... in this case, the edge lengths. (In using this methodology on this problem, I hit a couple of complications I won't discuss here. Choosing a different elimination order may have helped --it often does-- but I didn't give it much thought since the &quot;manual&quot; method turned out to be pretty straightforward.)&lt;/p&gt;&#xA;" OwnerUserId="409" LastEditorUserId="409" LastEditorDisplayName="" LastEditDate="2011-02-15T13:31:54.960" LastActivityDate="2011-02-15T13:31:54.960" CommentCount="1" />
  <row Id="22158" PostTypeId="1" AcceptedAnswerId="22283" CreationDate="2011-02-15T11:52:22.787" Score="0" ViewCount="96" Body="&lt;p&gt;As &lt;a href=&quot;http://mathoverflow.net/questions/53579/a-question-on-permutations-of-coefficients-of-polynomials&quot; rel=&quot;nofollow&quot;&gt;this question&lt;/a&gt;  suggests, I quite like the notion of permuting the coefficients of polynomials.&lt;br&gt;&#xA;And, moreover, I have another question on this direction:If L|F is a finite normal field extension, then it must be normal, then my question is: is there a name for the field extension L|F such that L contains &lt;em&gt;all roots of polynomials obtained by permuting the coefficients of p(x)&lt;/em&gt; where p(x) is a polynomial one of whose roots lie in L.&lt;br&gt;&#xA;In any case, thanks for paying attention.  &lt;/p&gt;&#xA;" OwnerUserId="6792" LastEditorUserId="6792" LastEditorDisplayName="" LastEditDate="2011-02-16T04:55:47.033" LastActivityDate="2011-02-16T05:03:17.783" Title="Is there a specific name for this notion of extensions of fields?" Tags="&lt;polynomials&gt;&lt;field-theory&gt;" AnswerCount="1" CommentCount="9" FavoriteCount="1" />
  <row Id="22159" PostTypeId="1" AcceptedAnswerId="22220" CreationDate="2011-02-15T11:54:22.213" Score="7" ViewCount="140" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm interested in the number of binary matrices of a given size that are distinct with regard to row and column permutations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If ~ is the equivalence relation on n⨯m binary matrices such that A ~ B iff one can obtain B from applying a permutation matrix to A, I'm interested in the number of ~-equivalence classes over all n⨯m binary matrices.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know there are 2^(nm) binary matrices of size n⨯m, and n!m! possible permutations, but somehow I fail to get an intuition on what this implies for the equivalence classes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Best regards,&#xA;P.&lt;/p&gt;&#xA;" OwnerUserId="7111" LastActivityDate="2011-02-15T20:58:42.647" Title="How many n⨯m binary matrices are there, up to row and column permutations?" Tags="&lt;linear-algebra&gt;&lt;combinatorics&gt;&lt;matrices&gt;&lt;permutations&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="0" />
  <row Id="22160" PostTypeId="1" AcceptedAnswerId="22168" CreationDate="2011-02-15T12:00:58.307" Score="1" ViewCount="67" Body="&lt;p&gt;I have a query below, and want to understand the reason of this in detail.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is the graph of $x^2+(y-\sqrt{x^2})^2=1$ &lt;a href=&quot;http://www.wolframalpha.com/input/?i=x%5E2%2B%28y-sqrt%28x%5E2%29%29%5E2%3D1&quot; rel=&quot;nofollow&quot;&gt;drawn by Wolfram Alpha&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/oUdgp.gif&quot; alt=&quot;graph of $x^2+(y-\sqrt{x^2})^2=1$&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this equation if you replace the term, $\sqrt{x^2}$ by $x$, which is a valid substitution, the graph &lt;a href=&quot;http://www.wolframalpha.com/input/?i=x%5E2%2B%28y-x%29%5E2%3D1&quot; rel=&quot;nofollow&quot;&gt;seems to become an ellipse&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/Fif7q.gif&quot; alt=&quot;graph of $x^2+(y-x)^2=1$&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Shouldn't the curve remain same?  What is the reason for this? Am I seriously missing some fundamental point here?&lt;/p&gt;&#xA;" OwnerUserId="1249" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2011-02-15T14:15:52.270" LastActivityDate="2011-02-15T14:15:52.270" Title="How come this curve changes if equation remains same?" Tags="&lt;geometry&gt;&lt;algebra-precalculus&gt;&lt;graph&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="22161" PostTypeId="1" AcceptedAnswerId="22170" CreationDate="2011-02-15T12:10:50.183" Score="10" ViewCount="89" Body="&lt;p&gt;Let $A$ be a commutative (unital) ring, and $A[x_1,\ldots,x_n]$ a polynomial ring over it in some finite number of variables. The inclusion $i\colon A \hookrightarrow A[x_1,\ldots,x_n]$ induces (by contraction) a continuous surjection $\mathrm{Spec}(i)\colon \mathrm{Spec}(A[x_1,\ldots,x_n]) \twoheadrightarrow \mathrm{Spec}(A)$ on the prime spectra. Is $\mathrm{Spec}(i)$ a closed map of topological spaces? Does this become the case if $A$ is assumed to be Noetherian and/or an integral domain or a field? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it's not closed, (under whatever assumptions on $A$), could someone provide a simple counterexample?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I realize this is probably a very stupid question. It seems like the map should be obviously be closed or obviously not be, but I've vacillated as to which. I seem finally to have devised a proof it is closed, but I am suspicious of this quasi-proof, because it seems to make an exercise I've been working on easier than the hint provided would indicate, and also fails to use some of the hypotheses granted for the exercise. Also, if it were true, I would expect to have seen some mention of it on the Internet or in some text, and so far I haven't.&lt;/p&gt;&#xA;" OwnerUserId="7112" LastEditorUserId="7112" LastEditorDisplayName="" LastEditDate="2011-02-15T12:16:30.553" LastActivityDate="2011-02-15T16:45:09.137" Title="Does inclusion of a ring into a polynomial ring induce a closed map on prime spectra?" Tags="&lt;commutative-algebra&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="2" />
  <row Id="22162" PostTypeId="2" ParentId="22121" CreationDate="2011-02-15T12:22:14.103" Score="-2" ViewCount="" Body="&lt;p&gt;With a calculator trying some random numbers ! seriously that is the most easiest way to prove it !!!&lt;/p&gt;&#xA;" OwnerUserId="4987" LastActivityDate="2011-02-15T12:22:14.103" CommentCount="3" />
  <row Id="22163" PostTypeId="2" ParentId="22161" CreationDate="2011-02-15T12:29:30.313" Score="4" ViewCount="" Body="&lt;p&gt;Nope. Let $A = \mathbb{Z}_{(p)}, n = 1$. Then $\text{Spec } A[x]$ has a closed point given by the morphism $\phi : A[x] \to \mathbb{Q}$ sending $x$ to $\frac{1}{p}$, and this closed point maps to the generic point in $\text{Spec } A$, which is not closed. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$A$ is Noetherian and an integral domain, so neither of those hypotheses help. If $A$ is a field then $\text{Spec } A$ consists of a single point.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2011-02-15T12:29:30.313" CommentCount="1" />
  <row Id="22164" PostTypeId="1" CreationDate="2011-02-15T06:10:12.020" Score="0" ViewCount="330" Body="&lt;p&gt;I have two vectors of matching lengths. They are readings from two different sensors (one is from a smartphone and the other is from a wiimote) of the same hand movement. I am trying to find the time offset between them to synchronise the readings for further processing. The readings I get are of the format (Time(ms) Value) for accelerations in the X,Y and Z direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the synchronization, I plotted the cross-correlation function &lt;code&gt;xcorr2()&lt;/code&gt; between the two sets. I am getting the same graph (a weird triangle peak and a straight line at the bottom) for Accelerations along the x, y and z directions (which I guess is good) but I don't know how to interpret it. What do the axes in the graph represent?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone explain to me what &lt;code&gt;xcorr2()&lt;/code&gt; means in a qualitative sense. From the correlation function, how do I determine the offset (i.e. how many seconds is sensor1 behind sensor2)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imelza&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/n7pKZ.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="7101" LastEditorUserId="1543" LastEditorDisplayName="" LastEditDate="2011-03-08T16:48:32.433" LastActivityDate="2011-03-10T11:43:34.567" Title="MATLAB interpretation of Xcorr2 - Cross Correlation function" Tags="&lt;signal-processing&gt;&lt;matlab&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="22165" PostTypeId="2" ParentId="22148" CreationDate="2011-02-15T12:50:58.493" Score="10" ViewCount="" Body="&lt;p&gt;Since $f$ is holomorphic it has a unique power series representation about origin,&#xA;$$&#xD;&#xA;f(z)=\sum_{k=0}^{\infty} c_k z^k.&#xD;&#xA;$$&#xA;Since $f\left(\dfrac{z}{2}\right)=\dfrac{1}{2} f(z)$ we obtain&#xA;$$&#xD;&#xA;\frac{c_k}{2^k}=\frac{1}{2}c_k,\quad k=0,1,\ldots.&#xD;&#xA;$$&#xA;From this we obtain $c_k=0$ if $k\neq 1$. Therefore $f(z)=c_1z$. We know that $f(1)=1$. Thus $c_1=1$ and $f(z)=z$.&lt;/p&gt;&#xA;" OwnerUserId="5418" LastActivityDate="2011-02-15T12:50:58.493" CommentCount="1" />
  <row Id="22166" PostTypeId="2" ParentId="22117" CreationDate="2011-02-15T13:02:32.220" Score="2" ViewCount="" Body="&lt;p&gt;In the second model, the locus of points at a fixed distance $d$ from a point $A$ is an axis-aligned square of side length $2d$ centered on $A$.  In the first model, the locus is a square of side length $d\sqrt{2}$ and rotated $45^{\circ}$ from the axes, again centered on $A$.  There is a combined rotation and dilation that maps the first model to the second.&lt;/p&gt;&#xA;" OwnerUserId="5546" LastActivityDate="2011-02-15T13:02:32.220" CommentCount="1" />
  <row Id="22168" PostTypeId="2" ParentId="22160" CreationDate="2011-02-15T13:10:23.553" Score="2" ViewCount="" Body="&lt;p&gt;For each positive real number $y$, there are two numbers whose square is $y$, one positive and one negative. $\sqrt{y}$ is defined as the positive of those two numbers. The two numbers whose square is $x^2$ are $x$ and $−x$, so $\sqrt{x^2}$ is $x$ or $−x$, depending on which of those two is positive. If $x$ is negative, it is $−x$ that's positive, and thus $\sqrt{x^2}=−x$ for $x$ negative.&lt;/p&gt;&#xA;" OwnerUserId="6622" LastActivityDate="2011-02-15T13:10:23.553" CommentCount="1" />
  <row Id="22169" PostTypeId="1" CreationDate="2011-02-15T13:12:53.547" Score="3" ViewCount="108" Body="&lt;p&gt;Hi Guys,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm hoping that such a calculation as below will be very easy for the users of this website:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a 3D Carousel on my website and a colleague is going to render out some 3D components to sit on the carousel. However, he needs the angle that the &quot;camera&quot; is at on my 3D carousel. The issue is that my 3D carousel is faux 3D, meaning the images at the rear of the carousel do not travel along a Z plane but they just reduce in size to give the impression that they are.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've added 2 black lines to the carousel and worked out (through a pixel measurement) that there is 243 pixels between the front and back of the carousel. In theory this carousel is perfectly circular.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The radius of the circle is:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;r = 2.3 (a number provided by the component to work out the radius)&#xA;w = 1000px (the width of the carousel's container)&#xA;w / r = 434.782609&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I know that there are 243px between the two black lines and that the radius of the circle is 434.78px so how would I work out the angle of the camera using these measurements? I've provided a screenshot to help if I've not explained it properly - on the diagram, I need to find the blue angle:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/piMo4.png&quot; alt=&quot;Screenshot of 3D Carousel&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/kQbKD.png&quot; alt=&quot;View of what I need&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need to know &lt;em&gt;how&lt;/em&gt; as opposed to just giving the answer because if we decide to change the angle, I need to work out how to reflect this in the 3D render of each component.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you need a little more information, just comment saying what you need and I'll do my best to trawl through the javascript file to find the answer.&lt;/p&gt;&#xA;" OwnerUserId="7115" LastEditorUserId="7115" LastEditorDisplayName="" LastEditDate="2011-02-15T14:17:55.927" LastActivityDate="2011-02-15T14:17:55.927" Title="Working out a &quot;Camera&quot; Angle from distance markers" Tags="&lt;geometry&gt;" AnswerCount="1" />
  <row Id="22170" PostTypeId="2" ParentId="22161" CreationDate="2011-02-15T13:17:23.947" Score="11" ViewCount="" Body="&lt;p&gt;Another good example to think about, more geometric and less arithmetic than that of Qiaochu, is obained by taking $A = k[x]$ ($k$ an algebraically closed field) and considering $k[x] \to k[x,y]$.  The induced map on Spectra is the map $\mathbb A^2 \to \mathbb A^1$ from the affine plane to the affine line given by the projection&#xA;$(x,y) \mapsto x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a (perhaps the most!) famous example of a non-closed map in algebraic geometry, which motivates the defintion of projective spaces, properness, complete varieties, and so on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To see that it is non-closed, consider the hyperbola $xy = 1$, which is a closed subset of $\mathbb A^2$.  Its image is the subset $x \neq 0$ of $\mathbb A^1$, which is not closed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is the geometric picture: to see if a given point $x_0$ of $\mathbb A^1$ is in the image of this map, we have to take the vertical line $x = x_0$ and intersect with hyperbola $x y = 1$, and ask whether or not this intersection is non-empty.  What we see is that the intersection &lt;em&gt;is&lt;/em&gt; non-empty if $x_0 \neq 0$,&#xA;but as we pass to the limit $x_0 = 0$, the intersection suddenly becomes empty.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This illustrates the general phenomenon that in affine varieties, there is no &quot;conservation of intersection number&quot; when we make continuous deformations of the varieties being intersected.  Rectifying this problem is one of the main motivations for introducing projective spaces, or, more generally, complete varieties.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Technically, if you look at the basic definitions related to projective varieties, or more generally, complete varieties, you will see that &quot;conservation of intersection number&quot; is not explicitly mentioned, but that the property of &lt;em&gt;properness&lt;/em&gt; (which has to do with the closedness of certain maps) is what looms large.  This may seem a little mysterious, but in fact it turns out that the failure of certain maps to be closed is more or less equivalent to the failure of conservation of intersection number.   The example of the map&#xA;$\mathbb A^2 \to \mathbb A^1$ above illustrates how the two issues are connected, and this is one reason why it is worth thinking about this example very carefully. &lt;/p&gt;&#xA;" OwnerUserId="221" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2011-02-15T16:45:09.137" LastActivityDate="2011-02-15T16:45:09.137" CommentCount="1" />
  <row Id="22171" PostTypeId="1" CreationDate="2011-02-15T08:33:41.980" Score="1" ViewCount="96" Body="&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/hb7HU.png&quot; alt=&quot;alt&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See diagram for more info.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This problem been bugging me for the past few days.&#xA;Not sure what I did wrong although I follow thoroughly every&#xA;tutorials and source code on the web.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.codezealot.org/archives/55&quot; rel=&quot;nofollow&quot;&gt;http://www.codezealot.org/archives/55&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7150" LastActivityDate="2011-03-17T16:33:35.030" Title="Something bugging me for the past few days, question on Separating Axis Theorem..." Tags="&lt;physics&gt;&lt;algorithms&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="22172" PostTypeId="2" ParentId="22171" CreationDate="2011-02-15T12:01:05.370" Score="1" ViewCount="" Body="&lt;p&gt;I'm not sure what result you are expecting, or why, but in your diagram both b and c are the same distance from the axis you've drawn, so of course you get the same result for both...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems a bit strange to see a separating axis with only one object - generally speaking you need two things in order for there to be an axis separating them!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps you don't really need or want a separating axis? Otherwise, I think you need to post some more detail on what you're trying to do.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want the distance between the points, just calculate the length of the vector between them (i.e. subtract one point from the other, square the components, sum them, and then take the square root - basic pythagoras).&lt;/p&gt;&#xA;" LastActivityDate="2011-02-15T12:01:05.370" CommentCount="1" />
  <row Id="22173" PostTypeId="1" AcceptedAnswerId="22174" CreationDate="2011-02-15T13:37:21.570" Score="0" ViewCount="58" Body="&lt;p&gt;It asks me to graph the region in the complex plane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$Re(z+iz) \le 1$&lt;/p&gt;&#xA;" OwnerUserId="2637" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2011-02-15T13:40:20.193" LastActivityDate="2011-02-15T13:55:15.260" Title="How can I proceed with this exercise?" Tags="&lt;complex-analysis&gt;&lt;plane-curves&gt;" AnswerCount="1" CommentCount="0" />
  <row Id="22174" PostTypeId="2" ParentId="22173" CreationDate="2011-02-15T13:39:52.643" Score="3" ViewCount="" Body="&lt;p&gt;Hint: If you write $z=x+yi$, $x$ and $y$ real, you should recognize it.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2011-02-15T13:55:15.260" LastActivityDate="2011-02-15T13:55:15.260" CommentCount="6" />
  <row Id="22175" PostTypeId="2" ParentId="22169" CreationDate="2011-02-15T13:53:52.330" Score="1" ViewCount="" Body="&lt;p&gt;Am I understanding correctly?  You have an area 1000 pixels wide by 243 pixels deep that you show objects in, representing (part of) the outer annulus of a 435 radius circle.  The deeper an object, the smaller it is rendered.  Is the back line the outer edge of the carousel, and if so shouldn't it be curved?  The question would then be to find a relation between depth and size of the rendering.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I have read it right and the camera is at the center, the distance to the front line is 435-243=192 pixels and the distance to the back line is 435 pixels.  So an object on the back line should be 192/435 as large as one on the front line.  Taking d as the depth from the front line, the size would be $\frac{192}{d+192}$ of the size on the front line.  But I am not sure this was the question.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-02-15T13:53:52.330" CommentCount="11" />
  <row Id="22176" PostTypeId="2" ParentId="7392" CreationDate="2011-02-15T14:14:47.020" Score="1" ViewCount="" Body="&lt;p&gt;Two more examples I think work...&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Let $k$ be a field, and $A = k[x_1,x_2,\ldots]$&#xA;a polynomial ring over $k$ in countably many indeterminates.&#xA;Let $\mathfrak{b}$ be the ideal generated by $x_1^2$ and &#xA;$x_n - x_{n+1}^2$ for all $n \geq 1$.&#xA;Write $y_n = \bar x_n$ in $B = A/\mathfrak{b}$.&#xA;Then $y_1^2 = 0$ and $y_n = y_{n+1}^2$ for all $n \geq 1$,&#xA;so $y_n^{2^n} = 0$.&#xA;If $\mathfrak{p} = (y_1,y_2,\ldots)$,&#xA;we have $B/\mathfrak{p} \cong k$, so $\mathfrak{p}$ is maximal.&#xA;On the other hand, the generators of $\mathfrak{p}$ are nilpotent, so&#xA;$\mathfrak{p}$ is contained in the nilradical of $B$, and hence is the unique minimal prime as well.&#xA;Since all primes then contain the maximal ideal $\mathfrak{p}$&#xA;$\mathfrak{p}$ is the only prime of $B$. &#xA;Thus $\mathrm{Spec}(B) = \{\mathfrak{p}\}$ is obviously Noetherian.&#xA;But $(y_1) \subsetneq (y_2) \subsetneq (y_3) \subsetneq \cdots$ &#xA;is an infinite ascending chain of ideals, so $B$ is not Noetherian.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Let $k$ be a field, $\mathbb{Q}$ the additive group of rational numbers, and $k[\mathbb{Q}]$ the group algebra. If $K$ is its field of fractions, there is a naturally associated valuation $v\colon K^\times \twoheadrightarrow \mathbb{Q}$; let $A = \{0\} \cup \{x \in K : v(x) \geq 0\}$ be the associated valuation ring. It has an ideal $\mathfrak{a}_q = \{x \in K : v(x) \geq q\}$ for each rational $q &amp;gt; 0$, with $\mathfrak{a}_q \subsetneq \mathfrak{a}_r$ if $r &amp;lt; q$. Thus, taking an infinite decreasing sequence $(q_j)$, we see $A$ is not Noetherian. However, the only prime ideal of $A$ is $\mathfrak{p} = \{x \in K : v(x) &amp;gt; 0\}$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="7112" LastActivityDate="2011-02-15T14:14:47.020" />
  <row Id="22177" PostTypeId="1" AcceptedAnswerId="22178" CreationDate="2011-02-15T14:24:53.600" Score="1" ViewCount="63" Body="&lt;p&gt;The expression is this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$|3z+2| \leq 1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;then, I did&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$|3x+2+3iy| \leq 1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and I got stuck.&lt;/p&gt;&#xA;" OwnerUserId="2637" LastActivityDate="2011-02-15T14:32:23.077" Title="How can I graph this in the complex plane?" Tags="&lt;homework&gt;&lt;complex-analysis&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="22178" PostTypeId="2" ParentId="22177" CreationDate="2011-02-15T14:32:23.077" Score="1" ViewCount="" Body="&lt;p&gt;Hint: How do you find the modulus of a complex number?&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-02-15T14:32:23.077" CommentCount="3" />
  <row Id="22179" PostTypeId="1" AcceptedAnswerId="22189" CreationDate="2011-02-15T14:54:44.317" Score="3" ViewCount="292" Body="&lt;p&gt;I am just starting out with vector spaces and I am having a hard time understanding it. One of the requirements state that 1v=v where 1 is the multiplicative identity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does 1 have to be the identity? Or is that whatever is the multiplicative identity is labelled 1?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I ran across a question where &lt;em&gt;a&lt;/em&gt;(x,y) was defined as ( &lt;em&gt;a&lt;/em&gt; x/3, &lt;em&gt;a&lt;/em&gt; y/3). It was not specified but I guess the question also implied that scalar &lt;em&gt;a&lt;/em&gt; belongs to real numbers. &#xA;Why has 1 got to be the identity here? Why can't I define 3 to be the identity here?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: I realise that I didn't describe the question in enough detail.  V here is the set of all ordered pairs (x,y) where x y are real numbers.&#xA;Addition is defined as (x1,y1) + (x2,y2) = (2x1-3x2,y1-y2).  Now addition obviously violates many axioms but I am interested in scalar multiplication,&#xA;which is defined as given above. The question is a simple example question asking us to list down all axioms violated.&lt;/p&gt;&#xA;" OwnerUserId="7118" LastEditorUserId="7118" LastEditorDisplayName="" LastEditDate="2011-02-15T16:25:49.980" LastActivityDate="2011-02-15T16:27:35.333" Title="Does the multiplicative identity have to be 1?" Tags="&lt;linear-algebra&gt;&lt;vector-space&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="22180" PostTypeId="2" ParentId="22112" CreationDate="2011-02-15T15:03:36.647" Score="0" ViewCount="" Body="&lt;p&gt;Usually, envelope is connected with a curve families (I imagine you know it), your situation is different.. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can write what you said, that is tipical in Signal Theory: modulation of a sinuoidal signal witha generic function. In your situation the envelop is a sort of a continous extension of the succession of local maxim values for your function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mathematically (and here I wish this can help you) we have a swinging signal modulated from a unknowed function f(t): how can we obtain this envelop?&#xA;In your situation, we suppose the Fourier transform of f(t) is different from 0 in a limited domain D, and frequency of sinusoidal function $\omega_0&amp;gt; \frac{\max_{\omega}D-\min_{\omega}D}{2}$&#xA;$$&#xD;&#xA;f(t)cos(\omega_0 t + \phi)= f(t)\frac{e^{i(\omega_0 t+\phi)}+e^{-i(\omega_0 t+\phi)}}{2}&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know that a product in x-variables is a convolution in coniugated variables for Fourier Transform:&#xA;$$&#xD;&#xA;\frac{1}{2\pi}\int_{-\infty}^{+\infty}e^{i\omega t}f(t)cos(\omega_0 t + \phi) dt= \frac{1}{2}\int_{-\infty}^{+\infty}\hat{f}(\omega-\omega&amp;#39;)\Bigl[e^{i\phi} \delta(\omega&amp;#39;+\omega_0)+e^{-i\phi}\delta(\omega&amp;#39;-\omega_0)\Bigl]d\omega&amp;#39; $$&#xA;$$&#xD;&#xA;=\frac{1}{2}\Bigl[e^{i\phi}\hat f(\omega+\omega_0)+e^{-i\phi}\hat f(\omega-\omega_0)\Bigl]&#xD;&#xA;$$&#xA;that is proportional to 2 summed Fourier transform of f(t) translated by $\mp\omega_0$. &#xA;Fourier transform of f(t) is different from 0 in a limited domain D, and frequency of sinusoidal function $\omega_0$ is higher than $\frac{max_{\omega}D-\min_{\omega}D}{2}$ for hyp. So the product of this with Heaviside function $\Theta(\omega)=0 $if $\omega&amp;lt;0$ and 1 if $\omega&amp;gt;0$ ($\omega=0$ has got zero measure) is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;\frac{1}{2}e^{-i\phi}\hat f(\omega-\omega_0)&#xD;&#xA;$$&#xA;So we obtaine Fourier transform translated of $\omega_0$ of envelopping function, that's FT of $\frac{1}{2} e^{-i(\omega_0 t+\phi)} f(t)$.&#xA;Fourier distributional anti-transform of $\Theta(\omega)$ is $\hat \Theta(t)=\frac{1}{i\ t}$. In \omega domain the product $=\frac{1}{2}\Theta(\omega)\Bigl[e^{i\phi}\hat f(\omega+\omega_0)+e^{-i\phi}\hat f(\omega-\omega_0)\Bigl]$ is the Fourier transform of convolution; then:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA; f(t) = -2i e^{i(\omega_0 t+\phi)} {\int_{-\infty}^{+\infty}f(\tau)cos(\omega_0 \tau + \phi)\frac{1}{t-\tau} d\tau}&#xD;&#xA;$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;that connects enveloping functions to envelopped sinusoid. This is about the physical process that happens in a lock-in, for filter signal from rumor. Remember that this integral has distributional meaning.&lt;/p&gt;&#xA;" OwnerUserId="3871" LastEditorUserId="3871" LastEditorDisplayName="" LastEditDate="2011-02-15T15:18:21.783" LastActivityDate="2011-02-15T15:18:21.783" />
  <row Id="22181" PostTypeId="2" ParentId="22002" CreationDate="2011-02-15T15:08:44.170" Score="6" ViewCount="" Body="&lt;p&gt;Let's first restrict the study to a simple class of continuous functions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A continuous function $f : X \rightarrow X$ is &quot;simple&quot; if $\forall y \in [0;1], \exists y&amp;#39; \in [0;1]$ such that :&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$f(0,y) = (0,y&amp;#39;)$&lt;/li&gt;&#xA;&lt;li&gt;$\forall q \in [0;1] \cap \mathbb{Q}, f(-qy,y) = (-qy&amp;#39;,y&amp;#39;)$ or $\forall q \in [0;1] \cap \mathbb{Q}, f(-qy,y) = (0,y&amp;#39;)$&lt;/li&gt;&#xA;&lt;li&gt;$\forall q \in [0;1] \cap \mathbb{Q}, f(q(1-y),y) = (q(1-y&amp;#39;),y&amp;#39;)$ or $\forall q \in [0;1] \cap \mathbb{Q}, f(q(1-y),y) = (0,y&amp;#39;)$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This means that the image of a point $x$ in the central segment is a point $f(x)$ on the central segment, the points to its left are either sent to the corresponding points on the left of $f(x)$ or all equal to $f(x)$, and the same with points on the right.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Call $\hom_S(X,X)$ the set of simple maps.&#xA;I could define a more general class of simple maps for example by allowing points on the left to be sent to points on the right, but it would only make things more complicated, and it's unneeded because this will be enough to separate the identity map from a constant map.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The identity map $id_X$, and the constant map $k : x \rightarrow (0,1/2)$ are in $\hom_S(X,X)$&#xA;Surely, if I want to prove that $X$ is not contractible, it is at least as hard as proving that it is not contractible with a homotopy that only uses simple maps. So, let's prove that there is no homotopy of simple maps between $id_X$ and $k$.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;In order to do that, I show that $\hom_S(X,X)$ is &quot;homeomorphic&quot; to a subset of $\hom([0;1],S^1)$ :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $f : [0;1] \rightarrow S^1$, define $\phi(f) \in \hom_S(X,X)$ with :&#xA;$$\phi(f)(0,y) = \left\{\begin{array}{ll}(0,2f(y)/pi) &amp;amp; \text{for} f(y) \in [0; \pi/2] \\&#xD;&#xA; (0,2-2f(y)/pi) &amp;amp; \text{for} f(y) \in [\pi/2; \pi] \\&#xD;&#xA; (0,2f(y)/pi-2) &amp;amp; \text{for} f(y) \in [\pi; 3\pi/2] \\&#xD;&#xA; (0,4-2f(y)/pi) &amp;amp; \text{for} f(y) \in [3\pi/2; 2\pi]\end{array} \right. $$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;the points on the left of $(0,y)$ are taken to the points on the left of $f(0,y)$ if $\theta \in [0; \pi]$, are sent to $f(0,y)$ otherwise ;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the points on the right of $(0,y)$ are taken to the points on the right of $f(0,y)$ if $\theta \in [-\pi/2; \pi/2]$, are sent to $f(0,y)$ otherwise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is easy to check that $\phi(f)$ is continuous if and only if $f(0) \notin [0; \pi]$ and&#xA; $f(1) \notin [-\pi/2; \pi/2]$. If I call $\hom_S([0;1],S^1)$ the subset of functions with this property, $\phi$ is a bijection from $\hom_S([0;1],S^1)$ into $\hom_S(X,X)$ that conserves homotopies.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Now we have to study homotopies in $\hom_S([0;1],S^1)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $f \in \hom_S([0;1],S^1)$, define $w(f) = $ the number of times the function winds up from $0$ to $\pi/2$. More precisely, define &#xA;$\alpha : S^1 \rightarrow S^1 : \theta \rightarrow 4\theta$ for $\theta \in [0;\pi/2]$ and $0$ otherwise.&#xA;$\forall f \in \hom_S([0;1],S^1), (\alpha \circ f)(0) = (\alpha \circ f)(1) = 0$, so we can quotient that map and get a function $\hat{f} : S^1 \rightarrow S^1$.&#xA;Then define $w(f)$ as the winding number of $\hat{f}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$w$ is a homotopic invariant (in fact it completely describes homotopy inside $\hom_S(X,X)$), but $w(\phi^{-1}(id_X)) = 1$ while $w(\phi^{-1}(k)) = 0$. Therefore $id_X$ and $k$ are not homotopic inside $\hom_S(X,X)$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Now what is left is to show that if there is a homotopy using any continuous maps, then there is one that only uses simple maps.&#xA;To do that, I need an approximation map $\psi : \hom(X,X) \rightarrow \hom_S (X,X)$ that push forward homotopies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, define $\psi(f)(0,y) = (0,y&amp;#39;)$ where $(x&amp;#39;,y&amp;#39;) = f(0,y)$.&#xA;If $x&amp;#39; \neq 0$, send all the points on the left/right side of $(0,y)$ into $(0,y&amp;#39;)$ as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Otherwise, we need to decide when to send the points on the left/right side of $(0,y)$ on the left/right side of $f(0,y)$. There are uncountably many ways to do it, so let's arbitrarily settle it with this :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $y \in [0;1]$, define $P_l(y) = \exists x_1,x_2,\ldots x_n \ldots &amp;lt; 0, \lim x_n = 0$, and $f(x_n,y)$ are on the (strictly) left half of $X$,&#xA;and similarly $P_r(y) = \exists x_1,x_2,\ldots x_n \ldots &amp;gt; 0, \lim x_n = 0$, and $f(x_n,y)$ are on the (strictly) right half of $X$.&#xA;Then, send the points on the left (resp. right) side of $(0,y)$ on the left (resp. right) of $\psi(f)(0,y)$ if and only if $P_l(y)$ (resp. $Pr(y)$) is true.&#xA;It is important to note that $P_l(0)$ and $P_r(1)$ are always false.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I have to prove that not only $\psi(f)$ is continuous, but that for any homotopy $h : [0;1] \times X \rightarrow X$, the map $\psi(h) : [0;1] \times X \rightarrow X$ is a homotopy of simple maps.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose $(t_n,y_n) \rightarrow (t,y)$ in $[0;1]^2$. &#xA;Call $(x&amp;#39;_n,y&amp;#39;_n) = h(t_n,(0,y_n))$ and $(x&amp;#39;,y&amp;#39;) = h(t(0,y))$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, $h(t_n,(0,y_n)) \rightarrow (x&amp;#39;,y&amp;#39;) = h(t,(0,y))$.&#xA;This shows that $\psi(h)(t_n,(0,y_n)) \rightarrow (0,y&amp;#39;) = \psi(h)(t,(0,y))$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $x&amp;#39; \neq 0$, then eventually neither are the $x&amp;#39;_n$ so $\psi(h)(t_n,(x_n,y_n)) = (0,y&amp;#39;_n) \rightarrow (0,y&amp;#39;) = \psi(h)(t,(x,y))$, for any sequence $x_n \rightarrow x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $y&amp;#39; \notin \{0;1\}$ then there is a neighbourhood of $(0,y&amp;#39;)$ in $X$ where the diagonal lines are all disconnected from each other, so for $n$ large enough, the points on the diagonal lines near $(0,y)$ who are sent on the diagonal lines near $(0,y&amp;#39;)$ are stuck there, so eventually $P_l$ and $P_r$ will not change.&#xA;This also means that if $y=0$ (resp. $y=1$), $P_l$ (resp. $P_r$) must eventually be false.&#xA;So there is no obstruction and $\psi(h)(t_n,(x_n,y_n)) \rightarrow \psi(h)(t,(x,y))$ for any sequence $x_n \rightarrow x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only troublesome points are when $y&amp;#39; = 0$ or $y&amp;#39; = 1$.&#xA;For example, if $y&amp;#39; = 0$, then $P_r$ will not change for the same reason as stated above, so $\psi(h)(t_n,(x_n,y_n)) \rightarrow (0,y&amp;#39;) = \psi(h)(t,(x,y))$ for any positive sequence $x_n \rightarrow x$.&#xA;For the points on the left, since $y&amp;#39;_n \rightarrow 0$, whatever $P_l(y_n)$ is and for any negative sequence $x_n \rightarrow x$, $\psi(h)(t_n,(x_n,y_n)) \rightarrow (0,0) = \psi(h)(t,(x,y))$ .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The case $y&amp;#39; = 1$ is similar, and this proves that the approximation map $\phi$ transport homotopies in $\hom(X,X)$ into homotopies in $\hom_S(X,X)$ (though it doesn't pull back homotopies at all).&#xA;Since $id_X$ and $k$ were not homotopic in $\hom_S(X,X)$, they are not homotopic in $\hom(X,X)$, which shows that $X$ is not contractible.&lt;/p&gt;&#xA;" OwnerUserId="6270" LastEditorUserId="6270" LastEditorDisplayName="" LastEditDate="2011-03-01T09:31:55.497" LastActivityDate="2011-03-01T09:31:55.497" CommentCount="3" />
  <row Id="22182" PostTypeId="2" ParentId="22164" CreationDate="2011-02-15T15:20:22.407" Score="1" ViewCount="" Body="&lt;p&gt;The peak of the triangle represents your offset.  If the two vectors were identical, the peak would appear exactly in the center.  If they are offset from each other, the peak will occur offset from the center.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd do an experiment and correlate a vector with itself.  Then use a function like argmax() to find the index of the maximum position.  It should be len(vector)/2 plus or minus 1.  Then do the correlation of the two different signals, and do argmax on that.  Subtract the value you just found for identical vectors, and you will have your offset.  If you reverse the order of the signals, the offset will be negative.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remember that there are different implementations of correlation, like a circular cross-correlation, where the signals are wrapped around.  You don't want that.  If the Matlab function is a circular cross-correlation (FFT-enhanced), then you need to zero pad first.  Read into the different implementations and options of &lt;code&gt;xcorr2&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://gist.github.com/376572&quot; rel=&quot;nofollow&quot;&gt;Here's a simple example in Python&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2206" LastEditorUserId="2206" LastEditorDisplayName="" LastEditDate="2011-02-15T15:57:07.277" LastActivityDate="2011-02-15T15:57:07.277" CommentCount="9" />
  <row Id="22183" PostTypeId="1" AcceptedAnswerId="22185" CreationDate="2011-02-15T15:30:52.697" Score="0" ViewCount="68" Body="&lt;p&gt;I have a value generated from complex circle geometry. That value is about 1 for the bottom edge of the circle and 0.2 for the top edge. I don't have the luxury of changing that variable. However, I can make an additional variable calculation from it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wish to apply an effect to the circle where the top edge has a value of 5 and the bottom edge has a value of 0 - see diagram to illustrate - the blue values are what I need to achieve and the red values are all I have to work with:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/14TDx.png&quot; alt=&quot;Calculation of reversing values&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do I make a calculation that given the top edge is 0.2 the answer is 5, and given that the bottom edge is 1 then the answer is 0 (or close enough to it). I apologise if I've not explained myself properly. The smallest number has to achieve the greatest output and the largest number has to achieve the the smallest output&lt;/p&gt;&#xA;" OwnerUserId="7115" LastEditorUserId="7115" LastEditorDisplayName="" LastEditDate="2011-02-15T15:39:34.280" LastActivityDate="2011-02-15T15:39:34.280" Title="Way to reverse numbers in calculation?" Tags="&lt;calculus&gt;&lt;geometry&gt;" AnswerCount="1" />
  <row Id="22184" PostTypeId="2" ParentId="9416" CreationDate="2011-02-15T15:37:48.033" Score="2" ViewCount="" Body="&lt;p&gt;Generally you need to apply a &lt;a href=&quot;http://en.wikipedia.org/wiki/Window_function&quot; rel=&quot;nofollow&quot;&gt;windowing function&lt;/a&gt; to your signal before performing the FFT, to get clean spikes for the frequency components (without &quot;skirts&quot;).  (The only exception is if your frequency components &lt;a href=&quot;https://gist.github.com/236567&quot; rel=&quot;nofollow&quot;&gt;fit &lt;em&gt;exactly&lt;/em&gt; into the length of the FFT&lt;/a&gt;, so that they each fill exactly one bin.)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you use a Gaussian windowing function, and then fit a parabola to the highest three points in your FFT, you can get a theoretically exact result for the frequency:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://ccrma.stanford.edu/~jos/sasp/Quadratic_Interpolation_Spectral_Peaks.html&quot; rel=&quot;nofollow&quot;&gt;Quadratic Interpolation of Spectral Peaks&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2206" LastActivityDate="2011-02-15T15:37:48.033" />
  <row Id="22185" PostTypeId="2" ParentId="22183" CreationDate="2011-02-15T15:38:00.483" Score="2" ViewCount="" Body="&lt;p&gt;$$\text{blue}=6.25(1-\text{red})$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Plotting the three pairs $(\text{red},\text{blue})$ from your diagram shows that all three points lie on a single line.  From any two points, the slope (change in blue over change in red) can be computed ($-6.25$).  An equation for the line is then $\text{blue}-\text{known blue}=\text{slope}(\text{red}-\text{known red})$, where $(\text{known red},\text{known blue})$ is any one of the known points.  Using $(1,0)$, $\text{blue}-0=-6.25(\text{red}-1)$, which can be rewritten as I did above.&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2011-02-15T15:38:00.483" CommentCount="1" />
  <row Id="22186" PostTypeId="2" ParentId="22056" CreationDate="2011-02-15T15:43:16.830" Score="0" ViewCount="" Body="&lt;p&gt;I can't get it, i have do the following.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Point 1(current position): 36.837147 -2.469118&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Point 2: 36.837437 -2.469040&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Point 3: 36.837452 -2.469162&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Point 4: 36.837547 -2.469532&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The angle given by the iPhone is 155º.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;What i do is:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;float x1=point1.longitude; &#xA;&#xA;float y1=point1.latitude;&#xA;float x2=point2.longitude;&#xA;float y2=point2.latitude;&#xA;&#xA;float x = x2 - x1;&#xA;float y = y2 - y1;&#xA;&#xA;float angle = atan2(y, x);&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Then the result is: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Angle between:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P1, P2=1.307330&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P1, P3=1.713568&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P1, P4=2.372586&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;But... Now you say to substract from the phone angle the obtained angle.For the three cases, it is something like 154.xx.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What i'm doing wrong? The true situation says that the P2 and P3 are inside 20º, but not the P3.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank You!!!&lt;/p&gt;&#xA;" OwnerUserId="7081" LastActivityDate="2011-02-15T15:43:16.830" CommentCount="4" />
  <row Id="22187" PostTypeId="1" AcceptedAnswerId="22188" CreationDate="2011-02-15T15:50:23.740" Score="1" ViewCount="53" Body="&lt;p&gt;I have a bus that does &lt;code&gt;A-------X------------------------B&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;it goes at 10 hours from A, having:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Speed at A &lt;code&gt;sA = 10 m/s&lt;/code&gt;&lt;br&gt;&#xA;constant acceleration &lt;code&gt;a = -0,5 m/s2&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At what hour it will be at X, knowing the distance AX &lt;code&gt;dAX = 50m&lt;/code&gt; ?&lt;/p&gt;&#xA;" OwnerUserId="4050" LastEditorUserId="4050" LastEditorDisplayName="" LastEditDate="2011-02-15T16:37:00.877" LastActivityDate="2011-02-15T16:37:00.877" Title="Knowing the time of arrival to point X" Tags="&lt;homework&gt;&lt;recreational-mathematics&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="22188" PostTypeId="2" ParentId="22187" CreationDate="2011-02-15T16:15:18.147" Score="2" ViewCount="" Body="&lt;p&gt;Being $s = \frac{a}{2} t^2 + v t + s_0$ you might be able to work something out.&lt;/p&gt;&#xA;" OwnerUserId="5870" LastActivityDate="2011-02-15T16:15:18.147" />
  <row Id="22189" PostTypeId="2" ParentId="22179" CreationDate="2011-02-15T16:27:35.333" Score="5" ViewCount="" Body="&lt;p&gt;Let's ignore the axiom that requires $1\mathbf{v}=\mathbf{v}$ to hold for a moment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's call an element $\alpha$ of the field with the property that $\alpha\mathbf{v}=\mathbf{v}$ for every $\mathbf{v}$ an &quot;identity&quot;. First, let's note that we really want to have at most one of them, not multiple ones. Otherwise, we could just define scalar multiplcation by &quot;do nothing&quot;, and we don't &lt;em&gt;really&lt;/em&gt; have a vector space structure (the scalars are just dummys, not doing anything). And also, because otherwise we're going to lose the notion of linear independence (which is very important later on): if $\alpha$ and $\beta$ both &quot;work&quot; as the identity and $\alpha\neq\beta$, then for every vector you are going to have &#xA;$$(\alpha-\beta)\mathbf{v} = \alpha\mathbf{v}-\beta\mathbf{v}=\mathbf{v}-\mathbf{v}=\mathbf{0},$$ &#xA;but with $\alpha\neq\beta$. This will mean that &lt;em&gt;no&lt;/em&gt; collection of vectors is linearly independent. So we don't want more than one identity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One axiom is the associativity of scalar multiplication:&#xA;$$(\alpha\beta)\mathbf{v} = \alpha(\beta\mathbf{v}).$$&#xA;Why can't you define the &quot;identity&quot; to be $3$? Because then you would have $3\mathbf{v}=\mathbf{v}$, hence &#xA;$$9\mathbf{v} = 3(3\mathbf{v}) = 3\mathbf{v}=\mathbf{v},$$ &#xA;so that now $9$ is &lt;em&gt;also&lt;/em&gt; an identity, even though $3\neq 9$, and we already said that is not a good idea.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, whatever the identity is, you want it to equal its square, for precisely this reason. Otherwise, you're going to have &lt;em&gt;lots&lt;/em&gt; of identities, and hence lots of troubles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But in a field, there are only two solutions to $x^2=x$: namely, $x=0$ and $x=1$. So the &quot;identity&quot; has to be either $0$ or $1$. But the fact that $0\mathbf{v}=\mathbf{0}$ holds follows without the identity axiom (&quot;$1\mathbf{v}=\mathbf{v}$&quot;), so unless &lt;em&gt;every&lt;/em&gt; vector is $\mathbf{0}$, you cannot have $0$ be the identity, because then we would have&#xA;$$\mathbf{v} = 0\mathbf{v}=\mathbf{0}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, really, the only reasonable choice for identity is $1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, then, why do we make it an axiom? Because otherwise, we can define scalar multiplication by &quot;any scalar times any vector is $\mathbf{0}$&quot;. This definition satisfies &lt;strong&gt;all&lt;/strong&gt; the axioms of a vector space &lt;em&gt;except&lt;/em&gt; the existence of the identity, but it's a &quot;silly&quot; vector space (and one where the notion of linear independence is also ultimately completely screwed up). So we want to exclude it. And the way to exclude it is precisely by requiring that there be an element of the field that does &lt;em&gt;not&lt;/em&gt; map everything to $\mathbf{0}$ through scalar multiplication, so it makes sense to kill both birds with one stone and ask for $1$ to not send everyone to $\mathbf{0}$ by multiplying each vector to itself.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2011-02-15T16:27:35.333" CommentCount="1" />
  <row Id="22190" PostTypeId="1" AcceptedAnswerId="38166" CreationDate="2011-02-15T16:36:18.677" Score="6" ViewCount="206" Body="&lt;p&gt;I've read the section &quot;Serre duality&quot; in Hartshorne's book and have several questions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) In Remark 7.1.1 it is claimed that on $X = \mathbb{P}^n$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\alpha = \frac{x_0^n}{x_1 \cdot ... \cdot x_n} d(\frac{x_1}{x_0}) \wedge ... \wedge d(\frac{x_n}{x_0})$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is a Cech cocycle which generates $H^n(X,\omega_X)$ (I have checked this) and that this is independent from the choice of the basis of $\mathbb{P}^n$. Well it's easy to check that it's independent from the order of $x_0,...,x_n$. But it does not seem to be invariant under other automorphisms of $\mathbb{P}^n$, which are typically given by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x_0 \mapsto x_0 + \lambda x_1 , x_1 \mapsto x_1, ..., x_n \mapsto x_n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or am I wrong? What is the fix group in $\text{Aut}(\mathbb{P}^n) = PGL(n)$ of $\omega_X$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) In Remark 7.12.1 it is shown that Serre duality implies that for a smooth projective variety $X$ of dimension $n$ we have that $H^n(X,\omega_X) \cong k$. It's quite fascinating to get such a concrete result with this abstract machinery of Ext-functors. What can be said about this isomorphism? Is there a &quot;canonical&quot; one? Is it possible to distinguish a generator? What examples do you know?&lt;/p&gt;&#xA;" OwnerUserId="1650" LastActivityDate="2011-05-10T16:09:18.467" Title="Questions about Serre duality" Tags="&lt;algebraic-geometry&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="22191" PostTypeId="1" CreationDate="2011-02-15T17:21:35.003" Score="1" ViewCount="120" Body="&lt;p&gt;Here's the question, from &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0071508619&quot; rel=&quot;nofollow&quot;&gt;my $10 mathbook&lt;/a&gt; that everyone is growing to love and hate:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;A saltwater solution initially contains 5lb of salt in 10 gal of fluid.  If water flows in at 0.5 gal/min and the mixture flows out at the same rate, how much salt is present after 20 mins?&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I get that this is a differential equation problem, and that the quantity we are observing is the &lt;strong&gt;concentration&lt;/strong&gt; of the salt in the water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rate of change of concentration ($\frac{dSolute}{dt}$) is proportional to:  the current concentration, and the rate at which the fluid is (being replaced) by fresh water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And so I of course have $y =y_0 e^{kt}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do I think about this problem?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a solution given, which is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ \frac{dS}{dt} = -\frac{1}{2}(\frac{S}{10})$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At $t=20$, $S=5e^{-1}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can guess where the constants came from ($-\frac{1}{2}$, 10) but I don't see what's going on here, especially with the $S=5e^{-1}$ quantity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Working backwards, it says&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\mbox{change in concentration}}{\mbox{unit of time}} = \mbox{(rate at which we are losing solution)}\left(\frac{\mbox{current salt concentration}}{\mbox{volume of water}}\right).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is that the formula?&lt;/p&gt;&#xA;" OwnerUserId="1069" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-15T21:20:10.030" LastActivityDate="2011-02-15T21:20:10.030" Title="Salty water question" Tags="&lt;differential-equations&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="22192" PostTypeId="1" AcceptedAnswerId="22353" CreationDate="2011-02-15T17:26:38.587" Score="4" ViewCount="108" Body="&lt;p&gt;I don't have much of any awareness about the representation theory of non-compact Lie groups but I bumped into it for my work. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there some idea that the representations of a non-compact group are labeled by those of its maximal compact subgroup? If yes then I would like to know of explanations for the above and of references from where I can pick this up.  &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;The conformal group of $3+1$ dimensional space-time is $SO(4,2)$ and apparently any representation of it can be written as a direct sum over representations of $SO(4) \times SO(2)$. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The N=2 superconformal group for $2+1$ dimensions is $SO(3,2)\times SO(2)$ and its maximal compact subgroup $SO(2) \times SO(3) \times SO(2)$ ? If yes then I would like to know how can this be proven. (In physics contexts these two $SO(2)$ factors are distinguished by physical generators of different meanings.)  &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I would like to know of the general framework in which the above fits in. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To quote from a paper a typical argument where such a thing seems to get used,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Any irreducible representation of the superconformal algebra may be decomposed into a finite number of distinct irreducible representations of the conformal algebra...which are in turn labeled by their own primary states...hence the state content of an irreducible representation of the superconformal algebra is completely specified by the quantum numbers of its conformal primaries&quot; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would be very glad if someone can also give expository references or explanations specific to the above argument. &lt;/p&gt;&#xA;" OwnerUserId="965" LastEditorUserId="274" LastEditorDisplayName="" LastEditDate="2011-02-16T15:00:30.843" LastActivityDate="2011-02-16T15:00:30.843" Title="Representations of a non-compact group are labeled by its maximal compact subgroup?" Tags="&lt;physics&gt;&lt;lie-groups&gt;&lt;representation-theory&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="22193" PostTypeId="2" ParentId="22191" CreationDate="2011-02-15T17:29:37.087" Score="1" ViewCount="" Body="&lt;p&gt;The differential equation is $\frac{dS}{dt} = -\frac{S}{20}$, which has solution $S=Ce^{\frac{-t}{20}}$.  S(0) is given as 5, so $S(t)=5e^{\frac{-t}{20}}$  At t=20, this is $5e^{-1}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Added:  I just took the equation from your post.  In a small amount of time, $\delta t$ you have $\frac{1}{2}\delta t$ pure water added and the same quantity of salty water removed.  This represents $\frac{1}{20}\delta t$ of the volume of the salty water, so you should remove that much salt.  The change in salt is $-\frac{1}{20}S\delta t$.  This gives $\frac{dS}{dt} = -\frac{S}{20}$&lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2011-02-15T17:43:45.120" LastActivityDate="2011-02-15T17:43:45.120" CommentCount="1" />
  <row Id="22194" PostTypeId="1" CreationDate="2011-02-15T17:31:53.590" Score="0" ViewCount="81" Body="&lt;p&gt;3 colors of balls, blue,red and yellow, there are a large number of each color of balls which are distributed evenly. Ask what's the probability to get 3 balls which are all red color? What is probability to get 3 balls which are red blue and yellow?&#xA;Thanks&lt;/p&gt;&#xA;" OwnerUserId="7122" LastActivityDate="2011-02-15T18:12:26.693" Title="probability question" Tags="&lt;probability&gt;" AnswerCount="1" CommentCount="7" />
  <row Id="22195" PostTypeId="1" CreationDate="2011-02-15T17:41:03.583" Score="5" ViewCount="528" Body="&lt;p&gt;Let $A$ be $n\times n$ matrix. Prove that if $A^2=\mathbf{0}$ then $A$ is not invertible.&lt;/p&gt;&#xA;" OwnerUserId="7123" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-15T19:53:28.423" LastActivityDate="2011-02-16T08:19:12.003" Title="Prove that if $A^2=0$ then $A$ is not invertible." Tags="&lt;linear-algebra&gt;" AnswerCount="8" CommentCount="5" />
  <row Id="22196" PostTypeId="2" ParentId="22195" CreationDate="2011-02-15T17:43:54.263" Score="6" ViewCount="" Body="&lt;p&gt;$0=\det (A^2) = (\det A)^2$. Hence $\det A=0$.&lt;/p&gt;&#xA;" OwnerUserId="589" LastActivityDate="2011-02-15T17:43:54.263" CommentCount="1" />
  <row Id="22197" PostTypeId="2" ParentId="22195" CreationDate="2011-02-15T17:47:27.693" Score="3" ViewCount="" Body="&lt;p&gt;If $A$ is invertible then $I=AB$ for some $B$. Then $A=AI=A^2B=0$, and so $A=0$, but $0$ is not invertible.&lt;/p&gt;&#xA;" OwnerUserId="589" LastActivityDate="2011-02-15T17:47:27.693" />
  <row Id="22198" PostTypeId="2" ParentId="22195" CreationDate="2011-02-15T17:49:19.403" Score="9" ViewCount="" Body="&lt;p&gt;This is simply a special case of the fact that, in &lt;em&gt;every&lt;/em&gt; ring, a unit is not a zero-divisor, $\:$ since $\rm\ A&amp;#39; A = 1,\ AB = 0\ \Rightarrow\ B = (A&amp;#39;A)B = A&amp;#39;(AB) = 0\:.\: $ The problem is the special case $\rm\ B = A\:.$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Informally: a ring collapses to the zero ring if one adjoins an inverse of $0$ (or a zero-divisor).&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2011-02-15T21:11:49.293" LastActivityDate="2011-02-15T21:11:49.293" />
  <row Id="22199" PostTypeId="2" ParentId="22195" CreationDate="2011-02-15T17:49:33.900" Score="10" ViewCount="" Body="&lt;p&gt;With no determinant. Assume that $A$ is invertible on the right, thus there exists $B$ such that $AB=I$, hence $A^2B=A$; since $A^2=0$, $A^2B=0$, hence $A=0$. But the matrix $0$ is not invertible on the right, hence the hypothesis was false: this proves that $A$ is not invertible on the right. Likewise if one assumes that $A$ is invertible on the left, that is, that there exists $B$ such that $BA=I$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As explained in a comment, in the infinite dimensional case, the fact that $A$ is not injective (as proved by Andres) prevents $A$ to be invertible on the left but not on the right.&lt;/p&gt;&#xA;" OwnerUserId="6179" LastEditorUserId="6179" LastEditorDisplayName="" LastEditDate="2011-02-16T08:19:12.003" LastActivityDate="2011-02-16T08:19:12.003" />
  <row Id="22200" PostTypeId="1" CreationDate="2011-02-15T18:02:36.410" Score="4" ViewCount="116" Body="&lt;p&gt;What are some examples of functions on the Power Set of the Reals? Is this an abuse of terminology - functions on the reals can be thought of as functions on the power set of the naturals with a specific ordering. I was hoping someone would kindly refer me to a text or article where explicit (not necessarily 'useful') examples of functions with the domain P(R) are given; or if this is confused idea why there is nothing to it. Thanks!&lt;/p&gt;&#xA;" OwnerUserId="7124" LastActivityDate="2011-02-15T18:35:35.337" Title="Functions on P(R) - are there examples?" Tags="&lt;analysis&gt;&lt;set-theory&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="22201" PostTypeId="2" ParentId="22195" CreationDate="2011-02-15T18:06:09.890" Score="1" ViewCount="" Body="&lt;p&gt;If $A,B \in \mathbb{R}^{n \times n}$, then $\text{rank}(A) + \text{rank}(B) - n \leq \text{rank}(AB) \leq \text{min}(\text{rank}(A),\text{rank}(B))$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here $A=B$ and $A^2 =0$. Hence $\text{rank}(A^2) = 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence if $\text{rank}(A) = k$, then we have $$k+k - n\leq 0 \leq k$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From which we get that $0 \leq k \leq \frac{n}{2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So this implies not only is the determinant $0$ and there is no inverse but also there cannot be more than $\frac{n}{2}$ linearly independent rows/columns of the matrix $A$.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastActivityDate="2011-02-15T18:06:09.890" />
  <row Id="22202" PostTypeId="2" ParentId="22194" CreationDate="2011-02-15T18:12:26.693" Score="1" ViewCount="" Body="&lt;p&gt;Hint 1:  If the chance of pulling a red ball is 1/3, what is the chance of doing it three times in a row?&#xA;Hint 2: What are the chances of pulling specifically a red ball first, a blue ball second, and a yellow ball third?  How many different ways are there to get one of each in three draws?&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-02-15T18:12:26.693" CommentCount="3" />
  <row Id="22203" PostTypeId="2" ParentId="22200" CreationDate="2011-02-15T18:22:10.597" Score="8" ViewCount="" Body="&lt;p&gt;A function on $\mathcal{P}(\mathbb{R})$ essentially means a &quot;rule&quot; of assigning to each subset of $\mathbb{R}$ an element of some set $S$, the codomain.  So e.g., there is the identity function $\mathcal{P}(\mathbb{R})\to\mathcal{P}(\mathbb{R})$ that sends a set $A$ to itself.  There is the $\sup$ map from $\mathcal{P}(\mathbb{R})$ to the extended reals $[-\infty,\infty]$ that sends $A\subseteq\mathbb{R}$ to $\sup A$, and similarly with $\inf$.  You could also define functions like $f(A)=1$ if $A$ is open and $f(A)=0$ if $A$ is not open, or other such maps indicating topological properties of subsets of $\mathbb{R}$.  You could define the function $c:\mathcal{P}(\mathbb{R})\to \{0,1,2,\ldots,2^{\aleph_0}\}$ such that $c(A)$ is the cardinality of $A$.  Or $C:\mathcal{P}(\mathbb{R})\to \{0,1,2,\ldots,2^{\aleph_0}\}$ such that $C(A)$ is the cardinality of the set of connected components of $A$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I see that Zhen Lin has indicated a couple of other useful examples in the comments.  The complement in $\mathbb{R}$ defines a bijection on $\mathcal{P}(\mathbb{R})$.  Each outer measure on $\mathcal{P}(\mathbb{R})$ defines a function from $\mathcal{P}(\mathbb{R})$ to $[0,\infty]$.  The closure and interior maps are other functions from $\mathcal{P}(\mathbb{R})$ to itself, mapping onto the set of closed and open subsets of $\mathbb{R}$ respectively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So yes, there are lots of explicit examples, but I don't know exactly what you're looking for.  The set of functions from $\mathcal{P}(\mathbb{R})$ to any fixed set $S$ is $S^{\mathcal{P}(\mathbb{R})}$, with cardinality $\displaystyle{|S|^{2^\mathfrak{c}}}$, which is at least $\displaystyle{2^{2^\mathfrak{c}}=2^{2^{2^{\aleph_0}}}}$ if $S$ has more than $1$ element.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2011-02-15T18:22:10.597" CommentCount="2" />
  <row Id="22204" PostTypeId="2" ParentId="22200" CreationDate="2011-02-15T18:35:35.337" Score="6" ViewCount="" Body="&lt;p&gt;Every set can be a domain of a function (in fact, a proper class can also be the domain of a function if one is careful enough). In particular the powerset of the real numbers can be.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some examples that might be used are measures in measure theory, granted the Lebesgue Measure is not defined for every subset of the real numbers (unless the axiom of choice is not assumed) but you can define the outer measure which is defined on $P(\mathbb{R})$, or some other measure that is not limited by the requirements of the Borel/Lebesgue measures.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other examples are negation, union and intersection (functions in two variables), as well symmetric difference.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming the axiom of choice, the Stone-Cech compactification of the natural numbers with the discrete topology is of cardinality $\beth_2$ (i.e. $|P(\mathbb{R})|$) and you can look at it as if you are assigning each subset of the real numbers an ultrafilter over the natural numbers.&lt;/p&gt;&#xA;" OwnerUserId="622" LastActivityDate="2011-02-15T18:35:35.337" />
  <row Id="22205" PostTypeId="1" AcceptedAnswerId="22212" CreationDate="2011-02-15T18:40:57.810" Score="1" ViewCount="76" Body="&lt;p&gt;$(X_n), n \in \mathbb{N}$ is a stochastic process.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I saw in one definition of Martingale that&#xA;$$E [X_{n+1} |X_0 , X_1 , . . . , X_n ] = X_n \quad a.s., \forall n \geq 0.$$&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;I understand what &quot;almost surely&quot; itself means. But I was wondering how to interpret the above usage of &quot;almost surely&quot; in the definition of&#xA;Martingale? Are $X_0 , X_1 , . . . , X_n$ seen as random variables or their given deterministic values?&lt;/li&gt;&#xA;&lt;li&gt;Plus, I don't see the definition of&#xA;Martingale on &lt;a href=&quot;http://en.wikipedia.org/wiki/Martingale_%28probability_theory%29#Definitions&quot; rel=&quot;nofollow&quot;&gt;its Wikipedia article&lt;/a&gt; relies on &quot;almost&#xA;surely&quot;. Is it really required?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Thanks and regards!&lt;/p&gt;&#xA;" OwnerUserId="1281" LastEditorUserId="1281" LastEditorDisplayName="" LastEditDate="2011-02-15T18:57:14.063" LastActivityDate="2011-02-15T19:11:59.957" Title="Almost surely in definition of Martingale" Tags="&lt;probability-theory&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="22206" PostTypeId="2" ParentId="22195" CreationDate="2011-02-15T18:52:17.417" Score="10" ViewCount="" Body="&lt;p&gt;I think that for a problem like this, using determinants (or invoking the rank-nullity theorem) ends up hiding what really goes on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instead, I suggest that you simply argue about this directly: $A^2=0$ means that $A^2v=0$ for any $v$. Then $A(Av)=0$ for any $v$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If for some $v$ we have $w=Av\ne 0$, then $A$ is not injective, because $$Aw=A(Av)=A^2v=0,$$ and we are done. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Else, $Av=0$ for all $v$, and then $A=0$, which of course is not invertible.&lt;/p&gt;&#xA;" OwnerUserId="462" LastActivityDate="2011-02-15T18:52:17.417" CommentCount="4" />
  <row Id="22207" PostTypeId="1" CreationDate="2011-02-15T18:57:54.717" Score="0" ViewCount="109" Body="&lt;p&gt;1) Let $p$ be an odd prime, $n$ an integer such that $n \geq 3p$ and $a$ a $2p$-cycle in $S_n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $b$ be a $p$-cycle in $S_n$ and assume that $a$ and $b$ are disjoint. Suppose $K$ is a subgroup of $S_n$ that contains $a$ and $b$. Prove that $K$ is not cyclic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(we may assume $a = (1\;\: 2\;\: 3\;\: \dotsb \;\:2p)$ and $b = (2p+1\;\: 2p+2\;\: \dotsb \;\:3p)$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Let $H_n = \lbrace a \in S_n \;|\; |a|$ is odd$\rbrace$. For which values of $n\geq 2$ is $H_n$ a subgroup of $S_n$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have trouble starting number 1 because of its ambiguous definition of $H$. Let's start by a proof by contradiction approach and assume that $H$ is cyclic. Then do we define $H = \langle ab \rangle = \{e,ab,a^2b^2,a^3b^3,\dotsc\}$ and so forth? Or are they defined as $H = \langle a \rangle\langle b \rangle$? Does $H$ contain any other elements besides $a$ and $b$? If so how do we treat that?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For number 2 let's first note that every element in $H_n$ is in $A_n$ (proved in other part). Then $H_n$ is a &quot;subset&quot; of $A_n$. I suppose I can go about to prove this by saying every element of $H_n$ is in $A_n$ and get something out of that, but I can't quite piece it all together.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance.&lt;/p&gt;&#xA;" OwnerUserId="6525" LastEditorUserId="1269" LastEditorDisplayName="" LastEditDate="2011-02-15T19:17:51.087" LastActivityDate="2011-02-15T19:21:25.157" Title="Permutation group proofs" Tags="&lt;homework&gt;&lt;group-theory&gt;&lt;permutations&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" />
  <row Id="22208" PostTypeId="1" AcceptedAnswerId="22215" CreationDate="2011-02-15T19:01:45.803" Score="1" ViewCount="48" Body="&lt;p&gt;Hello &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have an ODE of the form $y&amp;#39;&amp;#39;(x)=F(x,y&amp;#39;)$ which has the initial conditions, $y(\mu)=\mu$ and $y&amp;#39;(\mu)=1$. Now I have seen that an equation which is equidimensional in x, can be made autonomous by the change of variable $x=e^t$, ie now we must solve an ODE for the unknown $z(t)$ ($=y(x)$). But now what happens to the initial conditions?? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I often see change of variables of some sort in books. But no one really talks about what happens to the initial conditions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I guess people dont need to transform the initial conditions, since if a geneneral solution $z(t)$ to the transformed (and simplified) equation can be found. Then we can resolve any arbitrary constants, using the fact $z(ln(t))=y(x)$. But I'm trying to solve my equation through power series techniques so I dont think this method applies. I need the conditions in advance to compute the coefficients of the terms in the power series. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the chain rule may be applied to evaluate one of the conditions. That is by application of the chain rule we have $\frac{dy}{dx}=\frac{dy}{dt}\frac{1}{t}$, so $\frac{dy}{dt}(ln(\mu))=\mu$ since $\frac{dy}{dx}(\mu)=1$. But what about the first condition? how is that transformed?  &lt;/p&gt;&#xA;" OwnerUserId="6386" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2011-02-15T20:00:59.540" LastActivityDate="2011-02-15T20:00:59.540" Title="How are the initial conditions transformed with a change of variable?" Tags="&lt;calculus&gt;&lt;differential-equations&gt;" AnswerCount="1" />
  <row Id="22209" PostTypeId="2" ParentId="22115" CreationDate="2011-02-15T19:02:19.250" Score="2" ViewCount="" Body="&lt;p&gt;I think the answer is negative, you can't derive $x^d \ne y $ from $x^{f(n)/g(n)}\not \equiv y \pmod{4} $ for every $n$ and &#xA;$$\lim_{n\to \infty} \frac{f(n)}{g(n)} \to d $$&#xA;Look at this example&#xA;$$ x= 2, y = 3, d = \frac{\log 3}{\log 2} $$&#xA;as $d$ is irrational (see Hardy and Wright p. 162), we can find rational numbers $f(n)/g(n)$ with odd numerator and denominator that converge to $d$, (for example if $n$ is odd and&#xA;  $$ \frac{a_n}{n} &amp;lt; \frac{\log 3}{\log 2} &amp;lt; \frac{a_n+1}{n} $$&#xA;then pick $g(n) = n$ and $f(n) = a_n$ if $a$ is odd or $f(n)=a_n+1$ otherwise). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have obviously &#xA;$$ 2^{f(n)} \not\equiv 3^{g(n)} \pmod{4} $$&#xA;because the left hand side is even and the right hand side is odd however&#xA;$$ 2^d = 3$$&lt;/p&gt;&#xA;" OwnerUserId="3274" LastActivityDate="2011-02-15T19:02:19.250" CommentCount="2" />
  <row Id="22210" PostTypeId="2" ParentId="22190" CreationDate="2011-02-15T19:10:56.763" Score="0" ViewCount="" Body="&lt;p&gt;Here are the answers to your questions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) $PGL(n)$ acts trivially on $H^n(\mathbb{P}^n, \omega)$. This will follow from my answer to (2), which is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For any $n$ dimensional projective variety $X$, the isomorphism $\mathrm{Tr}: H^n(X, \omega_X) \to k$ is completely canonical. More precisely, if $g$ is an automorphism of $X$, and $\eta$ a cohomology class, then $\mathrm{Tr}(g^* \eta) = \mathrm{Tr}(\eta)$. More generally, if $g : X \to Y$ is a degree $d$ map, then $\mathrm{Tr}_Y(g^* \eta) = d \cdot \mathrm{Tr}_X(\eta)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried to write up a short proof, and failed miserably. I hope someone else will succeed.&lt;/p&gt;&#xA;" OwnerUserId="448" LastActivityDate="2011-02-15T19:10:56.763" CommentCount="10" />
  <row Id="22211" PostTypeId="2" ParentId="22205" CreationDate="2011-02-15T19:11:12.077" Score="3" ViewCount="" Body="&lt;p&gt;A careful look at the definition of $E [X_{n+1} |X_0 , X_1 , . . . , X_n ]$ on the left hand side of the equation shows that conditional expectation is defined only &lt;em&gt;almost surely&lt;/em&gt;. This is true for any conditional expectation $E[X | {\cal F}]$, whether conditioned on a $\sigma$-algebra or a set of random variables. This has nothing to do with martingales, per se. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Generally, the martingale equation only makes sense &lt;em&gt;almost surely&lt;/em&gt;, even though the random variable $X_n$ on the right hand side is defined everywhere on the probability space.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, if the $\sigma$-algebra generated by $X_0,X_1,\dots,X_n$ contains no non-trivial $P$-null sets, then the left hand side is also defined everywhere on the probability space, and the martingale identity must hold identically.  &lt;/p&gt;&#xA;" OwnerUserId="940" LastActivityDate="2011-02-15T19:11:12.077" />
  <row Id="22212" PostTypeId="2" ParentId="22205" CreationDate="2011-02-15T19:11:59.957" Score="3" ViewCount="" Body="&lt;p&gt;This is because conditional expectations can only be defined up to almost sure events. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Recall that $Z=E[X|Y]$ is defined as any random variable $Z$ such that (1) $Z$ is $\sigma(Y)$-measurable, and (2) $E[Z\varphi(Y)]=E[X\varphi(Y)]$ for every bounded measurable function $\varphi$. Hence, if $Z$ fulfills (1) and (2) and $P(Z&amp;#39;=Z)=1$, then $Z&amp;#39;$ fulfills (1) and (2) as well and $Z$ and $Z&amp;#39;$ have an equal right to be called $E[X|Y]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To sum up, $E[X|Y]$ may be seen as a whole class of random variables, each one equal to any other &lt;em&gt;up to a null event&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="6179" LastActivityDate="2011-02-15T19:11:59.957" />
  <row Id="22213" PostTypeId="2" ParentId="22207" CreationDate="2011-02-15T19:21:25.157" Score="2" ViewCount="" Body="&lt;p&gt;For (1), since every subgroup of a cyclic group is itself cyclic, it is enough to show that the subgroup $K_0$ of $S_n$ generated by $a$ and $b$ is not cyclic. You &lt;em&gt;should&lt;/em&gt; do some computations with a small example now, to see that $K_0$ is not cyclic.&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2011-02-15T19:21:25.157" />
  <row Id="22214" PostTypeId="2" ParentId="22195" CreationDate="2011-02-15T19:26:37.920" Score="0" ViewCount="" Body="&lt;p&gt;Assume there is an inverse denoted A&lt;sup&gt;-1&lt;/sup&gt;. Then you have AA = &lt;code&gt;0&lt;/code&gt; and then multiply it by your inverse A&lt;sup&gt;-1&lt;/sup&gt; you have AAA&lt;sup&gt;-1&lt;/sup&gt; = 0A&lt;sup&gt;-1&lt;/sup&gt; = A = &lt;code&gt;0&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now given the definition of &lt;code&gt;0&lt;/code&gt; :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given a Matrix M and the&lt;code&gt;0&lt;/code&gt;vector M&lt;code&gt;0&lt;/code&gt;=&lt;code&gt;0&lt;/code&gt;=&lt;code&gt;0&lt;/code&gt;M.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore by definition of&lt;code&gt;0&lt;/code&gt;it cannot have an inverse.&lt;/p&gt;&#xA;" OwnerUserId="7126" LastActivityDate="2011-02-15T19:26:37.920" />
  <row Id="22215" PostTypeId="2" ParentId="22208" CreationDate="2011-02-15T19:49:00.113" Score="0" ViewCount="" Body="&lt;p&gt;It is much like changing variables in definite integrals.  You can either change the old conditions to the new variables, or change the solution back to the old variables and use the original conditions.  Say your initial conditions are $y(1)=\mu, y&amp;#39;(1)=1$ (you shouldn't have a variable as the argument of $y$ for initial conditions-it is given).  Then if you set $x=e^t$ you have $y(t=0)=\mu, \frac{dy}{dt}(t=0)=\frac{dy}{dx}\frac{dx}{dt}(t=0)=1\cdot 1=1$ if you consider $y$ as a function of $t$&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-02-15T19:49:00.113" />
  <row Id="22216" PostTypeId="2" ParentId="21388" CreationDate="2011-02-15T20:22:57.433" Score="2" ViewCount="" Body="&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/y4oDa.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="130" LastActivityDate="2011-02-15T20:22:57.433" CommentCount="2" />
  <row Id="22217" PostTypeId="2" ParentId="22148" CreationDate="2011-02-15T20:24:06.077" Score="1" ViewCount="" Body="&lt;p&gt;Another way of looking at this is that you are given that $f(z) - z = 0$ for $z = 1$. Then use the condition that $f(z/2) = {1 \over 2}f(z)$ to inductively show that $f(z) - z$ also has zeroes at $z = 2^{-n}$ for all positive integers $n$. Thus the zeroes of $f(z) - z$ have an accumulation point at $z = 0$, which is only possible if $f(z) - z = 0$ for all $z$ since nonzero analytic functions can only have isolated zeroes. &lt;/p&gt;&#xA;" OwnerUserId="3035" LastActivityDate="2011-02-15T20:24:06.077" CommentCount="0" />
  <row Id="22219" PostTypeId="1" CreationDate="2011-02-15T20:57:02.380" Score="0" ViewCount="87" Body="&lt;p&gt;Apologies if this is too basic, but given a permutation matrix $M$, is there any parameter or formula based on $M$ that gives the disjoint cycle decomposition, or at least the conjugacy class, of the corresponding permutation?&lt;/p&gt;&#xA;" OwnerUserId="689" LastActivityDate="2011-02-15T21:07:41.060" Title="Conjugacy class of a permutation from its matrix representation" Tags="&lt;matrices&gt;&lt;permutations&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="22220" PostTypeId="2" ParentId="22159" CreationDate="2011-02-15T20:58:42.647" Score="4" ViewCount="" Body="&lt;p&gt;This is solved &lt;a href=&quot;http://oeis.org/A005748/a005748.PDF&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; using Pólya enumeration theory. For the square case ($n=m$), see &lt;a href=&quot;http://oeis.org/A002724&quot; rel=&quot;nofollow&quot;&gt;this sequence&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Comment: I found these by searching for $1,2,7$ in the OEIS.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2011-02-15T20:58:42.647" />
  <row Id="22221" PostTypeId="2" ParentId="22219" CreationDate="2011-02-15T21:07:41.060" Score="5" ViewCount="" Body="&lt;p&gt;&lt;em&gt;I don't really see what exactly you expect, as «parameter or formula based on $M$» is a pretty vague description. Here is one option...&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can compute the number of cycles of a specific length from the multiplicities of other eigenvalues---which are all roots of unity. For example, the multiplicity of $1$ as an eigenvalue is the number of cycles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More generally, if we call $c_\ell$ the number of cycles of length $\ell$ in the permutation, and $\mu_n$ the multiplicity of $e^{2\pi i/n}$ as an eigenvalue of the matrix, then&#xA;$$\mu_n = \sum_{n\mid\ell}c_\ell.$$ This relation be inverted using &lt;a href=&quot;http://en.wikipedia.org/wiki/M%C3%B6bius_inversion_formula&quot; rel=&quot;nofollow&quot;&gt;Moebius inversion&lt;/a&gt; to a formula exactly expressing the $c_\ell$ in terms of the multiplicities $\mu_n$.&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2011-02-15T21:07:41.060" CommentCount="6" />
  <row Id="22222" PostTypeId="2" ParentId="22063" CreationDate="2011-02-15T21:12:07.330" Score="3" ViewCount="" Body="&lt;p&gt;First, let me provide the context: Given are a unitary operator $U \in L(H)$ (actually $\|U\| \leq 1$ suffices) on a Hilbert space and $P$ is the orthogonal projection onto the subspace $V = \{x \in H\,:\, Ux=x\}$ of fixed vectors of $U$. The von Neumann mean ergodic theorem asserts that for all $x \in H$ there is convergence of the Ces&amp;agrave;ro averages of $U^{k}x$ to $Px$ in the norm:&#xA;\[&#xA;\left\Vert Px - \frac{1}{n+1} \sum_{k = 0}^{n} U^{k}x\right\Vert \; \xrightarrow{n \to \infty} \; 0.&#xA;\]&#xA;In other words, the sequence of operators $\frac{1}{n+1} \sum_{k = 0}^{n} U^{k}$ converges to $P$ in the strong operator topology on $L(H)$. Of course, convergence in the strong operator topology is the same as pointwise convergence of the operators on $H$, but I prefer not to speak of pointwise convergence because this might be confusing when speaking of a function space like $H = L^{2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt; &lt;strong&gt;Spoiler:&lt;/strong&gt; Note that if you have that, the exercise is already solved because you can multiply with $U^{M}$ inside the norm and write $N = n+M$ (or preferably do this backwards).&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I'm not sure what kind of answer you're looking for. Formally, the result looks stronger but it is equivalent, as you noticed and I argued in the spoiler above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think you're already pretty close to the intuition I have about this.&#xA;Let $S_{m}^{n} = \frac{1}{n-m+1}\sum_{k = m}^{n} U^{k}$ with $n \geq m$ (the $+1$ is rather immaterial but it should become clear from the stuff I say below why I prefer to add it). This is a triangular double sequence of operators in $L(H)$. As you say, taking $m = \text{const}$ you have convergence $\|S_{m}^{n}f - Pf\| \xrightarrow{n \to \infty} 0$. Now just requiring $(n-m) \to \infty$ means that only the increasing length of the tails matters eventually.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;For me a slightly more abstract stance is helpful, but this may be due to the fact that I've thought about &lt;em&gt;amenability&lt;/em&gt; too much. I'll give a brief account of this point of view anyway:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Ces&amp;agrave;ro averages are probability measures on the abelian semi-group $\mathbb{N}$, namely $\mu_{n} = \frac{1}{n+1}\sum_{k=0}^{n} \delta_{k}$. The shift $Sk = k+1$ on $\mathbb{N}$ acts on the probability measures and&#xA;\[&#xA;\|S\mu_{n} - \mu_{n}\| = \|\frac{1}{n+1} (\delta_{n+1} - \delta_{0})\| \leq \frac{2}{n+1} \to 0,&#xA;\]&#xA;where the norm is understood to be the total variation norm (or $\ell^{1}$-norm).&#xA;Now in general, a sequence $(\lambda_{n})$ (or even a net) of probability measures on $\mathbb{N}$ is called &lt;em&gt;approximately invariant&lt;/em&gt; (or a &lt;em&gt;Reiter sequence&lt;/em&gt;) if &#xA;$\|S\lambda_{n} - \lambda_{n}\| \to 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, the semigroup $\mathbb{N}$ acts on $H$ by $n \ast x = U^{n}x$. By pushing probability measures forward via the orbit map, we get an action on $H$ by the convolution semigroup $P(\mathbb{N})$ of probability measures on $\mathbb{N}$. Explicitly, $\mu \ast x = \sum_{n \in \mathbb{N}} \mu(n) U^{n} x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $(\lambda_{n})$ be an approximately invariant sequence (or net) of probability measures. Then&#xA;\[&#xA;\|\lambda_{n} \ast (Ux - x)\| = \| (S\lambda_{n}) \ast x - \lambda_{n} \ast x\| \leq \|S\lambda_{n} - \lambda_{n}\|\,\| x\|\;\xrightarrow{n \to \infty} \;0&#xA;\]&#xA;for all $x \in H$. Let $W$ be the closed linear span of the vectors $Ux - x$. We have just seen that $\lambda_{n} \ast w \to 0$ for all $w$ in a dense subspace of $W$, hence for all $w \in W$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moreover, it is not difficult to check that $W^{\perp} = V = \{x \,:\,x = Ux\}$. Indeed, if $y \in W^{\perp}$ then $0 = \langle y, x - Ux \rangle = \langle y - U^{\ast}y, x\rangle$ for all $x \in H$, so $y = U^{\ast}y$ and hence $Uy = y$ because $U$ is unitary. Therefore $V \supset W^{\perp}$ and the other inclusion is clear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For every $x \in H$ we have $x = Px + (1-P)x$ with $Px \in V$ and $(1-P)x \in W$. Therefore $\lambda_{n} \ast x = \lambda_{n} \ast Px + \lambda_{n} \ast (1-P)x = Px + \lambda_{n} \ast (1-P)x \to Px$ and we have proved the following version of the mean ergodic theorem:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Theorem.&lt;/strong&gt; If $(\lambda_{n})$ is an approximately invariant sequence (or net) of probability measures on $\mathbb{N}$ then $\lambda_{n} \ast x = \sum_{k \in \mathbb{N}} \lambda_{n}(k) U^{k}x$ converges to $Px$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Finally, I come back to your specific question. The operators $S_{m}^{n}$ are easily seen to arise from the probability measures $\lambda_{m}^{n} = \frac{1}{n-m+1} \sum_{k=m}^{n} \delta_{k}$, which is a net using lexicographic ordering and is approximately invariant &lt;em&gt;provided&lt;/em&gt; that $(n-m) \to \infty$.&lt;/p&gt;&#xA;" OwnerUserId="5363" LastEditorUserId="5363" LastEditorDisplayName="" LastEditDate="2011-02-16T11:34:14.347" LastActivityDate="2011-02-16T11:34:14.347" CommentCount="2" />
  <row Id="22223" PostTypeId="1" AcceptedAnswerId="22230" CreationDate="2011-02-15T21:15:42.863" Score="4" ViewCount="103" Body="&lt;p&gt;Hi&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've been struggling with this for a bit&#xA;and was wondering if anyone can give me a hint:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose $\{ T_n\}_1^\infty\subset \mathcal{L}\{ X,Y\}$ is a sequence of bounded operators from a banach space $X$ to a normed space $Y$, such that $\forall x\in X:\lim_n T_n x=0$. &#xA;Prove that $\lim_n(\sup\{\Vert T_n x\Vert:x\in K\})=0$ on any compact set $K\subset X$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as I can see, the fact that $T_n$ attains it's maximum on $K$ for any $n$ is not enough to prove this one, since it is possible that there exists an $\epsilon&amp;gt;0$ such that for any $N$ $\exists n&amp;gt;N$ such that $\Vert T_nx\Vert&amp;gt;\epsilon$ for some $x\in K$, and diminishes to 0 later on. Plus, this kind of solution doesn't employ the fact the $X$ is complete.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've been trying to work this question out as a varient of the Principle of Uniform Boundedness, but I keep getting stuck. Anyway I would be very thankful if anyone could tell me if I'm on the right track with this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cheers&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S&#xA;This is homework&lt;/p&gt;&#xA;" OwnerUserId="7129" LastActivityDate="2011-02-15T21:37:50.457" Title="Does strong convergence to 0 implies convergence to zero on compact sets?" Tags="&lt;homework&gt;&lt;functional-analysis&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="22224" PostTypeId="2" ParentId="22191" CreationDate="2011-02-15T21:17:49.197" Score="1" ViewCount="" Body="&lt;p&gt;Ross's answer was good, but here's a bit more clarity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is &quot;how much salt is lost after 20 minutes (of above described loss scenario)?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In $y = y_0 e^{kt}$, we have to find $k$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So how much salt is lost?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;You lose $\frac{1}{2}$ gallons of water per minute&lt;/li&gt;&#xA;&lt;li&gt;The concentration of salt per gallon is $\frac{S}{10}$, because there is S salt in the 10 gallons of water (at any instant of time)&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Thus the net rate of change of &lt;strong&gt;salt&lt;/strong&gt; in the vat is&#xA;$ -\frac{1}{2} $ [gallons/min] $ \frac{S}{10} $ [salt/gallon]&#xA;$ = -\frac{S}{20} $ [salt/min]&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Thus the differential equation is &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;$ \frac{dS}{dt} = -\frac{S}{20} $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Matching with&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ \frac{dy}{dt} = ky $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for a differential equation, this means k = $-\frac{1}{20}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$y = 5 e^{-\frac{1}{20}t}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$t = 20$ so&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$y = 5 e^{-1}$&lt;/p&gt;&#xA;" OwnerUserId="1069" LastActivityDate="2011-02-15T21:17:49.197" />
  <row Id="22225" PostTypeId="1" CreationDate="2011-02-15T21:17:52.407" Score="0" ViewCount="68" Body="&lt;p&gt;Hi all, i'm currently developing a FPS Game.&#xA;I need to trace a ray from the camera through the camera &quot;target&quot; (that I need to find), knowing the camera rotation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I make a code like this (by the theory of the Spherical Coordinates):&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;(rx,ry,rz) [Camera Rotation]&#xA;(cpx,cpy,cpz) [Current Player Position]&#xA;posx = (100 * sin(rz) * cos(rx)) + cpx&#xA;posy= (100 * sin(rz) * sin(rx)) + cpy&#xA;posz= (100*  cos(rz)) + cpz&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;It seems that something goes wrong in this operations (except for orthogonal angles), so i need an help!&#xA;Thank you very much&lt;/p&gt;&#xA;" OwnerUserId="7128" LastActivityDate="2011-02-15T21:17:52.407" Title="Spherical Coordinates Question" Tags="&lt;geometry&gt;" CommentCount="3" />
  <row Id="22226" PostTypeId="1" AcceptedAnswerId="22228" CreationDate="2011-02-15T21:20:54.663" Score="1" ViewCount="107" Body="&lt;p&gt;I have to find the domain of&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{z^2-1}{z^2+z+1}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, there can't be any $z$ that makes null the denominator,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$z^2+z+1$ = 0&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and after decomposing $z$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x^2-y^2+x+2xyi+yi=0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How to solve that quadratic complex equation?&lt;/p&gt;&#xA;" OwnerUserId="2637" LastActivityDate="2011-02-15T21:43:48.783" Title="How can I find the roots of this complex equation?" Tags="&lt;complex-analysis&gt;&lt;complex-numbers&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="22227" PostTypeId="2" ParentId="22226" CreationDate="2011-02-15T21:24:45.213" Score="3" ViewCount="" Body="&lt;p&gt;The maximal domain of definition is,as you imply, the complement in $\mathbb C$ of the set of roots of the polynomial $z^2+z+1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore to find the &lt;em&gt;bad&lt;/em&gt; points, we need to solve the equation $$z^2+z+1=0.$$ I am pretty sure, now, that you know how to solve quadratic equations, no?&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2011-02-15T21:24:45.213" CommentCount="1" />
  <row Id="22228" PostTypeId="2" ParentId="22226" CreationDate="2011-02-15T21:25:57.293" Score="5" ViewCount="" Body="&lt;p&gt;Three points:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;You did the computation wrong. Writing $z=x+iy$, then&#xA;$$z^2+z+1 = (x+iy)^2 + (x+iy) + 1 = (x^2-y^2 + x + 1) + i(2xy+y).$$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If $(x^2-y^2+x+1) + i(2xy+y) = 0$, with $x,y$ real numbers, then you need $x^2-y^2+x+1=0$ and $2xy+y=0$. You solve these the way you usually solve equations for real numbers. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, for example, you have $0=2xy+y = (2x+1)y$. So either $y=0$ or $2x+1=0$. If $y=0$, then the first equation reduces to $x^2+x+1=0$, which has no real solutions, so there are no solutions with $y=0$. If $y\neq 0$, then $2x+1=0$, so $x=-\frac{1}{2}$. Plugging into the first equation, we get &#xA;$$0 = \frac{1}{4}-y^2 -\frac{1}{2} + 1 = -y^2 +\frac{3}{4},$$&#xA;so you get that $y^2 = \frac{3}{4}$, or $y = \pm\frac{\sqrt{3}}{2}$. So the two solutions are $z=-\frac{1}{2}+i\frac{\sqrt{3}}{2}$ and $z=-\frac{1}{2}-i\frac{\sqrt{3}}{2}$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The acrobatics from step 2 are unnecessary. You don't have to decompose into real and imaginary parts, because the quadratic formula works for complex numbers! (Provided you take complex square roots). Since&#xA;$$z^2 + z + 1 = \left(z+\frac{1}{2}\right)^2 + \frac{3}{4}$$&#xA;(by completing the square), then this is zero if and only if &#xA;$$z+\frac{1}{2} = \sqrt{-\frac{3}{4}},$$&#xA;if and only if &#xA;$$z = -\frac{1}{2} + \frac{\sqrt{-3}}{2}.$$&#xA;You may recognize this as exactly what you get from the quadratic formula applied to $z^2+z+1$, and you may also recognize them as the solutions you get if you go through the contorsions of step 2 above. So just find the two complex square roots of $-3$, and rejoice! (The quadratic formula works even if the coefficients of the quadratic are complex numbers, instead of real numbers).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-15T21:43:48.783" LastActivityDate="2011-02-15T21:43:48.783" CommentCount="1" />
  <row Id="22229" PostTypeId="2" ParentId="22223" CreationDate="2011-02-15T21:31:31.883" Score="3" ViewCount="" Body="&lt;p&gt;Hint: a) uniform boundedness principle is the right track. b) If $||x - x_0|| &amp;lt; \delta$ then $||T_n(x)|| \leq ||T_n(x_0)|| + ||T_n||\delta $. &lt;/p&gt;&#xA;" OwnerUserId="3035" LastActivityDate="2011-02-15T21:31:31.883" />
  <row Id="22230" PostTypeId="2" ParentId="22223" CreationDate="2011-02-15T21:37:50.457" Score="3" ViewCount="" Body="&lt;p&gt;Since $T_nx\to 0$ for each $x$, $(T_nx)_n$ is bounded for each $x$ (convergent sequences are bounded), so you can apply uniform boundedness.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose the limit is not zero.  It either doesn't exist or is greater than $0$, but in either case there is an $\varepsilon\gt0$ and a subsequence $(T_{n_k})_k$ such that $\sup\{\|T_{n_k}x\|:x\in K\}\gt\varepsilon$ for each $k$.  This implies that there is a sequence $x_1,x_2,\ldots$ in $K$ such that $\|T_{n_k}x_k\|\gt\varepsilon$.  You can then use sequential compactness of $K$ and the result of uniform boundedness to derive a contradiction.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2011-02-15T21:37:50.457" CommentCount="2" />
  <row Id="22231" PostTypeId="1" AcceptedAnswerId="22233" CreationDate="2011-02-15T21:41:19.020" Score="1" ViewCount="112" Body="&lt;p&gt;I'm having some troubles computing this limit.&#xA;$$\lim_{M\to\infty}M\left(\frac{1}{s} - \frac{e^{\frac{-s}{M}}}{s}\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that the answer should be 1, but I can't seem to figure out the steps to get there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are the steps I've tried taking:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The exponential approaches 1 as $M\to\infty$,&#xA;$$\lim_{M\to\infty}M\left(\frac{1}{s} - \frac{1}{s}\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Simplifying the statement leaves $\lim\limits_{M\to\infty}M * 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where do I go from here, or have I made a mistake above?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a side note, this is for deriving the laplace transform of the dirac-delta function by approximating the dirac-delta function as a finite rectangular pulse starting a t=0 and ending at $t=\frac{1}{M}$ with a magnitude of M, then taking the limit as $M\to\infty$&lt;/p&gt;&#xA;" OwnerUserId="5747" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-16T03:41:21.223" LastActivityDate="2011-02-16T03:41:21.223" Title="Computing the limit $\lim\limits_{M\to\infty}M\left(\frac{1}{s} - \frac{\exp(-s/M)}{s}\right)$" Tags="&lt;homework&gt;&lt;limit&gt;" AnswerCount="2" CommentCount="7" />
  <row Id="22232" PostTypeId="1" CreationDate="2011-02-15T21:44:01.280" Score="3" ViewCount="70" Body="&lt;p&gt;Find the matrix of rotation in $\mathbb{R}^3$ by the angle $\theta$ around the column vector $(1, 2, 3)^T$. We assume that rotation is counterclockwise if we sit at the tip of the vector and look at the origin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could I use the product of three 3x3 matrices to rotate vectors about the x,y,and z points individually? If so, what would they look like?&lt;/p&gt;&#xA;" OwnerUserId="7123" LastEditorUserId="7123" LastEditorDisplayName="" LastEditDate="2011-02-15T22:15:32.283" LastActivityDate="2011-02-16T00:04:09.467" Title="How to find the matrix of rotation in $\mathbb{R}^3$ by the angle a around the vector column $(1, 2, 3)^T$." Tags="&lt;linear-algebra&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="22233" PostTypeId="2" ParentId="22231" CreationDate="2011-02-15T21:46:17.737" Score="3" ViewCount="" Body="&lt;p&gt;Try rewriting the equation as &#xA;$$\lim_{t\to 0^+}\left(\frac{1-e^{-t}}{0-(-t)}\right)$$&#xA;with the appropriate substitution.&#xA;From there on it should probably be quite easy to figure out why the answer is 1.&lt;/p&gt;&#xA;" OwnerUserId="7129" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-15T22:19:02.187" LastActivityDate="2011-02-15T22:19:02.187" CommentCount="5" />
  <row Id="22234" PostTypeId="2" ParentId="22232" CreationDate="2011-02-15T21:55:58.737" Score="1" ViewCount="" Body="&lt;p&gt;First normalize the column vector to a unit vector $\textbf{u}$. Then the rotation matrix $R$ is given by the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$R = \textbf{u} \otimes \textbf{u}+ \cos \theta(I- \textbf{u} \otimes \textbf{u})+\sin \theta[\textbf{u}]_{\times}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $[\textbf{u}]_{\times}$ is the skew-symmetric form of $\textbf{u}$, $\otimes$ is the tensor product, and $I$ is the identity matrix.&lt;/p&gt;&#xA;" OwnerUserId="2548" LastEditorUserId="2548" LastEditorDisplayName="" LastEditDate="2011-02-16T00:04:09.467" LastActivityDate="2011-02-16T00:04:09.467" CommentCount="1" />
  <row Id="22235" PostTypeId="2" ParentId="22232" CreationDate="2011-02-15T21:58:45.927" Score="1" ViewCount="" Body="&lt;p&gt;Do you know anything about quaternions?  In particular, in the standard representation of rotations by quaternions, a rotation of $\theta$ degrees about the vector $\vec{v} = (v_0, v_1, v_2) = v_0{\bf i}+v_1{\bf j}+v_2{\bf k}$ is represented by the quaternion ${\bf q} = \mathrm{cos}(\theta/2) + \mathrm{sin}(\theta/2)\hat{v}$, where $\hat{v}$ is the normalization of $\vec{v}$.  From there it's easy to find the rotation matrix; see the Wikipedia page on &lt;a href=&quot;http://en.wikipedia.org/wiki/Quaternions_and_spatial_rotation&quot; rel=&quot;nofollow&quot;&gt;quaternions and rotation&lt;/a&gt; for more details.&lt;/p&gt;&#xA;" OwnerUserId="785" LastActivityDate="2011-02-15T21:58:45.927" />
  <row Id="22236" PostTypeId="5" CreationDate="2011-02-15T22:14:49.383" Score="0" ViewCount="" Body="&lt;p&gt;For questions on congruences, linear diophantine equations, greatest common divisors, etc., please use the &lt;a href=&quot;http://math.stackexchange.com/questions/tagged/elementary-number-theory&quot;&gt;elementary-number-theory&lt;/a&gt; tag.&lt;/p&gt;&#xA;" OwnerUserId="-1" LastEditorUserId="1778" LastEditorDisplayName="" LastEditDate="2011-02-23T18:40:40.463" LastActivityDate="2011-02-23T18:40:40.463" />
  <row Id="22237" PostTypeId="4" CreationDate="2011-02-15T22:14:49.383" Score="0" ViewCount="" Body="Questions on more advanced topics of number theory." OwnerUserId="-1" LastEditorUserId="1778" LastEditorDisplayName="" LastEditDate="2011-02-23T18:40:29.510" LastActivityDate="2011-02-23T18:40:29.510" />
  <row Id="22238" PostTypeId="5" CreationDate="2011-02-15T22:18:41.393" Score="0" ViewCount="" Body="&lt;p&gt;Questions on elementary, or &quot;classic&quot;, number theory: congruences, greatest common divisor identities, divisibility, etc. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For more advanced questions, please use the [number-theory], [analytic-number-theory], or [algebraic-number-theory] tags, as appropriate.&lt;/p&gt;&#xA;" OwnerUserId="-1" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-15T22:18:41.393" LastActivityDate="2011-02-15T22:18:41.393" />
  <row Id="22239" PostTypeId="4" CreationDate="2011-02-15T22:18:41.393" Score="0" ViewCount="" Body="Questions on congruences, linear diophantine equations, greatest common divisor, divisibility, etc." OwnerUserId="-1" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-15T22:18:41.393" LastActivityDate="2011-02-15T22:18:41.393" />
  <row Id="22240" PostTypeId="2" ParentId="8332" CreationDate="2011-02-15T22:57:36.960" Score="1" ViewCount="" Body="&lt;p&gt;The question has been answered on MathOverflow: &lt;a href=&quot;http://mathoverflow.net/questions/49913/factoring-a-field-extension-into-one-which-adds-no-roots-of-unity-followed-by-on/49914#49914&quot; rel=&quot;nofollow&quot;&gt;http://mathoverflow.net/questions/49913/factoring-a-field-extension-into-one-which-adds-no-roots-of-unity-followed-by-on/49914#49914&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="264" LastActivityDate="2011-02-15T22:57:36.960" />
  <row Id="22241" PostTypeId="5" CreationDate="2011-02-15T23:18:11.637" Score="0" ViewCount="" Body="&lt;p&gt;Large cardinals are such cardinals whose existence cannot be proved within ZFC, and requires stronger axioms to be added to $ZFC$, they are often used to measure the consistency strength of a certain statement in the language of set theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For further reading:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;Kanamori, Akihiro&lt;/strong&gt;. &lt;em&gt;The Higher Infinite : Large Cardinals in Set Theory from Their Beginnings&lt;/em&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Kanamori, Akihiro; Magidor, M.&lt;/strong&gt;, &lt;em&gt;&lt;a href=&quot;http://math.bu.edu/people/aki/e.pdf&quot; rel=&quot;nofollow&quot;&gt;The evolution of large cardinal axioms in set theory&lt;/a&gt;&lt;/em&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="-1" LastEditorUserId="622" LastEditorDisplayName="" LastEditDate="2011-02-15T23:18:34.640" LastActivityDate="2011-02-15T23:18:34.640" />
  <row Id="22242" PostTypeId="4" CreationDate="2011-02-15T23:18:11.637" Score="0" ViewCount="" Body="Large cardinals are such cardinals whose existence cannot be proved within ZFC, and requires stronger axioms to be added to $ZFC$, they are often used to measure the consistency strength of a certain statement in the language of set theory." OwnerUserId="-1" LastEditorUserId="153" LastEditorDisplayName="" LastEditDate="2011-02-19T21:37:05.583" LastActivityDate="2011-02-19T21:37:05.583" />
  <row Id="22243" PostTypeId="1" AcceptedAnswerId="22245" CreationDate="2011-02-15T23:21:03.890" Score="0" ViewCount="58" Body="&lt;p&gt;Hi. Consider the the following problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/VLWE0.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Define $u&amp;#39;$ and $v&amp;#39;$ to be the projections of $u$ and $v$ onto the plane perpendicular to the axis of $\xi$. If $\omega \in \mathbb{R}^{3 \times 1}$ is a unit vector in the direction of the axis of $\xi$, then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$u&amp;#39; = u -\omega \omega ^{T} u$ and $v&amp;#39; = v -\omega \omega ^{T} v$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is something wrong with these equations?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I thought that $\omega \omega ^{T}u = u$ (dot product) so it becomes $u&amp;#39; = u - u = 0$&lt;/p&gt;&#xA;" OwnerUserId="6984" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2011-02-15T23:35:55.657" LastActivityDate="2011-02-15T23:46:47.270" Title="Vector projection" Tags="&lt;vector-calculus&gt;" AnswerCount="1" />
  <row Id="22244" PostTypeId="1" AcceptedAnswerId="28871" CreationDate="2011-02-15T23:24:49.923" Score="6" ViewCount="83" Body="&lt;p&gt;Suppose $dX_t = \mu(X_t)dt + \sigma(X_t)dW_t$ is a diffusion. Is there a sense in which the dynamics are &quot;dominated&quot; locally by the diffusion term, and dominated globally by the drift term?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $\mu$ and $\sigma$ are constants, then the law of the iterated logarithm says that the contribution from the diffusion term is slightly greater than $\sqrt{t}$, whereas the contribution from the drift term is linear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, over small timescales, small variations in the noise dominate any estimate one can make of the drift term.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does a similar principle apply to more general diffusions?&lt;/p&gt;&#xA;" OwnerUserId="3162" LastEditorUserId="3162" LastEditorDisplayName="" LastEditDate="2011-02-15T23:34:30.150" LastActivityDate="2011-03-24T20:46:43.930" Title="Diffusions - global and local" Tags="&lt;probability&gt;&lt;intuition&gt;&lt;stochastic-processes&gt;" AnswerCount="1" />
  <row Id="22245" PostTypeId="2" ParentId="22243" CreationDate="2011-02-15T23:26:06.490" Score="1" ViewCount="" Body="&lt;p&gt;Note that $\omega$ is a column unit vector and hence the inner product (or dot product) is $\omega^T \omega = 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The outer product $\omega \omega^T$ is a $3 \times 3$ matrix (which is called the projection matrix) and hence $\omega \omega^T u \neq u$ unless $u$ and $\omega$ are in the same direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The outer product $\omega \omega^T$ takes any vector and projects it onto the space spanned by the vector $\omega$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\omega^T u$ is the scalar projection of the vector $u$ onto the $\xi$ axis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\omega (\omega^T u)$ is the vector projection of the vector $u$ onto the $\xi$ axis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you subtract of the vector projection of $u$ onto the $\xi$ axis to get $u&amp;#39;$, the vector projection of $u$ onto the plane perpendicular to the $\xi$ axis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same thing holds for $v$&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2011-02-15T23:46:47.270" LastActivityDate="2011-02-15T23:46:47.270" />
  <row Id="22246" PostTypeId="1" AcceptedAnswerId="22247" CreationDate="2011-02-15T23:32:49.107" Score="0" ViewCount="84" Body="&lt;p&gt;It is easy to see that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$p:\mathbb{R} \rightarrow S^1: p(t)=(\cos t, \sin t)$ is a covering map of $S^1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Indeed take a point $x_0$ in $S^1$, take $U=S^1\setminus \{-x_0\}$ as an open neighborhood and then $p^{-1}(U)=\cup_{n \in Z} J_n$ where $J_n=\{t\in\mathbb{R} : t_0-n-1/2&amp;lt;t&amp;lt;t+0+n+1/2\}$. $p$ then maps $J_n$ homeomorphically onto $U$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can't see why we can't take a similar $p:(a,b) \to S^1$. For example, Massey states that if we take the open interval $(0,10)$ onto the circle, then some points in $S^1$ fail to have an elementary neighborhood. &lt;/p&gt;&#xA;" OwnerUserId="2219" LastActivityDate="2011-02-15T23:37:39.407" Title="Covering Space of Circle" Tags="&lt;algebraic-topology&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="1" />
  <row Id="22247" PostTypeId="2" ParentId="22246" CreationDate="2011-02-15T23:37:39.407" Score="2" ViewCount="" Body="&lt;p&gt;Look at a preimage of a small neighborhood $U$ of $p(10)$. It will have one component which doesn't map homeomorphically onto $U$.&lt;/p&gt;&#xA;" OwnerUserId="2941" LastActivityDate="2011-02-15T23:37:39.407" />
  <row Id="22248" PostTypeId="2" ParentId="22195" CreationDate="2011-02-16T00:10:30.810" Score="0" ViewCount="" Body="&lt;p&gt;Well I've heard that the more ways you can prove something, the merrier. :) So here's a sketch of the proof that immediately came to mind, although it may not be as snappy as some of the other good ones here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's prove the contrapositive, that is if $A$ is invertible then $A^2 \neq 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $A$ is invertible then we can write it as a product of elementary matrices,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$A = E_n...E_1I$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then $A^2$ can be written as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$AA = (E_n...E_1I)(E_n...E_1I) = (E_n...E_1 E_n...E_1)I$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is a sequence of elementary row operations on the identity matrix. But this will never produce the zero matrix $0$. QED.&lt;/p&gt;&#xA;" OwnerUserId="7090" LastEditorUserId="7090" LastEditorDisplayName="" LastEditDate="2011-02-16T05:18:55.983" LastActivityDate="2011-02-16T05:18:55.983" />
  <row Id="22249" PostTypeId="2" ParentId="22231" CreationDate="2011-02-16T00:12:15.067" Score="1" ViewCount="" Body="&lt;p&gt;You could also use the power series expansion of the exponential function. More specifically, $$e^x=1+x+O(x^2)$$ as $x\to 0$.&lt;/p&gt;&#xA;" OwnerUserId="1403" LastActivityDate="2011-02-16T00:12:15.067" />
  <row Id="22250" PostTypeId="1" AcceptedAnswerId="22252" CreationDate="2011-02-16T00:14:09.573" Score="0" ViewCount="50" Body="&lt;p&gt;$y$ is a function of $t$.  Simple differential equations are written to make&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ \frac{dy}{dt} = ky(t) $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The function $y(t)$ that fits this is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$y(t) = y(0) e^{kt} $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $y(0)$ is some initial condition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ \frac{d}{dt} y(t) = k y(0) e^{kt} $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Achieving our constraint $ \frac{dy}{dt} = ky(t) $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is all fine and dandy, but I'm wondering if anyone ever uses $2^{kt}, 3^{kt}$ or $n^{kt}$ in any problems.  There is an accumulating $ln(2)$ factor on each derivative, but I'm wondering if that's ever useful.&lt;/p&gt;&#xA;" OwnerUserId="1069" LastActivityDate="2011-02-16T00:22:50.890" Title="Using a different base for differential equations?" Tags="&lt;differential-equations&gt;" AnswerCount="2" />
  <row Id="22251" PostTypeId="2" ParentId="22250" CreationDate="2011-02-16T00:20:00.430" Score="0" ViewCount="" Body="&lt;p&gt;One can always change bases and get back the exponential function and exploit the fact that $f(x) = f&amp;#39;(x)$.&lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2011-02-16T00:20:00.430" />
  <row Id="22252" PostTypeId="2" ParentId="22250" CreationDate="2011-02-16T00:22:50.890" Score="2" ViewCount="" Body="&lt;p&gt;It depends on $k$. If your $k$ is $\log(2)$, then you could write $y(t) = 2^t$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any exponential can be rewritten in terms of other exponential.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a^{k_1 t} = b^{k_2 t}$, where $k_2 = k_1 \log_b(a)$ and equivalently $k_1 = k_2 \log_a(b)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But $f(t) = c e^t$ is the only &quot;nice&quot; function which satisfies $f&amp;#39;(t) = f(t)$&lt;/p&gt;&#xA;" OwnerUserId="3302" LastActivityDate="2011-02-16T00:22:50.890" />
  <row Id="22253" PostTypeId="2" ParentId="17405" CreationDate="2011-02-16T00:26:56.063" Score="6" ViewCount="" Body="&lt;p&gt;Apparently the same exercise can be done for the lemniscate with the same result. For instance, see &lt;a href=&quot;http://www.jstor.org/stable/2321821&quot; rel=&quot;nofollow&quot;&gt;http://www.jstor.org/stable/2321821&lt;/a&gt; where Theorem 2 states that&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If the lemniscate can be divided in n parts with ruler and compass, then n is a power of two times a product of distinct Fermat primes.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The main difficulty, when compared to the better known theorem about the circle, appears to be the shift from circular functions (sin, cos) to elliptic functions. For instance one requires some sort of addition theorem for these functions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is only one more curve, but one that can be associated to the important elliptic integral $\int \frac{dt}{\sqrt{1-t^4}}$, making an appearance as the arc-length of the lemniscate. I'm guessing there is a wide class of curves that are associated to elliptics integrals this way, but I doubt that any of them would naturally be as interesting as the circle or the lemniscate.&lt;/p&gt;&#xA;" OwnerUserId="5189" LastActivityDate="2011-02-16T00:26:56.063" />
  <row Id="22255" PostTypeId="1" AcceptedAnswerId="22256" CreationDate="2011-02-16T00:36:01.283" Score="1" ViewCount="124" Body="&lt;p&gt;I need to confirm that i am approaching this problem correctly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Problem:&#xA;Let $f$ and $g$ be functions $f,g:\mathbb{R}\to \mathbb{R}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Prove that if $f^2 + g^2$ is bounded then $f$ is bounded.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Attempted Solution:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Definition of bounded: $|f(x)| \leq M$ for all $x \in \mathbb{R}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using that definition we can say&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$|f^2 + g^2| \leq M$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then:&#xA;$|f| = \sqrt{|f^2|} \leq \sqrt{M}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore we can conclude that $|f|$ is bounded when $|f^2 + g^2|$ is bounded.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this true? Or have I gone horrible wrong? Any help is appreciated. Oh and if someone can point me to a good place to learn latex commands that would be sweet, thanks!&lt;/p&gt;&#xA;" OwnerUserId="5505" LastEditorUserId="4284" LastEditorDisplayName="" LastEditDate="2011-02-16T05:59:26.713" LastActivityDate="2011-02-16T05:59:26.713" Title="Bounded and Unbounded Functions" Tags="&lt;algebra-precalculus&gt;" AnswerCount="1" CommentCount="6" />
  <row Id="22256" PostTypeId="2" ParentId="22255" CreationDate="2011-02-16T00:44:56.487" Score="7" ViewCount="" Body="&lt;p&gt;There is an easier way, given that squares of real numbers are non-negative, so $f^2 \ge 0$, $g^2 \ge 0$ and $f^2 +g^2 \ge 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $ f^2 + g^2  \le M$ then $ f^2 \le M$, so $-\sqrt{M} \le f \le \sqrt{M}$.&lt;/p&gt;&#xA;" OwnerUserId="6460" LastActivityDate="2011-02-16T00:44:56.487" CommentCount="1" />
  <row Id="22257" PostTypeId="1" AcceptedAnswerId="22260" CreationDate="2011-02-16T00:51:03.637" Score="2" ViewCount="56" Body="&lt;p&gt;I'm lost on how to solve this for z. Can anyone help?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\left(7^\left(-2z\right)\right)^\left(7-z\right) = 49^\left(-12\right)$ &lt;/p&gt;&#xA;" OwnerUserId="6843" LastActivityDate="2011-02-16T01:00:21.863" Title="Exponents and Logarithms" Tags="&lt;homework&gt;&lt;logarithms&gt;" AnswerCount="3" CommentCount="0" />
  <row Id="22258" PostTypeId="2" ParentId="22257" CreationDate="2011-02-16T00:52:55.710" Score="2" ViewCount="" Body="&lt;p&gt;Take $\log_{7}$ of both sides and then simplify.&lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2011-02-16T00:52:55.710" />
  <row Id="22259" PostTypeId="2" ParentId="22257" CreationDate="2011-02-16T00:53:52.220" Score="1" ViewCount="" Body="&lt;p&gt;Write $7^{-2z}$ as $(7^2)^{-z} = 49^{-z}$&lt;/p&gt;&#xA;" OwnerUserId="3302" LastActivityDate="2011-02-16T00:53:52.220" />
  <row Id="22260" PostTypeId="2" ParentId="22257" CreationDate="2011-02-16T01:00:21.863" Score="4" ViewCount="" Body="&lt;p&gt;The equation is equivalent to $7^{(-2x)(7-x)}=7^{2\cdot(-12)}$.&lt;/p&gt;&#xA;" OwnerUserId="589" LastActivityDate="2011-02-16T01:00:21.863" />
  <row Id="22261" PostTypeId="1" AcceptedAnswerId="22281" CreationDate="2011-02-16T01:24:31.023" Score="3" ViewCount="94" Body="&lt;p&gt;Let $X_i$ be a sequence of probability spaces and define $\displaystyle X=\prod_{i=1}^\infty X_i$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $A$ be the algebra on $X$ generated by the sets of the form $$\displaystyle \prod_{i=1}^{n-1} X_i \times E_n \times \prod_{i=n+1}^{\infty} X_i, E_n \in P(X_n)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Show that every element in $A$ can be written as $$a= \displaystyle \cup_{j=1}^m (\prod_{i=1}^{n_j} E_{j,i}                                                                                                                                 \times \prod_{i=n_j}^{\infty} X_i)$$ where $E_{j,i} \in P(X_i)$ and the union is a disjoint union&lt;/p&gt;&#xA;" OwnerUserId="6556" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-16T03:49:56.157" LastActivityDate="2011-03-18T05:33:37.547" Title="Describing all elements in the set algebra generated by given sets in an infinite product" Tags="&lt;set-theory&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="22262" PostTypeId="1" CreationDate="2011-02-16T01:33:29.550" Score="3" ViewCount="170" Body="&lt;p&gt;How to find a differentiable map $T : \mathbb{R} \rightarrow \mathbb{R}$ whose fixed points are exactly integers?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a)Find points $|T&amp;#39;(x)| &amp;gt; 1$ (if any)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and prove that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;b)There must exist at least one $x$ such that $|T&amp;#39;(x)| &amp;gt; 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the functions &#xA;$$x - Tx = x(x^2 - 1)(x^2 - 4)(x^2 - 9)...$$&#xA;and&#xA;$$x - Tx = x(e^x - e^1)(e^{-x} - e^1)(e^x - e^2)(e^{-x} - e^2)(e^x - e^3)...$$ will satistfy the criteria.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any closed form expressions for $T&amp;#39;(x)$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I proceed further?&lt;/p&gt;&#xA;" OwnerUserId="5049" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2011-02-16T02:29:24.617" LastActivityDate="2011-02-16T17:06:21.093" Title="How to find a differentiable map $T : \mathbb{R} \rightarrow \mathbb{R}$ whose fixed points are exactly integers" Tags="&lt;real-analysis&gt;" AnswerCount="4" CommentCount="5" FavoriteCount="1" />
  <row Id="22263" PostTypeId="1" CreationDate="2011-02-16T01:37:42.120" Score="2" ViewCount="60" Body="&lt;p&gt;I have defined this simple sheaf. Take $E:set, B:set, p:E\to B$. Let $P(B)$ be a set of subsets of $B$. Let $S$ be the set of &lt;a href=&quot;http://en.wikipedia.org/wiki/Section_(fiber_bundle)&quot; rel=&quot;nofollow&quot;&gt;sections&lt;/a&gt; of $p$. Let $F$ be a contravariant functor from $P(B)$ as a poset category to $Set$ which sends:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;objects: $B&amp;#39;$ to ($S$ with elements restricted to $B'$);&lt;/li&gt;&#xA;&lt;li&gt;morphisms: an inclusion to the restriction of functions;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;and the discrete topology on $P(B)$ and obvious gluing. I call $F$ the “sheaf of sections”, is there a standard name? Was anybody studying it? (Perhaps it is too simple.) A motivation is that &lt;a href=&quot;http://ncatlab.org/nlab/show/dependent+type&quot; rel=&quot;nofollow&quot;&gt;sections can be identified with functions of a dependent type&lt;/a&gt; which generalize typed tuples.&lt;/p&gt;&#xA;" OwnerUserId="7011" LastActivityDate="2011-04-21T19:42:19.063" Title="a function of a dependent type, a section, a sheaf" Tags="&lt;soft-question&gt;&lt;elementary-set-theory&gt;&lt;computer-science&gt;&lt;definition&gt;&lt;sheaf-theory&gt;" AnswerCount="1" />
  <row Id="22264" PostTypeId="2" ParentId="22262" CreationDate="2011-02-16T01:39:46.307" Score="5" ViewCount="" Body="&lt;p&gt;You're on the right track but that product does not work. Find a function that has zeros exactly at the integers. Then add $x$ to it.&lt;/p&gt;&#xA;" OwnerUserId="589" LastActivityDate="2011-02-16T01:39:46.307" />
  <row Id="22265" PostTypeId="1" AcceptedAnswerId="22273" CreationDate="2011-02-16T01:49:02.393" Score="6" ViewCount="176" Body="&lt;p&gt;According to wikipedia, &lt;a href=&quot;http://en.wikipedia.org/wiki/Strassen_algorithm&quot; rel=&quot;nofollow&quot;&gt;Strassen's Algorithm&lt;/a&gt; runs in $O(N^2.807)$ time.  Has anyone seen a more rigorous analysis displaying constants, possibly in a specific language such as C or Java?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I realize this will vary from language to language, machine to machine etc, but does anyone know of an approximate input size where Strassen's algorithm starts to outperform regular matrix multiplication?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think this may belong on the computer science stack exchange but since it is somewhat mathematical I thought I would post it here.&lt;/p&gt;&#xA;" OwnerUserId="2533" LastEditorUserId="1016" LastEditorDisplayName="" LastEditDate="2011-02-16T02:26:03.267" LastActivityDate="2011-02-16T16:26:32.857" Title="Has there been a rigorous analysis of Strassen's algorithm?" Tags="&lt;analysis&gt;&lt;soft-question&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="22266" PostTypeId="2" ParentId="19562" CreationDate="2011-02-16T01:53:10.810" Score="1" ViewCount="" Body="&lt;p&gt;I'd like to give an answer which is different from the accepted one. If you consider “expression” and &lt;a href=&quot;http://en.wikipedia.org/w/index.php?title=Term_(first-order_logic)&quot; rel=&quot;nofollow&quot;&gt;“term”&lt;/a&gt; being synonyms, then $x\cdot x+x$ does not include constants. But it is still a polynomial because your definition does not say “at least one constant”. Compare to this example: if “a list is a sequence of elements”, the list may contain &lt;strong&gt;no elements&lt;/strong&gt; as well.&lt;/p&gt;&#xA;" OwnerUserId="7011" LastActivityDate="2011-02-16T01:53:10.810" CommentCount="3" />
  <row Id="22268" PostTypeId="1" AcceptedAnswerId="22275" CreationDate="2011-02-16T02:01:28.057" Score="1" ViewCount="52" Body="&lt;p&gt;Suppose that we have a family of linear functions $L(\alpha) : \mathbb{R}^n \rightarrow \mathbb{R}^n$, where $\alpha$ is a positive real number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For each $\alpha$, it is given that $L(\alpha)$ is a symmetric matrix, and so it has a basis of eigenvectors, and $n$ eigenvalues (not necessarily distinct). We define a function $f(\alpha)$ to be the smallest eigenvalue of $L(\alpha)$. For example, if $L(\alpha)$ is the matrix&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ \left( \begin{array}{cc}&#xD;&#xA;\alpha &amp;amp; 0  \\&#xD;&#xA;0 &amp;amp; 2\alpha&#xD;&#xA;\end{array} \right) $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;then $f(\alpha) = \alpha$ because $\alpha &amp;gt; 0$. It is also given that the elements of $L(\alpha)$ are smooth functions of $\alpha$ for each $\alpha$. Therefore, as the eigenvalues are the roots of a polynomial whose coefficients are also smooth functions of $\alpha$, $f(\alpha)$ is a smooth function of $\alpha$ for all but a few values of $\alpha$ (this follows from the implicit function theorem).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem I am having is to try to find the global minimum of $f(\alpha)$. Usually what one would do is to take its derivative with respect to alpha, and set that equal to 0, but that is actually not possible in the present case because there is no way to get an explicit formula for $f(\alpha)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A possible idea that I had was to get some kind of sequence which converges to the smallest eigenvalue of $L(\alpha)$, and hope that this sequence converges uniformly in $\alpha$, but I have not yet found such a thing.&lt;/p&gt;&#xA;" OwnerUserId="1089" LastActivityDate="2011-02-16T03:18:51.143" Title="Eigenvalues of a Parametrized Family of Linear Functions" Tags="&lt;linear-algebra&gt;&lt;analysis&gt;&lt;functional-analysis&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="22269" PostTypeId="2" ParentId="22262" CreationDate="2011-02-16T02:02:31.557" Score="3" ViewCount="" Body="&lt;p&gt;Since you are interested in infinite products, I recommend taking a look at the &lt;a href=&quot;http://en.wikipedia.org/wiki/Infinite_product&quot; rel=&quot;nofollow&quot;&gt;Wikipedia article&lt;/a&gt; for a little orientation on these.  I believe you will find it useful in finding the example you want.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(b) is a consequence of the mean value theorem. Consider $T(\frac{1}{2})$, for example.  It is either less than or greater than $\frac{1}{2}$.  If less, then the average rate of change from $\frac{1}{2}$ to $1$ is greater than $1$.  If greater, then the average rate of change from $0$ to $\frac{1}{2}$ is greater than $1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The difficulty of (a) could depend on the example, and that must wait until you have an explicit example.  Of course by (b) you know that there will be such points.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2011-02-16T02:02:31.557" />
  <row Id="22270" PostTypeId="2" ParentId="22031" CreationDate="2011-02-16T02:04:22.830" Score="4" ViewCount="" Body="&lt;p&gt;I agree compatibility is the main thing to look for. Since we're assuming the student is talented and ambitious, the question to ask is &quot;which professor thinks the most highly of me?&quot; This can be hard to measure, because some people are naturally more or less bubbly or expressive, so spend time getting to know a prospective advisor mathematically through a reading course or mini-research project. Wait at least a full semester or so, and see how the meetings are going. Are they productive? Is the material interesting? Does the professor seem to like how you work? Is he or she becoming engaged and interested in continuing working with you?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next, don't overthink it. As with all important life decisions, the outcome of this one depends largely on factors you can't hope to predict. Don't obsess too much about the status of your advisor, first because this is harder to gauge than you might think, and second because it's even harder to gauge how much this status will be used in your favor. Mathematical brilliance is a similar issue.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, here is one major pitfall I've seen a number of friends fall into. Beware the charismatic lecturer! This is why I recommended a reading course. Of course many people are wonderful lecturers and also wonderful advisors, but many are also great on stage and then distant in person. Watch out for the advisor who keeps you at an arm's length, even if you loved the course you took with him or her.&lt;/p&gt;&#xA;" OwnerUserId="7134" LastActivityDate="2011-02-16T02:04:22.830" CommunityOwnedDate="2011-02-16T02:04:22.830" />
  <row Id="22271" PostTypeId="2" ParentId="10697" CreationDate="2011-02-16T02:17:07.077" Score="0" ViewCount="" Body="&lt;p&gt;Abstract mathematics is not about the real world. Like, “If it is just taken out of someone's head, why you guys can not make it simple and understandable? Is it a problem with your head? Why you can't get rid of those obscure definitions and horrible proofs?” (Surely I exasperated it. :) ) Actually, can anyone?&lt;/p&gt;&#xA;" OwnerUserId="7011" LastActivityDate="2011-02-16T02:17:07.077" CommunityOwnedDate="2011-02-16T02:17:07.077" />
  <row Id="22272" PostTypeId="1" AcceptedAnswerId="22750" CreationDate="2011-02-16T02:17:50.697" Score="2" ViewCount="76" Body="&lt;p&gt;Starting with a given simple, directed Graph G, I define a two-edge swap as:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;select two edges u-&gt;v and x-&gt;y such that (u!=x) and (v!=y) and (u!=y) and (x!=v)&lt;/li&gt;&#xA;&lt;li&gt;delete the two edges u-&gt;v and x-&gt;y&lt;/li&gt;&#xA;&lt;li&gt;add edges u-&gt;y and x-&gt;v&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Is it guaranteed that I can reach any simple directed graph with the original (in- and out-) degree sequence in some finite number of 2-edge swaps? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we need some sort of 3-edge swaps, what are they?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Background: I intend to use this as MCMC steps to sample random graphs, but over at the &lt;a href=&quot;https://networkx.lanl.gov/trac/ticket/498&quot; rel=&quot;nofollow&quot;&gt;Networkx Developer site&lt;/a&gt;, there is a discussion that Theorem 7 of the paper &lt;a href=&quot;http://www.combinatorics.org/Volume_17/PDF/v17i1r66.pdf&quot; rel=&quot;nofollow&quot;&gt;P Erdos et al., &lt;em&gt;&quot;A simple Havel–Hakimi type algorithm to realize graphical degree sequences of directed graphs&quot;&lt;/em&gt;, Combinatorics 2010&lt;/a&gt; implies that we need 3-edge swaps to sample the whole space. &lt;/p&gt;&#xA;" OwnerUserId="2532" LastEditorUserId="2532" LastEditorDisplayName="" LastEditDate="2011-02-16T02:24:47.803" LastActivityDate="2011-02-19T06:10:00.933" Title="Reaching all possible simple directed graphs with a given degree sequence with 2-edge swaps" Tags="&lt;combinatorics&gt;&lt;graph-theory&gt;&lt;monte-carlo&gt;" AnswerCount="1" />
  <row Id="22273" PostTypeId="2" ParentId="22265" CreationDate="2011-02-16T02:37:51.800" Score="4" ViewCount="" Body="&lt;p&gt;You're probably better off asking such a question at the CS-site, but I suggest you have a look at the easily readable original article&#xA;&lt;a href=&quot;http://dx.doi.org/10.1007/BF02165411&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; (probably you need a university subscription), and the MR review is &lt;a href=&quot;http://www.ams.org/mathscinet-getitem?mr=248973&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;. Since Strassen's algorithm is quite famous, I would be surprised if it were difficult to find a detailed analysis in various dialects by googling a little.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The roughly is in fact precisely $\log_{2}{7} \approx 2.807$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to multiply two $n \times n$ matrices in the usual way you need $n^{3}$ multiplications and $n^{2}(n-1)$ additions, so in total you need $n^2 (2n-1)$ arithmetical operations. Strassen's algorithm (in his own analysis) gets away with less than $4.7 n^{\log_{2} 7}$ arithmetical operations (provided $n \geq 16$ if I'm not completely mistaken), and this should be better for all such $n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't see any other &quot;costly&quot; thing than addition and multiplication in an implementation. The amount of memory needed should also be rather easy to figure out from Strassen's description, but it should be of order $4 n^2$.&lt;/p&gt;&#xA;" OwnerUserId="5363" LastEditorUserId="5363" LastEditorDisplayName="" LastEditDate="2011-02-16T02:45:57.853" LastActivityDate="2011-02-16T02:45:57.853" />
  <row Id="22274" PostTypeId="2" ParentId="21411" CreationDate="2011-02-16T02:48:31.950" Score="1" ViewCount="" Body="&lt;p&gt;Let $C$ be a &lt;a href=&quot;http://en.wikipedia.org/wiki/Singleton_(mathematics)&quot; rel=&quot;nofollow&quot;&gt;singleton&lt;/a&gt;. Then $C^A$, $C^B$, $C^A$, and hence $(C^B)^{C^A}$ are singletons. Take $A$ and $B$ such that $\mathrm{card}(A\to B)\gt 1$, e.g. $A=B=\{0,1\}$. (“card” is a number of elements in a set, &lt;a href=&quot;http://en.wikipedia.org/wiki/Cardinality&quot; rel=&quot;nofollow&quot;&gt;cardinality&lt;/a&gt;.) Then $\mathrm{card}(B^A) \gt \mathrm{card}((C^B)^{C^A})$, so  there is no injection.&lt;/p&gt;&#xA;" OwnerUserId="7011" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-16T04:27:05.690" LastActivityDate="2011-02-16T04:27:05.690" CommentCount="1" />
  <row Id="22275" PostTypeId="2" ParentId="22268" CreationDate="2011-02-16T03:18:51.143" Score="0" ViewCount="" Body="&lt;p&gt;Always a fan of the cheap solution, suppose I take $n=1$, then $f(\alpha)$ has just one eigenvalue.  And $\frac{df(\alpha)}{d\alpha}$ can be anything.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-02-16T03:18:51.143" CommentCount="2" />
  <row Id="22276" PostTypeId="1" AcceptedAnswerId="22293" CreationDate="2011-02-16T03:47:58.387" Score="1" ViewCount="100" Body="&lt;p&gt;How can I compute this limit:&#xA;$$\lim_{z\rightarrow i} (z\cdot \overline{z})$$&#xA;(where $\overline{z}$ is the conjugate of $z$)?&lt;/p&gt;&#xA;" OwnerUserId="2637" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-16T04:00:15.927" LastActivityDate="2011-02-16T05:44:56.997" Title="The limit of $z\cdot\overline{z}$ as $z\to i$" Tags="&lt;complex-analysis&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="22277" PostTypeId="2" ParentId="22276" CreationDate="2011-02-16T03:54:13.500" Score="3" ViewCount="" Body="&lt;p&gt;Hint: Is $z&amp;#39;$ the conjugate of $z$ and the lower dot a multiplication?  If so, what is the problem?  You might try expressing $z=a+bi$, with $a,b$ real and trying a real limit.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-02-16T03:54:13.500" CommentCount="4" />
  <row Id="22278" PostTypeId="2" ParentId="8512" CreationDate="2011-02-16T03:59:11.433" Score="1" ViewCount="" Body="&lt;p&gt;Here is a second approach.  A matrix is diagonalizable if and only if it has a basis of eigenvectors.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose $v$ is an eigenvector of a matrix $A$ with eigenvalue $\lambda$. Then&#xA;$$A^nx = \lambda^n v.$$  Hence any eigenvector of $A$ is an eigenvector of $A^n$.  Hence, any basis of eigenvalues of a matrix $A$ is a basis of eigenvalues of $A^n$ for any nonnegative integer n.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More generally if $f$ is a polynomial in the scalar field of the matrix, $f(A)$ is diagonalizable too.  This is pretty easy to see.&lt;/p&gt;&#xA;" OwnerUserId="4154" LastActivityDate="2011-02-16T03:59:11.433" />
  <row Id="22279" PostTypeId="1" CreationDate="2011-02-16T04:26:07.997" Score="1" ViewCount="33" Body="&lt;p&gt;I asked &lt;a href=&quot;http://math.stackexchange.com/questions/22115/are-limits-on-exponents-in-moduli-possible&quot;&gt;a similar question to this recently&lt;/a&gt;. Here, I consider an arbitrary, but fixed, modulus m, which is relatively prime to x and y.  Can anybody extend the answer given in the previous question?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, my question is the following. Suppose I show that:&#xA;$$x^{f(z)/g(z)} = y \pmod{m}$$&#xA;is impossible for some given positive integers $x$ and $y$, and integer $m&amp;gt;4$, where m is relatively prime to both x and y, and $f(z)$ and $g(z)$ are defined by,&#xA;\begin{align*}&#xA;f(z) &amp;amp;= \phi(m)  k_1(z) + 1 \\&#xA;g(z) &amp;amp;= \phi(m) k_2(z) + 1 \\&#xA;\end{align*}&#xA;and $k_1(z)$ and $k_2(z)$ are integer functions, that approach infinity, such that $f(z)/g(z)$ approaches some irrational number. Can I then say, the equation:&#xA;$$x^{f(z)/g(z)} = y$$&#xA;has no solutions integer solutions, with the same $x$ and $y$, as $z$ goes to infinity as well?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is, if I let,&#xA;$$d = \lim_{z-&amp;gt;\infty}\frac{f(z)}{g(z)}$$&#xA;be the irrational number in the limit, then would it be true that,&#xA;$$x^d \neq y$$&#xA;for the same $x$ and $y$ ?&lt;/p&gt;&#xA;" OwnerUserId="7105" LastEditorUserId="7105" LastEditorDisplayName="" LastEditDate="2011-02-16T04:37:11.653" LastActivityDate="2011-02-16T04:37:11.653" Title="Are limits on exponents in moduli possible, if the modulus is relatively prime?" Tags="&lt;number-theory&gt;&lt;limit&gt;&lt;irrational-numbers&gt;&lt;powers&gt;" CommentCount="4" />
  <row Id="22280" PostTypeId="1" CreationDate="2011-02-16T04:29:11.333" Score="4" ViewCount="128" Body="&lt;p&gt;I asked this on mathoverflow, and was suggested to ask it here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Someone mentioned, in passing, to me that $u \mapsto \nabla \cdot ( c^2 \nabla u)$ is a Laplace-Beltrami operator. Does anyone have some insight into this? From my understanding, the Laplace-Beltrami operator generalizes the Laplacian to Riemannian manifolds, by taking the trace of the Hessian. I don't really see the connection to that and the above operator, unless c=1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This operator comes from the wave equation, where $\partial^2_t u -\nabla \cdot ( c^2 \nabla u) = f$. There may or not be some smoothness conditions on $c$, and all these are functions on subsets of $\mathbb{R}^n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for any help,&#xA;-Nick&lt;/p&gt;&#xA;" OwnerUserId="5564" LastActivityDate="2011-05-17T18:58:22.960" Title="is $\nabla \cdot ( c^2 \nabla)$ a Laplace-Beltrami operator?" Tags="&lt;differential-geometry&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="22281" PostTypeId="2" ParentId="22261" CreationDate="2011-02-16T04:35:54.633" Score="0" ViewCount="" Body="&lt;p&gt;You can find this in Paul Halmos' &lt;em&gt;Measure theory&lt;/em&gt;, Chapter I, §5. He talks about generated &lt;em&gt;rings&lt;/em&gt; but in your case the distintion is irrelevant---this is the content of one of the exercises in that section, in fact.&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2011-02-16T04:35:54.633" />
  <row Id="22282" PostTypeId="1" CreationDate="2011-02-16T05:02:10.203" Score="2" ViewCount="90" Body="&lt;p&gt;Statement: If E is a bounded set of real numbers, there exists a G-delta set G such that G contains E and has the same Lebesgue outer measure with E.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I completed the proof for the case that E is countable, but how should I prove this for the case that E is uncountable?&lt;/p&gt;&#xA;" OwnerUserId="7136" LastActivityDate="2011-02-16T06:15:42.567" Title="G-delta set with the same Lebesgue outer measure" Tags="&lt;measure-theory&gt;" AnswerCount="1" />
  <row Id="22283" PostTypeId="2" ParentId="22158" CreationDate="2011-02-16T05:03:17.783" Score="4" ViewCount="" Body="&lt;p&gt;Such a field is algebraically closed! Let $p \in K[x]$ be any polynomial, and consider $xp$; then $xp$ has a root in $K$ (namely, zero); so all the roots of $xp$ lie in $K$, and that includes all roots of $p$.&lt;/p&gt;&#xA;" OwnerUserId="536" LastActivityDate="2011-02-16T05:03:17.783" />
  <row Id="22284" PostTypeId="1" AcceptedAnswerId="22286" CreationDate="2011-02-16T05:04:49.327" Score="3" ViewCount="220" Body="&lt;p&gt;$$\int (2x^2+4x-2)^{-\frac{3}{2}} \ dx$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Complete the square?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int \frac{1}{(2(x+1)^2-4)^\frac{3}{2}} \ dx$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Not sure what do do next, or if I should try something else?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Big help if you can show as &quot;step-by-step&quot; possible as you can.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance!&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;@Chandru1:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm confused at what happens to the 2 that gets factored out when completing the square (the 2 in the denominator) because it looks like Arturo left it out and instead substituted: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\int \frac {1}{(2(x+1)^2-4)^\frac{3}{2}} \ dx$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$u = x+1$&#xA; $du =dx$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;which changes the integral to  $\int \frac {1}{(2(u^2-4))^\frac{3}{2}}$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would I be best factoring it out as a 1/2 in front of the integral  or leaving it in and then using trig. substitution like this?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$u= \sqrt{2}sec(t)$&#xA; $du = \sqrt{2}sec(t)*tan(t) dt$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which changes it to:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\int \frac {\sqrt{2}sec(t)*tan(t)}{(2(2sec^2(t)-4))^\frac{3}{2}} \ dt$ = $\int \frac {\sqrt{2}sec(t)*tan(t)}{(2(tan^2(t)))^\frac{3}{2}} \ dt$ = $\int \frac {\sqrt{2}sec(t)*tan(t)}{(2tan^3(t))} \ dt$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My $\sqrt{2}$ factors out like yours, but I have no 1/8 so figure I factor it out as a 1/2 earlier like I thought?  But then the constant in front of my integral would be $\frac{\sqrt{2}}{2}$  and you have $\frac{\sqrt{2}}{8}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know what's wrong, just not how I went wrong.&lt;/p&gt;&#xA;" OwnerUserId="6423" LastEditorUserId="6423" LastEditorDisplayName="" LastEditDate="2011-02-16T10:48:41.300" LastActivityDate="2011-02-16T15:38:39.157" Title="Help solving another integral $\int (2x^2+4x-2)^{-\frac{3}{2}} \ dx$" Tags="&lt;calculus&gt;&lt;homework&gt;&lt;integral&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="22286" PostTypeId="2" ParentId="22284" CreationDate="2011-02-16T05:13:35.143" Score="8" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Hint.&lt;/strong&gt; $(2x^{2}+4x-2) = 2(x^{2}+2x +1) -4 = 2(x+1)^{2}-4$. Now subsitute $x+1 = \sqrt{2} \sec{t}$. Then you have $dx = \sqrt{2} \sec{t} \cdot \tan{t} \ dt$. Now, $$2 \cdot \Bigl[(x+1)^{2} -2 \Bigr]= 2 \cdot \Bigl[ 2 \cdot (\sec^{2}{t}-1)\Bigr] = 4 \cdot (\sec^{2}{t}-1)$$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now what will you get if you power them by $\frac{3}{2}$. You will get $4^{3/2} \cdot \Bigl[ \tan^{2}{t}\Bigr]^{3/2} = 8 \cdot \tan^{3}{t}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And try to evaluate the integral. Simplyfying you should get &lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align*}&#xA;\int \frac{1}{(2x^{2}+4x-2)^{-\frac{3}{2}}} \ \textrm{dx} &amp;amp;= \frac{\sqrt{2}}{8} \int\frac{\sec{t} \cdot \tan{t}}{\tan^{3}{t}} \ \textrm{dt} \\ &amp;amp;= \frac{\sqrt{2}}{8} \int\frac{\cos{t}}{\sin^{2}{t}} \ \textrm{dt}&#xA;\end{align*}&lt;/p&gt;&#xA;" OwnerUserId="9413" LastEditorUserId="9413" LastEditorDisplayName="anonymous" LastEditDate="2011-02-16T10:56:52.600" LastActivityDate="2011-02-16T10:56:52.600" CommentCount="6" />
  <row Id="22287" PostTypeId="2" ParentId="22284" CreationDate="2011-02-16T05:16:00.963" Score="8" ViewCount="" Body="&lt;p&gt;First, factor out a $2$:&#xA;$$(2x^2 + 4x - 2) = 2(x^2+2x-1)\quad\text{so}\quad (2x^2+4x-2)^{-3/2} = 2^{-3/2}(x^2+2x-1)^{-3/2}.$$&#xA;Do this because you can pull it out of the integral and it will make things easier.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, &lt;strike&gt;up to a constant&lt;/strike&gt;, this is the same as doing the integral&#xA;$$\frac{1}{2^{3/2}}\int\frac{dx}{(x^2+2x-1)^{3/2}}.$$&#xA;Completing the square is a good first step. $x^2+2x-1 = (x^2+2x+1) - 2 = (x+1)^2 - 2$. So we get the integral&#xA;$$\frac{1}{2^{3/2}}\int\frac{dx}{\bigl( (x+1)^2 - 2\bigr)^{3/2}}.$$&#xA;Doing a change of variable $u=x+1$ changes it to&#xA;$$\frac{1}{2^{3/2}}\int\frac{du}{(u^2-2)^{3/2}}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now try a trigonometric substitution to get rid of that pesky square root in the exponent. Use $\tan^2\theta + 1 = \sec^2\theta$, or $\sec^2\theta - 1 =\tan^2\theta$. So set $u = \sqrt{2}\sec\theta$ to get&#xA;$$u^2-2 = 2\sec^2\theta - 2 = 2(\sec^2\theta - 1) = 2\tan^2\theta$$&#xA;and $du = \sqrt{2}\sec\theta\tan\theta\,d\theta$, so&#xA;$$\frac{1}{2^{3/2}}\int\frac{du}{(u^2-2)^{3/2}} = \frac{1}{2^{3/2}}\int\frac{\sqrt{2}\sec\theta\tan\theta\,d\theta}{(2\tan^2\theta)^{3/2}} = \frac{1}{2^{3/2}}\int\frac{\sqrt{2}\sec\theta\tan\theta\,d\theta}{2^{3/2}\tan^3\theta}.$$&#xA;Now work the trigonometric integral.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-16T15:38:39.157" LastActivityDate="2011-02-16T15:38:39.157" CommentCount="8" />
  <row Id="22288" PostTypeId="1" CreationDate="2011-02-16T05:20:22.427" Score="4" ViewCount="85" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/P-adic_number#Algebraic_approach&quot; rel=&quot;nofollow&quot;&gt;As we all know&lt;/a&gt;, the ring $Z_p$ can be constructed as the projective limit of the rings $Z/p^{n}Z$.&lt;br&gt;&#xA;Now is there any generalization such as the p-adic completions of a Dedekind Domain?&lt;br&gt;&#xA;This might be said to be inspired by the general treatment of extensions of Dedekind Domains from the treatment of algebraic number fields.&lt;br&gt;&#xA;In any case, thanks very much.  &lt;/p&gt;&#xA;" OwnerUserId="6792" LastEditorUserId="6792" LastEditorDisplayName="" LastEditDate="2011-02-16T09:35:41.130" LastActivityDate="2011-02-16T09:35:41.130" Title="Is there a notion of *p-adic Dedekind Domains* ?" Tags="&lt;reference-request&gt;&lt;algebraic-number-theory&gt;&lt;p-adic-number-theory&gt;" CommentCount="6" FavoriteCount="2" />
  <row Id="22289" PostTypeId="1" CreationDate="2011-02-16T05:25:57.400" Score="0" ViewCount="65" Body="&lt;p&gt;The table on the left gives the birth and death rates (per million) by region. The table on the right gives the populations (in millions) in each region for a number of years. Use matrix multiplication to get a matrix showing the total number of births and deaths in each year.&#xA;                Births          Deaths&#xA;Africa          .036             .014&#xA;Asia            .019             .008&#xA;Latin America   .021             .006&#xA;North America   .014             .008&#xA;Europe          .011             .011 years   Africa  Asia    L.America  N.America  Europe&#xA;                                      1960    283    1628      218       199         425&#xA;                                      2002    839    3518      539       320         513&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you&lt;/p&gt;&#xA;" OwnerUserId="7137" LastActivityDate="2011-03-18T06:33:37.447" Title="How do I label the solution of a word problem using matrix multiplication?" Tags="&lt;matrices&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="22290" PostTypeId="1" CreationDate="2011-02-16T05:34:23.737" Score="10" ViewCount="363" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Let $f\colon\mathbb{R}\to\mathbb{R}$. Prove that the set &#xA;  $$\{x \mid \mbox{if $y$ converges to $x$, then $f(y)$ converges to $\infty$}\}$$ is countable.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My book told me to consider  $g(x)=\arctan(f(x))$, then it said &quot;it is easy to see the set is countable.&quot; But I still can't understand what it mean.  &lt;/p&gt;&#xA;" OwnerUserId="7020" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-16T05:37:36.680" LastActivityDate="2011-02-17T06:19:20.307" Title="Proving the set of points at which a function diverges to $\infty$ is countable" Tags="&lt;real-analysis&gt;" AnswerCount="5" CommentCount="11" FavoriteCount="3" />
  <row Id="22291" PostTypeId="1" AcceptedAnswerId="22494" CreationDate="2011-02-16T05:34:46.370" Score="1" ViewCount="52" Body="&lt;p&gt;There's a way to represent proper colorings as graph homomorphisms, is there a similar construction for spanning trees of a graph?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For proper $m$-colorings of graph $G$, the relevant homomorphism is the mapping that takes adjacent vertices of $G$ to adjacent vertices of $K_m$&lt;/p&gt;&#xA;" OwnerUserId="998" LastActivityDate="2011-02-17T13:43:55.167" Title="spanning trees as graph homomorphisms" Tags="&lt;graph-theory&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="22292" PostTypeId="2" ParentId="22289" CreationDate="2011-02-16T05:35:34.880" Score="1" ViewCount="" Body="&lt;p&gt;Edit: Didn't notice there were several years on the &quot;table&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose there are $R$ regions (in your case, $R = 5$) and $Y$ years (in your case, $Y = 2$). You have an $R \times 2$ matrix $A_{rs}$, where $A_{r1}$ is the birth rate in region $r$, and $A_{r2}$ is the death rate. You also have the region population $Y \times R$ matrix $B_{yr}$, which represents the population of region $r$ at year $y$. You want the total number of deaths and births for each year. This is a summation over the different regions of something, which corresponds to a multiplication of the two matrices (perhaps transposed).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where are you stuck?&lt;/p&gt;&#xA;" OwnerUserId="1277" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2011-02-16T06:29:11.463" LastActivityDate="2011-02-16T06:29:11.463" CommentCount="2" />
  <row Id="22293" PostTypeId="2" ParentId="22276" CreationDate="2011-02-16T05:43:27.093" Score="4" ViewCount="" Body="&lt;p&gt;As $\mathbb{C}$ comes nicely equipped with a metric, limits behave as nice as ever. Therefore, the limit of the product is the product of the limits. And the limit of the conjugate is the conjugate of the limit (this can be proved using a standard epsilon-delta argument). Hence, in your example:&#xA;$$\lim_{z \to i} (z \cdot \overline{z})=\lim_{z \to i}(z) \lim_{z \to i}(\overline{z})= i \cdot(-i) = 1.$$&lt;/p&gt;&#xA;" OwnerUserId="4284" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-16T05:44:56.997" LastActivityDate="2011-02-16T05:44:56.997" CommentCount="1" />
  <row Id="22294" PostTypeId="1" AcceptedAnswerId="22301" CreationDate="2011-02-16T05:44:43.167" Score="2" ViewCount="64" Body="&lt;p&gt;We have a function $f$ of $N$ variables which is the product of $M$ polynomials: &#xA;$$f(x_1,x_2,\ldots, x_N) = P_1 \cdot P_2 \cdots  P_M.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each $P_i$ is a polynomial of at most three variables ($x_j$s) with degree 1, and each variable appears in at most two of the $P_i$ 's, so the highest degree of each variable in $f$ is limited by $2$. &#xA;Now, the question is that if we expand $f$ and then replace every $x_i^2$ with $x_i$, would the result be absolute zero or not? in other words: &#xA;$$g(x_1,x_2,\ldots, x_N) = f(x_1,x_2,\ldots, x_N) \bmod (x_1^2-x_1) \bmod (x_2^2-x_2) \cdots \bmod (x_N^2-x_N)$$&#xA;is $g == 0$ or not ? (by $\bmod$ I mean polynomial remainder)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;$f = (x+y - 2xy)(y+z - 2yz)(z+x - 2zx)$. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$g = \mathrm{remainder}(\mathrm{remainder}(\mathrm{remainder}(f , x^2-x), y^2-y), z^2-z) == 0$. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;i.e. if we expand f, and then replace every $x^2$ by $x$ , $y^2$ by $y$, and $z^2$ by $z$, all the terms will cancel out and the result will be absolute zero. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The challenge is that, expanding $f$ may result in an exponential number of intermediate terms, so I'm looking for an algorithm that can test whether $g==0$ without the need to fully expand $f$. For example by assigning some numerical values to $x_i$ and evaluating $f$ or through some transformation that can simplify calculation of $g$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks, &#xA;MG&lt;/p&gt;&#xA;" OwnerUserId="7138" LastEditorUserId="7138" LastEditorDisplayName="" LastEditDate="2011-02-16T06:55:14.743" LastActivityDate="2011-02-16T06:55:14.743" Title="An efficient way to check whether a polynomial (under certain condition) is absolutely equal to zero or not" Tags="&lt;number-theory&gt;&lt;linear-algebra&gt;&lt;algorithms&gt;&lt;polynomials&gt;&lt;computer-science&gt;" AnswerCount="1" FavoriteCount="2" />
  <row Id="22295" PostTypeId="1" AcceptedAnswerId="22543" CreationDate="2011-02-16T05:45:32.560" Score="3" ViewCount="140" Body="&lt;p&gt;Let $I$ be an uncountable set of indices and $U_\alpha\subset\mathbb{R}^3$ be an open subset for each $\alpha\in I$. We know that $U=\bigcup_{\alpha\in I} U_\alpha$ is an open subset.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is: does there exist a countable subset $J \subset I$ such that  $U=\bigcup_{\alpha\in J} U_\alpha$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general it is not true without the openness condition. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;As suggested by Chandru, this is a `named' property: Lindelöf space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Every $\sigma$--compact set is Lindelöf: Let $K_0\subset\cdots K_n\to\subset U$ be an increasing sequence of compact subsets with $\bigcup_{n\ge0}K_n=U$. Each $K_n$ has a finite open cover. So $U$ has a countable open cover.&lt;/p&gt;&#xA;" OwnerUserId="6145" LastEditorUserId="6145" LastEditorDisplayName="" LastEditDate="2011-02-16T06:04:23.043" LastActivityDate="2011-02-17T19:35:40.430" Title="When could an uncountable union be reduced to a countable union?" Tags="&lt;topology&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="22296" PostTypeId="2" ParentId="22262" CreationDate="2011-02-16T05:51:53.347" Score="0" ViewCount="" Body="&lt;p&gt;I think this problem is equivalent to the problem of finding a curvature on a plane which intersect with $y=x$ only at interger points .Of course ,this curvature must be the graph of a function.but it is not difficult to find.For example,you can piece some circle curves together.(I don't know how to post the picture.)&lt;/p&gt;&#xA;" OwnerUserId="7020" LastActivityDate="2011-02-16T05:51:53.347" />
  <row Id="22297" PostTypeId="1" CreationDate="2011-02-16T06:04:14.070" Score="2" ViewCount="82" Body="&lt;p&gt;&lt;strong&gt;Simple but Stuck: How do I find the point of intersection of two lines in Vector Calculus? Given symmetric equations and deriving parametric equations.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Find the point of intersection of the lines &lt;strong&gt;L1&lt;/strong&gt;: x/2 = (y+1)/3 = (z-1)/2 and &lt;strong&gt;L2&lt;/strong&gt;: x/4 = y/5 = z/5&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The parametric forms should be $x_1$ = 2t, $y_1$ = -1+3t, $z_1$=1+2t and $x_2$=4t, $y_2$=5t, $z_2$ = 5t&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The final solution is (4, 5, 5). It almost looks like someone just grabbed the coefficients from the parametric form of the second line to make the point. Working backwards, how do I get this answer?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sincerely,&lt;br&gt;&#xA;In need of the process.&lt;br&gt;&#xA;Thank you very much.&lt;/p&gt;&#xA;" OwnerUserId="7140" LastActivityDate="2011-02-16T06:07:29.980" Title="Simple but Stuck: How do I find the point of intersection of two lines in Vector Calculus?" Tags="&lt;calculus&gt;&lt;geometry&gt;&lt;vector-calculus&gt;" AnswerCount="1" />
  <row Id="22298" PostTypeId="2" ParentId="22297" CreationDate="2011-02-16T06:07:29.980" Score="1" ViewCount="" Body="&lt;p&gt;Let the parameter of the second line be $s$ and solve $2t=4s, -1+3t=5s, 1+2t=5s$.  If they are not consistent, there is no intersection.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-02-16T06:07:29.980" CommentCount="4" />
  <row Id="22299" PostTypeId="2" ParentId="22282" CreationDate="2011-02-16T06:15:42.567" Score="3" ViewCount="" Body="&lt;p&gt;How are you defining the Lebesgue outer measure $\tau$ of a set $E$? It should be the infimum of the set $M$ consisting of all those numbers $r$ that are the measures of certain open sets that cover $E$, right?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that this infimum is finite because $E$ is bounded: To be bounded means that for some $l$, $E\subseteq(-l,l)$, so $2l\in M$ and $\tau=\inf M\le 2l$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take a sequence $r_n$ of elements of $M$ that is decreasing and converges to $\tau$. This sequence exists because $\tau$ is an infimum: For each $n$ there must be an $r\in M$ with $\tau\le r&amp;lt;\tau+(1/n)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For each $r_n$ pick a &lt;em&gt;witnessing&lt;/em&gt; open set $E_n$ covering $E$ with measure $r_n$. We may as well assume that $E_n$ is bounded. If it is not, replace it with $E_n&amp;#39;=E_n\cap(-l,l)$. Of course, this may replace $r_n$ with a smaller number. That's fine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The point, of course, is that $F=\bigcap_nE_n$ is a $G_\delta$ set that contains $E$ (since each $E_n$ is open and contains $E$). To check that $F$ has the same outer measure as $E$, note that any open set covering $F$ also covers $E$, so the outer measure $\rho$ of $F$ satisfies $\rho\ge\tau$. On the other hand, $E_n$ covers $F$, so $\rho\le\tau+(1/n)$. Therefore, $\rho\le\tau$.&lt;/p&gt;&#xA;" OwnerUserId="462" LastActivityDate="2011-02-16T06:15:42.567" CommentCount="1" />
  <row Id="22300" PostTypeId="2" ParentId="22290" CreationDate="2011-02-16T06:20:06.783" Score="-1" ViewCount="" Body="&lt;p&gt;Maybe one of these two will help:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Every uncountable subset of R^n has a limit point --seems this limit point may&#xA;    prevent your function from being real-valued ( this follows from the plane being&#xA;    Lindeloff)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) The set of points of continuity of a function f:R--&gt;R is a G-delta set( of course, if f&#xA;   was continuous at x, then, by sequential continuity, we would necessarily have&#xA;   f(y)-&gt;f(x) )&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I cannot think of an actual proof, and I don't have enough points to post comments&lt;/p&gt;&#xA;" OwnerUserId="7141" LastActivityDate="2011-02-16T06:20:06.783" CommentCount="3" />
  <row Id="22301" PostTypeId="2" ParentId="22294" CreationDate="2011-02-16T06:24:20.577" Score="1" ViewCount="" Body="&lt;p&gt;This is a &lt;a href=&quot;http://en.wikipedia.org/wiki/Constraint_satisfaction_problem&quot; rel=&quot;nofollow&quot;&gt;constraint satisfaction problem&lt;/a&gt; with the additional requirement that every variable appear in at most two clauses. This is dealt with in a &lt;a href=&quot;http://www.google.ca/url?sa=t&amp;amp;source=web&amp;amp;cd=1&amp;amp;ved=0CBkQFjAA&amp;amp;url=http%3A%2F%2Fciteseerx.ist.psu.edu%2Fviewdoc%2Fdownload%3Fdoi%3D10.1.1.113.3236%26rep%3Drep1%26type%3Dps&amp;amp;rct=j&amp;amp;q=fanout%20limitations%20on%20constraint%20systems&amp;amp;ei=cWlbTaChKImUtweMg5CfDA&amp;amp;usg=AFQjCNG1c6Rln6D4rZZS48SyOnsLutelaA&amp;amp;cad=rja&quot; rel=&quot;nofollow&quot;&gt;paper&lt;/a&gt; by Feder and another &lt;a href=&quot;http://www.dtic.upf.edu/%7Evdalmau/mfcs6.ps.gz&quot; rel=&quot;nofollow&quot;&gt;paper&lt;/a&gt; by Dalmau and Ford.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you really allow all possible clauses involving three variables, then you can encode 3SAT, as Feder shows: if a variable $x$ is used $n$ times, replace every use of it by $x_i$, and add consistency clauses:&#xA;$$ (x_1=x_2=y_1) \land (y_1=x_3=y_2) \land (y_2=x_4=y_3) \land \cdots \land (y_{n-3} = x_{n-1} = x_n). $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Feder and Dalmau and Ford are trying to characterize the restrictions on the clauses which result in polytime-solvable problems; the latter (attributing the result to the former) have shown that if every variable is allowed to occur three times, then the problem is always NP-complete. Related work is &lt;a href=&quot;http://www.google.ca/url?sa=t&amp;amp;source=web&amp;amp;cd=1&amp;amp;sqi=2&amp;amp;ved=0CCAQFjAA&amp;amp;url=http%3A%2F%2Fwww.ccs.neu.edu%2Fhome%2Flieber%2Fcourses%2Fcsg260%2Ff06%2Fmaterials%2Fpapers%2Fmax-sat%2Fp216-schaefer.pdf&amp;amp;rct=j&amp;amp;q=schaeffer%20the%20complexity%20of%20sati&amp;amp;ei=bWxbTeZvgvjwBtb-2bQO&amp;amp;usg=AFQjCNG_f2JcF9lMFLruW8u5IBJGSiMSBA&amp;amp;cad=rja&quot; rel=&quot;nofollow&quot;&gt;Schaefer&lt;/a&gt; which does just that without the &quot;maximum-use&quot; constraint, and &lt;a href=&quot;http://www.google.ca/url?sa=t&amp;amp;source=web&amp;amp;cd=2&amp;amp;ved=0CBwQFjAB&amp;amp;url=http%3A%2F%2Fportal.acm.org%2Fcitation.cfm%3Fid%3D152385&amp;amp;rct=j&amp;amp;q=kratochvil%20One%20more%20occurrence%20of%20variables%20makes%20satisfiability%20jump%20from%20trivial%20to%20np-complete&amp;amp;ei=3mdbTbm9HpSgtwfZm42DDA&amp;amp;usg=AFQjCNHWuHNQOVTdwXm8b0cZw3SpUR9ogQ&amp;amp;cad=rja&quot; rel=&quot;nofollow&quot;&gt;Kratochvíl&lt;/a&gt; which shows that for $k$-SAT, there is a jump from polytime-solvable to NP-complete, depending on the allowed number of occurrences (I'm not sure whether they give the exact number).&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2011-02-16T06:24:20.577" CommentCount="1" />
  <row Id="22302" PostTypeId="2" ParentId="21949" CreationDate="2011-02-16T06:31:48.930" Score="7" ViewCount="" Body="&lt;p&gt;Question (2) is a known problem; it's called the &lt;em&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Change-making_problem&quot; rel=&quot;nofollow&quot;&gt;change-making problem&lt;/a&gt;&lt;/em&gt;.  The Wikipedia page lists integer programming and &lt;a href=&quot;http://en.wikipedia.org/wiki/Dynamic_programming&quot; rel=&quot;nofollow&quot;&gt;dynamic programming&lt;/a&gt; as solution approaches.  The change-making problem is also a variation of the famous &lt;a href=&quot;http://en.wikipedia.org/wiki/Knapsack_problem&quot; rel=&quot;nofollow&quot;&gt;knapsack problem&lt;/a&gt;.  One of the implications of that is that since the standard version of the knapsack problem is NP-complete, it's likely that the change-making problem is as well, dashing hopes for an easy-to-find fast algorithm for solving it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the Wikipedia page also says that for the US and most other coin systems, the greedy algorithm will give the optimal solution.  Thus for the nickel/dime/quarter example you give, the optimal solution is to use as many quarters as you can, then account for what's left by using as many dimes as you can, then make up the rest with nickels.  So 3 is the minimum number of nickels, dimes, and quarters required to make 40 cents.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;HR&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's the integer program formulation for Question (2).  If you let $x_i$ be the number of coins of type $i$ to be used, $d_i$ be the denomination of coin type $i$, and $A$ be the target amount, Question (2) entails solving &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\min \sum_i x_i $$&#xA;subject to&#xA;$$\sum_i d_i x_i = A,$$&#xA;$$x_i \geq 0, x_i \in \mathbb{Z}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While integer programs are hard to solve in general, for small problems like the example you give a good IP solver will return a solution very quickly. For instance, if you take coins of denominations 31, 37, and 41 cents (since nickels, dimes, and quarters are too easy :) ), and want to know the minimum number required to be equivalent to 2011 cents, the solver &lt;a href=&quot;http://www.lindo.com/index.php?option=com_content&amp;amp;view=article&amp;amp;id=2&amp;amp;Itemid=10&quot; rel=&quot;nofollow&quot;&gt;Lingo&lt;/a&gt; requires only the following code to find the answer:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;MIN = x1 + x2 + x3;&#xA;31*x1 + 37*x2 + 41*x3 = 2011;&#xA;@GIN(x1); @GIN(x2); @GIN(x3);&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Lingo outputs the optimal solution as $x_1 = 0, x_2 = 20, x_3 = 31,$ with only 7 solver iterations and &quot;00:00:00&quot; time elapsed.  So the minimum number of coins is $51$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You did ask specifically for an algorithm.  Many IP solvers (including Lingo) use the &lt;a href=&quot;http://mat.gsia.cmu.edu/orclass/integer/node13.html#SECTION00042000000000000000&quot; rel=&quot;nofollow&quot;&gt;branch and bound algorithm&lt;/a&gt;, which has solving a linear program (usually with the &lt;a href=&quot;http://en.wikipedia.org/wiki/Simplex_algorithm&quot; rel=&quot;nofollow&quot;&gt;simplex method&lt;/a&gt;) embedded in it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;HR&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For Question (1), you could use the following integer program to find a single solution or determine that one does not exist.  (In the latter case, the IP solver will report that the problem is infeasible.)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\max 1$$&#xA;subject to&#xA;$$\sum_i d_i x_i = A,$$&#xA;$$\sum_i x_i = n,$$&#xA;$$x_i \geq 0, x_i \in \mathbb{Z}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again, for small problems a good IP solver will return a solution almost instantaneously.   I'm not sure how (or even if it's possible) to modify an IP approach to find all solutions like you request.  The generating function techniques mentioned by cardinal and in the question linked to by Derek Jennings will tell you &lt;em&gt;how many&lt;/em&gt; solutions there are, but I'm not sure if they will tell you exactly what those solutions are.  &lt;/p&gt;&#xA;" OwnerUserId="2370" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2011-02-16T21:57:10.383" LastActivityDate="2011-02-16T21:57:10.383" />
  <row Id="22303" PostTypeId="1" CreationDate="2011-02-16T06:45:10.977" Score="2" ViewCount="165" Body="&lt;p&gt;$\forall x \in Z \exists y \in Z(x *y = x + y)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I'm understanding this correctly this is FALSE.&lt;/p&gt;&#xA;" OwnerUserId="1040" LastActivityDate="2011-03-18T17:33:39.983" Title="Universal Quantification and Sets $\forall x \in Z \exists y \in Z(x *y = x + y)$" Tags="&lt;logic&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="22304" PostTypeId="2" ParentId="22295" CreationDate="2011-02-16T06:52:53.700" Score="1" ViewCount="" Body="&lt;p&gt;As suggested by Chandru, this is a `named' property: Lindelöf space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Every $\sigma$--compact set is Lindelöf: Let $K_0\subset\cdots K_n\to\subset U$ be an increasing sequence of compact subsets with $\bigcup_{n\ge0}K_n=U$. Each $K_n$ has a finite open cover. So $U$ has a countable open cover.&lt;/p&gt;&#xA;" LastActivityDate="2011-02-16T06:52:53.700" CommunityOwnedDate="2011-02-16T07:56:37.173" />
  <row Id="22305" PostTypeId="1" AcceptedAnswerId="22306" CreationDate="2011-02-16T06:54:35.433" Score="4" ViewCount="129" Body="&lt;p&gt;Let S be the collection of vectors $\begin{pmatrix} x \\ y \\ z \end{pmatrix}$ in $\mathbb{R}^3$ such that $|x-y|=|y-z|$. Either prove that S forms a subspace of $\mathbb{R}^3$ or give a counterexample to show that it does not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(from Poole, Linear Algebra - A Modern Introduction 2nd ed., exercise 3.5.8)&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I'm working on this book by myself, so if you can just give me a hint to show me the right direction, that would be great. I have been trying to solve this by going through the three requirements of a subspace. &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The zero vector is in S.&lt;/li&gt;&#xA;&lt;li&gt;If &lt;b&gt;u&lt;/b&gt; and &lt;b&gt;v&lt;/b&gt; are in S, then &lt;b&gt;u&lt;/b&gt; + &lt;b&gt;v&lt;/b&gt; is in S.&lt;/li&gt;&#xA;&lt;li&gt;If &lt;b&gt;u&lt;/b&gt; is in S and &lt;i&gt;c&lt;/i&gt; is a scalar, then &lt;i&gt;c&lt;/i&gt;&lt;b&gt;u&lt;/b&gt; is in S.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;It's easy to see the first one holds. For the third one I noticed that by adding &lt;i&gt;c&lt;/i&gt; as multiplier on both sides of $|x-y|=|y-z|$ and fiddling a bit it becomes $|$&lt;i&gt;c&lt;/i&gt;$u_x$-&lt;i&gt;c&lt;/i&gt;$u_y|$ = $|$&lt;i&gt;c&lt;/i&gt;$u_y$-&lt;i&gt;c&lt;/i&gt;$u_z|$, which is the form I need to show it is closed under scalar multiplication.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But how to show the second one I have no idea. Stuff I tried tended to split into multiple cases because absolute values are taken of the left and right sides, it didn't feel like the right direction. Is there something really easy I'm missing here?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also tried imagining what the subspace would look like, but it's too difficult to visualize.&lt;/p&gt;&#xA;" OwnerUserId="7139" LastEditorUserId="3731" LastEditorDisplayName="" LastEditDate="2011-02-16T08:23:05.567" LastActivityDate="2011-02-16T09:08:50.970" Title="Prove $[x,y,z]$ such that $|x-y|=|y-z|$ is a subspace of $\mathbb{R}^3$" Tags="&lt;linear-algebra&gt;" AnswerCount="3" />
  <row Id="22306" PostTypeId="2" ParentId="22305" CreationDate="2011-02-16T06:58:55.813" Score="5" ViewCount="" Body="&lt;p&gt;What about vectors $\begin{bmatrix} 3\\ 7\\ 11 \end{bmatrix}$ and $\begin{bmatrix} 2\\ 8\\ 2 \end{bmatrix}$? These are both in $S$, but $\begin{bmatrix} 5\\ 15\\ 13 \end{bmatrix}$ is not in $S$.&lt;/p&gt;&#xA;" OwnerUserId="1583" LastEditorUserId="1583" LastEditorDisplayName="" LastEditDate="2011-02-16T07:13:50.647" LastActivityDate="2011-02-16T07:13:50.647" CommentCount="2" />
  <row Id="22307" PostTypeId="2" ParentId="22305" CreationDate="2011-02-16T07:10:53.560" Score="5" ViewCount="" Body="&lt;p&gt;An easy way to find a counter-example is to take a vector of form $\begin{bmatrix}&#xD;&#xA;a\\ &#xD;&#xA;b\\ &#xD;&#xA;c&#xD;&#xA;\end{bmatrix}$ where $|a-b|=|b-c|$ holds, but, for example, $a=b+c$ (or $c=a+b$). Then add some vector of form, say, $\begin{bmatrix}&#xD;&#xA;x\\ &#xD;&#xA;0\\ &#xD;&#xA;x&#xD;&#xA;\end{bmatrix}$ to your first vector to derive a counterexample.&lt;/p&gt;&#xA;" OwnerUserId="3731" LastActivityDate="2011-02-16T07:10:53.560" />
  <row Id="22309" PostTypeId="2" ParentId="22290" CreationDate="2011-02-16T08:09:31.453" Score="10" ViewCount="" Body="&lt;p&gt;There was some discussion about the terminology in comments to the question -- I interpret the statement &quot;if $y$ converges to $x$, then $f(y)$ converges to $\infty$&quot; to mean &quot;for all $M\in\mathbb{R}$ there is $\delta &amp;gt; 0$ such that $f(y)&amp;gt;M$ for all $y\neq x$ with $\lvert y-x\rvert &amp;lt; \delta$&quot;. Let $S$ be the set of all such points $x$, and call a set that contains uncountably many points of $S$ &quot;uncountably divergent&quot; for short.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't think that the $\arctan$ transformation is of any help, since it's just a homeomorphism between $\mathbb{R}$ and $(-\pi,\pi)$ and between $\mathbb{R}\cup\{\infty\}$ and $(-\pi,\pi]$ and doesn't change the substance of the required proof. [Edit, after seeing Nick Kirby's answer and following his links: The transformation does make a difference, in that the limit $\infty$ is not allowed as a function value here, whereas it is if it is finite. This difference is relevant with respect to the example of the function with removable discontinuities at all rationals given &lt;a href=&quot;http://math.stackexchange.com/questions/3777/is-there-a-function-with-a-removable-discontinuity-at-every-point&quot;&gt;here&lt;/a&gt;, since this uses the limit at the rationals as the function value at the irrationals, which we can't do in the present case, so that this example doesn't yield an example of a function &quot;converging to $\infty$&quot; at the rationals -- in fact there is no such function; see the remark about nowhere dense sets below.]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume that $S$ is uncountable. First, observe that $[n,n+1]$ must then be uncountably divergent for some $n\in\mathbb{N}$. Without loss of generality, assume $n=0$, so $[0,1]$ is uncountably divergent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we divide the interval in half. Either $[0,1/2]$ or $[1/2,1]$ or both are uncountably divergent. If only one of them is, we discard the other half and replace the interval by the uncountably divergent half and continue dividing it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume that after each subdivision only one half is uncountably divergent. These halves form a chain of nested closed intervals whose intersection contains exactly one point. Thus, except possibly for one point, all points of $S$ in $[0,1]$ lie in one of the discarded halves that each contain at most countably many points of $S$. But there are only countably many of these halves, contradicting the fact that there are uncountably many points of $S$ in $[0,1]$. Hence the assumption is false and the subdivision must encounter an interval of which both halves are uncountably divergent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once we encounter such an interval, we denote it by $I$ and recursively apply the entire halving procedure to it (including the part where we discard intervals of which only one half is uncountably divergent). We call the two resulting intervals $I_0$ and $I_1$ and continue applying the halving procedure to them, resulting in intervals $I_{00}$, $I_{01}$, $I_{10}$, $I_{11}$, and so on, where each resulting interval has the property that both its halves are uncountably divergent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We discard any points of $S$ lying in the discarded halves, as well as all points lying on a boundary of one of the intervals. Since there are only countably many such points, each interval that was uncountably divergent remains so. Note that each remaining point of $S$ in $[0,1]$ now lies in a chain of nested intervals $I$, $I_{i_1}$, $I_{i_1i_2}$, ... with lengths converging to $0$, and the intersection of that chain contains exactly that point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We now begin with the interval $I$, choose some point $s$ of $S$ in it and find $\delta$ such that $f(y) &amp;gt; 1$ for all $y\neq s$ with $y \in (s-\delta,s+\delta)$. Now $s$ lies in a chain of nested closed intervals with lengths converging to $0$, so one of these intervals is entirely contained in $(s-\delta,s+\delta)$. Denote this interval by $I_i$, where $i$ is a string of binary digits, and choose the one of $I_{i0}$ and $I_{i1}$ which does not contain $s$, so that $f&amp;gt;1$ on this entire subinterval. We now apply the same procedure to this interval, choosing a point of $S$ in it but now choosing $\delta$ such that $f(y) &amp;gt; 2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Iterating this procedure, we obtain a chain of nested closed intervals whose lengths converge to $0$, and the intersection of these intervals contains exactly one point $t$. But now $f(t)&amp;gt;n$ for all $n\in\mathbb{N}$, which is impossible. Hence $S$ must be countable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[Edit: Note that uncountability was only used to make sure that we have points in both halves of an interval at every level of subdivision -- thus, we can use the same proof to show that $S$ is nowhere dense (except for the &quot;removable technicality&quot; that we discarded the set of boundary points, i.e. the set of points with finite binary representation, which is itself dense -- this was just to ensure that each point of an interval is contained in exactly one of the closed subintervals, but we can alternatively ensure that by choosing one of the four sub-subintervals, at most two of which contain the given point.). The same is not true for the case of finite limits (i.e. limits which are allowed as function values elsewhere), as the example of the function with removable discontinuities at all rationals shows. The crucial difference is in the last step of the proof, since $f(t)&amp;gt;n$ for all $n\in\mathbb{N}$ is impossible, whereas the corresponding statement for the limit $0$, $\lvert f(t)\rvert&amp;lt;1/n$ for all $n\in\mathbb{N}$, is consistent.]&lt;/p&gt;&#xA;" OwnerUserId="6622" LastEditorUserId="6622" LastEditorDisplayName="" LastEditDate="2011-02-17T01:08:00.440" LastActivityDate="2011-02-17T01:08:00.440" CommentCount="6" />
  <row Id="22310" PostTypeId="1" CreationDate="2011-02-16T08:56:10.183" Score="1" ViewCount="125" Body="&lt;p&gt;$\int (\frac {x-1}{3-x})^\frac{1}{2} dx$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i'm stuck in this part &lt;/p&gt;&#xA;&#xA;&lt;p&gt;let $u$$=$$\frac {x-1}{3-x}$ them $du$$=$$\frac {2}{(x-3)^2}$ $dx$ which can be represented as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$du$$=$$  \frac{1}{3-x} -  \frac {1-x}{(3-x)^2}$ $dx$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i can't &quot;see&quot; how to get to this $2$ $\int \frac{(u)^\frac{1}{2}}{(u+1)^2}dx$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;after this part i know how to solve it ;&#xA;I just wish someone would show me &quot;step by step&quot; this part &#xA;It seems it involves some sort of &quot;leap&quot; of thought; or there is a systematic way using basic algebra?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="7143" LastActivityDate="2011-02-16T12:39:42.923" Title="help solving an integral $\int (\frac {x-1}{3-x})^\frac{1}{2} dx$" Tags="&lt;calculus&gt;&lt;homework&gt;" AnswerCount="5" />
  <row Id="22311" PostTypeId="2" ParentId="22305" CreationDate="2011-02-16T09:08:50.970" Score="2" ViewCount="" Body="&lt;p&gt;This may be the geometric angle for why you don't get a subspace:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Break down your problem into the 4 possible cases for absolute values:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One case would be:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;x-y&gt;=0&#xA;   y-z&gt;=0&lt;/p&gt;&#xA;&#xA;&lt;p&gt;etc. And check what happens where the 4 regions overlap&lt;/p&gt;&#xA;" OwnerUserId="7141" LastActivityDate="2011-02-16T09:08:50.970" />
  <row Id="22312" PostTypeId="1" AcceptedAnswerId="22343" CreationDate="2011-02-16T09:09:33.490" Score="1" ViewCount="84" Body="&lt;p&gt;I'm hating these variable resistance questions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A body of mass $m$ falls from rest in a medium that produces a resistance of magnitude $m\cdot k \cdot v$. where $k$ is a constant, where the speed of the particle is $v$. Show that when the body has reached a speed $V$ it will have fallen for a time&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{1}{k} \ln \left(\frac{g}{g-k\cdot V}\right) \; .$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Help much appreciated.&lt;/p&gt;&#xA;" OwnerUserId="2874" LastEditorUserId="3567" LastEditorDisplayName="" LastEditDate="2011-02-16T11:44:00.703" LastActivityDate="2011-02-16T13:11:27.377" Title="A body falls through a medium." Tags="&lt;integral&gt;&lt;differential-equations&gt;&lt;physics&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="22313" PostTypeId="1" AcceptedAnswerId="22316" CreationDate="2011-02-16T09:10:00.633" Score="0" ViewCount="37" Body="&lt;p&gt;limit: [(12x)/(9-8x)] as x --&gt; (+)9/8  and (-)9/8.&#xA; I am confused how to do this. L'hospital, substitution and conjugates will not solve this as they result in the division of zero. I tried factoring out the largest degree:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(x/x)(12/[9/x]-8)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;integer over variable results in 0:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;12/ (0 - 8) =&lt;/p&gt;&#xA;&#xA;&lt;p&gt;12/-8 =&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-3/2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, upon further investigation this is the limit as it approaches infinity. I have tried a few other things, but none have resulted in a correct answer, this is as close as I have gotten. What do I have to do to get x --&gt; (+)9/8 instead of x --&gt; (+)inf? Thanks.&lt;/p&gt;&#xA;" OwnerUserId="7144" LastActivityDate="2011-02-16T09:25:12.823" Title="Limits of positive and negatives?" Tags="&lt;limit&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="22314" PostTypeId="2" ParentId="22310" CreationDate="2011-02-16T09:16:15.953" Score="0" ViewCount="" Body="&lt;p&gt;Notice the $u+1$ in the denominator. What is $u+1$ in terms of $x$? In particular, what is $\frac{1}{(u+1)^2}$?&lt;/p&gt;&#xA;" OwnerUserId="5523" LastActivityDate="2011-02-16T09:16:15.953" />
  <row Id="22315" PostTypeId="2" ParentId="22310" CreationDate="2011-02-16T09:22:11.937" Score="2" ViewCount="" Body="&lt;p&gt;First note&#xA;$$&#xD;&#xA;du=\frac{2}{(x-3)^2}dx&#xD;&#xA;$$&#xA;which implies &#xA;$$&#xD;&#xA;\frac{(x-3)^2}{2}du=dx.&#xD;&#xA;$$&#xA;Also, &#xA;$$&#xD;&#xA;u+1=\frac{x-1}{3-x}+\frac{3-x}{3-x}=\frac{2}{3-x}&#xD;&#xA;$$&#xA;which implies&#xA;$$&#xD;&#xA;(u+1)^2=\left(\frac{2}{3-x}\right)^2=\frac{4}{(x-3)^2}.&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Putting it together you have&#xA;$$&#xD;&#xA;\int \left(\frac {x-1}{3-x}\right)^\frac{1}{2} dx = \int u^{1/2}\cdot\frac{(x-3)^2}{2}du = 2\int u^{1/2}\cdot\frac{(x-3)^2}{4}du&#xD;&#xA;= 2\int \frac{u^{1/2}}{(u+1)^2}du.&#xD;&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="1583" LastActivityDate="2011-02-16T09:22:11.937" />
  <row Id="22316" PostTypeId="2" ParentId="22313" CreationDate="2011-02-16T09:25:12.823" Score="2" ViewCount="" Body="&lt;p&gt;If the limit of the denominator is 0 and the limit of the numerator is nonzero, the limit does not exist (is $\pm\infty$). You can easily determine the sign by plugging in numbers close to $\frac{9}{8}$ on either side.&lt;/p&gt;&#xA;" OwnerUserId="5523" LastActivityDate="2011-02-16T09:25:12.823" />
  <row Id="22317" PostTypeId="1" CreationDate="2011-02-16T09:25:18.817" Score="1" ViewCount="85" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Find two different 5-sylow subgroups in S16.&lt;br&gt;&#xA;  hint: use group multiplication.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;any hints?&lt;br&gt;&#xA;thanks ahead.&lt;/p&gt;&#xA;" OwnerUserId="6535" LastActivityDate="2011-02-16T10:36:17.880" Title="sylow 5-subgroup in S16" Tags="&lt;abstract-algebra&gt;&lt;group-theory&gt;&lt;algebra&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="2" />
  <row Id="22318" PostTypeId="2" ParentId="22310" CreationDate="2011-02-16T09:35:21.703" Score="1" ViewCount="" Body="&lt;p&gt;You can write the fraction as $\frac{(x-1)^{1/2}}{(3-x)^{1/2}}$. Then put $3-x =t$. So you have $dt= -dx$ and $x-1 =3-t-1=2-t$. So you have to now evaluate the integral $$\int \sqrt{ \frac{2-t}{t}}\ -\rm{dt}$$ which can be evaluated by sing the subsitution $t = 2 \cos^{2}{v}$.&lt;/p&gt;&#xA;" LastActivityDate="2011-02-16T09:35:21.703" />
  <row Id="22319" PostTypeId="2" ParentId="22310" CreationDate="2011-02-16T09:45:30.193" Score="1" ViewCount="" Body="&lt;p&gt;You probably reached this point after inserting u:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\int u^\frac{1}{2} \frac{(x-3)^2}{2}du$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The leap of thought you would need here is to realize that the remaining x must be substituted by u and so what you want to find is a function which in terms of u can replace x, in this case we can go for $(x-3)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x-3 = f(u)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A trial and error approach would be to inspect $u$ and see that (as you realized) $u$ can be written as: $u = -1 + \frac{2}{3-x}$, here we already see the term $3-x$ so we just need to manipulate the equation to let $x-3$ be on one side. step by step:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$u+1 = \frac{2}{3-x}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{2}{u+1} = 3-x$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$-\frac{2}{u+1} = x-3$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I.e. $f(u) = -\frac{2}{u+1}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we can substitute $(x-3)$ with $-\frac{2}{u+1}$ in the integral to get  $2$ $\int \frac{(u)^\frac{1}{2}}{(u+1)^2}dx$&lt;/p&gt;&#xA;" OwnerUserId="5464" LastEditorUserId="5464" LastEditorDisplayName="" LastEditDate="2011-02-16T12:39:42.923" LastActivityDate="2011-02-16T12:39:42.923" />
  <row Id="22320" PostTypeId="1" CreationDate="2011-02-16T09:50:08.077" Score="3" ViewCount="108" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Possible Duplicate:&lt;/strong&gt;&lt;br&gt;&#xA;  &lt;a href=&quot;http://math.stackexchange.com/questions/8943/computing-gcd-of-all-permutations-of-the-digits-of-a-given-number&quot;&gt;Computing GCD of all permutations (of the digits) of a given number.&lt;/a&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;p&gt;How can we find the greatest common divisor (GCD) of all numbers that can be obtained by permuting the digits in the given number?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please suggest a feasible algorithm.&lt;/p&gt;&#xA;" OwnerUserId="3744" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-18T13:43:39.667" LastActivityDate="2011-03-18T13:43:39.667" ClosedDate="2011-03-18T13:49:50.323" Title="Finding GCD of all permutations of a decimal number." Tags="&lt;number-theory&gt;&lt;algorithms&gt;&lt;contest-math&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" />
  <row Id="22321" PostTypeId="1" AcceptedAnswerId="22346" CreationDate="2011-02-16T09:52:26.470" Score="5" ViewCount="199" Body="&lt;p&gt;Sorry for this whole bunch of questions. Please note, that I know what a commutative diagram is, and that I can somehow read them, at least the simpler ones. But often enough the diagrams are labelled and/or explained verbally, thus my question:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;How are commutative diagrams to be&#xA;  read in general? Can they always be&#xA;  deterministically translated into&#xA;  first-order sentences or properties?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Which &quot;background assumptions&quot; are needed for a commutative diagram to make specific sense? (A minimal assumption should be that they tell something about a category. What else?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which conventions are used? (e.g. for existence and uniqueness of arrows)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When is it important to label the objects and arrows?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When are verbal explanations necessary?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does every conceivable diagram mean something?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there (graphical) analogues of commutative diagrams for other (relational or algebraic) structures than categories?&lt;/p&gt;&#xA;" OwnerUserId="1792" LastEditorUserId="1792" LastEditorDisplayName="" LastEditDate="2011-02-16T10:12:59.767" LastActivityDate="2011-02-18T12:48:26.040" Title="Reading commutative diagrams?" Tags="&lt;soft-question&gt;&lt;category-theory&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" />
  <row Id="22322" PostTypeId="2" ParentId="22317" CreationDate="2011-02-16T09:54:46.060" Score="0" ViewCount="" Body="&lt;p&gt;Does conjugation work here?&lt;br&gt;&#xA;In this case we just need to find a 5-sylow subgroup of $S_{16}$ which is not normal which will fulfill the condition, don't we?&lt;br&gt;&#xA;If I miss something, please tell me, thanks.&lt;br&gt;&#xA;By the way, to type $S_{16}$ is better to use S_{16} instead of S_16 as far as I am concerned.&lt;/p&gt;&#xA;" OwnerUserId="6792" LastActivityDate="2011-02-16T09:54:46.060" />
  <row Id="22323" PostTypeId="1" AcceptedAnswerId="22330" CreationDate="2011-02-16T09:57:05.510" Score="1" ViewCount="62" Body="&lt;p&gt;I'm super rusty on my real analysis and I'm stuck on this easy question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Show that for a semi-continuous function $ g: \mathbb{R}^n \rightarrow \mathbb{R}$, the set $\{x \in \mathbb{R}^n: g(x) \leq 0\}$ is closed&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have found some proofs online though I am working with somewhat different definitions of semi-continuous functions and closed sets and am trying to see whether sticking to the definitions I know.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A closed set is a set such that all limits points are within the set.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A point $x \in S$ is a limit point of $S$ if $\forall \delta &amp;gt; 0, \exists y \in S$ such that $y \neq x$ and  $|x-y| &amp;lt; \delta$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A function $g$ is lower semicontinuous at $x$ if $\forall \epsilon &amp;gt;0, \space \exists \delta &amp;gt; 0$ such that $g(x) - g(y) &amp;lt; \epsilon \space \forall y \in N_{\delta}(x)$ &lt;/p&gt;&#xA;" OwnerUserId="7145" LastActivityDate="2011-02-16T10:42:06.470" Title="Show that: for a semi-continuous function $g: \mathbb{R}^n \rightarrow \mathbb{R}$, the set $\{x \in \mathbb{R}^n: g(x) \leq 0\}$ is closed" Tags="&lt;real-analysis&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="22324" PostTypeId="1" AcceptedAnswerId="22389" CreationDate="2011-02-16T09:59:55.053" Score="8" ViewCount="239" Body="&lt;p&gt;In my book, Galois group is defined to mean the set of automorphisms on $E/F$ that &quot;leave alone&quot; the elements in $F$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On Wikipedia it says:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;If $E/F$ is a Galois extension, then $Aut(E/F)$ is called the Galois group of (the extension) $E$ over $F$, $\dots$&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And Wikipedia's definition of Galois:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;An algebraic field extension $E/F$ is Galois if it is normal and separable. Equivalently, the extension $E/F$ is Galois if and only if it is algebraic, and the field fixed by the automorphism group $Aut(E/F)$ is precisely the base field $F$.&quot; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in one case, Wikipedia, the extension is restricted to be algebraic. So the set of automorphisms on $\mathbb{Q}(\pi) / \mathbb{Q}$ is not a Galois group. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question: How is it possible to have two different definitions of what a Galois group is? Do these not conflict? Or what am I missing here?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many thanks for your help. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm using J. Gallian, Contemporary Abstract Algebra and Allan Clark, Elements of Abstract Algebra. Both use the same terminology, not the same as Wikipedia. &lt;/p&gt;&#xA;" OwnerUserId="5798" LastEditorUserId="5798" LastEditorDisplayName="" LastEditDate="2011-02-17T09:39:46.330" LastActivityDate="2011-02-17T09:39:46.330" Title="A question regarding the definition of Galois group" Tags="&lt;abstract-algebra&gt;&lt;field-theory&gt;&lt;galois-theory&gt;" AnswerCount="3" CommentCount="7" FavoriteCount="1" />
  <row Id="22325" PostTypeId="2" ParentId="22310" CreationDate="2011-02-16T10:07:42.660" Score="1" ViewCount="" Body="&lt;p&gt;You only came unstuck at the point where you needed to set $\text{d}x$ equal to&#xA;$\frac{(3-x)^2}{2} \text{d}u,$ an equality you already had. So your problem reduces to expressing $3-x$ in terms of $u.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From your expression for $u$ you have $u(3-x)=x-1,$ and adding $3-x$ to both sides gives&#xA;$(3-x)(u+1)=2$ and hence $3-x=2/(1+u).$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sure, but slightly longer approach, is to solve $u(3-x)=x-1$ for $x$ and subtract from $3.$&lt;/p&gt;&#xA;" OwnerUserId="1301" LastActivityDate="2011-02-16T10:07:42.660" />
  <row Id="22326" PostTypeId="2" ParentId="22320" CreationDate="2011-02-16T10:08:34.503" Score="0" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://mathoverflow.net/questions/53579/a-question-on-permutations-of-coefficients-of-polynomials&quot; rel=&quot;nofollow&quot;&gt;This&lt;/a&gt; has a weaker version which might give some hint as far as I am concerned.&lt;br&gt;&#xA;In any case, thanks for paying attention and if I make some mistakes, please inform me.&lt;/p&gt;&#xA;" OwnerUserId="6792" LastActivityDate="2011-02-16T10:08:34.503" CommentCount="2" />
  <row Id="22327" PostTypeId="2" ParentId="22324" CreationDate="2011-02-16T10:16:00.987" Score="0" ViewCount="" Body="&lt;p&gt;It is supposed to be proved in your Galois theory book as an important theorem, if not, firstly, consider the degree of L|F, and secondly, consider the fixed field, or to take a look at &lt;em&gt;Van Der Wareden&lt;/em&gt;'s book &lt;strong&gt;Algebra&lt;/strong&gt; is to be of some help.&lt;br&gt;&#xA;Edit:Here I mean that the fixed field of Aut(E|F) is exactly equal to F when E|F is&lt;br&gt;&#xA;(1)Finite&lt;br&gt;&#xA;(2)Separable&lt;br&gt;&#xA;(3)Normal&lt;br&gt;&#xA;and the converse is also true, if this isn't your question, I will delete my post immediately, thanks. &lt;/p&gt;&#xA;" OwnerUserId="6792" LastEditorUserId="6792" LastEditorDisplayName="" LastEditDate="2011-02-16T13:31:17.030" LastActivityDate="2011-02-16T13:31:17.030" CommentCount="2" />
  <row Id="22328" PostTypeId="2" ParentId="22317" CreationDate="2011-02-16T10:36:17.880" Score="3" ViewCount="" Body="&lt;p&gt;What does a permutation of order 5 look like? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you give a condition when two permutations of order 5 commute?&lt;/p&gt;&#xA;" OwnerUserId="360" LastActivityDate="2011-02-16T10:36:17.880" CommentCount="2" />
  <row Id="22329" PostTypeId="1" AcceptedAnswerId="22354" CreationDate="2011-02-16T10:36:30.450" Score="3" ViewCount="173" Body="&lt;p&gt;the formula is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;from $$\sum\limits_{k=1}^{2n} (-1)^{k} \cdot k^{2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://img641.imageshack.us/img641/707/capturetnr.png&quot; rel=&quot;nofollow&quot;&gt;more detailed in this image&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How to find finite sum of this formula?&lt;/p&gt;&#xA;" OwnerUserId="7148" LastActivityDate="2011-02-16T13:42:36.487" Title="Help to find finite sum" Tags="&lt;sequences-and-series&gt;" AnswerCount="6" FavoriteCount="1" />
  <row Id="22330" PostTypeId="2" ParentId="22323" CreationDate="2011-02-16T10:42:06.470" Score="2" ViewCount="" Body="&lt;p&gt;We have to show that if $x$ is a limit point of $\{x \in \mathbb{R}^n: g(x) \leq 0\}$ then $x$ is in that set, i.e. $g(x) \leq 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First let's write what being a limit point tells us about $x$. It says that there are points $y$ such that $g(y) \leq 0$ as close to $x$ as you want :&#xA;$$\forall \delta &amp;gt; 0 \exists y \in N_\delta(x), g(y) \leq 0$$&#xA;Then write what lower semicontinuity of $g$ at $x$ means about the values of $g$ near $x$ . It says that if $y$ is sufficiently close to $x$ then $g(y)$ is at least as high as $g(x)- \varepsilon$:&lt;br&gt;&#xA;$$\forall \epsilon &amp;gt; 0, \exists \delta &amp;gt; 0, \forall y \in N_\delta(x), g(x)-g(y) &amp;lt; \varepsilon$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You see how they seem made for each other ?&#xA;Suppose $g(x) &amp;gt; 0$, and plug $\epsilon = g(x)$ in the lower semicontinuity property.&#xA;You get :&#xA;$$\forall \delta &amp;gt; 0 \exists y \in N_\delta(x), g(y) \leq 0$$&#xA;$$\exists \delta &amp;gt; 0 \forall y \in N_\delta(x), g(y) &amp;gt; 0$$&#xA;These two are the negation of each other so this is a contradiction.&#xA;The semicontinuity tells you about some $\delta &amp;gt; 0$, if you give it to the limit point property you learn about some $y \in N_\delta(x)$, such that $0 &amp;lt; g(y) \leq 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So $g(x) &amp;gt; 0$ is contradictory, therefore $g(x) \leq 0$.&lt;/p&gt;&#xA;" OwnerUserId="6270" LastActivityDate="2011-02-16T10:42:06.470" CommentCount="1" />
  <row Id="22331" PostTypeId="1" AcceptedAnswerId="22337" CreationDate="2011-02-16T11:01:35.340" Score="0" ViewCount="85" Body="&lt;p&gt;Hello! I know this is fairly basic stuff compared to most of the questions here, but I have to start somewhere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How would I find out whether the following lines intersect the same point, and if so, where that is?&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;y - 2x + 3=0&lt;/li&gt;&#xA;&lt;li&gt;2x + y - 53=0&lt;/li&gt;&#xA;&lt;li&gt;y - x - 11=0&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I've read the answers to &quot;&lt;a href=&quot;http://math.stackexchange.com/q/5267/4823&quot;&gt;How to find where 3 lines intersect.&lt;/a&gt;&quot;, but the answers were too advanced for me, partly because I'm not learning math in English.&lt;br&gt;&#xA;It would be great if someone could &lt;strong&gt;show&lt;/strong&gt; me how they approached this problem.&lt;/p&gt;&#xA;" OwnerUserId="4823" LastActivityDate="2011-02-16T11:17:59.387" Title="Do these three lines intersect the same point, and if so, where?" Tags="&lt;geometry&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="22332" PostTypeId="2" ParentId="22329" CreationDate="2011-02-16T11:02:11.537" Score="4" ViewCount="" Body="&lt;p&gt;Try working with the sequence $S_n=\sum_1^{2n}(-1)^kk^2$. If you look at the sequence of differences (sidrat hefreshim in hebrew, i'm not sure if that the correct translation. any hebrew speaker- please correct me here) of $A_n=S_{n+1}-S_n$ you should be able to get a first degree formula for $A_n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now- Notice that $S_n-S_1=\sum_1^{n-1}A_n$, and since $S_1$ is easy to calculate, and $\sum A_n$ is not to tricky- this should give you the solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;good luck!&lt;/p&gt;&#xA;" OwnerUserId="7129" LastActivityDate="2011-02-16T11:02:11.537" />
  <row Id="22333" PostTypeId="2" ParentId="22329" CreationDate="2011-02-16T11:03:09.823" Score="10" ViewCount="" Body="&lt;p&gt;Hint: What is $(2i)^2-(2i-1)^2$? Next, what are $\displaystyle\sum_{i=1}^n1$ and $\displaystyle\sum_{i=1}^ni$?&lt;/p&gt;&#xA;" OwnerUserId="6179" LastActivityDate="2011-02-16T11:03:09.823" />
  <row Id="22334" PostTypeId="2" ParentId="22329" CreationDate="2011-02-16T11:04:28.363" Score="2" ViewCount="" Body="&lt;p&gt;Sorry, I am in a bit of a hurry, but it should be &lt;a href=&quot;http://oeis.org/A014105&quot; rel=&quot;nofollow&quot;&gt;this sequence&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To see this, just remark that you are summing up difference of consecutive squares (-1+4) + (-9+16) + ..., and these difference are the odd integers congruent to 3 mod 4  (i.e., 3, 7, 11...).&lt;/p&gt;&#xA;" OwnerUserId="6507" LastActivityDate="2011-02-16T11:04:28.363" />
  <row Id="22335" PostTypeId="2" ParentId="22329" CreationDate="2011-02-16T11:05:02.247" Score="1" ViewCount="" Body="&lt;p&gt;You can separate this sum into two, one for $k$ odd and another for $k$ even, writing $k=2t+1$ and $k=2t$, respectively. You can also try finding $\sum k^2 z^k$ and then set $k=-1$.&lt;/p&gt;&#xA;" OwnerUserId="589" LastActivityDate="2011-02-16T11:05:02.247" />
  <row Id="22336" PostTypeId="2" ParentId="22329" CreationDate="2011-02-16T11:05:36.030" Score="1" ViewCount="" Body="&lt;p&gt;Yet another way (although not as simple as those already suggested) is to write it as&#xA;$$-\sum_{k=1}^{2n} k^2 + 2 \sum_{m=1}^{n} (2m)^2$$&#xA;and use the known formula for a sum of squares.&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2011-02-16T11:05:36.030" CommentCount="2" />
  <row Id="22337" PostTypeId="2" ParentId="22331" CreationDate="2011-02-16T11:14:34.863" Score="2" ViewCount="" Body="&lt;p&gt;If you add the first two equations, you will find that these lines intersect at $y=25$. What is the corresponding $x$ value? Try to do similar things for the second and third equation, and the first and third equation. If these lines intersect at the same point, you should find the same answers.&lt;/p&gt;&#xA;" OwnerUserId="5882" LastActivityDate="2011-02-16T11:14:34.863" CommentCount="1" />
  <row Id="22338" PostTypeId="2" ParentId="22331" CreationDate="2011-02-16T11:17:59.387" Score="2" ViewCount="" Body="&lt;p&gt;If you at first look at two of the equations and solve them as a set of two equations with two unknowns, you will find out where two of the lines intersect. Then all you have to do is to check if the last line also goes through the same point. (Does the x and y values from the first set of equations fit in the last equation?)&lt;/p&gt;&#xA;" OwnerUserId="7044" LastActivityDate="2011-02-16T11:17:59.387" />
  <row Id="22339" PostTypeId="1" AcceptedAnswerId="22345" CreationDate="2011-02-16T11:20:46.153" Score="6" ViewCount="78" Body="&lt;p&gt;Let $X$ and $Y$ be two Poisson variables with different mean. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a better (as in more concise or numerically faster) way to compute $P(X\leq Y)$ than using&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ P(X\leq Y) = \sum_{y=0}^\infty P(Y=y) P(X\leq y) $$&lt;/p&gt;&#xA;" OwnerUserId="505" LastActivityDate="2011-02-16T13:08:12.413" Title="Comparing Poisson variables" Tags="&lt;probability&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="22340" PostTypeId="1" AcceptedAnswerId="22342" CreationDate="2011-02-16T12:06:16.593" Score="3" ViewCount="272" Body="&lt;p&gt;Prove that for all natural numbers statement n, statement is dividable by 7 &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$15^n+6$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Base.&lt;/strong&gt; We prove the statement for $n = 1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;15 + 6 = 21 it is true&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Inductive step.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Induction Hypothesis.&lt;/em&gt; We assume the result holds for $k$. That is, we assume that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$15^k+6$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is divisible by 7&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;To prove:&lt;/em&gt; We need to show that the result holds for $k+1$, that is, that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$15^{k+1}+6=15^k\cdot 15+6$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and I don't know what to do &lt;/p&gt;&#xA;" OwnerUserId="7149" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-17T03:03:21.357" LastActivityDate="2011-02-21T17:56:39.620" Title="Prove 7 divides $15^n+6$ with mathematical induction" Tags="&lt;induction&gt;" AnswerCount="3" CommentCount="6" />
  <row Id="22341" PostTypeId="2" ParentId="22340" CreationDate="2011-02-16T12:09:25.800" Score="5" ViewCount="" Body="&lt;p&gt;By induction hypothesis, you have $15^k=7t-6$.&lt;/p&gt;&#xA;" OwnerUserId="589" LastActivityDate="2011-02-16T12:09:25.800" CommentCount="5" />
  <row Id="22342" PostTypeId="2" ParentId="22340" CreationDate="2011-02-16T12:10:58.277" Score="8" ViewCount="" Body="&lt;p&gt;Observe that $14$ is divisible by 7. Then let $15^k\cdot 15+6=15^k\cdot 14+ 15^k+6$. &lt;/p&gt;&#xA;" OwnerUserId="1772" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-17T03:05:49.280" LastActivityDate="2011-02-17T03:05:49.280" />
  <row Id="22343" PostTypeId="2" ParentId="22312" CreationDate="2011-02-16T12:25:23.153" Score="3" ViewCount="" Body="&lt;p&gt;Well the force on the particle due to gravity is $mg,$ where $g$ is the acceleration due to gravity, so the net downward force on the particle is $m(g-kv).$ So its acceleration is&#xA;$g-kv.$ So you need to solve $dv/dt = g-kv$ given the initial conditions. That is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_0^V \frac{ \text{d}v}{g-kv} = \int_0^T \text{d}t.$$&lt;/p&gt;&#xA;" OwnerUserId="1301" LastActivityDate="2011-02-16T12:25:23.153" />
  <row Id="22344" PostTypeId="2" ParentId="22095" CreationDate="2011-02-16T12:33:48.747" Score="2" ViewCount="" Body="&lt;p&gt;Let $R=k[x,y]/(x^2-xy, y^2-xy)$, $E_1=(x), E_2=(y)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then $Ass(E_1) = Ass (E_2) = (x-y)$ but $E_1+E_2$ contains the element $x-y$...&lt;/p&gt;&#xA;" OwnerUserId="223" LastActivityDate="2011-02-16T12:33:48.747" />
  <row Id="22345" PostTypeId="2" ParentId="22339" CreationDate="2011-02-16T12:40:04.450" Score="3" ViewCount="" Body="&lt;p&gt;&lt;em&gt;OK, this is how I've done it for now. It's very likely that this is not the best way to do it, but I just decided to post it, for the fun of it. There are probably some mistakes, so go ahead and point them out.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, for the sake of definiteness, I will posit $X \sim Poiss(\lambda)$ and $Y \sim Poiss(\mu)$. In particular, this means that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \mathbb{P}(X=k) = \frac{e^{-\lambda} \lambda^k}{k!} \; .$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, notice that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \frac{d}{d\lambda} \mathbb{P}(X=k) = \mathbb{P}(X=k-1) - \mathbb{P}(X=k) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with special case&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \frac{d}{d\lambda} \mathbb{P}(X=0) = - \mathbb{P}(X=0) \; . $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This implies that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \frac{d}{d\lambda} \mathbb{P}(X \leq y) = - \mathbb{P}(X=y) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;by summing our first identity over $k$ from $0$ to $y$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, differentiating the sought after probability with respect to $\lambda$ gives us&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \frac{d}{d\lambda}\mathbb{P}(X\leq Y) = - \sum_{y=0}^{\infty} \mathbb{P}(X=y)\mathbb{P}(Y=y) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sum on the right hand side can be expressed as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ - \sum_{y=0}^{\infty} \mathbb{P}(X=y)\mathbb{P}(Y=y) = - e^{-(\lambda + \mu)} \sum_{y=0}^{\infty} \frac{\lambda^y \mu^y}{(y!)^2} \; . $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The last part can be rewritten noting that the modified Bessel function $I_0(x)$ has the following Taylor expansion around $0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$I_0(x) = \sum_{n=0}^{\infty} \frac{(x/2)^{2n}}{(n!)^2} \; .$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We then obtain&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \frac{d}{d\lambda}\mathbb{P}(X\leq Y) = - e^{-(\lambda + \mu)} I_0(2 \sqrt{\lambda \mu})$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, integrating this we get the following expression&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \mathbb{P}(X\leq Y) - \mathbb{P}(0\leq Y)= - \int_0^{\lambda} e^{-(t + \mu)} I_0(2 \sqrt{t \mu}) dt $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \mathbb{P}(X\leq Y) = 1 - e^{-\mu}\int_0^{\lambda} e^{-t} I_0(2 \sqrt{\mu t}) dt \; . $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This integral should be in principle more numerically tractable than the previous sum, provided you have modified Bessel functions implemented in whatever program you are using to solve this.&lt;/p&gt;&#xA;" OwnerUserId="3567" LastEditorUserId="3567" LastEditorDisplayName="" LastEditDate="2011-02-16T12:55:05.180" LastActivityDate="2011-02-16T12:55:05.180" CommentCount="1" />
  <row Id="22346" PostTypeId="2" ParentId="22321" CreationDate="2011-02-16T12:44:32.277" Score="5" ViewCount="" Body="&lt;p&gt;Commutative diagrams are, to put it simply, a (very) handy way of writing systems of equations in categories. Instead of having rows of symbols and do symbol-pushing you reason diagramatically and geometrically.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, about your questions, I am not sure I understand all of them so my answer may be way off, but here it goes anyway.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Which &quot;background assumptions&quot; are needed for a commutative diagram to make specific sense?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Without going (too) technical here, diagrams in categories are oriented graphs where the vertices are labeled with objects and the edges with arrows and to every path we associate the composite arrow. The problem is that the composite is not well defined in the presence of loops, e.g. think of a single vertex with two loops labeled f and g: does it denote the composite fg or gf? If your labeled graph does not have loops then you are all set.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: making this all formal and rigorous is something of a chore and for a little gain. One possible way is to define diagrams of shape $G$ (a graph, for example a square) in a category $A$ as functors $FG\to A$ where $FG$ is the free category on the graph. Then a commutative diagram of a shape $G$ is a functor $FG/R \to A$ where $FG/R$ denotes the quotient of $FG$ by a congruence $R$ that &quot;forces&quot; commutativity, that is, the equality of certain (all) parallel pairs of paths. You can then define the gluing of diagrams along a common diagram as a certain pushout, which allows you to prove that if the smaller pieces of a diagram are commutative then the whole thingy is also commutative, etc.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Which conventions are used? (e.g. for existence and uniqueness of arrows)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am not sure understand what you are trying to ask here. What problem of existence and uniqueness are you referring to? The existence and uniqueness of a composite associated to a path was treated in the previous answer.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;When is it important to label the objects and arrows?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Unless there is no ambiguity about what arrows and objects you are mentioning (or you simply do not need them), feel free to not label the graph. In my experience, this happens very rarely so I tend to label just about everything.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;When are verbal explanations necessary?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No general rule here. The best that I can offer is use common sense and know your target audience.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Does every conceivable diagram mean something?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No, see my first answer.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Are there (graphical) analogues of commutative diagrams for other (relational or algebraic) structures than categories?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes, there is a fairly sophisticated graphical calculus for braided monoidal categories with duals involving tangles. There are also higher dimensional generalizations of diagrams to n-categories but here the problems of assigning a consistent composite to the corresponding notion of graph are much harder and their computational usefulness is rather limited for reasons that should be obvious. Even 2-diagrams in 2-categories are already very cumbersome. This is actually one of the reasons why higher dimensional category is inherently harder: you do not have a handy graphical calculus and computing anything can range from a pain in the butt to a pain in the (insert ultra-sensitive bodily region here).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope it helps, regards.&#xA;G. Rodrigues&lt;/p&gt;&#xA;" OwnerUserId="1033" LastEditorUserId="1033" LastEditorDisplayName="" LastEditDate="2011-02-18T12:48:26.040" LastActivityDate="2011-02-18T12:48:26.040" CommentCount="9" />
  <row Id="22347" PostTypeId="2" ParentId="22339" CreationDate="2011-02-16T13:08:12.413" Score="2" ViewCount="" Body="&lt;p&gt;Rasholnikov has shown you a method using modified Bessel functions.  You could also get that sort of result from a &lt;a href=&quot;http://en.wikipedia.org/wiki/Skellam_distribution&quot; rel=&quot;nofollow&quot;&gt;Skellam distribution&lt;/a&gt;, which is the difference between two Poisson distributions.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want a quick and simple approximation for $X \sim \text{Poiss}(\lambda)$ and $Y \sim \text{Poiss}(\mu)$ then &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\text{Pr}(X \leq Y) \approx \Phi\left( \frac{1/2 - (\lambda - \mu )}{\sqrt{\lambda + \mu}} \right)$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\Phi()$ is the cumulative distribution function of a standard normal distribution.  The $1/2$ is there as a continuity adjustment to deal with the possibility that $X=Y$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example with $\lambda =5$ and  $\mu =10$, the true result is about 0.9256 while the approximation gives 0.9222. &lt;/p&gt;&#xA;" OwnerUserId="6460" LastActivityDate="2011-02-16T13:08:12.413" CommentCount="2" />
  <row Id="22348" PostTypeId="1" CreationDate="2011-02-16T13:10:17.003" Score="0" ViewCount="55" Body="&lt;p&gt;Say I have 100 numbers that are averaged:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;number of values = 100&#xA;total sum of values = 2000&#xA;mean = 2000 / 100 =&amp;gt; 20&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;If I want to add a value and find out the new average:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;total sum of values = 2000 + 100&#xA;mean = 2100 / 101 =&amp;gt; 20.79&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;If I want to subtract a value and find out the new average:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;total sum of values = 2100 - 100&#xA;mean = 2000 / 100 =&amp;gt; 20&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;It seems to work, but is the above correct?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this the proper way to add/subtract values from a average without having to re-sum all the 100 numbers first?&lt;/p&gt;&#xA;" OwnerUserId="3603" LastEditorUserId="3603" LastEditorDisplayName="" LastEditDate="2011-02-16T13:15:43.673" LastActivityDate="2011-02-16T13:19:23.043" Title="How to add and subtract values from an average?" Tags="&lt;average&gt;" AnswerCount="1" />
  <row Id="22349" PostTypeId="2" ParentId="22312" CreationDate="2011-02-16T13:11:27.377" Score="1" ViewCount="" Body="&lt;p&gt;Okay here is my noob way of solving this  (I only learned this less than a year ago but I have forgotten everything and semester starts in a few weeks and I'll probably be required to do this stuff EEEEEEEEEEEEEEEEEEEEEEP),&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The forces on the &quot;body&quot; are the resistance and its weight (gravity) as far as I can tell from your description&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So &#xA;$$ \vec{F}_{net} = m\mathbf{a} =mg-mk\mathbf{v} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which can be simplified to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$  \mathbf{a} = g-k\mathbf{v}  $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which can be written in a slightly different form as because acceleration is derivative of velocity&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$  v&amp;#39;(t)+k v(t)=g  $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This looks type of equation looks familiar to me and can be solved as it is a &lt;a href=&quot;http://en.wikipedia.org/wiki/Linear_differential_equation#First_order_equation&quot; rel=&quot;nofollow&quot;&gt;&quot;linear first order differential equation&quot;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now here is where I may start to make errors so please forgive me&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First we multiple everything an &quot;integrating factor&quot; equal to $e^{\int f(t)\,dt}$ where $f(t)=k,$ to make it all work so&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$e^{\int f(t)\,dt}=e^{\int k\,dt}=e^{kt}$$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We don't need a &quot;plus C&quot; (I'm not entirely clear on this but I don't think it makes a difference to anything in the long run, SOMEONE PLEASE CONFIRM)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So when we multiply everything by $e^{kt} $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$  v&amp;#39;(t)e^{kt}+k v(t)e^{kt}=ge^{kt} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is &quot;equivalent&quot; to this (look at the product rule)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$  (v(t)e^{kt})&amp;#39;=ge^{kt} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now if we integrate both sides&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$  v(t)e^{kt}={\int ge^{kt}\,dt}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$  v(t)e^{kt}=\frac{ge^{kt}}{k}+C$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$  v(t)=\frac{C}{e^{kt}}+\frac{g}{k}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To get rid of the C, you said it starts from rest so $v(0)=0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$  v(0)=0=\frac{C}{e^{k0}}+\frac{g}{k} = \frac{C}{1}+\frac{g}{k}$$&#xA;$$ 0= C+\frac{g}{k}$$&#xA;$$ C=-\frac{g}{k}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$  v(t)=\frac{-\frac{g}{k}}{e^{kt}}+\frac{g}{k}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and I would simplify this to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$  v(t)=g\frac{1-e^{-kt}}{k}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now Wolfram Alpha would have done &lt;a href=&quot;http://www.wolframalpha.com/input/?i=v%27%28t%29%2Bk+v%28t%29%3Dg&quot; rel=&quot;nofollow&quot;&gt;this for you&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now to the second part of your question when v(t)=V what is t?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$  v(t)=V=g\frac{1-e^{-kt}}{k}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$  \frac{kV}{g}={1-e^{-kt}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$  e^{-kt}=1-\frac{kV}{g}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$  e^{-kt}=\frac{g-kV}{g}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$  e^{kt}=\frac{g}{g-kV}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$  kt=\log_e {\frac{g}{g-kV}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$  t=\frac{1}{k}\log_e ({\frac{g}{g-kV}})$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wolfram Alpha would do this too but is down for me now&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope this is correct&lt;/p&gt;&#xA;" OwnerUserId="7152" LastActivityDate="2011-02-16T13:11:27.377" CommentCount="1" />
  <row Id="22350" PostTypeId="1" AcceptedAnswerId="22356" CreationDate="2011-02-16T13:15:00.053" Score="4" ViewCount="118" Body="&lt;p&gt;I'm having trouble seeing why this (it's stated as obvious in a proof of Sylow's first theorem): if $n=mp^r$ ($p$ is prime), and $p^t\mid m$ but $p^{t+1}\nmid m$, then $p^t\mid\binom{n}{p^r}$ but $p^{t+1}\nmid\binom{n}{p^r}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also wonder if a combinatorial proof of this fact can be given.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you in advance for any help given.&lt;/p&gt;&#xA;" OwnerUserId="1313" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-16T18:09:34.010" LastActivityDate="2011-02-16T21:00:33.760" Title="$p^t$ divides $C(n,p^r)$, $p^{t+1}$ doesn't." Tags="&lt;combinatorics&gt;&lt;elementary-number-theory&gt;" AnswerCount="3" CommentCount="4" FavoriteCount="1" />
  <row Id="22351" PostTypeId="2" ParentId="22348" CreationDate="2011-02-16T13:19:23.043" Score="3" ViewCount="" Body="&lt;p&gt;$s=\frac{a_1+...+a_n}{n}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want the average of $a_1,...,a_n$ and $a_{n+1}$, then $s&amp;#39;=\frac{a_1+...+a_n+a_{n+1}}{n+1}=\frac{ns+a_{n+1}}{n+1}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want the average of $a_1,...,a_{n-1}$ then $s&amp;#39;&amp;#39;=\frac{a_1+...+a_{n-1}}{n-1}=\frac{ns-a_n}{n-1}$.&lt;/p&gt;&#xA;" OwnerUserId="1313" LastActivityDate="2011-02-16T13:19:23.043" CommentCount="1" />
  <row Id="22352" PostTypeId="1" AcceptedAnswerId="22392" CreationDate="2011-02-16T13:26:00.387" Score="4" ViewCount="191" Body="&lt;p&gt;Can there be such a thing as a classification of classification theorems?&lt;/p&gt;&#xA;" OwnerUserId="7485" LastActivityDate="2011-02-17T13:44:09.990" Title="Can there be such a thing as a classification of classification theorems?" Tags="&lt;soft-question&gt;" AnswerCount="2" CommentCount="6" FavoriteCount="2" />
  <row Id="22353" PostTypeId="2" ParentId="22192" CreationDate="2011-02-16T13:42:32.077" Score="10" ViewCount="" Body="&lt;p&gt;Suppose that we have (to fix ideas) a unitary irreducible representation of a semi-simple Lie group $G$ (such as the non-compact Lie groups that you write down) on a Hilbert space.  (Here I mean irreducible in the Hilbert space sense, i.e. there are no proper invariant &lt;em&gt;closed&lt;/em&gt; subspaces.)  Let's call the Hilbert space $V$ (just to give it a name).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A theorem of Harish-Chandra then says that $V$ is &lt;em&gt;admissible&lt;/em&gt;, which means the following: fix a maximal compact subgroup $K$ of $G$.  Then each irreducible representation $W$ of $K$ appears with finite multiplicity as a subrepresentation&#xA;of $V$.  If we call this multiplicity $m_W$, then we may write&#xA;$V = \hat{\oplus}_W W^{m_W}$, i.e. as the Hilbert space direct sum (i.e. the completed direct sum) of the various $W$, each appearing with multiplicity $m_W$.  (This is a consequence of the Peter--Weyl theorem, and is true&#xA;for any unitary representation of a compact group in which each irrep. appears &#xA;with finite multiplicity.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now inside $\hat{\oplus} W^{m_W}$ we have the actual algebraic direct sum&#xA;$\oplus_W W^{m_W}$, and this has an intrinsic characterization as a subspace&#xA;of $V$, as the &lt;em&gt;$K$-finite vectors&lt;/em&gt;.  (A vector $v$ is called $K$-finite if the&#xA;linear span of all its translates by elements of $K$ is finite-dimensional.)&#xA;Let's denote it by $V_K \subset V$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It turns out that $V_K$, although it is not invariant under the action of $G$&#xA;(typically, unless $V$ happens to be finite-dimensional, which it usually won't be), is invariant under $\mathfrak g$, the Lie algebra of $G$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One calls $V_K$ a $(\mathfrak{g},K)$-module, or also a Harish-Chandra module&#xA;(because it has actions of $\mathfrak{g}$ and $K$).  It turns out that&#xA;$V_K$ determines $V$, and the basis of Harish-Chandra's approach to the study&#xA;of unitary reps. of $G$ is to work instead with the underlying $(\mathfrak g,K)$-modules.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now in principle, to recover $V$, one really needs $V_K$ &lt;em&gt;as a $(\mathfrak g, K)$-module&lt;/em&gt;; i.e. forgetting $\mathfrak g$ and just remembering the $K$-action is throwing away a lot of information.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But in practice (at least in the examples that I know) non-isomorphic irreducible $V$ have different list of multiplicities $m_W$, and so just knowing&#xA;$V_K$ as a $K$-rep. may well already pin down $V$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, often one doesn't even have to know &lt;em&gt;all&lt;/em&gt; the $m_W$, but just&#xA;the first non-zero vanishing value. (If I think of the reps. $W$ as being&#xA;labelled by their highest weights lying in some choice of dominant Weyl chamber for $K$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A good place to read about this (it is not short, but I found it very good for dipping into) is Knapp's book &lt;em&gt;Representation theory of semisimple groups: an &#xA;overview based on examples&lt;/em&gt;.  He gives the basic definitions, a lot of examples, and goes on to develop various aspects of the theory (e.g. the theory of&#xA;the relationship between $V$ and the multiplicities $m_W$: this is known as&#xA;the theory of &lt;em&gt;$K$-types&lt;/em&gt;).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Incidentally, Harish-Chandra was a student of Dirac, and (as far as I know) his study of unitary reps. of semisimple groups was inspired by Bargmann's treatment of the special case of $SL_2(\mathbb R)$, which was in turn inspired in part by the role of this group in physics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, I don't know of a treatment of the theory which directly relates it to the physics literature, and I can't parse the physics argument that you wrote down in detail.&lt;/p&gt;&#xA;" OwnerUserId="221" LastEditorUserId="5363" LastEditorDisplayName="" LastEditDate="2011-02-16T14:11:16.260" LastActivityDate="2011-02-16T14:11:16.260" CommentCount="9" />
  <row Id="22354" PostTypeId="2" ParentId="22329" CreationDate="2011-02-16T13:42:36.487" Score="5" ViewCount="" Body="&lt;p&gt;As I'm a fan of &lt;a href=&quot;http://demonstrations.wolfram.com/ProofWithoutWords12N1NChoose2/&quot; rel=&quot;nofollow&quot;&gt;proofs without words,&lt;/a&gt; here is my pictorial effort of why&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{k=1}^{2n} (-1)^k k^2 = \sum_{i=1}^{2n} i = n(2n+1).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/9b7MP.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1301" LastActivityDate="2011-02-16T13:42:36.487" />
  <row Id="22355" PostTypeId="2" ParentId="22324" CreationDate="2011-02-16T13:51:50.320" Score="12" ViewCount="" Body="&lt;p&gt;I don't know which book you are using or what the precise statement therein is, but here is how the terminology is used (at least within the English speaking mathematical world) [Edit: This may be too categorical a statement; see Arturo Magidin's answer]:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $E/F$ is any field extension, then $Aut(E/F)$ denotes the field automorphisms of $E$ that leave each element of $F$ fixed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;If&lt;/em&gt; $E/F$ is a Galois extension (i.e. finite, separable, and normal) then one&#xA;writes $Gal(E/F)$ for $Aut(E/F)$ and calls it the Galois group of $E$ over $F$.&#xA;So Galois groups are a special case of automorphism groups. The reason for introducing this new terminology (i.e. $Gal(E/F)$) for an existing more general&#xA;concept (i.e. $Aut(E/F)$) is in part historical, and also to serve as a reminder that&#xA;in the particular case when $E/F$ is Galois, there is a rich theory (the Galois correspondence) relating $Gal(E/F)$ and the field theoretic structure of the extension $E/F$, which does not hold in more general contexts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that there is also a notion of an infinite degree Galois extension $E/F$&#xA;(if $E/F$ is an infinite extension of $F$, one says that it is Galois if&#xA;$E$ is the union of finite subextensions of $F$; in particular, $E$ is still necessarily algebraic over $F$), and in this context one also uses the&#xA;notation $Gal(E/F)$ for $Aut(E/F)$.  But in this case one also equips $Gal(E/F)$ &#xA;with extra structure (one makes it not just a group, but a topological group).&#xA;Also, this case is typically not treated in undergraduate textbooks and courses (at least in the U.S.), so if you are learning this material for the first time, you won't want to worry about it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, I want to say that studying $Aut(\mathbb Q(\pi)/\mathbb Q)$ (just to give one non-Galois theoretic example) is interesting and important; it just isn't part of Galois theory.&lt;/p&gt;&#xA;" OwnerUserId="221" LastEditorUserId="221" LastEditorDisplayName="" LastEditDate="2011-02-17T05:57:20.763" LastActivityDate="2011-02-17T05:57:20.763" CommentCount="7" />
  <row Id="22356" PostTypeId="2" ParentId="22350" CreationDate="2011-02-16T13:56:10.750" Score="3" ViewCount="" Body="&lt;p&gt;This is the same idea which &lt;em&gt;I.N.Herstein&lt;/em&gt; uses to prove the First Sylow theorem. So i would recommend you to have a peep into his, &lt;em&gt;Topics in Algebra&lt;/em&gt; book, Sec &lt;strong&gt;2.12,&lt;/strong&gt; Group Theory, &lt;strong&gt;Page 92&lt;/strong&gt;&#xA;&lt;img src=&quot;http://i.imgur.com/Fa5tB.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" LastActivityDate="2011-02-16T16:22:33.213" />
  <row Id="22357" PostTypeId="1" AcceptedAnswerId="22363" CreationDate="2011-02-16T13:57:21.047" Score="5" ViewCount="163" Body="&lt;p&gt;Does the principle of mathematical induction extend to a cardinality larger than that of the countably infinite?&lt;/p&gt;&#xA;" OwnerUserId="7485" LastEditorUserId="622" LastEditorDisplayName="" LastEditDate="2011-02-16T14:10:36.810" LastActivityDate="2011-02-16T14:22:08.297" Title="Does the principle of mathematical induction extend to higher cardinalities?" Tags="&lt;set-theory&gt;&lt;induction&gt;" AnswerCount="2" />
  <row Id="22358" PostTypeId="1" CreationDate="2011-02-16T13:59:11.560" Score="2" ViewCount="83" Body="&lt;p&gt;I have an algorithm and I would like to know how many times each line is called.&#xA;There I wrote which lines I understand and some lines is left.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;j := 1   ---------------------- 1 time&#xA;while j &amp;lt; n do  --------------- n times&#xA;  x := B[j]    ---------------- n-1 times&#xA;  k := j    ------------------- n-1 times&#xA;  for i := j + 1 to n do  ----- ?&#xA;    if B[i] &amp;lt; x then     ------ ?&#xA;      x := B[i]     ----------- ?&#xA;      k := i    --------------- ?&#xA;  B[l] := B[j]   -------------- n-1 times&#xA;  B[j] := x     --------------- n-1 times&#xA;  j := j + 1   ---------------- n-1 times&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Note: &lt;code&gt;k := i&lt;/code&gt; means that line is called once but &lt;code&gt;for i := j + 1 to n do&lt;/code&gt; - twice: assign and comparison.&lt;/p&gt;&#xA;" OwnerUserId="7148" LastActivityDate="2011-02-16T17:13:13.797" Title="Algorithm complexity in for loop" Tags="&lt;algorithms&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="22359" PostTypeId="2" ParentId="22357" CreationDate="2011-02-16T14:01:22.493" Score="3" ViewCount="" Body="&lt;p&gt;Yes see &lt;a href=&quot;http://en.wikipedia.org/wiki/Transfinite_induction&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; (site about transfinite induction).&lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2011-02-16T14:01:22.493" />
  <row Id="22360" PostTypeId="1" CreationDate="2011-02-16T14:03:28.090" Score="1" ViewCount="116" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The definition for the two processes&#xA;to be independent is given by&#xA;&lt;a href=&quot;http://planetmath.org/encyclopedia/IndependentStochasticProcesses.html&quot; rel=&quot;nofollow&quot;&gt;PlanetMath&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Two stochastic processes $\lbrace X(t)\mid t\in T \rbrace$ and $\lbrace Y(t)\mid t\in T \rbrace$ are said to&#xA;  be independent, if for any positive&#xA;  integer $n&amp;lt;\infty$, and any sequence&#xA;  $t_1,\ldots,t_n\in T$, the random&#xA;  vectors&#xA;  $\boldsymbol{X}:=(X(t_1),\ldots,X(t_n))$&#xA;  and&#xA;  $\boldsymbol{Y}:=(Y(t_1),\ldots,Y(t_n))$&#xA;  are independent.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I was wondering if according to the&#xA;definition, for any positive integer&#xA;$n,m&amp;lt;\infty$, and any sequence&#xA;$t_1,\ldots,t_n\in T$ and any&#xA;sequence $s_1,\ldots,s_m\in T$, the&#xA;random vectors&#xA;$\boldsymbol{X}:=(X(t_1),\ldots,X(t_n))$&#xA;and&#xA;$\boldsymbol{Y}:=(Y(s_1),\ldots,Y(s_m))$&#xA;are also independent?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Some related questions are for two&#xA;independent random vectors $V$ and&#xA;$W$:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Will any subvector of $V$ and any subvector of $W$ (the two subvectors&#xA;do not necessarily have the same indices in the original random vectors) be independent?&lt;/li&gt;&#xA;&lt;li&gt;For any two subvectors $V_1$ and    $V_2$ of $V$ and any two subvectors &#xA;$W_1$ and $W_2$ of $W$, will the conditional random vectors $V_1|V_2$&#xA;and $W_1|W_2$ also be independent?&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Thanks and regards!&lt;/p&gt;&#xA;" OwnerUserId="1281" LastEditorUserId="1281" LastEditorDisplayName="" LastEditDate="2011-02-16T14:08:45.353" LastActivityDate="2011-02-16T14:55:30.927" Title="Independent stochastic processes and independent random vectors" Tags="&lt;probability-theory&gt;&lt;stochastic-processes&gt;" AnswerCount="1" />
  <row Id="22361" PostTypeId="1" CreationDate="2011-02-16T14:15:10.923" Score="2" ViewCount="88" Body="&lt;p&gt;Do there exists permutations $\pi_1,\pi_2$ and polynomial size CFG that describe the finite language {$w \pi_1(w) \pi_2(w)$} over alphabet {0,1}?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Polynomial size in $|w|=n$&lt;/p&gt;&#xA;" OwnerUserId="4750" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2011-02-17T06:19:39.840" LastActivityDate="2011-02-17T06:19:39.840" Title="Do there exists permutations $\pi_1,\pi_2$ and polynomial size CFG that describe the finite language $\{w \pi_1(w) \pi_2(w)\}$ over alphabet {0,1}?" Tags="&lt;computer-science&gt;&lt;formal-systems&gt;" AnswerCount="1" CommentCount="6" />
  <row Id="22362" PostTypeId="2" ParentId="22358" CreationDate="2011-02-16T14:19:51.363" Score="0" ViewCount="" Body="&lt;p&gt;Hint: You enter the for loop n-1 times as shown from the line above.  Then how many loops to you do?  It should be something like n-j, but you have to figure out the ends-there may be a +1 or -1 or something.  Does your definition of for do the loop with i=n or not?  so the for will be executed 2*(n-1)*(something like n-j) times.  Given the if statement, you can't say for sure how many times the lines within it are executed.  All you can do is find a maximum, assuming the if always evaluates as true.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-02-16T14:19:51.363" />
  <row Id="22363" PostTypeId="2" ParentId="22357" CreationDate="2011-02-16T14:22:08.297" Score="7" ViewCount="" Body="&lt;p&gt;Mathematical induction can be extended, one of the usual ways to extend it is &lt;a href=&quot;http://en.wikipedia.org/wiki/Transfinite_induction&quot; rel=&quot;nofollow&quot;&gt;transfinite induction&lt;/a&gt;, as mentioned by PEV.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A more general way requires slightly more terms:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $R$ be a partial order. We say that $R$ is well-founded if for every non-empty class $A$ there is an $R$-minimal element in $A$. We say that $R$ is set-like if for every $x$ the class $\{y | yRx\}$ is a set.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose that $A$ is a class, $R$ is a well-founded partial order which is set like (at least on the elements of $A$), then if $B\subseteq A$ such that for every $x\in A$ we have $(yRx \rightarrow y\in B) \rightarrow x\in B$ then $A=B$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the same way induction is defined on the natural numbers, even if it isn't always clear how from this clumsy definition one can prove things by induction as usually.&lt;/p&gt;&#xA;" OwnerUserId="622" LastActivityDate="2011-02-16T14:22:08.297" />
  <row Id="22364" PostTypeId="1" AcceptedAnswerId="22370" CreationDate="2011-02-16T14:31:40.257" Score="2" ViewCount="87" Body="&lt;p&gt;Hello everyone,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here I have an interesting problem on linear algebra. It looks very simple, but not so easy to solve for me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $r_i, i=1,…,n$ be unit vectors in $R^n$, find a unit vector $x$ to minimize&#xA;$\sum \| r_i\times x \|^2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remark: if let $\theta$ be the angle between $r_i$ and $x$, then $\sum \| r_i\times x \|^2 = \sum sin^2 \theta _i$. But I don't like sinusoid functions, I think they make the problem more complex especially for high dimensional cases. Is it possible to solve the problem using linear algebra or matrix analysis?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you very much.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Shiyu&lt;/p&gt;&#xA;" OwnerUserId="7156" LastEditorUserId="7156" LastEditorDisplayName="" LastEditDate="2011-02-16T14:42:09.263" LastActivityDate="2011-02-16T15:22:22.377" Title="Minimize sum of the norm of cross products" Tags="&lt;linear-algebra&gt;&lt;cross-product&gt;" AnswerCount="1" CommentCount="14" />
  <row Id="22365" PostTypeId="2" ParentId="22360" CreationDate="2011-02-16T14:37:22.260" Score="1" ViewCount="" Body="&lt;p&gt;The answer to all your questions is yes. And they can be deduced from the following :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If two random vectors $\boldsymbol{X}:=(X_1, \ldots,X_n)$ &#xA;and $\boldsymbol{Y}:=(Y_1, \ldots,Y_m)$ are independent, any pair of &quot;marginalized&quot; random vectors $\boldsymbol{X_A}$,  $\boldsymbol{Y_B}$ (each formed by arbitrary subsets of the originals) are independent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This property (basically your second question) can be readily deduced from the definition of independence (factorization of joint densities) and marginalization. From this the other two follow.&lt;/p&gt;&#xA;" OwnerUserId="312" LastEditorUserId="312" LastEditorDisplayName="" LastEditDate="2011-02-16T14:55:30.927" LastActivityDate="2011-02-16T14:55:30.927" />
  <row Id="22366" PostTypeId="1" CreationDate="2011-02-16T14:39:03.027" Score="2" ViewCount="108" Body="&lt;p&gt;First of all, I would like to call a group immaculate provided that the orders of G and $\Sigma$(the order of N) where N varies over all normal subgroups of G are equal.&lt;br&gt;&#xA;From &lt;a href=&quot;http://arxiv.org/abs/math/0104012v1&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; it has been said that &lt;em&gt;any abelianizer of a group whose order is less than that of D(G), where D(G) is the latter function defined above, is cyclic.&lt;/em&gt;&lt;br&gt;&#xA;Then my question is: Is there any characterization of groups G such that $G^{ab}$ is an &lt;em&gt;imamculate group&lt;/em&gt; where $G^{ab}$ is the abelianizer of G which coincides with the previous one in the case that G is a cyclic group of the order a perfect number.&lt;br&gt;&#xA;This may be referenced to &lt;a href=&quot;http://mathoverflow.net/questions/54851/is-there-an-odd-order-group-whose-order-is-the-sum-of-the-orders-of-the-proper-no&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;, in any case, thanks.  &lt;/p&gt;&#xA;" OwnerUserId="6792" LastEditorUserId="6792" LastEditorDisplayName="" LastEditDate="2011-02-16T14:56:36.317" LastActivityDate="2011-02-16T14:56:36.317" Title="A question on a generalization of perfect numbers" Tags="&lt;number-theory&gt;&lt;group-theory&gt;&lt;number&gt;" CommentCount="4" FavoriteCount="1" />
  <row Id="22367" PostTypeId="1" AcceptedAnswerId="22376" CreationDate="2011-02-16T15:00:22.887" Score="2" ViewCount="271" Body="&lt;p&gt;Hello,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given two Martingale processes $(X_t)$ and $(Y_t)$, are their sum $(X_t+Y_t)$ and their product $(X_t \times Y_t)$ also Martingale?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If not, will the two $(X_t)$ and $(Y_t)$ being independent grant their sum and product Martingale? Thanks!&lt;/p&gt;&#xA;" OwnerUserId="3348" LastActivityDate="2011-04-23T08:05:31.730" Title="Sum and product of Martingale processes" Tags="&lt;stochastic-processes&gt;" AnswerCount="2" CommentCount="8" />
  <row Id="22368" PostTypeId="2" ParentId="12472" CreationDate="2011-02-16T15:12:45.663" Score="1" ViewCount="" Body="&lt;p&gt;Your existing formula does not produce integers all the time:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider number of types, k=1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe that means f(k,n)=n&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your f(1,n)=&#xA;$ \sum_{i=0}^n 1/i!  $&#xA; which is definitely not generally an integer.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you want is  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ \sum_{i=0}^n g(k,i)  $  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;where g(k,i) is the number of ways you can select exactly i objects from upto k types.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which can be similarly expanded to  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ \sum_{i=0}^n \sum_{j=0}^k h(j,i)  $  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;where h(j,i) is the number of ways you can select exactly i objects from exactly j types.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note, because h(j,i)=0,i&amp;lt;j, we can swap the summations:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ \sum_{j=0}^k \sum_{i=j}^n  h(j,i)  $  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;[  and l(j,n)=$ \sum_{i=j}^n  h(j,i)  $ is the number of ways you can select between j and n objects from exactly j types  &#xD;&#xA;&#xD;&#xA;which is the same as selecting between 0 and n-j objects from up to j types  &#xD;&#xA;&#xD;&#xA;thus l(j,n)=$ \sum_{i=0}^{n-j}  g(j,i)  $  &#xD;&#xA;&#xD;&#xA;so f(k,n)=$ \sum_{i=0}^n g(k,i) =  \sum_{j=0}^k \sum_{i=0}^{n-j}  g(j,i)  $  ]  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of g or h (or l) should be definable using combinations as suggested by Yuval Filmus's comment.&lt;/p&gt;&#xA;" OwnerUserId="820" LastActivityDate="2011-02-16T15:12:45.663" />
  <row Id="22369" PostTypeId="1" AcceptedAnswerId="22375" CreationDate="2011-02-16T15:15:49.703" Score="2" ViewCount="143" Body="&lt;p&gt;Hi there,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The equations for getting the expected value of the binomial distribution in one of my textbooks is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sum_{x=i}  x_i P(X=x_i)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and in another is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sum_x  x P(x)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and both texts say that this actually equals:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\mu = np$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and from what I can see the 'p' means the probability of success in the binomial equation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As usual, I find the math notation of the summation equations is difficult to understand. If p = 0.6 and the number of trials is 4 then I from the above equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\mu = 4 times 0.6 = 2.4$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which seems simple, but when I look at the summation equations, I can't see that they are telling me to add up p 4 times because:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;there is a value $x_i$ to multiply a value $P(X = x_i)$&lt;/li&gt;&#xA;&lt;li&gt;What is this value $x_i$? &lt;/li&gt;&#xA;&lt;li&gt;Is $P(X = x_i)$ the same as 'p' ie 0.6 in the above example, or does it have a different value for each value of $x_i$ whatever that is?&lt;/li&gt;&#xA;&lt;li&gt;I tried summing the values of the discrete probability distribution and dividing by the number of different discrete values but didn't get the same answer.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I'm sure I'm not too far off getting this, but I don't see how the summation notation is saying the same thing as $\mu = np$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance for any clarification.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regards&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Joe&lt;/p&gt;&#xA;" OwnerUserId="1094" LastActivityDate="2011-02-16T16:14:54.463" Title="Expected Value of Binomial Probability Distribution - Issue with Notation" Tags="&lt;probability&gt;" AnswerCount="1" />
  <row Id="22370" PostTypeId="2" ParentId="22364" CreationDate="2011-02-16T15:22:22.377" Score="0" ViewCount="" Body="&lt;p&gt;You want to maximize $\sum_i (r_i \cdot x)^2$ over unit vectors $x$.  To get rid of the constraint, this is $$\frac{\sum_i (r_i \cdot x)^2}{|x|^2}=\frac{\sum_i (r_i \cdot x)^2}{\sum_j x_j^2}=\frac{\sum_{ij} (r_{ij} x_j)^2}{\sum_j x_j^2}$$ where $r_{ij}$ is the $j^{th}$ component of $r_i$.  Now you can differentiate with respect to $x_j$ and set to zero without any trig functions getting in the way.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-02-16T15:22:22.377" CommentCount="1" />
  <row Id="22371" PostTypeId="2" ParentId="22303" CreationDate="2011-02-16T15:28:36.120" Score="1" ViewCount="" Body="&lt;p&gt;Let $x\in\mathbb{Z}$. Firstly, if $x\in\{0,2\}$ then $y=x$. Suppose if so $x\notin\{0,2\}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have that $x=\frac{x+y}{y}=\frac{x}{y}+1$. Since both sides are integers we have $\frac{x}{y}\in\mathbb{Z}$. Let $x$ be some prime number which is not $2$, then $y=\{\pm 1,\pm x\}$. Since $\frac{x}{y} = x-1 \neq \pm 1$ (because $x\neq 0,2$) we have that $y\neq\pm x$ therefore $y=\pm 1$ but then we have $x-1 = \pm x$ which is a contradiction.&lt;/p&gt;&#xA;" OwnerUserId="622" LastActivityDate="2011-02-16T15:28:36.120" />
  <row Id="22372" PostTypeId="2" ParentId="22367" CreationDate="2011-02-16T15:36:13.943" Score="2" ViewCount="" Body="&lt;p&gt;The sum is a martingale by linearity of conditional expectation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The product is not a martingale in general - if $X_t$ is a martingale, then $X_t \times X_t = X_t^2$ is a submartingale by Jensen's inequality. &lt;/p&gt;&#xA;" OwnerUserId="3162" LastActivityDate="2011-02-16T15:36:13.943" CommentCount="1" />
  <row Id="22373" PostTypeId="2" ParentId="12472" CreationDate="2011-02-16T15:36:49.097" Score="8" ViewCount="" Body="&lt;p&gt;Your reasoning is incorrect. $\frac{k^i}{i!}$ is not even an integer in general, so it can't count what you're claiming it counts. While $k^i$ is the number of ways to choose $i$ objects from $k$ types of objects &lt;em&gt;in order&lt;/em&gt;, you can't just divide by $i!$ to get rid of the ordering because some of the orderings are the same when you rearrange them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, if $k = 2, i = 3$, then the possible ways to choose $3$ objects from $2$ types of objects in order are $AAA, AAB, ABA, ABB, BAA, BAB, BBA, BBB$, and the possible ways to choose $3$ objects from $2$ types are $AAA, AAB, ABB, BBB$, which is $4$ and not $\frac{2^3}{3!} = \frac{4}{3}$. When you are dividing by $3!$ you are pretending that the different orderings group evenly into groups of $6$ but, as you can see, they don't.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The actual number you want, for fixed $i$, is the number of &lt;a href=&quot;http://en.wikipedia.org/wiki/Multiset&quot; rel=&quot;nofollow&quot;&gt;multisets&lt;/a&gt; of $i$ objects on a $k$-element set, which is actually&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$${k+i-1 \choose i}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This can be proven using what's called the &quot;stars and bars&quot; argument, as well as &lt;a href=&quot;http://en.wikipedia.org/wiki/Burnside%27s_lemma&quot; rel=&quot;nofollow&quot;&gt;Burnside's lemma&lt;/a&gt;. Note that if $i$ is fixed and $k$ tends to infinity this is asymptotically, but not exactly, $\frac{k^i}{i!}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you say, if you want the number of multisets of $n$ or less objects, then you just add in a dummy element of the set, so the final answer is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{i=0}^n {k+i-1 \choose i} = {k+n \choose n}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This identity is sometimes known as the hockey-stick identity because it has an interpretation in terms of summing along a hockey-stick-like shape on Pascal's triangle. &lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2011-02-17T09:02:48.603" LastActivityDate="2011-02-17T09:02:48.603" CommentCount="4" />
  <row Id="22374" PostTypeId="2" ParentId="22350" CreationDate="2011-02-16T16:10:00.947" Score="5" ViewCount="" Body="&lt;p&gt;If we expand ${ mp^r \choose p^r }$ in the usual way we obtain&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$${ mp^r \choose p^r } = m \prod_{k=1}^{p^r-1} \frac{mp^r-k}{p^r-k}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and since $mp^r -k= (m-1)p^r + (p^r-k)$ the power of $p$ dividing $mp^r-k$ is the same as the power of $p$ dividing $p^r-k.$ Hence the powers of $p$ in the numerator and denominator of the above product cancel, which gives us the result due to the $m$ before the product sign.&lt;/p&gt;&#xA;" OwnerUserId="1301" LastEditorUserId="1301" LastEditorDisplayName="" LastEditDate="2011-02-16T16:24:19.030" LastActivityDate="2011-02-16T16:24:19.030" />
  <row Id="22375" PostTypeId="2" ParentId="22369" CreationDate="2011-02-16T16:14:54.463" Score="3" ViewCount="" Body="&lt;p&gt;If the number of trials is 4, then $X$ can take the values 0, 1, 2, 3 and 4. Then $x_1 = 0, x_2=1, \cdots, x_5 = 4$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since this is a binomial distribution, $P(X=x_i) = \binom{n}{x_i}p^{x_i}(1-p)^{n-x_i}$, so $P(X=x_i)$ does not equal $p$ (at least not usually), and $P(X=x_i)$ has different values for different $x_i$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the formula for finding $P(X=x_i)$ (provided above), you will be able to multiply $P(X=x_i)$ with $x_i$ for $i = 1, 2, 3, 4, 5$. When you add these five products together you will find the sum (at the top of your question) for the expectation value. If you do it right, I think you will find out that the sum is equal to $np$.&lt;/p&gt;&#xA;" OwnerUserId="7044" LastActivityDate="2011-02-16T16:14:54.463" CommentCount="2" />
  <row Id="22376" PostTypeId="2" ParentId="22367" CreationDate="2011-02-16T16:22:36.947" Score="6" ViewCount="" Body="&lt;p&gt;The product of two independent martingales is a martingale--or rather it is or it is not, depending on the precise formulation of the hypothesis! When it is, one says that the martingales are &lt;em&gt;orthogonal&lt;/em&gt;. This is explained, for example, by &lt;a href=&quot;http://mech.math.msu.su/~cherny/&quot; rel=&quot;nofollow&quot;&gt;Alexander Cherny&lt;/a&gt; in the chapter &lt;a href=&quot;http://mech.math.msu.su/~cherny/pmt.pdf&quot; rel=&quot;nofollow&quot;&gt;Some Particular Problems of Martingale Theory&lt;/a&gt; of the &lt;a href=&quot;http://books.google.com/books?id=r7hcwexRIkcC&amp;amp;printsec=frontcover&amp;amp;source=bl&amp;amp;ots=8v4BBzh2rF&amp;amp;sig=mSqgPU-ZXhaD6LmSYMhNn8LuIUo&amp;amp;hl=fr&amp;amp;ei=OIeyTZ_WMsPX8gPAxZjRAQ&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=6&amp;amp;ved=0CEAQ6AEwBQ#v=onepage&quot; rel=&quot;nofollow&quot;&gt;Shiryaev Festschrift&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And yes, the sum of two independent martingales is a martingale but, here again, it might be wise to state the result with some care and, first of all, as mentioned by steveO in a comment, to specify the filtration(s) one is considering. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The trivial version is that if $X$ and $Y$ are two martingales (independent or not) with respect to a given filtration $\mathcal{G}$, then the sum $X+Y$ is also a martingale with respect to $\mathcal{G}$. But what happens if one assumes that $X$ is a martingale with respect to its own filtration $\mathcal{F}^X$ and that $Y$ is a martingale with respect to its own filtration $\mathcal{F}^Y$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Recall that the filtration $\mathcal{F}^Z$ of a process $Z$ is defined by $\mathcal{F}^Z_n=\sigma(\{Z_k;k\le n\})$ for every $n$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, &lt;em&gt;if $X$ and $Y$ are independent&lt;/em&gt;, $X+Y$ is a martingale with respect to its own filtration $\mathcal{F}^{X+Y}$ but, first, the proof, while not terribly difficult, requires to be careful (and uses more than the linearity of conditional expectations), and, second, for non independent martingales, this becomes horribly wrong.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To get an idea of the problem, consider a given integrable random variable $\xi$ and two $\sigma$-algebras $\mathcal{A}$ and $\mathcal{B}$ and try to find conditions guaranteeing that $E(\xi|\mathcal{A}\vee\mathcal{B})=E(\xi|\mathcal{A})$. Is $\xi$ independent of $\mathcal{B}$ enough? No, one has to assume that  $\mathcal{B}$ is independent of $\sigma(X)\vee\mathcal{A}$.&lt;/p&gt;&#xA;" OwnerUserId="6179" LastEditorUserId="6179" LastEditorDisplayName="" LastEditDate="2011-04-23T08:05:31.730" LastActivityDate="2011-04-23T08:05:31.730" CommentCount="2" />
  <row Id="22377" PostTypeId="1" CreationDate="2011-02-16T16:23:35.977" Score="1" ViewCount="73" Body="&lt;p&gt;For a function $f(x,y)$ of two independent variables we have an incomplete specification of its partial derivatives as follows: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac {\partial f(x,y)} {\partial x} = \frac {1} {g(x,y) \sqrt {1 - (\frac {k y} {x^{(1/3)}})^2}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac {\partial f(x,y)} {\partial y} = \left(\frac {3 x} {4}\right)\left (\frac {k} {x^{(1/3)}}\right)^2 (2 y) \frac {1} {g(x,y) \sqrt {1 - (\frac {k y} {x^{(1/3)}})^2}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Problem: finding a suitable $g(x,y)$ that makes the partial derivatives converge to a single function $f(x,y)$ that fulfills the condition $f(x,0) = x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I will be grateful if people with many flight hours can offer suggestions for $g(x,y)$.  Needless to say, I am &lt;strong&gt;not&lt;/strong&gt; asking that they verify those suggestions,&#xA;but in case someone would like, these are the inputs to Wolfram integrator:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1 / ( g(x,y as r) sqrt(1 - (k r / x^(1/3))^2) )&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(3 t / 4) (k / t^(1/3))^2 (2 x) / ( g(x as t,y as x) sqrt(1 - (k x / t^(1/3))^2) )&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance for your help.&lt;/p&gt;&#xA;" OwnerUserId="7161" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2011-02-16T17:29:18.287" LastActivityDate="2011-02-16T18:54:34.313" Title="Completing partial derivatives to make them converge" Tags="&lt;differential-equations&gt;" AnswerCount="1" CommentCount="6" />
  <row Id="22378" PostTypeId="2" ParentId="22265" CreationDate="2011-02-16T16:26:32.857" Score="2" ViewCount="" Body="&lt;p&gt;From a google search ([crossover strassen matrix multiplication]), I've found that experiments have found the crossover point to be from n = 8 to n = 20.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At some point in analysis of algorithms you jump from manipulating mathematical concepts straight to just running things experimentally and recording values, at which point it is not really called analysis of algorithms, but instead engineering where you manipulate things other than the basic algorithm (the machine the cooling system, the speed of memory components, caching strategies, etc). At that point the variables to consider are too many and underspecified to make an analytic (on paper) calculation, and you just have to compare using runtime data.&lt;/p&gt;&#xA;" OwnerUserId="1919" LastActivityDate="2011-02-16T16:26:32.857" />
  <row Id="22380" PostTypeId="2" ParentId="22303" CreationDate="2011-02-16T16:47:24.143" Score="2" ViewCount="" Body="&lt;p&gt;To prove it false, you just have to exhibit a single x for which there is no y ...&#xA;This is because the negation of &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\forall x \in Z \exists y \in Z(x *y = x + y)$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\exists x \in Z \forall y \in Z(x *y \ne x + y)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;yunone has done just that.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-02-16T16:47:24.143" />
  <row Id="22381" PostTypeId="2" ParentId="22262" CreationDate="2011-02-16T17:06:21.093" Score="1" ViewCount="" Body="&lt;p&gt;$f(x)=\sin(2\pi x)+x$ satisfies $f(n)=n$ for all $n\in\mathbb{Z}$  I don't know what you want with the rest...&lt;/p&gt;&#xA;" OwnerUserId="6925" LastActivityDate="2011-02-16T17:06:21.093" CommentCount="3" />
  <row Id="22383" PostTypeId="2" ParentId="22358" CreationDate="2011-02-16T17:13:13.797" Score="4" ViewCount="" Body="&lt;p&gt;Since B[i] can change depending on some as yet unknown thing (maybe some values of B[1], B[2], B[3],.. will behave differently than others, at this point e don't know maybe it will maybe it won't), let's start off by simplifying, getting rid of the stuff we don't know about and leaving unconditional operations:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;j := 1   ---------------------- 1 time&#xA;while j &amp;lt; n do  --------------- n times&#xA;  k := j    ------------------- n-1 times&#xA;  for i := j + 1 to n do  ----- ?&#xA;    k := i    ----------------- ?&#xA;  j := j + 1   ---------------- n-1 times&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;(This is the point of worst-case analysis, because of the conditional and the data in B, things may go much faster, but at least it'll be no worse than the analysis we're about to do.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The inner loop will occur altogether $n-1$ times (think of it just like the other assignments). But each operation in that inner loop (incrementing the $i$, comparing it to $n$, setting $k$ to $i$) how many times will they take place? Well, from $j+1$ to $n$ times. Which is $n - j - 1$. So as $j$ goes from 1 to $n$, $i$ goes from $j+1$ to $n$. This can be expressed as a summation:&#xA;$$\sum^n_{j=1} \ \sum^n_{i=j+1} ???$$&#xA;What is the question mark? What is the inner thing being summed? Well, let's be simple, ignore the extra comparisons and additions in the 'for' command' and just count the 'k := i ' assignment or 1 operation. So the summation looks like this:&#xA;$$\sum^n_{j=1} \ \sum^n_{i=j+1} 1.$$&#xA;Now you should be able to find it easier to manipulate this sum than it is to manipulate the programming language operations. For example, summing 1 from $a$ to $b$ is the same as summing 1 from 0 to $b-a$ which is equal to $b-a+1$ (why the '$+1$'?). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you can manipulate the summation, then you can go back to adding in complexities, like multiple assignments. You'll realize that those simplifications I made were not really important,they only change a constant in front of the final answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since for this algorithm (insertion sort), it ends up being dependent on the actual data $B$ being sorted, the worst case is the most exact answer you'll get (you really don't want to worry about the conditional) unless you know something about what $B$ looks like (is it already sorted, in reverse, some interesting subset of permutations, etc.&lt;/p&gt;&#xA;" OwnerUserId="1919" LastActivityDate="2011-02-16T17:13:13.797" />
  <row Id="22384" PostTypeId="1" CreationDate="2011-02-16T17:15:34.530" Score="-1" ViewCount="95" Body="&lt;p&gt;A dietitian has 3 supplement products that can be used. Each dose of Supplement I contains 5 units of Vitamin B, 4 units of calcium and 4 units of iron. Each dose of Supplement II contains 2 units of Vitamin B, 3 units of calcium and 2 unites of iron. Each dose of Supplement III contains 1 unit of vitamin B, 5 units of calcium and 2 units of iron. If the patient needs 13 units of Vitamin B, 16 units of calcium and 12 units of iron each week, how many doses of each supplement should the dietitian prescribe?&lt;/p&gt;&#xA;" OwnerUserId="7137" LastActivityDate="2011-02-16T17:50:43.153" Title="How to put the following information into a matrix?" Tags="&lt;matrices&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="22385" PostTypeId="1" CreationDate="2011-02-16T17:22:17.207" Score="12" ViewCount="214" Body="&lt;p&gt;Let $B$ and $F$ be compact Hausdorff spaces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $E\to B$ be a fiber bundle with fibre $F$ and structure group $\mathrm{Homeo}(F)$, the group of homeomorphisms of $F$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think this induces a fiber bundle $E&amp;#39;$ over $B$ with fiber $C(F,\mathbb C)$, the C*-algebra of continuous functions on $F$, and with structure group $\mathrm{Aut}(C(F,\mathbb C))\cong\mathrm{Homeo}(F)$, the group of *-automorphisms of $C(F,\mathbb C)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(To be more explicit about what happens here: my idea is: take a covering of $B$ which trivialises $E$. The transition functions give me a cocycle with values in the structure group $\mathrm{Homeo}(F)$. But, since $\mathrm{Homeo}(F)\cong\mathrm{Aut}(C(F,\mathbb C))$, I get a cocycle with values in $\mathrm{Aut}(C(F,\mathbb C))$, which I'd like to use to glue my new bundle.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $\Gamma(B,E&amp;#39;)$ denote the continuous sections of $E&amp;#39;$. I think pointwise operations turn this into a C*-algebra. Since the fiber $C(F,\mathbb C)$ is commutative, $\Gamma(B,E&amp;#39;)$ is commutative as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Question:&lt;/em&gt; What is the spectrum of $\Gamma(B,E&amp;#39;)$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example: If $E\cong B\times F$ is the trivial bundle, then $E&amp;#39;\cong B\times C(F,\mathbb C)$ and thus&#xA;$$\Gamma(B,E&amp;#39;)\cong C(B,C(F,\mathbb C))\cong C(B\times F,\mathbb C).$$&#xA;This suggests that the spectrum of $\Gamma(B,E&amp;#39;)$ is actually $E$.&lt;/p&gt;&#xA;" OwnerUserId="367" LastEditorUserId="367" LastEditorDisplayName="" LastEditDate="2011-05-05T14:53:29.923" LastActivityDate="2011-05-05T14:53:29.923" Title="What is the spectrum of the commutative C*-algebra I have constructed here?" Tags="&lt;topology&gt;&lt;functional-analysis&gt;&lt;c-star-algebras&gt;&lt;bundles&gt;" CommentCount="2" FavoriteCount="3" />
  <row Id="22386" PostTypeId="2" ParentId="22384" CreationDate="2011-02-16T17:45:13.767" Score="3" ViewCount="" Body="&lt;p&gt;Tell us more about your problem.  This may not address what you are struggling with, because you never told us.  In any case, do you know how to solve 3 equations and 3 unknowns?  &lt;a href=&quot;http://www.themathpage.com/alg/simultaneous-equations-3.htm&quot; rel=&quot;nofollow&quot;&gt;This&lt;/a&gt; and &lt;a href=&quot;http://www.tutorvista.com/math/solve-3-equations-3-unknowns&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; provide tutorials.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Try to find a way to get a system of 3 equations and 3 unknowns to solve this question.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint:  Assign variables $x,y,z$ to be how much of Supplement I, II, III is used respectively.  Then by looking at Iron we see that we must have $4x+2y+2z=12$.  Looking at Calcium and Vitamin B gives two more similar equations.  Now we can solve the system.&lt;/p&gt;&#xA;" OwnerUserId="6075" LastEditorUserId="6075" LastEditorDisplayName="" LastEditDate="2011-02-16T17:50:43.153" LastActivityDate="2011-02-16T17:50:43.153" />
  <row Id="22387" PostTypeId="2" ParentId="12507" CreationDate="2011-02-16T17:46:37.887" Score="1" ViewCount="" Body="&lt;p&gt;Here a fancy proof that uses operators and differentiation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $T_a$ be the translation operator which maps polynomials to their translates&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ T_a p(x) = p(x + a) .$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Restricting attention to polynomials of degree $n$ or lower, this is a map from an $n+1$-dimensional vector space to itself. The question to be resolved is whether the $n+1$ translation operators $T_{a_0}, T_{a_1}\dots, T_{a_n}$ are linearly independent (when applied to polynomials of degree $n$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is well-known that the translation operator can be written as the exponentiation of the differentiation operator $Dp(x) := p&amp;#39;(x)$ as follows&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ T_a = e^{aD} = 1 + \frac1{1!} aD + \frac1{2!} a^2D^2 + \dots + \frac1{n!} a^n D^n .$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is Taylor's theorem. Note that since $D$ is nilpotent on the space of polynomials of degree $n$, $D^{n+1} = 0$, the series is actually a finite sum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the point is that for any polynomial $p(x)$ of degree $n$, we know that its derivatives $p(x), Dp(x), D^2p(x), \dots, D^n p(x)$ are &lt;em&gt;linearly independent&lt;/em&gt; because the degree always decreases by 1. Hence, they form a basis of our vector space of polynomials. The expansion of the $T_a$ in terms of this new basis is given by the expression above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To show that the translates $T_{a_k} p(x)$ are linearly independent, we only have to check that their matrix with respect to the new basis is nonsingular. It reads&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{pmatrix} 1 &amp;amp; 1 &amp;amp; \dots &amp;amp; 1 \\&#xD;&#xA;a_0 &amp;amp; a_1 &amp;amp; \dots &amp;amp; a_n \\&#xD;&#xA;\frac1{2!} a_0^2 &amp;amp; \frac1{2!} a_1^2 &amp;amp; \dots &amp;amp; \frac1{2!} a_n^2 \\&#xD;&#xA;\vdots \\&#xD;&#xA;\frac1{n!} a_0^n &amp;amp; \frac1{n!} a_1^n &amp;amp; \dots &amp;amp; \frac1{n!} a_n^n \\&#xD;&#xA;\end{pmatrix}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But its determinant is clearly a non-zero multiple of the Vandermonde determinant, which is non-zero if and only if the values $a_0,a_1,\dots,a_n$ are pairwise distinct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Another way to see that this matrix is non-singular is to note that the Vandermonde determinant is related to the interpolation problem, whose solution is unique.)&lt;/p&gt;&#xA;" OwnerUserId="434" LastActivityDate="2011-02-16T17:46:37.887" />
  <row Id="22388" PostTypeId="2" ParentId="22280" CreationDate="2011-02-16T17:56:51.980" Score="0" ViewCount="" Body="&lt;p&gt;The Laplace-Beltrami operator, as far as I am aware, is `unique'.  Perhaps you mean that this operator is &lt;strong&gt;a&lt;/strong&gt; Laplacian---that is certainly true.  The idea is that it is still a divergence, and thus (with some assumptions on $c$) hopefully elliptic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most common generalisation of the Laplacian in this direction is the $p$-Laplacian, on which you can learn more here: &lt;a href=&quot;http://en.wikipedia.org/wiki/P-Laplacian&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/P-Laplacian&lt;/a&gt;.  A simple google search for &quot;p laplacian&quot; also turns up plenty of elementary material.&lt;/p&gt;&#xA;" OwnerUserId="4322" LastActivityDate="2011-02-16T17:56:51.980" CommentCount="2" />
  <row Id="22389" PostTypeId="2" ParentId="22324" CreationDate="2011-02-16T18:05:56.417" Score="11" ViewCount="" Body="&lt;p&gt;There is a slight divergence of nomenclature. Everyone agrees on what $\mathrm{Aut}(E/F)$ &lt;em&gt;is&lt;/em&gt;. The question is what to call it.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Some books (e.g., Hungerford, Rotman's &lt;em&gt;Galois Theory&lt;/em&gt;), always refer to $\mathrm{Aut}(E/F)$ as the &quot;&lt;em&gt;Galois group&lt;/em&gt;&quot; of $E$ over $F$ (or of the extension), whether or not the extension is a Galois extension.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Other books (e.g., Lang), use the generic term &quot;&lt;em&gt;automorphism group&lt;/em&gt;&quot; to refer to $\mathrm{Aut}(E/F)$ in the general case, and reserve the term &lt;em&gt;Galois group&lt;/em&gt; &lt;strong&gt;exclusively&lt;/strong&gt; for the situation in which $E$ is a Galois extension of $F$. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;So, in Lang, even just saying &quot;Galois group&quot; already implies that the extension must be a Galois extension, that is, normal and separable. In Hungerford, just saying &quot;Galois group&quot; does not imply anything beyond the fact that we are looking at the automorphism of the extension.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wikipedia is following Convention 2; your book is following convention 1. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is also the question of whether to admit infinite extensions or not. A lot of introductory books only consider only finite extensions when dealing with Galois Theory, and define an extension to be Galois if and only if $|\mathrm{Aut}(E/F)| = [E:F]$. This definition does not extend to the infinite extension, so the definitions are restricted to finite (algebraic) extensions, with infinite extensions not considered at all. Other characterizations of an extension being Galois (e.g., normal and separable) generalize naturally to infinite extensions, so no restriction is placed. Likewise, some books explicitly restrict to algebraic extensions, others do not; but note that most define &quot;normal&quot; to require algebraicity, because it is defined in terms of embeddings into the algebraic closure of the base field, so even if you don't explicitly require the extension to be algebraic in order to be Galois, in reality this restriction is (almost) always in place.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is not such a big deal as it might appear, because one can show that an arbitrary (possibly infinite) Galois extension $E/F$ is completely characterized in a very precise sense by the &lt;em&gt;finite&lt;/em&gt; Galois extensions $K/F$ with $F\subseteq K\subset E$ with $[K:F]\lt\infty$, as the automorphism group $\mathrm{Aut}(E/F)$ is the inverse limit of the corresponding finite automorphism groups.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-16T18:23:30.233" LastActivityDate="2011-02-16T18:23:30.233" />
  <row Id="22390" PostTypeId="1" AcceptedAnswerId="22393" CreationDate="2011-02-16T18:11:57.393" Score="6" ViewCount="194" Body="&lt;p&gt;Given a polynomial $y=C_0+C_1 x+C_2 x^2+C_3 x^3 + \ldots$ of some order $N$, I can easily calculate the polynomial of reduced order $M$ by taking only the first $M+1$ terms. This is equivalent to doing a Taylor series expansion with $M&amp;lt;=N$ around $x=0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But what if I want to take the Taylor series expansion around a different point $x_c$. In the end, I want the polynomial coefficients of $y_2=K_0+K_1 x + K_2 x^2 + K_3 x^3 + \ldots$ which represents the Taylor's expansion of $y$ around point $x_c$ such that $y(x_c)=y_2(x_c)$ including the first $M$ derivatives.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So given the coefficients $C_i$ with $i=0 \ldots N$, and a location $x_c$ I want to calculate the coefficients $K_j$ with $j=0 \ldots M$.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Example&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Given $y=C_0+C_1 x+C_2 x^2$ ( $N=2$ ) then the tangent line ($M=1$) through $x_c$ is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ y_2 = (C_0-C_2 x_c^2) + (C_1+2 C_2 x_c) x $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or $K_0 = C_0-C_2 x_c^2$, and $K_1 =C_1+2 C_2 x_c$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There must be a way to construct a ($M+1$ by $N+1$ ) matrix that transforms the coefficients $C_i$ into $K_j$. For the above example this matrix is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \begin{bmatrix}K_{0}\\&#xD;&#xA;K_{1}\end{bmatrix}=\begin{bmatrix}1 &amp;amp; 0 &amp;amp; -x_{c}^{2}\\&#xD;&#xA;0 &amp;amp; 1 &amp;amp; 2\, x_{c}\end{bmatrix}\begin{bmatrix}C_{0}\\&#xD;&#xA;C_{1}\\&#xD;&#xA;C_{2}\end{bmatrix} $$&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Example #2&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The reduction of a $5$-th order polynomial to a $3$-rd order around $x_c$ is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \begin{bmatrix}K_{0}\\&#xD;&#xA;K_{1}\\&#xD;&#xA;K_{2}\\&#xD;&#xA;K_{3}\end{bmatrix}=\left[\begin{array}{cccc|cc}&#xD;&#xA;1 &amp;amp;  &amp;amp;  &amp;amp;  &amp;amp; -x_{c}^{4} &amp;amp; -4\, x_{c}^{5}\\&#xD;&#xA; &amp;amp; 1 &amp;amp;  &amp;amp;  &amp;amp; 4\, x_{c}^{3} &amp;amp; 15\, x_{c}^{4}\\&#xD;&#xA; &amp;amp;  &amp;amp; 1 &amp;amp;  &amp;amp; -6\, x_{c}^{2} &amp;amp; -20\, x_{c}^{3}\\&#xD;&#xA; &amp;amp;  &amp;amp;  &amp;amp; 1 &amp;amp; 4\, x_{c} &amp;amp; 10\, x_{c}^{2}\end{array}\right]\begin{bmatrix}C_{0}\\&#xD;&#xA;C_{1}\\&#xD;&#xA;C_{2}\\&#xD;&#xA;C_{3}\\&#xD;&#xA;C_{4}\\&#xD;&#xA;C_{5}\end{bmatrix} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is a block matrix, and not an upper diagonal one as some of the answers have indicated.&lt;/p&gt;&#xA;" OwnerUserId="3301" LastEditorUserId="3301" LastEditorDisplayName="" LastEditDate="2011-02-16T19:09:32.837" LastActivityDate="2011-02-19T20:46:27.393" Title="Taylor series of a polynomial" Tags="&lt;linear-algebra&gt;&lt;polynomials&gt;&lt;taylor-expansion&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="2" />
  <row Id="22391" PostTypeId="1" CreationDate="2011-02-16T18:26:07.433" Score="2" ViewCount="158" Body="&lt;p&gt;hi all,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am having an NxN matrix . I want to read the elements of that NxN matrix diagonally and need to store it in an array.How?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, I am having one 3x3 matrix&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A B C&lt;/p&gt;&#xA;&#xA;&lt;p&gt;D E F&lt;/p&gt;&#xA;&#xA;&lt;p&gt;G H I&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to read this matrix diagonally(starting from A and downwards) and output should be&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A D B C E G H F I&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Update:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can i do it in reverse manner?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to fill the above read array(A D B C E G H F I) into an NxN (N should be given as input) HOW?&lt;/p&gt;&#xA;" OwnerUserId="7164" LastEditorUserId="7164" LastEditorDisplayName="" LastEditDate="2011-02-16T19:48:43.820" LastActivityDate="2011-02-16T19:56:08.333" Title="How to read an NxN matrix diagonally? After this how to write it diagonally?" Tags="&lt;matrices&gt;" AnswerCount="2" CommentCount="5" FavoriteCount="1" />
  <row Id="22392" PostTypeId="2" ParentId="22352" CreationDate="2011-02-16T18:27:40.313" Score="10" ViewCount="" Body="&lt;p&gt;In a sense, yes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The object of &lt;em&gt;descriptive set theory&lt;/em&gt; is to understand &quot;definable&quot; sets of reals (as opposed to arbitrary sets). So, for example, we concern ourselves with Borel sets, or their continuous images, or the complements of those images, or countable unions of such things, etc. Also, we study other spaces, not just the set of reals. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The basic setting is that of &lt;em&gt;Polish spaces&lt;/em&gt;, i.e., complete metrizable spaces with a countable dense subset. This of course includes ${\mathbb R}$, but many other spaces that appear in practice are here. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It turns out that many classification problems that occur in practice have the form: We have a Polish space $X$ and an equivalence relation $E$ on $X$ (typically, $E$ is either Borel as a subset of $X\times X$, or the continuous image of a Borel set). We then study the complexity of the quotient space $X/E$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This can be measured in several ways. For example: Can we pick in a &quot;Borel fashion&quot; a representative of each equivalence class? If not, can we in a sense approximate the graphs of &quot;choice functions&quot; even if we cannot actually single out one of them? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It turns out that we can prove results that say that certain classification problems are strictly harder than certain others, and we can study the &quot;partial ordering&quot; of classification problems according to complexity. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To be a bit more precise, consider the problem: Given two $5\times 5$ matrices with real entries, are they similar? This is a classification problem: We want to pair matrices that are similar, and the question is how hard is it to decide whether they are paired. It turns out that the Jordan canonical form of two matrices is the same iff they are similar, and the Jordan form can be handily coded by a real number, so the problem reduces to &quot;can we identify if two reals are equal?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are two ways of measuring the complexity of this problem. One is in terms of complexity theory, and then we need to talk about how the reals are &quot;given&quot; to us. The other way is the descriptive set theoretic one: We have a Polish space: ${\mathbb R}$; and an equivalence relation: equality. This is as simple as a problem gets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another problem is: Given two auto-homeomorphisms of the unit square, when are they conjugate? This is a harder problem, meaning, there is no &quot;Borel&quot; map that to an auto-homeomorphism assigns a real number so that two homeomorphisms are conjugate iff they have been assigned the same real.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A good place to learn about this (and about the wide variety of examples that this approach covers) is &quot;A survey of current and recent work on the theory of Borel equivalence relations&quot; by G. Hjorth. It is available from his webpage, at &lt;a href=&quot;http://www.math.ucla.edu/~greg/research.html&quot; rel=&quot;nofollow&quot;&gt;http://www.math.ucla.edu/~greg/research.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="462" LastActivityDate="2011-02-16T18:27:40.313" CommentCount="6" />
  <row Id="22393" PostTypeId="2" ParentId="22390" CreationDate="2011-02-16T18:36:03.883" Score="8" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Note:&lt;/strong&gt; Rewritten essentially from the ground up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sorry for the somewhat obfuscated previous version. I did not realize at first that you were trying to write the end result as a polynomial in $x$ instead of $(x-x_c)$, and then I was trying to bootstrap that into what I had already written.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm going to use $a$ instead of $x_c$, because it is a little simpler to type and harder to confuse with $x$, if you don't mind.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The simplest way to think about what you are doing is that it is the composition of three linear transformations.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;First you take a polynomial written in terms of the basis $\beta=[1,x,x^2,\ldots,x^N]$ of the space of polynomials degree at most $N$, and you rewrite it in terms of the basis $\gamma=[1,(x-a),(x-a)^2,\ldots,(x-a)^N]$. That is, a change of coordinates.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Then you take the polynomial written in terms of $\gamma$, and you project down to the subspace generated by $[1,(x-a),\ldots,(x-a)^M]$. That is, you erase the terms of degree $M+1,\ldots,N$. Call this $P^N_M$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Finally, you take the resulting polynomial, which is written in terms of the basis $\gamma$, and you translated it back into the basis $\beta$; another change of coordinates.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Being a composition of three linear transformations, it can be represented by a matrix which is the product of three matrices, the matrices that represent the three operations. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The matrix for $P^N_M$ is the simplest one: it is a block-diagonal matrix that has the $(M+1)\times (M+1)$ identity in the first block, and the $(N-M)\times(N-M)$ zero matrix as the second block.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The matrix for step 3 is just the inverse of the matrix for step 1. To each real number $r$, we have a matrix whose columns represent the binomial expansion of $(x+r)$:&#xA;$$B_r = \left(\begin{array}{cccccc}&#xD;&#xA;1 &amp;amp; r &amp;amp; r^2 &amp;amp; r^3 &amp;amp; \cdots &amp;amp; r^N\\\&#xD;&#xA;0 &amp;amp; 1 &amp;amp; 2r &amp;amp; 3r^2 &amp;amp; \cdots &amp;amp; \binom{N}{1}r^{N-1}\\\&#xD;&#xA;0 &amp;amp; 0 &amp;amp; 1 &amp;amp; 3r &amp;amp; \cdots &amp;amp; \binom{N}{2}r^{N-2}\\\&#xD;&#xA;0 &amp;amp; 0 &amp;amp; 0 &amp;amp; 1 &amp;amp; \cdots &amp;amp; \binom{N}{3}r^{N-3}\\\&#xD;&#xA;\vdots &amp;amp; \vdots &amp;amp; \vdots &amp;amp; \ddots &amp;amp; \vdots\\\&#xD;&#xA;0 &amp;amp; 0 &amp;amp; 0 &amp;amp; 0 &amp;amp; \cdots &amp;amp; \binom{N}{N-1}r\\\&#xD;&#xA;0 &amp;amp; 0 &amp;amp; 0 &amp;amp; 0 &amp;amp; \cdots &amp;amp; 1&#xD;&#xA;\end{array}\right).$$&#xA;When you go from having the basis given by powers of $x$ to the basis given by powers of $x-a$, you do this by multiplying by $B_a$. In order to go from the basis given by the powers of $(x-a)$ to the powers of $x$, the change of variable $u=x-a$ shows that you are going form powers of $u$ to powers of $u+a = u-(-a)$, so you achieve this by multiplying by $B_{(-a)}$. Thus, your operation is given by the matrix&#xA;$$B_{(-a)}P^N_MB_a.$$&#xA;Now, this is a product of three $(N+1)\times (N+1)$ matrices, so you may wonder why I have an $(N+1)\times (N+1)$ matrix and you have an $(M+1)\times (N+1)$ matrix. The reason is that because the last $N-M$ rows of $P_M$ are all zero and all matrices are upper triangular, the last $N-M$ rows of this product will likewise be zero; you are just omitting them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, if you have the polynomial $C_0 + C_1x + \cdots +C_Nx^N$, then your end result is the polynomial $K_0 + K_1x+\cdots +K_Nx^N$, where&#xA;$$\left(\begin{array}{c}&#xD;&#xA;K_0\\&#xD;&#xA;K_1\\&#xD;&#xA;\vdots\\&#xD;&#xA;K_N\end{array}\right) = B_{(-a)}P_MB_a\left(\begin{array}{c}&#xD;&#xA;C_0\\&#xD;&#xA;C_1\\&#xD;&#xA;\vdots\\&#xD;&#xA;C_N\end{array}\right).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For your second example, you have $N=5$ and $M=3$. For $N=5$, you have&#xA;\begin{align*}&#xA;B_{a} &amp;amp;= \left(\begin{array}{cccccc}&#xA;1 &amp;amp; a &amp;amp; a^2 &amp;amp; a^3 &amp;amp; a^4 &amp;amp; a^5\\&#xA;0 &amp;amp; 1 &amp;amp; 2a &amp;amp; 3a^2 &amp;amp; 4a^3 &amp;amp; 5a^4\\&#xA;0 &amp;amp; 0 &amp;amp; 1 &amp;amp; 3a &amp;amp; 6a^2 &amp;amp; 10a^3\\&#xA;0 &amp;amp; 0 &amp;amp; 0 &amp;amp; 1 &amp;amp; 4a &amp;amp; 10a^2\\&#xA;0 &amp;amp; 0 &amp;amp; 0 &amp;amp; 0 &amp;amp; 1 &amp;amp; 5a\\&#xA;0 &amp;amp; 0 &amp;amp; 0 &amp;amp; 0 &amp;amp; 0 &amp;amp; 1&#xA;\end{array}\right)\\&#xA;P^5_3 &amp;amp;= \left(\begin{array}{cccccc}&#xA;1 &amp;amp; 0 &amp;amp; 0 &amp;amp; 0 &amp;amp; 0 &amp;amp; 0\\&#xA;0 &amp;amp; 1 &amp;amp; 0 &amp;amp; 0 &amp;amp; 0 &amp;amp; 0\\&#xA;0 &amp;amp; 0 &amp;amp; 1 &amp;amp; 0 &amp;amp; 0 &amp;amp; 0\\&#xA;0 &amp;amp; 0 &amp;amp; 0 &amp;amp; 1 &amp;amp; 0 &amp;amp; 0\\&#xA;0 &amp;amp; 0 &amp;amp; 0 &amp;amp; 0 &amp;amp; 0 &amp;amp; 0\\&#xA;0 &amp;amp; 0 &amp;amp; 0 &amp;amp; 0 &amp;amp; 0 &amp;amp; 0&#xA;\end{array}\right)\\&#xA;B_{-a} &amp;amp;=\left(\begin{array}{rrrrrr}&#xA;1 &amp;amp; -a &amp;amp; a^2 &amp;amp; -a^3 &amp;amp; a^4 &amp;amp; -a^5\\&#xA;0 &amp;amp; 1 &amp;amp; -2a &amp;amp; 3a^2 &amp;amp; -4a^3 &amp;amp; 5a^4\\&#xA;0 &amp;amp; 0 &amp;amp; 1 &amp;amp; -3a &amp;amp; 6a^2 &amp;amp; -10a^3\\&#xA;0 &amp;amp; 0 &amp;amp; 0 &amp;amp; 1 &amp;amp; -4a &amp;amp; 10a^2\\&#xA;0 &amp;amp; 0 &amp;amp; 0 &amp;amp; 0 &amp;amp; 1 &amp;amp; -5a\\&#xA;0 &amp;amp; 0 &amp;amp; 0 &amp;amp; 0 &amp;amp; 0 &amp;amp; 1&#xA;\end{array}\right).&#xA;\end{align*}&#xA;If you multiply out $B_{-a}P^5_3B_a$ you get&#xA;\begin{align*}&#xA;B_{-a}P^5_3B_a &amp;amp;= \left(\begin{array}{rrrrrr}&#xA;1 &amp;amp; -a &amp;amp; a^2 &amp;amp; -a^3 &amp;amp; a^4 &amp;amp; -a^5\\&#xA;0 &amp;amp; 1 &amp;amp; -2a &amp;amp; 3a^2 &amp;amp; -4a^3 &amp;amp; 5a^4\\&#xA;0 &amp;amp; 0 &amp;amp; 1 &amp;amp; -3a &amp;amp; 6a^2 &amp;amp; -10a^3\\&#xA;0 &amp;amp; 0 &amp;amp; 0 &amp;amp; 1 &amp;amp; -4a &amp;amp; 10a^2\\&#xA;0 &amp;amp; 0 &amp;amp; 0 &amp;amp; 0 &amp;amp; 1 &amp;amp; -5a\\&#xA;0 &amp;amp; 0 &amp;amp; 0 &amp;amp; 0 &amp;amp; 0 &amp;amp; 1&#xA;\end{array}\right)\left(\begin{array}{cccccc}&#xA;1 &amp;amp; a &amp;amp; a^2 &amp;amp; a^3 &amp;amp; a^4 &amp;amp; a^5\\&#xA;0 &amp;amp; 1 &amp;amp; 2a &amp;amp; 3a^2 &amp;amp; 4a^3 &amp;amp; 5a^4\\&#xA;0 &amp;amp; 0 &amp;amp; 1 &amp;amp; 3a &amp;amp; 6a^2 &amp;amp; 10a^3\\&#xA;0 &amp;amp; 0 &amp;amp; 0 &amp;amp; 1 &amp;amp; 4a &amp;amp; 10a^2\\&#xA;0 &amp;amp; 0 &amp;amp; 0 &amp;amp; 0 &amp;amp; 0 &amp;amp; 0\\&#xA;0 &amp;amp; 0 &amp;amp; 0 &amp;amp; 0 &amp;amp; 0 &amp;amp; 0&#xA;\end{array}\right)\\&#xA;&amp;amp;= \left(\begin{array}{rrrrrr}&#xA;1 &amp;amp; 0 &amp;amp; 0 &amp;amp; 0 &amp;amp; -a^4 &amp;amp; -4a^5\\&#xA;0 &amp;amp; 1 &amp;amp; 0 &amp;amp; 0 &amp;amp; 4a^3 &amp;amp; 15a^4\\&#xA;0 &amp;amp; 0 &amp;amp; 1 &amp;amp; 0 &amp;amp; -6a^2 &amp;amp; -20a^3\\&#xA;0 &amp;amp; 0 &amp;amp; 0 &amp;amp; 1 &amp;amp; 4a &amp;amp; 10a^2\\&#xA;0 &amp;amp; 0 &amp;amp; 0 &amp;amp; 0 &amp;amp; 0 &amp;amp; 0\\&#xA;0 &amp;amp; 0 &amp;amp; 0 &amp;amp; 0 &amp;amp; 0 &amp;amp; 0&#xA;\end{array}\right)&#xA;\end{align*}&#xA;exactly what you have except for the two extra rows of zeros at the bottom.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added.&lt;/strong&gt; If you think about it, you'll see that this process will always produce a block-upper triangular matrix, with an $(M+1)\times(M+1)$ identity in the upper left and an $(N-M)\times(N-M)$ zero matrix in the bottom right; because $B_{-a}B_a = I$, the bottom rows you zero out in $B_a$ do not affect the products in the first $M+1$ columns, and they are all that matters in the last $N-M$ rows. So the only parts that need computation is the block on the upper right. This shows how it relates to your non-square matrices.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added 2.&lt;/strong&gt; I should have added this as soon as you said you were programming it...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is an alternative way of computing this matrix, which may be faster when $M$ is larger than half of $N$. The idea is just to note that the matrix $P^N_M$ can be written as a difference of the identity matrix with the matrix that has the $(N-M)\times(N-M)$ identity on the bottom right and zeros elsewhere; call it $Q^N_M$; that is,&#xA;$$P^N_M = I_{N+1} - Q^N_M = \left(\begin{array}{ccccc}&#xD;&#xA;1 &amp;amp; 0 &amp;amp; \cdots &amp;amp; 0 &amp;amp; 0\\\&#xD;&#xA;0 &amp;amp; 1 &amp;amp; \cdots &amp;amp; 0 &amp;amp; 0\\\&#xD;&#xA;\vdots &amp;amp; \vdots &amp;amp; \ddots &amp;amp; \vdots &amp;amp; \vdots \\\&#xD;&#xA;0 &amp;amp; 0 &amp;amp; \cdots &amp;amp; 1 &amp;amp; 0\\\&#xD;&#xA;0 &amp;amp; 0 &amp;amp; \cdots &amp;amp; 0 &amp;amp; 1&#xD;&#xA;\end{array}\right) - \left(\begin{array}{cccccc}&#xD;&#xA;0 &amp;amp; 0 &amp;amp; \cdots &amp;amp; 0 &amp;amp; 0\\\&#xD;&#xA;0 &amp;amp; 0 &amp;amp; \cdots &amp;amp; 0 &amp;amp; 0\\\&#xD;&#xA;\vdots &amp;amp; \vdots &amp;amp; \ddots &amp;amp; \vdots &amp;amp; \vdots\\\&#xD;&#xA;0 &amp;amp; 0 &amp;amp; \cdots &amp;amp; 1 &amp;amp; 0\\\&#xD;&#xA;0 &amp;amp; 0 &amp;amp; \cdots &amp;amp; 0 &amp;amp; 1&#xD;&#xA;\end{array}\right).$$&#xA;Then you have that&#xA;$$B_{-a}P^{N}_MB_a = B_{-a}(I-Q^N_M)B_a = (B_{-a}IB_a) - B_{-a}Q^N_MB_a = I-B_{-a}Q^{N}_MB_a.$$&#xA;If $M$ is more than half of $N$, then there will be fewer operations to perform in $B_{-a}Q^N_MB_a$ than in $B_{-a}P^N_MB_a$, because $Q^N_MB_a$ has only $N-M$ nonzero rows, whereas $P^N_MB_a$ has $M+1$ nonzero rows. So you can pick which of the two computations to do: either $B_{-a}P^N_MB_a$, or $I-B_{-a}Q^N_MB_a$, and for $N$ and $M$ large, with $M$ close to $N$, using $Q^N_M$ will probably be faster than using $P^N_M$. &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-19T20:46:27.393" LastActivityDate="2011-02-19T20:46:27.393" CommentCount="5" />
  <row Id="22394" PostTypeId="2" ParentId="22377" CreationDate="2011-02-16T18:54:34.313" Score="3" ViewCount="" Body="&lt;p&gt;Hint: From your data one obtains&#xA;$$-{f_x\over f_y}={-2\over 3 k^2 y x^{1/3}}.$$&#xA;It follows that the curves $f(x,y)=$const. satisfy the separable differential equation $$y&amp;#39;={-2\over 3 k^2 y x^{1/3}}.$$&lt;/p&gt;&#xA;" OwnerUserId="1303" LastActivityDate="2011-02-16T18:54:34.313" CommentCount="1" />
  <row Id="22395" PostTypeId="1" AcceptedAnswerId="22397" CreationDate="2011-02-16T19:06:13.980" Score="3" ViewCount="118" Body="&lt;p&gt;I have a question related to this: &lt;a href=&quot;http://math.stackexchange.com/questions/5329/projective-modules&quot;&gt;Projective modules&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm trying to understand the &quot;philosophy&quot; of the statement, because it seems too similar to the statement &quot;a module is free iff every element can be written uniquely as a finite linear combination of elements of a basis&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this &quot;projective basis&quot; property saying this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a module P is projective iff every element in P can be written as a finite linear combination of some elements of P?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We lose uniqueness in the expression as a sum: in the elements of P, in the elements of R, and in the number of terms (so the concept of &quot;rank&quot; wouldn't make sense). Is this all, or am I misunderstanding the statement?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any other intuition related to that property is also appreciated.&lt;/p&gt;&#xA;" OwnerUserId="2614" LastActivityDate="2011-02-16T20:30:06.387" Title="A module is projective iff it has a projective basis" Tags="&lt;abstract-algebra&gt;&lt;modules&gt;" AnswerCount="3" FavoriteCount="1" />
  <row Id="22396" PostTypeId="2" ParentId="22391" CreationDate="2011-02-16T19:16:14.703" Score="4" ViewCount="" Body="&lt;p&gt;The array is made in such a way that the elements whose row position + column posiiton are clumped together, in an increasing order. That is, all the elements $a_{ij}$ such that $i + j = k$ will form a &quot;group&quot; of elements in the array.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{pmatrix}&#xD;&#xA;a_{11} &amp;amp; a_{12} &amp;amp; a_{13} &amp;amp; a_{14} \\&#xD;&#xA;a_{21} &amp;amp; a_{22} &amp;amp; a_{23} &amp;amp; a_{24} \\&#xD;&#xA;a_{31} &amp;amp; a_{32} &amp;amp; a_{33} &amp;amp; a_{34} \\&#xD;&#xA;a_{41} &amp;amp; a_{42} &amp;amp; a_{43} &amp;amp; a_{44}&#xD;&#xA;\end{pmatrix}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;will give the list&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(&#xD;&#xA;\underbrace{a_{11}}_{i+j = 2}&#xD;&#xA;, \underbrace{a_{21} , a_{12}}_{i+j = 3}&#xD;&#xA;, \underbrace{a_{13} , a_{22} , a_{31}}_{i+j=4}&#xD;&#xA;, \underbrace{a_{41} , a_{32} , a_{23} , a_{14}}_{i+j = 5}&#xD;&#xA;, \underbrace{a_{24} , a_{33} , a_{42} }_{i+j = 6}&#xD;&#xA;, \underbrace{a_{43} , a_{34}}_{i+j = 7}&#xD;&#xA;, a_{44}&#xD;&#xA;)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This shows a systematic way you can do it. Iterate through the $k$. When you get to a new $k$, increase the largest of the $i,j$ if you can, otherwise increase the smallest, so you get $i+j = k$. Then decrease the largest of the two and increase the smallest of the two until you have the reverse (that is, the one that was the largest to begin with is now  1). Restart.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note also that you have a symmetry, you will have two groups with the same size for every size except $k = n+1$, which might be used to improve the algorithm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT:&#xA;To reverse the process, start with $p = 1$ and iterate $p$ to $n$, then decrease it back to $1$ in a loop (or two loops maybe).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Every time you go through the loop, take $p$ elements from the list, let's call them $a_1, a_2, \dots, a_p$, and put them in the matrix. To do this, use two variables $i$ and $j$. At the start of every iteration of the loop, $i$ and $j$ have the values of the element last inserted. Increase the largest value by 1, if you can (when $p$ is increasing), otherwise increase the smallest value. Stepwise reverse the numbers as before, inserting the elements $a_1, a_2, \dots, a_p$.&lt;/p&gt;&#xA;" OwnerUserId="5966" LastEditorUserId="5966" LastEditorDisplayName="" LastEditDate="2011-02-16T19:56:08.333" LastActivityDate="2011-02-16T19:56:08.333" CommentCount="6" />
  <row Id="22397" PostTypeId="2" ParentId="22395" CreationDate="2011-02-16T19:19:37.803" Score="5" ViewCount="" Body="&lt;p&gt;The statement &quot;every element in P can be written as a finite linear combination of some elements of P.&quot;, where &quot;some&quot; means a finite set, just says that the module is &lt;a href=&quot;http://en.wikipedia.org/wiki/Finitely_generated_module&quot; rel=&quot;nofollow&quot;&gt;finitely generated&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This has nothing to do with being projective.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take for instance the $\mathbb Z$-module $\mathbb Z/2$. Here every element can be written as a multiple of $[1]$. So $\mathbb Z/2$ is finitely generated but not projective.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand the infinite direct sum $\bigoplus_{\mathbb N}\mathbb Z$ is a projective $\mathbb Z$-module which is not finitely generated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The crucial thing in the definition of a projective basis is really that you have &lt;em&gt;homomorphisms out of $P$ into the ring (regarded as module over itself)&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="367" LastActivityDate="2011-02-16T19:19:37.803" CommentCount="4" />
  <row Id="22398" PostTypeId="2" ParentId="20102" CreationDate="2011-02-16T19:28:23.853" Score="0" ViewCount="" Body="&lt;p&gt;The Dixon's algorithm is. It does a inversion of the matrix in modular arithmetic, based on some prime modulus P (this step is easy and exact). Then it solves the trivial system for each x to get a p-adic number representation of the solution. It can be converted to a rational number. Method was described in &quot;Dixon J. D. Exact solution of linear equations using P-adic expansions // Numer. Math., 40, 1982, P. 137-141.&quot;, but this article is offline.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a article about Dixon's algorithm and its modifications online: &lt;a href=&quot;http://www2.isye.gatech.edu/~dsteffy/papers/OSLifting.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www2.isye.gatech.edu/~dsteffy/papers/OSLifting.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Author also have all described algorithms implemented and published at &lt;a href=&quot;http://www2.isye.gatech.edu/~dsteffy/rational/&quot; rel=&quot;nofollow&quot;&gt;http://www2.isye.gatech.edu/~dsteffy/rational/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another interesting approach is Montante's method, which is a gauss variant for integer coefficient matrices. This variant don't create rational numbers in the matrix while converting it to triangular. Take a look at the last picture in link &lt;a href=&quot;http://en.wikipedia.org/wiki/Montante&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Montante&lt;/a&gt;'s_method&lt;/p&gt;&#xA;" OwnerUserId="6545" LastEditorUserId="6545" LastEditorDisplayName="" LastEditDate="2011-02-16T19:34:14.893" LastActivityDate="2011-02-16T19:34:14.893" />
  <row Id="22399" PostTypeId="1" CreationDate="2011-02-16T19:35:17.167" Score="4" ViewCount="211" Body="&lt;p&gt;I also have to prove this for $$2^2 *4^2 *6^2 ...(p-1)^2 \equiv (-1)^{(p+1)/2} \pmod{p}$$&#xA;I made some progress so far and got stuck. I said that since p is odd, $(p+1)/2$ is even. Then we can say that $(-1)^{\mathrm{even}}= 1$, so $(-1)^{(p+1)/2}\pmod{p}$ can be written as $1\pmod{p}$. I know that the product of odds is odd and the product of evens is even, but I cant prove that the left side of this equation in either case is congruent to $1 \pmod{p}$. Any help here would be greatly appreciated. &lt;/p&gt;&#xA;" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-16T19:54:56.170" LastActivityDate="2011-02-17T09:32:51.250" Title="If p is an odd prime, prove that $1^2*3^2 *5^2 ...(p-2)^2 \equiv (-1)^{(p+1)/2}\pmod{p}$" Tags="&lt;elementary-number-theory&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="1" />
  <row Id="22400" PostTypeId="2" ParentId="22391" CreationDate="2011-02-16T19:40:59.863" Score="2" ViewCount="" Body="&lt;p&gt;You can do something like the following Haskell code:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;walk (n, m) = step (1,-1) (1,1)&#xA;  where step (dx, dy) (x, y) &#xA;          | (x, y) == (n, m)                    = [(x, y)]&#xA;          | (x' == 0 &amp;amp;&amp;amp; y' &amp;lt;= m) || x' == n + 1 = (x, y) : step (-dx, -dy) (x, y + 1)&#xA;          | (y' == 0 &amp;amp;&amp;amp; x' &amp;lt;= n) || y' == m + 1 = (x, y) : step (-dx, -dy) (x + 1, y)&#xA;          | otherwise                           = (x, y) : step (dx, dy) (x', y')&#xA;          where x' = x + dx&#xA;                y' = y + dy&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;with that definition, we have for example&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;*Main&amp;gt; walk (3,4)&#xA;[(1,1),(2,1),(1,2),(1,3),(2,2),(3,1),(3,2),(2,3),(1,4),(2,4),(3,3),(3,4)]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="274" LastActivityDate="2011-02-16T19:40:59.863" CommentCount="1" />
  <row Id="22401" PostTypeId="2" ParentId="22395" CreationDate="2011-02-16T19:49:01.090" Score="3" ViewCount="" Body="&lt;p&gt;The phrasing of your statement &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;a module P is projective iff every element in P can be written as a finite linear combination of some elements of P&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;is pretty bad, because it is not at all evident from that those &lt;em&gt;some elements&lt;/em&gt; are fixed: what you mean, but not what you wrote, probably is&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;a module is projective iff there is a set $X\subseteq P$ such that every element of $P$ is a finite combination of elements of $X$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now, this last statement is false: you can always pick $X=P$, so its truth would imply that all modules are projective---and there are rings which have non-projective modules!&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2011-02-16T19:49:01.090" CommentCount="1" />
  <row Id="22402" PostTypeId="2" ParentId="22399" CreationDate="2011-02-16T20:03:08.170" Score="4" ViewCount="" Body="&lt;p&gt;You can start with Wilson's theorem, $(p-1)! \equiv -1 \pmod{p}.$ Rearrange the factorial thus, with $p=2m+1,$&#xA;$$-1 \equiv 1(p-1)2(p-2) \ldots m(p-m) \equiv 1(-1)2(-2) \ldots m(-m) \equiv 1^22^2 \ldots m^2(-1)^m&#xD;&#xA;\pmod{p}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we have shown&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$1^22^2 \ldots m^2 \equiv (-1)^{m+1} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or, noting that $m=(p-1)/2,$ in terms of $p$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$1^12^2 \ldots \left( \frac{p-1}{2} \right)^2 \equiv (-1)^{(p+1)/2}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we construct the product on the LHS, so consider first $2^24^2 \ldots (p-1)^2$ and factor out $2^{2p-2}$ then use Fermat,&#xA;$2^{p-1} \equiv 1 \pmod{p}.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You should now have the result for the product with the even numbers, use that along with Wilson's theorem again to prove the desired result for the product with&#xA;the odd numbers.&lt;/p&gt;&#xA;" OwnerUserId="1301" LastEditorUserId="1301" LastEditorDisplayName="" LastEditDate="2011-02-17T09:32:51.250" LastActivityDate="2011-02-17T09:32:51.250" CommentCount="3" />
  <row Id="22403" PostTypeId="2" ParentId="22395" CreationDate="2011-02-16T20:12:50.177" Score="2" ViewCount="" Body="&lt;p&gt;The statement you're linking to is: A module $P$ is projective if and only if there is a family $\{x_{i}\}_{i \in I} \subset P$ and morphisms $f_{i}: P \to R$ such that for each $x \in P$ we have $x = \sum_{i \in I} f_{i}(x) x_{i}$. The last statement says three things:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;In order for the sum to make sense we must have that for all $x$ the set  $\{i\,:\,f_{i}(x) \neq 0 \}$ is finite. Or, as stated there: for all $x$ we have $f_{i}(x) = 0$ for almost all $i$.&lt;/li&gt;&#xA;&lt;li&gt;The set $(x_{i})_{i \in I}$ &lt;em&gt;generates&lt;/em&gt; $P$. In other words, the map $g: \bigoplus_{i \in I} R \to M$ sending $(r_{i})$ to $\sum_{i \in I} r_{i} x_{i}$ is an epimorphism (it suffices to take $r_{i} = f_{i}(x)$ to see that this map is onto).&lt;/li&gt;&#xA;&lt;li&gt;The epimorphism $g$ &lt;em&gt;splits&lt;/em&gt;: there is a right inverse $f: P \to \bigoplus_{i \in I} R$ of $g$, i.e. $gf = \operatorname{id}_{P}$ (this morphism $f$ is of the form $(f_{i})_{i \in I}$ with morphisms $f_{i}: P \to R$ and by the definition of a direct sum we have $f_{i}(x) \neq 0$ only for finitely many $i$).&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;In my opinion its just an extremely explicit way of phrasing the much more catchy &quot;a module is projective if and only if it is a direct summand of a free module&quot;. &lt;/p&gt;&#xA;" OwnerUserId="5363" LastEditorUserId="5363" LastEditorDisplayName="" LastEditDate="2011-02-16T20:30:06.387" LastActivityDate="2011-02-16T20:30:06.387" CommentCount="2" />
  <row Id="22404" PostTypeId="2" ParentId="20102" CreationDate="2011-02-16T20:16:47.523" Score="0" ViewCount="" Body="&lt;p&gt;Use Gaussian ellimination with only integers.&#xA;Don't do the division.&#xA;Instead to zero a paritucar cell in the matrix use euclid algorithm.&#xA;You may need bignum library.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is almost equivalent to using rational number gaussian ellimination but simpler.&lt;/p&gt;&#xA;" OwnerUserId="5552" LastActivityDate="2011-02-16T20:16:47.523" CommentCount="1" />
  <row Id="22405" PostTypeId="1" CreationDate="2011-02-16T20:31:56.793" Score="4" ViewCount="78" Body="&lt;p&gt;I am looking for any reference on Wigner's classification of irreducible representations of Poincare group. I know the classification, but is there any reference where the representations are constructed and explained. This classification gives the different spin particles in Quantum mechanics.&#xA;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="7167" LastActivityDate="2011-02-16T20:42:07.920" Title="Irreducible representations of Poincare group" Tags="&lt;lie-groups&gt;&lt;representation-theory&gt;" AnswerCount="1" />
  <row Id="22406" PostTypeId="2" ParentId="22405" CreationDate="2011-02-16T20:42:07.920" Score="2" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://www.worldscibooks.com/etextbook/5081/5081_chap1.pdf&quot; rel=&quot;nofollow&quot;&gt;Here&lt;/a&gt; is an article that explains Wigner's classification (but not in the exact same way as Wigner himself). To see Wigner's classification explicitly, you probably should check out the &lt;a href=&quot;http://www.jstor.org/pss/1968551&quot; rel=&quot;nofollow&quot;&gt;following&lt;/a&gt;[E.P. Wigner, Ann. Math., &lt;strong&gt;40&lt;/strong&gt;, 149, (1939)].&lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2011-02-16T20:42:07.920" />
  <row Id="22407" PostTypeId="1" AcceptedAnswerId="22480" CreationDate="2011-02-16T20:57:38.560" Score="3" ViewCount="171" Body="&lt;p&gt;For a family of random variables, I was wondering if independence and Conditional Independence under any condition among them imply each other? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If not, can these two concepts imply one another under some special cases? Thanks!&lt;/p&gt;&#xA;" OwnerUserId="1281" LastActivityDate="2011-02-17T13:44:24.153" Title="Independence and Conditional Independence between random variables" Tags="&lt;probability-theory&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="1" />
  <row Id="22408" PostTypeId="2" ParentId="20102" CreationDate="2011-02-16T20:58:45.613" Score="0" ViewCount="" Body="&lt;p&gt;The Gaussian elimination algorithm is simple to program.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Define a rational number object which can be a list of numerator factors and a list of denominator factors. Write functions for doing the four arithmetic operations using the rational number objects. In your Gaussian elimination program, substitute an array of the rational number objects for the array of integers, and use your functions for the calculations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You may or may not want the number objects to automatically reduce to simplest terms after a computation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't think it would be fast, but it's pretty straight forward, and I don't think the execution time would be long enough to be a bother, maybe a few seconds.&lt;/p&gt;&#xA;" OwnerUserId="1312" LastActivityDate="2011-02-16T20:58:45.613" CommentCount="1" />
  <row Id="22409" PostTypeId="2" ParentId="22350" CreationDate="2011-02-16T21:00:33.760" Score="2" ViewCount="" Body="&lt;p&gt;As others have mentioned, this follows easily by &lt;em&gt;aligning&lt;/em&gt; the numerator and denominator factors modulo $\rm\ p^r \ $ in order to obtain maximal cancellation of powers of $\rm\:p\:.\:$ This is a special case of the simple arithmetical method I described in my &lt;a href=&quot;http://math.stackexchange.com/questions/2158/division-of-factorials/2192#2192&quot;&gt;post here&lt;/a&gt; to prove that binomial coefficients are integers. See also this very &lt;a href=&quot;http://math.stackexchange.com/questions/5818/proving-n-choose-p-equiv-bigl-fracnp-bigr-textmod-p/5829#5829&quot;&gt;simple example&lt;/a&gt; for motivation. I think that you'll find that the key conceptual idea is clearer after studying these further examples.&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2011-02-16T21:00:33.760" />
  <row Id="22411" PostTypeId="1" AcceptedAnswerId="22435" CreationDate="2011-02-16T22:12:21.173" Score="15" ViewCount="221" Body="&lt;p&gt;I'd like to calculate, or find a reasonable estimate for, the Mertens-like product&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\prod_{2&amp;lt;p\le n}\frac{p}{p-2}=\left(\prod_{2&amp;lt;p\le n}1-\frac{2}{p}\right)^{-1}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, how does this behave asymptotically?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Hmm... trying to think this one out, I get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left(\prod_{2&amp;lt;p\le n}1-\frac{2}{p}\right)^{-1}=\exp\log\left(\left(\prod_{2&amp;lt;p\le n}1-\frac{2}{p}\right)^{-1}\right)=\exp-\log\left(\prod_{2&amp;lt;p\le n}1-\frac{2}{p}\right)$$&#xA;which is&#xA;$$\exp-\sum_{2&amp;lt;p\le n}\log\left(1-\frac{2}{p}\right)=\exp\sum_{2&amp;lt;p\le n}\left(\frac{2}{p}+\frac12\left(\frac{2}{p}\right)^2+\frac13\left(\frac{2}{p}\right)^3+\cdots\right)$$&#xA;which, with P(s) the prime zeta function and f(s)=P(s)-2^s, is less than&#xA;$$\exp\left(\frac42f(2)+\frac83f(3)+\cdots+\sum_{2&amp;lt;p\le n}\frac{2}{p}\right)$$&#xA;which might not be a bad approximation for n large.  But I can't immediately find a series for P(s) with $s\to+\infty$ and I'm not sure if there's a better way.  Help?&lt;/p&gt;&#xA;" OwnerUserId="1778" LastEditorUserId="6075" LastEditorDisplayName="" LastEditDate="2011-03-26T17:08:27.207" LastActivityDate="2011-03-26T17:08:27.207" Title="Computing the product of p/(p - 2) over the odd primes" Tags="&lt;number-theory&gt;&lt;analysis&gt;&lt;prime-numbers&gt;&lt;analytic-number-theory&gt;" AnswerCount="3" FavoriteCount="3" />
  <row Id="22412" PostTypeId="2" ParentId="22411" CreationDate="2011-02-16T22:23:03.657" Score="5" ViewCount="" Body="&lt;p&gt;Not sure if this is what you were looking for, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Taking logs,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We get $\sum (\log p - \log (p-2)) = \sum (\frac{2}{p} + \mathcal{O}(\frac{1}{p^2}))$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now it is well known that $ \sum_{p \le n} \frac{1}{p}  = \log \log n + A + \mathcal{O}(\frac{1}{\log^2n})$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And thus&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your product is asymptotically&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$K \log^2 n$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;for some constant $K$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you noticed in your question, you can compute the $K$ using the prime-zeta function (and knowing the $A$ above).&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2011-02-16T22:30:39.890" LastActivityDate="2011-02-16T22:30:39.890" CommentCount="2" />
  <row Id="22413" PostTypeId="1" AcceptedAnswerId="22420" CreationDate="2011-02-16T22:39:35.843" Score="2" ViewCount="79" Body="&lt;p&gt;Write the following sequence as a recurrence relation (with sufficient initial values specified):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$b_n=1-\frac{1}{2^n} \forall n\in\mathbb{N}^*$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think I am suppose to use induction (e.g. n-1) to prove the recurrence, but I am confused, how?&lt;/p&gt;&#xA;" OwnerUserId="7144" LastEditorUserId="524" LastEditorDisplayName="" LastEditDate="2011-02-16T23:06:59.360" LastActivityDate="2011-02-16T23:07:33.347" Title="Write the following sequence as a recurrence relation..." Tags="&lt;homework&gt;&lt;sequences-and-series&gt;&lt;discrete-mathematics&gt;&lt;induction&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="22414" PostTypeId="1" AcceptedAnswerId="22418" CreationDate="2011-02-16T22:41:02.607" Score="1" ViewCount="58" Body="&lt;p&gt;As often happens I'm not sure of the name of what I'm trying to do, so I'm having difficulty searching it. I know this is basic algebra, but my math is null so I'm struggling with it greatly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a coordinate system that starts from 0,0 in the bottom left corner, and ends in 1,1 in the top right corner. I must &quot;translate&quot; from coordinates to pixels and viceversa. I'm including a diagram to clarify what I'm trying to do.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm trying to do this with Javascript, but I guess it's irrelevant. A formula is enough to get me started.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/dv6Jb.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="7169" LastEditorUserId="7169" LastEditorDisplayName="" LastEditDate="2011-02-16T22:55:22.140" LastActivityDate="2011-02-16T22:59:25.750" Title="How do I map pixels to a coordinate system?" Tags="&lt;coordinate-systems&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="22415" PostTypeId="2" ParentId="21931" CreationDate="2011-02-16T22:47:37.667" Score="3" ViewCount="" Body="&lt;p&gt;For a discrete random variable $X$ with support $\mathbb{Z}$, the Fourier transform of the probability distribution $P_x \equiv P[X=x]$ is given by&#xA;$$&#xD;&#xA;\tilde{P}(k) = \sum_{x=-\infty}^{\infty} e^{ikx} P_x = E\left[e^{ikx}\right] = e^{h(k)},&#xD;&#xA;$$&#xA;where&#xA;$$&#xD;&#xA;h(k) = \sum_{n=1}^{\infty} \kappa_{n} \frac {(ik)^{n}}{n!}&#xD;&#xA;$$&#xA;is the natural logarithm of the characteristic function of $X$, and $\kappa_{n}$ is the $n$th cumulant of $X$.  Recall that $\kappa_{1} = \mu$ is the mean and $\kappa_{2} = \sigma^2$ is the variance.  The probability that a sum of $M$ independent variables $X_i$ with the same distribution is exactly $x \in {\mathbb{Z}}$ is then&#xA;$$&#xD;&#xA;\begin{eqnarray}&#xD;&#xA;P\left[\sum_{i=1}^{M} X_i = x\right] &amp;amp;=&amp;amp; \int_{-\pi}^{\pi}\frac{dk}{2\pi} e^{-ikx}\tilde{P}(k)^M \\&#xD;&#xA;&amp;amp;=&amp;amp; \int_{-\pi}^{\pi}\frac{dk}{2\pi} e^{Mh(k)-ikx} \\&#xD;&#xA;&amp;amp;=&amp;amp; \int_{-\pi}^{\pi}\frac{dk}{2\pi} e^{ik(M\mu - x) - \frac{1}{2}M\sigma^2 k^2} \exp\left(\sum_{n=3}^{\infty}M\kappa_{n}\frac{(ik)^{n}}{n!}\right).&#xD;&#xA;\end{eqnarray}&#xD;&#xA;$$&#xA;Considering the desired case where $x = M\mu \in {\mathbb{Z}}$, and making the change of variable $k \rightarrow k/(\sigma\sqrt{M})$, we have&#xA;$$&#xD;&#xA;P\left[\sum_{i=1}^{M} X_i = M\mu\right] = \frac{1}{\sigma\sqrt{2\pi M}}\int_{-\pi\sigma\sqrt{M}}^{\pi\sigma\sqrt{M}} d\Phi(k) \exp\left(\sum_{n=3}^{\infty} \sigma^{-n}M^{1-\frac{1}{2}n}\kappa_{n}\frac{(ik)^{n}}{n!}\right),&#xD;&#xA;$$&#xA;where $d\Phi(k) = \phi(k) dk$ is the standard normal distribution (with mean $0$ and variance $1$).  Here we assume that exponential decays rapidly away from $k=0$, so we may replace the limits of integration by $\pm\infty$.  Then, expanding the exponential in inverse powers of $M$, and using the fact that the $n$th central moment of the standard normal distribution vanishes for odd $n$ and is equal to $(n-1)!!$ for even $n$, we obtain the following:&#xA;$$&#xD;&#xA;P\left[\sum_{i=1}^{M} X_i = M\mu\right] = \frac{1}{\sigma\sqrt{2\pi M}}\left(1 + \frac{\kappa_4}{8M\sigma^4} - \frac{5\kappa_3^2}{24M\sigma^6} + O(M^{-2})\right).&#xD;&#xA;$$&#xA;This is essentially the Edgeworth expansion.  If $X$ is the Bernoulli distribution with probability of success $p = \frac{1}{2}(1+a)$ (and of failure $q=\frac{1}{2}(1-a)$), then it is straightforward to verify that&#xA;$$&#xD;&#xA;\begin{eqnarray}&#xD;&#xA;\kappa_2 &amp;amp;=&amp;amp; \sigma^2 = pq = \frac{1}{4}(1-a^2) \\&#xD;&#xA;\kappa_3 &amp;amp;=&amp;amp; \frac{1}{4}(1-a^2)(-a) = -\frac{1}{4}a(1-a^2) \\&#xD;&#xA;\kappa_4 &amp;amp;=&amp;amp; \frac{1}{8}(1-a^2)(3a^2-1),&#xD;&#xA;\end{eqnarray}&#xD;&#xA;$$&#xA;and hence&#xA;$$&#xD;&#xA;\begin{eqnarray}&#xD;&#xA;\frac{5\kappa_3^2}{24\sigma^6} &amp;amp;=&amp;amp; \frac{5a^2}{6(1-a^2)} \\&#xD;&#xA;\frac{\kappa_4}{8\sigma^4} &amp;amp;=&amp;amp; \frac{3a^2 - 1}{4(1-a^2)},&#xD;&#xA;\end{eqnarray}&#xD;&#xA;$$&#xA;for a total correction term proportional to&#xA;$$&#xD;&#xA;-\frac{5\kappa_3^2}{24M\sigma^6} + \frac{\kappa_4}{8M\sigma^4} = \frac{9a^2-3-10a^2}{12M(1-a^2)} = -\frac{3+a^2}{12M(1-a^2)} = -\frac{1-pq}{12Mpq},&#xD;&#xA;$$&#xA;which agrees with the Stirling approximation to the exact result.&lt;/p&gt;&#xA;" OwnerUserId="5546" LastActivityDate="2011-02-16T22:47:37.667" CommentCount="1" />
  <row Id="22416" PostTypeId="2" ParentId="22290" CreationDate="2011-02-16T22:49:41.840" Score="4" ViewCount="" Body="&lt;p&gt;Notice that $g(x)=\arctan(f(x))$ has a removable discontinuity at points $x$ such that $\lim_{t\to x} f(t)=\infty$. Therefore, the countability of the set of interest can be seen as a consequence of the fact discussed here: &lt;a href=&quot;http://math.stackexchange.com/questions/3777/is-there-a-function-with-a-removable-discontinuity-at-every-point&quot;&gt;Is there a function with a removable discontinuity at every point?&lt;/a&gt;  and particularly, Chandru1's link: &lt;a href=&quot;http://books.google.com/books?id=6l_E9OTFaK0C&amp;amp;pg=PA249&quot; rel=&quot;nofollow&quot;&gt;Chandru1's link to Theorem 5.63&lt;/a&gt;, which proves that the set of removable discontinuities is at most countable.&lt;/p&gt;&#xA;" OwnerUserId="5253" LastActivityDate="2011-02-16T22:49:41.840" />
  <row Id="22417" PostTypeId="2" ParentId="22340" CreationDate="2011-02-16T22:53:42.153" Score="2" ViewCount="" Body="&lt;p&gt;Often textbook solutions to induction problems like this are magically &quot;pulled out of a hat&quot; - completely devoid of intuition. Below I explain the intuition behind the induction in this proof. Namely, I show that the proof easily reduces to the completely trivial induction that $\rm\ 1^n\ \equiv\ 1\:$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $\rm\ 15^n + 6 = 15^n-1 + 7\:,\: $ it suffices to show that $\rm\ 7\ |\ 15^n - 1\:.\: $ The base case $\rm\ n=1\ $ is clear. The inductive step, slightly abstracted, is simply the following&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\quad\quad\quad\quad\quad\rm\ \ \ \ 7\ |\ \ c\ -1,\ \ \ d\ \:-\ 1\ \ \Rightarrow\ \ 7\ |\ cd-1 = (c-1)\ d + (d-1)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\quad$  thus $\quad\ \ \ 7\ |\ 15-1,\ 15^n-1\ \ \Rightarrow\ \ 7\ |\ 15^{n+1}-1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Said $\rm\ mod\ 7,\ \ 15\equiv 1\ \Rightarrow\ 15^n\equiv 1^n\equiv 1\ $ by inductively multiplying (&quot;powering&quot;) using this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;LEMMA&lt;/strong&gt; $\rm\ \ A\equiv a,\ B\equiv b\ \Rightarrow\ AB\equiv ab\ \ (mod\ m)\quad\quad$ (&lt;strong&gt;product rule for congruences&lt;/strong&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof&lt;/strong&gt; $\rm\ \ m\: |\: A-a,\:\:\ B-b\ \Rightarrow\ m\ |\ (A-a)\ B + a\ (B-b)\ =\ AB - ab $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice how this transformation from divisibility form to congruence arithmetic form has served to reduce the induction to the triviality $\rm\: 1^n \equiv 1$. Many induction problems can similarly be reduced to trivial inductions by appropriate conceptual preprocessing. Always think before calculating!&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2011-02-21T17:56:39.620" LastActivityDate="2011-02-21T17:56:39.620" />
  <row Id="22418" PostTypeId="2" ParentId="22414" CreationDate="2011-02-16T22:56:13.590" Score="2" ViewCount="" Body="&lt;p&gt;Your coordinates look odd, but going from co-ordinates to pixels, you probably want something like &#xA;$$f(x,y) = (190y+5,195-190x)$$&#xA;and going from pixels to coordinates &#xA;$$g(a,b) = \left(\frac{195-b}{190},\frac{a-5}{190},\right)$$&#xA;but you may need to adjust this slightly to deal with what happens on the edges&lt;/p&gt;&#xA;" OwnerUserId="6460" LastActivityDate="2011-02-16T22:56:13.590" CommentCount="3" />
  <row Id="22420" PostTypeId="2" ParentId="22413" CreationDate="2011-02-16T23:04:10.923" Score="6" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\ $ Eliminate $\rm\ 1/2^n\ $ from the equations for $\rm\ b_n,\ b_{n+1}\ $ &lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2011-02-16T23:04:10.923" />
  <row Id="22421" PostTypeId="2" ParentId="22413" CreationDate="2011-02-16T23:07:33.347" Score="2" ViewCount="" Body="&lt;p&gt;If you wrote &#xA;$$\frac{1}{2^{n}} = 1 - b_{n}$$&#xA;and similarly for the $n-1$ case then you could probably see a relationship between $b_{n-1}$ and $b_n$&lt;/p&gt;&#xA;" OwnerUserId="6460" LastActivityDate="2011-02-16T23:07:33.347" />
  <row Id="22422" PostTypeId="2" ParentId="22290" CreationDate="2011-02-16T23:49:55.267" Score="5" ViewCount="" Body="&lt;p&gt;For completeness, here is a direct proof without the $\arctan$ transformation using the ideas in the book cited in Nick Kirby's answer (where the proof of the part of the theorem relating to removable discontinuities is left as an exercise).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $S_{mn}$ with $m,n\in \mathbb{N}$ be the set of all points $x$ such that $f(x) &amp;lt; m$ and $f(y) &amp;gt; m$ for all $y\neq x$ with $x-1/n&amp;lt;y&amp;lt;x+1/n$. For each point $x$ at which $f$ &quot;converges to $\infty$&quot; (in the sense explicated in my other answer), we can choose $m &amp;gt; f (x)$, and then choose $n$ such that $f(y) &amp;gt; m$ for all $y\neq x$ with $x-1/n&amp;lt;y&amp;lt;x+1/n$. Thus each point at which $f$ &quot;converges to $\infty$&quot; is contained in one of the sets $S_{mn}$. But none of these sets has an accumulation point, since that would imply that there are two points which are less than $1/n$ apart, both of whose function values are below $m$ and both of which have a neighbourhood of radius $1/n$ where the function values are above $m$, which is impossible. Hence each $S_{mn}$ is countable, and the set of points at which $f$ &quot;converges to $\infty$&quot;, being the countable union of these countable sets, is also countable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[Edit: Note that this proof substantially uses countability and can't be adapted to nowhere-denseness as immediately as the proof in my other answer. This is because although each $S_{mn}$ is not only countable but also nowhere dense, this property, unlike countability, is not necessarily inherited in a countable union. Thus, we cannot exclude on the basis of this proof alone that the union of all $S_{mn}$ might be dense in $\mathbb{R}$, as demonstrated by the example of the function with removable discontinuities at all rationals. However, the union of all $S_{mn}$ with bounded $m$ cannot be dense anywhere, and it is this necessary unbounded growth of $m$ that in the present case precludes denseness, since we cannot have $\infty &amp;gt; m$ for all $m\in \mathbb{N}$ as a function value, whereas we can have $0 &amp;lt; 1/m$ for all $m\in \mathbb{N}$ as a function value in the finite case.]&lt;/p&gt;&#xA;" OwnerUserId="6622" LastEditorUserId="6622" LastEditorDisplayName="" LastEditDate="2011-02-17T06:19:20.307" LastActivityDate="2011-02-17T06:19:20.307" CommentCount="1" />
  <row Id="22423" PostTypeId="1" AcceptedAnswerId="22425" CreationDate="2011-02-17T00:08:55.000" Score="7" ViewCount="389" Body="&lt;p&gt;Prove that any positive real number $r$ satisfying: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$r - \frac{1}{r} = 5$ must be irrational. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the contradiction that the equation must be rational, we set $r= a/b$, where a,b are positive integers and substitute:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\begin{align*}&#xD;&#xA;&amp;amp;\frac{a}{b} - \frac{1}{a/b}\\&#xD;&#xA;&#xD;&#xA;&amp;amp;\frac{a}{b} - \frac{b}{a}\\&#xD;&#xA;&#xD;&#xA;&amp;amp;\frac{a^2}{ab} - \frac{b^2}{ab}\\&#xD;&#xA;&#xD;&#xA;&amp;amp;\frac{a^2-b^2}{ab}&#xD;&#xA;\end{align*}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am unsure what to do next?&lt;/p&gt;&#xA;" OwnerUserId="7144" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-17T06:00:43.917" LastActivityDate="2011-02-28T04:02:44.203" Title="Proof by contradiction: $r - \frac{1}{r} =5\Longrightarrow r$ is irrational ?" Tags="&lt;discrete-mathematics&gt;&lt;proof-strategy&gt;&lt;contradiction&gt;" AnswerCount="4" CommentCount="3" FavoriteCount="1" />
  <row Id="22424" PostTypeId="2" ParentId="22423" CreationDate="2011-02-17T00:22:24.023" Score="7" ViewCount="" Body="&lt;p&gt;To complete your solution, note that you can, without loss of generality, set $a$ and $b$ to be coprime. So you have $a^2=b^2+5ab=b(b+5a)$. Hence $a$ divides $b(b+5a)$. Euclid's lemma now tells you that $a$ divides $b+5a$ (because $a$ and $b$ are coprime). But then $a$ must divide $b$, which is contradiction with the fact that $a$ and $b$ are coprime.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's an alternative: transform it into $r^2-5r-1=0$. What are the real (if any) roots of this equation? The quadratic formula gives you: $r_{12}=5/2\pm \sqrt{29}/2$. Since there are at most two different roots for a quadratic polynomial in $\mathbb{R}$, these are the roots. So your problem comes down to showing that $\sqrt{29}$ is irrational. (In fact, you can prove that the square root of any non-perfect-square number is irrational. This is the number-theoretic part).&lt;/p&gt;&#xA;" OwnerUserId="1313" LastEditorUserId="1313" LastEditorDisplayName="" LastEditDate="2011-02-17T00:29:35.840" LastActivityDate="2011-02-17T00:29:35.840" CommentCount="1" />
  <row Id="22425" PostTypeId="2" ParentId="22423" CreationDate="2011-02-17T00:22:26.877" Score="10" ViewCount="" Body="&lt;p&gt;I think your method is sound:&#xA;if we assume by contradiction that $r=\frac{a}{b}$ with $a,b$ integers&#xA;we get that&#xA;$$\frac{a}{b}-\frac{b}{a}=5$$&#xA;since neither $b$ nor $a$ is 0, we can multiply by $ab$ and get&#xA;$$(a-b)(a+b)=5ab$$&#xA;now there are three cases to investigate, i'll just point them out so you can finish it off:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;If $a$ and $b$ are both odd- look at both sides of the equation (mod2).&lt;/li&gt;&#xA;&lt;li&gt;If either one of them is odd and the other even- try the same trick of (mod 2)&lt;/li&gt;&#xA;&lt;li&gt;lastly- if both are even- you can divide the entire equation by 2 and continue.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;hope that helps&lt;/p&gt;&#xA;" OwnerUserId="7129" LastEditorUserId="5363" LastEditorDisplayName="" LastEditDate="2011-02-17T00:31:41.190" LastActivityDate="2011-02-17T00:31:41.190" CommentCount="4" />
  <row Id="22426" PostTypeId="1" CreationDate="2011-02-17T00:30:12.263" Score="2" ViewCount="73" Body="&lt;p&gt;Does anyone know of an online probability course that uses Ross' &quot;A First Course in Probability&quot; as the text?&lt;/p&gt;&#xA;" OwnerUserId="7173" LastActivityDate="2011-02-17T00:30:12.263" CommunityOwnedDate="2011-02-17T01:23:46.537" Title="online probability course using Ross" Tags="&lt;probability&gt;" CommentCount="1" />
  <row Id="22427" PostTypeId="2" ParentId="22290" CreationDate="2011-02-17T00:48:11.490" Score="4" ViewCount="" Body="&lt;p&gt;If we allow $f$ to be a partial function, that is, where the domain is only a subset of $\mathbb{R}$ rather than all of $\mathbb{R}$, then there is an interesting counterexample. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Namely, let $C\subset\mathbb{R}$ be the usual Cantor set, and let $U=\mathbb{R}-C$ be the complement, an open set. For any real $x\in U$, let $$f(x)=\frac{1}{d(x)},$$ where $d(x)$ is the distance from $x$ to $C$, which is nonzero for all $x\in U$. Note that for any sequence $y_n\in U$ with $y_n\to x\in C$, we have $d(y_n)\to 0$ and hence $f(y_n)\to\infty$. (Note also that every $x\in C$ is a limit of points in $U$.) Since the Cantor set has uncountable size continuum, we have therefore an uncountable set of points where $f$ diverges to infinty. Furthermore, $f$ is continuous on its domain $U$.&lt;/p&gt;&#xA;" OwnerUserId="413" LastEditorUserId="413" LastEditorDisplayName="" LastEditDate="2011-02-17T01:11:53.387" LastActivityDate="2011-02-17T01:11:53.387" CommentCount="2" />
  <row Id="22428" PostTypeId="1" CreationDate="2011-02-17T00:48:24.570" Score="3" ViewCount="66" Body="&lt;p&gt;I know that I can use P(AB)=P(A)P(B) to prove whether sets are independent. But how can I use this to say that P(A)=0.3 and P(B)=0.4 and P(AUB)=0.6 are or aren't independent?&#xA;It's fine not to give an outright solution, an explanation would be much more helpful!&lt;/p&gt;&#xA;" OwnerUserId="7174" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2011-02-17T01:26:06.330" LastActivityDate="2011-02-17T01:26:06.330" Title="Proving sets to be independent" Tags="&lt;probability&gt;" AnswerCount="1" CommentCount="0" />
  <row Id="22429" PostTypeId="2" ParentId="22428" CreationDate="2011-02-17T00:52:34.713" Score="3" ViewCount="" Body="&lt;p&gt;Use the fact that $P(A\cup B)=P(A)+P(B)-P(A\cap B)$.&lt;/p&gt;&#xA;" OwnerUserId="1313" LastActivityDate="2011-02-17T00:52:34.713" CommentCount="3" />
  <row Id="22430" PostTypeId="1" AcceptedAnswerId="22432" CreationDate="2011-02-17T00:53:14.073" Score="5" ViewCount="124" Body="&lt;p&gt;Hi again!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is another homework question I can't figure out.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;For $|G|$ even, $\forall x\in G\exists b\in G\setminus{\{x^{-1}\}}$ such that $bxb = x^{-1}$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I tried to toy with associativity but to no avail. Also, I can't see the relevance of $|G|$ being even. Any hint (I'm not here for the solution) is appreciated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for your attention.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update&lt;/strong&gt;: I thought I would show you the proof I've written since that's the least I can trade for the effort you took to help me. I'm eager for any kind of feedback so, if you're about to comment on it, show no mercy. I'm trying to improve.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Let $x \in G$. There's left to prove that there exists $b \in G\setminus{\{x^{-1}\}}$ such that:&lt;br&gt;&#xA;  $bxb = x^{-1}$&lt;br&gt;&#xA;  $\Longleftrightarrow (bxb)x = x(bxb) = e$&lt;br&gt;&#xA;  $\Longleftrightarrow (bx)bx = xb(xb) = e$&lt;br&gt;&#xA;  $\Longleftrightarrow (bx)^{2} = (xb)^{2} = e$&lt;br&gt;&#xA;  $\Longleftrightarrow bx = (bx)^{-1}$ and $xb = (xb)^{-1}$&lt;br&gt;&#xA;  $\Longleftrightarrow bx = x^{-1}b^{-1}$ and $xb = b^{-1}x^{-1}$  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Let $a \in G\setminus{\{e\}}$ such that $a = a^{-1}$. (We know such $a$ exists by a previous result). Let $b = ax^{-1}$. Then:&lt;br&gt;&#xA;  $bx = x^{-1}b^{-1}$ and $xb = b^{-1}x^{-1}$&lt;br&gt;&#xA;  $\Longleftrightarrow ax^{-1}x = x^{-1}(ax^{-1})^{-1}$ and $xax^{-1} = (ax^{-1})^{-1}x^{-1}$&lt;br&gt;&#xA;  $\Longleftrightarrow a = x^{-1}xa^{-1}$ and $xax^{-1} = xa^{-1}x^{-1}$&lt;br&gt;&#xA;  $\Longleftrightarrow a = a^{-1}$ and $xax^{-1} = xax^{-1}$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What do you think of it?&lt;/p&gt;&#xA;" OwnerUserId="6538" LastEditorUserId="6538" LastEditorDisplayName="" LastEditDate="2011-02-17T15:51:13.307" LastActivityDate="2011-02-17T16:46:47.360" Title="For $|G|$ even, $\forall x\in G\exists b\in G\setminus{\{x^{-1}\}}$ such that $bxb = x^{-1}$" Tags="&lt;homework&gt;&lt;group-theory&gt;" AnswerCount="3" CommentCount="5" FavoriteCount="1" />
  <row Id="22431" PostTypeId="2" ParentId="22430" CreationDate="2011-02-17T01:02:50.243" Score="0" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Hint.&lt;/strong&gt; $bxb = x^{-1}\Longleftrightarrow bx = x^{-1}b^{-1}\Longleftrightarrow bx = (bx)^{-1}\Longleftrightarrow bx$ has order $1$ or $2$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Excluding $b=x^{-1}$ guarantees that $bx\neq e$.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2011-02-17T01:02:50.243" CommentCount="1" />
  <row Id="22432" PostTypeId="2" ParentId="22430" CreationDate="2011-02-17T01:02:58.210" Score="2" ViewCount="" Body="&lt;p&gt;Here are some hints, read them one at a time if you want to keep looking yourself :-)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1 - Saying |G| is even means the group contains some $a$ that satisfies $a^2 = 1$ but $a\neq 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2 - For given $x$ you should try to find $b$ such that $(bx)^2 = 1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3 - So you could attempt find $b$ such that $bx = a$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4 - This means $b = ax^{-1}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-edit&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Say $|G|$ is even. There exists $a$ such that $a^2 =1$, $a\neq 1$. (See arguments in the comments.)&#xA;Now take any $x\in G$ and let $b := ax^{-1}$. Indeed, this $b$ satisfies&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$b\neq x^{-1}$, because  $b = x^{-1}$ would yield the contradiction $bx = a = 1$&lt;/li&gt;&#xA;&lt;li&gt;$ bxb = ax^{-1}xax^{-1} = aax = x $&#xA;as required.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="5189" LastEditorUserId="5189" LastEditorDisplayName="" LastEditDate="2011-02-17T16:46:47.360" LastActivityDate="2011-02-17T16:46:47.360" CommentCount="7" />
  <row Id="22433" PostTypeId="2" ParentId="22430" CreationDate="2011-02-17T01:03:51.723" Score="1" ViewCount="" Body="&lt;p&gt;What the claim tells you is that for any element $x$ of the group $G$, there is some $b$ in $G$ (other than $x^{-1}$) such that $bx$ is its own inverse. Try to see what happens if you &quot;try&quot; all $b$'s (other than $x^{-1}$) in $G$ and never obtain, in the product $bx$, an element which is its own inverse. (Also note that when you run through all possible $b$'s in the product $bx$, you're actually covering the whole group -- this is because of the cancellation law). Think about the relationship between some elements being their own inverses (or, equivalently, having order 2) and the order of the group being even.&lt;/p&gt;&#xA;" OwnerUserId="1313" LastEditorUserId="1313" LastEditorDisplayName="" LastEditDate="2011-02-17T01:21:19.090" LastActivityDate="2011-02-17T01:21:19.090" />
  <row Id="22434" PostTypeId="2" ParentId="22411" CreationDate="2011-02-17T01:05:47.493" Score="3" ViewCount="" Body="&lt;p&gt;Do you know &lt;a href=&quot;http://projecteuclid.org/DPubS?service=UI&amp;amp;version=1.0&amp;amp;verb=Display&amp;amp;handle=euclid.ijm/1255631807&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; paper by Rosser and Schoenfeld?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It contains a number of bounds that might be useful for you, including (3.29 in the paper)&#xA;$$\prod_{p\le x} \frac{p}{p-1} &amp;lt; e^C (\log x)\left(1+\frac{1}{2 \log^2 x} \right)$$&#xA;for $286 \le x$, where $C$ is Euler's constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Their proof of this result might be useful, and/or other results in the paper might be.&lt;/p&gt;&#xA;" OwnerUserId="2937" LastActivityDate="2011-02-17T01:05:47.493" />
  <row Id="22435" PostTypeId="2" ParentId="22411" CreationDate="2011-02-17T01:13:04.877" Score="12" ViewCount="" Body="&lt;p&gt;Lets do this as explicitly as possible (I would like to find the constant and error term)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;1 The Original Problem&lt;/strong&gt;&#xA;First, consider the identity $\log\left(1-\frac{2}{p}+\frac{1}{p^2}\right) +\log\left(1- \frac{1}{(p-1)^2} \right)=\log\left(1- \frac{2}{p} \right)$.  (I spent a little time working that out, but for now lets just leave it as a black box since the computation yields nothing new.  I encourage you to multiply everything out and check for yourself) Then &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{2&amp;lt;p\leq n}\log\left(1-\frac{2}{p}+\frac{1}{p^2}\right) +\sum_{2&amp;lt;p\leq n} \log\left(1- \frac{1}{(p-1)^2} \right) = \sum_{2&amp;lt;p\leq n}\log\left(1-\frac{2}{p}\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Multiplying by negative one and exponentiating both sides yields&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left(\prod_{2&amp;lt;p\leq n}1-\frac{1}{p}\right)^{-2} \cdot \prod_{2&amp;lt;n&amp;lt;p} \left(1- \frac{1}{(p-1)^2} \right)^{-1} = \left(\prod_{2&amp;lt;p\leq n}1-\frac{2}{p}\right)^{-1} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Recall $ \Pi_2=\prod_{2&amp;lt;p} \left(1- \frac{1}{(p-1)^2} \right)$ is the &lt;a href=&quot;http://mathworld.wolfram.com/TwinPrimesConstant.html&quot; rel=&quot;nofollow&quot;&gt;Twin Prime Constant&lt;/a&gt;, and that the one product on the left hand side converges to the reciprocal of this.  It is then by one of Mertens formulas that we know $$\left(\prod_{2&amp;lt;p\leq n}1-\frac{1}{p}\right)^{-1}=\frac{1}{2}e^\gamma \log n + O(1)$$ where $\gamma$ is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Euler%E2%80%93Mascheroni_constant&quot; rel=&quot;nofollow&quot;&gt;Euler Mascheroni Constant&lt;/a&gt;. (Specifically this is Theorem 2.7 (e) in Montgomery &lt;strong&gt;Multiplicative Number Theory I. Classical Theory&lt;/strong&gt;)  Upon squaring this assymptotic result, we are able to conclude:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left(\prod_{2&amp;lt;p\leq n}1-\frac{2}{p}\right)^{-1} = \frac{1}{4}e^{2\gamma}\Pi_2^{-1} \log^2n + O(\log n)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope that helps,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note:  The reason I substituted the twin prime constant in for that other product is because it converges &lt;strong&gt;very&lt;/strong&gt; fast in comparison with the error term.  I can give more details if desired, but I'll leave it as an exercise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;2 What is best possible?&lt;/strong&gt;&#xA;Can the error term be made better?  Yes.  It turns out we can make that error term a lot better.  By using the Prime Number Theorem we find &#xA;$$\prod_{2&amp;lt;p\leq n} \left( 1-\frac{1}{p}\right)^{-1}=\frac{1}{2}e^\gamma \log n + e^{-c\sqrt{\log n}}$$ where $c$ is the constant used in the proof of the Zero Free Region.  Since $e^{-\sqrt{\log n}}$ decreases faster than any power of $\log$, we obtain a much better result upon squaring this estimate.  Precisely we have:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left(\prod_{2&amp;lt;p\leq n}1-\frac{2}{p}\right)^{-1} = \frac{1}{4}e^{2\gamma}\Pi_2^{-1} \log^2n + O\left( e^{-c\sqrt{\log n}} \right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(again the convergence to $\Pi_2$ is much to rapid to interfere with the error term)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would be willing to bet that this is the &lt;em&gt;best&lt;/em&gt; we can do, and that any better would imply stronger results regarding the error term for $\pi(x)$, the prime counting function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;3 Numerics&lt;/strong&gt;&#xA;Just for fun, the exact constant in front of the $\log^2n$ is:  1.201303.  How close is this?  Well for:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$n=10$ we get an error of $0.630811$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$n=50$ we get an error of $1.22144$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$n=100$ we get an error of $0.63493$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$n=1000$ we get an error of $0.438602$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$n=10^4$ we get an error of $0.250181$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$n=10^5$ we get an error of $0.096783$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$n=10^6$ we get an error of $0.017807$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where each time the error is positive.  That is the product seems to be slightly larger than the asymptotic (but converging fairly rapidly).  However, my intuition tells me it is almost certain (I will not prove it here) that the error term oscillates between negative and positive infinitely often.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;4 Under Riemann Hypothesis&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we assume the Riemann Hypothesis, the error term is bounded by $$\frac{C\log^2 x}{\sqrt{x}}$$ for some constant C. By analyzing the above data with numerical methods, the error seems to be best fitted by $\frac{C\log x}{\sqrt {x}}$.&lt;/p&gt;&#xA;" OwnerUserId="6075" LastEditorUserId="6075" LastEditorDisplayName="" LastEditDate="2011-03-26T17:08:03.743" LastActivityDate="2011-03-26T17:08:03.743" CommentCount="3" />
  <row Id="22436" PostTypeId="1" CreationDate="2011-02-17T01:14:06.627" Score="3" ViewCount="96" Body="&lt;p&gt;How would you go about proving the $(p, q, r)$ -pretzel knot is equivalent to the $(p, r, q)$ -pretzel knot?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By &quot;equivalent&quot; I mean you can change one knot into the other by elementary deformations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've found this question/example in several books and papers on knot theory where they state the proof as obvious.&lt;/p&gt;&#xA;" OwnerUserId="7175" LastActivityDate="2011-02-17T11:44:48.013" Title="Pretzel knot equivalence" Tags="&lt;knot-theory&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" />
  <row Id="22437" PostTypeId="1" AcceptedAnswerId="22441" CreationDate="2011-02-17T02:12:06.703" Score="3" ViewCount="104" Body="&lt;p&gt;Every rotation in 3D space can be defined by a rotation axis and an angle.&#xA;Now let's say we have two rotations R1(axis1, angle1), R2(axis2, angle2).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I remember that Rotation operator is closed under composition, so R1(R2(object)) will be a rotation again.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anybody know how to find the new axis and angle for R1 o R2?&lt;/p&gt;&#xA;" OwnerUserId="2939" LastActivityDate="2011-02-17T04:03:01.110" Title="Combining Two 3D Rotations" Tags="&lt;transformation&gt;&lt;3d&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="22438" PostTypeId="2" ParentId="22352" CreationDate="2011-02-17T02:21:56.713" Score="8" ViewCount="" Body="&lt;p&gt;Allow me to supplement Andres's excellent answer by copying over the following answer that I gave over at &lt;a href=&quot;http://mathoverflow.net/questions/10481/when-is-a-classification-problem-wild/10494#10494&quot; rel=&quot;nofollow&quot;&gt;this MO question&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can we understand in a precise general way the idea that a given classification problem is complicated or simple? How are we to compare the relative difficulty of two classification problems? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;These questions form the central motivation for the emerging subject known as Borel equivalence relation theory (see &lt;a href=&quot;http://www.math.ucla.edu/~greg/223b.1.06s/hjorth.dvi&quot; rel=&quot;nofollow&quot;&gt;Greg Hjorth's survey article&lt;/a&gt;, greatly missed after his recent death). The main idea is that many of the most natural equivalence relations arising in many parts of mathematics turn out to be Borel relations on a standard Borel space. To give one example, the isomorphism problem on finitely generated groups, but of course, there are hundreds of other examples. A classification problem for an equivalence relation E is really the problem of finding a way to describe the E-equivalence classes, of finding an E-invariant function that distinguishes the classes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Harvey Friedman defined that one equivalence relation E is &lt;em&gt;Borel-reducible&lt;/em&gt; to another relation F if there is a Borel function f such that x E y if and only if f(x) F f(y). That is, the function f maps E classes to F classes in such a way that different E classes get mapped to different F classes. This provides a classification of the E classes by using the F classes. The concept of reducibility provides a precise, robust way to say that one relation F is at least as complex as another E. Two relations are Borel equivalent if they reduce to each other, and we are led to the hierarchy of equivalence relations under Borel reducibility. By placing an equivalence relation into this hierarchy, we come to understand how complex it is in comparision with other equivalence relations. In particular, we say that one equivalence relation E is strictly simpler than F, if E reduces to F but not conversely.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It sometimes happens that one has a classification problem E and is able to provide a classification by assigning to each structure a countable list of data, such that two structures are equivalent iff they have the same data. This amounts to a reduction of E to the equality relation =, for two structures are E equivalent iff their data is equal. Such relations that reduce to equality are called &lt;em&gt;smooth&lt;/em&gt;, and lay near the bottom of the hierarchy of Borel equivalence relations. These are the simplest equivalence relations. Thus, one way of showing that a relation is comparatively simple, is to show that it is smooth, and to show it is comparatively hard, show that it is not smooth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The subject of Borel equivalence relation theory, as now developed by A. Kechris, G. Hjorth, S. Thomas and many others, is focused on placing many of the natural classification problems of mathematics into this hierarchy. Some of the main early results are the following interesting dichotomies:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Theorem.&lt;/b&gt;(Silver dichotomy) Every Borel equivalence relation E either has only countably many equivalence classes or = reduces to E.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The relation E&lt;sub&gt;0&lt;/sub&gt; says that two binary sequences are equivalent iff they agree from some point onward. It is easy to see that = reduces to E&lt;sub&gt;0&lt;/sub&gt;, and an elementary argument shows that E&lt;sub&gt;0&lt;/sub&gt; does not reduce to =. Thus, E&lt;sub&gt;0&lt;/sub&gt; is strictly harder than equality. Moreover, it is a kind of next-step up in the hiearchy, in light of the following.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Theorem.&lt;/b&gt;(Glimm-Effros dichotomy) Every Borel equivalence relation E either reduces to = or E&lt;sub&gt;0&lt;/sub&gt; reduces to E. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The subject continues with many interesting results that gradually illuminate more and more of the hierarchy of Borel equivalence relations. For example, the Feldman-Moore theorem shows that every Borel equivalence relation E having every equivalence class countable is the orbit equivalence of a countable group of Borel bijections of the space. The relation E&lt;sub&gt;oo&lt;/sub&gt; is the orbit equivalence of the left-translation action of the free group F&lt;sub&gt;2&lt;/sub&gt; on its power set. This relation is complete for the countable Borel equivalence relations, in the sense that every countable Borel equivalence relation reduces to it. It's great stuff!&lt;/p&gt;&#xA;" OwnerUserId="413" LastActivityDate="2011-02-17T02:21:56.713" />
  <row Id="22439" PostTypeId="1" AcceptedAnswerId="22450" CreationDate="2011-02-17T02:55:21.520" Score="3" ViewCount="113" Body="&lt;p&gt;I'm trying to solve the problem of showing that&#xA;$$\lim_{x\to6}\left(\frac{x}{4}+3\right) = \frac{9}{2}$$&#xA;using the $\epsilon$-$\delta$ definition of a limit.&lt;/p&gt;&#xA;" OwnerUserId="7176" LastEditorUserId="7176" LastEditorDisplayName="" LastEditDate="2011-02-17T03:26:04.333" LastActivityDate="2011-02-17T04:57:12.307" Title="Proving the value of a limit using the $\epsilon$-$\delta$ definition." Tags="&lt;calculus&gt;&lt;homework&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="22440" PostTypeId="2" ParentId="22439" CreationDate="2011-02-17T03:01:34.433" Score="7" ViewCount="" Body="&lt;p&gt;Given an $\epsilon\gt 0$, you want to show that provided $x$ is close enough to $6$, without being equal to $6$, then $\frac{x}{4}+3$ will be $\epsilon$-close to $\frac{9}{2}$. Well, first thing is to figure out how close $\frac{x}{4}+3$ is to $\frac{9}{2}$:&#xA;$$\left|\left(\frac{x}{4}+3\right) - \frac{9}{2}\right| = \left|\frac{x}{4}+3-\frac{9}{2}\right| = \left|\frac{x + 12 - 18}{4}\right| = \frac{|x-6|}{4}.$$&#xA;So, how can you make sure that $\displaystyle \left|\left(\frac{x}{4}+3\right) - \frac{9}{2}\right|$ is smaller than $\epsilon$, by placing conditions on how close $x$ is to $6$, that is, on the value of $|x-6|$?&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-17T03:34:48.323" LastActivityDate="2011-02-17T03:34:48.323" CommentCount="11" />
  <row Id="22441" PostTypeId="2" ParentId="22437" CreationDate="2011-02-17T03:01:43.247" Score="5" ViewCount="" Body="&lt;p&gt;Once again, the easiest way of doing this (IMHO, at least) turns out to be quaternions: a rotation of $\theta$ about the (normalized) vector $\hat{v} = (v_0, v_1, v_2)$ is represented by the quaternion ${\bf q}=\mathrm{cos}(\theta/2) + \mathrm{sin}(\theta/2)*(v_0{\bf i}+v_1{\bf j}+v_2{\bf k})$; and quaternion multiplication is easy to compute (all you need are the base axioms ${\bf i}^2 = {\bf j}^2 = {\bf k}^2 = {\bf i}{\bf j}{\bf k} = -1$ - note that the latter implies for instance that ${\bf i}{\bf j} = {\bf k}$ by simply multiplying both sides by ${\bf k}$); once you have your resultant quaternion, the axis can be extracted as the (normalized) imaginary part of the result and the angle can be found by taking the arc-cosine of the scalar part.  See the Wikipedia page on &lt;a href=&quot;http://en.wikipedia.org/wiki/Quaternions_and_spatial_rotation&quot; rel=&quot;nofollow&quot;&gt;quaternions and rotations&lt;/a&gt; for more details on just how this works.&lt;/p&gt;&#xA;" OwnerUserId="785" LastActivityDate="2011-02-17T03:01:43.247" CommentCount="3" />
  <row Id="22442" PostTypeId="2" ParentId="7774" CreationDate="2011-02-17T03:14:20.770" Score="5" ViewCount="" Body="&lt;p&gt;So, the exercise is incorrect as stated, as the nice example in the question shows.  They probably meant to say that the matrix is invertible &lt;em&gt;in the subalgebra of upper triangular matrices&lt;/em&gt; if and only if the diagonal entries are invertible.  This is the version given on page 16 in a set of &lt;a href=&quot;http://toknotes.mimuw.edu.pl/sem1/files/K_theory.pdf&quot; rel=&quot;nofollow&quot;&gt;lecture notes&lt;/a&gt; by Matthes and Szymański based primarily on the same book.  They also give a counterexample to the original statement.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2011-02-17T05:02:44.560" LastActivityDate="2011-02-17T05:02:44.560" CommunityOwnedDate="2011-02-17T03:14:20.770" CommentCount="1" />
  <row Id="22443" PostTypeId="2" ParentId="22361" CreationDate="2011-02-17T03:19:05.377" Score="1" ViewCount="" Body="&lt;p&gt;Without loss of generality, we can assume that a grammar for $L_n$ (such a language with specific $\pi_1,\pi_2 \in S_n$) is in Chomsky Normal Form. The language $L_n$ consists of the words $w(x) = x\pi_1(x)\pi_2(x)$ for all $x \in \{0,1\}^n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the Subword Lemma (see my previous answers), for each $w(x)$ we can find a substring $s(x)$ of length $$\frac{n}{2} \leq |s(x)| &amp;lt; n$$ generated by some symbol $A(x)$ and occurring at position $p(x)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose that $p(x) = p(y)$ and $A(x) = A(y)$. Since $|s(x)|&amp;lt;n$, the subword $s(x)$ cannot intersect both the $x$ part and the $\pi_2(x)$ part of $w(x)$; we can assume it is disjoint from the $x$ part. Thus $w(x)$ is of the form $x \alpha s(x) \beta$. This implies that $A(x)$ generates exactly one string, namely $s(x)$. Therefore $s(x) = s(y)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now $s(y)$ intersects either $\pi_1(y)$ or $\pi_2(y)$ in at least $n/4$ places, and thus determines at least $n/4$ bits of $y$. Therefore at most $2^{3n/4}$ strings $y \in \{0,1\}^n$ can have $p(x) = p(y)$ and $A(x) = A(y)$. Since there are at most $3n$ possibilities for $p(y)$, we get that there are at least $$\frac{2^{n/4}}{3n}$$ different non-terminals in the grammar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Comment: The same proof works if $\pi_1,\pi_2 \in S_{\{0,1\}^n}$, i.e. are arbitrary permutations on the set of all $n$-bit words. Given $n/4$ bits of $\pi_i(y)$, there are exactly $2^{3n/4}$ preimages $y$.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2011-02-17T03:19:05.377" CommentCount="6" />
  <row Id="22444" PostTypeId="2" ParentId="22423" CreationDate="2011-02-17T03:25:59.973" Score="6" ViewCount="" Body="&lt;p&gt;Here are a five methods. $\ $ (1) $\: $ Complete your proof: show that $\rm\ (a,b) = 1\ \Rightarrow\ (ab,\:a^2-b^2) = 1\:.\:$ Hint: $\ $ prime $\rm\ p\ |\ a,\ aa-bb\ \ \Rightarrow\ \ p\ |\ bb\ \Rightarrow\ p\ |\ b\:.\ $ Or: $\ $ since $\rm\ a,\:b\ $ are coprime to $\rm\ a-b,\:a+b\ $ then their products $\rm\ a\:b,\ a^2-b^2\:,\: $ are also coprime, by Euclid's Lemma.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) $\ $ The rational root test $\:\Rightarrow\:$ the only possible rational roots of $\rm\ x^2 -5\ x - 1\ $ are $\rm\ x = \pm 1\:.$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(3) $\ $ Suppose it has a rational root $\rm \ R = A/B\:.\ $ Suppose this is in lowest terms, i.e. $\rm\: B\:$ is minimal.   $\rm\ R = 5 + 1/R\ \Rightarrow\ A/B = (5\:A+B)/A\:.\:$ Taking fractional parts yields $\rm\ b/B = a/A\ $ for $\rm 0\le b &amp;lt; B\:.\:$ But $\rm\ b\ne0\ \Rightarrow\ A/B = a/b\ $ contra minimality of $\rm\:B\:.\:$ So $\rm\:b = 0\:,\:$ i.e. $\rm\ A/B\ $ has fractional part $ = 0\:,\:$ so $\rm\ R = A/B\ $ is an integer. Then so too is $\rm\ 1/R = R-5\:.\:$ So $\rm\ R = \pm 1\:,\:$ contra $\rm\ R^2 - 1 = 5\:R\:.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(4) $\ $ As in (3), $\rm\ A/B = C/A\ $ with $\rm\:B\:$ least implies $\rm\:B\ |\ A\:$ by &lt;a href=&quot;http://math.stackexchange.com/questions/5/how-can-you-prove-that-the-square-root-of-two-is-irrational/16544#16544&quot;&gt;unique fractionization,&lt;/a&gt; which concludes the proof as in (3).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For more on the relationship between (3) and (4), follow the above link, where you'll find my analysis of the analogous general square-root irationality proofs, and links to an interesting discussion of such between John Conway and I.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(5) $\ $ As Euclid showed a very long time ago, the Euclidean gcd algorithm works also for rationals, so they too have gcds, and such gcds enjoy the same laws as for integers, e.g. the distributive law. Thus $\rm\ (r,1)^2 = (r^2,r,1) = (5r+1,r,1) = (r,1)\ $ so $\rm\ (r,1) = 1\:,\:$ so $\rm\ 1\ |\ r \ $ in $\rm\:\mathbb Z\:,\:$ i.e. $\rm\ r\in\mathbb Z\:,\:$ and the proof concludes as above. This is - at the heart - the same proof alluded to by Moron using non-termination of the continued-fraction algorithm (a variant of the Euclidean gcd algorithm). Alternatively, scaling the prior equations by $\rm\:b^2\:,\:$ where $\rm\ r = a/b\:,\:$ converts it to one using only integer gcd's, namely $\rm\ 1 = (a,b)^2 = (a^2,ab,b^2) = (5ab+b^2,ab,b^2) = b\:(a,b)\:$ therefore $\rm\ b\ |\ 1\:.\:$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;These are essentially special cases of gcd/ideal-theoretic ways to prove that $\rm\: \mathbb Z\:$ is &lt;em&gt;integrally-closed&lt;/em&gt;, i.e. the monic case of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Rational_root_theorem&quot; rel=&quot;nofollow&quot;&gt;rational root test&lt;/a&gt; - any fractional root of a monic polynomial must be integral. Perhaps the slickest way to prove such results is the elegant one-line proof using Dedekind's notion of &lt;a href=&quot;http://math.stackexchange.com/questions/2883/show-that-3p2-q2-implies-3p-and-3q/2941#2941&quot;&gt;conductor ideal&lt;/a&gt;. Follow that link (and its links) for much further discussion (both elementary and advanced).&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2011-02-28T04:02:44.203" LastActivityDate="2011-02-28T04:02:44.203" CommentCount="5" />
  <row Id="22445" PostTypeId="1" AcceptedAnswerId="22447" CreationDate="2011-02-17T03:49:20.710" Score="5" ViewCount="96" Body="&lt;p&gt;If I understand correctly, the analytic multiplicity of a linear operator say $T:V\to V$ is the amount of times $\lambda$ shows up as a root in the characteristic polynomial (Assuming you have a matrix $A$ representing $T$ with respect to some basis of $V$, then the characteristic equation is $\det(A-\lambda I)$).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand how to find this, but what exactly does this mean for the linear operator?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, how do we find the geometric multiplicity of $T$?  What is it's significance?  I tried looking it up on Wikipedia but they are using some notation and words that I am not familiar with; the definitions there are usually quite formal.  Am I correct in thinking that it is the greatest amount of linearly independent vectors of the eigenvalues of $T$?  Do all the eigenvalues of a linear operator have the same amount of linearly independent vectors?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; Here is a specific example.&#xA;We have the matrix, say $A$, with $1$s all along the diagonal and above, and $0$s below.  Since this matrix is upper triangular, the eigenvalues of $A$ are $1$.  Looking at the equations we have to satisfy to find the eigenvector(s) of an eigenvalue $1$, we get:&#xA;\begin{align*}&#xA;x_1 + x_2 + \cdots + x_n &amp;amp;= \lambda x_1\\&#xA;x_2 + \cdots + x_n &amp;amp;= \lambda x_2\\&#xA;&amp;amp;\vdots\\&#xA;x_n &amp;amp;=\lambda x_n&#xA;\end{align*}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems like we need to do some inspection here, it would be hard (or at least I can't see how) to show this formally.  It seems to me like there are two possible linear independant eigenvectors here,&lt;br&gt;&#xA;$$\displaystyle \vec 0$$ and $$\displaystyle (1,0,...,0)$$&lt;br&gt;&#xA;So the geometric multiplicity is $1$?&lt;/p&gt;&#xA;" OwnerUserId="2533" LastEditorUserId="2533" LastEditorDisplayName="" LastEditDate="2011-02-17T04:47:15.103" LastActivityDate="2011-02-17T16:30:40.203" Title="Geometric and analytic multiplicity of a linear operator." Tags="&lt;linear-algebra&gt;" AnswerCount="1" CommentCount="7" FavoriteCount="1" />
  <row Id="22446" PostTypeId="2" ParentId="22437" CreationDate="2011-02-17T04:03:01.110" Score="0" ViewCount="" Body="&lt;p&gt;Take a look at: &lt;a href=&quot;http://www.siggraph.org/education/materials/HyperGraph/modeling/mod_tran/3drota.htm&quot; rel=&quot;nofollow&quot;&gt;http://www.siggraph.org/education/materials/HyperGraph/modeling/mod_tran/3drota.htm&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1369" LastActivityDate="2011-02-17T04:03:01.110" />
  <row Id="22447" PostTypeId="2" ParentId="22445" CreationDate="2011-02-17T04:03:11.813" Score="8" ViewCount="" Body="&lt;p&gt;The geometric multiplicity of $\lambda$ tells you how big a subspace of $V$ you can find where $T$ acts simply as &quot;multiplication by $\lambda$&quot; (that is, how big, dimensionally speaking, the subspace spanned by the eigenvectors of $\lambda$ is).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The analytic/algebraic multiplicity of $\lambda$ tells you how big that space &quot;should&quot; be for $V$ and $T$ to have a &quot;nice&quot; decomposition of the following kind: you can express $V$ as a direct sum of subspaces $E_{\lambda_1}$, $E_{\lambda_2},\ldots,E_{\lambda_k}$ (where $\lambda_1,\ldots,\lambda_k$ are the &lt;em&gt;distinct&lt;/em&gt; roots of the characteristic polynomial) so that on each $E_{\lambda_i}$, $T$ acts just by &quot;multiplication by $\lambda_i$&quot;. For $V$ to really be equal to the sum of these spaces, you need $\dim(E_{\lambda_i})$, which is the geometric multiplicity of $\lambda_i$, to equal the algebraic multiplicity of $\lambda_i$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It has other properties, but I think that's a good place to start.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yes, the geometric multiplicity is the largest possible number of linearly independent &lt;strong&gt;eigenvectors&lt;/strong&gt; of $T$ associated to $\lambda$ (vectors $\mathbf{v}$, $\mathbf{v}\neq\mathbf{0}$, such that $T(\mathbf{v}) = \lambda\mathbf{v}$; that is, vectors on which $T$ acts just by &quot;multiplication by $\lambda$). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;No, not all eigenvalues have the same geometric multiplicity; for example, in the matrix&#xA;$$\left(\begin{array}{cccc}&#xD;&#xA;2 &amp;amp; 1 &amp;amp; 0 &amp;amp; 0\\\&#xD;&#xA;0 &amp;amp; 2 &amp;amp; 0 &amp;amp; 0\\\&#xD;&#xA;0 &amp;amp; 0 &amp;amp; 2 &amp;amp; 0\\\&#xD;&#xA;0 &amp;amp; 0 &amp;amp; 0 &amp;amp; 1&#xD;&#xA;\end{array}\right),$$&#xA;the characteristic polynomial is $(2-\lambda)^3(1-\lambda)$, so the two eigenvalues are $\lambda_1=1$ and $\lambda_2=2$. The eigenvalue $\lambda_1=1$ has algebraic and geometric multiplicities both equal to $1$; $\lambda_2=2$ has algebraic multiplicity $3$, and geometric multiplicity $2$. (You can check the geometric multiplicity by finding the nullity of $A-2I$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Added.&lt;/em&gt; Since the geometric multiplicity of an eigenvalue $\lambda_i$ is the dimension of the subspace $E_{\lambda_i}$, your first task in finding that dimension is to identify the vectors $\mathbf{v}$ for which $T(\mathbf{v})=\lambda_i\mathbf{v}$. This is equivalent to finding the vectors for which $(T-\lambda_i I)(\mathbf{v})=\mathbf{0}$. The reason this is a better problem to tackle is that it is easier to solve a system that looks like $B\mathbf{v}=\mathbf{0}$, than one that looks like $A\mathbf{v}=\lambda\mathbf{v}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, you find the nullspace of $T-\lambda_iI$, that is, the collection of all vectors $\mathbf{v}$ for which $(T-\lambda_iI)(\mathbf{v})=\mathbf{0}$. Its dimension is precisely the geometric multiplicity of $\lambda_i$, so the geometric multiplicity of $\lambda_i$ is found by computing $\mathrm{nullity}(T-\lambda_iI)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For your specific example: we begin with the matrix corresponding to the standard basis:&#xA;$$A =\left(\begin{array}{cccc}&#xD;&#xA;1 &amp;amp; 1 &amp;amp; \cdots &amp;amp; 1\\\&#xD;&#xA;0 &amp;amp; 1 &amp;amp; \cdots &amp;amp; 1\\\&#xD;&#xA;\vdots &amp;amp; \vdots &amp;amp; \ddots &amp;amp; \vdots \\\&#xD;&#xA;0 &amp;amp; 0 &amp;amp; \cdots &amp;amp; 1&#xD;&#xA;\end{array}\right).$$&#xA;The characteristic polynomial is $\det(A-tI) = (1-t)^n$, so the only eigenvalue is $\lambda=1$, with algebraic multiplicity $n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To find the geometric multiplicity, take $A-1I$ (&quot;$1I$&quot; because we are taking $\lambda = 1$), and find its nullspace. Since&#xA;$$A - I = \left(\begin{array}{ccccc}&#xD;&#xA;0 &amp;amp; 1 &amp;amp; 1 &amp;amp;\cdots &amp;amp; 1\\\&#xD;&#xA; 0 &amp;amp; 0 &amp;amp; 1 &amp;amp; \cdots &amp;amp; 1\\\&#xD;&#xA;\vdots &amp;amp; \vdots &amp;amp; \vdots &amp;amp; \ddots &amp;amp; \vdots \\\&#xD;&#xA;0 &amp;amp; 0 &amp;amp; 0 &amp;amp; \cdots &amp;amp; 0&#xD;&#xA;\end{array}\right),$$&#xA;finding the reduced row-echelon form will give you the solutions to $(A-I)\mathbf{x}=\mathbf{0}$. The reduced row-echelon form of $A-I$ is&#xA;$$\left(\begin{array}{ccccc}&#xD;&#xA;0 &amp;amp; 1 &amp;amp; 0 &amp;amp; \cdots &amp;amp; 0\\\&#xD;&#xA;0 &amp;amp; 0 &amp;amp; 1 &amp;amp; \cdots &amp;amp; 0\\\&#xD;&#xA;\vdots &amp;amp; \vdots &amp;amp; \vdots &amp;amp; \ddots &amp;amp; \vdots\\\&#xD;&#xA;0 &amp;amp; 0 &amp;amp; 0 &amp;amp; \cdots &amp;amp; 1\\\&#xD;&#xA;0 &amp;amp; 0 &amp;amp; 0 &amp;amp; \cdots &amp;amp; 0&#xD;&#xA;\end{array}\right),$$&#xA;so $\mathbf{x}=(x_1,x_2,\ldots,x_n)$ is in the nullspace if and only if $x_2=x_3=\cdots=x_n=0$. So the eigenvectors of $\lambda=1$ are all vectors of the form $(a,0,0,\ldots,0)$ for arbitrary $a$; however, because $\mathbf{0}$ is &lt;em&gt;always&lt;/em&gt; a solution to $A\mathbf{x}=\lambda \mathbf{x}$ for any $\lambda$, we declare by fiat that an eigenvector has to be nonzero (this has no bearing on the geometric multiplicity of $\lambda$, because $\mathbf{0}$ can never be in a linearly independent set). A &lt;em&gt;basis&lt;/em&gt; for this nullspace is given by $(1,0,\ldots,0)$, so the nullspace has dimension $1$. This dimension is the geometric multiplicity of $\lambda=1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, in summary: $\lambda=1$ is the only eigenvalue; it has algebraic multiplicity $n$, and geometric multiplicity $1$. The eigenvectors are all nonzero multiples of $(1,0,0,\ldots,0)$. &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-17T16:30:40.203" LastActivityDate="2011-02-17T16:30:40.203" CommentCount="1" />
  <row Id="22448" PostTypeId="2" ParentId="21696" CreationDate="2011-02-17T04:36:43.367" Score="1" ViewCount="" Body="&lt;p&gt;Actually, your guess is right with a little more assumption. Namely, in a Noetherian local ring $A$, TFAE for an ideal $I$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) $I$ has finite projective dimension and $I/I^2$ is a free $A/I$-module.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) $I$ is generated by a regular sequence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(this is a result by Vasconcelos, I just found it as Exercise 20.23 in Eisenbud's book)&lt;/p&gt;&#xA;" OwnerUserId="223" LastActivityDate="2011-02-17T04:36:43.367" />
  <row Id="22449" PostTypeId="1" AcceptedAnswerId="22452" CreationDate="2011-02-17T04:39:57.057" Score="2" ViewCount="72" Body="&lt;p&gt;$\log(a - b) - \log(a - c)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does this have a simpler form? Perhaps one where the $a$s have cancelled out? I know it can also be expressed as a log of the fraction: $\log\frac{(a-b)}{(a-c)}$, but the same question applies.&lt;/p&gt;&#xA;" OwnerUserId="7178" LastEditorUserId="7178" LastEditorDisplayName="" LastEditDate="2011-02-17T04:45:09.563" LastActivityDate="2011-02-17T05:23:08.397" Title="difference of logs" Tags="&lt;arithmetic&gt;&lt;logarithms&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="22450" PostTypeId="2" ParentId="22439" CreationDate="2011-02-17T04:44:59.733" Score="6" ViewCount="" Body="&lt;p&gt;Since you said you're still lost after Arturo's post, I'll try to start earlier.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Q.  What do you mean intuitively by $\lim\limits_{x\rightarrow 6} \frac{x}{4} + 3$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A.  Intuitively, you keep plugging in particular $x$ values really close to $6$ (but never actually plugging in $6$ - things like 5.99999 and 6.0000001 - into $\frac{x}{4}+3$ and you record the outputs.  Now, as you keep plugging in things closer and closer to $6$, you expect the outputs to hone in on one number.  The limit, then, is that one number.  By looking at the graph of $\frac{x}{4}+3$, you'd probably guess that the output is $\frac{6}{4} + 3 = \frac{9}{2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, let me resay this answer in a way that will lead into the official math definition for a limit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I come along and say the limit is $\frac{9}{2}$, how would you test me?  Well, you could think to yourself &quot;if the values are honing in on $\frac{9}{2}$, eventually they must get and stay within $.1$ of $\frac{9}{2}$, and so you challenge me by asking me to show that this is indeed the case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then I could respond by saying, &quot;Once $x$ is within .01 of 6, then $\frac{x}{4} + 3$ will be within $.1$ of $\frac{9}{2}$.  For if $|x-6|&amp;lt;.01$, then $|\frac{x}{4}+3 - \frac{9}{2}| = |\frac{x}{4} - \frac{3}{2}| = |\frac{x-6}{4}| = \frac{|x-6|}{4} &amp;lt; \frac{.01}{4} &amp;lt; .1$.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If every time you come up with a tolerance (like $.1$), I can pass your test by making up a tolerance of my own ($.01$), then mathematically we'd say the limit is $\frac{9}{2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the official math definition is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\lim\limits_{x\rightarrow 6} \frac{x}{4} + 3 = \frac{9}{2}$ means for all $\epsilon &amp;gt; 0$, there is a $\delta &amp;gt; 0$ such that if $|x-6|&amp;lt;\delta$, then $|\frac{x}{4}+3 - \frac{9}{2}|&amp;lt; \epsilon$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In our previous &quot;conversation&quot;.  The $.1$ played the role of $\epsilon$ while the $.01$ played the role of $\delta$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After reading (and  possibly rereading, and rerereading) all of the above, I'd encourage you to reread Arturo's response and see if you can turn what he said into a full fledged answer.&lt;/p&gt;&#xA;" OwnerUserId="331" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-17T04:57:12.307" LastActivityDate="2011-02-17T04:57:12.307" />
  <row Id="22452" PostTypeId="2" ParentId="22449" CreationDate="2011-02-17T04:47:41.290" Score="1" ViewCount="" Body="&lt;p&gt;It depends what you think is simpler.  $\log(a - b) - \log(a - c)=\log \frac{a-b}{a-c}=\log \left(1-\frac{b-c}{a-c}\right)$&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-02-17T04:47:41.290" />
  <row Id="22453" PostTypeId="2" ParentId="8943" CreationDate="2011-02-17T05:01:24.393" Score="3" ViewCount="" Body="&lt;p&gt;First, if all the digits are the same, there is only one number and that is the GCD.  As was pointed out before, if 3 or 9 is a factor of one permutation it will be a factor of them all.  Otherwise, imagine swapping just the ones and tens digit when they are different.  The GCD of these two has to divide $100a+10b+c-100a+10c+b=9(b-c)$ where $b$ and $c$ are single digits.  For the GCD of all the numbers to have a factor 2, all the digits must be even.  For the GCD to have a factor 4, all the digits must be 0, 4, or 8 and for 8 they must be 0 or 8.  Similarly for 5 and 7.  Finally, the GCD will be 27 if all the digits are 0,3,6, or 9 and 27 divides one permutation and 81 if all the digits are 0 or 9 and 81 divides one permutation.  Can you prove the last assertion?  &lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-02-17T05:01:24.393" CommentCount="1" />
  <row Id="22454" PostTypeId="2" ParentId="22449" CreationDate="2011-02-17T05:23:08.397" Score="0" ViewCount="" Body="&lt;p&gt;Like Ross said, it depends what you think is simpler.&#xA;You can try out something like,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \log(a-b) - \log(a-c) = \log(a) + \log(1-\frac{b}{a}) - \log(a) - \log(1 - \frac{c}{a}) = \log(1-\frac{b}{a}) - \log(1 - \frac{c}{a})$$ where $\log(a)$ is cancelled out. Then expand it to get the results that you want. But since your question is not clear as to what results you are seeking, nor does it say anything about a,b and c, I am not sure if this might be of help.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyways, this expression wont have a simpler form, like you said, where the 'a's are cancelled out. It needs some extra condition to actually cancel out 'a' from the expression.&lt;/p&gt;&#xA;" OwnerUserId="2320" LastActivityDate="2011-02-17T05:23:08.397" />
  <row Id="22455" PostTypeId="2" ParentId="22423" CreationDate="2011-02-17T05:57:53.843" Score="3" ViewCount="" Body="&lt;p&gt;Since you have already seen answers which complete your proof, here is another proof using &lt;a href=&quot;http://en.wikipedia.org/wiki/Continued_fraction&quot; rel=&quot;nofollow&quot;&gt;continued fractions&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $r$ was rational, it will have a finite continued fraction, say $[a_1, a_2, \dots, a_n]$, but that can be extended to $[5, a_1, a_2, \dots , a_n]$ and $[5,5,a_1, a_2, \dots a_n]$ etc, because&#xA;$$ r = 5 + \frac{1}{r} = 5 + \frac{1}{[a_1, a_2, \dots, a_n]} = [5, a_1, a_2, \dots, a_n]$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, you can easily give the infinite continued fraction of $r$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$[5,5,5,5,\dots]$$&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2011-02-17T05:57:53.843" CommentCount="4" />
  <row Id="22456" PostTypeId="1" AcceptedAnswerId="22457" CreationDate="2011-02-17T06:12:50.887" Score="1" ViewCount="47" Body="&lt;p&gt;Hello,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have several BNF defined as follow:  &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;1. &amp;lt;S&amp;gt; -&amp;gt; 0&amp;lt;S&amp;gt; | 1 | 0  &#xA;2. &amp;lt;S&amp;gt; -&amp;gt; 1&amp;lt;S&amp;gt; | 1 | 0&#xA;3. &amp;lt;S&amp;gt; -&amp;gt; 0&amp;lt;S&amp;gt; | 1&amp;lt;S&amp;gt; | 1 | 0&#xA;4. &amp;lt;S&amp;gt; -&amp;gt; &amp;lt;S&amp;gt;0 | 1  | 0&#xA;5. &amp;lt;S&amp;gt; -&amp;gt; &amp;lt;S&amp;gt;1 | 0&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This is the solution from my friend, however, I doubt the correctness of some of them&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;1. { 0^n: n &amp;gt;= 1 } V { 0^n1: n &amp;gt;= 0 } -&amp;gt; This is fine.&#xA;2. { 1^n0^k: n &amp;gt; 0, k = 0, 1 } -&amp;gt; I don't quite get this one, is the same with grammar 1?&#xA;3. { 0 } V { 1^n0^k: n &amp;gt; 0 and k = 0, 1 } -&amp;gt; I think this one is incorrect.&#xA;4. { 0^n: n &amp;gt; 0 } V { 10^n: n &amp;gt;= 0 } -&amp;gt; I think this is also incorrect&#xA;5. { 01^n: n &amp;gt;= 0 } -&amp;gt; This is fine.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Could anyone help me explain this?   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks,&lt;br&gt;&#xA;Chan&lt;/p&gt;&#xA;" OwnerUserId="5711" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2011-02-17T06:19:45.830" LastActivityDate="2011-02-17T06:25:04.250" Title="Need help explain BNF" Tags="&lt;computer-science&gt;&lt;formal-systems&gt;" AnswerCount="1" />
  <row Id="22457" PostTypeId="2" ParentId="22456" CreationDate="2011-02-17T06:17:26.143" Score="1" ViewCount="" Body="&lt;p&gt;Questions $1$,$2$ and $4$ are very similar. If you get one of them you should get all of them. Question $5$ is even simpler.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For question $3$, it is easy to see that every non-empty string is generated by the BNF. For example, to generate 0110, use $$S \rightarrow 0S \rightarrow 01S \rightarrow 011S \rightarrow 0110.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As to your question on &quot;how to write all non-empty strings in $\{\}$ notation&quot;, this depends on the conventions used in your course. The corresponding regular expression is $(0+1)^+$, but I guess you haven't learned these just yet.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2011-02-17T06:25:04.250" LastActivityDate="2011-02-17T06:25:04.250" CommentCount="3" />
  <row Id="22458" PostTypeId="1" CreationDate="2011-02-17T06:37:42.853" Score="2" ViewCount="99" Body="&lt;p&gt;What is the formula for the derivative of the product of two matrices when they are different sizes? Further, what is the formula for the derivative of a hadamard product when the derivatives of the inputs aren't the same size as the inputs?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's what I'm thinking: assume an nX1 vector x, a nXn symmetric matrix S, f(x)=Sw, g(x)=w'Sw, h(x)=f(x)/g(x), and j(x)=had(x, h(x)) where had(a, b) is the hadamard product between the two vectors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For f(x) and g(x) the derivatives are S and 2*Sw, respectively. One would think that the derivative of h(x) would be (g(x)f'(x)-g'(x)f(x))/(g(x)^2) from the quotient rule. However, the numerator is nXn and the second part is multiplying a nX1 vector by an nX1 vector, which doesn't work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, I found a formula for the derivative of hadamard product and applying it to this would give had(x, h'(x) + had(I, h(x)) = had(x, h'(x)) + h(x). Since h(x) is nX1, h'(x) must be nXn. However, the hadamard product of an nX1 and nXn matrix doesn't work. Similarly, the opposite is the case in the second term. So far as I can tell, the terms that are nX1 need to be copied over to become nXn. This would indicate that the proper formula is something like had(kron(x,ones),h'(x))+had(I,kron(h(x),ones)), where I is an nXn identity matrix and ones is a 1Xn vector of ones.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think that's on the right track since I found a matrix calculus guide that connects the problem to Kronecker products: &#xA;&lt;a href=&quot;http://www4.ncsu.edu/~pfackler/MatCalc.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www4.ncsu.edu/~pfackler/MatCalc.pdf&lt;/a&gt;&#xA;However, I still had the problem of conformability when using their product rule formula.&lt;/p&gt;&#xA;" OwnerUserId="5844" LastEditorUserId="5844" LastEditorDisplayName="" LastEditDate="2011-02-17T14:33:39.937" LastActivityDate="2011-05-18T23:58:59.387" Title="Conformability and Matrix Derivatives" Tags="&lt;calculus&gt;&lt;matrices&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="22459" PostTypeId="1" AcceptedAnswerId="22460" CreationDate="2011-02-17T07:00:48.807" Score="1" ViewCount="40" Body="&lt;p&gt;For a p-primary group $G$, if  $\langle y\rangle$ is a cyclic group of order $p^i$ for any natural integer $i$, then $ng \in \langle y\rangle$ implies $g \in \langle y\rangle$ for any $g \in G$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, in the terms of group theory, what does the symbol $G^p$ means for a group $G$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wanted to ask this because I was referring to this post:&lt;a href=&quot;http://math.stackexchange.com/questions/18719/example-of-intersection-of-pure-subgroup-which-is-not-pure&quot;&gt;Example of Intersection of Pure Subgroup which is not Pure&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that $G$ is abelian. Thanks in advance.&lt;/p&gt;&#xA;" OwnerUserId="1683" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-17T07:22:38.323" LastActivityDate="2011-02-17T14:47:34.290" Title="A Question about p-primary Group." Tags="&lt;group-theory&gt;" AnswerCount="1" />
  <row Id="22460" PostTypeId="2" ParentId="22459" CreationDate="2011-02-17T07:19:59.340" Score="3" ViewCount="" Body="&lt;p&gt;The answer to your first question is &quot;no&quot;. Take $G=C_p\times C_{p^2}$, a product of a cyclic group of order $p$ and a cyclic group of order $p^2$, and let $y=(1,1)$, which has order $p^2$. Then let $g=(0,1)$. Then $pg = (0,p) = p(1,1) = py$, so $pg\in \langle y\rangle$, but $g\notin \langle y\rangle$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Added for completeness:&lt;/em&gt; As mentioned in the comment below, if you also have that $\gcd(n,p)=1$, however, then the conclusion does follow: since $G$ is a $p$-group, there exists $k$ such that $p^kg = 0$, and from $\gcd(n,p)=1$ we know there exist integers $r,s$ such that $1 = rn + sp^k$. Then, since $ng\in\langle y\rangle$, we have:&#xA;$$g = 1g = (rn+sp^k)g = (rn)g + (sp^k)g = r(ng) + s(p^kg) = r(ng)+s0 = r(ng)\in\langle y\rangle.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the above works in &lt;em&gt;any&lt;/em&gt; group: if $H$ is a subgroup of $G$, $g\in G$ is of order $k$, and $g^n\in H$ with $\gcd(n,k)=1$, then $g\in H$, essentially  because $\langle g\rangle = \langle g^n\rangle$ if $\gcd(n,k)=1$, so $$g\in\langle g\rangle = \langle g^n\rangle \subseteq H.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For your second question: for a group $G$, $G^p$ is the subgroup of $G$ generated by all elements of the form $x^p$ with $x\in G$; that is, the subgroup generated by the $p$th powers of all elements of $G$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $G$ is abelian, then this is simply the subgroup that consists of all $p$-powers, since $a^pb^p = (ab)^p$; for nonabelian groups, it may in general be larger.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your first sentence is not very well-written, by the way: &quot;any&quot; is incorrect there, you mean &quot;some&quot; (otherwise, you are saying that for all natural numbers $i$, the subgroup $\langle y\rangle$ has order $p^i$; this is impossible, since the subgroup can only have one order).&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-17T14:47:34.290" LastActivityDate="2011-02-17T14:47:34.290" CommentCount="6" />
  <row Id="22461" PostTypeId="1" AcceptedAnswerId="22462" CreationDate="2011-02-17T07:21:10.307" Score="3" ViewCount="126" Body="&lt;p&gt;Hey there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm trying to understand a solution I was given in a tutorial regarding a problem with Carmichael numbers and I was wondering if you guys can help clarify things:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A composite number $m$ is called a Carmichael number if the congruence $a^{m-1} \equiv 1 \pmod{m}$ is true for every number a with $\gcd(a,m) = 1$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Verify that $m = 561 = 3 \times 11 \times 17$ is a Carmichael number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solution given:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Apply Fermat's Little Theorem to each prime divisor of $m$:&#xA;\begin{align*}&#xA;a^2 &amp;amp;\equiv 1 \pmod{3}\\&#xA;a^{10} &amp;amp;\equiv 1 \pmod{11}\\&#xA;a^{16} &amp;amp;\equiv 1 \pmod{17}&#xA;\end{align*}&#xA;This somehow then implies that $a^{80} \equiv 1 \pmod{561}$ then accordingly $a^{560} \equiv 1 \pmod{561}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am lost as to how the 3 congruences imply $a^{80} \equiv 1 \pmod{561}$ ($80 = \mathrm{LCM}(2,10,16)$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can somebody clarify this for me? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="6523" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-17T16:06:23.917" LastActivityDate="2011-02-17T16:20:57.430" Title="Verifying Carmichael numbers" Tags="&lt;elementary-number-theory&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="22462" PostTypeId="2" ParentId="22461" CreationDate="2011-02-17T07:26:35.760" Score="6" ViewCount="" Body="&lt;p&gt;Note that $80 = \mathrm{lcm}(2,10,16)$. So you can write $a^{80} = (a^2)^{40} = (a^{10})^{8} = (a^{16})^5$. So, &#xA;\begin{align*}&#xA;a^{80}= (a^2)^{40} &amp;amp;\equiv 1^{40} = 1\pmod{3},\\&#xA;a^{80}= (a^{10})^{8} &amp;amp;\equiv 1^8 = 1 \pmod{11},\\&#xA;a^{80}= (a^{16})^5 &amp;amp;\equiv 1^5 = 1\pmod{17}.&#xA;\end{align*}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the Chinese Remainder Theorem, the system of congruences &#xA;\begin{align*}&#xA;x&amp;amp;\equiv 1\pmod{3}\\&#xA;x&amp;amp;\equiv 1\pmod{11}\\&#xA;x&amp;amp;\equiv 1\pmod{17}&#xA;\end{align*}&#xA;has a unique solution modulo $3\times 11\times 17 = 561$. But both $x=1$ and $x=a^{80}$ are solutions. Since the solution is unique modulo $561$, then &#xA;the two solutions we found must be congruent. That is,&#xA;$$a^{80}\equiv 1\pmod{561}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(&lt;em&gt;Added.&lt;/em&gt; Or, more simply, as Andres points out, since $3$, $11$, and $17$ each divide $a^{80}-1$, and are pairwise relatively prime, then their product divides $a^{80}-1$). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you have that $a^{80}\equiv 1\pmod{561}$, then any power of $a^{80}$ is also congruent to $1$ modulo $561$. In particular,&#xA;$$a^{560} = (a^{80})^{7} \equiv 1^7 = 1 \pmod{561}$$&#xA;as desired.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-17T14:42:17.937" LastActivityDate="2011-02-17T14:42:17.937" CommentCount="3" />
  <row Id="22463" PostTypeId="1" AcceptedAnswerId="22467" CreationDate="2011-02-17T07:36:28.123" Score="2" ViewCount="113" Body="&lt;p&gt;Hello,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a BNF defined as follow:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;&amp;lt;S&amp;gt; -&amp;gt; 0&#xA;&amp;lt;S&amp;gt; -&amp;gt; 1&#xA;&amp;lt;S&amp;gt; -&amp;gt; &amp;lt;S&amp;gt;&amp;lt;S&amp;gt;&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I think this grammar is not ambiguous, but the solution was 'yes'. Any idea?&lt;br&gt;&#xA;The book solution is odd to me, because I think it is the same as the one above.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;&amp;lt;S&amp;gt;-&amp;gt; 0|1|&amp;lt;S&amp;gt;&amp;lt;S&amp;gt;&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Thanks,&lt;br&gt;&#xA;Chan&lt;/p&gt;&#xA;" OwnerUserId="5711" LastActivityDate="2011-02-17T08:35:21.190" Title="Is this BNF grammar ambiguous?" Tags="&lt;computer-science&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="22464" PostTypeId="2" ParentId="22463" CreationDate="2011-02-17T07:48:20.587" Score="2" ViewCount="" Body="&lt;p&gt;Consider the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$S \to S S \to S S S \to 1 S S \to 1 1 S \to 1 1 1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$S \to S S \to 1 S \to 1 S S \to 1 1 S\to 1 1 1$&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2011-02-17T08:33:51.480" LastActivityDate="2011-02-17T08:33:51.480" CommentCount="12" />
  <row Id="22465" PostTypeId="1" AcceptedAnswerId="22470" CreationDate="2011-02-17T08:24:18.530" Score="1" ViewCount="157" Body="&lt;p&gt;Show that among any 52 integers, there are two whose sum or difference is divisible by 100.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume all the integers are placed into boxes based on their remainders, then there must be one box y that contains two integers. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If, (a - b) = (100x + y) - (100z + y), where x,z are integers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;= 100(x - z), which means 100*integer = (a - b).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this valid? I did some other examples, but the number of given integers was odd, which meant that there must be two integers in one box...so when given 52 that changes things up and I am wondering how that affects the problem and how I solve it? Any further explanation would be helpful. Thanks!&lt;/p&gt;&#xA;" OwnerUserId="7144" LastEditorUserId="7144" LastEditorDisplayName="" LastEditDate="2011-02-17T08:40:55.067" LastActivityDate="2011-02-17T15:39:19.550" Title="Proving by PigeonHole Principle?" Tags="&lt;discrete-mathematics&gt;&lt;proof-strategy&gt;" AnswerCount="2" FavoriteCount="2" />
  <row Id="22466" PostTypeId="1" CreationDate="2011-02-17T08:33:12.083" Score="1" ViewCount="94" Body="&lt;p&gt;Assume R is commutative. Show that an R-module M is irreducible if and only if M is isomorphic (as an R-Module) to R/I where I is a maximal ideal of R. &#xA;If I remove the hypothesis that R is communtative, is it result still true?&lt;/p&gt;&#xA;" OwnerUserId="2177" LastActivityDate="2011-05-27T11:03:04.940" Title="A question about module" Tags="&lt;modules&gt;" AnswerCount="1" FavoriteCount="0" />
  <row Id="22467" PostTypeId="2" ParentId="22463" CreationDate="2011-02-17T08:35:21.190" Score="5" ViewCount="" Body="&lt;p&gt;A general technique to check for ambiguity is this. Transform the grammar in an equation system:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$S(x) = S(x)^2 + 2x$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solving for $S(x)$ yields $S(x) = \frac{1}{2}(1 - \sqrt{1 - 8x})$ which is the ordinary generating function enumerating the grammar's left-derivation trees.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The generated language is $\{0,1\}^+$ which has the generating function $\frac{1}{1-2x} - 1$. This is obviously different from the above, therefore there are more left-derivation trees (for words of length $n$) than words (of length $n$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This technique goes back to Chomsky and Schützenberger, 1963.&lt;/p&gt;&#xA;" OwnerUserId="3330" LastActivityDate="2011-02-17T08:35:21.190" CommentCount="4" />
  <row Id="22468" PostTypeId="1" AcceptedAnswerId="22529" CreationDate="2011-02-17T08:37:50.047" Score="3" ViewCount="74" Body="&lt;p&gt;This is asserted in Exercise 1.15 of Joe Harris's algebraic geometry book (&lt;em&gt;Algebraic Geometry: A First Course&lt;/em&gt;, Pg. 11 in my copy). This result struck my fancy but I'm unable to solve it myself or find references to it elsewhere. The closest reference I've found is the following document, which asserts something weaker: &lt;a href=&quot;http://www.mast.queensu.ca/~tony/kn+1.ps&quot; rel=&quot;nofollow&quot;&gt;http://www.mast.queensu.ca/~tony/kn+1.ps&lt;/a&gt; Does anyone know of any references or see a solution off the top of their head? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(A rational normal curve is any curve equivalent to the Veronese image $v_n(\mathbb{P^1}) \subset \mathbb{P}^n$.)&lt;/p&gt;&#xA;" OwnerUserId="251" LastActivityDate="2011-04-07T09:45:48.220" Title="A polynomial of degree $k$ vanishing at $kd+1$ points on a rational normal curve in $\mathbb{P}^d$ must vanish on the whole curve" Tags="&lt;algebraic-geometry&gt;" AnswerCount="2" />
  <row Id="22469" PostTypeId="2" ParentId="22465" CreationDate="2011-02-17T08:38:03.520" Score="4" ViewCount="" Body="&lt;p&gt;It is not valid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have 100 boxes and 52 integers. Pigeon hole does not apply with the integers being the pigeons...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a proof which keeps the number of pigeons the same:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the &quot;boxes&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(0,100), \ (1,99), \ (2, 98), \dots, (50,50)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are $52$ pigeons (the remainders of the integers modulo 100) and $51$ boxes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At least one box must have two integers. If both are $y$, then their difference is divisible by 100. If one is $y$ and the other $100-y$, then their sum is divisible.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2011-02-17T15:39:19.550" LastActivityDate="2011-02-17T15:39:19.550" CommentCount="8" />
  <row Id="22470" PostTypeId="2" ParentId="22465" CreationDate="2011-02-17T08:39:38.753" Score="4" ViewCount="" Body="&lt;p&gt;Here are soms hints that should lead you to the answer:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Consider the set $S$ with these numbers: $a_1,a_2,...,a_{51},a_{52},-a_{1},-a_{2},...,-a_{51},-a_{52}$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;There are 104 numbers to two of them have the same remainder, note that the problem is solved unless these numbers are $a_i$ and $-a_i$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;But if $a_i \equiv -a_i$ then 50 divides $a_i$. If there are two $a_i$ that are divisible by $50$ then their sum (and difference) is divisible by 100. Therefore we may assume there is only one such number.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;So one can leave $a_i$ and $-a_i$ out of the set of 104 numbers, there are 102 numbers remaining. The same argument as before holds, but this time it cannot occur that the numbers found are $a_i$ and $-a_i$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="5189" LastActivityDate="2011-02-17T08:39:38.753" CommentCount="4" />
  <row Id="22471" PostTypeId="2" ParentId="22466" CreationDate="2011-02-17T08:43:48.980" Score="2" ViewCount="" Body="&lt;p&gt;Yes, this is true, if you replace ideal by &lt;strong&gt;right ideal&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See &lt;a href=&quot;http://en.wikipedia.org/wiki/Irreducible_module#cite_ref-0&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Irreducible_module#cite_ref-0&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="367" LastActivityDate="2011-02-17T08:43:48.980" />
  <row Id="22472" PostTypeId="1" AcceptedAnswerId="22484" CreationDate="2011-02-17T09:19:49.030" Score="6" ViewCount="142" Body="&lt;p&gt;hi all&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This seems to be a fairly easy question&#xA;but I'm looking for new points of view on it&#xA;and was wondering if anyone might be able to help&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(by the way- this question does come from home-work,&#xA;but i've already solved and handed it, and i'm posting this out of interest, so no HW tag)&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Let $B_n=B(x_n,r_n)$ be a sequence of nested closed balls in a banach space $X$ prove, that $\bigcap_1^\infty B_n\neq\emptyset$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As I said before, it should be rather simple. When the radii decrease to 0, it's just a matter of selecting any sequence of points in $B_n$, and it must be Cauchy- and the limit is in the intersection.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is what to do when the radii do not decrease to 0? I got some tips about multiplying the balls by a sequence of decreasing scalars, or reducing the radii so that they decrease to 0, but found too many pathological cases for both methods.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally- I used a geometric arguemnt (which i've shown to work in any normed space) that if $B(x_1,r_1)\subset B(x_2,r_2)$ then $\| x_1-x_2\|\leq|r_1-r_2|$.&#xA;This turned out to be some kind of technical catastrophe, but it worked...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Still, if anyone knows of a more elegant solution, I'd love to hear about it&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="7129" LastEditorUserId="7129" LastEditorDisplayName="" LastEditDate="2011-02-17T10:12:47.957" LastActivityDate="2011-02-20T04:53:56.403" Title="Nested sequences of balls in a Banach space" Tags="&lt;geometry&gt;&lt;functional-analysis&gt;&lt;banach-spaces&gt;" AnswerCount="1" CommentCount="6" FavoriteCount="2" />
  <row Id="22473" PostTypeId="1" AcceptedAnswerId="22505" CreationDate="2011-02-17T09:22:47.313" Score="4" ViewCount="149" Body="&lt;p&gt;Definition: F is a function iff F is a relation and $(x,y) \in F$ and $(x,z) \in F \implies y=z$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm reading Introduction to Set Theory by Monk, J. Donald (James Donald), 1930 and i came across a theorem 4.10.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Theorem 4.10  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(ii)$0:0 \to A$, if $F : 0 \to A$, then $F=0$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(iii) If $F:A\to 0$, then $A=F=0$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where the book just explain the concept of function and now is stating its function property. I am stuck on what actually does it mean and how to prove it. May be can give me a hint.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks ahead.  &lt;/p&gt;&#xA;" OwnerUserId="1683" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-17T16:36:18.123" LastActivityDate="2011-02-17T16:39:43.577" Title="How to Prove the following Function Properties." Tags="&lt;elementary-set-theory&gt;" AnswerCount="2" CommentCount="5" FavoriteCount="1" />
  <row Id="22474" PostTypeId="1" AcceptedAnswerId="22476" CreationDate="2011-02-17T09:41:48.430" Score="1" ViewCount="135" Body="&lt;p&gt;How would I go about calculating the total number of possible triangles where the perimeter is given, and all sides are integers?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Exempli gratia:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;a + b + c = 15&lt;/li&gt;&#xA;&lt;li&gt;d + e + f = 16&lt;/li&gt;&#xA;&lt;li&gt;x + y + z = 100&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="4823" LastActivityDate="2011-02-18T04:13:18.460" Title="How to calculate the total number of possible triangles where the perimeter is given, and all sides are integers?" Tags="&lt;geometry&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="22475" PostTypeId="1" CreationDate="2011-02-17T10:12:46.473" Score="0" ViewCount="38" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;why a torsion free flat module over a discrete valuation ring is Cohen-Macaulay?&lt;/p&gt;&#xA;" OwnerUserId="6583" LastActivityDate="2011-02-17T10:12:46.473" Title="cohen-macaulay implication" Tags="&lt;algebraic-geometry&gt;&lt;arithmetic-geometry&gt;" CommentCount="3" />
  <row Id="22476" PostTypeId="2" ParentId="22474" CreationDate="2011-02-17T10:13:15.663" Score="1" ViewCount="" Body="&lt;p&gt;It rather depends on whether you regard 7,5,4 as the same triangle as 5,7,4 (edges in a different order), and whether you allow the triangles 8,8,0 (with a zero edge) or 8,5,3 (with a zero area).  Let's suppose not in what follows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So start with the longest side: it cannot be longer than or equal to half the perimeter, but it must be at least a third of the perimeter.  So that gives you a limited set of values. Now consider the middle sized side: it must be at least half the difference between the perimeter and the longest side, but cannot be longer than the longest side. Together these will tell you the shortest side.  So just add up the number of different shortest sides for each possible longest side.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example with a perimeter of 15, the longest side must be 5, 6 or 7.  If it is 5 then the middle side can be 5; if it is 6 then the middle side can be 5 or 6; if it is 7 then the middle side can be 4, 5, 6 or 7.  Adding those up gives 1+2+4 = 7.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similar results can be found for other small perimeters. It is also possible to derive a general formula which has the curious property that $a(2n-3) = a(2n)$, or in other words, starting with a triangle with an odd perimeter, we can find a related triangle with a perimeter 3 more just by adding 1 to each side.&lt;/p&gt;&#xA;" OwnerUserId="6460" LastActivityDate="2011-02-17T10:13:15.663" CommentCount="3" />
  <row Id="22477" PostTypeId="1" CreationDate="2011-02-17T10:48:03.367" Score="1" ViewCount="135" Body="&lt;p&gt;I created a certain algorithm and now I try to prove it's convergence.&#xA;Here is the essence of the algorithm:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given are finite sets $A_i$, positive constants $\alpha_a$, and a sequence $     \vec{\delta}^{(k)} \in \mathbb{R}^n$:&#xA;$$ A_1,\; \ldots,\; A_n \subset \mathcal{A} $$&#xA;$$ B_a = \{i : a \in A_i\}$$&#xA;$$ \sum_{a\in \mathcal{A}}\alpha_a = n, \quad 0 &amp;lt; \alpha_a &amp;lt; |B_a| $$&#xA;$$\delta_i^{(0)} = 1 \quad i = 1,2,\; \ldots, \;n$$&#xA;$$ \beta_a^{(k)} = \sum_{i \in B_a} \delta_i^{(k)} $$&#xA;$$ \delta_i^{(k+1)} = \left(\sum_{a\in A_i} \frac{\alpha_a}{\beta_a^{(k)}}\right)^{-1} $$&#xA;Prove that $\vec{\delta}^{(k)}$ converges.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I did some numerical experiments but I am not sure that the bounds on $\alpha_a$ are sufficient. You can make stronger assumptions if you want.&#xA;I'm just lookgin for an idea how to prove this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt;&#xA;This may help: in experiments this function is always increases:&#xA;$$ \sum_{a \in A} \alpha_a \log\frac{\alpha_a}{\beta_a} - \sum^n_{i=1} \log \sum_{a\in A_i} \frac{\alpha_a}{\beta_a}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit 2:&lt;/strong&gt;&#xA;You can assume $|A_i| \geq 2 $&lt;/p&gt;&#xA;" OwnerUserId="5552" LastEditorUserId="5552" LastEditorDisplayName="" LastEditDate="2011-02-17T12:46:04.887" LastActivityDate="2011-02-17T12:46:04.887" Title="How to prove that this sequence converges?" Tags="&lt;sequences-and-series&gt;&lt;algorithms&gt;&lt;convergence&gt;" AnswerCount="1" CommentCount="9" />
  <row Id="22478" PostTypeId="1" AcceptedAnswerId="22525" CreationDate="2011-02-17T11:05:16.003" Score="0" ViewCount="248" Body="&lt;p&gt;Philosophical questions concerning the difference between equality, isomorphism, equality upto (unique) isomorphism, undistinguishability, and the like are not very popular among practicing mathematicians: there's a difference between equality upto isomorphism and equality upto unique isomorphism, and that's it (not to forget about isomorphism and natural isomorphism).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But personally, I'm not totally satisfied with this stance when looking at truly abstract structures like unlabelled graphs (finite or infinite, countable or uncountable), conceived as nothing-but-dots-and-arrows. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Are two abstract structures not to be considered&#xA;  &lt;strong&gt;equal&lt;/strong&gt; in the strongest sense - being one and the same - as soon&#xA;  as there is an isomorphism between&#xA;  them, regardless of being unique, natural, and/or&#xA;  not?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The existence of isomorphisms in turn tells us something about the symmetries of the abstract structure, but of one and only one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe there are no such abstract structures per se, but only concrete structures (models) and/or concrete presentations of them (like adjacency matrices). Then the question misses a subject. But if so?&lt;/p&gt;&#xA;" OwnerUserId="1792" LastEditorUserId="1792" LastEditorDisplayName="" LastEditDate="2011-02-21T20:21:56.710" LastActivityDate="2011-02-21T20:21:56.710" Title="Equality of abstract structures" Tags="&lt;set-theory&gt;&lt;graph-theory&gt;&lt;philosophy&gt;" AnswerCount="1" CommentCount="23" />
  <row Id="22479" PostTypeId="1" AcceptedAnswerId="22481" CreationDate="2011-02-17T11:13:06.827" Score="2" ViewCount="140" Body="&lt;p&gt;Hello,&#xA;I have an equation like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$te^{t} = \int_0^t e^\tau u(\tau)d\tau$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't really know how to solve it.. Would it be possible to differentiate both sides of the equation? If so, how can I do it - is it like differentiate one side and then the other or is it more complex? And if it is possible what conditions need to be satisfied in order to do it?&lt;/p&gt;&#xA;" OwnerUserId="6485" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2011-02-17T11:16:29.180" LastActivityDate="2011-02-17T12:57:37.590" Title="Equation with a definite integral - can I differentiate it?" Tags="&lt;integral&gt;&lt;differential&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="22480" PostTypeId="2" ParentId="22407" CreationDate="2011-02-17T11:13:44.860" Score="1" ViewCount="" Body="&lt;p&gt;Independence does not imply conditional independence: for instance, independent random variables are rarely independent conditionally on their sum or on their maximum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conditional independence does not imply independence: for instance, conditionally independent random variables uniform on $(0,u)$ where $u$ is uniform on $(0,1)$ are not independent.&lt;/p&gt;&#xA;" OwnerUserId="6179" LastEditorUserId="6179" LastEditorDisplayName="" LastEditDate="2011-02-17T11:24:28.300" LastActivityDate="2011-02-17T11:24:28.300" />
  <row Id="22481" PostTypeId="2" ParentId="22479" CreationDate="2011-02-17T11:33:07.257" Score="4" ViewCount="" Body="&lt;p&gt;If we differentiate the left hand side we get $e^t+t e^t$, if we differentiate the right hand side we get $e^t u(t)$ by the fundamental theorem of calculus. Then we see&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$u(t)=t+1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An we are done. I think we need to demand that $u(t)$ is continuous to get uniqueness. In general $u(t)=t+1$ almost everywhere.&lt;/p&gt;&#xA;" OwnerUserId="5882" LastEditorUserId="5882" LastEditorDisplayName="" LastEditDate="2011-02-17T12:57:37.590" LastActivityDate="2011-02-17T12:57:37.590" CommentCount="4" />
  <row Id="22482" PostTypeId="2" ParentId="22436" CreationDate="2011-02-17T11:44:48.013" Score="1" ViewCount="" Body="&lt;p&gt;You can change the $(p,q,r)$ pretzel link into the $(r,q,p)$ pretzel link by a 180 degree rotation around the midpoint of the central twist box (the one containing $q$ twists).   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is also a way to change the $(p,q,r)$ link into the $(q,r,p)$ link.  This is a bit three-dimensional, so takes a bit more work to see.  Draw the diagram on a sphere $S$ in $R^3$ instead of on a plane in $R^3$.  Now you can drag the twist box containing $r$ twists around the back of $S$ (&quot;past infinity&quot;) until it becomes the first twist box.  (The arcs connecting the twist boxes rearrange themselves in the most convenient way!)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Composing these two operations correctly will move you between the two diagrams.  To get a sequence of Reidemeister moves, project all of these movements to a fixed plane. &lt;/p&gt;&#xA;" OwnerUserId="1307" LastActivityDate="2011-02-17T11:44:48.013" />
  <row Id="22483" PostTypeId="1" AcceptedAnswerId="22515" CreationDate="2011-02-17T12:07:53.460" Score="0" ViewCount="43" Body="&lt;p&gt;I have:&lt;br&gt;&#xA;A1: {3, 5, 6} and A2: {3, 4, 6, 8}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Avg A1: 4.(6)&lt;br&gt;&#xA;Std.Deviation A1: 1.247 &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Avg A2: 5.25&lt;br&gt;&#xA;Std. Deviation A2: 1.92  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Merging A1 and A2, we get&lt;br&gt;&#xA;A12: {3, 3, 4, 5, 6, 6, 8}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Avg A12: &lt;strong&gt;5&lt;/strong&gt;&lt;br&gt;&#xA;Std. Deviation A12: &lt;strong&gt;1.69&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to be able to obtain those values (avg and std. deviation for A12), having only the following information about A1 and A2:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;A1&lt;/strong&gt;&lt;br&gt;&#xA;No. Elements: 3&lt;br&gt;&#xA;Avg A1: 4.(6)&lt;br&gt;&#xA;Std.Deviation A1: 1.247 &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;A2&lt;/strong&gt;&lt;br&gt;&#xA;No. Elements: 4&lt;br&gt;&#xA;Avg A2: 5.25&lt;br&gt;&#xA;Std.Deviation A2: 1.92 &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Average is easy, &#xA;(4.(6)*3 + 5.25*4)/7 = &lt;strong&gt;5&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I seem to be unable to calculate the correct std. deviation value.&#xA;I can only use the above information.&#xA;Any ideas?&lt;/p&gt;&#xA;" OwnerUserId="7182" LastEditorUserId="7182" LastEditorDisplayName="" LastEditDate="2011-02-17T14:30:37.603" LastActivityDate="2011-02-17T15:20:30.273" Title="Calculate standard deviation of merged sets with limited information (full details inside)" Tags="&lt;standard-deviation&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="22484" PostTypeId="2" ParentId="22472" CreationDate="2011-02-17T12:08:46.183" Score="4" ViewCount="" Body="&lt;p&gt;I don't know if this is more elegant, but that's about the best I can come up with at the moment and probably essentially the same as your argument.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Consider first the situation $B_{\leq r}(x) \subset B_{\leq s}(y)$. It is easy to see that $r \leq s$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Claim.&lt;/em&gt; $\|y - x\| \leq s - r$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Proof.&lt;/em&gt; If $x = y$ there is nothing to prove, so let's assume $x \neq y$. The point $z = x - r \frac{y-x}{\|y - x\|}$ belongs to $B_{\leq r}(x)$ and hence also to $B_{\leq s}(y)$. Therefore $\|y - z\| \leq s$. On the other hand,&#xA;\[&#xA;y - z = y - x + \frac{r}{\|y - x\|} (y - x) = \underbrace{\left(1 + \frac{r}{\|y - x\|}\right)}_{\lambda} (y - x),&#xA;\]&#xA;so $s \geq \lambda \|y - x\| = \|y - x\| + r$ and hence $\|y - x\| \leq s - r$. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;This means that a nested sequence of closed balls $B_{\leq r_{n}}(x_{n})$ has the following properties:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The sequence $r_{n}$ is monotonically decreasing, hence converges to some $r$.&lt;/li&gt;&#xA;&lt;li&gt;If $N$ is such that $r_{N} \leq r + \varepsilon$ then the above claim implies that for all $n\geq m \geq N$ we have $r_m - r_n \leq \varepsilon$, so $\|x_{m} - x_{n}\| \leq \varepsilon$ because $B_{\leq r_{n}}(x_{n}) \subset B_{\leq r_{m}}(x_{m})$.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;In other words, the centers $x_{n}$ form a Cauchy sequence and their limit point $x$ must belong to $\bigcap_{n = 1}^{\infty} B_{\leq r_{n}}(x_{n})$.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added:&lt;/strong&gt; As Jonas pointed out, the argument can be made even simpler and doesn't need completeness: Suppose $r_{n} \to r \gt 0$. Then there is $N$ such that $r_{N} \leq 2r$. Then for all $n \geq N$ we have $r \leq r_{n} \leq r_{N} \leq 2r$, so $r_{N} - r_{n} \leq r$ and the claim implies that $\|x_{n} - x_{N}\| \leq r \leq r_{n}$, so $x_{N} \in \bigcap_{n = 1}^{\infty} B_{\leq r_{n}} (x_{n})$.&lt;/p&gt;&#xA;" OwnerUserId="5363" LastEditorUserId="5363" LastEditorDisplayName="" LastEditDate="2011-02-20T04:53:56.403" LastActivityDate="2011-02-20T04:53:56.403" CommentCount="3" />
  <row Id="22485" PostTypeId="1" AcceptedAnswerId="22490" CreationDate="2011-02-17T12:10:53.067" Score="0" ViewCount="68" Body="&lt;p&gt;Suppose I have a singular matrix given by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$A = \begin{pmatrix}&#xD;&#xA;a_{11} &amp;amp; a_{12} &amp;amp; a_{13} &amp;amp; a_{14}\\&#xD;&#xA;a_{12} &amp;amp; a_{11} &amp;amp; a_{14} &amp;amp; a_{13}\\&#xD;&#xA;a_{31} &amp;amp; a_{32} &amp;amp; a_{33} &amp;amp; a_{34}\\&#xD;&#xA;a_{32} &amp;amp; a_{31} &amp;amp; a_{34} &amp;amp; a_{33}&#xD;&#xA;\end{pmatrix}$$&#xA;Which is a homogeneous system in $(x_1, x_2, x_3, x_4)^T$ where there is a variable $\Lambda$ in the coefficients that makes the matrix singular if chosen such that the determinant is zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I have the matrices:&#xA;$$B = \begin{pmatrix}&#xD;&#xA;a_{11} + a_{12} &amp;amp; a_{13} + a_{14}\\&#xD;&#xA;a_{31} + a_{32} &amp;amp; a_{33} + a_{34}&#xD;&#xA;\end{pmatrix}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$C = \begin{pmatrix}&#xD;&#xA;a_{11} - a_{12} &amp;amp; a_{13} - a_{14}\\&#xD;&#xA;a_{31} - a_{32} &amp;amp; a_{33} - a_{34}&#xD;&#xA;\end{pmatrix}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which are homogeneous systems in $(x_1 + x_2, x_3 + x_4)^T$ and $(x_1 - x_2, x_3 - x_4)^T$ respectively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I make the matrices $B$ and $C$ out of $A$, and I determine the $\Lambda$ such that those determinants are zero, can I say anything about the determinant of $A$ with those values of $\Lambda$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The idea is that I won't have a fourth degree polynomial to solve (which Maple or Mathematica can seem to do in this case) and then I can obtain the original result by summing or subtracting and such.&lt;/p&gt;&#xA;" OwnerUserId="1120" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2011-05-12T10:16:39.017" LastActivityDate="2011-05-12T10:16:39.017" Title="Singular matrix." Tags="&lt;linear-algebra&gt;&lt;matrices&gt;" AnswerCount="1" />
  <row Id="22486" PostTypeId="1" CreationDate="2011-02-17T12:30:17.410" Score="1" ViewCount="79" Body="&lt;p&gt;I'm searching for a formula to create &lt;a href=&quot;http://oeis.org/A057537&quot; rel=&quot;nofollow&quot;&gt;http://oeis.org/A057537&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="7184" LastActivityDate="2011-02-17T13:05:25.077" Title="Formula for oeis A057537" Tags="&lt;sequences-and-series&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="22487" PostTypeId="2" ParentId="22477" CreationDate="2011-02-17T12:32:16.503" Score="1" ViewCount="" Body="&lt;p&gt;It's not quite clear what you mean by &quot;$\vec{\delta}^{(k)}$ converges to a unique point&quot;, since you've specified initial values, so if $\vec{\delta}^{(k)}$ converges this will by definition be to a unique point. I assume that you mean that it converges to a point which is independent of the initial values? If so, that is not the case, since multiplying the initial values by a constant factor results in the entire sequence being multiplied by a constant factor, so the limit could only be &quot;unique&quot; in the above sense up to a constant factor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems that a useful way to approach these equations is to eliminate $\beta_a^{(k)}$ by substituting it from the upper iteration step into the lower iteration step, and then bringing the reciprocal on the right-hand side over to the left-hand side, yielding&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{a\in A_i} \frac{\alpha_a \delta_i^{(k+1)}}{\sum_{j\in B_a} \delta_j^{(k)}} = 1\;.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we drop the superscripts labeling the iterations steps, we get the equation that determines the fix point(s). Note a) that the sum of these equations over all $i$ is automatically satisfied, and b) all $n$ equations would be satisfied if $\sum_{j\in B_a} \delta_j$ were independent of $a$. So the iteration can be viewed as solving the problem of choosing weights $\delta_i$ for the sets such that for each point $a$ the sum over the weights of the sets in which it is contained is the same. This also makes sense of the fact that a constant factor in the weights is irrelevant.&lt;/p&gt;&#xA;" OwnerUserId="6622" LastActivityDate="2011-02-17T12:32:16.503" CommentCount="3" />
  <row Id="22489" PostTypeId="1" AcceptedAnswerId="22492" CreationDate="2011-02-17T12:37:19.217" Score="1" ViewCount="241" Body="&lt;p&gt;For the level set method, $\phi(\vec{x},t)$ is the level set function in 3D and the level set $\phi(\vec{x},t) = 0$ forms the interface. For evolving $\phi$ the derivation says to imagine a particle $\textbf{x}(t)$ on the surface, then we differentiate with respect to t&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{d}{dt}(\phi(\vec{x}, t)=0)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using chain rule we get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{\partial \phi}{\partial t} + \triangledown \phi \cdot \frac{d\vec{x}}{dt} = 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then this equation is turned into&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{\partial \phi}{\partial t} + \triangledown \phi \cdot \vec{V} = 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I guess $\vec{V}$ is the speed, and its elements are to replace $dx$, $dy$, .. that are elements of $d{\vec{x}}$. Hence we can substitute our own displacement and receive a result. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I do not understand is they continue like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Seperate $\vec{V}$ into normal and tangential components&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{\partial \phi}{\partial t} + \triangledown \phi \cdot (\vec{V_N}\vec{N} + \vec{V_T}\vec{T}) = 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then since $\vec{N} = \frac{\triangledown \phi}{|\triangledown \phi |}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{\partial \phi}{\partial t} + V_N \cdot |\triangledown \phi| = 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What happened to $V_T$? Also how can the dot product of $\triangledown \phi$ with  $\frac{\triangledown \phi}{|\triangledown \phi |}$ result in $|\triangledown \phi|$ ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Link: &lt;a href=&quot;http://www.cs.au.dk/~bang/smokeandwater2006/Lecture9_IntroToWaterAndLS.ppt&quot; rel=&quot;nofollow&quot;&gt;http://www.cs.au.dk/~bang/smokeandwater2006/Lecture9_IntroToWaterAndLS.ppt&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="6786" LastActivityDate="2011-02-17T13:08:37.800" Title="Derivation of Basic Level Set Equations" Tags="&lt;calculus&gt;&lt;differential-geometry&gt;&lt;vector-calculus&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="22490" PostTypeId="2" ParentId="22485" CreationDate="2011-02-17T13:03:19.210" Score="4" ViewCount="" Body="&lt;p&gt;Yes. If the $a_{44}$ in the lower right corner of your $A$ is a typo (which I am pretty sure about) and should be an $a_{33}$ instead, then $\det A=\det B\cdot \det C$. There are some different ways to see this; here is one:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your matrix $A$ can be written as the block matrix $\left(\begin{array}{cc} X&amp;amp;Y\\ U&amp;amp;W\end{array}\right)$, where $X$, $Y$, $U$, $W$ are the following $2\times 2$ matrices:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$X=\left(\begin{array}{cc} a_{11}&amp;amp;a_{12}\\ a_{12}&amp;amp;a_{11}\end{array}\right)$;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$Y=\left(\begin{array}{cc} a_{13}&amp;amp;a_{14}\\ a_{14}&amp;amp;a_{13}\end{array}\right)$;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$Z=\left(\begin{array}{cc} a_{31}&amp;amp;a_{32}\\ a_{32}&amp;amp;a_{33}\end{array}\right)$;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$W=\left(\begin{array}{cc} a_{33}&amp;amp;a_{34}\\ a_{34}&amp;amp;a_{33}\end{array}\right)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, these matrices $X$, $Y$, $U$, $W$ are &lt;a href=&quot;http://en.wikipedia.org/wiki/Circulant_matrix&quot; rel=&quot;nofollow&quot;&gt;circulant matrices&lt;/a&gt;, and thus can be diagonalized by the unitary discrete Fourier transform matrix&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$F_2=\frac{1}{\sqrt 2}\left(\begin{array}{cc} 1&amp;amp;1\\ 1&amp;amp;-1\end{array}\right)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$X=F_2\mathrm{diag}\left(a_{11}+a_{12},a_{11}-a_{12}\right)F_2^{-1}$;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$Y=F_2\mathrm{diag}\left(a_{13}+a_{14},a_{13}-a_{14}\right)F_2^{-1}$;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$Z=F_2\mathrm{diag}\left(a_{31}+a_{32},a_{31}-a_{32}\right)F_2^{-1}$;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$W=F_2\mathrm{diag}\left(a_{33}+a_{34},a_{33}-a_{34}\right)F_2^{-1}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a consequence, the block matrix $A=\left(\begin{array}{cc} X&amp;amp;Y\\ U&amp;amp;W\end{array}\right)$ can be written as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$A=\left(\begin{array}{cc} F_2&amp;amp;0\\ 0&amp;amp;F_2\end{array}\right)\left(\begin{array}{cc} \mathrm{diag}\left(a_{11}+a_{12},a_{11}-a_{12}\right) &amp;amp; \mathrm{diag}\left(a_{13}+a_{14},a_{13}-a_{14}\right) \\ \mathrm{diag}\left(a_{31}+a_{32},a_{31}-a_{32}\right) &amp;amp; \mathrm{diag}\left(a_{33}+a_{34},a_{33}-a_{34}\right) \end{array}\right) \left(\begin{array}{cc} F_2&amp;amp;0\\ 0&amp;amp;F_2\end{array}\right)^{-1}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(check this!), so that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\det A = \det \left(\begin{array}{cc} \mathrm{diag}\left(a_{11}+a_{12},a_{11}-a_{12}\right) &amp;amp; \mathrm{diag}\left(a_{13}+a_{14},a_{13}-a_{14}\right) \\ \mathrm{diag}\left(a_{31}+a_{32},a_{31}-a_{32}\right) &amp;amp; \mathrm{diag}\left(a_{33}+a_{34},a_{33}-a_{34}\right) \end{array}\right) $.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the determinant on the right hand side can be even simplified by transposing the second row with the third row and transposing the second column with the third column:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\det A = \det \left(\begin{array}{cccc} a_{11}+a_{12} &amp;amp; a_{13}+a_{14} &amp;amp; 0 &amp;amp; 0 \\ a_{31}+a_{32} &amp;amp; a_{33}+a_{34} &amp;amp; 0 &amp;amp; 0 \\ 0 &amp;amp; 0 &amp;amp; a_{11}-a_{12} &amp;amp; a_{13}-a_{14} \\ 0 &amp;amp; 0 &amp;amp; a_{31}-a_{32} &amp;amp; a_{33}-a_{34} \end{array}\right)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the matrix on the right hand side is obviously just the block matrix $\left(\begin{array}{cc} B&amp;amp;0\\ 0&amp;amp;C\end{array}\right)$, so its determinant is $\det B\cdot \det C$.&lt;/p&gt;&#xA;" OwnerUserId="586" LastEditorUserId="586" LastEditorDisplayName="" LastEditDate="2011-02-17T13:08:29.517" LastActivityDate="2011-02-17T13:08:29.517" CommentCount="1" />
  <row Id="22491" PostTypeId="2" ParentId="22486" CreationDate="2011-02-17T13:05:25.077" Score="1" ViewCount="" Body="&lt;p&gt;I would use a recursion.  With [the coins and notes] $c(1)$, $c(2)$, $c(3)$,... $c(m)$ in order, smallest first, I would start with &#xA;$$b(i,j) = 0 \textrm{ for } j&amp;lt;0 \textrm{ or } i&amp;lt;0$$&#xA;$$b(i,0) = 1 $$&#xA;$$b(0,j) = 0 \textrm{ for } j&amp;gt;0$$&#xA;$$b(i,j) = b(i-1,j)+b(i,j-c(j)) \textrm{ for } i,j&amp;gt;0$$&#xA;then $b(i,j)$ is the number of ways of making $i$ where the largest part is no bigger than $c(j)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So A057537($n$) and similar sequences are simply $b(n,m)$.  If the $c(j)$ represent a very large or infinite collection, then you only need to take $m$ &lt;em&gt;large enough&lt;/em&gt;, i.e. so $c(m+1)$ is greater than $n$. &lt;/p&gt;&#xA;" OwnerUserId="6460" LastActivityDate="2011-02-17T13:05:25.077" />
  <row Id="22492" PostTypeId="2" ParentId="22489" CreationDate="2011-02-17T13:08:37.800" Score="0" ViewCount="" Body="&lt;p&gt;Since $\nabla \phi$ is normal to the surface, its inner product with a vector tangential to the surface is zero. Furthermore $\nabla \phi \cdot \nabla \phi=|\nabla\phi|^2$, hence the second identity: $$\nabla\phi\cdot\frac{\nabla\phi}{|\nabla\phi|}=\frac{|\nabla\phi|^2}{|&#xD;&#xA;\nabla\phi|}=|\nabla\phi|$$.&lt;/p&gt;&#xA;" OwnerUserId="5882" LastActivityDate="2011-02-17T13:08:37.800" />
  <row Id="22493" PostTypeId="1" CreationDate="2011-02-17T13:34:27.673" Score="3" ViewCount="168" Body="&lt;p&gt;&lt;a href=&quot;http://mathoverflow.net/questions/25983/intuitive-crutches-for-higher-dimensional-thinking/26095#26095&quot; rel=&quot;nofollow&quot;&gt;On mathoverflow&lt;/a&gt;, Terry Tao says the following:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;For instance, one can view a high-dimensional vector space as a state space for a system with many degrees of freedom. A megapixel image, for instance, is a point in a million-dimensional vector space; by varying the image, one can explore the space, and various subsets of this space correspond to various classes of images.&#xA;&lt;br /&gt;&lt;br /&gt;&#xA;One can similarly interpret sound waves, a box of gases, an ecosystem, a voting population, a stream of digital data, trials of random variables, the results of a statistical survey, a probabilistic strategy in a two-player game, and many other concrete objects as states in a high-dimensional vector space, and various basic concepts such as convexity, distance, linearity, change of variables, orthogonality, or inner product can have very natural meanings in some of these models (though not in all).&#xA;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This paragraph is saying something that helps a little but I can't quite grasp it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The picture as a million-dimensional space.. &lt;strong&gt;why&lt;/strong&gt; would you consider it $10^6$-space and not simply 5-space (with x,y,R,G,B as the basis vectors).  Shouldn't you go for the simpler representation?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand that you can define orthogonality this way, by the inner product of two $10^6$ pictures equalling 0.  But what does it mean for 2 pictures to be orthogonal, and why would you want to define that?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why do we want to use multidimensional space to handle something like the Fourier Transform for example.  A soundwave.  You can decompose it into 22050-dimensional space using a 22050 point FFT, and MATLAB seems to have a rocking good time with that, and filters work.  But mathematically I'm unsure of the reason.&lt;/p&gt;&#xA;" OwnerUserId="1069" LastEditorUserId="1069" LastEditorDisplayName="" LastEditDate="2011-02-17T16:46:32.650" LastActivityDate="2011-02-17T16:46:32.650" Title="Why do we use n-dimensional spaces?" Tags="&lt;vector-space&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="3" />
  <row Id="22494" PostTypeId="2" ParentId="22291" CreationDate="2011-02-17T13:38:52.987" Score="2" ViewCount="" Body="&lt;p&gt;No. Let me assume you are working in the category of (simple, undirected) graphs such that a morphism $f : G \to H$ of graphs is a function sending vertices to vertices preserving the edge relation, since this is the category that makes the coloring fact you mentioned work out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose $G$ is a graph such that either the functor $\text{Hom}(-, G)$ or the functor $\text{Hom}(G, -)$ counts spanning trees. Let $1$ denote the one-element vertex. In the first case, since $|\text{Hom}(1, G)| = 1$ it follows that $G$ has one vertex, but this is absurd. In the second case, since $|\text{Hom}(G, 1)| = 1$ it follows that $G$ has no edges, but this is also absurd.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I had a more sophisticated argument in mind which should also handle any reasonable generalization, but it is unnecessary in this case: first for any particular generalization it should not be hard to argue that $G$ must be finite, but then $|\text{Hom}(G, H)| \le |H|^{|G|}$ and $|\text{Hom}(H, G)| \le |G|^{|H|}$, so the sequence of the number of spanning trees of $K_n$ is either polynomially or exponentially bounded in $n$. But this sequence is $n^{n-2}$; contradiction.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might be interested in the &lt;a href=&quot;http://en.wikipedia.org/wiki/Tutte_polynomial&quot; rel=&quot;nofollow&quot;&gt;Tutte polynomial&lt;/a&gt;, though. There is also the notion of the &lt;strong&gt;critical group&lt;/strong&gt; of a graph, which is a finite group with order the number of spanning trees, and I think its construction can be made functorial. I haven't checked this, though.&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2011-02-17T13:43:55.167" LastActivityDate="2011-02-17T13:43:55.167" CommentCount="2" />
  <row Id="22496" PostTypeId="2" ParentId="22407" CreationDate="2011-02-17T13:44:24.153" Score="1" ViewCount="" Body="&lt;p&gt;For the first question (independence does not imply conditional independence), one of my favorite simple counterexamples works.  Flip two fair coins.  Let $A$ be the event that the first coin is heads, $B$ the event that the second coin is heads, $C$ the event that the two coins are the same (both heads or both tails).  Clearly $A$ and $B$ are independent, but they are not conditionally independent given $C$.  If you want an example with random variables, consider the indicators $1_A, 1_B, 1_C$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of interest here is that $A,B,C$ are pairwise independent but not mutually independent (since any two determine the third).&lt;/p&gt;&#xA;" OwnerUserId="822" LastActivityDate="2011-02-17T13:44:24.153" />
  <row Id="22497" PostTypeId="2" ParentId="22493" CreationDate="2011-02-17T13:45:00.530" Score="1" ViewCount="" Body="&lt;p&gt;One reason to think about a million pixel picture in $10^6$ (or, more likely, $3\cdot 10^6$) dimensional space rather than RGB is that each point in the space corresponds to a different picture.  Pictures that are close in coordinates are close in content.  In RGB space (3 dimensional) you don't have that, in fact I don't know how to locate a picture in 3 space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you say, defining two pictures to be orthogonal doesn't seem very useful, but many of the operations we do in picture processing can be represented by matrices.  Sharpening and blurring filters are good examples.  So are &quot;lighten shadows&quot; and &quot;darken highlights&quot;&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-02-17T13:45:00.530" CommentCount="1" />
  <row Id="22498" PostTypeId="1" CreationDate="2011-02-17T13:55:26.293" Score="0" ViewCount="54" Body="&lt;p&gt;I have a truncated cone with the smaller perimeter = 70cm and the greater perimeter = 64cm. The edge length is 42cm&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I now calculate the angle alpha?&lt;/p&gt;&#xA;" OwnerUserId="5791" LastActivityDate="2011-02-17T14:20:15.573" Title="Angle in a truncated cone?" Tags="&lt;geometry&gt;" AnswerCount="1" />
  <row Id="22499" PostTypeId="1" CreationDate="2011-02-17T14:00:37.790" Score="2" ViewCount="73" Body="&lt;p&gt;I'm note sure if this is mathematical, however I know that there's an area of maths about folding. And regardless, this seems like an interesting question that I have no idea how to solve.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am designing a 5 day paper timetable for university that I will keep in my pocket so I know where and when my classes are. My current timetable is a table with days as columns, hours as rows and cells being classes. However when I came to fold this I found that it was awkward. In order to fold it small enough to fit in my pocket I needed to fold it many times, which in turn meant that to view some parts of the time table I needed to unfold it many times.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;How can I fold a timetable so that to view any part requires the least amount of unfolding?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="1099" LastActivityDate="2011-02-17T15:07:12.710" Title="How can I fold a timetable?" Tags="&lt;geometry&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="22500" PostTypeId="2" ParentId="22474" CreationDate="2011-02-17T14:15:24.160" Score="3" ViewCount="" Body="&lt;p&gt;This is given in &lt;a href=&quot;http://oeis.org/A005044&quot; rel=&quot;nofollow&quot;&gt;A005044&lt;/a&gt;.  The generating function is $\frac{x^3}{(1-x^2)(1-x^3)(1-x^4)}$  Some other formulas and references are given as well&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-02-17T14:15:24.160" />
  <row Id="22501" PostTypeId="2" ParentId="22498" CreationDate="2011-02-17T14:20:15.573" Score="1" ViewCount="" Body="&lt;p&gt;I presuming it is a right circular cone and $\alpha$ is the angle between the base and the slant (details which should have been included in the question).  Hint:  Think about a trapezoid formed with the axis of the cylinder, two parallel radii (one on each circular surface) and the slant between the ends of the radii.  You should be able to calculate the length of the radii and the angles at the ends of the axis are right angles.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-02-17T14:20:15.573" />
  <row Id="22502" PostTypeId="1" AcceptedAnswerId="22504" CreationDate="2011-02-17T14:20:45.217" Score="0" ViewCount="155" Body="&lt;p&gt;I'm looking for a calculator that shows a full answer regardless of it's length.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example the assignment &lt;code&gt;2^1000&lt;/code&gt; will output: &lt;code&gt;1,07150860718626732094842504906e+301&lt;/code&gt;.&#xA;Windows calculator shows me that amount because else it won't fit the display.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the question is:&#xA;&lt;strong&gt;Is there a calculator out there that shows the full answer instead of shortening it?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="7188" LastEditorUserId="7188" LastEditorDisplayName="" LastEditDate="2011-02-17T14:26:18.080" LastActivityDate="2011-06-01T14:53:30.277" Title="Is there a calculator out there that shows the full answer instead of shortening it?" Tags="&lt;calculator&gt;" AnswerCount="5" CommentCount="7" FavoriteCount="1" />
  <row Id="22503" PostTypeId="1" CreationDate="2011-02-17T14:22:23.997" Score="0" ViewCount="82" Body="&lt;p&gt;Given a topological space that is T3 (i.e. regular) and has a sigma-discrete base (where these conditions allow the topological space to be biconditionally metrizable), is there a way to understand what all the possible admittable metrics are? And do all these possibly admittable metrics form a space?&lt;/p&gt;&#xA;" OwnerUserId="7485" LastActivityDate="2011-02-17T14:22:23.997" Title="space of metric spaces of a topological space" Tags="&lt;topology&gt;" CommentCount="2" />
  <row Id="22504" PostTypeId="2" ParentId="22502" CreationDate="2011-02-17T14:27:54.580" Score="8" ViewCount="" Body="&lt;p&gt;If you don't mind a web-based solution, wolframalpha.com will do that for you:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;http://www.wolframalpha.com/input/?i=2^1000&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;outputs: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;10715086071862673209484250490600018105614048117055336074437503883703510511249361224931983788156958581275946729175531468251871452856923140435984577574698574803934567774824230985421074605062371141877954182153046474983581941267398767559165543946077062914571196477686542167660429831652624386837205668069376&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="6449" LastActivityDate="2011-02-17T14:27:54.580" CommentCount="2" />
  <row Id="22505" PostTypeId="2" ParentId="22473" CreationDate="2011-02-17T14:28:39.603" Score="4" ViewCount="" Body="&lt;p&gt;Consider the function $F\colon 0\to A$, suppose there is some $\langle x,y\rangle\in F$. This means that $x\in dom F$, since we have $dom F = 0$ then $x\in 0$ which is a contradiction. Therefore there are no ordered pairs in $F$, from the fact that it is a function we know that there are not other elements in $F$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If so, we proved $F=0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same proof goes for the other statement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt;&lt;br&gt;&#xA;An alternative method is by cardinal arithmetic: $|F|=|dom F| \le |dom F|\times|rng F|$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first equality is simply by projection $\langle x,F(x)\rangle \mapsto x$, where the second is by the identity map.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From this, suppose $dom F = 0$ then $F=0$ and suppose $rng F=0$ then $F=0$ and $dom F=0$.&lt;/p&gt;&#xA;" OwnerUserId="622" LastEditorUserId="622" LastEditorDisplayName="" LastEditDate="2011-02-17T16:39:43.577" LastActivityDate="2011-02-17T16:39:43.577" />
  <row Id="22506" PostTypeId="2" ParentId="22493" CreationDate="2011-02-17T14:33:27.607" Score="4" ViewCount="" Body="&lt;p&gt;The basic reason to think of waves as sitting inside vector spaces is that the &lt;a href=&quot;http://en.wikipedia.org/wiki/Wave_equation&quot; rel=&quot;nofollow&quot;&gt;wave equation&lt;/a&gt; is linear. That is, a linear combination of two waves is another wave. This is not structure that we arbitrarily impose on the problem (as one could argue about representations of a picture): it is structure that is already present which we are recognizing. The Fourier transform diagonalizes differentiation, so it is a natural way to think about differential equations like the wave equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These ideas turn out to be so useful that we try to use them even in contexts where it is not necessarily natural, at first glance, to use vector spaces (for example in additive combinatorics). It is all part of one of the most important principles in all of mathematics: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Linear algebra is easy, so reduce as much of the difficulty of a problem as you can to linear algebra.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It is astonishing how much of modern mathematics can be said to be an application of this principle.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2011-02-17T14:33:27.607" />
  <row Id="22507" PostTypeId="2" ParentId="22502" CreationDate="2011-02-17T14:36:02.697" Score="0" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://wims.unice.fr/wims/wims.cgi?session=PKC0C0E5B6.2&amp;amp;lang=en&amp;amp;module=tool%2Fanalysis%2Ffunction.en&quot; rel=&quot;nofollow&quot;&gt;Wims Function Calculator&lt;/a&gt; also gives 10715086071862673209484250490600018105614048117055336074437503883703510511249361224931983788156958581275946729175531468251871452856923140435984577574698574803934567774824230985421074605062371141877954182153046474983581941267398767559165543946077062914571196477686542167660429831652624386837205668069376.&lt;/p&gt;&#xA;" OwnerUserId="2810" LastActivityDate="2011-02-17T14:36:02.697" CommentCount="0" />
  <row Id="22508" PostTypeId="1" AcceptedAnswerId="22509" CreationDate="2011-02-17T14:39:12.167" Score="1" ViewCount="64" Body="&lt;p&gt;I initially asked this question &lt;a href=&quot;http://stats.stackexchange.com/questions/7330/rotation-matrices-and-prior-invariance-for-arbitrary-dimensions&quot;&gt;here&lt;/a&gt;, and someone suggested this may be a better place to get an answer.&#xA;I have a question about a rotation matrix, which can be represented in 2 dimensions as:&#xA;$$R_{2}(\theta)=\begin{bmatrix} cos\theta &amp;amp; sin\theta \\ -sin\theta &amp;amp; cos\theta\end{bmatrix}$$&#xA;For some arbitrary angle $\theta$. This can be extended to an arbitrary number of dimensions by adding an identity matrix:&#xA;$$R_{n}(\theta)=\begin{bmatrix} R_{2}(\theta) &amp;amp; 0 \\ 0 &amp;amp; I_{n-2}\end{bmatrix}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is: can any rotation in arbitrary dimensions be represented by a sequence of 2-D rotations? (i.e. a product of row/column permuted matrices of the above type?) It doesn't matter if the sequence is unique or not for my purposes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If this is true (which I think it would be), does adding a translation affect this representation, or does the rotation need to be carried about the origin for the representation to hold?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$X&amp;#39;=R_{n}^T(X-x_{0})$$&lt;/p&gt;&#xA;" OwnerUserId="7189" LastActivityDate="2011-02-17T14:42:33.053" Title="Rotation matrices for arbitrary dimensions" Tags="&lt;matrices&gt;&lt;transformation&gt;&lt;orthonormal&gt;" AnswerCount="1" />
  <row Id="22509" PostTypeId="2" ParentId="22508" CreationDate="2011-02-17T14:42:33.053" Score="3" ViewCount="" Body="&lt;p&gt;Yes. This is a consequence of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Spectral_theorem&quot; rel=&quot;nofollow&quot;&gt;spectral theorem&lt;/a&gt; and is proven, for example, in &lt;a href=&quot;http://qchu.wordpress.com/2011/02/05/so3-and-su2/&quot; rel=&quot;nofollow&quot;&gt;this blog post&lt;/a&gt;. Translations don't affect anything.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2011-02-17T14:42:33.053" CommentCount="2" />
  <row Id="22510" PostTypeId="2" ParentId="22473" CreationDate="2011-02-17T14:52:13.180" Score="2" ViewCount="" Body="&lt;p&gt;I assume that by $0$ you mean the empty set ($\varnothing$). I don't know how the book defines a relation (the usual definition is that it's a subset of the Cartesian product of two sets). But unless it mentions that the domain of a relation $R\subset A\times B$ is $A$ then the definition of a function &lt;em&gt;as you present it&lt;/em&gt; is different from the standard set theoretic definition of the function and furthermore (iii) is not correct. Under the standard definition of a function (namely that if $F\subset A\times B$ then the $dom(F)=A$) both (ii) and (iii) are correct. To see this you have to carefully examine whether they fall into the definition of a function:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Observe that $\varnothing\subset A\times B$. Also note that for any set $A$, $\varnothing\times A= A\times\varnothing=\varnothing$ and thus its only subset (and possible function) is the empty set. So $\varnothing$ is by definition a relation of $A\times B$ for any $A$ and $B$ and furthermore the only relation if one of the $A$ or $B$ is empty.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now if $F\subset \varnothing\times B$ then $F=\varnothing$ and there cannot be $(x,y)\in F$, $(x,z)\in F$ and $y\neq z$ (since $F$ is empty). Thus $F$ is a function. Note here that if we assume that for a function $F\subset A\times B$ we have $dom(F)\subset A$, then with a similar argument we show that given arbitrary sets $A$ and $B$ the empty set is always a function between $A$ and $B$ (and thus (iii) is wrong). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now for (iii) under the standard definition, if $A$ is not empty then $F$ has to be non-empty since its domain is not empty, but on the other hand $F=\varnothing$ (as a subset of the empty set). Thus $F=A=\varnothing$.&lt;/p&gt;&#xA;" OwnerUserId="1772" LastEditorUserId="1772" LastEditorDisplayName="" LastEditDate="2011-02-17T15:04:36.507" LastActivityDate="2011-02-17T15:04:36.507" CommentCount="3" />
  <row Id="22511" PostTypeId="2" ParentId="22499" CreationDate="2011-02-17T15:07:12.710" Score="3" ViewCount="" Body="&lt;p&gt;If you can make the overall form a horizontal strip that only needs to be folded vertically, then accordion folding (score at equal intervals, fold starting at one end, alternately forward and backward) will allow you to see any panel with a single unfolding.&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2011-02-17T15:07:12.710" CommentCount="0" />
  <row Id="22512" PostTypeId="1" AcceptedAnswerId="22514" CreationDate="2011-02-17T15:14:41.403" Score="1" ViewCount="53" Body="&lt;p&gt;Hi there,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need to be able to find the eigenvalues and eigenvectors of a matrix (sticking to $2 \times 2$ matrices at the moment).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm fine with finding the eigenvalues, but my answers for the eigenvectors don't match up with those on my problem sheet. However the &lt;strong&gt;proportion&lt;/strong&gt; between my values seem to be the same, but I'm wondering why, and especially how, the writer of the problem sheet came up with his answers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Matrix $A$:&#xA;$$A = \left(\begin{array}{rr}&#xD;&#xA;2 &amp;amp; -1\\&#xD;&#xA;-1 &amp;amp; 1&#xD;&#xA;\end{array}\right).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Eigenvalues found: $2.618$ and $0.382$&#xA;as $(A - \lambda I)X = 0$&#xA;$$\left(\begin{array}{cc}&#xD;&#xA;2 - 2.618 &amp;amp; -1\\&#xD;&#xA;-1 &amp;amp; 1-2.618&#xD;&#xA;\end{array}\right)\left(\begin{array}{c}&#xD;&#xA;x\\&#xD;&#xA;y&#xD;&#xA;\end{array}\right) = \left(\begin{array}{c}&#xD;&#xA;0\\0&#xD;&#xA;\end{array}\right).$$&#xA;so:&#xA;$$\begin{array}{rcccl}&#xD;&#xA;-0.618x &amp;amp; - &amp;amp; y &amp;amp; = &amp;amp; 0\\\&#xD;&#xA;-x &amp;amp; - &amp;amp; 1.618y &amp;amp; = &amp;amp; 0.&#xD;&#xA;\end{array}.$$&#xA;So:&#xA;\begin{align*}&#xA;x &amp;amp;= -1.618y\&#xA;y &amp;amp;= -0.618x.&#xA;\end{align*}&#xA;so this should give an eigenvector of:&#xA;$$\left(\begin{array}{c}&#xD;&#xA;1\\-1.618\end{array}\right)\quad\text{or}\quad&#xD;&#xA;\left(\begin{array}{c}&#xD;&#xA;0.618\\&#xD;&#xA;-1&#xD;&#xA;\end{array}\right).$$&#xA;Or can the signs be reversed?&#xA;$$\left(\begin{array}{c}&#xD;&#xA;-1\\&#xD;&#xA;1.618&#xD;&#xA;\end{array}\right)\quad&#xD;&#xA;\text{and}\quad&#xD;&#xA;\left(\begin{array}{c}&#xD;&#xA;-0.618\\ 1\end{array}\right)\quad???$$&#xA;The eigenvector given in my answers sheet for the eigenvalue $2.618$ is:&#xA;$$\left(\begin{array}{c}&#xD;&#xA;-0.85\\&#xD;&#xA;0.53&#xD;&#xA;\end{array}\right)$$&#xA;which if you look at the proportion and take into account the fact that these numbers have been rounded seem to be at least proportional to my answers, but what I'd like to know is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How come the answer above was given on the question sheet?&#xA;Does the sign actually matter? I suspect yes, but can't see why or how one chooses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many thanks in advance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Joe&lt;/p&gt;&#xA;" OwnerUserId="1094" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-17T19:06:08.887" LastActivityDate="2011-02-17T19:06:08.887" Title="Difficulty with last part of finding eigenvector" Tags="&lt;linear-algebra&gt;&lt;matrices&gt;&lt;eigenvectors&gt;" AnswerCount="1" />
  <row Id="22513" PostTypeId="1" CreationDate="2011-02-17T15:18:49.683" Score="22" ViewCount="565" Body="&lt;p&gt;I dread mathematics, and I believe it's because I have come to associate mathematics with the experience of terrible teachers. All of my math teachers have been grumpy, but one in particular was the epitome of evil. She would take any opportunity to yell and scream at me when I struggled to comprehend the problems given. She approached the kids in my class as if their struggling wasn't a result of a misunderstanding, but rather from a lack of discipline, one that she could solve by being some sort of mathematics drill sergeant. This was when I was a small child, which obviously left an impression on my mind that probably wouldn't have existed had I been older.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now that I am older, however, I need mathematics. I also have a growing curiosity and interest for it. Right now, I am planning to move out of my parent's house and live elsewhere. However, I have a fear in the back of my mind that my understanding of mathematics is not sufficient enough to do all of this; to handle a job; to handle expenses; to handle day-to-day life. Even the idea of becoming a cashier and having to handle money frightens me into avoiding those jobs; which leaves me (having no formal education) with virtually no options for employment. It's pretty intimidating. I seem to have some problem grasping even the simplest mathematic questions. Come up to me and ask me &quot;What's 8 + 6?&quot; and my mind wanders for the answer as if blindfolded. I would probably resort to sticking my hands behind my back and counting it off on my fingers or counting one at a time in my head. This just doesn't seem normal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to conquer my fear of mathematics and educate myself. I want to approach the field as an absolute beginner, and by that, I mean go back to the very basics and work my way up, no matter how reassured I am of my abilities at times. I know it's impossible to conquer the entire mathematic field, but I want to conquer what is necessary and then some. I &lt;em&gt;need&lt;/em&gt; the bare minimum, though I &lt;em&gt;want&lt;/em&gt; a sufficient understanding. Are there any approaches or, what I am specifically requesting, books or courses, that would allow me to to teach myself in such a manner?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sorry for the history lesson and/or if this is not a legitimate question.&lt;/p&gt;&#xA;" OwnerUserId="7191" LastActivityDate="2011-05-24T22:39:58.533" Title="Learning mathematics as if an absolute beginner?" Tags="&lt;self-learning&gt;" AnswerCount="3" CommentCount="7" FavoriteCount="9" />
  <row Id="22514" PostTypeId="2" ParentId="22512" CreationDate="2011-02-17T15:20:01.243" Score="3" ViewCount="" Body="&lt;p&gt;Your examples have $y=-1.618x$, not the other way around.  Any non-zero multiple (including -1) of an eigenvector is an eigenvector, so you can flip the signs on both components (but not only one).  As you say, the book answer is proportional to yours once you get x and y right.  I don't know how they chose the normalization.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2011-02-17T18:07:30.490" LastActivityDate="2011-02-17T18:07:30.490" CommentCount="3" />
  <row Id="22515" PostTypeId="2" ParentId="22483" CreationDate="2011-02-17T15:20:30.273" Score="2" ViewCount="" Body="&lt;p&gt;If $\sigma$ is the (population) standard deviation then &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sigma^2 = \frac{\sum (x_i - \mu)^2}{n} = \frac{\sum x_i^2}{n} - \mu^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so using the subscripts $1$, $2$ and $C$ for the first part, second part and combination&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sigma_C^2 = \frac{n_1 (\sigma_1^2 + \mu_1^2) + n_2 (\sigma_2^2 + \mu_2^2)}{n_1+n_2} - \mu_C^2$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\mu_C = \frac{n_1 \mu_1 + n_2 \mu_2}{n_1+n_2}$ as you have.  Then take the square root.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think this is easier to understand than &lt;a href=&quot;http://en.wikipedia.org/wiki/Standard_deviation#Combining_standard_deviations&quot; rel=&quot;nofollow&quot;&gt;Wikipedia's version&lt;/a&gt; though theirs is better if rounding errors are an issue (e.g. with large means close to each other and small standard deviations).&lt;/p&gt;&#xA;" OwnerUserId="6460" LastActivityDate="2011-02-17T15:20:30.273" CommentCount="1" />
  <row Id="22516" PostTypeId="2" ParentId="22493" CreationDate="2011-02-17T15:24:37.823" Score="10" ViewCount="" Body="&lt;p&gt;I think the question is related to why the space of images is 1-million dimensional, and not 3 dimensional, and why studying these spaces might be useful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, think about the 3 dimensional space of RGB colors: what is a point in that space? It's simply a vector with 3 coordinates: $&amp;lt;r, g, b&amp;gt;$. So that's just the information for one pixel, not for a whole image. Any point in that space carries only the information about how much red, green and blue &lt;strong&gt;one&lt;/strong&gt; pixel has.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now think about what would you do if you needed to store the information of color for TWO pixels: you would need one axis to represent the Red color of pixel 1 and one axis to represent the Red color of pixel 2; one for the Green color of pixel 1, one for Green color of pixel 2; and one for Blue color of pixel 1, and one for Blue color of pixel 2. So you would have 6 axis: 3 for each pixel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to represent the color of 1 million pixels, you need 3 * 1 million axis. These axis are orthogonal in the sense that if you want, you can change the color of just one pixel without having to adjust the color of any other pixels. So you would have a space with 3 * 1 million dimensions. Each point in that space now corresponds to an image: specifically, the coordinates along each of the 3 * 1 million axes gives you the value of R, G, or B for a given pixel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, in effect, if you want to have a space of IMAGES, and not a space of PIXELS, you need a 3 * 1 million-dimensional space, not a 3D space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now think about what would happen if you could take, for instance, 50 pictures of human faces and plot them in this 3 million-dimensional space, and see where they lie. Of course you can't visualize this, but you might expect that these pictures have SOMETHING in common (after all, they are all pictures of human faces, not arbitrary pictures of anything, or crazy combinations of pixel colors). If you &lt;em&gt;could&lt;/em&gt; see how these images are spread over that space (where the points corresponding to those pictures are, that is), you would see that they usually are not located ANYWHERE. For instance, human faces have typical colors -- they tend not to be green, for example. That means that the regions of the space where you would expect to see green pixels are probably sort of empty. That's what they mean when they talk about identifying the subspace of human faces. It's just the &quot;surface&quot;, or sub-regions of that 3*1 million-dimensional space where you'd expect to find points corresponding to human faces. Typically that sub-region might be described with fewer information than 3*1 million coordinates, if you just find a better representation for your image, instead of one that stores the value of each and all r,g,b component for all pixels. That is why image compression is possible: if you just find the right way of representing your information (like the value of your 3*1 million RGB components), you might do that with LESS than 3*1 million numbers; specifically, since this numbers have a pattern.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is possible to try to identify the FORM of the sub-region of the 3-million-dimensional space where human faces tend to appear. Then given another image which you don't know if it's a human face or not, you could try to GUESS if it is a human face. How? Well, check if the point corresponding to that new image, when plotted on that 3*1 million-dimensional space, is close to the sub-region where the points of human faces usually are. Sometimes they call the process of identifying this subregions by the name of &quot;manifold learning&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ok, lots of information. Just think about it for a while. It's hard (actually impossible) to visualize spaces with more than 3 dimensions, but once you get the idea of what's going on, often you'll see that your intuitions about what  happens in 2D or 3D carry on. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Try this exercise: imagine a black-and-white image with just 3 pixels; each pixel is then just a value between 0 and 1 (0 being completely black and 1 being completely white, and values in between being shades of gray). Now imagine the set of images where the first pixel is darker than the second one, and the second one is darker than the third one. That is, images like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$&amp;lt;  0.3,  0.8, 1.0 &amp;gt;$&#xA;or &#xA;$&amp;lt; 0.12, 0.53, 0.7 &amp;gt;$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now generate a bunch of those (say, 10.000 images like that) and plot them in the 3D where they lie. Notice that there is a pattern in these images: the value of the 2nd pixel is always larger than the value of the 1st pixel; the value of the 3rd pixel is always larger than the value of the 2nd. Clearly we shouldn't expect points corresponding to images like this to occupy ANY place in the 3D space. For example, we would most certainly not see points in the space close to &amp;lt; 0.5, 0.3, 0.1 &gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can actually see how 10.000 of such images in this plot, where I show 10.000 images like the ones I described. Since each axis corresponds to the value of one of the 3 pixels in a given image, we have 3 axes. Each point in the plot is thus a 3-pixel image.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/tZvcQ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice how the points are occupying just a small part of the whole 3D space. That happens because there is a relation between the values of the pixels. Images of that type all have something in common, so they occupy similar portions of the whole 3D space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same way, if you could plot 1-million pixel images (which would lie in a 3*1-million dimensional space, as mentioned before), and all of those images corresponded to images of human faces, you would see some pattern like the one I showed above. Specifically, the points corresponding to images of human faces would most likely NOT occupy the whole 3-million dimensional space. We could actually try to estimate the &quot;shape&quot; of the sub-region where human faces are by using techniques called &quot;manifold learning&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, notice that you could use the same ideas as above to analyze any other kind of data. Results of a statistical survey? Imagine that you have 50 questions, each one being a value from 0 to 100. You ask those questions to one person and get 50 numbers back. You ask them to another person and get another 50 numbers. How to &quot;visualize&quot; them? Plot them in a 50-dimensional space where each axis corresponds to the value of one given answer. A point in that space then corresponds to 50 numbers (specifically, the answers given in one specific survey). If you plot, say, 1000 of these surveys in this space, you would get 1000 50-dimensional points. Maybe there is some pattern can be found; maybe there isn't.  If there is, it might be the case that these 1000 50-dimensional points lie in a subspace (or sub-region) of the 50-D space. That is what Terrence Tao was saying when he said that it is useful to study these subspaces, or sub-regions, and when he said that the &quot;subsets of this space correspond to various classes of images.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope that helps!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bruno&lt;/p&gt;&#xA;" OwnerUserId="2368" LastActivityDate="2011-02-17T15:24:37.823" />
  <row Id="22517" PostTypeId="2" ParentId="22502" CreationDate="2011-02-17T16:00:06.843" Score="1" ViewCount="" Body="&lt;p&gt;Have a look at &lt;a href=&quot;http://alpertron.com/ECM.HTM&quot; rel=&quot;nofollow&quot;&gt;this site&lt;/a&gt;: it is dedicated to large number factorizations, but you can also make some integer based operations. You can input &quot;2^1000&quot; and try.&lt;/p&gt;&#xA;" OwnerUserId="354" LastActivityDate="2011-02-17T16:00:06.843" CommentCount="1" />
  <row Id="22518" PostTypeId="1" CreationDate="2011-02-17T16:15:47.980" Score="0" ViewCount="31" Body="&lt;p&gt;I'm looking for a way to mathematically combine two concepts: &lt;a href=&quot;http://en.wikipedia.org/wiki/LFSR&quot; rel=&quot;nofollow&quot;&gt;LFSRs&lt;/a&gt;, and &lt;a href=&quot;http://en.wikipedia.org/wiki/Barrel_shifter&quot; rel=&quot;nofollow&quot;&gt;Barrel Shifters&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm looking for a way, in O(1) time, to shift an LFSR loop a given number of shifts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I'm hoping to find is a simple process where I have the current state of the LFSR and the number of times I want to shift from that state as parameters to a quick/easy process.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At first I just thought about looking at all the taps, then moving the taps over by 1 and looking at them again, finding the shift in bit each time and appending it onto the end but of course this isn't O(1) and it gets complicated if I want to shift so many times that a tap would &quot;slide off&quot; the original LFSR state.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I may be in the wrong place to ask this because I'm looking for a way to do this programmatically. Although any incite into how it's possible (or why it's impossible) would be very helpful.&lt;/p&gt;&#xA;" OwnerUserId="757" LastActivityDate="2011-02-17T16:15:47.980" Title="Shifting an LFSR loop in O(1) time?" Tags="&lt;lfsr&gt;" CommentCount="2" />
  <row Id="22519" PostTypeId="2" ParentId="22461" CreationDate="2011-02-17T16:20:57.430" Score="0" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\: $ For primes $\rm\ p\neq q\:$ coprime to $\rm\:a\:,\:$ if $\rm\ p-1,q-1\ |\ m\ $ then $\rm\ p,q\ |\ a^m - 1\ \Rightarrow\ pq\ |\ a^m - 1$ since lcm = product for coprime integers (here distinct primes).&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2011-02-17T16:20:57.430" />
  <row Id="22520" PostTypeId="1" AcceptedAnswerId="22524" CreationDate="2011-02-17T16:31:00.057" Score="2" ViewCount="60" Body="&lt;p&gt;It is known that a string $s$ is actually made up of repetitions of another string $s_1$ of length $L_1$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also $s$ can be thought of as made up of repetitions of another string $s_2$ of length $L_2$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example the string $s = abababab$ is made by repeating the substring $s_1$ &quot;$ab$&quot; of length $L_1= 2$ or by repeating the substring $s_2$ &quot;$abab$&quot; of length $L_2=4$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe that in such a case $s$ can be made by repeating substring of length $\gcd(L_1,L_2)$. For example if $L_1=6$ and $L_2=10$ then $s$ should be repetitive in a substring of length $2$  (= $\gcd(6,10)$) too. This seems intuitive but how do I show this formally ?&lt;/p&gt;&#xA;" OwnerUserId="6445" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-17T16:38:35.233" LastActivityDate="2011-02-17T16:43:51.513" Title="Periodicity in strings" Tags="&lt;modular-arithmetic&gt;" AnswerCount="2" />
  <row Id="22521" PostTypeId="1" CreationDate="2011-02-17T16:31:12.310" Score="3" ViewCount="103" Body="&lt;p&gt;I have a $N\times N$ symmetric positive semidefinite matrix $Q$, and am considering a class of symmetric positive definite matrices having all eigenvalues in a given bounded interval $[a, b]$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it possible to find a matrix in this class such that the sum of the smallest $k$ elements of $\text{diag}(XQ)$ is minimized?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am welcoming any suggestion, either in the form &quot;not a trivial problem&quot;, or &quot;here is the solution, or a pointer to it&quot;.&lt;/p&gt;&#xA;" OwnerUserId="7192" LastEditorUserId="7192" LastEditorDisplayName="" LastEditDate="2011-02-17T17:05:05.543" LastActivityDate="2011-02-17T21:28:25.340" Title="Minimizing the sum of the $k$ smallest elements of the diagonal of a matrix" Tags="&lt;matrices&gt;&lt;optimization&gt;&lt;eigenvalues&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="22522" PostTypeId="1" AcceptedAnswerId="22569" CreationDate="2011-02-17T16:36:12.850" Score="2" ViewCount="87" Body="&lt;p&gt;I have stumbled upon a remarkable similarity between the proof of Baer's criterion and an extension theorem in field theory. Here are the statements:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Baer's criterion: Let $R$ be a ring. A left $R$-module $E$ is injective iff every $R$-map $f:I \to E$, where $I$ is a left ideal of $R$, can be extended to a map $R\to E$.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Extension theorem: let $F\subset K$ be an algebraic field extension, and $L$ an algebraically closed field. Then every field homomorphism $\sigma: F \to L$ can be extended to a field homomorphism $K\to L$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now, these two theorems don't seem to have anything to do with each other, but the proofs are strikingly similar. Here's how they go:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For Baer's criterion (the non-trivial statement, i.e. the sufficiency of the ideal condition): to prove injectivity, we let $A\subset B$ be a submodule and $f:A\to E$ an $R$-map, and define $X=\{(A&amp;#39;,g&amp;#39;): A\subset A&amp;#39;\subset B,\, g&amp;#39;: A&amp;#39; \to E,\, g&amp;#39;|_A=f\}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $\leq$ be the partial order in $X$ given by $(A&amp;#39;,g&amp;#39;)\leq (A&amp;#39;&amp;#39;,g&amp;#39;&amp;#39;) \iff A&amp;#39;\subset A&amp;#39;&amp;#39; \text{ and } g&amp;#39;&amp;#39;|_{A&amp;#39;}=g&amp;#39;$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By Zorn's lemma there's $(A_0,g_0)$ a maximal element. If $A_0=B$ we're done, if not, let $b\in B\setminus A_0$, define $I=\{r\in R: rb\in A_0\}$ and $h:I\to E$, $r \mapsto g_0(rb)$. Apply the hypothesis to find an extension $\tilde{h}$ of $h$ to $R$. Let $A_1=A_0 + Rb$ and $g_1:A_1 \to E$, $a_0+rb \mapsto g_0(a_0)+r\tilde{h}(1)$, and $(A_1,g_1)$ contradict the maximality of $(A_0,g_0)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the extension theorem: define $M=\{(A,\tau):F\subset A \subset K,\, \tau:A\to L,\, \tau|_F=\sigma\}$. Let $\leq$ be the partial order on $M$ given by $(A,\tau)\leq (A&amp;#39;,\tau&amp;#39;) \iff A\subset A&amp;#39; \text{ and } \tau&amp;#39;|_A=\tau$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By Zorn's lemma there's $(A,\tau)$ a maximal element. If $A=K$ we're done, if not let $\alpha \in K\setminus A$, I claim $\tau$ extends to $A(\alpha)\to L$, contradiction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $p$ be the minimal polynomial for $\alpha$ over $A$. The polynomial $\tau p\in L[X]$ has a root $r\in L$ by hypothesis. Since $A(\alpha)= \frac{A[X]}{\langle p \rangle}$, define&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\tilde{\tau}: \frac{A[X]}{\langle p \rangle} \to L$ as $\tilde{\tau}|_A=\tau$, $\tilde{\tau}(X)=r$, and $(A(\alpha),\tilde{\tau})$ contradicts the maximality of $(A,\tau)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The resemblance of both proofs shouldn't come as such a surprise since both are about &lt;em&gt;extending morphisms&lt;/em&gt;, so maybe the only observation to make is &quot;it's a useful technique, remember it's useful for extending morphisms&quot;. If that's it, then this question is useless. But I'm intrigued. Is this technique used to prove other extending theorems? Is it possible to generalize it and write a single categorical proof for situations of this kind? Any other observations are appreciated (or perhaps it's just a dumb observation that doesn't serve any purpose).&lt;/p&gt;&#xA;" OwnerUserId="2614" LastActivityDate="2011-02-17T22:37:01.617" Title="Extending morphisms with Zorn's lemma" Tags="&lt;field-theory&gt;&lt;modules&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="22523" PostTypeId="2" ParentId="22520" CreationDate="2011-02-17T16:42:39.970" Score="2" ViewCount="" Body="&lt;p&gt;Replace your original string $s$ by infinitely many copies of $s$, going both ways: $$S = ... sss ... $$&#xA;Now $s$ has period $n$ iff $S(x) = S(x+n)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $S$ has periods $n,m$ then $S$ also has period $(n,m)$ since $(n,m) = an + bm$ for some integers $a,b$.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2011-02-17T16:42:39.970" CommentCount="2" />
  <row Id="22524" PostTypeId="2" ParentId="22520" CreationDate="2011-02-17T16:43:51.513" Score="3" ViewCount="" Body="&lt;p&gt;This follows from the following Theorem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Theorem.&lt;/strong&gt; Let $w \in \Gamma^{*}$ be a word with periodicities $p_1$ and $p_2$. If $|w| \geq p_1+p_2- \gcd(p_1,p_2)$, then $w$ has periodicity $\gcd(p_1, p_2)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is known as &lt;a href=&quot;http://planetmath.org/encyclopedia/FineAndWilfsTheoremOnWords.html&quot; rel=&quot;nofollow&quot;&gt;Fine and Wilf's Theorem&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2011-02-17T16:43:51.513" CommentCount="1" />
  <row Id="22525" PostTypeId="2" ParentId="22478" CreationDate="2011-02-17T16:51:24.267" Score="2" ViewCount="" Body="&lt;p&gt;In practice, God does not hand you abstract structures: you construct them from other structures, and the point of caring about isomorphisms is that you want to &lt;em&gt;keep track&lt;/em&gt; of this construction process. For example, you are almost never handed a vector space $V$ and its dual $V^{\ast}$. Usually you are performing some construction (e.g. on the tangent space $T_p(M)$ of a manifold $M$ at a point $p$) which naturally involves elements of $T_p(M)$ as well as elements of the cotangent space $T_p(M)^{\ast}$, and if you are foolish enough to think that they are the same space then you will literally not be capable of doing calculations in this setting (e.g. changing coordinates).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One way to say this is that abstract structures often arise &lt;em&gt;functorially&lt;/em&gt;, and even if $F(c), G(c)$ are isomorphic where $F, G : C \to D$ are functors and $c \in C$ is an object, the functors $F, G$ need not be naturally isomorphic, and usually we actually care about the functors, not the objects $F(c), G(c)$ in isolation. In the above example taking duals is a contravariant functor $\text{Vect} \to \text{Vect}$, and since it is contravariant it does not behave at all like the identity functor. &lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2011-02-17T16:51:24.267" CommentCount="6" />
  <row Id="22526" PostTypeId="2" ParentId="22513" CreationDate="2011-02-17T16:59:51.440" Score="13" ViewCount="" Body="&lt;p&gt;This is a difficult question to answer, mainly because any advice must be very personal to be useful for you. I'll try anyway.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Before learning mathematics, one has to &lt;strong&gt;learn how to learn&lt;/strong&gt; mathematics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Concerning the contemporary &lt;strong&gt;school system&lt;/strong&gt;, it simply does &lt;strong&gt;not do a good job&lt;/strong&gt; at what it is supposed to do. The range of children who learn successfully from it, or rather in spite of it, is quite narrow. You simply fell through the cracks, as many others, but that does need to have any relation to your general ability to learn mathematics. (I once instructed someone who taught himself programming after dropping out of college. He was very capabable, though he lacked methods to learn efficiently.) You are simply outside the way too narrow scope of the school system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Still, the best advice for learning mathematics is to &lt;strong&gt;find a good teacher&lt;/strong&gt;, but these are very hard to come by, so that this advice may be useless. (That's basically the same advice as for learning to play the piano.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, there is the issue that school does not even teach &lt;strong&gt;proper mathematics&lt;/strong&gt;, only some nonsense that they label &quot;mathematics&quot; but which is better described as &quot;rote arithmetics&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Two of the sins of the rote arithmetics as taught in school are the notions that 1) every problem has a definite answer and 2) that you should be able to come up with that answer instantly or you're stupid. Both are wrong. The book &lt;a href=&quot;http://books.google.com/books?id=nCNVQgAACAAJ&quot; rel=&quot;nofollow&quot;&gt;Thinking Mathematically&lt;/a&gt; dispels these myths and presents a &lt;strong&gt;proper way to learn mathematics&lt;/strong&gt;. However, note that learning how to learn mathematics from a book is quite difficult. It is easy to miss the point, for instance by not doing the exercises. A good teacher would be preferable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That said, &quot;simple&quot; questions like &quot;what is 8+6&quot; are &lt;strong&gt;also&lt;/strong&gt; a matter of &lt;strong&gt;rote learning&lt;/strong&gt;. This is something that can be learned in the school system at least, but again, it's equally easy to pass the school system and never learn how to do rote learning properly. Learning a foreign language, in particular vocabulary, is usually a good way to exercise rote learning.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another sin of school teaching is 3) the notion that learning is no fun. It baffles me how the key ingredient to learning, &lt;strong&gt;fun&lt;/strong&gt;, is entirely absent from the school system. (I don't know any mathematician who is doing mathematics for reasons other than that he/she loves it and that it's a lot of fun.) Fortunately, since you are motivated to learn mathematics by yourself, this should be no problem for you. Every moment of understanding is a tremendous amount of fun, and if you have enough of these moments in the beginning, you'll be hooked for life.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Concerning the process of learning, there is another problem that might or might not apply to you, namely the issue of &lt;strong&gt;genuine medical conditions&lt;/strong&gt; that can inhibit learning. (&quot;Inability to concentrate&quot;, undiagnosed food allergies, ...) However, given all the things above that can go wrong before that, I'm hesitant to put blame on medical conditions. Still, it may be worth keeping in mind, though personally, I'm not convinced that current treatment options actually help (and don't just accidentally fix some of the 3 problems mentioned above). In case you have a medical condition that can be fixed, fix it, otherwise you're kinda screwed anyway, though not necessarily when it comes to learning mathematics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After these general but very important considerations of the process of learning itself, there is the problem of finding good &lt;strong&gt;materials for the actual mathematics&lt;/strong&gt;. Fortunately, this is a much easier problem, largely solved by the videos from the &lt;a href=&quot;http://www.khanacademy.org/&quot; rel=&quot;nofollow&quot;&gt;KhanAcademy&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="434" LastActivityDate="2011-02-17T16:59:51.440" CommentCount="9" />
  <row Id="22527" PostTypeId="1" CreationDate="2011-02-17T17:05:26.550" Score="1" ViewCount="92" Body="&lt;p&gt;Does anybody know of a good way to figure out if the variety defined by the ideal $I=(a_1y_2-a_2y_1,b_1y_2-b_2y_1)$ in $\mathbb A^2_{y_1,y_2}\times\mathbb P^1_{a_1,a_2}\times \mathbb P^1_{b_1,b_2}$ is Toric?&lt;/p&gt;&#xA;" OwnerUserId="7194" LastActivityDate="2011-05-18T17:58:54.773" Title="Is the following a toric variety?" Tags="&lt;algebraic-geometry&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="22528" PostTypeId="1" AcceptedAnswerId="22531" CreationDate="2011-02-17T17:11:54.250" Score="4" ViewCount="189" Body="&lt;p&gt;The following is possibly true but I can't find a corresponding theorem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $E/F$ is the splitting field of some polynomial in $F$ and $F \subset K \subset E$ then:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$Gal(E/K)$ normal subgroup of $Gal(E/F)$  $\Leftrightarrow$ $K/F$ is a normal field extension&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this true? I think saying that $E/F$ is the splitting field of some polynomial in $F$ is the same as saying $E/F$ is Galois and therefore the Galois correspondence theorem applies. But I'm not sure I see how the meaning of normality of a subgroup relates to the meaning of normality of a field extension. But maybe the above is wrong altogether?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In any case, many thanks for your help to clarify this.&lt;/p&gt;&#xA;" OwnerUserId="5798" LastEditorUserId="5798" LastEditorDisplayName="" LastEditDate="2011-02-17T21:41:37.607" LastActivityDate="2011-02-17T21:54:47.337" Title="A question regarding normal field extensions and Galois groups" Tags="&lt;abstract-algebra&gt;&lt;field-theory&gt;&lt;galois-theory&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="2" />
  <row Id="22529" PostTypeId="2" ParentId="22468" CreationDate="2011-02-17T17:24:56.330" Score="4" ViewCount="" Body="&lt;p&gt;Parameterize the curve with $[s:t]\to [s^d:s^{d-1}t:\ldots :t^d]$ and plug it into the polynomial $F$ of degree $k$. The result is a homogeneous polynomial of degree $kd$ in $s,t$. But any homogeneous polynomial in two variables factors into linear factors over an algebraically closed field, and each linear factor corresponds to a point. So, there are exactly $kd$ points in the curve that vanish on $F$ counting multiplicities, unless the polynomial vanishes identically which just means that $F$ vanishes on the whole curve.&lt;/p&gt;&#xA;" OwnerUserId="7194" LastActivityDate="2011-02-17T17:24:56.330" />
  <row Id="22530" PostTypeId="2" ParentId="22527" CreationDate="2011-02-17T17:30:09.100" Score="1" ViewCount="" Body="&lt;p&gt;The torus $(\mathbb C^\times)^2$ acts on your variety $X$, with $(s,t)\in (\mathbb C^\times)^2$ sending the point $(y_1,y_2,a_1:a_2,b_1:b_2)\in X$ to $(sy_1,ty_2,t^{-1}a_1:s^{-1}a_2,t^{-1}b_1:s^{-1}b_2)$. There is an open faithful and dense orbit. That's enough.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Of course&lt;/em&gt;, &quot;enough&quot; depends on what definition of toric varieties you have in mind...&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2011-02-17T17:30:09.100" CommentCount="2" />
  <row Id="22531" PostTypeId="2" ParentId="22528" CreationDate="2011-02-17T17:47:14.260" Score="6" ViewCount="" Body="&lt;p&gt;You are confused about the Galois correspondence - normal subgroups $H$ &lt;strong&gt;of the Galois group&lt;/strong&gt; $\text{Gal}(E/F)$ correspond to normal extensions $E^H/F$, where $E^H$ denotes the subfield of $E$ fixed by $H$. Note that $E$ being the splitting field of a polynomial in $F$ does not guarantee that $E/F$ is Galois. This is due to the fact that $E/F$ is Galois only when it is &lt;strong&gt;normal&lt;/strong&gt; (i.e., is a compositum of some splitting fields) and &lt;strong&gt;separable&lt;/strong&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I imagine the statement you intended was:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If $E/F$ is normal, then   $$H\triangleleft \text{Aut}(E/F) \iff E^H/F \text{ is a normal field extension.}$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is actually still true. It can be regarded as a salvaging of the Fundamental Theorem of Galois Theory in the case that $E/F$ is not necessarily separable. Here is my reasoning: Let $E/F$ be normal and let $G=\text{Aut}(E/F)$. Then $E^G/F$ is purely inseparable, and $E/E^G$ is separable. We have that $\text{Aut}(E/E^G)=\text{Aut}(E/F)$. Because $E/F$ is normal, we have that $E/E^G$ is normal and hence $E/E^G$ is Galois, and therefore a normal subgroup $H\triangleleft\text{Aut}(E/E^G)=\text{Aut}(E/F)$ corresponds to a normal extension $E^H/E^G$. It is known that if $C\subseteq B\subseteq A$ is a tower of field extensions and $A/C$ is normal and $B/C$ is purely inseparable, then $A/B$ is normal. Thus $E^H/F$ is normal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conversely, given a normal subextension $L/F$ of $E/F$ that is the fixed field $L=E^H$ of some subgroup $H\subseteq\text{Aut}(E/F)=\text{Aut}(E/E^G)$, then $L$ in fact contains $E^G$, and $L/F$ normal implies $L/E^G$ normal, hence the subgroup of $G$ that fixes $L$, namely $H$, is normal in $G$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please take note of this question: it is &lt;strong&gt;not&lt;/strong&gt; an obvious one!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strike&gt;However, constructing a counterexample is evading me. Here is a try: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $\mathbb{F}_p$ be a finite field where $3\nmid p-1$ (so that there are no cube roots of unity), let $F=\mathbb{F}_p(T)$, let $f=x^{3p}-T\in F[x]$, let $E$ be the splitting field of $f$ over $F$. Then $E=F(\sqrt[3p]{T},\sqrt[3]{1})$. This has as a subfield $M=F(\sqrt[3]{T})$, which is separable, but not normal, over $F$. EDIT: Nevermind, this doesn't work. $M$ isn't the fixed field of a normal subgroup of $\text{Aut}(E/F)$.&lt;/strike&gt;&lt;/p&gt;&#xA;" OwnerUserId="264" LastEditorUserId="264" LastEditorDisplayName="" LastEditDate="2011-02-17T19:02:04.260" LastActivityDate="2011-02-17T19:02:04.260" CommentCount="3" />
  <row Id="22532" PostTypeId="1" AcceptedAnswerId="22534" CreationDate="2011-02-17T18:03:57.313" Score="2" ViewCount="62" Body="&lt;p&gt;If you have an odd number of data values {39, 40, 42, 44, 47, 48, 49, 51, 53}, what is the score that value that corresponds to Quartile 1?  In general, I would like to know if you include the median's value (47) in figuring the position of quartile 1. &lt;/p&gt;&#xA;" OwnerUserId="6243" LastActivityDate="2011-02-17T18:11:18.237" Title="How do you find the number that corresponds to Quartile 1, when given an even number of scores?" Tags="&lt;statistics&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="22533" PostTypeId="2" ParentId="22532" CreationDate="2011-02-17T18:09:41.940" Score="0" ViewCount="" Body="&lt;p&gt;$Q_1$ would be $41$ (i.e. the median of the lower half of the data). One can include the median value or not. You just have to be consistent. I think most calculators do not include the median.&lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2011-02-17T18:09:41.940" />
  <row Id="22534" PostTypeId="2" ParentId="22532" CreationDate="2011-02-17T18:11:18.237" Score="3" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Quartile#Computing_methods&quot; rel=&quot;nofollow&quot;&gt;Quoting from Wikipedia&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;There is no universal agreement on choosing the quartile values.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;[...]&lt;/p&gt;&#xA;  &#xA;  &lt;h3&gt;Method 1&lt;/h3&gt;&#xA;  &#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;Use the median to divide the ordered data set into two halves. Do not include the median into the halves.&lt;/li&gt;&#xA;  &lt;li&gt;The lower quartile value is the median of the lower half of the data. The upper quartile value is the median of the upper half of the data.&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;  &#xA;  &lt;p&gt;This rule is employed by the TI-83 calculator boxplot and 1-Var Stats functions.&lt;/p&gt;&#xA;  &#xA;  &lt;h3&gt;Method 2&lt;/h3&gt;&#xA;  &#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;Use the median to divide the ordered data set into two halves. Include the median into both halves.&lt;/li&gt;&#xA;  &lt;li&gt;The lower quartile value is the median of the lower half of the data. The upper quartile value is the median of the upper half of the data.&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="72" LastActivityDate="2011-02-17T18:11:18.237" CommunityOwnedDate="2011-02-17T18:11:18.237" />
  <row Id="22535" PostTypeId="1" CreationDate="2011-02-17T18:18:44.940" Score="1" ViewCount="22" Body="&lt;p&gt;Hello all you sexy math people.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm kind of stuck with two math problems, in static equilibrium analysis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, here we go:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$Y=C+G+I$&lt;/li&gt;&#xA;&lt;li&gt;$C=C(Y)$&lt;/li&gt;&#xA;&lt;li&gt;$I=I(r)$&lt;/li&gt;&#xA;&lt;li&gt;$M=M(Y,r)$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;And I am supposed to find the effect when $M_r \rightarrow \infty$, on $\frac{dY}{dG}$ and $\frac{dY}{dM}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the problem is that I would say that $G$ and $R$ are the exogenous variables, but how can then $M$ have an effect on $Y$? I just don't understand anything! All he other tasks have gone pretty well, but this one which is supposed to be easy, no can do! &lt;/p&gt;&#xA;" OwnerUserId="7197" LastEditorUserId="5105" LastEditorDisplayName="" LastEditDate="2011-02-17T23:26:24.587" LastActivityDate="2011-02-17T23:26:24.587" Title="Comparativ-static equilibrium and endo/exo-genous variables." Tags="&lt;analysis&gt;" />
  <row Id="22536" PostTypeId="2" ParentId="22513" CreationDate="2011-02-17T18:39:56.097" Score="2" ViewCount="" Body="&lt;p&gt;If you really have to start from the very basics, then maybe following an &lt;a href=&quot;http://en.wikipedia.org/wiki/Adult_education&quot; rel=&quot;nofollow&quot;&gt;educational program for adults&lt;/a&gt; is the solution. It's still the rote learning, but for the very basic stuff, provided you are motivated, this is not such a bad approach. It gets bad if you want to go into real mathematics. On the other hand, since you are already an adult, the teachers will approach the subject matter differently than they would with kids. Try to find out what the possibilities are for adult school education near your living place.&lt;/p&gt;&#xA;" OwnerUserId="3567" LastActivityDate="2011-02-17T18:39:56.097" />
  <row Id="22537" PostTypeId="1" CreationDate="2011-02-17T18:59:19.597" Score="3" ViewCount="156" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am taking a probability course and we have been of course learning the classic problems of men with hats or variations of that in some other form. The problems being of the type if n men have n hats and mix them all together, whats the probability 0, 1, 2, or k men choose there hat. But from what I can gather is that they are all problems of the same type, for any given random permutation of k objects how many fixed points are there (we weren't taught this way, just my observations). My professor taught us to solve this using conditional probabilities (conditioning on whether the first person matched and then generating a system of equations) and using the inclusion-exclusion principle. But is there any other way to reason this? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for reading..&lt;/p&gt;&#xA;" OwnerUserId="7198" LastActivityDate="2011-02-19T00:13:43.637" Title="How many fixed points in a permutation" Tags="&lt;probability&gt;&lt;combinatorics&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="22538" PostTypeId="1" AcceptedAnswerId="22545" CreationDate="2011-02-17T19:08:20.937" Score="2" ViewCount="112" Body="&lt;p&gt;I already know the harmonic series ($1/n$, which diverges) and following the link, $1/n^2$ that converges.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What other useful series could you teach me, or perhaps some general advice? Any website/resource is very welcome also, since most of my search for them reveals only the harmonic one and not much more. (I know how to execute the test!).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Have a nice day!&lt;/p&gt;&#xA;" OwnerUserId="2950" LastEditorUserId="2950" LastEditorDisplayName="" LastEditDate="2011-02-17T19:57:28.463" LastActivityDate="2011-02-17T19:57:28.463" Title="Useful series for limit comparison test" Tags="&lt;sequences-and-series&gt;&lt;convergence&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="22539" PostTypeId="2" ParentId="22537" CreationDate="2011-02-17T19:16:10.947" Score="4" ViewCount="" Body="&lt;p&gt;I believe these are called &lt;a href=&quot;http://en.wikipedia.org/wiki/Rencontres_numbers&quot; rel=&quot;nofollow&quot;&gt;Rencontres Numbers&lt;/a&gt; (the number of permutations with exactly $k$ fixed points). These are also called Partial Derangements, I believe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The wiki page I linked to gives you a recursive/generating function approach I believe (which I suppose could possibly be the conditional probability approach you mention).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope that helps.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2011-02-18T21:37:10.970" LastActivityDate="2011-02-18T21:37:10.970" CommentCount="2" />
  <row Id="22540" PostTypeId="2" ParentId="22528" CreationDate="2011-02-17T19:25:06.603" Score="3" ViewCount="" Body="&lt;p&gt;It is not true in general that saying that $E$ is a splitting field over $F$ implies that the extension is Galois: the missing ingredient is &lt;em&gt;separability&lt;/em&gt;. The implication holds for characteristic zero and in more generality for perfect fields, but not always. For an example, take $F=\mathbb{F}_p(x)$, the field of rational functions over the field of $p$ elements, and let $E$ be the splitting field over $F$ of $t^p - x$. If $\alpha$ is a root of $t^p-x$ in $E$, then $t^p - x = t^p - \alpha^p = (t-\alpha)^p$ in $E$, since $E$ has characteristic $p$. So $E=F[\alpha]$ is a splitting field of $f(t)=t^p-x$ over $F$. But we also conclude that $\mathrm{Aut}(E/F)$ consists only of the identity (since $\sigma\in\mathrm{Aut}(E/F)$ is completely determined by its value in $\alpha$, but $\alpha$ must map to itself, since it must map to a root of $t^p - x$, and $\alpha$ is the only possibility). However, since $f(t)$ is irreducible over $F$ of degree $p$, then $[E:F]=p$. Hence, $|\mathrm{Aut}(E/F)|\lt[E:F]$, and the extension &lt;strong&gt;cannot&lt;/strong&gt; be a Galois extension.  The reason it fails to be a Galois extension is that the extension is not &lt;em&gt;separable&lt;/em&gt;, since it is given by an irreducible polynomial with multiple roots.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-17T21:54:47.337" LastActivityDate="2011-02-17T21:54:47.337" CommentCount="5" />
  <row Id="22541" PostTypeId="1" AcceptedAnswerId="22544" CreationDate="2011-02-17T19:25:49.003" Score="1" ViewCount="69" Body="&lt;p&gt;Hello!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $Q$ denote the group of quaternions. Why is every subgroup of $Q \times \mathbb{Z}_{2}$ normal?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$Q= \{I,A,A^{2},A^{3},B,BA,BA^{2},BA^{3}\}$, I the identity matrix and $A=\begin{bmatrix} 0 &amp;amp; 1 \\ -1 &amp;amp; 0\end{bmatrix}$ and $B=\begin{bmatrix} 0 &amp;amp; i\\ i &amp;amp; 0\end{bmatrix}$ where $i=\sqrt{-1}$.&lt;/p&gt;&#xA;" OwnerUserId="6495" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-17T19:39:43.917" LastActivityDate="2011-02-17T20:47:39.993" Title="Quaternions and normality of subgroups of $Q \times \mathbb{Z}_{2}$" Tags="&lt;group-theory&gt;" AnswerCount="2" CommentCount="11" />
  <row Id="22542" PostTypeId="2" ParentId="22538" CreationDate="2011-02-17T19:28:16.737" Score="1" ViewCount="" Body="&lt;p&gt;General advice: consider the &lt;a href=&quot;http://en.wikipedia.org/wiki/Integral_test_for_convergence&quot; rel=&quot;nofollow&quot;&gt;integral test for convergence&lt;/a&gt;. As an exercise, apply it to the series $\sum\limits_{n = 2}^\infty  {\frac{1}{{n(\log n)^\alpha  }}}$, $\alpha &amp;gt; 0$ fixed. For which $\alpha$ does the series coverge?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another very useful test is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Alternating_series_test&quot; rel=&quot;nofollow&quot;&gt;alternating series test&lt;/a&gt;. See also &lt;a href=&quot;http://en.wikipedia.org/wiki/Convergence_tests&quot; rel=&quot;nofollow&quot;&gt;convergence tests&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2810" LastEditorUserId="2810" LastEditorDisplayName="" LastEditDate="2011-02-17T19:42:24.307" LastActivityDate="2011-02-17T19:42:24.307" CommentCount="1" />
  <row Id="22543" PostTypeId="2" ParentId="22295" CreationDate="2011-02-17T19:35:40.430" Score="2" ViewCount="" Body="&lt;p&gt;If $X$ is a topological space, then if this property holds for every open subset of $X$, then this space is called hereditarily Lindelöf. All separable metric spaces have this property, and for metric spaces it's equivalent to separability. &lt;/p&gt;&#xA;" OwnerUserId="4280" LastActivityDate="2011-02-17T19:35:40.430" />
  <row Id="22544" PostTypeId="2" ParentId="22541" CreationDate="2011-02-17T19:39:13.183" Score="4" ViewCount="" Body="&lt;p&gt;I assume you mean the group $Q_8$, the 8-element group of quaternions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Lemma.&lt;/strong&gt; Let $G$ be a group. Then every &lt;em&gt;cyclic&lt;/em&gt; subgroup of $G$ is normal if and only if &lt;em&gt;every&lt;/em&gt; subgroup of $G$ is normal. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Proof.&lt;/em&gt; Clearly, if every subgroup is normal, then every cyclic subgroup is normal. For the converse, suppose that every cyclic subgroup is normal, and let $H$ be any subgroup of $G$; we must show that $H$ is normal. To show that $H$ is normal, let $h\in H$ and $g\in G$; we must show that $ghg^{-1}\in H$. But $h\in\langle h\rangle$, hence $ghg^{-1}\in g\langle h\rangle g^{-1} = \langle h\rangle\subseteq H$ by our hypothesis that all cyclic subgroups are normal, so $ghg^{-1}\in H$, as desired. Thus, $H\triangleleft G$. $\Box$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So it suffices to show that every cyclic subgroup of $Q_8\times \mathbb{Z}_2$ is normal. Let $(a,b)\in Q_8\times\mathbb{Z}_2$, and let $(x,y)\in Q_8\times\mathbb{Z}_2$ be an arbitrary element of $Q_8\times\mathbb{Z}_2$. We want to show that $(x,y)(a,b)(x,y)^{-1}\in\langle(a,b)\rangle$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice that $(x,y)(a,b)(x,y)^{-1} = (xax^{-1},yby^{-1}) = (xax^{-1},b)$, because $\mathbb{Z}_2$ is abelian. Also, since every subgroup of $Q_8$ is normal, then $xax^{-1}\in\langle a\rangle$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, notice that $xax^{-1} = a^k$ for some $k$, and that $xax^{-1}$ has the same order as $a$ (conjugation is an automorphism). Therefore, $k$ is relatively prime to the order of $a$. Since $|Q_8| = 8$, the order of $a$ is a power of $2$, and we may assume that $k$ is odd. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, $k=2r+1$ for some $r$, so we have&#xA;$$(x,y)(a,b)(x,y)^{-1} = (xax^{-1},b) = (a^k,b) = (a^{2r+1},b).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you show now that $(x,y)(a,b)(x,y)^{-1}$ lies in $\langle(a,b)\rangle$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added.&lt;/strong&gt; For an encore, prove the &quot;easy half&quot; of Dedekind's characterization of the Hamiltonian groups (nonabelian groups in which every subgroup is normal):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Theorem.&lt;/strong&gt; If $G=Q_8\times F\times D$, where $F$ is a direct sum of (any number of) copies of $\mathbb{Z}_2$, and $D$ is an abelian group in which every element has (finite) odd order, then every subgroup of $G$ is normal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added 2.&lt;/strong&gt; A few people seem to have jumped to the conclusion that the result follows because every subgroup of $Q_8$ is normal in $Q_8$, and $\mathbb{Z}_2$ is abelian so every subgroup is normal. Intuitively, they are thinking that if every subgroup of $G$ is normal, and every subgroup of $K$ is normal, then every subgroup of $G\times K$ &quot;should&quot; be normal, since conjugation is acting on each coordinate separately. The problem with the argument is that not every subgroup of $G\times K$ is of the form $A\times B$ with $A\leq G$ and $B\leq K$, so while conjugating $(a,b)$ will map $a$ to another element of $\langle a\rangle$, and $b$ to another element of $\langle b\rangle$, this element may not be a power of $(a,b)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, $G=Q_8\times Q_8$ already fails to have the property that every subgroup is normal even though each subgroup of each factor is normal in that factor: if $H=\langle (i,i)\rangle$, then $H$ is contained in the diagonal subgroup of $G$, but conjugating by $(j,1)$ moves $(i,i)$ outside the diagonal, so $H$ is not normal. The same problem arises with the counterexample I gave in comments, using $G=Q_8\times\mathbb{Z}_4$ and the subgroup generated by $(i,1)$. &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-17T20:47:39.993" LastActivityDate="2011-02-17T20:47:39.993" CommentCount="4" />
  <row Id="22545" PostTypeId="2" ParentId="22538" CreationDate="2011-02-17T19:41:47.413" Score="3" ViewCount="" Body="&lt;p&gt;There is the generalization of what you put above, the so called $p$-series.  Let $p\in \mathbb{R}$, then $$\sum_{n=1}^\infty \frac{1}{n^p}$$ converges if and only if $p&amp;gt;1$.  This can be proven using the integral test mentioned by Shai Covo.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also of interest are &lt;a href=&quot;http://en.wikipedia.org/wiki/Alternating_series&quot; rel=&quot;nofollow&quot;&gt;alternating series&lt;/a&gt;, such as $$\sum_{n=1}^\infty \frac{(-1)^n}{\sqrt{n}}$$ since there are good tests to see if they converge.  (The above sum converges) Specifically, the &lt;a href=&quot;http://en.wikipedia.org/wiki/Alternating_series_test&quot; rel=&quot;nofollow&quot;&gt;alternating series test&lt;/a&gt; tells us that if we have a sequence $a_n$ with &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) $a_n\cdot a_{n+1} &amp;lt;0$ for every $n$ (it alternates signs)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) $|a_{n+1}|\leq |a_n|$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(3) $\lim_{n\rightarrow \infty} a_n =0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then $\sum_{n=1}^\infty a_n$ converges.  This then brings up the topic of &lt;a href=&quot;http://en.wikipedia.org/wiki/Conditional_convergence&quot; rel=&quot;nofollow&quot;&gt;Conditional&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/Absolute_convergence&quot; rel=&quot;nofollow&quot;&gt;Absolute&lt;/a&gt; convergence.  For a generalization of the alternating series test, see &lt;a href=&quot;http://en.wikipedia.org/wiki/Dirichlet%27s_test&quot; rel=&quot;nofollow&quot;&gt;Dirichlets Test&lt;/a&gt;.  (This test allows us to give the conditions of convergence for series such as $\sum_{n=1}^\infty a_n \sin (n)$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope that helps,&lt;/p&gt;&#xA;" OwnerUserId="6075" LastActivityDate="2011-02-17T19:41:47.413" />
  <row Id="22546" PostTypeId="2" ParentId="22541" CreationDate="2011-02-17T19:43:48.293" Score="0" ViewCount="" Body="&lt;p&gt;i, j, and k pairwise anti-commute and $\mathbb{Z}/(2)$ (or 2-adic integers whichever you mean) is abelian.  So $xyx^{-1}=\pm y$ for all $x,y$.  so given a subgroup $H$, $xHx^{-1}\subseteq\pm H=H$&lt;/p&gt;&#xA;" OwnerUserId="6925" LastActivityDate="2011-02-17T19:43:48.293" CommentCount="1" />
  <row Id="22547" PostTypeId="2" ParentId="18131" CreationDate="2011-02-17T19:54:09.967" Score="0" ViewCount="" Body="&lt;p&gt;always draw a picture!  you'll see the volume lying between the $xy$-plane and the plane $x+2y+3z=1$, a tetrahedron.  It's projection on the $xy$-plane is a right triangle with hypotenuse $x+2y=1$ (obtained by setting $z=0$).  IMHO, the geometric approach is better than messing with a bunch of inequalities!&lt;/p&gt;&#xA;" OwnerUserId="6925" LastActivityDate="2011-02-17T19:54:09.967" />
  <row Id="22548" PostTypeId="1" CreationDate="2011-02-17T20:02:50.327" Score="0" ViewCount="103" Body="&lt;p&gt;... Or perhaps, what are some interesting examples of simple graphs that are not known to be planar or non-planar?&lt;/p&gt;&#xA;" OwnerUserId="1358" LastActivityDate="2011-02-17T20:02:50.327" Title="What is the simplest graph that is not know to be planar or non-planar?" Tags="&lt;graph-theory&gt;" CommentCount="9" />
  <row Id="22549" PostTypeId="2" ParentId="1747" CreationDate="2011-02-17T20:08:54.877" Score="0" ViewCount="" Body="&lt;p&gt;I'm also not a math student, but a software engineer student, I can tell you that I do not know the exact answer, but I might know enough to get an upvote for my effort, so:&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Assuming it is a fair coin&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Fair coin means, that probability to get result 'head' in a single toss is 50%.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sample questions you could ask at search engine &lt;a href=&quot;http://www.wolframalpha.com&quot; rel=&quot;nofollow&quot;&gt;wolfraalpha.com&lt;/a&gt; to get answer :&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.wolframalpha.com/input/?i=probability+12+heads+8+tails&quot; rel=&quot;nofollow&quot;&gt;32 coin tosses&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.wolframalpha.com/input/?i=probability+12+heads+8+tails&quot; rel=&quot;nofollow&quot;&gt;probability 12 heads 8 tails&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.wolframalpha.com/input/?i=fewer+than+18+heads+in+40+tosses&quot; rel=&quot;nofollow&quot;&gt;fewer than 18 heads in 40 tosses&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;h2&gt;Assuming it is a unfair coin&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Fair coin means, that probability to get result 'head' in a single toss is less or greater then 50% and if you add probability to get result 'tail' you still get 100%. So basically you still have a fair coin with a tilted probability, for example {64% to get head;36% to get tail} - you can use same equations, just shift the probability. &lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Assuming it is a unfair coin with undetermined probability&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;I do not even understand, how anyone could determine that coin has a fixed probability fixed number of from samples of coin tosses. For example, I could make a function, that would change coin toss probability by after each toss or some number of tosses - you as the user, can not determine that by any number of sample sequence tosses.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Conclusion - Answer to &quot;How 'off' should the result be, for it to be probable, that the coin is not 'fair'&quot;?&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;From definition of fair coin toss, it is more probable, that coin is 'unfair' as soon as: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;'head results'!='nr. tail results'&lt;/code&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, you do 1 million coin tosses with a 'fair' coin (but you do not know that), and end up with all heads, then what could you tell me about how 'fair'/'unfair' is the coin? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can conclude indirectly, that the probability of the coin being 'fair' is 1.0100 x 10^-301030. To get probability that coin is unfair, you solve &lt;code&gt;100%-'fair coin toss probability'&lt;/code&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If this is a physics problem, then addition to measurement you also need to add combined uncertainty (Uc). So, if combined uncertainty exceeds the probability of the coin being 'fair', you can conclude that it is 'unfair' or at least - acts as one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To get combined uncertainty there are at least 3 ways I am aware of:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;multiple measurements &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&#xA;&#xA; &#xA;&#xA;  &#xA;&#xA;   &#xA;&#xA;    U&#xA;&#xA;    c&#xA;&#xA;   &#xA;&#xA;   &amp;#8290;&#xA;&#xA;   &#xA;&#xA;    (&#xA;&#xA;    x&#xA;&#xA;    )&#xA;&#xA;   &#xA;&#xA;  &#xA;&#xA;  =&#xA;&#xA;  &#xA;&#xA;   &#xA;&#xA;    &#xA;&#xA;     &#xA;&#xA;      U&#xA;&#xA;      A&#xA;&#xA;      2&#xA;&#xA;     &#xA;&#xA;     &amp;#8290;&#xA;&#xA;     &#xA;&#xA;      (&#xA;&#xA;      x&#xA;&#xA;      )&#xA;&#xA;     &#xA;&#xA;    &#xA;&#xA;    +&#xA;&#xA;    &#xA;&#xA;     &#xA;&#xA;      U&#xA;&#xA;      B&#xA;&#xA;      2&#xA;&#xA;     &#xA;&#xA;     &amp;#8290;&#xA;&#xA;     &#xA;&#xA;      &#xA;&#xA;       (&#xA;&#xA;       x&#xA;&#xA;       )&#xA;&#xA;      &#xA;&#xA;      m&#xA;&#xA;     &#xA;&#xA;    &#xA;&#xA;   &#xA;&#xA;  &#xA;&#xA; &#xA;&#xA;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;1-measurement&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&#xA;&#xA; &#xA;&#xA;  &#xA;&#xA;   &#xA;&#xA;    U&#xA;&#xA;    c&#xA;&#xA;   &#xA;&#xA;   &amp;#8290;&#xA;&#xA;   &#xA;&#xA;    (&#xA;&#xA;    x&#xA;&#xA;    )&#xA;&#xA;   &#xA;&#xA;  &#xA;&#xA;  =&#xA;&#xA;  &#xA;&#xA;   &#xA;&#xA;    &#xA;&#xA;     &#xA;&#xA;      U&#xA;&#xA;      B&#xA;&#xA;      2&#xA;&#xA;     &#xA;&#xA;     &amp;#8290;&#xA;&#xA;     &#xA;&#xA;      &#xA;&#xA;       (&#xA;&#xA;       x&#xA;&#xA;       )&#xA;&#xA;      &#xA;&#xA;      M&#xA;&#xA;     &#xA;&#xA;    &#xA;&#xA;    +&#xA;&#xA;    &#xA;&#xA;     &#xA;&#xA;      U&#xA;&#xA;      B&#xA;&#xA;      2&#xA;&#xA;     &#xA;&#xA;     &amp;#8290;&#xA;&#xA;     &#xA;&#xA;      &#xA;&#xA;       (&#xA;&#xA;       x&#xA;&#xA;       )&#xA;&#xA;      &#xA;&#xA;      b&#xA;&#xA;     &#xA;&#xA;    &#xA;&#xA;   &#xA;&#xA;  &#xA;&#xA; &#xA;&#xA;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;indirect measurements&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&#xA;&#xA; &#xA;&#xA;  &#xA;&#xA;   &#xA;&#xA;    U&#xA;&#xA;    c&#xA;&#xA;   &#xA;&#xA;   &amp;#8290;&#xA;&#xA;   &#xA;&#xA;    (&#xA;&#xA;    y&#xA;&#xA;    )&#xA;&#xA;   &#xA;&#xA;  &#xA;&#xA;  =&#xA;&#xA;  &#xA;&#xA;   &#xA;&#xA;    &#xA;&#xA;     &#xA;&#xA;      &#xA;&#xA;       &amp;#8706;&#xA;&#xA;       y&#xA;&#xA;      &#xA;&#xA;      &#xA;&#xA;       &amp;#8706;&#xA;&#xA;       &#xA;&#xA;        x&#xA;&#xA;        1&#xA;&#xA;       &#xA;&#xA;      &#xA;&#xA;     &#xA;&#xA;     &amp;#8290;&#xA;&#xA;     &#xA;&#xA;      U&#xA;&#xA;      C&#xA;&#xA;     &#xA;&#xA;     &amp;#8290;&#xA;&#xA;     &#xA;&#xA;      &#xA;&#xA;       (&#xA;&#xA;       x&#xA;&#xA;       )&#xA;&#xA;      &#xA;&#xA;      2&#xA;&#xA;     &#xA;&#xA;    &#xA;&#xA;    +&#xA;&#xA;    &amp;#8230;&#xA;&#xA;    +&#xA;&#xA;    &#xA;&#xA;     &#xA;&#xA;      &#xA;&#xA;       &amp;#8706;&#xA;&#xA;       y&#xA;&#xA;      &#xA;&#xA;      &#xA;&#xA;       &amp;#8706;&#xA;&#xA;       &#xA;&#xA;        x&#xA;&#xA;        n&#xA;&#xA;       &#xA;&#xA;      &#xA;&#xA;     &#xA;&#xA;     &amp;#8290;&#xA;&#xA;     &#xA;&#xA;      U&#xA;&#xA;      C&#xA;&#xA;     &#xA;&#xA;     &amp;#8290;&#xA;&#xA;     &#xA;&#xA;      &#xA;&#xA;       (&#xA;&#xA;       x&#xA;&#xA;       )&#xA;&#xA;      &#xA;&#xA;      2&#xA;&#xA;     &#xA;&#xA;    &#xA;&#xA;   &#xA;&#xA;  &#xA;&#xA; &#xA;&#xA;&#xA;&#xA;&lt;p&gt;Where Ua and Ub are uncertainty components: Type A and B. &lt;/p&gt;&#xA;" OwnerUserId="2317" LastActivityDate="2011-02-17T20:08:54.877" />
  <row Id="22550" PostTypeId="1" CreationDate="2011-02-17T20:16:10.287" Score="2" ViewCount="51" Body="&lt;p&gt;I've been reading about Riemannian manifolds, and have come across a comment that says that for a metric $g$ on an $N$-dimensional manifold $M$, considered as a bilinear map&#xA;$$&#xD;&#xA;g:\Omega^1(M) \times \Omega^1(M) \to C^{\infty}(M),&#xD;&#xA;$$&#xA;there exists a canonically induced bilinear map&#xA;$$&#xD;&#xA;g_k:\Omega^k(M) \times \Omega^k(M) \to C^{\infty}(M),&#xD;&#xA;$$&#xA;for all $2 \geq k \leq N$. What is this canonically induced $g_k$ defined?&lt;/p&gt;&#xA;" OwnerUserId="2099" LastEditorUserId="2099" LastEditorDisplayName="" LastEditDate="2011-02-17T20:28:50.307" LastActivityDate="2011-02-17T20:28:50.307" Title="Extension of Riemannian Metric to Higher Forms" Tags="&lt;differential-geometry&gt;&lt;geometric-construction&gt;" CommentCount="2" />
  <row Id="22551" PostTypeId="1" AcceptedAnswerId="22552" CreationDate="2011-02-17T20:19:59.627" Score="1" ViewCount="91" Body="&lt;p&gt;Is there a better way to deal with the series expansion, in particular?  I'm interested in attempting to use the series expansion to estimate the value, which I will eventually combine with more complicated expressions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, to begin, I calculated the following expansion for this series (note that $n$ is an integer, $i$ is the imaginary number, and this is a series of complex $t$):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{1}{1-e^{int}}$ = $-\frac{1}{int}\sum_{k=0}^\infty{\left(-\sum_{j=0}^\infty\frac{(int)^{j+1}}{(j+2)!}\right)^k}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem occurs when I attempt to expand the series by summing $k$ and $j$ through values less than infinity.  For instance, if I run through the values 0 to 2 for both variables, I get a maximum power of $t$ when both are 2.  This gives a term for $t^6$.  Unfortunately, I'm missing the rest of the coefficient values for this term, and so it's inaccurate.  In fact, I've found that the expansions will always be inaccurate because the series takes a term to the $k$th power.  It's like I have to find a way to run through each term of the series expansion, instead of simply going through and pluuging in a value for infinity.  In other words, I need a way to &quot;traverse&quot; through the series term by term, instead of by a simple expansion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm really just looking for an accurate way to partially expand this series, but I want to actually work with the series.  To make this clear - I want to attempt to use a series expansion.  The one I have is ackward to work with.  So I'd like to find another way to deal with series expansions, if possible.  Any help would be greatly appreciated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I'm asking is for a better way to deal with the series expansion.  Perhaps a new series? ...or perhaps a new way to get terms of the existing series.&lt;/p&gt;&#xA;" OwnerUserId="2626" LastActivityDate="2011-02-17T20:48:39.743" Title="Series Expansion of $\frac{1}{1-e^{int}}$." Tags="&lt;power-series&gt;" AnswerCount="1" />
  <row Id="22552" PostTypeId="2" ParentId="22551" CreationDate="2011-02-17T20:23:03.870" Score="3" ViewCount="" Body="&lt;p&gt;Using that $$\frac{t}{e^t-1}=\sum_{k\geq0}B_k\frac{t^k}{k!},$$ where $B_k$ is a &lt;a href=&quot;http://en.wikipedia.org/wiki/Bernoulli_number&quot; rel=&quot;nofollow&quot;&gt;Bernoulli number&lt;/a&gt;, you can get a Taylor series for your function by doing a little massaging to this series. Now, it &lt;em&gt;will&lt;/em&gt; contain Bernoulli numbers...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To actually compute, it may be simpler and better to just compute $e^{int}$ and then use the geometric series $(1-e^{int})^{-1}=\sum_{k\geq0}e^{inkt}$.&lt;/p&gt;&#xA;" OwnerUserId="274" LastEditorUserId="274" LastEditorDisplayName="" LastEditDate="2011-02-17T20:48:39.743" LastActivityDate="2011-02-17T20:48:39.743" CommentCount="2" />
  <row Id="22553" PostTypeId="1" AcceptedAnswerId="22602" CreationDate="2011-02-17T20:28:06.250" Score="2" ViewCount="176" Body="&lt;p&gt;Hello! Can anyone show some hints to this? If $\gcd(a,b)=1$, then $a^{\phi(b)}+b^{\phi(a)}=1 \pmod{ab}$.&#xA;I know that $a^{\phi(b)}=1 \pmod{b}$, and similarly, $b^{\phi(a)}=1 \pmod{a}$, but then how do I combine the work so that I get the result I need?&#xA;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="6912" LastEditorUserId="6912" LastEditorDisplayName="" LastEditDate="2011-02-18T04:48:14.493" LastActivityDate="2011-02-18T06:06:15.170" Title="Phi function application" Tags="&lt;homework&gt;&lt;elementary-number-theory&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="1" />
  <row Id="22554" PostTypeId="2" ParentId="22553" CreationDate="2011-02-17T20:31:50.387" Score="11" ViewCount="" Body="&lt;p&gt;Verify the congruence holds modulo $a$ and modulo $b$. Then use the fact that if $a|k$, $b|k$, and $\gcd(a,b)=1$, then $ab|k$. &lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2011-02-17T20:31:50.387" />
  <row Id="22555" PostTypeId="2" ParentId="22521" CreationDate="2011-02-17T20:48:16.840" Score="1" ViewCount="" Body="&lt;p&gt;I don't know if there's a quick and easy solution (I doubt it), but if you want to approach the problem numerically, I'd suggest using the spectral decomposition of $X$. Then you could alternatingly solve the minimization problem for the eigenvalues alone and for the eigenvectors alone. For each of these minimization problems, you could keep the set of diagonal elements that are summed fixed in each step; then if it turns out that after the step different diagonal elements have become the smallest ones, the target value can only decrease when you switch to summing the right elements, so this shouldn't cause oscillations. With fixed diagonal elements being summed, the minimization problem for the eigenvalues would be a trivial linear problem with solution either $a$ or $b$ for each eigenvalue. (That seems a bit too easy -- am I making a mistake there?) Since all eigenvalues will be either $a$ or $b$ in the end, you could also consider doing $N+1$ separate eigenvector optimizations with all possible combinations of eigenvalues; then you don't have to worry about the eigenvalues at all -- that would probably be less efficient but easier to code. I wouldn't be surprised to find that one could show that they will all end up $a$ anyway -- there is clearly a bias towards $a$, and they can certainly not all be $b$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regarding the minimization problem for the eigenvectors, I've done that sort of thing before, using rotation matrices that reproduce the correct gradient to linear order; I could write more about the details of that but that will get a bit more involved so I'll stop here and wait to see if you think this is at all a promising approach.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S.: Note that if it's true that all the eigenvalues will be either $a$ or $b$, then you wouldn't in fact be optimizing a full set of eigenvectors, but just a decomposition into two orthogonal eigenspaces of given dimensions. This might also lead to ideas for an analytical solution after all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.P.S.: In fact I think it does :-). I just realized that for summation over fixed diagonal elements (say, with index set $I$) you can write the target function as the sum over a quadratic form evaluated at each of the eigenvectors and multiplied by the corresponding eigenvalue:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{i\in I}\mathrm{diag}(XQ)_i = \sum_{i\in I} \vec{e}_i^\mathrm{T}\left(\sum_{j=1}^N\lambda_j x_jx_j^\mathrm{T}\right)Q\vec{e}_i=\sum_{j=1}^N\lambda_jx_j^\mathrm{T}&#xD;&#xA;\left(\sum_{i\in I}Q\vec{e}_i\vec{e}_i^\mathrm{T}\right)x_j\;.&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd be surprised if the solution for minimizing this weren't that the eigenvectors are the eigenvectors of that quadratic form and the eigenvalues are $a$ or $b$ according as the eigenvalues of the quadratic form are positive or negative.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If that's true, then you could find the global optimum by trying out all $\left(N\atop k\right)$ subsets of $k$ diagonal elements and diagonalizing the corresponding quadratic forms. But I still feel I may have made some gigantic mistake along the way -- it seems simpler than it should be now...&lt;/p&gt;&#xA;" OwnerUserId="6622" LastEditorUserId="6622" LastEditorDisplayName="" LastEditDate="2011-02-17T21:28:25.340" LastActivityDate="2011-02-17T21:28:25.340" />
  <row Id="22556" PostTypeId="1" CreationDate="2011-02-17T21:05:19.597" Score="0" ViewCount="195" Body="&lt;p&gt;So from the fourier series, we can simplify it further and use trig identities to get the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ f(t) = \frac{a_0}{2} + \sum^{\infty}_{n=1}&#xD;&#xA;\left(\frac{a_n}{2}+\frac{b_n}{2i}\right)e^{i n \omega t} +&#xD;&#xA;\left(\frac{a_n}{2}+\frac{b_n}{2i}\right)e^{-i n \omega t} $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So how do you go from the above line to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ f(t) = \sum^{\infty}_{n=-\infty} C_n e^{in\omega t} $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ f(t) = \frac{1}{\sqrt{2\pi}} \int^{\infty}_{-\infty} F(\omega) e^{-i \omega t} d\omega $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="4789" LastEditorUserId="4789" LastEditorDisplayName="" LastEditDate="2011-02-17T23:01:38.127" LastActivityDate="2011-05-19T02:59:00.800" Title="Derivation of Fourier Transform?" Tags="&lt;fourier-analysis&gt;" AnswerCount="1" CommentCount="6" />
  <row Id="22557" PostTypeId="1" AcceptedAnswerId="22636" CreationDate="2011-02-17T21:14:31.060" Score="2" ViewCount="85" Body="&lt;p&gt;Given an expression, for example, &#xA;$$f(x,y)=\frac{x}{(x-1)(1-y)(xy+x-1)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can we find the general expression involving $m$ and $n$ for $$\lim_{x\to 0, y\to 0}{\frac{\partial^{n+m}{f}}{\partial{x}^{m}\partial{y}^{n}}}$$&lt;/p&gt;&#xA;" OwnerUserId="6016" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-17T21:19:33.107" LastActivityDate="2011-02-18T06:14:44.333" Title="How to get the expression for higher order derivatives systematically" Tags="&lt;calculus&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="1" />
  <row Id="22559" PostTypeId="1" AcceptedAnswerId="22562" CreationDate="2011-02-17T21:24:19.667" Score="1" ViewCount="71" Body="&lt;p&gt;Hello. I was reading the proof that $A_{4}$ is the unique subgroup of order $12$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the author counts the number of conjugates:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1 cycle of type () , 6 cycles of type $(1 \ 2)$, 8 cycles of type $(1 \ 2 \ 3)$, 6 cycles of type $(1 \ 2 \ 3 \ 4)$ and 3 cycles of type $(1 \ 2)(3 \ 4)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now it says, the only possible way to get 12 elements is 1 + 3 + 8. Why is this? can't we have 1 cycle of type (), 2 cycles of type $(1 \ 2)$, 3 cycles of type  $(1 \ 2 \ 3)$, 2 cycles of type $(1 \ 2)(3 \ 4)$ and 4 cycles of type $(1 \ 2 \ 3 \ 4)$. This also gives you 12. Why is this impossible though? I don't really understand why the only possibility is 1 + 3 + 8.&lt;/p&gt;&#xA;" OwnerUserId="6495" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-17T21:26:37.323" LastActivityDate="2011-02-17T21:33:12.887" Title="$A_{4}$ unique subgroup of $S_4$ of order 12" Tags="&lt;group-theory&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="22560" PostTypeId="1" AcceptedAnswerId="22564" CreationDate="2011-02-17T21:31:45.727" Score="4" ViewCount="76" Body="&lt;p&gt;This is the equation that I use to calculate a percentage margin between cost and sales prices, where x = sales price and y = cost price:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;z=\frac{x-y}{x}*100&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This can be solved for x to give the following equation, which calculates sales price based on cost price and margin percentage:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;x=\frac{y}{1-(\frac{z}{100})}&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is, what are the steps involved in solving the first equation for x? It's been 11 years since I last did algebra at school and I can can't seem to figure it out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm guessing the first step is to divide both sides by 100 like so:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;\frac{z}{100}=\frac{x-y}{x}&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then what? Do I multiply both sides by x? If so how to I reduce the equation down to a single x?&lt;/p&gt;&#xA;" OwnerUserId="7199" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2011-02-17T21:38:45.953" LastActivityDate="2011-02-17T21:53:57.523" Title="What are the steps to solve this simple algebraic equation?" Tags="&lt;algebra&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="22561" PostTypeId="1" AcceptedAnswerId="29252" CreationDate="2011-02-17T21:31:46.367" Score="1" ViewCount="49" Body="&lt;p&gt;Does a local minimum  of a function always satify the Armijo rule?&lt;/p&gt;&#xA;" OwnerUserId="5951" LastActivityDate="2011-03-27T07:11:02.477" Title="Does a local minimum of a function always satify the Armijo rule" Tags="&lt;nonlinear-optimization&gt;" AnswerCount="1" />
  <row Id="22562" PostTypeId="2" ParentId="22559" CreationDate="2011-02-17T21:33:12.887" Score="8" ViewCount="" Body="&lt;p&gt;A subgroup of index 2 is always normal. Two elements of $S_n$ are conjugate iff they have the same cycle type. Thus, if your subgroup contains one element of a given cycle type, it contains all of them. Since the subgroup must contain the identity, and there's only one way of getting 11 as a sum from ${3,6,6,8}$, $A_4$ is the only possibility.&lt;/p&gt;&#xA;" OwnerUserId="5203" LastActivityDate="2011-02-17T21:33:12.887" />
  <row Id="22564" PostTypeId="2" ParentId="22560" CreationDate="2011-02-17T21:38:37.153" Score="7" ViewCount="" Body="&lt;p&gt;First, clear the denominator by multiplying both sides by $x$:&#xA;\begin{align*}&#xA;z &amp;amp;= \frac{100(x-y)}{x}\\&#xA;zx &amp;amp;= 100(x-y)&#xA;\end{align*}&#xA;Then move all the terms that have an $x$ in it to one side of the equation, all other terms to the other side, and factor out the $x$:&#xA;\begin{align*}&#xA;zx &amp;amp;= 100x - 100y\\&#xA;zx - 100x &amp;amp;= -100y\\&#xA;x(z-100) &amp;amp;= -100y&#xA;\end{align*}&#xA;Now divide through by $z-100$ to solve for $x$; you have to worry about dividing by $0$, but in order for $z-100$ to be $0$, you need $z=100$; the only way for $z$ to be equal to $100$ is if $\frac{x-y}{x}=1$, that is, if $x-y=x$, that is, if $y=0$. Since, presumably, you don't get the things for free, you can assume that $y\neq 0$ so this division is valid. You get:&#xA;$$x = \frac{-100y}{z-100}.$$&#xA;Now, to get it into nicer form, use the minus sign in the numerator to change the denominator from $z-100$ to $100-z$. Then divide both the numerator and the denominator by $100$ to get it into the form you have:&#xA;\begin{align*} &#xA;x &amp;amp; = \frac{-100y}{z-100}\\&#xA;x &amp;amp;= \frac{100y}{100-z}\\&#xA;x &amp;amp;= \frac{\frac{1}{100}\left(100 y\right)}{\frac{1}{100}(100-z)}\\&#xA;x &amp;amp;= \frac{y}{1 - \frac{z}{100}}.&#xA;\end{align*}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Added:&lt;/em&gt; Alternatively, following Myself's very good point, you can go &quot;unsimplify&quot; $\frac{x-y}{x}$ to $1 - \frac{y}{x}$, to go from &#xA;$$\frac{z}{100} = \frac{x-y}{x} = 1 - \frac{y}{x}$$&#xA;to&#xA;$$\frac{y}{x} = 1 - \frac{z}{100}.$$&#xA;Taking reciprocals and multiplying through by $y$ gives&#xA;\begin{align*}&#xA;\frac{x}{y} = \frac{1}{1 - \frac{z}{100}}\\&#xA;x = \frac{y}{1-\frac{z}{100}}&#xA;\end{align*}&#xA;which is probably how the particular expression you had (as opposed to $\frac{100y}{100-z}$) arose in the first place. &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-17T21:53:57.523" LastActivityDate="2011-02-17T21:53:57.523" CommentCount="1" />
  <row Id="22565" PostTypeId="2" ParentId="22560" CreationDate="2011-02-17T21:40:01.713" Score="0" ViewCount="" Body="&lt;p&gt;$$ z = 100 \cdot \frac{x-y}{x}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ zx = 100(x-y)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$zx - 100x = -100y$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x(z-100) = -100y$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x = -\frac{100y}{z-100}$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then divide both numerator and denominator by $-100$ to get $$x = \frac{y}{1-(\frac{z}{100})}$$&lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2011-02-17T21:40:01.713" />
  <row Id="22566" PostTypeId="1" CreationDate="2011-02-17T21:54:37.813" Score="5" ViewCount="120" Body="&lt;p&gt;I am trying to learn differential geometry (i.e., teach myself!)&#xA;So here is a question that came up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For some $h &amp;gt; 0$, consider the cone &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$C_h = \{ (x,y,z) \; : \;  0 \le z = \sqrt{x^2 + y^2} &amp;lt; h \} \subset \mathbb{R}^3$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;endowed with subspace topology. It seems that we can cover this with a single chart $(U,\phi)$ where $U = C_h$ and $\phi$ is the projection $\phi(x,y,z) = (x,y)$. So it seems that this defines a differentiable structure and we get a smooth ($C^\infty$) 2-dimensional manifold. (Is it correct?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now consider the inclusion map $i : C_h \to \mathbb{R}^3$, is this maps smooth? It doesn't seem to me that it is. The expression of $i$ in the chart above is not smooth at $(0,0)$ and I don't seem to be able to find any other compatible chart around zero which has a smooth representation. (Haven't given it  much thought though). If this is true how one shows that this map is not smooth. (Also, if this is true, a vague question is whether removing the origin is the only way to fix this problem)&lt;/p&gt;&#xA;" OwnerUserId="7202" LastActivityDate="2011-02-17T22:56:57.023" Title="Elementary question in differential geometry" Tags="&lt;differential-geometry&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="2" />
  <row Id="22567" PostTypeId="1" AcceptedAnswerId="22571" CreationDate="2011-02-17T22:07:50.830" Score="2" ViewCount="39" Body="&lt;p&gt;The following is a step in a proof: $f$ is a Lipschitz function from $E$ to $F$ where $E$ is a finite-dimensional Banach space and $F$ an arbitrary Banach space.  $\phi\geq 0$ is a $C^\infty$ function with compact support, $\int\phi=1$, $\phi(x)=\phi(-x)$. The function $g(z)=\int f(z+x)\phi(x)dx$ is defined.  The claim is that $||g||_{Lip}\leq||f||_{Lip}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I see why this would be true with some additional condition on $\phi$ (say, $||\phi||_{L^2}\leq 1$), but it's not clear to me why the given conditions are sufficient to control the Lipschitz constant of $g$.&lt;/p&gt;&#xA;" OwnerUserId="3763" LastActivityDate="2011-02-17T22:34:53.880" Title="Preservation of Lipschitz Constant by Convolutions" Tags="&lt;convolution&gt;&lt;banach-spaces&gt;" AnswerCount="1" />
  <row Id="22568" PostTypeId="2" ParentId="22458" CreationDate="2011-02-17T22:10:20.467" Score="0" ViewCount="" Body="&lt;p&gt;Someone has provided the answers to me.&#xA;He noted that g'(x) is 1xn not nx1 so there is no conformability problem with h'(x). Second, the derivative of the Hadamard product f(x)og(x) is diag(g(x))f'(x)+diag(f(x))g'(x).&lt;/p&gt;&#xA;" OwnerUserId="5844" LastActivityDate="2011-02-17T22:10:20.467" />
  <row Id="22569" PostTypeId="2" ParentId="22522" CreationDate="2011-02-17T22:19:24.017" Score="3" ViewCount="" Body="&lt;p&gt;I do see some nontrivial commonality here, but it is not precisely between the two theorems you suggest.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Namely, just as the injective modules are (obviously!) the injective objects in the category of $R$-modules, the algebraically closed fields are the injective objects in &#xA;the category of fields.  Indeed, the latter statement is precisely your Extension Theorem.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus your Extension Theorem doesn't strike me as a &quot;Baer criterion&quot; &lt;em&gt;per se&lt;/em&gt;.  Perhaps it is better the other way around: if we define an injective field as being a field which satisfies the conclusion of the Extension Theorem, then maybe &quot;Baer's Criterion&quot; is that a field is injective iff it is algebraically closed?  (To be clear, this is definitely a true fact; it's not clear whether it is worthy of this name.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me try this: to check whether a field $K$ is injective, it suffices to check that for every maximal ideal $I$ of $K[t]$ there exists a $K$-algebra map $K[t]/I \rightarrow K$.  This is perhaps &quot;Baeresque&quot;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that I came to the analogy here by trying to understand why the injective envelope $M \rightarrow E(M)$ is not a natural [in the categorical sense] construction: it is unique up to nonunique isomorphism over $M$.  This is just the same situation as the algebraic closure of a field.  And then I found &lt;a href=&quot;http://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.8.194&quot; rel=&quot;nofollow&quot;&gt;this paper&lt;/a&gt; by Adamek, Herrlich, Rosicky and Tholen exploring this issue in a more general categorical context.&lt;/p&gt;&#xA;" OwnerUserId="299" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2011-02-17T22:37:01.617" LastActivityDate="2011-02-17T22:37:01.617" CommentCount="2" />
  <row Id="22570" PostTypeId="2" ParentId="7002" CreationDate="2011-02-17T22:27:04.477" Score="0" ViewCount="" Body="&lt;p&gt;While studying today, I think I did a very similar exercise. It is presented as:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Being $G$ a finite group and $H \leq K \leq G$, prove that $|G:H| = |G:K|\cdot|K:H|$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I answered the following:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$|G:K|\cdot|K:H| = \frac{|G|}{|K|}\cdot\frac{|K|}{|H|} = \frac{|G|\cdot|K|}{|K|\cdot|H|} = |G:H|$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Is it the same thing or am I mistaking anything?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you for the attention.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I know this isn't an answer to your question, and if you find this post spam, please let me know.)&lt;/p&gt;&#xA;" OwnerUserId="6538" LastActivityDate="2011-02-17T22:27:04.477" CommentCount="2" />
  <row Id="22571" PostTypeId="2" ParentId="22567" CreationDate="2011-02-17T22:34:53.880" Score="3" ViewCount="" Body="&lt;p&gt;Recall Bochner's theorem: A measurable function $h: \Omega \to F$ is integrable if and only if the scalar function $\|h\|$ is integrable. This is a straightforward consequence of the usual dominated convergence theorem (approximate $h$ by simple functions).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This readily implies that $\|\int h \| \leq \int \|h\|$, see e.g. Section 2.3 in Ryan, &lt;em&gt;Introduction to tensor products of Banach spaces&lt;/em&gt;, Springer Monographs in Mathematics, 2002.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using this, you have&#xA;\[&#xA;\Vert g(z) - g(z')\Vert \leq \int \Vert f(z+x) - f(z'+x)\Vert \phi(x)\,dx \leq \|f\|_{\text{Lip}} \cdot \|z - z'\| \int \phi \, dx,&#xA;\]&#xA;so $\Vert g\Vert_{\text{Lip}} \leq \Vert f \Vert_{\text{Lip}}$ as desired.&lt;/p&gt;&#xA;" OwnerUserId="5363" LastActivityDate="2011-02-17T22:34:53.880" CommentCount="2" />
  <row Id="22572" PostTypeId="1" AcceptedAnswerId="22585" CreationDate="2011-02-17T22:38:40.337" Score="4" ViewCount="224" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Let $f$ and $g$ be functions such that $f\colon A\to B$ and $g\colon B\to C$. Prove or disprove the following&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;a) If $g\circ f$ is injective, then $g$ is injective&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here's my proof that this is true.&#xA;Let $A = \{4,5\}$, $B = \{3,9\}$, $C = \{1,2\}$&#xA;$f(4) = 3$; $f(5) = 9$&#xA;$g\circ f(4) = 1$ and $g\circ f(5) = 2$&#xA;Since no 2 elements map to the same element in the range $g\circ f$ and $g$ are both injective.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;b) If $f$ and $g$ are surjective, then $g\circ f$ is surjective. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here's my proof that this is true.&#xA;Let $A = \{1,2\}$, $B = \{4\}$, $C = \{5\}$,&#xA;$f(1) = 4$; $f(2) = 4$&#xA;$g\circ f(1) = 5$ and $g\circ f(2) = 5$.&#xA;Since no elements are left unmapped then $g\circ f$ and $f$ and $g$ are surjective&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are these proofs valid?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;**Edit:&#xA;Ok so I revised my proof for a. I will disprove it using a counterexample. &#xA;Let A = {4}, B = {3,9}, C={1}. &#xA;g∘f(4) = 1 and g(3) = 1 and g(9) = 1&#xA;g∘f is injective but g is not injective because 3 and 9 both map to 1. &#xA;Is that a valid disproof?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I'm just having some trouble disproving or proving b. If anyone could lend some tips.&#xA;Also this is not homework. Just reviewing for an exam. &lt;/p&gt;&#xA;" OwnerUserId="6331" LastEditorUserId="6331" LastEditorDisplayName="" LastEditDate="2011-02-17T23:14:09.400" LastActivityDate="2011-02-17T23:56:20.343" Title="Injective and Surjective Functions" Tags="&lt;functions&gt;&lt;elementary-set-theory&gt;" AnswerCount="3" CommentCount="5" FavoriteCount="1" />
  <row Id="22573" PostTypeId="2" ParentId="22572" CreationDate="2011-02-17T22:51:19.950" Score="1" ViewCount="" Body="&lt;p&gt;Let us consider statement a).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is interpreted as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $f: A \to B $ and $g: B \to C$ are functions such $ g \circ f$ is injective, then $g$ is injective.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If this statement were to be true,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;it means that for &lt;em&gt;any&lt;/em&gt; functions $f,g$ such that $g \circ f$ is injective it &lt;em&gt;must&lt;/em&gt; be the case that $g$ is injective.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you have done is shown a specific example of $f$ and $g$ for which this is true.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So basically you have proved:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are &lt;em&gt;some&lt;/em&gt; $f$, $g$ such that $g \circ f$ and $g$ are injective.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the statement:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;If a dog has four legs then it has a tail&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To prove this, you have to show that &lt;em&gt;every&lt;/em&gt; dog which has four legs also has a tail.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you have done is you have just taken a specific example of a dog...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now if the statement had been&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;No dog with four legs has a tail&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then by demonstrating a four-legged dog with a tail, you have disproven that statement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if you can find a specific case of $f$, $g$ such that $g \circ f$ is injective, but $g$ is not, then you have a valid proof (called proof by counter-example) that the statement is false.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2011-02-17T22:57:20.227" LastActivityDate="2011-02-17T22:57:20.227" />
  <row Id="22574" PostTypeId="2" ParentId="22572" CreationDate="2011-02-17T22:55:17.450" Score="2" ViewCount="" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Theorem. If $n$ is a number, then $n^2 = 2n$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Proof. Set $n = 2$. Then $n^2 = 4$ and $2n = 4$. Therefore $n^2 = 2n$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;... Hold on a second. For $n=3$ we have that $3^2 = 9$, but $2\cdot 3 = 6$, and $6\neq 9$. So the statement isn't valid after all! Can you see what went wrong?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same thing applies for your statements. Instead of proving them in one particular case, you should proof them for &lt;em&gt;every possible&lt;/em&gt; $f$ and $g$ you can think of.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[edit]&#xA;Here's some help with case (b), I suggest you stop reading after every step and try to complete the proof yourself. If it doesn't work, read the next hint.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;We have to show that $g\circ f$ is surjective. Give any $c\in C$, we must show there is some $x\in A$ such that $g(f(x)) = (g\circ f)(x) = c$.&lt;/li&gt;&#xA;&lt;li&gt;We know that $g$ is surjective, therefore there exists some $b\in B$ such that $g(b) = c$.&lt;/li&gt;&#xA;&lt;li&gt;We know that $f$ is surjective, therefore there exists some $a\in A$ such that $f(a) = b$.&lt;/li&gt;&#xA;&lt;li&gt;Now $c = g(b) = g(f(a))$.&lt;/li&gt;&#xA;&lt;li&gt;We conclude that such $x$ exists indeed, it is precisely the element $a$ that&#xA;we have constructed.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="5189" LastEditorUserId="5189" LastEditorDisplayName="" LastEditDate="2011-02-17T23:56:20.343" LastActivityDate="2011-02-17T23:56:20.343" />
  <row Id="22575" PostTypeId="1" AcceptedAnswerId="22579" CreationDate="2011-02-17T22:55:31.627" Score="1" ViewCount="100" Body="&lt;p&gt;Given a measure space $(\Omega, \mathbb{F},\mu)$ and any two measurable real-valued functions $f,g$ defined on $\Omega$, I was wondering if there is an inequality like&#xA;$$&#xD;&#xA;\int_\Omega |f*g| d\mu \leq \int_\Omega |f| d\mu * \int_\Omega |g| d\mu?&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or what are the correct relations between product and integral? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What if the measure is a probability measure?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks and regards!&lt;/p&gt;&#xA;" OwnerUserId="1281" LastActivityDate="2011-02-18T02:26:38.267" Title="Is there such an inequality between product and integral for functions" Tags="&lt;measure-theory&gt;" AnswerCount="3" />
  <row Id="22576" PostTypeId="2" ParentId="22566" CreationDate="2011-02-17T22:56:57.023" Score="3" ViewCount="" Body="&lt;p&gt;You've endowed $C_h$ with the structure of an &lt;em&gt;abstract manifold&lt;/em&gt; but $C_h$ is not a &lt;em&gt;submanifold&lt;/em&gt; of $\mathbb R^3$.  The fact that your set isn't a submanifold boils down to two observations: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) The fact your map $i$ is not differentiable at the origin&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) An application of the implicit function theorem gives the proof by contradiction. The implicit function theorem says that if your set was a submanifold, $i$ would have to be smooth -- technically you have to consider the two other coordinate projections $(x,y,z) \to (y,z)$, $(x,y,z) \to (x,z)$ but $C_h$ does not satisfy the &quot;vertical line rule&quot; so it can't be a graph of a function of $(y,z)$ or $(x,z)$. &lt;/p&gt;&#xA;" OwnerUserId="642" LastActivityDate="2011-02-17T22:56:57.023" CommentCount="4" />
  <row Id="22577" PostTypeId="2" ParentId="22575" CreationDate="2011-02-17T23:03:52.997" Score="1" ViewCount="" Body="&lt;p&gt;&lt;em&gt;Edit:&lt;/em&gt; I realise that we are apparently talking about the pointwise product of functions. Then the statement is false:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $f$ and $g$ be constant with value $1$. Let $\mu(\Omega)=1/2$. Then the left hand side is $1/2$ while the right hand side is $1/4$. That's a counterexample.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;First of all: Convolution only makes sense if $\Omega$ has a group structure like for instance $\mathbb R$.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you assume that, then, yes, your statement is true.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A reformulation of your estimation is that the $L^1$-norm on $L^1(G,\mu)$ is sub-multiplicative, see&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Convolution_algebra#The_convolution_algebra_L1.28G.29&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Convolution_algebra#The_convolution_algebra_L1.28G.29&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="367" LastEditorUserId="367" LastEditorDisplayName="" LastEditDate="2011-02-17T23:09:19.013" LastActivityDate="2011-02-17T23:09:19.013" />
  <row Id="22578" PostTypeId="2" ParentId="22575" CreationDate="2011-02-17T23:04:57.280" Score="1" ViewCount="" Body="&lt;p&gt;Assuming that the $*$ means product we have Cauchy-Schwarz (a special case of Hölder)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int |fg| \, d\mu = \sqrt{\int |f|^2 \, d\mu \int |g|^2 \, d\mu}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it is convolution look up the Youngs inequality for convolutions. In the common setting your RHS doesn't make any sense.&lt;/p&gt;&#xA;" OwnerUserId="1120" LastActivityDate="2011-02-17T23:04:57.280" />
  <row Id="22579" PostTypeId="2" ParentId="22575" CreationDate="2011-02-17T23:05:46.113" Score="2" ViewCount="" Body="&lt;p&gt;Below, assume that all integrals/expectations are finite.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the measure is a probability measure, then your question reads&#xA;$$&#xD;&#xA;{\rm E}|XY| \le {\rm E}|X|{\rm E}|Y|?&#xD;&#xA;$$&#xA;(Here $X$ and $Y$ are random variables.)&#xA;It does hold&#xA;$$&#xD;&#xA;|{\rm E}(XY)|^2  \le {\rm E}(X^2 ){\rm E}(Y^2 ).&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that $X$ and $Y$ (which are measurable functions from $\Omega$ to $\mathbb{R}$) correspond to $f$ and $g$. That is, the correct inequality is&#xA;$$&#xD;&#xA;\bigg|\int_\Omega  {fg\,d\mu } \bigg|^2  \le \bigg(\int_\Omega  {f^2 \,d\mu } \bigg)\bigg(\int_\Omega  {g^2 \,d\mu } \bigg)&#xD;&#xA;$$&#xA;(generalized below), where $\mu$ is the probability measure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your inequality, however, is evidently false (in general), for if $X$ is nonnegative, it gives ${\rm E}(X^2 ) \le {\rm E}^2 (X)$, or ${\rm Var}(X) \leq 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: More (but not most) generally, if $p,q \in (1,\infty)$ satisfy $1/p + 1/q = 1$, then&#xA;$$&#xD;&#xA;{\rm E}|XY| \le ({\rm E}|X|^p )^{1/p} ({\rm E}|Y|^q)^{1/q}, &#xD;&#xA;$$&#xA;or&#xA;$$&#xD;&#xA;\int_\Omega  {|fg|\,d\mu }  \le \bigg(\int_\Omega  {|f|^p \,d\mu } \bigg)^{1/p} \bigg(\int_\Omega  {|g|^q \,d\mu } \bigg)^{1/q}. &#xD;&#xA;$$&#xA;For further details, see &lt;a href=&quot;http://en.wikipedia.org/wiki/H%C3%B6lder_inequality&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; (note the section &quot;Generalization for probability measures&quot;).&lt;/p&gt;&#xA;" OwnerUserId="2810" LastEditorUserId="2810" LastEditorDisplayName="" LastEditDate="2011-02-18T02:26:38.267" LastActivityDate="2011-02-18T02:26:38.267" CommentCount="1" />
  <row Id="22580" PostTypeId="1" AcceptedAnswerId="22609" CreationDate="2011-02-17T23:14:35.067" Score="5" ViewCount="114" Body="&lt;p&gt;Given a integer $n$, we want to know if $n=m^k$ for some $m$ and $k&amp;gt;1$. What is the fastest method? (Suppose the factors of $n$ are not given).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The best method I can think of is to try to calculate $n^\frac{1}{k}$ to a certain precision, for $k$ from $2$ to $\log_2 n$. Determine if $n^\frac{1}{k}$ is a integer by test if $\lfloor n^\frac{1}{k} \rfloor ^k = n$.&lt;/p&gt;&#xA;" OwnerUserId="96" LastActivityDate="2011-02-18T01:53:37.573" Title="How to check if a integer is a power of some integer?" Tags="&lt;number-theory&gt;&lt;algorithms&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="22581" PostTypeId="1" AcceptedAnswerId="22626" CreationDate="2011-02-17T23:15:37.337" Score="6" ViewCount="134" Body="&lt;p&gt;Hi, this is a little exercise I've been fiddling with for a while now.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Let $f\colon X\to Y$ and $g\colon Y\to X$ be functions. I want to show that there are subsets $A\subseteq X$ and $B\subseteq Y$ such that $f(A)=B$ and $g(Y-B)=X-A$. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Of course, if $f$ is surjective, then taking $A=X$, one has $f(A)=B=Y$ and $g(Y-Y)=g(\emptyset)=\emptyset=X-X=X-A$, and you're done. So I suppose $f$ is not surjective.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried to approach it by contradiction. It seems that for any $A\subseteq X$, there is obviously a $B\subseteq Y$ such that $f(A)=B$, so if the result is not the case, for each pair of subsets $A$ and $f(A)$, we must have $g(Y-f(A))\neq X-A$. Then for every $f(A)\subseteq Y$, there exists a $y\in Y-f(A)$ such that $g(y)\notin X-A$. This would imply $g(y)\notin X \lor g(y)\in A$, but since $g(y)\in X$, we must have $g(y)\in A$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only thing I can glean from this is that $g$ is surjective, since for any singleton $\{x\}\subseteq X$, we could then find a $y\in Y-\{f(x)\}$ such that $g(y)=x$. But I don't quite see how to get a contradiction. What direction should I head? I attempted to apply the fact that a monotone function on power sets has a fixed point, but that only seems to apply then the function maps from a power set into itself. Thanks!&lt;/p&gt;&#xA;" OwnerUserId="6742" LastActivityDate="2011-02-18T05:00:28.260" Title="Do there exist sets $A\subseteq X$ and $B\subseteq Y$ such that $f(A)=B$ and $g(Y-B)=X-A$?" Tags="&lt;homework&gt;&lt;elementary-set-theory&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="2" />
  <row Id="22582" PostTypeId="2" ParentId="8415" CreationDate="2011-02-17T23:17:37.920" Score="1" ViewCount="" Body="&lt;p&gt;The function $f(n)$ must be $\Theta (\log n)$.  &lt;strong&gt;Update:&lt;/strong&gt; As Didier Piau points out in the comments to the &lt;a href=&quot;http://mathoverflow.net/questions/46777/asymptotic-difference-between-a-function-and-its-binomial-average&quot; rel=&quot;nofollow&quot;&gt;question after I asked it on Math Overflow&lt;/a&gt;, we can say something stronger: $\frac{f(n)}{\log_2 n} \to L$ as $n \to \infty$.  See the update at the end of the argument.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose, for some positive $L$ (the negative case is similar), $$\lim_{n \to \infty} \left(f(n) - 2^{-n} \sum_{k=0}^n \binom{n}{k} f(k) \right) = L.$$&#xA;Thus $$f(n) = L + r(n) + 2^{-n} \sum_{k=0}^n \binom{n}{k} f(k)$$ for some function $r(n)$ such that $r(n) \to 0$ as $n \to \infty$.  This gives us a recurrence relation for $f(n)$.  Since $r(n) \to 0$, $L + r(n) &amp;gt; 0$ for all sufficiently large $n$.  So, for all sufficiently large $n$, there exist positive constants $C$ and $D$ such that $C &amp;lt; L + r(n) &amp;lt; D$.  Since the initial terms in the function eventually become negligible in determining the value of $f(n)$ via the recurrence relation, there exist functions $g(n)$ and $h(n)$ such that for all sufficiently large $n$, $g(n) \leq f(n) \leq h(n)$ and $g(n)$ and $h(n)$ satisfy &#xA;$$g(n) = C + 2^{-n} \sum_{k=0}^n \binom{n}{k} g(k),$$&#xA;$$h(n) = D + 2^{-n} \sum_{k=0}^n \binom{n}{k} h(k).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the problem reduces to showing that $g(n)$ and $h(n)$ are $\Theta (\log n)$.  The argument is the same for both.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are some different ways to do this; my favorite is to interpret the $g(n)$ recurrence probabilistically.  Suppose we start at time $g(0)$, we flip a set of $n$ coins simultaneously each round, and it takes $C$ time units to do one round of flips.  When a coin turns up heads for the first time, we cease flipping it.  Let $T(n)$ be the time at which the last coin to achieve its first head does so.  To find $E[T(n)]$, condition on the number of coins that achieve heads in the first round of flips.  This yields $$E[T(n)] = C + 2^{-n} \sum_{k=0}^n \binom{n}{k} E[T(n-k)] = C + 2^{-n} \sum_{k=0}^n \binom{n}{k} E[T(k)].$$  Thus $g(n) = E[T(n)]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another way to view $T(n)$ is that it is $g(0) + C M_n$, where $M_n = \max\{X_1, X_2, \ldots, X_n\}$ and the $X_i$'s are independent and identically distributed geometric $(1/2)$ random variables.  (Each geometric random variable models the first time a head appears.)  Thus $g(n) = g(0) + C E[M_n]$.  It is known that $\frac{H_n}{\log 2} \leq E[M_n] \leq \frac{H_n}{\log 2} + 1$ and, more precisely, that $E[M_n]$ is logarithmically summable to $\frac{H_n}{\log 2} + \frac{1}{2}$.  (See, for example, Bennett Eisenberg's paper &quot;On the expectation of the maximum of IID geometric random variables,&quot; &lt;em&gt;Statistics and Probability Letters&lt;/em&gt;  78 (2008) 135-143.  See also this MO question, &quot;&lt;a href=&quot;http://mathoverflow.net/questions/41604/what-is-the-expected-maximum-out-of-a-sample-size-n-from-a-geometric-distributi/&quot; rel=&quot;nofollow&quot;&gt;What is the Expected Maximum out of a Sample (size $n$) from a Geometric Distribution&lt;/a&gt;?&quot;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus $g(n) = \frac{C}{\log 2} \log n + O(1)$, which means that $h(n) = \frac{D}{\log 2} \log n + O(1)$ and $f(n) = \Theta (\log n)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update:&lt;/strong&gt; Given $\epsilon &amp;gt; 0$, if we take take $C &amp;gt; 0$ such that $L - \epsilon \leq C &amp;lt; L$ and $D = L + \epsilon$, this argument shows that &#xA;$$L - \epsilon + O\left(\frac{1}{\log n}\right) \leq \frac{f(n)}{\log_2 n} \leq L + \epsilon + O\left(\frac{1}{\log n}\right).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, as $n \to \infty$, $\frac{f(n)}{\log_2 n} \to L.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For some other ideas that pertain to this result, including what are effectively some alternative derivations, see Pradipta's recent MO question, &quot;&lt;a href=&quot;http://mathoverflow.net/questions/55625/coin-flipping-and-a-recurrence-relation&quot; rel=&quot;nofollow&quot;&gt;Coin Flipping and a Recurrence Relation&lt;/a&gt;&quot;.  In fact, reading Pradipta's question and some of its answers gave me the ideas needed to construct this argument!  So, thanks go to Pradipta, Didier Piau, Emil Jeřábek, and Louigi Addario-Berry.&lt;/p&gt;&#xA;" OwnerUserId="2370" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2011-02-21T05:48:22.920" LastActivityDate="2011-02-21T05:48:22.920" />
  <row Id="22583" PostTypeId="2" ParentId="13575" CreationDate="2011-02-17T23:20:06.027" Score="7" ViewCount="" Body="&lt;p&gt;If you want to have an overall knowledge Physics-flavored the best books are &lt;strong&gt;Nakahara&lt;/strong&gt;'s &lt;em&gt;&quot;Geometry, Topology and Physics&quot;&lt;/em&gt; and above all: &lt;strong&gt;Frankel&lt;/strong&gt;'s &lt;em&gt;&quot;The Geometry of Physics&quot;&lt;/em&gt; (great book, but sometimes his notation can bug you a lot compared to standards).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to learn Differential Topology study these in this order: &lt;strong&gt;Milnor&lt;/strong&gt;'s &lt;em&gt;&quot;Topology from a Differentiable Viewpoint&quot;&lt;/em&gt;, &lt;strong&gt;Jänich/Bröcker&lt;/strong&gt;'s &lt;em&gt;&quot;Introduction to Differential Topology&quot;&lt;/em&gt; and &lt;strong&gt;Madsen&lt;/strong&gt;'s &lt;em&gt;&quot;From Calculus to Cohomology&quot;&lt;/em&gt;. Although it is always nice to have a working knowledge of general point set topology which you can quickly learn from &lt;strong&gt;Jänich&lt;/strong&gt;'s &lt;em&gt;&quot;Topology&quot;&lt;/em&gt; and more rigorously with &lt;strong&gt;Runde&lt;/strong&gt;'s &lt;em&gt;&quot;A Taste of Topology&quot;&lt;/em&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To start Algebraic Topology these two are of great help: &lt;strong&gt;Croom&lt;/strong&gt;'s &lt;em&gt;&quot;Basic Concepts of Algebraic Topology&quot;&lt;/em&gt; and &lt;strong&gt;Sato/Hudson&lt;/strong&gt; &lt;em&gt;&quot;Algebraic Topology an intuitive approach&quot;&lt;/em&gt;. Graduate level standard references are &lt;strong&gt;Hatcher&lt;/strong&gt;'s &lt;em&gt;&quot;Algebraic Topology&quot;&lt;/em&gt; and &lt;strong&gt;Bredon&lt;/strong&gt;'s &lt;em&gt;&quot;Topology and Geometry&quot;&lt;/em&gt;, &lt;strong&gt;tom Dieck&lt;/strong&gt;'s &lt;em&gt;&quot;Algebraic Topology&quot;&lt;/em&gt; along with &lt;strong&gt;Bott/Tu&lt;/strong&gt; &lt;em&gt;&quot;Differential Forms in Algebraic Topology.&quot;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To really understand the classic and intuitive motivations for modern differential geometry you should master curves and surfaces from books like &lt;strong&gt;Toponogov&lt;/strong&gt;'s &lt;em&gt;&quot;Differential Geometry of Curves and Surfaces&quot;&lt;/em&gt; and make the transition with &lt;strong&gt;Kühnel&lt;/strong&gt;'s &lt;em&gt;&quot;Differential Geometry - Curves, Surfaces, Manifolds&quot;&lt;/em&gt;. Other nice classic texts are &lt;strong&gt;Kreyszig&lt;/strong&gt; &lt;em&gt;&quot;Differential Geometry&quot;&lt;/em&gt; and &lt;strong&gt;Struik&lt;/strong&gt;'s &lt;em&gt;&quot;Lectures on Classical Differential Geometry&quot;&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For modern differential geometry I cannot stress enough to study carefully the books of &lt;strong&gt;Jeffrey M. Lee&lt;/strong&gt; &lt;em&gt;&quot;Manifolds and Differential Geometry&quot;&lt;/em&gt; and &lt;strong&gt;Livio Nicolaescu&lt;/strong&gt;'s &lt;em&gt;&quot;Geometry of Manifolds&quot;&lt;/em&gt;. Both are deep, readable, thorough and cover a lot of topics with a very modern style and notation. In particular, Nicolaescu's is my favorite. For Riemannian Geometry I would recommend &lt;strong&gt;Jost&lt;/strong&gt;'s &lt;em&gt;&quot;Riemannian Geometry and Geometric Analysis&quot;&lt;/em&gt; and &lt;strong&gt;Petersen&lt;/strong&gt;'s &lt;em&gt;&quot;Riemannian Geometry&quot;&lt;/em&gt;. A nice introduction for Symplectic Geometry is &lt;strong&gt;Cannas da Silva&lt;/strong&gt; &lt;em&gt;&quot;Lectures on Symplectic Geometry&quot;&lt;/em&gt; or &lt;strong&gt;Berndt&lt;/strong&gt;'s &lt;em&gt;&quot;An Introduction to Symplectic Geometry&quot;&lt;/em&gt;. If you need some Lie groups and algebras the book by &lt;strong&gt;Kirilov&lt;/strong&gt; &lt;em&gt;&quot;An Introduction to Lie Groops and Lie Algebras&quot;&lt;/em&gt; is nice; for applications to geometry the best is &lt;strong&gt;Helgason&lt;/strong&gt;'s &lt;em&gt;&quot;Differential Geometry - Lie Groups and Symmetric Spaces&quot;&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;FOR TONS OF SOLVED PROBLEMS ON DIFFERENTIAL GEOMETRY the best book by far is the recent volume by &lt;strong&gt;Gadea/Muñoz - &quot;Analysis and Algebra on Differentiable Manifolds: a workbook for students and teachers&quot;&lt;/strong&gt;. From manifolds to riemannian geometry and bundles, along with amazing summary appendices for theory review and tables of useful formulas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are interested in learning Algebraic Geometry I recommend the books of my Amazon list. They are in recommended order to learn from the beginning by yourself:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.amazon.com/lm/RHQS8Y3V7LJRQ/ref=cm_pdp_lm_title_1&quot; rel=&quot;nofollow&quot;&gt;http://www.amazon.com/lm/RHQS8Y3V7LJRQ/ref=cm_pdp_lm_title_1&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, from that list, a quick path to understand basic Algebraic Geometry would be to read &lt;strong&gt;Bertrametti&lt;/strong&gt; et al. &lt;em&gt;&quot;Lectures on Curves, Surfaces and Projective Varieties&quot;&lt;/em&gt;, &lt;strong&gt;Shafarevich&lt;/strong&gt;'s &lt;em&gt;&quot;Basic Algebraic Geometry&quot;&lt;/em&gt; vol. 1, 2 and &lt;strong&gt;Perrin&lt;/strong&gt;'s &lt;em&gt;&quot;Algebraic Geometry an Introduction&quot;&lt;/em&gt;. But then you are entering the world of abstract algebra.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are interested in Complex Geometry (Kähler, Hodge...) I recommend &lt;strong&gt;Moroianu&lt;/strong&gt;'s &lt;em&gt;&quot;Lectures on Kähler Geometry&quot;&lt;/em&gt;, &lt;strong&gt;Ballmann&lt;/strong&gt;'s &lt;em&gt;&quot;Lectures on Kähler Manifolds&quot;&lt;/em&gt; and &lt;strong&gt;Huybrechts&lt;/strong&gt;' &lt;em&gt;&quot;Complex Geometry&quot;&lt;/em&gt;. To connect this with Analysis of Several Complex Variables I recommend trying &lt;strong&gt;Fritzsche/Grauert&lt;/strong&gt; &lt;em&gt;&quot;From Holomorphic Functions to Complex Manifolds&quot;&lt;/em&gt; and also &lt;strong&gt;Wells&lt;/strong&gt;' &lt;em&gt;&quot;Differential Analysis on Complex Manifolds&quot;&lt;/em&gt;. Afterwards, to connect this with algebraic geometry, try, in this order, &lt;strong&gt;Miranda&lt;/strong&gt;'s &lt;em&gt;&quot;Algebraic Curves and Riemann Surfaces&quot;&lt;/em&gt;, &lt;strong&gt;Mumford&lt;/strong&gt;'s &lt;em&gt;&quot;Algebraic Geometry - Complex Projective Varieties&quot;&lt;/em&gt;, &lt;strong&gt;Voisin&lt;/strong&gt;'s &lt;em&gt;&quot;Hodge Theory and Complex Algebraic Geometry&quot;&lt;/em&gt; vol. 1 and 2, and &lt;strong&gt;Griffiths/Harris&lt;/strong&gt; &lt;em&gt;&quot;Principles of Algebraic Geometry&quot;&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can see their table of contents at Amazon.&#xA;Hope this helps... good luck!&lt;/p&gt;&#xA;" OwnerUserId="4058" LastEditorUserId="4058" LastEditorDisplayName="" LastEditDate="2011-03-02T02:10:12.243" LastActivityDate="2011-03-02T02:10:12.243" CommentCount="2" />
  <row Id="22584" PostTypeId="1" AcceptedAnswerId="23041" CreationDate="2011-02-17T23:22:38.863" Score="6" ViewCount="159" Body="&lt;p&gt;It is well known that the solution to the classical Dido problem is a semicircle, and that the solution to the classical isoperimetric problem is a circle. It's also reasonably obvious that the solution to the following variant is a circular arc:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Let $A$ and $B$ be fixed points on a plane, and let $l$ be a length greater than $\overline{AB}$. Which (smooth) curve through $A$ and $B$, of length $l$, maximises the area between itself and the line $AB$?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It's a straightforward exercise to verify extremality using the calculus of variations, but are there alternative proofs that do not invoke e.g. the Euler-Lagrange equations? This was originally a homework problem with the isoperimetric inequality given as a hint, and I'm just wondering what the intended solution was...&lt;/p&gt;&#xA;" OwnerUserId="5191" LastActivityDate="2011-02-21T21:12:17.997" Title="The Dido problem with an arclength constraint" Tags="&lt;differential-geometry&gt;&lt;euclidean-geometry&gt;&lt;plane-curves&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="22585" PostTypeId="2" ParentId="22572" CreationDate="2011-02-17T23:24:36.270" Score="2" ViewCount="" Body="&lt;p&gt;The idea of a mathematical proof is that without dependence of a specific input, if certain properties are true, then the conclusion is also true.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case, if the composition is injective then it has to be that the &quot;outer&quot; function is injective, and that if the functions are surjective then their composition is as such.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Showing a specific case is a valid method for disproving a claim, as it shows that at a certain time the properties hold but the conclusion is false.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, consider $A=\{0\}$ and $B=\{1,2\}$ and $C=\{0\}$ and $f(0)=1$ and $g(1)=0, g(2)=0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The composition $g\circ f$ is a function from $A$ to $C$, since $A$ is a singleton every function is injective. However $g$ is clearly not injective.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a counterexample to the claim, thus disproving it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The proof for the second one, if both functions are surjective then so is their composition, let us check this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $A,B,C$ any sets, and $f,g$ functions as required. To show that $g\circ f$ is surjective we want to show that every element of $C$ is in the range of $g\circ f$. Assume $c\in C$ then there is some element $b\in B$ such that $g(b)=c$. Since $f$ is surjective we have some $a\in A$ such that $f(a)=b$. It follows that $g\circ f (a) = c$ and therefore the composition of surjective functions is surjective.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, consider this proof. There was nothing to limit ourself. I did not require anything from $A,B,C$ other than them being sets, and I did not require anything from $f,g$ other than them being functions from/onto the required set in the premise of the claim. Furthermore, I did not check the surjective-ness of the composition by choosing a certain element. I picked and &lt;strong&gt;arbitrary&lt;/strong&gt; element, from a set which was also &lt;strong&gt;arbitrary&lt;/strong&gt;. Using this sort of argument assures us that the property is not dependent on any of the characteristics of the set (for example, some things are only true on finite, or infinite, or other certain sets). This allows us to have a &lt;em&gt;very general&lt;/em&gt; theorem. Whenever we have three sets, and two functions which are surjective between them, then the composition is also surjective!&lt;/p&gt;&#xA;" OwnerUserId="622" LastActivityDate="2011-02-17T23:24:36.270" CommentCount="2" />
  <row Id="22591" PostTypeId="2" ParentId="22584" CreationDate="2011-02-18T00:12:12.053" Score="0" ViewCount="" Body="&lt;p&gt;While I'm not sure that this will perfectly solve the matter for arbitrary $l$, it's easy to get this in special cases, in a way that may point the way to a general solution (in particular, by continuity) : build an isosceles triangle on $AB$ with the other two sides both equal to the radius of the appropriate circle, and consider the 'ice-cream-cone' with your curve and that the two radii, and piece together however many of these it takes to form a circle (note that this only works for arcs that are integral divisions of the appropriate circle!).  The isoperimetric inequality then says that the total area-maximizing curve is a circle, which implies that the section of curve between $AB$ that maximizes the wedge area is a circular arc, and since the area of the triangle is constant you can just subtract it out...&lt;/p&gt;&#xA;" OwnerUserId="785" LastActivityDate="2011-02-18T00:12:12.053" CommentCount="1" />
  <row Id="22592" PostTypeId="1" AcceptedAnswerId="22595" CreationDate="2011-02-18T00:17:57.207" Score="0" ViewCount="134" Body="&lt;p&gt;I want to find parametric and symmetric equations for the line of intersection of the planes $x+y+z=1$ and $x+z$=0. Here's what I have so far:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need to find a point on this line of intersection. So by setting z=0 in the equation of both planes, I get $x=0, y=1, z=0$. So $(0,1,0)$ is on the line. Now since this line is intersecting both planes, it must be perpendicular to both planes. So to find a line parallel to this line, I take the cross product of the normal vectors of both planes. I get $&amp;lt;1,-1,0&amp;gt;$. And now I'm stuck. I dont know what to do next. Any ideas?&lt;/p&gt;&#xA;" OwnerUserId="2985" LastActivityDate="2011-02-18T01:08:37.227" Title="Finding parametric and symmetric equations for a line" Tags="&lt;linear-algebra&gt;" AnswerCount="2" />
  <row Id="22593" PostTypeId="1" AcceptedAnswerId="22598" CreationDate="2011-02-18T00:17:58.690" Score="5" ViewCount="126" Body="&lt;p&gt;I first saw this on &lt;a href=&quot;http://people.missouristate.edu/lesreid/Adv92.html&quot; rel=&quot;nofollow&quot;&gt;the Missouri State problem page&lt;/a&gt; and it has never been solved there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the group generated by a,b,c, and d subject to the relations&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ab = c, bc = d, cd = a, and da = b&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the first relation, the second relation becomes bab = d. Using this expression and the first relation, we obtain&lt;/p&gt;&#xA;&#xA;&lt;p&gt;abbab = a and baba = b&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Taking the second relation above and multiplying both sides on the left by a^-1b^-1 and on the right by a^-1, we have b = a^-2. Now first relation above becomes aa^-2a^-2aa^-2 = a or a^-4 = a, hence i = a^5. Therefore our group consists of the five elements i, a, a^2, a^3, a^4. The other elements can be expressed in terms of a as follows: b = a^3, c = a^4, and d = a^2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, we get to his month's problems. How many elements are there in the groups given by the following generators and relations?&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;* Generators:a,b,c&#xA;  Relations: ab = c, bc = a, ca = b&#xA;&#xA;* Generators:a,b,c,d,e&#xA;  Relations: ab = c, bc = d, cd = e, de = a, ea = b&#xA;&#xA;* Generators:a,b,c,d,e,f&#xA;  Relations: ab = c, bc = d, cd = e, de = f, ef = a, fa = b &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Source: John H. Conway &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I recognize the first one as the quaternions but have made no progress on the other two.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-02-18T10:17:57.140" Title="A Conway problem on identifying groups" Tags="&lt;group-theory&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" />
  <row Id="22594" PostTypeId="2" ParentId="21548" CreationDate="2011-02-18T00:21:31.667" Score="4" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\displaystyle\rm\quad \frac{n^5}5\: +\: \frac{n^3}3\: +\: \frac{7\:n}{15}\ =\ \frac{n^5-n}5\: +\: \frac{n^3-n}3\: +\: n\ \in \mathbb Z\ $ by Fermat's Little Theorem.&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2011-02-18T00:21:31.667" />
  <row Id="22595" PostTypeId="2" ParentId="22592" CreationDate="2011-02-18T00:30:38.003" Score="1" ViewCount="" Body="&lt;p&gt;The line of intersection is in both planes, not perpendicular to them.  You have two equations in three unknowns, which should have a solution depending on one parameter, just like a line.  So can you solve the two equations to give something like that?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Added:  If you subtract the two equations, you get y=1.  If you plug that into the first, they become identical, x+z=0.  So all points on both planes have to satisfy both of these.  A parameterization of the line would then be (t,1,-t)&lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2011-02-18T00:43:19.113" LastActivityDate="2011-02-18T00:43:19.113" CommentCount="1" />
  <row Id="22596" PostTypeId="1" CreationDate="2011-02-18T00:34:39.557" Score="5" ViewCount="146" Body="&lt;p&gt;Hello everyone,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am trying to refresh my knowledge (and hopefully learn more) about Algorithm Analysis. I took a course on this two years ago but I am trying to catch up on what I had learned back then. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The way I am going about it is by doing exercises from the classic text Introduction to Algorithms, 2nd Edition - CLRS. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem that I am trying to solve is this: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;hr&gt;&#xA;&lt;strong&gt;Exercise 7.2-2&lt;/strong&gt; &lt;br&gt;&#xA;What is the running time of QUICKSORT when all elements of array A have the same value? &lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My Solution&lt;/strong&gt; &lt;br&gt;&#xA;The running time of QUICKSORT when all elements of array A have the same value will be equivalent to the worst case running of QUICKSORT since no matter what pivot is picked, QUICKSORT will have to go through all the values in A. And since all values are the same, each recursive call will lead to unbalanced partitioning. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus the recurrence will be:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$T(n) = T(n-1) + \theta(n)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The above recurrence has the solution (I will prove this later):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$T(n) = \theta(n^2)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence the running time of QUICKSORT in this case is $$\theta(n^2)$$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Please verify if I have answered the question appropriately :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please let me know if this question should be posted elsewhere in the SE network.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Jake Clawson&lt;/p&gt;&#xA;" OwnerUserId="7205" LastActivityDate="2011-02-18T00:34:39.557" Title="Quicksort Running Time" Tags="&lt;analysis&gt;&lt;algorithms&gt;&lt;self-learning&gt;" CommentCount="1" />
  <row Id="22597" PostTypeId="1" AcceptedAnswerId="22651" CreationDate="2011-02-18T00:34:49.470" Score="11" ViewCount="314" Body="&lt;p&gt;Sophie Germain proved Fermat's Last Theorem $x^p+y^p \neq z^p$ for the special case where p is a Sophie Germain prime and $p\not|xyz$.  Does any one know of a proof for the other case, where $p|xyz$? Note: I am looking for a proof restricted to the Sophie Germain primes, as of course, Wiles proved this generally.&lt;/p&gt;&#xA;" OwnerUserId="2505" LastEditorUserId="2505" LastEditorDisplayName="" LastEditDate="2011-02-21T12:54:23.397" LastActivityDate="2011-02-27T07:51:38.027" Title="Fermat's Last Theorem - Special Case of Sophie Germain Primes" Tags="&lt;number-theory&gt;&lt;reference-request&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="2" />
  <row Id="22598" PostTypeId="2" ParentId="22593" CreationDate="2011-02-18T00:36:31.633" Score="8" ViewCount="" Body="&lt;p&gt;These are the so called &quot;Fibonacci groups&quot; F(2,n), for varying n. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In D.L. Johnsons's &quot;Presentations of Groups&quot;, Johnson gives as exercise that&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;F(2,5) is finite and cyclic&lt;/li&gt;&#xA;&lt;li&gt;F(2,6) is infinite, because there exists a homomorphism $\chi \colon F(2,6)\to \mathrm{Sym}{\mathbb Z}$ such that $(\chi(a))(n) = n+1$, $(\chi(b))(n)=-n$ for $n\in \mathbb Z$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In a preceeding example, Johnson shows that F(2,4) is cyclic by making use of van Kampen diagrams, maybe that helps for F(2,5).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I haven't tried it myself, but they seem quite tedious exercises anyway.&lt;/p&gt;&#xA;" OwnerUserId="5189" LastActivityDate="2011-02-18T00:36:31.633" CommentCount="1" />
  <row Id="22599" PostTypeId="1" CreationDate="2011-02-18T00:37:08.963" Score="4" ViewCount="92" Body="&lt;p&gt;How do I prove this by induction?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Prove that for every natural number n, $ 2^0 + 2^1 + ... + 2^n = 2^{n+1}-1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is my attempt.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Base Case: let $ n = 0$ Then, $2^{0+1} - 1 = 1$ Which is true.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Inductive Step to prove is: $ 2^{n+1} = 2^{n+2} - 1$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Our hypothesis is: $2^n = 2^{n+1} -1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is where I'm getting off track. Lets look at the right side of the last equation: $2^{n+1} -1$ I can rewrite this as the following.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$2^1(2^n) - 1$ But, from our hypothesis $2^n = 2^{n+1} - 1$ Thus:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$2^1(2^{n+1} -1) -1$ This is where I get lost. Because when I distribute through I get this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$2^{n+2} -2 -1$ This is wrong is it not? Am I not applying the rules of exponents correctly here? I have the solution so I know what I'm doing is wrong. Here is the correct proof.&#xA;&lt;img src=&quot;http://i.imgur.com/6R3ik.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1040" LastActivityDate="2011-02-18T01:19:47.350" Title="How do I prove this by induction?" Tags="&lt;induction&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="22600" PostTypeId="2" ParentId="22599" CreationDate="2011-02-18T00:44:43.020" Score="7" ViewCount="" Body="&lt;p&gt;Both &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Inductive Step to prove is: $ 2^{n+1} = 2^{n+2} - 1$&lt;/li&gt;&#xA;&lt;li&gt;Our hypothesis is: $2^n = 2^{n+1} -1$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;are wrong and should be &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Inductive Step to prove is: $ 2^0 + 2^1 + ... + 2^n + 2^{n+1} = 2^{n+2} - 1$&lt;/li&gt;&#xA;&lt;li&gt;Our hypothesis is: $ 2^0 + 2^1 + ... + 2^n = 2^{n+1}-1$ &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Add $2^{n+1}$ to both sides of the hypothesis and you have the step to prove since $2^{n+1}-1 +2^{n+1} =  2^{n+2} - 1$&lt;/p&gt;&#xA;" OwnerUserId="6460" LastActivityDate="2011-02-18T00:44:43.020" />
  <row Id="22601" PostTypeId="1" AcceptedAnswerId="22604" CreationDate="2011-02-18T00:46:45.840" Score="2" ViewCount="80" Body="&lt;p&gt;Hi again!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This time I'm having trouble with the following exercise:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Being $|G| = 20$ and $H$ and $K$ subgroups of $G$ whose order is 5, prove that $K = H$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm also recommended to start by proving that $|H \cap K| = 5$.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My draft:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Let $x \in (H \cap K)\setminus{\{e\}}$. Then $|x|$ divides $|H| = |K| = 5$. Because $x \neq e$, $|x|$ can only be 5. So $|&amp;lt;x&amp;gt; = H \cap K| = 5$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Is this part OK?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But now, on to prove that $H = K$, I've been wandering around unsuccessfully (my rubber is already half the size it was earlier today :p). Can you drop any hint on this matter?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for taking the time to read!&lt;/p&gt;&#xA;" OwnerUserId="6538" LastEditorUserId="6538" LastEditorDisplayName="" LastEditDate="2011-02-18T01:10:15.773" LastActivityDate="2011-02-18T01:10:15.773" Title="Being $|G| = 20$ and $H$ and $K$ subgroups of $G$ whose order is 5, prove that $K = H$" Tags="&lt;homework&gt;&lt;group-theory&gt;" AnswerCount="1" CommentCount="9" FavoriteCount="1" />
  <row Id="22602" PostTypeId="2" ParentId="22553" CreationDate="2011-02-18T00:47:43.053" Score="5" ViewCount="" Body="&lt;p&gt;It's just $\rm\ (0,1) + (1,0)\: =\: (1,1)\ $ in $\rm\ \mathbb Z/a \times \mathbb Z/b\:,\ $ i.e. $\rm\ a^{\phi(b)}\to (0,1)\ $ being $\ 0\:$ mod $\rm\:a\:,\ 1\: $ mod $\rm\: b$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the special case $\rm\ \alpha = \beta = 1\ $ in the following form of the Chinese Remainder Theorem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;THEOREM (CRT)&lt;/strong&gt; $\rm\ \ $ If $\rm\ a,\:b\:$ are coprime integers then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\displaystyle\quad\quad\quad\quad\quad x\equiv \alpha\ (mod\ a),\ \ x\equiv \beta\ (mod\ b)\ \ \iff\ \ x\ \equiv\ \alpha\ b^{\phi(a)}+ \beta\ a^{\phi(b)} \ \ (mod\ a\:b)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof&lt;/strong&gt; $\rm\ (\Leftarrow)\ \ $ Immediate from $\rm\ b^{\phi(a)}\equiv 1\ (mod\ a),\ \ a^{\phi(b)}\equiv 1\ (mod\ b)\:$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\ (\Rightarrow)\ \ $ The solution is unique $\rm\ (mod\ ab)\ $ since if $\rm\ x&amp;#39;,\:x\ $ are solutions then $\rm\ x&amp;#39;\equiv x\ $ mod $\rm\:a,b\:$ therefore $\rm\ a,\:b\ |\ x&amp;#39;-x\ \Rightarrow\ a\:b\ |\ x&amp;#39;-x\ \ $ since $\rm\ \:a,\:b\:$ coprime $\rm\:\Rightarrow\ lcm(a,b) = a\:b\:.\quad\quad$ &lt;strong&gt;QED&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2011-02-18T06:06:15.170" LastActivityDate="2011-02-18T06:06:15.170" CommentCount="1" />
  <row Id="22603" PostTypeId="1" CreationDate="2011-02-18T00:54:33.307" Score="4" ViewCount="99" Body="&lt;p&gt;On p.3 of the first volume of Spivak's Comprehensive Introduction to Differential Geometry, he says that it is an &quot;easy exercise&quot; to show that the invariance of domain theorem (if $f:U\subset\mathbb{R}^n\rightarrow\mathbb{R}^n$ is one-to-one and continuous and $U$ is open then $f(U)$ is open) implies that in his definition of a manifold&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;a metric space $M$ such that every&#xA;  point $x\in M$ has a neighborhood $U$&#xA;  of $x$ and some integer $n\geq0$ such&#xA;  that $U$ is homeomorphic to&#xA;  $\mathbb{R}^n$,&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;the neighborhood $U$ in fact must be open. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question: My proof seems to require a bit of set up, as well as two appeals two the invariance of domain theorem, including once to first prove that the dimension of a manifold is well defined (which Spivak discusses on p.4). In any case, I feel like the argument is longer than what Spivak calls a &quot;complicated little argument&quot; on the previous page. Am I missing something obvious?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps my real question is whether this kind of comment is to be expected from Spivak, since I am reading this book on my own.&lt;/p&gt;&#xA;" OwnerUserId="3656" LastEditorUserId="1543" LastEditorDisplayName="" LastEditDate="2011-04-21T22:09:42.237" LastActivityDate="2011-04-21T22:09:42.237" Title="Spivak and Invariance of Domain" Tags="&lt;differential-geometry&gt;" CommentCount="1" />
  <row Id="22604" PostTypeId="2" ParentId="22601" CreationDate="2011-02-18T00:54:39.533" Score="0" ViewCount="" Body="&lt;p&gt;I'm afraid your draft isn't quite complete, because you're assuming such $x$ exists. In other words, you're assuming $H\cap K$ is non-trivial. However,  if you could exlude that case then you would be done.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a complete reasoning, don't read it if you want to keep looking :-)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You should note that $H\cap K\leq H$, therefore $ |K\cap H|\mid |H| = 5$, because of Lagrange's theorem. In other words $|H\cap K| = 1$ or $|H\cap K| = 5$, because $5$ is prime.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now assume $|H\cap K| = 1$ indeed, (we will try to exclude this case, by deriving a contradiction). Then $|HK| = \frac{|H|\cdot |K|}{|H\cap K|} = \frac{5\cdot 5}{1} = 25$, which is impossible in a group of $20$ elements. (I'm assuming you are familiar with that formula?) Therefore this situation is not possible and $|H\cap K| = 5$ indeed.&lt;/p&gt;&#xA;" OwnerUserId="5189" LastActivityDate="2011-02-18T00:54:39.533" CommentCount="7" />
  <row Id="22605" PostTypeId="2" ParentId="22592" CreationDate="2011-02-18T01:08:37.227" Score="0" ViewCount="" Body="&lt;p&gt;You have said a wrong thing (the line is not perpendicolar to planes, but ot them normal vectors...) but your idea is right. Maybe it was a lapsus. ;-)&#xA;I think that you calculus of the vector is wrong, because the two normal vectors are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(1,1,1), \ \  (1,0,1)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and your vector (1,-1,0) is not perpendicular to them. You can easily find a perpendicular vector by eyes: (1,0,-1), so you don't need to do cross product.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your line so is parametrically $t(1,0,-1)$ but traslated from the origin (0,0,0) to (0,1,0): that's $(0,1,0)+t(1,0,-1)$. Or, if you prefer this writing:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;\left\{ \begin{aligned}&#xD;&#xA;&amp;amp;x=t \\&#xD;&#xA;&amp;amp;y=1 \\&#xD;&#xA;&amp;amp;z=-t&#xD;&#xA;\end{aligned} \right.&#xD;&#xA;&#xD;&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="3871" LastActivityDate="2011-02-18T01:08:37.227" />
  <row Id="22606" PostTypeId="2" ParentId="22599" CreationDate="2011-02-18T01:19:47.350" Score="1" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\ $ Here's the inductive proof for summing a general geometric series.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;THEOREM&lt;/strong&gt; $\rm\quad 1 + x + \cdots + x^{n-1}\ =\ \dfrac{x^{n}-1}{x-1}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof&lt;/strong&gt; $\ $ Base case: It is true for $\rm\ n = 1\:,\:$ viz. $\rm\ 1 = (x-1)/(x-1)\:$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Inductive step: Suppose it is true for $\rm\ n = k\:.\ $ Then we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\rm\ x^k + (x^{k-1} +\: \cdots\: + 1)\:\ =\:\ x^k +\frac{x^k-1}{x-1}\ =\  \frac{x^{k+1}-1}{x-1}$$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;which implies it is true for $\rm\: n = k+1\:,\:$ thus completing the inductive proof.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The proof you seek is just the special case $\rm\ x = 2\ $. &lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2011-02-18T01:19:47.350" />
  <row Id="22607" PostTypeId="1" CreationDate="2011-02-18T01:25:30.557" Score="3" ViewCount="141" Body="&lt;p&gt;I know how to handle the 2d case: &lt;a href=&quot;http://www.proofwiki.org/wiki/Set_Difference_of_Cartesian_Products&quot; rel=&quot;nofollow&quot;&gt;http://www.proofwiki.org/wiki/Set_Difference_of_Cartesian_Products&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I am having trouble simplifying the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $X=\prod_{1}^\infty X_i, A_i \subset X_i$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I simplify/rewrite $X - (A_1 \times A_2 \times \cdots A_n \times X_{n+1} \times X_{n+2} \cdots)$ with unions/intersections?&lt;/p&gt;&#xA;" OwnerUserId="7209" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2011-02-19T18:39:22.050" LastActivityDate="2011-02-19T18:45:30.607" Title="Cartesian product set difference" Tags="&lt;elementary-set-theory&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="22608" PostTypeId="2" ParentId="22556" CreationDate="2011-02-18T01:37:14.520" Score="0" ViewCount="" Body="&lt;p&gt;There's a mistake. Sin formula is $\sin(\omega nt)=\frac{e^{i \omega nt}-e^{-i\omega nt}}{2i}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so the serie become:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ f(t) = \frac{a_0}{2} + \sum^{\infty}_{n=1}&#xD;&#xA;\left(\frac{a_n}{2}+\frac{b_n}{2i}\right)e^{i n \omega t} +&#xD;&#xA;\left(\frac{a_n}{2}-\frac{b_n}{2i}\right)e^{-i n \omega t} $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, you call $C_{n}=\left(\frac{a_n}{2}+\frac{b_n}{2i}\right)$ and $C_{-n}=\left(\frac{a_n}{2}-\frac{b_n}{2i}\right)$ and obtain:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ f(t) = \frac{a_0}{2} + \sum^{\infty}_{n=1}&#xD;&#xA;C_{n}e^{i n \omega t} +&#xD;&#xA;C_{-n}e^{-i n \omega t} $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;that can be re-written in&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ f(t) = \sum^{\infty}_{n=-\infty} C_n e^{in\omega t} $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;because $e^{i\omega0t}=1$ and you can rename $\frac{a_0}{2}$ in ${c_0}e^{i\omega0t}$.&#xA;Notice that if f(t) is real $a_n$ and $b_n$ are reals. so $c_n=\overline c_{-n}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For Fourier Transform the question is a little more complicated, because Fourier Transform is an operator that you define in $L^1$, you extend it on $L^2$ and in temperate distributions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fourier serie is connected with  a function $L^1$on a interval (that become periodic on $\mathbb{R}$) or an hyperrectangle on $\mathbb{R}^n$.  Imagine you can extend this to all $R^n$...you obtain Fourier transform. The factor  $\frac{1}{2\pi}$ is arbitrary, there are other choices, but all they must be coherent with the choice of the factor of the inverse Fourier Transform. $\frac{1}{\sqrt {2\pi}}$ is the most natural, because it's very similar to normalized complex exponencials that you use in Fourier Serie for the finite interval.&#xA;And this choice makes Fourier Transform an Unitary Operator, because $F^{+}=F^{-1}$.&lt;/p&gt;&#xA;" OwnerUserId="3871" LastEditorUserId="3871" LastEditorDisplayName="" LastEditDate="2011-02-18T01:44:37.087" LastActivityDate="2011-02-18T01:44:37.087" />
  <row Id="22609" PostTypeId="2" ParentId="22580" CreationDate="2011-02-18T01:53:37.573" Score="1" ViewCount="" Body="&lt;p&gt;Like Jonas Meyer said, the &lt;a href=&quot;http://mathoverflow.net/questions/13843/how-to-quickly-determine-whether-a-given-natural-number-is-a-power-of-any-other-n&quot; rel=&quot;nofollow&quot;&gt;mathoverflow&lt;/a&gt; post contains all the information.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem can be solved in almost linear time by &lt;a href=&quot;http://cr.yp.to/papers/powers.pdf&quot; rel=&quot;nofollow&quot;&gt;this paper&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="96" LastActivityDate="2011-02-18T01:53:37.573" />
  <row Id="22610" PostTypeId="1" AcceptedAnswerId="22622" CreationDate="2011-02-18T02:05:33.527" Score="4" ViewCount="143" Body="&lt;p&gt;Hi everyone, I have got a proof of $det(AB)$=$(detA)(detB)$ in my book.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It goes as follows (for invertible A): we know that rref[A|AB]=[$I_{n}$|B]&#xA;We also know that det(A)=$(-1)^{s}k_{1}k_{2}...k_{r}$ where s is the number of row swaps needed to get to the rrefA, and $k_{i}$ are coefficients by which we divide rows of A to get to rrefA. Hence, the book concludes, det(AB)=$(-1)^{s}k_{1}k_{2}...k_{r}$(detB)=(detA)(detB), but I don't see how we go from det(AB) to $(-1)^{s}k_{1}k_{2}...k_{r}$(detB). Could you please explain the logic behind this step? (I see how (detA)(detB)=$(-1)^{s}k_{1}k_{2}...k_{r}$(detB), obviously).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The book then also uses the fact that if A is not invertible, neither will be AB (because image of AB is contained in A), so (detA)(detB)=0(detB)=0=det(AB). My question is, how would we prove that this equation holds if B is non-invertible, and hence detB=0? I could think of saying that since B is not invertible then it can't be represented as a product of elementary matrices, while A can, so AB can't be represented as such either, but that sounds hand-wavy to me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks a lot!&lt;/p&gt;&#xA;" OwnerUserId="5157" LastEditorUserId="5157" LastEditorDisplayName="" LastEditDate="2011-02-18T04:39:12.023" LastActivityDate="2011-02-18T04:39:12.023" Title="Two questions with respect to the determinants" Tags="&lt;linear-algebra&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" />
  <row Id="22611" PostTypeId="2" ParentId="8415" CreationDate="2011-02-18T02:08:04.653" Score="0" ViewCount="" Body="&lt;p&gt;A very quick and rough approach: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems probable to me that the particular weights of the averaging are not important, that things would be similar with a (eg) uniform averaging. Assuming this, in the continuous, a simple analogy would be&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle \lim_{x \to \infty } f(x) -\frac{\int_0^x f(x) dx }{x} = \lim_{x \to \infty }f(x) - \frac{F(x)}{x} = k \hspace{1cm}$  with $f(x) &amp;gt;0 $ and  $0 &amp;lt;k &amp;lt;\infty$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And $f(x) = a \log(x) + b  $  (or any function that grows logarithmically) fits the bill. And, in particular, the harmonic numbers, grows in this way.&lt;/p&gt;&#xA;" OwnerUserId="312" LastActivityDate="2011-02-18T02:08:04.653" CommentCount="3" />
  <row Id="22612" PostTypeId="2" ParentId="22607" CreationDate="2011-02-18T02:08:29.000" Score="1" ViewCount="" Body="&lt;p&gt;Try writing&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\prod_{k=n+1}^{\infty} X_k = X&amp;#39;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;then you want the difference of&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(X_1  \times X_2  \times \cdots  \times X_n  \times X&amp;#39;) - (A_1  \times A_2  \times\cdots  \times A_n  \times X&amp;#39;)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can use the rule that you linked inductively to this difference. Then note that in some parts of the expression you will get $X&amp;#39; - X&amp;#39; = \emptyset$.&lt;/p&gt;&#xA;" OwnerUserId="5105" LastEditorUserId="5105" LastEditorDisplayName="" LastEditDate="2011-02-19T08:20:25.847" LastActivityDate="2011-02-19T08:20:25.847" CommentCount="3" />
  <row Id="22613" PostTypeId="1" CreationDate="2011-02-18T02:25:36.900" Score="4" ViewCount="133" Body="&lt;p&gt;I'm a very, very logical person -- but I've never taken any math classes; really, I know.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, on that note -- how do I teach myself math? As my question might suggest, I see the answer as being an online reference that chains mathematical knowledge from simple to complex -- but I'm open to other suggestions. One thing that would be of interest is if the resource had a rating on the complexity/depth of knowledge required to reach an understanding of the given concept; yes, unlikely, but doesn't hurt to ask.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Questions, feedback, comments -- just comment, thanks!!&lt;/p&gt;&#xA;" OwnerUserId="2453" LastActivityDate="2011-02-18T02:45:22.063" Title="Is there a online reference that chains mathematical knowledge from simple to complex?" Tags="&lt;education&gt;&lt;self-learning&gt;" AnswerCount="1" />
  <row Id="22614" PostTypeId="1" CreationDate="2011-02-18T02:31:15.690" Score="2" ViewCount="58" Body="&lt;p&gt;Hello,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The BNF is defined as followed:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;S -&amp;gt; aAb | bBA &#xA;A -&amp;gt; ab | aAB&#xA;B -&amp;gt; bB | b&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The sentence is:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;aaAbBb&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;And this is the parse tree:&#xA;&lt;img src=&quot;http://i.imgur.com/gpdeo.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Phrases:&lt;/strong&gt; aaAbBb, aAbB, bB&lt;br&gt;&#xA;&lt;strong&gt;Simple Phrases:&lt;/strong&gt; bB&lt;br&gt;&#xA;&lt;strong&gt;Handle:&lt;/strong&gt; ?    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the book, &lt;code&gt;handle&lt;/code&gt; is defined as followed:&#xA;&lt;code&gt;B&lt;/code&gt; is the handle of the right sentential from y = aBw if and only if:&lt;br&gt;&#xA;$S -&amp;gt;_{rm} \cdot aAw -&amp;gt;_{rm} aBw$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in my case, what's the handle? Any idea?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks,&lt;br&gt;&#xA;Chan&lt;/p&gt;&#xA;" OwnerUserId="5711" LastEditorUserId="2736" LastEditorDisplayName="" LastEditDate="2011-02-18T03:38:34.150" LastActivityDate="2011-02-18T03:38:34.150" Title="Help understand $\text{handle}$ in parsing problem" Tags="&lt;computer-science&gt;&lt;formal-systems&gt;" CommentCount="14" />
  <row Id="22615" PostTypeId="2" ParentId="22613" CreationDate="2011-02-18T02:45:22.063" Score="2" ViewCount="" Body="&lt;p&gt;Learning maths is like building, all foundations must be down before the next level.  So the first question you need to answer well is: &lt;strong&gt;&quot;What do I understand already?&quot;&lt;/strong&gt; (note the use of &lt;em&gt;understand&lt;/em&gt;, not &lt;em&gt;know&lt;/em&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you know where you are at, then you can find the next level up, if your level is under that taught in regular schools you should be able to identify your level roughly by grade (probably per domain: number, geometry, etc.) and move up from there.  If you are already up to a degree level, then finding a natural path ahead is more subjective matter, there is much to cover but many routes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps expand your question with your current opinion of your level to get more accurate help.&lt;/p&gt;&#xA;" OwnerUserId="3726" LastActivityDate="2011-02-18T02:45:22.063" CommentCount="6" />
  <row Id="22616" PostTypeId="1" AcceptedAnswerId="22619" CreationDate="2011-02-18T02:46:58.047" Score="4" ViewCount="179" Body="&lt;p&gt;Lets say we have:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\mathbf{A=BX}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where A and B are known matrices, X is unknown. In case B was square, a solution can be found by $\mathbf{B^{-1}A=X}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But how do you attempt to solve for X when B is not square, i.e. $n\neq m$?&lt;/p&gt;&#xA;" OwnerUserId="5877" LastActivityDate="2011-02-18T10:03:12.440" Title="Solving non square matrix equations" Tags="&lt;linear-algebra&gt;&lt;matrices&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="22618" PostTypeId="1" AcceptedAnswerId="22623" CreationDate="2011-02-18T03:19:34.810" Score="2" ViewCount="53" Body="&lt;p&gt;This is for a graduate level course on computational linear systems. I want to do it myself, without help, but it's been ten years since I took linear algebra and I'm not sure I understand what the first part is asking for. The problem states:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Find a solution $x = x^* = (x_1^*, x_2^*, x_3^*, x_4^*)^T$ to the LLS problem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left\|&#xD;&#xA;\begin{pmatrix}&#xD;&#xA;1 &amp;amp; 1 &amp;amp; 1 &amp;amp; 1 \\&#xD;&#xA;0 &amp;amp; 0 &amp;amp; -e &amp;amp; 0 \\&#xD;&#xA;e &amp;amp; 0 &amp;amp; 0 &amp;amp; 0 \\&#xD;&#xA;0 &amp;amp; 0 &amp;amp; 0 &amp;amp; e \\&#xD;&#xA;0 &amp;amp; e &amp;amp; 0 &amp;amp; 0 \\&#xD;&#xA;-e &amp;amp; 0 &amp;amp; 0 &amp;amp; 0 \\&#xD;&#xA;0 &amp;amp; -e &amp;amp; 0 &amp;amp; 0 \\&#xD;&#xA;0 &amp;amp; 0 &amp;amp; 0 &amp;amp; e \\&#xD;&#xA;0 &amp;amp; 0 &amp;amp; e &amp;amp; 0&#xD;&#xA;\end{pmatrix}&#xD;&#xA;\begin{pmatrix}&#xD;&#xA;x_1 \\ x_2 \\ x_3 \\ x_4 &#xD;&#xA;\end{pmatrix}&#xD;&#xA;-&#xD;&#xA;\begin{pmatrix}&#xD;&#xA;1 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0&#xD;&#xA;\end{pmatrix}&#xD;&#xA;\right\|_2 \to \min&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;as a function of $e$.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Find a function $f(x) = f(x_1, x_2, x_3, x_4)$ such that $x$ minimizes $f(x)$ is solution $x^*$ to problem&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;In case it isn't clear from the formatting, we have an (4x9) matrix with a variable e (not the log base) multiplied by a vertical vector of x1,x2,x3,x4, take the difference of that and another vertical vector with one 1 and 8 zeros and minimize. At one point in the assignment, he calls them Normal Equations. I've looked those up in Wolfram Alpha, but I don't think it helps. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm stumped by the question labeled 1. What is the teacher asking for? This is just the beginning part. I think once I understand the question he's asking here I can move on to solving Cholesky's algorithm, etc. &lt;/p&gt;&#xA;" OwnerUserId="7206" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-18T04:17:48.580" LastActivityDate="2011-02-18T04:34:30.560" Title="How to find a representative equation for a linear system" Tags="&lt;homework&gt;&lt;linear-algebra&gt;&lt;matrices&gt;" AnswerCount="1" CommentCount="8" />
  <row Id="22619" PostTypeId="2" ParentId="22616" CreationDate="2011-02-18T03:58:11.117" Score="5" ViewCount="" Body="&lt;p&gt;Let $\mathbf{a}_1,\ldots,\mathbf{a}_k$ be the columns of $\mathbf{A}$ (so $\mathbf{A}$ is $n\times k$ for some $k$). Notice that we will have $\mathbf{B}$ is $n\times m$ and $\mathbf{X}$ is $m\times k$ for some integer $m$ (for $\mathbf{BX}=\mathbf{A}$ to work out). Let $\mathbf{x}_1,\ldots,\mathbf{x}_k$ be the columns of $\mathbf{X}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice that $\mathbf{a}_i$ depends &lt;em&gt;only&lt;/em&gt; on $\mathbf{B}$ and $\mathbf{x}_i$, since $\mathbf{Bx}_i = \mathbf{a}_i$. So to determine the $i$th column of $\mathbf{X}$, it suffices to solve the system of linear equations&#xA;$$\mathbf{B}\mathbf{x}_i = \mathbf{a}_i.$$&#xA;So finding $\mathbf{X}$ is equivalent to solving $k$ systems of linear equations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, you can just do them all at the same time. Simply take a matrix that is made up of $\mathbf{B}$ followed by $\mathbf{A}$:&#xA;$$\left(\mathbf{B}|\mathbf{A}\right)$$&#xA;and use Gauss-Jordan elimination on $\mathbf{B}$. The solutions you find for each column corresponding to $\mathbf{A}$ yield the columns of $\mathbf{X}$.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-18T04:56:55.223" LastActivityDate="2011-02-18T04:56:55.223" />
  <row Id="22620" PostTypeId="1" CreationDate="2011-02-18T04:12:00.367" Score="2" ViewCount="157" Body="&lt;p&gt;I understand what it means for a prime number to ramify in a ring of integers of a number field. However, an infinite prime is an archimedean valuation, what does it mean for an archimedean valuation to ramify in a number field?&lt;/p&gt;&#xA;" OwnerUserId="7212" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-18T04:20:54.053" LastActivityDate="2011-02-19T15:54:57.280" Title="what does it mean for a prime at infinity to ramify?" Tags="&lt;algebraic-number-theory&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="22621" PostTypeId="2" ParentId="22474" CreationDate="2011-02-18T04:13:18.460" Score="1" ViewCount="" Body="&lt;p&gt;Let ${\cal S}_p$ be the set of ordered triples $(a,b,c) \in \mathbb{N}^3$ such that $c \ge b \ge a \ge 1$, $a+b+c=p$, and $a+b &amp;gt; c$ (the triangle inequality).  For instance, ${\cal S}_3 = \{ (1,1,1) \}$ and ${\cal S}_4 = \{ \}$.  Each triple in ${\cal S}_{p-2}$ gives rise to a triple in ${\cal S}_{p}$ under the transformation $(a,b,c)\rightarrow(a,b+1,c+1)$, and all triples in ${\cal S}_{p}$ are produced by this transformation except for those of the form $(a,a,c)$.  These latter triples can be counted directly: there is one for each value of $a$ for which $3a \le p &amp;lt; 4a$, and so there are exactly $I_p = \lfloor{p/3}\rfloor - \lceil{(p+1)/4}\rceil + 1$ of them.  So we have the recurrence $S_p = S_{p-2} + I_p$ for the values of $S_p = |{\cal S}_p|$, with initial conditions $S_0=S_1=0$.  The addend has period $12$, and takes on values&#xA;$$&#xD;&#xA;I_{12k+(0,1,2,...,11)} = k + (0,0,0,1,0,0,1,1,0,1,1,1).&#xD;&#xA;$$&#xA;We then have $S_{12(k+1)} = S_{12k} + 6k + 3$ and $S_{12(k+1)+1} = S_{12k+1} + 6k + 5$, leading to quadratic solutions $S_{12k} = 3k^2$ and $S_{12k+1} = 3k^2 + 2k$.  The full set of solutions is&#xA;$$&#xD;&#xA;\begin{eqnarray}&#xD;&#xA;S_{12k} &amp;amp;=&amp;amp; 3k^2 \\&#xD;&#xA;S_{12k+1} &amp;amp;=&amp;amp; 3k^2 + 2k \\&#xD;&#xA;S_{12k+2} &amp;amp;=&amp;amp; 3k^2 + k \\&#xD;&#xA;S_{12k+3} &amp;amp;=&amp;amp; 3k^2 + 3k + 1\\&#xD;&#xA;S_{12k+4} &amp;amp;=&amp;amp; 3k^2 + 2k \\&#xD;&#xA;S_{12k+5} &amp;amp;=&amp;amp; 3k^2 + 4k + 1 \\&#xD;&#xA;S_{12k+6} &amp;amp;=&amp;amp; 3k^2 + 3k + 1 \\&#xD;&#xA;S_{12k+7} &amp;amp;=&amp;amp; 3k^2 + 5k + 2 \\&#xD;&#xA;S_{12k+8} &amp;amp;=&amp;amp; 3k^2 + 4k + 1 \\&#xD;&#xA;S_{12k+9} &amp;amp;=&amp;amp; 3k^2 + 6k + 3 \\&#xD;&#xA;S_{12k+10} &amp;amp;=&amp;amp; 3k^2 + 5k + 2 \\&#xD;&#xA;S_{12k+11} &amp;amp;=&amp;amp; 3k^2 + 7k + 4.&#xD;&#xA;\end{eqnarray}&#xD;&#xA;$$&#xA;Note that, as noted elsewhere, for all odd $p$ we have $S_{p} = S_{p+3}$.&lt;/p&gt;&#xA;" OwnerUserId="5546" LastActivityDate="2011-02-18T04:13:18.460" />
  <row Id="22622" PostTypeId="2" ParentId="22610" CreationDate="2011-02-18T04:16:28.567" Score="4" ViewCount="" Body="&lt;p&gt;If $\det(A)=0$, then you have $\det(A)\det(B) = 0 = \det(AB)$, so there is nothing to do. So all you need to do is consider the case where $A$ is invertible, i.e., where $\det(A)\neq 0$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To show the result holds if $B$ is not invertible (not necessary, but you ask about it) notice that if $B$ is not invertible then neither is $AB$ (since there exists a nontrivial solution to $B\mathbf{x}=\mathbf{0}$, and this nontrivial solution is also a nontrivial solution to $AB\mathbf{x}=\mathbf{0}$, so $AB$ is not invertible either). So $\det(AB)=0$, and equality holds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The key to the argument the book uses is that if $A$ is invertible, then it can be written as a product of elementary matrices: $A = E_1\cdots E_t$. These elementary matrices correspond to the elementary row operations performed to get rrefA (in the order of the indices). But when you left-multiply any matrix by an elementary matrix, then you perform the corresponding elementary row operation on the matrix.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, what happens if $A$ is just an elementary matrix, and you look at $\det(AB)$? If $A$ is an elementary matrix corresponding to a swap of rows, then $AB$ is the result of swapping rows of $B$, so $\det(AB) = -\det(B)$. But if $A$ is an elementary matrix corresponding to a swap of rows, then $\det(A) = -1$. So in this case, $\det(AB) = -\det(B) = \det(A)\det(B)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $A$ corresponds to adding a multiple of one row to another row, then $\det(A)=1$ and $\det(AB) = \det(B) = \det(A)\det(B)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, if $A$ corresponds to multiplying a row by $k$, then $AB$ is the result of multiplying a row of $B$ by $k$, so by the properties of the determinant you have that $\det(AB) = k\det(B)$; but again, $\det(A) = k$, so $\det(AB)=k\det(B)$ = \det(A)\det(B)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the result works if $A$ is an elementary matrix. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, how do the $E_i$ correspond to the operations you do when you compute the rref of $A$? They are the &lt;em&gt;opposite&lt;/em&gt; of what you do to $A$ to get the rref: because if $F_1,\ldots,F_t$ are what you do to $A$ to get the rref of $A$, then $F_t\cdots F_1 A = I$, so $A = (F_t\cdots F_1)^{-1} = F_1^{-1}\cdots F_t^{-1}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The inverse of an elementary matrix corresponding to a swap of rows is an elementary matrix corresponding to a swap of rows; the inverse of an elementary matrix corresponding to adding a multiple of one row to another is an elementary matrix corresponding to adding a multiple of one row to another; and the inverse of an elementary matrix that corresponds to &lt;em&gt;dividing&lt;/em&gt; a row by $k$ is the elementary matrix that corresponds to &lt;strong&gt;multiplying&lt;/strong&gt; a row by $k$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, to see how $E_1$, $E_2,\ldots,E_t$ affect $\det(B)$ when you compute $\det(E_1\cdots E_tB)$, you only need to keep track of (i) the number of times you swapped rows; and (ii) the constants you divided the rows of $A$ by to get the rref of $A$. Each row swap multiplies the determinant by $-1$, and dividing a row of $A$ by $k$ &lt;em&gt;multiplies&lt;/em&gt; the determinant by $k$. So, if $s$ is the number of times you swapped rows, and $k_1,\ldots,k_r$ are the constants by which you divided rows of $A$, then&#xA;$$\det(AB) = \det(E_1\cdots E_t B) = (-1)^sk_1\cdots k_r\det(B),$$&#xA;which is the step you were having trouble with.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope that helps.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2011-02-18T04:16:28.567" />
  <row Id="22623" PostTypeId="2" ParentId="22618" CreationDate="2011-02-18T04:18:56.970" Score="1" ViewCount="" Body="&lt;p&gt;Okay, let's call the big matrix $A$ and the vector with 9 elements $b$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, of course, you want a solution $x$ such that $Ax = b$, because then $Ax - b = 0$ and $\|Ax-b\| = 0$. My guess is that this is not possible for all $e$ (it is possible for $e = 0$, easy to check).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, what you then have is a &lt;a href=&quot;http://en.wikipedia.org/wiki/Linear_least_squares_%28mathematics%29&quot; rel=&quot;nofollow&quot;&gt;linear least squares problem&lt;/a&gt;. That is, you can &quot;solve&quot; for the $x$ such that $Ax$ is as close as it gets to $b$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In question 1, the teacher probably wants you to write down a function $f$, such that if $f(x)$ reaches it minimum at $x$, then $\|Ax - b\|$ is minimal too.&lt;/p&gt;&#xA;" OwnerUserId="5966" LastEditorUserId="5966" LastEditorDisplayName="" LastEditDate="2011-02-18T04:34:30.560" LastActivityDate="2011-02-18T04:34:30.560" CommentCount="1" />
  <row Id="22624" PostTypeId="1" AcceptedAnswerId="22635" CreationDate="2011-02-18T04:19:16.640" Score="2" ViewCount="45" Body="&lt;p&gt;I'm having trouble with this real analysis/optimization problem. The result seems intuitively obvious, but I don't know how I could possibly formalize it. Any advice is appreciated:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $X \subset \mathbb{R}^n$,  $Y \subset \mathbb{R}$, $f:x \rightarrow Y$ and $g:: Y \rightarrow \mathbb{R}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose $X$ is a compact set, $f$ and $g$ are continuous functions on their domains, and $g$ is a strictly monotonically decreasing function on its domain. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Argue that the sets referred to exist and are equivalent:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\text{argmax}_{\textbf{x} \in X} f(\textbf{x}) = \text{argmin}_{\textbf{x} \in X} g \circ f(\textbf{x})$&lt;/p&gt;&#xA;" OwnerUserId="7145" LastEditorUserId="7145" LastEditorDisplayName="" LastEditDate="2011-02-18T05:25:16.443" LastActivityDate="2011-02-18T06:05:22.373" Title="If $f$ and $g$ are cts and $g$ is monotonically decreasing , prove that $\text{argmax}_{x \in X} f(x) = \text{argmin}_{x \in X} g \circ f(x)$" Tags="&lt;real-analysis&gt;&lt;optimization&gt;" AnswerCount="1" />
  <row Id="22625" PostTypeId="2" ParentId="22610" CreationDate="2011-02-18T04:32:16.160" Score="2" ViewCount="" Body="&lt;p&gt;First, in case anyone else, like me, doesn't know the abbreviation &quot;rref&quot;: it means &quot;&lt;a href=&quot;http://en.wikipedia.org/wiki/Row_echelon_form&quot; rel=&quot;nofollow&quot;&gt;reduced row echelon form&lt;/a&gt;&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The structure of the proof in the book isn't entirely clear from your exposition. It is clear that the first part assumes that $A$ is invertible. (You didn't mention that assumption.) It's not quite clear to me whether the first part also assumes that $B$ is invertible. (If it does not, your question in the second part wouldn't affect the validity of the proof.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I) To the first part of the question (&quot;Could you please explain the logic behind this step?&quot;):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a) If we assume that $B$ is invertible, we can proceed as follows: Since $\mathrm{rref}[A|AB]=[I_n|B]$, we know that applying row operations that transform $A$ into reduced row echelon form to $AB$ yields $B$. But that means that whatever row operations we perform to get $B$ into reduced row echelon form, we can perform the same operations on $AB$ after first performing $s$ row swaps and dividing rows of $AB$ by the $k_i$. Thus, applying $\det(A)=(-1)^{s}k_{1}k_{2}...k_{r}$ to $AB$, we see that this product for $AB$ must be the corresponding product for $A$ multiplied by the corresponding product for $B$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;b) If we don't assume that $B$ is invertible in the first part, I think we need a bit more than what you've written. For instance, it would be enough to know how row operations affect the determinant, namely that a row swap multiplies the determinant by $-1$, dividing a row by $k_i$ divides the determinant by $k_i$, and adding a multiple of a row to another row leaves the determinant unchanged. Applying this to $\mathrm{rref}[A|AB]=[I_n|B]$ yields the desired equation, since $AB$ is transformed into  $B$ by $s$ row swaps and divisions by the $k_i$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;II) To the second part of the question (&quot;how would we prove that this equation holds if B is non-invertible&quot;):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As mentioned, this is only required if we assume in the first part that $B$ is invertible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your &quot;hand-waving&quot; proof seems fine to me -- more explicitly: We've dealt with the case where $A$ is not invertible. So assume that $A$ is invertible. Then if $AB$ were invertible, that would allow us to express both $A$ and $AB$ as a product of elementary matrices, which in turn would allow us to represent $B$ as such a product, which can't be if $B$ is not invertible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other ways of proving the same thing would be: If $A$ is invertible (assumed as above), we can write $B=(A^{-1}A)B=A^{-1}(AB)$. So if $AB$ were invertible, then $(AB)^{-1}A$ would be an inverse for $B$, which can't be if $B$ is not invertible. Or, in the same vein as what the book does in the case where $A$ isn't invertible: If $B$ isn't invertible, its kernel is not empty, but the kernel of $AB$ contains the kernel of $B$, and hence the kernel of $AB$ is not empty, and thus $AB$ isn't invertible. (Note that here we don't have to assume that $A$ is invertible.)&lt;/p&gt;&#xA;" OwnerUserId="6622" LastActivityDate="2011-02-18T04:32:16.160" />
  <row Id="22626" PostTypeId="2" ParentId="22581" CreationDate="2011-02-18T04:48:46.093" Score="3" ViewCount="" Body="&lt;p&gt;There's a proof &lt;a href=&quot;http://books.google.de/books?id=kvoaoWOfqd8C&amp;amp;lpg=PP1&amp;amp;ots=Jp46YyLHec&amp;amp;dq=mathematical%20foundations%20of%20computational%20engineering&amp;amp;pg=PA152#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; (where &quot;isotonic&quot; means &quot;order-preserving&quot;). Theorem (F1) there is a corollary of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Knaster-Tarski_theorem&quot; rel=&quot;nofollow&quot;&gt;Knaster–Tarski theorem&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="6622" LastEditorUserId="6622" LastEditorDisplayName="" LastEditDate="2011-02-18T05:00:28.260" LastActivityDate="2011-02-18T05:00:28.260" CommentCount="8" />
  <row Id="22627" PostTypeId="1" AcceptedAnswerId="22645" CreationDate="2011-02-18T05:07:30.810" Score="1" ViewCount="160" Body="&lt;p&gt;Let $f$ be a degree $n$ univariate polynomial in $Z[x]$, defined as $f = \sum\limits_{i=0}^{i=n} f_i x^i$. Let $$g = f^2 = f \cdot f = \sum\limits_{k=0}^{i=2n} g_k x^k$$ be the square of $f$. I am trying to find an expression for the coefficients $g_k$. So far, I have the standard expression: $$g_k = \sum\limits_{i+j=k,0 \le i,j \le n} f_i f_j.$$&#xA;But while playing with polynomials in Maple, I noticed that there is (seemingly) more structure in the coefficients. Is there a nicer formula for raising a polynomial to arbitrary powers? (and more specifically to power 2?)&lt;/p&gt;&#xA;" OwnerUserId="2468" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-18T05:08:48.110" LastActivityDate="2011-02-18T08:43:45.297" Title="Raising a polynomial to a power." Tags="&lt;polynomials&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="0" />
  <row Id="22628" PostTypeId="1" AcceptedAnswerId="22630" CreationDate="2011-02-18T05:26:02.863" Score="2" ViewCount="58" Body="&lt;p&gt;Is there any way to solve the inequation: $${e^{2x - 1}} \ge 2x$$ without plotting the graph?&lt;/p&gt;&#xA;" OwnerUserId="7018" LastActivityDate="2011-02-18T05:38:13.443" Title="How to solve this inequation without plotting the graph?" Tags="&lt;inequality&gt;" AnswerCount="1" />
  <row Id="22629" PostTypeId="1" AcceptedAnswerId="22631" CreationDate="2011-02-18T05:27:20.903" Score="5" ViewCount="188" Body="&lt;p&gt;Denote by $M_{n*n}(k)$ the ring of n by n matrices with coefficients in the field k.&lt;br&gt;&#xA;Then why does this ring not contain any two-sided ideal?&lt;br&gt;&#xA;Thanks for any clarification, and this is an exercise from the notes of commutative algebra by @Pete L Clark, of which I thought as simple but I cannot figure it out now.&lt;/p&gt;&#xA;" OwnerUserId="6792" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-18T05:57:04.493" LastActivityDate="2011-02-18T06:08:02.873" Title="Why is the ring of matrices over a field simple?" Tags="&lt;linear-algebra&gt;&lt;homework&gt;&lt;algebra&gt;" AnswerCount="1" CommentCount="6" FavoriteCount="2" />
  <row Id="22630" PostTypeId="2" ParentId="22628" CreationDate="2011-02-18T05:29:53.170" Score="4" ViewCount="" Body="&lt;p&gt;All you need to know is $e^x \geq 1 + x$, $\forall x \in \mathbb{R}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Equality holds when $x=0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To prove $e^x \geq 1 + x$, $\forall x \in \mathbb{R}$, consider $f(x) = e^x - 1 -x$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f&amp;#39;(x) = e^x - 1 = 0 \Rightarrow$, $x=0$ is an extremum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Further $f&amp;#39;&amp;#39;(x) = e^x &amp;gt; 0$, $\forall x \in \mathbb{R}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, $x=0$ is a minimum&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, we get $f(x) \geq f(0)$, $\forall x \in \mathbb{R}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, $e^x - 1 - x \geq 0$, $\forall x \in \mathbb{R}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, $e^x \geq 1 + x$, $\forall x \in \mathbb{R}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Replacing $x$ by $2x-1$, we get $e^{2x-1} \geq 1 + (2x-1) = 2x$, $\forall x \in \mathbb{R}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Equality holds when $2x-1=0$ i.e. when $x = \frac{1}{2}$&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2011-02-18T05:38:13.443" LastActivityDate="2011-02-18T05:38:13.443" />
  <row Id="22631" PostTypeId="2" ParentId="22629" CreationDate="2011-02-18T05:35:52.290" Score="5" ViewCount="" Body="&lt;p&gt;Suppose that you have an ideal $\mathfrak{I}$ which contains a matrix with a nonzero entry $a_{ij}$. Multiplying by the matrix that has $0$'s everywhere except a $1$ in entry $(i,i)$, kill all rows except the $i$th row; multiplying by a suitable matrices on the right, kill all columns except the $j$th column; now you have a matrix, necessarily in  $\mathfrak{I}$, which contains exactly one nonzero entry, namely $a_{ij}$ in position $(i,j)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now show that $\mathfrak{I}$ must contain &lt;em&gt;all&lt;/em&gt; matrices in $M_{n\times n}(k)$. This will show that a $2$-sided ideal consists either of &lt;em&gt;only&lt;/em&gt; the $0$ matrix, or must be equal to the entire ring.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Added.&lt;/em&gt; Now that you have a matrix that has a single nonzero entry, can you get a matrix that has a single nonzero entry on whatever coordinate you specify, and such that this nonzero entry is whatever element of $k$ you want, by multiplying this matrix (on either left, or right, or both) by suitable elementary matrices? Will they all be in $\mathfrak{I}$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left(\begin{array}{cc}&#xD;&#xA;a&amp;amp;b\\&#xD;&#xA;c&amp;amp;d&#xD;&#xA;\end{array}\right) = \left(\begin{array}{cc}&#xD;&#xA;a &amp;amp; 0\\&#xD;&#xA;0 &amp;amp; 0&#xD;&#xA;\end{array}\right) + \cdots$$&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-18T06:08:02.873" LastActivityDate="2011-02-18T06:08:02.873" CommentCount="5" />
  <row Id="22632" PostTypeId="2" ParentId="22627" CreationDate="2011-02-18T05:35:55.983" Score="3" ViewCount="" Body="&lt;p&gt;At least you can do it faster, using discrete Fourier transform.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $a$ be a vector with $2n$ elements whose first $n$ elements are the coefficients of the polynomial $f$, the rest zero. Let $c$ be the vector with the coefficients for $g$. Then $c$ is the cyclic convolution of $a$ with itself:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$c = a * a$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But convolution turns to multiplication under the Fourier transform $\mathcal F$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathcal F (c) = \mathcal F(a) \mathcal F(a)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;using the inverse Fourier transform you get:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$c = \mathcal F^{-1} \left( \mathcal F(a) \mathcal F(a) \right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This can of course be generalized to arbitrary powers and polynomials which are not equal.&lt;/p&gt;&#xA;" OwnerUserId="5966" LastEditorUserId="5966" LastEditorDisplayName="" LastEditDate="2011-02-18T06:10:03.723" LastActivityDate="2011-02-18T06:10:03.723" CommentCount="3" />
  <row Id="22633" PostTypeId="1" CreationDate="2011-02-18T05:39:48.913" Score="3" ViewCount="115" Body="&lt;p&gt;If two matrices from GL(2,Z) have the same determinant, will there always be a matrix from SL(2,Z) which transforms one matrix to the other?&lt;/p&gt;&#xA;" OwnerUserId="7485" LastActivityDate="2011-02-19T05:02:50.130" Title="If two matrices from GL(2,Z) have the same determinant, will there always be a matrix from SL(2,Z) which transforms one matrix to the other?" Tags="&lt;linear-algebra&gt;" AnswerCount="1" CommentCount="8" FavoriteCount="1" />
  <row Id="22634" PostTypeId="2" ParentId="22633" CreationDate="2011-02-18T05:43:32.653" Score="6" ViewCount="" Body="&lt;p&gt;If by &quot;transform&quot; you mean conjugate, no. Both&#xA;$$\left(\begin{array}{cc}&#xD;&#xA;1 &amp;amp; 1\\&#xD;&#xA;0 &amp;amp; 1&#xD;&#xA;\end{array}\right)\qquad\text{and}\qquad \left(\begin{array}{cc}1 &amp;amp; 0\\0 &amp;amp; 1&#xD;&#xA;\end{array}\right)$$&#xA;are in $\mathrm{GL}(2,\mathbb{Z})$, have determinant $1$, but they aren't conjugate even in $\mathrm{GL}(2,\mathbb{R})$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But you don't mean &quot;conjugate&quot;; you mean, given $A$ and $B$ with same determinant and in $\mathrm{GL}(2,\mathbb{Z})$, is there a matrix $C$ from $\mathrm{SL}(2,\mathbb{Z})$ that will multiply $A$ into $B$, i.e., $CA=B$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, if $CA=B$, and $C$, $A$, and $B$ are all in invertible, what is $C$? Is it in $\mathrm{SL}(2,\mathbb{Z})$?&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-18T05:55:06.867" LastActivityDate="2011-02-18T05:55:06.867" />
  <row Id="22635" PostTypeId="2" ParentId="22624" CreationDate="2011-02-18T06:05:22.373" Score="2" ViewCount="" Body="&lt;p&gt;Suppose $x \in \mathrm{argmax} f$. Equivalently, for all $y \in X$, $f(x) \geq f(y)$. This is equivalent to $g(f(x)) \leq g(f(y))$. Which is the same as $x \in \mathrm{argmin} g \circ f$.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2011-02-18T06:05:22.373" />
  <row Id="22636" PostTypeId="2" ParentId="22557" CreationDate="2011-02-18T06:14:44.333" Score="1" ViewCount="" Body="&lt;p&gt;Suppose $f = \sum_{i,j \geq 0} c_{ij} x^i y^j$. The limit you're looking for is $m!n! c_{mn}$. Your function is better written as $$f(x,y) = \frac{x}{(1-x)(1-y)(1-x-xy)},$$ where the most interesting factor is $$\frac{1}{1-x-xy} = \sum_{t \geq 0} x^t(1 + y)^t = \sum_{t,s \geq 0} \binom{t}{s} x^t y^s.$$ Multiplying by a variable $1/(1-z)$ is like taking &quot;running sums&quot; across $z$. So $$\frac{x}{(1-x)(1-y)(1-x-xy)} = \sum_{t,s \geq 0} x^{t+1} y^s \left[ \sum_{T \leq t, S \leq s} \binom{T}{S} \right]. $$ So the value you're looking for is $$m!n!\sum_{T \leq m-1,S \leq n} \binom{T}{S}.$$&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2011-02-18T06:14:44.333" />
  <row Id="22637" PostTypeId="1" AcceptedAnswerId="22638" CreationDate="2011-02-18T06:36:12.140" Score="5" ViewCount="115" Body="&lt;p&gt;Hello,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is: write 101! in canonical form.&lt;br&gt;&#xA;I solved it by try to find the highest power of each factor prime factor less than 101. When running factor() function using TI-89, I realized that the power of each prime factor decrease. &#xA;So I guess I have two questions:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;How can write this idea using Math notation? I'm think of $\prod$ and $\sum$ notation, but I don't know how to express this idea.&lt;/li&gt;&#xA;&lt;li&gt;As a programmer, I want to write a program to handle this situation. I could check every prime factor, however, I'm curious about the decrease of these prime factor. Is there any pattern behind the scene? If there is, it would increase my algorithm a bit.   &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update&lt;/strong&gt;&lt;br&gt;&#xA;Based on Moron answer, my attempt is:&#xA;$$\prod_{i=1}^{\infty}p_i^{\sum_{k=1}^{\infty} \left\lfloor\frac{n}{p^k}\right\rfloor}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does it make sense?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks,&lt;br&gt;&#xA;Chan&lt;/p&gt;&#xA;" OwnerUserId="5711" LastEditorUserId="5711" LastEditorDisplayName="" LastEditDate="2011-02-18T06:55:32.513" LastActivityDate="2011-02-18T06:55:32.513" Title="Question regarding to canonical factorization of n!?" Tags="&lt;elementary-number-theory&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="3" />
  <row Id="22638" PostTypeId="2" ParentId="22637" CreationDate="2011-02-18T06:39:43.037" Score="7" ViewCount="" Body="&lt;p&gt;You can use Legendre's formula that the highest power of a prime $p$ which divides $n!$ is given by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\text{ord}(n,p) = \sum_{k=1}^{\infty} \left\lfloor\frac{n}{p^k}\right\rfloor$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note, even though the upper limit says $\displaystyle \infty$, the summation is finite: $\displaystyle \left\lfloor\frac{n}{p^k}\right\rfloor$ is zero when $\displaystyle p^k \gt n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your case, the only primes you need to consider are all in the range $\displaystyle 1 \lt p \le 101$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can thus write $n!$ as &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$n! =  \prod_{p \le n} p^{\text{ord}(n,p)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\displaystyle \text{ord}(n,p)$ is defined as above and $\displaystyle p$ runs through the primes.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2011-02-18T06:47:13.870" LastActivityDate="2011-02-18T06:47:13.870" CommentCount="3" />
  <row Id="22639" PostTypeId="2" ParentId="22637" CreationDate="2011-02-18T06:42:47.040" Score="3" ViewCount="" Body="&lt;p&gt;First note that the largest prime dividing $N!$ has to be less than or equal to $N$. Let this prime be $p$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you observe this, write $N!$ as a product of primes $N! = 2^{\alpha_2} 3^{\alpha_2} 5^{\alpha_5} 7^{\alpha_7} 11^{\alpha_{11}} \ldots p^{\alpha_p}$ where $\alpha_i \in \mathbb{N}$ and $p$ is the largest prime not exceeding $N$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now if you want to find the maximum power of a prime $q$ dividing $N!$, it is given by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\alpha_q = \left \lfloor \frac{N}{q} \right \rfloor + \left \lfloor \frac{N}{q^2} \right \rfloor + \left \lfloor \frac{N}{q^3} \right \rfloor + \cdots$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first term appears since you want to count the number of terms less than $N$ and are multiples of $q$ and each of these contribute one $q$ to $N!$. But then when you have multiples of $q^2$ you are not adding just one $q$ but you are adding two of these primes $q$ to the product. So you now count the number of multiple of $q^2$ less than $N$ and add and so on and so forth for higher powers of $q$ less than $N$.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastActivityDate="2011-02-18T06:42:47.040" />
  <row Id="22640" PostTypeId="1" AcceptedAnswerId="22641" CreationDate="2011-02-18T07:19:01.040" Score="2" ViewCount="52" Body="&lt;p&gt;I need to find a counterexample to disprove the following statement:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for upper semi-continuous function $g: \mathbb{R}^n \rightarrow \mathbb{R}$, the set $\{x \in \mathbb{R}^n: g(x) \leq 0\}$ is closed.&lt;/p&gt;&#xA;" OwnerUserId="7145" LastActivityDate="2011-02-18T07:30:12.160" Title="Dispove: for an upper semi-continuous function $g:$, the set $\{x \in \mathbb{R}^n: g(x) \leq 0\}$ is closed" Tags="&lt;homework&gt;&lt;real-analysis&gt;" AnswerCount="1" />
  <row Id="22641" PostTypeId="2" ParentId="22640" CreationDate="2011-02-18T07:30:12.160" Score="1" ViewCount="" Body="&lt;p&gt;For the case of n=1 you can take the function $f(x)=x$ when $x&amp;lt;0$ and $f(x)=1+x$ when $x\geq 0$ then it is continuous everywhere except in $x=0$ where it is upper semi continuous. The set $f^{-1}(y\leq 0)= (-\infty,0)$ which is not close.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;you can extend this for every n, by taking a function which is continuous everywhere except a hyper plane (which in n=1 is just a point). So in n=2 do the same thing just the jump will be on the line $x=0$&lt;/p&gt;&#xA;" OwnerUserId="2125" LastActivityDate="2011-02-18T07:30:12.160" />
  <row Id="22642" PostTypeId="1" AcceptedAnswerId="22643" CreationDate="2011-02-18T07:39:27.550" Score="5" ViewCount="92" Body="&lt;p&gt;I've found some time to read a little more on set theory, and I've come across the following question. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose I have four sets $X$, $Y$, $Z$, and $W$ such that $Y\subseteq W$ and $Z\subseteq X$. Suppose also that $X\cup Y\sim Y$, where by $\sim$ I mean that the two sets $X\cup Y$ and $Y$ are equinumerous. How can I show that $Z\cup W\sim W$? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I thought the Bernstein-Schroeder theorem might be applicable. The identity function maps $W$ into $Z\cup W$ injectively, so I figured it suffices to show that there is an injection from $Z\cup W$ into $W$. From $X\cup Y\sim Y$, there is an injection $f\colon X\cup Y\to Y$, and thus $f|_X$ is an injection from $X$ into $Y$. Since $Z\subseteq X$ there is an injection from $Z$ to $X$, and likewise from $Y$ into $W$. Composing all these would give an injection from $Z$ into $W$. Those were my thoughts, but I don't think I can use them to show that $Z\cup W$ maps injectively into $W$. There must be a better way. Thanks for any help.&lt;/p&gt;&#xA;" OwnerUserId="1583" LastActivityDate="2011-02-18T08:11:43.663" Title="How can $Z\cup W\sim W$ for sets $Z$ and $W$?" Tags="&lt;elementary-set-theory&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" />
  <row Id="22643" PostTypeId="2" ParentId="22642" CreationDate="2011-02-18T07:54:48.387" Score="3" ViewCount="" Body="&lt;p&gt;$Z \cup W \subseteq X \cup W = X \cup (Y \cup (W \;\backslash Y)) = (X \cup Y) \cup (W \;\backslash Y) \sim Y \cup (W \;\backslash Y) = W \subseteq Z \cup W$&lt;/p&gt;&#xA;" OwnerUserId="6622" LastActivityDate="2011-02-18T07:54:48.387" CommentCount="5" />
  <row Id="22644" PostTypeId="2" ParentId="22642" CreationDate="2011-02-18T08:11:43.663" Score="1" ViewCount="" Body="&lt;p&gt;If $f\colon X\cup Y\to Y$ is an injection and and $Z\subseteq X$, then the restriction to $f\colon Z\cup Y\to Y$ is also an injection.  With $Y\subseteq W$ and using the identity injection $i\colon W \backslash Y \to W \backslash Y$ restricted to the injection $i\colon W \backslash (Z \cup Y) \to W \backslash Y$, you can combine $f|_{Z\cup Y}$ and $i|_{W \backslash (Z \cup Y)}$ to provide an injection from $Z\cup W$ to $W$   &lt;/p&gt;&#xA;" OwnerUserId="6460" LastActivityDate="2011-02-18T08:11:43.663" CommentCount="3" />
  <row Id="22645" PostTypeId="2" ParentId="22627" CreationDate="2011-02-18T08:31:13.653" Score="1" ViewCount="" Body="&lt;p&gt;I've written this for the more general case of the p'th power of a powerseries $ f(x)=1+ax+bx^2+... $ You can just set all non-wanted coefficients to zero. Here is a screenshot:&#xA;&lt;img src=&quot;http://i.imgur.com/6O5ad.png&quot; alt=&quot;p'th power of a power series&quot;&gt;&#xA;The numerical coefficients at some term $ a^{e_1}*b^{e_2}*c^{e_3} $depends on the occurences of the exponents and the number of factors (multinomials)&lt;/p&gt;&#xA;" OwnerUserId="1714" LastEditorUserId="1714" LastEditorDisplayName="" LastEditDate="2011-02-18T08:43:45.297" LastActivityDate="2011-02-18T08:43:45.297" />
  <row Id="22646" PostTypeId="1" AcceptedAnswerId="22665" CreationDate="2011-02-18T08:50:55.767" Score="9" ViewCount="216" Body="&lt;p&gt;I know about the &lt;a href=&quot;http://en.wikipedia.org/wiki/Finitely-generated_abelian_group#Classification&quot; rel=&quot;nofollow&quot;&gt;structure theorem for finitely generated abelian groups&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm wondering whether there exists a similar structure theorem for abelian groups that are not finitely generated. In particular, I'm interested in &lt;em&gt;torsion groups&lt;/em&gt;. Maybe having a &lt;em&gt;finite exponent&lt;/em&gt; helps?&lt;/p&gt;&#xA;" OwnerUserId="1683" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-18T16:11:56.347" LastActivityDate="2011-02-19T21:43:29.740" Title="Structure Theorem for abelian torsion groups that are not finitely generated." Tags="&lt;group-theory&gt;&lt;abelian-groups&gt;" AnswerCount="2" FavoriteCount="2" />
  <row Id="22647" PostTypeId="1" AcceptedAnswerId="22662" CreationDate="2011-02-18T09:35:40.103" Score="1" ViewCount="72" Body="&lt;p&gt;&lt;strong&gt;Premise:&lt;/strong&gt; I'm a bit unsure about the terminology most frequently used, so I'll try to be as clear as possible, please tell me if something isn't clear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Talking about regular smooth manifolds, by which I mean function $P: E \subseteq \mathbb{R}^2 \rightarrow \mathbb{R}^3$ such that $P_u \times P_v \not\equiv 0 \; \forall (u,v)\in E$ (by &quot;$\times$&quot; I mean vector product) and $P \in C^{\infty}(E)$, we can define something which I'd like to call &quot;cartesian&quot; manifolds, manifolds with parametrization that looks like $(u,v,f(u,v))$, which coincide with the graph of some function $f: \mathbb{R}^2 \rightarrow \mathbb{R}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is: is it true that if any umbilical point exists on the manifold, then it must be a maximum or a minimum of the function $f$? If this condition is too strong, would it suffice for the point to be a stationary point (namely, the gradient of $f$ vanishes)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the viceversa, would it be sufficient for a point to be umbilical on a cartesian manifold that the gradient of $f$ vanished in that point, and that such point was not flat? (A flat point is a point where the matrix of the second fundamental form is the zero matrix)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question arises in the context of my geometry class (where we covered both algebraic and differential geometry from an introductory point of view)...my professor probably said something about this (he showed us an example of a elliptic paraboloid which has its &quot;vertex&quot; as a umbilical point), but I can't remember well. Thanks in advance for any insight.&lt;/p&gt;&#xA;" OwnerUserId="1415" LastActivityDate="2011-02-18T14:10:13.907" Title="Umbilical points and maxima/minima of a function" Tags="&lt;differential-geometry&gt;" AnswerCount="1" />
  <row Id="22648" PostTypeId="1" AcceptedAnswerId="22657" CreationDate="2011-02-18T09:59:54.533" Score="1" ViewCount="34" Body="&lt;p&gt;Perhaps this is a very basic question, but I can't find a fast solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a cyclic interval [0,n] (distance from n to 0 is 1). I need a function that for a given value x returns the point in the interval where it maximizes the distance to all previous points (x-1, x-2...)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, if n=100 and f(0) = 0&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;f(1) = 50&#xA;f(2) = 25&#xA;f(3) = 75&#xA;f(4) = 12'5&#xA;f(5) = 37'5&#xA;f(6) = 62'5&#xA;f(7) = 87'5&#xA;f(8) = 6'25&#xA;f(100) = ??&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Note that f(0)=0 is for convinience, but you may start at f(0)=50.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the end, what I need is that an uniform distribution for {f(0), f(1)... f(n)}&lt;/p&gt;&#xA;" OwnerUserId="7216" LastEditorUserId="367" LastEditorDisplayName="" LastEditDate="2011-02-18T10:30:30.567" LastActivityDate="2011-02-18T12:33:30.067" Title="Maximum distance in cycle interval" Tags="&lt;algorithms&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="22649" PostTypeId="2" ParentId="22616" CreationDate="2011-02-18T10:03:12.440" Score="1" ViewCount="" Body="&lt;p&gt;In addition to Arturo's answer, maybe you could take a look at mine to this &lt;a href=&quot;http://math.stackexchange.com/questions/1822/is-reducing-a-matrix-to-row-echelon-form-useful-at-all/1831#1831&quot;&gt;question&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="664" LastActivityDate="2011-02-18T10:03:12.440" />
  <row Id="22650" PostTypeId="1" AcceptedAnswerId="22654" CreationDate="2011-02-18T10:16:57.343" Score="6" ViewCount="166" Body="&lt;p&gt;I wonder why $qvq^{-1}$ does give the rotated vector of $v$?&#xA;Are there any easy-to-understand proof for it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was on &lt;a href=&quot;http://en.wikipedia.org/wiki/Quaternions_and_spatial_rotation&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;, but I could not understand the proof there&#xA;because of the conversions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why $uv-vu$ will be $2(u \times v)$ and why $uvu$ will be $v(uu)-2(uv)u$?&lt;/p&gt;&#xA;" OwnerUserId="7217" LastEditorUserId="1543" LastEditorDisplayName="" LastEditDate="2011-02-18T10:25:55.847" LastActivityDate="2011-02-18T11:11:20.577" Title="Why quaternions represent rotations?" Tags="&lt;quaternions&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="22651" PostTypeId="2" ParentId="22597" CreationDate="2011-02-18T10:17:01.173" Score="9" ViewCount="" Body="&lt;p&gt;These two cases are traditionally called ${\bf Case \,\, 1}$ and ${\bf Case \,\, 2}$, and you are after a proof of Case 2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The essence of Dirichlet's proof of Case 2 when $p=5$ can be found in&#xA;Fermat's Last Theorem: A Genetic Introduction to Algebraic Number Theory, by Harold M. Edwards. It's in chapter three, at about page 70.&lt;/p&gt;&#xA;" OwnerUserId="1301" LastActivityDate="2011-02-18T10:17:01.173" CommentCount="3" />
  <row Id="22652" PostTypeId="2" ParentId="22593" CreationDate="2011-02-18T10:17:57.140" Score="3" ViewCount="" Body="&lt;p&gt;For your second group, we can write it as:&#xA;$$ \langle a,b\ |\ babab^2ab=a, ab^2aba=b\rangle. $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first relation gives $baba=ab^{-1}a^{-1}b^{-2}$, and plugging that into the second relation gives $ab(ab^{-1}a^{-1}b^{-2})=b$, or $aba=b^3ab$.  Plugging that again into the second relation gives $ab^2(b^3ab)=b$, or $ab^5a=1$, so $b^5=a^{-2}$.  Thus $a^2$ is central; combining the first and second relation gives $bab^2=a^2$, and plugging that into the first relation gives $ba^4b=a$.  Since $a^2$ is central, this is the same as saying $b^2=a^{-3}$. This is enough to abelianize your group, so it was abelian to begin with; a simple check then shows it is $C_{11}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is easy to see the last group is infinite: the presentation can be given as&#xA;$$ \langle a,b\ |\ b^2ab^2abab^2ab, abab^2aba\rangle;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Quotienting out by the normal subgroup $\langle a^2,b^2\rangle$ (you can check it is normal), gives the group $C_2\ast C_2$, the infinite dihedral group.&lt;/p&gt;&#xA;" OwnerUserId="641" LastActivityDate="2011-02-18T10:17:57.140" />
  <row Id="22653" PostTypeId="1" AcceptedAnswerId="22655" CreationDate="2011-02-18T10:58:45.360" Score="1" ViewCount="73" Body="&lt;p&gt;Suppose $h \in \mathbb{N}$, is there a known non trivial upper bound for&#xA;$$\left| \frac{1}{n} \sum_{m=1}^n e^{2 \pi i h (2 \pi m)} \right|?$$&lt;/p&gt;&#xA;" OwnerUserId="899" LastActivityDate="2011-02-18T11:16:39.320" Title="Non trivial upper bound for an exponential sum" Tags="&lt;diophantine-approximation&gt;&lt;exponential-sum&gt;" AnswerCount="1" />
  <row Id="22654" PostTypeId="2" ParentId="22650" CreationDate="2011-02-18T10:59:56.890" Score="8" ViewCount="" Body="&lt;p&gt;A (real) quaternion is a &quot;number&quot; of the form $q=a+bi+cj+dk$ where the coefficients $a$, $b$, $c$ and $d$ are real numbers and $i^2=j^2=-1$, $ij=k$, $jk=i$, $ki=j$, $ji=-k$ and so on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The conjugate quaternion is $\overline{q}=a-bi-cj-dk$, and the (reduced) quaternionic norm of $q$ is the real number $q\overline{q}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pure quaternions, i.e. those for which $a=0$ or equivalently $\overline{q}=-q$ form a 3-dimensional real space, an obvious basis being $\{i,j,k\}$. The quaternionic norm restricted to the space of pure quaternions turns out to be simply the euclidean norm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The transformation $u\mapsto quq^{-1}$ preserves the space of pure quaternions and preserves the norm, thus can be read as an orthogonal tranformation of ${\Bbb R}^3$ which moreover preserves orientation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One concludes recalling that an orientation-preserving orthogonal tranformation of ${\Bbb R}^3$ is rotation around some axis.&lt;/p&gt;&#xA;" OwnerUserId="688" LastActivityDate="2011-02-18T10:59:56.890" CommentCount="3" />
  <row Id="22655" PostTypeId="2" ParentId="22653" CreationDate="2011-02-18T11:06:28.727" Score="3" ViewCount="" Body="&lt;p&gt;This is a partial sum of a geometric series:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left| \frac{1}{n} \sum_{m=1}^n e^{2 \pi i h (2 \pi m)} \right|&#xD;&#xA;=\left| \frac{1}{n} \sum_{m=1}^n q^m \right| &#xD;&#xA;=\left| \frac{q}{n} \frac{1-q^n}{1-q}\right| \le \frac{2}{n\left|1-q\right|}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with $q:=e^{(2 \pi)^2 i h}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If your question is whether there is an upper bound for this independent of $h$, the answer is no -- $2 \pi h$ comes arbitrarily close to integers as $h$ increases, and hence the denominator comes arbitrarily close to $0$. For such values of $h$, the summands remain arbitrarily close to $1$, and the only bound is the trivial one $(1/n)\cdot n=1$.&lt;/p&gt;&#xA;" OwnerUserId="6622" LastEditorUserId="6622" LastEditorDisplayName="" LastEditDate="2011-02-18T11:16:39.320" LastActivityDate="2011-02-18T11:16:39.320" />
  <row Id="22656" PostTypeId="2" ParentId="22650" CreationDate="2011-02-18T11:11:20.577" Score="4" ViewCount="" Body="&lt;p&gt;A proof is outlined &lt;a href=&quot;http://qchu.wordpress.com/2011/02/12/su2-and-the-quaternions/&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;, although I skipped a few computations you should verify. &lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2011-02-18T11:11:20.577" />
  <row Id="22657" PostTypeId="2" ParentId="22648" CreationDate="2011-02-18T12:33:30.067" Score="1" ViewCount="" Body="&lt;p&gt;Suppose $f(0)=0$ and $n=1$.  For positions $2^{a} \le i &amp;lt; 2^{a+1}$, the sequence takes on values of the form $k 2^{-(a+1)}$ for odd $k$ less than $2^{a+1}$.  For a given $a$, these values can be covered in any order, since all the distances are equal; but concretely we may take $k=2(i-2^{a})+1$, so&#xA;$$&#xD;&#xA;f(i; 0, 1) = \frac{2(i-2^{a})+1}{2^{a+1}} = \frac{i-2^{a}}{2^a} + \frac{1}{2^{a+1}}&#xD;&#xA;$$&#xA;for all $i&amp;gt;0$, where $a = \lfloor\log_2(i)\rfloor$.  More generally, for $f(0)=\alpha n$ and arbitrary $n$, we have&#xA;$$&#xD;&#xA;f(i; \alpha, n) = \left(\frac{2\pi^{(a)}(i - 2^{a}) + 1}{2^a} + \frac{1}{2^{a+1}} + \alpha\right)n \mod n&#xD;&#xA;$$&#xA;where each $\pi^{(a)}$ is an arbitrary permutation of $0,1,2,...,2^{a}-1$.&lt;/p&gt;&#xA;" OwnerUserId="5546" LastActivityDate="2011-02-18T12:33:30.067" />
  <row Id="22659" PostTypeId="1" CreationDate="2011-02-18T12:55:57.560" Score="6" ViewCount="270" Body="&lt;p&gt;Is it adequate to expect to accomplish some amount of original work in a masters thesis?&lt;/p&gt;&#xA;" OwnerUserId="7485" LastActivityDate="2011-02-19T05:17:15.100" CommunityOwnedDate="2011-02-18T17:04:01.007" ClosedDate="2011-02-19T09:27:04.403" Title="Is it adequate to expect to accomplish some amount of original work in a masters thesis?" Tags="&lt;soft-question&gt;" AnswerCount="2" CommentCount="6" FavoriteCount="2" />
  <row Id="22660" PostTypeId="1" AcceptedAnswerId="22668" CreationDate="2011-02-18T13:35:06.833" Score="2" ViewCount="118" Body="&lt;p&gt;Hello everyone,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A matrix $A \in R^{n\times n}$ is said to be orthogonally equivalent to $B\in R^{n\times n}$ if there is an orthogonal matrix $U\in R^{n\times n}$, $U^T U=I$, such that $A=U^T B U$. My question is what kind of matrices are orthogonally equivalent to themselves? i.e.,&#xA;$A=U^T A U$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A similar interesting question is: if &#xA;$$U^T \Lambda U=\Lambda $$&#xA;where $\Lambda$ is a diagonal matrix and $U$ is a orthogonal matrix, are the diagonal entries of $\Lambda$ equal? That is whether $\Lambda=kI$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Look forward to your opinion. Thank you very much.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Shiyu&lt;/p&gt;&#xA;" OwnerUserId="7156" LastEditorUserId="7156" LastEditorDisplayName="" LastEditDate="2011-02-18T14:41:55.997" LastActivityDate="2011-02-18T15:15:11.083" Title="What kind of matrices are orthogonally equivalent to themselves?" Tags="&lt;linear-algebra&gt;&lt;orthonormal&gt;" AnswerCount="1" CommentCount="8" FavoriteCount="1" />
  <row Id="22661" PostTypeId="1" CreationDate="2011-02-18T13:45:47.967" Score="1" ViewCount="89" Body="&lt;p&gt;transport equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \frac{d}{dt}u +cu = 0\qquad    \mbox{ in } \mathbb{R}^n\times (0,\infty)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ u(x,0) = 0     \qquad       \mbox{ on }\mathbb{R}^n\times \{t=0\} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;b-const&lt;/p&gt;&#xA;" OwnerUserId="7224" LastEditorUserId="7224" LastEditorDisplayName="" LastEditDate="2011-02-20T18:38:59.027" LastActivityDate="2011-02-20T18:38:59.027" Title="PDE transport equation please help" Tags="&lt;differential-equations&gt;" AnswerCount="1" />
  <row Id="22662" PostTypeId="2" ParentId="22647" CreationDate="2011-02-18T14:10:13.907" Score="1" ViewCount="" Body="&lt;p&gt;Generically, an umbilic is an isolated point on the surface $S$ where $K=H^2$, $K$ and $H$ being the Gaussian and the mean curvature, respectively. What this implies for the partial derivatives of your height function $f$ you have to find out by yourself by doing the calculations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In any case, for an umbilic it is not necessary that the tangent plane to $S$ is horizontal there. See it this way: If $p$ is an umbilic on an arbitrary surface $S\subset{\mathbb R}^3$ and if the tangent plane to $S$ at $p$ is not parallel to the $z$-axis then you can always parametrize $S$ in a neighbourhood of $p$  in the form $(u,v)\to(u,v, f(u,v))$ for a suitable $f$.&lt;/p&gt;&#xA;" OwnerUserId="1303" LastActivityDate="2011-02-18T14:10:13.907" />
  <row Id="22663" PostTypeId="1" CreationDate="2011-02-18T14:29:44.640" Score="4" ViewCount="229" Body="&lt;p&gt;Is it possible to create an &quot;antigroup&quot;? What I mean by this is, given some group G, and some &quot;antigroup&quot; H, then the &quot;free product&quot; of G and H, G*H will equal the &quot;group&quot; (vacuously a group) of no generators and no relations. Can I construct such a thing? It makes me think that I need &quot;anti-generators&quot; and &quot;anti-relations&quot;, somehow the &quot;union&quot; of the generators and anti-generators causes cancellation, and same goes for the relations. Which now makes me think that I need to generalise the notion of &quot;union&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And is it possible to construct an &quot;anti-manifold&quot;? Where if given some manifold J and some &quot;anti-manifold&quot; K, then the &quot;connected-sum&quot; J#K will equal the empty set?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that these sorts of constructions are external to what groups or manifolds are. That is an &quot;anti-group&quot; won't be a group, because the free product of groups will always add complexity, as does the connected sum of manifolds, unless the connected sum is with an n-sphere (of appropriate dimension).&lt;/p&gt;&#xA;" OwnerUserId="7485" LastEditorUserId="7485" LastEditorDisplayName="" LastEditDate="2011-02-18T15:06:55.683" LastActivityDate="2011-06-04T11:11:05.083" Title="Can &quot;anti-groups&quot; and &quot;anti-manifolds&quot; be constructed? (and other &quot;anti-objects&quot;)" Tags="&lt;geometry&gt;&lt;group-theory&gt;" AnswerCount="1" CommentCount="7" FavoriteCount="1" />
  <row Id="22664" PostTypeId="2" ParentId="22661" CreationDate="2011-02-18T14:32:08.303" Score="1" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;hint&lt;/strong&gt; Consider the integral curve $\gamma: (0,\infty)\to\mathbb{R}^n$ given by $$\gamma(s) = (sb_1,sb_2,\ldots, sb_n)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;then &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{d}{ds} u(s,\gamma(s)) = \left(\frac{\partial}{\partial t}u + b\cdot\nabla u\right)(s,\gamma(s))$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and your equation becomes the ordinary differential equation along the curves $\gamma$ given by $\frac{d}{ds} u(s,\gamma(s)) = - cu$. &lt;/p&gt;&#xA;" OwnerUserId="1543" LastActivityDate="2011-02-18T14:32:08.303" CommentCount="2" />
  <row Id="22665" PostTypeId="2" ParentId="22646" CreationDate="2011-02-18T14:34:53.583" Score="20" ViewCount="" Body="&lt;p&gt;A torsion abelian group can always be decomposed into its $p$-parts: given an abelian group $A$, and a prime $p$, let&#xA;$$A_p = \{a\in A\mid \mbox{the order of $a$ is a power of $p$}\}.$$&#xA;Then $A_p$ is a subgroup of $A$, each has trivial intersection with the subgroup generated by all the others, so the subgroup they generate is isomorphic to their direct sum;  and so &#xA;$$\Bigl\langle A_p\;\Bigm|\; \mbox{$p$ prime}\Bigr\rangle = \mathop{\oplus}\limits_{p}A_p \leq A,$$&#xA;where $p$ ranges over all primes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given any $a\in A$, since $a$ is torsion, $a$ can be written as a sum of elements of $A$, each of which has order a prime power, so we have&#xA;$$A = \mathop{\oplus}\limits_{p} A_p,$$&#xA;where $p$ ranges over all primes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Two torsion abelian groups are isomorphic if and only if their $p$-components are isomorphic. So the problem reduces to classifying abelian $p$-groups. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can then decompose the group into its divisible and its reduced part, $A = A_{\mathrm{div}}\oplus A_{\mathrm{red}}$ (recall that an abelian group $A$ is &lt;em&gt;divisible&lt;/em&gt; if and only if for every $a\in A$ and every positive integer $n$, there exists an element $x\in A$ such that $nx=a$; an abelian group $A$ is &lt;em&gt;reduced&lt;/em&gt; if its largest divisible subgroup is the trivial group). Since every divisible abelian group is a direct sum of copies of $\mathbb{Q}$ and copies of the Prufer groups $\mathbb{Z}_{p^{\infty}}$ for different primes, in the $p$-group torsion case your divisible part will just be a direct sum of (possibly infinitely many) copies of $\mathbb{Z}_{p^{\infty}}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Two abelian groups are isomorphic if and only if their divisible parts are isomorphic and their reduced parts are isomorphic. So the problem is down to reduced abelian $p$-groups.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(The following can be found in Rotman's &lt;em&gt;Introduction to the Theory of Groups&lt;/em&gt;, 4th Edition, Chapter 10).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of bounded exponent, there is no problem: Prufer (1923 for countable groups) and Baer (1934 for the general case) proved that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Theorem.&lt;/strong&gt; (Prufer, Baer) If $A$ is an abelian group of bounded exponent (that is, there exists $n\gt 0$ such that $na=0$ for all $a\in A$), then $A$ is a direct sum of cyclic groups.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if the exponent of the reduced part is bounded, it's just a sum of cyclic groups (necessarily of $p$-power order). And then, counting with cardinality, you get that two are isomorphic if and only if they have the same number of direct summands of each order.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, things get more complicated at this stage.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some results: for nonnegative integer $n$, define $U\{n,A\}$ to be&#xA;$$U\{n,A\}= \dim_{\mathbf{F}_p}\left(\frac{p^nA\cap A[p]}{p^{n+1}A\cap A[p]}\right)$$&#xA;where $A[p]$ is the subgroup of elements such that $pa=0$. If $A$ is a sum of cyclic groups, then $U\{n,A\}$ is the number of cyclic summands of order $p^{n+1}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In 1933, Ulm proved that there is a transfinite version of these numbers, with $n$ ranging over ordinal numbers, called the &lt;em&gt;Ulm invariants&lt;/em&gt;. Two countable torsion abelian $p$-groups are isomorphic if and only if their Ulm invariants are equal. (The result does not extend to uncountable groups). Prufer proved that countable abelian $p$-group is a direct sum of cyclic groups if and only if $\cap_{n=1}^{\infty}p^nA = 0$ (again, false for uncountable groups). Kulikov characterized the groups that are direct sums of cyclic groups.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strike&gt;Note that Prufer's result gives you that in the countable reduced case you get a direct sum of cyclic groups (an element of the intersection would necessarily have a $p^i$th root for every $i$, and being of order a power of $p$, it would have $n$th roots for all $n$, hence taking it and its $n$th roots you get a divisible subgroup, so since $A$ is reduced the element must be trivial). So really, the problems start appearing when you go to the uncountable case. &lt;/strike&gt; That argument is incorrect; taking $x$ and its $p^i$th roots does not necessarily lead to a divisible subgroup, because you may not be able to pick a system of $p^i$th roots that goes &quot;all the way to $\infty$&quot;. E.g., you could have countably many different $p$th roots, such that the $i$th $p$th root has a $p^{i-1}$th root but no $p^i$th root; you could end up with an element that has $n$th roots for all $n$, but the roots themselves don't have $n$th roots for all $n$, so the subgroup is not actually divisible. Sorry about that, and thanks to Jack for the heads up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added.&lt;/strong&gt; There is also a theorem of Kulikov:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Theorem&lt;/strong&gt; (Kulikov, 1945) Every torsion abelian group $A$ is an extension of a direct sum of cyclic groups by a divisible group.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might be tempted to jump to the conclusion that this means that the reduced part of the $p$-component of $A$ will necessarily be a direct sum of cyclic groups (since it has no nontrivial divisible subgroup), but the problem is that a reduced group may have divisible quotients (even if it is torsion), so this is not the case. For an example of a reduced group with a divisible quotient, take&#xA;$$A = \mathop{\oplus}\limits_{n=1}^{\infty}C_{p^n},$$&#xA;where $C_{p^i}$ is a cyclic group of order $p^i$, generated by $x_i$. Note that $A$ is reduced. But if we let&#xA;$$B = \bigl\langle x_1-px_2,\ x_2-px_3,\ \ldots, x_n-px_{n+1},\ldots\bigr\rangle,$$&#xA;then $A/B$ is isomorphic to the Prufer group $\mathbb{Z}_{p^{\infty}}$, which is divisible. So Kulikov's theorem gives you a &lt;em&gt;different&lt;/em&gt; way of thiking about the reduced part, as an extension.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-19T21:05:12.980" LastActivityDate="2011-02-19T21:05:12.980" CommentCount="11" />
  <row Id="22667" PostTypeId="1" AcceptedAnswerId="22671" CreationDate="2011-02-18T15:14:07.590" Score="2" ViewCount="141" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Given a function series, what are&#xA;the ways often used to find the&#xA;function that the series converges&#xA;to?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, how do you construct&#xA;the function on the right hand side&#xA;from the series on the left hand&#xA;side: $$\sum_{n \in \mathbb{N} \cup \{ 0\}} (\begin{array}{c} 2n \\ n&#xD;&#xA;    \end{array}) x^{n}=(1-4x)^{-1/2}$$ I suppose one&#xA;may try to go the other way around,&#xA;i.e. verify that the Taylor&#xA;expansion of the function on RHS around 0 is&#xA;the series on the LHS, but I was&#xA;wondering if this is one of the&#xA;acceptable way to construct the RHS&#xA;from the LHS?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Related questions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given a sequence of real numbers,&#xA;what are the ways to find the number&#xA;the sequence converges to? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are the ways to find the number&#xA;a given series converges to? For&#xA;example,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{n=0}^\infty \frac{1}{n!}=e$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and the series that converges to&#xA;$\pi$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="1281" LastEditorUserId="1281" LastEditorDisplayName="" LastEditDate="2011-02-18T15:22:49.887" LastActivityDate="2011-02-18T16:22:34.120" Title="How to find what a given series/sequence converges to" Tags="&lt;real-analysis&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="22668" PostTypeId="2" ParentId="22660" CreationDate="2011-02-18T15:15:11.083" Score="1" ViewCount="" Body="&lt;p&gt;The family of matrices $U^{T}BU$, where $B$ is a fixed, positive definite matrix $\mathbb{R}^{n\times n}$, and $U$ varies over the orthogonal group $O(n)$, is obtaining by rigidly rotating and reflecting the eigenvectors of $B$.  The matrix $B$ is invariant under such a transformation iff its eigenspaces are preserved.  Even if there are $n$ distinct eigenvalues (so that all eigenspaces are $1$-dimensional), there are $2^n$ discrete choices for $U$ that preserve $B$: namely, reflections of any subset of the eigenvectors.  Note that these form a discrete subgroup of $O(n)$ under matrix multiplication: it can be represented as $O(1)^n$.  When eigenvalues are degenerate, then additional orthogonal transformations of the higher-dimensional eigenspaces will preserve the matrix $B$.  In general, if the eigenspaces of $B$ associated with eigenvalues $\lambda_1 &amp;lt; \lambda_2 &amp;lt; ... &amp;lt; \lambda_k$ have dimensions $d_1,d_2,...d_k$, with $d_1+d_2+...+d_k=n$, then the subgroup of $O(n)$ that preserves $B$ is isomorphic to $O(d_1)\times O(d_2) \times ... \times O(d_k)$.&lt;/p&gt;&#xA;" OwnerUserId="5546" LastActivityDate="2011-02-18T15:15:11.083" CommentCount="2" />
  <row Id="22669" PostTypeId="1" AcceptedAnswerId="22672" CreationDate="2011-02-18T15:55:21.673" Score="5" ViewCount="82" Body="&lt;p&gt;Matsumura mentions this as if it is obvious, and I can't find this result anywhere. Am I missing something obvious here?&lt;/p&gt;&#xA;" OwnerUserId="4341" LastActivityDate="2011-02-18T20:14:42.430" Title="Why is UFD a Krull domain?" Tags="&lt;abstract-algebra&gt;&lt;commutative-algebra&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="22670" PostTypeId="1" AcceptedAnswerId="22700" CreationDate="2011-02-18T15:55:27.587" Score="1" ViewCount="55" Body="&lt;p&gt;Hello,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am working through L.C.Evans' &lt;em&gt;Partial Differential Equations&lt;/em&gt; -- the chapter on second-order elliptic equations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have got a general question on symmetric vs. non-symmetric elliptic operators. Consider an operator of the form&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle Lu = \sum_{i,j}a^{ij} u_{,ij} + \sum_i b^i u_{,i} + cu$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In his book, Evans mainly treats the case of a symmetric highest order part (i.e. the coefficients for the second order derivatives form a symmetric matrix, $a^{ij}=a^{ji}$). The author frequently tells this restriction is 'without loss of generality'. However, I don't see what he means by this. Hence I wonder&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;You can you generalize to the case of $(a^{ij})_{ij}$ being non-symmetric?&lt;/li&gt;&#xA;&lt;li&gt;How does this relate with the operator as a whole being symmetric?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;It would be great help if you could clarify this picture.&lt;/p&gt;&#xA;" OwnerUserId="3557" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2011-02-18T20:11:23.250" LastActivityDate="2011-02-18T20:11:23.250" Title="Question on Evans' treatment of elliptic 2n order equations" Tags="&lt;differential-equations&gt;&lt;pde&gt;" AnswerCount="1" />
  <row Id="22671" PostTypeId="2" ParentId="22667" CreationDate="2011-02-18T16:04:15.253" Score="2" ViewCount="" Body="&lt;p&gt;In general most power series and most sequences do not have closed forms. The specific series you mention is &lt;a href=&quot;http://en.wikipedia.org/wiki/Hypergeometric_series&quot; rel=&quot;nofollow&quot;&gt;hypergeometric&lt;/a&gt; and there are known algorithms for working with these, e.g. guessing or proving identities or algebraic or differential equations. I don't know much about this subject, but &lt;a href=&quot;http://www.math.upenn.edu/~wilf/AeqB.html&quot; rel=&quot;nofollow&quot;&gt;Petkovsek, Wilf, and Zeilberger&lt;/a&gt; might be a good place to start. (If you just want to get from the RHS to the LHS, use the &lt;a href=&quot;http://en.wikipedia.org/wiki/Binomial_theorem#Newton.27s_generalized_binomial_theorem&quot; rel=&quot;nofollow&quot;&gt;generalized binomial theorem&lt;/a&gt;.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I do not understand your second example. $\sum \frac{1}{n!} = e$ is more or less a definition. If you define $e = \lim_{n \to \infty} \left( 1 + \frac{1}{n} \right)^n$, then you can prove this by proving that $e^x = \sum \frac{x^n}{n!} = \lim_{n \to \infty} \left( 1 + \frac{x}{n} \right)^n$, which is a nice exercise. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are special techniques which work on series that have a particular form, but again, there is no reason to expect that an arbitrary sum has a reasonable expression in terms of known constants. &lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2011-02-18T16:04:15.253" />
  <row Id="22672" PostTypeId="2" ParentId="22669" CreationDate="2011-02-18T16:05:54.560" Score="6" ViewCount="" Body="&lt;p&gt;Just for completeness: A Krull domain is an integral domain $D$ with field of fractions $K$ for which there is a family $\mathcal{F}=\{R_{\lambda}\}_{\lambda\in\Lambda}$ of discrete valuation rings of $K$ such that:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$D = \mathop{\cap}\limits_{\lambda\in\Lambda}R_{\lambda}$; and&lt;/li&gt;&#xA;&lt;li&gt;For every $x\in K$, $x\neq 0$, there are at most a finite number of $\lambda\in\Lambda$ such that $v_{\lambda}(x)\neq 0$. &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;So, assume that $D$ is a UFD. For each (class of associated) irreducible element $\pi$ of $D$, localizing away from $\pi$ (only allowable denominators are prime to $\pi$) gives you a DVR with valuation given by $(\pi)$. The intersection of all of these DVRs, viewed as subrings of the field of fractions of $D$, is exactly $D$ (because the only elements in the field of fractions that can be written with denominators prime to all irreducibles are the elements of $D$), and every element of the field of fractions has nonzero $\pi$-valuation only at finitely many $\pi$ (write the fraction in reduced terms: only those $\pi$ that show up in the numerator or the denominator give you nonzero valuation). So $D$ is a Krull ring.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-18T20:14:42.430" LastActivityDate="2011-02-18T20:14:42.430" />
  <row Id="22673" PostTypeId="2" ParentId="22659" CreationDate="2011-02-18T16:11:17.010" Score="2" ViewCount="" Body="&lt;p&gt;The answer depends on the standards of your institute, but even in good institutes, it is often acceptable to just present a summary of a difficult paper, with all details filled in. While certainly not original work, it can be difficult for the student since in math often results are presented in a very cryptic way, as part of the conventions of the subject (this is very different in, say, computer science).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is always better if you do some original research, but you may not be mature enough for that, or versed enough in the subject you're hoping to work on. Original work comes later, during your PhD, which is significantly longer, and having an appreciably higher starting point.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2011-02-18T16:11:17.010" CommunityOwnedDate="2011-02-18T17:04:01.007" CommentCount="4" />
  <row Id="22674" PostTypeId="2" ParentId="18114" CreationDate="2011-02-18T16:22:02.010" Score="0" ViewCount="" Body="&lt;p&gt;I think you should read the wiki page on the &lt;a href=&quot;http://en.wikipedia.org/wiki/Stress_%28mechanics%29&quot; rel=&quot;nofollow&quot;&gt;classical stress tensor&lt;/a&gt;, especially the section on change of basis.  Perhaps what you want to do is use your sensors to measure the stress tensor, and then find the basis which diagonalizes it.  Also you might want to see Feynman Lectures on Physics, volume 2 chapter 39, and volume 3 chapter 5, section 5-7.&lt;/p&gt;&#xA;" OwnerUserId="1417" LastActivityDate="2011-02-18T16:22:02.010" CommentCount="1" />
  <row Id="22675" PostTypeId="2" ParentId="22667" CreationDate="2011-02-18T16:22:34.120" Score="1" ViewCount="" Body="&lt;p&gt;It depends on the function.  In your first question you have the generating function for central binomial coefficients and a variety of techniques such as those described in &lt;a href=&quot;http://www.math.upenn.edu/~wilf/DownldGF.html&quot; rel=&quot;nofollow&quot;&gt;generatingfunctionology&lt;/a&gt;.  Your equation appears as (2.5.11)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your second question, it is unclear exactly what you are asking, as the first equation is for me the definition of $e$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that $\sum_{n=0}^\infty \frac{x^n}{n!}$ is its own derivative and so a multiple of $\exp (x)$, and looking at $x=0$ it must be $\exp (x)$, so back to your question with $x=1$ the value is $\exp(1)$ or $e$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or perhaps you want an approximation to the decimal value: for example the terms up to $n=9$ add up to 2.7182815... and it is easy to show that the others add up to less than 0.000001 so the sum is about 2.71828... &lt;/p&gt;&#xA;" OwnerUserId="6460" LastActivityDate="2011-02-18T16:22:34.120" />
  <row Id="22676" PostTypeId="1" AcceptedAnswerId="22694" CreationDate="2011-02-18T16:29:24.643" Score="0" ViewCount="58" Body="&lt;p&gt;using sum of geometric progression and inverse DFT we can show:&#xA;$${\sum_{j=0}^{k-1}\frac{\exp(2\pi ij\frac{u+1}{k})}{\exp(2\pi i\frac{j}{k})-a}=k\frac{a^{u(mod\ k)}}{1-a^{k}}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is how to compute&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$${\sum_{j=0}^{k-1}\frac{\exp(2\pi ij\frac{u+1}{k})}{(\exp(2\pi i\frac{j}{k})-a)^m}}$$&#xA;here $m,u,k$ are natural numbers&lt;/p&gt;&#xA;" OwnerUserId="6052" LastEditorUserId="6052" LastEditorDisplayName="" LastEditDate="2011-02-18T16:52:51.110" LastActivityDate="2011-02-18T18:39:46.387" Title="DFT of a certain function" Tags="&lt;fourier-analysis&gt;" AnswerCount="1" />
  <row Id="22677" PostTypeId="1" CreationDate="2011-02-18T16:31:14.590" Score="0" ViewCount="91" Body="&lt;p&gt;Let $g$ a Lie algebra and $V$ a finite-dimensional irreducible $g$-module, then each generalized eigenspace of $V$ is actually an eigenspace? If not, what is a condition to guarantee this fact?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="2764" LastEditorUserId="2764" LastEditorDisplayName="" LastEditDate="2011-02-18T16:48:54.700" LastActivityDate="2011-02-18T18:44:30.607" Title="Eigenspaces of a representation" Tags="&lt;representation-theory&gt;" AnswerCount="1" CommentCount="6" />
  <row Id="22678" PostTypeId="1" CreationDate="2011-02-18T16:37:11.730" Score="4" ViewCount="137" Body="&lt;p&gt;Let $f(t) = \sum_{i = 0}^{n} c_{i} \ t^{i}$ be a degree $n$ polynomial. In the binomial basis, we can write $f(t) = \sum_{k = 0}^{n} h_{k} \ \binom{t + n - k}{n}$. Using the calculus of finite differences, I can prove that the  coefficient $c_{i}$ can be computed from $\{ h_{0}, \dots, h_{n} \}$ by the equation&#xA;\begin{eqnarray}&#xA;c_i = \frac{1}{n!} \sum_{k = 0}^{n} \left( \sum_{\ell = i}^{n} s(n,\ell) \binom{\ell}{i} (n-k)^{\ell - i} \right) h_{k},&#xA;\end{eqnarray}&#xA;where $s(n,\ell)$ is the (signed) Stirling Number of the first kind. I'd like to have the inverse formula. That is, how does one compute $h_{k}$ as a function of $\{ c_{0}, \dots, c_{n} \}$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Update:&lt;/em&gt; Thanks to the suggestions below, I can prove &#xA;\begin{eqnarray}&#xA;f(t) = \sum_{i = 0}^{n} \sum_{k = 0}^{i} \sum_{j = k}^{i} c_{i} k! \binom{i}{j} r^{i-j} S(j,k) \binom{t - r}{k}&#xA;\end{eqnarray}&#xA;for any positive integer $r$. I'm not sure how to convert this into an expansion in the basis $\{ \binom{t + n - k}{k} \}$ or $\{ \binom{t + n -k}{n} \}$, since I believe that $r$ cannot depend on $k$. Any ideas are certainly appreciated!&lt;/p&gt;&#xA;" OwnerUserId="2720" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2011-05-29T02:46:23.837" LastActivityDate="2011-05-29T02:46:23.837" Title="Computing the Coefficients of a Polynomial in the Binomial Basis" Tags="&lt;algebra&gt;&lt;binomial-coefficients&gt;&lt;stirling-numbers&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="2" />
  <row Id="22679" PostTypeId="1" AcceptedAnswerId="22680" CreationDate="2011-02-18T16:47:02.180" Score="1" ViewCount="74" Body="&lt;p&gt;When solving an equation with irrational (or algebraic in the case) powers, are the roots likely to be transcendental or algebraic, or does it vary?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an example, I was trying to figure out if $(x + 1)^{\sqrt{2}}=x^2 - 2 x + 2$ had an exact solution. I tried it myself and then let Wolfram Alpha work on it, only getting approximate solutions (0.31375... and 3. something)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was curious if these solutions are algebraic, and thus the solution of some polynomial, or transcendental.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, are there any known techniques for solving this kind of thing?&lt;/p&gt;&#xA;" OwnerUserId="2247" LastActivityDate="2011-02-18T17:07:37.213" Title="Polynomials in irrational powers and their roots" Tags="&lt;algebra&gt;&lt;roots&gt;&lt;transcendental-numbers&gt;" AnswerCount="1" />
  <row Id="22680" PostTypeId="2" ParentId="22679" CreationDate="2011-02-18T17:07:37.213" Score="4" ViewCount="" Body="&lt;p&gt;Probably the solutions are almost always transcendental, but I don't know if this is provable in general with current technology. Known results along these lines include the &lt;a href=&quot;http://en.wikipedia.org/wiki/Gelfond%E2%80%93Schneider_theorem&quot; rel=&quot;nofollow&quot;&gt;Gelfond-Schneider&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/Lindemann%E2%80%93Weierstrass_theorem&quot; rel=&quot;nofollow&quot;&gt;Lindemann-Weierstrass&lt;/a&gt; theorems.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2011-02-18T17:07:37.213" CommentCount="2" />
  <row Id="22681" PostTypeId="1" AcceptedAnswerId="22682" CreationDate="2011-02-18T17:20:56.373" Score="1" ViewCount="91" Body="&lt;p&gt;How do you determine the different odds in Blackjack? For example, what would be the difference in odds from using 1, 2, or 3 decks? Also, what would be the difference in odds if you shuffle the deck before each hand instead of after the cards are used up?&lt;/p&gt;&#xA;" OwnerUserId="3284" LastActivityDate="2011-02-18T18:00:08.763" Title="Determining odds in Blackjack?" Tags="&lt;statistics&gt;&lt;recreational-mathematics&gt;" AnswerCount="2" />
  <row Id="22682" PostTypeId="2" ParentId="22681" CreationDate="2011-02-18T17:30:39.580" Score="2" ViewCount="" Body="&lt;p&gt;Generally, as you use more decks you are approaching drawing with replacement instead of without.  The chance of getting a blackjack with one deck is 2(4/52)(16/51), with the 2 being whether you draw the ace first, 4/52 drawing an ace, and 16/51 drawing a 10 count.  If you have a huge number of decks this becomes 2(4/52)(16/52).  If you don't shuffle before the hand, it is impacted by the cards you have seen go by.  If all the aces have already been drawn, the probability is zero.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-02-18T17:30:39.580" />
  <row Id="22683" PostTypeId="1" CreationDate="2011-02-18T17:38:58.860" Score="5" ViewCount="135" Body="&lt;p&gt;The most basic example of a topologically non-trivial real line bundle is the well-known &lt;em&gt;Möbius strip&lt;/em&gt;. Everyone who learns about vector bundles will be confronted by it, if only because it has the distinguished advantage that we can draw a picture of it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/AlO5R.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to &lt;em&gt;draw pictures&lt;/em&gt; of other line bundles, too. In particular, I have a complex line bundle which I would like to visualize somehow. How do I do that?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To be more specific:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The base manifold is the torus $M = S^1\times S^1$. It should be fine to visualize it as a rectangle, though.&lt;/li&gt;&#xA;&lt;li&gt;The complex line bundle has structure group $U(1)$.&lt;/li&gt;&#xA;&lt;li&gt;It is given as a direct summand of the trivial bundle $M \times L^2(\mathbb R^3)$. In other words, it is embedded in an infinite dimensional Hilbert space bundle. In particular, there is an induced connection coming from the hermitian form (scalar product).&lt;/li&gt;&#xA;&lt;li&gt;(The bundle arises from an analysis of the Quantum Hall Effect.)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;My questions:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;1) Are there any example drawings of complex line bundles?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I imagine that one attaches a plane to every point of the base manifold, but it is not clear how to me how to arrange them such that one obtains a qualitative picture of the fact that they represent complex numbers.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;2) Is there a minimal dimension $N$ such that every complex line bundle can be embedded into $\mathbb R^N$ in a suitable fashion?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It is probably the case that $N \geq 4$, so this won't be of much use, but it might still shed some insight on the problem, in particular because we are also given a connection.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;3a) Any ideas of how one might go about drawing a complex line bundle?&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;3b) Any ideas on how to best visualize the connection coming from a hermitian form?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="434" LastActivityDate="2011-02-19T12:42:50.173" Title="How to draw a complex line bundle?" Tags="&lt;differential-topology&gt;&lt;visualization&gt;&lt;vector-bundles&gt;" AnswerCount="2" CommentCount="6" />
  <row Id="22684" PostTypeId="2" ParentId="22681" CreationDate="2011-02-18T18:00:08.763" Score="1" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Edward_O._Thorp&quot; rel=&quot;nofollow&quot;&gt;Edward Oakley Throp&lt;/a&gt; went into depth on this subject with his book ${\bf Beat \,\, the \,\, Dealer,}$ which demonstrated mathematically how the house advantage can be overcome in blackjack by &lt;a href=&quot;http://en.wikipedia.org/wiki/Card_counting&quot; rel=&quot;nofollow&quot;&gt;card counting.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1301" LastActivityDate="2011-02-18T18:00:08.763" />
  <row Id="22685" PostTypeId="1" AcceptedAnswerId="22687" CreationDate="2011-02-18T18:06:34.403" Score="0" ViewCount="158" Body="&lt;p&gt;I need some help understanding why $\frac{x}{x+1}=\frac{1}{x^{-1}+1}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would be grateful if someone could explain. Be explicit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you!&lt;/p&gt;&#xA;" OwnerUserId="5877" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2011-02-18T20:11:12.370" LastActivityDate="2011-02-18T20:11:12.370" Title="Why $\frac{x}{x+1}=\frac{1}{x^{-1}+1}$" Tags="&lt;algebra-precalculus&gt;" AnswerCount="3" CommentCount="5" FavoriteCount="1" />
  <row Id="22686" PostTypeId="1" CreationDate="2011-02-18T18:13:38.537" Score="10" ViewCount="542" Body="&lt;p&gt;The ancient discipline of construction by straightedge and compass is both fascinating and entertaining. But what is its significance in a mathematical sense? It is still taught in high school geometry classes even today. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I'm getting at is this: Are the rules of construction just arbitrarily imposed restrictions, like a form of poetry, or is there a meaningful reason for prohibiting, say, the use of a protractor?&lt;/p&gt;&#xA;" OwnerUserId="130" LastEditorUserId="1055" LastEditorDisplayName="" LastEditDate="2011-05-19T13:50:26.067" LastActivityDate="2011-05-19T13:50:26.067" Title="What is the (mathematical) point of geometric constructions?" Tags="&lt;geometry&gt;&lt;education&gt;&lt;geometric-construction&gt;" AnswerCount="5" CommentCount="15" FavoriteCount="2" />
  <row Id="22687" PostTypeId="2" ParentId="22685" CreationDate="2011-02-18T18:14:10.377" Score="6" ViewCount="" Body="&lt;p&gt;$$\frac{x}{x+1} = \frac{x \cdot 1}{x \cdot (1+\frac{1}{x})}= \frac{x}{x} \cdot \frac{1}{1+\frac{1}{x}} = 1 \cdot \frac{1}{1+\frac{1}{x}} = \frac{1}{1+x^{-1}}$$&lt;/p&gt;&#xA;" OwnerUserId="4216" LastActivityDate="2011-02-18T18:14:10.377" CommentCount="1" />
  <row Id="22688" PostTypeId="2" ParentId="22685" CreationDate="2011-02-18T18:14:13.947" Score="1" ViewCount="" Body="&lt;p&gt;Divide both the numerator and the denominator by $x$.&lt;/p&gt;&#xA;" OwnerUserId="689" LastActivityDate="2011-02-18T18:14:13.947" />
  <row Id="22689" PostTypeId="1" CreationDate="2011-02-18T18:21:56.860" Score="2" ViewCount="161" Body="&lt;p&gt;My question is the following&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/DLSBm.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;how can i get the x3,y3 point from x1,y1 and x2,y2 points? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;the x3,y3 point distance from x1,y1 is 300.&lt;/p&gt;&#xA;" LastEditorUserId="3731" LastEditorDisplayName="" LastEditDate="2011-02-19T00:22:34.257" LastActivityDate="2011-04-09T14:16:50.830" Title="Getting the third point from two points on one line" Tags="&lt;geometry&gt;&lt;euclidean-geometry&gt;&lt;analytic-geometry&gt;" AnswerCount="1" />
  <row Id="22690" PostTypeId="2" ParentId="22689" CreationDate="2011-02-18T18:25:40.390" Score="6" ViewCount="" Body="&lt;p&gt;Assume&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;dx = x2 - x1&#xA;dy = y2 - y1&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Then,&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;x3 = x1 + dx*k&#xA;y3 = y1 + dy*k&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The square of distance between &lt;code&gt;p1&lt;/code&gt; and &lt;code&gt;p3&lt;/code&gt; is&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;(dx^2 + dy^2)*k^2 = 300^2&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Now you can find &lt;code&gt;k&lt;/code&gt; and then &lt;code&gt;x3&lt;/code&gt; and &lt;code&gt;y3&lt;/code&gt;&lt;/p&gt;&#xA;" OwnerUserId="1349" LastActivityDate="2011-02-18T18:25:40.390" />
  <row Id="22691" PostTypeId="2" ParentId="22685" CreationDate="2011-02-18T18:33:02.867" Score="3" ViewCount="" Body="&lt;p&gt;Your equation is not correct.  Note that the expression on the left-hand side is defined for $x=0$, but the right-hand expression is not.  Thus, the two expressions are not equivalent.  However, assuming $x\neq 0$, we may transform the left-hand expression into the right-hand expression as tpv has shown.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular to tpv's explanation, note that $\frac{x}{x}=1$ only if $x \neq 0$.&lt;/p&gt;&#xA;" OwnerUserId="2937" LastActivityDate="2011-02-18T18:33:02.867" CommentCount="2" />
  <row Id="22692" PostTypeId="2" ParentId="22683" CreationDate="2011-02-18T18:38:10.240" Score="1" ViewCount="" Body="&lt;p&gt;In any case... try making pictures of the bundle of unit circles associated to your complex line bundle, which has one dimension less. If the base space is the torus $S^1\times S^1$, then the total space of the circle bundle can be constructed by identifying faces of a cube, and &lt;em&gt;that&lt;/em&gt; is something one can make pictures of.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you described the actual bundle of which you want a picture, then someone could come up with something sensible. The general problem of «drawing complex line bundles» is pretty hopeless.&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2011-02-18T18:38:10.240" CommentCount="3" />
  <row Id="22693" PostTypeId="2" ParentId="11214" CreationDate="2011-02-18T18:39:23.793" Score="2" ViewCount="" Body="&lt;p&gt;I've never thought about it as a generalization of the Taylor series, but what you're describing sounds like the &lt;a href=&quot;http://en.wikipedia.org/wiki/Mellin_transform&quot; rel=&quot;nofollow&quot;&gt;Mellin transform&lt;/a&gt;, which is defined as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\hat{f}(s) = \int_0^{\infty}x^{s-1} f(x)\mathrm{d}x$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's used a lot in certain branches of high-energy physics theory (which is how I happen to know about it). The Wikipedia article linked above describes some other uses.&lt;/p&gt;&#xA;" OwnerUserId="1190" LastActivityDate="2011-02-18T18:39:23.793" />
  <row Id="22694" PostTypeId="2" ParentId="22676" CreationDate="2011-02-18T18:39:46.387" Score="2" ViewCount="" Body="&lt;p&gt;This is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{1}{n!} \frac{\mathrm{d}^{n}}{\mathrm{d}a^{n}} S (a)\;,$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $n=m-1$ and $S(a)$ is the sum you computed.&lt;/p&gt;&#xA;" OwnerUserId="6622" LastActivityDate="2011-02-18T18:39:46.387" />
  <row Id="22695" PostTypeId="1" AcceptedAnswerId="22706" CreationDate="2011-02-18T18:40:20.253" Score="2" ViewCount="104" Body="&lt;p&gt;Assume that I have a large, non-symmetric matrix $\bf{A}$ of zeros and ones. If I want the &lt;em&gt;exact&lt;/em&gt; diagonal elements of $\bf{A}^{k}$ (not the trace, but the elements themselves) where $k$ is some large whole number, I can&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Multiply the matrices out using with exact integers, possibly speeding the computation by using some &lt;a href=&quot;http://en.wikipedia.org/wiki/Addition-chain_exponentiation&quot; rel=&quot;nofollow&quot;&gt;Addition-chain exponentiation&lt;/a&gt;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Diagonalize the matrix as $\bf A = V \Gamma V^{-1}$ and compute $\bf A^{k} = V \Gamma^{k} V^{-1}$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The first method is slow but accurate, while the second method can introduce numerical errors (since the exact answer is required). Is there some decomposition I am missing that does what I'm looking for?&lt;/p&gt;&#xA;" OwnerUserId="196" LastActivityDate="2011-02-18T19:45:58.967" Title="Exact powers of an integer matrix" Tags="&lt;matrices&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" />
  <row Id="22696" PostTypeId="2" ParentId="22677" CreationDate="2011-02-18T18:44:30.607" Score="2" ViewCount="" Body="&lt;p&gt;I am not sure I get exactly what you mean, but the answer is probably &lt;em&gt;no&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $g$ is one-dimensional and we fix a non-zero element $X_0\in g$, then providing a $g$-module structure to a vector space $V$ is the same thing as picking an endomorphism $T\in\gl(V)$, corresponding to the action of $X_0$ on the module (&lt;em&gt;i.e.&lt;/em&gt;, what you call $\rho(X_0)$ is a comment above).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you pick a linear map $T:V\to V$ which is not semisimple (that is, whose Jordan canonical form is not diagonal), then your «generalized eigenspaces of $V$» are surely the same thing as the generalized eigenspaces of $T$, and these are certainly not actual eigenspaces, because of the choice of $T$.&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2011-02-18T18:44:30.607" />
  <row Id="22697" PostTypeId="1" AcceptedAnswerId="22699" CreationDate="2011-02-18T18:54:27.900" Score="2" ViewCount="150" Body="&lt;p&gt;Hello&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $\pi: X \times Y \to X$ be a projection map where $Y$ is compact. Prove that $\pi$ is a closed map.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;First i would like to see a proof of this claim. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I want to know that here why compactness is necessary or do we have any other weaker condition other than compactness for the same result to hold.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1252" LastActivityDate="2011-02-19T13:22:50.430" Title="Projection map being a closed map" Tags="&lt;topology&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="22698" PostTypeId="2" ParentId="21614" CreationDate="2011-02-18T18:58:14.200" Score="0" ViewCount="" Body="&lt;p&gt;Determinants are best understood in the context of &lt;a href=&quot;http://en.wikipedia.org/wiki/Exterior_algebra&quot; rel=&quot;nofollow&quot;&gt;exterior algebra&lt;/a&gt;, which goes back to the work of &lt;a href=&quot;http://www-history.mcs.st-and.ac.uk/Biographies/Grassmann.html&quot; rel=&quot;nofollow&quot;&gt;Hermann Grassmann&lt;/a&gt;. Here a down-to-earth description of the intuition behind it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider an $n$-dimensional vector space $V = K^n$ where $K$ is some field. Assume that the dimension $n$ is as large as we need it for our examples. Let $e_1,\dots,e_n$ denote a basis of the space $V$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you know, any (say) three linearly independent vectors $a,b,c$ span a 3-dimensional subspace of $V$. Wouldn't it be nice if we could somehow &lt;em&gt;calculate&lt;/em&gt; with subspaces? I.e. if we had some notion of algebraic product that is geometrically meaningful? Like, if the product of two 3-dimensional subspaces is zero, then something geometric has happened. It turns out that — with some limitations — there is something like that, namely the &lt;strong&gt;exterior algebra&lt;/strong&gt; or &lt;strong&gt;Grassmann algebra&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The key point that the &lt;strong&gt;exterior product&lt;/strong&gt; of two &quot;subspaces&quot; corresponds to their &lt;em&gt;geometric span&lt;/em&gt;. The drawback of the exterior product is that it's definition is entirely formal, but it's usefulness shall serve as motivation. Namely, the exterior product $\wedge$ of three vectors $a,b,c,$ is defined as the formal expression&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ a \wedge b \wedge c$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is subject to the following three rules&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The product is linear in each factor, $(\alpha x + y) \wedge z = \alpha(x \wedge z) + (y \wedge z)$.&lt;/li&gt;&#xA;&lt;li&gt;The two factors are the same, then the product vanishes, $x \wedge x = 0$.&lt;/li&gt;&#xA;&lt;li&gt;The product is antisymmetric, $x \wedge y = - y \wedge x$. (This actually follows from 1 and 2)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;For instance, we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ (a+b) \wedge b \wedge d = (a \wedge b \wedge d) + (b \wedge b \wedge d) = a \wedge b \wedge d + 0 = a \wedge b \wedge d$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you can see, we also consider sums of these products, though they don't necessarily carry geometric meaning.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The beauty of these rules is that they allow us to represent the subspace spanned by the three vectors $a,b,c$ by their exterior product $a \wedge b \wedge c$. In particular, it is an easy exercise to show that the rules imply that&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;We have $a \wedge b \wedge c  = \lambda (a&amp;#39; \wedge b&amp;#39; \wedge c&amp;#39;)$ for some scalar multiple $\lambda$ if and only if the three vectors $a,b,c$ span the &lt;strong&gt;same subspace&lt;/strong&gt; as the three vectors $a&amp;#39;,b&amp;#39;,c&amp;#39;$.&lt;/li&gt;&#xA;&lt;li&gt;We have $a \wedge b \wedge c = 0$ if and only if the three vectors are &lt;strong&gt;linearly dependent&lt;/strong&gt;.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Isn't that awesome? The question of whether vectors are linearly dependent has been transformed into a question about an algebraic product being zero, something that is easy to calculate with!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, if you have a matrix $A \in K^{n\times n}$, the exterior product of the column vectors&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ Ae_1 \wedge Ae_2 \wedge \dots \wedge Ae_n $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is zero if and only if the matrix is singular. It is not difficult to show that in an $n$-dimensional vectors space, the only non-zero products of $n$ factors must be of the form $\lambda(e_1\wedge e_2 \wedge \dots \wedge e_n)$. The scalar multiple $\lambda$ can be interpreted as the &lt;em&gt;volume&lt;/em&gt; of the parallelepiped. The determinant of a matrix is defined to be this factor:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ (\det A)(e_1\wedge e_2 \wedge \dots \wedge e_n) := Ae_1 \wedge Ae_2 \wedge \dots \wedge Ae_n .$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is an instructive exercise to show that this coincides with the standard formulas for the determinant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For more about using the exterior product to do calculations with subspaces, consider a book by &lt;a href=&quot;http://sites.google.com/site/grassmannalgebra/&quot; rel=&quot;nofollow&quot;&gt;John Browne&lt;/a&gt;, though I think it's a bit long-winded.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These techniques feature prominently in differential geometry as &lt;strong&gt;differential forms&lt;/strong&gt;. The cup-product in &lt;strong&gt;cohomology&lt;/strong&gt; is another instance of these ideas. &lt;a href=&quot;http://en.wikipedia.org/wiki/Schubert_calculus&quot; rel=&quot;nofollow&quot;&gt;Schubert calculus&lt;/a&gt; is a natural extension of these ideas to problems from algebraic geometry.&lt;/p&gt;&#xA;" OwnerUserId="434" LastActivityDate="2011-02-18T18:58:14.200" />
  <row Id="22699" PostTypeId="2" ParentId="22697" CreationDate="2011-02-18T18:58:15.400" Score="6" ViewCount="" Body="&lt;p&gt;There is a standard example for why some hypothesis on $Y$ is necessary: let $X=Y=\mathbb R$, and consider the closed subset $F=\{(x,y)\in \mathbb R\times\mathbb R:xy=1\}\subset\mathbb R\times\mathbb R$. What is its projection to the first factor?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, one can prove that a space $Y$ is compact iff for all spaces $X$ the projection $X\times Y\to X$ is closed. So while compactness is not necessary (I think...) for the closedness of the projection for one $X$, it is necessary if you want &lt;em&gt;all&lt;/em&gt; such projections to be closed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the proof you want in the first bullet point... this is a standard exercise in topology: what have you tried?&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2011-02-18T18:58:15.400" CommentCount="2" />
  <row Id="22700" PostTypeId="2" ParentId="22670" CreationDate="2011-02-18T19:00:41.527" Score="3" ViewCount="" Body="&lt;p&gt;The reason for this generalization is actually simple. If you assume your function is sufficiently smooth, then you have $\frac{\partial^2 u}{\partial x_i \partial x_j} = \frac{\partial^2 u}{\partial x_j \partial x_i}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance if $n=2$, the highest order term in the PDE is of the form&#xA;$$a_{11} \frac{\partial^2 u}{\partial x_1^2} + a_{12} \frac{\partial^2 u}{\partial x_1 x_2} + a_{21} \frac{\partial^2 u}{\partial x_2 x_1} + a_{22} \frac{\partial^2 u}{\partial x_2^2}$$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that as said before is $u$ is smooth enough, then $$\frac{\partial^2 u}{\partial x_1 x_2} = \frac{\partial^2 u}{\partial x_2 x_1}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, the leading order term in the PDE becomes&#xA;$$a_{11} \frac{\partial^2 u}{\partial x_1^2} + (a_{12} + a_{21}) \frac{\partial^2 u}{\partial x_1 x_2} + a_{22} \frac{\partial^2 u}{\partial x_2^2}$$&#xA;which could be re-written as&#xA;$$a_{11} \frac{\partial^2 u}{\partial x_1^2} + \frac{(a_{12} + a_{21})}{2} \frac{\partial^2 u}{\partial x_1 x_2} + \frac{(a_{21} + a_{12})}{2} \frac{\partial^2 u}{\partial x_2 x_1} + a_{22} \frac{\partial^2 u}{\partial x_2^2}$$&#xA;Hence, the leading order term in the PDE becomes&#xA;$$a&amp;#39;_{11} \frac{\partial^2 u}{\partial x_1^2} + a&amp;#39;_{12} \frac{\partial^2 u}{\partial x_1 x_2} + a&amp;#39;_{21} \frac{\partial^2 u}{\partial x_2 x_1} + a&amp;#39;_{22} \frac{\partial^2 u}{\partial x_2^2}$$&#xA;where&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle&#xD;&#xA;\begin{align*}&#xD;&#xA;a&amp;#39;_{11} &amp;amp; = &amp;amp; a_{11}\\&#xD;&#xA;a&amp;#39;_{22} &amp;amp; = &amp;amp; a_{22}\\&#xD;&#xA;a&amp;#39;_{12} &amp;amp; = \frac{(a&amp;#39;_{12} + a&amp;#39;_{21})}{2} = \frac{(a&amp;#39;_{21} + a&amp;#39;_{12})}{2} = &amp;amp; a&amp;#39;_{21}&#xD;&#xA;\end{align*}&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, you can always write the leading order term of the PDE as a symmetric part.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastActivityDate="2011-02-18T19:00:41.527" />
  <row Id="22701" PostTypeId="2" ParentId="21614" CreationDate="2011-02-18T19:09:48.373" Score="1" ViewCount="" Body="&lt;p&gt;In Artin's book, &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0132413779&quot; rel=&quot;nofollow&quot;&gt;Algebra&lt;/a&gt;, the determinant is defined as the only function $\det: \mathbb{R}^{n \times n} \rightarrow \mathbb{R}$ such that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1 - $\det(I) = 1$ &lt;br&gt;&#xA;2 - $\det$ is an $n$-multilinear function&lt;br&gt;&#xA;3 - if two rows (columns) of $A$ are equal, then $\det(A) = 0$&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which, I think, can be proven using Qiaochu's answer (but I'm not sure).&lt;/p&gt;&#xA;" OwnerUserId="1415" LastActivityDate="2011-02-18T19:09:48.373" CommentCount="2" />
  <row Id="22703" PostTypeId="2" ParentId="22686" CreationDate="2011-02-18T19:32:21.693" Score="9" ViewCount="" Body="&lt;p&gt;I personally think straightedge and compass is one of the most important tools i acquired during high school, regarding mathematics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other then it being, as you said, a fascinating and entertaining subject- &#xA;It was the first time before college I ever had to work under severe restriction- and got unbelievable results. Later on, when I discovered Galois and Field theory the whole thing became even cooler.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I'm getting at, is that I think subjects like Construtions are extremely important for math education.&#xA;A few days ago someone posted Lockharts Lement- It mentions the fact that math became a subject of industry- kids learn math as a technical and terrifying subject. Telling them to use protractors and calculators and other stuff that make &quot;short cuts&quot; to solutions of problems makes math into paper-work. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Putting on restrictions, and teaching how to come up with beautiful stuff under them- makes it into a game. I remember when I was a kid I wasn't very much interested in paperworks. But games, I liked games. And that kind of things is what got me to enroll in a mathematics degree. At least that's my opinion on the matter :)&lt;/p&gt;&#xA;" OwnerUserId="7129" LastEditorUserId="7129" LastEditorDisplayName="" LastEditDate="2011-02-20T07:52:22.400" LastActivityDate="2011-02-20T07:52:22.400" CommentCount="4" />
  <row Id="22704" PostTypeId="1" AcceptedAnswerId="22713" CreationDate="2011-02-18T19:34:45.633" Score="3" ViewCount="214" Body="&lt;p&gt;$\begin{align*}&#xD;&#xA; \lim_{x\to 0}\frac{\frac{1}{\sqrt{4+x}}-\frac{1}{2}}{x}&#xD;&#xA; &amp;amp;=\lim_{x\to 0}\frac{\frac{2}{2\sqrt{4+x}}-\frac{\sqrt{4+x}}{2\sqrt{4+x}}}{x}\\&#xD;&#xA; &amp;amp;=\lim_{x\to 0}\frac{\frac{2-\sqrt{4+x}}{2\sqrt{4+x}}}{x}\\&#xD;&#xA; &amp;amp;=\lim_{x\to 0}\frac{2-\sqrt{4+x}}{2x\sqrt{4+x}}\\&#xD;&#xA; &amp;amp;=\lim_{x\to 0}\frac{(2-\sqrt{4-x})(2+\sqrt{4-x})}{(2x\sqrt{4+x})(2+\sqrt{4-x})}\\&#xD;&#xA; &amp;amp;=\lim_{x\to 0}\frac{2 \times 2 + 2\sqrt{4-x}-2\sqrt{4-x}-((\sqrt{4-x})(\sqrt{4-x})) }{2 \times 2x\sqrt{4+x} + 2x\sqrt{4+x}\sqrt{4-x}}\\&#xD;&#xA; &amp;amp;=\lim_{x\to 0}\frac{4-4+x}{4x\sqrt{4+x} + 2x\sqrt{4+x}\sqrt{4-x}}\\&#xD;&#xA; &amp;amp;=\lim_{x\to 0}\frac{x}{x(4\sqrt{4+x} + 2\sqrt{4+x}\sqrt{4-x})}\\&#xD;&#xA; &amp;amp;=\lim_{x\to 0}\frac{1}{(4\sqrt{4+x} + 2\sqrt{4+x}\sqrt{4-x})}\\&#xD;&#xA;&amp;amp;=\frac{1}{(4\sqrt{4+0} + 2\sqrt{4+0}\sqrt{4-0})}\\&#xD;&#xA;&amp;amp;=\frac{1}{16} &#xD;&#xA;\end{align*}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;wolframalpha says it's negative. What am I doing wrong? &lt;/p&gt;&#xA;" OwnerUserId="2230" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-18T19:51:56.443" LastActivityDate="2011-02-18T21:41:34.030" Title="Simple limit, wolframalpha doesn't agree, what's wrong? (Just the sign of the answer that's off)" Tags="&lt;calculus&gt;" AnswerCount="5" CommentCount="5" />
  <row Id="22705" PostTypeId="2" ParentId="22704" CreationDate="2011-02-18T19:43:07.317" Score="1" ViewCount="" Body="&lt;p&gt;Certainly for $x \gt 0,\frac{1}{\sqrt{4+x}}-\frac{1}{2} \lt 0$ so the limit should be negative.  Between the  fifth and sixth limit you flipped a sign under the sqrt in the numerator and that changes the sign of the total thing&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-02-18T19:43:07.317" CommentCount="0" />
  <row Id="22706" PostTypeId="2" ParentId="22695" CreationDate="2011-02-18T19:45:58.967" Score="3" ViewCount="" Body="&lt;p&gt;(edit : this is for large k, so it may not be useful for k comparable to n)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A faster way to compute high powers of a matrix is by computing in the $\mathbb{Z}$-algebra $\mathbb{Z}[M]$.&#xA;The space $(I_n,M,M^2, \ldots)$ is a subspace of $M_n(\mathbb{Z})$, and in fact the Cayley-Hamilton theorem tells you that it has dimension at most $n$. For example, sometimes, the $n$ coefficients on the diagonal of the matrix (or the the coefficients on the first row, or column, etc) are enough to know what the rest of the coefficients are.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $P$ is the minimal polynomial of $M$ over $\mathbb{Z}$ and has degree $d = d_P$, the basis the best suited in order to compute products, however, is the basis $(1,M,\ldots M^{d-1})$ : we want to use the isomorphism $\mathbb{Z}[M] = \mathbb{Z}[X]/P$.&#xA;Doing one multiplication in there requires $O(d^2)$ operations :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Keep a table of the coordinates of $(1,M,\ldots M^{2d-2})$ in the basis $(1,M,\ldots M^{d-1})$, then compute the coefficients $c_k$ in&#xA;$$ (\Sigma_{i=0}^{d-1} a_i M^i)(\Sigma_{j=0}^{d-1} b_j M^j) = \Sigma_{k=0}^{2d-2} (c_k = \Sigma_{i+j=k} a_i b_j) M^k$$&#xA;Then use that table to convert this sum into a vector $\Sigma_{k=0}^{d-1} d_k M^k$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can further optimize this algorithm if the polynomial $P$ is reducible.&#xA;If $P = QR$ with $Q$ and $R$ having no factor in common, then $\mathbb{Z}[M] = \mathbb{Z}[X]/P \simeq \mathbb{Z}[X]/Q \times \mathbb{Z}[X]/R$. The isomorphism here sends $X$ onto the pair $X,X$. you will need to compute the inverse of the isomorphism, but once you do that, you gain in complexity because $d_Q^2 + d_R^2 &amp;lt; (d_Q + d_R)^2 = d_P^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can also improve if you have $P = Q^k$ for some $k$, because then, $\mathbb{Z}[X]/P \simeq \mathbb{Z}[X,Y]/(Q(X),Y^k)$, where the isomorphism sends $X$ onto $X+Y$. Once again, the computations there are shorter since you have $k$ polynomials of degree $d/k$ to compute, it will take $O(k(d/k)^2 = O(d^2/k)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have to compute $P$, its factorization, and the isomorphisms only once. Once you have done all this preliminary work you get a very good algorithm. &#xA;If the matrix is diagonalizable (in $M_n(\mathbb{Z})$), you will get that $P$ factors completely into distincts factors of degree 1, and this will give you an algorithm equivalent to the one you gave.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even if you don't want to factorize $P$, simply working in $\mathbb{Z}[X]/P$ gives a good optimisation. You can compute $P$ by taking some element $x$ in $\mathbb{Z}^n$, and checking when $M^k x$ becomes linearly dependant on $1, x, \ldots x^{k-1}$. If you are lucky you get some dependancy for $k&amp;lt;n$, which means you have found a factor of $P$ of degree $k&amp;lt;n$. If you know of any subspace of $\mathbb{Z}^n$ stable by $M$, be sure to make use of it, picking $x$ in it will give you such a factor of $P$.&lt;/p&gt;&#xA;" OwnerUserId="6270" LastActivityDate="2011-02-18T19:45:58.967" CommentCount="1" />
  <row Id="22707" PostTypeId="2" ParentId="22704" CreationDate="2011-02-18T19:46:29.917" Score="0" ViewCount="" Body="&lt;p&gt;Some of your $4+x$ terms under the radical were misprinted as $4-x$, leading to the sign error.  In particular, after &quot;rationalizing&quot; the numerator you should have $4-(4+x)=-x$ in the numerator.  Ross has already given the reason the limit cannot be positive; the quotient is negative for all $x\neq 0$.  The case he didn't explicitly mention is $x\lt 0$, in which case $\sqrt{4+x}&amp;lt;2$ implies $\frac{1}{\sqrt{4+x}}-\frac{1}{2}\gt0$.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2011-02-18T19:46:29.917" CommunityOwnedDate="2011-02-18T19:46:29.917" />
  <row Id="22708" PostTypeId="2" ParentId="22704" CreationDate="2011-02-18T19:46:30.167" Score="0" ViewCount="" Body="&lt;p&gt;An argument why it should be negative.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When $x&amp;gt;0$, we have $\sqrt{4+x} &amp;gt; 2$ and hence $\frac{1}{\sqrt{4+x}} &amp;lt; \frac{1}{2}$ and hence&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\frac{1}{\sqrt{4+x}} - \frac{1}{2}}{x} &amp;lt; 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, When $x&amp;lt;0$, we have $\sqrt{4+x} &amp;lt; 2$ and hence $\frac{1}{\sqrt{4+x}} &amp;gt; \frac{1}{2}$ and hence&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\frac{1}{\sqrt{4+x}} - \frac{1}{2}}{x} &amp;lt; 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So either way, as $x \rightarrow 0$, the value is negative.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastActivityDate="2011-02-18T19:46:30.167" />
  <row Id="22709" PostTypeId="2" ParentId="22704" CreationDate="2011-02-18T19:47:19.510" Score="1" ViewCount="" Body="&lt;p&gt;In between the fourth and fifth steps, you go from &#xA;$$\lim_{x\to 0}\frac{2-\sqrt{4+x}}{2x\sqrt{4+x}} \text{ to } \lim_{x\to 0}\frac{(2-\sqrt{4-x})(2+\sqrt{4-x})}{(2x\sqrt{4+x})(2+\sqrt{4-x})}$$&#xA;which is not correct. It should be &#xA;$$\lim_{x\to 0}\frac{(2-\sqrt{4+x})(2+\sqrt{4+x})}{(2x\sqrt{4+x})(2+\sqrt{4+x})}$$&lt;/p&gt;&#xA;" OwnerUserId="1016" LastEditorUserId="1016" LastEditorDisplayName="" LastEditDate="2011-02-18T21:41:34.030" LastActivityDate="2011-02-18T21:41:34.030" CommentCount="2" />
  <row Id="22710" PostTypeId="1" AcceptedAnswerId="22712" CreationDate="2011-02-18T19:57:24.707" Score="3" ViewCount="123" Body="&lt;p&gt;Let $A$ and $B$ be two groups and let $G=A*B$ be their free product. Suppose both $A$ and $B$ are nontrivial. I want now to prove that $G$ is infinite and that $Z(G)$ is trivial.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think I can prove that $G$ is infinite using this argument: two reduced words of $A*B$ are equal if and only if they have the same length and the ith letter of the first word is the same as the ith letter of the second one. Hence one can find a reduced word of $G$ of length arbitrarily big (since we are sure it won't be equal to a shorter word). Do you think it works? Has anyone another suggestion?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, I don't really have a clue for the second part except trying to analyze a long list of cases of reduced words and prove for each one that the only element with which they commute is the identity. Does anyone have a smarter idea?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance!&lt;/p&gt;&#xA;" OwnerUserId="5465" LastActivityDate="2011-02-18T20:58:12.283" Title="Center of a free product" Tags="&lt;homework&gt;&lt;group-theory&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="22711" PostTypeId="1" CreationDate="2011-02-18T20:06:58.370" Score="4" ViewCount="104" Body="&lt;p&gt;Let $\mathfrak {so}_{n}$ denote the skew-symmetric complex $n \times n$-matrices and let $M$ denote the symmetric $n \times n$-matrices of trace 0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As I understand, $M$ is a module over $\mathfrak {so}_n$. What then is its decomposition into irreducibles? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The standard representation of $\mathfrak {so}_n$ has dimension $n$, the adjoint representation dimension $\frac 1 2 n \cdot (n-1)$ and there are two spin representations of small dimension. But I don't see a way how these, together with the trivial representation, should add up to the dimension of $M$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: This comes from trying to understand the Cartan decomposition $\mathfrak g=\mathfrak k \oplus \mathfrak p$, where $[\mathfrak k,\mathfrak p] \subseteq \mathfrak p$, cf. the &#xA;wikipedia article on &lt;a href=&quot;http://en.wikipedia.org/wiki/Cartan_decomposition&quot; rel=&quot;nofollow&quot;&gt;Cartan decomposition&lt;/a&gt;. As the associated symmetric should be irreducible, the representation should be irreducible, but my numbers just don't add up.&lt;/p&gt;&#xA;" OwnerUserId="2436" LastEditorUserId="2436" LastEditorDisplayName="" LastEditDate="2011-02-18T22:27:13.727" LastActivityDate="2011-02-19T01:05:20.010" Title="Symmetric matrices as module over the skewsymmetric ones" Tags="&lt;representation-theory&gt;" AnswerCount="1" CommentCount="9" FavoriteCount="2" />
  <row Id="22712" PostTypeId="2" ParentId="22710" CreationDate="2011-02-18T20:07:49.987" Score="6" ViewCount="" Body="&lt;p&gt;Well, you would have a hard time proving that a reduced word only commutes with the identity, because every word commutes with every power of itself, and of course, any element $a\in A$ will commute with all elements in $C_A(a)$, and any $b\in B$ with all elements in $C_B(b)$. So no element of $A*B$ has trivial centralizer! But you don't need a nontrivial element that doesn't commute with anything for the center to be trivial (good thing, since there is no such thing in any nontrivial group); you just need every nontrivial element to not commute with &lt;em&gt;something&lt;/em&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You don't really need to consider a lot of cases: the only two things you have to worry a bit about are what to do if both of your groups have only two elements, and what the reduced word starts and ends with. For example, if $\mathbf{w}$ is a reduced word that starts and ends with an element of $A$, then just take $b\in B$, $b\neq e$ to get something it does not commute with, so $\mathbf{w}$ cannot be central. If it starts with an element of $A$ and ends with an element of $B$, then take a word $\mathbf{w}&amp;#39;$ that starts with an element of $A$ and ends with an element of $B$ (so that $\mathbf{w}\mathbf{w}&amp;#39;$ and $\mathbf{w}&amp;#39;\mathbf{w}$ are already reduced), and make $\mathbf{w}&amp;#39;$ either begin with something different from what $\mathbf{w}$ begins with, or end with something different from what $\mathbf{w}$ ends with; etc. You should have just 4 cases, and only two that can give you trouble if your groups are too small, in which case you can just stare at them directly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your argument for why $A*B$ is infinite is fine. You can even produce, explicitly, words of arbitrary length: just take $a\in A$, $b\in B$ with $a\neq e\neq b$, and consider $(ab)^n$, $n\in\mathbb{Z}$.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-18T20:58:12.283" LastActivityDate="2011-02-18T20:58:12.283" CommentCount="4" />
  <row Id="22713" PostTypeId="2" ParentId="22704" CreationDate="2011-02-18T20:59:34.593" Score="13" ViewCount="" Body="&lt;p&gt;Others have already pointed out a sign error. One way to avoid such is to first simplify the problem by changing variables. Let $\rm\ z = \sqrt{4+x}\ $ so $\rm\ x = z^2 - 4\:.\:$ Then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\rm \frac{\frac{1}{\sqrt{4+x}}-\frac{1}{2}}{x}\ =\ \frac{\frac{1}z - \frac{1}2}{z^2-4}\ =\ \frac{-(z-2)}{2\:z\:(z^2-4)}\ =\ \frac{-1}{2\:z\:(z+2)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this form it is very easy to compute the limit as $\rm\ z\to 2\:$.&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2011-02-18T20:59:34.593" />
  <row Id="22714" PostTypeId="1" AcceptedAnswerId="22715" CreationDate="2011-02-18T21:04:01.800" Score="4" ViewCount="108" Body="&lt;p&gt;We usually say that a graph is planar if it can be embedded into 2-space s.t. no edges intersect.  Here's a different way to describe the same situation: a graph is planar if it can be embedded into 2-space s.t. the edges form the boundaries of cells which nowhere overlap.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My question:&lt;/strong&gt; Can you take this second definition and extend the idea to higher dimensions?  For instance, allow the cells to be volumes in 3-space - like soap bubbles where the graph edges are the intersection of soap film walls.&lt;/p&gt;&#xA;" OwnerUserId="1358" LastActivityDate="2011-02-19T00:21:33.103" Title="Higher dimensional analog to planar graphs?" Tags="&lt;graph-theory&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="1" />
  <row Id="22715" PostTypeId="2" ParentId="22714" CreationDate="2011-02-18T21:09:56.173" Score="4" ViewCount="" Body="&lt;p&gt;A planar graph is nothing other than a &lt;a href=&quot;http://en.wikipedia.org/wiki/Triangulation_(topology)&quot; rel=&quot;nofollow&quot;&gt;triangulation&lt;/a&gt; of the $2$-sphere (via &lt;a href=&quot;http://en.wikipedia.org/wiki/Stereographic_projection&quot; rel=&quot;nofollow&quot;&gt;stereographic projection&lt;/a&gt;), at least if one allows arbitrary polygons instead of just triangles, so a natural generalization to three dimensions is triangulations of the $3$-sphere where one allows arbitrary (convex) polyhedra instead of just simplices. One can of course talk about triangulations of arbitrary manifolds. &lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2011-02-18T21:09:56.173" CommentCount="5" />
  <row Id="22716" PostTypeId="2" ParentId="22678" CreationDate="2011-02-18T21:13:29.230" Score="3" ViewCount="" Body="&lt;p&gt;Since $\binom{t+n-k}{k} = \frac{(t+n-k)!}{k! (t+n-2k)!} = \frac{(t+n-k)^{\underline{k}}}{k!}$, your problem reduces to how to find the coefficients $s_0, s_1, \ldots, s_n$ such that $$t^n = \sum_{k=0}^n s_k (t+n-k)^{\underline{k}}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you had $n-k = 0$, this would involve the Stirling numbers of the second kind $S(n,k)$ via the formula &#xA;$$t^n = \sum_{k=0}^n S(n,k) t^{\underline{k}}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instead, what you want are called the &lt;em&gt;non-central Stirling numbers of the second kind&lt;/em&gt;.  Unfortunately, I can't seem to find a good web reference.  Charalambides's &lt;em&gt;&lt;a href=&quot;http://books.google.com/books?id=PDMGA-v5G54C&amp;amp;pg=PA315&quot; rel=&quot;nofollow&quot;&gt;Enumerative Combinatorics&lt;/a&gt;&lt;/em&gt;, pp. 314-319, has a decent discussion, but not all of it appears on Google books.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The relevant formulas are&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$t^n = \sum_{k=0}^n S(n,k;r) (t-r)^{\underline{k}},$$&#xA;and &#xA;$$S(n,k;r) = \sum_{j=k}^n \binom{n}{j} r^{n-j} S(j,k),$$&#xA;where $S(j,k)$ is (again) a Stirling number of the second kind.  Note that the second equation is dual (as it should be!) to the expression inside the parentheses in your question. &lt;/p&gt;&#xA;" OwnerUserId="2370" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2011-05-10T11:03:42.787" LastActivityDate="2011-05-10T11:03:42.787" CommentCount="0" />
  <row Id="22717" PostTypeId="2" ParentId="22714" CreationDate="2011-02-18T21:18:15.597" Score="1" ViewCount="" Body="&lt;p&gt;I'm not clear about your definition, for a graph below, what are cells and boundaries?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://yaroslavvb.com/blog/cholesky/kary-orig.png&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="998" LastActivityDate="2011-02-18T21:18:15.597" CommentCount="2" />
  <row Id="22718" PostTypeId="2" ParentId="22537" CreationDate="2011-02-18T21:36:56.023" Score="2" ViewCount="" Body="&lt;p&gt;The paper &quot;&lt;a href=&quot;http://www.jstor.org/stable/2689812&quot; rel=&quot;nofollow&quot;&gt;Matchings, Derangements, Rencontres&lt;/a&gt;,&quot; by Hanson, Seyffarth, and Weston, has a nice discussion of the rencontre and related problems, including a generalization of the rencontre problem.  They also prove an interesting recursive formula not given in the Wikipedia page linked to in Moron's answer: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$D_{n,k} = n D_{n-1,k} + (-1)^{n-k} \binom{n}{k}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This, of course, contains the famous derangement recurrence as a special case: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$D_n = n D_{n-1} + (-1)^n,$$&#xA;where $D_n$ is the number of derangements (no fixed points) on $n$ elements.&lt;/p&gt;&#xA;" OwnerUserId="2370" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2011-02-19T00:13:43.637" LastActivityDate="2011-02-19T00:13:43.637" />
  <row Id="22720" PostTypeId="1" AcceptedAnswerId="22818" CreationDate="2011-02-18T22:42:00.440" Score="1" ViewCount="111" Body="&lt;p&gt;Hello,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let the heat kernel on $(0,\infty)\times \mathbb R^n$ be given by &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Psi(x,t) = (4\pi t )^{-\frac{n}{2}} e^{ -\dfrac{|x|^2}{4t} }$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for $t&amp;gt;0$, otherwise $0$ except at the origin of space-time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is clear that the heat kernel is smooth everywhere except at the origin. Let $\delta &amp;gt; 0$. How can you show that ( the absolute values of ) the heat kernel and all of its derivatives are uniformly bounded on $[\delta, \infty ) \times \mathbb R^n$, say, by some constant $C_\delta &amp;gt; 0$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems clear as each derivative does not change the relation between the exponents in $x$ and $t$. However, I am stuck and don't know how to proceed. Search on the web provided me only with statements of this fact without a proof.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you.&lt;/p&gt;&#xA;" OwnerUserId="3557" LastEditorUserId="3557" LastEditorDisplayName="" LastEditDate="2011-02-19T04:37:29.027" LastActivityDate="2011-02-19T18:46:18.797" Title="Heat kernel has uniformly bounded derivatives" Tags="&lt;analysis&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="2" />
  <row Id="22721" PostTypeId="1" AcceptedAnswerId="22723" CreationDate="2011-02-18T22:52:51.520" Score="3" ViewCount="265" Body="&lt;p&gt;Hello,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't need to list all proper divisors, I just want to get its sum. Because for a small number, checking all proper divisors and adding them up is not a big deal. However, for a large number, this would run extremely slow. Any idea?   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks,&lt;br&gt;&#xA;Chan Nguyen&lt;/p&gt;&#xA;" OwnerUserId="5711" LastActivityDate="2011-02-19T02:42:35.907" Title="Is there a formula to calculate the sum of all proper divisors of a number?" Tags="&lt;elementary-number-theory&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="22723" PostTypeId="2" ParentId="22721" CreationDate="2011-02-18T23:06:33.943" Score="7" ViewCount="" Body="&lt;p&gt;If the prime factorization of $n$ is $$n=\prod_k p_k^{a_k}$$ where the $p_k$ are the distinct prime factors and the $a_k$ are the positive integer exponents, the sum of &lt;em&gt;all&lt;/em&gt; the positive integer factors is $$\prod_k\left(\sum_{i=0}^{a_k}p_k^i\right).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, the sum of all of the factors of $120=2^3\cdot3\cdot5$ is $$(1+2+2^2+2^3)(1+3)(1+5)=15\cdot4\cdot6=360.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For &lt;em&gt;proper&lt;/em&gt; factors, subtract $n$ from this sum.  This may or may not be faster, depending on the number and how you'd get the prime factorization, but this is the typical technique for high school contest problems of this sort.&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2011-02-18T23:06:33.943" CommentCount="4" />
  <row Id="22725" PostTypeId="1" AcceptedAnswerId="22730" CreationDate="2011-02-19T00:08:27.727" Score="3" ViewCount="198" Body="&lt;p&gt;When considering an infinitesimal distance/interval/in calculus, what is the intuitive interpretation? Is it too small to be measurable but still has some distance on an unattainable scale? Are there different interpretations? If so, what I am considering for the time being is the interpretation in calculus, but I'm still glad to hear of all views.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: I may not be talking about &quot;measurable&quot; in the same sense as measure theory. Sorry about that.&lt;/p&gt;&#xA;" OwnerUserId="4820" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2011-02-19T00:11:26.367" LastActivityDate="2011-02-22T20:13:29.223" Title="Infinitesimals - what's the intuition?" Tags="&lt;calculus&gt;&lt;intuition&gt;" AnswerCount="3" CommentCount="7" />
  <row Id="22726" PostTypeId="2" ParentId="22720" CreationDate="2011-02-19T00:14:43.747" Score="0" ViewCount="" Body="&lt;p&gt;Even with the exponent on $t$ being negaitve I don't think it is true.  At the origin, you just need to show that the $\exp(\frac{-1}{t})$ term dominates every polynomial.  But for $t$ a little bit positive, $\frac {d}{dt}t^{-n}\exp(\frac{-1}{t})\approx t^{-n-2}\exp(\frac{-1}{t})$ which grows unboundedly with $n$.  Similarly taking more derivatives puts more factors of $t$ in the denominator, which increases the derivative.  So it is not uniformly bounded.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2011-02-19T18:12:39.587" LastActivityDate="2011-02-19T18:12:39.587" CommentCount="1" />
  <row Id="22728" PostTypeId="2" ParentId="10761" CreationDate="2011-02-19T00:18:48.333" Score="1" ViewCount="" Body="&lt;p&gt;It is a fractal much like the cantor set with dimension 2/r. That is Jarniks theorem. You can find a proof in the Falconer book Fractal Geometry: Mathematical Foundations and Applications.&lt;/p&gt;&#xA;" OwnerUserId="4290" LastActivityDate="2011-02-19T00:18:48.333" CommentCount="1" />
  <row Id="22729" PostTypeId="2" ParentId="22714" CreationDate="2011-02-19T00:21:33.103" Score="2" ViewCount="" Body="&lt;p&gt;Another generalization is along the genus axis, i.e., consider which graphs can be embedded into a 2d compact surface, such as an $n$-torus. This area is called &lt;a href=&quot;http://en.wikipedia.org/wiki/Topological_graph_theory&quot; rel=&quot;nofollow&quot;&gt;topological graph theory&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="589" LastActivityDate="2011-02-19T00:21:33.103" CommentCount="3" />
  <row Id="22730" PostTypeId="2" ParentId="22725" CreationDate="2011-02-19T00:28:15.837" Score="4" ViewCount="" Body="&lt;p&gt;The intuitive interpretation (and one construction in non-standard analysis) is a sequence of distances converging to zero. So it's a process rather than one single distance. Every &quot;real&quot; distance $x$ can be thought of as the sequence $$x,x,x,\ldots,$$ whereas infinitesimals are sequences like $$1,\frac{1}{2},\frac{1}{3},\ldots.$$ Getting all this to work is Robinson's non-standard analysis.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2011-02-19T00:28:15.837" CommentCount="2" />
  <row Id="22731" PostTypeId="1" AcceptedAnswerId="22735" CreationDate="2011-02-19T00:32:07.807" Score="4" ViewCount="97" Body="&lt;p&gt;Hello,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How could I find all the pairs (n, k) for this equation. The most obvious pair solution that I can see is (1, 1).&lt;br&gt;&#xA;Using summation identity, I have:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{n(n+1)}{2} = \frac{k(k + 1)(2k + 1)}{6}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then I thought of using cubic formula for k-equation, but it involved many variables. Any idea?   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks,&lt;br&gt;&#xA;Chan&lt;/p&gt;&#xA;" OwnerUserId="5711" LastEditorUserId="5711" LastEditorDisplayName="" LastEditDate="2011-02-19T00:57:05.280" LastActivityDate="2011-02-19T01:17:42.387" Title="Find a pair (n, k) such that $\sum_{i=1}^{n} i = \sum_{i=1}^{k} i^2$?" Tags="&lt;elementary-number-theory&gt;" AnswerCount="2" CommentCount="9" FavoriteCount="1" />
  <row Id="22732" PostTypeId="1" AcceptedAnswerId="22736" CreationDate="2011-02-19T00:38:55.810" Score="5" ViewCount="208" Body="&lt;p&gt;Hello,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there exists such an n then n must be the highest of power of 2 that divides n!. But to find out the highest power we need to increase n for each step until it reaches n. Any idea?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks,&lt;br&gt;&#xA;Chan&lt;/p&gt;&#xA;" OwnerUserId="5711" LastActivityDate="2011-02-19T03:50:39.030" Title="Is there an n such that $2^n|n!$?" Tags="&lt;elementary-number-theory&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="22733" PostTypeId="1" AcceptedAnswerId="33434" CreationDate="2011-02-19T00:43:37.813" Score="9" ViewCount="260" Body="&lt;p&gt;Let $H$ be the &lt;a href=&quot;http://en.wikipedia.org/wiki/Hawaiian_earring&quot; rel=&quot;nofollow&quot;&gt;Hawaiian earring&lt;/a&gt; and let $H&amp;#39;$ be the reflection of the Hawaiian earring across the $y$-axis (in the Wikipedia picture). There is a canonical homomorphism from the free product $\pi_1(H) * \pi_1(H&amp;#39;)$ to $\pi_1(H \cup H&amp;#39;)$ (with basepoint their intersection), but it is not an isomorphism.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This was intended to be a recent homework problem of mine, but &lt;em&gt;as stated&lt;/em&gt; the problem actually asked whether the two groups are abstractly isomorphic. I don't know the answer to this question, and neither does my professor. My guess is that they are not isomorphic, but I don't have good intuitions about such large groups. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: to be clear, I know how to do the intended problem, and I also know that $H \cup H&amp;#39;$ is homeomorphic to $H$. &lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2011-02-19T01:24:09.447" LastActivityDate="2011-04-17T07:52:46.217" Title="The fundamental group of a pair of Hawaiian earrings" Tags="&lt;algebraic-topology&gt;" AnswerCount="2" CommentCount="7" FavoriteCount="3" />
  <row Id="22734" PostTypeId="2" ParentId="22731" CreationDate="2011-02-19T00:51:22.557" Score="1" ViewCount="" Body="&lt;p&gt;Fix the variable $k$. Let &#xA;$$k&amp;#39; = \dfrac{k(k+1)(2k+1)}{6}.$$ &#xA;Then you get the quadratic equation &#xA;$$n^2+n-2k&amp;#39; = 0$$&#xA;with the solutions&#xA;$$n_{1/2} = -\dfrac{1}{2} \pm \sqrt{(\dfrac{1}{2})^2+2k&amp;#39;}.$$&#xA;Now you can generate your solution pairs.&lt;/p&gt;&#xA;" OwnerUserId="4857" LastActivityDate="2011-02-19T00:51:22.557" CommentCount="1" />
  <row Id="22735" PostTypeId="2" ParentId="22731" CreationDate="2011-02-19T00:53:16.047" Score="5" ViewCount="" Body="&lt;p&gt;There are only two variables involved.  If you want to search, you can write it as a quadratic in $n$, just try values of $k$, solve for $n$, and see if it comes out integral.  I find k=5, n=10, k=6, n=13 and k=85, n=645 as solutions as well with no more under k=200.  Then &lt;a href=&quot;http://oeis.org/A053612&quot; rel=&quot;nofollow&quot;&gt;OEIS&lt;/a&gt; has no more and asserts the series is finite.  There are references for this claim in &lt;a href=&quot;http://oeis.org/A053611&quot; rel=&quot;nofollow&quot;&gt;A053611&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2011-02-19T01:17:42.387" LastActivityDate="2011-02-19T01:17:42.387" CommentCount="3" />
  <row Id="22736" PostTypeId="2" ParentId="22732" CreationDate="2011-02-19T00:57:45.650" Score="12" ViewCount="" Body="&lt;p&gt;The power of 2 that divides n! is $\lfloor \frac{n}{2} \rfloor + \lfloor \frac{n}{4} \rfloor +\lfloor \frac{n}{8} \rfloor + \ldots \lfloor \frac{n}{2^k} \rfloor$  This can never be more than $n-1$, which occurs when $n$ is a power of 2.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-02-19T00:57:45.650" CommentCount="2" />
  <row Id="22738" PostTypeId="2" ParentId="22732" CreationDate="2011-02-19T01:00:36.357" Score="7" ViewCount="" Body="&lt;p&gt;No. Consider the formula due to Legendre: the power of the prime $p$ in the factorization of $n!$ is $\sum_{k=1}^\infty \lfloor n/p^k \rfloor$.  In the case $p=2$ you have $\sum_{k=1}^\infty \lfloor n/2^k \rfloor$. But&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \sum_{k=1}^\infty \lfloor n/2^k \rfloor &amp;lt; \sum_{k=1}^\infty n/2^k  $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;since we can't have $n/2^k$ an integer for &lt;i&gt;all&lt;/i&gt; $k$ simultaneously, so at least one term in the left-hand sum is less than the corresponding term in the right-hand sum. And the right-hand sum is just $n$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, as you might imagine from the fact that $\lfloor n/2^k \rfloor$ is close to $n/2^k$, the exponent of 2 in $n!$ is pretty close to $n$; in fact it's $n$ minus the number of 1s in the binary expansion of $n$.  In fact, if $n$ is a power of 2, then $2^{n-1}|n!$.&lt;/p&gt;&#xA;" OwnerUserId="173" LastEditorUserId="173" LastEditorDisplayName="" LastEditDate="2011-02-19T03:50:39.030" LastActivityDate="2011-02-19T03:50:39.030" CommentCount="3" />
  <row Id="22739" PostTypeId="2" ParentId="22711" CreationDate="2011-02-19T01:05:20.010" Score="1" ViewCount="" Body="&lt;p&gt;I just worked out the weights by hand for the case $n=3$, except I dropped the trace = 0 condition (since it allows for a nicer basis to do computations in).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case, we find that the 6d representation is 5-d irreducible rep + Trivial.  Of course, the trivial rep is the scalar multiples of the identity, and the complement is the traceless matrices, so the action on $M$ is the unique irreducible 5-d rep.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Without computing higher things by hand, I can't say more than this - but it at least shows you that neither the standard rep nor the adjoint need to show up.&lt;/p&gt;&#xA;" OwnerUserId="331" LastActivityDate="2011-02-19T01:05:20.010" CommentCount="2" />
  <row Id="22740" PostTypeId="2" ParentId="22733" CreationDate="2011-02-19T01:15:52.590" Score="2" ViewCount="" Body="&lt;p&gt;This morphism is not surjective, for the exact same reason $\pi(H)$ is not the free group with countably many generators.&#xA;Just as there are loops in $\pi_1(H)$ that go through an infinite sequence of circles,&#xA;here you have loops in $\pi_1(H \cup H&amp;#39;)$ that go through an infinite sequence of circles, and who change whose circle they're using an infinite amount of times (for example, just pick the $nth$ circle from the left if $n$ is even, and from the right if $n$ is odd).&#xA;So for the same reason, the free product doesn't give you all the cycles, and that morphism is not an isomorphism.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I'm pretty sure you can find an isomorphism between $\pi_1(H \cup H&amp;#39;)$ and $\pi_1(H)$, because they are homeomorphic.&#xA;For example, send the circle of radius $1/n$ of the $H$ from $H \cup H&amp;#39;$ onto the circle of radius $1/2n$ of $H$ ; and the circle of radius $1/n$ of the $H&amp;#39;$ from $H \cup H&amp;#39;$ onto the circle of radius $1/(2n-1)$ of $H$.&lt;/p&gt;&#xA;" OwnerUserId="6270" LastEditorUserId="367" LastEditorDisplayName="" LastEditDate="2011-02-19T03:36:01.123" LastActivityDate="2011-02-19T03:36:01.123" CommentCount="1" />
  <row Id="22741" PostTypeId="2" ParentId="4680" CreationDate="2011-02-19T01:19:33.433" Score="2" ViewCount="" Body="&lt;p&gt;Pardon my skepticism, but has anyone so much as breadboarded Blömer '92 or Landau '93 in all these 18 years?  For lack of same, people still publish ugly surdballs, &lt;em&gt;e.g.&lt;/em&gt;,&#xA;$$\vartheta _3\left(0,e^{-6 \pi }\right)=\frac{\sqrt[3]{-4+3 \sqrt{2}+3 \sqrt[4]{3}+2 \sqrt{3}-3^{3/4}+2 \sqrt{2}\, 3^{3/4}} \sqrt[4]{\pi }}{2\&#xD;&#xA;   3^{3/8} \sqrt[6]{\left(\sqrt{2}-1\right) \left(\sqrt{3}-1\right)} \Gamma \left(\frac{3}{4}\right)}$$&#xA;(J. Yi / J. Math. Anal. Appl. 292 (2004) 381–400, Thm 5.5 vi) instead of&#xA;$$\vartheta _3\left(0,e^{-6 \pi }\right)=\frac{\sqrt{2+\sqrt{2}+\sqrt{2} \sqrt[4]{3}+\sqrt{6}} \,\sqrt[4]{\pi }}{2\ 3^{3/8} \Gamma&#xD;&#xA;   \left(\frac{3}{4}\right)}\quad .$$&#xA;And why do both papers trot out the same old Ramanujan denestings instead of new and interesting ones?  &lt;em&gt;E.g.&lt;/em&gt;,&#xA;$$\sqrt{2^{6/7}-1}=\frac{2^{8/7}-2^{6/7}+2^{5/7}+2^{3/7}-1}{\sqrt{7}}$$&#xA;or&#xA;$$\sqrt[3]{3^{3/5}-\sqrt[5]{2}}=\frac{2^{2/5}+\sqrt[5]{3}+2^{3/5} 3^{2/5}-\sqrt[5]{2}\, 3^{3/5}}{5^{2/3}}$$&#xA;or&#xA;$$\frac{\sqrt[3]{1+\sqrt{3}+\sqrt{2}\, 3^{3/4}}}{\sqrt[6]{\sqrt{3}-1}}=\frac{\sqrt{1+\sqrt{3}+\sqrt{2} \sqrt[4]{3}}}{\sqrt[6]{2}}\quad ?$$&#xA;These results were found by two young students of mine who would very much like to know values of &lt;em&gt;q&lt;/em&gt; and &lt;em&gt;b&lt;/em&gt; in Bill Dubuque's structure theorem which effect the denesting&#xA;$$\sqrt[3]{-\frac{106}{25}-\frac{369 \sqrt{3}}{125}+\frac{3 \sqrt{3} \left(388+268 \sqrt{3}\right)}{100 \sqrt[3]{2}\,&#xD;&#xA;   5^{2/3}}}=\frac{3}{5^{2/3}}-\frac{1+\sqrt{3}}{\sqrt[3]{10}}+\frac{1}{5} \sqrt[3]{2} \left(3+2 \sqrt{3}\right)\quad.$$&#xA;Thanks in advance.&lt;/p&gt;&#xA;" OwnerUserId="7234" LastEditorUserId="7234" LastEditorDisplayName="" LastEditDate="2011-02-19T23:08:28.830" LastActivityDate="2011-02-19T23:08:28.830" CommentCount="1" />
  <row Id="22742" PostTypeId="2" ParentId="22686" CreationDate="2011-02-19T02:12:19.320" Score="4" ViewCount="" Body="&lt;p&gt;Throughout the history of mathematics there has been a tension between the issue of showing that something exists and actually constructing what one may know is there. One can, using Euler's polyhedral formula show that there are 5 combinatorially regular planar graphs, and since these are planar and 3-connected graphs, there have to exist convex polyhedra that correspond to these graphs. This does not, however, show that these polyhedra exist as &quot;regular polyhedra.&quot; Euclid actually constructs the 5 regular convex polyhedra though since no mention is made of convexity the claim that these are the only &quot;regular polyhedra&quot; is problematical. Using a broader collection of &quot;rules&quot; one gets the four Kepler-Poinsot solids, which are not convex but are &quot;regular.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are also other rules for &quot;construction.&quot; For example, one can carry out constructions using folding rules in the spirit of origami constructions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.langorigami.com/science/hha/origami_constructions.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.langorigami.com/science/hha/origami_constructions.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One can also look at constructions with compass alone or ruler alone. All of these lead to interesting mathematics.&lt;/p&gt;&#xA;" OwnerUserId="1369" LastActivityDate="2011-02-19T02:12:19.320" />
  <row Id="22743" PostTypeId="2" ParentId="22686" CreationDate="2011-02-19T02:35:58.710" Score="2" ViewCount="" Body="&lt;p&gt;I think it is safe to say that the practical use of straightegde-compass constructions is close to zero, just like that of playing or listening to music, painting, or reading literature. So those who suggest that playful mathematics should be dropped from the curriculum should also advocate the abolition of music and arts lessons at school (see also Lockhart's Lament linked to by Qiaochu above).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why specifically straightedge-compass constructions? I guess the main reason for why these classical problems are still there is because it is an area in which one can quickly get to problems that require a mix of creative thinking and training without having a lot of technical baggage. It can also be regarded as a piece of history education, since these problems were so important to people in the past (Greeks, motivation for Galois theory, etc.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am sure different people have had different experiences with maths lessons at school, but for me, teachers that were not focused on mindless training of techniques and on practical applications were a welcome exception, rather than the rule. It is clear that a balance between different requirements on mathematics education must be found, and that will surely include practical applications &lt;em&gt;as well as&lt;/em&gt; playful and creative mathematics, such as straightedge-compass constructions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A side note: showing students that maths is fun and creative, as well as developing their problem solving skills, is a good practical investment even if these students don't become mathematicians.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, let me address the argument put forward e.g. by Arturo Magidin, that since students are not taught enough basic skills in mathematics, one should sacrifice some of the more playful elements of maths education in favour of those basic skills (numeracy). I claim that doing that will result in students that learn &lt;em&gt;less&lt;/em&gt; numeracy, not more. The reason is that the playful elements are fun and can help persuade the students that it's worth listening and participating in the maths lessons. If the entire maths education consisted of &quot;numeracy&quot;, students would just doze off, and I wouldn't blame them.&lt;/p&gt;&#xA;" OwnerUserId="3212" LastEditorUserId="3212" LastEditorDisplayName="" LastEditDate="2011-02-20T03:28:26.287" LastActivityDate="2011-02-20T03:28:26.287" CommunityOwnedDate="2011-02-20T03:28:26.287" CommentCount="3" />
  <row Id="22744" PostTypeId="2" ParentId="22721" CreationDate="2011-02-19T02:37:13.167" Score="3" ViewCount="" Body="&lt;p&gt;Just because it is interesting:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is actually a (less known) recursive formula for calculating $\sigma(n)$, the sum of the divisors of $n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sigma(n) = \sigma(n-1) + \sigma(n-2) - \sigma(n-5) - \sigma(n-7) + \sigma(n-12) +\sigma(n-15) + ..$$&#xA;Here  $1,2,5,7,...$ is the sequence of &lt;em&gt;generalized pentagonal numbers&lt;/em&gt; $\frac{3n^2-n}{2}$ for $n = 1,-1,2,-2,...$ and the signs are repetitions of $1,1,-1,-1$. The summation continues until you try to calculate $\sigma$ of something negative. However, if $\sigma(0)$ occurs in the summation (this happens precisely when $n$ is a generalized pentagonal number), it should be replaced by $n$ itself. In other words&#xA;$$ \sigma(n) = \sum_{i\in \mathbb Z_0} (-1)^i\left( \sigma(n - \tfrac{3i^2-i}{2}) + \delta(n,\tfrac{3i^2-i}{2}) \right), $$&#xA;where we set $\sigma(i) = 0$ for $i\leq 0$ and $\delta(\cdot,\cdot)$ is the Kronecker delta.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that  calculating $\sigma(n)$ requires $\sigma(n-1)$ already, therefore its complexity is at least $\mathcal O(n)$, which makes it kind of useless for practical purposes. Note however the lack of reference to divisibility in this formula, which makes it a bit miraculous and therefore worth mentioning.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.math.dartmouth.edu/~euler/pages/E175.html&quot; rel=&quot;nofollow&quot;&gt;Here&lt;/a&gt;'s a reference to the Euler's paper from 1751.&lt;/p&gt;&#xA;" OwnerUserId="5189" LastEditorUserId="5189" LastEditorDisplayName="" LastEditDate="2011-02-19T02:42:35.907" LastActivityDate="2011-02-19T02:42:35.907" CommentCount="1" />
  <row Id="22745" PostTypeId="1" CreationDate="2011-02-19T04:15:12.373" Score="0" ViewCount="96" Body="&lt;p&gt;I have recently developed a type of number that I have never seen before (possibly due to my limited experience with mathematics).  The number is essentially a Directed Acyclic Graph, where each node is assigned a single digit, and carry-over for each digit is determined by the connections between nodes.   This allows addition, subtraction, multiplication, division, powers, etc... to be defined on these &quot;graph numbers&quot;.  (I am pretty sure the term &quot;graph number&quot; is already taken by mathematics, but I didn't know what else to call it.)  Just as the most trivial example, one can define an integer as a graph, somewhat like:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \ldots \longleftarrow d_4 \longleftarrow d_3 \longleftarrow d_2 \longleftarrow d_1 \longleftarrow d_0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, I've never seen anything like this in mathematics, but again, that may be due to my limited experience.  Does anyone know if such numbers already exist, and if so, what kind of mathematics has been done with them?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;public GraphNumber add(GraphNumber b) {&#xA;  if (!GraphNumber.checkSameType(this, b)) {&#xA;    System.out.println(&quot;\nNot the same type!&quot;);&#xA;    System.exit(-1);&#xA;  }&#xA;  GraphNumber tmp = this.cpShape();&#xA;  int k;&#xA;  for (int i=0, cnstsize = size, cnstsize2 = conn[0].length;&#xA;       i&amp;lt;cnstsize; i++) { &#xA;    for (int j=0; j&amp;lt;cnstsize2; j++) {&#xA;      k = conn[i][j];&#xA;      tmp.nodes[k] += (nodes[i] + b.nodes[i] + tmp.nodes[i]) &amp;gt;&amp;gt; 1;&#xA;    }&#xA;    tmp.nodes[i] = (nodes[i] + b.nodes[i] + tmp.nodes[i]) % 2;&#xA;  }&#xA;return tmp;&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;}&lt;/p&gt;&#xA;" OwnerUserId="7237" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-19T05:54:57.060" LastActivityDate="2011-02-19T05:54:57.060" ClosedDate="2011-02-19T09:07:42.200" Title="Numbers defined on graph structures" Tags="&lt;number&gt;&lt;graph&gt;" CommentCount="26" />
  <row Id="22746" PostTypeId="1" CreationDate="2011-02-19T04:40:21.017" Score="3" ViewCount="195" Body="&lt;p&gt;I'd like to know why inner products in Reproducing kernel Hilbert spaces are (linear) evaluation functionals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand that inner products are linear functionals, and I know what an evaluation functional is, I just can't explain why an inner product (in a RKHS) is evaluation functional, and vise-versa.&lt;/p&gt;&#xA;" OwnerUserId="1752" LastActivityDate="2011-02-20T18:54:28.240" Title="Why are inner products in RKHS linear evaluation functionals?" Tags="&lt;functional-analysis&gt;" AnswerCount="1" FavoriteCount="2" />
  <row Id="22748" PostTypeId="2" ParentId="22659" CreationDate="2011-02-19T05:17:15.100" Score="2" ViewCount="" Body="&lt;p&gt;Your question is abstract and needs to be made concrete.  Much like finding a husband or wife, you don't need to know how to charm every person in the world, just one.  In this case, you only need to satisfy one relatively small group of people at a particular institution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Identify the people who will directly need to approve your work.  For instance, it might depend on a particular adviser. Alternatively, there may also be a committee.  As people have said, there are no real rules here.  There is no central authority that forces every mathematics department in the world to do the same thing; and so, as far as I know, they don't.  After you have figured out the main players, figure out their relative importance.  It may turn out that if you adviser approves, then everybody else will rubber stamp it.  This may be the case even if the rules on paper say otherwise.  Bureaucratic institutions can be like that and universities are quite bureaucratic. It may turn out that you need to have some committee's approval.  There may be some people who will not bother to read it.  There may be others who will nitpick everything and make you rewrite everything a dozen times.  The best way to avoid nasty surprises is to get some intel on the people involved.  The easiest way to do this is to just go and ask directly, &quot;Prof. X, I am planning to do Y for my thesis.  Would that be what you are looking for?&quot;.  Later you might have to say, &quot;Prof. X, I tried to do Y, but I only managed to do Z.  Is that enough?&quot;  If you keep him informed, then it shouldn't be a surprise to you when you turn in the final copy whether he will approve it or not.  You can also get information from people who have been through the same process, other graduate students.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This all involves a lot of chatting and people skills and you may not want to do this.  However, the alternative is you are wasting time and money.  It's okay to approach this in a professional way.  I think most professors will respond professionally and provide you with the information you need to make informed decisions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you don't have an adviser yet, try to schedule a few meetings and try to specify as much as possible what you might be working on.  Pick the project that has the best chance of allowing you to complete everything adequately and on time.&lt;/p&gt;&#xA;" OwnerUserId="1004" LastActivityDate="2011-02-19T05:17:15.100" CommunityOwnedDate="2011-02-19T05:17:15.100" />
  <row Id="22749" PostTypeId="1" AcceptedAnswerId="22761" CreationDate="2011-02-19T05:48:28.773" Score="2" ViewCount="97" Body="&lt;p&gt;I am unfamiliar with the general process of converting NFA to DFA. I have general understanding of the theory, but I don't have the method established. Please help explain the process required to transform an NFA to DFA. Thank you.&lt;/p&gt;&#xA;" OwnerUserId="7240" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2011-03-21T05:32:35.323" LastActivityDate="2011-03-21T05:32:35.323" Title="Nondeterministic Finite Automata to Deterministic Finite Automata?" Tags="&lt;computer-science&gt;&lt;automata&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="22750" PostTypeId="2" ParentId="22272" CreationDate="2011-02-19T05:56:49.883" Score="1" ViewCount="" Body="&lt;p&gt;On page 6 of the &lt;a href=&quot;http://www.combinatorics.org/Volume_17/PDF/v17i1r66.pdf&quot; rel=&quot;nofollow&quot;&gt;linked paper&lt;/a&gt;, a triple swap is defined. Basically, you transform $v_n \rightarrow v_k \rightarrow v_i \rightarrow v_m$ into $v_n \rightarrow v_i \rightarrow v_k \rightarrow v_m$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/rKe3e.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="2643" LastEditorUserId="2643" LastEditorDisplayName="" LastEditDate="2011-02-19T06:10:00.933" LastActivityDate="2011-02-19T06:10:00.933" />
  <row Id="22751" PostTypeId="1" AcceptedAnswerId="22752" CreationDate="2011-02-19T06:00:42.647" Score="2" ViewCount="90" Body="&lt;p&gt;Hello,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given an odd prime $p$, how does one find the highest power of $p$ that divides&#xA;$$\displaystyle\prod_{i=0}^n(2i+1)?$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wrote it down all paper and realized that the highest power of $p$ that divides this product will be the same as the highest power of $p$ that divides $(\lceil\frac{n}{2}\rceil - 1)!$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since &#xA;$$10! = 1\times 2\times 3\times 4\times 5\times 6\times 7\times 8\times 9\times 10$$ while &#xA;$$\prod_{i=0}^{4} (2i+1) = 1\times 3\times 5\times 7\times 9$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Am I in the right track?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks,&lt;br&gt;&#xA;Chan&lt;/p&gt;&#xA;" OwnerUserId="5711" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-19T06:35:11.213" LastActivityDate="2011-02-19T06:35:40.607" Title="How to find the highest power of a prime $p$ that divides $\prod_{i=0}^{n} 2i+1$?" Tags="&lt;elementary-number-theory&gt;" AnswerCount="1" CommentCount="15" />
  <row Id="22752" PostTypeId="2" ParentId="22751" CreationDate="2011-02-19T06:35:40.607" Score="5" ViewCount="" Body="&lt;p&gt;Note that $\displaystyle \prod_{i=1}^{n} (2i-1) = \frac{(2n)!}{2^n n!}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clearly, the highest power of $2$ dividing the above product is $0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For odd primes $p$, we proceed as follows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the highest power of $p$ dividing $\frac{a}{b}$ is nothing but the highest power of $p$ dividing $a$ - highest power of $p$ dividing $b$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i.e. if $s_p$ is the highest power of $p$ dividing $\frac{a}{b}$ and $s_{p_a}$ is the highest power of $p$ dividing $a$ and $s_{p_b}$ is the highest power of $p$ dividing $b$, then $s_p = s_{p_a}-s_{p_b}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the highest power of $p$ dividing $\displaystyle \frac{(2n)!}{2^n n!}$ is nothing but $s_{(2n)!}-s_{2^n}-s_{n!}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that $s_{2^n} = 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now if you want to find the maximum power of a prime $q$ dividing $N!$, it is given by&#xA;$$s_{N!} = \left \lfloor \frac{N}{q} \right \rfloor + \left \lfloor \frac{N}{q^2} \right \rfloor + \left \lfloor \frac{N}{q^3} \right \rfloor + \cdots$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(&lt;a href=&quot;http://math.stackexchange.com/questions/17916/number-of-zeros-at-the-end-of-a-factorial/17917#17917&quot;&gt;Look up this stackexchange thread for the justification of the above claim&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, the highest power of a odd prime $p$ dividing the product is $$\left ( \left \lfloor \frac{2N}{p} \right \rfloor + \left \lfloor \frac{2N}{p^2} \right \rfloor + \left \lfloor \frac{2N}{p^3} \right \rfloor + \cdots \right ) - \left (\left \lfloor \frac{N}{p} \right \rfloor + \left \lfloor \frac{N}{p^2} \right \rfloor + \left \lfloor \frac{N}{p^3} \right \rfloor + \cdots \right)$$&lt;/p&gt;&#xA;" OwnerUserId="3302" LastActivityDate="2011-02-19T06:35:40.607" CommentCount="5" />
  <row Id="22753" PostTypeId="1" CreationDate="2011-02-19T06:37:56.470" Score="10" ViewCount="264" Body="&lt;p&gt;I've learned the process of orthogonal diagonalisation in an algebra course I'm taking...but I just realised I have no idea what the point of it is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The definition is basically this: &quot;A matrix A is orthogonally diagonalisable if there exists a matrix P which is orthogonal and D = P&lt;sup&gt;t&lt;/sup&gt;AP where D is diagonal&quot;. I don't understand the significance of this though...what is special/important about this relationship?&lt;/p&gt;&#xA;" OwnerUserId="7241" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-19T06:45:04.400" LastActivityDate="2011-02-20T12:49:37.587" Title="What's the point of orthogonal diagonalisation?" Tags="&lt;linear-algebra&gt;&lt;matrices&gt;" AnswerCount="5" CommentCount="2" FavoriteCount="3" />
  <row Id="22754" PostTypeId="2" ParentId="22753" CreationDate="2011-02-19T06:44:22.983" Score="14" ViewCount="" Body="&lt;p&gt;When you have a matrix $A$ that is diagonalisable, the matrix $U$ such that $U^{-1}AU=D$ is diagonal is the matrix whose columns are a basis of eigenvectors of $A$. Having a basis of eigenvectors makes understanding $A$ much easier, and makes computations with $A$ easy as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What the basis of eigenvectors does not do, however, is preserve the usual notions of size of vectors, distance between vectors, or angles between vectors, that come from the inner/dot product. The standard basis for $\mathbb{R}^n$ is particularly nice because it is not only a basis, it is an &lt;em&gt;orthonormal&lt;/em&gt; basis. Orthonormal bases make all sorts of things easy to compute, such as how to express vectors as linear combinations of them, least squares, the size of vectors, angles between vectors, etc. For example, if you think about it, it's very easy to express a vector as a linear combination of the vectors in the standard basis; it's usually not so easy to do it in terms of some other bases. With orthonormal bases, all you need to do is take the dot product with the basis elements to get the coefficients of the linear transformation. &lt;em&gt;Very&lt;/em&gt; easy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So now you have two notions that make life simpler: a basis of eigenvectors makes your life easy relative to the linear transformation/basis; an orthonormal basis makes your life easy relative to the vector space itself. Unfortunately, usually you have to choose one or the other, and you cannot have both.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When a matrix $P$ is orthogonal, then its columns are not just a basis for $\mathbb{R}^n$, they are an &lt;strong&gt;orthonormal&lt;/strong&gt; basis. The fact that $A$ is orthogonally diagonalisable means that, for $A$, &lt;em&gt;you don't have to choose!&lt;/em&gt; The columns of $P$ are &lt;em&gt;both&lt;/em&gt; an orthonormal basis, &lt;em&gt;and&lt;/em&gt; a basis of eigenvectors for $A$. So you get the best of both worlds. And being an orthonormal basis, you still keep a good handle on the notions of distance, angles between vectors, and so on.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-19T20:44:26.813" LastActivityDate="2011-02-19T20:44:26.813" />
  <row Id="22755" PostTypeId="1" AcceptedAnswerId="22756" CreationDate="2011-02-19T06:46:34.693" Score="16" ViewCount="285" Body="&lt;p&gt;eg. $1000$ has $16$ factors $(1,  2,  4,  5,  8,  10,  20,  25,  40,  50,  100,  125,  200,  250,  500,  1000)$&lt;/p&gt;&#xA;" OwnerUserId="7242" LastEditorUserId="7242" LastEditorDisplayName="" LastEditDate="2011-02-19T07:04:25.920" LastActivityDate="2011-02-19T07:04:25.920" Title="Why does $10^x$ have $(x+1)^2$ factors?" Tags="&lt;elementary-number-theory&gt;" AnswerCount="3" FavoriteCount="4" />
  <row Id="22756" PostTypeId="2" ParentId="22755" CreationDate="2011-02-19T06:50:13.817" Score="19" ViewCount="" Body="&lt;p&gt;Assuming $x$ is a positive integer, of course...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How many factors does a positive integer $n$ have? If you factor $n$ into primes,&#xA;$$n = p_1^{a_1}\times p_2^{a_2}\times\cdots\times p_r^{a_r}$$&#xA;then every factor of $n$ is of the form&#xA;$$m = p_1^{b_1}\times p_2^{b_2}\times\cdots\times p_r^{b_r}$$&#xA;with $0\leq b_i\leq a_i$. So the number of factors of $n$ is&#xA;$$(a_1+1)(a_2+1)\cdots(a_r+1)$$&#xA;(because you have $a_1+1$ choices for the exponent of $p_1$, $(a_2+1)$ choices for the exponent of $p_2$, etc.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $10 = 2\times 5$, then $10^x = 2^x\times 5^x$, so the number of factors is $(x+1)(x+1)=(x+1)^2$. &lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2011-02-19T06:50:13.817" CommentCount="1" />
  <row Id="22757" PostTypeId="2" ParentId="22755" CreationDate="2011-02-19T06:50:56.603" Score="2" ViewCount="" Body="&lt;p&gt;${\displaystyle 10^x = 2^x5^x}$. A factor of $10^x$ is a number of the form ${\displaystyle 2^y5^z}$ where $0 \leq y \leq x$ and $0 \leq z \leq x$, $y$ and $z$ integers. There are $(x+1)$ integers from $0$ to $x$, so the total number of possible $(y,z)$ is $(x+1)*(x+1) = (x+1)^2$. &lt;/p&gt;&#xA;" OwnerUserId="3035" LastActivityDate="2011-02-19T06:50:56.603" CommentCount="1" />
  <row Id="22758" PostTypeId="2" ParentId="22755" CreationDate="2011-02-19T06:53:35.743" Score="4" ViewCount="" Body="&lt;p&gt;The short answer to your question is $10^x = 2^x 5^x$. We shall now enumerate the number of divisors. Let $d$ be one such divisor. Then $d = 2^y 5^z$, where $0 \leq y \leq x$ and $0 \leq z \leq x$. Now $y$ has $x+1$ choices and similarly $z$ has $x+1$ choices and each of these choices are independent and will give rise to an unique divisor $d$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, the total number of divisors of $10^x$ is $(x+1) \times (x+1) = (x+1)^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, to find the number of divisors of $n$, write $n = p_1^{\alpha_1} \times p_2^{\alpha_2} \times \ldots p_k^{\alpha_k}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A divisor $d$ of $n$ must be of the form  $d = p_1^{\beta_1} \times p_2^{\beta_2} \times \ldots p_k^{\beta_k}$ where $0 \leq \beta_i \leq \alpha_i$, $i \in \{0,1,2,\ldots,k\}$. Hence, each $\beta_i$ has $(1+\alpha_i)$ choices and each of these choices are independent and will give rise to an unique divisor $d$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, the total number of divisors of $n$ is $$\displaystyle \prod_{i=1}^{k} (1+\alpha_i)$$&lt;/p&gt;&#xA;" OwnerUserId="3302" LastActivityDate="2011-02-19T06:53:35.743" CommentCount="1" />
  <row Id="22759" PostTypeId="1" AcceptedAnswerId="22762" CreationDate="2011-02-19T07:15:14.770" Score="2" ViewCount="90" Body="&lt;p&gt;Hello. I want to compute the dimension of $\mathbb{A}_{\mathbb{C}}^{1}$, that is the dimension of the affine space in 1 dimension over the field $\mathbb{C}$ but with respect the $\textbf{Euclidean}$ topology.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My definition of dimension of a topological space $X$: is the greatest upper bound of the set of all ascending chains of length n that consists of irreducible non-empty and closed subsets of X ( $C_{0} \subset C_{1} \ .... \subset C_{n}$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $A \subset \mathbb{C}$ with say two points, then $A = \{x,y\} = \{x\} \cup \{y\}$. But $\mathbb{C}$ is a metric space hence every singleton is closed. Thus if A has at least two points then A is reducible. So the only irreducible sets are singletons because $\mathbb{C}$ with the Euclidean topology is not irreducible. Therefore the dimension is equal to $0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is the above correct? Also, another question, in general is there a quick or easy way to figure out the dimension of a given topological space? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="6495" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2011-02-19T09:58:50.920" LastActivityDate="2011-02-19T09:58:50.920" Title="Elementary question: dimension of the 1- affine space over the complex numbers" Tags="&lt;algebraic-geometry&gt;&lt;general-topology&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" />
  <row Id="22760" PostTypeId="2" ParentId="22753" CreationDate="2011-02-19T07:15:15.943" Score="1" ViewCount="" Body="&lt;p&gt;Arturo has given a nice answer. Here are couple of my additions to Arturo's answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any symmetric matrix which can be diagonalized can be re-written in an orthogonal diagonalized form. Another aspects of this orthogonal diagonalization is that this essentially means that the left and right singular vectors are the same (except probably for a change of sign). The singular values for such a matrix are nothing but the magnitude of the eigen values. So in essence, the orthogonal diagonalization gives the Singular Value Decomposition as well and knowing the SVD is all you need to know about any matrix.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastActivityDate="2011-02-19T07:15:15.943" CommentCount="1" />
  <row Id="22761" PostTypeId="2" ParentId="22749" CreationDate="2011-02-19T07:19:17.887" Score="3" ViewCount="" Body="&lt;p&gt;Suppose the original NFA had state set $S$, initial state $q \in S$, and accepting states $F \subset S$. The DFA is going to keep track of &lt;i&gt;what possible states the NFA could get into reading the input so far&lt;/i&gt;. Therefore each state of the DFA corresponds to a subset of $S$, viz. the possible states the NFA could get into. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The initial state is composed of $q$ and all possible states reachable from $q$ via epsilon transitions. The accepting states are all those containing a state from $F$. The transitions are defined in such a way that the interpretation of states in the DFA conforms to what's written above. In order to see what happens in state $\sigma$ upon reading input $a$, consider for all $s \in \sigma$ all states (if any) reachable by following $a$ and then epsilon transitions; collect all of these, for all $s \in \sigma$, in a set $\tau$, which is the target of the arrow labeled $a$ emanating from $\sigma$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For examples and more formal definitions, check the various textbooks and lecture notes detailing this topic (the latter are available online, just google &quot;NFA to DFA&quot;).&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2011-02-19T07:19:17.887" />
  <row Id="22762" PostTypeId="2" ParentId="22759" CreationDate="2011-02-19T09:03:51.983" Score="6" ViewCount="" Body="&lt;p&gt;The notion of dimension of a topological space you are dealing with is not the one (or any of the ones) that topologists usually deal with.  For instance, for most reasonable notions of topological dimension (e.g. &lt;a href=&quot;http://en.wikipedia.org/wiki/Lebesgue_covering_dimension&quot; rel=&quot;nofollow&quot;&gt;covering dimension&lt;/a&gt;) the Euclidean space $\mathbb{R}^n$ has dimension $n$ (and thus $\mathbb{C}^n \cong \mathbb{R}^{2n}$ has dimension $2n$).  Let us call the invariant you have defined the &lt;strong&gt;Krull dimension&lt;/strong&gt; of a topological space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Edit&lt;/b&gt;: in this answer I want to make reference to two different topologies on $\mathbb{C}$.  It seems best to use separate notation for them.  Let me define&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\mathbb{C}_E$ to be $\mathbb{C}$ endowed with its standard (Euclidean) topology, and&lt;br&gt;&#xA;$\mathbb{C}_Z$ to be $\mathbb{C}$ endowed with the Zariski topology, or equivalently the &#xA;cofinite topology: the proper closed sets are precisely the finite sets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are a couple of problems with your computation of the Krull dimension of $\mathbb{C}_E$.  First there is an issue with the indexing: you seem to be claiming that the only nonempty irreducible closed subsets of $\mathbb{C}_E$ are the singleton sets.  But if this is true, this means the maximal length of a chain is $0$, not $1$, and thus the Krull dimension of $\mathbb{C}_E$ would be zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second: your claim is true, but your argument for it does not seem correct to me.  What you have said is that in a separated space -- i.e., a space in which all singletons are closed (others say &quot;$T_1$&quot;) -- no $2$-element subset is irreducible.  That's true, but there could be larger irreducible closed subsets.  Indeed, the space $\mathbb{C}_Z$ is separated and irreducible: it cannot be the union of two proper closed subsets, since it is infinite and every proper closed subset is finite.  Your argument works equally well to show that a finite closed set in a separated space is only irreducible if it consists of a single point, so the maximal chains of irreducible closed subsets in $\mathbb{C}_Z$ are $\{x\} \subset \mathbb{C}$ and thus the Krull dimension is $1$.  (More generally, if you endow $\mathbb{C}^n$ with the Zariski topology, it has Krull dimension $n$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So separatedness does not distinguish between $\mathbb{C}_E$ and $\mathbb{C}_Z$.  You need to use some stronger property of metric topologies here, and the next most obvious guess turns out to be correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Exercise: If $X$ is a Hausdorff space, then the only irreducible closed subsets are the singleton sets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You should try to prove this on your own and ask if you have any difficulties.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The exercise implies that the Krull dimension of any Hausdorff space is zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a little more information on irreducible spaces, see $\S 13.4$ of &lt;a href=&quot;http://math.uga.edu/~pete/integral.pdf&quot; rel=&quot;nofollow&quot;&gt;these notes&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for your last question: well, most spaces one meets in conventional topology are Hausdorff, so have Krull dimension zero.  Probably the most important class of spaces which need not be Hausdorff (especially for an algebraic geometer!) are the spaces of the form $\operatorname{Spec} R$ for $R$ a commutative ring (i.e., Zariski topologies).  Here the name &quot;Krull dimension&quot; is justified by the fact that the Krull dimension of &#xA;$\operatorname{Spec} R$ is equal to the Krull dimension of $R$ in the sense of ring theory, i.e., computed in terms of lengths of chains of prime ideals.  This follows from the correspondence between irreducible closed subsets and prime ideals.  Finally, if you have a space $X$ which is a finite union of open subspaces $X_1,\ldots,X_n$, then the Krull dimension of $X$ is equal to the maximum of the Krull dimensions of the $X_i$'s.  This helps to compute the Krull dimension of the underlying topological space of an algebraic variety or a sufficiently reasonable scheme.  In practice, if you are given a scheme then its Krull dimension is one of the easiest things to compute: if you don't know that, then you really don't know much about your scheme.&lt;/p&gt;&#xA;" OwnerUserId="299" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2011-02-19T09:23:31.310" LastActivityDate="2011-02-19T09:23:31.310" CommentCount="3" />
  <row Id="22763" PostTypeId="2" ParentId="22753" CreationDate="2011-02-19T09:17:34.117" Score="4" ViewCount="" Body="&lt;p&gt;Mathematics is all about equivalence relations. Often we are not really interested in objects on-the-nose, we are interested in objects up to some natural &lt;a href=&quot;http://en.wikipedia.org/wiki/Equivalence_relation&quot; rel=&quot;nofollow&quot;&gt;equivalence relation&lt;/a&gt;. Matrices really define &lt;em&gt;linear transformations&lt;/em&gt;, so in order to talk about linear transformations using matrices we need to specify when two matrices give the same abstract linear transformation. At the level of a first course in linear algebra, there are at least two natural choices, which are often not distinguished.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Up to change of basis. This is a reasonable choice if your vector space has no natural norm or inner product (e.g. if it is a space of polynomials). The nicest matrices here are the ones that can be diagonalized (by an arbitrary invertible matrix). &lt;/li&gt;&#xA;&lt;li&gt;Up to change of &lt;em&gt;orthonormal&lt;/em&gt; basis. This is a reasonable choice if your vector space has an inner product / Euclidean norm, since you want to be able to compute inner products in the easiest possible way with respect to your basis. Here the nicest matrices are the ones that can be orthogonally diagonalized.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;One of the most important examples of the latter is the use of &lt;a href=&quot;http://en.wikipedia.org/wiki/Fourier_series&quot; rel=&quot;nofollow&quot;&gt;Fourier series&lt;/a&gt;, which orthogonally diagonalize differentiation. This is an important technique for solving differential equations.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Perhaps you are also not familiar with the point of &lt;em&gt;diagonalization.&lt;/em&gt; The point of diagonalization is to change coordinates so that the linear transformation you're interested in is as simple as possible (it doesn't get simpler than diagonal matrices). That makes it easy to analyze, as in the Fourier series example above. &lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2011-02-19T09:43:16.030" LastActivityDate="2011-02-19T09:43:16.030" CommentCount="1" />
  <row Id="22764" PostTypeId="2" ParentId="22697" CreationDate="2011-02-19T09:18:32.757" Score="3" ViewCount="" Body="&lt;p&gt;Suppose $Z \subset X \times Y$ is closed, and suppose $x_0 \in X \setminus \pi[Z]$.&#xA;For any $y \in Y, (x_0, y) \notin Z$, and as $Z$ is closed we find a basic open subset $U(y) \times V(y)$ of $X \times Y$ that contains $(x_0, y)$ and misses $Z$.&#xA;The $V(y)$ cover $Y$, so finitely many of them cover $Y$ by compactness, say $V(y_1),\ldots,V(y_n)$ do. Now define $U = \cap_{i=1}^{n} U(y_i)$, and note that $U$ is an open neighbourhood of $x_0$ that misses $\pi[Z]$. So $\pi[Z]$ is closed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To see that the closed projection property implies compactness (sketch): suppose $X$ has the closed projection property along $X$, and let $\cal{F}$ be a filter on $X$. Define a space $Y$ that is as a set $X \cup \{\ast\}, \ast \notin X$, where $X$ has the discrete topology and a neighbourhood of $\ast$ is of the form $A \cup \{\ast\}$ with $A \in \cal{F}$. Then $D = \{(x,x): x \in X\}$ is a subset $X \times Y$ and closedness of the projection $p: X \times Y \rightarrow Y$ implies that some point $(x,\ast)$ is in its closure, and this $x$ is an adherence point of the filter.&lt;/p&gt;&#xA;" OwnerUserId="4280" LastEditorUserId="4280" LastEditorDisplayName="" LastEditDate="2011-02-19T13:22:50.430" LastActivityDate="2011-02-19T13:22:50.430" />
  <row Id="22765" PostTypeId="1" AcceptedAnswerId="22766" CreationDate="2011-02-19T10:01:01.790" Score="1" ViewCount="138" Body="&lt;p&gt;Certainly is not a circle; it looks like an ellipse but I don't think it is.&#xA;Thanks&lt;/p&gt;&#xA;" OwnerUserId="7243" LastEditorUserId="7243" LastEditorDisplayName="" LastEditDate="2011-05-12T19:13:11.830" LastActivityDate="2011-05-12T19:13:11.830" Title="What conic curve is the graph of y=sin(x) from 0 to pi?" Tags="&lt;calculus&gt;&lt;trigonometry&gt;" AnswerCount="1" />
  <row Id="22766" PostTypeId="2" ParentId="22765" CreationDate="2011-02-19T10:04:24.523" Score="5" ViewCount="" Body="&lt;p&gt;Unsurprisingly, it's called a &lt;a href=&quot;http://en.wikipedia.org/wiki/Sine_wave&quot; rel=&quot;nofollow&quot;&gt;sinusoid&lt;/a&gt;. It's &lt;em&gt;not&lt;/em&gt; a conic.&lt;/p&gt;&#xA;" OwnerUserId="589" LastEditorUserId="589" LastEditorDisplayName="" LastEditDate="2011-02-19T12:49:27.637" LastActivityDate="2011-02-19T12:49:27.637" CommentCount="1" />
  <row Id="22767" PostTypeId="2" ParentId="21068" CreationDate="2011-02-19T10:17:46.690" Score="2" ViewCount="" Body="&lt;p&gt;Because there is a group homomorphism into Dirichlet series.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This can be used to relate zeta values e.g. since $\varphi \star 1 = I$ leads to $\sum \varphi(n)/n^s = \zeta(s-1)/\zeta(s)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It can probably be used to prove arithmetic statements using zeta functions but I don't have any examples of that.&lt;/p&gt;&#xA;" OwnerUserId="4290" LastActivityDate="2011-02-19T10:17:46.690" />
  <row Id="22768" PostTypeId="2" ParentId="21366" CreationDate="2011-02-19T10:29:37.227" Score="2" ViewCount="" Body="&lt;p&gt;That's probably a typo. Try $p_x f(x)= f(x) p_{x_0}$ or, alternatively, $f(x)=1-p_{x_0}-p_x +2 p_{x_0} p_{x}$ in the rest of the proof (which I can't see).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As given, the equality is not true. Think of 2 lines, say of slope 2 and 1/2, in R^2 and the projections along vectors (0,1) and (1,0). Then the image of the first composition is all of the first line,  and of the second one all of the second line. No way such maps can be equal.&lt;/p&gt;&#xA;" OwnerUserId="2633" LastActivityDate="2011-02-19T10:29:37.227" CommentCount="1" />
  <row Id="22769" PostTypeId="1" AcceptedAnswerId="22909" CreationDate="2011-02-19T10:51:56.940" Score="4" ViewCount="213" Body="&lt;p&gt;Hello, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What does it mean to say a function is differentiable with respect to the lebesgue measure almost everywhere. A definition would be helpfull.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do I need to learn about the Radon Nikodym derivative to understand this.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="6386" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2011-02-19T18:40:51.877" LastActivityDate="2011-02-20T08:37:55.387" Title="What does it mean to say a function is differentiable with respect to lebesgue measure?" Tags="&lt;integral&gt;&lt;measure-theory&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="22770" PostTypeId="2" ParentId="22769" CreationDate="2011-02-19T11:04:02.833" Score="1" ViewCount="" Body="&lt;p&gt;I wonder what part of your question is not covered by &lt;a href=&quot;http://en.wikipedia.org/wiki/Lebesgue_differentiation_theorem&quot; rel=&quot;nofollow&quot;&gt;this page&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="6179" LastActivityDate="2011-02-19T11:04:02.833" CommentCount="4" />
  <row Id="22771" PostTypeId="1" CreationDate="2011-02-19T11:20:57.847" Score="0" ViewCount="37" Body="&lt;p&gt;Hi, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to do some image processing using the log-Gabor wavelet transform. I have managed to get it working as long as I Fourier-transform and back the entire signal. But I don't have any idea about how to make it with small windows at a time and still with a continuous response. Is that possible at all? How is normally done?&lt;/p&gt;&#xA;" OwnerUserId="7246" LastActivityDate="2011-02-19T11:20:57.847" Title="Windowing and the log-Gabor wavelet transform..." Tags="&lt;signal-processing&gt;&lt;wavelet-transform&gt;" FavoriteCount="1" />
  <row Id="22772" PostTypeId="1" AcceptedAnswerId="22804" CreationDate="2011-02-19T11:29:41.947" Score="2" ViewCount="127" Body="&lt;p&gt;Lets say we construct a list as follows.&#xA;It has f(n) 1's, and f(2) 2's and f(3) number of three's in it etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let L(n,f) be this list, so if f(n)=n^2 we get L(3,n^2)=(1,2,2,2,2,3,3,3,3,3,3,3,3,3)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If f(n) is such that f(m)&gt;(sum f(n) n=1 to n=m-1). Then the probability to pick the number n from the list L(n,f) is above 1/2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if we take that f(n), but reverse is such that the smallest number always has the most appearances in the list, we make it into g(f) (we can do this since it was arbitrary), then the probability to pick the nmuber &quot;1&quot; from L(n,g(f)) is above 1/2 for all n, but then we let n-&gt;inf, then L(inf,g(f)) contains all the distinc integers, but the probability if you pick a random numebr from the list that it is 1, is still 1/2, is it possible ?&lt;/p&gt;&#xA;" OwnerUserId="4929" LastActivityDate="2011-02-19T15:49:11.657" Title="Probability to pick &quot;1&quot; from a list of infinite disistnct integers." Tags="&lt;number-theory&gt;&lt;probability&gt;&lt;sequences-and-series&gt;&lt;soft-question&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="22773" PostTypeId="1" AcceptedAnswerId="22802" CreationDate="2011-02-19T11:38:05.130" Score="4" ViewCount="59" Body="&lt;p&gt;I've been thinking about this for awhile now (I am trying to find a method of proving the Borsuk-Ulam theorem in 2 dimensions without resorting to the usual, and not so intuitive to non-mathematicians topological arguments).  I saw the question &lt;a href=&quot;http://math.stackexchange.com/questions/5065/does-a-continuous-scalar-field-on-a-sphere-have-continuous-loop-of-isothermic-an&quot;&gt;Does a continuous scalar field on a sphere have continuous loop of &quot;isothermic antipodes&quot;&lt;/a&gt;&#xA;posted here, but am curious about a stronger statement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Say you have a continuous map $f$ from $S^2$ to $R$.  Let $U$ be the set of points such that $f(x)=f(-x)$ for $-x$ being the antipode of $x$.  I;ve been able to show that if $U$ is not the entire sphere, then every pair of antipodal points which are not mapped to the same value by $f$ are disconnected in $S^2-U$ (using an argument similar to one posted in the above question).  So; the set $U$ must then contain at least one continuous loop. What I am trying to show is that $U$ must contain a loop $C$ such that $C=-C$ (equivelantly, that there is a loop $C$ in $U$ for which $S^2-C$ is two disjoint &quot;hemispheres&quot; of equal areas). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems to me that if this were not the case then you could construct a second map $g$ on $S^2$ such that its set of antipodes mapping to the same value would be completely disjoint from $U$ (since two loops on a sphere which are not &quot;equators&quot; in the sense that they divide the area of the sphere in half do not necessarily need to intersect), and then you could construct a map from $S^2$ to $R^2$ which had no antipodes mapping to the same value by $F=(f,g)$, which would contradict the theorem I know is true but would like to prove.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe I'm going about this in a difficult way, so any suggestions on how to study the nature of the set $U$ would be appreciated.&lt;/p&gt;&#xA;" OwnerUserId="4772" LastActivityDate="2011-02-19T15:30:13.880" Title="Continuous scalar fields on spheres." Tags="&lt;topology&gt;&lt;intuition&gt;" AnswerCount="1" />
  <row Id="22774" PostTypeId="1" CreationDate="2011-02-19T11:40:48.513" Score="2" ViewCount="156" Body="&lt;p&gt;$w$ is a complex number that satisfy:&#xA;$$5{w^3} - 3i{\left| w \right|^2} - 2i=0$$&#xA;Can we find $|w|$ without finding $w$ or easy root?&lt;/p&gt;&#xA;" OwnerUserId="7018" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2011-02-19T18:37:50.190" LastActivityDate="2011-02-19T18:37:50.190" Title="Find $|w|$ without finding $w$?" Tags="&lt;complex-numbers&gt;" AnswerCount="3" FavoriteCount="1" />
  <row Id="22775" PostTypeId="1" AcceptedAnswerId="22903" CreationDate="2011-02-19T12:16:03.277" Score="0" ViewCount="66" Body="&lt;p&gt;Hello&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need a formula which will fill an array of X slots with values (from $0$ to $1$). The sum of all array values should be $1$. The values should look like this:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;^&#xA;|&#xA;|&#xA;|&#xA;0.2&#xA;|&#xA;|&#xA;|         .....&#xA;|      ..       ..&#xA;|    .             .&#xA;|  .                 .&#xA;| .                   .&#xA;|.                     .&#xA;|.                     .&#xA;+------------------------array-slot-&amp;gt;&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Which means that for an array of size 7 I'd approximatelly want something like:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;i[0]=0.09&#xA;i[1]=0.14&#xA;i[2]=0.17&#xA;i[3]=0.20 &#xA;i[4]=0.17&#xA;i[5]=0.14&#xA;i[6]=0.09&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;which add to exactly $1$ and look like gaussian (or something).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Accuracy is not very important but I'd like it to sum to $1$ and not look linear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;thanks&lt;/p&gt;&#xA;" OwnerUserId="7249" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2011-02-20T03:35:34.087" LastActivityDate="2011-02-20T07:25:54.963" Title="Filling an array with values which sum to $1$ and look like gaussian" Tags="&lt;statistics&gt;" AnswerCount="3" />
  <row Id="22776" PostTypeId="1" CreationDate="2011-02-19T12:16:40.483" Score="0" ViewCount="79" Body="&lt;p&gt;I am having trouble understanding part of a lecture in class.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $G$ be a group, $\phi$ a representation of $G$ into $GL_n(C)$, and let $E^{ij}$ denote the matrix with 1 in the $i$th row and $j$th column and zero everywhere else. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Define the matrix $M = \sum_{g \in G} \phi(g) E^{ij} \phi(g)^{-1}$. We showed that $M$ commutes with $\phi(g)$ for every $g$. So by Schur's lemma, we know that $M = \alpha I_n$ for some nonzero complex number $\alpha$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if you compute the trace of $M$, it comes out to be $|G| \delta_{ij}$, where $\delta_{ij}$ is the Kronecker delta function. Since $M = \alpha I_n$, the trace should simply be a fixed number $\alpha n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where is the mistake?&lt;/p&gt;&#xA;" OwnerUserId="7251" LastActivityDate="2011-02-19T12:33:57.530" Title="Consequence of Schur's lemma" Tags="&lt;representation-theory&gt;" AnswerCount="1" CommentCount="0" />
  <row Id="22777" PostTypeId="1" AcceptedAnswerId="22785" CreationDate="2011-02-19T12:19:02.267" Score="6" ViewCount="348" Body="&lt;p&gt;This is probably a very stupid question, but I just learned about integrals so I was wondering what happens if we calculate the integral of $\sqrt{1 - x^2}$ from $-1$ to $1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We would get the surface of the semi-circle, which would equal to $\pi/2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would it be possible to calculate $\pi$ this way?&lt;/p&gt;&#xA;" OwnerUserId="5905" LastEditorUserId="1301" LastEditorDisplayName="" LastEditDate="2011-02-21T14:00:56.437" LastActivityDate="2011-02-23T09:51:39.123" Title="Calculate $\pi$ precisely using integrals?" Tags="&lt;calculus&gt;&lt;integral&gt;&lt;pi&gt;&lt;circle&gt;" AnswerCount="4" CommentCount="3" />
  <row Id="22778" PostTypeId="2" ParentId="22774" CreationDate="2011-02-19T12:25:36.750" Score="1" ViewCount="" Body="&lt;p&gt;Since the second and third terms are pure imaginary, the angle of the solution is either $\pi/6$, $5\pi/6$, or $3\pi/2$.  To solve for the magnitudes, just solve the real equation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$5{w^3} - 3 w^2 - 2=0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which clearly has a real positive root at $1$.&lt;/p&gt;&#xA;" OwnerUserId="4726" LastActivityDate="2011-02-19T12:25:36.750" />
  <row Id="22779" PostTypeId="2" ParentId="22777" CreationDate="2011-02-19T12:26:11.660" Score="1" ViewCount="" Body="&lt;p&gt;Yes, this integral converges to $\pi/2$. If you evaluate the integral numerically, with your favorite integration scheme, you can compute digits of $\pi$.&lt;/p&gt;&#xA;" OwnerUserId="5882" LastActivityDate="2011-02-19T12:26:11.660" CommentCount="4" />
  <row Id="22780" PostTypeId="2" ParentId="22774" CreationDate="2011-02-19T12:26:30.540" Score="5" ViewCount="" Body="&lt;p&gt;Take $w = r e^{i \phi}$. Then you get:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$5 r^3 e^{i 3 \phi} - 3i r^2 - 2i = 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which can be rearranged to:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$5r^3 e^{i 3 \phi} = i(3r^2 + 2)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;divide by $i$ on both sides:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$5r^3 e^{i 3 \phi - i\frac{\pi}{2}} = \underbrace{3r^2 + 2}_{\in \mathbb R}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;since RHS is real and positive, LHS is real and positive, i.e. $e^{ i 3 \phi - i \frac{\pi}{2}} = 1$. Thus we get:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$5r^3 - 3r^2 - 2 = 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and we want the positive real roots to this equations, which is $r = 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Fixed an error.&lt;/p&gt;&#xA;" OwnerUserId="5966" LastEditorUserId="5966" LastEditorDisplayName="" LastEditDate="2011-02-19T12:33:29.333" LastActivityDate="2011-02-19T12:33:29.333" />
  <row Id="22782" PostTypeId="1" AcceptedAnswerId="22806" CreationDate="2011-02-19T12:31:52.867" Score="2" ViewCount="82" Body="&lt;p&gt;All functions are smooth and continuous.&lt;br&gt;&#xA;I use ' to signfy $d/dt$.&lt;br&gt;&#xA;Given:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$q_i&amp;#39;(0)=0$, for i=1,2,3,4.&lt;br&gt;&#xA;$q_2(t)q_3(t)&amp;gt;q_1(t)q_4(t)$ for $0&amp;lt;=t&amp;lt;T$. T is finite. At t=T the inequality breaks.&lt;br&gt;&#xA;$q_4(0)*q_3&amp;#39;&amp;#39;(0)&amp;gt;q_4&amp;#39;&amp;#39;(0)*q_3(0)$&lt;br&gt;&#xA;$(q_4&amp;#39;&amp;#39;(0)+q_2&amp;#39;&amp;#39;(0))(q_1(0)+q_3(0)&amp;gt;(q_4(0)+q_2(0))(q_1&amp;#39;&amp;#39;(0)+q_3&amp;#39;&amp;#39;(0))$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This also holds but I suspect it is superfluous:&lt;br&gt;&#xA;$q_2(0)*q_1&amp;#39;&amp;#39;(0)&amp;lt;q_2&amp;#39;&amp;#39;(0)*q_1(0)$&lt;br&gt;&#xA;One can also add this without loss of generality:&lt;br&gt;&#xA;$q_1(0)=1, q_2(0)=-1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Prove or disprove that:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$q_4(t)*q_3&amp;#39;&amp;#39;(t)&amp;gt;q_4&amp;#39;&amp;#39;(t)*q_3(t)$&lt;br&gt;&#xA;$(q_4&amp;#39;&amp;#39;(t)+q_2&amp;#39;&amp;#39;(t))(q_1(t)+q_3(t)&amp;gt;(q_4(t)+q_2(t))(q_1&amp;#39;&amp;#39;(t)+q_3&amp;#39;&amp;#39;(t))$&lt;br&gt;&#xA;for $0&amp;lt;=t&amp;lt;T$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Counterexample will do.&lt;/p&gt;&#xA;" OwnerUserId="5904" LastEditorUserId="5904" LastEditorDisplayName="" LastEditDate="2011-02-19T15:56:30.743" LastActivityDate="2011-02-19T16:16:47.573" Title="System of inequalities" Tags="&lt;real-analysis&gt;&lt;differential-geometry&gt;&lt;differential-equations&gt;" AnswerCount="1" />
  <row Id="22783" PostTypeId="2" ParentId="22776" CreationDate="2011-02-19T12:33:57.530" Score="4" ViewCount="" Body="&lt;p&gt;Schur's lemma doesn't tell you that $\alpha$ is nonzero, and you have also neglected to make it clear with your notation that $M = M^{ij}$ and $\alpha = \alpha^{ij}$ both depend on $i$ and $j$. You then get $n \alpha^{ij} = |G| \delta_{ij}$ and there is no contradiction here. &lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2011-02-19T12:33:57.530" />
  <row Id="22784" PostTypeId="2" ParentId="22683" CreationDate="2011-02-19T12:42:50.173" Score="0" ViewCount="" Body="&lt;p&gt;Apparently, Mario Serna has produced pictures of $U(1)$-bundles on his &lt;a href=&quot;http://www-thphys.physics.ox.ac.uk/people/MarioSerna/GaugeGeometryResearch/&quot; rel=&quot;nofollow&quot;&gt;webpage&lt;/a&gt; and in his paper &lt;a href=&quot;http://arxiv.org/abs/hep-th/0205250&quot; rel=&quot;nofollow&quot;&gt;&quot;Riemannian Gauge Theory and Charge Quantization&quot;&lt;/a&gt;. Here an example&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/Os5Q8.png&quot; width=&quot;600&quot; /&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The image represents a trivial $\mathbb R^3$ over some rectangular base manifold. The $U(1)$ bundle which we want to visualize is shown as an $\mathbb R^2$-sub-bundle: the disks indicate the 2-dimensional fibers at each point, to be understood as subspaces of small 3-dimensional boxes at each point (not shown).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems that the disks are also meant to give an impression of the connection, but I don't fully understand how parallel transport is supposed to work here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;He cites a result by Narasimhan and Ramanan which says that every $U(1)$ bundle can be embedded into a trivial $(2d+1)$-dimensional complex vector bundle where $d = \text{dim} M$ is the dimension of the base manifold. Fortunately, the dimension is lower in the cases drawn.&lt;/p&gt;&#xA;" OwnerUserId="434" LastActivityDate="2011-02-19T12:42:50.173" />
  <row Id="22785" PostTypeId="2" ParentId="22777" CreationDate="2011-02-19T12:43:30.577" Score="4" ViewCount="" Body="&lt;p&gt;In fact, the indefinite integral of $\sqrt{1-x&amp;#178;}$ is $\frac12(\sqrt{1-x&amp;#178;}x + \arcsin{x}) + C$, so you are actually &quot;using&quot; $\pi$ in the arcsine if you solve this somehow symbolically, as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_{-1}^1 \sqrt{1-x&amp;#178;} dx = \arcsin(1) = \frac{\pi}{2}$$&lt;/p&gt;&#xA;" OwnerUserId="7247" LastActivityDate="2011-02-19T12:43:30.577" />
  <row Id="22786" PostTypeId="2" ParentId="22775" CreationDate="2011-02-19T12:49:20.520" Score="4" ViewCount="" Body="&lt;p&gt;Just use the &lt;a href=&quot;http://en.wikipedia.org/wiki/Binomial_distribution&quot; rel=&quot;nofollow&quot;&gt;binomial probability density&lt;/a&gt; for $p=1/2$ and $n$ equal to the size of your array minus 1.&lt;/p&gt;&#xA;" OwnerUserId="3567" LastActivityDate="2011-02-19T12:49:20.520" CommentCount="3" />
  <row Id="22787" PostTypeId="2" ParentId="22775" CreationDate="2011-02-19T12:55:00.743" Score="1" ViewCount="" Body="&lt;p&gt;Let's f(x) = a . exp((x-b)²/(2.c²)) with :&lt;br&gt;&#xA; - a having any value (explained later)&lt;br&gt;&#xA; - b = (n-1)/2&lt;br&gt;&#xA; - c having any value (proportional to n would make sense though)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then :&lt;br&gt;&#xA;    int sum=0;&lt;br&gt;&#xA;    for (int i=0; i smaller than n; i++)&lt;br&gt;&#xA;    {&lt;br&gt;&#xA;      a[i] = f(i);&lt;br&gt;&#xA;      sum+=a[i];&lt;br&gt;&#xA;    }&lt;br&gt;&#xA;    for (int i=0; i smaller than n; i++)&lt;br&gt;&#xA;    {&lt;br&gt;&#xA;      a[i]/=sum;&lt;br&gt;&#xA;    }  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I haven't tried but it seems to make sense. Maybe there'll be a problem of offset but you'll see it when testing.&lt;/p&gt;&#xA;" OwnerUserId="7252" LastActivityDate="2011-02-19T12:55:00.743" CommentCount="1" />
  <row Id="22788" PostTypeId="1" AcceptedAnswerId="22989" CreationDate="2011-02-19T13:11:08.567" Score="2" ViewCount="79" Body="&lt;p&gt;Hello,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am aware that if there is a bifurcating tree with N leaves, then there are (N-1) internal nodes (branching points) with a single root node. How is this relationship proved?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Best,&lt;/p&gt;&#xA;" OwnerUserId="1587" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2011-02-19T18:41:59.473" LastActivityDate="2011-02-21T15:10:44.503" Title="Number of inner nodes in relation to the leaf number N" Tags="&lt;graph-theory&gt;&lt;graph&gt;&lt;trees&gt;" AnswerCount="2" />
  <row Id="22789" PostTypeId="1" AcceptedAnswerId="22790" CreationDate="2011-02-19T13:24:51.163" Score="4" ViewCount="100" Body="&lt;p&gt;What does &quot;M&quot; stand for in Weierstrass M-Test?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just asking...&lt;/p&gt;&#xA;" OwnerUserId="6838" LastActivityDate="2011-02-19T13:34:10.193" Title="Weierstrass M-Test" Tags="&lt;math-history&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="0" />
  <row Id="22790" PostTypeId="2" ParentId="22789" CreationDate="2011-02-19T13:34:10.193" Score="9" ViewCount="" Body="&lt;p&gt;The M stands for majorant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In german it is also called &lt;a href=&quot;http://de.wikipedia.org/wiki/Weierstra%C3%9Fsches_Majorantenkriterium&quot; rel=&quot;nofollow&quot;&gt;Weierstraßsches Majorantenkriterium&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="367" LastActivityDate="2011-02-19T13:34:10.193" CommentCount="4" />
  <row Id="22791" PostTypeId="2" ParentId="22777" CreationDate="2011-02-19T13:41:30.867" Score="9" ViewCount="" Body="&lt;p&gt;If you want to calculate $\pi$ in this way, note that the expansion of&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sqrt{1-x^2} = 1 - \sum_{n=1}^\infty \frac{(2n)!}{(2n-1)2^{2n}(n!)^2} x^{2n} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and so if we integrate term by term and evaluate from $-1$ to $1$ we will end up with the following formula for $\pi$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \pi = 4 \left\lbrace 1 - \sum_{n=1}^\infty \frac{(2n)!}{(4n^2-1)2^{2n}(n!)^2} \right\rbrace .$$&lt;/p&gt;&#xA;" OwnerUserId="1301" LastEditorUserId="1301" LastEditorDisplayName="" LastEditDate="2011-02-19T14:54:11.863" LastActivityDate="2011-02-19T14:54:11.863" />
  <row Id="22792" PostTypeId="2" ParentId="22788" CreationDate="2011-02-19T13:58:36.863" Score="3" ViewCount="" Body="&lt;p&gt;I think the easiest way it to regard this a knock-out competition with $n$ teams, each inner node is a match, and there is one winner. So there are $n-1$ teams to knock out and so $n-1$ nodes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or you could do it by induction noting that replacing a pair of leaves and an inner node by a leaf reduces the number of inner nodes and leaves by 1, and that if there is only one leaf then there is no inner node.&lt;/p&gt;&#xA;" OwnerUserId="6460" LastActivityDate="2011-02-19T13:58:36.863" CommentCount="1" />
  <row Id="22793" PostTypeId="1" AcceptedAnswerId="22795" CreationDate="2011-02-19T14:15:02.010" Score="2" ViewCount="141" Body="&lt;p&gt;Please help in to understand why/how $$|\vec{a} +\vec{b}|^2 = |\vec{a}|^2 +|\vec{b}|^2 + 2 \vec{a}\cdot \vec{b}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My precise confusion is why the modulus is not included in $2 \vec{a}\cdot \vec{b}$ ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since if we do &#xA;$(|\vec{a}| +|\vec{b}|)^2$, we have modulus sign in every $\vec{a}$ and $\vec{b}$.&lt;/p&gt;&#xA;" OwnerUserId="2109" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2011-02-19T19:13:13.873" LastActivityDate="2011-02-19T19:13:13.873" Title="How $|\vec{a} +\vec{b}|^2 = |\vec{a}|^2 +|\vec{b}|^2 + 2\vec{a}\cdot \vec{b}$ ?" Tags="&lt;linear-algebra&gt;&lt;inner-product-space&gt;" AnswerCount="3" FavoriteCount="1" />
  <row Id="22794" PostTypeId="2" ParentId="22793" CreationDate="2011-02-19T14:25:51.720" Score="2" ViewCount="" Body="&lt;p&gt;Because...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$|\vec{a} +\vec{b}|^2 = (\vec{a} +\vec{b}) \cdot  (\vec{a} +\vec{b})$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(|\vec{a}| +|\vec{b}|)^2 = (|\vec{a}| +|\vec{b}|)\cdot (|\vec{a}| +|\vec{b}|)$&lt;/p&gt;&#xA;" LastActivityDate="2011-02-19T14:25:51.720" />
  <row Id="22795" PostTypeId="2" ParentId="22793" CreationDate="2011-02-19T14:27:36.010" Score="6" ViewCount="" Body="&lt;p&gt;The problem is that $|\vec a+ \vec b|^2 \neq (|\vec a| + |\vec b|)^2$, because $|\vec a + \vec b| \neq |\vec a| + |\vec b|$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Suppose $\vec a$ and $\vec b$ both have length 1 but have opposite directions; then you can see why it's not equal!)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have to use the fact $|\vec a|^2 = \vec a \cdot \vec a$ and properties of the dot product to prove the identity you have above.  Using the fact that the dot product of a vector with itself is the square of the modulus, your identity is the same as $(\vec a + \vec b) \cdot (\vec a + \vec b) = \vec a \cdot \vec a + 2 \vec a \cdot \vec b + \vec b \cdot \vec b$.  Now can you see why it's true?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Note: I use the LaTeX code &lt;code&gt;\cdot&lt;/code&gt; to write the dot product.  Even though they have some of the same properties, the dot product is not the same as multiplication.&lt;/p&gt;&#xA;" OwnerUserId="1461" LastEditorUserId="1461" LastEditorDisplayName="" LastEditDate="2011-02-19T14:33:38.650" LastActivityDate="2011-02-19T14:33:38.650" CommentCount="1" />
  <row Id="22796" PostTypeId="1" AcceptedAnswerId="22800" CreationDate="2011-02-19T14:34:39.003" Score="3" ViewCount="104" Body="&lt;p&gt;In order to learn about vector bundles, I would like to draw the tautological vector bundle over the complex projective line&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ E = \{(x,v) \in \mathbb{CP}^1 \times \mathbb{C}^2 : v \in x \} .$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Identifying the complex projective line with the Riemann sphere, $\mathbb{CP}^1 \cong S^2$, I hope that it might be possible to visualize this bundle by attaching small planes to each point of the sphere, similar to how one can visualize the tangent bundle of sphere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, I'm looking for an embedding $E \hookrightarrow S^2 \times \mathbb{R}^3$ into a trivial bundle. (Obviously, $E$ has to be viewed as a 2-dimensional real vector bundle for this to make sense.) I am aware that such a thing might not exist, in which case I would like to learn why.&lt;/p&gt;&#xA;" OwnerUserId="434" LastActivityDate="2011-02-19T15:17:42.667" Title="Trying to draw the tautological line bundle ($\subseteq \mathbb{CP}^1\times \mathbb{C}^2$)" Tags="&lt;differential-topology&gt;&lt;visualization&gt;&lt;vector-bundles&gt;" AnswerCount="2" />
  <row Id="22797" PostTypeId="1" CreationDate="2011-02-19T14:35:13.950" Score="4" ViewCount="72" Body="&lt;p&gt;Let $R$ be a commutative ring with unit. Let $R_p$ be a domain for all $p\in SpecR$ and let $SpecR$ be connected. Is it true that $R$ is a domain or can someone provide a counterexample. Note here that $R$ is not necessarily a Noetherian ring. For a Noetherian ring this is easy.&lt;/p&gt;&#xA;" OwnerUserId="7254" LastActivityDate="2011-02-19T18:12:35.420" Title="Locally a domain and connected implies a domain" Tags="&lt;commutative-algebra&gt;" AnswerCount="1" />
  <row Id="22798" PostTypeId="2" ParentId="22796" CreationDate="2011-02-19T14:48:23.793" Score="1" ViewCount="" Body="&lt;p&gt;I think you can't do much better than to visualize the Hopf fibration $S^{3} \to S^{2}$, which can be defined by restricting the tautological line bundle to $S^{3} \subset \mathbb{C}^{2}$ (in other words, the fiber over each point $x \in \mathbb{CP}^{1}$ is the circle of points of length $1$ in $x$). Most of the pictures are obtained by stereographically projecting $S^{3}\smallsetminus \{N\}$ to $\mathbb{R}^{3}$. Just google for Hopf fibration and you'll get lots of nice pictures.&lt;/p&gt;&#xA;" OwnerUserId="5363" LastEditorUserId="5363" LastEditorDisplayName="" LastEditDate="2011-02-19T14:55:40.587" LastActivityDate="2011-02-19T14:55:40.587" />
  <row Id="22799" PostTypeId="2" ParentId="22774" CreationDate="2011-02-19T15:09:15.443" Score="4" ViewCount="" Body="&lt;p&gt;The answer is &lt;em&gt;yes we can&lt;/em&gt;... &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For any solution $w$, the two complex numbers $5w^3$ and $(3\left| w \right|^2+2)\mathrm{i}$ are the same hence their modulus are the same: $5\left|w\right|^3=3\left| w \right|^2+2$. This means that the modulus $r$ of any solution $w$ is such that $5r^3-3r^2-2=0$. Here ends the answer to the OP question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Naturally one can go further since an obvious solution of the $r$ equation is $r=1$. Hence $5r^3-3r^2-2=(r-1)(5r^2+2r+2)$ and, since $5r^2+2r+2=0$ has no real solution, every solution of the $w$ equation is such that $\left| w \right|=1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All this does not say that there exists any solution $w$ at all but one can go further and note that a complex number $w$ such that $\left| w \right|=1$ is a solution iff the argument of $w^3$ is $\pi/2$ hence the argument of $w$ is $\pi/6$ or $5\pi/6$ or $3\pi/2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally the three solutions are $\mathrm{e}^{\mathrm{i}\pi/6}=(\sqrt{3}+\mathrm{i})/2$, $\mathrm{e}^{\mathrm{i}5\pi/6}=(-\sqrt{3}+\mathrm{i})/2$ and $\mathrm{e}^{\mathrm{i}3\pi/2}=-\mathrm{i}$.&lt;/p&gt;&#xA;" OwnerUserId="6179" LastActivityDate="2011-02-19T15:09:15.443" />
  <row Id="22800" PostTypeId="2" ParentId="22796" CreationDate="2011-02-19T15:17:42.667" Score="1" ViewCount="" Body="&lt;p&gt;I claim that there is no bundle embedding of the realification of the tautological bundle $\mathcal{O}(-1)$ into the trivial (real) bundle $S^2 \times \mathbb{R}^3$. Suppose there were; then we could take the orthogonal complement $N$, and we would obtain a decomposition&#xA;$$ N \oplus \mathcal{O}(-1)_{\mathbb{R}} = S^2 \times \mathbb{R}^3$$&#xA;where $N$ is a real line bundle. But real line bundles on any compact CW complex $X$ are classified by the first Chern class, which lives in $H^1(X, \mathbb{Z}/2)$ (since the infinite 1-Grassmannian is a $K(\mathbb{Z}/2, 1)$). However, $H^1(S^2, \mathbb{Z}/2)=0$, and so $N$ is trivial.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It follows that if such an embedding existed, then $\mathcal{O}(-1)_{\mathbb{R}}$ would be &lt;em&gt;stably&lt;/em&gt; trivial. This is, however, not the case. Stable trivialty would imply that the Stiefel-Whitney classes were trivial, as the product formula for them shows. However, we know that the top Chern class in $H^2(\mathbb{CP}^1, \mathbb{Z})$ generates the group, and also that (Proposition 3.8 in Hatcher's &lt;em&gt;Vector Bundles and K-theory&lt;/em&gt;, available &lt;a href=&quot;http://www.math.cornell.edu/~hatcher/VBKT/VBpage.html&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;) implies that the top Stiefel-Whitney class is the image  of the top Chern class. But the image of a generator in $H^2(\mathbb{CP}^1, \mathbb{Z})$ in $H^2(S^2, \mathbb{Z}/2)$ is nonzero. &lt;/p&gt;&#xA;" OwnerUserId="536" LastActivityDate="2011-02-19T15:17:42.667" CommentCount="3" />
  <row Id="22801" PostTypeId="2" ParentId="22793" CreationDate="2011-02-19T15:24:48.660" Score="6" ViewCount="" Body="&lt;p&gt;Let us write the norm in terms of inner product, namely $|x| = \sqrt{\langle x,x\rangle}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now: &#xA;$$\begin{align*}&#xD;&#xA;|a+b|^2 &amp;amp;=\langle a+b,a+b\rangle\\&#xD;&#xA; &amp;amp;= \langle a,a+b\rangle + \langle b,a+b\rangle \\&#xD;&#xA; &amp;amp;= \langle a,a\rangle + \langle a,b\rangle + \langle b,a\rangle +\langle b,b\rangle\\&#xD;&#xA; &amp;amp;= |a|^2 + |b|^2 + \langle a,b\rangle +\overline{\langle a,b\rangle}\\&#xD;&#xA; &amp;amp;= |a|^2 + |b|^2 + 2Re(\langle a,b\rangle)&#xD;&#xA;\end{align*}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $Re$ is taking the real part of a complex number, and of course when the vector space is real then $2Re(\langle a,b\rangle) = 2\langle a,b\rangle$.&lt;/p&gt;&#xA;" OwnerUserId="622" LastActivityDate="2011-02-19T15:24:48.660" />
  <row Id="22802" PostTypeId="2" ParentId="22773" CreationDate="2011-02-19T15:30:13.880" Score="1" ViewCount="" Body="&lt;p&gt;Since &lt;a href=&quot;http://math.stackexchange.com/questions/5065/does-a-continuous-scalar-field-on-a-sphere-have-continuous-loop-of-isothermic-an&quot;&gt;the question you refer to&lt;/a&gt; was answered negatively (such a loop does not necessarily exist), you should not assume that the set $U$ contains at least one continuous loop.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even if $U$ does contain a continuous loop, it is not necessarily the case that there exists a loop $C$ so that $C = -C$.  For example, there exists a function on the sphere whose $U$ region is the union of two antipodal small circles on the sphere connected by a pair of antipodal topologist's sine curves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, if the function $S^2\to\mathbb{R}$ is sufficiently well-behaved, the set $U$ will be a graph on the sphere.  In this case, it is relatively easy to show that $U$ must contain a loop $C$ such that $C = -C$.  Specifically, since $U=-U$, the antipode of each vertex of $U$ is again a vertex of $U$.  If we choose a path from a vertex to its antipode, then the union of this path with the antipodal path is a closed loop.  (This argument works whenever $U$ is path-connected.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is possible that the loop created in this fashion intersects itself, since the original path may contain more than one pair of antipodal points.  However, it is always possible to shorten the path so that the only pair of antipodal points are the endpoints.  Specifically, given a path $\gamma\colon[0,1]\to S^2$ between two antipodal points, let&#xA;$$&#xD;&#xA;a = \max\{t \in [0,1] \mid \gamma(t) = -\gamma(s)\text{ for some }s&amp;gt;t\}&#xD;&#xA;$$&#xA;and let&#xA;$$&#xD;&#xA;b = \min\{s \in [a,1] \mid \gamma(s) = -\gamma(a)\}&#xD;&#xA;$$&#xA;Then the restriction of $\gamma$ to $[a,b]$ is a path whose only pair of antipodal points are its endpoints, so the union of $\gamma$ and $-\gamma$ is a simple closed curve $C$ for which $C=-C$.&lt;/p&gt;&#xA;" OwnerUserId="1726" LastActivityDate="2011-02-19T15:30:13.880" CommentCount="2" />
  <row Id="22804" PostTypeId="2" ParentId="22772" CreationDate="2011-02-19T15:49:11.657" Score="2" ViewCount="" Body="&lt;p&gt;By integer in your comment, it seems you mean positive integer.  A probability distribution that meets your request would be $P(n)=\frac{1}{2^n}$.  It sums to 1 and has positive value for all $n \gt 0$ with $P(1) \gt P(2) \gt \ldots$  If you really want all integers it can easily be reworked, but you would have to specify which ones you want more probable.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-02-19T15:49:11.657" CommentCount="2" />
  <row Id="22805" PostTypeId="2" ParentId="22620" CreationDate="2011-02-19T15:54:57.280" Score="3" ViewCount="" Body="&lt;p&gt;On Alex's request I'm leaving my comment as an answer. Let $L/K$ be a finite extension of number fields. A real place $v$ of $K$ (which can be thought of as an embedding of $K$ into $\mathbb{C}$ with image contained in $\mathbb{R}$) is said to ramify in $L$ if it extends to an embedding of $L$ into $\mathbb{C}$ with non-real image. If all extensions of $v$ to places of $L$ are real (the associated embeddings have real image), then $v$ is unramified (also said to be split) in $L$. A complex place of $K$ (an embedding into $\mathbb{C}$ with non-real image) is always unramified. So, the extension $L/K$ is unramified at $\infty$ if all the real places stay real. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, in the extension $\mathbb{Q}(\zeta_p)/\mathbb{Q}(\zeta_p+\zeta_p^{-1})$, where $\zeta_p$ is a primitive $p$-th root of unity for some odd prime $p$, the base field is totally real (all its Archimedean places are real) while the top field is totally imaginary, so all real places ramify in the extension. &lt;/p&gt;&#xA;" OwnerUserId="628" LastActivityDate="2011-02-19T15:54:57.280" />
  <row Id="22806" PostTypeId="2" ParentId="22782" CreationDate="2011-02-19T16:16:47.573" Score="2" ViewCount="" Body="&lt;p&gt;It can't be true as you have nothing except smoothness to restrict the functions away from zero.  Note that you are trying to prove $(q_1(t)q_2&amp;#39;(t)-q_1&amp;#39;(t)q_2(t))&amp;#39;\gt 0$  So start them off with $q_1(t)=a+bt^2, q_2(t)=c+dt^2, q_3(t)=e+ft^2, q_4(t)=g+ht^2$ for $t\le 1$.  Your second equation can be satisfied if $c,e \gt \gt a,g$  Then for $t\gt 1$ add something smooth with positive derivative to $q_1$ like $\exp(\frac{a}{x^2-1})$  If you pick $a$ large enough, the derivatives will be larger than the function.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-02-19T16:16:47.573" />
  <row Id="22807" PostTypeId="1" AcceptedAnswerId="22812" CreationDate="2011-02-19T16:33:10.670" Score="5" ViewCount="78" Body="&lt;p&gt;Let $V$ be a vector space over a field $k$ (not necessarily finite-dimensional) and $T : V \to V$ a linear operator. Is there an accepted term for the following condition on $T$?&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;For any $v \in V$ the subspace $\text{span}(v, Tv, T^2 v, ...)$ is finite-dimensional, and $T$ is nilpotent on any such subspace. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For example, the differential operator $\frac{d}{dx}$ acting on $k[x]$ satisfies this condition but is not nilpotent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Motivation:&lt;/strong&gt; When $\text{char}(k) = 0$, this condition ensures that the exponential $e^T : V \to V$ is well-defined without giving $V$ any additional structure, since $e^T v$ is a finite sum for any particular $v$. &lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2011-02-19T17:00:13.310" Title="Is there an accepted term for &quot;locally&quot; nilpotent linear operators?" Tags="&lt;linear-algebra&gt;&lt;terminology&gt;" AnswerCount="1" CommentCount="0" />
  <row Id="22808" PostTypeId="1" AcceptedAnswerId="22810" CreationDate="2011-02-19T16:44:57.567" Score="2" ViewCount="86" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a way to estimate the reciprocal of a decimal number? e.g. 1/0.388. Are there any formulas for this sort or thing, or do I just have to calculate it by hand?&lt;/p&gt;&#xA;" OwnerUserId="6815" LastActivityDate="2011-02-19T16:50:37.363" Title="Is there a way to quickly estimate the reciprocal of a number?" Tags="&lt;number-theory&gt;" AnswerCount="2" />
  <row Id="22809" PostTypeId="2" ParentId="22808" CreationDate="2011-02-19T16:49:37.350" Score="2" ViewCount="" Body="&lt;p&gt;$ 1/0.388 = 1/(388/1000)=1000/388=250/97$. Now do it by hand, as precise as you need.&lt;/p&gt;&#xA;" OwnerUserId="589" LastActivityDate="2011-02-19T16:49:37.350" CommentCount="2" />
  <row Id="22810" PostTypeId="2" ParentId="22808" CreationDate="2011-02-19T16:50:37.363" Score="4" ViewCount="" Body="&lt;p&gt;It depends what facts you have in your head.  Probably you know that 1/2=0.5, 1/3=0.3333, 1/4=0.25, 1/5=0.2.  It is less common to know 1/6=0.16666, 1/7=0.142857, 1/8=0.125, 1/9=0.11111.  Leading zeros just make a factor of 10.  You can just &quot;look down the list&quot; to see 1/.388 is between 2 and 3.  Then 0.388 is about 15% bigger than 0.333, so 1/0.388 is about 15% less than 3, or about 2.55.  Here we are using that $(1+x)^{-1}\approx 1-x$ for $x$ small compared with 1.  This is only 1% off&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-02-19T16:50:37.363" />
  <row Id="22812" PostTypeId="2" ParentId="22807" CreationDate="2011-02-19T17:00:13.310" Score="4" ViewCount="" Body="&lt;p&gt;The standard name is &lt;em&gt;locally nilpotent&lt;/em&gt;. Thus one hears about locally nilpotent derivations, for example, like $\frac{\mathrm d}{\mathrm dt}$ in $k[t]$.&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2011-02-19T17:00:13.310" CommentCount="1" />
  <row Id="22813" PostTypeId="2" ParentId="22746" CreationDate="2011-02-19T17:06:49.600" Score="5" ViewCount="" Body="&lt;p&gt;On a Hilbert space, all continuous linear functionals are inner product functionals (Riesz), and conversely (Cauchy-Schwarz).  In a RKHS, evaluation functionals are continuous, which is equivalent to being inner product functionals.  The converse is usually not true.  That is, inner product functionals on a RKHS need not be point evaluations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $H$ be a Hilbert space consisting of complex-valued functions on a set $X$ such that for each $x\in X$, the evaluation functional $f\mapsto f(x)$ is continuous.  Then for each $x\in X$ there is a $k_x\in H$ such that for all $f\in H$, $f(x)=\langle f,k_x\rangle$.  (The function $K:X\times X\to \mathbb{C}$ defined by $K(x,y)=k_y(x)=\langle k_y,k_x\rangle$ is the reproducing kernel of the RKHS $H$, and some authors start with $K$ in defining a RKHS.)  Only the inner products with the elements $k_x$ are evaluation functionals.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, consider the Hardy space $H^2$ of holomorphic functions on the open unit disk whose sequences of Maclaurin series coefficients are in $\ell^2$, with inner product $\displaystyle{\left\langle \sum_{k=0}^\infty a_kz^k,\sum_{k=0}^\infty b_kz^k\right\rangle=\sum_{k=0}^\infty a_k\overline{b_k}}$.  Evaluations on the open disk are continuous, as can be seen directly by writing down the element $k_w$ of $H^2$ whose inner product functional is evaluation at $w$, $k_w(z)=\sum_{k\geq0} \overline{w}^kz^k=\frac{1}{1-\overline{w}z}$.  So a necessary and sufficient condition for an inner product functional $f\mapsto\langle f,g\rangle$ to be an evaluation functional is the existence of a $w$ in the open unit disk such that $g=k_w$, a condition which typical $g\in H^2$ will not satisfy.  Note that the set of evaluation functionals is not closed under scalar multiplication, nor addition.  In fact, it is linearly independent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A simpler but in some ways less interesting example is $\ell^2$ thought of as a space of functions on the nonnegative integers, where the evaluation functionals are just the inner products with elements of $\ell^2$ that have the value $1$ at one point and vanish elsewhere.  An even simpler example would be a finite dimensional Hilbert space thought of as functions on a finite set.  In these cases, cardinality is enough to see that most inner product functionals are not point evaluations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The inner product functionals and evaluation functionals would be identical if you were considering a Hilbert space $H$ as a space of functions on its dual space, in the usual isomorphism of $H$ with its double dual.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added:&lt;/strong&gt; Here is some elaboration on the first 2 sentences.  If $H$ is a Hilbert space and $g\in H$, then the function $T_g :H\to \mathbb{C}$ defined by $T_g(f)=\langle f,g\rangle$ is a linear functional on $H$ called an &quot;inner product functional&quot; above.  Each inner product functional is continuous.  The &lt;a href=&quot;http://en.wikipedia.org/wiki/Operator_norm&quot; rel=&quot;nofollow&quot;&gt;operator norm&lt;/a&gt; of $T_g$ is equal to $\|g\|$.  The Cauchy-Schwarz inequality gives $|T_g(f)|\leq \|f\|\|g\|$ for all $f$, which means $\|T_g\|\leq\|g\|$.  Plugging $g$ into $T_g$ gives $|T_g(g)|=\|g\|^2$, showing that $\|T_g\|\geq \|g\|$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So inner product functionals are continuous, and this would be true in any inner product space.  The Riesz representation theorem (for Hilbert space, sometimes also called Riesz's lemma) says that the converse is true for a Hilbert space.  You can see this for example in &lt;a href=&quot;http://en.wikipedia.org/wiki/Riesz_representation_theorem#The_Hilbert_space_representation_theorem&quot; rel=&quot;nofollow&quot;&gt;the Wikipedia article&lt;/a&gt;, and in many textbooks including the basics of Hilbert spaces, such as Rudin's &lt;em&gt;Real and complex analysis&lt;/em&gt;.  That is, if $T:H\to\mathbb{C}$ is any continuous linear functional, then there is a $g\in H$ such that $T=T_g$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hopefully the first sentence is clearer now.  As for the second sentence, it follows directly from the first sentence and the definition of RKHS, and the second paragraph elaborates on this.  There is more than one way to characterize RKHS, and if continuity of point evaluations isn't clear from your definition, perhaps you could provide the definition to make it easier to answer your questions.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2011-02-20T18:54:28.240" LastActivityDate="2011-02-20T18:54:28.240" CommentCount="18" />
  <row Id="22815" PostTypeId="1" CreationDate="2011-02-19T18:09:54.780" Score="0" ViewCount="549" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Possible Duplicate:&lt;/strong&gt;&lt;br&gt;&#xA;  &lt;a href=&quot;http://math.stackexchange.com/questions/12906/is-value-of-pi-4&quot;&gt;Is value of $\pi$ = 4 ?&lt;/a&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;p&gt;I've seen an interesting math puzzle today, purporting that $\pi$ = 4:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/agBz5.jpg&quot; alt=&quot;puzzle image&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't really know where the mistake is. Clearly, the perimeter of the polygon at each step is exactly 4, so it should be the same in the limit as well. Also, in the limit the polygon becomes the circle, because each point outside the circle gets removed in a finite number of steps.&lt;/p&gt;&#xA;" OwnerUserId="7258" LastEditorUserId="4423" LastEditorDisplayName="" LastEditDate="2011-05-23T00:10:35.250" LastActivityDate="2011-05-23T00:10:35.250" CommunityOwnedDate="2011-05-21T23:11:40.290" ClosedDate="2011-02-19T18:14:38.897" Title="Pi = 4? How to disprove?" Tags="&lt;geometry&gt;&lt;pi&gt;" CommentCount="4" />
  <row Id="22816" PostTypeId="2" ParentId="22797" CreationDate="2011-02-19T18:12:35.420" Score="2" ViewCount="" Body="&lt;p&gt;See &lt;a href=&quot;http://mathoverflow.net/questions/7477/non-integral-scheme-having-integral-local-rings&quot; rel=&quot;nofollow&quot;&gt;http://mathoverflow.net/questions/7477/non-integral-scheme-having-integral-local-rings&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2011-02-19T18:12:35.420" CommentCount="1" />
  <row Id="22817" PostTypeId="2" ParentId="22607" CreationDate="2011-02-19T18:45:30.607" Score="0" ViewCount="" Body="&lt;p&gt;You cannot describe the indicated set $S$ without using some negation. So let $A_j&amp;#39;:=X_j \setminus A_j$ and $\hat A_j:=\pi_j^{-1}(A_j&amp;#39;)=\lbrace x=(x_i)_{i\geq1}\in X | x_j\in A_j&amp;#39;\rbrace$. Then the set $S$ can be written as $S=\bigcup_{1\leq j\leq n} \hat A_j$, because an $x\in X$ is a member of $S$ iff at least one of the first $n$ &quot;coordinates&quot; $x_j$ of $x$ does not belong to the corresponding $A_j$.&lt;/p&gt;&#xA;" OwnerUserId="1303" LastActivityDate="2011-02-19T18:45:30.607" />
  <row Id="22818" PostTypeId="2" ParentId="22720" CreationDate="2011-02-19T18:46:18.797" Score="4" ViewCount="" Body="&lt;p&gt;The Fourier transform in the $x$-variables of ${\displaystyle h(x,t) = {1 \over (4\pi t)^{n \over 2}}e^{-{\vert x\vert ^2 \over t}}}$ is given by ${\displaystyle 2^{-n}e^{-\pi|\xi|^2t}}$ (The $2^{-n}$ factor may not be there depending on your normalization).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Taking an $x_i$-derivative of $h(x,t)$ corresponds to multiplying by $i\xi_i$ on the Fourier transform side, while taking a $t$ derivative corresponds on the Fourier transform side to once again taking a $t$-derivative. Thus the $x$-Fourier transform of the partial derivative $\partial_x^{\alpha}\partial_t^{\beta}h(x,t)$ is of the form ${\displaystyle p(\xi)e^{-\pi|\xi|^2t}}$ where $p(\xi)$ is a monomial. One can evaluate $\partial_x^{\alpha}\partial_t^{\beta}h(x,t)$ at any point by using the Fourier inversion formula on ${\displaystyle p(\xi)e^{-\pi|\xi|^2t}}$. This implies that $|\partial_x^{\alpha}\partial_t^{\beta}h(x,t)|$ is at most the $L^1$ norm of ${\displaystyle p(\xi)e^{-\pi|\xi|^2t}}$ (in the $\xi$-variables). Since this function is decreasing in $t$, $|\partial_x^{\alpha}\partial_t^{\beta}h(x,t)|$ is going to be bounded by &#xA;$${\displaystyle \int_{R^n} |p(\xi)|e^{-\pi|\xi|^2\delta}}\,d\xi$$&#xA;Thus one has a uniform bound for a given derivative. But the bounds will not be uniform over all derivatives as equality above is achieved for $x = 0$.&lt;/p&gt;&#xA;" OwnerUserId="3035" LastActivityDate="2011-02-19T18:46:18.797" CommentCount="1" />
  <row Id="22819" PostTypeId="2" ParentId="21068" CreationDate="2011-02-19T19:27:41.813" Score="1" ViewCount="" Body="&lt;p&gt;I saw a talk by John Thompson, not that I knew what it was about, looking at a copy of SL(2,Z) in dirichlet series under convolution. This paper &lt;a href=&quot;http://arxiv.org/abs/0803.1121&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/0803.1121&lt;/a&gt; might be something...&lt;/p&gt;&#xA;" OwnerUserId="6925" LastActivityDate="2011-02-19T19:27:41.813" />
  <row Id="22820" PostTypeId="1" CreationDate="2011-02-19T20:18:59.387" Score="1" ViewCount="97" Body="&lt;p&gt;Hello,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the direct formula for the following variation of Josephus problem?&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;There are $n$ persons, numbered $1$ to $n$, around a circle.. Starting from $kth$ person, every second person is eliminated. Given the $n$ and $k$, determine the index of the $xth$ person who is eliminated.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;ADDED:In the first step the $k+1$th is eliminated.For n = 16,k = 6, elimination={7,9,...}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="7262" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-19T21:39:37.897" LastActivityDate="2011-05-28T14:47:05.337" Title="Direct formula for a variation of Josephus problem:" Tags="&lt;contest-math&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="22821" PostTypeId="1" AcceptedAnswerId="22834" CreationDate="2011-02-19T20:21:10.493" Score="0" ViewCount="125" Body="&lt;p&gt;I &lt;a href=&quot;http://math.stackexchange.com/questions/22745/numbers-defined-on-graph-structures&quot;&gt;asked a question yesterday&lt;/a&gt; regarding numbers defined on graph structures that I call graph numbers.  I posted the algorithm I was using to define graph numbers, which are simply a natural extension of linear numbers we use all the time.  Nolion's response was &lt;em&gt;close&lt;/em&gt; but not quite right.  The equations are actually:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$d_{m+n}(v) = \biggl (d_{n}(v) + d_{m}(v) + \sum_{(u,v)\in{E}}{d_{{m+n},{carry over}}(u)} \biggr ) \mod {2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$d_{{m+n},carryover}(v) = \biggl \lfloor \frac {d_{n}(v) + d_{m}(v) + \sum_{(u,v)\in{E}}{d_{{m+n},{carry over}}(u)} } {2} \biggr \rfloor$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After reading nolion's response, I was able to come up with these equations. Notice, they are added in $Z$, which is why I need the $\mod 2$ to bring them back to $Z/2Z$.  However, the digits of m and n are chosen from $Z/2Z$ (just as in normal binary addition of more than 2 numbers).  But the order of evaluation of $d_{m+n}$ matters (which, again, I am not sure how to describe mathematically) starting from the &quot;root&quot; nodes and evaluating as in a Breadth First Search.  Notice, that if the Directed Acyclic Graph is a linear connection of nodes, this becomes normal arithmetic in base 2.  My question is, has there been any work done on such numbers?&lt;/p&gt;&#xA;" OwnerUserId="7263" LastEditorUserId="7263" LastEditorDisplayName="" LastEditDate="2011-02-19T21:23:09.770" LastActivityDate="2011-02-19T22:14:20.703" Title="Numbers defined on graph structures" Tags="&lt;number&gt;&lt;graph&gt;" AnswerCount="1" CommentCount="14" />
  <row Id="22822" PostTypeId="2" ParentId="18231" CreationDate="2011-02-19T20:28:36.993" Score="1" ViewCount="" Body="&lt;p&gt;By the simple argument given by Jonas and explained in the Wiki page Theo referred to, there can exist no such &lt;em&gt;positive&lt;/em&gt; measure $\mu$ since, using the notation below, $M_2M_4&amp;lt;(M_3)^2$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As regards the existence of a &lt;em&gt;signed&lt;/em&gt; measure $\mu$ of bounded variation on the interval $[0,1]$ with a given sequence of moments &#xA;$$&#xD;&#xA;M_n=\displaystyle\int_0^1x^n\mathrm{d}\mu(x),&#xD;&#xA;$$&#xA;it might be worthwhile, following Jonas's hint, to recall the result stated in chapter 3 section 2 of Widder's &lt;em&gt;The Laplace transform&lt;/em&gt;: such a measure $\mu$ exists iff the sequence $(S_n)$ is bounded, where for every nonnegative $n$,&#xA;$$&#xD;&#xA;S_n=\sum_{k=0}^n{n\choose k}\left|\sum_{i=0}^k(-1)^i{k\choose i}M_{n-i}\right|.&#xD;&#xA;$$&#xA;Hence to determine if $(S_n)$ is bounded or not when $M_n=\mathrm{e}^{-n^2}$ for every $n\ge1$ would give the answer.&lt;/p&gt;&#xA;" OwnerUserId="6179" LastEditorUserId="6179" LastEditorDisplayName="" LastEditDate="2011-02-19T21:41:46.467" LastActivityDate="2011-02-19T21:41:46.467" />
  <row Id="22823" PostTypeId="2" ParentId="22820" CreationDate="2011-02-19T20:31:19.860" Score="1" ViewCount="" Body="&lt;p&gt;If you go to &lt;a href=&quot;http://en.wikipedia.org/wiki/Josephus_problem&quot; rel=&quot;nofollow&quot;&gt;the Wikipedia page,&lt;/a&gt; there is &lt;a href=&quot;http://www.cs.man.ac.uk/~shamsbaa/Josephus.pdf&quot; rel=&quot;nofollow&quot;&gt;a reference&lt;/a&gt; at the bottom&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-02-19T20:31:19.860" />
  <row Id="22824" PostTypeId="1" AcceptedAnswerId="22826" CreationDate="2011-02-19T20:43:00.193" Score="5" ViewCount="191" Body="&lt;p&gt;Von Neumann's Minimax theorem (quoted from &lt;a href=&quot;http://en.wikipedia.org/wiki/Minimax#Minimax_theorem&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;For every two-person, zero-sum game with finite strategies, there exists a value V and a mixed strategy for each player, such that (a) Given player 2's strategy, the best payoff possible for player 1 is V, and (b) Given player 1's strategy, the best payoff possible for player 2 is −V.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Nash's theorem:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Every finite game has a mixed strategy equilibrium.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now, to me, it seems that the Minimax theorem is simply a specific instance of the Nash theorem, for a two-player zero-sum  game (the moment an equilibrium is established, the results concerning the game value follow immediately). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But in my Game Theory course, we studied these as two separate theorems, with entirely different proofs. Some exams even had proving both theorems as two exam questions - making it seem like a claim that &quot;Minimax follows immediately from Nash's theorem&quot; would be suspect. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Am I misunderstanding some fundamental difference between these two theorems? Or did we simply learn two different proofs of the same thing?&lt;/p&gt;&#xA;" OwnerUserId="7261" LastActivityDate="2011-02-19T21:36:51.573" Title="Difference between Minimax theorem and Nash Equilibrium existence" Tags="&lt;game-theory&gt;&lt;famous-theorems&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="22825" PostTypeId="1" AcceptedAnswerId="22832" CreationDate="2011-02-19T20:50:34.677" Score="1" ViewCount="182" Body="&lt;p&gt;If I have only the upper triangular part of a symmetric matrix A&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;How could I compute SVD, having this&#xA;upper triangle makes easier the&#xA;computing?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;$$\begin{pmatrix} 1 &amp;amp; 22 &amp;amp; 13 &amp;amp; 14 \\  &amp;amp; 1 &amp;amp; 45 &amp;amp; 24 \\  &amp;amp;  &amp;amp; 1 &amp;amp; 34 \\  &amp;amp;  &amp;amp;  &amp;amp; 1\end{pmatrix}$$&lt;/p&gt;&#xA;" OwnerUserId="7265" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2011-04-07T05:59:38.580" LastActivityDate="2011-04-07T05:59:38.580" Title="How to compute  SVD (Singular Value Decomposition) of a symmetric matrix" Tags="&lt;linear-algebra&gt;&lt;matrices&gt;&lt;eigenvalues&gt;" AnswerCount="2" CommentCount="5" />
  <row Id="22826" PostTypeId="2" ParentId="22824" CreationDate="2011-02-19T21:36:51.573" Score="3" ViewCount="" Body="&lt;p&gt;Wikipedia agrees with you, saying &quot;In zero-sum games, the minimax solution is the same as the Nash equilibrium&quot; (second statement of contents of the article about Minimax). So the existence of Nash equilibrium is a generalization of the Minimax theorem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Presumably, the proof of the minimax theorem is much simpler than the proof of the general theorem. Another crucial difference is that the proof of the minimax theorem is &lt;i&gt;constructive&lt;/i&gt; (it amounts to linear programming), whereas &lt;a href=&quot;http://people.csail.mit.edu/costis/journal_ver10.pdf&quot; rel=&quot;nofollow&quot;&gt;finding a Nash equilibrium is PPAD-complete&lt;/a&gt;, even for &lt;a href=&quot;http://citeseerx.ist.psu.edu/viewdoc/download;jsessionid=B5B67EAE78F22B6D3E416EFC88473B35?doi=10.1.1.93.8285&amp;amp;rep=rep1&amp;amp;type=pdf&quot; rel=&quot;nofollow&quot;&gt;two player games&lt;/a&gt;. It is even hard to &lt;a href=&quot;http://people.csail.mit.edu/costis/appxhardness.pdf&quot; rel=&quot;nofollow&quot;&gt;find an approximate Nash equilibrium&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2011-02-19T21:36:51.573" CommentCount="3" />
  <row Id="22827" PostTypeId="2" ParentId="22646" CreationDate="2011-02-19T21:43:29.740" Score="6" ViewCount="" Body="&lt;p&gt;To complement Arturo's excellent survey of the foundations of abelian p-groups:  A useful survey of the later (1960-2000) development of the theory of abelian p-groups is given in:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hill, Paul. &quot;The development of the theory of p-groups.&quot;&#xA;Rocky Mountain J. Math. 32 (2002), no. 4, 1135–1151.&#xA;MR&lt;a href=&quot;http://www.ams.org/mathscinet-getitem?mr=1987598&quot; rel=&quot;nofollow&quot;&gt;1987598&lt;/a&gt;&#xA;DOI:&lt;a href=&quot;http://dx.doi.org/10.2307/1999950&quot; rel=&quot;nofollow&quot;&gt;10.2307/1999950&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I personally enjoyed the earlier theory a lot more (Kulakoff, Prüfer, Zippin, Ulm) which was gathered up into textbook form by Fuchs and Kaplansky.  However, lots of positive and negative results were achieved after then, so if you aren't daunted by the several hundred pages of the textbooks by Fuchs, Kaplansky, and Griffith, then the survey has a lot of important papers beyond the textbook level.&lt;/p&gt;&#xA;" OwnerUserId="583" LastActivityDate="2011-02-19T21:43:29.740" />
  <row Id="22828" PostTypeId="2" ParentId="22825" CreationDate="2011-02-19T22:00:04.253" Score="1" ViewCount="" Body="&lt;p&gt;A symmetric matrix always has an eigenvalue decomposition (for which there a specialized routines). I'm not really sure for what application one would need the SVD.&lt;/p&gt;&#xA;" OwnerUserId="7266" LastActivityDate="2011-02-19T22:00:04.253" />
  <row Id="22829" PostTypeId="1" CreationDate="2011-02-19T22:01:13.330" Score="5" ViewCount="122" Body="&lt;p&gt;The symmetric group $S_n$ acts naturally on the set $\{1,2,\dots, n\}$. Is this a left or a right action?&lt;/p&gt;&#xA;" OwnerUserId="7267" LastEditorUserId="1583" LastEditorDisplayName="" LastEditDate="2011-02-19T22:23:35.127" LastActivityDate="2011-02-19T22:37:46.997" Title="Is the natural action of $S_n$ on $\{1, 2, \dots, n\}$ a left or a right action?" Tags="&lt;group-theory&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="22830" PostTypeId="5" CreationDate="2011-02-19T22:04:26.930" Score="0" ViewCount="" Body="&lt;p&gt;The axiom of choice is an axiom usually added to the Zermelo-Frankael theory of sets, it states that given a set $I$ and $a_i$, for $i\in I$, non-empty sets there exists a function $f\colon I\to \bigcup a_i$ such that $f(i)\in a_i$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The axiom of choice is equivalent to the statement that every set can be well-ordered, as well to the Zorn's lemma which asserts that if a partial order has the property that every chain is bounded from above, then there is a maximal element.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The axiom of choice is generally independent of the ZF theory, and if ZF is consistent then it is consistent with both the axiom of choice as well its negation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some theorems which follow from the axiom of choice:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The product of compact spaces is compact (equivalent)&lt;/li&gt;&#xA;&lt;li&gt;Countable union of countable sets is countable&lt;/li&gt;&#xA;&lt;li&gt;Every infinite set has a countable subset&lt;/li&gt;&#xA;&lt;li&gt;Every field has an algebraic closure&lt;/li&gt;&#xA;&lt;li&gt;There are sets of real numbers which are not Lebesgue measurable&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Most of the mathematicians nowadays assume the axiom of choice when they deal with their mathematics. Mostly because without it infinite processes in mathematics become much harder to handle, and one can construct models without the axiom of choice in which continuity of sequences is not equivalent to $\varepsilon-\delta$ continuity, or some fields have no algebraic closure.&lt;/p&gt;&#xA;" OwnerUserId="-1" LastEditorUserId="622" LastEditorDisplayName="" LastEditDate="2011-02-19T22:09:15.880" LastActivityDate="2011-02-19T22:09:15.880" />
  <row Id="22831" PostTypeId="4" CreationDate="2011-02-19T22:04:26.930" Score="0" ViewCount="" Body="The axiom of choice is an axiom usually added to the Zermelo-Frankael theory of sets, it states that given a set $I$ and $a_i$, for $i\in I$, non-empty sets there exists a function $f\colon I\to \bigcup a_i$ such that $f(i)\in a_i$." OwnerUserId="-1" LastEditorUserId="622" LastEditorDisplayName="" LastEditDate="2011-02-19T22:09:53.897" LastActivityDate="2011-02-19T22:09:53.897" />
  <row Id="22832" PostTypeId="2" ParentId="22825" CreationDate="2011-02-19T22:09:49.510" Score="3" ViewCount="" Body="&lt;p&gt;The SVD for a symmetric matrix $A = U \Sigma V^T$, where $U$ and $V$ are unitary matrices with $U = \left[u_1 | u_2 | \ldots | u_n \right]$, $V = \left[v_1 | v_2 | \ldots | v_n \right]$ and $\Sigma$ is a diagonal matrix with non-negative diagonal entries and $v_i = \pm u_i$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a symmetric matrix the following decompositions are equivalent to SVD. (Well, almost equivalent if you do not worry about the signs of the vectors).&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Eigen-value decomposition&lt;/strong&gt; i.e. $A = X \Lambda X^{-1}$. When $A$ is symmetric, the eigen values are real and the eigenvectors can be chosen to be orthonormal and hence $X^TX = XX^T = I$ i.e. $X^{-1} = X^T$. The only difference is that the singular values are the magnitudes of the eigen values and hence the column of $X$ needs to be multiplied by a negative sign if the eigen value turns out to be negative to get the singular value decomposition. Hence, $U = X$ and $\sigma_i = |\lambda_i|$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Orthogonal decomposition&lt;/strong&gt; i.e. $A = PDP^T$, where $P$ is a unitary matrix and $D$ is a diagonal matrix. This exists only when matrix $A$ is symmetric and is the same as eigen value decomposition.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Schur decomposition&lt;/strong&gt; i.e. $A = Q S Q^T$, where $Q$ is a unitary matrix and $S$ is an upper triangular matrix. This can be done for any matrix. When $A$ is symmetric, then $S$ is a diagonal matrix and again is the same as the eigen value decomposition and orthogonal decomposition.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I do not remember the cost for each of these operations i.e. I don't remember the coefficients before the leading order $n^3$ term. If my memory is right, the typical algorithm for orthogonal decomposition is slightly cheaper than the other two, though I cannot guarantee.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2011-02-19T22:25:36.863" LastActivityDate="2011-02-19T22:25:36.863" CommentCount="15" />
  <row Id="22833" PostTypeId="2" ParentId="22829" CreationDate="2011-02-19T22:12:41.290" Score="10" ViewCount="" Body="&lt;p&gt;First of all, there is no essential difference between a left or right action. A left action $\lambda: G \times S \to S$ becomes a right action $\rho: S \times G \to S$ by setting $\rho(s,g) = \lambda(g^{-1},s)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next, it really depends on how you think of $S_{n}$, how you realize it and on your conventions. If you think of $S_{n}$ as bijections $\pi: \mathbf{n} \to \mathbf{n}$ with composition, and &lt;em&gt;if&lt;/em&gt; you adhere to the convention that functions take arguments on the right, then the action is on the left: $(\sigma \circ \pi)(k) = \sigma(\pi(k))$.&lt;/p&gt;&#xA;" OwnerUserId="5363" LastActivityDate="2011-02-19T22:12:41.290" />
  <row Id="22834" PostTypeId="2" ParentId="22821" CreationDate="2011-02-19T22:14:20.703" Score="0" ViewCount="" Body="&lt;p&gt;I had a nice, long answer written out that has been subsumed by all the comments above.  Oh well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only part that isn't redundant is that for multiplication, you want to look at whether the set of endomorphisms of your DAG acts vertex transitively.  Ideally, it would act simply vertex-transitively, because otherwise, you would have to include choices of endomorphisms explicitly in the structure.  Anyway, that's the mathematical terminology for your &quot;fractal structure&quot;.&lt;/p&gt;&#xA;" OwnerUserId="7268" LastActivityDate="2011-02-19T22:14:20.703" CommentCount="6" />
  <row Id="22835" PostTypeId="2" ParentId="22686" CreationDate="2011-02-19T22:16:11.063" Score="18" ViewCount="" Body="&lt;p&gt;Okay, I seem to be ranting too much in comments, so let me try to put forth my points and opinion here (as a community wiki, since it is opinion). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, to address the question: What is the significance of straightedge and compass constructions within mathematics? Historically, they played an important role and led to a number of interesting material (the three famous impossibilities lead very naturally to transcendental numbers, theories of equations, and the like). They are related to fascinating stuff (numbers constructible by origami, etc). But I would say that their significance parallels a bit the &lt;a href=&quot;http://math.stackexchange.com/questions/10029/10034#10034&quot;&gt;significance of Cayley's Theorem&lt;/a&gt; in Group Theory: although important historically, and relevant to understand the development of many areas of mathematics, they are not that particularly important today. As you can see from the responses, many find them &quot;fascinating&quot;, many find them &quot;boring&quot;, but nobody seems to have come forth with an important application.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, addressing the issue of teaching it at K-12. Let me preface this by saying that I am a &quot;survivor&quot; of the &lt;a href=&quot;http://en.wikipedia.org/wiki/New_Math&quot; rel=&quot;nofollow&quot;&gt;New Math&lt;/a&gt;, which came to Mexico (where I grew up) in the 70s. I would say I became a mathematician in large part &lt;em&gt;despite&lt;/em&gt; having been taught with New Math, rather than &lt;em&gt;because&lt;/em&gt; of it. Also, I did not attend K-12 education or undergraduate in the U.S.; I am a bit more familiar with undergraduate at the level of precalculus and above thanks to my job, but not very much with the details of curricula in sundry states in the U.S. So I may very well have a wrong impression of details in what follows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, one problem, in my view, is that when we talk about &quot;math education&quot;, we are really talking about &lt;em&gt;two&lt;/em&gt; different things: numeracy and mathematics. This is the same phenomenon we see when we think about &quot;English class&quot;. I suspect that English Ph.D.s are nonplussed at people who think they spend their time dealing with grammar, spelling, punctuation, etc, just as mathematicians are nonplussed that people think we spend our time multiplying really big numbers by hand. English education has two distinct components, which we might call &quot;Literacy&quot; and &quot;Literature&quot;. Literacy is the component where we try to teach students to read and write effectively, spelling, punctuation, etc. Making themselves understood in written form, and understanding the written word. On the other hand, Literature is the component where they are introduced to Shakespeare, novelists, book reading, short stories, poetry, creative writing, historical and world literature, etc. We consider English education at our schools a failure when it fails in its mission with regards to &lt;strong&gt;literacy&lt;/strong&gt;: we don't consider it a failure if students come out not being particularly enthused with reading classic novels or don't become professional poets, or even if they don't care or like poetry (or Shakespeare).  The professional English Ph.D. engages in literature, not in literacy. Literacy is the domain of the grammarian.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mathematics education likewise has two components; numeracy (to borrow the term from John Allen Paulos) and mathematics. Numeracy is the parallel of literacy: we want children to be able to handle and understand numbers and basic algebra, percentages, etc., because they are necessary to function in the world. Mathematics is the stuff that mathematicians do, and which we all find so interesting and beautiful. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Mathematics&quot; also includes advanced topics that are necessary for someone who is going to go on to study areas that require mathematics: so the physicists and engineers need to know trigonometry; business and economists need to know advanced statistics; computer science needs to know discrete mathematics; etc. Much like someone going on to college likely needs more than simple command of grammar and spelling. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Part of the problem with math education is that it so often conflates numeracy with mathematics; another part of the problem is that mathematics was included in the &quot;classical&quot; curriculum for much the same reason as Latin and Greek were included: historical reasons, because every &quot;gentleman&quot; was expected to know some Latin, some Greek, and some mathematics (and by &quot;mathematics&quot;, people meant Euclid). To some extent, we still teach geometry in K-12 because we've always taught geometry. But it is not part of the numeracy curriculum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It would be wonderful if we had the time in K-12 to teach students &lt;em&gt;both&lt;/em&gt; numeracy and mathematics; it is a fact that today we are failing at both. We don't perform better in teaching students numeracy by teaching them beautiful mathematics, even if they understand and appreciate them, just like making them really understand and care for Shakespeare's plays (through performance, say) will make them able to understand a set of written instructions, or write a coherent argument.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Trying to excite students about mathematics is all well and good; but numeracy should come first. Trying to excite students about the wonderful world of books, plays, and poetry is all well and good, but we need them to be able to read and write first, because that's part of what they will need to function in society.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Constructions by straightedge and compass can become an interesting part of &lt;em&gt;mathematics&lt;/em&gt; education, just like geometry. I just don't think they have a place in &lt;em&gt;numeracy&lt;/em&gt; education. But they are taking the time we need for that numeracy education. Trigonometry &lt;em&gt;used to be&lt;/em&gt; part of the numeracy education that people needed; this is no longer the case today. Trigonometry, today, is a foundational science for more advanced studies, not part of numeracy. But we are still teaching trigonometry as a numeracy subject (hence the rules, recipes, mnemonics, and the like). We could try to turn trigonometry into a &lt;em&gt;mathematical&lt;/em&gt; subject, sure; or we could postpone it until later and only teach it to those for whom it is an important foundation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Added.&lt;/em&gt; To clarify: I don't mean to say that the &quot;solution&quot; is to do less math and more numeracy. I think the solution is likely to be complicated, but the first step is to identify exactly what parts of what is currently branded as &quot;mathematics&quot; are really numeracy, and which parts are mathematics. Trigonometry is taught as part of &quot;advanced mathematics&quot;, but it is taught as rote and rules because it was really numeracy. We don't need to teach trigonometry as numeracy any more, so we shouldn't. If it is to be taught, it needs to be taught in the right context. Geometry is similar: geometry used to be taught as basic numeracy because &quot;every educated person should know geometry&quot;; (of course, &quot;educated&quot; at the time meant &quot;rich and land owner, or with aspirations in that direction&quot;). We don't need most of geometry as basic numeracy, we want it now as mathematics. So teaching geometry &lt;em&gt;as&lt;/em&gt; numeracy is a waste of time, and it takes up the &lt;em&gt;numeracy&lt;/em&gt; time that should be spent in other things. (It can still be taught during the &lt;em&gt;mathematics&lt;/em&gt; time). I certainly don't say &quot;drop all the math, concentrate on the numeracy&quot;. I say, &quot;when dealing with numeracy, concentrate on the numeracy, not the math, and don't confuse the two.&quot; Nobody seems to confuse spelling rules with reading novels, because we separate literacy from literature. Too many people confuse arithmetic with mathematics, because we don't separate them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason I talk about dropping trigonometry and doing some basic statistics is precisely that: trigonometry is being taught as part of the &lt;em&gt;numeracy&lt;/em&gt; curriculum, when it shouldn't. Basic statistics, say at the level of the wonderful &lt;strong&gt;How to Lie With Statistics&lt;/strong&gt;, is not taught as part of numeracy. But in today's world, there is a far better case for statistics being part of the basic numeracy education than trigonometry. Everyone coming out of High School &lt;em&gt;should know&lt;/em&gt; that taking a 10% pay-cut and then getting a 5% raise does not mean you are now at 95% of your old salary (go do a spot check, see how many people think you &lt;em&gt;are&lt;/em&gt;). They need to know the difference between &lt;em&gt;average&lt;/em&gt; and &lt;em&gt;median&lt;/em&gt;, so they are not misled by statements about &quot;the average salary of the American worker&quot;. They should understand what &quot;false positive&quot; and &quot;false negative&quot; means. They should be able to interpret graphs (even the silly ones on the cover of every &lt;em&gt;USA Today&lt;/em&gt; issue) and be able to spot the distortions created by chopping axes, etc. These are &lt;em&gt;numeracy&lt;/em&gt; issues. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Likewise high school geometry: it is trying to be both numeracy and mathematics, and I think it generally fails at both. There are some components of geometry that are part of numeracy, they ought to be treated that way, but the parts that are mathematics should be separate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One problem I have with Lockhart's lament is that he does not make clear the distinction between numeracy and mathematical education. The nightmare he paints for the musician and the artist is precisely that musical education is being turned into the equivalent of numeracy/literacy education, thus doing a disservice to music-as-an-art. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The science of mathematics that we all know and love has some intersection with numeracy and arithmetic, but we all know it is a limited intersection; just as the study of literature has some intersection with the study of grammar an spelling, but the intersection is limited. The main purpose of K-12 education (or at the very least, K-6 or K-8) should be &lt;em&gt;numeracy&lt;/em&gt;, with some limited forays into mathematics (just as the main purpose will be &lt;em&gt;literacy&lt;/em&gt; with some limited forays into &lt;em&gt;literature&lt;/em&gt;). The way to teach numeracy is necessarily different from the way to teach mathematics. Numeracy requires that we memorize multiplication tables, for all the horror this will cause to modern education people; this of course is a far cry from teaching the &lt;em&gt;mathematics&lt;/em&gt; of multiplication, which may very well be very interesting and awaken the child's curiosity and wonder at the world. That can be done within the context of &lt;em&gt;mathematical&lt;/em&gt; education, but it shouldn't be done in the context, and at the expense of, numeracy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Trigonometry used to be part of numeracy; it no longer is. It is now either mathematical or foundational for advanced studies, so it should be treated as such. Geometry used to be taught for reasons which no longer hold, and to some extent we continue teaching it as a historical legacy; we shouldn't. Those components which are numeracy should be taught as that, and we could move the rest (including constructions with compass and straightedge) to the more creative, mathematical education side of the equation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, I've ranted long enough, and probably made myself a few detractors along the way...&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2011-05-12T10:23:04.190" LastActivityDate="2011-05-12T10:23:04.190" CommunityOwnedDate="2011-02-19T22:16:11.063" CommentCount="11" />
  <row Id="22836" PostTypeId="2" ParentId="22753" CreationDate="2011-02-19T22:22:41.770" Score="1" ViewCount="" Body="&lt;p&gt;&lt;b&gt;If a matrix $A$ is unitarily diagonalizable, they one can define a &quot;Fourier transform&quot; for which $A$ is a &quot;convolution&quot; matrix.&lt;/b&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is an example. We have a family $F$ of subsets of some finite set $S$, i.e. $F \subset 2^S$, such that any two sets in $F$ agree in some coordinate, i.e. for any two $A,B \in F$, there is some element $x \in S$ that either belongs to both $A,B$ or to neither; we call such a family $F$ &lt;i&gt;agreeing&lt;/i&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How big can an agreeing family $F$ be? Clearly, it can contain at most half the sets, since $A$ and $S \setminus A$ cannot both belong to $F$. On the other hand, for any $x \in S$, the family $$F = \{ A \subset S : x \in A \}$$ is an agreeing family containing exactly half the sets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a different proof. Let $F$ be an agreeing family, and $f$ its characteristic function. i.e. $f(A) = 1$ iff $A \in F$. Its Fourier transform is defined by $$ \hat{f}(B) = 2^{-|S|} \sum_A f(A) (-1)^{|A \cap B|}. $$ The Fourier transform is really presenting the function $f$ in the orthonormal basis $$\chi_B(A) = (-1)^{|A\cap B|},$$ which is orthonormal with respect to the inner product $$\langle g,h \rangle = 2^{-|S|} \sum_A g(A) h(A). $$ The Fourier transform is defined so that $$f = \sum_B \hat{f}(B) \chi_B, $$ and so the formula for the transform can also be written $$\hat{f}(B) = \langle f,\chi_B \rangle;$$ this works since the basis is orthonormal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An easy calculation gives that $\hat{f}(\emptyset) = |F|/2^{|S|}$. Moreover, $$\langle f,f \rangle = \sum_{B,C} \langle \hat{f}(B) \chi_B, \hat{f}(C) \chi_C \rangle = \sum_B \hat{f}(B)^2,$$ again from orthonormality. Note that $$\langle f,f\rangle = \langle f^2, 1 \rangle = \langle f,\chi_\emptyset \rangle = \hat{f}(\emptyset),$$ where we used the fact that $f$ is $\{0,1\}$-valued.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider now the operator $X$ which corresponds to complementation, i.e. $$Xe_A = e_{S \setminus A},$$ where $e_A$ is the vector which is $1$ only for the set $A$. Another way to look at the operator $X$ is that it is convolution with $S$, where the group operation is symmetric difference (i.e. $A \triangle S = S \setminus A$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A straightforward computation shows that the eigenvectors of $X$ are exactly the Fourier basis vectors $\chi_B$:&#xA;$$ X \chi_B(A) = (-1)^{|(S\setminus A)\cap B|} = (-1)^{|S \cap B|} (-1)^{|A \cap B|} = (-1)^{|B|} \chi_B(A). $$&#xA;This is not surprising since $X$ is a convolution operator for the group $\mathbb{Z}_2^{|S|}$, and the Fourier basis is a basis of characters for that abelian group.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the family $f$ is agreeing, $f(A)f(S\setminus A) = 0$, and so $$0 = \langle f,Xf \rangle = \sum_B (-1)^{|B|} \hat{f}(B)^2,$$ where we again used orthonormality of the Fourier basis. Denoting $|f| = |F|/2^{|S|}$, we have seen above that $$\sum_B \hat{f}(B)^2 = |f|, \quad \hat{f}(\emptyset)^2 = |f|^2.$$ So the even and odd squared Fourier coefficients balance; their total weight is $|f|$, and the weight of one of them is $|f|^2$. Evidently, $|f|^2$ can be at most half the total weight, and so $|f| \leq 1/2$.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;This proof might seem silly (since we presented a &quot;one-line&quot; proof preceding it), but the same method can be used to prove much more difficult theorems. For example, we can look at a family of &quot;colored&quot; sets, i.e. generalize the two colors above (corresponding to &quot;not in the set&quot; and &quot;in the set&quot;) to an arbitrary number of colors. The largest family is still obtained by fixing one coordinate, but the combinatorial proof is more difficult (takes about a page); the Fourier proof is almost the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are some more difficult examples:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Families of graphs over a fixed vertex set, the intersection of any two of which contains a triangle. The &quot;best&quot; family is obtained by taking all supergraphs of a fixed triangle. The only known proof is using very similar Fourier methods.&lt;/li&gt;&#xA;&lt;li&gt;Families of permutations, any two of which have a common &quot;input/output&quot; pair. The &quot;best&quot; family is in general obtained by fixing the image of some element (all permutations taking $i$ to $j$). There is a simple combinatorial proof of that along the lines above. But it's much harder to prove that these are the unique optimal families, whereas it follows relatively easily from the &quot;Fourier&quot; proof. Moreover, the latter can be extended to the case where the permutations are required to have $t$ matching input/output pairs. The &quot;best&quot; families take $t$ fixed inputs to $t$ fixed outputs. The only known proof is spectral (it requires the representation theory of $S_n$).&lt;/li&gt;&#xA;&lt;li&gt;Families of subsets of $S$ of size $k &amp;lt; |S|/2$, any two of which intersect. The maximal families are the same as above (supersets of a fixed elements). This is the celebrated Erdős-Ko-Rado theorem. There is a Fourier proof with some extra benefits over some of the combinatorial proofs, viz. it can describe the structure of almost optimal families.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The Fourier proof of the last example actually optimizes some skewed measure of a general family of subsets, in other words, instead of limiting the sets to be of size $k$, we just give more &quot;relevance&quot; to sets of size roughly $k$. The proof goes as follows:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Find some inner product so that $\langle f,1 \rangle_p$ is the required skewed measure (which is $\mu_p(A) = p^{|A|} (1-p)^{|S\setminus A|}$ for $p \approx k/n$).&lt;/li&gt;&#xA;&lt;li&gt;Find some &quot;convolution operator&quot; $X$ such that $\langle f,Xf \rangle_p = 0$ for every intersecting family, and moreover the eigenvectors of $X$ are orthogonal with respect to the inner product.&lt;/li&gt;&#xA;&lt;li&gt;Follow the same steps as above to conclude that $\mu_p(f) \leq p$.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The construction of the convolution operator $X$ crucially uses the fact that symmetric matrices are unitarily diagonalizable (the symmetric matrix in question is obtained from a reversible Markov chain). The operator $X$ is not strictly unitarily diagonalizable, since its eigenvectors are only orthogonal with respect to the skewed inner product (in which the stationary distribution of the Markov chain crops up), but it is obtained from such a matrix through scaling of rows.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The material is taken from a series of papers by Ehud Friedgut and friends. I will refer to them by their current numbers on his &lt;a href=&quot;http://www.ma.huji.ac.il/~ehudf/PPP&quot; rel=&quot;nofollow&quot;&gt;list&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The general method (including the bound on &quot;agreeing families of colored sets&quot;) is #16.&lt;/li&gt;&#xA;&lt;li&gt;The application to permutations is #2 (there are some follow-up papers by David Ellis).&lt;/li&gt;&#xA;&lt;li&gt;The application to graphs is #1.&lt;/li&gt;&#xA;&lt;li&gt;The spectral proof of Erdős-Ko-Rado is #7.&lt;/li&gt;&#xA;&lt;li&gt;The general construction (using the crucial property that symmetric matrices are unitarily diagonalizable) is #6.&lt;/li&gt;&#xA;&lt;li&gt;The connection between #7 and #6 is explained in the last section of #5.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2011-02-19T22:22:41.770" />
  <row Id="22837" PostTypeId="2" ParentId="22829" CreationDate="2011-02-19T22:28:44.870" Score="3" ViewCount="" Body="&lt;p&gt;If you think of permutations as functions, then the action is on the left as indicated by Theo.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There &lt;em&gt;is&lt;/em&gt;, however, a slight wrinkle when you write your permutations as products of cycles. Some books/people like to perform the multiplication of cycles &quot;left to right&quot;, as you are reading the cycles, while others do it right to left. For example, if $\rho=(1,2,3,4)$, $\sigma=(1,2,3)$, and $\tau=(1,4)(2,3)$, and you write $\rho\sigma\tau$, then there are two ways of computing it:&#xA;$$(1,2,3,4)(1,2,3)(1,4)(2,3)$$&#xA;Working &quot;left-to-right&quot;, you'd say: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$1$ maps to $2$, then $2$ maps to $3$, then $3$ maps to $2$, so $1\mapsto 2$. $2$ maps to $3$, $3$ maps to $1$, $1$ maps to $4$, so $2\mapsto 4$. $3$ maps to $4$, then $4$ maps to $1$, so $3\mapsto 1$; and finally $4$ maps to $1$, $1$ to $3$, and $3$ to $2$, so $4\mapsto 2$. Thus, $\rho\sigma\tau = (1,2,4,3).$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm not in my office, so I cannot confirm, but I believe Herstein's &quot;Topics in Algebra&quot; uses this convention.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, you could read them right-to-left, as a composition of functions. In that case, you would say:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$1$ maps to $4$, then $4$ is fixed, then $4$ to $1$, so $1$ is fixed. $2$ maps to $3$, then $3$ to $1$, then $1$ to $2$, so $2$ is fixed. $3$ maps to $2$, then $2$ to $3$, then $3$ to $4$, so $3\mapsto 4$. And $4$ maps to $1$, then $1$ to $2$, then $2$ to $3$, so $4\mapsto 3$. Therefore, $\rho\sigma\tau = (3,4)$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If compute your products of cycles in the former manner, then your action is a &lt;em&gt;right action&lt;/em&gt;. If you compute them in the latter, then your action is a &lt;em&gt;left action&lt;/em&gt;, just like Theo described it.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-19T22:37:46.997" LastActivityDate="2011-02-19T22:37:46.997" CommentCount="3" />
  <row Id="22838" PostTypeId="1" AcceptedAnswerId="22843" CreationDate="2011-02-19T22:50:50.910" Score="4" ViewCount="125" Body="&lt;p&gt;Hello, here is a pair of quick questions that I am wondering about, I would appreciate any help:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it always the case that any matrix $A$ can be represented as $E_{1}...E_{n}$ * rref($A$) (and in special case when the matrix is invertible, the representation becomes simply $E_{1}...E_{n}$) where $E_{i}$ is an elementary matrix?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And is it always the case that we can write the determinant of any matrix $A$ in the form $(-1)^{s}k_{1}...k_{n}$ * det(rref($A$)) where $s$ signifies the number of row swaps needed to get to the rref form, and $k_{i}$ are the scalars by which we divided rows of $A$ to get to its rref form (and in the special case when the matrix is not invertible, det(rref($A$)) is equal to zero, so the product of the whole expression equals to zero)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you!&lt;/p&gt;&#xA;" OwnerUserId="5157" LastActivityDate="2011-02-19T23:22:58.230" Title="Determinant and matrix representation questions" Tags="&lt;linear-algebra&gt;" AnswerCount="1" CommentCount="8" />
  <row Id="22839" PostTypeId="1" AcceptedAnswerId="22840" CreationDate="2011-02-19T23:07:34.677" Score="4" ViewCount="142" Body="&lt;p&gt;For a real-valued function $f$ defined on $\mathbb{R}$ or its subset,&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;is it possible that it is&#xA;differentiable at one point and not&#xA;in one of its neighbourhoods except&#xA;the point itself?&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;is it possible that it is&#xA;differentiable over an interval, but&#xA;its derivative over the interval is&#xA;not continuous?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found on &lt;a href=&quot;http://mathforum.org/kb/thread.jspa?messageID=3210909&amp;amp;tstart=0&quot; rel=&quot;nofollow&quot;&gt;this link&lt;/a&gt; that for&#xA;$f(x) = x^2 \sin(1/x)$, when $x$ is not&#xA;0, the derivative is $2x \sin(1/x) -&#xD;&#xA;    \cos(1/x)$ which does not have a limit&#xA;as x approaches 0, but the&#xA;derivative of $f$ does exist at 0:&#xA;$$\lim_{h \rightarrow 0} ( h^2&#xD;&#xA;    \sin(1/h) - 0)/(h-0) = 0.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was wondering how $f$ is&#xA;differentiable at 0? Doesn't it&#xA;require $f$ to be defined on 0?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Thanks and regards!&lt;/p&gt;&#xA;" OwnerUserId="1281" LastActivityDate="2011-02-19T23:19:52.657" Title="Questions about derivative and differentiation" Tags="&lt;real-analysis&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" />
  <row Id="22840" PostTypeId="2" ParentId="22839" CreationDate="2011-02-19T23:10:38.443" Score="5" ViewCount="" Body="&lt;p&gt;Here is the answer to the third question&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let us take a look at the function $f(x) = x^2 \sin(\frac{1}{x})$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first question is &quot;Is the function even in $C^{(0)}$&quot;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer is not yet since the function is ill-defined at the origin. However if we define $f(0) = 0$, then yes the function is in $C^0$. This can be seen from the fact that $\sin(\frac{1}{x})$ is bounded and hence the function is bounded above by $x^2$ and below by $-x^2$. So as we go towards $0$, the function is bounded by functions which themselves tend to $0$. And the limit is $0$ and thereby the function is continuous.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the next question &quot;Is the function differentiable everywhere?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is obvious that the function is differentiable everywhere except at $0$. At $0$, we need to pay little attention. If we were to blindly differentiate $f(x)$ using the conventional formulas, we get $g(x) = f&amp;#39;(x) = 2x \sin(\frac{1}{x}) + x^2 \times \frac{-1}{x^2} \cos(\frac{1}{x})$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now $g(x)$ is ill-defined for $x=0$. Further $\displaystyle \lim_{x \rightarrow 0} g(x)$ doesn't exist. This is what we get if we use the formula. So can we say that $f(x)$ is not differentiable at the origin. Well no! All we can say is $g(x)$ is discontinuous at $x=0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what about the derivative at $x=0$? Well as I always prefer to do, get back to the definition of $f&amp;#39;(0)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f&amp;#39;(0) = \displaystyle \lim_{\epsilon \rightarrow 0} \frac{f(\epsilon) - f(0)}{\epsilon} = \displaystyle \lim_{\epsilon \rightarrow 0} \frac{\epsilon^2 \sin(\frac{1}{\epsilon})}{\epsilon} = \displaystyle \lim_{\epsilon \rightarrow 0} \epsilon \sin(\frac{1}{\epsilon}) = 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Since $|\sin(\frac{1}{\epsilon})| \leq 1$ so it is bounded).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we find that the function $f(x)$ has a derivative at the origin whereas the function $g(x) = f&amp;#39;(x)$, $\forall x \neq 0$ is not continuous or even well-defined at the origin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we have this function whose derivative exists everywhere but then $f(x) \notin C^{(1)}$ since the derivative is not continuous at the origin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Look up &lt;a href=&quot;http://en.wikipedia.org/wiki/Volterra%27s_function&quot; rel=&quot;nofollow&quot;&gt;Volterra's function&lt;/a&gt; as an answer to your second question.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2011-02-19T23:19:34.850" LastActivityDate="2011-02-19T23:19:34.850" CommentCount="2" />
  <row Id="22841" PostTypeId="2" ParentId="22839" CreationDate="2011-02-19T23:19:52.657" Score="5" ViewCount="" Body="&lt;ol&gt;&#xA;&lt;li&gt;Let $f(x)=x^2$ if $x$ is rational, $f(x)=0$ if $x$ is irrational.  If $x\neq 0$, then $f$ is not continuous at $x$, and hence not differentiable at $x$.  If $h\neq0$ is rational, then $\frac{f(h)}{h}=h$, while if $h$ is irrational, then $\frac{f(h)}{h}=0$.  Therefore $f&amp;#39;(0)=\lim_{h\to0}\frac{f(h)}{h}=0$, and in particular it exists.   If you wanted the example to be continuous, take a &lt;a href=&quot;http://en.wikipedia.org/wiki/Weierstrass_function&quot; rel=&quot;nofollow&quot;&gt;continuous function that is nowhere differentiable&lt;/a&gt; and multiply by $x$ to get a continuous function differentiable only at $0$.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2011-02-19T23:19:52.657" CommentCount="1" />
  <row Id="22842" PostTypeId="2" ParentId="20282" CreationDate="2011-02-19T23:22:19.200" Score="0" ViewCount="" Body="&lt;p&gt;Additionally, Hatchers book provides another way of answering this problem. We can realize X (the space in question) as a CW complex, with 0-skeleton a point, 1-skeleton S^1 v S^1, say with fundamental group free on a and b, and attach 2 2-cells along ab, (or any other pair, depending on direction of paths and such). This leads to an isomorphism of the fundamental group of the 0-skeleton mod (ab) with the fundamental group of X. Despite choice of how to glue the 2-cells, we get (a)*(b)/(ab) which is infinite cyclic.&lt;/p&gt;&#xA;" OwnerUserId="7270" LastActivityDate="2011-02-19T23:22:19.200" />
  <row Id="22843" PostTypeId="2" ParentId="22838" CreationDate="2011-02-19T23:22:58.230" Score="3" ViewCount="" Body="&lt;p&gt;Yes to both.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Multiplying on the left by an elementary matrix corresponds to doing an elementary row operation. Conversely, every elementary row operation can be &quot;performed&quot; by multiplying on the left by an appropriate elementary matrix. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you go from $A$ to the reduced row-echelon form of $A$, $\mathrm{rref}A$, you do so by performing a (finite) sequence of elementary row operations; these can be represented instead by performing multiplications on the left by appropriate elementary matrices. So, you can find a sequence of elementary matrices $F_1,F_2,\ldots,F_n$, corresponding to the actions you took, that will make the following true:&#xA;$$F_nF_{n-1}\cdots F_1A = \mathrm{rref}A.$$&#xA;Now, every elementary matrix is invertible, and its inverse is an elementary matrix (of the same type). So if we let $E_i = F_i^{-1}$, then multiplying on the left both sides of the equation by $E_n$, $E_{n-1},\ldots, E_1$, you get&#xA;$$A = E_1E_2E_3\cdots E_n\mathrm{rref}A,$$&#xA;with each $E_i$ an elementary matrix.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An elementary matrix that corresponds to &quot;add a multiple of one row to another row&quot; has an inverse of the same type, and in both cases the determinant is just $1$, so you don't need to keep track of them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An elementary matrix that corresponds to &quot;swap two rows&quot; has an inverse of the same type (in fact, it is its own inverse), and has determinant $-1$, so you only need to keep track of &lt;em&gt;how many&lt;/em&gt; of these matrices are around.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But an elementary matrix that corresponds to &quot;divide one row by $k_i$&quot; has as inverse the elementary matrix that corresponds to &quot;&lt;strong&gt;multiply&lt;/strong&gt; one row by $k_i$&quot;. The former has determinant $\frac{1}{k_i}$, the latter has determinant $k_i$. So here you need to keep track of exactly what you divded by, and the taking of inverses is why for the determinant computation you &lt;em&gt;multiply&lt;/em&gt; by the $k_i$ whereas you originally &lt;em&gt;divided&lt;/em&gt; by them. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the determinant of the product is the product of the determinant (when all matrices are square), from&#xA;$$A = E_1\cdots E_n\mathrm{rref}A$$&#xA;you get the formula you write for the determinant.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2011-02-19T23:22:58.230" />
  <row Id="22844" PostTypeId="1" AcceptedAnswerId="22848" CreationDate="2011-02-19T23:56:57.410" Score="1" ViewCount="80" Body="&lt;p&gt;(Context: polynomial multiplication using DFT/FFT)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $f = \sum\limits_{i=0}^{n-1} f_i x^i$ and $g = \sum\limits_{j=0}^{n-1} g_j x^j$ be polynomials in $F[x]$ for some field $F.$ The convolution of $f$ and $g$ is given by $$f \ast g = \sum\limits_{k=0}^{n-1} c_k x^k$$ where $$c_k = \sum\limits_{i+j \equiv k \mod n} f_i g_j.$$ The $k$th coefficient in the product $fg$ is given by $$\sum\limits_{i+j = k} f_i g_j.$$ Why is convolution equivalent to multiplication in $F[x]/(x^n-1)?$ Shouldn't that be mod $x^n$?&lt;/p&gt;&#xA;" OwnerUserId="2468" LastActivityDate="2011-02-20T00:10:33.550" Title="Why is polynomial convolution equivalent to multiplication in $F[x]/(x^n-1)$?" Tags="&lt;polynomials&gt;" AnswerCount="2" />
  <row Id="22845" PostTypeId="1" CreationDate="2011-02-20T00:03:05.967" Score="0" ViewCount="106" Body="&lt;p&gt;The answer in the book is $e-1$, and but I can't figure out how to go from $x(\ln(x) - 1)$ to that answer...&lt;/p&gt;&#xA;" OwnerUserId="6392" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2011-02-20T00:55:33.320" LastActivityDate="2011-02-20T00:55:33.320" Title="Find the area bounded by $y = \ln(x)$, $y = 1$, the x-axis and the y-axis." Tags="&lt;calculus&gt;&lt;homework&gt;&lt;integral&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="22846" PostTypeId="2" ParentId="12906" CreationDate="2011-02-20T00:04:46.273" Score="3" ViewCount="" Body="&lt;p&gt;(non-rigorous:)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is simply another example of why the &quot;limit of the sum&quot; is not the &quot;sum of the limit.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Length of curves are a subset of Sums/Integrals which are really the same thing in my mind.  If you like, in this case &quot;the limit of the lengths of the curves &quot; is not the &quot;length of the limit curve&quot;)&lt;/p&gt;&#xA;" OwnerUserId="6075" LastActivityDate="2011-02-20T00:04:46.273" />
  <row Id="22847" PostTypeId="2" ParentId="22844" CreationDate="2011-02-20T00:06:07.817" Score="2" ViewCount="" Body="&lt;p&gt;This is a special case of the isomorphism between the &lt;a href=&quot;http://en.wikipedia.org/wiki/Group_algebra&quot; rel=&quot;nofollow&quot;&gt;group algebra&lt;/a&gt; $K[G]$ of a group $G$ under multiplication and the algebra of finitely supported functions $G \to K$ under convolution. The isomorphism sends an element $g \in K[G]$ to the indicator function $1_g$ of $g$ and is not hard to verify explicitly. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For polynomials, take $G = \mathbb{Z}$ (and ignore negative terms); for polynomials in the context of DFT, take $G = \mathbb{Z}/n\mathbb{Z}$. &lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2011-02-20T00:06:07.817" />
  <row Id="22848" PostTypeId="2" ParentId="22844" CreationDate="2011-02-20T00:10:33.550" Score="3" ViewCount="" Body="&lt;p&gt;Note that the product of the polynomials is given by&#xA;$$&#xD;&#xA;\begin{align}&#xD;&#xA;fg&amp;amp;=\sum_{k=0}^{2n-2}\sum_{i+j=k}f_ig_jx^k\\&#xD;&#xA;&amp;amp;=\sum_{k=0}^{n-1}\left(\sum_{i+j=k}f_ig_j+\sum_{i+j=k+n}f_ig_jx^n\right)x^k\\&#xD;&#xA;&amp;amp;=f*g +(x^n-1)\sum_{k=0}^{n-1}\sum_{i+j=k+n}f_ig_jx^k&#xD;&#xA;\end{align}&#xD;&#xA;$$&#xA;So, $fg=f*g$ mod $x^n-1$ as claimed. In mod $x^n-1$ arithmetic the terms $x^k$ and $x^{k+n}$ become the same, which is why you have a sum over $i+j=k$ mod $n$ in the expression for $c_k$.&lt;/p&gt;&#xA;" OwnerUserId="1321" LastActivityDate="2011-02-20T00:10:33.550" CommentCount="2" />
  <row Id="22849" PostTypeId="2" ParentId="22845" CreationDate="2011-02-20T00:14:00.160" Score="4" ViewCount="" Body="&lt;p&gt;Your area is nothing but $\displaystyle \int_{y=0}^{y=1} x dy$ where $x = e^y$. Plug this in to get the answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/C4E2j.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2011-02-20T00:29:45.580" LastActivityDate="2011-02-20T00:29:45.580" />
  <row Id="22850" PostTypeId="1" AcceptedAnswerId="22853" CreationDate="2011-02-20T00:17:03.113" Score="4" ViewCount="61" Body="&lt;p&gt;I believe the following is true, but I do not know how to prove it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given two sets of distinct non-zero complex numbers $z_1, \dots , z_n$ and $w_1, \dots , w_n$, if &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \begin{bmatrix}&#xD;&#xA;  z_{1}  &amp;amp; \cdots &amp;amp; z_{n} \\&#xD;&#xA;  z_{1}^{2} &amp;amp; \cdots &amp;amp; z_{n}^{2} \\&#xD;&#xA;  \vdots  &amp;amp; \ddots &amp;amp; \vdots  \\&#xD;&#xA;  z_{1}^n  &amp;amp; \cdots &amp;amp; z_{n}^n&#xD;&#xA; \end{bmatrix}&#xD;&#xA;\begin{bmatrix}&#xD;&#xA;  1 \\&#xD;&#xA;  1 \\&#xD;&#xA;  \vdots  \\&#xD;&#xA;  1&#xD;&#xA; \end{bmatrix}&#xD;&#xA;=&#xD;&#xA;\begin{bmatrix}&#xD;&#xA;  w_{1}  &amp;amp; \cdots &amp;amp; w_{n} \\&#xD;&#xA;  w_{1}^{2} &amp;amp; \cdots &amp;amp; w_{n}^{2} \\&#xD;&#xA;  \vdots  &amp;amp; \ddots &amp;amp; \vdots  \\&#xD;&#xA;  w_{1}^n  &amp;amp; \cdots &amp;amp; w_{n}^n&#xD;&#xA; \end{bmatrix}&#xD;&#xA;\begin{bmatrix}&#xD;&#xA;  1 \\&#xD;&#xA;  1 \\&#xD;&#xA;  \vdots  \\&#xD;&#xA;  1&#xD;&#xA; \end{bmatrix}&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;then $w_1, \dots , w_n$ is a permutation of $z_1, \dots , z_n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any ideas?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Background&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again given distinct non-zero complex numbers $z_1, \dots , z_n$, consider the matrix equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \begin{bmatrix}&#xD;&#xA;  1  &amp;amp; z_{1}  &amp;amp; z_{1}^2 &amp;amp; \cdots &amp;amp; z_{1}^{n-1} \\&#xD;&#xA;  1  &amp;amp; z_{2}  &amp;amp; z_{2}^2 &amp;amp; \cdots &amp;amp; z_{2}^{n-1} \\&#xD;&#xA;  \vdots  &amp;amp; \vdots &amp;amp;  \vdots &amp;amp; \ddots  &amp;amp; \vdots  \\&#xD;&#xA;  1  &amp;amp; z_{n}  &amp;amp; z_{n}^2 &amp;amp; \cdots &amp;amp; z_{n}^{n-1}&#xD;&#xA; \end{bmatrix}&#xD;&#xA;\begin{bmatrix}&#xD;&#xA;  a_0 \\&#xD;&#xA;  a_1 \\&#xD;&#xA;  \vdots  \\&#xD;&#xA;  a_{n-1}&#xD;&#xA; \end{bmatrix}&#xD;&#xA;=&#xD;&#xA;\begin{bmatrix}&#xD;&#xA;  0 \\&#xD;&#xA;  0 \\&#xD;&#xA;  \vdots  \\&#xD;&#xA;  0&#xD;&#xA; \end{bmatrix}&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the fundamental theorem of algebra, this equation can only be satisfied when the $a_1, \dots , a_n$'s are all zero, and so the matrix is non-singular. Since the complex numbers $z_1, \dots , z_n$ are non-zero, the matrix&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA; \begin{bmatrix}&#xD;&#xA;  z_{1}  &amp;amp; z_{1}^2 &amp;amp; \cdots &amp;amp; z_{1}^{n} \\&#xD;&#xA;  z_{2}  &amp;amp; z_{2}^2 &amp;amp; \cdots &amp;amp; z_{2}^{n} \\&#xD;&#xA;  \vdots &amp;amp;  \vdots &amp;amp; \ddots  &amp;amp; \vdots  \\&#xD;&#xA;  z_{n}  &amp;amp; z_{n}^2 &amp;amp; \cdots &amp;amp; z_{n}^{n}&#xD;&#xA; \end{bmatrix}&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is also non-singular, and so its columns form a linearly independent basis for $\mathbb{C}^n$. Considered in this light, the claim above compares two bases formed in this way and says, in effect, if the dot-products of each member of the bases with $[ 1,1, \dots ,1]$ are the same, then the two bases are the same. &lt;/p&gt;&#xA;" OwnerUserId="6272" LastActivityDate="2011-02-20T00:29:32.343" Title="Linear Algebra of Symmetric Sums" Tags="&lt;linear-algebra&gt;&lt;complex-numbers&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="22851" PostTypeId="2" ParentId="22845" CreationDate="2011-02-20T00:18:58.037" Score="2" ViewCount="" Body="&lt;p&gt;We have the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \int_{y=0}^{y=1} e^y \ dy$$ which is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$e-1$$&lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2011-02-20T00:18:58.037" />
  <row Id="22852" PostTypeId="2" ParentId="22850" CreationDate="2011-02-20T00:22:32.680" Score="2" ViewCount="" Body="&lt;p&gt;Yes, this is true. It follows from &lt;a href=&quot;http://en.wikipedia.org/wiki/Newton%27s_identities&quot; rel=&quot;nofollow&quot;&gt;Newton's identities&lt;/a&gt;, which show that the first $n$ symmetric sums of $n$ numbers determine the coefficients of a polynomial with those numbers as its roots (in characteristic $0$). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A non-recursive formula follows from the &lt;a href=&quot;http://qchu.wordpress.com/2009/06/24/gila-vi-the-cycle-index-polynomials-of-the-symmetric-groups/&quot; rel=&quot;nofollow&quot;&gt;exponential formula&lt;/a&gt; in combinatorics, which shows that the identity&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\prod_{i=1}^n (1 + tz_i) = \exp \left( p_1 - \frac{p_2 t^2}{2} + \frac{p_3 t^3}{3} \mp ... \right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is equivalent to the identity&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$e_k = \frac{1}{k!} \sum_{\pi \in S_k} \text{sgn}(\pi) p_{\pi}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$p_k = \sum_{i=1}^n z_i^k$,&lt;/li&gt;&#xA;&lt;li&gt;$e_k$ is the $k^{th}$ coefficient of $\prod (1 + tz_i)$,&lt;/li&gt;&#xA;&lt;li&gt;$\text{sgn}(\pi)$ is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Parity_of_a_permutation&quot; rel=&quot;nofollow&quot;&gt;sign&lt;/a&gt; of the permutation $\pi$, and&lt;/li&gt;&#xA;&lt;li&gt;$p_{\pi} = p_{\lambda_1} ... p_{\lambda_m}$ where the permutation $\pi$ has cycle type $(\lambda_1, ... \lambda_n)$. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;A short proof of the first identity is as follows: let $M$ be a matrix whose eigenvalues are $z_1, ... z_n$. Then the LHS is $\det (I + tM)$ and the RHS is $\exp \text{tr } \log (1 + tM)$. But the identity $\det \exp N = \exp \text{tr } N$ where $N$ is a matrix is straightforward to prove (e.g. by proving it for $N$ diagonalizable and extending by continuity). &lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2011-02-20T00:29:32.343" LastActivityDate="2011-02-20T00:29:32.343" />
  <row Id="22853" PostTypeId="2" ParentId="22850" CreationDate="2011-02-20T00:25:21.610" Score="4" ViewCount="" Body="&lt;p&gt;The result is true. Look at &lt;a href=&quot;http://en.wikipedia.org/wiki/Symmetric_polynomial&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Symmetric_polynomial&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/Newton%27s_identities&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Newton%27s_identities&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A &lt;em&gt;symmetric polynomial&lt;/em&gt; in $n$ variables $p(x_1,\dots,x_n)$ is one that does not change under any permutation of the variables. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As explained in the first Wikipedia page linked to above, any symmetric polynomial in $x_1,\dots, x_n$ can be expressed as a polynomial expression with rational coefficients in the power sum symmetric polynomials $p_i(x_1, \dots, x_n)$ for $i=1,\dots,n$, where $$p_i(x_1,\dots,x_n)=x_1^i+x_2^i+\dots+x_n^i. $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The point is then that the coefficients of $$q(x)=(x-z_1)(x-z_2)\dots(x-z_n)$$ are symmetric polynomials of $z_1,\dots,z_n$ and therefore expressible in terms of the $p_i(z_1,\dots,z_n)$. Since these coefficients completely determine $q(x)$, we have the result you are after.&lt;/p&gt;&#xA;" OwnerUserId="462" LastActivityDate="2011-02-20T00:25:21.610" />
  <row Id="22854" PostTypeId="2" ParentId="12906" CreationDate="2011-02-20T00:28:26.133" Score="2" ViewCount="" Body="&lt;p&gt;Correct answer: Nothing is wrong with this, as long as your space is defined using a Manhattan metric.  Normal Euclidean space is defined using a Euclidean metric. &lt;/p&gt;&#xA;" OwnerUserId="7263" LastActivityDate="2011-02-20T00:28:26.133" CommentCount="2" />
  <row Id="22855" PostTypeId="1" AcceptedAnswerId="22856" CreationDate="2011-02-20T00:29:03.173" Score="4" ViewCount="87" Body="&lt;p&gt;Hello. I'm trying to learn by myself some algebraic geometry, since I will probably be taking an elementary course next semester.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have some trouble understanding the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $Y=Z(y-x^{2})$ then by definition $Y$ is algebraic set in $\mathbb{A}^{2}$. The author finds the coordinate ring $K[x,y]/I(Y)$ where $I(Y)=\{f \in K[x,y]: f=0 \ \textrm{in Y}\}$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Claim: $K[x,y]/I(y) \cong K[t]$ where $t$ is an indeterminate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Define a map $\phi: K[x,y] \rightarrow K[t]$ by $x \mapsto t$ and $y \mapsto t^{2}$. Then $f$ is a ring morphism by the universal property.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Questions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Why it suffices to say where to send $x$ and $y$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) How do you check using the universal property that this is indeed a ring morphism? I don't know how to apply this property&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) What is the reason of sending $x$ to $t$ and $y \mapsto t^{2}$,i.e how do you figure out the images of $x$ and $y$ under $\phi$ ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would really appreciate if you can please explain as much as possible, hope that is not too much to ask.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance&lt;/p&gt;&#xA;" OwnerUserId="6495" LastActivityDate="2011-02-20T00:37:52.680" Title="Elementary question about coordinate ring" Tags="&lt;algebraic-geometry&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" />
  <row Id="22856" PostTypeId="2" ParentId="22855" CreationDate="2011-02-20T00:37:52.680" Score="3" ViewCount="" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;It suffices to say where $x$ and $y$ go because they generate $K[x, y]$ (by the universal property).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The universal property tells you that a $K$-algebra homomorphism out of a polynomial ring over $K$ is determined by the images of the indeterminates.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The targets of $x$ and $y$ are chosen so that the ideal of the homomorphism is $I(Y)$. A different way of seeing this is to note that there is a (surjective) morphism $\mathbb{A}^1 \to Y$ sending $t \mapsto (t, t^2)$ — a morphism of affine varieties corresponds to a $K$-algebra homomorphism of the coordinate rings (but in the opposite direction!) and vice-versa.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="5191" LastActivityDate="2011-02-20T00:37:52.680" CommentCount="2" />
  <row Id="22857" PostTypeId="1" CreationDate="2011-02-20T00:54:40.810" Score="0" ViewCount="72" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Find the area bounded by these two functions:&#xA;  $$y = \frac{\ln x}{x}\quad\mbox{and}\quad y = \frac{1}{e} + \frac{(e^2+1)(x-e)}{e^2-1}.$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="6392" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2011-02-20T01:17:19.120" LastActivityDate="2011-02-20T01:19:34.850" Title="Find the area bounded by these two functions?" Tags="&lt;calculus&gt;&lt;homework&gt;&lt;integral&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="22858" PostTypeId="1" CreationDate="2011-02-20T00:59:15.043" Score="2" ViewCount="79" Body="&lt;p&gt;Question: Given a twist of the projective space, how do I find unit quaternions that represent it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Backgroud and what do I mean: Following Conway &amp;amp; Smith's &lt;em&gt;On Quaternions and Octonions&lt;/em&gt;, every pair of unitary quaternions $l,r$ gives a map $[l,r]:x \mapsto \bar{l}xr$, and a map $\ast[l,r]:x \mapsto \bar{l} \bar{x} r$, and every element of $\operatorname{GO}(4)$ can be seen as one of these two mappings (and $\operatorname{SO}(4)$ consists of those of the first type). When working projectively, the four expressions $[\pm l, \pm r]$ are all the same projective map.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, the elements of $\operatorname{PSO}(4)$ can be thought, in a geometric way, as twists, that is, the product of two rotations on polar lines with different angles (in particular a rotation happens when one of the two angles is zero).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to see, geometrically, what is the twist obtained by the product of two twists that I know (in terms of their lines). Using quaternions that would be straightforward, if I knew, given a (geometric) twist (ie. its two polar lines and the angles) how can we find the unit quaternions $l,r$ that represent it?&lt;/p&gt;&#xA;" OwnerUserId="7272" LastEditorUserId="5363" LastEditorDisplayName="" LastEditDate="2011-02-20T01:28:28.607" LastActivityDate="2011-02-20T01:28:28.607" Title="The geometry of $\operatorname{PSO}(4)$ and the quaternions" Tags="&lt;quaternions&gt;&lt;projective-geometry&gt;" CommentCount="1" />
  <row Id="22859" PostTypeId="1" CreationDate="2011-02-20T01:05:54.530" Score="5" ViewCount="134" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Second_derivative_test&quot; rel=&quot;nofollow&quot;&gt;Second derivative test&lt;/a&gt; in&#xA;Wikipedia says that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a real function of one variable:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If the function f is twice&#xA;  differentiable at a stationary point&#xA;  x, meaning that $\ f^{\prime}(x) = 0$&#xA;  , then:&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;If $ f^{\prime\prime}(x) &amp;lt; 0$ then $f$&#xA;  has a local maximum at $x$. If $f^{\prime\prime}(x) &amp;gt; 0$ then $f$&#xA;  has&#xA;  a local minimum at $x$. If $f^{\prime\prime}(x) = 0$, the&#xA;  second&#xA;  derivative test says nothing about the&#xA;  point $x$, a possible inflection&#xA;  point.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For a function of more than one&#xA;variable:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Assuming that all second order partial&#xA;  derivatives of $f$ are continuous on a&#xA;  neighbourhood of a stationary point $x$,&#xA;  then:&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;if the eigenvalues of the Hessian at $x$&#xA;  are all positive, then $x$ is a local&#xA;  minimum. If the eigenvalues are all&#xA;  negative, then $x$ is a local maximum,&#xA;  and if some are positive and some&#xA;  negative, then the point is a saddle&#xA;  point. If the Hessian matrix is&#xA;  singular, then the second derivative&#xA;  test is inconclusive.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My question is why in the&#xA;multivariate case, the test requires&#xA;the second order partial derivatives&#xA;of $f$ to be continuous on a&#xA;neighbourhood of $x$, while in the&#xA;single variable case, it does not&#xA;need the second derivative to be&#xA;continuous around $x$? Do both also&#xA;require that the first derivative&#xA;and the function itself to be&#xA;continuous around $x$?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;Similarly, does &lt;a href=&quot;http://en.wikipedia.org/wiki/First_derivative_test#Precise_statement_of_first_derivative_test&quot; rel=&quot;nofollow&quot;&gt;first derivative&#xA;test&lt;/a&gt; for $f$ at $x$ need $f$ to&#xA;be continuous and differentiable in&#xA;a neighbourhood of $x$?&lt;/li&gt;&#xA;&lt;li&gt;For &lt;a href=&quot;http://en.wikipedia.org/wiki/Higher-order_derivative_test&quot; rel=&quot;nofollow&quot;&gt;higher order derivative&#xA;test&lt;/a&gt;, it doesn't mention if $f$&#xA;is required to be continuous and&#xA;differentiable in a neighbourhood of&#xA;some point $c$ up to some order. So&#xA;does it only need that $f$ is&#xA;differentiable at $c$ up to order $n$?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Thanks for clarification!&lt;/p&gt;&#xA;" OwnerUserId="1281" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2011-02-20T01:14:31.397" LastActivityDate="2011-03-02T23:59:21.200" Title="Questions about determining local extremum by derivative" Tags="&lt;real-analysis&gt;&lt;optimization&gt;" AnswerCount="4" />
  <row Id="22860" PostTypeId="2" ParentId="22857" CreationDate="2011-02-20T01:12:53.043" Score="4" ViewCount="" Body="&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/DzKPe.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The area is the integral $\displaystyle \int_{x_1}^{x_2} y dx$ where $y=y_{\text{upper}}-y_{\text{lower}}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$y_{\text{upper}} = \frac{\ln(x)}{x}$ and $y_{\text{lower}} = \frac{1}{e} + \frac{e^2+1}{e^2-1}(x-e)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(x_1,y_1)$ and $(x_2,y_2)$ are obtained by equating $y_{\text{lower}}$, $y_{\text{upper}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: It is easy to obtain $(x_1,y_1)$ and $(x_2,y_2)$ by guessing.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2011-02-20T01:19:34.850" LastActivityDate="2011-02-20T01:19:34.850" CommentCount="3" />
  <row Id="22861" PostTypeId="1" CreationDate="2011-02-20T01:20:08.573" Score="13" ViewCount="610" Body="&lt;p&gt;How do you effectively study mathematics? How does one read a maths book instead or just staring at it for hours? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Apologies in advance if the question is ill-posed or too subjective in its current form to meet the requirements of the FAQ; I'd certainly appreciate any suggestions for its modification if need be.)&lt;/p&gt;&#xA;" OwnerUserId="7273" LastActivityDate="2011-04-23T23:17:23.750" CommunityOwnedDate="2011-02-20T01:48:34.140" Title="How to effectively and efficiently learn mathematics" Tags="&lt;soft-question&gt;&lt;education&gt;" AnswerCount="8" CommentCount="1" FavoriteCount="11" />
  <row Id="22862" PostTypeId="2" ParentId="22859" CreationDate="2011-02-20T01:29:04.087" Score="4" ViewCount="" Body="&lt;p&gt;For your second question: Yes: you need the derivative of $f$ to be defined in a puncture neighborhood of $x$ for the first derivative test to even make sense: you need to know the sign of $f&amp;#39;$ on $(x-\delta,x)$ and on $(x,x+\delta)$ for some $\delta\gt 0$ to be able to apply the test; this automatically requires continuity as well on those intervals. And of course, if $f$ is not continuous at $x$, then you cannot derive a conclusion about the value of $f$ at $x$. Take&#xA;$$f(x) =\left\{\begin{array}{ll}&#xA;-x &amp;amp; \mbox{if $x\lt 0$,}\\&#xA;\frac{1}{2} &amp;amp; \mbox{if $x=0$,}\\&#xA;x+\frac{3}{4} &amp;amp; \mbox{if $x\gt 0$.}&#xA;\end{array}\right.$$&#xA;Then the derivative is negative on $(-\infty,0)$, and positive on $(0,\infty)$; it is undefined on $x=0$. But $f$ has neither a local maximum nor a local minimum at $0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, the First Derivative Test is usually stated for functions that are continuous.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2011-02-20T01:29:04.087" CommentCount="1" />
  <row Id="22863" PostTypeId="2" ParentId="22861" CreationDate="2011-02-20T01:30:47.967" Score="10" ViewCount="" Body="&lt;p&gt;First, besides the book, you need paper and a pen. Second, you must do the exercices of the book. Third, you must do the exercises of the book. Fourth, you must do the exercices of the book. Fifth..., did I mention you must do the exercises of the book? -Do them!&lt;/p&gt;&#xA;" OwnerUserId="664" LastActivityDate="2011-02-20T01:30:47.967" CommunityOwnedDate="2011-02-20T01:48:34.140" CommentCount="9" />
  <row Id="22864" PostTypeId="2" ParentId="22861" CreationDate="2011-02-20T01:31:26.313" Score="2" ViewCount="" Body="&lt;p&gt;I like your question. For me, I prefer to get a big picture first. If books that I have do not work for me in this way, perhaps because of my weak math background, I will search online for a comprehensive overview and mostly be led to Wikipedia. More details will be pursued when needed.&lt;/p&gt;&#xA;" OwnerUserId="1281" LastActivityDate="2011-02-20T01:31:26.313" CommunityOwnedDate="2011-02-20T01:48:34.140" />
  <row Id="22865" PostTypeId="2" ParentId="22861" CreationDate="2011-02-20T01:43:17.140" Score="18" ViewCount="" Body="&lt;p&gt;The only way is to put in the time, not give up, and keep studying it.  I don't believe in &quot;shortcuts.&quot;  One must carefully, and diligently read through the books, and &lt;strong&gt;do the excercises&lt;/strong&gt;  (as Agusti Roig mentions!!), and spend hour after hour working at it.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For me, one of the most important things is to ask &lt;em&gt;yourself&lt;/em&gt; questions.  Why does the theorem have theses hypothesis?  Can this be generalized?  Why this definition?  What was the key idea in the proof? Can I apply this idea, this method of proof to other questions?  For what kinds of questions will this method fail, and why does it fail?  Is there another proof?  etc... The list of possibilities goes on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The more time you spend thinking about a subject, the better you will understand it.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Or in some cases, you will begin understand how little you understand....)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT:  I remembered a post on Terence Tao's blog which says something similar.  The post is &quot;&lt;a href=&quot;http://terrytao.wordpress.com/career-advice/ask-yourself-dumb-questions-%E2%80%93-and-answer-them/&quot; rel=&quot;nofollow&quot;&gt;Ask yourself the dum questions - and answer them!&lt;/a&gt;, it is worth a read.&lt;/p&gt;&#xA;" OwnerUserId="6075" LastEditorUserId="6075" LastEditorDisplayName="" LastEditDate="2011-04-23T23:17:23.750" LastActivityDate="2011-04-23T23:17:23.750" CommunityOwnedDate="2011-02-20T01:48:34.140" CommentCount="2" />
  <row Id="22866" PostTypeId="1" AcceptedAnswerId="22870" CreationDate="2011-02-20T01:49:38.850" Score="3" ViewCount="116" Body="&lt;p&gt;Hello, I'm trying to go through the section of Newer Results in Hartshorne's Geometry: Euclid and Beyond.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This particular exercise has been bugging me for a good while:&#xA;&lt;img src=&quot;http://i.imgur.com/gs1BS.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My first approach was to show that the perpendicular bisectors of $A&amp;#39;B&amp;#39;$ and $A&amp;#39;D&amp;#39;$ intersect at the same point as the perpendicular bisectors of $A&amp;#39;B&amp;#39;$ and $B&amp;#39;C&amp;#39;$. I figured since the center of the circle circumscribed around a triangle has its center at the intersection of the three perpendicular bisectors, this would show that $A&amp;#39;$, $B&amp;#39;$, $C&amp;#39;$ and $D&amp;#39;$ would all be on the same circle. However, I didn't see any way to show this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also tried extending $A&amp;#39;D&amp;#39;$ and $A&amp;#39;C&amp;#39;$ down into the circle containing $D$, $D&amp;#39;$ and $C$ with hopes that it $\angle D&amp;#39;A&amp;#39;C&amp;#39;\cong\angle D&amp;#39;B&amp;#39;C&amp;#39;$, but this seemed like a dead end also. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm going a little mad, if any one sees a possible solution, I'd be quite grateful. Thanks.&lt;/p&gt;&#xA;" OwnerUserId="1583" LastActivityDate="2011-02-20T02:36:14.753" Title="From Hartshorne's Geometry: Euclid and Beyond: intersections of four circles forming a cyclic quadrilateral?" Tags="&lt;geometry&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" />
  <row Id="22867" PostTypeId="1" CreationDate="2011-02-20T02:06:20.947" Score="0" ViewCount="199" Body="&lt;p&gt;Is it true to say that if you enjoy applications it means that you didn't learn the applications within mathematics? &lt;/p&gt;&#xA;" OwnerUserId="7273" LastActivityDate="2011-02-20T03:10:43.597" CommunityOwnedDate="2011-02-20T03:26:27.950" ClosedDate="2011-02-20T12:26:02.060" Title="&quot;Pure&quot; Maths vs. &quot;Applied&quot; Maths" Tags="&lt;soft-question&gt;&lt;education&gt;&lt;applications&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="1" />
  <row Id="22869" PostTypeId="2" ParentId="22861" CreationDate="2011-02-20T02:27:17.113" Score="4" ViewCount="" Body="&lt;p&gt;A math book usually represents a way of thinking about a topic, a perspective on that topic. Hence, you have to agree with the authors opinion on how to present the topic in order to get a useful learning approach. (Contrary, if you completely disagree with the book's perspective, you may regard it as a challenge.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In my experience, learning achievements are enhanced if you let the knowledge flow through you own hands. This means you have to put down the content of the book in a way which fits your way of thinking best. The author will probably have a (slightly) different perspective than you have, due to taste and ability.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A good approach towards a book is: &quot;The author is lying.&quot; - each line of the book has to be justified. If you can't do so, you do not understand the topic in full.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore, I have made the experience it is inevitable to spend lots of time with the matter. Difficult and inaccessible proofs may unveil if you read it over and over again (imo, good examples of these are Hörmander's books on linear pde).&lt;/p&gt;&#xA;" OwnerUserId="3557" LastActivityDate="2011-02-20T02:27:17.113" CommunityOwnedDate="2011-02-20T02:27:17.113" />
  <row Id="22870" PostTypeId="2" ParentId="22866" CreationDate="2011-02-20T02:36:14.753" Score="1" ViewCount="" Body="&lt;p&gt;I see two ways of doing it. The first is probably more in spirit of classical geometry, the second uses the cross ratio.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Let $\beta = \angle A&amp;#39;B&amp;#39;C&amp;#39;$ and $\delta = \angle A&amp;#39;D&amp;#39;C&amp;#39;$. We want to show that $\beta + \delta = \pi$. Draw the sides of the five quadrilaterals $AA&amp;#39;B&amp;#39;B$, $BB&amp;#39;C&amp;#39;C$, $CC&amp;#39;D&amp;#39;D$ and $DD&amp;#39;A&amp;#39;A$ as well as $ABCD$ into the picture. By assumption, these quadrilaterals are cyclic, hence opposite angles are supplementary (add up to $\pi$). We have&#xA;\begin{align*}&#xA;\beta &amp;amp; = 2\pi - \angle A'B'B - \angle BB'C' = (\pi - \angle A'B'B) + (\pi - \angle BB'C') \\&#xA;&amp;amp; = \angle A'AB + \angle C'CB.&#xA;\end{align*}&#xA;Similarly, $\delta = \angle A&amp;#39;AD + \angle C&amp;#39;CD$. Therefore&#xA;\[&#xA;\beta+\delta = (\angle A'AB + \angle A'AD) + (\angle C'CB + \angle C'CD) =&#xA;\angle DAB + \angle DCB = \pi&#xA;\]&#xA;as we wanted.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Compute the cross ratio $[A&amp;#39;,B&amp;#39;,C&amp;#39;,D&amp;#39;]$ and show that it is real by using that the cross ratios $[A,B,C,D]$ and $[A,A&amp;#39;,B&amp;#39;,B]$, $[B,B&amp;#39;,C&amp;#39;,C]$, $[C,C&amp;#39;,D&amp;#39;,D]$ and $[D,D&amp;#39;,A&amp;#39;,A]$ are all real.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="5363" LastActivityDate="2011-02-20T02:36:14.753" CommentCount="2" />
  <row Id="22871" PostTypeId="1" AcceptedAnswerId="22921" CreationDate="2011-02-20T02:40:47.107" Score="1" ViewCount="78" Body="&lt;p&gt;I know there is an easy way to sample standard mixture models if the mixture components can be easily sampled. But, I have a slightly different scenario and I'm wondering if there is an analogous method (or any method at all that is nice). I'll first explain the standard scenario so there is no confusion, and then explain how my problem is different. Sorry if some of this is too remedial, I just want my question to be completely clear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A standard mixture model has pdf&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f_X(x) = \sum_i w_i f^i_X(x)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where the $w_i$'s are non-negative and sum to one and the $f^i$'s are pdfs. A pretty important mixture model is mixture of Gaussians, where the $f^i$'s are normal distributions with different means and variances, i.e.,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f_X(x) = \sum_i w_i \mathcal{N}(\mu_i, \sigma^2_i; x)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can easily sample this distribution with two samples. I'll assume there is a latent variable that governs which of the components generates the value of the random variable. Thus, I can choose $i$ with probability $w_i$ and then sample $\mathcal{N}(\mu_i, \sigma^2_i; \cdot)$ to generate my sample $\bar{x}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In my case, I have a slightly different mixture model &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f_X(x) = \sum_i w_i f^i_X(x) - \sum_j w_j f^j_X(x)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically, some of the weights are negative so the components $w_j f^j$ &lt;strong&gt;subtract&lt;/strong&gt; from the density at any $x$. Assume that $f_X$ is a pdf, so it is non-negative and integrates to one. Of course, there are methods for generating samples from arbitrary distributions, but it seems that computing the cdf for this one may be hard. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the best way to draw a sequence of samples from this distribution?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance for any help you can provide.&lt;/p&gt;&#xA;" OwnerUserId="5799" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2011-02-20T03:23:59.940" LastActivityDate="2011-02-21T02:07:11.527" Title="Efficient sampling of a mixture model" Tags="&lt;probability&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="22872" PostTypeId="2" ParentId="22867" CreationDate="2011-02-20T02:53:13.637" Score="0" ViewCount="" Body="&lt;p&gt;I think the answer to this is yes and no.  Yes, because, at least for me, with every new concept in mathematics that I master, there is this point where I fully grasp the concept and absolutely hate it at the same time.  And no, because, over time when you truly master the concept and can work with it fluidly, it becomes enjoyable.  It is like speaking a language.  But this is really a subjective question.&lt;/p&gt;&#xA;" OwnerUserId="7263" LastEditorUserId="7263" LastEditorDisplayName="" LastEditDate="2011-02-20T03:10:43.597" LastActivityDate="2011-02-20T03:10:43.597" CommunityOwnedDate="2011-02-20T03:26:27.950" CommentCount="2" />
  <row Id="22873" PostTypeId="1" AcceptedAnswerId="22878" CreationDate="2011-02-20T02:54:38.537" Score="4" ViewCount="126" Body="&lt;p&gt;The proof in my text is as such:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $a_1,a_2,a_3,\ldots$ and $b_1, b_2, b_3,\ldots$ be the labels of the left- and right- hand endpoints respectively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the set A of the left-hand endpoints of the intervals, and let x = sup A. Since x is an upper bound for A, we have $a_n \leq x$. Since each $b_n$ is an upper bound for A, we have $x\leq b_n$. Then since $a_n\leq x \leq b_n$, we can conclude that $x\in I_n$ for every choice of $n\in \mathbb{N}$. Hence x is in the infinite intersection of nested intervals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is.. could the proof work with $x = a_n$ instead? It seems that all the key properties would still hold -- $a_n \leq a_n \leq b_n$ for all n. This does not seem to require the AoC to be true.&lt;/p&gt;&#xA;" OwnerUserId="4633" LastActivityDate="2011-02-20T03:06:48.857" Title="Why does the following proof of the Nested Interval Property require the Axiom of Completeness?" Tags="&lt;real-analysis&gt;&lt;analysis&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="22874" PostTypeId="1" CreationDate="2011-02-20T02:58:40.450" Score="0" ViewCount="140" Body="&lt;p&gt;I'm talking about these specific relations, where g is the determinant of the metric tensor (so it's symmetric spscific), which is a function of $x^k$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{1}{2g}\frac{\partial g}{\partial x^k}=\frac{1}{2}\frac{\partial \ln g}{\partial x^k}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've tried making discrete variations of the matrix, but I can't reach this particular relation.&lt;/p&gt;&#xA;" OwnerUserId="6153" LastActivityDate="2011-03-22T03:33:49.270" Title="How to prove matrix identities involving the determinant, logarithm and derivatives" Tags="&lt;matrices&gt;" AnswerCount="2" />
  <row Id="22875" PostTypeId="2" ParentId="22861" CreationDate="2011-02-20T02:59:11.763" Score="3" ViewCount="" Body="&lt;p&gt;It depends on what level you are at.  But Schaum's outlines can introduce you to the basics of topology, group theory, abstract algebra, and several other areas in mathematics.  Just do the exercises.&lt;/p&gt;&#xA;" OwnerUserId="7263" LastActivityDate="2011-02-20T02:59:11.763" CommunityOwnedDate="2011-02-20T02:59:11.763" />
  <row Id="22876" PostTypeId="1" AcceptedAnswerId="22882" CreationDate="2011-02-20T03:01:54.360" Score="2" ViewCount="39" Body="&lt;p&gt;If I have 5 people classify an image as offensive or not, how can I calculate my confidence that the answer that 3 or more people agree on is the correct answer?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have past accuracy rates, but am not sure how to incorporate my confidence in those accuracy rates. For example, Person A may be 90% accurate, but with only 10 sample points. Person B may be 85% accurate with 1000 sample points.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've gotten as far as knowing the probability that all 5 people will choose the correct answer is, assuming P(X) is the probability that Person X will choose the correct answer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P(A)*P(B)*P(C)*P(D)*P(E)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've been looking at confidence intervals and multinomial models, but haven't been able to piece everything together.&lt;/p&gt;&#xA;" OwnerUserId="7274" LastActivityDate="2011-02-20T18:28:41.493" Title="Confidence in classifying data" Tags="&lt;probability&gt;&lt;statistics&gt;" AnswerCount="1" />
  <row Id="22877" PostTypeId="2" ParentId="22874" CreationDate="2011-02-20T03:01:59.117" Score="1" ViewCount="" Body="&lt;p&gt;This formula is a direct consequence of the chain rule. &lt;/p&gt;&#xA;" OwnerUserId="5882" LastActivityDate="2011-02-20T03:01:59.117" />
  <row Id="22878" PostTypeId="2" ParentId="22873" CreationDate="2011-02-20T03:06:48.857" Score="3" ViewCount="" Body="&lt;p&gt;Take some irrational number, e.g. $\sqrt{2}$. Consider its decimal expansion&#xA;$$ 1.4142\ldots $$&#xA;Now construct a sequence of nested intervals:&#xA;$$ [1,2] \supset [1.4,1.5] \supset [1.41,1.42] \supset [1.414,1.415] \supset [1.4142,1.4143] \supset \cdots $$&#xA;These are nested intervals with rational endpoints, but their intersection contains no rational number. So without completeness the Nested Interval Property is false.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2011-02-20T03:06:48.857" CommentCount="2" />
  <row Id="22879" PostTypeId="2" ParentId="22874" CreationDate="2011-02-20T03:18:52.117" Score="0" ViewCount="" Body="&lt;p&gt;This follows from chain rule. In general, if you want some identities involving derivatives of matrices, I would recommend you to look at &quot;&lt;a href=&quot;http://matrixcookbook.com/&quot; rel=&quot;nofollow&quot;&gt;The Matrix Cookbook&lt;/a&gt;&quot;&lt;/p&gt;&#xA;" OwnerUserId="3302" LastActivityDate="2011-02-20T03:18:52.117" CommentCount="1" />
  <row Id="22880" PostTypeId="1" AcceptedAnswerId="22891" CreationDate="2011-02-20T03:57:56.990" Score="11" ViewCount="193" Body="&lt;p&gt;This question comes from computer security, but I'll distill it into a probability question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a biased die with 96 sides.  95 sides are equiprobable, each having a 1% chance of landing up.  The remaining side, side X, has a 5% chance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All sides &lt;strong&gt;look&lt;/strong&gt; identical; I want to identify side X.  My best strategy is obviously to roll it a bunch of times and take the majority, but my question is this: how many rolls are needed until Pr{X is majority result} &gt; p for any given p &gt; 1/2?&lt;/p&gt;&#xA;" OwnerUserId="7162" LastActivityDate="2011-02-20T07:26:47.527" Title="How many times do I roll an unfair die to determine its bias?" Tags="&lt;probability&gt;&lt;dice&gt;" AnswerCount="2" FavoriteCount="2" />
  <row Id="22881" PostTypeId="1" AcceptedAnswerId="22883" CreationDate="2011-02-20T04:02:34.807" Score="1" ViewCount="51" Body="&lt;p&gt;Why is it that for quaternions, $u*v = \mathrm{cross}(u,v)-\mathrm{dot}(u,v)$? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wonder for what reason they are equal to each others.&lt;/p&gt;&#xA;" OwnerUserId="7217" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-20T06:31:58.507" LastActivityDate="2011-02-20T06:31:58.507" Title="Why u*v = cross(u,v)-dot(u,v) at quaternions?" Tags="&lt;quaternions&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="22882" PostTypeId="2" ParentId="22876" CreationDate="2011-02-20T04:12:25.317" Score="1" ViewCount="" Body="&lt;p&gt;There are a lot of different tools you can use, but they all depend on a slightly different formulation (assumptions) of your problem. Let me offer one based on a Bayesian approach to the problem. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume each item is either offensive $X=1$ or not $X=0$. Then, define the random variable $Y_i$ to be the label assigned by the $i^\mathrm{th}$ person. You mentioned you had a model for each person, i.e., $p_{Y_i|X}(Y_i = 1 | X = 1)$ and $p_{Y_i|X}(Y_i = 1 | X = 0)$. So, if person $i$ labeled the item offensive, the probability the item is actually offensive is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ p_{X|Y_i}(1|1) = \frac{p_{Y_i|X}(1|1)p_X(1)}{p_{Y_i}(1)} = \frac{p_{Y_i|X}(1|1)p_X(1)}{p_{Y_i|X}(1|1)p_X(1)+p_{Y_i|X}(1|0)p_X(0)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;from Bayes law. If we assume the responses are conditionally independent (when conditioned on $X$) then we get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ p_{X|Y_i, Y_j}(1|1,1) = \left(\frac{p_{Y_i|X}(1|1)}{p_{Y_i}(1)} \times \frac{p_{Y_j|X}(1|1)}{p_{Y_j}(1)}\right) p_X(1)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where the denominators can be computed as above. This just becomes a product over all the observations of the hidden variable you have (for $Y_k$, etc.). By the way, this is exactly a &lt;a href=&quot;http://en.wikipedia.org/wiki/Naive_Bayes_classifier&quot; rel=&quot;nofollow&quot;&gt;Naive Bayes classifier&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the question is what is the probability the next person will label it offensive? Let $Z_k$ denote the unknown response. Then,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$p_{Z_k|Y_i,Y_j} = p_{Z_k|Y_i,Y_j,X}(1|y_i,y_j,1)p_{X|Y_i,Y_j}(1|y_i,y_j) + p_{Z_k|Y_i,Y_j,X}(1|y_i,y_j,0)p_{X|Y_i,Y_j}(0|y_i,y_j)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;from total law of probability. Basically, its the probability (given previous responses) that the item is offensive and the new response is correct plus the probability that the item is not offensive and the new response is wrong. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Good luck. &lt;/p&gt;&#xA;" OwnerUserId="5799" LastEditorUserId="5799" LastEditorDisplayName="" LastEditDate="2011-02-20T18:28:41.493" LastActivityDate="2011-02-20T18:28:41.493" CommentCount="3" />
  <row Id="22883" PostTypeId="2" ParentId="22881" CreationDate="2011-02-20T04:18:04.173" Score="4" ViewCount="" Body="&lt;p&gt;Assume that $u = bi + cj + dk$ and $v = xi + yj + zk$ are imaginary quaternions (no real part). Then a straightforward computation using the identities $i^2 = j^2 = k^2 = ijk = -1$ gives &#xA;\[&#xA;uv = -(bx + cy + dz) + (cz - dy)i + (dx - bz)j + (by - cx)k = - \langle u,v \rangle + u \times v&#xA;\]&#xA;with the usual identification $\mathbb{R}^{3} = \operatorname{Im}\mathbb{H}$.&lt;/p&gt;&#xA;" OwnerUserId="5363" LastActivityDate="2011-02-20T04:18:04.173" />
  <row Id="22884" PostTypeId="1" CreationDate="2011-02-20T04:18:07.807" Score="5" ViewCount="111" Body="&lt;p&gt;I have an exercise scribbled down, and I am not sure what it is asking. It is somewhat similar to Burnside's lemma.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have a finite group $G$ acting on a set $X$. For each $g \in G$, let $X^g$ denote the set of elements in $X$ fixed by $g$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_g |X^g|^2 = |G| \cdot \text{(number of orbits of a stabilizer)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not sure what it means by &quot;orbit of a stabilizer&quot;. I am guessing that it refers to the action of $G$ on cosets of a stabilizer by multiplication. But this really doesn't make sense to me since this action is transitive and the orbit is just the entire set.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anyone know of such an exercise and can someone explain what the precise statement of the problem should be?&lt;/p&gt;&#xA;" OwnerUserId="7276" LastEditorUserId="7276" LastEditorDisplayName="" LastEditDate="2011-02-20T10:33:56.177" LastActivityDate="2011-02-20T16:51:37.037" Title="Similar result to Burnside's lemma" Tags="&lt;group-theory&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="22885" PostTypeId="2" ParentId="22884" CreationDate="2011-02-20T04:33:12.170" Score="5" ViewCount="" Body="&lt;p&gt;Consider $G$ acting on $X^2$ component-wise. The number of elements of $X^2$ fixed by $G$ is exactly $|X_g|^2$. So the average of $|X_g|^2$ is the number of orbits of $G$'s action on $X^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This &quot;exercise&quot; (replacing $2$ with an arbitrary natural number) can be used to show that the distribution of the number of fixed points of a permutation is roughly Poissonian. It has lots of other uses in enumeration theory.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2011-02-20T04:33:12.170" />
  <row Id="22886" PostTypeId="1" AcceptedAnswerId="22897" CreationDate="2011-02-20T04:51:43.487" Score="8" ViewCount="215" Body="&lt;p&gt;My brother has a 30 year fixed mortgage. He pays monthly. Every month my brother doubles his principal payment (so every month, he pays a little bit more, according to how much more principal he's paying).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;He told me he'd pay his mortgage off in 15 years this way. I told him I though it'd take more than 15 years. Who's right? If I'm right (it'll take more than 15 years) how would I explain this to him?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;CLARIFICATION: He doubles his principal by looking at his statement and doubling the &quot;amount applied to principal this payment&quot; field.&lt;/p&gt;&#xA;" OwnerUserId="7278" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-20T07:20:33.753" LastActivityDate="2011-04-29T09:25:19.323" Title="Paying off a mortgage twice as fast?" Tags="&lt;finance&gt;" AnswerCount="6" CommentCount="7" />
  <row Id="22887" PostTypeId="1" AcceptedAnswerId="22888" CreationDate="2011-02-20T04:55:53.063" Score="4" ViewCount="198" Body="&lt;p&gt;I'm reading a statistics textbook which defines the mean of a random variable $X$ with CDF $F$ as a &lt;em&gt;statistical function&lt;/em&gt; $t(\centerdot)$, where&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ t(F) = \int x \, dF(x).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can someone explain this definition? I'm familiar with the definition of the mean as an integral over the PDF $f$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \int x \, f(x) \, dx.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But what does it mean to have the function $F(x)$ in the variable of integration?&lt;/p&gt;&#xA;" OwnerUserId="7277" LastActivityDate="2011-02-20T13:03:46.460" Title="Definition of mean as an integral over the CDF" Tags="&lt;statistics&gt;" AnswerCount="2" />
  <row Id="22888" PostTypeId="2" ParentId="22887" CreationDate="2011-02-20T05:17:20.323" Score="6" ViewCount="" Body="&lt;p&gt;The former integral is a Stieltjes integral. See &lt;a href=&quot;http://en.wikipedia.org/wiki/Riemann%E2%80%93Stieltjes_integral&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt;, for example, in particular the section &quot;Application to probability theory&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In brief, the integral $I_1 = \int {xdF(x)} $ is a generalization of $I_2 = \int {xf(x)dx} $. $I_2$ can be used only when the distribution is absolutely continuous, that is, has a probability density function. $I_1$ can be used for any distribution, even if it is discrete or continuous singular (provided the expectation is well-defined, that is $\int {|x|dF(x)}  &amp;lt; \infty $). Note that if $F&amp;#39;(x) = f(x)$, then $\frac{d}{{dx}}F(x) = f(x)$ gives rise to $dF(x) = f(x)dx$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a thorough account of this topic, see, for example, &lt;a href=&quot;http://homepages.nyu.edu/~eo1/Book-PDF/pChapterDDD.pdf&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An example. Suppose that a random variable $X$ has CDF $F$ such that $F&amp;#39;(x) = f_1 (x)$ for $x &amp;lt; a$, $F&amp;#39;(x) = f_2 (x)$ for $x &amp;gt; a$, and $F(a) - F(a-) = p$, where $f_1$ and $f_2$ are nonnegative continuous functions, and $0 &amp;lt; p \leq 1$. Note that $F$ is everywhere differentiable except at $x=a$ where it has a jump discontinuity of size $p$. In particular, $X$ does not have a PDF $f$, hence you cannot compute ${\rm E}(X)$ as the integral $I_2$. However, ${\rm E}(X)$ can be computed as follows:&#xA;$$&#xD;&#xA;{\rm E}(X) = \int_{ - \infty }^\infty  {xdF(x)}  = \int_{ - \infty }^a {xf_1 (x)dx}  + a[F(a) - F(a - )] + \int_a^\infty  {xf_2 (x)dx} .&#xD;&#xA;$$&#xA;In case $f_1$ and $f_2$ are identically zero and, hence, $p=1$, this gives&#xA;$$&#xD;&#xA;{\rm E}(X) = \int_{ - \infty }^\infty  {xdF(x)}  =  a[F(a) - F(a - )] = a,&#xD;&#xA;$$&#xA;which is obvious since $X$ has the $\delta_a$ distribution (that is, ${\rm P}(X=a)=1$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Exercise 1. Suppose that $X$ takes finitely many values $x_1,\ldots,x_n$, with probabilities $p_1,\ldots,p_n$, respectively. Conclude that ${\rm E}(X) = \int_{ - \infty }^\infty  {xdF(x)}  = \sum\nolimits_{k = 1}^n {x_k p_k } $.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Exercise 2. Suppose that with probability $0 &amp;lt; p &amp;lt; 1$ a random variable $X$ takes the value $a$, and with probability $1-p$ it is uniform$[0,1]$. Find the CDF $F$ of $X$, and compute its expectation (note that $X$ is neither discrete nor continuous random variable).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: The integral $I_1$ is, of course, a special case of $\int {h(x)dF(x)} $ (say, for $h$ a continuous function). In particular, letting $h=1$, we have $\int {dF(x)}  = 1$ (which is a generalization of $\int {f(x)dx}  = 1$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;: Further details.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that if $h$ is continuous, then ${\rm E}[h(X)]$, if exists, is given by &#xA;$$&#xD;&#xA;{\rm E}[h(X)] = \int {h(x)dF(x)}.&#xD;&#xA;$$&#xA;In particular, if the $n$-th moment exists, it is given by&#xA;$$&#xD;&#xA;{\rm E}[X^n] = \int {x^n dF(x)}.&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In principle, the limits of integration range from $-\infty$ to $\infty$. In this context, consider the following important example. Suppose that $X$ is any nonnegative random variable. Then its Laplace transform is &#xA;$$&#xD;&#xA;{\rm E}[e^{ - sX} ] = \int {e^{ - sx} dF(x)}  = \int_{ 0^- }^\infty  {e^{ - sx} dF(x)}, \;\; s \geq 0,&#xD;&#xA;$$&#xA;where $0^-$ can be replaced by $-\varepsilon$ for any $\varepsilon &amp;gt; 0$. While the $n$-th moment of (the nonnegative) $X$ is given, for any $n \geq 1$, by&#xA;$$&#xD;&#xA;{\rm E}[X^n] = \int_{ 0^- }^\infty  {x^n dF(x)}  = \int_0^\infty  {x^n dF(x)}, &#xD;&#xA;$$&#xA;it is not true in general that also&#xA;$$&#xD;&#xA;{\rm E}[e^{ - sX} ] = \int_{0 }^\infty  {e^{ - sx} dF(x)}.&#xD;&#xA;$$&#xA;Indeed, following the definition of the integral,&#xA;$$&#xD;&#xA;{\rm E}[e^{ - sX} ] = \int_{ 0^- }^\infty  {e^{ - sx} dF(x)} = e^{-s0}[F(0)-F(0-)] + \int_{ 0 }^\infty  {e^{ - sx} dF(x)},&#xD;&#xA;$$&#xA;hence the jump of $F$ at zero should be added (if positive, of course). In the $n$-th moment case, on the other hand, the corresponding term is $0^n[F(0)-F(0-)] = 0$, hence a jump of $F$ at zero does not affect the overall integral.&lt;/p&gt;&#xA;" OwnerUserId="2810" LastEditorUserId="2810" LastEditorDisplayName="" LastEditDate="2011-02-20T13:03:46.460" LastActivityDate="2011-02-20T13:03:46.460" />
  <row Id="22889" PostTypeId="2" ParentId="22887" CreationDate="2011-02-20T05:19:15.400" Score="6" ViewCount="" Body="&lt;p&gt;The cumulative density function $F$ is related to the probability density function $f$ by $dF(x)/dx=f(x)$. The equation you have in terms of $F$ can be re-expressed in terms of $f$ by substituting in $dF(x) = f(x)\,dx$. In fact, for many purposes, you can take this as the definition of the differential term $dF$. However, in more general circumstances where $F$ is not differentiable and the PDF $f$ is not well-defined, the form involving $dF$ still holds (interpreting it as a &lt;a href=&quot;http://en.wikipedia.org/wiki/Riemann%E2%80%93Stieltjes_integral&quot; rel=&quot;nofollow&quot;&gt;Riemann-Stieltjes&lt;/a&gt; integral). For example, if the distribution is discrete, so that $F$ is piecewise constant, then $dF$ becomes a sum over &lt;a href=&quot;http://en.wikipedia.org/wiki/Dirac_measure&quot; rel=&quot;nofollow&quot;&gt;Dirac&lt;/a&gt; distributions.&lt;/p&gt;&#xA;" OwnerUserId="1321" LastEditorUserId="1321" LastEditorDisplayName="" LastEditDate="2011-02-20T11:18:30.000" LastActivityDate="2011-02-20T11:18:30.000" CommentCount="0" />
  <row Id="22890" PostTypeId="2" ParentId="22884" CreationDate="2011-02-20T05:28:34.063" Score="4" ViewCount="" Body="&lt;p&gt;The &quot;number of orbits of a stabilizer&quot;: given $x \in X$, the stabilizer $G_x$ acts on $X$. The number of orbits of the stabilizer means the number of orbits of $G_x$ on $X$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you add the extra condition that the action of $G$ is transitive on $X$. If that is so, you first not that $\Sigma_{g \in G_x} |X_g| = \Sigma_{g \in G_y} |X_g|$, for every $x, y \in X$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the action of $G$ is transitive on $X$, then $|G|=|X||G_x|$, implying that&#xA;$$k|G|=k|X||G_x|=(|X|)(k|G_x|)=\Sigma_{x\in X} \Sigma_{g \in G_x} |X_g| = \Sigma_{g\in G} \Sigma_{x \in X_g} |X_g| = \Sigma_{g\in G}|X_g|^2 ,$$&#xA;where $k$ is the number of orbits of $G_x$ on $X$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if the action of $G$ on $X$ is not transitive, the stabilizers can have different number of orbits, depending on the point that is stabilized. For example if $1, 2, 3, 4$ represent the vertices of a regular square, and $G$ consists of the identity and the reflection $R$ in the line through $1$ and $3$, then $\Sigma_{g\in G}|X_g|^2 = |X_{id}|^2 + |X_R|^2 = 4^2+2^2 =20$. On one hand the stabilizer of 1, $G_1$, has three orbits on the vertices (the one containing 1, the one containing 3 and the one containing 2 and 4), and hence (&quot;number of orbits of the stabilizer&quot;)|G|=3*2=6. &lt;/p&gt;&#xA;" OwnerUserId="7272" LastEditorUserId="7272" LastEditorDisplayName="" LastEditDate="2011-02-20T16:51:37.037" LastActivityDate="2011-02-20T16:51:37.037" CommentCount="2" />
  <row Id="22891" PostTypeId="2" ParentId="22880" CreationDate="2011-02-20T05:39:09.283" Score="9" ViewCount="" Body="&lt;p&gt;There's a way to approximate the answer using information theory. Let $X$ be the variable representing the side, and let $D$ be an outcome of a roll. Each roll tells you that much information about $X$:&#xA;$$ I(X;D) = H(D) - H(D|X) = \log 96 + 95 \cdot 0.01 \log 0.01 + 0.05 \log 0.05. $$&#xA;As an approximation, the information gained from different rolls is additive; it's certainly at most additive. Since $H(X) = \log 96$ (since you have no a-priori belief about $X$), the number of rolls needed is at least&#xA;$$ \frac{H(X)}{I(X;D)} \approx 115. $$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Another way to approach the same question is as follows. Suppose the die was thrown $N$ times. The good side came up approximately $N/20$ times, and all the others came up $\mathrm{Bin}(N,0.01) \approx \mathcal{N}(0.01N,0.0099N)$. So we want the probability that such a normal variable be larger than $N/20$ to be smaller than $1/95$ (since there are $95$ such variables; we're ignoring their dependence). This is $$\frac{0.05N - 0.01N}{\sqrt{0.0099N}} \approx 0.4\sqrt{N}$$ standard deviations. Looking at a table, we need roughly $2.3$ standard deviations to get a probability of $1/95$, so we need $$N \approx (2.3/0.4)^2 = 33. $$ Of course, this estimate is pretty rough (it doesn't even match our previous supposed lower bound).&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The estimate in the previous paragraph is particularly problematic since in this range the convergence to the normal distribution is pretty bad (the expectation is very small, roughly $1$). In this regime, the probability is actually roughly Poissonian. Let's estimate the probability of success when $N=100$ using this approximation. The correct side is going to show up roughly $5$ times. Each other side is going to show up $P(1)$ times, so it's going to be as large as $4$ with probability&#xA;$$ 1 - e^{-1} \sum_{t=0}^4 \frac{1^t}{t!} \approx 0.00366. $$&#xA;So the expected number of &quot;contenders&quot; is&#xA;$$ 0.00366 \cdot 95 \approx 0.35.$$&#xA;Assuming independence, the probability of no contenders is roughly&#xA;$$ (1 - 0.00366)^{95} \approx 0.706. $$&#xA;This approximation is only roughly vindicated by the experiments below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to find a better approximation, let's take into account the roughly $P(5)$ distribution of the histogram of $X$:&#xA;$$ e^{-5} \sum_{H=0}^\infty \frac{5^H}{H!} \left(e^{-1} \sum_{t=0}^{H-1} \frac{1}{t!} \right)^{95} \approx 0.527. $$&#xA;This conforms much better to the results below.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Finally, here are experimental results from one million trials, along with the Poisson approximation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \begin{array}{r|c|c}&#xD;&#xA;N &amp;amp; \text{Experiment} &amp;amp; \text{Approx} \\\hline&#xD;&#xA;50 &amp;amp; 0.272323 &amp;amp; 0.274900 \\&#xD;&#xA;100 &amp;amp; 0.526765 &amp;amp; 0.527424 \\&#xD;&#xA;150 &amp;amp; 0.718004 &amp;amp; 0.714956 \\&#xD;&#xA;200 &amp;amp; 0.840143 &amp;amp; 0.836808 \\&#xD;&#xA;250 &amp;amp; 0.912917 &amp;amp; 0.910115 \\&#xD;&#xA;300 &amp;amp; 0.954097 &amp;amp; 0.951953&#xD;&#xA;\end{array}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The results support the information-theoretic bound, as well as our estimates obtained using the Poisson approximation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: When the success probability is $p$ and there are $T$ trials, the standard deviation is $$\sqrt{\frac{p(1-p)}{T}} \leq \frac{1}{2\sqrt{T}}.$$ In our case, $T = 10^6$ and so a standard deviation is at most $0.5\cdot 10^{-3}$. With probability $95\%$, the error is at most two standard deviations, i.e. $\pm 0.001$. So the experimental results are correct up to roughly the third digit (which may be slightly off itself).&lt;/p&gt;&#xA;" OwnerUserId="1277" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2011-02-20T07:08:43.697" LastActivityDate="2011-02-20T07:08:43.697" />
  <row Id="22892" PostTypeId="1" AcceptedAnswerId="22896" CreationDate="2011-02-20T05:39:22.463" Score="0" ViewCount="44" Body="&lt;p&gt;Let's say I have a 2D coordinate space between (minX,minY,maxX,maxY) = -5,-5 and 5,5.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then let's say I have coordinates (x1,y1,x2,y2) for a line that will run from 4,4 to 6,7.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do I truncate that line so that it ends at the edge of the coordinate space (but maintains the same slope)?  In other words, how do I get the values for x2,y2 so that y2 is 5 (the maximum it could be)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I originally thought that just using the percentage change in the highest of the out of boundary coordinates would do the trick (e.g. (maxX/y2)*x2), but I get a different slope.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If things change when x or y goes negative I need to account for that too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;thank you for any help!!&lt;/p&gt;&#xA;" OwnerUserId="7279" LastActivityDate="2011-02-20T08:34:02.800" Title="how to truncate a line along the same slope at a given boundary?" Tags="&lt;geometry&gt;&lt;algebra&gt;&lt;coordinate-systems&gt;" AnswerCount="2" />
  <row Id="22893" PostTypeId="2" ParentId="22886" CreationDate="2011-02-20T05:41:19.783" Score="3" ViewCount="" Body="&lt;p&gt;I put together an Excel spreadsheet.  It depends upon what you mean by doubling the principal.  For a 5% loan the 30 year payment is 5.3682/1000.  If you look at the amortization schedule of the 30 year loan and increase the payment by the principal amount (with the additional applied to principal) it will take 20 years, not 15.  But if you look at the principal amount based on the current balance and double it, you do pay off in 15 years.  I would suggest you make your own spreadsheet and play with it.  Each month you charge interest (in my case .05/12 of the current balance) then subtract the payment made to find the new balance.  Excel has the PMT function to determine the payment.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-02-20T05:41:19.783" />
  <row Id="22894" PostTypeId="2" ParentId="22886" CreationDate="2011-02-20T05:42:25.953" Score="3" ViewCount="" Body="&lt;p&gt;The short answer to your question is if your brother pays double the monthly payment then he will pay off the mortgage in &lt;strong&gt;less&lt;/strong&gt; than $15$ years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The right way to analyze such problems from a mathematical frame-point is to look at the present value of the total amount he is paying.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Say your brother pays $\$x$ per month for $30$ years starting from the end of the first month. Let $r \%$ be the rate of interest per annum. We will compare the present value of the mortgage/loan.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Note: that $\$1$ today is worth $\$\left(1+\frac{r}{12}\right)$ at the end of the first month).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The present value of the total amount he pays in $30$ years is $$\frac{x}{\left(1+\frac{r}{12}\right)} + \frac{x}{\left(1+\frac{r}{12}\right)^2} + \cdots + \frac{x}{\left(1+\frac{r}{12}\right)^{12 \times 30}} = \frac{12x}{r} \left (1- \left (\frac{12}{12+r}\right)^{360} \right )$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instead of $\$x$ if he were to pay $\$(2x)$ starting from the end of the first month, and let him pay for $n$ years, the present value of the total amount he pays in $n$ years is $$\frac{2x}{\left (1+\frac{r}{12} \right )} + \frac{2x}{\left (1+\frac{r}{12} \right)^2} + \cdots + \frac{2x}{\left (1+\frac{r}{12} \right)^{12 \times n}} = \frac{24x}{r} \left(1- \left (\frac{12}{12+r}\right)^{12n} \right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the mortgage to be fair, both the present values must be the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So find $n$ such that $$2 \left(1- \left (\frac{12}{12+r}\right)^{12n} \right) = \left(1- \left (\frac{12}{12+r}\right)^{360} \right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solving for $n$, we get $$n = \frac{\log \left (1+\left (\frac{12}{12+r} \right )^{360} \right)-\log \left(2 \right)}{12 \log \left(\frac{12}{12+r} \right)}$$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So say plugging in $r = 6\%$ i.e. $r = 0.06$, we get $n = 9.01466$ years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whatever be the rate of interest, the total amount, the amount he is paying per month, if he decides to double the monthly payment he will pay it off in less than $15$ years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, if a mortgage is for $N$ years and a person needs to pay $x$ per month, if the person decides to pay $k x$ per month instead of $x$ per month and $k&amp;gt;1$, then the person will pay the mortgage in less than $\frac{N}{k}$ years.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2011-02-20T06:08:00.583" LastActivityDate="2011-02-20T06:08:00.583" CommentCount="2" />
  <row Id="22896" PostTypeId="2" ParentId="22892" CreationDate="2011-02-20T05:49:15.440" Score="0" ViewCount="" Body="&lt;p&gt;The two point formula for a line is $y-y_1=(x-x_1)\frac{y_2-y_1}{x_2-x_1}$ where the points are $(x_1,y_1)$ and $(x_2,y_2)$  You can plug your boundaries into this to find where the line intersects the boundaries.  So in your case you have $y-4=\frac{3}{2}(x-4)$ and you can put in $y=5$ to solve for $x$ where you hit the boundary, getting $4\frac{2}{3}$.  If you put in all four boundaries, two of the points will be &quot;in play&quot; and two will be &quot;out of bounds&quot;, so just draw the line between the ones &quot;in play&quot;.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-02-20T05:49:15.440" CommentCount="6" />
  <row Id="22897" PostTypeId="2" ParentId="22886" CreationDate="2011-02-20T05:50:27.397" Score="8" ViewCount="" Body="&lt;p&gt;Suppose you have a $\$ $100 debt at 10% monthly interest, and you pay $\$ $15 a month.  The amortization for this payment is:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Month     Principal   Interest    Payment   Applied to Principal&#xA;  1         100          10         15              5&#xA;  2         95            9.5       15              5.50&#xA;  3         89.50         8.95      15              6.05&#xA;  4         83.45         8.35      15              6.65&#xA;  5         76.80         7.68      15              7.32&#xA;  6         69.48         6.95      15              8.05&#xA;  7         62.43         6.24      15              8.76&#xA;  8         53.67         5.37      15              9.63&#xA;  9         44.04         4.40      15             10.60&#xA; 10         33.44         3.34      15             11.66&#xA; 11         21.78         2.18      15             12.82&#xA; 12          8.96         0.90       9.86&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;So you pay it off in one year.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, I'm not clear from your description if your brother is planning to pay off double what the original amortization table would indicate, or double what that month's principal would have been (given that he already paid off more than he was expected to), so let me do both.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose first that he simply pays pays his original 15, plus the column &quot;applied to principal&quot; from the original amortization: so the first month, instead of paying 15 with 5 towards principal, he pays 20 (10 towards principal). Next month, instead of 15, he pays 20.50 (so, whatever it will be towards principal, plus an extra 5.50 corresponding to the original amortization table &quot;applied to principal&quot; column; etc.) We have:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Month     Principal   Interest    Payment   Applied to Principal&#xA;  1         100          10        15+5            10&#xA;  2          90           9        15+5.50         11.50&#xA;  3          78.5         7.85     15+6.05         13.20&#xA;  4          65.3         6.53     15+6.65         15.12&#xA;  5          50.18        5.02     15+7.32         17.30&#xA;  6          32.88        3.29     15+8.05         19.76&#xA;  7          13.12        1.31     14.43&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;so it takes 7 months, more than half. Here, what I add each  month to the payment is what the &quot;Applied to Principal&quot; column indicated for that month in the original amortization table.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if each month he pays each again the amount that would &lt;em&gt;now&lt;/em&gt; apply to the principal, we have:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Month     Principal   Interest    Payment   Applied to Principal&#xA;  1         100          10         15+5          10&#xA;  2          90           9         15+6          12&#xA;  3          78           7.80      15+7.20       14.40&#xA;  4          63.60        6.36      15+8.64       17.38&#xA;  5          45.62        4.56      15+10.44      20.88&#xA;  6          24.74        2.47      15+9.74       24.74&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;so you pay it off in six month (the last month because 15-2.47 = 12.53, and 15+12.53 is more than what is owed). Here, what I add to the payment each month is equal to the difference between the basic payment of $\$ $15 and the interest that is being paid off. So for example, in month 4, you have to pay down $\$ $4.56 interest; that means your $\$ $15 payment will pay down 15-4.56=10.44 principal, so you add another $\$ $10.44 to the payment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Added.&lt;/em&gt; And then there seems to be a third option; from what youo describe, ti seems to me that he would look at the &quot;Applied to Principal&quot; line in the previous month, and add that amount to his payment fo the next month. If we do that, we get the following table:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Month     Principal   Interest    Payment   Applied to Principal&#xA;  1         100          10         15              5&#xA;  2          95           9.50      15+5           10.50&#xA;  3          84.50        8.45      15+10.50       12.05&#xA;  4          72.45        7.25      15+12.05       19.80&#xA;  5          52.65        5.27      15+19.80       29.53&#xA;  6          23.12        2.31      15+10.24       25.24&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;so it's 6 months this way, with a slightly larger final payment than the previous version, because he is not really doubling the amount of principal he would have paid down that particular month, but rather something slightly smaller. Here, what I add to the payment is the previous month's &quot;Applied to Principal&quot; column; so, since in month 4 there was $\$ $19.80 applied to principal, that's how much is added to the payment of Month 5. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, in the first scenario (double what he &lt;em&gt;would&lt;/em&gt; have paid towards principal in the original amortization), it takes him more than half the time. In the second, where he pays again over what he would have paid off that month in principal, it takes him about half the time. In the final scenario, where he pays again over what he paid &lt;em&gt;last&lt;/em&gt; time towards principal, it's a bit more than half the time, but by very little compared to the first method. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;These are just examples, of course, but they give you an indication of how things go over 30 years. It would seem &lt;em&gt;he&lt;/em&gt; is correct, and it will take him just a bit over 15 years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This agrees with the results reported by Ross using Excel: my &quot;first scenario&quot; is what Ross reports as taking 20 years; the second what he reports as taking 15 years. My third scenario is very close to the second, but he's a bit off from &quot;doubling&quot; because he is applying the &lt;em&gt;previous&lt;/em&gt; amount that was used to pay down principal, not the current one.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-20T07:49:27.240" LastActivityDate="2011-02-20T07:49:27.240" />
  <row Id="22898" PostTypeId="2" ParentId="22886" CreationDate="2011-02-20T05:52:12.647" Score="0" ViewCount="" Body="&lt;p&gt;Let's look at two scenarios: two months of payment $P$ vs. one month of payment $2P$. Start with the second scenario. Assume the total amount to be payed is $X$ and the rate is $r &amp;gt; 1$, the total amount to be payed after one month would be $$r(X-2P).$$ Under the first scheme, the total amount to be payed after two months would be $$r(r(X-P)-P) = r(rX - (1+r)P).$$ Most of the time, $X$ is much larger than $P$, and so $X-2P$ is significantly smaller than $rX - (1+r)P$ (remember $r \approx 1$). So it should take your brother &lt;i&gt;less&lt;/i&gt; than 15 years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note I first subtract the payment and then take interest, but it shouldn't really matter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This all assumes the payments are fixed, but looking at the other answers this is not really the case... I guess my banking skills are lacking. Too young to take loans.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2011-02-20T05:52:12.647" CommentCount="1" />
  <row Id="22899" PostTypeId="1" AcceptedAnswerId="22900" CreationDate="2011-02-20T06:42:48.407" Score="3" ViewCount="322" Body="&lt;p&gt;Hi everyone, this is a problem from my textbook I am working on. I am not sure how to approach it -- I would like to solve it using a 'matrix' way if possible -- that is, to set up a matrix with coefficients etc. and solve for the basis of the kernel, which would be the basis of what I am looking for (not sure if that's possible).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How to find a basis for all polynomials $f(t)$ of degree 3 such that $f(1)=0$ and $\int_{-1}^{1} f(t) dt$=0&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, my book defines polynomials of degree n as subspaces/vector spaces, but Mariano in the following question argues that it is not such because it does not contain the zero element (though I think if we set all the coefficients to 0, it does), so now I am confused as to whether the problem even makes any sense:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://math.stackexchange.com/questions/1350/are-the-polynomials-of-degree-n-a-vector-space&quot;&gt;Are the polynomials of degree n a vector space?&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5157" LastActivityDate="2011-02-20T11:27:54.447" Title="Finding a basis for all polynomials $f(t)$ of degree 3 such that $f(1)=0$ and $\int_{-1}^{1} f(t) dt$=0" Tags="&lt;linear-algebra&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="22900" PostTypeId="2" ParentId="22899" CreationDate="2011-02-20T06:54:51.710" Score="6" ViewCount="" Body="&lt;p&gt;Almost certainly, your textbook is refering to the set of all polynomials of degree &lt;strong&gt;at most&lt;/strong&gt; $n$, rather than the polynomials of degree &lt;strong&gt;exactly&lt;/strong&gt; $n$. The latter is indeed not a vector space under the usual addition of polynomials: it doesn't contain the $0$ polynomial (which either has no degree, or has some degree smaller than $0$), and it's not closed under sums: both $x^n+x^{n-1}$ and $-x^n$ are polynomials of degree $n$, but their sum is not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I suspect your textbook is asking for the vector space that consists of all polynomials of degree &lt;strong&gt;at most&lt;/strong&gt; $3$, which further satisfy $f(1)=0$ and $\int_{-1}^1 f(t)\,dt = 0$. That is, if we let $\mathbf{P}_3$ denote the vector space of polynomials of degree at most $3$, then it wants you to find a basis for $\mathbf{W}$, where&#xA;$$\mathbf{W} = \left\{ f(t)\in\mathbf{P}_3\;\left|\; f(1)=0\text{ and }\int_{-1}^1f(t)\,dt = 0\right.\right\}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You say you want to solve this via a matrix; well: the map $\mathbf{P}_3\to\mathbb{R}$ given by &quot;evaluation at $1$&quot; ($p(t)\longmapsto p(1)$) is a linear transformation. And the map $\mathbf{P}_3\to\mathbb{R}$ given by &quot;integrate over $[-1,1]$ ($p(t)\longmapsto \int_{-1}^1p(t)\,dt$) is &lt;em&gt;also&lt;/em&gt; a linear transformation. Call the first linear transformation $E$ and the second $T$. The polynomials that are in the nullspace/kernel of the first map are precisely the ones that satisfy $f(1) = 0$. The ones that are in the kernel of the second one are precisely the ones that satisfy $\int_{-1}^1f(t)\,dt = 0$. You want the ones that are in the kernel of &lt;em&gt;both&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The standard way of achieving this is to map to the product: take the map $L\colon\mathbf{P_3}\to \mathbb{R}\times\mathbb{R}=\mathbb{R}^2$, given by:&#xA;$$L(p(t)) = \Bigl( E(p),\; T(p)\Bigr) = \left(p(1),\quad\int_{-1}^1p(t)\,dt\right).$$&#xA;Then the subspace you want is &lt;em&gt;precisely&lt;/em&gt; the kernel of $L$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To find the kernel of $L$, one possibility is to pick a basis for $\mathbf{P}_3$ (say, the standard basis $[1,t,t^2,t^3]$), a basis for $\mathbb{R}^2$ (say, the standard basis $[(1,0), (0,1)]$), construct the matrix representation for $L$, and find the nullspace of that matrix. Interpreting this nullspace in terms of the basis $[1,t,t^2,t^3]$ will give you the kernel you want, and finding a basis for that nullspace will yield a basis for the subspace of $\mathbf{P}_3$ that you are looking at.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-20T07:06:44.533" LastActivityDate="2011-02-20T07:06:44.533" CommentCount="4" />
  <row Id="22901" PostTypeId="2" ParentId="22859" CreationDate="2011-02-20T06:54:53.690" Score="2" ViewCount="" Body="&lt;p&gt;For question (3), the idea is to take the Taylor expansion in order to see which type the critical point is. If the Taylor approximation is&#xA;$$ f(x + \epsilon) \approx f(x) + C \epsilon^n, \quad 0 \neq C = \frac{f^{(n)}}{n!} $$ then if $n$ is odd, it's not an optimum at all, whereas if it is even, it's a minimum if $C &amp;gt; 0$ and a maximum if $C &amp;lt; 0$ (it's enough to notice that $x^2$ has a minimum at $0$). In order for the Taylor approximation to work, you need the $n+1$ derivative of $f$ to be continuous, to ensure that the remainder term is continuous in some neighborhood of $x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In practice, functions are usually infinitely differentiable, so you shouldn't worry so much about the existence of derivatives.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For question (1), we do the same Taylor business and get an expansion&#xA;$$f(\vec{x} + \vec{\epsilon}) \approx f(\vec{x}) + \vec{\epsilon}&amp;#39; \nabla f(\vec{x}) + \frac{1}{2} \vec{\epsilon}&amp;#39; Hf(\vec{x}) \vec{\epsilon}, $$&#xA;where $\nabla f$ is the gradient and $Hf$ is the Hessian, the matrix of second derivatives. If $Hf(\vec{x})$ is positive definite with smallest eigenvalue $\lambda &amp;gt; 0$, then $$\vec{\epsilon}&amp;#39; Hf(\vec{x}) \vec{\epsilon} \geq \lambda \|\vec{\epsilon}\|^2, $$ and we get a minimum (since the error is $O(\|\vec{\epsilon}\|^3)$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some funky math allows you somehow to deduce all of this even without having the extra derivative required for having a bounded Taylor's remainder term, but I wouldn't worry too much about it.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2011-02-20T06:54:53.690" />
  <row Id="22902" PostTypeId="2" ParentId="20522" CreationDate="2011-02-20T07:07:25.593" Score="1" ViewCount="" Body="&lt;p&gt;If the property $\mbox{P}$  is expressible by an equivalent first-order formula $\varphi(x)$ of set theory in the free variable $x$, an exact (i.e. sufficient and necessary) condition for the existence of some set $A$ of all sets $x$, which satisfy $\mbox{P}$ in $\mbox{ZFC}$, i.e. for &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$\models_{ZFC} \exists A \forall x (x \in A \leftrightarrow \varphi(x))$,&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;is by some axiom of separation the condition &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$\models_{ZFC} \exists B \forall x (\varphi(x) \rightarrow x \in B)$. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I.e. you need  only by the completeness theorem of first-order logic to &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;express $\mbox{P}$ as an equivalent first-order formula $\varphi(x)$ of set theory, &lt;/li&gt;&#xA;&lt;li&gt;find some suitable set $B$, and &lt;/li&gt;&#xA;&lt;li&gt;prove $\vdash_{ZFC} \forall x (\varphi(x) \rightarrow x \in B)$ in your metatheory.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="7280" LastEditorUserId="7280" LastEditorDisplayName="" LastEditDate="2011-02-20T11:02:26.237" LastActivityDate="2011-02-20T11:02:26.237" CommentCount="2" />
  <row Id="22903" PostTypeId="2" ParentId="22775" CreationDate="2011-02-20T07:25:54.963" Score="0" ViewCount="" Body="&lt;p&gt;I used a good old sine implementation for this which solved my problem.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;public static double[] getProbabilities(int size) {&#xA;    double[] result = new double[size];&#xA;    double sum = 0;&#xA;    for (int i = 0; i &amp;lt; size; i++) {&#xA;        result[i] = Math.sin((i+1)/((double)size+1) * Math.PI);&#xA;        sum+=result[i];&#xA;    }&#xA;    for (int i = 0; i &amp;lt; size; i++) {&#xA;        result[i] /= sum;&#xA;    }&#xA;    return result;&#xA;}&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The first loop assigns values and also calculates the sum. The second loop normalizes all values so the end sum is 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For size 7 it produces:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;0.07612046748871325&#xA;0.140652283836026&#xA;0.18377106498543178&#xA;0.198912367379658&#xA;0.18377106498543178&#xA;0.140652283836026&#xA;0.07612046748871326&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Which is symetrical, adds up to 1 and looks like the graph I mentioned in my question.&lt;/p&gt;&#xA;" OwnerUserId="7249" LastActivityDate="2011-02-20T07:25:54.963" />
  <row Id="22904" PostTypeId="2" ParentId="22880" CreationDate="2011-02-20T07:26:47.527" Score="4" ViewCount="" Body="&lt;p&gt;As the other answer shows, the answer is not simple. This is because it depends on the exact sequence of rolls. You can see this using Bayesian reasoning. Let $X$ be a r.v. that indicates the biased side. Denote the value of the $i^\mathrm{th}$ role by $Y_i$. Then, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$p(Y_1, \cdots, Y_n|X=x) =  \left(\frac{1}{20}\right)^{m_x} \left(\frac{1}{100}\right)^{n-m_x}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$m_x = \sum_{i=1}^n \mathbf{1}\left(Y_i = x\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and $\mathbf{1}$ is the indicator function. Then, we can use Bayes rule to calculate the probability of $X=x$ given a sequence of rolls.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$p(X=x|Y_1, \cdots, Y_n) = \frac{p(Y_1, \cdots, Y_n|X=x)p(X=x)}{P(Y_1, \cdots, Y_n)} = \frac{p(Y_1, \cdots, Y_n|X=x)p(X=x)}{\sum_{x&amp;#39;} P(Y_1, \cdots, Y_n | X=x&amp;#39;)p(X=x&amp;#39;)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Substituting from above and assuming all sides are uniformly equal &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$=\frac{\left(\frac{1}{20}\right)^{m_x} \left(\frac{1}{100}\right)^{n-m_x} \frac{1}{96}}{\sum_{x&amp;#39;=1}^{96}\left[ \left(\frac{1}{20}\right)^{m_{x&amp;#39;}} \left(\frac{1}{100}\right)^{n-m_{x&amp;#39;}}\right] \frac{1}{96}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, basically, its hard to know a priori how many rolls because it depends on how many &quot;repeats&quot; on all sides comes up. But, you can easily compute the probability that $X=x$ where $x$ is any side you want and stop after enough rolls.&lt;/p&gt;&#xA;" OwnerUserId="5799" LastActivityDate="2011-02-20T07:26:47.527" CommentCount="1" />
  <row Id="22905" PostTypeId="1" AcceptedAnswerId="22906" CreationDate="2011-02-20T07:58:21.083" Score="4" ViewCount="67" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This problem came up in my conversation with a friend -- not sure how basic it is, but it seems quite interesting:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Prove that If $f_n$ where $n&amp;gt;3$ is prime, then $n$ is prime for a Fibonacci sequence where $f_1$=$f_2$=1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can this be tackled? Thanks a bunch for any ideas.&lt;/p&gt;&#xA;" OwnerUserId="3731" LastActivityDate="2011-02-20T08:02:09.577" Title="Prove that If $f_n$ where $n&gt;3$ is prime, then $n$ is prime for a Fibonacci series where $f_1$=$f_2$=1" Tags="&lt;number-theory&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="22906" PostTypeId="2" ParentId="22905" CreationDate="2011-02-20T08:02:09.577" Score="2" ViewCount="" Body="&lt;p&gt;The answer may be found at this link:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.dfcd.net/articles/fibprime.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.dfcd.net/articles/fibprime.pdf&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" LastActivityDate="2011-02-20T08:02:09.577" />
  <row Id="22907" PostTypeId="1" CreationDate="2011-02-20T08:14:32.970" Score="1" ViewCount="318" Body="&lt;p&gt;I want to solve transcendental equations by Mathematica 7. &lt;strong&gt;Solve&lt;/strong&gt; does not help me, &lt;strong&gt;reduce&lt;/strong&gt; - as well. Seams that I shoulkd use &lt;strong&gt;FindRoot&lt;/strong&gt;, but it does not give all solutions. It gives only one solution. So how can I solve transcendental equations?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is the equation I want to solve:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/rQedG.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So for a fixed value of L, A1, A2 (L &gt;= 0 and is an integer) parameters I want to find all values for k expressed with A1 and A2, for which the determinant is 0. In addition I must say that A1 and A2 &gt;0. &lt;/p&gt;&#xA;" OwnerUserId="7281" LastEditorUserId="7281" LastEditorDisplayName="" LastEditDate="2011-02-20T14:44:35.453" LastActivityDate="2011-02-20T20:17:55.437" Title="How to solve transcendental equations with Mathematica 7?" Tags="&lt;mathematica&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="22908" PostTypeId="2" ParentId="22892" CreationDate="2011-02-20T08:34:02.800" Score="0" ViewCount="" Body="&lt;p&gt;I'd suggest you to use vector formulas and at the end unfold them to X and Y coordinates. So, the function of your line is $r_0+t\cdot (r_1-r_0)$ with a scalar parameter $t$, where $r_0=(4,4)$ and $r_1=(6,7)$. Then boundaries of your rectangle comes into play. The top boundary has equation $y=5$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$r_{0y}+t\cdot (r_{1y}-r_{0y})=5$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$t=\frac{5-r_{0y}}{r_{1y}-r_{0y}}$ (found $t$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x$ of the intersection point = $r_{0x}+t\cdot (r_{1x}-r_{0x})$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For 4 boundaries you will have 4 intersection points. But your line obviously has 2 points of intersection with the rectangle. You chose intersection points based on the values of $t$. (Do not forget that every intersection point has a $t$ parameter.) Namely, trash 2 intersection points with the least $t$ and the greatest $t$. (I hope it is clear what I'm saying here.)&lt;/p&gt;&#xA;" OwnerUserId="7011" LastActivityDate="2011-02-20T08:34:02.800" />
  <row Id="22909" PostTypeId="2" ParentId="22769" CreationDate="2011-02-20T08:37:55.387" Score="0" ViewCount="" Body="&lt;p&gt;After some thought and since I didn't really get any satisfactory answers. This is what I make of the statement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The definition of differentiation in the statement is taken in the usual sense (the standard defintion from calculus, for some reason I thought the statement was referring to some new definition of differentiation!). So the statemtnt simply says, The set of points at which the function is not differentiable is a null set w.r.t. the lebesgue measure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the statement would have been clearer I think if the author had written &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f$ is differentiable almost everywhere with respect to the lebesgue measure ... I think thats right!!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So no new definition of differentiation, phew! just some confusing terminology. &lt;/p&gt;&#xA;" OwnerUserId="6386" LastActivityDate="2011-02-20T08:37:55.387" CommentCount="8" />
  <row Id="22910" PostTypeId="1" CreationDate="2011-02-20T08:49:01.497" Score="7" ViewCount="199" Body="&lt;p&gt;A series arising in perturbation theory in quantum mechanics:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sum_{m\neq n} \frac{1}{n^2 - m^2}$, where $n$ is a given positive odd integer and $m$ runs through all odd positive integers different from $n$. I have a hunch that residue methods are applicable here, but I don't know complex analysis.&lt;/p&gt;&#xA;" OwnerUserId="7282" LastEditorUserId="1301" LastEditorDisplayName="" LastEditDate="2011-02-20T10:49:27.817" LastActivityDate="2011-02-20T15:46:17.967" Title="Computing $\sum_{m \neq n} \frac{1}{n^2-m^2}$" Tags="&lt;sequences-and-series&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" />
  <row Id="22911" PostTypeId="2" ParentId="22861" CreationDate="2011-02-20T08:55:35.317" Score="5" ViewCount="" Body="&lt;p&gt;If “staring” means “I have no thoughts”, then you are probably stuck at some tough place. Mathematics was developed for centuries, do not expect of yourself to develop it by your own in minutes. Put in words what you are stuck at, ask people.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd found revealing to switch between textbooks on the same subject, as textbooks have different strengths and weaknesses. However, this method has its own disadvantage, because you should integrate knowledge from different books on your own. They may use different notations etc.&lt;/p&gt;&#xA;" OwnerUserId="7011" LastActivityDate="2011-02-20T08:55:35.317" CommunityOwnedDate="2011-02-20T08:55:35.317" />
  <row Id="22913" PostTypeId="1" CreationDate="2011-02-20T09:42:22.733" Score="0" ViewCount="107" Body="&lt;p&gt;What is combinatorial distance?&lt;/p&gt;&#xA;" OwnerUserId="7485" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2011-02-20T17:11:12.113" LastActivityDate="2011-02-20T17:11:12.113" Title="What is combinatorial distance?" Tags="&lt;combinatorics&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="22914" PostTypeId="2" ParentId="22910" CreationDate="2011-02-20T09:59:04.210" Score="2" ViewCount="" Body="&lt;p&gt;Use the cotangent indentity at the &lt;a href=&quot;http://mathworld.wolfram.com/Cotangent.html&quot; rel=&quot;nofollow&quot;&gt;very end of the page&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are several ways to derive this identity, some use residues, other the Euler product formula for the sine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: So we have the identity:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\pi z \cot (\pi z) - 1}{2z^2} = \sum_{m=1}^{\infty} \frac{1}{z^2-m^2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, we want to compute the sum $\sum_{n\neq m} \frac{1}{n^2-m^2}$, so we rewrite the identity as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{\substack{m=1, \\ n\neq m}}^{\infty} \frac{1}{z^2-m^2} = \frac{\pi z \cot (\pi z) - 1}{2z^2} - \frac{1}{z^2-n^2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And we take the limit for $z \to n$ for both sides to get the desired sum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT 2: After some work rearranging, I arrived at the following identity&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$-\frac{\pi \tan \left(\pi \frac{z}{2}\right)}{4z} = \sum_{n \; odd} \frac{1}{z^2-n^2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which can be used along the lines explained in the previous edit.&lt;/p&gt;&#xA;" OwnerUserId="3567" LastEditorUserId="3567" LastEditorDisplayName="" LastEditDate="2011-02-20T11:47:03.577" LastActivityDate="2011-02-20T11:47:03.577" CommentCount="3" />
  <row Id="22915" PostTypeId="2" ParentId="22910" CreationDate="2011-02-20T10:33:54.873" Score="13" ViewCount="" Body="&lt;p&gt;You can write&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \frac{1}{n^2 - m^2} = \frac{1}{2n}&#xD;&#xA;\left\lbrace \frac{1}{m+n} - \frac{1}{m-n} \right\rbrace . \quad (1)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now if we sum up both sides over all odd $m \ne n ,$ taking into account that $n$ is odd, lots of cancelling goes on and we obtain&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{m \ne n} \frac{1}{n^2 - m^2} = -\frac{1}{4n^2}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At first sight it appears the cotangent identity could be useful but it's not actually needed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a numerical check try summing the following with wolframAlpha&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$1/24 + 1/16 + \sum_{k=3}^\infty 1/(5^2 - (2k+1)^2),$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;you will see that it is $-1/100,$ as expected.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or try this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$1/48 + 1/40 + 1/24 + \sum_{k=4}^\infty 1/(7^2 - (2k+1)^2).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You will get $-1/196.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: To clarify the cancellation taking place when we sum the RHS of $(1).$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{m \ne n, \,\, m \textrm{ odd} }&#xD;&#xA;\left\lbrace \frac{1}{m+n} - \frac{1}{m-n} \right\rbrace&#xD;&#xA;&#xD;&#xA;= \sum_{m \ne n, \,\, m \textrm{ odd} }&#xD;&#xA;\left\lbrace \frac{1}{n+m} + \frac{1}{n-m} \right\rbrace $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$= \left\lbrace \left( \frac{1}{n+1} + \frac{1}{n-1} \right) +&#xD;&#xA;\left( \frac{1}{n+3} + \frac{1}{n-3} \right) +&#xD;&#xA;\left( \frac{1}{n+5} + \frac{1}{n-5} \right) + \cdots&#xD;&#xA;+ \left( \frac{1}{2n-2} + \frac{1}{2} \right) \right\rbrace $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$+ \left( \frac{1}{2n+2} - \frac{1}{2} \right)&#xD;&#xA;+ \left( \frac{1}{2n+4} - \frac{1}{4} \right)&#xD;&#xA;+ \left( \frac{1}{2n+6} - \frac{1}{6} \right) + \cdots $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and rearranging all the terms in the braces&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$= \left\lbrace \frac{1}{2} + \frac{1}{4} + \cdots + \frac{1}{2n-2} \right\rbrace&#xD;&#xA;+ \left( \frac{1}{2n+2} - \frac{1}{2} \right)&#xD;&#xA;+ \left( \frac{1}{2n+4} - \frac{1}{4} \right) + \cdots $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$=\left\lbrace \frac{1}{2} + \frac{1}{4} + \cdots + \frac{1}{2n-2} \right\rbrace&#xD;&#xA;- \left\lbrace \frac{1}{2} + \frac{1}{4} + \cdots + \frac{1}{2n} \right\rbrace&#xD;&#xA;= - \frac{1}{2n}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and hence the result.&lt;/p&gt;&#xA;" OwnerUserId="1301" LastEditorUserId="1301" LastEditorDisplayName="" LastEditDate="2011-02-20T15:46:17.967" LastActivityDate="2011-02-20T15:46:17.967" CommentCount="2" />
  <row Id="22916" PostTypeId="2" ParentId="22859" CreationDate="2011-02-20T10:46:51.907" Score="1" ViewCount="" Body="&lt;p&gt;Here's a somewhat handwaving answer to question (1):&#xA;To check whether $f(a,b)$ is a local minimum, you want to compare the value $f(a+h,b+k)$ to the value $f(a,b)$.&#xA;The $x$-derivatives at $(a,b)$ tell you something about how $f(a+h,b)$ compares to $f(a,b)$,&#xA;but, unless you assume continuity, the $y$-derivatives at $(a,b)$ don't tell you anything about how $f(a+h,b+k)$ compares to $f(a+h,b)$ (since that comparison involves the $y$-derivatives at the point $(a+h,b)$ rather than at $(a,b)$).&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2011-02-20T10:46:51.907" />
  <row Id="22917" PostTypeId="1" CreationDate="2011-02-20T10:49:16.023" Score="1" ViewCount="56" Body="&lt;p&gt;Let R be a ring. If a direct sum of R-modules is projective, can we imply that the direct sum is also a free module over R ?&lt;/p&gt;&#xA;" OwnerUserId="2177" LastActivityDate="2011-02-20T11:27:28.607" Title="A question about free module" Tags="&lt;modules&gt;" AnswerCount="1" />
  <row Id="22918" PostTypeId="2" ParentId="22913" CreationDate="2011-02-20T11:07:38.863" Score="2" ViewCount="" Body="&lt;p&gt;In a graph $G,$ I understand combinatorial distance to be the distance defined by the number of edges connecting two vertices. The distance between two vertices being the number of edges of the shortest path connecting them.&lt;/p&gt;&#xA;" OwnerUserId="1301" LastEditorUserId="1301" LastEditorDisplayName="" LastEditDate="2011-02-20T11:22:20.073" LastActivityDate="2011-02-20T11:22:20.073" />
  <row Id="22919" PostTypeId="2" ParentId="22899" CreationDate="2011-02-20T11:10:46.470" Score="1" ViewCount="" Body="&lt;p&gt;A basis of the linear subspace of polynomials $f$ of degree at most $3$ such that $f(1)=0$ is $(p_1,p_2,p_3)$ where&#xA;$$&#xD;&#xA;p_1(t)=2(1-t),\quad&#xD;&#xA;p_2(t)=3(1-t)^2,\quad&#xD;&#xA;p_3(t)=4(1-t)^3.&#xD;&#xA;$$&#xA;(Proof: Let $p_0(t)=1$. Then $(p_0,p_1,p_2,p_3)$ is a basis of the linear subspace of polynomials $f$ of degree at most $3$, as every family of polynomials with degree $0$, $1$, $2$ and $3$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the integral of $a_1p_1+a_2p_2+a_3p_3$ over $[-1,1]$ is zero iff $4a_1+8a_2+8a_3=0$ iff $a_1=-2a_2-4a_3$, hence a basis of the subspace the OP is interested in is $(2p_1-p_2,4p_1-p_3)$, or equivalently, $(q_2,q_3)$ with&#xA;$$&#xD;&#xA;q_2(t)=(1-t)(1+3t),\quad&#xD;&#xA;q_3(t)=(1-t)(1+2t-t^2).&#xD;&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="6179" LastEditorUserId="6179" LastEditorDisplayName="" LastEditDate="2011-02-20T11:27:54.447" LastActivityDate="2011-02-20T11:27:54.447" />
  <row Id="22920" PostTypeId="2" ParentId="22917" CreationDate="2011-02-20T11:27:28.607" Score="3" ViewCount="" Body="&lt;p&gt;I assume you're excluding trivial summands; otherwise the answer would trivially be &quot;no&quot;. But it's also &quot;no&quot; with non-trivial summands. Take for instance $\mathbb{Z}$ as a module over $\mathbb{Z}\oplus\mathbb{Z}$, with multiplication defined by $(a,b)\cdot c= a\cdot c$. This is a projective module, but its direct sum with itself is not a free module (which you can see e. g. by noting that many elements of $\mathbb{Z}\oplus\mathbb{Z}$ leave this module invariant whereas only the identity element would leave a free module invariant).&lt;/p&gt;&#xA;" OwnerUserId="6622" LastActivityDate="2011-02-20T11:27:28.607" />
  <row Id="22921" PostTypeId="2" ParentId="22871" CreationDate="2011-02-20T12:11:36.867" Score="2" ViewCount="" Body="&lt;p&gt;Since $$\int f_X^i(x) dx = 1 \forall i$$ and $$\int f_X^j(x) dx = 1 \forall j,$$ it is also required that $$\sum_i w_i -  \sum_j w_j = 1.$$ Let us call $\sum_i w_i = a$, so we know that $\sum w_j = a-1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is the algorithm:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Pick $i$ based on weights $w_i$.&lt;/li&gt;&#xA;&lt;li&gt;Sample y from $f_X^i(x)$.&lt;/li&gt;&#xA;&lt;li&gt;Calculate $$p_a = \frac{\sum_i w_i f^i_X(x)- \sum_j w_j f^j_X(x)}{\sum_i w_i f_X^i(x)}$$&lt;/li&gt;&#xA;&lt;li&gt;Sample $u$ from $Uniform(0,1)$.&lt;/li&gt;&#xA;&lt;li&gt;If $u &amp;lt; p_a$, then emit y, else reject this sample and go back to 1 to try another.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;This is called acceptance-rejection sampling. For more information see the &lt;a href=&quot;https://secure.wikimedia.org/wikipedia/en/wiki/Rejection_sampling&quot; rel=&quot;nofollow&quot;&gt;wikipedia article&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2532" LastEditorUserId="2532" LastEditorDisplayName="" LastEditDate="2011-02-21T02:07:11.527" LastActivityDate="2011-02-21T02:07:11.527" CommentCount="6" />
  <row Id="22922" PostTypeId="2" ParentId="12906" CreationDate="2011-02-20T12:23:24.557" Score="-2" ViewCount="" Body="&lt;p&gt;The fallacy with this approach is as follows: The perimeter of the circle is clearly less than the perimeter of the square (we accept this as fact because we know $c = \pi d$).  One would expect that the sequence of estimates of $\pi$ should converge (i.e. start at $4$ and gradually converge to the true value) as we have decreasing granularity in our estimates.  The fact that this sequence does not converge (or even move away from 4) should point out that we are not making any progress with subsequent steps in this approach.&lt;/p&gt;&#xA;" OwnerUserId="4726" LastEditorUserId="4726" LastEditorDisplayName="" LastEditDate="2011-05-27T23:23:33.890" LastActivityDate="2011-05-27T23:23:33.890" CommentCount="3" />
  <row Id="22923" PostTypeId="2" ParentId="20522" CreationDate="2011-02-20T12:45:18.657" Score="11" ViewCount="" Body="&lt;p&gt;In ZFC, the following are equivalent:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The class $\{x\mid P(x)\}$ is a set.&lt;/li&gt;&#xA;&lt;li&gt;There is an ordinal $\alpha$ such that whenever $P(x)$, then $x$ has rank at most $\alpha$. &lt;/li&gt;&#xA;&lt;li&gt;There is a set $B$ such that whenever $P(x)$, then $x\in B$.&lt;/li&gt;&#xA;&lt;li&gt;There is no class function $F$ from the class $\{x\mid P(x)\}$ onto the ordinals.&lt;/li&gt;&#xA;&lt;li&gt;There is an ordinal $\theta$ such that there is no function mapping $\{x\mid P(x)\}$ onto $\theta$. &lt;/li&gt;&#xA;&lt;li&gt;There is a set $C$ such that there is no function mapping $\{x\mid P(x)\}$ onto $C$.&lt;/li&gt;&#xA;&lt;li&gt;There is an ordinal $\theta$ that does not map injectively into $\{x\mid P(x)\}$. &lt;/li&gt;&#xA;&lt;li&gt;There is a set $D$ that does not map injectively into $\{x\mid P(x)\}$. &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Proof. (1 iff 2) If the class is a set, then it must be contained in some $V_\alpha$, and so every element will have rank at most $\alpha$. The converse is the Separation axiom.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2 iff 3) Use that every $V_\alpha$ is a set, and every set is contained in some $V_\alpha$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1 implies 4) The ordinals are not a set, so this follows by the Replacement axiom. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(4 implies 2) Map each $x$ for which $P(x)$ holds to its rank.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1 implies 5) For every set, there is an ordinal onto which it does not map, namely, the successor of its cardinality. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(5 iff 6) Every set is bijective with an ordinal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(5 implies 7) If a class does not map surjectively onto $\theta$, then $\theta$ cannot map injectively into the class.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(7 iff 8) Every set is bijective with an ordinal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(7 implies 2) If $\theta$ does not map injectively into $\{x\mid P(x)\}$, then that class cannot contain sets of arbitrarily large rank. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;QED&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Meanwhile, the following notions are strictly weaker in ZFC, if ZFC is consistent:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;There is a map from the ordinals onto $\{x \mid P(x)\}$. &lt;/li&gt;&#xA;&lt;li&gt;There is a bijection of $\{x\mid P(x)\}$ with the ordinals.&lt;/li&gt;&#xA;&lt;li&gt;There is a bijection of${x\mid P(x)}$ with $V$, the entire set-theoretic universe.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;There reason that they are weaker in general is that it is relatively consistent with ZFC that there is no definable (from parameters) well-ordering of the universe. In such a model $V$, there is no class surjection or bijection from the ordinals to $V$, since this would provide the desired well-ordering, but $V$ is not a set. Similarly, in such a model, there is no bijection from the class of ordinals to the entire universe, but the class of ordinals is not a set. Such a model can be constructed using the forcing technique, by an Easton support iteration that adds a Cohen subset to unboundedly many regular cardinals. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Addendum.&lt;/b&gt; Let me add that there can be no purely syntactic characterization of the properties $P$ for which $\{x\mid P(x)\}$ is a set. The reason is that some properties determine sets in some models of ZFC, but not in others. So the question of whether this class is a set depends not just on the syntactic features of $P$, but on the properties of the universe in which the class is to be formed. An example of this is the property $P(x)\iff CH\wedge x=x$, which determines a set just in case $\neg CH$. &lt;/p&gt;&#xA;" OwnerUserId="413" LastEditorUserId="413" LastEditorDisplayName="" LastEditDate="2011-02-20T15:05:46.347" LastActivityDate="2011-02-20T15:05:46.347" />
  <row Id="22924" PostTypeId="1" AcceptedAnswerId="22926" CreationDate="2011-02-20T12:49:15.857" Score="3" ViewCount="168" Body="&lt;p&gt;This is given in my module as a part of a problem's solution:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$m^4 + 4 = 0 $$ $$\Rightarrow m = 1 \pm i,-1\pm i$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not getting how this conversion is taking place,could somebody explain?&lt;/p&gt;&#xA;" OwnerUserId="2109" LastEditorUserId="536" LastEditorDisplayName="" LastEditDate="2011-02-20T23:27:23.270" LastActivityDate="2011-02-20T23:54:43.420" Title="How $m^4+4 = 0 \Rightarrow m = 1 \pm i,-1\pm i$?" Tags="&lt;complex-numbers&gt;" AnswerCount="4" CommentCount="2" />
  <row Id="22925" PostTypeId="2" ParentId="22753" CreationDate="2011-02-20T12:49:37.587" Score="2" ViewCount="" Body="&lt;p&gt;Matrices are complicated objects.  At first glance they are rectangular arrays of numbers with a complicated multiplication rule.  Diagonalization helps us reduce the the matrix multiplication operation to a sequence of simple steps which make sense.  In your case you are asking about orthogonal diagonalization, so I will limit my comments to that.  Note that I normally think about diagonalization as a factorization of the matrix $\mathbf A$ as &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathbf A = \mathbf P \mathbf D \mathbf P^T$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know that $\mathbf D$ contains the eigenvalues of $\mathbf A$ and $\mathbf P$ contains the corresponding eigenvectors.  Consider the multiplication $\mathbf{y=Ax}$.  We can perform the multiplication $\mathbf y = \mathbf P \mathbf D \mathbf P^T \mathbf x$ step-by-step as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First: $\mathbf x&amp;#39; = \mathbf P^T \mathbf x$.  This step projects $\mathbf x$ onto the eigenvectors because the eigenvectors are in the rows of $\mathbf P^T$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second: $\mathbf y&amp;#39; = \mathbf D \mathbf x&amp;#39;$.  This step stretches the resultant vector independently in each direction which corresponds to an eigenvector.  This is the key.  A matrix does a simple scaling (in general we may also shear or rotate) operation in independent directions (orthogonal case only!) in a particular basis or representation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Third:  $\mathbf y = \mathbf P \mathbf y&amp;#39;$.  We take our resulting stretched vector and linearly combine the eigenvectors to get back to our original space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in summary, diagonalization tells us that under matrix multiplication, as vector can be represented in a special basis under which the actual operation of the matrix is a simple diagonal matrix (the simplest possible) and then represented back in our original space.&lt;/p&gt;&#xA;" OwnerUserId="4726" LastActivityDate="2011-02-20T12:49:37.587" />
  <row Id="22926" PostTypeId="2" ParentId="22924" CreationDate="2011-02-20T13:06:53.727" Score="7" ViewCount="" Body="&lt;p&gt;One way to look at it is like this:&#xA;$$&#xD;&#xA;m^4+4=0\implies m^4=-4\implies m^2=\pm 2i.&#xD;&#xA;$$&#xA;But&#xA;$$&#xD;&#xA;m^2=2i\implies m=\pm\sqrt{2}\sqrt{i}.&#xD;&#xA;$$&#xA;Also,&#xA;$$&#xD;&#xA;m^2=-2i\implies m=\pm i\sqrt{2}\sqrt{i}.&#xD;&#xA;$$&#xA;Now you can use that fact that there are two square roots of $i$, &#xA;$$&#xD;&#xA;\frac{1+i}{\sqrt{2}}\ \text{and}\ \frac{1+i}{-\sqrt{2}}&#xD;&#xA;$$ &#xA;to simplify each possibility, and get the $4$ values of $m$ you mentioned.&lt;/p&gt;&#xA;" OwnerUserId="1583" LastActivityDate="2011-02-20T13:06:53.727" />
  <row Id="22927" PostTypeId="2" ParentId="22861" CreationDate="2011-02-20T13:24:53.290" Score="4" ViewCount="" Body="&lt;p&gt;Two points :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Work the proofs. Work them &lt;strong&gt;hard&lt;/strong&gt;. I said work, and not learn, for a specific reason. Many students just think if you are able to recite the proof, it's ok.&#xA;What you have to do is (in my opinion) :&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Understand which part of the proof is a &lt;em&gt;key idea&lt;/em&gt;. Not all statements are equally important.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Understand why there is this condition, and not another one. Aka find counter-examples. It's one of the first exercice you should do. It can be really hard for some theorem, but it is very instructive. As already stated, ask yourself questions !&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;2) Do exercices. A lot of them. And by a lot, if you are undergrad, I mean &lt;strong&gt;a lot&lt;/strong&gt;. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Never skip a correction even if you have the good results. You should always try to see differences between your answer and the book one. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Learn to be good in mental calculus. Yeah, it's annoying to work, but you have to do it.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Do not work only short exercises. Working problems helps to get the &quot;big picture&quot;, especially when you mix tools like analysis and algebra together.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If you are stuck in an exercise after some time, ask help, dont put it away. Not be able to do an exercise on your own is the way to make some progress. Even the best mathematicians ask help to their collaborators.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;But the most important point (third one) is to be &lt;strong&gt;patient&lt;/strong&gt;, and to &lt;strong&gt;enjoy&lt;/strong&gt; working. Math is not a quickly rewarding field, but it's worth it.&lt;/p&gt;&#xA;" OwnerUserId="6582" LastActivityDate="2011-02-20T13:24:53.290" CommunityOwnedDate="2011-02-20T13:24:53.290" CommentCount="2" />
  <row Id="22928" PostTypeId="1" CreationDate="2011-02-20T13:37:38.047" Score="1" ViewCount="271" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Find&#xA;  $$ \lim_{x \to \infty} \sqrt[3]{1 + x^{2} + x^{3}} -x$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7284" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2011-02-20T19:13:39.150" LastActivityDate="2011-02-20T21:20:00.100" Title="Limit of algebraic function avoiding l'Hopital's rule" Tags="&lt;analysis&gt;&lt;limit&gt;" AnswerCount="5" CommentCount="11" />
  <row Id="22929" PostTypeId="2" ParentId="22924" CreationDate="2011-02-20T13:40:38.363" Score="3" ViewCount="" Body="&lt;p&gt;You can also factor $x^4+4=(x^2-2x+2)(x^2+2x+2)$.&lt;/p&gt;&#xA;" OwnerUserId="589" LastActivityDate="2011-02-20T13:40:38.363" CommentCount="1" />
  <row Id="22930" PostTypeId="2" ParentId="22928" CreationDate="2011-02-20T13:46:44.330" Score="0" ViewCount="" Body="&lt;p&gt;For such problems, quite often it pays off to use the identity&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a^b = \exp[ b * \log(a)]$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;because then one can apply all his/her knowledge about exp and log.&lt;/p&gt;&#xA;" OwnerUserId="7266" LastActivityDate="2011-02-20T13:46:44.330" CommentCount="2" />
  <row Id="22932" PostTypeId="2" ParentId="22907" CreationDate="2011-02-20T13:51:56.037" Score="2" ViewCount="" Body="&lt;p&gt;Finding all solutions of a general transcendental equation is a nontrivial task. As you have more than one equation this even looks hopeless ;-) So unless you know more about the equations and the allowed range for the variables (real, complex, from a compact set, ...), Mathematica (and this forum) cannot help.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For your specific problem, Mathematica can help. Bessel functions of half integer arguments can be reduced to trigonometric functions. Mathematica will help you with this task if you write 1/2 instead of 0.5. So for L=0, you obtain that the determinant is proportional to&#xA; $\sin[ (A_1 -A_2)k]$&#xA;and therefore the solutions are at $k= n \pi/(A_1-A_2)$ with $n\in \mathbb{Z}$. For $L=1,2,…$ the busyness gets more and more tricky, but numerical solutions are always obtainable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;E.g., for $L=1$ the determinant is zero whenever $(1+A_1 A_2 k^2) \sin[(A_1-A_2)k] - (A_1 -A_2) k \cos[(A_1-A_2)k]=0$. There is a trivial solution for $k=0$. For $|k| \geq (A_1 -A_2)/A_1 A_2$ this equation has no solutions. So you only need to find the solutions in a finite interval (except for $A_1 = A_2$ where the determinant is always zero). For this FindRoot helps.&lt;/p&gt;&#xA;" OwnerUserId="7266" LastEditorUserId="7266" LastEditorDisplayName="" LastEditDate="2011-02-20T18:13:44.420" LastActivityDate="2011-02-20T18:13:44.420" />
  <row Id="22934" PostTypeId="2" ParentId="22924" CreationDate="2011-02-20T14:15:25.660" Score="4" ViewCount="" Body="&lt;p&gt;Here $m = (-4)^{1/4}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now $-4 = 4(\cos(\theta) + i\sin(\theta))$ in polar form where $\theta = (2n-1)\pi$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By de Moivre's theorem&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$m = (-4)^{1/4} = 4^{1/4}(\cos(\frac{\theta}{4}) + i\sin(\frac{\theta}{4}))$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $n = 0$, $\displaystyle m = \sqrt{2}\left(\frac{1}{\sqrt{2}} - \frac{i}{\sqrt{2}}\right) = 1 - i$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $n = 1$, $\displaystyle m = \sqrt{2}\left(\frac{1}{\sqrt{2}} + \frac{i}{\sqrt{2}}\right) = 1 + i$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $n = 2$, $\displaystyle m = \sqrt{2}\left(-\frac{1}{\sqrt{2}} + \frac{i}{\sqrt{2}}\right) = -1 + i$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $n = 3$, $\displaystyle m = \sqrt{2}\left(-\frac{1}{\sqrt{2}} - \frac{i}{\sqrt{2}}\right) = -1 - i$&lt;/p&gt;&#xA;" OwnerUserId="5049" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-20T23:54:43.420" LastActivityDate="2011-02-20T23:54:43.420" />
  <row Id="22935" PostTypeId="2" ParentId="22928" CreationDate="2011-02-20T14:24:32.800" Score="4" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;hint&lt;/strong&gt;: you can rewrite the terms inside the limit as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ x \cdot \left( \sqrt[3]{\frac{1}{x^3} + \frac{1}{x} + 1} - 1\right) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for the term underneath the cube-root, use the Taylor expansion of the cube-root function near the value 1:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \sqrt[3]{1 + y} = 1 + \frac{1}{3}y - \frac{1}{9} y^2 + \ldots $$&lt;/p&gt;&#xA;" OwnerUserId="1543" LastActivityDate="2011-02-20T14:24:32.800" CommentCount="4" />
  <row Id="22937" PostTypeId="1" AcceptedAnswerId="22941" CreationDate="2011-02-20T14:32:46.270" Score="2" ViewCount="161" Body="&lt;p&gt;Hi there,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In reading &lt;a href=&quot;http://news.discovery.com/space/milky-way-stuffed-with-50-billion-alien-worlds.html&quot; rel=&quot;nofollow&quot;&gt;this article&lt;/a&gt; about updated estimates for the number of exoplanets in the Milky Way, I am curious how to get an estimate of the mean distance between them.  The Milky Way is ~50,000 light years in radius and an average of ~1000 ly in thickness, and the article estimates there to be ~500 million exoplanets in the habitable zone around their stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To simplify this greatly (grotesquely!), I think the galaxy can be thought of as a plane, since I want to know the method of doing this for a disc anyway, and there are so many estimates here, it shouldn't affect the results greatly.  So, the area of this disc would be:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\pi (50000ly)^2 \approx 7853981634 ly^2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, beyond this, to find an even distribution of 500,000,000 points, I'm not really sure how to go about that.  Given this, I can later apply it to an oblate disk, but I'd like to know the general method of evenly distributing points in a circle, and possibly finding the distance between them if it's not straightforward.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you in advance.&lt;/p&gt;&#xA;" OwnerUserId="7285" LastEditorUserId="7285" LastEditorDisplayName="" LastEditDate="2011-02-20T16:46:15.033" LastActivityDate="2011-02-20T16:46:15.033" Title="Finding the mean distance between n points evenly distributed in a disc of radius r" Tags="&lt;geometry&gt;&lt;probability-distributions&gt;" AnswerCount="2" CommentCount="5" />
  <row Id="22939" PostTypeId="1" CreationDate="2011-02-20T16:00:11.567" Score="2" ViewCount="86" Body="&lt;p&gt;Let $R$ be a commutative ring with identity. Let $I$ be an ideal of $R$. Suppose, we give a topology on $R$ where a set is open if and only if it is a union of cosets of powers of $I$. Then, is $R$ a topological ring?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: The question has been edited in the light of comments below.&lt;/p&gt;&#xA;" OwnerUserId="7288" LastEditorUserId="7288" LastEditorDisplayName="" LastEditDate="2011-02-20T21:42:56.437" LastActivityDate="2011-02-20T23:53:06.207" Title="For a ring R with an ideal I, the I-adic topology makes R into a topological ring" Tags="&lt;topology&gt;&lt;commutative-algebra&gt;" AnswerCount="2" CommentCount="6" />
  <row Id="22940" PostTypeId="2" ParentId="22937" CreationDate="2011-02-20T16:21:31.693" Score="1" ViewCount="" Body="&lt;p&gt;It seems a rough approximation is useful here, as the distance from one planet to the nearest will vary considerably.  I would just calculate the volume of interest (taking account of the bulge and arms as best you can), then divide by the number of planets to get the volume each one &quot;occupies&quot;.  The expected distance will then be something like 1-2 times the radius of a sphere of this volume.  Very rough, but doable.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-02-20T16:21:31.693" CommentCount="4" />
  <row Id="22941" PostTypeId="2" ParentId="22937" CreationDate="2011-02-20T16:25:35.050" Score="4" ViewCount="" Body="&lt;p&gt;Insofar as the number of points is practically infinite it is enough to calculate the average distance $a(D)$ of two independently chosen and uniformly distributed points in the unit disc $D$. The result is $$a(D)= {128\over 45\pi}.$$ I found this on page 1 of the following document:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.math.uni-muenster.de/reine/u/burgstal/d18.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.math.uni-muenster.de/reine/u/burgstal/d18.pdf&lt;/a&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;where also references are given.&lt;/p&gt;&#xA;" OwnerUserId="1303" LastActivityDate="2011-02-20T16:25:35.050" CommentCount="1" />
  <row Id="22942" PostTypeId="2" ParentId="22939" CreationDate="2011-02-20T16:27:42.357" Score="1" ViewCount="" Body="&lt;p&gt;That cannot work: the topology you want has as a basis the set of all translations of all powers of I, while not all sets you describe in the first paragraph are unions of such things.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, suppose $I$ is such that $I^2=0$. Then the open sets are just the cosets of $I$ in $R$. But there are sets which &lt;em&gt;contain&lt;/em&gt; a coset but are not a union of cosets. We can construct a concrete example as follows: Let $k$ be a field, and let $R=k[x]/(x^2)$. Let $\varepsilon$ be the image of $x\in k[x]$ in the quotient $R$. Then $I=(\varepsilon)$ is an ideal which squares to zero. And $I\cup(k\setminus\{0\})$ is a set which contains a power of $I$, but it is not a union of cosets of $I$.&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2011-02-20T16:27:42.357" CommentCount="8" />
  <row Id="22943" PostTypeId="2" ParentId="22928" CreationDate="2011-02-20T16:33:16.987" Score="6" ViewCount="" Body="&lt;p&gt;Here is a more elementary way using the difference of cubes identity.  (It is not as elegant as the Taylor series presented by Willie Wong, but requires less background)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $x=\sqrt[3]{x^{3}}$, we are looking at $\lim_{x\rightarrow\infty}\sqrt[3]{1+x^{2}+x^{3}}-\sqrt[3]{x^{3}}$.Recall that the cubic identity $a^{3}-b^{3}=\left(a-b\right)\left(a^{2}+ab+b^{2}\right)$, which tells us that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left(\sqrt[3]{1+x^{2}+x^{3}}\right)^{3}-\left(\sqrt[3]{x^{3}}\right)^{3}$$&#xA;$$=\left(\sqrt[3]{1+x^{2}+x^{3}}-\sqrt[3]{x^{3}}\right)\cdot \left(\left(1+x^{2}+x^{3}\right)^{\frac{2}{3}}+\sqrt[3]{x^{3}}\cdot\left(1+x^{2}+x^{3}\right)^{\frac{1}{3}}+\left(x^{3}\right)^{\frac{2}{3}}\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and hence &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{1+x^{2}}{\left(1+x^{2}+x^{3}\right)^{\frac{2}{3}}+x\cdot\left(1+x^{2}+x^{3}\right)^{\frac{1}{3}}+x^{2}}=\sqrt[3]{1+x^{2}+x^{3}}-\sqrt[3]{x^{3}}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Divide the top and bottom of the left hand side by $x^{2}$ to find $$\frac{1+\frac{1}{x^{2}}}{\left(1+\frac{1}{x^{2}}+\frac{1}{x^{3}}\right)^{\frac{2}{3}}+\left(1+\frac{1}{x^{2}}+\frac{1}{x^{3}}\right)^{\frac{1}{3}}+1}=\sqrt[3]{1+x^{2}+x^{3}}-\sqrt[3]{x^{3}}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $$\lim_{x\rightarrow\infty}1+\frac{1}{x^{2}}=1$$ and $$\lim_{x\rightarrow\infty}\left(1+\frac{1}{x^{2}}+\frac{1}{x^{3}}\right)^{\frac{2}{3}}+\left(1+\frac{1}{x^{2}}+\frac{1}{x^{3}}\right)^{\frac{1}{3}}+1=3$$ &#xA;we see by the quotient rule for limits $$\lim_{x\rightarrow\infty}\sqrt[3]{1+x^{2}+x^{3}}-\sqrt[3]{x^{3}}=\frac{1}{3}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope that helps,&lt;/p&gt;&#xA;" OwnerUserId="6075" LastActivityDate="2011-02-20T16:33:16.987" CommentCount="2" />
  <row Id="22944" PostTypeId="1" AcceptedAnswerId="22946" CreationDate="2011-02-20T16:50:52.933" Score="1" ViewCount="73" Body="&lt;p&gt;Suppose $w \in L^{1}(G)$ and $w \geq 0$. The geometric mean of $w$ is defined by $$ \Delta(w) = \exp \int_{G} \log w(x) \ dx$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\Delta(w) = 0$ if the integral is $-\infty$. What is the intuition behind this definition? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Source: &lt;em&gt;Fourier Analysis on Groups&lt;/em&gt; by Walter Rudin&lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2011-02-20T17:07:17.653" Title="Intuition Behind Geometric Means" Tags="&lt;intuition&gt;" AnswerCount="1" />
  <row Id="22945" PostTypeId="1" CreationDate="2011-02-20T16:56:41.707" Score="5" ViewCount="151" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Snell%27s_law&quot; rel=&quot;nofollow&quot;&gt;Snell's law&lt;/a&gt; from geometrical optics states that the ratio of the angles of incidence $\theta_1$ and of the angle of refraction $\theta_2$ as shown in figure1, is the same as the opposite ratio of the indices of refraction $n_1$ and $n_2$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;\frac{\sin\theta_1}{\sin \theta_2} = \frac{n_2}{n_1}&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/fYyJb.png&quot; alt=&quot;figure 1&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(figure originally from &lt;a href=&quot;http://upload.wikimedia.org/wikipedia/commons/thumb/3/3f/Snells_law2.svg/220px-Snells_law2.svg.png&quot; rel=&quot;nofollow&quot;&gt;wikimedia&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let $P$ be a point in one medium (with refraction index $n_1$) and $Q$ a point in the other one as in the figure. My question is, is there is a nice geometrical construction (at best using only ruler and compass) to find the point $O$ in the figure such that Snell's law is satisfied. (Suppose you know the interface and $n_2/n_1$)?&lt;/p&gt;&#xA;" OwnerUserId="3445" LastEditorUserId="3445" LastEditorDisplayName="" LastEditDate="2011-02-20T17:35:26.760" LastActivityDate="2011-02-20T20:39:36.933" Title="Geometrical construction for Snell's law?" Tags="&lt;euclidean-geometry&gt;&lt;physics&gt;&lt;mathematical-modeling&gt;&lt;mathematical-physics&gt;&lt;elementary-geometry&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="2" />
  <row Id="22946" PostTypeId="2" ParentId="22944" CreationDate="2011-02-20T17:01:02.100" Score="4" ViewCount="" Body="&lt;p&gt;Isn't this an extension of the geometric mean in discrete setting?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $a$ is the geometric mean of $a_1,a_2,a_3,\ldots,a_n$ then $$a=\sqrt[n]{a_1 \times a_2 \times \cdots \times a_n}$$ which implies $$\log(a) = \frac{\log(a_1) + \log(a_2) + \cdots + \log(a_n)}{n}$$ i.e. $\log(a)$ is the arithmetic mean of $\log(a_i)$'s.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When we talk of average in continuous sense, we replace the summation by integrals and hence&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle \log(\Delta(w)) = \frac{\int_G \log(w(x)) dx}{\int_G dx}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Delta(w) = \exp \left(\frac{\int_G \log(w(x)) dx}{\int_G dx} \right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and if $l \rightarrow -\infty$, then $\displaystyle \lim_{l \rightarrow -\infty} e^l = 0$ and hence $\Delta(w) = 0$ if the integral is $-\infty$&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2011-02-20T17:07:17.653" LastActivityDate="2011-02-20T17:07:17.653" />
  <row Id="22947" PostTypeId="2" ParentId="22913" CreationDate="2011-02-20T17:09:07.323" Score="1" ViewCount="" Body="&lt;p&gt;It is possible to find a &quot;distance&quot; - function which obeys the same abstract properties that Euclidean distance obeys, see  &lt;a href=&quot;http://en.wikipedia.org/wiki/Metric_%28mathematics%29&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Metric_%28mathematics%29&lt;/a&gt; between combinatorial objects of different kinds: trees, strings, graphs, etc. There are many such distances but a good place to get started for this circle of ideas is with edit distance and Levenshtein distance:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Edit_distance&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Edit_distance&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Levenshtein_distance&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Levenshtein_distance&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1369" LastActivityDate="2011-02-20T17:09:07.323" />
  <row Id="22948" PostTypeId="1" AcceptedAnswerId="22951" CreationDate="2011-02-20T17:35:59.433" Score="2" ViewCount="63" Body="&lt;p&gt;Define the rank of a group $G$ as the minimal size of a subset $S$ such that $S$ generates $G$. I want to prove that, given $G$ and $H$ two groups, if there exists a surjective homomorphism $f$ between $G$ and $H$ then rank$(G)$ $\geq$ rank$(H)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I thought of this: let $S$ be a minimal generating set of $G$. Let us consider $f(S)$. Then clearly $\vert S \vert \geq \vert f(S) \vert$. It would then suffice to prove that $f(S)$ generates $H$. However, I see no good reason for this... Does anyone see a way to prove it?&lt;/p&gt;&#xA;" OwnerUserId="5465" LastEditorUserId="5465" LastEditorDisplayName="" LastEditDate="2011-02-20T17:45:26.947" LastActivityDate="2011-02-20T17:45:26.947" Title="$G \twoheadrightarrow H \Rightarrow$ rank$(G) \geq$ rank$(H)$" Tags="&lt;group-theory&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="22949" PostTypeId="1" AcceptedAnswerId="23002" CreationDate="2011-02-20T17:37:45.883" Score="0" ViewCount="94" Body="&lt;p&gt;Given an integer N and a smooth base B; what is the (approximate) probability that N is completely divisible by primes &amp;lt;= B.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I assume there is some nice connection to the de Bruijn or Dickman function but I can't see it.&lt;/p&gt;&#xA;" OwnerUserId="7290" LastActivityDate="2011-02-21T00:42:46.260" Title="Probability of being B smooth" Tags="&lt;number-theory&gt;&lt;probability&gt;" AnswerCount="1" CommentCount="9" />
  <row Id="22950" PostTypeId="2" ParentId="22945" CreationDate="2011-02-20T17:38:35.653" Score="0" ViewCount="" Body="&lt;p&gt;Yes. I guess there is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Draw perpendicular from point P to interface say PP'.&#xA;Draw perpendicular from point Q to interface say QQ'.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Divide segment P'Q' in ratio n_2/n_1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;O is the point that divides P'Q' in n_2/n_1.&lt;/p&gt;&#xA;" OwnerUserId="705" LastActivityDate="2011-02-20T17:38:35.653" CommentCount="1" />
  <row Id="22951" PostTypeId="2" ParentId="22948" CreationDate="2011-02-20T17:41:20.247" Score="1" ViewCount="" Body="&lt;p&gt;It is true that $H$ is generated by $f(S)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $h\in H$. Then there exists some $g\in G$ such that $h = f(g)$, because $f$ is a surjection. Now write $g = s_1 s_2 \dots s_\ell$ with $s_i \in S\cup S^{-1}$. Now apply $f$ to both sides of the equation and we find that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ h = f(g) = f(s_1 \dots s_\ell) = f(s_1)f(s_2) \dots f(s_\ell),$$&#xA;because $f$ is a homomorphism. Also note that $f(x^{-1})=f(x)^{-1}$, therefore indeed $h$ is generated by $f(S)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that $|S| \geq |f(S)|$, because $f$ is a surjection, but they might not be equal.&lt;/p&gt;&#xA;" OwnerUserId="5189" LastActivityDate="2011-02-20T17:41:20.247" CommentCount="1" />
  <row Id="22952" PostTypeId="2" ParentId="22945" CreationDate="2011-02-20T18:05:09.823" Score="2" ViewCount="" Body="&lt;p&gt;If you know the interface, then drop perpendiculars from $P$ and $Q$ to the interface. Let the points of intersection be $P&amp;#39;$ and $Q&amp;#39;$. Let $PP&amp;#39; = y_P$ and $QQ&amp;#39; = y_Q$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now consider the line segment $P&amp;#39;Q&amp;#39;=x$. You need to find a point $O$ inside $P&amp;#39;Q&amp;#39;$ such that $OP&amp;#39; + OQ&amp;#39; = x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $OP&amp;#39; = x_P$ and $OQ&amp;#39; = x_Q$.&#xA;&lt;img src=&quot;http://i.imgur.com/84Swl.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We now have two equations to solve for $\theta_1$ and $\theta_2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x_P + x_Q = x$ i.e. $$y_P \tan(\theta_1) + y_Q \tan(\theta_2) = x$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\sin(\theta_1)}{\sin(\theta_2)} = \frac{n_2}{n_1}$$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the problem is well-defined and hence solving for $\theta_1$ and $\theta_2$ gives $x_1$ and $x_2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I shall post the geometric construction later.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2011-02-20T20:39:36.933" LastActivityDate="2011-02-20T20:39:36.933" CommentCount="1" />
  <row Id="22953" PostTypeId="2" ParentId="22928" CreationDate="2011-02-20T18:16:08.977" Score="6" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\ $ It's simply a first derivative: $\: $ changing variables $\rm\ x\to 1/x\ $ transforms it to &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\rm\displaystyle\ \lim_{x\to\ 0^{+}}\ \frac{f(x)-f(0)}x\ =\ f\:&amp;#39;(0) \ \ \ for\ \ \ f(x) = (1+x+x^3)^{1/3}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now it is easy to calculate $\rm\ f\:&amp;#39;(0)\ =\ 1/3\ $ by direct evaluation (it's not indeterminate). Namely&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\rm f\:&amp;#39;(x)\ =\ \frac{d}{dx}\ (1+x+x^3)^{1/3}\ =\ \frac{1+3\ x^2}{3\ (1+x+x^3)^{2/3}}\ \ \Rightarrow\ \ f\:&amp;#39;(0)\ =\  \frac{1}3$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that this method employs only knowledge of the &lt;em&gt;definition&lt;/em&gt; of the derivative and basic &lt;em&gt;rules&lt;/em&gt; for calculating derivatives of polynomial and powers. It does not employ more advanced techniques such as L'Hospital's rule, or (binomial) power series expansions, etc.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2011-02-20T21:20:00.100" LastActivityDate="2011-02-20T21:20:00.100" CommentCount="3" />
  <row Id="22954" PostTypeId="1" CreationDate="2011-02-20T18:16:18.867" Score="3" ViewCount="384" Body="&lt;p&gt;I am to prove that being $x$ a string and $|x|$ its length, one should have the following property hold true for any two strings $x$ and $y$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ |xy| = |x| + |y| $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with $x, y \in \Sigma^*$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To prove this, I am expected to make use of the following two definitions:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;for $x=\epsilon$ we have $|x|=0$&lt;/li&gt;&#xA;&lt;li&gt;for $x=au$, being $u \in \Sigma^*$ and $a \in \Sigma$ we have $|x|=1+|u|$&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Intuitively it is easy to understand what is being asked here. The idea is to prove that the length of the concatenation of any two strings is equal to the sum of their individual lenghts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I am failing to realise is how to tackle the problem. Should I make use of induction? Just algebraic manipulation?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any pointers would be welcome.&lt;/p&gt;&#xA;" OwnerUserId="148" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2011-03-26T15:26:38.180" LastActivityDate="2011-03-27T08:47:12.767" Title="Proving that $|xy| = |x| + |y|$ being $x$ and $y$ two strings" Tags="&lt;combinatorics&gt;&lt;computer-science&gt;&lt;induction&gt;" AnswerCount="3" CommentCount="17" />
  <row Id="22955" PostTypeId="1" CreationDate="2011-02-20T18:21:52.433" Score="0" ViewCount="108" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;If $X$ is inductive, then the set $\{x\in X\mid x\text{ is transitive}\}$ is inductive. Hence&#xA;every $n\in\mathbb{N}$ is transitive.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If $X$ is inductive, then the set $\{x\in X\mid x\text{ is transitive and }x\notin x\}$ is inductive. Hence, $n\notin n$ and $n\neq n+1$ for each $n\in\mathbb{N}$. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="7224" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-20T20:26:53.570" LastActivityDate="2011-02-20T20:39:47.827" Title="Some set theory, inductive sets" Tags="&lt;elementary-set-theory&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="22956" PostTypeId="2" ParentId="22263" CreationDate="2011-02-20T18:25:46.017" Score="1" ViewCount="" Body="&lt;p&gt;If I read that right, you're just looking at the sheaf of sections of E-&gt;B, with E, B sets, B with topology P(B) [If by P(B) you meant the power set, the discrete topology].  I don't think these examples are particularly popular, but you can treat them like any other sheaves of sets (see, perhaps, MacLane-Moerdijk, Sheaves in Geometry and Logic).&lt;/p&gt;&#xA;" OwnerUserId="7292" LastActivityDate="2011-02-20T18:25:46.017" />
  <row Id="22957" PostTypeId="1" CreationDate="2011-02-20T18:26:25.877" Score="0" ViewCount="81" Body="&lt;p&gt;What are the open sets in a topology specified by a fundamental system of neighbourhoods of $0$ of a topolgical group? Also, how is this topology unique. I searched this online, but the books I found only mention these statements, without specifying what the open sets are and why the topology is unique. Any help will be appreciated.&lt;/p&gt;&#xA;" OwnerUserId="7288" LastActivityDate="2011-03-22T18:33:55.370" Title="Topology defined by a fundamental system of neighbourhoods of zero in a topological group" Tags="&lt;topology&gt;" AnswerCount="1" />
  <row Id="22958" PostTypeId="2" ParentId="22928" CreationDate="2011-02-20T18:28:24.360" Score="3" ViewCount="" Body="&lt;p&gt;A very similar question was asked recently &lt;a href=&quot;http://math.stackexchange.com/questions/22093/limit-involving-as-x-to-infty-of-sqrt5x5-3x4-17-x/22104#22104&quot;&gt;here&lt;/a&gt;...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The limit follows immediately upon showing, using the mean value theorem, that&#xA;$$&#xD;&#xA;\sqrt[3]{{\bigg(x + \frac{1}{3}\bigg)^3 }} - \sqrt[3]{{\bigg(x + \frac{1}{3}\bigg)^3  - \bigg(\frac{x}{3} - \frac{{26}}{{27}}\bigg)}} \to 0&#xD;&#xA;$$&#xA;as $x \to \infty$.&lt;/p&gt;&#xA;" OwnerUserId="2810" LastActivityDate="2011-02-20T18:28:24.360" CommentCount="3" />
  <row Id="22959" PostTypeId="1" AcceptedAnswerId="22976" CreationDate="2011-02-20T18:29:05.457" Score="1" ViewCount="126" Body="&lt;p&gt;Hi!&#xA;I am using quaternions to rotate my camera.&#xA;I only rotating the camera around it's x and y axises. &#xA;But for some reason, after some rotation, y axis also gets rotated(looks like around z axis).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I could avoid the problem with fixing y to (0,1,0), but I am curious where I am wrong.&lt;/p&gt;&#xA;" OwnerUserId="7217" LastActivityDate="2011-02-21T02:43:37.873" Title="Problem with my camera rotation" Tags="&lt;quaternions&gt;" AnswerCount="2" />
  <row Id="22960" PostTypeId="2" ParentId="22957" CreationDate="2011-02-20T18:32:23.767" Score="1" ViewCount="" Body="&lt;p&gt;Basically, because to define a topology we need to define a local base at every point in a consistent way. In a topological group $G$, for every $x$ and $y$ we have a homeomorphism of $G$ that maps $x$ to $y$, just use $h(z) = y*x^{-1}*z$. So we can transport a neighbourhood base of $e$ (the identity of $G$) to any other point of $G$, using such $h$. One then checks this is a consistent assignment and so determines the topology on $G$.&lt;/p&gt;&#xA;" OwnerUserId="4280" LastActivityDate="2011-02-20T18:32:23.767" CommentCount="2" />
  <row Id="22961" PostTypeId="2" ParentId="22686" CreationDate="2011-02-20T18:40:45.220" Score="4" ViewCount="" Body="&lt;p&gt;The beauty of straightedge and compass constructions, as opposed to the use of, say, a protractor, is that you don't measure anything. With ruler and compass you can bisect an angle without knowing its size, whereas with a protractor, you would have to measure the angle and then calculate the result.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, the point of this form of geometry is that it can be done independently of calculations and numbers. I think this is an important idea to teach: mathematics is not about numbers, but about objects adhering to certain rules (axioms).&lt;/p&gt;&#xA;" OwnerUserId="434" LastActivityDate="2011-02-20T18:40:45.220" CommentCount="1" />
  <row Id="22962" PostTypeId="2" ParentId="22924" CreationDate="2011-02-20T18:40:45.500" Score="3" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\rm\ -4 = m^4\ \Rightarrow\ \pm 2\ i = m^2 = (a+b\ i)^2 = a^2-b^2 + 2\:a\:b\ i\ \Rightarrow\ |a| = |b| = 1\:$.&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2011-02-20T18:40:45.500" />
  <row Id="22963" PostTypeId="2" ParentId="22093" CreationDate="2011-02-20T18:49:49.600" Score="4" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\ $ It's simply a first derivative: $\: $ changing variables $\rm\ x\to 1/x\ $ yields &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\rm\displaystyle\ \lim_{x\to\ 0^{+}}\ \frac{f(x)-f(0)}x \ \ \ for\ \ \ f(x)\ =\ (17\ x^5-3\ x+1)^{1/5}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now it is very easy to calculate $\rm\ f\:&amp;#39;(0)\ =\: -3/5\:$.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2011-02-20T18:59:04.093" LastActivityDate="2011-02-20T18:59:04.093" />
  <row Id="22964" PostTypeId="2" ParentId="8750" CreationDate="2011-02-20T18:55:57.090" Score="3" ViewCount="" Body="&lt;p&gt;EDIT: It has been brought to my attention by the author of &lt;a href=&quot;http://whistleralley.com/reuleaux/reuleaux.htm&quot; rel=&quot;nofollow&quot;&gt;this webpage&lt;/a&gt; that the solution presented below is incomplete/incorrect. It provides a formula for the height of the road as a function of $\theta$. However, to be useful, we need a formula for the height as a function of &lt;em&gt;horizontal displacement&lt;/em&gt; (e.g.: we want $h = f(x)$, not $h = f(\theta)$). To easily see why the solution below is insufficient, note that the curve below spans a (horizontal) distance of $\pi \over 3$, which means the arc length of the curve itself must be larger. Yet, the arc length of one segment of the Reuleaux triangle is exactly $\pi \over 3$. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I'll provide a solution for the triangle. It is easily generalizable to other Reuleaux polygons. From &lt;a href=&quot;http://www.geogebra.org/en/upload/files/piman/WheelsRoads/reuleaux_road.html&quot; rel=&quot;nofollow&quot;&gt;this javascript demo&lt;/a&gt; and &lt;a href=&quot;http://whistleralley.com/reuleaux/reuleaux.htm&quot; rel=&quot;nofollow&quot;&gt;this webpage&lt;/a&gt;, we see that a road amenable to Reuleaux wheels consists of a series of bumps. Our goal here is to provide a formula for one of those bumps.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/OXzD9.png&quot; alt=&quot;Reuleaux diagram&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $ABC$ be an equilateral triangle of side length 1. Let $D$ be the centroid of $ABC$. Let $AEFB$ be a circular arc centered on $C$ and passing through $A$ and $B$, where $CE$ is chosen to bisect $AB$ and $F$ is an arbitrarily chosen point on the arc. Finally, let $G$ be a point on $CE$ so that $FG$ is perpendicular to $CE$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The idea is to measure the length $\overline{DF}$ as a function of $\theta = \angle FCE$. We proceed as follows:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$\overline{CD} = {1\over{\sqrt{3}}}$ (exercise left to the reader)&lt;/li&gt;&#xA;&lt;li&gt;$\overline{CF} = 1$ (by definition)&lt;/li&gt;&#xA;&lt;li&gt;$\overline{FG} = \sin(\theta)$&lt;/li&gt;&#xA;&lt;li&gt;$\overline{CG} = \cos(\theta)$&lt;/li&gt;&#xA;&lt;li&gt;$\overline{DG} = \overline{CG} - \overline{CD} = \cos(\theta) - {1\over{\sqrt{3}}}$&lt;/li&gt;&#xA;&lt;li&gt;$\overline{DF} = \sqrt{\sin(\theta)^2 + \cos(\theta)^2 - {2\over{\sqrt{3}}}\cos(\theta) + {1\over 3}}$&lt;/li&gt;&#xA;&lt;li&gt;$\overline{DF} = \sqrt{{4\over 3} - {2\over{\sqrt{3}}}\cos(\theta)}$&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;We observe that $\overline{DF}$ is maximized when $\theta$ is greatest; in this case, when $\theta = {\pi \over 6}$, where $\overline{DF} = \sqrt{1 \over 3}$. From this, we conclude that we can model one of the Reuleaux road bumps by:$$\sqrt{1 \over 3} - \sqrt{{4\over 3} - {2\over{\sqrt{3}}}\cos(\theta)}\quad,\qquad {-\pi \over 6} \leq \theta \leq {\pi \over 6}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can go to &lt;a href=&quot;http://www.wolframalpha.com/input/?i=Plot%28+sqrt%281%2F3%29-sqrt%284%2F3+-+2%2Fsqrt%283%29%2acos%28t%29%29%2C+t%3D-Pi%2F6..Pi%2F6%2C+y%3D0..1%29&quot; rel=&quot;nofollow&quot;&gt;Wolfram Alpha&lt;/a&gt; for a plot of this curve:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/6UPQb.gif&quot; alt=&quot;Wolfram plot&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="2643" LastEditorUserId="2643" LastEditorDisplayName="" LastEditDate="2011-02-21T16:34:02.763" LastActivityDate="2011-02-21T16:34:02.763" />
  <row Id="22965" PostTypeId="2" ParentId="22954" CreationDate="2011-02-20T18:59:07.897" Score="0" ViewCount="" Body="&lt;p&gt;So, I will only cover here how to solve this for the inductive step, as the other is rather easy (not that the inductive step one isn't!).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Considering a string $x&amp;#39;=\lambda x$ we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$|x&amp;#39;y|=|x&amp;#39;|+|y|$$&#xA;$$|\lambda xy|=|\lambda x| + |y|$$&#xA;$$|\lambda z| = 1 + |x| + |y|$$&#xA;$$1 + |z| = 1 + |x| + |y|$$&#xA;$$1+|xy| = 1 + |xy|$$&lt;/p&gt;&#xA;" OwnerUserId="148" LastActivityDate="2011-02-20T18:59:07.897" CommentCount="5" />
  <row Id="22966" PostTypeId="2" ParentId="22093" CreationDate="2011-02-20T19:11:21.227" Score="1" ViewCount="" Body="&lt;p&gt;If you want to solve ab initio, then the way to go about is the way Arturo suggested.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that $a^5 - b^5 = (a-b)(a^4 + ba^3+ b^2a^2 + b^3a + b^4)$ and hence $$(a-b)= \frac{a^5 - b^5}{a^4 + ba^3+ b^2a^2 + b^3a + b^4}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now take $a=\sqrt[5]{x^5-3x^4+17}$ and $b=x$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sqrt[5]{x^5-3x^4+17}-x = \frac{(x^5-3x^4+17)-x^5}{\left(\sqrt[5] {x^5-3x^4+17} \right)^4 + x \left(\sqrt[5] {x^5-3x^4+17} \right)^3 + x^2 \left(\sqrt[5] {x^5-3x^4+17}\right)^2 + x^3 \left(\sqrt[5] {x^5-3x^4+17}\right)+ x^4}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Simplifying, we get,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sqrt[5]{x^5-3x^4+17}-x = \frac{-3x^4+17}{x^4 \left(\sqrt[5] {1-\frac{3}{x}+\frac{17}{x^5}} \right)^4 + x^4 \left(\sqrt[5] {1-\frac{3}{x}+\frac{17}{x^5}} \right)^3 + x^4 \left(\sqrt[5] {1-\frac{3}{x}+\frac{17}{x^5}} \right)^2 + x^4 \left(\sqrt[5] {1-\frac{3}{x}+\frac{17}{x^5}} \right) + x^4}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sqrt[5]{x^5-3x^4+17}-x = \frac{-3+\frac{17}{x^4}}{\left(\sqrt[5] {1-\frac{3}{x}+\frac{17}{x^5}} \right)^4 + \left(\sqrt[5] {1-\frac{3}{x}+\frac{17}{x^5}} \right)^3 + \left(\sqrt[5] {1-\frac{3}{x}+\frac{17}{x^5}} \right)^2 + \left(\sqrt[5] {1-\frac{3}{x}+\frac{17}{x^5}} \right) + 1}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, we have&#xA;$$\sqrt[5]{x^5-3x^4+17}-x = \frac{Nr(x)}{Dr(x)}$$ where $Nr(x) = -3+\frac{17}{x^4}$ and $Dr(x) = \left(\sqrt[5] {1-\frac{3}{x}+\frac{17}{x^5}} \right)^4 + \left(\sqrt[5] {1-\frac{3}{x}+\frac{17}{x^5}} \right)^3 + \left(\sqrt[5] {1-\frac{3}{x}+\frac{17}{x^5}} \right)^2 + \left(\sqrt[5] {1-\frac{3}{x}+\frac{17}{x^5}} \right) + 1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle \lim_{x \rightarrow \infty} Nr(x) = -3$ and $\displaystyle \lim_{x \rightarrow \infty} Dr(x) = 5$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, we have $\displaystyle \lim_{x \rightarrow \infty} Nr(x)$ and $\displaystyle \lim_{x \rightarrow \infty} Dr(x)$ exists as a real number and hence&#xA;$$\displaystyle \lim_{x \rightarrow \infty} \sqrt[5]{x^5-3x^4+17}-x = \frac{\displaystyle \lim_{x \rightarrow \infty} Nr(x)}{\displaystyle \lim_{x \rightarrow \infty} Dr(x)} = \frac{-3}{5} = - \frac{3}{5}$$&lt;/p&gt;&#xA;" OwnerUserId="3302" LastActivityDate="2011-02-20T19:11:21.227" />
  <row Id="22967" PostTypeId="1" CreationDate="2011-02-20T19:37:43.597" Score="2" ViewCount="394" Body="&lt;p&gt;I'm having trouble getting my head arround the concept. &#xA;The folks at mathoverflow were too clever to answer me.&#xA;Can someone explain it to me?&lt;/p&gt;&#xA;" OwnerUserId="7293" LastActivityDate="2011-02-20T23:42:22.340" Title="What is a Lie Group in layman's terms?" Tags="&lt;lie-groups&gt;" AnswerCount="4" CommentCount="4" FavoriteCount="3" />
  <row Id="22968" PostTypeId="2" ParentId="22907" CreationDate="2011-02-20T20:11:07.317" Score="2" ViewCount="" Body="&lt;p&gt;Not directly on-topic, but Mathematica 8's &lt;code&gt;Solve&lt;/code&gt; gives the following solutions&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{array}{l}&#xD;&#xA; \cos \left(A_2 k\right)\to \sin \left(A_2 k\right) \cot \left(A_1 k\right) \\&#xD;&#xA; \cos \left(A_1 k\right)\to 0 \\&#xD;&#xA; \sin \left(A_1 k\right)\to 0 \\&#xD;&#xA; \sin \left(A_2 k\right)\to 0 \\&#xD;&#xA; \sin \left(A_1 k\right)\to 0&#xD;&#xA;\end{array}$$&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;L = 0;&#xA;mat = {{BesselJ[L + 1/2, k A1],BesselJ[-(L + 1/2), k A1]},&#xA;      {BesselJ[L + 1/2, k A2], BesselJ[-(L + 1/2), k A2]}};&#xA;Solve[Det[mat] == 0]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="998" LastEditorUserId="998" LastEditorDisplayName="" LastEditDate="2011-02-20T20:17:55.437" LastActivityDate="2011-02-20T20:17:55.437" />
  <row Id="22970" PostTypeId="1" AcceptedAnswerId="22973" CreationDate="2011-02-20T20:32:40.070" Score="0" ViewCount="54" Body="&lt;p&gt;Suppose I have to solve the following boundary value problem: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$-u''=f\quad\text{and}\quad u(0)=0,\ u(1)=0.\qquad \mbox{Where $f \colon [0,1] \to \mathbb{R}$.}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the solution to the exercise, the professor integrates as follows to solve the equation: &#xA;$$u(x)=-\int_{0}^{x}\int_{1}^{y}f(s)dsdy+c_{1}x+c_{2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could someone explain to me why does he use these limits of integration specifically? Why from $1$ to $y$ then from $0$ to $x$?&lt;/p&gt;&#xA;" OwnerUserId="7053" LastEditorUserId="7053" LastEditorDisplayName="" LastEditDate="2011-02-20T20:39:21.020" LastActivityDate="2011-02-20T20:48:18.130" Title="Limits of integration of a 2nd order differential equation" Tags="&lt;differential-equations&gt;" AnswerCount="1" />
  <row Id="22971" PostTypeId="2" ParentId="22955" CreationDate="2011-02-20T20:34:04.387" Score="2" ViewCount="" Body="&lt;p&gt;Please ask questions, don't just quote your assingment/problem. It's better if you specify what you have tried or where you are stuck.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To prove that a set $S$ is inductive, you need to prove that:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$\emptyset\in S$; and&lt;/li&gt;&#xA;&lt;li&gt;If $a\in S$, then $a\cup\{a\} \in S$.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Conversely, if $S$ is inductive, then 1 and 2 will hold.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, for your first problem: to show the set you are given is inductive, you must show that $\emptyset$ is in the set. That is, you need to show that $\emptyset \in X$ and that $\emptyset$ is transitive. Since $X$ is assumed inductive, then $\emptyset\in X$. And $\emptyset$ is transitive, so $\emptyset$ is in the set.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, assume $a$ is in the set; you need to show that $a\cup\{a\}$ is in the set. That is, assume that $a$ is an element of $X$ and is transitive; use that to show that $a\cup \{a\}$ is in $X$ and is transitive. That will show your set is inductive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the final clause: $\mathbb{N}$ is inductive, so that means that $\{n\in\mathbb{N}\mid n\text{ is transitive}\}$ is an inductive set by what you will have proven by this point. It is also a subset of $\mathbb{N}$. What subsets of $\mathbb{N}$ are inductive?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Part 2 is proven the same way: show $\emptyset$ is in the given set, then show that if $a$ is in the set, then so is $a\cup\{a\}$. The first part of the &quot;Hence&quot; is immediate from the first clause of $2$. for the second part of the &quot;Hence&quot;, let me ask you a question: is $n\in n+1$?&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-20T20:39:47.827" LastActivityDate="2011-02-20T20:39:47.827" CommentCount="0" />
  <row Id="22972" PostTypeId="2" ParentId="21933" CreationDate="2011-02-20T20:35:55.430" Score="3" ViewCount="" Body="&lt;p&gt;It's actually even more shocking than what Arturo's answer might suggest.  Consider the following algorithm.  Given a group presentation $G=\langle S|R \rangle$, enumerate all possible proofs (say, using the axioms of ZFC), and check whether or not they give a proof that $G$ is trivial or not.  If $G$ is trivial, then this algorithm must finish.  Indeed, if $G$ is trivial, then we can write each element of $S$ as a product of conjugates of elements of $R$, and this constitutes a proof.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, we know that there are inputs for which this algorithm does not terminate.  This implies that there are finite presentations that define nontrivial groups, but for which there is no proof that they are nontrivial.  This gives a concrete and down-to-earth example of Goedel's incompleteness theorem!&lt;/p&gt;&#xA;" OwnerUserId="7296" LastActivityDate="2011-02-20T20:35:55.430" CommentCount="1" />
  <row Id="22973" PostTypeId="2" ParentId="22970" CreationDate="2011-02-20T20:48:18.130" Score="5" ViewCount="" Body="&lt;p&gt;Call $u_1(x)=-\displaystyle\int_{0}^{x}\int_{1}^{y}f(s)dsdy$ and $u_0(x)=-\displaystyle\int_{0}^{x}\int_{0}^{y}f(s)dsdy$. Then $u_1(x)=u_0(x)+cx$ with $c=\displaystyle\int_{0}^{1}f(s)ds$ hence both $u(x)=u_0(x)+c_1x+c_2$ (as you probably have in mind) or $u(x)=u_1(x)+c_1x+c_2$ (as your professor did) are valid forms of the general solution to the equation $-u&amp;#39;&amp;#39;=f$.&lt;/p&gt;&#xA;" OwnerUserId="6179" LastActivityDate="2011-02-20T20:48:18.130" CommentCount="1" />
  <row Id="22974" PostTypeId="1" CreationDate="2011-02-20T20:58:13.857" Score="1" ViewCount="113" Body="&lt;p&gt;A Fourier series arising in perturbation theory in quantum mechanics is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{m\neq n} \frac{1}{n^2 - m^2} \cos \frac{m\pi x}{2a} \, .$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $n$ is an odd positive integer and $n$ runs through all odd positive integers other than $n$. (The numbers are odd so that the Fourier terms are zero at $\pm a$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have no idea what kind of function produces this series. Is it familiar to anyone?&lt;/p&gt;&#xA;" OwnerUserId="7282" LastEditorUserId="10557" LastEditorDisplayName="" LastEditDate="2011-06-06T01:15:06.157" LastActivityDate="2011-06-06T01:49:31.570" Title="The fourier series $\sum_{m\neq n} \frac{1}{n^2 - m^2} \cos \frac{m\pi x}{2a}$" Tags="&lt;fourier-series&gt;" AnswerCount="2" CommentCount="0" />
  <row Id="22975" PostTypeId="2" ParentId="22974" CreationDate="2011-02-20T21:12:01.363" Score="1" ViewCount="" Body="&lt;p&gt;Taking $n=1$ and $a= \frac{1}{2}$, wolframalpha gives me $$\displaystyle \sum_{m = \text{odd}}^{\infty} \frac{\cos(m \pi x)}{1-m^2} = \frac{1}{4}(\pi(1-2x)\sin(\pi x) - \cos(\pi x))$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Taking $n=3$ and $a= \frac{1}{2}$, wolframalpha gives me $$\displaystyle \sum_{m = \text{odd}}^{\infty} \frac{\cos(m \pi x)}{9-m^2} = \frac{1}{36}(3\pi(1-2x)\sin(3\pi x) - \cos(3\pi x))$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Following this my hunch would be for $a= \frac{1}{2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle \sum_{m = \text{odd}}^{\infty} \frac{\cos(m \pi x)}{n^2-m^2} = \frac{1}{4n^2}(n\pi(1-2x)\sin(n\pi x) - \cos(n\pi x))$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Playing around a bit more with wolfram alpha, my new guess is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $x \in [0,a]$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle \sum_{m\neq n} \frac{1}{n^2 - m^2} \cos \frac{m\pi x}{2a} = \frac{1}{4n^2} \left(n\pi \left(1-\frac{x}{a} \right) \sin \left(\frac{n\pi x}{2a} \right) - \cos \left(\frac{n\pi x}{2a} \right) \right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and For $x \in (-a,0]$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle \sum_{m\neq n} \frac{1}{n^2 - m^2} \cos \frac{m\pi x}{2a} = \frac{1}{4n^2} \left(n\pi \left(-1-\frac{x}{a} \right) \sin \left(\frac{n\pi x}{2a} \right) - \cos \left(\frac{n\pi x}{2a} \right) \right)$$&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2011-02-20T21:46:07.820" LastActivityDate="2011-02-20T21:46:07.820" CommentCount="0" />
  <row Id="22976" PostTypeId="2" ParentId="22959" CreationDate="2011-02-20T21:17:48.483" Score="2" ViewCount="" Body="&lt;p&gt;How are you handling rotations?  If you're composing rotations then you might be running into the fundamental 'trick' with rotations: their non-commutativity.  In particular, if you have some rotation $R$ that's built (e.g.) as 'rotate around the $x$ axis $10^{\circ}$, then rotate around the $y$ axis $10^{\circ}$', then composing $R$ with itself (i.e., 'rotate around the $x$ axis $10^{\circ}$, then rotate around the $y$ axis $10^{\circ}$, then rotate around $x$ $10^{\circ}$ again, then rotate around $y$ $10^{\circ}$ again) is &lt;em&gt;not&lt;/em&gt; the same as 'rotate around the $x$ axis $20^{\circ}$, then rotate around the $y$ axis $20^{\circ}$'!  (This happens because 'rotate around $x$ by $10^{\circ}$, then rotate around $y$ by $10^{\circ}$' isn't the same as 'rotate around $y$ by $10^{\circ}$, then rotate around $x$ by $10^{\circ}$'.)  If you want to keep your camera upright and you're not allowing arbitrary camera rotations, then it's a good idea to explicitly consider your camera in terms of the classic Yaw/Pitch/Roll Euler angles and manipulate those rather than by composing rotations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Incidentally, for questions like this (practical questions about the implementation of rotation of objects, rather than abstract questions about the space of rotations) I'd recommend &lt;a href=&quot;http://gamedev.stackexchange.com&quot;&gt;the gamedev StackExchange site&lt;/a&gt;; this is exactly the sort of thing they handle on a regular basis.&lt;/p&gt;&#xA;" OwnerUserId="785" LastEditorUserId="785" LastEditorDisplayName="" LastEditDate="2011-02-21T02:43:37.873" LastActivityDate="2011-02-21T02:43:37.873" />
  <row Id="22977" PostTypeId="2" ParentId="22886" CreationDate="2011-02-20T21:48:13.103" Score="2" ViewCount="" Body="&lt;p&gt;It seems your brother is essentially right.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a standard amortization schedule, the amount applied to principal&#xA;each month increases geometrically, at the interest rate.&#xA;Doubling these amounts (or increasing them by any constant factor&#xA;over the amounts in the original amortization schedule) corresponds&#xA;to making payments at a higher &lt;em&gt;constant&lt;/em&gt; level that amortizes&#xA;the loan over a shorter total time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's the math: For month $j=0,1,2\ldots$ of the loan, let $P_j$ be&#xA;the principal remaining at the start of the month,&#xA;and $Y$ the payment, paid at the end of the month.&#xA;The amount paid toward interest is $I_j=rP_j$ with $r=0.05/12$,&#xA;and the amount paid toward principal is $A_j=Y-I_j=Y-rP_j$.&#xA;Then the new principal is&#xA;$$&#xD;&#xA;P_{j+1}= P_j-A_j = P_j(1+r)-Y,&#xD;&#xA;$$&#xA;so&#xA;$$&#xD;&#xA;I_{j+1}= rP_{j+1}= (1+r)I_j-rY = (1+r)(I_j-Y)+Y.&#xD;&#xA;$$&#xA;Hence $(1+r)A_j=A_{j+1}$ and therefore $A_j=(1+r)^jA_0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The standard payment $Y$ is rigged to make $P_N=I_N=0$ with&#xA;$N=360$ months. A higher (constant) payment $\hat Y$ corresponds to&#xA;principal payments $\hat A_j$ that are larger than $A_j$&#xA;by always the same proportion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a 30 year loan at 5 percent, the standard monthly payment&#xA;is \$5.3692 per \$1000.  Doubling the principal payment results&#xA;in a monthly \$6.5697 per \$1000, which amortizes the loan over&#xA;about 20 years.  Increasing the principal payments by 200 percent&#xA;(tripling them) amortizes the loan over a bit more than 15 years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But from your description it seems your brother is doing something&#xA;different, something that increases his payments each month.&#xA;A spreadsheet calculation shows that he would indeed pay off&#xA;the loan in 15 years, if he adds, to the standard 30-year payment $Y$,&#xA;the amount of principal that the payment $Y$ &lt;em&gt;would pay off this month&lt;/em&gt;.&#xA;(This amount may be listed on his statement.)&#xA;This means his payment at the end of month $j$&#xA;is $$&#xD;&#xA;Y_j=Y+ (Y-rP_j).&#xD;&#xA;$$&#xA; As above, now his remaining principal satisfies&#xA;$$&#xD;&#xA;P_{j+1}= P_j+I_j - Y_j = P_j(1+2r)-2Y.&#xD;&#xA;$$&#xA;So effectively his principal is reduced as if he makes the constant&#xA;payment $2Y$ on a loan with interest rate $2r$.  As it happens,&#xA;with $Y$ being the original 30-year standard payment, $2Y$ is almost just the right value to amortize this loan over 15 years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any way you do it, paying principal off early is a great way to&#xA;save lots on interest later.&lt;/p&gt;&#xA;" OwnerUserId="2947" LastActivityDate="2011-02-20T21:48:13.103" />
  <row Id="22978" PostTypeId="2" ParentId="22974" CreationDate="2011-02-20T21:52:31.073" Score="0" ViewCount="" Body="&lt;p&gt;You can find a differential equation for the expression by differentiating it twice. Just to get rid of the annoying factors, let's define $z={\pi x}/{2a}$. Then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{d^2}{dz^2}\sum_{\substack{m \; odd \\ m\neq n}} \frac{1}{n^2 - m^2} \cos (mz) = \sum_{\substack{m \; odd \\ m\neq n}} \frac{-m^2}{n^2 - m^2} \cos (mz) = $$&#xA;$$= \sum_{\substack{m\;  odd \\ m\neq n}} \cos (mz) - n^2 \sum_{\substack{m \; odd \\ m\neq n}} \frac{1}{n^2 - m^2} \cos (mz)\; .$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f&amp;#39;&amp;#39;(z) + n^2 f(z) = \sum_{\substack{m \; odd \\ m\neq n}} \cos (mz) = -\cos (nz) \; .$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sum of cosines should be easy to perform (replace the cosines by complex exponentials, use the geometric summation formula and take the real part), and solving the differential equation is also not difficult.&lt;/p&gt;&#xA;" OwnerUserId="3567" LastEditorUserId="10557" LastEditorDisplayName="" LastEditDate="2011-06-06T01:49:31.570" LastActivityDate="2011-06-06T01:49:31.570" CommentCount="4" />
  <row Id="22979" PostTypeId="1" CreationDate="2011-02-20T22:03:36.973" Score="3" ViewCount="133" Body="&lt;p&gt;I'm an undergrad student fairly keen on algebra. Over the different algebra courses I've taken, I've often encountered the so-called $p$-Prüfer group on exercises but somehow never got around to them. Now I'm trying to take care of that, but there are some statements I've seen about this group which I don't know how to prove (maybe because I lack some more background in group theory, especially in the study of infinite abelian groups?)&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Definition&lt;/strong&gt; A $p$-group is a &lt;em&gt;$p$-Prüfer group&lt;/em&gt; if it is isomorphic to $$C_{p^\infty}=\{e^{\frac{2k\pi i}{p^n}}:k\in \mathbb{Z}, n\in\mathbb{Z}^+\} \subset (\mathbb{C}^\times, \cdot)$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What I'm having trouble to prove is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The following are $p$-Prüfer groups:&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;1) An infinite $p$-group whose subgroups are totally ordered by inclusion,&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;2) An infinite $p$-group such that every finite subset generates a cyclic group,&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;3) An infinite abelian $p$-group such that $G$ is isomorphic to every proper quotient,&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;4) An infinite abelian $p$-group such that every subgroup is finite&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Just for the record, what I (think I) could prove was that the following are $p$-Prüfer groups:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;5) An injective envelope of $C_{p^n}$, for any $n\geq 1$,&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;6) A Sylow $p$-subgroup of $\frac{\mathbb{Q}}{\mathbb{Z}}$,&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;7) The direct limit of $0\subset C_p \subset C_{p^2}\subset ...$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here $C_{p^n}$ denotes a cyclic group of order $p^n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any other characterizations of the $p$-Prüfer group are welcome.&lt;/p&gt;&#xA;" OwnerUserId="2614" LastActivityDate="2011-03-21T10:50:35.697" Title="Characterizations of the $p$-Prüfer group" Tags="&lt;group-theory&gt;&lt;abstract-algebra&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="22980" PostTypeId="1" AcceptedAnswerId="22981" CreationDate="2011-02-20T22:06:38.840" Score="0" ViewCount="181" Body="&lt;p&gt;Question 5.33 from &quot;Topology and its Applications&quot;  by Baesner is to compute the fundamental group of the torus ($T^2$) with $n$ points removed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can &quot;see&quot; in my mind that if we remove one point we get a bouquet of two circles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Less clear is what happens when we remove 2 (and more) points.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any hints?&lt;/p&gt;&#xA;" OwnerUserId="2219" LastActivityDate="2011-02-21T08:20:07.910" Title="$T^2$ with $n$ points removed - Question from Basener" Tags="&lt;algebraic-topology&gt;" AnswerCount="2" />
  <row Id="22981" PostTypeId="2" ParentId="22980" CreationDate="2011-02-20T22:18:14.593" Score="1" ViewCount="" Body="&lt;p&gt;Hint: removing $n$ points will also give something that only consists of 1-dimensional things.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can also use van Kampen's theorem to calculate the fundamental group directly. EDIT: Err, you can't, at least not in the way I thought. Let $T_n$ be a torus with $n$ holes. Then, van Kampen's theorem gives a short exact sequence of groups&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ 1 \to \pi_1(S^1) \to \pi_1(T_n) \to \pi_1(T_{n-1}) \to 1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but that's not enough information to deduce $\pi_1(T_n)$.&lt;/p&gt;&#xA;" OwnerUserId="434" LastEditorUserId="434" LastEditorDisplayName="" LastEditDate="2011-02-21T08:20:07.910" LastActivityDate="2011-02-21T08:20:07.910" CommentCount="2" />
  <row Id="22982" PostTypeId="1" AcceptedAnswerId="23117" CreationDate="2011-02-20T22:24:34.437" Score="3" ViewCount="228" Body="&lt;p&gt;I have been looking for some good material covering Markov Chains but everything seems so difficult to me...&#xA;After reading about the subject, I figured out that there is basically three kinds of processes: discrete-time, continuous-time and decision Markov Processes.&#xA;I started reading Introduction to Probability Models, Tenth Edition, from Sheldon M. Ross, about discrete-time processes and then, after judging myself introduced to the subject tried to read about continuous-time processes and got confused.&#xA;Could someone give me an explanation about the subject (like the other Markov Chains topic: What is a Markov Chain?), focusing on differences and at least one simple application ? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also would be glad if you could suggest books that don't use a very complicated language to explain the subject. The book I cited above seems cool (although I have lots of doubts over the formulas/proofs and the math notation) but there are too many people telling the book doesn't explain the subject properly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you very much!&lt;/p&gt;&#xA;" OwnerUserId="7298" LastActivityDate="2011-02-22T16:38:01.237" Title="What is the difference between all types of Markov Chains?" Tags="&lt;statistics&gt;&lt;stochastic-processes&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="0" />
  <row Id="22983" PostTypeId="2" ParentId="22980" CreationDate="2011-02-20T22:25:23.687" Score="2" ViewCount="" Body="&lt;p&gt;its a bouquet of $n+1$ circles, so the free group on $n+1$ generators.  Think of a rectangle with $n-1$ horizontal lines across it.  Roll it up into a tube, so you have a line segment with $n+1$ circles attached to it.  Then identify the top and bottom circles.  So you have a circle with $n$ circles attached to it which is homotopic to a bouquet of $n+1$ circles.&lt;/p&gt;&#xA;" OwnerUserId="6925" LastActivityDate="2011-02-20T22:25:23.687" />
  <row Id="22984" PostTypeId="2" ParentId="22959" CreationDate="2011-02-20T22:27:36.570" Score="2" ViewCount="" Body="&lt;p&gt;An answer you would probably receive in a game dev forum: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It sounds like your using a Y up system and pitching around the world X axis. Typically, for a game camera where you only want yaw &amp;amp; pitch, usually the yaw is desired about the world's vertical axis but pitch is desired about the camera's local horizontal axis(local X axis).  When the camera orientation is in the identity orientation, the camera's local horizontal axis happens to be aligned with the world horizontal axis (usually the X axis) but when yaw is applied the local horizontal axis rotates away from the world X axis. If you then apply pitch around the world X axis, it can cause a rotation with a Z axis component to it. But if you apply the pitch rotation first (before the yaw) while the local and world X axes are aligned, there will be no Z component to the rotation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This pitch rotation causes the local Up axis to rotate away from the world Up axis too but you really don't care because you always yaw around the world up axis anyway.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So try applying pitch first, then yaw and see if it helps.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or... calculate the local horizontal axis after yawing and use that axis to pitch around instead of the world X axis.&lt;/p&gt;&#xA;" OwnerUserId="2915" LastActivityDate="2011-02-20T22:27:36.570" />
  <row Id="22985" PostTypeId="2" ParentId="22967" CreationDate="2011-02-20T22:35:51.193" Score="1" ViewCount="" Body="&lt;p&gt;it's a group thats a smooth manifold.  Technically, the group structure and topology should be related, i.e. $X\times X\to X, (a,b)\mapsto ab^{-1}$ should be smooth.  For instance, $\mathbb{R}^n$, the circle, a torus (surface of a doughnut) the three sphere ($\mathbb{R}^3$ with a point at infinity) can all be given the structure of a lie group.&lt;/p&gt;&#xA;" OwnerUserId="6925" LastActivityDate="2011-02-20T22:35:51.193" CommentCount="1" />
  <row Id="22986" PostTypeId="2" ParentId="22967" CreationDate="2011-02-20T22:56:05.003" Score="3" ViewCount="" Body="&lt;p&gt;Group theory is the study of symmetry.  Lie groups were invented by Lie to study the symmetries of differential equations.  You might like the article:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Starrett, John. &quot;Solving differential equations by symmetry groups.&quot;&#xA;Amer. Math. Monthly 114 (2007), no. 9, 778–792.&#xA;MR&lt;a href=&quot;http://www.ams.org/mathscinet-getitem?mr=2360906&quot; rel=&quot;nofollow&quot;&gt;2360906&lt;/a&gt;&#xA;&lt;sup&gt;(author's &lt;a href=&quot;http://euler.nmt.edu/%7Ejstarret/05-649LieGroupODEFinalVersion.pdf&quot; rel=&quot;nofollow&quot;&gt;preprint&lt;/a&gt;)&lt;/sup&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It describes how to solve some reasonably calc-2 or calc-4 problems using symmetry and can serve as introduction to what Lie groups do.  It also has pretty pictures of smoothly flowing curves.&lt;/p&gt;&#xA;" OwnerUserId="583" LastActivityDate="2011-02-20T22:56:05.003" />
  <row Id="22987" PostTypeId="2" ParentId="22979" CreationDate="2011-02-20T22:57:22.350" Score="4" ViewCount="" Body="&lt;p&gt;How one procedes to prove these probably depends a bit on which description of the $p$-Prüfer group one is most comfortable with. So my particular arguments may not seem very natural to you, in which case this will probably not help much.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Because the subgroups are ordered by inclusion, given any $x,y\in G$, either $x\in\langle y\rangle$ or $y\in\langle x\rangle$. Take an element $x_1\in G$. The subgroup it generates cannot be all of $G$ (it is finite, since $G$ is a $p$-group), so there exists $x_2\in G$ with $x_2\notin\langle x_1\rangle$. Therefore, $x_1\in\langle x_2\rangle$. But $x_2$ cannot generate all of $G$. So there exists $x_3\in G$, $x_3\notin \langle x_2\rangle$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Continuing this way, you obtain a (countably infinite) collection of elements of $G$, $x_1,x_2,\ldots,$ such that $x_n\in\langle x_{n+1}\rangle$, $x_{n+1}\notin\langle x_n\rangle$. By introducing suitable elements into the sequence, we may assume that $[\langle x_{n+1}\rangle\colon\langle x_n\rangle]=p$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now prove that the subgroup generated by all the $x_i$ is isomorphic to the $p$-Prüfer group, and that it is equal to $G$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Let $x_1\in G$. It does not generate all of $G$, so let $y\in G$ that is not in $\langle x_1\rangle$. Then $\langle x_1,y\rangle$ is cyclic, strictly larger than $\langle x_1\rangle$; let $x_2$ be a generator of this cyclic group. Since $x_2$ does not generate all of $G$, there exists $y\in G$ not in $\langle x_2\rangle$; but $\langle x_2,y\rangle$ is cyclic, so let $x_3$ be a generator. Lather, rinse, repeat, to get an infinite sequence as above, adding elements if necessary to get each step to increase the size of the cyclic group by $p$; that is, $x_n$ is of order $p^n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again, prove the group generated by the $x_i$ is the $p$-Prüfer group and equal to all of $G$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I'd written suitable arguments for 3 and 4, but Jack has posted them as well and they are essentially the same, so I'll defer to him on them.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-20T23:24:15.423" LastActivityDate="2011-02-20T23:24:15.423" />
  <row Id="22989" PostTypeId="2" ParentId="22788" CreationDate="2011-02-20T23:10:17.997" Score="0" ViewCount="" Body="&lt;p&gt;I did it myself, the base case &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;n1 is n=2 2-1=1, &#xA;&#xA;n2 is n=3 3-1=2. &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt; &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;n(k) is n=k k-1=k-1; n&#xA;&#xA;n(k+1) is n=k+1 (k+1)-1=(k-1)+1=n(k)+1&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Since the k+1 step relies and follows from the k step the base case can be reached.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;best,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;blockquote&gt;&#xA;    &lt;blockquote&gt;&#xA;      &lt;p&gt;based on a comment below (by  Steven Stadnicki), some insigt into the process the induction is based upon is needed to make this complete which is correct. The process is that with the basic tree n1, to add another internal node, a leaf node is replaced by an internal node (-1) and has then two leaves attached to the new internal node (+2) for a total of (+1) from the alteration of adding another internal node. So the addition of an internal node and leaf node is 1 to 1, but the initial offset with there being 1 internal to 2 leaf nodes remains. Therefore we can easily see how the individual induction steps work and n-1 is the relationship that hold.&gt;&gt;&gt;&gt;&lt;/p&gt;&#xA;    &lt;/blockquote&gt;&#xA;  &lt;/blockquote&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1587" LastEditorUserId="1587" LastEditorDisplayName="" LastEditDate="2011-02-21T15:10:44.503" LastActivityDate="2011-02-21T15:10:44.503" CommentCount="1" />
  <row Id="22990" PostTypeId="1" AcceptedAnswerId="22997" CreationDate="2011-02-20T23:15:36.977" Score="7" ViewCount="260" Body="&lt;p&gt;In Lebesgue integration we usually approximate the function we want to integrate with step-functions on measurable sets. How much &quot;power&quot; do we take away if we require that the step functions are on intervals instead? What functions are left that are integrable?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm asking this because I want some integral to converge but I only know the values on $1_{(x_1,x_2)}$. Maybe we can get to Lipschitz functions that way?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit to be more specific.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Usually we define an integral for a positive function in this sense. First if we have&#xA;$$f = \sum a_i 1_{A_i} \text{ then } \int f \, d\mu = \sum a_i \mu(A_i)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now if $f$ is a positive measurable function we then define&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int f \, d\mu := \sup \left \{\int g : g \leq f \text{ and $g$ is a simple function} \right \}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question now is: what is left of the theory if we require the $A_i$ to be intervals instead of elements of the whole $\sigma$-algebra?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My apologies for the unclear question. I shouldn't ask questions in the middle of the night.&lt;/p&gt;&#xA;" OwnerUserId="1120" LastEditorUserId="1120" LastEditorDisplayName="" LastEditDate="2011-02-21T00:28:37.107" LastActivityDate="2011-02-21T00:29:29.133" Title="Pull the teeth out of Lebesgue integration." Tags="&lt;soft-question&gt;&lt;measure-theory&gt;" AnswerCount="2" CommentCount="11" FavoriteCount="2" />
  <row Id="22991" PostTypeId="2" ParentId="22979" CreationDate="2011-02-20T23:24:06.317" Score="4" ViewCount="" Body="&lt;p&gt;Arturo already explained the hint 1⇒7, 2⇒7 (and 1⇒2 is easy enough, I trust you can do it).  I'll explain how to use pG = { pg : g in G } to understand G for 3 and 4.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If every nonzero quotient of G is isomorphic to G, then what does G/pG look like?  Well, it is an elementary abelian p-group, so a vector space over Z/pZ.  If it is nonzero, then it has a one-dimensional quotient: Z/pZ.  By the hypothesis on G, that would mean G itself is Z/pZ.  Unfortunately such a group is not infinite, and so is not a group as in 3.  Hence a group as in 3 must have G=pG.  As it is a p-group, it is divisible, so a direct sum of Prüfer p-groups.  However, such a direct sum always has a single Prüfer p-group as a quotient, and so G must itself be the Prüfer p-group.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If every proper subgroup of G is finite, then what does pG look like?  If it is finite, then G/pG is infinite, and so infinite dimensional.  Take a proper subspace.  The preimage of that subspace in G is a proper subgroup that is infinite. Oops. So pG cannot be finite!  So again G=pG, and G is a direct sum of Prüfer p-groups.  How many?  Well each summand is a subgroup, and so if there is more than one, then one has an infinite proper subgroup.  So G must itself be a single Prüfer p-group.&lt;/p&gt;&#xA;" OwnerUserId="583" LastActivityDate="2011-02-20T23:24:06.317" CommentCount="7" />
  <row Id="22992" PostTypeId="2" ParentId="22967" CreationDate="2011-02-20T23:35:11.980" Score="5" ViewCount="" Body="&lt;p&gt;Consider the set of $(n\times n)$ matrices that have non-zero determinant. Such a matrix corresponds to a system of linear equations ($n$ equations in $n$ unknowns) that has a unique solution. You can think of the solution as the unique point of intersection between the graph of a function and a horizontal hyperplane. Here it is helpful to think of $n=1$. In other words, the coefficients of the system correspond to a transformation of space: the variables $x_1, \ldots x_n$ are transformed to $\sum a_{ij} x_i$. The set of such transformations form a group: the matrices can be multiplied, each has an inverse, the multiplication is associative, and the identity transformation fixes each point of space. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Intuitively, it is easy to see which transformations are close to one another. They are close if they move points that are nearby to points that are nearby. Arithmetically, if the entries in the matrix are close, then the transformations are close: thus $0.14x + .33y$, is a reasonable approximation to $x/7+y/3$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus the  &lt;em&gt;set&lt;/em&gt; of invertible $(n\times n)$ matrices is a &lt;em&gt;space&lt;/em&gt; of invertible $(n\times n)$ matrices. What is not easy for a layman to see is that its spacial characteristics are defined via the determinant since as a set, the $(n\times n)$ matrices are a subset of $n^2$-space. The non-singular matrices are the pull-back of a regular value of the determinant function. [There is a small lie here: this is true for for matrices of determinant 1, but all non-zero determinant matrices deform onto that smaller space]. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One important spacial characteristic is that these matrices form a smooth manifold. This is something that is analogous to the surface of a sphere (which is NOT a lie group), the surface of a torus (which is) or the $3$-dimensional sphere that consists of the set of $(x,y,z,w)$ such that $x^2+y^2+z^2+w^2=1$ (which also happens to be a Lie group). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From these examples, we abstract the idea of a Lie group which is a group (that can be thought of as a set of transformations or symmetries) that has the structure of a smooth manifold --- at small scales it is indistinguishable from ordinary Euclidean space. The multiplication and inversion maps are a differentiable functions. And these multiplications occur between pairs of symmetries --- they should not be confused with the action of the matrices on the vector space which is where I started the discussion. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Examples include the real line, the non-zero real numbers, the circle, the torus, the $3$-sphere, the set of rotations of 3-dimensional space, and the special unitary groups representations of which determine particles in physics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are some small problems with the definition that I gave. A smooth manifold is a topological space which is paracompact and Hausdorff (neither definition will play a role in the layman's understanding), and that is covered by coordinate charts with specific properties. I imagine that Wikipedia has the relevant definitions articulated carefully. &lt;/p&gt;&#xA;" OwnerUserId="722" LastActivityDate="2011-02-20T23:35:11.980" />
  <row Id="22993" PostTypeId="2" ParentId="22967" CreationDate="2011-02-20T23:42:22.340" Score="4" ViewCount="" Body="&lt;p&gt;I think that understanding comes through examples. The most fundamental example I believe to be the rotation group. Consider the sphere $S^2\subset \mathbb{R}^3$. The sphere has rotational symmetries. If we rotate the sphere by any angle, the sphere doesn't change. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The collection of all rotations forms a Lie group. The group property basically means that&#xA;if we rotate the sphere over any angle $\alpha$, after this over an angle $\beta$, it is the same if we would have rotated it in one go (over some different angle). Also any rotation has an inverse (rotating it over the opposite angle). This makes the rotations a group. The &quot;Lie&quot; in Lie group means that these rotations can be done arbitrary small. Many small rotations makes for a big rotation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lie groups capture the concept of &quot;continuous symmetries&quot;.&lt;/p&gt;&#xA;" OwnerUserId="5882" LastActivityDate="2011-02-20T23:42:22.340" />
  <row Id="22994" PostTypeId="1" AcceptedAnswerId="23004" CreationDate="2011-02-20T23:44:58.547" Score="2" ViewCount="40" Body="&lt;p&gt;Let $F$ be a local field, $G=GL_2(F)$, $\chi_1,\chi_2$ quasicharacters, and $B(\chi_1,\chi_2)$ be the principal series (we are assuming $\chi_1,\chi_2$ to be such that the representation is irreducible).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is $B(\chi_1,\chi_2)$ a constituent of $L^2(Z\backslash G)$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm pretty sure it never is, but can't prove it. Am familiar with Bump up to 4.5.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, say $\chi_1\chi_2^{-1} = |\cdot |^{\pm 1}$ (so not a principal series). Same question?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This time I think it is...&lt;/p&gt;&#xA;" OwnerUserId="6404" LastActivityDate="2011-02-21T01:22:31.127" Title="When are the principal series of GL_2(F) square integrable?" Tags="&lt;representation-theory&gt;" AnswerCount="1" />
  <row Id="22995" PostTypeId="2" ParentId="22939" CreationDate="2011-02-20T23:53:06.207" Score="0" ViewCount="" Body="&lt;p&gt;In another, now deleted, venue you mentioned a more specific question of just showing the operations are continuous.  You mentioned you could handle negation, but had trouble with addition and multiplication.  Here is one way to verify them:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let J = I&lt;sup&gt;n&lt;/sup&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The inverse image of x+J under addition is { (y,z) : y+z-x in J }. This is a union of A(y) = (y+J)⊕(x-y+J) ≤ R⊕R as y varies over R. Each A(y) is open (being a direct product of open sets), and so the union is open. In other words, the preimage under + of an open set is open.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The inverse image x+J under multiplication is { (y,z) : yz-x in J }. This is a union of B(y) = (y+J)⊕(∪{ z+J : yz-x in J }). Each B(y) is open (being a direct product of an open set and a union of open sets). In other words the preimage under multiplication of an open set is open.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only trick to it is (1) knowing some open sets of a direct product and (2) noticing that you could have just worked in a quotient ring R/J where every set (every!) is open, since it is a union of singletons, x+J. In plainer language, (y+j)*(z+j') = yz + jz+yj'+jj' = yz + (something in J) is still in yz+J.&lt;/p&gt;&#xA;" OwnerUserId="583" LastActivityDate="2011-02-20T23:53:06.207" />
  <row Id="22996" PostTypeId="1" AcceptedAnswerId="22998" CreationDate="2011-02-21T00:05:15.700" Score="12" ViewCount="225" Body="&lt;p&gt;I'm trying to prove that any group $G$ of infinite order has an infinite number of subgroups.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think that if the group has an element of infinite order, then it's easy because I can take the groups generated by the powers of this element.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What if it doesn't? Every element generates a cyclic subgroup. Every element belongs to at least one cyclic subgroup (that generated by itself). So the group is the union of its cyclic subgroups. If all these are finite, we would have to have an infinite collection of subgroups anyway.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is that correct?&lt;/p&gt;&#xA;" OwnerUserId="1313" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-21T00:14:12.630" LastActivityDate="2011-02-21T01:06:26.250" Title="Finite number of subgroups $\Rightarrow$  finite group." Tags="&lt;group-theory&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="0" />
  <row Id="22997" PostTypeId="2" ParentId="22990" CreationDate="2011-02-21T00:06:29.280" Score="8" ViewCount="" Body="&lt;p&gt;Terminologically speaking, &lt;em&gt;step function&lt;/em&gt; usually means linear combination of characteristic functions of intervals, whereas &lt;em&gt;simple function&lt;/em&gt; is used for functions taking on finitely many values.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're talking about Lebesgue measure, you &lt;em&gt;can&lt;/em&gt; approximate arbitrary $L^1$ functions with step functions in $L^1$ norm.  It's just that in defining the Lebesgue integral, it's easiest to do so first for arbitrary simple functions, because then all measurable functions can be approximated in a particularly nice way.  In particular, if $f$ is nonnegative, then $\int f$ can be defined to be the supremum of the integrals of nonnegative simple functions dominated by $f$.  This would not work for step functions, as the characteristic function of the irrationals in $[0,1]$ shows; or, less trivially, the characteristic function of a fat Cantor set.   Thus, your proposed change in definitions would break down for relatively nice Borel functions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the problem of $L^1$ approximation by step functions could be settled if you could find, for each finite measure set $E$ and for each $\varepsilon\gt0$, a finite union of intervals $F$ such that $\int |\chi_E - \chi_F |\lt \varepsilon$.  That is, you want the measure of the symmetric difference of $E$ and $F$ to be less than $\varepsilon$.  That this is true for Lebesgue measure on the line is (a version of) one of &lt;a href=&quot;http://en.wikipedia.org/wiki/Littlewood%27s_three_principles_of_real_analysis&quot; rel=&quot;nofollow&quot;&gt;Littlewood's 3 principles&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2011-02-21T00:29:29.133" LastActivityDate="2011-02-21T00:29:29.133" CommentCount="7" />
  <row Id="22998" PostTypeId="2" ParentId="22996" CreationDate="2011-02-21T00:13:15.407" Score="10" ViewCount="" Body="&lt;p&gt;Yes, it's correct, though perhaps somewhat disorganized.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fact that a group is a union of its cyclic subgroups holds regardless of any hypothesis on the group. So you can state and prove that first.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then: an infinite cyclic group has infinitely many (cyclic) subgroups. So, a group is infinite if and only if it has infinitely many cyclic subgroups: If the group is finite, then it has only finitely many subsets, so finitely many subgroups, so finitely many cyclic subgroups. If the group is infinite, then being a union of its cyclic subgroups, either one of the cyclic subgroups is infinite (and hence there will be infinitely many cyclic subgroups, since &lt;em&gt;that&lt;/em&gt; subgroup has infinitely many cyclic subgroups), or else there are infinitely many finite cyclic subgroups in the union. Either way you are done.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is your argument, just organized a bit more. &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-21T00:30:11.300" LastActivityDate="2011-02-21T00:30:11.300" />
  <row Id="22999" PostTypeId="1" AcceptedAnswerId="23001" CreationDate="2011-02-21T00:18:49.347" Score="10" ViewCount="188" Body="&lt;p&gt;My interest in combinatorics was recently sparked when I read about the many things that the Catalan numbers count, as found by Richard Stanley. I picked up a copy of Brualdi's Combinatorics, and while browsing the section on counting sequences I found a nice little puzzle that has definitely puzzled me.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Let $m$ and $n$ be nonnegative integers with $n\geq m$. There are $m+n$ people in line to get into a theater for which admission if $50$ cents. Of the $m+n$ people, $n$ have a $50$-cent piece and $m$ have a $\$ 1$ dollar bill. The box offices opens with an empty cash register. Show that the number of ways the people can line up so that change is available when needed is&#xA;  $$&#xD;&#xA;\frac{n-m+1}{n+1}\binom{m+n}{m}.&#xD;&#xA;$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I first noted that the first person to enter must be one of the $n$ with a half-dollar. Now the register has a half-dollar change. The second person can be either a person with a half-dollar or a dollar. In the first case, the register will now have two half-dollars, in the second case, the register will now have one dollar bill. So it seems to me that when one of the $n$ people with a half-dollar enters, the number of half-dollars in the register increases by $1$, and when one of the $m$ people with a bill enters, the number of half-dollars decreases by $1$ but the number of bills increases by $1$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried to model this by looking at paths in $\mathbb{Z}^2$. The $x$-axis is like the number of half-dollars, and the $y$-axis is the number of bills. You start at $(0,0)$, and you can take steps forward $(1,0)$ or backwards diagonally $(-1,1)$ corresponding to who enters, but you must always stay in the first quadrant of the plane without crossing over the axes. The goal is to make $m+n$ moves, and I figured maybe the number of such paths is counted by $\frac{n-m+1}{n+1}\binom{m+n}{m}$, but I'm not sure how to show this. I don't know if this observation simplifies the problem at all, as I don't know how to finish up. I'd be happy to see how this problem is done, thank you.&lt;/p&gt;&#xA;" OwnerUserId="1583" LastActivityDate="2011-02-21T03:26:12.910" Title="How many ways can we let people into a movie theater if they only have half-dollars and dollars?" Tags="&lt;combinatorics&gt;&lt;puzzle&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" />
  <row Id="23000" PostTypeId="2" ParentId="22990" CreationDate="2011-02-21T00:21:05.133" Score="6" ViewCount="" Body="&lt;p&gt;There is a way to use step functions, and only step functions, to set up the Lebesgue integral due to Mikusinski. It requires something more subtle than a supremum. We say that a function $f$ is &lt;strong&gt;integrable&lt;/strong&gt; if there exists a sequence $f_i$ of step functions such that $\sum \int |f_i|$ converges and such that $f(x) = \sum f_i(x)$ pointwise whenever $\sum |f_i|$ converges, and we define its &lt;strong&gt;integral&lt;/strong&gt; to be $\int f = \sum \int f_i$. &lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2011-02-21T00:21:05.133" CommentCount="2" />
  <row Id="23001" PostTypeId="2" ParentId="22999" CreationDate="2011-02-21T00:28:48.197" Score="10" ViewCount="" Body="&lt;p&gt;You are correct that you can think of this as a problem of counting (restricted) paths on the $\mathbb{Z}^2$, and that this is probably a good way to think about it. But I think it is easier if you think of a square array, and you are trying to get from the bottom left, $(0,0)$, to the upper right $(n,m)$, and your steps must be from $(k,\ell)$ to $(k+1,\ell)$ or from $(k,\ell)$ to $(k,\ell+1)$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $n$ is the number of people with $50$ cent pieces, and $m$ is the number of people with dollar bills. When a person with a 50 cent piece enters, you take a step right. When a person with a dollar bill enters, you take a step up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you try to take a step from $(k,\ell)$ to $(k,\ell+1)$, you will only have enough change in the till if $\ell+1\leq k$. Otherwise, you'll be out of luck (because you need one person with 50 cent piece for every person with a dollar that has managed to come in).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, you must always stay at or below the diagonal. So the paths you want to count are the paths from $(0,0)$ to $(n,m)$ that stay at or below the main diagonal. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Notice that $\binom{n+m}{m}$ is the number of &lt;em&gt;total&lt;/em&gt; paths from $(0,0)$ to $(n,m)$ taking only steps right or up: you must take $n+m$ steps total, and of those $m$ will be steps up; so $\binom{n+m}{m}$ picks which of the $n+m$ steps will be steps up. So the factor $\frac{n+1-m}{n+1}$ must be the fraction of the paths that stay at or below the main diagonal.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does that help sufficiently, or should I expand more?&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="6460" LastEditorDisplayName="" LastEditDate="2011-02-21T00:33:58.987" LastActivityDate="2011-02-21T00:33:58.987" CommentCount="6" />
  <row Id="23002" PostTypeId="2" ParentId="22949" CreationDate="2011-02-21T00:42:46.260" Score="3" ViewCount="" Body="&lt;p&gt;Let $p_1, ... p_n$ be the first $n$ primes and let $\pi_n(N)$ be the number of positive integers less than or equal to $N$ divisible by only the primes $p_i$. This is precisely the number of non-negative integer solutions to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{i=1}^n e_i \log p_i \le \log N$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which defines an $n$-dimensional region with volume $V = \frac{(\log N)^n}{n! \log p_1 ... \log p_n}$. So this is a first-order estimate of the number of solutions; in other words, asymptotically we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\pi_n(N) \sim \frac{(\log N)^n}{n! \log p_1 ... \log p_n}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(The error term is $O(( \log N)^{n-1})$.) A reasonable guess for the probability that $N$ itself is divisible only by the primes $p_i$ is then roughly the derivative of this, or &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{n (\log N)^{n-1}}{N n! \log p_1 ... \log p_n}.$$&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2011-02-21T00:42:46.260" CommentCount="1" />
  <row Id="23003" PostTypeId="2" ParentId="22996" CreationDate="2011-02-21T01:06:26.250" Score="0" ViewCount="" Body="&lt;p&gt;I might be oversimplifying, but it seems to me that if G has infinite elements, it has an infinite number of cyclic subgroups, one for each g in G.&lt;/p&gt;&#xA;" OwnerUserId="7270" LastActivityDate="2011-02-21T01:06:26.250" CommentCount="2" />
  <row Id="23004" PostTypeId="2" ParentId="22994" CreationDate="2011-02-21T01:22:31.127" Score="4" ViewCount="" Body="&lt;p&gt;A principal series representation is never square integrable.  Probably in the case of $GL_2$ one can compute matrix coefficients explicitly to verify this, but it also follows from the (somewhat more theoretical) computation of matrix coefficients via the Jacquet module.   A good reference for this is Casselman's notes (available on his web-page), which develop the theory of the Jacquet module and its application to matrix coefficients essentially from scratch.  (Casselman work's in the context of an arbitrary reductive group, and so uses some basic Lie theoretic results as input; but in the case of $GL_2$ one can just replace that&#xA;input by explicit facts and computations about $GL_2$ and its Borel subgroups.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case when the induced representation is trivial, the infinite dimensional&#xA;constituent (which I would call special; but I don't know what Bump's terminology is) will be square integrable (up to a twist; one may need to apply a twist to ensure that the central character is unitary, if it is not already).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again, this can be seen via a Jacquet module computation: the Jacquet module of&#xA;an induced representation for $GL_2$ is two-dimensional, and so there are two exponents, one of which gives rise to square integrable matrix coefficients, but &#xA;the other of which doesn't.  In the reducible case, the one-dimensional constituent is contributing the &quot;bad&quot; exponent, and so when we remove that constituent, the infinite-dimensional piece that's left has just a one-dimensional Jacquet module, whose exponent is good enough to ensure square integrability.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My memory is that Casselman discusses this example specifically in his notes.&lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2011-02-21T01:22:31.127" />
  <row Id="23005" PostTypeId="1" AcceptedAnswerId="23010" CreationDate="2011-02-21T01:23:48.813" Score="1" ViewCount="98" Body="&lt;ol&gt;&#xA;&lt;li&gt;Suppose there are three measurable spaces&#xA;$(\Omega, \mathbb{F})$, $(S_i,&#xD;&#xA;    \mathbb{S}_i), i=1,2$, and two&#xA;measurable mappings $f_i: \Omega&#xD;&#xA;    \rightarrow S_i, i=1,2$. Is the &#xA;mapping $f$ defined as&#xA;$f(\omega):=(f_1(\omega),&#xD;&#xA;    f_2(\omega))$ a measurable mapping&#xA;from $(\Omega, \mathbb{F})$ to&#xA;$(\prod_{i=1}^2 S_i, \prod_{i=1}^2&#xD;&#xA;    \mathbb{S}_i)$, where $\prod_{i=1}^2&#xD;&#xA;    \mathbb{S}_i$ is the product sigma&#xA;algebra of $\mathbb{S}_i, i=1,2$?&lt;/li&gt;&#xA;&lt;li&gt;Suppose there are four measurable spaces&#xA;$(\Omega_i, \mathbb{F}_i), i=1,2$,&#xA;$(S_i, \mathbb{S}_i), i=1,2$, and&#xA;two measurable mappings $f_i:&#xD;&#xA;    \Omega_i \rightarrow S_i, i=1,2$. Is&#xA;the mapping $f$ defined as&#xA;$f(\omega_1,&#xD;&#xA;    \omega_2):=(f_1(\omega_1),&#xD;&#xA;    f_2(\omega_2))$ a measurable mapping&#xA;from $(\prod_{i=1}^2 \Omega_i,&#xD;&#xA;    \prod_{i=1}^2 \mathbb{F}_i)$ to&#xA;$(\prod_{i=1}^2 S_i, \prod_{i=1}^2&#xD;&#xA;    \mathbb{S}_i)$?&lt;/li&gt;&#xA;&lt;li&gt;In Part 1 and Part 2, conversely, if&#xA;$f$ is a measurable mapping, will&#xA;$f_i, i=1,2$ be measurable mappings?&lt;/li&gt;&#xA;&lt;li&gt;Can the statements in Part 1,2 and&#xA;3 be generalized to any&#xA;collection of $(S_i, \mathbb{S}_i) i&#xD;&#xA;    \in I$ and $(\Omega_i, \mathbb{F}_i)&#xD;&#xA;    i \in I$?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Thanks and regards! Are there some websites or books that address these questions?&lt;/p&gt;&#xA;" OwnerUserId="1281" LastEditorUserId="1281" LastEditorDisplayName="" LastEditDate="2011-02-21T01:36:25.587" LastActivityDate="2011-02-21T05:17:05.980" Title="Questions about measurable mapping and product sigma algebra" Tags="&lt;measure-theory&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="23006" PostTypeId="1" CreationDate="2011-02-21T01:23:59.163" Score="4" ViewCount="148" Body="&lt;p&gt;One learns in a probability course that a (real) random variable is a measurable mapping of some probability space $(\Omega,\mathcal{A},\mathbf{P})$ into $(\mathbb{R},\mathcal{B}(\mathbb{R}))$. But as soon as one gets into topics that are a little advanced, the space $(\Omega,\mathcal{A},\mathbf{P})$ is not mentioned unless it is absolutely necessary. After a long time of frustration, I have become quite comfortable with this language. But some things still trouble me. The following kind of reasoning comes up in the book I'm reading: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The author says that $(X_i)_{i\in I}$ is a family of random variables and specifies the distribution of each random variable. Then he phrases some (random) proposition $A((X_i)_{i_\in I})$ (this is a little imprecise, I hope you get the meaning) and talks about $\mathbf{P}[A((X_i)_{i_\in I}) \text{ holds}]$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question: Let $(\Omega&amp;#39;,\mathcal{A}&amp;#39;,\mathbf{P}&amp;#39;)$ be another probability space and $(Y_i)_{i\in I}$ random variables such that, for each $i\in I$, the distribution of $Y_i$ is the same as the distribution of $X_i$. Is it then obvious that $\mathbf{P}[(A(X_i)_{i_\in I}) \text{ holds}]=\mathbf{P}&amp;#39;[(A(Y_i)_{i_\in I}) \text{ holds}]$? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now my guess is that this is true, but needs a proof, which is not completely trivial in case $I$ is infinite, at least not for a beginner. However, in the book this problem isn't discussed at all. So did I miss something?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure whether the question was correctly understood, so I'll rephrase it a little.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $(\Omega,\mathcal{A},\mathbf{P})$ and $(\Omega&amp;#39;,\mathcal{A}&amp;#39;,\mathbf{P}&amp;#39;)$ be two probability spaces, $I$ a set, and $(X_i)_{i\in I}$ and $(Y_i)_{i\in I}$ families of random variables on $(\Omega,\mathcal{A},\mathbf{P})$ and $(\Omega&amp;#39;,\mathcal{A}&amp;#39;,\mathbf{P}&amp;#39;)$ respectively such that, for each $i\in I$, the distribution of $X_i$ is equal to the distribution of $Y_i$. Let $J$ be a countable subset of $I$ and $B_j$ a Borel set for each $j\in J$. The question is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it obvious that $\mathbf{P}\left[\bigcup_{j\in J}\{X_j\in B_j\}\right]=\mathbf{P}&amp;#39;\left[\bigcup_{j\in J}\{Y_j\in B_j\}\right]$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sets $B_j$ and the union over $J$ are just an example. What I mean, but cannot formalize: Let $A\in\mathcal{A}$ and $A&amp;#39;\in\mathcal{A}&amp;#39;$ such that there is an expression for $A$ in terms of the $X_i$, and $A&amp;#39;$ is given by the same expression replacing $X_i$ by $Y_i$ for each $i$. Is it obvious that $\mathbf{P}[A]=\mathbf{P}&amp;#39;[A&amp;#39;]$?&lt;/p&gt;&#xA;" OwnerUserId="1878" LastEditorUserId="1878" LastEditorDisplayName="" LastEditDate="2011-02-21T11:51:03.307" LastActivityDate="2011-02-21T12:39:59.737" Title="The role of the &quot;hidden&quot; probability space on which random variables are defined" Tags="&lt;probability-theory&gt;" AnswerCount="2" CommentCount="11" FavoriteCount="1" />
  <row Id="23007" PostTypeId="1" CreationDate="2011-02-21T01:33:22.027" Score="10" ViewCount="196" Body="&lt;p&gt;I recently learned that the only parallelizable spheres are $\mathbb{S}^1$, $\mathbb{S}^3$, and $\mathbb{S}^7$.  This led me to wonder:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is $T\mathbb{S}^2$?  Is it diffeomorphic to a more familiar space?  What about $T\mathbb{S}^n$ for $n \neq 1, 3, 7$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT (for precision): Is $T\mathbb{S}^2$ diffeomorphic to some finite product, connected sum, and/or quotient of spheres, projective spaces, euclidean spaces, and linear groups?&lt;/p&gt;&#xA;" OwnerUserId="640" LastEditorUserId="640" LastEditorDisplayName="" LastEditDate="2011-02-21T02:27:24.513" LastActivityDate="2011-02-21T07:37:08.740" Title="What is $T\mathbb{S}^2$?" Tags="&lt;differential-geometry&gt;&lt;differential-topology&gt;" AnswerCount="3" FavoriteCount="3" />
  <row Id="23008" PostTypeId="2" ParentId="23007" CreationDate="2011-02-21T01:59:02.140" Score="7" ViewCount="" Body="&lt;p&gt;Your question is perhaps too vague.  The tangent bundle has a definition and that's what it is -- presumably you're asking for more than this but you don't specify with any precision.  It would be good to edit your question to make it more precise. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One description of $TS^2$ is you take the mapping cylinder $SO_3 \to S^2$ where this map is the orbit of a single point in $SO_3$'s natural action on $S^2$ by isometries.  Technically, $TS^2$ is the above mapping cylindre after you erase the $SO_3$ boundary (to make it non-compact).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another description of $TS^2$ would be the configuration space of two points in $S^2$.  Precisely, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$C_2 S^2 = \{ (x,y) \in S^2 \times S^2 : x \neq y \}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can identify the two by a stereographic projection map construction.  There's many more such constructions.  But you really ought to say what you're looking for because the list is endless. &lt;/p&gt;&#xA;" OwnerUserId="642" LastActivityDate="2011-02-21T01:59:02.140" CommentCount="1" />
  <row Id="23009" PostTypeId="1" AcceptedAnswerId="23017" CreationDate="2011-02-21T02:00:04.683" Score="3" ViewCount="133" Body="&lt;p&gt;Given a $\sigma$-algebra on a Cartesian product of a collection of sets, do there always exist a $\sigma$-algebra on each set, so that their product $\sigma$-algebra on their Cartesian product will be the same as the given $\sigma$-algebra?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If yes, how to construct the $\sigma$-algebras on the sets from the $\sigma$-algebra on their Cartesian product?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If no, when they exist, how to construct the component $\sigma$-algebras? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks and regards! Any reference is appreciated as well.&lt;/p&gt;&#xA;" OwnerUserId="1281" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-21T02:51:45.570" LastActivityDate="2011-03-03T03:31:22.670" Title="From $\sigma$-algebra on product space to $\sigma$-algebra on component space" Tags="&lt;measure-theory&gt;" AnswerCount="2" CommentCount="6" />
  <row Id="23010" PostTypeId="2" ParentId="23005" CreationDate="2011-02-21T02:00:51.157" Score="2" ViewCount="" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The product $\sigma$-algebra on $S_1\times S_2$ is generated by sets of the form $A\times B$, with $A\in \mathbb{S}_1$ and $B\in\mathbb{S}_2$. The function $f$ is measurable if and only if the inverse image of a measurable set is measurable, and it suffices to check that the inverse image of a measurable set in a specific generating set for the $\sigma$-algebra is measurable. So it suffices to see if $f^{-1}(A\times B)$, with $A$ and $B$ as above, is measurable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $x\in f^{-1}(A\times B)$, then $f(x)\in A\times B$, so $f_1(x)\in A$ and $f_2(x)\in B$. Thus, $x\in f_1^{-1}(A)\cap f_2^{-1}(B)$. Conversely, if $x\in f_1^{-1}(A)\cap f_2^{-1}(B)$, then $f(x)\in A\times B$. So $f^{-1}(A\times B) = f_1^{-1}(A)\cap f_2^{-1}(B)$. Is this set in $\mathbb{F}$?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Same idea: what is $f^{-1}(A\times B)$? $(x,y)\in f^{-1}(A\times B)$ if and only if $x\in f_1^{-1}(A)$ and $y\in f_2^{-1}(B)$, so $f^{-1}(A\times B) = f_1^{-1}(A)\times f_2^{-1}(B)$. Is this set in the product $\sigma$-algebra $\mathbb{F}_1\times\mathbb{F}_2$?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Are the projection maps $\pi_i\colon (S_1\times S_2,\,\mathbb{S}_1\times\mathbb{S}_2)$ measurable? Is the composition of measurable functions measurable? What are $\pi_i\circ f$?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I'll leave 4 to you.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-21T05:17:05.980" LastActivityDate="2011-02-21T05:17:05.980" CommentCount="3" />
  <row Id="23011" PostTypeId="2" ParentId="23006" CreationDate="2011-02-21T02:01:00.983" Score="0" ViewCount="" Body="&lt;p&gt;The question is not very clear, but here is some idea. Let $U$ be a uniform$[0,1]$ random variable, and define a random process $X=\{X_t: t \in [0,1]\}$ by $X_t = \mathbf{1}(t=U)$, where $\mathbf{1}$ is the indicator function. Then, $X$ is identical in law to the zero process $Y$ defined by $Y_t = 0$ for all $t \in [0,1]$; that is ${\rm P}[X_{t_1} = 0, X_{t_2}=0,\ldots,X_{t_n}=0]=1$ for any choice of $n \geq 1$ and $0 \leq t_1 &amp;lt; \ldots \leq t_n \leq 1$. However, $X$ and $Y$ are quite different: $X$ is not continuous, its $\sup$ is equal to $1$, etc.&lt;/p&gt;&#xA;" OwnerUserId="2810" LastActivityDate="2011-02-21T02:01:00.983" CommentCount="1" />
  <row Id="23012" PostTypeId="1" AcceptedAnswerId="23021" CreationDate="2011-02-21T02:07:35.823" Score="10" ViewCount="350" Body="&lt;p&gt;This question is based on a homework exercise:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Let $m$ be a negative, square-free integer with at least two prime factors. Show that $\mathbb{Z}[\sqrt{m}]$ is not a &lt;a href=&quot;http://en.wikipedia.org/wiki/Principal_ideal_domain&quot; rel=&quot;nofollow&quot;&gt;PID&lt;/a&gt;.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In an aside comment in the book we're using (Milne), it was noted that a lot was known about quadratic extension with $m$ negative, but very little with $m$ positive. Why is this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have not solved the exercise, mainly because I can't think of a propery of negative numbers that positive numbers don't possess ($|m|\neq m$, but that's trivial).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seem there should be some relatively straightforward way to calculate class numbers for quadratic extensions with $m$ negative and composite. Or maybe the way to do this is to produce an ideal that is not principal - but then again I must find this property of negative numbers that separates them from positive.&lt;/p&gt;&#xA;" OwnerUserId="4284" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2011-02-22T02:14:37.483" LastActivityDate="2011-02-26T04:52:40.867" Title="What is so special about negative numbers $m$, $\mathbb{Z}[\sqrt{m}]$? " Tags="&lt;homework&gt;&lt;algebraic-number-theory&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="2" />
  <row Id="23013" PostTypeId="1" AcceptedAnswerId="23016" CreationDate="2011-02-21T02:14:37.783" Score="0" ViewCount="222" Body="&lt;p&gt;Hello everyone,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I got this problem within others as homework, and I don't know how to do it. Does anyone know how to start solving it? Thank you! Any help would be appreciated. Sorry for my writing in LaTex because for some reason I can't install mathjax, and so what I see is in Latex format/display. Thanks for understanding.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let x_1 and x_2 be independent exponentially distributed random variables with parameters $\lambda_1$ and $\lambda_2$ so that $Pr{X_i&amp;gt;t}=e^{-\lambda_it}$ for $t&amp;gt;=0$.&#xA;Let $N=1$ if $X_1&amp;lt;X_2$, and $N=2$ if $X_2&amp;lt;=X_1$, $U=\min\{X_1, X_2\}=X_N$, $V=\max\{X_1,X_2\}$ and $W=V-U=|X_1-X_2|$.&#xA;Show:&#xA;(a). $Pr\{N=1\}=\frac{\lambda_1}{\lambda_1+\lambda_2}$ and $Pr\{N=2\}=\frac{\lambda_2}{\lambda_1+\lambda_2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(b). $Pr\{U&amp;gt;t\}=e^{-(\lambda_1+\lambda_2)t}$ for $t&amp;gt;=0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(c). $N$ and $U$ are independent random variables&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(d). $Pr\{W&amp;gt;t|N=1\}=e^{-\lambda_2t}$ and $Pr\{W&amp;gt;t|N=2\}=e^{-\lambda_1t}$ for $t&amp;gt;=0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(e). $U$ and $W=V-U$ are independent random variables.&lt;/p&gt;&#xA;" OwnerUserId="6912" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-21T02:20:24.657" LastActivityDate="2011-02-22T16:39:09.800" Title="Independent exponentially distributed random variables" Tags="&lt;homework&gt;&lt;probability-theory&gt;&lt;stochastic-processes&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="23014" PostTypeId="2" ParentId="23009" CreationDate="2011-02-21T02:20:07.247" Score="2" ViewCount="" Body="&lt;p&gt;Suppose $A$ and $B$ are each 2-element sets.  It's easy to create a measure on the 4-element set $A\times B$ which is not the product of measures on $A$ and $B$.  Or perhaps I misunderstood you?&lt;/p&gt;&#xA;" OwnerUserId="2579" LastActivityDate="2011-02-21T02:20:07.247" CommentCount="5" />
  <row Id="23015" PostTypeId="2" ParentId="23007" CreationDate="2011-02-21T02:26:33.417" Score="7" ViewCount="" Body="&lt;p&gt;Here's how I think about it.  (Ryan Budney posted his answer while I was typing this.  One can think of this as a fleshing out of his answer).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, we need to understand the unit tangent bundle $T^1 S^2$.  Once we know this, we product this with $[0,\infty)$ and then quotient all points of the form $(u, 0)\in T^1S^2 \times\mathbb{R}$ somehow to get the 0 section $S^2$.  (This is precisely the mapping cylinder construction Ryan mentions).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Before we can talk about the &quot;unit tangent bundle&quot;, we must have a notion of length of vectors.  So in the background, pretend like I picked a Riemannian metric so lengths make sense.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I claim $T^1 S^2$ is diffeomorphic to $SO(3)$ (the collection of 3 x 3 orthogonal matrices of determinant 1) which is diffeomorphic to $\mathbb{R}P^3$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The map from $T^1 S^2$ to $SO(3)$ sends $(u,v)$ to the matrix with columns $u, v, u\times v$.  Here, I'm thinking of a unit tangent vector $v\in T_u S^2$ as a vector in $\mathbb{R}^3$ orthogonal to the vector $u$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The easiest way to see $SO(3)$ and $\mathbb{R}P^3$ are diffeomorphic is to note they are both quotients of $S^3 = SU(2)$ by the same quotienting map.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, we understand $T^1 S^2$, the unit length vectors in $TS^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To allow for length, we product with $[0,\infty)$.  Now, the only problem is the 0 section should be an $S^2$, and it's currently an $\mathbb{R}P^3$, so some quotienting must happen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What quotienting must happen?  Well, all the unit vectors at a given point must collapse to the point.  Well, there is the action of a circle on $T^1S^2$ given by rotation vectors clockwise (say) as seen from the normal vector to the sphere.  This action is clearly free.  Now, it's a fact that if you translate this circle action into the $SO(3)$ picture, the circle action is the Hopf action.  This implies that we identify $\mathbb{R}P^3$ with $S^2$ by quotienting by the Hopf action:  Two points in $\mathbb{R}P^3$ iff they are in the same Hopf orbit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Incidentally, I just learned a few days ago that $TS^2$ is not homeomorphic to $S^2\times \mathbb{R}^2$, though I'm still not sure how to prove it ;-).  (Of course, it's clear that they are not bundle isomorphic, but they could still be abstractly homeomorphic).  I don't know about the other nonparallizable tangent bundles, though.&lt;/p&gt;&#xA;" OwnerUserId="331" LastActivityDate="2011-02-21T02:26:33.417" CommentCount="5" />
  <row Id="23016" PostTypeId="2" ParentId="23013" CreationDate="2011-02-21T02:26:48.303" Score="2" ViewCount="" Body="&lt;p&gt;For a), use the law of total probability:&#xA;$$&#xD;&#xA;{\rm P}(X_1  &amp;lt; X_2 ) = \int_0^\infty  {{\rm P}(X_1  &amp;lt; X_2 |X_2  = t)f_{X_2 } (t)\,{\rm d}t} ,&#xD;&#xA;$$&#xA;where $f_{X_2}$ is the PDF of $X_2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For b), notice that $\min \{ X_1 ,X_2 \}  &amp;gt; t$ if and only if $X_1 &amp;gt; t$ and $X_2 &amp;gt; t$ (and use the fact that $X_1$ and $X_2$ are independent).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For c), calculate ${\rm P}(\min \{X_1,X_2 \}&amp;gt;t , X_1 &amp;gt; X_2 )$ using the law of total probability, conditioning on $X_2$. You should easily find that&#xA;$$&#xD;&#xA;{\rm P}(\min \{X_1,X_2 \}&amp;gt;t , X_1 &amp;gt; X_2 ) = {\rm P}(\min \{X_1,X_2 \}&amp;gt;t ){\rm P}(X_1 &amp;gt; X_2 ) = \frac{{\lambda _2 }}{{\lambda _1  + \lambda _2 }}e^{ - (\lambda _1  + \lambda _2 )t}. &#xD;&#xA;$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For d), note that&#xA;$$&#xD;&#xA;{\rm P}(|X_1  - X_2 | &amp;gt; t|N = 1) = \frac{{{\rm P}(X_2  - X_1  &amp;gt; t,X_1  &amp;lt; X_2 )}}{{{\rm P}(X_1  &amp;lt; X_2 )}} = \frac{{{\rm P}(X_2  &amp;gt; X_1  + t)}}{{{\rm P}(X_1  &amp;lt; X_2 )}},&#xD;&#xA;$$&#xA;and you should easily show using the law of total probability, conditioning on $X_1$, that &#xA;$$&#xD;&#xA;{P(X_2  &amp;gt; X_1  + t)} = \frac{{\lambda _1 }}{{\lambda _1  + \lambda _2 }}e^{ - \lambda _2 t}. &#xD;&#xA;$$&#xA;Note: The calculation for ${\rm P}(|X_1  - X_2 | &amp;gt; t|N = 2)$ is completely analogous.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;NOTE&lt;/strong&gt;: Since question e) is not so easy, I give more than hints. However, try solving a significant part of it by yourself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For e), it is straightforward to show, using that $N$ and $U$ are independent, that&#xA;$$&#xD;&#xA;{\rm P}(W &amp;gt; t | U=u) = {\rm P}(W &amp;gt; t | N=1, U=u){\rm P}(N=1) + {\rm P}(W &amp;gt; t | N=2, U=u){\rm P}(N=2).&#xD;&#xA;$$&#xA;For this purpose, you may replace $U=u$ by $U \in [u,u+{\rm d}u]$, where ${\rm d}u \to 0$, in order to condition on events with positive probability. Now, given $U=u$ and $N=1$, we have that $X_1 = u$ and that $X_2 - X_1$, by standard property of the exponential distribution, is exponential$(\lambda_2)$. Analogously, given $U=u$ and $N=2$, we have that $X_2 = u$ and that $X_1 - X_2$ is exponential$(\lambda_1)$. From this you should find that&#xA;$$&#xD;&#xA;{\rm P}(W &amp;gt; t | U=u) = \frac{{\lambda _1 }}{{\lambda _1  + \lambda _2 }}e^{ - \lambda _2 t}  + \frac{{\lambda _2 }}{{\lambda _1  + \lambda _2 }}e^{ - \lambda _1 t} .&#xD;&#xA;$$&#xA;Now we are done by&#xA;$$&#xD;&#xA;{\rm P}(W &amp;gt; t ) = {\rm P}(W &amp;gt; t ,N = 1) + {\rm P}(W &amp;gt; t ,N = 2),&#xD;&#xA;$$&#xA;as it gives us, by virtue of a) and d),&#xA;$$&#xD;&#xA;{\rm P}(W &amp;gt; t) = \frac{{\lambda _1 }}{{\lambda _1  + \lambda _2 }}e^{ - \lambda _2 t}  + \frac{{\lambda _2 }}{{\lambda _1  + \lambda _2 }}e^{ - \lambda _1 t} .&#xD;&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="2810" LastEditorUserId="2810" LastEditorDisplayName="" LastEditDate="2011-02-22T15:22:35.127" LastActivityDate="2011-02-22T15:22:35.127" CommentCount="6" />
  <row Id="23017" PostTypeId="2" ParentId="23009" CreationDate="2011-02-21T02:32:06.343" Score="3" ViewCount="" Body="&lt;p&gt;Let $A=\{0,1\}$, $B=\{0,1\}$, $\mathbb{S}=\{\emptyset, \{(0,0),(1,1)\}, \{(1,0),(0,1)\}, A\times B\}$. Then $\mathbb{S}$ is a $\sigma$-algebra on $A\times B$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there $\sigma$-algebras on $A$ and on $B$ whose product is $\mathbb{S}$? Well, there aren't that many $\sigma$-algebras on the two-element set. There's the total $\sigma$-algebra, $\mathcal{P}(\{0,1\})$, and the $\sigma$-algebra $\{\emptyset,\{0,1\}\}$, and that's it (if it contains any proper subset, then it contains its complement, and it is the entire thing). Call these $\mathbb{S}_1$ and $\mathbb{S}_2$, respectively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is $\mathbb{S}$ the product of two of these? No.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's not equal to $\mathbb{S}_1\times\mathbb{S}_i$, because this $\sigma$-algebra contains $\{1\}\times \{0,1\}$, but $\mathbb{S}$ does not. Symmetrically, it's not equal to $\mathbb{S}_i\times \mathbb{S}_1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the only possibility would be that it is equal to &#xA;$$\mathbb{S}_2\times\mathbb{S}_2 = \Bigl\{\emptyset,\{0,1\}\Bigr\} \times \Bigl\{\emptyset,\{0,1\}\Bigr\} = \Bigl\{ \emptyset, A\times B\Bigr\}$$&#xA;which is not equal to $\mathbb{S}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So $\mathbb{S}$ is a $\sigma$-algebra on $A\times B$ which is not a product of a $\sigma$-algebra on $A$ and a $\sigma$-algebra on $B$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it doesn't work for two, it has no hope of working for an infinite number of factors, so the potential distinction I suggested in the comments is in fact irrelevant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Added.&lt;/em&gt; To answer the added question: suppose that there &lt;em&gt;do&lt;/em&gt; exist $\sigma$-algebras $\mathbb{S}_i$ on $X_i$ such that $\mathbb{S}=\prod\mathbb{S}_i$. Any $A_i\in\mathbb{S}_i$ will necessarily satisfy that&#xA;$$A\times\prod_{j\neq i} X_j \in\mathbb{S}.$$&#xA;Conversely, suppose that $A\subseteq X_i$ is such that $A\times \prod_{j\neq i}X_j\in\mathbb{S}$. &lt;strike&gt;Looking at the projection from $\prod X_i$ to &#xA;$X_i$, you get that $A\in\mathbb{S}_i$.&lt;/strike&gt; The projection does not work; I'm fairly sure it is the case that such an $A$ will lie in $\mathbb{S}_i$, but I find myself unable to prove it right now. &lt;strong&gt;Edit:&lt;/strong&gt; This gap was filled by Morning in &lt;a href=&quot;http://math.stackexchange.com/questions/23227/further-questions-about-from-product-sigma-algebra-to-component-ones&quot;&gt;his answer to this question.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming this is the case, then $\mathbb{S}_i$ would consist exactly of the subsets of $X_i$ such that $A\times\prod_{j\neq i}X_j\in\mathbb{S}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can see how this fails in the example above. The collection of all $A\subseteq \{0,1\}$ such that $A\times\{0,1\}\in\mathbb{S}$ is just $\{\emptyset, \{0,1\}\}$, but the product of these does not exhaust the $\sigma$-algebra. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you define the $\mathbb{S}_i$ this way, then the product $\sigma$-algebra $\prod\mathbb{S}_i$ is always contained in $\mathbb{S}$, but need not be equal to it. &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-03T03:31:22.670" LastActivityDate="2011-03-03T03:31:22.670" CommentCount="7" />
  <row Id="23018" PostTypeId="1" AcceptedAnswerId="23019" CreationDate="2011-02-21T02:45:23.710" Score="6" ViewCount="293" Body="&lt;p&gt;I am working on this practice problem, and I was wondering if I could get some help.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a $T$:$\mathbb{R^{2x2}}\to \mathbf{P_{2}}$, that is, from 2x2 matrices to polynomials of degree at most 2. The transformation is given as following: &#xA;$$T\left(\begin{bmatrix}&#xD;&#xA;a &amp;amp; b\\ &#xD;&#xA;c &amp;amp; d&#xD;&#xA;\end{bmatrix}\right)&#xD;&#xA;= a-c+2d+(b+2c-d)t+(a-c+3d)t^{2}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To get the basis of kernel of $T$, I solved a system of equations needed to get the 'O' element in the $\mathbf{P_{2}}$ -- $a-c+2d=0$, $b+2c-d=0$ and $a-c+3d=0$. As a result, I got the basis of the kernel equal to $$\begin{bmatrix}&#xD;&#xA;1 &amp;amp; -2\\ &#xD;&#xA;1 &amp;amp; 0\\ &#xD;&#xA;\end{bmatrix}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When it comes to image, if I understand correctly, I need to factor out all the variables separately, to see what is it that they span. So I got $a(1+t^{2})+b(t)+c(-t^{2}+2t-1)+d(3t^{2}-t+2)$. So would I be correct in saying that these three polynomials (without the coefficients $a$, $b$, $c$, and $d$) form the basis for the image $T$? Thank you!&lt;/p&gt;&#xA;" OwnerUserId="5157" LastEditorUserId="5157" LastEditorDisplayName="" LastEditDate="2011-02-23T18:00:54.257" LastActivityDate="2011-02-23T18:00:54.257" Title="Basis for Kernel and Image of the following T" Tags="&lt;linear-algebra&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="23019" PostTypeId="2" ParentId="23018" CreationDate="2011-02-21T03:03:49.257" Score="4" ViewCount="" Body="&lt;p&gt;Your answers are correct, but for the image part a) you need some additional reasoning and b) you can considerably simplify the result. a) These polynomials form a generating set for the image of $T$; to show that they form a basis, you have to check whether they're linearly independent (they are). b) You've correctly described the image space, but in quite a complicated way. What's the dimension of the space spanned by your basis of three polynomials?  And what's the dimension of the entire codomain $\mathbf{P_2}$? What does that tell you about which space these elements actually span?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the way, the basis element of the kernel should be written as a $2\times2$ matrix, not as a column vector, and the basis is the set containing that element, not the element itself.&lt;/p&gt;&#xA;" OwnerUserId="6622" LastActivityDate="2011-02-21T03:03:49.257" CommentCount="2" />
  <row Id="23020" PostTypeId="1" AcceptedAnswerId="23112" CreationDate="2011-02-21T03:07:11.120" Score="6" ViewCount="106" Body="&lt;p&gt;I am in the process of learning about Mapping class groups. At this point it seems like most of what I've read involves very low dimensional (surfaces and 3-manifolds) applications.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was wondering if they were studied (or arise naturally) in higher dimensional settings?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, any references to their uses in homotopy theory would be appreciated.&lt;/p&gt;&#xA;" OwnerUserId="6851" LastActivityDate="2011-02-21T20:18:27.383" Title="Mapping class groups in higher dimensions" Tags="&lt;reference-request&gt;&lt;algebraic-topology&gt;" AnswerCount="2" />
  <row Id="23021" PostTypeId="2" ParentId="23012" CreationDate="2011-02-21T03:07:27.063" Score="4" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\ $ Consider factorizations like $\rm\ \ \sqrt{-pq}\ \sqrt{-pq}\ =\: -p\:q\ $  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note too $\rm\ p\ |\ d\ \Rightarrow\ (p,\:\sqrt{d})^2\ =\ p\ (p,d/p,\sqrt{d})\ =\ (p)\ $ since $\rm\:d\:$ squarefree $\rm \Rightarrow\ (p,d/p) = 1\:$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More generally, one can show that a quadratic number ring is a PID (or investigate its class group structure) by analysing the splitting or ramification of rational primes below the Minkowski bound. For example, this explains Euler's famous &lt;a href=&quot;http://math.stackexchange.com/questions/2561/behaviour-of-polynomials-in-a-pid/2590#2590&quot;&gt;prime-producing polynomials.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One explanation as to why this is simpler for complex vs. real quadratic number fields has to do with the fact that norms reflect &lt;em&gt;much&lt;/em&gt; of the multiplicative structure, and complex quadratic fields have simpler norms, being positive definite vs. indefinite sign for real quadratic fields.  Thus, in the first example I mention above, it is very simple to show that $\rm\ \sqrt{-pq}\ $ is irreducible since otherwise $\rm\ r\ |\ \sqrt{-pq}\ \Rightarrow\ rr&amp;#39;\ |\ pq\ $ in $\mathbb Z\:,\:$ so $\rm\ rr&amp;#39; = p\:$ or $\rm\:q\:,\:$ contra $\rm\ r\:r&amp;#39; = m^2 + pq\ n^2\ $ takes only &lt;em&gt;square&lt;/em&gt; values below $\rm\:pq\ (\Rightarrow n = 0)\:$. Notice how this proof depends crucially on the &lt;em&gt;positivity&lt;/em&gt; of the norm form. In a real field we have instead $\rm\ rr&amp;#39; = m^2 - pq\:n^2\ $ which, having indefinite sign, thwarts the proof. For example, in $\rm\ \mathbb Z[\sqrt 14]\ $ we have $\rm\ 2 = N(4 + \sqrt{14}),\ -7 = N(7 + 2\ \sqrt{14})\:,\:$ where $\rm\:N\:$ is the norm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact non-UFD number rings are characterized by this property, viz. $\rm\ \sqrt{-pq}\ $ being irreducible but having norm splitting into two or more primes. Indeed, it's not difficult to show that a number ring is a UFD iff its irreducibles have prime power norms, i.e. exactly one rational prime $\rm\:p\:$ occurs in norms of irreducible elements $\rm\ N(\pi) = p^n\:.\:$ More generally one can show that in many favorable contexts (e.g. Galois) that a number ring enjoys unique factorization iff its monoid of norms does. For references (Bumby and Dade, Lettl, Coykendall) see my &lt;a href=&quot;http://groups.google.com/group/sci.math/msg/e2d24bd27e600942&quot; rel=&quot;nofollow&quot;&gt;sci.math post on 19 Dec 2007.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2011-02-26T04:52:40.867" LastActivityDate="2011-02-26T04:52:40.867" />
  <row Id="23022" PostTypeId="2" ParentId="23018" CreationDate="2011-02-21T03:10:03.923" Score="3" ViewCount="" Body="&lt;p&gt;The simplest way to compute the image, abstractly, is to take a basis of the domain, and see what its image spans. So here, you could take the standard basis of $\mathbb{R}^{2\times      2}$, to wit,&#xA;$$\left(\begin{array}{cc}&#xD;&#xA;1 &amp;amp; 0\\&#xD;&#xA;0 &amp;amp; 0&#xD;&#xA;\end{array}\right),\quad&#xD;&#xA;\left(\begin{array}{cc}&#xD;&#xA;0 &amp;amp; 1\\&#xD;&#xA;0 &amp;amp; 0&#xD;&#xA;\end{array}\right),\quad&#xD;&#xA;\left(\begin{array}{cc}&#xD;&#xA;0 &amp;amp; 0\\&#xD;&#xA;1 &amp;amp; 0&#xD;&#xA;\end{array}\right),\quad&#xD;&#xA;\left(\begin{array}{cc}&#xD;&#xA;0 &amp;amp; 0\\&#xD;&#xA;0 &amp;amp; 1&#xD;&#xA;\end{array}\right).$$&#xA;Here, you have that the images of these basis vectors are, respectively,&#xA;$$1+t^2,\quad t,\quad, -1+2t-t^2,\quad 2-t+3t^2.$$&#xA;So you would want to find out the span of these four polynomials in $\mathbb{P}^2$ (note that, of course, these four polynomials must be linearly dependent; after all, $\dim(\mathbf{P}_2) = 3$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively, you can think about the matrix representation of $T$. Convince yourself that the image of $T$ corresponds to the columnspace of the matrix that represents $T$. One way to find the columnspace is to find a row-echelon form of the matrix, and then take the columns &lt;em&gt;in the original matrix&lt;/em&gt; that correspond to the columns that contain the pivots of the row-echelon form. Since you may have already computed the row-echelon form of the matrix in order to find a basis for the null-space, it's likely you have already done all the needed work and can just exploit it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, as you noted in comments, if you already know that the dimension of the image is $3$ (by the Dimension Theorem, aka the Rank-Nullity Theorem, say), and you know that $\mathbf{P}_2$ has dimension $3$, then you know the image is all of $\mathbf{P}_2$ will do. The descriptions above are the general procedures you might use with any linear transformation.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="6622" LastEditorDisplayName="" LastEditDate="2011-02-21T03:16:10.093" LastActivityDate="2011-02-21T03:16:10.093" />
  <row Id="23023" PostTypeId="2" ParentId="22999" CreationDate="2011-02-21T03:26:12.910" Score="0" ViewCount="" Body="&lt;p&gt;There is one nice one-to-one correspondence that may be set up here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First of all, there is clearly ${n+m \choose{n}}$ different arrangements that people can enter in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $(i,j)$ denote the moment at which $i+j$ people have paid for their ticket, where $i$ represents the number of people paying with a $50$ cent piece, $j$ the number of people paying with a $\$1$. Here it does not matter if everyone receives change or not. Every arrangement of people ends with $(n,m)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To make sure that change is available for everyone, it must always be the case that $i \ge j$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We count all the arrangements that are bad, i.e. that at some point someone does not recieve their change. When the first person who does not get change back enters, we have $(k,k+1)$ for some $k$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here, do the following trick: suppose all the people that had $\$1$'s, now wish to pay $50$ cents, and vice versa. So out of the remaining people, $m - (k+1)$ pay $50$ cents and $n - k$ pay $\$1$ bills. In total, we have $(k+m-(k+1),k+1+n-k) = (m-1, n+1)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We are given that $n \ge m$, so $n+1 &amp;gt; m-1$. So in this case, the number of people paying $\$1$ bills is strictly greater than the number of peple paying otherwise. Now notice, that &lt;strong&gt;in any arrangement of people entering&lt;/strong&gt; that ends with $(m-1,n+1)$ there must be a a first $k$, when we have $(k,k+1)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can now switch &lt;em&gt;&quot;who pays what&quot;&lt;/em&gt; as before, and we get back an arrangement that ends with $(n,m)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This creates a nice bijection between all bad arrangements (where someone does not get their change) that end with $(n,m)$ and &lt;strong&gt;all possible&lt;/strong&gt; arrangements that end with $(m-1,n-1)$. Counting as before, there are ${n+1+m-1 \choose{n+1}} = {n+m \choose{n+1}}$ possible arrangements ending with $(m-1, n-1)$. So the number of arrangements where everyone gets their change is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$${n+m \choose{n}} - {n+m \choose{n+1}} = {{n+m \choose{n}}} \dfrac{n+1-m}{n+1}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note also that the probability of a &lt;strong&gt;good&lt;/strong&gt; arrangement occuring comes out nicely to &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\dfrac{n-m+1}{n+1}$$&lt;/p&gt;&#xA;" OwnerUserId="5105" LastActivityDate="2011-02-21T03:26:12.910" />
  <row Id="23024" PostTypeId="2" ParentId="23012" CreationDate="2011-02-21T03:48:09.513" Score="8" ViewCount="" Body="&lt;p&gt;One way to think about this is in terms of the reduction theory of positive definite binary quadratic forms over $\mathbb{Z}$: see for instance Cox's book &lt;em&gt;Primes of the form $x^2 + ny^2$&lt;/em&gt;, or, for a short introduction, &lt;a href=&quot;http://math.uga.edu/~pete/8430notes3.pdf&quot; rel=&quot;nofollow&quot;&gt;these notes&lt;/a&gt;.  The key result here is that for each &lt;em&gt;positive definite&lt;/em&gt; primitive binary quadratic form there is a &lt;strong&gt;unique&lt;/strong&gt; Minkowski-reduced form equivalent to it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are looking at forms $x^2 + ab y^2$ with $1 \leq a \leq b$, $\operatorname{gcd}(a,b) = 1$, then you find that both&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$q_1(x,y) = x^2 + ab y^2$ and&lt;br&gt;&#xA;$q_2(x,y) = a x^2 + by^2$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;are Minkowski reduced, so the class number of the quadratic order of discriminant $-4ab$ must be at least $2$.  This is not the whole story, because you also have to deal with discriminants of the form $D \equiv 1 \pmod 4$, but a similar analysis can be done here.  Anyway, this gives you some feel for what is different between imaginary and real quadratic fields: the associated binary quadratic forms behave very differently.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Added&lt;/b&gt;: After seeing Weaam's (correct) answer, I looked back at the question and saw that it is asking something easier than I had thought: the question asks why $\mathbb{Z}[\sqrt{-m}]$ is not a PID for certain integers $m &amp;gt; 0$.  But as Weaam points out, it is easy to show that $\mathbb{Z}[\sqrt{-m}]$ is not a UFD for any $m &amp;gt; 2$: this also occurs in my lecture notes for the course I taught (more or less) based on Cox's book: see Corollary 6 of &lt;a href=&quot;http://math.uga.edu/~pete/8430notes1.pdf&quot; rel=&quot;nofollow&quot;&gt;these notes&lt;/a&gt; (the first handout in the course).  What I say above is also a complete solution to this question -- well, at least as long $m$ is not a square; in that case one could just argue that the ring is not integrally closed -- but is more complicated than is necessary.  What I was indicating is that for squarefree composite $m &amp;gt; 0$, the class number of the &lt;em&gt;full ring of integers&lt;/em&gt; of the imaginary quadratic field $\mathbb{Q}(\sqrt{-m})$ is greater than $1$.  This does seem to be most easily handled by elementary reduction theory / genus theory, so far as I know.  For yet another closely related result, see Landau's Theorem at the end of the latter set of notes.&lt;/p&gt;&#xA;" OwnerUserId="299" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2011-02-21T08:07:28.370" LastActivityDate="2011-02-21T08:07:28.370" CommentCount="1" />
  <row Id="23025" PostTypeId="1" CreationDate="2011-02-21T03:54:35.330" Score="5" ViewCount="136" Body="&lt;p&gt;Is $(\mathbf{V} \cap \mathbf{W})^{\bot}=(\mathbf{V}^{\bot} \cap \mathbf{W}^{\bot})$? I tried element-chasing, but I am getting confused when trying to determine mutual containment.&lt;/p&gt;&#xA;" OwnerUserId="7304" LastActivityDate="2011-02-21T04:19:27.533" Title="Is $(\mathbf{V} \cap \mathbf{W})^{\bot}=(\mathbf{V}^{\bot} \cap \mathbf{W}^{\bot})$?" Tags="&lt;vector-space&gt;" AnswerCount="2" CommentCount="5" />
  <row Id="23026" PostTypeId="2" ParentId="23025" CreationDate="2011-02-21T04:04:10.750" Score="7" ViewCount="" Body="&lt;p&gt;No.  Try $W=\{0\}$, $V\neq\{0\}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is true is that the perp of the span of two spaces is the intersection of the perps: $(V\vee W)^\perp=V^\perp\cap W^\perp$.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2011-02-21T04:04:10.750" />
  <row Id="23027" PostTypeId="2" ParentId="23025" CreationDate="2011-02-21T04:19:27.533" Score="4" ViewCount="" Body="&lt;p&gt;If the ambient space has dimension $n$, then taking the orthogonal complement takes a subspace of dimension $k$ to a subspace of dimension $n-k$. Intersection in general reduces the dimension. So your equation looks suspect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A nice analogy is from propositional logic: the orthogonal complement is similar to negation, and intersection is similar to logical and. So according to de Morgan's laws, if in one side of the equation you have $\cap$, in the other side you should have the dual operator, viz. $+$ (what Jonas uses $\lor$ for).&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2011-02-21T04:19:27.533" />
  <row Id="23028" PostTypeId="2" ParentId="23012" CreationDate="2011-02-21T04:24:37.060" Score="4" ViewCount="" Body="&lt;p&gt;By finding an irreducible that's not prime for d&amp;lt;-2 should suffice as a prime is the same as an irreducible in a PID.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A good candidate is 2 which is not prime in $Z[\sqrt{d}]$, since it divides $d(d-1)$ but not $\frac{d \pm \sqrt{d}}{2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To show that 2 is irreducible assume to the contrary that $2 = \alpha \beta$  where neither a units. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take the norm, but note that the $N(a + b\sqrt{d}) = a^2 - db^2$ is multiplicative, always nonnegative, and N(u)=1 iff u is unit, and units are +1, -1 which wouldn't be the case if d were positive, as Arturo Magidin suggested in a previous comment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then $N(\alpha)=N(\beta)=2$, i.e $a^2 - bd^2 = \pm 2$. Then you may deduce that if d&amp;lt;-2 then no such a or b exists, and the result follows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Addendum:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let d&amp;lt;-5 be squarefree such that d = pr, where p prime.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The ideal $(p, \sqrt{d})$ is not principal in $Z[\sqrt{d}]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose $Z[\sqrt{d}]$ is PID, then $(p, \sqrt{d}) = (n)$ for some $n = a + b\sqrt{d}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Taking the norm, then $N(n) = a^2 - db^2$ divides d (inclusion of ideals).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If b = |1|. Then $a^2 \pm d$ divides d and a = 0. But then $N(n)=d$ divides $N(p)=p^2$, but d = pq. Case b &gt;|1| is also trivial.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Case b = 0. Then $a^2$ divides d, and then $a=1$. Hence (n) = (1) = $Z[\sqrt{d}] = (p, \sqrt{d})$. Then $1 = px + \sqrt{d}y$ for x, y in $Z[\sqrt{d}]$, but then $ps - dt = 1$, which is impossible since d = pr.&lt;/p&gt;&#xA;" OwnerUserId="1746" LastEditorUserId="1746" LastEditorDisplayName="" LastEditDate="2011-02-21T06:41:14.097" LastActivityDate="2011-02-21T06:41:14.097" CommentCount="0" />
  <row Id="23029" PostTypeId="1" AcceptedAnswerId="23053" CreationDate="2011-02-21T04:44:23.480" Score="1" ViewCount="57" Body="&lt;p&gt;Hello,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I used computer program to generate numbers of this form, then verify if they are amicable. However, it took too long to finish, is there any special technique to verify if a pair is amicable?    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One pair that I found is $m = 284 = 2^{2}.71, n = 220 = 2^{2}.5.11$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks,&lt;br&gt;&#xA;Chan&lt;/p&gt;&#xA;" OwnerUserId="5711" LastActivityDate="2011-02-21T11:36:07.787" Title="Find amicable pair numbers of the from $m = 2^{k}a, n = 2^{k}bc$, where $k &lt; 35$ and a, b, c are odd primes?" Tags="&lt;elementary-number-theory&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="23030" PostTypeId="1" AcceptedAnswerId="23073" CreationDate="2011-02-21T05:43:32.080" Score="3" ViewCount="122" Body="&lt;p&gt;This is a continuation of &lt;a href=&quot;http://math.stackexchange.com/questions/21945/infinite-product-of-measurable-spaces&quot;&gt;my previous question&lt;/a&gt;. I admit I still don't fully understand the concept of infinite product of $\sigma$-algebras.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $(E_i, \mathbb{B}_i)$ be measurable spaces, where $i \in I$ is an index set, possibly infinite.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understood the definition of product sigma algebra. However, I am not sure after complement and/or countable union/intersection on those generating subsets, if $\prod_{i \in I} B_i$, $\forall B_i \in \mathbb{B}_i$, is  measurable wrt the product sigma algebra $\prod_{i \in I} \mathbb{B}_i$? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was expecting a &quot;not always&quot; answer to my first question. In this scenario, I was wondering if the product of all proper measurable subsets is always nonmeasurable wrt the product sigma algebra, i.e. is $\prod_{i \in I} B_i$, $\forall B_i \in \mathbb{B}_i$ and $B_i \neq E_i$, always nonmeasurable wrt the product sigma algebra $\prod_{i \in I} \mathbb{B}_i$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could you explain why? Thanks and regards!&lt;/p&gt;&#xA;" OwnerUserId="1281" LastEditorUserId="1281" LastEditorDisplayName="" LastEditDate="2011-02-21T06:11:18.503" LastActivityDate="2011-04-11T10:46:58.717" Title="Is product of measurable subsets measurable wrt infinite product $\sigma$-algebra?" Tags="&lt;measure-theory&gt;" AnswerCount="1" CommentCount="15" />
  <row Id="23031" PostTypeId="1" CreationDate="2011-02-21T06:17:03.520" Score="2" ViewCount="102" Body="&lt;p&gt;At the moment, I'm studying Ergodic Theory and I find myself a little stuck. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Weyl Equidistribution Theorem states that the following are equivalent:&#xA;1. For any $f \in L^{1}([0,1])$ and sequence $\{ u_{k} \}_{k \geq 1}$ on $[0,1]$,&#xA;\begin{eqnarray}&#xA;\lim_{n \to \infty} \frac{1}{n} \sum_{k = 1}^{n} f( u_{k} ) = \int_{0}^{1} f(x) \, dx.&#xA;\end{eqnarray}&#xA;2. The sequence $\{ u_{k} \}_{k \geq 1}$ is uniformly distributed modulo 1, i.e., equidistributed on $[0,1]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For fixed real $b, t &amp;gt; 0$, how does one compute the following limit?&#xA;\begin{eqnarray}&#xA;\lim_{a \to \infty} \tfrac{1}{a t} \sum_{i = 1}^{at} \lbrace b ( t - \tfrac{i}{a} ) \rbrace&#xA;\end{eqnarray}&#xA;where $\lbrace \cdot \rbrace$ denotes the fractional part function. It is known that the sequence of fractional parts of the integral multiples of the irrational $\alpha$, i.e., $\{ \{ k \alpha \} \}_{k \geq 1}$, is equidistributed on $[0,1]$. I'm not sure how to work with the fractional part in the sum above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Update&lt;/em&gt;: For fixed real $a, b &amp;gt; 0$, how does one compute the following limit?&#xA;\begin{eqnarray}&#xA;\lim_{t \to \infty} \tfrac{1}{a t} \sum_{i = 1}^{at} \lbrace b ( t - \tfrac{i}{a} ) \rbrace&#xA;\end{eqnarray}&#xA;Unfortunately, the Riemann summation trick doesn't work in this case. (I believe the answer is $\frac{1}{2}$ in the general case, and I think it follows from the above theorem and the average value of $\{ \cdot \}$ on $[0,1]$. However, if $a$ divides $b$, then the limit is $0$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="2720" LastEditorUserId="2720" LastEditorDisplayName="" LastEditDate="2011-02-22T14:42:57.527" LastActivityDate="2011-02-22T14:42:57.527" Title="Weyl Equidistribution Theorem and a Limit" Tags="&lt;number-theory&gt;&lt;real-analysis&gt;&lt;ergodic-theory&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" />
  <row Id="23032" PostTypeId="1" AcceptedAnswerId="23034" CreationDate="2011-02-21T06:32:55.293" Score="2" ViewCount="78" Body="&lt;p&gt;Next question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $X$ be a wedge of two circles, $X=A \vee B$ and let $x_0$ denote the point of tangency.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $q:S^1 \to X$ be the identification map that identifies -1 and 1 $\left(q\left(-1\right)=x_0=q\left(1\right)\right)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is to show that $(S^1,q)$ is not a covering space of $X$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question: Does this &quot;fail&quot; at the vertex $x_0$? &lt;/p&gt;&#xA;" OwnerUserId="2219" LastActivityDate="2011-02-21T06:46:32.747" Title="Covering space of a wedge of two circles" Tags="&lt;algebraic-topology&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="23034" PostTypeId="2" ParentId="23032" CreationDate="2011-02-21T06:46:32.747" Score="3" ViewCount="" Body="&lt;p&gt;Yes, indeed!  It fails exactly at $x_0$ (and nowhere else).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To show this, look at the inverse image of a small open neighbourhood $U$ of $x_0$.  Intuitively, it should be clear that this can’t be expressed as a disjoint union of copies of $U$.  To show this rigorously, first do a warm-up exercise (or recall it, if you’ve seen it before): show that the letter X is not homeomorphic to an interval.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That approach shows very concretely how the definition of covering space fails here.  If you have already seen the path lifting property for covering spaces, then that can be used to give a nice, less direct proof that this example is not a covering space.  Just look at a path going through $x_0$ the “wrong” way…&lt;/p&gt;&#xA;" OwnerUserId="2439" LastActivityDate="2011-02-21T06:46:32.747" CommentCount="1" />
  <row Id="23035" PostTypeId="2" ParentId="23007" CreationDate="2011-02-21T07:37:08.740" Score="4" ViewCount="" Body="&lt;p&gt;A slightly random answer, but if we concretely identify $TS^{n-1}$ as the embedded real manifold $\{ (x, v) \in \mathbb{R}^n \times \mathbb{R}^n : \| x \| = 1, \langle x, v \rangle = 0 \}$, then it is diffeomorphic to the complex affine quadric $\{ (z_1, \ldots, z_n) \in \mathbb{C}^n : z_1^2 + \cdots + z_n^2 = 1 \}$. (This was an amusing homework exercise I did yesterday.)&lt;/p&gt;&#xA;" OwnerUserId="5191" LastActivityDate="2011-02-21T07:37:08.740" />
  <row Id="23036" PostTypeId="1" CreationDate="2011-02-21T07:41:12.190" Score="2" ViewCount="140" Body="&lt;p&gt;In Wikipedia, bilinear mapping to the underlying field $f \in \operatorname{Hom}(V \otimes V, k)$ is defined as non-degenerate iff $X \mapsto f(-,X)$ is an isomorphism to the continuous (?) dual.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried to prove the proposition that $X \mapsto f(-,X)$ is an iso iff $X \mapsto f(X,-)$ is an iso, but only managed to prove that this proposition is equivalent to the proposition that $\varphi := \lbrace(f(-,X), f(X,-))\rbrace$ is an automorphism of $V^*$ (and a mapping, of course).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is the proposition true? I assume it's likely because there is no mention of left vs. right degeneracy on the wiki.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm asking because I'm trying to study orthogonal complements in the most general setting, so I wanted to see what conditions on $f$ are required for it to define complement, when $V \cong U \oplus U^\perp$ and when left and right complements coincide.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: I am specifically concerned about the infinite-dimensional case. For the finite-dimensional case, the proposition is easy to prove using a variety of strategies (e.g. via determinants).&lt;/p&gt;&#xA;" OwnerUserId="3793" LastEditorUserId="3793" LastEditorDisplayName="" LastEditDate="2011-02-21T08:22:58.063" LastActivityDate="2011-05-22T09:00:27.387" Title="Left vs. right degeneracy" Tags="&lt;linear-algebra&gt;" AnswerCount="1" CommentCount="6" />
  <row Id="23037" PostTypeId="2" ParentId="23036" CreationDate="2011-02-21T08:01:56.447" Score="0" ViewCount="" Body="&lt;p&gt;Let $f:V\otimes V\to k$ be a bilinear form, and suppose that $k$ is a field and that $V$ is finite dimensional.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Define $f_l$, $f_r:V\to V^*$ by setting $f_l(v)(w)=f(v\otimes w)$ and $f_r(v)(w)=f(w\otimes v)$ for all $v$, $w\in V$. Suppose moreover that $f_l$ is not injective, so that there exists a $v_0\in V\setminus0$ such that $f_l(v_0)=0$, that is, for all $w\in V$ we have $f(v_0\otimes w)=0$. &#xA;In other words, for all $w\in V$ we have that $\langle v_0\rangle\subseteq\ker f_r(w)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $\bar V=V/\langle v_0\rangle$ and let $\pi:V\to\bar V$ be the quotient map.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This implies that for each $w\in V$ there is a map $g(w)\in\bar V^*$ such that $f_r(w)=g(w)\circ\pi$. One easily sees that $g:V\to\bar V^*$ is linear, so we have a factorization $f_r=\pi^t\circ g$ where $\pi^t:\bar V^*\to V^*$ is the transposed of $\pi$. Since the rank of $g$ is at most $\dim V-1$, the rank of $f_r$ is almost $\dim V-1&amp;lt;\dim V^*$. It follows that $f_r$ is not surjective.&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2011-02-21T08:01:56.447" CommentCount="3" />
  <row Id="23038" PostTypeId="2" ParentId="23031" CreationDate="2011-02-21T08:10:50.473" Score="2" ViewCount="" Body="&lt;p&gt;First make the substitution $n = a t$, then your sum becomes $$\frac{1}{n} \sum_{i=1}^n \{(b t) (1 - i/n) \}.$$ Now this is really just a Riemann sum which converges (as $n \rightarrow \infty$) to&#xA;$$ \int_0^1 \{(bt) (1-i) \} di$$&#xA;Putting $u = (bt)(1-i)$ the integral becomes&#xA;$$ \frac{1}{b t} \int_0^{bt} \{u\} du $$&#xA;and this is equal to&#xA;$$\frac{1}{2 bt} ( \lfloor{b t\rfloor} + \{ b t \}^2 ).$$&#xA;(Because there are $\lfloor{b t\rfloor}$ triangles of area $1/2$ and one triangle of area $\{ b t \}^2 / 2$.)&lt;/p&gt;&#xA;" OwnerUserId="899" LastActivityDate="2011-02-21T08:10:50.473" CommentCount="4" />
  <row Id="23039" PostTypeId="1" CreationDate="2011-02-21T08:34:08.280" Score="1" ViewCount="108" Body="&lt;p&gt;How can I show that the set of rational numbers with denominator a power of two form a dense subset of the reals?&lt;/p&gt;&#xA;" OwnerUserId="7485" LastActivityDate="2011-02-21T19:59:44.367" Title="How can I show that the set of rational numbers with denominator a power of two form a dense subset of the reals?" Tags="&lt;real-analysis&gt;" AnswerCount="4" CommentCount="1" />
  <row Id="23040" PostTypeId="2" ParentId="23039" CreationDate="2011-02-21T08:40:01.410" Score="1" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT:&lt;/strong&gt; Given $x \in \mathbb{R}$, how can you use $\frac{\left \lfloor 2^kx \right \rfloor}{2^k}$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here $\lfloor \cdot \rfloor$ is the floor function, which in this case sends $2^kx$ to the nearest integer less than or equal to $2^kx$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rational numbers whose denominator is a power of two are called &lt;em&gt;dyadic rationals&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="5105" LastEditorUserId="5105" LastEditorDisplayName="" LastEditDate="2011-02-21T09:55:40.610" LastActivityDate="2011-02-21T09:55:40.610" CommentCount="3" />
  <row Id="23041" PostTypeId="2" ParentId="22584" CreationDate="2011-02-21T08:43:36.533" Score="4" ViewCount="" Body="&lt;p&gt;This diagram might help:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/jXEhS.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For any length $l &amp;gt; \overline{AB}$, we can find a circle passing through $AB$ such that the length of a circular arc between $A$ and $B$ is equal to $l$. In the diagram above, suppose that the red, dotted arc $ADB$ and the upper part of the circle both have length $l$. Note that the region bounded by $ADBC$ (i.e.: the pink area) has the same perimeter length as the circle. &lt;/p&gt;&#xA;" OwnerUserId="2643" LastEditorUserId="2643" LastEditorDisplayName="" LastEditDate="2011-02-21T21:12:17.997" LastActivityDate="2011-02-21T21:12:17.997" CommentCount="2" />
  <row Id="23042" PostTypeId="1" CreationDate="2011-02-21T08:46:23.160" Score="1" ViewCount="88" Body="&lt;p&gt;Does knowing the exact number of $\mathbb{Z}^{n}$-lattice points in an $n$-sphere of arbitrary radius help one compute $n$-sphere packing density and kissing number?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="2720" LastEditorUserId="2720" LastEditorDisplayName="" LastEditDate="2011-02-21T08:53:12.590" LastActivityDate="2011-02-21T08:53:12.590" Title="Counting Lattice Points in an $n$-sphere and Sphere Packing" Tags="&lt;number-theory&gt;&lt;geometry&gt;" CommentCount="4" FavoriteCount="1" />
  <row Id="23043" PostTypeId="2" ParentId="23013" CreationDate="2011-02-21T09:38:16.850" Score="2" ViewCount="" Body="&lt;p&gt;This is to show that one can solve (e) and similar questions in a fully automatized way. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One is asked to show that $U$ and $W$ are independent, let us be more ambitious and try to compute the distribution of $(U,W)$. If this distribution is a product, we are done. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Question&lt;/strong&gt; How to compute the distribution of any random variable $Z$? &lt;br&gt;&lt;strong&gt;Answer&lt;/strong&gt; By writing $E(\varphi(Z))$ as the integral of $\varphi$ with respect to a measure $\mu$, for every (bounded measurable) function $\varphi$. Then $\mu$ is the distribution of $Z$. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And this is actually often quite easy to do...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let us see what happens for $Z=(U,W)$. The first step is to replace $Z$ by a function of $(X_1,X_2)$, in the case at hand,&#xA;$$&#xD;&#xA;\varphi(Z)=\varphi(\min\{X_1,X_2\},\max\{X_1,X_2\}-\min\{X_1,X_2\}).&#xD;&#xA;$$&#xA;Now the RHS is a (quite ugly) function of $(X_1,X_2)$ but this does not matter. The only important thing is that the distribution of $(X_1,X_2)$ has density $f_1(x_1)f_2(x_2)$, hence, like for any function of $(X_1,X_2)$, by definition of the distribution of $(X_1,X_2)$, one has&#xA;$$&#xD;&#xA;E(\varphi(Z))=\int\varphi(\min\{x_1,x_2\},\max\{x_1,x_2\}-\min\{x_1,x_2\})f_1(x_1)f_2(x_2)\mathrm{d}x_1\mathrm{d}x_2.&#xD;&#xA;$$&#xA;Up to this point, everything is general. Now one begins to use the max/min thing. This forces us to decompose the integral into two parts, one for the domain where $x_1\le x_2$ and the other for the domain $x_1&amp;gt;x_2$. This decomposition yields $E(\varphi(Z))$ as $(*)+(**)$ with&#xA;$$&#xD;&#xA;(*)=\int\varphi(x_1,x_2-x_1)f_1(x_1)f_2(x_2)\mathbf{1}_{x_1&amp;lt;x_2}\mathrm{d}x_1\mathrm{d}x_2,&#xD;&#xA;$$&#xA;and&#xA;$$&#xD;&#xA;(**)=\int\varphi(x_2,x_1-x_2)f_1(x_1)f_2(x_2)\mathbf{1}_{x_1&amp;gt;x_2}\mathrm{d}x_1\mathrm{d}x_2.&#xD;&#xA;$$&#xA;Recall that our goal is to write $E(\varphi(Z))$ as&#xA;$$&#xD;&#xA;(o)=\int\varphi(u,w)\mathrm{d}\mu(u,w),&#xD;&#xA;$$&#xA;for a given measure $\mu$. Let us rewrite $(*)$ and $(**)$ with this goal in mind. The changes of variables $[u=x_1,w=x_2-x_1]$ in $(*)$ and $[u=x_2,w=x_1-x_2]$ in $(**)$ lead to&#xA;$$&#xD;&#xA;(*)=\int\varphi(u,w)f_1(u)f_2(u+w)\mathbf{1}_{w&amp;gt;0}\mathrm{d}u\mathrm{d}w,&#xD;&#xA;$$&#xA;and to&#xA;$$&#xD;&#xA;(**)=\int\varphi(u,w)f_1(u+w)f_2(u)\mathbf{1}_{w&amp;gt;0}\mathrm{d}u\mathrm{d}w.&#xD;&#xA;$$&#xA;Comparing $(o)$ with $(*)+(**)$, one sees that the only way to make them equal for every $\varphi$ is that $\mathrm{d}\mu(u,w)=g(u,w)\mathrm{d}u\mathrm{d}w$ with&#xA;$$&#xD;&#xA;g(u,w)=[f_1(u)f_2(u+w)+f_1(u+w)f_2(u)]\mathbf{1}_{w&amp;gt;0}.&#xD;&#xA;$$&#xA;We are done and two things are to be noted: first, all these steps are fully automatic, and second, the formula for $g$ is valid for any $(U,W)$ based on independent $X_1$ and $X_2$ with densities $f_1$ and $f_2$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case at hand, $f_i(x)=\lambda_i\mathrm{e}^{-\lambda_i x}$ for $x&amp;gt;0$, hence, for every $u&amp;gt;0$ and $w&amp;gt;0$,&#xA;$$&#xD;&#xA;g(u,w)=\lambda_1\lambda_2\mathrm{e}^{-(\lambda_1+\lambda_2)u}[\mathrm{e}^{-\lambda_1w}+\mathrm{e}^{-\lambda_2w}].&#xD;&#xA;$$&#xA;The function $g(u,w)$ is a product $g_1(u)g_2(w)$ hence $(U,W)$ is independent.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And naturally, this proves simultaneously that the functions $g_1$ and $g_2$ are the densities of the distributions of $U$ and $W$, up to multiplicative positive constants.&lt;/p&gt;&#xA;" OwnerUserId="6179" LastEditorUserId="6179" LastEditorDisplayName="" LastEditDate="2011-02-22T16:39:09.800" LastActivityDate="2011-02-22T16:39:09.800" CommentCount="1" />
  <row Id="23044" PostTypeId="1" AcceptedAnswerId="23530" CreationDate="2011-02-21T09:50:09.057" Score="2" ViewCount="164" Body="&lt;p&gt;Suppose we divide the the interval $[0,1]$ into $t$ equal intervals labeled $i_1$ upto $i_t$, then we make a function $f(t,x)$ that returns $1$ if $x$ is in $i_n$ and $n$ is odd, and $0$ if $n$ is even.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is $\lim_{t \rightarrow \infty} f(t,1/3)$?&lt;br&gt;&#xA;What is $\lim_{t \rightarrow \infty} f(t,1/2)$?&lt;br&gt;&#xA;What is $\lim_{t \rightarrow \infty} f(t,1/\pi)$?&lt;br&gt;&#xA;What is $\lim_{t \rightarrow \infty} f(t,x)$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;joriki clarification in comments is correct, does $\lim_{t \rightarrow \infty} f(t,1/\pi)$ exist, is it 0 or 1 or (0 or 1) or undefined?&#xA;Is it incorrect to say that is (0 or 1)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a way to express this:&lt;br&gt;&#xA;$K=\lim_{t \rightarrow \infty} f(t,x)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;K, without limit operator ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think to say K is simply undefined is an easy way out. Something undefined cant have properties. Does K have any properties? Is K a concept?&lt;/p&gt;&#xA;" OwnerUserId="4929" LastEditorUserId="4929" LastEditorDisplayName="" LastEditDate="2011-02-24T10:29:07.257" LastActivityDate="2011-02-24T17:07:57.833" Title="Limit of function of a set of intervals labeled i_n in [0,1]" Tags="&lt;functions&gt;&lt;limit&gt;&lt;convergence&gt;" AnswerCount="3" CommentCount="11" />
  <row Id="23045" PostTypeId="1" CreationDate="2011-02-21T10:04:33.170" Score="0" ViewCount="55" Body="&lt;p&gt;How can I show for some transcendental number $\alpha \in \mathbb{C}$, that two distinct polynomials of the form $b_{n}\alpha^n + \cdot \cdot \cdot + b_{2}\alpha^2 + b_{1}\alpha$ and $a_{m}\alpha^m + \cdot \cdot \cdot + a_{2}\alpha^2 + a_{1}\alpha$ can never equal each other?&lt;/p&gt;&#xA;" OwnerUserId="7485" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2011-02-22T23:49:48.203" LastActivityDate="2011-02-22T23:49:48.203" Title="&quot;Non-equality&quot; of distinct polynomials with the same transcendental argument" Tags="&lt;polynomials&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="23047" PostTypeId="2" ParentId="23039" CreationDate="2011-02-21T10:13:30.993" Score="4" ViewCount="" Body="&lt;p&gt;Just expand the reals in base 2. With this convention, the rationals of the form $m/2^r$ will be identified with those reals whose expansion's digit is definitely $0$, i.e. those which have a &quot;finite&quot; expansion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is evident that you can get a finitely expanded real as close as you want to any fixed real $r$, just cut away all the digits after the $n$-th, whatever $n$ you choose (exactly as you do with decimal expansions!)&lt;/p&gt;&#xA;" OwnerUserId="688" LastActivityDate="2011-02-21T10:13:30.993" />
  <row Id="23048" PostTypeId="2" ParentId="23039" CreationDate="2011-02-21T10:18:25.380" Score="0" ViewCount="" Body="&lt;p&gt;A nice sequence that I can think of would be by looking at the convergents of the continued fraction for any given $x \in \mathbb{R}$, and associating a diadic number to each successive convergent, such that the power of two in the denominator is the &quot;closest&quot; value to the denominator of the convergent. &lt;/p&gt;&#xA;" OwnerUserId="7485" LastActivityDate="2011-02-21T10:18:25.380" CommentCount="1" />
  <row Id="23050" PostTypeId="2" ParentId="23020" CreationDate="2011-02-21T10:39:24.470" Score="3" ViewCount="" Body="&lt;p&gt;Dear Patrick,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you mentioned, people have been studying the mapping class groups mostly in connection with (two-dimensional) surfaces, and also some work have been done for 3-manifolds - here the most notable result is Kojima's theorem which states that every finite group is the  mapping class group of a compact hyperbolic 3-manifold.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know only a very few cases, where mapping class groups appear in higher dimensions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One is a recent generalization of Kojima's theorem for higher dimensional hyperbolic manifolds by Belolipetsky and Lubotsky. See: M. Belolipetsky and A. Lubotzky, Finite Groups and Hyperbolic Manifolds, Invent. Math. 162 (2005) 459–472.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another one is the use of mapping class groups (of 2d surfaces) in the study of symplectic 4-manifolds (and even 6-manifolds) in &lt;a href=&quot;http://math.berkeley.edu/~auroux/papers/mcg-farb.pdf&quot; rel=&quot;nofollow&quot;&gt;http://math.berkeley.edu/~auroux/papers/mcg-farb.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All the best,&#xA;Zoltan&lt;/p&gt;&#xA;" OwnerUserId="7317" LastActivityDate="2011-02-21T10:39:24.470" CommentCount="1" />
  <row Id="23052" PostTypeId="2" ParentId="23045" CreationDate="2011-02-21T11:08:23.420" Score="3" ViewCount="" Body="&lt;p&gt;That follows directly from the definition of transcendental number: if $p(\alpha)=q(\alpha)$ then $p(\alpha)-q(\alpha)=0$.&lt;/p&gt;&#xA;" OwnerUserId="589" LastActivityDate="2011-02-21T11:08:23.420" />
  <row Id="23053" PostTypeId="2" ParentId="23029" CreationDate="2011-02-21T11:36:07.787" Score="1" ViewCount="" Body="&lt;p&gt;If you are looking for a nice program for finding amicable pairs, there is a related question on &lt;a href=&quot;http://stackoverflow.com/questions/4218056/how-to-optimize-code-for-finding-amicable-pairs&quot;&gt;stackoverflow&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://tutoree7.pastebin.com/wKvMAWpT&quot; rel=&quot;nofollow&quot;&gt;Here&lt;/a&gt; is a code from this question. Hope that helps.&lt;/p&gt;&#xA;" OwnerUserId="5105" LastActivityDate="2011-02-21T11:36:07.787" CommentCount="1" />
  <row Id="23054" PostTypeId="1" CreationDate="2011-02-21T11:37:14.383" Score="2" ViewCount="227" Body="&lt;p&gt;How do I calculate the distance between the line joining the two points on a spherical surface and another point on same surface? I have illustrated my problem in the image below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/BdHxo.png&quot; alt=&quot;Sphere&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the above illustration, the points A, B and X lies on a spherical surface, I need to find the distance between points (A,B) and X. I am not a mathematics guy. If possible please illustrate me the solution as non-mathematics guys could understand.  Thanks.&lt;/p&gt;&#xA;" OwnerUserId="7319" LastEditorUserId="3305" LastEditorDisplayName="" LastEditDate="2011-02-21T18:08:28.030" LastActivityDate="2011-02-24T21:59:55.650" Title="How to find the distance between a point and line joining two points on a sphere?" Tags="&lt;analytic-geometry&gt;&lt;spherical-geometry&gt;&lt;spherical-coordinates&gt;" AnswerCount="4" CommentCount="6" FavoriteCount="1" />
  <row Id="23055" PostTypeId="1" CreationDate="2011-02-21T11:47:20.847" Score="3" ViewCount="130" Body="&lt;p&gt;I need to express the value of $\pi$ as numerator/denominator. What is the best pair considering that the numerator should be less than or equal to $2^{62}$? Or how to get this pair?  &lt;/p&gt;&#xA;" OwnerUserId="5403" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2011-02-21T16:10:16.093" LastActivityDate="2011-02-21T16:10:16.093" Title="What is the best numerator and denominator couple to get the value of $\pi$?" Tags="&lt;pi&gt;" CommentCount="3" />
  <row Id="23056" PostTypeId="1" CreationDate="2011-02-21T12:02:19.297" Score="3" ViewCount="154" Body="&lt;p&gt;Hello,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am trying to calculate the touching point of two circles. I have the following information.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Circle $1$:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Centre $(h,k)$&lt;/li&gt;&#xA;&lt;li&gt;Radius $r_1$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Circle $2$:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Point on Circumference $(x_1,y_1)$&lt;/li&gt;&#xA;&lt;li&gt;Radius $r_2$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;From this I would like to calculate the centre point of circle $2$: $(s,r)$ and the point where they touch: $(x,y)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have worked out that to find the second centre point I can express the formulas like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(x_1-s)^2+(y_1-r)^2=r_2^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(s-h)^2+(r-k)^2=(r_1+r_2)^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unforunately my maths skills are not up to the task of solving these equations for $s$ and $r$, and ultimately for $x$ and $y$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would appreciate any help that can be provided.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regards&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Martin.&lt;/p&gt;&#xA;" OwnerUserId="7322" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2011-02-21T20:25:56.547" LastActivityDate="2011-02-23T13:20:22.237" Title="Calculation of the touching point of two circles given limited data." Tags="&lt;geometry&gt;&lt;algebra-precalculus&gt;&lt;analytic-geometry&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="23057" PostTypeId="1" AcceptedAnswerId="23060" CreationDate="2011-02-21T12:07:21.787" Score="3" ViewCount="254" Body="&lt;p&gt;Suppose f(m,n) is a double sequence in $\mathbb R$. Under what condition do we have $\lim\limits_{n\to\infty}\sum\limits_{m=1}^\infty f(m,n)=\sum\limits_{m=1}^\infty \lim\limits_{n\to\infty} f(m,n)$? Thanks!&lt;/p&gt;&#xA;" OwnerUserId="7321" LastActivityDate="2011-02-21T12:39:28.987" Title="Under what condition can we interchange lim and \sum?" Tags="&lt;calculus&gt;&lt;real-analysis&gt;" AnswerCount="2" FavoriteCount="2" />
  <row Id="23059" PostTypeId="2" ParentId="23057" CreationDate="2011-02-21T12:29:14.757" Score="0" ViewCount="" Body="&lt;p&gt;Denote $h(m)=\lim_{n\to\infty}f(m,n)$, and suppose that $\sum_{m=1}^\infty h(m)&amp;lt;\infty$. Then my guess is that the sufficient condition is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;\limsup_{n,m}\left|\sum_{k=1}^mf(k,n)-\sum_{k=1}^m h(k)\right|=0.&#xA;\end{align}&lt;/p&gt;&#xA;" OwnerUserId="4742" LastEditorUserId="4742" LastEditorDisplayName="" LastEditDate="2011-02-21T12:35:59.053" LastActivityDate="2011-02-21T12:35:59.053" />
  <row Id="23060" PostTypeId="2" ParentId="23057" CreationDate="2011-02-21T12:39:28.987" Score="7" ViewCount="" Body="&lt;p&gt;A fairly general set of conditions, sufficient for many applications, is given by the hypotheses of &lt;a href=&quot;http://en.wikipedia.org/wiki/Dominated_convergence_theorem&quot; rel=&quot;nofollow&quot;&gt;dominated convergence&lt;/a&gt;. (Note that sums are just integrals with respect to the counting measure on $\mathbb{N}$, so dominated convergence applies with no modification.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Without domination, the idea is that lumps of positive mass can &quot;escape to infinity&quot; when one attempts to interchange sum and limit. Here is a basic example: let $f_{m,n} = 1$ if $m = n$ and $0$ otherwise. Then $\sum_{m=1}^{\infty} f(m, n) = 1$ for all $n$, so the LHS is $1$, but $\lim_{n \to \infty} f(m, n) = 0$, so the RHS is $0$. The point of domination is to prevent these lumps of mass from escaping. &lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2011-02-21T12:39:28.987" CommentCount="3" />
  <row Id="23061" PostTypeId="2" ParentId="20813" CreationDate="2011-02-21T12:39:48.363" Score="1" ViewCount="" Body="&lt;p&gt;“Heyting. Intuitionism: An introduction.” (a Russian translation is also available) constructs real numbers.&lt;/p&gt;&#xA;" OwnerUserId="7011" LastActivityDate="2011-02-21T12:39:48.363" />
  <row Id="23062" PostTypeId="2" ParentId="23006" CreationDate="2011-02-21T12:39:59.737" Score="2" ViewCount="" Body="&lt;p&gt;The distributions of each $X_i$ and each $Y_i$ are far from being sufficient to decide anything about the families $(X_i)_i$ and $(Y_i)_i$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume for instance that $X_1$, $X_2$, $Y_1$ and $Y_2$ are all uniform $\pm1$ Bernoulli random variables and that $X_1=X_2=Y_1=-Y_2$. Then the event $[X_1=1\ \mbox{or}\ X_2=1]$ has probability $\frac12$ while the event $[Y_1=1\ \mbox{or}\ Y_2=1]$ has probability $1$.&lt;/p&gt;&#xA;" OwnerUserId="6179" LastActivityDate="2011-02-21T12:39:59.737" />
  <row Id="23063" PostTypeId="1" AcceptedAnswerId="23065" CreationDate="2011-02-21T13:09:04.220" Score="6" ViewCount="138" Body="&lt;p&gt;Hey there,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Today we had a probational exam in analysis. I wasn't able to solve one of the exercises and I have no idea what theorem to apply in order to solve it:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $I=[0,1]$ and $f: I \rightarrow \mathbb{R}$ be continuously differentiable. Assuming that $f$ has a root. Show: $$\max_{x \in I} |f(x)| \leq \max_{x \in I} |f&amp;#39;(x)|$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does this theorem have a name? What other theorem will I need in order to prove it? I'm sure the fact that the function has a root is important, but I don't see why and how to make use of it...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for your help!&lt;/p&gt;&#xA;" OwnerUserId="3787" LastActivityDate="2011-02-21T13:19:24.600" Title="How to prove $\max_{x \in I} |f(x)| \leq \max_{x \in I} |f'(x)|$?" Tags="&lt;analysis&gt;" AnswerCount="3" FavoriteCount="1" />
  <row Id="23064" PostTypeId="1" AcceptedAnswerId="23165" CreationDate="2011-02-21T13:16:52.733" Score="1" ViewCount="147" Body="&lt;p&gt;Hello,&#xA;recently I've come across a few cases where I had to evaluate $e^{At}$ (differential equations) where A is in Frobenius canonical form. The algorithm looks like:&#xA;$$e^{At} = Pe^{Jt}P^{-1}$$ so I need to do Jordan decomposition. From Frobenius canonical form I can easily write the characteristic equation =&gt; compute the roots which are eigenvalues and write the Jordan canonical form.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This was the easy part. Now the other part: compute matrices $P$ and $P^{-1}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any standard form of those matrices if matrix A is in a Frobenius canonical form?&lt;/p&gt;&#xA;" OwnerUserId="6485" LastEditorUserId="1543" LastEditorDisplayName="" LastEditDate="2011-04-18T12:13:44.360" LastActivityDate="2011-04-22T06:32:59.370" Title="Convert Frobenius canonical form to Jordan canonical form" Tags="&lt;linear-algebra&gt;&lt;differential-equations&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="23065" PostTypeId="2" ParentId="23063" CreationDate="2011-02-21T13:19:01.560" Score="6" ViewCount="" Body="&lt;p&gt;The example $f(x)\equiv 1$ shows that having a root is important.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $x_0\in I$ be such that $|f(x_0)|=\max_{x\in I}|f(x)|$, and let $r\in I$ be such that $f(r)=0$.  Then $\left|\frac{f(x_0)-f(r)}{x_0-r}\right|\geq |f(x_0)|$, and you can apply the Mean Value Theorem to finish.  Continuity of the derivative isn't necessary, just continuity of $f$ on $I$ and differentiability on $(0,1)$, except that with continuity of $f&amp;#39;$ you can really say that the right-hand side of the inequality is a max rather than a sup.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2011-02-21T13:19:01.560" CommentCount="1" />
  <row Id="23066" PostTypeId="2" ParentId="23063" CreationDate="2011-02-21T13:19:15.350" Score="5" ViewCount="" Body="&lt;p&gt;I don't know if this theorem has a name, but you're guessing right that it is important that $f$ has a root (otherwise adding a constant allows you to make the values of $f$ arbitrarily large without changing $f'$). One proof is a straightforward application of the fundamental theorem of analysis and a standard estimate on the integral. Indeed, if $x_{0}$ is such that $f(x_{0}) = 0$ then $f(x) = \int_{x_{0}}^{x} f&amp;#39;(t)\,dt$. Now combine this with the fact that $|\int_{a}^{b} h| \leq \int_{a}^{b}|h|$ for all $a,b$ and all integrable $h$.&lt;/p&gt;&#xA;" OwnerUserId="5363" LastActivityDate="2011-02-21T13:19:15.350" />
  <row Id="23067" PostTypeId="2" ParentId="23063" CreationDate="2011-02-21T13:19:24.600" Score="4" ViewCount="" Body="&lt;p&gt;Hints: Mean value theorem between a root and any point + the fact that the diameter of $I$ is $1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the way, continuity on the closed interval $I$ and differentiability on the interior of $I$ are sufficient.&lt;/p&gt;&#xA;" OwnerUserId="6179" LastActivityDate="2011-02-21T13:19:24.600" />
  <row Id="23068" PostTypeId="1" AcceptedAnswerId="23110" CreationDate="2011-02-21T13:36:51.613" Score="1" ViewCount="97" Body="&lt;p&gt;As we all know, Sturm's axioms have completely solved the problem for finding the number of roots in an arbitrary interval $[a,b]$, using the derivative and forms a Sturm set.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now my question follows naturally that is there any other use of this axiom.&lt;br&gt;&#xA;When I learned the Sturm set, I thought it will be difficult to find any such set for an arbitrary polynomial, nonetheless, the use of derivatives is really marvelous, and from then on, this question has occupied a part of my mind, shouting at me, asking me to treat it seriously.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any response is well appreciated, thanks.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: The four axioms are as follows as far as I am concerned:&lt;br&gt;&#xA;Given a set of polynomials ordered by natural numbers and the 0-th is the desired polynomial.&lt;br&gt;&#xA;(1):$\forall i \in$ {1,...,n}, $f_{i}$ and $f_{i+1}$ do not share the same root.&lt;br&gt;&#xA;(2):$f_n$(x) does not have one root.&lt;br&gt;&#xA;(3):If $\mathfrak{a}$ is a root of $f_k$, then $f_{k-1}(\mathfrak{a})$ and $f_{k+1}(\mathfrak{a})$ do not have the same sign.&lt;br&gt;&#xA;(4):If $\mathfrak{b}$ is a root of f(x), then in the interval $(-infinity, \mathfrak{b})$, $f_0(\mathfrak{b})$ and $f_1(\mathfrak{b})$  do not have the same sign; and in the interval $(\mathfrak{b},infinity)$, they share the same sign.&lt;br&gt;&#xA;After reading the answer by @Bill Dubuque, since I am not familiar with the theory of algorithms, I found the sturm's set to be full of unfamiliar things.&lt;br&gt;&#xA;In any case, thanks very much.  &lt;/p&gt;&#xA;" OwnerUserId="6792" LastEditorUserId="6792" LastEditorDisplayName="" LastEditDate="2011-02-22T09:02:48.340" LastActivityDate="2011-02-22T09:02:48.340" Title="Are there any  sets other than the usual in which we can apply Sturm's axioms?" Tags="&lt;algebra&gt;&lt;polynomials&gt;&lt;applications&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="1" />
  <row Id="23069" PostTypeId="1" CreationDate="2011-02-21T13:39:26.327" Score="2" ViewCount="120" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Possible Duplicate:&lt;/strong&gt;&lt;br&gt;&#xA;  &lt;a href=&quot;http://math.stackexchange.com/questions/16175/on-sort-of-linear-functions&quot;&gt;On sort-of-linear functions&lt;/a&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;p&gt;It is an interesting exercise to show that $Gal(\mathbb R ~\colon \mathbb Q)$ is trivial. The only solution I know hinges on the fact that the automorphism is order-preserving, which in turn depends on the fact that $\theta(xy)=\theta(x)\theta(y)$ for $\theta \in Aut~ \mathbb R$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, a function $L:\mathbb R \to \mathbb R$ with just the property that $L(x+y)=L(x)+L(y)$ can be shown to preserve multiplication on the rationals. And, I have been unsuccessful in trying to extend this fact to the reals. This could be because such a function could be discontinous, an example of which I also have failed to construct (it's a bad day).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Can you give me an example of a discontinuous function $L:\mathbb R \to \mathbb R$ with the property that $L(x+y)=L(x)+L(y)$ and $L|_{\mathbb Q}=$identity. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;An idea: Perhaps we could consider a function that preserves order on rationals but reverses it on irrationals.&lt;/p&gt;&#xA;" OwnerUserId="4806" LastActivityDate="2011-02-21T15:20:01.057" ClosedDate="2011-02-21T15:22:25.677" Title="A discontinuous linear function over the rationals" Tags="&lt;analysis&gt;&lt;abstract-algebra&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="23070" PostTypeId="1" CreationDate="2011-02-21T13:46:40.290" Score="1" ViewCount="130" Body="&lt;ol&gt;&#xA;&lt;li&gt;Given a random vector $X: (\Omega,&#xD;&#xA;    \mathbb{F}, P) \rightarrow (\prod_{i&#xD;&#xA;    \in I} S_i, \prod_{i \in I}&#xD;&#xA;    \mathbb{S}_i)$, is each component&#xA;variable $X_i, \forall i \in I$ of&#xA;the random vector $X$ always a&#xA;random variable from $(\Omega,&#xD;&#xA;    \mathbb{F}, P)$ to $(S_i,&#xD;&#xA;    \mathbb{S}_i)$?&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If yes, I guess there are two ways&#xA;to define the distributions for each&#xA;component variable $X_i, i \in I$:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;First the random vector $X$ induces a    probability measure&#xA;$P_X$ from the    its domain to its&#xA;codomain. Then    define $P_{X_i}&#xD;&#xA;    (A): = P_{X}(A \times    \prod_{j&#xD;&#xA;    \in I, j \neq i} S_j),    \forall A&#xD;&#xA;    \in \mathbb{S}_i$.&lt;/li&gt;&#xA;&lt;li&gt;$X_i$ can induce a probability    measure $P&amp;#39;_{X_i}$ from its domain&#xA;to    its codmain.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I was wondering if $P_{X_i}$ and&#xA;$P&amp;#39;_{X_i}$ are always the same on&#xA;$\mathbb{S}_i$? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Will the first definition of marginal probability measures make the product of all the marginal probability measures to be the same as the joint measure? &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Can the above definitions and their&#xA;relation be generalized to arbitrary&#xA;measure space $(\Omega, \mathbb{F},&#xD;&#xA;    \mu)$ and mearuable mapping&#xA;$f:(\Omega, \mathbb{F}) \rightarrow&#xD;&#xA;    (\prod_{i \in I} S_i, \prod_{i \in&#xD;&#xA;    I} \mathbb{S}_i)$, to define the&#xA;distribution of each component&#xA;mapping $f_i $ of $f$? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does the first way of definition&#xA;require that the measure $\mu$ must&#xA;be a probability measure? How to&#xA;adjust the definition when $\mu$ can&#xA;be any measure? Shall one define $\mu_{f_i}$ from $\mu_f$ as&#xA;$\mu_{f_i}(A_i):=&#xD;&#xA;        \frac{\mu_f(A_i \times \prod_{j \in&#xD;&#xA;        I, j\neq i} S_i)}{\prod_{j \in I,&#xD;&#xA;        j\neq i} \mu_{f_j}(S_i)}$, or $\mu_{f_i}(A_i):=&#xD;&#xA;        \mu_f(A_i \times \prod_{j \in&#xD;&#xA;        I, j\neq i} S_i)$, $\forall A_i \in \mathbb{S}_i $ or something&#xA;else?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When $\mu$ can be any measure, if it&#xA;is not possible to define in the&#xA;first way, or it is possible but the&#xA;two definitions are not the same,&#xA;how about when $\mu$ is finite, i.e.&#xA;$\mu(\Omega) &amp;lt; \infty$?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Thanks and regards!&lt;/p&gt;&#xA;" OwnerUserId="1281" LastEditorUserId="1281" LastEditorDisplayName="" LastEditDate="2011-02-22T15:55:32.507" LastActivityDate="2011-02-22T23:31:52.737" Title="How to understand marginal distribution" Tags="&lt;probability-theory&gt;&lt;measure-theory&gt;" AnswerCount="1" CommentCount="8" FavoriteCount="1" />
  <row Id="23071" PostTypeId="2" ParentId="23069" CreationDate="2011-02-21T13:48:48.433" Score="2" ViewCount="" Body="&lt;p&gt;This is impossible to do in ZF alone, but possible with a Hamel basis for $\mathbb{R}$ as a $\mathbb{Q}$ vector space.  Hence no &quot;explicit&quot; example can be expected.  Complete $\{1\}$ to a Hamel basis $B$, let $x\in B\setminus\{1\}$, and let $f:\mathbb{R}\to\mathbb{R}$ be the unique linear function such that $f(x)=0$ and $f(b)=b$ for all $b\in B\setminus\{x\}$; in particular, $f(1)=1$ implies that $f$ is the identity on the rationals.  Because $x$ can be approximated by rationals (not to mention with rational multiples of other elements of $B$), $f$ is not continuous at $x$.  In fact, $f$ is not continuous anywhere.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2011-02-21T15:20:01.057" LastActivityDate="2011-02-21T15:20:01.057" CommunityOwnedDate="2011-02-21T15:14:48.523" />
  <row Id="23072" PostTypeId="2" ParentId="22777" CreationDate="2011-02-21T14:14:31.663" Score="2" ViewCount="" Body="&lt;p&gt;You can also refer this thread:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://math.stackexchange.com/questions/1956/is-there-an-integral-that-proves-pi-333-106/1959#1959&quot;&gt;Is there an integral that proves $\pi &gt; 333/106$?&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;$$ \int\limits_{0}^{1} \frac{x^{5}(1-x)^{6}(197+462x^{2})}{530(1+x^{2})} + \frac{333}{106}= \pi$$&lt;/p&gt;&#xA;" LastActivityDate="2011-02-23T09:51:39.123" />
  <row Id="23073" PostTypeId="2" ParentId="23030" CreationDate="2011-02-21T14:17:03.357" Score="5" ViewCount="" Body="&lt;p&gt;Let $B_i\in\mathbb{B}_i$ be nonempty for each $i\in I$, and consider the set $I_0 = \{i\in I\mid B_i\neq E_i\}$. If $I_0$ is countable, then $\prod B_i\in\prod\mathbb{B}_i$. If the set $I_0$ is uncountable, then $\prod B_i\notin\prod\mathbb{B}_i$. In particular, if $I$ is countable then you get all the &quot;box&quot; sets, and if it is uncountable, you will never have a set $\prod B_i$ where $\emptyset\neq B_i\neq E_i$ for all $i\in I$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The key is that the product $\sigma$-algebra is the smallest $\sigma$-algebra that makes all the canonical projections measurable; that is, such that for every $i\in I$, every $B_i\mathbb{B}_i$, $\pi_i^{-1}(B_i)\in\prod\mathbb{B}_i$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $X=\prod\limits_{i\in I}E_i$, and $\mathbb{B}=\prod\limits_{i\in I}\mathbb{B}_i$, for simplicity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, assume that $I_0$ is countable, and let $B_i\in\mathbb{B}_i$ for each $i$; I claim that $\prod\limits{i\in I}B_i\in\mathbb{B}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To see this, for each $i\in I_0$ let $S_i =\pi_i^{-1}(E_i-B_i)= (E_i-B_i)\times\prod\limits_{j\neq i}E_j$. Then $S_i\in\mathbb{B}$, and hence so is $\mathop{\cup}\limits_{i\in I_0}S_i$. Since $I_0$ is countable, this is an element of $\mathbb{B}$, being a countable union of elements of $\mathbb{B}$. An element $(e_i)\in X$ lies in $\cup S_i$ if and only if $e_i\notin B_i$ for some $i\in I_0$. Therefore, $(e_i)\in X-\cup S_i$ if and only if $e_i\in B_i$ for all $i\in I_0$. That is, $X-\cup S_i = \mathop{\prod}\limits_{i\in I_0}B_i\times\mathop{\prod}\limits_{i\notin I_0}E_i = \prod\limits_{i\in I}B_i$. This is the complement of an element of $\mathbb{B}$, hence lies in $\mathbb{B}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note however that $\mathbb{B}$ will usually contain other stuff besides these sets. Even in the case where $I=\{1,2\}$, you usually have more than just the &quot;box&quot; sets in the $\sigma$ algebra. For example, the product of the Borel $\sigma$-algebra on $[0,1]$ with itself contains the set $([\frac{1}{4},\frac{1}{2}]\times[\frac{1}{4},\frac{1}{2}])\cup([\frac{3}{4},1]\times[\frac{3}{4},1])$, which is not of the form $A\times B$ with $A$ and $B$ Borel subsets of $[0,1]$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now assume that $I_0$ is uncountable. I will construct a $\sigma$-algebra that necessarily contains $\mathbb{B}$ (but need not be equal to it), and which does &lt;em&gt;not&lt;/em&gt; contain any set of the form $\prod\limits{i\in I}B_i$ in which the subset $I_0 = \{i\in I\mid B_i\neq E_i\}$ is uncountable (in particular, when it is equal to all of $I$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $S$ be the $\sigma$-algebra on $\prod\limits_{i\in I}E_i$ that consists of all sets $A$ of the following form: there exists a &lt;em&gt;countable&lt;/em&gt; set $J\subseteq I$, and an element $B\in \mathcal{P}\left(\prod\limits_{j\in J}E_i\right)$, such that $A = B\times\prod\limits_{i\notin J}E_i.$ That is, $A$ is an arbitrary thing in the $J$-coordinates, but is all of $E_i$ in the non-$J$ coordinates. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I claim that $S$ is a $\sigma$-algebra on $X$. It contains the empty set by selecting $B=\emptyset$. It is closed under complements, as $X-A = \left((\prod\limits_{j\in J}E_j) - B\right)\times\prod\limits_{i\notin J}E_i$ is of the same form. And it is closed under countable unions: if $A_n$ is a subset of the desired form, with associated countable $J_n\subseteq I$, letting $J=\cup J_n$ we have that $\cup A_n$ is of the form $C\times\prod\limits_{i\notin J}E_i$ with $C$ a subset of $\prod_{j\in J}E_j$. Since each $J_n$ is countable, $J$ is countable, so this set is in $S$. Thus, $S$ is a $\sigma$-algebra on $X$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note also that $S$ contains the $\sigma$-algebra $\mathbb{B}$, since the generators of $\mathbb{B}$ all lie in $S$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if $B_i\in\mathbb{B}_i$ for each $i$, and the set $J=\{i\in I\mid B_i\neq E_i\}$ is uncountable, then $\prod B_i\notin S$, and hence does not lie in $\mathbb{B}$ either. Thus, $\mathbb{B}$ does not contain any set which is a product of elements of the different $\mathbb{B}_i$ in which uncountably many components are not the entire space.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-21T16:23:01.920" LastActivityDate="2011-02-21T16:23:01.920" />
  <row Id="23074" PostTypeId="1" AcceptedAnswerId="36485" CreationDate="2011-02-21T15:00:49.220" Score="7" ViewCount="229" Body="&lt;p&gt;Can one state and prove the Pythagorean theorem using Hilbert's axioms of geometry, without any reference to arithmetic?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: Here is a possible motivation for this question (and in particular for the &quot;state&quot; part of this question). It is known that the theory of Euclidean geometry is complete. Every true statement in this theory is provable. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, it is known that the axioms of (Peano) arithmetic cannot be proven to be consistent. So, basically, I ask if there is a reasonable theory which is known to be consistent and complete, and in which the Pythagorean theorem can be stated and proved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In summary, I guess I am asking - can we be sure that the Pythagorean theorem is true? :)&lt;/p&gt;&#xA;" OwnerUserId="5491" LastEditorUserId="5491" LastEditorDisplayName="" LastEditDate="2011-02-21T16:30:28.897" LastActivityDate="2011-05-02T22:04:58.143" Title="The Pythagorean theorem and Hilbert axioms" Tags="&lt;euclidean-geometry&gt;" AnswerCount="1" CommentCount="10" />
  <row Id="23075" PostTypeId="1" CreationDate="2011-02-21T15:08:22.713" Score="1" ViewCount="45" Body="&lt;p&gt;I have an equation:&#xA;$(1-\epsilon)x^2 -2x +1=0$ (regularly perturbed problem, we anticipate all roots to remain bounded when $\epsilon$ goes to $0$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I substitute $x=\displaystyle \sum_{n\geq0}C_n \epsilon^n$ and have to work out the equations for $C_0,C_1,C_2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I get: $C_0^2-2C_0+1=0$ so $C_0=1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and there a problem starts: as the equation for $C_1$ is: $2C_0C_1-C_0^2-2C_1=0$ so we get $1=0$. &#xA;What is the reason this approach fails and how to devise a way to fix it?&lt;/p&gt;&#xA;" OwnerUserId="7320" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2011-02-21T15:41:35.187" LastActivityDate="2011-02-21T16:03:19.020" Title="Perturbation problem equation" Tags="&lt;perturbation-theory&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="23077" PostTypeId="1" AcceptedAnswerId="23080" CreationDate="2011-02-21T15:38:59.607" Score="2" ViewCount="63" Body="&lt;p&gt;Suppose $T_1(X)$ is a sufficient statistic for $\theta$ for a family of distributions indexed by $\theta$ and $T_2(X)$ is another statistic that induces a “finer”(what is meant by a finer ? ) partition than $T_1$. How do I know that $T_2$ is also sufficient  ? &lt;/p&gt;&#xA;" OwnerUserId="6335" LastActivityDate="2011-02-21T16:23:15.650" Title="Sufficient statistics" Tags="&lt;statistics&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="23078" PostTypeId="2" ParentId="23075" CreationDate="2011-02-21T15:49:29.420" Score="4" ViewCount="" Body="&lt;p&gt;Let us actually solve and see what exactly happens and what is the reason for this erroneous result you get.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solving we get $$\displaystyle x = \frac{2 \pm \sqrt{4-4(1-\epsilon)}}{2(1-\epsilon)} = \frac{1 \pm \sqrt{\epsilon}}{1-\epsilon} = \frac{1}{1 \mp \sqrt{\epsilon}} = 1 \pm \sqrt{\epsilon} + \epsilon \pm \sqrt{\epsilon^3} + \epsilon^2 \pm \sqrt{\epsilon^5} + \cdots$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When we assume the perturbed solution is $x = \displaystyle \sum_{n\geq 0} c_n \epsilon^n$, we are not accounting for the fractional powers of $\epsilon$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem essentially stems from our inability to represent $\frac{1}{1 \pm \sqrt{\epsilon}}$ as a taylor series about origin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The error $1=0$ arises when you cancel off $2C_1$ with $-2C_1$. If we try to express $\frac{1}{1 \pm \sqrt{\epsilon}}$ as a taylor series, then you will find that the derivative at the origin $C_1$ is unbounded. Hence, you are canceling of $+\infty$ and $-\infty$ and this is where the error creeps in and hence you get $1=0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To avoid this, a better way to deal with the problem would be as follows. If we know that the perturbation is positive, we can replace $\epsilon$ by $\epsilon^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the problem becomes $$(1-\epsilon^2)x^2 + 2x + 1 = 0$$ Solving, we get $$x=\frac{1}{1 \mp \epsilon} = 1 \pm \epsilon + \epsilon^2 \pm \epsilon^3 + \epsilon^4 \pm \epsilon^5 + \cdots$$&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2011-02-21T16:03:19.020" LastActivityDate="2011-02-21T16:03:19.020" />
  <row Id="23079" PostTypeId="1" AcceptedAnswerId="23113" CreationDate="2011-02-21T16:01:35.570" Score="4" ViewCount="130" Body="&lt;p&gt;For some formulae $\phi(x)$ it can be proved from the axioms of ZFC, that there is no set $X$ with $(x)x\in X \equiv \phi(x)$. Thus the collection $\lbrace x\ |\ \phi(x)\rbrace$ is a proper class.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore, there is no set $Y$ that is in bijection with such a collection, right? Thus, every proper class must be in some sense larger than any set, right? (&quot;too large to be a set&quot;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For two formulae $\phi(x)$, $\psi(x)$, one of them defining a proper class, it may be possible two show that $(x) \phi(x) \equiv \psi(x)$. This means a) that also the second collection is a proper class, and b) that the two have equal size in some sense. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Can the notion of size of proper classes (or equipollency) be made precise?  Is it possible that two proper classes&#xA;  do not have equal size in this sense?&#xA;  Would it thus follow, that there are&#xA;  smaller and larger proper classes? Or&#xA;  do all proper classes have the same&#xA;  size (of $V$)?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And - finally - can the size of $V$ be &lt;em&gt;characterized&lt;/em&gt;? To talk at large, maybe like this (if sense of $P(V)$ could be made): &lt;em&gt;$V$ is so large such that $|V| = |P(V)|$&lt;/em&gt; (which never holds for sets)? (This would admittedly be counter-intuitive, since the discrepancy between $|X|$ and $|P(X)|$ grows with $|X|$.)&lt;/p&gt;&#xA;" OwnerUserId="1792" LastEditorUserId="1792" LastEditorDisplayName="" LastEditDate="2011-02-21T23:10:56.093" LastActivityDate="2011-02-21T23:10:56.093" Title="Some questions concerning the size of proper classes in ZFC" Tags="&lt;logic&gt;&lt;set-theory&gt;&lt;elementary-set-theory&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="23080" PostTypeId="2" ParentId="23077" CreationDate="2011-02-21T16:04:51.147" Score="2" ViewCount="" Body="&lt;p&gt;Consider the following partitions: $\textbf{P} = \{[1,2], (2,4] \}$ and $\textbf{P&amp;#39;} = \{[1,2), \{2 \}, (2,3), [3,4] \}$. Then $\textbf{P&amp;#39;}$ is finer than $\textbf{P}$. In other words, for every $J$ in $\textbf{P&amp;#39;}$ there exists a $K$ in $\textbf{P}$ such that $J \subseteq K$. Sufficiency is a property of partitions. So if we have sufficiency over $[1,2]$ we have sufficiency over $[1,2)$ and $\{2 \}$. Likewise, if we have sufficiency over $(2,4]$, we have sufficiency over $(2,3)$ and $[3,4]$. If we look at sufficiency over the coarsest partition $\textbf{Q} = \{[1,4] \}$ then this is the minimal sufficient statistic. &lt;/p&gt;&#xA;" OwnerUserId="2548" LastEditorUserId="2548" LastEditorDisplayName="" LastEditDate="2011-02-21T16:23:15.650" LastActivityDate="2011-02-21T16:23:15.650" />
  <row Id="23082" PostTypeId="1" CreationDate="2011-02-21T16:13:39.090" Score="2" ViewCount="39" Body="&lt;p&gt;At time $t_{0}$ I have $N$ server instances running. Each instance is at $100\%$ CPU utilization. Collectively, the instances are attempting to respond to $R_{0}$ web requests. Some unknown percent of $R_{0}$ requests are being ignored because all the instances are under too much load.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At time $t_{1}$, the load lessens, and all $N$ instances are now at only $50\%$ CPU utilization. Collectively, the instances are now successfully responding to $100\%$ of $R_{1}$ web requests. Note $R_{1}$ &amp;lt; $R_{0}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How many additional server instances ($N_{\text{additional}}$ so that $N_{\text{new}} = N + N_{\text{additional}}$) should you create so that should $R_{0}$ traffic occur again, CPU utilization on each instance will be no more than $75\%$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My approach is to use the performance under $50\%$ load to calculate the predicted maximum number of requests per instance, and then divide the $R_{0}$ by this amount to get the estimated number of instances needed, multiplying by some amount to over-estimate to get the $75\%$ max CPU utilization.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;e.g.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maximum $R$ PerInstance $= 100 \times \frac{\frac{R_1}{N}}{50}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Predicted Number of Instances $= \frac{\text{Predicted R}}{\text{Max } R \text{ Per Instance}} \times 1.25$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does that seem like the right approach?&lt;/p&gt;&#xA;" OwnerUserId="2114" LastEditorUserId="2114" LastEditorDisplayName="" LastEditDate="2011-02-22T21:44:57.673" LastActivityDate="2011-02-22T21:44:57.673" Title="Calculating the Optimal Number of Server Instances" Tags="&lt;algebra&gt;" AnswerCount="1" />
  <row Id="23083" PostTypeId="2" ParentId="23079" CreationDate="2011-02-21T16:17:08.200" Score="3" ViewCount="" Body="&lt;p&gt;The answer seems to depend on your set theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Specifically, in an &lt;a href=&quot;http://groups.google.com/group/sci.math/msg/88e0bac0c4d39038&quot; rel=&quot;nofollow&quot;&gt;August 2006 post in &lt;code&gt;sci.math&lt;/code&gt;&lt;/a&gt;, Herman Rubin said that if $\mathbf{V}=\mathbf{L}$ (Goedel's Constructible Universe), then every proper class is equipollent to (bijectable with) the class of all ordinals, hence equipollent with one another, so you only have a single &quot;size&quot; of proper classes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, he also said that there are models of ZFC with different versions of the Axiom of Choice for proper classes and in some of these not all proper classes are equipollent (even if you assume the Axiom of Choice for sets).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For completeness: what do we mean by equipollency of proper classes? Goedel defined equivalence of proper classes as follows: For proper classes $X$ and $Y$, $X\sim Y$ if and only if there exists $Z$ such that:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$Z$ is a relation (contained in $\mathbf{V}^2$);&lt;/li&gt;&#xA;&lt;li&gt;$Z$ is twice unary ($Z$ is single-valued, and so is its inverse);&lt;/li&gt;&#xA;&lt;li&gt;The domain of $Z$ is $X$; and&lt;/li&gt;&#xA;&lt;li&gt;The domain of the inverse of $Z$ is $Y$.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;One model where all proper classes are equipollent is when we assume the &quot;Axiom of Global Choice&quot;, which states that there exists a unary relation $A$ such that for all nonempty sets $X$, there exists $y\in X$ such that $(X,y)\in A$. This is given by Goedel in his paper on the consistency of AC and GCH with ZF. It implies that any class is equipollent to an initial segment of the ordinals, and since any proper initial segment of the ordinals is a set, a proper class must be equipollent to the proper class of all ordinals, giving the result that all proper classes are equipollent. &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-21T19:11:27.920" LastActivityDate="2011-02-21T19:11:27.920" CommentCount="8" />
  <row Id="23084" PostTypeId="2" ParentId="23054" CreationDate="2011-02-21T16:55:56.327" Score="0" ViewCount="" Body="&lt;p&gt;Assuming you mean the geodesic line $AB$.&lt;br&gt;&#xA;&lt;strong&gt;Hint:&lt;/strong&gt; Calculate the angle $\theta =\angle XCP$ where $P$ is a point on the line and $C$ is the center of the sphere. For instance with $\cos \theta =\frac{\vec{CX}\cdot \vec{CP}}{|\vec{CX}|\cdot |\vec{CP}|}=\frac{\vec{CX}\cdot \vec{CP}}{r^2}$, where $r$ is the length of the sphere's radius and $\vec{CX}, \vec{CP}$ are vectors in $\mathbb{R}^3$. This angle is proportional to the geodesic distance between $P$ and $X$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you have to do then, is to find the point $P$ for which the angle is smallest.  &lt;/p&gt;&#xA;" OwnerUserId="7044" LastEditorUserId="7044" LastEditorDisplayName="" LastEditDate="2011-02-21T17:17:19.170" LastActivityDate="2011-02-21T17:17:19.170" />
  <row Id="23085" PostTypeId="1" CreationDate="2011-02-21T17:16:13.413" Score="2" ViewCount="66" Body="&lt;p&gt;Hello,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;this will probably be an easy-to-answer and a not-well-posed question, since I'm a total beginner in the field, but here goes: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $V(F)$ and $V(G)$ be two projective curves in $\mathbb{P}^2$ ($F,G\in\mathbb{F}[x_0,x_1,x_2]$ homogenous polynomials) and $P\in V(F)\cap V(G)$ a point. Let $\mu_P(F,G)$ denote the intersection number (multiplicity), i.e. the number, defined in &lt;a href=&quot;http://www.math.lsa.umich.edu/~wfulton/CurveBook.pdf&quot; rel=&quot;nofollow&quot;&gt;Fultons book&lt;/a&gt; on p. 36, section 3.3 and also chapter 5. &lt;strong&gt;Is there any relationship between $\mu_P(F,G)$ and the possible equality of partial derivatives&lt;/strong&gt; &#xA;$$\frac{\partial F}{\partial x_i}\!\!\!(P)\overset{?}{=}\frac{\partial G}{\partial x_i}\!\!\!(P),\;\ldots,\;\frac{\partial^k F}{\partial x_i^k}\!\!\!(P)\overset{?}{=}\frac{\partial^k G}{\partial x_i^k}\!\!\!(P)\;?$$&#xA;&lt;strong&gt;Do the equalities hold for $k=\mu_P(F,G)$?&lt;/strong&gt; If not, what is the correspondence? I assume degrees of $F$ and $G$ must be involved?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S. Some references from (preferably recent) books are highly desirable.&lt;/p&gt;&#xA;" OwnerUserId="7325" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2011-02-21T17:31:41.647" LastActivityDate="2011-02-21T17:31:41.647" Title="intersection multiplicity and partial derivatives of algebraic curves" Tags="&lt;algebraic-geometry&gt;&lt;commutative-algebra&gt;&lt;algebraic-curves&gt;" />
  <row Id="23086" PostTypeId="1" AcceptedAnswerId="23094" CreationDate="2011-02-21T17:28:38.970" Score="2" ViewCount="82" Body="&lt;p&gt;How does one prove that the class number of $\mathbb{Z}[\zeta_3]$ is $1$?&lt;/p&gt;&#xA;" OwnerUserId="2212" LastActivityDate="2011-02-21T20:23:28.300" Title="Prove that the class number of $\mathbb{Z}[\zeta_3]$ is $1$." Tags="&lt;number-theory&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="23087" PostTypeId="1" AcceptedAnswerId="23089" CreationDate="2011-02-21T17:33:23.633" Score="1" ViewCount="47" Body="&lt;p&gt;If &lt;strong&gt;P=NP&lt;/strong&gt;, is there a polynomial-time algorithm $A$ that can decide the $\text{Independent Set}$ decision problem?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is, with an undirected graph $G = (V, E)$ and a positive integer $k$, does $G$ contain a set of $k$ independent vertices (i.e. a set of vertices no two of which are joined by an edge)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am unsure of whether to look at $\text{Independent Set for trees}$ or algorithms that could perhaps solve problems equivalent to $\text{Clique, Vertex Cover}$ etc. in polynomial-time. I would appreciate some guidance on my options and the following development...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Upon choosing $A$, how could I use it to create a further polynomial-time algorithm that &lt;em&gt;finds&lt;/em&gt; an $\text{Independent Set}$ of size $k$, rather than just returning a Boolean truth value.&lt;/p&gt;&#xA;" OwnerUserId="6796" LastActivityDate="2011-02-21T17:52:01.477" Title="Independent Set decision problem in P" Tags="&lt;homework&gt;&lt;algorithms&gt;&lt;computer-science&gt;&lt;np-complete&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="23088" PostTypeId="1" CreationDate="2011-02-21T17:40:53.153" Score="3" ViewCount="109" Body="&lt;p&gt;What is the best software for helping a 10-year-old learn about advanced mathematics?&lt;/p&gt;&#xA;" OwnerUserId="7328" LastActivityDate="2011-02-21T20:47:01.413" CommunityOwnedDate="2011-02-21T17:47:32.360" Title="Software for helping kids learn advanced mathematics?" Tags="&lt;math-software&gt;&lt;education&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="1" />
  <row Id="23089" PostTypeId="2" ParentId="23087" CreationDate="2011-02-21T17:52:01.477" Score="2" ViewCount="" Body="&lt;p&gt;Well, Independent Set is certainly in NP (pick $k$ vertices and check if they form an independent set).  So, if P=NP, then yes, there is a polynomial time algorithm for Independent Set.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For your second question, first use $A$ to ensure that there is an independent set of size $k$.  Then, remove the first vertex and it's neighbors.  Use $A$ to see if the resulting graph has a $k-1$ independent set.  If yes, mark the first vertex as in your independent set and recurse.  Otherwise, put back everything you took out and try with the second vertex, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's a really informal sketch, but you get the idea. &lt;/p&gt;&#xA;" OwnerUserId="7268" LastActivityDate="2011-02-21T17:52:01.477" />
  <row Id="23090" PostTypeId="2" ParentId="23088" CreationDate="2011-02-21T17:56:04.330" Score="3" ViewCount="" Body="&lt;p&gt;(A tad tongue in cheek) If you mean advanced mathematics the same way I do, then &lt;a href=&quot;http://djvu.org/&quot; rel=&quot;nofollow&quot;&gt;DejaVu viewer&lt;/a&gt;. (It is even a great software for helping 25-year-olds learn advanced mathematics.)&lt;/p&gt;&#xA;" OwnerUserId="1543" LastActivityDate="2011-02-21T17:56:04.330" CommunityOwnedDate="2011-02-21T17:56:04.330" />
  <row Id="23091" PostTypeId="1" CreationDate="2011-02-21T18:02:35.643" Score="2" ViewCount="109" Body="&lt;p&gt;I'm having trouble comprehending the following.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.stat.berkeley.edu/users/pitman/s205s03/lecture28.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.stat.berkeley.edu/users/pitman/s205s03/lecture28.pdf&lt;/a&gt; in the proof of theorem 28.3.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It says that $H_t(x,y)=H_t\delta_y(x)$, where $\delta_y(x)=1$ if $x=y$ and $0$ if $x\neq y$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm stuck with understanding this. Please help?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also why is $E(H_t f)=Ef$ ? I particularly don't know why $K(c)=c$ if $c$ is a constant and $K$ is a Markov operator, because I thought $K$ could only operate on functions. Could we consider $c$ as a function too although it really is just a scalar?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;EDIT&lt;/em&gt;&#xA;I know this is another late question but about the same paper, there's an equality I don't understand.  Near the end, it says&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$|h_t(x,y)-1|=\left|\sum_{z}(h_{t/2}(x,z)-1)(h_{t/2}(z,y)-1)\pi(z)\right|$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the RHS it looks like I should get $|\sum_{z} (h_{t/2}(x,z)h_{t/2}(z,y)-h_{t/2}(x,z)-h_{t/2}(z,y)+1) \pi(z)|.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However I can't simplify that to be the LHS.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And lastly, the inequality $$\left\|H_t-E\right\|_{2\to\infty}\leq \left\|H_{t_1}\right\|_{2\to\infty} \left\|H_{t_2}-E\right\|_{2\to 2}$$ for $t=t_1+t_2.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These have been two questions I have been unable to solve and they would help me greatly in understanding Markov bounds.  Thanks!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;pr.probability &lt;/p&gt;&#xA;" OwnerUserId="7329" LastEditorUserId="7329" LastEditorDisplayName="" LastEditDate="2011-02-23T14:24:05.917" LastActivityDate="2011-02-23T14:24:05.917" Title="Eigenfunction expansion with Markov operators" Tags="&lt;probability&gt;" AnswerCount="1" CommentCount="6" />
  <row Id="23092" PostTypeId="1" CreationDate="2011-02-21T18:04:55.263" Score="9" ViewCount="239" Body="&lt;p&gt;The Collatz function $T$ is defined on the set $\mathbb{Z}^+$ of positive integers&#xA;as: $T(n)=n/2$ if $n$ is even, and $T(n)=3n+1$ if $n$ is odd. Let $T^k$ be the $k$th iteration&#xA;of $T$. We say $n$ terminates if $T^k(n)=1$ for some $k$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $n$ be an integer of the form $$3^{2^k(j-1)}+3^{2^k(j-2)}+\cdots +3^{2^k}+1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $k,j\in \mathbb{Z}^+$ and $j$ is odd.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Question: Will $n$ terminate for all such $k,j$?&lt;/p&gt;&#xA;" OwnerUserId="3249" LastActivityDate="2011-02-23T13:35:00.310" Title="A question regarding Collatz conjecture" Tags="&lt;number-theory&gt;" AnswerCount="0" CommentCount="8" FavoriteCount="2" />
  <row Id="23093" PostTypeId="1" AcceptedAnswerId="23100" CreationDate="2011-02-21T18:06:18.017" Score="3" ViewCount="115" Body="&lt;p&gt;My understanding right now is that an example of conditional independence would be:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If two people live in the same city, the probability that person &lt;strong&gt;A&lt;/strong&gt; gets home in time for dinner, and the probability that person &lt;strong&gt;B&lt;/strong&gt; gets home in time for dinner are independent; that is, we wouldn't expect one to have an affect on the other. But if a snow storm hits the city and introduces a probability &lt;strong&gt;C&lt;/strong&gt; that traffic will be at a stand still, you would expect that the probability of both &lt;strong&gt;A&lt;/strong&gt; getting home in time for dinner and &lt;strong&gt;B&lt;/strong&gt; getting home in time for dinner, would change.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If this is a correct understanding, I guess I still don't understand what exactly conditional independence &lt;em&gt;is&lt;/em&gt;, or what it does for us (why does it have a separate name, as opposed to just compounded probabilities), and if this isn't a correct understanding, could someone please provide an example with an explanation?&lt;/p&gt;&#xA;" OwnerUserId="1389" LastActivityDate="2011-02-21T18:55:26.137" Title="Could someone explain conditional independence?" Tags="&lt;probability&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="23094" PostTypeId="2" ParentId="23086" CreationDate="2011-02-21T18:08:08.733" Score="3" ViewCount="" Body="&lt;p&gt;Here's a nice geometric argument, which you can find essentially in Klein's lecture on Ideal Numbers (Lecture VIII in &lt;strong&gt;Lectures of Mathematics&lt;/strong&gt;, by Felix Klein, AMS Chelsea Publishing, AMS, 2000).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the lattice $\mathbb{Z}[\zeta_3]$ in $\mathbb{C}$ ($\zeta_3 = \frac{-1+\sqrt{-3}}{2}$). Given $a,b\in\mathbb{Z}[\zeta_3]$, $a\neq 0$. Consider the points in the lattice of the form $(r+s\zeta_3)a$, and use them to tessellate the complex plane. Then locate $b$, and the closest corner of one of the squares to $b$, $qa$. Then let $r=b-qa$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a picture which is slightly off, since my lattice here is $\mathbb{Z}[i]$, not $\mathbb{Z}[\zeta_3]$, but it should give you an idea; it's taken from the slides for a talk I gave some years ago. It's a bit hard to see, but the small dots are the lattice points; the big dots represent the $\mathbb{Z}[\zeta_3]$ multiples of $a$, with $a$ the first large dot to the right of the one labeled $0$ and above the horizontal line. (Well, as I said,t he small dots are really in the position of the lattice points of $\mathbb{Z}[i]$, but you get the idea, I hope). The three labeled dots above zero are (counterclockwise) $a$, $(1+\zeta_3)a$, and $\zeta_3 a$. The dot corresponding to $b$ is circled in red. The blue dot is $qa$, the blue arrow is the vector corresponding to $qa$, and the green arrow is the vector corresponding to $r  = b-qa$. (The &quot;1&quot; at the very bottom of the picture is the page number, so ignore it...)&#xA;&lt;img src=&quot;http://i.imgur.com/05XE4.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How big can $r$ be? The tessellation is by parallelograms whose sides have length $|a|$ and $|\zeta_3 a|=|a|$, so the furthest that $b$ can be from the corner it is closest to is $\frac{|a|\sqrt{2}}{2}$. That is, $|r|\leq\frac{|a|\sqrt{2}}{2}\lt|a|$.  From this it follows that $0\leq N(r)\lt N(a)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, for all $a,b\in\mathbb{Z}[\zeta_3]$, there exists $q,r\in\mathbb{Z}[\zeta_3]$ such that $b = qa+r$, and $0\leq N(r)\lt N(a)$. So $N$ is a Euclidean function on $\mathbb{Z}[\zeta_3]$, the latter is Euclidean, hence a PID, hence the class number is $1$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same geometric argument can be used to show that $\mathbb{Z}[i]$ and $\mathbb{Z}[\sqrt{-2}]$ are Euclidean, but it breaks down when you get to $\mathbb{Z}[\sqrt{-5}]$, because the size of the rectangles now allows the distance from $b$ to a corner to be larger than the length of $a$ (good thing, too, since $\mathbb{Z}[\sqrt{-5}]$ is not a UFD). &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-21T20:23:28.300" LastActivityDate="2011-02-21T20:23:28.300" CommentCount="1" />
  <row Id="23095" PostTypeId="1" AcceptedAnswerId="23098" CreationDate="2011-02-21T18:09:16.617" Score="1" ViewCount="80" Body="&lt;p&gt;I am a student taking a &quot;discrete maths&quot; course. Teacher seems to jump from one subject to another rapidly and this time he covered ring theory, Z/nZ, and polynomial rings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is hard for me to understand anything in his class, and so the reports he gives become very hard.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I did my best to find answers using google, but I just couldn't find it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Specifically he asked us to find all ideals of Z/6Z, and prove that these are in fact all of them.&#xA;He also asked us to find all ideals of F[X]/(X^3-1) where F stands for Z/2Z.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand the idea behind ideals, like I can see why {0,3} is ideal of Z/6Z, but how do I find ALL the ideals?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And regarding polynomials, is there some kind of a mapping between polynomials and Z/nZ?&#xA;Because otherwise I have no idea how to find ideals of polynomials.&lt;/p&gt;&#xA;" OwnerUserId="7289" LastActivityDate="2011-02-21T18:36:16.483" Title="A basic question about finding ideals of rings and proving that these are all the ideals" Tags="&lt;ring-theory&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="23096" PostTypeId="2" ParentId="23082" CreationDate="2011-02-21T18:12:11.410" Score="1" ViewCount="" Body="&lt;p&gt;Your idea/approach is correct but from where do you get the factor $1.25$? You need to multiply by a factor $\frac{4}{3}$ since you now want them to be working at $75\%$ utilization. ($75\%$ of $\frac{4}{3}$ is $1$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume that each web request has the same CPU utilization.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$R_1$ web requests corresponds to $0.5 \times N$ instances.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$1$ web request corresponds to $\frac{N}{2R_1}$ instances.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$R_0$ web requests corresponds to $\frac{NR_0}{2R_1}$ instances.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And this should be $0.75 \times N_{\text{new}}$ instances.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, $N_{\text{new}} = \frac{4}{3} \frac{NR_0}{2R_1} = \frac{2}{3} \frac{NR_0}{R_1}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the number of additional server instances you need is $\frac{2}{3} \frac{NR_0}{R_1} - N$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Note that you have $R_0 &amp;gt; 2R_1$ and hence $\frac{2}{3} \frac{NR_0}{R_1} - N &amp;gt; \frac{4}{3} N - N = \frac{N}{3}$. Hence the additional server instance you need is greater than $\frac{N}{3}$).&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2011-02-21T18:19:28.763" LastActivityDate="2011-02-21T18:19:28.763" />
  <row Id="23097" PostTypeId="2" ParentId="23054" CreationDate="2011-02-21T18:12:16.137" Score="6" ViewCount="" Body="&lt;p&gt;This assumes that everything is on the surface of the sphere.  Furthermore I assume the sphere has radius 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Change the coordinates so that A and B are both on the equator of the sphere. For definiteness, move $A$ to $(1,0,0)$ and move $B$ to $(\cos \theta, \sin \theta, 0)$ where $\theta$ is the angle between the vectors from the center of the sphere to $A$ and to $B$. This is a linear transformation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then what you care about is the latitude of $X$. If $X$ is in the sector of the sphere immediately north or south of the line $AB$, then the answer is just $2\pi \phi$ where $\phi$ is the latitude of $X$. If $X$ is not in that sector then the answer is just the distance to either $A$ or $B$, whichever is closer.&lt;/p&gt;&#xA;" OwnerUserId="173" LastActivityDate="2011-02-21T18:12:16.137" />
  <row Id="23098" PostTypeId="2" ParentId="23095" CreationDate="2011-02-21T18:23:05.203" Score="3" ViewCount="" Body="&lt;p&gt;Since $\mathbb{Z}/6\mathbb{Z}$ is finite, it is not difficult to try to find all ideals: you've got $\{0\}$ and you've got $\mathbb{Z}/6\mathbb{Z}$. Suppose the ideal contains $a\neq 0$. Then it must also contain $a+a$, $a+a+a$, and so on. Check the possibilities. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;No, there generally is no mapping between $F[X]/(p(x))$ and a $\mathbb{Z}/n\mathbb{Z}$. But first, notice that by doing long division, every polynomial $p(x)$ in $F[X]$ can be written as $p(x) = q(x)(x^3-1) + r(x)$, where $r(x)=0$ or else $\deg(r)\lt 3$. That means that every element of $F[x]/(x^3-1)$ corresponds to one of the &quot;remainders&quot;, and there are only $8$ possible remainders (the remainder must be of the form $a+bx+cx^2$, with $a,b,c\in\mathbb{Z}/2\mathbb{Z}$), so again $F[x]/(x^3-1)$ is finite, and you can check the possibilities. Here adding an element to itself is not going to help much (because $p+p=0$ for all $p$) but you can instead consider a given $p(x)$ and all $8$ multiples of it that you get when you multiply by elements of $F[x]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively, the ideals of $R/I$ correspond to ideals of $R$ that contain $I$. So the ideals of $\mathbb{Z}/6\mathbb{Z}$ correspond to ideals of $\mathbb{Z}$ that contain $6\mathbb{Z}$, and ideals of $F[X]/(x^3-1)$ correspond to ideals of $F[x]$ that contain $(x^3-1)$. Notice that $(a)$ contains $(b)$ if and only if $a$ divides $b$. &lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2011-02-21T18:23:05.203" CommentCount="3" />
  <row Id="23099" PostTypeId="1" AcceptedAnswerId="23115" CreationDate="2011-02-21T18:25:24.987" Score="0" ViewCount="59" Body="&lt;p&gt;I'm exploring the use of FFTs for multiplication, but even with simple examples it seems to go wrong. For example, here I'm trying to multiply $1$ by $2x$ (code is in matlab, but I think you can probably understand it even if you don't know the language):&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;# x and y are two polynomials, x(n) = 1; y(n) = 2n&#xA;x = [0 0 0 1];&#xA;y = [0 0 2 0];&#xA;# Evaluate both polynomials at roots of unity&#xA;xval = fft(x);&#xA;yval = fft(y);&#xA;# zval(n) = x(n)*y(n)&#xA;zval = xval .* yval;&#xA;# interpolate to get back prodvec(n) = x(n)*y(n)&#xA;prodvec = ifft(zval) / length(zval);&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This gives an output of &lt;code&gt;[0 0 0 .5]&lt;/code&gt; instead of &lt;code&gt;[0 0 2 0]&lt;/code&gt;. Any idea what I'm doing wrong? It seems obvious enough that I should be doing $Wx$ to find the values of $x(n)$, and then $W^{-1}z$ will give me back the polynomial I want. But it's not working.&lt;/p&gt;&#xA;" OwnerUserId="2549" LastActivityDate="2011-02-21T19:06:26.180" Title="Multiplication using an FFT" Tags="&lt;fourier-analysis&gt;" AnswerCount="1" />
  <row Id="23100" PostTypeId="2" ParentId="23093" CreationDate="2011-02-21T18:26:14.623" Score="6" ViewCount="" Body="&lt;p&gt;The scenario you describe provides a good example for conditional independence, though you haven't quite described it as such. As &lt;a href=&quot;http://en.wikipedia.org/wiki/Conditional_independence&quot; rel=&quot;nofollow&quot;&gt;the Wikipedia article&lt;/a&gt; puts it, &quot;$R$ and $B$ are conditionally independent [given $Y$] if and only if, given knowledge of whether $Y$ occurs, knowledge of whether $R$ occurs provides no information on the likelihood of $B$ occurring, and knowledge of whether $B$ occurs provides no information on the likehood of $R$ occurring&quot;. In this case, $R$ and $B$ are the events of persons &lt;strong&gt;A&lt;/strong&gt; and &lt;strong&gt;B&lt;/strong&gt; getting home in time for dinner, and $Y$ is the event of a snow storm hitting the city. Certainly the probabilities of $R$ and $B$ will depend on whether $Y$ occurs. However, just as it's plausible to assume that if these two people have nothing to do with each other their probabilities of getting home in time are independent, it's also plausible to assume that, while they will both have a lower probability of getting home in time if a snow storm hits, these lower probabilities will nevertheless still be independent of each other. That is, if you already know that a snow storm is raging and I tell you that person &lt;strong&gt;A&lt;/strong&gt; is getting home late, that gives you no new information about whether person &lt;strong&gt;B&lt;/strong&gt; is getting home late. You're getting information on that from the fact that there's a snow storm, but given that fact, the fact that &lt;strong&gt;A&lt;/strong&gt; is getting home late doesn't make it more or less likely that &lt;strong&gt;B&lt;/strong&gt; is getting home late, too. So conditional independence is the same as normal independence, but restricted to the case where you know that a certain condition is or isn't fulfilled. Not only can you not find out about &lt;strong&gt;A&lt;/strong&gt; by finding out about &lt;strong&gt;B&lt;/strong&gt; in general (normal independence), but you also can't do so under the condition that there's a snow storm (conditional independence).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An example of events that are independent but not conditionally dependent would be: You randomly sample two people &lt;strong&gt;A&lt;/strong&gt; and &lt;strong&gt;B&lt;/strong&gt; from a large population and consider the probabilities that they will get home in time. Without any further knowledge, you might plausibly assume that these probabilities are independent. Now you introduce event $Y$, which occurs if the two people live in the same neighbourhood (however that might be defined). If you know that $Y$ occurred and I tell you that &lt;strong&gt;A&lt;/strong&gt; is getting home late, then that would tend to increase the probability that &lt;strong&gt;B&lt;/strong&gt; is also getting home late, since they live in the same neighbourhood and any traffic-related causes of &lt;strong&gt;A&lt;/strong&gt; getting home late might also delay &lt;strong&gt;B&lt;/strong&gt;. So in this case the probabilities of &lt;strong&gt;A&lt;/strong&gt; and &lt;strong&gt;B&lt;/strong&gt; getting home in time are not conditionally independent given $Y$, since once you know that $Y$ occurred, you are able to gain information about the probability of &lt;strong&gt;B&lt;/strong&gt; getting home in time by finding out whether &lt;strong&gt;A&lt;/strong&gt; is getting home in time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Strictly speaking, this scenario only works if there's always the same amount of traffic delay in the city overall and it just moves to different neighbourhoods. If that's not the case, then it wouldn't be correct to assume independence between the two probabilities, since the fact that one of the two is getting home late would already make it somewhat likelier that there's heavy traffic in the city in general, even without knowing that they live in the same neighbourhood.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To give a precise example: Say you roll a blue die and a red die. The two results are independent of each other. Now you tell me that the blue result isn't a $6$ and the red result isn't a $1$. You've given me new information, but that hasn't affected the independence of the results. By taking a look at the blue die, I can't gain any knowledge about the red die; after I look at the blue die I will still have a probability of $1/5$ for each number on the red die except $1$. So the probabilities for the results are conditionally independent given the information you've given me. But if instead you tell me that the sum of the two results is even, this allows me to learn a lot about the red die by looking at the blue die. For instance, if I see a $3$ on the blue die, the red die can only be $1$, $3$ or $5$. So in this case the probabilities for the results are not conditionally independent given this other information that you've given me. This also underscores that conditional independence is always relative to the given condition -- in this case, the results of the dice rolls are conditionally independent with respect to the event &quot;the blue result is not $6$ and the red result is not $1$&quot;, but they're not conditionally independent with respect to the event &quot;the sum of the results is even&quot;.&lt;/p&gt;&#xA;" OwnerUserId="6622" LastEditorUserId="6622" LastEditorDisplayName="" LastEditDate="2011-02-21T18:55:26.137" LastActivityDate="2011-02-21T18:55:26.137" CommentCount="1" />
  <row Id="23101" PostTypeId="1" CreationDate="2011-02-21T18:32:58.373" Score="2" ViewCount="184" Body="&lt;p&gt;Could you help me to show: $g(x,\epsilon)+f(x,\epsilon)=O(|\phi(x,\epsilon)|+|\psi(x,\epsilon)|)$ &#xA;but $g(x,\epsilon)+f(x,\epsilon)\neq O(\phi(x,\epsilon)+\psi(x,\epsilon))$  (both when $\epsilon\to0$), where $O$ stands for the big Oh notation?&lt;/p&gt;&#xA;" OwnerUserId="7320" LastEditorUserId="3567" LastEditorDisplayName="" LastEditDate="2011-02-21T19:00:45.470" LastActivityDate="2011-02-22T12:03:54.323" Title="How to prove asymptotic property with big O notation?" Tags="&lt;asymptotics&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="23102" PostTypeId="2" ParentId="18508" CreationDate="2011-02-21T18:34:49.057" Score="0" ViewCount="" Body="&lt;p&gt;I am not sure what you mean by system of differential equations. I would recommend this &lt;a href=&quot;http://books.google.com/books?id=uvpwKK7ZVwMC&amp;amp;printsec=frontcover&amp;amp;dq=fundamentals+of+engineering+numerical+analysis&amp;amp;source=bl&amp;amp;ots=LnG-pgD66w&amp;amp;sig=dj5VGdHzFlYStPoTarRzKNwSttQ&amp;amp;hl=en&amp;amp;ei=nq9iTcTMI5OosAOBv7HNCA&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=3&amp;amp;ved=0CCwQ6AEwAg#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;book&lt;/a&gt;. It covers a good amount of numerical analysis for ODE's and PDE's. It has a good amount of problems and illustrations and is fairly rigorous.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastActivityDate="2011-02-21T18:34:49.057" />
  <row Id="23103" PostTypeId="2" ParentId="23095" CreationDate="2011-02-21T18:36:16.483" Score="0" ViewCount="" Body="&lt;p&gt;They're both products of fields: $\mathbb{Z}/6\cong\mathbb{Z}/2\times\mathbb{Z}/3$ and $F[x]/(x^3-1)\cong GF(2)\times GF(2^2)$ (if you've covered this kind of thing) which makes it easy to see the idea structure.&lt;/p&gt;&#xA;" OwnerUserId="6925" LastActivityDate="2011-02-21T18:36:16.483" />
  <row Id="23104" PostTypeId="1" CreationDate="2011-02-21T18:38:40.847" Score="7" ViewCount="346" Body="&lt;p&gt;There is an ancient problem, I remember reading once in a book while I was a kid, that says: there was a father who had $N$ sons and $T$ cows. He divided the cows among the sons in the following order: for the first son, $1 + 1/7$ of the remaining; for the second son $2+ 1/7$ of the remaining; ..., for the $i$'th son $i+1/7$ of the remaining... the question asked for the numbers T and N. In fact $N=6$ and $T=36$ is a solution. I found that it can be generalized to any given ratio $1/M$ and $N=M-1$ and $T=N^2$ is a solution. I was wondering if there is any other solution to the general problem in $\mathbb{Z}$. I have formalized the problem as following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have a system of equations described by the following $N+1$ equations defined in $\mathbb{Z}$:&#xA;\begin{align*}&#xA;n_1 &amp;amp;= 1+ \frac{T-1}{M}\\&#xA;n_2 &amp;amp;= 2+ \frac{T-2-n_1}{M}\\&#xA;&amp;amp;\vdots\\&#xA;n_i &amp;amp;= i+ \frac{T-i-\sum\limits_{j=1}^{i-1} n_j}{M}\\&#xA;T&amp;amp;=\sum\limits_{i=1}^{N} n_i.&#xA;\end{align*}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One set of solutions is given by:&#xA;\begin{align*}&#xA;T &amp;amp;= N^2\\&#xA;n_i&amp;amp;=N\\&#xA;M&amp;amp;=N+1&#xA;\end{align*}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, when $N=6$, $T=36$, $n_1 = n_2 = \cdots = n_6 = 6$ is a solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm wondering if there is any other non-trivial integer solutions to the above system of equations. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;MG&lt;/p&gt;&#xA;" OwnerUserId="7138" LastEditorUserId="7138" LastEditorDisplayName="" LastEditDate="2011-04-01T15:01:15.647" LastActivityDate="2011-04-06T03:17:39.853" Title="Cattle problem -- all solutions in $\mathbb{Z}$ " Tags="&lt;number-theory&gt;&lt;diophantine-equations&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="23105" PostTypeId="1" AcceptedAnswerId="23108" CreationDate="2011-02-20T21:30:02.807" Score="2" ViewCount="505" Body="&lt;ul&gt;&#xA;&lt;li&gt;Suppose $s_{1}$ are the numbers with 10-digits that have $1$ prime factor.&lt;/li&gt;&#xA;&lt;li&gt;Suppose $s_{2}$ are the numbers with 10-digits that have $2$ prime factors.&lt;/li&gt;&#xA;&lt;li&gt;Suppose $s_{n}$ are the numbers with 10-digits that have $n$ prime factors.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Leading $0$s are allowed. Which of $s$ would have the smallest expected value of tries to guess the right $10$ digit number?&lt;/p&gt;&#xA;" OwnerUserId="5902" LastEditorUserId="1778" LastEditorDisplayName="" LastEditDate="2011-03-07T21:14:32.140" LastActivityDate="2011-03-07T21:14:32.140" Title="What set of $10$-digit numbers with $k$ prime factors has largest cardinality?" Tags="&lt;number-theory&gt;&lt;prime-numbers&gt;" AnswerCount="3" CommentCount="5" FavoriteCount="2" />
  <row Id="23108" PostTypeId="2" ParentId="23105" CreationDate="2011-02-21T04:33:14.833" Score="9" ViewCount="" Body="&lt;p&gt;I don't know how to do this mathematically, but you can do it programmatically: use the Sieve of Eratosthenes to generate factor counts from 1 to $10^{10}-1$.  Note that $10^{10}$ is a bit over $2^{33}$ so you won't be able to store this in memory on a conventional computer (one byte per entry means you'd need 9.3 GB of RAM just for the array).  There are some time-memory trade-offs here that could keep you off the disk, and there are ways to parallelize some of this, but it's still not something you can easily figure out on a commodity computer (unless there is a better method).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I ran a C program for 10-digit numbers and obtained the following counts:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&#xA;01: 455052511&#xA;02: 1493776443&#xA;03: 2227121996&#xA;04: 2139236881&#xA;05: 1570678136&#xA;06: 977694273&#xA;07: 550454756&#xA;08: 291646797&#xA;09: 148930536&#xA;10: 74342563&#xA;11: 36585097&#xA;12: 17836903&#xA;13: 8641282&#xA;14: 4167745&#xA;15: 2002277&#xA;16: 959377&#xA;17: 458176&#xA;18: 218163&#xA;19: 103657&#xA;20: 49031&#xA;21: 23133&#xA;22: 10837&#xA;23: 5091&#xA;24: 2349&#xA;25: 1089&#xA;26: 499&#xA;27: 224&#xA;28: 102&#xA;29: 44&#xA;30: 19&#xA;31: 7&#xA;32: 3&#xA;33: 1&#xA;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This table gives the counts for each $s_i$ for $1 \leq i \leq 33$ where integers are between 2 and $10^{10} - 1$.  Prime multiplicities are counted, as you specified in a comment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This ran in about 12 minutes on an x86 using 10 GB of RAM.  So the answer to your question is that $s_3$ is largest.  Code posted below.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;&#xA;#include &amp;lt;stdio.h&gt;&#xA;#include &amp;lt;stdlib.h&gt;&#xA;&#xA;#define MAX 10000000000       // sieve size&#xA;#define PPOWER 255            // marker for prime powers&#xA;&#xA;// allocate MAX bytes of memory&#xA;unsigned char sieve[MAX];&#xA;&#xA;// use # of factors as index, # of ints is value&#xA;unsigned long tally[100];&#xA;&#xA;&#xA;// sieve[] is initialized to all 0's; then for each i, sieve[i] will contain&#xA;// the number of prime factors (including multiplicity) for each i, except&#xA;// sieve[p^j] for prime p, j &gt;1 will contain the special marker PPOWER.&#xA;// We do this to properly account for prime multiplicities.&#xA;&#xA;main() {&#xA;  unsigned long i, j;&#xA;  int c;&#xA;&#xA;  for (i=2; i &amp;lt; MAX; i++) {&#xA;&#xA;    if (sieve[i] != PPOWER) {&#xA;      // if composite, tally and continue&#xA;      if (sieve[i] &gt; 0) {&#xA;        tally[sieve[i]]++;&#xA;        continue;&#xA;      }&#xA;&#xA;      // ok, i is prime; tally and mark all prime powers as PPOWER&#xA;      // (takes some thought to see why we do this)&#xA;      tally[1]++;&#xA;      j = i*i;&#xA;      c = 2;&#xA;      while (j &amp;lt; MAX) {&#xA;        sieve[j] = PPOWER;&#xA;        tally[c]++;&#xA;        j = j * i;&#xA;        c++;&#xA;      }&#xA;    }&#xA;&#xA;    // now sieve as usual&#xA;    for (j=i*2; j &amp;lt; MAX; j += i) {&#xA;      if (sieve[j] != PPOWER)&#xA;          sieve[j]++;&#xA;    }&#xA;  }&#xA;&#xA;  for (c=1; c &amp;lt; 100; c++)&#xA;    if (tally[c]) printf(&quot;%.2d: %ld\n&quot;, c, tally[c]);&#xA;}&#xA;&lt;/code&gt;&#xA;&lt;/pre&gt;&#xA;" OwnerUserId="7162" LastEditorUserId="7162" LastEditorDisplayName="" LastEditDate="2011-02-24T16:44:46.887" LastActivityDate="2011-02-24T16:44:46.887" CommentCount="18" />
  <row Id="23109" PostTypeId="2" ParentId="23104" CreationDate="2011-02-21T18:51:31.750" Score="6" ViewCount="" Body="&lt;p&gt;Are you interested in solutions in $\mathbb{Z}$ or solutions in $\mathbb{Z}^+$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another trivial solution in integers is $$M=1$$ $$n_1=T$$ $$n_i = 0, \forall i \neq 1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So setting $M=1$, you can choose $T$ to be any integer and you will get only integer solutions since the condition $T = \displaystyle \sum_{i=1}^N n_i$ is trivially satisfied and $n_i$ are defined recursively as integer combinations of previous $n_i$'s and $T$ with $n_1 = T$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;: Lets analyze further to see other possible integer solutions. (My analysis is incomplete will try to do this later tonight/tomorrow.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$n_N = N + \frac{T-N - \displaystyle \sum_{j=1}^{N-1} n_j}{M}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the constraint, $T = \displaystyle \sum_{j=}^N n_j$, we get $n_N = T - \displaystyle \sum_{j=1}^{N-1} n_j$ and hence&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$n_N = N + \frac{n_N - N}{M}$$&#xA;$$\left( n_N-N \right) \left(1-\frac{1}{M} \right) = 0$$ which gives us that $$n_N = N \text{ or } M=1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $M=1$, then we already have the solution $$n_1 = T \text{ and } n_i = 0 \text{ } \forall \text{ }i&amp;gt;1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $M \neq 1$, then $n_N = N$, and we have $$n_{N-1} = N-1 + \frac{T-(N-1) - (n_1+n_2+ \cdots n_{N-2})}{M} = N - 1 + \frac{N + n_{N-1}-(N-1)}{M}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(n_{N-1} - N + 1)(M-1) = N$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From which we get that $M \neq 1$ and since we are interested only in integer solutions we have $(M-1)|N$ i.e. $M = 1 + a$ where $a|N$. Then $n_{N-1} = N + \frac{N}{a} - 1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$n_{N-2} = N-2 + \frac{n_N + n_{N-1} + n_{N-2} - (N-2)}{1+a}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$n_{N-2} - (N-2) - \frac{n_{N-2} - (N-2)}{1+a} = \frac{N + N + \frac{N}{a}-1}{1+a}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$a(n_{N-2} - (N-2)) = 2N + \frac{N}{a} - 1$$&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2011-02-21T21:01:20.793" LastActivityDate="2011-02-21T21:01:20.793" CommentCount="0" />
  <row Id="23110" PostTypeId="2" ParentId="23068" CreationDate="2011-02-21T18:55:35.257" Score="5" ViewCount="" Body="&lt;p&gt;Presumably by &quot;four axioms&quot; you refer to the properties that define a polynomial Sturm sequence. Sturm's polynomial root-counting algorithm algorithm works in any real-closed-field. Recall that a field is &lt;em&gt;real&lt;/em&gt; if $-1$ is not a sum of squares, and a &lt;em&gt;real-closed&lt;/em&gt; field is a real field with no proper real algebraic extension. There are many equivalent ways to characterize real-closed fields $\rm\:R\:,\:$ e.g. either $\rm\:r\:$ or $\rm\:-r\:$ has a square-root $\rm\in R$ and every odd-degree polynomial $\rm\ f(x)\in R[x]$ has root $\rm\in R\:.\:$ Alternatively: $\rm\ R\ $ is an ordered field wrt the order $\rm\ r\ge 0\iff r\ $ is a square, and $\rm\ R\ $ has the intermediate value property wrt this order.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sturm's algorithm plays a fundamental role in the theory of real-closed fields. For example, it easily yields the uniqueness of a real-closure fixing a given order. Further it can be generalized to prove that the projection of a semi-algebraic set is semi-algebraic - a geometric form of Tarski's celebrated quantifier elimination for real-closed fields. This leads to an effective  algorithm for deciding the truth of first order statements in real-closed fields - by cylindric algebraic decomposition (CAD). The algorithm simply decomposes $\rm\ R^n\ $ into a &lt;em&gt;finite&lt;/em&gt; number of constant-sign &quot;cylinders&quot; and then simply inspects the sign on each component (this is completely obvious in the one-dimensional case, e.g. &lt;a href=&quot;http://math.stackexchange.com/questions/13654/problems-with-inequalities/13681#13681&quot;&gt;see here&lt;/a&gt;). You can read about this in any textbook on real algebraic geometry.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2011-02-21T20:07:38.460" LastActivityDate="2011-02-21T20:07:38.460" />
  <row Id="23111" PostTypeId="1" AcceptedAnswerId="23136" CreationDate="2011-02-21T18:57:27.437" Score="6" ViewCount="237" Body="&lt;p&gt;I've encountered this whilst writing an optimisation benchmark for some heuristic search algorithms. Feels like there should be a basic solution out there!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A square-grid graph is constructed from $n \times m$ vertices embedded into the cartesian plane, such that each vertex corresponds to a pair of integers. Say that a vertex is on a boundary if the $x$ or $y$ coordinate is minimal or maximal (i.e. it is at the lower/upper bound of the grid).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a general expression for the number of simple paths (paths that do not revisit the same vertex) between a pair of vertices $(a,b)$ on a square-grid graph. Where $a$ is on a boundary and $b$ is not?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="7330" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2011-02-22T02:06:44.887" LastActivityDate="2011-02-22T20:52:48.690" Title="Number of simple paths between two vertices on an $n \times m$ square-grid graph?" Tags="&lt;combinatorics&gt;&lt;graph-theory&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="23112" PostTypeId="2" ParentId="23020" CreationDate="2011-02-21T19:01:03.860" Score="3" ViewCount="" Body="&lt;p&gt;In high dimensions there are several variants that are all distinct (which for surfaces they all agree).  There's mapping class groups in the &quot;homotopy category&quot; meaning the homotopy-classes of homotopy equivalences of a topological space, with composition giving the group structure.  This is a &quot;core&quot; object of study of classical algebraic topology.  In the topological/pl/smooth categories there are isotopy classes of homeomorphisms/pl automorphisms/diffeomorphisms of a manifold.  The smooth category gets quite a bit of attention -- for example the smooth category mapping class group of $S^n$ (if you restrict to orientation-preserving diffeomorphisms) is the group of homotopy $(n+1)$-spheres, provided $n \geq 5$.   There has been some work on stable high-dimensional mapping class groups by people like Giansiracusa (Swansea).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have to head out but I can add more later. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Giansiracusa reference is this: &lt;a href=&quot;http://www.arxiv.org/abs/math.gt/0510599&quot; rel=&quot;nofollow&quot;&gt;http://www.arxiv.org/abs/math.gt/0510599&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Modulo some qualifiers the statement is that the stable mapping class group of a 4-manifold is the automorphisms of homology that preserve the intersection form. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mapping class groups of a products of circles $(S^1)^n$ in the topological, PL and smooth categories were computed by Hatcher in his &quot;Higher simple homotopy theory&quot; paper.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there anything in particular you're interested in?&lt;/p&gt;&#xA;" OwnerUserId="642" LastEditorUserId="642" LastEditorDisplayName="" LastEditDate="2011-02-21T20:18:27.383" LastActivityDate="2011-02-21T20:18:27.383" CommentCount="2" />
  <row Id="23113" PostTypeId="2" ParentId="23079" CreationDate="2011-02-21T19:06:22.903" Score="5" ViewCount="" Body="&lt;p&gt;Hi Hans. This question was &lt;a href=&quot;http://mathoverflow.net/questions/44303/cardinality-of-proper-classes&quot; rel=&quot;nofollow&quot;&gt;asked&lt;/a&gt; at MO a while ago. Below I quote the relevant part of my answer. But let me add some comments.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In ZFC (proper) classes are not actual objects, we only treat them formally, and they are really just short-cuts for formulas (with parameters), i.e., we identify formulas $\phi(x)$ with the collection of sets that satisfy $\phi$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This makes talking of relations between classes somewhat awkward. Arturo's nice answer, for example, indicates how one can make sense of, say, a bijection between two classes. A bit more formally, if $X$ is the class of $x$ such that $\phi(x)$ and $Y$ is the class of $y$ such that $\psi(y)$, then a bijection between $X$ and $Y$ is a class $Z$ given by a formula $\rho(x,y)$ such that:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;For any $x,y,z$, if $\rho(x,y)$ and $\rho(x,z)$ then $y=z$. &lt;/li&gt;&#xA;&lt;li&gt;Similarly, for any $x,y,z$, if$ \rho(x,y)$ and $\rho(z,y)$, then $x=z$.&lt;/li&gt;&#xA;&lt;li&gt;For any $x,y$, $\rho(x,y)$ implies that $\phi(x)$ and $\psi(y)$ holds.&lt;/li&gt;&#xA;&lt;li&gt;For any $x$ such that $\phi(x)$ there is a $y$ such that $\rho(x,y)$.&lt;/li&gt;&#xA;&lt;li&gt;And, for any $y$ such that $\psi(y)$, there is an $x$ such that $\rho(x,y)$.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;This awkwardness also makes it impossible to develop an appropriate theory of classes. For example, the natural question: &quot;Are $X$ and $Y$ pf the same size?&quot; cannot even be &lt;em&gt;asked&lt;/em&gt; in ZFC. Of, course, if there is a $\rho$ as above, then we can say that they have the same size, as witnessed by $Z$, and if there is no such $\rho$, outside of ZFC we can say that thy don't have the same size, but all we can do in ZFC is to say, of any given formula $\rho$, that $\rho$ does not define a bijection between $X$ and $Y$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, equipotency is only an example of what we cannot discuss freely about classes. Thus when interested in proper classes, we move from ZFC to other theories. There are (at least) two natural extensions of ZFC where classes can be treated as formal objects. One is NBG (Von Neumann-Bernays-Goedel). Here, the objects are classes, and sets are classes that belong to other classes. $Z$ is a bijection between $X$ and $Y$ iff $A$ is a function, its domain is $X$ and its range is $Y$ (just as with sets). NBG is usually preferred because it is a &lt;em&gt;conservative&lt;/em&gt; extension of ZFC; in a sense, all we have done is to allow references to classes without changing what we mean by &quot;set&quot; in any way. Formally, any theorem of NBG that only mentions sets is a theorem of ZFC. And our interpretation of classes as given by formulas gives us a way of extending any model of ZFC to one of NBG.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fact that NBG is conservative over ZFC is in a sense a serious limitation, as we limit the notion of class somewhat artificially. When discussing elementary embedding formulations of large cardinal axioms (something very common in modern set theory), for example, the discussion can be carried out in NBG, but not in the smoothest possible fashion. In technical terms, the comprehension axiom in NBG is predicative, but it is hard to justify this when we are allowing proper classes at all. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most natural extension of ZFC to treat proper classes is Morse-Kelley (MK). Here, comprehension is unrestricted, so it is more natural to combine classes into new ones by usual operations. The cost of this is that MK is not a conservative extension of ZFC. In fact, MK can prove the consistency of NBG (and therefore of ZFC). That being said, to discuss equipotency of classes, MK seems the appropriate framework.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is what I said in the MO post mentioned above: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In extensions of set theory where classes are allowed (not just formally as in ZFC, but as actual objects as in MK or GB), sometimes it is suggested to add an axiom (due to Von Neumann, I believe) stating that any two classes are in bijection with one another. Under this axiom, the &quot;cardinality&quot; of a proper class would be ORD, the class of all ordinals. (By the way, by class forcing, given any proper class, one can add a bijection between the class and ORD without adding sets, so this assumption bears no implications for set theory proper.)&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Without assuming Von Neumann's axiom, or the axiom of choice, I know of no sensible way of making sense of this notion, as now we could have some proper classes that are &quot;thinner&quot; than others, or even incomparable. Of course, we could study models where this happens (for example, work in ZF, assume there is a strong inaccessible $\kappa$, and consider  $V_\kappa$ as the universe of sets, and $Def(V_\kappa)$ in Gödel's sense (or even  $V_{\kappa+1}$) as the collection of classes). &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Let me expand on this a bit. I argue in ZFC as this is the best known framework of the three mentioned above:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, class forcing allows us to add a bijection between $V$ and $ORD$ without adding sets. Essentially, what we do is to &quot;thread&quot; through the class of bijections between sets and ordinals. In the resulting extension, we have added no sets, but we have a new class $G$. The resulting structure $(V,G,\in)$ is a model of the strong version of ZFC where we allow $G$ to appear as a predicate in instances of the replacement axiom. Also, any proper class $A$ here (in the sense of &quot;definable by a formula&quot;) is also in bijection with the ordinals, and such a bijection is easily definable from $A$ and $G$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This shows that the assumption that all classes have the same size is completely harmless: No new theorems of ZFC can be proved by adding this assumption.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The model we obtain is not a &quot;natural&quot; model of NBG, though, since G is not definable. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Arturo's suggestion ($V=L$) gives us models where bijections between classes and the ordinals are definable. It has the disadvantage that V=L is quite limiting. As I mentioned in the comments to his answer, there is an alternative: We can assume $G$ definable. This is because, over any model of ZFC, we can force to obtain a new model where $V=HOD$ (on the other hand, once $V\ne L$ we cannot force to have the equality back).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$HOD$ is the class of hereditarily ordinal definable sets. This means that $x\in HOD$ implies that $x\subset HOD$, and that to be in HOD, $x$ must be definable from ordinal parameters. Of course $V=L$ implies $V=HOD$, but $HOD$ is compatible with all known large cardinals, while $L$ is not. Moreover, $HOD$ carries a definable well-ordering (in order type ORD), so if $V=HOD$, then any proper class is in bijection with the ordinals. Moreover, $V=HOD$ is equivalent to the statement that there is a (definable) bijection between $V$ and $ORD$. So, at least in ZFC, this characterizes in a natural way when all proper classes have the same size (and it is really the only way that &quot;sizes of proper classes&quot; can be freely discussed in ZFC). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, it &lt;em&gt;is&lt;/em&gt; consistent that $V\ne HOD$, in which case $V$ and $ORD$ have different cardinality in the ZFC sense defined above, and in fact one can arrange that there are incomparable &quot;sizes&quot; of proper classes. &lt;/p&gt;&#xA;" OwnerUserId="462" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-21T19:12:20.930" LastActivityDate="2011-02-21T19:12:20.930" CommentCount="2" />
  <row Id="23114" PostTypeId="1" AcceptedAnswerId="23138" CreationDate="2011-02-21T19:06:26.263" Score="6" ViewCount="139" Body="&lt;p&gt;I am familiar with using cross products in physics to answer questions about force and torque, but are there applications to other scientific fields?  The examples that I've seen in most calculus textbooks deal only with physics or engineering questions.&lt;/p&gt;&#xA;" OwnerUserId="5727" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2011-02-21T21:41:53.443" LastActivityDate="2011-02-21T21:41:53.443" CommunityOwnedDate="2011-02-22T10:23:12.073" Title="Applications of cross product in sciences other than physics" Tags="&lt;calculus&gt;&lt;vector-calculus&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="23115" PostTypeId="2" ParentId="23099" CreationDate="2011-02-21T19:06:26.180" Score="2" ViewCount="" Body="&lt;p&gt;There are two things you're doing wrong:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Vectors in matlab are read left-to-right, so you should have $x,y$ reversed.&lt;/li&gt;&#xA;&lt;li&gt;The FFT and inverse FFT operations are inverses, so there's no need to divide by the normalization factor.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Try this instead:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;x = [1 0 0 0];&#xA;y = [0 2 0 0];&#xA;ifft(fft(x).*fft(y))&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2011-02-21T19:06:26.180" CommentCount="4" />
  <row Id="23116" PostTypeId="1" AcceptedAnswerId="23127" CreationDate="2011-02-21T19:07:19.180" Score="4" ViewCount="85" Body="&lt;p&gt;You need to arrange a list of participants - four administrators and four students - who will sit behind a table in the order listed.  In how many ways can you list them if you must alternate students and administrators?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This question is from &lt;em&gt;Discrete Mathematics for Computer Scientists&lt;/em&gt; by Stien, Drysdale and Bogart.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is what I am thinking they be arrange two ways&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) SASASASA&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) ASASASAS&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In both 1 &amp;amp; 2, you can arrange by (4*3*2*1) students and (4*3*2*1) admins.  My solution would be:  (4! + 4!) * 2 = 4*4!.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The text does not provide a solution, so I hope someone here can confirm.  &lt;/p&gt;&#xA;" OwnerUserId="6391" LastEditorUserId="6391" LastEditorDisplayName="" LastEditDate="2011-02-21T19:17:21.250" LastActivityDate="2011-02-21T22:11:59.777" Title="How many ways can you list them if you must alternate students and administrators?" Tags="&lt;combinatorics&gt;" AnswerCount="2" />
  <row Id="23117" PostTypeId="2" ParentId="22982" CreationDate="2011-02-21T19:07:49.780" Score="2" ViewCount="" Body="&lt;p&gt;A Markov chain is basically a stochastic process in which the past history of the process is irrelevant if you know the current system state.  In other words, all information about the past and present that would be useful in saying something about the future is contained in the present state.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A discrete-time Markov chain is one in which the system evolves through discrete time steps.  So changes to the system can only happen at one of those discrete time values.  An example is a board game like &lt;a href=&quot;http://en.wikipedia.org/wiki/Snakes_and_ladders&quot; rel=&quot;nofollow&quot;&gt;Chutes and Ladders&lt;/a&gt; (apparently called &quot;Snakes and Ladders&quot; outside the U.S.) in which pieces move around on the board according to a die roll.  If you are looking at the board at the beginning of someone's turn and wondering what the board will look like at the beginning of the next person's turn, it doesn't matter how the pieces arrived at their current positions (the past history of the system).  All that matters is that the pieces are where they currently are (the current system state) and the upcoming die roll (the probabilistic aspect).  This is discrete because changes to the system state can only happen on someone's turn.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A continuous-time Markov chain is one in which changes to the system can happen at any time along a continuous interval.  An example is the number of cars that have visited a drive-through at a local fast-food restaurant during the day.  A car can arrive at any time $t$ rather than at discrete time intervals.  Since arrivals are basically independent, if you know the number of cars that have gone through by 10:00 a.m., what happened before 10:00 a.m. doesn't give you any additional information that would be useful in predicting the number of cars that will have visited the drive-through by, say, noon.  (This is under the usual but reasonable assumption that the arrivals to the drive-through follow a &lt;a href=&quot;http://en.wikipedia.org/wiki/Poisson_process&quot; rel=&quot;nofollow&quot;&gt;Poisson process&lt;/a&gt;.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A Markov decision process is just a Markov chain that includes an agent that makes decisions that affect the evolution of the system over time.  (So I don't think of it as a separate kind of Markov chain, since the usual Markov chain definition doesn't include such an agent.)  A (continuous-time) example would be the potato chip inventory for a local grocery store.  If you know the inventory level at 10:00 a.m. and are trying to predict the inventory level at noon, the inventory levels before 10:00 a.m. don't tell you anything beyond what you already know about the level at 10:00 a.m.  The decision aspect arises because the manager can decide when to place orders so that bags arrive at certain times.  Thus the inventory level at any time $t$ depends not just on (the probabilistic aspect of) customers arriving randomly and taking bags off the shelf but also on the manager's (deterministic) decisions.  (An example of a discrete-time Markov decision process is the board game &lt;a href=&quot;http://en.wikipedia.org/wiki/Parcheesi&quot; rel=&quot;nofollow&quot;&gt;Parcheesi&lt;/a&gt;.  The board position at the beginning of the next player's turn depends only on the current board position, the current player's dice roll (the Markov chain aspect) and the current player's decision as to which pieces to move based on the dice roll (the decision process aspect).)  &lt;/p&gt;&#xA;" OwnerUserId="2370" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2011-02-22T16:38:01.237" LastActivityDate="2011-02-22T16:38:01.237" CommentCount="2" />
  <row Id="23118" PostTypeId="2" ParentId="23105" CreationDate="2011-02-21T19:09:28.180" Score="6" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://www.wolframalpha.com/input/?i=PrimePi%2810%5E10%29&quot; rel=&quot;nofollow&quot;&gt;Wolfram Alpha &lt;/a&gt;gives the number of up to 10 digit primes as 455,052,511.  When you ask for 2 prime factors, do repeated factors count?  So how many prime factors does $2^{31}*3=6442450944$ have?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Added:  &lt;a href=&quot;http://oeis.org/A066265&quot; rel=&quot;nofollow&quot;&gt;OEIS&lt;/a&gt; gives the number of up to 10 digit semiprimes as 1493776443  I don't find a result for other numbers of factors easily.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edited to allow numbers with leading zeros in the 10 digits&lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2011-02-21T21:51:22.987" LastActivityDate="2011-02-21T21:51:22.987" CommentCount="5" />
  <row Id="23119" PostTypeId="1" CreationDate="2011-02-21T19:14:05.700" Score="0" ViewCount="22" Body="&lt;p&gt;It is known that unimodular transformations are bijections of the lattice $\mathbb{Z}^{n}$. Suppose I have a (not necessarily integral) polytope $P$ of the form $\textbf{conv}(\mathbf{0}, a_{1} \mathbf{e}_{1}, \dots, a_{n} \mathbf{e}_{n})$, a so-called simplicial $n$-polytope with $a_{i} \in \mathbb{R}_{+}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Question:&lt;/em&gt; How does an arbitrary unimodular transformation act on such polytopes, i.e., let $T$ be such a transformation, what is $T(\mathbf{0}, a_{1} \mathbf{e}_{1}, \dots, a_{n} \mathbf{e}_{n})$? I understand that this transformation must leave the number of intersection lattice points invariant, so the lattice point counting function $L_{P}(t) = |tP \cap \mathbb{Z}^{n}|$ satisfies $L_{TP}(t) = L_{P}(t)$. &lt;/p&gt;&#xA;" OwnerUserId="2720" LastActivityDate="2011-02-21T19:14:05.700" Title="Unimodular Transformations of Simplicial $n$-Polytopes" Tags="&lt;number-theory&gt;&lt;linear-algebra&gt;&lt;combinatorics&gt;" CommentCount="1" FavoriteCount="1" />
  <row Id="23120" PostTypeId="1" CreationDate="2011-02-21T19:20:28.657" Score="3" ViewCount="109" Body="&lt;p&gt;Is it a mistake &lt;em&gt;ab initio&lt;/em&gt; to think of categories as of &lt;em&gt;models&lt;/em&gt; of category theory, just as we think of (inner) models-of-set-theory as of models of set theory, graphs as of models of graph theory, groups as of models of group theory, topologies as of models of topology, and so on?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is one big difference, and maybe the essence lies within: all &quot;normal&quot; models have to been based on &lt;em&gt;sets&lt;/em&gt;, while categories may be based on &lt;em&gt;proper classes&lt;/em&gt;. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Why would it be a problem for &quot;normal&quot;&#xA;  models of &quot;normal&quot; theories to be based on proper classes, but not&#xA;  for categories as models of category theory?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;&lt;strong&gt;Side remark:&lt;/strong&gt; Forgive me for talking at large: &quot;Normal&quot; models of &quot;normal&quot; theories are required to be based on sets, because only then one can apply the machinery of set theory. For categories one definitely wants to apply another machinery, the machinery of category theory. But if this machinery can handle class models in principle, why not apply this machinery on class models of &quot;normal&quot; theories, too - and in a direct way?&lt;/sup&gt;&lt;/p&gt;&#xA;" OwnerUserId="1792" LastEditorUserId="1792" LastEditorDisplayName="" LastEditDate="2011-02-22T06:33:22.967" LastActivityDate="2011-02-22T06:33:22.967" Title="Class models in set theory and category theory" Tags="&lt;set-theory&gt;&lt;category-theory&gt;&lt;model-theory&gt;" CommentCount="3" />
  <row Id="23121" PostTypeId="2" ParentId="23091" CreationDate="2011-02-21T19:20:50.437" Score="1" ViewCount="" Body="&lt;p&gt;The first is an application of the definition of $H_t$ as an operator which works on functions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$H_t f(x) = e^{-t} \sum_{i=0}^{\infty} \frac{t^i K^i f(x)}{i!} \; .$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Apply it on the function $\delta_y$ to give&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$H_t \delta_y(x) = e^{-t} \sum_{i=0}^{\infty} \frac{t^i K^i \delta_y(x)}{i!} \; .$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the crucial thing to work out is what is the action of $K$ on $\delta_y$. Again using the definition, this becomes&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$K \delta_y (x) = \sum_{z \in \mathcal{X}} K(x,z)\delta_y(z)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $\delta_y(z)=1$ when $z=y$ and $0$ elsewhere this simplifies to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$K \delta_y (x) = K(x,y) \; .$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Applying K $i$ times to $\delta_y(x)$ should give you the $i$^th power of the stochastic matrix $K(x,y)$ which is denoted in the text by $K^i(x,y)$. (This confused me at first, since I though they meant an ordinary power of the $(x,y)$-component.) So for instance, for power two, this means&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$K^2 \delta_y(x) \equiv K^2(x,y) = \sum_{z \in \mathcal{X}} K(x,z)K(z,y) \; .$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Putting this all together, we get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$H_t \delta_y(x) = e^{-t} \sum_{i=0}^{\infty} \frac{t^i K^i(x,y)}{i!} \; .$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But this is exactly how $H_t(x,y)$ is defined. So up till there, much ado about nothing, just some definitions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT 1&lt;/strong&gt; Purely using the definition &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$h_t(x,y)=\frac{H_t(x,y)}{\pi(y)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;one can show that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{eqnarray}&#xD;&#xA;h_t(x,y) &amp;amp; = &amp;amp;\frac{H_t(x,y)}{\pi(y)}=\frac{\sum_z H_{t/2}(x,z)H_{t/2}(z,y)}{\pi(y)} \\&#xD;&#xA;&amp;amp; = &amp;amp;\sum_z \frac{H_{t/2}(x,z)}{\pi(z)}\frac{H_{t/2}(z,y)}{\pi(y)}\pi(z) = \sum_z h_{t/2}(x,z)h_{t/2}(z,y)\pi(z)\end{eqnarray}$$&lt;/p&gt;&#xA;" OwnerUserId="3567" LastEditorUserId="3567" LastEditorDisplayName="" LastEditDate="2011-02-22T18:20:41.470" LastActivityDate="2011-02-22T18:20:41.470" CommentCount="0" />
  <row Id="23122" PostTypeId="2" ParentId="23116" CreationDate="2011-02-21T19:26:09.280" Score="6" ViewCount="" Body="&lt;p&gt;For each arrangement of students you have 4! arrangements of admins, so you have 2(whether you start with A or S) *4! (arrangements of students) *4! (arrangements of admins) = 1152 &lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-02-21T19:26:09.280" />
  <row Id="23123" PostTypeId="1" CreationDate="2011-02-21T19:28:34.817" Score="0" ViewCount="264" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Find parametric equations for the tangent line at the point $(\cos(-\frac{4 \pi}{6}), \sin(-\frac{4 \pi}{6}), -\frac{4 \pi}{6}))$ on the curve $x = \cos(t), y = \sin(t), z=t$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I understand that in order to find the solution, I need to use partial derivatives. However, the method in my textbook works for simpler problems -- I seem to be making a calculation error when I try to apply the method to this problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone suggest how to approach this problem?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found a &lt;em&gt;very&lt;/em&gt; &lt;a href=&quot;http://answers.yahoo.com/question/index?qid=20070403182912AAd4PCr&quot; rel=&quot;nofollow&quot;&gt;similar problem and solution here&lt;/a&gt;, but the solution by the person who answered is hard for me to follow. Unfortunately, I get stuck at the line where he subtracts $\frac{\pi}{6}$ from $\pi$ within the trigonometric functions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.math.ucla.edu/~ronmiech/Calculus_Problems/32A/chap11/section7/732d45/732_45.html&quot; rel=&quot;nofollow&quot;&gt;Here is the &quot;simple&quot; method that I was originally using.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any sincere help would be appreciated. Thank you.&lt;/p&gt;&#xA;" OwnerUserId="3800" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2011-02-21T21:46:02.657" LastActivityDate="2011-02-21T21:46:02.657" Title="Finding parametric equations for the tangent line at a point on a curve" Tags="&lt;calculus&gt;&lt;geometry&gt;&lt;trigonometry&gt;&lt;vector-calculus&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="23124" PostTypeId="1" AcceptedAnswerId="23179" CreationDate="2011-02-21T19:33:06.280" Score="6" ViewCount="233" Body="&lt;p&gt;For $\alpha\in\mathbb{R}$, it seems that we have $\displaystyle \sum_{k \ge 0}\frac{1}{k^2+k-\alpha} = \frac{\pi}{\sqrt{4\alpha+1}}\tan\left(\frac{1}{2}\pi\sqrt{4\alpha+1}\right).$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've tried many techniques and ways -- partial fractions, turning it into an integral, finding the partial sum and so on -- to show it, but all ended in failure. Many thanks for the help in advance. &lt;/p&gt;&#xA;" OwnerUserId="4558" LastEditorUserId="4558" LastEditorDisplayName="" LastEditDate="2011-02-22T01:24:10.913" LastActivityDate="2011-02-22T13:12:48.160" Title="Show that $\displaystyle \sum_{k \ge 0}\frac{1}{k^2+k-\alpha} = \frac{\pi}{\sqrt{4\alpha+1}}\tan\left(\frac{1}{2}\pi\sqrt{4\alpha+1}\right)$" Tags="&lt;calculus&gt;&lt;complex-analysis&gt;&lt;sequences-and-series&gt;" AnswerCount="1" CommentCount="8" FavoriteCount="1" />
  <row Id="23125" PostTypeId="1" AcceptedAnswerId="23133" CreationDate="2011-02-21T19:33:13.420" Score="4" ViewCount="135" Body="&lt;p&gt;Hi, I'd like to know how to derive these functions (I know the answers, I want to know how to get there)&#xA;\begin{align*}&#xA;f(x) &amp;amp;= \arcsin\left(\frac{x}{3}\right)\\&#xA;f(x) &amp;amp;= \arccos(2x+1)\\&#xA;f(x) &amp;amp;= \arctan(x^2)\\&#xA;f(x) &amp;amp;= \mathrm{arcsec}(x^7)\\&#xA;\end{align*}&#xA;etc.&lt;/p&gt;&#xA;" OwnerUserId="6350" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-21T19:42:09.537" LastActivityDate="2011-02-21T20:16:49.257" Title="Trying to derive an inverse trigonometric function..." Tags="&lt;calculus&gt;&lt;trigonometry&gt;" AnswerCount="4" CommentCount="4" />
  <row Id="23126" PostTypeId="2" ParentId="23125" CreationDate="2011-02-21T19:39:41.303" Score="2" ViewCount="" Body="&lt;p&gt;When you say &quot;derive these functions&quot; do you mean &quot;take the derivative of these functions?&quot;  If so, the &lt;a href=&quot;http://en.wikipedia.org/wiki/Chain_rule&quot; rel=&quot;nofollow&quot;&gt;chain rule&lt;/a&gt; is your friend.  So for $f(x)=\arctan(x^2)$, $f&amp;#39;(x)=\frac{1}{1+(x^2)^2}2x$&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-02-21T19:39:41.303" CommentCount="2" />
  <row Id="23127" PostTypeId="2" ParentId="23116" CreationDate="2011-02-21T19:40:30.193" Score="8" ViewCount="" Body="&lt;p&gt;Your answer is incorrect for several reasons. First, you don't add the totals: since you must order &lt;em&gt;both&lt;/em&gt; the students and the administrators, you should &lt;em&gt;multiply&lt;/em&gt; the totals; you multiply when you are counting how many ways you can do both $A$ and $B$; you add when you are counting how many ways you can do &lt;em&gt;either&lt;/em&gt; $A$ or $B$. See the opening remarks in &lt;a href=&quot;http://math.stackexchange.com/questions/11307/how-do-i-do-combinatorics-counting/11312#11312&quot;&gt;this previous answer&lt;/a&gt;. And you are not taking into account that you have both possibilities: either begin with an administrator, or begin with a student (which would require you to add the results you got for each case). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's two different ways of counting them, both leading to the same final answer, both giving the same answer Ross gave, but with details explained.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Along the lines you describe, but using the Sum and Product rules correctly: You can &lt;em&gt;either&lt;/em&gt; begin with an administrator or with a student. No matter who begins, you order the administrators, which as you correctly point out can be done in $4!$ ways, and you order the students, which again as you correctly point out can be done in $4!$ ways. Since you must to both, in each case (whether an administrator goes first or a student goes first) you have $4!\times 4!$ ways of ordering them (you must order the administrators &lt;em&gt;and&lt;/em&gt; order the students, so the Product Rule applies). This describes the full ordering, since once you order the students, order the administrators, and decide who goes first, you've got the entire order. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since you can &lt;em&gt;either&lt;/em&gt; begin with a student &lt;strong&gt;or&lt;/strong&gt; with an administrator, you want to add the total number of ways to do it with a student first (which is $4!\times 4!$), with the number of ways to do it with an administrator first (also $4!\times 4!$). So the total number is $(4!\times 4!) + (4!\times 4!) = 2\times 4!\times 4!$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;You can pick anyone to go first; that's $8$ possible ways. Once you decide who goes first, you've got $4$ possibilities for who goes second (must be someone from the other group). Then you've got $3$ for go who goes third (only three people left in the group to which the first person belongs), and $3$ for who goes fourth (any of the three remaining people in the group to which the person in the second position belongs); then $2$ for who goes fifth, $2$ for who goes sixth, and the 7th and 8th place are now forced. So that gives $$8\times 4\times 3\times 3\times 2\times 2 = 2\times(4\times 3\times 2)\times(4\times 3\times 2) = 2\times4!\times 4!$$ &#xA;ways of ordering them, same as before.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-21T22:11:59.777" LastActivityDate="2011-02-21T22:11:59.777" CommentCount="1" />
  <row Id="23128" PostTypeId="2" ParentId="23039" CreationDate="2011-02-21T19:59:44.367" Score="2" ViewCount="" Body="&lt;p&gt;This follows immediately from the &lt;strong&gt;Archimedian&lt;/strong&gt; property. To find a dyadic $\rm\ m/2^n \in (r,s)\ $ start at at any dyadic $\rm\ k &amp;lt; r\ $ (e.g. an integer) and keep taking dyadic step sizes smaller than the interval, say $\rm\ 1/2^j\ &amp;lt;\ s-r\:.\: $ By the Archimedean property eventually you'll land in the interval, necessarily at some dyadic rational (being a sum of such). This is a special case of the proof I &lt;a href=&quot;http://math.stackexchange.com/questions/21435/equivalency-of-archimedian-fields-properties/21457#21457&quot;&gt;explained here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2011-02-21T19:59:44.367" />
  <row Id="23129" PostTypeId="1" AcceptedAnswerId="23140" CreationDate="2011-02-21T20:05:00.353" Score="18" ViewCount="518" Body="&lt;p&gt;In Euclidean geometry, I frequently use concepts related to invariance under scaling.  For example, I know that if two squares have different side lengths, the ratio of their side lengths is the square root of the ratio of their areas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I became curious about the justification of this principle recently when I was asked to explain pi, the ratio of circumference to diameter of a circle, to a young student.  It occurred to me half way through the explanation that I do not know a reason why pi exists; I'm not sure why all circles have the same ratio.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I haven't studied non-Euclidean geometry, but I presume this scale-invariance is a special feature of Euclidean geometry.  On the surface of a sphere, for example, the sum of the angles of a triangle is related to the ratio of the area of the triangle to the surface area of the sphere, whereas Euclidean triangles have the same sum of angles regardless of size.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that Euclidean geometry is distinguished from other geometries by the parallel postulate, the statement that for any line and any point not on the line, there exists a unique line parallel to the original and through the point.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is, &quot;How can I go from this parallel postulate and other relevant features of geometry to understanding why Euclidean geometry has no innate length scale involved?&quot;&lt;/p&gt;&#xA;" OwnerUserId="3305" LastActivityDate="2011-05-31T23:24:23.423" Title="Why is Euclidean geometry scale-invariant?" Tags="&lt;euclidean-geometry&gt;" AnswerCount="3" CommentCount="7" FavoriteCount="8" />
  <row Id="23130" PostTypeId="2" ParentId="23123" CreationDate="2011-02-21T20:07:00.190" Score="1" ViewCount="" Body="&lt;p&gt;So $\textbf{r}(t) = \left&amp;lt; \cos t, \sin t, t \right&amp;gt;$. Then $\textbf{r}&amp;#39;(t) = \left&amp;lt;-\sin t, \cos t, 1 \right&amp;gt;$. So $t = -4 \pi/6$. So $\textbf{r&amp;#39;}(-\frac{4 \pi}{6}) = \left&amp;lt;-\sin( -\frac{4 \pi}{6}), \cos \left( -\frac{4 \pi}{6} \right), 1 \right&amp;gt;$. So the equation of the tangent line would be &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x = -\frac{4 \pi}{6}+t\left(-\sin\left( -\frac{4 \pi}{6}\right)\right)$$ &#xA;$$y = -\frac{4 \pi}{6}+t\left(\cos \left(-\frac{4 \pi}{6} \right) \right)$$ &#xA;and $$z = -\frac{4 \pi}{6}+t$$&lt;/p&gt;&#xA;" OwnerUserId="2548" LastEditorUserId="2548" LastEditorDisplayName="" LastEditDate="2011-02-21T20:18:28.430" LastActivityDate="2011-02-21T20:18:28.430" CommentCount="2" />
  <row Id="23131" PostTypeId="2" ParentId="23123" CreationDate="2011-02-21T20:08:18.537" Score="0" ViewCount="" Body="&lt;p&gt;I'm pretty sure that the answer to which you linked on Yahoo! Answers is just wrong.  The parametric equations $x=\cos t$, $y=\sin t$, $z=t$ describe a spiral on a cylinder of radius 1 coaxial with the $z$-axis.  Since there is only 1 parameter, these parametric equations cannot describe a 2-dimensional surface.  The method used in your second link seems appropriate—the direction vector of the tangent line at any point on $\langle x(t),y(t),z(t)\rangle=\langle\cos t,\sin t,t\rangle$ is $\langle x&amp;#39;(t),y&amp;#39;(t),z&amp;#39;(t)\rangle=\cdots$ (no partial derivatives needed) and you know a point on the line, so you can write a parametric equation for the tangent line.&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2011-02-21T20:08:18.537" />
  <row Id="23132" PostTypeId="2" ParentId="23125" CreationDate="2011-02-21T20:13:40.230" Score="4" ViewCount="" Body="&lt;p&gt;I'll walk you through a derivation (how to get there) of the derivative of the arcsine function.  The same idea can be applied to the other inverse trigonometric functions. If $y=\arcsin x$, then $\sin y=\sin(\arcsin x)=x$.  $$\sin y=x$$ Take the derivative of both sides with respect to $x$ (remember the chain rule!). $$\cos y\cdot\frac{dy}{dx}=1$$  Isolate $\frac{dy}{dx}$, which is what we're trying to find. $$\frac{dy}{dx}=\frac{1}{\cos y}$$ Since we want $\frac{dy}{dx}$ in terms of $x$, substitute for $y$. $$\frac{dy}{dx}=\frac{1}{\cos (\arcsin x)}$$ $\cos(\arcsin x)=\sqrt{1-x^2}$ (see &lt;a href=&quot;http://math.stackexchange.com/questions/15514/how-can-i-find-the-derivative-of-y-sin-arctan-x-tan-arcsin-x/15517#15517&quot;&gt;this answer of mine&lt;/a&gt; for a technique for simplifying a trig function of an inverse trig function), so $$\frac{dy}{dx}=\frac{1}{\sqrt{1-x^2}}.$$&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2011-02-21T20:13:40.230" />
  <row Id="23133" PostTypeId="2" ParentId="23125" CreationDate="2011-02-21T20:14:01.633" Score="10" ViewCount="" Body="&lt;p&gt;The formulas you need are the derivatives of $\arcsin(u)$, $\arccos(u)$, $\arctan(u)$, $\mathrm{arcsec}(u)$, and presumably $\mathrm{arccot}(u)$ and $\mathrm{arccsc}(u)$. Once you know these, you can apply the Chain Rule.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And how do you find these derivatives? Well, the Inverse Function Theorem is your first friend. If $y = g(x)$ has an inverse, is differentiable at $x=a$, $g(a)=b$, and $g&amp;#39;(a)\neq 0$, then $(g^{-1})&amp;#39;(b) = \frac{1}{g&amp;#39;(a)}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, consider $y=\sin(\theta)$. Since the derivative of $\sin(\theta)$ is $\cos(\theta)$, you have that&#xA;$$\frac{d}{du}\arcsin(u) = \frac{1}{\cos(\arcsin(u))}.$$&#xA;But... what is $\cos(\arcsin(u))$? Suppose $\arcsin(u)=\theta$. That means that $\sin(\theta) = u$, and since $\sin^2(\theta)+\cos^2(\theta)=1$, then $\cos^2(\theta) = 1 - \sin^2(\theta) = 1-u^2$. Therefore, $|\cos(\theta)|=\sqrt{\cos^2\theta} = \sqrt{1-u^2}$; and because in order to talk about the inverse of $\sin \theta$ we must have $-\frac{\pi}{2}\leq \theta\leq \frac{\pi}{2}$, then $\cos\theta\geq 0$, so $|\cos\theta|=\cos\theta$. That is, $\cos\theta = \sqrt{1-u^2}$. So, plugging into the formula for the derivative of $\arcsin(u)$, we have:&#xA;$$\frac{d}{du}\arcsin(u) = \frac{1}{\cos(\arcsin u)} = \frac{1}{\sqrt{1-u^2}}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Performing the same kind of analysis for $\arccos(u)$, we get&#xA;$$\frac{d}{du}\arccos(u) = \frac{1}{-\sin(\arccos u)} = -\frac{1}{\sqrt{1-u^2}}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $\arctan u$, using the fact that $(\tan\theta)&amp;#39; = \sec^2\theta$, we have&#xA;$$\frac{d}{du}\arctan u = \frac{1}{\sec^2(\arctan u)}.$$&#xA;Now, if $\arctan u = \theta$, then $\tan(\theta) = u$. Using the fact that $\tan^2\theta + 1 = \sec^2\theta$, we get that $sec^2(\arctan u) = \sec^2(\theta) = 1 + \tan^2(\theta) = 1+u^2$, so&#xA;$$\frac{d}{du}\arctan u = \frac{1}{\sec^2(\arctan u)} = \frac{1}{1+u^2}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $\mathrm{arccot u}$, the same analysis works, provided you remember that $(\cot\theta)&amp;#39; = -\csc^2\theta$ and that $1 + \cot^2\theta = \csc^2\theta$, so&#xA;$$\frac{d}{du}\mathrm{arccot}(u) = \frac{1}{-\csc^2(\mathrm{arccot}(u))} = -\frac{1}{1+u^2}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With $\mathrm{arcsec}u$, we have $(sec\theta)&amp;#39; = sec\theta\tan\theta$, so&#xA;$$\frac{d}{du}\mathrm{arcsec}(u) = \frac{1}{\sec(\mathrm{arcsec} (u))\tan(\mathrm{arcsec} u)}.$$&#xA;Here, $\sec(\mathrm{arcsec} (u)) = u$; if $\mathrm{arcsec}(u)=\theta$, then $\sec\theta = u$, and from $\tan^2\theta + 1 = \sec^2\theta$, we get $|\tan\theta| = \sqrt{u^2 - 1}$. You get:&#xA;$$\frac{d}{du}\mathrm{arcsec}(u) = \frac{1}{\sec(\mathrm{arcsec}(u))\tan(\mathrm{arcsec}(u))} = \frac{1}{u\sqrt{u^2-1}}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And finally, using the fact that $(\csc\theta)&amp;#39; = -\csc\theta\cot\theta$, you get&#xA;$$\frac{d}{du}\mathrm{arccsc}(u) = \frac{1}{-\csc(\mathrm{arccsc}(u))\cot(\mathrm{arccsc}(u))} = -\frac{1}{u\sqrt{u^2-1}}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you have these formulas, the Chain Rule takes care of the rest. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you have:&#xA;\begin{align*}&#xA;\frac{d}{du}\arcsin(u) &amp;amp;= \frac{1}{\sqrt{1-u^2}}, &amp;amp;\qquad \frac{d}{du}\arccos u &amp;amp;= -\frac{1}{\sqrt{1-u^2}},\\&#xA;\frac{d}{du}\arctan(u) &amp;amp;=\frac{1}{1+u^2}, &amp;amp;\frac{d}{du}\mathrm{arccot}(u) &amp;amp;= -\frac{1}{1+u^2},\\&#xA;\frac{d}{du}\mathrm{arcsec}(u) &amp;amp;=\frac{1}{u\sqrt{u^2-1}}, &amp;amp;\frac{d}{du}\mathrm{arccsc}(u) &amp;amp;= - \frac{1}{u\sqrt{u^2-1}}.&#xA;\end{align*}&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2011-02-21T20:14:01.633" CommentCount="3" />
  <row Id="23134" PostTypeId="2" ParentId="23129" CreationDate="2011-02-21T20:16:10.707" Score="12" ViewCount="" Body="&lt;p&gt;Maybe this reason isn't &quot;deep&quot; enough, but it's because the Euclidean distance formula (the Pythagorean identity) is scale invariant:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$a^2 + b^2 = c^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(Xa)^2 + (Xb)^2 = X^2(a^2+b^2) = (Xc)^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;whereas the distance formulas in spherical geometry:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\cos a \cos b = \cos c$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and hyperbolic geometry:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\cosh a \cosh b = \cosh c$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;lack this property.&lt;/p&gt;&#xA;" OwnerUserId="4519" LastActivityDate="2011-02-21T20:16:10.707" CommentCount="1" />
  <row Id="23135" PostTypeId="2" ParentId="23125" CreationDate="2011-02-21T20:16:49.257" Score="1" ViewCount="" Body="&lt;p&gt;if $y=\arcsin(x)$ then $\sin(y)=x$ and differentiating wrt $x$ we get $cos(y)y&amp;#39;=1$.  So, $y&amp;#39;=\frac{1}{\cos(\arcsin(x))}$.  draw a triangle (with sides 1, $x$, $\sqrt{1-x^2}$) to see that $\cos(\arcsin(x))=\sqrt{1-x^2}$. Hence $\frac{d}{dx}\arcsin(x)=\frac{1}{\sqrt{1-x^2}}$.  This kind of reasoning works for various inverse function (using implicit differentiation).  see any calc textbook under implicit differentiation.&lt;/p&gt;&#xA;" OwnerUserId="6925" LastActivityDate="2011-02-21T20:16:49.257" />
  <row Id="23136" PostTypeId="2" ParentId="23111" CreationDate="2011-02-21T20:25:02.837" Score="5" ViewCount="" Body="&lt;p&gt;This is a variation of the problem of enumerating &lt;a href=&quot;http://en.wikipedia.org/wiki/Self-avoiding_walk&quot; rel=&quot;nofollow&quot; title=&quot;self avoiding walks&quot;&gt;self avoiding walks&lt;/a&gt;. In general there are no closed form expressions for the number of self avoiding walks between two arbitrary points in a grid of arbitrary dimension; although I have not seen the problem considered with the particular constraints in question, I would be surprised if there was a closed form solution in this case. However, reliable approximation methods do exist for such enumeration problems, and such an approximation may be suitable for your purposes.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: Using a rather naive approach, I've worked out the following lower bound for the number of self-avoiding walks on an  $n\times m$  rectangular grid from a given boundary point to a given interior point. With more mathematical sophistication, it is likely that significant improvements could be made on this bound.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let  $A$ be an $n\times m$ rectangular grid, with $n&amp;gt;2$ and $m&amp;gt;2$. That is, let $A$ be the collection of integer points in an $xy$ cartesian plane satisfying the conditions $0 \leq x \leq n$ and $0 \leq y \leq m$. Let $\left(a,b\right)$ be a point in the interior of $A$. In the wikipedia article it discusses the case of walks from one diagonal to another with moves only in the positive direction. I shall call such walks positive walks. There is a simple formula for the number of positive walks from one point to another in a rectangular grid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We claim that the number of self-avoiding walks from a given point on the boundary to $\left(a,b\right)$ is bounded below, independently of the choice of the point on the boundary by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \binom{a+b}{b} + \binom{a+m-b}{m-b} + \binom{n-a+b}{b} + \binom{n-a+m-b}{m-b}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is, by the number of positive walks from any of the four corners of the grid to the point $\left(a,b\right)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $\alpha$ be a point on the boundary of $A$. We first bound the number of self-avoiding walks from $\alpha$ to $\left(a,b\right)$ below by the number of positive walks from any point on the boundary of an $\left(n-2\right)\times\left(m-2\right)$ rectangular grid $B$ to the point $\left(a-1,b-1\right)$ within $B$. Our requirement that $n&amp;gt;2$ and $m&amp;gt;2$ is to allow this step to work with a minimum of headache. There is probably an exact solution in the case where $n \leq 2$ or $m \leq 2$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We may consider those walks which first move $k$ steps in the counter-clockwise direction around the boundary, then move one step into the interior, and whose successive moves follow a positive walk directly to $\left(a,b\right)$. We may do this for $k$ from 0 to $2n+2m-1$, excluding those values of $k$ which place the walker upon a corner, where there is no move into the interior. It is clear that we have produced a disjoint collection of self-avoiding walks equal in number to the number of positive walks from the boundary of an $\left(n-2\right)\times\left(m-2\right)$ $B$ to the point $\left(a-1,b-1\right)$ in $B$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we use the result that the number of positive walks from one diagonal to another on an $j\times k$ grid equals&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\binom{j+k}{j} = \binom{j+k}{j,k} = \binom{j+k}{k} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using this result, we may express the number of positive walks from any point on the boundary of $B$ to the point $\left(a-1,b-1\right)$ as the sum&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{i=0}^{b-1}\binom{a-1+i}{a-1} + \sum_{i=0}^{m-b-1}\binom{a-1+i}{a-1} + \sum_{i=0}^{b-1}\binom{n-a-1+i}{n-a-1} +  $$&#xA;$$\sum_{i=0}^{m-b-1}\binom{n-a-1+i}{n-a-1} + \sum_{i=0}^{a-1}\binom{b-1+i}{b-1} + \sum_{i=0}^{n-a-1}\binom{b-1+i}{b-1} + $$&#xA;$$ \sum_{i=0}^{a-1}\binom{m-b-1+i}{m-b-1} + \sum_{i=0}^{n-a-1}\binom{m-b-1+i}{m-b-1} - 4$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where we subtract 4 because the above expression involving binomial coefficients counts each straight line path from the boundary of $B$ to $\left(a-1,b-1\right)$ twice. Since we have specified that $n&amp;gt;2$ and $m&amp;gt;2$, we may safely ignore this 4 by adding in some walks that first move $k$ steps in the clock-wise direction, move away from the boundary, and then follow a positive walk to the finish point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the well known binomial coefficient identity&#xA;$$ \sum_{i=0}^{k}\binom{r+i}{r} = \binom{r+k+1}{r+1}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We may express the above sum, without the addition of the constant $-4$, as&#xA;$$\binom{a+b-1}{a}+\binom{a-1+m-b}{a}+\binom{n-a-1+b}{n-a}+\binom{n-a-1+m-b}{n-a} + $$&#xA;$$\binom{b-1+a}{b}+\binom{b-1+n-a}{b}+\binom{m-b-1+a}{m-b}+\binom{m-b-1+n-a}{m-b}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After some simple manipulations, we may pair up the terms and combine each pair using the identity $\binom{n}{r}=\binom{n-1}{r}+\binom{n-1}{r-1}$ to get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \binom{a+b}{b} + \binom{a+m-b}{m-b} + \binom{n-a+b}{b} + \binom{n-a+m-b}{m-b}$$&#xA;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The result we were attempting to prove. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure how useful this result is, but I thought it was interesting how easily things lined up to be simplified by standard binomial coefficient identities. I also thought it was interesting how the final formula has a straightforward combinatorial interpretation.&lt;/p&gt;&#xA;" OwnerUserId="7323" LastEditorUserId="7323" LastEditorDisplayName="" LastEditDate="2011-02-22T20:52:48.690" LastActivityDate="2011-02-22T20:52:48.690" CommentCount="1" />
  <row Id="23137" PostTypeId="2" ParentId="23056" CreationDate="2011-02-21T20:25:26.003" Score="1" ViewCount="" Body="&lt;p&gt;For the two equations you have, expand the squares on the left sides.  You'll have $s^2$ and $r^2$ on the left side in both equations.  Subtracting the two equations gives an equation that is linear in $s$ and $r$.  Solve the linear equation for one variable and substitute into one of your original two equations to solve for the other variable.  I believe that there are two solutions (by symmetry over the line through the known center and the known circumference point).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&lt;strong&gt;edit&lt;/em&gt;&lt;/strong&gt;  Let me redefine your variables, then go through some of the algebra.  I'll continue to call the known center $(h,k)$ and the radius of that circle $r_1$ and the radius of the other circle $r_2$; I'll call the known point on the circumference of the other circle $(a,b)$, the desired point of tangency $(c,d)$, and the unknown center of the other circle $(m,n)$.  The system of equations you had, written in my variables, is:&#xA;$$\begin{align}&#xD;&#xA;(a-m)^2+(b-n)^2&amp;amp;=r_2^2&#xD;&#xA;\\&#xD;&#xA;(m-h)^2+(n-k)^2&amp;amp;=(r_1+r_2)^2&#xD;&#xA;\end{align}$$&#xA;We want to solve for $(m,n)$.  Expand the squares (use $(x+y)^2=x^2+2xy+y^2$) on the left sides of both equations:&#xA;$$\begin{align}&#xD;&#xA;a^2-2am+m^2+b^2-2bn+n^2&amp;amp;=r_2^2&#xD;&#xA;\\&#xD;&#xA;m^2-2mh+h^2+n^2-2nk+k^2&amp;amp;=(r_1+r_2)^2&#xD;&#xA;\end{align}$$&#xA;Subtract the two equations (I'm subtracting the first from the second):&#xA;$$(m^2-2mh+h^2+n^2-2nk+k^2)-(a^2-2am+m^2+b^2-2bn+n^2)=(r_1+r_2)^2-r_2^2$$&#xA;$$m(2a-2h)+n(2b-2k)+(h^2+k^2-a^2-b^2)=r_1^2+2r_1r_2$$&#xA;I've grouped the terms on the left side into $m$ times some constant, $n$ times some constant, and a constant term (only $m$ and $n$ are variables here); I'll continue to group terms in this way.  Solve for one of the variables ($m$):&#xA;$$\begin{align}&#xD;&#xA;m(2a-2h)&#xD;&#xA;&amp;amp;=r_1^2+2r_1r_2-n(2b-2k)-(h^2+k^2-a^2-b^2)&#xD;&#xA;\\&#xD;&#xA;&amp;amp;=(r_1^2+2r_1r_2-h^2-k^2+a^2+b^2)-n(2b-2k)&#xD;&#xA;\end{align}$$&#xA;$$m=\frac{(r_1^2+2r_1r_2-h^2-k^2+a^2+b^2)-n(2b-2k)}{2a-2h}$$&#xA;Now, substitute this expression for $m$ in one of the two original equations (I'll use the first):&#xA;$$(a-\frac{(r_1^2+2r_1r_2-h^2-k^2+a^2+b^2)-n(2b-2k)}{2a-2h})^2+(b-n)^2=r_2^2$$&#xA;This is a (very messy) quadratic equation in $n$ (it can be written in the form $()n^2+()n+()=0$; $n$ is the only variable; the rest are constants).  While it's possible to continue to solve by hand, the symbolic manipulation is messy (though I can fill in more detail if necessary).  The two solutions for $n$ (note the $\pm$) that result are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$n=\scriptstyle\frac{a^2 (b+k)-2 a h (b+k)+h^2 (b+k)+(b-k) \left(b^2-k^2+r_1^2+2 r_1r_2\right)\pm\sqrt{-(a-h)^2 \left((a-h)^2+(b-k)^2-r_1^2\right) \left(a^2-2 a h+(b-k)^2+(h-r_1-2 r_2) (h+r_1+2 r_2)\right)}}{2 \left((a-h)^2+(b-k)^2\right)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The next step would be to take this, put it back into the expression for $m$ to get two values of $m$.  Here are the results I get using Mathematica:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(m,n)=$&#xA;$\begin{matrix}\scriptstyle{\left(&#xD;&#xA;  \frac{1}{2 (a-h) \left((a-h)^2+(b-k)^2\right)}&#xD;&#xA;  \left(\scriptstyle{\begin{align}&#xD;&#xA;    a^4&amp;amp;-2 a^3 h+2 a h^3-h^2 \left(h^2+(b-k)^2\right)+a^2 (b-k)^2+(a-h)^2 r_1 \left(r_1+2 r_2\right)\\&#xD;&#xA;    &amp;amp;+b \sqrt{-(a-h)^2 \left((a-h)^2+(b-k)^2-r_1^2\right) \left((a-h)^2+(b-k)^2-\left(r_1+2 r_2\right){}^2\right)}\\&#xD;&#xA;    &amp;amp;-k \sqrt{-(a-h)^2 \left((a-h)^2+(b-k)^2-r_1^2\right) \left((a-h)^2+(b-k)^2-\left(r_1+2 r_2\right){}^2\right)}&#xD;&#xA;  \end{align}}\right)\right.},\\ \quad\quad\scriptstyle{\left.&#xD;&#xA;  \frac{\left((a-h)^2+(b-k)^2\right) (b+k)+(b-k) r_1 \left(r_1+2 r_2\right)-\sqrt{-(a-h)^2 \left((a-h)^2+(b-k)^2-r_1^2\right) \left((a-h)^2+(b-k)^2-\left(r_1+2 r_2\right){}^2\right)}}{2 \left((a-h)^2+(b-k)^2\right)}\right)}\end{matrix}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or $(m,n)=$&#xA;$\begin{matrix}\scriptstyle{\left(&#xD;&#xA;  \frac{1}{2 (a-h) \left((a-h)^2+(b-k)^2\right)}&#xD;&#xA;  \left(\scriptstyle{\begin{align}&#xD;&#xA;    a^4&amp;amp;-2 a^3 h+2 a h^3-h^2 \left(h^2+(b-k)^2\right)+a^2 (b-k)^2+(a-h)^2 r_1 \left(r_1+2 r_2\right)\\&#xD;&#xA;    &amp;amp;-b \sqrt{-(a-h)^2 \left((a-h)^2+(b-k)^2-r_1^2\right) \left((a-h)^2+(b-k)^2-\left(r_1+2 r_2\right){}^2\right)}\\&#xD;&#xA;    &amp;amp;+k \sqrt{-(a-h)^2 \left((a-h)^2+(b-k)^2-r_1^2\right) \left((a-h)^2+(b-k)^2-\left(r_1+2 r_2\right){}^2\right)}&#xD;&#xA;  \end{align}}\right)\right.},\\ \quad\quad\scriptstyle{\left.&#xD;&#xA;  \frac{\left((a-h)^2+(b-k)^2\right) (b+k)+(b-k) r_1 \left(r_1+2 r_2\right)+\sqrt{-(a-h)^2 \left((a-h)^2+(b-k)^2-r_1^2\right) \left((a-h)^2+(b-k)^2-\left(r_1+2 r_2\right){}^2\right)}}{2 \left((a-h)^2+(b-k)^2\right)}\right)}\end{matrix}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since we know the two circles are tangent at the point $(c,d)$, the point $(c,d)$ is on the line connecting the centers of the circles and $\frac{r_1}{r_1+r_2}$ of the way from $(h,k)$ to $(m,n)$, so $$(c,d)=\left(\frac{r_2}{r_1+r_2}h+\frac{r_1}{r_1+r_2}m,\frac{r_2}{r_1+r_2}k+\frac{r_1}{r_1+r_2}n\right).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mathematica isn't generating any nice simplification of that expression with $m$ and $n$ substituted in, so I'll stop here.&lt;/p&gt;&#xA;" OwnerUserId="72" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2011-02-22T19:02:50.887" LastActivityDate="2011-02-22T19:02:50.887" CommentCount="2" />
  <row Id="23138" PostTypeId="2" ParentId="23114" CreationDate="2011-02-21T20:32:59.750" Score="3" ViewCount="" Body="&lt;p&gt;To &quot;questions about force and torque&quot; I would add the purely geometrical field of ${\it kinematics}$ which deals with the movement of three-dimensional bodies in three-space, and there are applications in elementary three-dimensional geometry, as determining the distance between two lines in three-space. Then there is a simple rule for solving two homogeneous linear  equations in three unknowns: The general solution of the system $$a_1 x_1+a_2 x_2+a_3 x_3=0, \quad b_1 x_1+b_2 x_2+b_3 x_3=0$$&#xA;is given by $\ {\bf x}=\lambda\ ({\bf a}\times {\bf b})$, $\ \lambda\in{\mathbb R}$. But otherwise no application of the cross product comes to mind, the reason being that this concept is strictly confined to euclidean spaces of dimension three. Only for $d=3$ a skew bilinear form can be represented by a vector, i.e. an element of the base space. &lt;/p&gt;&#xA;" OwnerUserId="1303" LastActivityDate="2011-02-21T20:32:59.750" CommunityOwnedDate="2011-02-22T10:23:12.073" />
  <row Id="23139" PostTypeId="1" AcceptedAnswerId="23143" CreationDate="2011-02-21T20:39:30.337" Score="4" ViewCount="154" Body="&lt;p&gt;Given $\left\{ a_{i}\right\} _{i=0}^{n}\subset\mathbb{R}$ which are&#xA;distinct, show that $\left\{ e^{a_{i}t}\right\} \subset C^{0}\left(\mathbb{R},\mathbb{R}\right)$,&#xA; form a linearly independent set of functions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any tips on how to go about this proof. I tried a working from the&#xA;definition of an exponential and combining sums but that didn't seem&#xA;to get me anywhere. I saw a tip on the internet that said write it&#xA;in the form&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\mu_{1}e^{a_{1}t}+\dots+\mu_{n}e^{a_{n}t}=0$ to try to show $\mu_{1}=\dots=\mu_{n}=0$&#xA;considering each term of the left hand side must be positive, but&#xA;I can't get my head around that because while I understand $e^{x}&amp;gt;0\forall x\in\mathbb{R}$&#xA;I cannot see why $\mu_{i}$ must be positive in any case. I have thought&#xA;about differentiating but that doesn't seem to help. The question&#xA;did originally ask for a &quot;rigourous&quot; proof but I'll take any hints right now and the provided the solution of 'is obvious' is most unhelpful to me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any input would be fantastic. Thank you.&lt;/p&gt;&#xA;" OwnerUserId="7333" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-21T20:47:32.963" LastActivityDate="2011-02-22T00:38:31.830" Title="Proof of linear independence of $e^{at}$" Tags="&lt;functions&gt;&lt;linear-algebra&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="23140" PostTypeId="2" ParentId="23129" CreationDate="2011-02-21T20:39:50.757" Score="11" ViewCount="" Body="&lt;p&gt;Similarity &lt;em&gt;is&lt;/em&gt; a special feature of Euclidean geometry. It follows from the Euclidean parallel postulate:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;For each line and each point not on&#xA;  the line, there is a unique second&#xA;  line, parallel to the first, that&#xA;  contains the point.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As a matter of fact, the following claim, called Wallis's postulate, is equivalent to the parallel postulate:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Given any triangle $\triangle ABC$ and any scale factor, r, there exists&#xA;  a triangle $\triangle DEF$ similar to $\triangle ABC$ &#xA;  where r is the ratio of the lengths of&#xA;  corresponding sides.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There is a list of 15 claims all logically equivalent to the Euclidean parallel postulate on the &lt;a href=&quot;http://en.wikipedia.org/wiki/Parallel_postulate&quot; rel=&quot;nofollow&quot;&gt;Wikipedia Parallel Postulate article&lt;/a&gt;. One of which, not surprisingly, is the Pythagorean theorem. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Perhaps the following will provide some insight. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;em&gt;AAA Similarity Theorem:&lt;/em&gt; If two triangles have all three pairs of angles congruent, then the triangles are similar.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;the proof of which does not require the parallel postulate but is rather involved. Nonetheless, with it, we can show:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;em&gt;Similar Triangle Construction Theorem:&lt;/em&gt; If $\triangle ABC$ is triangle and $\overline{DE}$ any segment, then there is a point $F$ such that $\triangle ABC$ is similar to $\triangle DEF$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Proof:&lt;/em&gt; We can construct a unique ray $\overrightarrow{DP}$ so that $\angle EDP \cong \angle A$. We can similarly construct a unique ray $\overrightarrow{EQ}$ on the same side of $\overleftrightarrow{DE}$ so that $\angle DEQ \cong \angle B$. Since the angle measures of the constructed angles, $\angle EDP$ and $\angle DEQ$, sum to less than $180$, by Euclid's fifth postulate (equivalent to Euclidean parallel postulate), the two rays must intersect at some point $F$ forming a triangle $\triangle DEF$.  By the angle-sum theorem (also logically equivalent to the Euclidean parallel postulate), $\angle C \cong \angle F$. Now by the AAA similarity theorem, the two triangles are similar. $\square$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/tAU3b.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then given any scale factor $r$, We can construct a segment $\overline {DE}$ so that&#xA;$$r=\frac{DE}{AB}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From this, Wallis's postulate follows straight forwardly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conversely, let us assume Wallis's postulate and from it derive the Euclidean parallel postulate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Proof of EPP from Wallis's Postulate:&lt;/em&gt; Suppose we have a line $\ell$ and a point $P$ not on the line. We can drop a perpendicular from $P$ to line $\ell$ and let the perpendicular intersect $\ell$ at point $F$. Then construct a new line $m$ at $P$ at $90$ degrees to our perpendicular. This new line will be parallel $\ell$. If it were not, $m$ and $\ell$ would intersect forming a triangle which would violate the exterior angle theorem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It remains to show the line $m$ is unique. Suppose there were a different line $n$ also through $P$. We must show $n$ is not parallel, that is, line $n$ intersects $\ell$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $n$ is our perpendicular $\overleftrightarrow{PF}$ obviously it intersects $\ell$. So if $n$ is not $\overleftrightarrow{PF}$, lets choose a point $Q$ of $n$ between lines $\ell$ and $m$ but not on $\overleftrightarrow{PF}$. Let $R$ be the point of intersection of a perpendicular dropped from $Q$ onto line $\overleftrightarrow{PF}$. Let $r=\frac{PF}{PR}$. Wallis's theorem assures us that there is somewhere a triangle $\triangle ABC$ that is similar to $\triangle PQR$ with scale factor $r$. We can construct a new triangle $\triangle PSF$ on the segment $\overline {PF}$ congruent to $\triangle ABC$ by placing $S$ on $\ell$ at distance $BC$ from $F$ and on the same side of $\overleftrightarrow{PF}$ as $Q$. This triangle is congruent to $\triangle ABC$ by the side-angle-side congruence theorem and so is similar to $\triangle PQR$. So the angles $\angle FPS$ and $\angle RPQ$ are the same angle. Hence the lines $n$, $\overleftrightarrow{PQ}$, and $\overleftrightarrow{PS}$ are the same line, and $n$ does indeed intersect $\ell$. $\square$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/40OOW.png&quot; alt=&quot;Wallis to EPP diagram&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="6272" LastEditorUserId="6272" LastEditorDisplayName="" LastEditDate="2011-02-22T00:54:50.477" LastActivityDate="2011-02-22T00:54:50.477" CommentCount="7" />
  <row Id="23141" PostTypeId="2" ParentId="23088" CreationDate="2011-02-21T20:47:01.413" Score="0" ViewCount="" Body="&lt;p&gt;There are various &quot;edutainment&quot; video games (e.g. JumpStart, I think it was; see &lt;a href=&quot;http://en.wikipedia.org/wiki/Educational_game#Video_games&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;) that I played in elementary school, but I don't really know whether they had any positive effect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Teaching him/her a programming language such as Python, if you think they can handle it, is probably a good way to go. For a 10 year old, there are also things like &lt;a href=&quot;http://scratch.mit.edu/&quot; rel=&quot;nofollow&quot;&gt;Scratch&lt;/a&gt; and &lt;a href=&quot;http://mindstorms.lego.com/en-us/Default.aspx&quot; rel=&quot;nofollow&quot;&gt;Lego Mindstorms&lt;/a&gt; to give an introduction to programming.&lt;/p&gt;&#xA;" OwnerUserId="2954" LastActivityDate="2011-02-21T20:47:01.413" CommunityOwnedDate="2011-02-21T20:47:01.413" />
  <row Id="23142" PostTypeId="1" AcceptedAnswerId="23196" CreationDate="2011-02-21T20:47:19.313" Score="3" ViewCount="75" Body="&lt;p&gt;My solution concerning a problem about Ergodic Recurrence requires me to prove that $\|P_T 1_B\| &amp;gt; 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $P_T$ is the projection onto the space $I := \{f \in L^2 : f \circ T = f\}$, $T$ is a measure preserving mapping (so for all $B$ measurable $\mu(T^{-1} B) = \mu(B)$) and $B$ is of positive measure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can someone hint my why $\|P_T 1_B\|$ should be strictly positive? If $1_B$ would be in $I$ then I would have that $T^{-1} B = B$ which is probably not the case.&lt;/p&gt;&#xA;" OwnerUserId="1120" LastActivityDate="2011-02-22T06:14:15.237" Title="Ergodic Recurrence." Tags="&lt;functional-analysis&gt;&lt;ergodic-theory&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="1" />
  <row Id="23143" PostTypeId="2" ParentId="23139" CreationDate="2011-02-21T20:50:50.660" Score="4" ViewCount="" Body="&lt;p&gt;If you put $t=0$, the sum of the $\mu_i$ is zero, so at least one of them is negative (unless they are all zero).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Intuitively, assume $a_n$ is greater than all of the other $a$'s.  Then if $t$ gets very large, that term will dominate over all the others, so $\mu_n$ must be zero.  Then argue the same about the next largest.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-02-21T20:50:50.660" CommentCount="2" />
  <row Id="23145" PostTypeId="1" AcceptedAnswerId="23148" CreationDate="2011-02-21T21:17:50.883" Score="3" ViewCount="81" Body="&lt;p&gt;I've found in my solution book multiple references to this &quot;dimension theorem&quot; but have been unlucky looking for a more extensive description of this. When does this hold? Are there any other hypothesis needed?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="6427" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-21T21:18:57.153" LastActivityDate="2011-02-21T21:32:14.297" Title="For a linear transformation $f \colon V \to W$, $\dim \mathrm{Ker}\, f + \dim \mathrm{Im}\, f = \dim V$" Tags="&lt;linear-algebra&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="23146" PostTypeId="1" CreationDate="2011-02-21T21:21:15.293" Score="0" ViewCount="92" Body="&lt;p&gt;With apologies for the lack of mathematical sophistication, I would like an answer to the following problem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are six bins from which I can draw &lt;strong&gt;3&lt;/strong&gt;, &lt;strong&gt;3&lt;/strong&gt;, &lt;strong&gt;5&lt;/strong&gt;, &lt;strong&gt;3&lt;/strong&gt;, &lt;strong&gt;11&lt;/strong&gt;, and &lt;strong&gt;7&lt;/strong&gt; objects respectively giving me a total of &lt;strong&gt;10395&lt;/strong&gt; combinations. Is there a standard analytic method or algorithm for making multiple drawings from these bins that would be less than or equal to the original combination in cost assuming a unit value of &lt;strong&gt;$100&lt;/strong&gt; per object?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Assume an approximately equivalent toy staffing problem, using a stratified random selection process, I have chosen staff members from six departments (for example, three from finance (a), three from marketing (b), and so on) to cover the work week. Each contract employee costs me 100 per use and I need a minimum of six workers to meet customer demand. Union regulations require that I distribute the workload across as many employees as possible. In their ideal world, I would pay out $3,200 = (100 \times (3 + 3 + 5 + 3 + 11 + 7))$ by employing the full quota for the week. However, I would like to minimize my total cost while honoring the union obligation. &#xA;In other words, I want to give an equal opportunity to every employee within each department by varying the numbers across the weekdays while minimizing the total cost. &lt;/p&gt;&#xA;" OwnerUserId="7335" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2011-02-22T05:09:16.610" LastActivityDate="2011-05-23T06:00:47.693" Title="Multiple Subset Combination Equivalent to Original Single Combination" Tags="&lt;combinatorics&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="23147" PostTypeId="2" ParentId="23145" CreationDate="2011-02-21T21:24:23.957" Score="0" ViewCount="" Body="&lt;p&gt;No, there is no other hypothesis needed. In particular, the dimension of $V$ may be infinite.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See &lt;a href=&quot;http://en.wikipedia.org/wiki/Rank-nullity_theorem&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Rank-nullity_theorem&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="367" LastActivityDate="2011-02-21T21:24:23.957" />
  <row Id="23148" PostTypeId="2" ParentId="23145" CreationDate="2011-02-21T21:26:36.270" Score="7" ViewCount="" Body="&lt;p&gt;The Dimension Theorem is often called the &quot;Rank Nullity Theorem&quot;, so you may have more luck finding it under that name.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The result holds for &lt;strong&gt;any&lt;/strong&gt; linear transformation and vector spaces; it's not even restricted to finite dimensional spaces, so long as your sum is taken to be a sum of cardinals and you assume the Axiom of Choice (so that all vector spaces have bases). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Dimension Theorem, aka Rank Nullity Theorem, states:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Dimension Theorem.&lt;/strong&gt; &lt;em&gt;If&lt;/em&gt; $\mathbf{V}$ &lt;em&gt;and&lt;/em&gt; $\mathbf{W}$ &lt;em&gt;are&lt;/em&gt; &lt;strong&gt;any&lt;/strong&gt; &lt;em&gt;vector spaces (over the same field), and&lt;/em&gt; $T$ &lt;em&gt;is&lt;/em&gt; &lt;strong&gt;any&lt;/strong&gt; &lt;em&gt;linear transformation from&lt;/em&gt; $\mathbf{V}$ &lt;em&gt;to&lt;/em&gt; $\mathbf{W}$, &lt;em&gt;then&lt;/em&gt; &#xA;$\dim(\mathrm{ker} T) + \dim(\mathrm{Im} T) = \dim(\mathbf{V})$&#xA;&lt;em&gt;where the sum is a sum of cardinals.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The idea is to start with a basis $\beta$ for $\mathrm{ker} T$, and then extend it to a basis $\beta\cup\gamma$ for all of $\mathbf{V}$. Then you show that the image of $\gamma$ under $T$ is linearly independent: if $\mathbf{v}_1,\ldots,\mathbf{v}_n\in\gamma$ are such that &#xA;$$\alpha_1T(\mathbf{v}_1)+\cdots\alpha_nT(\mathbf{v}_n) = \mathbf{0}$$&#xA;in $\mathbf{W}$, then you can rewrite this as&#xA;$$T(\alpha_1\mathbf{v}_1+\cdots+\alpha_n\mathbf{v}_n) = \mathbf{0}$$&#xA;hence $\alpha_1\mathbf{v}_1+\cdots+\alpha_n\mathbf{v}_n\in\mathrm{ker}(T)$. That means that this linear combination of vectors of $\gamma$ lies in the span of $\beta$, but since $\beta\cup\gamma$ is linearly independent, the only way this can happen is if $\alpha_1\mathbf{v}_1+\cdots+\alpha_n\mathbf{v}_n=\mathbf{0}$; this is a linear combination of vectors in $\gamma$, which is linearly independent, so $\alpha_1=\cdots=\alpha_n=0$. This proves that&#xA;$$T(\gamma) = \{T(\mathbf{v})\mid \mathbf{v}\in\gamma\}$$&#xA;is a linearly independent subset of $\mathbf{W}$. It is now easy to show that $\mathrm{Im}(T) = \mathrm{span}T(\beta\cup\gamma) = \mathrm{span}(T(\gamma))$, so that $\dim(\mathrm{Im}(T)) = |\gamma|$. So you have that&#xA;$$\dim\mathbf{V} = |\beta\cup\gamma| = |\beta|+|\gamma| = \dim(\mathrm{ker}\; T) + \dim(\mathrm{Im}\; T),$$&#xA;as desired.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-21T21:32:14.297" LastActivityDate="2011-02-21T21:32:14.297" />
  <row Id="23149" PostTypeId="2" ParentId="23139" CreationDate="2011-02-21T21:31:07.497" Score="4" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\rm\ e^{a_0\:t} = c_1\ e^{a_1\:t}\:+\:\cdots\:+ c_n\ e^{a_n\:t}\ \Rightarrow\ e^{a_0\:t}\ $ is killed by $\rm\ D-a_0\ $ and by  $\rm\ (D-a_1)\:\cdots\:(D-a_n)\: $.&#xA;But applying the latter to $\rm\ f = e^{a_0\:t}\ $ yields $\rm\ (a_0-a_1)\cdots (a_0-a_n)\ f \ne 0\ $ since $\rm\ a_{\:i}\ne a_{\:0}\:$ for $\rm\: i\ne 0\:$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, they are unequal because their characteristic polynomials share no roots.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2011-02-21T21:38:03.557" LastActivityDate="2011-02-21T21:38:03.557" CommentCount="2" />
  <row Id="23150" PostTypeId="2" ParentId="23146" CreationDate="2011-02-21T21:38:39.327" Score="0" ViewCount="" Body="&lt;p&gt;If there are a,b,c,d,e, and f objects in each bin, there are $\binom{a}{3}\binom{d}{3}\binom{c}{5}\binom{d}{3}\binom{e}{11}\binom{f}{7}$ combinations that you can draw.  Presumably this is what you mean by 10395.  Please clarify the second question-is there a specific cost for drawing from each bin?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second try:  If you draw one object from each bin, there are 10395 combinations.  There are other combinations of numbers to draw from each bin that give 10395 combinations (like 2 from the 11, 1 from each other except the 5; 3 from the 11, 2 from the 7 and 1 from a 3), but all I find involve drawing six total objects.  Is this in the right direction?&lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2011-02-21T23:28:29.283" LastActivityDate="2011-02-21T23:28:29.283" CommentCount="4" />
  <row Id="23151" PostTypeId="1" AcceptedAnswerId="23171" CreationDate="2011-02-21T22:07:55.900" Score="2" ViewCount="137" Body="&lt;p&gt;If $z=f(x,y)$, then total derivative is $\mathrm{d}z=\frac{\partial f}{\partial x}\mathrm{d}x+\frac{\partial f}{\partial y}\mathrm{d}y$. If $\mathrm{d} z=0$, how do you show that $z$ is a constant?&lt;/p&gt;&#xA;" OwnerUserId="7340" LastActivityDate="2011-02-22T01:04:14.253" Title="Question about total derivative" Tags="&lt;multivariable-calculus&gt;" AnswerCount="2" />
  <row Id="23152" PostTypeId="2" ParentId="23151" CreationDate="2011-02-21T22:12:40.370" Score="2" ViewCount="" Body="&lt;p&gt;If $df = 0$, then $\frac{\partial f}{\partial x} \ dx = -\frac{\partial f}{\partial y} \ dy$. I guess one could solve for $f(x,y)$ to get $f(x,y) = g(x-y)$ since $(f_x+f_y)g(x-y) = g&amp;#39;(x-y)+(-1)g&amp;#39;(x-y) = 0$ identically for some $g \in C^1$. &lt;/p&gt;&#xA;" OwnerUserId="2548" LastEditorUserId="2548" LastEditorDisplayName="" LastEditDate="2011-02-21T22:59:27.243" LastActivityDate="2011-02-21T22:59:27.243" CommentCount="1" />
  <row Id="23153" PostTypeId="1" CreationDate="2011-02-21T22:24:03.393" Score="0" ViewCount="101" Body="&lt;p&gt;Here I have a question about time scales in dynamic systems - for reference you can look at a previous question that spurs this one:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://math.stackexchange.com/questions/8575/minimizing-the-cost-of-a-path-in-a-dynamic-system&quot;&gt;Minimizing the cost of a path in a dynamic system&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That question was about finding the minimum cost path from $0$ to $c$ on the real line, where cost is quadratic in the size of the step at each time $t$, and there was a drift towards 0 of $-P(t)$, $P(t)$ being your position at time $t$.  The question is, what is the minimum cost path, when you  minimize over size of step each period, and number of steps, subject only to the constraint of eventually getting to $c$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The solution in this discrete time problem is to see that the problem is essentially linear, with linear constraints, and a solution easy to characterize for a given length of path $T$, and whose cost is decreasing in $T$, so that the minimum cost path is to take a infinite amount of time to get from o to $c$.  Taking very very small steps is the way to go.  Well and good, the question is answered.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;It may then seem natural to ask, &quot;What about the continuous time analog to that problem?&quot;  That is, if the original dynamics are given by &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;P(t) = (1-\gamma) P(t-1) + \gamma x(t) &#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\gamma&amp;lt;1$ limits fast I can move, and which can be written&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;\frac{P(t)- P(t-1)}{\gamma} = - P(t-1) +x(t),&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What if we send $\gamma$ to zero, essentially discretizing the time interval finer and finer?  The problem then becomes to solve the following problem:  Minimize the cost:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;\int _1^T ( x(t))^2 dt,&#xD;&#xA;$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;subject to the constraint that P(0) = 0, P(T) = c, and P goes from 0 to c, and $P$ follows the following dynamics:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;\dot{P} = -P(t) + x(t) &#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The constraint is essentially  that $P$ solves a particular ODE.  I ask mathematica about it, and am informed that $P$ has the following form:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;P(z) = e^{-z} ( K + \int_1^{z} e^t x(t) dt),&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $K$ is a constant of integration.  So, adding in the boundary conditions, the constraint on the minimization problem is that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;e^{-T} \int_1^{T} e^t x(t) dt = c,&#xD;&#xA;$$&#xA;that is, whatever $x(t)$ is, $P$ must end up at $c$ at time $T$.  So the lagrangian is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;L = \int _1^T ( x(t))^2 dt - \lambda (e^{-T} \int_1^{T} e^t x(t) dt - c)&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But note, just as before, this lagrangian is simply quadratic in $x(t)$, and can be solved for a fixed $T$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I ask mathematica to do so, I find that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;x(t) = \frac{e^t c}{e^{t-1} - e},&#xD;&#xA;$$&#xA;Which is very simple, linear in distance to be moved $c$.  Note that this doesn't depend on $T$, the total time length of the path;  indeed, when we minimize cost with respect to $T$, we find that cost in minimized for &#xA;$$&#xD;&#xA;T = \frac{1}{2} (1 - 2 ProductLog[-1, -(1/(2 \sqrt{e}))]),&#xD;&#xA;$$&#xA;where &quot;ProductLog(-1,z)&quot; is mathematica's way of calculating Lambert's W function.  So a finite $T$ is cost minimizing, and it is about 2.25643.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, this isn't necessarily a contradiction to the discrete case; after all, a finite length of time in the continuous model is an infinite number of periods in the discrete model, so the answers are not in conflict.  However, the answer in the continuous model might easily have been &quot;infinite amount of (continuous) time,&quot; but it wasn't.  My questions are &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Why not?&lt;/li&gt;&#xA;&lt;li&gt;How to interpret the value of cost-minimizing $T$ - what does it mean?  How much &quot;time&quot; is that?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The point of all this is to &quot;justify&quot; some simluations of the discrete system we are doing;  we have a problem where we are simulating random shocks to a dynamic system, and we have some theorems that say, mean time to escape of from some point is a function of this cost minimization problem, IE the cost minimizing way to escape is the dominant way to escape, the most likely way to escape.  But we are having a hard time interpreting the cost minimizing escape; it says that in the discrete case, the dominant escape should take an infinite amount of time, while in the continuous case(which the theorems are actually written for), it takes a finite amount of time, a time which seems to have nothing to do with the distance attempting to be escaped!&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Well, obviously no one had anything to say on this issue - is it due to a poorly framed question?  Would this question be more appropriate  at mathoverflow?&lt;/p&gt;&#xA;" OwnerUserId="2464" LastEditorUserId="2464" LastEditorDisplayName="" LastEditDate="2011-03-10T18:07:11.943" LastActivityDate="2011-05-10T20:56:22.867" Title="How to Interpret Time Scales in a Dynamic System" Tags="&lt;differential-equations&gt;&lt;optimization&gt;&lt;dynamical-systems&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="23154" PostTypeId="1" CreationDate="2011-02-21T22:47:06.703" Score="0" ViewCount="99" Body="&lt;p&gt;I have this eq.&#xA;$$&#xD;&#xA;\frac{x}{x-1} + e^{(x-1)/\epsilon} - \epsilon = 0.&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need to derive an asymptotic expansion for any bounded roots $x(\epsilon)$ this equation might have, to find the coefficient of the nth (‘general’) term and show that asymptotic expansion converges for all $\epsilon$ in a neighborhood of zero. Does this asymptotic expansion converge to $x(\epsilon)$ for $\epsilon\neq 0$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some1 please help :( &lt;/p&gt;&#xA;" OwnerUserId="7320" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2011-02-22T00:46:31.913" LastActivityDate="2011-02-22T00:46:31.913" Title="Solving an equation" Tags="&lt;analysis&gt;" AnswerCount="1" />
  <row Id="23155" PostTypeId="2" ParentId="23154" CreationDate="2011-02-21T23:19:37.890" Score="1" ViewCount="" Body="&lt;p&gt;For $\epsilon$ near zero, the term $e^{(x-1)/\epsilon}$ will be extremely small unless $x \gt 1$.  So if we ignore it, we have $\epsilon=1+\frac{1}{x-1}$ or $x=\frac{\epsilon}{\epsilon -1}$, safely less than 1.  Numerically you can iterate this, plugging this $x$ into the exponential term and solving again.  Convergence will be quite rapid.  To prove there is a root, you should be able to find points of opposite signs on either side of this $x$.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2011-02-22T00:06:30.967" LastActivityDate="2011-02-22T00:06:30.967" />
  <row Id="23157" PostTypeId="1" AcceptedAnswerId="23250" CreationDate="2011-02-21T23:36:02.247" Score="2" ViewCount="62" Body="&lt;p&gt;This is a question on how to make sense of an equation (specifically, a description of a distribution) that, no matter how I slice it and dice it, always has some left over pieces that don't fit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The equation in question describes a random variable $G$ drawn from a Dirichlet process $\textrm{DP}(\alpha_0,G_0)$:&#xA;\begin{align}&#xA;G = \sum_{k=1}^\infty \beta_k \delta_{\theta_k}&#xA;\end{align}&#xA;where $\alpha_0 &amp;gt; 0$ and $G_0$ is some base measure. The $\beta_k$ are stick breaking weights that depend on $\alpha_0$ and $\theta_k$ are atoms drawn from $G_0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The full description can be found at the top of page 4 of &lt;a href=&quot;http://ftp.stat.berkeley.edu/tech-reports/653.pdf&quot; rel=&quot;nofollow&quot;&gt;this paper&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My first question is whether I can read $G$ like a function. Say, if I have some &quot;atom&quot; (which term I don't fully understand) $\theta_j$, can I plug it in like:&#xA;\begin{align}&#xA;G(\theta_j) &amp;amp;= \sum_{k=1}^\infty \beta_k \delta_{\theta_k} (\theta_j) \newline&#xA;&amp;amp;= \sum_{k \:\textrm{where} \: \theta_k = \theta_j} \beta_k&#xA;\end{align}&#xA;meaning if I happened to draw an atom that was identical to more than one $\theta_k$ I am just adding up the relevant weights $\beta_k$. And that the random variable $G$ itself can be loosely understood as very long vector of weights $(\ldots,G_j = \sum_{k \:\textrm{where} \: \theta_k = \theta_j} \beta_k,\ldots)$ such that it is just a repartitioning of the infinite dimension vector $(\ldots,\beta_k,\ldots)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if I interpret it this way, I get into trouble because apparently, you're supposed to use the $G$ to draw the so-called &quot;atoms&quot; again. So I revise my understanding to think of $G$ as a multinomial distribution. But then the equation that sums over the infinitely many values of $\beta_k$ doesn't make any sense. A sum is a sum. How do you draw things from a sum, which has just a fixed scalar value?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is my interpretation wrong? Whether it's wrong at the foundations or wrong in subtle ways, please correct me regardless.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm trying to make the leap from the prob/statistics in DeGroot's book to understanding measure theory and nonparametric processes and it's things like these that completely throw me for a loop.&lt;/p&gt;&#xA;" OwnerUserId="6690" LastActivityDate="2011-02-22T18:07:40.740" Title="Notation question with Dirichlet processes" Tags="&lt;probability&gt;&lt;statistics&gt;&lt;probability-distributions&gt;" AnswerCount="1" CommentCount="6" />
  <row Id="23158" PostTypeId="1" AcceptedAnswerId="23160" CreationDate="2011-02-21T23:43:51.017" Score="0" ViewCount="80" Body="&lt;p&gt;Q: Find the volume of the solid of revolution obtained by revolving the region bounded by the line y = 2 and the curve y = sec^2x, -90 &amp;lt; x &amp;lt; 90, around the x-axis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Attempted solution: 1/3 tanx ( sec^2(x) + 2)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm sure I did the integral correctly, but now when I need to sub in for 90, tan(90) does not exist?  (The answer is 2(pi)^2 - 8pi/3)&lt;/p&gt;&#xA;" OwnerUserId="6392" LastActivityDate="2011-02-22T00:04:54.057" Title="How to find the volume of the solid of revolution?" Tags="&lt;calculus&gt;&lt;homework&gt;&lt;integral&gt;" AnswerCount="1" />
  <row Id="23159" PostTypeId="1" AcceptedAnswerId="23162" CreationDate="2011-02-21T23:58:31.273" Score="1" ViewCount="60" Body="&lt;p&gt;Hello. Let $\mathbb{A}_{2}$ denote the 2-affice space over $\mathbb{C}$. Now let $Y=V(y-x^{2})$, here $V$ denotes the zero-locus. Now by definition the coordinate ring is the quotient ring $k[x,y]/I(Y)$ where $I(Y)$ is the ideal generated by all polynomials vanishing on $Y$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the author computes $k[x,y]/I(Y)$, then it computes $k[x,y]/&amp;lt;y-x^{2}&amp;gt;$, so my question is: why is $I(V(y-x^{2}))=&amp;lt;y-x^{2}&amp;gt;$?&lt;/p&gt;&#xA;" OwnerUserId="6495" LastActivityDate="2011-02-22T00:16:52.263" Title="Clarification about zero locus and ideal generated" Tags="&lt;algebraic-geometry&gt;" AnswerCount="2" />
  <row Id="23160" PostTypeId="2" ParentId="23158" CreationDate="2011-02-21T23:58:44.970" Score="1" ViewCount="" Body="&lt;p&gt;Presumably your arguments are degrees, not radians.  The volume element of the solid of revolution formed by the rectangle of width $dx$ and extending from $y_1$ to $y_2$ is $\pi(y_2^2-y_1^2)dx$  You integrated $\sec^4(x)dx$ correctly, but that is not quite what you need to integrate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sec^2(x)$ crosses 2-are you supposed to just consider the part of it below y=2?  This will eliminate the problem of the tangent blowing up.  If you are supposed to consider the area outboard of y=2 all the way to 90 degrees, then the volume is infinite, as your calculation shows.  &lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2011-02-22T00:04:54.057" LastActivityDate="2011-02-22T00:04:54.057" CommentCount="1" />
  <row Id="23161" PostTypeId="1" CreationDate="2011-02-22T00:05:48.967" Score="0" ViewCount="68" Body="&lt;p&gt;Hello,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The BNF is defined as follows:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;&amp;lt;S&amp;gt; -&amp;gt; &amp;lt;S&amp;gt;a&amp;lt;S&amp;gt;a&amp;lt;S&amp;gt; | b&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This is my review question for a quiz, and according to my teacher, this grammar is ambiguous. However, I realized this grammar is not.&lt;br&gt;&#xA;For example, given a sentence:  &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;babababab&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;There are three ways to generate this sentence depending on which &lt;code&gt;&amp;lt;S&amp;gt;&lt;/code&gt; we use, but its parse-tree is unique.&lt;br&gt;&#xA;Am I right in this case? Any feedback would be greatly appreciated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks,&lt;br&gt;&#xA;Chan&lt;/p&gt;&#xA;" OwnerUserId="5711" LastActivityDate="2011-05-19T06:10:39.697" Title="Question about ambiguity of BNF" Tags="&lt;computer-science&gt;" AnswerCount="1" />
  <row Id="23162" PostTypeId="2" ParentId="23159" CreationDate="2011-02-22T00:12:11.317" Score="3" ViewCount="" Body="&lt;p&gt;By the Nullstellensatz, $I(V(J))=\sqrt{J}$ for any ideal $J$. Because $y-x^2$ is an irreducible polynomial in $k[x,y]$, we have that $J=(y-x^2)$ is a prime ideal, and hence a radical ideal (i.e. $J=\sqrt{J}$). Thus $I(V(y-x^2))=(y-x^2)$.&lt;/p&gt;&#xA;" OwnerUserId="264" LastActivityDate="2011-02-22T00:12:11.317" CommentCount="1" />
  <row Id="23163" PostTypeId="2" ParentId="23139" CreationDate="2011-02-22T00:14:05.070" Score="1" ViewCount="" Body="&lt;p&gt;Suppose that for some reason you do not want to differentiate or let $t$ tend to $\infty$.  Evaluating $\mu_1e^{a_1t}+\cdots+\mu_ne^{a_nt}$ at $t=0,1,\ldots,n-1$ yields the system of equations&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\begin{matrix} &#xD;&#xA;\mu_1 &amp;amp; + &amp;amp; \mu_2 &amp;amp; + &amp;amp; \cdots &amp;amp; + &amp;amp; \mu_n &amp;amp;=&amp;amp;0\\&#xD;&#xA;\mu_1e^{a_1} &amp;amp;+&amp;amp; \mu_2e^{a_2} &amp;amp;+&amp;amp; \cdots &amp;amp;+&amp;amp; \mu_ne^{a_n}&amp;amp;=&amp;amp;0\\&#xD;&#xA;\mu_1\left(e^{a_1}\right)^2 &amp;amp;+&amp;amp; \mu_2\left(e^{a_2}\right)^2 &amp;amp;+&amp;amp; \cdots &amp;amp;+&amp;amp; \mu_n\left(e^{a_n}\right)^2&amp;amp;=&amp;amp;0\\&#xD;&#xA; \vdots&amp;amp; &amp;amp;\vdots &amp;amp; &amp;amp;\vdots&amp;amp; &amp;amp;\vdots &amp;amp; &amp;amp;\vdots\\&#xD;&#xA;\mu_1\left(e^{a_1}\right)^{n-1} &amp;amp;+&amp;amp; \mu_2\left(e^{a_2}\right)^{n-1} &amp;amp;+&amp;amp; \cdots &amp;amp;+&amp;amp; \mu_n\left(e^{a_n}\right)^{n-1}&amp;amp;=&amp;amp;0,&#xD;&#xA;\end{matrix}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or in other words, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\begin{bmatrix}&#xD;&#xA;1&amp;amp;1&amp;amp;\cdots&amp;amp;1\\&#xD;&#xA;e^{a_1}&amp;amp;e^{a_2}&amp;amp;\cdots&amp;amp;e^{a_n}\\&#xD;&#xA;\left(e^{a_1}\right)^2&amp;amp;\left(e^{a_2}\right)^2&amp;amp;\cdots&amp;amp;\left(e^{a_n}\right)^2\\&#xD;&#xA;\vdots&amp;amp;\vdots&amp;amp;\ddots&amp;amp;\vdots\\&#xD;&#xA;\left(e^{a_1}\right)^{n-1}&amp;amp;\left(e^{a_2}\right)^{n-1}&amp;amp;\cdots&amp;amp;\left(e^{a_n}\right)^{n-1}&#xD;&#xA;\end{bmatrix}&#xD;&#xA;\begin{bmatrix}&#xD;&#xA;\mu_1\\&#xD;&#xA;\mu_2\\&#xD;&#xA;\vdots\\&#xD;&#xA;\mu_n&#xD;&#xA;\end{bmatrix}&#xD;&#xA;=\vec 0.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because the $e^{a_i}$ are distinct, the &lt;a href=&quot;http://en.wikipedia.org/wiki/Vandermonde_matrix&quot; rel=&quot;nofollow&quot;&gt;Vandermonde matrix&lt;/a&gt; $A$ above is invertible, and therefore the unique solution $\vec x$ to $A\vec x=\vec 0$ is $\vec x=\vec 0$.  Thus all $\mu_i$ are zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similar reasoning would apply if you evaluated at any arithmetic progression of $n$ terms.  Alternatively, evaluating the $0^\text{th}$ through $(n-1)^\text{st}$ derivatives at $0$ (or another point) would give a Vandermonde system with $a_i$ in place of $e^{a_i}$.  The latter approach is finding the &lt;a href=&quot;http://en.wikipedia.org/wiki/Wronskian&quot; rel=&quot;nofollow&quot;&gt;Wronskian&lt;/a&gt;, as referred to in Qiaochu's comment.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2011-02-22T00:38:31.830" LastActivityDate="2011-02-22T00:38:31.830" />
  <row Id="23164" PostTypeId="2" ParentId="23159" CreationDate="2011-02-22T00:16:52.263" Score="4" ViewCount="" Body="&lt;p&gt;You don't need the Nullstellensatz. A polynomial $p(x, y) \in k[x, y]$ vanishes on the set of all points $(t, t^2)$ if and only if $p(t, t^2)$ is the zero polynomial when $k$ is infinite, and one can verify directly that this is true if and only if $p(x, y)$ is divisible by $y - x^2$ by replacing all occurrences of $y$ with $x^2$. In particular you do not need $k$ algebraically closed. &lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2011-02-22T00:16:52.263" />
  <row Id="23165" PostTypeId="2" ParentId="23064" CreationDate="2011-02-22T00:26:34.737" Score="4" ViewCount="" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;The matrix P is the matrix of coefficients of certain divisors of the characteristic polynomial of the Frobenius canonical form.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In fact this works for any companion matrix, not just the companion matrix of a power of an irreducible polynomial.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose a matrix A has 1s above the diagonal, and the last row is [ v0, v1, v2, v3, ..., vn ], so&#xA;$$ A = \begin{bmatrix}&#xD;&#xA;0 &amp;amp; 1 &amp;amp; 0 &amp;amp; 0 &amp;amp; \ldots &amp;amp; 0 &amp;amp; 0 \\&#xD;&#xA;0 &amp;amp; 0 &amp;amp; 1 &amp;amp; 0 &amp;amp; \ldots &amp;amp; 0 &amp;amp; 0 \\&#xD;&#xA;0 &amp;amp; 0 &amp;amp; 0 &amp;amp; 1 &amp;amp; \ldots &amp;amp; 0 &amp;amp; 0 \\&#xD;&#xA;\vdots &amp;amp; \vdots &amp;amp; \vdots &amp;amp; \vdots &amp;amp; \ddots &amp;amp; \vdots \\&#xD;&#xA;0 &amp;amp; 0 &amp;amp; 0 &amp;amp; 0 &amp;amp; \ldots &amp;amp; 1 &amp;amp; 0 \\&#xD;&#xA;0 &amp;amp; 0 &amp;amp; 0 &amp;amp; 0 &amp;amp; \ldots &amp;amp; 0 &amp;amp; 1 \\&#xD;&#xA;v_0 &amp;amp; v_1 &amp;amp; v_2 &amp;amp; v_3 &amp;amp; \ldots &amp;amp; v_{n-1} &amp;amp; v_n \\&#xD;&#xA;\end{bmatrix}.$$  Then&#xA;$$ A^{(n+1)} = v_0 A^0 + v_1 A^1 + v_2 A^2 + \ldots + v_n A^n$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose $x-\alpha$ is a factor of&#xA;$$&#xD;&#xA;f(x) = x^{(n+1)} - v_n x^n - \ldots - v_1 x^1 - v_0 x^0&#xD;&#xA;$$&#xA;and write&#xA;$$&#xD;&#xA;f(x)/(x-\alpha) = w_n x^n + w_{n-1} x^{(n-1)} + \ldots + w_1 x^1 + w_0 x^0&#xD;&#xA;$$&#xA;where of course $w_n = 1$.  Then&#xA;$w = [ w_0, w_1, w_2, \ldots, w_n ]$ is a left eigenvector of A, since $wA = \alpha w$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In more detail, $$wA = [ v_0, w_0 + v_1, w_1 + v_2, \ldots, w_{n-1} + v_n ]$$&#xA;and&#xA;$$f(x) = (x-\alpha)( w_n x^n + w_{n-1} x^{(n-1)} + \ldots + w_1 x^1 + w_0 x^0)&#xD;&#xA;= x^{n+1} - (-\alpha+w_{n-1}) x^{n} \ldots$$&#xA;so by equating coefficients of $x^n$, one gets $v_n = \alpha - w_{n-1}$ and $v_n + w_{n-1} = \alpha w_n$.  Similarly, $v_{n-1} = \alpha w_{n-1} - w_{n-2}$, and in general $v_i = \alpha w_i - w_{i-1}$ so $v_i + w_{i-1} = \alpha w_i$ and $wA = \alpha w$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Something similar works with repeated roots even, to give you Jordan blocks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So to find your matrix P, find each root α of f(x) with multiplicity k, and then the rows of P are the coefficients (in increasing order of power of x) of f(x)/(x−α)&lt;sup&gt;i&lt;/sup&gt;, from i=1 to k, with corresponding diagonal entry of the matrix just being α, and the run from 1 to k forming the Jordan block.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Example 1&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Find the JCF of&#xA;$$A = \begin{bmatrix}0&amp;amp;1&amp;amp;0&amp;amp;0\\0&amp;amp;0&amp;amp;1&amp;amp;0\\0&amp;amp;0&amp;amp;0&amp;amp;1\\-1&amp;amp;4&amp;amp;-6&amp;amp;4\end{bmatrix}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The characteristic polynomial is just $f(x) = x^4-4x^3+6x^2-4x+1$ since it is a companion matrix of $f$.  Of course, $f(x) = (x-1)^4$ is easy to factor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We calculate &#xA;$$f(x)/(x-1) = 1x^3 - 3x^2 + 3x - 1,$$&#xA;$$f(x)/(x-1)^2 = 0x^3+1x^2-2x+1,$$&#xA;$$f(x)/(x-1)^3 = 0x^3+0x^2+x-1,\text{ and}$$&#xA;$$f(x)/(x-1)^4 = 0x^3+0x^2+0x+1.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We then write down the coefficients (in reverse order, I guess):&#xA;$$P = \begin{bmatrix} -1 &amp;amp; 3 &amp;amp; -3 &amp;amp; 1 \\ 1 &amp;amp; -2 &amp;amp; 1 &amp;amp; 0 \\ -1 &amp;amp; 1 &amp;amp; 0 &amp;amp; 0 \\ 1 &amp;amp; 0 &amp;amp; 0 &amp;amp; 0 \end{bmatrix}$$&#xA;so that&#xA;$$ P A P^{-1} = \begin{bmatrix} 1 &amp;amp; 0 &amp;amp; 0 &amp;amp; 0 \\ 1 &amp;amp; 1 &amp;amp; 0 &amp;amp; 0 \\ 0 &amp;amp; 1 &amp;amp; 1 &amp;amp; 0 \\ 0 &amp;amp; 0 &amp;amp; 1 &amp;amp; 1 \end{bmatrix}$$&lt;/p&gt;&#xA;" OwnerUserId="583" LastEditorUserId="583" LastEditorDisplayName="" LastEditDate="2011-02-22T00:43:55.847" LastActivityDate="2011-02-22T00:43:55.847" CommentCount="1" />
  <row Id="23166" PostTypeId="2" ParentId="23146" CreationDate="2011-02-22T00:34:15.160" Score="0" ViewCount="" Body="&lt;p&gt;Assume an approximately equivalent toy staffing problem, using a stratified random selection process, I have chosen staff members from six departments (for example, three from finance (a), three from marketing (b), and so on) to cover the work week. Each contract employee costs me 100 per use and I need a minimum of six workers to meet customer demand. Union regulations require that I distribute the workload across as many employees as possible. In their ideal world, I would pay out 3,200 = &#xA;(100 x (3 + 3 + 5 + 3 + 11 + 7)) by employing the full quota for the week. However, I would like to minimize my total cost while honoring the union obligation. &#xA;In other words, I want to give an equal opportunity to every employee within each department by varying the numbers across the weekdays while minimizing the total cost. &lt;/p&gt;&#xA;" OwnerUserId="7343" LastEditorUserId="7343" LastEditorDisplayName="" LastEditDate="2011-02-22T00:49:54.343" LastActivityDate="2011-02-22T00:49:54.343" CommentCount="2" />
  <row Id="23167" PostTypeId="1" CreationDate="2011-02-22T00:38:23.580" Score="2" ViewCount="38" Body="&lt;p&gt;If $f:X\rightarrow Y$ is a surjective morphism of schemes and $g:X&amp;#39;\rightarrow Y$ is another morphism of schemes, one can show that $p_{2}:X\times_{Y}X&amp;#39;\rightarrow X&amp;#39;$ is also surjective.  The proof that I have seen (in Mumford's Red Book, for instance) goes as follows:  Given $P\in X&amp;#39;$, we can find points $R\in Y$ and $P\in X$ and inclusions $k\left(R\right)\hookrightarrow k\left(P\right)$ and $k\left(R\right)\hookrightarrow k\left(Q\right)$.  The desired result can then be established if we can say that there exists a field $F$ containing each of $k\left(P\right)$ and $k\left(Q\right)$.  However, I don't see how we can conclude that such a field exists.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is easy to just give a different proof without looking at a field extensions of $k\left(P\right)$ and $k\left(Q\right)$, but I wanted to know how the sketch of a proof above can be used.  Thanks!&lt;/p&gt;&#xA;" OwnerUserId="7344" LastActivityDate="2011-02-22T04:04:07.467" Title="&quot;Surjectivity is stable under base change&quot; and field compositums" Tags="&lt;field-theory&gt;&lt;schemes&gt;" AnswerCount="1" />
  <row Id="23168" PostTypeId="1" AcceptedAnswerId="23170" CreationDate="2011-02-22T00:39:41.123" Score="0" ViewCount="53" Body="&lt;p&gt;How to find the real valued sufficient statistic of $\theta$ for a random sample from a distribution with the probability density function written below?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f(x;\theta) = \theta ax^{a−1} \exp(−\theta x^a), x &amp;gt; 0, \theta &amp;gt; 0, a &amp;gt; 0$$&lt;/p&gt;&#xA;" OwnerUserId="6335" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2011-02-22T01:05:44.823" LastActivityDate="2011-02-22T01:05:44.823" Title="Real valued sufficient statistic" Tags="&lt;statistics&gt;&lt;probability-distributions&gt;" AnswerCount="2" />
  <row Id="23169" PostTypeId="2" ParentId="23168" CreationDate="2011-02-22T00:58:17.907" Score="2" ViewCount="" Body="&lt;p&gt;Use the factorization theorem. In particular we have the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Theorem.&lt;/strong&gt; $T$ is a sufficient statistic for $\theta$ if the liklihood factorizes in the following form: $$L(x_1, \dots, x_n| \theta) = g(\theta, T(x_1, \dots, x_n)) \cdot h(x_1, \dots, x_n)$$ for some functions $g$ and $h$. &lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2011-02-22T00:58:17.907" />
  <row Id="23170" PostTypeId="2" ParentId="23168" CreationDate="2011-02-22T01:01:57.480" Score="3" ViewCount="" Body="&lt;p&gt;$$f(x,\theta) = \theta a x^{a-1} \exp(-\theta x^{a}) = a x^{a-1} \exp(-\theta x^a + \log(\theta))$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle \prod_{i=1}^n f(x_i;\theta) = a^n \left( \prod_{i=1}^n x_i^{(a-1)} \right) \exp(-\theta \sum_{i=1}^n x_i^a + n \log(\theta))$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given $a$, define $$h(x_1,x_2,\ldots,x_n) = a^n \left( \prod_{i=1}^n x_i^{(a-1)} \right)$$ $$T(x_1,x_2,\ldots,x_n) = \sum_{i=1}^n x_i^a$$ $$g(\theta, T(x_1,x_2,\ldots,x_n)) = \exp(-\theta T + n \log(\theta))$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, by &lt;a href=&quot;http://en.wikipedia.org/wiki/Sufficient_statistic#Fisher.E2.80.93Neyman_factorization_theorem&quot; rel=&quot;nofollow&quot;&gt;Fisher factorization theorem&lt;/a&gt; $\displaystyle T(x_1,x_2,\ldots,x_n) = \sum_{i=1}^n x_i^a$ is a sufficient statistic.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastActivityDate="2011-02-22T01:01:57.480" CommentCount="1" />
  <row Id="23171" PostTypeId="2" ParentId="23151" CreationDate="2011-02-22T01:04:14.253" Score="3" ViewCount="" Body="&lt;p&gt;Since $dz=0$, $\frac{\partial z}{\partial x} = \frac{\partial z}{\partial y}=0$.&#xA;Hence, on the one hand $z(x,y)=f(y)$ and on the other $z(x,y)=g(x)$ by integrating. &#xA;So $f(y)=g(x)$. Differentiating both sides with respect to $x$, we have $0=\frac{\partial g}{\partial x}$. Therefore $z(x,y)=g(x)=C$.&lt;/p&gt;&#xA;" OwnerUserId="5253" LastActivityDate="2011-02-22T01:04:14.253" CommentCount="2" />
  <row Id="23172" PostTypeId="1" AcceptedAnswerId="23178" CreationDate="2011-02-22T01:18:13.030" Score="7" ViewCount="144" Body="&lt;p&gt;Hello,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Prove that if $p^{a}$ is a factor of the canonical factorization of ${{2n}\choose{n}}$ then $p^{a} &amp;lt; 2n$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My attempt:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$${{2n}\choose{n}} = \frac{(2n)!}{n!n!}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $a_1$ be the highest of power of $(2n)!$&lt;br&gt;&#xA;Let $a_2$ be the highest of power of $n!$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the highest power of $\frac{(2n)!}{n!n!}$ = $a_1 - 2a_2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $a_1 &amp;lt;= 2n - 1$ and $2a_2 &amp;lt;= 2n - 2$&lt;br&gt;&#xA;Therefore the highest power of $p$ that divides $\frac{(2n)!}{n!n!}$ is $2n - 1 - 2n + 2 = 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $a &amp;lt;= 1 \implies p^{a} &amp;lt; 2n$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Am I in the right track? Any idea?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Following Ross Millikan's hint:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $a$ be the highest power of $p$ such that $p^{a}|n!$&#xA;Then, $a$ =  $\lfloor \frac{n}{p} \rfloor + \lfloor \frac{n}{p^2} \rfloor +\lfloor \frac{n}{p^3} \rfloor + \ldots \lfloor \frac{n}{p^k} \rfloor$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $b$ be the highest power of $p$ such that $p^{b}|(2n)!$&#xA;Then, $b$ =  $\lfloor \frac{2n}{p} \rfloor + \lfloor \frac{2n}{p^2} \rfloor +\lfloor \frac{2n}{p^3} \rfloor + \ldots \lfloor \frac{2n}{p^q} \rfloor$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Longrightarrow b - 2a$ is the highest power of $p$ such that $p|\frac{(2n)!}{n!n!}$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $a, b \in N \implies b - 2a &amp;lt; b$&lt;br&gt;&#xA;Besides, $p^{b} &amp;lt; 2n$&lt;br&gt;&#xA;$\therefore p^{b - 2a} &amp;lt; p^{b} &amp;lt; 2n$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Am I in the right track now?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks,&lt;br&gt;&#xA;Chan&lt;/p&gt;&#xA;" OwnerUserId="5711" LastEditorUserId="5711" LastEditorDisplayName="" LastEditDate="2011-02-22T05:38:48.470" LastActivityDate="2011-02-22T05:38:48.470" Title="Prove that if $p^{a}$ is a factor of the canonical factorization of ${{2n}\choose{n}}$ then $p^{a} &lt; 2n$?" Tags="&lt;elementary-number-theory&gt;" AnswerCount="2" CommentCount="7" />
  <row Id="23173" PostTypeId="1" AcceptedAnswerId="23175" CreationDate="2011-02-22T01:37:10.183" Score="2" ViewCount="78" Body="&lt;p&gt;Hi, I want to find the irreducible components of the algebraic set $Y$, in $\mathbb{A}^{3}$ given by the zero-locus of the equations $x^{2}-yz$ and $xz-z$. I also want to compute the dimension of $Y$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well if $xz-z=0$ then $x=1$ or $z=0$. If $z=0$ then $x=0$ so we obtain the $y-axis$.&#xA;Now if $x=1$ then $yz-1=0$ hence:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$Y = V_{1}(x,z) \cup V_{2}(x-1,yz-1)$. (Here $V_{i}$ denotes the locus-set).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now $k[x,y,z]/(x,z) \cong k[y]$ so $V_{1}(x,z)$ is irreducible because $k[y]$ is a integral domain. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm stuck in showing $V_{2}(x-1,yz-1)$ is irreducible. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you please help? (Also in general how do you compute the dimension?)&lt;/p&gt;&#xA;" OwnerUserId="6495" LastEditorUserId="6495" LastEditorDisplayName="" LastEditDate="2011-02-22T02:43:57.197" LastActivityDate="2011-02-22T02:43:57.197" Title="Help with irreducible components" Tags="&lt;algebraic-geometry&gt;" AnswerCount="1" />
  <row Id="23175" PostTypeId="2" ParentId="23173" CreationDate="2011-02-22T02:10:56.927" Score="5" ViewCount="" Body="&lt;p&gt;We have that $k[V_2]=k[x,y,z]/(x-1,yz-1)\cong k[y,z]/(yz-1)$. You can show that the map from $V_2$ to $\mathbb{A}^1-\{0\}$ defined by $(x,y,z)\mapsto y$ is an isomorphism by showing that the corresponding morphism of coordinate rings $k[\mathbb{A}^1-\{0\}]=k[y,y^{-1}]\rightarrow k[y,z]/(yz-1)$ (i.e., the morphism defined by $y\mapsto y$, $y^{-1}\mapsto z$) is an isomorphism. Because $\mathbb{A}^1-\{0\}$ is a non-empty open set of the irreducible set $\mathbb{A}^1$, it is also irreducible, and hence $V_2$ is irreducible as well. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I feel like this may be a bit roundabout, but I think it works.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The dimension of $Y$ is 1, because a maximal-length chain of irreducible subsets of $Y$ will start at either of its irreducible components, each of which is dimension 1. The irreducible components are a line, which is clearly of dimension 1 (also: its coordinate ring is $k[y]$, which is a ring of dimension 1), and a hyperbola $yz-1$, which (as we showed above) is isomorphic to $\mathbb{A}^1-\{0\}$, and by Prop 1.10 in Hartshorne, $\dim(Y)=\dim(\overline{Y})$ for any quasi-affine variety $Y$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I often find it helpful to try to visualize what's going on - here is the output from Mathematica. The two planes are the zero locus of $xz-z$, and the cone is the zero locus of $x^2-yz$ - their intersection is, as expected, the $y$-axis and a hyperbola which lies on the plane $x=1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/siPJo.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="264" LastEditorUserId="264" LastEditorDisplayName="" LastEditDate="2011-02-22T02:43:40.863" LastActivityDate="2011-02-22T02:43:40.863" CommentCount="4" />
  <row Id="23176" PostTypeId="1" CreationDate="2011-02-22T02:32:57.647" Score="4" ViewCount="107" Body="&lt;p&gt;I'm working on an integration by parts problem, and I'm trying to substitute to simplify the equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_\sqrt{\frac{\pi}{2}}^\sqrt{\pi} \theta^3 \cos(\theta^2) d\theta$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the substitution rule for definite integrals, I substitute $\theta^2 = t$ and apply the same to the limits of integration:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_\frac{\pi}{2}^\pi t^\frac{3}{2} \cos(t) dt$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, Wolfram|Alpha tells me that I have done something wrong, as these two integrals are not equivalent. Where did I screw up?&lt;/p&gt;&#xA;" OwnerUserId="451" LastActivityDate="2011-02-22T03:13:17.940" Title="Substitution Rule for Definite Integrals" Tags="&lt;calculus&gt;&lt;integral&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="23177" PostTypeId="2" ParentId="23176" CreationDate="2011-02-22T02:34:27.333" Score="0" ViewCount="" Body="&lt;p&gt;We have $dt = 2 \theta \ d \theta$. So you have to multiply by $\frac{1}{2}$.&lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2011-02-22T02:34:27.333" CommentCount="3" />
  <row Id="23178" PostTypeId="2" ParentId="23172" CreationDate="2011-02-22T02:42:52.660" Score="2" ViewCount="" Body="&lt;p&gt;Hint: The highest power of a prime,$p$, that divides $n!$ is $\lfloor \frac{n}{p} \rfloor + \lfloor \frac{n}{p^2} \rfloor +\lfloor \frac{n}{p^3} \rfloor + \ldots \lfloor \frac{n}{p^k} \rfloor$.  This is your $a_2$.  Can you compare twice this with the expression for $2n$, your $a_1$?&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-02-22T02:42:52.660" CommentCount="1" />
  <row Id="23179" PostTypeId="2" ParentId="23124" CreationDate="2011-02-22T02:49:07.047" Score="9" ViewCount="" Body="&lt;p&gt;First we note that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ S = \sum_{k \ge 0} \frac{1}{k^2+k-\alpha} =&#xD;&#xA;\sum_{k \ge 0} \frac{1}{(k+1/2)^2 - (4\alpha+1)/4} =&#xD;&#xA;4\sum_{k \ge 0} \frac{1}{(2k+1)^2 -a^2},$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $a=\sqrt{4\alpha+1},$ and so&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ S = 4 \sum_{m \textrm{ odd}} \frac{1}{m^2 - a^2}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we use the well-known &lt;a href=&quot;http://mathworld.wolfram.com/Cotangent.html&quot; rel=&quot;nofollow&quot;&gt;cotangent identity&lt;/a&gt; (at the bottom of the page)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\pi\cot(a\pi) = \frac{1}{a} + \sum_{m=1}^\infty \frac{2a}{a^2-m^2} . \quad (1)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Replacing $a$ by $a/2$ and dividing by $2$ gives&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \frac12 \pi\cot \left( \frac{a\pi}{2}\right) = \frac{1}{a} + \sum_{m=1}^\infty \frac{2a}{a^2-(2m)^2} . \quad (2)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Subtracting $(2)$ from $(1)$ gives &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \pi \cot(a\pi) - \frac12 \pi\cot \left( \frac{a\pi}{2}\right) =&#xD;&#xA;\sum_{m=1, \textrm{ odd } m}^\infty \frac{2a}{a^2-m^2} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but $\cot(\theta) -\frac12 \cot(\theta/2)= -\frac12 \tan(\theta/2)$ and so&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \frac{\pi}{4a} \tan \left( \frac{a \pi}{2} \right) = &#xD;&#xA;\sum_{m \textrm{ odd}} \frac{1}{m^2 - a^2},$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;from which the result follows setting $a = \sqrt{4\alpha + 1}.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: The cotangent identity is proven &lt;a href=&quot;http://people.reed.edu/~jerry/311/cotan.pdf&quot; rel=&quot;nofollow&quot;&gt;here.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT2: An easy way to discover the cotangent identity is to take logarithms&#xA;of the following product formula for $\sin  \pi x$ and differentiate wrt $x.$&#xA;$$\sin \pi x = \pi x \prod_{n=1}^{\infty} \left( 1 - \frac{x^2}{n^2} \right)$$&lt;/p&gt;&#xA;" OwnerUserId="1301" LastEditorUserId="1301" LastEditorDisplayName="" LastEditDate="2011-02-22T13:12:48.160" LastActivityDate="2011-02-22T13:12:48.160" CommentCount="2" />
  <row Id="23181" PostTypeId="1" CreationDate="2011-02-22T02:52:45.480" Score="4" ViewCount="110" Body="&lt;p&gt;I'm having trouble recasting the following question in a form amenable to the calculus of variations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Question: Given a set of $n$ points $P=\{(x_1,y_1),..(x_n,y_n)\}$ what is the curve passing through these points such that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(C1) its length is minimal&#xA;&lt;p&gt;&#xA;(C2) the maximum curvature it takes on, as it interpolates the above points, is as small as possible?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Note: this is related to my earlier MO question &lt;a href=&quot;http://mathoverflow.net/questions/39554/interpolation-splines-of-bounded-curvature&quot; rel=&quot;nofollow&quot;&gt;here  &lt;/a&gt;, I wasn't sure if this was good enough for MO.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My partial answer: Suppose we express the curve in the form $y=f(x)$; suppose further that $L(y)$ and $\kappa(y)$ denote length and curvature as a function of $y$. Then, since we need to reduce &lt;i&gt; both&lt;/i&gt; length and curvature, we need to minimize $L(y) + \lambda \kappa(y)$ over the interval $[x_1,x_n]$. This reduces to minimizing $F(y) = \int_{x1}^{x_n} \sqrt{1+y&amp;#39;^2} dx  + \lambda \frac{ y&amp;#39;&amp;#39;} { (1+y&amp;#39;^2)^{3/2}}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Doubts:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) Is $F(y)$ the correct functional? It seems not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) How do I alter $F(y)$ to ensure that it passes through all points in $P$?&lt;/p&gt;&#xA;" OwnerUserId="1585" LastEditorUserId="1585" LastEditorDisplayName="" LastEditDate="2011-02-22T03:06:15.240" LastActivityDate="2011-02-22T03:06:15.240" Title="Extremal curve passing through a set of points" Tags="&lt;optimization&gt;&lt;multivariable-calculus&gt;" CommentCount="2" />
  <row Id="23183" PostTypeId="2" ParentId="23176" CreationDate="2011-02-22T03:13:17.940" Score="3" ViewCount="" Body="&lt;p&gt;Here's how I tend to think about substitution in integrals:  you've decided to let $t=\theta^2$.  The corresponding relationship in differentials is $dt=2\theta d\theta$.  Now, back to the integral, try to rearrange the parts of the integral into recognizable chunks.&#xA;$$\begin{align}&#xD;&#xA;\int_\sqrt{\frac{\pi}{2}}^\sqrt{\pi} \theta^3 \cos(\theta^2) d\theta&#xD;&#xA;&amp;amp;=\int_{\theta=\sqrt{\frac{\pi}{2}}}^{\theta=\sqrt{\pi}} \theta^2 \cos(\theta^2)\frac{1}{2}\cdot 2\theta d\theta&#xD;&#xA;\\&#xD;&#xA;&amp;amp;=\frac{1}{2}\int_{t=\frac{\pi}{2}}^{t=\pi} (\theta^2) \cos(\theta^2)(2\theta d\theta)&#xD;&#xA;\\&#xD;&#xA;&amp;amp;=\frac{1}{2}\int_{\frac{\pi}{2}}^{\pi} t \cos(t)dt&#xD;&#xA;\end{align}$$&#xA;In particular, notice that the $d\theta$ doesn't just magically become $dt$; rather, $dt$ replaces the equivalent expression $2\theta d\theta$.&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2011-02-22T03:13:17.940" />
  <row Id="23184" PostTypeId="1" AcceptedAnswerId="23185" CreationDate="2011-02-22T03:18:00.003" Score="0" ViewCount="205" Body="&lt;p&gt;I'm having difficulty understanding how to find Hydrostatic Force on various shapes. One problem is a right triangle with a height of 3ft and a width of 7ft. Its submerged 1ft below the surface with the 7ft side facing the surface. What do I do?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand the idea. My problem is more about setting up the integral. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;so I don't need to take the integral? There is no thickness. the way we've been doing this in class is that dF=pgdA where F=force on the triangular plate, pg=62.5 and dA=is the change in area that somehow includes dx OR dy. &lt;/p&gt;&#xA;" OwnerUserId="7347" LastEditorUserId="7347" LastEditorDisplayName="" LastEditDate="2011-02-24T13:12:44.433" LastActivityDate="2011-02-24T14:28:13.327" Title="Hydrostatic Force/Pressure" Tags="&lt;calculus&gt;&lt;physics&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="23185" PostTypeId="2" ParentId="23184" CreationDate="2011-02-22T03:23:26.493" Score="2" ViewCount="" Body="&lt;p&gt;The force is simply the weight of the water displaced.  You should have a theorem that proves that.  So integrate to find the volume and multiply by the density of water.  But if you have a triangle, the volume is zero unless it has some thickness.  Then it's easy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Added:  if it is completely submerged, the force is simply the volume of the shape.  For a right triangle with legs of 3 and 7, the area is 21/2 and the volume would be $\frac{21t}{2}$, where $t$ is the thickness.  Then the force would be $\frac{21t}{2}62.4$ pounds upward.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Added2:  If there is no thickness, you are considering everything to have unit thickness.  The 62.5 is the density of water in lb/ft^3 (though it is really 62.43 max unless you have salt water).  The reason you don't need an integral is that you said it was submerged 1 foot below the surface.  It's like if you asked in integral calculus for the area under $y=1-x$ from 0 to 2. If the upper limit were below 1, I might do an integral, but with limit above 1 I would just find the area of the triangle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But maybe you meant the lower corner is 1 foot down, so the triangle is partially submerged.  You still need the area of what is in the water and your expression integrating $dA$ is to do exactly that.  You can either use similar triangles to see that the submerged shape has height 1 foot and width $\frac{7}{3}$ feet for an area of $\frac{7}{6}$ foot, or integate $\int_0^1\frac{7}{3}xdx$ get the same as the volume element is $dx$ high and $\frac{7}{3}x$ wide (determined by the same similar triangles I used in the first).&lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2011-02-24T14:28:13.327" LastActivityDate="2011-02-24T14:28:13.327" />
  <row Id="23186" PostTypeId="1" CreationDate="2011-02-22T03:47:31.727" Score="0" ViewCount="69" Body="&lt;p&gt;I'm reading this book and cannot make sense of the proof presented here: &lt;a href=&quot;http://bit.ly/h8qAOJ&quot; rel=&quot;nofollow&quot;&gt;http://bit.ly/h8qAOJ&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have some understanding of abstract algebra but I'm not even sure if that's what we're dealing with here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some questions:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;What is a homogeneous morphism?  Is it the same as a homomorphism?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;What does the notation &quot;f [element of] Hom(v,u)&quot; mean?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;What does the notation &quot;f = (f_1,...f_n)&quot; mean?  Is this a permutation group?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;What is the degree of a homogeneous morphism?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;What is A^p?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;reddit xref: &lt;a href=&quot;http://www.reddit.com/r/math/comments/fpzja/homogeneous_morphism_proof_from_perecs_life_a/&quot; rel=&quot;nofollow&quot;&gt;http://www.reddit.com/r/math/comments/fpzja/homogeneous_morphism_proof_from_perecs_life_a/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7348" LastActivityDate="2011-02-22T04:15:32.840" Title="What is a homogenous morphism? (related to proof from Perec's &quot;Life A User's Manual&quot;)" Tags="&lt;abstract-algebra&gt;" AnswerCount="1" />
  <row Id="23187" PostTypeId="2" ParentId="23172" CreationDate="2011-02-22T03:54:02.027" Score="4" ViewCount="" Body="&lt;p&gt;Here's a way I like a lot, I have outlined the steps:&#xA;(Each step is a one line proof)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(i) There are $\lfloor N/q \rfloor$ integers less than or equal to $N$ that are divisible by $q$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(ii) Deduce that the difference in the number of integers in the numerator and denominator of $\left({2N\atop N}\right)$ which are divisible by $q$ is $\lfloor 2N/q\rfloor -2\lfloor N/q\rfloor$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(iii) Show that this quantity equals either 0 or 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(iv) Deduce that if $p^r$ divides $\left({2N\atop N}\right)$ then $p^r\leq 2N$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope that helps,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remark:  This is an exercise I did a while ago from a book by Dr. Andrew Granville.  It had similar outline, and was not my invention.  &lt;/p&gt;&#xA;" OwnerUserId="6075" LastActivityDate="2011-02-22T03:54:02.027" CommentCount="1" />
  <row Id="23188" PostTypeId="2" ParentId="23167" CreationDate="2011-02-22T04:04:07.467" Score="2" ViewCount="" Body="&lt;p&gt;Form the tensor product $k(P)\otimes_{k(R)} k(Q)$.  Since $k(R)$ is a field, and&#xA;each of $k(P)$ and $k(Q)$ is non-zero, this tensor product is non-zero.  Thus it contains at least one maximal ideal, say $\mathfrak m$.  Write $F :=(k(P)\otimes_{k(R)} k(Q))/\mathfrak m.$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;By construction $F$ is a field extension of $k(R)$, and it admits $k(R)$-algebra morphisms $k(P)\to F$ and $k(Q) \to F$, which must be embeddings, since their domains are fields.  Thus $F$ is the desired field.&lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2011-02-22T04:04:07.467" />
  <row Id="23189" PostTypeId="2" ParentId="23186" CreationDate="2011-02-22T04:05:24.380" Score="3" ViewCount="" Body="&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;$f\in\mathrm{Hom}(\nu,\mu)$ means that $f$ is a homomorphism with domain $\nu$ and codomain $\mu$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;It is clear later that the arguments of $f$ are $p$-tuples of the form $(a_1,\ldots,a_p)$. Being &quot;homogeneous&quot; means that there exists a natural number $k$ such that for all scalars $\alpha$, $f(\alpha a_1,\alpha a_2,\ldots,\alpha a_p) = \alpha^k f(a_1,a_2,\ldots,a_p)$.  The natural number $k$ is the degree of this homogeneous map.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Just like the &quot;inputs&quot; to $f$ are tuples, the outputs are meant to be $n$-tuples. Each component is computed separately, so you can think of each component as consisting of a function $f_i(a_1,\ldots,a_p)$. For example, if you have a function $f\colon\mathbb{R}^2\to\mathbb{R}^3$ given by $f(x,y) = (x^2-y, x+y, y-2x)$, then you can think of $f$ as consisting of three &quot;component&quot; functions, $f_1$, $f_2$, and $f_3$, with $f_1(x,y)=x^2-y$, $f_2(x,y)=x+y$, and $f_3(x,y)=y-2x$. Then we write $f=(f_1,f_2,f_3)$. That's what the notation $f=(f_1,\ldots,f_m)$ means. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$A^p$ is the collection of all $p$-tuples of elements of $A$. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Frankly, I don't think you are meant to make sense of that quote. For one thing, the &quot;quote&quot; misuses &quot;$o$&quot; to stand for $\circ$, composition, writing $f_iog$ instead of $f_i\circ g$. Which suggests the author of the book didn't really know what he was quoting either.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-22T04:15:32.840" LastActivityDate="2011-02-22T04:15:32.840" />
  <row Id="23191" PostTypeId="1" AcceptedAnswerId="23195" CreationDate="2011-02-22T05:19:53.103" Score="0" ViewCount="71" Body="&lt;p&gt;I am trying to understand how $\epsilon$ transitions work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From what I've read, when you &quot;go&quot; to a state S that has arrows pointing outwards with $\epsilon$'s in it, you automatically go to those states too. That is, in the following diagram, when you are about to give the first digit as input to this FSA, you must consider you are not just in state &lt;em&gt;S1&lt;/em&gt; but in &lt;em&gt;S1&lt;/em&gt; AND &lt;em&gt;S2&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/UiTVi.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this correct? Following this reasoning, when converting this NDFA to a DFA, one should have as first state &lt;em&gt;S13&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The $\delta (S13, b) = S123$, as&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;if I consider the path starting at&#xA;&lt;em&gt;S1&lt;/em&gt;, I will instantly go to &lt;em&gt;S3&lt;/em&gt;&lt;/li&gt;&#xA;&lt;li&gt;if I consider the path starting at &lt;em&gt;S3&lt;/em&gt;,&#xA;as the input is &lt;em&gt;b&lt;/em&gt;, this path dies&lt;/li&gt;&#xA;&lt;li&gt;if I consider the path starting at &lt;em&gt;S1&lt;/em&gt; with a &lt;em&gt;b&lt;/em&gt; as input,&#xA;I will get to &lt;em&gt;S2&lt;/em&gt;.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I have a vague impression my understanding of $\epsilon$ transitions is wrong, as in some exercises I seem to get it done right, while in others wrong.&lt;/p&gt;&#xA;" OwnerUserId="148" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2011-03-21T05:32:54.277" LastActivityDate="2011-03-21T05:32:54.277" Title="Understanding $\epsilon$ transitions in a finite state automaton" Tags="&lt;soft-question&gt;&lt;computer-science&gt;&lt;automata&gt;" AnswerCount="1" />
  <row Id="23192" PostTypeId="1" AcceptedAnswerId="23290" CreationDate="2011-02-22T05:22:57.010" Score="0" ViewCount="149" Body="&lt;p&gt;following &lt;a href=&quot;http://math.stackexchange.com/questions/22825/how-to-compute-svd-singular-value-decomposition-of-upper-triangular-matrix&quot;&gt;@Sivaram Ambikasaran's answer&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for SVD, I get the computing (using MATLAB)of:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;A =  2     1     3&#xA;     1     2     5&#xA;     3     5     4    &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;[U,S,V]            = svd(A)&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;U = -0.3793   -0.2964   -0.8765&#xA;    -0.5571   -0.6832    0.4721&#xA;    -0.7387    0.6674    0.0941&#xA;&#xA;S = 9.3111         0         0&#xA;         0    2.4506        0&#xA;         0         0    1.1395    &#xA;&#xA;V =-0.3793    0.2964   -0.8765&#xA;   -0.5571    0.6832    0.4721&#xA;   -0.7387   -0.6674    0.0941    &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;for eigenvalues I get:&lt;br&gt;&#xA;[eigVector,lambda] = eig(A)&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;eigVector =    &#xA;    0.2964    0.8765    0.3793&#xA;    0.6832   -0.4721    0.5571&#xA;   -0.6674   -0.0941    0.7387&#xA;&#xA;lambda =    &#xA;   -2.4506         0         0&#xA;         0    1.1395         0&#xA;         0         0    9.3111&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I know understand that the singular values are the magnitudes of the eigen values &#xA;so:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;lambda's diagonal magnitudes are equal to S, in SVD,&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;    |-2.4506| =  2.4506&#xA;    |1.1395 | =  1.1395&#xA;    |9.3111 | =  9.3111&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The question here is, as the other values&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;How to get U from eigenvector in that&#xA;example&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;what criteria is used to change sign?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;Could you tell me if computing&#xA;eigenvalue decomposition takes less&#xA;time than computing SVD (it's order)??&lt;/li&gt;&#xA; - &#xA;&lt;/ul&gt;&#xA;" OwnerUserId="7265" LastEditorUserId="7265" LastEditorDisplayName="" LastEditDate="2011-02-22T05:45:57.560" LastActivityDate="2011-02-22T21:58:52.610" Title="Eigenvalue decomposition Singular value decomposition" Tags="&lt;matrices&gt;&lt;eigenvalues&gt;&lt;eigenvectors&gt;" AnswerCount="1" />
  <row Id="23193" PostTypeId="1" AcceptedAnswerId="23597" CreationDate="2011-02-22T05:34:59.460" Score="12" ViewCount="501" Body="&lt;p&gt;The question I am asking has basically already been asked.  Please see &lt;a href=&quot;http://math.stackexchange.com/questions/21851/connecting-a-n-n-point-grid/21930#21930&quot;&gt;this&lt;/a&gt; MSE thread.  There are a few questions brought up on that thread, and a smaller number were answered.  The reason I am asking this question is that I want to focus explicitly on one of the unanswered parts: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does there exist $n$, and $r&amp;lt;2n-2$, such that the $n\times n$ square grid can be connected with an unbroken path of $r$ straight lines?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice that $2n-1$ is the trivial solution.  This is obtained by either spiralling towards the center, or by zig-zagging up and down.  On the &lt;a href=&quot;http://math.stackexchange.com/questions/21851/connecting-a-n-n-point-grid/21930#21930&quot;&gt;other&lt;/a&gt; MSE thread, I posted an answer showing $2n-2$ was possible for $n\geq 3$. (this is obtained by reducing to the $3\times 3$ case)  I felt $2n-2$ should be optimal, but I couldn't think of a proof.  (Also, it is not necessarily clear that for a 100 billion by 100 billion grid you can't use some kind of cleverness to eliminate &lt;em&gt;one&lt;/em&gt; single line)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This has been bothering me since I saw it, and I would like to know if anyone has any ideas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit:  This is regarding an question/answer I received:  Not only are 45 degree lines allowed, but lines in &lt;em&gt;any&lt;/em&gt; direction.  Naturally 0,45,90 seem like the best since those angles will intersect the most lattice points, and we ask ourselves &quot;Will any optimal solution consist only of lines at these angles?&quot;.  The answer is a definite &lt;em&gt;no&lt;/em&gt;.  Consider the &lt;a href=&quot;http://www.mathpuzzle.com/dots10x10.gif&quot; rel=&quot;nofollow&quot;&gt;following solution&lt;/a&gt; to the $10\times 10$ grid which uses 18 lines (so $2n-2$, i.e. currently optimal).  Also there is &lt;a href=&quot;http://www.mathpuzzle.com/JK8x8.GIF&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; solution to the $8\times 8 $ grid using 14 lines.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Interestingly, this solution to the $10\times 10 $ generalizes to give another $2n-2$ line solution to the $n\times n$ grid when $n$ is even.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I found this $8\times8$ and $10\times 10$ solution on a link posted by user3123 on the &lt;a href=&quot;http://math.stackexchange.com/questions/21851/connecting-a-n-n-point-grid/21930#21930&quot;&gt;other&lt;/a&gt; question page.  Specifically it was &lt;a href=&quot;http://www.mathpuzzle.com/dots.html&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; link.)&lt;/p&gt;&#xA;" OwnerUserId="6075" LastEditorUserId="6075" LastEditorDisplayName="" LastEditDate="2011-02-24T16:56:42.097" LastActivityDate="2011-02-26T23:23:26.097" Title="Not lifting your pen on the $n\times n$ grid." Tags="&lt;combinatorics&gt;&lt;graph-theory&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="3" />
  <row Id="23194" PostTypeId="2" ParentId="21243" CreationDate="2011-02-22T05:45:14.780" Score="0" ViewCount="" Body="&lt;p&gt;I'm also interested on the convergence of Markov chains on a graph and found your questions very interesting. &#xA;As per my knowledge, for adjacancy matrices, eigenvalues of AM implies many things. It implies the connectivity of the graph. If the difference between the largest eigenvalue and the second largest eigenvalue (this is known as spectral gap) is large, it means that the graph is a well-connected graph. In fact in spectral graph theory expansion parameter of a graph (for a d-regular graph) is lower bounded by the spectral gap of the adjacency matrix. You can read more on this in &lt;a href=&quot;http://www.math.ias.edu/~boaz/ExpanderCourse/allnotes.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.math.ias.edu/~boaz/ExpanderCourse/allnotes.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a question on one thing you mentioned.&#xA;It's about what you mentioned as &quot;In a chain, the matrix of transition probabilities P (for discrete time markov chains), or the matrix of transition frequencies Q (for cont-time markov chains), can be evaluated (their eigenvalues) in order to inspect whether the chain is ergodic or not (with a parallelism to Control Theory: eigenvalues in the unitary circle or in the negative half-plane).&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you please elaborate more on this and help me to get an idea about which properties can be implied using the eigenvalues of the transition probability matrix of a markov chain on graph.?&lt;/p&gt;&#xA;" OwnerUserId="7351" LastActivityDate="2011-02-22T05:45:14.780" />
  <row Id="23195" PostTypeId="2" ParentId="23191" CreationDate="2011-02-22T05:54:48.477" Score="3" ViewCount="" Body="&lt;p&gt;$\epsilon$ transitions are optional - you can always take them, but you don't have to. That's why the starting state of the automaton in question is indeed $\{1,3\}$. On input $b$, you mention correctly that from state $1$ you go to $2$, whereas from state $3$ there is no arrow labeled $b$. Therefore the automaton goes to state $\{2\}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, there can be more than one output arrow having the same label: for example, if the automaton reads $a$ while in state $\{2\}$, it goes to state $\{2,3\}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, $\epsilon$ transitions can be taken in chain. There's no such opportunity in the automaton in question, but supposing that there was one more state $4$ connected to $3$ via an $\epsilon$ transition, the starting state would be $\{1,3,4\}$ - you can take as many $\epsilon$ transitions as you want (but never have to take one).&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2011-02-22T05:54:48.477" CommentCount="3" />
  <row Id="23196" PostTypeId="2" ParentId="23142" CreationDate="2011-02-22T06:03:46.963" Score="2" ViewCount="" Body="&lt;p&gt;It doesn't matter whether $B$ is invariant or not, it only has to be of positive measure so that its characteristic function $[B]$ is nonzero in $L^{2}$. I assume that $X$ is a finite measure space (so $[X] \in L^{2}$) and that $T: X \to X$ is a measure-preserving transformation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clearly, $P_{T}[X] = [X]$, so &#xA;\[&#xA;0 &amp;lt; \mu(B) = \langle [B],[X] \rangle = &#xA;\langle [B], P_{T}[X] \rangle = \langle P_{T}[B],[X]\rangle&#xA;\]&#xA;using $P_{T}^{\ast} = P_{T}$ in the last equation, hence $P_{T}[B] \neq 0$ in $L^{2}$.&lt;/p&gt;&#xA;" OwnerUserId="5363" LastEditorUserId="5363" LastEditorDisplayName="" LastEditDate="2011-02-22T06:14:15.237" LastActivityDate="2011-02-22T06:14:15.237" />
  <row Id="23197" PostTypeId="1" CreationDate="2011-02-22T06:04:44.307" Score="0" ViewCount="164" Body="&lt;p&gt;The question I'm trying to figure out states that I have 3 points &lt;em&gt;P1, P2&lt;/em&gt; and &lt;em&gt;P3&lt;/em&gt; in space.  In one frame (Frame A I called it) those points are: &lt;em&gt;Pa1, Pa2&lt;/em&gt; and &lt;em&gt;Pa3&lt;/em&gt;, same story for Frame B (namely: &lt;em&gt;Pb1, Pb2&lt;/em&gt; and &lt;em&gt;Pb3&lt;/em&gt;).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whats the rotation matrix from one to the other?  That's literally all the information I have. What was suggested was make an intermediate coordinate frame and align one of it's axis' with a point.   However I don't see how this will help with the problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone offer some advice how to tackle this problem? I'm stumped. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks,&#xA;Edit: Both frames have the same origin, so there is no translation component.&lt;/p&gt;&#xA;" OwnerUserId="7353" LastActivityDate="2011-02-25T10:19:11.340" Title="Finding a Rotation Transformation from two Coordinate Frames in 3-Space" Tags="&lt;geometry&gt;&lt;matrices&gt;&lt;transformation&gt;&lt;rigid-transformation&gt;" AnswerCount="4" CommentCount="1" />
  <row Id="23198" PostTypeId="2" ParentId="23197" CreationDate="2011-02-22T06:18:32.250" Score="0" ViewCount="" Body="&lt;p&gt;Think the general problem is referred to as the absolute orientation problem and suggest to read below paper.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://cis.jhu.edu/software/lddmm-similitude/umeyama.pdf&quot; rel=&quot;nofollow&quot;&gt;http://cis.jhu.edu/software/lddmm-similitude/umeyama.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5312" LastActivityDate="2011-02-22T06:18:32.250" />
  <row Id="23199" PostTypeId="1" AcceptedAnswerId="23201" CreationDate="2011-02-22T06:18:45.220" Score="2" ViewCount="102" Body="&lt;p&gt;How can I prove that $\displaystyle{\lim_{(x, y) \to (0, 0)} \frac{x^3 - y^3}{x^2 + y^2}}=0$?  The method I've been taught is the pinching one, where you compare the absolute value of the function to greater limits that are known to equal zero, but I haven't managed to find a comparison that works.  Could someone point me in the right direction?&lt;/p&gt;&#xA;" OwnerUserId="7354" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2011-02-22T06:37:36.330" LastActivityDate="2011-02-22T06:37:36.330" Title="Finding the limit of a 2-dimensional function" Tags="&lt;limit&gt;" AnswerCount="2" />
  <row Id="23201" PostTypeId="2" ParentId="23199" CreationDate="2011-02-22T06:28:29.687" Score="5" ViewCount="" Body="&lt;p&gt;$$\left|\frac{x^3 - y^3}{x^2 + y^2}\right|\leq \left|\frac{x^3}{x^2 + y^2}\right|+\left|\frac{y^3}{x^2 + y^2}\right|\leq |x| + |y|.$$&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2011-02-22T06:28:29.687" CommunityOwnedDate="2011-02-22T06:28:29.687" />
  <row Id="23202" PostTypeId="2" ParentId="23199" CreationDate="2011-02-22T06:29:13.397" Score="3" ViewCount="" Body="&lt;p&gt;Hint: If $(x,y)$ is such that both $|x|\le\varepsilon$ and $|y|\le\varepsilon$, then your function $f$ is such that $|f(x,y)|\le2\varepsilon$.&lt;/p&gt;&#xA;" OwnerUserId="6179" LastActivityDate="2011-02-22T06:29:13.397" />
  <row Id="23203" PostTypeId="1" CreationDate="2011-02-22T06:37:07.297" Score="0" ViewCount="86" Body="&lt;p&gt;I recently read this pdf: &lt;a href=&quot;http://www.math.princeton.edu/~nelson/papers/warn.pdf&quot; rel=&quot;nofollow&quot;&gt;Warning Signs of a Possible Collapse of Contemporary Mathematics&lt;/a&gt;, and I'm having some trouble understanding the issues it raises.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The author says that the consistency of Peano Arithmetic cannot be proved within the system, which is a direct application of Gödel's second incompleteness theorem. It can, however, be proved using transfinite induction, but this is a problem in his opinion, because one has to trust the existence of infinite cardinals, contrary to his monotheistic faith. My first question is whether this is what he meant, or if I misunderstood some part of his argument.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nelson then goes on to define some classes of numbers which are contained in one of them, that of &quot;counting numbers&quot;, and proves that these counting numbers are closed under addition and multiplication. But he implies that it is not possible to prove that exponentiation and superexponentiation and, in general, faster-growing recursive functions, whose definition is completely permitted by the axiom of induction, are guaranteed to terminate, i.e., to be computable. My other questions are: can all recursive functions be shown to terminate after a finite number of steps, assuming the existence of some transfinite cardinals? And what exactly is the finitist objection to this assumption? Finally, what would be a finistically acceptable model for mathematics? Is it usable, in a general sense?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="6427" LastActivityDate="2011-02-22T06:37:07.297" Title="Finitistic objections to the current mathematical model" Tags="&lt;philosophy&gt;" CommentCount="3" />
  <row Id="23204" PostTypeId="1" AcceptedAnswerId="23205" CreationDate="2011-02-22T06:48:12.573" Score="1" ViewCount="63" Body="&lt;p&gt;It is known the following equivalence: Let $Y \subset \mathbb{A}^{n}$ be an algebraic set. Then $Y$ is irreducible if and only if $I(Y)$ is prime, where:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$I(Y) = \{f \in K[x_{1},x_{2},..,x_{n}]: \textrm{for all p in Y}, \ f(p)=0 \}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So as an example, the author considers $J = \langle (xz-y^2,x^3-yz) \rangle \subset k[x,y,z]$. Then let $Y=V(J)$ (the locus set).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if we want to find whether $Y$ is irreducible we must study $I(Y)=I(V(J))$ and show&#xA;this ideal is prime (or show it is not).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The author then shows that $J$ is not a prime ideal. But why is this? don't we need to show that $I(V(J))$ is a prime ideal? why we must check $J$? or do we always have $J=I(V(J))$?&lt;/p&gt;&#xA;" OwnerUserId="6495" LastActivityDate="2011-02-22T07:27:52.650" Title="Clarification about equivalence irreducibility and prime ideals" Tags="&lt;algebraic-geometry&gt;" AnswerCount="1" />
  <row Id="23205" PostTypeId="2" ParentId="23204" CreationDate="2011-02-22T07:27:52.650" Score="3" ViewCount="" Body="&lt;p&gt;No, we don't always have $I(V(J))=J$ - the Nullstellensatz says that we always have $I(V(J))=\sqrt{J}$, where $\sqrt{J}$ is the radical of the ideal $J$, but there are many ideals that are not equal to their radical. For example, the radical of the non-prime ideal $J=(x^2)$ in $k[x]$ is the prime ideal $(x)$, so that $V(J)$ is irreducible even though $J$ is not prime. So your complaint is valid; it sounds like the book's argument for why $V(J)$ is not irreducible is in error. Could you specify the author/title?&lt;/p&gt;&#xA;" OwnerUserId="264" LastActivityDate="2011-02-22T07:27:52.650" CommentCount="4" />
  <row Id="23207" PostTypeId="1" CreationDate="2011-02-22T08:34:13.950" Score="1" ViewCount="141" Body="&lt;p&gt;If two topological spaces have the same homotopy type, how do I prove that their complements in an ambient space are homeomorphic?&lt;/p&gt;&#xA;" OwnerUserId="7485" LastActivityDate="2011-02-23T17:00:34.587" Title="If two topological spaces have the same homotopy type, how do I prove that their complements in an ambient space are homeomorphic?" Tags="&lt;topology&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="23208" PostTypeId="1" CreationDate="2011-02-22T08:37:09.733" Score="1" ViewCount="155" Body="&lt;p&gt;Given is a second-order tensor $T$, and three arbitrary vectors, $u$, $v$ and $w$, defined in Euclidean point space $\mathcal{E}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Prove that the determinant of the tensor $T$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\det T=\frac{Tu.(Tv \times Tw)}{u.(v \times w)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is independent of $u$, $v$ and $w$, as long as they form a base in the 3-dimensional vector space $\mathcal{V}$ defined on $\mathcal{E}$ (that is, vectors $u$, $v$ and $w$ are linearly independent).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is basically proving that the determinant of a tensor is invariant of the particular basis it happens to be resolved in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've tried to work through this using index notation, but it got pretty cumbersome quickly, so I thought there might be another way to do it. I'm not looking for the full treatment here, but rather as a general advice how to tackle this. I think I can manage the details myself.&lt;/p&gt;&#xA;" OwnerUserId="3022" LastActivityDate="2011-02-22T16:16:22.727" Title="Prove the determinant of a tensor is invariant" Tags="&lt;homework&gt;&lt;tensors&gt;" AnswerCount="2" />
  <row Id="23209" PostTypeId="1" CreationDate="2011-02-22T08:51:36.653" Score="0" ViewCount="92" Body="&lt;p&gt;Relations of finite arity $n$ over a base set $A$ can be seen as unary predicates of functions $f:[n] \rightarrow A$. Nothing prohibits us to consider more general functions $f:\mathbb{N} \rightarrow A$ or even $f:\mathbb{R}^+_0 \rightarrow A$.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Is there a model theory that allows for&#xA;  relations and functions of infinite&#xA;  and even uncountable arity?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Or is this general case implicitly covered by &quot;classical&quot; model theory?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;&lt;strong&gt;Side question:&lt;/strong&gt; What about considering functions $f:\mathbb{Q}^+_0 \rightarrow A$? Couldn't one allow for functions with arbitrary order types as domain.&lt;/sup&gt;&lt;/p&gt;&#xA;" OwnerUserId="1792" LastEditorUserId="1792" LastEditorDisplayName="" LastEditDate="2011-02-22T09:26:28.027" LastActivityDate="2011-02-22T09:26:28.027" Title="Relations of infinite arity" Tags="&lt;logic&gt;&lt;model-theory&gt;" />
  <row Id="23210" PostTypeId="2" ParentId="23207" CreationDate="2011-02-22T09:29:11.070" Score="7" ViewCount="" Body="&lt;p&gt;I don't know how your question can be fixed so as to get a true statement, but as asked this is simply wrong. Take the $x$-axis in $\mathbb{R}^{2}$. It is contractible and its complement has two connected components. However, the complement of a point only has one connected component, so this is an easy example of two subspaces of the same homotopy type whose complements aren't homeorphic or even homotopic.&lt;/p&gt;&#xA;" OwnerUserId="5363" LastActivityDate="2011-02-22T09:29:11.070" CommentCount="3" />
  <row Id="23212" PostTypeId="2" ParentId="19318" CreationDate="2011-02-22T09:56:08.653" Score="2" ViewCount="" Body="&lt;p&gt;Group theory is used extensively in chemistry.  It's what determines whether or not two molecules will bond (it's based on the symmetry of their orbitals).  Group theory is essentially the basis of the Molecular Orbital Theory which is the basis of modern chemistry.&lt;/p&gt;&#xA;" OwnerUserId="7364" LastActivityDate="2011-02-22T09:56:08.653" />
  <row Id="23213" PostTypeId="1" CreationDate="2011-02-22T10:03:59.903" Score="0" ViewCount="66" Body="&lt;p&gt;$f$ is uni-variate polynomial of degree $d$. I am interested in lower bounding modulus first derivative of $f$ (i.e. $|f'|$)in interval $[a,b]$ given the promise that in interval $[a,b]$ $f&amp;#39;$ don't have any root.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Like in &lt;strong&gt;Markoff theorem&lt;/strong&gt;..to upper bound first derivative of polynomial in open interval (a,b) is&#xA;$|f&amp;#39;(x)|\leq \frac{2*M*d*d}{b-a}$ where M is upper bound of $f$ in $[a,b]$,but I don't know proof of this theorem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know polynomial $f$ i.e. its coefficient,degree and upper bound as well as lower bound of f in interval [a,b], Now I want to lower bound f' in [a,b] in terms of these terms..say like Markoff theorem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pls let me know whether this time I able to put my question clearly &lt;/p&gt;&#xA;" OwnerUserId="7363" LastActivityDate="2011-02-22T10:03:59.903" Title="lower bounding first derivative of polynomial" Tags="&lt;polynomials&gt;" CommentCount="3" />
  <row Id="23214" PostTypeId="2" ParentId="23208" CreationDate="2011-02-22T10:10:26.057" Score="2" ViewCount="" Body="&lt;p&gt;Since you use the cross and dot products in your question, you are working over an inner product space, therefore $\mathcal{V}$ is canonically identified with its dual. In particular this means that we can treat $T$, a rank-two tensor, as a linear transformation on $\mathcal{V}$. Therefore $T$ extends to a linear transformation on the exterior algebra of $\mathcal{V}$ and in particular you have a map $\tilde{T}:\Lambda^3V \to\Lambda^3V$. Using the inner-product, you can take the Hodge dual identifying $*: \Lambda^3V \to \mathbb{R}$ satisfying $** = 1$. So pre- and post-composing with $*$, you can extend $T$ to a linear map $\hat{T}:\mathbb{R}\to\mathbb{R}$ (which is just multiplication by a real number).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can check that your expression is in fact&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \det T = \frac{*[\tilde{T}(u\wedge v \wedge w)]}{*(u\wedge v \wedge w)} = \frac{\hat{T}s}{s}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is clearly independent of the chosen basis of $\mathcal{V}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The statement in the end reflects the fact that the determinant of a linear transformation on a vector space represents the numerical factor by which the &lt;em&gt;volume element&lt;/em&gt; in the vector space changes under the linear transformation. &lt;/p&gt;&#xA;" OwnerUserId="1543" LastActivityDate="2011-02-22T10:10:26.057" />
  <row Id="23215" PostTypeId="2" ParentId="8750" CreationDate="2011-02-22T10:25:52.817" Score="0" ViewCount="" Body="&lt;p&gt;&lt;img src=&quot;http://www.math.ethz.ch/~blatter/Reul.tiff&quot; alt=&quot;Reuleaux figure&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I refer to the above figure. The triangle has side length $1$, and the culmination point of the road is at the origin $O$. Let the road $s\to (x(s),y(s))$ be parametrized by arc length and let $\theta:=-\arg(\dot x,\dot y)$. Then the point $A$ in the figure is given by $A=(x+\sin\theta, y+\cos\theta)$, and the centroid $Z$ of the triangle by $Z=(\ldots,\  y+\cos\theta-\cos(\theta + s)/\sqrt{3})$. As $Z$ should keep its second coordinate constant we get the condition&#xA;$$-\sin\theta-\sin\theta\cdot\dot \theta +\sin(\theta + s)\cdot(\dot\theta+1)/\sqrt{3} =0$$&#xA;(note that $\dot y=-\sin\theta$) from which we obtain&#xA;$$(\dot\theta+1)(\sin(\theta+s)-\sqrt{3} \sin\theta)=0.$$&#xA;It follows that here the second factor has to vanish identically. This leads to&#xA;$$\tan\theta(s)={\sin s\over\sqrt{3} -\cos s}$$  resp. to&#xA;$$\dot x(s)=\cos\theta={\sqrt{3}-\cos s\over\sqrt{4-2\sqrt{3}\cos s}},\qquad &#xD;&#xA;\dot y(s)=-\sin\theta={-\sin s\over \sqrt{4-2\sqrt{3}\cos s}}.$$&#xA;Unfortunately $x(s)$ is not an elementary function.&lt;/p&gt;&#xA;" OwnerUserId="1303" LastActivityDate="2011-02-22T10:25:52.817" />
  <row Id="23216" PostTypeId="2" ParentId="23197" CreationDate="2011-02-22T10:28:43.970" Score="0" ViewCount="" Body="&lt;p&gt;I give an example using the matrix-calculator-language MatMate. We have a matrix &lt;strong&gt;A&lt;/strong&gt; which represents three points in a 3D-coordinate system; the columns represent the x,y,z-axes. Then we have a rotation, its coordinates in matrix &lt;strong&gt;B&lt;/strong&gt;. We try to find the roationmatrix &lt;strong&gt;t&lt;/strong&gt; which provides &lt;strong&gt;A * t = B&lt;/strong&gt;. To provide valid matrices I generate one random matrix &lt;strong&gt;A&lt;/strong&gt; and by a random-rotation &lt;strong&gt;tx&lt;/strong&gt; I generate a valid/meaningful matrix &lt;strong&gt;B&lt;/strong&gt; The key is to find rotations to a &quot;normalized&quot; position, for instance a triangular layout of the coordinate-matrices. Then with &lt;strong&gt;t1&lt;/strong&gt; rotating &lt;strong&gt;A-&gt;triangular&lt;/strong&gt; and &lt;strong&gt;t2&lt;/strong&gt; rotating &lt;strong&gt;B-&gt;triangular&lt;/strong&gt; we get &lt;strong&gt;B = A * t1 * inverse(t2)&lt;/strong&gt;&lt;br&gt;&#xA;Example:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; A = randomu(3,3)   // provide a random matrix with 3 points in 3 axes&#xA;      A : &#xA;   28.78       28.00       27.32&#xA;    9.36       87.72       54.37&#xA;   62.89       16.12       96.08&#xA;&#xA; tx = gettrans(randomu(3,3),&quot;drei&quot;)   // provide a random rotation-matrix&#xA; B = A*tx  // generate a valid matrix B which is a rotation of A&#xA;      B : &#xA;   46.98        8.03       -9.36&#xA;   86.42       50.39       27.03&#xA;  108.94      -38.80       -8.57&#xA;&#xA;===========================================================================&#xA;&#xA;t1 = gettrans(A,&quot;drei&quot;)  // find a &quot;normalized&quot; position of A, for example the     &#xA;                         // rotation to the &quot;triangular&quot; position/layout. &#xA;                         // keep the required rotation-matrix in t1&#xA;A1 = A * t1              // rotate A to A1&#xA;      t1 : &#xA;    0.59       -0.74       -0.32&#xA;    0.58        0.66       -0.47&#xA;    0.56        0.10        0.82&#xA;&#xA;      A1 : &#xA;   48.57        0.00       -0.00&#xA;   86.70       56.75        0.00&#xA;  100.61      -26.37       51.27&#xA;&#xA;&#xA;t2=gettrans(B,&quot;drei&quot;) // now find the rotation which rotates B to the&#xA;                      // same &quot;normalized&quot; (=triangular) position/layout&#xA;B1 = B * t2&#xA;      t2 : &#xA;    0.97        0.04        0.25&#xA;    0.17        0.64       -0.75&#xA;   -0.19        0.77        0.61&#xA;&#xA;      B1 : // we see, that A1 and B1 are equal&#xA;   48.57        0.00        0.00&#xA;   86.70       56.75       -0.00&#xA;  100.61      -26.37       51.27&#xA;&#xA;&#xA; chk = A * t1 * t2'  // check whether the combined rotation t1 and &#xA;                     // inverse(t2)=transpose(t2) rotates A to B&#xA;      chk : &#xA;   46.98        8.03       -9.36&#xA;   86.42       50.39       27.03&#xA;  108.94      -38.80       -8.57&#xA; // We see, that this is equal to B and the combined rotation is tx=t1*t2' &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="1714" LastEditorUserId="1714" LastEditorDisplayName="" LastEditDate="2011-02-22T10:35:59.210" LastActivityDate="2011-02-22T10:35:59.210" />
  <row Id="23217" PostTypeId="1" AcceptedAnswerId="23225" CreationDate="2011-02-22T10:34:20.760" Score="2" ViewCount="317" Body="&lt;p&gt;It goes like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $G$ be a group, $H &amp;lt; G$, and $N \lhd G$ be the smallest normal subgroup containing $H$. Then any $f \in \operatorname{Hom}(G,G&amp;#39;)$ such that $H$ is in its kernel uniquely factors through $G/N$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's a simple proposition, but it's so weirdly formulated that I spent the whole day picking it apart.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically, it says two things:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) $H &amp;lt; N \lhd G$, and $N &amp;lt; \operatorname{Ker}f$ (so $N \lhd \operatorname{Ker}f$), which is trivial to prove.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Any hom uniquely factors through a quotient by any normal group that is a subgroup of its kernel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So why combine two natural propositions into a single one that is so weird? Is it used in some important place or does it have any significance in the group theory by itself? Or am I mistaken and the second proposition I put forward is generally false?&lt;/p&gt;&#xA;" OwnerUserId="3793" LastEditorUserId="3793" LastEditorDisplayName="" LastEditDate="2011-02-22T12:18:43.710" LastActivityDate="2011-05-28T19:49:27.413" Title="A weird proposition in Lang's &quot;Algebra&quot;" Tags="&lt;group-theory&gt;" AnswerCount="2" CommentCount="5" FavoriteCount="1" />
  <row Id="23218" PostTypeId="1" CreationDate="2011-02-22T10:47:58.907" Score="0" ViewCount="42" Body="&lt;p&gt;Hello, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $G = (V, E)$ and $G&amp;#39; = (V&amp;#39;, E&amp;#39;)$ be two graphs, and let $f: V \rightarrow V&amp;#39;$ be a surjection. Then $f$ is a &lt;em&gt;covering map&lt;/em&gt; from $G$ to $G&amp;#39;$ if for each $v \in G$, the restriction of $f$ to the neighbourhood of $v$ is a bijection onto the neighbourhood of $f(v) \in V&amp;#39;$ in $G&amp;#39;$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question (homework) is how to easily prove that if there exists a cycle in $G$, there also exists a cycle in $G&amp;#39;$? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a proof based on the size of the preimage of each vertex of $G&amp;#39;$. But, it seems to complicate. I would like to know your point of view.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks a lot in advance. &lt;/p&gt;&#xA;" OwnerUserId="7365" LastActivityDate="2011-02-22T12:31:04.117" Title="Covering and Cycles" Tags="&lt;graph-theory&gt;" AnswerCount="1" />
  <row Id="23219" PostTypeId="2" ParentId="23217" CreationDate="2011-02-22T10:57:07.153" Score="1" ViewCount="" Body="&lt;p&gt;Well, I don't know Lang's reason, so I can only guess. First of all, it is a concise and easy enough statement and I don't see what's so particularly &lt;em&gt;weird&lt;/em&gt; about it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That said, I think the main reason is that the proposition as stated by Lang accurately describes the categorical image $\langle\langle H \rangle\rangle_{G}$ (the &lt;em&gt;normal closure&lt;/em&gt; of $H$ in $G$, or as your proposition states the kernel of the cokernel) of the inclusion $H \hookrightarrow G$ and its categorical cokernel $G/\langle\langle H \rangle\rangle_{G}$ when the inclusion is viewed as a morphism in the category of groups, see e.g. &lt;a href=&quot;http://en.wikipedia.org/wiki/Cokernel&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for your last question, your second proposition is correct and follows from the usual &lt;a href=&quot;http://en.wikipedia.org/wiki/Fundamental_theorem_on_homomorphisms&quot; rel=&quot;nofollow&quot;&gt;homomorphism theorem&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added much later:&lt;/strong&gt; It seems that $\langle H^{G} \rangle$ is the usual notation for the normal closure which I denoted $\langle\langle H \rangle\rangle_{G}$ &amp;mdash; some people insisted on that by editing it in &amp;mdash; but in my opinion this is &lt;em&gt;really ugly&lt;/em&gt; and, much worse, quite ambiguous.&lt;/p&gt;&#xA;" OwnerUserId="5363" LastEditorUserId="5363" LastEditorDisplayName="" LastEditDate="2011-05-28T19:49:27.413" LastActivityDate="2011-05-28T19:49:27.413" CommentCount="8" />
  <row Id="23220" PostTypeId="1" AcceptedAnswerId="23237" CreationDate="2011-02-22T11:17:17.227" Score="1" ViewCount="234" Body="&lt;p&gt;I could not understand the concept while googling. can anybody provide help?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;what will be the determinant of the following matrix?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;\left[\begin{array}{cccc}&#xD;&#xA;1 &amp;amp; 2 &amp;amp; 3 &amp;amp; 4 \\&#xD;&#xA;5 &amp;amp; 6 &amp;amp; 7 &amp;amp; 8 \\&#xD;&#xA;9 &amp;amp; 10 &amp;amp; 11 &amp;amp; 12 \\&#xD;&#xA;13 &amp;amp; 14 &amp;amp; 15 &amp;amp; 16&#xD;&#xA;\end{array} \right]$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.... :)&lt;/p&gt;&#xA;" OwnerUserId="7366" LastEditorUserId="8476" LastEditorDisplayName="" LastEditDate="2011-04-21T13:44:53.340" LastActivityDate="2011-04-21T14:04:32.923" Title="How to find Determinant of a matrix" Tags="&lt;matrices&gt;" AnswerCount="4" CommentCount="8" />
  <row Id="23221" PostTypeId="2" ParentId="23197" CreationDate="2011-02-22T11:37:18.757" Score="1" ViewCount="" Body="&lt;p&gt;Consider the matrix $M_a$ that gives you the first frame representation of vectors $P_1, P_2, P_3$: $P_{a1}=M_a \cdot P_1, P_{a2}=M_a \cdot P_2, P_{a3}=M_a \cdot P_3$. Similarly, you've got some matrix $M_b$ that gives you the second frame representation: $P_{b1}=M_b \cdot P_1, P_{b2}=M_b \cdot P_2, P_{b3}=M_b \cdot P_3$. Since any $P_i$ is a column-vector, the equations above allow to combine the columns into matrices: $P=(P_1 P_2 P_3), P_a=(P_{a1} P_{a2} P_{a3}), P_b=(P_{b1} P_{b2} P_{b3})$, then $P_a=M_a \cdot P$ and $P_b=M_b \cdot P$. We want to find a matrix, $X$, that would transform $M_a$ into $M_b$: $M_b=M_a \cdot X$, then $X=M_a^{-1} \cdot M_b$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We derive $M_a$ and $M_b$: $P_a \cdot P^{-1} = M_a$, $P_b \cdot P^{-1} = M_b$. Then $X = P \cdot P_a^{-1} \cdot P_b \cdot P^{-1}$. That obviously requires non-degenerate (non-planar) initial set of $P$ and non-degenerate frames $M_a$ and $M_b$.&lt;/p&gt;&#xA;" OwnerUserId="3806" LastActivityDate="2011-02-22T11:37:18.757" />
  <row Id="23222" PostTypeId="2" ParentId="23101" CreationDate="2011-02-22T12:03:54.323" Score="2" ViewCount="" Body="&lt;p&gt;So, assuming $g(x,\epsilon)=O(\phi(x,\epsilon))$ and $f(x,\epsilon)=O(\psi(x,\epsilon))$ this means that $\exists \delta_1,\delta_2,M_1,M_2$ such that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\forall |\epsilon|&amp;lt;\delta_1 : |g(x,\epsilon)| &amp;lt; M_1 |\phi(x,\epsilon)|$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\forall |\epsilon|&amp;lt;\delta_2 : |f(x,\epsilon)| &amp;lt; M_2 |\psi(x,\epsilon)| \; .$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This implies that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{eqnarray}&#xD;&#xA;\forall |\epsilon| &amp;lt; \min(\delta_1,\delta_2) : &amp;amp; |g(x,\epsilon) + f(x,\epsilon)| &amp;amp; \leq |g(x,\epsilon)| + |f(x,\epsilon)| \\&#xD;&#xA;&amp;amp; &amp;amp; &amp;lt; \max(M_1,M_2)(|\phi(x,\epsilon)| + |\psi(x,\epsilon)|)  \; , \end{eqnarray} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;using the triangle inequality. Or in other words $g(x,\epsilon)+f(x,\epsilon)=O(|\phi(x,\epsilon)|+|\psi(x,\epsilon)|)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, using Yuval Filmus' suggestion, you can see that for the choice $f(x)=g(x)=\phi(x)=x$ and $\psi(x)=-x$, $2x=O(2|x|)$, which is correct. The other identity can not be correct however since $2x \neq O(0)$.&lt;/p&gt;&#xA;" OwnerUserId="3567" LastActivityDate="2011-02-22T12:03:54.323" />
  <row Id="23223" PostTypeId="2" ParentId="23218" CreationDate="2011-02-22T12:31:04.117" Score="1" ViewCount="" Body="&lt;p&gt;If $u,v\in V$ and $(u,v)\in E$ then since $u$ is in the neighborhood of $v$ then the condition on the local bijection gives you that $(f(v),f(u))\in E&amp;#39;$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose C is a cycle in G. Take the subgraph C' of G' with the vertices $f(v)$ such that $v\in C$ and the edges $(f(v),f(u))$ such that $(v,u)\in C$. If $u\in C$ then it has two different neighbors $v,w\in C$ so by the local bijection $f(u)$ has two different neighbors $f(v),f(w)$ in C'. By the definition of C' these are all the neighbors $f(u)$ has. So C' must be a cycle. (you can show that C' is connected, but even if it isn't you can say that every connected component is a cycle)&lt;/p&gt;&#xA;" OwnerUserId="2125" LastActivityDate="2011-02-22T12:31:04.117" />
  <row Id="23224" PostTypeId="1" CreationDate="2011-02-22T12:33:23.053" Score="1" ViewCount="126" Body="&lt;p&gt;How to integrate $x^{1/2}e^{-x}$ using integration by parts?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Answer should be $\left(-\sqrt{x} e^{-x}+(1/2)\sqrt{\pi} \mbox{erf}(\sqrt{x})\right)+c$&lt;/p&gt;&#xA;" OwnerUserId="2723" LastEditorUserId="3567" LastEditorDisplayName="" LastEditDate="2011-02-22T13:09:06.890" LastActivityDate="2011-02-22T13:15:08.797" Title="Integrate $x^ {1/2}e^{-x}$ using integration by parts." Tags="&lt;integral&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="23225" PostTypeId="2" ParentId="23217" CreationDate="2011-02-22T12:45:43.890" Score="8" ViewCount="" Body="&lt;p&gt;Dear Alexei,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suspect you find Lang's statement weird because you are thinking of it as a technique for &lt;em&gt;generating&lt;/em&gt; homomorphisms, and then wondering why Lang is putting a strange emphasis on these particular kinds of normal subgroups.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But to appreciate the proposition, you should think from a different point of view.  E.g. suppose that you are &lt;em&gt;given&lt;/em&gt; a homomorphism $f: G \to G&amp;#39;$, and that&#xA;you are told that $f(h) = 1$ for every $h \in H$.  Then what can you conclude about $ker(f)$?   Well, it certainly contains $H$, since this is what you are told.  But since the kernel is normal, it in fact contains the normal subgroup of $G$ generated by $H$, which is what Lang calls $N$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a side remark: although $N$ is correctly described as the smallest normal subgroup containing $H$, it is probably better (from a psychological point of view) to think of $N$ as the normal subgroup of $G$ generated by $H$; this helps bring out the role of $H$.   &lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2011-02-22T12:45:43.890" CommentCount="1" />
  <row Id="23226" PostTypeId="2" ParentId="23224" CreationDate="2011-02-22T13:15:08.797" Score="4" ViewCount="" Body="&lt;p&gt;I am not familiar with using &lt;strong&gt;e.r.f&lt;/strong&gt; but i shall tell you the procedure. Take $u= \sqrt{x}$ and $dv = e^{-x}$. So you have $du = \frac{1}{2 \cdot \sqrt{x}} \ \rm{dx}$ and $v = -e^{-x}$. Now using the integration formula we have,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align*}&#xA; \int \sqrt{x} e^{-x} \ \textrm{dx} &amp;amp;= \Bigl[ u \cdot v ] - \int v \cdot \ \textrm{du} \\ &amp;amp; = \Bigl [ -\sqrt{x} \cdot e^{-x}\Bigr] + \frac{1}{2} \int \frac{e^{-x}}{\sqrt{x}} \ \textrm{dx}&#xA;\end{align*}&lt;/p&gt;&#xA;" LastActivityDate="2011-02-22T13:15:08.797" CommentCount="4" />
  <row Id="23227" PostTypeId="1" AcceptedAnswerId="23249" CreationDate="2011-02-22T13:41:12.830" Score="2" ViewCount="79" Body="&lt;p&gt;This is a continuation of &lt;a href=&quot;http://math.stackexchange.com/q/23009/1281&quot;&gt;my previous question&lt;/a&gt; and inspired by &lt;a href=&quot;http://math.stackexchange.com/questions/23009/from-sigma-algebra-on-product-space-to-sigma-algebra-on-component-space/23017#23017&quot;&gt;Arturo Margidin's reply&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose there are a collection of measurable spaces $(X_i, \mathbb{S}_i), i \in I$. Let $\mathbf{X}=\prod_{i\in I}X_i$, and let $\mathbb{S}=\prod_{i\in I} \mathbb{S}_i$ be the product $\sigma$-algebra.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Suppose $A_{i_0}\subseteq X_{i_0}$ is such that $A_{i_0}\times\prod_{j\neq i_0}X_j\in\mathbb{S}$. Does it follow that $A_{i_0}\in\mathbb{S}_{i_0}$?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;More generally, suppose we have a family of subsets, $A_i\subseteq X_i$, and we know that $A_j\in\mathbb{S}_j$ for all $j\neq i_0$ and that $\prod_{i\in I}A_i\in\mathbb{S}$. Does it follow that $A_{i_0}\in \mathbb{S}_{i_0}$?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Basically, I would like to reconstruct each $\mathbb{S}_i$&#xA;from $\prod_{i \in I} \mathbb{S}_i$.&#xA;If you have other approaches, please&#xA;don't hesitate to reply. In particular, Part 1 and Part 2 are attempts based on projection of measurable rectangles in the product $\sigma$-algebra. I was wondering if it is possible to go from sectioning a measurable set in the product $\sigma$-algebra?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Thanks and regards!&lt;/p&gt;&#xA;" OwnerUserId="1281" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-22T14:22:04.150" LastActivityDate="2011-03-24T18:37:58.557" Title="Further questions about from product $\sigma$-algebra to component ones" Tags="&lt;measure-theory&gt;" AnswerCount="1" CommentCount="7" FavoriteCount="1" />
  <row Id="23228" PostTypeId="1" AcceptedAnswerId="23235" CreationDate="2011-02-22T14:57:52.983" Score="18" ViewCount="710" Body="&lt;p&gt;In my classes I sometimes have a contest concerning who can write the largest number in ten symbols. It almost never comes up, but I'm torn between two &quot;best&quot; answers: a stack of ten 9's (exponents) or a 9 followed by nine factorial symbols. Both are undoubtedly huge, but I haven't been able to produce an argument that one is larger (surely they aren't equal). Any insight into which of these two numbers is bigger would be greatly appreciated. Thanks!&lt;/p&gt;&#xA;" OwnerUserId="7372" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2011-02-22T15:21:30.120" LastActivityDate="2011-05-14T14:03:39.863" Title="Which is bigger?" Tags="&lt;recreational-mathematics&gt;" AnswerCount="6" CommentCount="7" FavoriteCount="3" />
  <row Id="23230" PostTypeId="1" CreationDate="2011-02-22T15:19:38.010" Score="0" ViewCount="78" Body="&lt;p&gt;This is inspired by &lt;a href=&quot;http://math.stackexchange.com/questions/23009/from-sigma-algebra-on-product-space-to-sigma-algebra-on-component-space/23014#23014&quot;&gt;Carl Offner's reply to one of my previous questions&lt;/a&gt; and &lt;a href=&quot;http://math.stackexchange.com/questions/23070/how-to-understand-marginal-distribution&quot;&gt;my previous question about marginal and joint measures&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Given a measure $\mu$ on product&#xA;$\sigma$-algebra $\prod_{i \in I}&#xD;&#xA;    \mathbb{S}_i$ of a collection of&#xA;measurable spaces $(X_i,&#xD;&#xA;    \mathbb{S}_i), i \in I$, does there&#xA;exist a measure $\mu_i$ on each&#xA;component $\sigma$-algebra&#xA;$\mathbb{S}_i$, s.t. their product&#xA;$\prod_{i \in I} \mu_i$ is the given&#xA;measure $\mu$ on the product&#xA;$\sigma$-algebra?&lt;/li&gt;&#xA;&lt;li&gt;If no, what are some necessary and/or&#xA;sufficient conditions for the given&#xA;measure $\mu$ to have such a&#xA;decomposition?&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;When they exist, how to construct&#xA;the component measures $\mu_i$ from&#xA;$\mu$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, is this a viable way&#xA;by defining  $$\mu_i(A_i):=&#xD;&#xA;    \frac{\mu(A_i \times \prod_{j \in&#xD;&#xA;    I, j\neq i} X_i)}{\prod_{j \in I,&#xD;&#xA;    j\neq i} \mu_j(X_i)}, \forall A_i \in \mathbb{S}_i ?$$ If not,&#xA;when will it become viable?  ADDED: I asked this question, because obviously, the product and the division may not make sense in some cases. Also I actually made a mistake of circular definition, where I define $\mu_i$ in terms of $\mu_j, j\neq i$ which have to be defined in similar ways.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Thanks and regards!&lt;/p&gt;&#xA;" OwnerUserId="1281" LastEditorUserId="1281" LastEditorDisplayName="" LastEditDate="2011-02-22T19:18:05.157" LastActivityDate="2011-02-22T19:18:05.157" Title="Measure from on product $\sigma$-algebra to on component $\sigma$-algebras" Tags="&lt;measure-theory&gt;" AnswerCount="1" CommentCount="7" />
  <row Id="23231" PostTypeId="2" ParentId="23228" CreationDate="2011-02-22T15:23:48.753" Score="-3" ViewCount="" Body="&lt;p&gt;&amp;alefsym;&lt;sub&gt;9!!!!!!!!&lt;/sub&gt; is a lot bigger than either of those.&lt;/p&gt;&#xA;" OwnerUserId="5656" LastActivityDate="2011-02-22T15:23:48.753" CommentCount="5" />
  <row Id="23232" PostTypeId="1" AcceptedAnswerId="28314" CreationDate="2011-02-22T15:17:35.080" Score="0" ViewCount="154" Body="&lt;p&gt;I'm supposed to construct a DFA which accepts&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;{ w | w is a word except 'aa' and 'aaa' }&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Is this the correct &lt;a href=&quot;https://docs.google.com/drawings/pub?id=1jebzHyD265xhFif1OUr8l3YqscQTgGpM2ZiepCs8ZDs&amp;amp;w=1411&amp;amp;h=700&quot; rel=&quot;nofollow&quot;&gt;solution&lt;/a&gt;? The &lt;strong&gt;thick&lt;/strong&gt; line state is supposed to be the end state.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sry, somehow mixed two different exercises. Corrected.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT 2&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's the corrected &lt;a href=&quot;https://docs.google.com/drawings/pub?id=1jebzHyD265xhFif1OUr8l3YqscQTgGpM2ZiepCs8ZDs&amp;amp;w=1411&amp;amp;h=700&quot; rel=&quot;nofollow&quot;&gt;solution&lt;/a&gt;(?)!&lt;/p&gt;&#xA;" OwnerUserId="7374" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2011-03-21T05:31:41.567" LastActivityDate="2011-03-21T15:42:41.840" Title="Is this DFA correct" Tags="&lt;computer-science&gt;&lt;automata&gt;" AnswerCount="2" CommentCount="7" />
  <row Id="23233" PostTypeId="2" ParentId="23228" CreationDate="2011-02-22T15:39:06.660" Score="16" ViewCount="" Body="&lt;p&gt;Wolfram Alpha shows $$9^{9^9}\approx 10^{10^{8.56}}$$ while $$(9!)! \approx 10^{10^{6.26}}$$  Adding another character adds another 10 to the bottom of the stack without changing the upper exponent much at all.  It will even do the full stack of 10.  The exponents have 8.5678 atop the tower of 10's, while the factorials have 6.26949.  So the exponents win.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Added:  In Douglas Hofstadter's May, 1982 column &quot;On Number Numbness&quot; he declares these essentially equal.  For numbers of this size, the first thing you should look at is how many times you have to take a log to make it reasonable, which is 9 for both of them.  Then look at the number on top of the stack, which is the only one that matters.  So it is like 9.85678 compared with 9.626949, which are very close.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2011-05-14T14:03:39.863" LastActivityDate="2011-05-14T14:03:39.863" CommentCount="6" />
  <row Id="23234" PostTypeId="2" ParentId="19318" CreationDate="2011-02-22T15:44:14.827" Score="2" ViewCount="" Body="&lt;p&gt;In mathematics, we study structures of various sorts, and the relationships (or morphisms) between them, which we usually encode as functions of a special type. More generally, the objects and relationships or morphisms form what is called a category, which consists of the objects and morphisms, and the rule for how morphisms compose with each other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are many many important examples of categories, at least one for every branch of mathematics, and often quite a few. (Sets and functions, vector spaces and linear maps, topological spaces and continuous maps, metric spaces and short maps, rings and ring homomorphisms, graphs and graph homomorphisms, and so on.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we take any category at all, and examine a single object in it, then the morphisms from that object to itself which are invertible form a group, called the automorphism group of the object. This group abstractly measures the &quot;symmetries&quot; of that object, under the sort of transformations we're considering important. Symmetry has shown itself to be a powerful tool for reasoning in many circumstances -- it allows us to take an argument about one part of a structure and carry it throughout the structure to many other places where it applies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because essentially every object we study in mathematics has an automorphism group, it makes sense to study groups in general, so that we don't have to start afresh in each branch of mathematics when we want to consider symmetries of whatever it is that we're studying.&lt;/p&gt;&#xA;" OwnerUserId="3869" LastActivityDate="2011-02-22T15:44:14.827" />
  <row Id="23235" PostTypeId="2" ParentId="23228" CreationDate="2011-02-22T15:57:06.810" Score="22" ViewCount="" Body="&lt;p&gt;$n!$ grows more rapidly than $9^n$ (it grows approximately with $n^n$), so eventually it wins out; in fact, a few moments with Wolfram Alpha suggests that for all $n\gt 21$, $n! \gt 9^n$.  This means that the best solution is mixed: after the first exponentiation (since $9^9$ &lt;em&gt;is&lt;/em&gt; greater than $9!$), you're better off using factorials, giving the answer $9^9!!!!!!!!$.  (Also, note that we're all assuming that the correct parentheses here are implicit, since there's a big difference between $\left(9^9\right)^9$ and $9^{\left(9^9\right)}$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, if you're allowed to use more-or-less stock mathematical notation, you may want to have a look at &lt;a href=&quot;http://en.wikipedia.org/wiki/Knuth%27s_up-arrow_notation&quot; rel=&quot;nofollow&quot;&gt;Knuth's arrow notation&lt;/a&gt;; $9\uparrow\uparrow\uparrow\uparrow\uparrow\uparrow\uparrow\uparrow9$ is bigger by far than the rest of these.  (And numbers even larger than this have come up in mathematics - have a look, for instance, at &lt;a href=&quot;http://en.wikipedia.org/wiki/Graham%27s_number&quot; rel=&quot;nofollow&quot;&gt;Graham's Number&lt;/a&gt;, which most conveniently uses the arrow notation in its definition.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt; In fact it's pretty straightforward to prove that the answer in the first paragraph, $9^9!!!!!!!!$, is the best that can be done (with these operations).  First, as long as $a$ is at least two symbols (for this group of symbols), then $a^9 \lt 9^a\lt a!$, so adding a factorial symbol is always going to be better than adding a $9$ onto your tower.  But what about structures like $9!^{9!}$?  Well, if both $a$ and $b$ are at least two symbols long, then $a^b\lt \mathrm{max}(a,b)!!$; assuming for the moment that $b$ is larger (because if $a$ were larger, then $b^a\gt a^b$), then $b! \approx b^b$, and while this isn't enough to ensure that it's greater than $a^b$ (consider the case where $a$ = $b$ = $9!$), it's close enough that we &lt;em&gt;can&lt;/em&gt; be certain taking a second factorial will be larger - so those extra symbols you spent on exponentiation should retroactively have just gone into more exclamation points, and this is enough to guarantee that $9^9!!!!!!!!$ is the largest possible combination of these particular operations.&lt;/p&gt;&#xA;" OwnerUserId="785" LastEditorUserId="785" LastEditorDisplayName="" LastEditDate="2011-02-22T16:51:01.050" LastActivityDate="2011-02-22T16:51:01.050" CommentCount="3" />
  <row Id="23236" PostTypeId="2" ParentId="23232" CreationDate="2011-02-22T16:01:09.383" Score="3" ViewCount="" Body="&lt;p&gt;Your automaton is not a DFA (consider fourth state from the left on the top row).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, your automaton rejects other words besides $aa$ and $aaa$. Can you see why?&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2011-02-22T16:01:09.383" CommentCount="4" />
  <row Id="23237" PostTypeId="2" ParentId="23220" CreationDate="2011-02-22T16:05:46.237" Score="3" ViewCount="" Body="&lt;p&gt;By the way, if you've continued your determinant (in order to draw a &lt;em&gt;real&lt;/em&gt; one) in the same vein:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;D =&#xD;&#xA;\begin{vmatrix}&#xD;&#xA;1 &amp;amp; 2 &amp;amp; 3 &amp;amp; 4  \\\&#xD;&#xA;5 &amp;amp; 6 &amp;amp; 7 &amp;amp; 8  \\\&#xD;&#xA;9 &amp;amp; 10 &amp;amp; 11 &amp;amp; 12   \\\&#xD;&#xA;13 &amp;amp; 14 &amp;amp; 15 &amp;amp; 16&#xD;&#xA;\end{vmatrix} \ ,&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the answer would have been easy:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;D =&#xD;&#xA;\begin{vmatrix}&#xD;&#xA;1 &amp;amp; 2 &amp;amp; 3 &amp;amp; 4  \\\&#xD;&#xA;5 &amp;amp; 6 &amp;amp; 7 &amp;amp; 8  \\\&#xD;&#xA;4 &amp;amp; 4 &amp;amp; 4 &amp;amp; 4   \\\&#xD;&#xA;4 &amp;amp; 4 &amp;amp; 4 &amp;amp; 4&#xD;&#xA;\end{vmatrix}&#xD;&#xA;= 0 \ .&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;:-)&lt;/p&gt;&#xA;" OwnerUserId="664" LastEditorUserId="664" LastEditorDisplayName="" LastEditDate="2011-02-23T08:33:32.477" LastActivityDate="2011-02-23T08:33:32.477" CommentCount="3" />
  <row Id="23238" PostTypeId="2" ParentId="23208" CreationDate="2011-02-22T16:16:22.727" Score="0" ViewCount="" Body="&lt;p&gt;Well, one should observe that $u \cdot (v \times w)$ is equal to the determinant, in an oriented orthonormal basis, of the matrix U with $u, v, w$ as its columns. So, if we assume the determinant is multiplicative, we have $Tu \cdot (Tv \times Tw) = \det TU = \det T \det U = (\det T) u \cdot (v \times w)$, which is what we wanted to prove.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, perhaps the point of the question is to prove that the determinant &lt;em&gt;is&lt;/em&gt; multiplicative. This is a little more difficult, and essentially amounts to proving certain facts about alternating multilinear forms.&lt;/p&gt;&#xA;" OwnerUserId="5191" LastActivityDate="2011-02-22T16:16:22.727" />
  <row Id="23239" PostTypeId="2" ParentId="6863" CreationDate="2011-02-22T16:44:22.967" Score="0" ViewCount="" Body="&lt;p&gt;The Kalman filter can be used as is for two separate sensors, just consider a measurement at time $t$ to be&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$y_t = [y^\mathrm{accel}_t, y^\mathrm{gyro}_t]&amp;#39;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and write the appropriate sensor noise covariance matrix ($H$). Your levels of confidence of the &quot;good and bad&quot; parts of the sensors, encoded in $H$, will automatically &quot;fuse&quot; the sensor measurements in the sense the Kalman filter will give the optimal linear estimator of the state, given your measurements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, the farther your measurement model deviates from a linear function with white Gaussian noise, the less successful the Kalman filter will be.&lt;/p&gt;&#xA;" OwnerUserId="5799" LastActivityDate="2011-02-22T16:44:22.967" />
  <row Id="23240" PostTypeId="1" CreationDate="2011-02-22T16:57:35.220" Score="0" ViewCount="42" Body="&lt;p&gt;Let $X$ be a complex manifold equipped with a smooth hermitian metric $h$. We can define a sub-fibration $B \to X$ of the tangent bundle $T_X$ by requiring that the fiber over a point be the unit ball in $T_X$ at that point, i.e.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ B_x = \{ \xi \in T_{X,x} \, | \, \| \xi \|_{h(x)} = 1 \}. $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to see that if $X$ is compact, then the fibration $B$ is compact in the total space of $T_X$. This is just a global version of the usual fact that the unit ball in a normed finite dimensional vector space is compact.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can prove this by using the (horrible) definition of $T_X$ as the disjoint union of the stalks $T_{X,x}$ modulo an equivalence relation. As each $B_x$ is compact, their disjoint union is compact in the disjoint union of the stalks by Tychonoff, and dividing by the equivalence relation is a continuous map so $B$ is compact in the total space of $T_X$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Isn't there a nicer way to see this? I ask both because I really don't like the definition of the tangent space as a disjoint union of stalks, and because I have to prove a similar lemma for the relative tangent space associated to deformations of a manifold $X$. A similar proof works in that case, but it is quite dirty.&lt;/p&gt;&#xA;" OwnerUserId="3225" LastActivityDate="2011-02-22T17:54:29.777" Title="The unit ball fibration in a tangent bundle" Tags="&lt;complex-analysis&gt;&lt;differential-geometry&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="23241" PostTypeId="1" AcceptedAnswerId="23243" CreationDate="2011-02-22T17:04:39.927" Score="1" ViewCount="96" Body="&lt;p&gt;Let $p$ be a prime and let $F=\mathbb{Z}/p\mathbb{Z}$. Can a nonzero multivariate polynomial $f\in F[x_1,....,x_n]$ such that $\mathrm{deg}_if&amp;lt; p$ for all $i=1,\ldots,n$ be identically zero as a polynomial function $F^n\to F$, where $\mathrm{deg}_if$ stands for the degree of $f$ in $x_i$?&lt;/p&gt;&#xA;" OwnerUserId="2261" LastEditorUserId="2261" LastEditorDisplayName="" LastEditDate="2011-02-22T21:06:48.560" LastActivityDate="2011-02-22T21:06:48.560" Title="Identically zero multivariate polynomial function" Tags="&lt;commutative-algebra&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="23242" PostTypeId="1" AcceptedAnswerId="23252" CreationDate="2011-02-22T17:18:12.777" Score="0" ViewCount="94" Body="&lt;p&gt;Variable z varies from -50.0 to 18.0. Find the equation that gives d such that when z is equal to -50 d is equal to 0.1 and when z is 18.0 d is 1.0 and d values varies from 0.1 to 1 uniformly depending on the value of z with the boundary conditions given above.&lt;/p&gt;&#xA;" OwnerUserId="7378" LastActivityDate="2011-02-22T18:10:20.267" Title="What is the math equation that satisfies conditions given below?" Tags="&lt;algebra&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="23243" PostTypeId="2" ParentId="23241" CreationDate="2011-02-22T17:39:22.907" Score="4" ViewCount="" Body="&lt;p&gt;No.  The phenomenon of polynomials over fields which are not the zero polynomial but evaluate to the zero function is studied in loving detail in $\S 2.1$ of &lt;a href=&quot;http://math.uga.edu/~pete/4400ChevalleyWarning.pdf&quot; rel=&quot;nofollow&quot;&gt;these notes&lt;/a&gt;: in particular, the result you want is Theorem 8.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also agree with Qiaochu's comment: it would be possible to make much less of a production of this by simply inducting on the number of variables, beginning with the basic fact that no polynomial with coefficients in an integral domain can have more roots than its degree (except the polynomial which has all coefficients zero, if you take the convention that this polynomial has degree $-\infty$).  I have my reasons for the more elaborate treatment given in my notes: in particular, an at least somewhat careful analysis of the situation leads directly to a proof of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Chevalley%E2%80%93Warning_theorem&quot; rel=&quot;nofollow&quot;&gt;Chevalley-Warning Theorem&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="299" LastActivityDate="2011-02-22T17:39:22.907" />
  <row Id="23244" PostTypeId="2" ParentId="23228" CreationDate="2011-02-22T17:41:19.190" Score="14" ViewCount="" Body="&lt;p&gt;A thorough study of this question is given in the article &quot;&lt;a href=&quot;http://www.jstor.org/stable/27642031&quot; rel=&quot;nofollow&quot;&gt;Exponential vs. Factorial&lt;/a&gt;&quot; by Velleman (&lt;em&gt;American Mathematical Monthly&lt;/em&gt; Vol. 113, No. 8, Oct. 2006, pp. 689-704).  In the motivating example there are 5 characters instead of 10.  Of course, the conclusion is the same as in Steven Stadnicki's answer, that $9^9!!!!!!!!$ (or $9^9!!!$ in the case of 5 characters) is the largest possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The article also addresses your original question, proving that the exponential tower of $n$ $9$s is always larger than a $9$ with $n-1$ factorials applied (and many more general statements).&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2011-02-22T17:54:22.430" LastActivityDate="2011-02-22T17:54:22.430" CommentCount="2" />
  <row Id="23245" PostTypeId="2" ParentId="23240" CreationDate="2011-02-22T17:54:29.777" Score="1" ViewCount="" Body="&lt;p&gt;&lt;b&gt;Hint&lt;/b&gt;: Here is an outline of a possible proof.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Do the problem in the special case that the ball fibration is trivial, i.e., isomorphic to a product.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Convince yourself that the ball fibration is locally trivial.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) Since the base is compact, you can find a finite open cover $\{U_i\}$ such that the restriction of the fibration to each $U_i$ is trivial.  Now use the fact that a finite union of compact sets is compact.&lt;/p&gt;&#xA;" OwnerUserId="299" LastActivityDate="2011-02-22T17:54:29.777" CommentCount="10" />
  <row Id="23246" PostTypeId="1" AcceptedAnswerId="23253" CreationDate="2011-02-22T17:55:45.503" Score="2" ViewCount="53" Body="&lt;h2&gt;Background&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;I am conducting a quantitative synthesis (meta-analysis) that includes previously published data. For the model that I am using, an exact or inflated estimate of SE is acceptable.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Question&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Given a 95% Confidence Interval (measured from mean to upper or lower confidence limit), $\alpha$, and $n$, is it correct to estimate the standard error thus:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$SE = \frac{CI}{t_{(1-\alpha/2,n)}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is based on the calculation by &lt;a href=&quot;http://eskes.psychiatry.dal.ca/Files/2003_-_Saville_-_Basic_statistics_and_the_inconsistency_of_m.pdf&quot; rel=&quot;nofollow&quot;&gt;Saville, 2003&lt;/a&gt;, but most calculations of CI using the Z score. Unfortunately, $Z=f(SE)$, and I don't think it can be otherwise estimated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;note: this question is a subset of one that has remained unanswered on &lt;a href=&quot;http://stats.stackexchange.com/questions/2917/are-these-formulas-for-transforming-p-lsd-msd-hsd-ci-to-se-as-an-exact-or-in&quot;&gt;stats.stackexchange&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3733" LastEditorUserId="6460" LastEditorDisplayName="" LastEditDate="2011-02-22T18:19:36.927" LastActivityDate="2011-02-22T18:32:34.873" Title="How to estimate $SE$ from $95\%$CI, $\alpha $ and $n$ ?" Tags="&lt;probability&gt;&lt;statistics&gt;" AnswerCount="1" />
  <row Id="23248" PostTypeId="2" ParentId="23220" CreationDate="2011-02-22T17:56:09.933" Score="0" ViewCount="" Body="&lt;p&gt;Determinants are defined only for squared matrix. If you are sure about your question then simply the determinant for your matrix does not exist. &lt;/p&gt;&#xA;" OwnerUserId="6335" LastActivityDate="2011-02-22T17:56:09.933" CommentCount="2" />
  <row Id="23249" PostTypeId="2" ParentId="23227" CreationDate="2011-02-22T17:57:56.043" Score="1" ViewCount="" Body="&lt;p&gt;Consider $I$ to be countable as suggested by Arturo Magidin. We show part (2). Define&#xA;$$&#xD;&#xA;{\cal G} = \{\pi_i^{-1}(A):A\in \mathbb S_i, i\in I\}.&#xD;&#xA;$$&#xA;The product $\sigma$-field $\mathbb S$ is generated by $\cal G$. Fix $i_0$, and pick arbitrary $x_i\in A_i$ for every $i\in I, i\neq i_0$. (Note that one needs to assume $\mathbb S_i\ni A_i\neq \emptyset$, otherwise the statement is obviously not correct.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Define a mapping &#xA;$$&#xD;&#xA;T_{i_0} : X_{i_0}\ni x_{i_0} \mapsto \{x_i\}_{t\in I}\in {\bf X}.&#xD;&#xA;$$&#xA;Observe that if $T_{i_0}$ is $\mathbb S_{i_0}/\mathbb S$ measurable, then $\prod_{i\in I}A_i\in\mathbb S$ implies that $T_{i_0}^{-1}(\prod_{i\in I}A_i) = A_{i_0}\in\mathbb S_{i_0}$, and we are done. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It remains to show that $T_{i_0}$ is measurable and it suffices to show that $T_{i_0}^{-1}(G)\in\mathbb S_{i_0}$ for all $G\in \cal G$. This is true since for all $i\in I, A\in \mathbb S_i$, $T_{i_0}^{-1}(\pi_i^{-1}(A))$ equals (1) $A$ if $i=i_0$, (2) $X_{i_0}$ if $i\neq i_0, x_i\in A$, and (3) $\emptyset$ if $i\neq i_0, x_i\notin A$.&lt;/p&gt;&#xA;" OwnerUserId="5836" LastActivityDate="2011-02-22T17:57:56.043" CommentCount="2" />
  <row Id="23250" PostTypeId="2" ParentId="23157" CreationDate="2011-02-22T18:07:40.740" Score="1" ViewCount="" Body="&lt;p&gt;For a discrete probability measure, each point that has positive measure is an atom. For example, a Bernoulli distribution has atoms 0 and 1. Then, $G$ can be seen as a random discrete probability measure with atoms at $\theta_k$, with (random) mass $\beta_k$. To draw a sample according to $G$, it means to draw a point $\theta_k$ according to the weight $\beta_k$. &lt;/p&gt;&#xA;" OwnerUserId="5836" LastActivityDate="2011-02-22T18:07:40.740" />
  <row Id="23251" PostTypeId="1" AcceptedAnswerId="23257" CreationDate="2011-02-22T18:09:25.103" Score="0" ViewCount="91" Body="&lt;p&gt;Is it true that every inner product in $\ l_2 $ is of the form $\langle x,y\rangle_a =\sum_{n=1} ^ {\infty} {a_n x_n y_n}$ ? (Of course $\ x=(x_n) , y=(y_n) $ are in $\ l_2 $ .)&lt;/p&gt;&#xA;" OwnerUserId="5422" LastEditorUserId="5422" LastEditorDisplayName="" LastEditDate="2011-03-01T18:38:33.603" LastActivityDate="2011-03-01T18:38:33.603" Title="Form of the inner product in $\ l_2$." Tags="&lt;functional-analysis&gt;&lt;inner-product-space&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="23252" PostTypeId="2" ParentId="23242" CreationDate="2011-02-22T18:10:20.267" Score="1" ViewCount="" Body="&lt;p&gt;As $d$ changes from $0.1$ to $1$ (a change of $1-0.1=0.9$), $z$ changes from $-50$ to $18$ (a change of $18--50=68$), so for every change of $1$ in $z$, $d$ changes by $\frac{0.9}{68}$.  When $z=-50$, $d=0.1$.  So, an equation relating $z$ and $d$ is $$d-0.1=\frac{0.9}{68}(z+50).$$&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2011-02-22T18:10:20.267" />
  <row Id="23253" PostTypeId="2" ParentId="23246" CreationDate="2011-02-22T18:12:22.947" Score="5" ViewCount="" Body="&lt;p&gt;I would say yes.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If your sample isn't large enough you should be using $t$ scores to find the confidence interval rather than $z$ scores, anyway.  (Some folks argue that you should &lt;em&gt;always&lt;/em&gt; use $t$-scores, since the assumption required for using $z$-scores - that you know the population standard deviation $\sigma$ - is rarely satisfied.)  With $t$ scores, and using your notation, the formula for $CI$, half of the width of the confidence interval for a population mean, is known to be&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$CI = t_{(1 - \alpha/2,n)} SE$, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;so solving that for $SE$ yields the expression you have in your question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update:&lt;/strong&gt; Now that Henry has fixed the link, I see that the expression I give is exactly that in Saville's paper.  It's a well-known expression for the confidence interval using $t$-scores, and so, again, I don't see why it would be invalid to use it.  For example, see just about any introductory statistics text, such as Moore, McCabe, and Craig, &lt;em&gt;Introduction to the Practice of Statistics&lt;/em&gt;, 6th edition, p. 420.&lt;/p&gt;&#xA;" OwnerUserId="2370" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2011-02-22T18:32:34.873" LastActivityDate="2011-02-22T18:32:34.873" />
  <row Id="23254" PostTypeId="2" ParentId="23230" CreationDate="2011-02-22T18:14:43.817" Score="2" ViewCount="" Body="&lt;p&gt;The answer to (1) is certainly no; not every measure on a product space is a product measure, not even for a finite product.  Consider for example the following: let $m$ be Lebesgue measure on $[0,1]$, let $F : [0,1] \to [0,1]^2$ be given by $F(x) = (x,x)$, and let $\mu = m \circ F^{-1}$ be the pushforward measure on the product $[0,1]^2$ (with its product $\sigma$-algebra of course).  $\mu$ then is effectively 1-dimensional Lebesgue measure on the diagonal of $[0,1]$.  Now $\mu$ cannot be a product measure.  For suppose $\mu = \mu_1 \times \mu_2$.  Let $0 &amp;lt; a &amp;lt; 1$; it's clear that we have $\mu((0,a)^2) = a &amp;gt; 0$ and $\mu((a,1)^2) = 1-a &amp;gt; 0$.  Thus $\mu_i((0,a)) &amp;gt; 0$ and $\mu_i((a,1)) &amp;gt; 0$ for $i=1,2$.  But $\mu((0,a) \times (a,1)) = \mu((a,1) \times (0,a)) = 0$, a contradiction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is best thought of in terms of probability theory: a probability measure $\mu$ on $\mathbb{R}^d$ gives the (joint) distribution of a random vector $(X_1, \dots, X_d)$.  If $\pi_i : \mathbb{R}^d \to \mathbb{R}$, $i=1,\dots,d$ is the projection onto the $i$'th component, then $\mu_i = \mu \circ \pi_i^{-1}$ is the marginal distribution of $X_i$.  But $\mu$ is a product measure iff $\{X_1, \dots, X_d\}$ are independent.  In our example, $\mu$ is the joint distribution of $(U,U)$, where $U \sim U(0,1)$; obviously $U$ is not independent of itself.&lt;/p&gt;&#xA;" OwnerUserId="822" LastActivityDate="2011-02-22T18:14:43.817" />
  <row Id="23255" PostTypeId="2" ParentId="23228" CreationDate="2011-02-22T18:19:37.617" Score="-2" ViewCount="" Body="&lt;p&gt;Define an operation facexp where facexp(9) is equal to: 9^8^7^6^5^4^3^2^1.&#xA;The hats should be read as &quot;to the power&quot; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Devise a positional notation for the operation, so it doesn't need a '!' or similar to take up our allotment of characters. Subscripting would work for that, just as superscripting works for powers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Make a string consisting of ten nines, each, after the first, subscripted from the position of its predecessor.&lt;/p&gt;&#xA;" OwnerUserId="7380" LastActivityDate="2011-02-22T18:19:37.617" CommentCount="4" />
  <row Id="23256" PostTypeId="2" ParentId="23241" CreationDate="2011-02-22T18:33:46.663" Score="0" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\ $ By induction the leading coefficient of $\rm\ f(x_n) \ $ is nonzero for some  $\rm (x_1,\cdots,x_{n-1})\in  F^{n-1}$ Evaluating $\rm\:f\:$ at this point produces a &lt;em&gt;univariate&lt;/em&gt; polynomial $\rm\ 0 \ne \bar f(x_n)\ \in\ \mathbb F_p[x_n]\ $ of degree $\rm&amp;lt; p\:.\ $  So some $\rm\:x_n\:$ isn't a root of $\rm\:\bar f\:,\:$ since a nonzero poly over a domain has no more roots than its degree.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Remark&lt;/strong&gt;  $\ $ In fact the above root-bound property completely characterizes integral domains: the ring $\rm\: D\:$ is a domain $\iff$ every nonzero polynomial $\rm\ f(x)\in D[x]\ $ has at most $\rm\ deg\ f\ $ roots in $\rm\:D\:.\:$ For the simple proof see my &lt;a href=&quot;http://math.stackexchange.com/questions/4597/simple-explanation-and-examples-of-the-miller-rabin-primality-test/4604#4604&quot;&gt;post here&lt;/a&gt;, where I illustrate it constructively in $\rm\: \mathbb Z/m\: $ by showing that, given any $\rm\:f(x)\:$ with more roots than its degree, we can quickly compute a nontrivial factor of $\rm\:m\:$ via a $\rm\:gcd\:$. The quadratic case of this result is at the heart of many integer factorization algorithms, which attempt to factor $\rm\:m\:$ by searching for a nontrivial square root in $\rm\: \mathbb Z/m\:,\:$ e.g. a square root of $1$ that is not $\:\pm 1$.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2011-02-22T18:56:00.803" LastActivityDate="2011-02-22T18:56:00.803" />
  <row Id="23257" PostTypeId="2" ParentId="23251" CreationDate="2011-02-22T18:34:47.037" Score="4" ViewCount="" Body="&lt;p&gt;I'm not sure I understand the question. First of all, every bounded sequence $a = (a_{n})$ with $0 \lt a_{n}$ for all $n$ will give a scalar product on $\ell^{2}$ by $\langle x,y \rangle_{a} = \sum_{n = 1}^{\infty} a_{n} x_{n} y_{n}$. Not every bounded sequence is square-summable. For instance, the usual scalar product is not of the form you're asking about.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moreover, for every bounded and injective operator $A : \ell^{2} \to \ell^{2}$ you get a scalar product by setting $\langle x,y \rangle_{A} = \langle Ax, Ay \rangle$.&lt;/p&gt;&#xA;" OwnerUserId="5363" LastActivityDate="2011-02-22T18:34:47.037" CommentCount="1" />
  <row Id="23258" PostTypeId="1" AcceptedAnswerId="23275" CreationDate="2011-02-22T19:03:00.450" Score="3" ViewCount="125" Body="&lt;p&gt;Hello,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;let $H$ be a separable Hilbert space. Let $(e_i)_i$ be an orthonormal basis. For any bounded linear map $T$ we write, whenever possible&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$tr \, T := \sum_{i}^{\infty} \langle T e_i, e_i \rangle$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let $L(H)$ be the set of bounded operators, $N(H)$ the set of nuclear operators. We want to show&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$i: L(H) \rightarrow N(H)&amp;#39; , T \mapsto ( S \mapsto tr(TS))$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is an isometric isomorphism.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the ideal property of the nuclear operators, it is easy to see that $i$ is well-defined, linear and bounded with $|i| \leq 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Recall that the operators $\langle \cdot, e_i \rangle e_j$ form a basis of $N(H)$. You easily see the operator is injective (if the induced functional is zero, it's preimage must have been zero,too) and surjective (simply build up an operator $T \in L(H)$ which induces a given functional in $N(H)$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Question:&#xA;How can I finish the proof with showing that $i$ is an isometry?&lt;/p&gt;&#xA;" OwnerUserId="3557" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-22T19:58:33.783" LastActivityDate="2011-02-22T20:43:16.850" Title="On the isometry between bounded linear operators and the dual of nuclear linear operators " Tags="&lt;functional-analysis&gt;&lt;operator-theory&gt;" AnswerCount="2" />
  <row Id="23259" PostTypeId="1" AcceptedAnswerId="23261" CreationDate="2011-02-22T19:11:31.370" Score="1" ViewCount="191" Body="&lt;p&gt;Or, in general, what does the magnitude of the cross product mean? How would you prove or disprove this?&lt;/p&gt;&#xA;" OwnerUserId="7382" LastActivityDate="2011-02-23T11:15:20.223" Title="Is the cross product of two unit vectors itself a unit vector?" Tags="&lt;linear-algebra&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="23260" PostTypeId="2" ParentId="23228" CreationDate="2011-02-22T19:13:39.523" Score="-2" ViewCount="" Body="&lt;p&gt;Former one is bigger 9 with exponents&lt;/p&gt;&#xA;" OwnerUserId="7383" LastActivityDate="2011-02-22T19:13:39.523" CommentCount="1" />
  <row Id="23261" PostTypeId="2" ParentId="23259" CreationDate="2011-02-22T19:13:43.387" Score="7" ViewCount="" Body="&lt;p&gt;No - for example, the cross product of any unit vector with itself is 0. In general, the magnitude of the cross product of vectors $\vec{a}$ and $\vec{b}$ is $$|\vec{a}\times\vec{b}|=|\vec{a}||\vec{b}|\sin(\theta)$$&#xA;where $\theta$ is the angle between the vectors $\vec{a}$ and $\vec{b}$. Thus, the cross product of two unit vectors $\vec{u}$ and $\vec{v}$ is itself a unit vector if and only if $\vec{u}$ and $\vec{v}$ are orthogonal, i.e. meet at right angles (this makes $\sin(\theta)=\sin(\frac{\pi}{2})=1$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As to the general interpretation of the magnitude of the cross product, see &lt;a href=&quot;http://en.wikipedia.org/wiki/Cross_product#Geometric_meaning&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The magnitude of the cross product can be interpreted as the positive area of the parallelogram having $a$ and $b$ as sides.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="264" LastEditorUserId="264" LastEditorDisplayName="" LastEditDate="2011-02-22T19:20:04.330" LastActivityDate="2011-02-22T19:20:04.330" CommentCount="2" />
  <row Id="23262" PostTypeId="1" AcceptedAnswerId="23282" CreationDate="2011-02-22T19:16:38.727" Score="6" ViewCount="209" Body="&lt;p&gt;Similar questions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://math.stackexchange.com/questions/17117/chance-of-7-of-a-kind-with-10-dice&quot;&gt;Chance of 7 of a kind with 10 dice&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://math.stackexchange.com/questions/17208/probability-of-getting-exactly-k-of-a-kind-in-n-rolls-of-m-sided-dice-wher&quot;&gt;Probability of getting exactly $k$ of a kind in $n$ rolls of $m$-sided dice, where $k\leq n/2$.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Probability was never my thing, so please bear with me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've reviewed the threads above to the best of my ability, but I still wonder how to go about finding a match of 3 from 7 dice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;At least three match, but no more (two sets of three is okay, a set of three and a set of four is not):&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(a) : $ \frac{6  \binom{7}{3} 5^4}{6^7}  $ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the other discussions, this wasn't desired since it would allow for a second triple to occur, or even a quadruple. &#xA;Odds of a quadruple with the remaining 4 dice:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(b) : $(1/5)^4 $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, the probability that from rolling 7 dice that there is at least three that match, and no more than three, would be:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(c) : $ \frac{6  \binom{7}{3} * 5^4}{6^7}- (1/5)^4 $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Exactly two sets of three:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(d) : $ \frac{6  \binom{7}{3}  \binom{4}{3} \binom{1}{1}}{ 6^7} $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe?  My thought process was that if $\binom{7}{3}$ will give me a set of three, then with the remaining 4, I could pick 3 yielding $\binom{4}{3}$ with 1 leftover.  I realize this is probably wrong.  Why? What would be the proper way to go about this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Exactly one set of three:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then to find the probability that there is one and only one set of three from 7 dice, we could take the probability of one or more sets of three (c) and subtract the probability of exactly two sets (d), for:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ \frac{6  \binom{7}{3}  5^4}{ 6^7} - (1/5)^4 - \frac{6  \binom{7}{3}  \binom{4}{3} \binom{1}{1} }{ 6^7} $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(e) : $  \left(\frac{6  \binom{7}{3}}{ 6^7}\right) \left( 5^4 -  \binom{4}{3} \binom{1}{1} \right) - (1/5)^4 $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this at all on the right path?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS.&#xA;Sorry about the syntax, but I couldn't figure out how to make the standard nCr() symbol with MathJaX.&lt;/p&gt;&#xA;" OwnerUserId="7339" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-22T20:42:17.707" LastActivityDate="2011-02-23T01:22:05.143" Title="Probability of 3 of a kind with 7 dice" Tags="&lt;probability&gt;&lt;combinatorics&gt;" AnswerCount="2" CommentCount="5" />
  <row Id="23263" PostTypeId="2" ParentId="23259" CreationDate="2011-02-22T19:16:48.460" Score="-3" ViewCount="" Body="&lt;p&gt;yeap! The cross product of two unit vectors is a unit vector&lt;/p&gt;&#xA;" OwnerUserId="7383" LastActivityDate="2011-02-22T19:16:48.460" CommentCount="3" />
  <row Id="23264" PostTypeId="1" AcceptedAnswerId="23268" CreationDate="2011-02-22T19:17:03.200" Score="0" ViewCount="31" Body="&lt;p&gt;Let's say that I have a variable $j$ defined by the following formula:&#xA;$$j=\frac{n(n+2) + m}{2}$$&#xA;where $n$ and $m$ are two parameters, both integers, that satisfy the following conditions:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$n\in \left[0,1,2,...\right]$&lt;/li&gt;&#xA;&lt;li&gt;$m\in\left[-n,n\right]$&lt;/li&gt;&#xA;&lt;li&gt;$\left(n-m\right)$ even&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;We know that, subject to the above conditions, there exists a bijection between $j$ and the tuples $\left(m,n\right)$. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Is there a way of  expressing $m$ and $n$ as (hopefully compact) explicit functions of (only) $j$?  &lt;/li&gt;&#xA;&lt;li&gt;How would one proceed to find such a formula?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="5519" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-22T19:26:31.170" LastActivityDate="2011-02-22T20:13:14.870" Title="Reversing bijections defined via conditional expressions" Tags="&lt;functions&gt;&lt;functional-equations&gt;" AnswerCount="2" />
  <row Id="23266" PostTypeId="2" ParentId="22725" CreationDate="2011-02-22T19:22:49.270" Score="0" ViewCount="" Body="&lt;p&gt;Imagine the classic distance vs time graph typically used to teach derivatives in schools. Speed = change of distance/time taken for the change or s = &amp;Delta;d/&amp;Delta;t . Now what is more interesting is to know the speed at a particular point t , ie; function s(t). Now using the above definition blindly it appears like speed at a point should be infinite, since s=&amp;Delta;d/&amp;Delta;t and &amp;Delta;t=0 at a point. Now that is absurd since you know there is a finite speed at each point if you are driving.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's where infinitesimal or limits comes into picture, instead of the earlier definition let s(t) be &amp;Delta;d/&amp;Delta;t as &amp;Delta;t -&gt; 0, or it as t becomes infinitesimally small. Think of this being a very small time interval that you cannot really measure the size of it , the moment you measure the size of it you can make it smaller than that (That is a property of real numbers , purposely skipping the math part here). Now here &amp;Delta;t != 0 since it is never the same point. Also the only value of speed which can satisfy the above condition is speed at the point t for if it not the speed at time t and is the average of the speed on the interval then I can make the interval still smaller by definition of infinitesimally small and contradict that. Thus the introduction of 'infinitesimally small interval' concept suddenly gives us a tool to define the value of a function at a point and we can solve for limits to get those values. This also defines fundamentally what a derivative is.&lt;/p&gt;&#xA;" OwnerUserId="7379" LastActivityDate="2011-02-22T19:22:49.270" />
  <row Id="23267" PostTypeId="1" CreationDate="2011-02-22T19:26:49.517" Score="1" ViewCount="34" Body="&lt;p&gt;Suppose I have a computer application which does the following&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Receives requests and forwards them to an outside service&lt;/li&gt;&#xA;&lt;li&gt;Processes responses from the outside service (one response per request)&lt;/li&gt;&#xA;&lt;li&gt;Sends periodic heartbeat messages to the outside service, and the outside service response to each message.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I would like to answer the following question.  In a given interval of time (t0, t1), what is the probability that the application has two things to process (any combination of requests from the client or responses from the service).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm assuming that the requests are generated by some stochastic random process &lt;em&gt;G&lt;/em&gt;, and that the round trip time from the application to the service follows some distribution &lt;em&gt;RTT&lt;/em&gt;.  Additionally, the heartbeats are generated by the application according to some uniform distribution &lt;em&gt;H&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure how to put this together, though.  I had started by not considering the heartbeats or the round trip times, and just assuming that requests and responses were generated by two independent Poisson processes.  In this simplified case, answering my question is straightforward.  I would like to see if the answer changes much with the more complicated approach, or if what I'm already doing is a fair approximation.&lt;/p&gt;&#xA;" OwnerUserId="7384" LastActivityDate="2011-02-22T19:26:49.517" Title="Estimating resource contention with stochastic proccesses?" Tags="&lt;statistics&gt;&lt;stochastic-processes&gt;" />
  <row Id="23268" PostTypeId="2" ParentId="23264" CreationDate="2011-02-22T19:34:14.577" Score="2" ViewCount="" Body="&lt;p&gt;Since $2j$ depends quadratically on $n$, you basically need the square root of $2j$ to get $n$. To get the limits exactly right, I'd try $n=\lfloor \sqrt{2j+a}+b\rfloor$ and then determine $a$ and $b$ so that everything comes out right. Once you have $n$, it's easy to get $m$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Update: You can systematically determine $a$ and $b$ by looking at what happens at the boundaries. You need&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\lfloor \sqrt{n(n+2)-n+a}+b\rfloor=n$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\lfloor \sqrt{(n-1)((n-1)+2)+(n-1)+a}+b\rfloor=n-1\;.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The difference between the two arguments of the square root is $2$. To avoid rounding problems you can choose $a$ and $b$ such that the argument of the floor function is the integer $n$ when the argument of the square root is in the middle between the two:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sqrt{n^2+n-1 + a} + b = n\;,$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$n^2+n-1+a=(n-b)^2\;,$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$n-1+a=-2bn+b^2\;.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To make this true for all $n$, you can equate the constant terms and the coefficients of $n$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$1=-2b\;,$$&#xA;$$-1+a=b^2\;.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That yields $a=\frac{5}{4}$ and $b=-\frac{1}{2}$, so the formula for $n$ is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$n=\lfloor \sqrt{2j+\frac{5}{4}}-\frac{1}{2}\rfloor\;.$$&lt;/p&gt;&#xA;" OwnerUserId="6622" LastEditorUserId="6622" LastEditorDisplayName="" LastEditDate="2011-02-22T20:13:14.870" LastActivityDate="2011-02-22T20:13:14.870" />
  <row Id="23269" PostTypeId="1" AcceptedAnswerId="23272" CreationDate="2011-02-22T19:44:29.307" Score="3" ViewCount="272" Body="&lt;p&gt;Hello,&#xA;I'm having some trouble while trying to understand one task..&#xA;The task is as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ddot{x}(t) + \dot{x}(t) + 2x(t) = \sin(\omega t)$$&#xA;where $x(0) = 7, t\geq 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The solution is in the following form:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x(t) = f(t) + A\sin(\omega t + \varphi)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And the task is: find $\omega$ so that $A$ is max.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My understanding of this is that $f(t)$ is the solution of the homogeneous differential equation and the rest is the special solution of the nonhomogeneous  equation. Still that does not give me any clue about how to evaluate the relationship between $A$ and $\omega$.&#xA;Any clues?&lt;/p&gt;&#xA;" OwnerUserId="6485" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-22T19:58:58.307" LastActivityDate="2011-02-23T00:36:10.753" Title="Differential equation, quite weird task.." Tags="&lt;differential-equations&gt;" AnswerCount="2" CommentCount="0" />
  <row Id="23270" PostTypeId="1" AcceptedAnswerId="23486" CreationDate="2011-02-22T19:54:55.337" Score="0" ViewCount="71" Body="&lt;p&gt;This is a question about the proof of Proposition 1.4 in Farb and Margalit's &quot;Primer on Mapping Class Groups&quot; (in v. 5.0, it is on page 37 in the PDF, which you can download &lt;a href=&quot;http://www.math.uchicago.edu/~margalit/mcg/mcgv50.pdf&quot; rel=&quot;nofollow&quot; title=&quot;here&quot;&gt;here&lt;/a&gt;).  The proposition states&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Let $\alpha$ be a non-nullhomotopic simple closed curve on the (hyperbolic) surface $S$; then $[\alpha]\in\pi_1(S)$ is primitive.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Most of the proof I'm OK with, except right in the beginning, when they write&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;...let $\phi\in\text{Isom}^+(\mathbb{H}^2)$ be the hyperbolic isometry corresponding to some element of the conjugacy class of $\alpha$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Two questions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) What do they mean by &lt;em&gt;the&lt;/em&gt; hyperbolic isometry $\phi$?  Don't different elements of $\pi_1(S)$ correspond to different elements of $\text{Isom}^+(\mathbb{H}^2)$? (Here $\pi_1(S)$ is acting as deck transformations on $\mathbb{H}^2$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Why is there a &lt;em&gt;hyperbolic&lt;/em&gt; isometry corresponding to $\alpha$? For example, if $\alpha$ is a simple loop around a puncture point, then shouldn't any such $\phi$ be parabolic?&lt;/p&gt;&#xA;" OwnerUserId="641" LastActivityDate="2011-02-24T04:19:24.157" Title="Question on proof in &quot;Primer on MCGs&quot;" Tags="&lt;group-theory&gt;&lt;hyperbolic-geometry&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="23271" PostTypeId="2" ParentId="23264" CreationDate="2011-02-22T19:58:09.507" Score="1" ViewCount="" Body="&lt;p&gt;You can take $2j = n(n+2)+m = (n+1)^2 + (m-1)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider $k^2$ and $(k+1)^2$. The difference between them is $2k+1$; if you allow yourself to add at most $k$ to $k^2$, you &quot;get&quot; all the way to $k^2+k$. If you allow yourself to subtract at most $k$ from $(k+1)^2$, you can &quot;get down&quot; all the way to $k^2+k+1$. So that is why the map $(m,n)\mapsto n(n+1)+m$ is bijective under the given conditions (the parity condition is just to ensure that this number is even). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So look at $k = \lfloor \sqrt{2j}\rfloor$. This is either $n+1$ or $n$, depending on whether $m$ is positive or negative, and once you determine if $k=n$ or if $k=n+1$, you can compute $m$ from $m = 2j - n(n+2)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, since $k=n$ or $k=n+1$, either $2j-k(k+2)$ will lie between $-k$ and $k$ or not. If it does, then $k=n$ and $m=2j-k(k+2)$. If it does not lie between $-k$ and $k$, then $k=n+1$, so $m = 2j-(k-1)(k+1)$.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2011-02-22T19:58:09.507" />
  <row Id="23272" PostTypeId="2" ParentId="23269" CreationDate="2011-02-22T20:04:32.807" Score="4" ViewCount="" Body="&lt;p&gt;Since $f$ is the (generic) homogeneous solution, you can choose to set it equal to zero; then you just have the special solution $x(t) = A\mathrm{sin}(\omega t+\phi)$.  You should be able to find $\ddot{x}(t)$ and $\dot{x}(t)$, then plug them all into your core differential equation and see what it means for both sides to be equal - if you've done it right, this will give you the relation between $\omega$ and $A$ that you're looking for.&lt;/p&gt;&#xA;" OwnerUserId="785" LastActivityDate="2011-02-22T20:04:32.807" CommentCount="3" />
  <row Id="23273" PostTypeId="2" ParentId="23045" CreationDate="2011-02-22T20:08:38.043" Score="0" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\ \ $ By definition, $\rm\ \alpha\:$ is transcendental over $\rm\:F\ \iff\ F[x] \cong F[\alpha]\ $ via evaluation $\rm\ x\to \alpha\:.\ $ For example, $\rm\ \mathbb Q[\pi] \cong \mathbb Q[x]\:.\:$ Therefore, in particular, transcendental evaluation is always injective.  Thus &lt;em&gt;any&lt;/em&gt; element transcendental over a field $\rm\:F\:$ serves equally well as an &quot;indeterminate&quot; over $\rm\:F\:$.&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2011-02-22T20:08:38.043" />
  <row Id="23274" PostTypeId="2" ParentId="22725" CreationDate="2011-02-22T20:13:29.223" Score="1" ViewCount="" Body="&lt;p&gt;First think about these properties of the number zero. Let $r$ be any positive real number, then...&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;$0 &amp;lt; r$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$0 &amp;gt; -r$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$0 \cdot r = 0$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$0 + r = r$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;You can think of infinitesimals as &quot;zero-like numbers&quot;. Let $r$ be any positive real number. Then if $\epsilon$ is an infinitesimal, it has these properties:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;$\epsilon &amp;lt; r$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$\epsilon &amp;gt; -r$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$\epsilon \cdot r$ is an infinitesimal&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$\epsilon + r$ is &quot;infinitely close&quot; to the real number r&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;They are useful in calculus for many reasons, but the major reason that someone in Calculus 1 will appreciate is that they simplify your derivative. The traditional definition of the derivative is that $f&amp;#39;(x) = lim_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}$, which we must do because simply plugging in $h = 0$ as you would normally do to &quot;solve the limit&quot; yields the undefined algebraic expression $\frac{0}{0}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Infinitesimally, we define $f&amp;#39;(x) = ($ &quot;the real number infinitely close to&quot; $\frac{f(x+\epsilon) - f(x)}{\epsilon})$. This definition prevents you from having to do complicated limit arguments and lets you focus on the algebra of the situation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another good example where infinitesimals simplify calculus is the definition of continutiy. Usually, you define a function $f$ to be continuous at $a$ if and only if $lim_{x \rightarrow a} f(x) = f(a)$ which unpacks by the definition of limit to $\forall \epsilon &amp;gt; 0 \exists \delta &amp;gt; 0$ such that $|x - a| &amp;lt; \delta$ implies $|f(x) - f(a)| &amp;lt; \epsilon$. That is, the definition requires two universal quantifiers (&quot;for all&quot;) and one existential (&quot;there exists&quot;). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The infinitesimal definition of continuous is much easier to understand: &quot;if $x$ is infinitely close to $a$, then $f(x)$ is infinitely close to $f(a)$&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The difficulty in using infinitesimals is not necessarily understanding their properties as much as it is difficulty in CONSTRUCTING them. Understanding how &quot;infinitely close to&quot; works is similarly not very difficult for the functions in any first calculus course. Fortunately, we usually don't worry about constructing real numbers in an introductory calculus course, so it should be reasonable to use infinitesimals in a similar way!&lt;/p&gt;&#xA;" OwnerUserId="1796" LastActivityDate="2011-02-22T20:13:29.223" />
  <row Id="23275" PostTypeId="2" ParentId="23258" CreationDate="2011-02-22T20:18:23.070" Score="1" ViewCount="" Body="&lt;p&gt;I would do it as follows:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;For $(x,y) \in H \times H$ define an element of $N(H)$ by $x \,\tilde{\otimes}\, y = \langle \cdot , y \rangle x$ and note that $\|x \, \tilde{\otimes}\,y\|_{1} = \|x\|\,\|y\|$.&lt;/li&gt;&#xA;&lt;li&gt;For a functional $\varphi \in N(H)&amp;#39;$ define a sesquilinear form on $H$ by $B_{\varphi}(x,y) = \varphi(x \,\tilde{\otimes}\, y)$. The map $\varphi \mapsto B_{\varphi}$ is clearly linear and of norm $\leq 1$ since $|B_{\varphi}(x,y)| = |\varphi(x \, \tilde{\otimes}\,y)| \leq \|\varphi\|\,\|x\|\,\|y\|$ by 1.&lt;/li&gt;&#xA;&lt;li&gt;Use the version of the Riesz representation theorem stating that every bounded sesquilinear form $B$ is of the form $B(x,y) = \langle x,T y \rangle$ for a unique $T$. Moreover, $\|B\| = \|T\|$. In other words, the map $B \mapsto T$ from bounded sesquilinear forms to $L(H)$ is a linear isometry.&lt;/li&gt;&#xA;&lt;li&gt;Define $j(\varphi)$ by $\varphi \mapsto B_{\varphi} \mapsto T_{\varphi}$ by combining the maps from 2. and 3. Verify that $ji = 1_{L(H)}$ and $ij = 1_{N(H)&amp;#39;}$. Since both $i$ and $j$ have norm $\leq 1$, they must be isometries.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="5363" LastActivityDate="2011-02-22T20:18:23.070" CommentCount="2" />
  <row Id="23276" PostTypeId="1" AcceptedAnswerId="23277" CreationDate="2011-02-22T20:18:35.707" Score="7" ViewCount="151" Body="&lt;p&gt;Since $\varphi(p)=p-1$ is even the p'th cyclotomic field contains some quadratic field. Hecke says that in fact every quadratic field is contained by some cyclotomic field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is this theorem called and how is it proved?&lt;/p&gt;&#xA;" OwnerUserId="4290" LastActivityDate="2011-02-22T20:38:42.640" Title="Relationship between Cyclotomic and Quadratic fields" Tags="&lt;algebraic-number-theory&gt;&lt;galois-theory&gt;" AnswerCount="2" />
  <row Id="23277" PostTypeId="2" ParentId="23276" CreationDate="2011-02-22T20:21:43.643" Score="10" ViewCount="" Body="&lt;p&gt;This is a special case of the Kronecker-Weber theorem, which says that any abelian extension of $\mathbb{Q}$ is contained inside some cyclotomic field; any quadratic extension of $\mathbb{Q}$ is automatically abelian. I don't believe the special case of the theorem for quadratic fields has a separate name.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, one does not need the full power of this (very advanced) theorem. The following two steps are used in exercise 8 of Chapter 2 in Marcus's &lt;em&gt;Number Fields&lt;/em&gt; to prove precisely the case of quadratic fields:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;strong&gt;Show that $\mathbb{Q}(\zeta_p)$ contains $\sqrt{p}$ if $p\equiv 1\bmod 4$ and $\sqrt{-p}$ if $p\equiv 3\bmod 4$.&lt;/strong&gt;  (Note: This step follows from results in the preceding chapter in Marcus about the discriminant of $\mathbb{Q}(\zeta_p)$ being &#xA;$$\prod_{1\leq r&amp;lt;s\leq p-1}(\zeta_p^r-\zeta_p^s)^2=\left(\prod_{1\leq r&amp;lt;s\leq p-1}(\zeta_p^r-\zeta_p^s)\right)^2=\pm p^{p-2}$$&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Show that $\mathbb{Q}(\zeta_8)$ contains $\sqrt{2}$.&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Obviously $\mathbb{Q}(i)$ contains $\sqrt{-1}$, so to get a cyclotomic field containing $\mathbb{Q}(\sqrt{m})$, we just need to take the compositum of the cyclotomic fields corresponding to each of $m$'s prime factors (note that we can assume $m$ is squarefree).&lt;/p&gt;&#xA;" OwnerUserId="264" LastEditorUserId="264" LastEditorDisplayName="" LastEditDate="2011-02-22T20:33:50.273" LastActivityDate="2011-02-22T20:33:50.273" />
  <row Id="23278" PostTypeId="2" ParentId="23276" CreationDate="2011-02-22T20:38:42.640" Score="2" ViewCount="" Body="&lt;p&gt;This result of Gauss is very well-known in algebraic number theory. The obvious web searches will easily locate proofs and much more, e.g. here's a proof from &lt;a href=&quot;http://books.google.com/books?id=-RZDK6N4j8wC&amp;amp;pg=PA114&quot; rel=&quot;nofollow&quot;&gt;Weintraub: Galois Theory&lt;/a&gt;&#xA;&lt;img src=&quot;http://i.imgur.com/ZVn50.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2011-02-22T20:38:42.640" />
  <row Id="23279" PostTypeId="2" ParentId="23262" CreationDate="2011-02-22T20:39:51.857" Score="4" ViewCount="" Body="&lt;p&gt;Carrying around the $6^7$ is just complicating your life. Instead, just &lt;em&gt;count&lt;/em&gt; how many distinct rolls have exactly one triple; the end probability will be that count, divided by the total number of possible rolls (namely, $6^7$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is also the issue of distinguishable and non-distinguishable dice. But let's assume you roll a single dice in sequence and write down the results.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proceeding along the lines you have, and hoping I'm not making a mistake:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;At least 3-of-a-kind of &lt;strong&gt;x&lt;/strong&gt;, with the other four dice being anything except &lt;strong&gt;x&lt;/strong&gt;. This one you compute correctly: pick what is &lt;strong&gt;x&lt;/strong&gt; (six possibilities), then pick which three dice fall come up &lt;strong&gt;x&lt;/strong&gt; ($\binom{7}{3}$ ways), then choose any of five possible outcomes for each of the remaining four dice, $5^4$. So you have $\binom{6}{1}\binom{7}{3}5^4$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Edit.&lt;/em&gt; However, this overcounts! (&lt;em&gt;Thanks to Henry for spotting it&lt;/em&gt;) If a roll has &lt;em&gt;two&lt;/em&gt; different three-of-a-kind, then I count that roll twice; once when I'm counting the lower three-of-a-kind, and again when I'm counting the higher one. So we need to subtract those rolls that have two three-of-a-kinds. We select the two ranks by using $\binom{6}{2}$, then select the dice that show the lower rank with $\binom{7}{3}$, and then select the dice, out of the remaining $4$, that have the higher rank three-of-a-kind with $\binom{4}{3}$. Then the remaining die can be any of the remaining $4$ ranks; so we overcounted by $\binom{6}{2}\binom{7}{3}\binom{4}{3}4$. So the &lt;em&gt;real&lt;/em&gt; number of rolls that have at least one three-of-a-kind is&#xA;$$\binom{6}{1}\binom{7}{3}5^4 - \binom{6}{2}\binom{7}{3}\binom{4}{3}4.$$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Discard the case in which we get 3-of-a-kind of &lt;strong&gt;x&lt;/strong&gt;, but the other four dice are equal to &lt;strong&gt;y&lt;/strong&gt;, with &lt;strong&gt;x&lt;/strong&gt;$\neq$ &lt;strong&gt;y&lt;/strong&gt;. There are $\binom{6}{1}$ ways of picking &lt;strong&gt;x&lt;/strong&gt;, and then you need to pick what the other $4$ are; there are $\binom{5}{1}$ ways of doing it; then pick the three dice that will come up &lt;strong&gt;x&lt;/strong&gt;, $\binom{7}{3}$ ways. So you have $\binom{6}{1}\binom{5}{1}\binom{7}{3}$ outcomes that you &lt;em&gt;don't&lt;/em&gt; want.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Discard the case in which you get two sets of 3-of-a-kind. We already figured out, when we computed the overcount, that there are&#xA;$$4\binom{6}{2}\binom{7}{3}\binom{4}{3}$$&#xA;rolls with two 3-of-a-kind. So we must subtract this total again.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;So: the total number of rolls in which the largest number of dice that come up the same is $3$, and there is one and only one set that does this, is:&#xA;$$\binom{6}{1}\binom{7}{3}5^4 - \binom{6}{1}\binom{7}{3}\binom{5}{1} - 2\times\left(\binom{6}{2}\binom{7}{3}\binom{4}{3}4\right) = 113400.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given that there are $6^7$ sequential rolls possible, you get a probability of&#xA;$$\frac{\mbox{number of rolls with exactly one three-of-a-kind, no 4-of-a-kind}}{6^7} = \frac{113400}{6^7} \approx 0.4050925926.$$&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-23T01:22:05.143" LastActivityDate="2011-02-23T01:22:05.143" CommentCount="6" />
  <row Id="23280" PostTypeId="2" ParentId="23258" CreationDate="2011-02-22T20:43:16.850" Score="1" ViewCount="" Body="&lt;p&gt;I think it helps to think of the commutative analogue of this, which is the fact that $\ell^\infty$ is the dual of $\ell^1$.  If $(a_n)_n\in\ell^\infty$, then the corresponding functional is $(x_n)_n\mapsto\sum_n a_nx_n$.  To show that the map from $\ell^\infty$ to the dual of $\ell^1$ does not decrease norm, let $\varepsilon\gt0$, take $m$ such that $|a_m|\gt \|(a_n)_n\|_\infty-\varepsilon$, and let $(x_n)_n\in\ell^1$ be such that $x_m=1$ and $x_n=0$ for $n\neq m$.  Then $|\sum_n a_nx_n|=|a_m|=|a_m|\|(x_n)_n\|_1 \geq(\|(a_n)_n\|_\infty-\varepsilon)\|(x_n)_n\|_1$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the noncommutative case, you have a bounded operator $T$, and given small $\varepsilon$ you can find a unit vector $e\in H$ such that $\|Te\|\geq\|T\|-\varepsilon$, and a unit vector $f\in H$ such that $\langle Te,f\rangle = \|Te\|$.  Consider the rank one operator $S=\langle\cdot,f\rangle e$, which has norm $1$ in $N(H)$.  If my calculations are correct, the trace of TS is $\langle Te,f\rangle=\|Te\|$, showing that the norm of $i(T)$ is at least $\|T\|-\varepsilon$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that on Hilbert space, nuclear operators are often called &lt;a href=&quot;http://en.wikipedia.org/wiki/Trace_class&quot; rel=&quot;nofollow&quot;&gt;&quot;trace class&quot;&lt;/a&gt;, and that a proof that $L(H)$ is the dual of the trace class operators can be found in standard texts on operator theory.  &lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2011-02-22T20:43:16.850" />
  <row Id="23282" PostTypeId="2" ParentId="23262" CreationDate="2011-02-22T20:53:52.293" Score="6" ViewCount="" Body="&lt;p&gt;You are on the right track with $6^7 = 279936$ as the denominator. To find how many cases have three but no more matching, I would start by looking at the four partitions of 7 into up to 6 parts where the largest is 3: 3+3+1, 3+2+2, 3+2+1+1, 3+1+1+1+1.  You can then work out each systematically, taking account both the numbers that appear and the order they appear in.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the first (which has two sets of three) is $$\frac{6!}{2!\;1!\;3!} \times \frac{7!}{3!\;3!\;1!} = 8400$$ which is ten times what you have in (c).  The others (just one three-of-a-kind) are &#xA;$$\frac{6!}{1!\;2!\;3!} \times \frac{7!}{3!\;2!\;2!} + \frac{6!}{1!\;1!\;2!\;2!} \times \frac{7!}{3!\;2!\;1!\;1!} + \frac{6!}{1!\;4!\;1!} \times \frac{7!}{3!\;1!\;1!\;1!\;1!} = 113400$$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I get about 0.405 for the probability of exactly one three-of-a-kind (but no four or more) and about 0.435 for the probability of the one or more threes-of-a-kind (but no four or more) &lt;/p&gt;&#xA;" OwnerUserId="6460" LastEditorUserId="6460" LastEditorDisplayName="" LastEditDate="2011-02-22T21:16:36.580" LastActivityDate="2011-02-22T21:16:36.580" CommentCount="5" />
  <row Id="23283" PostTypeId="1" CreationDate="2011-02-22T21:07:06.427" Score="1" ViewCount="181" Body="&lt;p&gt;I'm completely new with the sum notation and Leva-Civita-Symbols. Can somebody tell me if what I did was correct?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $a,b \in \mathbb{R}^3$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to prove $$ (b \wedge a) = -(a \wedge b) $$ so I did $$(b \wedge a)_i = \epsilon_{ijk}b_ja_k = - \epsilon_{ikj}b_ja_k = -\epsilon_{ikj}a_kb_j = - (a \wedge b)_i $$&lt;/p&gt;&#xA;" OwnerUserId="5450" LastActivityDate="2011-02-22T21:07:06.427" Title="(very simple) Vector identity proof" Tags="&lt;homework&gt;&lt;vector-calculus&gt;" CommentCount="5" />
  <row Id="23284" PostTypeId="1" AcceptedAnswerId="23288" CreationDate="2011-02-22T21:16:17.733" Score="1" ViewCount="72" Body="&lt;p&gt;Player A, picks $n_1$ integers $a_1$,...,$a_{n_1}$ uniformly at random from $1$..$N$, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;and player B picks $n_2$ integers $b_1$,...,$b_{n_2}$ the same way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given $N$, $n_1$, $n_2$, and $d$, what is the expected number of pairs ($a_i$,$b_j$) where |$a_i$-$b_j$| $=&amp;lt;$ $d$, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A. if A and B are allowed to select duplicates in their lists.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;B. if duplicate numbers are not allowed. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks, &#xA;MG&lt;/p&gt;&#xA;" OwnerUserId="7138" LastActivityDate="2011-02-22T22:05:27.123" Title="Expected number of close pairs" Tags="&lt;probability&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" />
  <row Id="23285" PostTypeId="2" ParentId="23105" CreationDate="2011-02-22T21:25:44.530" Score="7" ViewCount="" Body="&lt;p&gt;There are about&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{10^{10}(\log\log 10^{10})^{k-1}}{(k - 1)!\log 10^{10}}$$&#xA;10-digit numbers with k prime factors, allowing leading zeros, or&#xA;$$\frac{1}{(k - 1)!}\left( \frac{10^{10}(\log\log 10^{10})^{k-1}}{\log 10^{10}}-\frac{10^9(\log\log 10^9)^{k-1}}{\log 10^9} \right)$$&#xA;without leading zeros.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Based on this you'd expect those with 3 or 4 prime factors to be most likely.  The ones with two and five are pretty far behind, with the others not even close.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Counts (with leading 0s):&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;1 455052511 (predicted: 434294482)&#xA;2 1493776443 (predicted: 1362215689)&#xA;3 2227121996 (predicted: 2136374810)&#xA;4 2139236881 (predicted: 2233663566)&#xA;5 1570678136 (predicted: 1751537079)&#xA;6 977694273 (predicted: 1098780384)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;So those with three prime factors win.  Note: the exact counts assume that you want to count repeated prime factors more than once (the predictions are basically the same either way).  If you look at only the number of distinct prime factors the totals vary but 3 remains in first place.&lt;/p&gt;&#xA;" OwnerUserId="1778" LastEditorUserId="1778" LastEditorDisplayName="" LastEditDate="2011-02-22T21:42:57.583" LastActivityDate="2011-02-22T21:42:57.583" CommentCount="2" />
  <row Id="23286" PostTypeId="1" AcceptedAnswerId="23289" CreationDate="2011-02-22T21:26:23.940" Score="3" ViewCount="173" Body="&lt;p&gt;Let $\tau_a=\inf\{t: B_t=a\}$, the hitting time of the standard Brownian motion to reach the boundary $a$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is easily derived&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$E(e^{-\lambda \tau_a})=e^{-|a|\sqrt{2\lambda}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I am having a problem of using this formula to get the moments of $\tau_a$ by matching the coefficients of terms for each power of $\lambda$, specifically, the LHS, if expanded, has all integer powers of $\lambda$; however, the RHS has the various terms of $\sqrt{\lambda}$. How to reconcile the difference here? Could somebody please help? Thanks.&lt;/p&gt;&#xA;" OwnerUserId="6016" LastActivityDate="2011-02-22T23:55:12.327" Title="A question regarding the hitting time formula in brownian motion" Tags="&lt;probability&gt;&lt;stochastic-processes&gt;" AnswerCount="2" />
  <row Id="23287" PostTypeId="1" CreationDate="2011-02-22T21:26:59.133" Score="8" ViewCount="166" Body="&lt;p&gt;Let $\gamma: I \rightarrow \mathbb{R}^n$ be a smooth curve. We define a Frenet frame to be an orthonormal frame $X_1, \ldots X_n$ such that for all $1 \leq k \leq n$, $\gamma^{(k)}(t)$ is contained in the linear span of $X_1(t), \ldots, X_k(t)$, $t \in I$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is an example of a smooth curve $\gamma$ with no Frenet frame?&lt;/p&gt;&#xA;" OwnerUserId="7387" LastActivityDate="2011-04-19T17:54:00.957" Title="Smooth curve with no Frenet frame" Tags="&lt;differential-geometry&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="3" />
  <row Id="23288" PostTypeId="2" ParentId="23284" CreationDate="2011-02-22T21:28:32.673" Score="6" ViewCount="" Body="&lt;p&gt;In both cases, the expected number is $n_1n_2p$ where $p$ is the probability that $|a_1-b_1|\le d$. There are $n(d)=N+2(N-1)+\cdots+2(N-d)$ such couples $(a,b)$ hence $p=n(d)/N^2$. Finally $n(d)=(2d+1)N-d(d+1)$, hence the expected number you ask for is &#xA;$$n_1n_2[(2d+1)N-d(d+1)]/N^{2}.$$&lt;/p&gt;&#xA;" OwnerUserId="6179" LastEditorUserId="6179" LastEditorDisplayName="" LastEditDate="2011-02-22T21:35:36.267" LastActivityDate="2011-02-22T21:35:36.267" />
  <row Id="23289" PostTypeId="2" ParentId="23286" CreationDate="2011-02-22T21:34:06.660" Score="3" ViewCount="" Body="&lt;p&gt;Well, in fact $E(\tau_a)$ is infinite...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A classic argument to see that $\tau_a$ does not have finite mean is to consider $\tau_{2a}$. On the one hand, to hit $2a$, one has to hit $a$ and then, starting from $a$, to hit $2a$. By the strong Markov property, this second duration is independent of $\tau_a$ and, by invariance by translations, distributed like $\tau_a$. On the other hand, by the scale invariance of Brownian motion, $(2B_t)$ is distributed like $(B_{4t})$, hence $\tau_{2a}$ is distributed like $4\tau_a$. Finally, $\tau_a$ solves a distribution equality: the distribution of $T+T&amp;#39;$ and the distribution of $4T$ coincide, where $T$ and $T&amp;#39;$ are independent copies of $\tau_a$. This is enough to prove that $\tau_a$ is not integrable (and almost enough to deduce its Laplace transform, for that matter).&lt;/p&gt;&#xA;" OwnerUserId="6179" LastEditorUserId="6179" LastEditorDisplayName="" LastEditDate="2011-02-22T22:25:24.157" LastActivityDate="2011-02-22T22:25:24.157" CommentCount="1" />
  <row Id="23290" PostTypeId="2" ParentId="23192" CreationDate="2011-02-22T21:58:52.610" Score="1" ViewCount="" Body="&lt;p&gt;Arrange the eigen values in decreasing order of magnitude. Arrange the eigen vectors also based on decreasing values of magnitude of eigenvalues. Now do you see what $U$ and $V$ are. Notice that $U$ and $V$ are almost same except for a change in sign of the column corresponding to the negative eigenvalue.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastActivityDate="2011-02-22T21:58:52.610" CommentCount="3" />
  <row Id="23291" PostTypeId="2" ParentId="23286" CreationDate="2011-02-22T22:03:44.690" Score="2" ViewCount="" Body="&lt;p&gt;To add on Didier's answer, note that the PDF of $\tau_a$, for $a &amp;gt; 0$, is given by&#xA;$$&#xD;&#xA;f_{\tau _a } (t) = \frac{a}{{\sqrt {2\pi } }}t^{ - 3/2} e^{ - a^2 /(2t)} ,\;\; t &amp;gt; 0.&#xD;&#xA;$$&#xA;From this you immediately see that ${\rm E}(\tau_a) = \infty$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;: The distribution function of $\tau_a$ can be derived as follows (cf. p.1 &lt;a href=&quot;http://www.maths.leeds.ac.uk/uk_dsgs/lecture3b_gl.pdf&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;).&#xA;First note that, for any $t &amp;gt; 0$,&#xA;$$&#xD;&#xA;{\rm P}(\tau _a  &amp;lt; t) = {\rm P}(\tau _a  &amp;lt; t,B_t  &amp;lt; a) + {\rm P}(\tau _a  &amp;lt; t,B_t  &amp;gt; a).&#xD;&#xA;$$&#xA;The right-hand side is equal to $2{\rm P}(\tau _a  &amp;lt; t,B_t  &amp;gt; a)$, hence to $2{\rm P}(B_t  &amp;gt; a)$.&#xA;Thus, &#xA;$$&#xD;&#xA;{\rm P}(\tau _a  &amp;lt; t) = 2 {\rm P}(B_t  &amp;gt; a) = \frac{2}{{\sqrt {2\pi t} }}\int_a^\infty  {e^{ - \frac{{y^2 }}{{2t}}} \,{\rm d}u}  = \frac{2}{{\sqrt \pi  }}\int_{\frac{a}{{\sqrt {2t} }}}^\infty  {e^{ - y^2 } \,{\rm d}y}.&#xD;&#xA;$$&#xA;Differentiating with respect to $t$ gives the above formula for $f_{\tau _a } (t)$. (For arbitrary $a \in \mathbb{R}$, just put $|a|$ instead of $a$ in that formula.)&lt;/p&gt;&#xA;" OwnerUserId="2810" LastEditorUserId="2810" LastEditorDisplayName="" LastEditDate="2011-02-22T23:55:12.327" LastActivityDate="2011-02-22T23:55:12.327" />
  <row Id="23292" PostTypeId="2" ParentId="23284" CreationDate="2011-02-22T22:05:27.123" Score="2" ViewCount="" Body="&lt;p&gt;If $d\ge N-1$, then all pairs must be within $d$ of each other, and the total number of close pairs must be exactly $n_1 n_2$.  Now assume $d &amp;lt; N-1$.  For any $i,j$, the random variables $a_i$ and $b_j$ are independent and uniformly distributed over $[N]=\{1,2,...,N\}$.  The points $(x,y)$ in $[N]^2$ with $|x-y|&amp;gt;d$ make up two right triangles with side length $N-1-d$, and the total number of such points is $(N-1-d)(N-d)$, or $(N-d)^2 - (N-d)$.  So the probability that a particular pair $(a_i,b_j)$ is close is given by&#xA;$$&#xD;&#xA;\begin{eqnarray}&#xD;&#xA;P\left[|a_i-b_j|\le d\right] &amp;amp;=&amp;amp; \frac{1}{N^2}\left(N^2 - (N-d)^2 + (N-d)\right) \\&#xD;&#xA;&amp;amp;=&amp;amp; \frac{1}{N^2}\left(N(2d+1) - d(d+1)\right).&#xD;&#xA;\end{eqnarray}&#xD;&#xA;$$&#xA;The expected number of close pairs is just $n_1 n_2$ times this probability:&#xA;$$&#xD;&#xA;E\left[\text{# pairs }(a_i, b_j)\text{ s.t. }|a_i-b_j|\le d\right] = \frac{n_1 n_2}{N^2}\left(N(2d+1) - d(d+1)\right).&#xD;&#xA;$$&#xA;This is true whether or not $A$ and $B$ are allowed to have duplicates within their individual lists of integers.  It does, however, make a difference whether $A$ and $B$ may have duplicates &lt;em&gt;between&lt;/em&gt; their lists.  If no duplicates are to be allowed even between $A$ and $B$, then $N$ points $(x,x)$ are removed from consideration for each pair $(a_i, b_j)$: the final result will have its $N^2$ denominator replaced by $N(N-1)$.&lt;/p&gt;&#xA;" OwnerUserId="5546" LastActivityDate="2011-02-22T22:05:27.123" />
  <row Id="23293" PostTypeId="1" AcceptedAnswerId="23401" CreationDate="2011-02-22T23:15:46.523" Score="1" ViewCount="279" Body="&lt;p&gt;I've an confession to make. I've been using pdf's and pmf's without actually knowing what they are. The idea that I've been having so long is that density = area under the curve but if I look at it that way then a lot of times it didn't make sense when we refer to mass of a random variable in discrete distributions. How can I interpret this? Why do we call mass or density instead of something else?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;p.s.  Please feel free to change the question itself in a more understandable way if you feel this is a logically wrong question.&lt;/p&gt;&#xA;" OwnerUserId="6335" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2011-02-23T17:25:54.550" LastActivityDate="2011-02-23T17:25:54.550" Title="Probability density function vs. probability mass function" Tags="&lt;probability-theory&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="23294" PostTypeId="2" ParentId="23293" CreationDate="2011-02-22T23:27:31.663" Score="1" ViewCount="" Body="&lt;p&gt;Probability mass functions are used for discrete distributions. It assigns a probability to each point in the sample space. Whereas the integral of a probability density function gives the probability that a random variable falls within some interval. &lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2011-02-22T23:27:31.663" CommentCount="3" />
  <row Id="23295" PostTypeId="2" ParentId="23070" CreationDate="2011-02-22T23:31:52.737" Score="2" ViewCount="" Body="&lt;p&gt;First, $\prod_{i\in I}\mathbb S_i$ should be understood as the product $\sigma$-algebra, i.e., the smallest $\sigma$-algebra on $\prod_{i\in I}S_i$ such that the projection $\pi_i:\prod_{j\in I}S_j \ni (s_j)_{j\in I}\mapsto s_i\in \mathbb S_i$ is measurable. In this case, then the answers to the first two questions are Yes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For part (3), as explained in the comments and answers in another closely related post &lt;a href=&quot;http://math.stackexchange.com/q/23230/5836&quot;&gt;Measure from on product $\sigma$-algebra to on component $\sigma$-algebras&lt;/a&gt;, one needs to work with finite measures on the product space. In this case, without loss of generality assuming $\mu = \mathbb P$ is a probability measure and using $X$ instead of $f$, one recovers part (2).&lt;/p&gt;&#xA;" OwnerUserId="5836" LastActivityDate="2011-02-22T23:31:52.737" />
  <row Id="23296" PostTypeId="1" AcceptedAnswerId="23300" CreationDate="2011-02-22T23:36:22.143" Score="2" ViewCount="233" Body="&lt;ul&gt;&#xA;&lt;li&gt;Polynomials of degree 2, if the first term is positive, starts descreasing.&lt;/li&gt;&#xA;&lt;li&gt;Polynomials of degree 3, if the first term is positive, starts increasing.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;is this constant?&#xA;like..&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Polynomials of degree 4, if the first term is positive, starts descreasing.&lt;/li&gt;&#xA;&lt;li&gt;Polynomials of degree 5, if the first term is positive, starts increasing.&lt;/li&gt;&#xA;&lt;li&gt;...&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="4501" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2011-02-23T00:56:56.927" LastActivityDate="2011-02-23T13:43:06.233" Title="Do all polynomials of even degree start by decreasing as you plot from $-\infty$ upward?" Tags="&lt;polynomials&gt;&lt;graph&gt;" AnswerCount="3" CommentCount="4" />
  <row Id="23297" PostTypeId="2" ParentId="23296" CreationDate="2011-02-22T23:42:31.193" Score="2" ViewCount="" Body="&lt;p&gt;If by &quot;starts&quot;, you mean the behaviour in a $(-\infty, -M)$ where $M$ some large positive real number, then yes.&lt;/p&gt;&#xA;" OwnerUserId="7367" LastEditorUserId="4501" LastEditorDisplayName="" LastEditDate="2011-02-23T00:21:02.643" LastActivityDate="2011-02-23T00:21:02.643" />
  <row Id="23298" PostTypeId="2" ParentId="23269" CreationDate="2011-02-22T23:43:57.057" Score="4" ViewCount="" Body="&lt;p&gt;The correct value of $\omega$ is not $\frac{\sqrt{7}}{2} \approx 1.32$.  It's $\sqrt{\frac{3}{2}} \approx 1.22$.  We are looking for a near resonance effect, but you can't actually have resonance when the harmonic oscillator is damped.  This makes the analysis is a little different.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A reference is the section on &lt;a href=&quot;http://en.wikipedia.org/wiki/Harmonic_oscillator#Sinusoidal_driving_force&quot; rel=&quot;nofollow&quot;&gt;sinusoidal forcing in the Wikipedia page on the harmonic oscillator&lt;/a&gt;.  From the formulas there you can see that the relationship between $A$ and $\omega$ is given by &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$A = \frac{1}{\sqrt{\omega^2 + (2 - \omega^2)^2}}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solving $\frac{dA}{d \omega} = 0$ yields $\omega = \sqrt{\frac{3}{2}}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;HR&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update:&lt;/strong&gt; Here's the derivation.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The trick is to generalize, and solve the differential equation $x&amp;#39;&amp;#39; + x&amp;#39; + 2x = e^{i \omega t}$.  The resulting solution will have a real part and an imaginary part.  Since $e^{i \omega t} = \cos \omega t + i \sin \omega t$, you actually want the &lt;em&gt;imaginary part&lt;/em&gt; of the solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the driving force is an exponential, we know that the particular solution must be of the form $x_p(t) = c e^{i \omega t}$.  Subbing that into the differential equation produces the auxiliary equation $-c \omega^2 + i c\omega + 2c = 1$.  Solving that for $c$ yields $$c = \frac{1}{a + i b} = \frac{a - i b}{a^2 + b^2},$$&#xA;where $a = 2 - \omega^2$ and $b = \omega$.  Thus the particular solution to the complex differential equation is &#xA;$$x_p(t) = \frac{a - i b}{a^2 + b^2} (\cos \omega t + i \sin \omega t),$$ of which the imaginary part is $$-\frac{b}{a^2 + b^2} \cos \omega t + \frac{a}{a^2+b^2} \sin \omega t.$$  Since $A$ is just the magnitude of this solution (you're doing a rotation to the vertical axis when converting to $A \sin (\omega t + \phi)$), we get $$A = \frac{1}{\sqrt{a^2+b^2}} = \frac{1}{\sqrt{\omega^2 + (2 - \omega^2)^2}}.$$&lt;/p&gt;&#xA;" OwnerUserId="2370" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2011-02-23T00:36:10.753" LastActivityDate="2011-02-23T00:36:10.753" CommentCount="1" />
  <row Id="23299" PostTypeId="2" ParentId="9846" CreationDate="2011-02-22T23:55:57.167" Score="5" ViewCount="" Body="&lt;p&gt;There are several possible ways of defining &quot;density on a sphere&quot;, each one giving somewhat different results.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alas, most of them have some &quot;maximum number of vertices&quot; that give exactly equal density.&#xA;Above that maximum number, further tessellation can at best &lt;em&gt;approximate&lt;/em&gt; constant density.&#xA;(That approximation is more than adequate for many purposes).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Unfortunately, it is a well-known group theoretical result that there are no completely regular point distributions on the sphere for N &gt; 20.&quot; -- &lt;a href=&quot;http://space.mit.edu/home/tegmark/icosahedron.html&quot; rel=&quot;nofollow&quot;&gt;Max Tegmark&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Equal density as equal areas of the triangles formed by the vertices:&#xA;You can tessellate a sphere to give a geodesic sphere such that every triangle has exactly equal area, to any desired resolution, using any equal-area projection such as the Snyder equal area projection.&#xA;(A few people use &lt;a href=&quot;http://en.wikipedia.org/wiki/geodesic_grid&quot; rel=&quot;nofollow&quot;&gt;geodesic grids&lt;/a&gt; based on this principle).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Equal density as congruent triangles formed by the vertices:&#xA;When a person builds a geodesic dome out of panels, it would be super-convenient if every panel were identically the same size and shape.&#xA;Alas, the maximum &quot;size&quot; is the 120 identical faces of the&#xA;hexakis icosahedron (aka disdyakis triacontahedron).&#xA;Any convex solid with more than 120 faces must necessarily have 2 or more kinds of faces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Equal density as minimum-energy configurations of charged particles:&#xA;&lt;a href=&quot;http://mathpages.com/home/kmath005/kmath005.htm&quot; rel=&quot;nofollow&quot;&gt;Min-Energy Configurations of Electrons On A Sphere&lt;/a&gt;.&#xA;You can put &lt;em&gt;any&lt;/em&gt; integer number of repelling particles on a sphere, and calculate some minimum-energy configuration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Equal density as equal distance from every vertex to the N nearby vertices:&#xA;When a person builds a geodesic dome out of struts, it would be super-convenient if all the struts were the same length.&#xA;Most &quot;naive&quot; methods of dividing the large triangles of the icosahedron into smaller triangles generates lots of different edge lengths;&#xA;but there are ways to &quot;tweak&quot; the tessellation subdivision in order to minimize the number of different lengths of edges.&#xA;(Fewer unique lengths requires fewer jigs in manufacturing and fewer spares needed to replace any damaged strut).&#xA;Alas, the maximum &quot;size&quot; of a strictly convex polyhedron made entirely of equilateral triangles (convex deltahedron) is the 30 edges of the icosahedron.&#xA;(You could try to make the pentakis dodecahedron out of 60 equilateral triangles, giving 90 equal-length edges, but then it would be slightly concave).&#xA;Any strictly convex solid made of triangles with more than 30 edges must necessarily have 2 or more lengths of edges.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A few more &lt;a href=&quot;http://david.carybros.com/html/3d_design.html#sphere_approximation&quot; rel=&quot;nofollow&quot;&gt;notes on sphere approximation&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1705" LastActivityDate="2011-02-22T23:55:57.167" />
  <row Id="23300" PostTypeId="2" ParentId="23296" CreationDate="2011-02-23T00:12:35.777" Score="3" ViewCount="" Body="&lt;p&gt;Assuming you mean if we start drawing it from $-\infty$ and proceed towards $+\infty$, then yes that is correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is because $a_m x^m$ is the dominating term in the polynomial $P(x) = \sum_{k=0}^{m} a_k x^k$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now for even degree $Q(x)$ (with positive leading coefficient), it's derivative $P(x) = Q&amp;#39;(x)$ is of odd degree and so is negative as we start out from $-\infty$, and thus $Q(x)$ is decreasing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For odd degree $Q(x)$, the derivative has even degree and so is positive near $-\infty$ (and also $+\infty$) and is thus increasing.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2011-02-23T00:12:35.777" CommentCount="4" />
  <row Id="23301" PostTypeId="1" AcceptedAnswerId="23302" CreationDate="2011-02-23T00:15:59.623" Score="2" ViewCount="63" Body="&lt;p&gt;Suppose I have a sequence of real numbers such that AGM $(a_{n},a_{n+1}) = 1/2^n$. Would that sequence be unique and how would I go about finding the individual $a_{n}$? (AGM is the arithmetic geometric mean)&lt;/p&gt;&#xA;" OwnerUserId="890" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2011-02-23T00:21:31.597" LastActivityDate="2011-02-23T01:41:59.010" Title="Implicitly determined agm sequence" Tags="&lt;classical-analysis&gt;&lt;agm&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="23302" PostTypeId="2" ParentId="23301" CreationDate="2011-02-23T00:53:44.680" Score="3" ViewCount="" Body="&lt;p&gt;One series that works is generated by noting that $AGM(x,\frac{x}{2})=\frac {3 \pi x}{8K(\frac{1}{9})}$ where $K(x)$ is the complete elliptic integral of the first kind.  So we can take $a_n=\frac{8K(\frac{1}{9})}{3\pi 2^n}$  Numerically, $K(\frac{1}{9})\approx 1.617386$  This yields $AGM(x,\frac{x}{2})\approx 0.7283955 x$ and $a_n\approx \frac{1.37288}{2^n}$.  I suspect you could apply a dither, raising the odd $n$ and lowering the even $n$, but haven't checked.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2011-02-23T01:41:59.010" LastActivityDate="2011-02-23T01:41:59.010" CommentCount="2" />
  <row Id="23303" PostTypeId="1" AcceptedAnswerId="23350" CreationDate="2011-02-23T00:54:55.583" Score="3" ViewCount="76" Body="&lt;p&gt;Let $\pi: X\to C$ be a fibration in curves where $C$ is a non-singular curve and $X$ a regular, integral surface and the generic fiber $X_\eta$ is a non-singular curve over $k(C)$ (these hypotheses might be stronger than necessary, but I just threw a bunch on to make it as nice as possible).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now a point on $X_\eta$, say $p$ is also a point on $X$ itself. Note the generic point of $X$ is the generic point of $X_\eta$, say $\zeta$. All the other points on $X_\eta$ are closed in the curve and are height $1$ points on $X$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I read in a paper that for any point on the generic fiber, $p$, we have $\mathcal{O}_{X,p}\simeq \mathcal{O}_{X_\eta, p}$, and at first I just thought to myself that it's obvious, but when I tried to actually think of a reason it wasn't so obvious.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $X_\eta$ were open in $X$, then this would be clear since restricting to an open and then taking a stalk doesn't cause problems, but why should this still be true for the generic fiber which is neither open nor closed?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One noted consequence is that $k(X)=\mathcal{O}_{X,\zeta}=\mathcal{O}_{X_\eta, \zeta}=k(X_\eta)$.&lt;/p&gt;&#xA;" OwnerUserId="2000" LastActivityDate="2011-02-23T08:22:46.047" Title="Local ring on generic fiber" Tags="&lt;algebraic-geometry&gt;" AnswerCount="1" />
  <row Id="23304" PostTypeId="1" CreationDate="2011-02-23T01:21:07.023" Score="0" ViewCount="83" Body="&lt;p&gt;Everyone is familiar with distributivity of multiplication over addition of real numbers. The distributivity of two binary operations sometimes goes both ways (e.g. max and min, or for lattices in general.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Out of curiosity, I looked at the set of real numbers for which &lt;strong&gt;addition distributes over multiplication.&lt;/strong&gt; A simple computation shows that this set is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\{a,b,c\in \mathbb{R}\, |\, a+b+c=1\}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this just a meaningless fluke, or is there reason to expect something like this?&lt;/p&gt;&#xA;" LastActivityDate="2011-02-23T01:40:23.050" Title="When addition distributes over multiplication" Tags="&lt;soft-question&gt;&lt;algebra&gt;" AnswerCount="1" CommentCount="9" />
  <row Id="23305" PostTypeId="1" CreationDate="2011-02-17T20:49:29.210" Score="1" ViewCount="169" Body="&lt;p&gt;Can someone point me to the full derivation of the Fourier Series? I'm having problems understanding how the a's and b's coeffients are worked out.&lt;/p&gt;&#xA;" OwnerUserId="4789" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2011-02-23T01:44:53.673" LastActivityDate="2011-03-25T02:38:07.033" Title="Derivation of Fourier Series?" Tags="&lt;sequences-and-series&gt;&lt;fourier-analysis&gt;&lt;fourier-series&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="23306" PostTypeId="2" ParentId="23305" CreationDate="2011-02-17T21:37:03.930" Score="0" ViewCount="" Body="&lt;p&gt;If you already believe that any function can be written as an infinite series of sines and cosines with appropriate coefficients, and all you are wondering about is how we go about actually computing the coefficients, then you're most of the way there. The tricky part is proving that the Fourier series works at all; the derivation of the coefficients is pretty straightforward. It's still not exactly intuitive and involves some steps that you would never think of, but it's easy enough to follow. A good derivation is here: &lt;a href=&quot;http://cnx.org/content/m10733/latest/&quot; rel=&quot;nofollow&quot;&gt;http://cnx.org/content/m10733/latest/&lt;/a&gt;&lt;/p&gt;&#xA;" LastActivityDate="2011-02-17T21:37:03.930" />
  <row Id="23307" PostTypeId="2" ParentId="23304" CreationDate="2011-02-23T01:33:38.863" Score="9" ViewCount="" Body="&lt;p&gt;If you mean, &quot;for which real numbers $a$, $b$, and $c$ do we have $a+(bc) = (a+b)(a+c)$?&quot; then you get &#xA;$$a+bc = a^2 + ab+ac + bc,$$ &#xA;or $a=a(a+b+c)$. This means that &lt;em&gt;either&lt;/em&gt; $a=0$, or else $a+b+c=1$, which is probably what you were trying to say (though you forgot the possibility that $a=0$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But this is not a &quot;set of real numbers for which addition distributes over multiplication&quot;. Such a set would be a collection of real numbers $X$, such that for all $r,s,t\in X$ you have $r+st = (r+s)(r+t)$, which would require that given &lt;em&gt;any&lt;/em&gt; three elements in the set, either the first one you picked is $0$, or else they add up to $1$. In particular, any number you pick, picking it three times, would have to be either $0$ or $\frac{1}{3}$; but you cannot have both $\frac{1}{3}$ and $0$, because then picking $\frac{1}{3}$ for $r$ and $s$, and $0$ for $t$, you would not have $r+(st) = (r+s)(r+t)$. So the only collections $X$ that satisfy &lt;em&gt;that&lt;/em&gt; condition are $X=\{0\}$ and $X=\{\frac{1}{3}\}$. Not very interesting at all...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So it's probably better to think about what you are looking for as the collection of all $3$-tuples of real numbers $(a,b,c)$ such that $a+bc = (a+b)(a+c)$, which consists exactly of all $3$-tuples with either $a=0$ or $a+b+c=1$. Geometrically you get the union of two planes in $3$-space: the $yz$-plane, and the $x+y+z=1$ plane. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nothing terribly exciting, I think, or particularly significant. But if you are interested in structures in which &quot;addition&quot; distributes over &quot;multiplication&quot; and vice-versa, then consider looking at &lt;a href=&quot;http://en.wikipedia.org/wiki/Boolean_algebra_%28structure%29&quot; rel=&quot;nofollow&quot;&gt;boolean algebras&lt;/a&gt;, or more generally &lt;a href=&quot;http://en.wikipedia.org/wiki/Lattice_%28order%29&quot; rel=&quot;nofollow&quot;&gt;lattices&lt;/a&gt;, where $\wedge$ distributes over $\vee$ and vice-versa (like $\cap$ and $\cup$ do for sets, and the logical operators AND and OR do for logic). &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-23T01:40:23.050" LastActivityDate="2011-02-23T01:40:23.050" />
  <row Id="23308" PostTypeId="1" CreationDate="2011-02-23T02:10:09.790" Score="0" ViewCount="77" Body="&lt;p&gt;I am given a 2D &lt;em&gt;convex&lt;/em&gt; polygon ( given in terms of coordinate of the vertices), and I want to lay it on a grid with an undetermined uniform square grid size, $x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, my task is to choose $x$ such that, regardless of the size and the shape of the polygon, the number of the grid, $G_n$, inside, or intersecting the polygon must remain &lt;strong&gt;&lt;em&gt;roughly&lt;/em&gt;&lt;/strong&gt; the same. $G_n$ is a constant that is given.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some additional info: $x$ should be relatively small compare to the dimension of the polygon, and so, $G_n$ should be quite big ( $&gt;100$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the algorithm to do this? Note that the polygon can be of any size and shape, as long as it is convex. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand that the definition is a bit loose here, but I wonder whether anyone has done any research on this, to help to clarify all the concepts?&lt;/p&gt;&#xA;" OwnerUserId="550" LastEditorUserId="550" LastEditorDisplayName="" LastEditDate="2011-02-23T05:53:46.673" LastActivityDate="2011-02-23T12:17:10.573" Title="Given a polygon, and the fix number of grid, find the grid size" Tags="&lt;geometry&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="23310" PostTypeId="2" ParentId="23305" CreationDate="2011-02-23T02:28:31.690" Score="1" ViewCount="" Body="&lt;p&gt;Imagine that $f(x) = \sum_n a_n \cos n x + \sum_n b_n \sin n x$ (for $x \in [0,2\pi]$, say).  The idea for computing the $a_n$s and $b_n$s is that when&#xA;you write down integrals of the form $\int_0^{2\pi} \cos m x \cos n x,$&#xA;or $\int_0^{2\pi} \sin m x \sin n x$, or $\int_0^{2 \pi} \cos m x \sin n x$,&#xA;then the integrals vanish (just compute them!) unless $m = n$ and the functions&#xA;coincide; and in the cases when they don't vanish, their values are easily computed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So taking $f$, and then computing $\int_0^{2\pi} f(x) \cos n x$ or $\int_0^{2 \pi}&#xD;&#xA;f(x) \sin n x$, one exactly reads off $a_n$ or $b_n$ (for the value of $n$ you&#xA;chose).  This is where the formulas come from.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The way people normally think about this is as a kind of orthogonal projection:&#xA;the functions $\cos n x$ and $\sin n x$ are like orthogonal basis vectors in &#xA;a vector space, and the integral is like an inner product.  So to find the &#xA;coefficient of a given basis vector (i.e. an $a_n$ or a $b_n$) one takes the&#xA;inner product against that particular basis vector.&lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2011-02-23T02:28:31.690" />
  <row Id="23311" PostTypeId="2" ParentId="23308" CreationDate="2011-02-23T02:30:53.787" Score="1" ViewCount="" Body="&lt;p&gt;I think I'm a little confused what you mean by 'must remain the same' - must remain the same under what transformations (or other operations)?  In general (assuming that you want $G_n$ to be invariant under translational displacement with respect to the grid), this is impossible; choose as the given polygon a (convex) quadrilateral in general position with all vertex positions (relative to the 'top' vertex) and all line slopes transcendental.  Then whatever the grid size, if the top vertex is positioned precisely on some reference grid point, none of the edges of the polygon and none of the other vertices can pass through a grid point.  This means that the polygon can be shifted vertically some small amount to bring the reference grid point either inside or outside without changing the status of any other grid points, thus changing your value of $G_n$.&lt;/p&gt;&#xA;" OwnerUserId="785" LastActivityDate="2011-02-23T02:30:53.787" CommentCount="1" />
  <row Id="23312" PostTypeId="1" AcceptedAnswerId="23325" CreationDate="2011-02-23T02:31:25.610" Score="11" ViewCount="369" Body="&lt;p&gt;What is the importance of eigenvalues/eigenvectors?  &lt;/p&gt;&#xA;" OwnerUserId="6764" LastActivityDate="2011-02-23T10:58:30.603" Title="What is the importance of eigenvalues/eigenvectors?" Tags="&lt;linear-algebra&gt;&lt;eigenvalues&gt;&lt;eigenvectors&gt;" AnswerCount="4" CommentCount="5" FavoriteCount="6" />
  <row Id="23313" PostTypeId="2" ParentId="23312" CreationDate="2011-02-23T02:42:01.510" Score="2" ViewCount="" Body="&lt;p&gt;The behaviour of a linear transformation can be obscured by the choice of basis. For some transformations, this behaviour can be made clear by choosing a basis of eigenvectors: the linear transformation is then a (non-uniform in general) scaling along the directions of the eigenvectors. The eigenvalues are the scale factors.&lt;/p&gt;&#xA;" OwnerUserId="589" LastActivityDate="2011-02-23T02:42:01.510" />
  <row Id="23314" PostTypeId="1" CreationDate="2011-02-23T02:50:31.417" Score="1" ViewCount="164" Body="&lt;p&gt;I was wondering what would you classify the damping state (under damped, over-damped, critically damped) a second-order ode system with a negative damping ratio? To me it doesn't make much sense since a negative damping ratio results in an unstable system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The system I have is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$G(s) = \frac{5}{s^2-6s+16}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So solving for the natural frequency and damping ratio:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\omega_n = \sqrt{16} = 4$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$2 \zeta \omega_n = -6$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\zeta = \frac{-6}{2 \omega_n} = -0.75$&lt;/p&gt;&#xA;" OwnerUserId="5747" LastActivityDate="2011-03-25T08:38:07.500" Title="Negative damping ratio for second order system?" Tags="&lt;homework&gt;&lt;differential-equations&gt;" AnswerCount="1" />
  <row Id="23315" PostTypeId="2" ParentId="23296" CreationDate="2011-02-23T03:05:51.297" Score="3" ViewCount="" Body="&lt;p&gt;As already noted, the answer is &quot;yes&quot;.  Here is one way to see it:&#xA;factor out the leading term, so the polynomial has the form&#xA;$P(x) = a x^n( 1 + \text{ expression in powers of } \frac{1}{x}).$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $x$ is very positive or very negative, i.e. if $|x| \gg 0$, then the expression in powers of &#xA;$\frac{1}{x}$ will be negligible, and so &#xA;$P(x) \sim a x^n$ for $| x | \gg 0.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So now just look at the behaviour of $a x^n$.  If $n$ is odd this is very negative&#xA;when $x$ is very negative, becoming more so as $x\to -\infty$, while it is very positive when $x$ is very positive, becoming more so as $x \to \infty$.  If $n$ is even then it is very positive when $x$ is either very negative or very positive, and increases both as&#xA;$x \to -\infty$ and as $x \to \infty$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: As Didier Piau notices, this argument with crude asymptotics is not precise enough to conclude true monotonicity for $|x| \gg 0$; Moron's argument with derivatives is better for that.  However,  it does give an explanation for the rough behaviour of $P(x)$ for $|x| \gg 0$.  &lt;/p&gt;&#xA;" OwnerUserId="221" LastEditorUserId="221" LastEditorDisplayName="" LastEditDate="2011-02-23T13:43:06.233" LastActivityDate="2011-02-23T13:43:06.233" CommentCount="4" />
  <row Id="23316" PostTypeId="1" CreationDate="2011-02-23T03:06:02.283" Score="2" ViewCount="133" Body="&lt;p&gt;Hey all,&#xA;I know that gcd(a,b) divides a and b, and must also then divide (a)(t) (t being some integer).  This makes sense to me, but how do I prove it?  It seems that the addition of (a)(t) is a continuation of the linear combination of gcd(a,b) = av + bu for some v, u being elements of Z.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any help?&lt;/p&gt;&#xA;" OwnerUserId="7395" LastEditorUserId="462" LastEditorDisplayName="" LastEditDate="2011-02-23T05:11:33.280" LastActivityDate="2011-02-23T09:19:38.533" Title="Prove: gcd(a,b) = gcd(a, b + at)" Tags="&lt;elementary-number-theory&gt;" AnswerCount="5" />
  <row Id="23317" PostTypeId="1" AcceptedAnswerId="23340" CreationDate="2011-02-23T03:10:11.420" Score="2" ViewCount="117" Body="&lt;p&gt;I'm having a terrible time trying to understand Fourier transforms.  I'm very visual so leaving the X,Y,Z,t domain is not working form me :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm trying to figure out the basics at the moment.  Like, taking a Sine wave (they're odd right?) and converting it into its real and imaginary numbers.  I'm pretty sure I got that working, but to make sure, what should the plotted data look like?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, how do I find the power spectrum of a transform?  How do I use FT to identify the n most significant frequencies in a signal?  That last question shows how lost I am!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I have:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know how to get the real and imaginary numbers from a signal.  I know how to get the phase and the magnitude.  What I need to get is the power spectrum and the most significant frequencies.  Also any dumbed down explaination of what's going on would be very helpful!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="7396" LastActivityDate="2011-02-23T05:56:06.713" Title="Fourier Transforms." Tags="&lt;homework&gt;&lt;analysis&gt;&lt;fourier-analysis&gt;&lt;signal-processing&gt;" AnswerCount="1" FavoriteCount="2" />
  <row Id="23318" PostTypeId="2" ParentId="23316" CreationDate="2011-02-23T03:11:26.077" Score="1" ViewCount="" Body="&lt;p&gt;Let $d = \gcd(a,b)$ and $d&amp;#39; = \gcd(a, b+at)$. Then $d$ divides $a$ and $d$ divides $b$. So $d$ divides $at$. So $d$ divides $b+at$. Likewise, $d&amp;#39;$ divides $a$ and $d&amp;#39;$ divides $b+at$. So $d&amp;#39;$ divides $b$.&lt;/p&gt;&#xA;" OwnerUserId="2548" LastEditorUserId="2548" LastEditorDisplayName="" LastEditDate="2011-02-23T03:53:50.047" LastActivityDate="2011-02-23T03:53:50.047" CommentCount="1" />
  <row Id="23319" PostTypeId="2" ParentId="23316" CreationDate="2011-02-23T03:45:17.003" Score="0" ViewCount="" Body="&lt;p&gt;You're asking for a proof of the below.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;let c = GCD(a,b); then c | a.t for integer t.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Let's refactor:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;if c | a then c | a.t for integer t.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This is fundamental (it's very close to the definition of 'divisibility'), but to prove it:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;if c | a then cx = a, for some integer x (this is the definition of divisibility)&#xA;if cx = a then cxt = at, for all integer t&#xA;c | cxt (again, the definition of divisibility)&#xA;if c | cxt, then c | at&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="662" LastActivityDate="2011-02-23T03:45:17.003" CommentCount="9" />
  <row Id="23320" PostTypeId="2" ParentId="23312" CreationDate="2011-02-23T03:53:58.897" Score="1" ViewCount="" Body="&lt;p&gt;I think if you want a better answer, you need to tell us more precisely what you&#xA;  may have in mind: are you interested in theoretical aspects of eigenvalues; do&#xA;  you have a specific application in mind? Matrices by themselves are just arrays of&#xA;  numbers, which take meaning once you set up a context. Without the context, it seems&#xA;  difficult to give you a good answer. If you use matrices to describe adjacency relations,&#xA;  then eigenvalues/vectors may mean one thing; if you use them to represent linear maps&#xA;  something else, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One possible application: In some cases, you may be able to diagonalize your&#xA;  matrix M using the eigenvalues, which gives you a nice expression for M^k. &#xA;  Specifically, you may be able to decompose your matrix into a product&#xA;  SDS^-1 , where D is diagonal, with entries the eigenvalues, and S is the&#xA;  matrix with the associated respective eigenvectors.&#xA;    I hope it is not a problem to post this as a comment. I got a couple of Courics&#xA;  here last time for posting a comment in the answer site.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mr. Arturo:&#xA;   Interesting approach!. This seems to connect with the theory of characteristic&#xA;   curves in PDE's(who knows if it can be generalized to dimensions higher than 1), which&#xA;   are curves along which a PDE becomes an ODE, i.e., as you so brilliantly said,&#xA;   curves along which the PDE decouples. &lt;/p&gt;&#xA;" OwnerUserId="7398" LastEditorUserId="7398" LastEditorDisplayName="" LastEditDate="2011-02-23T05:54:27.593" LastActivityDate="2011-02-23T05:54:27.593" CommentCount="2" />
  <row Id="23321" PostTypeId="2" ParentId="23316" CreationDate="2011-02-23T03:54:05.413" Score="3" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\rm\ \ $ If $\rm\ c\ |\ a\ $ then $\rm\ c\ |\ b + a\ t\ \iff\ c\ |\ b\:.\ $ This implies that  $\rm\ \{a\:,\:b+a\ t\}\ $ and $\rm\ \{a\:,\: b\}\ $ have the same set of common divisors $\rm\:c\:,\:$ hence they have the same &lt;em&gt;greatest&lt;/em&gt; common divisor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Modly: $\:$ if $\rm\ a\equiv 0\ $ then $\rm\ b+a\ t\equiv 0\: \iff\: b\equiv 0\ \ \ (mod\ c)$&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2011-02-23T05:32:36.930" LastActivityDate="2011-02-23T05:32:36.930" CommentCount="3" />
  <row Id="23322" PostTypeId="1" CreationDate="2011-02-23T03:56:48.610" Score="3" ViewCount="85" Body="&lt;p&gt;Is it possible, given a point in the interior of an angle to construct a circle through that point tangent to the rays forming the angle?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose you have a point $A$ in the interior of an angle with vertex $O$. You can bisect the angle, take a point $I$ on the angle bisector, and drop perpendiculars $IP$ and $IQ$ to each side of the angle. The circle with center $I$ and radius $IP$ is then tangent to the rays of the angle. Call this circle $\Gamma$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can then draw a line $OA$, which will intersect $\Gamma$ at some point $A&amp;#39;$ on the far side of $\Gamma$ from $O$. You can then draw lines $EA$ parallel to $PA&amp;#39;$ and $FA$ parallel to $QA&amp;#39;$, where $E$ is the intersection with the ray containing $P$, and $F$ is the intersection with the ray containing $Q$. It seems intuitively obvious that the circle through $EAF$ will be a circle through $A$ tangent to the rays of the angle. But is there a way to prove it? I don't see how to do so. Thanks.&lt;/p&gt;&#xA;" OwnerUserId="7399" LastActivityDate="2011-02-23T05:37:07.450" Title="Constructing a circle through a point in the interior of an angle." Tags="&lt;geometry&gt;" AnswerCount="3" />
  <row Id="23323" PostTypeId="1" CreationDate="2011-02-23T04:13:10.817" Score="0" ViewCount="38" Body="&lt;p&gt;I ran into this problem with my abstract algebra homework the other night, and I think I am really close to solving it.  By the definition of modulo n, I know that&#xA;c=a + rm, m being some integer&#xA;c=b + sn, n being some integer&#xA;I want to prove a=b + (rm + sn)&#xA;I have tried algebraically manipulating these equalities, but to no avail.  Is this the right track, or do I need a different approach?  I have a quiz tomorrow, so an answer would be much appreciated. :D &lt;/p&gt;&#xA;" OwnerUserId="7395" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-23T04:39:00.053" LastActivityDate="2011-02-23T04:39:00.053" Title="If $c\equiv a \pmod{r}$ and $c\equiv b\pmod{s}$,  prove $a\equiv b\pmod{\gcd(a,b)}$" Tags="&lt;elementary-number-theory&gt;" AnswerCount="3" CommentCount="3" />
  <row Id="23324" PostTypeId="1" AcceptedAnswerId="23382" CreationDate="2011-02-23T04:15:09.267" Score="0" ViewCount="114" Body="&lt;p&gt;The quaternion group of order 8 has an irreducible two dimensional representation over $\mathbb{C}$ but how does one show that this representation cannot be defined over $\mathbb{R}$?&lt;/p&gt;&#xA;" OwnerUserId="6369" LastEditorUserId="6369" LastEditorDisplayName="" LastEditDate="2011-02-23T04:31:33.003" LastActivityDate="2011-02-23T14:39:31.603" Title="representation of quaternion group over $\mathbb{C}$ and $\mathbb{R}$" Tags="&lt;abstract-algebra&gt;&lt;representation-theory&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="23325" PostTypeId="2" ParentId="23312" CreationDate="2011-02-23T04:17:32.403" Score="15" ViewCount="" Body="&lt;p&gt;Short answer: Eigenvectors make understanding linear transformations easy. They are the &quot;axes&quot; (directions) along which a linear transformation acts simply by &quot;stretching/compressing&quot; and/or &quot;flipping&quot;; eigenvalues give you the factors by which this compression occurs. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The more directions you have along which you understand the behavior of a linear transformation, the easier it is to understand the linear transformation; so you want to have as many linearly independent eigenvectors as possible associated to a single linear transformation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Slightly longer answer: there are a lot of problems that can be modeled with linear transformations, and the eigenvectors give very simply solutions. For example, consider the system of linear differential equations&#xA;\begin{align*}&#xA;\frac{dx}{dt} &amp;amp;= ax + by\\&#xA;\frac{dy}{dt} &amp;amp;= cx + dy.&#xA;\end{align*}&#xA;This kind of system arises when you describe, for example, the growth of population of two species that affect one another. For example, you might have that species $x$ is a predator on species $y$; the more $x$ you have, the fewer $y$ will be around to reproduce; but the fewer $y$ that are around, the less food there is for $x$, so fewer $x$s will reproduce; but then fewer $x$s are around so that takes pressure off $y$, which increases; but then there is more food for $x$, so $x$ increases; and so on and so forth. It also arises when you have certain physical phenomena, such a particle on a moving fluid, where the velocity vector depends on the position along the fluid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solving this system directly is complicated. But suppose that you could do a change of variable so that instead of working with $x$ and $y$, you could work with $z$ and $w$ (which depend linearly on $x$; that is, $z=\alpha x+\beta y$ for some constants $\alpha$ and $\beta$, and $w=\gamma x + \delta y$, for some constants $\gamma$ and $\delta$) and the system transformed into something like&#xA;\begin{align*}&#xA;\frac{dz}{dt} &amp;amp;= \kappa z\\&#xA;\frac{dw}{dt} &amp;amp;= \lambda w&#xA;\end{align*}&#xA;that is, you can &quot;decouple&quot; the system, so that now you are dealing with two &lt;em&gt;independent&lt;/em&gt; functions. Then solving this problem becomes rather easy: $z=Ae^{\kappa t}$, and $w=Be^{\lambda t}$. Then you can use the formulas for $z$ and $w$ to find expressions for $x$ and $y$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can this be done? Well, it amounts &lt;em&gt;precisely&lt;/em&gt; to finding two linearly independent eigenvectors for the matrix $\left(\begin{array}{cc}a &amp;amp; b\\c &amp;amp; d\end{array}\right)$! $z$ and $w$ correspond to the eigenvectors, and $\kappa$ and $\lambda$ to the eigenvalues.  By taking an expression that &quot;mixes&quot; $x$ and $y$, and &quot;decoupling it&quot; into one that acts independently on two different functions, the problem becomes a lot easier. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is the essence of what one hopes to do with the eigenvectors and eigenvalues: &quot;decouple&quot; the ways in which the linear transformation acts into a number of independent actions along separate &quot;directions&quot;, that can be dealt with independently. A lot of problems come down to figuring out these &quot;lines of independent action&quot;, and understanding them can really help you figure out what the matrix/linear transformation is &quot;really&quot; doing. &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-23T04:42:08.270" LastActivityDate="2011-02-23T04:42:08.270" CommentCount="1" />
  <row Id="23328" PostTypeId="2" ParentId="23323" CreationDate="2011-02-23T04:23:17.437" Score="1" ViewCount="" Body="&lt;p&gt;I am &lt;em&gt;guessing&lt;/em&gt; that the question you are trying to ask is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If $c$ is congruent to $a$ modulo $r$, and $c$ is congruent to $b$ modulo $s$, then $a$ is congruent to $b$ modulo $\gcd(r,s)$ (not modulo $\gcd(a,b)$).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If so: you know $r$ divides $c-a$, so $\gcd(r,s)$, which divides $r$, also divides $c-a$. And $s$ divides $c-b$, so $\gcd(r,s)$ divides $c-b$. So $\gcd(r,s)$ divides &lt;em&gt;both&lt;/em&gt; $c-a$ and $c-b$, which means it divides their difference. Which just happens to be...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or along what you write: express $r$ and $s$ as multiples of $\gcd(r,s)=d$. Since $r=dk$ and $s=d\ell$ for some $k$ and $\ell$, you get&#xA;\begin{align*}&#xA;c &amp;amp;= a+rm = a+d(km).\\&#xA;c &amp;amp;= b+sn = b+d(\ell n).&#xA;\end{align*}&#xA;That means $a+d(km) = b+d(\ell n)$. Now, can you write $a = b+dt$ for some integer $t$?&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2011-02-23T04:23:17.437" />
  <row Id="23329" PostTypeId="2" ParentId="23323" CreationDate="2011-02-23T04:23:41.413" Score="1" ViewCount="" Body="&lt;p&gt;Suppose $c \equiv a \pmod r$ and $c \equiv b \pmod s$. We want to show that $a \equiv b \pmod{t}$ where $t = \gcd(r,s)$. So $c-a = rl$ and $c-b = sm$ for $l,m \in \mathbb{Z}$. So $a-b = sm-rl$. &lt;/p&gt;&#xA;" OwnerUserId="2548" LastEditorUserId="2548" LastEditorDisplayName="" LastEditDate="2011-02-23T04:32:20.560" LastActivityDate="2011-02-23T04:32:20.560" />
  <row Id="23330" PostTypeId="2" ParentId="23323" CreationDate="2011-02-23T04:27:55.637" Score="1" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\rm\ \ \ d\ |\ r\ |\ c-a,\ \ \ d\ |\ s\ |\ c-b\ \ \Rightarrow\ \ d\ | \ c-b-(c-a)\ =\ a-b\:.\ $ Now put $\rm\ d = gcd(r,s)$&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2011-02-23T04:36:37.763" LastActivityDate="2011-02-23T04:36:37.763" />
  <row Id="23331" PostTypeId="2" ParentId="23322" CreationDate="2011-02-23T04:53:03.983" Score="1" ViewCount="" Body="&lt;p&gt;Let $\Phi$ be the homothety centered at $O$ mapping $A&amp;#39;$ to $A$.  Then $\Phi$ fixes the lines $\overline{OP}$ and $\overline{OQ}$, so it maps $P$ to $E$ and $Q$ to $F$.  Then it maps the circle $\Gamma$ to the circle through $EAF$, so the circle through $EAF$ must be tangent to $\overline{OP}$ and $\overline{OQ}$&lt;/p&gt;&#xA;" OwnerUserId="1726" LastActivityDate="2011-02-23T04:53:03.983" />
  <row Id="23332" PostTypeId="2" ParentId="23322" CreationDate="2011-02-23T04:55:16.773" Score="1" ViewCount="" Body="&lt;p&gt;Consider $\triangle OAE$ and $\triangle OA&amp;#39;P$.  Because they share vertex $O$ and $AE\parallel A&amp;#39;P$, the triangles are similar.  Similar triangles are dilation images of one another, so $\triangle OAE$ is the image of $\triangle OA&amp;#39;P$ under the dilation with center $O$ and ratio $\frac{OA}{OA'}$—call this dilation $T_d$.  Likewise, $\triangle OAF$ is the image of $\triangle OA&amp;#39;Q$ under $T_d$.  That is, $E$, $A$, and $F$ are the images of $P$, $A&amp;#39;$, and $Q$, respectively, under $T_d$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since three points uniquely determine a circle, the circle through $E$, $A$, and $F$ is the image of the circle you called $\Gamma$ under $T_d$, so its center—call it $X$—is the image of $I$ under $T_d$.  Now, $T_d$ maps $\triangle OIP$ to $\triangle OXE$, so $\overline{XE}\perp\overline{OE}$, and similarly $\overline{XF}\perp\overline{OF}$, so the circle through $E$, $A$, and $F$ is tangent to the rays of the angle.&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2011-02-23T04:55:16.773" />
  <row Id="23333" PostTypeId="2" ParentId="23207" CreationDate="2011-02-23T05:14:34.857" Score="1" ViewCount="" Body="&lt;p&gt;Alexander duality is the closest to what you seem to want: &lt;a href=&quot;http://en.wikipedia.org/wiki/Alexander_duality&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Alexander_duality&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1138" LastActivityDate="2011-02-23T05:14:34.857" />
  <row Id="23334" PostTypeId="1" AcceptedAnswerId="23368" CreationDate="2011-02-23T05:20:57.817" Score="0" ViewCount="41" Body="&lt;p&gt;I have been doing some casual internet research on biclusters.  (I have read the Wiki article several times.)  So far, it seems as if there are few definitions or standard terminology.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;I was wondering if there were any standard papers or books that anybody who is interested in algorithms for finding biclusters should read.  &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Is it possible to say what is the state of the art in the field?  I was intrigued by the notion of finding biclusters using genetic algorithms, so I would appreciate comments on that approach in particular in the context of other approaches.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Usually in clustering, the goal is to partition the data-set into groups where each element is in some group.  Do bicluster algorithms also seek to put all elements in a particular group?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Note: Also posted &lt;a href=&quot;http://stats.stackexchange.com/questions/7419/seeking-advice-on-getting-started-with-biclustering&quot;&gt;here&lt;/a&gt;.  I thought perhaps the larger number of mathematically-inclined eyeballs on here might help.&lt;/p&gt;&#xA;" OwnerUserId="1004" LastActivityDate="2011-02-23T11:52:20.410" Title="Getting started with biclustering" Tags="&lt;clustering&gt;" AnswerCount="1" FavoriteCount="0" />
  <row Id="23335" PostTypeId="2" ParentId="23316" CreationDate="2011-02-23T05:26:37.933" Score="1" ViewCount="" Body="&lt;p&gt;$$\displaystyle ax+by = a(x-ty)+(b+at)y$$&lt;/p&gt;&#xA;" OwnerUserId="3302" LastActivityDate="2011-02-23T05:26:37.933" />
  <row Id="23336" PostTypeId="1" AcceptedAnswerId="23337" CreationDate="2011-02-23T05:27:25.303" Score="0" ViewCount="38" Body="&lt;p&gt;Suppose $z = x + y$ and in addition, there is a function $f(z)=F(x,y)$ for all values.  Is there a way to relate,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{df}{dz}, \frac{\partial F}{\partial x}\text{ and }\frac{\partial F}{\partial y}\textrm{?}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, suppose $z = g(x,y)$, what is the relationship (if any)? &lt;/p&gt;&#xA;" OwnerUserId="1004" LastEditorUserId="1004" LastEditorDisplayName="" LastEditDate="2011-02-23T06:03:57.280" LastActivityDate="2011-02-23T06:03:57.280" Title="Derivatives with respect to a variable that is a function of other variables." Tags="&lt;real-analysis&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="23337" PostTypeId="2" ParentId="23336" CreationDate="2011-02-23T05:33:31.687" Score="2" ViewCount="" Body="&lt;p&gt;We're given $F(x,y)=f(x+y)$. So $\frac{\partial F}{\partial x} (x,y)= f&amp;#39;(x+y)\cdot 1=\frac{\partial F}{\partial y}(x,y)$, by the Chain Rule.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, if $F(x,y)=f(g(x,y))$, we only have the statement of the Chain Rule. $\frac{\partial F}{\partial x}(x,y) = f&amp;#39;(g(x,y)) \frac{\partial g}{\partial x} (x,y)$ and $\frac{\partial F}{\partial y}(x,y) = f&amp;#39;(g(x,y)) \frac{\partial g}{\partial y} (x,y)$.&lt;/p&gt;&#xA;" OwnerUserId="5253" LastActivityDate="2011-02-23T05:33:31.687" />
  <row Id="23338" PostTypeId="2" ParentId="23322" CreationDate="2011-02-23T05:37:07.450" Score="1" ViewCount="" Body="&lt;p&gt;This isn't exactly a construction, but this might prove that such a circle exists. The basic idea is to slide a point $p$ on the angle bisector until the length of the dropped perpendicular is equal to the distance between $p$ and $A$. Then $p$ will be the center of a circle tangent to the two rays (and the circle will include $A$ by design). Here goes:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can assume one of the rays determining the angle is the positive x-axis. So &lt;em&gt;O&lt;/em&gt; is the origin, and let $x_0$ be the x-coordinate of your point &lt;em&gt;A&lt;/em&gt;. We want to find a point $p=(x_1, y_1)$ on the angle bisector such that $d(A,p)=x_1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So let $p=(x_1,y_1)$ be an arbitrary point on the bisector. We have $x_1=0$ when $p=O$ (in particular $x_1&amp;lt;d(A,p)$ ), and we have $x_1&amp;gt;d(A,p)$ when $x_0=x_1$. Since these distances are varying continuously for $x_1\in [0, x_0]$, the IVT implies there is a point $p$ on the bisector with $x_1=d(A,p)$. This is the point you want.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;NOTE: I would post a nice picture, but my reputation won't allow it.&lt;/p&gt;&#xA;" LastActivityDate="2011-02-23T05:37:07.450" />
  <row Id="23339" PostTypeId="2" ParentId="23056" CreationDate="2011-02-23T05:46:49.943" Score="0" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;EDIT: This is not entirely correct since the radius of the second circle is given.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I'm reading your question correctly, your second circle is not uniquely determined. Here are some examples:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/Z1vwl.png&quot; alt=&quot;non-uniqueness&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The point on the circumference of your (non-unique) second circle is labeled A, but there are infinitely many circles through A that are tangent to the first circle (centered at &lt;em&gt;Q&lt;/em&gt;).&lt;/p&gt;&#xA;" LastActivityDate="2011-02-23T06:02:30.870" CommentCount="2" />
  <row Id="23340" PostTypeId="2" ParentId="23317" CreationDate="2011-02-23T05:56:06.713" Score="1" ViewCount="" Body="&lt;p&gt;I'll try and give some intuition from an Electrical engineering perspective. It is useful to think of a Fourier transform as giving you the frequency domain picture of a given signal, i.e., it gives you a picture of both the amplitude and the phase of the different frequency components that make up the signal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are familiar with Fourier series representation of periodic signals, the Fourier transform can be viewed as an extension to non periodic signals. Of course, not all continuous functions have a well defined Fourier transform, but if you look at typical engineering applications, that is not an issue.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When dealing with deterministic signals, the power spectrum is given by the squared magnitude of the Fourier transform. If $f(t)$ is the signal in the (continuous) time domain and $F(\omega)$ is the frequency domain representation, then the Power spectrum is $|F(\omega)|^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When dealing with random signals, the power spectral density is the Fourier transform of the autocorrelation function of the process (Provided the process is wide sense stationary).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The above explanations carry over to the discrete domain as well. So, if you are working with DFTs of signal samples, things are not significantly different. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the N most significant frequencies, I think you need to take the DFT of your signal and then look at the frequencies in increasing order of amplitudes.&lt;/p&gt;&#xA;" OwnerUserId="2641" LastActivityDate="2011-02-23T05:56:06.713" />
  <row Id="23341" PostTypeId="2" ParentId="150" CreationDate="2011-02-23T06:11:54.507" Score="0" ViewCount="" Body="&lt;p&gt;I don't mean to nitpick but I think we should be specific about the&#xA;  norm/topology we are using when we state that these functions are a&#xA;  dense subset. When we work in [a,b], I think we use the sup norm.&#xA;    Sorry, all the info about norm/topology, etc. is included in the&#xA;    above link.&lt;/p&gt;&#xA;" OwnerUserId="7398" LastActivityDate="2011-02-23T06:11:54.507" />
  <row Id="23342" PostTypeId="1" AcceptedAnswerId="23482" CreationDate="2011-02-23T06:13:17.060" Score="1" ViewCount="37" Body="&lt;p&gt;Let $f_C$ be the convex envelope of $f$ on a non-empty convex set $C \subset \mathbb{R}^n$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need to show that $\{ x^∗ \in C : f(x^∗) \leq f(x), \forall x \in C \} \subset \{x^∗ \in C : f_C(x^∗) \leq f_C(x), \forall x \in C \}.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Right now, I can show that: $\{ x^∗ \in C : f(x^∗) \leq f(x), \forall x \in C \} \subseteq \{x^∗ \in C : f_C(x^∗) \leq f_C(x) \forall x \in C\}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I need a counterexample to show that the equality in the subset relation does not hold. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For reference: The convex envelope of $f$ on $C$, denoted $f_C : C \rightarrow R$ is a convex function such that $f_C(x) \leq f(x) \forall x \in C$. The definition also requires that if $g$ is any other convex function on $C$ for which $g(x) \leq f(x), \forall x \in C$ then $f_C(x) \geq g(x),  \forall x \in C$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: &lt;strong&gt;Sorry for this, turns out that there was a typo and the $\subset$ should have actually been a $\subseteq$.&lt;/strong&gt; After all, f could be a convex function in which case, it would be the same as it's convex hull. &lt;/p&gt;&#xA;" OwnerUserId="7145" LastEditorUserId="7145" LastEditorDisplayName="" LastEditDate="2011-02-24T03:56:07.207" LastActivityDate="2011-02-24T03:56:57.707" Title="Counterexample to show that the set of global minima of a function $f$ is a strict subset of the set of minima of the convex envelope of $f$" Tags="&lt;homework&gt;&lt;convex-optimization&gt;" AnswerCount="2" />
  <row Id="23343" PostTypeId="1" AcceptedAnswerId="23354" CreationDate="2011-02-23T06:26:07.470" Score="1" ViewCount="77" Body="&lt;p&gt;This is a problem from Borevich and Shafarevich (p.180).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $\mathfrak{O} = k[x]$, the problem is to find the set of valuations $A$ of the field $K = k(x)$ which satisfies the conditions (1),(2),(3) below.  I think I want to find the largest such set.  I also want to do the same when $\mathfrak{O} = k[1/x]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can someone please give me some guidance on how to do this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is the relevant terminology.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A function $v$ on a field $K$ is called a valuation of $K$ if it satisfies the following:&lt;br&gt;&#xA;(i) $v(\alpha)$ takes on all rational integral values as $\alpha$ ranges through the nonzero elements of $K$, $v(0)=\infty$&lt;br&gt;&#xA;(ii) $v(\alpha \beta) = v(\alpha) + v(\beta)$&lt;br&gt;&#xA;(iii) $v(\alpha + \beta) \geq \min(v(\alpha),v(\beta))$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $\mathfrak{O}$ be a integral domain with quotient field $K$ and let $A$ be a set of valuations of $K$.  Consider the following properties that $A$ may possess.&lt;br&gt;&#xA;(1) For any $\alpha \neq 0$ of $\mathfrak{O}$, $v(\alpha) = 0$ for all but finitely many valuations $v \in A$&lt;br&gt;&#xA;(2) An element $\alpha$ of $K$ belongs to $\mathfrak{O}$ if and only if $v(\alpha) \geq 0$ for all $v \in A$&lt;br&gt;&#xA;(3) For any finite set of distinct valuations $v_1, \ldots, v_m$ of $A$ and for any set of nonnegative integers $k_1,\ldots,k_m$, there is an element $\alpha \in \mathfrak{O}$ for which $v_i(\alpha) = k_i$, $(i=1,\ldots,m)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that $A$ satisfying the conditions (1),(2),(3) is equivalent to $A$ inducing a theory of divisors on $\mathfrak{O}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A theory of divisors on a domain $\mathfrak{O}$ consists of a commutative semigroup $D$ (semigroup = associative multiplication with identity) having the property of unique factorization into irreducibles along with a homomorphism $\alpha \rightarrow (\alpha)$ of the semigroup $\mathfrak{O}^{\times}$ into $D$ satisfying&lt;br&gt;&#xA;(i) An element $\alpha \in \mathfrak{O}^{\times}$ is divisible by $\beta \in \mathfrak{O}^{\times}$ in the ring $\mathfrak{O}$ if and only if $(\alpha)$ is divisible by $(\beta)$ in $D$.&lt;br&gt;&#xA;(ii) If $\alpha$ and $\beta$ of $\mathfrak{O}$ are divisible by $a \in D$, then $\alpha \pm \beta$ are also divisible by $a$.&lt;br&gt;&#xA;(iii) If $a$ and $b$ are two elements of $D$ and the set of all elements $\alpha \in \mathfrak{O}$ which are divisible by $a$ coincides with the set of all elements $\beta \in \mathfrak{O}$ which are divisible by $b$, then $a = b$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: I have one clarification question about condition (i) in the definition of a valuation.  Is the intention that $v$ should assume all integers values and also only integer values?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I ask because another problem is to show that an algebraically closed field $K$ has no valuations, and I have only been able to prove this when $v$ assumes only integer values.  My proof is that for any given $\alpha \in K$ with valuation $v(\alpha) = m \neq 0$ and every positive integer $n$, the equation $x^n = \alpha$ has a solution $x \in K$, and so the equation $ny = m$ has a solution $y = v(\alpha) \in \mathbb{Z}$, which says that every integer $n$ divides the given nonzero integer $m$.  Contradiction.&lt;/p&gt;&#xA;" OwnerUserId="6887" LastEditorUserId="6887" LastEditorDisplayName="" LastEditDate="2011-02-23T14:39:09.750" LastActivityDate="2011-02-23T16:09:30.983" Title="How to find the set of valuations of the field $k(x)$ that satisfy certain properties?" Tags="&lt;number-theory&gt;&lt;algebraic-number-theory&gt;" AnswerCount="1" />
  <row Id="23344" PostTypeId="2" ParentId="150" CreationDate="2011-02-23T06:27:37.147" Score="1" ViewCount="" Body="&lt;p&gt;A natural class of examples would be paths of Brownian motion. These are continuous but non-differentiable everywhere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You may also be interested in fractal curves such as the Takagi function, which is also continuous but nowhere differentiable. (I think Wikipedia calls it the &quot;Blancmange curve&quot;.) I like this one better than the Weierstrass function, but this is personal preference.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Brownian_motion&quot; rel=&quot;nofollow&quot;&gt;Brownian Motion&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Blancmange_curve&quot; rel=&quot;nofollow&quot;&gt;Takagi function&lt;/a&gt;&lt;/p&gt;&#xA;" LastActivityDate="2011-02-23T06:27:37.147" />
  <row Id="23345" PostTypeId="2" ParentId="23342" CreationDate="2011-02-23T06:31:04.477" Score="0" ViewCount="" Body="&lt;p&gt;Take the function $f(x)=x^2 (x-1)^2$ on $I=[-2,2]$, then the global minimum of f are $0,1$. If $f_C$ is its convex envelope, then you showed that 0,1 are global minimum for $f_C$ but then all the points in $[0,1]$ must be global minimum. (or in the general case, the global minimum of $f_C$ will contain the convex hull of the global minimum of $f$).&lt;/p&gt;&#xA;" OwnerUserId="2125" LastEditorUserId="2125" LastEditorDisplayName="" LastEditDate="2011-02-23T06:50:07.977" LastActivityDate="2011-02-23T06:50:07.977" />
  <row Id="23346" PostTypeId="2" ParentId="21677" CreationDate="2011-02-23T07:06:45.973" Score="1" ViewCount="" Body="&lt;p&gt;The answer is yes. Three lines in general position determine a unique hyperboloid of one sheet (since they determine a &lt;strong&gt;ruling&lt;/strong&gt;, or a one parameter family of lines on a hyperbola). But there is a second ruling of the same hyperboloid; every line in this ruling meets the original three lines. So Ross is correct, there is a one parameter family of solutions (in the shape of a hyperboloid).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I'm ever not up to my eyeballs in homework, I'll make a picture of this and post it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;NOTE: Questions of this sort can be addressed in generality using the &lt;strong&gt;Schubert calculus&lt;/strong&gt;, which is roughly a part of enumerative geometry.&lt;/p&gt;&#xA;" LastActivityDate="2011-02-23T07:06:45.973" CommentCount="0" />
  <row Id="23347" PostTypeId="1" AcceptedAnswerId="23366" CreationDate="2011-02-23T07:28:36.097" Score="0" ViewCount="95" Body="&lt;p&gt;As a step in proving that the union of intervals $B_E \subset [0,1]$ (where $E$ is the set of infinite Bernoulli sequences, e.g. 0.0110...., such that some finite pattern repeats infinitely often) is a Borel set, the authors in &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0817638849&quot; rel=&quot;nofollow&quot;&gt;this book&lt;/a&gt; build the set as follows. First, they define $E_n$ as the set of sequences where the pattern occurs beginning at position $n$ and $B_{E_n}$ to be the corresponding union of intervals. Then they state without proof that&#xA;\begin{align}&#xA;B_E = \bigcap^\infty_{k=1} \; \bigcup_{n \geq k} \; B_{E_n}&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After thinking about this statement for a few days, I have now convinced myself that this does indeed represent the set of Bernoulli sequences (and the corresponding intervals) where a given finite pattern repeats infinitely often.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, if someone asked me to prove that this infinite intersection of infinite unions does represent $B_E$, I'm not sure where I should start (I think I could do a hand waving argument, though). So my question is, how do you formally prove equivalence between the LHS and RHS? &lt;/p&gt;&#xA;" OwnerUserId="6690" LastActivityDate="2011-02-23T11:37:50.243" Title="Building a Bernoulli sequence where finite patterns repeat infinitely often" Tags="&lt;measure-theory&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="23350" PostTypeId="2" ParentId="23303" CreationDate="2011-02-23T08:22:46.047" Score="1" ViewCount="" Body="&lt;p&gt;Take affine affine open sets $U\subseteq X$ and $V\subseteq C$ such that $\pi|_U: U\rightarrow V$.Then $B:=O_X(U)$ is an $A:=O_C(V)$-algebra and $O_{X,p}=B_p$, where $p\in\mathrm{Spec} (B)$ satisfies $p\cap A=0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The generic fibre $X_\zeta\cap U$ equals $\mathrm{Spec}((A\setminus 0)^{-1}B)$. Hence&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$&#xD;&#xA;O_{X_\zeta ,p} =((A\setminus 0)^{-1}B)_{p(A\setminus 0)^{-1}B}=B_p&#xD;&#xA;$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;since $B\setminus p$ contains $A\setminus 0$.&lt;/p&gt;&#xA;" OwnerUserId="2479" LastActivityDate="2011-02-23T08:22:46.047" CommentCount="1" />
  <row Id="23351" PostTypeId="1" AcceptedAnswerId="23423" CreationDate="2011-02-23T03:21:41.693" Score="4" ViewCount="172" Body="&lt;p&gt;I was reading &lt;a href=&quot;http://en.wikipedia.org/wiki/Honeycomb&quot; rel=&quot;nofollow&quot;&gt;this paragraph&lt;/a&gt; and it got me thinking:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The closed ends of the honeycomb cells&#xA;  are also an example of geometric&#xA;  efficiency, albeit three-dimensional&#xA;  and little-noticed. The ends are&#xA;  trihedral (i.e., composed of three&#xA;  planes) sections of rhombic&#xA;  dodecahedra, with the dihedral angles&#xA;  of all adjacent surfaces measuring&#xA;  120°, the angle that minimizes surface&#xA;  area for a given volume. (The angle&#xA;  formed by the edges at the pyramidal&#xA;  apex is approximately 109° 28' 16&quot; (=&#xA;  180° - arccos(1/3)).)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is hardly intuitive; is there a proof of this somewhere?&lt;/p&gt;&#xA;" OwnerUserId="550" LastActivityDate="2011-02-23T21:00:08.740" Title="Prove that Honeycomb Structures are the Most Geometrically Efficient Structure" Tags="&lt;geometry&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="23353" PostTypeId="2" ParentId="23351" CreationDate="2011-02-23T06:44:26.403" Score="1" ViewCount="" Body="&lt;p&gt;It is called geometrically efficient because it is  &lt;a href=&quot;http://mathworld.wolfram.com/CirclePacking.html&quot; rel=&quot;nofollow&quot;&gt;&lt;strong&gt;densely packed&lt;/strong&gt;&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also read:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Sphere_packing&quot; rel=&quot;nofollow&quot;&gt;Sphere Packing Wikipedia&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="705" LastActivityDate="2011-02-23T06:44:26.403" />
  <row Id="23354" PostTypeId="2" ParentId="23343" CreationDate="2011-02-23T08:37:19.470" Score="1" ViewCount="" Body="&lt;p&gt;In the sequel let $v$ be a valuation of $k(x)$, let $R$ be the valuation ring of $v$ and $M$ its maximal ideal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $v$ satisfies (2) then $k[x]\subseteq R$; in particular $v$ is trivial on $k$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The prime ideal $P:=M\cap k[x]$ is non-zero, thus has the form $P=pk[x]$ with a monic prime polynomial $p$ (monic means that the coefficient of the leading term of $p$ is $1$, not that $p$ has degree $1$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The localication $k[x]_P$ is contained in $R$. Since $k[x]$ is factorial we can write every element of $k(x)$ in the form $f/g$, $f,g\in k[x]$ and not both divisible by $p$. Condition (2) implies that $f/g\in R$ is equivalent to $p$ does not divide $g$. The latter implies $f/g\in k[x]_P$, hence $k[x]_P=R$. Moreover one sees that the valuation $v$ is the $p$-adic valuation of $k(x)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the polynomial ring $k[x]$ these valuations are in bijection with the monic prime polynomials of $k[x]$. The valuation $v$ associated to such a polynomial $p$ is defined in the same way in which the p-adic valuations of the rationals are defined.&#xA;The set $A$ of valuations you get actually coincides with the set of all valuations of $K(x)$ that are non-negative on $k[x]$. So it remains to prove that (1) and (3) are satisfied. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proof of (1) for the family $A$ of all $p$-adic valuations of $k(x)$: an element $\alpha\in k[x]$ has a positive value at a $p$-adic valuation if and only if $\alpha$ is divisible by $p$. Thus the relevant $p$ are the prime factors of $\alpha$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proof of (2): take the prime elements $p_i$ corresponding to the $v_i$, raise them to the powers $k_i$ and multiply these powers to get $\alpha$.&lt;/p&gt;&#xA;" OwnerUserId="2479" LastEditorUserId="2479" LastEditorDisplayName="" LastEditDate="2011-02-23T16:09:30.983" LastActivityDate="2011-02-23T16:09:30.983" CommentCount="6" />
  <row Id="23355" PostTypeId="2" ParentId="23314" CreationDate="2011-02-23T08:37:45.657" Score="1" ViewCount="" Body="&lt;p&gt;I assume the G(s) which you give is the Laplace transform of the Green's function (please indicate next time what your symbols mean otherwise, a question can be hard to understand). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A negative damping rate is not damping but driving. Physically, it results from an external system which constantly pumps energy into your oscillator (I assume your 2nd order ode is an oscillator). In your case with $G(s) =\frac{5}{s^2-6s+16}$, the real-time Green's function is $G(t) = \frac{5}{\sqrt{7}} \sin(\sqrt{7} t) e^{3t}$, i.e., an oscillatory solution whose amplitude is exponentially growing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the way, why do you get that the frequency is 4 (I have $\sqrt{7}$). Do you use a different convention for the Laplace transform. My convention is $G(s) = \int_0^\infty \! dt \, G(t) e^{-s t}$. &lt;/p&gt;&#xA;" OwnerUserId="7266" LastActivityDate="2011-02-23T08:37:45.657" CommentCount="3" />
  <row Id="23356" PostTypeId="2" ParentId="21677" CreationDate="2011-02-23T09:02:52.157" Score="3" ViewCount="" Body="&lt;p&gt;Label the three lines $\ell_1$, $\ell_2$, and $\ell_3$. They cannot intersect for then they would be co-planar. Since lines $\ell_1$ and $\ell_2$ do not intersect, they have points where they are closest and the line $m$ connecting those two points will be perpendicular to both $\ell_1$ and $\ell_2$. In fact, perpendicular to this line $m$ are two parallel planes: $\mathcal{P}_1$ which contains $\ell_1$ and $\mathcal{P}_2$ which contains $\ell_2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now pick any point $A$ of line $\ell_3$ not in either plane. The point $A$ together with the line $\ell_2$ defines a plane $\mathcal{Q}$ that contains them. This plane, since it intersects plane $\mathcal{P}_2$, must intersect the parallel plane $\mathcal{P}_1$. Moreover, the line $r$ formed by the intersection of planes $\mathcal{Q}$ and $\mathcal{P}_1$ is parallel to line $\ell_2$. Note that lines $r$ and $\ell_1$ are both on $\mathcal{P}_1$ and cannot be parallel. If they were, $\ell_2$ would also be parallel to $\ell_1$, and $\ell_1$ and $\ell_2$ would, therefore, be co-planar. So, lines $r$ and $\ell_1$ must intersect at some point $B$. If you connect points $A$ and $B$ with a line, it will connect point $A$ on $\ell_3$ passes through $\ell_2$ and connect to $B$ on $\ell_1$.&lt;/p&gt;&#xA;" OwnerUserId="6272" LastEditorUserId="6272" LastEditorDisplayName="" LastEditDate="2011-02-23T09:10:11.327" LastActivityDate="2011-02-23T09:10:11.327" CommentCount="1" />
  <row Id="23357" PostTypeId="2" ParentId="23316" CreationDate="2011-02-23T09:19:38.533" Score="0" ViewCount="" Body="&lt;p&gt;Suppose gcd(a, b) = k - (1) &#xA;        gcd(a, b+at)=k' - (2) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If k' != k - &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are two cases, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1/ k' &amp;lt; k : Contradiction because k|(b+at).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2/ k' &gt; k : Since k'|a again contradiction due to (1).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore k = k'.&lt;/p&gt;&#xA;" OwnerUserId="7402" LastActivityDate="2011-02-23T09:19:38.533" />
  <row Id="23358" PostTypeId="1" AcceptedAnswerId="23380" CreationDate="2011-02-23T10:12:25.907" Score="10" ViewCount="412" Body="&lt;p&gt;A very basic ring theory question, which i am not able to solve. How does one show that &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;$\mathbb{Z}[i]/(3-i) \cong \mathbb{Z}/10\mathbb{Z}$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Extending the result: $\mathbb{Z}[i]/(a-ib) \cong \mathbb{Z}/(a^{2}+b^{2})\mathbb{Z}$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;My attempt was to define a map, $\varphi :\mathbb{Z}[i] \to \mathbb{Z}/10\mathbb{Z}$ and show that the kernel is the ideal generated by $&amp;lt;3-i&amp;gt;$. But i couldn't think of such a map. Anyhow, any ideas would be helpful. &lt;/p&gt;&#xA;" OwnerUserId="9413" LastEditorUserId="9413" LastEditorDisplayName="" LastActivityDate="2011-03-21T16:32:05.253" Title="How $\mathbb{Z}[i]/(3-i) \cong \mathbb{Z}/10\mathbb{Z}$" Tags="&lt;abstract-algebra&gt;&lt;ring-theory&gt;" AnswerCount="5" CommentCount="4" FavoriteCount="4" />
  <row Id="23359" PostTypeId="2" ParentId="11499" CreationDate="2011-02-23T10:16:42.830" Score="0" ViewCount="" Body="&lt;p&gt;I've been told there are are only 3 ways to subdivide the starting triangles to make a &lt;a href=&quot;http://www.adventurewagen.net/MEN/009/009_074.HTM&quot; rel=&quot;nofollow&quot;&gt;geodesic dome&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;class 1 (alternative method): Starting with one triangle, divide each edge into u segments, dividing the entire triangle into u^2 smaller triangles, using the original edges and new edges more or less parallel to the 3 original edges. This gives &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Starting with a 20 sided icosahedron, the class 1 subdivision gives 20*u^2 triangles that cover the entire sphere:&#xA;20, 80, 180, 320, 500, 720, 980, 1280, 1620, 2000, etc. triangular faces.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;class 2 (triacon method): Starting with one triangle, pick some even integer u, divide each edge into u segments, and divide the entire triangle into 3*(u/2)^2 smaller triangles with new edges at right angles to the 3 original edges (erasing the original edges; the non-equilateral pieces along the edges count as as &quot;half a triangle&quot;).&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Starting with a 20 sided icosahedron, the class 2 subdivision gives 20*3*(u/2)^2 triangles that cover the entire sphere:&#xA;60, 240, 540, 960, 1500, 2160, etc. triangular faces.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.cjfearnley.com/fuller-faq-4.html&quot; rel=&quot;nofollow&quot;&gt;class 3&lt;/a&gt;: ... alas, I don't seem to have the formula at hand.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So you won't get a geodesic sphere with exactly 1000 faces, but you can get one with around 1000 faces, with a frequency of 8 or 9.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While sub-dividing appears to create congruent equilateral triangles, stopping there gives small flat coplanar triangles -- the next design steps to &quot;inflate&quot; those vertices so they touch the surface of a sphere result in lots of triangles that are approximately equilateral, but not exactly -- even worse, they are not exactly congruent to each other, either.&#xA;A &lt;a href=&quot;http://www.simplydifferently.org/Geodesic_Dome_Notes?page=3#3V%205/9%20Icosahedron%20Dome&quot; rel=&quot;nofollow&quot;&gt;3v Icosahedron Dome&lt;/a&gt; uses 2 kinds of faces and 3 kinds of strut lengths.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is possible to tweak the edge lengths to make &lt;a href=&quot;http://space.mit.edu/home/tegmark/icosahedron.html&quot; rel=&quot;nofollow&quot;&gt;thousands of triangles have exactly equal areas&lt;/a&gt;, even though they have slightly different shapes.&lt;/p&gt;&#xA;" OwnerUserId="1705" LastActivityDate="2011-02-23T10:16:42.830" />
  <row Id="23360" PostTypeId="1" CreationDate="2011-02-23T10:18:28.923" Score="0" ViewCount="150" Body="&lt;p&gt;Let n be a natural number with unique prime factorization $p^m$... $q^k$ . Show that n can be written as a square if and only if  all (m, ...k) are even&lt;/p&gt;&#xA;" OwnerUserId="7404" LastEditorUserId="1778" LastEditorDisplayName="" LastEditDate="2011-02-23T15:44:20.823" LastActivityDate="2011-02-23T15:44:20.823" Title="Prime factorization of square numbers" Tags="&lt;elementary-number-theory&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="23362" PostTypeId="2" ParentId="23312" CreationDate="2011-02-23T10:58:30.603" Score="2" ViewCount="" Body="&lt;p&gt;A short explanation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider a matrix $A$, for an example one representing a physical transformation. When this matrix is used to transform a given vector $x$ the result is $y = A x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Now an interesting question is are there any vectors $x$ which does not change it's direction under this transformation? However allow the vector magnitude to vary by scalar $ \lambda $.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Such a question is of the form $A x = \lambda x $.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So such special $x$ are called eigen vectors and the change in magnitude depends on the eigen value $ \lambda $.&lt;/p&gt;&#xA;" OwnerUserId="5312" LastActivityDate="2011-02-23T10:58:30.603" />
  <row Id="23363" PostTypeId="2" ParentId="23360" CreationDate="2011-02-23T11:08:01.170" Score="1" ViewCount="" Body="&lt;p&gt;Assuming you mean written as a square, if $m,\dots,k$ are all even, $m=2m&amp;#39;,\dots,k=2k&amp;#39;$ for some $m&amp;#39;,\dots, k&amp;#39;\in\mathbb{Z}$ then &#xA;$$&#xD;&#xA;n=p^m\cdots q^k=p^{2m&amp;#39;}\cdots q^{2k&amp;#39;}=(p^{m&amp;#39;}\cdots q^{k&amp;#39;})^2.&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $n$ can be written as a square, then for some $m$ with factorization $r_1^{a_1}\cdots r_n^{a_n}$,&#xA;$$&#xD;&#xA;n=m^2=(r_1^{a_1}\cdots r_n^{a_n})^2=(r_1^{2a_1}\cdots r_n^{2a_n})=p^m\cdots q^k.&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then for any prime $s$ in the factorization of $n$, $s|r_1^{2a_1}\cdots r_n^{2a_n}$, which implies there is a unique $r_i$ such that $s|r_i^{2a_i}\implies s|r_i\implies s=r_i$, since $s$ and $r_i$ are both prime. By uniqueness of the factorization, if $s$ has power $t$ in the factorization of $n$, $s^t=r_i^{2a_i}$, which implies $t=2a_i$, so all the powers in $p^m\dots q^k$ are even.&lt;/p&gt;&#xA;" OwnerUserId="1583" LastActivityDate="2011-02-23T11:08:01.170" />
  <row Id="23364" PostTypeId="2" ParentId="23259" CreationDate="2011-02-23T11:15:20.223" Score="2" ViewCount="" Body="&lt;p&gt;$|\vec{a}\times\vec{b}|=|\vec{a}||\vec{b}||\sin(\theta)|$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $a,b$ unit vectors, so we have $|a| = |b| = 1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$|\vec{a}\times\vec{b}|=|\sin(\theta)| \le 1$ (equality is when $|\sin(\theta)| = 1$ i.e. when a and b are perpendicular)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore in general the result won't be a unit vector.&lt;/p&gt;&#xA;" OwnerUserId="5312" LastActivityDate="2011-02-23T11:15:20.223" />
  <row Id="23365" PostTypeId="2" ParentId="23351" CreationDate="2011-02-23T11:18:50.840" Score="0" ViewCount="" Body="&lt;p&gt;This question can be explained and understood with the aid of &#xA;the &lt;strong&gt;physics&lt;/strong&gt; of soap bubbles. :=)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One problem when thinking on such questions is that one &#xA;often thinks the walls as &quot;rubber sheets&quot;. For the surface &#xA;(and in arbtrary cross sections ways) minimisation the surface &#xA;tension has to be thought constant. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So always when three lamellae join in a common &quot;corner&quot; &#xA;the orthogonal cross section is 120 degrees. (thee identical &#xA;forces in one point) Excuse this &quot;corner&quot; and orthogonal,&#xA;I am not aware of geometry in English. &lt;/p&gt;&#xA;" OwnerUserId="7407" LastEditorUserId="7407" LastEditorDisplayName="" LastEditDate="2011-02-23T19:12:35.857" LastActivityDate="2011-02-23T19:12:35.857" />
  <row Id="23366" PostTypeId="2" ParentId="23347" CreationDate="2011-02-23T11:37:50.243" Score="3" ViewCount="" Body="&lt;p&gt;If you can understand that &quot;the pattern appears infinitely often in the sequence $x$&quot; is the same as the formal mathematical sentence &quot;$\forall k, \exists n \ge k$ such that the pattern appears at index $n$ in the sequence $x$&quot;, then the RHS is just a traduction in terms of sets of this sentence :&#xA;$$ x \in \bigcap^\infty_{k=1} \; \bigcup_{n \geq k} \; B_{E_n} \Leftrightarrow&#xD;&#xA; \forall k, \; x \in \bigcup_{n \geq k} \; B_{E_n} \Leftrightarrow&#xD;&#xA; \forall k \; \exists n \ge k, \; x \in B_{E_n}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you need to formally prove equivalence between the LHS and the RHS, you need a formal description of what the LHS is.&#xA;In fact, if $B_E$ isn't formally defined before, this equality &lt;i&gt;is&lt;/i&gt; its definition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have a predicate $\phi$ that depends on an integer parameter $n$ (like $\phi(n)$ = $x_n$ is even, or $\phi(n)$ = the pattern appears at index $n$ in the sequence $x$), we mean that :&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&quot;$\phi$ is sometime true&quot; means &quot;$\exists n \phi(n)$&quot;&lt;/li&gt;&#xA;&lt;li&gt;&quot;$\phi$ is always true&quot; means &quot;$\forall n \phi(n)$&quot;&lt;/li&gt;&#xA;&lt;li&gt;&quot;$\phi$ is eventually true&quot; means &quot;$\exists k, \forall n \ge k, \phi(n)$&quot;&lt;/li&gt;&#xA;&lt;li&gt;&quot;$\phi$ is true infinitely many times (or infinitely often)&quot; means &quot;$\forall k, \exists n \ge k, \phi(n)$&quot;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In order to see why it is the right translation, it's best to understand that &quot;$\phi$ is true infinitely often&quot;, is the same as &quot;$\phi$ is not eventually false&quot;,&#xA;since the translation of &quot;eventually true&quot; is easier to understand.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $\phi$ is true infinitely often and eventually false, you have a contradiction.&#xA;If it's not the case that $\phi$ is true infinitely often, then it is eventually false.&lt;/p&gt;&#xA;" OwnerUserId="6270" LastActivityDate="2011-02-23T11:37:50.243" CommentCount="1" />
  <row Id="23367" PostTypeId="2" ParentId="23358" CreationDate="2011-02-23T11:43:28.317" Score="15" ViewCount="" Body="&lt;p&gt;Go back one step and add the defining equation for $i$ to the ideal. In other words, consider your ring as a quotient of the ring of polynomials $\mathbb Z[x]$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \mathbb Z[i] / (3-i) = \mathbb Z [x] / (3-x,x^2+1) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Manipulating the ideal $(3-x,x^2+1)$ a bit, you will find that the quotient is indeed equal to $\mathbb Z/10\mathbb Z$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This makes sense because setting $i^2=-1$ and $i=3$ implies that $9 = -1$, which is true in $\mathbb Z/10\mathbb Z$.&lt;/p&gt;&#xA;" OwnerUserId="434" LastActivityDate="2011-02-23T11:43:28.317" CommentCount="0" />
  <row Id="23368" PostTypeId="2" ParentId="23334" CreationDate="2011-02-23T11:52:20.410" Score="1" ViewCount="" Body="&lt;p&gt;I guess a good starting point would be &lt;a href=&quot;http://www.google.com/url?sa=t&amp;amp;source=web&amp;amp;cd=1&amp;amp;ved=0CBcQFjAA&amp;amp;url=http%3A%2F%2Fciteseerx.ist.psu.edu%2Fviewdoc%2Fdownload%3Fdoi%3D10.1.1.118.8302%26rep%3Drep1%26type%3Dpdf&amp;amp;rct=j&amp;amp;q=biclustering%20survey%20shamir&amp;amp;ei=nvRkTeerJsiu8QP0i5XqBg&amp;amp;usg=AFQjCNFx0RLoztVM3pLIiakOx0QvYo694A&amp;amp;sig2=N0fhYThDoa4c8Nia8dbf2A&amp;amp;cad=rja&quot; rel=&quot;nofollow&quot;&gt;Tanay, Shamir and Sharan's survey&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="689" LastActivityDate="2011-02-23T11:52:20.410" />
  <row Id="23370" PostTypeId="1" AcceptedAnswerId="23371" CreationDate="2011-02-23T12:40:03.250" Score="0" ViewCount="52" Body="&lt;p&gt;Consider a simple linear equation of the form:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$n=\frac{2x+2}{3}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $n$ and $x$ represent something that comes in whole positive quantities (for example physical objects).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Define the equation &lt;em&gt;only&lt;/em&gt; for $n$ and $x$ that are a part of natural numbers (whole numbers $&gt;0$)&lt;/li&gt;&#xA;&lt;li&gt;Solve the equation satisfying the above restriction (without for instace graphing it and looking for $n$ and $x$ that work)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="5877" LastActivityDate="2011-02-23T12:47:56.520" Title="Linear equation for a special domain and range" Tags="&lt;linear-algebra&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="23371" PostTypeId="2" ParentId="23370" CreationDate="2011-02-23T12:47:56.520" Score="0" ViewCount="" Body="&lt;p&gt;If you want only integer solutions, then you have a &lt;a href=&quot;http://en.wikipedia.org/wiki/Diophantine_equation#Linear_Diophantine_equations&quot; rel=&quot;nofollow&quot;&gt;Diophantine problem&lt;/a&gt;. Your equation can be written $3n=2x+2$, from which you deduce that $n$ is even. From there, it's easy to find all integer solutions. You can then select the positive ones, if any.&lt;/p&gt;&#xA;" OwnerUserId="589" LastActivityDate="2011-02-23T12:47:56.520" CommentCount="2" />
  <row Id="23372" PostTypeId="1" CreationDate="2011-02-23T13:09:07.273" Score="2" ViewCount="152" Body="&lt;p&gt;Lusin's theorem states that for every $\varepsilon$, for every borel measure $\mu$, for every function $f:\mathbb{R}^n\to\mathbb{R}^m$, for every open set $A$ of finite measure, there exists a compact set $K$ such that $\mu(A-K)&amp;lt;\varepsilon$ and $f$ restricted to $K$ is continue.&#xA;Now, I'm wondering about what's the form of the compact $K$ in the case of the Dirichlet's function. Can you help me?&lt;/p&gt;&#xA;" OwnerUserId="5885" LastEditorUserId="5885" LastEditorDisplayName="" LastEditDate="2011-02-23T15:20:13.133" LastActivityDate="2011-02-23T15:20:13.133" Title="Lusin's theorem" Tags="&lt;real-analysis&gt;" AnswerCount="1" CommentCount="6" />
  <row Id="23375" PostTypeId="2" ParentId="5921" CreationDate="2011-02-23T13:31:10.873" Score="1" ViewCount="" Body="&lt;p&gt;The assertion is false, even for independent random variables $X$ and $Y$. Fix $u$ in $(0,\frac12)$ and consider i.i.d. Bernoulli random variables $X$ and $Y$ with&#xA;$$P(X=0)=P(Y=0)=1-u,\quad&#xD;&#xA;P(X=1)=P(Y=1)=u.&#xD;&#xA;$$ &#xA;Then $[Y&amp;lt;X]=[Y=0,X=1]$, hence $P(Y&amp;lt;X)=u(1-u)$, which is at most $\frac14$ and can be as close to $0$ as desired. In particular $P(Y&amp;lt;X)&amp;lt;\frac12$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let us now consider the probability that $[Y\le X]$. Then the independence hypothesis is crucial. To see this, consider $u$ in $(0,1)$, $X$ uniformly distributed on the interval $(u,1+u)$ and $Y$ uniformly distributed on the interval $(0,1)$. Then $F_Y(x)\ge F_X(x)$ for every $x$. One can realize $(X,Y)$ in at least three ways. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) If $Y=X-u$, then $P(Y\le X)=1$ hence $P(Y\le X)\ge\frac12$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) If $(X,Y)$ are independent, $P(Y\le X)=\frac12+u-\frac12u^2$ hence $P(Y\le X)\ge\frac12$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(3) If $Y=\varphi(X)$ with $\varphi(x)=x+u$ if $x\le1-u$ and $\varphi(x)=x+u-1$ otherwise, one can check that $Y$ is uniformly distributed on $(0,1)$ and that $[Y\le X]=[X&amp;gt;1-u]$ hence $P(Y\le X)=2u$ and $P(Y\le X)$ is as close to $0$ as desired. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This shows that the result $P(Y\le X)\ge\frac12$ cannot hold in full generality.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, for independent $X$ and $Y$, indeed $P(Y\le X)\ge\frac12$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A proof which does not assume the existence of densities is as follows. Note that $$P(Y\le X)=E(P(Y\le X|X))=E(F_Y(X))\ge E(F_X(X))=P(X&amp;#39;\le X),&#xD;&#xA;$$ &#xA;where $X&amp;#39;$ and $X$ are i.i.d. Now, by symmetry $[X&amp;#39;\le X]$ and $[X\le X&amp;#39;]$ have the same probability and $P(X&amp;#39;\le X)+P(X\le X&amp;#39;)=1+P(X=X&amp;#39;)\ge1$ hence $P(X&amp;#39;\le X)\ge\frac12$ and we are done. &lt;/p&gt;&#xA;" OwnerUserId="6179" LastEditorUserId="6179" LastEditorDisplayName="" LastEditDate="2011-02-23T13:39:31.353" LastActivityDate="2011-02-23T13:39:31.353" />
  <row Id="23377" PostTypeId="2" ParentId="23054" CreationDate="2011-02-23T13:37:29.627" Score="1" ViewCount="" Body="&lt;p&gt;There is a detailed formula in &lt;a href=&quot;http://en.wikipedia.org/wiki/Great-circle_distance&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;, with a warning about rounding errors and alternate versions to avoid them.  This gets you the distance between any two points, say A and X or B and X.  I am not sure what you mean by the distance from (A,B) to X unless it is this.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-02-23T13:37:29.627" />
  <row Id="23378" PostTypeId="2" ParentId="23358" CreationDate="2011-02-23T13:57:24.730" Score="6" ViewCount="" Body="&lt;p&gt;In general, one knows that if $\alpha$ is an integer in the number field $K$, then &#xA;$$&#xD;&#xA;{\rm N}_{K/{\Bbb Q}}(\alpha)=\left|\frac{A}{A\alpha}\right|&#xD;&#xA;$$&#xA;Here $\rm N$ is the norm and $A$ denotes the ring of integers. In the question's situation, $A={\Bbb Z}[i]$ is the ring of Gaussian integer and since $K$ is quadratic imaginary, ${\rm N}_{K/{\Bbb Q}}(\alpha)=\alpha\bar\alpha$ where the bar denotes complex conjugation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When $\alpha=3-i$, ${\rm N}_{K/{\Bbb Q}}(\alpha)=(3-i)(3+i)=10$, thus &#xA;${\Bbb Z}[i]/(3-i)$ is a ring with 10 elements, whose representatives are&#xA;$$&#xD;&#xA;\left\{0,1,2,i,i+1,i+2,2i,2i+1,2i+2,-1 \right\}.&#xD;&#xA;$$&#xA;A minute of thought and a brief inspection of these representatives should convince that the ring is indeed isomorphic to the ring of classes modulo 10.&lt;/p&gt;&#xA;" OwnerUserId="688" LastActivityDate="2011-02-23T13:57:24.730" CommentCount="3" />
  <row Id="23379" PostTypeId="2" ParentId="23358" CreationDate="2011-02-23T14:07:11.293" Score="12" ViewCount="" Body="&lt;p&gt;Firstly: it is not true in general that $\mathbb Z[i]/(a - ib) \cong \mathbb Z/(a^2 + b^2).$  (Consider the case of $3 - 0\cdot i$.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The claimed isomorphism &lt;em&gt;does&lt;/em&gt; hold if $a$ and $b$ are coprime.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a sketch of how to see this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To begin with, note that it is much easier to consider maps &lt;em&gt;from&lt;/em&gt; $\mathbb Z$ to other rings, rather than maps in the opposite direction (as you suggested in your answer), because $\mathbb Z$ maps to &lt;em&gt;any&lt;/em&gt; ring with unity in a canonical way, by sending $1$ to $1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So consider the canonical map $\mathbb Z \to \mathbb Z[i]/(a - i b).$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The target is finite of order $a^2 + b^2$, and so this map factors to give an&#xA;injection&#xA;$\mathbb Z/(n) \hookrightarrow \mathbb Z[i]/(a - i b)$&#xA;for some $n$ dividing $a^2 + b^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now if $a$ and $b$ are coprime then $b$ is coprime to $a^2 + b^2$, hence coprime to $n$, and so $b$ is invertible in $\mathbb Z/(n)$.  Combining this observation with the equation $a - i b = 0$ (which holds in $\mathbb Z[i]/(a - i b)$) one finds (and I leave this as an exercise!) that the map&#xA;$\mathbb Z/(n) \hookrightarrow \mathbb Z[i]/(a - ib)$ contains $i$ in its image, and hence is surjective as well as injective, and so we are done.&lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2011-02-23T14:07:11.293" CommentCount="1" />
  <row Id="23380" PostTypeId="2" ParentId="23358" CreationDate="2011-02-23T14:35:37.727" Score="7" ViewCount="" Body="&lt;p&gt;Define&#xA;$$\phi: \mathbb{Z} \rightarrow \mathbb{Z}[i]/(3-i) \text{ where } \phi(z) = z + (3-i)\mathbb{Z}[i]$$&#xA;It follows simply that&#xA;$\ker \phi = \{z \in (3-i)\mathbb{Z}[i] \cap \mathbb{Z}\}$.\&#xA;So for any such $z \in \ker \phi$, we have $z = (3-i)(a+bi)$ for some $a,b \in \mathbb{Z}.&#xD;&#xA;(3-i)(a+bi) \in \mathbb{Z}$ which happens if and only if $3b-a=0$.&#xA;So $$\begin{align*}&#xD;&#xA;\ker \phi = (3-i)\mathbb{Z}[i] \cap \mathbb{Z} &amp;amp;= \{(3-i)(3b+bi)|b \in \mathbb{Z}\}\\&#xD;&#xA;&amp;amp;= \{(9b + b) + i(3b-3b)|b \in \mathbb{Z}\}\\&#xD;&#xA;&amp;amp;= \{10b|b \in \mathbb{Z}\}\\&#xD;&#xA;&amp;amp;= 10\mathbb{Z}&#xD;&#xA;\end{align*}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To see $\phi$ is surjective, let $(a+bi) + (3-i)\mathbb{Z}[i] \in \mathbb{Z}[i]/(3-i)$. Then $a+bi=a+3b-3b+bi=(a+3b)-b(3-i)$, so $\phi(a+3b) = (a+bi) + (3-i)\mathbb{Z}[i]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence $\mathbb{Z}[i]/(3-i) \cong \mathbb{Z}/10\mathbb{Z}$.&lt;/p&gt;&#xA;" OwnerUserId="1016" LastEditorUserId="1016" LastEditorDisplayName="" LastEditDate="2011-03-21T16:32:05.253" LastActivityDate="2011-03-21T16:32:05.253" CommentCount="3" />
  <row Id="23381" PostTypeId="1" AcceptedAnswerId="23383" CreationDate="2011-02-23T14:36:31.767" Score="0" ViewCount="49" Body="&lt;p&gt;Hi,&#xA;I know a similar question was &lt;a href=&quot;http://math.stackexchange.com/questions/3987/how-to-calculate-the-number-of-decimal-digits-for-a-binary-number&quot;&gt;asked before&lt;/a&gt;, but I wanted to know if this can be extended to any number system by a generic formula.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, given a number X in number system A, how many digits would it have in number system B?&#xA;I'm looking for a sort of universal formula in terms of X, A and B.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="7412" LastActivityDate="2011-02-23T15:09:19.230" Title="Number of digits in different number systems?" Tags="&lt;number-theory&gt;&lt;arithmetic&gt;&lt;logarithms&gt;" AnswerCount="1" />
  <row Id="23382" PostTypeId="2" ParentId="23324" CreationDate="2011-02-23T14:39:31.603" Score="2" ViewCount="" Body="&lt;p&gt;There are sophisticated ways you can do this, but I think its best to push it through in a straightforward way. Here are two approaches; I'm deliberately not giving the details because this is a good homework problem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Algebraic:&lt;/strong&gt; Let $C$ be the subgroup $\{ 1, i, -1, -i \}$ of the quaternions. Let $V$ be the real representation of $C$ where $i$ acts by $\left( \begin{smallmatrix} 0 &amp;amp; 1 \\ -1 &amp;amp; 0 \end{smallmatrix} \right)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose that the two dimensional representation of the quaternions could be defined over $\mathbb{R}$. Show that the restriction of this representation to $C$ is isomorphic to $V$. In other words, you can always choose a basis where $i$ acts by $\left( \begin{smallmatrix} 0 &amp;amp; 1 \\ -1 &amp;amp; 0 \end{smallmatrix} \right)$. Now write down the equations $ij=ji^3$ and $j^2=i^2$ and try to solve them&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Geometric:&lt;/strong&gt; Any representation on $\mathbb{R}^2$ can be conjugated to preserve the standard inner product; meaning that the action is by rotations and reflections. A case by case analysis should show you pretty quickly that you can't find rotations and reflections of the plane that give an action of the quaternions.&lt;/p&gt;&#xA;" OwnerUserId="448" LastActivityDate="2011-02-23T14:39:31.603" CommentCount="1" />
  <row Id="23383" PostTypeId="2" ParentId="23381" CreationDate="2011-02-23T14:45:55.843" Score="1" ViewCount="" Body="&lt;p&gt;As J. M. said in a comment to the question you ask, an integer $X$ has $\lfloor 1+\log_B X\rfloor$ base-$B$ digits.  Going from base $A$ to base $B$ the number of digits is multiplied by $\frac{\log A}{\log B}$ but that is approximate due to the 1 and the floor function.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2011-02-23T15:09:19.230" LastActivityDate="2011-02-23T15:09:19.230" CommentCount="2" />
  <row Id="23384" PostTypeId="1" CreationDate="2011-02-23T14:54:20.427" Score="0" ViewCount="180" Body="&lt;p&gt;Hello!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need your help to solve this question &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A subset $X \subset k^n$ ($k$ a field) is called algebraic if there exist polynomials $f_1, \dots, f_m \in k[t_1,\dots,t_n]$ such that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$X = \{x \in K^n | f_1(x) = ...=f_m(x)=0\}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The coordinate ring $k[X]$ of $X$ is the ring of all functions $f: X\to k$ that can be represented by some polynomial. That is, there exists a polynomial $g \in k[t_1,\dots,t_n]$ such that $f(x) = g(x)$ for all $x \in X$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1- Show that for two different points $x,y \in X$ there exists $f \in k[X]$ with $f(x) \neq f(y)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2- Let $k=\mathbb{Z}_2 = \mathbb{Z}/2\mathbb{Z}$ and $X \subset k^2$ be the set $\{(0,1),(1,0)\}$. Is $X$ algebraic? Determine its coordinate ring.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3- Let $g_i$ in $N$ be a sequence of polynomials and define $Y:= \{ x \in k^n | g_i (x) = 0~\forall~i\}$. Show that $Y$ is algebraic set.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="7413" LastEditorUserId="367" LastEditorDisplayName="" LastEditDate="2011-02-23T16:04:57.773" LastActivityDate="2011-02-26T21:57:12.450" Title="coordinate ring questions" Tags="&lt;algebraic-geometry&gt;" AnswerCount="0" CommentCount="5" />
  <row Id="23385" PostTypeId="2" ParentId="23372" CreationDate="2011-02-23T15:12:53.970" Score="3" ViewCount="" Body="&lt;p&gt;I found an answer by myself. First, let's order the rationals, say $\{q_n\}$. Then consider the covering $O_n=(q_n-\frac{\varepsilon}{2^n},q_n+\frac{\varepsilon}{2^n})$ for $n=1,2,\ldots$. It's straightforward to check that the measure of $\cup O_n$ is less than $\epsilon$. Now it suffices to choose the complementary $K$ of $\cup O_n$ in $A$ which is obviously compact. The Dirichlet's function on $K$ is $0$.&lt;/p&gt;&#xA;" OwnerUserId="5885" LastActivityDate="2011-02-23T15:12:53.970" />
  <row Id="23386" PostTypeId="1" AcceptedAnswerId="23424" CreationDate="2011-02-23T15:18:21.933" Score="3" ViewCount="89" Body="&lt;p&gt;This is a problem from Borevich and Shafarevich (p.180).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, a definition:&lt;br&gt;&#xA;A theory of divisors on a domain $\mathfrak{O}$ consists of a commutative semigroup $D$ (semigroup = associative multiplication with identity) having the property of unique factorization into irreducibles along with a homomorphism $\alpha \rightarrow (\alpha)$ of the semigroup $\mathfrak{O}^{\times}$ into $D$ satisfying&lt;br&gt;&#xA;(i) An element $\alpha \in \mathfrak{O}^{\times}$ is divisible by $\beta \in \mathfrak{O}^{\times}$ in the ring $\mathfrak{O}$ if and only if $(\alpha)$ is divisible by $(\beta)$ in $D$.&lt;br&gt;&#xA;(ii) If $\alpha$ and $\beta$ of $\mathfrak{O}$ are divisible by $a \in D$, then $\alpha \pm \beta$ are also divisible by $a$.&lt;br&gt;&#xA;(iii) If $a$ and $b$ are two elements of $D$ and the set of all elements $\alpha \in \mathfrak{O}$ which are divisible by $a$ coincides with the set of all elements $\beta \in \mathfrak{O}$ which are divisible by $b$, then $a = b$.&lt;br&gt;&#xA;The elements of $D$ are called divisors and elements of the form $(\alpha)$ for $\alpha \in \mathfrak{O^{\times}}$ are called principal divisors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is to show that every element $a$ of $D$ is the greatest common divisor of two principal divisors $(\alpha),(\beta)$ [which means that $a$ is a common divisor of $(\alpha)$ and $(\beta)$ that is divisible by every other common divisor of these two elements].&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My first thought was to mimic the proof that every ideal $a$ (in a Dedekind domain) is generated by two elements $\alpha, \beta \in a$, which means exactly that $a$ is the gcd of the principal ideals $(\alpha),(\beta)$.  However, this proof seems to rely critically on the addition of ideals and the notion of the inverse of an ideal, neither of which seem to transfer over to the setting of divisors.  Most basic books in algebraic number theory have this proof; see for instance Stewart and Tall.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have been told that (iii) implies that $a$ divides at least one principal divisor $(\alpha)$ [though I don't see how].&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $D$ has unique factorization, I know that $(\alpha) = \prod_{i}^{n} p_{i}^{r_i}$ for irreducibles $p_i$ and positive integers $r_i$, and that $a = \prod_{i}^{n} p_{i}^{s_i}$ where $s_i \leq r_i$.  Hence, if $(\beta)$ exists it must have the form $(\beta) = \prod_{i}^{n} p_{i}^{t_i} \cdot q_1 \cdots q_m$ such that $r_i = \min(s_i,t_i)$ and the $q_i$'s are irreducibles not equal to the $p_i$'s.  This means that the $t_i$ are restricted as &#xA;$t_i = r_i$ if $s_i &amp;lt; r_i$ and $t_i =$ any integer $\geq s_i$ if $s_i = r_i$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I can make a valid choice of the $t_i$'s and $q_i$'s that makes $\prod_{i}^{n} p_{i}^{t_i} \cdot q_1 \cdots q_m $ principal, then I will be done.  However, I am not sure how to do this.  I am not even sure if this is the right approach.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can someone please help me?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: What I call a semigroup above is actually a monoid.&lt;/p&gt;&#xA;" OwnerUserId="6887" LastEditorUserId="6887" LastEditorDisplayName="" LastEditDate="2011-02-25T18:38:08.320" LastActivityDate="2011-02-25T18:38:08.320" Title="Show that every divisor is the gcd of two principal divisors (in the abstract setting of a semigroup $D$ of &quot;divisors&quot;)" Tags="&lt;number-theory&gt;&lt;algebraic-number-theory&gt;" AnswerCount="1" />
  <row Id="23387" PostTypeId="1" AcceptedAnswerId="23389" CreationDate="2011-02-23T15:30:58.097" Score="3" ViewCount="163" Body="&lt;p&gt;It's easy to show $C[0,1]$ is not dense in $L^{\infty}[0,1]$ in the norm topology, but $C[0,1]$ is dense in $L^{\infty}[0,1]$ in the weak*-topology when take $L^{\infty}$ as the dual of $L^{1}$. how to proof?&lt;/p&gt;&#xA;" OwnerUserId="5593" LastActivityDate="2011-02-23T19:47:46.307" Title="On the density of $C[0,1]$ in the space $L^{\infty}[0,1]$" Tags="&lt;functional-analysis&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="23388" PostTypeId="1" CreationDate="2011-02-23T15:37:59.870" Score="1" ViewCount="70" Body="&lt;p&gt;what is the solution for max &lt;strong&gt;y$^T$&lt;/strong&gt; $\Sigma$ &lt;strong&gt;y&lt;/strong&gt; in terms of maximum eigen value and the associated eigen vectors ?&lt;/p&gt;&#xA;" OwnerUserId="7415" LastActivityDate="2011-06-02T13:44:39.843" Title="maximum value of a quadratic problem" Tags="&lt;linear-algebra&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="23389" PostTypeId="2" ParentId="23387" CreationDate="2011-02-23T15:51:52.343" Score="3" ViewCount="" Body="&lt;p&gt;Given $f\in L^\infty$, apply Lusin's theorem infinitely many times to get a sequence $(f_n)_n$ in $C[0,1]$ uniformly bounded by $\|f\|_\infty$ such that the measure of the set where $f$ and $f_n$ are unequal is less than $\frac{1}{n}$.  If $g\in L^1$, then given $\varepsilon\gt 0$ there exists $N$ such that $\int_A|g|&amp;lt;\varepsilon$ whenever the measure of $A$ is less than $\frac{1}{N}$.  Then $|\int_0^1(f-f_k)g|\leq 2\|f\|_\infty\varepsilon$ for all $k\geq N$.  This implies that $f_n\to f$ in the weak-$*$ topology, because $\int_0^1 f_ng\to\int_0^1fg$ for all $g\in L^1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively, you can show that the sequence of Cesàro means of the Fourier series of $f$ converges to $f$ in the weak-$*$ topology.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2011-02-23T16:02:55.040" LastActivityDate="2011-02-23T16:02:55.040" />
  <row Id="23390" PostTypeId="1" AcceptedAnswerId="23397" CreationDate="2011-02-23T15:59:33.583" Score="3" ViewCount="155" Body="&lt;p&gt;It is well known that the symmetric property is $d(x,y)=d(y,x)$ is not necessary in the definition of distance if the triangle inequality is carefully stated. On the other hand there are examples of functions satifying&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) $d(x,y)\geq 0$ and $d(x,y)=0$ if and only if $x=y$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) $d(x,y)\leq d(x,z)+d(z,y)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which are not symmetric: in the three point space $(a,b,c)$ take the non-zero values of $d$ as $1=d(a,b)=d(c,b)$, $2=d(b,a)=d(b,c)=d(a,c)=d(c,a)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do you know other examples of &quot;non symmetric distances&quot;? Are there examples on the real numbers, etc.? Are there examples of spaces were every function satisfying (1) and (2) is symmetric?&lt;/p&gt;&#xA;" OwnerUserId="7416" LastActivityDate="2011-02-23T17:57:30.310" Title="Examples of non symmetric distances" Tags="&lt;metric-spaces&gt;" AnswerCount="6" CommentCount="2" />
  <row Id="23391" PostTypeId="1" CreationDate="2011-02-23T16:26:00.587" Score="1" ViewCount="89" Body="&lt;p&gt;Find $g$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a)&#xA;If $g∘(2f)=f+h$  where $f(x)=2x + 5$ and $h(x)=x^3 -2x$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;b)&#xA;If $(2f)∘g=f+h$ where $f(x)=\ln(x+2)$ and $h(x)=\sin(x^2)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="7143" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-23T19:41:24.847" LastActivityDate="2011-02-23T19:41:24.847" Title="Function composition:Find $g$" Tags="&lt;homework&gt;&lt;functions&gt;" AnswerCount="1" />
  <row Id="23392" PostTypeId="2" ParentId="23391" CreationDate="2011-02-23T16:51:21.287" Score="2" ViewCount="" Body="&lt;p&gt;(a) Write out $f+h$ (that you can do directly). Then write out $2f$. Then try to expresss $f+h$ in terms of $2f$. Your expression is what $g$ does.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, if we have $f(x) = 3x+1$ and $h(x) = 36x^2+21x+3$, then $f+h = 36x^2 + 24x+4$. On the other hand, $2f = 6x+2$. Can we write $f+h$ in terms of $2f$? Yes: $(6x+2)^2 = 36x^2 + 24x + 4$, so $g(u) = u^2$ would satisfy $g\circ(2f)=f+h$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(b) Similar to (a), but now you are doing $(2f)\circ g$ with the given $f$ and $h$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In short: Write it out, figure it out.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2011-02-23T16:51:21.287" CommentCount="5" />
  <row Id="23393" PostTypeId="1" CreationDate="2011-02-23T16:54:55.317" Score="0" ViewCount="29" Body="&lt;p&gt;I have a markov-chain type model which I compute its state transition function P, and then need to find the probabilities after n steps. Problem is, each P(X|Y) is a normally distributed variable by itself because it has a normally distributed error (thats the assumption anyway). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Right now I use the standard stuff to find out state transition probabilities after n steps by using the mean, and get an approximated result, but ofcourse what I have is a joint distribution which is a result of random variable sums and products. So basically what I really want is to get some features out of this joint distribution, for example:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Right now I have an approximation for nth step P, but what i really want instead is the mean of that distribution. This is different since product of two normal variables is not normal (and more over multiplying that variable with another normal variable which I could't find any analytic solution for)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I would also like to know my error for the nth step, in the form of standard deviation, or better - say 95% confidence interval.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;one obvious way to do this would be to sample the variables and then numerically do an approximated integral for each sum or product, but this takes way too much computations.&#xA;Another is to run a monte-carlo on the model and get an approximation for the entire distribution, but thats also not really feasable if I need say 10000s of iterations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there's a better way to approximate specific features, or even a smart way to assess the number of iterations needed in a monte carlo simulation?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S Since this is an error, and for simplicity, I currently assume all variables are independent.&lt;/p&gt;&#xA;" OwnerUserId="7417" LastEditorUserId="7417" LastEditorDisplayName="" LastEditDate="2011-02-23T21:30:00.183" LastActivityDate="2011-02-23T21:30:00.183" Title="Approximating features of joint distribution of a model that makes arithmetics with normally distributed independent variables" Tags="&lt;approximation&gt;&lt;probability-distributions&gt;&lt;monte-carlo&gt;" />
  <row Id="23394" PostTypeId="2" ParentId="23207" CreationDate="2011-02-23T17:00:34.587" Score="4" ViewCount="" Body="&lt;p&gt;Indeed, the whole subject of knot theory is interesting because different subsets of $R^3$ that are homeomorphic to each other (namely different knots, each homeomorphic to a circle) have non-homeomorphic complements in $R^3$.  And knots are homeomorphic, not just homotopy equivalent.&lt;/p&gt;&#xA;" OwnerUserId="6071" LastActivityDate="2011-02-23T17:00:34.587" />
  <row Id="23395" PostTypeId="2" ParentId="23390" CreationDate="2011-02-23T17:08:20.767" Score="3" ViewCount="" Body="&lt;p&gt;For any directed graph $G=(V,E)$, we may define a non-symmetric distance as follows: for $x,y\in V$, the &lt;em&gt;directed graph distance&lt;/em&gt; is the length of the shortest directed path from $x$ to $y$, or $\infty$ if there is no such path.  This clearly satisfies the triangle inequality as stated above.  An even more general family of examples is generated by allowing a positive weight, in which case we may take the distance to be the smallest weight of any directed path from $x$ to $y$ (or, if $G$ is infinite, the greatest lower bound of all such weights; in this case we must additionally require all edge weights to be bounded away from zero).&lt;/p&gt;&#xA;" OwnerUserId="5546" LastActivityDate="2011-02-23T17:08:20.767" />
  <row Id="23396" PostTypeId="1" CreationDate="2011-02-23T17:08:46.650" Score="1" ViewCount="43" Body="&lt;p&gt;According to Batyrev, there are exactly 18 types of smooth toric Fano threefolds. As projective toric varieties, these are defined by the normal fan of certain 3-dimensional polytopes. Does anyone know of a reference where these polytopes are listed, e.g., as convex hulls of points?&lt;/p&gt;&#xA;" OwnerUserId="6064" LastActivityDate="2011-02-23T17:31:38.120" Title="Toric fano threefolds" Tags="&lt;algebraic-geometry&gt;" AnswerCount="1" />
  <row Id="23397" PostTypeId="2" ParentId="23390" CreationDate="2011-02-23T17:09:54.387" Score="7" ViewCount="" Body="&lt;p&gt;A classic example is the circle $S^1$ with metric the length of the shortest clockwise path between $x$ and $y$, but let me say some things in general.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://ncatlab.org/nlab/show/metric+space&quot; rel=&quot;nofollow&quot;&gt;Lawvere&lt;/a&gt; once made the point that this is really a much more natural definition of a metric space, since it allows them to be interpreted as a type of enriched category. The triangle inequality then becomes a consequence of composition of morphisms, which is extremely reasonable if you think of distances in a metric space as measuring &quot;the best way to get from $a$ to $b$&quot;: clearly the best way to get from $a$ to $b$ to $c$ is at most as good at the best way to get from $a$ to $c$, and this is precisely the (asymmetric) triangle inequality. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, there is no reason in general that the best way to get from $a$ to $b$ has to look anything like the best way to get from $b$ to $a$. This is the sense in which the symmetry requirement is unnatural. There is also no reason in general that it should be possible to get from $a$ to $b$ at all! This is the sense in which the requirement that distances be finite is unnatural. Finally, there is no reason in general that it should not be possible to instantaneously get from $a$ to $b$ (in other words, that $a$ and $b$ be isomorphic); this is the sense in which the requirement that distances be positive-definite is unnatural. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is how to use that idea to generate a large class of examples. Let $(M, d)$ be a metric space and let $h : M \to \mathbb{R}$ be a function. Define a new metric&#xA;$$d&amp;#39;(x, y) = \begin{cases} d(x, y) + h(y) - h(x) \text{ if } h(y) \ge h(x) \\\&#xD;&#xA; d(x, y) \text{ otherwise} \end{cases}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Intuitively, $h$ is a potential (e.g. a height if one is thinking of gravitational potential), and the new metric $d&amp;#39;$ penalizes you for going against the potential (e.g. going uphill). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The directed graph example given by mjqxxxx is also a good illustration of this philosophy about metric spaces. &lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2011-02-23T17:09:54.387" CommentCount="1" />
  <row Id="23398" PostTypeId="2" ParentId="23390" CreationDate="2011-02-23T17:12:08.343" Score="0" ViewCount="" Body="&lt;p&gt;What about the following distance function on the reals?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{array}{rcl}&#xD;&#xA;d(x,y) &amp;amp; = 1, &amp;amp; \mbox{if} \;\; x&amp;lt;y \\&#xD;&#xA; &amp;amp; = 2, &amp;amp; \mbox{if} \;\; x&amp;gt;y \\&#xD;&#xA; &amp;amp; = 0, &amp;amp; \mbox{if} \;\; x=y &#xD;&#xA;\end{array}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A bit artificial, but it seems to fulfill the requirements, unless I overlooked something.&lt;/p&gt;&#xA;" OwnerUserId="3567" LastActivityDate="2011-02-23T17:12:08.343" CommentCount="2" />
  <row Id="23399" PostTypeId="1" AcceptedAnswerId="23462" CreationDate="2011-02-23T17:22:11.747" Score="5" ViewCount="228" Body="&lt;p&gt;The recurrence $a_{n+1}=a_n(n-1/2)$ is related  to $\Gamma(n+1/2)$ ( not difficult to prove) and it could be represented in a way like $\frac {(2n-1)!!} {2^n}$&#xA;Also I know that &#xA;$(2n-1)!!$ is  the number of permutations of 2n whose cycle type consists of n parts equal to 2; these are the involutions without fixed points (A).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, for each $n \in N$, let $f(n)$ is  the number of subsets of set $[n]=\ {1,2,...,n}$. Then $f(n)=2^n$ (B)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wonder about the understanding of the  meaning ( sense) of the ration: A/B?&#xA;What could be the meaning of $\frac {(2n-1)!!} {2^n}$ in terms of  sets?&lt;/p&gt;&#xA;" OwnerUserId="7418" LastEditorUserId="413" LastEditorDisplayName="" LastEditDate="2011-02-24T05:20:12.160" LastActivityDate="2011-02-28T14:29:36.440" Title="The ratio in terms of sets" Tags="&lt;combinatorics&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" />
  <row Id="23400" PostTypeId="2" ParentId="23390" CreationDate="2011-02-23T17:22:24.350" Score="2" ViewCount="" Body="&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Hausdorff_distance&quot; rel=&quot;nofollow&quot;&gt;Hausdorff distance&lt;/a&gt; is a symmetric version of a natural non-symmetric distance.&lt;/p&gt;&#xA;" OwnerUserId="589" LastActivityDate="2011-02-23T17:22:24.350" />
  <row Id="23401" PostTypeId="2" ParentId="23293" CreationDate="2011-02-23T17:25:19.637" Score="2" ViewCount="" Body="&lt;p&gt;(This answer takes as its starting point the OP's question in the comments, &quot;Let me understand mass before going to density. Why do we call a point in the discrete distribution as mass? Why can't we just call it a point?&quot;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We could certainly call it a point.  The utility of the term &quot;probability mass function,&quot; though, is that it tells us something about how the function in the discrete setting relates to the function in the continuous setting because of the associations we already have with &quot;mass&quot; and &quot;density.&quot;  And I think to understand why we use these terms in the first place we have to start with what we call the density function.  (In fact, I'm not sure we would even be using &quot;probability mass&quot; without the corresponding &quot;probability density&quot; function.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's say we have some function $f(x)$ that we haven't named yet but we know that $\int_a^b f(x) dx$ yields the probability that we see an outcome between $a$ and $b$.  What should we call $f(x)$?  Well, what are its properties?  Let's start with its units.  We know that, in general, the units on a definite integral $\int_a^b f(x) dx$ are the units of $f(x)$ times the units of $dx$.  In our setting, the integral gives a probability, and $dx$ has units in say, length.  So the units of $f(x)$ must be probability per unit length.  This means that $f(x)$ must be telling us something about how much probability is concentrated per unit length near $x$; i.e., how &lt;em&gt;dense&lt;/em&gt; the probability is near $x$.  So it makes sense to call $f(x)$ a &quot;probability density function.&quot;  (In fact, one way to view $\int_a^b f(x) dx$ is that, if $f(x) \geq 0$, $f(x)$ is &lt;em&gt;always&lt;/em&gt; a density function.  From this point of view, height is area density, area is volume density, speed is distance density, etc.  One of my colleagues uses an approach like this when he discusses applications of integration in second-semester calculus.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now that we've named $f(x)$ a density function, what should we call the corresponding function in the discrete setting?  It's not a density function; its units are probability rather than probability per unit length.  So what is it?  Well, when we say &quot;density&quot; without a qualifier we are normally talking about &quot;mass density,&quot; and when we integrate a density function over an object we obtain the mass of that object.  With this in mind, the relationship between the probability function in the continuous setting to that of the probability function in the discrete setting is exactly that of density to mass.  So &quot;probability mass function&quot; is a natural term to grab to apply to the corresponding discrete function.  &lt;/p&gt;&#xA;" OwnerUserId="2370" LastActivityDate="2011-02-23T17:25:19.637" CommentCount="3" />
  <row Id="23402" PostTypeId="2" ParentId="23396" CreationDate="2011-02-23T17:31:38.120" Score="2" ViewCount="" Body="&lt;p&gt;Google immediately finds &lt;a href=&quot;http://magma.maths.usyd.edu.au/users/kasprzyk/research/pdf/GAELTalk.pdf&quot; rel=&quot;nofollow&quot;&gt;Smooth toric fano varieties and their polytopes&lt;/a&gt; by A. Kasprzyk, which has &lt;em&gt;pictures&lt;/em&gt; of the 18 polytopes. Surely following the references therein should get you coordinates of the vertices.&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2011-02-23T17:31:38.120" CommentCount="4" />
  <row Id="23403" PostTypeId="2" ParentId="23390" CreationDate="2011-02-23T17:32:18.747" Score="3" ViewCount="" Body="&lt;p&gt;Most distances in real life are going to be more or less asymmetric due to one-way roads, going uphill resp. downhill, different public transportation schedules, congestion, etc. &lt;/p&gt;&#xA;" OwnerUserId="677" LastActivityDate="2011-02-23T17:32:18.747" />
  <row Id="23404" PostTypeId="2" ParentId="23390" CreationDate="2011-02-23T17:57:30.310" Score="1" ViewCount="" Body="&lt;p&gt;This may or may not be what you're looking for, but I thought the idea was fun, so I'll share:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This will be a non-symmetric metric on $\mathbb{Z}^+$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose that one is only allowed to move left one unit, and in addition, one is allowed to jump right from one power of two to the next.  I.e., to find a sequence of steps to get from 5 to 11, one must travel as follows: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;5 \to 4 \to8 \to16 \to15 \to 14 \to13 \to12\to11.&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a (positive) integer $n$, define $T(n)$ to be largest power of $2$ which is less than or equal to n.  Similarly, define $S(n)$ to be the smallest power of $2$ greater than or equal to $n$.  Then, if $|\cdot|$ denotes the usual absolute value, we have:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;d(x,y) = &#xD;&#xA;\left\{&#xD;&#xA;\begin{array}{lcc}&#xD;&#xA;|x - y| &amp;amp; &amp;amp; x \geq y \\&#xD;&#xA;|x - T(x)| + \log_2(S(y)/T(x)) + |S(y) - y| &amp;amp;  &amp;amp; x &amp;lt; y&#xD;&#xA;\end{array}&#xD;&#xA;\right.&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $x \leq y$, one just moves left $|y - x|$ units.  If not, one moves from $x$ to the largest power of $x$ less than or equal to $x$.  One then jumps $\log_2(S(y)) - \log_2(T(x))$ powers of $2$, and finally moves the remaining $|S(y) - y|$ units.  It is not hard to show that this a metric (It clear satisfies condition (i), and to show it satisfies (ii), one just needs to consider the case of $d(x,y)$ when $x &gt; y$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This metric came to mind in lieu of Cauchy's Proof of the Arithmetic-Geometric mean inequality (&lt;a href=&quot;http://en.wikipedia.org/wiki/Inequality_of_arithmetic_and_geometric_means#Proof_by_Cauchy&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Inequality_of_arithmetic_and_geometric_means#Proof_by_Cauchy&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $P(n)$ denote the statement that the AG-mean inequality is true for $n$ numbers.  Cauchy's idea was to prove the inequality by first inducting along powers of two (i.e., P($2^k$) is true implies P($2^{k+1}$) is true.  Then, he showed that $P(n)$ is true implies $P(n-1)$ is true.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming that we know the arithmetic-geometric mean inequality is true for some step $n$, $d(n,m)$ is the number of steps needed using Cauchy's proof to show $P(m)$ is true, assuming that we know $P(n)$ is true.&lt;/p&gt;&#xA;" OwnerUserId="6491" LastActivityDate="2011-02-23T17:57:30.310" />
  <row Id="23405" PostTypeId="1" AcceptedAnswerId="23408" CreationDate="2011-02-23T18:09:07.590" Score="1" ViewCount="59" Body="&lt;p&gt;Given a social choice function $F$, a subset $B\subset A$ of the candidates and a coalition $S\subset N$ of the voters, $\beta$-effectiveness of $S$ for $B$ is equivalent to $N\setminus S$ not being $\beta$-effective for $A\setminus B$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, for a social choice correspondence it is not true - the fact that $N\setminus S$ is not $\beta$-effective for $A\setminus B$ does not, in general, lead to $S$ being $\beta$-effective for $B$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could anyone give an example that shows why it is so? What is special about social choice functions that gives this symmetry?&lt;/p&gt;&#xA;" OwnerUserId="766" LastEditorUserId="766" LastEditorDisplayName="" LastEditDate="2011-02-23T18:19:56.303" LastActivityDate="2011-02-23T19:47:07.947" Title="$N\setminus S$ is not $\beta$-effective for $A\setminus B$, and $S$ is not $\beta$-effective for $B$." Tags="&lt;game-theory&gt;&lt;social-choice-theory&gt;" AnswerCount="1" />
  <row Id="23406" PostTypeId="1" CreationDate="2011-02-23T18:29:18.153" Score="0" ViewCount="175" Body="&lt;p&gt;Specifically, how to show that an affine variety over complex number is never compact in Euclidean topology unless it is a single point.  I got a hint on this qiestion: Given an affine variety X, show that the image of X under the projection map onto the first coordinate is either a point or an open subset (in the Zariski topology).&lt;/p&gt;&#xA;" OwnerUserId="7419" LastActivityDate="2011-02-24T06:23:16.610" Title="An affine space of positive dimension is not complete" Tags="&lt;algebraic-geometry&gt;" AnswerCount="3" CommentCount="3" />
  <row Id="23407" PostTypeId="1" CreationDate="2011-02-23T18:40:31.037" Score="3" ViewCount="157" Body="&lt;p&gt;I want to learn discrete mathematics by reading book by myself but I find sometime it's very hard to understand what author trying to say.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to know, are there any good online video that teach discrete mathematic?&lt;/p&gt;&#xA;" OwnerUserId="725" LastActivityDate="2011-02-23T19:53:08.283" Title="Are there any good Discrete Mathematics video online?" Tags="&lt;discrete-mathematics&gt;&lt;online-resources&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" />
  <row Id="23408" PostTypeId="2" ParentId="23405" CreationDate="2011-02-23T18:49:15.993" Score="2" ViewCount="" Body="&lt;p&gt;The core difference is that a social choice function selects only one candidate - the fact that $N - S$ is not $\beta$-effective for $A - B$ with regards to a s.c.f F means that there exists $P^S \in L^S$ such that for every $Q^{N - S} \in L^{N - S}$, it does not hold that $F(P^S,Q^{N-S}) \in A - B$. Since F is a s.c.f, this means that $F(P^S, Q^{N-S}) \in B$ and therefore that $S$ is $\beta$-effective for $B$ with regards to F. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if we talk about $\beta$-effectiveness with regards to a s.c.c. $H$, the fact that it does not hold that $H(P^S, Q^{N-S}) \subset  A-B$ does &lt;em&gt;not&lt;/em&gt; necessarily mean that $H(P^S,Q^{N-S}) \subset B$ - we could have that H selects both members of $B$ and $A-B$. Is this clear enough for you to come up with a s.c.c that does not satisfy this condition?&lt;/p&gt;&#xA;" OwnerUserId="5573" LastEditorUserId="5573" LastEditorDisplayName="" LastEditDate="2011-02-23T19:47:07.947" LastActivityDate="2011-02-23T19:47:07.947" CommentCount="1" />
  <row Id="23409" PostTypeId="1" AcceptedAnswerId="23412" CreationDate="2011-02-23T19:07:15.220" Score="11" ViewCount="217" Body="&lt;p&gt;I hope this isn't off topic - sorry if I'm wrong.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In 1976, Kenneth Appel and Wolfgang Haken proved the claim that a map can always be coloured with four colours, with no adjacent regions given the same colour. This was controversial because the proof process required a computer to evaluate many different cases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From what I've read, the controversy was (1) that no human could reasonably confirm the proof, and (2) that being a computer proof, all it did was give a yes/no answer to the question - it didn't contribute to the understanding of the problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first issue seems to be a non-issue now - the proof has been replicated using different software. But the second issue seems to stand - and I don't really understand that issue.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Quite a few proofs require a number of particular cases of the problem to be separately checked. Having eliminated every possible case where the answer might have been &quot;no&quot; is a routine way of proving that the answer is &quot;yes&quot;. Does it really make any difference in principle whether the number of cases is 2 or 2 million? Does the scale of that number make any difference to the degree of human understanding of the problem?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And, given that people wrote the software that evaluated all the cases, all the ideas underlying the proof seem to be understandable by, and to have been understood by, people.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To me, having evaluated a thousand different variations on the same kind of a problem shows no more understanding than evaluating one. Solving a thousand quadratic equations, for instance, is much the same as solving one - all the repetitions are just plugging different numbers into the same formula, or repeating the same completing-the-square procedure, or whatever.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, I am very much impressed that Appel and Haken were able to understand the problem in sufficient depth that they could write a program to derive what all the special cases are and check them. Writing software to reliably determine all the cases often shows even deeper understanding than manual derivation of all the cases, where deeper understanding can be bypassed to a degree by trial-and-error.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Getting the computer to run the program, once written, seems irrelevant to me. The program presumably could have been (eventually) executed by a person, in the same way that it could have been executed by a Turing machine, but doing the mechanical follow-the-steps stuff seems irrelevant to depth of understanding.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there something I'm missing?&lt;/p&gt;&#xA;" OwnerUserId="510" LastActivityDate="2011-02-23T20:02:45.820" Title="Did the Appel/Haken graph colouring (four colour map) proof really not contribute to understanding?" Tags="&lt;graph-theory&gt;&lt;math-history&gt;&lt;coloring&gt;" AnswerCount="1" />
  <row Id="23410" PostTypeId="1" CreationDate="2011-02-23T19:10:56.107" Score="1" ViewCount="105" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a Computer Science background and not very knowledgeable in Probability and Statistics. So excuse me if my question,notation, or language is flawed. Anyways, the problems is that we have two Bernoulli variables $X_1,X_2$ that generate sequences of values for $n$ consecutive Bernoulli trials. Now we have $m$ independent observations of those trials in discrete independent time-slots. For example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;let:&#xA; $n=4, m=3$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The result of the aforementioned trials is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for $X_1: ((1,0,0,1),(0,0,1,1),(0,0,0,1))$&lt;br&gt;&#xA;for $X_2: ((0,0,1,0),(0,1,1,0),(0,1,0,0))$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where each of &quot;inner-sequences&quot; is one of $m$ independent sequence of Bernoulli trials. Now we aggregate each of those &quot;inner-sequences&quot; into a Beta distribution function to represent the &lt;em&gt;posterior probability&lt;/em&gt; of success/failure of each variable in different observation. For instance the above sequence transforms to the following&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$X_1:(B_{11},B_{12},B_{13})$&#xA;$X_2:(B_{21},B_{22},B_{23})$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where each $B_{ij}$ is a Beta distribution function associated with the corresponding sequence of trials in the previous part of the example.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we a have two sequences Beta Distribution where we want to use in order to find the correlation between $X_1$ and $X_2$ preferably producing a final beta curve that shows the degree of correlation between $X_1$ and $X_2$ incorporating the factor of uncertainty, or a Gaussian curve. A very simple approach is to find the correlation based on the mean of the curves and using the Pearson's correlation method. However, this is not precise enough. Is there any method to find the correlation between two Beta distribution functions. An easier question is where can I find useful information about the detection of correlation based on two distribution functions of any kind (easiest should be Gaussian functions). Thank you so much in advance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Amir&lt;/p&gt;&#xA;" OwnerUserId="7420" LastEditorUserId="7420" LastEditorDisplayName="" LastEditDate="2011-02-23T20:22:24.283" LastActivityDate="2011-02-23T20:22:24.283" Title="Correlation between Beta distributions" Tags="&lt;probability&gt;&lt;statistics&gt;&lt;probability-theory&gt;&lt;probability-distributions&gt;&lt;correlation&gt;" />
  <row Id="23411" PostTypeId="1" AcceptedAnswerId="23415" CreationDate="2011-02-23T19:20:24.113" Score="3" ViewCount="144" Body="&lt;p&gt;I'm trying to figure out how I can work out the number of possible valid hand combinations of a poker game that a player (opponent) could possibly have when the flop has been dealt on the table.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I'm guessing that I have to take the probability that each hand can occur and then see which one is the highest one that could possibly be?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's say that the opponent has bet $10 and now I have to make my next move.  Is there some math way, (with use of computer) that I can work out the number of valid combinations of hands that this opponent could still have and which he'll most likely pick?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If anyone could give some pointers as to what I should look into? &lt;/p&gt;&#xA;" OwnerUserId="807" LastActivityDate="2011-02-23T20:36:20.013" Title="Poker combinations &amp; probability" Tags="&lt;probability&gt;&lt;combinatorics&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="23412" PostTypeId="2" ParentId="23409" CreationDate="2011-02-23T19:22:49.003" Score="19" ViewCount="" Body="&lt;p&gt;When people say that Appel-Haken did not really contribute understanding, they aren't necessarily talking about the four-color problem itself. They mean that the proof did not really contribute any understanding of &lt;em&gt;mathematics.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A famous example when the proof of a long-standing conjecture really did contribute a lot of understanding is Andrew Wiles' proof of Fermat's Last Theorem. People weren't excited about this proof because they cared directly about the answer to FLT; they were excited because the proof brought together and integrated ideas from many disciplines and represented progress on the much larger &lt;a href=&quot;http://en.wikipedia.org/wiki/Langlands_program&quot; rel=&quot;nofollow&quot;&gt;Langlands program&lt;/a&gt;. Similarly, Perelman's proof of the Poincare Conjecture introduced important new ideas and tools. This was why many experts were convinced that Perelman's proof was valid even before it was formally checked; the high-level summary contained enough nontrivial ideas that they already saw that there was something extremely interesting going on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, the problem with Appel-Haken is that it is &lt;em&gt;boring.&lt;/em&gt; It was more or less an application of an already-understood technique, just on a larger scale, and so has led to very little interesting new mathematics. People are still looking for a conceptual proof of the Four-Color Theorem analogous to the two proofs above, for example people working in quantum topology; see &lt;a href=&quot;http://sbseminar.wordpress.com/2009/10/07/how-to-almost-prove-the-4-color-theorem/&quot; rel=&quot;nofollow&quot;&gt;this blog post&lt;/a&gt; by Noah Snyder and Kainen's &lt;a href=&quot;http://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.110.6627&quot; rel=&quot;nofollow&quot;&gt;Quantum interpretations of the four-color theorem&lt;/a&gt;. A proof along these lines would be much more interesting, as it would likely shed light on a number of other issues in quantum topology. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;This principle applies to more than just long-standing open problems. Many problems you might encounter in an undergraduate course might be solvable by a tedious computation along the lines of the Appel-Haken proof, but often there exist much more interesting conceptual proofs along the lines of Wiles' proof. If you stopped after finding the tedious calculation you might never find the conceptual argument, which often turns out to be much more interesting (e.g. it naturally suggests generalizations, interesting concepts, ties to other branches of mathematics...). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I will take your comment about quadratic equations as an example. It's true that the quadratic formula allows you to solve quadratic equations mechanistically. The question, then, is whether the quadratic formula leads to any significant conceptual understanding of polynomials. For example, does it suggest a natural route to the &lt;em&gt;cubic&lt;/em&gt; formula?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you think of the quadratic equation in terms of completing the square, then you quickly run into a problem: you cannot, in general, complete the cube. So completing the square does not generalize to cubic equations. If you want to understand cubic equations, it follows that you need to think about the quadratic formula more conceptually.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The conceptual breakthrough is the following: what the quadratic formula really shows you is that there is a &lt;em&gt;symmetry&lt;/em&gt; to the roots of a quadratic polynomial. The roots&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x_1 = \frac{-b + \sqrt{b^2 - 4ac}}{2a}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x_2 = \frac{-b - \sqrt{b^2 - 4ac}}{2a}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;are &lt;em&gt;conjugate&lt;/em&gt;: they are related by a symmetry which flips the sign of the square root. This symmetry manifests itself concretely in the fact that the sum&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x_1 + x_2 = - \frac{b}{a}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is completely invariant under flipping the sign of the square root, whereas the sum&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x_1 - x_2 = \frac{ \sqrt{b^2 - 4ac} }{a}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is completely negated under flipping the sign of the square root. Taking this idea seriously leads you to the method of &lt;a href=&quot;http://en.wikipedia.org/wiki/Cubic_function#Lagrange_s_method&quot; rel=&quot;nofollow&quot;&gt;Lagrange resolvents&lt;/a&gt;, and now if you were born in the right century you would be well on your way to inventing group theory, Galois theory, and (if you were really observant) representation theory. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Isn't that way more interesting than using the quadratic formula? &lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2011-02-23T20:02:45.820" LastActivityDate="2011-02-23T20:02:45.820" CommentCount="7" />
  <row Id="23413" PostTypeId="1" AcceptedAnswerId="23414" CreationDate="2011-02-23T19:33:41.807" Score="0" ViewCount="48" Body="&lt;p&gt;Hello,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have the following question $f(x) = -x^2 + 4x$ when $x = -3$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Should the structure of the equation be as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$-3^2 + 4 x -3 = -21$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The issue I'm having is with the double negative in the variable $x$ and in the function formula.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If anyone can help me, be much appreciated. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="6771" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-23T19:45:15.957" LastActivityDate="2011-02-23T19:50:22.267" Title="structure of an algebraic equation" Tags="&lt;homework&gt;&lt;algebra&gt;" AnswerCount="2" />
  <row Id="23414" PostTypeId="2" ParentId="23413" CreationDate="2011-02-23T19:36:16.410" Score="2" ViewCount="" Body="&lt;p&gt;It should be $-(-3)^2+4(-3)$ which is $-9-12 = -21$.&lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2011-02-23T19:36:16.410" CommentCount="1" />
  <row Id="23415" PostTypeId="2" ParentId="23411" CreationDate="2011-02-23T19:45:18.947" Score="2" ViewCount="" Body="&lt;p&gt;The number of hands is 47*46/2=1081 as you see five cards (your two plus the flop) so there are 47 left to choose from.  If you are looking for the highest, there is no probability involved, just look at all of them and find the highest.  Ignoring any information from the betting, the chance that an opponent has three of a kind is just (number of holdings that make three of a kind)/1081.  For this example, if none of the five cards you see match in value, there are nine hands that make your opponent three of a kind (three pairs of each value that matches one of the flop cards).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;None of this speaks to which of the hands are reasonable for your opponent to behave as you have seen so far.  There are many books on the subject (most of which give similar advice) that your opponent may well have read, so you can take some guidance there.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-02-23T19:45:18.947" CommentCount="4" />
  <row Id="23416" PostTypeId="2" ParentId="23387" CreationDate="2011-02-23T19:47:46.307" Score="5" ViewCount="" Body="&lt;p&gt;Let $\psi_n$ be a sequence of standard mollifiers, symmetric about 0.  If $f \in L^\infty$, $g \in L^1$, it is a quick application of Fubini's theorem to see that $\int (f * \psi_n) g = \int f (\psi_n * g)$, where $*$ denotes convolution.  Since $f * \psi_n$ is continuous and $\psi_n * g \to g$ in $L^1$, we are done.&lt;/p&gt;&#xA;" OwnerUserId="822" LastActivityDate="2011-02-23T19:47:46.307" />
  <row Id="23417" PostTypeId="2" ParentId="23413" CreationDate="2011-02-23T19:50:22.267" Score="1" ViewCount="" Body="&lt;p&gt;Several comments:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;You don't give a question, so I don't know what &quot;question&quot; you have. You wrote down a formula for a function, and a value for $x$, but no &lt;em&gt;question&lt;/em&gt;. Is your question &quot;What is the value of $f(x)$ when $x=-3$, if $f(x)=-x^2 + 4x$?&quot; If so, please write out the question explicitly.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Be careful with the signs and the operations. When you write down a bunch of numbers with operations between them, there is a &lt;em&gt;precedence&lt;/em&gt; of operations that is universally agreed upon. Some operations have to happen first. For example, if you write $2+3\times 5$, then the convention is that you do the &lt;em&gt;product&lt;/em&gt; first, and the sum second, so that $2+3\times 5$ equals $17$ (not $25$). In particular, one does the exponentials first, then the products (including a minus sign, which is really multiplication by $-1$), and sums at the end.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Now: to evaluate $f(x)$ at $x=-3$, you have to take the formula for $f(x)$, and &lt;em&gt;plug in&lt;/em&gt; the value of $x$. That is, take the formula for $f$, which is&#xA;$$f(x) = -x^2 + 4x$$&#xA;and wherever you see an $x$, you have to put in $-3$ &lt;strong&gt;instead&lt;/strong&gt;. Putting parentheses so that it is clear what we are doing, you would have&#xA;$$f(-3) = -(-3)^2 + 4(-3).$$&#xA;When you evaluate this expression, the &lt;em&gt;first&lt;/em&gt; thing you need to do is the square. Then  you have to do the products (the $-$ times the $(-3)^2$, and the $4$ times $(-3)$); and finally you do the sum of the two totals you have.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your attempt is incorrect in sundry ways: you should not have an $x$ left after evaluating; the end result must be a &lt;em&gt;number&lt;/em&gt;. You plugged in $3$ for $x$ in the $x^2$ instead of $-3$. And what you wrote, $-3^2+4x-3$, is not equal to $-21$. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="742" LastActivityDate="2011-02-23T19:50:22.267" CommentCount="1" />
  <row Id="23418" PostTypeId="1" CreationDate="2011-02-23T19:50:33.077" Score="4" ViewCount="98" Body="&lt;p&gt;I'm trying to derive a taxonomy from sparse &quot;is-a&quot; relationships and am looking for a linear algebraic solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More specifically, the data is noisy and I want to detect relations that violate anti-transitivity, remove the least offenders, and then fill-in the transitive relationships, inheriting all relations from parents to children.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is anyone aware of mathematical methods to:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;compute a general statistic of &quot;transitivity&quot; of some graph matrix?&lt;/li&gt;&#xA;&lt;li&gt;compute a matrix of relations in violation of anti-transitivity?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I could remove the offending rows from (2) and see how they affect (1).  Then, I could compute transitive closure, which I already know how to do [1].&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[1] &lt;a href=&quot;http://www.cs.nmsu.edu/~ipivkina/TransClosure/index.html&quot; rel=&quot;nofollow&quot;&gt;http://www.cs.nmsu.edu/~ipivkina/TransClosure/index.html&lt;/a&gt; Warning: contains god-forsaken Java Applet.&lt;/p&gt;&#xA;" OwnerUserId="7422" LastActivityDate="2011-02-28T04:49:47.093" Title="How to detect antitransitivity from an adjacency matrix?" Tags="&lt;linear-algebra&gt;&lt;matrices&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="23419" PostTypeId="2" ParentId="23407" CreationDate="2011-02-23T19:53:08.283" Score="2" ViewCount="" Body="&lt;p&gt;Check out the free audio/video courses in mathematics available through MIT OpenCourseWare: &lt;a href=&quot;http://ocw.mit.edu/courses/audio-video-courses/#mathematics&quot; rel=&quot;nofollow&quot;&gt;http://ocw.mit.edu/courses/audio-video-courses/#mathematics&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7423" LastActivityDate="2011-02-23T19:53:08.283" />
  <row Id="23420" PostTypeId="1" AcceptedAnswerId="23421" CreationDate="2011-02-23T20:02:37.347" Score="3" ViewCount="84" Body="&lt;p&gt;While working on the problem $g(x)= \sqrt{x^{2}-4x+4}$, I got stuck and plugged the formula into Wolfram Alpha.  It simplified alot of things for me, but there is one section I'm having trouble with, since one of my weaknesses is radicals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to the Chain Rule, $\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}$, and therefore $u= x^{2}-4x+4$ and $y= \sqrt{u}$, resulting in $\frac{d\sqrt{u}}{du}$.  That then transforms to $\frac{1}{2\sqrt{u}}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How does that work?&lt;/p&gt;&#xA;" OwnerUserId="7006" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2011-02-23T20:27:53.443" LastActivityDate="2011-02-23T20:27:53.443" Title="Differentiating function with radicals" Tags="&lt;calculus&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="23421" PostTypeId="2" ParentId="23420" CreationDate="2011-02-23T20:10:10.917" Score="2" ViewCount="" Body="&lt;p&gt;I sometimes find that working with abstract functions is easier than specific ones, especially when powers and radicals are involved as they are very confusing for me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The chain rule, as you quote it, is correct. Your implementation, however, is not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Define $f(u) = \sqrt u$ and $g(x) = x^2 -4x +4$, then you wish to derive $f(g(x))$, which should give &#xA;$$(f(g(x))&amp;#39; = f&amp;#39;(g(x))\cdot g&amp;#39;(x)=\frac{1}{2\sqrt{g(x)}}\cdot g&amp;#39;(x) = \frac{1}{2\sqrt{x^2-4x+4}}\cdot(2x-4)$$&lt;/p&gt;&#xA;" OwnerUserId="622" LastActivityDate="2011-02-23T20:10:10.917" />
  <row Id="23422" PostTypeId="2" ParentId="23411" CreationDate="2011-02-23T20:31:19.693" Score="0" ViewCount="" Body="&lt;p&gt;There is no software that can accurately take into consideration what move your opponent has done (i.e., assuming you were in position and he bet before you, take into account his bet or check) and tell you what likely has he has.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are way too many variables to account for here, including but not limited to, how aggressive, how tight he is playing pre-flop, how likely he is bluffing given the cards if they look scary or not, etc.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That said, the best one can do after the flop is rely on previous experience and their own judgement calls.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is software that takes into account number of players at the table, chip counts, etc, that will give you the mathematically correct play late in tournaments.  This is usually when you are in push or fold mode.  If you wish to see how this works, please see some software such as sit and go wizard (sngwiz.com)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As to answering your original question, I don't think the hands a player could possibly has would be useful to you at all, but of course, it is not hard to see what these are based on the cards you see.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt;&lt;br&gt;&#xA;In answer to your question about where you should look into, if you are looking to improve your game, which seems to be the case, the best thing you can do in my opinion is sign up at a forum such as 2 plus 2 or any really, that has some good players willing to help you out.  Post hand histories and see the players will be more than happy to comment on your play.&lt;/p&gt;&#xA;" OwnerUserId="2533" LastEditorUserId="2533" LastEditorDisplayName="" LastEditDate="2011-02-23T20:36:20.013" LastActivityDate="2011-02-23T20:36:20.013" />
  <row Id="23423" PostTypeId="2" ParentId="23351" CreationDate="2011-02-23T20:35:14.810" Score="5" ViewCount="" Body="&lt;p&gt;If you want to divide space up into uniform volume cells with minimum surface area, the honeycomb is not optimal. Look at the &lt;a href=&quot;http://en.wikipedia.org/wiki/Weaire%E2%80%93Phelan_structure&quot; rel=&quot;nofollow&quot;&gt;Weaire–Phelan structure&lt;/a&gt;. While honeycombs are not quite optimal, they are certainly close enough for bees -- they're suboptimal by only 0.3%. &lt;/p&gt;&#xA;" OwnerUserId="2501" LastEditorUserId="2501" LastEditorDisplayName="" LastEditDate="2011-02-23T21:00:08.740" LastActivityDate="2011-02-23T21:00:08.740" />
  <row Id="23424" PostTypeId="2" ParentId="23386" CreationDate="2011-02-23T20:36:24.137" Score="1" ViewCount="" Body="&lt;p&gt;If  $\rm\:A|a\:\Rightarrow\: a=0\ $ then ditto for $\rm\:A^2\:,\:$ since $\rm\:A|A^2|a\:.\:$ Thus by (iii) $\rm\ A^2 = A\ $ so $\rm\: A = (1)\:\Rightarrow\: A|1\ \Rightarrow\Leftarrow\:$ Therefore there exists an $\rm\ a\ne 0\ $ such that $\rm\: A\:|\:a\:,\:$ say $\rm\ AB = (a)\:.\:$ Applying the lemma below yields $\rm\ (b) = AC,\ (B,C) = 1\:.$ So $\rm\ (a,b) = (AB,AC) = A\:(B,C) = A\:.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;LEMMA&lt;/strong&gt; $\ $ For any $\rm\:A,B\:$ there exists $\rm\:b,C\:$ with $\rm (b) = AC,\ (B,C)=1\:.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof&lt;/strong&gt; $\ $ Let $\rm\: B = P_1^{e_1}\cdots P_n^{e_n},\ P_i\:$ prime, $\rm\ Q = P_1\cdots P_n,\ \:Q_i = Q/P_i\:.\ Q_i|Q\ $ proper $\rm\:\Rightarrow\ AQ_i|AQ$ proper, so there exists $\rm\ b_i:\ AQ_i|b_i,\ AQ\nmid b_i\:.\:$ Put $\rm\ b = b_1+\cdots+b_n\:.\:$ Note $\rm\ A|b\ $ by $\rm\ A|AQ_i|b_i\:.\:$ $\rm\ i\ne j\ \Rightarrow\ P_i|Q_j\ \Rightarrow\ AP_i|AQ_j|b_j\:,\:$ but $\rm\ AP_i\nmid b_i\ $ (else $\rm\ AP_i,AQ_i|b_i $ $\rm \Rightarrow$ $\rm\: lcm(AP_i,AQ_i)=AQ|b_i\:$). Thus $\rm\ AP_i\nmid b\ $ so $\rm\ P_i\nmid (b)/A =: C\:,\: $ hence $\rm\ (B,C) = 1,\ \ (b) = AC\:.\ \ $ &lt;strong&gt;QED&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2011-02-23T22:02:58.227" LastActivityDate="2011-02-23T22:02:58.227" CommentCount="5" />
  <row Id="23425" PostTypeId="1" CreationDate="2011-02-23T20:36:42.493" Score="1" ViewCount="224" Body="&lt;p&gt;I know how to solve linear homogeneous ordinary differential equations with constant coefficients using the differential operator D, by using &lt;a href=&quot;http://math.ucsd.edu/~dmeyer/teaching/20Dspring07/diffop2.pdf&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; method.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it possible to use a similar method (using the differential operator) to solve more advanced ODEs? I'm thinking of both more advanced linear ODEs, such as Euler-Cauchy differential equations, as well as non-linear ODEs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there any articles on the web on this topic, or even textbooks that use this method to solve ODEs?&lt;/p&gt;&#xA;" OwnerUserId="4576" LastActivityDate="2011-02-23T22:15:00.597" Title="Solving ordinary differential equations using the differential operator D" Tags="&lt;calculus&gt;&lt;differential-equations&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="23426" PostTypeId="1" CreationDate="2011-02-23T20:41:18.023" Score="3" ViewCount="49" Body="&lt;p&gt;I am interested in the relationship between non-linear differential equations and the Hausdorff, or fractal, dimension of the set of solutions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, the &lt;a href=&quot;http://en.wikipedia.org/wiki/Lorenz_attractor&quot; rel=&quot;nofollow&quot;&gt;Lorenz Attractor&lt;/a&gt;, which is a standard example in the theory of dynamical systems, apparently has a &quot;correlation dimension&quot; of 2.05, whatever that means.  &lt;a href=&quot;http://adsabs.harvard.edu/cgi-bin/nph-bib_query?bibcode=1983PhyD....9..189G&amp;amp;db_key=PHY&quot; rel=&quot;nofollow&quot;&gt;According to the authors&lt;/a&gt; who derived this result, the correlation dimension &quot;is closely related to the fractal dimension and the information dimension, but its computation is considerably easier.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I really want is the following: does anyone know of a system of coupled differential equations, preferably with smooth coefficients, for which a non-integer Hausdorff dimension of the solution set can be derived in an elegant way?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="1417" LastEditorUserId="1417" LastEditorDisplayName="" LastEditDate="2011-02-23T20:48:37.053" LastActivityDate="2011-02-23T20:48:37.053" Title="The Hausdorff dimension of the set of solutions of a system of  coupled differential equations." Tags="&lt;dynamical-systems&gt;&lt;fractals&gt;" />
  <row Id="23427" PostTypeId="5" CreationDate="2011-02-23T20:49:08.020" Score="0" ViewCount="" Body="&lt;p&gt;For questions about the theory of probability, such as questions about sigma-algebras or probability measures, the &quot;probability-theory&quot; tag is more appropriate.&lt;/p&gt;&#xA;" OwnerUserId="-1" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2011-02-23T20:49:54.773" LastActivityDate="2011-02-23T20:49:54.773" />
  <row Id="23428" PostTypeId="4" CreationDate="2011-02-23T20:49:08.020" Score="0" ViewCount="" Body="This tag is for basic questions about probability and for questions in which one wants a probability or expected value calculated. " OwnerUserId="-1" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2011-02-23T20:49:52.933" LastActivityDate="2011-02-23T20:49:52.933" />
  <row Id="23429" PostTypeId="1" CreationDate="2011-02-23T20:49:34.820" Score="4" ViewCount="49" Body="&lt;p&gt;Hello,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have the following function &#xA;$$f(x) = 3x^3 - 5x^2 - 4x + 4.$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to find the value of $f(x)$ when $x = -3$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have ordered the equation in the following way.&#xA;$$f(-3) = (3 * -3^3) - (5 * -3^2) - (4*-3) + 4 = -28.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone tell me if the following syntax is correct and if not, where have I gone wrong.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many Thanks&lt;/p&gt;&#xA;" OwnerUserId="6771" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-23T21:04:10.513" LastActivityDate="2011-02-23T21:04:10.513" Title="Order of operation - polynomial equation" Tags="&lt;homework&gt;&lt;algebra&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="23430" PostTypeId="5" CreationDate="2011-02-23T20:50:36.507" Score="0" ViewCount="" Body="&lt;p&gt;For basic questions about probability or questions about calculating a probability or an expected value, please use the &quot;probability&quot; tag.&lt;/p&gt;&#xA;" OwnerUserId="-1" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2011-02-23T20:54:08.887" LastActivityDate="2011-02-23T20:54:08.887" />
  <row Id="23431" PostTypeId="4" CreationDate="2011-02-23T20:50:36.507" Score="0" ViewCount="" Body="This tag is for questions about the theory of probability, such as questions about sigma-algebras or probability measures." OwnerUserId="-1" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2011-02-23T20:51:08.563" LastActivityDate="2011-02-23T20:51:08.563" />
  <row Id="23432" PostTypeId="2" ParentId="23429" CreationDate="2011-02-23T20:56:09.073" Score="4" ViewCount="" Body="&lt;p&gt;When you write $5*-3^2$, the precedence of operations means that you first compute $3^2$, then you multiply the result by $-1$, then you multiply the entire thing by $5$. This is &lt;em&gt;not&lt;/em&gt; what you want.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Don't be scared of parentheses! The best way to write this, if you want to insist on putting the $*$ in it (we normally just write multiplication by juxtaposition) would be:&#xA;$$f(-3) = (3*(-3)^3) - (5*(-3)^2) - (4*(-3)) + 4.$$&#xA;More succinctly, you can write it as:&#xA;$$f(-3) = 3(-3)^3 - 5(-3)^2 - 4(-3) + 4.$$&#xA;By the precedence order, you would first compute the exponentials, then the products, then the sums.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That said, I don't think you computed $f(-3)$ correctly. &lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2011-02-23T20:56:09.073" CommentCount="1" />
  <row Id="23433" PostTypeId="1" AcceptedAnswerId="23441" CreationDate="2011-02-23T21:07:57.013" Score="8" ViewCount="226" Body="&lt;p&gt;How do I solve the following?&#xA;$$\lim_{x\to 0} \int_0^1 \cos\left(\frac{1}{xt}\right)\, dt$$&lt;/p&gt;&#xA;" OwnerUserId="6556" LastEditorUserId="6075" LastEditorDisplayName="" LastEditDate="2011-02-24T00:33:55.090" LastActivityDate="2011-02-24T18:08:13.163" Title="having trouble with limit of integral" Tags="&lt;analysis&gt;&lt;integral&gt;" AnswerCount="2" CommentCount="6" FavoriteCount="1" />
  <row Id="23435" PostTypeId="1" AcceptedAnswerId="23442" CreationDate="2011-02-23T21:52:26.387" Score="6" ViewCount="70" Body="&lt;p&gt;Hi! I'm starting a mini-course on Combinatorics and, although I can &quot;see&quot; the results, I'm having difficulties proving them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Being $\#A = n \in \mathbb{N}$, prove that $\#A^{k} = n^{k}$, for $k \in \mathbb{N}$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I understand that $A^{k} = \{(x_1, \ldots, x_k) : x_i \in A\}$, and that we have $n \times n \ldots \times n$ ($k$ times) possible layouts for a sequence like $(x_1, \ldots, x_k)$, but I don't know what to use to prove the result.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you suggest some plan? If possible, could it be an advice that would give me traction to prove the following related results?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for the time of took to read my question. I highly appreciate it.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update:&lt;/strong&gt; Since this question is so straightforward (see comments), and was only presented with the intent of getting acquainted with the tools of the trade, I shall move on to another result I wish I can prove. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Prove that the number of elements in $\{(x_1, \ldots, x_k) : x_i \in A, i \in \{1, \ldots, k\}, x_i \neq x_j \Longleftrightarrow i \neq j, j \in \{1, \ldots, k\}\}$ is given by the formula $\frac{n!}{(n-k)!}$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Thanks for your replies so far!&lt;/p&gt;&#xA;" OwnerUserId="6538" LastEditorUserId="6538" LastEditorDisplayName="" LastEditDate="2011-02-23T22:34:51.483" LastActivityDate="2011-02-23T22:35:42.287" Title="How to prove that if $\#A = n$, then $\#A^{k} = n^{k}$? And what about the formula $\frac{n!}{(n-k)!}$?" Tags="&lt;homework&gt;&lt;combinatorics&gt;&lt;elementary-set-theory&gt;" AnswerCount="1" CommentCount="10" />
  <row Id="23436" PostTypeId="1" CreationDate="2011-02-23T21:56:48.343" Score="0" ViewCount="135" Body="&lt;p&gt;Let $a\sin^{2}x + b\cos^{2}x=a\sin^{2}y + b\cos^{2}y=1$&#xA;and $a\tan x=b\tan y$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Prove $(a-b)(b-1)=0$&lt;/p&gt;&#xA;" OwnerUserId="7425" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-23T22:43:12.980" LastActivityDate="2011-02-23T22:43:12.980" ClosedDate="2011-02-23T23:07:00.843" Title="Prove trigonometry" Tags="&lt;trigonometry&gt;" CommentCount="1" />
  <row Id="23437" PostTypeId="1" CreationDate="2011-02-23T22:02:18.507" Score="-2" ViewCount="133" Body="&lt;p&gt;Let $H$ be a Hilbert space and assume that $\{e_k\}_{k\in\mathbb{Z}}$ is an orthonormal basis of $H$.  Define $V_j$ for $j\in\mathbb Z$ as the smallest closed subspace of $H$ containing $\{e_k:k\leq j\}$.  Show that $(V_j)_{j\in\mathbb Z}$ is a weak multiresolution.&lt;/p&gt;&#xA;" OwnerUserId="7428" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2011-02-26T03:16:03.283" LastActivityDate="2011-02-26T04:04:22.980" Title="Weak multiresolution" Tags="&lt;hilbert-spaces&gt;" AnswerCount="1" CommentCount="6" />
  <row Id="23438" PostTypeId="1" CreationDate="2011-02-23T22:05:11.353" Score="2" ViewCount="79" Body="&lt;p&gt;I've a $m \times n$ column matrix $A$ that can contain values {0,0.5,1}. i.e $\Omega = \{0,0.5,1\}$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$A$ = &#xA;$\begin{bmatrix} &#xD;&#xA;0 &amp;amp; 0.5 &amp;amp; 0.5 &amp;amp; 1 &amp;amp; 1 &amp;amp; \cdots &amp;amp; 0 \\&#xD;&#xA;0.5 &amp;amp; 0 &amp;amp; 1 &amp;amp; 1 &amp;amp; 0 &amp;amp; \cdots &amp;amp; 1 \\&#xD;&#xA;\vdots &amp;amp; \vdots &amp;amp; \vdots &amp;amp; \vdots &amp;amp; \vdots  &amp;amp; \vdots \\ &#xD;&#xA;0 &amp;amp; 0 &amp;amp; 0.5 &amp;amp; 0.5 &amp;amp; 1 &amp;amp; \cdots &amp;amp; 0 &#xD;&#xA;\end{bmatrix}$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;let the column matrix $x$ be equal to $\mathbb{R_0^{+ n \times 1}}$ (non-negative real numbers including 0)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x$ = &#xA;$\begin{bmatrix} &#xD;&#xA; 1 \\ 2 \\ 0 \\ 0.5 \\ 3 \\ \vdots \\ 0.3 &#xD;&#xA;\end{bmatrix}$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know that $Ax = b$. So given $b$ we can find $x$ by $x = A \backslash b$ (performing matrix left division). 99.9% the matrix A is full rank (because it can contain only 3 values and $m$ is very large) but if not I could always use $x$ = pseudo-inverse$(A) \times b$.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is now if we are given $b$ and a matrix $A^{\prime}$ such that $A^{\prime} \supseteq A$ where $A^{\prime} \in \mathbb{R^{m \times \textbf{N}}}$ and $\textbf{N} \supseteq n$ and belong to the same domain as $A$ (pardon my math explanation. $A$ and $A^{\prime}$ has same rows but the columns of $A$ are $\subseteq$ of $A^{\prime}$  ), is it possible to find $x$ ? (Note that we do not need $A$, if a column in $A^{\prime}$ is involved in the linear combination it gets a value or else it can just be $0$).&lt;/p&gt;&#xA;" OwnerUserId="6335" LastEditorUserId="6335" LastEditorDisplayName="" LastEditDate="2011-02-24T13:02:03.893" LastActivityDate="2011-02-24T13:02:03.893" Title="Deducing linear combinations of column matrices." Tags="&lt;linear-algebra&gt;&lt;matrices&gt;" CommentCount="3" />
  <row Id="23439" PostTypeId="2" ParentId="23406" CreationDate="2011-02-23T22:05:41.153" Score="1" ViewCount="" Body="&lt;p&gt;The hint is a consequence of Chevalley's theorem on constructible sets: &lt;a href=&quot;http://en.wikipedia.org/wiki/Constructible_set_%28topology%29&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Constructible_set_%28topology%29&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A non-empty Zariski-open susbset of the affine line is clearly not compact, so the image has to be a finite set.&#xA;Projecting on each coordinate, you get that the variety is finite.&lt;/p&gt;&#xA;" OwnerUserId="2660" LastActivityDate="2011-02-23T22:05:41.153" />
  <row Id="23440" PostTypeId="2" ParentId="23425" CreationDate="2011-02-23T22:15:00.597" Score="1" ViewCount="" Body="&lt;p&gt;There is a large literature on &quot;&lt;a href=&quot;http://en.wikipedia.org/wiki/Operational_calculus&quot; rel=&quot;nofollow&quot;&gt;operational calculus&lt;/a&gt;&quot; dating back to Heaviside's pioneering work. One particularly powerful operator factorization technique is the Infeld - Hull ladder method - which plays a big role in unifying many classes of special functions that arise in physics (mainly via separation of variables in various coordinate systems). Willard Miller showed that this method is equivalent to the representation theory of four local Lie groups. This lie-theoretic approach served to powerfully unify and &quot;explain&quot; all prior similar attempts to provide a unfied theory of such classes of special functions. See &lt;a href=&quot;http://math.stackexchange.com/questions/622/importance-of-representation-theory/2803#2803&quot;&gt;my post here&lt;/a&gt; for further details and references.&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2011-02-23T22:15:00.597" />
  <row Id="23441" PostTypeId="2" ParentId="23433" CreationDate="2011-02-23T22:19:11.950" Score="7" ViewCount="" Body="&lt;p&gt;If it's defined, it must be equal to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\lim_{x\to 0}\;\lim_{a\to 0} \int_a^1 \cos\left(\frac{1}{xt}\right)\, \mathrm{d}t\;.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the integral is well-defined, and you can substitute $u = 1 / (xt)$, yielding&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\lim_{x\to 0}\;\lim_{a\to 0} \frac{1}{x}\int_{1/x}^{1/(ax)} \frac{\cos u}{u^2}\, \mathrm{d}u\;.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then integration by parts leads to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\lim_{x\to 0}\;\lim_{a\to 0} \frac{1}{x}\left(\left[\frac{\sin  u}{u^2}\right]_{1/x}^{1/(ax)}+2\int_{1/x}^{1/(ax)} \frac{\sin u}{u^3}\, \mathrm{d}u\right)\;.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now you can bound the absolute value of the integral by taking the absolute value of the integrand and dropping the sine, and both terms vanish in the limit(s).&lt;/p&gt;&#xA;" OwnerUserId="6622" LastActivityDate="2011-02-23T22:19:11.950" />
  <row Id="23442" PostTypeId="2" ParentId="23435" CreationDate="2011-02-23T22:27:26.103" Score="2" ViewCount="" Body="&lt;p&gt;To address the second question, the set in question is the number of elements of $A^k$ such that no two entries are the same. Since $A$ has $n$ elements, you then have $n$ choices for the first entry, and $n-1$ choices for the second entry, as you cannot choose the element which you placed in the first entry. You then have $n-2$ choices for the third entry, $\dots$, $n-(k-1)=n-k+1$ choices for the the $k^{\text{th}}$ entry, for the same reasoning. But notice&#xA;$$&#xD;&#xA;n(n-1)(n-2)\cdots(n-k+1)=\frac{n(n-1)\cdots(n-k+1)(n-k)\cdots 1}{(n-k)\cdots 1}=\frac{n!}{(n-k)!}.&#xD;&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="1583" LastEditorUserId="1583" LastEditorDisplayName="" LastEditDate="2011-02-23T22:35:42.287" LastActivityDate="2011-02-23T22:35:42.287" CommentCount="1" />
  <row Id="23443" PostTypeId="1" AcceptedAnswerId="23444" CreationDate="2011-02-23T22:35:28.527" Score="6" ViewCount="209" Body="&lt;p&gt;I found an example of a function $f: \mathbb{R}\to\mathbb{R}$ that is continuous and bounded, but is not uniformly continuous. It is $\sin(x^2)$. I think it's not uniformly continuous because the derivative is bigger and bigger as $x$ increases. But I dont know how to prove this is uniformly continous. Is $\sin(x^2)$ uniformly continuous then? if it isn't, can you guys think of any other examples? thanks&lt;/p&gt;&#xA;" OwnerUserId="7429" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-23T22:40:57.133" LastActivityDate="2011-02-23T23:32:42.917" Title="function $\mathbb{R}\to\mathbb{R}$ that is continuous and bounded, but not uniformly continuous" Tags="&lt;real-analysis&gt;" AnswerCount="4" />
  <row Id="23444" PostTypeId="2" ParentId="23443" CreationDate="2011-02-23T22:46:42.057" Score="5" ViewCount="" Body="&lt;p&gt;To prove that $f(x)=\sin(x^2)$ is not uniformly continuous, let $\epsilon=\frac{1}{2}$. You want to show that for every $\delta\gt 0$ there exists $x$ and $y$ (which may depend on $\delta$) such that $|x-y|\lt \delta$, but $|f(x)-f(y)|\geq \frac{1}{2}$. Suggestion: try to pick a $y$ where $f(y)=0$, and a very nearby $x$ where $f(1)=\pm 1$. You want $y^2=n\pi$ for some integer $n$, and $x^2$ to be $\frac{1}{2}\pi$ away.  If you cannot produce a pair explicitly, maybe you can show that you can pick them as close as you want anyway, provided $x$ and $y$ are large enough (which agrees with your observation that the derivative is unbounded). &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-23T23:04:28.617" LastActivityDate="2011-02-23T23:04:28.617" CommentCount="3" />
  <row Id="23445" PostTypeId="2" ParentId="23443" CreationDate="2011-02-23T22:46:57.370" Score="0" ViewCount="" Body="&lt;p&gt;One option is to consider a simpler function, like $f(x)=x^2$. Can you prove it is not uniformly continuous?&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2011-02-23T22:46:57.370" CommentCount="2" />
  <row Id="23446" PostTypeId="1" AcceptedAnswerId="23447" CreationDate="2011-02-23T22:52:30.893" Score="3" ViewCount="77" Body="&lt;p&gt;To solve for the $\sin^{-1} z$ ($z$ element of $\mathbb{C}$), the book reads that &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;$w = \sin^{-1} z$ when $z = \sin w$ implies:&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$w = \sin^{-1} z$ when $z = (e^{iw} - e^{-iw})/2i$  implies:&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$(e^{iw})^2 - 2iz(e^{iw}) - 1 = 0$ *&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I don't see (2) implies (3). I mean it looks like the first step they used was mutiply out the $2i$, then subtract $2zi$ from both sides, but I'm lost after that.&lt;/p&gt;&#xA;" OwnerUserId="6758" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-23T22:55:28.393" LastActivityDate="2011-02-24T00:54:44.330" Title="What was used to establish this equivalence? [Textbook]" Tags="&lt;algebra-precalculus&gt;&lt;complex-analysis&gt;" AnswerCount="2" />
  <row Id="23447" PostTypeId="2" ParentId="23446" CreationDate="2011-02-23T22:57:42.477" Score="5" ViewCount="" Body="&lt;p&gt;If $z = \frac{e^{iw}-e^{-iw}}{2i}$, then $2iz = e^{iw}-e^{-iw}$. Multiplying through by $e^{iw}$ you get&#xA;$$2ize^{iw} = (e^{iw}-e^{-iw})e^{iw} = e^{iw}e^{iw}-e^{-iw}e^{iw} = (e^{iw})^2 - 1.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Why would you multiply through by $e^{iw}$? Because &#xA;$$e^{iw} - e^{-iw} = e^{iw} - \frac{1}{e^{iw}} = \frac{e^{iw}e^{iw} - 1}{e^{iw}}$$&#xA;so you &quot;really&quot; have $2iz = \frac{e^{iw}e^{iw}-1}{e^{iw}}$, so multiplying through by $e^{iw}$ just clears denominators, just like the initial multiplication by $2i$). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So $-2ize^{iw} - 1 = -(e^{iw})^2 + 1 - 1 = -(e^{iw})^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Adding $(e^{iw})^2$ will give you $0$.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2011-02-23T22:57:42.477" />
  <row Id="23448" PostTypeId="2" ParentId="23443" CreationDate="2011-02-23T23:02:36.413" Score="2" ViewCount="" Body="&lt;p&gt;Let $T_n(x) = 1-n|x|$ for x in $[-1/n, 1/n]$.  As $T_n$ is linear, the required $\delta$ is $\epsilon/n$.  Try to construct a continuous function by using combinations of these $T_n$'s and constant (zero) functions.  I'm imagining a graph where every once in a while there is a spike with graph $T_n$, but as you go out to infinity, the spikes get steeper.&lt;/p&gt;&#xA;" OwnerUserId="4009" LastActivityDate="2011-02-23T23:02:36.413" />
  <row Id="23449" PostTypeId="2" ParentId="23443" CreationDate="2011-02-23T23:05:23.010" Score="0" ViewCount="" Body="&lt;p&gt;As another example, consider the sequence $S_n  = \sum\limits_{k = 1}^n {\frac{1}{k}} $, and a function $f$ such that $f(S_n)=1$ and $f\big(\frac{{S_n  + S_{n + 1} }}{2}\big) = 0$. Further note that &#xA;$$&#xD;&#xA;\frac{{S_n  + S_{n + 1} }}{2} - S_n  = \frac{1}{{2(n + 1)}} \to 0.&#xD;&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="2810" LastEditorUserId="2810" LastEditorDisplayName="" LastEditDate="2011-02-23T23:32:42.917" LastActivityDate="2011-02-23T23:32:42.917" CommentCount="1" />
  <row Id="23450" PostTypeId="2" ParentId="23418" CreationDate="2011-02-23T23:38:09.023" Score="1" ViewCount="" Body="&lt;p&gt;I think what you're looking for is &quot;transitive reduction&quot;, which gives the smallest relation whose transitive closure is the same as the transitive closure of the given relation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And this makes me think, why even bother with the transitive reduction when you can just take the transitive closure to begin with...that is, why worry about the 'noise' and just add in all the missing transitive arcs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If your noise is such that you have -cycles-... well, that's a whole nother problem (how do you know which edge is the 'bad' one?)&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;From the comments, I'm adding to my answer. I think the question really is something like this: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Given a directed acyclic graph plus some additional edges (which may induce cycles), return some kind of poset (a transitive directed acyclic graph)&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once the cycles are broken, one can then take a transitive closure (algebraically, as you like by $A^* = I + A + A^2 + ... + A^n = (I - A)^{-1}$ (OK, I threw in reflexive there too).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the difficulty is to get rid of the cycles. There is no way to know which ones are spurious, so you have to arbitrarily throw away edges to get rid of cycles. A naive approach would be to DFS repeatedly (this isn't particularly algebraic), removing the edge on which a cycle is recognized, until no more cycles are found. This isn't particularly elegant because it may remove more edges than necessary (to induce acyclicity).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The next idea is to try to get rid of the -fewest- edges possible, which off the top of my head sounds so very NP-complete that I wouldn't bother (can somebody confirm or not?).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When that happens, we can take off the theoretician hat and just be practical...how many spurious edges do you expect? If very few, I'd say go with the DFS approach, there won't be many intertwining cycles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You may be concerned that such a thing will result in a taxonomy that is not ... right. Say, a dog is an animal is a beagle (you wanted the algorithm to throw out the spurious 'an animal is a beagle' and not 'a beagle is a dog'). But if your algorithm is to be anonymous, it can't (it's not possible to) know which is the best edge to remove (in a cycle, no edge is distinguished).  That is, the bad example I gave is unavoidable in a general setting.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS: since we moved to dealing with cycles rather than transitivity (since any acyclic graph has an acyclic transitive closure), I realize now that an appropriate 'statistic' might by 'girth', the length of the largest cycle (with an acyclic graph having infinite girth). And ...I don't know of a fractional analog to girth. See Scheinerman, Fractional Graph Theory for ideas. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;PPS: I'm sure there are heuristics that might help choose bad edges (edges that you want to remove), for example, to make the hierarchy more ... pyramidal... try to minimize the number of 'top' nodes (nodes with no in-edges) or try to maximize the number of descendants of these top nodes. This might be expensive (for each edge in a cycle, compute descendants for everybody, then compare these results and pick the best one). &lt;/p&gt;&#xA;" OwnerUserId="1919" LastEditorUserId="1919" LastEditorDisplayName="" LastEditDate="2011-02-28T04:49:47.093" LastActivityDate="2011-02-28T04:49:47.093" CommentCount="2" />
  <row Id="23451" PostTypeId="1" AcceptedAnswerId="23452" CreationDate="2011-02-23T23:38:21.700" Score="5" ViewCount="91" Body="&lt;p&gt;I have the following inequality:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{2x-3}{x+1}\leq1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so, considering $x \neq -1$, I started multiplying $x+1$ both sides:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$2x-3\leq x+1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;then I subtracted $x$ both sides:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x-3\leq1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and then sum $3$ both sides:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x\leq4$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, my solution for $x\neq-1$ is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(-\infty,4]$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the book solution is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(-1,4]$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I did wrong?&lt;/p&gt;&#xA;" OwnerUserId="4501" LastEditorUserId="6075" LastEditorDisplayName="" LastEditDate="2011-02-24T02:43:54.887" LastActivityDate="2011-02-24T02:58:38.667" Title="How to find the solution for $\frac{2x-3}{x+1} \leq 1$?" Tags="&lt;algebra&gt;&lt;algebra-precalculus&gt;&lt;polynomials&gt;&lt;inequality&gt;" AnswerCount="4" CommentCount="3" />
  <row Id="23452" PostTypeId="2" ParentId="23451" CreationDate="2011-02-23T23:43:53.210" Score="4" ViewCount="" Body="&lt;p&gt;Note that if we have $\frac{a}{b} \leq 1$, this means that if $b&amp;gt;0$, then $a \leq b$ and if $b&amp;lt;0$, then $a \geq b$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you will need to split this into cases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First note that you can &lt;strong&gt;multiply without changing&lt;/strong&gt; the $\leq$ sign only when $x+1 &amp;gt; 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So when $x+1 &amp;gt; 0$, we have $2x-3 \leq x+1$ which gives us $x \leq 4$. Hence, when $x+1&amp;gt;0$, we have $x \leq 4$ and hence $x \in (-1,4]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $x+1 &amp;lt; 0$, the $\leq$ gets reversed to $\geq$ when you multiply by $x+1$ and we get $2x-3 \geq x+1$ when $x+1 &amp;lt; 0$. This gives us $x \geq 4$ when $x &amp;lt; -1$ which is not possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, the solution is $x \in (-1,4]$&lt;/p&gt;&#xA;" OwnerUserId="3302" LastActivityDate="2011-02-23T23:43:53.210" />
  <row Id="23453" PostTypeId="2" ParentId="23451" CreationDate="2011-02-23T23:44:42.803" Score="5" ViewCount="" Body="&lt;p&gt;The problem is when you multiplied both sides by $x+1$.  Remember that when you multiply an inequality by a negative number it changes signs.  This means we have to split into cases:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Case 1:  $x+1&amp;gt;0$.  Then we get $$2x-3\leq x+1$$&#xA;By the same reasoning that you present above we find $x\leq 4$.  Then rewrite the inequality $x+1&amp;gt;0$ as $x&amp;gt;-1$.  Combining this with $x\leq 4$ we see $x\in (-1,4]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Case 2: $x+1&amp;lt;0$.  Then we get $$2x-3\geq x+1$$ (notice that since $x+1&amp;lt;0$ the sign had to switch directions)  We then solve to find $x\geq 4$.  Since for this case we also had $x+1&amp;lt;0$, which is the same as $x&amp;lt;-1$ we conclude no such $x$ exists.  (A number cannot be less than -1 and greater than 4)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope that helps,&lt;/p&gt;&#xA;" OwnerUserId="6075" LastEditorUserId="6075" LastEditorDisplayName="" LastEditDate="2011-02-23T23:56:42.800" LastActivityDate="2011-02-23T23:56:42.800" CommentCount="1" />
  <row Id="23454" PostTypeId="1" AcceptedAnswerId="23457" CreationDate="2011-02-23T23:45:49.043" Score="2" ViewCount="167" Body="&lt;p&gt;In an exercise asking to mark true or false, it shows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{1}{a/x-b/x}=\frac{1}{a-b}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It really look like &lt;strong&gt;false&lt;/strong&gt; to me. But the answer is &lt;strong&gt;true&lt;/strong&gt;! How can it be?&lt;/p&gt;&#xA;" OwnerUserId="4501" LastEditorUserId="6075" LastEditorDisplayName="" LastEditDate="2011-02-24T02:43:29.410" LastActivityDate="2011-02-24T03:49:51.063" Title="How can $\frac{1}{a/x-b/x}$ be equal to $\frac{1}{a-b}$?" Tags="&lt;algebra&gt;&lt;algebra-precalculus&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="23455" PostTypeId="2" ParentId="23454" CreationDate="2011-02-23T23:49:01.227" Score="1" ViewCount="" Body="&lt;p&gt;Are you sure it is not $\frac{1}{a/x-b/x}=\frac{x}{a-b}$?  Otherwise I agree with you.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-02-23T23:49:01.227" />
  <row Id="23456" PostTypeId="2" ParentId="23454" CreationDate="2011-02-23T23:51:38.497" Score="1" ViewCount="" Body="&lt;p&gt;$\frac{1}{\frac{a}{x}-\frac{b}{x}} = \frac{1}{\frac{a-b}{x}} = \frac{x}{a-b} \not\equiv \frac{1}{a-b}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So this is &lt;em&gt;not&lt;/em&gt; true in the general case.&lt;/p&gt;&#xA;" OwnerUserId="37" LastActivityDate="2011-02-23T23:51:38.497" />
  <row Id="23457" PostTypeId="2" ParentId="23454" CreationDate="2011-02-23T23:51:43.457" Score="2" ViewCount="" Body="&lt;p&gt;Suppose $x\neq 0$, and $a\neq b$.  Multiplying top and bottom of the left hand side by $x$ shows $$\frac{1}{a/x-b/x}=\frac{x}{a-b}$$ and this equals $\frac{1}{a-b}$ if and only if $x=1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In short, it can't be true, but my guess is that the book meant to have an $x$ as the numerator of the right hand side of the original equation.&lt;/p&gt;&#xA;" OwnerUserId="6075" LastEditorUserId="1321" LastEditorDisplayName="" LastEditDate="2011-02-24T03:49:51.063" LastActivityDate="2011-02-24T03:49:51.063" CommentCount="3" />
  <row Id="23458" PostTypeId="2" ParentId="23451" CreationDate="2011-02-24T00:03:19.447" Score="3" ViewCount="" Body="&lt;p&gt;As others have mentioned, multiplying by $x+1$ forces you to consider cases at the outset. Instead, you can write $\frac{2x-3}{x+1} \leq 1$ as $\frac{2x-3}{x+1} -1 \leq 0$. Simplify this into $\frac{p(x)}{q(x)} \leq 0$ and consider when a fraction is negative.&lt;/p&gt;&#xA;" OwnerUserId="589" LastActivityDate="2011-02-24T00:03:19.447" />
  <row Id="23459" PostTypeId="2" ParentId="23451" CreationDate="2011-02-24T00:10:46.850" Score="4" ViewCount="" Body="&lt;p&gt;To preserve the $\:\le\:$ you must multiply by $\rm\ (x+1)^2\ $ not $\rm\ x+1\:,\:$ namely&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\quad\quad\quad\quad\quad\  \displaystyle\frac{2x-3}{x+1}\ \le\ 1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\quad\iff\quad \displaystyle\frac{x-4}{x+1}\ \ \le\ 0 $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\quad\iff\quad  (x+1)\ (x - 4)\ \le\ 0,\quad x\ne -1 $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\quad\iff\quad\ x\ \in\ (-1,4\:] $&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2011-02-24T01:10:31.703" LastActivityDate="2011-02-24T01:10:31.703" />
  <row Id="23460" PostTypeId="2" ParentId="23446" CreationDate="2011-02-24T00:54:44.330" Score="0" ViewCount="" Body="&lt;p&gt;$\sin(z)=\frac{e^{iz}-e^{-iz}}{2i}$&lt;/p&gt;&#xA;" OwnerUserId="6925" LastActivityDate="2011-02-24T00:54:44.330" CommentCount="1" />
  <row Id="23461" PostTypeId="1" AcceptedAnswerId="23463" CreationDate="2011-02-24T00:57:01.470" Score="4" ViewCount="121" Body="&lt;p&gt;Hello,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My attempt:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $a^{k} \equiv b^{k}( \text{mod}\ \ m ) \implies m|( a^{k} - b^{k} )$ and $a^{k+1} \equiv b^{k+1}( \text{mod}\ \ m ) \implies m|( a^{k+1} - b^{k+1} ) $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using binomial identity, we have:&#xA;$$a^{k} - b^{k} = ( a - b )( a^{k - 1} + a^{k - 2}b + a^{k - 3}b + ... ab^{k - 2} + b^{k - 1} )$$&#xA;$$a^{k + 1} - b^{k + 1} = ( a - b )( a^{k} + a^{k - 1}b + a^{k - 2}b + ... ab^{k - 1} + b^{k} )$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now there are two cases:&lt;br&gt;&#xA;1. If $m|(a - b)$, we're done.&lt;br&gt;&#xA;2. Else $m|( a^{k - 1} + a^{k - 2}b + a^{k - 3}b + ... ab^{k - 2} + b^{k - 1} )$ and $m|( a^{k} + a^{k - 1}b + a^{k - 2}b + ... ab^{k - 1} + b^{k} )$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And I was stuck from here, since I could not deduce anything from these two observations. I still have $(a, m) = 1$, and I guess this condition is used to prevent $m$ divides by the two right hand side above.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A hint would be greatly appreciated.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks,&lt;br&gt;&#xA;Chan&lt;/p&gt;&#xA;" OwnerUserId="5711" LastEditorUserId="6075" LastEditorDisplayName="" LastEditDate="2011-02-24T01:20:54.210" LastActivityDate="2011-02-24T02:46:30.523" Title="Prove that if $a^{k} \equiv b^{k}  \pmod m $ and $a^{k+1} \equiv b^{k+1}  \pmod m $ and $\gcd( a, m ) = 1$ then $a \equiv b \pmod m $" Tags="&lt;elementary-number-theory&gt;" AnswerCount="4" CommentCount="6" FavoriteCount="1" />
  <row Id="23462" PostTypeId="2" ParentId="23399" CreationDate="2011-02-24T01:06:59.473" Score="3" ViewCount="" Body="&lt;p&gt;This is a good question, although its statement may need to be cleaned up a little bit. It occasionally happens that looking for patterns between seemingly different types of problems and seemingly different branches of mathematics produces real results. The issue with this particular problem is not that there aren't any interpretations that this quantity can be given, but that there are too many. The solution is not to throw this problem away, but to refine it to the point that it admits a clear answer. This is a natural process in mathematical research.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suspect that your looking for a combinatorial interpretation to the formula&#xA;&#xA;$\frac{\left(2n-1\right)!!}{2^n}$. Since $\mbox{gcd}\left(2^n,\left(2n-1\right)!!\right) = 1$ for all $n \geq 1$, this formula cannot be interpreted as enumerating the points in some specified finite set. Since $2^n &amp;lt; \left(2n-1\right)!!$ for all $n\geq 3$, this formula cannot be interpreted as a probability of some kind. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The formula $\frac{\left(2n-1\right)!!}{n!2^n}$ can be interpreted as giving the probability that if two people each flip two separate fair coins $n$ times, then each person gets heads the same number of times. I worked this out by unraveling $\frac{A}{B}$ using binomial identities until I got something that looked like the probability of some easily described random event. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit:&#xA;In the light of this &lt;a href=&quot;http://math.stackexchange.com/questions/23808/coin-toss-question&quot;&gt;question&lt;/a&gt; and its answers, it seems that there is a connection between this simple game I've described and the numbers you are interested in. It's strange how these things happen.&lt;/p&gt;&#xA;" OwnerUserId="7323" LastEditorUserId="7323" LastEditorDisplayName="" LastEditDate="2011-02-26T22:24:59.260" LastActivityDate="2011-02-26T22:24:59.260" CommentCount="2" />
  <row Id="23463" PostTypeId="2" ParentId="23461" CreationDate="2011-02-24T01:10:06.460" Score="4" ViewCount="" Body="&lt;p&gt;Here is the idea:  Show that (2) is impossible.  Suppose that $$m|\left( a^{k-1}+a^{k-2}\cdot b+\cdots + b^{k-1}\right)$$ and &#xA;$$m|\left(a^{k}+a^{k-1}\cdot b+\cdots + b^{k}\right)$$&#xA;Then subtract $b$ times the first integer from the second integer.  That is consider $$\left( a^{k}+a^{k-1}\cdot b+\cdots + b^{k} \right) -b\cdot \left( a^{k-1}+a^{k-2}\cdot b+\cdots + b^{k-1} \right)= a^k$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It then follows that $m$ divides $a^k$, but that is impossible.  (why?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope that helps.&lt;/p&gt;&#xA;" OwnerUserId="6075" LastActivityDate="2011-02-24T01:10:06.460" CommentCount="2" />
  <row Id="23464" PostTypeId="2" ParentId="23461" CreationDate="2011-02-24T01:11:30.300" Score="2" ViewCount="" Body="&lt;p&gt;We have $$m|(a^{k}+a^{k-1}b+a^{k-2}b^2 + \cdots + ab^{k-1}+b^k)$$ $$-t(a^{k-1}+a^{k-2}b+a^{k-3}b^2+ \cdots + ab^{k-2} + b^{k-1}) = a^k$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is impossible.&lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2011-02-24T01:11:30.300" CommentCount="1" />
  <row Id="23465" PostTypeId="2" ParentId="23461" CreationDate="2011-02-24T01:18:13.007" Score="5" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\rm\ \ b^k \equiv a^k\ \Rightarrow\ b^{k+1} \equiv\: b\ a^k \equiv\: a^{k+1}\ \Rightarrow\ b\equiv a\ $ by cancelling $\rm\ a^k\:.\: $ Recall that if $\rm\ (a,m) = 1\ $ then $\rm\:a\:$ is invertible, so cancellable $\rm\ (mod\ m)$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same proof works in any ring where $\rm\:a\:$ is a unit, i.e. invertible. The proof cancels the units $\rm\ a^k = b^k\ $ from both sides of $\rm\ a^{k+1} = b^{k+1}\ $ to obtain $\rm\ a = b\:.\:$ Here we've used that unit $\rm\:a\ \Rightarrow\: $ unit $\rm\: b\ $ since $\rm\:b|b^k|a^k\:$, and units are closed under products/divisors, i.e. $\rm\ x\ y\:$ unit $\rm\iff\ x\:$ unit and $\rm\:y\:$ unit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's clearer dehomogenized using $\rm\ c = b/a\:,\: $ viz. $\rm\ 1 = c^k\ \Rightarrow\ c = c^{k+1} = 1\:$.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2011-02-24T02:46:30.523" LastActivityDate="2011-02-24T02:46:30.523" CommentCount="3" />
  <row Id="23466" PostTypeId="1" AcceptedAnswerId="23468" CreationDate="2011-02-24T01:35:38.490" Score="5" ViewCount="78" Body="&lt;p&gt;I am doing calculus integration, and need to show my work for Horizontal slicing (even though Vertical slicing is far easier). The equation is y= x/$\sqrt{2x^2+1}$&lt;br&gt;&#xA;I need to rewrite the equation so that it is x=... in order to horizontally slice it (in other words, it should be rewritten so that it is dependent on y). &#xA;This isn't exactly a calculus question, although it is being used for calculus. I'm probably missing something that is pretty obvious. &#xA;Any help would be greatly appreciated!&lt;/p&gt;&#xA;" OwnerUserId="7432" LastActivityDate="2011-02-24T04:33:31.857" Title="Rewrite equation to solve for x, not y" Tags="&lt;calculus&gt;&lt;algebra-precalculus&gt;" AnswerCount="2" />
  <row Id="23467" PostTypeId="2" ParentId="23466" CreationDate="2011-02-24T01:42:33.980" Score="1" ViewCount="" Body="&lt;p&gt;If you are working on an interval where $x\ge 0$, $$\begin{align}&#xD;&#xA;\frac{x}{\sqrt{2x^2+1}}&amp;amp;=\sqrt{\frac{x^2}{2x^2+1}}&#xD;&#xA;\\&#xD;&#xA;&amp;amp;=\sqrt{\frac{2x^2+1-1}{2(2x^2+1)}}&#xD;&#xA;\\&#xD;&#xA;&amp;amp;=\sqrt{\frac{1}{2}\left(1-\frac{1}{2x^2+1}\right)}&#xD;&#xA;\end{align}$$&#xA;Now that there's only one instance of $x$ on the right side, you should be able to more easily solve for $x$.&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2011-02-24T01:42:33.980" />
  <row Id="23468" PostTypeId="2" ParentId="23466" CreationDate="2011-02-24T01:50:39.360" Score="6" ViewCount="" Body="&lt;p&gt;Heres a tip:  Take reciprocals.&#xA;Notice $$\left( \frac{1}{y}\right)^2=\left( \frac{\sqrt{2x^2+1}}{x}\right)^2=\frac{2x^2+1}{x^2}=2+\frac{1}{x^2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then we get $$\frac{1}{x^2}=\frac{1}{y^2}-2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take reciprocals again and we find&#xA;$$x^2=\frac{1}{\frac{1}{y^2}-2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take square roots, and we are finished.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope that helps&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit:  Just to make things complete I decided to add the final line: $$x=\pm \sqrt{\frac{y^2}{1-2y^2}}$$&lt;/p&gt;&#xA;" OwnerUserId="6075" LastEditorUserId="6075" LastEditorDisplayName="" LastEditDate="2011-02-24T04:33:31.857" LastActivityDate="2011-02-24T04:33:31.857" CommentCount="3" />
  <row Id="23469" PostTypeId="2" ParentId="23461" CreationDate="2011-02-24T01:58:33.240" Score="2" ViewCount="" Body="&lt;p&gt;This is related to what Bill Dubuque said but is more elementary:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $a^k = b^k$ mod $m$, one has $m | (a^k - b^k)$, and similarly $m | (a^{k+1} - b^{k+1})$. Thus by the first equation $m$ divides $b(a^k - b^k) = (a^kb - b^{k+1})$. Subtracting, we have that $m | (a^{k+1} - b^{k+1}) - (a^kb - b^{k+1}) = a^{k+1} - a^kb = a^k(a - b)$. But since $(a,m) = 1$, the fact that $m | a^k(a - b)$ means $m | a - b$, or equivalently $a = b$ mod $m$ which is what you needed to show. &lt;/p&gt;&#xA;" OwnerUserId="3035" LastActivityDate="2011-02-24T01:58:33.240" CommentCount="1" />
  <row Id="23470" PostTypeId="1" AcceptedAnswerId="23529" CreationDate="2011-02-24T01:59:48.197" Score="3" ViewCount="91" Body="&lt;p&gt;Hello,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am interested in a good comprehensive resource on realification and complexification of vector spaces over the reals or complexes (and the interplay of these structures on the 'same' space in general).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, understanding of the basic theory is necessary and useful for a more intuitive approach towards functional analysis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you give me a tip? For example, Serge Lang's classical book does not explicitly work this part out. I am aware of a few pages in Arnold's book on ODE, but there should be something more comprehensive and neat somewhere out there.&lt;/p&gt;&#xA;" OwnerUserId="3557" LastActivityDate="2011-02-24T14:30:30.240" Title="Realification and Complexification of vector spaces" Tags="&lt;linear-algebra&gt;&lt;reference-request&gt;&lt;vector-space&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="23471" PostTypeId="1" CreationDate="2011-02-24T02:03:02.373" Score="1" ViewCount="348" Body="&lt;p&gt;What is the difference between a field and a group in algebra?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could someone briefly explain it?&lt;/p&gt;&#xA;" OwnerUserId="7433" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-24T03:04:36.867" LastActivityDate="2011-04-25T17:08:39.713" Title="what is the difference between field and group in algebra" Tags="&lt;abstract-algebra&gt;" AnswerCount="4" CommentCount="1" FavoriteCount="3" />
  <row Id="23472" PostTypeId="2" ParentId="23471" CreationDate="2011-02-24T02:05:41.307" Score="0" ViewCount="" Body="&lt;p&gt;As Arturo points out, &quot;a group has three operations with signature (2,1,0) (binary, unary, zeroary), while a field has two binary, one unary, two zeroary operations and one unary partial operation&quot; (5 operations and 1 partial operation). A group has to satisfy (i) closure (ii) associativity, (iii) identity and (iv) invertibility. A field has to satisfy these as well and also have distributivity of multiplication over addition and commutativity. We can also talk about &lt;a href=&quot;http://en.wikipedia.org/wiki/Field_extension&quot; rel=&quot;nofollow&quot;&gt;field extensions&lt;/a&gt; and &lt;a href=&quot;http://terrytao.wordpress.com/2010/01/23/some-notes-on-group-extensions/&quot; rel=&quot;nofollow&quot;&gt;group extensions&lt;/a&gt;. As Tao notes, one usually talks about field extensions as embedding a field $F$ into a larger field $K$ that has $F$ as a subspace. Group extensions, on the other hand, are defined by covering. That is, we cover a group $G$ by another group $G&amp;#39;$ that has $G$ has a quotient. &lt;/p&gt;&#xA;" OwnerUserId="2548" LastEditorUserId="2548" LastEditorDisplayName="" LastEditDate="2011-02-24T03:07:54.320" LastActivityDate="2011-02-24T03:07:54.320" CommentCount="4" />
  <row Id="23473" PostTypeId="2" ParentId="23471" CreationDate="2011-02-24T02:07:15.623" Score="4" ViewCount="" Body="&lt;p&gt;I feel this question is too broad for MSE, and could easily be answered by using a common search engine.  Briefly: A field has two operations and is always commutative.  See &lt;a href=&quot;http://en.wikipedia.org/wiki/Group_%28mathematics%29#Definition&quot; rel=&quot;nofollow&quot;&gt;Definiton of Group&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/Field_%28mathematics%29#Definition_and_illustration&quot; rel=&quot;nofollow&quot;&gt;Definition of Field&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="6075" LastActivityDate="2011-02-24T02:07:15.623" CommentCount="1" />
  <row Id="23475" PostTypeId="1" AcceptedAnswerId="23483" CreationDate="2011-02-24T02:57:35.437" Score="4" ViewCount="141" Body="&lt;p&gt;I'm trying to prove divergence of the integral $\int_1^{\infty}\frac{\left|\cos{x^2}\right|}{x^q}dx$, where $0&amp;lt;q&amp;lt;1$, by applying the mean-value theorem to $\int_{n\pi}^{(n+1)\pi}\frac{\left|\cos{x^2}\right|}{x^q}dx$: there exists $c_n\in [n\pi,(n+1)\pi]$ such that $\int_{n\pi}^{(n+1)\pi}\frac{\left|\cos{x^2}\right|}{x^q}dx=\left|\cos{c_n^2}\right|\int_{n\pi}^{(n+1)\pi}\frac{1}{x^q}dx$. So if I can show that $\inf{\left|\cos{c_n^2}\right|}\neq 0$, we would have the divergence, by comparison with $\int_1^{\infty}\frac{1}{x^q}dx$. It isn't apparent to me how to prove that bit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that the divergence can also be proved by noting that $\left|\cos{x^2}\right|\geq\cos^2{x^2}=\frac{1+\cos{2x^2}}{2}$ and using a test akin to Dirichlet's test for series to prove the convergence of $\int_1^{\infty}\frac{\cos{2x^2}}{x^q}dx$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I'd like to know if my idea can be finished.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you.&lt;/p&gt;&#xA;" OwnerUserId="1313" LastEditorUserId="1313" LastEditorDisplayName="" LastEditDate="2011-02-24T03:34:38.857" LastActivityDate="2011-02-24T08:35:13.427" Title="Divergence of an improper integral - |cos(x^2)|/x^q." Tags="&lt;calculus&gt;&lt;improper-integrals&gt;" AnswerCount="3" CommentCount="0" />
  <row Id="23477" PostTypeId="2" ParentId="23471" CreationDate="2011-02-24T03:15:19.363" Score="13" ViewCount="" Body="&lt;p&gt;This is more of a rejoinder to PEV's response as it existed before the edit. One can think of &quot;subtraction&quot; as an operation separate from addition, though rather than consider the binary operation, one usually considers the &lt;em&gt;unary&lt;/em&gt; operation that maps an element to its additive inverse (it has nicer properties than subtraction, which is not associative, and it automatically captures the idea that the additive inverse is unique).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, from the point of view of Universal Algebra, we want a group to have &lt;strong&gt;three&lt;/strong&gt; operations: a binary operation $\cdot$ (that corresponds to the group product), a unary operation ${}^{-1}$ (that maps every element to its inverse), and a zeroary operation $e$ (distinguished element). Then the &quot;axioms&quot; of a group can be stated as identities of the operations:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$(a\cdot b)\cdot c = a\cdot(b\cdot c)$;&lt;/li&gt;&#xA;&lt;li&gt;$e\cdot a = a\cdot e = a$;&lt;/li&gt;&#xA;&lt;li&gt;$a\cdot a^{-1} = a^{-1}\cdot a = e$.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Being expressible as identities has all sorts of nice consequences. Consequently, universal algebraists will consider a group as a set together with three operations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fields are more problematical, though. You start the same way: with two binary operations $+$ and $\times$; a unary operation $-$; two zeroary operations $0$ and $1$, and the axioms that describe these operations. Multiplicative inverses, however, are not defined via an operation, because $0$ does not have a multiplicative inverse. Instead, they must be defined via a &lt;em&gt;partial&lt;/em&gt; operations, so fields are not &quot;algebras&quot; within the sense of universal algebra, but rather they are &lt;em&gt;partial algebras&lt;/em&gt;. We know that fields cannot be defined as algebras with operations, because if they could then the category of fields (or even of fields of a given characteristic) with field morphisms would necessarily have categorical products, and they do not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The theory of partial algebras is far more difficult than the theory of algebras. I seem to recall Brian Davey and David Clark, in their book &lt;strong&gt;Natural Dualities for the Working Mathematician&lt;/strong&gt; mentioning that one can prove a lot of very nice and powerful &quot;theorems&quot; about partial algebras if one forgets some of the problems (e.g., if $f\colon A\to B$ is a homomorphism of partial algebras, then $f(A)$ need not be a partial-subalgebra of $B$; though this is particularly problematical when one consider relational algebras and relational partial algebras, where we have relations in addition to operations).&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-24T03:21:59.137" LastActivityDate="2011-02-24T03:21:59.137" CommentCount="15" />
  <row Id="23478" PostTypeId="2" ParentId="23406" CreationDate="2011-02-24T03:43:01.363" Score="1" ViewCount="" Body="&lt;p&gt;As Plop states, the hint follows from Chevalley's theorem.  However, in this context one shouldn't need to appeal to the full strength of that theorem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, Chevalley's theorem is a variation on Noether normalization (and both are variations on the Nullstellensatz --- see &lt;a href=&quot;http://mathoverflow.net/questions/15611/to-prove-the-nullstellensatz-how-can-the-general-case-of-an-arbitrary-algebraica/15719#15719&quot; rel=&quot;nofollow&quot;&gt;this MO answer&lt;/a&gt;), but Noether normalization is usually taught at an ealier stage than Chevalley's theorem, so &#xA;you might consider using it instead.  (Regard this as an alternative hint.)&lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2011-02-24T03:43:01.363" CommentCount="2" />
  <row Id="23480" PostTypeId="2" ParentId="23475" CreationDate="2011-02-24T03:51:33.857" Score="1" ViewCount="" Body="&lt;p&gt;Careful.  Showing $\cos{c_n^2}\neq 0 $ does &lt;em&gt;not&lt;/em&gt; show divergence.  (Consider $\cos{c_n^2}=n^{-(n!)}$)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For your idea to work you need to show that there exists $c_n$ such that $c_n\leq \int_{n\pi}^{(n+1)\pi}|\cos x^2|dx$ and $c_n$ does not go to zero too &lt;em&gt;quickly&lt;/em&gt;.  That is we need to bound the integral from below.  (Using inequalities rather than the equality given by the mean value theorem will suffice.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then &#xA;$$\int_{n\pi}^{(n+1)\pi}\frac{|\cos x^2|}{x^q}dx\geq c_n\cdot \int_{n\pi}^{(n+1)\pi}\frac{1}{((n+1)\pi)^q}dx=\frac{c_n\pi}{((n+1)\pi)^q}$$ &#xA;Where the inequality follows since $\frac{1}{((n+1)\pi)^q}\leq x^q $ on this interval, and the average of $\cos x^2$ is greater than $c_n$. &lt;/p&gt;&#xA;" OwnerUserId="6075" LastActivityDate="2011-02-24T03:51:33.857" />
  <row Id="23481" PostTypeId="1" CreationDate="2011-02-24T03:52:52.310" Score="2" ViewCount="52" Body="&lt;p&gt;Say I have 2 LPs, LP_1 and LP_2 which have real and integer variables and a staircase structure (i.e. the solution and feasible region of LP_2 depends on the solution of LP_1).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$LP_1$ has the form&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\min c_1 (x_1 + z_1)$&lt;br&gt;&#xA;$ s.t.$&lt;br&gt;&#xA;$ A_1 (x_1 + z_1) = b_1$&lt;br&gt;&#xA;$ x_1, z_1 \geq 0 $&lt;br&gt;&#xA;$ x_1, \in \mathbb{R^{n_1^{real}}}, z_1 \in \mathbb{Z^{n_1^{int}}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and the second LP $LP_2$ has the form&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\min c_2 (x_2 z_2)$&lt;br&gt;&#xA;$s.t.$&lt;br&gt;&#xA;$ A_2 (x_2 + z_2) + E_2 (x_1 + z_1) = b_2$&lt;br&gt;&#xA;$ x_2, z_2 \geq 0 $&lt;br&gt;&#xA;$ x_2, \in \mathbb{R^{n_2^{real}}}, z_2 \in \mathbb{Z^{n_2^{int}}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given a trial solution $x_1, z_1$, how can I solve $LP_2$ and then generate a cutting plane using Bender's Decomposition to add to $LP_1$? Any advice is appreciated - I believe the answer involves &quot;lifting&quot; though I'm not sure on what it is and why it works. &lt;/p&gt;&#xA;" OwnerUserId="7145" LastActivityDate="2011-02-24T03:52:52.310" Title="Bender's Decomposition for Mixed Integer Programs" Tags="&lt;optimization&gt;&lt;linear-programming&gt;" />
  <row Id="23482" PostTypeId="2" ParentId="23342" CreationDate="2011-02-24T03:56:57.707" Score="0" ViewCount="" Body="&lt;p&gt;EDIT: Sorry for this, turns out that there was a typo and the ⊂ should have actually been a ⊆. After all, f could be a convex function in which case, it would be the same as it's convex hull. &lt;/p&gt;&#xA;" OwnerUserId="7145" LastActivityDate="2011-02-24T03:56:57.707" />
  <row Id="23483" PostTypeId="2" ParentId="23475" CreationDate="2011-02-24T03:57:02.830" Score="3" ViewCount="" Body="&lt;p&gt;You can avoid the issues with the mean value theorem by proceeding like this: Change variables from $x$ to $x^2$ as one of the answerers suggested, so that one is attempting to show divergence of ${\displaystyle \int_1^{\infty} {|\cos(x)| \over x^r}\,dx}$, where now ${1 \over 2} &amp;lt; r &amp;lt; 1$. Next, note that one has &#xA;$$\int_{2\pi}^{\infty} {|\cos(x)| \over x^r}\,dx = \sum_{n=1}^{\infty} \int_{2n\pi}^{2(n+1)\pi}{|\cos(x)| \over x^r}\,dx$$&#xA;Since the denominator is monotone increasing, this is bounded below by&#xA;$$\sum_{n=1}^{\infty} \int_{2n\pi}^{2(n+1)\pi}{|\cos(x)| \over (2\pi(n+1))^r}\,dx$$&#xA;$$= {1 \over (2\pi)^r}\sum_{n=1}^{\infty} {1 \over (n+1)^r}\int_{2n\pi}^{2(n+1)\pi}|\cos(x)|\,dx$$&#xA;Since $\cos(x)$ has period $2\pi$, the integral above has some fixed value $C &amp;gt; 0$, so the above is equal to&#xA;$$C {1 \over (2\pi)^r}\sum_{n=1}^{\infty} {1 \over (n+1)^r}$$&#xA;Since $r &amp;lt; 1$ this diverges.&lt;/p&gt;&#xA;" OwnerUserId="3035" LastEditorUserId="3035" LastEditorDisplayName="" LastEditDate="2011-02-24T04:03:18.060" LastActivityDate="2011-02-24T04:03:18.060" />
  <row Id="23484" PostTypeId="1" AcceptedAnswerId="23487" CreationDate="2011-02-24T04:08:30.297" Score="6" ViewCount="286" Body="&lt;p&gt;$\int {\frac{8x^4+15x^3+16x^2+22x+4}{x(x+1)^2(x^2+2)}dx}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I used partial fractions, solved A = 2, C = 3.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;$$\frac{A}{x} + \frac{B}{x+1} + \frac{C}{(x+1)^2} +\frac{(Dx+E)}{(x^2+2)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align*}&#xA;&amp;amp;8x^4+15x^3+16x^2+22x+4\\&#xA;&amp;amp;\quad = A(x+1)^2(x^2+2)+B(x)(x+1)(x^2+2)+C(x)(x^2+2)+(Dx+E)(x)(x+1)^2&#xA;\end{align*}&#xA;Substitute in $x=0$ to get $4=A(1)(2)$, so  $A = 2$&#xA;$$6x^4+11x^3+10x^2+14x = B(x)(x+1)(x^2+2)+C(x)(x^2+2)+(Dx+E)(x)(x+1)^2$$&#xA;Substitute in $x=-1$ to get&#xA;$$6-11+10-14 = C(-1)(1+2)$$&#xA;so $-9=-3C$, thus $C=3$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Leaving me what I have below:&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Which brings me to where I am currently stuck.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$6x^4 +8x^3 +10x^2+8x = B(x)(x+1)(x^2+2) + (Dx + E) (x) (x+1)^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is the next best move to use substitution to solve for $B$?&lt;/p&gt;&#xA;" OwnerUserId="6423" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-24T05:04:34.923" LastActivityDate="2011-02-24T17:09:15.027" Title="Help solving $\int {\frac{8x^4+15x^3+16x^2+22x+4}{x(x+1)^2(x^2+2)}dx}$" Tags="&lt;calculus&gt;&lt;homework&gt;&lt;integral&gt;" AnswerCount="4" CommentCount="7" />
  <row Id="23485" PostTypeId="2" ParentId="23484" CreationDate="2011-02-24T04:14:53.907" Score="2" ViewCount="" Body="&lt;p&gt;Given what you have, what happens when you set $x = -1$?&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2011-02-24T04:14:53.907" CommunityOwnedDate="2011-02-24T04:14:53.907" CommentCount="7" />
  <row Id="23486" PostTypeId="2" ParentId="23270" CreationDate="2011-02-24T04:19:24.157" Score="1" ViewCount="" Body="&lt;p&gt;For (1):  They mean pick any element of the conjugacy class, and look at the corresponding $\phi$.  It doesn't matter which one you look at because being primitive is a conjugacy invariant.&lt;/p&gt;&#xA;" OwnerUserId="6071" LastActivityDate="2011-02-24T04:19:24.157" CommentCount="1" />
  <row Id="23487" PostTypeId="2" ParentId="23484" CreationDate="2011-02-24T04:42:10.797" Score="6" ViewCount="" Body="&lt;p&gt;First, I trust you used the correct partial fraction decomposition:&#xA;$$\frac{8x^4+15x^3+16x^2+22x+4}{x(x+1)^2(x^2+2)} = \frac{A}{x} + \frac{B}{x+1} + \frac{C}{(x+1)^2} + \frac{Dx+E}{x^2+2}.$$&#xA;This leads to&#xA;\begin{align*}&#xA;&amp;amp;8x^4 + 15x^3 + 16x^2 + 22x + 4\\&#xA;&amp;amp;\qquad = A(x+1)^2(x^2+2) + Bx(x+1)(x^2+2) + Cx(x^2+2) + (Dx+E)x(x+1)^2.&#xA;\end{align*}&#xA;A useful &quot;trick&quot; is to evaluate at the zeros of the linear factors to get some information; I suspect you evaluated at $x=0$ to get $2A = 4$, from which you got $A=2$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can then evaluate at $x=-1$ to get $-3C = -9$, which is how you got $C=3$. Looking good. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then you used that to simplify.&#xA;$$2(x^2+2x+1)(x^2+2) +3x(x^2+2) = 2x^4 + 7x^3 + 6x^2 + 14x + 4,$$&#xA;which subtracted from $8x^4 + 15x^3 + 16x^2 + 22x + 4$ gave you&#xA;$$6x^4 + 8x^3 + 10x^2 + 8x = Bx(x+1)(x^2+2) + (Dx+E)x(x+1)^2.$$&#xA;Hmmm... Which is not quite what you have. Did you use the correct decomposition, or did you forget about being careful with that $(x+1)^2$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway: here's where you go stuck because you are used to being able to solve the partial fractions problems using only the evaluation trick. But when you have irreducible quadratic factors or powers of linear factors (or worse, &lt;em&gt;both&lt;/em&gt;), the trick doesn't get you all the way there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here, we can factor out $x$ from both sides to get&#xA;$$6x^3 + 8x^2 + 10x + 8 = B(x+1)(x^2+2) + (Dx+E)(x+1)^2.$$&#xA;(We factored out $x$ from both sides and cancelled; that's how we dropped from fourth power to cube).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Edit.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can further factor out $x+1$ from both sides:&#xA;$$(x+1)(6x^2 + 2x + 8) = (x+1)(B(x^2+2) + (Dx+E)(x+1))$$&#xA;to get&#xA;$$6x^2 + 2x + 8 = B(x^2+2) + (Dx+E)(x+1).$$&#xA;Contrary to your claim before, now that we had all the right terms, we &lt;em&gt;cannot&lt;/em&gt; simply conclude that $D=6$, because there are &lt;em&gt;two&lt;/em&gt; quadratic terms: $Bx^2$ and $Dx^2$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can, however, evaluate at $x=-1$ to get $12 = 3B$, or $B=4$; from this you go to&#xA;$$6x^2 + 2x + 8 = 4x^2 + 8 + (Dx+E)(x+1)$$&#xA;or &#xA;$$2x^2 + 2x = (Dx+E)(x+1).$$&#xA;Noting that the constant term on the right is $E$, and $0$ on the left, you get $E=0$. This gives&#xA;$$2x(x+1) = Dx(x+1)$$&#xA;which, cancelling $x(x+1)$ yields $D=2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively, from $2x^2+2x = (Dx+E)(x+1)$, we can factor the left hand side completely to get&#xA;$$2x(x+1) = (Dx+E)(x+1)$$&#xA;from which we immediately get $Dx+E = 2x$, so $D=2$ and $E=0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, in summary, $A=2$, $B=4$, $C=3$, $D=2$, $E=0$. &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-24T05:38:40.390" LastActivityDate="2011-02-24T05:38:40.390" CommentCount="8" />
  <row Id="23488" PostTypeId="1" AcceptedAnswerId="23489" CreationDate="2011-02-24T04:50:38.483" Score="2" ViewCount="246" Body="&lt;p&gt;Suppose you're given a circle with center $O$, I'm curious, how can one construct with ruler and compass three circles inside the larger circle such that each is tangent to the larger circle as well as to the other two?&lt;/p&gt;&#xA;" OwnerUserId="7440" LastEditorUserId="7440" LastEditorDisplayName="" LastEditDate="2011-02-24T05:06:11.547" LastActivityDate="2011-02-24T16:24:52.970" Title="Three tangent circles inside a larger circle." Tags="&lt;geometry&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="23489" PostTypeId="2" ParentId="23488" CreationDate="2011-02-24T04:59:59.153" Score="0" ViewCount="" Body="&lt;p&gt;If the radius of the circle is 1, the radius of three circles that will be internally tangent is $2\sqrt{3}-3\approx 0.464$ by Soddy's formula, given in &lt;a href=&quot;http://en.wikipedia.org/wiki/Problem_of_Apollonius&quot; rel=&quot;nofollow&quot;&gt;the Soddy's circles section of this&lt;/a&gt;.  You can construct this length, then mark it off on a diameter of the circle to find one of the centers, and finish constructing the equilateral triangle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Added:  For the construction, make a right angle with 1 on one side, swing 2 as a hypotenuse, and you have $\sqrt{3}$.  Then you can mark a line to get $2\sqrt{3}-3$.  Draw a diameter of the circle you are given and mark off $2\sqrt{3}-3$ from the circumfrence to find the center of one circle.  Construct an equilateral triangle of $2(2\sqrt{3}-3)$ bisected by the diameter and you have the three centers.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2011-02-24T16:24:52.970" LastActivityDate="2011-02-24T16:24:52.970" CommentCount="4" />
  <row Id="23491" PostTypeId="1" CreationDate="2011-02-24T05:37:16.803" Score="2" ViewCount="51" Body="&lt;p&gt;Figured you math dudes (and dudettes) could knock this one out of the park!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've got a website that stores users' game scenario results. Each scenario record can have one of three possible results: Side 1 wins, Draw, or Side 2 wins.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To this end, I'm currently representing balance by showing the simple percentage of total victories for each result. Example:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Result Totals&#xA;  Side 1 - 5&#xA;  Draw - 1&#xA;  Side 2 - 4&#xA;&#xA;Display&#xA;  Side 1    Draw    Side 2&#xA;   50%       10%     40%&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;but the astute reader will immediately note that using this methods means a scenario with 5 plays split 2-1-2 and a scenario with 100 plays split 40-20-40 will have the same apparent &quot;balance&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I would like to do is take into account the total plays for a scenario. Thereby, a scenario with a split of 41-20-39 will rank much higher than one splitting 2-1-2, even though the simple percentage is actually less balanced.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does that make sense? Ideally, this would map to a scale of some sort, so scenarios could be given discrete balance ratings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for any help you can provide.&lt;/p&gt;&#xA;" OwnerUserId="7444" LastActivityDate="2011-02-24T05:51:36.067" Title="Calculating the &quot;most balanced&quot; scenarios of a game?" Tags="&lt;algorithms&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" />
  <row Id="23493" PostTypeId="2" ParentId="23491" CreationDate="2011-02-24T05:51:36.067" Score="4" ViewCount="" Body="&lt;p&gt;In fact, your measurement of balance with 2-1-2 and 40-20-40 are the same-each player has the same chance of winning.  What is different is the assurance that the measurement is reasonably accurate.  Under the usual (worrisome) statistical assumptions you can calculate a standard deviation on the number of player 1 wins as $\sqrt{Np(1-p)}$ where $N$ is the number of games and $p$ is the fraction won by player 1.  So for 2-1-2 the standard deviation on the number of player 1 wins is $\sqrt{5\cdot0.4\cdot0.6}\approx 1$  So you would say player 1 wins $(40 \pm 20)\%$.  In the case of 40-20-40 you would have a standard deviation of about 5, so you would say player 1 wins $(40 \pm 5)\%$.  It's not that the second measurement shows more balance, but you have better reason to think it is close.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-02-24T05:51:36.067" CommentCount="1" />
  <row Id="23494" PostTypeId="2" ParentId="23406" CreationDate="2011-02-24T06:23:16.610" Score="0" ViewCount="" Body="&lt;p&gt;Matt E, all I know about Noether normalization is the existence of an algebraically independent elements of a finitely generated commutative algebra.&lt;/p&gt;&#xA;" OwnerUserId="7445" LastActivityDate="2011-02-24T06:23:16.610" CommentCount="7" />
  <row Id="23495" PostTypeId="1" AcceptedAnswerId="23497" CreationDate="2011-02-24T06:27:19.960" Score="3" ViewCount="85" Body="&lt;p&gt;Today I came across two possible canonical maps in the context of representation theory from reading stackexchange and a friend's email. However, I don't really know what they are (and have thus far not found satisfactory information about them):&#xA;1) $hom_k(V \otimes V,k ) \rightarrow hom_k(V, V^*)$&#xA;2) $(V^*)^* \otimes V \rightarrow hom_k(V, V^*)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In both cases $V$ is a finite dimensional representation of some (finite) group $G$. In 2) $k$ maybe assumed algebraically closed. Perhaps 1) is a result of exterior derivatives (my guess). Any pointer is welcomed.&lt;/p&gt;&#xA;" OwnerUserId="6369" LastActivityDate="2011-02-24T07:23:54.240" Title="a couple of canonical maps in representation theory" Tags="&lt;linear-algebra&gt;&lt;representation-theory&gt;" AnswerCount="1" />
  <row Id="23496" PostTypeId="1" AcceptedAnswerId="23499" CreationDate="2011-02-24T06:50:28.270" Score="2" ViewCount="128" Body="&lt;p&gt;I'm having trouble figuring out how to evaluate the integral $\int_{|z|=\rho} \frac{|dz|}{|z-a|^2}$ where $|a| \neq \rho$. This is a problem in Ahlfors in the section on Cauchy's Integral Formula, and I think by convention when he says $|z|=\rho$, he means the parametrization $z=\rho e^{it}, \; 0 \leq t \leq \pi$ (so that the winding number of a point inside this circle would be 1). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm guessing there is some smart way to apply the integral formula (since it's in this section), and I naively tried to expand the integrand. However, you end up with $\frac{1}{(z-a)(\bar{z}-\bar{a})}$, and I don't believe $\bar{z}-\bar{a}$ is an analytic function, so I'm not sure how to proceed from here.&lt;/p&gt;&#xA;" OwnerUserId="1736" LastActivityDate="2011-02-25T03:44:09.983" Title="Evaluating a complex integral" Tags="&lt;complex-analysis&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="23497" PostTypeId="2" ParentId="23495" CreationDate="2011-02-24T06:56:14.910" Score="4" ViewCount="" Body="&lt;p&gt;You know there is a natural isomorphism $\hom_k(U\otimes V,W)\to\hom_k(U, \hom_k(V,W))$. For &lt;em&gt;(1)&lt;/em&gt;, take $U=V$ and $W=k$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the second one, there is no such natural isomorphism: on one hand, $(V^*)^*$ is naturally isomorphic to $V$, so $(V^*)^*\otimes V$ is naturally isomorphic to $V\otimes V$. On the other hand, $\hom(V,V^*)$ is naturally isomorphic to $\hom(V\otimes V,k)$, or, in other words, to $(V\otimes V)^*$. Now there is no natural isomorphism from $V\otimes V$ to $(V\otimes V)^*$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;NB:&lt;/strong&gt; In fact, $(V\otimes V)^*$ and $V\otimes V$ are in general not even isomorphic, much less naturally isomorphic (naturality is problematic, as the two depend with different variances on $V$ anyways...) One way to see this is to let $G$ be a cyclic group generated by an element $g$ of order $100$, and letting $V$ be the one-dimensional representation of $G$ where $g$ acts by a primitive $100$th root of unity $\omega$. Then $V\otimes V$ is also one-dimensional, with $g$ acting by $\omega^2$, and you can surely describe $(V\otimes V)^*$ and see that it is not isomorphic to $V\otimes V$.&lt;/p&gt;&#xA;" OwnerUserId="274" LastEditorUserId="274" LastEditorDisplayName="" LastEditDate="2011-02-24T07:23:54.240" LastActivityDate="2011-02-24T07:23:54.240" CommentCount="0" />
  <row Id="23498" PostTypeId="2" ParentId="23484" CreationDate="2011-02-24T07:16:23.560" Score="1" ViewCount="" Body="&lt;p&gt;Below I show how the &lt;a href=&quot;http://en.wikipedia.org/wiki/Heaviside_cover-up_method&quot; rel=&quot;nofollow&quot;&gt;Heaviside cover-up method&lt;/a&gt; generalizes to handle nonlinear denominators. With the numerator $\rm\:f(x)\ =\ 8x^4+15x^3+16x^2+22x+4\:,\: $ the undetermined partial fraction is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\rm\frac{f(x)}{x(x+1)^2(x^2+2)}\ =\ \frac{a}{x}\ +\ \frac{b\ (x+1) + c}{(x+1)^2}\ +\ \frac{d\ x+e}{x^2+2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To find the numerator of the $\rm\: x^2+2\ $ fraction, clear denominators and collect factors of $\rm\: x^2 + 2\: $ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\rm f(x)\ \ =\ \ x\ (x+1)^2\ (d\ x +\: e)\ \ +\ \ (x^2+\:2)\ g(x)\ ,\quad\quad some\ \ g(x) \in \mathbb Q[x]$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Evaluating this mod $\rm\ x^2 + 2\:,\ $ i.e.$\:$ iteratively applying the rewrite rule $\rm\ x^2 \to -2\:,\:\: $ yields&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\rm - 8\ x + 4\ =\: -(4\ d +\: e)\ x + 2\ d - 4\ e\quad \Rightarrow\quad d=2,\ e=0 $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that this method amounts to ignoring (&quot;covering&quot;) all the undetermined partial fractions having denominator different (i.e. coprime) from the current denominator $\rm\:p(x) = x^2+2\:$ (i.e. having different roots) then evaluating what remains at the roots of $\rm\:p(x)\:$ or, $\:$ equivalently, $\:$ evaluating mod $\rm\:p(x)\:$. To avoid computing inverses mod $\rm\:p(x)\:$ we scale to clear denominators &lt;em&gt;before&lt;/em&gt; evaluating. This is simply the higher-degree analog of the classical Heaviside method - where covering up and evaluating at $\rm\: x = r\:$ is equivalent to evaluating modulo $\rm\:x-r\:$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the same method we can solve for the numerator of the $\rm\ (x+1)^2$ fraction&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\rm f(x)\ \ =\ \ x\ (x^2+2)\ (b\ (x+1) + c)\ \ +\ \ (x+1)^2\ h(x)\ ,\quad\quad some\ \ h(x) \in\mathbb Q[x]$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Evaluating this mod $\rm\: (x+1)^2\:,\ $ i.e. $\:$ iteratively applying the rewrite rule $\rm\: x^2 \to -2\ x - 1\:,\: $ yields&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\rm 3\ x - 6\ =\ (5\ c - 3\ b)\ x + 2\ c - 3\ b\quad \Rightarrow\quad c = 3,\ b = 4 $$&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2011-02-24T17:09:15.027" LastActivityDate="2011-02-24T17:09:15.027" />
  <row Id="23499" PostTypeId="2" ParentId="23496" CreationDate="2011-02-24T07:20:45.397" Score="2" ViewCount="" Body="&lt;p&gt;Hint: Try to write everything in terms of $z$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since we're working on the circle, we have $|z|^2 = z\overline{z} = \rho^2$.  Also, $|dz| = \rho\,dt$ and $dz = i\rho e^{it}\,dt = iz\,dt$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might then want to use partial fractions.&lt;/p&gt;&#xA;" OwnerUserId="640" LastActivityDate="2011-02-24T07:20:45.397" CommentCount="2" />
  <row Id="23500" PostTypeId="1" CreationDate="2011-02-24T08:27:13.407" Score="-1" ViewCount="94" Body="&lt;p&gt;For $b$ in $\mathbb{R}$ and a function $f:\mathbb{R} \to \mathbb{R}$ define the translate $T_{b}f:\mathbb{R} \to \mathbb{R}$ of $f$ by $(T_{b}f) (x):=f(x-b)$. Show that for $f$ in $L^2(\mathbb{R})$ the function $T_b f$ is again in $L^2(\mathbb{R})$, moreover show that the translation operator $T_bf:L^2(R)\to L^2(R)$ is an isometry and the inverse operator is $T_{-b}$.&lt;/p&gt;&#xA;" OwnerUserId="7428" LastEditorUserId="5363" LastEditorDisplayName="" LastEditDate="2011-02-24T08:49:35.383" LastActivityDate="2011-02-24T10:11:57.263" Title="Translation operator" Tags="&lt;analysis&gt;" AnswerCount="2" />
  <row Id="23501" PostTypeId="2" ParentId="23475" CreationDate="2011-02-24T08:35:13.427" Score="0" ViewCount="" Body="&lt;p&gt;Like Zarrax has pointed out, we could look at ${\displaystyle \int_1^{\infty} {|\cos(x)| \over x^r}\,dx}$ where ${1 \over 2} &amp;lt; r &amp;lt; 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We could argue that if ${\displaystyle \int_1^{\infty} {\cos^2(x) \over x^r}\,dx}$ diverges then by the comparison test our integral diverges, since ${|\cos(x)| \over x^r} \geq {\cos^2(x) \over x^r}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Showing divergence of ${\displaystyle \int_1^{\infty} {\cos^2(x) \over x^r}\,dx}$ is simple if we note the fact that: $$\frac{\cos^2(x)}{x^r}=\frac{1}{2x^r}+\frac{\cos(2x)}{2x^r}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But ${\displaystyle \int_1^{\infty} \frac{\cos(2x)}{2x^r}\,dx}$ converges by Dirichlets test and ${\displaystyle \int_1^{\infty} \frac{1}{2x^r}\,dx}$ obviously diverges hence $$\int_1^{\infty} {\cos^2(x) \over x^r}\,dx$$ diverges as needed.&lt;/p&gt;&#xA;" OwnerUserId="3237" LastActivityDate="2011-02-24T08:35:13.427" />
  <row Id="23502" PostTypeId="2" ParentId="23500" CreationDate="2011-02-24T08:37:26.000" Score="1" ViewCount="" Body="&lt;p&gt;Assuming that this is homework: Solve the respective integrals by substitution.&lt;/p&gt;&#xA;" OwnerUserId="3148" LastActivityDate="2011-02-24T08:37:26.000" CommentCount="1" />
  <row Id="23503" PostTypeId="1" AcceptedAnswerId="23506" CreationDate="2011-02-24T08:51:58.620" Score="5" ViewCount="350" Body="&lt;p&gt;Hi, is there some way to create unique number from 2 positive integer numbers? Result must be unique even for these pairs: 2 and 30, 1 and 15, 4 and 60. In general, if I take 2 random numbers result must be unique(or with very high probability unique)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks a lot&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: calculation is for computer program,so computational complexity is important&lt;/p&gt;&#xA;" OwnerUserId="7449" LastEditorUserId="7449" LastEditorDisplayName="" LastEditDate="2011-02-24T10:56:15.813" LastActivityDate="2011-03-08T22:16:07.843" Title="Create unique number from 2 numbers" Tags="&lt;functions&gt;" AnswerCount="3" CommentCount="7" FavoriteCount="2" />
  <row Id="23505" PostTypeId="2" ParentId="23503" CreationDate="2011-02-24T09:18:48.837" Score="12" ViewCount="" Body="&lt;p&gt;if the numbers are $a$ and $b$, take $2^a3^b$. This method works for any number of numbers (just take different primes as the bases), and all the numbers are distinct.&lt;/p&gt;&#xA;" OwnerUserId="2538" LastActivityDate="2011-02-24T09:18:48.837" />
  <row Id="23506" PostTypeId="2" ParentId="23503" CreationDate="2011-02-24T09:22:02.937" Score="22" ViewCount="" Body="&lt;p&gt;The sort of function you are looking for is often called a pairing function. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A standard example is the Cantor pairing function $\mathbb{N}\times\mathbb{N}\rightarrow\mathbb{N}$, given by:&#xA;$\pi(a,b) = \frac{1}{2}(a+b)(a+b+1) + b$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can find more information here: &lt;a href=&quot;http://en.wikipedia.org/wiki/pairing_function&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/pairing_function&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7062" LastActivityDate="2011-02-24T09:22:02.937" CommentCount="2" />
  <row Id="23509" PostTypeId="1" CreationDate="2011-02-24T09:52:41.117" Score="2" ViewCount="111" Body="&lt;p&gt;All rings and algebra in this question are commutative and contains unity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose $M$ is an $A$ module and $A$ a $R$ algebra. If $pd_R(M) &amp;lt; \infty$, then will that imply $pd_A(M) &amp;lt; \infty$? &#xA;  In particular if $M = \frac{R[X_1, X_2, \ldots,X_n]_N}{I}$, $A = R[X_1, X_2, \ldots,X_n]_N$ where $R[X_1, X_2, \ldots,X_n]$ is polynomial algebra , $N$ a multiplicative closed set and $I$ an ideal, is the answer affirmative?&lt;/p&gt;&#xA;" OwnerUserId="7451" LastActivityDate="2011-02-25T01:38:09.970" Title="projective dimension" Tags="&lt;commutative-algebra&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="23510" PostTypeId="2" ParentId="23500" CreationDate="2011-02-24T10:11:57.263" Score="3" ViewCount="" Body="&lt;p&gt;This is straightforward from the observation that the measure on $\Bbb R$ is invariant by translation, or equivalently that if $y=x+c$ then $dy=dx$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, you have the same isometry whenever you consider the, say, $L^2$ space over a space endowed with a measure that satisfies the same property (in order for the translation to make sense, the &quot;space&quot; should be a group or a space that supports a group action).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, consider the punctured line ${\Bbb R}^\times$, i.e. the multiplicative group of real numbers, with the measure&#xA;$$&#xD;&#xA;d^\times x=\frac{dx}{|x|}.&#xD;&#xA;$$&#xA;Mind that here translation is actually multiplication: $x\mapsto cx$.&#xA;An important theorem of Analysis says that a measure with this property exists, and in fact is essentially unique, in a fairly general situation, namely when the space is a locally compact group.&lt;/p&gt;&#xA;" OwnerUserId="688" LastActivityDate="2011-02-24T10:11:57.263" CommentCount="1" />
  <row Id="23511" PostTypeId="1" CreationDate="2011-02-24T10:39:07.690" Score="2" ViewCount="97" Body="&lt;p&gt;Last time I did any maths was A-Level (some time ago!).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a programming/layout problem where I need to display 7 items across a page, with 6 gaps between them. I have a fixed width, but need to determine values of x and y that are integers where x &gt; y.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This 'feels' like something I could plot on a graph, hence solve with calculus but I need a pointer in the right direction. Ideally I'd like to solve it programatically but I need an understanding of how to solve it first. Further down the line, I'd like to be able to vary the number of items to get different values.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="7452" LastEditorUserId="1778" LastEditorDisplayName="" LastEditDate="2011-02-25T00:54:38.197" LastActivityDate="2011-02-25T10:46:18.577" Title="How to solve 700 = 7x + 6y for a range of values?" Tags="&lt;calculus&gt;&lt;algebra&gt;&lt;diophantine-equations&gt;" AnswerCount="2" />
  <row Id="23512" PostTypeId="1" AcceptedAnswerId="23549" CreationDate="2011-02-24T10:59:59.287" Score="2" ViewCount="81" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wanted to know if it was true that if we are given a one-dimensional Feller process taking values in $\mathbb{R}$ and a hitting time $\tau_A=\inf\{t&amp;gt;0 s.t. X_t\in A\}$ with $A$ a open set (this to avoid measurability complexities but I would be also interested in the general case) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;then if $X_0\in \bar{A}^c$ we have :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\lim_{t\to 0}P[\tau_A&amp;lt;t]=0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regards&lt;/p&gt;&#xA;" OwnerUserId="4437" LastEditorUserId="5363" LastEditorDisplayName="" LastEditDate="2011-02-24T15:36:44.023" LastActivityDate="2011-02-24T16:15:33.647" Title="Limit probability for some Hitting time of a Feller Process" Tags="&lt;stochastic-processes&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="23513" PostTypeId="1" AcceptedAnswerId="23517" CreationDate="2011-02-24T11:01:44.993" Score="1" ViewCount="95" Body="&lt;p&gt;This is a fairly simple, if not a bit stupid question, but I'm a little rusty on my permutations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If anyone here is familiar with the Lowest Common Subsequence problem, they probably know that the number of posibble subsequences in a sequence in 2^n; n being the length of the sequence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A{a,b,c,d}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;here, n=4;&#xA;thus number of possible subsequences = 2^4 = 16.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know it sounds a fairly basic question, but can anybody please explain why is this so?&lt;/p&gt;&#xA;" OwnerUserId="7453" LastActivityDate="2011-02-24T15:17:07.247" Title="Why is the number of possible subsequences 2^n?" Tags="&lt;algorithms&gt;&lt;probability-distributions&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="23515" PostTypeId="1" CreationDate="2011-02-24T11:15:28.557" Score="5" ViewCount="224" Body="&lt;p&gt;When a mathematical notion is said to exist, it seems that there is a lot of freedom in what this particularly means, a freedom which accounts I think for why some people are finitists or ultra-finitists. How would you define mathematical existence?&lt;/p&gt;&#xA;" OwnerUserId="7485" LastActivityDate="2011-03-26T21:38:15.410" Title="mathematical existence" Tags="&lt;soft-question&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="23516" PostTypeId="2" ParentId="23511" CreationDate="2011-02-24T11:16:18.533" Score="3" ViewCount="" Body="&lt;p&gt;First, find one solution. Here you can say 700 = 7*100 + 6*0, or 700 = 700*1 = 700*(7-6) = 7*700 + 6*(-700).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One you have one solution $(x_1,y_1)$, another solution $(x_2,y_2)$ has to satisfy&#xA;$7x_1+6y_1 = 700 = 7x_2+6y_2$, so $7(x_1-x_2) + 6(y_1-y_2) = 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here this means that $7(x_1-x_2) = 6(y_2-y_1)$. Since 6 and 7 are coprime, this number is a multiple of 42, so $7(x_1-x_2) = 6(y_2-y_1) = 42k$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now you get $x_2 = x_1 - 6k, y_2 = y_1+7k$, and that gives you the set of solutions &#xA;$S = \{(100-6k, 0+7k), k \in \mathbb{Z}\}$.&#xA;If you only want to have solutions with positive $0&amp;lt;y&amp;lt;x$, this means you get constraints on $k$ : $0 &amp;lt; k$ and $k &amp;lt; 100/13$, which means $k \in \{1, 2, \ldots 7 \}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to solve $a = nx + (n-1)y$, since $n$ and $n-1$ are coprime, the same method works :&#xA;start with the particular solution $a = a*1 = n*a + (n-1)*(-a)$, so $(x=a, y= -a)$.&#xA;the solutions in $\mathbb{Z}^2$ are $\{(x=a-k(n-1), y= kn-a)), k \in \mathbb{Z}\}$.&#xA;Then you write the constraints $0&amp;lt;y&amp;lt;x$ in terms of $k$ and you get $a/n &amp;lt; k &amp;lt; 2a/(2n-1)$&lt;/p&gt;&#xA;" OwnerUserId="6270" LastEditorUserId="6270" LastEditorDisplayName="" LastEditDate="2011-02-25T10:46:18.577" LastActivityDate="2011-02-25T10:46:18.577" CommentCount="2" />
  <row Id="23517" PostTypeId="2" ParentId="23513" CreationDate="2011-02-24T11:23:23.453" Score="1" ViewCount="" Body="&lt;p&gt;If the order of the elements in each subsequence does not matter, then every subsequence is an element of the powerset of the given sequence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your example, the powerset of your sequence A contains&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\{\}, \{a\}, \{b\}, \{c\}, \{d\}, \{a, b\}, \{a, c\}, \{a, d\}, \{b, c\}, \{b, d\}, \{c, d\}, \{a, b, c\}, \{a, b, d\},$ &#xA;$\{a, c, d\}, \{b, c, d\}, \{a, b, c, d\}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;each one being considered as a subsequence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider then the cardinality of the powerset:&#xA;if a set A contains n elements, then the powerset of A contains $2^n$ elements.&lt;/p&gt;&#xA;" OwnerUserId="5672" LastEditorUserId="5672" LastEditorDisplayName="" LastEditDate="2011-02-24T11:35:50.957" LastActivityDate="2011-02-24T11:35:50.957" CommentCount="6" />
  <row Id="23518" PostTypeId="2" ParentId="23484" CreationDate="2011-02-24T11:34:59.663" Score="0" ViewCount="" Body="&lt;p&gt;Following your original method from the point where you pose your question, here are a couple of tricks you can use, which were not mentioned in previous answers. (1) Set $x=\varepsilon -1$ and neglect $\varepsilon^2$, to get $B=4$ immediately. (2) Set $x=\mathrm{i}\sqrt2$ and equate real and imaginary parts to find $D$ and$E$ (independently of $B$). &lt;/p&gt;&#xA;" OwnerUserId="875" LastActivityDate="2011-02-24T11:34:59.663" />
  <row Id="23519" PostTypeId="2" ParentId="23513" CreationDate="2011-02-24T11:46:54.407" Score="1" ViewCount="" Body="&lt;p&gt;The proof can be given by induction. For set of n elements, the number of subsets are $2^n$. For the set of n + 1 elements, you can take the previous $2^n$ elements and add the (n + 1)th element in each of these sets. So now you have $2^n$ (original subsets) + $2^n$ (new subsets). It is simple to see the result now. I haven't mentioned base case as it is elementary. &lt;/p&gt;&#xA;" OwnerUserId="5403" LastEditorUserId="5403" LastEditorDisplayName="" LastEditDate="2011-02-24T14:46:40.170" LastActivityDate="2011-02-24T14:46:40.170" />
  <row Id="23520" PostTypeId="1" CreationDate="2011-02-24T12:14:28.383" Score="0" ViewCount="63" Body="&lt;p&gt;I had seen a video lecture of Gilbert Strang online, which was about chaos and Newton's method.  I could not find it again online, could someone help me locate it?  Thanks in advance. &lt;/p&gt;&#xA;" OwnerUserId="7455" LastActivityDate="2011-02-24T12:14:28.383" Title="Gilbert Strang's video lecture about chaos and Newton's method" Tags="&lt;reference-request&gt;&lt;chaos-theory&gt;" />
  <row Id="23522" PostTypeId="1" CreationDate="2011-02-24T12:52:16.680" Score="1" ViewCount="159" Body="&lt;p&gt;How many proofs of the Cauchy-Schwarz inequality are there? Is there some kind of reference that lists all of these proofs?&lt;/p&gt;&#xA;" OwnerUserId="7340" LastEditorUserId="5363" LastEditorDisplayName="" LastEditDate="2011-02-24T12:55:35.287" LastActivityDate="2011-02-24T12:55:35.287" Title="Proofs of the Cauchy-Schwarz Inequality?" Tags="&lt;real-analysis&gt;&lt;inequality&gt;" CommentCount="10" />
  <row Id="23523" PostTypeId="2" ParentId="23509" CreationDate="2011-02-24T13:11:58.503" Score="2" ViewCount="" Body="&lt;p&gt;In general, the answer is &lt;em&gt;no&lt;/em&gt;.  E.g. suppose that $R$ is a field.  Every module over a field is free, and so has projective dimension $0$.  But if we take $A = k[x]/(x^2),$ then the $A$-module $M := k$ (with $x$ acting via zero) has infinite projective dimension over $A$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, your particular case has $A$ being a localization of $R$.  In this case the answer is &lt;em&gt;yes&lt;/em&gt;:&#xA;let $P^{\bullet} \to M$ be a finite length projective resolution of $M$ as an $R$-module.&#xA;Then using the facts (two basic properties of localizations) that (a) $M \cong A\otimes_R M$, and (b) $A$ is flat over $R$, we see that $A\otimes_R P^{\bullet}$ is a finite length projective resolution of $M$ as an $A$-module.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: This &quot;answer&quot; doesn't actually answer the question, because in the question there are really three rings involved, $R$, $R[X_1,\ldots,X_n]$, and then the localization $A := R[X_1,\ldots,X_n]_A$.    So in the above answer,&#xA;the ring called $R$ should actually be $R[X_1,\ldots,X_n]$, and the question becomes: if $M$ is an $R[X_1,\ldots,X_n]$-module which is of finite projective dimension as an $R$-module, is it of finite projective dimension as an&#xA;$R[X_1,\ldots,X_n]$-module?   I believe that the answer is again &lt;em&gt;yes&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If this is true, it is surely well-known (and if it's false, I apologize for the blunder!); in any case, here is an attempt at a proof.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Firstly, by induction on $n$ (i.e. adjoin variables one at a time) we reduce to&#xA;the case $n = 1$, so $M$ is an $R[x]$-module, of finite projective dimension as an $R$-module.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now take a projective resolution of $M$ as an $R[x]$-module; since $R[x]$ is projective as an $R$-module, this is also a projective resolution of $M$ as an $R$-module, and after some point in this resolution, the objects become projective as $R$-modules.  So it suffices to show that an $R[x]$-module which is &lt;em&gt;projective&lt;/em&gt; as an $R$-module has finite projective dimension over $R[x]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now adding a complement to this projective $R$-module, we can make it free&#xA;over $R$, and we can make it an $R[x]$-module just by having $x$ act via zero on the complement that we added.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we reduce to showing that an $R[x]$-module that is &lt;em&gt;free&lt;/em&gt; over $R$ has finite projective dimension over $R$.  (This step is probably superfluous, but it helps me psychologically to really be working with a free $R$-module.) &#xA;Let $F$ be our free $R$-module; the action of $x$ then corresponds to an&#xA;element $\phi \in End_R(F).$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now write $F[x] := R[x]\otimes_R F$, and consider the complex of $R[x]$-modules&#xA;$$ 0 \to F[x] \buildrel x - \phi \over\longrightarrow F[x] \to F \to 0.$$&#xA;This is a presentation of $M$, and is short exact, so indeed we see that $F$ has finite projective dimension over $R[x]$ (in fact it has projective dimension equal to one), so we are done. (In fact the proof shows that the projective dimension of our original $M$ over $R[x]$ is at most one greater than&#xA;its projective dimension over $R$.)&lt;/p&gt;&#xA;" OwnerUserId="221" LastEditorUserId="221" LastEditorDisplayName="" LastEditDate="2011-02-25T01:38:09.970" LastActivityDate="2011-02-25T01:38:09.970" />
  <row Id="23524" PostTypeId="1" AcceptedAnswerId="23525" CreationDate="2011-02-24T13:33:41.967" Score="4" ViewCount="144" Body="&lt;p&gt;I have been looking on the net for a textbook that only is about &quot;Generating Functions&quot; but did not found much. Is there any textbook just about &quot;Generating Functions&quot;?&lt;/p&gt;&#xA;" OwnerUserId="7340" LastActivityDate="2011-02-24T13:36:51.877" Title="Textbook on &quot;Generating Functions&quot;" Tags="&lt;analysis&gt;&lt;generating-functions&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" />
  <row Id="23525" PostTypeId="2" ParentId="23524" CreationDate="2011-02-24T13:36:51.877" Score="16" ViewCount="" Body="&lt;p&gt;Try  &lt;a href=&quot;http://www.math.upenn.edu/~wilf/DownldGF.html&quot; rel=&quot;nofollow&quot;&gt;generatingfunctionology&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="589" LastActivityDate="2011-02-24T13:36:51.877" />
  <row Id="23526" PostTypeId="2" ParentId="23503" CreationDate="2011-02-24T13:38:10.837" Score="3" ViewCount="" Body="&lt;p&gt;Google &lt;a href=&quot;http://en.wikipedia.org/wiki/Pairing_function&quot; rel=&quot;nofollow&quot;&gt;pairing function&lt;/a&gt;. As I mentioned in the &lt;a href=&quot;http://math.stackexchange.com/questions/5587/related-integer-combination/5590#5590&quot;&gt;similar question,&lt;/a&gt; there are also other pairing functions besides the well-known one due to Cantor. For example, see this &lt;a href=&quot;http://szudzik.com/ElegantPairing.pdf&quot; rel=&quot;nofollow&quot;&gt;&quot;elegant&quot; pairing function,&lt;/a&gt; which has the useful property that it orders many expressions by depth.&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2011-02-24T13:38:10.837" />
  <row Id="23527" PostTypeId="1" AcceptedAnswerId="23546" CreationDate="2011-02-24T13:44:43.600" Score="1" ViewCount="52" Body="&lt;p&gt;In many places I see the entropy definition as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$H(X) = \sum_{i=1}^n {p(x_i)\,I(x_i)} = -\sum_{i=1}^n {p(x_i) \log_b p(x_i)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In Wikipedia I saw: $H(X)  =  \operatorname{E}(I(X))$ where E is expected value and I is information content.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Today I was discussing an issue with my friends and couldn't decide if chessboard's entropy is high or low. Because I can define chessboard as &quot;1 black, 1 white square tiled into $8\times8$ grid&quot;. I don't need to send all the cell values. The difference is more distinct if the board size is large, say $1000\times1000$. I think the second definition is better since it includes definition of &lt;em&gt;information content&lt;/em&gt;, it is more intuitive for people who comes from different backgrounds. I cannot utilize the first definition for the chessboard case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same thing goes for lots of things. Take digits of $\pi$. It takes too much resource to send the first 1M digits compared to sending just the $\pi$ generating formula. Is the entropy of $\pi$ high or low?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; Because I don't know the terms in the field, my title may not be very descriptive. Any title suggestions are welcome.&lt;/p&gt;&#xA;" OwnerUserId="4776" LastActivityDate="2011-02-24T16:05:06.350" Title="Entropy of generatable(?) structures" Tags="&lt;information-theory&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="23528" PostTypeId="1" CreationDate="2011-02-24T14:11:19.480" Score="0" ViewCount="95" Body="&lt;p&gt;Dear colegues, I need your help with the following issue:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need to prove that the p-adic topology is a group (that it has group properties) and analyse the conection, compactness, metric, and whatever related with the p-adic topology.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you very much for your help and attention, I look forward to hearing your answers.&lt;/p&gt;&#xA;" OwnerUserId="7458" LastActivityDate="2011-02-24T14:11:19.480" Title="p-adic topology" Tags="&lt;topology&gt;" CommentCount="4" />
  <row Id="23529" PostTypeId="2" ParentId="23470" CreationDate="2011-02-24T14:30:30.240" Score="3" ViewCount="" Body="&lt;p&gt;As others have said, one can find (often brief) treatments of these issues in many places.  However, an unusually thorough and insightful account is given in &lt;a href=&quot;http://www.math.uconn.edu/~kconrad/blurbs/linmultialg/complexification.pdf&quot; rel=&quot;nofollow&quot;&gt;this exposition&lt;/a&gt; of Keith Conrad.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: &quot;unusually&quot; means &quot;unusually for treatments of complexification&quot;, not &quot;unusually for Keith Conrad&quot;.  In fact I give his entire page of expository writing -- namely &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.math.uconn.edu/~kconrad/blurbs/&quot; rel=&quot;nofollow&quot;&gt;http://www.math.uconn.edu/~kconrad/blurbs/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;my highest recommendation: it is a treasure trove of intermediate level math.  And when I say &quot;intermediate level&quot;, I don't mean that grown up mathematicians can't profit from reading them.  We sure can -- I have learned a lot, and I have not hesitated to flatter them in the most sincere way.  &lt;/p&gt;&#xA;" OwnerUserId="299" LastActivityDate="2011-02-24T14:30:30.240" />
  <row Id="23530" PostTypeId="2" ParentId="23044" CreationDate="2011-02-24T14:42:24.770" Score="4" ViewCount="" Body="&lt;p&gt;There is no limit for any $0&amp;lt;x&amp;lt;1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(a) $f(t, 1/3) = 1$ for $t$ of the form $6n+1$ or $6n+2$ and $0$ for $t$ of the form $6n+4$ or $6n+5$, and on the boundary in other cases.  For example $\frac{2n}{6n+1} &amp;lt; \frac{1}{3} &amp;lt; \frac{2n+1}{6n+1}$ and $\frac{2n+1}{6n+4} &amp;lt; \frac{1}{3} &amp;lt; \frac{2n+2}{6n+4} .$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(b) $f(t, 1/2) = 1$ for $t$ of the form $4n+1$ and $0$ for $t$ of the form $4n+3$ and on the boundary in other cases&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(c) If $f(t, 1/\pi) = 1$ then $f(t+3, 1/\pi) = 0$ or $f(t+4, 1/\pi) = 0$ and similarly if $f(t, 1/\pi) = 0$ then $f(t+3, 1/\pi) = 1$ or $f(t+4, 1/\pi) = 1$.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(d) If $f(t, x) = 1$ then $f\left(t+\lfloor{1/x1}\rfloor , x \right) = 0$ or $f\left(t+\lceil{1/x1}\rceil , x \right) = 0$  and similarly if $f(t, x) = 0$ then $f\left(t+\lfloor{1/x1}\rfloor , x \right) = 1$ or $f\left(t+\lceil{1/x1}\rceil , x \right) = 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So there is no convergence and so no limit.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If instead you explicitly gave boundary cases the value $1/2$ (only necessary for rational $x$) and took the partial average of $f(s,x)$ over $1 \le s \le t$, then the limit of the average as $t$ increases would be $1/2$.   &lt;/p&gt;&#xA;" OwnerUserId="6460" LastActivityDate="2011-02-24T14:42:24.770" />
  <row Id="23532" PostTypeId="1" CreationDate="2011-02-24T14:53:15.527" Score="0" ViewCount="57" Body="&lt;p&gt;I want to show that the product of smooth quasi-projective varieties is smooth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Call my varieties $A$ and $B$. Since both are smooth, at every point the dimension of the tangent space equals the dimension of the variety. Will the dimension of $A$x$B$ be the sum of the dimensions of $A$ and $B$? If so, how can I show that the same thing happens to the dimension of the tangent space at any point?&lt;/p&gt;&#xA;" OwnerUserId="7461" LastActivityDate="2011-02-24T14:53:15.527" Title="product of smooth quasi-projective varieties" Tags="&lt;algebraic-geometry&gt;" CommentCount="2" />
  <row Id="23533" PostTypeId="1" AcceptedAnswerId="23536" CreationDate="2011-02-24T14:54:03.687" Score="3" ViewCount="94" Body="&lt;p&gt;I'm reviewing past assignments and am still having trouble formulating a proof for this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the sequence $(x_n)$, where $x_n = (1,\frac{1}{2}, \frac{1}{3}, \ldots, \frac{1}{n}, 0, 0, \ldots)$. Determine whether $(x_n)$ converges in $l_1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's simple to show that the coordinate-wise limit does not converge, but how can I show that the coordinate-wise limit is the only possible limit? Alternatively, I'm trying to show that $(x_n)$ is unbounded, which I think should be straightforward also, but is giving me trouble.&lt;/p&gt;&#xA;" OwnerUserId="4518" LastEditorUserId="5363" LastEditorDisplayName="" LastEditDate="2011-02-24T15:33:18.820" LastActivityDate="2011-02-24T15:33:18.820" Title="How to prove that the sequence of $x_n = (1,\frac{1}{2}, \frac{1}{3}, ... \frac{1}{n}, 0, 0...)$ does not converge under $\|\cdot\|_1$?" Tags="&lt;functional-analysis&gt;&lt;convergence&gt;&lt;metric-spaces&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="23534" PostTypeId="1" AcceptedAnswerId="23540" CreationDate="2011-02-24T14:58:36.680" Score="0" ViewCount="71" Body="&lt;p&gt;I know about algebraic numbers and transcendental numbers.&#xA;How the roots of a polynomial with irrational coefficients are classified.&#xA;Are they transcendental?&lt;/p&gt;&#xA;" OwnerUserId="5049" LastActivityDate="2011-02-24T15:32:51.297" Title="Roots of a polynomial with irrational coefficients" Tags="&lt;polynomials&gt;&lt;roots&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="23536" PostTypeId="2" ParentId="23533" CreationDate="2011-02-24T15:05:35.947" Score="8" ViewCount="" Body="&lt;p&gt;Both your ideas can be made into arguments, let me show you how:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The standard trick for showing divergence is to estimate&#xA;\[&#xA;\|x_{2n} - x_{n}\| = \|(0,\ldots,0,\frac{1}{n+1},\frac{1}{n+2},\ldots,\frac{1}{2n},0,\ldots)\| = \sum_{k=n+1}^{2n} \frac{1}{k} \geq n \cdot \frac{1}{2n} = \frac{1}{2},&#xA;\]&#xA;so the sequence $(x_{n})$ is not a Cauchy sequence, hence it can't converge. On the other hand, this also shows that $\|x_{2^k}\| \geq \frac{k}{2}$, so the sequence is unbounded.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Note that the map $\ell^{1} \to \mathbb{R}$ sending a sequence $(y_{n})_{n \in \mathbb{N}} \in \ell^{1}$ to its $n$th coordinate $y_{n}$ is continuous because $|y_{n}| \leq \|(y_{n})_{n \in \mathbb{N}}\|$. So an $\ell^{1}$-limit must be a pointwise limit. So if the pointwise limit is outside $\ell^{1}$, there can't be an $\ell^{1}$-limit.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="5363" LastEditorUserId="5363" LastEditorDisplayName="" LastEditDate="2011-02-24T15:29:01.260" LastActivityDate="2011-02-24T15:29:01.260" CommentCount="1" />
  <row Id="23537" PostTypeId="1" AcceptedAnswerId="23541" CreationDate="2011-02-24T15:06:24.553" Score="1" ViewCount="84" Body="&lt;p&gt;Recent news piqued my interest in integer partitions again.  I'm working my way back through an old text and I'm completely hung up on this problem:&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Recall that $p_k(n)$ is the number of partitions of $n$ into exactly $k$ parts.  Prove that for all positive integers $k\leq n$, the inequality $p_k(n) \leq (n-k+1)^{k-1}$ holds.&lt;/p&gt;&#xA;" OwnerUserId="7463" LastActivityDate="2011-02-24T15:28:26.047" Title="Upper bound on integer partitions of n into k parts." Tags="&lt;combinatorics&gt;&lt;partitions&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="23538" PostTypeId="2" ParentId="23533" CreationDate="2011-02-24T15:06:54.543" Score="2" ViewCount="" Body="&lt;p&gt;Unboundedness should be straightforward: what is $||x_n||_1$?  What does it do as $n \to \infty$?&lt;/p&gt;&#xA;" OwnerUserId="822" LastActivityDate="2011-02-24T15:06:54.543" />
  <row Id="23539" PostTypeId="2" ParentId="23513" CreationDate="2011-02-24T15:17:07.247" Score="1" ViewCount="" Body="&lt;p&gt;Another way of thinking about this problem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For any given sub-sequence a specific element can either be included or excluded.  This means you have 2 possibilities for each element and a total of n-elements.  Thus there are $2^n$ possible ways to generate a sub-sequence.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Another possible way to wrap your head around the concept.  Think of every sub-sequence as containing $n$ blank spaces, each corresponding to one of the elements of your original sequence.  You can then fill each of these blanks with either a 1 to denote the inclusion of that element or a 0 to denote the exclusion of that element.  That gives 2 possibilities for each blank.  There are n blanks, thus $2^n$ possible sub sequences.&lt;/p&gt;&#xA;" OwnerUserId="7463" LastActivityDate="2011-02-24T15:17:07.247" />
  <row Id="23540" PostTypeId="2" ParentId="23534" CreationDate="2011-02-24T15:24:18.673" Score="3" ViewCount="" Body="&lt;p&gt;The roots of a polynomial with algebraic coefficients are all algebraic, and a monic polynomial whose roots are all algebraic has algebraic coefficients.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So a monic polynomial with some transcendental coefficient must have at least one transcendental root (and vice versa), but it can also have algebraic roots (for example, $0$ is a non transcendental root of $X^2- \pi X = 0$).&lt;/p&gt;&#xA;" OwnerUserId="6270" LastEditorUserId="6270" LastEditorDisplayName="" LastEditDate="2011-02-24T15:32:51.297" LastActivityDate="2011-02-24T15:32:51.297" CommentCount="2" />
  <row Id="23541" PostTypeId="2" ParentId="23537" CreationDate="2011-02-24T15:28:26.047" Score="2" ViewCount="" Body="&lt;p&gt;Each of the $k$ parts must have a minimum of 1 item, leaving $n-k$ to distribute into the $k$ bins.  When you choose how many to put in the first $k-1$ bins, the number going into the last bin is fixed.  So you have $k-1$ choices, each of which is within the range $[0,n-k]$, so there are $n-k+1$ of them.  This is a generous upper bound.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-02-24T15:28:26.047" />
  <row Id="23542" PostTypeId="2" ParentId="23509" CreationDate="2011-02-24T15:31:03.920" Score="0" ViewCount="" Body="&lt;p&gt;@Matt //&#xA;Are you claiming that $R[X_1, X_2 \ldots X_n]_N \otimes_R \frac{R[X_1, X_2 \ldots X_n]_N}{I} = \frac{R[X_1, X_2 \ldots X_n]_N}{I}$?&#xA;Soppose R is a field and N=1 then $R[X_1, X_2, \ldots, X_n] \otimes_R \frac{R[X_1, X_2 \ldots X_n]}{I} = \frac{R[X_1, X_2 \ldots X_n]}{I}$ does not seem to be true as vector space dimensions over $R$ are different for both sides for suitable choice of I. &lt;/p&gt;&#xA;" OwnerUserId="7465" LastActivityDate="2011-02-24T15:31:03.920" CommentCount="3" />
  <row Id="23543" PostTypeId="2" ParentId="23044" CreationDate="2011-02-24T15:33:15.417" Score="2" ViewCount="" Body="&lt;p&gt;Let $n(t, x)$ be the index of the interval into which $x$ falls when $[0,1)$ is divided into $t$ identical intervals,&#xA;$$&#xD;&#xA;[0,1) = \bigcup_{i=1}^{t} \left[\frac{i-1}{t}, \frac{i}{t}\right).&#xD;&#xA;$$&#xA;Then $(n-1)/t \le x &amp;lt; n/t$, so&#xA;$$&#xD;&#xA;n(t,x) = \lfloor{tx + 1}\rfloor.&#xD;&#xA;$$&#xA;Clearly $n(t+1,x)-n(t,x) \le 1$ for $x&amp;lt;1$; that is, $n(t,x)$ cannot skip any values.  On the other hand, for $x&amp;gt;0$, $n(t,x)$ grows without bound as $t\rightarrow\infty$.  Combining these two facts, we see that for $x&amp;gt;0$, $n(t,x)$ is both even and odd infinitely often, hence $f(t,x)$ is equal to both $0$ and $1$ infinitely often, and hence $\lim_{t\rightarrow\infty}f(t,x)$ does not exist.  At the remaining point, $x=0$, the limit does exist: $n(t,0)$ is identically equal to $1$, which is odd, so $f(t,0)$ is identically equal to $1$, and $\lim_{t\rightarrow\infty}f(t,0)=1$.&lt;/p&gt;&#xA;" OwnerUserId="5546" LastActivityDate="2011-02-24T15:33:15.417" />
  <row Id="23544" PostTypeId="1" CreationDate="2011-02-24T15:37:56.487" Score="10" ViewCount="215" Body="&lt;p&gt;Let $R$ be a commutative ring. The Cartesian square $A=R\times R$ is endowed with the operation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(a_1,b_1)\circ(a_2,b_2)=(a_1+a_2,b_1+b_2+a_1a_2^2+a_1^2a_2)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which turns $A$ into a commutative group. I have two questions concerning this group.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question 1:&lt;/strong&gt; For what $R$ is $(A,\circ)$ isomorphic to $R\oplus R$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I managed to show that if $3$ is invertible in $R$ then the isomorphism holds.&#xA;It &lt;strike&gt;also holds for R=F_9, but&lt;/strike&gt; does not hold for $R=\mathbb{F}_3$, $\mathbb{F}_9$, or $\mathbb{F}_{27}$.&#xA;Other rings $R$ in which $3$ is not invertible (including $R=\mathbb{Z}$) remain a mystery to me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question 2:&lt;/strong&gt; When is $(A,\circ)$ generated by the elements of the form $(a,0)$, $a\in R$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again, only partial results here. Let $B$ be the subgroup of $A$ generated by $(a,0)$, $a\in R$. Then $B$ contains $(0,a_1a_2^2+a_1^2a_2)$ for all $a_1,a_2\in R$. Hence,&#xA;we have $B=A$ if $R$ is additively generated by the elements of the form $a_1a_2^2+a_1^2a_2$. This is so for $R=\mathbb{Z}/p\mathbb{Z}$ with $p$ odd. &lt;/p&gt;&#xA;" OwnerUserId="2261" LastEditorUserId="2261" LastEditorDisplayName="" LastEditDate="2011-04-06T13:23:10.807" LastActivityDate="2011-04-06T13:23:10.807" Title="A commutative group structure on $R\times R$ for a ring $R$" Tags="&lt;group-theory&gt;&lt;commutative-algebra&gt;" AnswerCount="2" CommentCount="7" FavoriteCount="2" />
  <row Id="23545" PostTypeId="1" AcceptedAnswerId="23548" CreationDate="2011-02-24T15:58:25.243" Score="3" ViewCount="84" Body="&lt;p&gt;Hello. Let $x_{1},x_{2},...,x_{m}$ be elements of $\mathbb{A}^{n}$, where $\mathbb{A}^{n}$ is the n-affine space over an algebraically closed field $k$. Now define $X=\{x_{1},x_{2},...,x_{m}\}$. Why is the coordinate ring $A(X)$, isomorphic to $\oplus_{j=1}^{m} k = k^{m}$?&lt;/p&gt;&#xA;" OwnerUserId="6495" LastActivityDate="2011-02-24T16:21:11.893" Title="Isomorphism in coordinate ring" Tags="&lt;algebraic-geometry&gt;" AnswerCount="2" />
  <row Id="23546" PostTypeId="2" ParentId="23527" CreationDate="2011-02-24T16:05:06.350" Score="2" ViewCount="" Body="&lt;p&gt;Entropy is defined for a probability distribution; it measures how much information must be provided, on average, to identify a randomly chosen element from that distribution.  To measure the information required to generate a &lt;em&gt;specific&lt;/em&gt; string, on the other hand, one can use the &lt;a href=&quot;http://en.wikipedia.org/wiki/Kolmogorov_complexity&quot; rel=&quot;nofollow&quot;&gt;Kolmogorov complexity&lt;/a&gt; or a related complexity measure.  The colors of the $10^6 \times 10^6$ chessboard and the first trillion digits of $\pi$ can be expressed using far less than a trillion bits, and this is a consequence of the low complexity of those objects.&lt;/p&gt;&#xA;" OwnerUserId="5546" LastActivityDate="2011-02-24T16:05:06.350" CommentCount="1" />
  <row Id="23547" PostTypeId="2" ParentId="23545" CreationDate="2011-02-24T16:07:55.703" Score="4" ViewCount="" Body="&lt;p&gt;For each $i$, $A(\{x_i\}) = k[x]/I(x_i) \cong k$, so each $I(x_i)$ is a maximal ideal of &#xA;$k[x]$.  I assume the points $x_1,\ldots,x_n$ are distinct, from which it follows easily &#xA;that the ideals $I(x_1),\ldots,I(x_n)$ are distinct maximal ideals.  Thus they are &lt;strong&gt;pairwise comaximal&lt;/strong&gt; and the &lt;strong&gt;Chinese Remainder Theorem&lt;/strong&gt; -- see e.g. $\S 4.3$ of &lt;a href=&quot;http://math.uga.edu/~pete/integral.pdf&quot; rel=&quot;nofollow&quot;&gt;these notes&lt;/a&gt; -- applies.  I leave it to you to check that it gives the conclusion you want.&lt;/p&gt;&#xA;" OwnerUserId="299" LastActivityDate="2011-02-24T16:07:55.703" CommentCount="1" />
  <row Id="23548" PostTypeId="2" ParentId="23545" CreationDate="2011-02-24T16:12:16.553" Score="4" ViewCount="" Body="&lt;p&gt;The coordinate ring $A(X)$ is defined to be the quotient of $k[z_1,\ldots,z_n]$ (I'm using $z_i$ because you are using $x_i$ to denote the points of affine space) by the ideal $I(X)$. The ideal $I(X)$ is the ideal of all elements of $k[z_1,\ldots,z_n]$ that are zero at $X$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the Nullstellensatz, the ideal corresponding to a single point $x_i = (x_{i1},\ldots,x_{in})$ is given by $(z_1-x_{i1}, z_2-x_{i2},\ldots,z_n-x_{in})$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The ideal of a union is the intersection of the ideals; so you are trying to mod out by&#xA;$$\bigcap_{i=1}^m (z_1-x_{i1},\ldots,z_n-x_{in}).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the ideals $(z_1-x_{i1},\ldots,z_n-x_{in})$ are maximal ideals, since $k[z_1,\ldots,z_n]/(z_1-x_{i1},\ldots,z_n-x_{in}) \cong k$. If they are all distinct, then they are pairwise comaximal, hence pairwise coprime. By the Chinese Remainder Theorem, we know that if $\mathfrak{p}_1,\ldots,\mathfrak{p}_k$ are pairwise coprime ideals in $R$, then&#xA;$$\frac{R}{\mathfrak{p}_1\cap\cdots\cap\mathfrak{p}_k} \cong \frac{R}{\mathfrak{p}_1}\oplus\cdots\oplus\frac{R}{\mathfrak{p}_k}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we have that&#xA;$$A(X) = \frac{k[z_1,\ldots,z_n]}{\cap_{i=1}^m(z_1-x_{i1},\ldots,z_n-x_{in})} \cong \mathop{\bigoplus}_{i=1}^m \frac{k[z_1,\ldots,z_n]}{(z_1-x_{i1},\ldots,z_m-x_{im})} \cong \mathop{\bigoplus}_{i=1}^m k.$$&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-24T16:21:11.893" LastActivityDate="2011-02-24T16:21:11.893" CommentCount="1" />
  <row Id="23549" PostTypeId="2" ParentId="23512" CreationDate="2011-02-24T16:15:33.647" Score="3" ViewCount="" Body="&lt;p&gt;Some useful facts about Feller processes are at &lt;a href=&quot;http://almostsure.wordpress.com/2010/07/19/properties-of-feller-processes/&quot; rel=&quot;nofollow&quot;&gt;http://almostsure.wordpress.com/2010/07/19/properties-of-feller-processes/&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, every Feller process admits a cadlag modification.  If you are willing to assume you are dealing with such a modification, then your question becomes pretty easy, since by continuity of probability $\lim_{t \to 0} P(\tau_A &amp;lt; t) = P(\tau_A = 0)$.  But $\bar{A}^c$ is open and $X_t$ is a.s. continuous at 0 so this has probability 0.  It doesn't much matter that $A$ be open or even measurable, since if you replace $A$ by $\bar{A}$ the answer is the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you insist on using a non-cadlag modification, I am not sure that you can say much.&lt;/p&gt;&#xA;" OwnerUserId="822" LastActivityDate="2011-02-24T16:15:33.647" CommentCount="1" />
  <row Id="23550" PostTypeId="1" AcceptedAnswerId="23552" CreationDate="2011-02-24T16:23:35.960" Score="0" ViewCount="80" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Possible Duplicate:&lt;/strong&gt;&lt;br&gt;&#xA;  &lt;a href=&quot;http://math.stackexchange.com/questions/20418/how-to-find-the-sum-of-the-following-series&quot;&gt;How to find the sum of the following series&lt;/a&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;p&gt;Hey there,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do I (generally) calculate values of series? What tricks and theorems are there to make my life easier? E.g.:&#xA;$$\sum_{n=1}^\infty \frac{n^2}{2^n} = ?$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know perfectly well how to show whether a series converges or diverges, but (at least in this and a lot of other cases) I have no idea how to calculate the value to which it converges. I'm looking forward to your help.&lt;/p&gt;&#xA;" OwnerUserId="3787" LastActivityDate="2011-02-24T16:36:08.503" ClosedDate="2011-02-24T16:50:45.943" Title="How do I calculate values of series?" Tags="&lt;sequences-and-series&gt;" AnswerCount="1" CommentCount="11" />
  <row Id="23551" PostTypeId="2" ParentId="23358" CreationDate="2011-02-24T16:31:17.757" Score="9" ViewCount="" Body="&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/QLjSF.png&quot; alt=&quot;Gaussian integers modulo 3-i&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This diagram shows the Gaussian integers modulo 3-i.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The red points shown are all considered to be 0 but their locations in Z[i] are 0, 3-i, i(3-i) and 3-i + i(3-i). Every congruence class must be inside that box once and you can see there are 10 of them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The arrows show adding by 1 each time. Doing that takes you through every equivalence class and then back to the start.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So $\mathbb{Z}[i]/(3-i) \cong \mathbb{Z}/10\mathbb{Z}$.&lt;/p&gt;&#xA;" OwnerUserId="4290" LastEditorUserId="4290" LastEditorDisplayName="" LastEditDate="2011-02-24T16:36:47.093" LastActivityDate="2011-02-24T16:36:47.093" />
  <row Id="23552" PostTypeId="2" ParentId="23550" CreationDate="2011-02-24T16:36:08.503" Score="0" ViewCount="" Body="&lt;p&gt;Consider the more general problem $\sum {n^2}{z^n}$. To handle this, start with $\sum {z^n}$ and $\sum {n}{z^n}$. The first of these should be easy.&lt;/p&gt;&#xA;" OwnerUserId="589" LastActivityDate="2011-02-24T16:36:08.503" />
  <row Id="23553" PostTypeId="1" AcceptedAnswerId="23561" CreationDate="2011-02-24T16:54:59.743" Score="-1" ViewCount="128" Body="&lt;p&gt;I am writing a scientific article with a few mathematical equations. Can I assume that my audience will know what lhs and rhs mean?&lt;/p&gt;&#xA;" OwnerUserId="3733" LastActivityDate="2011-02-24T17:31:52.127" ClosedDate="2011-02-24T17:42:44.770" Title="Can I assume that a biologist will know what &quot;lhs&quot; and &quot;rhs&quot; mean? Or what are some other ways of indicating the left/right hand sides of an equation?" Tags="&lt;notation&gt;" AnswerCount="4" CommentCount="7" />
  <row Id="23554" PostTypeId="2" ParentId="23553" CreationDate="2011-02-24T16:58:01.410" Score="0" ViewCount="" Body="&lt;p&gt;I think it should be pretty obvious. Or you can simply say that LHS means &quot;left hand side&quot; and RHS means &quot;right hand side.&quot; But many biologists have probably studied some probability or statistics. So they might have seen this notation before. &lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2011-02-24T16:58:01.410" CommentCount="2" />
  <row Id="23555" PostTypeId="1" AcceptedAnswerId="23560" CreationDate="2011-02-24T17:01:42.813" Score="2" ViewCount="144" Body="&lt;p&gt;&lt;strong&gt;Background&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As background, I have found that taylor expansion provides poor estimates of a function at extreme parameter values. Indeed, the approximation at extreme values can get worse (more rapid exponential increase) as the order of the taylor series increases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This seems intuitive, but I don't know that it is a rule ... or if there is a proof. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Questions&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;do all polynomials with order &gt; 1 go to $\pm$ infinity?&lt;/li&gt;&#xA;&lt;li&gt;Is there a good reference where I can find answers to a qustion such as this?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="3733" LastEditorUserId="3733" LastEditorDisplayName="" LastEditDate="2011-02-24T18:33:26.467" LastActivityDate="2011-02-24T21:37:44.977" Title="do all polynomials with order $&gt; 1$ go to $\pm$ infinity?" Tags="&lt;algebra&gt;&lt;polynomials&gt;&lt;taylor-expansion&gt;" AnswerCount="3" CommentCount="3" />
  <row Id="23556" PostTypeId="2" ParentId="23553" CreationDate="2011-02-24T17:02:40.737" Score="2" ViewCount="" Body="&lt;p&gt;I think abbreviations like that rarely pay.&lt;/p&gt;&#xA;" OwnerUserId="589" LastActivityDate="2011-02-24T17:02:40.737" CommentCount="2" />
  <row Id="23557" PostTypeId="1" AcceptedAnswerId="31488" CreationDate="2011-02-24T12:27:45.550" Score="1" ViewCount="126" Body="&lt;p&gt;How to prove the Energy function of a Hopfield net is monotonically decreasing?&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$E = -1/2 \sum_{i,j} {w_{ij}}{s_i}{s_j} + \sum_{i}^{}s_{i} \theta_i$ &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'll assume a proof involves the standard definition(s) of the activation function:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;(1) $a_i \leftarrow \left\{ \begin{matrix} 1 &amp;amp; \mbox {if }\sum_{j}{w_{ij}s_j}&amp;gt;\theta_i, \\&#xD;&#xA;-1 &amp;amp; \mbox {otherwise.}\end{matrix} \right\}$&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;(2) $a_i \leftarrow \left\{ \begin{matrix} 1 &amp;amp; \mbox {if }\sum_{j}{w_{ij}s_j}&amp;gt;\theta_i, \\&#xD;&#xA;0 &amp;amp; \mbox {otherwise.}\end{matrix} \right\}$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And the normal rules for weights:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$w_{ij}=0, \forall i=j$ (not reflexive)&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$w_{ij} = w_{ji}, \forall i,j$ (symmetric)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If someone could set me on the right path, it would be appreciated.&lt;/p&gt;&#xA;" OwnerUserId="4509" LastEditorUserId="4509" LastEditorDisplayName="" LastEditDate="2011-03-16T05:12:29.977" LastActivityDate="2011-04-07T07:40:34.023" Title="How to prove the Energy function of a Hopfield net is monotonically decreasing?" Tags="&lt;machine-learning&gt;&lt;descent&gt;" AnswerCount="1" CommentCount="7" />
  <row Id="23558" PostTypeId="2" ParentId="23044" CreationDate="2011-02-24T17:07:57.833" Score="1" ViewCount="" Body="&lt;p&gt;@Eric Why is there an enormous bounty on this question? It has been answered correctly...?&#xA;Yea that was a misclick, can I have it back? Now I cant even comment&lt;/p&gt;&#xA;" OwnerUserId="5904" LastActivityDate="2011-02-24T17:07:57.833" CommentCount="1" />
  <row Id="23559" PostTypeId="2" ParentId="23555" CreationDate="2011-02-24T17:10:08.457" Score="2" ViewCount="" Body="&lt;p&gt;Even linear polynomials get larger (in absolute value) than any fixed number at the variable goes to $\pm \infty$  Please see the discussion cited by lhf in the comment.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-02-24T17:10:08.457" CommentCount="1" />
  <row Id="23560" PostTypeId="2" ParentId="23555" CreationDate="2011-02-24T17:10:18.567" Score="7" ViewCount="" Body="&lt;p&gt;Yes all non-constant polynomials are unbounded.  You could see &lt;a href=&quot;http://en.wikipedia.org/wiki/Liouville%27s_theorem_%28complex_analysis%29&quot; rel=&quot;nofollow&quot;&gt;Liouville's Theorem&lt;/a&gt;, or just notice that the leading coefficient of the polynomial will dominate.  That is if I look at $$p(x)=a_n x^n +a_{n-1}x^{n-1}+\cdots+a_1x+a_0$$ then as $x\rightarrow\infty$ we have $p(x)\sim a_nx^n$ which goes to infinity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit:  I will elaborate, but it may not be needed, and may not clarify things further.  Factoring out $x^n$, we see $$p(x)=x^n\cdot \left( a_n+\frac{a_{n-1}}{x}+\frac{a_{n-2}}{x^2}+\cdots+\frac{a_0}{x^n} \right)$$ taking $x\rightarrow \infty$ we find $$ \left( a_n+\frac{a_{n-1}}{x}+\frac{a_{n-2}}{x^2}+\cdots+\frac{a_0}{x^n} \right)\rightarrow a_n$$ so that $p(x)\sim a_nx^n$.  As $x^n\rightarrow \infty$ when $x\rightarrow \infty$ we see that for nonzero $n$ the polynomial will go to infinity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remark:  The symbol $g(x)\sim f(x)$ is short hand for $\lim_{x\rightarrow \infty}\frac{f(x)}{g(x)}=1$, and reads &quot;f(x) is asymptotic to g(x).&quot;&lt;/p&gt;&#xA;" OwnerUserId="6075" LastEditorUserId="6075" LastEditorDisplayName="" LastEditDate="2011-02-24T17:18:52.913" LastActivityDate="2011-02-24T17:18:52.913" CommentCount="2" />
  <row Id="23561" PostTypeId="2" ParentId="23553" CreationDate="2011-02-24T17:31:23.440" Score="8" ViewCount="" Body="&lt;p&gt;If you even have to think if your audience knows what a particular abbreviation means, then you &lt;em&gt;must&lt;/em&gt; explain it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In any case, writing LHS/RHS in anything but very informal contexts seems simply unacceptable to me.&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2011-02-24T17:31:23.440" />
  <row Id="23562" PostTypeId="2" ParentId="23553" CreationDate="2011-02-24T17:31:52.127" Score="1" ViewCount="" Body="&lt;p&gt;I like to define acronyms like:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;The left hand side (LHS) of Equation 1...&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And then use LHS throughout the rest of the paper if it is used frequently. If you only use &quot;left hand side&quot; a few times, it's best to just spell it out.&lt;/p&gt;&#xA;" OwnerUserId="592" LastActivityDate="2011-02-24T17:31:52.127" />
  <row Id="23563" PostTypeId="2" ParentId="23555" CreationDate="2011-02-24T17:42:30.620" Score="3" ViewCount="" Body="&lt;p&gt;By scaling by $\rm\:x^k\:,\:$ a nonzero rational function (polynomial fraction) $\rm\:f(x)\:$ can be written uniquely in the form $\rm\ c\ x^n\: \bar f(x)\ $ where $\rm\ \lim_{x\to\infty}\:\bar f(x) = 1\:.\:$ The degree $\rm\:n\:$ is known as the order of $\rm\:f\:$ at $\rm\infty\:.\:$ These integer exponents serve as an (archimedean) scale to compare the asymptotic growth of rational functions in a neighborhood of $\rm\:\infty\:.\:$ More generally one can extend to wider classes of asymptotically well-behaved functions such as $\rm\ log(x),\ exp(x)\ $ etc to obtain more general (non-archimedean) scales that measure so-called &quot;orders of infinity&quot;. For example see Hardy's classic textbook by that name, google &quot;transseries&quot;, and see my post on &lt;a href=&quot;http://math.stackexchange.com/questions/9911/displaystyle-sum-n-1-infty-frac1n-logsn/9915#9915&quot;&gt;boundaries of convergence.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2011-02-24T21:37:44.977" LastActivityDate="2011-02-24T21:37:44.977" />
  <row Id="23564" PostTypeId="1" AcceptedAnswerId="23570" CreationDate="2011-02-24T17:42:48.110" Score="8" ViewCount="204" Body="&lt;p&gt;In my question about the convergence/divergence of &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;\sum_{n=2}^\infty \frac{1\cdot 3\cdot 5\cdot 7\cdots (2n-3)}{2^nn!}.&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;here:  &lt;a href=&quot;http://math.stackexchange.com/questions/9372/why-doesnt-this-series-converge&quot;&gt;Why Doesn't This Series Converge?&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Zarrax gave the answer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;You can use Taylor approximations here. Note that the ratio between consecutive terms is ${2n - 3 \over 2n} = \exp(\ln(1 - 3/2n)) = \exp(-{3 \over 2n} + O(1/n^2))$. So the product is comparable to $\exp(-{3 \over 2} \sum_{i = 2}^n {1 \over n} + O(1/n))$, which in turn is comparable to $\exp(-{3 \over 2} \ln(n))$ or $n^{-{3 \over 2}}$. Thus the series converges.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've decided to give a talk in a graduate seminar about the danger of coming up with examples off the top of your head and now wish to understand what this answer means.  The problem is that I have no exposure to big O notation and am not having luck online.  Basically, I don't understand the answer at all.  I can break it into a few questions:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;How does Zarrax pass from $\exp(\ln(1 - 3/2n))$ to $\exp(-{3 \over 2n} + O(1/n^2))$?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;To which product is Zarrax referring in his third sentence?  And how is it comparable to $\exp(-{3 \over 2} \sum_{i = 2}^n {1 \over n} + O(1/n))$?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;How does Zarrax pass from that to $\exp(-{3 \over 2} \ln(n))$?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Thank you for your help.&lt;/p&gt;&#xA;" OwnerUserId="2838" LastActivityDate="2011-02-24T18:35:56.810" Title="How to Use Big O Notation" Tags="&lt;asymptotics&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="23566" PostTypeId="1" AcceptedAnswerId="23603" CreationDate="2011-02-24T18:04:19.143" Score="19" ViewCount="6810" Body="&lt;p&gt;I love math, and I used to be very good at it. The correct answers came fast and intuitively. I never studied, and redid the demonstration live for the tests (sometimes inventing new ones). I was the one who answered the tricky questions in class (8 hours of math/week in high school)... You get the idea.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As such I used to have a lot of confidence in my math abilities, and didn't think twice about saying the first idea that came to mind when answering a question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That was more than 10 years ago, and I (almost) haven't done any math since then. I've graduated in a scientific field that requires little of it (I prefer not to give details) and worked for some time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I'm back at school (master of statistics) and I need to do math, once again. I make mistakes upon blunders with the same confidence I used to have when I was good, which is extremely embarrassing when it happens in class. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I feel like a tone deaf musician and an ataxic painter at the same time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One factor that probably plays a role is that I've learnt math in my mother tongue, and I'm now using it English, but I wouldn't expect it to make such a difference. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that it will require practice and hard work, but I need direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any help is welcome.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Kind regards,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-- Mathemastov&lt;/p&gt;&#xA;" OwnerUserId="7466" LastActivityDate="2011-04-08T20:12:36.893" Title="Formerly good at math, but after 12 years I've lost most of my skills. Now I need them once again. Any advice to grow them back?" Tags="&lt;soft-question&gt;&lt;intuition&gt;&lt;self-learning&gt;&lt;learning&gt;" AnswerCount="13" CommentCount="1" FavoriteCount="15" />
  <row Id="23567" PostTypeId="2" ParentId="23433" CreationDate="2011-02-24T18:08:13.163" Score="4" ViewCount="" Body="&lt;p&gt;Here's another take.  Since $\cos \left(\frac{1}{xt}\right)$ is an even function of $x$, we can take the limit from the right or from the left, and the result is the same.  So assume $x &amp;gt; 0$.  Then apply the $u = 1/(xt)$ substitution as in joriki's answer, followed by integration by parts in the opposite direction from joriki's answer.  Rewriting the result in terms of the sine integral produces the output from Wolfram Alpha mentioned by Eivind and Sivaram: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \lim_{x \to 0} \left(\frac{1}{x} \int_0^{1/x} \frac{\sin u}{u} du  - \frac{\pi}{2x} + \cos \left(\frac{1}{x}\right)\right).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, apply the following known &lt;a href=&quot;http://en.wikipedia.org/wiki/Trigonometric_integral#Asymptotic_series_.28for_large_argument.29&quot; rel=&quot;nofollow&quot;&gt;series expansion for the sine integral&lt;/a&gt; $\int_0^x \frac{\sin u}{u} du$  (valid for large values of $x$):&#xA;$$\int_0^x \frac{\sin u}{u} du = \frac{\pi}{2} -\frac{\cos x}{x} \left(1 - \frac{2!}{x^2} + \frac{4!}{x^4} \pm \cdots \right) - \frac{\sin x}{x} \left(\frac{1}{x} - \frac{3!}{x^3} \pm \cdots \right).$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;After simplifying, we are left with &#xA;$$\lim_{x \to 0} \left(-\cos \left(\frac{1}{x}\right) \left(- 2! x^2 + 4! x^4 \pm \cdots \right) - \sin\left(\frac{1}{x}\right) \left(x - 3! x^3 \pm \cdots \right) \right).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $\cos \left(\frac{1}{x}\right)$ and $\sin \left(\frac{1}{x}\right)$ are both bounded by $-1$ and $1$, the limit is $0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In addition, we can see that the dominant term in the limit is $- x\sin\left(\frac{1}{x}\right)$, which is clearly bounded by $|x|$, as conjectured by Sivaram in the comments.&lt;/p&gt;&#xA;" OwnerUserId="2370" LastActivityDate="2011-02-24T18:08:13.163" />
  <row Id="23568" PostTypeId="2" ParentId="23566" CreationDate="2011-02-24T18:10:40.283" Score="1" ViewCount="" Body="&lt;p&gt;Revisit all of your High school books. It will be easier to grasp and remind you of many stuff you forgot.&lt;/p&gt;&#xA;" OwnerUserId="6335" LastActivityDate="2011-02-24T18:10:40.283" />
  <row Id="23570" PostTypeId="2" ParentId="23564" CreationDate="2011-02-24T18:22:45.250" Score="8" ViewCount="" Body="&lt;p&gt;To answer your questions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) He is using $\log(1-x) = -x-x^2/2-x^3/3 - \cdots = -x + O(x^2).$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) The product he's referring to is&#xA;$$\frac{1\cdot 3\cdot 5\cdot 7\cdots (2n-3)}{2^nn!}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and it's comparable to $\exp(-{3 \over 2} \sum_{i = 2}^n {1 \over n} + O(1/n))$ because this is the product of $n-1$ versions of $\exp(-{3 \over 2n} + O(1/n^2)).$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) He uses $\sum_{i=2}^n 1/i = \log n + \text{constant} + O(1/n).$&lt;/p&gt;&#xA;" OwnerUserId="1301" LastActivityDate="2011-02-24T18:22:45.250" />
  <row Id="23571" PostTypeId="2" ParentId="23564" CreationDate="2011-02-24T18:23:15.083" Score="6" ViewCount="" Body="&lt;p&gt;When an expression uses $\mathcal{O}(1/n^2)$ it means that this is actually a function $f$ such that $f \in \mathcal{O}(1/n^2)$. This is a very convenient abuse of notation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that $\mathcal{O}(1/n^2)$ is actually a &lt;em&gt;set&lt;/em&gt; of functions. A function $f \in \mathcal{O}(1/n^2)$ iff there is a constant $C$ (dependent on $f$) such that $|f(n)| \le \frac{C}{n^2}$ for all sufficiently large $n$, i.e. as $n \to \infty$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: we are talking about positive integers for now, the definitions are also valid when $n$ is real or when $n \to A$, where $A$ need not be $\infty$, in which case, we talk about the inequality holding in a neighbourhood of $A$. Usually, what $A$ is, is clear from the context and is left out (another convenience).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This symbol is called BigOh (as you seem to know already). You can find more information about that here: &lt;a href=&quot;http://en.wikipedia.org/wiki/Big_O_notation&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Big_O_notation&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So when you use $\mathcal{O}(1/x^2)$, in the expression &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ln(1-\frac{1}{x}) = \frac{1}{x} + \mathcal{O}(1/x^2)$ what we really mean is that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for some $f \in \mathcal{O}(1/x^2)$ we have that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ln(1-\frac{1}{x}) = \frac{1}{x} + f(x)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now you can read Derek's &lt;a href=&quot;http://math.stackexchange.com/questions/23564/how-to-use-big-o-notation/23570#23570&quot;&gt;answer&lt;/a&gt; :-) I guess I don't have to repeat it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that when Derek says&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ln(1- x) = x + \mathcal{O}(x^2)$, he is talking about $x$ in the neigbourhood of $0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are plenty of other answers here which use the BigOh.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are a few of them (I especially recommend you read the first one)&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://math.stackexchange.com/questions/3215/convergence-of-sqrtnx-n-where-x-n1-sinx-n/3220#3220&quot;&gt;Convergence of $\sqrt{n}x_{n}$ where $x_{n+1} = \sin(x_{n})$&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://math.stackexchange.com/questions/13300/big-mathcalo-notation-question-while-estimating-sum-frac-log-nn&quot;&gt;Big $\mathcal{O}$ Notation question while estimating $\sum \frac{\log n}{n}$&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://math.stackexchange.com/questions/12566/proving-sum-limits-p-leq-x-frac1-sqrtp-geq-frac12-logx-log/12588#12588&quot;&gt;Proving $\sum\limits_{p \leq x} \frac{1}{\sqrt{p}} \geq \frac{1}{2}\log{x} -\log{\log{x}}$&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://math.stackexchange.com/questions/14515/what-is-the-expression-of-n-that-equals-to-sum-i-1n-frac1i2/14518#14518&quot;&gt;What is the expression of $n$ that equals to $\sum_{i=1}^n \frac{1}{i^2}$?&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2011-02-24T18:35:56.810" LastActivityDate="2011-02-24T18:35:56.810" />
  <row Id="23572" PostTypeId="1" AcceptedAnswerId="23578" CreationDate="2011-02-24T18:41:18.543" Score="1" ViewCount="76" Body="&lt;p&gt;That is to ask, is $e^{\ln(q_0)}$ = $\ln(e^{q_0})$?&lt;/p&gt;&#xA;" OwnerUserId="7382" LastActivityDate="2011-02-24T19:17:48.407" Title="For quaternions, is the natural log the inverse of the exponential function?" Tags="&lt;quaternions&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="23573" PostTypeId="1" AcceptedAnswerId="23576" CreationDate="2011-02-24T19:00:35.533" Score="0" ViewCount="64" Body="&lt;p&gt;$$\frac{\ln(T+1)-\ln(V-S+1)}{\ln(V-S+1)}=\frac{1}{K}-1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are the steps if I want to solve for $V$?&lt;/p&gt;&#xA;" OwnerUserId="7468" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2011-02-24T19:15:11.053" LastActivityDate="2011-02-24T20:31:49.863" Title="How do I solve for $V$ in this equation?" Tags="&lt;algebra&gt;&lt;algebra-precalculus&gt;&lt;logarithms&gt;" AnswerCount="3" />
  <row Id="23574" PostTypeId="1" AcceptedAnswerId="23577" CreationDate="2011-02-24T19:04:07.867" Score="2" ViewCount="68" Body="&lt;p&gt;Hello,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is the beginning of a proof that&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Every abelian finite group $(B,+)$ with $|B|=p^r$ ($p$ prime, $r\in\mathbb{N}$) is isomorphic to a direct product of cyclic groups.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;By induction on $r$. If $r=1$, then $B$ is cyclic. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Else let $b\in B$ of maximal order and $B_1$ the cyclic group generated by $b$. We use the induction hypothesis on $\bar B=B/B&amp;#39;$:&#xA;$$\bar B\cong \bar B_1\times \bar B_2\times\cdots \times \bar B_k$$&#xA;where $\bar B_1, \dots, \bar B_k$ are cyclic groups.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then the proof uses that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\bar B_1+ \bar B_2+\cdots+ \bar B_k\cong \bar B_1\times \bar B_2\times\cdots\times \bar B_k \ \small (1)$$&#xA;to prove that there exists cyclic sub-groups of $B$, $B_1, \dots, B_k$ such that $$B_1+\cdots+B_k\cong B_1\times\cdots\times B_k$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The isomorphism $\small (1)$ should be obvious as it is given without any details, but I am unable to understand why it holds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could someone enlighten me ?&lt;/p&gt;&#xA;" OwnerUserId="524" LastEditorUserId="524" LastEditorDisplayName="" LastEditDate="2011-02-24T20:20:54.993" LastActivityDate="2011-02-24T20:20:54.993" Title="Decomposition of abelian finite $p-$groups" Tags="&lt;abstract-algebra&gt;&lt;abelian-groups&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="23575" PostTypeId="2" ParentId="23573" CreationDate="2011-02-24T19:12:38.290" Score="6" ViewCount="" Body="&lt;p&gt;Set $x = \ln(V-S+1)$ and solve the equation (which is linear in $x$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then use $V = e^x + S - 1$.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2011-02-24T19:12:38.290" CommentCount="1" />
  <row Id="23576" PostTypeId="2" ParentId="23573" CreationDate="2011-02-24T19:13:06.900" Score="4" ViewCount="" Body="&lt;p&gt;Hint:  Get rid of the second term of the numerator on the LHS as -1 and move to RHS.  A little algebra will leave $\ln(V-S+1)$ on the RHS and all else on the LHS.  Then exponentiate both sides.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-02-24T19:13:06.900" />
  <row Id="23577" PostTypeId="2" ParentId="23574" CreationDate="2011-02-24T19:15:10.443" Score="1" ViewCount="" Body="&lt;p&gt;You are identifying the cyclic group $\overline{B}_i$ with the subgroup&#xA;$$\{(\overline{b}_1,\ldots,\overline{b}_k)\in \overline{B}\mid b_j=0\mbox{ for }j\neq i\}.$$&#xA;That is, you are going from the &quot;external&quot; direct product to the &quot;internal&quot; direct product.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The corresponding subgroups of $\overline{B}$ are disjoint ($\overline{B}_i\cap\mathop{\oplus}\limits_{j\neq i}\overline{B}_j = {0}$), and span $\overline{B}$. So you get that the sum equals $\overline{B}$, and hence is isomorphic to the direct product/sum of the groups.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2011-02-24T19:15:10.443" CommentCount="5" />
  <row Id="23578" PostTypeId="2" ParentId="23572" CreationDate="2011-02-24T19:17:48.407" Score="2" ViewCount="" Body="&lt;p&gt;$e^{\ln(q_0)} = q_0$ always holds for all branches of the logarithm and even for square matrices (because this is essentially the definition of the log).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ln(e^{q_0}) = q_0$ is already for complex numbers not always true (because the logarithm is multivalued)&lt;/p&gt;&#xA;" OwnerUserId="7266" LastActivityDate="2011-02-24T19:17:48.407" />
  <row Id="23579" PostTypeId="1" AcceptedAnswerId="23585" CreationDate="2011-02-24T19:28:08.063" Score="4" ViewCount="121" Body="&lt;p&gt;The following theorem is in Rudin's &lt;em&gt;Principles of Mathematical Analysis&lt;/em&gt;, 8.2: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose $c_n \geq 0$ and $S = \sum c_n$ converges. Put $f(x) = \sum_{n=0}^{\infty} c_nx^n$. Show that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;\lim_{x \rightarrow 1} f(x) = \sum_{n=0}^{\infty} c_n.&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rudin goes on to give a somewhat involved proof using Abel Summation. I'm wondering why this is necessary; why not just this argument? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Proof:&quot; Pick $N$ such that $\sum_{n &amp;gt; N} c_n &amp;lt; \epsilon$. Pick $r$ such that $1-r^N &amp;lt; \epsilon/S$. Then we have &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;\sum_{n=0}^{\infty} c_n - \sum_{n=0}^{\infty} c_nr^n = \sum_{n=0}^N c_n(1-r^n) + \sum_{n&amp;gt;N} c_n(1-r^n) &amp;lt; 2 \epsilon.&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Briefly, Rudin's method involves using Abel summation and then bounding the new sequence of partial sums, which seems like a less natural approach to me. Does anyone have a reason why he did that?&lt;/p&gt;&#xA;" OwnerUserId="1491" LastActivityDate="2011-02-25T05:00:26.620" Title="Proving this convergence of series" Tags="&lt;sequences-and-series&gt;&lt;convergence&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="0" />
  <row Id="23580" PostTypeId="2" ParentId="23566" CreationDate="2011-02-24T19:30:46.467" Score="16" ViewCount="" Body="&lt;p&gt;I think the key to your problem is in your first paragraph. You say, &quot;The correct answers came fast and intuitively. I never studied.&quot; This is the classic high school con that can lead one to doubt one's own abilities as soon as the going gets more challenging.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;No matter what your abilities, to do worthwhile work in mathematics you will need to study and work hard. You say in your last paragraph that you know it will require practice and hard work, but I do not think you have fully taken this on board.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you do, you will stop worrying about making mistakes. They are an essential part of the learning process and not something to beat yourself up about. Forget the belief, instilled in high school, that you should be able to come up with the correct answer straight away. Those questions were crafted to have short, snappy answers within the reach of rote learning.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your mathematical career has now gone beyond this stage. So get down to study and make as many mistakes as you like on the path to a deeper understanding, and enjoy!&lt;/p&gt;&#xA;" OwnerUserId="1301" LastActivityDate="2011-02-24T19:30:46.467" CommentCount="1" />
  <row Id="23581" PostTypeId="1" AcceptedAnswerId="23582" CreationDate="2011-02-24T19:34:06.710" Score="0" ViewCount="58" Body="&lt;p&gt;I am trying to write an equation to solve my problem of figuring out which post is most popular. While one would think the one with the highest stars would be, if the stars are equal then the one with 61 votes should rise higher than the one with 5 for example (since more people are downloading that)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Star system is 1-5&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I thought of something like numberofvotes*star&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would that work?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The issue comes in that something with less stars should never rank higher, so the number of stars is always the first determining factor.&lt;/p&gt;&#xA;" OwnerUserId="1595" LastEditorUserId="5191" LastEditorDisplayName="" LastEditDate="2011-02-24T19:38:45.657" LastActivityDate="2011-02-24T19:46:02.353" Title="Equation for Ranking Top Posts" Tags="&lt;statistics&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="23582" PostTypeId="2" ParentId="23581" CreationDate="2011-02-24T19:38:38.077" Score="1" ViewCount="" Body="&lt;p&gt;No. But assume that N is a large number larger than the maximum number of votes you ever expect. Then &quot;numberofvotes + N*star&quot;&#xA;would work.&lt;/p&gt;&#xA;" OwnerUserId="7266" LastActivityDate="2011-02-24T19:38:38.077" CommentCount="2" />
  <row Id="23584" PostTypeId="2" ParentId="23581" CreationDate="2011-02-24T19:46:02.353" Score="1" ViewCount="" Body="&lt;p&gt;Your formula will not work if you think 5 five-star votes (5 in your formula) is better than 61 one-star votes (12.2 in your formula)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An alternative might be something like $\frac{20 + \textrm{total number of stars}}{10 +  \textrm{number of voters}}$ which might say that an unvoted post is assumed to be equivalent in quality to one with lots of two-star votes.  If it then gets lots of five-star votes it will increase, and with lots of one-star votes will decrease.  Going back to your example, 5 five-star votes would be roughly the same as a very large number of three-star votes.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can adjust the 20 and 10 to suit your particular needs.   &lt;/p&gt;&#xA;" OwnerUserId="6460" LastActivityDate="2011-02-24T19:46:02.353" />
  <row Id="23585" PostTypeId="2" ParentId="23579" CreationDate="2011-02-24T19:48:15.300" Score="4" ViewCount="" Body="&lt;p&gt;In my copy of Rudin there is no assumption that $c_n \ge 0$, and your proof fails if you can't assume that $\sum c_n$ converges absolutely. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Indeed, the power of this theorem is precisely that it applies in situations where $\sum c_n$ does &lt;em&gt;not&lt;/em&gt; converge absolutely. For example it applies to $\arctan x = x - \frac{x}{3} + \frac{x}{5} \mp ... $ to give the classic formula $\frac{\pi}{4} = 1 - \frac{1}{3} + \frac{1}{5} \mp ...$. &lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2011-02-24T19:55:30.113" LastActivityDate="2011-02-24T19:55:30.113" CommentCount="2" />
  <row Id="23586" PostTypeId="2" ParentId="23544" CreationDate="2011-02-24T19:55:25.447" Score="4" ViewCount="" Body="&lt;p&gt;With $R=\mathbb{Z}$, you get an isomorphism. Note that $(\mathbb{Z}\times\mathbb{Z},\circ)$ is torsionfree abelian: given $(a,b)\in\mathbb{Z}\times\mathbb{Z}$, if $a\neq 0$, then $n(a,b) = (na,x)$ for some $x$, so $n(a,b)=(0,0)$ requires $n=0$. If $a=0$, then $n(0,b) = (0,nb)$, so again $n(0,b)=0$ requires $n=0$ or $b=0$. (I'm using $m(x,y)$ to mean $(x,y)$ $\circ$-added to itself $m$ times if $m\gt 0$, and the $\circ$-inverse if $m\lt 0$, as usual for an abelian group).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, $(\mathbb{Z}\times\mathbb{Z},\circ)$ is $2$-generated: $(1,0)$ and $(0,1)$ certainly generate: to get an arbitrary $(a,b)$ using $(1,0)$ and $(0,1)$, just take $a(1,0)$, which will give an element of the form $(a,x)$ for some $x$, and then take $(a,x)\circ(0,b-x)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So $(\mathbb{Z}\times\mathbb{Z},\circ)$ is free abelian and $2$-generated, hence either cyclic or isomorphic to $\mathbb{Z}\times\mathbb{Z}$. Moding out by the infinite cyclic normal subgroup $\{(0,b)\mid b\in\mathbb{Z}\}$ we get a group isomorphic to $\mathbb{Z}$, so the group $(\mathbb{Z}\times\mathbb{Z},\circ)$ cannot be infinite cyclic (because the quotient of an infinite cyclic group by a nontrivial subgroup is finite). Thus, with $R=\mathbb{Z}$ you get an isomorphism.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More generally, if you have an isomorphism for $R$ and one for $S$, then you also get an isomorphism for $R\oplus S$, since the addition will just be &quot;coordinate wise&quot;, so you can take the isomorphism $f\colon (R\times R,\circ)\to R\oplus R$ and $g\colon (S\times S,\circ)\to S\oplus S$, and obtain an isomorphism $f\times g\colon (R\oplus S\times R\oplus S,\circ)\to (R\oplus R)\times (S\oplus S)\cong (R\oplus S)\times (R\oplus S)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This will give you isomorphisms for all rings of the form&#xA;$$\mathbb{Z}^t \times\frac{\mathbb{Z}}{p_1^{a_1}\mathbb{Z}}\times\cdots\times\frac{\mathbb{Z}}{p_r^{a_r}\mathbb{Z}}$$&#xA;where $p_1,\ldots,p_r$ are primes, different from $3$, $a_i\gt 0$, $t,r\geq 0$, and the ring structure is the obvious one.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Added&lt;/em&gt;. (Well that was silly). For rings of characteristic $3$, you get an isomorphism if and only if $a^3=0$ for all $a\in R$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To see that this is necessary, notice that &#xA;\begin{align*}&#xA;3(a,0) &amp;amp;= (a,0)\circ(a,0)\circ(a,0) = (2a,2a^3)\circ(a,0)\\&#xA;&amp;amp;= (3a,2x^3 + 4a^3+2a^3) = (0,8a^3) = (0,-a^3).&#xA;\end{align*}&#xA;Thus, $(R\times R,\circ)$ has characteristic $3$ if and only if $a^3=0$ for all $a$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To see that the condition is sufficient, if $R$ is of characteristic $3$ and $a^3=0$ for all $a\in R$,then $(R\times R,\circ)$ is an abelian $3$-group, hence a vector space over $\mathbb{F}_3$. Since it has the same cardinality as $R\oplus R$, which is also a vector space over $\mathbb{F}_3$, they are isomorphic as $\mathbb{F}_3$-vector spaces, and hence as abelian groups. &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-25T00:44:47.227" LastActivityDate="2011-02-25T00:44:47.227" CommentCount="5" />
  <row Id="23587" PostTypeId="1" CreationDate="2011-02-24T19:56:10.860" Score="2" ViewCount="126" Body="&lt;p&gt;$$ \sum_{i=0}^{n-2}\left(\sum_{j=i+1}^{n-1} i\right) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Formulas in my book give me equations to memorize and solve simple questions like $$ \sum_{i=0}^{n} i $$ ... However, For the question on top, how would I go about solving it by hand without a calculator? WolfRamAlpha seems to give the equation of 1/6[(n-2)(n-1)n].&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any suggestions would be appreciated. It's not a homework question, but I am studying for a test. I wrote the mathematical version of two nested for-loops for code that checks to see if a number in an array is unique or not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you.&lt;/p&gt;&#xA;" OwnerUserId="3800" LastActivityDate="2011-02-24T20:19:52.033" Title="Question on Solving a Double Summation" Tags="&lt;algorithms&gt;&lt;sequences-and-series&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="23588" PostTypeId="1" AcceptedAnswerId="23592" CreationDate="2011-02-24T20:00:46.683" Score="0" ViewCount="150" Body="&lt;p&gt;Is it true that, in a Dedekind Domain, all maximal ideals are prime? &lt;/p&gt;&#xA;" OwnerUserId="2212" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-24T21:16:57.297" LastActivityDate="2011-02-25T05:52:02.263" Title="Is it true that, in a Dedekind Domain, all maximal ideals are prime?" Tags="&lt;commutative-algebra&gt;&lt;ring-theory&gt;" AnswerCount="3" />
  <row Id="23589" PostTypeId="1" CreationDate="2011-02-24T20:01:25.050" Score="3" ViewCount="195" Body="&lt;p&gt;I know that the recurrence $\displaystyle a(n+1)=a(n)(n-1/2)$ can be represented like $\displaystyle \frac{(2n-1)!!}{2^n}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Actually the initial recurrence is slightly different:&#xA;$$\displaystyle a(n+1)=a(n)(n-1/2)+o(1/n)$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am looking  for the new representation of it in terms of $\displaystyle \frac{(2n-1)!!}{2^n}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Probably the question has to be refined. The recurrence is very close to the property of Kendall-Mann numbers &lt;a href=&quot;http://oeis.org/A181609&quot; rel=&quot;nofollow&quot;&gt;http://oeis.org/A181609&lt;/a&gt;&#xA;that is why I am trying to find an interval of $\displaystyle n$'s when the $\displaystyle \frac{(2n-1)!!}{2^n}$ or similar representation works fine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand that $\displaystyle o(1/n)$ is quite big error for the representation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am also interested in the case the recurrence is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle a(n+1)=a(n)(n-1/2+o(1/n))$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is closer to the recurrence satisfied by the Kendall-Mann numbers.&lt;/p&gt;&#xA;" OwnerUserId="7418" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2011-02-27T23:35:27.790" LastActivityDate="2011-02-27T23:38:08.210" Title="Recurrence representation(s): $a(n+1)=a(n)(n-1/2)+o(1/n)$ and $a(n+1)=a(n)(n-1/2+o(1/n))$" Tags="&lt;number-theory&gt;&lt;combinatorics&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="23590" PostTypeId="2" ParentId="23588" CreationDate="2011-02-24T20:04:10.157" Score="1" ViewCount="" Body="&lt;p&gt;It is true in every ring that maximal ideals are prime (because the quotient is a field, hence also an integral domain).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is true in Dedekind domains however (by definition) is that every nonzero prime ideal is maximal. See also the Wikipedia article. &lt;a href=&quot;http://en.wikipedia.org/wiki/Dedekind_domain&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Dedekind_domain&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt;&#xA;To add a little more substance to this answer, here are two examples of Dedekind domains. First, the obvious one: $\mathbb{Z}$. Indeed, for every prime ideal $(p)$, we have $\mathbb{Z}/(p)$ is a finite integral domain, hence a field. So every non-zero prime ideal is maximal. It is easy to see that $\mathbb{Z}$ is Noetherian (in fact, of dimension 1), and integrally closed (as is any PID).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second example: $\mathbb{Z}[i]\cong \mathbb{Z}[x]/(x^2+1)$. By Hilbert's basis theorem, this is Noetherian (overkill-argument!), it is integrally closed (since it is the number field of $\mathbb{Q}(i)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the algebraic closure of $\mathbb{Z}$, namely $\bar{\mathbb{Z}}$ is &lt;strong&gt;not&lt;/strong&gt; a Dedekind domain. Unique factorization fails miserably: $a=\sqrt{a} \sqrt{a}=\sqrt{\sqrt{a}}\sqrt{\sqrt{a}}$ and so on. Another way to see it is to consider the ideals $(a)^{\frac{1}{i}}$ for $i=1,2,3,\ldots$. This is a non-terminating increasing chain of ideals, hence $\bar{\mathbb{Z}}$ is not Noetherian.&lt;/p&gt;&#xA;" OwnerUserId="4284" LastEditorUserId="4284" LastEditorDisplayName="" LastEditDate="2011-02-25T05:52:02.263" LastActivityDate="2011-02-25T05:52:02.263" CommentCount="5" />
  <row Id="23591" PostTypeId="2" ParentId="23588" CreationDate="2011-02-24T20:04:47.293" Score="10" ViewCount="" Body="&lt;p&gt;In &lt;em&gt;any&lt;/em&gt; commutative ring with identity, all maximal ideals are prime. An ideal is maximal if and only if the quotient ring is a field; an ideal is prime if and only if the quotient ring is an integral domain. Fields are always integral domains. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(In fact, in any ring with identity, commutative or not, maximal ideals are prime ideals: if $\mathfrak{A}\mathfrak{B}\subseteq \mathfrak{M}$, with $\mathfrak{M}$ maximal, and $\mathfrak{A}\not\subset\mathfrak{M}$, then $\mathfrak{M}+\mathfrak{A}=R$; hence &#xA;$$\mathfrak{B} = R\mathfrak{B} = (\mathfrak{A}+\mathfrak{M})\mathfrak{B} = \mathfrak{A}\mathfrak{B} + \mathfrak{MB}\subseteq \mathfrak{M},$$&#xA;so $\mathfrak{B}\subseteq\mathfrak{M}$, proving that $\mathfrak{M}$ is prime.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Did you perhaps mean it the other way? &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Is it true that in a Dedekind Domain, all &lt;em&gt;nonzero&lt;/em&gt; prime ideals are maximal?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The answer there is &quot;yes&quot; as well, but the reason will depend on which definition of Dedekind domain you have. One definition is that they are domains in which every ideal is finitely generated, every nonzero prime ideal is maximal, and the domain is integrally closed in its field of fractions, which would make the answer true &quot;by definition&quot;, but there are other, equivalent, definitions. So if this is what you meant to ask, you'll have to specify what your definition is if you want elucidation. &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-24T20:45:28.957" LastActivityDate="2011-02-24T20:45:28.957" CommentCount="2" />
  <row Id="23592" PostTypeId="2" ParentId="23588" CreationDate="2011-02-24T20:05:04.157" Score="3" ViewCount="" Body="&lt;p&gt;Maximal ideals are prime in every ring. Probably you mean the converse, which is true since &quot;divides = contains&quot; in Dedekind domains. So an ideal properly containing a prime ideal would be one properly dividing it, so it can only be 1.&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2011-02-24T20:05:04.157" CommentCount="2" />
  <row Id="23593" PostTypeId="2" ParentId="23587" CreationDate="2011-02-24T20:06:48.090" Score="2" ViewCount="" Body="&lt;p&gt;Since the sum $$\sum_{j=i+1}^{n-1} i $$ does not depend on $j$ we see $$\sum_{j=i+1}^{n-1} i = i\cdot\sum_{j=i+1}^{n-1} 1= i(n-1-i) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then you have to find $$\sum_{i=0}^{n-2} \left( i(n-1-i)\right)=\sum_{i=0}^{n-2} \left( (n-1)i-i^2)\right)=(n-1) \sum_{i=0}^{n-2} i- \sum_{i=0}^{n-2} i^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you solve it from here?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope that helps,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit:  Perhaps you wanted a $j$.  In other words, lets evaluate $$ \sum_{i=0}^{n-2}\left(\sum_{j=i+1}^{n-1} j\right) $$&#xA;Reversing the order of summation yields:&#xA;$$ \sum_{j=1}^{n-1}\left(\sum_{i=j-1}^{n-2} j\right) $$ which can be solved by the exact same method presented above.&lt;/p&gt;&#xA;" OwnerUserId="6075" LastEditorUserId="6075" LastEditorDisplayName="" LastEditDate="2011-02-24T20:19:52.033" LastActivityDate="2011-02-24T20:19:52.033" />
  <row Id="23594" PostTypeId="2" ParentId="7195" CreationDate="2011-02-24T20:18:58.073" Score="1" ViewCount="" Body="&lt;p&gt;Using the substitution $x=\sin(u)$, you get your problem in the form&#xA;$\int\!dx\, \frac{\sqrt{(1-mx^2)(1-x^2)}}{x^2 (1-x^2)}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Legendre studied the problem of solving integrals which involve  rational functions of $x$ and the square root of a polynomial of degree less or equal four in $x$. You will find a recipe on the page &lt;a href=&quot;http://everything2.com/title/elliptic+integral+standard+forms&quot; rel=&quot;nofollow&quot;&gt;http://everything2.com/title/elliptic+integral+standard+forms&lt;/a&gt;. Following the argument, noting that in your case&#xA;$A=0, B=1, C=x^2-x^4, D=1, R=(1-mx^2)(1-x^2)$, you will obtain the result as a function of the three elliptic integrals.&lt;/p&gt;&#xA;" OwnerUserId="7266" LastActivityDate="2011-02-24T20:18:58.073" />
  <row Id="23595" PostTypeId="2" ParentId="23589" CreationDate="2011-02-24T20:19:34.373" Score="2" ViewCount="" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;For the recurrence $\displaystyle a_{n+1} = (n - 1/2) a_n + o(1/n)$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you set $\displaystyle a_n = \frac{(2n-1)!!}{2^n} - e_n$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;you get $\displaystyle e_{n+1} - e_n = o\left(\frac{1}{n}\right)$ and thus $e_n = o(\log n)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using Stirling's approximation, I believe we have that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle \frac{(2n-1)!!}{2^n} \approx \frac{Kn!}{\sqrt{n}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus we get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\lim_{n \to \infty} \frac{2^n a_n}{(2n-1)!!} = 1 $$&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Now if the recurrence was &#xA;  $\displaystyle a_{n+1} = a_{n} (n - 1/2 + o(1/n))$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Set $\displaystyle a_{n+1} = e_n\frac{(2n-1)!!}{2^n}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle \frac{e_{n+1}}{e_n} = 1 + o(1/n^2)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And thus&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle  e_n = e_1 \prod_{k=1}^{n-1} (1 + o(1/k^2))$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now using $1 + x \le e^x$ we get that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle e_n = \mathcal{O}(1)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and we can easily show that $\displaystyle e_n$ is convergent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$a_n \sim \frac{C(2n-1)!!}{2^n}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for some constant $\displaystyle C = \lim_{n \to \infty} e_n$ .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We could chose your $\displaystyle f(n) = o(1/n)$ as we like, to make this constant different from $\displaystyle 1$.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2011-02-27T23:38:08.210" LastActivityDate="2011-02-27T23:38:08.210" CommentCount="13" />
  <row Id="23596" PostTypeId="1" AcceptedAnswerId="29230" CreationDate="2011-02-24T20:25:22.530" Score="5" ViewCount="241" Body="&lt;p&gt;If I have a covariance matrix for a data set and I multiply it times one of it's eigenvectors.  Let's say the eigenvector with the highest eigenvalue.  The result is the eigenvector or a scaled version of the eigenvector.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What does this really tell me?  Why is this the principal component?  What property makes it a principal component?  Geometrically, I understand that the principal component (eigenvector) will be sloped at the general slope of the data (loosely speaking).  Again, can someone help understand why this happens?  &lt;/p&gt;&#xA;" OwnerUserId="6764" LastActivityDate="2011-03-27T02:51:39.703" Title="Why is the eigenvector of a covariance matrix equal to a principal component?" Tags="&lt;linear-algebra&gt;&lt;eigenvectors&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="23597" PostTypeId="2" ParentId="23193" CreationDate="2011-02-24T20:26:37.223" Score="11" ViewCount="" Body="&lt;p&gt;I believe this is a proof that you can't do better than $2n-2$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $m$ be the number of horizontal and vertical lines in the path. Then, since there are $2n$ rows and columns, there are $2n-m$ rows and columns that don't contain a line along their length; say, $k$ rows and $l$ columns with $k+l=2n-m$. Imagine the grid points where these rows and columns intersect taken out of the grid and arranged in a rectangle of $k$ rows and $l$ columns. Since each of the points in this rectangle is both in a row and in a column that doesn't contain a line along its length, each of these points must be on some line that is neither horizontal nor vertical.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $k=0$, then we have $n$ horizontal lines, and these must be connected by at least $n-1$ non-horizontal lines to make an unbroken path. Thus, for $k=0$ there must be $2n-1$ lines, so we can exclude that case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $k=1$, then we have $n-1$ horizontal lines, and we have to connect each of the $n$ grid points on the remaining row with a separate line, so again we must have at least $2n-1$ lines, so we can exclude that case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus $k\ge2$, and likewise $l\ge2$. Thus we may consider the border of the rectangle, i.e. the two outermost rows and the two outermost columns. This border consists of $2k+2l-4=2(k+l-2)=2(2n-m-2)$ grid points. Every line that doesn't run along the length of one of these rows or columns can hit at most two of these points. That means there must be at least $2n-m-2$ lines in addition to the $m$ horizontal and vertical lines, for a total of $2n-2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S.: You can check this out in the $8\times8$ and $10\times10$ solutions linked to in the question -- in each case, the &quot;rectangle&quot; constructed in the proof is actually a contiguous rectangle in the grid (with $k=6$, $l=8$ and $k=10$, $l=2$, respectively), and you can see that the path optimally connects two points of that rectangle's border with each line that isn't horizontal or vertical.&lt;/p&gt;&#xA;" OwnerUserId="6622" LastEditorUserId="6622" LastEditorDisplayName="" LastEditDate="2011-02-24T20:38:15.033" LastActivityDate="2011-02-24T20:38:15.033" CommentCount="4" />
  <row Id="23598" PostTypeId="2" ParentId="23573" CreationDate="2011-02-24T20:31:49.863" Score="1" ViewCount="" Body="&lt;p&gt;Note that the equation involves the composition of three invertible functions $\rm\ f(g(h(x))) = 0\ $ where $\rm\:f\:$ has fractional linear form $\rm\ f(x) = (a\ x + b)/(c\ x + d)\:,\: $ and $\rm\ g(x) = \ln\: x\:,\:$ and $\rm\ h(x) = x - S - 1\:.\:$ Therefore, upon  applying inverses we conclude that$\rm\ x = h^{-1}(g^{-1}(f^{-1}(0)))\:.\:$ That's effectively what the other solutions do. The same method solves (inverts) any composition of invertible functions.&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2011-02-24T20:31:49.863" />
  <row Id="23599" PostTypeId="1" AcceptedAnswerId="23604" CreationDate="2011-02-24T20:39:52.140" Score="3" ViewCount="167" Body="&lt;p&gt;Studying some integral table, I came across the following definite integral&#xA;$$\int_0^{\pi} \log [ a^2 + b^2 -2 a b \cos \phi ]\,d\phi$$ for $a,b \in \mathbb{R}$. Does somebody know a nice way to get the results?&lt;/p&gt;&#xA;" OwnerUserId="7266" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-24T20:44:38.443" LastActivityDate="2011-02-24T21:57:05.607" Title="How to compute the following definite integral" Tags="&lt;calculus&gt;" AnswerCount="3" CommentCount="5" />
  <row Id="23600" PostTypeId="1" CreationDate="2011-02-24T20:42:39.510" Score="2" ViewCount="225" Body="&lt;p&gt;I'm struggling with the concept of conditional expectation. First of all, if you have a link to any explanation that goes beyond showing that it is a generalization of elementary intuitive concepts, please let me know. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me get more specific. Let $(\Omega,\mathcal{A},P)$ be a probability space and $X$ an integrable real random variable defined on $(\Omega,\mathcal{A},P)$. Let $\mathcal{F}$ be a sub-$\sigma$-algebra of $\mathcal{A}$. Then $E[X|\mathcal{F}]$ is the a.s. unique random variable $Y$ such that $Y$ is $\mathcal{F}$-measurable and for any $A\in\mathcal{F}$, $E[X1_A]=E[Y1_A]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The common interpretation seems to be: &quot;$E[X|\mathcal{F}]$ is the expectation of $X$ given the information of $\mathcal{F}$.&quot; I'm finding it hard to get any meaning from this sentence.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;In elementary probability theory, expectation is a real number. So the sentence above makes me think of a real number instead of a random variable. This is reinforced by $E[X|\mathcal{F}]$ sometimes being called &quot;conditional expected value&quot;. Is there some canonical way of getting real numbers out of $E[X|\mathcal{F}]$ that can be interpreted as elementary expected values of something?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;In what way does $\mathcal{F}$ provide information? To know that some event occurred, is something I would call information, and I have a clear picture of conditional expectation in this case. To me $\mathcal{F}$ is not a piece of information, but rather a &quot;complete&quot; set of pieces of information one could possibly acquire in some way. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Maybe you say there is no real intuition behind this, $E[X|\mathcal{F}]$ is just what the definition says it is. But then, how does one see that a martingale is a model of a fair game? Surely, there must be some intuition behind that!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope you have got some impression of my misconceptions and can rectify them.&lt;/p&gt;&#xA;" OwnerUserId="1878" LastActivityDate="2011-02-24T23:08:39.260" Title="Intuition behind Conditional Expectation" Tags="&lt;probability-theory&gt;&lt;probability&gt;" AnswerCount="5" CommentCount="3" FavoriteCount="3" />
  <row Id="23601" PostTypeId="2" ParentId="23600" CreationDate="2011-02-24T20:48:21.683" Score="1" ViewCount="" Body="&lt;p&gt;I happened to read an article on Wikipedia today on &lt;a href=&quot;http://en.wikipedia.org/wiki/Conditional_expectation&quot; rel=&quot;nofollow&quot;&gt;Conditional Expectation&lt;/a&gt;. That clarified a lot of my questions. Hope it helps!&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;For your first question, in the&#xA;linked article, there is the&#xA;definition for conditional&#xA;expectation of a r.v. $X: \Omega&#xD;&#xA;    \rightarrow \mathbb{R}$ given a sub&#xA;sigma algebra $\mathcal{F}$ of the&#xA;one $\mathcal{A}$ on domain&#xA;$\Omega$. It is a&#xA;$\mathcal{F}$-measurable function $:&#xD;&#xA;    \Omega \rightarrow \mathbb{R}$,&#xA;denoted as $E(X \vert \mathcal{F})$.&#xA;If you evaluate this conditional&#xA;expectation at a point $\omega \in&#xD;&#xA;    \Omega$, you will get a value $E(X&#xD;&#xA;    \vert \mathcal{F})(\omega)$, which&#xA;is called the conditional&#xA;expectation of $X$ given&#xA;$\mathcal{F}$ at $\omega$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the r.v. $X$ is an indicator&#xA;function on some measurable subset&#xA;say $A \in \mathcal{A}$, its&#xA;conditional expectation given the&#xA;sub sigma algebra is called the&#xA;conditional probability of the&#xA;subset $A$ given the sub sigma&#xA;algebra $\mathcal{F}$, denoted as&#xA;$P( A \vert \mathcal{F})$. It  is a&#xA;mapping: $\Omega \rightarrow&#xD;&#xA;    \mathbb{R}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we let $A$ vary within&#xA;$\mathcal{A}$, the conditional&#xA;probability $P( \cdot \vert&#xD;&#xA;    \mathcal{F})$ is a mapping:&#xA;$\mathcal{A} \times \Omega&#xD;&#xA;    \rightarrow \mathbb{R}$. In some&#xA;cases,  $\forall \omega \in \Omega$,&#xA;$P( \cdot \vert&#xD;&#xA;    \mathcal{F})(\omega)$ is a&#xA;probability measure on $(\Omega,&#xD;&#xA;    \mathcal{F})$, in which case&#xA;$P(\cdot \vert \mathcal{F})$ is&#xA;called a regular conditional&#xA;probability.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When $\mathcal{F}$ is generated by&#xA;another r.v. $Y$, then the&#xA;conditional expectation and&#xA;conditional probability will be&#xA;called the ones given the r.v. $Y$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;For your second question, I am still&#xA;wondierng what kind of information a&#xA;sigma algebra (of a r.v.) can&#xA;provide, and how it is provided?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="1281" LastEditorUserId="1281" LastEditorDisplayName="" LastEditDate="2011-02-24T22:16:33.947" LastActivityDate="2011-02-24T22:16:33.947" />
  <row Id="23602" PostTypeId="2" ParentId="23515" CreationDate="2011-02-24T21:06:10.680" Score="1" ViewCount="" Body="&lt;p&gt;The short answer is that mathematical existence is always relative to a system. If you are working in the realm of positive integers, then only such integers exist. But mathematical systems can always be extended to a larger system in which more things exist, and often these systems are developed to prove results about the original, smaller, system which includes just those things that we are primarily concerned about. The things peculiar to the larger system may be variously regarded as nonexistent abominations, convenient fictions, or perfectly good entities as long as their definition is consistent. Most mathematicians take the latter view. Then the question arises: how far can this system-extending process be taken? We can systematize this question by considering that most of mathematics can be formalized within set theory (I would say all mathematics, but that's not universally accepted). The standard formulation of set theory is ZFC. Some people think that's already too much (in particular, they are not happy about the &quot;C&quot; part), most people go along with it, and some are intrigued by the possibilities of extending it. A major direction of this extension research is in positing the existence of what are called &quot;large cardinals&quot;. Although each class of large cardinals was conceived originally by specific properties rather than by reference to the others, it turns out amazingly that, as far as is known, the cardinals form a linear hierarchy of existence: if you believe in one, then you can be assured of the existence of the cardinals below it, and you can consistently disbelieve, should you be so inclined, in the existence of those in a higher class. Large cardinals are not all as fanciful as they may seem: the existence of some of them is equivalent to the truth of some relatively down-to-earth-looking mathematical statements (Harvey Friedman is the big name in this area.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The next step is: how far can the large-cardinal hierarchy be extended? Here I think it all gets blurred. And perhaps it's not reasonable to always expect finite answers to questions about the infinite.  &lt;/p&gt;&#xA;" OwnerUserId="875" LastActivityDate="2011-02-24T21:06:10.680" />
  <row Id="23603" PostTypeId="2" ParentId="23566" CreationDate="2011-02-24T21:08:40.020" Score="12" ViewCount="" Body="&lt;p&gt;I'd like to emphasize a remark that Eric made in a comment to his answer. Introspection is essential when learning mathematics - not only to analyze problem solving techniques - but also in many other ways. The web of mathematics is connected in many mysterious and marvelous ways. Spending a little effort attempting to discover these links can go a long way towards better understanding the essence of the matter. After you solve (or fail to solve) a problem you should spend some time trying to abstract a bit from the specific problem. Can it easily be generalized from a specific trick to a general method that you can add to your tools for later reuse? For example, if it's a problem about numbers can it be generalized to one about functions? If so, can the proof be simplified by appealing to properties unique to functions such as derivatives? For a simple example &lt;a href=&quot;http://math.stackexchange.com/questions/6310/n-mid-an-bn-longrightarrow-n-mid-fracan-bna-b/6448#6448&quot;&gt;see here,&lt;/a&gt; or consider the trivial proof of the $\rm\:abc$ theorem for polynomials (vs. difficult conjecture for numbers), which proof exploits to the hilt the power of derivatives (viz. wronskians). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Eventually, with enough training, you will be able to effortlessly move back and forth between the general and the specific, and better recognize the essence of the matter when sizing up problems - the same way a chess grandmaster can evaluate a chessboard in a single glance. Don't be frustrated if this doesn't come easy - or if it has atrophied - because - just like chess - it takes much practice to remain proficient. Unlike vision, language, etc. these mental faculties were not programmed into our minds by evolution, so one must continually reprogram these faculties for other purposes - whether they be chess or mathematical reasoning (it's no accident that one frequently sees strong correlation between chess and math abilities - both depend strongly upon pattern-matching, e.g. see the famous studies by the psychologist de Groot: &lt;a href=&quot;http://www.google.com/#hl=en&amp;amp;sugexp=ldymls&amp;amp;xhr=t&amp;amp;q=de+groot+though+choice+chess&quot; rel=&quot;nofollow&quot;&gt;Thought and choice in chess&lt;/a&gt;).&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2011-02-24T21:08:40.020" CommentCount="4" />
  <row Id="23604" PostTypeId="2" ParentId="23599" CreationDate="2011-02-24T21:15:09.713" Score="8" ViewCount="" Body="&lt;p&gt;This is Gauss' law in two dimensions in disguise. In two dimensions, the Green's function of the Poisson equation is $\ln r$, so since the argument of the logarithm is the distance between two points at distances $a$ and $b$ from the center with an angle $\phi$ between their vectors, you're calculating the potential between a point charge and a uniformly charged circle, which, since the potential solves the Poisson equation, depends only on the outer radius and is the same as the potential in the case where the charge of the circle is concentrated at its centre.&lt;/p&gt;&#xA;" OwnerUserId="6622" LastActivityDate="2011-02-24T21:15:09.713" CommentCount="1" />
  <row Id="23605" PostTypeId="2" ParentId="23600" CreationDate="2011-02-24T21:17:49.927" Score="3" ViewCount="" Body="&lt;p&gt;An example. Suppose that $X \sim {\rm binomial}(m,p)$ and $Y \sim {\rm binomial}(n,p)$ are independent ($0 &amp;lt; p &amp;lt; 1$). For any integer $0 \leq s \leq m+n$, it holds&#xA;$$&#xD;&#xA;{\rm E}[X|X + Y = s] = \frac{{m }}{{m  + n }}s.&#xD;&#xA;$$&#xA;This means that&#xA;$$&#xD;&#xA;{\rm E}[X|X + Y] = \frac{{m }}{{m  + n }}(X+Y).&#xD;&#xA;$$&#xA;Note that ${\rm E}[X|X + Y]$ is a random variable which is a function of $X+Y$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that, in general, the conditional expectation of $X$ given $Z$, denoted ${\rm E}[X|Z]$, is defined as ${\rm E}[X|\sigma(Z)]$, where $\sigma(Z)$ is the $\sigma$-algebra generated by $Z$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;. In response to the OP's request, I note that the binomial distribution (which is discrete) plays no special role in the above example. For completely analogous results for the normal and gamma distributions (both are continuous) see &lt;a href=&quot;http://math.stackexchange.com/questions/11487/y-xn-what-is-exy-y/11521#11521&quot;&gt;this&lt;/a&gt; and &lt;a href=&quot;http://math.stackexchange.com/questions/11673/ex-1-x-1-x-2-for-independent-gamma-random-variables&quot;&gt;this&lt;/a&gt;, respectively; for a substantial generalization, see &lt;a href=&quot;http://mathoverflow.net/questions/47168/ex-1-x-1-x-2-where-x-i-are-integrable-independent-infinitely-divisib/47204#47204&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2810" LastEditorUserId="2810" LastEditorDisplayName="" LastEditDate="2011-02-24T23:07:18.080" LastActivityDate="2011-02-24T23:07:18.080" CommentCount="13" />
  <row Id="23607" PostTypeId="1" AcceptedAnswerId="23632" CreationDate="2011-02-24T21:26:56.207" Score="4" ViewCount="152" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am trying to upper bound the following sum:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{k=1}^{n/2} \frac{\binom{n-2}{k-1}k^{k-2}(n-k)^{n-k-2}}{n^{n-3}}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Based on numerical computations, it seems like the upper bound is a constant (there is also another complicated proof that suggested the upper bound should be a constant). Any idea how to prove this? Stirling's approximation does not seem to help: using Stirling's (in a loose way) I can show that the sum is $O(log n)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A related bound that would imply the bound on the above sum is to show that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \frac{\binom{n-2}{k-1}k^{k-2}(n-k)^{n-k-2}}{n^{n-3}} \leq \frac{c}{k^2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for some constant $c$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="7469" LastEditorUserId="6075" LastEditorDisplayName="" LastEditDate="2011-02-25T01:08:34.110" LastActivityDate="2011-02-25T01:33:59.627" Title="Upper bound on summation involving binomial coefficients" Tags="&lt;combinatorics&gt;&lt;binomial-coefficients&gt;&lt;sequences-and-series&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="0" />
  <row Id="23609" PostTypeId="2" ParentId="23544" CreationDate="2011-02-24T21:38:55.413" Score="2" ViewCount="" Body="&lt;p&gt;Ok, so inspired by Arturo's answer, here is another partial answer that includes the Gaussian integers and the 3-adic integers:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Define τ:R→R:n↦2*(n+1 choose 3) and notice the formal equality&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$2\cdot\binom{n+m+1}{3} = 2\cdot\binom{n+1}{3}+2\cdot\binom{m+1}{3} + (mn^2+m^2n)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then define [ a, b ] = ( a, b + τ(a) ) to be a different coordinate system on (R⊕R,∘).  Then [ a1, b1 ]∘[ a2, b2 ] = [ a1+a2, b1+b2 ] is clearly isomorphic to R⊕R.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose R is a domain with field of fractions K.  Then τ:R→K:n↦2*(n+1 choose 3) definitely exists, so one just needs some condition for τ(R) ≤ R.  I thought this was at least sort of common, but I can't think of any real examples other than Z.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Certainly R=Z[x] doesn't work for τ, though both R⊕R and A will be free abelian of countably infinite rank.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I guess the only improvement is that we don't need every element of R to be divisible by 3, only the elements of the form xxx−x.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe the ring R=Z[i] works for this, since Z[i]/(3) ≅ Z/3Z × Z/3Z satisfies xxx-x ≡ 0.  This ring is less exciting in some ways though, since the additive group is free abelian.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rings that are even nicer, where all binomial coefficients exist, are called binomial rings.  For example the p-adic integers for any prime p (and where p=3 is the interesting one for us) also work.  The additive group of p-adic integers is not free abelian, so this is a real gain.&lt;/p&gt;&#xA;" OwnerUserId="583" LastActivityDate="2011-02-24T21:38:55.413" CommentCount="5" />
  <row Id="23610" PostTypeId="2" ParentId="23599" CreationDate="2011-02-24T21:41:42.307" Score="4" ViewCount="" Body="&lt;p&gt;Note that by symmetry we can assume $|b| \geq |a|$. Since $\cos(\phi)$ is even, your integral is half of the integral from $-\pi$ to $\pi$, or &#xA;$${1 \over 2}\int_{-\pi}^{\pi} \log(a^2 + b^2 - 2ab\cos(\phi))\,d\phi$$&#xA;Note that $|a^2 + b^2 - 2ab\cos(\phi)|$ is the same as $|ae^{i\phi} - b|^2$, so your integral becomes &#xA;$$\int_{-\pi}^{\pi}\log|ae^{i\phi} - b|\,d\phi$$&#xA;Thinking of $ae^{i\phi}$ as a complex number $z$, the integrand is the real part of the function $\ln(az - b)$, which is analytic on the interior of the unit disk (we use that $|b| \geq |a|$). Thus the integrand is harmonic and therefore satisfies the mean-value property: the integral over the unit circle is $2\pi$ times the value at the center of the circle, which is $\log|a0 - b| = \log|b|$ in this case. So your answer is $2\pi\log|b|$, which is the same as $2\pi\max(\log|a|,\log|b|)$ since we assume $|b| \geq |a|$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Technical point: we glossed over applying the mean-value property when $|b| = |a|$ and thus the integrand has a singularity at the boundary; but any of a number of limiting techniques will deal with this case as well.&lt;/p&gt;&#xA;" OwnerUserId="3035" LastActivityDate="2011-02-24T21:41:42.307" CommentCount="1" />
  <row Id="23611" PostTypeId="2" ParentId="23599" CreationDate="2011-02-24T21:47:17.140" Score="7" ViewCount="" Body="&lt;p&gt;Here is a different way, and is strangely the first thing that came to my mind.  It uses generating series, and power series as well as other various techniques:&#xA;(Warning!: It is significantly more complicated)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume that $a\geq b\geq0$ without loss of generality.  Notice that our integral is $$\pi \log(a^2+b^2)+\int_{0}^{\pi}\log\left( 1-\frac{2ab\cos\phi}{a^2+b^2}\right). $$ Since $\log(1-x)=-\sum_{i=1}^{\infty}\frac{x^{i}}{i}$ we have &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_{0}^{\pi}\log\left( 1-\frac{2ab\cos\phi}{a^2+b^2}\right)d\phi=-\int_{0}^{\pi}\sum_{i=1}^{\infty} \frac{1}{i}\left(\frac{2ab\cos\phi}{a^2+b^2}\right)^i d\phi$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Switch the order or integration and summation to find:  (More on why this is legal at the end)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$-\int_{0}^{\pi}\sum_{i=1}^{\infty} \frac{1}{i}\left(\frac{2ab\cos\phi}{a^2+b^2}\right)^i d\phi=-\sum_{i=1}^{\infty} \frac{1}{i}\left(\frac{2ab}{a^2+b^2}\right)^i \int_{0}^{\pi} \cos^i\phi d\phi$$&#xA;For $n$ odd we have $\int_0^\pi \cos^n xdx=0$.  For even $n=2r$ we see by integration by parts that $\int_{0}^{\pi}\cos(x)^{2r}dx=\frac{2r-1}{2r}\int_{0}^{\pi}\cos(x)^{2r-2}dx$, so that induction yields $$\int_{0}^{\pi}\cos(x)^{2r}dx=\pi\prod_{i=1}^{r}\frac{2i-1}{2i}=\pi\frac{\left({2r\atop r}\right)}{4^{r}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus our sum becomes &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$-\pi \sum_{r=1}^{\infty} \frac{1}{2r}\left(\frac{ab}{a^2+b^2}\right)^{2r} \left({2r\atop r}\right ) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let $$f(z)=\sum_{r=1}^{\infty} \frac{1}{r} \left({2r\atop r}\right) \frac{z^r}{4^r}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then we need to find $-\frac{\pi}{2} f\left(\left( \frac{2ab}{a^2+b^2} \right)^2\right)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $Y=\left( \frac{2ab}{a^2+b^2} \right)^2$.&#xA;Recall the generating series $$\sum_{n=0}^{\infty}\left({2n\atop n}\right)x^{n}=\frac{1}{\sqrt{1-4x}}. $$  Differentiating yields $1+xf^{&amp;#39;}(x)=\frac{1}{\sqrt{1-x}}$ and hence $f(Y)=\int_{0}^{Y}\frac{1-\sqrt{1-x}}{x\sqrt{1-x}}dx$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Make the substitution $x=\sin^{2}(u)$, and we get $$f(Y)=\int_{0}^{\theta}\frac{1-\cos(u)}{\sin^{2}(u)\cos(u)}2\sin(u)\cos(u)du=2\int_{0}^{\theta}\csc(u)-\cot(u)du$$ where $\theta$ is in the first quadrant and satisfies $\sin (\theta)=\frac{2ab}{a^2+b^2}.$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then since $\int\csc x\, dx=-\log\left|\csc x+\cot x\right|$, and $\int\cot x dx=\log\left|\sin x\right|$, we see that $\int\csc(u)-\cot(u)du=-\log|\sin(u)\csc(u)+\sin(u)\cot u|=-\log|1+\cos(u)|$. Thus $$f(Y)=-2\log|1+\cos(u)|\biggr|_{u=0}^{u=\theta}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Drawing the triangle with sides $2ab$, $a^2-b^2$ and $a^2+b^2$ tells us that $\cos(\theta)=\frac{a^2-b^2}{a^2+b^2}$, and hence $$f(Y)=2\log 2- 2\log \left(1+\frac{a^2-b^2}{a^2+b^2}\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus the final answer is  $$\pi \log(a^2+b^2)-\frac{\pi}{2}f(Y)=\pi\log(a^2+b^2)-\pi \log 2 +\pi\log\left(1+\frac{a^2-b^2}{a^2+b^2}\right)=2\pi\log(a)$$&lt;/p&gt;&#xA;" OwnerUserId="6075" LastEditorUserId="6075" LastEditorDisplayName="" LastEditDate="2011-02-24T21:57:05.607" LastActivityDate="2011-02-24T21:57:05.607" CommentCount="1" />
  <row Id="23612" PostTypeId="2" ParentId="23054" CreationDate="2011-02-24T21:59:55.650" Score="0" ViewCount="" Body="&lt;p&gt;The question is a little ambiguous: the three previous answers used three different interpretations. If the OP wants the surface distance from the point $X$ to the geodesic line $\overleftrightarrow{AB}$, the answer is straightforward. If the desired distance is between $X$ and the segment $\overline{AB}$, a bit more work is required.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using longitude ($\theta$) and latitude ($\phi$), let $A=(\theta_A, \phi_A)$, $B=(\theta_B, \phi_B)$, and $X=(\theta_X, \phi_X)$. The direction vectors for these points are &#xA;$$\hat A = (\cos \phi_A \cos \theta_A, \cos \phi_A \sin \theta_A, \sin \phi_A),$$ $$ \hat B = (\cos \phi_B \cos \theta_B, \cos \phi_B \sin \theta_B, \sin \phi_B), $$&#xA;$$\hat X = (\cos \phi_X \cos \theta_X, \cos \phi_X \sin \theta_X, \sin \phi_X).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $\Phi$ be the  distance on the unit sphere between $\hat X$ and the geodesic line passing through $\hat A$ and $\hat B$. Imagine the plane $\mathcal{P}$ passing through $\hat A$, $\hat B$, and the origin, which cuts the unit sphere in half. Then the Euclidean distance of $\hat X$ from  plane $\mathcal{P}$ is $\sin \Phi$. Now let $\hat n$ be a unit normal vector for $\mathcal{P}$, and we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\hat n = \hat A \times \hat B$$&#xA;$$\sin \Phi = | \hat n \cdot \hat X |$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, if the radius of the original sphere is $R$, then the surface distance from the point $X$ to the geodesic line $\overleftrightarrow{AB}$ is $R \Phi$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To determine the distance to the segment $\overline{AB}$, we need to determine whether or not the point of line $\overleftrightarrow{ A B}$ that $ X$ is closest to is between $A$ and $B$. If the closest point is between $A$ and $B$, then the surface distance to the segment is $R \Phi$. Otherwise, the distance to the segment is the distance to the closest endpoint, which is best resolved though the methods described in the &lt;a href=&quot;http://en.wikipedia.org/wiki/Great-circle_distance&quot; rel=&quot;nofollow&quot;&gt;Wikipedia article&lt;/a&gt; referenced by Ross Millikan. One way to make this determination is to find the point $\hat{X}_{\textrm{proj}}$, the projection of $\hat X$ onto plane $\mathcal{P}$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\hat{X}_{\textrm{proj}} = \hat X - (\hat n \cdot \hat X) \hat n,$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and then normalize $\hat{X}_{\textrm{proj}}$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\hat x = \frac{\hat{X}_{\textrm{proj}} }{| \hat{X}_{\textrm{proj}} |},$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So determining whether the point of line $\overleftrightarrow{AB}$ that $X$ is closest to is between $A$ and $B$ reduces to determining whether $\hat x$ is between $\hat A$ and $\hat B$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now consider the mid-point of $\hat A$ and $\hat B$, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$M=\frac{\hat A + \hat B}{2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the projection of $\hat x$ on the ray $\overrightarrow{OM}$ is further along than the projection of $\hat A$ or $\hat B$, then $\hat x$ is between $\hat A$ and $\hat B$, that is, if $\; \hat x \cdot M &amp;gt; \hat A \cdot M \; \; (=\hat B \cdot M)$, then $\hat x$ is between $\hat A$ and $\hat B$, otherwise not.&lt;/p&gt;&#xA;" OwnerUserId="6272" LastActivityDate="2011-02-24T21:59:55.650" />
  <row Id="23613" PostTypeId="2" ParentId="23600" CreationDate="2011-02-24T22:24:24.313" Score="4" ViewCount="" Body="&lt;p&gt;Maybe this simple example will help. I use it when I teach &#xA;conditional expectation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) The first step is to think of ${\mathbb E}(X)$ in a new way:&#xA;as the best estimate for the value of a random variable $X$ in the absence of any information. &#xA;To minimize the squared error&#xA;$${\mathbb E}[(X-e)^2]={\mathbb E}[X^2-2eX+e^2]={\mathbb E}(X^2)-2e{\mathbb E}(X)+e^2,$$&#xA;we differentiate to obtain $2e-2{\mathbb E}(X)$, which is zero at $e={\mathbb E}(X)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, if I throw a fair die and you have to&#xA;estimate its value $X$, according to the analysis above, your best bet is to guess ${\mathbb E}(X)=3.5$.&#xA;On specific rolls of the die, this will be an over-estimate or an under-estimate, but in the long run it minimizes the mean square error. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) What happens if you &lt;em&gt;do&lt;/em&gt; have additional information?&#xA;Suppose that I tell you that $X$ is an even number.&#xA;How should you  modify your estimate to take this new information into account?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The mental process may go something like this: &quot;Hmmm, the possible values &lt;em&gt;were&lt;/em&gt; $\lbrace 1,2,3,4,5,6\rbrace$&#xA;but we have eliminated $1,3$ and $5$, so the remaining possibilities are $\lbrace 2,4,6\rbrace$. &#xA;Since I have no other information, they should be considered equally likely and hence  the revised expectation is $(2+4+6)/3=4$&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, if I were to tell you that $X$ is odd, your revised (conditional) expectation is 3. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(3) Now imagine that I will roll the die and I will tell you the parity of $X$; that is, I will &#xA;tell you whether the die comes up odd or even. You should now see that a single numerical response &#xA;cannot cover both cases. You would respond &quot;3&quot; if I tell you &quot;$X$ is odd&quot;, while you would respond &quot;4&quot; if I tell you &quot;$X$ is even&quot;.&#xA;A single numerical response is not enough because the particular piece of information that I will give you is &lt;strong&gt;itself random&lt;/strong&gt;.&#xA;In fact, your response is necessarily a function of this particular piece of information. &#xA;Mathematically, this is reflected in the requirement that ${\mathbb E}(X\ |\ {\cal F})$ must be $\cal F$ measurable. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think this covers point 1 in your question, and tells you why a single real number is not sufficient. &#xA;Also concerning point 2, you are correct in saying that the role of $\cal F$ in ${\mathbb E}(X\ |\ {\cal F})$&#xA;is not a single piece of information, but rather tells what possible specific pieces of (random) information may occur. &lt;/p&gt;&#xA;" OwnerUserId="940" LastActivityDate="2011-02-24T22:24:24.313" CommentCount="3" />
  <row Id="23614" PostTypeId="2" ParentId="23600" CreationDate="2011-02-24T22:26:18.193" Score="1" ViewCount="" Body="&lt;p&gt;You can think of the conditional expectation as the orthogonal projection onto the closed subspace of $\mathcal F$-measurable random variables in the Hilbert space of square integrable random variables.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.ee.nthu.edu.tw/cschang/Talk01142008.pdf&quot; rel=&quot;nofollow&quot;&gt;This&lt;/a&gt; is a detailed and elementary discussion of this viewpoint.&lt;/p&gt;&#xA;" OwnerUserId="367" LastEditorUserId="367" LastEditorDisplayName="" LastEditDate="2011-02-24T23:08:39.260" LastActivityDate="2011-02-24T23:08:39.260" />
  <row Id="23615" PostTypeId="2" ParentId="23596" CreationDate="2011-02-24T22:40:16.580" Score="1" ViewCount="" Body="&lt;p&gt;Some informal explanation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Covariance matrix $C_y$ (it is symmetric) encodes the correlations between variables of a vector. In general a covariance matrix is non-diagonal (i.e. have non zero correlations with respect to different variables).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;But it's interesting to ask, is it possible to diagonalize the covariance matrix by changing basis of the vector?&lt;/strong&gt;. In this case there will be no (i.e. zero) correlations between different variables of the vector. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Diagonalization of this symmetric matrix is possible with eigen value decomposition.&#xA;You may read 'A Tutorial on Principal Component Analysis'by Jonathon Shlens to get a good understanding. (pages 6-7)&lt;/p&gt;&#xA;" OwnerUserId="5312" LastEditorUserId="5312" LastEditorDisplayName="" LastEditDate="2011-02-24T22:45:26.507" LastActivityDate="2011-02-24T22:45:26.507" />
  <row Id="23617" PostTypeId="2" ParentId="23600" CreationDate="2011-02-24T22:55:00.843" Score="3" ViewCount="" Body="&lt;p&gt;I think a good way to answer question 2 is as follows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am performing an experiment, whose outcome can be described by an element $\omega$ of some set $\Omega$.  I am not going to tell you the outcome, but I will allow you to ask certain questions yes/no questions about it.  (This is like &quot;20 questions&quot;, but infinite sequences of questions will be allowed, so it's really &quot;$\aleph_0$ questions&quot;.) We can associate a yes/no question with the set $A \subset \Omega$ of outcomes for which the answer is &quot;yes&quot;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, one way to describe some collection of &quot;information&quot; is to consider all the questions which could be answered with that information.  (For example, the 2010 Encyclopedia Britannica is a collection of information; it can answer the questions &quot;Is the dodo extinct?&quot; and &quot;Is the elephant extinct?&quot; but not the question &quot;Did Justin Bieber win a 2011 Grammy?&quot;)  This, then, would be a set $\mathcal{F} \subset 2^\Omega$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I know the answer to a question $A$, then I also know the answer to its negation, which corresponds to the set $A^c$ (e.g. &quot;Is the dodo not-extinct?&quot;).  So any information that is enough to answer question $A$ is also enough to answer question $A^c$.  Thus $\mathcal{F}$ should be closed under taking complements.  Likewise, if I know the answer to questions $A,B$, I also know the answer to their disjunction $A \cup B$ (&quot;Are either the dodo or the elephant extinct?&quot;), so $\mathcal{F}$ must also be closed under (finite) unions.  Countable unions require more of a stretch, but imagine asking an infinite sequence of questions &quot;converging&quot; on a final question.  (&quot;Can elephants live to be 90? Can they live to be 99? Can they live to be 99.9?&quot; In the end, I know whether elephants can live to be 100.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think this gives some insight into why a $\sigma$-field can be thought of as a collection of information.&lt;/p&gt;&#xA;" OwnerUserId="822" LastActivityDate="2011-02-24T22:55:00.843" />
  <row Id="23618" PostTypeId="1" AcceptedAnswerId="23623" CreationDate="2011-02-24T22:57:54.273" Score="1" ViewCount="154" Body="&lt;p&gt;So the problem was to integrate $\overline{z}$ over the half circle of radius 2, $ z = 2e^{i \theta}$ from $z = -2i$ to $z = 2i$. After going through the math, they say &#xA;this integral is equal to the integral over the curve of $\frac{dz}{z}$ which is equal to $\pi i $. I don't follow how this equivalence was established. I hope I provided enough detail, still learning how to use MathJax. Thanks!&lt;/p&gt;&#xA;" OwnerUserId="6758" LastEditorUserId="6296" LastEditorDisplayName="" LastEditDate="2011-02-24T23:13:53.370" LastActivityDate="2011-02-25T00:30:11.623" Title="What does this statement in my text mean? (Complex Analysis)" Tags="&lt;complex-analysis&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="23620" PostTypeId="1" CreationDate="2011-02-24T23:13:59.400" Score="1" ViewCount="126" Body="&lt;p&gt;Suppose $(\Omega, \mathcal{F}, P)$ is a probability space and $(U, \mathcal{\Sigma})$ is a measurable space. There seem to be two ways of defining the conditional expectation of a r.v. $X: \Omega \rightarrow \mathbb{R}$ given another r.v. $Y: \Omega \rightarrow U$, denoted as $E(X\vert Y)$:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;As a $\sigma(Y)$-measurable&#xA;mapping from $\Omega$ to&#xA;$\mathbb{R}$, defined as: $$E(X\vert Y) = E(X \vert \sigma(Y)). $$&#xA;where $\sigma(Y)$ is the sigma algebra of r.v. $Y$, which I think is also denoted as $Y^{-1}(\mathcal{\Sigma})$?&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;As a $\mathcal{\Sigma}$-measurable&#xA;mapping from $U$ to $\mathbb{R}$,&#xA;defined as follows (from &lt;a href=&quot;http://en.wikipedia.org/wiki/Conditional_expectation#Conditioning_as_factorization&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Define measure Q on U to be the&#xA;probability measure induced by $Y$&#xA;on $(U, \mathcal{\Sigma})$, as $Q(B)&#xD;&#xA;    = P(Y^{−1}(B)), \forall B \in \mathcal{\Sigma}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Define $E(X \vert Y)$ to be the&#xA;integrable function $g:U \rightarrow&#xD;&#xA;    \mathbb{R}$ such that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \int_{Y^{-1}(B)} X(\omega) \ d&#xD;&#xA;    \operatorname{P} = \int_{B}&#xD;&#xA;    g(u) \ d \operatorname{Q},&#xD;&#xA;    \forall B \in \mathcal{\Sigma}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I am correct, this definition is related to the first one as:&#xA;$$E(X \mid Y) \circ Y= E\left(X \mid Y^{-1} \left(\Sigma\right)\right). $$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I was wondering which of the above two is the definition of $E(X \mid Y)$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks and regards! References (links or books) will also be appreciated! &lt;/p&gt;&#xA;" OwnerUserId="1281" LastEditorUserId="1281" LastEditorDisplayName="" LastEditDate="2011-02-25T06:57:33.177" LastActivityDate="2011-02-25T14:00:38.247" Title="Definition of conditional expectation of a random variable given another one" Tags="&lt;probability-theory&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="23621" PostTypeId="2" ParentId="21872" CreationDate="2011-02-24T23:38:09.160" Score="1" ViewCount="" Body="&lt;p&gt;Standard curves are usually used in the field of biology to help you identify how much &quot;stuff&quot; (in many cases, this is the concentration of some kind of protein) you have in your samples.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Simply put, a standard curve is a plot of known x and y values.  Given just y from your experiment, you should be able to determine x from the plot of known values (standard curve).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To get into a bit more detail, a standard curve for immunoassays such as ELISAs is usually fit to the &lt;a href=&quot;http://www.miraibio.com/blog/2010/08/the-4-parameter-logistic-4pl-nonlinear-regression-model/&quot; rel=&quot;nofollow&quot;&gt;4 parameter logistic (4PL)&lt;/a&gt; or &lt;a href=&quot;http://www.miraibio.com/blog/2009/02/5-pl-logistic-regression/&quot; rel=&quot;nofollow&quot;&gt;5 parameter logistic (5PL)&lt;/a&gt; nonlinear regression model equations.  &lt;/p&gt;&#xA;" OwnerUserId="7474" LastActivityDate="2011-02-24T23:38:09.160" />
  <row Id="23622" PostTypeId="1" AcceptedAnswerId="23626" CreationDate="2011-02-24T23:49:43.677" Score="4" ViewCount="118" Body="&lt;p&gt;Consider the collection $$\left\{e^{2\pi ikx/(b-a)}\,:\, k=0,\pm 1, \pm2,\ldots \right\}\,,$$ on $(a,b)$. Verify that this system is orthogonal in $L^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have attempted the two most obvious things: (1) integrating with exponentials, (2) integrating with trig functions. Both are messy, and neither are working out. I don't want to write out a lot of work, because my notes are a tedious nightmarescape, but I basically keep ending up with the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Set $\theta=\frac{2\pi}{b-a}$. Take $e^{2\pi ikx/(b-a)}$ and $e^{2\pi ikx/(b-a)}$. Taking their inner product, I get something like $$\frac{i}{\theta(k-n)}\left[e^{\theta(k-n)bi}-e^{\theta(k-n)ai}\right]\,.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I feel like I'm going crazy, but it is not at all apparent that this will end being zero....&lt;/p&gt;&#xA;" OwnerUserId="2313" LastActivityDate="2011-02-25T08:18:41.663" Title="Verifying a system is orthogonal in $L^2$." Tags="&lt;homework&gt;&lt;real-analysis&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="23623" PostTypeId="2" ParentId="23618" CreationDate="2011-02-25T00:21:10.443" Score="4" ViewCount="" Body="&lt;p&gt;For $z$ on this half circle $z\overline{z}=|z|^2=4$ so that $\overline{z}=\frac{4}{z}$.  Let $C$ be the contour in question.  Then we have&#xA; $$\int_C \overline{z}=\int_C \frac{4}{z}dz=\int_{-\pi/2}^{\pi/2}\frac{4}{2e^{i\theta}}d(2e^{i\theta})=\int_{-\pi/2}^{\pi/2}4id\theta=4\pi i$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have an extra factor of $4$.  Are you sure that is not correct?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope that helps,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Alternative:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just put in your parametrization directly, that is $z=2e^{i\theta}$.  Then we get $$\int_C \overline{z}dz=\int_{-\pi/2}^{\pi/2}2e^{-i\theta}d(2e^{i\theta})=\int_{-\pi/2}^{\pi/2}4id\theta=4\pi i$$ &lt;/p&gt;&#xA;" OwnerUserId="6075" LastEditorUserId="6075" LastEditorDisplayName="" LastEditDate="2011-02-25T00:30:11.623" LastActivityDate="2011-02-25T00:30:11.623" CommentCount="1" />
  <row Id="23624" PostTypeId="1" AcceptedAnswerId="23630" CreationDate="2011-02-25T00:35:53.323" Score="1" ViewCount="52" Body="&lt;p&gt;How can we show that the projective dimension of the $\mathbb{Z}/p^2 \mathbb{Z}$-module $\mathbb{Z}/p \mathbb{Z}$ is infinite?&lt;/p&gt;&#xA;" OwnerUserId="7475" LastActivityDate="2011-02-25T01:15:23.147" Title="How to show that the projective dimension is infinite" Tags="&lt;commutative-algebra&gt;" AnswerCount="1" />
  <row Id="23625" PostTypeId="1" AcceptedAnswerId="23666" CreationDate="2011-02-25T00:45:30.503" Score="4" ViewCount="157" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Let's say we have a real, continuous,&#xA;  positive function f(x) for which we&#xA;  define the quantity:&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$$\pi(f,a) = \frac{\int_0^a f(x)&#xD;&#xA; dx}{\int_0^a \sqrt{1+\left(\frac{df(x)}{dx}&#xD;&#xA; \right)^2 }dx}$$&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;we want to &lt;em&gt;find the function f&lt;/em&gt; that&#xA;  &lt;strong&gt;maximizes&lt;/strong&gt; $\pi$ for a given $a$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In general how do we attack problems of this kind: find $f$ such that  $\mathrm{F}(f)$ is maximum? Are there any constrains that guarantee that there is an analytical solution? How could the problem above be modified to have a solution?&lt;/p&gt;&#xA;" OwnerUserId="7476" LastEditorUserId="6075" LastEditorDisplayName="" LastEditDate="2011-02-25T01:43:17.107" LastActivityDate="2011-02-26T08:53:16.620" Title="Function that maximizes a function." Tags="&lt;functions&gt;&lt;optimization&gt;" AnswerCount="4" CommentCount="11" />
  <row Id="23626" PostTypeId="2" ParentId="23622" CreationDate="2011-02-25T00:48:12.447" Score="2" ViewCount="" Body="&lt;p&gt;Note that $\displaystyle{\frac{b}{b-a} = \frac{a}{b-a} + 1}$, so those exponential terms differ by a factor of $e^{2\pi(k-n)i}$.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2011-02-25T00:48:12.447" CommunityOwnedDate="2011-02-25T00:48:12.447" CommentCount="3" />
  <row Id="23627" PostTypeId="2" ParentId="23625" CreationDate="2011-02-25T00:54:49.193" Score="3" ViewCount="" Body="&lt;p&gt;It looks like you are wanting to maximize the integral of $f$ with respect to its arclength. (there appears to be a typo on the bottom, but I could be wrong)  However, this quantity is unbounded.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the constant function $f(x)=c$, and let $a&amp;gt;0$.  Then $$\pi(f,a)=\frac{\int_0^a cdx}{\int_0^a\sqrt{1+0^2}dx}=\frac{ac}{a}=c$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Taking $c$ to be as large as we want we see there is no maximum, and $\pi(f,a)$ is unbounded.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope that helps&lt;/p&gt;&#xA;" OwnerUserId="6075" LastActivityDate="2011-02-25T00:54:49.193" CommentCount="1" />
  <row Id="23628" PostTypeId="2" ParentId="23625" CreationDate="2011-02-25T00:55:32.720" Score="4" ViewCount="" Body="&lt;p&gt;You are computing the ratio of area under the graph of $f(x)$ to arc length, from $x=0$ to $x=a$. The arc length is invariant under up-shifts, but the area under the graph is not. For any $g(x)$ and any $r\gt 0$, the difference between $\pi(g+r,a)$ and $\pi(g,a)$ is proportional to $r$, and so you have that $\pi(f,a)$ is unbounded. There is no maximum for any class of functions that is closed under adding constants.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2011-02-25T00:55:32.720" CommentCount="2" />
  <row Id="23629" PostTypeId="2" ParentId="23511" CreationDate="2011-02-25T01:01:33.463" Score="0" ViewCount="" Body="&lt;p&gt;If you want integer solutions, you have a &lt;a href=&quot;http://en.wikipedia.org/wiki/Diophantine_equation#Linear_Diophantine_equations&quot; rel=&quot;nofollow&quot;&gt;linear diophantine equation&lt;/a&gt;. You can start by observing that $y$ must be a multiple of $7$, then simplify the equation using this.&lt;/p&gt;&#xA;" OwnerUserId="589" LastActivityDate="2011-02-25T01:01:33.463" />
  <row Id="23630" PostTypeId="2" ParentId="23624" CreationDate="2011-02-25T01:15:23.147" Score="2" ViewCount="" Body="&lt;p&gt;If $M=\mathbb Z/p\mathbb Z$ had finite projective dimension, the kernel of every surjective map $P\to \mathbb Z/p\mathbb Z$ from a $\mathbb Z/p^2\mathbb Z$-projective module $P$ onto $M$ would have strictly smaller projective dimension. Yet there is a short exact sequence $$0\to\mathbb Z/p\mathbb Z\to \mathbb Z/p^2\mathbb Z\to\mathbb Z/p\mathbb Z\to 0$$ &lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2011-02-25T01:15:23.147" />
  <row Id="23631" PostTypeId="1" AcceptedAnswerId="23635" CreationDate="2011-02-25T01:18:44.720" Score="3" ViewCount="62" Body="&lt;p&gt;Hello Everyone,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I do not have enough complexity theory background, but I was wondering about the kind of reductions that we normally do to show NP-Completeness. I think all of the reductions that I have seen are one-to-one; in that, given an instance $I_1$ of problem $P_1$, they map it to instance $I&amp;#39;_1$ of problem $P&amp;#39;_1$ such that $I&amp;#39;_1$ is a yes instance of $P&amp;#39;_1$ iff $I_1$ is a yes instance of $P_1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I have never seen any reduction which can, say map an instance $I_1$ of $P_1$ to say $2$ or more instances of $P_2$ (in some sense, the mapping should involve creating some structure in $I_2$ where we have $2$ or more choices and we could pick any one arbitrarily). This will be kind of surprising (to me) if it exists, but nevertheless the following question bothers me even more.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Leaving that aside, I have never seen any reduction which might map $2$ or more (say YES instances) of a problem $P$ to the same instance of some problem $Q$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would appreciate if you can exhibit these reductions for NP-Complete problems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;NOTE: I understand that this is more of a cstheory question. But since this is certainly not research level, I decided to post it here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-Akash&lt;/p&gt;&#xA;" OwnerUserId="5391" LastActivityDate="2011-02-25T04:16:45.507" Title="Many one and One many reductions" Tags="&lt;computer-science&gt;&lt;np-complete&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="23632" PostTypeId="2" ParentId="23607" CreationDate="2011-02-25T01:33:59.627" Score="3" ViewCount="" Body="&lt;p&gt;Lets do some rearranging.  First split up the binomial notation: $$\frac{\binom{n-2}{k-1}k^{k-2}(n-k)^{n-k-2}}{n^{n-3}}=\frac{(n-2)!k^{k-2}(n-k)^{n-k-2}}{(k-1)!(n-k-1)!n^{n-3}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Multiply the numerators and denominators to remove the $-1$'s and $-2$'s:&#xA;$$=\frac{n!}{(n-1)n}\cdot\frac{k}{k!}\cdot\frac{n-k}{(n-k)!}\cdot\frac{k^{k}}{k^{2}}\cdot\frac{(n-k)^{n-k}}{(n-k)^{2}}\frac{n^{3}}{n^{n}}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now rearrange again so that Sterlings formula jumps out at us:&#xA;$$=\frac{n}{(n-1)}\frac{n}{k\cdot(n-k)}\cdot\frac{n!}{n^{n}}\cdot\frac{k^{k}}{k!}\cdot\frac{(n-k)^{(n-k)}}{(n-k)!}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Applying Sterlings formula &lt;em&gt;roughly&lt;/em&gt;, this becomes&#xA;$$\approx\frac{n}{(n-1)}\frac{n}{k\cdot(n-k)}\cdot\left(\sqrt{2\pi n}e^{-n}\right)\cdot\left(\frac{1}{\sqrt{2\pi k}}e^{k}\right)\cdot\left(\frac{1}{\sqrt{2\pi(n-k)}}e^{n-k}\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$=\frac{n}{(n-1)\sqrt{2\pi}}\cdot\left(\frac{n}{k(n-k)}\right)^{3/2}$$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now compare the last piece to the integral&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_{1}^{n-1}\left(\frac{n}{x(n-x)}\right)^{3/2}dx.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This integral is bounded by a constant for every $n$ so the proof is finished. (The bound can be placed on the integral by a partition trick that yields an infinite geometric series.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope that helps,&lt;/p&gt;&#xA;" OwnerUserId="6075" LastActivityDate="2011-02-25T01:33:59.627" CommentCount="4" />
  <row Id="23633" PostTypeId="1" CreationDate="2011-02-25T02:08:24.707" Score="1" ViewCount="55" Body="&lt;p&gt;Recently I've started looking at building a 6 degree of freedom motion simulator based on a Stewart Platform.  Although it's unlikely to be built due to time, money and space constraints, I'm still looking at writing some software for it as software development is my day job.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've been able to get motion data from Flight Simulator such as acceleration and pitch and roll amount, but have no idea on where to go to get an idea of the math needed to translate these into actuator movements.  My aim is to get a 3d model working first then maybe build a small scale platform as a proof of concept.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've been reading through a few papers on the subject but since I don't really have a engineering background they're not making much sense.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If anyone could provide some first steps or some more layman type papers to start building an algorithm up, it would be much appreciated.&lt;/p&gt;&#xA;" OwnerUserId="7477" LastActivityDate="2011-02-25T02:08:24.707" Title="Building a Motion Cue Algorithm" Tags="&lt;algorithms&gt;" CommentCount="2" />
  <row Id="23635" PostTypeId="2" ParentId="23631" CreationDate="2011-02-25T02:11:29.727" Score="2" ViewCount="" Body="&lt;p&gt;For an example of one to many, from my answer here: &lt;a href=&quot;http://math.stackexchange.com/questions/7130/reduction-from-hamiltonian-cycle-to-hamiltonian-path/7141#7141&quot;&gt;Reduction from Hamiltonian cycle to Hamiltonian path&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a reduction from Hamiltonian Cycle to Hamiltonian Path.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Given a graph $G$ of which we need to&#xA;  find Hamiltonian Cycle, for a single&#xA;  edge $e = \{u,v\}$ add new vertices&#xA;  $u&amp;#39;$ and $v&amp;#39;$ such that $u&amp;#39;$ is&#xA;  connected only to $u$ and $v&amp;#39;$ is&#xA;  connected only to $v$ to give a new&#xA;  graph $G_e$.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$G_e$ has a hamiltonian path if and&#xA;  only if $G$ has a hamiltonian cycle&#xA;  with the edge $e=\{u,v\}$.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Run the Hamiltonian path algorithm on&#xA;  each $G_e$ for each edge $e \in G$. If&#xA;  all graphs have no hamiltonian path,&#xA;  then $G$ has no hamiltonian cycle. If&#xA;  at least one $G_e$ has a hamiltonian&#xA;  path, then $G$ has a hamiltonian cycle&#xA;  which contains the edge $e$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2011-02-25T02:11:29.727" CommentCount="3" />
  <row Id="23636" PostTypeId="1" AcceptedAnswerId="23642" CreationDate="2011-02-25T02:15:04.710" Score="2" ViewCount="71" Body="&lt;p&gt;EDIT:Let $S$ be any finite set and suppose $x \notin S$. Let $K=S \cup \{x\}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1.Prove that $P(K)$ is the disjoint union of $P(S)$ and $X=\{T \subset K : x \in T\}$. That is, show that $P(K) = P(S) \cup X$.&lt;/p&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2.Prove that every element of $X$ is the union of a subset of $S$ with $\{x\}$, and that if you take different subsets of $S$ you get different elements of $X$. Argue that, therefore, $X$ has the same number of elements as $P(S)$.&lt;/p&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3.Argue that the previous two parts allow you to conclude that if $S$ is a finite set, then $P(K)$ has twice as many elements as $P(S)$.&lt;/p&gt;&lt;/p&gt;&#xA;" OwnerUserId="7478" LastEditorUserId="5105" LastEditorDisplayName="" LastEditDate="2011-02-25T02:54:35.053" LastActivityDate="2011-02-25T02:54:35.053" Title="Prove that $P(K)$ is the disjoint union of $P(S)$ and $X=\{T \subset K : x \in T\}$" Tags="&lt;discrete-mathematics&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="23637" PostTypeId="2" ParentId="23566" CreationDate="2011-02-25T02:20:48.437" Score="3" ViewCount="" Body="&lt;p&gt;One thing I remember realizing in high school was that I would often see something, or read something, and think to myself, &quot;yes, that makes sense&quot;.  But then a day later I wouldn't have a clue about it.  That was my first introduction to the difference between what I later learned was &quot;active understanding&quot; as opposed to &quot;passive understanding&quot;.  It's something that is not obvious to anyone without a certain amount of introspection.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another thing I learned (much later, in graduate school) was that if I wanted to learn something, it was a good idea to get at least three books on the subject.  I always found that reading just one book, I would inevitably run across something that would stop me dead, completely unable to continue.  But if I had one or two other books available, almost always there would be some alternative explanation that made sense to me and made it possible for me to go on.&lt;/p&gt;&#xA;" OwnerUserId="2579" LastActivityDate="2011-02-25T02:20:48.437" CommentCount="1" />
  <row Id="23638" PostTypeId="1" CreationDate="2011-02-25T02:28:39.583" Score="3" ViewCount="99" Body="&lt;p&gt;I have $n$ coins and each of them shows head with probability $p$ if I throw them separately. Now I throw them one-by-one, and each current throw has a correlation $\rho$ with the previous result. (I hope this is rigorously defined through the definition of correlation coefficient.) After one round of tossing, what is the probability that I got $k$ heads in total?&lt;/p&gt;&#xA;" OwnerUserId="6016" LastActivityDate="2011-02-25T04:20:22.917" Title="correlated coin problem" Tags="&lt;probability&gt;" AnswerCount="0" CommentCount="2" FavoriteCount="1" />
  <row Id="23639" PostTypeId="1" AcceptedAnswerId="23640" CreationDate="2011-02-25T02:29:32.970" Score="3" ViewCount="157" Body="&lt;p&gt;Prove that for every positive integer x of exactly four digits, if the sum of digits is divisible by 3, then x itself is divisible by 3 (i.e. consider x = 6132, the sum of digits of x is 6+1+3+2 = 12, which is divisible by 3, so x is also divisible by 3)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How could I approach this proof? &#xA;I'm not sure where I would even begin. &lt;/p&gt;&#xA;" OwnerUserId="6331" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2011-02-25T04:42:07.337" LastActivityDate="2011-02-25T04:42:07.337" Title="Number Theory - Proof of divisibility by 3" Tags="&lt;number-theory&gt;&lt;number&gt;" AnswerCount="3" CommentCount="8" />
  <row Id="23640" PostTypeId="2" ParentId="23639" CreationDate="2011-02-25T02:34:05.047" Score="1" ViewCount="" Body="&lt;p&gt;Suppose you have a number whose decimal digits are represented $a$, $b$, $c$, and $d$, so $x=abcd$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In base $10$, this means&#xA;$$&#xD;&#xA;x=abcd=a\cdot 10^3+b\cdot 10^2+c\cdot 10^1+d\cdot 10^0.&#xD;&#xA;$$&#xA;Try looking at $x$ modulo $3$, and remember that $10\equiv 1\pmod{3}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This concept is easily extended to an integer $x$ of any number of digits.&lt;/p&gt;&#xA;" OwnerUserId="1583" LastEditorUserId="1583" LastEditorDisplayName="" LastEditDate="2011-02-25T02:40:54.697" LastActivityDate="2011-02-25T02:40:54.697" />
  <row Id="23641" PostTypeId="2" ParentId="23639" CreationDate="2011-02-25T02:39:11.173" Score="1" ViewCount="" Body="&lt;p&gt;Actually, this is true for an integral number with any digits. The proof is quite easy. Let's denote the integral number by $\overline{a_n a_{n-1} \ldots a_1}$. If the sum of its digits $\sum_{i=1}^n{a_i}$ is divisible by 3, then $\sum_{i=1}^n{(1+\overline{9...9}_{i-1})*a_i}$ is too. Here $\overline{9...9}_{i-1}$ denotes the integer with $i-1$ 9's. But this second sum is just the original number $\overline{a_n a_{n-1} \ldots a_1}$ expanded.&lt;/p&gt;&#xA;" OwnerUserId="6016" LastEditorUserId="6016" LastEditorDisplayName="" LastEditDate="2011-02-25T02:46:03.063" LastActivityDate="2011-02-25T02:46:03.063" />
  <row Id="23642" PostTypeId="2" ParentId="23636" CreationDate="2011-02-25T02:41:40.047" Score="1" ViewCount="" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Show that $P(K) \subset (P(S) \cup X)$ and $(P(S) \cup X) \subset P(K)$. To show this, argue that if $A \in P(K)$ is some set, then so must $A \in (P(S) \cup X)$. And conversely. To see that $P(S)$ and $X$ are disjoint, let $A \in P(S)$ and $B \in X$. Why cannot it ever be the case that $A=B$? Try to find an element of one set that is not an element of the other.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Pick some subset $A \in X$. How does the construction of $X$ allow you to prove that $A$ is the same set as some subset of $S$ union the singleton $\{x\}$?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Can you find a one-to-one correspondence between $P(S)$ and $X$? How does this allow you to count the number of subsets of $K$?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="5105" LastEditorUserId="5105" LastEditorDisplayName="" LastEditDate="2011-02-25T02:48:54.537" LastActivityDate="2011-02-25T02:48:54.537" CommentCount="2" />
  <row Id="23643" PostTypeId="2" ParentId="23620" CreationDate="2011-02-25T03:09:47.580" Score="2" ViewCount="" Body="&lt;p&gt;I've only seen $E[X|Y]$ used to denote the first one.  The second is a function $g$ such that $g(Y) = E[X|\sigma(Y)]$. (by the way, $g$ is only unique up to $Q$-null sets).  Informally, you might write $g(y) = E[X|Y=y]$, which is actually correct when $Y$ is discrete (i.e. when $Q$ is atomic).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By definition, $E[X|\sigma(Y)]$ is a $\sigma(Y)$-measurable random variable that is supposed to answer the following: if I told you the value of $Y$, what would be your best estimate of the value of $X$?  Your estimate would be different depending on the value of $Y$, so it should be some function $g$ of $Y$.  Your second definition is referring to that function $g$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to prove that in the discrete case $g(y) = E[X|Y=y]$, call the right side $h(y)$.  Then show that the random variable $h(Y)$ satisfies the conditions that uniquely define $E[X|\sigma(Y)]$: namely, $h(Y)$ is $\sigma(Y)$-measurable and for any $A \in \sigma(Y)$, $E[h(Y);A] = E[X;A]$.  Note that in the discrete case, $A$ is necessarily a countable union of events of the form $\{Y=y_i\}$.&lt;/p&gt;&#xA;" OwnerUserId="822" LastEditorUserId="822" LastEditorDisplayName="" LastEditDate="2011-02-25T14:00:38.247" LastActivityDate="2011-02-25T14:00:38.247" CommentCount="3" />
  <row Id="23644" PostTypeId="2" ParentId="21969" CreationDate="2011-02-25T03:11:02.833" Score="13" ViewCount="" Body="&lt;p&gt;The answer is &quot;no&quot;. Here's a proof:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The main idea is to consider the two sums as approximations of the same integral over $1/x$ with the rectangle rule and then show that the approximation errors can't be equal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It will be more convenient to work with the centre of each sum rather than the beginning. So let $k_i$ be the number of terms in the $i$th sum (as in the question), let $n_i=m_i + (k_i-1)/2$ be the central denominator (which may or may not actually occur as the denominator of a term in the sum depending on whether it's an integer or a half-integer), and let $\sigma_i = \sigma(m_i,k_i)$. Without loss of generality, assume $n_2 &amp;gt; n_1$, and as Henry observed, we can also assume that the sums don't overlap.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now $\sigma_i$ is roughly $k_i/n_i$, so the quantity $\Delta:=k_1n_2-k_2n_1$ would need to be small in order for the sums to be equal. This quantity is either an integer or a half-integer. Note that if $k_1$, $k_2$ and $n_2$ are fixed, then $\sigma_2$ is fixed, $n_1$ depends monotonically on $\Delta$, and $\sigma_1$ in turn depends monotonically on $n_1$, so the difference between the two sums can change sign at most once as $\Delta$ changes. Thus, if we can show that the difference has opposite signs for $\Delta = 0$ and $\Delta = 1/2$, it will follow that it cannot be zero for any possible value of $\Delta$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now consider $1/l$ as the first term in the following expansion:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_{l-\frac{1}{2}}^{l+\frac{1}{2}}\frac{\mathrm{d}x}{x} = \int_{-\frac{1}{2}}^{+\frac{1}{2}}\frac{\mathrm{d}x}{l-x}=\int_{-\frac{1}{2}}^{+\frac{1}{2}}\frac{1}{l}\sum_{i=0}^\infty\left(\frac{x}{l}\right)^i\mathrm{d}x=2\sum_{j=0}^\infty\frac{\left(2l\right)^{-(2j+1)}}{2j+1}\;.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then $\sigma_1$ is the first term of the following expansion:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_{n_1-k_1/2}^{n_1+k_1/2}\frac{\mathrm{d}x}{x}=\sum_{l=n_1-(k_1-1)/2}^{n_1+(k_1-1)/2}\int_{l-\frac{1}{2}}^{l+\frac{1}{2}}\frac{\mathrm{d}x}{x}=2\sum_{j=0}^\infty\frac{1}{2j+1}\sum_{l=n_1-(k_1-1)/2}^{n_1+(k_1-1)/2}\left(2l\right)^{-(2j+1)}\;.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now consider first the case $\Delta=0$. Then $n_2=\frac{k_2}{k_1}n_1$, so by substituting $u=\frac{k_2}{k_1}x$ we can write the two sums as approximations of the same integral:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_{n_1-k_1/2}^{n_1+k_1/2}\frac{\mathrm{d}x}{x}=\int_{n_2-k_2/2}^{n_2+k_2/2}\frac{\mathrm{d}u}{u}=2\sum_{j=0}^\infty\frac{1}{2j+1}\sum_{l=n_2-(k_2-1)/2}^{n_2+(k_2-1)/2}\left(2l\right)^{-(2j+1)}\;,$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where the $j=0$ term of this expansion is now $\sigma_2$. Since these are two expansions of the same integral, to show $\sigma_1&amp;lt;\sigma_2$ it suffices to show that for each $j&amp;gt;0$ the term in the first expansion is greater than the term in the second expansion. (This is plausible, since in going from the first approximation of the integral to the second, we've increased the number of intervals by a factor $k_2/k_1$ but decreased the $j$th error term in each interval roughly by a factor $(n_1/n_2)^{2j+1}$, which is $(k_1/k_2)^{2j+1}$ since $\Delta=0$.) In fact, we only need to treat the $j=1$ case; the inequalities for higher $j$ then follow since all additional factors of $l^{-2}$ are greater in the first expansion than in the second expansion, since we assumed that the two sums of reciprocals don't overlap. (We could have avoided the terms with higher $j$ altogether by writing the error term as a derivative at an intermediate value, but that would have introduced cumbersome shifts of $1/2$ to account for the intermediate values.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $j=1$, we use the convexity of $l^{-3}$ in both directions, decreasing the greater one of the sums by collapsing it onto its center:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{l=n_1-(k_1-1)/2}^{n_1+(k_1-1)/2}l^{-3}\ge\sum_{l=n_1-(k_1-1)/2}^{n_1+(k_1-1)/2}n_1^{-3}=\frac{k_1}{n_1^{3}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and increasing the lesser one by smearing it out over an integral:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{l=n_2-(k_2-1)/2}^{n_2+(k_2-1)/2}l^{-3}&amp;lt;\int_{n_2-k_2/2}^{n_2+k_2/2}l^{-3}\mathrm{d}l=\frac{(n_2-k_2/2)^{-2}-(n_2+k_2/2)^{-2}}{2}=$$&#xA;$$=\frac{n_2k_2}{(n_2-k_2/2)^2(n_2+k_2/2)^2}&amp;lt;\frac{n_2k_2}{n_1^2n_2^2}=\frac{k_1}{n_1^{3}}\;.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This establishes that $\sigma_1&amp;lt;\sigma_2$ when $\Delta=0$. Now consider $\Delta \neq 0$. Then $\frac{k_1}{k_2}n_2 = n_1+\frac{\Delta}{k_2}$, and we need to shift the integrand to make the integral limits match:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_{n_2-k_2/2}^{n_2+k_2/2}\frac{\mathrm{d}x}{x}=\int_{n_1-k_1/2+\Delta/k_2}^{n_1+k_1/2+\Delta/k_2}\frac{\mathrm{d}u}{u}=\int_{n_1-k_1/2}^{n_1+k_1/2}\frac{\mathrm{d}t}{t+\Delta/k_2}\;.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since all the error terms in the expansions for the integrals have the same sign, their difference is less than the larger of the two, which, as we showed above, is the one for $\sigma_1$. Thus, to show that the difference changes sign when $\Delta$ goes from $0$ to $1/2$, it suffices to show that the change due to the shift in the integrand is larger than the approximation error in the expansion for $\sigma_1$. To do this, we can expand the shifted integral in the same form as the error expansion, again using convexity to bound each integral by the central value of the integrand:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_{n_1-k_1/2}^{n_1+k_1/2}\frac{\mathrm{d}t}{t+\Delta/k_2}=\sum_{l=n_1-(k_1-1)/2}^{n_1+(k_1-1)/2}\int_{l-\frac{1}{2}}^{l+\frac{1}{2}}\frac{\mathrm{d}t}{t+\Delta/k_2}&amp;gt;\sum_{l=n_1-(k_1-1)/2}^{n_1+(k_1-1)/2}\frac{1}{l+\Delta/k_2}\;.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To show that for each $l$ the summand differs from $1/l$ by more than the corresponding approximation error, we estimate the latter so that we can sum the series over $j$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$2\sum_{j=0}^\infty\frac{1}{2j+1}\left(2l\right)^{-(2j+1)}&amp;lt;&#xD;&#xA;\frac{2}{3}\sum_{j=0}^\infty\left(2l\right)^{-(2j+1)}=\frac{2}{3}\left(2l\right)^{-3}\frac{1}{1-1/(2l)}=\frac{2}{3}\frac{1}{(2l)^2}\frac{1}{2l-1}\;.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, the difference from the shift in the reciprocals is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{1}{l}-\frac{1}{l+\Delta/k_2}=\frac{\Delta/k_2}{l(l+\Delta/k_2)}\;.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we can estimate the quotient of these two values (using $\Delta=1/2$):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{2}{3}\frac{1}{(2l)^2}\frac{1}{2l-1}\frac{l(l+\Delta/k_2)}{\Delta/k_2}&amp;lt;&#xD;&#xA;\frac{1}{12}\frac{1}{l}\frac{1}{l-1/2}\frac{l+1/2}{\Delta/k_2}&amp;lt;&#xD;&#xA;\frac{k_2}{6(l-1)}\;.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, the change due to the shift in the integrand is larger than the approximation error provided $k_2&amp;lt;6(l-1)$. This we can derive by considering the powers of $2$ in the denominators of the two sums.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The highest power of $2$ that divides one of the denominators in the sum cannot be cancelled and hence divides the reduced denominator of the sum. This is because between any two numbers containing the same number of factors of $2$, there is one containing at least one more factor of $2$, and hence each sum contains a unique denominator $d$ with the highest power of $2$ in that sum. If we add all the other reciprocals and reduce them to a common denominator, that denominator will necessarily contain fewer powers of $2$ than $d$, and hence these powers cannot cancel if we then add $1/d$. Thus, the two sums can only be equal if the denominator with the most factors of $2$ has the same number of factors of $2$ in both sums.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For this to be the case, $k_2$ must be at least $1$ less than twice the largest denominator in $\sigma_1$; otherwise any interval of $k_2$ numbers would necessarily contain a number with one factor of $2$ more than the highest power of $2$ in the denominators of $\sigma_1$. Thus we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$k_2 \le 2\left(n_1+\frac{k_1-1}{2}\right)-1=2n_1+k_1-2\;.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the sums don't overlap, we also have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$n_1+\frac{k_1-1}{2}\le n_2-\frac{k_2-1}{2}-1\;,$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and hence&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{k_2}{2}\le \left(n_1+\frac{k_1-1}{2}\right)-\frac{1}{2} \le n_2-\frac{k_2-1}{2}-1-\frac{1}{2}\;,$$&#xA;$$k_2\le n_2-1\;.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With $\Delta=k_1n_2-k_2n_1= 1/2$, this yields&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$k_1=\frac{k_2}{n_2}n_1+\frac{\Delta}{n_2}\le n_1-\frac{n_1}{n_2}+\frac{\Delta}{n_2}&amp;lt;n_1\;,$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and hence $k_1\le n_1 - 1/2$, since $k_1$ is an integer and $n_1$ an integer or a half-integer. Thus&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$k_2\le2n_1+k_1-2&amp;lt;3n_1-2&amp;lt;6(\frac{n_1}{2}-\frac{1}{4})\le 6\left(n_1-\frac{k_1-1}{2}-1\right)\le 6(l-1)\;.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This completes the proof.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S.: In case you're surprised that we get exactly the factor of $6$ that we need from considering the powers of $2$: I am, too :-) If there's any deeper reason behind this, I have no idea what it might be. It's possible to derive a slightly tighter bound using &lt;a href=&quot;http://en.wikipedia.org/wiki/Bertrand%27s_postulate#Better_results&quot; rel=&quot;nofollow&quot;&gt;Jitsuro Nagura's improvement on Bertrand's postulate&lt;/a&gt;, since, as Henry observed, any prime in the second sum that is greater than half its endpoint is enough to make the reduced denominators differ. But I think such improvements in bounding $k_2$ with respect to $n_2$ can't make the factor in the final inequality lower than $4$, since they only improve the estimate of $k_1$ with respect to $n_1$, but not that of $k_2$ with respect to $n_1$ from the powers of $2$, which is $k_2 &amp;lt; 2 n_1$ even for $k_1=1$. &lt;/p&gt;&#xA;" OwnerUserId="6622" LastEditorUserId="6622" LastEditorDisplayName="" LastEditDate="2011-02-25T18:09:33.147" LastActivityDate="2011-02-25T18:09:33.147" CommentCount="1" />
  <row Id="23645" PostTypeId="1" CreationDate="2011-02-25T03:20:13.710" Score="3" ViewCount="54" Body="&lt;p&gt;Given a lie algebra $g$, how does one approach finding the Gel'fand Models. For clarity, by this I mean &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$\bigoplus_{\lambda\in P^+}V(\lambda)$ where $P^+$ are the dominant weights, and $V(\lambda)$ is the highest weight representation of weight $\lambda$. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;One can calculate the weight modules and just take their sum, however I would like something more succinct. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, consider the simple case of $sl_2(C)$. This Gel'fand model is simply the complex two variable polynomials. One sees this by writing the highest weight representations of $sl_2(C)$ as homogeneous polynomials in variables $x,y$ by considering the Liebniz action of $sl_2(C)$ on $C&amp;lt;x,y&amp;gt;$. By summing these you get the polynomials in two variables.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I find this particularly intuitive. However, in the more general situation of $sl_n$, I don't see how to do this. Note, I am particularly interested in showing they are isomorphic to rings with nicer forms(I don't care to argue about what I mean by nicer, I think we both know).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&lt;strong&gt;What I am even more interested in, is this question for quantized universal enveloping algebras, and again a nice simple case would be $U_q(sl_n)$.&lt;/em&gt;&lt;/strong&gt; Again, our simple case, $U_q(sl_2)$ I know and like: the quantum plane, two variable polynomials quotient $xy-qyx$ for parameter $q$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know of a paper or two that mention some of these, but none that I have explain how to see this for the general type A case. In particular, papers about the quantum version are especially rare. References are appreciated. I would also appreciate proofs for other specific cases, they might be enlightening. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Note:This coincides with the homogeneous coordinate ring for $sl_n$. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: A large discussion has taken place with Mariano below.  He pointed out that my previous language was incorrect, and has helped me identify the correct question that I wished to ask. Hail to the chief!(I hope he doesn't mind I call him chief. :/)&lt;/p&gt;&#xA;" OwnerUserId="16" LastEditorUserId="16" LastEditorDisplayName="" LastEditDate="2011-03-15T05:02:33.033" LastActivityDate="2011-03-15T05:02:33.033" Title="How to compute the Gel'fand Models for a (quantum) Lie Algebra" Tags="&lt;reference-request&gt;&lt;representation-theory&gt;&lt;lie-algebra&gt;&lt;quantum-groups&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="23646" PostTypeId="1" CreationDate="2011-02-25T03:25:36.060" Score="3" ViewCount="146" Body="&lt;p&gt;Given a mobius transformation that maps one pair of concentric circles to another pair of concentric circles, why is the ratio of the radii preserved through the map? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I thought about how mobius transformations are compositions of rotations, scaling, inversion, and translation, and that intuitively, these types of maps shouldn't change the ratio of radii between two circles. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would it be correct to just say that if $\frac{r_1}{r_2}$ is the ratio of radii between the two circles, then &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) The radii are invariant under translation, $z \mapsto z+a$, so $\frac{r_1}{r_2}$ stays the same&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Under scaling by a factor $z \mapsto az$, $\frac{ar_1}{ar_2} = \frac{r_1}{r_2}$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) Under inversion, $z \mapsto \frac{1}{z}$, $\frac{1/r_1}{1/r_2} = \frac{r_2}{r_1}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or is there a different/better way to think about this problem? &lt;/p&gt;&#xA;" OwnerUserId="1736" LastEditorUserId="1736" LastEditorDisplayName="" LastEditDate="2011-02-25T03:55:43.073" LastActivityDate="2011-02-25T09:11:10.077" Title="Mobius transformations and concentric circles" Tags="&lt;complex-analysis&gt;" AnswerCount="2" CommentCount="5" />
  <row Id="23647" PostTypeId="2" ParentId="23645" CreationDate="2011-02-25T03:32:30.803" Score="5" ViewCount="" Body="&lt;p&gt;For a semisimple Lie algebra, the representation ring is a polynomial ring, and can be described quite concretely as the invariant ring $\mathbb Z[\Lambda]^W$ of the group algebra $\mathbb Z[\Lambda]$ of the weight lattice $\Lambda$ under the natural action of the Weyl group $W$. In the quantum case with $q$ not a root of unity, the ring has a similar description, as the deformation is not strong enough to mess much with it, in a sense; if $q$ is a root of unity, things are considerably more complicated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The classical case is treated in pretty much any good text on representation theory of semisimple algebras, in one form or another. For example, the ever great &lt;em&gt;Representation Theory&lt;/em&gt; by Fulton and Harris. The quantum case is treated in the corresponding quantum books :)&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2011-02-25T03:32:30.803" CommentCount="10" />
  <row Id="23648" PostTypeId="2" ParentId="23496" CreationDate="2011-02-25T03:44:09.983" Score="0" ViewCount="" Body="&lt;p&gt;To use Zarrax and Jesse's hints, it's possible to see that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_{|z|=\rho} \frac{|dz|}{|z-a|^2} = \int_{|z|=\rho} \frac{1}{(z-a)(\bar{z}-\bar{a})}(-i\rho) \frac{dz}{z}$$&#xA;$$ =\int_{|z|=\rho} \frac{-i \rho}{(z-a)(\frac{\rho^2}{z}-\bar{a})(z)} dz$$&#xA;$$ =\int_{|z|=\rho} \frac{-i \rho}{(z-a)(\rho^2 - \bar{a}z)} dz$$&#xA;$$ =\frac{i \rho}{\bar{a}} \int_{|z|=\rho} \frac{1}{(z-a)(z-\frac{\rho^2}{\bar{a}})} dz$$&#xA;$$ =\frac{i \rho}{\bar{a}} 2 \pi i n(\gamma,a) \frac{1}{a-\frac{\rho^2}{\bar{a}}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\gamma = \{|z|=\rho\}$, $n(\gamma,a)$ is the winding number of $a$ along $\gamma$, and where we assume that $|a| &amp;lt; \rho$ such that $a$ lies inside the circle $|z|=\rho$.\&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $|a|&amp;gt;\rho$, then we can reverse the role of the two functions, since then $\frac{1}{z-a}$ will be analytic inside $\gamma$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does that look correct?&lt;/p&gt;&#xA;" OwnerUserId="1736" LastActivityDate="2011-02-25T03:44:09.983" />
  <row Id="23649" PostTypeId="1" AcceptedAnswerId="23650" CreationDate="2011-02-25T03:51:54.113" Score="5" ViewCount="116" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know this is a simple question for most of you, but I am currently studying for a Calculus exam and was just wondering why an online calculator I am using to double-check my work was disagreeing with me on this question:&#xA;$$\lim_{x\to 0} \cot(x)\sec(x)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I reduce this down to $\frac{1}{\sin(x)}$, and in that case $x\to 0^-$ the limit is equal to negative infinity; and if $x\to 0^+$, the limit is equal to positive infinity. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Doesn't this mean that the limit as $x\to 0$ does not exist? I use the calculator (linked below), and while it verifies that the two sides approach opposite infinity, it solves the entire limit as approaching &quot;infinity&quot;. What does this mean?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.numberempire.com/limitcalculator.php&quot; rel=&quot;nofollow&quot;&gt;http://www.numberempire.com/limitcalculator.php&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7484" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-25T03:53:53.390" LastActivityDate="2011-02-25T03:58:12.540" Title="Limit approaches infinity on one side and negative infinity on other side" Tags="&lt;calculus&gt;&lt;infinity&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="23650" PostTypeId="2" ParentId="23649" CreationDate="2011-02-25T03:58:12.540" Score="6" ViewCount="" Body="&lt;p&gt;Your analysis is correct. Alternatively, $\sec(x)\to 1$ as $x\to 0$, and you can deal with $\cot(x)$, which goes to $\infty$ as $x\to 0^+$ and to $-\infty$ as $x\to 0^-$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note, though, the fact that each one-sided limit does not exist is already enough to tell you the limit does not exist. (Saying that the limit equals $\infty$ or $-\infty$ is &lt;em&gt;not&lt;/em&gt; saying that the limit exists, it is saying that the limit does not exist &lt;em&gt;and explaining why&lt;/em&gt;: because the values of the function grow without bound, either in the positive direction or in the negative direction, respectively). Even though we write things like&#xA;$$\lim_{x\to 0}\frac{1}{x^2} = \infty$$&#xA;this limit &lt;em&gt;does not&lt;/em&gt; exist.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As to the limit calculator at your link, I don't know what it means when it says as two-sided limit is $\infty$, since it says the same thing for $\lim\limits_{x\to 0}\frac{1}{x}$. In other words, it means that the on-line calculator is either not giving the correct answer, or else it means something other than what we think it means.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2011-02-25T03:58:12.540" />
  <row Id="23651" PostTypeId="2" ParentId="23639" CreationDate="2011-02-25T04:07:29.330" Score="1" ViewCount="" Body="&lt;p&gt;It's due to radix representation being &lt;em&gt;polynomial form&lt;/em&gt; in the radix, e.g. $\rm\ n = 4321 = p(10)\ $ for $\rm\ p(x) = 4\: x^3 + 3\: x^2 + 2\: x + 1\:.\:$ Thus mod $\rm\:3\::\ 10\equiv 1\ \Rightarrow\ p(10)\equiv p(1)\ =\ \sigma(n) :=$ sum of digits. Aternatively one may simply put $\rm\ x = 10\ $ in the $\:$ &lt;strong&gt;Factor Theorem&lt;/strong&gt; $\rm\ \ x-1\ |\ p(x)-p(1)\:,\: $ hence $\rm\ 3\ |\ 9\ |\ p(10)-p(1) = n - \sigma(n)\:.\:$ This is a special case of &lt;a href=&quot;http://math.stackexchange.com/search?q=user:242+casting&quot;&gt;casting out nines.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2011-02-25T04:07:29.330" />
  <row Id="23652" PostTypeId="2" ParentId="23631" CreationDate="2011-02-25T04:16:45.507" Score="2" ViewCount="" Body="&lt;p&gt;The easiest way to get a many-to-one reduction is to partially solve the original problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, suppose we want to reduce CLIQUE to 3COLORABILITY. We get as input a pair $(G,k)$ and need to output $G&amp;#39;$ where $G&amp;#39;$ is 3-colorable iff $G$ has a clique of size exactly $k$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We will proceed as follows.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;If $k\leq 10$, solve CLIQUE by brute force in time O(n^10). If $G$ contains a clique of size $k$ output the trivial graph. Otherwise, output $K_4$.&lt;/li&gt;&#xA;&lt;li&gt;If $k&amp;gt;10$, then use the standard reduction.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Every graph of the form $(G,k)$, $k\leq 10$, reduces to the same 2 graphs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I realize that this seems artificial (because it is). The reason that you rarely see such reductions in practice is that it's often easier to reason about a 1-1 reduction. After all, you're trying to show that $x\in P \Leftrightarrow f(x)\in Q$ and so the reverse implication is often easier that way.&lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2011-02-25T04:16:45.507" CommentCount="1" />
  <row Id="23653" PostTypeId="1" AcceptedAnswerId="23805" CreationDate="2011-02-25T04:17:59.140" Score="4" ViewCount="159" Body="&lt;p&gt;This is about problem $5$ in Section IV.$5$ of Hungerford's Algebra book. The question is the following:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;blockquote&gt;&#xA;    &lt;p&gt;If $A&amp;#39;$ is a submodule of the right $R$-module $A$ and $B&amp;#39;$ is a submodule of the left $R$-module $B$, then $$\frac{A}{A&amp;#39;} \otimes _R \frac{B}{B&amp;#39;} \cong \frac{A \otimes _R B}{C}$$ where $C$ is the subgroup of $A \otimes _R B$ generated by all elements $a&amp;#39; \otimes b$ and $a \otimes b&amp;#39;$ with $a \in A$, $a&amp;#39; \in A&amp;#39;$, $b \in B$, $b&amp;#39; \in B&amp;#39;$.&lt;/p&gt;&#xA;  &lt;/blockquote&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm just starting to somehow grasp the tensor product concept and I'm having a lot of trouble when trying to prove that some tensor product is isomorphic to a certain group. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular what I feel that's causing me problems is the fact that in the tensor product $A \otimes _R B$ not every element is an elementary tensor of the form $a \otimes b$ so that it is really hard for me sometimes to be able to define an appropriate map to try to produce an explicit isomorphism.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure if the universal property could be of help here, since I can try to define a map $$\varphi : \frac{A}{A&amp;#39;} \times \frac{B}{B&amp;#39;} \longrightarrow \frac{A \otimes _R B}{C} $$ by $$\varphi(a + A&amp;#39;, b + B&amp;#39;) := a \otimes b + C$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I've been trying without getting anywhere. Any hints or advice on how to approach this exercise would be very much appreciated. Thank you.&lt;/p&gt;&#xA;" OwnerUserId="900" LastActivityDate="2011-02-26T05:25:39.053" Title="A question about tensor products" Tags="&lt;homework&gt;&lt;abstract-algebra&gt;&lt;modules&gt;" AnswerCount="1" CommentCount="11" />
  <row Id="23654" PostTypeId="2" ParentId="23646" CreationDate="2011-02-25T04:18:31.170" Score="2" ViewCount="" Body="&lt;p&gt;If you think of the two concentric circles as bounding an annulus $A_1$, and then of their images as bounding an annulus $A_2$, then the Mobius transformation is inducing a conformal transformation between $A_1$ and $A_2$, and it is a general fact the ratio of the inner and outer radii of an annulus is a conformal invariant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This places the property you are asking about in a broader context, although I haven't answered the question as to &lt;em&gt;why&lt;/em&gt; this general fact is true.  For the moment, let me defer to &lt;a href=&quot;http://en.wikipedia.org/wiki/Extremal_length&quot; rel=&quot;nofollow&quot;&gt;wikipedia&lt;/a&gt; for one proof of this fact.&lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2011-02-25T04:18:31.170" />
  <row Id="23655" PostTypeId="1" AcceptedAnswerId="23657" CreationDate="2011-02-25T04:19:11.330" Score="4" ViewCount="131" Body="&lt;p&gt;So I am trying to understand what a coset is and what a quotient ring is. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I am going to tell you guys what I know. And please let me know if my thinking is right or wrong, and if I am missing something. For the rest of this post, assume R is a ring and I is an ideal of that ring. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So $I = (m)$ is a principal ideal generated by $m$ where $m \in R$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the congruence class of $a$ and $I$ is denoted by $[a]_m$ but this congruence class can also be written as $a + (m)$ or simply $a + I$. &#xA;now this congruence class is obviously a set. So is this the coset of it? So for any $a$ that you choose in $R$ and you &quot;add&quot; the ideal to it (generated by m which is also in R) you get a coset. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The quotient ring R/I just means ALL the cosets of I in R. So does this say that if hypothetically speaking m = 3, then 1 + (3) is one coset, 2 + (3) is another coset and hence the quotient ring is ALL the cosets for every possible a? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope I make sense. If someone could send me a link to an easy (introduction to algebra) article or a &quot;tutorial&quot;, that would be appreciated. I am using Hungerford Algebra. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If anyone can explain this to me in easy english, that is appreciated. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="287" LastActivityDate="2011-02-25T04:38:45.547" Title="What is a quotient ring and cosets?" Tags="&lt;abstract-algebra&gt;" AnswerCount="1" CommentCount="6" />
  <row Id="23657" PostTypeId="2" ParentId="23655" CreationDate="2011-02-25T04:38:45.547" Score="3" ViewCount="" Body="&lt;p&gt;Remember that every equivalence relation induces a partition on the set on which you have defined the relation. An ideal $I$ defines an equivalence relation on the set $R$ by saying that $a\sim b$ if and only if $a-b\in I$; we express this by saying that $a$ is &lt;em&gt;congruent&lt;/em&gt; to $b$ modulo $I$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That means that $R$ is partitioned into equivalence class under this &quot;congruent to modulo $I$&quot; relation. The equivalence classes are called &quot;cosets&quot; (I claimed some time ago this is short for &quot;congruence set&quot;, but have been unable to substantiate this; but you can surely imagine that it is). The cosets are the equivalence classes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, what is the equivalence class of an $a\in R$? it is the set of all things that are congruent to $a$ modulo $I$; this consists exactly of all elements of the form $a+x$ with $x\in I$, so we write it as&#xA;$$ a + I = \{ a+x \mid x\in I\}.$$&#xA;This is its description as a &lt;em&gt;set&lt;/em&gt;. If we want to think of it in terms of the equivalence relation and remember that it is the equivalence class of $a$, then we use the standard notation for equivalence classes and write $[a]$ (or $[a]_I$, or $[a]_m$, to remind us also of &lt;em&gt;which&lt;/em&gt; ideal $I$ we are dealing with). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you ask if $a+I$ is &quot;the coset of it&quot;, it is unclear to me who &quot;it&quot; is. But, $a+I$ is the equivalence class of $a$, so it is the coset of $a$ (since &quot;the coset of $x$&quot; just means &quot;the equivalence class of $x$ under the equivalence relation 'congruent modulo $I$'&quot;). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason that when you &quot;add&quot; the ideal you get the coset is just because of what the definition of the equivalence relation is: every element of the form $a+i$ is congruent to $a$ modulo $I$, because $a-(a+i) = -i\in I$; and if $b$ is congruent to $a$ modulo $I$, then $a-b=i$ for some $i\in I$, so we get that $b=a-i$. That is, every element of the form $a+x$ with $x\in I$ is in the coset, and everything in the coset is of the form $a+x$ with $x\in I$. So the notation $a+I$ is both suggestive and useful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the &lt;strong&gt;set&lt;/strong&gt; $R/I$ is just the set of equivalence classes; as a &lt;em&gt;set&lt;/em&gt;, the elements are the cosets. Each coset has many different names, since $[a]_I = [b]_I$ whenever $a\sim b$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a &lt;strong&gt;ring&lt;/strong&gt;, $R/I$ is the ring whose elements are the cosets/equivalence classes, and whose operations $\oplus$ and $\odot$ are defined by&#xA;\begin{align*}&#xA;[x]_I \oplus [y]_I &amp;amp;= [x+y]_I,\\&#xA;[x]_I\odot[y]_I &amp;amp;= [x\cdot y]_I&#xA;\end{align*}&#xA;where $+$ and $\cdot$ are the operations in $R$. (We later drop the distinction between $+$ and $\oplus$, but the point here is that they are operations defined on different sets, so they are really different functions, though very closely related). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your example, yes: $R/I$ is the collection of &lt;em&gt;all&lt;/em&gt; cosets, but remember that the same coset may have many different names. So, for instance, if $R=\mathbb{Z}$ and $I=(3)$, then $R/I$ consists of &lt;em&gt;all&lt;/em&gt; cosets, which are sets of the form&#xA;$$ a + (3) = \{ \ldots, a+3(-2), a+3(-1) , a+3(0), a+3(1), a+3(2), a+3(4),\ldots\}$$&#xA;for all $a$. But as it happens, every single coset is equal to either $0+(3)$, $1+(3)$, or $2+(3)$, so in fact $R/I$ has only three elements, even though each of those elements has infinitely many names:&#xA;\begin{align*}&#xA;0+(3) &amp;amp;= 3+(3) = 6+(3) = 9+(3) =\cdots = 3k+(3),\\&#xA;1+(3) &amp;amp;= 4+(3) = 7+(3) = 10+(3) = \cdots = 3k+1 + (3)\\&#xA;2+(3) &amp;amp;= 5+(3) = 8+(3) = 11+(3) = \cdots = 3k+2 + (3)&#xA;\end{align*}&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2011-02-25T04:38:45.547" CommentCount="8" />
  <row Id="23658" PostTypeId="1" CreationDate="2011-02-25T04:42:29.913" Score="8" ViewCount="178" Body="&lt;p&gt;What are the various sets of postulates that can used to derive Euclidean geometry?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It might be nice to have several different approaches together for comparison purposes and for ready reference.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It might also be interesting to include an axiomatization (or two) of elliptical geometry.&lt;/p&gt;&#xA;" OwnerUserId="6272" LastEditorUserId="6272" LastEditorDisplayName="" LastEditDate="2011-02-25T05:04:33.927" LastActivityDate="2011-03-01T22:37:29.077" Title="What are the postulates that can be used to derive geometry?" Tags="&lt;geometry&gt;&lt;big-list&gt;&lt;euclidean-geometry&gt;&lt;axioms&gt;" AnswerCount="5" CommentCount="1" FavoriteCount="3" />
  <row Id="23659" PostTypeId="2" ParentId="23658" CreationDate="2011-02-25T04:43:40.800" Score="4" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Axioms used in David Hilbert's &lt;em&gt;The Foundations of Geometry&lt;/em&gt; (1899), as translated by E. J. Townsend in 1902&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The undefined terms in Hilbert's axiomatization are &lt;em&gt;point&lt;/em&gt;, &lt;em&gt;line&lt;/em&gt;, &lt;em&gt;plane&lt;/em&gt;, &lt;em&gt;lies upon&lt;/em&gt;, &lt;em&gt;between&lt;/em&gt; and &lt;em&gt;congruent&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&lt;strong&gt;Group I: Axioms of Connection.&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The axioms of this group establish a connection&#xA;between the concepts indicated above; namely, points,&#xA;straight lines, and planes. These axioms are as follows:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Two distinct points $A$ and $B$ always completely&#xA;determine a straight line $a$. We write $AB = a$&#xA;or $BA = a$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Any two distinct points of a straight line completely&#xA;determine that line; that is, if $AB = a$ and&#xA;$AC=a$, where $B \neq C$, then is also $BC=a$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Three points $A$, $B$, $C$ not situated in the same&#xA;straight line always completely determine a plane&#xA;$\alpha$. We write $ABC=a$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Any three points $A$, $B$, $C$ of a plane $\alpha$, which&#xA;do not lie in the same straight line, completely determine&#xA;that plane.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If two points $A$, $B$ of a straight line $a$ lie in&#xA;a plane $\alpha$, then every point of $a$ lies in $a$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If two planes $\alpha$, $\beta$ have a point $A$ in common,&#xA;then they have at least a second point $B$ in common.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Upon every straight line there exist at least two&#xA;points, in every plane at least three points not&#xA;lying in the same straight line, and in space there&#xA;exist at least four points not lying in a plane.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&lt;strong&gt;Group II: Axioms of Order.&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The axioms of this group define the idea expressed&#xA;by the word &lt;em&gt;between&lt;/em&gt;, and make possible, upon the&#xA;basis of this idea, an order of sequence of the points&#xA;upon a straight line, in a plane, and in space. The&#xA;points of a straight line have a certain relation to one&#xA;another which the word &lt;em&gt;between&lt;/em&gt; serves to describe.&#xA;The axioms of this group are as follows:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;If $A$, $B$, $C$ are points of a straight line and&#xA;$B$ lies between $A$ and $C$, then $B$ lies also between&#xA;$C$ and $A$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If $A$ and $C$ are two points of a straight line,&#xA;then there exists at least one point $B$ lying between&#xA;$A$ and $C$ and at least one point $D$ so situated that&#xA;$C$ lies between $A$ and $D$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Of any three points situated on a straight line,&#xA;there is always one and only one which lies between&#xA;the other two.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Any four points $A$, $B$, $C$, $D$ of a straight line&#xA;can always be so arranged that $B$ shall lie between&#xA;$A$ and $C$ and also between $A$ and $D$, and, furthermore,&#xA;that $C$ shall lie between $A$ and $D$ and also&#xA;between $B$ and $D$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Let $A$, $B$, $C$ be three points not lying in the&#xA;same straight line and&#xA;let $a$ be a straight line lying in the plane $ABC$ and not passing through any of the&#xA;points $A$, $B$, $C$. Then, if the straight line $a$ passes through a point&#xA;of the segment $AB$, it will also pass through either a point of the segment&#xA;$BC$ or a point of the segment $AC$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&lt;strong&gt;Group III: Axiom of Parallels (The Axiom of Euclid).&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The introduction of this axiom simplifies greatly&#xA;the fundamental principles of geometry and facilitates&#xA;in no small degree its development. This axiom may&#xA;be expressed as follows:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;In a plane $\alpha$ there can be drawn through any&#xA;point $A$, lying outside of a straight line $a$, one and&#xA;only one straight line which does not intersect the&#xA;line $a$. This straight line is called the parallel to&#xA;$a$ through the given point $A$.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&lt;strong&gt;Group IV. Axioms of Congruence.&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The axioms of this group define the idea of congruence&#xA;or displacement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Segments stand in a certain relation to one another&#xA;which is described by the word &lt;em&gt;congruent&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;If $A$, $B$ are two points on a straight line $a$,&#xA;and if $A&amp;#39;$ is a point upon the same or another&#xA;straight line $a&amp;#39;$, then, upon a given side of $A&amp;#39;$ on&#xA;the straight line $a&amp;#39;$, we can always find one and&#xA;only one point $B&amp;#39;$ so that the segment $AB$ (or $BA$)&#xA;is congruent to the segment $A&amp;#39;B&amp;#39;$. We indicate&#xA;this relation by writing $AB\equiv A&amp;#39;B&amp;#39;$.&#xA;Every segment is congruent to itself; &#xA;that is, we always have $AB\equiv AB$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If a segment $AB$ is congruent to the segment&#xA;$A&amp;#39;B&amp;#39;$ and also to the segment $A&amp;#39;&amp;#39;B&amp;#39;&amp;#39;$, then the segment&#xA;$A&amp;#39;B&amp;#39;$ is congruent to the segment $A&amp;#39;&amp;#39;B&amp;#39;&amp;#39;$; that&#xA;is, if $AB \equiv A&amp;#39;B$ and $AB \equiv A&amp;#39;&amp;#39;B&amp;#39;&amp;#39;$, then&#xA;$A&amp;#39;B&amp;#39; \equiv A&amp;#39;&amp;#39;B&amp;#39;&amp;#39;$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Let $AB$ and $BC$ be two segments of a straight&#xA;line $a$ which have no points in common aside from&#xA;the point $B$, and, furthermore, let $A&amp;#39;B&amp;#39;$ and $B&amp;#39;C&amp;#39;$&#xA;be two segments of the same or of another straight&#xA;line $a&amp;#39;$ having, likewise, no point other than $B&amp;#39;$ in&#xA;common.  Then, if $AB \equiv A&amp;#39;B&amp;#39;$ and $BC \equiv B&amp;#39;C&amp;#39;$,&#xA;we have $AC \equiv A&amp;#39;C&amp;#39;$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Let an angle $(h,k)$ be given in the plane&#xA;$\alpha$ and let a straight line $a&amp;#39;$ be given in a plane $\alpha&amp;#39;$.&#xA;Suppose also that, in the plane $\alpha$, a definite side&#xA;of the straight line $a&amp;#39;$ be assigned. Denote by $h&amp;#39;$ a&#xA;half-ray of the straight line $a&amp;#39;$ emanating from a&#xA;point $O&amp;#39;$ of this line. Then in the plane $\alpha&amp;#39;$ there&#xA;is one and only one half-ray $k&amp;#39;$ such that the angle&#xA;$(h,k)$, or $(k,h)$, is congruent to the angle $(h&amp;#39;,k&amp;#39;)$&#xA;and at the same time all interior points of the angle&#xA;$(h&amp;#39;,k&amp;#39;)$ lie upon the given side of $a&amp;#39;$. We express&#xA;this relation by means of the notation&#xA;$\angle (h,k) \equiv \angle (h&amp;#39;,k&amp;#39;)$ &#xA;Every angle is congruent to itself; that is, &#xA;$\angle (h,k) \equiv \angle (h,k)$&#xA;or $\angle (h,k) \equiv \angle (k,h)$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;f the angle $(h,k)$ is congruent to the angle&#xA;$(h&amp;#39;,k&amp;#39;)$ and to the angle $(h&amp;#39;&amp;#39;,k&amp;#39;&amp;#39;)$, then the angle&#xA;$(h&amp;#39;,k&amp;#39;)$ is congruent to the angle $(h&amp;#39;&amp;#39;,k&amp;#39;&amp;#39;)$; that is&#xA;to say, if $\angle (h, k) \equiv \angle (h&amp;#39;, k&amp;#39;)$ and&#xA;$\angle (h, k) \equiv \angle (h&amp;#39;&amp;#39;,k&amp;#39;&amp;#39;)$, then&#xA;$\angle (h&amp;#39;,k&amp;#39;) \equiv \angle (h&amp;#39;&amp;#39;,k&amp;#39;&amp;#39;)$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If, in the two triangles $ABC$ and $A&amp;#39;B&amp;#39;C&amp;#39;$&#xA;the congruences&#xA;$AB \equiv A&amp;#39;B&amp;#39;, \: AC \equiv A&amp;#39;C&amp;#39;, \: \angle BAC \equiv \angle B&amp;#39;A&amp;#39;C&amp;#39;$&#xA;hold, then the congruences $\angle ABC \equiv \angle A&amp;#39;B&amp;#39;C&amp;#39; \:\mbox{and}\; \angle ACB \equiv \angle A&amp;#39;C&amp;#39;B&amp;#39;$&#xA;also hold.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Some definitions relevant to the axioms of congruence:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $\alpha$ be any arbitrary plane and $h$,&#xA;$k$ any two distinct half-rays lying in $\alpha$ and emanating&#xA;from the point $O$ so as to form a part of two different&#xA;straight lines. We call the system formed by these&#xA;two half-rays $h$, $k$ an &lt;em&gt;angle&lt;/em&gt; and represent it by the&#xA;symbol $\angle(h, k)$ or $\angle(k, h)$. From axioms II, 1--5, it&#xA;follows readily that the half-rays $h$ and $k$, taken together&#xA;with the point $O$, divide the remaining points&#xA;of the plane a into two regions having the following&#xA;property: If $A$ is a point of one region and $B$ a point&#xA;of the other, then every broken line joining $A$ and $B$&#xA;either passes through $O$ or has a point in common&#xA;with one of the half-rays $h$, $k$. If, however, $A$, $A&amp;#39;$&#xA;both lie within the same region, then it is always possible&#xA;to join these two points by a broken line which&#xA;neither passes through $O$ nor has a point in common&#xA;with either of the half-rays $h$, $k$. One of these two&#xA;regions is distinguished from the other in that the segment&#xA;joining any two points of this region lies entirely&#xA;within the region. The region so characterised is&#xA;called the &lt;em&gt;interior of the angle $(h,k)$&lt;/em&gt;. To distinguish&#xA;the other region from this, we call it the &lt;em&gt;exterior of&#xA;the angle $(h,k)$&lt;/em&gt;. The half rays $h$ and $k$ are called the&#xA;&lt;em&gt;sides of the angle&lt;/em&gt;, and the point $O$ is called the &lt;em&gt;vertex&#xA;of the angle&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&lt;strong&gt;Group V. Axiom of Continuity (The Axiom of Archimedes).&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This axiom makes possible the introduction into&#xA;geometry of the idea of continuity. In order to state&#xA;this axiom, we must first establish a convention concerning&#xA;the equality of two segments. For this purpose,&#xA;we can either base our idea of equality upon the&#xA;axioms relating to the congruence of segments and&#xA;define as &lt;em&gt;equal&lt;/em&gt; the correspondingly congruent segments,&#xA;or upon the basis of groups I and II, we may&#xA;determine how, by suitable constructions, &#xA;a segment is to be laid off from a point of a&#xA;given straight line so that a new, definite segment is&#xA;obtained &lt;em&gt;equal&lt;/em&gt; to it. In conformity with such a&#xA;convention, the axiom of Archimedes may be stated&#xA;as follows:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Let $A_1$ be any point upon a straight line between&#xA;the arbitrarily chosen points $A$ and $B$. Take the&#xA;points $A_2$, $A_3$, $A_4,\ldots$ so that $A_1$ lies between $A$&#xA;and $A_2$, $A_2$ between $A_1$ and $A_3$, $A_3$ between $A_2$ and&#xA;$A_4$ etc. Moreover, let the segments&#xA;$  A A_1, \; A_1 A_2, \; A_2 A_3, \; A_3 A_4, \;\ldots$&#xA;be equal to one another. Then, among this series&#xA;of points, there always exists a certain point $A_n$&#xA;such that $B$ lies between $A$ and $A_n$.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="6272" LastEditorUserId="6272" LastEditorDisplayName="" LastEditDate="2011-02-25T05:47:02.767" LastActivityDate="2011-02-25T05:47:02.767" CommunityOwnedDate="2011-02-25T04:44:22.610" />
  <row Id="23660" PostTypeId="2" ParentId="23658" CreationDate="2011-02-25T04:45:21.797" Score="3" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Postulates used in John M. Lee's &lt;em&gt;Axiomatic Geometry&lt;/em&gt; (Draft, 2011)&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Used with author's permission&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&lt;strong&gt;Postulates of Neutral Geometry&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The undefined terms in Jack Lee's axiomatization are &lt;em&gt;point&lt;/em&gt;, &lt;em&gt;line&lt;/em&gt;, &lt;em&gt;distance&lt;/em&gt; (between points) and &lt;em&gt;measure&lt;/em&gt; (of an angle).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Postulate 1&lt;/strong&gt; (The Set Postulate). Every line is a set of points, and there is a set of point called the plane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Postulate 2&lt;/strong&gt; (The Existence Postulate). There exist at least two distinct points.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Postulate 3&lt;/strong&gt; (The Unique Line Postulate). Given any two points, there is a unique line that contains both of them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Postulate 4&lt;/strong&gt; (The Distance Postulate). For every pair of points $A$ and $B$, the &lt;em&gt;distance between $A$ and $B$&lt;/em&gt; is a non-negative real number determined by $A$ and $B$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Postulate 5&lt;/strong&gt; (The Ruler Postulate). For every line $\ell$, there is a bijective function $\mathcal{f}:\ell \rightarrow \mathbb{R}$ with the property that for any two points $A, B \in \ell$, we have &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ AB = | \mathcal{f}(B) - \mathcal{f}(A)|.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Postulate 6&lt;/strong&gt; (The Plane Separation Postulate). For any line $\ell$, the set of all points not on $\ell$ is the union of two disjoint, non-empty subsets called &lt;em&gt;sides of $\ell$&lt;/em&gt;. If $A$ anf $B$ are distinct points not on $\ell$, then both of the following conditions are satisfied:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;$A$ and $B$ are on the same side of line $\ell$ if and only if $\overline{AB} \cap \ell= \oslash$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$A$ and $B$ are on the opposite side of line $\ell$ if and only if $\overline{AB} \cap \ell \neq \oslash$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Postulate 7&lt;/strong&gt; (The Angle Measure Postulate). For every angle $\angle ab$, the &lt;em&gt;measure of $\angle ab$&lt;/em&gt;, written as $m\angle ab$, is a real number in the closed interval $[0,180]$ determined by $\angle ab$.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Definition: If $\overrightarrow r$ is a ray starting at point $O$, and $P$ is a point not on the line $\overleftrightarrow r$, the &lt;em&gt;half-rotation of rays determined by $\overrightarrow r$ and $P$&lt;/em&gt;, denoted by $HR(\overrightarrow r, P )$, is the set whose elements are all the rays that start at $O$ and whose points are either in line $\overleftrightarrow r$ or on the same side of $\overleftrightarrow r$ as $P$. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Postulate 8&lt;/strong&gt; (The Protractor Postulate). For every ray $\overrightarrow r$ and every point $P$ not on line $\overleftrightarrow{r}$, there is a bijective function $g:HR(\overrightarrow{r}, P) \rightarrow [0,180]$ such that the following two conditions are satisfied:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;the function $g$ assigns the number $0$ to $\overrightarrow{r}$ and the number $180$ to the ray opposite $\overrightarrow{r}$. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If $\overrightarrow{a}$ and $\overrightarrow{b}$ are any two rays in $HR(\overrightarrow{r}, P)$ then $$ m\angle ab = |g(\overrightarrow b)-g(\overrightarrow b)|.$$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Postulate 9&lt;/strong&gt; (The SAS Postulate). If there is a correspondence between the vertices of two triangles such that two sides and the included angle of one triangle are congruent to the corresponding sides and angle of the other triangle, then the triangles are congruent under that correspondence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&lt;strong&gt;Postulates of Euclidean Geometry&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Postulates for Euclidean geometry are the postulates are Postulates 1–9 of neutral geometry, plus the following: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Postulate 10E&lt;/strong&gt; (The Euclidean Parallel Postulate). For each line $\ell$ and each point $A$ that does not lie on $\ell$, there is a unique line that contains $A$ and is parallel to $\ell$.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Definition: Lines $\ell$ and $m$ are called parallel if they do not intersect.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;There is one additional undefined term for Euclidean geometry: &lt;em&gt;area&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Postulate 11E&lt;/strong&gt; (The Euclidean Area Postulate).   For every polygonal region $\mathcal{R}$, there is a positive real number $\alpha(\mathcal{R})$ called the &lt;em&gt;area of $\mathcal{R}$&lt;/em&gt;, which satisfies the following three conditions:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;(Area Congruence Property) If $\mathcal{R}_1$ and $\mathcal{R}_2$ are congruent simple regions, then $$ \alpha(\mathcal{R}_1)=\alpha(\mathcal{R}_2).$$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;(Area Addition Property) If $\mathcal{R}_1, \dots , \mathcal{R}_n$ are non-overlapping simple regions, then $$\alpha(\mathcal{R}_1 \cup \cdots \cup \mathcal{R}_n)= \alpha(\mathcal{R}_1) + \cdots + \alpha( \mathcal{R}_n).$$  &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;(Unit Area Property) If $\mathcal{R}$ is a square region with sides of length $1$, then $\alpha(\mathcal{R})=1$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Some useful definitions: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;A &lt;em&gt;polygon&lt;/em&gt; is a broken line segment that is simple closed and proper.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;A &lt;em&gt;broken line segment&lt;/em&gt; is the union of finitely many segments $\overline{A_1 A_2},\, \overline{A_2 A_3}, \dots ,\, \overline{A_{n} A_{n+1}}$ determined by points $A_1, \dots , A_{n+1}$, not necessarily distinct. We denote this $\overline{A_1, \dots , A_{n+1}}$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Given a broken line segment $\overline{A_1, \dots , A_{n+1}}$, we call it &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;simple&lt;/em&gt; if the first $n$ vertices are distinct and no two of its constituent segments, edges, intersect except at a common end point.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;closed&lt;/em&gt; if the first and last points points, $A_1$ and $A_{n+1}$, are the same.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;proper&lt;/em&gt; if for each $i=1, \dots , n-1$, the three end points $A_i, A_{i+1}, A_{i+2}$ are non-collinear and if the broken line segment is closed, we require that $A_n, A_1,  A_2$ are non-collinear as well.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;We call two polygons &lt;em&gt;congruent&lt;/em&gt; if there is a correspondence between their vertices such that consecutive vertices correspond to consecutive vertices, corresponding edges are congruent, and corresponding interior angle measures are equal.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Given a polygon $\mathcal{P}$, a point $Q$ not on $\mathcal{P}$ is said to be an &lt;em&gt;interior point of $\mathcal{P}$&lt;/em&gt; if a ray starting with $Q$ and not containing any vertices of the polygon $\mathcal{P}$ has an odd number of intersections with the polygon.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;We call a set of points $\mathcal{R}$ a &lt;em&gt;simple region&lt;/em&gt; if, for some  polygon $\mathcal{P}$, $\mathcal{R}$ is the union of $\mathcal{P}$ and its interior.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Two regions are &lt;em&gt;congruent&lt;/em&gt; if their associated polygons are congruent.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Two simple regions are said to be &lt;em&gt;non-overlapping&lt;/em&gt; if their interiors are disjoint.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;A polygon is called a &lt;em&gt;square&lt;/em&gt; if it has four sides, all four sides are congruent and all four angles have measure 90.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;We describe a region as a &lt;em&gt;square region&lt;/em&gt; if its associated polygon is a square.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&lt;strong&gt;Postulates of Hyperbolic Geometry&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Postulates for hyperbolic geometry are the postulates are Postulates 1–9 of neutral geometry, plus the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Postulate 10H&lt;/strong&gt; (The Hyperbolic Parallel Postulate). For each line $\ell$ and each point $A$ that does not lie on $\ell$, there are at least two distinct lines that contain $A$ and are parallel to $\ell$.&lt;/p&gt;&#xA;" OwnerUserId="6272" LastEditorUserId="6272" LastEditorDisplayName="" LastEditDate="2011-02-25T05:19:59.427" LastActivityDate="2011-02-25T05:19:59.427" CommunityOwnedDate="2011-02-25T04:45:21.797" />
  <row Id="23661" PostTypeId="2" ParentId="23658" CreationDate="2011-02-25T04:46:48.070" Score="3" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Postulates used in George D. Birkhoff's &lt;em&gt;A set of postulates for plane geometry, based on scale and protractor&lt;/em&gt; (1932)&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The undefined terms in Birkhoff's axiomatization are &lt;em&gt;point&lt;/em&gt;, &lt;em&gt;line&lt;/em&gt;, &lt;em&gt;distance&lt;/em&gt; and &lt;em&gt;angle&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Postulate I&lt;/strong&gt; (The Postulate of Line Measure). The points $A, B, \dots$ of any line can be put into $1:1$ correspondence with the real numbers $x$ so that $|x_B-x_A| = d(A,B)$ for all points $A$ and $B$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Postulate II&lt;/strong&gt; (The Point-line Postulate). One and only one line, $\ell$, contains any two distinct points $P$ and $Q$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Postulate III&lt;/strong&gt; (The Postulate of Angle Measure). The half-lines, or rays, $\ell, m, \dots$ through any point $O$ can be put into $1:1$ correspondence with the real numbers $a \; ( \textrm{ mod }2 \pi)$ so that if $A \neq 0$ and $B \neq 0$ are points of $\ell$ and $m$, respectively, the difference $a_m - a_l \; ( \textrm{ mod }2 \pi)$ of the numbers associated with lines $\ell$ and $m$ is $\angle AOB$. Furthermore, if the point $B$ on $m$ varies continuously in a line $r$ not containing the vertex $O$, the number $a_m$ varies continuously also.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Postulate IV&lt;/strong&gt; (Postulate of Similarity). If in two triangles $\triangle ABC$ and $\triangle A&amp;#39;B&amp;#39;C&amp;#39;$ and for some constant $k &amp;gt; 0,\; d(A&amp;#39;, B&amp;#39;) = kd(A, B),\; d(A&amp;#39;, C&amp;#39;) = kd(A, C)$, and $\angle B&amp;#39;A&amp;#39;C&amp;#39; = \pm \angle BAC$, then also $d(B&amp;#39;, C&amp;#39;) = kd(B, C),\; \angle C&amp;#39;B&amp;#39;A&amp;#39; = \pm \angle CBA$, and $\angle A&amp;#39;C&amp;#39;B&amp;#39; = \pm \angle ACB$.&lt;/p&gt;&#xA;" OwnerUserId="6272" LastEditorUserId="6272" LastEditorDisplayName="" LastEditDate="2011-02-25T06:43:56.453" LastActivityDate="2011-02-25T06:43:56.453" CommunityOwnedDate="2011-02-25T04:46:48.070" CommentCount="3" />
  <row Id="23662" PostTypeId="2" ParentId="23579" CreationDate="2011-02-25T04:49:45.110" Score="3" ViewCount="" Body="&lt;p&gt;I agree heartily with Qiaochu: the case in which all terms are non-negative is a trivial case of Abel's Theorem.  This comes out in my take on Abel's Theorem in &lt;a href=&quot;http://math.uga.edu/~pete/243series7.pdf&quot; rel=&quot;nofollow&quot;&gt;these notes&lt;/a&gt; on an undergraduate analysis class I taught a few years ago.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that I begin by giving exactly the proof from Rudin (why not?).  After that I have some discussion of &quot;Abel summation&quot; and applications of this result.  Note in particular Exercise 40: a series with non-negative terms is Abel-summable iff it is convergent in the usual sense.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It might also help to see some typical applications of Abel's Theorem, which are to non-convergent series.  In my notes, I first use it to prove that &lt;em&gt;if&lt;/em&gt; the &lt;a href=&quot;http://en.wikipedia.org/wiki/Cauchy_product&quot; rel=&quot;nofollow&quot;&gt;Cauchy product&lt;/a&gt; of two convergent series converges at all, then it converges to the expected value, namely the product of the two sums.  Note that earlier on in my notes I had proved that when at least one of the two series is absolutely convergent the Cauchy product necessarily converges to the expected value, whereas there are examples of divergent Cauchy products of two nonabsolutely convergent series.  Then I use Abel's Theorem to justify the &quot;calculus identity&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sum_{n=0}^{\infty} \frac{(-1)^n}{2n+1} = \frac{\pi}{4}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In both cases the setting is of non-absolute convergence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I haven't looked back at my copy of Rudin, but I am willing to believe that he does not explicitly mention the significance of the non-absolute convergence here.  If so, this is an instance of his book being a little too laconic for contemporary eyes.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Added&lt;/b&gt;: Okay, I just looked back at Rudin's book.  First, he also follows the statement and proof of the theorem with the application to Cauchy products (what a coincidence...).  Second -- it's true: he really should hit a little harder the point that the content comes when the terms are not all non-negative, and moreover when the series is not absolutely convergent.  (Nevertheless Chapter 8 is by far my favorite in Rudin's book: a lot of classical content is covered in a very clean, efficient way.)&lt;/p&gt;&#xA;" OwnerUserId="299" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2011-02-25T05:00:26.620" LastActivityDate="2011-02-25T05:00:26.620" />
  <row Id="23664" PostTypeId="1" CreationDate="2011-02-25T04:52:45.837" Score="10" ViewCount="246" Body="&lt;p&gt;Are there any areas of mathematics that are known to be impossible to formalise in terms of set theory?&lt;/p&gt;&#xA;" OwnerUserId="7485" LastActivityDate="2011-02-25T04:52:45.837" Title="Are there any areas of mathematics that are known to be impossible to formalise in terms of set theory?" Tags="&lt;set-theory&gt;" CommentCount="7" FavoriteCount="1" />
  <row Id="23665" PostTypeId="1" CreationDate="2011-02-25T05:51:08.373" Score="2" ViewCount="89" Body="&lt;p&gt;I'm looking for a way to project a vector (in this case a function on the real line) onto a basis for that space (in this case the set of N-degree polynomials over the domain of a closed interval) with respect to minimizing the L-infinity norm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another way you could phrase this is that I'm looking to find the polynomial of degree N which approximates a function with the smallest absolute value distance from that function on a given interval.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When looking to approximate the function using the L2 norm, it was pretty straightforward.  I created an orthonormal basis of polynomials, and then I could simply take the dot product with each of these basis vectors to do the projection.  But that minimized the mean squared error, which is not as important as the absolute value error for my purposes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem with the L-Infinity norm is that you can't get an orthonormal basis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also tried looking at it as a minimization problem, with respect to partial derivatives of the coefficients, but because it involves an absolute value and boundary conditions, you end up with a whole bunch of corner cases, not to mention a larger and larger set of equations to solve for higher degree polynomials.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've considered just using some machine learning technique to approximate the values, but that's nasty and I'm interested in the math behind it anyway.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT:I have found out that what I am looking for is called the &quot;minimax polynomial&quot;, but I'd still be interested in some of the math behind it.&lt;/p&gt;&#xA;" OwnerUserId="7480" LastEditorUserId="7480" LastEditorDisplayName="" LastEditDate="2011-02-25T08:53:40.043" LastActivityDate="2011-02-25T08:53:40.043" Title="Projecting onto vector space with L-Infinity norm/ minimum absolute value polynomial fitting" Tags="&lt;polynomials&gt;&lt;optimization&gt;&lt;norm&gt;" FavoriteCount="1" />
  <row Id="23666" PostTypeId="2" ParentId="23625" CreationDate="2011-02-25T06:32:48.403" Score="2" ViewCount="" Body="&lt;p&gt;I don't know very much about this, but it seems like the relevant field is &lt;a href=&quot;http://en.wikipedia.org/wiki/Calculus_of_variations&quot; rel=&quot;nofollow&quot;&gt;Calculus of Variations&lt;/a&gt;.  For a fixed $a$, $\pi(f,a)$ is a functional of f.  The continuity of the functional would depend on what space of functions you are optimizing over.  I read a few chapters of &lt;em&gt;Calculus of Variations&lt;/em&gt; by Gelfand a while back and I found it quite accessible.&lt;/p&gt;&#xA;" OwnerUserId="1004" LastActivityDate="2011-02-25T06:32:48.403" />
  <row Id="23667" PostTypeId="1" CreationDate="2011-02-25T06:34:34.967" Score="1" ViewCount="56" Body="&lt;p&gt;In &lt;a href=&quot;http://en.wikipedia.org/wiki/Conditional_expectation&quot; rel=&quot;nofollow&quot;&gt;the Wikipedia article for conditional expectation&lt;/a&gt;,  conditional probability is defined in terms of conditional expectation.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Given a sub sigma algebra of the one&#xA;on a probability space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given a probability space $(\Omega,&#xD;&#xA;    \mathcal{F}, P)$, a conditional&#xA;probability $P(A \mid \mathcal{B})$&#xA;of a measurable subset $A \in&#xD;&#xA;    \mathcal{F}$ given a sub sigma&#xA;algebra $\mathcal{B}$ of&#xA;$\mathcal{F}$, is defined as the&#xA;conditional expectation $E(A \mid&#xD;&#xA;    \mathcal{B})$ of indicator function&#xA;$i_A$ of $A$ given $\mathcal{B}$,&#xA;i.e. $$ P(A \mid \mathcal{B}): = E(A&#xD;&#xA;    \mid \mathcal{B}), \forall A \in&#xD;&#xA;    \mathcal{F}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So actually the conditional&#xA;probability $P(\cdot \mid&#xD;&#xA;    \mathcal{B})$ is a mapping $: \Omega&#xD;&#xA;    \times \mathcal{F} \rightarrow&#xD;&#xA;    \mathbb{R}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A conditional probability $P(\cdot&#xD;&#xA;    \mid \mathcal{B})$ is called regular if&#xA;$P(\cdot|\mathcal{B})(\omega),&#xD;&#xA;    \forall \omega \in \Omega$ is also a&#xA;probability measure. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;what are some&#xA;necessary and/or sufficient&#xA;conditions for a conditional&#xA;probability $P(\cdot \mid&#xD;&#xA;    \mathcal{B})$ to be regular?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Given a r.v. on a probability space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose $(\Omega, \mathcal{F}, P)$&#xA;is a probability space and $(U,&#xD;&#xA;    \mathcal{\Sigma})$ is a measurable&#xA;space. There seem to be two ways of&#xA;defining the conditional expectation&#xA;$E(X\mid Y)$ of a r.v. $X: \Omega&#xD;&#xA;    \rightarrow \mathbb{R}$ given&#xA;another r.v. $Y: \Omega \rightarrow&#xD;&#xA;    U$, either as a&#xA;$\sigma(Y)$-measurable mapping $:&#xD;&#xA;    \Omega \rightarrow \mathbb{R}$, or&#xA;as a $\Sigma$-measurable mapping $:&#xD;&#xA;    U \rightarrow \mathbb{R}$,  as &lt;a href=&quot;http://math.stackexchange.com/questions/23620/definition-of-conditional-expectation-of-a-random-variable-given-another-one&quot;&gt;in&#xA;my previous post&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If one let $X$ to be the indicator&#xA;function $1_A$ for some $A \in&#xD;&#xA;    \mathcal{F}$, one can similarly&#xA;define  $E(1_A \mid Y)$ to be&#xA;conditional probability of $A$ given&#xA;$Y$, denoted as $P(A\mid Y)$.&#xA;Therefore $P(\cdot \mid Y)$ is a&#xA;mapping $: \Omega \times \mathcal{F}&#xD;&#xA;    \rightarrow \mathbb{R}$ or a mapping&#xA;$: U \times \mathcal{F} \rightarrow&#xD;&#xA;    \mathbb{R}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Questions:&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1). What are some necessary and/or&#xA;    sufficient conditions for $P(\cdot&#xD;&#xA;        \mid Y)$ to be regular, i.e. to be a&#xA;    mapping $: \Omega \rightarrow \{&#xD;&#xA;        \text{probability measures on&#xD;&#xA;        }(\Omega, \mathcal{F}) \}$ or a&#xA;    mapping $: U \rightarrow \{&#xD;&#xA;        \text{probability measures on&#xD;&#xA;        }(\Omega, \mathcal{F}) \}$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2). Under what kinds of conditions, will&#xA;    $P(X \mid Y)$ defined as above be&#xA;    equal to the ratio $\frac{P(X, Y)}{P(Y)}$, the&#xA;    definition used in elementary&#xA;    probability?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Thanks and regards! References (links or books) will also be appreciated!&lt;/p&gt;&#xA;" OwnerUserId="1281" LastEditorUserId="1281" LastEditorDisplayName="" LastEditDate="2011-02-25T07:20:37.653" LastActivityDate="2011-02-25T08:52:02.307" Title="When a conditional probability becomes a mapping to probability measures" Tags="&lt;probability-theory&gt;" AnswerCount="1" />
  <row Id="23668" PostTypeId="1" AcceptedAnswerId="23672" CreationDate="2011-02-25T07:05:17.867" Score="4" ViewCount="74" Body="&lt;p&gt;Hi all, i hope that you can help.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the Introduction to Set Theory by Monk, J. Donald (James Donald)... he defined ordinal to be:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Definition (1):$A$ is an ordinal iff $A$ is ∈-transitive and each member of $A$ is ∈-transitive.$A$ is ∈-transitive iff for all $x$ and $y$ , $x \in y \in A \implies x \in A$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And using definition (1), i have a problem of showing  $0 \in $Ord.     &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My solution: let $x \in y \in 0$, then show that $x \in 0$, but in the theorem shown before this, there cannot exist an $x \in 0$ for any $x$, for if $x \in 0$, then $x$ is a set and $x \neq x$, which is absurd. Hence  there cannot be an $x$ such that $x\in 0$ .But if this is the case, we cannot shown the above theorem by definition. Any ideas?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;thanks for your help.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit:Initially i have 2 question to ask. But later on, i have found the answer to my first question. That is why i deleted it and the question post above is my second question.&lt;/p&gt;&#xA;" OwnerUserId="1683" LastEditorUserId="1683" LastEditorDisplayName="" LastEditDate="2011-02-25T07:31:18.107" LastActivityDate="2011-02-25T08:02:37.350" Title="About Ordinal Number." Tags="&lt;set-theory&gt;&lt;elementary-set-theory&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="23669" PostTypeId="1" AcceptedAnswerId="23698" CreationDate="2011-02-25T07:34:54.077" Score="7" ViewCount="132" Body="&lt;p&gt;hi&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $f,g$ be continuous functions from $X$ to $Y$, $X$ is a topological space and $Y$ a topological space under ordered topology. Then prove that the set $\{x \in X \ | \ f(x) &amp;lt; g(x)\}$ is open.&#xA;  I want to know that what intrisic property of order makes it possible.&lt;/p&gt;&#xA;" OwnerUserId="1252" LastActivityDate="2011-02-25T13:34:44.093" Title="property about topological space." Tags="&lt;general-topology&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" />
  <row Id="23670" PostTypeId="1" AcceptedAnswerId="23836" CreationDate="2011-02-25T07:36:52.623" Score="3" ViewCount="103" Body="&lt;ol&gt;&#xA;&lt;li&gt;When there is no hash function used, $s = k-x \cdot m \mod {q}$ &lt;strong&gt;instead of&lt;/strong&gt; $s = k-x \cdot H(m||r) \mod{q}$?&lt;/li&gt;&#xA;&lt;li&gt;When a hash function is defined as $H(m)$ &lt;strong&gt;instead of&lt;/strong&gt; $H(m||r)$?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Ref: &lt;a href=&quot;http://en.wikipedia.org/wiki/Schnorr_signature&quot; rel=&quot;nofollow&quot;&gt; Schnorr Signature &lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7488" LastEditorUserId="7488" LastEditorDisplayName="" LastEditDate="2011-02-26T09:37:16.380" LastActivityDate="2011-02-26T15:34:13.127" Title="How is the Schnorr Signature insecure in the following 2 scenarios:" Tags="&lt;homework&gt;&lt;cryptography&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="23671" PostTypeId="1" AcceptedAnswerId="23676" CreationDate="2011-02-25T08:01:19.463" Score="1" ViewCount="61" Body="&lt;p&gt;I saw this expression in a book and I cannot understand how did he get this expression.&#xA;Suppose $Z_t$ and $D_t$ are some stochastic processes and we have these expressions,&#xA;$Z_{t_k} - Z_{t_{k-1}} = (t_k - t_{k-1})(D_1 + ... + D_{n-k+1}) - t_{k-1}D_{n-k+2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$Z_{t_i} - Z_{t_{i-1}} = (t_i - t_{i-1})(D_1 + ... + D_{n-i+1}) - t_{i-1}D_{n-i+2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How did they get this covariance equation?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$Cov(Z_{t_k} - Z_{t_{k-1}}, Z_{t_i} - Z_{t_{i-1}}) = (t_k - t_{k-1})(t_i - t_{i-1}) \sum_{m=1}^{n-i+1} Var(D_m) - t_{i-1}((t_k - t_{k-1})Var(D_{n-i+2})$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any hint appreciated as I cannot figure out which formula they are using here.&lt;/p&gt;&#xA;" OwnerUserId="957" LastActivityDate="2011-02-25T08:27:17.137" Title="example on variance of stochastic processes" Tags="&lt;stochastic-processes&gt;&lt;correlation&gt;" AnswerCount="1" />
  <row Id="23672" PostTypeId="2" ParentId="23668" CreationDate="2011-02-25T08:02:37.350" Score="2" ViewCount="" Body="&lt;p&gt;Seoral: A statement of the form &quot;for all $x\in A$ blah&quot; literally means &quot;for all $x$, if it is the case that $x\in A$, then blah&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now: As you mentioned, $y\in 0$ is false for any $y$. We need to show that if $x\in y\in 0$ then $x\in 0$. More precisely, this means: For all $x$ and $y$, if $y\in 0$ and $x\in y$, then $x\in 0$. Since $y\in 0$ is false, we need to show that for all $x$ and $y$, (false and $x\in y$) implies $x\in 0$. Recall that &quot;false and blah&quot; is false, so this is just: For all $x$ and $y$, false implies $x\in 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But false implies anything! (More precisely, statements $p\to q$ are true whenever $p$ is false. They are also true when $q$ is true, but that doesn't matter here.) So, we have &quot;for all $x$ and $y$, true&quot;, which is the same as &quot;true&quot;. I.e., you have proved what you needed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One usually doesn't express situations as the one above this way; instead, people talk of statements being &quot;vacuously true&quot;, as pointed out in a comment above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's review: 0 is transitive, because any element of 0 is a subset. (Precisely, because there are no elements of 0.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, any element of 0 is transitive. Again, because there are no elements of 0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is somewhat strange the first time one encounters it, since it is also the case that any element of 0 fails to be transitive (and that any element of 0 is blue, etc).&lt;/p&gt;&#xA;" OwnerUserId="462" LastActivityDate="2011-02-25T08:02:37.350" CommentCount="6" />
  <row Id="23673" PostTypeId="2" ParentId="23669" CreationDate="2011-02-25T08:12:15.917" Score="1" ViewCount="" Body="&lt;p&gt;Hey i found this somewhere on the net. See &lt;strong&gt;problem 5&lt;/strong&gt; in the given link:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.math.uiowa.edu/~jsimon/COURSES/M132Fall07/M132Fall07_Exam1_Solutions.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.math.uiowa.edu/~jsimon/COURSES/M132Fall07/M132Fall07_Exam1_Solutions.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" LastActivityDate="2011-02-25T13:34:44.093" />
  <row Id="23674" PostTypeId="1" AcceptedAnswerId="23678" CreationDate="2011-02-25T08:14:13.577" Score="4" ViewCount="124" Body="&lt;p&gt;Let $v$ be the 3-adic valuation on $\mathbb{Q}$ and consider the subring $\mathbb{Z}_{(3)}$ of $\mathbb{Q}$ defined by&#xA;$$&#xD;&#xA;\mathbb{Z}_{(3)} = \{ x \in \mathbb{Q} : v(x) \geq 0  \}.&#xD;&#xA;$$&#xA;That is, $\mathbb{Z}_{(3)}$ is the ring of rational numbers that are integral with respect to $v$.  $\mathbb{Z}_{(3)}$ is also the localization of $\mathbb{Z}$ at the prime ideal $(3)$.  I know $\mathbb{Z}_{(3)}$ is integrally closed in $\mathbb{Q}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to find the integral closure of $\mathfrak{O}$ in the field $\mathbb{Q}(\sqrt{-5})$:&#xA;$$&#xD;&#xA;\overline{\mathbb{Z}_{(3)}} = \{x \in  \mathbb{Q}(\sqrt{-5})  : x \text{ is a root of a monic irreducible polynomial with coefficients in } \mathbb{Z}_{(3)} \}&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I do this?  What should I be thinking about?&lt;/p&gt;&#xA;" OwnerUserId="6887" LastActivityDate="2011-02-25T14:10:53.053" Title="How to find the integral closure of $\mathbb{Z}_{(3)}$ in the field $\mathbb{Q}(\sqrt{-5})$?" Tags="&lt;number-theory&gt;&lt;algebraic-number-theory&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="2" />
  <row Id="23675" PostTypeId="2" ParentId="23622" CreationDate="2011-02-25T08:18:41.663" Score="1" ViewCount="" Body="&lt;p&gt;You should consider the cases $n=k$ and $n\not=k$ separately (before you integrate).&lt;/p&gt;&#xA;" OwnerUserId="5882" LastActivityDate="2011-02-25T08:18:41.663" />
  <row Id="23676" PostTypeId="2" ParentId="23671" CreationDate="2011-02-25T08:27:17.137" Score="1" ViewCount="" Body="&lt;p&gt;Assume $n-i+1 &amp;lt; n-k+1$ and that $D_i$ are independent. Substitute for $Z_{t_k}-Z_{t_{k-1}}$ and $Z_{t_i}-Z_{t_{i-1}}$ in&#xA;${\rm Cov}(Z_{t_k}-Z_{t_{k-1}},Z_{t_i}-Z_{t_{i-1}})$. Use linearity of covariance. Use the fact that independent random variables are uncorrelated. Use that fact that ${\rm Cov}(X,X) = {\rm Var}(X)$. &lt;/p&gt;&#xA;" OwnerUserId="2810" LastActivityDate="2011-02-25T08:27:17.137" CommentCount="2" />
  <row Id="23677" PostTypeId="1" CreationDate="2011-02-25T08:43:34.267" Score="2" ViewCount="54" Body="&lt;ol&gt;&#xA;&lt;li&gt;Given random variables $X, Y, Z$, is&#xA;joint distribution of $X$ and &quot;$Y|Z$&quot; &#xA;same as distribution of $(X, Y)|Z$?&lt;/li&gt;&#xA;&lt;li&gt;Given random variables $X_1, X_2,&#xD;&#xA;    Y_1, Y_2$, is joint distribution of&#xA;&quot;$X_1 | X_2$&quot; and &quot;$Y_1|Y_2$&quot;  same as&#xA;distribution of $(X_1, Y_1) | (X_2, Y_2)$?&lt;/li&gt;&#xA;&lt;li&gt;For two independent random vectors&#xA;$X$ and&#xA;    $Y$, and any two subvectors $X_1$ and $X_2$ of $X$ and any two&#xA;subvectors &#xA;    $Y_1$ and $Y_2$ of $Y$, will the conditional random vectors &quot;$X_1|X_2$&quot;&#xA;    and &quot;$Y_1|Y_2$&quot; also be independent?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Why? Thanks and regards!&lt;/p&gt;&#xA;" OwnerUserId="1281" LastEditorUserId="1281" LastEditorDisplayName="" LastEditDate="2011-02-25T09:02:49.827" LastActivityDate="2011-02-25T11:43:55.647" Title="Is joint distribution of $X$ and &quot;$Y|Z$&quot; same as distribution of $(X, Y)|Z$" Tags="&lt;probability-theory&gt;" AnswerCount="1" CommentCount="7" FavoriteCount="1" />
  <row Id="23678" PostTypeId="2" ParentId="23674" CreationDate="2011-02-25T08:43:35.180" Score="4" ViewCount="" Body="&lt;p&gt;The integral closure of $\Bbb Z$ in $K={\Bbb Q}\sqrt{-5}$ is the ring of integers ${\cal O}_K={\Bbb Z}\sqrt{-5}$, the latter equality because $-5\equiv 3\bmod 4$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you observed, ${\Bbb Z}_{(3)}$ is the localization of $\Bbb Z$ at the ideal $(3)$. Then, it follows from general properties (see Atiyah-Macdonald, Ch 5) that the integral closure $A$ of ${\Bbb Z}_{(3)}$ in $K$ is the localization at the ideal $(3)$ in ${\cal O}_K$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $3$ splits in $K$, $A$ has two maximal ideals, namely ${\frak m}_{\pm}=(1\pm\sqrt{-5})A$.&lt;/p&gt;&#xA;" OwnerUserId="688" LastActivityDate="2011-02-25T08:43:35.180" CommentCount="4" />
  <row Id="23679" PostTypeId="2" ParentId="23677" CreationDate="2011-02-25T08:49:48.647" Score="1" ViewCount="" Body="&lt;p&gt;(1) No if one random variable is $X$ and the other is $Y|Z$. Take for example six equally probable cases for $(X,Y,Z)$: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1, 1, 2),&#xA;  (2, 1, 1),&#xA;  (3, 2, 1),&#xA;  (1, 2, 2),&#xA;  (2, 3, 1),&#xA;  (3, 3, 2)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then $Y|Z$ is 1, 2 or 3 with equal probability, no matter what $Z$ is: they are pairwise independent.  So the joint distribution of $X$ and of $Y|Z$ is the same six pairs as the joint distribution of $X$ and of $Y$. But $(X,Y)$ is not independent of $Z$ since if $Z=1$ then $X$ is not $2$.    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) No. Not if $X_1|X_2$ is not independent of $Y_2$ , or $Y_1|Y_2$ is not independent of $X_2$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(3) If I understand your use of &lt;em&gt;vectors&lt;/em&gt; and &lt;em&gt;subvectors&lt;/em&gt;, then yes&lt;/p&gt;&#xA;" OwnerUserId="6460" LastEditorUserId="6460" LastEditorDisplayName="" LastEditDate="2011-02-25T11:43:55.647" LastActivityDate="2011-02-25T11:43:55.647" CommentCount="4" />
  <row Id="23680" PostTypeId="2" ParentId="23667" CreationDate="2011-02-25T08:52:02.307" Score="1" ViewCount="" Body="&lt;p&gt;For question 1,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know if CNS conditions exists but here are 2 sufficient conditions I know :&lt;br&gt;&#xA;-the space is a standard probability space look &lt;a href=&quot;http://en.wikipedia.org/wiki/Standard_probability_space&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;&lt;br&gt;&#xA;-the space is a Radon Space look &lt;a href=&quot;http://en.wikipedia.org/wiki/Radon_space&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think that there is anothers but can't remember the exact form except that I know that it implie a polish space and a countably generated sigma field and maybe another condtition. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regards&lt;/p&gt;&#xA;" OwnerUserId="4437" LastActivityDate="2011-02-25T08:52:02.307" CommentCount="1" />
  <row Id="23681" PostTypeId="1" CreationDate="2011-02-25T09:06:51.803" Score="2" ViewCount="50" Body="&lt;p&gt;I've never seen the following comparison before. Let me start with a specific example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given a finite structure with two symmetric binary relations, i.e. a graph $G$ with one vertex set $V$ and two edge sets $E_1$, $E_2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Giving an explicit description of $G$ in a formal language can be seen as a special case of defining a function $f$ from a set $T$ with a betweenness relation $B$ to the set $V\ \cup \big( E_1 \times \lbrace E_1  \rbrace \big) \cup \big( E_2 \times \lbrace E_2 \rbrace \big)$ such that $(v,w) \in E_i$ iff &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(\exists x,y,z) f(x) = v \wedge f(y) = ((v,w),E_i) \wedge f(z) = w \wedge B(x,y,z)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Such a tuple $(x,y,z)$ represents the &quot;sentence&quot; that $(v,w) \in E_i$. The sentences may overlap and reuse &quot;symbols&quot;, but this can be avoided. For example ($T = \mathbb{N}$):&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;|u|v|vwE|uwE|w| | | |...&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;represents the sentences $(u,w) \in E$ and $(v,w) \in E$. But also does:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;|u|uwE|w|v|vwE|w| | | |...&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;In fact, it's enough to consider functions $f$ from $T$ to $V\ \cup \lbrace E_1 \rbrace \cup \lbrace E_2 \rbrace$ such that $(v,w) \in E_i$ iff &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(\exists x,y,z) f(x) = v \wedge f(y)=E_i \wedge f(z) = w \wedge B(x,y,z)$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This comes closer to the normal usage of formal languages. So&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;|u|E|w|v|E|w| | | |...&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;represents the sentences $(u,w) \in E$ and $(v,w) \in E$, which are oftenly written as $uEw$ and $vEw$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is what &quot;symbolic&quot; description essentially is: a &quot;structure preserving&quot; function &lt;strong&gt;from a medium to the structure&lt;/strong&gt;. (Using intermediate symbols from an alphabet isn't essential: the elements of the structure may symbolize themselves.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In contrast, &quot;analog&quot; description essentially is a &quot;structure preserving&quot; function &lt;strong&gt;from the structure to a medium&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider a function $f$ from the set $V\ \cup \big( E_1 \times \lbrace E_1  \rbrace \big) \cup \big( E_2 \times \lbrace E_2 \rbrace \big)$ to a set $T$ with a betweenness relation $B$ such that $(v,w) \in E_i$ iff $f((v,w),E_i))$ is between $f(v)$ and $f(w)$, or:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(\exists x,y,z) f(v) = x \wedge f(((v,w),E_i)) = y  \wedge f(w) = z \wedge B(x,y,z)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example ($T = \mathbb{N}^2$):&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;|u|uwE|w|&#xA;| |vwE| |&#xA;|v|   | |&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Again, we can ignore the specific edges and write/draw for short:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;|u|E|w|&#xA;| |E| |&#xA;|v| | |&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;We even can ignore the specific vertices and write/draw for short (note, how this &lt;em&gt;looks&lt;/em&gt; like a unlabelled graph!):&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;|V|E|V|&#xA;| |E| |&#xA;|V| | |&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Analog description can be generalized for relations of arbitrary arity, but that's a bit tedious. In principle, it works.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Is this comparison in any sense&#xA;  enlightening, or is it just baublery? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As I said before, I've never seen this comparison made explicit. So can any references be given?&lt;/p&gt;&#xA;" OwnerUserId="1792" LastEditorUserId="1792" LastEditorDisplayName="" LastEditDate="2011-02-25T09:27:27.690" LastActivityDate="2011-02-25T09:27:27.690" Title="Comparing symbolic and analog descriptions" Tags="&lt;soft-question&gt;&lt;graph-theory&gt;&lt;model-theory&gt;" CommentCount="3" />
  <row Id="23682" PostTypeId="2" ParentId="23646" CreationDate="2011-02-25T09:11:10.077" Score="1" ViewCount="" Body="&lt;p&gt;You may assume that all four circles are centered at the origin. The $x$-axis $l_1$ and the $y$-axis $l_2$ can be considered as circles that intersect the first pair of circles at $90^\circ$. They will be mapped by $f$ onto two circles that intersect the second pair of circles at $90^\circ$, and it is easy to see that this is only possible if the circles $f(l_i)$ are again lines through the origin. As $l_1$ and $l_2$, as well as $f(l_1)$ and $f(l_2)$, intersect at $0$ and $\infty$ it follows that $f$ either keeps $0$ and $\infty$ fixed or interchanges these two points. This in turn implies $f(z)=c z$ or $f(z)=c/z$ for a suitable $c\ne 0$. In both cases the ratio between the larger and the smaller radius of the two circles stays the same.&lt;/p&gt;&#xA;" OwnerUserId="1303" LastActivityDate="2011-02-25T09:11:10.077" />
  <row Id="23683" PostTypeId="1" AcceptedAnswerId="23688" CreationDate="2011-02-25T09:11:31.737" Score="1" ViewCount="74" Body="&lt;p&gt;$\frac{256}{2^S}=64$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How would you solve for S?&lt;/p&gt;&#xA;" OwnerUserId="7490" LastEditorUserId="7339" LastEditorDisplayName="" LastEditDate="2011-02-25T10:13:22.723" LastActivityDate="2011-02-25T10:13:22.723" Title="Missing exponent?" Tags="&lt;homework&gt;" AnswerCount="2" />
  <row Id="23684" PostTypeId="2" ParentId="23683" CreationDate="2011-02-25T09:16:36.697" Score="1" ViewCount="" Body="&lt;p&gt;Hint: 256=2^8, 64=2^6, and remember the rule $2^x/2^y = 2^{x-y}$.&lt;/p&gt;&#xA;" OwnerUserId="5491" LastActivityDate="2011-02-25T09:16:36.697" CommentCount="1" />
  <row Id="23685" PostTypeId="1" AcceptedAnswerId="23750" CreationDate="2011-02-25T09:48:23.953" Score="2" ViewCount="160" Body="&lt;p&gt;This question stems from &lt;a href=&quot;http://math.stackexchange.com/questions/23262/probability-of-3-of-a-kind-with-7-dice/23279#23279&quot;&gt;Probability of 3 of a kind with 7 dice&lt;/a&gt; as a means to check my understanding.  I'm not sure whether this is general practice on this website, so I'm sorry if creating a new topic is frowned upon.  I figured the question is large enough on its own to warrant a new topic.  (Note: I intend to go back later to choose &quot;best answer&quot; on the previous one.  It's a tough choice because I've used both extensively.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now without further ado:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With a set of three desired from 9 dice, there are &lt;/p&gt;&#xA;&#xA;&lt;p&gt;a) $ \binom{6}{1}\binom{9}{3}5^6 $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;possible outcomes which contain one set of three.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These outcomes also contain other undesirables such as 111222222, 111222223, 111222333, etc. which must be accounted for.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Number of possibilities which contains a set of six in the remaining dice:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;b) $ \binom{6}{2}\binom{9}{6}4^0 $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Number of possibilities which contains a set of five in the remaining dice:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;c) $ \binom{6}{2}\binom{9}{5}4^1 $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Number of possibilities which contains a set of four in the remaining dice:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;d) $ \binom{6}{2}\binom{9}{4}4^2 $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Number of possibilities which contains a set of three in the remaining dice:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;e) $ \binom{6}{2}\binom{9}{3}\binom{6}{3}4^3 $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Number of possibilities which contain two sets of three in the remaining dice (three sets of three in the nine):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;f) $ \binom{6}{3}\binom{9}{3}\binom{6}{3}\binom{3}{3}3^0 $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, this last set of three option (e) (ie. 111222345) may contain something that looks like 111222333 due to the randomness of the three remaining dice, so we need to subtract (f) from (e) in order to find the number of options which contain only 2 sets of three.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;g) e - f: $ \binom{6}{2}\binom{9}{3}\binom{6}{3}4^3 - \binom{6}{3}\binom{9}{3}\binom{6}{3}\binom{3}{3}3^0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because (a) computes &lt;strong&gt;all&lt;/strong&gt; combinations where there is one set of three, it will pull certain things twice, and certain other things even more. ex. 111222344 is the same as 222111344 AND 111222333 ~ 222111333 ~ 333222111: there are 3! different arrangements of this latter scenario. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I was initially planning to write out everything as I understood it, but as I try, I find myself going in circles and getting massively confused.  To get the final solution of &quot;how many ways to get exactly one set of 3 in 9 dice,&quot; there should be something like the following unless I'm mistaken:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ a - b - c - d - ??? $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How many e's, f'g, and/or g's need to be taken out to account for both the initial duplicates counted in (a) (as described two paragraphs up) AND how many e's, f's, and/or g's need to be removed to ensure only one set of three remains confuses me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is what I have so far along the right path? If not, why?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If so, or even if just the overall idea is okay but the specifics are off, is there an easy/managable way to keep track of duplicates and items that need to be excluded in situations like this one?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was also wondering if there's a good way to make use of the knowledge of that 111222333 ~ 333222111 is really just abc ~ cba; or 3! as opposed to working out more specifically what to include or exclude.  Any ideas? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As always, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many thanks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I got pulled away too many times while writing thanks to a dog who wanted to run around the house with a bloody toe, so I hope it all flows well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solution:&#xA;I'm sad to say I basically gave up.  I thought I had it, I can figure out specific scenarios, but in my chart the probabilities weren't adding up to 1 when all scenarios should have been accounted for. Therefore something someplace went horribly wrong.  I took programmer approach and wrote a short script to count what I wanted out of a million rolls and filled in the chart that way (no, not exact, but close enough for what I want from it).  I need a bath.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the bright side all this grueling let me know why, when the program turned out unexpected results, it was right and could be explained. &lt;/p&gt;&#xA;" OwnerUserId="7339" LastEditorUserId="7339" LastEditorDisplayName="" LastEditDate="2011-03-03T19:16:24.483" LastActivityDate="2011-03-03T19:16:24.483" Title="Number of ways to get exactly one 3 of a kind from 9 dice." Tags="&lt;combinatorics&gt;" AnswerCount="2" CommentCount="8" />
  <row Id="23686" PostTypeId="1" CreationDate="2011-02-25T09:52:28.410" Score="6" ViewCount="103" Body="&lt;p&gt;If I have two exact triangles $X \to Y \to Z \to X[1]$ and $X&amp;#39; \to Y&amp;#39; \to Z&amp;#39; \to X&amp;#39;[1]$ in a triangulated category, and I have morphisms $X \to X&amp;#39;$, $Y \to Y&amp;#39;$ which 'commute' (i.e. such that $X \to Y \to Y' = X \to X' \to Y'$), thene there exists a (not-necessarily unique) map $Z \to Z&amp;#39;$ which completes what we've got to a morphism of triangles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a criterion which ensures the uniqueness of this cone-map?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd like something along the lines of: if $Ext^{-1}(X,Y&amp;#39;)=0$ then yes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I might be too optimistic, cfr. Prop 10.1.17 of Kashiwara-Schapira Categories and Sheaves: in addition to $Hom(X[1],Y&amp;#39;)=0$ they also assume $Hom(Y,X&amp;#39;)=0.$ I really don't have this second assumption.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(In the case I'm interested in $X=X&amp;#39;, Y=Y&amp;#39;$ and $X\to X&amp;#39;$, $Y \to Y&amp;#39;$ are the identity maps)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(If it makes things easier, although I doubt it, you can take the category to be the bounded derived category of coherent sheaves on some, fairly nasty, scheme)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the context I have in mind $X, Y, X&amp;#39;, Y&amp;#39;$ are all objects of the heart of a bounded t-structure.&#xA;If we assumed $Hom(Z,Y&amp;#39;)=0$ or $Hom(X[1],Z&amp;#39;)=0$ then the result easily follows. I don't think I'm happy making those assumptions though.&lt;/p&gt;&#xA;" OwnerUserId="1007" LastEditorUserId="1007" LastEditorDisplayName="" LastEditDate="2011-02-25T15:23:05.243" LastActivityDate="2011-02-25T15:23:05.243" Title="cones in the derived category" Tags="&lt;algebraic-geometry&gt;&lt;category-theory&gt;&lt;homological-algebra&gt;" CommentCount="4" />
  <row Id="23687" PostTypeId="1" AcceptedAnswerId="23697" CreationDate="2011-02-25T09:56:26.143" Score="4" ViewCount="191" Body="&lt;p&gt;I was wondering what is the relation between &lt;a href=&quot;http://en.wikipedia.org/wiki/Borel%E2%80%93Cantelli_lemma&quot; rel=&quot;nofollow&quot;&gt;the first and second Borel–Cantelli lemmas&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/Kolmogorov%27s_zero-one_law&quot; rel=&quot;nofollow&quot;&gt;Kolmogorov's zero-one law&lt;/a&gt;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The former is about limsup of a sequence of events, while the latter is about tail event of a sequence of independent sub sigma algebras or &lt;a href=&quot;http://planetmath.org/encyclopedia/TailEvent.html&quot; rel=&quot;nofollow&quot;&gt;of independent random variables&lt;/a&gt;. Both have results for limsup/tail event to have either probability 0 or 1. I guess there are relations between but cannot identify them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can the former be viewed as a special case of the latter? How about in reverse direction?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks and regards!&lt;/p&gt;&#xA;" OwnerUserId="1281" LastActivityDate="2011-02-25T16:47:57.640" Title="Relation between Borel–Cantelli lemmas and Kolmogorov's zero-one law" Tags="&lt;probability-theory&gt;&lt;measure-theory&gt;" AnswerCount="1" FavoriteCount="2" />
  <row Id="23688" PostTypeId="2" ParentId="23683" CreationDate="2011-02-25T10:02:49.560" Score="0" ViewCount="" Body="&lt;p&gt;FURTHER HINT: $\mathrm{log}_2(2^{x}) = x$&lt;/p&gt;&#xA;" OwnerUserId="7493" LastActivityDate="2011-02-25T10:02:49.560" />
  <row Id="23689" PostTypeId="1" CreationDate="2011-02-25T10:13:39.477" Score="6" ViewCount="142" Body="&lt;p&gt;I would be glad if someone can give me some (hopefully easy to understand!) references for learning about these groups Osp, USp and PSU and their representations. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I run into these mostly while studying about supersymmetry and I wonder if PSU groups are the same as what are called ``supergroups&quot;.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I frequently run into $Osp(3\vert4)$, $USp(2N)$, $PSU(2,2\vert 4)$, $SU(2,2)$ and $SU(2\vert 4)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I don't understand this notation of having a &quot;$\vert$&quot; and I don't know the definition of any of these groups) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some of the concepts about them that I face are,&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;The commuting subalgebra of the lie algebra of $SU(2,2,\vert m)$ is that of $SU(2,1\vert m-1)$. The generators of $SU(2,1\vert m-1)$ are related to those of $SU(2,2,\vert m)$ via some &quot;obvious reduction&quot;. I don't know what that means. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The &quot;bosonic subgroup&quot; of $SU(2,1\vert m-1)$ is apparently $SU(2,1) \times U(m-1)$. I wonder what this means.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The supersymmetry generators are supposed to transform as &quot;bifundamentals&quot;  under the bosonic subalgebra of $SU(2,2)$ and $SU(m)$. (I don't know what is a &quot;bifundamental&quot;) &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I am very confused about these concepts and would be grateful if someone one help traverse this subject.    &lt;/p&gt;&#xA;" OwnerUserId="965" LastEditorUserId="965" LastEditorDisplayName="" LastEditDate="2011-02-25T10:19:46.610" LastActivityDate="2011-03-02T16:39:43.987" Title="Osp, USp, SU(,) and PSU" Tags="&lt;reference-request&gt;&lt;representation-theory&gt;&lt;lie-groups&gt;&lt;lie-algebra&gt;" CommentCount="2" />
  <row Id="23690" PostTypeId="1" CreationDate="2011-02-25T10:15:07.133" Score="3" ViewCount="68" Body="&lt;p&gt;I read about how Hecke groups are a particular generalisation of the modular group, generalising one of the generators of the modular group to $z\mapsto z + \lambda$. Have people studied generalisations of the other generator of the modular group, perhaps generalising to $z \mapsto \dfrac{-1}{z^n}$  ?&lt;/p&gt;&#xA;" OwnerUserId="7485" LastActivityDate="2011-02-25T10:15:07.133" Title="generalisations of the modular group" Tags="&lt;group-theory&gt;" CommentCount="2" />
  <row Id="23691" PostTypeId="2" ParentId="23197" CreationDate="2011-02-25T10:19:11.340" Score="0" ViewCount="" Body="&lt;p&gt;The way the question is worded, it sounds like you already have two frames A and B (which are orthonormal bases) readily available, which would make the points redundant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[ If you only have the points, as long as they're not collinear, you can always construct an ONB from either of them: have vector u going from P&lt;sub&gt;1&lt;/sub&gt; to P&lt;sub&gt;2&lt;/sub&gt; and v from P&lt;sub&gt;1&lt;/sub&gt; to P&lt;sub&gt;3&lt;/sub&gt;, then i = u/|u|; j = uxv, j = j/|j|; k = ixj; and your base is (i,j,k). ]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since A is a transform from the common &quot;neutral&quot; coordinate system to the A coordinate system, multiplying with the inverse of A will be a transform back to the common coordinate system. In this case, A is an ONB, so &quot;inverse&quot; and &quot;transpose&quot; are synonyms, which makes your life a lot easier.&lt;br&gt;&#xA;Similarly, B transforms from the neutral coordinate system to the B coordinate system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, to transform any point from A to B, you multiply it with A&lt;sup&gt;T&lt;/sup&gt; and then with B. You can multiply the matrices together to do the transform in one step if you have to transform many points.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that if you do this in a computer program, the transpose actually &lt;em&gt;saves&lt;/em&gt; you computional effort rather than adding some, if you are wise while multiplying the matrices together (not the naive route of calling transpose() first and then multiply()).&lt;br&gt;Multiplying two matrices can be seen as taking dot products of rows and columns. Or, if you look at it this way, dot products of rows and transposed rows. Which makes multiplying a transposed matrix look a lot nicer, because now it maps perfectly to SIMD instructions and memory layout.&lt;/p&gt;&#xA;" OwnerUserId="7491" LastActivityDate="2011-02-25T10:19:11.340" />
  <row Id="23692" PostTypeId="1" AcceptedAnswerId="23713" CreationDate="2011-02-25T10:48:43.507" Score="2" ViewCount="90" Body="&lt;p&gt;As we know,a hyperplane can seem as a divisor,and a divisor can become a linear bundle,I want to know what the structure of linear bundle is.&#xA;For example, the hyperplane is given by $a_0 z_0+a_1 z_1+\ldots+a_n z_n=0$,what is the projective map、transform function and so on?&lt;/p&gt;&#xA;" OwnerUserId="4706" LastEditorUserId="536" LastEditorDisplayName="" LastEditDate="2011-02-25T14:54:36.490" LastActivityDate="2011-02-26T20:24:41.907" Title="How does a hyperplane become a linear bundle?" Tags="&lt;algebraic-geometry&gt;" AnswerCount="1" />
  <row Id="23695" PostTypeId="1" AcceptedAnswerId="23756" CreationDate="2011-02-25T12:27:56.933" Score="3" ViewCount="120" Body="&lt;p&gt;I am poking on a proof of the subject theorem. Given sets $A_i$ and injections $f_i:A_i\to A_{1-i}$, $i\in \{0,1\}$, theorem defines a bijection $b$ between $A_0$ and $A_1$. $b$ uses an auxilary function&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$degree (x:A_i) := \cases{&#xD;&#xA;1+degree(f_{1-i}^{-1}(x)), x\in f_{1-i}(A_{1-i}) \cr 0}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;“degree” may loop, denote this $\bot$. Then $b$ returns some value$\neq\bot$ on $\bot$ and is not constant in general. $b$ solves the halting problem. So, it is possible? (I assume that the codomain of “degree” is a flat domain.) (I have not posted this on &lt;a href=&quot;http://cstheory.stackexchange.com/&quot;&gt;cstheory&lt;/a&gt; because this is not a research-level question.)&lt;/p&gt;&#xA;" OwnerUserId="7011" LastActivityDate="2011-02-25T21:24:49.653" Title="Cantor–Bernstein–Schröder theorem and recursion" Tags="&lt;computer-science&gt;&lt;elementary-set-theory&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="23696" PostTypeId="2" ParentId="23685" CreationDate="2011-02-25T12:31:27.817" Score="1" ViewCount="" Body="&lt;p&gt;What was wrong with my previous method?  9 can be partitioned into up to 6 parts with one 3 and other parts smaller, in three ways, namely $3+2+2+2(+0+0)$, $3+2+2+1+1(+0)$, and $3+2+1+1+1+1$.  So I think the answer is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \frac{6!}{1!\;3!\;2!} \times \frac{9!}{3!\;2!\;2!\;2!} + \frac{6!}{1!\;2!\;2!\;1!} \times \frac{9!}{3!\;2!\;2!\;1!\;1!} + \frac{6!}{1!\;1!\;4!} \times \frac{9!}{3!\;2!\;1!\;1!\;1!\;1!} = 4082400 $$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;i.e. just over 4 million.&lt;/p&gt;&#xA;" OwnerUserId="6460" LastActivityDate="2011-02-25T12:31:27.817" CommentCount="5" />
  <row Id="23697" PostTypeId="2" ParentId="23687" CreationDate="2011-02-25T12:58:02.797" Score="7" ViewCount="" Body="&lt;p&gt;The first Borel-Cantelli is simply the fact that the probability of a union is at most the sum of the probabilities; it has nothing in common with Kolmogorov zero-one law.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What Kolmogorov zero-one law tells you in the setting of the second Borel-Cantelli lemma is that the probability of the limsup is $0$ or $1$, because (1) the limsup is always in the tail $\sigma$-algebra, (2) you are considering &lt;strong&gt;independent&lt;/strong&gt; events, hence their tail $\sigma$-algebra is trivial. Then the second Borel-Cantelli lemma itself tells you that this probability is in fact $1$ under the non summability condition which you know.&lt;/p&gt;&#xA;" OwnerUserId="6179" LastEditorUserId="6179" LastEditorDisplayName="" LastEditDate="2011-02-25T16:47:57.640" LastActivityDate="2011-02-25T16:47:57.640" CommentCount="4" />
  <row Id="23698" PostTypeId="2" ParentId="23669" CreationDate="2011-02-25T13:01:30.927" Score="7" ViewCount="" Body="&lt;p&gt;Look at the map $(f,g)\colon X\to Y\times Y$. This map is continuous because $f$ and $g$ are.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your set $\{x \in X \mid f(x) &amp;lt; g(x)\}$ is the preimage under $(f,g)$ of the set $\{(a,b) \in Y\times Y \mid a &amp;lt; b\}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the only thing that remains to prove is that $\{(a,b) \in Y\times Y \mid a &amp;lt; b\}\subseteq Y\times Y$ is open.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This follows from the definition of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Order_topology&quot; rel=&quot;nofollow&quot;&gt;Order topology&lt;/a&gt; on $Y$. Do you need help with that?&lt;/p&gt;&#xA;" OwnerUserId="367" LastActivityDate="2011-02-25T13:01:30.927" CommentCount="1" />
  <row Id="23699" PostTypeId="1" CreationDate="2011-02-25T13:34:06.683" Score="2" ViewCount="78" Body="&lt;p&gt;Everywhere I look in the literature, Sobolev spaces are defined on an open subset of the real line. What are the technical issues with defining a Sobolev space on a closed subset, i.e. are there problems at the boundary, and does anyone know any good references that cover this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My main purpose is to prove $H^1([0,T];\mathbb{R}) = &#xD;&#xA;\{ x \in L^2([0,T];\mathbb{R}) : ||x&amp;#39;||_{L^2} + \gamma^{2}||x||_{L^2} &amp;lt; \infty \}$ is a reproducing kernel Hilbert space. I can do this for $(0,T)$ and want to know if the proof is transferable to the case of the closed interval $[0,T]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many thanks,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Matthew.&lt;/p&gt;&#xA;" OwnerUserId="7498" LastActivityDate="2011-04-26T18:42:42.060" Title="Sobolev space on closed subset of the real line" Tags="&lt;analysis&gt;" AnswerCount="1" />
  <row Id="23700" PostTypeId="1" CreationDate="2011-02-25T13:36:51.673" Score="1" ViewCount="131" Body="&lt;p&gt;i need a mathematical function F which is defined as F(x) = 0 if x equal to zero and F(x) = 1 if x is not equal to zero. can F(x) be represented in the form of a single mathematical expression in terms of x?&lt;/p&gt;&#xA;" OwnerUserId="7500" LastActivityDate="2011-02-27T17:24:34.707" Title="Mathematical function to check whether its parameter is zero or not" Tags="&lt;functions&gt;&lt;zeros&gt;" AnswerCount="3" CommentCount="5" />
  <row Id="23701" PostTypeId="1" CreationDate="2011-02-25T13:53:42.800" Score="1" ViewCount="23" Body="&lt;p&gt;Hi Guys,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have an application that stores data in database. I need search functionality to work on this database.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For this to work I need a &quot;relevance&quot; score, a score that is calculated based on a set of criteria to output as a value that can be then used to order a set of data.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Say for instance the user enters three keywords: X, Y and Z -  I need to generate a score based on a database entry. I wish the criteria to be related to how many times each appears.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Database Entry A - X appears 8 times Y appears once and Z appears once. Giving a collective score of 10.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Database Entry B - X appears 24 times Y does not appear and Z does not appear. Giving a collective score of 24.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's my problem. Database Entry A IS more relevant based on the search of XYZ because it has all three database entries, not just one, yet a standard calculation would class Database Entry B as more relevant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need to figure out a way to calculate the results and give an number score to the result based on not just how many of each keyword appears, but also giving higher scores for those results that have more than one keyword displayed, exponentially (i.e. entering 10 keywords would show results where all 10 appear above ones with large amounts of one).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The solution needs to be something that can be easily calculated using PHP ideally.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for your help guys&lt;/p&gt;&#xA;" OwnerUserId="7115" LastActivityDate="2011-02-25T13:53:42.800" ClosedDate="2011-05-13T15:24:15.583" Title="Working out a &quot;relevance&quot; score for a search application" Tags="&lt;calculus&gt;&lt;logic&gt;" />
  <row Id="23702" PostTypeId="2" ParentId="23674" CreationDate="2011-02-25T14:10:53.053" Score="3" ViewCount="" Body="&lt;p&gt;Here is a different and (perhaps) somewhat more elementary approach than Andrea's. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $R$ be the integral closure of $\mathbb{Z}_{(3)}$ in $\mathbb{Q}(\sqrt{-5})$.  Clearly $R$ is a subring of $\mathbb{Q}(\sqrt{-5})$, and thus we are looking for a necessary and sufficient condition on $a,b \in \mathbb{Q}$ such that $a+b\sqrt{-5} \in R$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $P(a,b)$ be the minimal polynomial of $a+b\sqrt{-5}$ over $\mathbb{Q}$, i.e., the  unique monic polynomial with $\mathbb{Q}$-coefficients of least degree satisfied by $a+b\sqrt{-5}$.  Certainly if $P(a,b)$ has coefficients lying in $\mathbb{Z}_{(3)}$ then $a+b\sqrt{-5}$ lies in $R$.  In fact the converse is true because the ring $\mathbb{Z}_{(3)}$ is integrally closed (e.g. it is a PID and PID $\implies$ UFD $\implies$ integrally closed).  For a proof of this fact, see e.g. the section on Integrally Closed Domains in &lt;a href=&quot;http://math.uga.edu/~pete/integral.pdf&quot; rel=&quot;nofollow&quot;&gt;these notes&lt;/a&gt;.  (Currently this is Theorem $260$ in $\S 14.5$, but that more precise information is subject to change.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Try out this computation for yourself: you should get that $a+b\sqrt{-5} \in R \iff a,b \in \mathbb{Z}_{(3)}$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now in my notes I also prove the compatibility of integral closure with localization which Andrea uses in his answer: in fact it comes about five pages earlier than the fact about minimal polynomials mentioned above.  So it is certainly arguable which of these facts is &quot;more basic&quot;.  My reason for choosing the latter is because I think it is a little more concrete and amenable to computation: indeed, if you don't know this fact about minimal polynomials, it seems to me that you're going to have a hard time computing any nontrivial examples of integral closures whatsoever.&lt;/p&gt;&#xA;" OwnerUserId="299" LastActivityDate="2011-02-25T14:10:53.053" />
  <row Id="23703" PostTypeId="1" CreationDate="2011-02-25T14:11:31.517" Score="2" ViewCount="111" Body="&lt;p&gt;Can someone please share a link or source where I can find the derivation of FFT(base-2) from the DFT. I need to put this in latex for my thesis and am finding so many different explanations that I don't know which one to use.&lt;/p&gt;&#xA;" OwnerUserId="7501" LastActivityDate="2011-03-30T09:38:43.410" Title="Derivation of FFT" Tags="&lt;fourier-analysis&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" />
  <row Id="23704" PostTypeId="1" AcceptedAnswerId="23707" CreationDate="2011-02-25T14:14:14.123" Score="3" ViewCount="152" Body="&lt;p&gt;I hit a snag whilst revising some log rules, could anyone confirm my suspicion:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\log _b \left( x \right) = \log _b \left( y \right) \rightarrow x = y ?$$&lt;/p&gt;&#xA;" OwnerUserId="6409" LastActivityDate="2011-02-25T16:00:20.847" Title="Does $\log _b \left( x \right) = \log _b \left( y \right) \rightarrow x = y$ ?" Tags="&lt;logarithms&gt;" AnswerCount="4" CommentCount="1" />
  <row Id="23705" PostTypeId="1" CreationDate="2011-02-25T14:17:05.663" Score="4" ViewCount="178" Body="&lt;p&gt;Cheers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if I don't make sense, I have a value for $y$, I need to know what $x$ is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$y = a \exp(bx) + c \exp(dx)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a =       12.85$,&#xA;$b =    0.001857$,&#xA;$c =      -54.24$,&#xA;$d =    -0.05316$&lt;/p&gt;&#xA;" OwnerUserId="7502" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2011-02-25T15:20:47.733" LastActivityDate="2011-02-25T23:27:13.280" Title="How to invert this exponential function to solve for x:  $y = a \exp(bx) + c \exp(dx)$?" Tags="&lt;functions&gt;&lt;inverse&gt;" AnswerCount="3" />
  <row Id="23706" PostTypeId="2" ParentId="23704" CreationDate="2011-02-25T14:19:19.200" Score="1" ViewCount="" Body="&lt;p&gt;By definition, $b^{\log_b t} = t$.&lt;/p&gt;&#xA;" OwnerUserId="589" LastActivityDate="2011-02-25T14:19:19.200" />
  <row Id="23707" PostTypeId="2" ParentId="23704" CreationDate="2011-02-25T14:20:40.743" Score="4" ViewCount="" Body="&lt;p&gt;Yes because $b^{\log_{b} y} = b^{\log_{b} x} \Longleftrightarrow y = x$. &lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2011-02-25T14:20:40.743" />
  <row Id="23709" PostTypeId="2" ParentId="23700" CreationDate="2011-02-25T14:31:56.157" Score="3" ViewCount="" Body="&lt;p&gt;Knuth suggests using the &lt;a href=&quot;http://en.wikipedia.org/wiki/Iverson_bracket&quot; rel=&quot;nofollow&quot;&gt;Iverson bracket notation&lt;/a&gt; $[x\ne0]$ for that function. You can also try $|\mbox{sgn}(x)|$. See &lt;a href=&quot;http://en.wikipedia.org/wiki/Sign_function&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Sign_function&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="589" LastEditorUserId="4290" LastEditorDisplayName="" LastEditDate="2011-02-25T14:42:56.260" LastActivityDate="2011-02-25T14:42:56.260" />
  <row Id="23711" PostTypeId="2" ParentId="23704" CreationDate="2011-02-25T14:32:59.027" Score="5" ViewCount="" Body="&lt;p&gt;Yes it does, by the following argument: Suppose that $\mathrm{log}_b(x) = \mathrm{log}_b(y)$. Then $b^{\mathrm{log}_b(x)} = b^{\mathrm{log}_b(y)}$. But now (by the definition of $\mathrm{log_b(\cdot)}$) we know that $b^{\mathrm{log}_b(x)} = x$, so we conclude that $x = y$ as required.&lt;/p&gt;&#xA;" OwnerUserId="7493" LastActivityDate="2011-02-25T14:32:59.027" />
  <row Id="23712" PostTypeId="1" CreationDate="2011-02-25T14:49:19.100" Score="0" ViewCount="93" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Consider conditional expectation of&#xA;a real-valued r.v. $X$ given a sub&#xA;sigma algebra $\mathcal{B}$ of the&#xA;probability space $(\Omega,&#xD;&#xA;    \mathcal{F}, P)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Will independence between&#xA;$\sigma(X)$, i.e. the sigma algebra&#xA;of the r.v., and the given sub sigma&#xA;algebra $\mathcal{B}$ make $E(X \mid&#xD;&#xA;    \mathcal{B}) \equiv E(X)$ ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is independence between $\sigma(X)$&#xA;and the given sub sigma algebra&#xA;$\mathcal{B}$ is the only way to&#xA;define independence between $X$ and&#xA;$\mathcal{B}$? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there are other ways, will they&#xA;make $E(X \mid \mathcal{B}) \equiv&#xD;&#xA;    E(X)$ ?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Consider conditional expectation of&#xA;a real-valued r.v. $X$ given another&#xA;$S$-valued r.v. $Y$ on the same&#xA;probability space $(\Omega,&#xD;&#xA;    \mathcal{F}, P)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Will independence between $X$ and&#xA;$Y$ make $E(X \mid Y) \equiv E(X)$ ?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Why? References are also appreciated!&#xA;Thanks and regards!&lt;/p&gt;&#xA;" OwnerUserId="1281" LastActivityDate="2011-02-25T15:04:07.950" Title="Conditional expectation and independence" Tags="&lt;probability-theory&gt;" AnswerCount="1" />
  <row Id="23713" PostTypeId="2" ParentId="23692" CreationDate="2011-02-25T14:54:16.117" Score="2" ViewCount="" Body="&lt;p&gt;I'm going to assume your question is in the context of algebraic geometry. First, if we are working over affine space, then the hyperplane is cut out by a global function, so the divisor is principal; in particular, this line bundle is trivial.&#xA;(In fact, any vector bundle over affine space is trivial, though this is a hard theorem, the &lt;a href=&quot;http://en.wikipedia.org/wiki/Quillen-Suslin_theorem&quot; rel=&quot;nofollow&quot;&gt;Quillen-Suslin theorem&lt;/a&gt;.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is more interesting when one is working with projective space $\mathbb{P}^n_k$ over a field $k$. In that case, one obtains the standard line bundle $\mathcal{O}(1)$. See section 4.3 of &lt;a href=&quot;http://people.fas.harvard.edu/~amathew/linebund.pdf&quot; rel=&quot;nofollow&quot;&gt;http://people.fas.harvard.edu/~amathew/linebund.pdf&lt;/a&gt;.  The basic idea is that given a line bundle, one obtains the associated Weil divisor by picking a rational section and taking its divisor. So there is a global section $x_0$&#xA;of $\mathcal{O}(1)$ (which is basically the structure sheaf twisted by a  homogeneous degree), and where this section does not vanish (and does not vanish with multiplicity one) is just a hyperplane. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, since one can show directly that the Weil class group of $\mathbb{P}^n_k$ is isomorphic to $\mathbb{Z}$ (any hypersurface of degree $d$ being equivalent to $d$ times the hyperplane $\{x_0 = 0\}$, as any homogeneous polynomial of degree $d$ divided by $x_0^d$ is a rational function on $\mathbb{P}^n$), so the Picard group of line bundles on $\mathbb{P}^n$ is precisely $\mathbb{Z}$, generated by a copy of $\mathcal{O}(1)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The line bundle $\mathcal{O}(1)$ is the &lt;em&gt;dual&lt;/em&gt; of the so-called tautological line bundle over $\mathbb{P}^n$ consisting of pairs $(\ell, p)$ where $\ell$ is a line and $p$ is a point in that line (this is more basic in the topological category).&lt;/p&gt;&#xA;" OwnerUserId="536" LastActivityDate="2011-02-25T14:54:16.117" CommentCount="6" />
  <row Id="23714" PostTypeId="1" CreationDate="2011-02-25T14:54:53.413" Score="2" ViewCount="56" Body="&lt;p&gt;How to calculate the distance between a convex polytope and a point?&lt;br&gt;&#xA;Polytope is specified as the solution to the system of linear inequalities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm looking for the method that is computationally efficient.&lt;/p&gt;&#xA;" OwnerUserId="4811" LastActivityDate="2011-02-25T14:54:53.413" Title="Distance between a polytope and a point" Tags="&lt;geometry&gt;&lt;computational-geometry&gt;" CommentCount="3" />
  <row Id="23715" PostTypeId="2" ParentId="23712" CreationDate="2011-02-25T15:04:07.950" Score="3" ViewCount="" Body="&lt;p&gt;The answer to the first question in 1. is yes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a special case, the answer to 2. is also yes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Concerning the second question in 1.: It's the only way I can think of. One definition is enough, isn't it? You can of course spell out what indepence actaully means in this case and take what you get as your definition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Concerning the third question in 1.: They should, otherwise they cannot be equivalent defitions.&lt;/p&gt;&#xA;" OwnerUserId="367" LastActivityDate="2011-02-25T15:04:07.950" CommentCount="6" />
  <row Id="23716" PostTypeId="2" ParentId="23705" CreationDate="2011-02-25T15:05:40.543" Score="2" ViewCount="" Body="&lt;p&gt;I doubt there is an analytic solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With those values of $a,b,c,d$, it appears that $y$ is smoothly increasing and bijective function of $x$ so given any value of $y$ you should be able to find a very close approximation $x$ by standard numerical methods.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, if $y=0$ then $x \approx 26.175\ldots$, if $y=100$ then $x \approx 1104.9\ldots$, if $y=-100$ then $x \approx -13.7278\ldots$.&lt;/p&gt;&#xA;" OwnerUserId="6460" LastActivityDate="2011-02-25T15:05:40.543" />
  <row Id="23717" PostTypeId="1" AcceptedAnswerId="23718" CreationDate="2011-02-25T15:06:27.613" Score="0" ViewCount="54" Body="&lt;p&gt;Hi there, I've got a few criteria and I need to find a simple (as simple as possible) function which fits those criteria:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\lim_{x \to -\infty }{f(x)} = C_1$$&#xA;$$f(0) = C_2$$&#xA;$$f&amp;#39;(0) = C_3$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $C_1,C_2,C_3$ are constants... ($C_1$ will be $1$, and the other constants are determined from another function so the function is continuous).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The function only has to be defined to $x = 0$. (As for the positive $x$ another function would be used).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm thinking along the line of:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f(x) = \frac{ 1 }{ax + b} + c$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However I'm more or less struck on &quot;using&quot; the limit constraint to solve this problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f(0) = \frac{1}{b} + c = C_2$$&#xA;$$f&amp;#39;(0) = - \frac{a}{b^2} = C_3$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance,&#xA;paul23&lt;/p&gt;&#xA;" OwnerUserId="7504" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2011-02-25T15:34:10.620" LastActivityDate="2011-02-25T16:21:42.403" Title="In search of a formula which fullfills criteria" Tags="&lt;limit&gt;&lt;generating-functions&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="23718" PostTypeId="2" ParentId="23717" CreationDate="2011-02-25T15:19:00.327" Score="0" ViewCount="" Body="&lt;p&gt;If you choose, $f(x) = c + \frac{1}{ax+b}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$C_1 = \lim_{x \rightarrow -\infty} f(x) = c$$&#xA;provided $a \neq 0$. Hence $c = C_1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$C_2 = f(0) = C_1 + \frac{1}{b} \Rightarrow b = \frac{1}{C_2-C_1}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$C_3 = f&amp;#39;(0) = -\frac{a}{b^2} \Rightarrow a = -\frac{C_3}{(C_2-C_1)^2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, $a = -\frac{C_3}{(C_2-C_1)^2}$, $b = \frac{1}{C_2-C_1}$ and $c=C_1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f(x) = \frac{(C_2-C_1)^2}{-C_3 x + (C_2 - C_1)} + C_1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An other possible candidate is $$f(x) = (C_2-C_1) e^{\frac{C_3}{C_2-C_1}x} + C_1$$ provided $C_3 \times(C_2 - C_1) &amp;gt; 0$.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2011-02-25T15:28:05.307" LastActivityDate="2011-02-25T15:28:05.307" />
  <row Id="23719" PostTypeId="2" ParentId="23717" CreationDate="2011-02-25T15:20:41.067" Score="1" ViewCount="" Body="&lt;p&gt;If you define $f(x)=\frac{(C_1-C_2)x}{x+a}+C_2$ you get the proper limit at $-\infty$ and value at 0.  So then $f&amp;#39;(0)=-\frac{(C_1-C_2)a}{a^2}=C_3$ or $a=\frac{-(C_1-C_2)}{C_3}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: parentheses added in the last after original posting&lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2011-02-25T16:21:42.403" LastActivityDate="2011-02-25T16:21:42.403" />
  <row Id="23720" PostTypeId="1" AcceptedAnswerId="23721" CreationDate="2011-02-25T15:24:39.603" Score="0" ViewCount="50" Body="&lt;p&gt;How would I write a summation for the average speed of an object, given $10$ speeds (which were calculated $r=\frac{d}{t}$)?&#xA;Would it be:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \sum_{i = 1}^{10} \frac{d_i}{t_i}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or am I wrong? I'm not sure about the subscript i's after distance and time, and if they are actually number sequences that I could use. Thanks.&lt;/p&gt;&#xA;" OwnerUserId="7505" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2011-02-25T15:35:29.137" LastActivityDate="2011-02-25T17:36:16.070" Title="Summation for Average Speed" Tags="&lt;physics&gt;" AnswerCount="2" />
  <row Id="23721" PostTypeId="2" ParentId="23720" CreationDate="2011-02-25T15:29:33.517" Score="2" ViewCount="" Body="&lt;p&gt;No. Generally average speed is defined as the total distance divided by the total time. That would be&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \frac{\sum_{i = 1}^{10} d_i}{\sum_{j = 1}^{10} t_j} $$&lt;/p&gt;&#xA;" OwnerUserId="1543" LastActivityDate="2011-02-25T15:29:33.517" />
  <row Id="23722" PostTypeId="2" ParentId="23625" CreationDate="2011-02-25T15:35:42.397" Score="0" ViewCount="" Body="&lt;p&gt;Add the condition $f(0)=f(a)=0$. The graph of the optimal $f$ has to be a circular arc of a certain angle $2\alpha$, $\ 0\leq\alpha\leq{\pi\over2}$, for otherwise one could increase the area $A$ without changing the length $L$. Expressing $A$ and $L$ using the variable $\alpha$ we get&#xA;$${A\over L}={a\over 4} ({1\over\sin\alpha}-{\cos\alpha\over\alpha}),$$&#xA; and this assumes its maximal value ${a\over 4}$ for $\alpha={\pi\over2}$, so that we again arrive at a semicircle.&lt;/p&gt;&#xA;" OwnerUserId="1303" LastEditorUserId="1303" LastEditorDisplayName="" LastEditDate="2011-02-26T08:53:16.620" LastActivityDate="2011-02-26T08:53:16.620" CommentCount="2" />
  <row Id="23723" PostTypeId="1" CreationDate="2011-02-25T15:37:02.803" Score="6" ViewCount="81" Body="&lt;p&gt;&lt;a href=&quot;http://math.stackexchange.com/q/23129/2949&quot;&gt;This question&lt;/a&gt; explains that scale-invariance (or more accurately, similarity) is an important property of Euclidean geometry. Are there any other ways to define scale-invariant geometries in any sense? And how do they differ from Euclidean geometry?&lt;/p&gt;&#xA;" OwnerUserId="2949" LastActivityDate="2011-04-18T21:34:02.080" Title="What types of geometries are scale-invariant?" Tags="&lt;geometry&gt;&lt;euclidean-geometry&gt;" AnswerCount="1" />
  <row Id="23724" PostTypeId="1" AcceptedAnswerId="23732" CreationDate="2011-02-25T15:44:00.583" Score="2" ViewCount="112" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Let $A \subseteq P(\omega)$, where $\omega$ is the set of all natural numbers and $P(\omega)$ is the power set of $\omega$. If $\langle A,\subseteq\rangle$ is a well ordered set how can you prove that $A$ is a countable set.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Let $A$ be a set which elements are closed sets of real numbers. If $\langle A,\subseteq\rangle$ is a well ordered set how can you prove that $A$ is a countable set.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;How can you prove that Zorn's lemma (so and the axiom of choice) is equivalent to that for every partially ordered set $\langle A,\le\rangle$ that satisfies Zorn's condition, for every $b\in A$ there exists a maximum element $a$ for which $b\le a$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="7507" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-25T16:25:53.253" LastActivityDate="2011-02-25T16:39:47.510" Title="set theory problems" Tags="&lt;elementary-set-theory&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="23725" PostTypeId="2" ParentId="23699" CreationDate="2011-02-25T15:55:04.587" Score="1" ViewCount="" Body="&lt;p&gt;Well, I suspect you defined the Sobolev space on the close interval &lt;em&gt;as&lt;/em&gt; the one on the open interval (i.e. using the same functional equation taking the test functions on $\mathscr{C}^{\infty}_0((0,T))$), and then extend every function of it by choosing arbitrary values on the boundary. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In that case $H^1((0,T))=H^1([0,T])$ and your problem is of little concern, because every function of $W^{1,1}((0,T))$ is also continue on the closed interval $[0,T]$ (i.e. they admit finite limit on the boundary). So you can apply, by limit, the same technique you used in the case of the open interval $(0,T)$ (for example using the continuous inclusion of $W^{1,1}(I)$ in $L^{\infty}(I)$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, I think the main reason beyond defining Sobolev space on open sets lies in this: $\mathscr{C}^{\infty}_0(\bar{I})=\mathscr{C}^{\infty}(\bar{I})$.&lt;/p&gt;&#xA;" OwnerUserId="5885" LastActivityDate="2011-02-25T15:55:04.587" />
  <row Id="23726" PostTypeId="2" ParentId="23704" CreationDate="2011-02-25T16:00:20.847" Score="1" ViewCount="" Body="&lt;p&gt;$\log(x)$ is increasing, for instance its derivative $1/x$ is positive.&lt;/p&gt;&#xA;" OwnerUserId="6925" LastActivityDate="2011-02-25T16:00:20.847" />
  <row Id="23727" PostTypeId="1" AcceptedAnswerId="23738" CreationDate="2011-02-25T16:09:15.663" Score="5" ViewCount="69" Body="&lt;p&gt;We've just started this today in first year applied maths at university. Today we were given the problem of deriving the formula for the area of an ellipse. I've got as far as saying there's some relationship between the semi-major axis and the semi-minor axis on the one hand, and the area of the ellipse on the other, but now I'm stuck.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I do not want a direct answer to the problem. I'm looking for a good online tutorial for these sorts of problems that might get me thinking in the right way. I've tried doing a search myself, but keep getting stuff on performing conversions. Can anyone point me to a good source for my purposes?&lt;/p&gt;&#xA;" OwnerUserId="7509" LastActivityDate="2011-02-25T17:16:31.477" Title="Anyone know a clear, useful online tutorial on dimensional analysis" Tags="&lt;conic-sections&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" />
  <row Id="23728" PostTypeId="1" CreationDate="2011-02-25T16:09:40.713" Score="2" ViewCount="308" Body="&lt;p&gt;This is a task from an old programming contest the task is as follows, &lt;strong&gt;A list of system of linear equations is given in the inputs.If the system is solvable we have to output the solution in the form of 'A + Bk',else we have to output &quot;no solutions&quot;.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now lets consider the inputs:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;x = 3 (mod 4)&#xA;&#xA;x = 4 (mod 3)&#xA;&#xA;x = 1 (mod 2)&#xA;x = 2 (mod 3)&#xA;x = 3 (mod 4)&#xA;x = 4 (mod 5)&#xA;x = 5 (mod 6)&#xA;x = 0 (mod 7)&#xA;&#xA;x = 2 (mod 2)&#xA;x = 4 (mod 3)&#xA;x = 6 (mod 5)&#xA;x = 8 (mod 7)&#xA;x = 10 (mod 11)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Using Chinese remainder theorem,I guess the outputs should be:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;3 + 4k&#xA;1 + 3k&#xA;119 + 420k&#xA;736 + 2310k&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;But according to the Judges I/O the output is :&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;3 + 4k&#xA;no solutions&#xA;119 + 420k&#xA;no solutions&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Now I can see many had solved this task could anybody explain how this output holds valid for the inputs?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS:There is a note from the problem author that &quot;&lt;strong&gt;Not all are solvable using the Chinese Remainder Theorem, and some aren't solvable at all!&lt;/strong&gt;&quot;&lt;/p&gt;&#xA;" OwnerUserId="2109" LastEditorUserId="2109" LastEditorDisplayName="" LastEditDate="2011-02-25T16:15:13.850" LastActivityDate="2011-02-25T16:20:14.223" Title="Solving systems of linear congruences (modular equations)." Tags="&lt;number-theory&gt;&lt;algorithms&gt;&lt;contest-math&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="23729" PostTypeId="1" AcceptedAnswerId="23752" CreationDate="2011-02-25T16:18:25.480" Score="4" ViewCount="158" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Algebraic_closure&quot; rel=&quot;nofollow&quot;&gt;As we all know&lt;/a&gt;, the algebraic closure often has an infinite degree.&lt;br&gt;&#xA;Also, &lt;a href=&quot;http://math.uga.edu/~pete/FieldTheory.pdf&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; shows the necessary and sufficient condition for a Galois extension to be a finite extension of fields.&lt;br&gt;&#xA;However, we may want to characterize the cases of the case where the extension is not Galois, which is actually my question.&lt;br&gt;&#xA;And this question is related to &lt;a href=&quot;http://math.stackexchange.com/questions/12082/field-reductions&quot;&gt;this question&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="6792" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-25T16:20:15.977" LastActivityDate="2011-02-25T20:01:36.250" Title="What is the condition for a field to make the degree of its algebraic closure over it infinite?" Tags="&lt;abstract-algebra&gt;&lt;field-theory&gt;&lt;galois-theory&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="2" />
  <row Id="23730" PostTypeId="2" ParentId="23728" CreationDate="2011-02-25T16:20:14.223" Score="1" ViewCount="" Body="&lt;p&gt;The author believes that things can only equal 0, 1, or 2 mod 3 and nothing can be equal to 4 mod 3.  Others would accept that 1 and 7 equal 4 mod 3 (and might use the equivalence sign instead of the equals sign).&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-02-25T16:20:14.223" />
  <row Id="23731" PostTypeId="2" ParentId="23729" CreationDate="2011-02-25T16:23:41.370" Score="10" ViewCount="" Body="&lt;p&gt;The &lt;a href=&quot;http://www.math.uconn.edu/~kconrad/blurbs/galoistheory/artinschreier.pdf&quot; rel=&quot;nofollow&quot;&gt;Artin-Schreier theorem&lt;/a&gt; asserts that, if the field $k$ is not algebraically closed and $[\bar{k} : k]$ is finite, then in fact it equals $2$ and $\bar{k} = k[i]$ &lt;em&gt;and&lt;/em&gt; $k$ is &lt;a href=&quot;http://en.wikipedia.org/wiki/Real_closed_field&quot; rel=&quot;nofollow&quot;&gt;real closed&lt;/a&gt;. So I guess the condition you're looking for is: if and only if $k$ is not algebraically closed and also not real closed. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am answering the title question, not the body question. The body question is not equivalent to the title question; the answer to it is that the algebraic closure is Galois if and only if it's separable, hence if and only if $k$ is perfect. Perhaps you have only seen the definition of Galois extension for finite extensions. This is not the definition which works in maximal generality. The definition which does work in maximal generality is the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Definition:&lt;/strong&gt; An algebraic extension $k \to L$ is &lt;em&gt;Galois&lt;/em&gt; if the fixed field of $\text{Aut}_k(L)$ is $k$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Algebraic_closure#Separable_closure&quot; rel=&quot;nofollow&quot;&gt;separable closure&lt;/a&gt; $k_s$ of $k$ is always Galois (in fact it is the maximal Galois extension of $k$), and agrees with the algebraic closure if and only if $k$ is perfect. &lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2011-02-25T16:30:56.930" LastActivityDate="2011-02-25T16:30:56.930" CommentCount="1" />
  <row Id="23732" PostTypeId="2" ParentId="23724" CreationDate="2011-02-25T16:39:47.510" Score="7" ViewCount="" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Since $A$ is well-ordered, we can order-biject it with an ordinal $\beta$, $f\colon \beta\to A$. Now, you can use $f$ to define an injection from $\beta$ to $\omega$ by letting $g(\alpha)$ be the least element of $f(\alpha+1)-f(\alpha)$. Show that $g$ is an injection, and therefore $\beta$ is countable.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Consider the set of complements, which are open, and well ordered by $\supseteq$. Show that there is always a rational that is in one set but not in its $\supseteq$-successor, and that you can pick distinct rationals at each stage (that is, define an injection from $A$ to $\mathbb{Q}$).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Replacing &quot;maximum element $a$&quot; with &quot;maximal element $a$&quot;, assuming Zorn's Lemma and given $\langle A,\leq\rangle$ and $b$, look at $\mathcal{C}=\{x\in A\mid x\geq b\}$ and apply Zorn's Lemma to it. For the converse, pick an arbitrary $b\in A$ and conclude the original set has maximal elements.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="742" LastActivityDate="2011-02-25T16:39:47.510" />
  <row Id="23733" PostTypeId="2" ParentId="23705" CreationDate="2011-02-25T16:42:47.133" Score="3" ViewCount="" Body="&lt;p&gt;Write $\displaystyle b = B/M$ and $\displaystyle d = D/M$, where $\displaystyle M \gt 0$ and $\displaystyle B,D$ are integers (I am assuming $\displaystyle b,d$ are rational).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Set $\displaystyle z = e^{x/M}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle y = az^B + c z^D$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a polynomial in $\displaystyle z$. You could try using polynomial root finding methods now.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2011-02-25T16:42:47.133" CommentCount="3" />
  <row Id="23734" PostTypeId="1" CreationDate="2011-02-25T16:42:49.033" Score="11" ViewCount="279" Body="&lt;p&gt;Some definite integrals, such as $\int_0^\infty e^{-x^2}\,dx$, are known despite the fact that there is no closed-form antiderivative. However, the method I know of calculating this particular integral (square it, and integrate over the first quadrant in polar coordinates) is not dependent on the Riemann sum definition. What I thought might be interesting is a definite integral $\int_a^bf(x)\,dx$ for which the limit of the Riemann sums happens to be calculable, but for which no closed-form antiderivative of $f$ exists. Of course there are some obvious uninteresting examples, like integrating odd functions over symmetric intervals, but one doesn't need Riemann sums to calculate these uninteresting examples.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; To make this a bit clearer, it would be nice to have a &quot;natural&quot; continuous function $f(x)$ where by some miracle $\lim_{n\to\infty} \sum_{i=1}^nf(x_i)\Delta x$ is computable (for some interval $[a,b]$) using series trickery, but for which no antiderivative exists composed of elementary functions.&lt;/p&gt;&#xA;" OwnerUserId="2941" LastEditorUserId="2941" LastEditorDisplayName="" LastEditDate="2011-02-25T19:25:25.723" LastActivityDate="2011-02-26T17:44:49.503" Title="Is there a definite integral for which the Riemann sum can be calculated but for which there is no closed-form antiderivative?" Tags="&lt;calculus&gt;" AnswerCount="2" CommentCount="8" />
  <row Id="23735" PostTypeId="2" ParentId="23734" CreationDate="2011-02-25T16:49:43.927" Score="3" ViewCount="" Body="&lt;p&gt;Dirichlet function defined on $\displaystyle [0,1]$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle D(x) = 0$ is x is irrational.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle D(x) = \frac{1}{q}$ if $\displaystyle x = \frac{p}{q}$ in a rational in reduced form, $\displaystyle q \gt 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle F(x) = \int_{0}^{x} \ D(t) \text{d}t = 0 \ \forall \ x \in [0,1]$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that there is no function $\displaystyle f$ such that $\displaystyle f&amp;#39;(x) = D(x)$, because derivatives have the intermediate value property (see here for instance: &lt;a href=&quot;http://www.math.wvu.edu/~kcies/teach/Fall03Spr04/451NadlerText/451Nadler101-120.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.math.wvu.edu/~kcies/teach/Fall03Spr04/451NadlerText/451Nadler101-120.pdf&lt;/a&gt;) and $\displaystyle D(x)$ takes only rational values.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Not really sure if this is what you are looking for, though.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If one can prove that an integral is some value, you have proven that the limit of Riemann Sums is the same value.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, hope that helps.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2011-02-25T17:21:23.497" LastActivityDate="2011-02-25T17:21:23.497" CommentCount="4" />
  <row Id="23738" PostTypeId="2" ParentId="23727" CreationDate="2011-02-25T17:16:31.477" Score="3" ViewCount="" Body="&lt;p&gt;There is an &lt;a href=&quot;http://ocw.mit.edu/courses/mathematics/18-098-street-fighting-mathematics-january-iap-2008/&quot; rel=&quot;nofollow&quot;&gt;MIT OCW course&lt;/a&gt; and &lt;a href=&quot;http://mitpress.mit.edu/books/full_pdfs/Street-Fighting_Mathematics.pdf&quot; rel=&quot;nofollow&quot;&gt;textbook&lt;/a&gt; called &lt;em&gt;Street-Fighting Mathematics&lt;/em&gt; by Mahajan which discusses dimensional analysis among other things that you might find useful. For this problem the key observation is to think about what happens to an ellipse when you stretch or shrink along either axis. (Hint: you get another ellipse. What is the relationship between their areas? Between their semimajor / semiminor axes?)&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2011-02-25T17:16:31.477" CommentCount="1" />
  <row Id="23739" PostTypeId="1" AcceptedAnswerId="23747" CreationDate="2011-02-25T17:21:30.157" Score="7" ViewCount="104" Body="&lt;p&gt;Suppose I define a principal $G$-bundle as a map $\pi: P \to M$ with a smooth right action of $G$ on $P$ that acts freely and transitively on the fibers of $\pi$.  Does it follow that $P$ is locally isomorphic to $M \times G$ with the obvious right action of $G$ on $M \times G$?  Let's suppose $M$ is a manifold.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that fiber bundles over a contractible set are trivial and a manifold is locally contractible, but I believe this statements refers to locally trivial fiber bundles and so will not apply to this case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A related question is: if we have a fibration such that the base space is contractible and all fibers are homeomorphic, does it follow that the fibration is just the product of the base with the fiber?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="9" LastActivityDate="2011-02-25T19:34:49.187" Title="Local triviality of principal bundles" Tags="&lt;differential-geometry&gt;&lt;algebraic-topology&gt;&lt;fiber-bundles&gt;" AnswerCount="1" CommentCount="6" />
  <row Id="23740" PostTypeId="1" CreationDate="2011-02-25T17:34:36.460" Score="4" ViewCount="134" Body="&lt;p&gt;I'm gonna take a Calculus course next year, my professor suggest me to review high school algebra. I want to know, which book is good for refresh knowledge on high school algebra?&lt;/p&gt;&#xA;" OwnerUserId="725" LastActivityDate="2011-02-25T21:01:16.287" Title="Good book for high school algebra" Tags="&lt;algebra&gt;&lt;algebra-precalculus&gt;&lt;books&gt;" AnswerCount="2" />
  <row Id="23741" PostTypeId="2" ParentId="23720" CreationDate="2011-02-25T17:36:16.070" Score="0" ViewCount="" Body="&lt;p&gt;This type of thing leads to some amusing puzzles.  Like if you take a trip and for the first half you go 30 mph and for the last half you go 70 mph, is your average speed 50 mph?  Answer:  depends what you mean by half.  If &quot;half&quot; means half of the time, then the answer is yes, but if &quot;half&quot; means half of the distance, then the answer is no.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly:  if you average 30 mph on an outbound trip, how fast do you have to go on the return trip in order to average 60 mph for the whole trip?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly:  construct a scenario where there are two baseball players, and on each day player 1 has a higher batting average than player 2, but after a week player 2 has a higher (cumulative) average than player 1.&lt;/p&gt;&#xA;" OwnerUserId="6071" LastActivityDate="2011-02-25T17:36:16.070" />
  <row Id="23742" PostTypeId="2" ParentId="23740" CreationDate="2011-02-25T18:06:02.473" Score="2" ViewCount="" Body="&lt;p&gt;This &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0817636773&quot; rel=&quot;nofollow&quot;&gt;book&lt;/a&gt;, by Israel Gelfand, is perhaps the best book on high school algebra I have ever seen. Written by a leading mathematician, it is short, clear, and concise, yet provides a depth of understanding that you cannot find in many other books on the subject.&lt;/p&gt;&#xA;" OwnerUserId="7323" LastActivityDate="2011-02-25T18:06:02.473" CommentCount="1" />
  <row Id="23743" PostTypeId="2" ParentId="23740" CreationDate="2011-02-25T18:09:45.537" Score="0" ViewCount="" Body="&lt;p&gt;See also Lang's &lt;em&gt;Basic Mathematics&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="589" LastActivityDate="2011-02-25T18:09:45.537" />
  <row Id="23744" PostTypeId="1" AcceptedAnswerId="23745" CreationDate="2011-02-25T18:22:39.083" Score="1" ViewCount="117" Body="&lt;p&gt;is $x \in H$ then does it follow that $||x|| &amp;lt; \infty$?&lt;/p&gt;&#xA;" OwnerUserId="6556" LastActivityDate="2011-02-25T19:56:35.890" Title="question about hilbert space" Tags="&lt;hilbert-spaces&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="23745" PostTypeId="2" ParentId="23744" CreationDate="2011-02-25T18:31:38.557" Score="3" ViewCount="" Body="&lt;p&gt;Short answer (with some extra text to fill it out): Yes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;:)&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2011-02-25T18:31:38.557" />
  <row Id="23746" PostTypeId="1" CreationDate="2011-02-25T18:59:41.960" Score="12" ViewCount="416" Body="&lt;p&gt;This question is related to my bath towel, which I hang on a rope, so let's have fun (you can use your own towel to do this experiment in bath-o).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is this rectangle with sides $a&amp;lt;b$. The rectangle is bent along a line that passes through the center of the rectangle. At which angle $\alpha$ (&lt;em&gt;i. e.&lt;/em&gt;, angle $\angle RAU$ – see the picture) should we bend the rectangle in order to get the minimum area of crossing intersection?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is obvious that if $|a-b| \gg 0$, then $\alpha\approx\pi/4$. Also, we can look at the pentagon NOPQR. In addition to that, for a certain $a/b$ we get the triangle ($a/b&amp;lt;0.8150237, \alpha =\pi/4$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All in all, I am looking for a graph: $a/b$ as in terms of $\alpha$ and the area of the crossing intersection, which is $\displaystyle S = [2ab - (a^2+b^2)t + 2abt^2 - (a^2+b^2)t^3]/[4(1-t^2)]$, where $t=\tan \alpha$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any help is highly welcome.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/j18po.jpg&quot; alt=&quot;MAth towel&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="7418" LastEditorUserId="6622" LastEditorDisplayName="" LastEditDate="2011-05-14T23:05:00.947" LastActivityDate="2011-05-15T01:31:44.300" Title="Bath towel on the rope" Tags="&lt;classical-analysis&gt;&lt;elementary-geometry&gt;" AnswerCount="0" CommentCount="20" FavoriteCount="1" />
  <row Id="23747" PostTypeId="2" ParentId="23739" CreationDate="2011-02-25T19:34:49.187" Score="3" ViewCount="" Body="&lt;p&gt;For the first question - yes, at least if you suppose $P$ is a smooth manifold and, say, $G$ is a Lie group. For the principal $G$-bundles with your definition, just as with the regular one, being trivial is the same as admitting a section (in the section is $s$, map $s(x)$ to ($x$, unit of $G$), and use the action to define the rest of the trivializing map). Now to construct a section locally, near $x_0$ take any $s(x_0)$ in the fiber above $x_0$. Pick an auxiliary Riemann metric near  take  the orthogonal subspace to the tangent of the fiber at $s(x)$. Exponential map will give you a section locally. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other categories you would need to construct the section $s$ in a different manner. I think this can be done in the category of topological manifolds. Not sure about more general cases, but it would seem ok. Maybe for CW complexes you can go cell by cell?&lt;/p&gt;&#xA;" OwnerUserId="2633" LastActivityDate="2011-02-25T19:34:49.187" CommentCount="1" />
  <row Id="23748" PostTypeId="1" AcceptedAnswerId="23753" CreationDate="2011-02-25T19:47:10.730" Score="5" ViewCount="238" Body="&lt;p&gt;Does there exist a property which is known to be satisfied by only one integer, but such that this property does not provide a means by which to compute this number? I am asking because this number could be unfathomably large. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was reading &lt;a href=&quot;http://math.stackexchange.com/questions/514/conjectures-that-have-been-disproved-with-extremely-large-counterexamples&quot;&gt;Conjectures that have been disproved with extremely large counterexamples?&lt;/a&gt; , does there exist a conjecture that is known to have a counterexample, but which has not been found, and where there is no &quot;bound&quot; on the expected magnitude of this integer?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there known something about how the largest integer that is expressable in n symbols, grows with n?&lt;/p&gt;&#xA;" OwnerUserId="4929" LastActivityDate="2011-02-25T20:28:29.620" Title="Largest known integer" Tags="&lt;number-theory&gt;&lt;intuition&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" />
  <row Id="23749" PostTypeId="2" ParentId="23744" CreationDate="2011-02-25T19:56:35.890" Score="4" ViewCount="" Body="&lt;p&gt;On any real or complex vector space $X$ for which a norm $\|\cdot\|$ is defined, part of the definition is that $\|x\|$ is a real number for each $x\in X$.  The norm on a real or complex inner product space $H$ fits into this context, because part of the definition of the inner product is that $\langle x,y\rangle$ is a real or complex number for each $x$ and $y$ in $H$, and that $\langle x,x\rangle$ is nonnegative for each $x\in H$, and hence $\langle x,x\rangle$ is a nonnegative real number (excluding the possibility of $\langle x,x\rangle=\infty$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In some contexts there is notational abuse of $\|\cdot\|$, which may be the source of the question here.  For example, suppose that $\mu$ is a positive measure on $X$, and $1\leq p\lt \infty$.  Some authors will say that for a measurable real or complex-valued function $f$ on $X$, $\|f\|_p$ is defined to be the $p^\text{th}$ root of $\int_X |f|^pd\mu$, before defining $L^p(\mu)$ to be the set of such $f$ for which $\|f\|_p$ is finite.  With this convention, $\|\cdot\|_p$ is a norm &lt;em&gt;when restricted to $L^p(\mu)$&lt;/em&gt;, but the extended notation allows $\|f\|_p=\infty$ to also be a meaningful statement; it is equivalent to saying that $f$ is not in $L^p(\mu)$.  So for example, $\|f\|_2$ can be infinite for some measurable $f$, but $\|\cdot\|_2$ is a norm on the Hilbert space $L^2(\mu)$, meaning in part that $\|f\|_2$ is a nonnegative real number for all $f\in L^2(\mu)$.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2011-02-25T19:56:35.890" />
  <row Id="23750" PostTypeId="2" ParentId="23685" CreationDate="2011-02-25T19:56:39.270" Score="1" ViewCount="" Body="&lt;p&gt;Henry's method is really superior here, because the number of things you need to keep track of by direct counting (both to account for inclusion/exclusion, and to get rid of &quot;undesirables&quot;) gets more and more complicated as you increase the number of dice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But let's discuss your counts, so you can see the errors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One way to think about these counts is that you are giving instructions to someone on how to write down these rolls; the binomial coefficients just count how many ways that person can follow the instructions. So for example, when you want to count how many rolls have three 3-of-a-kinds, you can tell someone: &quot;first pick what the 3-of-a-kinds will be; then to write them down, pick which dice the largest one shows up in, then which dice the smallest one shows up in; the middle one will be in the remaining dice&quot;. This explains how to write down any particular instance of three 3-of-a-kind. So you can say, &quot;Okay, I'll pick 3, 5, and 2. Then I pick the dice for the 5s, which I will be 8-2-9; then from the remaining ones I'll pick the three dice for the 2s, which I choose to be 1-5-7. Then remaining dice, 3-4-6, that's where the 3s show up. That is, this roll is 253323255.&quot; Then the binomial coefficients just count the number of ways in which you can make each choice. But then you have to be careful with overcounting, not to count the same exact roll more than once by focusing on different parts of it (that's why here I said that your first choice of three dice would correspond to the &lt;em&gt;largest&lt;/em&gt; of the ranks; if I had just said &quot;the first rank you picked&quot;, then we could have ended up repeating this particular roll if we selected the ranks and the dice in a different order).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I will use $n$-oak to mean &quot;$n$ of a kind&quot;...&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;As I mentioned, your count for the total numbers is off because you are overcounting. First we select the rank for the 3-oak. Then we select which dice it shows up in. Finally, we assign the remaining 6 dice to any of the remaining 5 ranks. However, this will overcount any roll that has several 3-oaks, because we will count it once for each 3-oak. So 111222345 will be counted once when we select 1 for the 3-oak and positions 1-2-3; and will be counted &lt;em&gt;again&lt;/em&gt; when we select 2 for the 3-oak in positions 4-5-6. Likewise, rolls with three 3-oaks will be counted several times, once for each choice. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How many times? Rolls with exactly two 3-oaks will be counted twice; rolls with three 3-oaks will be counted 3 times. So we want to count the number of rolls with two 3-oaks and subtract them from the previous total; and the number of rolls with three 3-oaks and subract them &lt;em&gt;twice&lt;/em&gt; from the previous count.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Okay, so let's count the number of rolls with two 3-oaks. We first select the two ranks; then we select where the higher rank shows up; then where the lower rank shows up; then we assign the remaining three dice to any of the remaining 4 ranks. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, this &lt;em&gt;also&lt;/em&gt; counts rolls with three 3-oaks. In fact, it will count each one of them &lt;em&gt;three&lt;/em&gt; times, because there are three ways in which we can specify two of the three 3-oaks. So if just subtract the total above, we will be subtracting three times the number of rolls with three 3-oaks, but we only wanted to subtract it &lt;em&gt;twice&lt;/em&gt;. So we will need to count the rolls with three 3-oaks and add them back in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To count the number of rolls with three 3-oaks, select the three ranks, then decide where the highest rank goes, then where the lowest rank goes, the rest is now forced.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(The process above is called &quot;inclusion/exclusion&quot;; first we included too much, then we subtract but we subtract too much, so then we include back in what need to; as the number of dice increases, you would then probably have to take out some stuff that you added back extra, then add back in some of them, etc. Complicated).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So: &quot;pick a rank for the 3-oak, pick the dice it shows up in, remaining six dice can be anything except the rank of the 3-oak&quot; is&#xA;$$\binom{6}{1}\binom{9}{3}5^6=7875000.$$&#xA;Then &quot;select two ranks for two 3-oaks, select where the highest rank goes, select where the lowest rank goes, and the remaining three dice can be anything other than the ranks of these two&quot; is&#xA;$$\binom{6}{2}\binom{9}{3}\binom{6}{3}4^3=1612800.$$&#xA;And then &quot;select the three ranks for the three 3-oaks; select where the highest rank goes, select where the lowest rank goes&quot; is&#xA;$$\binom{6}{3}\binom{9}{3}\binom{6}{3}=33600.$$&#xA;So you $A$ should have been&#xA;$$A = 7875000-1612800+33600 = 6295800.$$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;A 3-oak and a 6-oak: first select the rank for the 3-oak, then the rank for the 6-oak, then select where the 3-oak is. Note that the order of choice matters, because the &lt;em&gt;first&lt;/em&gt; choice is the 3-oak. So we get&#xA;$$\binom{6}{1}\binom{5}{1}\binom{9}{3} = 2520.$$&#xA;You got fewer, because instead of selecting the rank for the 3-oak and the rank for the 6-oak, you just selected two ranks without specifying which one would be the 3-oak and which one would be the 6-oak. So here, you counted 111222222 the same as 222111111, which is incorrect. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;A 3-oak and a 5-oak. First select the rank for the 3-oak, then the rank for the 5-oak. Then decide where the 3-oak goes, then where the 5-oak goes. Finally, choose what the remaining die is, from the other four ranks:&#xA;$$\binom{6}{1}\binom{5}{1}\binom{9}{3}\binom{6}{5}4=60480.$$&#xA;Again, your error was not to distinguish between which rank will be the 3-oak and which one will be the 5-oak. Selecting them with $\binom{6}{2}$ does not distinguish which one you pick first and which one you pick second.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;A 3-oak and a 4-oak. Here we have two possibilities: a 3-oak, a 4-oak, and a 2-oak; or a 3-oak, a 4-oak, and nothing else. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the first kind: select the rank for the 3-oak; then the rank for the 4-oak; then the rank for the 2-oak. Then select where the 3-oak goes; then select where the 4-oak goes. The pair is then forced into the remaining dice. This is&#xA;$$\binom{6}{1}\binom{5}{1}\binom{4}{1}\binom{9}{3}\binom{6}{4} = 151200.$$&#xA;For the second kind: select rank for the $3$-oak, the rank for the $4$-oak; then select where the 3-oak goes, then where the 4-oak goes from among the remaining 6 positions. Finally, pick two ranks from the remaining 4, and select which dice will have the larger one of the two (this allows us to count them exactly); the remaining dice shows the smaller one. This is&#xA;$$\binom{6}{1}\binom{5}{1}\binom{9}{3}\binom{6}{4}\binom{4}{2}\binom{2}{1} = 453600.$$&#xA;So we get a total of $151200+453600 = 604800$ rolls with a 3-oak and a 4-oak (accounting for both 3-oak, 4-oak, and 2-oak; and 3-oak, 4-oak, and two singletons).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Two 3-oaks but not three. We counted them above: first select the two ranks; then the dice where the larger one goes; then select the dice for the lower rank. Then anything from the remaining 4 ranks to the other three dice. However, this will overcount, because we are including the rolls with 3-oaks, counting each of them three times. So we need to subtract the rolls with three 3-oaks three times. The number of ways of rolling three 3-oaks: select the three ranks, select where the highest ranking one goes, select where the lowest ranking one goes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the first count is given by&#xA;$$\binom{6}{4}\binom{9}{3}\binom{6}{3}4^3 = 1612800;$$&#xA;and then the ones we overcounted are given by&#xA;$$\binom{6}{3}\binom{9}{3}\binom{6}{3} = 33600.$$&#xA;So we get $1612800 - 3(33600) = 1512000$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You tried to to take into account those three 3-oak rolls, but you did not compensate quite right, because you only subtracted the three 3-oak rolls once; you needed to subtract it more times, because you overcounted each of them more than once.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Finally, three 3-oaks, which we figured out to be $33600$ above.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The totals above match up with Henry's in the comments.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the total you would get from this would &#xA;$$6295800 - (2520+60480 + 604800+1512000+33600) = 4082400,$$&#xA;same as Henry's answer, but after a lot more computations and having to keep track of a lot more things (and compensating for overcounts, then overcompensating for undercounts, etc.)&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2011-02-25T19:56:39.270" CommentCount="1" />
  <row Id="23751" PostTypeId="1" CreationDate="2011-02-25T19:58:15.350" Score="2" ViewCount="86" Body="&lt;p&gt;I have been given the following definition of simplicial approximation in lectures:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Let $K, L$ be simplicial complexes and $f : |K| \to |L|$ be a continuous map of their polyhedra. A &lt;em&gt;simplicial approximation&lt;/em&gt; of $f$ is a map $g$ of vertices of $K$ to vertices of $L$ such that $$f(\mathrm{st}_K(v)) \subseteq \mathrm{st}_L(g(v))$$ for each vertex $v$ in $K$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;However, elsewhere I find the definition that a simplicial approximation is any simplicial map which is homotopic to the original map. This seems, to me, to be considerably more general than the definition I have, since, for example, if $f$ is homotopic to a constant map, then I can have very trivial simplicial approximations in this sense. So my question is, which is the more common / &quot;morally&quot; correct / better definition?&lt;/p&gt;&#xA;" OwnerUserId="5191" LastActivityDate="2011-02-26T02:32:03.917" Title="Definition of simplicial approximation" Tags="&lt;algebraic-topology&gt;&lt;definition&gt;&lt;homology&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="23752" PostTypeId="2" ParentId="23729" CreationDate="2011-02-25T20:01:36.250" Score="6" ViewCount="" Body="&lt;p&gt;Let me summarize the results of $\S 12.4$ of my notes.  (This involves making explicit some things which were left as &quot;exercises&quot;, but I'm okay with that.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $K$ be any field, let $K^{\operatorname{sep}}$ be any separable closure and let $\overline{K}$ be any algebraic closure containing $K^{\operatorname{sep}}$.  Then:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Suppose first that $K = K^{\operatorname{sep}}$, i.e., $K$ is separably closed.  Then either:&lt;br&gt;&#xA;1a) $K$ is algebraically closed, or&lt;br&gt;&#xA;1b) It isn't, i.e., $K$ has positive characteristic $p$ and there exists $a \in K$ such that &#xA;the polynomial $t^p - a$ is irreducible.  Then $t^{p^n}-a$ is irreducible for all $n$, so &#xA;$[\overline{K}:K]$ is infinite.  (For this, see Lemma 32 in $\S 6.1$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Suppose that $K$ is not separably closed, so $G = \operatorname{Aut}(K^{\operatorname{sep}}/K)$ is nontrivial.  Then by the Artin-Schreier theorem, either&lt;br&gt;&#xA;2a) $\# G = 2$, in which case $K$ is formally real and $K(\sqrt{-1})$ is algebraically closed, or&lt;br&gt;&#xA;2b) $\# G &amp;gt; 2$, in which case $G$ is infinite, which implies $[\overline{K}:K]$ is infinite.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that some of the exercises outline further extensions of the Artin-Schreier Theorem, i.e., if $\overline{K}/K$ is &quot;small&quot; in other ways then $K$ is either real closed or algebraically closed.&lt;/p&gt;&#xA;" OwnerUserId="299" LastActivityDate="2011-02-25T20:01:36.250" CommentCount="2" />
  <row Id="23753" PostTypeId="2" ParentId="23748" CreationDate="2011-02-25T20:28:29.620" Score="5" ViewCount="" Body="&lt;p&gt;One can easily generate &quot;conjectures&quot; with large counterexamples using Goodstein's theorem and related results. For example, if we conjecture that the Goodstein sequence $\rm\:4_k\:$ never converges to $0$ then the least counterexample is $\rm\ k = 3\ (2^{402653211}-1) \approx 10^{121210695}\:$. For much further discussion of Goodstein's theorem see my &lt;a href=&quot;http://groups.google.com/groups?selm=WGD.95Dec11023450%40martigny.ai.mit.edu&quot; rel=&quot;nofollow&quot;&gt;sci.math post of Dec 11 1995&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2011-02-25T20:28:29.620" CommentCount="5" />
  <row Id="23754" PostTypeId="1" CreationDate="2011-02-25T20:43:32.083" Score="2" ViewCount="105" Body="&lt;p&gt;&lt;i&gt;[Context: I'm working through &lt;a href=&quot;http://see.stanford.edu/materials/aimlcs229/problemset1.pdf&quot; rel=&quot;nofollow&quot;&gt;problem set one of Andrew Ng's Machine Learning course, question 2&lt;/a&gt;, trying to derive the Hessian myself. My trouble, though, is with the simple linear algebra. I'll ignore the regularization parameters in my question to keep things simple.]&lt;/i&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can find $H_{jk}$ by taking partial derivatives of the original function with respect to $j$ and $k$. I've worked it through to the point where I know that $H_{jk} = \sum_{i=1}^m d_i X_{ij} X_{ik}$, and I know the answer is that $H = X^TDX$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;($d$ is a vector of reals, $H$, $X$ and $D$ are $m \times m$ matrices, $D$ diagonal with $D_{ii} = d_i$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's easy for me to verify that $H = X^TDX$ gives $H_{jk} = \sum_{i=1}^m d_i X_{ij} X_{ik}$. But I don't &quot;see&quot; how to go easily in the opposite direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Is there a method, or a set of rules to memorize, or a way of talking through it that would make it easy, or do I just need to do enough basic linear algebra homework to start to recognize patterns?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Am I doing it wrong by breaking it down and deriving $H_{jk}$? Are there tricks for taking these kind of derivatives (which involve logs and exponentials) over the whole matrix at once?&lt;/p&gt;&#xA;" OwnerUserId="7512" LastActivityDate="2011-05-28T06:04:01.017" Title="Going from individual elements back to to matrix/vector notation" Tags="&lt;linear-algebra&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="23756" PostTypeId="2" ParentId="23695" CreationDate="2011-02-25T21:04:31.580" Score="6" ViewCount="" Body="&lt;p&gt;I believe that you are asking for a computable version of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Schroeder-Bernstein_theorem&quot; rel=&quot;nofollow&quot;&gt;Cantor-Schroeder-Bernstein theorem&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The traditional answer to this question is to point to &lt;a href=&quot;http://en.wikipedia.org/wiki/Myhill_isomorphism_theorem&quot; rel=&quot;nofollow&quot;&gt;Myhill's theorem&lt;/a&gt;, which asserts that if $A$ and $B$ are sets of natural numbers and you have a computable one-reduction of $A$ to $B$ (that is, a computable one-to-one function taking $A$ to $B$ and the complement of $A$ to the complement of $B$) and a computable one-reduction of $B$ to $A$, then there is a computable permutation of the natural numbers taking $A$ exactly to $B$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The proof of Myhill's theorem follows the idea of the classical proof of the Cantor-Schroeder-Bernstein theorem, but gets around the problem you mention of not knowing the nature of the loop that a given input is in with a clever switching method, which periodically closes the bijection off into finite loops.&lt;/p&gt;&#xA;" OwnerUserId="413" LastEditorUserId="413" LastEditorDisplayName="" LastEditDate="2011-02-25T21:24:49.653" LastActivityDate="2011-02-25T21:24:49.653" />
  <row Id="23757" PostTypeId="1" CreationDate="2011-02-25T21:13:34.413" Score="0" ViewCount="91" Body="&lt;p&gt;Is there any common solution how to perform deformation of the open region on the plane (simply connected 2-d closed manifold) with non-Euclidian metric to get Euclidian metric? Thanks.&lt;/p&gt;&#xA;" OwnerUserId="7513" LastActivityDate="2011-02-25T21:13:34.413" Title="How to make a plane deformation with non-Euclidean metrics into plane with Euclidean?" Tags="&lt;differential-geometry&gt;&lt;differential-topology&gt;&lt;manifold&gt;" CommentCount="1" />
  <row Id="23758" PostTypeId="1" AcceptedAnswerId="23761" CreationDate="2011-02-25T21:39:53.977" Score="5" ViewCount="126" Body="&lt;p&gt;I recently stumbled across the wikipedia page on equivalents to the Axiom of Choice. I noticed that every infinite Cartesian product of a non-empty family of non-empty sets being non-empty was equivalent to the axiom of choice. Which isn't hard to see, since any element from such a product is essentially a choice function. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This led me to wonder under what circumstances can you construct such infinite Cartesian products without the axiom of choice. I'm curious about general conditions required, presumably if there's some well-ordering on each set in the family this would be sufficient. In particular I'm curious about $\mathbb R^{\mathbb R}$ and $\mathbb R^{\mathbb N}$ that is the sets of real-valued functions on the real line and all real-valued sequences. Can we still claim that these sets exist without the axiom of choice? &lt;/p&gt;&#xA;" OwnerUserId="1930" LastEditorUserId="1930" LastEditorDisplayName="" LastEditDate="2011-02-25T22:00:53.590" LastActivityDate="2011-02-25T22:00:53.590" Title="Constructing Infinite Cartesian Products without AC" Tags="&lt;set-theory&gt;&lt;axiom-of-choice&gt;" AnswerCount="2" CommentCount="0" />
  <row Id="23759" PostTypeId="1" AcceptedAnswerId="23766" CreationDate="2011-02-25T21:41:24.127" Score="6" ViewCount="98" Body="&lt;p&gt;Is it possible that a finite combinatorial problem may admit a closed form solution, and for it to be impossible in practice to prove the validity of this solution? I'm not sure if a rigorous definition can be given to the notion of a finite combinatorial problem, but I mean problems of the following nature:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&#xA;&lt;em&gt;Given a set finite set $X$, which may or may not have additional structure, enumerate the number of elements in $X$ which satisfy some constraint defined in terms of the primitives of set theory and the additional structure of $X$.&lt;/em&gt;&#xA;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&#xA;Since we are speaking of closed form solutions, we are really interested in families of finite combinatorial problems, parametrized by $\mathbb{N}$, which&#xA;scale in some natural way with increasing $n\in \mathbb{N}$. I'm not sure if the notion of a closed form solution can be given a rigorous definition, but I mean something along the lines of the following definition from Graham, Knuth, and Patashnik's &lt;em&gt;Concrete Mathematics&lt;/em&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&#xA;&lt;em&gt;An expression for a quantity $f\left(n\right)$ is in closed form if we can compute it using at most a fixed number of &quot;well known&quot; standard operations, independent of $n$.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&#xA;I understand that this question is vague and open ended, so I would be happy with answers or partial answers to any of the following subquestions.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Are there examples of finite combinatorial problems for which empirical evidence suggests there is a closed form solution, but for which significant effort has failed to produce any proofs of the validity of this solution?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Is it possible to give a rigorous axiomatization that captures the notion of of a finite combinatorial problem that working combinatorialists work with, and then bring to bear ideas along the lines of Godel's Theorems and Turing's work on the Halting Problem to produce an existence proof for such families of combinatorial problems? Can any rigorous formulation be given to the notion of a family of finite combinatorial problems scaling naturally with $n\in\mathbb{N}$?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Are there examples of finite combinatorial problems which display a kind of regularity for large $n$. That is, are there any known families of finite combinatorial problems that scale naturally with $n\in\mathbb{N}$, such that $f\left(n\right)$, the answer to the problem associated to $n$, behaves haphazardly for small $n$, but can then be given in terms of a closed form expression for sufficiently large $n$? Is there anything to suggest that difficult combinatorial problems that have been studied, but for which little is known, may display this kind of regularity?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Finally, are there any results in mathematical logic, set theory, or proof theory of which I am unaware, that render my question trivial or foolish?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I would appreciate any help with this question that can be given. As someone with a decent background in combinatorics, but no deep knowledge of logic or set theory, I don't know where to begin with this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit:(In response to a comment of Qiaochu Yuan)&#xA;$n$ need not be the size of $X$. I hope this example should clarify what I'm trying to get at. Consider the problem of enumerating the permutations of the elements of a finite set $X$ of cardinality $n$. This problem may be cast as the following problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&#xA;Enumerate the elements of $X^n$, $\left(x_0,\ldots ,x_{n-1}\right)$, for which&#xA;$x_i = x_j$ if and only if $i = j$. The problem has solution $n!$, which may or may not be considered closed, depending on what you mean by &quot;well known&quot; in Knuth's definition. The answer to this problem is dependent only on the size of $X$, not on any interpretation of what the elements might be. In a sense, the problems of this manner could be said to scale naturally with $n$. Part of my question is to provide a rigorous definition of what I mean by scale naturally.&lt;/p&gt;&#xA;" OwnerUserId="7323" LastEditorUserId="7323" LastEditorDisplayName="" LastEditDate="2011-02-25T22:42:38.220" LastActivityDate="2011-02-26T02:08:00.473" Title="On the existence of closed form solutions to finite combinatorial problems." Tags="&lt;combinatorics&gt;&lt;logic&gt;&lt;set-theory&gt;&lt;proof-theory&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" />
  <row Id="23760" PostTypeId="2" ParentId="23758" CreationDate="2011-02-25T21:44:35.463" Score="1" ViewCount="" Body="&lt;p&gt;Sure, they exist always, by Comprehension, and they are non-empty in this case, because $X^X$ always contains the identity and constant functions. The same holds for $X^N$: we already have constant functions and many other ones. Powers are not really the issue, but product of indexed sets, of different types, that have no &quot;structure&quot; like a well-order to exploit, roughly speaking.&lt;/p&gt;&#xA;" OwnerUserId="4280" LastActivityDate="2011-02-25T21:44:35.463" />
  <row Id="23761" PostTypeId="2" ParentId="23758" CreationDate="2011-02-25T21:46:03.157" Score="8" ViewCount="" Body="&lt;p&gt;What is equivalent to the Axiom of Choice is the assertion that &lt;em&gt;every&lt;/em&gt; cartesian product of &lt;em&gt;any&lt;/em&gt; nonempty family of nonempty sets is itself nonempty (not merely that &quot;an infinite Cartesian product of a non-empty family of nonempty sets&quot;). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are plenty of families for which one can prove, without needing to invoke the Axiom of Choice, that their cartesian product is nonempty. For example, as you note, a product of any nonempty family of nonempty well-ordered sets is nonempty, regardless of the cardinality of the family. (However, proving that a denumerably infinite product of denumerably infinite sets is nonempty requires at least &lt;em&gt;some&lt;/em&gt; Choice; you can think of this as a weakening of the previous case, since here we are saying that the sets are well-order &lt;strong&gt;able&lt;/strong&gt; but not necessarily well-order &lt;strong&gt;ed&lt;/strong&gt;. That is, rather than coming already provided with a well-ordering, we are just told that a well ordering exists). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, any family in which all the sets are the same also has nonempty product: if $\{A_i\}_{i\in I}$ is a nonempty family, and $A_i=A_j\neq\emptyset$ for all $i,j\in I$, then since $A_i$ is nonempty, there exists $a\in A_i$. Then the function $f\colon I\to\cup A_i$ given by $f(i) = a$ for all $i\in I$ is a choice function for the family (and an element of $\times_{i\in I} A_i$). In particular, both the sets $\mathbb{R}^\mathbb{R}$ and $\mathbb{R}^{\mathbb{N}}$ are nonempty, and we can prove this without invoking choice. Just the $f\colon\mathbb{R}\to\mathbb{R}$ given by $f(r)=0$ for all $r$; this is an element of $\mathbb{R}^{\mathbb{R}}$; and $g(n)=0$ for all $n\in\mathbb{N}$, this is an element of $\mathbb{R}^{\mathbb{N}}$. Neither one requires AC. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Likewise, any family $\{A_i\}_{i\in I}$ in which there is a cofinite $J\subseteq I$ with $\cap_{j\in J}A_j\neq\emptyset$ will have a choice function whose existence does not depend on AC: take any $J$ with the given property, take any $x\in \cap_{j\in J}A_j$, and letting $I-J = \{i_1,\ldots,i_k\}$, pick $a_t\in A_{i_t}$, $t=1,\ldots,k$. Then $f\colon J\to\cup A_i$ given by&#xA;$$f(i) = \left\{\begin{array}{ll}&#xA;x &amp;amp; \mbox{if $i\in J$;}\\&#xA;a_1 &amp;amp;\mbox{if $i=i_1$;}\\&#xA;\vdots\\&#xA;a_k &amp;amp;\mbox{if $i=i_k$.}&#xA;\end{array}\right.$$&#xA;is a choice function for the family, whose existence can be established without invoking the Axiom of Choice. This is of course not necessary, merely sufficient.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-25T21:53:37.470" LastActivityDate="2011-02-25T21:53:37.470" CommentCount="5" />
  <row Id="23763" PostTypeId="1" CreationDate="2011-02-25T22:39:21.350" Score="4" ViewCount="252" Body="&lt;p&gt;In my recent &lt;em&gt;explorations&lt;/em&gt; I stumbled upon the following series &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;\vartheta_{4}(0,e^{\alpha \cdot z})=1+2\sum_{k=1}^{\infty} (-1)^{k}\cdot e^{\alpha \cdot z\cdot k^{2}} ; \alpha \in  \mathbb{R}, z \in \mathbb{C}&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is one of the well known  Jacobi theta functions/series with the peculiarity of having the variable $z \in \mathbb{C}$ in a different place, i.e. $e^{\alpha \cdot \mathbf{z} \cdot k^{2}}$!!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The usual form of the theta function is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align*}&#xA;\vartheta_{4}(z,e^{\alpha })=1+2\sum_{k=1}^{\infty} (-1)^{k}\cdot e^{\alpha \cdot k^{2}}\cos(2kz) ; &#xA;\end{align*}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but not in the case I have in hands. Does the former formula make any sense? Where are this kind of series &lt;strong&gt;used&lt;/strong&gt;  or &lt;strong&gt;analysed&lt;/strong&gt;? (Apart from the well known case of &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\psi(x)=\sum_{n=1}^{\infty}e^{-n^{2}\pi x}=\frac{1}{2} \left[ \vartheta_{3}(0,e^{-\pi x})-1 \right]$$&#xA;used in the context of the Riemann zeta-function.)&lt;/p&gt;&#xA;" OwnerUserId="1747" LastEditorUserId="1208" LastEditorDisplayName="" LastEditDate="2011-04-11T19:28:47.703" LastActivityDate="2011-04-12T08:25:26.767" Title="Where are this kind of series used, $\vartheta_{4}(0,e^{\alpha \cdot z})$?" Tags="&lt;number-theory&gt;&lt;sequences-and-series&gt;&lt;complex-analysis&gt;" AnswerCount="1" />
  <row Id="23764" PostTypeId="1" AcceptedAnswerId="23767" CreationDate="2011-02-25T22:48:00.817" Score="7" ViewCount="174" Body="&lt;p&gt;It's a well-known theorem (Corollary 8.10 in Lee Smooth) that given a smooth map of manifolds  $\phi:M\rightarrow N$ and a regular point $p\in N$ of $\phi$, the level set $\phi^{-1}(p)\subset M$ is a closed embedded submanifold. Is the converse true? That is, given an embedded submanifold $S\subset M$, is there necessarily a manifold $N$, smooth map $\phi:M\rightarrow N$, and regular point $p\in N$ of $\phi$ such that $S=\phi^{-1}(p)$? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Prop. 8.12 in Lee Smooth shows that this is true locally; specifically,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Let $S$ be a subset of a smooth $n$-manifold $M$. Then $S$ is an embedded $k$-submanifold of $M$ if and only if every point $p\in S$ has a neighborhood $U\subset M$ such that $U\cap S$ is a level set of a submersion $\phi:U\rightarrow\mathbb{R}^{n-k}$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(and any level set of a submersion is of course the level set of a regular point). I feel like this is the kind of question where, if there is a counterexample, it probably is very simple, but I wasn't able to come up with one. &lt;/p&gt;&#xA;" OwnerUserId="264" LastActivityDate="2011-02-26T09:11:21.323" Title="Is every embedded submanifold globally a level set?" Tags="&lt;differential-topology&gt;&lt;manifold&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="23765" PostTypeId="1" AcceptedAnswerId="23773" CreationDate="2011-02-25T22:55:52.413" Score="3" ViewCount="199" Body="&lt;p&gt;I am using Sipser terminology here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone give examples of languages A and B such that we can prove B is undecidable using A in a proof by contradiction but we A is not $\leq_m$ B.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An example of this is $E_{TM}$ as defined on page 89 of Sipser's &lt;em&gt;Introduction to the Theory of Computation 2nd ed&lt;/em&gt;. We use $A_{TM}$ in the proof by contradiction but $A_{TM}$ is not mapping reducible to $E_{TM}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance.&lt;/p&gt;&#xA;" OwnerUserId="6299" LastEditorUserId="6299" LastEditorDisplayName="" LastEditDate="2011-02-26T00:03:14.650" LastActivityDate="2011-03-17T01:23:48.507" Title="Undecidable languages and mapping reducibility" Tags="&lt;computability&gt;" AnswerCount="2" />
  <row Id="23766" PostTypeId="2" ParentId="23759" CreationDate="2011-02-25T23:10:38.903" Score="4" ViewCount="" Body="&lt;p&gt;There's some work on complexity of combinatorial counting of objects definable in Monadic Second Order Logic (MSOL).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, you can define &quot;X is an independent set of G&quot; using MSOL. Time to count number of independent sets is independent of size of the graph for family of graphs of bounded tree width. For independent sets it is independent for family of bounded clique width, and one of Makowksi's papers gives conditions beyond MSOL which allow tractability with bounded clique width.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See page 8 of Makowski/Godlin's &lt;a href=&quot;http://yaroslavvb.com/upload/save/godlin-graph.pdf&quot; rel=&quot;nofollow&quot;&gt;Graph Polynomials: From Recursive Definitions to Subset Expansion Formulas&lt;/a&gt; for examples of properties definable using MSOL on a graph.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's a number of papers from Makowski and co-authors on this line of work, for instance his student's &lt;a href=&quot;http://www.cs.technion.ac.il/~tkotek/pubfiles/YG70.pdf&quot; rel=&quot;nofollow&quot;&gt;Definability of Combinatorial Functions and their Linear Recurrence Relations.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="998" LastEditorUserId="998" LastEditorDisplayName="" LastEditDate="2011-02-26T02:08:00.473" LastActivityDate="2011-02-26T02:08:00.473" CommentCount="1" />
  <row Id="23767" PostTypeId="2" ParentId="23764" CreationDate="2011-02-25T23:14:30.150" Score="5" ViewCount="" Body="&lt;p&gt;If $M$ is compact, you can find a tubular neighborhood $U$ of $M$, which is diffeomorphic to the normal bundle $\mathcal N$ of $M$ in $N$, by some diffeo $h:\mathcal N\to U$. Now pick a metric on $\mathcal N$, and let $q:\mathcal N\to\mathbb R$ be the square of the corresponding norm. If $\phi:\mathcal N\to\mathbb R$ is a smooth function sich that $\phi(v)\sim\tfrac1{q(v)}$ outside of the unit balls in each fiber, with the aproximation better and better as $q(v)$ grows, and constantly equal to $1$ inside the balls of radius $\tfrac12$ in each fiber, then the product $\phi\cdot q$ composed with $h$ gives you a map $U\to\mathbb R$, which extends to a smooth map $N\to\mathbb R$ (with value $1$ outside of $U$), whose zero-set is precisely $N$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $M$ is not compact, you can have problems: for example, consider the map $t\in\mathbb R\mapsto ((1+e^t)\cos t,(1+e^t)\sin t)\in\mathbb R^2$. ANy function vanishing in its image also vanishes on the unit circle.&lt;/p&gt;&#xA;" OwnerUserId="274" LastEditorUserId="274" LastEditorDisplayName="" LastEditDate="2011-02-25T23:24:55.627" LastActivityDate="2011-02-25T23:24:55.627" />
  <row Id="23768" PostTypeId="1" AcceptedAnswerId="23837" CreationDate="2011-02-25T23:22:21.117" Score="6" ViewCount="135" Body="&lt;p&gt;Hello,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that we can assume that formulas are objects in $V_\omega$, and that notions such as formula and satisfiability for a standard model (when the universe is a set) are definable and absolute. Because of that it appears to me that we can define an absolute (metamathematical) formula that &quot;decides&quot; when a formula (as an object in the universe) belongs in ZFC and furthermore through it we can write a formula $S(A)$ that states $(A,\in)\models ZFC$ , which again will be absolute. Is this correct? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore, given sets $A$ and $E\subset A\times A$ it looks to me that the satisfiability of the model $(A,E)$ can be described with a $\Delta_1$ formula $S(A,E,\ulcorner\phi\urcorner)$ and thus is absolute. Is this true? And does this mean that &quot;$(A,E)\models ZFC$&quot; is absolute as well?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance.&lt;/p&gt;&#xA;" OwnerUserId="1772" LastEditorUserId="1772" LastEditorDisplayName="" LastEditDate="2011-02-26T00:30:01.703" LastActivityDate="2011-02-26T15:13:21.560" Title="Is the sentence &quot;$(A,\in)\models ZFC$&quot; absolute?" Tags="&lt;logic&gt;&lt;set-theory&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="23769" PostTypeId="2" ParentId="16066" CreationDate="2011-02-25T23:25:31.090" Score="2" ViewCount="" Body="&lt;p&gt;Suggestions&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;11&lt;/strong&gt;: &lt;strong&gt;$e^{\pi i}$&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;4&lt;/strong&gt;: &lt;strong&gt;${\sqrt 5}^{\sqrt 3}$&lt;/strong&gt;   (aprox. 4.03019240+ )&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While, it's no match for the previous entry, I can't resist giving one more for &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;11&lt;/strong&gt;: $\frac{\sqrt[4]{104 + e^{\pi \sqrt{58}}}}{6^2}$&lt;/p&gt;&#xA;" OwnerUserId="7367" LastEditorUserId="7367" LastEditorDisplayName="" LastEditDate="2011-02-26T12:41:58.520" LastActivityDate="2011-02-26T12:41:58.520" CommunityOwnedDate="2011-02-25T23:25:31.090" CommentCount="1" />
  <row Id="23770" PostTypeId="2" ParentId="23705" CreationDate="2011-02-25T23:27:13.280" Score="2" ViewCount="" Body="&lt;p&gt;You have $y=ae^{bx} + ce^{dx}$ with $a, b$ positive and $c, d$ negative. So this means:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;if $x$ is large and positive, then $y \approx ae^{bx}$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;if $x$ is large and negative, then $y \approx ce^{dx}$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So for $x$ large and positive, $x \approx (1/b) \log (y/a)$; for $x$ large and negative, $x \approx (1/d) \log (y/c)$ -- these come from solving the above approximations for $x$. If you need to go further I'd say start with these approximations and then use something like Newton's method.&lt;/p&gt;&#xA;" OwnerUserId="173" LastActivityDate="2011-02-25T23:27:13.280" />
  <row Id="23771" PostTypeId="2" ParentId="16066" CreationDate="2011-02-25T23:36:27.497" Score="4" ViewCount="" Body="&lt;p&gt;Some that I like:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;I'm rather fond of $3 \approx \log 20$. Every so often I find myself taking large powers of $e$ in my head; knowing this is helpful.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$7 \approx 12 \log_2 (3/2)$. This expresses the fact that, in music, a fifth is seven-twelfths of an octave.  &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$6 \approx \log (\pi^4 + \pi^5)$; I've actually seen T-shirts claiming that $\pi^4 + \pi^5 = e^6$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$\pi + \pi^2 \approx 13$, although unless we're in an Orwell book your clocks have no thirteen.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Mathematical_coincidence&quot; rel=&quot;nofollow&quot;&gt;wikipedia article on mathematical coincidences&lt;/a&gt; may have more ideas.&lt;/p&gt;&#xA;" OwnerUserId="173" LastActivityDate="2011-02-25T23:36:27.497" CommunityOwnedDate="2011-02-25T23:36:27.497" CommentCount="5" />
  <row Id="23772" PostTypeId="1" AcceptedAnswerId="23774" CreationDate="2011-02-25T23:49:55.250" Score="1" ViewCount="78" Body="&lt;p&gt;The operators in this question are in Hilbert space, but I'm going to word it as a discrete spectrum and a finite dimensional (linear algebra) answer is fine by me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Operators $A$ and $\Delta$ are bounded, self-adjoint and non-commutative, and $B\equiv e^{- A} \Delta e^{ A}$. Some of the eigenvalues of $A$ or $\Delta$ may be negative. I want to prove that the $B \Delta$ is positive semi-definite and that the eigenvalue $0$ has the same multiplicity in $B\Delta$ as it does in $\Delta$ (I'm not sure that this second part is true). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$B$ is a similarity transformation, so the eigenvalues, $\Delta_i$, of $\Delta$ are the same as those of $B$, but the eigenvectors are different, so the eigenvalues of $B\Delta$ are not $\Delta_i^2$. The $e^A$ form (positive eigenvalues) is important, since the product of general similar operators doesn't have a non-negative spectrum in general.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm tempted to expand in bases of the eigenvectors &#xA;$|x\rangle = \sum_i b_i |B_i\rangle&#xD;&#xA; = \sum_j d_j |\Delta_j\rangle$, &#xA;$|B_i\rangle = \sum_j c_{i,j} |\Delta_j\rangle \implies d_j = \sum_i b_ic_{i,j}$, &#xA;to find&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;\begin{aligned}&#xD;&#xA;\langle x | B\Delta | x \rangle &#xD;&#xA;&amp;amp; = \sum_{i,j} b_i^* d_j \langle B_i | B\Delta | \Delta_j \rangle&#xD;&#xA;= \sum_{i,j} b_i^* d_j \Delta_i \Delta_j \langle B_i| \Delta_j \rangle&#xD;&#xA;=\sum_{i,j,k} b_i^* d_j c_{i,k}^* \Delta_i\Delta_j \delta_{kj}  \\&#xD;&#xA;&amp;amp; = \sum_{i,j} b_i^* d_j c_{i,j}^* \Delta_i\Delta_j &#xD;&#xA;\end{aligned}&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which doesn't help much, because I haven't used that $e^A$ is positive. If I expand in the basis of eigenvectors of $A$ and $\Delta$, I get a similar but slightly more complicated expression, and I think there must be a simpler approach. &lt;/p&gt;&#xA;" OwnerUserId="7515" LastActivityDate="2011-02-26T01:51:30.730" Title="Eigenvalues of a product of similar operators" Tags="&lt;linear-algebra&gt;&lt;functional-analysis&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="23773" PostTypeId="2" ParentId="23765" CreationDate="2011-02-26T00:30:55.040" Score="4" ViewCount="" Body="&lt;p&gt;There are plenty of them, for example take two languages A and B such that A is non-recursive (that is, undecidable), $A \leq_{T} B$ (where $\leq_{T}$ denotes Turing reducibility) but not $A \leq_{m} B$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a rather trivial example (since it can be solved considerably easier), take $A=HALT$ and $B=\overline{HALT}$. Then, clearly $A \leq_{T} B$ (to compute if $x \in B$, just run the oracle to ask if $x \in A$ and say yes if the answer of the oracle is no, and say no if the answer of the oracle is yes). But if $A \leq_{m} B$, we have a recursive function f such that $x \in A \iff f(x) \in B$, and then we can show that B is recursively enumerable: to check if $x \in B$ just compute f(x) and generate HALT, when (if it happens) f(x) appears in the generation of HALT, stop. This procedure is a semidecision procedure for $B=\overline{HALT}$, hence B is recursively enumerable. Since we know HALT itself is recursively enumerable, we would have HALT recursive, a contradiction.&#xA;Thus, A does not m-reduces to B, but nonetheless, since A T-reduces to B, and we know that A is undecidable, we can immediately conclude that B is undecidable.&lt;/p&gt;&#xA;" OwnerUserId="2926" LastActivityDate="2011-02-26T00:30:55.040" CommentCount="4" />
  <row Id="23774" PostTypeId="2" ParentId="23772" CreationDate="2011-02-26T00:35:42.320" Score="1" ViewCount="" Body="&lt;p&gt;Consider the similarity transformation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$e^{A/2}B\Delta e^{-A/2}=e^{-A/2}\Delta e^{A/2} e^{A/2} \Delta e^{-A/2}=M^\dagger M$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with $M:=e^{A/2}\Delta e^{-A/2}$. An operator of the form $M^\dagger M$ is positive semi-definite, since $\langle x \vert M^\dagger M \vert x \rangle=\langle x M \vert M x \rangle \ge 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That the eigenvalue $0$ has the same multiplicity in $B\Delta$ as it does in $\Delta$ follows because the eigenvalues of $M^\dagger M$ are the squared absolute values of the singular values of $M$, the number of non-zero singular values of $M$ is the rank of $M$, the number of non-zero eigenvalues of $\Delta$ is the rank of $\Delta$, and these two ranks are equal since $M$ is obtained from $\Delta$ by a similarity transformation.&lt;/p&gt;&#xA;" OwnerUserId="6622" LastEditorUserId="6622" LastEditorDisplayName="" LastEditDate="2011-02-26T01:51:30.730" LastActivityDate="2011-02-26T01:51:30.730" CommentCount="1" />
  <row Id="23775" PostTypeId="1" AcceptedAnswerId="23785" CreationDate="2011-02-26T00:44:33.607" Score="3" ViewCount="212" Body="&lt;p&gt;Hello, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm having trouble showing the following:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If $f_n$ is a sequence of measurable functions such that $f_n$ converges to $f$ almost everywhere, then $f$ is measurable.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I was thinking of using $\limsup$ since I know that $\limsup f_n$ is measurable. But now I'm not sure how to continue my argument.&lt;/p&gt;&#xA;" OwnerUserId="6556" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2011-02-26T04:35:08.777" LastActivityDate="2011-05-07T21:55:50.657" Title="Limit a.e. of a sequence measurable functions is measurable" Tags="&lt;measure-theory&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="23776" PostTypeId="1" CreationDate="2011-02-26T00:49:49.730" Score="2" ViewCount="259" Body="&lt;p&gt;My question is how to interpret sigma algebra, especially in the context of probability theory (stochastic processes included). I would like to know if there is some clear and general way to interpret sigma algebra, which can unify various ways of saying it as history, future, collection of information, size/likelihood-measurable etc? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Specifically,I hope to know how to interpret the following in some consistent way:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;being given/conditional on a sigma algebra&lt;/li&gt;&#xA;&lt;li&gt;a subset being measurable or nonmeasurable w.r.t. a sigma&#xA;algebra&lt;/li&gt;&#xA;&lt;li&gt;a mapping being measurable or nonmeasurable w.r.t. a&#xA;sigma algebra in domain and another&#xA;sigma algebra in codomain&lt;/li&gt;&#xA;&lt;li&gt;a collection of increasing sigma algebras, i.e. a filtration of sigma algebras&lt;/li&gt;&#xA;&lt;li&gt;...&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Following are a list of examples that I have met. They are nice examples, but I feel their ways of interpretation are not clear and consistent enough for me to apply in practice. Even if there is no unified way to interpret all the examples, I would like to know what some different ways of interpretation are. &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Stopping_time&quot; rel=&quot;nofollow&quot;&gt;Stopping time&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Let $(I, \leq)$ be an ordered index&#xA;  set, and let $(\Omega, \mathcal{F},\mathcal{F}_t, \mathbb{P})$ be a&#xA;  filtered probability space. &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Then a random variable $\tau : \Omega \to I$ is called a stopping time if&#xA;  $\{ \tau \leq t \} \in \mathcal{F}_{t} \forall t \in I$. &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Speaking concretely, for τ to be a&#xA;  stopping time, it should be possible&#xA;  to decide whether or not $\{ \tau \leq t \}$ has occurred on the basis of the&#xA;  knowledge of $\mathcal{F}_t$, i.e.,&#xA;  event $\{ \tau \leq t \}$ is&#xA;  $\mathcal{F}_t$-measurable.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I was still wondering how exactly to &quot;decide whether or not $\{ \tau \leq t \}$ has occurred on the basis of the knowledge of $\mathcal{F}_t$, i.e., event $\{ \tau \leq t \}$ is  $\mathcal{F}_t$-measurable.&quot; &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Martingale_%28probability_theory%29&quot; rel=&quot;nofollow&quot;&gt;Martingale process&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If a stochastic process $Y : T \times \Omega \rightarrow S$ is a martingale&#xA;  with respect to a filtration $\{ \Sigma_t\}$ and probability measure&#xA;  $P$, then  for all s and t with $s &amp;lt; t$ and all $F \in \Sigma_s$,&#xA;          $$Y_s = \mathbf{E}_{\mathbf{P}} ( Y_t | \Sigma_s ),$$ &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;where $\Sigma_s $ is interpreted as &quot;history&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was also wondering how $\Sigma_s, s &amp;lt; t$ can act as history, $\Sigma_s, s=t$ as present, and $\Sigma_s, s &amp;gt; t$ as future?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;I originally interpret a measurable&#xA;subset wrt a sigma algebra as a&#xA;subset whose &quot;size&quot;/&quot;likelihood&quot;  is measurable,&#xA;and the class of such&#xA;size-measurable subsets must be&#xA;closed under complement and&#xA;countable union.&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;In &lt;a href=&quot;http://math.stackexchange.com/questions/23600/intuition-behind-conditional-expectation/23617#23617&quot;&gt;a post by Nate Eldredge&lt;/a&gt;, a&#xA;measurable subset wrt a sigma&#xA;algebra is interpreted by analogy of questions being answered:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If I know the answer to a question&#xA;  $A$, then I also know the answer to&#xA;  its negation, which corresponds to the&#xA;  set $A^c$ (e.g. &quot;Is the dodo&#xA;  not-extinct?&quot;).  So any information&#xA;  that is enough to answer question $A$&#xA;  is also enough to answer question&#xA;  $A^c$.  Thus $\mathcal{F}$ should be&#xA;  closed under taking complements. &#xA;  Likewise, if I know the answer to&#xA;  questions $A,B$, I also know the&#xA;  answer to their disjunction $A \cup B$&#xA;  (&quot;Are either the dodo or the elephant&#xA;  extinct?&quot;), so $\mathcal{F}$ must also&#xA;  be closed under (finite) unions. &#xA;  Countable unions require more of a&#xA;  stretch, but imagine asking an&#xA;  infinite sequence of questions&#xA;  &quot;converging&quot; on a final question. &#xA;  (&quot;Can elephants live to be 90? Can&#xA;  they live to be 99? Can they live to&#xA;  be 99.9?&quot; In the end, I know whether&#xA;  elephants can live to be 100.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance for sharing your views, and any reference that has related discussion is also appreciated!&lt;/p&gt;&#xA;" OwnerUserId="1281" LastEditorUserId="1281" LastEditorDisplayName="" LastEditDate="2011-02-26T17:20:28.890" LastActivityDate="2011-03-13T22:19:19.260" Title="Interpretation of sigma algebra" Tags="&lt;probability-theory&gt;&lt;measure-theory&gt;&lt;stochastic-processes&gt;" AnswerCount="1" CommentCount="6" FavoriteCount="1" />
  <row Id="23777" PostTypeId="1" AcceptedAnswerId="23781" CreationDate="2011-02-26T00:56:18.370" Score="4" ViewCount="93" Body="&lt;p&gt;Let $A$ be a 3x3 real symmetric matrix, and let $v_1, v_2$ be orthogonal eigenvectors of $A$.  Suppose $v_3$ is orthogonal to both $v_1$ and $v_2$.  Is $v_3$ necessarily an eigenvector of $A$?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To put it another way: by the spectral theorem we know that $\mathbb{R}^3$ has an orthonormal basis consisting of eigenvectors of $A$, but can we extend any two orthogonal eigenvectors of $A$ into such a basis?&lt;/p&gt;&#xA;" OwnerUserId="4904" LastActivityDate="2011-02-26T01:13:27.907" Title="Eigenvectors of real symmetric matrices?" Tags="&lt;linear-algebra&gt;" AnswerCount="2" CommentCount="0" />
  <row Id="23778" PostTypeId="2" ParentId="23777" CreationDate="2011-02-26T01:01:46.863" Score="3" ViewCount="" Body="&lt;p&gt;Yes: If we denote by $W\subset\mathbb{R}^3$ the span of $\{v_1,v_2\}$, then the possible choices of such $v_3$ are exactly the nonzero elements of the line $W^\perp$. Given that $A$ is real symmetric and $v_1,v_2$ are orthogonal eigenvectors, there must exist some such choice of $v_3$ which is an eigenvector, but then all other possible choices of $v_3$ are nonzero multiples of this one, hence also eigenvectors.&lt;/p&gt;&#xA;" OwnerUserId="5116" LastActivityDate="2011-02-26T01:01:46.863" />
  <row Id="23780" PostTypeId="1" AcceptedAnswerId="23786" CreationDate="2011-02-26T01:12:46.360" Score="1" ViewCount="110" Body="&lt;p&gt;Hello,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the book, it said:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $n = a_{k}10^{k} + a_{k-1}10^{k-1} + a_{k-2}10^{k-2} + ... + a_110 + a_0$&lt;br&gt;&#xA;Then, because $10 \equiv 0 \pmod{2}$ it follows that $10^j \equiv 0 \pmod{2^j}$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What congruence property did they use in this case? Is that:&lt;br&gt;&#xA;If $a \equiv b \pmod{k_1}$ and $c \equiv d \pmod{k_2}$ then,&#xA;$ab \equiv cd \pmod{k_1k_2}$ ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I saw one property in the book, which is:&lt;br&gt;&#xA;$a \equiv b \pmod{k}$ and $c \equiv d \pmod{k}$then,&#xA;$ab \equiv cd \pmod{k}$&#xA;But I really don't understand how this property relates to the one above it. Any idea?  &lt;/p&gt;&#xA;" OwnerUserId="5711" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-26T03:59:57.860" LastActivityDate="2011-02-26T03:59:57.860" Title="How does divisibility test using congruence work?" Tags="&lt;elementary-number-theory&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="23781" PostTypeId="2" ParentId="23777" CreationDate="2011-02-26T01:13:27.907" Score="3" ViewCount="" Body="&lt;p&gt;You don't need the spectral theorem to prove this. We are given that $\langle Av_1, v_3 \rangle = \langle Av_2, v_3 \rangle = 0$, and it follows directly by symmetry that $\langle v_1, Av_3 \rangle = \langle v_2, Av_3 \rangle = 0$. To make it really obvious, $A$ must be symmetric in the basis $\{ v_1, v_2, v_3 \}$, and in that basis we are given that it looks like&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left[ \begin{array}{ccc}&#xD;&#xA; \lambda_1 &amp;amp; 0 &amp;amp; * \\\&#xD;&#xA; 0 &amp;amp; \lambda_2 &amp;amp; * \\\&#xD;&#xA; 0 &amp;amp; 0 &amp;amp; * \end{array} \right]$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and we can complete the rest of the matrix by symmetry. &lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2011-02-26T01:13:27.907" />
  <row Id="23782" PostTypeId="1" CreationDate="2011-02-26T01:13:29.900" Score="6" ViewCount="114" Body="&lt;p&gt;In reading journal articles (in physics), I often come across recurring equations in the Introduction sections. Sometimes they don't mention its name. For example, I come across&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{eqnarray*}&#xD;&#xA;&#xD;&#xA;E\Psi &amp;amp;=&amp;amp; {\skew{6}{\hat}{H}}\Psi\\&#xD;&#xA;&#xD;&#xA;i\hbar\frac{\partial}{\partial t} &amp;amp;=&amp;amp; {\skew{6}{\hat}{H}}\Psi\\&#xD;&#xA;&#xD;&#xA;\end{eqnarray*}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yes, that's the Schrödinger equation (not that I actually understand it, just got it from Wikipedia). But what if I didn't know what it is? Is there some place I can type in the equation, then I'll know what its name is? Then from there, I'll know where to start looking for more information.&lt;/p&gt;&#xA;" OwnerUserId="1323" LastActivityDate="2011-02-26T01:49:42.650" Title="Have equation, want its name" Tags="&lt;math-software&gt;" AnswerCount="2" />
  <row Id="23783" PostTypeId="2" ParentId="23782" CreationDate="2011-02-26T01:20:22.703" Score="4" ViewCount="" Body="&lt;p&gt;I've been reasonably successful typing the latex code on google. Try googling &quot;\hbar i \frac{\partial \psi}{\partial t}=H\psi&quot; to see what I mean.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: try without the quotes first..&lt;/p&gt;&#xA;" OwnerUserId="6153" LastActivityDate="2011-02-26T01:20:22.703" CommentCount="2" />
  <row Id="23785" PostTypeId="2" ParentId="23775" CreationDate="2011-02-26T01:31:55.353" Score="2" ViewCount="" Body="&lt;p&gt;Based on your comment, it looks like you have a handle on the argument.  A limsup of a sequence of measurable functions is an inf of a sequence of functions each of which is a sup of a sequence of measurable functions, so it reduces to showing that both infs and sups of sequences of measurable functions are measurable.  Assuming you're working with a complete measure, it doesn't matter what happens on the set of measure zero where $(f_n)$ does not converge to $f$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just for fun, here's another way to think of this, assuming the functions are all defined on $[0,1]$ with Lebesgue measure.  By Egoroff's theorem, off of a set $A$ of arbitrarily small measure, $f_n\to f$ uniformly.  By Lusin's theorem, off of a set $B$ of arbitrarily small measure each $f_n$ is continuous (you can apply Lusin to each function with progressively smaller exceptional sets and take $B$ to be the union of these sets). Off of $A\cup B$, $f$ is a uniform limit of a sequence of continuous functions, hence continuous.  As came up at &lt;a href=&quot;http://math.stackexchange.com/questions/19146/thoughts-about-measurable-functions&quot;&gt;another question&lt;/a&gt;, a function that is continuous off of sets of arbitrarily small measure is measurable.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2011-02-26T02:11:00.540" LastActivityDate="2011-02-26T02:11:00.540" />
  <row Id="23786" PostTypeId="2" ParentId="23780" CreationDate="2011-02-26T01:46:30.923" Score="1" ViewCount="" Body="&lt;p&gt;Maybe it will help to notice that $10^j=2^j5^j$, so clearly $2^j|10^j$, and thus $10^j\equiv 0\pmod{2^j}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Essentially for that particular case, you have $10\equiv 0\pmod{2}$, which says $2|10$. It follows that $10^j\equiv 0\pmod{2^j}$ because $10^j$ has $j$ factors of $10$, each of which is divisible by $2$, and thus you can divide by $j$ factors of $2$. That is $2^j|10^j$, or $10^j\equiv 0\pmod{2^j}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does that make it more clear?&lt;/p&gt;&#xA;" OwnerUserId="1583" LastActivityDate="2011-02-26T01:46:30.923" CommentCount="5" />
  <row Id="23787" PostTypeId="1" CreationDate="2011-02-26T01:47:18.717" Score="0" ViewCount="65" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Possible Duplicate:&lt;/strong&gt;&lt;br&gt;&#xA;  &lt;a href=&quot;http://math.stackexchange.com/questions/23384/coordinate-ring-questions&quot;&gt;coordinate ring questions&lt;/a&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;p&gt;Hello!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need your help to solve this question&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A subset X⊂k^n (k a field) is called algebraic if there exist polynomials f1,…,fm∈k[t1,…,tn] such that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;X={x∈Kn|f1(x)=...=fm(x)=0}.&#xA;The coordinate ring k[X] of X is the ring of all functions f:X→k that can be represented by some polynomial. That is, there exists a polynomial g∈k[t1,…,tn] such that f(x)=g(x) for all x∈X.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1- Show that for two different points x,y∈X there exists f∈k[X] with f(x)≠f(y).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2- Let k=Z_2=Z/2Z and X⊂k^2 be the set {(0,1),(1,0)}. Is X algebraic? Determine its coordinate ring.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3- Let g_i in N be a sequence of polynomials and define Y:={x∈k^n|gi(x)=0 ∀ i}. Show that Y is algebraic set.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="7517" LastActivityDate="2011-02-26T01:47:18.717" ClosedDate="2011-02-26T02:27:32.850" Title="coordinate ring" Tags="&lt;ring-theory&gt;" CommentCount="1" />
  <row Id="23788" PostTypeId="2" ParentId="23780" CreationDate="2011-02-26T01:47:36.447" Score="1" ViewCount="" Body="&lt;p&gt;You you need to use this property $j$ times, since:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$10 \equiv 0 \mod{2} $ and $10 \equiv 0 \mod{2}$, then $10 \cdot 10 \equiv 10^2 \equiv 0 \mod{2^2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You know that $2|10$ so it must be that $2^2 | 10^2$ (factorization). Repeat again:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$10 \equiv 0   \mod{2}$ and $ 10^2 \equiv 0  \mod{2^2}$, then $10 \cdot 10^2 \equiv 10^3 \equiv 0  \mod{2^3}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in the end you get:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$10 \equiv 0  \mod{2}$ and $ 10^{j-1} \equiv 0 \mod{2^{j-1}}$, then $10 \cdot 10^{j-1} \equiv 10^j \equiv 0  \mod{2^j}$&lt;/p&gt;&#xA;" OwnerUserId="5105" LastActivityDate="2011-02-26T01:47:36.447" CommentCount="3" />
  <row Id="23789" PostTypeId="2" ParentId="23782" CreationDate="2011-02-26T01:49:42.650" Score="6" ViewCount="" Body="&lt;p&gt;You could try &lt;a href=&quot;http://uniquation.com/&quot; rel=&quot;nofollow&quot;&gt;http://uniquation.com/&lt;/a&gt;. This is basically a tex searcher, but it is better than a full text search. For instance search for &lt;code&gt;\frac{a}{b}&lt;/code&gt; returns results which contain &lt;code&gt;\frac{e}{m}&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe this is still in Beta though.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Caveat: I haven't used this site much.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2011-02-26T01:49:42.650" />
  <row Id="23790" PostTypeId="1" AcceptedAnswerId="23889" CreationDate="2011-02-26T01:53:15.003" Score="3" ViewCount="229" Body="&lt;p&gt;So the question states that the intersection of two prime ideals is always a prime ideal. Well this is false but I need an example to counter it. I looked online and found one &quot;For example, inside Z, 2Z and 3Z are prime, but there intersection, 6Z is not prime&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so I just need some explanation to what a prime ideal is and how you can determine that an ideal is prime. The definition I know of is &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Let $R$ be a comm. ring with identity. An ideal P is prime iff $R \neq P$ and whenever $bc \in P$ then $b \in P$ or $c \in P$ &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I dont know how to apply this definition to the example above.&lt;/p&gt;&#xA;" OwnerUserId="287" LastActivityDate="2011-02-27T03:16:13.310" Title="prime ideals and examples of them" Tags="&lt;abstract-algebra&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="1" />
  <row Id="23791" PostTypeId="2" ParentId="23764" CreationDate="2011-02-26T02:08:01.727" Score="0" ViewCount="" Body="&lt;p&gt;Obvious nessessary condition is $S$ closed . But it is not sufficient even if $S$ is compact because you have obstruction : fiber of submersion $\phi:M\to N$ at $n \in N$ has trivial normal bundle with fiber equal to $T_n (N)$. So for example if you take nontrivial line bindle on circle (=Mobius bundle) then circle can not be fiber of any submersion defined on bundle. Of course if $S$ is closed it is zero set of smooth function on $M$ by Whitney theorem; but function is not submersion .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Variation on same theme: if you take orientable manifold and submersion to orientable manifold, all fibers are orientable. So if you inbed any not orientable manifold in open subset of $\mathbb R^n$ (always possible by other Whitney theorem) it cannot be fiber of submersion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit for to take account Mariano S-A remark: Same proof show that $S$ is not fiber of regular value either:  Because points on $M$ where $\phi$ has maximal rank is open subset $M_1 \subset M$ containing $S$.And image of $M_1$ is open subset $N_1 \subset N$ because submersion is allways open . Now restrict to $\phi_1:M_1\to N_1$ . Normal bundle does not change and apply preceded result for submersions.&lt;/p&gt;&#xA;" OwnerUserId="2872" LastEditorUserId="2872" LastEditorDisplayName="" LastEditDate="2011-02-26T09:11:21.323" LastActivityDate="2011-02-26T09:11:21.323" CommentCount="1" />
  <row Id="23793" PostTypeId="2" ParentId="23751" CreationDate="2011-02-26T02:32:03.917" Score="1" ViewCount="" Body="&lt;p&gt;Mariano's comment is dead on. It might be helpful to think also of cellular approximation. Given a map of CW complexes $f: X \to Y$, $g: X \to Y$ is a cellular approximation of $f$ if they are homotopic and $g$ is cellular, that is it restricts to a map between the n-skeletons of $X$ and $Y$. I personally don't think a lot about simplicial approximations or cellular approximations because I know they always exist in the settings I care about. The point of having these approximations is then you can make standard types of arguments (like inductive arguments) that are more familiar without losing any generality. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;An interesting question is whether or not cellular/simplicial maps are cofibrant objects in a particular/obvious model category structure on the category of morphisms (if such a model structure exists). Maybe I should ask such a question on some sort of math question site...&lt;/p&gt;&#xA;" OwnerUserId="627" LastActivityDate="2011-02-26T02:32:03.917" CommentCount="3" />
  <row Id="23794" PostTypeId="2" ParentId="23790" CreationDate="2011-02-26T02:51:38.553" Score="1" ViewCount="" Body="&lt;p&gt;An equivalent definition of prime ideals is the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An ideal I in a ring R is a &lt;strong&gt;prime ideal&lt;/strong&gt; if the quotient ring R/I is an integral domain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proof of equivalence: Assume $I$ is a prime ideal in &quot;your&quot; definition. And assume $xy=0$ in $R/I$. This says precisely that $xy \in I$. But this means that either $x$ or $y$ is in $I$, since $I$ is prime, hence, say, $x=0$ in $R/I$, hence $R/I$ is an integral domain. The converse is similar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For an example of a prime ideal, consider $(x)$ in the polynomial ring $k[x,y]$ (all polynomials in $x,y$ over a field $k$). The quotient $k[x,y]/(x)$ is isomorphic to $k[y]$ which is an integral domain, hence $(x)$ is prime. However, the intersection of the two prime ideals $(x)$ and $(y)$, $(x) \cap (y)$ is $(xy)$, which is not prime. For consider the quotient $k[x,y]/(xy)$. The product $xy=0$ zero, but neither $x$ nor $y$ is zero. So the quotient is not an integral domain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In $\mathbb{Z}$, the prime ideals correspond to the principal ideal $(p)$ generated by prime numbers. Every non-zero prime ideal in $\mathbb{Z}$ is also maximal. Therefore, the intersection of any two of them is properly contained in one of them, and so the intersection is not maximal and so the intersection is not prime. (this is a bit circular though!)&lt;/p&gt;&#xA;" OwnerUserId="4284" LastEditorUserId="4284" LastEditorDisplayName="" LastEditDate="2011-02-26T03:01:29.493" LastActivityDate="2011-02-26T03:01:29.493" CommentCount="1" />
  <row Id="23795" PostTypeId="2" ParentId="23763" CreationDate="2011-02-26T03:03:46.040" Score="2" ViewCount="" Body="&lt;p&gt;I've seen this series in Ono's book &quot;The web of modularity&quot; in Theorem 1.60 on page 17 where he gives the identity:&#xA;$$\frac{\eta(z)^2}{\eta(2z)} = \sum_{n= -\infty}^\infty (-1)^n q^{n^2}$$&#xA;where $q = e^{\pi i z}$ and &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\eta(z) := q^{1/24}\prod_{n=1}^\infty (1-q^n).$$&#xA;is the Dedekind Eta function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suppose one reason that this kind of thing could be useful is that it is very sparse, and therefore fast to compute lots of coefficients very quickly. For example you could use it to calculate the $q$-expansion of the Eisenstein series $E_4$ using the identity (equation 1.28 in Ono)&#xA;$$E_4 = \frac{\eta(z)^{16}}{\eta(2z)^8} + 2^8 \frac{\eta(2z)^{16}}{\eta(z)^8}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(However I think there are faster ways to calculate $E_4$?!)&lt;/p&gt;&#xA;" OwnerUserId="668" LastActivityDate="2011-02-26T03:03:46.040" CommentCount="0" />
  <row Id="23796" PostTypeId="2" ParentId="23790" CreationDate="2011-02-26T03:10:18.340" Score="1" ViewCount="" Body="&lt;p&gt;By definition, $\rm\ 6\ \mathbb Z\ $ isn't prime $\rm\iff$ there are $\rm\:b,c\ $ such that $\rm\ bc \in 6\ \mathbb Z,\ b,c \not\in 6\ \mathbb Z\:.\:$ Equivalently, $\rm\ 6\ |\ bc,\ 6\ \nmid b,c\:.\ $ So it suffices to exhibit a &lt;em&gt;proper&lt;/em&gt; factorization of $6$. Hence the argument shows that the generator of a principal prime ideal must be irreducible (= prime in a UFD). That explains why a minimal polynomial is sometimes called an &lt;em&gt;irreducible polynomial&lt;/em&gt;. Beware that minimal polynomials needn't be irreducible over algebras with zero-divisors (e.g. matrix rings).&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2011-02-26T19:45:54.917" LastActivityDate="2011-02-26T19:45:54.917" />
  <row Id="23797" PostTypeId="1" CreationDate="2011-02-26T03:15:36.010" Score="1" ViewCount="184" Body="&lt;p&gt;I am having trouble with the following problem and would appreciate some help.  It is a piece of a proof of Cauchy's integral formula for a circle that appears in Dieudonne's &lt;em&gt;Treatise on Analysis, Vol. I&lt;/em&gt;.  I am interested in this proof because it proceeds directly from the assumption that $f$ is complex differentiable in an open neighborhood of the disk.  All other proofs that I have seen take a more indirect route, often going through some form of Cauchy's theorem first.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $f$ be complex differentiable on an open neighbourhood of the closed unit disc.&#xA;Assume that, for each given $\lambda \in (0,1)$ and $\epsilon &amp;gt; 0$, there are points $t_0 = 0 &amp;lt; t_1 &amp;lt; \cdots &amp;lt; t_m = 2\pi$, a number $\delta &amp;gt; 0$, and in each interval $[t_k,t_{k+1}]$ a point $\theta_k$ such that &#xA;$$&#xD;&#xA;| f(\zeta_k + x) - f(\zeta_k) -  f^{\prime}(\zeta_k)x | \leq \epsilon |x|&#xD;&#xA;$$&#xA;whenever $|h| \leq \delta$ and $t_k \leq t \leq t_{k+1}$.&#xA;Here $\zeta_k = z + \lambda(e^{i\theta_k} - z)$ and $\zeta_k + x = z + (\lambda+h)(e^{it} - z)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Define&#xA;$$&#xD;&#xA;g(\lambda) = \int_{0}^{2\pi} \frac{ f(z + \lambda(e^{it} - z)) - f(z)}{e^{it} - z} e^{it} dt.&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By comparing each integral &#xA;$$&#xD;&#xA;A_k = \int_{t_k}^{t_{k+1}} \frac{ f( z + (\lambda + h)(e^{it} - z) ) - f( z + \lambda(e^{it} - z) ) }{e^{it} - z} e^{it} dt&#xD;&#xA;$$&#xA;to the corresponding expression &#xA;$$&#xD;&#xA;B_k = \frac{h}{i \lambda} (   f( z + \lambda(e^{i t_{k+1}} - z) ) - f( z + \lambda(e^{i t_k} - z) )   )&#xD;&#xA;$$&#xA;show that, for each given $\lambda \in (0,1)$ and $\epsilon &amp;gt; 0$, there is a $\delta &amp;gt; 0$ such that&#xA;$$&#xD;&#xA;| g(\lambda + h) - g(\lambda) - 0 \cdot h | \leq \epsilon |h|&#xD;&#xA;$$&#xA;for $|h| \leq \delta$.  That is, show that $g^{\prime}(\lambda) = 0$ for $\lambda \in (0,1)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note&#xA;$$&#xD;&#xA;\sum_{k=0}^{m-1} (A_k - B_k) = g(\lambda + h) - g(\lambda) - 0 \cdot h.&#xD;&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="6887" LastEditorUserId="6887" LastEditorDisplayName="" LastEditDate="2011-03-02T05:00:08.700" LastActivityDate="2011-03-02T05:00:08.700" Title="An estimate needed in a proof of Cauchy's integral formula" Tags="&lt;analysis&gt;&lt;complex-analysis&gt;" CommentCount="2" />
  <row Id="23798" PostTypeId="2" ParentId="23780" CreationDate="2011-02-26T03:32:47.767" Score="2" ViewCount="" Body="&lt;p&gt;It's no coincidence that the congruence sign $\equiv$ resembles the equality sign $=\:$. This notation was explicitly devised to help remind you of the fact that congruence relations share many of the same properties as the equality relation. In particular, just like equations in the ring of integers, ring congruences can be added, multiplied, scaled, etc. Thus, considering this analogy, how would you prove that $\rm\ n = 0\ \Rightarrow\ n^{\:j} = 0\ $ for $\rm\:n\:$ an integer? Precisely the same proof works for congruences.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For completeness, here is a proof of the &lt;strong&gt;congruence product rule&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;LEMMA&lt;/strong&gt; $\rm\ \ A\equiv a,\ B\equiv b\ \Rightarrow\ AB\equiv ab\ \ (mod\ m)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof&lt;/strong&gt; $\rm\ \ m\: |\: A-a,\:\:\ B-b\ \Rightarrow\ m\ |\ (A-a)\ B + a\ (B-b)\ =\ AB - ab $&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2011-02-26T03:37:58.947" LastActivityDate="2011-02-26T03:37:58.947" />
  <row Id="23799" PostTypeId="1" AcceptedAnswerId="23802" CreationDate="2011-02-26T03:35:44.173" Score="11" ViewCount="455" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Second-order_logic&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt; describes the first-order vs. second-order logic as follows:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;First-order logic uses only variables that range over &lt;strong&gt;individuals&lt;/strong&gt; (elements of the domain of discourse); second-order logic has these variables as well as additional variables that range over &lt;strong&gt;sets of individuals&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It gives $\forall P\,\forall x (x \in P \lor x \notin P)$ as an SO-logic formula, which makes perfect sense to me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, in a &lt;a href=&quot;http://cstheory.stackexchange.com/q/5117/873&quot;&gt;post at CSTheory&lt;/a&gt;, the poster claimed that $\forall x\forall y(x=y\leftrightarrow\forall z(z\in x\leftrightarrow z\in y))$ is an FO formula. I think this must not be the case, since in the above formula, $x$ and $y$ are &lt;strong&gt;sets of individuals&lt;/strong&gt;, while $z$ is an &lt;strong&gt;individual&lt;/strong&gt; (and therefore this must be an SO formula).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I mentioned this as a comment, but two users commented that ZF can be entirely described by FO logic, and therefore $\forall x\forall y(x=y\leftrightarrow\forall z(z\in x\leftrightarrow z\in y))$ is an FO formula.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm confused. Could someone explain this please?&lt;/p&gt;&#xA;" OwnerUserId="2420" LastActivityDate="2011-04-07T20:40:17.370" Title="First-Order Logic vs. Second-Order Logic" Tags="&lt;logic&gt;&lt;set-theory&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="6" />
  <row Id="23800" PostTypeId="2" ParentId="23799" CreationDate="2011-02-26T03:54:14.153" Score="9" ViewCount="" Body="&lt;p&gt;The statement is indeed a first order statement in standard Set Theory. But it is no wonder you are a bit confused.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most people think of sets and elements as different things. However, in standard set theory this is not the case. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In standard set theory, &lt;em&gt;everything&lt;/em&gt; is a set (there are no &quot;ur-elements&quot;, elements that are not sets). The &lt;em&gt;objects&lt;/em&gt; of set theory are sets themselves. The primitive relation $\in$ is a relation between &lt;em&gt;sets&lt;/em&gt;, not between ur-elements and sets. So, for example, the Axiom of the Power Set in ZFC states that&#xA;$$\forall x\exists y(\forall z(z\in y \leftrightarrow z\subseteq x))$$&#xA;which is a first order statement, because all the things being quantified over are objects in the theory (namely, &quot;sets&quot;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you need to forget the notion that &quot;elements&quot; are things &lt;em&gt;in&lt;/em&gt; sets and sets are things that contain elements. In ZF, &lt;em&gt;everything&lt;/em&gt; is a set.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's a little hard to see the distinction between first and second order statements in ZF precisely because &quot;sets&quot; are the objects, and moreover, given any set, there is a &lt;em&gt;set&lt;/em&gt; that contains all the subsets (the power set). In a way, the Axioms are set up precisely to allow you to talk about collections of sets without having to go to second order logic. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In ZF, to get to second order you need to start talking about &quot;proper classes&quot; or &quot;properties&quot;. For instance, that's why Comprehension is not a single axiom, but an entire infinite family of axioms. Comprehension essentially says that for  &lt;em&gt;every&lt;/em&gt; property $P$ and every object $x$ of the theory (i.e., every set $x$), $\{y\mid y\in x\wedge P(y)\}$ is a set. But trying to quantify over &lt;em&gt;all propositions&lt;/em&gt; would be a second order statement. Instead, you have an &quot;Axiom Schema&quot; which says that for each property $P$, you have an axiom that says&#xA;$$\forall x\exists y\Bigl(z\in y\leftrightarrow\bigl(z\in x\wedge P(z)\bigr)\Bigr).$$&#xA;If you try quantifying over &quot;all $P$&quot;, &lt;strong&gt;then&lt;/strong&gt; you get a second order statement in ZF. &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-26T19:30:43.750" LastActivityDate="2011-02-26T19:30:43.750" CommentCount="4" />
  <row Id="23802" PostTypeId="2" ParentId="23799" CreationDate="2011-02-26T04:03:29.960" Score="11" ViewCount="" Body="&lt;p&gt;It seems to me that you're confusing the first order formulas with their intended interpretations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The language of set theory consists of just a single 2 place predicate symbol, usually denoted $\in$.  The statement you quote is a first order statement - it means just what is says: &quot;for all $x$ and for all $y$, ($x = y$ iff for all $z$,  ($z \in x$ iff $z \in  y$))&quot;, but it does &lt;em&gt;not&lt;/em&gt; tell you what $x$ is.  When you say &quot;but $x$ is a set and $z$ is an individual, so this statement looks second order!&quot;, you're adding an interpretation to the picture which is not specified by the first order formula alone - namely that &quot;for all x&quot; means &quot;for all sets x&quot; and &quot;$\in$&quot; means &quot;the usual $\in$ in set theory&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In first order logic, this &quot;adding of interpretation&quot; is usually called &quot;exhibiting a model&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's another way of looking at this same first order statement.  Suppose I reinterpret things - I say &quot;for all $x$&quot; means &quot;for all real numbers $x$&quot; and $\in$ means $&amp;lt;$.  Then, $\forall x \forall y$ ($x=y$ iff $\forall z$, ($z \in x$ iff $z \in y$)) is a true statement: it says two real numbers are equal iff they have the same collection of smaller things.  Notice that in this model, nothing looks second order.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By contrast, in second order logic, you are directly referring to subsets, so that in any model (that is, interpretation), $\forall S$ means &quot;for all subsets of whatever set the variable $x$ ranges over&quot;.&lt;/p&gt;&#xA;" OwnerUserId="331" LastEditorUserId="2420" LastEditorDisplayName="" LastEditDate="2011-02-26T05:59:20.423" LastActivityDate="2011-02-26T05:59:20.423" CommentCount="2" />
  <row Id="23803" PostTypeId="2" ParentId="23437" CreationDate="2011-02-26T04:04:22.980" Score="1" ViewCount="" Body="&lt;p&gt;I'm not 100% sure what is meant by weak multiresolution, but I have a somewhat educated guess.  My understanding is that &lt;a href=&quot;http://en.wikipedia.org/wiki/Multiresolution_analysis#Definition&quot; rel=&quot;nofollow&quot;&gt;multiresolutions&lt;/a&gt; in Hilbert spaces are usually defined not for abstract Hilbert spaces, but for $L^2$ spaces, and the hypotheses generally include properties described in terms of functions in the space.  However, at least &lt;a href=&quot;http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.127.5106&amp;amp;rep=rep1&amp;amp;type=pdf&quot; rel=&quot;nofollow&quot;&gt;one source&lt;/a&gt; (page 228, or page 5 of the pdf) uses the term &quot;multiresolution&quot; in an abstract Hilbert space $H$ to refer to a collection of subspaces $(V_j)_{j\in\mathbb{Z}}$ such that $V_j\subseteq V_{j+1}$ for all $j\in\mathbb Z$, $\cap_{j\in\mathbb Z}V_j=\{0\}$, and $\cup_{j\in\mathbb{Z}}V_j$ is dense in $H$ (and a further reference is given there as a source of this definition).  My best guess is that that is what is meant in this question.  If so, this is a straightforward exercise, and I hope that the OP will elaborate on where he or she is having trouble so that we have a better idea of how we can help.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2011-02-26T04:04:22.980" CommunityOwnedDate="2011-02-26T04:04:22.980" />
  <row Id="23804" PostTypeId="2" ParentId="23734" CreationDate="2011-02-26T04:52:22.350" Score="3" ViewCount="" Body="&lt;p&gt;This may not answer your question, but I read of another example of an integral which can be calculated without finding an anti-derivative in Halmos's &lt;em&gt;Problems for Mathematicians, Young and Old&lt;/em&gt; is the integral:&#xA;$$ \int_{0}^{1} \int_{0}^{1} \frac{1}{1-xy} dx dy.$$&#xA;The trick here is to look at the integrand at each $x,y$ as a number $\frac{1}{1-r}$ where $0&amp;lt;r=xy&amp;lt;1$ and follow your heart. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Edit:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On second thought, this answer is probably not to the point, since the solution will require a convergence theorem (monotone convergence works), and then we need an anti-derivative-less calculation of integrals of the form $\int_{0}^{1} x^k dx$. There are such (found &lt;a href=&quot;http://math2.org/math/integrals/more/x%5En.htm&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;, for instance). In any case, the method for finding this integral invites the generalizations$$\int_{0}^{1} \cdots \int_{0}^{1} \int_{0}^{1} \frac{1}{1- x_1 x_2 \cdots x_n} dx_1 dx_2 \cdots dx_n$$&#xA;where $n&amp;gt;1$.&lt;/p&gt;&#xA;" OwnerUserId="5253" LastEditorUserId="5253" LastEditorDisplayName="" LastEditDate="2011-02-26T17:44:49.503" LastActivityDate="2011-02-26T17:44:49.503" />
  <row Id="23805" PostTypeId="2" ParentId="23653" CreationDate="2011-02-26T05:25:39.053" Score="1" ViewCount="" Body="&lt;p&gt;Given a bilinear map $A/A&amp;#39; \times B/B&amp;#39; \to M$ to some abelian group $M$, lift it to a bilinear map $A \times B \to M$. It is easy to check that &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$A \times B \to M$ factors through $A \otimes B/C$ uniquely,&lt;/li&gt;&#xA;&lt;li&gt;$A/A&amp;#39; \times B/B&amp;#39; \to M$ can be factored as $\phi$ and the map $A \otimes B/C \to M$ in 1 uniquely.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;This verifies the universal property of tensor product.&lt;/p&gt;&#xA;" OwnerUserId="325" LastActivityDate="2011-02-26T05:25:39.053" />
  <row Id="23806" PostTypeId="1" AcceptedAnswerId="23818" CreationDate="2011-02-26T05:27:18.397" Score="17" ViewCount="341" Body="&lt;p&gt;I'm currently a Calculus III student. I enjoy math a lot, but only when I understand its beauty and meaning. However, so many times I have no idea what it is I am learning about, althought I am still able to solve problems pertaining to those topics. I'm memorizing the sounds this language makes, but I dont understand them. I think a big reason why most children and teenagers really dread math over any other subject is because the only thing that is taught to them is equations and numbers, and to them is not being explained its significance. For if people really knew what it all meant, then they'd realize it's probably the most important think they should ever study. Even in college, at this relatively high level of math, they &lt;em&gt;still&lt;/em&gt; do not preach its meaning, but rather scare you into cramming pages after pages of material so you can pass the next exam. When you have passed the exam, then it is safe for you to forget the material you just absorbed. This is the reason I often find myself bored of studying my current topics. For some things, I see the intuition behind it, and those are the things that keep me interested in calculus, but its often so hard to come up with a good meaning of what I'm learning all by myself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It took mankind hundreds and thousands of years to come to where we are with math, so I dont expect to understand its true meaning in an hour, but I'd really like to. The school curriculum, here in America at least, doesnt teach meaning or utility. This is the reason so many youngsters always ask &quot;when will we ever need to use this stuff?&quot;-what a naive question this I learned to be. So I guess what I'm trying to say is that I've grown bored of this material, as we often cram 2 full sections of a topic in one day. Its impossible to keep up with its meaning, but if I am to survive this course along with more advanced courses to come, I must be able to understand its meaning. My textbook doesn't help me with this issue though- no math book can really teach intuition, but they dont even attempt to. They barely even go into the history of the current topic, and thats one of my favorite parts-I like to read about how a normal man came up with this theory that revolutionized the world. It makes math feel human to me, and that I too can understand it like men before me. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most of the question answerers on this site are mathematicians, if not at least one in training. This means you have come to where you are by understanding what you have learned in the past. How have you done this? How have you been able to connect all the pieces? What are some good resources that will help in what I hope to do? How do I stop being a robot, and actually connect with what I'm learning?&lt;/p&gt;&#xA;" OwnerUserId="2985" LastActivityDate="2011-02-27T06:40:31.990" CommunityOwnedDate="2011-02-26T09:23:42.963" Title="Understanding the intuition behind math" Tags="&lt;intuition&gt;&lt;math-history&gt;" AnswerCount="4" CommentCount="4" FavoriteCount="10" />
  <row Id="23807" PostTypeId="2" ParentId="23806" CreationDate="2011-02-26T05:38:03.630" Score="5" ViewCount="" Body="&lt;p&gt;Great teachers, great books, and (more recently) some great bloggers like Terry Tao and &lt;a href=&quot;http://www.dpmms.cam.ac.uk/~wtg10/mathsindex.html&quot; rel=&quot;nofollow&quot;&gt;Tim Gowers&lt;/a&gt;. There are, in fact, some books that do a great job at giving context, rationale and intuition; Silverman and Tate's &quot;Rational Points on Elliptic Curves&quot; is one example, and some of it should be accessible to a motivated high-schooler. You should definitely read &quot;Mathematics: A Very Short Introduction&quot; by Gowers, and maybe browse the Princeton Companion to Mathematics from time to time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, study physics. &lt;/p&gt;&#xA;" OwnerUserId="308" LastActivityDate="2011-02-26T05:38:03.630" CommunityOwnedDate="2011-02-26T09:23:42.963" CommentCount="1" />
  <row Id="23808" PostTypeId="1" AcceptedAnswerId="23825" CreationDate="2011-02-26T05:43:12.283" Score="17" ViewCount="382" Body="&lt;p&gt;Two players A and B each has a fair coin and they start to toss simultaneously (counted as one round). They toss in $n$ ($\ge 1$) rounds and stop because they have accumulated the same number of heads (could be 0, the case that both of them did not get any head) for the first time. What is the distribution of $n$ and its expectation?&lt;/p&gt;&#xA;" OwnerUserId="6016" LastActivityDate="2011-02-27T08:54:14.970" Title="coin toss question" Tags="&lt;probability&gt;" AnswerCount="3" FavoriteCount="5" />
  <row Id="23809" PostTypeId="1" AcceptedAnswerId="23811" CreationDate="2011-02-26T05:58:30.017" Score="3" ViewCount="137" Body="&lt;p&gt;Hello,  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone share me the trick to solve this problem using congruence?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks,&lt;br&gt;&#xA;Chan&lt;/p&gt;&#xA;" OwnerUserId="5711" LastActivityDate="2011-02-26T06:48:24.140" Title="What is the remainder of $16!$ is divided by 19?" Tags="&lt;elementary-number-theory&gt;" AnswerCount="2" />
  <row Id="23810" PostTypeId="2" ParentId="23809" CreationDate="2011-02-26T06:10:57.243" Score="4" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\ $ By &lt;a href=&quot;http://en.wikipedia.org/wiki/Wilson%27s_theorem&quot; rel=&quot;nofollow&quot;&gt;Wilson's theorem&lt;/a&gt; $\ $ mod $19\::\ -1 \ \equiv\ 18\:!\ \equiv\ 18\cdot17\cdot 16\:!\ \equiv\ (-1)\ (-2)\cdot 16\:! $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly we have the Wilson reflection formula $\rm\displaystyle\ (p-1-k)\:!\ \equiv\ \frac{(-1)^{k+1}}{k!}\ \ (mod\ p)\:,\ $ $\rm\:p\:$ prime&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2011-02-26T06:48:24.140" LastActivityDate="2011-02-26T06:48:24.140" CommentCount="7" />
  <row Id="23811" PostTypeId="2" ParentId="23809" CreationDate="2011-02-26T06:13:07.607" Score="5" ViewCount="" Body="&lt;p&gt;If you've seen &lt;a href=&quot;http://en.wikipedia.org/wiki/Wilson%27s_theorem&quot; rel=&quot;nofollow&quot;&gt;Wilson's theorem&lt;/a&gt; then you know the remainder when $18!$ is divided by $19$.  To get $16!$ mod $19$, then, you can multiply by the multiplicative inverses of $17$ and $18$ mod $19$.  Do you know how to find these?  (Edit: Referring to both inverses might be a bit misleading, because you really only need to invert their product.  See also Bill Dubuque's answer.)&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2011-02-26T06:36:11.277" LastActivityDate="2011-02-26T06:36:11.277" CommentCount="4" />
  <row Id="23812" PostTypeId="2" ParentId="20441" CreationDate="2011-02-26T06:21:20.690" Score="5" ViewCount="" Body="&lt;p&gt;Here's a generalization of the dual identity described by the OP.  It requires $n \geq m$ for the integral.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(1) \hspace{1cm} x^m e^x = \sum_{n=0}^{\infty} n^{\underline{m}}\frac{x^n}{n!},$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(2) \hspace{0.6cm} \frac{n!}{n^{\underline{m}}} = \int_0^{\infty} \frac{x^n}{x^m e^x} dx.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Equation (1) can be rewritten $$x^m e^x = \sum_{n=m}^{\infty} \frac{x^n}{(n-m)!} = \sum_{n=0}^{\infty} \frac{x^{n+m}}{n!},$$&#xA;which follows immediately from the sum in the OP's question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Equation (2) is just $$(n-m)! = \int_0^{\infty} \frac{x^{n-m}}{e^x} dx,$$ which is also straightforward.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;HR&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;OP asked in the comments how I came up with this.  It's a bit roundabout, actually. I was thinking about the question &quot;&lt;a href=&quot;http://math.stackexchange.com/questions/23734/is-there-a-definite-integral-for-which-the-riemann-sum-can-be-calculated-but-for&quot;&gt;Is there a definite integral for which the Riemann sum can be calculated but for which there is no closed-form antiderivative?&lt;/a&gt;&quot; and wondering if one way to construct an answer would be to find an interesting sum that we know how to evaluate and that could also be expressed as a Riemann sum but for which the corresponding integral had no closed form.  That eventually led me to the Wikipedia page on mathematical series, especially &lt;a href=&quot;http://en.wikipedia.org/wiki/List_of_mathematical_series#Factorial_denominators&quot; rel=&quot;nofollow&quot;&gt;this section on series with factorial denominators&lt;/a&gt;, which has your infinite sum as the first example.  I remembered your question (I had starred it as a favorite) and wondered if some of the other examples there could be used to construct answers to your question.  The sum $\sum_{m=0}^{\infty} m x^m = xe^x$ worked, and I thought I might have a generalization of your identities, but then the sum $\sum_{m=0}^{\infty} m^2 x^m = (x+x^2) e^x$ did not.  The page mentions that some of the examples there can be thought of as moments of a Poisson distribution.  I then remembered that the &lt;a href=&quot;http://en.wikipedia.org/wiki/Factorial_moment&quot; rel=&quot;nofollow&quot;&gt;factorial moments of a Poisson ($x$) distribution&lt;/a&gt; are much simpler than the usual moments; the factorial moments are just powers of $x$.  I checked it, and that turned out to be the right generalization.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As to the meaning, I'm not sure.  However, if you read &lt;a href=&quot;http://math.stackexchange.com/questions/3444/intuition-for-the-definition-of-the-gamma-function/3498#3498&quot;&gt;George Lowther's answer&lt;/a&gt; or &lt;a href=&quot;http://math.stackexchange.com/questions/3444/intuition-for-the-definition-of-the-gamma-function/20239#20239&quot;&gt;Benja's answer&lt;/a&gt; to the second question linked to in Qiaochu Yuan's answer to your question you see the Poisson distribution and its relationship to the gamma distribution figuring prominently.  That might be the source of the relationship for the generalization as well as for your original pair of identities.&lt;/p&gt;&#xA;" OwnerUserId="2370" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2011-02-27T22:10:32.737" LastActivityDate="2011-02-27T22:10:32.737" CommentCount="2" />
  <row Id="23814" PostTypeId="1" AcceptedAnswerId="23827" CreationDate="2011-02-26T06:29:20.950" Score="12" ViewCount="281" Body="&lt;p&gt;I recently showed my Algorithms class how to bound $\ln n! = \sum \ln n$ by integrals, thereby obtaining the simple factorial approximation &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ e \left(\frac{n}{e}\right)^{n} \leq n! \leq en\left(\frac{n}{e}\right)^{n} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But one student, having seen Stirling's approximation, asked where the $\sqrt{2 \pi n}$&#xA;term comes from in&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ n! \sim \left(\frac{n}{e}\right)^n \sqrt{2 \pi n} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which I couldn't readily answer.  Looking around, I can't find a straightforward and intuitive explanation.  Can someone explain this in terms understandable to a Computer Science undergrad (ie, no analysis beyond first-year calculus)?&lt;/p&gt;&#xA;" OwnerUserId="7162" LastActivityDate="2011-03-06T17:53:37.330" Title="How best to explain the $\sqrt{2\pi n}$ term in Stirling's?" Tags="&lt;approximation&gt;&lt;factorial&gt;" AnswerCount="5" CommentCount="1" FavoriteCount="2" />
  <row Id="23815" PostTypeId="2" ParentId="23814" CreationDate="2011-02-26T06:34:04.087" Score="4" ViewCount="" Body="&lt;p&gt;I'm not exactly sure how much analysis is covered by first-year calculus, but &lt;a href=&quot;http://www.math.ucdavis.edu/~romik/home/Publications_files/stirling.pdf&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; is probably the shortest proof I know of this fact. It doesn't go very far in terms of making this intuitive, though, and I don't know that this can be done at all.&lt;/p&gt;&#xA;" OwnerUserId="308" LastActivityDate="2011-02-26T06:34:04.087" />
  <row Id="23816" PostTypeId="2" ParentId="23814" CreationDate="2011-02-26T06:44:39.930" Score="3" ViewCount="" Body="&lt;p&gt;I think that the most intuitive explanation of this factor is as the total mass of a Gaussian integral in the evaluation of the stirling formual by means of steepest descent on the Gamma function defining integral. See, for example, section 2.4 of the following &lt;a href=&quot;http://ocw.mit.edu/courses/mathematics/18-238-geometry-and-quantum-field-theory-fall-2002/lecture-notes/sec2.pdf&quot; rel=&quot;nofollow&quot;&gt;lecture note&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1430" LastActivityDate="2011-02-26T06:44:39.930" CommentCount="1" />
  <row Id="23817" PostTypeId="1" AcceptedAnswerId="23842" CreationDate="2011-02-26T06:44:47.887" Score="2" ViewCount="73" Body="&lt;p&gt;Let $W^{m,p}(\Omega) = \{ f \in L^p(\Omega): D^\alpha f \in L^p(\Omega) \text{ for multi-indices } |\alpha| \leq m\}$, where $D$ denotes the weak derivative. Let $W_0^{m,p}$ denote the closure of $C_c^\infty(\Omega)$ in $W^{m,p}(\Omega)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is it true that $W_0^{m,p}(\mathbb{R}^d) = W^{m,p}(\mathbb{R}^d)$, but in general $W_0^{m,p}(\Omega) \subsetneq W^{m,p}(\Omega)$? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am trying to understand why there is a need to consider $W_0^{m,p}(\mathbb{R}^d)$. I'm guessing it's because the elements in $W^{m,p}(\Omega)$ can get really messy, but I don't have very good intuition about both spaces.&lt;/p&gt;&#xA;" OwnerUserId="1736" LastEditorUserId="1120" LastEditorDisplayName="" LastEditDate="2011-02-26T13:51:59.150" LastActivityDate="2011-02-26T13:51:59.150" Title="Some basics of Sobolev spaces" Tags="&lt;real-analysis&gt;&lt;functional-analysis&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="23818" PostTypeId="2" ParentId="23806" CreationDate="2011-02-26T06:47:31.937" Score="12" ViewCount="" Body="&lt;p&gt;I like this question a lot and I think that it's an important one. So here goes a (necessarily incomplete) attempt at answering such a broad and personal question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, &quot;motivation&quot; and &quot;understanding for the essence&quot; can mean very different things. There is of course physical motivation and intuition, and that probably applies most immediately to the Calculus III course that you are talking about. E.g. for the concept of derivatives of vector valued functions, you can think of the vector valued function of time that gives the position of an object as a vector. Of course, its derivative with respect to time will be the velocity (also a vector, since it described the speed and the direction of the movement) and the second derivative will be the acceleration. A good course in such an applicable subject will not just ask question like &quot;compute the derivative of such and such a function&quot;, but will actually confront the student with real life examples.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But there is also intuition for less physical and more platonic concepts, such as that of a group, or of a prime number. Again, examples help. Also, you should always try to ask yourself the question &quot;Could I have invented this?&quot;. If you see a new definition, ask yourself &quot;What concrete problem might have prompted someone to define such a thing?&quot;. If you see a new result, ask yourself &quot;Why was this to be expected, why would it be at least a reasonable conjecture?&quot;. Then try to convert your intuition into a proof. When you see a proof, ask yourself &quot;Why is this a natural approach to try? Could I have proven this?&quot;. I agree with you that knowing the historical development can be very helpful in this and you should invest time in researching it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to contradict you in your assertion that intuition, motivation and historical context are black magic secrets that mathematicians acquire and then keep to themselves. It is true of some books and some teachers. So, you just have to find the right books. For that, you could ask for a specific recommendation here, including the area you want to learn and the books you have looked at, together with the reason you found them deficient. Of course, you can also ask specific &quot;intuition&quot; type questions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To learn to appreciate mathematics, it is important to think about mathematics in your &quot;spare time&quot;. Go out into nature and think about what your lecturer just told you in the last lecture. Or just think about whatever you find interesting. Then come back home with specific questions and look them up or ask them here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, something that I preach my students all the time is that they should develop a critical approach to what they are taught: if I give them a definition, they should try to come up with as many examples as possible. If a state a theorem of the type &quot;A implies B&quot;, they should go home and find an example that &quot;B does not necessarily imply A&quot;. If they do find such an example, they should ask themselves what additional hypotheses they need to impose to get the converse. If they don't, they should come back to me and ask me &quot;but you haven't told us the whole story. What about the converse?&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In short, don't expect your lecturers to tell you everything you need to know. You should expect to have to think, to investigate yourself, to ask questions, and, above all, to think about mathematics because you can't help it, rather than because you are told to. This is not something, most people are born with, it's something that you have to cultivate.&lt;/p&gt;&#xA;" OwnerUserId="3212" LastEditorUserId="3212" LastEditorDisplayName="" LastEditDate="2011-02-26T12:29:44.390" LastActivityDate="2011-02-26T12:29:44.390" CommunityOwnedDate="2011-02-26T09:23:42.963" CommentCount="2" />
  <row Id="23819" PostTypeId="1" CreationDate="2011-02-26T06:55:36.593" Score="1" ViewCount="170" Body="&lt;p&gt;What is the formula to get sum of series &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$1+(1+x)+(1+x)^2+.... +(1+x)^n,$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $n$ is Integer and $x$ is Rational&lt;/p&gt;&#xA;" OwnerUserId="7520" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2011-02-26T18:16:41.920" LastActivityDate="2011-02-27T15:43:46.117" Title="Formula for sum of series $1+(1+x)+(1+x)^2+.... +(1+x)^n$" Tags="&lt;sequences-and-series&gt;" AnswerCount="3" />
  <row Id="23820" PostTypeId="2" ParentId="23819" CreationDate="2011-02-26T06:58:28.760" Score="3" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\ $ It's a &lt;a href=&quot;http://en.wikipedia.org/wiki/Geometric_series&quot; rel=&quot;nofollow&quot;&gt;geometric series&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2011-02-26T06:58:28.760" />
  <row Id="23821" PostTypeId="2" ParentId="23819" CreationDate="2011-02-26T07:03:50.513" Score="2" ViewCount="" Body="&lt;p&gt;It's a geometric series with common ration $(1+x)$. The sum for a geometric series is given by $$ S = a \frac{ r^{n}-1}{r-1}$$ where $a$ is the first term and $r$ is the common ratio.&lt;/p&gt;&#xA;" LastActivityDate="2011-02-26T07:03:50.513" CommentCount="1" />
  <row Id="23822" PostTypeId="2" ParentId="23814" CreationDate="2011-02-26T07:22:33.487" Score="7" ViewCount="" Body="&lt;p&gt;One way is to apply &lt;a href=&quot;http://en.wikipedia.org/wiki/Euler%E2%80%93Maclaurin_formula&quot; rel=&quot;nofollow&quot;&gt;Euler-McLaurin Summation&lt;/a&gt; (which can be viewed as a clever application of repeated integration by parts) to $\displaystyle \log x$ to show that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$n! \sim C \sqrt{n} \left(\frac{n}{e}\right)^n$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then use &lt;a href=&quot;http://crypto.stanford.edu/pbc/notes/pi/wallis.xhtml&quot; rel=&quot;nofollow&quot;&gt;Wallis formula&lt;/a&gt; (another clever application of integration by parts)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\pi}{2} = \frac{2}{1} \times \frac{2}{3} \times \frac{4}{3} \times \frac{4}{5} \dots $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;to show that $\displaystyle C = \sqrt{2\pi}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I recall correctly, this is how this was derived historically.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, this does not give an intuitive reason as to why the constant is $\displaystyle \sqrt{2\pi}$&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2011-02-26T07:22:33.487" />
  <row Id="23823" PostTypeId="1" AcceptedAnswerId="23824" CreationDate="2011-02-26T07:30:56.020" Score="2" ViewCount="178" Body="&lt;p&gt;Hello,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm really confused about congruence. I tried hard, but I kept failing :(  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$30! \equiv -1 \pmod{31} \text{ by Wilson&amp;#39;s Theorem}$$ &#xA;$$ \Longleftrightarrow 30.29.28.27.26.25! \equiv -1 \pmod{31}$$&#xA;$$ \Longleftrightarrow (-1).15.10.(-8).6.25! \equiv -1 \pmod{31}$$&#xA;$$ \Longleftrightarrow 15.4.5!.25! \equiv -1 \pmod{31}$$&#xA;$$ \Longleftrightarrow 60.5!.25! \equiv -1 \pmod{31}$$&#xA;$$ \Longleftrightarrow 15.5!.25! \equiv -1 \pmod{31}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And I was stuck here :( ? Furthermore, I have to use computer to find a pair of solution of the Diophantine equation $ax + 31y = 1$ for each number: $30, 29, 28, 27, 26 ...  $&#xA;I wonder is there an easier way to do this? Because I think this way is very time consuming. Any idea?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks,&lt;br&gt;&#xA;Chan&lt;/p&gt;&#xA;" OwnerUserId="5711" LastActivityDate="2011-02-26T18:06:44.747" Title="Need help to find remainder when $5!25!$ is divided by $31$?" Tags="&lt;elementary-number-theory&gt;" AnswerCount="3" />
  <row Id="23824" PostTypeId="2" ParentId="23823" CreationDate="2011-02-26T07:40:43.730" Score="6" ViewCount="" Body="&lt;p&gt;So by Wilson's Theorem, you have &#xA;$$&#xD;&#xA;30\cdot 29\cdot 28\cdot 27\cdot 26\cdot 25!\equiv -1\pmod{31}.&#xD;&#xA;$$&#xA;But notice this implies&#xA;$$&#xD;&#xA;(-1)(-2)(-3)(-4)(-5)25!\equiv (-1)^5 5!25!\equiv -1\pmod{31},&#xD;&#xA;$$&#xA;since $30\equiv -1\pmod{31}$, $29\equiv -2\pmod{31}$,$\dots$, and $26\equiv -5\pmod{31}$. &lt;/p&gt;&#xA;" OwnerUserId="1583" LastActivityDate="2011-02-26T07:40:43.730" CommentCount="6" />
  <row Id="23825" PostTypeId="2" ParentId="23808" CreationDate="2011-02-26T07:42:07.827" Score="15" ViewCount="" Body="&lt;p&gt;The problem can be mapped on a &lt;a href=&quot;http://en.wikipedia.org/wiki/Random_walk#One-dimensional_random_walk&quot; rel=&quot;nofollow&quot;&gt;one-dimensional discrete random walk&lt;/a&gt;. Imagine that you start at 0 and you make a step +1 (for the outcome HT) with probability $1/4$, a step -1 (for the outcome TH) with probability $1/4$, and a step 0 (for the outcomes TT and HH) with probability $1/2$. The question about the number of coin tosses after which both have accumulate the same number of heads is then equivalent to asking after how many steps the random walk returns to the origin. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let us introduce the random variable $Z_i$ at step $i$ with value +1 (for HT), -1 (for TH), and 0 in the other cases. As described above, it holds that $P(Z_i = 1) =1/4$, $P(Z_i = -1) =1/4$, and $P(Z_i = 0) =1/2$. Furthermore, introduce $S_n = \sum_{i=1}^n Z_i$. $S_n$ denotes the number of heads which player 1 has obtained minus the number of heads which player 2 has obtained after $n$ steps. The sequence $\{S_n\}$ is called random walk. The question about both players having the same number of heads corresponds to the problem of the &lt;em&gt;first&lt;/em&gt; return to the origin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let us first solve a simpler problem: what is the probability $P_n$ to return after $n$ steps? We should have the same number of $Z$'s having +1 and -1. Each step we have 3 outcomes $Z= \pm 1,0$. In order to return to the origin, we need to have $m$ times $+1$, $m$ times $-1$, and $n-2m$ times $0$ ($0 \leq m \leq n/2$); this event has probability $$ \frac{n!}{m! m! (n-2m)!} P(Z_i = 1)^m P(Z_i = 1)^m P(Z_i = 0)^{n-2m} = \frac{n!}{(m!)^2 (n-2m)! 2^{n+2m}} .$$ The probability to return after $n$ steps is thereby given by&#xA;$$ P_n = \sum_{m=0}^{\lfloor n/2 \rfloor} \frac{n!}{(m!)^2 (n-2m)! 2^{n+2m} }= {2n \choose n}2^{-2n}.$$ What remains to be done is to obtain the probability for the &lt;em&gt;first&lt;/em&gt; return.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a quite general relation between the probability $P_n$ for a return after $n$ steps and the probability $\pi_n$ for the &lt;em&gt;first&lt;/em&gt; return,&#xA;$$ P_n= \pi_1 P_{n-1} + \cdots + \pi_n P_0.$$ The interpretation is simple: to return at step $n$ either you have the first return after one step and then return again after $n-1$ steps, or you return the first time after the second step and then again after $n-2$ steps, ..., or you return the first time after $n$ steps ($P_0$ = 1). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The right hand side is a convolution of two probability distributions. Therefore, it is convenient to introduce the generating functions &#xA; $P(x) = \sum_{n=0}^\infty P_n x^n$ and $\Pi (x) = \sum_{n=0}^\infty \pi_n x^n$ (we set $\pi_0=0$). Multiplying the aforementioned equation by $x^n$ and summing over $n$, we obtain the relation $P(x) = 1 + P(x) \Pi(x)$ (remember $\pi_0 =0$ and $P_0 =1$). We can solve this equation for $$\Pi(x) = \frac{ P(x) -1 }{P(x)}.$$ All what remains to be done is to evaluate $P(x)$. This can be done (binomial theorem) with the result $$ P(x) = \frac{1}{\sqrt{1-x}}$$ and therefore we obtain $$\Pi(x) = 1- \sqrt{1-x}.$$ Expanding this result in a Taylor series, we obtain the probability for the &lt;em&gt;first&lt;/em&gt; return (the probability that the game stops after $n$ rounds)&#xA;$$ \pi_n = \frac{{2n \choose n} }{(2n -1) 2^{2n}} \quad n\geq1.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The expected number of rounds can be also calculated $$&#xD;&#xA;\bar n  = \sum_{n=1}^\infty \pi_n n  \to \infty,$$ i.e., it diverges. The reason is that the asymptotic behavior of $\pi_n$ is given by&#xA;$$\pi_n \sim \frac{1}{2 \sqrt{\pi} n^{3/2}}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question has arisen about the evaluate the sum leading to $P_n$. Borrowing from the solution of Mike there is an alternative way to get $P_n$ (and thereby proving that the sum really is what I wrote). Instead of looking at the game as having $n$ rounds, we can think that the game has $2n$ rounds where we split the tosses of player 1 and player 2. If player 1 has H then we move $+1/2$ for tail $-1/2$. If player 2 has T then we move $+1/2$ for tail $-1/2$. After two rounds this leads to the same random walk as I described above (HH and TT no move, HT move $+1$, TH move $-1$). The probability $P_n$ to return after $n$ rounds (of the original game) is therefore equivalent to return after $2n$ rounds in the new game. There are $2n \choose n$ possible paths which return after $2n$ rounds because we can choose which $n$ steps which move up (H for player 1 and T for player 2) and in the rest of the steps we have to move down. In total there are $2^{2n}$ paths of length $2n$. The probability to return at the origin after $2n$ steps ($n$ steps in the original game) is given by&#xA;$$ P_n = {2n \choose n} 2^{-2n}.$$&lt;/p&gt;&#xA;" OwnerUserId="7266" LastEditorUserId="7266" LastEditorDisplayName="" LastEditDate="2011-02-27T08:54:14.970" LastActivityDate="2011-02-27T08:54:14.970" CommentCount="4" />
  <row Id="23826" PostTypeId="2" ParentId="23823" CreationDate="2011-02-26T07:46:51.913" Score="1" ViewCount="" Body="&lt;p&gt;If you are looking for $x$ such that $15x = 1 \mod 31$, notice that $15 \times 2 = 30 = -1 \mod 31$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence $15 \times (-2) = 1 \mod 31$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, as yunone's answer shows, you have a mistake somewhere.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2011-02-26T07:46:51.913" CommentCount="1" />
  <row Id="23827" PostTypeId="2" ParentId="23814" CreationDate="2011-02-26T09:04:47.697" Score="13" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://terrytao.wordpress.com/2010/01/02/254a-notes-0a-stirlings-formula/&quot; rel=&quot;nofollow&quot;&gt;This blog post&lt;/a&gt; by Terence Tao explains how to get this constant from the central limit theorem (equivalently, from the normalization factor we use to define the normal distribution). The question remains where it came from in the central limit theorem, and I think the most honest answer to that question is Fourier analysis; $\frac{1}{\sqrt{2\pi}}$ is the normalization factor which makes the Fourier transform and its inverse look the same, and the Gaussian distribution is its own Fourier transform. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(That is, I don't think you should look for a straightforward and intuitive explanation because I think this is actually a rather deep fact which is not really explained in most courses. On the other hand, just the $\sqrt{n}$ term is straightforward to explain: instead of using the left Riemann sum or the right Riemann sum, you use the trapezoid rule.)&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2011-02-26T16:16:59.577" LastActivityDate="2011-02-26T16:16:59.577" CommentCount="3" />
  <row Id="23828" PostTypeId="1" AcceptedAnswerId="23835" CreationDate="2011-02-26T09:32:13.047" Score="1" ViewCount="150" Body="&lt;p&gt;Here is the original formula:&#xA;$$\frac{256}{2^x}=y$$&#xA;In order to solve for $x$, I've done this:&#xA;$$\log_{2}\left(\frac{256}{y}\right)=x$$&#xA;The problem is that $y$ can be zero. What should I do to solve for $x$?&lt;/p&gt;&#xA;" OwnerUserId="7490" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-26T20:23:12.390" LastActivityDate="2011-02-26T20:23:12.390" Title="Problems with the number $0$" Tags="&lt;homework&gt;" AnswerCount="4" CommentCount="5" />
  <row Id="23829" PostTypeId="2" ParentId="23828" CreationDate="2011-02-26T09:35:18.193" Score="0" ViewCount="" Body="&lt;p&gt;The original equation does not have a solution for i, so why is this an issue?&lt;/p&gt;&#xA;" OwnerUserId="7155" LastActivityDate="2011-02-26T09:35:18.193" CommentCount="1" />
  <row Id="23830" PostTypeId="2" ParentId="23828" CreationDate="2011-02-26T09:37:10.013" Score="0" ViewCount="" Body="&lt;p&gt;If i is zero, then s has no solution at all. Therefore it should have a limitation.&lt;/p&gt;&#xA;" OwnerUserId="3066" LastActivityDate="2011-02-26T09:37:10.013" />
  <row Id="23832" PostTypeId="1" AcceptedAnswerId="23838" CreationDate="2011-02-26T09:57:16.653" Score="1" ViewCount="190" Body="&lt;p&gt;In Cryptography, I find it commonly mentioned: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Let G be cyclic group of Prime order q and with a generator g.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you please exemplify this with a trivial example please!&#xA;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="7488" LastActivityDate="2011-02-26T13:30:24.330" Title="Example for Cyclic Groups and Selecting a generator" Tags="&lt;elementary-number-theory&gt;&lt;cryptography&gt;" AnswerCount="3" />
  <row Id="23833" PostTypeId="2" ParentId="23832" CreationDate="2011-02-26T10:11:13.013" Score="0" ViewCount="" Body="&lt;p&gt;A common example would be the integers modulo $5$, $\mathbb{Z}_5$. This a cyclic group under addition with a possible generator $1$, and has prime order $5$. &lt;/p&gt;&#xA;" OwnerUserId="1583" LastActivityDate="2011-02-26T10:11:13.013" CommentCount="3" />
  <row Id="23834" PostTypeId="2" ParentId="23832" CreationDate="2011-02-26T10:17:36.097" Score="1" ViewCount="" Body="&lt;p&gt;In fact if you take the group $(\mathbb{Z}_{p},+)$ for a prime number $p$, then every element is a generator. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Take $G= \{a^{q}=e, a ,a^{2}, \cdots, a^{q-1}\}$. Now $|G|=q$ and $G = &amp;lt;a&amp;gt;$, which means that $G$ is generated by $a$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" LastActivityDate="2011-02-26T10:17:36.097" CommentCount="6" />
  <row Id="23835" PostTypeId="2" ParentId="23828" CreationDate="2011-02-26T10:19:32.413" Score="2" ViewCount="" Body="&lt;p&gt;If you rewrite $\dfrac{256}{2^x} = y$ as $256 = y \times 2^x$, then it is clearly impossible to have $y=0$ while remaining within the real numbers, as 256 is not a multiple of 0.  Since $2^x$ is positive for all real $x$, you must have $y&amp;gt;0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So $x = \log_2 \left( \dfrac{256}{y} \right) = 8 - \log_2 \left( y \right)$ is fine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, it &lt;em&gt;is&lt;/em&gt; possible to have $x=0$, in which case $2^x = 1$ and so $y = 256$ and $\log_2 \left( y \right) = 8$. &lt;/p&gt;&#xA;" OwnerUserId="6460" LastActivityDate="2011-02-26T10:19:32.413" CommentCount="2" />
  <row Id="23836" PostTypeId="2" ParentId="23670" CreationDate="2011-02-26T10:33:14.997" Score="0" ViewCount="" Body="&lt;p&gt;As to 2: if $e= H(m)$ and $s = k - x \cdot H(m)$, the $k$ serves no purpose any more.&#xA;Anyone can create $H(m)$ and there is no way to verify that $s$ has been created by the one possessing the private key $x$. In the original scheme, $g^s \cdot y^e$ yields $g^k = r$, so that we can verify that $e = H(m \| r)$ is correct. In variation 2, the $r$ has been eliminated, and cannot be computed by a verifier. This is not even a signature scheme, really, as there is no verifying beyond checking $e = H(m)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To address variation 1, as in the comment: if we use $(r,s)$ where $s = k - x \cdot m$, then the verifying equation becomes $g^s \cdot y^m = r$, given $(r,s)$ and the message $m$ and public key $y$; if it holds it's a valid signature (no need for hashing). But then from a valid signature $(r,s)$ for message $m$ we generate one for $m+1$ by multiplying $r$ by $y$, and keeping $s$. The check then becomes $$g^s \cdot y^{m+1} = g^s \cdot y^m \cdot y = r \cdot y$$ which is valid too. So then also the hashing stays necessary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And if we use $e = H(m)$ and so $(r,s) = (g^k, k - x \cdot H(m))$ as the signature, we can do a similar thing: the verifying equation becomes $g^s \cdot y^{H(m)} = r$ (valid iff true)&#xA;and then $(g \cdot r, s+1)$ is also a valid signature for $m$: verifying gives&#xA;$$g^{s+1} \cdot y^{H(m)} = g \cdot (g^s \cdot y^{H(m)}) = g \cdot r$$ as $(r,s)$ was valid, and so the new one is valid too. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note also that if we re-use the same $k$ (and thus $r = g^k$) for 2 signatures for different messages (in the secure, normal scheme), and the opponent knows or guesses this, then we can compute $x$, the secret key: we then have $(e_1, s_1)$ and $(e_2, s_2)$ where $e_1 = H(m_1 \|  r)$, $e_2 = H(m_2 \| r)$, $s_1 = k - x \cdot e_1$, $s_2 = k - x \cdot e_2$; the opponent computes $$s_1 - s_2 = x \cdot (e_2 - e_1)$$ and $s_1, s_2, e_1, e_2$ are known, allowing us to solve for $x$, in most cases. Of course, for a large enough group, and good random choices, this is very unlikely to happen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you see, it's quite intricate to see why we need certain stuff.&lt;/p&gt;&#xA;" OwnerUserId="4280" LastEditorUserId="4280" LastEditorDisplayName="" LastEditDate="2011-02-26T15:34:13.127" LastActivityDate="2011-02-26T15:34:13.127" CommentCount="3" />
  <row Id="23837" PostTypeId="2" ParentId="23768" CreationDate="2011-02-26T10:44:24.687" Score="4" ViewCount="" Body="&lt;p&gt;Here is a way to see directly that the satisfaction relation $M\models\varphi[a]$ has complexity $\Delta_1$ in set theory. This relation is defined (by Tarski) by induction on the complexity of $\varphi$. Thus, we can say that $M\models\varphi[a]$ if and only if there is a function mapping formulas and points in $M$ to the set {satisfied, not satisfied}, which satisfies the inductive properties of Tarski's definition (so that atomic formulas are correctly satisifed, the satisfaction relation on negation inverts the answer, the relation on Boolean combinations works correctly and the relation on quantifiers works correctly). These requirements on the satisfaction relation S have complexity $\Delta_0$, since we need only quantifiy over $S$, over $M$ and over the natural numbers (codes for formulas). Furthermore, one can prove that there is a unique satisfaction relation, so $M\model\varphi[a]$ if and only if there is a satisfaction relation showing this, if and only if all satisfaction relations show this. So the satisfaction relation is $\Delta_1$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It works the same for any theory, such as ZFC as in your question, since this just adds an additional quantifier over the natural numbers (for every axiom, it is satisfied), which is bounded and therefore doesn't increase the complexity beyond $\Delta_1$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, being a model of any particular assertion or theory is $\Delta_1$ and therefore absolute in the sense you may have had in mind. For example, any model of set theory will agree with all its forcing extensions and inner models about whether a given set structure is a model of ZFC.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Addendum.&lt;/b&gt; But there is another subtle sense in which being a model of ZFC is not absolute. For example, suppose that $M$ is a model of $ZFC+\neg Con(ZFC)$, so that $M$ thinks there are no models of ZFC at all. So this model $M$ must have nonstandard natural numbers, since otherwise it would have the same proofs from ZFC as we do and so it would realize that ZFC is consistent. Inside $M$, we may enumerate the axioms of what $M$ thinks of as ZFC (this includes nonstandard instances of the axioms). In $M$, consider the largest initial segment of the axioms that $M$ thinks is true in some rank initial segment $(V_\alpha)^M$. By the Reflection theorem, this includes all the standard initial segments. Thus, what $M$ thinks is the longest such realizable initial segment of the axioms has nonstandard length. And so the corresponding set $(V_\alpha)^M$ satisfies all the standard instances of axioms of ZFC. That is to say, it really is a model of ZFC, but $M$ doesn't think it is, because $M$ thinks that some of the nonstandard axioms are not true there. So this is an example of non-absoluteness, where two universes of set-theory can disagree about whether a given structure is a model of ZFC or not.&lt;/p&gt;&#xA;" OwnerUserId="413" LastEditorUserId="413" LastEditorDisplayName="" LastEditDate="2011-02-26T15:13:21.560" LastActivityDate="2011-02-26T15:13:21.560" CommentCount="1" />
  <row Id="23838" PostTypeId="2" ParentId="23832" CreationDate="2011-02-26T11:13:18.877" Score="1" ViewCount="" Body="&lt;p&gt;A better example (because the log problem is hard, which is not the case for the additive groups): the multiplicative group of $\mathbb{Z}_{p} = \{ 1,2,\ldots,p-1\}$ under multiplication. Say for $p=11$, we have $G = \{1,2,\ldots,10\}$, and not all elements are generators, e.g. $1$ is not. Let's try 2: $$2^1 = 2, 2^2 = 4, 2^3 = 8, &#xD;&#xA;2^4 = 16 = 5, 2^5 = 10 = -1, $$&#xA;$$ 2^6 = -2 = 9, 2^7 = -4 = 7, 2^8 = -8 = 3, 2^9 = 6, 2^{10} = 12 = 1$$ again. So 2 is a generator, and the discrete log for 3 equals 8, as $2^8 = 3 \mod 11$ etc. For large primes, this gets harder and harder (we cannot make a table like this anymore). In general, you need to work a bit to find a generator, but they always exist. E.g. see &lt;a href=&quot;http://en.wikipedia.org/wiki/Primitive_root_modulo_n&quot; rel=&quot;nofollow&quot;&gt;the wikipedia page on this&lt;/a&gt;, e.g if we choose $p = 2q+1$, where $q$ is also prime (a &quot;safe prime&quot; in cryptography) then $g$ is a multiplicative primitive in $\mathbb{Z}_p$ iff $g^{\frac{p-1}{2}} = -1 \mod p$, and this is easy to test (fast modular exponentiation algorithms exist). As there are $\phi(\phi(n)) = \phi(2q) = q-1$ many primitive elements, a picking a few random numbers and testing them will give a primitive element. &#xA;In practice, a set of fixed primes, and corresponding generators, have already been fixed by a committee, like in the TLS/SSL standard, so that a user need not generate them himself, but can just say &quot;use group x from that standard&quot;; but like I said, it's not that hard for a PC to find primitive elements.&lt;/p&gt;&#xA;" OwnerUserId="4280" LastEditorUserId="4280" LastEditorDisplayName="" LastEditDate="2011-02-26T13:30:24.330" LastActivityDate="2011-02-26T13:30:24.330" CommentCount="3" />
  <row Id="23840" PostTypeId="2" ParentId="23808" CreationDate="2011-02-26T11:57:59.557" Score="11" ViewCount="" Body="&lt;p&gt;The probability of the game ending after $n$ rounds is $\frac{2}{4^n}C_{n-1}$, where $C_n = \frac{1}{n+1} \binom{2n}{n}$, the $n$th &lt;a href=&quot;http://en.wikipedia.org/wiki/Catalan_number&quot; rel=&quot;nofollow&quot;&gt;Catalan number&lt;/a&gt;, and the expected number of rounds is infinite.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;HR&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;To see that the probability is $\frac{2}{4^n}C_{n-1}$&lt;/strong&gt;, we use the interpretation of the $n-1$ Catalan number as the number of &lt;a href=&quot;http://mathworld.wolfram.com/DyckPath.html&quot; rel=&quot;nofollow&quot;&gt;Dyck paths&lt;/a&gt; from $(0,0)$ to $(n,n)$ that lie strictly below and do not touch the diagonal $y=x$ except at $(0,0)$ and $(n,n)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume Player 1 has more heads except at the beginning and at the end.  Let $P1(k)$ denote the number of heads Player 1 has after $k$ rounds and $P2(k)$ denote the number of heads Player 2 has after $k$ rounds.  Let $Z_k = P1(k) - P2(k)$.  Let $(x_k,y_k)$ denote the position on the lattice after $2k$ steps in a Dyck path.  Let $X_k = x_k - k$.    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Associate the outcomes of the coin flips (H = heads, T = tails, with Player 1's outcome first) with two steps in a Dyck path (R = right, U = up) in the following fashion:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;HT $\Leftrightarrow$ RR&lt;/p&gt;&#xA;&#xA;&lt;p&gt;HH $\Leftrightarrow$ RU&lt;/p&gt;&#xA;&#xA;&lt;p&gt;TT $\Leftrightarrow$ UR&lt;/p&gt;&#xA;&#xA;&lt;p&gt;TH $\Leftrightarrow$ UU&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have the following:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;If the flips are HT, then $Z_k$ increments by 1.  With an RR move, $X_k$ increments by 1. &lt;/li&gt;&#xA;&lt;li&gt;If the flips are HH, then $Z_k$ doesn't change. With an RU move, $X_k$ doesn't change. &lt;/li&gt;&#xA;&lt;li&gt;If the flips are TT, then $Z_k$ doesn't change.  With a UR move, $X_k$ doesn't change. &lt;/li&gt;&#xA;&lt;li&gt;If the flips are TH, then $Z_k$ decrements by 1.  With a UU move, $X_k$ decrements by 1.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;$Z_0 = X_0 = 0$, and both the game and the Dyck path end when $Z_n = X_n = 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we see that there is a bijection between the set of ways for the game to end after $n$ rounds with Player 1 ahead except at the beginning and the end and the number of Dyck paths from $(0,0)$ to $(n,n)$ that lie strictly below and do not touch the diagonal $y=x$ except at $(0,0)$ and $(n,n)$.  Thus the number of ways for the game to end after $n$ rounds with Player 1 ahead except at the beginning and the end is $C_{n-1}$.  Double that to account for Player 2 being ahead except at the beginning and the end, and then obtain the probability of ending after $n$ rounds by dividing by $4^n$, the number of possible sequences of flips after $n$ rounds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;HR&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;To see that the expected number of rounds is infinite&lt;/strong&gt;, let $X_n$ denote the absolute value of the difference between the number of heads and the number of tails obtained by the two players by the $n$th round.  Then $X_n$ is a discrete-time Markov chain in which the transition probabilities are $p_{00} = \frac{1}{2}$, $p_{01} = \frac{1}{2}$, and, for $i \geq 1$, $p_{i,i+1} = p_{i, i-1} = \frac{1}{4}$, $p_{ii} = \frac{1}{2}$.  We want $M_0$, the mean recurrence time for state 0.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Such Markov chains are well-studied; see, for example, Example 5 in &lt;a href=&quot;http://www.ifp.illinois.edu/~srikant/ECE534/Spring09/DTMC.pdf&quot; rel=&quot;nofollow&quot;&gt;these notes&lt;/a&gt;.  There the author solves the more general case in which $p_{00} = 1- \lambda$, $p_{01} = \lambda$, and, for $i \geq 1$, $p_{i,i+1} = \lambda(1-\mu), p_{i, i-1} = \mu (1  - \lambda)$, $p_{ii} = \lambda \mu + (1 - \lambda)(1 - \mu)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The solution is that if $\lambda &amp;lt; \mu$, then &#xA;$$M_0 = \frac{1}{1 - \frac{(1 - \mu)\lambda}{(1- \lambda) \mu}},$$&#xA;and if $\lambda \geq \mu$, then $M_0 = \infty$.  (As joriki points out in the comments, the author's expression for $\pi_0$ should actually be the expression for $M_0$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since we have $\lambda = \mu = \frac{1}{2}$, $M_0 = \infty$.&lt;/p&gt;&#xA;" OwnerUserId="2370" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2011-02-27T00:44:07.527" LastActivityDate="2011-02-27T00:44:07.527" CommentCount="7" />
  <row Id="23841" PostTypeId="1" AcceptedAnswerId="23860" CreationDate="2011-02-26T12:25:51.147" Score="2" ViewCount="75" Body="&lt;p&gt;let $a_n$ be a series and +5, -5 partial limits of it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;as well note that $\left|a_{n+1}-a_{n}\right| \leq \frac{1}{n}$ for all $n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;prove that 0 is also a partial limit of $a_n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you.&lt;/p&gt;&#xA;" OwnerUserId="6163" LastEditorUserId="1415" LastEditorDisplayName="" LastEditDate="2011-02-26T13:52:03.033" LastActivityDate="2011-02-26T16:07:19.633" Title="partial limits of a series" Tags="&lt;sequences-and-series&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="23842" PostTypeId="2" ParentId="23817" CreationDate="2011-02-26T12:53:28.990" Score="2" ViewCount="" Body="&lt;p&gt;The difference between $W^{m,p}$ and $W_0^{m,p}$ is not merely a technical one. The idea of the space $W_0^{1,p}(\Omega)$ is that it consists of those functions in $W^{1,p}(\Omega)$ which take the value zero at the boundary of $\Omega$. For general $W_0^{m,p}$ ($m&gt;1$) the derivatives up to order $m-1$ have to be zero at the boundary as well. Since the boundary of $\mathbb{R}^n$ is empty it makes sense that $W_0^{m,p}(\mathbb{R}^n)=W^{m,p}(\mathbb{R}^n)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now to define what it means to restrict a Sobolev function to the boundary is not straightforward, because the boundary has measure zero and $L^p$-functions are in fact equivalence classes of functions, that is, they are only defined up to sets of measure zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The simplest way out of this is to use the definition you gave (closure of compactly supported smooth functions). It has the advantage that it is easy to work with. The drawback is of course that it is not obvious that the definition really captures the intended notion of zero boundary values.&lt;/p&gt;&#xA;" OwnerUserId="1609" LastActivityDate="2011-02-26T12:53:28.990" CommentCount="3" />
  <row Id="23843" PostTypeId="2" ParentId="23819" CreationDate="2011-02-26T12:57:51.557" Score="2" ViewCount="" Body="&lt;p&gt;&lt;em&gt;Simplified after the comment below by Yuval Filmus&lt;/em&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$S=1+(1+x)+(1+x)^{2}+(1+x)^{3}+\ldots +(1+x)^{n}$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;is the sum of a  &lt;a href=&quot;http://en.wikipedia.org/wiki/Geometric_progression&quot; rel=&quot;nofollow&quot;&gt;geometric progression&lt;/a&gt; with ratio $1+x$ and $n+1$ terms, the first of which is $1$. The usual way to derive $S$ is to multiply it by the ratio&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(1+x)S=(1+x)+(1+x)^{2}+(1+x)^{3}+\ldots +(1+x)^{n}+(1+x)^{n+1}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and subtract from $S$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{eqnarray*}&#xD;&#xA;S-(1+x)S &amp;amp;=&amp;amp;1+\left( (1+x)-(1+x)\right) +\left(&#xD;&#xA;(1+x)^{2}-(1+x)^{2}\right)  \\&#xD;&#xA;&amp;amp;&amp;amp;+\ldots +\left( (1+x)^{n}-(1+x)^{n}\right) -(1+x)^{n+1} \\&#xD;&#xA;&amp;amp;=&amp;amp;1-(1+x)^{n+1}.&#xD;&#xA;\end{eqnarray*}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solving for $S$, we get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$S=\frac{(1+x)^{n+1}-1}{x}\qquad (x\neq 0).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $x=0$ the sum of the series is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$S=1+1+1^{2}+1^{3}+\ldots +1^{n}=1+n.$$ &lt;/p&gt;&#xA;" OwnerUserId="752" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2011-02-27T15:43:46.117" LastActivityDate="2011-02-27T15:43:46.117" CommentCount="2" />
  <row Id="23844" PostTypeId="1" AcceptedAnswerId="23848" CreationDate="2011-02-26T12:58:04.680" Score="4" ViewCount="229" Body="&lt;p&gt;Let $\alpha = \frac{1+\sqrt{-19}}{2}$. Let $A = \mathbb Z[\alpha]$. Let's assume that we know that its invertibles are $\{1,-1\}$. During an exercise we proved that:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Lemma:&lt;/strong&gt; If $(D,g)$ is an Euclidean Domain such that its invertibles are $\{1,-1\}$, and $x$ is an element of minimal degree among the elements that are not invertible, then $D/(x)$ is isomorphic to $\mathbb Z/2\mathbb Z$ or $\mathbb Z/3\mathbb Z$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now the exercise asks:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Prove that $A$ is not an Euclidean Domain.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Everything hints to an argument by contradiction: were $(A, d)$ be an ED and $x$ an element of minimal degree among the non invertibles we'd like to show that $A/(x)$ is not isomorphic to $\mathbb Z/2\mathbb Z$ or $\mathbb Z/3\mathbb Z$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strike&gt;How do we do that? My problem is that, since I don't know what this degree function looks like, I don't know how to choose this $x$!&lt;/strike&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that the elements of $A/(x)$ are of the form $a+(x)$, with $a$ of degree less than $x$ or zero. By minimality of $x$ this means that $a\in \{0, 1, -1\}$. Now I'm lost: how do we derive a contradiction from this?&lt;/p&gt;&#xA;" OwnerUserId="4471" LastEditorUserId="6296" LastEditorDisplayName="" LastEditDate="2011-03-03T04:34:19.207" LastActivityDate="2011-03-03T04:34:19.207" Title="A ring that is not an Euclidean domain" Tags="&lt;homework&gt;&lt;abstract-algebra&gt;&lt;ring-theory&gt;" AnswerCount="3" CommentCount="0" />
  <row Id="23845" PostTypeId="2" ParentId="23808" CreationDate="2011-02-26T13:13:43.583" Score="11" ViewCount="" Body="&lt;p&gt;The problem is symmetric with the respect to the two players. Treating the first round separately, we have probability $1/2$ to stop at $n=1$ (both $0$ heads or both $1$ head), and probability $1/2$ to continue with a difference of $1$ in the head counts. It will be convenient to denote by $q_n$ the conditional probability that we end &lt;em&gt;after&lt;/em&gt; round $n$ if the first step left us with a difference of $1$. The overall probability to end in round $n$ will then be given by $1/2$ for $n=1$ and by $q_{n-1}/2$ for $n&amp;gt;1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sign of the difference cannot change since we stop when it becomes zero, so we can assume without loss of generality that it's non-negative. Thus we are interested in the probability distribution $p_n(k)$, where $n$ is the round and $k$ is the difference in head counts. This is a Markov chain with an absorbing state at $k=0$. The evolution of the probabilities is given by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$p_{n+1}(k)=\frac{1}{4}p_n(k-1)+\frac{1}{2}p_n(k)+\frac{1}{4}p_n(k+1)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for $k&amp;gt;1$ and $p_{n+1}(1)=\frac{1}{2}p_n(1) + \frac{1}{4}p_n(2)$ for $k=1$. Let's ignore the complication at the origin for a bit and just treat the recursion relation for general $k$. This is a linear operator being applied to the sequence $p_n (k)$ to obtain the sequence $p_{n+1}(k)$, and the eigenfunctions of this linear operator are the sequences $e^{ik\phi}$ with $-\pi &amp;lt; \phi \le \pi$ (since other values of $\phi$ just reproduce the same sequences). We can obtain the corresponding eigenvalue $\lambda(\phi)$ from&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\lambda(\phi) e^{ik\phi}=\frac{1}{4}e^{i(k-1)\phi}+\frac{1}{2}e^{ik\phi}+\frac{1}{4}e^{i(k+1)\phi}\;,$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\lambda(\phi) =\frac{1}{4}e^{-i\phi}+\frac{1}{2}+\frac{1}{4}e^{i\phi}\;,$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\lambda(\phi)=\frac{1+\cos\phi}{2}=\cos^2\frac{\phi}{2}\;.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have to combine the sequences $e^{ik\phi}$ and $e^{-ik\phi}$ into sines and cosines to get real sequences. Here's where the boundary at $k=0$ comes into play again. The equation $\lambda(\phi) p_n(1)=\frac{1}{2}p_n(1) + \frac{1}{4}p_n(2)$ provides an additional condition that selects a particular linear combination of the sines and cosines. In fact, it selects the sines, since this equation differs only by the term $\frac{1}{4}p_n(0)$ from the general recursion relation, and they can only both be satisfied if this term would have been zero anyway, which is the case for the sines.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since we know the time evolution of the eigenfunctions (they are multiplied by the corresponding eigenvalue in each round), we can now decompose our initial sequence, $p_1(1)=1$ and $p_1(k)=0$ for $k\neq1$, into sines and write its time evolution as the sum of the time evolution of the sines. Thus,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$p_1(k)=\int_0^\pi f(\phi) \sin (k\phi)\mathrm{d}\phi$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for $k \ge 1$, and we obtain our initial sequence by taking $f(\phi)=(2/\pi)\sin\phi$. Then the time evolution is given by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$p_n(k)=\frac{2}{\pi}\int_0^\pi \sin\phi \sin (k\phi)\left(\frac{1+\cos\phi}{2}\right)^{n-1}\mathrm{d}\phi\;.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the probability $q_n$ that we end after round $n$ is just the probability $p_n(1)$ times the probability $1/4$ that we move from $k=1$ to $k=0$, and so&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$q_n=\frac{1}{2\pi}\int_0^\pi \sin^2\phi \left(\frac{1+\cos\phi}{2}\right)^{n-1}\mathrm{d}\phi\;.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.wolframalpha.com/input/?i=%5Cint_0%5E%5Cpi+%281%2F%282%5Cpi%29%29%28sin%28phi%29%29%5E2%28%281%2Bcos%5Cphi%29%2F2%29%5E%28n-1%29&quot; rel=&quot;nofollow&quot;&gt;According to Wolfram&lt;/a&gt;, this is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$q_n=\frac{\Gamma(n+1/2)}{\sqrt{\pi}\,\Gamma(n+2)}\;,$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which I presume could be simplified for integer $n$. We can check that the $q_n$ sum up to $1$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{n=1}^\infty q_n=\frac{1}{2\pi}\int_0^\pi \sin^2\phi \sum_{n=1}^\infty\left(\frac{1+\cos\phi}{2}\right)^{n-1}\mathrm{d}\phi$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$=\frac{1}{2\pi}\int_0^\pi \frac{\sin^2\phi}{1-\frac{1+\cos\phi}{2}}\mathrm{d}\phi$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$=\frac{1}{\pi}\int_0^\pi \frac{\sin^2\phi}{1-\cos\phi}\mathrm{d}\phi$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$=\frac{1}{\pi}\int_0^\pi \frac{\sin^2\phi}{1-\cos^2\phi}\left(1+\cos\phi\right)\mathrm{d}\phi$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$=\frac{1}{\pi}\int_0^\pi 1+\cos\phi\,\mathrm{d}\phi$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$=1\;.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can also try to find the first moment of this probability distribution:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{n=1}^\infty n q_n=\frac{1}{2\pi}\int_0^\pi \sin^2\phi \sum_{n=0}^\infty n\left(\frac{1+\cos\phi}{2}\right)^{n-1}\mathrm{d}\phi$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$=-\frac{1}{\pi}\int_0^\pi \sin\phi \sum_{n=0}^\infty \frac{\mathrm{d}}{\mathrm{d}\phi}\left(\frac{1+\cos\phi}{2}\right)^n\mathrm{d}\phi$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$=\frac{1}{\pi}\int_0^\pi \cos\phi \sum_{n=0}^\infty \left(\frac{1+\cos\phi}{2}\right)^n\mathrm{d}\phi$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$=\frac{2}{\pi}\int_0^\pi \frac{\cos\phi}{1-\cos\phi}\mathrm{d}\phi\;.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This integral diverges at the limit $\phi=0$ (which corresponds to the sequence wandering off to large $k$), and thus as Mike had already pointed out the expected number of rounds is infinite.&lt;/p&gt;&#xA;" OwnerUserId="6622" LastEditorUserId="6622" LastEditorDisplayName="" LastEditDate="2011-02-26T23:25:16.673" LastActivityDate="2011-02-26T23:25:16.673" CommentCount="10" />
  <row Id="23846" PostTypeId="1" CreationDate="2011-02-26T13:13:56.927" Score="0" ViewCount="147" Body="&lt;p&gt;I stuck with this algebra question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I try to prove that the exterior algebra $R$ over $K^d$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;the k-algebra that generated by $x_1,\ldots,x_d$ and $x_ix_j=- x_jx_i$  for each $i,j$, has just one simple module which is not faithful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the only simple module is $k$ but I am not really sure if my idea does work or not&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can I use the fact $(x_i)^2=0$ for all $i$, then $k$ has cyclic subrings.If yes, then HOW?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, one more question if $k$ is finitely generated, is that enough to say that  $R$ is Artinian? &#xA;Thank you&lt;/p&gt;&#xA;" OwnerUserId="7517" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-26T20:10:34.170" LastActivityDate="2011-02-26T20:10:34.170" Title="exterior algebra" Tags="&lt;homework&gt;&lt;ring-theory&gt;" AnswerCount="1" />
  <row Id="23847" PostTypeId="2" ParentId="23806" CreationDate="2011-02-26T13:14:38.597" Score="5" ViewCount="" Body="&lt;p&gt;I can offer a few things that help me gain an intuition for new material, in addition to the more mathematically-related advice given by others:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Talk about the mathematics, either with other students, or with your professor during office hours.  I've noticed that while lots of times lectures can be dry and fast-paced, visiting after hours and asking about the concepts underlying the material can be very rewarding.  Talking with other students in the class can also expose the meaning as long as you find a group of other students who are also interested in learning the &quot;why&quot; behind the &quot;how.&quot;  Ask each other questions (&quot;what do you get out of this theorem?&quot;) and challenge the material.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Find books in the subject you are learning that are a bridge between the typical undergraduate text book and the more advanced (graduate level) books and read them as a supplement to the course.  For instance, Tom Apostol's Calculus (Vol. 1 &amp;amp; 2) is a great companion book for the calculus series, and will answer many of the mathematical &quot;why&quot; questions that the typical undergrad books leave out (plus, you'll be a leg up when you get to advanced calculus!).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lastly, read lots of mathematics books that aren't textbooks.  I started learning about mathematics through books like Euclid's Window, Fermat's Enigma, and Prime Obsession.  Journey Through Genius was also a good one, as it runs through the motivation behind several important results in mathematics.  Since then I have switched to reading biographies and autobiographies to get to know the mathematicians who have come before me, how they thought, how they lived, and how their lives are all intertwined (some good ones: The Man who Love Only Numbers (Paul Erdos), The Man who Knew Infinity (Ramanujan), I Want to be a Mathematician (Paul Halmos)).  In my experience, learning this side of mathematics has been invaluable for my understanding of the discipline in general.&lt;/p&gt;&#xA;" OwnerUserId="7051" LastActivityDate="2011-02-26T13:14:38.597" CommunityOwnedDate="2011-02-26T13:14:38.597" />
  <row Id="23848" PostTypeId="2" ParentId="23844" CreationDate="2011-02-26T13:40:27.527" Score="3" ViewCount="" Body="&lt;p&gt;What a coincidence, this was a recent homework problem for me as well. Here's an additional hint: Show that $X^2 + X + 5$ does not split over $\mathbb{F}_2$ or $\mathbb{F}_3$. Deduce a contradiction to the minimality of the degree of $x$.&lt;/p&gt;&#xA;" OwnerUserId="5191" LastActivityDate="2011-02-26T13:40:27.527" CommentCount="2" />
  <row Id="23849" PostTypeId="2" ParentId="23844" CreationDate="2011-02-26T13:48:03.540" Score="0" ViewCount="" Body="&lt;p&gt;You don't need to know what the degree function looks like to choose $x$: you already chose it when you said &quot;$x$ an element of minimal degree.&quot; (By the well-ordering principle, the set of degrees of nonzero, non-unital elements has a least element.)&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2011-02-26T13:48:03.540" CommentCount="2" />
  <row Id="23850" PostTypeId="1" AcceptedAnswerId="24059" CreationDate="2011-02-26T14:10:15.820" Score="5" ViewCount="486" Body="&lt;p&gt;Are there any known barriers to show the following invariant (perhaps by some sort of induction)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $\Sigma$ be some finite alphabet with $|\Sigma| \geq 2$, let $M$ be some (deciding) deterministic Turing machine with input alphabet $\Sigma$, and let $L_0 \subseteq \Sigma^{\star}$ be some non-sparse, $\mbox{NP}$-complete language.&lt;br&gt;&#xA;Then at least one of the following properties hold:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$M$ doesn't terminate always.&lt;/li&gt;&#xA;&lt;li&gt;$M$ has superpolynomial time complexity.&lt;/li&gt;&#xA;&lt;li&gt;$L(M) \triangle L_0$ is non-sparse.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Concise problem description: $\mbox{NP} \not\subseteq \mbox{P-close}$ (according to Tsuyoshi Ito, see his answer).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Caution: This problem is equivalent to $\mbox{P} \neq \mbox{NP}$.&lt;/p&gt;&#xA;" OwnerUserId="7280" LastEditorUserId="7280" LastEditorDisplayName="" LastEditDate="2011-03-04T09:48:03.063" LastActivityDate="2011-03-04T09:48:03.063" CommunityOwnedDate="2011-02-28T21:40:12.917" Title="Are there any known barriers to some approach for solving P vs. NP?" Tags="&lt;computer-science&gt;" AnswerCount="1" CommentCount="10" />
  <row Id="23851" PostTypeId="1" AcceptedAnswerId="23857" CreationDate="2011-02-26T14:36:09.550" Score="1" ViewCount="117" Body="&lt;p&gt;Hello all,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've started learning about measure theory and I'm trying to get some intuitive grasp of the basic concepts. This is only succeeding partially so far.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is an exercise which I don't quite understand. Here it is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Let $\Omega$ be the set of all sequences $\omega = (\omega_1,\omega_2,\ldots)$  where $\omega_n \in \{0,1\}$  $\forall n \geq 1$. Define for all $n$ the projections $p_n:\Omega \rightarrow \{0,1\}$ and let $\mathcal{F}_n = \sigma(p_1,\ldots,p_n)$. Prove that $\mathcal{F}_1 \subseteq \mathcal{F}_2 \subseteq \ldots$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The course material that I'm using defines the $\sigma$-algebra of a function in the context of borel sets $B \in \mathcal{B}(\mathbb{R})$, as in $\sigma(f) = \{\{f \in B\} : B \in \mathcal{B}(\mathbb{R})\}$ where $\{f \in B \} = \{\omega \in \Omega : f(\omega) \in B\}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have two questions about this exercise:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Is a $\sigma$-algebra generated by some function $f$ always defined in the context of Borel-algebras? Ie, in the case of our functions $p_i$, should we think of $p_i$ as a function mapping some sequence $\omega$ to $\{0,1\} \subseteq (a,b)$ for some $a,b \in \mathbb{R}$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) How should I read $\sigma(p_1)$? Because I'm having trouble connecting the nature of $p_i$ with the aformentioned definition of $\sigma(f)$.&lt;/p&gt;&#xA;" OwnerUserId="7155" LastActivityDate="2011-02-26T16:03:04.727" Title="Question on the notion of a $\sigma$-algebra generated by a function" Tags="&lt;homework&gt;&lt;sigma-algebra&gt;" AnswerCount="3" />
  <row Id="23852" PostTypeId="1" CreationDate="2011-02-26T14:51:16.940" Score="-3" ViewCount="105" Body="&lt;p&gt;So is a magnetic fractal something that describe something indirect in nature, or actual fractal growth readjusting toward something with an affinity for it. For example like a large magnet changing the shape of a heavy chain of paper clips (fractal)?&lt;/p&gt;&#xA;" OwnerUserId="7526" LastActivityDate="2011-02-26T16:44:06.027" Title="3D Magnetic Fractal approximating morphs" Tags="&lt;differential-geometry&gt;&lt;graph-theory&gt;&lt;fractals&gt;&lt;mathematical-modeling&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="23853" PostTypeId="2" ParentId="23846" CreationDate="2011-02-26T15:07:07.457" Score="1" ViewCount="" Body="&lt;p&gt;&lt;em&gt;(I wll be assuming $k$ is a field; if it is not, then you will need some hypothesis on it for your statement to be true)&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose $R$ is a ring and that $S$ is a non-zero simple left $R$-module. Pick a non-zero element $s_0\in S$; then the map $\phi:r\in R\mapsto sr_0\in S$ is a surjective map of left $R$-modules. Let $I\subseteq R$ be its kernel, a left ideal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now suppose $t\in R$ is an homogeneous element of positive degree. It is easy to see that the subset $tS=\{ts:s\in S\}$ is a submodule of $S$. Since $S$ is simple, then either $tS=0$, and in that case $t\in T$, or $tS=S$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to check that the second cannot occur. Suppose otherwise: then there exists an $s\in S$ such that $ts=s_0$ and, since $s_0$ generates $S$, there also exists an $r\in R$ such that $s=rs_0$; it follows that $s_0=rts_0$ or, in other words, that $(1-rt)s_0=0$. Now it turns out that the element $1-rt$ is invertible in $R$ (since $t$ has positive degree, $rt$ is a sum of homogeneous elements of positive degree, and then $(rt)^d=0$: it follows that $\sum_{n\geq0}(rt)^n$ is a finite sum and the usual argument shows that it is the inverse of $1-rt$), so we see that $s_0=0$, which is absurd.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We thus conclude that ever homogeneous element of positive degree is $R$ is in fact contained in $I$. Now we observe that there is exactly &lt;em&gt;one&lt;/em&gt; left ideal in $R$ which contains all the homogeneous elements of positive degree, so there is in fact, as we wanted, exactly one isomorphism class of simple modules.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;NB:&lt;/strong&gt; if you know a little more about ring theory---specifically, of Artinian algebras---we can recast this argument as follows: let $R$ be your exterior algebra on $d$ generators, and let $I$ be the ideal generated by all homogeneous elements of positive degree. It is pretty obvious that $I^d=0$, so that $I$ is a nilpotent ideal; on the other hand, it is also clear that $R/I$ is isomorphic to $k$ as a ring, so that it is in particular semisimple. It follows from a well-known characterization of the Jacobson radical that  $I$ &lt;em&gt;is&lt;/em&gt; the Jacobson radical of $R$. As a consequence, there is a bijection between the isoclasses of simple $R$-modules and the isoclasses of simple $R/I$-modules, and there is oly one of the latter.&lt;/p&gt;&#xA;" OwnerUserId="274" LastEditorUserId="274" LastEditorDisplayName="" LastEditDate="2011-02-26T16:47:19.403" LastActivityDate="2011-02-26T16:47:19.403" CommentCount="1" />
  <row Id="23854" PostTypeId="2" ParentId="23851" CreationDate="2011-02-26T15:28:44.767" Score="0" ViewCount="" Body="&lt;p&gt;1) By default, yes to the first question. As for the second one, note that $\{0,1\}\subset\mathbb R$, hence $p_i$ is a mapping from $\Omega$ to $\mathbb R$. No need to introduce $(a,b)$ here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) If you write out $p_i:\Omega\ni \omega  = \{\omega_n\}_{n\in\mathbb N} \mapsto \omega_i\in\{0,1\}\subset\mathbb R$, then $\sigma(p_i)$ should be clearly defined as before. Note that here, $\Omega$ is usually equipped with the product $\sigma$-algebra. For example, $p_1^{-1}(\{0\}) = \{0\}\times\{0,1\}^{\mathbb N}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Actually, you don't need to write $\sigma(p_i)$ out to solve the exercise, since by definition, $\sigma(f)\subset\sigma(f,g)$ for arbitrary measurable functions $f,g$.&lt;/p&gt;&#xA;" OwnerUserId="5836" LastActivityDate="2011-02-26T15:28:44.767" CommentCount="1" />
  <row Id="23855" PostTypeId="2" ParentId="23841" CreationDate="2011-02-26T15:29:00.273" Score="5" ViewCount="" Body="&lt;p&gt;The idea is this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine a child draws the real line on the ground and then starts hopping around on it in such a way that again and again, they land very close to &quot;5&quot; and to &quot;-5&quot;. They can do this without ever landing close to &quot;0&quot;: for example, they can just hop back and forth between &quot;5&quot; and &quot;-5&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But now imagine that as time goes on, the child gets tired and their hops become smaller and smaller (actually: arbitrarily small). Then if they still go back and forth between &quot;5&quot; and &quot;-5&quot;, then on the way they'll eventually have to land on or very close to &quot;0&quot;: once their hops get smaller than $\epsilon$, if you divide the interval from &quot;5&quot; to &quot;-5&quot; into boxes of size $\epsilon$, then on their way from the one to the other, they will have to land in every box, including the box containing &quot;0&quot;, since their hops aren't large enough to jump over an entire box in one hop. And if you wait long enough, you can do this for arbitrarily small $\epsilon$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you turn that little story into a formal proof? :-)&lt;/p&gt;&#xA;" OwnerUserId="6578" LastActivityDate="2011-02-26T15:29:00.273" CommentCount="1" />
  <row Id="23856" PostTypeId="1" AcceptedAnswerId="23871" CreationDate="2011-02-26T15:35:05.970" Score="1" ViewCount="141" Body="&lt;p&gt;Hi,&#xA; I am trying to understand how definite integral works when the partitions of the function are of unequal length. I found this &lt;a href=&quot;http://usma387.wikidot.com/project%3aii%3agroup-1&quot; rel=&quot;nofollow&quot;&gt;link&lt;/a&gt; and am stuck here:&#xA;$$I = \int_a^b f(x)dx =  \lim_{mesh(P) \to 0} R(f,P,T)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand that $\int_a^b f(x)dx = \text{Area under the curve}$ but not how that is equal to $\lim\limits_{mesh(P) \to 0} R(f,P,T)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The way I see this, in the case of equal partitions, the mesh is $\Delta x_{i} = \frac{1}{n}(b-a)$ so a 0 mesh means that there will be infinitely many thin partitions so that the sum of the area of each partition equals the area under the curve.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of unequal partitions, since the mesh is $\Delta x_{i} = x_i - x_{i-1}$, if each mesh goes to 0, it appears that the number of partitions $n$ remains fixed so that the sum of the area of each partition definitely won't cover the area under the curve. Am I visualizing this correctly? If the number of partitions increase, how is $n$ related to $\Delta x_{i}$?&lt;/p&gt;&#xA;" OwnerUserId="2973" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-26T20:11:18.507" LastActivityDate="2011-02-26T20:11:18.507" Title="Definite Integral with unequal partitions" Tags="&lt;calculus&gt;&lt;integral&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="23857" PostTypeId="2" ParentId="23851" CreationDate="2011-02-26T15:57:38.370" Score="1" ViewCount="" Body="&lt;p&gt;If $f: X \to (Y,\Sigma)$ is a function from a set to any measurable space (a space equipped with a $\sigma$-algebra) then it is easy to check that $\{f^{-1}(S)\,:\,S \in S\} = \sigma(f)$ is a $\sigma$-algebra (sorry, I can't bring myself to writing $\{f \in S\}$ for the pre-image $f^{-1}(S)$ of $S$ even if there's some justification). This is because taking pre-images commutes with unions, intersections and complements. By definition of measurability, $\sigma(f)$ is the smallest $\sigma$-algebra $\mathcal{A}$ on $X$ making $f$ measurable, i.e. $\sigma(f)$ is the smallest $\sigma$-algebra $\mathcal{A}$ such that $f^{-1}(S) \in \mathcal{A}$ for all $S \in \Sigma$. So the answer to your first question is no.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now if $f,g: X \to (Y,\Sigma)$ are two functions, then $\sigma(f,g)$ is the smallest $\sigma$-algebra making both $f$ and $g$ measurable. It doesn't have such an easy description in general, but it is clear that $\sigma(f), \sigma(g) \subset \sigma(f,g)$ and actually $\sigma(f,g)$ is the intersection of all $\sigma$-algebras containing both $\sigma(f)$ and $\sigma(g)$. In fact, if $\mathcal{S}$ is any collection of subsets of $X$ then it generates a $\sigma$-algebra $\sigma(\mathcal{S}) = \bigcap_{\mathcal{S} \subset \mathcal{A}} \mathcal{A}$, where the intersection is taken over all $\sigma$-algebras containing $\mathcal{S}$. Since the power set of $X$ contains $\mathcal{S}$ and is a $\sigma$-algebra, this intersection is non-empty. You should convince yourself that an arbitrary intersection of $\sigma$-algebras is again a $\sigma$-algebra.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your example, you have $X = Y^{\mathbb{N}}$ and in this situation the $\sigma$-algebra $\sigma(p_{1},\ldots,p_{n})$ has a semi-concrete description. Note that a pre-image of $p_{i}$ is of the form $\underbrace{Y \times \cdots \times Y}_{(i-1)\;\text{times}} \times S_{i} \times Y \times Y \times \cdots$, where $S_{i} \subset Y$ is arbitrary. By taking the intersection of the sets $p_{i}^{-1}(S_{i})$ this means that all the sets of the form $S_{1} \times \cdots \times S_{n} \times Y \times Y \times \cdots$ with $S_{1},\ldots,S_{n} \in \Sigma$ must belong to $\sigma(p_{1},\cdots,p_{n})$ and in fact, these so-called &lt;em&gt;cylinder sets&lt;/em&gt; generate the $\sigma$-algebra $\sigma(p_{1},\cdots,p_{n})$.&lt;/p&gt;&#xA;" OwnerUserId="5363" LastEditorUserId="5363" LastEditorDisplayName="" LastEditDate="2011-02-26T16:03:04.727" LastActivityDate="2011-02-26T16:03:04.727" />
  <row Id="23858" PostTypeId="2" ParentId="23851" CreationDate="2011-02-26T15:58:38.963" Score="1" ViewCount="" Body="&lt;p&gt;The answer to (1) is no (sorry, Morning... :-)) since the only structure one needs is a $\sigma$-algebra on the image set. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;More in details, let $\Omega$ denote any set and $(E,\mathcal{E})$ any measurable space (this means only that $E$ is a set and that $\mathcal{E}$ is a $\sigma$-algebra on $E$). For any function $f:\Omega\to E$, the $\sigma$-algebra generated by $f$ is defined as $\sigma(f)=\{\{f\in A\},A\in\mathcal{E}\}$, where $\{f\in A\}=f^{-1}(A)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, if $\Omega$ is itself a measurable space, that is, if one is given a $\sigma$-algebra $\mathcal{F}$ on $\Omega$, then $f$ is a random variable from $(\Omega,\mathcal{F})$ to $(E,\mathcal{E})$ iff  $\sigma(f)\subset\mathcal{F}$. This condition means exactly (surprise, surprise...) that $f^{-1}(A)\in\mathcal{F}$ for every $A\in\mathcal{E}$.&lt;/p&gt;&#xA;" OwnerUserId="6179" LastActivityDate="2011-02-26T15:58:38.963" CommentCount="3" />
  <row Id="23859" PostTypeId="1" AcceptedAnswerId="23873" CreationDate="2011-02-26T16:05:08.807" Score="1" ViewCount="58" Body="&lt;p&gt;This is an attempt to refine my previous &lt;a href=&quot;http://math.stackexchange.com/questions/23759/on-the-existence-of-closed-form-solutions-to-finite-combinatorial-problems&quot;&gt;question&lt;/a&gt; into something precise enough to admit a resolution. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&#xA;Let $P\;$ be a graph property, let $T_n\left(P\;\right)$ be the set of isomorphism classes of graphs on $n$ vertices which satisfy $P\;$, and let $f\left(n\right) = \left|T_n\left(P\; \right)\right|$, the number of elements in $T_n\left(P\;\right)$. Is it possible for $f\left(n\right)$ to satisfy a simple recurrence $f\left(n\right)=\mathcal{R}\left(f\left(1\right),f\left(2\right),\ldots,f\left(n-1\right)\right)$, yet for it be an intractable problem to determine this recurrence given a formulation of $P\;$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This problem as stated is trivial. Let $X$ be some collection of graphs for which $g\left(n\right)$, the number of graphs in $X$ with $n$ vertices, satisfies some simple recurrence. Then let $P\;$ be given as $x\in X$. In this case $P\;$ tells us nothing of how to find this recurrence. Is there some property in logic or set theory that identifies such examples? It seems that there should be some way to consider a large class of formulations of graph properties for which this question is not trivial, but at the moment I am too ignorant to formalize this notion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: (In response to Qiaochu's comment)&#xA;&#xA;I suspect that it doesn't matter too much what class of recurrences we consider. For now lets just consider linear recurrences with integer coefficients, $f\left(n\right)= \sum_{i=n-k}^{n-1}\alpha_{n-i} f\left(i\right)$, where $k$ is some constant. Then the problem I am interested in is whether or not there exists an algorithm which given a formula $\mathcal{F}$ describing a graph property $P\;$, outputs a proof that $f\left(n\right)=\left|T_n\left(P\right)\right|$ satisfies a linear recurrence with integer coefficients and produces the recurrence in all cases&#xA;in which $f\left(n\right)$ satisfies such a recurrence, or otherwise runs forever. Of course, we must consider some restricted class of such formulas $\mathcal{F}$, but as I stated above I am unsure at the moment how to define such a class. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: I have come to believe that much of the above is irrelevant, and that my line of reasoning has been confused. This is because everything seems to hang upon the class of graph properties (actually formula describing graph properties) we consider, and I have left this completely open. It now seems apparent that we should consider finitely encoded formula describing graph properties which:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Provide all of the information which is needed to check if a given graph possesses the property.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Describe something which is intrinsic to graphs.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I believe that the latter condition can be formalized in some way, but I am not sure of this. If a formula describing a graph property does not satisfy the first condition, then I do not see how it could be possible to enumerate the graphs which have $n$ vertices and possess the property, given only the formula. Allowing graph properties which do not satisfy the second condition allows us to encode a wide class of difficult mathematical problems into our framework. For example, we could let our property be described by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;$X$ is a graph for which the number of edges equals the $kth$ digit in the decimal expansion for the Euler-Mascheroni constant $\lambda$, where $k$ is the number of vertices of $X$.&lt;/em&gt; &#xA;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $\lambda$ is rational, then one could say that $f\left(n\right)$, which for each $n\in \mathbb{N}$, returns the number of such graphs with $n$ vertices, has a simple closed form solution. It is however not known whether or not $\lambda$ is rational, and this could conceivably be an undecidable proposition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I cannot be sure, but I suspect that if we limit the graph properties  under consideration in this way, then this question is no longer terribly interesting.&#xA;Perhaps someone may have something interesting to say on this topic, but I don't think I'm going to think about this any longer.&#xA;&lt;/p&gt;&#xA;" OwnerUserId="7323" LastEditorUserId="7323" LastEditorDisplayName="" LastEditDate="2011-02-27T21:44:17.553" LastActivityDate="2011-02-27T21:44:17.553" Title="Graph enumeration problems which admit recursive solutions." Tags="&lt;graph-theory&gt;&lt;computability&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="23860" PostTypeId="2" ParentId="23841" CreationDate="2011-02-26T16:07:19.633" Score="4" ViewCount="" Body="&lt;p&gt;It might be easiest to do a proof by contradiction.. suppose $0$ is not a partial limit. Then there is an $\epsilon &amp;gt; 0$ such that $|a_n| &amp;gt; \epsilon$ for all $n$. So if $n &amp;gt; {1 \over \epsilon}$, $a_{n+1}$ must have the same sign as $a_n$ since $|a_{n+1} - a_n| \leq {1 \over n} &amp;lt; \epsilon$. As a result, all $a_n$ for $n &amp;gt; {1 \over \epsilon}$ have the same sign, preventing both a positive and negative number from being a partial limit.&lt;/p&gt;&#xA;" OwnerUserId="3035" LastActivityDate="2011-02-26T16:07:19.633" CommentCount="2" />
  <row Id="23861" PostTypeId="1" AcceptedAnswerId="23864" CreationDate="2011-02-26T16:32:03.910" Score="4" ViewCount="96" Body="&lt;p&gt;Theorem:&#xA;If $x,y \in \mathbb{R}$ and $x &amp;gt; 0$, $\exists$ a positive integer $n$ such that $nx &amp;gt; y$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I read the proof by Rudin and understood it. I think it is very elegant and uses the LUB property of $\mathbb{R}$. But the way I tried to prove it is the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $x &amp;gt; y$, then $n = 1$. If $x &amp;lt; y$, then $y &amp;gt; 0$ and it makes sense to define $m = \frac{y}{x}$. $m \in \mathbb{R}$ using one of the field axioms. Since the set of natural numbers is unbounded $\exists n \in \mathbb{N}$ such that $n &amp;gt; m$. This implies $nx &amp;gt; y$. This proves the Archimedean property.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there something wrong with the proof or am I assuming something that is not obvious? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is something I frequently wonder while I am trying to learn analysis. I really like the way Rudin proves the theorems in his book. But I find it really difficult to adapt that kind of thinking and rigor. I think a big reason why I find this difficult is because I have an engineering background and I am never used to being so rigorous with proofs. Can anyone suggest some tips (or your experience) that deal with improving the ability to write rigorous proofs (and the ability to find logical gaps in proofs).&lt;/p&gt;&#xA;" OwnerUserId="3050" LastActivityDate="2011-02-26T16:40:52.327" Title="Archimedean property of $\mathbb{R}$" Tags="&lt;analysis&gt;&lt;soft-question&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" />
  <row Id="23862" PostTypeId="1" CreationDate="2011-02-26T16:32:52.630" Score="1" ViewCount="97" Body="&lt;p&gt;The diagram shows the graph of $y=|mx+n|$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/1m9In.png&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(i tried my best to do the same thing as my exercise book, actually 1 is propotional to 1 and 3 is propotional to 3, but 2 is not propotional to 2)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Find the value of $m$ and $n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyone can help me solve this problem? Thanks&lt;/p&gt;&#xA;" OwnerUserId="7527" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2011-02-26T18:59:45.537" LastActivityDate="2011-02-26T19:06:39.503" Title="Understanding $y=|mx+n|$" Tags="&lt;functions&gt;&lt;absolute-value&gt;" AnswerCount="4" CommentCount="7" />
  <row Id="23864" PostTypeId="2" ParentId="23861" CreationDate="2011-02-26T16:35:40.867" Score="8" ViewCount="" Body="&lt;p&gt;But you don't know that the set of natural numbers is unbounded in the reals. In fact that statement is &lt;em&gt;equivalent&lt;/em&gt; to the statement that $\mathbb{R}$ has the Archimedean property.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I presume Rudin shows it in pretty much the same way as suggested in Exercise 3 in another &lt;a href=&quot;http://math.stackexchange.com/questions/11923/completion-of-rational-numbers-via-cauchy-sequences/12009#12009&quot;&gt;answer of mine&lt;/a&gt; (man, I've gotten a lot of mileage out of that!)&lt;/p&gt;&#xA;" OwnerUserId="1269" LastEditorUserId="1269" LastEditorDisplayName="" LastEditDate="2011-02-26T16:40:52.327" LastActivityDate="2011-02-26T16:40:52.327" CommentCount="2" />
  <row Id="23865" PostTypeId="2" ParentId="21693" CreationDate="2011-02-26T16:38:21.657" Score="0" ViewCount="" Body="&lt;p&gt;I got help elsewhere about this in the end. Here is the answer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Decidable sets allow one to determine if a given element is in&#xA;the set.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Enumerable sets have an effective procedure that will produce all&#xA;elements in the set.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some sets are both decidable and enumerable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are various relationships between decidability and enumerability&#xA;such as that every decidable set is theoretically if not practically&#xA;enumerable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are no general non brute force ways of converting between&#xA;decidable and enumerable definitions of sets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are however lots of specific ways, for example there is a method for conic equations which could be used between &quot;x^2 + y^2 = 1&quot; and &quot;x = sin(n), y = cos(n)&quot;.&lt;/p&gt;&#xA;" OwnerUserId="6982" LastActivityDate="2011-02-26T16:38:21.657" />
  <row Id="23866" PostTypeId="1" AcceptedAnswerId="23869" CreationDate="2011-02-26T16:43:01.667" Score="1" ViewCount="72" Body="&lt;p&gt;How do I show that on a projective hypersurface of homogenous equation F = 0, the singular&#xA;points arise as common zeros of the partial derivatives dx_i(F)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have an assignment to calculate the singular points of the Steiner surface and it is not so hard using the above theorem. But I would like to understand why this fact holds in general.&lt;/p&gt;&#xA;" OwnerUserId="7528" LastActivityDate="2011-02-26T17:05:48.367" Title="Singular points and projective hypersurfaces." Tags="&lt;algebraic-geometry&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="23867" PostTypeId="2" ParentId="23852" CreationDate="2011-02-26T16:44:06.027" Score="0" ViewCount="" Body="&lt;p&gt;The theme of self-similarity across scale seems to be abundant in nature. The question, as &#xA;I read it, asks whether this is an organizing principle in and of itself or if it is simply a consequence of physical law. A parallel way of viewing this would be does physical law emerge from a self-similar organizing principle or vice-versa. And I think that when asked like this answer is that the fractal constructions we see in nature are the result of the step by step applications of the physical laws that assemble such constructions. So while the fractal nature of these objects may indicate that such assemblage occurs deterministically(according to a rule), in increments of time, and with units building upon themselves- it does not necessarily indicate something deeper beneath the law itself (much how the fractal generated by given algorithm doesn't say much about number theory and logic- only the way it which it is implemented)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hopefully this is along the lines of what you're looking for.&#xA;Also this is definitely not the last word on this as developing theories may incorporate a fractal definition of space and time:  &lt;a href=&quot;http://www.physorg.com/news157203574.html&quot; rel=&quot;nofollow&quot;&gt;http://www.physorg.com/news157203574.html&lt;/a&gt;   &lt;/p&gt;&#xA;" OwnerUserId="6160" LastActivityDate="2011-02-26T16:44:06.027" CommentCount="5" />
  <row Id="23868" PostTypeId="2" ParentId="23862" CreationDate="2011-02-26T17:00:54.517" Score="4" ViewCount="" Body="&lt;p&gt;Hints: What value inside the absolute value sign can make $y=0$?  If you set $x=0$, what is $y$?&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-02-26T17:00:54.517" />
  <row Id="23869" PostTypeId="2" ParentId="23866" CreationDate="2011-02-26T17:05:48.367" Score="2" ViewCount="" Body="&lt;p&gt;As is often the case for such basic questions, it depends on your precise definition for &quot;singular point&quot; on an algebraic variety.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One standard definition is via the &lt;strong&gt;Jacobian condition&lt;/strong&gt;: in an affine coordinate chart, the matrix of partial derivatives should have the largest possible rank: here, one.  Your condition is simply a projectivized version of this basic condition: it is equivalent to their being an affine coordinate chart $f(x,y) = 0$ for which at least one of the two partial derivatives is non-vanishing.  It may seem that it is easier for a projective curve $F(x,y,z) = 0$ to be smooth than an affine curve $f(x,y) = 0$ since in the projective case it is enough for any one of three partial derivatives to be nonzero, whereas in the affine case there are only two.  However this is illusory, because of course we only care about the vanishing of the partial derivatives &lt;em&gt;at points lying on the curve&lt;/em&gt;, and it turns out that in the projective case $\frac{\partial F}{\partial x}(P) = \frac{\partial F}{\partial y}(P) = \frac{\partial F}{\partial z}(P) = 0$ implies $F(P) = 0$, whereas this does not hold in the affine case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other common definition of nonsingularity of a variety at a point $P$ is that the &lt;strong&gt;local ring&lt;/strong&gt; at that point is a regular local ring.  The equivalence of this (more intrinsic) definition with the Jacobian condition is shown, for instance, in $\S I.5$ of Hartshorne's &lt;em&gt;Algebraic Geometry&lt;/em&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(As an &lt;em&gt;arithmetic&lt;/em&gt; geometer, I should point out that Hartshorne is rather unnecessarily working over an algebraically closed base field throughout his book -- despite spending a lot of time talking about &quot;general schemes&quot;.  This equivalence continues to hold over any &lt;strong&gt;perfect&lt;/strong&gt; ground field $k$.  For imperfect ground fields, smoothness is the stronger and more geometric condition: $X$ is smooth over $k$ iff the base extension of $X$ to an algebraic closure $\overline{k}$ is regular.  But if $\overline{k}/k$ is not separable, funny things can happen...)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Much later on, in $\S III.10$, Hartshorne gives yet another equivalent condition for smoothness in terms of relative differentials.  All three turn out to be important, but in my experience at least the Jacobian condition is by far the easiest to check computationally.  &lt;/p&gt;&#xA;" OwnerUserId="299" LastActivityDate="2011-02-26T17:05:48.367" />
  <row Id="23870" PostTypeId="2" ParentId="23862" CreationDate="2011-02-26T17:10:26.930" Score="2" ViewCount="" Body="&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/ilmmK.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;Hint.&lt;/strong&gt; This is the graph of $y=|x|$. The figure which you have has shifted the bend by $\frac{3}{2}$ units on the right hand side. So i think your graph should $y= | x - \frac{3}{2}|$. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" LastActivityDate="2011-02-26T17:16:49.610" CommentCount="2" />
  <row Id="23871" PostTypeId="2" ParentId="23856" CreationDate="2011-02-26T17:25:47.997" Score="1" ViewCount="" Body="&lt;p&gt;The &quot;convergence&quot; of partitions, associated with defining a Riemann integral through a limiting process, is a bit trickier than convergence of a sequence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We want two notions: that one partition &quot;refines&quot; another partition, and that a partition of an interval [a,b] produces subintervals of width at most h &gt; 0.  You seem to have a good grasp on the second of these, namely that even if the partition is not constructed to have n equal length subintervals, the subintervals may still have some &quot;small&quot; bound on their maximum width.  We can specify that &quot;mesh size&quot; is small in this way, setting the stage for proving something about the limit as mesh size tends to zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is the first concept, that of one partition refining another, which typically causes more difficulty.  The notes you linked to do not seem to treat that topic.  Here is &lt;a href=&quot;http://math.asu.edu/~kuiper/370files/Riemann_1.pdf&quot; rel=&quot;nofollow&quot;&gt; a PDF on the Riemann integral&lt;/a&gt; which does.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a given function f on interval [a,b] we define the &lt;strong&gt;lower Riemann integral&lt;/strong&gt; as the sup over all partitions of their lower Riemann sums, and the &lt;strong&gt;upper Riemann integral&lt;/strong&gt; as the inf over all partitions of their upper Riemann sums.  If these agree, then we say the function f is Riemann integrable on [a,b], and the common value of the upper and lower Riemann integrals is the value of the Riemann integral.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A partition $Q$ of [a,b] is said to refine partition $P$ when the endpoints of $P$ are also endpoints of $Q$.  The theory of Riemann integration uses a construction that given two partitions $P_1$ and $P_2$ produces a common refinement of both $Q$ by taking the union of the two sets of endpoints.  Just having a smaller mesh size is not enough to guarantee that one partition refines another.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See the notes of the link above for more details of how the theory of Riemann integration is developed using the concept of partition refinement.&lt;/p&gt;&#xA;" OwnerUserId="3111" LastActivityDate="2011-02-26T17:25:47.997" CommentCount="6" />
  <row Id="23872" PostTypeId="2" ParentId="23844" CreationDate="2011-02-26T17:33:04.913" Score="8" ViewCount="" Body="&lt;p&gt;I think that you will find enlightening the following sketched proof that $\rm\: \mathbb Z[w],\ w = (1 + \sqrt{-19})/2\ $ is a non-Euclidean PID -- based on a sketch by the eminent number-theorist Hendrik W. Lenstra.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the the proof in Dummit &amp;amp; Foote, uses the &lt;a href=&quot;http://groups.google.com/group/sci.math/msg/db5b65f820ffd142&quot; rel=&quot;nofollow&quot;&gt;Dedekind-Hasse theorem&lt;/a&gt;&#xA;to prove it is a PID, and the &lt;a href=&quot;http://groups.google.com/group/sci.math/msg/3944f591098651cc&quot; rel=&quot;nofollow&quot;&gt;universal side divisor criterion&lt;/a&gt;&#xA;to prove it is not Euclidean, is probably the simplest known.&#xA;The so-called universal side divisor criterion is essentially a&#xA;special case of research of Lenstra, Motzkin, Samuel, Williams &#xA;et al. that apply in much wider generality to Euclidean domains. &#xA;You can obtain a deeper understanding of Euclidean domains from&#xA;the excellent surveys by Lenstra in Mathematical Intelligencer&#xA;1979/1980 (Euclidean Number Fields 1,2,3) and Lemmermeyer's superb survey &lt;a href=&quot;http://www.fen.bilkent.edu.tr/~franz/publ/survey.pdf&quot; rel=&quot;nofollow&quot;&gt;The Euclidean algorithm in algebraic number fields&lt;/a&gt;. Below is said sketched proof of Lenstra, excerpted from &lt;a href=&quot;http://math.berkeley.edu/~gbergman/grad.hndts/nonEucPID.ps&quot; rel=&quot;nofollow&quot;&gt;George Bergman's web page&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $\rm\:w\:$ denote the complex number $\rm\ (1 + \sqrt{-19}/2\:,\:$ and $\rm\:R\:$ the ring $\rm\: Z[w]\:.$ &#xA;We shall show that $\rm\:R\:$ is a principal ideal domain, but not a Euclidean&#xA;ring. This is Exercise III.3.8 of Hungerford's &lt;em&gt;Algebra&lt;/em&gt; (2nd edition),&#xA;but no hints are given there; the proof outlined here was sketched for&#xA;me (Bergman) by H. W. Lenstra, Jr.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) Verify that $\rm\  w^2 - w + 5 = 0\:,\:$ that  $\rm\ R = \{m + n\ a\ :\ m, n \in \mathbb Z\} = \{m + n\ \bar a\ :\ m, n \in \mathbb Z\}\:,\:$ where the bar denotes complex conjugation, and that the map  $\rm\ x \to |x|^2 = x \bar x\ $  is nonnegative integer-valued &#xA;and respects multiplication.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) Deduce that $\rm\ |x|^2 = 1\ $  for all units of $\rm\:R\:,\:$ and using a lower bound on the absolute value of the imaginary part of any noneal member of $\rm\:R\:,\:$ conclude that the only units of $\rm\:R\:$ are $\pm 1\:.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(3) Assuming $\rm\:R\:$ has a Euclidean function $\rm\:h\:,\:$ let $\rm\:x\:$ be a nonzero nonunit of R minimizing $\rm\: h(x)\:.\:$  Show that $\rm\:R/xR\:$ consists of the images in this ring of $\:0\:$ and the units of $\rm\:R\:,\:$ hence has cardinality at most $3$. What nonzero rings are there of such cardinalities?  Show  $\rm\ w^2 - w + 5 = 0 \ $ has no solution in any of these rings, and deduce a contradiction, showing that R is not Euclidean.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We shall now show that $\rm\:R\:$ is a principal ideal domain. To do this, let&#xA;$\rm\:I\:$ be any nonzero ideal of $\rm\:R\:,\:$ and  $\rm\:x\:$  a nonzero element of $\rm\:I\:$ of least absolute value, i.e., minimizing the integer $\rm\ x \bar x\:.\:$ We shall prove $\rm\ I = x\:R\:.\:$ (Thus, we are using the function $\rm\ x \to x \bar x\ $  as a substitute for a Euclidean function, even though it doesn't enjoy all such properties.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For convenience, let us &quot;normalize&quot; our problem by taking  $\rm\ J = x^{-1}\ I\:.\:$ Thus, $\rm\:J\:$ is an $\rm\:R$-submodule of $\:\mathbb C\:,\:$ containing $\rm\:R\:$ and having no nonzero element of absolute value $&amp;lt; 1\:.\:$ We shall show from these properties that $\rm J - R = \emptyset\:,\:$ i.e., that $\rm\ J = R\:.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(4) Show that any element of $\rm\:J\:$ that has distance less than $1$ from some element of $\rm\:R\:$ must belong to $\rm\:R\:.\:$  Deduce that in any element of  $\rm\ J - R\:,\:$ the imaginary part must differ from any integral multiple of  $\:\sqrt{19}/2\:$  by at least $\:\sqrt{3}/2\:.\:$ (Suggestion: draw a picture showing the set of complex numbers which the preceding observation excludes. However, unless you are told the contrary, this picture does not replace a proof; it is merely to help you find a proof.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(5) Deduce that if $\rm\: J - R\:$  is nonempty, it must contain an element  $\rm\:y\:$  with imaginary part in the range $\rm\ [\sqrt{3}/2,\ \sqrt{19}/2 - \sqrt{3}/2]\:,\:$ and real part in the range $\rm\: (-1/2,\ 1/2]\:.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(6) Show that for such a $\rm\: y\:,\:$ the element $\rm\: 2\:y\:$  will have imaginary part too close to  $\:\sqrt{19}/2\:$  to lie in $\rm\: J - R\:.\:$  Deduce that  $\rm\ y = w/2\ $  or  $\rm- \bar w/2\:,\:$ and hence that $\rm\ w\ \bar w/2\ \in J\:.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(7) Compute  $\rm\: w\ \bar w/2\:,\:$ and obtain a contradiction. Conclude that $\rm\:R\:$ is a principal ideal domain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(8) What goes wrong with these arguments if we replace $19$ throughout by $17$?&#xA;By $23$?&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2011-03-03T04:19:37.307" LastActivityDate="2011-03-03T04:19:37.307" />
  <row Id="23873" PostTypeId="2" ParentId="23859" CreationDate="2011-02-26T17:38:33.593" Score="1" ViewCount="" Body="&lt;p&gt;A somewhat similar situation is where the graph minor theorem puts you in. If you take any minor-closed family of graphs, you know that it has a finite list of forbidden minors, but presumably it's hard to find &quot;in general&quot; (I don't know if anyone tried to make that precise, but it's conceivable that in some setting it's not computable). Given the list of forbidden minors, you have an $O(n^3)$ algorithm for testing membership in the family. Whether that gives you an efficient enumeration algorithm, I'm not sure; but the resulting situation is that there is an $O(n^3)$ algorithm, but it might be hard to find.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2011-02-26T17:38:33.593" CommentCount="1" />
  <row Id="23874" PostTypeId="1" AcceptedAnswerId="23877" CreationDate="2011-02-26T17:56:34.677" Score="7" ViewCount="211" Body="&lt;p&gt;I have been thinking about this for quite sometime.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;Eisentein Criterion for Irreducibility:&lt;/strong&gt; &lt;em&gt;Let $f$ be a primitive polynomial over a commutative unique factorization domain $R$, say&#xA;$$f(x)=a_0 + a_1x + a_2x^2 + \ldots + a_nx^n \;.$$&#xA;If $R$ has an irreducible element $p$ such that&#xA;$$p\mid a_m \qquad 0\le m\le n-1$$&#xA;$$p^2 \nmid a_0$$&#xA;$$p \nmid a_n$$&#xA;then $f$ is irreducible.&lt;/em&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Can anyone give me an &lt;strong&gt;explanation&lt;/strong&gt; of how one &lt;em&gt;might have conjectured this problem&lt;/em&gt;? Thinking along, the same lines the first polynomial which came to my mind was $x^{2}+1 \in \mathbb{R}[x]$ which is irreducible. But there are lot's of polynomials and it's very difficult to think of a condition, which would make them irreducible. &lt;/p&gt;&#xA;" OwnerUserId="9413" LastEditorUserId="9413" LastEditorDisplayName="" LastActivityDate="2011-03-03T04:04:24.853" Title="Motivation for Eisenstein Criterion" Tags="&lt;abstract-algebra&gt;&lt;polynomials&gt;&lt;ring-theory&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="5" />
  <row Id="23875" PostTypeId="2" ParentId="23823" CreationDate="2011-02-26T18:06:44.747" Score="0" ViewCount="" Body="&lt;p&gt;As I said in your &lt;a href=&quot;http://math.stackexchange.com/questions/23809/what-is-the-remainder-of-16-is-divided-by-19/23810#23810&quot;&gt;prior question,&lt;/a&gt; Wilson's theorem yields the &lt;strong&gt;reflection formula&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\rm (p-1-k)\:!\ \equiv\ \frac{(-1)^{k+1}}{k!}\ \ (mod\ p),\ \ for\ p\ prime $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence $\rm\displaystyle\ 25!\equiv \frac{1}{5!}\ (mod\ 31)\:.\:$ But $\rm\ 5!\equiv 4\ (5\cdot3\cdot2)\equiv -4,\:\:$ and $\rm\ 4\cdot 8\equiv 1\ \Rightarrow\ 1/4\equiv 8\ (mod\ 31)\:.\:$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore we conclude that $\rm\ 1/5!\equiv -1/4 \equiv -8\ (mod\ 31)\:.\ $ Indeed $\rm -8\cdot 5!  = 1 - 31^2$   &lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2011-02-26T18:06:44.747" />
  <row Id="23876" PostTypeId="2" ParentId="23874" CreationDate="2011-02-26T18:12:00.510" Score="9" ViewCount="" Body="&lt;p&gt;The point is this: if $R \to S$ is a homomorphism and the image of $f$ is irreducible in $S[x]$, then $f$ must also be irreducible in $R[x]$ (since any factorization of $f$ in $R[x]$ maps to a factorization of $f$ in $S[x]$). This idea gets to the heart of what the point of homomorphisms is: they are maps that send algebraic statements ($f$ has a factorization) to other algebraic statements. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I guess the best motivated way to approach this is to make two choices for $S$. First we choose $S = \mathbb{Z}/p\mathbb{Z}$. This is a surprisingly useful choice for specific polynomials, especially of low degree (it works for some prime $p$ if and only if the Galois group of the polynomial $f$ contains a $\deg f$-cycle, which always happens in degree $2, 3$ and is still very likely in higher degrees). For example the polynomial $x^3 - x + 1$ must be irreducible because it has no roots $\bmod 2$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;At some point we might encounter a polynomial like $x^3 - 2x - 2$. Now this polynomial is reducible $\bmod 2$ since it factors as $x^3 = x \cdot x \cdot x$, but this implies that if $x^3 - 2x - 2$ had an actual factorization $fg$ in the integers then $f, g$ would both have to have all their non-leading terms divisible by $2$, hence the constant term of $fg$ is divisible by $4$; contradiction. Eisenstein's criterion is a simple generalization of this example; it corresponds to choosing $S = \mathbb{Z}/p^2\mathbb{Z}$ or $S = \mathbb{Z}_p$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nowadays it is understood that Eisenstein's criterion is really a statement about &lt;a href=&quot;http://www.math.uconn.edu/~kconrad/blurbs/gradnumthy/totram.pdf&quot; rel=&quot;nofollow&quot;&gt;total ramification&lt;/a&gt;, but my guess is that this was not known when it was first written down. My guess is that Eisenstein wrote down this criterion for the sole purpose of proving that the cyclotomic polynomials $\Phi_p(x) = x^{p-1} + ... + 1$ are irreducible, and here it is very natural to think about what happens $\bmod p$ since $(x - 1) \Phi_p(x) = x^p - 1$ splits into linear factors $\bmod p$ by Fermat's little theorem. It is not a big leap from here to the general Eisenstein criterion. &lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2011-02-26T20:46:46.990" LastActivityDate="2011-02-26T20:46:46.990" />
  <row Id="23877" PostTypeId="2" ParentId="23874" CreationDate="2011-02-26T18:27:06.217" Score="8" ViewCount="" Body="&lt;p&gt;As Qiaochu hinted, Eisenstein's Criterion arises from exploiting factorization properties in simpler image rings. Here the property exploited is simply that the image reduces to a prime power, and prime products always factor &lt;em&gt;uniquely&lt;/em&gt; in domains. With that in mind one can easily generalize Eisenstein's criterion to other polynomials of similar shape. However, there is a limit to how far one can go this way since the criterion corresponds to totally ramified primes, but there are number fields without such primes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to find fruitful generalizations one may apply valuation-theoretic ideas. The Eisenstein and related irreducibility tests  are all special cases of much more general techniques exploiting &lt;em&gt;Newton polygons&lt;/em&gt;. They yield the  &lt;em&gt;master theorem&lt;/em&gt; behind all these related results. An excellent comprehensive introduction can be found in Filaseta's lecture notes - see the links below. See also this &lt;a href=&quot;http://mathoverflow.net/questions/15460/variants-of-eisenstein-irreducibility&quot; rel=&quot;nofollow&quot;&gt;MathOverflow question.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[1] &lt;a href=&quot;http://www.math.sc.edu/~filaseta/gradcourses/Math788F/latexbook/&quot; rel=&quot;nofollow&quot;&gt;http://www.math.sc.edu/~filaseta/gradcourses/Math788F/latexbook/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[2] &lt;a href=&quot;http://www.math.sc.edu/~filaseta/gradcourses/Math788F/NewtonPolygonsTalk.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.math.sc.edu/~filaseta/gradcourses/Math788F/NewtonPolygonsTalk.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[3] Newton Polygon Applet&#xA;&lt;a href=&quot;http://www.math.sc.edu/~filaseta/newton/newton.html&quot; rel=&quot;nofollow&quot;&gt;http://www.math.sc.edu/~filaseta/newton/newton.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[4] Abhyankar, Shreeram S.&lt;br&gt;&#xA;&lt;em&gt;Historical ramblings in algebraic geometry and related algebra.&lt;/em&gt;&lt;br&gt;&#xA;Amer. Math. Monthly 83 (1976), no. 6, 409-448.&lt;br&gt;&#xA;&lt;a href=&quot;http://mathdl.maa.org/mathDL/22/?pa=content&amp;amp;sa=viewDocument&amp;amp;nodeId=2964&amp;amp;pf=1&quot; rel=&quot;nofollow&quot;&gt;http://mathdl.maa.org/mathDL/22/?pa=content&amp;amp;sa=viewDocument&amp;amp;nodeId=2964&amp;amp;pf=1&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2011-03-03T04:04:24.853" LastActivityDate="2011-03-03T04:04:24.853" />
  <row Id="23878" PostTypeId="2" ParentId="23828" CreationDate="2011-02-26T18:32:00.907" Score="0" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\ $ Rewrite it as $\rm\ Y = 2^{\:8 - X}$&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2011-02-26T18:32:00.907" />
  <row Id="23879" PostTypeId="1" AcceptedAnswerId="23906" CreationDate="2011-02-26T18:43:03.810" Score="3" ViewCount="74" Body="&lt;p&gt;X is a projective variety, W is a quasi-projective variety over the algebraically closed ﬁeld k.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to construct a k-algebra isomorphism between O(XxW) and O(W) (the rings of regular functions). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the level of varieties I know that the projection XxW-&gt;W maps closed sets to closed sets. I feel like this should lead me to the algebra, but I have no idea how I would define the map going the other way.&lt;/p&gt;&#xA;" OwnerUserId="7528" LastActivityDate="2011-02-26T21:58:10.350" Title="Projective, Quasi-Projective" Tags="&lt;algebraic-geometry&gt;" AnswerCount="1" />
  <row Id="23880" PostTypeId="2" ParentId="23862" CreationDate="2011-02-26T18:51:14.003" Score="0" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\ $ Inspecting the diagram reveals the slope $\rm\:m\:$ and $\rm\:y$-intercept $\rm\:n\:$ of the line segment(s).&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2011-02-26T18:51:14.003" />
  <row Id="23881" PostTypeId="1" CreationDate="2011-02-26T19:02:05.097" Score="1" ViewCount="193" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Possible Duplicates:&lt;/strong&gt;&lt;br&gt;&#xA;  &lt;a href=&quot;http://math.stackexchange.com/questions/23129/why-is-euclidean-geometry-scale-invariant&quot;&gt;Why is Euclidean geometry scale-invariant?&lt;/a&gt; &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;a href=&quot;http://math.stackexchange.com/questions/3198/proof-that-pi-is-constant-the-same-for-all-circles-without-using-limits&quot;&gt;Proof that Pi is constant (the same for all circles), without using limits&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;p&gt;The answer with the most up votes will be selected as the answers after 72-hours has passed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Questions, feedback, requests -- just comment, thanks!!&lt;/p&gt;&#xA;" OwnerUserId="2453" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2011-05-31T02:43:39.287" LastActivityDate="2011-05-31T02:43:39.287" ClosedDate="2011-02-26T19:38:33.010" Title="Why is the ratio of the circumference of a circle to its diameter independent of the circle? " Tags="&lt;geometry&gt;&lt;pi&gt;" AnswerCount="1" CommentCount="13" />
  <row Id="23882" PostTypeId="2" ParentId="23862" CreationDate="2011-02-26T19:06:39.503" Score="0" ViewCount="" Body="&lt;p&gt;If it was without modulus graph will go through Oy at (0;-3) so n=-3, I assume we know that it touch Ox on (1.5;0) so &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;y=mx+n&#xA;0=1.5m-3&#xA;m=2&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;y=|2x-3|&lt;/p&gt;&#xA;" OwnerUserId="7149" LastActivityDate="2011-02-26T19:06:39.503" />
  <row Id="23883" PostTypeId="2" ParentId="23881" CreationDate="2011-02-26T19:39:48.560" Score="1" ViewCount="" Body="&lt;p&gt;Because when you change the scale, both the diameter and the  circumference change by multiplying by the same scale factor, since they are both of dimension 1. Thus, their ratio is independent of the scale factor. Since all circles are similar, the ratio is the same for all circles.&lt;/p&gt;&#xA;" OwnerUserId="589" LastActivityDate="2011-02-26T19:39:48.560" CommentCount="3" />
  <row Id="23884" PostTypeId="2" ParentId="23193" CreationDate="2011-02-26T19:41:09.353" Score="2" ViewCount="" Body="&lt;p&gt;Since I have to explain Joriki's solution to all the people I mentioned the  problem to, I decided to provide a typed solution written up in my own words.  Although there are no new ideas, there doesn't seem to be any good reason not to just share it anyway:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Lemma&lt;/strong&gt; Let  $a\geq 1$, $b\geq 1$.  Given an $a\times b$ rectangular grid, without using horizontal or vertical lines it requires at least $a+b-2$ lines to cover all the points.  (Here the path need not be connected)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;proof:&lt;/em&gt; Consider the exterior of the grid.  If $a\geq 2$ and $b\geq 2$ then each diagonal line can cover at most two points, but there are $2a+2b-4$ points that must be covered.  If $b=1$ then each diagonal line covers at most 1 point, and hence we need $a=a+b-1&amp;gt;a+b-2$ lines.  Similarly if $a=1$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given a solution to the $n\times n$ grid, let $h$ be the number of horizontal lines, and $v$ be the number of vertical lines used.  If $h=n$, we must have at least $2n-1$ lines since it will take at least $n-1$ lines to connect all of these horizontal parts.  Similarly for $v=n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, let $v&amp;lt;n$, $h&amp;lt;n$ and remove all the rows and columns where a horizontal or vertical line appears.  Then we have an $n-h\times n-v$ grid left over, which by the lemma requires at least $(n-h)+(n-v)-2=2n-2-h-v$ lines to cover.  Since we already used $h+v$ lines, the entire grid requires $(h+v)+(2n-2-h-v)=2n-2$ lines, and the proof is complete. &lt;/p&gt;&#xA;" OwnerUserId="6075" LastEditorUserId="6075" LastEditorDisplayName="" LastEditDate="2011-02-26T23:23:26.097" LastActivityDate="2011-02-26T23:23:26.097" CommentCount="2" />
  <row Id="23885" PostTypeId="1" AcceptedAnswerId="23890" CreationDate="2011-02-26T19:44:05.577" Score="2" ViewCount="103" Body="&lt;p&gt;In probability,  events are considered to be closed under countable union and complement, so mathematically they are modeled by $\sigma$-algebra.  I was wondering why events  are considered to be closed under countably union and complement?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In &lt;a href=&quot;http://math.stackexchange.com/questions/23600/intuition-behind-conditional-expectation/23617#23617&quot;&gt;Nate Eldredge's post&lt;/a&gt;, he has done an excellent job on explaining this, by using whether questions are answered or not as an analogy to whether events occur or not, if I understand his post correctly. However, if someone could explain plainly without analogy, it could be clearer to me. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was particularly curious why events are not considered to be closed under infinite (possibly uncountably) union, but instead just under countably union? So possibly to model events using the power set? I think this is not addressed in &lt;a href=&quot;http://math.stackexchange.com/questions/23600/intuition-behind-conditional-expectation/23617#23617&quot;&gt;Nate Eldredge's post&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My guess would be that the reason might be related to the requirement on the likelihood of any event to occur to be &quot;measurable&quot; in some sense. But how exactly to understand this requirement is unclear to me. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS: This post is related to my previous one &lt;a href=&quot;http://math.stackexchange.com/questions/23776/interpretation-of-sigma-algebra&quot;&gt;Interpretation of sigma algebra&lt;/a&gt;, but the questions asked in these two are not the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks and regards!&lt;/p&gt;&#xA;" OwnerUserId="1281" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2011-02-26T20:17:37.047" LastActivityDate="2011-02-26T20:17:37.047" Title="Why events in probability are closed under countable union and complement?" Tags="&lt;probability-theory&gt;&lt;measure-theory&gt;" AnswerCount="2" />
  <row Id="23886" PostTypeId="1" AcceptedAnswerId="23891" CreationDate="2011-02-26T19:45:17.427" Score="3" ViewCount="102" Body="&lt;p&gt;There are $n$ symmetric polynomials in the eigenvalues of a square matrix. Two of these are the determinant and the trace, each of which have countless applications and interpretations in algebra and geometry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What about the other symmetric polynomials? They are also similarity invariants, yet I've never seen them used or referenced. Are there any geometric interpretations, or applications, for these other invariants?&lt;/p&gt;&#xA;" OwnerUserId="7530" LastActivityDate="2011-02-26T20:19:57.620" Title="generalizations of determinant and trace" Tags="&lt;linear-algebra&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="23887" PostTypeId="2" ParentId="23886" CreationDate="2011-02-26T19:58:34.037" Score="0" ViewCount="" Body="&lt;p&gt;I think you  are thinking of the coefficients of the characteristic polynomial -the trace and the determinant are the &quot;extremes&quot;. &#xA;See eg &#xA;&lt;a href=&quot;http://alert.comule.com/?p=7741&quot; rel=&quot;nofollow&quot;&gt;http://alert.comule.com/?p=7741&lt;/a&gt;&#xA;and &#xA;&lt;a href=&quot;http://mathoverflow.net/questions/33478/geometric-interpretation-of-characteristic-polynomial&quot; rel=&quot;nofollow&quot;&gt;http://mathoverflow.net/questions/33478/geometric-interpretation-of-characteristic-polynomial&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="312" LastActivityDate="2011-02-26T19:58:34.037" CommentCount="1" />
  <row Id="23888" PostTypeId="2" ParentId="23885" CreationDate="2011-02-26T20:00:35.283" Score="2" ViewCount="" Body="&lt;p&gt;If events were closed under arbitrary unions, and if singletons are events, then everything would be an event, because every set is the union of the singletons it contains.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Trying to define probabilities or measures on all sets can lead to &lt;a href=&quot;http://en.wikipedia.org/wiki/Non-measurable_set&quot; rel=&quot;nofollow&quot;&gt;problems&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2011-02-26T20:00:35.283" />
  <row Id="23889" PostTypeId="2" ParentId="23790" CreationDate="2011-02-26T20:06:43.550" Score="1" ViewCount="" Body="&lt;p&gt;I'm going to concentrate on commutative rings, because those are the ones closest to what you might be familiar with. To deal with the notions in noncommutative rings takes a bit more work, but is certainly doable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ideals, at least at first, are meant to generalize the notion of &quot;is a multiple of&quot; (turns out that there is a different motivation for singling out ideals among the subrings, which is essentially the same reason for singling out normal subgroups among all subgroups of a group; this is not relevant right now, but you might be interested in taking a look at &lt;a href=&quot;http://math.stackexchange.com/questions/14282/why-do-we-define-quotient-groups-for-normal-subgroups-only/14315#14315&quot;&gt;this answer about normal subgroups&lt;/a&gt; later). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you consider the integers, you can characterize a number $n$ (up to sign) by describing all elements that are multiples of $n$. If you know exactly who the multiples of $n$ are, then you know exactly who $n$ is (except that you might confuse it with $-n$). So, instead of looking at the number $n$, we can look at the collection of all its multiples, &#xA;$$n\mathbb{Z} = (n) = \{a\in\mathbb{Z}\mid a\text{ is a multiples of }n\}.$$&#xA;What properties do the collections of &quot;all multiples of a given number&quot; have? Well:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The collection always contains $0$.&lt;/li&gt;&#xA;&lt;li&gt;If $a$ and $b$ are in the collection, so are $a+b$ and $a-b$.&lt;/li&gt;&#xA;&lt;li&gt;If $a$ is in the collection, and $r$ is any element of the ring, then $ra$ is also in the collection.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In the integers, and also in many other rings (for example, $\mathbb{R}[x]$, the polynomials over $\mathbb{R}$), &lt;em&gt;every&lt;/em&gt; collection that satisfies these three properties is in fact the collection of all multiples of some $a\in R$. But there are other rings where this does not happen. For example, if you consider the ring $\mathbb{Z}[x]$ of all polynomials with integer coefficients, you can take&#xA;$$I = \{ p(x)\in\mathbb{Z}[x]\mid p(0)\text{ is even}\}.$$&#xA;This collection satisfies all three properties: $0$ is in $I$; if $p(x)$ and $q(x)$ are in $I$, then so is $p(x)+q(x)$, because $(p+q)(0) = p(0)+q(0)$ is a sum of two even numbers, hence even; and if $p(x)$ is in the collection and $q(x)$ is any polynomial with integer coefficients, when $pq(0) = p(0)q(0)$ is even, because $p(0)$ is even and $q(0)$ is an integer. So $I$ is an ideal. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this $I$ the collection of &quot;all multiples of $a$&quot; for some $a\in\mathbb{Z}[x]$? No. If there were such an $a$, then since $2\in I$, then $2$ would have to be a multiple of $a$. That means that $a$ must be a constant polynomial, and must be either $\pm 1$ or $\pm 2$ (the only elements of $\mathbb{Z}[x]$ that divide $2$). But $x\in I$ as well, since evaluating $x$ at $0$ is even; and none of $\pm 1$ or $\pm 2$ divide $x$ in $\mathbb{Z}[x]$. So even though $I$ is an ideal, it is not &quot;all multiples of&quot; someone. So the notion of &quot;ideal&quot;, even though it starts up as &quot;all multiples of&quot; someone, is actually more general. This is the distinction between &lt;strong&gt;principal&lt;/strong&gt; ideals (ideals which &lt;em&gt;are&lt;/em&gt; &quot;all multiples of $a$&quot; for some $a$), and more general ideals (which need not be made up of &quot;all multiples of $a$&quot; for some $a$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nonetheless, ideals are closely connected to the notions of divisibility; as Dedekind noted when he introduced them in the 19th century, if you want to try to do &quot;modulo arithmetic&quot; as in the integers (working modulo $n$ is &quot;really&quot; working in $\mathbb{Z}/(n)$) then the conditions you need on a collection are &lt;em&gt;precisely&lt;/em&gt; the conditions that are needed to have ideals. That is, ideals are exactly the things for which you can do &quot;modulo arithmetic&quot;. And modulo arithmetic is all about divisibility (after all, $a\equiv b\pmod{n}$ means that $n$ divides $a-b$). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we want to also keep track of a few of the other special properties that some numbers have, and &quot;translate&quot; them into what they mean for ideals. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Prime numbers play a major role in divisibility issues in the integers. How does the &quot;prime&quot; property translate into the setting of ideals? A prime is a number $p$ such that:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$p\neq \pm 1$; and&lt;/li&gt;&#xA;&lt;li&gt;If $p$ divides a product, then it divides at least one of the factors.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Okay, how does that translate into ideals? If you think of an ideal $I$ as &quot;the collection of all multiples of some number&quot; (again, not really that in the general setting, but that's where the intuition and some of the definitions come from), then when do the multiples correspond to a prime? We need the prime not to divide &lt;em&gt;everything&lt;/em&gt;; so we require the ideal to not be the entire ring, $I\neq R$. And the second condition: if $ab$ is a multiple, then either $a$ or $b$ is a multiple. In other words: if $ab\in I$, then either $a\in I$ or $b\in I$. So we define:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;An ideal $I$ is a &lt;em&gt;prime ideal&lt;/em&gt; if and only if $I\neq R$, and whenever $ab\in I$, either $a\in I$ or $b\in I$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Going back to the intuition for ideals: what does the intersection of ideals correspond to? If $I$ is sort of like &quot;all multiples of $a$&quot;, and $J$ is sort of like &quot;all multiples of $b$&quot;, then what is $I\cap J$? All things that are multiples of &lt;em&gt;both&lt;/em&gt; $a$ and $b$!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, if $P$ and $Q$ are both prime ideals, would $P\cap Q$ be a prime ideal? Generally no: in general, you don't expect things that are multiples of &lt;em&gt;two&lt;/em&gt; different primes to be themselves prime. And so you get to your example. $(2)$ is a prime ideal in $\mathbb{Z}$, precisely because $2$ is a prime number: if $ab\in(2)$, then $ab$ is a multiple of $2$, so either $a$ is a multiple of $2$ or $b$ is a multiple of $2$ (because $2$ is a prime &lt;em&gt;number&lt;/em&gt;), so either $a\in(2)$ or $b\in(2)$. Similarly with $(3)$. But $(2)\cap(3)$ will be all numbers that are multiples of &lt;em&gt;both&lt;/em&gt; $2$ and $3$; this corresponds to &quot;all multiples of $6$&quot;, as we know from elementary number theory: $(2)\cap(3)=(6)$. But $6$ is not a prime &lt;em&gt;number&lt;/em&gt;, so there is no reason to expect $(6)$ to be a prime &lt;em&gt;ideal&lt;/em&gt;. In fact, a witness to the fact that $6$ is not a prime number should also work as a witness to the fact that $(6)$ is not a prime ideal. And indeed it does.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Caveat.&lt;/strong&gt; The analogy of ideals as &quot;set of all multiples of &lt;em&gt;something&lt;/em&gt;&quot; works reasonably well in very familiar settings, but breaks down very quickly once you get beyond the most basic of rings. For instance, in the integers, you cannot have two nonzero prime ideals $(p)$ and $(q)$ with $p\neq 0$, $q\neq 0$, $p\neq \pm q$, and $(p)\subseteq (q)$: that would mean that $p$ is a multiple of $q$, and with prime numbers that can only happen if $p=\pm q$. But in other rings it is certainly possible for it to happen. For instance, in $R=\mathbb{R}[x,y]$, the ring of polynomials in two variables, both&#xA;\begin{align*}&#xA;(x) &amp;amp;= \{ p(x,y)\in R\mid p(0,y) = 0\text{ for all }y\};\\&#xA;(x,y) &amp;amp;= \{ p(x,y)\in R\mid p(0,0) = 0\}&#xA;\end{align*}&#xA;are ideals; clearly $(x)\subseteq (x,y)$, $(x)\neq (0)$, $(x,y)\neq (0)$, and $(x)\neq (x,y)$. Yet both $(x)$ and $(x,y)$ are prime ideals. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the analogy can only take you so far, and it can be misleading if you try to take it all the way. But at least at first you might find it a useful hook for thinking about possible examples and possible counterexamples. &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-27T03:16:13.310" LastActivityDate="2011-02-27T03:16:13.310" CommentCount="1" />
  <row Id="23890" PostTypeId="2" ParentId="23885" CreationDate="2011-02-26T20:13:14.537" Score="4" ViewCount="" Body="&lt;p&gt;As Jonas mentioned, allowing arbitrary unions is not &quot;consistent&quot;, in the sense that there is no proper definition of probability. This is also related to the fact that infinite sums make much more sense when countable, since it's not clear how to attach a finite number to an uncountable sum of positive reals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, many desirable events are describable using countable unions and intersections. For example, events like &quot;the random walk returns to the origin&quot; is a union of countably many events &quot;the random walk returns to the origin at time $t$&quot;, and any one of those is a finite union of &quot;basic&quot; events.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, first order properties always correspond to taking countable unions and intersections; this means that if you have a statement of the form &quot;$\forall x \exists y \cdots P(x,y,\ldots)$&quot;, where $x,y,\ldots$ are integers, and the $P$s are basic events (e.g. for a random walk, depend on finitely many times), then the corresponding event is guaranteed to be in the $\sigma$-algebra, i.e. is guaranteed to have assigned to it a &quot;probability&quot;.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2011-02-26T20:13:14.537" CommentCount="1" />
  <row Id="23891" PostTypeId="2" ParentId="23886" CreationDate="2011-02-26T20:19:57.620" Score="6" ViewCount="" Body="&lt;p&gt;The trace and the determinant are the most useful invariants because the trace is additive and the determinant is multiplicative. The other coefficients of the characteristic polynomial are neither. The determinant also has a clear geometric interpretation. In addition, all of the coefficients of the characteristic polynomial of an operator $T$ can be computed from the traces of the operators $T^n$; this is one reason why it is not so surprising that traces of group elements in group representations carry a lot of information. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is not to say that people never &lt;em&gt;use&lt;/em&gt; the other invariants, although they don't tend to have special names. For example, my understanding is that the &lt;a href=&quot;http://en.wikipedia.org/wiki/Killing_form&quot; rel=&quot;nofollow&quot;&gt;Killing form&lt;/a&gt; in Lie theory, an important tool, was discovered by messing around with characteristic polynomials. And the construction underlying the coefficients of the characteristic polynomial, the &lt;a href=&quot;http://en.wikipedia.org/wiki/Exterior_algebra&quot; rel=&quot;nofollow&quot;&gt;exterior algebra&lt;/a&gt;, is enormously useful. &lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2011-02-26T20:19:57.620" />
  <row Id="23893" PostTypeId="1" CreationDate="2011-02-26T20:27:07.023" Score="-2" ViewCount="107" Body="&lt;p&gt;Given a continuous increasing function $f(x)$, $x \in \mathbb{R}$, and a sequence of real numbers $\{a_{n}\}$. Is there any relation between $f &amp;#39;(a_{n})$, and $f &amp;#39;(a_{n}+c_{n})$ for some constants $c_{n}$, $n\in \mathbb{N}$ ? I mean:&#xA;$|f &amp;#39;(a_{n}) - f &amp;#39;(a_{n}+c_{n})|\leq something$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="7531" LastEditorUserId="7531" LastEditorDisplayName="" LastEditDate="2011-02-26T20:50:51.650" LastActivityDate="2011-02-26T21:06:29.860" Title="Increasing functions" Tags="&lt;calculus&gt;" AnswerCount="3" CommentCount="7" />
  <row Id="23894" PostTypeId="2" ParentId="23893" CreationDate="2011-02-26T20:30:51.217" Score="1" ViewCount="" Body="&lt;p&gt;If you take any positive function and integrate it then you get a continuous increasing function. So in particular there's no reason that your $f&amp;#39;$ will satisfy a property like unto the one you've mentioned.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2011-02-26T20:30:51.217" CommentCount="3" />
  <row Id="23895" PostTypeId="2" ParentId="23874" CreationDate="2011-02-26T20:36:17.637" Score="2" ViewCount="" Body="&lt;p&gt;If you don't know about ramification or valuations and things, there is still a great easy way to &quot;see&quot; Eisenstein from the $\mathbb{Z}[x]$ case. This is presented in &lt;em&gt;Integers, Polynomials, and Rings&lt;/em&gt; by Ron Irving as a series of exercises to get the student to guess Eisenstein on their own.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First examine the case $x^n-p$ where $p$ is a prime. If this factors, then there are $f(x)=\sum_{k=0}^m a_kx^k$ and $g(x)=\sum_{k=0}^r b_kx^k$ for which $f(x)g(x)=x^n-p$. First, you get that $a_0b_0=p$ and hence without loss of generality $a_0=p$ and $p$ does not divide $b_0$ (we'll merely use this fact rather than $b_0=1$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now since $a_0b_1+a_1b_0=0$ we get that $p|(a_0b_1+a_1b_0)$ and $p|a_0$ so $p|a_1b_0$. This means $p|a_1$ or $p|b_0$, but $p$ does not divide $b_0$, so $p|a_1$. You can keep bootstrapping this argument up all the coefficients $a_k$ to get that $p|a_m$ the top coefficient, which is a contradiction since $a_mb_r=1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But what was really used here? Not much. So you could repeat the exact same argument with $x^n-pm$ where $(p,m)=1$. But that still isn't as general as you could go. You didn't need that $\sum_{j+k=l} a_jb_k=0$, you only needed that $p$ divided that sum to make the argument work, which is the same as checking that the polynomial $x^n+c_{n-1}x^{n-1}+\cdots +c_0$ has all $c_i$ divisible by $p$ (and $c_0$ not divisible by $p^2$). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yet again, this still isn't as general as the argument allows you to go, because the contradiction only came from the fact that $p$ did not divide the leading coefficient, it wasn't that it was $1$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So merely extrapolating what made the proof that $x^n-p$ is irreducible in $\mathbb{Z}[x]$ work gives you the full Eisenstein in $\mathbb{Z}[x]$. Lastly, if you want to go to $R[x]$, you need to figure out which facts about division you needed about $\mathbb{Z}$. But Eisenstein really is a criterion invented artificially in $\mathbb{Z}$ that could be guessed from examining what the key properties of one proof were.&lt;/p&gt;&#xA;" OwnerUserId="2000" LastActivityDate="2011-02-26T20:36:17.637" />
  <row Id="23896" PostTypeId="2" ParentId="23893" CreationDate="2011-02-26T20:41:58.740" Score="1" ViewCount="" Body="&lt;p&gt;No; the difference in the derivatives can be arbitrarily large, even if $c_n\to 0$. That is, the rate of change of $f&amp;#39;(x)$ can be very large, or even unbounded.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Fixed to make $f$ increasing on $[0,\infty)&lt;/em&gt;: The following example works on $[0,\infty)$. Since the value is $0$ at $0$, you can get an example that works and is increasing over all real numbers by letting $g(x) = f(x)$ for $x\geq 0$, and $g(x) = -f(-x)$ for $x\lt 0$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, take $f(x) = 2x^2+\sin(x^2)$ with $x\geq 0$; $f&amp;#39;(x) = 4x+2x\cos(x^2) = 2x(2+\cos(x^2))$ (positive on $[0,\infty)$, so $f(x)$ is increasing), and $f&amp;#39;&amp;#39;(x) = 2(2+\cos(x^2)) + 4x^2\cos(x^2)$. The second derivative can become very large for $x$ very large, which tells you that $f&amp;#39;(x)$ will have very large increases/decreases over very small intervals. So even if $c_n\to 0$, you can pick $a_n$ so that $f&amp;#39;(a_n) - f&amp;#39;(a_n+c_n)$ is unbounded.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To see this exlicitly, note that if $a_n=\sqrt{2n\pi}$, then $f&amp;#39;(a_n) = &#xD;&#xA;6\sqrt{2n\pi}$. If we now let $c_n = \sqrt{2n\pi + \frac{\pi}{2}} - \sqrt{2n\pi}$, then $a_n+ c_n = \sqrt{2n\pi + \frac{\pi}{2}}$, so $f&amp;#39;(a_n+c_n) = 4\sqrt{2n\pi+\frac{\pi}{2}}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then $|f&amp;#39;(a_n) - f&amp;#39;(a_n+c_n)| = \left|6\sqrt{2n\pi} - 4\sqrt{2n\pi+\frac{\pi}{2}}\right|$, which I claim goes to $\infty$ as $n\to\infty$. Also, $c_n\to 0$ as $n\to\infty$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, both claims will be established if we establish the latter one, since $|f&amp;#39;(a_n)-f&amp;#39;(a_n+c_n)| \geq 2\sqrt{2n\pi} - 4|c_n|$, and $2\sqrt{2n\pi}\to \infty$ as $n\to\infty$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And indeed, we have:&#xA;\begin{align*}&#xA;\lim_{n\to\infty} c_n &amp;amp; =  \lim_{n\to\infty}\left(\sqrt{2n\pi} - \sqrt{2n\pi+\frac{\pi}{2}}\right)\\&#xA;&amp;amp;=\lim_{n\to\infty}\frac{2n\pi - 2n\pi - \frac{\pi}{2}}{\sqrt{2n\pi} + \sqrt{2n\pi+\frac{\pi}{2}}} \\&#xA;&amp;amp; =  \lim_{n\to\infty}\frac{-\pi}{2\left(\sqrt{2n\pi} + \sqrt{2n\pi+\frac{\pi}{2}}\right)}\\&#xA;&amp;amp;= 0.&#xA;\end{align*}&#xA;So we have $c_n\to 0$ and $|f&amp;#39;(a_n)-f&amp;#39;(a_n+c_n)|\to\infty$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, if you set $c_n = 0$ for all $n\in\mathbb{N}$, then the answer is &quot;yes&quot;...&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-26T21:06:27.043" LastActivityDate="2011-02-26T21:06:27.043" CommentCount="3" />
  <row Id="23897" PostTypeId="1" AcceptedAnswerId="23900" CreationDate="2011-02-26T20:50:27.427" Score="1" ViewCount="59" Body="&lt;p&gt;I'm trying to understand the proof of lemma 3.2 of &lt;a href=&quot;http://www.math.uconn.edu/~kconrad/blurbs/grouptheory/Ansimple.pdf&quot; rel=&quot;nofollow&quot;&gt;this article&lt;/a&gt; by Keith Conrad on proofs of the simplicity of $A_n$. There is a step I don't follow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When he says &quot;Relabelling, we may write $\sigma=(1 2 \dots r) \pi$&quot;, I don't see why we could do that. Relabelling what?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As I tried to understand the statement, I thought of this: if $(i_1 \dots i_r)$ is a cycle appearing in $\sigma$, then we can conjugate $\sigma$ to $\sigma&amp;#39;=(1 \dots r) \pi$. If we show that $\sigma&amp;#39;$ coincides with $\sigma$ at some index, then we're fine. But why would it coincide, and where? &lt;/p&gt;&#xA;" OwnerUserId="2614" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-26T21:24:27.143" LastActivityDate="2011-02-26T23:11:01.440" Title="Relabelling a permutation" Tags="&lt;group-theory&gt;&lt;permutations&gt;" AnswerCount="1" />
  <row Id="23898" PostTypeId="2" ParentId="23893" CreationDate="2011-02-26T21:06:29.860" Score="1" ViewCount="" Body="&lt;p&gt;So it really depends on the sequences in question.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is just one example:  Consider $f(x)=x^n$.  Then for $c_n\geq0$, $a_n\geq 0$ $$f^&amp;#39;(a_n+c_n)-f^&amp;#39;(a_n)=na_n^{n-1}c_n+\cdots\geq na_n^{n-1}c_n$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then given any $c_n\rightarrow 0$, and any $M\in \mathbb{R}$ we can choose $a_n$ such that $f^&amp;#39;(a_n+c_n)-f^&amp;#39;(a_n)\geq M$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But what do you mean?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Heres looking at the more restrictive case where $a_n\rightarrow a$ for some $a\in \mathbb{R}$?  Without loss of generality $a_n\rightarrow 0$, so consider $f(x)=x^{1/n}$ on the interval $(0,1)$.  Again given any $c_n\rightarrow 0$ we can choose $a_n$ so that the difference is arbitrarily large.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And when $c_n$ does not converge to $0$ things blow up even more.&lt;/p&gt;&#xA;" OwnerUserId="6075" LastActivityDate="2011-02-26T21:06:29.860" />
  <row Id="23899" PostTypeId="1" AcceptedAnswerId="23901" CreationDate="2011-02-26T21:21:59.783" Score="2" ViewCount="87" Body="&lt;p&gt;The characteristic polynomial of a matrix A is defined as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\chi(A) = \det(xI-A) = \sum_{i=0}^n (-1)^i\cdot \operatorname{tr}^{(i)}(A) \cdot x^{(n-i)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The trace is the sum of the eigenvalues of a matrix, tr(A) = tr&lt;sup&gt;(1)&lt;/sup&gt;(A).  It is also the sum of the diagonal entries:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\operatorname{tr}(A) = \sum_{i=1}^n A_{ii}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sum of the products of pairs of eigenvalues is like the next trace.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this formula valid?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\operatorname{tr}^{(2)}(A) = \sum_{1 \leq i &amp;lt; j \leq n } A_{ii} A_{jj} - A_{ij} A_{ji}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What about this one?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\operatorname{tr}^{(2)}(A) = \tfrac12(\operatorname{tr}(A)^2 - \operatorname{tr}(A^2))$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there corresponding formulas for the next one, tr&lt;sup&gt;(3)&lt;/sup&gt;?&lt;/p&gt;&#xA;" OwnerUserId="583" LastActivityDate="2011-02-27T12:46:56.843" Title="Formulas for the (top) coefficients of the characteristic polynomial of a matrix" Tags="&lt;linear-algebra&gt;" AnswerCount="3" />
  <row Id="23900" PostTypeId="2" ParentId="23897" CreationDate="2011-02-26T21:23:18.517" Score="6" ViewCount="" Body="&lt;p&gt;Two explanations:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Informal one.&lt;/strong&gt; You are thinking of the elements of $A_n$ as acting on the set $\{x_1,x_2,\ldots,x_n\}$, by letting $\sigma$ map $x_i$ to $x_{\sigma(i)}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So say that $\sigma$ maps $x_3$ to $x_5$, and $x_5$ to $x_2$, and $x_2$ to $x_3$. Stick new labels on your elements, so that what you used to call &quot;$x_3$&quot; is now called $x_{1_{\mathrm{new}}}$, what you used to call $x_5$ is now called $x_{2_{\mathrm{new}}}$, and what you used to call $x_2$ is now called $x_{3_{\mathrm{new}}}$. Of course, you'll also need a new identity for $x_1$, say $x_{5_{\mathrm{new}}}$, since $5$ is no longer being used; and keep the other elements with their old names, so that $x_{i_{\mathrm{new}}} = x_i$ if $i\notin\{1,2,3,5\}$. Then $\sigma$ now acts on $\{x_{1_{\mathrm{new}}},x_{2_{\mathrm{new}}},\ldots,x_{n_{\mathrm{new}}}\}$ by looking at their &quot;old&quot; identities and acting as before. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But now $\sigma$ maps $x_{1_{\mathrm{new}}}\mapsto x_{2_{\mathrm{new}}}\mapsto x_{3_{\mathrm{new}}}\mapsto x_{1_{\mathrm{new}}}$, so if we write its cycle decomposition in terms of the &quot;new&quot; labels, we have that&#xA;$$\sigma = (1_{\mathrm{new}},2_{\mathrm{new}},3_{\mathrm{new}})\pi$$&#xA;for some $\pi$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Formal one.&lt;/strong&gt; Suppose the longest cycle of $\sigma$ is $(i_1,i_2,\ldots,i_r)$. Consider a permutation $\tau\in S_n$ with $1\mapsto i_1$, $2\mapsto i_2,\ldots,r\mapsto i_r$, and which acts in some (unspecified but irrelevant) way on the remaining elements of $\{1,\ldots,n\}$. Then consider the automorphism of $S_n$ given by conjugation by $\tau$, $\psi\mapsto \tau\psi\tau^{-1}$. You've probably seen already that if $\psi$ sends $k$ to $\ell$, then $\tau\psi\tau^{-1}$ sends $\tau(k)$ to $\tau(\ell)$. So we can look at $\tau A_n\tau^{-1}$ and $\tau\sigma\tau^{-1}$ instead of $A_n$ and $\sigma$ (they &lt;em&gt;are&lt;/em&gt; isomorphic copies, after all, so we can just &quot;translate back&quot; to $A_n$ by using the inverse automorphism when needed). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But in $\tau A_n\tau^{-1}$, the longest cycle of $\tau\sigma\tau^{-1}$ is just $(\tau(i_1),\tau(i_2),\ldots,\tau(i_r)) = (1,2,\ldots,r)$. So, by switching to this isomorphic copy of $A_n$, we may assume that the longest cycle of $\sigma$ is in fact $(1,2,\ldots,r)$. &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-26T23:11:01.440" LastActivityDate="2011-02-26T23:11:01.440" CommentCount="1" />
  <row Id="23901" PostTypeId="2" ParentId="23899" CreationDate="2011-02-26T21:29:52.220" Score="3" ViewCount="" Body="&lt;p&gt;The first guess is correct and follows directly from the permutation expansion of the determinant. The second guess is also correct. The generalization is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\text{tr}^{(n)}(A) = \text{tr } \Lambda^n(A) = \frac{1}{n!} \sum_{\pi \in S_n} \text{sgn}(\pi) \text{tr}_{\pi}(A)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\Lambda^n(A)$ is the action of $A$ on the &lt;a href=&quot;http://en.wikipedia.org/wiki/Exterior_algebra&quot; rel=&quot;nofollow&quot;&gt;exterior power&lt;/a&gt; $\Lambda^n$, $\text{tr}_n(A) = \text{tr}(A^n)$ and, for a permutation $\pi$ with cycle type $(\lambda_1, ... \lambda_k)$, we define $\text{tr}_{\pi} = \text{tr}_{\lambda_1} ... \text{tr}_{\lambda_k}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, $S_3$ has one element with cycle type $(1, 1, 1)$, three elements with cycle type $(1, 2)$, and two elements with cycle type $(3)$. It follows that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\text{tr}^{(3)}(A) = \frac{1}{6} \left( \text{tr}(A)^3 - 3 \text{tr}(A^2) \text{tr}(A) + 2 \text{tr}(A^3) \right).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This identity is really an identity of symmetric functions expressing the &lt;a href=&quot;http://en.wikipedia.org/wiki/Elementary_symmetric_polynomial&quot; rel=&quot;nofollow&quot;&gt;elementary symmetric functions&lt;/a&gt; in terms of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Power_sum_symmetric_polynomial&quot; rel=&quot;nofollow&quot;&gt;power symmetric functions&lt;/a&gt;. It can be deduced from &lt;a href=&quot;http://en.wikipedia.org/wiki/Newton%27s_identities&quot; rel=&quot;nofollow&quot;&gt;Newton's identities&lt;/a&gt; or the &lt;a href=&quot;http://qchu.wordpress.com/2009/06/24/gila-vi-the-cycle-index-polynomials-of-the-symmetric-groups/&quot; rel=&quot;nofollow&quot;&gt;exponential formula&lt;/a&gt;. You can find a more thorough exposition in Stanley's &lt;em&gt;Enumerative Combinatorics&lt;/em&gt;, Vol. II, Chapter 7. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Some exposition is in order regarding what this has to do with exterior powers. Suppose that $A$ is diagonalizable with eigenvalues $\lambda_1, ... \lambda_n$ and eigenvectors $e_1, ... e_n$. Then of course $\text{tr}(A) = \sum \lambda_i$. More generally $\text{tr}^{(k)}(A) = \sum_{i_1 &amp;lt; ... &amp;lt; i_k} \lambda_{i_1} ... \lambda_{i_k}$ is the $k^{th}$ elementary symmetric polynomial in the eigenvalues.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if $A$ acts on a vector space $V$, then the exterior power $\Lambda^k(V)$ inherits an action of $A$ by functoriality. In this basis, it is particularly easy to describe: it is given by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Lambda^k(A)(e_{i_1} \wedge ... \wedge e_{i_k}) = \lambda_{i_1} ... \lambda_{i_k} e_{i_1} \wedge ... \wedge e_{i_k}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence the wedges of all possible ordered $k$-tuples of distinct eigenvectors of $A$ forms an eigenbasis for $\Lambda^k(A)$ acting on $\Lambda^k(V)$. It follows that $\text{tr } \Lambda^k(A) = \text{tr}^{(k)}(A)$ for all diagonalizable $A$, hence for all $A$ by continuity. Of course you can ignore all of this if you want to, but I thought I would mention it because it is a precise sense in which the other coefficients of the characteristic polynomial encode &quot;higher traces.&quot; &lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2011-02-27T12:46:56.843" LastActivityDate="2011-02-27T12:46:56.843" CommentCount="12" />
  <row Id="23902" PostTypeId="1" CreationDate="2011-02-26T21:44:44.397" Score="10" ViewCount="268" Body="&lt;p&gt;I ask because, as a first-year calculus student, I am running into the fact that I didn't quite get this down when understanding the derivative:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, a derivative is the rate of change of a function with respect to changes in it's variable, this much I get.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thing is, definitions of 'differential' tend to be in the form of defining the derivative and calling the differential 'an infinitesimally small change in x', which is fine as far it goes, but then why bother even defining it formally outside of needing it for derivatives?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And THEN, the bloody differential starts showing up as a function in integrals, where it appears to be ignored part of the time, then functioning as a variable the rest.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why do I say 'practical'? Because when I asked for an explanation from other mathematician parties, I got one involving the graph of the function and how, given a right-angle triangle, a derivative is one of the other angles, where the differential is the line opposite the angle. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm sure that explanation is correct as far it goes, but it doesn't tell me what the differential DOES, or why it's useful, which are the two facts I need in order to really understand it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any assistance?&lt;/p&gt;&#xA;" OwnerUserId="7532" LastActivityDate="2011-02-26T23:04:06.097" Title="What is the practical difference between a differential and a derivative?" Tags="&lt;calculus&gt;&lt;differential&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="2" />
  <row Id="23903" PostTypeId="1" CreationDate="2011-02-26T21:49:57.467" Score="6" ViewCount="140" Body="&lt;p&gt;Is quantum logic producing interesting/different mathematics?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it different from the intuitionist approach to mathematics? How?&lt;/p&gt;&#xA;" OwnerUserId="4226" LastActivityDate="2011-03-06T11:38:12.040" Title="Is quantum logic producing interesting/different mathematics?" Tags="&lt;logic&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="23904" PostTypeId="2" ParentId="23899" CreationDate="2011-02-26T21:51:31.957" Score="2" ViewCount="" Body="&lt;p&gt;I thought I might add a bit on the background of what we are talking about.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider a vector space $V$ with a linear operator $A$, and lets choose a basis $e_i$.  Let $A\otimes A$ be the linear operator on $V\otimes V$ given by $A\otimes A (e_i\otimes e_j)=Ae_i\otimes Ae_j$.  It is easy to see that $tr(A\otimes A)=tr(A)^2$.&#xA;Let $\theta$ be the automorphism of $V\otimes V$ given by $\theta(e_i\otimes e_j)=e_j\otimes e_i$.  We can define $Sym^2(V)$, the Symmetric Square to be the set of all elements in $V\otimes V$ which are annihilated by $\theta -Id$, and $Alt^2(V)$, the Alternating Square, to be the set of all elements that are annihilated by $\theta + Id$.  Then, $V\otimes V$ decomposes into $Sym^2(V)\oplus Alt^2 (V)$ and what you are then looking for is the Trace of $A\otimes A$ when restricted to the alternating square.  (Note: restriction to an invariant subspace gives a linear operator, so we can take its Trace)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is this Trace the sum of $\lambda_i \lambda_j$?  First, the set of all $e_i\otimes e_j+e_j\otimes e_i$ is a basis for $Sym^2(V)$ and the set of all $e_i\otimes e_j - e_j\otimes e_i$ is a basis for $Alt^2(V)$.  To see this, write a vector in either subspace in the basis for $V\otimes V$ and consider the fact that it is annihilated by either $\theta +Id$ or $\theta -Id$.  Applying these operators gives and equation, and allows us to write the original vector in one of the bases I mentioned above.&lt;br&gt;&#xA;Now lets do the case where $A$ is diagonalizable, and the $e_i$ above form an eigenbasis.  Then $$A\otimes A (e_i\otimes e_j - e_j\otimes e_i)= \lambda_i\lambda_j(e_i\otimes e_j - e_j\otimes e_i)$$  Hence $\lambda_i\lambda_j$ are exactly the eigenvalues of $A\otimes A$ when restricted to $Alt^2(V)$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In more generality, the sum of the products of $n$ distinct eigenvalues will be the Trace of $A\otimes \cdots \otimes A$ restricted to the intersection of all subspaces annihilated by $\theta_\pi +Id$ where $\theta_\pi $ is some automorphism given by an odd permutation.  This subspace is sometimes called the &lt;a href=&quot;http://en.wikipedia.org/wiki/Exterior_algebra&quot; rel=&quot;nofollow&quot;&gt;Exterior Power&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, how do we actually calculate a formula for these Traces based on the original matrix $A$?  That is supplied in Qiaochu's Answer.&lt;/p&gt;&#xA;" OwnerUserId="6075" LastEditorUserId="6075" LastEditorDisplayName="" LastEditDate="2011-02-26T22:06:33.067" LastActivityDate="2011-02-26T22:06:33.067" CommentCount="5" />
  <row Id="23906" PostTypeId="2" ParentId="23879" CreationDate="2011-02-26T21:58:10.350" Score="3" ViewCount="" Body="&lt;p&gt;Any regular function on a projective variety (assumed irreducible) is constant. Recall that this is  a consequence of the valuative criterion for properness (among other things; alternatively, the image would be proper, and no non-finite subset of the affine line is proper). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So given any regular function $f$ on $X \times W$, it is constant on every fiber over $W$. Thus if $W \to X \times W$ is any section of the projection, the pull-back of $f$ to $W$ pulls back to $f$ via the projection $X \times W \to W$. Thus the map $\mathcal{O}(W) \to \mathcal{O}(X \times W)$ is surjective. Injectivity is clear because $X \times W \to W$ is surjective.&lt;/p&gt;&#xA;" OwnerUserId="536" LastActivityDate="2011-02-26T21:58:10.350" />
  <row Id="23907" PostTypeId="1" AcceptedAnswerId="23908" CreationDate="2011-02-26T21:59:36.990" Score="2" ViewCount="80" Body="&lt;p&gt;Hello,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fermat's little theorem says:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If $p$ is prime, and $a$ is an integer with $p \nmid a$, then $a^{p - 1 } \equiv 1 \pmod{p}$.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And this is a small part that I extracted from the book:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $n = 4k + r$ with $0 \leq r \leq 3$. Then by Fermat's little Theorem, we have&#xA;$$b^n \equiv b^{4k + r} \equiv (b^4)^kb^r \equiv 1^kb^r \equiv b^r\pmod{5} \text{for any integer b}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And I guess the author applied FLT for $b^4 \equiv 1 \pmod{5}$&#xA;But we don't know that $b$ divides 5 or not, how could the statement above be true? Any idea?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks,&lt;br&gt;&#xA;Chan&lt;/p&gt;&#xA;" OwnerUserId="5711" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-26T22:08:32.833" LastActivityDate="2011-02-26T22:11:19.550" Title="Question regarding Fermat's little theorem." Tags="&lt;elementary-number-theory&gt;&lt;modular-arithmetic&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="23908" PostTypeId="2" ParentId="23907" CreationDate="2011-02-26T22:11:19.550" Score="1" ViewCount="" Body="&lt;p&gt;The exact statement of what you wrote is slightly incorrect.  If $r=0$ and $b=5$ then it does not hold.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most likely the condition $\gcd (b,5)=1$ is missing.&lt;/p&gt;&#xA;" OwnerUserId="6075" LastActivityDate="2011-02-26T22:11:19.550" CommentCount="1" />
  <row Id="23909" PostTypeId="1" AcceptedAnswerId="23917" CreationDate="2011-02-26T22:11:27.973" Score="6" ViewCount="161" Body="&lt;p&gt;While helping a friend out for an exam (last year of high school), I found an exercise that neither of us could solve. I've tried a couple of different approaches but nothing seemed to work. Could anyone tell me how to solve it, or at least some hints? This is the exercise:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Knowing that&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$$\lim_{x \to a}\frac{x^2-\sqrt{a^3x}}{\sqrt{ax}-a}=12$$&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Find out $a$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We can assume that $a$ exists and is real.&lt;/p&gt;&#xA;" OwnerUserId="2757" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-26T22:42:48.840" LastActivityDate="2011-02-26T23:38:13.600" Title="Help solving a limit" Tags="&lt;limit&gt;" AnswerCount="4" />
  <row Id="23910" PostTypeId="2" ParentId="23909" CreationDate="2011-02-26T22:20:40.053" Score="9" ViewCount="" Body="&lt;p&gt;First note that $a&amp;gt;0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $a&amp;lt;0$, then $$\lim_{x \rightarrow a} \frac{x^2-\sqrt{a^3x}}{\sqrt{ax}-a} = \frac{a^2 - a^2}{-a-a} = 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $a=0$, then $\lim_{x \rightarrow a} \frac{x^2-\sqrt{a^3x}}{\sqrt{ax}-a}$ doesn't exist.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, $a&amp;gt;0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $\sqrt{ax}=y$ i.e. $x = \frac{y^2}{a}$. Note that as $x \rightarrow a$, $y \rightarrow a$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, $$\lim_{x \rightarrow a} \frac{x^2-\sqrt{a^3x}}{\sqrt{ax}-a} = \lim_{y \rightarrow a} \frac{y}{a^2} \frac{y^3-a^3}{y-a} = \frac{3a^2}{a} = 3a$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, $a=4$&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2011-02-26T22:37:15.647" LastActivityDate="2011-02-26T22:37:15.647" CommentCount="2" />
  <row Id="23911" PostTypeId="2" ParentId="23909" CreationDate="2011-02-26T22:23:14.817" Score="3" ViewCount="" Body="&lt;p&gt;Using &lt;a href=&quot;http://en.wikipedia.org/wiki/L%27H%C3%B4pital%27s_rule&quot; rel=&quot;nofollow&quot;&gt;L'hopital Rule&lt;/a&gt; we have $$ \lim_{x \to a} \frac{2x- \frac{a^3}{2 \sqrt{a^{3}x}}}{\frac{a}{2 \sqrt{ax}}} = 12$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Longrightarrow \frac{2a-\frac{a^3}{2a^2}}{\frac{1}{2}} = 12$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Longrightarrow 2\left(2a- \frac{a^3}{2a^2}\right)= 12$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So $a = 4$.&lt;/p&gt;&#xA;" OwnerUserId="2548" LastEditorUserId="2548" LastEditorDisplayName="" LastEditDate="2011-02-26T22:29:43.423" LastActivityDate="2011-02-26T22:29:43.423" CommentCount="2" />
  <row Id="23912" PostTypeId="2" ParentId="23909" CreationDate="2011-02-26T22:26:04.270" Score="3" ViewCount="" Body="&lt;p&gt;One way is application of L'hoipitals rule.  Since both denominator and numerator in $$\lim_{x \to a}\frac{x^2-\sqrt{a^3x}}{\sqrt{ax}-a}$$ &#xA;go to zero, by l'hopitals rule this is the same as &#xA;$$\lim_{x \to a}\frac{2x-a^{3/2}\frac{1}{2}x^{-1/2}}{\frac{\sqrt{a}}{2}x^{-1/2}}$$  Simplifying the expression we get that this equals &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\lim_{x \to a}\frac{4x^{3/2}-a^{3/2}}{\sqrt{a}}=\frac{3a^{3/2}}{\sqrt{a}}=3a$$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus choose $a=4$.&lt;/p&gt;&#xA;" OwnerUserId="6075" LastActivityDate="2011-02-26T22:26:04.270" CommentCount="1" />
  <row Id="23913" PostTypeId="2" ParentId="23902" CreationDate="2011-02-26T22:34:19.367" Score="5" ViewCount="" Body="&lt;p&gt;The distinction between the two concepts is not really clarified until you move up to higher dimensions and start doing multivariable calculus. Derivatives become &lt;a href=&quot;http://en.wikipedia.org/wiki/Jacobian_matrix_and_determinant&quot; rel=&quot;nofollow&quot;&gt;Jacobians&lt;/a&gt;, which are matrices giving linear approximations to smooth functions at a point, while differentials become &lt;a href=&quot;http://en.wikipedia.org/wiki/Differential_form&quot; rel=&quot;nofollow&quot;&gt;differential forms&lt;/a&gt;, which are things you integrate over higher-dimensional regions. (This is why we bother with differentials.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In one dimension it is hard to tell the difference between these two concepts because the Jacobian (derivative) $f&amp;#39;(x)$ looks an awful lot like the differential form $f&amp;#39;(x) \, dx$. But in higher dimensions the distinction becomes clearer: the relationship between the Jacobian and differential forms is given by the multivariate &lt;a href=&quot;http://en.wikipedia.org/wiki/Change_of_variables_formula#Substitution_for_multiple_variables&quot; rel=&quot;nofollow&quot;&gt;change of variables formula&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: to be completely precise, the problem is that in one dimension there is only one space of differential forms and the action of the Jacobian on it doesn't look special at all. In $n$ dimensions there are $n$ spaces of differential forms, and the Jacobian acts &lt;em&gt;differently&lt;/em&gt; on each of them. In particular, on the highest-dimension forms, it acts by the determinant, which is where the change of variables formula comes from. &lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2011-02-26T23:04:06.097" LastActivityDate="2011-02-26T23:04:06.097" />
  <row Id="23914" PostTypeId="2" ParentId="23902" CreationDate="2011-02-26T22:40:28.093" Score="5" ViewCount="" Body="&lt;p&gt;Originally, &quot;differentials&quot; and &quot;derivatives&quot; were intimately connected, with derivative being &lt;em&gt;defined&lt;/em&gt; as the ratio of the differential of the function by the differential of the variable (see my &lt;a href=&quot;http://math.stackexchange.com/questions/21199/dy-dx-is-not-a-ratio/21209#21209&quot;&gt;previous discussion on the Leibnitz notation for the derivative&lt;/a&gt;). Differentials were simply &quot;infinitesimal changes&quot; in &lt;code&gt;whatever&lt;/code&gt;, and the derivative of $y$ with respect to $x$ was the ratio of the infinitesimal change in $y$ relative to the infinitesimal change in $x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For integrals, &quot;differentials&quot; came in because, in Leibnitz's way of thinking about them, integrals were the sums of infinitely many infinitesimally thin rectangles that lay below the graph of the function. Each rectangle would have height $y$ and base $dx$ (the infinitesimal change in $x$), so the area of the rectangle would be $y\,dx$ (height times base), and we would add them all up as $S\; y\,dx$ to get the total area (the integral sign was originally an elongated $S$, for &quot;summa&quot;, or sum). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Infinitesimals, however, cause all sorts of headaches and problems. A lot of the reasoning about infinitesimals was, well, let's say not entirely rigorous (or logical); some differentials were dismissed as &quot;utterly inconsequential&quot;, while others were taken into account. For example, the product rule would be argued by saying that the change in $fg$ is given by &#xA;$$(f+df)(g+dg) -fg = fdg + gdf + df\,dg,$$ &#xA;and then ignoring $df\,dg$ as inconsequential, since it was made up of the product of two infinitesimals; but if infinitesimals that are really small can be ignored, why do we &lt;em&gt;not&lt;/em&gt; ignore the infinitesimal change $dg$ in the first factor?  Well, you can wave your hands a lot of huff and puff, but in the end the argument essentially broke down into nonsense, or the problem was ignored because things worked out regardless (most of the time, anyway).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, there was a need of a more solid understanding of just what derivatives and differentials actually are so that we can really reason about them; that's where limits came in. Derivatives are no longer ratios, instead they are limits. Integrals are no longer infinite sums of infinitesimally thin rectangles, now they are limits of Riemann sums (each of which is finite and there are no infinitesimals around), etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The notation is left over, though, because it is very useful notation and is very suggestive. In the integral case, for instance, the &quot;dx&quot; is no longer &lt;em&gt;really&lt;/em&gt; a quantity or function being multiplied: it's best to think of it as the &quot;closing parenthesis&quot; that goes with the &quot;opening parenthesis&quot; of the integral (that is, you are integrating whatever is between the $\int$ and the $dx$, just like when you have $2(84+3)$, you are multiplying by $2$ whatever is between the $($ and the $)$ ). But it is very useful, because for example it helps you keep track of what changes need to be made when you do a change of variable. One can justify the change of variable without appealing at all to &quot;differentials&quot; (whatever they may be), but the &lt;em&gt;notation&lt;/em&gt; just leads you through the necessary changes, so we treat them as if they were actual functions being multiplied by the integrand because they help keep us on the right track and keep us honest. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But here is an ill-kept secret: we mathematicians tend to be lazy. If we've already come up with a valid argument for situation A, we don't want to have to come up with a new valid argument for situation B if we can just explain how to get from B to A, even if solving B directly would be easier than solving A (old joke: a mathematician and an engineer are subjects of a psychology experiment; first they are shown into a room where there is an empty bucket, a trashcan, and a faucet. The trashcan is on fire. Each of them first fills the bucket with water from the faucet, then dumps it on the trashcan and extinguishes the flames. Then the engineer is shown to another room, where there is again a faucet, a trashcan on fire, and a bucket, but this time the bucket is already filled with water; the engineer takes the bucket, empties it on the trashcan and puts out the fire. The mathematican, later, comes in, sees the situation, takes the bucket, and empties it &lt;em&gt;on the floor&lt;/em&gt;, and then says &quot;which reduces it to a previously solved problem.&quot;) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where were we? Ah, yes. Having to translate all those informal manipulations that work so well and treat $dx$ and $dy$ as objects in and of themselves, into formal justifications that don't treat them that way is a real pain. It can be done, but it's a real pain. Instead, we want to come up with a way of justifying all those manipulations that will be valid always. One way of doing it is by actually giving them a &lt;em&gt;meaning&lt;/em&gt; in terms of the new notions of derivatives. And that is what is done.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically, we want the &quot;differential&quot; of $y$ to be the infinitesimal change in $y$; this change will be closely approximated to the change along the tangent to $y$; the tangent has slope $y&amp;#39;(a)$. But because we don't have infinitesimals, we have to say &lt;em&gt;how much&lt;/em&gt; we've changed the argument. So we define &quot;the differential in $y$ at $a$ when $x$ changes by $\Delta x$&quot;, $d(y,\Delta x)(a)$, as $d(y,\Delta x)(a) = y&amp;#39;(a)\Delta x$. This is exactly the change along the tangent, rather than along the graph of the function.  If you take the limit of $d(y,\Delta x)$ over $\Delta x$ as $\Delta x\to 0$, you just get $y&amp;#39;$. But we tend to think of the limit of $\Delta x\to 0$ as being $dx$, so abuse of notation leads to &quot;$dy = \frac{dy}{dx}\,dx$&quot;; this is &lt;em&gt;suggestive&lt;/em&gt;, but not quite true literally; instead, one then can show that arguments that treat differentials as functions tend to give the right answer under mild assumptions. Note that under this definition, you get $d(x,\Delta x) = 1\Delta x$, leading to $dx = dx$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, notice an interesting reversal: originally, differentials came first, and they were used to define the derivative as a ratio. Today, derivatives come first (defined as limits), and differentials are defined &lt;em&gt;in terms&lt;/em&gt; of the derivatives.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the practical difference, though? You'll probably be disappointed to hear &quot;not much&quot;. Except one thing: when your functions represent actual quantities, rather than just formal manipulation of symbols, the derivative and the differential &lt;strong&gt;measure different things.&lt;/strong&gt; The derivative measures a &lt;strong&gt;rate&lt;/strong&gt; of change, while the differential measures &lt;strong&gt;the change itself.&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the units of measurement are different: for example, if $y$ is distance and $x$ is time, then $\frac{dy}{dx}$ is measured in distance over time, i.e., velocity. But the differential $dy$ is measured in distance units of distance, because it represents the change in distance (and the difference/change between two distances is still a distance, not a velocity any more). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is it useful to have the distinction? Because sometimes you want to know how something &lt;em&gt;is changing&lt;/em&gt;, and sometimes you want to know how much something &lt;em&gt;changed&lt;/em&gt;. It's all nice and good to know the rate of inflation (change in prices over time), but you might sometimes want to know how much more the loaf of bread is now (rather than the rate at which the price is changing). And because being able to manipulate derivatives as if they were quotients can be very useful when dealing with integrals, differential equations, etc, and differentials give us a way of making sure that these manipulations don't lead us astray (as they sometimes did in the days of infinitesimals). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure if that answers your question or at least gives an indication of where the answers lie. I hope it does. &lt;em&gt;Added.&lt;/em&gt; I see Qiaochu has pointed out that the distinction becomes much clearer once you go to higher dimensions/multivariable calculus, so the above may all be a waste. Still...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added.&lt;/strong&gt; As Qiaochu points out (and I mentioned in passing elsewhere), there &lt;em&gt;are&lt;/em&gt; ways in which one can give &lt;em&gt;formal&lt;/em&gt; definitions and meanings to infinitesimals, in which case we &lt;em&gt;can&lt;/em&gt; define differentials as &quot;infinitesimal changes&quot; or &quot;changes along infinitesimal differences&quot;; and then use them to define derivatives as integrals just like Leibnitz did. The standard example of being able to do this is Robinson's &lt;a href=&quot;http://en.wikipedia.org/wiki/Non-standard_analysis&quot; rel=&quot;nofollow&quot;&gt;non-standard analysis&lt;/a&gt; Or if one is willing to forgo looking at &lt;em&gt;all&lt;/em&gt; kinds of functions and only at some restricted type of functions, then you can also give infinitesimals, differentials, and derivatives substance/meaning which is much closer to their original conception. &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-26T22:56:26.790" LastActivityDate="2011-02-26T22:56:26.790" CommentCount="1" />
  <row Id="23915" PostTypeId="1" AcceptedAnswerId="23916" CreationDate="2011-02-26T23:23:56.800" Score="1" ViewCount="137" Body="&lt;p&gt;Hello,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reduced residue system modulo $10$ is: $1, 3, 7, 9$&lt;br&gt;&#xA;But how could we find these numbers?&lt;br&gt;&#xA;The only thing I know is they're relatively prime to $10$. &#xA;What does it mean by &quot;no two different elements of the set are congruent to modulo m&quot;?   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks,&lt;br&gt;&#xA;Chan&lt;/p&gt;&#xA;" OwnerUserId="5711" LastActivityDate="2011-02-26T23:52:26.347" Title="How to find a reduced residue system modulo of a number?" Tags="&lt;elementary-number-theory&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="23916" PostTypeId="2" ParentId="23915" CreationDate="2011-02-26T23:32:03.010" Score="2" ViewCount="" Body="&lt;p&gt;The line &quot;no two different elements of the set are congruent modulo $m$&quot; just means that all of your elements are distinct modulo $m$.  For example, $1,3,7,9,11,111,1111$ are all relatively prime to $10$, but they do &lt;strong&gt;not&lt;/strong&gt;  form a reduced residue system since $1,11,111,1111$ are all the same modulo $10$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another way to specify the condition is:  The reduced residue system modulo $N$ is the set of all integers $m$ with $\gcd(m,N)=1$ and $0\leq m\leq N$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope that helps,&lt;/p&gt;&#xA;" OwnerUserId="6075" LastActivityDate="2011-02-26T23:32:03.010" CommentCount="1" />
  <row Id="23917" PostTypeId="2" ParentId="23909" CreationDate="2011-02-26T23:38:13.600" Score="5" ViewCount="" Body="&lt;p&gt;Besides L'Hopital one can simply rationalize the denominator, after discarding case $\rm\:a \le 0\:,\:$ viz:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\rm \frac{x^2-a\sqrt{ax}}{\sqrt{ax}-a}\ =\ \frac{x^2-a\sqrt{ax}}{\sqrt{ax}-a} \ \frac{\sqrt{ax}+a}{\sqrt{ax}+a}\ =\ \frac{ax\:(x-a)+\sqrt{ax}\ (x^2-a^2) }{a\:(x-a) }\ =\ x+(x+a)\sqrt{\frac{x}{a}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the above $\rm\to 3\ a\ $ as $\rm\ x\to a\ $ the problem reduces to solving $\rm\ 3\ a = 12\:$.&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2011-02-26T23:38:13.600" CommentCount="4" />
  <row Id="23918" PostTypeId="1" AcceptedAnswerId="23919" CreationDate="2011-02-26T23:40:37.887" Score="1" ViewCount="128" Body="&lt;p&gt;Hello,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm thinking of modulo, but really don't know how to start? A hint would be sufficient.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks,&lt;br&gt;&#xA;Chan&lt;/p&gt;&#xA;" OwnerUserId="5711" LastActivityDate="2011-02-26T23:44:01.097" Title="How to find the last digit of $3^{1000}$?" Tags="&lt;elementary-number-theory&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="23919" PostTypeId="2" ParentId="23918" CreationDate="2011-02-26T23:43:21.113" Score="2" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\rm\ \ mod\:\ 10\::\ \ 3^2 \equiv -1\ \Rightarrow\ 3^4 \equiv 1\ $ so you need only consider the exponent $\rm\ (mod\ 4)$&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2011-02-26T23:43:21.113" CommentCount="1" />
  <row Id="23920" PostTypeId="2" ParentId="23918" CreationDate="2011-02-26T23:44:01.097" Score="6" ViewCount="" Body="&lt;p&gt;Lets look at the last digits of the first few powers:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The last digit of $3^0$ is $1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The last digit of $3$ is $3$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The last digit of $3^2$ is $9$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The last digit of $3^3$ is $7$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The last digit of $3^4$ is $1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The last digit of $3^5$ is $3$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The last digit of $3^6$ is $9$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice a pattern?  Why does this pattern exist?  What is going on when I multiply by three?  Based on this we could guess that it has a period of $4$ so that $3^{4n}\equiv 1$.&lt;br&gt;&#xA;Use this to find the last digit of $3^{1000}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Do you know modular arithmetic?  If so it is a lot easier)&lt;/p&gt;&#xA;" OwnerUserId="6075" LastActivityDate="2011-02-26T23:44:01.097" />
  <row Id="23921" PostTypeId="2" ParentId="23915" CreationDate="2011-02-26T23:52:26.347" Score="0" ViewCount="" Body="&lt;p&gt;Presumably you refer to the Wikipedia definition of &lt;a href=&quot;http://en.wikipedia.org/wiki/Reduced_residue_system&quot; rel=&quot;nofollow&quot;&gt;reduced residue system&lt;/a&gt;. The point of the definition is to specify a system of representatives for the $\rm\:phi(n)\:$ congruence classes that are units (invertible) $\rm\ (mod\ n)\:.\:$ This amounts to choosing a set  of $\rm\:\phi(n)\:$ integers coprime to $\rm\:n\:$ such they they are all distinct $\rm\ (mod\ m)\:$.&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2011-02-26T23:52:26.347" CommentCount="1" />
  <row Id="23922" PostTypeId="2" ParentId="23899" CreationDate="2011-02-26T23:58:57.387" Score="0" ViewCount="" Body="&lt;p&gt;We want a &quot;simple&quot; formula for the coefficients of the characteristic polynomial in terms of the entries of the matrix, at least for the top few coefficients.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The characteristic polynomial can be written in terms of the eigenvalues:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;\chi(A) = (x-\lambda_1)(x-\lambda_2)\cdots(x-\lambda_n)&#xD;&#xA;$$&#xA;$$= x^n - (\lambda_1 + \dots + \lambda_n) x^{(n-1)} + (\lambda_1 \lambda_2 + \lambda_1 \lambda_3 + \dots + \lambda_{n-1} \lambda_n) x^{(n-2)} - \dots + (-1)^n \lambda_1 \lambda_2 \cdots \lambda_n&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each coefficient tr&lt;sup&gt;(k)&lt;/sup&gt; is just (±) the sum of the product of all k-sets of the eigenvalues λ&lt;sub&gt;i&lt;/sub&gt;.  We'd like to use the regular trace to write this sum down, and so we try to find some crazy matrix Λ&lt;sup&gt;(k)&lt;/sup&gt;(A) whose eigenvalues are exactly those products!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For k=1 this is easy: just use A itself.  For k=0, this is degenerate: just use [1].  For k=n, there is a sneaky trick: use [ det(A) ].  That is basically cheating, but it turns out to be part of a larger pattern.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To define B = Λ&lt;sup&gt;(k)&lt;/sup&gt;(A) we consider all k-subsets K of {1,2,…,n} and use those to index the rows and columns of B.  For two such k-subsets K,L define the (K,L)th entry of B to be the determinant of the submatrix of A formed by taking only the rows from K and the columns from L.  In other words, B is the matrix formed of the &lt;b&gt;k-minors&lt;/b&gt; of A.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance if K={i,j} and L={i,j}, then the (K,L)th entry is A&lt;sub&gt;ii&lt;/sub&gt;A&lt;sub&gt;jj&lt;/sub&gt;−A&lt;sub&gt;ij&lt;/sub&gt;A&lt;sub&gt;ji&lt;/sub&gt;.  The trace of this matrix is just the sum of the (K,K)th entries as K varies over all k-subsets of {1,2,…,n}.  For k=2, this is exactly the first formula given:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\operatorname{tr}^{(2)}(A) = tr(B) = \sum_{1 \leq i &amp;lt; j \leq n} A_{ii}A_{jj}-A_{ij}A_{ji}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now for k=3, the formula is a little bit awful, but primarily because the formula for a 3×3 determinant is a little awful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\operatorname{tr}^{(3)}(A) = tr(B) = \sum_{1\leq i&amp;lt;j&amp;lt;k \leq n} \left|\begin{array}{rrr}A_{ii} &amp;amp; A_{ij} &amp;amp; A_{ik} \\ A_{ji} &amp;amp; A_{jj} &amp;amp; A_{jk} \\ A_{ki} &amp;amp; A_{kj} &amp;amp; A_{kk} \end{array}\right|$$&#xA;$$= \sum_{1 \leq i &amp;lt; j &amp;lt; k \leq n} A_{ii}A_{jj}A_{kk}+A_{ij}A_{jk}A_{ki}+A_{ik}A_{ji}A_{kj}-A_{ik}A_{jj}A_{ki}-A_{ij}A_{ji}A_{kk}-A_{ii}A_{jk}A_{kj}$$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Now relating Eric Naslund's answer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Kronecker product of two matrices, M⊗N is another related construction.  Its rows are indexed by pairs (i1,i2) of row indices (i1 from M, and i2 from N), its columns are indexed by pairs (j1,j2) of column indices (j1 from M, and j2 from N), and the ((i1,i2),(j1,j2))th entry is just the the product of the (i1,j1)st entry of the first matrix with (i2,j2)nd entry of the second matrix.  In other words, one gets a block matrix where one the first matrix decides a common multiplier on the (i1,j1)st block, and the second matrix simply provides that block.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{pmatrix} 0 &amp;amp; 1 \\ -1 &amp;amp; 0 \end{pmatrix} \otimes \begin{pmatrix} 5 &amp;amp; 6 &amp;amp; 7 \\ 8 &amp;amp; 9 &amp;amp; 10 \\ 11 &amp;amp; 12 &amp;amp; 13 \end{pmatrix} =&#xD;&#xA;\left(\begin{array}{rrr|rrr}0&amp;amp;0&amp;amp;0&amp;amp;5&amp;amp;6&amp;amp;7\\0&amp;amp;0&amp;amp;0&amp;amp;8&amp;amp;9&amp;amp;10\\0&amp;amp;0&amp;amp;0&amp;amp;11&amp;amp;12&amp;amp;13\\\hline&#xD;&#xA;-5 &amp;amp; -6 &amp;amp; -7 &amp;amp; 0&amp;amp;0&amp;amp;0 \\  -8 &amp;amp; -9 &amp;amp; -10 &amp;amp; 0&amp;amp;0&amp;amp;0 \\ -11&amp;amp;-12&amp;amp;-13&amp;amp;0&amp;amp;0&amp;amp;0 \end{array}\right)&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Kronecker product has the nice feature that its eigenvalues are the products of eigenvalues, one from the first matrix and one from the second.  In particular, for A⊗A, one gets &lt;em&gt;all&lt;/em&gt; products of eigenvalues, even λ&lt;sub&gt;1&lt;/sub&gt;λ&lt;sub&gt;1&lt;/sub&gt; and both λ&lt;sub&gt;1&lt;/sub&gt;λ&lt;sub&gt;2&lt;/sub&gt; and λ&lt;sub&gt;2&lt;/sub&gt;λ&lt;sub&gt;1&lt;/sub&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Kronecker product of the two matrices acts on V⊗W, the vector space whose basis consists of pairs of basis vectors, one from V and one from W, and where M acts on V and N acts on W.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This vector space V⊗V is too large, it gives not only the products of 2-sets of eigenvalues, but also any ordered 2-tuple, including (1,1) and both (1,2) and (2,1).  However, there is a nice subspace called the &lt;strong&gt;exterior power&lt;/strong&gt; which is the intersection of the kernels of all those coordinate switchers mentioned by Eric Naslund.  The matrix of A acting on the k'th exterior power Λ&lt;sup&gt;(k)&lt;/sup&gt;(V) is just the matrix of k-minors, Λ&lt;sup&gt;(k)&lt;/sup&gt;(A).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The eigenvalues divide up nicely in the second tensor power: the subspace spanned by x⊗x is called the symmetric power Sym&lt;sup&gt;2&lt;/sup&gt;(V) and it gets one copy of each of the products λ&lt;sub&gt;i&lt;/sub&gt;λ&lt;sub&gt;j&lt;/sub&gt; for i ≤ j, including all of the repeats λ&lt;sub&gt;i&lt;/sub&gt;λ&lt;sub&gt;i&lt;/sub&gt;, so it has dimension n(n+1)/2.  The space spanned by all x⊗y−y⊗x is called the exterior power Alt&lt;sup&gt;2&lt;/sup&gt;(V) and it gets the leftover eigenvalues, one copy each of the λ&lt;sub&gt;i&lt;/sub&gt;λ&lt;sub&gt;j&lt;/sub&gt; for i &amp;lt; j, so it has dimension n(n-1)/2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For higher tensor powers the eigenvalues still have to be divided up, and one still has spaces Sym&lt;sup&gt;k&lt;/sup&gt;(V) and Alt&lt;sup&gt;k&lt;/sup&gt;(V)=Λ&lt;sup&gt;(k)&lt;/sup&gt;(V), and these get their corresponding eigenvalues: all products of non-decreasing k-tuples of eigenvalues λ&lt;sub&gt;i&lt;/sub&gt;⋯λ&lt;sub&gt;j&lt;/sub&gt; with 1≤i≤…≤j≤n for Sym&lt;sup&gt;k&lt;/sup&gt;(V) and all products of increasing k-tuples (or just plain k-sets) of eigenvalues for Alt&lt;sup&gt;k&lt;/sup&gt;(V).  Weirdly though, that doesn't add up to the total dimension, and there still other subspaces left!  These are called Schur functors and apparently don't have too direct a relation on this question, but you can see they have to exist by looking at the third tensor power of a three dimensional space (a vector space of dimension 3×3×3=27): Sym3 has (1,1,1), (1,1,2), (1,2,2), (1,2,3), (1,3,3), (2,2,2), (2,2,3), (2,3,3), (3,3,3) for a total of 9.  Alt3 has only (1,2,3) for a total of 1.  Where did the other 17 dimensions go?  For instance, who got λ&lt;sub&gt;3&lt;/sub&gt;λ&lt;sub&gt;2&lt;/sub&gt;λ&lt;sub&gt;1&lt;/sub&gt;?&lt;/p&gt;&#xA;" OwnerUserId="583" LastEditorUserId="583" LastEditorDisplayName="" LastEditDate="2011-02-27T01:08:10.447" LastActivityDate="2011-02-27T01:08:10.447" />
  <row Id="23923" PostTypeId="1" AcceptedAnswerId="23926" CreationDate="2011-02-27T00:07:51.297" Score="0" ViewCount="78" Body="&lt;p&gt;I have a problem that is has the notation &lt;code&gt;lg^2 n&lt;/code&gt; and I just want to verify that it actually means / is the same as &lt;code&gt;(lg n)^2&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it is not the same please tell me how to evaluate &lt;code&gt;lg^2 n&lt;/code&gt; (always assuming &lt;code&gt;lg&lt;/code&gt; is base 2).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I do not have a source to quote from, I was just given a bunch of algorithms to sort out.  Sorting them isn't a problem for me as long as I know what they are saying :)  I retyped the notation exactly how it looks and provided it in the picture below.  I have included some other formulas too so you can compare the notations.  I imagine it was written as &lt;code&gt;lg^2 n&lt;/code&gt; just so he didn't have to use parenthesis, but I wanted to verify.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/k6Irf.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="6096" LastEditorUserId="6096" LastEditorDisplayName="" LastEditDate="2011-02-27T00:50:22.873" LastActivityDate="2011-02-27T00:50:22.873" Title="Are the notations lg^2 n and (lg n)^2 the same thing?" Tags="&lt;calculus&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="23924" PostTypeId="1" AcceptedAnswerId="23928" CreationDate="2011-02-27T00:13:49.293" Score="1" ViewCount="79" Body="&lt;p&gt;Hello,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Show that if $c_1, c_2,\ldots, c_{\phi(m)}$ is a reduced residue system modulo $m$, $m \neq 2$, and $m$ is a positive integer, then $c_1 +\cdots+ c_{\phi(m)} \equiv 0 \pmod{m}$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;From the problem statement, I only know that $\gcd(c_i, m ) = 1$.&lt;br&gt;&#xA;Is there any related theorem that I missed?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A hint would be greatly appreciated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks,&lt;br&gt;&#xA;Chan&lt;/p&gt;&#xA;" OwnerUserId="5711" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-27T00:45:22.880" LastActivityDate="2011-02-27T01:30:42.467" Title="Show that if $c_1, c_2, \ldots, c_{\phi(m)}$ is a reduced residue system modulo m, $m \neq 2$ then $c_1 + \cdots+ c_{\phi(m)} \equiv 0 \pmod{m}$" Tags="&lt;elementary-number-theory&gt;" AnswerCount="2" />
  <row Id="23925" PostTypeId="2" ParentId="23924" CreationDate="2011-02-27T00:40:49.137" Score="2" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt;: If $c_i$ is a reduced residue class, then so is $m-c_i$. (Why?) and $\phi(m)$ is even $\forall m &amp;gt;2$&lt;/p&gt;&#xA;" OwnerUserId="3302" LastActivityDate="2011-02-27T00:40:49.137" CommentCount="1" />
  <row Id="23926" PostTypeId="2" ParentId="23923" CreationDate="2011-02-27T00:48:24.533" Score="3" ViewCount="" Body="&lt;p&gt;It depends on the source. Usually, the notation $f^n(x)$ to mean &quot;$(f(x))^n$&quot; is restricted to trigonometric and hyperbolic functions, and for other functions $f^n(x)$ means $$\underbrace{f\circ\dots\circ f}_{\mbox{$n$ times}}(x).$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you need to know what convention your source is using. It could mean either $\mathrm{lg}(\mathrm{lg}(n))$, following the usual conventions, or $(\mathrm{lg}(n))^2$, if the author is using the notation as in the case of trigonometric functions.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2011-02-27T00:48:24.533" CommentCount="1" />
  <row Id="23927" PostTypeId="1" CreationDate="2011-02-27T00:54:05.213" Score="6" ViewCount="139" Body="&lt;p&gt;Is there an explicit well-ordering of $\mathbb{N}^{\mathbb{N}}:=\{g:\mathbb{N}\rightarrow \mathbb{N}\}$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've been thinking about that for awhile but nothing is coming to my mind. My best idea is this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Denote by $&amp;lt;$ the usual &quot;less than&quot; relation on $\mathbb{N}$. Since $\mathbb{N}^{\mathbb{N}}$ is the set of infinite&#xA;sequences ${\{x_{n}\}}_{n\in \mathbb{N}}$ with $x_{n}\in \mathbb{N}$,&#xA;we can define ${\{x_{n}\}}_{n\in \mathbb{N}}\leq ^{\prime }{\{y_{n}\}}_{n\in&#xD;&#xA;\mathbb{N}}$ as follows. If $x_{0}&amp;lt;y_{0}$, then ${\{x_{n}\}}_{n\in&#xD;&#xA;\mathbb{N}}\leq ^{\prime }{\{y_{n}\}}_{n\in \mathbb{N}}$. If $x_{k-1}=y_{k-1}$,&#xA;for $k\in \mathbb{N}\setminus \{0\}$, then ${\{x_{n}\}}_{n\in \mathbb{N}}\leq&#xD;&#xA;^{\prime }{\{y_{n}\}}_{n\in \mathbb{N}}$ if and only if $x_{k}&amp;lt;y_{k}$.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I think that under this relation not every subset of $\mathbb{N}^{\mathbb{N}}$ has a least element.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any ideas?&lt;/p&gt;&#xA;" OwnerUserId="5956" LastActivityDate="2011-02-27T01:24:53.273" Title="Explicit well-ordering of $\mathbb{N}^{\mathbb{N}}$." Tags="&lt;set-theory&gt;" AnswerCount="2" CommentCount="11" />
  <row Id="23928" PostTypeId="2" ParentId="23924" CreationDate="2011-02-27T00:59:15.460" Score="2" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\ $ It's a special case of Wilson's theorem for groups - see &lt;a href=&quot;http://math.stackexchange.com/questions/9311/product-of-all-elements-in-an-odd-finite-abelian-group-is-1/9345#9345&quot;&gt;my answer here&lt;/a&gt; - which highlights the key role played by symmetry (here a negation reflection / involution). Namely, since your set is closed under negation, its non-fixed points $\rm -k\not\equiv k\:$ pair up and contribute zero to the sum, leaving only the sum of the fixed points $\rm - k\equiv k\ \iff\ 2\:k\equiv 0\:,\ $ so $\rm\ k\equiv 0\ $ if $\rm\: m\:$ is odd, else $\rm\ k \equiv 0,\ m/2\:$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: this was my answer to a &lt;a href=&quot;http://math.stackexchange.com/questions/9877/prove-the-sum-of-all-numbers-that-do-not-have-a-multiplicative-inverse-mod-n/9903#9903&quot;&gt;similar question.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2011-02-27T01:30:42.467" LastActivityDate="2011-02-27T01:30:42.467" CommentCount="5" />
  <row Id="23929" PostTypeId="2" ParentId="23927" CreationDate="2011-02-27T00:59:21.527" Score="4" ViewCount="" Body="&lt;p&gt;If you had a well-ordering of $\mathbb N^{\mathbb N}$ it wouldn't be too hard to construct a well-ordering of $\mathbb R$ from that. However, it is believed that there is no explicit well-ordering of $\mathbb R$, so I'm afraid there won't be one for $\mathbb N^{\mathbb N}$ either. JDH is the expert on this!&lt;/p&gt;&#xA;" OwnerUserId="2941" LastEditorUserId="2941" LastEditorDisplayName="" LastEditDate="2011-02-27T01:20:04.783" LastActivityDate="2011-02-27T01:20:04.783" CommentCount="1" />
  <row Id="23930" PostTypeId="1" AcceptedAnswerId="23960" CreationDate="2011-02-27T01:08:02.040" Score="3" ViewCount="69" Body="&lt;p&gt;consider the simple function &#xA;$$f(x) = \left\{&#xA;     \begin{array}{ll}&#xA;       1 &amp;amp; \mbox{if $x \in [0,1]$}\\&#xA;       10 &amp;amp; \mbox{if $x \in \mathbb{Q} \cap [1,2]$}\\&#xA;      -10 &amp;amp; \mbox{if  $x \in \mathbb{Q^c} \cap [1,2]$}\\&#xA;       2 &amp;amp; \mbox{if  $x \in [2,3]$}\&#xA; \end{array} \right.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is it correct to say that f is continuous on $[0,1] \cup [2,3]$?&lt;/p&gt;&#xA;" OwnerUserId="6556" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-27T04:28:08.787" LastActivityDate="2011-02-27T05:21:42.327" Title="continuity question" Tags="&lt;calculus&gt;" AnswerCount="4" />
  <row Id="23933" PostTypeId="2" ParentId="23930" CreationDate="2011-02-27T01:19:00.620" Score="2" ViewCount="" Body="&lt;p&gt;Hint: Your function is not well defined at 1 and 2, as those $x$ values each meet two of your criteria.  Having fixed that, say by making the interval in the second line $(1,2)$, what happens near 1 and 2?&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-02-27T01:19:00.620" />
  <row Id="23934" PostTypeId="2" ParentId="23930" CreationDate="2011-02-27T01:19:28.113" Score="2" ViewCount="" Body="&lt;p&gt;This function is not well-defined at some points. For example $f(1) = 1$ since $1 \in [0,1]$, but also $f(1) = 10$ because $1 \in \mathbb{Q} \cap [1,2]$ as well (similarly for $x=2$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you correct this, note that if you define $f(1)=1$, then arbitrarily close to $1$ you can find numbers that are rational (or irrational) and greater than one. Here the function evaluates to $10$ (or $-10$), and so it will not be continuous at $1$ (nor at $2$ for the same reason).&lt;/p&gt;&#xA;" OwnerUserId="5105" LastActivityDate="2011-02-27T01:19:28.113" />
  <row Id="23935" PostTypeId="2" ParentId="23927" CreationDate="2011-02-27T01:23:51.457" Score="0" ViewCount="" Body="&lt;p&gt;What is the least element of $\mathbb{N}^{\mathbb{N}} \setminus (0,0,0,\dots)$?  I don't think one is defined.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-02-27T01:23:51.457" CommentCount="2" />
  <row Id="23936" PostTypeId="1" AcceptedAnswerId="23939" CreationDate="2011-02-27T01:25:31.277" Score="1" ViewCount="92" Body="&lt;p&gt;I read a problem to determine the integral $\int_1^{100}x^xdx$ with error at most 5% from the book &quot;Which way did the bicycle go&quot;. I was a bit disappointed to read the solution which used computer or calculator. I was wondering whether there is a solution to the problem which does not use computers or calculators. In particular, is there way to prove that the solution given in the book has a mistake because it claims that&#xA;$$\frac{99^{99}-1}{1+\ln 99}+\frac{100^{100}-99^{99}}{1+\ln 100}\leq \int_1^{100}x^xdx$$&#xA;gives a bound $1.78408\cdot 10^{199}\leq \int_1^{100}x^xdx$ but I think the LHS should be $1.78407\cdot 10^{199}\leq \int_1^{100}x^xdx$? I checked this by Sage and Wolfram Alpha but I was unable to do it by pen and paper.&lt;/p&gt;&#xA;" OwnerUserId="7537" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2011-02-27T01:37:45.487" LastActivityDate="2011-02-27T17:10:54.090" Title="Numerical approximation of an integral" Tags="&lt;numerical-methods&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="23937" PostTypeId="1" AcceptedAnswerId="23941" CreationDate="2011-02-27T01:29:08.683" Score="4" ViewCount="139" Body="&lt;p&gt;I can't figure this problem out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $b_1, b_2, \cdots, b_{\phi(m)}$ be the integers between $1$ and $m$ that are relatively prime to $m$. $B$ is the product of these integers. Prove that either $B$ is congruent to $1$ or $B$ is congruent to $-1 \pmod{m}$. Could you give me some hint where to start?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Please avoid use of group theory. I am an undergrad in my first semester of number theory.&lt;/p&gt;&#xA;" OwnerUserId="7435" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-27T04:12:30.733" LastActivityDate="2011-02-27T19:18:52.770" Title="Prove that the product of coprimes must be congruent to $-1$ or $1 \pmod{m}$" Tags="&lt;elementary-number-theory&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="0" />
  <row Id="23938" PostTypeId="1" CreationDate="2011-02-27T01:31:18.370" Score="2" ViewCount="119" Body="&lt;p&gt;Hi, I realize that this is not a typical programing question but its still related. If anyone could help me out I would really appreciate it because I have a midterm coming up and this is the part that I don't understand. This is not a homework problem so don't worry about me trying to get out of my work. I just need someone to explain how to do this is normal plain english instead of whatever my professor is using.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $p(n) = \sum_{i=0}^d a_i n^i$ where $a_i,d &amp;gt; 0$ be a polynomial in $n$ of degree $d$. Use the definitions of the asymptotic notations to prove the following properties:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a) If $k \geq d$, then $p(n) = O(n^k)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are also 4 more correspoding to the Omega, theta small o and small omega properties but if I could get an idea on how to start I can figure the other ones out on my own. Thanks so Much!&lt;/p&gt;&#xA;" OwnerUserId="7538" LastEditorUserId="2548" LastEditorDisplayName="" LastEditDate="2011-02-27T02:14:17.957" LastActivityDate="2011-02-27T14:17:03.000" Title="Big-O Notation and Asymptotics" Tags="&lt;asymptotics&gt;" AnswerCount="2" />
  <row Id="23939" PostTypeId="2" ParentId="23936" CreationDate="2011-02-27T01:44:01.017" Score="0" ViewCount="" Body="&lt;p&gt;$x^x$ grows really fast.  Notice &#xA;$$n^n&amp;gt;\frac{1}{n}\sum_{i=1}^{n-1} i^i.$$&lt;br&gt;&#xA;In short $100^{100}$ is a &lt;em&gt;lot&lt;/em&gt; bigger then $99^{99}$, so $$\frac{99^{99}-1}{1+\ln99}+\frac{100^{100}-99^{99}}{1+ln100}\sim \frac{100^{100}}{1+\ln100}=10^{99}\frac{10}{1+\ln100}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By calculator $$\frac{10}{1+\ln100}=1.78407,$$ (notice the same as above) but we can get bascially that by hand.  First show $4&amp;lt;\ln100&amp;lt;5$.  Then $$1.6=\frac{10}{6}\leq  \frac{10}{1+\ln100}\leq \frac{10}{5}=2.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But lets do better.  Use the fact that $$\sum_{i=1}^n\frac{1}{i}=\ln n +\gamma +E$$ where $|E|\leq \frac{1}{2n}$, and $n\geq 4$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then $$\ln(100)=1+\frac{1}{2}+\frac{1}{3}+\cdots +\frac{1}{100}-0.577+E$$ and diligently adding this together (under 5 minutes) gives $\ln(100)\sim 4.6$.  From here we get the coefficient $\frac{25}{14}$ with an error bounded by $\frac{2}{100}$ or 2%.  (We gained error because I cut off 4.61 to 4.6)&lt;/p&gt;&#xA;" OwnerUserId="6075" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2011-02-27T17:10:54.090" LastActivityDate="2011-02-27T17:10:54.090" CommentCount="3" />
  <row Id="23940" PostTypeId="2" ParentId="23937" CreationDate="2011-02-27T01:49:28.373" Score="2" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\ $  It's a special case of Wilson's theorem for groups, i.e. exploit the innate inversion symmetry to pair up inverses - see my &lt;a href=&quot;http://math.stackexchange.com/questions/23924/show-that-if-c-1-c-2-ldots-c-phim-is-a-reduced-residue-system-modulo-m/23928#23928&quot;&gt;answer here.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2011-02-27T01:49:28.373" />
  <row Id="23941" PostTypeId="2" ParentId="23937" CreationDate="2011-02-27T01:49:38.447" Score="2" ViewCount="" Body="&lt;p&gt;The numbers $b_i$ you have are precisely the invertible elements modulo $m$, then you can pair them up with their inverses, which are also among the $b_i$. I think that this may help you figure out the rest. Try some numerical examples first so that you can see what's actually going on.&lt;/p&gt;&#xA;" OwnerUserId="900" LastActivityDate="2011-02-27T01:49:38.447" CommentCount="10" />
  <row Id="23942" PostTypeId="2" ParentId="23937" CreationDate="2011-02-27T02:00:39.747" Score="6" ViewCount="" Body="&lt;p&gt;The result in question is a generalization of Wilson's Theorem which was established by Gauss in his &lt;em&gt;Disquisitiones Arithmeticae&lt;/em&gt;.  &lt;a href=&quot;http://arxiv.org/PS_cache/arxiv/pdf/0711/0711.3879v1.pdf&quot; rel=&quot;nofollow&quot;&gt;This paper&lt;/a&gt; of C.S. Dalawat gives a nice treatment of this question and its number field analogues, suitable for students having taken a basic course in graduate level algebraic number theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(In the section on the Chinese Remainder Theorem in my commutative algebra notes, I pose this result as a problem and suggest that the reader try to prove a function field analogue of Dalawat's results.  It seems like this should be rather straightforward, but I haven't actually tried it myself...)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Added&lt;/b&gt;: I decided that the group-theoretic version of Wilson's Theorem should appear in my &lt;a href=&quot;http://www.math.uga.edu/~pete/4400algebra2point5.pdf&quot; rel=&quot;nofollow&quot;&gt;handout on commutative groups&lt;/a&gt;, which is an algebraic supplement to notes for an advanced undergraduate number theory course I have taught twice at UGA.  (It may help to explain that the prerequisite for this course is one semester of abstract algebra, but the standard first semester algebra course at UGA does not include any treatment of groups whatsoever!)  So I wrote it up and included as an application the proof of Gauss's Theorem that the OP is asking about.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I realize that by using a little group theory I am not meeting the request of the OP.  Sorry about that.  After thinking about it a little more, it is not so obvious to me how to remove the group theory from the argument!  One could look in Gauss's &lt;em&gt;Disquisitiones Arithmeticae&lt;/em&gt;...but I don't recommend that.  Those who have read this book closely tell me that it shows that Gauss most definitely knew about finite commutative groups when he wrote it -- he just doesn't use any explicitly group-theoretic language.  This makes some of his arguments rather difficult to read.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then I got curious and searched on MathSciNet for the sources for this folkloric &quot;Wilson's Theorem in a Finite Commutative Group&quot;.  My search began very auspiciously: &lt;a href=&quot;http://www.math.uga.edu/~pete/Miller03.pdf&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; is a 1903 paper by the early American group theorist G.A. Miller which proves the result.  In fact, comparing my two-page writeup to Miller's two-and-a-half-page paper I found the resemblance to be uncanny.  The only significant difference is that he does not mention Gauss at all (nor in fact does he reference anything that does not appear in an American journal; perhaps he was making a point by doing so).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What about after 1903?  The result appears in the middle of a long, strange 1958 &lt;em&gt;Monthly&lt;/em&gt; article of Vandiver and Weaver.  (Check out MathReview MR0105386 for a bemused and somewhat critical review; my reaction was similar.)  And then it appears (though it is not stated as a theorem or given a name) in the first paragraph of Dalawat's 2009 paper linked to above.  &lt;a href=&quot;http://www.isibang.ac.in/~sury/genwilsonrev.pdf&quot; rel=&quot;nofollow&quot;&gt;This note&lt;/a&gt; of B. Sury gives an interesting partial generalization to finite &lt;em&gt;noncommutative&lt;/em&gt; groups (in which case one must consider the issue of all possible ways to take the product, making the problem significantly more complicated).  Also Bill Dubuque has written informally on this subject several times in recent years.  And that's all I found!&lt;/p&gt;&#xA;" OwnerUserId="299" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2011-02-27T19:18:52.770" LastActivityDate="2011-02-27T19:18:52.770" CommentCount="5" />
  <row Id="23943" PostTypeId="2" ParentId="23938" CreationDate="2011-02-27T02:25:52.237" Score="1" ViewCount="" Body="&lt;p&gt;For (a) we want to show that $p(n) = O(n^k)$ for $k \geq d$. This is the same thing as showing that there exists a positive number $M$ and a real number $n_0$ such that $|p(n)| \leq M|n^k|$ for all $n&amp;gt;n_0$. So we have &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$|a_{0}+a_{1}n+a_{2}n^2 + \cdots +a_{d}n^d| \leq a_{d}n^{d} +a_{d-1}n^d + \cdots+ a_{0}n^d$$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\leq (a_0+ a_1 + \cdots + a_d)n^d$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \ \leq (a_0+ a_1 + \cdots + a_d)|n^d|$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So $f(n) = O(n^d) \implies f(n) = O(n^k)$ for $k \geq d$.&lt;/p&gt;&#xA;" OwnerUserId="2548" LastEditorUserId="2548" LastEditorDisplayName="" LastEditDate="2011-02-27T02:33:33.450" LastActivityDate="2011-02-27T02:33:33.450" CommentCount="1" />
  <row Id="23944" PostTypeId="1" CreationDate="2011-02-27T02:55:57.143" Score="1" ViewCount="231" Body="&lt;p&gt;The goal is to maximize the following function:&#xA;\begin{align}&#xA;K_p(q) = q\log \frac{q}{p} + (1-q)\log \frac{1-q}{1-p}&#xA;\end{align}&#xA;where&#xA;\begin{align}&#xA;0 \leq q \leq 1&#xA;\end{align}&#xA;and $p \in (0,0.5)$ and is some constant. In other words, I'm maximizing the KL divergence in terms of one Bernoulli variable with success parameter $q$ with regard to another Bernoulli with success parameter $p$. The way I've set it up, I can see without formal calculations that it's maximized when $q=1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I'd like to use the KKT conditions to solve this problem formally and I'm running into problems doing that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, I set up the Lagrangian to minimize with the KKT conditions:&#xA;\begin{align}&#xA;\mathcal{L}(q,\mu,\lambda) &amp;amp; = -K_p(q) - \mu(1-q) - \lambda q \newline&#xA;\textrm{where} \quad&#xA;&amp;amp; \mu, \lambda \geq 0 \newline&#xA;&amp;amp; 0 \leq q \leq 1 \newline&#xA;&amp;amp; \mu(1-q) = 0 \newline&#xA;&amp;amp; \lambda q = 0&#xA;\end{align}&#xA;I then set the derivative in terms of $q$ to zero and plug it back in and get the following dual form:&#xA;\begin{align}&#xA;D(\mu,\lambda) &amp;amp; = -\mu + \log(1-p+pe^{\mu-\lambda}) \newline&#xA;&amp;amp; = \log(e^{-\mu}-pe^{-\mu}+pe^{-\lambda})&#xA;\end{align}&#xA;which I'm supposed to maximize.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But again, I run into the problem where the maximum of the dual is attained when $\mu,\lambda=0$ (and I got these values again by looking at the equation because setting derivatives to zero was infeasible). Which, according to the KKT conditions, means that q is neither zero nor one. But that can't be right, because the maximum of $K$ is attained when $q=1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And my obvious question is: what am I doing wrong? And how is this done correctly?&lt;/p&gt;&#xA;" OwnerUserId="6690" LastActivityDate="2011-05-30T07:06:07.237" Title="Using KKT conditions to maximize function" Tags="&lt;nonlinear-optimization&gt;&lt;convex-optimization&gt;&lt;information-theory&gt;" AnswerCount="2" FavoriteCount="2" />
  <row Id="23945" PostTypeId="1" CreationDate="2011-02-27T02:58:29.960" Score="5" ViewCount="58" Body="&lt;p&gt;&lt;a href=&quot;http://garden.irmacs.sfu.ca/?q=op/reconstruction_conjecture&quot; rel=&quot;nofollow&quot;&gt;Reconstruction conjecture&lt;/a&gt; says that graphs (with at least three vertices) are determined uniquely by their vertex deleted subgraphs. This conjecture is five decades old.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Searching relevant literature, I found that the following classes of graphs are known to be reconstructible :&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;trees&lt;/li&gt;&#xA;&lt;li&gt;disconnected graphs, graphs whose complement is disconnected&lt;/li&gt;&#xA;&lt;li&gt;regular graphs&lt;/li&gt;&#xA;&lt;li&gt;Maximal Outerplanar Graphs&lt;/li&gt;&#xA;&lt;li&gt;maximal planar graphs&lt;/li&gt;&#xA;&lt;li&gt;outerplanar graphs&lt;/li&gt;&#xA;&lt;li&gt;Critical blocks&lt;/li&gt;&#xA;&lt;li&gt;Separable graphs without end vertices&lt;/li&gt;&#xA;&lt;li&gt;unicyclic graphs (graphs with one cycle)&lt;/li&gt;&#xA;&lt;li&gt;non-trivial cartesian product graphs&lt;/li&gt;&#xA;&lt;li&gt;squares of trees&lt;/li&gt;&#xA;&lt;li&gt;bidegreed graphs&lt;/li&gt;&#xA;&lt;li&gt;unit interval graphs&lt;/li&gt;&#xA;&lt;li&gt;threshold graphs&lt;/li&gt;&#xA;&lt;li&gt;nearly acyclic graphs (i.e., G-v is acyclic)&lt;/li&gt;&#xA;&lt;li&gt;cacti graphs&lt;/li&gt;&#xA;&lt;li&gt;graphs for which one of the vertex deleted graph is a forest.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I recently proved that a special case of partial 2-trees are reconstructible. I am wondering if partial 2-trees (a.k.a &lt;a href=&quot;http://en.wikipedia.org/wiki/Series-parallel_graph&quot; rel=&quot;nofollow&quot;&gt;series-parallel graphs&lt;/a&gt;) are known to be reconstructible. Partial 2-trees do not seem to fall into any of the above mentioned categories.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Am I missing any other known classes of reconstructible graphs in the above list ?&lt;/li&gt;&#xA;&lt;li&gt;In particular, are partial 2-trees known to be reconstructible ?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I asked this &lt;a href=&quot;http://cstheory.stackexchange.com/questions/5155/reconstruction-conjecture-and-partial-2-trees&quot;&gt;question at cstheory&lt;/a&gt; website also.&lt;/p&gt;&#xA;" OwnerUserId="7540" LastActivityDate="2011-02-27T02:58:29.960" Title="Reconstruction Conjecture and Partial 2-trees" Tags="&lt;combinatorics&gt;&lt;graph-theory&gt;" CommentCount="1" />
  <row Id="23946" PostTypeId="1" CreationDate="2011-02-27T02:59:01.040" Score="1" ViewCount="61" Body="&lt;p&gt;I am trying to solve a problem which I have reduced down to find out one of integer solutions to a number of nonhomogeneous linear inequalities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can this be done efficiently ? If so: how?; if not: why?&lt;/p&gt;&#xA;" OwnerUserId="4210" LastActivityDate="2011-03-29T11:38:35.123" Title="Integer solutions of nonhomogeneous linear inequalities" Tags="&lt;algorithms&gt;&lt;inequality&gt;" AnswerCount="1" />
  <row Id="23947" PostTypeId="1" AcceptedAnswerId="23977" CreationDate="2011-02-27T03:02:21.570" Score="5" ViewCount="199" Body="&lt;p&gt;Hello, &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Let $\phi(n) $ is the numbers of number that are relatively prime to n.&lt;br&gt;&#xA;  Then, how could we solve the equation $\phi(n) = k, k &amp;gt; 0?$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For example:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\phi(n) = 8 $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can use computer program to check all numbers that are relatively prime to $n$, but I think there must be an easier way to approach this problem.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Base on this formula:&#xA;$$\prod_{i=0}^{k} p_{i}a^{a_i} $$&lt;br&gt;&#xA;The only thing I can see is n must not have a prime factor &gt; 9, otherwise $\phi(n) &amp;gt; 8 $.&#xA;I really don't know where to start :( ? A hint would be greatly appreciated.&lt;/p&gt;&#xA;" OwnerUserId="5711" LastActivityDate="2011-02-27T08:04:51.730" Title="How to solve the equation $\phi(n) = k$?" Tags="&lt;elementary-number-theory&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="1" />
  <row Id="23948" PostTypeId="1" AcceptedAnswerId="23952" CreationDate="2011-02-27T03:06:26.293" Score="3" ViewCount="119" Body="&lt;p&gt;I understand the process in calculating a simple series like this one&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{n=0}^4 2n$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but I do not understand the steps to calculate a series like this one&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{n=1}^x n^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have an awesome calculator and know how to use it, so I know the solution&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{x^3}{3} + \frac{x^2}{2} + \frac{x}{6}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but I don't know the steps.  All the explanations I find (like those on &lt;a href=&quot;http://www.purplemath.com/modules/series2.htm&quot; rel=&quot;nofollow&quot;&gt;Purple Math&lt;/a&gt;) are for summations like the first one I listed.  If someone can provide a detailed explanation on how to calculate a summation like mine or provide a link to one it would be much appreciated.&lt;/p&gt;&#xA;" OwnerUserId="6096" LastEditorUserId="1208" LastEditorDisplayName="" LastEditDate="2011-02-28T21:45:30.113" LastActivityDate="2011-02-28T21:45:30.113" Title="Help calculating a series" Tags="&lt;sequences-and-series&gt;" AnswerCount="3" CommentCount="3" />
  <row Id="23950" PostTypeId="2" ParentId="23948" CreationDate="2011-02-27T03:24:13.820" Score="7" ViewCount="" Body="&lt;p&gt;$$(k+1)^3-k^3 = 3k^2 + 3k + 1$$ Now sum it up from $k=1$ to $n$. Notice the telescopic summation on the left side and use $\displaystyle \sum_{k=1}^n k = \frac{n(n+1)}{2}$ to get the answer. This is a standard technique to compute such sums.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastActivityDate="2011-02-27T03:24:13.820" CommentCount="5" />
  <row Id="23952" PostTypeId="2" ParentId="23948" CreationDate="2011-02-27T03:29:48.847" Score="5" ViewCount="" Body="&lt;p&gt;Yes, you can use differences to find such formulas. Many times there are also easier ways (involving tricks). For this one you can use that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{i=0}^n i^3 = \left ( \sum_{i=0}^n (i+1)^3 \right ) - (n+1)^3$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From here it is an easy computation to find the answer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{i=0}^n i^3 = \sum_{i=0}^n i^3 + \sum_{i=0}^n 3i^2 + \sum_{i=0}^n 3i + \sum_{i=0}^n 1 -(n+1)^3$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which gives&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{i=0}^n i^2 = \frac{-\frac{3n(n+1)}{2} - n -1 + n^3 +3n^2 +3n +1}{3} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{i=0}^n i^2 = \frac{n^3}{3} + \frac{n^2}{2} + \frac{n}{6}$$&lt;/p&gt;&#xA;" OwnerUserId="5105" LastEditorUserId="5105" LastEditorDisplayName="" LastEditDate="2011-02-27T03:37:31.990" LastActivityDate="2011-02-27T03:37:31.990" CommentCount="4" />
  <row Id="23953" PostTypeId="2" ParentId="23947" CreationDate="2011-02-27T03:32:06.750" Score="1" ViewCount="" Body="&lt;p&gt;You might take a look into this:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.jstor.org/stable/2308462&quot; rel=&quot;nofollow&quot;&gt;http://www.jstor.org/stable/2308462&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" LastActivityDate="2011-02-27T03:32:06.750" CommentCount="2" />
  <row Id="23954" PostTypeId="1" CreationDate="2011-02-27T03:36:38.490" Score="0" ViewCount="90" Body="&lt;p&gt;The following is a quote from &lt;a href=&quot;http://www.math.ucla.edu/~pak/papers/stoc2.pdf&quot; rel=&quot;nofollow&quot;&gt;Lifting Markov Chains to Speed up Mixing&lt;/a&gt;, by Chen, Lovasz, and Pak:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;...Thus we have described a (randomized) stopping rule that, for any starting node, stops in an expected number $O(\sqrt{n})$ steps, and the probability of stopping at each node is at least 1/2 its stationary probability. By standard results (see e.g. [9]), this implies that the mixing time is $O(\sqrt{n})$. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My question: I would like a precise reference for this fact, especially one which has a proof I can read. The reference [9] is &lt;a href=&quot;http://www.cs.elte.hu/~lovasz/mixtimes.ps&quot; rel=&quot;nofollow&quot;&gt;Mixing Times&lt;/a&gt; by Lovasz and Winkler, and I was unable to find this statement in there. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moreover, there is some ambiguity of what a mixing time of means; what I want it to mean in this case is that except from the eigenvalue at $1$, every other eigenvalue of the chain has modulus at most $1-c/\sqrt{n}$ for some constant $c&amp;gt;0$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can someone provide a precise reference? &lt;/p&gt;&#xA;" OwnerUserId="2120" LastActivityDate="2011-05-28T07:04:01.330" Title="Stopping rules for Markov Chains" Tags="&lt;probability-theory&gt;" AnswerCount="1" />
  <row Id="23955" PostTypeId="2" ParentId="23947" CreationDate="2011-02-27T03:38:54.383" Score="1" ViewCount="" Body="&lt;p&gt;See &lt;a href=&quot;http://mathoverflow.net/questions/31691/inverting-the-totient-function&quot; rel=&quot;nofollow&quot;&gt;http://mathoverflow.net/questions/31691/inverting-the-totient-function&lt;/a&gt; and &lt;a href=&quot;http://arxiv.org/abs/math/0404116&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/math/0404116&lt;/a&gt; ; also &lt;a href=&quot;http://www.ams.org/journals/mcom/2006-75-254/S0025-5718-06-01826-6/home.html&quot; rel=&quot;nofollow&quot;&gt;http://www.ams.org/journals/mcom/2006-75-254/S0025-5718-06-01826-6/home.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="589" LastActivityDate="2011-02-27T03:38:54.383" CommentCount="1" />
  <row Id="23956" PostTypeId="1" AcceptedAnswerId="23959" CreationDate="2011-02-27T04:13:14.337" Score="2" ViewCount="53" Body="&lt;p&gt;I'm trying to do a piece-wise affine transform in Python.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have one image with a set of points hand marked and another set of points where I wish to &quot;move&quot; my current points and the texture between.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is my understanding of the transformation:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;Find the delaunay triangulation of the moved points  &lt;/li&gt;&#xA;  &lt;li&gt;For each pixel in the image, find the corresponding triangle  &lt;/li&gt;&#xA;  &lt;li&gt;&lt;p&gt;Find the barycentric coordinates of the pixel for the triangle found in 2.&lt;/p&gt;&lt;/li&gt;&#xA;  &lt;li&gt;&lt;p&gt;Multiply the barycentric coordinates for the pixel in 3 and you&#xA;  have the coordinates of the pixel in&#xA;  the new image.&lt;/p&gt;&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I ran some tests and I believe my understanding is correct. I have two bottlenecks in my code currently. First, finding out if a point is inside a triangle. For that I am using the algorithm from the &lt;a href=&quot;http://en.wikipedia.org/wiki/Barycentric_coordinate_system_%28mathematics%29&quot; rel=&quot;nofollow&quot;&gt;wikipedia page&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second bottleneck, and this is the real problem, is finding out the triangle a point belongs to (hence the title :)) I tried finding the distance from the points to the centers, then ordering the triangles in each iteration by the distance from its center to the point, but that doesn't always work. I believe the problem lies in the fact that I have to use the triangulation of the image I want and not from the image I have.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I figure the displacement of the points/triangles comes into play then, but I'm not sure how.&lt;/p&gt;&#xA;" OwnerUserId="7542" LastActivityDate="2011-02-27T04:45:23.530" Title="Given a point and a set of triangles, what's would be a fast way to find which triangle the point belongs to?" Tags="&lt;geometry&gt;&lt;trigonometry&gt;&lt;image-processing&gt;" AnswerCount="1" />
  <row Id="23957" PostTypeId="2" ParentId="23930" CreationDate="2011-02-27T04:42:02.863" Score="1" ViewCount="" Body="&lt;p&gt;(Assume the definition of the function is corrected so that $f(1)=1$, $f(2)=2$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f$ is a function on $[1,3]$.  The points where $f$ is continuous are $[0,1) \cup (2,3]$.  In that sense, one would not say that $f$ is continuous on $[0,1] \cup [2,3]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;However&lt;/em&gt;, what is true is that the &lt;em&gt;restriction&lt;/em&gt; $f|_{[0,1] \cup [2,3]}$ of $f$ to $[0,1] \cup [2,3]$ is a continuous function on all of its domain.  But strictly speaking this is a different function from $f$.&lt;/p&gt;&#xA;" OwnerUserId="822" LastActivityDate="2011-02-27T04:42:02.863" />
  <row Id="23958" PostTypeId="1" CreationDate="2011-02-27T02:15:25.587" Score="5" ViewCount="215" Body="&lt;p&gt;I've always used the method of Lagrange multipliers with blind confidence that it will give the correct results when optimizing problems with constraints. But I would like to know if anyone can provide or recommend a derivation of the method at physics undergraduate level that can highlight its limitations, if any.&lt;/p&gt;&#xA;" OwnerUserId="7544" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-27T04:46:05.637" LastActivityDate="2011-03-03T02:32:54.523" Title="Derivation of the method of Lagrange multipliers?" Tags="&lt;optimization&gt;&lt;multivariable-calculus&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="1" />
  <row Id="23959" PostTypeId="2" ParentId="23956" CreationDate="2011-02-27T04:45:23.530" Score="0" ViewCount="" Body="&lt;p&gt;Two things:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) For finding out what triangle a point belongs to, I'd recommend a gridding system - overlay a small (something in the range of 20x20 to 50x50, depending on how many points are in your triangulation) grid on top of your destination triangulation and bin all the triangles into each grid cell they overlap (this should be fairly quick; with the right resolution, many of your triangles will fall into a single cell and most of the rest will only straddle a couple).  Then to determine which triangle a point belongs to, figure out which bin the point is in and test only the triangles that overlap that bin.  There are other, similar schemes available but this is probably the simplest and it should be more than good enough for what you're doing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Since you don't make it clear which way you're iterating in your description, I want to emphasize that you should be looping over the points in your destination image, not the points in your source image; if you loop over the points in the source image and forward-transform then you may wind up with holes in your destination image; by looping over your destination image and back-transforming to find the corresponding point in the source image, you eliminate this problem.  Also, you almost certainly want to do some measure of antialiasing; the easiest thing to do is probably to subsample your destination image and do 4 or 9 samples for each destination pixel.&lt;/p&gt;&#xA;" OwnerUserId="785" LastActivityDate="2011-02-27T04:45:23.530" CommentCount="3" />
  <row Id="23960" PostTypeId="2" ParentId="23930" CreationDate="2011-02-27T04:48:19.563" Score="1" ViewCount="" Body="&lt;p&gt;As others have pointed out, it seems reasonable to assume that you meant to define $f$ like this:&#xA;$$f = \chi_{[0,1]} + 10\chi_{\mathbb{Q}\cap (1,2)} -10\chi_{(1,2)\setminus\mathbb{Q}} +2\chi_{[2,3]}, $$&#xA;where $\chi_A$ is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Indicator_function&quot; rel=&quot;nofollow&quot;&gt;indicator function&lt;/a&gt; on $A$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And it is true that $f|_{[0,1]\cup [2,3]}$ is continuous since it's constant on each of its domain's &lt;a href=&quot;http://en.wikipedia.org/wiki/Connected_component_%28topology%29#Connected_components&quot; rel=&quot;nofollow&quot;&gt;connected components&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1269" LastEditorUserId="1269" LastEditorDisplayName="" LastEditDate="2011-02-27T05:21:42.327" LastActivityDate="2011-02-27T05:21:42.327" CommentCount="2" />
  <row Id="23961" PostTypeId="2" ParentId="23754" CreationDate="2011-02-27T04:51:09.877" Score="0" ViewCount="" Body="&lt;p&gt;I believe I have found the answer. The key insight is that multiplication by diagonal matrices multiplies each &lt;em&gt;row&lt;/em&gt; $i$ by $D_{ii}$ when the diagonal matrix is on the left, and each &lt;em&gt;column&lt;/em&gt; $i$ by $D_{ii}$ when the diagonal matrix is on the right.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus $\sum_{i=1}^m d_i X_{ij} X_{ik}$ can be rewritten as $\sum_{i=1}^m X_{ij} (DX)_{ik}$ = $\sum_{i=1}^m X_{ji}^T (DX)_{ik}$, which is a simple product of $X^T$ and $(DX)$.&lt;/p&gt;&#xA;" OwnerUserId="7512" LastActivityDate="2011-02-27T04:51:09.877" />
  <row Id="23962" PostTypeId="2" ParentId="23954" CreationDate="2011-02-27T04:51:22.013" Score="0" ViewCount="" Body="&lt;p&gt;We know that (on page 12) the mixing time is $$\mathcal{H} = \max_{i} \ \max_{j}(H(i,j)-H(\pi,j))$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems that $H(i,j) = O(\sqrt{n})$. &lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2011-02-27T04:51:22.013" CommentCount="1" />
  <row Id="23964" PostTypeId="1" CreationDate="2011-02-27T04:57:15.993" Score="1" ViewCount="129" Body="&lt;p&gt;I am a little muddled and am hoping I can get some clarification about forms in a complex manifold.  Since I am only concerned with local issues, consider $M = \mathbb C^n$ as a complex manifold.  So I have complex coordinates $z_1,\ldots,z_n$ and corresponding real coordinates $x_1, y_1,\ldots, x_n, y_n$ with $z_j = x_j + i y_j$.  Now I have a canonical complex structure $J$ on $M$ given by $J \partial_{x_j} = \partial_{y_j}, J \partial_{y_j} = -\partial_{x_j}$.  Then I have an isomorphism of the complex bundles between $TM$ and the holomorphic tangent bundle $T_{(1,0)} M = span_{\mathbb C} \{\partial_{z_j}\}$.  This isomorphism is given by&#xA;$$&#xD;&#xA;(a+ib) \partial_{z_j} \mapsto a\partial_{x_j} + b\partial_{y_j}.&#xD;&#xA;$$&#xA;Now let $\omega = \sum_i dx_i \wedge dy_i$ be the standard symplectic form on $M$.  What does this correspond to under the above isomorphism.  At first glance it is&#xA;$$&#xD;&#xA;\tilde \omega = \frac{i}{2}\sum_j dz_j \wedge d\bar z_j&#xD;&#xA;$$&#xA;where $dz_j = dx_j + i dy_j, d\bar{z_j} = dx_j - i dy_j$.  At second glance this can't be correct since this is zero on the holomorphic tangent bundle.  On third glance, everything seems to work out if I think of&#xA;$$&#xD;&#xA;d\bar{z_j}(c\partial_{z_j}) = \bar c.&#xD;&#xA;$$&#xA;But this is unsettling to me.  Is this the right way to think about it though?  Can I get in any trouble by thinking of it this way?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem seems to be that the differentials of the real coordinates give real dual vectors, but the covectors $dz_j$, $d\bar z_j$ are inherently complex (their real span is not the real dual to $T_{(1,0)} M$ unless I interpret $d\bar z_j$ as mentioned above).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe what is usually done is $\tilde\omega$ is considered to be a form on $TM \otimes \mathbb C$ and is a real symplectic form on&#xA;$$&#xD;&#xA;\mathbb Rspan\{\partial_{x_j}, \partial_{z_j}\} = \mathbb Rspan\{\partial_{z_j} + \partial_{\bar z_j}, i(\partial_{z_j} - \partial_{\bar z_j})\}&#xD;&#xA;$$&#xA;but I'd prefer to not complexify if I don't have to (since I am just concerned with the symplectic geometry and thus the real bundle, but I would like the convenience of the complex notation).&lt;/p&gt;&#xA;" OwnerUserId="9" LastEditorUserId="9" LastEditorDisplayName="" LastEditDate="2011-02-27T16:27:27.470" LastActivityDate="2011-03-24T18:05:40.223" Title="Symplectic form on a complex manifold" Tags="&lt;differential-geometry&gt;&lt;complex-geometry&gt;&lt;symplectic-geometry&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="23965" PostTypeId="2" ParentId="23958" CreationDate="2011-02-27T04:58:15.713" Score="11" ViewCount="" Body="&lt;p&gt;Lagrange multipliers are used to obtain the maximum of a function $f(\mathbf{x})$ on a surface $\{ \mathbf{x}\in\mathbb{R}^n\mid g(\mathbf{x}) = 0\}$ (I use &quot;surface&quot;, but whether it is a 2-dimensional, 1-dimensional, or whatever-dimensional object will depend on the $g$ and the $\mathbb{R}^n$ we are dealing with).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The gradient of $f$, $\nabla f$, points in the direction of greatest increase for $f$. If we want to find the largest value of $f$ along $g$, then we need the direction of greatest increase to be &lt;em&gt;orthogonal&lt;/em&gt; to $g$; otherwise, moving along $g$ will &quot;capture&quot; some of that increase and $f$ will not achieve its maximum among $g$ at that point (this is akin to the fact that in one-variable calculus, the derivative should be $0$ at the maximum, otherwise moving a bit will increase in one direction will increase the value of the function).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order for $\nabla f$ to be perpendicular to the surface, it must be parallel to the &lt;em&gt;gradient&lt;/em&gt; of $g$; so $\nabla f$ must be a scalar multiple of $\nabla g$. So this amounts to finding a solution to the system&#xA;\begin{align*}&#xA;\nabla f(\mathbf{x}) &amp;amp;= \lambda \nabla g(\mathbf{x})\\&#xA;g(\mathbf{x}) &amp;amp;= 0&#xA;\end{align*}&#xA;for both $\mathbf{x}$ and $\lambda$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Added.&lt;/em&gt; Such a point is not &lt;em&gt;guaranteed&lt;/em&gt; to be a maximum or a minimum; it could also be a saddle point, or nothing at all, much as in the one-variable case, points where $f&amp;#39;(x)=0$ are not guaranteed to be extremes of the function. Another obvious limitation is that if the surface $g$ is not differentiable (does not have a well-defined gradient) then you cannot even set up the system.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-27T05:21:34.200" LastActivityDate="2011-02-27T05:21:34.200" CommentCount="3" />
  <row Id="23966" PostTypeId="1" CreationDate="2011-02-27T05:10:30.083" Score="1" ViewCount="67" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;If $\displaystyle L = \frac{3+\sqrt{-3}}{2}$,&#xA;  and if $x\equiv 1\pmod{L}$,&lt;br&gt;&#xA;  show that $x^3\equiv 1\pmod{L^4}$. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have already shown that if $x\equiv 1\pmod{L}$,&lt;br&gt;&#xA;then $x^3\equiv 1\pmod{L^3}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="7543" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-27T05:15:08.557" LastActivityDate="2011-02-27T05:28:22.840" Title="A congruence in a quadratic number field" Tags="&lt;number-theory&gt;&lt;modular-arithmetic&gt;" AnswerCount="1" />
  <row Id="23967" PostTypeId="1" CreationDate="2011-02-27T05:22:31.893" Score="2" ViewCount="112" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Show that if a set $A$ in a metric space is bounded, so is each subset&#xA;  $B \subseteq A$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7545" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2011-02-27T05:58:07.943" LastActivityDate="2011-02-27T06:20:47.667" Title="Real Analysis question - hw" Tags="&lt;homework&gt;&lt;real-analysis&gt;&lt;topology&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="23968" PostTypeId="2" ParentId="23967" CreationDate="2011-02-27T05:26:41.497" Score="2" ViewCount="" Body="&lt;p&gt;If $A$ is bounded then there exists $x \in M$ and $r&amp;gt;0$ such that for all $a \in A$, we have $d(x,a) &amp;lt;r$. So how would you show that each subset $B \subseteq A$ is bounded?&lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2011-02-27T05:26:41.497" />
  <row Id="23970" PostTypeId="2" ParentId="23966" CreationDate="2011-02-27T05:28:22.840" Score="1" ViewCount="" Body="&lt;p&gt;Let $w = \frac{-1 + \sqrt{-3}}{2}$. Your question is to show that if $x \equiv 1 \pmod{2+w}$, then $x^3 \equiv 1 \pmod{(2+w)^4}$.&#xA;By direct expansion, we see that $(2+w)^4 = 9w$. So it suffices to show $x^3 \equiv 1 \pmod{9}$. Now check the following,&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$x \equiv 1 \pmod{2+w}$ is equivalent to $x = a+bw$, such that $a+b \equiv 1 \pmod{3}$, where $a,b$ are integers.&lt;/li&gt;&#xA;&lt;li&gt;$(a+bw)^3 = a^3+b^3-3ab^2 + 3ab(a-b)w$.&lt;/li&gt;&#xA;&lt;li&gt;Check that each coefficient is divisible by 9, by looking at two cases: either a or b are divisible by 3, or that $a \equiv b \equiv -1 \pmod{3}$.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="325" LastActivityDate="2011-02-27T05:28:22.840" />
  <row Id="23971" PostTypeId="2" ParentId="23948" CreationDate="2011-02-27T05:36:28.397" Score="2" ViewCount="" Body="&lt;p&gt;Another approach is to know (guess) that $\sum_{n=1}^xn^2$ is cubic in $x$.  Similar to an integral, the sum adds one degree.  Then you can just say $\sum_{n=1}^xn^2=Ax^3+Bx^2+Cx+D$.  If you calculate that the sum up to 1,2,3,4 is 1,5,14,30 (or start with 0 and sum to 0 makes it a bit easier) you can just solve the simultaneous equations for A,B,C, and D.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-02-27T05:36:28.397" CommentCount="1" />
  <row Id="23972" PostTypeId="2" ParentId="23958" CreationDate="2011-02-27T05:37:44.573" Score="4" ViewCount="" Body="&lt;p&gt;An algebraic way of looking at this is as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From an algebraic view point, we know how to find the extremum of a function of many variables. Say we want to find the extremum of $f(x_1,x_2,\ldots,x_n)$, we set the gradient to zero and look at the definiteness of the Hessian.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We would like to extend this idea, when we want to find the extremum of a function along with some constraints. Say the problem is:&#xA;$$\begin{align}&#xD;&#xA;\text{Minimize }f(x_1,x_2,\ldots,x_n)\\\&#xD;&#xA;\text{subject to: }g_k(x_1,x_2,\ldots,x_n) = 0\\\&#xD;&#xA;\text{where }k \in \{1,2,\ldots,m\}\\\&#xD;&#xA;\end{align}&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we find the extremum of $f$ just by setting the gradient of $f$ to zero, these extremum need not satisfy the constraints.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, we would like to include the constraints in the previous idea. One way to it is as follows. Define a new function:&#xA;$$F(\vec{x},\vec{\lambda}) = f(\vec{x}) - \lambda_1 g_1(\vec{x}) - \lambda_2 g_2(\vec{x}) - \cdots  - \lambda_m g_m(\vec{x})$$&#xA;where&#xA;$\vec{x} = \left[ x_1,x_2,\ldots,x_n \right], \vec{\lambda} = \left[\lambda_1,\lambda_2,\ldots,\lambda_m \right]$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that when the constraints are enforced, we have $F(\vec{x},\vec{\lambda}) = f(\vec{x})$ since $g_j(x) = 0$ when the constraints are enforced.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let us find the extremum of $F(\vec{x},\vec{\lambda})$. This is done by setting $\frac{\partial F}{\partial x_i} = 0$ and $\frac{\partial F}{\partial \lambda_j} = 0$ where $i \in \{1,2,\ldots,n\}$ and $j \in \{1,2,\ldots,m\}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Setting $\frac{\partial F}{\partial x_i} = 0$ gives us $$\vec{\nabla}f = \vec{\nabla}g \cdot \vec{\lambda}$$ where $\vec{\nabla}g = \left[\vec{\nabla} g_1(\vec{x}),\vec{\nabla} g_2(\vec{x}),\ldots,\vec{\nabla} g_m(\vec{x}) \right]$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Setting $\frac{\partial F}{\partial \lambda_j} = 0$ gives us $$g_j(x) = 0$$ where $j \in \{1,2,\ldots,m\}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, we find that when we find the extremum of $F$, the constraints are automatically enforced. This means that the extremum of $F$ corresponds to extremum of $f$ with the constraints enforced.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To decide, if the extremum is a minimum (or) maximum (or) if the point we obtain by solving the system is a saddle point, we need to look at the definiteness of the Hessian of $F$ and decide.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastActivityDate="2011-02-27T05:37:44.573" CommentCount="1" />
  <row Id="23973" PostTypeId="2" ParentId="23967" CreationDate="2011-02-27T06:20:47.667" Score="1" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Definition:&lt;/strong&gt; &lt;em&gt;A metric space $M$ is called bounded if there exists some number $r&amp;gt;0$, such that $d(x,y) \leq r$ for all $x, y \in M$.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since you are assuming $A$ to be &lt;strong&gt;bounded&lt;/strong&gt; then there is some number $k$ such that $d(x,a) \leq k$ for all $x,a\in A$. Since $B \subset A$ and this condition happens for all elements in $A$, therefore it should happen for elements in $B$ as well.&lt;/p&gt;&#xA;" LastActivityDate="2011-02-27T06:20:47.667" />
  <row Id="23974" PostTypeId="1" AcceptedAnswerId="23980" CreationDate="2011-02-27T06:33:15.303" Score="1" ViewCount="98" Body="&lt;p&gt;Background:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $V$ be a vector space over a field $\Bbbk$. Let $V^*=Hom_\Bbbk(V,\Bbbk)$ be the vector space of linear functionals on $V$. Consider the map $V^*\times V\to End_\Bbbk(V)$ given by $(f,v_0)\to\phi$ where $\phi(v)=f(v)\cdot v_0$. This map is obviously bilinear and thus induces a linear map $V^*\otimes_\Bbbk V\to End_\Bbbk V$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $V$ is finite-dimensional, then a choice of basis $\{e_1,\dots,e_n\}$ and a dual basis $\{e_1^*,\dots,e_n^*\}$ establishes that the map is injective since evaluations at $e_i$ for the image of each basis element $e_i^*\otimes e_j$ of $V^*\otimes V$ shows that the corresponding endomorphisms are linearly independent. Identifying $End_\Bbbk(V)$ with the $n^2$-dimensional space of $n\times n$ matrices and the fact that $\dim V^*\otimes V$ is $n^2$ shows that the map is in fact an isomorphism.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Now onto my question. If $V$ is an infinite-dimensional vector space over $\Bbbk$, then:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;is the above (obvious?) map $V^*\otimes_\Bbbk V\to End_\Bbbk(V)$ still injective? If not, then under what conditions is it?&lt;/li&gt;&#xA;  &lt;li&gt;given a $\phi\in End(V)$, is there a criterion to determine whether $\phi$ comes from an element of $V^*\otimes V$?&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="400" LastActivityDate="2011-02-27T07:05:06.800" Title="What does $V^*\otimes_\Bbbk V$ look like inside $End_\Bbbk V$ for infinite-dimensional $V$?" Tags="&lt;linear-algebra&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="23975" PostTypeId="1" CreationDate="2011-02-27T06:34:49.340" Score="4" ViewCount="140" Body="&lt;p&gt;[disclaimer: I've studied a lot of logic but never been good at analysis, so that's the angle I'm coming from below]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;in my attempt to find a precise version of the 'definitions' usually given when first introducing leibniz notation in single or multivariable calculus or analysis, wherein there is no appeal to differentials or infinitesimals,  I've discovered that I'm confused about a few interellated low-level issues around formalization and notation, etc.  I don't know which questions are the more basic ones here, so I'll just ask them as go along explaining what I think I &lt;em&gt;do&lt;/em&gt; understand about proposed formal defintions of the sort in question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm concerned with both the dy/dx notation and the 'del y/del x' notation for partial derivatives, but just the real-valued case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since I suspect that my confusions stem from use-mention confusions, and the confusion of functions, variables, and their values, I will use, and assume familiarity with lambda notation, metavariables, and quasi-quotation throughout.  Where not stated, ⌜λx.φ⌝ refers to the function on the largest &lt;em&gt;real&lt;/em&gt; domain on which φ is a real number. Assume only definitions for real variables are sought after below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, on with the definitions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These short articles by the author Thurston (the first five results) all give roughly the same formal definition of leibniz notation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://scholar.google.ca/scholar?hl=en&amp;amp;as_sdt=0,5&amp;amp;q=thurston+leibniz&quot; rel=&quot;nofollow&quot;&gt;http://scholar.google.ca/scholar?hl=en&amp;amp;as_sdt=0,5&amp;amp;q=thurston+leibniz&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;whereas the top results for this search are by the author Harrison, and each give one of a few slight variants on a different definition:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.google.ca/search?q=%22leibniz+notation%22+%22lambda+term%22&amp;amp;ie=utf-8&amp;amp;oe=utf-8&amp;amp;aq=t&amp;amp;rls=org.mozilla:en-GB:official&amp;amp;client=firefox-a&quot; rel=&quot;nofollow&quot;&gt;http://www.google.ca/search?q=%22leibniz+notation%22+%22lambda+term%22&amp;amp;ie=utf-8&amp;amp;oe=utf-8&amp;amp;aq=t&amp;amp;rls=org.mozilla:en-GB:official&amp;amp;client=firefox-a&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;here is my understanding of their definitions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;HARRISON:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;⌜dφ/dψ⌝ is a shorthand for ⌜D(λψ.φ)(ψ)⌝ i.e.&#xA;&quot;dy/dx&quot; stands for &quot;D(λx.y)(x)&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ψ must be a variable of the underlying logic (and it has a free and a bound occurence here) and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;φ must be a string such that ⌜∀x, φ = f(x)⌝ is true for some function f:S→R where S⊆R.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Q1. does anyone have a simpler way to state the restriction on φ?&#xA;Q1.1 what is the proper name for the sort of string φ must be?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;THURSTON:&#xA;⌜dφ/dψ⌝ is a shorthand for ⌜ψ'/φ'⌝ i.e.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;dy/dx&quot; is short for &quot;y'/x'&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means that φ and ψ must be names of functions from the reals to the reals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Q1.2 Are there other formal definitions in the literature I should compare to these? I haven't found any yet...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Harrison's defintion does not return a value because a free variable is uninstntiated, in the same sense in which wffs which are not sentences do not return a truth value.    Thurston's version, however returns a function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;df/dx = f'(x) for harrison,&#xA;df/dx = f' for thurston&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More concretely&lt;/p&gt;&#xA;&#xA;&lt;p&gt;d(x²+x)/dx =&#xA;2x+1 for harrison&#xA;λx.2x+1 for thurston&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Q2 is the value '2x+1' which contains a free variable being returned, or are we implicitly quantifying over x, or, let's call it 'ξ' for the purposes of quasi quotation, and saying: ∀ξ, ⌜2ξ+1⌝ refers to (not 'is') the value returned?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, consider the following 'typical' calculus problem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;&#xA;y = f(x)&#xA;x = g(u)&#xA;g(x) = x³ - 7&lt;/p&gt;&#xA;&#xA;&lt;p&gt;find df/dx&#xA;&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thurston gets us df/dx = λx.1/(3x²)&#xA;This is undefined on harison's approach, since x is not a variable of the logic, but the name of a function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Q3 should I be considering the possibility that the logic allows for variables ranging over functions?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And if we pretended that it was we'd still get df/dx = D(λx.f)(x) but this is mal-formed since λx needs something like 'f(x)' rather than 'f' as input.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And if we added a case to harrison's definition to append '(x)' or the like when it's missing, we'd still get df/dx = 1, which is not equal to the result we got with thurston's definition--but more strikingly, the function we evaluated to get 1 was, unike the f' vs f'(x) case above, not even the same function as was returned by thurston's definition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Q4 What should I conclude from the fact that these definitions diverge in this manner?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now consider:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;&#xA;y = f(x)&#xA;f(x) = x⁹&lt;/p&gt;&#xA;&#xA;&lt;p&gt;find dy/dx&#xA;&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On harrison's account, we could view y as a metavariable, so that f(x) is placed substitutionally into the defining string, but I have a feeling that is not the right way to understand it. However, if y is merely a variable of the logic, it is a &lt;em&gt;free&lt;/em&gt; variable in the result, and we end up with one free variable too many..&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On thurston's account, y must be the name of a function, but &quot;y = f(x)&quot; sets y equal to an expression with a free variable, not equal to the name of a function&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Q5 should i view statements like &quot;y=f(x)&quot; as involving a supressed &quot;(x)&quot; and &quot;∀&quot; so that we get &quot;∀x,y(x)=f(x)&quot; ? Or should I see y as a metavariable? Or, should I imagine the logic extended to allow some new syntactic category of 'dependent' variables, while thinking of the usual variables in the logic as 'independent' variables--i.e. those whose value does not depend on others? I think I am very confused about what happends when one variable depends on another.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Q6 On a related note, I saw a passage recently that spoke in terms like &quot;x(u) is the inverse function of u(x)&quot;--how should this be understood more precisely? I've come to discover that I don't understand expressions of this sort at all!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Q7 Does either of these defintions clearly capture 'practice' better than another?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Q8 How should similar attempts be made for the 'del' notation for patial derivatives?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Q9 Can someone give me an example of where d/dx and del/delx return different values on the same input?  If I'm not mistaken, in some formalizations this never happens, and in other formalizations it does--I think harrison's would not allow for this since it just returns an 'expression' rather than one of the various functions that can be formed by an expression when you apply a lambda operator to it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I started also trying to read this article on revising patial derivative notation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[I've hit my link limit as I'm new here, but google &quot;revised notation for partial derivatives&quot;  (with quotes).  It's by WC Hassenpflug]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but I got stuck on the sentence:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;If we have a function u = f(x,y) and the transformation y=g(x∩) is made, then it is not clear whether del u / dx means del u / dx |y or del y / dx | n&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can someone explain that one to me?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Q10 This all bears some superficial similarity to the relationship between so-called 'random variables', which are actually functions, and what are called 'variables' in the underlying logic--this has also confused me, and I see many operations done in text on random variables where the operators have only been defined for values in the random variable's domain, and not on functions. Can anyone comment on this? I would be nice if I could dismiss two long-standing confusions with one stone :p&lt;/p&gt;&#xA;" OwnerUserId="7546" LastActivityDate="2011-02-27T13:31:23.247" Title="Formalizing Those Readings of Leibniz Notation that Don't Appeal to Infinitesimals/Differentials" Tags="&lt;calculus&gt;&lt;real-analysis&gt;&lt;logic&gt;&lt;notation&gt;&lt;definition&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="23976" PostTypeId="2" ParentId="23806" CreationDate="2011-02-27T06:40:31.990" Score="0" ViewCount="" Body="&lt;p&gt;I just read this paper: &lt;a href=&quot;http://arxiv.org/abs/math.HO/9404236&quot; rel=&quot;nofollow&quot;&gt;On proof and progress in mathematics&lt;/a&gt;, it could help you.&lt;/p&gt;&#xA;" OwnerUserId="1938" LastActivityDate="2011-02-27T06:40:31.990" CommunityOwnedDate="2011-02-27T06:40:31.990" />
  <row Id="23977" PostTypeId="2" ParentId="23947" CreationDate="2011-02-27T06:46:50.323" Score="3" ViewCount="" Body="&lt;p&gt;This is too long to be comment and hence the post.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\phi(n) = 8$. Note that $\sqrt{n} \leq \phi(n) \leq n-1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This implies $n$ is at most $64$. So you could write a brute force computer and compute $\phi(n)$ when $n \in [9,64]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A better way would be as follows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $n=\displaystyle \prod_{i=1}^k p_i^{\alpha_i} \Rightarrow \phi(n) = \displaystyle \prod_{i=1}^k p_i^{\alpha_i-1} (p_i-1)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First note that $n$ can be of the form $\displaystyle 2^\alpha \left( \prod_{i=1}^k p_i \right)$ i.e. the exponent of the odd primes in the prime factorization of $n$ is $1$. This is so, because if not these primes will then divide $\phi(n) = 8$ which is not possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $k=0$, then we have $n=2^{\alpha}$, $\displaystyle 2^{\alpha-1} = \phi(n) = 2^3 \Rightarrow \alpha=4$. Hence, $k=1 \Rightarrow n=16$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $k=1$. Then we have $n=2^{\alpha} p_1$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $\alpha = 0,1$, then $\displaystyle (p_1-1) = \phi(n) = 2^3 \Rightarrow p_1 = 9 \Rightarrow \text{ Not possible}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $\alpha = 2$, then $\displaystyle 2 (p_1 - 1) = \phi(n) = 2^3 \Rightarrow p_1=5$. Hence, $n=20$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $\alpha = 3$, then $\displaystyle 2^2 (p_1 - 1) = \phi(n) = 2^3 \Rightarrow p_1=3$. Hence, $n=24$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let $k=2$. Then we have $n=2^{\alpha} p_1 p_2$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $\alpha = 0,1$, then $\displaystyle (p_1-1)(p_2-1) = \phi(n) = 2^3 \Rightarrow p_1 = 3, p_2 = 5$. Hence, $n=15$ when $\alpha = 0$ and $n=30$ when $\alpha = 1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $\alpha = 2$, then $\displaystyle 2(p_1-1)(p_2-1) = \phi(n) = 2^3 \Rightarrow (p_1-1)(p_2-1) = 4 \Rightarrow \text{ Not Possible}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$k=3$ is not possible since $(3-1) \times (5-1) \times (7-1) &amp;gt; 8$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, the only solutions (hope I have not missed any case) are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$n=15,16,20,24,30$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similar idea extends to other problems where we want to find the inverse of the totient function.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2011-02-27T08:04:51.730" LastActivityDate="2011-02-27T08:04:51.730" CommentCount="1" />
  <row Id="23978" PostTypeId="1" AcceptedAnswerId="23979" CreationDate="2011-02-27T06:47:42.037" Score="2" ViewCount="120" Body="&lt;p&gt;My textbook says that solutions for the equation $y&amp;#39;=-y^2$ must always be 0 or decreasing. I don't understand—if we're solving for y', then wouldn't it be more accurate to say it must always be 0 or negative. Decreasing seems to imply that we're looking at a full graph, even though the book is talking about individual solutions. Can someone explain this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Secondly, it gives that the family $y=\frac{1}{x+C}$ as solutions for the equation. It then tells me that 0 is a solution for y in the original equation that doesn't match the family, but I don't quite understand that. How can we know that y' will equal 0 if we're specifically looking outside of the family of solutions it gives?&lt;/p&gt;&#xA;" OwnerUserId="7547" LastActivityDate="2011-02-27T07:05:18.107" Title="How exactly do differential equations work?" Tags="&lt;calculus&gt;&lt;homework&gt;&lt;differential-equations&gt;" AnswerCount="3" />
  <row Id="23979" PostTypeId="2" ParentId="23978" CreationDate="2011-02-27T06:59:24.180" Score="3" ViewCount="" Body="&lt;p&gt;If a function is differentiable and is decreasing, then the derivative is non-positive. In your case, $y^2 \geq 0 \Rightarrow y&amp;#39; = -y^2 \leq 0$. Hence, the solution you obtain must be decreasing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that we obtain the family $y = \frac{1}{x+c}$ assuming $y \neq 0$. This can be seen from the way we solve the equation. We have $$\frac{dy}{dx} = -y^2$$ &lt;strong&gt;Assuming&lt;/strong&gt; $y \neq 0$, we can divide by $y^2$ to get $\frac{dy}{y^2} = -dx$. Now we can integrate to get, $y = \frac{1}{x+\frac{1}{y_0}}$. Note that for each $y_0$, we get a different solution. So we have a one parameter family of solutions. Also, note that this family of $y$'s is nowhere $0$ and hence it is fine when we divided by $y^2$ in the original equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the process of solving &lt;strong&gt;assumes&lt;/strong&gt; that $y \neq 0$. We also observe that $y=0$ is a solution by plugging it into the differential equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, the solutions to the differential equation are $$y=0 \text{ and } y=\frac{1}{x+\frac{1}{y_0}}, \text{ where } y_0 \in \mathbb{R}$$&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2011-02-27T07:04:46.593" LastActivityDate="2011-02-27T07:04:46.593" />
  <row Id="23980" PostTypeId="2" ParentId="23974" CreationDate="2011-02-27T06:59:48.970" Score="5" ViewCount="" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Yes.  If $\sum_{j=1}^n f_j\otimes e_j$ is an element of $V^*\otimes_k V$, you can assume WLOG that $\{e_1,\ldots,e_n\}$ is linearly independent.  If $\sum_{j=1}^n f_j(v)e_j=0$ for all $v\in V$, then for each $j$, $f_j(v)=0$ for all $v\in V$, so that $f_1=f_2=\cdots=f_n=0$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;They are the linear transformations with finite dimensional range.  If the range of $\phi\in \mathrm{End}(V)$ has basis $e_1,\ldots,e_n$, then you can write $\phi(v)=\sum_{j=1}^n c_j(v)e_j$, and each coefficient map $c_j:V\to k$ is linear.  Thus $\phi$ comes from $\sum_{j=1}^n c_j\otimes e_j$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2011-02-27T07:05:06.800" LastActivityDate="2011-02-27T07:05:06.800" CommentCount="1" />
  <row Id="23981" PostTypeId="2" ParentId="23978" CreationDate="2011-02-27T07:00:12.027" Score="2" ViewCount="" Body="&lt;p&gt;You're confusion can be resolved by clarifying one thing (I think). It seems you think that to &quot;solve&quot; a differential equation, you are looking for $y&amp;#39;$? Really, a solution is a function, maybe to decrease confusion we'll call it $f(x)$, with the property that $f&amp;#39;(x)=-[f(x)]^2$. That is what is meant by $y&amp;#39;=-y^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So when we think of the graph of $y=f(x)$ we know that it has the property that $\frac{dy}{dx}=-[f(x)]^2$, or since a square is always positive, and then we take the negative, we know the derivative of $y=f(x)$ is negative and hence the function is decreasing.&lt;/p&gt;&#xA;" OwnerUserId="2000" LastActivityDate="2011-02-27T07:00:12.027" />
  <row Id="23982" PostTypeId="2" ParentId="23978" CreationDate="2011-02-27T07:05:18.107" Score="2" ViewCount="" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Remember if $y&amp;#39;(x) &amp;gt; 0$, $y(x)$ is increasing; if $y&amp;#39;(x) &amp;lt; 0$, $y(x)$ is decreasing; if $y&amp;#39;(x) = 0$ then $y(x)$ is constant. In our case $y&amp;#39;(x) \le 0$ which means $y(x)$ is always constant or decreasing.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;You can verify yourself: if $y(x) = 0$ for all $x$, then $y&amp;#39;(x) = 0$ so it is true that $y&amp;#39; = -y^2$ therefore $y(x) = 0$ is a solution but it isn't in the form $\frac{1}{x + C}$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="7548" LastActivityDate="2011-02-27T07:05:18.107" />
  <row Id="23983" PostTypeId="1" CreationDate="2011-02-27T07:10:23.617" Score="1" ViewCount="166" Body="&lt;p&gt;What are resources for &lt;code&gt;Calculus&lt;/code&gt; you recommend for beginners such that it describes the topics easily?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="7549" LastEditorUserId="7549" LastEditorDisplayName="" LastEditDate="2011-02-27T07:43:23.447" LastActivityDate="2011-02-27T14:17:49.430" Title="Calculus for beginners" Tags="&lt;calculus&gt;" AnswerCount="3" CommentCount="4" FavoriteCount="1" />
  <row Id="23984" PostTypeId="2" ParentId="23983" CreationDate="2011-02-27T07:32:26.647" Score="0" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Books:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Calculus by Michael Spivak&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Calculus by Thomas Finney&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Introduction to Calculus and Analysis, Volume 1 by Richard Courant and Fritz John&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Calculus, Vol. 1: by Tom M. Apostol&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;A Course of Pure Mathematics by G.H. Hardy&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Elementary Analysis: The Theory of Calculus by Kenneth Ross&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Online Resources:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;There are lot of resources online like MIT OCW which provide material as well as problem sheets which you can work and hone your skills.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;I got this video from youtube: &lt;a href=&quot;http://www.youtube.com/watch?v=jbIQW0gkgxo&quot; rel=&quot;nofollow&quot;&gt;http://www.youtube.com/watch?v=jbIQW0gkgxo&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Hope this helps.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Happy Reading!!&lt;/p&gt;&#xA;" LastActivityDate="2011-02-27T07:32:26.647" CommentCount="1" />
  <row Id="23985" PostTypeId="2" ParentId="23944" CreationDate="2011-02-27T07:33:32.913" Score="0" ViewCount="" Body="&lt;p&gt;Is there any specific reason, you need to use KKT conditions to solve this problem? I have written a formal way of solving it below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f(q) = q \log \left( \frac{q}{p} \right) + \left( 1-q \right) \log \left( \frac{1-q}{1-p} \right)$$&#xA;$$\frac{\partial f}{\partial q} = \log \left( \frac{q}{p} \right) + q \times \frac{p}{q} \times \frac{1}{p} - \log \left( \frac{1-q}{1-p} \right) + \left( 1-q \right) \times \frac{1-p}{1-q} \times \frac{-1}{1-p} = \log \left( \frac{q}{p} \frac{1-p}{1-q} \right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now $q = p$ gives the minimum of $0$ since $\frac{\partial^2 f}{\partial q^2} = \frac{1}{q} + \frac{1}{1-q} = \frac{1}{q (1-q)} &amp;gt; 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that when $q&amp;gt;p$, $f&amp;#39;(q) &amp;gt; 0$ and when $q&amp;lt;p$, $f&amp;#39;(q) &amp;lt;0$. This means that the function starts at $q=0$ decreases to a minimum value of $0$ at $q=p$ and then increases and ends at $q=1$. So the maximum can occur either at $q=1$ or $q=0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$q=0$ gives us $f(0) = - \log(1-p)$ and $q=1$ gives us $f(1) = - \log(p)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if $p \in (0,0.5]$, then the maximum occurs at $q=1$ with the maximum being $- \log(p)$ and if $p \in [0.5,1)$, then the maximum occurs at $q=0$ with the maximum being $- \log(1-p)$&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2011-02-27T07:44:39.300" LastActivityDate="2011-02-27T07:44:39.300" CommentCount="7" />
  <row Id="23986" PostTypeId="1" AcceptedAnswerId="23996" CreationDate="2011-02-27T07:45:39.340" Score="3" ViewCount="99" Body="&lt;p&gt;These are a series of questions about Turing machines.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, are the number of a given Turing machine configurations (state + tape) countable?&#xA;Secondly, given that a computation history is a sequence of configurations, are the number of possible computation histories for a given Turing machine also countable?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My intuition on this question is that since configurations have to be finite (I think), then a configuration can be lexicographically ordered and are therefore countable. Computation histories also have to be finite (I think) so are also countable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks. &lt;/p&gt;&#xA;" OwnerUserId="6299" LastEditorUserId="6299" LastEditorDisplayName="" LastEditDate="2011-02-27T07:55:30.877" LastActivityDate="2011-02-27T16:04:30.227" Title="Turing machine configuration and computation history" Tags="&lt;computer-science&gt;&lt;computability&gt;" AnswerCount="1" />
  <row Id="23987" PostTypeId="2" ParentId="23983" CreationDate="2011-02-27T07:46:35.110" Score="1" ViewCount="" Body="&lt;p&gt;Here is the first calculus resource I ever used: &lt;a href=&quot;http://www.calculus-help.com/tutorials&quot; rel=&quot;nofollow&quot;&gt;http://www.calculus-help.com/tutorials&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I watched all of them around 4-5 years ago, and remember liking them.&#xA;It explains limits and the derivative in a very simple way.  I mean in a &lt;em&gt;really&lt;/em&gt; simple way.  I doubt there is a more basic explanation around.  If this is your first time ever looking at calculus, I would definitely give this site a try. (If you feel your mathematics is University level, maybe just skip it)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope that helps,&lt;/p&gt;&#xA;" OwnerUserId="6075" LastActivityDate="2011-02-27T07:46:35.110" />
  <row Id="23989" PostTypeId="1" CreationDate="2011-02-27T08:23:21.323" Score="1" ViewCount="76" Body="&lt;p&gt;Does a binary operation * exist such that (N;+) and (Z;*) are isomorphic?&#xA;N : set of natural numbers, Z : set of integers and + is the addition operation. &#xA;If yes, please give me an isomorphism function?&lt;/p&gt;&#xA;" OwnerUserId="7550" LastActivityDate="2011-02-27T08:42:28.843" Title="Isomorphic of (N,+)?" Tags="&lt;linear-algebra&gt;" AnswerCount="1" FavoriteCount="0" />
  <row Id="23990" PostTypeId="1" AcceptedAnswerId="23992" CreationDate="2011-02-27T08:41:29.077" Score="2" ViewCount="104" Body="&lt;p&gt;$\dot{\theta} \equiv \frac{d \theta(t)}{dt}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I encountered this ODE&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ddot{\theta} + \mu \dot{\theta}^2=0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do I find a solution for $\theta(t)$. I know $\dot{\theta}(t=0)=\omega_0$&lt;/p&gt;&#xA;" LastActivityDate="2011-02-27T08:56:28.033" Title="How to solve $\ddot{\theta} + \mu \dot{\theta}^2=0$" Tags="&lt;differential-equations&gt;" AnswerCount="2" />
  <row Id="23991" PostTypeId="2" ParentId="23989" CreationDate="2011-02-27T08:42:28.843" Score="11" ViewCount="" Body="&lt;p&gt;Whenever you have two sets that are in bijection you can &quot;transport&quot; an operation, or any other algebraic structure, from one to the other. Namely, if $A$ is set endowed with an operation $+$ and $f:A\rightarrow B$ is a bijection of sets, you can define an operation $\ast$ on $B$ simply as&#xA;$$&#xD;&#xA;b\ast b^\prime=f(a+a^\prime)&#xD;&#xA;$$&#xA;where $a$ and $a^\prime$ are the unique elements in $A$ such that $f(a)=b$ and $f(a^\prime)=b^\prime$. Then $(B,\ast)$ is isomoprphic to $(A,+)$ by construction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This can be applied to the question's case since there are surely bijections ${\Bbb N}\rightarrow{\Bbb Z}$. The problem is that the operations on $\Bbb Z$ that you construct in this way are very artificial and quite uninteresting. &lt;/p&gt;&#xA;" OwnerUserId="688" LastActivityDate="2011-02-27T08:42:28.843" />
  <row Id="23992" PostTypeId="2" ParentId="23990" CreationDate="2011-02-27T08:51:18.227" Score="2" ViewCount="" Body="&lt;p&gt;Let $u = \dot{\theta}$. The equation now becomes&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\dot{u}+mu^2=0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This can now simply be solved as a straightforward separable differential equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\dot{u}=-mu^2 \rightarrow \frac{\dot{u}}{u^2}=-m$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which yields&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{1}{u}=mt+C \rightarrow u=\frac{1}{mt+C}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\dot{\theta}$ is now an explicit function of t, so $\theta$ can be integrated directly (with your boundary conditions, so $C$ in this case is $\frac{1}{\omega_0}$).&lt;/p&gt;&#xA;" OwnerUserId="7551" LastEditorUserId="7551" LastEditorDisplayName="" LastEditDate="2011-02-27T08:56:28.033" LastActivityDate="2011-02-27T08:56:28.033" />
  <row Id="23993" PostTypeId="2" ParentId="23990" CreationDate="2011-02-27T08:54:52.297" Score="2" ViewCount="" Body="&lt;p&gt;Let $y = \dot{\theta}$, then we get $\dot{y} + \mu y^2 = 0$. Assuming $y \neq 0$, we get $\frac{dy}{y^2} + \mu dt = 0 \Rightarrow -\frac{1}{y(t)} + \frac{1}{y(0)} + \mu t = 0 \Rightarrow y(t) = \frac{1}{\mu t + \frac{1}{y(0)}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, we have $\frac{d \theta}{dt} = \frac{1}{\mu t + \frac{1}{y(0)}} \Rightarrow \theta(t) = \frac{1}{\mu} \log(\mu t+ \frac{1}{y(0)}) + \theta(0) = \frac{1}{\mu} \log(\mu t+ \frac{1}{\omega_0}) + \theta(0)$&lt;/p&gt;&#xA;" OwnerUserId="3302" LastActivityDate="2011-02-27T08:54:52.297" CommentCount="2" />
  <row Id="23994" PostTypeId="1" AcceptedAnswerId="23995" CreationDate="2011-02-27T08:55:41.317" Score="2" ViewCount="115" Body="&lt;p&gt;For which values of m does the equation:&#xA;$$3 \ln x+m x^3 = 17$$&#xA;have $1$ solution? $2$ solutions? $0$ solution?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="6163" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2011-02-27T17:27:54.147" LastActivityDate="2011-02-27T18:07:19.620" Title="Solutions to equation $3 \ln x+m x^3 = 17$" Tags="&lt;calculus&gt;" AnswerCount="2" />
  <row Id="23995" PostTypeId="2" ParentId="23994" CreationDate="2011-02-27T09:09:26.333" Score="3" ViewCount="" Body="&lt;p&gt;If $m\geq 0$, note that $f(x)=3\ln x +mx^3$ is always increasing, goes to positive infinity as $x$ does, and goes to minus infinity as $x$ decreases to zero.  Also, it is continuous. This is enough information to answer the question in this case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $m\lt 0$, then $f&amp;#39;(x)=3(\frac{1}{x}+mx^2)$ has a unique positive root at $r=\sqrt[3]{-\frac{1}{m}}$.  Thus $f$ is increasing on $(0,r)$ and decreasing on $(r,\infty)$.  Also, $f$ goes to $-\infty$ as $x$ decreases to zero or as $x$ goes to $\infty$.  The number of solutions in this case will be determined by the value of $f(r)$.  The $3$ cases are $f(r)&amp;lt;17$, $f(r)=17$, and $f(r)&amp;gt;17$.  To determine the condition on $m$ when each occurs, you can solve the logarithmic equation you get by plugging in $r$, $f(r)=17$.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2011-02-27T09:09:26.333" CommentCount="2" />
  <row Id="23996" PostTypeId="2" ParentId="23986" CreationDate="2011-02-27T09:23:32.800" Score="3" ViewCount="" Body="&lt;p&gt;Yes. The number of states of a Turing machine are countable (finite in fact), and since the number of arrangements of symbols (which are of finite variety) on the tape are also countable, the cartesian product of these two sets will also be countable. Note: the cartesian product is an upper bound on the number of possible configurations, some combinations of state and tape will be inaccessible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As to your next question, are the number of computational histories countable, the answer is yes, since the histories, as you point out, are a finite number of steps. As a result, you can list them by listing all histories that can be achieved in one step from all strings of length one, and then all histories that can be achieved in two steps from all strings of length one, and then all histories that can be achieve in one step from strings of length 2, and so on. You probably saw this with proof that the rationals are countable, by zigzagging along the infinite square of pairs.&lt;/p&gt;&#xA;" OwnerUserId="7551" LastEditorUserId="7551" LastEditorDisplayName="" LastEditDate="2011-02-27T16:04:30.227" LastActivityDate="2011-02-27T16:04:30.227" CommentCount="1" />
  <row Id="23997" PostTypeId="2" ParentId="23936" CreationDate="2011-02-27T09:26:46.810" Score="1" ViewCount="" Body="&lt;p&gt;Because your integrand grows so fast the whole integral is dominates for $x\approx 100$. We can write $x^x = \exp[ x \ln(x)]$ and then expanding $x \ln(x) = 100 \ln(100) + [1+ \ln(100)] (x- 100) + \cdots$ around $x = 100$ (note that it is important to expand inside the exponent). The integral can therefore be estimated as&#xA;$$\int_1^{100} dx \, x^x \approx 100^{100} \int_{-\infty}^{100} dx\, e^{[1+ \ln(100)] (x- 100) }&#xD;&#xA;= \frac{100^{100}}{1 + \ln (100)}.$$ Numerics shows that this result is off by $3\times 10^{-4}$.&lt;/p&gt;&#xA;" OwnerUserId="7266" LastActivityDate="2011-02-27T09:26:46.810" />
  <row Id="23998" PostTypeId="2" ParentId="23658" CreationDate="2011-02-27T09:48:10.100" Score="4" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Axioms used in Afred Tarski's and Steven Givant's &lt;em&gt;Tarski's System of Geometry&lt;/em&gt; (1991).&lt;/strong&gt;&lt;br&gt;&#xA;Link: &lt;a href=&quot;http://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.27.9012&quot; rel=&quot;nofollow&quot;&gt;http://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.27.9012&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(This is basically a summary of the first part of Tarski's and Givant's paper.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alfred Tarski proved around 1930 that his system of geometry is complete, decidable and that there is a constructive concistency proof for the theory. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;History&lt;/strong&gt;&lt;br&gt;&#xA;Through out the years, the set of Tarski's axioms have changed. At first, there were 20 (1-4, 5$_1$, 6, 7$_2$, 8$^{(2)}$, 9$_1^{(2)}$, 10, 12-21), as well as all instances of axiom schema 11. This made it an axiom system for elementary 2-dimensional geometry. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It turned out that axiom 13 and 19 was derivable from the remaining axioms, and they where omitted. Axiom 20 was replaced by a more precise variant, 20$_1$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In 1956-57 a substantial simplification of the axiom set was obtained by the efforts of Eva Kallin, Scott Taylor and Tarski. The axioms 5$_1$, 7$_2$, 9$_1^{(2)}$ and 10 was respectively replaced by 5, 7$_1$, 9$^{(2)}$ and 10$_1$. In the modified axiom set, the axioms 12, 14, 16, 17, 20$_1$ and 21 are shown to be derivable from the remaining ones. Then we have only the twelve axioms 1-6, 7$_1$, 8$^{(2)}$, 9$^{(2)}$, 10$_1$, 15 and 18, and all instances of the axiom schema 11. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In Tarski's course on the foundations of geometry at the University of California (1956-57), he pointed out that by replacing the axiom schema (11) with the second-order sentence 11 for the full (non-elementary) 2-dimensional Euclidean geometry. It was also pointed out that the axioms 8$^{(2)}$, 9$_1^{(2)}$ could be changed in order to obtain an axiom set for $n$-dimensional geometry. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gupta then showed that axioms 6 and 18 can be derived from the remaining axioms. So what people now usually mean when they talk about Tarski's axioms of geometry, is the axioms 1-5, 7$_1$, 8$^{(n)}$, 9$^{(n)}$ ($n=2,3,...$), 10$_1$, 15 and 11 (either the first-order axiom schema or the second-order single sentence). &lt;em&gt;These are the ones marked with &lt;strong&gt;&lt;/em&gt;*&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Definitions&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The ternary relation $B(abc)$&#xA;expresses that betweenness holds&#xA;between $a,b$ and $c$, which&#xA;intuitively means that $b$ lies on&#xA;the line segment joining $a$ and $d$.&lt;/li&gt;&#xA;&lt;li&gt;The quarternary relation $ab \equiv&#xD;&#xA;   cd$ expresses that the relation of&#xA;equidistance (or congruence of&#xA;segments) holds among the points&#xA;$a,b,c,d$, which intuitively means&#xA;that the distance from $a$ to $b$ is&#xA;the same as the distance from $c$ to&#xA;$d$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The only primitive geometrical object in Tarski's axiom system, are points. All variables $a,b,c,...$  are assumed to range over points.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;*&lt;strong&gt;Axiom 1&lt;/strong&gt; Reflexivity Axiom for Equidistance.&#xA;$$ab\equiv ba$$&#xA;The distance between the points $a$ and $b$ is the same as the distance between $b$ and $a$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;*&lt;strong&gt;Axiom 2&lt;/strong&gt; Transitivity Axiom for Equidistance.&#xA;$$ab \equiv pq \land ab \equiv rs \to pq \equiv rs$$&#xA;If the distance between the points $a$ and $b$ is the same as the distance between $p$ and $q$ and also the same as the distance between $r$ and $s$, then the distance between $p$ and $q$ is the same as the distance between $r$ and $s$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;*&lt;strong&gt;Axiom 3&lt;/strong&gt; Identity Axiom for Equidistance.&#xA;$$ab\equiv cc\to a = b$$&#xA;If the distance between $a$ and $b$ is the same as the distance between $c$ and $c$, then $a$ and $b$ are the same point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;*&lt;strong&gt;Axiom 4&lt;/strong&gt; Axiom of Segment Construction.&#xA;$$\exists x (B(qax)\land ax \equiv bc)$$&#xA;There is a point $x$ such that $a$ lies between $q$ and $x$ and the distance between $a$ and $x$ is equal to the distance between $b$ and $c$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Intuitively this means that, given any line segment $bc$, it is possible to construct a line segment congruent to it (of equal length), starting on any point $a$ and going in the direction of ray that is determined by $a$ and the endpoint $q$ on the ray.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;*&lt;strong&gt;Axiom 5&lt;/strong&gt; Five-Segment Axiom.&#xA;$$[a \neq b \land B(abc)\land B(a&amp;#39;b&amp;#39;c&amp;#39;)\land ab \equiv a&amp;#39;b&amp;#39;\land bc \equiv b&amp;#39;c&amp;#39; \land ad \equiv a&amp;#39;d&amp;#39; \land bd \equiv b&amp;#39;d&amp;#39;]\to cd \equiv c&amp;#39;d&amp;#39;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$a$ and $b$ is not the same point and&lt;/li&gt;&#xA;&lt;li&gt;$b$ lies between $a$ and $c$ and &lt;/li&gt;&#xA;&lt;li&gt;$b&amp;#39;$ lies between $a&amp;#39;$ and $c&amp;#39;$ and &lt;/li&gt;&#xA;&lt;li&gt;the distance between $a$ and $b$ is the same as the distance between $a&amp;#39;$ and $b&amp;#39;$ and&lt;/li&gt;&#xA;&lt;li&gt;the distance between $b$ and $c$ is the same as the distance between $b&amp;#39;$ and $c&amp;#39;$ and&lt;/li&gt;&#xA;&lt;li&gt;the distance between $a$ and $d$ is the same as the distance between $a&amp;#39;$ and $d&amp;#39;$ and&lt;/li&gt;&#xA;&lt;li&gt;the distance between $b$ and $d$ is the same as the distance between $b&amp;#39;$ and $d&amp;#39;$,&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;then the distance between $c$ and $d$ is the same as the distance between $c&amp;#39;$ and $d&amp;#39;$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;The Five-Segment Axiom asserts (in the non-degenerate case) that, given two triangles $\triangle acd$ and $\triangle a&amp;#39;c&amp;#39;d&amp;#39;$, and given interior points $b$ and $b&amp;#39;$ of the sides $ac$ and $a&amp;#39;c&amp;#39;$, from the congruences of certain corresponding pairs of line segments, one can conclude the congruence of another pair of corresponding line segments. Thus, this axiom is similar in character to the well-known theorems of Euclidean geometry that allow one to conclude, from hypotheses about the congruence of certain corresponding sides and angles in two triangles, the congruence of other corresponding sides and angles.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(There is another variant of this axiom at page 179 in the paper i linked to.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Axiom 6&lt;/strong&gt; Identity Axiom for Betweenness.&#xA;$$B(aba) \to a = b$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Axiom 7&lt;/strong&gt; First (or Inner) form of the Pasch Axiom.&#xA;$$B(apc) \land B(bqc) \to \exists x [B(pxb)\land B(qxa)]$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;*&lt;strong&gt;Axiom 7$_1$&lt;/strong&gt; Second (or Outer) form of Pasch Axiom.&#xA;$$B(apc) \land B(qcb) \to \exists x [B(axq)\land B(bpx)]$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $p$ lies between $a$ and $c$, and $c$ lies between $q$ and $b$, then there is an $x$ such that $x$ lies between $a$ and $q$ and $p$ lies between $b$ and $x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;In the outer form of the Pasch Axiom [...] the point $b$ lies on the extension of the side $cq$ in the direction from $q$ to $c$, and the line is assumed to intersect the “inner” side of the triangle (from the perspective of $bp$). The conclusion is that it must intersect the side $aq$ in some point $x$ on the extension of the side $bp$; this is expressed by the assertion $B(bpx)$. In other words, it intersects the &quot;outer&quot; side of the triangle.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Page 180 contains another variant of this axiom.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Axiom 7$_3$&lt;/strong&gt; Weak Pasch Axiom.&#xA;$$B(atd) \land B(bdc) \to \exists x \, \exists y [B(axb) \land B(ayc)\land B(ytx)]$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Axiom 8$^{(1)}$&lt;/strong&gt; Lower 1-Dimensional Axiom.&#xA;$$\exists a \,\exists b \, (a \neq b)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Axiom 8$^{(2)}$&lt;/strong&gt; Lower 2-Dimensional Axiom.&#xA;$$\exists a\, \exists b\, \exists c\, [\neg B(abc) \land \neg B(bca) \land \neg B(cab)]$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;The Lower 2-Dimensional Axiom asserts that there exist three non-collin-&#xA;ear points.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;*&lt;strong&gt;Axiom 8$^{(n)}$&lt;/strong&gt; Lower $n$-Dimensional Axiom for $n = 3, 4, ...$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\exists a \, \exists b \, \exists c \, \exists p_1 \, \exists p_2 \,\cdots \, \exists p_{n-1} \,$ $$\left[ \bigwedge_{1\leq i &amp;lt; j &amp;lt; n} p_i \neq p_j \land \bigwedge_{i=2}^{n-1} a p_1 \equiv a p_i \land \bigwedge_{i=2}^{n-1} b p_1 \equiv b p_i \land \bigwedge_{i=2}^{n-1} c p_1 \equiv c p_i \land [\neg B(abc)\land \neg B(bca) \land \neg B(cab)]\right]$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;The Lower $n$-Dimensional Axiom for $n = 3, 4, . . .$ asserts that there exist&#xA;$n − 1$ distinct points $p_1 , p_2 , . . . , p_{n−1}$ and three points $a, b, c$ such that each of the three points is equidistant from each of the $n-1$ points, but the three points are not collinear.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Axiom 9$^{(0)}$&lt;/strong&gt; Upper 0-Dimensional Axiom.&#xA;$$a=b$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Axiom 9$^{(1)}$&lt;/strong&gt; Upper 1-Dimensional Axiom.&#xA;$$B(abc) \lor B(bca) \lor B(cab)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Axiom 9$_1^{(2)}$&lt;/strong&gt; Upper 2-Dimensional Axiom.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\exists y \,$ {$([B(xya) \lor B(yax) \lor B(axy)] \land B(byc))$&lt;br&gt;&#xA;$([B(xyb) \lor B(ybx) \lor B(bxy)] \land B(cya))$&lt;br&gt;&#xA;$([B(xyc) \lor B(ycx) \lor B(cxy)] \land B(ayb))$}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Page 183 contains another variant of this axiom.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;*&lt;strong&gt;Axiom 9$^{(n)}$&lt;/strong&gt; Upper $n$-Dimensional Axiom (for $n=2,3,...$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left[ \bigwedge_{1\leq i &amp;lt; j \leq n} p_i \neq p_j \land \bigwedge_{i=2}^{n} a p_1 \equiv a p_i \land \bigwedge_{i=2}^{n} b p_1 \equiv b p_i \land \bigwedge_{i=2}^{n} c p_1 \equiv c p_i\right]\to [B(abc)\lor B(bca) \lor B(cab)]$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;The Upper $n$-Dimensional Axiom for $n = 2, 3, . . .$ asserts that any three&#xA;points $a$, $b$, $c$ which are equidistant from each of $n$ distinct points $p_1, p_2, ..., p_n$ must be collinear.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;*&lt;strong&gt;Axiom 10$_1$&lt;/strong&gt; First Form of Euclid’s Axiom.&#xA;$B(adt) \land B(bdc) \land a \neq d \to \exists x \, \exists y \, [B(abx) \land B(acy) \land B(ytx)]$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;The First Form of Euclid’s Axiom says that through any point $t$ in the&#xA;interior of an angle $\triangle bac$ there is a line—here, the line $xy$—that intersects both sides of the angle—here, in the points $x$ and $y$.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Page 183 contains another variant of this axiom.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Axiom 10$_2$&lt;/strong&gt; Second Form of Euclid’s Axiom.&#xA;$$B(abc) \lor B(bca) \lor B(cab) \lor \exists x \, [ax \equiv bx\land ax \equiv cx]$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Axiom 10$_3$&lt;/strong&gt; Third Form of Euclid’s Axiom.&#xA;$$[B(abf) \land ab \equiv bf \land B(ade) \land ad\equiv de \land B(bdc) \land bd \equiv dc] \to bc \equiv fe$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;*&lt;strong&gt;Axiom 11&lt;/strong&gt; Axiom of Continuity.&#xA;$$\exists a\, \forall x \, \forall y [x\in X \land y \in Y \to B(axy)] \to \exists b\, \forall x \, \forall y [x\in X \land y \in Y \to B(xby)]$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;The Axiom of Continuity asserts: any two sets $X$ and $Y$ such that the&#xA;elements of $X$ precede the elements of $Y$ with respect to some point $a$ (that&#xA;is, $B(axy)$ whenever $x$ is in $X$ and $y$ is in $Y$ ) are separated by a point $b$.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;*&lt;strong&gt;Axiom Schema 11(alternative to axiom 11.)&lt;/strong&gt; Axiom Schema of Continuity.&#xA;$$\exists a\, \forall x \, \forall y [\alpha \land \beta \to B(axy)] \to \exists b\, \forall x \, \forall y [\alpha \land \beta \to B(xby)]$$&#xA;where $\alpha, \beta$ are first-order formulas, the first of which does not contain any free occurrences of $a, b, y$ and the second any free occurrences of $a, b, x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can use the Axiom Schema in stead of axiom 11 to keep the axioms within the framework of first-order logic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Axiom 12&lt;/strong&gt; Reflexivity Axiom for Betweenness.&#xA;$$B(abb)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Axiom 13&lt;/strong&gt;&#xA;$$a=b \to B(aba)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Axiom 14&lt;/strong&gt; Symmetry Axiom for Betweenness.&#xA;$$B(abc) \to B(cba)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;*&lt;strong&gt;Axiom 15&lt;/strong&gt; Inner Transitivity Axiom for Betweenness.&#xA;$$B(abd) \land B(bcd) \to B(abc)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Axiom 16&lt;/strong&gt; Outer Transitivity Axiom for Betweenness.&#xA;$$B(abc) \land B(bcd) \land b \neq c \to B(abd)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Axiom 17&lt;/strong&gt; Inner Connectivity Axiom for Betweenness.&#xA;$$B(abd) \land B(acd) \to [B(abc) \lor B(acb)]$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Axiom 18&lt;/strong&gt; Outer Connectivity Axiom for Betweenness.&#xA;$$B(abc) \land B(abd) \land a \neq b \to [B(acd) \lor B(adc)]$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Axiom 19&lt;/strong&gt;&#xA;$$a=b \to ac \equiv bc$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Axiom 20&lt;/strong&gt; Uniqueness Axiom for Triangle Construction.&#xA;$$[ac\equiv ac&amp;#39;\land bc \equiv bc&amp;#39;\land B(adb) \land B(ad&amp;#39;b)\land B(cdx)\land B(c&amp;#39;d&amp;#39;x) \land d \neq x\land d&amp;#39; \neq x] \to c=c&amp;#39;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Page 187 contains another variant of this axiom.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Axiom 21&lt;/strong&gt; Existence Axiom for Triangle Construction.&#xA;$$ab \equiv a&amp;#39;b&amp;#39; \to \exists c \, \exists x \, (ac\equiv a&amp;#39;c&amp;#39; \land bc \equiv b&amp;#39;c&amp;#39; \land B(cxp) \land [B(abx) \lor B(bxa) \lor B(xab)])$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Axiom 22&lt;/strong&gt; Density Axiom for Betweenness.&#xA;$$x \neq z \to \exists y[x\neq y \land z \neq y \land B(xyz)]$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Axiom 23&lt;/strong&gt; &#xA;$$[B(xyz) \land B(x&amp;#39;y&amp;#39;z&amp;#39;)\land xy \equiv x&amp;#39;y&amp;#39;\land yz \equiv y&amp;#39;z&amp;#39;] \to xz \equiv x&amp;#39;z&amp;#39;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Axiom 24&lt;/strong&gt; &#xA;$$[B(xyz) \land B(x&amp;#39;y&amp;#39;z&amp;#39;)\land xz \equiv x&amp;#39;z&amp;#39; \land yz \equiv y&amp;#39;z&amp;#39;] \to xy \equiv x&amp;#39;y&amp;#39;$$&lt;/p&gt;&#xA;" OwnerUserId="7044" LastEditorUserId="7044" LastEditorDisplayName="" LastEditDate="2011-02-27T23:10:12.540" LastActivityDate="2011-02-27T23:10:12.540" />
  <row Id="23999" PostTypeId="2" ParentId="23975" CreationDate="2011-02-27T09:49:23.923" Score="2" ViewCount="" Body="&lt;p&gt;That's a lot of questions! I agree with some of your points about notation, but unfortunately that's just how it's (ab)used in common practice. However, I also think that you're thinking &lt;em&gt;too&lt;/em&gt; much about what everything means. I suspect most authors at this level do not worry about using good notation and distinguishing between free variables and bound variables and instantiated variables and so on. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's start by discussing &lt;em&gt;Lagrange&lt;/em&gt; notation, as this is closest in form to pure-functional notation. If you have a function $f: U \to \mathbb{R}$, $U$ an open subset of $\mathbb{R}$, we say it is differentiable on $U$ if there is a function $f&amp;#39;: U \to \mathbb{R}$ such that $$\forall x. \forall \epsilon. \exists \delta. \forall h. [ \epsilon &amp;gt; 0 \land \delta &amp;gt; 0 \land |h| &amp;lt; \delta \land x \in U \land x + h \in U \implies |f(x) - f(x + h) - h f&amp;#39;(x)| &amp;lt; \epsilon ]$$&#xA;Note that there's no dependence on how functions are defined and that it depends purely on the extensional properties of the function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the problem with understanding Leibniz's notation in this framework is that often one wants to write $\dfrac{df}{dx}$ for $f&amp;#39;(x)$, but then this leads to absurdities such as $\dfrac{df}{d2}$ for $f&amp;#39;(2)$. So what some authors do is write $\dfrac{df}{dx}(2)$ for $f&amp;#39;(2)$... which is also troublesome, as it seems to say that $f$ inherently depends on $x$. I think Leibniz notation is best understood in a framework where one has expressions depending on some fixed set of variables, rather than in a framework where one is dealing with functions and numbers directly. (I personally think Leibniz notation should be abolished. A friend of mine — very competent in calculus — once misinterpreted $f&amp;#39;(2x)$ as meaning the same thing as $\dfrac{d}{dx}[f(2x)]$ when in fact the latter expression is double the former.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similar criticisms can be made about the notation for partial derivatives, but here there is a problem because there are no good alternatives. Mathematica generalises Lagrange notation for this purpose: for example, $f^{(0, 1, 0)}$ means the first partial derivative of $f$ with respect to its second variable. In this notation, the Laplacian of a function $f$ of three variables may be written as $f^{(2, 0, 0)} + f^{(0, 2, 0)} + f^{(0, 0, 2)}$, where addition is pointwise. The problem with this notation is that it assumes that partial differentiation with respect to different variables commutes, but does not hold when the function isn't smooth enough. Another alternative is Einstein's index notation, in which $f_{i,jk}$ means the $i$-th component of the vector-valued function $f$ differentiated with respect to the $j$-th variable, and then again with respect to the $k$-th variable. In Penrose's abstract index notation this instead means the second total derivative of $f$. There's also a somewhat non-standard reading of $\dfrac{\partial}{\partial x}$ where the whole symbol is regarded as a named differential operator. This is usually in the context of differential geometry, and fits in with the identification of tangent vectors as differential operators. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, regarding random variables: I think this is actually relatively easy to sort out compared to the Leibniz notation mess. In the Kolmogorov formalism, a random variable is a function on the underlying sample space. For simplicity of exposition I'll assume we work with real-valued random variables. We can define an &lt;em&gt;algebra&lt;/em&gt; of real-valued random variables by pointwise operations: if I have two random variables $X, Y : \Omega \to \mathbb{R}$, there are random variables $X + Y, XY: \Omega \to \mathbb{R}$, where $(X + Y)(\omega) = X(\omega) + Y(\omega)$ and $(XY)(\omega) = X(\omega) Y(\omega)$, and for each $\lambda \in \mathbb{R}$ there is a random variable which takes the constant value $\lambda$ at all sample points. But better than that, we can apply arbitrary measurable functions $\mathbb{R} \to \mathbb{R}$ to random variables to obtain new ones: so if $f: \mathbb{R} \to \mathbb{R}$ is measurable and $X : \Omega \to \mathbb{R}$ is a random variable, then $f(X) : \Omega \to \mathbb{R}$ is just the composite $f \circ X$. There are other formalisms where the sample space is not only suppressed but entirely absent; Terence Tao has a &lt;a href=&quot;http://terrytao.wordpress.com/2010/02/10/245a-notes-5-free-probability/&quot; rel=&quot;nofollow&quot;&gt;blog post&lt;/a&gt; about this point of view.&lt;/p&gt;&#xA;" OwnerUserId="5191" LastActivityDate="2011-02-27T09:49:23.923" CommentCount="2" />
  <row Id="24000" PostTypeId="1" AcceptedAnswerId="24006" CreationDate="2011-02-27T10:17:03.837" Score="10" ViewCount="392" Body="&lt;p&gt;I have no idea about the underlying theory from which the mathematical induction was derived.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How to prove the mathematical induction is true?&lt;/p&gt;&#xA;" OwnerUserId="3901" LastEditorUserId="3901" LastEditorDisplayName="" LastEditDate="2011-02-27T11:15:07.607" LastActivityDate="2011-02-28T14:01:31.323" Title="How to prove the mathematical induction is true?" Tags="&lt;proof-theory&gt;" AnswerCount="4" CommentCount="4" FavoriteCount="1" />
  <row Id="24001" PostTypeId="1" CreationDate="2011-02-27T10:20:28.033" Score="5" ViewCount="279" Body="&lt;p&gt;The method of Lagrange multipliers is used to find the extrema of $f(x)$ subject to the constraints $\vec g(x)=0$, where  $x=(x_1,\dots,x_n)$ and $\vec g=(g_1,\dots,g_m)$ for $m \leq n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Although many textbooks get the final equations by arguing that at an extrema, the variation of $f(x)$  must be orthogonal to the surface $g(x)=0$, the &quot;simpler&quot; approach (and that which is commonly seen in field theory / optimizing functionals) is to construct the Lagrange function&#xA;$$ L(x,\lambda) = f(x) + \vec\lambda\cdot\vec g(x) $$&#xA;and varying w.r.t. $x$ and $\lambda$ to get the vector equations&#xA;$$&#xD;&#xA;\begin{align}&#xD;&#xA;  &amp;amp;x:&amp;amp;              0 &amp;amp;= \nabla f(x) + \sum_i \lambda_i \nabla g_i(x) \,, \\&#xD;&#xA;  &amp;amp;\vec \lambda:&amp;amp;   0 &amp;amp;= \vec g(x) \ .&#xD;&#xA;\end{align}&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The method only works if the extremal point is a regular point of the constraint surface, i.e. if $\mathrm{rnk}(\nabla\vec g) = m$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the &lt;em&gt;best&lt;/em&gt; way of understanding what goes wrong when the extrema is not a regular point of the constraint?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And, &lt;strong&gt;most importantly&lt;/strong&gt; to me, how does this generalize to field theories (i.e. optimizing functionals) with &lt;strong&gt;local constraints&lt;/strong&gt;? What is the equivalent regularity condition for constraints in field theory?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instructive examples are more than welcome.&lt;/p&gt;&#xA;" OwnerUserId="954" LastEditorUserId="954" LastEditorDisplayName="" LastEditDate="2011-03-10T06:43:06.547" LastActivityDate="2011-03-10T06:43:06.547" Title="How does the method of Lagrange multipliers fail (in classical field theories with local constraints)?" Tags="&lt;functional-analysis&gt;&lt;optimization&gt;&lt;multivariable-calculus&gt;&lt;examples&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" />
  <row Id="24002" PostTypeId="1" AcceptedAnswerId="24007" CreationDate="2011-02-27T10:26:51.013" Score="1" ViewCount="117" Body="&lt;h2&gt;Hi&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;I want to integrate this integral and ask if my work is correct or not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int^\infty_0 dx x^{\alpha-1} e^{-x} (a+bx)^{-\alpha}$$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I want to integrate it by parts, so I have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(a+bx)^{-\alpha} = v$$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$-b\alpha(a+bx)^{-\alpha-1}dx = dv$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x^{\alpha-1} e^{-x} dx = du$$   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Gamma(\alpha) = u$$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;now the integral becomes&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left.\Gamma(\alpha)(a+bx)^{-\alpha}\right|_0^\infty + \int^\infty_0 \Gamma(\alpha) b\alpha(a+bx)^{-\alpha-1}dx = 0$$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;the problem is in integration by parts. Is it correct to put $$\Gamma(\alpha) = u$$. if it is not correct how can I compute this integral? please help.&lt;/p&gt;&#xA;" OwnerUserId="7553" LastEditorUserId="8246" LastEditorDisplayName="" LastEditDate="2011-05-26T09:27:58.473" LastActivityDate="2011-05-26T09:27:58.473" Title="Integration by parts" Tags="&lt;calculus&gt;&lt;analysis&gt;&lt;integral&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" />
  <row Id="24003" PostTypeId="1" AcceptedAnswerId="24008" CreationDate="2011-02-27T10:43:49.737" Score="5" ViewCount="191" Body="&lt;p&gt;On the interval $(0, \infty)$,the function $f \geq 0$,$f' \leq 0$, and $f&amp;#39;&amp;#39; \geq 0$.Prove that $\lim\limits_{x \to \infty} xf&amp;#39;(x) = 0$. &lt;/p&gt;&#xA;" OwnerUserId="4706" LastActivityDate="2011-02-27T22:56:18.003" Title="A calculus question" Tags="&lt;calculus&gt;&lt;functions&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="2" />
  <row Id="24004" PostTypeId="2" ParentId="23799" CreationDate="2011-02-27T10:50:48.550" Score="4" ViewCount="" Body="&lt;p&gt;Replace the non-logical relational symbol ∈ with R and the confusion goes away. Your problem arises from the fact that the symbol ∈ makes you confuse the neutral symbol of the language of set theory (which has no properties other than those expressed by axioms of the theory) with the intended interpretation of the symbol about real sets. The set theories like ZF have very strange models which has nothing to do with the intended model and interpretation of the relation $R$ has nothing to do with the real membership relation between sets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A first-order theory can have many sorts and the intended meaning of some of those sorts can be higher type objects, e.g. we can have a two sorted theory where the intended meaning of the objects of the first sort are natural numbers and the intended meaning of the objects of the second sort are sets of natural numbers. We can quantify over them and the theory is still first order, though it has two sorts. The models of this theory needs to have two sets, one for interpreting the objects of the first sort and the other for interpreting the objects of the second sort, but there does not need to be any relation between these two sets that are used for interpreting objects even if the intended interpretation for the second sort of sets of objects from the first sort. The only properties of that these two sets need to satisfy are those expressed by axioms of the theory. You can add axioms of a set theory like ZFC for the second sort and consider objects of the first sort as urelements of the set theory that satisfying some other axioms e.g. first order Peano Arithmetic. This is still a first order theory. And the theory will have models where the interpretations of the two sort are not related in the intended way, i.e. the objects of the second sort will not be the sets of objects of the first sort, and there are no way to enforce this syntactically, i.e. using axioms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A second- or higher-order theory enforces some restriction on possible interpretations of some sort which are called the higher sort. These restrictions are not syntactical, i.e. they are not axioms of theory, but semantical, i.e. we restrict the models of the theory to those that satisfy some certain conditions, e.g. the members of the set interpreting the second sort in the two sorted number theory I mentioned above are &lt;em&gt;really&lt;/em&gt; the sets of objects of the first sort. This is like assuming some amount of set theory semantically. Without these semantical restrictions about interpretation the theory is still first order.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The higher-order theories are interesting for studying things like natural numbers. The (complete/full) second-order Peano arithmetic will force the set interpreting the second sort to be exactly the powerset of the first sort. Note that this theory is not axiomatizable, i.e. there are no set of first order axioms that will capture exactly these models. (In fact even weaker version of this second-order set theory which do not require the existence of all subsets of the first sort are not axiomatizable.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, I don't know any higher order set theory, and in fact the concept itself seems a little bit unnatural (of course it can be due to my lack of knowledge).&lt;/p&gt;&#xA;" OwnerUserId="468" LastEditorUserId="2420" LastEditorDisplayName="" LastEditDate="2011-02-27T11:42:46.727" LastActivityDate="2011-02-27T11:42:46.727" CommentCount="3" />
  <row Id="24005" PostTypeId="2" ParentId="24000" CreationDate="2011-02-27T10:56:27.610" Score="8" ViewCount="" Body="&lt;p&gt;The principle of induction (on the natural numbers) is equivalent to the axiom of well-foundedness of the natural numbers. Wikipedia gives (half of) a proof &lt;a href=&quot;http://en.wikipedia.org/wiki/Mathematical_induction#Proof_of_mathematical_induction&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="5191" LastActivityDate="2011-02-27T10:56:27.610" />
  <row Id="24006" PostTypeId="2" ParentId="24000" CreationDate="2011-02-27T11:00:13.347" Score="11" ViewCount="" Body="&lt;p&gt;A &quot;proof&quot; in mathematics always means a proof in some &lt;em&gt;system/theory&lt;/em&gt;. You have to specify the system/theory that you want a proof for the induction axiom. (You should also formally specify what you mean by the induction axiom since there are various axioms that are called induction axiom.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The induction axiom in an arithmetical theory (like Peano arithmetic) is an &lt;em&gt;axiom&lt;/em&gt;, i.e. it is one of the axioms of the theory, and therefore the proof is just a single line stating the axiom.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a set theory like $ZFC$ we can prove the induction axiom for the set of natural numbers using the fact that the set of natural numbers is &lt;em&gt;defined&lt;/em&gt; as the smallest inductive set that contains zero, and the proof is almost trivial. (An inductive set means a set that contains the successor of $x$ whenever it contains $x$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In high school or undergraduate courses, when one is asked to prove induction axiom, they are usually asked to derive the induction axiom from some other axioms like the least number principle for natural numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another possible question is what are the justifications for believing that the induction axiom is &lt;em&gt;true&lt;/em&gt; (or for accepting it as an axiom), which is a question in philosophy of mathematics and might be more suitable for MathOverflow.&lt;/p&gt;&#xA;" OwnerUserId="468" LastEditorUserId="468" LastEditorDisplayName="" LastEditDate="2011-02-27T11:40:24.293" LastActivityDate="2011-02-27T11:40:24.293" />
  <row Id="24007" PostTypeId="2" ParentId="24002" CreationDate="2011-02-27T11:15:30.157" Score="0" ViewCount="" Body="&lt;p&gt;Just as an aside, this is the result from &lt;a href=&quot;http://www.wolframalpha.com/&quot; rel=&quot;nofollow&quot;&gt;Wolfram Alpha&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/2IrTU.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="954" LastActivityDate="2011-02-27T11:15:30.157" CommentCount="3" />
  <row Id="24008" PostTypeId="2" ParentId="24003" CreationDate="2011-02-27T11:24:25.613" Score="2" ViewCount="" Body="&lt;p&gt;If the limit is not zero, there is an $\epsilon&amp;gt;0$ such that $xf&amp;#39;(x) \le -\epsilon$ at arbitrarily large $x$. So we can construct a sequence $x_n$ such that $x_n\ge2^n$, $x_{n+1}\ge 2x_n$ and $x_nf&amp;#39;(x_n) \le -\epsilon$ for all $n$. But then the lower halves of the rectangles that the points $(x_n,-\epsilon/x_n)$ form with the origin all lie above the graph of $f&amp;#39;$ (i.e. between it and the $x$ axis), are all disjoint, and all have area $\epsilon/2$. It follows that the indefinite integral of $f&amp;#39;$, i.e. $f$, diverges to $-\infty$, in contradiction with $f\ge0$.&lt;/p&gt;&#xA;" OwnerUserId="6622" LastActivityDate="2011-02-27T11:24:25.613" CommentCount="3" />
  <row Id="24009" PostTypeId="2" ParentId="23946" CreationDate="2011-02-27T11:33:01.087" Score="1" ViewCount="" Body="&lt;p&gt;You are describing &lt;a href=&quot;http://en.wikipedia.org/wiki/Integer_programming&quot; rel=&quot;nofollow&quot;&gt;integer programming&lt;/a&gt;. It's an NP-hard problem, so there are no known efficient (i.e., polynomial-time) solutions in general. However, this does not mean that your particular problem cannot be solved in reasonable time. Try the &lt;a href=&quot;http://www.gnu.org/software/glpk/&quot; rel=&quot;nofollow&quot;&gt;GLPK&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="589" LastActivityDate="2011-02-27T11:33:01.087" />
  <row Id="24010" PostTypeId="2" ParentId="23700" CreationDate="2011-02-27T12:32:42.220" Score="0" ViewCount="" Body="&lt;p&gt;Sorry this is not another answer, but I didn't know how to comment on the answers to my question.&#xA;@mplungjan - thanks for inviting me over to this forum, but i guess this is not the right place for me.&#xA;@Joshua - precisely&#xA;@quanta and lhf - thanks a lot for the answer, i suppose now i have something that i can work on. (i can use the binary representation to extract the sign, depending on the computer's architecture of course)&#xA;@Qiaochu Yuan and Thomas Rot - I believe I wasn't accurate enough with the question, I am writing a program and I want to get the functionality i mentioned in as few instructions as possible. So if I want to define the square function, I can define it as&#xA;F(x) = 0 if x=0,&#xA;     = 1 if x=1,&#xA;     = 4 if x=2,&#xA;.&#xA;.&#xA;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or&lt;/p&gt;&#xA;&#xA;&lt;p&gt;F(x) = x^x&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is why i wanted to &quot;express&quot; F in terms of x. Otherwise I can even define F to compute seemingly impossible results for me&#xA;F(x) = largest prime number, if x=0;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps it is the word &quot;function&quot; that is the root of all misunderstanding here. For me, &quot;function&quot; is a procedure - a set of instructions, and in this context, I want to implement that procedure without using the if-else construct. Hope I cleared up the confusion caused earlier by my limited understanding of functions. Thanks for all the replies guys, you are one of the most responsive forums I've ever visited!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regards,&#xA;Anil Raj&lt;/p&gt;&#xA;" OwnerUserId="7555" LastActivityDate="2011-02-27T12:32:42.220" CommentCount="2" />
  <row Id="24011" PostTypeId="2" ParentId="24001" CreationDate="2011-02-27T12:40:48.103" Score="7" ViewCount="" Body="&lt;p&gt;Generically, the $m$ equations $g_i(x)=0$ define a manifold $S$ of dimension $d:=n-m$. At each point $p\in S$ the $m$ gradients $\nabla g_i(p)$ are orthogonal to the tangent space $S_p$ of $S$ at $p$. The condition rnk$(\nabla g(p))=m$ means that these $m$ gradients are linearly independent, so that they span the full orthogonal complement $S_p^\perp$ which has dimension $m=n-d$. At a conditionally stationary point $p$  of $f$ the gradient $\nabla f(p)$ is in $S_p^\perp$, and if the rank condition is fulfilled, there will be constants $\lambda_i$ such that $\nabla f(p)=\sum_{i=1}^m \lambda_i\nabla g_i(p)$. In this case the given &quot;recipe&quot; will find the point $p$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider now the following example where the rank condition is violated: The two constraints&#xA;$$g_1(x,y,z):=x^6-z=0,\qquad g_2(x,y,z):=y^3-z=0$$&#xA;define a curve $S\subset{\mathbb R}^3$ with the parametric representation $$S: \quad x\mapsto (x,x^2,x^6)\qquad (-\infty &amp;lt; x &amp;lt;\infty).$$&#xA;The function $f(x,y,z):=y$ assumes its minimum on $S$ at the origin $o$. But if we compute the gradients&#xA;$$\nabla f(o)=(0,1,0), \qquad \nabla g_1(o)=\nabla g_2(o)=(0,0,-1),$$&#xA;it turns out that $\nabla f(o)$ is not a linear combination of the $\nabla g_i(o)$. As a consequence Lagrange's method will not bring this conditionally stationary point to the fore.&lt;/p&gt;&#xA;" OwnerUserId="1303" LastEditorUserId="1303" LastEditorDisplayName="" LastEditDate="2011-02-27T18:53:12.127" LastActivityDate="2011-02-27T18:53:12.127" CommentCount="4" />
  <row Id="24012" PostTypeId="1" AcceptedAnswerId="24014" CreationDate="2011-02-27T12:49:18.097" Score="5" ViewCount="92" Body="&lt;p&gt;I have a homework question where we have to prove the following definitions of a binomial coefficient are equal, algebraically.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/6NJ6S.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is what I got so far, and it's getting pretty complicated. And I could use some directions on how to continue. At the moment I think I'm just doing it wrong from the start and I'm overcomplicating things. But I'm not quite sure how to continue, because I'm just staring at these numbers and can't continue.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have writting this in Microsoft Word, so I'll use images to show what I've got so far, because writing this all again in Latex is tedious.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/q7zRd.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also this is homework for me, so please don't give me the answer yet, because I know I could just look this up on the internet. I'd like to prove it myself for most of the part, but since I'm stuck I was wondering if someone could give me a slight hint on what to do.&lt;/p&gt;&#xA;" OwnerUserId="3605" LastActivityDate="2011-02-27T13:10:05.587" Title="Showing two definitions of a binomial coefficient are the same" Tags="&lt;homework&gt;&lt;binomial-coefficients&gt;&lt;proof-strategy&gt;" AnswerCount="1" CommentCount="9" FavoriteCount="0" />
  <row Id="24013" PostTypeId="1" AcceptedAnswerId="33439" CreationDate="2011-02-27T12:52:22.140" Score="2" ViewCount="212" Body="&lt;p&gt;Hello,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;how can I please calculate an arc length of $\dfrac{e^x-e^{-x}}{2}$. I tried to substitute $\dfrac{e^x-e^{-x}}{2}=\sinh x$, which leads to $\int\sqrt{1+\cosh^2x}dx$, which unfortunately I can't solve.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you very much.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;C.&lt;/p&gt;&#xA;" OwnerUserId="7557" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2011-02-27T15:56:51.760" LastActivityDate="2011-04-17T18:05:27.763" Title="arc length of $\dfrac{e^x - e^{-x}}{2}$" Tags="&lt;real-analysis&gt;&lt;analysis&gt;&lt;integral&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="24014" PostTypeId="2" ParentId="24012" CreationDate="2011-02-27T13:10:05.587" Score="4" ViewCount="" Body="&lt;p&gt;You were on the right track, but you haven't chosen the easiest/lowest common denominator.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you've written,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$${{n-1} \choose {k-1}}+{{n-1} \choose k}=\frac{(n-1)!}{(n-k)!(k-1)!}+\frac{(n-1)!}{(n-k-1)!k!}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Isn't one of the denominator a multiple of the other one ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remember that $(n+1)!=(n+1)\cdot n \cdots 2\cdot 1=(n+1)\cdot n!$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that you can also prove this using a &lt;a href=&quot;http://en.wikipedia.org/wiki/Combinatorial_proof&quot; rel=&quot;nofollow&quot;&gt;combinatorial proof&lt;/a&gt;, which will give you a more intuitive idea of why the equality holds.&lt;/p&gt;&#xA;" OwnerUserId="524" LastActivityDate="2011-02-27T13:10:05.587" CommentCount="3" />
  <row Id="24015" PostTypeId="2" ParentId="23975" CreationDate="2011-02-27T13:31:23.247" Score="0" ViewCount="" Body="&lt;p&gt;I do not propose this as a proper answer, but it is too long for a comment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same problem was troubling me some time ago. IMHO Leibniz notation may be put on a formal ground, but context where it works is vividly different. Something like: Calculus is done in a “diagram”. A diagram is a set of variables and assignments&lt;/p&gt;&#xA;&#xA;&lt;p&gt;variable→set&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(list variable)×variable→function&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Functions go between variables. The diagram is commuting in the category theory sense. So, given a variable v, we can get a function to v from the diagram. This allows to “differentiate a variable”. (We also can define a diagram as a product preserving functor from a finite poset category with products to Set.) Every expression involving operations like sum and product creates a new variable v and a new function …→v . Likewise the differential operator. This allows to speak formally about partial derivatives also, prove the chain rule etc. However, I did not pursued my idea further and never came across something in the literature.&lt;/p&gt;&#xA;" OwnerUserId="7011" LastActivityDate="2011-02-27T13:31:23.247" />
  <row Id="24016" PostTypeId="1" AcceptedAnswerId="24018" CreationDate="2011-02-27T13:36:14.973" Score="5" ViewCount="162" Body="&lt;p&gt;We know the Stone-Weierstrass theorem for locally compact Hausdorff spaces (LCH) which states the following:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Theorem:&lt;/strong&gt; Suppose $X$ is LCH. A subalgebra $\mathcal{A}$ of $C_0(X)$ is dense if and only if it &lt;em&gt;separates points&lt;/em&gt; ($\forall x,y \in X : x\neq y \implies \exists f \in \mathcal{A}: f(x) \neq f(y)$) and &lt;em&gt;vanishes nowhere&lt;/em&gt; $(\forall x \in X \exists f \in \mathcal{A} : f(x) \neq 0$) and is closed under complex conjugation&lt;sup&gt;1&lt;/sup&gt;. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It's also easy to show that $C_b(X)$, the continuous and bounded functions on a topological space $X$, is a Banach space&lt;sup&gt;2&lt;/sup&gt;. It would be obvious to try and show a variant of Stone-Weierstrass for $C_b(X)$ where $X$ is merely Hausdorff, and since it's obviously dense in itself, for such a theorem to exist it would be required that $C_b(X)$ separates points (that it vanishes nowhere is clear: it contains the constant functions). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've tried to come up with an example of a Hausdorff space where $C_b(X)$ fails to separate points (this could for example be a space where every non-constant continuous function is unbounded), but to no avail. So does anyone happen to have an instructive example lying around?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS: It's an interesting exercise to prove that given a LCH space $X$ the continuous functions with compact support, $C_{00}(X)$, separates points and vanishes nowhere. You get to use many theorems from topology in the process of constructing a continuous function such that $f(x)=1$ and $f(y)=0$ given distinct points $x,y \in X$ [Hint: Find a good compact subset and use Urysohn's lemma].&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[1]: In the case of real-valued functions this is essentially a no-op, so including it in the theorem statement doesn't hurt.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[2]: &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0521497566&quot; rel=&quot;nofollow&quot;&gt;[Car00]&lt;/a&gt; shows in &lt;a href=&quot;http://books.google.com/books?id=4VFDVy1NFiAC&amp;amp;lpg=PP1&amp;amp;dq=real%20analysis&amp;amp;pg=PA153#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;Lemma 10.8&lt;/a&gt;  that $B(X)$, the set of bounded functions on a set $X$ is a Banach space, and it's an immediate corollary from &lt;a href=&quot;http://books.google.com/books?id=4VFDVy1NFiAC&amp;amp;lpg=PP1&amp;amp;dq=real%20analysis&amp;amp;pg=PA150#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;Thm 10.4&lt;/a&gt; that $C_b(X)$ is closed in $B(X)$ for $X$ a metric space. Of course this  generalises readily to the case of $X$ a topological space.&lt;/p&gt;&#xA;" OwnerUserId="1269" LastActivityDate="2011-02-27T20:05:42.173" Title="Example of Hausdorff space $X$ s.t. $C_b(X)$ does not separate points?" Tags="&lt;real-analysis&gt;&lt;topology&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="24017" PostTypeId="2" ParentId="24013" CreationDate="2011-02-27T13:43:52.830" Score="2" ViewCount="" Body="&lt;p&gt;There is no solution for arbitrary integration limits in elementary terms: we face an &lt;a href=&quot;http://en.wikipedia.org/wiki/Elliptic_integral&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;elliptic integral&lt;/em&gt;&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.wolframalpha.com/input/?i=int%28sqrt%281%2Bcosh%28x%29%5E2%29%2Cx%29&quot; rel=&quot;nofollow&quot;&gt;Wolfram alpha output&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="367" LastEditorUserId="5363" LastEditorDisplayName="" LastEditDate="2011-02-27T14:18:32.860" LastActivityDate="2011-02-27T14:18:32.860" CommentCount="6" />
  <row Id="24018" PostTypeId="2" ParentId="24016" CreationDate="2011-02-27T13:53:55.800" Score="8" ViewCount="" Body="&lt;p&gt;Looking at the table at the back of &lt;a href=&quot;http://en.wikipedia.org/wiki/Counterexamples_in_Topology&quot; rel=&quot;nofollow&quot;&gt;Counterexamples in Topology&lt;/a&gt;, the first space which is Hausdorff but not Urysohn (which in that book means $C(X)$ doesn't separate points) is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Prime_integer_topology&quot; rel=&quot;nofollow&quot;&gt;relatively prime integer topology&lt;/a&gt;. That is, the topology on $\mathbb{N}\setminus \{0\}$ with basis all sets of the form $\{b+na|n \in \mathbb{N}\}$ for $a$, $b$ coprime.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hausdorff is easy: if $k, l \in \mathbb{N}\setminus \{0\}$, take a prime $p$ larger than both and consider the open neighbourhoods $\{k+np \}, \{l+np \}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$C(X)$ fails to separate points because any two basic open sets have nondisjoint closures, since the closure of  $\{b+na \}$ contains all multiplies of $a$.&lt;/p&gt;&#xA;" OwnerUserId="5203" LastEditorUserId="5203" LastEditorDisplayName="" LastEditDate="2011-02-27T14:06:14.230" LastActivityDate="2011-02-27T14:06:14.230" />
  <row Id="24020" PostTypeId="2" ParentId="23938" CreationDate="2011-02-27T14:17:03.000" Score="2" ViewCount="" Body="&lt;p&gt;All students have difficulties when they meet the $O$- and the $o$-notations for the first time. Whenever such an $O$ appears it refers to a pre-agreed limit process for the independent variable, in the case at stake to $n\to\infty$. The statement $p(n)=O(n^k)\ (n\to\infty)$  does not mean that the (usually &quot;complicated&quot;) function $p(n)$ is equal to some other function $O(\cdot)$, evaluated at $n^k$. Instead it expresses the (claimed or proven) fact that the function $p(n)$ under study,  after  division by $n^k$, stays bounded when $n\to\infty$; which is the same thing as saying that $p(n)=b(n)\cdot n^k$ where now $b(n)$ is a bounded function. In your example, each $n^i/n^k$ is $\leq 1$, so in fact $|p(n)|\leq C\cdot n^k$ with $C:=\sum_{i=1}^d |a_i|$.&lt;/p&gt;&#xA;" OwnerUserId="1303" LastActivityDate="2011-02-27T14:17:03.000" CommentCount="1" />
  <row Id="24021" PostTypeId="2" ParentId="23983" CreationDate="2011-02-27T14:17:49.430" Score="3" ViewCount="" Body="&lt;p&gt;Another very helpful site is &lt;a href=&quot;http://www.khanacademy.org/&quot; rel=&quot;nofollow&quot;&gt;http://www.khanacademy.org/&lt;/a&gt;. It has videos on almost any subject and they are only about 10minutes long. They are clear and easy to understand. Khan teaches the idea and does example problems. I use it all the time. &lt;/p&gt;&#xA;" OwnerUserId="7347" LastActivityDate="2011-02-27T14:17:49.430" />
  <row Id="24022" PostTypeId="2" ParentId="8814" CreationDate="2011-02-27T14:40:43.403" Score="2" ViewCount="" Body="&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/zBPEc.png&quot; alt=&quot;Parallelogram&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left|z+z&amp;#39;\right|^{2}+\left|z-z&amp;#39;\right|^{2}=2\times\left(\left|z\right|^{2}+\left|z&amp;#39;\right|^{2}\right)$$&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The sum of the squares of the sides&#xA;  equals the sum of the squares of the&#xA;  diagonals.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2380" LastActivityDate="2011-02-27T14:40:43.403" CommunityOwnedDate="2011-02-27T14:40:43.403" />
  <row Id="24023" PostTypeId="2" ParentId="24016" CreationDate="2011-02-27T14:59:15.273" Score="4" ViewCount="" Body="&lt;p&gt;I'll just point out that &lt;em&gt;boundedness&lt;/em&gt; is not the obstruction: any space that admits a nonconstant real-valued continuous function admits one which is bounded, and if $C(X,\mathbb{R})$ separates points then so does $C_b(X,\mathbb{R})$.  This is easy to see: if $f$ is continuous and $f(x_1)=a &amp;lt; b = f(x_2)$, then $g(x) = (f(x) \vee a) \wedge b$ is continuous and bounded and also has $g(x_1)=a&amp;lt;b=g(x_2)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, continuous functions with compact support may not be enough if $X$ is not locally compact, even if it is metrizable.  For instance, it's a nice exercise to check that if $X$ is an infinite-dimensional Banach space, the only real-valued continuous function on $X$ with compact support is the zero function, i.e. in your notation $C_{00}(X) = \{0\}$.  (I think most authors call this space either $C_0(X)$ or $C_c(X)$.)&lt;/p&gt;&#xA;" OwnerUserId="822" LastEditorUserId="822" LastEditorDisplayName="" LastEditDate="2011-02-27T20:05:42.173" LastActivityDate="2011-02-27T20:05:42.173" CommentCount="2" />
  <row Id="24024" PostTypeId="1" AcceptedAnswerId="24028" CreationDate="2011-02-27T15:16:47.013" Score="1" ViewCount="44" Body="&lt;p&gt;Okay, so my text required me to actually prove both sides; The non-empty subset $W$ is a subspace of a $V$ if and only if $a\mathbf{x}+b\mathbf{y} \in W$ for all scalars $a,b$ and all vectors $\mathbf{x},\mathbf{y} \in W$. I figured out one direction already (that if $W$ is a subspace, then $a\mathbf{x}+b\mathbf{y}$ is an element of $W$ since $a\mathbf{x}$ and $b\mathbf{y}$ are in $W$ and thus so is their sum), but I'm stuck on the other direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I got that if $a\mathbf{x}+b\mathbf{y} \in W$, then $c(a\mathbf{x}+b\mathbf{y}) \in W$ as well since we can let $a&amp;#39; = ca$ and $b&amp;#39; = cb$ and we're good, so $W$ is closed under scalar multiplication. But for closure under addition, my text states that I can &quot;cleverly choose specific values for $a$ and $b$&quot; such that $W$ is closed under addition as well but I cannot find any values that would work. What I'm mostly confused about is how choosing specific values for $a$ and $b$ would prove anything, since $a, b$ can be any scalars and $\mathbf{x},\mathbf{y}$ can be any vectors, so setting conditions like $a = b$, $a = -b$, $a = 0$ or $b = 0$ don't seem to prove anything.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also something I'm not sure about is if they're saying that $a\mathbf{x}+b\mathbf{y} \in W$, am I to assume that that is the only form? So if I'm testing for closure under addition, I have to do something like $(a\mathbf{x}+b\mathbf{y})+(c\mathbf{z}+d\mathbf{w})$?&lt;/p&gt;&#xA;" OwnerUserId="1562" LastActivityDate="2011-02-27T15:39:37.770" Title="If $a\mathbf{x}+b\mathbf{y}$ is an element of the non-empty subset $W$, then $W$ is a subspace of $V$." Tags="&lt;linear-algebra&gt;&lt;homework&gt;" AnswerCount="4" />
  <row Id="24025" PostTypeId="1" CreationDate="2011-02-27T15:23:44.430" Score="0" ViewCount="82" Body="&lt;p&gt;I recently found the primitive element theorem a little bit unnatural, and hence I was trying to replace the proofs of some theorems using the primitive element theorem, and here is one.&lt;br&gt;&#xA;In the book by Jurgen Neukirch, the theorem that states (1) and (2) are equivalent, and uses the primitive element theorem:&lt;br&gt;&#xA;$$\begin{array}{rcl}&#xD;&#xA;&amp;amp;L&amp;#39;&amp;amp;\\&#xD;&#xA;&amp;amp;/\backslash\\&#xD;&#xA;L &amp;amp;&amp;amp;K&amp;#39;\\&#xD;&#xA;&amp;amp;\backslash/ \\&#xD;&#xA;&amp;amp;K&amp;amp;&#xD;&#xA;\end{array}$$&#xA;In this diagram, $L&amp;#39;$ is just the compositum of $L$ and $K&amp;#39;$.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The following are equivalent:&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;(1) $L|K$ is unramified.  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;(2) $L&amp;#39;|K&amp;#39;$ is unramified. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And he uses the primitive element to show one implication, while the other is a direct consequence of the first.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I was trying to deal with this, the concept of value groups came across me, and I am wondering if the unit group of $L|K$ is contained in that of $L&amp;#39;|K&amp;#39;$; if this is the case, then since $L|K$ is unramified, the unit group of $L|K$ is just $\{1\}$, and hence $L&amp;#39;|K&amp;#39;$ is unramified as well.&lt;br&gt;&#xA;Nevertheless, this is how far I am now, thanks for any clarification or elaboration.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Edit:&lt;/em&gt; Here is a sketch of how the primitive element theorem works:&lt;br&gt;&#xA;Fisrtly, since $L|K$ is unramified, the residue class field extension is separable, hence there exists a primitive element of the residue class field extension.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then we lift it to an element of $L|K$, and use the minimal polynomial of that element to deduce that $[L&amp;#39;:K&amp;#39;]$ is in fact equal to the degree of the residue class field extension of $L&amp;#39;/K&amp;#39;$, therefore we have proved what was to be proved.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This only serves as a supplement, if anyone needs a full proof, in my opinion, this is contained in every book on the valuation and ramification theory.&lt;br&gt;&#xA;Thanks very much.   &lt;/p&gt;&#xA;" OwnerUserId="6792" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-28T02:56:39.970" LastActivityDate="2011-02-28T02:56:39.970" Title="Are the correspondences between value groups and field extensions order preserving?" Tags="&lt;abstract-algebra&gt;&lt;algebraic-number-theory&gt;" CommentCount="12" FavoriteCount="1" />
  <row Id="24026" PostTypeId="2" ParentId="24024" CreationDate="2011-02-27T15:26:38.920" Score="0" ViewCount="" Body="&lt;p&gt;Consider $a=b=1$ and $a=-b=1$.&lt;/p&gt;&#xA;" OwnerUserId="5253" LastActivityDate="2011-02-27T15:26:38.920" />
  <row Id="24027" PostTypeId="1" AcceptedAnswerId="24031" CreationDate="2011-02-27T15:27:07.320" Score="4" ViewCount="124" Body="&lt;p&gt;We define cardinality as an equivalence relation on sets. But the class of all sets is not a set, so how do we do that? In particular, I'm interested in the proposition that equivalence classes form a partition of the initial set. It seems like it can be translated to cardinality, but I do not know how, at least in ZFC (and I don't even know ZFC :))&lt;/p&gt;&#xA;" OwnerUserId="3793" LastActivityDate="2011-02-27T16:15:02.013" Title="Equivalence relation on a proper class" Tags="&lt;set-theory&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="24028" PostTypeId="2" ParentId="24024" CreationDate="2011-02-27T15:29:18.763" Score="3" ViewCount="" Body="&lt;p&gt;You seem to be misunderstanding $\forall x,y \in W, \forall a,b,  ax+by \in W$.&#xA;This doesn't say that for a specific pair $(x,y) \in W^2$, every element of $W$ would be of the form $ax+by$ for some scalars $a$ and $b$. It does not say anything like $&amp;quot;\ldots \forall w \in W \ldots w =ax+by&amp;quot;$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, when you proved that $\forall w \in W, \forall c, cw \in W$,&#xA;you implicitly supposed that $w$ was of this form, while you never stated who $x$ and $y$ were. If you thought that $x$ and $y$ were fixed members of $W$, then it's false.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, you can say that since $w \in W$, and $0$ is a scalar, your hypothesis says that if you &lt;i&gt;choose&lt;/i&gt; $x=w, y=w, a=c, b=0$, you get $cw+0w \in W$. Thus $cw \in W$.&#xA;This is a valid proof of $\forall w \in W, \forall c, cw \in W$, and if you understand it, you should be able to prove the rest.&lt;/p&gt;&#xA;" OwnerUserId="6270" LastActivityDate="2011-02-27T15:29:18.763" />
  <row Id="24029" PostTypeId="2" ParentId="24024" CreationDate="2011-02-27T15:30:20.710" Score="2" ViewCount="" Body="&lt;p&gt;It seems to me that your question arises out of a misunderstanding of the expression &quot;$a\mathbf{x}+b\mathbf{y} \in W$ for all scalars $a,b$ and all vectors $\mathbf{x},\mathbf{y} \in W$&quot;. You seem to be taking that to mean that all elements of $W$ are of the form $a\mathbf{x}+b\mathbf{y}$ (which they are, but that's irrelevant). It just means that if you pick &lt;em&gt;any&lt;/em&gt; scalars $a$, $b$ and &lt;em&gt;any&lt;/em&gt; vectors $\mathbf{x},\mathbf{y} \in W$, then $a\mathbf{x}+b\mathbf{y} \in W$. So you're not &quot;setting conditions&quot; if you pick particular values of $a$ and $b$, you're just using what's given -- and what's given is far more general than what you're trying to prove, so it makes sense that you have to pick specific values to prove it.&lt;/p&gt;&#xA;" OwnerUserId="6622" LastActivityDate="2011-02-27T15:30:20.710" CommentCount="1" />
  <row Id="24030" PostTypeId="2" ParentId="24024" CreationDate="2011-02-27T15:39:37.770" Score="0" ViewCount="" Body="&lt;p&gt;As you may be a little bit confused this, let me remember you that&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$W\subset$ is a subspace of $V$ if and&#xA;  only if:&lt;/p&gt;&#xA;  &#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;$W\not=\emptyset$&lt;/li&gt;&#xA;  &lt;li&gt;$w+w&amp;#39;\in W$ for all $w,w&amp;#39;\in W$&lt;/li&gt;&#xA;  &lt;li&gt;$aw&amp;#39;\in W$ for all $w&amp;#39;\in W, \ a\in\mathbb{R}$&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Thus if $W$ is a subspace of $V$, it is, as you say, easy to see that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$aw+bw&amp;#39;\in W \ \forall \ w,w&amp;#39;\in W, \ a,b\in \mathbb{R}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;if you combine 2 and 3.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the other side, suppose that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$aw+bw&amp;#39;\in W \ \forall \ w,w&amp;#39;\in W, \ a,b\in \mathbb{R}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and you have to prove that then the three points above are true.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$W$ is not empty by definition.&lt;/li&gt;&#xA;&lt;li&gt;Take $w, w&amp;#39;\in W$. Letting $a=1, \ b=1$ gives that $w+w&amp;#39;\in W$&lt;/li&gt;&#xA;&lt;li&gt;Take $w\in W$. Letting $a=1, \ b=0$ gives that $1w+0w&amp;#39;=w\in W$&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;So you see that they are simply particular cases of the relation.&lt;/p&gt;&#xA;" OwnerUserId="524" LastActivityDate="2011-02-27T15:39:37.770" />
  <row Id="24031" PostTypeId="2" ParentId="24027" CreationDate="2011-02-27T16:15:02.013" Score="4" ViewCount="" Body="&lt;p&gt;We can of course define cardinality as you say: Two sets are equipotent (or have the same cardinality) iff there is a bijection between them. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can prove directly that this notion is reflexive, symmetric and transitive. For example, the last statement is: For any sets $A,B,C$, if there is a bijection from $A$ to $B$ and a bijection from $B$ to $C$, then there is a bijection from $A$ to $C$. Note that this does not require that we reference directly the collection of equivalence classes or even that we consider a single equivalence class as a given object. What I mean is: We can do all that we need to do without talking about proper classes or collections of proper classes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What would be the statement corresponding to &quot;the equivalence classes of an equivalence relation on a non-empty set form a partition of the set into non-empty subsets&quot;? Simply the conjunction of the following two statements: &#xA; 1. &quot;Cardinality is defined for all sets&quot;, which simply means that given any two sets $A$ and $B$, the statement &quot;$A$ and $B$ have the same cardinality&quot; is meaningful, and is either true or false. But of course that is the case, since $A$ and $B$ have the same cardinality iff there is a bijection between them, this is the very definition. In fact, &quot;$A$ and $B$ have the same cardinality&quot; is simply a linguistic shortcut for &quot;there is a bijection between $A$ and $B$&quot;. &#xA; 2. &quot;Given two sets $A$ and $B$, if there is a set $C$ such that $A$ and $C$ have the same cardinality and also $B$ and $C$ have the same cardinality, then so do $A$ and $B$.&quot; And this can be easily proved in the expected way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All of this can be easily formalized in set theory (ZFC or even much weaker systems). Again, the point is that there is no need to directly argue about proper classes or collections of classes (but, if you want, then there are also appropriate set theories, such as MK, where this is possible).&lt;/p&gt;&#xA;" OwnerUserId="462" LastActivityDate="2011-02-27T16:15:02.013" CommentCount="5" />
  <row Id="24032" PostTypeId="1" AcceptedAnswerId="24041" CreationDate="2011-02-27T16:20:45.490" Score="21" ViewCount="323" Body="&lt;p&gt;My 7-year-old son was staring at the graph of tan() and its endlessly-repeating serpentine strokes on the number line between multiples of $\pi$ and he asked me the question in the title.  More precisely, is the following true or false?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For any $\epsilon &amp;gt; 0$, there exists some $N \in \mathbb{Z}^+$ such that &#xA;$|\tan(N)-\lfloor \tan(N) \rceil| &amp;lt; \epsilon$.&lt;/p&gt;&#xA;" OwnerUserId="7162" LastEditorUserId="7162" LastEditorDisplayName="" LastEditDate="2011-02-27T17:46:35.383" LastActivityDate="2011-03-02T07:05:32.630" Title="Can we make $\tan(x)$ arbitrarily close to an integer when $x\in \mathbb{Z}$?" Tags="&lt;calculus&gt;&lt;real-analysis&gt;" AnswerCount="2" CommentCount="5" FavoriteCount="4" />
  <row Id="24033" PostTypeId="1" AcceptedAnswerId="24036" CreationDate="2011-02-27T16:22:27.830" Score="0" ViewCount="57" Body="&lt;p&gt;What are good computational geometry journals? I'm also counting geometric modeling as part of computational geometry. Are there any good journals that are considered to be non-mainstream in the US?&lt;/p&gt;&#xA;" OwnerUserId="6299" LastEditorUserId="7676" LastEditorDisplayName="" LastEditDate="2011-03-20T19:11:12.810" LastActivityDate="2011-03-20T19:11:12.810" Title="Computational Geometry journals" Tags="&lt;reference-request&gt;&lt;computational-geometry&gt;" AnswerCount="2" />
  <row Id="24034" PostTypeId="2" ParentId="24033" CreationDate="2011-02-27T16:34:51.297" Score="2" ViewCount="" Body="&lt;p&gt;I highly recommend Discrete and Computational Geometry: &lt;a href=&quot;http://www.springer.com/mathematics/numbers/journal/454&quot; rel=&quot;nofollow&quot;&gt;http://www.springer.com/mathematics/numbers/journal/454&lt;/a&gt;  If what you want is an excellent survey of materials related to discrete and computational geometry I also recommend this Handbook: &lt;a href=&quot;http://cs.smith.edu/~orourke/books/discrete.html&quot; rel=&quot;nofollow&quot;&gt;http://cs.smith.edu/~orourke/books/discrete.html&lt;/a&gt;  edited by Joseph O'Rourke and Jacob Eli Goodman.&lt;/p&gt;&#xA;" OwnerUserId="1369" LastEditorUserId="237" LastEditorDisplayName="" LastEditDate="2011-03-20T16:54:42.450" LastActivityDate="2011-03-20T16:54:42.450" />
  <row Id="24035" PostTypeId="2" ParentId="24003" CreationDate="2011-02-27T16:38:56.600" Score="-1" ViewCount="" Body="&lt;p&gt;For $x&amp;gt;1$, $f(x)$ is bounded below by $0$ and above by $f(1)$.  Therefore $\lim_{x\to\infty} {f(x)\over \log x}=0$.  By l'Hopital's rule, $\lim_{x\to\infty} xf&amp;#39;(x) =0$ as well.&lt;/p&gt;&#xA;" OwnerUserId="7562" LastActivityDate="2011-02-27T16:38:56.600" CommentCount="2" />
  <row Id="24036" PostTypeId="2" ParentId="24033" CreationDate="2011-02-27T16:44:47.713" Score="1" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://www.elsevier.com/locate/comgeo&quot; rel=&quot;nofollow&quot;&gt;Computational Geometry: Theory and Applications&lt;/a&gt; and &lt;a href=&quot;http://ejournals.wspc.com.sg/ijcga/ijcga.shtml&quot; rel=&quot;nofollow&quot;&gt;International Journal of Computational Geometry and Applications&lt;/a&gt; are the ones that focus on algorithms. For geometric modeling, there's &lt;a href=&quot;http://www.elsevier.com/locate/cad&quot; rel=&quot;nofollow&quot;&gt;Computer Aided Design&lt;/a&gt;, &lt;a href=&quot;http://www.elsevier.com/locate/cagd&quot; rel=&quot;nofollow&quot;&gt;Computer Aided Geometric Design&lt;/a&gt;, &lt;a href=&quot;http://www.elsevier.com/locate/gmod&quot; rel=&quot;nofollow&quot;&gt;Graphical Models&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="589" LastActivityDate="2011-02-27T16:44:47.713" />
  <row Id="24037" PostTypeId="1" AcceptedAnswerId="24044" CreationDate="2011-02-27T16:44:47.857" Score="4" ViewCount="137" Body="&lt;p&gt;Hi, I posted this on mathOverflow previously which was the wrong place to post it and I was asked to try this forum instead. Can anyone explain this in simple terms:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I met projective space via a recent class on perspective drawing, believe it or not, but I didn't know that this was the &quot;space&quot; we were using. I came across a more detailed description trawling the net. In a book on point-set topology that I bought, it describes Euclidean n-space as a field made of (sorry I don't know how to write mathematical symbols yet):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[ {n-tuples of reals}, Op(&quot;+&quot;), Op(&quot;.&quot;) ]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what is the equivalent set-theoretic description for projective space? I haven't been able to find one anywhere. All I've found is that basically it is constructed by taking a regular plane and adding the 'horizon' line but I want to understand mathematically what it is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT(2): I guess it's the fact of being on the sphere that matters. I think I get it... The whole antipodal points minus origin thing is how you would &lt;strong&gt;&lt;em&gt;describe&lt;/em&gt;&lt;/strong&gt; the surface of a sphere, isn't it? Parallel lines in R3 would hit the spherical boundary and then NECESSARILY converge as they follow that new curved surface... Yay! :o)&lt;/p&gt;&#xA;" OwnerUserId="7563" LastEditorUserId="7563" LastEditorDisplayName="" LastEditDate="2011-02-28T03:01:20.327" LastActivityDate="2011-02-28T03:01:20.327" Title="Is there a set-theoretic definition of Projective Space?" Tags="&lt;topology&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="24038" PostTypeId="1" CreationDate="2011-02-27T16:19:29.187" Score="1" ViewCount="139" Body="&lt;p&gt;I'm a little confused as to the results of multiplying vectors and scalars. If I multiply a vector and a scalar, is it a vector? And if I multiply a vector and another vector, is the results also a vector? I looked at &lt;a href=&quot;http://mathworld.wolfram.com/VectorMultiplication.html&quot; rel=&quot;nofollow&quot;&gt;http://mathworld.wolfram.com/VectorMultiplication.html&lt;/a&gt; but still haven't figured it out.&lt;/p&gt;&#xA;" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2011-02-27T17:12:25.880" LastActivityDate="2011-02-27T17:16:57.430" Title="Vector multiplication with scalars" Tags="&lt;linear-algebra&gt;&lt;geometry&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="24039" PostTypeId="2" ParentId="24038" CreationDate="2011-02-27T16:36:54.743" Score="1" ViewCount="" Body="&lt;p&gt;If $\alpha$ is a scalar and $\mathbf{v}$ is a vector that can be decomposed into some basis $\mathbf{e}_i$, then&#xA;$$\alpha \mathbf{v} = \alpha v^i \mathbf{e}_i\, ,$$&#xA;thus remaining a vector.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other vector multiplications where you get out a vector as well (e.g. the cross product) can be seen as $$\mathbf{v}\times \mathbf{w} = A^i_{kl}v^k w^l \mathbf{e}_i$$ but there are also other ways you can define a map from some $\mathbb{R}^n$ to a $\mathbb{R}^m$ like a scalar product etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It depends on what you want to calculate what you will get out. You may consider some Wikipedia articles such as &lt;a href=&quot;http://en.wikipedia.org/wiki/Linear_algebra&quot; rel=&quot;nofollow&quot;&gt;linear algebra&lt;/a&gt; to specify your questions in the future.&lt;/p&gt;&#xA;" OwnerUserId="2671" LastActivityDate="2011-02-27T16:36:54.743" />
  <row Id="24040" PostTypeId="2" ParentId="24038" CreationDate="2011-02-27T16:41:49.360" Score="2" ViewCount="" Body="&lt;p&gt;Ok, this is a math question really, but a 'physics answer' could go as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Velocity is a vector: it has a size (speed) and a direction. If you multiply a velocity vector by - say - a factor of two, you go twice as fast in the same direction. So multiplying a (velocity) vector with a scalar gives again a (velocity) vector. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the multiplication of two vectors. There are several meaningful definitions of vector multiplications. Out of these, the inner product multiplication (or dot product) is most abundant in physics. We take force vectors and displacement vectors as examples. Dot multiplying the force vector exerted on a particle with its displacement vector yields the amount of work done. This amount of work is a scalar (a number without further attributes such as direction). So the dot product between two vectors gives a scalar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As I mentioned, other definitions of products between vectors are possible. For instance, the direct product between two vectors yields a (rank two) tensor. No need to be bothered about this in basic physics though. &lt;/p&gt;&#xA;" LastActivityDate="2011-02-27T16:41:49.360" CommentCount="1" />
  <row Id="24041" PostTypeId="2" ParentId="24032" CreationDate="2011-02-27T16:49:43.527" Score="27" ViewCount="" Body="&lt;p&gt;Yes, you can.  In fact, you can make $\tan(N)$ arbitrarily close to your favorite integer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pick your favorite integer, which is probably 17.  Consider $x=\arctan 17$.  Then $\tan(x)=17$ and also $\tan(x+n\pi)=17$ for all integers $n$.  If you could make $x+n\pi$ very close to an integer (say $K$), then since the tangent function is continuous, $\tan(K)$ will be very close to 17.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the problem is now to make $x+n\pi$ very close to an integer.  This can be done because the multiples of $\pi$ (or any irrational number) are dense mod 1, so you can make $n\pi$ arbitrarily close to $-x$ mod 1.  This takes proof but it is elementary:  think about a circle of circumference 1, and start walking around the circle, around and around, marking a point each time your total distance travelled is a multiple of $\pi$.  You'll see the marks start to fill in the circle, and as you go on and on, the largest interval without a mark in it keeps getting smaller.  Eventually you will mark a point that's within $\epsilon$ of $-x$.&lt;/p&gt;&#xA;" OwnerUserId="6071" LastEditorUserId="6071" LastEditorDisplayName="" LastEditDate="2011-02-27T18:08:50.243" LastActivityDate="2011-02-27T18:08:50.243" CommentCount="4" />
  <row Id="24043" PostTypeId="2" ParentId="24037" CreationDate="2011-02-27T16:56:06.053" Score="2" ViewCount="" Body="&lt;p&gt;See my answer from MO.  The point is that a line through the origin is determined by a point (actually a pair of points) on the sphere, so if you want an $n$-dimensional space then you need to take lines through 0 in $R^{n+1}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Think of a line through 0 as a &quot;direction.&quot;  For example the space of lines through 0 in $R^2$ is 1-dimensional.  (In fact it is a circle.)  In general projective space is not just the same as an $n$-dimensional sphere because two antipodal points (opposite ends of a diameter) represent the same line through 0, hence the same point of projective space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus another way to define $n$-dimensional projective space is as a quotient of an $n$-dimensional sphere by identifying antipodal points.  The $n$-dimensional sphere is defined as a subset of $R^{n+1}$.&lt;/p&gt;&#xA;" OwnerUserId="6071" LastActivityDate="2011-02-27T16:56:06.053" CommentCount="5" />
  <row Id="24044" PostTypeId="2" ParentId="24037" CreationDate="2011-02-27T16:58:27.953" Score="0" ViewCount="" Body="&lt;p&gt;Albraically a projective space is $\frac{V \setminus \underline{0}}{\sim}$, where $\sim$ is an equivalence relation defined as $\underline{x} \sim \underline{y} \Leftrightarrow \underline{x} = \rho \underline{y} \; \rho \in \mathbb{R}^{*}$, $V$ is a vector space, $\underline{0}$ is the null vector and $\mathbb{R}^* = \mathbb{R} \setminus \{ 0 \}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically, what the definition says is that a projective space is a vector space without the zero vector and where all the vectors that multiples of each other are identified. This leads to topological representations of projective spaces such as the projective line, which can be identified as a circle (if you add the &quot;point at infinity&quot; on a line, it closes).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for your intuition, it is right. What happens in a projective space is that you &quot;add&quot; some &quot;points at infinity&quot; which identify a direction (all parallel lines &quot;end&quot; in the same point).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Like aaron points out, since we qutient out all multiples, the projective space has a dimension less than the vector space we construct it from.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: re-reading the OP's comment, I see I answered only partially. The point that the horizon is a circle is derived from a topological model of the 2 dimensional projective space (projective plane). You can see it as your &quot;normal&quot; cartesian plane, where you add a &quot;circle&quot; at infinity, where all the antipodal points are identified. This means that the points lying on the circumference on the first quadrant are the same as the points on the circumference on the third (same for the second and fourth quadrant). This is just a description of what aaron said: you're taking all lines from the origin and identifying their directions (their &quot;points at infinity&quot;) so that all parallel lines end up at the same point. (I could draw something on the computer, but it would look really silly, because I can't program anything that would do it for me automatically, so I'd have to use MS Paint) So, yes, the horizon is actually a circle and &quot;in perspective&quot; parallel lines meet &quot;at infinity&quot;, which was the motivation for artists to start and think mathematically of perspective (this is, I think, one of the historical origins to the projective plane). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the projective line, you should try to think to add one single point to a regular line, let's say &quot;at infinity&quot;. What happens is that the line starts from this point very far away and ends in this same point (intuitively, a line has only &quot;one infinity&quot; to go to) so it closes in a circle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this more explanatory? :)&lt;/p&gt;&#xA;" OwnerUserId="1415" LastEditorUserId="1415" LastEditorDisplayName="" LastEditDate="2011-02-27T23:57:12.727" LastActivityDate="2011-02-27T23:57:12.727" CommentCount="4" />
  <row Id="24045" PostTypeId="1" AcceptedAnswerId="24052" CreationDate="2011-02-27T17:00:46.023" Score="2" ViewCount="107" Body="&lt;p&gt;$$y=\frac{x^{5}}{6}+\frac{1}{10x^{3}}\qquad 1\leq x\leq 2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{dy}{dx}=\frac{5}{6}x^{4}-\frac{3}{10x^{4}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;squaring this $$=\frac{25}{36}x^{8}+\frac{9}{100x^{8}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Plugging into the formula $$ds=\sqrt{1+\left( \frac{dy}{dx}\right) ^{2}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_{1}^{2}\sqrt{1+\frac{25}{36}x^{8}+\frac{9}{100x^{8}}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this correct so far? And how would I go about evaluating this integral.&lt;/p&gt;&#xA;" OwnerUserId="6331" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2011-02-27T17:49:43.413" LastActivityDate="2011-02-27T22:11:33.220" Title="Find the length of the curve: $y=\frac{x^{5}}{6}+\frac{1}{10x^{3}}\qquad 1\leq x\leq 2$" Tags="&lt;calculus&gt;" AnswerCount="2" CommentCount="10" />
  <row Id="24046" PostTypeId="2" ParentId="24038" CreationDate="2011-02-27T17:07:03.487" Score="1" ViewCount="" Body="&lt;p&gt;Multiplication on scalar is an operation that must be defined on every vector space. Multiplicaton of two vectors is trickier, however, there is no single right way of doing it. Indeed, for any vector space $V$ over a field $k$ you can fix any $k$-bilinear mapping $f: V \times V \to V$ and call it multiplication. Then (V, f) is called a $k$-algebra. You can further require $f$ to satisfy some properties and get associative algebra, unital associative algebra, commutative associative algebra, division algebra, or maybe Lie algebra which is much different :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or you can fix a $k$-bilinear mapping $f: V \times V \to k$ and call it a scalar product. Depending on what additional properties you require, it may turn, for example, into (pseudo)-euclidean form or into symplectic form, which is much different. You get the picture :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some concrete examples with $\mathbb{R}^n$ as the vector space:&lt;br&gt;&#xA;$(\mathbf{x}, \mathbf{y}) \mapsto \mathbf{x} \cdot \mathbf{y} = \sum_{i = 1}^n x^i y^i$ is a euclidean form,&lt;br&gt;&#xA;$(\mathbf{x}, \mathbf{y}) \mapsto \sum_{i = 1}^n k_i x^i y^i$ $(k_i \in \lbrace 1, -1 \rbrace)$ is a pseudo-euclidean form,&lt;br&gt;&#xA;$(\mathbf{x}, \mathbf{y}) \mapsto \mathbf{x} \times \mathbf{y}$ gives you a Lie algebra,&lt;br&gt;&#xA;$((x^1, \ldots, x^n), (y^1, \ldots, y^n)) \mapsto (x^1 y^1, \ldots, x^n y^n)$ gives you a division algebra (not sure about that, though).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, in $\mathbb{R}^2$ you can define a very well-behaved algebra :)&lt;/p&gt;&#xA;" OwnerUserId="3793" LastEditorUserId="3793" LastEditorDisplayName="" LastEditDate="2011-02-27T17:16:57.430" LastActivityDate="2011-02-27T17:16:57.430" />
  <row Id="24047" PostTypeId="1" CreationDate="2011-02-27T17:15:01.760" Score="0" ViewCount="45" Body="&lt;p&gt;Let $X$ and $Y$ be (irreducible, quasi-projective) varieties (over an algebraically closed field $k$), and $\phi: X \to Y$ be a rational map. I think I understand what it means for $\phi$ to be regular at a point $x \in X$, but I'm lost as to how to prove that $x$ &lt;em&gt;is&lt;/em&gt; (or &lt;em&gt;isn't&lt;/em&gt;) a regular point of $\phi$, when I am given a single representative of $\phi$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, suppose $X = Y = \mathbb{P}^2$ is the projective plane, and $\phi$ is the rational map determined by $\phi([x : y: z]) = [yz : zx : xy]$. This uniquely defines a rational map since $\mathbb{P}^2$ is irreducible and the equation makes sense away from $P = \{ [1 : 0 : 0], [0 : 1 : 0], [0 : 0 : 1] \}$, which is a closed set. I can see that $\phi$ is a birational equivalence (and even an involution!), and it seems intuitive that $\phi$ cannot be regular on $P$. But how do I prove this rigorously?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, for a twist, consider the line $L \subset \mathbb{P}^2$ which satisfies the equation $x + y + z = 0$. Clearly, $L \cap P = \emptyset$, and $\phi(L)$ is the conic $xy + yz + zx = 0$, and $P \subset \phi(L)$. It turns out $\phi$ restricted to $\phi(L)$ is a &lt;em&gt;morphism&lt;/em&gt; of varieties $\phi(L) \to L$, because we've thrown away enough of the obstructions to regularisation. So regularity seems to be something a bit more subtle than staring at polynomials... but the way I showed that this restriction did turn $\phi$ into an honest morphism was precisely that: I fiddled about with polynomials and equations until I got a consistent set which covered the whole curve. Is there a better / more general way of finding the points at which a a rational map is regular?&lt;/p&gt;&#xA;" OwnerUserId="5191" LastActivityDate="2011-02-27T17:15:01.760" Title="Finding the regular points of a rational map" Tags="&lt;algebraic-geometry&gt;" CommentCount="1" />
  <row Id="24048" PostTypeId="2" ParentId="23700" CreationDate="2011-02-27T17:24:34.707" Score="0" ViewCount="" Body="&lt;p&gt;Any algorithm that does what you want will be dependent upon the F(x) you are working with.  Some F's are easy-for x^2 you can take the integer sqrt of x, square it, and see if you get x back again.  For x^x you could just store a table of them up to the largest number of interest and search the table.  Unless you are doing arbitrary precision work it will be much less than 1000 entries.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-02-27T17:24:34.707" />
  <row Id="24049" PostTypeId="1" CreationDate="2011-02-27T17:25:50.480" Score="5" ViewCount="170" Body="&lt;p&gt;Let $(R,m)$ be commutative noetherian local ring with unity. Suppose $P$ is a finitely generated projective module over $R[X]$ of rank $n$ . Is $P$ free? If not,what is the counter example?&lt;/p&gt;&#xA;" OwnerUserId="7565" LastActivityDate="2011-02-28T06:26:29.407" Title="projective module" Tags="&lt;commutative-algebra&gt;" AnswerCount="1" CommentCount="6" />
  <row Id="24050" PostTypeId="2" ParentId="24045" CreationDate="2011-02-27T17:43:23.517" Score="3" ViewCount="" Body="&lt;p&gt;&quot;dy/dx = (5x^4)/6 + -3/(10x^{-4}) squaring this = (25x^8)/36 + 9/(100x^-8)&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Do not write two arithmetic operations one next to the other. It leads to confusion. Use brackets instead. For example instead of 3+-5, write 3+(-5)&lt;br&gt;&#xA;2) As for the square, as pointed out, use the binomial formula $(a+b)^2=a^2+2ab+b^2$ and its variant $(a-b)^2=a^2-2ab+b^2$&lt;br&gt;&#xA;3) Learn the basics of Latex. &lt;/p&gt;&#xA;" OwnerUserId="7243" LastActivityDate="2011-02-27T17:43:23.517" CommentCount="2" />
  <row Id="24052" PostTypeId="2" ParentId="24045" CreationDate="2011-02-27T18:04:20.217" Score="1" ViewCount="" Body="&lt;p&gt;As Américo Tavares pointed out, you are missing the term $-1/2$ due to the crossterm when squaring $d y/ dx$. The integral you want to solve is&#xA;$$\int_1^2 dx \,\sqrt{\frac{1}{2}+ \frac{25 x^8}{36} + \frac{9}{100 x^8}}&#xD;&#xA;= \int_1^2 dx \, \sqrt{\frac{(9+25 x^8)^2}{900 x^8}}$$&#xA;$$ = \int_1^2 dx \,\frac{9+25 x^8}{30 x^4} = \left[-\frac{1}{10 x^3} + \frac{x^4}{6} \right]_{x=1}^2 =  \frac{1261}{240}. $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope every step is reproducible.&lt;/p&gt;&#xA;" OwnerUserId="7266" LastEditorUserId="7266" LastEditorDisplayName="" LastEditDate="2011-02-27T22:11:33.220" LastActivityDate="2011-02-27T22:11:33.220" CommentCount="8" />
  <row Id="24053" PostTypeId="2" ParentId="23994" CreationDate="2011-02-27T18:07:19.620" Score="2" ViewCount="" Body="&lt;p&gt;Can't improve on Jonas Meyer's answer, but playing with &lt;a href=&quot;http://omnium-gatherum.appspot.com/pages/so23994.html&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; geogebra applet might help you get a feel for the equation. Move the slider to change $m$ in increments of 1/1000000.&lt;/p&gt;&#xA;" OwnerUserId="130" LastActivityDate="2011-02-27T18:07:19.620" CommentCount="1" />
  <row Id="24054" PostTypeId="1" CreationDate="2011-02-27T18:08:05.550" Score="1" ViewCount="201" Body="&lt;p&gt;OK, so I have the following polar equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$r = Θ/20$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And I would like to translate this a little to the right, and down from the polar origin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, I figure since I know cartesian coordinate equation translations are quite simple, the best way to do this would be to convert this polar equation to a rectangular equation, translate, and then convert back to polar again.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I was having some problems making the initial conversion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know we are supposed to use the following relationships between polar and rectangular equations:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$r^2 = x^2 + y^2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$y = rsin(Θ)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, I cannot seem to convert my equation correctly...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would someone be able to suggest a simpler way to make this translation?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or, assist me in converting to polar and back again?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm quite new at this, so I apologize if this question is remedial.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="7568" LastActivityDate="2011-02-28T09:10:04.383" Title="Converting polar equation to cartesian coordinate polar equation and back again?" Tags="&lt;algebra-precalculus&gt;&lt;functional-equations&gt;&lt;polar-coordinates&gt;" AnswerCount="2" />
  <row Id="24056" PostTypeId="1" AcceptedAnswerId="24058" CreationDate="2011-02-27T18:31:03.527" Score="3" ViewCount="147" Body="&lt;p&gt;From I.N.Herstein's &lt;em&gt;Topics in Algebra.&lt;/em&gt; &lt;strong&gt;Chap 5 Sec 5.3 Page 227 Problem 8&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;Problem 8&lt;/strong&gt;: If $n&amp;gt;1$ prove that the splitting field of $x^{n}-1$ over the field of rational numbers is of degree $\Phi(n)$ where $\Phi$ is the Euler $\Phi$-function. ( This is a well known theorem. I know of no easy soluion, so don't be disappointed if you fail to get it. If you get an easy proof, I would like to see it.)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;First, I would like to see a proof of this result. Next, I think I have seen this proof in Dummit and Foote's &lt;em&gt;Abstract Algebra&lt;/em&gt; book, but not sure. Anyway, next question is: Has an easy solution been found to this problem? If not, I would like to know what efforts have been taken to make the proof more simple. And why does Herstein think an easy solution can exist.&lt;/p&gt;&#xA;" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-27T19:15:11.997" LastActivityDate="2011-02-27T19:15:11.997" Title="Splitting field of $x^{n}-1$ over $\mathbb{Q}$" Tags="&lt;abstract-algebra&gt;&lt;algebra&gt;&lt;field-theory&gt;&lt;galois-theory&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="24057" PostTypeId="2" ParentId="24054" CreationDate="2011-02-27T18:43:14.507" Score="2" ViewCount="" Body="&lt;p&gt;The other equation you need is $\theta = \arctan \frac {y}{x}$ and you have to worry about the branches of arctan.  So now you have $\sqrt{x^2+y^2}=\frac{1}{20} \arctan \frac {y}{x}$.  To translate right and down, you replace $x$ with $x+a$ and $y$ with $y-b$.  And you have quite a mess.  Taking $\sqrt{(x+a)^2+(y-b)^2}=\frac{1}{20} \arctan \frac {y-b}{x+a}$ back to polar is a lot of work and probably not illuminating.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-02-27T18:43:14.507" CommentCount="3" />
  <row Id="24058" PostTypeId="2" ParentId="24056" CreationDate="2011-02-27T18:52:55.217" Score="5" ViewCount="" Body="&lt;p&gt;I think the difficulty is proving that the $n$th &lt;a href=&quot;http://en.wikipedia.org/wiki/Cyclotomic_polynomial&quot; rel=&quot;nofollow&quot;&gt;cyclotomic polynomial&lt;/a&gt; is irreducible. Wikipedia says it's a non-trivial result. This gives a factorization of $x^n-1$ as the product of all cyclotomic polynomials $\Phi_d$ for $d$ dividing $n$.&lt;/p&gt;&#xA;" OwnerUserId="589" LastActivityDate="2011-02-27T18:52:55.217" />
  <row Id="24059" PostTypeId="2" ParentId="23850" CreationDate="2011-02-27T19:05:01.047" Score="7" ViewCount="" Body="&lt;p&gt;This is a slightly more detailed version of some of my comments on your &lt;a href=&quot;http://cstheory.stackexchange.com/questions/5162/are-there-any-known-barriers-to-use-some-approach-for-solving-p-vs-np&quot;&gt;cross-posting on cstheory.stackexchange.com&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The statement which you described can be concisely written as NP ⊈ P-close.  Here &lt;a href=&quot;http://qwiki.stanford.edu/index.php/Complexity_Zoo%3aP#pclose&quot; rel=&quot;nofollow&quot;&gt;P-close&lt;/a&gt; is the class of decision problems for which there exists a polynomial-time algorithm A such that the set of instances on which A fails to answer correctly is sparse.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is easy to see that P ⊆ P-close ⊆ P/poly, from which it is easy to see the implications hold that NP ⊈ P/poly ⇒ NP ⊈ P-close ⇒ P≠NP.  Since NP ⊈ P-close implies P≠NP, any proof of NP ⊈ P-close must also overcome the relativization barrier and the algebrization barrier.  I do not know if the natural-proof barrier (which every proof of NP ⊈ P/poly must overcome) necessarily applies to NP ⊈ P-close.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;del&gt;I do not think that it is known that NP ⊈ P-close is equivalent to P≠NP as you claim.&lt;/del&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Edit&lt;/em&gt;: On the contrary to what I wrote in an earlier revision, I learned that NP ⊈ P-close is indeed equivalent to P≠NP.  Although I already answered your question about barriers above, I guess that writing down the proof of this equivalence may be useful.  The proof is based on what you described on cstheory.stackexchage.com with one modification (namely, I use the result by Ogihara and Watanabe instead of Mahaney’s theorem).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As stated above, we have the implication NP ⊈ P-close ⇒ P≠NP.  We will prove the converse: NP ⊆ P-close ⇒ P=NP.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A &lt;em&gt;polynomial-time &lt;i&gt;k&lt;/i&gt;-truth-table reduction&lt;/em&gt; from a language &lt;i&gt;L&lt;/i&gt;&lt;sub&gt;1&lt;/sub&gt; to a language &lt;i&gt;L&lt;/i&gt;&lt;sub&gt;2&lt;/sub&gt; is a Turing reduction from &lt;i&gt;L&lt;/i&gt;&lt;sub&gt;1&lt;/sub&gt; to &lt;i&gt;L&lt;/i&gt;&lt;sub&gt;2&lt;/sub&gt; which invokes the oracle at most &lt;i&gt;k&lt;/i&gt; times nonadaptively.  Note that a many-one reduction is a special case of a 1-truth-table reduction.  Ogihara and Watanabe [OW91] proved the following result:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Theorem&lt;/strong&gt; [OW91].  If some sparse language is NP-complete under polynomial-time &lt;i&gt;k&lt;/i&gt;-truth-table reducibility for some constant &lt;i&gt;k&lt;/i&gt;, then P=NP.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that this theorem generalizes &lt;a href=&quot;http://blog.computationalcomplexity.org/2006/04/favorite-theorems-small-sets.html&quot; rel=&quot;nofollow&quot;&gt;Mahaney’s theorem&lt;/a&gt;, which is the special case of the theorem where the reduction is restricted to a polynomial-time many-one reduction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume NP ⊆ P-close.  Then SAT ∈ P-close.  Equivalently, there exists a language &lt;i&gt;L&lt;/i&gt;∈P such that the symmetric difference &lt;i&gt;S&lt;/i&gt;=SAT△&lt;i&gt;L&lt;/i&gt; is sparse.  Then the following is a polynomial-time 1-truth-table reduction from SAT to &lt;i&gt;S&lt;/i&gt;: given an input &lt;i&gt;x&lt;/i&gt;, decide (in polynomial time) whether &lt;i&gt;x&lt;/i&gt;∈&lt;i&gt;L&lt;/i&gt; and decide (by invoking the oracle for &lt;i&gt;S&lt;/i&gt;) whether &lt;i&gt;x&lt;/i&gt;∈&lt;i&gt;S&lt;/i&gt;, and return the XOR of the two results.  Therefore, the sparse set &lt;i&gt;S&lt;/i&gt; is NP-complete under polynomial-time 1-truth-table reducibility.  This implies P=NP by the aforementioned theorem by Ogihara and Watanabe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[OW91] Mitsunori Ogihara and Osamu Watanabe.  On polynomial-time bounded truth-table reducibility of NP sets to sparse sets.  &lt;em&gt;SIAM Journal on Computing&lt;/em&gt;, 20(3):471–483, June 1991.  &lt;a href=&quot;http://dx.doi.org/10.1137/0220030&quot; rel=&quot;nofollow&quot;&gt;http://dx.doi.org/10.1137/0220030&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="926" LastEditorUserId="926" LastEditorDisplayName="" LastEditDate="2011-02-28T16:14:40.867" LastActivityDate="2011-02-28T16:14:40.867" CommentCount="11" />
  <row Id="24060" PostTypeId="1" AcceptedAnswerId="24065" CreationDate="2011-02-27T19:38:31.233" Score="2" ViewCount="139" Body="&lt;p&gt;I need to prove that if $f: (0,1) \rightarrow \mathbb{R}$ is Uniformly continuous then it is  bounded. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you.&lt;/p&gt;&#xA;" OwnerUserId="6163" LastEditorUserId="6075" LastEditorDisplayName="" LastEditDate="2011-02-27T20:07:19.100" LastActivityDate="2011-02-27T20:07:19.100" Title="Uniform continuity" Tags="&lt;real-analysis&gt;&lt;analysis&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="24062" PostTypeId="2" ParentId="24016" CreationDate="2011-02-27T19:48:42.367" Score="1" ViewCount="" Body="&lt;p&gt;I recommend &lt;a href=&quot;http://www.math.uga.edu/~pete/Gantner71.pdf&quot; rel=&quot;nofollow&quot;&gt;this short 1971 note of T.E. Gantner&lt;/a&gt;.  He first proves:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Theorem: Each regular space $Z$ can be embedded as a subspace of a regular space $Q(Z)$ such that every continuous, real-valued function on $Q(Z)$ is constant on $Z$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;He applies the theorem as follows: take $X_0$ to be a one-point space, and inductively, for all $n \in \mathbb{N}$, define $X_{n+1} = Q(X_n)$.  Let $X = \lim_{n \rightarrow \infty} X_n$, endowed with the direct limit (or &lt;a href=&quot;http://en.wikipedia.org/wiki/Final_topology&quot; rel=&quot;nofollow&quot;&gt;final&lt;/a&gt;) topology.  Then $X$ is an infinite regular space on which every continuous real-valued function is constant.&lt;/p&gt;&#xA;" OwnerUserId="299" LastActivityDate="2011-02-27T19:48:42.367" />
  <row Id="24063" PostTypeId="2" ParentId="24060" CreationDate="2011-02-27T19:51:35.827" Score="5" ViewCount="" Body="&lt;p&gt;Here's one approach.  You can use uniform continuity (with $\varepsilon=1$, say) to show that $f$ is bounded on $(0,\delta)$ and $(1-\delta,1)$.  You probably already know the theorem that implies that $f$ is bounded on $[\delta,1-\delta]$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you don't know about boundedness of continuous functions on $[a,b]$, then what you can do here is cover $(0,1)$ with a finite number of tiny intervals where you know (using uniform continuity) that $f$ can't vary by more than $1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Uniformly continuous functions can also be extended to the closure, so an approach that would actually do more would be to show that $\lim_{x\to1}f(x)$ and $\lim_{x\to 0}f(x)$ exist, so that you may consider $f$ to be a restriction of a continuous function on $[0,1]$.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2011-02-27T19:51:35.827" CommentCount="4" />
  <row Id="24065" PostTypeId="2" ParentId="24060" CreationDate="2011-02-27T19:53:49.913" Score="5" ViewCount="" Body="&lt;p&gt;Recall that if $f$ is uniformly continuous, than given $\epsilon&amp;gt;0$ we can find $\delta&amp;gt;0$ such that $|x-y|&amp;lt;\delta$ implies $|f(x)-f(y)|&amp;lt;\epsilon$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To show it is bounded, it doesn't really matter what $\epsilon$ is, so let it be some fixed constant.  Let $N=\lfloor\frac{1}{\delta}\rfloor$ and take&#xA;$$x_1=\delta,\  x_2=2\delta,\dots,\  x_n=n\delta,\dots, x_N=N\delta.$$  Notice every $y\in (0,1)$ satisfies $|y-x_i|&amp;lt;\delta$ for some $i$. Then $|f|$ will be bounded by $$\text{max}_{1\leq i\leq N} \{|f(x_i)|+\epsilon \}$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;which follows by applying the definition of Uniformly Continuous.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope that helps,&lt;/p&gt;&#xA;" OwnerUserId="6075" LastActivityDate="2011-02-27T19:53:49.913" CommentCount="2" />
  <row Id="24066" PostTypeId="2" ParentId="24037" CreationDate="2011-02-27T19:55:43.593" Score="0" ViewCount="" Body="&lt;p&gt;The projective plane, also called two-dimensional projective space, is by definition the set of all straight lines through the origin in three-dimensional Euclidean space. So a &quot;point&quot; in projective space is actually a line through the origin $O$ in the ordinary 3-space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you fix some plane in 3-space which doesn't pass through the origin (take the plane $z=1$ to be concrete), then each point $P=(x,y,1)$ in that plane corresponds to a unique element in the projective plane, namely the line in 3-space passing through $O$ and $P$. These lines (those that intersect the plane $z=1$) are the &quot;finite points&quot; in the projective plane, and they can be said to have finite coordinates $(x,y)$, so they look just like points in an ordinary Euclidean plane. But there are additional elements in the projective plane, namely the lines through $O$ which lie along the $xy$-plane (so that they never intersect the plane $z=1$). It's these lines that are called the &quot;points at infinity&quot; in the projective plane (and they, in turn, form a one-dimensional projective space, the &quot;projective line at infinity&quot;, also known as the &quot;horizon&quot;).&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2011-02-27T19:55:43.593" CommentCount="2" />
  <row Id="24067" PostTypeId="1" AcceptedAnswerId="24073" CreationDate="2011-02-27T20:06:59.957" Score="1" ViewCount="141" Body="&lt;p&gt;Let $A$ be the following matrix:&#xA;$$A = \left(\begin{array}{rrr}&#xD;&#xA;-1 &amp;amp; \hphantom{-}3 &amp;amp; \hphantom{-}0\\&#xD;&#xA;0  &amp;amp; 2 &amp;amp; 0\\&#xD;&#xA;-3 &amp;amp; 3 &amp;amp; 2&#xD;&#xA;\end{array}\right).$$&#xA;I've found that the eigenvalues are -1 and 2 (multiplicity 2). However, when I try to find the eigenvectors for the eigenvalue 2, I can only find one, as the augmented matrix $A-2I$ reduces to &#xA;$$\left(\begin{array}{rrr}&#xD;&#xA;1 &amp;amp; -1 &amp;amp; 0\\&#xD;&#xA;0 &amp;amp; 0 &amp;amp; 0\\&#xD;&#xA;0 &amp;amp; 0 &amp;amp; 0&#xD;&#xA;\end{array}\right).$$&#xA;So is there any other way to diagonalise this matrix, or have I made a mistake somewhere?&lt;/p&gt;&#xA;" OwnerUserId="7087" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-27T20:40:17.057" LastActivityDate="2011-02-27T20:40:17.057" Title="Is this matrix diagonalisable?" Tags="&lt;matrices&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="24069" PostTypeId="2" ParentId="24000" CreationDate="2011-02-27T20:16:39.510" Score="2" ViewCount="" Body="&lt;p&gt;The induction axiom in Peano Arithmetic says that for any predicate (statement about numbers) $\phi$, if you can prove $\phi(0)$ is true and you can also prove that for any number $n$, $\phi(n) \implies \phi(n+1)$ then $\phi(n)$ is true for all $n$.  So you have proven $\phi(0)$ and $\phi(0) \implies \phi(1)$ and $\dots$ $\phi(10) \implies \phi(11)$ and $\dots$.  Intuitively, if there is an $m$ such that $\phi(m)$ is not true, the base case of $\phi(0)$ or one of these implications must have broken down.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The usual structure of an inductive proof is designed to take advantage of this, so following one carefully may help.  Arturo Magidin gave an excellent discussion in his answer to &lt;a href=&quot;http://math.stackexchange.com/questions/19485/dominoes-and-induction&quot;&gt;this post&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-02-27T20:16:39.510" />
  <row Id="24070" PostTypeId="1" AcceptedAnswerId="24130" CreationDate="2011-02-27T20:18:24.347" Score="0" ViewCount="116" Body="&lt;p&gt;Let $f: U \rightarrow \mathbb{R}^3$ be an immersion that parametrizes a piece of a surface, and let $(h_{ij})$ be the matrix for the second fundamental form of that surface.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to pg. 70 of the text &lt;em&gt;Differential Geometry&lt;/em&gt; by Wolfgang Kuhnel, we can think of the $(h_{ij})$ as &quot;the Hessian of matrix of a function $h$, which represents the surface as a graph over its tangent plane&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a &quot;heuristic&quot; understanding of what's going on, but I'd like to be a bit more careful about this.  What exactly is the function $h$?  Can we write it down explicitly (perhaps in terms of the parametrization $f$, the unit normal $\nu$, and their derivatives), so that we can directly check that its Hessian is indeed the second fundamental form $(h_{ij})$?&lt;/p&gt;&#xA;" OwnerUserId="4904" LastActivityDate="2011-02-28T02:05:56.770" Title="Why can we think of the second fundamental form as a Hessian matrix?" Tags="&lt;differential-geometry&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="24071" PostTypeId="1" AcceptedAnswerId="24075" CreationDate="2011-02-27T20:23:07.057" Score="3" ViewCount="71" Body="&lt;p&gt;If language L* (Kleene Star) is regular, does it imply that L is also regular?&lt;/p&gt;&#xA;" OwnerUserId="7572" LastActivityDate="2011-02-27T22:14:48.250" Title="L* regular -&gt; L regular?" Tags="&lt;computer-science&gt;" AnswerCount="2" />
  <row Id="24072" PostTypeId="2" ParentId="13019" CreationDate="2011-02-27T20:27:35.513" Score="1" ViewCount="" Body="&lt;p&gt;A bit late to the game, but here is an answer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;1:&lt;/strong&gt; To show it is finite:  Write $e^{-x}=e^{-x/2}\cdot e^{-x/2}$.  For every $y&amp;gt;0$ there exists $N$ such that $e^{-x/2}x^{y-1}&amp;lt;1$ when $x&amp;gt;N$ so that $$\int_N^\infty e^-x x^{y-1}dx\leq \int_N^\infty e^-{x/2} &amp;lt;\infty.$$  For $x$ between $0,1$ compare to $\int x^{y-1}dx$ which converges whenever $y-1&amp;gt;-1$, so for all $y$.  Lastly bounding $\int_1^N e^{-x}x^{y-1}dx$ I leave to you.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;2:&lt;/strong&gt; For this part, what you have done so far is good.  All we need to do is switch the last integral with the limit.  To do this, notice that in some neighborhood of radius $\delta$ around $y_0$ we can bound $$\int_0^\infty e^{-x} x^{y-1}dx$$ by similar methods as in &lt;strong&gt;1&lt;/strong&gt;.  Then, applying the Dominated Convergence Theorem to that neighborhood, we can switch the limit and the integral, solving the problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remark:  For that last part, notice we cannot bound $$\int_0^\infty e^{-x} x^{y-1}dx$$ uniformely for all $y\in(0,\infty)$ since the function blows up at $0$ and at $\infty$.  But fortunately, we can bound it uniformely for all $y$ in any compact subset of $(0,\infty)$, allowing us to use the DCT.&lt;/p&gt;&#xA;" OwnerUserId="6075" LastActivityDate="2011-02-27T20:27:35.513" />
  <row Id="24073" PostTypeId="2" ParentId="24067" CreationDate="2011-02-27T20:29:58.937" Score="7" ViewCount="" Body="&lt;p&gt;First: not every matrix is diagonalizable; if the dimension of the eigenspace of a given eigenvalue is strictly smaller than the multiplicity of the eigenvalue, then the matrix is not diagonalizable (the condition is both necessary and sufficient for matrices whose characteristic polynomial splits: a matrix is diagonalizable if and only if the characteristic polynomial splits, and the geometric multiplicity of each eigenvalue equals the geometric multiplicity).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second: &lt;strong&gt;this&lt;/strong&gt; matrix &lt;em&gt;is&lt;/em&gt; diagonalizable. You did everything right, but misinterpreted what $A-2I$ was telling you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are correct that your matrix has characteristic polynomial $\chi(t) = (2-t)^2(-1-t)$, so that $\lambda=2$ is an eigenvalue with multiplicity $2$, and $\lambda=-1$ is an eigenvalue with multiplicity $1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you try to find the nullspace for $A-2I$ to find the eigenspace for $\lambda=2$, you get&#xA;$$\left(\begin{array}{rrr}&#xD;&#xA;-3 &amp;amp; 3 &amp;amp; 0\\&#xD;&#xA;0 &amp;amp; 0 &amp;amp; 0\\&#xD;&#xA;-3 &amp;amp; 3 &amp;amp; 0&#xD;&#xA;\end{array}\right) \rightarrow \left(\begin{array}{rrr}&#xD;&#xA;1 &amp;amp; -1 &amp;amp; 0\\&#xD;&#xA;0 &amp;amp; 0 &amp;amp; 0\\&#xD;&#xA;0 &amp;amp; 0 &amp;amp; 0&#xD;&#xA;\end{array}\right),$$&#xA;exactly what you described, so far so good.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But now you've misinterpreted what you found: this matrix has rank $1$, so the nullspace has dimension &lt;strong&gt;two&lt;/strong&gt;, exactly what you need! The matrix corresponds to the system that has the unique equation $x-y = 0$. This  means that $z$ is completely free, and $x$ must equal $y$. So the solutions to this system are:&#xA;$$\begin{array}{rcl}&#xA;x &amp;amp; = &amp;amp;s\\&#xA;y &amp;amp; = &amp;amp;s\\&#xA;z &amp;amp; = &amp;amp;t&#xA;\end{array}\qquad\quad\mbox{$s$ and $t$ arbitrary.}$$&#xA;So you get &lt;strong&gt;two&lt;/strong&gt; linearly independent vectors: one corresponding to $s=0$ and $t=1$, and one corresponding to $s=1$ and $t=0$. So you can find two linearly independent eigenvectors corresponding to $2$: $(0,0,1)$ and $(1,1,0)$. Multiply them by $A$ to verify that they are indeed eigenvectors of $\lambda=2$.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2011-02-27T20:29:58.937" />
  <row Id="24074" PostTypeId="1" AcceptedAnswerId="24085" CreationDate="2011-02-27T20:34:08.943" Score="1" ViewCount="167" Body="&lt;p&gt;I'm wondering if the method of finding a solution to a nonhomogeneous PDE by the method of eigenfunction expansion works if the nonhomogeneous term is a constant, rather than a function of the independent variables?  For example, in a hyperbolic PDE with x and t as the independent variables the eigenfunctions might be something like $\sum_{n=1}^\infty sin(n\pi x)$, and to create an eigenfunction expansion of the nonhomogeneous term I have to solve for the coefficients A of $\sum_{n=1}^\infty A sin(n\pi x) = B$ by using the orthogonality of sines, where B is the constant nonhomogeneous term. I guess I'm having trouble seeing how an infinite series of sines could converge to a constant - I know in a Fourier series one has to solve for the &quot;DC component&quot; separately.  &lt;/p&gt;&#xA;" OwnerUserId="7571" LastActivityDate="2011-06-03T09:48:51.210" Title="Eigenfunction expansion solution to a PDE with a constant non homogeneous term" Tags="&lt;differential-equations&gt;&lt;pde&gt;" AnswerCount="1" />
  <row Id="24075" PostTypeId="2" ParentId="24071" CreationDate="2011-02-27T20:34:12.537" Score="5" ViewCount="" Body="&lt;p&gt;No. Pick a nonregular language L over an alphabet A that contains every letter of A as one-letter words.&#xA;(for example, A = {0,1}, L = {w such that $|w|_0 - |w|_1 = \pm 1$}).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then L* = A* so L* is regular, but L is not.&lt;/p&gt;&#xA;" OwnerUserId="6270" LastActivityDate="2011-02-27T20:34:12.537" CommentCount="1" />
  <row Id="24076" PostTypeId="1" AcceptedAnswerId="24079" CreationDate="2011-02-27T20:41:23.857" Score="2" ViewCount="52" Body="&lt;p&gt;Let $A$ be a DVR in its field of fractions $F$, and $F&amp;#39;\subset F$ a subfield. Then is it true that $A\cap F&amp;#39;$ is a DVR in $F&amp;#39;$? I can see that it is a valuation ring of $F&amp;#39;$, but how do I I show, for example, that $A\cap F&amp;#39;$ is Noetherian?&lt;/p&gt;&#xA;" OwnerUserId="4341" LastActivityDate="2011-02-27T21:00:13.763" Title="Sub-discrete valuation ring?" Tags="&lt;abstract-algebra&gt;&lt;commutative-algebra&gt;" AnswerCount="2" />
  <row Id="24077" PostTypeId="1" AcceptedAnswerId="24183" CreationDate="2011-02-27T20:43:33.030" Score="1" ViewCount="62" Body="&lt;p&gt;Is following language context-free?&#xA;Alphabet: {a,b,c,d}&#xA;L = {w | w is not in {aabbc,abc,add}}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think it is: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;{aabbc},{abc},{add} are all regular.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because of closure properties(Union)  R = {w | w is in {aabbc,abc,add}} is also regular.&#xA;Because of closure of complement (L is complement of R) L is also regular.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So L is regular, and since all regular languages are context-free : L is context-free.&lt;/p&gt;&#xA;" OwnerUserId="7572" LastActivityDate="2011-02-28T06:45:29.540" Title="Is language L context-free?" Tags="&lt;computer-science&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="24078" PostTypeId="2" ParentId="24076" CreationDate="2011-02-27T20:52:33.297" Score="3" ViewCount="" Body="&lt;p&gt;This is not necessarily true. For instance, you could have $A = k[[t]]$ for a field $k$ and $F&amp;#39; \subset k((t))$ equal to $k$.  Then $A \cap F&amp;#39; = F&amp;#39;$ and a  field is often not considered a DVR. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But this is the only thing that can go wrong. In general, if a domain has a nontrivial valuation taking values in the nonnegative integers, such that the elements of valuation zero are units, then it is a DVR (and noetherianness follows --- for instance, one checks that the only ideals occur as ${x: x \ \mathrm{has \ valuation \ at \ least \ } n}.$)&lt;/p&gt;&#xA;" OwnerUserId="536" LastEditorUserId="536" LastEditorDisplayName="" LastEditDate="2011-02-27T20:59:23.513" LastActivityDate="2011-02-27T20:59:23.513" CommentCount="2" />
  <row Id="24079" PostTypeId="2" ParentId="24076" CreationDate="2011-02-27T20:54:52.240" Score="2" ViewCount="" Body="&lt;p&gt;Restrict the valuation corresponding to A to $F&amp;#39;$. Then the restriction is also a valuation with discreate value group( a subgroup of ring of integers) if it is non trivial. But if the value group is trivial, then it is field and therefore not a DVR.&lt;/p&gt;&#xA;" OwnerUserId="7565" LastEditorUserId="7565" LastEditorDisplayName="" LastEditDate="2011-02-27T21:00:13.763" LastActivityDate="2011-02-27T21:00:13.763" />
  <row Id="24080" PostTypeId="2" ParentId="505" CreationDate="2011-02-27T20:55:08.277" Score="6" ViewCount="" Body="&lt;p&gt;And one more proof, using the topological notion of continuity: Suppose for contradiction that there exists some $x$ with $f(x)-g(x) = k \neq 0$. Since $f$ and $g$ are continuous, $f-g$ is continuous, so we must have that $f(x) - g(x) \neq 0$ on a non-empty open set $S$ since the inverse image of the open interval $(3k/2, k/2)$ must be open. But since the rationals are dense in the reals, $S$ must contain a rational number $y$, with $f(y) \neq g(y)$, a contradiction.&lt;/p&gt;&#xA;" OwnerUserId="7530" LastEditorUserId="7530" LastEditorDisplayName="" LastEditDate="2011-02-27T21:05:26.510" LastActivityDate="2011-02-27T21:05:26.510" CommentCount="1" />
  <row Id="24081" PostTypeId="1" CreationDate="2011-02-27T21:04:41.480" Score="3" ViewCount="87" Body="&lt;p&gt;Question is the following language decidable:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;{(M)|given input &quot;aaaaa&quot; Turing machine M will perform at least 1295 steps}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would say, yes it is. Just let the Universal Turing Machine count each step, and accept if it reaches step 1295.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But is this not a case where Rice´s theorem applies? Which means that we cannot decide such a property?&lt;/p&gt;&#xA;" OwnerUserId="7572" LastActivityDate="2011-03-17T14:35:57.510" Title="Decidable? Turing machine runs X steps for certain input?" Tags="&lt;computer-science&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="24083" PostTypeId="1" AcceptedAnswerId="24084" CreationDate="2011-02-27T21:27:43.037" Score="2" ViewCount="64" Body="&lt;p&gt;Given an undirected graph I'd like to color each node either black or red such that at most half of every node's neighbors have the same color as the node itself. As a first step, I'd like to show that this is always possible. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe this problem is the essence of the math quiz #2 in &lt;em&gt;Communications of the ACM&lt;/em&gt; 02/2011 where I found it, so you might consider this a homework-like question. The quiz deals with partitions but I found it more natural to formulate the problem as a graph-coloring problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Coming from computer science with some math interest I'm not sure how to approach this and would be glad about some hints. One observation is that any node of degree 1 forces its neighbor to be of the opposite color. This could lead to a constructive proof (or a greedy algorithm) that provides a coloring. However, an existence proof would be also interesting.&lt;/p&gt;&#xA;" OwnerUserId="6678" LastActivityDate="2011-02-27T21:54:21.830" Title="Graph coloring problem (possibly related to partitions)" Tags="&lt;graph&gt;&lt;partitions&gt;&lt;coloring&gt;" AnswerCount="2" />
  <row Id="24084" PostTypeId="2" ParentId="24083" CreationDate="2011-02-27T21:32:24.937" Score="1" ViewCount="" Body="&lt;p&gt;Hint: Start with a random colouring and try to increase the number of edges which have differently coloured endpoints.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Spoiler:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Pick a node which has more than half of it's neighbour of the same colour as itself and flip it's colour. Now show that, as a result, the number of edges with different coloured endpoints increases by at least 1. Repeat.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2011-02-27T21:54:21.830" LastActivityDate="2011-02-27T21:54:21.830" />
  <row Id="24085" PostTypeId="2" ParentId="24074" CreationDate="2011-02-27T21:34:53.860" Score="0" ViewCount="" Body="&lt;p&gt;An infinite series of sines can indeed converge to a constant, in the interior of the interval in question (I guess you have Dirichlet boundary conditions at the endpoints). If you look at the sum of that Fourier series on the whole real line you will see a &lt;a href=&quot;http://en.wikipedia.org/wiki/Square_wave&quot; rel=&quot;nofollow&quot;&gt;square wave&lt;/a&gt;, where the part that you're interested in constitutes half a period.&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2011-02-27T21:34:53.860" CommentCount="2" />
  <row Id="24086" PostTypeId="2" ParentId="24083" CreationDate="2011-02-27T21:37:41.850" Score="1" ViewCount="" Body="&lt;p&gt;Perhaps related is the following well-known riddle: At each step, all the nodes (simultaneously) choose their color as the majority color of their neighbors (or black in case of a tie). Show that this process converges to either a fixed point or a $2$-cycle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This charming riddle is solved by comparing the state at time $t$ to the state at time $t+2$, and using a potential function. This solution was published, and generalized in a sequence of papers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your case, you want to take the anti-majority. Perhaps the same techniques work, though I'm not sure how a $2$-cycle will help you.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2011-02-27T21:37:41.850" />
  <row Id="24087" PostTypeId="2" ParentId="24071" CreationDate="2011-02-27T21:38:46.290" Score="1" ViewCount="" Body="&lt;p&gt;A nice exercise is to show that, for any $L \subseteq \{0\}^*$, $L^{*}$ is regular.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Picking $SQR = \{0^{n^2}\ |\ n \in \mathbb{N}\}$ provides a counterexample to your statement. That $SQR$ is not regular, can be proved using the Pumping Lemma.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Spoiler:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This follows from the &lt;a href=&quot;http://en.wikipedia.org/wiki/Coin_problem&quot; rel=&quot;nofollow&quot;&gt;Frobenius problem&lt;/a&gt; for two coins (which was discoverd by J.J Sylvester).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2011-02-27T22:14:48.250" LastActivityDate="2011-02-27T22:14:48.250" />
  <row Id="24088" PostTypeId="1" AcceptedAnswerId="24094" CreationDate="2011-02-27T21:40:36.867" Score="6" ViewCount="180" Body="&lt;p&gt;The inequality&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\zeta(s)^3 | \zeta(s + it)^4 \zeta(s + 2it)| \ge 1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;follows from&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$3 + 4 \cos(\theta) + \cos(2 \theta) \ge 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How is that done? What is the relationship between zeta and the trigonometry?&lt;/p&gt;&#xA;" OwnerUserId="4290" LastEditorUserId="6075" LastEditorDisplayName="" LastEditDate="2011-03-31T20:19:42.890" LastActivityDate="2011-03-31T20:19:42.890" Title="Trig identity to zeta function identity?" Tags="&lt;number-theory&gt;&lt;analytic-number-theory&gt;&lt;riemann-zeta&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="2" />
  <row Id="24089" PostTypeId="2" ParentId="24088" CreationDate="2011-02-27T21:48:38.790" Score="6" ViewCount="" Body="&lt;p&gt;The inequality you state does not seem to make too much sense.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, you can find a related discussion in the book Riemann's Zeta Function, by H.M. Edwards on pages 79/80.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Google books link: &lt;a href=&quot;http://books.google.com/books?id=ruVmGFPwNhQC&amp;amp;pg=PA79&quot; rel=&quot;nofollow&quot;&gt;http://books.google.com/books?id=ruVmGFPwNhQC&amp;amp;pg=PA79&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Snapshot:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/pIHZy.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope that helps.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2011-02-27T22:05:44.890" LastActivityDate="2011-02-27T22:05:44.890" />
  <row Id="24091" PostTypeId="1" AcceptedAnswerId="24345" CreationDate="2011-02-27T21:58:07.840" Score="3" ViewCount="104" Body="&lt;p&gt;I was recently trying to think of a simple example that demonstrates that the natural inclusion of an abelian group $G$ into $$\widehat{\widehat{G}}=\text{Hom}_{\mathsf{Ab}}(\text{Hom}_{\mathsf{Ab}}(G,\mathbb{C}^\times),\mathbb{C}^\times)$$&#xA;is not necessarily an isomorphism. Note that I'm looking at all homomorphisms; no topology on the group is involved (or, if you prefer, they are all discrete).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously, $G$ has to be infinite. However, I was having a bit of trouble finding an example satisfactorily simple - in fact, the only one I could prove worked was $G=\mathbb{Z}$, in which case $\widehat{G}=\text{Hom}_{\mathsf{Ab}}(\mathbb{Z},\mathbb{C}^\times)\cong\mathbb{C}^\times$, so that&#xA;$$\widehat{\widehat{G}}\cong\text{Hom}_{\mathsf{Ab}}(\mathbb{C}^\times,\mathbb{C}^\times)$$&#xA;which is uncountable due to the existence of uncountably many automorphisms of the field $\mathbb{C}$. However, that requires the axiom of choice, which seems like it ought not to be necessary. I'm sure I'm missing an obvious one - could someone provide a $G$ which doesn't require the axiom of choice to prove the map isn't surjective?&lt;/p&gt;&#xA;" OwnerUserId="264" LastEditorUserId="264" LastEditorDisplayName="" LastEditDate="2011-02-27T22:29:35.023" LastActivityDate="2011-03-01T03:00:09.243" Title="The natural inclusion of an infinite abelian group $G$ into $\widehat{\widehat{G}}$" Tags="&lt;group-theory&gt;&lt;characters&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="24092" PostTypeId="2" ParentId="24049" CreationDate="2011-02-27T22:18:05.960" Score="4" ViewCount="" Body="&lt;p&gt;Here is some elaboration on the wiki entry in George's comment. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose $R$ is a  domain. $R$ is called seminormal if whenever $b^2=c^3$ in $R$ one can find $t \in R$ such that $b=t^3, c=t^2$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The relevant thing here is the following fact: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;R is seminormal if and only if $Pic(R) \cong Pic(R[X])$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So if $R$ is local and not seminormal then there will be a projective, non-free $R[x]$-module of rank $1$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for an implicit example, take $R = k[t^2,t^3]_{(t^2,t^3)}$. One can check that $I = (1-tx, t^2x^2)$ is an invertible (fractional) ideal of $R[x]$ which is non-free. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;UPDATE: by request, a reference is &lt;a href=&quot;http://pages.uoregon.edu/vitulli/WeakAndSeminormality.pdf&quot; rel=&quot;nofollow&quot;&gt;this survey&lt;/a&gt;, see page 16. I am sure you can find more by googling the relevant terms. &lt;/p&gt;&#xA;" OwnerUserId="223" LastEditorUserId="223" LastEditorDisplayName="" LastEditDate="2011-02-28T06:26:29.407" LastActivityDate="2011-02-28T06:26:29.407" CommentCount="4" />
  <row Id="24093" PostTypeId="2" ParentId="8623" CreationDate="2011-02-27T22:34:31.837" Score="0" ViewCount="" Body="&lt;p&gt;I came across this on this website today purely by chance. I struggled with this particular problem for a long time before arriving at an answer that turned out to be wrong. After some more work I eventually found the following solution. I hope it is correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are a total of 11 letters, 3 of which are duplicated:&#xA;AA, MM, TT, HEICS&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are 4 mutually exclusive ways of permutation that add to the solution:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Permutation of all different letters: $\frac{8!}{4!}$=1680&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Permutation of  2 from 5 letters (excluding all 3 duplicates), HEICS, and then each of this permutated with each pair of duplicates. There are $\frac{5!}{2!}$  = 20 permutations of HEICS  and then permutations of these with each of the duplicate, e.g. HEMM, gives $\frac{4!}{2!}$ = 12, giving a total of 12 x 20 = 240 and with 3 pairs we get 240 x 3 = 720&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The 3 duplicate pairs with each other, e.g. AAMM, $\frac{4!}{2!2!}$ = 6 for each pair, giving 6 x 3 = 18&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;One pair of duplicates permutated with one each of the other 2 pairs, e.g. MMAT, $\frac{4!}{2!}$ = 12 and with 3 pairs there are 12 x 3 = 36 permutations.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;So altogether, there are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1680 + 720 + 18 + 36  = 2452 permutations of 4 letters from MATHEMATICA.&lt;/p&gt;&#xA;" OwnerUserId="7569" LastEditorUserId="-1" LastEditorDisplayName="" LastEditDate="2011-03-03T16:25:04.447" LastActivityDate="2011-03-03T16:25:04.447" CommentCount="1" />
  <row Id="24094" PostTypeId="2" ParentId="24088" CreationDate="2011-02-27T22:37:29.697" Score="10" ViewCount="" Body="&lt;p&gt;Just in case, everyone recall the notation $s=\sigma+it$ to denote complex numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The inequality you are looking for is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$|\zeta(\sigma)^3 \zeta(\sigma + it)^4 \zeta(\sigma + 2it)| \geq 1.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for $\sigma&amp;gt;1$.  From this we can prove the &lt;a href=&quot;http://en.wikipedia.org/wiki/Prime_number_theorem&quot; rel=&quot;nofollow&quot;&gt;Prime Number Theorem&lt;/a&gt; (by using a contour integral and some limits switching) because it shows that the zeta function has no zeros on the line $\sigma=1$.  To see why, count the zeros of the numerator and the zeros of the denominator when $\sigma\rightarrow 1$.  We see that a zero at $1+it$ would force a pole to exist at $1+2it$, but that is impossible since the only pole of $\zeta(s)$ is at $s=1$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So why does it follow from the trigonometric identity&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$3 + 4 \cos(\theta) + \cos(2 \theta) \ge 0?$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First take logarithms.  Then the above is equivalent to showing that $$3\log|\zeta(\sigma)|+4\log|\zeta(\sigma+it)|+\log|\zeta(\sigma+2it)|\geq 0.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Recall that $\log|z|=\Re(\log z)$ and $\Re(z)+\Re(w)=\Re(w+z)$ where $\Re(z)$ denotes the real part of $z$. Hence we need to show that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Re \left(3\log\zeta(\sigma)+4\log\zeta(\sigma+it)+\log\zeta(\sigma+2it)\right) \geq 0.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now since &#xA;$$\log(\zeta(s))=\sum_{n=1}^\infty \frac{\Lambda(n)}{\log n}n^{-s}$$ &#xA;when $\sigma&amp;gt;1$, where $\Lambda(n)$ is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Von_Mangoldt_function&quot; rel=&quot;nofollow&quot;&gt;Von Mangoldt Lambda Function&lt;/a&gt;, grouping the sums changes the left hand side into&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Re \left(\sum_{n=1}^\infty \frac{\Lambda(n)}{\log n}\left( 3n^\sigma+4n^{\sigma+it}+n^{\sigma+2it} \right) \right).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since everything is sufficiently convergent, we can bring $\Re$ inside the summation to get &#xA;$$\left(\sum_{n=1}^\infty \frac{\Lambda(n)}{\log n}\Re\left( 3n^\sigma+4n^{\sigma+it}+n^{\sigma+2it} \right) \right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, notice $\Re(n^{x+iy})= n^{x}\cos(y\log n)$, so that the above becomes&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left(\sum_{n=1}^\infty \frac{\Lambda(n)}{\log n} n^{-\sigma}\left(3+4\cos(t\log n)+\cos(2t\log n) \right) \right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lastly, the term above must be greater than or equal to zero since every term in the summation is non-negative.  Thus we conclude &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Re \left(3\log\zeta(\sigma)+4\log\zeta(\sigma+it)+\log\zeta(\sigma+2it)\right) \geq 0$$&#xA;and hence&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$|\zeta(\sigma)^3 \zeta(\sigma + it)^4 \zeta(\sigma + 2it)| \geq 1$$&#xA;as desired.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope that helps,&lt;/p&gt;&#xA;" OwnerUserId="6075" LastEditorUserId="6075" LastEditorDisplayName="" LastEditDate="2011-02-27T22:43:24.390" LastActivityDate="2011-02-27T22:43:24.390" />
  <row Id="24095" PostTypeId="1" AcceptedAnswerId="24096" CreationDate="2011-02-27T22:48:03.770" Score="2" ViewCount="64" Body="&lt;p&gt;Suppose that $p$ is a prime. Suppose further that $h$ and $k$ are non-negative integers such that&#xA;$h + k = p − 1$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to prove that $h!k! + (−1)^h \equiv 0 \pmod{p}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My first thought is that by Wilson's theorem, $(p-1)! \equiv -1 \pmod{p}$, and $h!k!$ divides $(p-1)!$ (definition of a binomial). Where would I go from here?&lt;/p&gt;&#xA;" OwnerUserId="7578" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2011-02-27T22:53:10.167" LastActivityDate="2011-02-28T02:07:14.273" Title="$h+k=p-1$, $p$ prime. Prove $h!k! + (-1)^h \equiv 0 \pmod{p}$?" Tags="&lt;elementary-number-theory&gt;&lt;number&gt;" AnswerCount="2" CommentCount="0" />
  <row Id="24096" PostTypeId="2" ParentId="24095" CreationDate="2011-02-27T22:50:56.707" Score="3" ViewCount="" Body="&lt;p&gt;Use the fact that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$h! = (-1)^h (p-1)(p-2) \dots (p-h) \mod p$$&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2011-02-27T22:50:56.707" CommentCount="1" />
  <row Id="24097" PostTypeId="2" ParentId="24003" CreationDate="2011-02-27T22:56:18.003" Score="1" ViewCount="" Body="&lt;p&gt;On the one hand, $f$ is monotone decreasing (since $f' \leq 0$) and $f \geq 0$; hence, for some $l \geq 0$,&#xA;$\lim _{x \to \infty } f(x) = l$.&#xA;On the other hand, $f&amp;#39;$ is monotone increasing (since $f'' \geq 0$) and $f&amp;#39; \leq 0$; hence &#xA;$$&#xD;&#xA;f(x) - f(x/2) = \int_{x/2}^x {f&amp;#39;(u)du}  \le \int_{x/2}^x {f&amp;#39;(x)du}  = \frac{{x f&amp;#39;(x)}}{2} \le 0.&#xD;&#xA;$$&#xA;Letting $x \to \infty$, the left-hand side converges to $0$; hence $x f&amp;#39;(x) \to 0$ too.&lt;/p&gt;&#xA;" OwnerUserId="2810" LastActivityDate="2011-02-27T22:56:18.003" />
  <row Id="24098" PostTypeId="1" CreationDate="2011-02-27T23:08:52.257" Score="10" ViewCount="214" Body="&lt;p&gt;I've just learned about topological quotient spaces and was wondering if anyone can help me with this example I thought of.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $(\mathbb{Q}, +)$ be the usual group of rational numbers for addition, likewise $(\mathbb{R}, +)$. Set $S$ to be the set of all cosets, t.i. $S=\mathbb{R}/\mathbb{Q}=\{x + \mathbb{Q} \mid x \in \mathbb{R} \}$. What is the quotient space $\mathbb{R} / S$ like? ($\mathbb{R}$ is equipped with the regular euclidian topology) What is it homeomorphic to? What does a typical open set look like?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="7580" LastActivityDate="2011-02-28T15:29:03.500" Title="Quotient Space $\mathbb{R} / \mathbb{Q}$" Tags="&lt;topology&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="3" />
  <row Id="24099" PostTypeId="1" CreationDate="2011-02-27T23:09:26.577" Score="0" ViewCount="97" Body="&lt;p&gt;$f$ is a linear map on a $k-$vector space $M$ with $f^n=0$ for some $n&amp;gt;0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$k$ is a field&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$k^d$ is $k-$algebra generated by $x_1,\ldots,x_d$ satisfy the relation $x_i x_j + x_j x_i =0$ , $\forall i,j=1,...,d$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Show that $\text{ker}(f)$ is non-zero&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you&lt;/p&gt;&#xA;" OwnerUserId="7517" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2011-02-27T23:32:35.677" LastActivityDate="2011-03-29T23:38:38.653" Title="algebra problem" Tags="&lt;homework&gt;&lt;algebra&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="24100" PostTypeId="1" AcceptedAnswerId="24103" CreationDate="2011-02-27T23:16:18.410" Score="5" ViewCount="175" Body="&lt;p&gt;I am proving that &#xA;$$5n^3 + 7n^5 \equiv 0 \pmod{12}$$&#xA;It would suffice to show &#xA;$$n^3 \equiv n^5 \pmod{12}$$&#xA;How would I go about doing that?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suppose I could just go through each $n \equiv r \pmod{12}$ with $r$ from $1$ to $11$ and show that $n^3 \equiv n^5 \pmod{12}$ for each, but that would be tedious.  Surely there's a better way.&lt;/p&gt;&#xA;" OwnerUserId="7578" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-28T00:33:02.217" LastActivityDate="2011-02-28T03:51:56.690" Title="$n^3 \equiv n^5 \pmod{12} $?" Tags="&lt;homework&gt;&lt;elementary-number-theory&gt;" AnswerCount="3" CommentCount="5" />
  <row Id="24101" PostTypeId="1" AcceptedAnswerId="24114" CreationDate="2011-02-27T23:17:44.477" Score="1" ViewCount="94" Body="&lt;p&gt;I'm trying to find two statistics T1, T2 such that (T1, T2) is jointly sufficient for (λ, θ) for a random sample $X_1,\ldots,X_n$ from a two parameter exponential distribution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f(x) = \begin{cases}&#xD;&#xA;\lambda e^{-\lambda (x-\theta)}, &amp;amp; \theta &amp;lt; x &amp;lt; \infty, \\&#xD;&#xA;0, &amp;amp; elsewhere.&#xD;&#xA;\end{cases}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="6335" LastEditorUserId="6335" LastEditorDisplayName="" LastEditDate="2011-02-27T23:49:23.887" LastActivityDate="2011-02-28T00:24:09.567" Title="Finding joint sufficient statistics." Tags="&lt;statistics&gt;&lt;probability-distributions&gt;" AnswerCount="1" CommentCount="0" />
  <row Id="24102" PostTypeId="2" ParentId="24100" CreationDate="2011-02-27T23:22:42.640" Score="2" ViewCount="" Body="&lt;ul&gt;&#xA;&lt;li&gt;n^2 = 0,1,4 or 9 mod 12 by checking each number 0,1,2,3,4,5,6 (we don't have to check 7,8,9,10 and 11 since they are negatives and x^2 = (-x)^2).&lt;/li&gt;&#xA;&lt;li&gt;n^4 = (n^2)^2 = n^2 mod 12 by checking each number 0,1,4 and 9.&lt;/li&gt;&#xA;&lt;li&gt;thus n^3 = n^5.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="4290" LastActivityDate="2011-02-27T23:22:42.640" CommentCount="1" />
  <row Id="24103" PostTypeId="2" ParentId="24100" CreationDate="2011-02-27T23:24:47.600" Score="9" ViewCount="" Body="&lt;p&gt;We need to show $$12|n^5-n^3$$ for each $n$.  Factor this as $$n^3(n-1)(n+1).$$ This quantity will always be divisible by 3 (why?) and will always be divisible by 4 (why?) so it is divisible by 12.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint:&lt;/strong&gt; If I have three consecutive integers one is divisible by 3.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint:&lt;/strong&gt; If $n$ is even then we are done.  If $n$ is odd both $n+1$ and $n-1$ are divisible by 2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Remark:&lt;/strong&gt; By this exact reasoning we actually get $24|n^5-n^3$ for each $n$.&lt;/p&gt;&#xA;" OwnerUserId="6075" LastActivityDate="2011-02-27T23:24:47.600" CommentCount="2" />
  <row Id="24104" PostTypeId="1" CreationDate="2011-02-27T23:26:40.413" Score="2" ViewCount="51" Body="&lt;p&gt;sorry for my bad English...&#xA;I've a system consists of many cuboids, many are adjacent to the others.&#xA;My problem is... I want to find the largest cube in this system (which may consists of many of some or all part of the cuboids)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a decent algorithm for finding this ?&lt;/p&gt;&#xA;" OwnerUserId="7581" LastActivityDate="2011-02-27T23:26:40.413" Title="find largest cube in lot of cuboids" Tags="&lt;geometry&gt;&lt;computational-geometry&gt;" CommentCount="5" />
  <row Id="24105" PostTypeId="2" ParentId="24099" CreationDate="2011-02-27T23:27:14.680" Score="1" ViewCount="" Body="&lt;p&gt;Hint: Choose $n$ in the question such that $f^{n-1}$ is not zero, and notice that $f(f^{n-1}(M))=\{0\}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively, prove that in general a composition of injective maps is injective.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2011-02-27T23:27:14.680" CommunityOwnedDate="2011-02-27T23:27:14.680" CommentCount="4" />
  <row Id="24108" PostTypeId="1" AcceptedAnswerId="24112" CreationDate="2011-02-27T23:48:32.173" Score="1" ViewCount="35" Body="&lt;p&gt;I have the following Lemma (from Rotman)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $X$ be a space and, for $i=0,1$, let $\lambda_i^X:X \to X \times I$ be defined by $x \mapsto (x,i)$. If $H_n \left(\lambda_0^X\right)=H_n\left(\lambda_1^X\right):H_n(x)\to H_n \left(X \times I\right)$, then $H_n(f)=H_n(g)$ whenever $f$ and $g$ are homotopic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With the proof of this fact, this step is used:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$H_n\left(F\lambda_0^X\right)=H_n(F)H_n(\lambda_0^X)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe (probably) I am missing something obvious, but it is not immediately clear to me why that is true&lt;/p&gt;&#xA;" OwnerUserId="2219" LastActivityDate="2011-02-28T00:18:00.490" Title="Lemma in the Homotopy Axiom theorem" Tags="&lt;algebraic-topology&gt;" AnswerCount="1" />
  <row Id="24109" PostTypeId="1" CreationDate="2011-02-27T23:57:26.927" Score="0" ViewCount="70" Body="&lt;p&gt;Let us define a one-to-one function $f$ that maps binary strings of length $n$ to ternary strings of length $n$ such that if $x$ is random then $f(x)$ must be random. My question&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Is there an algorithm that computes such function? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2459" LastEditorUserId="2459" LastEditorDisplayName="" LastEditDate="2011-02-28T00:14:34.580" LastActivityDate="2011-02-28T07:04:02.020" Title="Mapping between random strings?" Tags="&lt;computer-science&gt;&lt;random&gt;" AnswerCount="2" CommentCount="14" />
  <row Id="24110" PostTypeId="1" AcceptedAnswerId="24111" CreationDate="2011-02-28T00:01:48.253" Score="0" ViewCount="44" Body="&lt;p&gt;Hello, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I have:&lt;br&gt;&#xA;\begin{align*}&#xA;N &amp;amp;\equiv 1 &amp;amp;\pmod{2}\\&#xA;N &amp;amp;\equiv 2 &amp;amp;\pmod{3}\\&#xA;N &amp;amp;\equiv 3 &amp;amp;\pmod{4}\\&#xA;&amp;amp;\vdots\\ &#xA;N &amp;amp;\equiv n - 1 &amp;amp;\pmod{n}&#xA;\end{align*}&#xA;How could I solve for $N$? Is there any property relates to this problem?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update&lt;/strong&gt;&lt;br&gt;&#xA;Base on Moron hint, we have:&lt;br&gt;&#xA;\begin{align*}&#xA;N + 1 &amp;amp;\equiv 0 &amp;amp;\pmod{2}\\&#xA;N + 1 &amp;amp;\equiv 0 &amp;amp;\pmod{3}\\&#xA;N + 1 &amp;amp;\equiv 0 &amp;amp;\pmod{4}\\&#xA;\vdots\\&#xA;N + 1 &amp;amp;\equiv 0 &amp;amp;\pmod{n}&#xA;\end{align*}&#xA;Hence, &#xA;$$N + 1 \equiv 0 \pmod{\mathrm{lcm}(2\cdot 3\cdots n}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\therefore N + 1 = lcm(2.3.4...n) * k \text{ for some integers k } $$&#xA;$$\implies N = lcm(2.3.4...n) * k - 1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does it look right?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks,&lt;br&gt;&#xA;Chan&lt;/p&gt;&#xA;" OwnerUserId="5711" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-28T00:24:20.323" LastActivityDate="2011-02-28T00:24:20.323" Title="Question regarding modulo property" Tags="&lt;elementary-number-theory&gt;&lt;modular-arithmetic&gt;" AnswerCount="1" CommentCount="7" />
  <row Id="24111" PostTypeId="2" ParentId="24110" CreationDate="2011-02-28T00:07:53.317" Score="4" ViewCount="" Body="&lt;p&gt;Hint: Consider the possible values for $N+1$.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2011-02-28T00:07:53.317" CommentCount="1" />
  <row Id="24112" PostTypeId="2" ParentId="24108" CreationDate="2011-02-28T00:18:00.490" Score="1" ViewCount="" Body="&lt;p&gt;(Let's say you are working with singular homology.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have maps of topological spaces $X \stackrel{f}{\to} Y \stackrel{g}{\to} Z$, then you have $H_n(g \circ f) = H_n(g) \circ H_n(f)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To prove this, check where a cycle $\gamma: I^n \to X$ is being sent to under $H_n(g \circ f)$ and $H_n(g) \circ H_n(f)$. Here $I^n$ is the standard $n$-simplex.&lt;/p&gt;&#xA;" OwnerUserId="325" LastActivityDate="2011-02-28T00:18:00.490" CommentCount="1" />
  <row Id="24113" PostTypeId="1" AcceptedAnswerId="24116" CreationDate="2011-02-28T00:22:52.063" Score="3" ViewCount="142" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Possible Duplicate:&lt;/strong&gt;&lt;br&gt;&#xA;  &lt;a href=&quot;http://math.stackexchange.com/questions/2746/is-there-an-elementary-proof-that-k-1n-1-k-is-never-an-integer&quot;&gt;Is there an elementary proof that $∑_{k=1}^n 1/k$ is never an integer?&lt;/a&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;p&gt;Hello,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Prove that $1 + \frac{1}{2} + \frac{1}{3} + ... + \frac{1}{n}$ is not an integer.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I tried to prove by induction on $n$, but I was stuck :(&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume $1 + \frac{1}{2} + \frac{1}{3} + ... + \frac{1}{n} = \frac{a}{b}$ for some integers $a, b$ and $a \neq b \text{and} b \neq 0$&lt;br&gt;&#xA;Then $ 1 + \frac{1}{2} + \frac{1}{3} + ... + \frac{1}{n + 1} = \frac{a}{b} + \frac{1}{n + 1}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then how can I prove that this expression is not integer? A hint would be greatly appreciated.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks,&lt;br&gt;&#xA;Chan&lt;/p&gt;&#xA;" OwnerUserId="5711" LastActivityDate="2011-02-28T00:26:07.030" ClosedDate="2011-02-28T00:40:36.357" Title="Prove that $1 + \frac{1}{2} + \frac{1}{3} + ... + \frac{1}{n}$ is not an integer" Tags="&lt;elementary-number-theory&gt;" AnswerCount="2" CommentCount="9" FavoriteCount="1" />
  <row Id="24114" PostTypeId="2" ParentId="24101" CreationDate="2011-02-28T00:24:09.567" Score="2" ViewCount="" Body="&lt;p&gt;First note that&#xA;$$&#xD;&#xA;f(x_1 ,x_2 , \ldots ,x_n |\lambda ,\theta ) = \lambda ^n e^{\lambda \theta n} \exp \bigg( - \lambda \sum\limits_{i = 1 }^n {X_i } \bigg) \prod\limits_{i = 1}^n {{\mathbf 1}(x_i  &amp;gt; \theta )}, &#xD;&#xA;$$&#xA;where $\mathbf 1$ denotes the indicator function. Then note that&#xA;$$&#xD;&#xA;\prod\limits_{i = 1}^n {{\mathbf 1}(x_i  &amp;gt; \theta )} = {\mathbf 1}(\min \lbrace x_1 , \ldots ,x_n \rbrace  &amp;gt; \theta ),&#xD;&#xA;$$&#xA;and apply Theorem 2 given &lt;a href=&quot;http://math.arizona.edu/~tgk/466/sufficient.pdf&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2810" LastActivityDate="2011-02-28T00:24:09.567" CommentCount="7" />
  <row Id="24115" PostTypeId="2" ParentId="24113" CreationDate="2011-02-28T00:25:47.710" Score="2" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt;: There is always a prime between $\frac{n}{2}$ and $n$, $\forall n \geq 4$&lt;/p&gt;&#xA;" OwnerUserId="3302" LastActivityDate="2011-02-28T00:25:47.710" CommentCount="1" />
  <row Id="24116" PostTypeId="2" ParentId="24113" CreationDate="2011-02-28T00:26:07.030" Score="2" ViewCount="" Body="&lt;p&gt;Hint: look at the largest power of 2 less than $n$. Can it get canceled out from the denominator?&lt;/p&gt;&#xA;" OwnerUserId="264" LastActivityDate="2011-02-28T00:26:07.030" />
  <row Id="24117" PostTypeId="1" AcceptedAnswerId="24118" CreationDate="2011-02-28T00:30:38.810" Score="2" ViewCount="50" Body="&lt;p&gt;Let $V$ be a finite-dimensional irreducible representation of a finite group G over an algebraically closed field. If $V \otimes V$ has a fixed subspace acted by $G$, why should it be 1-dimensional? I really don't know how to think about this. I tried reasoning that there is a fixed space then $\phi(g)$ must have eigenvalue 1, but it's not contributive. &lt;/p&gt;&#xA;" OwnerUserId="6369" LastActivityDate="2011-02-28T00:40:32.163" Title="fixed space of tensor representation" Tags="&lt;representation-theory&gt;&lt;tensors&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="24118" PostTypeId="2" ParentId="24117" CreationDate="2011-02-28T00:40:32.163" Score="1" ViewCount="" Body="&lt;p&gt;There need not be a fixed G-invariant subspace of dimension 1.  Consider the cyclic group G of order 4 and the representation on V=C given by g&lt;sup&gt;n&lt;/sup&gt;↦sqrt(−1)&lt;sup&gt;n&lt;/sup&gt;.  The corresponding representation on V⊗V ≅ C is given by g&lt;sup&gt;n&lt;/sup&gt;↦(−1)&lt;sup&gt;n&lt;/sup&gt;.  The fixed subspace has dimension 0, not 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, if V is irreducible and V⊗V has a fixed G-invariant subspace of dimension larger than 0, then the dimension is exactly 1.  This follows from character theory:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$[1,\chi \cdot \chi] = [ \bar \chi, \chi ] = \begin{cases} 0 &amp;amp; \chi \neq \bar \chi \\ 1 &amp;amp; \chi = \bar \chi \end{cases}$$&lt;/p&gt;&#xA;" OwnerUserId="583" LastActivityDate="2011-02-28T00:40:32.163" CommentCount="1" />
  <row Id="24119" PostTypeId="1" CreationDate="2011-02-27T22:22:58.353" Score="2" ViewCount="242" Body="&lt;p&gt;I've come across the term &quot;pairwise distinct&quot; in many research papers. But, I don't understand how it differs from just saying that the elements of a set are unique instead of saying that they are pairwise distinct. &#xA;Can someone please explain the difference, if any, to me?&lt;/p&gt;&#xA;" OwnerUserId="5169" LastActivityDate="2011-02-28T01:04:29.723" Title="Difference between pairwise distinct and unique?" Tags="&lt;terminology&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="24120" PostTypeId="1" AcceptedAnswerId="24122" CreationDate="2011-02-28T00:59:06.383" Score="2" ViewCount="63" Body="&lt;p&gt;I am having a hard time understanding a condition in building a measurable function. This is an example of a measurable function from &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0817638849&quot; rel=&quot;nofollow&quot;&gt;this book&lt;/a&gt; (p. 55). Quoting verbatim:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Consider $R_n(\omega)$, the Rademacher functions. These are measurable because they are piecewise constant; namely, for any subset $A$ of $\mathbb{R} \cup \{+\infty\} \cup \{-\infty\}$, we have that $\{\omega \in I; R_n(\omega) \in A\}$ is a finite union of intervals.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;where $I = [0,1]$ and the $n$th Rademacher function&#xA;\begin{align}&#xA;R_n(\omega) = \begin{cases}&#xA;1 &amp;amp; \textrm{if } a_n = 1 \newline&#xA;-1 &amp;amp; \textrm{if } a_n = 0&#xA;\end{cases}&#xA;\end{align}&#xA;and $a_n$ is the $n$th number in the binary expansion of $\omega \in I$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Namely, I don't get the condition placed regarding&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;any subset $A$ of $\mathbb{R} \cup \{+\infty\} \cup \{-\infty\}$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If the only values $R_n$ can take are +1 and -1, what's the point in saying that $A$ can be any interval on the extended real line? Isn't it just as valid to state&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$\{\omega \in I; R_n(\omega) \in \{-1,1\}\}$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and argue that the Rademacher functions are measurable?&lt;/p&gt;&#xA;" OwnerUserId="6690" LastEditorUserId="6690" LastEditorDisplayName="" LastEditDate="2011-02-28T01:44:43.583" LastActivityDate="2011-02-28T01:44:43.583" Title="Measurability of a function" Tags="&lt;measure-theory&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="24121" PostTypeId="2" ParentId="24119" CreationDate="2011-02-28T01:04:29.723" Score="3" ViewCount="" Body="&lt;p&gt;The first problem is that, set-theoretically, what matters is only whether an element is in a set or not. That is, for sets $A$ and $B$,&#xA;$$A=B\Longleftrightarrow\mbox{for all $x$, $x\in A$ if and only if $x\in B$.}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What this means, among other things, is that the set $\{1,1\}$ is &lt;em&gt;equal&lt;/em&gt; to the set $\{1\}$, because every element in the first is in the second and vice-versa. The fact that $1$ shows up twice in the first set is completely immaterial and irrelevant, the two sets are &lt;em&gt;equal&lt;/em&gt;. So there is no way, set-theoretically, to say that $1$ only shows up once in the second set but shows up twice in the first.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So it doesn't really make sense to say that &quot;elements of a set are unique.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instead, you really want to talk about either multisets (which introduces its own complications), or else you want to talk about &lt;em&gt;ordered tuples&lt;/em&gt; and say that entries with distinct indices should be distinct. Something like:&#xA;$$(a_i)_{i\in I}\text{ and }a_i\neq a_j\text{ if }i\neq j.$$&#xA;But this also introduces complications of its own, such as having to introduce an index set, not to mention lots of extra words.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So instead what we want to say is that given any &lt;em&gt;pair&lt;/em&gt; of elements (and we want to say &quot;pair&quot;, because in mathematics, if you simply say &quot;any $x$ and $y$&quot;, you do &lt;em&gt;not&lt;/em&gt; exclude the possibility that you selected the same element twice), the two elements are different. And this is what &quot;pairwise distinct&quot; means: every pair of elements consists of two different things. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another potential problem is that the term &quot;unique&quot; is usually reserved for a different kind of meaning. For example,  the Chinese Remainder theorem says that a certain system of congruences has solutions, and that the solutions are &quot;unique modulo $M$&quot; (where $M$ is a certain integer defined in terms of the hypothesis). That use of &quot;unique&quot; means that if you find any two solutions, then they are the same solution modulo $M$. That usage would be at odds with using unique in order to say &quot;these list of things doesn't have any repeats&quot;. &lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2011-02-28T01:04:29.723" />
  <row Id="24122" PostTypeId="2" ParentId="24120" CreationDate="2011-02-28T01:26:37.630" Score="4" ViewCount="" Body="&lt;p&gt;If $(X,\mathcal{A})$ and $(Y,\mathcal{B})$ are measurable spaces, then the &lt;em&gt;definition&lt;/em&gt; of measurable function says that $f\colon X\to Y$ is measurable if and only if $f^{-1}(B) \in \mathcal{A}$ for every $B\in\mathcal{B}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(This is very similar/essentially identical to the definition of continuity for a function between topological spaces: if $(X,\tau)$ and $(Y,\sigma)$ are topological spaces, then $f\colon X\to Y$ is continuous if and only if $f^{-1}(B)\in\tau$ for every $B\in\sigma$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, if you have a $B\in\mathcal{B}$ such that $B\cap\mathrm{Im}(f)=\emptyset$, then you will automatically get that $f^{-1}(B)\in\mathcal{A}$; and if $\mathrm{Im}(f)\subseteq B$, then you will also automatically get $f^{-1}(B)\in \mathcal{A}$. So generally you are only concerned with elements of $\mathcal{B}$ that intersect, but don't contain, the image of $f$; not because you don't need to check the others, but rather because the other elements of $\mathcal{B}$ will trivially satisfy the condition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What they are showing is that the Rademacher function is measurable when mapping into $\mathbb{R}$ with &lt;strong&gt;any&lt;/strong&gt; $\sigma$-algebra, by showing that it is measurable relative to the $\sigma$-algebra $\mathcal{P}(\mathbb{R})$ of all subsets of $\mathbb{R}$. This is pretty much the strongest possible statement about measurability of these functions. In point of fact, it &lt;em&gt;would&lt;/em&gt; be enough to consider only the sets $A=\{1\}$ and $A=\{-1\}$; any $A$ that contains both will necessarily have inverse image $I$ (measurable); any set that contains &lt;em&gt;neither&lt;/em&gt; will necessarily have inverse image $\emptyset$ (measurable); and any set that contains exactly one of them will have inverse image equal to $R_n^{-1}(\{1\})$ or $R_n^{-1}(\{-1\})$ (depending on which one they contain). But there is no need to argue that way, and we can simply take an arbitrary subset $A$, even though the argument will just reduce to those two nontrivial cases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you &lt;em&gt;cannot&lt;/em&gt; do is simply restrict to the case $A=\{1,-1\}$; this does not show that $R_n$ is measurable, because it does not show that a measurable subset of $\mathbb{R}$ which contains $1$ but not $-1$ has an inverse image in the (Borel) $\sigma$-algebra of $I$. That is, what you propose would be to check merely that $f^{-1}(B)$, where $B$ contains both $1$ and $-1$, is measurable. This is not sufficient to establish measurability. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(What you &lt;em&gt;could&lt;/em&gt; do is talk about $A$ being any subset of $\{-1,1\}$, instead of saying that $A$ is any subset of $\mathbb{R}\cup\{+\infty\}\cup\{-\infty\}$, because the former suffice in this case.)&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2011-02-28T01:26:37.630" CommentCount="1" />
  <row Id="24123" PostTypeId="1" AcceptedAnswerId="24128" CreationDate="2011-02-28T01:29:46.507" Score="4" ViewCount="109" Body="&lt;p&gt;Prove by element-wise:&#xA;$$A \cup (B \cap C) = (A \cup B) \cap (A \cup C).$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is easy to prove by Venn Diagram, which I have already done. I do not, however, know how to prove using this technique. I know it has to do with proving each is a subset, but I do not understand the general method to prove this formally?&lt;/p&gt;&#xA;" OwnerUserId="7144" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-28T01:31:52.383" LastActivityDate="2011-02-28T12:08:38.107" Title="Element-Wise Proofs?" Tags="&lt;homework&gt;&lt;elementary-set-theory&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="24124" PostTypeId="2" ParentId="24100" CreationDate="2011-02-28T01:30:48.890" Score="1" ViewCount="" Body="&lt;p&gt;It's just case $\rm\ p^j = 2^2,\ q^k = 3,\ d = e = 2\ $ of this simple  generalization of Euler's little theorem&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;THEOREM&lt;/strong&gt; $\ $ For primes $\rm\:p \ne q\:,\:$ naturals $\rm\:e\:$ and $\rm\ j,\ k \:\le\: d\ $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\rm\quad\quad\ \phi(p^j),\ \phi(q^k)\ |\ e\ \ \Rightarrow\ \ p^j\ q^k\ |\ n^d\ (n^e - 1)\ \ \ \forall\ n\in \mathbb N $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof&lt;/strong&gt; $\ $ If $\rm\ p\ |\ n\ $ then $\rm\ p^j\ |\ n^d\ $ by $\rm\ j\le d\:.\:$ Else $\rm\:n\:$ is coprime to $\rm\: p\:,\:$ so by Euler's little theorem we have: $\ $ mod $\rm\ p^j\:,\ \ n^{\phi(p^j)}\equiv 1\ \Rightarrow\ n^e\equiv 1\ $ by $\rm\ \phi(p^j)\ |\ e\:.\ $  Thus $\rm\  n^d\ (n^e - 1)\ $ is divisible by $\rm\ p^j\ $ and, similarly it is divisible by $\rm\ q^k\:,\ $ hence it is also divisible by their lcm = product. $\quad$ &lt;strong&gt;QED&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact for $\rm\ p = 2,\ j &amp;gt; 2\ $ we can use $\rm\ \phi(2^j)/2\ $ vs. $\rm\ \phi(2^j)\ $ because $\rm\ \mathbb Z/2^j\ $ has multiplicative group $\rm\ C(2)\times C(2^{j-2})\ $ for $\rm\ j&amp;gt; 2\:$. Thus, in fact, $\rm\ 24\ |\ n^3\ (n^2 - 1)\:.$ For more see my post on the &lt;a href=&quot;http://math.stackexchange.com/questions/1626/modular-exponentiation-using-eulers-theorem/1726#1726&quot;&gt;Fermat-Euler-Carmichael theorem.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2011-02-28T01:37:58.977" LastActivityDate="2011-02-28T01:37:58.977" CommentCount="1" />
  <row Id="24125" PostTypeId="2" ParentId="24123" CreationDate="2011-02-28T01:33:48.670" Score="1" ViewCount="" Body="&lt;p&gt;$(\Rightarrow)$: Suppose $x \in A \cup (B \cap C)$. Then $x \in A$ or $x \in B$ and $x \in C$. So $x \in A$ or $x \in B$ and $x \in A$ or $x \in C$. So $x \in (A \cap B) \cap (A \cup C)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(\Leftarrow)$: Try this one.&lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2011-02-28T01:33:48.670" CommentCount="1" />
  <row Id="24126" PostTypeId="1" AcceptedAnswerId="24129" CreationDate="2011-02-28T01:33:51.513" Score="0" ViewCount="79" Body="&lt;p&gt;Hello,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm looking for a formula to calculate the product of all proper divisors of a number number $n$. Any idea?    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks,&lt;br&gt;&#xA;Chan&lt;/p&gt;&#xA;" OwnerUserId="5711" LastActivityDate="2011-02-28T01:49:23.053" Title="What does the product of all proper divisors equal to?" Tags="&lt;elementary-number-theory&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="24128" PostTypeId="2" ParentId="24123" CreationDate="2011-02-28T01:36:23.893" Score="4" ViewCount="" Body="&lt;p&gt;Proving &quot;by element-wise&quot; (I do hope you are translating; that's awful abuse of language) means showing that each side is a subset of the other. In other words, the argument looks something like:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Suppose $x\in A\cup(B\cap C)$; then &lt;code&gt;.... stuff ....&lt;/code&gt;, so $x\in (A\cup B)\cap (A\cup B)$; therefore, $A\cup(B\cap C) \subseteq (A\cup B)\cap (A\cup C)$.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Now suppose that $y\in (A\cup B)\cap (A\cup C)$. Then &lt;code&gt;... stuff...&lt;/code&gt;, so $y\in A\cup (B\cap C)$. Therefore, $(A\cup B)\cap (A\cup C) \subseteq A\cup(B\cap C)$.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Since each is contained in the other, $A\cup(B\cap C)=(A\cup B)\cap(A\cup C)$. QED.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Or you can try to do both inclusions at the same time using equivalences. Then you would have something like&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$x\in A\cup (B\cap C) \Longleftrightarrow$ &lt;code&gt;stuff&lt;/code&gt; $\Longleftrightarrow$ &lt;code&gt;more stuff&lt;/code&gt; $\Longleftrightarrow$ &lt;code&gt;even more stuff&lt;/code&gt; $\Longleftrightarrow x\in (A\cup B)\cap (A\cup C)$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In other words, &quot;element-wise&quot; means &quot;chasing an element&quot;: pick an element in one side, show it has to be in the other and vice versa. After all, equality of sets is defined in terms of them containing the same elements, not in terms of Venn diagrams.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2011-02-28T01:36:23.893" CommentCount="1" />
  <row Id="24129" PostTypeId="2" ParentId="24126" CreationDate="2011-02-28T01:41:32.220" Score="5" ViewCount="" Body="&lt;p&gt;Like Sivaram points out. If $n$ is &lt;em&gt;not&lt;/em&gt; a perfect square, then every divisor $d_1$ can be paired with the divisor $\frac{n}{d_1}$, which is distinct from $d_1$; the product of these two is $n$. So the product of all divisors is equal to $n^k$, where $2k$ is the number of divisors. (Note that a positive integer has an odd number of distinct divisors if and only if it is a square).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $n$ &lt;em&gt;is&lt;/em&gt; a perfect square, then it will have $2k+1$ divisors for some $k$; the product will be $n^k\times\sqrt{n} = n^{k+\frac{1}{2}}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Either way, the answer is $n^{d(n)/2}$, where $d(n)$ is the number of divisors of $n$ (sometimes written $\sigma_0(n)$ or $\tau(n)$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want the &lt;em&gt;proper&lt;/em&gt; divisors, then you are excluding $n$ from the product, so the answer becomes $n^{(d(n)/2) - 1}$.  &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-28T01:49:23.053" LastActivityDate="2011-02-28T01:49:23.053" CommentCount="2" />
  <row Id="24130" PostTypeId="2" ParentId="24070" CreationDate="2011-02-28T02:05:56.770" Score="0" ViewCount="" Body="&lt;p&gt;Thanks Ryan!  So in conclusion, if we choose coordinates such that the surface is locally described by $f(u,v) = (u,v,h(u,v))$, then the second fundamental form at the point $f(0,0)$ is precisely the Hessian of $h$ at $(0,0)$.&lt;/p&gt;&#xA;" OwnerUserId="4904" LastActivityDate="2011-02-28T02:05:56.770" CommunityOwnedDate="2011-02-28T02:05:56.770" />
  <row Id="24131" PostTypeId="2" ParentId="24095" CreationDate="2011-02-28T02:07:14.273" Score="0" ViewCount="" Body="&lt;p&gt;This is simply a reformulation of the Wilson reflection formula that I mentioned &lt;a href=&quot;http://math.stackexchange.com/questions/23823/need-help-to-find-remainder-when-525-is-divided-by-31/23875#23875&quot;&gt;&lt;strong&gt;twice&lt;/strong&gt; yesterday&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\rm (p-1-k)\:!\ \equiv\ \frac{(-1)^{k+1}}{k!}\ \ (mod\ p),\ \ for\ p\ prime $$&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2011-02-28T02:07:14.273" />
  <row Id="24132" PostTypeId="1" AcceptedAnswerId="24135" CreationDate="2011-02-28T02:09:40.560" Score="1" ViewCount="49" Body="&lt;p&gt;Hello,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example:&lt;br&gt;&#xA;$6$ has this property since proper divisors of $6$ are: $2$ and $3.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From this thread: &lt;a href=&quot;http://math.stackexchange.com/questions/24126/what-does-the-product-of-all-proper-divisors-equal-to&quot;&gt;What does the product of all proper divisors equal to?&lt;/a&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My attempt was:&lt;br&gt;&#xA;If $n = p_1^{a_1} \times p_2^{a_2} \times ... \times p_k^{a_k}$&lt;br&gt;&#xA;Then $n = n^{\frac{\tau(n)}{2} - 1}$. Where $\tau(n) = (a_1 + 1) \times (a_2 + 1) \times ... \times (a_k + 1)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So is it good enough to stop here, or we can express $n$ in a better formula?  &lt;/p&gt;&#xA;" OwnerUserId="5711" LastEditorUserId="5711" LastEditorDisplayName="" LastEditDate="2011-02-28T02:16:52.303" LastActivityDate="2011-02-28T02:51:27.650" Title="Find all numbers that are equal to the product of all their proper divisors" Tags="&lt;elementary-number-theory&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="24133" PostTypeId="2" ParentId="24132" CreationDate="2011-02-28T02:12:04.817" Score="3" ViewCount="" Body="&lt;p&gt;Any number of the form $n = p_1 \times p_2$ where $p_1,p_2$ are primes.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastActivityDate="2011-02-28T02:12:04.817" CommentCount="2" />
  <row Id="24134" PostTypeId="1" AcceptedAnswerId="24261" CreationDate="2011-02-28T02:13:29.567" Score="5" ViewCount="253" Body="&lt;p&gt;I came across this post lying dormant on &lt;a href=&quot;http://www.mathforum.org/kb/thread.jspa?threadID=383701&amp;amp;messageID=1183710#1183710&quot; rel=&quot;nofollow&quot;&gt;some online forum&lt;/a&gt;. I am putting it here verbatim, it seems to me worth a lot.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;By Prof. S. D. Agashe, IIT Bombay&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Source: Vector Calculus, by Durgaprasanna Bhattacharyya, University&#xA;Studies Series,Griffith Prize Thesis, 1918, published by the University&#xA;of Calcutta, India, 1920, 90 pp)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Chapter IV: The Linear Vector Function, article 15, p.24:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;The most general vector expression linear in $r$ can contain terms only&#xA;of three possible types, $r$, $(a.r)b$ and $c\times r$, $a$, $b$, $c$ being constant unit vectors. Since $r$, $(a.r)b$ and $c\times r$ are in general non-coplanar,it follows from the theorem of the parallelepiped of vectors that the most general linear vector expression can be written in the form $\lambda . r + \mu (a.r)b + \nu (c\times r)$, where $\lambda, \mu, \nu$ are scalar constants&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bhattacharyya does not prove this. Has anyone seen a similar result and its proof?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bhattacharyya uses this to show that the divergence of the linear&#xA;function is ($3 \lambda + a.b$), that the curl is ($a \times b + 2c$). He goes on to define div and curl of a differentiable function as the div and curl of the (linear) derivative function. The div and curl of a linear function are defined in terms of certain surface integrals. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am excited about this result because it seems to provide an excellent&#xA;route to div and curl, as Bhattacharyya himself remarks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sorry for a rather long and &quot;technical&quot; communication.&lt;/p&gt;&#xA;" OwnerUserId="1674" LastActivityDate="2011-02-28T17:04:16.830" Title="Div, curl and linear algebra" Tags="&lt;linear-algebra&gt;&lt;vector-calculus&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="24135" PostTypeId="2" ParentId="24132" CreationDate="2011-02-28T02:14:54.157" Score="4" ViewCount="" Body="&lt;p&gt;$n = n^{\tau(n)/2 - 1}$ implies $\tau(n)=4$ and so $n$ is a product of two primes or the cube of a prime.&lt;/p&gt;&#xA;" OwnerUserId="589" LastEditorUserId="589" LastEditorDisplayName="" LastEditDate="2011-02-28T02:51:27.650" LastActivityDate="2011-02-28T02:51:27.650" CommentCount="1" />
  <row Id="24136" PostTypeId="2" ParentId="24109" CreationDate="2011-02-28T02:19:44.340" Score="2" ViewCount="" Body="&lt;p&gt;If the person reading the ternary string knows your algorithm, then s/he will know that $f(x)$ is not random.  There are $2^n$ binary strings of length $n$, so it takes $n$ bits to select one.  But there are $3^n$ ternary strings of length $n$, so it takes $n\log_2 (3)$ bits to select one.  Your $f(x)$ can only have a range of $2^n$ strings, not $3^n$.  However, to somebody who doesn't know your algorithm, the output can look random.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2011-02-28T03:20:48.747" LastActivityDate="2011-02-28T03:20:48.747" />
  <row Id="24137" PostTypeId="1" AcceptedAnswerId="24138" CreationDate="2011-02-28T02:21:09.403" Score="3" ViewCount="116" Body="&lt;p&gt;this might be a dumb question but is the following true? $$&amp;lt;\sum_{n=1}^\infty x_n, y&amp;gt;=\sum_{n=1}^\infty &amp;lt;x_n, y&amp;gt;$$&lt;/p&gt;&#xA;" OwnerUserId="6556" LastActivityDate="2011-04-26T02:09:14.860" Title="does linearity of inner product hold for infinite sum?" Tags="&lt;inner-product-space&gt;" AnswerCount="1" CommentCount="6" FavoriteCount="2" />
  <row Id="24138" PostTypeId="2" ParentId="24137" CreationDate="2011-02-28T02:26:53.677" Score="6" ViewCount="" Body="&lt;p&gt;As Mariano points out, you need to have a topological vector space - otherwise there is no such thing as an infinite sum, because an &quot;infinite sum&quot; is really defined to be a limit of partial sums, and no topology means no limits. However, if there is a topology, and the inner product is continuous with respect to it, we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\langle\sum_{n=1}^\infty x_n,y\rangle=\langle\lim_{N\rightarrow\infty}\sum_{n=1}^N x_n, y\rangle=\lim_{N\rightarrow\infty}\langle\sum_{n=1}^N x_n,y\rangle=\lim_{N\rightarrow\infty}\sum_{n=1}^N\langle x_n,y\rangle=\sum_{n=1}^\infty \langle x_n,y\rangle$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where the second equality is justified by the assumption that the inner product is continuous and hence preserves limits, and the third equality is justified by the bilinearity of the inner product.&lt;/p&gt;&#xA;" OwnerUserId="264" LastEditorUserId="5363" LastEditorDisplayName="" LastEditDate="2011-04-26T02:09:14.860" LastActivityDate="2011-04-26T02:09:14.860" />
  <row Id="24140" PostTypeId="1" CreationDate="2011-02-28T02:35:49.270" Score="1" ViewCount="108" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;A 30-foot ladder rests vertically against a wall. A bug starts at the bottom of the ladder and climbs up at a rate of 3.5 feet per minute. At the same time, the foot of the ladder is being pulled along the ground at a rate of 1.5 feet per minute until the top of the ladder reaches the ground. Let $x$ be the distance of the bug from the wall at time $t$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I got $x(t)=-.175t^2 +1.5t$ to represent the horizontal distance of the bug to the wall as a function of time, $t$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question also asks:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;When the bug is farthest from the wall how high is it?&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;When the bug reaches the top of the ladder how high is it from the ground?&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;What speed does the bug need to climb in order to reach the top at the same time the top of the ladder reaches the ground?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What I am confused about is what equation/steps do I need to find in order to compare the equation I found with, which will allow me to find the answers. I looked up related rates problems, but was confused because this problem has the ladder moving down, but none have one with something going up. Thanks in advance!&lt;/p&gt;&#xA;" OwnerUserId="7584" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-15T19:09:15.867" LastActivityDate="2011-03-15T22:29:39.117" Title="Related Rates Problem With Ladder" Tags="&lt;homework&gt;&lt;algebra-precalculus&gt;" AnswerCount="1" />
  <row Id="24141" PostTypeId="1" CreationDate="2011-02-28T02:42:54.833" Score="2" ViewCount="74" Body="&lt;p&gt;Can anyone help me solve this problem. I have no idea where to even start on it. Link inside &lt;a href=&quot;http://imgur.com/U2Eg9&quot; rel=&quot;nofollow&quot;&gt;stock option problem&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/AFuMC.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="7585" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-28T04:45:06.637" LastActivityDate="2011-03-02T21:39:52.630" Title="Stock Option induction problem" Tags="&lt;homework&gt;&lt;probability&gt;&lt;induction&gt;" AnswerCount="1" CommentCount="6" />
  <row Id="24142" PostTypeId="2" ParentId="24140" CreationDate="2011-02-28T02:43:35.490" Score="1" ViewCount="" Body="&lt;p&gt;Let $x$ be horizontal and $y$ vertical.  I would first express the bug position along the ladder as a function of t: $z=3.5t$.  Then you need to find out where this point along the ladder is.  The foot of the ladder is at $(1.5t, 0)$, so the top of the ladder is at $(0,what?)$.  From these two points you should be able to find the $(x,y)$ coordinates of the bug as a function of time, which will answer all your questions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Added and corrected:  the foot of the ladder is at $(1.5t,0)$ and the top is at $(0,\sqrt{900-2.25t^2})$  The bug is $\frac{3.5t}{30}$ of the way along the ladder, so is at $(1.5t(1-\frac{3.5t}{30}),\frac{3.5t}{30}\sqrt{900-2.25t^2})$&lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2011-03-15T22:29:39.117" LastActivityDate="2011-03-15T22:29:39.117" CommentCount="2" />
  <row Id="24143" PostTypeId="1" AcceptedAnswerId="24145" CreationDate="2011-02-28T02:56:32.177" Score="5" ViewCount="126" Body="&lt;p&gt;Hello&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Prove that two distinct number of the form $a^{2^{n}} + 1$ and $a^{2^{m}} + 1$ are relatively prime if a is even and have $gcd=2$ if a is odd&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My attempt:&lt;br&gt;&#xA;If $a$ is even, let $a = 2^{s}k$ for some integers $k, s$&lt;br&gt;&#xA;Then, $$a^{2^{n}} + 1 = 2^{2^{n}s}\times k^{2^n} + 1$$ and $$a^{2^{m}}  + 1 = 2^{2^{m}s}\times k^{2^m} + 1$$&#xA;To prove that they're relatively prime, we need to show that their gcd = 1.&#xA;And I was stuck here, how could I prove that gcd of two numbers is $1$?   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A hint would be sufficient. Thanks.  &lt;/p&gt;&#xA;" OwnerUserId="5711" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-28T03:05:14.597" LastActivityDate="2011-02-28T05:44:51.803" Title="Prove that two distinct number of the form $a^{2^{n}} + 1$ and $a^{2^{m}} + 1$ are relatively prime if $a$ is even and have $gcd=2$ if $a$ is odd" Tags="&lt;elementary-number-theory&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="1" />
  <row Id="24144" PostTypeId="2" ParentId="24143" CreationDate="2011-02-28T03:00:04.383" Score="5" ViewCount="" Body="&lt;p&gt;If $x = -1 \mod p$, then $x^{2^n} = 1 \mod p$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume $n \gt m$. So if $p$ divides $x + 1  = a^{2^m} + 1$, then, $a^{2^n} + 1 = x^{2^{n-m}} + 1 = 1 + 1 = 2  \mod p$.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2011-02-28T03:08:04.750" LastActivityDate="2011-02-28T03:08:04.750" CommentCount="8" />
  <row Id="24145" PostTypeId="2" ParentId="24143" CreationDate="2011-02-28T03:01:38.740" Score="5" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Hint:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the following proof when $a=2$: &lt;a href=&quot;http://planetmath.org/encyclopedia/FermatNumbersAreCoprime.html&quot; rel=&quot;nofollow&quot;&gt;http://planetmath.org/encyclopedia/FermatNumbersAreCoprime.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Try adapting it to work for all $a$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint 2:&lt;/strong&gt; Factor $a^{2^n}-1$.&lt;/p&gt;&#xA;" OwnerUserId="6075" LastActivityDate="2011-02-28T03:01:38.740" CommentCount="8" />
  <row Id="24146" PostTypeId="1" AcceptedAnswerId="24147" CreationDate="2011-02-28T03:12:01.060" Score="3" ViewCount="117" Body="&lt;p&gt;This was a question on one of our practice midterms:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Find all invertible $n \times n$ matrices $A$ such that $$A^2 + A = 0.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was told to expand $A^2$ and then solve, but that seems like a really ugly (and hard-to-generalize) solution... are there any better ones?&lt;/p&gt;&#xA;" OwnerUserId="4890" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2011-02-28T03:29:43.127" LastActivityDate="2011-02-28T03:33:43.853" Title="Find all invertible $n\times n$ matrices $A$ such that $A^2 + A = 0$" Tags="&lt;linear-algebra&gt;&lt;matrices&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="1" />
  <row Id="24147" PostTypeId="2" ParentId="24146" CreationDate="2011-02-28T03:13:46.367" Score="6" ViewCount="" Body="&lt;p&gt;Note that&#xA;$$A^2+A=0\iff A^2=-A\iff A=-I$$&#xA;where $I$ is the identity matrix, and where we have used that $A$ is invertible in going from $A^2=-A$ to $A=-I$ (we multiplied by $A^{-1}$ on both sides). So the only solution is $A=-I$.&lt;/p&gt;&#xA;" OwnerUserId="264" LastActivityDate="2011-02-28T03:13:46.367" CommentCount="15" />
  <row Id="24148" PostTypeId="2" ParentId="24146" CreationDate="2011-02-28T03:14:23.307" Score="1" ViewCount="" Body="&lt;p&gt;It is given that $A$ is invertible and hence multiply by $A^{-1}$ to get $$A^{-1}A^2 + A^{-1}A = 0 \Rightarrow A + I =0 \Rightarrow A = -I$$&lt;/p&gt;&#xA;" OwnerUserId="3302" LastActivityDate="2011-02-28T03:14:23.307" />
  <row Id="24149" PostTypeId="2" ParentId="24146" CreationDate="2011-02-28T03:15:36.350" Score="9" ViewCount="" Body="&lt;p&gt;If $A^2 + A = 0$, then the minimal polynomial of $A$ divides $t^2+t = t(t+1)$. That means that the characteristic polynomial of $A$ must be of the form $(-1)^nt^k(t+1)^r$, where $k+r=n$. $\lambda=0$ cannot be a root, though (since you specify that $A$ is invertible, so $0$ is not an eigenvalue), so the characteristic polynomial is necessarily $(-1)^n(t+1)^n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But that means that the minimal polynomial of $A$ must be $t+1$ (since it must divide the characteristic polynomial, and also $t(t+1)$). This implies that $A+I=0$, hence $A=-I$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Added.&lt;/em&gt; If we drop the requirement that $A$ be invertible, then the invertible case proceeds as above. In the noninvertible case, the minimal polynomial is $t(t+1)$ or $t$. If it is $t$, then $A=0$. If the minimal polynomial is $t(t+1)$, then the matrix is diagonalizable (since the minimal polynomial splits and is square free), and the only eigenvalues are $0$ and $-1$. So the matrix $A$ is similar to a diagonal matrix in which every diagonal entry is $0$ or $-1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In summary, if we drop the requirement that $A$ be invertible, then there are $n+1$ similarity classes for possible $A$'s, each similarity class corresponding to a diagonal matrix that has $k$ diagonal entries equal to  $0$, followed by $n-k$ entries equal to $-1$s in the diagonal, $k=0,\ldots,n$. &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-28T03:33:43.853" LastActivityDate="2011-02-28T03:33:43.853" CommentCount="2" />
  <row Id="24151" PostTypeId="1" AcceptedAnswerId="24153" CreationDate="2011-02-28T03:25:05.357" Score="3" ViewCount="52" Body="&lt;p&gt;I was given the problem: find the area of the region bounded by $y=1/x$, $y=x^2$, $y=0$, and  $x=e$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My approach was to break it up into two integrals, $\displaystyle \int_0^1 (x^2-0)\,dx$ and $\displaystyle \int_1^e\left(\frac{1}{x}-0\right)\,dx$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Doing it this way resulted in an answer of 4/3. Simply by eyeballing it, the result seems reasonable, but I was wondering if there was a way to double check, or if someone else could verify that I am either doing this correctly or incorrectly.&lt;/p&gt;&#xA;" OwnerUserId="1389" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-28T03:31:17.617" LastActivityDate="2011-02-28T03:42:31.280" Title="Finding the area" Tags="&lt;integral&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="24152" PostTypeId="2" ParentId="24143" CreationDate="2011-02-28T03:27:16.807" Score="1" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\rm\ \: (A+1,\ A^{2\:K}+1)\ =\ (A+1,\ (-1)^{2\:K}+1)\ =\ (A+1,\:2\:)\:.\:$ Put $\rm\ A = a^{2^{N}}\ $ (wlog $\rm\ N &amp;lt; M\:)\:.$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ $ using $\rm\ \  (A+I,\ \ F(A)\ )\ \ \  =\ \ \: (A+I,\ \ F(-I)\ )\ \ $ for all polynomials $\rm\ F(X)\in \mathbb Z[X],\ \ A,\:I\in \mathbb Z$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ \ \ $ i.e. $\rm\ \ \ \ \ A\ \equiv\: -I\ \ \ \Rightarrow\ \ \ F(A)\ \equiv\ F(-I)\ \ \ \ (mod\ D)\:,\ $ if $\rm\ D\ |\ A+I$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ \ \ $ i.e. $\rm\ \ \ (B,\ C)\ =\ (B,\ C\ mod\ B)\ \ $ -- $\:$ the &lt;em&gt;modular property&lt;/em&gt; of GCDs.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2011-02-28T05:44:51.803" LastActivityDate="2011-02-28T05:44:51.803" CommentCount="3" />
  <row Id="24153" PostTypeId="2" ParentId="24151" CreationDate="2011-02-28T03:29:06.683" Score="3" ViewCount="" Body="&lt;p&gt;What you did is indeed correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Splitting up as above, the area is $$A=\int_0^1x^2dx+\int_1^e \frac{1}{x}dx=\frac{1}{3}-0+\ln(e)-\ln(1)=\frac{4}{3}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope that helps,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Eyeballing:&lt;/strong&gt;  So here is a way to see get be a bit more confident with the answer.  When I drew the picture, it fit inside a rectangle of height $1$ and length $e$.  (Vertices $(0,0)$, $(0,1)$, $(e,0)$, $(e,1)$).  It looked like our area was roughly half of this rectangle, so $\frac{e}{2}$ since the rectangle has area $e\times 1=e$.  On a calculator, $4/3=1.333$ and $e/2=1.359$ so we are pretty close, as expected.&lt;/p&gt;&#xA;" OwnerUserId="6075" LastEditorUserId="6075" LastEditorDisplayName="" LastEditDate="2011-02-28T03:42:31.280" LastActivityDate="2011-02-28T03:42:31.280" CommentCount="2" />
  <row Id="24154" PostTypeId="1" CreationDate="2011-02-28T03:35:08.267" Score="1" ViewCount="93" Body="&lt;p&gt;Given a uniformly continuous function $f(x)$ on the real numbers $\Bbb R$, then by the definition of uniform continuity this means: for any $\epsilon&amp;gt;0$ there exists $\delta &amp;gt;0$ such that $|f(x)-f(y)|&amp;lt;\epsilon$ whenever $|x-y|&amp;lt;\delta$. My question is: Given $\epsilon_{n} =\frac{1}{3^{n}}$ (for example), and $y_{n}=x_{n}+w_{n}$ so that $|x_{n} -y_{n}|=|w_{n}|$,  then we know that there is $\delta_{n}&amp;gt;0$, how can we find the corresponding $\delta_{n}$? I don't know if my question makes sense. &lt;/p&gt;&#xA;" OwnerUserId="7586" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-28T03:43:40.040" LastActivityDate="2011-02-28T03:43:40.040" Title="Uniformly continuous functions." Tags="&lt;calculus&gt;&lt;analysis&gt;" CommentCount="6" />
  <row Id="24155" PostTypeId="1" AcceptedAnswerId="24159" CreationDate="2011-02-28T03:40:34.050" Score="3" ViewCount="87" Body="&lt;p&gt;Hello,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm trying to formally show that if two circles $\Gamma$ and $\Gamma&amp;#39;$ are tangent at $A$, then $\Gamma&amp;#39;$ lies either entirely inside or entirely outside $\Gamma$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The one particular case I'm struggling with is when $\Gamma$ and $\Gamma&amp;#39;$ lie on the same side of the tangent line $l$ at $A$. I know that $O$, $O&amp;#39;$ and $A$ are collinear. I suppose the radius $r$ of $\Gamma&amp;#39;$ is smaller than radius $s$ of $\Gamma$, and I want to show that $\Gamma&amp;#39;$ is entirely inside $\Gamma$, except for $A$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/BFsEs.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know line $OO&amp;#39;$ meets $\Gamma&amp;#39;$ on the other side of $A$ at some point $C$. I've been able to show that $C$ is inside $\Gamma$. For if $C*O*O&amp;#39;$, then $CO&amp;lt;CO&amp;#39;=r&amp;lt;s$, so $C$ is inside $\Gamma$. If $O*C*O&amp;#39;$, then $CO&amp;lt;OO&amp;#39;&amp;lt;OA=s$, so again $C$ is inside $\Gamma$. But when I take an arbitrary point $B$ on $\Gamma&amp;#39;$, I have no idea how to show that $OB&amp;lt;s$, to prove that $B$ is inside $\Gamma$. Does anyone know a nice way to rigorously show this? Thank you.&lt;/p&gt;&#xA;" OwnerUserId="1583" LastActivityDate="2011-02-28T04:24:56.097" Title="Why are tangent circles entirely inside or outside the other?" Tags="&lt;geometry&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="24156" PostTypeId="2" ParentId="24134" CreationDate="2011-02-28T03:51:30.453" Score="1" ViewCount="" Body="&lt;p&gt;I'm not sure if it is such a good approach. The most general vector linear in $r$ is $M r$ where $M$ is a 3x3 matrix. The number of unfixed constants in your formula is 12, while you only need 9 in general. If you set $b=c$ it seems more sensible, assuming $r\times c\neq0$. Then it's essentially just the expansion of a vector in the basis $r$, $c$, $r\times c$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The matrix $M$ can be divided into 3 parts, the trace part, a traceless symmetric part and a antisymmetric part:&#xA;$$ M = \frac13\mathrm{tr}(M)\,I + \frac12(M+M^t-\frac23\mathrm{tr}(M)\,I) + \frac12(M-M^t) $$&#xA;where $I$ is the identity matrix and ${}^t$ denotes transpose.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the terms in $M\cdot r$ correspond to the terms&#xA;$$ x r + y (r\cdot a)c + z (r\times c)\ .$$ You can see exactly how by just taking the gradient of both sides to get&#xA;$$ M = x\, I + y\, c a^t + z\, \varepsilon\cdot c$$&#xA;where $\varepsilon$ is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Levi-Civita_symbol&quot; rel=&quot;nofollow&quot;&gt;totally antisymmetric tensor&lt;/a&gt; and $c a^t$ can be chosen to be traceless, ie $a \cdot c=0$. In terms of components, this reads&#xA;$$ M_{ij} = x\, \delta_{ij} + y\, c_i a_j + z\, \sum_k\varepsilon_{ijk} c_k \ ,$$&#xA;where $\delta$ is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Kronecker_delta&quot; rel=&quot;nofollow&quot;&gt;Kronecker delta symbol&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="954" LastEditorUserId="954" LastEditorDisplayName="" LastEditDate="2011-02-28T04:13:41.107" LastActivityDate="2011-02-28T04:13:41.107" CommentCount="1" />
  <row Id="24158" PostTypeId="2" ParentId="20948" CreationDate="2011-02-28T04:06:39.193" Score="0" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;NOTE&lt;/strong&gt; $\ $ This is precisely the same answer I gave a couple months ago in &lt;a href=&quot;http://math.stackexchange.com/questions/11477/showing-that-an-equation-holds-true-with-a-fibonacci-sequence-f-nm-f-n-1/11482#11482&quot;&gt;this duplicate thread&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2011-02-28T04:06:39.193" />
  <row Id="24159" PostTypeId="2" ParentId="24155" CreationDate="2011-02-28T04:16:59.487" Score="1" ViewCount="" Body="&lt;p&gt;I'm not sure what axiomatization of Euclidean Geometry you're using, and what results you are considering as given at this point, but are there any reasons why you cannot use the triangle inequality to show that $OB&amp;lt;OO^{&amp;#39;}+O^{&amp;#39;}B$? Then since $O^{&amp;#39;}B=O^{&amp;#39;}A$, you would have that $OB&amp;lt;OA$. (You have claimed that I can take as given that $O$, $O^{&amp;#39;}$, and $A$ are colinear). &lt;/p&gt;&#xA;" OwnerUserId="7323" LastEditorUserId="7323" LastEditorDisplayName="" LastEditDate="2011-02-28T04:24:56.097" LastActivityDate="2011-02-28T04:24:56.097" CommentCount="1" />
  <row Id="24160" PostTypeId="1" AcceptedAnswerId="24173" CreationDate="2011-02-28T04:21:08.837" Score="2" ViewCount="108" Body="&lt;p&gt;Let $p(z)$  be a polynomial and $C$ denote the circle $|z-a| = R$. I want to evaluate the integral over $C$ of $p(z)$ with respect to $\overline{z}$, the conjugate of $z$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think to start off, I should probably use a Taylor expansions centered at $a$, but I'm lost with what to do next. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any hints on how to approach this problem? &lt;/p&gt;&#xA;" OwnerUserId="6758" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-28T05:39:44.070" LastActivityDate="2011-02-28T05:54:59.937" Title="A Difficult Contour Integral?" Tags="&lt;homework&gt;&lt;complex-analysis&gt;" AnswerCount="1" />
  <row Id="24161" PostTypeId="1" AcceptedAnswerId="24163" CreationDate="2011-02-28T04:21:30.330" Score="7" ViewCount="128" Body="&lt;p&gt;This is a well known lemma that consistently appears in textbooks, either as a statement without proof, or as an exercise (see for example pp. 146 of Hatcher)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $0 \stackrel{id}{\to} A \stackrel{f}{\to} B \stackrel{g}{\to} C\stackrel{h}{\to} 0$ is a short exact sequence of finitely generated abelian groups, then rank B = rank A + rank C&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've been trying to prove this unsuccessfully. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What do we know? $f$ is injective, $g$ is surjective, $\mathrm{Im} f = \mathrm{ker} g$, $\mathrm{Im} g = \mathrm{ker} h$, $C\simeq B/A$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I start with a maximally linearly independent subset $\{ a_\alpha \}$ of $A$ such that the sum (with only finite non-zero entries)&#xA;$$\sum n_\alpha a_\alpha=0$$ for $n_\alpha \in \mathbb{Z}$, implies that $n_\alpha=0$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where to go from here is a puzzle? Any hints would be appreciated&lt;/p&gt;&#xA;" OwnerUserId="2219" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-28T04:32:45.457" LastActivityDate="2011-02-28T05:01:36.933" Title="Short Exact Sequences &amp; Rank Nullity" Tags="&lt;group-theory&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="24162" PostTypeId="1" AcceptedAnswerId="24233" CreationDate="2011-02-28T04:33:05.967" Score="1" ViewCount="122" Body="&lt;p&gt;In many articles and papers I find the phrase&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;...discrete logarithm problem in many groups of cryptographic interest...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;commonly used. I would like what groups are they exactly referring to when they mention &quot;groups of cryptographic interest&quot; when it comes to the discrete logarithm problem. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="7488" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-28T04:50:50.170" LastActivityDate="2011-02-28T14:14:55.307" Title="&quot;Discrete logarithm problem in many groups of cryptographic interest&quot;?" Tags="&lt;cryptography&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="24163" PostTypeId="2" ParentId="24161" CreationDate="2011-02-28T04:39:22.080" Score="7" ViewCount="" Body="&lt;p&gt;Pick a maximal linearly independent subset $\{c_{\beta}\}$ of $C$. Now push the $a_{\alpha}$ to $B$ using $f$, and for each $c_{\beta}$ pick $c&amp;#39;_{\beta}\in B$ such that $g(c&amp;#39;_{\beta}) = c_{\beta}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now suppose that you have a finite linear combination of the $a_{\alpha}$ and the $c&amp;#39;_{\beta}$ that is equal to $0$,&#xA;$$n_{\alpha_1}f(a_{\alpha_1}) + \cdots + n_{\alpha_k}f(a_{\alpha_k}) + m_{\beta_1}c&amp;#39;_{\beta_1} + \cdots + m_{\beta_{\ell}}c&amp;#39;_{\beta_{\ell}} = 0.$$&#xA;Use $g$ to get a conclusion about the $m_{\beta_j}$; then use $f$ to get a conclusion about the $n_{\alpha_i}$. This will give you that $\mathrm{rank}(B)\geq \mathrm{rank}(A)+\mathrm{rank}(C)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do you also need help with the converse inequality?&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2011-02-28T04:39:22.080" CommentCount="9" />
  <row Id="24164" PostTypeId="1" AcceptedAnswerId="24165" CreationDate="2011-02-28T04:54:20.577" Score="3" ViewCount="115" Body="&lt;p&gt;Hello,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to develop a divisible test for 13, we use $1000 \equiv -1 \pmod{1001}$.&lt;br&gt;&#xA;I understand the idea; however, why do we use $1001$, can we use any smaller number? For example, to test for divisible by $11$, we need to use only $10 \equiv -1 \pmod{11}$? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks,&lt;/p&gt;&#xA;" OwnerUserId="5711" LastActivityDate="2011-02-28T07:30:36.537" Title="Question regarding divisibility test of 13" Tags="&lt;elementary-number-theory&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="24165" PostTypeId="2" ParentId="24164" CreationDate="2011-02-28T04:58:46.167" Score="5" ViewCount="" Body="&lt;p&gt;There are many divisibility tests for $13$. But tests like that for $3$, $9$ and $11$ (among others) are particularly good because they not only test the number for divisibility, but they actually tell you the remainder when the number is not divisible by what you are testing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What the &quot;test in development&quot; is trying to do is use something similar to the tests for $3$ and $9$ (adding the digits) or for $11$ (alternating sums and differerences of digits). In order to do something like that, with groups of digits, you want to find the smallest power of $10$ for which $10^k \equiv 1$ or $10^k\equiv -1 \pmod{13}$. The smallest such power happens to be $10^3 = 1000$ (as $10\equiv 10\equiv -3 \pmod{13}$, and $100 \equiv 9\equiv -4\pmod{13}$). So $10^3$ is the smallest one that can be used to develop a test that follows the pattern of those for $3$, $9$, and $11$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Added.&lt;/em&gt; There are other tests, of course. For example, you can develop a test similar to the one for $7$: take the last digit, multiply it by $2$, and subtract it from the rest of the digits; the original number is divisible by $7$ if and only if the result is divisible by $7$; but if the result is not divisible by $7$, the remainder need not be the same as that for the original number).  A similar test for $13$ is: take last digit, multiply it by $4$, and add it to the rest; the original number is divisible by $13$ if and only if the result is divisible by $14$. &lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2011-02-28T04:58:46.167" CommentCount="3" />
  <row Id="24166" PostTypeId="2" ParentId="24161" CreationDate="2011-02-28T05:01:36.933" Score="6" ViewCount="" Body="&lt;p&gt;If you're not familiar with tensor products, then learning them for this problem is overkill. But if you are familiar with them, then I think the following solution is nice.&#xA;$\newcommand{\Q}{\mathbb{Q}}&#xD;&#xA;\newcommand{\Z}{\mathbb{Z}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a finitely generated abelian group $A$, note that $\mathrm{rank}(A) = \dim_\Q(A\otimes_\Z \Q)$. One can show (basically following Arturo's answer) that given a short exact sequence of abelian groups $0\to A\to B\to C\to 0$, the sequence obtained by tensoring with $\Q$ is also exact:&#xA;$$0\to A\otimes_\Z \Q \to B\otimes_\Z \Q \to C\otimes_\Z \Q \to 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the statement about ranks boils down to an easy statement about how dimensions of vector spaces add in a short exact sequence. &lt;/p&gt;&#xA;" OwnerUserId="60" LastActivityDate="2011-02-28T05:01:36.933" CommentCount="2" />
  <row Id="24167" PostTypeId="1" CreationDate="2011-02-28T05:02:00.277" Score="1" ViewCount="61" Body="&lt;p&gt;Let $X$ be a non-empty subset of $\mathbb{R}^n$. For $x=(x_1,x_2,\dotsc,x_n)$ and $(k_1,k_2,\dotsc,k_n)$ a finite sequence of non-negative integers, consider the map&#xA;$x \mapsto {x_1}^{k_1} + {x_2}^{k_2} + \dotsb + {x_n}^{k_n}$.&#xA;Prove that the linear combination of these are dense in $C^r(X)$. &lt;/p&gt;&#xA;" OwnerUserId="7588" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-28T05:13:08.380" LastActivityDate="2011-02-28T05:13:08.380" Title="When a linear combination of functions is dense" Tags="&lt;analysis&gt;" CommentCount="4" />
  <row Id="24168" PostTypeId="2" ParentId="24164" CreationDate="2011-02-28T05:33:48.567" Score="2" ViewCount="" Body="&lt;p&gt;Unless you're dividing by the radix $\rm\pm 1$, i.e. $9$ or $11$ in decimal, such divisibitity tests aren't really any quicker than simply evaluating the radix polynomial in &lt;em&gt;Horner form&lt;/em&gt; using modular arithmetic. Here's one of my old posts describing such (which, as a bonus, describes how repeated squaring arises from binary Horner form).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/VLdsU.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2011-02-28T05:33:48.567" CommentCount="1" />
  <row Id="24169" PostTypeId="1" AcceptedAnswerId="24170" CreationDate="2011-02-28T05:42:31.160" Score="6" ViewCount="129" Body="&lt;p&gt;Hello,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My attempt was:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By Fermat's little theorem:&lt;br&gt;&#xA;$$2^{22} \equiv 1 \pmod{23}$$&#xA;$$(2^{11})^2 \equiv 1 \pmod{23}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I checked with my calculator the remainder is actually $1$. However, I wonder if I can take the square root on both sides of congruence. Any idea?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks,&lt;/p&gt;&#xA;" OwnerUserId="5711" LastActivityDate="2011-03-01T07:36:02.743" Title="Find the remainder of $2^{11}$ by $23$" Tags="&lt;elementary-number-theory&gt;" AnswerCount="3" CommentCount="4" />
  <row Id="24170" PostTypeId="2" ParentId="24169" CreationDate="2011-02-28T05:47:32.003" Score="2" ViewCount="" Body="&lt;p&gt;Yes you can take the square root, the elements $\{0,1,2, \dots, 22\}$ form a finite field, when the operations are taken modulo $23$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That only tells you that it is $\pm 1$, though.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2011-02-28T05:47:32.003" CommentCount="7" />
  <row Id="24171" PostTypeId="1" AcceptedAnswerId="24175" CreationDate="2011-02-28T05:51:33.280" Score="1" ViewCount="98" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Let $f_n, f : [a,b] \to R.$ &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;If $f_n \to f$  uniformly then show that the lebesgue integrals are equal. ie. $\int f = lim \int f_n$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is clearly true for continuous functions, but how do I handle the case of non-continuous functions?&lt;/p&gt;&#xA;" OwnerUserId="6556" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2011-02-28T06:07:18.230" LastActivityDate="2011-02-28T06:07:18.230" Title="lebesgue integral uniform convergence" Tags="&lt;real-analysis&gt;&lt;integral&gt;" AnswerCount="2" />
  <row Id="24173" PostTypeId="2" ParentId="24160" CreationDate="2011-02-28T05:54:59.937" Score="1" ViewCount="" Body="&lt;p&gt;Imagine the circle shifted to the origin and scaled to the unit circle -- can you now express $\overline{z}$ in terms of $z$?&lt;/p&gt;&#xA;" OwnerUserId="6622" LastActivityDate="2011-02-28T05:54:59.937" />
  <row Id="24174" PostTypeId="2" ParentId="24171" CreationDate="2011-02-28T05:57:37.283" Score="3" ViewCount="" Body="&lt;p&gt;It follows from&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\left| \int f \, dm - \int f_n \, dm \right| \le \int \left|f - f_n\right| \, dm \le (b-a) \Vert f - f_n \Vert_\infty$&lt;/p&gt;&#xA;" OwnerUserId="3208" LastActivityDate="2011-02-28T05:57:37.283" />
  <row Id="24175" PostTypeId="2" ParentId="24171" CreationDate="2011-02-28T06:06:05.237" Score="2" ViewCount="" Body="&lt;p&gt;Do you know how to do this argument for Riemann integrable functions?  It's the same one for Lebesgue integrable functions.  Hint: if $f_n \rightarrow f$ uniformly, then for all $\epsilon &amp;gt; 0$ and all sufficiently large $n$, $|f_n(x) - f(x)|$ is measurable and less than $\epsilon$ for all $x \in [a,b]$.  What does that tell you about the Lebesgue &#xA;integral $\int_{[a,b]} f_n -f$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the hypothesis that the total measure of the space be finite is essential.  It is easy to construct a sequence of integrable continuous functions $f_n: [0,\infty) \rightarrow [0,\infty)$ which converges uniformly to $0$ but so that $\int_{[0,\infty)} f \rightarrow \infty$.   &lt;/p&gt;&#xA;" OwnerUserId="299" LastActivityDate="2011-02-28T06:06:05.237" CommentCount="2" />
  <row Id="24176" PostTypeId="2" ParentId="24169" CreationDate="2011-02-28T06:09:04.860" Score="3" ViewCount="" Body="&lt;p&gt;$\left( \frac{2}{23}\right) = 1 \Rightarrow 2^{11} \equiv 1 \left( \bmod 23 \right)$&lt;/p&gt;&#xA;" OwnerUserId="3302" LastActivityDate="2011-02-28T06:09:04.860" CommentCount="2" />
  <row Id="24177" PostTypeId="1" AcceptedAnswerId="24182" CreationDate="2011-02-28T06:14:12.697" Score="3" ViewCount="38" Body="&lt;p&gt;Suppose $u$ and $v$ are measures on a measurable space $E$. Further suppose $u$ is finite and absolutely continuous with respect to v ($v(S)=0 \implies u(S)=0$).&#xA;The problem is: Show that $\forall \epsilon&amp;gt;0 \;\;\;\exists \delta: v(S)&amp;lt;\delta \implies u(S)&amp;lt;\epsilon.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've tried the following: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fix $\epsilon&amp;gt;0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $S_n=\{measurable\;\; B\in E|u(B)&amp;gt;\epsilon, v(B)&amp;lt;1/n\}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $S_n$ is finite for some n we are done ($v(S)=0 \implies u(S)=0$). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now what I am trying to show is that the other case, $S_n$ is infinite $\forall n\in\mathbb{N}$, leads to a contradiction.&#xA;Any suggestions?&lt;/p&gt;&#xA;" OwnerUserId="7590" LastActivityDate="2011-02-28T06:31:08.453" Title="Absolute continuity of measures" Tags="&lt;homework&gt;" AnswerCount="1" />
  <row Id="24178" PostTypeId="1" AcceptedAnswerId="24597" CreationDate="2011-02-28T06:17:46.203" Score="10" ViewCount="177" Body="&lt;p&gt;Let $S$ be a finite group with operator + and $\pi$ be a permutation on $S$.  Then what is the probability that $\pi(x) + x$ is injective over choices of $\pi$?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The concrete instantiation I'm interested in is $S=$GF$(2^n)$ for fixed $n &amp;gt; 0$.  (Computer Scientists call this &quot;xor on $n$-bit strings.&quot;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $n=1$ we have two permutations, neither of which produce an injection. For $n=2$ we have 24 permutations, 8 of which induce an injection so the probability is 1/3.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is an example $\pi$ for $n=2$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\pi(0)=0,\ \pi(1) = z,\ \pi(z)=z+1,\ \pi(z+1)=1.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here notice that $\pi(x) + x$ produces $0$, $z+1$, $1$, and $z$, respectively, which is an injection.&lt;/p&gt;&#xA;" OwnerUserId="7162" LastActivityDate="2011-03-02T08:00:25.750" Title="What is the probability that $\pi(x) + x$ is injective?" Tags="&lt;probability&gt;&lt;discrete-mathematics&gt;&lt;counting&gt;" AnswerCount="2" FavoriteCount="3" />
  <row Id="24179" PostTypeId="1" AcceptedAnswerId="24180" CreationDate="2011-02-28T06:18:47.900" Score="2" ViewCount="95" Body="&lt;p&gt;Hello,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Definition:&#xA;  A complete system of residues modulo m is a set of integers such that every integer is congruent modulo m to exactly one integer of the set.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;Example&lt;/strong&gt;: The division algorithm shows that the set of $0, 1, 2, ..., m - 1$ is a complete system of residues modulo m.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What I did not understand is &quot;congruent modulo m to exactly one integer of the set&quot;. Could anyone give me a counter example to this? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And here is the problem:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Prove that the set $0, 1, 3, 3^2, 3^3, ...., 3^{15}$ is a complete system of residues modulo $17$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I really have no idea how to start and what to prove :(. To be a complete system of residues modulo, what property does this set have to have? A hint would be greatly appreciated.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks, &lt;/p&gt;&#xA;" OwnerUserId="5711" LastActivityDate="2011-02-28T14:29:06.323" Title="How to prove a set of numbers is a complete system of residues modulo of a number?" Tags="&lt;elementary-number-theory&gt;" AnswerCount="2" />
  <row Id="24180" PostTypeId="2" ParentId="24179" CreationDate="2011-02-28T06:21:28.567" Score="2" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT:&lt;/strong&gt;  $\left(\frac{3}{17}\right) = -1$. Essentially you want to prove all the remainders are distinct $\bmod 17$.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastActivityDate="2011-02-28T06:21:28.567" CommentCount="5" />
  <row Id="24181" PostTypeId="2" ParentId="24169" CreationDate="2011-02-28T06:22:06.263" Score="3" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\rm\ \ mod\ 23\::\ \ 2\ \equiv\ 5^{\:2}\ \ \Rightarrow\ \ 2^{\:11}\ \equiv\  5^{\:22}\ \equiv\ 1\ $ by Fermat's little Theorem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See &lt;a href=&quot;http://en.wikipedia.org/wiki/Euler%27s_criterion&quot; rel=&quot;nofollow&quot;&gt;Euler's Criterion&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/Quadratic_reciprocity&quot; rel=&quot;nofollow&quot;&gt;quadratic reciprocity&lt;/a&gt; to understand what happens generally.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regarding square-roots, $\rm\ x^2 = a^2\ \iff\ (x-a)\ (x+a) = 0\ \iff\ x = \pm\: a\ \ $ holds true in any integral domain, i.e. it's true in any ring without zero-divisors. More concretely, in $\rm\ \mathbb Z/p\:,\: $ we have prime $\rm\ p\ |\ (x-a)\ (x+a)\ \Rightarrow\ p\ |\ x-a\ $ or $\rm\ p\ |\ x+a\:,\: $ so $\rm\ x \equiv \pm\: a\ \ (mod\ p)\:.$&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2011-02-28T07:01:09.537" LastActivityDate="2011-02-28T07:01:09.537" CommentCount="10" />
  <row Id="24182" PostTypeId="2" ParentId="24177" CreationDate="2011-02-28T06:31:08.453" Score="1" ViewCount="" Body="&lt;p&gt;I think that if you change the 1/n into $1/(2^n)$ and choose a sequence $B_n\in S_n$ such that $v(B_n)&amp;lt;2^{-n}$, and $u(B_n)&amp;gt;\epsilon$, and consider $B=\limsup B_n=\bigcap(\bigcup B_n)$, you might get your contradiction.&lt;/p&gt;&#xA;" OwnerUserId="7129" LastActivityDate="2011-02-28T06:31:08.453" />
  <row Id="24183" PostTypeId="2" ParentId="24077" CreationDate="2011-02-28T06:45:29.540" Score="2" ViewCount="" Body="&lt;p&gt;You're right. In fact, every finite language is regular, and so is it's complement. They're context-free a fortiori.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2011-02-28T06:45:29.540" />
  <row Id="24184" PostTypeId="1" AcceptedAnswerId="24185" CreationDate="2011-02-28T06:47:28.570" Score="4" ViewCount="128" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Prove that if a and b are relatively prime, then&lt;br&gt;&#xA;  $$\sum_{n=1}^{a-1} \left\lfloor \frac{nb}{a}\right\rfloor = \frac{(a - 1)(b - 1)}{2}$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My attempt was:&lt;br&gt;&#xA;We have:&#xA;$$\sum_{i=1}^{n-1} i = \frac{n(n - 1)}{2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, &#xA;$$\sum_{n=1}^{a-1} \left\lfloor \frac{nb}{a}\right\rfloor = \left\lfloor \frac{a(a - 1)b}{2a}\right\rfloor$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could I apply the summation formula for floor function like above? Am I in the right track?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks,&lt;br&gt;&#xA;Chan&lt;/p&gt;&#xA;" OwnerUserId="5711" LastEditorUserId="1269" LastEditorDisplayName="" LastEditDate="2011-02-28T06:49:36.177" LastActivityDate="2011-02-28T08:51:40.393" Title="How does summation formula work with floor function?" Tags="&lt;elementary-number-theory&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="24185" PostTypeId="2" ParentId="24184" CreationDate="2011-02-28T06:50:16.213" Score="3" ViewCount="" Body="&lt;p&gt;No, that doesn't work - it's just not true, since in general $\lfloor x + y \rfloor \neq \lfloor x \rfloor + \lfloor y \rfloor$, e.g. try $x=y=0.6$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have the general formula $$\left\lfloor \frac{x}{y} \right\rfloor = \frac{x-(x \mod{y})}{y}.$$ You can calculate the same using this formula.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2011-02-28T06:50:16.213" CommentCount="1" />
  <row Id="24186" PostTypeId="2" ParentId="24000" CreationDate="2011-02-28T07:03:08.787" Score="2" ViewCount="" Body="&lt;p&gt;If you are having difficulties accepting mathematical induction and other of Peano's Axioms, I find it helps to visualize the natural numbers as directed graph (a network of nodes connected by one-way arrows) as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a node (call it 0) that has no arrows going into it (Axioms 1 and 3, below). Every node has an arrow leaving it (Axiom 2, below). Every node has at most one arrow going into it (Axiom 4). You can get to any node by starting at 0 and following the arrows; that is, there are no isolated portions of the graph (Axiom 5). You may also require that every node have at most one arrow leaving it (not explicitly covered in the axioms as presented below).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Peano's Axioms from from Wolfram MathWorld &lt;a href=&quot;http://mathworld.wolfram.com/PeanosAxioms.html&quot; rel=&quot;nofollow&quot;&gt;http://mathworld.wolfram.com/PeanosAxioms.html&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Zero is a number.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If x is a number, the successor of x is a number.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Zero is not the successor of a number.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Two numbers of which the successors are equal are themselves equal.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If a set  of numbers contains zero and also the successor of every number in , then every number is in .&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Added:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The following is a better presentation of Peano's axioms (from &lt;a href=&quot;http://www.ms.uky.edu/~lee/ma502/notes2/node7.html&quot; rel=&quot;nofollow&quot;&gt;http://www.ms.uky.edu/~lee/ma502/notes2/node7.html&lt;/a&gt;) It uses 1 instead of 0 as the first natural number. Unlike the above presentation, the fact that, graphically, there would be at most one arrow leaving each node is made explicit in Axiom 2, below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Axiom 1: 1 is a natural number. That is, our set is not empty; it contains an object called 1 (read &quot;one&quot;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Axiom 2: For each x there exists exactly one natural number, called the successor of x, which will be denoted by x'.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Axiom 3: We always have ~x'=1. That is, there exists no number whose successor is 1. That is, there exists no number whose successor is 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Axiom 4: If x'=y' then x=y. That is, for any given number there exists either no number or exactly one number whose successor is the given number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Axiom 5 (Axiom of Induction): Let there be given a set M of natural numbers, with the following properties:&#xA;I. 1 belongs to M.&#xA;II. If x belongs to M then so does x'.&#xA;Then M contains all the natural numbers.&lt;/p&gt;&#xA;" OwnerUserId="3515" LastEditorUserId="3515" LastEditorDisplayName="" LastEditDate="2011-02-28T14:01:31.323" LastActivityDate="2011-02-28T14:01:31.323" />
  <row Id="24187" PostTypeId="2" ParentId="24109" CreationDate="2011-02-28T07:04:02.020" Score="0" ViewCount="" Body="&lt;p&gt;Continuing Ross's answer, the correct way is to take $n$ bit binary numbers to $\lfloor (\log_3 2) n \rfloor$ ternary numbers. If you choose such an $n$ so that $(\log_3) 2 n$ is close to an integer, this will be &quot;relatively random&quot; How close to an integer can $(\log_3 2) n$ be? Dirichlet's theorem tells us that the error can be roughly $1/n$ (this is optimal by Roth's theorem), and so letting $m = \lfloor (\log_3 2) n \rfloor$, the range of your function will be $$3^{m-\Theta(1/m)} = 3^{-\Theta(1/m)} 3^m.$$ As $m$ gets bigger, the range of the function becomes more and more closer to $3^m$, which makes it more and more difficult to break the function. You will need to look for that many numbers:&#xA;$$ \frac{1}{1 - 3^{-\Theta(1/m)}} = \Theta(m). $$ In cryptography, when trying to break an $m$-digits generator, you can use polynomially many queries (in $m$). So in this sense, the function is not secure after all.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2011-02-28T07:04:02.020" />
  <row Id="24188" PostTypeId="1" AcceptedAnswerId="24191" CreationDate="2011-02-28T07:08:06.753" Score="4" ViewCount="195" Body="&lt;p&gt;Hello, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is $2^{340} - 1 \equiv 1 \pmod{341} $?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is one of my homework problem, prove the statement above. However, I believe it is wrong. Since $2^{10} \equiv 1 \pmod{341}$, so $2^{10 \times 34} \equiv 1 \pmod{341}$ which implies $2^{340} - 1 \equiv 0 \pmod{341}$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any idea?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks,  &lt;/p&gt;&#xA;" OwnerUserId="5711" LastEditorUserId="5711" LastEditorDisplayName="" LastEditDate="2011-03-01T00:56:19.777" LastActivityDate="2011-03-01T00:56:19.777" Title="Is $2^{340} - 1 \equiv 1 \pmod{341} $?" Tags="&lt;elementary-number-theory&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="1" />
  <row Id="24189" PostTypeId="2" ParentId="24184" CreationDate="2011-02-28T07:09:02.743" Score="2" ViewCount="" Body="&lt;p&gt;Notice for each $n$ $$\left\lfloor\frac{nb}{a}\right\rfloor=\frac{nb}{a}-\frac{r_n}{a}$$ where $r_n\equiv nb \pmod a$ and $0\leq r_n&amp;lt; a$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then since $a$ and $b$ are relatively prime, $nb\equiv mb \pmod a$ implies $n\equiv m \pmod a$ so that we conclude each remainder $r_n$ is unique.  But, we know $r_n\in \{1,2,\dots,a-1\}$ for each $n$, and since there are $a-1$ different values for $n$ we see that $r_n$ takes on each of these values.  Hence $$\sum_{i=1}^{a-1}r_i=\sum_{i=1}^{a-1}i=\frac{a(a-1)}{2}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then our original sum is &#xA;$$\sum_{n=1}^{a-1} \left\lfloor \frac{nb}{a}\right\rfloor = \sum_{n=1}^{a-1} \frac{nb}{a}-\sum_{n=1}^{a-1}\frac{r_n}{a}=\frac{a(a - 1)b}{2a}-\frac{a(a-1)}{2a}=\frac{(a - 1)(b-1)}{2}$$ as desired.&lt;/p&gt;&#xA;" OwnerUserId="6075" LastActivityDate="2011-02-28T07:09:02.743" CommentCount="2" />
  <row Id="24191" PostTypeId="2" ParentId="24188" CreationDate="2011-02-28T07:13:03.443" Score="5" ViewCount="" Body="&lt;p&gt;What you wrote is correct.  $$2^{340}\equiv 1\pmod {341}$$  This is smallest example of a pseudoprime to the base two.  See &lt;a href=&quot;http://en.wikipedia.org/wiki/Fermat_pseudoprime&quot; rel=&quot;nofollow&quot;&gt;Fermat Pseudo Prime&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="6075" LastActivityDate="2011-02-28T07:13:03.443" CommentCount="1" />
  <row Id="24192" PostTypeId="2" ParentId="24179" CreationDate="2011-02-28T07:14:43.987" Score="2" ViewCount="" Body="&lt;p&gt;By the pigeonhole principle, $\rm\:S\:$ is a CSR $\rm\ (mod\ m)\ $ if $\rm\:S\:$ has $\rm\:m\:$ elements that are distinct $\rm (mod\ m)\:.\:$ Hence, for your problem, you need to show that $\rm\ \ i\ne j\ \Rightarrow\ 3^{\:i}\: \not\equiv 3^{\:j}\ (mod\ 17)\:,\ $ for $\rm\ 0\le i,\:j\le 15\:,\ $ i.e. you need to show mod $\:17\:,\:$ that $\:3\:$ has order $\rm 16\:$. If its order $\rm\:n\:$ is $ &amp;lt; 16\: $ then $\rm\:n\: |\: 16\ $ so $\rm\: n\: |\: 8\:,\: $ but $\rm\ 3^4 \equiv -4\ \Rightarrow\ 3^8\equiv -1\:.\:$ So, indeed, $3\:$ has order $16\:$.  Combining this with $\rm\ 3^{\:i}\:\not\equiv 0\ (mod\ 17)\ $ (why?) completes the proof.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2011-02-28T14:29:06.323" LastActivityDate="2011-02-28T14:29:06.323" />
  <row Id="24194" PostTypeId="1" CreationDate="2011-02-28T07:28:07.147" Score="1" ViewCount="64" Body="&lt;p&gt;Use the method of cylindrical shells to find the volume generated by rotating the region bounded by the given curves about the y-axis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$y= 27e^x$, $y= 27e^{-x}$, $x = 1$.&lt;/p&gt;&#xA;" OwnerUserId="7591" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-28T14:52:33.650" LastActivityDate="2011-02-28T14:52:33.650" Title="solid of revolution" Tags="&lt;calculus&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="24195" PostTypeId="2" ParentId="24164" CreationDate="2011-02-28T07:30:36.537" Score="1" ViewCount="" Body="&lt;p&gt;My course handout at &lt;a href=&quot;http://www.math.uconn.edu/~kconrad/blurbs/ugradnumthy/universaldivtest.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.math.uconn.edu/~kconrad/blurbs/ugradnumthy/universaldivtest.pdf&lt;/a&gt; discusses a general method to derive such divisibility tests (by 7, 13, etc.).&lt;/p&gt;&#xA;" OwnerUserId="619" LastActivityDate="2011-02-28T07:30:36.537" />
  <row Id="24196" PostTypeId="1" CreationDate="2011-02-28T07:32:26.630" Score="5" ViewCount="201" Body="&lt;p&gt;Of course there are many ways to prove this.  However, I came across the following exercise (Ch. 0 #3).&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Prove that:&#xA;  (a) a regular surface&#xA;  $S\subset \mathbb{R}^3$ is an&#xA;  orientable manifold if and only if&#xA;  there exists a differentiable mapping&#xA;  of $N:S\rightarrow \mathbb{R}^3$ with&#xA;  $N(p)\perp T_p(S)$ and $|N(p)|=1$, for&#xA;  all $p\in S$. (b) the Mobius band&#xA;  (Example 4.9 (b)) is non-orientable.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In Example 4.9 (b), he constructs the Mobius band as the quotient by the antipodal map of the cylinder $C=\{(x,y,z)\in \mathbb{R}^3:x^2+y^2=1,|z|&amp;lt;1\}$.  The problem, of course, is that this isn't given as a surface in $\mathbb{R}^3$!  I was thinking for a second that maybe I should try and construct a map $C\rightarrow S^2$ with the right properties and check that it doesn't descend to a map on $M$, but that's stupid because if I &lt;i&gt;were&lt;/i&gt; to embed $M\subset \mathbb{R}^3$, I'm pretty sure it couldn't possibly have those tangent planes anyways.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anyone have any insight?  Presumably the solution to (b) should use the fact given in (a).&lt;/p&gt;&#xA;" OwnerUserId="2159" LastActivityDate="2011-02-28T11:08:11.170" Title="Exercise in Do Carmo's &quot;Riemannian Geometry&quot;: the Mobius band is nonorientable." Tags="&lt;differential-geometry&gt;" AnswerCount="1" CommentCount="6" />
  <row Id="24197" PostTypeId="2" ParentId="24188" CreationDate="2011-02-28T07:40:25.743" Score="1" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\rm\ \ 2^{340}\equiv 2\ \ (mod\ 11\cdot31)\ \Rightarrow\ mod\ 11\::\ 2^{339}\equiv 1\equiv 2^{10}\ \Rightarrow\ 2^{gcd(339,10)} \equiv 1\:,\:$ i.e. $\ 2\equiv 1\ \Rightarrow\Leftarrow$&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2011-02-28T07:40:25.743" />
  <row Id="24198" PostTypeId="2" ParentId="23399" CreationDate="2011-02-28T07:40:27.277" Score="0" ViewCount="" Body="&lt;p&gt;@Albert: The recurrence $M(n+1)/M(n)=n-1/2+o(1/n)$ is related to Kendall-Mann property&#xA;&lt;a href=&quot;http://oeis.org/A181609&quot; rel=&quot;nofollow&quot;&gt;http://oeis.org/A181609&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could you look at the answer from Moron please&#xA;&lt;a href=&quot;http://math.stackexchange.com/questions/23589/recurrence-representations-an1-ann-1-2o1-n-and-an1-ann-1-2&quot;&gt;Recurrence representation(s): $a(n+1)=a(n)(n-1/2)+o(1/n)$ and $a(n+1)=a(n)(n-1/2+o(1/n))$&lt;/a&gt;&#xA;It seems to me that your game is a good interpretation, am I right?&lt;/p&gt;&#xA;" OwnerUserId="7592" LastEditorUserId="7592" LastEditorDisplayName="" LastEditDate="2011-02-28T14:29:36.440" LastActivityDate="2011-02-28T14:29:36.440" CommentCount="4" />
  <row Id="24199" PostTypeId="1" CreationDate="2011-02-28T08:07:32.900" Score="1" ViewCount="105" Body="&lt;p&gt;N urns are assigned m balls in a stochastic process based on a Pareto distribution. The process is as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;X is a Pareto random variable (xminimum = 1, alpha is a parameter)&#xA;if X &gt; N, throw the ball out&#xA;Otherwise, put the ball in urn number floor(X)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do this until m balls have been put in urns.  Then rank the urns by number of balls.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What should the final distribution look like, as N and m approach infinity? Does it resemble any standard prob. distribution? Doing Monte Carlo simulations, it seems to be heavy tailed, but not a Pareto distribution.  But I'm new to this - just getting my feet wet with stochastic processes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: Primary question is: What is prob. mass. function for balls in k-th urn (after ordering them).  But I'm open to answers along the lines of: Look at it differently.  Basic goal is - given a selection of choices, with balls assigned to them, but some choices a priori more popular, what is relationship between stochastic process and pmf of results?&#xA;(And, yes, if you can't tell, I'm coming from an &lt;em&gt;applied&lt;/em&gt; perspective... this question is a simplification of a model encountered assigning awards to favorites).&lt;/p&gt;&#xA;" OwnerUserId="7593" LastEditorUserId="7593" LastEditorDisplayName="" LastEditDate="2011-02-28T08:27:13.343" LastActivityDate="2011-03-01T13:54:32.423" Title="Stochastic Urn Process using a Pareto distribution" Tags="&lt;probability&gt;&lt;statistics&gt;&lt;stochastic-processes&gt;&lt;probability-distributions&gt;&lt;simulation&gt;" AnswerCount="2" CommentCount="5" />
  <row Id="24200" PostTypeId="1" CreationDate="2011-02-28T08:07:36.063" Score="2" ViewCount="84" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Possible Duplicate:&lt;/strong&gt;&lt;br&gt;&#xA;  &lt;a href=&quot;http://math.stackexchange.com/questions/3283/why-do-i-always-get-1-when-i-keep-hitting-the-square-root-button-on-my-calculato&quot;&gt;&amp;ldquo;Why do I always get 1 when I keep hitting the square root button on my calculator?&amp;rdquo;&lt;/a&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;p&gt;Take any positive number above or below 1. Upon performing repeated square root operations the end result is always 1. Is there a theorem in Math that can prove this rather intriguing result? Thank you.&#xA;Klim&lt;/p&gt;&#xA;" OwnerUserId="7594" LastActivityDate="2011-02-28T09:26:01.350" ClosedDate="2011-02-28T12:53:02.863" Title="The theory of the result obtained from repeated square root operation of numbers" Tags="&lt;number-theory&gt;" AnswerCount="1" CommentCount="0" />
  <row Id="24201" PostTypeId="1" CreationDate="2011-02-28T08:08:10.843" Score="3" ViewCount="99" Body="&lt;p&gt;Let $k$ be a field and $R$ be the exterior ring over $k^d $, that is, $k$-algebra generated by elements  $$x_1,\ldots,x_d,$$&#xA;where $$\ x_ix_j= - x_jx_i?$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Is $R$ Artinian?&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="7517" LastEditorUserId="367" LastEditorDisplayName="" LastEditDate="2011-02-28T09:07:20.103" LastActivityDate="2011-02-28T15:12:56.597" Title="Artinian Ring Question" Tags="&lt;homework&gt;&lt;algebra&gt;&lt;ring-theory&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="24202" PostTypeId="2" ParentId="24200" CreationDate="2011-02-28T08:12:53.890" Score="6" ViewCount="" Body="&lt;p&gt;You have the sequence $x_{n+1} = \sqrt{x_n}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $0 \lt x_1 \lt 1$, we have that $x_{n+1} \gt x_n$ and $x_n \lt 1$. So this is a monotonically increasing sequence bounded above, and so has a limit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The limit $L  = \lim_{n \to \infty} x_n$ satisfies $0 \lt x_1 \le L \le 1$ and $L = \sqrt{L}$, thus $L = 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $x_1 \gt 1$, you get a monotonically decreasing sequence which is bounded below and you get $\lim_{n \to \infty} x_n = 1$, similarly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that, if you don't start with $1$ (or $0$), in a finite number of steps, you will get very close to $1$, but don't really attain $1$. The result of $1$ you see is likely due to the precision limitations of your calculator.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2011-02-28T08:12:53.890" />
  <row Id="24203" PostTypeId="1" CreationDate="2011-02-28T08:34:58.987" Score="1" ViewCount="104" Body="&lt;p&gt;I want some hints in how to solve these questions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I separate two different point by polynomial in the coordinate ring $\ k[X] $?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many thanks&lt;/p&gt;&#xA;" OwnerUserId="7517" LastEditorUserId="7517" LastEditorDisplayName="" LastEditDate="2011-03-05T00:24:09.347" LastActivityDate="2011-03-05T00:24:09.347" Title="A question .. need help" Tags="&lt;homework&gt;&lt;algebra&gt;&lt;ring-theory&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="24204" PostTypeId="1" CreationDate="2011-02-28T08:39:58.377" Score="3" ViewCount="140" Body="&lt;p&gt;The Pareto distribution is used to say, given a particular person X, what is the pdf of his wealth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to explore the reciprocal question: Given the total amount of wealth in a population, what portion does a random person have.  I conjecture that this is simply a constant times the Pareto distribution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More interestingly: What is the shape of the distribution curve, if the richest person would be at the 0 point on the x axis, the next richest person to the right, and so on - we would see a monotonically decreasing curve.  But what is its shape? What is its derivative?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's quite likely that I'm not phrasing that question properly.  Let me ask a more basic question: What is the appropriate terminilogy to explore the question? Give a probability distribution applied many times over, what is the shape of the resultant allocation curve?&lt;/p&gt;&#xA;" OwnerUserId="7593" LastActivityDate="2011-03-01T05:22:29.843" Title="Understanding the Pareto distribution as applied to wealth" Tags="&lt;probability&gt;&lt;statistics&gt;&lt;probability-distributions&gt;&lt;mathematical-modeling&gt;" AnswerCount="1" />
  <row Id="24205" PostTypeId="1" CreationDate="2011-02-28T08:44:18.173" Score="1" ViewCount="36" Body="&lt;p&gt;What is the difference between additive and multiplicative kalman filters, as well as some other kinds?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm also looking for reference texts and articles that describe the algorithms, so recommendations for those as well would be preferable.&lt;/p&gt;&#xA;" OwnerUserId="6955" LastActivityDate="2011-02-28T08:44:18.173" Title="What is the difference between various kalman filters?" Tags="&lt;stochastic-processes&gt;&lt;signal-processing&gt;&lt;control-theory&gt;" FavoriteCount="1" />
  <row Id="24206" PostTypeId="2" ParentId="24203" CreationDate="2011-02-28T08:48:01.993" Score="2" ViewCount="" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;This question needs clarification. What is $N(\ a)$?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;Hint&lt;/em&gt;: look at coordinate projections.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Edit&lt;/em&gt;: Concerning 1: If $b$ and $c$ are integers, then $b^2+15c^2$ is non-negative, so it can never be -2 or -5. If $c&amp;gt;0$ then $b^2+15c^2\geq 15$. If $c=0$ then $b^2+15c^2=b^2$ is a square number and thus cannot be 2 or 5 either.&lt;/p&gt;&#xA;" OwnerUserId="367" LastEditorUserId="367" LastEditorDisplayName="" LastEditDate="2011-02-28T10:55:07.593" LastActivityDate="2011-02-28T10:55:07.593" CommentCount="2" />
  <row Id="24207" PostTypeId="2" ParentId="23703" CreationDate="2011-02-28T08:49:59.557" Score="1" ViewCount="" Body="&lt;p&gt;I guess I found a better &lt;a href=&quot;ftp://www.comm-tec.com/oops/ismar/Benthos_chirp2_win2000_install/DSPDriver/web/pdfs/tech/fft-bench.pdf&quot; rel=&quot;nofollow&quot;&gt;link&lt;/a&gt;.&#xA;Thank you for your help.&lt;/p&gt;&#xA;" OwnerUserId="7501" LastActivityDate="2011-02-28T08:49:59.557" />
  <row Id="24208" PostTypeId="2" ParentId="24184" CreationDate="2011-02-28T08:51:40.393" Score="1" ViewCount="" Body="&lt;p&gt;A different Hint:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider a square of side $ab$, with one corner at $(0,0)$ and the opposite corner at $(ab, ab)$. Draw the vertical lines $x = b$, $x=2b$ etc and the horizontal lines $y = a$, $y = 2a$ etc.  Now try to count in two different ways, the number of intersection points of these lines, which lie on or below the diagonal of the square.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2011-02-28T08:51:40.393" />
  <row Id="24209" PostTypeId="1" AcceptedAnswerId="24212" CreationDate="2011-02-28T09:02:17.077" Score="1" ViewCount="100" Body="&lt;p&gt;I'm giving a presentation about orientable surfaces (as a student project) and I was wondering what I should talk about. The presentation should be 20-30 min long. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've been thinking maybe something like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) definition of a surface&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) what does orientable mean (orientable vs. oriented)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And then I'm a bit lost. In the general topology course last year, classification of compact surfaces and covering maps were mentioned but I don't much about either and I don't know if it has anything to do with orientability of surfaces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this class, algebraic topology, we're going to learn about cell complexes and homology, but I don't think that has anything to do with orientability either.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Question: what is an interesting fact or theorem that I should definitely mention in this presentation?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many thanks for your help!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I did &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) definition of surface and manifold&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) definition of orientable (with paper moebius strip)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) classification of compact orientable surfaces&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just in case anyone reads this post later, they might use this to get inspiration.&#xA;I had also prepared material about homology but the professor said homology was going to be discussed towards the very end of the lecture so I did not present what I had prepared.&lt;/p&gt;&#xA;" OwnerUserId="5798" LastEditorUserId="5798" LastEditorDisplayName="" LastEditDate="2011-03-04T10:10:20.540" LastActivityDate="2011-03-04T10:10:20.540" Title="Presentation about orientable surfaces" Tags="&lt;algebraic-topology&gt;" AnswerCount="1" CommentCount="6" />
  <row Id="24210" PostTypeId="2" ParentId="24054" CreationDate="2011-02-28T09:10:04.383" Score="0" ViewCount="" Body="&lt;p&gt;If you want to do a translation in polar coordinates by $(r_0,\theta_0)$, the following formulas hold&#xA;$$r&amp;#39; = \sqrt{r^2 + r_0^2 -2 r r_0 \cos(\theta - \theta_0)}$$&#xA;$$\theta&amp;#39; = \arctan([ r \sin(\theta) -r_0 \sin(\theta_0)]/[r \cos(\theta) -r_0 \cos(\theta_0)])$$&#xA;where $&amp;#39;$ denotes the new system. Note that there is a problem with the ambiguity of the $\arctan$ function. For this reason often &lt;a href=&quot;http://en.wikipedia.org/wiki/Atan2&quot; rel=&quot;nofollow&quot;&gt;arctan2&lt;/a&gt; is implemented (or you can implement it yourself). Having arctan2 the second equation reads&#xA;$$\theta&amp;#39; = \text{arctan2}( r \sin(\theta) -r_0 \sin(\theta_0),r \cos(\theta) -r_0 \cos(\theta_0)).$$&lt;/p&gt;&#xA;" OwnerUserId="7266" LastActivityDate="2011-02-28T09:10:04.383" />
  <row Id="24211" PostTypeId="1" CreationDate="2011-02-28T09:12:50.833" Score="1" ViewCount="88" Body="&lt;p&gt;I am learning some Cryptography and I came across this exercise where I have to make the following proof (translated from German, so I hope it is accurate).&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Proof the following assertion:&#xA;  Let $n \in \mathbb{Z}$ be odd and $a \in \mathbb{Z}_n^*$. If $a^{\frac{n-1}{2}} \neq \pm 1 \mod n$ , then $n$ is a composite. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now I don't really have any math background, so my proof goes as follows, but I am not sure if I really proofed the assertion:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Proof by contradiction. Assume that $n$ is a composite.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$$a^{\frac{n-1}{2}} \equiv 1 \mod n$$ &#xA;  $$\Rightarrow a^{\frac{n-1}{2}}\cdot a^{\frac{n-1}{2}} \equiv 1 \cdot 1 \mod n$$&#xA;  $$\Rightarrow a^{n-1} \equiv 1 \mod n$$&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Now since $a^{\varphi(n)} \equiv 1 \mod n$ for any finite group and with $\varphi$ being the Euler's totient function. But $\varphi(n) = n - 1$ if and only if $n$ is prime. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And I could do the same thing $a^{\frac{n-1}{2}} \equiv -1 \mod n$&lt;/p&gt;&#xA;" OwnerUserId="517" LastActivityDate="2011-02-28T11:07:06.880" Title="Proof of Primality Testing" Tags="&lt;prime-numbers&gt;&lt;cryptography&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="24212" PostTypeId="2" ParentId="24209" CreationDate="2011-02-28T09:22:43.600" Score="2" ViewCount="" Body="&lt;p&gt;Homology certainly has something to do with orientability, but if you are not yet familiar, I wouldn't recommend diving into the subject for this. You should definitely speak about non-orientable surfaces (e.g. the Klein bottle); this will really help understanding the idea of orientability. Cutting up the mobius band in front of class is also very insightful.&lt;/p&gt;&#xA;" OwnerUserId="5882" LastActivityDate="2011-02-28T09:22:43.600" CommentCount="5" />
  <row Id="24213" PostTypeId="1" CreationDate="2011-02-28T09:23:33.540" Score="0" ViewCount="28" Body="&lt;ol&gt;&#xA;&lt;li&gt;need a theorem that proves that a locally convex topological space is compact (infinite dimensional) &lt;/li&gt;&#xA;&lt;li&gt;need a theorem on fixed point that will establish existence of a fixed point for the above space&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="7595" LastActivityDate="2011-02-28T09:23:33.540" Title="A solution for non lipschitz quantum stochastic differential equation" Tags="&lt;functional-analysis&gt;" CommentCount="1" />
  <row Id="24215" PostTypeId="1" CreationDate="2011-02-28T10:13:08.407" Score="1" ViewCount="38" Body="&lt;p&gt;I have the following (directed) graph:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A --&gt; B&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the transition is labled C. Is there symbol to denote the inverse of the (A,B) transtion?&lt;/p&gt;&#xA;" OwnerUserId="4166" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2011-02-28T14:06:07.913" LastActivityDate="2011-02-28T22:29:08.203" Title="Indicate reverse of graph transition" Tags="&lt;graph-theory&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="24216" PostTypeId="2" ParentId="24211" CreationDate="2011-02-28T10:16:22.977" Score="2" ViewCount="" Body="&lt;p&gt;Consider the contrapositive. Suppose p an odd prime and $a \in \mathbb{Z}_p^*$. Then $a^\frac{p-1}{2} \equiv \pm 1 \mod p$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint. $(a^\frac{p-1}{2})^2 \equiv 1 \mod p$ since p is prime, i.e. $(a^\frac{p-1}{2} - 1)(a^\frac{p-1}{2}+1)$ is divisible by p.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note on your proof: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The implication &quot;A implies B&quot; is false only when A is true and B is false, which we assume in a proof by contradiction, and not A is False and B is true. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $a^{n-1} \equiv 1 \mod n$ for all $a \in \mathbb{Z}_p^*$, then n need not be prime, consider the composite number n = 561, $a \in \mathbb{Z}_{561}^*$ then $a^{560} \equiv 1 \mod 561$. These numbers are called Carmichael.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The euler-phi $\phi(n)$ is not the order of a, i.e. it is not the smallest such that $a^{\phi(n)} \equiv 1 \mod n$. Indeed, if n = 7, then $\phi(7) = 6$ and $2^3 \equiv 1 \mod 7$.&lt;/p&gt;&#xA;" OwnerUserId="1746" LastEditorUserId="1746" LastEditorDisplayName="" LastEditDate="2011-02-28T11:07:06.880" LastActivityDate="2011-02-28T11:07:06.880" />
  <row Id="24217" PostTypeId="2" ParentId="24196" CreationDate="2011-02-28T11:01:05.247" Score="1" ViewCount="" Body="&lt;p&gt;What doCarmo might have had in mind: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I didn't check, but I think the quotient map &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \pi: C \to M $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is precisely the &lt;a href=&quot;http://en.wikipedia.org/wiki/Orientability#Orientable_double_cover&quot; rel=&quot;nofollow&quot;&gt;orientation covering&lt;/a&gt; of $M$. And one can prove that this covering is connected iff $M$ is not orientable (see Lee &quot;Intro to Smooth Manifolds&quot;, p. 331 for example).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the above is not useful: Maybe you could try to pull back the orientation of $M$ to $C$ and get a contradiction or so. (although this really has not much to do with part a) of the exercise, so it might be that doCarmo wants you to imbed the Möbius strip)&lt;/p&gt;&#xA;" OwnerUserId="3208" LastEditorUserId="3208" LastEditorDisplayName="" LastEditDate="2011-02-28T11:08:11.170" LastActivityDate="2011-02-28T11:08:11.170" CommentCount="1" />
  <row Id="24218" PostTypeId="1" AcceptedAnswerId="24240" CreationDate="2011-02-28T11:09:54.913" Score="2" ViewCount="65" Body="&lt;p&gt;Σ:&lt;a href=&quot;http://en.wikipedia.org/wiki/Signature_(logic)&quot; rel=&quot;nofollow&quot;&gt;signature&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;T:Σ-theory&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For M:Σ-model which satisfies T, is it proper to name it “Σ,T-model” or “Σ-model satisfying T”? In comparison, in the context of Lawvere theories, any &lt;a href=&quot;http://en.wikipedia.org/wiki/Lawvere_theory&quot; rel=&quot;nofollow&quot;&gt;Lawvere theory&lt;/a&gt; L incorporates a signature as well as axioms. If we construct L from Σ and T, any L-model obviously satisfies T. Therefore “Σ,T-model” seems as an appropriate name.&lt;/p&gt;&#xA;" OwnerUserId="7011" LastEditorUserId="7011" LastEditorDisplayName="" LastEditDate="2011-02-28T15:37:12.803" LastActivityDate="2011-02-28T15:37:12.803" Title="does the definition of model depend on a theory or just a signature?" Tags="&lt;logic&gt;&lt;definition&gt;&lt;model-theory&gt;" AnswerCount="1" />
  <row Id="24219" PostTypeId="2" ParentId="24194" CreationDate="2011-02-28T11:13:48.473" Score="1" ViewCount="" Body="&lt;p&gt;Consider the approximately rectangular vertical strip running from the curve $y=27e^{-x}$ to the curve $y=27e^x$ at a distance $x$ from the $y$-axis and of width $\delta x.$ The height of the strip is about $27(e^x-e^{-x})$ and so its area is approximately&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$27(e^x-e^{-x}) \delta x.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now consider rotating this about the $y$-axis one complete revolution. The rectangle will sweep out a volume of approximately&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ 2 \pi x \times 27(e^x-e^{-x}) \delta x.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You then need to sum up all these elements of volume from $x=0$ to $x=1$ in the limit as $\delta x \rightarrow 0.$ This gives you the required integral.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Be sure to draw a diagram to see what is going on.&lt;/p&gt;&#xA;" OwnerUserId="1301" LastActivityDate="2011-02-28T11:13:48.473" />
  <row Id="24220" PostTypeId="1" AcceptedAnswerId="24255" CreationDate="2011-02-28T11:51:39.580" Score="3" ViewCount="98" Body="&lt;p&gt;Thank you~&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Show that if $G$ is a finite nilpotent group, then every Sylow subgroup is normal in $G$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I know that the normalizer of any proper subgroup of a nilpotent group contains this subgroup properly. So I think maybe I can prove the normality of the Sylow subgroup of $G$, say $P$, by showing that $N(P)=N(N(P))$, thus showing that $N(P)=G$. (Here, $N(P)$ represents the normalizer of $P$ in $G$.) But I don't know how to complete this step.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll appreciate your help. Many thanks.&lt;/p&gt;&#xA;" OwnerUserId="4736" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-28T14:50:29.907" LastActivityDate="2011-02-28T16:12:30.243" Title="Show that if $G$ is a finite nilpotent group, then every Sylow subgroup is normal in $G$." Tags="&lt;group-theory&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" />
  <row Id="24221" PostTypeId="1" CreationDate="2011-02-28T11:52:54.257" Score="0" ViewCount="95" Body="&lt;p&gt;What software do you use and would recommend for solving recurrence relations?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example mathematica has &lt;code&gt;RSolve&lt;/code&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ a(n+1)-2 a(n) = 1$ for $a(n)$:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;RSolve[a[n + 1] - 2 a[n] == 1, a[n], n]&#xA;-&amp;gt; a(n) = c 2^(n-1)+2^n-1&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Are there any good (preferably open) alternatives? &lt;/p&gt;&#xA;" OwnerUserId="7476" LastActivityDate="2011-03-30T12:38:46.433" Title="Software for recurrence relations." Tags="&lt;math-software&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="24222" PostTypeId="2" ParentId="24123" CreationDate="2011-02-28T12:03:36.473" Score="0" ViewCount="" Body="&lt;p&gt;“Element-wise” means using the definitions of an intersection and a union and an equality of sets:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\forall x. x\in(A\cup B) \leftrightarrow x\in A\lor x\in B$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\forall x. x\in(A\cap B) \leftrightarrow x\in A\land x\in B$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(A=B) \leftrightarrow (\forall x.x\in A \leftrightarrow x\in B)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reason about $x$ which is “an element”. Then the deduction rules of intuitionistic logic give you the desired theorem.&lt;/p&gt;&#xA;" OwnerUserId="7011" LastEditorUserId="7011" LastEditorDisplayName="" LastEditDate="2011-02-28T12:08:38.107" LastActivityDate="2011-02-28T12:08:38.107" />
  <row Id="24223" PostTypeId="1" CreationDate="2011-02-28T12:03:51.710" Score="1" ViewCount="90" Body="&lt;p&gt;The answer is 2454, apparently. I am getting 1446, however. Can anyone point out my error ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We need to partition the solution into the mutually exclusive cases of choices of 4 letters where we have no repetitions (e.g. MATH), where we have one repetition (e,g, MATT) and where we have two repetitions (e.g. MMTT). Then we compute the corresponding number of perms of each case, and add, thus:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have 8 unique letters, so we can make 8C4 selections with no repetitions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have 3 pairs of repetitions {MM, AA, TT} which gives us 3 * 7C2 choices with one repetition, since having chosen, say, MM, we need to select 2 more letters from the remaining 7 unique letters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have 3C2 choices with 2 repetitions taken from the set {MM, AA, TT}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So our total number of perms is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(#choices of 4, no reps * 4!) + (#choices of 4, 1 rep * 4!/2!) + (#choices of 4, 2 reps * 4!/(2!2!) ) = (8C4 * 24) + (3 * 7C2 * 12) + (3C2 * 6) = 672 + 756 + 18 = 1446&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where did I go wrong ?&lt;/p&gt;&#xA;" OwnerUserId="7597" LastActivityDate="2011-02-28T12:21:35.337" Title="How many perms of 4 from letters of MATHEMATICS ?" Tags="&lt;combinatorics&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="24224" PostTypeId="2" ParentId="24221" CreationDate="2011-02-28T12:07:54.297" Score="1" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://www.wolframalpha.com/input/?i=a%28n%2B1%29-2a%28n%29%3D1&quot; rel=&quot;nofollow&quot;&gt;WolframAlpha&lt;/a&gt; does an okay job.&lt;/p&gt;&#xA;" OwnerUserId="1301" LastActivityDate="2011-02-28T12:07:54.297" CommentCount="1" />
  <row Id="24225" PostTypeId="2" ParentId="24199" CreationDate="2011-02-28T12:09:04.373" Score="2" ViewCount="" Body="&lt;p&gt;Ignoring the reranking (as jorki says, it seems to makes things too complicated), I think you can say something like:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The probability that the any particular ball which goes into an urn turns out to go into the $k$th urn [$1 \le k &amp;lt; N $] is $$P_k = \frac{1/k^{\alpha} - 1/(k+1)^{\alpha}}{1-1/N^{\alpha}}$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;With $m$ balls in total in the urns, the probability that there are $m_k$ in the $k$th urn is &#xA;$$\Pr (M_k=m_k) = \binom{m}{m_k} P_k^{m_k} (1-P_k)^{m-m_k}$$&#xA;i.e. a binomial distribution $B(m,P_k)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Taking the distribution over the urns together, rather than individual urns, this is a &lt;a href=&quot;http://en.wikipedia.org/wiki/Multinomial_distribution&quot; rel=&quot;nofollow&quot;&gt;multinomial distribution&lt;/a&gt;, with the parameters $m$ and $P_1, P_2, \ldots$&lt;/p&gt;&#xA;" OwnerUserId="6460" LastEditorUserId="6460" LastEditorDisplayName="" LastEditDate="2011-02-28T14:23:53.857" LastActivityDate="2011-02-28T14:23:53.857" CommentCount="5" />
  <row Id="24226" PostTypeId="2" ParentId="24223" CreationDate="2011-02-28T12:17:31.547" Score="2" ViewCount="" Body="&lt;p&gt;$^8C_4 * 24$ is $1680$, not $672$.&lt;/p&gt;&#xA;" OwnerUserId="1508" LastActivityDate="2011-02-28T12:17:31.547" CommentCount="3" />
  <row Id="24227" PostTypeId="2" ParentId="20605" CreationDate="2011-02-28T12:21:16.993" Score="0" ViewCount="" Body="&lt;p&gt;Hi Wistful,&#xA;I would recommend reading &lt;code&gt;http://www.unige.ch/~gander/consulting/X/EigenUpdate.ps.gz&lt;/code&gt; and having a look at the cited work of &lt;strong&gt;Golub and Van Loan&lt;/strong&gt;.&#xA;They show howto update matrices with rank-one-changes. You can understand you update matrix $D \in\mathbb{R}^{n\times n}$ as a sum of $n$ rank-one-updates.&#xA;Good luck!&lt;/p&gt;&#xA;" OwnerUserId="7596" LastActivityDate="2011-02-28T12:21:16.993" />
  <row Id="24229" PostTypeId="2" ParentId="24220" CreationDate="2011-02-28T13:04:28.550" Score="2" ViewCount="" Body="&lt;p&gt;A fairly simple way to do this is to note that a normal Sylow-subgroup is always characteristic. Thus, since you can make a subnormal chain from the Sylow-subgroup to the group itself, this Sylow-subgroup must in fact be normal in each subgroup in the chain and thus in the entire group.&lt;/p&gt;&#xA;" OwnerUserId="2538" LastActivityDate="2011-02-28T13:04:28.550" CommentCount="1" />
  <row Id="24230" PostTypeId="1" AcceptedAnswerId="24231" CreationDate="2011-02-28T14:04:55.397" Score="1" ViewCount="119" Body="&lt;p&gt;$f$ is continuous between $[0,1]$, and $f(0)=f(1)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to prove that there is an $a \in [0,0.5]$ such that $f(a+0.5)=f(a)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ok, so Rolle's theorem can be useful here, but I can't see the connection to the derivative,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Weierstrass, Uniform continuity?) I'll be glad to instructions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="6163" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2011-03-01T02:43:47.417" LastActivityDate="2011-03-01T02:43:47.417" Title="For a continuous function, $f(0) = f(1) \Longrightarrow f(a) = f(a + 1/2)$." Tags="&lt;calculus&gt;" AnswerCount="3" CommentCount="4" />
  <row Id="24231" PostTypeId="2" ParentId="24230" CreationDate="2011-02-28T14:11:18.750" Score="4" ViewCount="" Body="&lt;p&gt;Consider the function $g(a) = f(a+0.5) - f(a)$ on the interval $[0,0.5]$, and use the intermediate value theorem.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2011-02-28T15:09:51.060" LastActivityDate="2011-02-28T15:09:51.060" CommentCount="2" />
  <row Id="24232" PostTypeId="2" ParentId="24230" CreationDate="2011-02-28T14:11:43.093" Score="3" ViewCount="" Body="&lt;p&gt;HINT: Work with the function $g(a)=f(a+0.5)-f(a)$. Consider $g(0)$ and $g(0.5)$ (and their sum).&lt;/p&gt;&#xA;" OwnerUserId="2810" LastActivityDate="2011-02-28T14:11:43.093" CommentCount="1" />
  <row Id="24233" PostTypeId="2" ParentId="24162" CreationDate="2011-02-28T14:14:55.307" Score="3" ViewCount="" Body="&lt;p&gt;Not all cyclic groups are created equal; some groups are used (currently) for cryptographic applications, while others are not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Groups of cryptographic interest&quot; refers to groups that are normally used in actual applications of cryptography that relate to the discrete log problem; these are Diffie-Hellman key exchange, ElGamal, and the like. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Usually, the groups that are used for these problems are the multiplicative group of integers modulo a very large prime $p$; the multiplicative group of a finite field (particularly of finite fields of characteristic $2$, because they tend to be easy to implement); or points of curves of elliptic curves (over finite or global fields).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some groups have very easy discrete logarithm problem (the &lt;strong&gt;additive&lt;/strong&gt; group of integers modulo $n$, for example) so they are not used in actual applications of cryptography; other groups are too hard to implement, so they are not used either. The discrete logarithm problem for these groups is irrelevant for cryptography, since they are not used for cryptography. So these groups are &lt;em&gt;not&lt;/em&gt; of &quot;cryptographic interest&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that being of &quot;cryptographic interest&quot; is both time-dependent (it depends on what is being used &lt;em&gt;now&lt;/em&gt;), and more importantly as noted by Qioachu, it is not invariant under isomorphism. The multiplicative group of a finite field of order $p^k$ is (abstractly) isomorphic to the additive group of integers modulo $n=p^k-1$; but while the discrete logarithm problem for the former is considered &quot;hard&quot;, the discrete logarithm problem for the latter is &quot;easy&quot;. The problem here is that finding an isomorphism &lt;em&gt;is&lt;/em&gt; essentially equivalent to constructing a full logarithm table for the multiplicative group of the finite field, so having an isomorphism is pretty much the same as solving the discrete logarithm problem.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2011-02-28T14:14:55.307" CommentCount="1" />
  <row Id="24234" PostTypeId="1" AcceptedAnswerId="24274" CreationDate="2011-02-28T14:17:00.327" Score="3" ViewCount="173" Body="&lt;p&gt;In my project I'm using the value of public exponent of 4451h. I thought it's safe and ok until I started to use one commercial RSA encryption library. If I use this exponent with this library, it throws exception.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I contacted developpers of this library and got the following reply: &quot;This feature is to prevent some attacks on RSA keys. The consequence is that the exponent value is limited to {3, 5, 17, 257 or 65537}. Deactivating this check is still being investigated, as the risks may be great.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's the first time in my life I hear that values other than {3, 5, 17, 257 or 65537} are used to break RSA. I knew only of using 3 with improper padding being vulnerable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is that really so? Surely, I can use another library, but after such answer I worried about security of my solution.&lt;/p&gt;&#xA;" OwnerUserId="7599" LastEditorUserId="7599" LastEditorDisplayName="" LastEditDate="2011-02-28T14:26:59.210" LastActivityDate="2011-02-28T18:14:07.833" Title="Should RSA public exponent be only in {3, 5, 17, 257 or 65537} due to security considerations?" Tags="&lt;cryptography&gt;" AnswerCount="3" CommentCount="7" FavoriteCount="1" />
  <row Id="24235" PostTypeId="2" ParentId="24230" CreationDate="2011-02-28T14:25:20.227" Score="1" ViewCount="" Body="&lt;p&gt;&lt;em&gt;Or&lt;/em&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;consider if there is a $b\in [0,1]$ such that $f(b) = f(0) = f(1)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What if there is no such $b$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What if $b = 0.5$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[ You don't need Rolle's theorem this way ]&lt;/p&gt;&#xA;" OwnerUserId="7476" LastActivityDate="2011-02-28T14:25:20.227" />
  <row Id="24236" PostTypeId="1" AcceptedAnswerId="24239" CreationDate="2011-02-28T14:31:44.523" Score="1" ViewCount="129" Body="&lt;p&gt;the fucnction y(x) implement  the Equation y''-xy=0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In addition, know that y(0)=0 and y'(o)=1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;find the value of y(0)^(n), which means the value of 0 in the nth derivative.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="6163" LastEditorUserId="6163" LastEditorDisplayName="" LastEditDate="2011-02-28T17:05:38.053" LastActivityDate="2011-02-28T17:05:38.053" Title="Another  question in calculus" Tags="&lt;calculus&gt;" AnswerCount="1" CommentCount="8" />
  <row Id="24237" PostTypeId="1" AcceptedAnswerId="24405" CreationDate="2011-02-28T14:32:10.713" Score="7" ViewCount="128" Body="&lt;p&gt;&lt;strong&gt;Update&lt;/strong&gt;: it is not possible to reply to this question without additional information. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My comment below: &quot;I have to agree with you that one &quot;cannot derive (2) from (1) alone&quot;. Now it seems to me that one must consider how the sequences $v_{h,n}^{\prime }$ and $v_{h,n}$ are constructed. That is described in the first part of the paper, but unfortunately it is not easy for me to summarize it. As I understand Apéry transformed repeatidely a continued fraction whose approximants are $\dfrac{u_{h,n}^{\prime }}{u_{h,n}}$, iterating on $h$. The above sequences are $v_{h,n}^{\prime }=\dfrac{u_{h,n}^{\prime }}{h!n!}, v_{h,n}=\dfrac{u_{h,n}}{h!n!}$.&quot;  &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;In &lt;em&gt;Irrationalité de Certaines Constantes&lt;/em&gt;, Bull. section des sciences du C.T.H.S., n.º3, p.37-53, Roger Apéry derives rational approximations $\dfrac{v_{h,n}^{\prime }}{v_{h,n}}$ for $\ln (1+t)$, &#xA;$\zeta (2)$ and $\zeta (3)$, the simplest being the one for the $\ln $. The&#xA;sequences $v_{h,n}^{\prime }$ and $v_{h,n}$, whose ratio converges&#xA;to $\ln (2)$, satisfy the recursive relation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(n+1)v_{h,n+1}-(2h+1)v_{h,n}-nv_{h,n-1}=0.\qquad (1)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The diagonal sequences $w_{n}^{\prime }=v_{n,n}^{\prime },w_{n}=v_{n,n}$&#xA;satisfy&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(n+1)w_{n+1}-3\left( 2n+1\right) w_{n}-nw_{n-1}=0.\qquad (2)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remarks: &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The initial conditions for $v_{h,n}^{\prime }$, $v_{h,n}$, $w_{n}^{\prime&#xD;&#xA;}$, $w_{n}$ are not indicated in the paper. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Recurrences $(1)$ and $(2)$ are the particular case for $t=1$ of, respectively,&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;$$(n+1)v_{h,n+1}-\left( \left( n+1\right) -nt+h\left( 1+t\right) \right)&#xD;&#xA;v_{h,n}-ntv_{h,n-1}=0\qquad (\ast)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(to simplify the notation the index $h$ was deleted in the original) and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(n+1)w_{n+1}-\left( 2n+1\right) \left( 2+t\right) w_{n}-nt^{2}w_{n-1}=0.\qquad (\ast\ast)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question&lt;/strong&gt;: How do you derive $(2)$ from $(1)$?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Copy of the mentioned paper&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/TeBGq.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="752" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2011-06-03T13:48:21.093" LastActivityDate="2011-06-03T13:48:21.093" Title="Diagonal  of the double sequence $(n+1)v_{h,n+1}-(2h+1)v_{h,n}-nv_{h,n-1}=0$" Tags="&lt;number-theory&gt;&lt;sequences-and-series&gt;&lt;recurrence-relations&gt;" AnswerCount="1" />
  <row Id="24238" PostTypeId="2" ParentId="24201" CreationDate="2011-02-28T14:32:21.603" Score="2" ViewCount="" Body="&lt;p&gt;To expand Pete's ḧint:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The key observation is that your $R$ is a &lt;em&gt;finite dimensional&lt;/em&gt; $k$-vector space, and each of its left ideals is a vector subspace. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now consider a decreasing chain of left ideals in $R$... It is in particular a decreasing chain of subspaces in a finite dimensional vector space: can it &lt;em&gt;not&lt;/em&gt; stop?&lt;/p&gt;&#xA;" OwnerUserId="274" LastEditorUserId="274" LastEditorDisplayName="" LastEditDate="2011-02-28T14:52:33.270" LastActivityDate="2011-02-28T14:52:33.270" />
  <row Id="24239" PostTypeId="2" ParentId="24236" CreationDate="2011-02-28T14:36:53.517" Score="4" ViewCount="" Body="&lt;p&gt;Add xy to both sides (To get $y'' = xy$), and start taking derivatives.  A pattern emerges.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;: You should get:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$y^{(3)} = y + xy&amp;#39;$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$y^{(4)} = 2y&amp;#39; + xy&amp;#39;&amp;#39;$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$y^{(5)} = 3y&amp;#39;&amp;#39; + xy&amp;#39;&amp;#39;&amp;#39;$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$y^{(6)} = 4y^{(3)} + xy^{(4)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$y^{(7)} = 5y^{(4)} + xy^{(5)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and so on...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, ask yourself what happens when you start evaluating each of these terms at $0$.&lt;/p&gt;&#xA;" OwnerUserId="6491" LastEditorUserId="6491" LastEditorDisplayName="" LastEditDate="2011-02-28T14:52:30.430" LastActivityDate="2011-02-28T14:52:30.430" CommentCount="4" />
  <row Id="24240" PostTypeId="2" ParentId="24218" CreationDate="2011-02-28T15:03:39.393" Score="4" ViewCount="" Body="&lt;p&gt;The usual terminology in model theory is that for a given language or signature $\Sigma$, one may speak of a &lt;em&gt;structure&lt;/em&gt; in the signature $\Sigma$, or a &lt;em&gt;$\Sigma$-structure&lt;/em&gt;, and this places no requirements on the theory of the model. When one has a theory $T$ in the language of $\Sigma$, one may have a &lt;em&gt;model&lt;/em&gt; of $T$, which is a structure in that signature satisfying that theory. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One ambiguity, as you notice, is that if a theory is regarded as a set of sentences, then one cannot fully tell the language from the theory, since perhaps some parts of the language were not used in the theory. This ambiguity is not actually ambiguous in practice if the signature $\Sigma$ is clear from context, and when it is not, then you are right that one must say what the language is, in the cases that this is a difference that matters. So in practice, this amounts essentially to the same as the Lawvere practice. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;An interesting example where the distinction is important is that it is possible for a theory $T$ to be decidable in a small language, but not in a larger language. For example, Presburger arithmetic is decidable in the language with just $+$, but not in the language having both plus and times, even when no additional axioms are added concerning times. Indeed, almost no theory will be decidable in a language that includes extra unmentioned binary or higher arity relation or function symbols, since the empty theory in the language with such relation or functions symbols is undecidable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In practice, researchers often use the term &lt;em&gt;model&lt;/em&gt; when no theory has been specified, and in such cases, they usually mean &lt;em&gt;structure&lt;/em&gt;, unless a particular theory and signature is implicit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, regarding your proposal for the term &quot;$\Sigma,T$-model&quot;, let me say that my own taste tends towards natural language formulations, where we could simply speak of a model of $T$ in the language $\Sigma$. &lt;/p&gt;&#xA;" OwnerUserId="413" LastEditorUserId="413" LastEditorDisplayName="" LastEditDate="2011-02-28T15:14:11.510" LastActivityDate="2011-02-28T15:14:11.510" CommentCount="2" />
  <row Id="24241" PostTypeId="1" CreationDate="2011-02-28T15:09:57.143" Score="39" ViewCount="1756" Body="&lt;p&gt;I have much more experience programming than I do with advanced mathematics, so perhaps this is just a comfort thing with me, but I often get frustrated trying to follow mathematical notation. Specifically, I get frustrated trying to keep track of what each variable signifies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a programmer, this would be completely unacceptable no matter how many comments you added explaining it:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;float A(float P, float r, float n, float t) {&#xA;  return P * pow(1 + r / n, n * t);&#xA;}&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Yet a mathematician would have no problem with this:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$A = P\ \Bigl(1+\frac{r}{n}\Bigr)^{nt}$&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Where,&lt;br&gt;&#xA;  A = final amount&lt;br&gt;&#xA;  P = principal amount (initial investment)&lt;br&gt;&#xA;  r = annual nominal interest rate (as a decimal)&lt;br&gt;&#xA;  n = number of times the interest is compounded per year&lt;br&gt;&#xA;  t = number of years&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So why don't I ever see the following?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$final\_amount = principal\ \Bigl(1+\frac{interest\_rate}{periods\_per\_yr}\Bigr)^{periods\_per\_yr\ \cdot\ years}$&lt;/p&gt;&#xA;" OwnerUserId="7600" LastActivityDate="2011-04-01T20:45:03.597" Title="Why do mathematicians use single-letter variables?" Tags="&lt;notation&gt;&lt;math-history&gt;" AnswerCount="12" CommentCount="8" FavoriteCount="10" />
  <row Id="24242" PostTypeId="2" ParentId="24241" CreationDate="2011-02-28T15:11:45.897" Score="26" ViewCount="" Body="&lt;p&gt;We are very, very lazy. I am very, very serious about this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;NB1:&lt;/strong&gt; The history is told in Florian Cajori's book on the history of notation. In very old times, there were no variables (and no formulas, really) and everything was incredibly verbose. Cajori's book beautifully shows the very long and tortuous way from that to modern day notation for variables; there are several sections regarding the notation of unknowns and of their powers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;NB2:&lt;/strong&gt; Additionally, we usually deal with very complicated expressions, so using verbose names for variables you render things almost impossible. Writing down the formula for Gaussian curvature in terms of $E$, $F$, $G$ and the Christoffel symbols if we wrote $\mathsf{Chistoffel}^i_jk$ instead of $\Gamma^{i}_{jk}$ would turn differential geometry into a dead subject very soon :P&lt;/p&gt;&#xA;" OwnerUserId="274" LastEditorUserId="274" LastEditorDisplayName="" LastEditDate="2011-02-28T15:42:44.060" LastActivityDate="2011-02-28T15:42:44.060" CommentCount="3" />
  <row Id="24243" PostTypeId="2" ParentId="24201" CreationDate="2011-02-28T15:12:56.597" Score="0" ViewCount="" Body="&lt;p&gt;Actually, I am not sure the question is true as stated if $1 = -1$ in $k$!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $2$ is invertible, then you get $x_i^2=0$ for all $i$. As suggested in the comments and Mariano's answer, the key point is $R$ has finite dimension as a $k$-vector space. Since the monomials span $R$, you want a lot of them to vanish. For example: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x_1x_2x_1 = -x_1x_1x_2= -x_1^2x_2 = 0$$&lt;/p&gt;&#xA;" OwnerUserId="223" LastActivityDate="2011-02-28T15:12:56.597" CommentCount="2" />
  <row Id="24244" PostTypeId="2" ParentId="24234" CreationDate="2011-02-28T15:14:48.710" Score="5" ViewCount="" Body="&lt;p&gt;Actually, a small exponent like $3$ could spell trouble, at least if you're doing your best to do things wrong (say if the message, as a number, is &quot;small&quot;) - something that can never be ruled out when cryptography is involved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason that these exponents are used is that they're fast to compute. They can be computed by repeated squaring plus one multiplication (we need the exponent to be odd).&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2011-02-28T15:14:48.710" />
  <row Id="24245" PostTypeId="2" ParentId="24241" CreationDate="2011-02-28T15:15:59.963" Score="12" ViewCount="" Body="&lt;p&gt;I can't imagine writing full words when solving equation and rewriting it 10 times, when programming you usually don't write it so much, you can use functions, classes, copy/paste, programs which automatically insert variables etc.&lt;/p&gt;&#xA;" OwnerUserId="7149" LastActivityDate="2011-02-28T15:15:59.963" CommentCount="1" />
  <row Id="24246" PostTypeId="2" ParentId="24241" CreationDate="2011-02-28T15:21:51.527" Score="15" ViewCount="" Body="&lt;p&gt;My linear algebra professor (J. Komlos) said something that has always stuck with me: we should always use the &lt;em&gt;same&lt;/em&gt; letters to denote certain variables, and different letters for different (math) subjects. In this way, our brains are able to build mental pathways so that when we see certain letters we can remember lots of other things we know about that subject because we associate those letters with certain facts, theorems, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I actually think it's mostly a cultural phenomenon, comp-sci people like to use acronyms in part because it's more clear to program with an acronym than a single letter. But similarly, chemists like to use Latin words, physicists make up descriptive names for things, and mathematicians make up new words no one has ever heard of before.&lt;/p&gt;&#xA;" OwnerUserId="1417" LastEditorUserId="457" LastEditorDisplayName="" LastEditDate="2011-03-01T11:26:10.007" LastActivityDate="2011-03-01T11:26:10.007" CommentCount="8" />
  <row Id="24247" PostTypeId="2" ParentId="24241" CreationDate="2011-02-28T15:23:50.783" Score="3" ViewCount="" Body="&lt;p&gt;It's a convention like any other; never underestimate the power of historical inertia! Different disciplines have different conventions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would also guess (and I have no evidence to support this) that a big reason it is currently still around is that typesetting equations was expensive and time-consuming before the rise of modern typesetting (e.g. TeX) and writers of textbooks were probably encouraged to keep their notation as concise as possible. This would explain the disparity between mathematicians and programmers since the latter, after all, have the benefit of a history of using computers. &lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2011-02-28T15:23:50.783" />
  <row Id="24248" PostTypeId="2" ParentId="24241" CreationDate="2011-02-28T15:25:30.787" Score="14" ViewCount="" Body="&lt;p&gt;A few thoughts:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;It is not true that one does not see formulae as you wrote in the very last line, something like $$ \mathrm{velocity} = \frac{\mathrm{displacement}}{\mathrm{time}} $$ In fact, I was &lt;em&gt;taught&lt;/em&gt; algebra that way: start with word problems and write down mathematical expressions with words and labels, instead of numbers. I believe this kind of notation is also often used in textbooks. &lt;/li&gt;&#xA;&lt;li&gt;Like Mariano and Templar said in their answers, one of the main advantage of the short single letter notation is ease of copying. For computations it is much easier to write (by hand) single letters than whole strings. &lt;/li&gt;&#xA;&lt;li&gt;Another issue similar to the above is that it is easier to visually recognize identical terms when you use single letters. Take the expression for the &quot;final amount&quot; you wrote down. I happen to think that using the string &quot;periods_per_yr&quot; makes it harder to notice that the same term appear both in the denominator inside the brackets and in the exponent outside. Algebraically it is perhaps easier to keep track of structures using single letters. &lt;/li&gt;&#xA;&lt;li&gt;Also, what about when performing abstract arguments where the variables do not refer to anything specific and concrete? Is the phrase &quot;take three points in the plane, point_one, point_two, and point_three...&quot; more easily to understand than &quot;take three points $P,Q,R$ in the plane...&quot;? &lt;/li&gt;&#xA;&lt;li&gt;Lastly, in my PhD dissertation I have several computations involving expressions that span close to half a page when printed. I shudder to think what would happen if I use verbose names for all the terms: is it worth it to make individual variables more &quot;legible&quot; at the expense of making each equation span 3 pages? Does that actually improve overall legibility? &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1543" LastActivityDate="2011-02-28T15:25:30.787" />
  <row Id="24249" PostTypeId="2" ParentId="24098" CreationDate="2011-02-28T15:29:03.500" Score="3" ViewCount="" Body="&lt;p&gt;Since stackexchange is being silly and I can't seem to comment on my own question - I'll post this as an answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm thinking the topology is trivial on the set $S$. Since if the set $U$ is open in $\mathbb{R} / S$ then it's preimage of $q$ (where $q$ is quotient mapping) must be open in $\mathbb{R}$, meaning there exists an open interval $J \subseteq q^{-1}(U)$. But $q(J)$ equals all of the cosets in $\mathbb{R} / S$. Am I right?&lt;/p&gt;&#xA;" OwnerUserId="7602" LastActivityDate="2011-02-28T15:29:03.500" />
  <row Id="24250" PostTypeId="2" ParentId="24234" CreationDate="2011-02-28T15:30:52.027" Score="4" ViewCount="" Body="&lt;p&gt;That sounds like rubbish to me. I think they are just trying to shut you up.&lt;/p&gt;&#xA;" OwnerUserId="1508" LastActivityDate="2011-02-28T15:30:52.027" />
  <row Id="24251" PostTypeId="1" AcceptedAnswerId="24260" CreationDate="2011-02-28T15:35:33.520" Score="2" ViewCount="141" Body="&lt;p&gt;Let there be given a set of $n$ real numbers, $\{r_i\} \subset (0,1)$; is it possible to find some conditions satisfied by a real number $m \in \mathbb{R}$ to ensure that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\hspace{6cm}r_1^m+r_2^m+r_3^m+...+r_n^m=1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, it might be possible to define such an $m$ by an infinite series.&lt;/p&gt;&#xA;" OwnerUserId="1417" LastEditorUserId="1417" LastEditorDisplayName="" LastEditDate="2011-02-28T20:39:40.153" LastActivityDate="2011-03-01T09:39:14.933" Title="A problem on summing real numbers all taken to the same exponent." Tags="&lt;analysis&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="1" />
  <row Id="24252" PostTypeId="2" ParentId="24241" CreationDate="2011-02-28T16:06:32.963" Score="2" ViewCount="" Body="&lt;p&gt;I don't think that substituting single-letter variables with more letters would do much positive in most mathematics, because most theorems (and such) are short enough so that remembering a few variables isn't a problem. And if you forget, it is just to look at the start of the theorem. Also, there is conventions for what variable-letters to use in different circumstances. And last, it is a way to make it easier to separate words from variables.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In programming, the ability to have more letters for one variable, is a huge advantage because a program often consists of thousands of lines, and hundreds of variables. &lt;/p&gt;&#xA;" OwnerUserId="7044" LastActivityDate="2011-02-28T16:06:32.963" />
  <row Id="24253" PostTypeId="2" ParentId="24241" CreationDate="2011-02-28T16:07:05.083" Score="30" ViewCount="" Body="&lt;p&gt;Perhaps the most compelling reason for using single character variables is that it enables the usual convention of omitting the multiplication sign in products. This enables great conciseness in notating polynomials - which is important since polynomials are ubiquitous in mathematics, so any conventuion that simplifies their notation, comprehension, etc is surely worthwhile. Thus we can write $\rm\ xyz\ $ to mean $\rm\ x\cdot y\cdot z\ $ without any worry that it will be mistaken for a variable name. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;While having to insert the multiplication signs doesn't reduce conciseness much for a &lt;em&gt;monomial&lt;/em&gt;, it can greatly increase complexity for a &lt;em&gt;polynomial&lt;/em&gt; of many terms. For it may cause equations to overflow the line/page length, etc, greatly hindering comprehension. Moreover, as many cognitive studies show, humans read read words by their shape (e.g. cover up the top/bottom half of a line of text and note how you can still easily read it), so any conventions that alter the shape (or increase its visual complexity) necessarily inhibit visual parsing, pattern-matching, and global inference of key structural characteristics.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2011-02-28T16:38:19.473" LastActivityDate="2011-02-28T16:38:19.473" CommentCount="5" />
  <row Id="24254" PostTypeId="2" ParentId="24241" CreationDate="2011-02-28T16:10:11.353" Score="3" ViewCount="" Body="&lt;p&gt;Because a variable is just a number and they all behave equally, different variables dont have any different meaning or interpretation in math, you dont gain any additional information by writing &quot;final_amount&quot; instead of x. Its only confusing and unproffesional. Even if you try to follow this path, there can be intermediate steps in solving an equation in which these words can not be interpreted in any sensible physical or informal way. For instance it can be convenient to introduce imaginary numbers even when solving equations with real solutions.&lt;/p&gt;&#xA;" OwnerUserId="4929" LastActivityDate="2011-02-28T16:10:11.353" />
  <row Id="24255" PostTypeId="2" ParentId="24220" CreationDate="2011-02-28T16:12:30.243" Score="2" ViewCount="" Body="&lt;p&gt;Let $G$ be any finite group, and let $P$ be a Sylow $p$-subgroup of $G$, and $N(P)$ the normalizer in $G$ of $P$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that $P$ is a Sylow $p$-subgroup of $N(P)$, and in fact is normal in $N(P)$; that means that $P$ is the &lt;em&gt;only&lt;/em&gt; Sylow $p$-subgroup of $G$ that is contained in $N(P)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, suppose $g\in G$ normalizes $N(P)$ (that is, $g\in N(N(P))$). Then $g^{-1}N(P)g = N(P)$. We also know that $g^{-1}Pg$ is a Sylow $p$-subgroup of $G$, and since $P\subseteq N(P)$, then $g^{-1}Pg\subseteq g^{-1}N(P)g = N(P)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What does that tell you about $g^{-1}Pg$? Conclusion?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that this particular result does not require you to assume $G$ is nilpotent.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2011-02-28T16:12:30.243" CommentCount="1" />
  <row Id="24256" PostTypeId="2" ParentId="24241" CreationDate="2011-02-28T16:16:07.313" Score="35" ViewCount="" Body="&lt;p&gt;I think one reason is that often one does &lt;em&gt;not&lt;/em&gt; want to remember what the variable names really represent. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an example, when we choose to talk about the matrix $(a_{ij})$ instead of the matrix $(\mathrm{TransitionProbability}_{ij})$, this expresses the important fact that once we have formulated our problem in terms of matrices, it is perfectly safe to forget where the problem came from originally -- in fact, remembering what the matrix &quot;really&quot; describes might only be unnecessary psychological baggage that prevents us from applying all linear-algebraic tools at our disposal. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(As an aside, have you ever seen code written by a mathematician? It very often looks exactly like your first example.)&lt;/p&gt;&#xA;" OwnerUserId="677" LastActivityDate="2011-02-28T16:16:07.313" CommentCount="1" />
  <row Id="24257" PostTypeId="1" AcceptedAnswerId="24264" CreationDate="2011-02-28T16:48:33.290" Score="3" ViewCount="70" Body="&lt;p&gt;I am curious if something like the Nine Lemma (http://en.wikipedia.org/wiki/Nine_lemma) is true in an arbitrary triangulated category. To be more explicit, suppose I have a map of cofiber sequences/distinguished triangles and I take the cofiber/mapping cone at each stage vertically (this gives a diagram like the diagram in the wikipedia link without the zeroes) then is the bottom row a cofiber sequence/distinguished triangle?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am particularly interested in the category of spectra if that makes things easier or harder.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, if the result is not true in general what about when one of the maps that we end up taking the cofiber of is the identity map?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I feel like this ought to be true but I did not see anything in the two references I checked and I am not sure how to make us of verdier's/octahedral axiom.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;thanks for your time.&lt;/p&gt;&#xA;" OwnerUserId="627" LastActivityDate="2011-02-28T17:20:50.653" Title="Nine lemma in Triangulated categories" Tags="&lt;algebraic-topology&gt;&lt;homotopy-theory&gt;&lt;triangulated-categories&gt;" AnswerCount="1" />
  <row Id="24259" PostTypeId="2" ParentId="24178" CreationDate="2011-02-28T16:51:47.713" Score="5" ViewCount="" Body="&lt;p&gt;My laptop has extended your results to $n=3$ and $n=4$ (but $n=5$ is completely out of reach).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $n=3$ we have $8!=40320$ permutations, of which $384$ induce an injection. $384/40320 = 1/105$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $n=4$ we have $16!=20922789888000$ permutations, of which $244744192$ induce an injection. $244744192/20922789888000 = 97/8292375$ in lowest terms, so I don't see a nice pattern here.&lt;/p&gt;&#xA;" OwnerUserId="1508" LastActivityDate="2011-02-28T16:51:47.713" CommentCount="1" />
  <row Id="24260" PostTypeId="2" ParentId="24251" CreationDate="2011-02-28T16:53:26.313" Score="6" ViewCount="" Body="&lt;p&gt;A hint: For given $r_i\in ]0,1[$ consider the function $f(x):=\sum_{i=1}^n r_i^x$ and see what happens for $x\to 0$ resp. $x\to\infty$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My response to the comments: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One has $f(0)=n&amp;gt;1$ and $\lim_{x\to\infty} f(x)=0$; furthermore $f$ is monotonically decreasing, so there is exactly one $x&amp;gt;0$ with $f(x)=1$. Unless all $r_i$ are equal (and maybe in some other special cases) it is not possible to solve the equation $f(x)=1$ explicitly. For a numerical solution assume $r_1\leq r_2\leq\ldots\leq r_n$ and put&#xA;$$a:={\log n\over \log(1/r_1)},\qquad b:={\log n\over \log(1/r_n)}.$$&#xA;Then $f(a)\geq 1$ and $f(b)\leq 1$. Now use binary search.&lt;/p&gt;&#xA;" OwnerUserId="1303" LastEditorUserId="1303" LastEditorDisplayName="" LastEditDate="2011-03-01T09:39:14.933" LastActivityDate="2011-03-01T09:39:14.933" CommentCount="5" />
  <row Id="24261" PostTypeId="2" ParentId="24134" CreationDate="2011-02-28T17:04:16.830" Score="3" ViewCount="" Body="&lt;p&gt;The claim is true.  Any $3\times3$ matrix can be expressed as &#xA;$$&#xD;&#xA;A= \lambda I+  a b^T + B&#xD;&#xA;$$&#xA;where $\lambda$ is real, $a$ and $b$ are 3-vectors and $B$ is skew&#xA;(so that $Bx=c\times x$ for some vector $c$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To prove this, choose an orthogonal matrix&#xA;$Q$ to diagonalize the symmetric part of $A$.&#xA;Then $Q^TAQ=D+K$ where $D$ is diagonal and $K$ is skew.&#xA;If the diagonal entries of $D$ are not all distinct then it is&#xA;easy to write&#xA;$D=\lambda I+\hat a \hat b^T$ and we finish as below.&#xA;If the entries are all distinct, &#xA;we can suppose that $Q$ was chosen so that &#xA;the largest eigenvalue of $D$ is first, the smallest &#xA;second and the middle last.  Then for some positive &#xA;$\mu$ and $\nu$, the matrix&#xA;$D$ can be written&#xA;$$&#xD;&#xA;D = \lambda I + \mu&#xD;&#xA;\begin{pmatrix}&#xD;&#xA;1 &amp;amp; 0 &amp;amp; 0\cr 0 &amp;amp;-\nu^2&amp;amp;0\cr&#xD;&#xA;0&amp;amp;0&amp;amp;0&#xD;&#xA;\end{pmatrix}&#xD;&#xA;=\lambda I + \hat a\hat b^T+\hat K,&#xD;&#xA;$$ with $$&#xD;&#xA; \hat a= \mu\begin{pmatrix}1\cr \nu\cr0\end{pmatrix}, &#xD;&#xA;\quad&#xD;&#xA;\hat b=&#xD;&#xA;\begin{pmatrix} 1 \cr -\nu\cr 0 \end{pmatrix},&#xD;&#xA;\quad \hat K=\mu\begin{pmatrix}&#xD;&#xA;0&amp;amp;\nu&amp;amp;0\cr -\nu &amp;amp; 0&amp;amp;0\cr 0&amp;amp;0&amp;amp;0&#xD;&#xA;\end{pmatrix} .&#xD;&#xA;$$&#xA;Let $a=Q\hat a$, $b=Q\hat b$, and $B=Q(\hat K+K)Q^T$,&#xA;and you're done.&lt;/p&gt;&#xA;" OwnerUserId="2947" LastActivityDate="2011-02-28T17:04:16.830" CommentCount="2" />
  <row Id="24262" PostTypeId="1" AcceptedAnswerId="24266" CreationDate="2011-02-28T17:09:29.197" Score="4" ViewCount="254" Body="&lt;p&gt;I'm am a little bit stuck on this question, any help is appreciated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Show that for every $n\in\mathbb{N}$, $3^{4n+2} + 1$ is divisible by $10$.&lt;/p&gt;&#xA;" OwnerUserId="5505" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2011-02-28T19:30:33.663" LastActivityDate="2011-03-02T13:25:59.597" Title="Show that $3^{4n+2} + 1$ is divisible by $10$" Tags="&lt;elementary-number-theory&gt;" AnswerCount="6" CommentCount="3" />
  <row Id="24263" PostTypeId="2" ParentId="24262" CreationDate="2011-02-28T17:12:37.257" Score="7" ViewCount="" Body="&lt;p&gt;hint : $3^{4n+2} = (10-1)^{2n+1}$.&lt;/p&gt;&#xA;" OwnerUserId="6270" LastActivityDate="2011-02-28T17:12:37.257" />
  <row Id="24264" PostTypeId="2" ParentId="24257" CreationDate="2011-02-28T17:13:35.400" Score="3" ViewCount="" Body="&lt;p&gt;Yes, a version of it is true (however, I don't think you can do it with &lt;em&gt;any&lt;/em&gt; morphism of distinguished triangles - since the map of the cones is not unique).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The statement I know is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Every commutative square sits inside a $4 \times 4$-diagram whose rows and colums are distinguished triangles, $8$ squares commute and one square is sign commutative.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is called &quot;Verdier's exercise&quot; in the folklore and can be found in Bernstein-Beilinson-Deligne, &lt;em&gt;faisceaux pervers&lt;/em&gt;, Proposition 1.1.11.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can also find a proof as Lemma 2.6 in May's, &lt;em&gt;The additivity of traces in tensor triangulated categories&lt;/em&gt;, available &lt;a href=&quot;http://www.math.uchicago.edu/~may/PAPERS/AddJan01.pdf&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to prove it yourself, here's an outline: Start with a commutative square $ABA&amp;#39;B&amp;#39;$ and draw the diagonal $A \to B&amp;#39;$. Build octahedra over the two ensuing commutative triangles. Only using these octahedra, you will then be able to build a diagram of the form&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; A  -&amp;gt; B -&amp;gt;  C -&amp;gt; A[1]&#xA; |     |     |     |&#xA; v     v     v     v&#xA; A' -&amp;gt; B' -&amp;gt; C'-&amp;gt; A'[1]&#xA; |     |&#xA; v     v&#xA; A''-&amp;gt; B''&#xA; |     |&#xA; v     v&#xA;A[1]-&amp;gt;B[1]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The morphism $C \to C&amp;#39;$ will be a composition of two morphisms and build yet another octahedron and complete the diagram. You'll have to rotate one triangle and that's the reason for a sign commutativity occurring in the bottom right square.&lt;/p&gt;&#xA;" OwnerUserId="5363" LastEditorUserId="5363" LastEditorDisplayName="" LastEditDate="2011-02-28T17:20:50.653" LastActivityDate="2011-02-28T17:20:50.653" CommentCount="5" />
  <row Id="24265" PostTypeId="1" CreationDate="2011-02-28T17:16:52.933" Score="1" ViewCount="70" Body="&lt;p&gt;I'm looking for a basic motivation for this topic. Meaning, why would one want to study the function field of a curve. The reason I usually give to this question is that one understands a space by understanding the functions that can be defined on it. Also, this object categorically classifies curves up to birational equivalence. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, to motivate a younger student who doesn't know anything about birational geometry nor indeed has yet developed the intuition about understanding geometry through functions, these reasons can seem arbitrary. Does anyone know of a way or analogy to see why one would come to study function fields? In particular, of curves?&lt;/p&gt;&#xA;" OwnerUserId="7606" LastActivityDate="2011-02-28T17:16:52.933" Title="Function field of a curve" Tags="&lt;algebraic-geometry&gt;" CommentCount="3" />
  <row Id="24266" PostTypeId="2" ParentId="24262" CreationDate="2011-02-28T17:19:09.093" Score="7" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Hint:&lt;/strong&gt; $$3^{4n+2}=3^{4n}\cdot 9$$ and $$3^{4n}\equiv 1 \pmod{10}$$ since $\phi(10)=4$.&lt;/p&gt;&#xA;" OwnerUserId="6075" LastActivityDate="2011-02-28T17:19:09.093" />
  <row Id="24267" PostTypeId="2" ParentId="24262" CreationDate="2011-02-28T17:21:04.053" Score="1" ViewCount="" Body="&lt;p&gt;$$9+1=10$$&#xA;$$81 \times \left( 3^{4n-2} +1 \right) = 3^{4n+2} + 1 +80$$&lt;/p&gt;&#xA;" OwnerUserId="6460" LastActivityDate="2011-02-28T17:21:04.053" CommentCount="1" />
  <row Id="24268" PostTypeId="1" AcceptedAnswerId="24270" CreationDate="2011-02-28T17:23:46.670" Score="4" ViewCount="158" Body="&lt;p&gt;I am aware of the &lt;em&gt;big theta&lt;/em&gt; notation $f = \Theta(g)$ if and only if there are &lt;em&gt;positive constants&lt;/em&gt; $A, B$ and $x_0 &amp;gt; 0$ such that for all $x &amp;gt; x_0$ we have&#xA;$$&#xD;&#xA;A|g(x)| \leq |f(x)| \leq B |g(x)|.&#xD;&#xA;$$&#xA;What if the condition is the following:&#xA;$$&#xD;&#xA;C_1 + A|g(x)| \leq |f(x)| \leq C_2 + B |g(x)|&#xD;&#xA;$$&#xA;where $C_1, C_2$ are possibly negative? Certainly more can be said than just $f = O(g)$. Is there a generalized $\Theta$ notation which allows shifts (by, say $C_1, C_2$)? In particular, I'm interested in the special case:&#xA;\begin{eqnarray}&#xA;-C  \leq f(x) - g(x) \leq C&#xA;\end{eqnarray}&#xA;for some positive $C$. How does $f$ compare to $g$ in this case? If $f$ and $g$ are positive functions of $x$ which both diverge to $\infty$, is it true that $f(x) = -C + g(x) + \Theta(1)$? What is the appropriate asymptotic notation in this case?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Update&lt;/em&gt; Thanks for the clarifying answers. Now here is a slightly harder question. Suppose $f$ is discrete and $g$ is continuous. Suppose further that as $x \to \infty$, the difference $f(x) - g(x)$ is asymptotically bounded in the interval $[-C,C]$ but does not necessarily converge to $0$. Does $f \sim g$ still make sense? Would it be more appropriate to use $\liminf_{x \to \infty} f(x) - g(x) = - C$ and $\limsup_{x \to \infty} f(x) - g(x) = C$?&lt;/p&gt;&#xA;" OwnerUserId="2720" LastEditorUserId="2720" LastEditorDisplayName="" LastEditDate="2011-02-28T23:54:04.817" LastActivityDate="2011-02-28T23:54:04.817" Title="Big $O$ vs Big $\Theta$" Tags="&lt;real-analysis&gt;&lt;asymptotics&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="4" />
  <row Id="24270" PostTypeId="2" ParentId="24268" CreationDate="2011-02-28T17:56:13.477" Score="2" ViewCount="" Body="&lt;p&gt;If $g(x)$ and $f(x)$ tends to $\infty$, then there is a value $x_0$ such that for $x &amp;gt; x_0$, $g(x)$ and $f(x)$ are strictly positive.  Therefore, if $-C \leq f(x) - g(x) \leq C$, then for $x &amp;gt; x_0$, we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \frac{-C}{g(x)} \leq \frac{f(x)}{g(x)} -1 \leq \frac{C}{g(x)}. $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Taking limits, you see that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \lim_{x \to \infty} \frac{f(x)}{g(x)} = 1, $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;if the limit exists.  In this case, you can write $f \sim g$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update:&lt;/strong&gt; To answer your second question, $f \sim g$ may not be appropriate here as $\displaystyle\lim_{x \to \infty} \frac{f(x)}{g(x)}$ may or may not exist.  If the limit does exist, then you can write $f \sim g$ as before.  If not, then the situation is trickier, and it must be dealt with individually, depending on the functions $f$ and $g$.  You should just make the statement that best exemplifies what you are trying to say between the relationship of $f(x)$ and $g(x)$.  The big-Oh (or big-Theta) notation may not be the best fit here.  Hope this is helpful.&lt;/p&gt;&#xA;" OwnerUserId="6491" LastEditorUserId="6491" LastEditorDisplayName="" LastEditDate="2011-02-28T19:18:06.763" LastActivityDate="2011-02-28T19:18:06.763" />
  <row Id="24271" PostTypeId="2" ParentId="24268" CreationDate="2011-02-28T17:57:12.883" Score="2" ViewCount="" Body="&lt;p&gt;If $\lim |f(x)| = \lim |g(x)| = \infty$ then there is no difference between your two concepts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $f$ is a $\Theta(g)$, then it is a &quot;shifted&quot; $\Theta(g)$ with $C_1 = C_2 = 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $f$ is a &quot;shifted $\Theta(g)$, then it is a $\Theta(g)$ :&#xA;Since $\lim |g(x)| = \infty$, there exists $x_1$ from which $C_2/B \le |g(x)|$ and $ -2C_1/A \le |g(x)|$. This shows that for $x \ge \max(x_0,x_1)$, $A|g(x)|/2 \le C_1 + A|g(x)| \le |f(x)| \le C_2 + B|g(x)| \le 2B|g(x)|$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $-C \le f(x)-g(x) \le C$, then this is exactly the same as saying $f-g = O(1)$.&#xA;In this case, you have $f = g+O(1)$, and if their limit is $\pm \infty$, $f \sim g$&lt;/p&gt;&#xA;" OwnerUserId="6270" LastActivityDate="2011-02-28T17:57:12.883" />
  <row Id="24272" PostTypeId="1" CreationDate="2011-02-28T17:58:32.257" Score="0" ViewCount="52" Body="&lt;p&gt;This question is two parts, the first part I've done the second part seems simple enough but I keep getting the wrong answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obtain the complete viewing transformation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a) maps a window with world co-ordinates $1 \leq x \leq 10$, $1 \leq y \leq 10$ onto the viewport with $0.25\leq x\leq 0.75$, $0 \leq y \leq 0.5$ in normalised device space, and then ...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;b) maps a window with $0.25 \leq x \leq 0.5$, $0.25 \leq y \leq 0.5$ in the normalised device space into a viewport with $1 \leq x \leq 10$, $1 \leq y \leq 10$ on the physical display device.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I got the answer to the first part&#xA;Tw.S.Tv = &#xA;$\begin{bmatrix}&#xD;&#xA;1/18 &amp;amp; 0 &amp;amp; 0 \\&#xD;&#xA;0 &amp;amp; 1/18 &amp;amp; 0 \\&#xD;&#xA;7/36 &amp;amp; -1/18 &amp;amp; 1&#xD;&#xA;\end{bmatrix}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;however the second part stumbles me a bit, I've been trying to do it the reverse way round but no luck. the answer is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\begin{bmatrix}&#xD;&#xA;2 &amp;amp; 0 &amp;amp; 0 \\&#xD;&#xA;0 &amp;amp; 2 &amp;amp; 0 \\&#xD;&#xA;-1 &amp;amp; 10 &amp;amp; 1&#xD;&#xA;\end{bmatrix}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but I can't get it right :(&lt;/p&gt;&#xA;" OwnerUserId="2874" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-28T18:05:58.303" LastActivityDate="2011-02-28T18:05:58.303" Title="Viewing Transformations." Tags="&lt;matrices&gt;&lt;transformation&gt;" />
  <row Id="24273" PostTypeId="2" ParentId="24262" CreationDate="2011-02-28T18:03:11.693" Score="1" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\rm\ \ \ A^K\: \equiv -1\ \ \Rightarrow\ \ A^{\:J+2\:K}\ \equiv\ A^{\:J}$&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2011-02-28T18:03:11.693" />
  <row Id="24274" PostTypeId="2" ParentId="24234" CreationDate="2011-02-28T18:14:07.833" Score="4" ViewCount="" Body="&lt;p&gt;Like Yuval said, there is a performance reason: if you use $e = 2^k + 1$ for some (small) $k$, then computing $x^e \mod n$ is going to be faster than for most other exponents. Taking $k=16$ is very common and gives $e = 65537$, while taking $k$ equal to 1,2,4 or 8 give the other values in your library. Also, as these are (Fermat) primes, all $n = pq$ with $p \mod e \neq 1$ and $q \mod e \neq 1$ will give a valid $(n,e)$-combination, and many libraries build their $n$ this way, for efficient testing and generation of $n$. A small $e$ is potentially dangerous because of so-called broadcast attacks (sending 3 times the same message with $e = 3$ to 3 different people (with different moduli) compromises the message), even though this can be thwarted by padding, and small $e$ are thus often avoided.&#xA;But in principle any large enough $e$ that is coprime with $\phi(n)$ can be used, provided that $d$ is not too small etc. So this library has chosen to optimize its generation and testing (take shortcuts that can be made for these choices of $e$) and disallow other $e$. There are standards that always take 65537 for $e$ (so you don't have to transmit that info), so this library is even flexible compared to that. &lt;/p&gt;&#xA;" OwnerUserId="4280" LastActivityDate="2011-02-28T18:14:07.833" />
  <row Id="24275" PostTypeId="2" ParentId="24204" CreationDate="2011-02-28T18:57:31.357" Score="1" ViewCount="" Body="&lt;p&gt;What you're looking for is called &lt;em&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Zipf%27s_law&quot; rel=&quot;nofollow&quot;&gt;Zipf's law&lt;/a&gt;&lt;/em&gt;.  This law says that many distribution curves in which the data values are placed in rank order on the horizontal axis by frequency (or, equivalently, percent) follow a power law.  The most famous use of Zipf's law is to describe the frequency of word usage in any given language, although the Wikipedia article specifically mentions income rankings as you ask for.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It can be thought of as a discrete version of the Pareto distribution, so you're right about that.  &lt;em&gt;Added:&lt;/em&gt; This is because Zipf's law is the discrete power law distribution, and Pareto is the continuous power law distribution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;HR&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(&lt;strong&gt;Update&lt;/strong&gt;, in response to the OP's request for more on the relationship between Zipf and Pareto.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm going to do this in the general case.  The argument will also be for numbers and amounts, rather than probabilities, with the understanding that we can convert the functions involved to pdfs or pmfs by scaling by the appropriate constants.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose we have the density function $p(x)$ for dollars (although it could be any good) allocated among people in a group, so that $\int_a^b p(x) dx$ gives the number of people in the group who have between $a$ and $b$ dollars.  Now, rank the people in the group by wealth, and let $z(y)$ denote the wealth that the person ranked $y$ has.  The question then is, &quot;What is the relationship between $p(x)$ and $z(y)$?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the number of people who have more than $M$ dollars.  Using $p(x)$, that is given by $\int_M^{\infty} p(x) dx$.  But this is also $R$, where $R$ is the largest rank of a person who has at least $M$ dollars.  (In other words, if the 34th person has at least $M$ but the 35th does not, then $R = 34$.)  So $z(R+1) &amp;lt; M \leq z(R)$.  If the population is large enough, we can say $z(R) = M$ without losing much accuracy.  Thus $R = z^{-1}(M)$.  So there's our relationship (approximately): $$\int_M^{\infty} p(x) dx = z^{-1}(M).$$&#xA;Thus the ranking function $z(y)$ is the inverse of the wealth tail cumulative distribution function $\int_M^{\infty} p(x) dx$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How does this relate to power laws?  Well, in this special case, if $p(x) = \frac{C}{x^{\alpha+1}}$ (i.e., a Pareto distribution) for some $\alpha &amp;gt; 0$ and constant $C$, then we have $$z^{-1}(M) = \int_M^{\infty} \frac{C}{x^{\alpha}} dx = \frac{C}{\alpha M^{\alpha}},$$&#xA;which means, for some constant $K$, $$z(y) = \frac{K}{y^{\frac{1}{\alpha}}}.$$ &#xA;Thus $z(y)$ is also a power law.  Thus a Pareto (power law) distribution function for some good produces a power law ranking function for people with that good (i.e., Zipf).  &lt;/p&gt;&#xA;" OwnerUserId="2370" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2011-03-01T05:22:29.843" LastActivityDate="2011-03-01T05:22:29.843" CommentCount="5" />
  <row Id="24276" PostTypeId="1" AcceptedAnswerId="24277" CreationDate="2011-02-28T19:02:57.247" Score="1" ViewCount="46" Body="&lt;p&gt;I have 3 vectors in a high-dimensional euclidean space.  I would like to express the length of their sum in terms of their lengths and inner products. For two vectors, I can do it with the law of cosines:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$c^2 = a^2 + b^2 - 2ab\cos\theta$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which in my context translates to:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$||c||^2 =||a||^2+||b||^2 -2 a \cdot b$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(where c = a+b and all are vectors, I project them into the span of a,b to get the triangle and then apply the law of cosines from trig)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do I do this for 3 vectors, d=a+b+c?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll keep working on this and answer my own question if I come up with a solution.&lt;/p&gt;&#xA;" OwnerUserId="7609" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-28T19:43:51.397" LastActivityDate="2011-02-28T19:43:51.397" Title="Length of sum of high-dimensional 3 vectors in terms of their lengths and inner products" Tags="&lt;linear-algebra&gt;&lt;euclidean-geometry&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="24277" PostTypeId="2" ParentId="24276" CreationDate="2011-02-28T19:06:00.193" Score="2" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Hint:&lt;/strong&gt; Use associativity:  $$\overline{a}+\overline{b}+\overline{c}=(\overline{a}+\overline{b})+\overline{c}$$&#xA;This yields the answer since you already know how to add two vectors.  As a check, notice that the answer has to be the same if I permute $a,b,c$ since addition is commutative.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Say you have vectors $a,b,c$.  Then $$\|a+b+c\|=\|(a+b)\|^2+\|c\|^2-2(a+b)\cdot c$$&#xA;$$=\|(a+b)\|^2+\|c\|^2 -2a\cdot c-2b\cdot c$$&#xA;$$=\|a\|^2+\|b\|^2+\|c\|^2 -2\left( a\cdot c+b\cdot c+a\cdot b\right)$$&lt;/p&gt;&#xA;" OwnerUserId="6075" LastEditorUserId="6075" LastEditorDisplayName="" LastEditDate="2011-02-28T19:11:26.960" LastActivityDate="2011-02-28T19:11:26.960" CommentCount="3" />
  <row Id="24278" PostTypeId="1" AcceptedAnswerId="24329" CreationDate="2011-02-28T19:31:41.983" Score="1" ViewCount="121" Body="&lt;p&gt;I have the following somewhat awkward definition of covariant differentiation along a curve:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Let $S \subseteq \mathbb{R}^N$ be a smoothly and isometrically embedded manifold, and $\alpha : I \to S$ be a smooth curve. The &lt;em&gt;covariant derivative along the curve&lt;/em&gt; $\alpha$ of a vector field $V: I \to \mathbb{R}^N$, $V(t) \in T_{\alpha(t)}S$ for each $t \in I$, is the orthogonal projection of $\dot{V}$ onto $T_{\alpha(t)}S$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I feel like it ought to be easy to show that if I have an isometry $f: S \to \tilde{S}$, then covariant differentiation and the differential $df$ ought to commute in the obvious way, namely&#xA;$$ df \left( \nabla_{\dot{\alpha}} V \right) = \nabla_{df(\dot{\alpha})} \left( df(V) \right) $$&#xA;but I spent the better part of the afternoon thinking about it without any success, at which point I resorted to taking charts and comparing coordinate expressions. That was easy but it felt like cheating. Is there a way to do this directly from the definition, exploiting the fact that $S$ and $\tilde{S}$ are embedded manifolds?&lt;/p&gt;&#xA;" OwnerUserId="5191" LastEditorUserId="5191" LastEditorDisplayName="" LastEditDate="2011-02-28T19:41:30.900" LastActivityDate="2011-03-01T00:32:57.793" Title="Proving some basic properties of covariant differentiation" Tags="&lt;differential-geometry&gt;&lt;manifold&gt;&lt;riemannian-geometry&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="24279" PostTypeId="2" ParentId="24251" CreationDate="2011-02-28T19:45:53.093" Score="1" ViewCount="" Body="&lt;p&gt;It is possible to tighten the range of possible values for $m$ to the interval&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$-\frac{\ln n }{\ln \min_i r_i} \leq m \leq -\frac{\ln n }{\ln \max_i r_i}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is not so good if the maximum and the minimum of the $r_i$ are very close to 1 and 0 respecively.&lt;/p&gt;&#xA;" OwnerUserId="3567" LastActivityDate="2011-02-28T19:45:53.093" />
  <row Id="24280" PostTypeId="1" CreationDate="2011-02-28T20:08:37.967" Score="2" ViewCount="118" Body="&lt;p&gt;I have the following differential equation:&#xA;$$y&amp;#39;&amp;#39;+y=\cos(t)\cos(2t)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe something can be done to $\cos(t)\cos(2t)$ to make it easier to solve. Any ideas?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance.&lt;/p&gt;&#xA;" OwnerUserId="7611" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-02-28T20:17:03.847" LastActivityDate="2011-03-01T03:49:34.487" Title="Second-Order Linear Differential Equation" Tags="&lt;differential-equations&gt;" AnswerCount="2" CommentCount="5" />
  <row Id="24281" PostTypeId="1" AcceptedAnswerId="24283" CreationDate="2011-02-28T20:32:42.477" Score="1" ViewCount="74" Body="&lt;p&gt;In a random drawing with 50 entries are the odds of winning better if I:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A.  Enter 1 contest 3 times&lt;br /&gt;&#xA;or &lt;br /&gt;&#xA;B.  Enter 3 contests 1 time&lt;br /&gt;&lt;/p&gt;&#xA;" OwnerUserId="7613" LastActivityDate="2011-02-28T20:46:44.510" Title="Odds of winning a contest" Tags="&lt;statistics&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="24282" PostTypeId="1" AcceptedAnswerId="24288" CreationDate="2011-02-28T20:40:36.963" Score="0" ViewCount="76" Body="&lt;p&gt;I have a bunch of data, and I'd like to find the closest exponential aproximation I can to fit the points.  I'm guessing there's a (relatively) straightforward way to do this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, if I have the data points (5, 9, 17, 33), the equation I'd be looking for would be f(n) = 2^x + 1, since that would fit the data points exactly.  Of course, if I had a similar set of points (5, 9, 16, 33), the equation wouldn't work quite right: f(n) = 2^x + 1 would still be &quot;close&quot;, but presumably some other numbers than 2 and 1 could get me ever so slightly &quot;closer&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a way to figure this out?  (I'm guessing &quot;yes&quot;).  Ideally, I'd like to have it be in the form described above (something^x + possible_constant), though methods for figuring this out in a more general way are welcome, so long as I can tie down some of the parameters (so that any other factors would be forced to be 0).&lt;/p&gt;&#xA;" OwnerUserId="175" LastActivityDate="2011-02-28T20:59:03.883" Title="How to find closest exponential approximation?" Tags="&lt;approximation&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="24283" PostTypeId="2" ParentId="24281" CreationDate="2011-02-28T20:46:44.510" Score="3" ViewCount="" Body="&lt;p&gt;In the first case, you have three entries out of 50, so your chance of winning assuming all drawings have equal chance of being picked is $$\frac{3}{50}=0.06.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you enter 3 contests 1 time, and I assume each contest is the same, then your chance of winning at least one of them is the complement to losing all of them. This probability is given by&#xA;$$&#xD;&#xA;1-\left(\frac{49}{50}\right)^3=\frac{7351}{125000}\approx 0.058808.&#xD;&#xA;$$&#xA;So it seems you have a slightly better chance of winning at least once if put all three of your bets in one contest.&lt;/p&gt;&#xA;" OwnerUserId="1583" LastActivityDate="2011-02-28T20:46:44.510" CommentCount="2" />
  <row Id="24284" PostTypeId="1" CreationDate="2011-02-28T20:51:13.640" Score="6" ViewCount="160" Body="&lt;p&gt;Consider the string 'Let y = f(x).&quot;  Suppose that it occurs in some elementary context, such as when graphing the function f using x/y coordinates.  How is this to be understood in predicate logic?  We can't have either x or y be free variables, for consider the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;let f:ℝ→ℝ, ∀x,f(x)=2x&lt;/p&gt;&#xA;&#xA;&lt;p&gt;let g:ℝ→ℝ, ∀x,g(x)=x+x&lt;/p&gt;&#xA;&#xA;&lt;p&gt;let y = f(x)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;let z = g(x)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;∴ y = z&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here the last line is clearly true, but would lack a truth value if either variable were a free variable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if both variables are bound, we're stuck with permutations of quantifiers that mean the wrong things:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;∀x,∀y,y=f(x)   [says the universe has cardinality 1]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;∀x,∃y,y=f(x)   [says f's domain is the universe]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;∃y,∀x,y=f(x)   [says f is a constant function]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;∃x,∀y,y=f(x)   [says the universe has cardinality 1 and f is nonempty]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;∀y,∃x,y=f(x)   [says f is onto the universe]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;∃x,∃y,y=f(x)   [says f is not the empty function]&lt;/p&gt;&#xA;" OwnerUserId="7546" LastActivityDate="2011-03-04T05:02:17.207" Title="Is the 'variable' in 'let y=f(x)' free, bound, or neither?" Tags="&lt;logic&gt;&lt;definition&gt;" AnswerCount="5" CommentCount="1" />
  <row Id="24285" PostTypeId="2" ParentId="24282" CreationDate="2011-02-28T20:52:02.700" Score="1" ViewCount="" Body="&lt;p&gt;You can solve for $f(x) = Ca^x$ using linear regression, since $\log f(x) = (\log a)x + \log C$. Here $x$ should be $1,2,\ldots$, and $y$ should be your data points.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2011-02-28T20:52:02.700" CommentCount="3" />
  <row Id="24286" PostTypeId="2" ParentId="24241" CreationDate="2011-02-28T20:57:16.300" Score="7" ViewCount="" Body="&lt;p&gt;A number of excellent explanations have been given, ranging from laziness of mathematicians to convention. I'd like to add one more that may be slightly more debatable than others.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I dare to say I experience some sort of &lt;em&gt;freedom&lt;/em&gt; when assigning variable names. This freedom is very typical to math: ideas are not bound by physical reality, the only constraints are your own imagination and the consistency of your thoughts. This freedom suffers if you're thinking about your math too concretely. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are a few remarks to explain my point of view:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Somewhere in the 15th century ($\pm$ a few centuries) people realised they didn't have to discribe their equations in words, they could just assign letters to their quantities and let the formulae speak for themselves. Suddenly people realised it was not necessary for algebraic equations to have a direct geometric meaning. If people had kept writing $\rm area = length\cdot width$ until the end of days, this might never have happened.&lt;/li&gt;&#xA;&lt;li&gt;Let's say we were thinking about distance. We might write&#xA;$$ \rm distance = \sqrt{horizontal\ displacement^2 + vertical\ displacement^2}.$$ However, would you ever dare to write $\rm distance = |horizontal\ displacement| + |vertical\ displacement|$? Maybe you would, but write this on the blackboard of an average high-school physics teacher (no offence meant) and they might try to hit you with something. This is because they're thinking about the physical reality, not about the abstract properties of such a formula or it's consequences in a more abstract setting. Therefore if you ever want to define the concept of a metric space and study it's properties in full generality, it's probably safer to denote distance simply by &quot;d&quot;. (Choosing 'd' as a small mnemonic to where it all started.)&lt;/li&gt;&#xA;&lt;li&gt;Let's say we're thinking about a curve this time. At some point you may come to realise there's actually an interesting group structure on your curve. If you're refering to your curve simply as &quot;curve&quot; everywhere, it's probably a giant leap to suddenly rebaptize your curve into &quot;curve that is actually a group as well&quot;. You'll probably still call your curve &quot;curve&quot; all the time and you may unconsciously make it harder for yourself to realise the importance and usefullness of it's group structure.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;In conclusion: math is an &lt;em&gt;abstract&lt;/em&gt; science, assigning the name all to concrete names to objects will constrain your imagination, which is always a bad thing in math.&lt;/p&gt;&#xA;" OwnerUserId="5189" LastActivityDate="2011-02-28T20:57:16.300" />
  <row Id="24287" PostTypeId="1" AcceptedAnswerId="24499" CreationDate="2011-02-28T20:58:14.397" Score="0" ViewCount="76" Body="&lt;p&gt;Are there any restrictions on $\pi_1(S^3\backslash K)$ for a (tame) knot $K$ besides having $\pi_1^{\text{ab}}(S^3\backslash K)=\mathbb{Z}$?&#xA;&lt;hr&gt;&#xA;So we have 1) finite presentation, 2) $H_1=\mathbb{Z}$, 3) $H_2=0$, and I saw somewhere 4) the knot group must be the normal closure of a single element.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Apparently this characterizes knot groups in dimensions $n\geq5$ (i.e. $\pi_1(S^n\backslash S^{n-2})$), but fails to do so in lower dimensions.  Basically I just wanted to know if there is simple characterization of knot groups (not &quot;it has a presentation that looks like a wirtinger presentation&quot;)&lt;/p&gt;&#xA;" OwnerUserId="6925" LastEditorUserId="6925" LastEditorDisplayName="" LastEditDate="2011-03-01T18:56:03.527" LastActivityDate="2011-03-01T22:26:28.853" Title="restrictions on the fundamental group of a knot complement?" Tags="&lt;knot-theory&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="24288" PostTypeId="2" ParentId="24282" CreationDate="2011-02-28T20:59:03.883" Score="2" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://www.xuru.org/rt/ExpR.asp#CopyPaste&quot; rel=&quot;nofollow&quot;&gt;Here&lt;/a&gt; is an online exponential regression calculator. &lt;/p&gt;&#xA;" OwnerUserId="2810" LastActivityDate="2011-02-28T20:59:03.883" CommentCount="2" />
  <row Id="24289" PostTypeId="2" ParentId="24284" CreationDate="2011-02-28T21:01:07.250" Score="2" ViewCount="" Body="&lt;p&gt;It's just a definition of $y$. The statement $y=f(x)$ defines $y$ to be $f(x)$. It doesn't correspond to any particular statement of predicate logic, since you're not claiming anything. Whenever you use $y$ in a statement $P(y)$, you really mean $$\forall y, y=f(x) \implies P(y).$$ You can then use your knowledge that $f$ is a function: $$\exists! y, y = f(x).$$ Using this, you can prove that the extension of the language by adding $y$ is conservative. That means that everything that you prove using this definition of $y$ and does not involve $y$ is actually true even without using $y$. The uniqueness of $y$ guarantees that $f(x) = y$; note that when you say $f(x)$, this is also shorthand for something. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now to your example. What you're really trying to prove is&#xA;$$\forall x, \forall y, \forall z, f(x) = y \land g(x) = z \implies y = z. $$ The proof reduces to $x+x=2x$ through the definition of &quot;$f(x)=y$&quot;.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2011-02-28T21:01:07.250" CommentCount="2" />
  <row Id="24290" PostTypeId="2" ParentId="24287" CreationDate="2011-02-28T21:12:56.547" Score="1" ViewCount="" Body="&lt;p&gt;If $\pi$ is a knot group, in addition to $H_1(\pi)=\mathbb Z$, you also have $H_2(\pi)=0$. This is because $H_2(S^3\setminus K)=0$, and $K(\pi,1)=(S^3\setminus K)$ union  cells of dimension $\geq 3$. So $H_2(S^3\setminus K)$ surjects onto $H_2(K(\pi,1))=H_2(\pi)$. &lt;/p&gt;&#xA;" OwnerUserId="2941" LastActivityDate="2011-02-28T21:12:56.547" />
  <row Id="24291" PostTypeId="2" ParentId="24262" CreationDate="2011-02-28T21:14:17.310" Score="3" ViewCount="" Body="&lt;p&gt;What is the formation law of the remainders of the division by $10$ of the&#xA;powers $3^{n}$? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(For the notation see &lt;a href=&quot;http://en.wikipedia.org/wiki/Modular_arithmetic&quot; rel=&quot;nofollow&quot;&gt;modular arithmetic&lt;/a&gt;.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left\{ &#xD;&#xA;\begin{array}{c}&#xD;&#xA;3\equiv 3\quad \pmod{10}  \\ &#xD;&#xA;3^{2}\equiv 9\quad \pmod{10}  \\ &#xD;&#xA;3^{3}\equiv 7\quad \pmod{10}  \\ &#xD;&#xA;3^{4}\equiv 1\quad \pmod{10} &#xD;&#xA;\end{array}\right. $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left\{ &#xD;&#xA;\begin{array}{c}&#xD;&#xA;3^{5}\equiv 3\quad \pmod{10}  \\ &#xD;&#xA;3^{6}\equiv 9\quad \pmod{10}  \\ &#xD;&#xA;\cdots  \\ &#xD;&#xA;\cdots &#xD;&#xA;\end{array}\right. $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, with respect to the divisor $10$ the remainders of $3^{n}$ are periodic ($3,9,7,1,3,9,7,1,\dots$) with period $4$. This together with $4n+2\equiv 2\quad \pmod{4}$ yields $3^{4n+2}\equiv 3^{2}\quad \pmod{10}$. Also $1\equiv 1\quad \pmod{10}$. Hence, for all $n\ge 1$ $$3^{4n+2}+1\equiv 3^{2}+1\equiv 0\quad \pmod{10},$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which means the remainders of the devision of $3^{4n+2}+1$ by $10$ are $0$.&lt;/p&gt;&#xA;" OwnerUserId="752" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2011-03-02T13:25:59.597" LastActivityDate="2011-03-02T13:25:59.597" />
  <row Id="24292" PostTypeId="1" AcceptedAnswerId="24293" CreationDate="2011-02-28T21:19:21.940" Score="3" ViewCount="127" Body="&lt;p&gt;Let $G$ be a group, $X$ a set. Defining an action $G\times X \to X$ is the same as defining a group morphism $\rho: G\to Sym(X)$, through the formula $g\cdot - = \rho(g)$. The morphism $\rho$ is called a &lt;em&gt;permutation representation&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, in what little representation theory I have studied one replaces the set $X$ with a vector space $V$, then defining a group morphism $\rho: G \to GL(V)$ is the same as defining an action $G\times V \to V$ such that $g\cdot-$ is linear for every g, through the same formula $g\cdot - = \rho(g)$. The morphism $\rho$ is called a &lt;em&gt;linear representation&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first question is: on what other objects can I make my group act? What are the conditions for the previous constructions to make sense when picking an object $X$ from a given category, and picking $Aut(X)$ the set of isomorphisms of $X$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Addenda: 1) and when is this &lt;em&gt;fruitful&lt;/em&gt;? On what categories have actions been studied?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Since we can always define through a morphism such a &lt;em&gt;categorial representation&lt;/em&gt;, when is it equivalent to the definition of the form $G\times X \to X$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the second question: the construction of the semidirect product of groups $N$, $Q$ involves a morphism $Q \to Aut(N)$. One says that $Q$ acts on $N$ by automorphisms. I just noticed &lt;em&gt;why&lt;/em&gt; one says so, and it's because it's just another case of an &quot;action with extra structure&quot;. In this case it's the same giving such a morphism than giving an action $Q \times N \to N$ such that $q\cdot -$ is a group morphism for every $q$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the semidirect product is linked to representations in this way. The second question is, are there analogue constructions with other type of representations? What I mean is: making a group act on another group yields a new group (the semidirect product) which has as underlying set the direct product of both groups. Do actions over different categories yield new objects in a similar fashion?&lt;/p&gt;&#xA;" OwnerUserId="2614" LastEditorUserId="2614" LastEditorDisplayName="" LastEditDate="2011-02-28T21:29:24.913" LastActivityDate="2011-02-28T23:24:52.707" Title="Defining representations; representations and semidirect products" Tags="&lt;abstract-algebra&gt;&lt;group-theory&gt;&lt;representation-theory&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" />
  <row Id="24293" PostTypeId="2" ParentId="24292" CreationDate="2011-02-28T21:23:03.120" Score="5" ViewCount="" Body="&lt;p&gt;The answer to your first question is essentially anything. Given any group $G$ and an object $X$ in a category $C$, a representation of $G$ on $X$ is precisely a group homomorphism $G \to \text{Aut}(X)$. For example, we can and do make groups act on&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;abelian groups,&lt;/li&gt;&#xA;&lt;li&gt;topological spaces such as manifolds,&lt;/li&gt;&#xA;&lt;li&gt;graphs,&lt;/li&gt;&#xA;&lt;li&gt;categories,&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;and so forth. The refinement of your first question is addressed by &lt;a href=&quot;http://mathoverflow.net/questions/56304/looking-for-interesting-actions-that-are-not-representations&quot; rel=&quot;nofollow&quot;&gt;this MO question&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $C$ is a concrete category (a category of sets and structure-preserving maps between sets) then giving a group action of $G$ on $X$ is the same as giving a map $G \times X \to X$ satisfying certain properties, but in general it's not possible to make sense of the object $G \times X$ if $G$ is just an ordinary group. If you want a general definition of group actions along these lines, then I think the right thing to do is to talk about &lt;a href=&quot;http://ncatlab.org/nlab/show/group+object&quot; rel=&quot;nofollow&quot;&gt;group objects&lt;/a&gt; in $C$ instead of groups and require that $C$ have, say, finite products. Then it ought to be true that an action of a group object in $C$ on an object in $C$ is the same as a morphism $G \times X \to X$ satisfying certain properties, where $\times$ is the product in $C$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, a group object in $\text{Top}$ is a topological group, and a group action of a topological group on a topological space is precisely a continuous map $G \times X \to X$ satisfying certain properties. In situations like this, this definition is actually preferable, since I think $\text{Aut}(X)$ cannot always be defined as a topological group in the correct way. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;As long as you are thinking about category theory at all, there are natural ways to generalize the notion of representation, namely to &lt;a href=&quot;http://en.wikipedia.org/wiki/Functor&quot; rel=&quot;nofollow&quot;&gt;functors&lt;/a&gt;. A group $G$ can be regarded as a category with one object and a morphism $g$ for every $g \in G$ which compose according to the way they compose in $G$, and then a permutation representation of $G$ is nothing more than a functor $G \to \text{Set}$, a linear representation is nothing more than a functor $G \to \text{Vect}$, and so forth. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So not only can one consider more general target categories, but one can also consider more general source categories, and this is an enormously fruitful idea even if we restrict ourselves to situations which look vaguely like group representations. For example it leads to &lt;a href=&quot;http://en.wikipedia.org/wiki/Quiver_(mathematics)&quot; rel=&quot;nofollow&quot;&gt;quiver representations&lt;/a&gt;, &lt;a href=&quot;http://ncatlab.org/nlab/show/Lawvere+theory&quot; rel=&quot;nofollow&quot;&gt;Lawvere theories&lt;/a&gt;, &lt;a href=&quot;http://ncatlab.org/nlab/show/topological+quantum+field+theory&quot; rel=&quot;nofollow&quot;&gt;topological quantum field theories&lt;/a&gt;... and of course the general idea of a functor is enormously fruitful as well.&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2011-02-28T22:35:00.140" LastActivityDate="2011-02-28T22:35:00.140" CommentCount="2" />
  <row Id="24294" PostTypeId="1" AcceptedAnswerId="24320" CreationDate="2011-02-28T21:26:40.313" Score="2" ViewCount="120" Body="&lt;p&gt;Lets say you have a Riemannian Manifold $(M,g)$, and you have some given chart where $g = g_{ij} dx_i dx_j$ and you wish to compute the Christoffel symbols for the Riemannian connection in this chart. To do this involves:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Calculating the inverse matrix of $g$, which is not too bad in dimensions 2 and 3, but becomes quite painful in higher dimensions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Using the formula&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Gamma_{ij}^k = \frac{1}{2} g^{kl}(\partial_i g_{jl} + \partial_j g_{il} - \partial_l g_{ij}) $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which if you look closely the index that is summed over on the right is $l$, so there are $n=dim M$ terms on the right that need to be evaluated, and inside each of these you must take a partial derivative of three terms in $g$, so in total on the right-side there are $3n$ terms to be computed. This is just for each value of $k$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Luckily $\Gamma_{ij}^k$ is symmetric in $ij$, and so you only have to compute about half of the $3n^3$ quantities involved, but it is still a very involved calculation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anyone know a simpler method to go about this? Or an easier way to look at the calculation of this? As $\Gamma_{ij}^k$ is an $n \times n$ symmetric matrix for each fixed $k$, maybe there is a way to write $(\Gamma_{ij}^k)$, the matrix, as a matrix product of some other matrices that would be easier to write down. That way you would not have to work with index notation for doing the calculation, which also slows things down.&lt;/p&gt;&#xA;" OwnerUserId="1089" LastActivityDate="2011-02-28T23:46:25.940" Title="Is there a good way to compute Christoffel Symbols" Tags="&lt;differential-geometry&gt;&lt;riemannian-geometry&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="24295" PostTypeId="2" ParentId="24262" CreationDate="2011-02-28T21:34:09.570" Score="0" ViewCount="" Body="&lt;p&gt;$3^{4n+2} = 3^{2(2n+1)} = 9^{2n+1}$.  So your question is really asking about odd powers of 9 and then adding 1.  Let's look at powers of 9, and focus on the last digit (ie, take them mod 10):&#xA;$9^1 = 9,\ 9^2 = 1,\ 9^3=9^1\cdot 9^2=9 \cdot 1 = 9, \cdots$.  So the last digit in consecutive powers of 9 goes 9, 1, 9, 1, 9, 1, ...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now consider adding 1 to only the odd terms above.&lt;/p&gt;&#xA;" OwnerUserId="7162" LastActivityDate="2011-02-28T21:34:09.570" />
  <row Id="24296" PostTypeId="2" ParentId="24287" CreationDate="2011-02-28T21:40:09.980" Score="4" ViewCount="" Body="&lt;p&gt;Famously, knot groups are torsion free.  One proof of this uses the fact that $S^3 - K$ is a $K(\pi, 1)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another nice fact - knots have Seifert surfaces and so knot groups contain free groups. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another one - knot complements have Heegaard splittings.  It follows that any knot group has a presentation with one more generator than relation.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think that combining Waldhausen's work with Gordon-Luecke should prove that the knot group (possibly equipped with a $Z^2$ &quot;peripheral&quot; subgroup) determines the knot in $S^3$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Neuwirth wrote a book entitled &quot;Knot groups&quot;, published in 1965. There is lots more to know since then!&lt;/p&gt;&#xA;" OwnerUserId="1307" LastActivityDate="2011-02-28T21:40:09.980" />
  <row Id="24297" PostTypeId="1" CreationDate="2011-02-28T21:43:04.063" Score="0" ViewCount="64" Body="&lt;p&gt;Let $M$ be the manifold in $R^3$ which is a plane defined by $z = x+y$. There is a global chart for this manifold $\phi(x,y) = (x,y,x+y)$ with inverse $\phi^{-1}(x,y,z)=(x,y)$. Then, we can define a metric tensor using the embedding in $R^3$ as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$g = dx^2 + dy^2 + (dx+dy)^2 = 2 dx^2 + 2 dy^2 + 2 dx dy$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus $g_{ij} = 2$, and $det(g) = 0$, which means that $g^{ij}$ does not exist, and in particular one cannot compute the Christoffel symbols for the Riemannian Connection. Why is this? What is special about this manifold or these coordinates that don't allow the Christoffel symbols to be computed?&lt;/p&gt;&#xA;" OwnerUserId="1089" LastActivityDate="2011-02-28T21:43:04.063" Title="What if the metric tensor has zero-determinant?" Tags="&lt;differential-geometry&gt;&lt;riemannian-geometry&gt;" CommentCount="3" />
  <row Id="24298" PostTypeId="1" CreationDate="2011-02-28T21:53:20.863" Score="0" ViewCount="85" Body="&lt;p&gt;I'm working on implementing &lt;a href=&quot;http://www.almaden.ibm.com/cs/projects/iis/hdb/Publications/papers/opes.pdf&quot; rel=&quot;nofollow&quot;&gt;Order Preserving Encryption for Numeric Data&lt;/a&gt;, and part of the algorithm includes approximating density of the distribution as a linear density function $f(p) = qp+r$ where $p$ exists in the distribution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the article:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If a distribution over $[0,ph)$ has the density function $qp + r$, where $p \in [0,ph)$, then for any constant $z &amp;gt; 0$, the mapping function $M(p) = z(\frac{q}{2r}p^2 + p)$ will yield a uniformly distributed set of values.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The algorithm splits up the distribution into a set of &quot;buckets&quot;, and a unique density function is needed for each bucket.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I'm working with a normal distribution over $[0,100)$ with $\mu=65$ and $\sigma=10$, split into buckets $[0,40)$, $[40, 65)$, $[65, 80)$, and $[80,100)$, how can I produce a density function for each bucket?&lt;/p&gt;&#xA;" OwnerUserId="7614" LastActivityDate="2011-02-28T21:53:20.863" Title="Approximate linear density function for a normal distribution." Tags="&lt;probability&gt;&lt;approximation&gt;&lt;probability-distributions&gt;&lt;normal-distribution&gt;" />
  <row Id="24299" PostTypeId="1" CreationDate="2011-02-28T21:57:03.637" Score="0" ViewCount="45" Body="&lt;p&gt;Let $a$ and $b$ be positive reals both greater than $1$. I'd like to compute the limit of the summation &#xA;\begin{eqnarray}&#xA;\lim_{x \to \infty} \frac{1}{\log_a x} \sum_{k = 0}^{\lfloor \log_{a} x \rfloor} \left \lbrace (\log_{b}  x) - k \log_b a \right \rbrace,&#xA;\end{eqnarray}&#xA;where $\{ \cdot  \}$ denotes the fractional part function and $\lfloor \cdot \rfloor$ denotes the floor function. I believe that the limit is $\frac{1}{2}$ for almost all $a$ and $b$, and I believe the limit doesn't exist if $b = a^r$ where $r$ is a positive rational. I also have a feeling that Weyl's theorem may be playing a role here but I'm not sure if $\lbrace (\log_{b}  x) - k \log_b a \rbrace$ is equidistributed on any interval.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any help is certainly appreciated!&lt;/p&gt;&#xA;" OwnerUserId="2720" LastEditorUserId="2720" LastEditorDisplayName="" LastEditDate="2011-02-28T22:40:15.440" LastActivityDate="2011-02-28T22:40:15.440" Title="Equidistribution and a Fractional Part Sum Limit" Tags="&lt;real-analysis&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="24300" PostTypeId="2" ParentId="24299" CreationDate="2011-02-28T22:09:59.013" Score="2" ViewCount="" Body="&lt;p&gt;Your notation is unclear, but I'll assume you meant&#xA;$$ \log_b(a^{-k}x) = \log_b(x) - k\log_b a. $$&#xA;Weyl's theorem shows that if $\log_b a^{-1}$ is irrational then $\{k\log_b a^{-1}\}$ is asymptotically equidistributed. Shifting it by $\{\log_b x\}$ shouldn't make a difference, and so the limit approaches $1/2$.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2011-02-28T22:09:59.013" CommentCount="1" />
  <row Id="24301" PostTypeId="2" ParentId="24280" CreationDate="2011-02-28T22:10:08.073" Score="4" ViewCount="" Body="&lt;p&gt;I don't see an elegant solution (and I fail to see how Arturo's comment helps immediately), but it can be done by brute force using the method of &lt;a href=&quot;http://en.wikipedia.org/wiki/Variation_of_parameters&quot; rel=&quot;nofollow&quot;&gt;variation of parameters&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The homogeneous differential equation $y&amp;#39;&amp;#39;(t) + y(t) = 0$ has the two fundamental solutions $y_{1}(t) = \sin{(t)}$ and $y_{2}(t) = \cos(t)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We must solve&#xA;\[&#xA;\begin{pmatrix}&#xA;  0 \\&#xA;  \cos(t)\cos(2t)&#xA;\end{pmatrix} =&#xA;\begin{pmatrix}&#xA;  y_{1} &amp;amp; y_{2} \\&#xA;  y_{1}^{\prime} &amp;amp; y_{2}^{\prime}&#xA;\end{pmatrix}&#xA;\begin{pmatrix}&#xA;  u_{1} \\&#xA;  u_{2}&#xA;\end{pmatrix} =&#xA;\begin{pmatrix}&#xA;  \sin{t} &amp;amp; \cos{t} \\&#xA;  -\cos{t} &amp;amp; \sin{t}&#xA;\end{pmatrix}&#xA;\begin{pmatrix}&#xA;  u_{1} \\&#xA;  u_{2}&#xA;\end{pmatrix}&#xA;\]&#xA;for $u_{1}$ and $u_{2}$ and find $U_{1}$ and $U_{2}$ such that $U_{1}^{\prime} = u_{1}$ and $U_{2}^{\prime} = u_{2}$.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A particular solution will then be given by&#xA;  \[y_{0}(t) = U_{1} y_{1} + U_{2}y_{2}\qquad\text{and}&#xA;  \qquad y=y_{0} + c_{1} y_{1} + c_{2}y_{2}\]&#xA;  will be the general solution.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Since the matrix $A(t) = \begin{pmatrix}&#xD;&#xA;  \sin{t} &amp;amp; \cos{t} \\\&#xD;&#xA;  -\cos{t} &amp;amp; \sin{t}&#xD;&#xA;\end{pmatrix}$ is orthogonal, its inverse is its transpose, so&#xA;\[&#xA;\begin{pmatrix}&#xA;  u_{1} \\&#xA;  u_{2}&#xA;\end{pmatrix}= &#xA;\begin{pmatrix}&#xA;  \sin{t} &amp;amp; -\cos{t} \\&#xA;  \cos{t} &amp;amp; \sin{t}&#xA;\end{pmatrix}&#xA;\begin{pmatrix}&#xA;  0 \\&#xA;  \cos(t)\cos(2t)&#xA;\end{pmatrix} =&#xA;\begin{pmatrix}&#xA;\cos^2(t)\cos(2t) \\&#xA;\sin(t)\cos(t)\cos(2t)&#xA;\end{pmatrix}&#xA;\]&#xA;Note that $u_{2}(t) = \sin(t)\cos(t)\cos(2t) = \frac{1}{2}\sin{(2t)}\cos{(2t)} = \frac{1}{4}\sin{(4t)}$, so we can take&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;\[U_{2}(t) = -\frac{1}{16}\cos{(4t)}.\]&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;On the other hand, using Arturo's comment three times, we get&#xA;\begin{align*}&#xA;u_{1}(t) &amp;amp; = \cos^2(t)\cos(2t) = \cos(t)\frac{1}{2}(\cos(3t)+\cos(t)) \\&#xA;&amp;amp; = \frac{1}{2} \cos{(t)}\cos(3t) + \frac{1}{2} \cos(t)\cos(t) = \frac{1}{4}&#xA;(\cos{(4t)} + \cos{(2t)}) + \frac{1}{4}(\cos{(2t)} + 1) \\&#xA;&amp;amp; = \frac{1}{4}\cos{(4t)} + \frac{1}{2} \cos{(2t)} + \frac{1}{4}.&#xA;\end{align*}&#xA;Integrating this gives&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;\[U_{1}(t) = \frac{1}{16} \sin(4t) + \frac{1}{4}\sin{(2t)} + \frac{1}{4}t.\]&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now we can simplify the particular solution &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;\begin{align*}&#xA;  y_{0} = U_{1} y_{1} + U_{2}y_{2} = \cdots = \frac{1}{4}t \sin(t) - \frac{1}{16} \cos(3t) + \frac{5}{16}\cos(t)&#xA;  \end{align*}&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;using some further trig identities (I'm really growing tired of this, sorry).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You should at least check yourself that $y_{0}^{\prime\prime} + y_{0} = \frac{1}{2}(\cos(t) + \cos(3t)) = \cos(t)\cos{(2t)}$ by Arturo's comment again.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Therefore the general solution is &#xA;  \[&#xA;  y(t) = y_{0} + c_{1} y_{1} + c_{2} y_{2} = \frac{1}{4}t \sin(t) - \frac{1}{16} \cos(3t) + \frac{5}{16}\cos(t) + c_{1}\sin{(t)} + c_{2} \cos(t)&#xA;  \]&#xA;  for some real constants $c_{1}$ and $c_{2}$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5363" LastEditorUserId="5363" LastEditorDisplayName="" LastEditDate="2011-02-28T22:34:58.580" LastActivityDate="2011-02-28T22:34:58.580" CommentCount="3" />
  <row Id="24302" PostTypeId="2" ParentId="18334" CreationDate="2011-02-28T22:19:33.203" Score="2" ViewCount="" Body="&lt;p&gt;I like most people don't read P=NP? proofs. There seems to be something missing in the current approaches.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have an approach that only lacks one step from resolving the issue. I only need to determine the complexity of determining membership in a fractal set. Decades more of work by me and I am sure I can resolve the final issues. Or not.&lt;/p&gt;&#xA;" OwnerUserId="5055" LastActivityDate="2011-02-28T22:19:33.203" />
  <row Id="24303" PostTypeId="2" ParentId="24215" CreationDate="2011-02-28T22:29:08.203" Score="1" ViewCount="" Body="&lt;p&gt;As Yuval mentioned there's no direct simple standard notation. But:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;a directed graph is a relation, and this  an be represented by a square adjacency matrix $A$, so the graph with all edges reversed has the transposed adjacency matrix $A^T$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;in category theory, for a given category $C$, the same objects and the reverse of all arrows as $C$ is called $C^{op}$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1919" LastActivityDate="2011-02-28T22:29:08.203" />
  <row Id="24304" PostTypeId="1" AcceptedAnswerId="24305" CreationDate="2011-02-28T22:41:27.470" Score="1" ViewCount="47" Body="&lt;p&gt;Let $F$ be an algebraic Number Field and $O_F$ its ring of integers.  If $\alpha \in O_F$ then is there a criterion for when $\alpha$, $\alpha&amp;#39;$ are relatively prime?\&#xA;\&#xA;I am most interested in $O_F=\mathbb{Z}[\zeta_3]$ for the time being.\&#xA;\&#xA;It seems like a possible criterion could be Let $a,b\in \mathbb{Z}$ and $\alpha=a+b\zeta$.  Then, $gcd(a,b)=1$ iff $\alpha, \alpha&amp;#39;$ are relatively prime.  Is this correct? The $\Rightarrow$ direction is most important to me.\&#xA;\&#xA;Thanks.\&#xA;\&lt;/p&gt;&#xA;" OwnerUserId="2212" LastEditorUserId="5191" LastEditorDisplayName="" LastEditDate="2011-02-28T23:05:57.970" LastActivityDate="2011-02-28T23:05:57.970" Title="Is there a criterion for when an algebraic integer and its conjugate are relatively prime?" Tags="&lt;number-theory&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="24305" PostTypeId="2" ParentId="24304" CreationDate="2011-02-28T22:48:04.967" Score="2" ViewCount="" Body="&lt;p&gt;In your case we have $\gcd(a + b \zeta, a + b \zeta^2) = \gcd(a + b \zeta, b(\zeta - 1))$. If $\gcd(a, b) = 1$ then this equals $\gcd(a + b \zeta, \zeta - 1) = \gcd(a + b, \zeta - 1)$ which equals $1$ if and only if $3 \nmid a + b$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general I am not sure what you mean by &quot;relatively prime&quot; if $\mathcal{O}_F$ is not a UFD. Do you mean that they generate the unit ideal? There are lots of ways to check this generalizing the method above. A sufficient condition is that $(\alpha, \alpha&amp;#39;) = (x, y)$ where the norms of $x, y$ are relatively prime. &lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2011-02-28T22:58:03.297" LastActivityDate="2011-02-28T22:58:03.297" />
  <row Id="24306" PostTypeId="1" AcceptedAnswerId="24313" CreationDate="2011-02-28T22:50:50.740" Score="2" ViewCount="68" Body="&lt;p&gt;Let $\mathcal{O}_K$ be the ring of integers of some number field $K$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It happens that $\mathcal{O}_K$ might not have unique factorization, but...&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;We can form the multiplicative group of ideals of $\mathcal{O}_K$&lt;/li&gt;&#xA;&lt;li&gt;It has unique factorization&lt;/li&gt;&#xA;&lt;li&gt;This construction doesn't seem to be a ring&lt;/li&gt;&#xA;&lt;li&gt;Each ideal can be put into the form $(\alpha,\beta)$ with both $\alpha,\beta \in \mathcal{O}_K$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I think the ideal $(\alpha,\beta)$ represents the gcd of $\alpha$ and $\beta$ (analogous to field of fractions) so why can't we build a new ring out of the algebraic integers which has gcd closed and unique factorization?&lt;/p&gt;&#xA;" OwnerUserId="4290" LastActivityDate="2011-02-28T23:08:48.377" Title="Ideals as an algebraic integer ring?" Tags="&lt;algebra&gt;&lt;algebraic-number-theory&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" />
  <row Id="24307" PostTypeId="2" ParentId="24306" CreationDate="2011-02-28T22:55:48.733" Score="2" ViewCount="" Body="&lt;p&gt;There's no &quot;additive inverses&quot; to ideals. However, the ideals of a ring do form a semiring - see this &lt;a href=&quot;http://mathoverflow.net/questions/26607&quot; rel=&quot;nofollow&quot;&gt;MO question&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="264" LastEditorUserId="264" LastEditorDisplayName="" LastEditDate="2011-02-28T23:04:17.933" LastActivityDate="2011-02-28T23:04:17.933" CommentCount="4" />
  <row Id="24308" PostTypeId="1" AcceptedAnswerId="24354" CreationDate="2011-02-28T22:55:56.520" Score="2" ViewCount="117" Body="&lt;p&gt;Hello,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;(1)&lt;/strong&gt; I have never studied any projective/affine geometry or algebraic curves. I'd like to see a &lt;strong&gt;clear definition of a line in the projective space $\mathbb{P}^n(k)$&lt;/strong&gt;, since I need it for my algebraic geometry study.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;(2)&lt;/strong&gt; I'm guessing a line is the variety $\mathcal{V}(f_1,\ldots,f_{n-1})$, where $f_i$ are linear homogenous polynomials, whose coefficients form a $n\times n\!-\!1$ matrix of full rank. &lt;strong&gt;Yes or no?&lt;/strong&gt;&#xA;Another guess would be, that a line in $\mathbb{P}^n(k)$ is uniquely determined by two points $a\!=\![a_0\!:\!\ldots\!:\!a_n]$, $b\!=\![b_0\!:\!\ldots\!:\!b_n]$, such that the matrix $\begin{bmatrix} a \\ b \end{bmatrix}$ is of rank $2$. But &lt;strong&gt;how is such a line parametrized&lt;/strong&gt;? Is any of my two attempts of a definition correct?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;(3)&lt;/strong&gt; What are the &lt;strong&gt;defining equations of two intersecting lines in $\mathbb{P}^3$&lt;/strong&gt;? And now, most importantly: how can I &lt;strong&gt;compute the Hilbert polynomial&lt;/strong&gt; of such a variety?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For such an elementary concept, one would expect it to be the first object defined, but to my annoyance and frustration, I have yet to see an official definition. I have the book Introduction to Algebraic Geometry (Hassett), as well as Algebraic Curves (Fulton) as my main source. Any references would be highly desirable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;thank you&lt;/p&gt;&#xA;" OwnerUserId="7325" LastActivityDate="2011-03-01T18:43:05.780" Title="what is the definition of a line in $\mathbb{P}^n(k)$ + how to compute the hilbert polynomial of two intersecting lines?" Tags="&lt;algebraic-geometry&gt;&lt;commutative-algebra&gt;&lt;projective-geometry&gt;&lt;projective-space&gt;" AnswerCount="3" />
  <row Id="24310" PostTypeId="1" AcceptedAnswerId="24330" CreationDate="2011-02-28T23:02:28.160" Score="3" ViewCount="83" Body="&lt;p&gt;I am giving a talk on Euler's proof that $X^3+Y^3=Z^3$ has no solutions in positive integers.  Some facts that I believe to be true are the following.  For some I give proof.  Please verify that my reasoning is correct and make any pertinent comments.  I use the notation $\zeta=\zeta_3$.&lt;br/&gt;&#xA;&lt;br/&gt;&#xA;(a) $\mathbb{Q}(\sqrt{-3})=\mathbb{Q}(\zeta)$&lt;br/&gt;&#xA;Proof:   Note that $\{1, \zeta\}$ and $\{1, \frac{1+\sqrt{-3}}{2}\}$ are bases for $\mathbb{Z}[\zeta]$, $\mathbb{Z}[\sqrt{-3}]$, respectively.  If $\alpha \in \mathbb{Z}[\zeta]$, then, $\alpha=a+b\zeta=a-b+b+b\zeta=(a-b)+b(1+\zeta)=(a-b)+b(\frac{1+\sqrt{-3}}{2})\in \mathbb{Z}[\sqrt{-3}]$.  Similarly, if $\alpha \in \mathbb{Z}[\sqrt{-3}]$, then $\alpha=a+b(\frac{1+\sqrt{-3}}{2})=a+b-b+b(\frac{1+\sqrt{-3}}{2})=(a+b)+b\zeta \in \mathbb{Z}[\zeta]$.&lt;br/&gt;&#xA;&lt;br/&gt;&#xA;(b) $O=\mathbb{Z}[\frac{1+\sqrt{-3}}{2}]$.  Proof:  This follows from (a).&lt;br/&gt;&#xA;&lt;br/&gt;&#xA;(c) Ord$_3(1-\zeta)=1/2$ makes sense and is well defined.&lt;br/&gt;&#xA;Here is my reasoning for this.  $1=$Ord$_3(3)=$Ord$_3((1-\zeta)(1-\zeta'))=$Ord$_3(1-\zeta)+$Ord$_3(1-\zeta')$.  There is a way around using this in the proof, but this is kinda cool so Im interested.&lt;br/&gt;&#xA;&lt;br/&gt;&#xA;(d) The conjugate of $a-\zeta b$ is $a-\zeta&amp;#39;b$.  Proof:  Since the conjugate of products is the product of conjugates and the conjugate of sums is the sum of the conjugates, this follows.&lt;br/&gt;&lt;/p&gt;&#xA;" OwnerUserId="2212" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2011-02-28T23:19:02.873" LastActivityDate="2011-03-01T00:44:13.103" Title="Some questions about $\mathbb{Z}[\zeta_3]$" Tags="&lt;number-theory&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="24312" PostTypeId="1" AcceptedAnswerId="24315" CreationDate="2011-02-28T23:03:39.143" Score="1" ViewCount="38" Body="&lt;p&gt;How can I express that if an expression evaluates to a negative number, I assign it a value of zero? Is it possible to do this without repeating the function twice?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is an example&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f(\vec{\phi})=\begin{cases}&#xD;&#xA; f(\vec{\phi}) &amp;amp; \text{for}\; f(\vec{\phi}) \gt 0 \\\&#xD;&#xA; 0  &amp;amp; \text{for}\; f(\vec{\phi})\lt 0.&#xD;&#xA;\end{cases}$$&lt;/p&gt;&#xA;" OwnerUserId="3733" LastEditorUserId="5363" LastEditorDisplayName="" LastEditDate="2011-02-28T23:26:06.353" LastActivityDate="2011-02-28T23:27:14.417" Title="Is there an clear way to state truncation?" Tags="&lt;functions&gt;&lt;notation&gt;" AnswerCount="1" />
  <row Id="24313" PostTypeId="2" ParentId="24306" CreationDate="2011-02-28T23:03:47.527" Score="4" ViewCount="" Body="&lt;p&gt;In some sense, we can; I think this is what the ring of integers in the &lt;a href=&quot;http://en.wikipedia.org/wiki/Hilbert_class_field&quot; rel=&quot;nofollow&quot;&gt;Hilbert class field&lt;/a&gt; does.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I don't think this is the right way to think about the move from elements to ideals in general. The point of passing to ideals is to abstract out the main property we want out of divisibility: $m | n$ if and only if the ideal $(m)$ contains the ideal $(n)$. So the natural structure on ideals is as a lattice ordered by inclusion, and it just happens to be a happy fact about Dedekind domains that this lattice is isomorphic to a product of copies of $\mathbb{N}$, one for each prime ideal. In general the order structure on ideals is much more complicated and the idea that one can think about ideals as generalized elements breaks down (e.g. try to apply this philosophy to $F[x, y]$ for $F$ a field). &lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2011-02-28T23:08:48.377" LastActivityDate="2011-02-28T23:08:48.377" CommentCount="0" />
  <row Id="24315" PostTypeId="2" ParentId="24312" CreationDate="2011-02-28T23:12:14.320" Score="4" ViewCount="" Body="&lt;p&gt;It sounds like you are looking for $f^+(x) = \frac{f(x) + |f(x)|}{2}$. Positive values will evaluate to $f(x)$, but negative values will be taken to $0$, but this also repeats the expression so I hope that you were merely looking to avoid a piecewise definition.&lt;/p&gt;&#xA;" OwnerUserId="557" LastEditorUserId="557" LastEditorDisplayName="" LastEditDate="2011-02-28T23:27:14.417" LastActivityDate="2011-02-28T23:27:14.417" CommentCount="2" />
  <row Id="24316" PostTypeId="2" ParentId="24292" CreationDate="2011-02-28T23:24:52.707" Score="2" ViewCount="" Body="&lt;p&gt;For the first question: one situation in which other representations have been very useful is in finite group theory and finite geometry.  There is a very close connection between certain finite groups and certain finite geometries, and so both directions are useful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many simple groups were discovered from their (finite) geometric properties.  For instance the &lt;a href=&quot;http://en.wikipedia.org/wiki/Mathieu_group&quot; rel=&quot;nofollow&quot;&gt;Matheiu groups&lt;/a&gt; can be understood as (derivatives of) automorphism groups of &lt;a href=&quot;http://en.wikipedia.org/wiki/Steiner_system&quot; rel=&quot;nofollow&quot;&gt;Steiner systems&lt;/a&gt;, the finite projective &lt;a href=&quot;http://en.wikipedia.org/wiki/Classical_group&quot; rel=&quot;nofollow&quot;&gt;classical groups&lt;/a&gt; are all automorphism groups of &lt;a href=&quot;http://en.wikipedia.org/wiki/Finite_geometry&quot; rel=&quot;nofollow&quot;&gt;finite geometries&lt;/a&gt; such as &lt;a href=&quot;http://en.wikipedia.org/wiki/Projective_geometry&quot; rel=&quot;nofollow&quot;&gt;projective spaces&lt;/a&gt;, the sporadic group &lt;a href=&quot;http://en.wikipedia.org/wiki/Higman-Sims_group&quot; rel=&quot;nofollow&quot;&gt;HS&lt;/a&gt; is the automorphism group of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Higman-Sims_graph&quot; rel=&quot;nofollow&quot;&gt;Higman–Sims graph&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To understand &lt;a href=&quot;http://en.wikipedia.org/wiki/Non-Desarguesian_projective_plane&quot; rel=&quot;nofollow&quot;&gt;some fancy projective planes&lt;/a&gt;, Dickson and Zassenhaus classified all finite &lt;a href=&quot;http://en.wikipedia.org/wiki/Near-field_(mathematics)&quot; rel=&quot;nofollow&quot;&gt;near-fields&lt;/a&gt;.  Later Suzuki classified all &lt;a href=&quot;http://en.wikipedia.org/wiki/Zassenhaus_group&quot; rel=&quot;nofollow&quot;&gt;Zassenhaus groups&lt;/a&gt;, which look a lot like the automorphism groups of projective planes, but also included an infinite family of new simple groups, the &lt;a href=&quot;http://en.wikipedia.org/wiki/Suzuki-Ree_groups&quot; rel=&quot;nofollow&quot;&gt;Suzuki groups&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many investigations on the structure of permutation groups quickly inherit a geometric feel, and many investigations in finite geometry require delicate classifications.  Many of the papers of William Kantor represent these two ideas quite well.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;For the second question, I think it depends on what you want to use the new construction for.  One nice effect of semi-direct products is it allows one to consider any group acting on any other group as if the first is acting by conjugation.  This means you can use strange things like Sylow's theorem in either the actor or the actee.  It seems like such an advantage is lost if the actor and the actee are different sorts of objects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, you can take a semi-direct product of a matrix group and a vector space, and this is quite powerful.  It allows you to use group theory to understand a representation, and representation theory to understand a group.  Again, it works out nicely because both the group and the vector space are matrices, so you have a common ground to work with both of them.&lt;/p&gt;&#xA;" OwnerUserId="583" LastActivityDate="2011-02-28T23:24:52.707" CommentCount="3" />
  <row Id="24317" PostTypeId="2" ParentId="24284" CreationDate="2011-02-28T23:34:16.510" Score="3" ViewCount="" Body="&lt;p&gt;When I was teaching predicate calculus in the introductory course last semester (Well, I was a TA but still...) we took the following approach:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Define &lt;strong&gt;terms&lt;/strong&gt; and &lt;strong&gt;formulae&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A term is practically a name for some element of the interpretation (the structure), either a free variable, or a constant in the language and if $F$ is a function and $t_1,\ldots ,t_k$ are terms then $F(t_1,\ldots ,t_k)$ is also a term.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, formulae have the familiar structure. You start with atomic formulae which are relations (either equality or within the language) and terms and you say that if you have two formulae then the conjugation, disjunction, negation and implication give you a new formula, as well quantification of $\exists$ and $\forall$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the formula $\varphi(x,y)\colon = y=f(x)$ is a formula which receives a TRUE value whenever $y$ is assigned to be $f^M(x)$, it is an atomic formula - namely the equality of two terms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now considering the $f$ and $g$ that you have presented above. You will need to have $2$ as a constant in your language, as well the functions of multiplication and addition. Then you want to require your structure to satisfy the axioms that $x+x = 2x$, namely the usual ring axioms (well, you want $\mathbb{R}$ to behave normally) and then you can deduce that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The lines &quot;let $y = f(x)$&quot; and &quot;let $z=g(x)$&quot; then &quot;$z=y$&quot; can be formulated as follow:&#xA;$$\forall x\forall y\forall z((y=f(x) \wedge z=g(x))\rightarrow z=y)$$&#xA;Or more explicitly:&#xA;$$\forall x\forall y\forall z((y=2\cdot x\wedge z=x+x)\rightarrow z=y)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Addendum&lt;/strong&gt;:&lt;br&gt;&#xA;Another way to look at that is that by definibility. Your language has the addition and multiplication, and $0$, and $1$ as constants.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With that you can define a formula $\varphi(x,y)$ which is only satisfied when $y=f(x)$. In this sense, we have enriched the language with new function symbols - $f$ and $g$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now when you say &quot;let $y=f(x)$ and $z=g(x)$&quot; you enrich the language with three other constants, named $x,y,z$, and you add an axiom to you theory &quot;$y=f(x)$&quot; and &quot;$z=g(x)$&quot;. Now you claim that from your theory you can deduce that the new constants $y$ and $z$ are interpreted as the same element in the structure.&lt;/p&gt;&#xA;" OwnerUserId="622" LastEditorUserId="622" LastEditorDisplayName="" LastEditDate="2011-03-01T12:55:43.617" LastActivityDate="2011-03-01T12:55:43.617" />
  <row Id="24318" PostTypeId="1" CreationDate="2011-02-28T23:34:44.293" Score="2" ViewCount="114" Body="&lt;p&gt;Given an algebraic monoid $M$ (by which I mean a monoid in the category of schemes), one has  the group $M^\times$ of invertible elements (which by general abstract nonsense is also a scheme, see my comment below), which comes with a map of algebraic monoids $M^\times \to M$. Is this map an open immersion?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have only the following evidence to work from: one knows that the general linear group over $\mathbb{C}$ is open in the monoid of matrices (with the usual topology), and more generally the general linear group $GL_n$ is Zariksi open in all matrices $M_n$. One can then consider subvarieties of $M_n$ which are also submonoids, and the invertible elements thereof should again be Zariski open.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unless I've got things terribly wrong, the inclusion $M^\times \hookrightarrow M$ is a monomorphism of schemes (it's described as a certain equaliser), and so I suppose we just need to show it's smooth. But my algebraic geometry is non-existent, so help, please!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: I've added a bounty, in case anyone can sharpen Matt E's result, in other words, work out if we need to restrict to smooth schemes or not.&lt;/p&gt;&#xA;" OwnerUserId="3835" LastEditorUserId="3835" LastEditorDisplayName="" LastEditDate="2011-03-15T08:10:15.437" LastActivityDate="2011-03-15T08:10:15.437" Title="Is the inclusion of the group of units in an algebraic monoid an open immersion?" Tags="&lt;algebraic-geometry&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="24319" PostTypeId="1" CreationDate="2011-02-28T23:38:25.110" Score="1" ViewCount="75" Body="&lt;p&gt;Okay, this is really bugging me:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My Math book has this practice where I need to get the area and perimeter of the next Circular Crown:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;R = 3cm , r = 1.75cm&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Well, I do it. But my results are simply different than the ones the book tells me (on the back of the book, there is an answers sheet). Why? The process to get the area and perimeter are relatively simple. The instructions are on the book. I follow them, and I fail!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, my Math teacher told me it was because &lt;em&gt;[put reason here]&lt;/em&gt;, but it was ages ago, and I can't remember. I remember the book &lt;em&gt;is right&lt;/em&gt;, but  I forgot why and how.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The book says:&#xA;Area is 5.94π^2&#xA;Perimeter is 9.5π&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While I get:&#xA;Area is (95/16)π&#xA;Perimeter is (19/2)π&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What am I doing wrong?&lt;/p&gt;&#xA;" OwnerUserId="5775" LastActivityDate="2011-02-28T23:38:25.110" Title="Getting a Circular Crown's area and perimeter." Tags="&lt;circle&gt;" CommentCount="5" />
  <row Id="24320" PostTypeId="2" ParentId="24294" CreationDate="2011-02-28T23:46:25.940" Score="2" ViewCount="" Body="&lt;p&gt;You can read off the Christoffel symbols from the Euler–Lagrange equations for the Lagrangian naturally defined by the metric, $L=\frac12 g_{ij} v^i v^j$. Still messy, but at least there's not much you have to memorize in order to do the computation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can't recall a good reference right off the bat, but Google found this, for example:&#xA;&lt;a href=&quot;http://count.ucsc.edu/~rmont/classes/121A/polarChristoffelE_Lag.pdf&quot; rel=&quot;nofollow&quot;&gt;http://count.ucsc.edu/~rmont/classes/121A/polarChristoffelE_Lag.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2011-02-28T23:46:25.940" />
  <row Id="24321" PostTypeId="1" CreationDate="2011-02-28T23:52:23.577" Score="2" ViewCount="57" Body="&lt;p&gt;I'm trying to figure out the probability of a 3rd failure occurring on the 5th attempt of doing something. Let's just call the probability of success of failure P(S) or P(F), I won't put numbers as I want to actually learn.&lt;/p&gt;&#xA;" OwnerUserId="7625" LastEditorUserId="7625" LastEditorDisplayName="" LastEditDate="2011-02-28T23:58:11.997" LastActivityDate="2011-03-01T00:11:36.910" Title="Probability of nth failure on nth attempt" Tags="&lt;homework&gt;&lt;probability&gt;" AnswerCount="2" />
  <row Id="24322" PostTypeId="1" AcceptedAnswerId="24331" CreationDate="2011-02-28T23:57:01.427" Score="3" ViewCount="34" Body="&lt;p&gt;Question as in title. I only really need the special case where one of the open embeddings is the identity, but the more general case would be useful to know.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit - By product I mean: given $U\to X$ and $V\to Y$ open embeddings, is the canonical map $U\times V \to X\times Y$ an open embedding?&lt;/p&gt;&#xA;" OwnerUserId="3835" LastEditorUserId="3835" LastEditorDisplayName="" LastEditDate="2011-03-01T00:38:03.373" LastActivityDate="2011-03-01T00:42:07.007" Title="Is the product of two open embeddings of schemes an open embedding?" Tags="&lt;algebraic-geometry&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="24323" PostTypeId="1" AcceptedAnswerId="24363" CreationDate="2011-02-28T23:57:16.777" Score="1" ViewCount="75" Body="&lt;p&gt;I realize this is a novice question, but I'm stuck here. Suppose you have an ordered basis $B = (v_1, v_2, v_3, v_4)$ of $R^4$. My problem is how to determine a system of equations that defines, relatively to the basis $B$, the linear subspace of $R^4$ generated by the vectors $v_1-v_2+v_4$ and $-v_1+3v_3$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again, I realize this is a novice question, but I'd appreciate some insight into it...&#xA;Thanks in advance.&lt;/p&gt;&#xA;" OwnerUserId="6666" LastActivityDate="2011-03-01T05:18:36.633" Title="System of equations for a subspace" Tags="&lt;linear-algebra&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="24324" PostTypeId="1" AcceptedAnswerId="24327" CreationDate="2011-03-01T00:01:02.907" Score="3" ViewCount="78" Body="&lt;p&gt;I'm trying to solve an equation for B where&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$2 B e^{(1-B)}=1$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I plugged it into wolfram alpha and got:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/n8t6w.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is that $W_n$ mean? How do I intrepret this answer?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Feel free to retag it, I wasn't entirely sure what it falls under!)&lt;/p&gt;&#xA;" OwnerUserId="2333" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2011-03-01T00:36:07.807" LastActivityDate="2011-03-01T00:36:07.807" Title="What is $W_n ?$" Tags="&lt;terminology&gt;" AnswerCount="1" CommentCount="11" />
  <row Id="24325" PostTypeId="2" ParentId="24321" CreationDate="2011-03-01T00:02:42.727" Score="1" ViewCount="" Body="&lt;p&gt;HINTS: &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Can you imagine a sequence of successes and failures which would involve the 3rd failure occurring on the 5th attempt.  What is the probability of this sequence?&lt;/li&gt;&#xA;&lt;li&gt;What is the probability of another sequence involving the 3rd failure occurring on the 5th attempt?&lt;/li&gt;&#xA;&lt;li&gt;How many sequences are there involving the 3rd failure occurring on the 5th attempt? (You could count them as it is certainly in single figures, or you might find an expression)&lt;/li&gt;&#xA;&lt;li&gt;How do you combine the probabilities of the different sequences?     &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="6460" LastActivityDate="2011-03-01T00:02:42.727" CommentCount="9" />
  <row Id="24326" PostTypeId="2" ParentId="24321" CreationDate="2011-03-01T00:05:13.637" Score="1" ViewCount="" Body="&lt;p&gt;See &lt;a href=&quot;http://en.wikipedia.org/wiki/Negative_binomial_distribution&quot; rel=&quot;nofollow&quot;&gt;Negative Binomial Distribution&lt;/a&gt; (the section &quot;Alternative formulations&quot;).&lt;/p&gt;&#xA;" OwnerUserId="2810" LastEditorUserId="2810" LastEditorDisplayName="" LastEditDate="2011-03-01T00:11:36.910" LastActivityDate="2011-03-01T00:11:36.910" />
  <row Id="24327" PostTypeId="2" ParentId="24324" CreationDate="2011-03-01T00:10:55.643" Score="6" ViewCount="" Body="&lt;p&gt;$W(z)$ is the &lt;a href=&quot;http://mathworld.wolfram.com/LambertW-Function.html&quot; rel=&quot;nofollow&quot;&gt;Lambert $W$ function&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$W_k(-\frac{1}{2e})$ where $k \in \mathbb{Z}$ denotes the $k^{th}$ root of the equation $xe^x = -\frac{1}{2e}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/zW3s9.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2011-03-01T00:27:13.447" LastActivityDate="2011-03-01T00:27:13.447" CommentCount="4" />
  <row Id="24328" PostTypeId="2" ParentId="24323" CreationDate="2011-03-01T00:11:53.600" Score="1" ViewCount="" Body="&lt;p&gt;Relative to $B$, the given vectors have coordinates $(1,-1,0,1)$ and $(-1,0,3,0)$.&lt;/p&gt;&#xA;" OwnerUserId="589" LastActivityDate="2011-03-01T00:11:53.600" />
  <row Id="24329" PostTypeId="2" ParentId="24278" CreationDate="2011-03-01T00:32:57.793" Score="2" ViewCount="" Body="&lt;p&gt;Thanks for clarifying your question. Given an isometry of only the surface (which say does not extend to the ambient space) it shouldn't be immediately clear that covariant differentiation, given by your formula, commutes with the isometry.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason for this is you're saying that covariant differentiation is &lt;em&gt;intrinsic&lt;/em&gt; -- this is a key step in Gauss's Theorema Egregium.   Once you know covariant differentiation is intrinsic, then you have a clear attack at proving the Riemann curvature tensor is intrinsic, so Gauss curvature is intrinsic.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Usually in a differential geometry course this is where you make a big deal about the fact that a Riemann manifold has a unique Levi-Civita connection and your formula for the covariant derivative is one manifestation of it. This sometimes goes under the name of &lt;em&gt;the fundamental theorem of differential geometry&lt;/em&gt; so it's not a trivial thing. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think there's a variety of alternative ways of proving what you want but they all require some kind of tool such as the above argument.  I suppose you could reduce it to a linear-algebra argument if you take the next step to view your formula for the covariant derivative as a cute formula for an Ehresmann connection (this starts to relate your covariant derivative to an intrinsic language).  You would then have to ensure that your connection and the pull-back connection (by your isometry) are the same -- in effect you'd be encoding the &quot;fundamental theorem&quot;'s proof but in the language of Ehresmann connections. &lt;/p&gt;&#xA;" OwnerUserId="642" LastActivityDate="2011-03-01T00:32:57.793" />
  <row Id="24330" PostTypeId="2" ParentId="24310" CreationDate="2011-03-01T00:37:57.243" Score="3" ViewCount="" Body="&lt;p&gt;b) does not follow from a). By definition, for $K$ a number field, $\mathcal{O}_K$ is the set of algebraic integers in $K$. You have shown, at best, that the ring of algebraic integers in $K = \mathbb{Q}(\zeta)$ contains $\mathbb{Z}[\zeta]$, but you have not shown that this is all of $\mathcal{O}_K$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a complete proof of b) although, again, this is unnecessary since all you need for this proof is that $\mathbb{Z}[\zeta]$ has unique prime factorization. Let $a + b \zeta$ be an algebraic integer in $K$. Then its conjugate is $a + b \zeta^2$, hence its trace is $a - b$, which must be an integer. After multiplying by $\zeta$ we get $a \zeta + b \zeta^2 = -b + (a-b) \zeta$. Since this is also an algebraic integer, its trace $-b - (a-b) = -a$ must also be an integer, hence $a, b$ are both integers. On the other hand any element of $K$ of the form $a + b \zeta, a, b \in \mathbb{Z}$ is an algebraic integer, so $\mathcal{O}_K = \mathbb{Z}[\zeta]$ as desired.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;c) is not a good idea. The correct definition is this: for a prime ideal $P$ in a Dedekind domain $D$, there is a &lt;a href=&quot;http://en.wikipedia.org/wiki/Discrete_valuation&quot; rel=&quot;nofollow&quot;&gt;discrete valuation&lt;/a&gt; $\nu_P$ defined as follows: if $x \in \mathcal{O}_K$, then let $\nu_P(x)$ be the greatest power of $P$ that divides the ideal $(x)$. So in this case the relevant prime ideal is $P = (1 - \zeta)$ since $(3)$ is not a prime ideal, and one has $\nu_P(1 - \zeta) = 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Associated to any discrete valuation is an &lt;a href=&quot;http://en.wikipedia.org/wiki/Absolute_value&quot; rel=&quot;nofollow&quot;&gt;absolute value&lt;/a&gt; $2^{-\nu_P(x)}$ on $K$, and the reason you do not want to try defining valuations with respect to non-prime ideals is that the corresponding absolute value will not satisfy the triangle inequality. In this case you are lucky because $3$ ramifies (it is the square of a prime ideal) but, for example, $7 = (2 - \zeta)(2 - \zeta^2)$ does not ramify and the valuations $\nu_{2 - \zeta}, \nu_{2 - \zeta^2}$ do not coincide, so the naive definition of $\nu_7$ will not satisfy the triangle inequality. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;d) depends on your definition of &quot;conjugate.&quot; &lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2011-03-01T00:44:13.103" LastActivityDate="2011-03-01T00:44:13.103" CommentCount="6" />
  <row Id="24331" PostTypeId="2" ParentId="24322" CreationDate="2011-03-01T00:42:07.007" Score="4" ViewCount="" Body="&lt;p&gt;Yes. In fact, this can be seen as follows. Open immersions are preserved under base-change; thus $U \times V \to X \times V$ is an open immersion; similarly so is $X \times V \to X \times Y$; thus the composition is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most of the nice properties of schemes (immersions, separatedness, properness, finite type, quasicompactness, quasiseparatedness, smoothness, flatness, quasifiniteness, surjectivity, radicialness) are preserved the basic operations of base-change and composition. This implies that they are preserved under products as you ask. (Much of made of this in EGA.)&lt;/p&gt;&#xA;" OwnerUserId="536" LastActivityDate="2011-03-01T00:42:07.007" CommentCount="1" />
  <row Id="24332" PostTypeId="1" AcceptedAnswerId="24334" CreationDate="2011-03-01T00:53:11.103" Score="5" ViewCount="115" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Prove that $gcd(a + a&amp;#39;, b + b&amp;#39;) = 1$ if $ab - a&amp;#39;b&amp;#39; = \pm 1$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My attempt was:&lt;br&gt;&#xA;Case 1:&lt;br&gt;&#xA;$ab - a&amp;#39;b&amp;#39; = 1 \implies \gcd(a, b&amp;#39;) = 1$ and $\gcd(a&amp;#39;, b) = 1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then is it sufficient to conclude that $\gcd(a + a&amp;#39;, b + b&amp;#39;) = 1$? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore, when we write $\pm 1$, does it mean &lt;code&gt;or&lt;/code&gt; or &lt;code&gt;and&lt;/code&gt;?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks,  &lt;/p&gt;&#xA;" OwnerUserId="5711" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-01T04:43:17.867" LastActivityDate="2011-03-01T05:06:25.310" Title="Prove that $\gcd( a + a', b + b' ) = 1$ if $ab - a'b' = \pm 1$" Tags="&lt;elementary-number-theory&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="2" />
  <row Id="24334" PostTypeId="2" ParentId="24332" CreationDate="2011-03-01T01:03:37.830" Score="7" ViewCount="" Body="&lt;p&gt;No, $\text{gcd}(a,b) = 1$ and $\text{gcd}(x,y) = 1$ does not imply $\text{gcd}(a+x, b+y) = 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, $\text{gcd}(a,b) = \text{gcd}(b,a) = 1$ does not mean $\text{gcd}(a+b, a+b) = 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The $\pm 1$ means OR.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint: to the question in title:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you find $x$ and $y$ so that $(a+a&amp;#39;)x - (b+b&amp;#39;)y = ab - a&amp;#39;b&amp;#39;$?&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2011-03-01T01:21:51.730" LastActivityDate="2011-03-01T01:21:51.730" CommentCount="4" />
  <row Id="24335" PostTypeId="2" ParentId="24308" CreationDate="2011-03-01T01:17:18.770" Score="4" ViewCount="" Body="&lt;p&gt;The line through $(a_0 : ... : a_n)$ and $(b_0 : ... : b_n)$ is uniquely parameterized by $(a_0 X + b_0 Y : a_1 X + b_1 Y : ... a_n X + b_n Y)$. Note that this precisely describes a morphism $\mathbb{P}^1 \to \mathbb{P}^n$ which is an isomorphism onto its image. I am surprised this is not given somewhere in Fulton. &lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2011-03-01T01:17:18.770" CommentCount="1" />
  <row Id="24336" PostTypeId="1" AcceptedAnswerId="24341" CreationDate="2011-03-01T01:49:14.607" Score="1" ViewCount="150" Body="&lt;p&gt;Explain $-\int_{b}^{a}\frac{Q}{4 \pi \epsilon \bar{r}^{2}} \cdot d \bar{r}= \left[\frac{Q}{4\pi \epsilon \bar{r}}\right]_{b}^{a}$. What is the term $\frac{1}{\bar{r}^{2}} \cdot d\bar{r}$? I find it hard to think about it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$r$ is a radius of a ball.&lt;/p&gt;&#xA;" OwnerUserId="7628" LastEditorUserId="7628" LastEditorDisplayName="" LastEditDate="2011-03-01T02:21:50.990" LastActivityDate="2011-03-01T13:42:28.357" Title="Explain $-\int^{a}_{b}\frac{Q}{4 \pi \epsilon \bar{r}^{2}} \cdot d \bar{r}= \left[\frac{Q}{4\pi \epsilon \bar{r}}\right]^{a}_{b}$" Tags="&lt;integral&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="24337" PostTypeId="1" AcceptedAnswerId="24340" CreationDate="2011-02-28T21:44:53.353" Score="3" ViewCount="96" Body="&lt;p&gt;hi, i'm trying to solve this question:&#xA;Given a turing machine that is decidable by at most 50 * n^4 steps, can we build a dif algorithm that can decide it in n^4 steps?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Me and my friends thought about it, and we coulnd't get it right.&#xA;Some points we had in mind:&#xA;1. if you can reduce the overall cost by 50 somehow, couldn't you reduce it over and over again?&#xA;2. we tried to think about algorithms that require 50*n^4 moves (at most) and we thought about this language: string str is in A if str.GetHashCode().GetHashCode()... 50 times is even. do you think this algorithm is unreducable?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;thanks for the help!&lt;/p&gt;&#xA;" OwnerUserId="7622" LastActivityDate="2011-03-01T02:19:08.840" Title="Reducing the complexity of a turing machine algorithm" Tags="&lt;computability&gt;" AnswerCount="1" CommentCount="14" />
  <row Id="24338" PostTypeId="1" CreationDate="2011-03-01T02:05:02.707" Score="1" ViewCount="136" Body="&lt;p&gt;If $H$ is a Hilbert space of entire functions with weighted norm $||f||^{2}=\int_{R} |\frac{f(t)}{g(t)}|^{2}dt$ for some entire function $g$ (not necessary in $H$). Can we find any relation between the norm of $f$ and the norm of it's derivative? Something like:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$||f&amp;#39;||\leq C ||f||$ for some constant $C$. (Note: so far we don't know whether $f&amp;#39;$ belongs to $H$ or not).&lt;/p&gt;&#xA;" OwnerUserId="7630" LastEditorUserId="2548" LastEditorDisplayName="" LastEditDate="2011-03-01T04:16:12.433" LastActivityDate="2011-03-01T10:34:11.910" Title="Hilbert Space - Norm of derivative" Tags="&lt;functional-analysis&gt;" AnswerCount="2" />
  <row Id="24339" PostTypeId="1" AcceptedAnswerId="24343" CreationDate="2011-03-01T02:09:04.233" Score="3" ViewCount="47" Body="&lt;p&gt;Let $D^n$ be the closed unit disk in $\mathbb{R}^n$. I'm looking for a homeomorphism of $D^n$ to itself which is the identity on the boundary, and takes the origin to a given point $z$ in its interior. I can visualize such a function as &quot;shifting around the mass&quot; within $D^n$, but I can't seem to write one down.&lt;/p&gt;&#xA;" OwnerUserId="5290" LastActivityDate="2011-03-01T02:47:05.993" Title="Autohomeomorphism of the unit disk which is constant on the boundary, and which takes 0 to a prescribed point" Tags="&lt;general-topology&gt;" AnswerCount="1" />
  <row Id="24340" PostTypeId="2" ParentId="24337" CreationDate="2011-03-01T02:19:08.840" Score="2" ViewCount="" Body="&lt;p&gt;You can find the answer on &lt;a href=&quot;http://en.wikipedia.org/wiki/Linear_speedup_theorem&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;. If you don't want to look it up, think how can you build a TM then executes $50$ steps of the old machine at once; you may need some bootstrapping phase.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2011-03-01T02:19:08.840" CommunityOwnedDate="2011-03-01T02:19:08.840" />
  <row Id="24341" PostTypeId="2" ParentId="24336" CreationDate="2011-03-01T02:24:38.447" Score="4" ViewCount="" Body="&lt;p&gt;In rigorous mathematics, one does not usually write this as you did with a multiplication dot, as the symbol $\mathrm{d}\bar{r}$ is considered as a formal part of the notation of the integral (indicating the variable of integration) and not as a factor to be multiplied. However, this integral comes from physics, and in physics it's quite usual to treat expressions like $\mathrm{d}\bar{r}$ as if they were &quot;infinitesimal quantities&quot; which can be manipulated algebraically much like ordinary quantities (though not exactly like them). In fact, this was also the viewpoint taken by many in mathematics before analysis was put on a rigorous basis with definitions that don't use such &quot;infinitesimal quantities&quot;. However, there's a branch of mathematics called &lt;a href=&quot;http://en.wikipedia.org/wiki/Non-standard_analysis&quot; rel=&quot;nofollow&quot;&gt;non-standard analysis&lt;/a&gt; that does deal with basing analysis on a formal definition of what such &quot;infinitesimal quantities&quot; mean.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Having said all that, in the field where this integral occurs, in physics, $\mathrm{d}\bar{r}$ is often considered as an infinitesimal change in the radius $\bar{r}$, and then $\frac{1}{\bar{r}^2}\mathrm{d}\bar{r}$ means the product of that change with the inverse square of the radius. That is, the integral is considered as &quot;summing&quot; over a large/infinite number of small/infinitesimal radial intervals, and for each interval the product of the change in radius on that interval, $\bar{r}$, with (some proportionality constant times) the inverse square $\frac{1}{\bar{r}^2}$ of the radius is added to the value of the integral. Does that answer your question?&lt;/p&gt;&#xA;" OwnerUserId="6622" LastEditorUserId="6622" LastEditorDisplayName="" LastEditDate="2011-03-01T02:29:50.393" LastActivityDate="2011-03-01T02:29:50.393" CommentCount="7" />
  <row Id="24342" PostTypeId="1" CreationDate="2011-03-01T02:37:59.260" Score="9" ViewCount="269" Body="&lt;p&gt;The upper bound for the number of moves required to solve a regular Rubik's cube has been shown to be &lt;a href=&quot;http://www.cube20.org/&quot; rel=&quot;nofollow&quot;&gt;20&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Two questions come to mind:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Does this result have more general significance?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;What are the most pressing issue with regards to Rubik's cube (or generalizations) or its group?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="2809" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2011-05-01T12:58:03.350" LastActivityDate="2011-05-31T14:07:15.727" Title="Rubik's cube interesting questions?" Tags="&lt;combinatorics&gt;&lt;group-theory&gt;&lt;recreational-mathematics&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" />
  <row Id="24343" PostTypeId="2" ParentId="24339" CreationDate="2011-03-01T02:47:05.993" Score="4" ViewCount="" Body="&lt;p&gt;Map the rays starting at $z$ to the rays starting at the origin. Each ray is identified by its point on the boundary.&lt;/p&gt;&#xA;" OwnerUserId="589" LastActivityDate="2011-03-01T02:47:05.993" CommentCount="1" />
  <row Id="24344" PostTypeId="1" AcceptedAnswerId="24367" CreationDate="2011-03-01T02:51:09.467" Score="4" ViewCount="129" Body="&lt;p&gt;Hello. I'm reading the proof of the following fact:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let A be a ring, a an ideal of A, M an A-module. Then $(A/a) \otimes_{A} M$ is isomorphic to M/aM.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the solution is to tensor the exact sequence $0 \rightarrow a \rightarrow A \rightarrow A/a \rightarrow 0$ with M.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This induces an exact sequence:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a \otimes_{A} M \rightarrow A \otimes_{A} M \rightarrow (A/a) \otimes_{A} M \rightarrow 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is $a \otimes_{A} M \cong aM$ ? Why is this &quot;clear and obvious&quot; i.e how do we see it quickly?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I just started reading about the tensor product of modules and have trouble seeing why such isomorphisms are natural, I'm sure there's a &quot;quick&quot; way to see them, so why this is clear? is there any other way to think about the tensor product besides been the quotient of a free module by a set of generators?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="6495" LastEditorUserId="325" LastEditorDisplayName="" LastEditDate="2011-03-01T03:39:20.977" LastActivityDate="2011-03-01T06:40:25.190" Title="Help understanding tensoring of exact sequences" Tags="&lt;modules&gt;" AnswerCount="4" CommentCount="5" FavoriteCount="1" />
  <row Id="24345" PostTypeId="2" ParentId="24091" CreationDate="2011-03-01T03:00:09.243" Score="0" ViewCount="" Body="&lt;p&gt;The question was answered on MO: &lt;a href=&quot;http://mathoverflow.net/questions/56955&quot; rel=&quot;nofollow&quot;&gt;http://mathoverflow.net/questions/56955&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="264" LastActivityDate="2011-03-01T03:00:09.243" />
  <row Id="24346" PostTypeId="2" ParentId="24344" CreationDate="2011-03-01T03:20:08.600" Score="4" ViewCount="" Body="&lt;p&gt;The isomorphism $A \otimes_A M \to M$ is by sending $a \otimes m \to am$. Under this isomorphism, the image of $\mathfrak{a} \otimes M$ would be $\mathfrak{a}M$.&lt;/p&gt;&#xA;" OwnerUserId="325" LastActivityDate="2011-03-01T03:20:08.600" />
  <row Id="24347" PostTypeId="2" ParentId="24332" CreationDate="2011-03-01T03:29:14.567" Score="10" ViewCount="" Body="&lt;p&gt;Here's one way of looking at these problems. In general, integers $m$ and $n$ are relatively prime if and only if there are integers $x$ and $y$ such that $mx - ny = \pm 1$. You can write this in matrix form as &#xA;$$det\pmatrix {m&amp;amp;n \cr y&amp;amp;x \cr} = \pm 1$$&#xA;Correspondingly, the condition you are given is that&#xA;$$det\pmatrix {a&amp;amp;b&amp;#39; \cr a&amp;#39;&amp;amp;b \cr} = \pm 1$$&#xA;Adding one row to another doesn't change the determinant, so you also have&#xA;$$det\pmatrix {a + a&amp;#39;&amp;amp;b + b&amp;#39; \cr a&amp;#39;&amp;amp;b \cr} = \pm 1$$&#xA;This gives that $a + a&amp;#39;$ and $b + b&amp;#39;$ are relatively prime. &lt;/p&gt;&#xA;" OwnerUserId="3035" LastEditorUserId="3035" LastEditorDisplayName="" LastEditDate="2011-03-01T05:06:25.310" LastActivityDate="2011-03-01T05:06:25.310" CommentCount="3" />
  <row Id="24348" PostTypeId="1" CreationDate="2011-03-01T03:32:32.027" Score="1" ViewCount="76" Body="&lt;p&gt;As I understand it, to convolve f and g means to find ∫f(a)g(t-a)da, which is also apparently commutative, and therefore ∫f(a)g(t-a)da = ∫f(t-a)g(a)da (definite integral from -Inf to +Inf)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That means, if f(g) = 1, then ∫g(t-a)da = ∫g(a)da. For g(t) = t, for example, is not true. So what am I misunderstanding?&lt;/p&gt;&#xA;" OwnerUserId="7634" LastEditorUserId="7634" LastEditorDisplayName="" LastEditDate="2011-03-01T03:39:24.487" LastActivityDate="2011-03-01T06:06:53.040" Title="Commutativity of convolutions implies ∫f(a)da = ∫f(t-a)da ?" Tags="&lt;convolution&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="24349" PostTypeId="2" ParentId="23944" CreationDate="2011-03-01T03:41:38.887" Score="0" ViewCount="" Body="&lt;p&gt;Setting the derivative of $\mathcal{L}$ to zero does not guarantee that you obtain the minimum of it. I believe $\mathcal{L}$ is a concave function of $q$ at least for some $p,\lambda,\mu$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And also, the function $K_p$ to maximize is a convex function; therefore, solving the dual problem may not give you an optimal solution to the primal problem.&lt;/p&gt;&#xA;" OwnerUserId="7583" LastActivityDate="2011-03-01T03:41:38.887" CommentCount="2" />
  <row Id="24350" PostTypeId="2" ParentId="24280" CreationDate="2011-03-01T03:49:34.487" Score="4" ViewCount="" Body="&lt;p&gt;There's a general way of doing these things. You solve the homogeneous equation $y&amp;#39;&amp;#39; + y = 0$, giving $C_1\cos(t) + C_2\sin(t)$ for constants $C_1$ and $C_2$, then add to it a single solution $y_p(t)$ to the inhomogeneous equation $y&amp;#39;&amp;#39; + y = \cos(t)\cos(2t) = {1 \over 2}\cos(3t) + {1 \over 2}\cos(t)$. The result will be the general solution to your differential equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To find $y_p(t)$, you try $y_p(t) = a_1\cos(3t) + a_2\sin(3t) + b_1t\cos(t) + b_2t\sin(t)$. You plug it in and solve for $a_1,a_2,b_1,$ and $b_2$. Normally you just try combinations of $\cos(3t), \sin(3t), \cos(t),$ and $\sin(t)$, but since the latter two solve the homogeneous equation you have to stick a $t$ in front. I will trust Theo Buehler is right and that $a_1 = -{1 \over 16}$, $b_2 = {1 \over 4}$ and $a_2 = b_1 = 0$. Thus your general solution will be&#xA;$$y(t) = C_1\cos(t) + C_2\sin(t) -{1 \over 16}\cos(3t) + {1 \over 4}t\sin(t)$$&#xA;(The ${5 \over 16}\cos(t)$ term gets absorbed into the solution to the homogeneous equation).&lt;/p&gt;&#xA;" OwnerUserId="3035" LastActivityDate="2011-03-01T03:49:34.487" CommentCount="2" />
  <row Id="24351" PostTypeId="1" CreationDate="2011-03-01T02:15:34.757" Score="1" ViewCount="192" Body="&lt;p&gt;Which is a good book to refresh discrete maths fundamentals for a grad student? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It would be great if the book has short and terse explanations of concepts with lots of worked out examples(/to be worked out exercises) to set the brain rolling.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can someone suggest a similar refresher for graduate level algorithms?&lt;/p&gt;&#xA;" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2011-03-01T04:42:17.543" LastActivityDate="2011-03-01T04:44:24.687" Title="Discrete maths fundamentals" Tags="&lt;big-list&gt;&lt;books&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="0" />
  <row Id="24352" PostTypeId="2" ParentId="24351" CreationDate="2011-03-01T02:55:45.143" Score="2" ViewCount="" Body="&lt;p&gt;Concrete Mathematics by Knuth is useful as a refresher AFAIK. But this is not a research level question so not really belongs here :).&lt;/p&gt;&#xA;" OwnerUserId="7642" LastActivityDate="2011-03-01T02:55:45.143" CommentCount="2" />
  <row Id="24353" PostTypeId="2" ParentId="24351" CreationDate="2011-03-01T03:10:58.297" Score="0" ViewCount="" Body="&lt;p&gt;I don't think it's a graduate textbook, but I really like the discrete math &lt;a href=&quot;http://books.google.com/books?id=I6wxfY2dUwQC&amp;amp;&quot; rel=&quot;nofollow&quot;&gt;book&lt;/a&gt; by Matousek and Nesetril.  It presents the topics more succinctly than the standard discrete math undergraduate texts and has much more interesting exercises.&lt;/p&gt;&#xA;" LastActivityDate="2011-03-01T03:10:58.297" />
  <row Id="24354" PostTypeId="2" ParentId="24308" CreationDate="2011-03-01T03:53:50.770" Score="4" ViewCount="" Body="&lt;p&gt;Regarding the third part of your question: The intersection of two lines in $\mathbb P^3$ is generically empty (if you write down two random lines, they will be skew), but sometimes the lines will be coplanar (i.e. lie in a common plane), and then they will meet in a point (as any two lines in $\mathbb P^2$ do).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you know that &lt;em&gt;a priori&lt;/em&gt; that the two lines meet in a point, then they must &#xA;be coplanar, and the problem is the same as for two intersecting lines &#xA;in $\mathbb P^2$.&lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2011-03-01T03:53:50.770" CommentCount="6" />
  <row Id="24355" PostTypeId="2" ParentId="6485" CreationDate="2011-03-01T04:00:30.710" Score="0" ViewCount="" Body="&lt;p&gt;I don't think the feasible space of the problem is a convex set. Why are you pretty sure that there is no dual gap?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know what's the dual problem of this one, but we know the solution is $(x^*,y^*)=(\infty,0)$, that is unbounded.&lt;/p&gt;&#xA;" OwnerUserId="7583" LastEditorUserId="7583" LastEditorDisplayName="" LastEditDate="2011-03-01T04:33:42.893" LastActivityDate="2011-03-01T04:33:42.893" />
  <row Id="24356" PostTypeId="2" ParentId="24344" CreationDate="2011-03-01T04:01:48.797" Score="4" ViewCount="" Body="&lt;p&gt;That the natural map $a \otimes_A M \rightarrow aM$ be an isomorphism is very far from being &quot;clear and obvious&quot;.  As others have noted, it is not generally true.  But even more, it may be viewed as the entire obstruction to the functor &quot;tensoring with $M$&quot; being exact.  In other words, if for all ideals $a$ of $A$ (in fact it suffices even to restrict to finitely generated ideals) the map $a \otimes_A M \rightarrow aM$ is an isomorphism, then the module $M$ is &lt;strong&gt;flat&lt;/strong&gt; and tensoring any injection with $M$ gives an injection.  See for instance $\S 3.11$ of &lt;a href=&quot;http://math.uga.edu/~pete/integral.pdf&quot; rel=&quot;nofollow&quot;&gt;my commutative algebra notes&lt;/a&gt;, where I state and prove this (well-known) result, which I call the &lt;strong&gt;Tensorial Criterion For Flatness&lt;/strong&gt;.&lt;/p&gt;&#xA;" OwnerUserId="299" LastActivityDate="2011-03-01T04:01:48.797" CommentCount="1" />
  <row Id="24357" PostTypeId="2" ParentId="24344" CreationDate="2011-03-01T04:03:19.337" Score="1" ViewCount="" Body="&lt;p&gt;As mentioned by curious, the induced sequence is in general not exact on the left, though it is exact at the other two places. But as both curious and Soarer point out, the image of the map on the left is $aM$ under the isomorphism $A\otimes_A M \cong M$, and then exactness at the other two places gives you the isomorphism that you want.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;is there any other way to think about the tensor product besides been the quotient of a free module by a set of generators?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes! In fact, there is a much more useful, if more abstract, way to think of tensor products. Let $A$ be a commutative ring, and let $M,N,Z$ be $A$-modules. (Analogous statements hold over noncommutative rings, but the formulation is not quite as clean.) An $A$-&lt;em&gt;bilinear map&lt;/em&gt; $b:M\times N\to Z$ is a map of sets such that for each $m\in M$ and each $n\in N$, the maps $b(m,-):N\to Z$ and $b(-,n):M\to Z$ are $A$-linear. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bilinear maps appear all over the place; you can think of them as &quot;generalized multiplications&quot; -- for any $A$-algebra $R$, its multiplication $R\times R\to R$ is $A$-bilinear.  Note that a bilinear map $M\times N\to Z$ is &lt;em&gt;not&lt;/em&gt; $A$-linear with respect to the usual $A$-module structure on $M\times N$. That's too bad, because it's always nice to have things be linear. To &quot;correct&quot; this, the tensor product $M\otimes_A N$ is exactly the $A$-module such that bilinear maps $M\times N\to Z$ are the same thing as linear maps $M\otimes_A N\to Z$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More precisely, $M\otimes_A N$ is characterized by the following universal property: if $Q$ is an $A$-module and $b:M\times N\to Q$ is an $A$-bilinear map, then there is a unique $A$-&lt;em&gt;linear&lt;/em&gt; map $\tilde b:M\otimes_A N\to Q$ such that $\tilde b(m\otimes n) = b(m,n)$ for all $(m,n)\in M\times N$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, for example, if $R$ is an $A$-algebra, then the multiplication of $R$ is an $A$-linear map $R\otimes_A R\to R$. Indeed, this lets you phrase the axioms of an associative algebra completely in terms of linear maps, which is a useful thing to do if, for example, you want to talk about &quot;algebra objects&quot; in categories other than the category of $A$-modules.&lt;/p&gt;&#xA;" OwnerUserId="5116" LastActivityDate="2011-03-01T04:03:19.337" />
  <row Id="24358" PostTypeId="2" ParentId="24338" CreationDate="2011-03-01T04:39:19.273" Score="1" ViewCount="" Body="&lt;p&gt;I think not.  In a small neighborhood $U$, $g$ will be almost constant, so your new norm will look very much like the standard $L^2$ norm.  And of course for the standard norm one can find functions with $||f||$ arbitrarily small and $||f&amp;#39;||$ arbitrarily large.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What makes you think this should hold?&lt;/p&gt;&#xA;" OwnerUserId="822" LastActivityDate="2011-03-01T04:39:19.273" />
  <row Id="24359" PostTypeId="2" ParentId="24351" CreationDate="2011-03-01T04:44:24.687" Score="0" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0321295358&quot; rel=&quot;nofollow&quot;&gt;Kleinberg and Tardos&lt;/a&gt; is a good book. I followed this book for an Introductory Graduate Discrete Mathematics course. It starts from basics and covers a good amount of material in depth. Kleinberg and Tardos has a lot of Exercises at the end of each chapter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.cs.berkeley.edu/~vazirani/algorithms.html&quot; rel=&quot;nofollow&quot;&gt;Algorithms&lt;/a&gt; book by S. Dasgupta, C.H. Papadimitriou, and U.V. Vazirani is an equally good book and is probably well-written than Kleinberg and Tardos. I have not read this book completely though.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastActivityDate="2011-03-01T04:44:24.687" />
  <row Id="24361" PostTypeId="2" ParentId="24308" CreationDate="2011-03-01T04:53:16.817" Score="1" ViewCount="" Body="&lt;p&gt;(2) Your guess is correct, except I think it is an $(n-1)\times (n+1)$ matrix. Think of your line as the intersection of $n-1$ hyperplanes in non-degenerate position.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(3) Assuming you meant the &lt;em&gt;union&lt;/em&gt; of two lines $L_1, L_2$, then the defining variety is given by $I_1I_2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the Hilbert poly., use:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$0 \to \mathcal O_{L_1\cup L_2} \to \mathcal O_{L_1}\oplus \mathcal O_{L_2} \to \mathcal O_{L_1\cap L_2} \to 0$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;and the fact that Hilbert poly. are additive. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Details added: the two terms on the right of the above sequence can be calculated easily. Think of a line as $Proj(k[x,y])$ and a point (since you know $L_1,L_2$ intersect at a point) as $Proj(k[x])$. I am sure you can find the Hilbert polynomials of those rings.  &lt;/p&gt;&#xA;" OwnerUserId="223" LastEditorUserId="223" LastEditorDisplayName="" LastEditDate="2011-03-01T18:43:05.780" LastActivityDate="2011-03-01T18:43:05.780" CommentCount="4" />
  <row Id="24362" PostTypeId="1" AcceptedAnswerId="24368" CreationDate="2011-03-01T05:11:11.520" Score="5" ViewCount="137" Body="&lt;p&gt;Hello,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In my notes: &#xA;$$\sum_{d|n} \mu(d)\sigma(d) = (-1)^{k} \prod_{i=1}^{k} p_i$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\mu(d)$ is Mobius function and $\sigma(d)$ is the sum of all positive divisors of d.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And I have no idea how they got the expression on the right hand side. Could anyone help me explain how this work? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks,&lt;br&gt;&#xA;Chan&lt;/p&gt;&#xA;" OwnerUserId="5711" LastEditorUserId="5711" LastEditorDisplayName="" LastEditDate="2011-03-01T05:36:09.633" LastActivityDate="2011-03-01T09:11:31.743" Title="How does Mobius function work?" Tags="&lt;elementary-number-theory&gt;" AnswerCount="3" CommentCount="5" FavoriteCount="1" />
  <row Id="24363" PostTypeId="2" ParentId="24323" CreationDate="2011-03-01T05:18:36.633" Score="1" ViewCount="" Body="&lt;p&gt;Since you are working relative to the basis $B$, you may as well imagine that you are working over the standard basis and that the vectors you have are, as lhf points out, $(1,-1,0,1)$ and $(-1,0,3,0)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is a system of equations that determines these vectors? To get a subspace, you'll want a &lt;em&gt;homogeneous&lt;/em&gt; system of equations. And these two vectors should generate all solutions. So, first of all, your system should have four unknowns and two equations, since you want two parameters. The solutions will all look like&#xA;$$\left(\begin{array}{c}&#xD;&#xA;x_1\\x_2\\x_3\\x_4\end{array}\right) = \left(\begin{array}{r}&#xD;&#xA;1\\-1\\0\\1\end{array}\right)r + \left(\begin{array}{r}-1 \\0\\3\\0\end{array}\right)s$$&#xA;with $r$ and $s$ arbitrary. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What system is this? Well,&#xA;\begin{align*}&#xA;x_1 &amp;amp; = r - s\\&#xA;x_2 &amp;amp;= -r\\&#xA;x_3 &amp;amp;= 3s\\&#xA;x_4 &amp;amp;= r.&#xA;\end{align*}&#xA;Equivalently, plugging in $x_4$ for $r$ and $\frac{1}{3}x_3$ for $s$, we get&#xA;\begin{align*}&#xA;x_1 +\frac{1}{3}x_3 - x_4 &amp;amp;= 0\\&#xA;x_2 +x_4 &amp;amp;= 0.&#xA;\end{align*}&#xA;You'll note that the two vectors you have, $(1,-1,0,1)$ and $(-1,0,3,0)$, are solutions. And any solution is a linear combination of these two. So this gives the system you want &lt;em&gt;in terms of coordinate vectors relative to $B$&lt;/em&gt;. &lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2011-03-01T05:18:36.633" />
  <row Id="24364" PostTypeId="2" ParentId="24362" CreationDate="2011-03-01T05:25:55.980" Score="3" ViewCount="" Body="&lt;p&gt;Hint: the Mobius function is only non-zero on squarefree numbers (&lt;a href=&quot;http://en.wikipedia.org/wiki/M%C3%B6bius_function#Definition&quot; rel=&quot;nofollow&quot;&gt;see here&lt;/a&gt;), and it is either $+1$ or $-1$ depending on the number of factors. Then consider: What do the squarefree divisors of $n=p_1^{a_1}\cdots p_r^{a_r}$ look like? What does $\sigma$ (the sum of divisors) of a squarefree number look like?&lt;/p&gt;&#xA;" OwnerUserId="264" LastActivityDate="2011-03-01T05:25:55.980" />
  <row Id="24365" PostTypeId="2" ParentId="24318" CreationDate="2011-03-01T05:31:01.497" Score="5" ViewCount="" Body="&lt;p&gt;I think that (assuming say that $M$ is finite type over a field $k$) the map&#xA;$M^{\times} \to M$ should be an open immersion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is an attempt at a proof, hopefully not too bogus:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First of all, let's begin by carefully defining $M^{\times}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have have the multiplication map $\mu:M \times M \to M,$ the identity&#xA;element $e: \mathrm{Spec} k \to M$, and the tranposition $\tau: M\times M&#xD;&#xA;\to M \times M$ (switching factors).  (Here I am assuming I have a monoid in the &#xA;category of $k$-schemes, for some field $k$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We may form the product $\nu:=\mu\times (\mu\circ \tau): M\times M \to M\times M,$&#xA;which in terms of $S$-valued points (for any $k$-scheme $S$) maps&#xA;the pair $(x,y)$ to the pair $(xy, yx).$&#xA;If we form the fibre product of $\nu$ and $e\times e: \mathrm{Spec} k \to M\times M$, we obtain a closed subscheme&#xA;$X \subset M\times M$, whose $S$-points consist of &#xA;pairs $(x,y)$ such that $x y = yx = e.$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now in fact we can make $X$ a group scheme, by defining (on $S$-valued points) the multiplication $(x,y)\cdot (u,v) =&#xD;&#xA;(xu,vy),$ the identity element $(e,e)$, and the inverse operation $(x,y)^{-1} = (y,x)$. Intuitively, we have that $y = x^{-1}$, and this is just the product&#xA;on invertible elements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus we should write $X = M^{\times}$; this is the formation of the group&#xA;of units in a categorical manner.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now projection onto the first factor gives a map $X \to M$,&#xA;which is a monomorphism (look at $S$-valued points, and observe by the usual proof for monoids in sets that $x$ determines $y$).    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Everything so far is purely categorical.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we have to do some algebraic geometry. I'll assume that $M$ is finite type&#xA;over $k$, and argue with Zariski tangent spaces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the Zariski tangent space to $M$ at $e$.  This is equal to&#xA;morphisms $\mathrm{Spec} k[\epsilon]/(\epsilon^2) \to M$ lying above&#xA;the map $e: \mathrm{Spec}k \to M$, and so &#xA;is naturally a monoid in the category of $k$-vector spaces,&#xA;with zero acting as the identity.  A standard argument then shows&#xA;that the monoid structure on the Zariski tangent space structure coincides&#xA;with the underlying addition (coming from the &lt;em&gt;a priori&lt;/em&gt; vector space&#xA;structure).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now this vector space is a group under addition, and so we can lift this map&#xA;$\mathrm{Spec} k[\epsilon]/(\epsilon^2)$ from a map to $M$ to a map to $M^{\times}$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In conclusion: the map $M^{\times} \to M$ induces an isomorphism of Zariski tangent spaces the identity.  Since $M^{\times}$ is homogeneous, it must do so&#xA;at every (say $\overline{k}$-valued) point of its domain.  (Alternatively, we could just repeat the above argument as these points.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So $M^{\times} \to M$ is a monomorphism that induces an isomorphism on Zariski tangent spaces at every $\overline{k}$-valued point of its domain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This should imply that $M^{\times} \to M$ is an etale monomorphism, and hence is an open immersion.  (Hopefully this last step is correct.  If $M$ is smooth it is okay; if $M$ is not smooth then it also seems okay to me at the moment, but perhaps I am blundering ... .)&lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2011-03-01T05:31:01.497" CommentCount="1" />
  <row Id="24366" PostTypeId="1" AcceptedAnswerId="24369" CreationDate="2011-03-01T05:32:32.227" Score="2" ViewCount="59" Body="&lt;p&gt;Hello&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Prove that$f(n) = 2^{\omega(n)}$ is multiplicative where $\omega(n)$ is the number of distinct primes.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My attempt:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $a = p_1p_2\cdots p_k$ and $b = q_1q_2\cdots q_t$&#xA;where $p_i$ and $q_j$ are prime factors, and $p_i \neq q_j$ for all $1 \leq i \leq k$ and $1 \leq j \leq t$.&#xA;We will show that $2^{\omega(ab)} = 2^{\omega(a)} \times 2^{\omega(b)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Indeed,&#xA;$\omega(a) = k$ and $\omega(b) = t$. Then $2^{\omega(a)} \times 2^{\omega(b)} = 2^{k + t}$&lt;br&gt;&#xA;Where $2^{\omega(ab)} = 2^{k + t}$&lt;br&gt;&#xA;$\therefore 2^{\omega(ab)} = 2^{\omega(a)} \times 2^{\omega(b)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Am I in the right track?   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks,  &lt;/p&gt;&#xA;" OwnerUserId="5711" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-01T05:51:14.980" LastActivityDate="2011-03-01T05:51:14.980" Title="Prove that$f(n) = 2^{\omega(n)}$ is multiplicative where $\omega(n)$ is the number of distinct primes" Tags="&lt;elementary-number-theory&gt;" AnswerCount="1" CommentCount="6" />
  <row Id="24367" PostTypeId="2" ParentId="24344" CreationDate="2011-03-01T05:37:12.973" Score="2" ViewCount="" Body="&lt;p&gt;Since $a \otimes_{A} M \rightarrow A \otimes_{A} M \rightarrow (A/a) \otimes_{A} M \rightarrow 0$ is exact,&lt;br&gt;&#xA; the second function in the sequence $g:A \otimes_{A} M \rightarrow (A/a) \otimes_{A} M $ is surjective. Moreover, the image of $f:a \otimes_{A} M \rightarrow A \otimes_{A} M $, that is $f(a \otimes_{A} M)$, is the kernel of $g$. &lt;br&gt;&#xA; By the first theorem of isomorphism for modules,&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$A \otimes_{A} M/f(a \otimes_{A} M) \cong (A/a) \otimes_{A} M $     (1)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if $h: S \rightarrow T$ is an isomorphism of modules, and $S&amp;#39;$ is a submodule of $S$,&#xA;we have $S/S&amp;#39;\cong T/h(S&amp;#39;)$  (2)&lt;br&gt;&#xA;which can be proved by applying again the theorem of isomorphism to the projection function $S\rightarrow T/h(S&amp;#39;)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $h:A \otimes_{A} M\rightarrow M$ , defined by $h(a\otimes_{A} m)=am$ is an isomorphism and $h(f(a \otimes_{A} M))=aM$, it follows by (2) that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$A \otimes_{A} M/f(a \otimes_{A} M) \cong M/aM$ (3)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, (1) and (3) prove your requirement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another proof can be given using the universal property of the tensor product mentioned above.&lt;/p&gt;&#xA;" OwnerUserId="7243" LastEditorUserId="7243" LastEditorDisplayName="" LastEditDate="2011-03-01T06:40:25.190" LastActivityDate="2011-03-01T06:40:25.190" CommentCount="3" />
  <row Id="24368" PostTypeId="2" ParentId="24362" CreationDate="2011-03-01T05:44:10.537" Score="5" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Caveat:&lt;/strong&gt; This may not be the most direct way of doing it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Write $n=p_1^{a_1}\cdots p_k^{a_k}$ with $p_i$ pairwise distinct primes. Since the Moebius function is zero at any non-square-free integer, the only divisors that matter are those which are products of some pairwise distinct $p_i$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $$\sigma(p_{i_1}\cdots p_{i_r}) = \prod_{j=1}^r \sigma(p_{i_j}) = \prod_{j=1}^{r}(1+p_{i_j}),$$&#xA;then the sum on the left hand side is the sum of these products, with a minus sign if the number of factors is odd, and a plus sign if the number of factors is even.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $x_i = 1 + p_i$. Let $S_r(x_1,\ldots,x_k)$ is the $r$th elementary symmetric polynomial on $x_1,\ldots,x_r$; that is, &#xA;\begin{align*}&#xA;S_0(x_1,\ldots,x_k) &amp;amp;= 1,\\&#xA;S_1(x_1,\ldots,x_k) &amp;amp;= x_1+\cdots + x_k,\\&#xA;S_2(x_1,\ldots,x_k) &amp;amp;= x_1x_2 + \cdots + x_1x_k + x_2x_3+\cdots + x_{k-1}x_k,\\&#xA;&amp;amp;\vdots\\&#xA;S_k(x_1,\ldots,x_k) &amp;amp;= x_1\cdots x_k.&#xA;\end{align*}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore:&#xA;$$\sum_{d|n}\mu(d)\sigma(d) = \sum_{r=0}^k (-1)^rS_r(x_1,\ldots,x_k).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now consider the polynomial $(t-x_1)\cdots(t-x_k)$. The coefficient of $t^i$ is precisely $(-1)^{k-i}S_{k-i}(x_1,\ldots,x_k)$. Thus, the sum on the right hand side is this polynomial evaluated at $t=1$. Therefore,&#xA;$$\sum_{d|n}\mu(d)\sigma(d) = \sum_{r=0}^k(-1)^rS_r(x_1,\ldots,x_k) = (1-x_1)\cdots(1-x_k) = \prod_{i=1}^k(1-x_i).$$&#xA;But $1-x_i = 1-(1+p_i) = -p_i$. Therefore,&#xA;$$\sum_{d|n}\mu(d)\sigma(d) = \prod_{i=1}^k(-p_i) = (-1)^k\prod_{i=1}^kp_i,$$&#xA;as claimed.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-01T05:49:36.553" LastActivityDate="2011-03-01T05:49:36.553" CommentCount="6" />
  <row Id="24369" PostTypeId="2" ParentId="24366" CreationDate="2011-03-01T05:45:55.183" Score="3" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Hint:&lt;/strong&gt; Notice $$2^{\omega(a)}\times 2^{\omega(b)}=2^{\omega(a)+\omega(b)}$$ so try to relate $\omega(a)+\omega(b)$ and $\omega(ab)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To simplify things, you need only show it holds for $a=p^r$, $b=q^t$ where $q,p$ are prime numbers. &lt;/p&gt;&#xA;" OwnerUserId="6075" LastActivityDate="2011-03-01T05:45:55.183" CommentCount="4" />
  <row Id="24370" PostTypeId="2" ParentId="24348" CreationDate="2011-03-01T06:06:53.040" Score="3" ViewCount="" Body="&lt;p&gt;I think the answers in the comments are slightly besides the point. It's true that the integrals aren't well-defined in the case of $g(t)=t$, but the more fundamental point here seems to me that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_{-\infty}^{+\infty} g(t-a)\mathrm{d}a=\int_{-\infty}^{+\infty} g(a)\mathrm{d}a$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is in fact true, and whenever one of the sides is well-defined the other is, too. It follows without any mention of convolutions from a simple change of variables $u=t-a$, which leaves the limits of integration at infinity unchanged. So it seems that to answer the question needs to involve addressing why Joey thought that these two were not equal in the first place, which probably had nothing to do with lack of convergence. Joey, I'm guessing that you didn't see that making the substitution from $a$ to $t-a$ leaves the limits at infinity unchanged?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another way of seeing this is that the area under $g(t-a)$ is just the shifted mirror image of the area under $g(a)$. &lt;/p&gt;&#xA;" OwnerUserId="6622" LastActivityDate="2011-03-01T06:06:53.040" />
  <row Id="24371" PostTypeId="1" AcceptedAnswerId="24375" CreationDate="2011-03-01T06:56:08.273" Score="2" ViewCount="78" Body="&lt;p&gt;This idea resulted while I heard an advertisement for Sonic, where they claim to have something like 300,000 different drinks they serve.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Essentially, what they are allowing you do to is mix any soda to create your own &quot;unique&quot; drink.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As such, it made me curious how to calculate the number of &quot;unique&quot; drinks you make.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's say you have 4 sodas (soda: A, B, C, and D) and you want to find out the total number of unique drinks you can make by using any number of those sodas mixed together.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is the answer just 4! (4 Factorial)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answers would include:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;A&lt;/li&gt;&#xA;&lt;li&gt;B&lt;/li&gt;&#xA;&lt;li&gt;C&lt;/li&gt;&#xA;&lt;li&gt;D&lt;/li&gt;&#xA;&lt;li&gt;AB&lt;/li&gt;&#xA;&lt;li&gt;AC&lt;/li&gt;&#xA;&lt;li&gt;AD&lt;/li&gt;&#xA;&lt;li&gt;BC&lt;/li&gt;&#xA;&lt;li&gt;BD&lt;/li&gt;&#xA;&lt;li&gt;CD&lt;/li&gt;&#xA;&lt;li&gt;ABC&lt;/li&gt;&#xA;&lt;li&gt;ABD&lt;/li&gt;&#xA;&lt;li&gt;ACD&lt;/li&gt;&#xA;&lt;li&gt;BCD&lt;/li&gt;&#xA;&lt;li&gt;ABCD&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="7638" LastActivityDate="2011-03-01T07:16:14.407" Title="Combination vs Permutation?" Tags="&lt;factorial&gt;" AnswerCount="2" />
  <row Id="24372" PostTypeId="1" AcceptedAnswerId="24374" CreationDate="2011-03-01T07:01:32.557" Score="3" ViewCount="42" Body="&lt;p&gt;Given a set of 2D points I have to find the points that when connected will form a polygon that contains all the points in the set.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A quick example: imagine you have a set &lt;code&gt;S={(1,1),(2,1),(1,2),(2,2),(4,4)}&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The solution would be &lt;code&gt;SolS={(1,1),(2,1),(1,2),(4,4)}&lt;/code&gt; as those points (when connected) form a polygon that cointains all the points in the set (&lt;code&gt;(2,2)&lt;/code&gt; is cointained inside the polygon).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a diagram describing the problem :&#xA;&lt;img src=&quot;http://i.imgur.com/o82k5.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And the Solution (Sorry for the quick scketeches):                &lt;img src=&quot;http://i.imgur.com/NcuWE.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, this obviously need to be extended to deal with &lt;code&gt;n&lt;/code&gt; 2D points.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have been cracking my head trying to work out an algorithm but I just cant find it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you so much for your help!&lt;/p&gt;&#xA;" OwnerUserId="7639" LastActivityDate="2011-03-01T07:05:10.463" Title="Given a set of 2D points (x,y) (cloud of points), find the points that, when connected, will contain all other points." Tags="&lt;geometry&gt;&lt;algorithms&gt;&lt;set-theory&gt;&lt;signal-processing&gt;&lt;coordinate-systems&gt;" AnswerCount="1" />
  <row Id="24373" PostTypeId="2" ParentId="24371" CreationDate="2011-03-01T07:04:33.553" Score="2" ViewCount="" Body="&lt;p&gt;It would be $2^n-1$ where $n$ possible drink flavors can be added.  Although, I think it's important to note that the proportions of the ingredients in recipes can have a huge effect on the actual taste so the potential could be much larger.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the simple case, where we ignore the proportions, the options are to include or not include each drink flavor and so this doubles the possible options for each drink added.  However, I subtract one because one of the options is to not include any of the drinks at all i.e. not to buy a drink (and this doesn't count as a possible drink flavor).&lt;/p&gt;&#xA;" OwnerUserId="1004" LastActivityDate="2011-03-01T07:04:33.553" />
  <row Id="24374" PostTypeId="2" ParentId="24372" CreationDate="2011-03-01T07:05:10.463" Score="4" ViewCount="" Body="&lt;p&gt;The word for what you are describing is the &lt;strong&gt;convex hull&lt;/strong&gt; of the points. &lt;a href=&quot;http://en.wikipedia.org/wiki/Convex_hull_algorithms&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt; has a list of algorithms for computing them.&lt;/p&gt;&#xA;" OwnerUserId="264" LastActivityDate="2011-03-01T07:05:10.463" CommentCount="3" />
  <row Id="24375" PostTypeId="2" ParentId="24371" CreationDate="2011-03-01T07:08:00.697" Score="2" ViewCount="" Body="&lt;p&gt;$4! = 24$ but you only have $15$ possibilities in your list so something is wrong.  In fact you have $2^4 -1$, and the reason is that each of the $4$ ingredients  can be in or out giving $2$ possibilities four times, hence $2 \times 2 \times 2 \times 2$.  But having no ingredients at all does not count, so you have to subtract $1$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This works in general, so with $n$ possible ingredients you would get $2^n - 1$ possibilities.  There might be more or fewer, if for example you could have double concentration of a particular ingredient, or if some mixtures were not allowed.  &lt;/p&gt;&#xA;" OwnerUserId="6460" LastActivityDate="2011-03-01T07:08:00.697" />
  <row Id="24377" PostTypeId="1" AcceptedAnswerId="24379" CreationDate="2011-03-01T07:13:05.517" Score="3" ViewCount="74" Body="&lt;p&gt;I apologize in advance for what is probably an obvious question but here goes:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What structure do functions on affine schemes respect?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If we consider other types of 'spaces' we have a notion of function on them which respect structure relevant to the type of 'space'. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a smooth manifold, we consider differential functions to $\mathbb{R}$(or $\mathbb{C}$ if you like) which respect the Differential structure. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of affine schemes, functions are simply elements of our ring. Why is this natural to think of functions on schemes as elements of the underlying ring, and what structure preservation does this emerge from?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope this is clear. &lt;/p&gt;&#xA;" OwnerUserId="16" LastActivityDate="2011-03-01T09:23:30.220" Title="What do functions on affine schemes preserve?" Tags="&lt;algebraic-geometry&gt;" AnswerCount="2" />
  <row Id="24378" PostTypeId="1" AcceptedAnswerId="24380" CreationDate="2011-03-01T07:19:26.823" Score="4" ViewCount="214" Body="&lt;p&gt;Hello,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Prove that two any consecutive terms of Fibonacci sequence are relatively prime&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My attempt:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have $f_1 = 1, f_2 = 1, f_3 = 2...$. So obviously $\gcd(f1, f2) = 1$.&lt;br&gt;&#xA;Suppose that $\gcd(f_n, f_{n+1}) = 1$, we will show that $\gcd(f_{n+1}, f_{n+2}) = 1$&lt;br&gt;&#xA;Consider, $\gcd(f_{n+1}, f_{n+2}) = \gcd(f_{n+1}, f_{n+1} + f_n)$&#xA;because $f_{n+2} = f_{n+1} + f_n.$&lt;br&gt;&#xA;Then $\gcd(f_{n+1}, f_{n+1} + f_n) = \gcd(f_{n+1}, f_{n}) = 1$ ( gcd property )&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, $\gcd(f_n, f_{n+1}) = 1$ for all $n &amp;gt; 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Am I in the right track?&lt;br&gt;&#xA;Any feedback would be greatly appreciated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks,  &lt;/p&gt;&#xA;" OwnerUserId="5711" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2011-05-01T13:44:38.427" LastActivityDate="2011-05-01T13:44:38.427" Title="Prove that two any consecutive terms of Fibonacci sequence are relatively prime" Tags="&lt;elementary-number-theory&gt;&lt;fibonacci-number&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="24379" PostTypeId="2" ParentId="24377" CreationDate="2011-03-01T07:22:43.430" Score="3" ViewCount="" Body="&lt;p&gt;I'm still very much a novice in algebraic geometry, but as far as I know, when one views the elements of a ring $R$ as &quot;functions&quot; on $\text{Spec}(R)$, this is not from the point of view of preserving any structure. In fact, they aren't actually functions per se, since the &quot;value&quot; of $f\in R$ at $p\in\text{Spec}(R)$ is $\overline{f}\in\kappa(p)$, and the field $\kappa(p)$ can vary as $p$ varies. One reason that viewing $R$ as &quot;functions&quot; on $\text{Spec}(R)$ is useful (besides the fact that it generalizes the case of a variety $X$, when all the $\kappa(p)$'s are $\mathbb{C}$, and the elements of the coordinate ring $A(X)$ are actually functions) is that it motivates the definition of a morphism of affine schemes $f:\text{Spec}(S)\rightarrow\text{Spec}(R)$ to be a map such that we can &quot;pull back&quot; &quot;functions&quot; on $\text{Spec}(R)$ to &quot;functions&quot; on $\text{Spec}(S)$ (i.e., send elements of $R$ to elements of $S$, i.e. have a homomorphism from $R$ to $S$, which we know is in fact equivalent to a map of schemes from $\text{Spec}(S)$ to $\text{Spec}(R)$). By analogy, a map of manifolds $f:M\rightarrow N$ is smooth iff for any smooth function $g:N\rightarrow\mathbb{R}$, the pullback $g\circ f$ is a smooth function on $M$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I think that $R$-as-functions-on-$\text{Spec}(R)$ should really be considered as something special, and inherent to the structure of $\text{Spec}(R)$ (specifically, this is the structure sheaf), and that this then motivates the definition of a morphism of schemes as something which preserves &lt;strong&gt;that&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Oh, I should also mention a neat idea from Eisenbud + Harris that also motivates the &quot;function&quot; analogy:&#xA;&lt;img src=&quot;http://i.imgur.com/ssGge.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="264" LastEditorUserId="264" LastEditorDisplayName="" LastEditDate="2011-03-01T08:05:02.673" LastActivityDate="2011-03-01T08:05:02.673" />
  <row Id="24380" PostTypeId="2" ParentId="24378" CreationDate="2011-03-01T07:27:28.863" Score="1" ViewCount="" Body="&lt;p&gt;Congratulations, you have solved it.&#xA;You have used the fact that gcd(a+b,b)=gcd(a,b)&lt;/p&gt;&#xA;" OwnerUserId="7243" LastActivityDate="2011-03-01T07:27:28.863" CommentCount="1" />
  <row Id="24381" PostTypeId="2" ParentId="24362" CreationDate="2011-03-01T07:28:21.520" Score="3" ViewCount="" Body="&lt;p&gt;This looks more like a special case of &lt;a href=&quot;http://en.wikipedia.org/wiki/M%C3%B6bius_inversion_formula&quot; rel=&quot;nofollow&quot;&gt;Mobius Inversion Formula&lt;/a&gt;. You need to choose an appropriate $f$ and $g$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that $n = p_1^{\alpha_1} p_2^{\alpha_2} \ldots p_k^{\alpha_k}$ and $n = p_1 p_2 \ldots p_k$ should give the same answer. (Essentially you are removing the $\mu(d) = 0$ terms on the left side).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, we could just deal with $n = p_1 p_2 \ldots p_k$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let $f(n) = n$ and $g(n) = \displaystyle \sum_{d|n} f(d)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$g(n)$ is nothing but the sum of divisors of $n$ i.e. $g(n) = \sigma(n)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that $\mu(d) = (-1)^k \mu \left( \frac{n}{d} \right)$ when $n = p_1 p_2 \ldots p_k$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;By &lt;a href=&quot;http://en.wikipedia.org/wiki/M%C3%B6bius_inversion_formula&quot; rel=&quot;nofollow&quot;&gt;Mobius Inversion Formula&lt;/a&gt;, $$n = \sum_{d|n} \mu(d) g\left( \frac{n}{d} \right) = (-1)^k \sum_{d|n} \mu\left( \frac{n}{d} \right) g \left( \frac{n}{d} \right) = (-1)^k \sum_{d|n} \mu\left( d \right) g \left( d \right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, $$\sum_{d|n} \mu\left( d \right) \sigma \left( d \right) = (-1)^k p_1 p_2 \ldots p_k$$&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2011-03-01T09:11:31.743" LastActivityDate="2011-03-01T09:11:31.743" CommentCount="2" />
  <row Id="24383" PostTypeId="1" CreationDate="2011-03-01T07:40:42.993" Score="2" ViewCount="62" Body="&lt;p&gt;Here is the question from Rotman, verbatim:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A sequence $S&amp;#39;_*\stackrel{f}{\to} S_* \stackrel{g}{\to} S&amp;#39;&amp;#39;_*$ is exact in Comp if and only if $S&amp;#39;_{n}*\stackrel{f_n}{\to} S_n \stackrel{g_n}{\to} S&amp;#39;&amp;#39;_{n}$ is exact in Ab for every $n\in \mathbb{Z}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Note here Comp and Ab are the categories of chain complexes and abelian groups respectively. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am slightly confused - I thought this is the definition of an exact sequence of chain complexes. Indeed this is what Massey appears to say (Definition 2.6)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What am I missing here? &lt;/p&gt;&#xA;" OwnerUserId="2219" LastEditorUserId="367" LastEditorDisplayName="" LastEditDate="2011-03-01T07:49:06.400" LastActivityDate="2011-03-01T07:49:06.400" Title="Exact sequences in the category of chain complexes" Tags="&lt;algebraic-topology&gt;&lt;homological-algebra&gt;" CommentCount="5" />
  <row Id="24384" PostTypeId="1" AcceptedAnswerId="24386" CreationDate="2011-03-01T07:40:58.360" Score="1" ViewCount="82" Body="&lt;p&gt;I'm confused by what my solution manual is telling me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Part a of a problem I'm doing is:&#xA;Sketch the plane curve with the given vector equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$r(t) = &amp;lt;t-2, t^2+1&amp;gt;, t = -1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The solution manual basically says:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $(x+2)^2 = t^2 = y-1 =&amp;gt; y = (x+2)^2 - 1$, the curve is a parabola.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How is $(x+2)^2 = t^2 = y-1?$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another problem $r(t) = sinti + 2costj, t = pi/4$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I thought I was supposed to sketch it by plugging in values for t, but the solution manual says it's a parabola by doing this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x = sint, y = 2cost$ so $x^2 + (y/2)^2 = 1$ and the curve is an ellipse.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How come you are allowed to just square both x and y to get the ellipse?&lt;/p&gt;&#xA;" OwnerUserId="3514" LastActivityDate="2011-03-01T07:55:22.543" Title="How do you sketch a vector/plane curve?" Tags="&lt;homework&gt;&lt;vector-space&gt;" AnswerCount="1" />
  <row Id="24385" PostTypeId="1" AcceptedAnswerId="24412" CreationDate="2011-03-01T07:44:37.310" Score="4" ViewCount="96" Body="&lt;p&gt;Let $K$ be a quadratic number field, and $\mathcal{O}_K$ the ring of integers of $K$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The map $\pi: Spec(\mathcal{O}_K) \rightarrow Spec(\mathbb{Z})$ that sends a prime ideal $\mathbb{p}$ to $\mathbb{p} \cap \mathbb{Z}$ is induced since $\mathcal{O}_K$ contains $\mathbb{Z}$. And the fiber $\pi^{-1}$ of the prime ideal $(p)$ of $\mathbb{Z}$ is then understood as the decomposition of $(p)$ in $\mathcal{O}_K$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We then obtain a geometric interpretation of how p factors in $\mathcal{O}_K$ using results obtained from Algebraic number theory. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm looking for hints to (major or minor!) results that can be proved regarding the behaviour of $(p)$ in $\mathcal{O}_K$, or other interesting aspects of $\mathcal{O}_K$ using &quot;as much as possible&quot; Algebraic geometry  (at level of a first course in Schemes, using, say first seven chapters of Liu's &quot;Algebraic Geometry and Arithmetic Curves&quot;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd also appreciate a recommendation of a textbook or notes that discusses these ideas in details.&lt;/p&gt;&#xA;" OwnerUserId="1746" LastEditorUserId="1746" LastEditorDisplayName="" LastEditDate="2011-03-01T07:58:58.390" LastActivityDate="2011-03-01T13:02:49.660" Title="Factorization of primes and $Spec(\mathcal{O}_K)$" Tags="&lt;algebraic-geometry&gt;&lt;algebraic-number-theory&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="1" />
  <row Id="24386" PostTypeId="2" ParentId="24384" CreationDate="2011-03-01T07:49:58.257" Score="0" ViewCount="" Body="&lt;p&gt;You probably mean $r(t) = &amp;lt;t-2, t^2 +1, -1&amp;gt;$. $r(t) = &amp;lt;x, y, z&amp;gt;$ where $x = t-2$, $y=t^2+1$ and $z = -1$ hence $(x+2)^2 = t^2$ etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See &lt;a href=&quot;http://en.wikipedia.org/wiki/Parametric_equation&quot; rel=&quot;nofollow&quot;&gt;parametric equations&lt;/a&gt; in Wikipedia.&lt;/p&gt;&#xA;" OwnerUserId="7476" LastEditorUserId="7476" LastEditorDisplayName="" LastEditDate="2011-03-01T07:55:22.543" LastActivityDate="2011-03-01T07:55:22.543" />
  <row Id="24387" PostTypeId="1" AcceptedAnswerId="24403" CreationDate="2011-03-01T08:08:17.900" Score="1" ViewCount="150" Body="&lt;p&gt;I am writing up an assignment with includes many identities of Fibonacci numbers. I have made up the following notation (here $f_n$ is the number of tilings of an $n$-board by dominoes and squares - a well known interpretation of Fibonacci numbers):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given an $n$-board, and some sequence $seq$ of dominos and squares, let $f_n^{seq}$ represent all tilings of an $n$-board that end with the sequence $seq$. So for example   $f_n^{DS}$ are all tilings of an $n$-board that end with a domino succeeded by a square. Furthermore, for repeating sequences write:&#xA;$$&#xD;&#xA;\underbrace{seq,seq, \cdots, seq}_{n-\text{times}} = n\cdot seq&#xD;&#xA;$$&#xA;So for example $f_n^{SSDSDSDSDS} = f_n^{2S4DS}$. Let the size of a sequence $|seq|$ be the number of tiles the sequence occupies, so  $|SS4DS| = 14$. It is clear that:&#xA;$$&#xD;&#xA;f_n^{seq} = f_{n - |seq|}&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How may this be improved? Are statements such as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA; f_{3n-3k} = f_{3n}^{kDS} = f_{3n}^{SSkDS} + f_{3n}^{DkDS} + f_{3n}^{(k+1)DS} = f_{3n-3k}^{SS} + f_{3n-3k}^{D} + f_{3n-3k}^{DS} =&#xD;&#xA;$$&#xA;$$&#xD;&#xA;= 2f_{3n-3k-2} +  f_{3n-3k-3} &#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a bit understandable without explanation? (I do offer an explanation of course).&lt;/p&gt;&#xA;" OwnerUserId="5105" LastActivityDate="2011-03-01T17:26:32.947" Title="Fibonacci Numbers: Is This Notation Clear? How Can It Be Improved?" Tags="&lt;combinatorics&gt;&lt;notation&gt;&lt;fibonacci-number&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="24388" PostTypeId="1" AcceptedAnswerId="24423" CreationDate="2011-03-01T08:12:13.557" Score="2" ViewCount="95" Body="&lt;p&gt;For programming purposes I want a function f(x,R) that given a certain seed x returns the same random value every time, in an arbitrary range R. But, I also want the output to be equally distributed. Is this mathematically possible?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, I can take the 1st - nth decimal places of sin(x) such that 10^n &gt; R and divide by R, but I'm sure that this isn't evenly distributed even for R = 10.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Otherwise (veering slightly into Stack Overflow territory here) are there existing well-established functions/algorithms for producing (pseudo)random numbers in this fashion?&lt;/p&gt;&#xA;" OwnerUserId="7643" LastEditorUserId="7643" LastEditorDisplayName="" LastEditDate="2011-03-02T06:37:37.523" LastActivityDate="2011-03-02T06:37:37.523" Title="What simple functions return equally distributed random values in an arbitrary given range?" Tags="&lt;random&gt;" AnswerCount="2" CommentCount="5" FavoriteCount="0" />
  <row Id="24389" PostTypeId="1" AcceptedAnswerId="24391" CreationDate="2011-03-01T08:31:00.113" Score="3" ViewCount="141" Body="&lt;p&gt;Let $f$ be a continuous function , with the intial  conditions: $f(0)=1$, $\frac{df}{dx} \geq f$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How does one show that $f(x) \geq e^x$ for every $x \geq 0$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried using Taylor series&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you.&lt;/p&gt;&#xA;" OwnerUserId="6163" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2011-03-01T08:39:36.700" LastActivityDate="2011-03-01T13:38:06.377" Title="Showing that $f(x) \geq e^x$" Tags="&lt;calculus&gt;" AnswerCount="3" CommentCount="5" />
  <row Id="24390" PostTypeId="2" ParentId="24336" CreationDate="2011-03-01T08:33:58.177" Score="1" ViewCount="" Body="&lt;p&gt;Maybe, I write up your equation in more conventional (physics) notation. You have the electric field $$\vec{E}(\vec{r}) = \frac{Q}{4 \pi \epsilon_0 r^2} \frac{\vec{r}}{r} \in \mathbb{R}^3$$ at the position $\vec{r}\in \mathbb{R}^3$ of a particle with charge $Q$ at the origin ($r=|\vec{r}|$). Integrating this field over a line gives the potential difference between the endpoints of the line. In your case you want to integrate the electric field from somewhere on the surface with $r=a$ to the surface with $r=b$ (along which path exactly you integrate the field does not matter as the system is radially symmetric and we are in a static situation  [mathematically $\nabla \wedge \vec{E}(\vec{r}) =0$]. So we want to integrate (this is a &lt;a href=&quot;http://en.wikipedia.org/wiki/Line_integral&quot; rel=&quot;nofollow&quot;&gt;line integral&lt;/a&gt;)$$\int_{r=a}^{r=b} \vec{E}(\vec{r}) \cdot d\vec{r} = [-U(r)]_{r=a}^{b} $$&#xA;where we introduced the potential $U(r) = \frac{Q}{4 \pi \epsilon_0 r}$ with the property that $E= - \nabla U$ and the &lt;a href=&quot;http://en.wikipedia.org/wiki/Gradient_theorem&quot; rel=&quot;nofollow&quot;&gt;gradient theorem&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="7266" LastEditorUserId="7266" LastEditorDisplayName="" LastEditDate="2011-03-01T13:42:28.357" LastActivityDate="2011-03-01T13:42:28.357" CommentCount="3" />
  <row Id="24391" PostTypeId="2" ParentId="24389" CreationDate="2011-03-01T08:43:14.060" Score="7" ViewCount="" Body="&lt;p&gt;$$\frac{d(e^{-x}f(x))}{dx} = e^{-x} \frac{df}{dx} - e^{-x} f = e^{-x} \left( \frac{df}{dx} - f \right) \geq 0$$ Hence, $e^{-x} f(x)$ is an increasing function and therefore $e^{-x}f(x) \geq e^{-0}f(0)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, $$f(x) \geq f(0) e^x = e^x$$&lt;/p&gt;&#xA;" OwnerUserId="3302" LastActivityDate="2011-03-01T08:43:14.060" CommentCount="1" />
  <row Id="24392" PostTypeId="2" ParentId="24389" CreationDate="2011-03-01T08:47:10.423" Score="1" ViewCount="" Body="&lt;p&gt;Try using the definition of derivative:&#xA;$$f&amp;#39;(a)=\lim_{h\to 0}{f(a+h)-f(a)\over h}$$&#xA;for which you know that $f&amp;#39;(a) \geq f(a)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Try using a helper function $h(x) = f(x) - e^x$ and prove that it's positive or zero.&lt;/p&gt;&#xA;" OwnerUserId="7476" LastActivityDate="2011-03-01T08:47:10.423" CommentCount="1" />
  <row Id="24393" PostTypeId="2" ParentId="24389" CreationDate="2011-03-01T08:51:42.333" Score="1" ViewCount="" Body="&lt;p&gt;Define $\psi(x)=\log f(x)$. Then $\psi(0)=0$ and $\psi&amp;#39;(x) = f&amp;#39;(x)/f(x) \geq 1$. Hence $\psi(x) \geq x$, that is $f(x) \geq e^x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remark: Note that $\psi(x)$ is well-defined (since $f(x) &amp;gt; 0$ for all $x &amp;gt; 0$; consider the mean value theorem).&lt;/p&gt;&#xA;" OwnerUserId="2810" LastEditorUserId="2810" LastEditorDisplayName="" LastEditDate="2011-03-01T13:38:06.377" LastActivityDate="2011-03-01T13:38:06.377" CommentCount="4" />
  <row Id="24394" PostTypeId="2" ParentId="20070" CreationDate="2011-03-01T09:19:00.293" Score="0" ViewCount="" Body="&lt;p&gt;It's $O(n)$, because $\lfloor \frac{n}{2} \rfloor \leq \frac{n}{2} = \frac{1}{2} n$.&lt;/p&gt;&#xA;" OwnerUserId="3793" LastActivityDate="2011-03-01T09:19:00.293" />
  <row Id="24395" PostTypeId="2" ParentId="24377" CreationDate="2011-03-01T09:23:30.220" Score="3" ViewCount="" Body="&lt;p&gt;If you like, &quot;affine structure.&quot; A scheme is a gadget that locally looks like an affine scheme, and a morphism of schemes is a gadget that locally looks like a morphism of affine schemes. Then an element of a ring $R$ is precisely a morphism $\mathbb{Z}[x] \to R$, or in the other direction a morphism $\text{Spec } R \to \text{Spec } \mathbb{Z}[x] \cong \mathbb{A}^1(\mathbb{Z})$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But it is also valid to argue that the functions &lt;em&gt;are&lt;/em&gt; the structure, in the sense that all of these categories (manifolds, smooth manifolds, schemes, etc.) are full subcategories of the category of locally ringed spaces. &lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2011-03-01T09:23:30.220" />
  <row Id="24396" PostTypeId="1" AcceptedAnswerId="24644" CreationDate="2011-03-01T09:27:31.000" Score="3" ViewCount="200" Body="&lt;p&gt;I want to find the complex fixpoint $t=b^t $ for real bases $b&amp;gt; \eta = \exp(\exp(-1))$.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;added remark: &lt;em&gt;I'm aware that there is a solution using branches of the Lambert-W-function, but I've no Maple/Mathematica and only a rough implementation in Pari/GP for real values. That motivated to try a solution via Newton/Raphson. And to understand and solve such an implementation (which has to deal with derivatives and complex values) is/was then my question here. See also my comment to Fabian's answer below. I seem to have solved it myself for the &quot;principal branch&quot; (see my own answer below) but it is still open for the general case of k'th branch.&lt;/em&gt;  [end of remark]  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I have is a function depending on a parameter $ \beta $ giving the auxiliary values&lt;br&gt;&#xA;$$  u = \frac{\beta}{ \sin(\beta) } *\exp( i * \beta) $$&#xA;$$   t=\exp(u) $$&#xA;$$   b= f(\beta) = \exp(u/t) = \exp(u * \exp(-u))  $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By this I can do an approximation given a base $B$ using binary search. I can find the bounds of an interval taking lower and upper-limit beta's $ \beta_l = \epsilon $ and $ \beta_u = \pi-\epsilon $ with small epsilons giving the lower and upper bases $b_l$ and $b_u$ respectively. Then comparing $b_m = f(\beta_m)$ where $ \beta_m = (\beta_l + \beta_u)/2 $ with my given base $B$ I can implement a binary search which approximates $b_m$ to $B$ arbitrarily well and having $\beta_m$ I can reconstruct &lt;em&gt;u&lt;/em&gt; and the fixpoint &lt;em&gt;t&lt;/em&gt; by the above formula.     &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, that binary search needs surprisingly many iterations and I thought, possibly a Newton-like method for that approximation would be more efficient. But since I have complex values involved I do not even see the derivative and even less the formula how to involve that derivative in such an approximation-formula and how to apply this finally to actually do the iterations...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&lt;strong&gt;[update 4]&lt;/em&gt;&lt;/strong&gt; moved my own findings into an own answer (as suggested in meta.&lt;em&gt;*&lt;/em&gt;)    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&lt;strong&gt;[update 1]&lt;/em&gt;&lt;/strong&gt;  (in this old plot I used the letter &lt;em&gt;s&lt;/em&gt; instead of &lt;em&gt;b&lt;/em&gt;) &lt;img src=&quot;http://i.imgur.com/AbPxl.png&quot; alt=&quot;plot of function f(beta)&quot;&gt; &lt;/p&gt;&#xA;" OwnerUserId="1714" LastEditorUserId="1714" LastEditorDisplayName="" LastEditDate="2011-03-03T12:43:02.003" LastActivityDate="2011-03-10T15:13:21.997" Title="How to find a Newton-like approximation for that function?" Tags="&lt;analysis&gt;&lt;functions&gt;&lt;approximation&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="24397" PostTypeId="1" AcceptedAnswerId="24399" CreationDate="2011-03-01T09:47:46.427" Score="1" ViewCount="133" Body="&lt;p&gt;Let f be a convex and  bounded function, meaning there is a constant $C$, such that $f(x) &amp;lt; C$ for every $x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need to prove that $f$ is a constant function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="6163" LastEditorUserId="2219" LastEditorDisplayName="" LastEditDate="2011-03-01T10:13:15.370" LastActivityDate="2011-03-01T10:35:36.647" Title="Convex and bounded function is constant" Tags="&lt;calculus&gt;" AnswerCount="2" />
  <row Id="24398" PostTypeId="2" ParentId="24397" CreationDate="2011-03-01T10:11:16.460" Score="1" ViewCount="" Body="&lt;p&gt;Hint: What can you say about derivatives of convex functions? Then compare the straight line joining two points (say $x$ and $x_1$) with the tangent line at $x$&lt;/p&gt;&#xA;" OwnerUserId="2219" LastEditorUserId="2219" LastEditorDisplayName="" LastEditDate="2011-03-01T10:21:56.987" LastActivityDate="2011-03-01T10:21:56.987" CommentCount="4" />
  <row Id="24399" PostTypeId="2" ParentId="24397" CreationDate="2011-03-01T10:28:56.403" Score="1" ViewCount="" Body="&lt;p&gt;Suppose f is not constant, i.e., $\exists x,y\in\mathbb{R}:f(x)&amp;gt;f(y)$.&#xA;Since f is convex, we have: $f(x)\leq\lambda f(\frac{x-(1-\lambda)y}{\lambda})+(1-\lambda)f(y)\;\;\;\forall\lambda\in(0,1).$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(This is just the definition of convexity, $f(\lambda x&amp;#39;+(1-\lambda)y&amp;#39;)\leq\lambda f(x&amp;#39;)+(1-\lambda)f(y&amp;#39;)\;\;\;\forall\lambda\in(0,1)$, with $x=\lambda x&amp;#39;+(1-\lambda)y&amp;#39;$ and $y=y&amp;#39;$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence $\frac{f(x)-(1-\lambda)f(y)}{\lambda}\leq f(\frac{x-(1-\lambda)y}{\lambda}).$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, since $f(x)&amp;gt;f(y)$, $\frac{f(x)-(1-\lambda)f(y)}{\lambda}=\frac{f(x)-f(y)}{\lambda}+f(y)\rightarrow \infty$ as $\lambda\rightarrow0^+.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence f is not bounded above. &lt;/p&gt;&#xA;" OwnerUserId="5959" LastEditorUserId="5959" LastEditorDisplayName="" LastEditDate="2011-03-01T10:35:36.647" LastActivityDate="2011-03-01T10:35:36.647" CommentCount="2" />
  <row Id="24400" PostTypeId="2" ParentId="24338" CreationDate="2011-03-01T10:34:11.910" Score="1" ViewCount="" Body="&lt;p&gt;An explicit counterexample: fix some $f$ and define &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\tilde{f}(t) = e^{iKt}f(t) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for some large, unspeficied $K$. $\tilde{f}$ is clearly also in $H$. Then $$\tilde{f}&amp;#39;(t) = e^{iKt}f&amp;#39;(t) + Ke^{iKt}f(t)$$ An inequality of the form you seek will imply that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$K\|f\| - \|f&amp;#39;\| \leq \| f&amp;#39; + K f\| \leq C\|f\| $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for all $K$, which is absurd. &lt;/p&gt;&#xA;" OwnerUserId="1543" LastActivityDate="2011-03-01T10:34:11.910" />
  <row Id="24401" PostTypeId="2" ParentId="24387" CreationDate="2011-03-01T11:15:33.327" Score="3" ViewCount="" Body="&lt;p&gt;I must say that I don't find your $n \cdot seq$ easy to read, and it can appear ambiguous. I'd be inclined to borrow from automata theory and write e.g. $SSDSDSDSDS$ as $S^2(DS)^4$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would be inclined to help the lazy reader by also reminding him with the notation that you're talking about the end of sequences, so $\ldots S^2(DS)^4$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure how aesthetic this is when superscripted, but you can see for yourself:&#xA;$f_{3n}^{\ldots S^2(DS)^4}$&lt;/p&gt;&#xA;" OwnerUserId="5676" LastActivityDate="2011-03-01T11:15:33.327" CommentCount="2" />
  <row Id="24402" PostTypeId="1" CreationDate="2011-03-01T11:28:28.500" Score="-1" ViewCount="92" Body="&lt;p&gt;Are there any algebraic structures with a binary operation that is commutative but not distributive?&lt;/p&gt;&#xA;" OwnerUserId="7649" LastActivityDate="2011-03-01T13:24:16.353" ClosedDate="2011-03-01T18:15:17.103" Title="Are there any algebraic structures with a binary operation that is commutative but not distributive?" Tags="&lt;algebra&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="1" />
  <row Id="24403" PostTypeId="2" ParentId="24387" CreationDate="2011-03-01T11:30:34.173" Score="4" ViewCount="" Body="&lt;p&gt;Putting all that into a superscript (especially with the exponential notation suggested by others) is hard on the eyes. How about:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f_{3n-3k} = f_{3n}(*(DS)^k) = f_{3n}(*SS(DS)^k) + f_{3n}(*D(DS)^k) + f_{3n}(*(DS)^{k+1})$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;etc. Here * represents any sequence (so the argument in parentheses is not the end of the sequence, it's the whole sequence).&lt;/p&gt;&#xA;" OwnerUserId="1508" LastActivityDate="2011-03-01T11:30:34.173" CommentCount="1" />
  <row Id="24404" PostTypeId="2" ParentId="24388" CreationDate="2011-03-01T11:34:42.303" Score="0" ViewCount="" Body="&lt;p&gt;To give a StackOverflowy answer, how about generating a random value in your given range for every new input and return the previously generated value if the input appears again? You'd have to store every pair, though.&#xA;I think this is the only way to introduce randomness into your setup. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The example given in your question is not random at all, it is deterministic. Are you sure that you are really looking for randomness instead of maybe a hash function?&lt;/p&gt;&#xA;" OwnerUserId="505" LastActivityDate="2011-03-01T11:34:42.303" CommentCount="2" />
  <row Id="24405" PostTypeId="2" ParentId="24237" CreationDate="2011-03-01T11:45:11.550" Score="3" ViewCount="" Body="&lt;p&gt;I hope I'm not misunderstanding your question; please correct me if I am.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the answer must be that you cannot derive (2) from (1) alone. For each $h$, (1) is a separate recurrence relation linking only values with the same $h$. If you assume only (1), you can freely choose two initial values $v_{h,0}$ and $v_{h,1}$ and use them to generate a two-dimensional vector space of sequences (for fixed $h$) that can be made to take any value for a given $n$. But (2) contains only members for different values of $h$, so each of the members occurring in (2) can separately be chosen arbitrarily by choosing suitable initial values for its value of $h$. But there can't be a linear relationship between three quantities that can all be independently made to take an arbitrary value.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(It's possible that for some values of $h$ and $n$ (1) forces $v_{h,n}$ to be zero, but that could only imply (2) if this were the case on the entire diagonal, i.e. if the diagonal sequence were in fact zero -- if it is non-zero somewhere, we can always take a multiple of the sequence for that $h$ and thereby disturb (2) but not (1).)&lt;/p&gt;&#xA;" OwnerUserId="6622" LastEditorUserId="6622" LastEditorDisplayName="" LastEditDate="2011-03-01T12:12:01.260" LastActivityDate="2011-03-01T12:12:01.260" CommentCount="2" />
  <row Id="24406" PostTypeId="1" CreationDate="2011-03-01T11:47:05.773" Score="3" ViewCount="153" Body="&lt;p&gt;Let $(R,m)$ be a D.V.R with field of fraction $K$ and $L$ any finite algebraic field extension of $K$. Suppose $\bar{R}$ is the integral closure of $R$ in $L$. Then it is well known that $\bar{R}$ is a Dedekind domain and for any nonzero ideal $J$ of $\bar{R}$ the ring $\bar{R}/J$ is a finite $R$ module. My question is if $\bar{R}$ is by itself a finite $R$ module. If $L$ is separable over $K$ or $(R,m)$ is essentially finite over a field, then the answer is affirmative. I think in general it is not true. But I am not getting any easy counter-example. &lt;/p&gt;&#xA;" OwnerUserId="7565" LastActivityDate="2011-03-02T09:20:53.683" Title="A question related to krull akizuki" Tags="&lt;commutative-algebra&gt;" AnswerCount="2" FavoriteCount="2" />
  <row Id="24407" PostTypeId="2" ParentId="24402" CreationDate="2011-03-01T11:58:36.560" Score="1" ViewCount="" Body="&lt;p&gt;If I assume that the algebraic structure has two binary operation then the answer is as follows.&#xA;Consider the algebraic structure of real numbers with addition $+$ and subtraction $-$. Note $+$ is commutative. Then $-$ is not distributive over $+$. For example $a-(b + c) \neq (a - b) + (a -c)$ in general. &lt;/p&gt;&#xA;" OwnerUserId="7565" LastActivityDate="2011-03-01T11:58:36.560" CommentCount="1" />
  <row Id="24408" PostTypeId="1" AcceptedAnswerId="24411" CreationDate="2011-03-01T12:10:37.610" Score="1" ViewCount="150" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Possible Duplicate:&lt;/strong&gt;&lt;br&gt;&#xA;  &lt;a href=&quot;http://math.stackexchange.com/questions/1748/undergraduate-algebraic-geometry-textbook-recomendations&quot;&gt;(undergraduate) Algebraic Geometry Textbook Recomendations&lt;/a&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;p&gt;I am interested in algebraic number theory and I am recently acquainted with the theory of valuations, which further leads to Riemann-Roch theory, and which is closely related to algebraic geometry, and the algebraic-K-theory.&lt;br&gt;&#xA;Therefore, my problem is:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Are there excellent introductory books of the theory of Algebraic Geometry to recommend?  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Since I know in general nothing about this theory, I may want a book which explains the ideas as clear as possible and which at the mean time contains as much material as possible.&lt;br&gt;&#xA;If I am asking too much, then any good book in your view suffices.&lt;br&gt;&#xA;Thanks very much.&lt;/p&gt;&#xA;" OwnerUserId="6792" LastActivityDate="2011-03-01T12:44:58.633" ClosedDate="2011-03-01T13:35:13.227" Title="Are there any good algebraic geometry books to recommend?" Tags="&lt;soft-question&gt;&lt;algebraic-geometry&gt;&lt;reference-request&gt;&lt;big-list&gt;&lt;algebraic-number-theory&gt;" AnswerCount="1" CommentCount="8" FavoriteCount="1" />
  <row Id="24409" PostTypeId="1" AcceptedAnswerId="24433" CreationDate="2011-03-01T12:29:03.137" Score="2" ViewCount="112" Body="&lt;p&gt;Let $f$ be a continuous function, with the initial condition: $f&amp;#39;&amp;#39;&amp;gt;0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need to prove that $f(2x)-f(x) &amp;lt; f(3x)-f(2x)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By $f&amp;#39;&amp;#39;&amp;gt;0$ I can learn that $f&amp;#39;$ is monotonous increasing, and so I tried to use Lagrange or Taylor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you.&lt;/p&gt;&#xA;" OwnerUserId="6163" LastEditorUserId="1120" LastEditorDisplayName="" LastEditDate="2011-03-01T12:36:45.603" LastActivityDate="2011-03-01T17:44:16.213" Title="$f(2x)-f(x)&lt;f(3x)-f(2x)$ using Lagrange theorem" Tags="&lt;calculus&gt;" AnswerCount="3" CommentCount="6" />
  <row Id="24411" PostTypeId="2" ParentId="24408" CreationDate="2011-03-01T12:44:58.633" Score="2" ViewCount="" Body="&lt;p&gt;As I mentioned at another post &lt;a href=&quot;http://math.stackexchange.com/questions/13575/teaching-myself-differential-topology-and-differential-geometry/22583#22583&quot;&gt;Teaching myself differential topology and differential geometry.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are interested in learning Algebraic Geometry I recommend the books of my Amazon list. They are in recommended order to learn from the beginning by yourself:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.amazon.com/lm/RHQS8Y3V7LJRQ/ref=cm_pdp_lm_title_1&quot; rel=&quot;nofollow&quot;&gt;http://www.amazon.com/lm/RHQS8Y3V7LJRQ/ref=cm_pdp_lm_title_1&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, from that list, a quick path to understand basic Algebraic Geometry would be to read &lt;strong&gt;Bertrametti et al&lt;/strong&gt;. &lt;em&gt;&quot;Lectures on Curves, Surfaces and Projective Varieties&quot;&lt;/em&gt;, &lt;strong&gt;Shafarevich&lt;/strong&gt;'s &quot;&lt;em&gt;Basic Algebraic Geometry&quot;&lt;/em&gt; vol. 1, 2 and &lt;strong&gt;Perrin&lt;/strong&gt;'s &lt;em&gt;&quot;Algebraic Geometry an Introduction&quot;&lt;/em&gt;. But then you are entering the world of abstract algebra.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is no a single complete book and much less explaining the ideas as clearly as possible. If you are starting from the very beginnig, I recommend these in this order: Karen &lt;strong&gt;Smith's, Beltrametti, Hulek, Safarechiv vol. 1, Perrin, Shafarevich vol. 2&lt;/strong&gt; and then scheme theory with &lt;strong&gt;Ueno&lt;/strong&gt;'s three volumes.... then you can jump with enough background to the bible by &lt;strong&gt;Hartshorne&lt;/strong&gt;, or &lt;strong&gt;Griffiths/Harris&lt;/strong&gt; for the more complex geometric side. &lt;/p&gt;&#xA;" OwnerUserId="4058" LastActivityDate="2011-03-01T12:44:58.633" CommentCount="4" />
  <row Id="24412" PostTypeId="2" ParentId="24385" CreationDate="2011-03-01T12:49:54.187" Score="3" ViewCount="" Body="&lt;p&gt;As suggested in a comment by Quiaochu Yuan, Neukirch's book is really good, but I'd recommend you to have a look also at the beautiful lecture notes on Arithmetic Geometry by Lucien Szpiro. You can find them in his webpage, &lt;a href=&quot;http://math.gc.cuny.edu/faculty/szpiro/notes_summary.html&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&#xA;These are the notes of a course given by Szpiro in Orsay and they are in French (&lt;a href=&quot;http://math.gc.cuny.edu/faculty/szpiro/CoursdeGeometrie.htm&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;Cours De Géométrie Arithmétique&lt;/em&gt;&lt;/a&gt;), but it looks like somebody TeX-ed and translated (alas, just part of) them (&lt;a href=&quot;http://math.gc.cuny.edu/faculty/szpiro/cagsep30.pdf&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;Basic Arithmetic Geometry Notes&lt;/em&gt;&lt;/a&gt;).&#xA;Szpiro starts by studying Picard groups and one-dimensional rings and then discusses in a very geometric flavour some classical results of algebraic number theory (finiteness of $Pic$, Dirichlet's unit theorem...).&lt;/p&gt;&#xA;" OwnerUserId="6314" LastEditorUserId="6314" LastEditorDisplayName="" LastEditDate="2011-03-01T13:02:49.660" LastActivityDate="2011-03-01T13:02:49.660" CommentCount="1" />
  <row Id="24413" PostTypeId="1" AcceptedAnswerId="24422" CreationDate="2011-03-01T12:57:04.743" Score="6" ViewCount="187" Body="&lt;p&gt;Could give some examples of nonnegative measurable function $f:\mathbb{R}\to[0,\infty)$, such that its integral over any bounded interval is infinite?&lt;/p&gt;&#xA;" OwnerUserId="5593" LastActivityDate="2011-03-01T14:09:23.973" Title="examples of measurable functions on $\mathbb{R}$" Tags="&lt;real-analysis&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="24414" PostTypeId="2" ParentId="24241" CreationDate="2011-03-01T13:05:07.180" Score="3" ViewCount="" Body="&lt;p&gt;Actually, I think the reason why single letter identifiers are popular in math, and not so popular in programming is Intelisense. Today, every profesional developer would flinch if shown something like &lt;code&gt;return P * pow(1 + r / n, n * t)&lt;/code&gt;, but we have been spoiled by the ease of use of automatic filling of long names.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As I remember, 10-15 years ago it was more common to use single or double letter names for any local variable, because the developer would have to write each character himself, by hand. Right now, I just press a few keys, and the IDE does the rest.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;BTW, the name of the function used in the example (&lt;em&gt;pow&lt;/em&gt; for Power) fits nicely with the explanation.&lt;/p&gt;&#xA;" OwnerUserId="4160" LastActivityDate="2011-03-01T13:05:07.180" CommentCount="4" />
  <row Id="24416" PostTypeId="2" ParentId="24402" CreationDate="2011-03-01T13:18:40.007" Score="0" ViewCount="" Body="&lt;p&gt;Well, this is half an answer. In a (left or right) &lt;a href=&quot;http://en.wikipedia.org/wiki/Near-ring&quot; rel=&quot;nofollow&quot;&gt;commutative near ring&lt;/a&gt; one has only one of the distributive laws.&lt;/p&gt;&#xA;" OwnerUserId="2614" LastEditorUserId="2614" LastEditorDisplayName="" LastEditDate="2011-03-01T13:24:16.353" LastActivityDate="2011-03-01T13:24:16.353" />
  <row Id="24417" PostTypeId="2" ParentId="24409" CreationDate="2011-03-01T13:25:30.500" Score="1" ViewCount="" Body="&lt;p&gt;I'll try to expand my comment a bit:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First assume $x&amp;gt;0$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $f$ is continuous everywhere, it is continuous on $[x,2x]$. By the mean value theorem, there is some $y_1\in [x,2x]$ such that $f(2x)-f(x) = xf&amp;#39;(y_1)$.&#xA;Similarly, there is some $y_2\in [2x,3x]$ suc that $f(3x)-f(2x) = xf&amp;#39;(y_2)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore it is sufficient to show that $xf&amp;#39;(y_2) &amp;gt; xf&amp;#39;(y_1)$, or (because $x&amp;gt;0)$ that $f&amp;#39;(y_2) &amp;gt; f&amp;#39;(y_1)$. We know that $y_2 \geq y_1$ and that $f&amp;#39;$ is strictly increasing so we'll just have to exclude the possibility that $y_1 = y_2$. But in that case $y_2 = y_1 = 2x$ which leads to a contradiction. ( $f(2x) = f(x) + \int_x^{2x} f&amp;#39;(t)dt &amp;lt; f(x) + \int_x^{2x} f&amp;#39;(2x) dt = f(x) + xf&amp;#39;(2x) = f(2x)$. Although there are probably neater ways to find a contradiction.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now assume $x&amp;lt;0$. Construct $g(x) = f(-x)$ and reason with $g(x)$; I haven't done it but it shouldn't be too hard.&lt;/p&gt;&#xA;" OwnerUserId="5189" LastActivityDate="2011-03-01T13:25:30.500" />
  <row Id="24418" PostTypeId="2" ParentId="24396" CreationDate="2011-03-01T13:30:00.483" Score="3" ViewCount="" Body="&lt;p&gt;It might be advantageous to reduce the problem to something which has been studied in the literature: the keyword is &lt;a href=&quot;http://en.wikipedia.org/wiki/Lambert_W_function&quot; rel=&quot;nofollow&quot;&gt;Lambert W function&lt;/a&gt;. The Lambert W function $W(z)$ is defined via $$z= W(z)e^{W(z)}.$$ &#xA;On many systems there are already implementations of this function. In terms of the Lambert W function your $t$ is given via the relation $$t = - \frac{W(-\log b)}{\log b}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We need to find a way to calculate $w=W(z)$ for $z&amp;lt;-1/e$. I follow an idea of &lt;a href=&quot;http://ioannis.virtualcomposer2000.com/math/LWCalculating.html&quot; rel=&quot;nofollow&quot;&gt;this page&lt;/a&gt;. Let $z=-r$ (with $r&gt;1/e$), $w=R e^{i \theta}$ ($R&gt;0$). Writing $w e^{w} =z$ in polar coordinates, reduces to the set of equation $$r= R e^{R \cos \theta}$$ and $$\pi =  \theta + R \sin \theta.$$ This set of equations can be solved for $R&amp;gt;0$ and $\theta \in [0,\pi]$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second equation can be rewritten as $R= (\pi - \theta)/\sin \theta$. Plugging this into the first equation yields&#xA;$$r \sin \theta= (\pi - \theta) e^{(\pi - \theta) \cot \theta}.$$ This last equation can be solved using bisection (we know that the solution is between 0 and $\pi$). &lt;/p&gt;&#xA;" OwnerUserId="7266" LastActivityDate="2011-03-01T13:30:00.483" CommentCount="3" />
  <row Id="24419" PostTypeId="2" ParentId="24199" CreationDate="2011-03-01T13:49:07.777" Score="0" ViewCount="" Body="&lt;p&gt;Got it! Unbelievable.  The process of accumulating and then ranking is identical to simply computing the CDF, then reflecting over the y=x axis (taking the inverse)!!!  P(Urn X has x balls) = M*Pareto(X).  Let f(z) = $CDF^{-1}(M*Pareto(z))$.  That means that (using frequentist terminology) n*z urns have less than f(z) balls.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the final answer - to the original question without simplifying - is $n * CDF^{-1}(M*Pareto(x))$.  (Or, to preserve the order I mentioned in the question, reflecting &lt;em&gt;that&lt;/em&gt; over the y-axis).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm still a novice here, so can I ask you experts to verify or disprove this?&lt;/p&gt;&#xA;" OwnerUserId="7593" LastEditorUserId="7593" LastEditorDisplayName="" LastEditDate="2011-03-01T13:54:32.423" LastActivityDate="2011-03-01T13:54:32.423" CommentCount="1" />
  <row Id="24420" PostTypeId="2" ParentId="24413" CreationDate="2011-03-01T13:57:53.810" Score="3" ViewCount="" Body="&lt;p&gt;See exercise 26 (c) on p. 327 &lt;a href=&quot;http://books.google.co.il/books?id=4VFDVy1NFiAC&amp;amp;pg=PA327&amp;amp;lpg=PA327&amp;amp;dq=%22not+integrable+on+any%22&amp;amp;source=bl&amp;amp;ots=l7cZisZp5g&amp;amp;sig=Bv-_cghd99DfBpFJtdqwSMIQnLo&amp;amp;hl=iw#v=onepage&amp;amp;q=%22not%20integrable%20on%20any%22&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2810" LastActivityDate="2011-03-01T13:57:53.810" CommentCount="1" />
  <row Id="24421" PostTypeId="1" AcceptedAnswerId="24452" CreationDate="2011-03-01T14:07:58.223" Score="8" ViewCount="251" Body="&lt;p&gt;How to prove that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;\left(\sum_{i=1}^n w_i n_i \sqrt{\dfrac{y_i(1-y_i)}{n_i+1}}\right)^2 \leq \dfrac{\left(\sum_{i=1}^n w_i n_i y_i\right)\left(\sum_{i=1}^n w_i n_i (1-y_i)\right)}{(\sum_{i=1}^n w_i n_i+1)}&#xD;&#xA;$$&#xA;where $w_i\geq0$, $\sum_{i=1}^n w_i=1$, $n_i&amp;gt;0$ and $y_i \in (0,1)$ for $i=1,\dots,n$, with $n&amp;gt;1$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have verified numerically that it should hold, but I cannot still find an elegant way to show it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The formula comes from an inequality for the variance of a convex combination of beta-distributed variables.&lt;/p&gt;&#xA;" OwnerUserId="2821" LastEditorUserId="2720" LastEditorDisplayName="" LastEditDate="2011-03-01T14:24:11.953" LastActivityDate="2011-03-01T20:46:56.530" Title="Upsetting inequality" Tags="&lt;real-analysis&gt;&lt;probability-theory&gt;&lt;inequality&gt;" AnswerCount="1" FavoriteCount="2" />
  <row Id="24422" PostTypeId="2" ParentId="24413" CreationDate="2011-03-01T14:09:23.973" Score="5" ViewCount="" Body="&lt;p&gt;The easiest example I know is constructed as follows. Let $q_{n}$ be an enumeration of the rational numbers in $[0,1]$. Consider $g(x) = \sum_{n=1}^{\infty} 2^{-n} \frac{1}{|x-q_{n}|^{1/2}}$. Since each function  $\frac{1}{|x-q_{n}|^{1/2}}$ is integrable on $[0,1]$, so is $g(x)$ [verify this!]. Therefore $g(x) &amp;lt; \infty$ almost everywhere, so we can simply set $g(x) = 0$ in the points where the sum is infinite. On the other hand, $f = g^{2}$ has infinite integral over each interval in $[0,1]$. Indeed, if $0 \leq a \lt b \leq 1$ then $(a,b)$ contains a number $q_{n}$, so $\int_{a}^{b} f(x)\,dx \geq \int_{a}^{b} \frac{1}{|x-q_{n}|}\,dx = \infty$. Now in order to get the function $f$ defined at every point of $\mathbb{R}$, simply define $f(n + x) = f(x)$ for $0 \leq x \lt 1$.&lt;/p&gt;&#xA;" OwnerUserId="5363" LastActivityDate="2011-03-01T14:09:23.973" CommentCount="3" />
  <row Id="24423" PostTypeId="2" ParentId="24388" CreationDate="2011-03-01T14:17:01.427" Score="0" ViewCount="" Body="&lt;p&gt;As per your comment, it seems that you actually want some deterministic function $f(N,x)$ that returns a integer in $[1..N]$, and which, for a fixed $N$ and for different values of $x$ returns values that appear random in that interval. Still, this requirement is not totally clear, because one should stipulate a distribution of $x$. But, if I understand your scenario, one can assume that $x$ is uniform in some large interval (say, the range the full range of integers in you platform-language). So, a simple recipe would be to use a standard pseudorandom generator and use $x$ as a seed. For example, in Java:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;static  int getrand(int N,long x) {&#xA;    Random r = new Random(x);  // random number generator, with x as seed&#xA;    return r.nextInt(N)+1;     // pseudoradom number in 1..N range &#xA;} &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Or you can directly implement your own pseudorandom routine, &lt;a href=&quot;http://en.wikipedia.org/wiki/Linear_congruential_generator#Parameters_in_common_use&quot; rel=&quot;nofollow&quot;&gt;eg&lt;/a&gt;,&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;static  int getrand2(int N,long x) {&#xA;    long a=1664525;  &#xA;    long c=1013904223;&#xA;    long res = (a*x+c)&amp;amp;0xFFFFFFFF; &#xA;    return (int)((res % N)+1);&#xA;}&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="312" LastEditorUserId="312" LastEditorDisplayName="" LastEditDate="2011-03-01T14:25:05.383" LastActivityDate="2011-03-01T14:25:05.383" CommentCount="1" />
  <row Id="24424" PostTypeId="1" AcceptedAnswerId="24440" CreationDate="2011-03-01T14:28:41.843" Score="0" ViewCount="30" Body="&lt;p&gt;Concept of &lt;strong&gt;Proxy Signatures&lt;/strong&gt; serving the main purpose of &lt;em&gt;delegation&lt;/em&gt; is clear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But what exactly is &lt;strong&gt;proxy re-signatures&lt;/strong&gt;? &#xA;How and why is it performed? Does it use the same warrant generating and proxy signing mechanisms?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="7488" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-01T14:49:55.860" LastActivityDate="2011-03-01T16:29:20.330" Title="What is the notion behind proxy re-signatures?" Tags="&lt;cryptography&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="24425" PostTypeId="1" AcceptedAnswerId="24426" CreationDate="2011-03-01T14:29:29.340" Score="1" ViewCount="81" Body="&lt;p&gt;Suppose we have an algebraically closed field $F$ and $n+1$ variables $X_0, \dots, X_n$, where $n &amp;gt; 1$. Does there exist an irreducible homogeneous polynomial in these variables of degree $d$ for any positive integer $d &amp;gt; 1$? In other words, does there always exist an irreducible hypersurface of arbitrary degree? Of course, I am also interested in constructions of these polynomials.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you.&lt;/p&gt;&#xA;" OwnerUserId="7651" LastActivityDate="2011-03-01T14:58:08.663" Title="Irreducible homogeneous polynomials of arbitrary degree" Tags="&lt;algebraic-geometry&gt;&lt;polynomials&gt;" AnswerCount="1" />
  <row Id="24426" PostTypeId="2" ParentId="24425" CreationDate="2011-03-01T14:40:56.487" Score="2" ViewCount="" Body="&lt;p&gt;Yes, this is straightforward. First note that a homogeneous polynomial $f(X_0, ... X_n)$ which is not divisible by $X_0$ is irreducible if and only if $f(1, ... X_n)$ is irreducible, so the problem reduces to constructing irreducible polynomials in $k[x_1, ... x_n]$ of degree $d$. To do this we can take $x_1^2 - x_2 h(x_3, ... x_n)$ for any polynomial $h$ of degree $d-1$. (If $n = 2$ then take $h = 1$). &lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2011-03-01T14:58:08.663" LastActivityDate="2011-03-01T14:58:08.663" />
  <row Id="24427" PostTypeId="1" AcceptedAnswerId="24458" CreationDate="2011-03-01T15:18:26.697" Score="2" ViewCount="99" Body="&lt;p&gt;If $A$ is a Krull domain in its field of fractions $F$, and if $F&amp;#39;\subset F$ is a subfield, then $A\cap F&amp;#39;$ is a Krull domain. But is it necessarily a Krull domain of $F&amp;#39;$? That is, is $F&amp;#39;$ necessarily the field of fractions of $A\cap F&amp;#39;$?&lt;/p&gt;&#xA;" OwnerUserId="4341" LastActivityDate="2011-03-01T20:48:22.510" Title="About Krull domain" Tags="&lt;abstract-algebra&gt;&lt;commutative-algebra&gt;" AnswerCount="1" />
  <row Id="24428" PostTypeId="2" ParentId="6845" CreationDate="2011-03-01T15:25:30.147" Score="0" ViewCount="" Body="&lt;p&gt;You may want to read up on Gosper's algorithm. It helps you find closed-form expressions for certain classes of sums. &lt;/p&gt;&#xA;" OwnerUserId="4623" LastActivityDate="2011-03-01T15:25:30.147" CommentCount="1" />
  <row Id="24429" PostTypeId="1" AcceptedAnswerId="24436" CreationDate="2011-03-01T15:34:26.930" Score="10" ViewCount="230" Body="&lt;p&gt;Suppose I have a function $f$ that is analytic on the unit disk $D = \{ z \in \mathbb{C} : |z| &amp;lt; 1 \}$ that is also continuous up to $\bar{D}$. If $f$ is identically zero on some segment of of the boundary (e.g. $\{ e^{it}, 0 \leq t \leq \pi/2 \}$ ), is it then true that $f$ is identically zero on the entire boundary? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that analytic functions that are zero at an accumulation point inside the domain of analyticity are identically zero throughout the entire domain, but I don't know what (if anything) can be said if something similar occurs on the boundary of the domain.  &lt;/p&gt;&#xA;" OwnerUserId="1736" LastActivityDate="2011-03-06T00:12:30.650" Title="on the boundary of analytic functions" Tags="&lt;complex-analysis&gt;" AnswerCount="3" CommentCount="5" FavoriteCount="5" />
  <row Id="24430" PostTypeId="1" CreationDate="2011-03-01T15:35:27.240" Score="0" ViewCount="57" Body="&lt;p&gt;We have 10 bricks, 3 red, 2 white, 2 yellow, 2 blue, 1 black. In how many ways can these be arranged such that only 2 red bricks are adjacent ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We want to distribute the elements in {RR, R} over the spaces in arrangements of the form:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;_B_B_B_B_B_B_B_&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where _ represents a space (of which there are 8), and B a brick (we have 7 bricks left after removing {RR, R}). So we have:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;8C2 * 7!(2!2!2!) = 7*7!/2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;perms in total, since we choose 2 of the 8 spaces to distribute {RR, R} and we have 7!(2!2!2!) perms for the remaining 7 bricks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the answer provided is 7*7!. Can anyone spot my error ?&lt;/p&gt;&#xA;" OwnerUserId="7597" LastActivityDate="2011-03-01T15:40:04.087" Title="an arrangement of 10 objects, with restriction" Tags="&lt;combinatorics&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="24431" PostTypeId="2" ParentId="24430" CreationDate="2011-03-01T15:40:04.087" Score="2" ViewCount="" Body="&lt;p&gt;It is not enough to just consider the 8C2 locations where the elements $RR$ and $R$ will go. The arrangement is also important. So, if you were to exchange $RR$ with $R$ in a given permutation, you would get a different permutation. So you have to multiply your answer by 2 to get $7*7!$.&lt;/p&gt;&#xA;" OwnerUserId="2641" LastActivityDate="2011-03-01T15:40:04.087" CommentCount="1" />
  <row Id="24432" PostTypeId="2" ParentId="24429" CreationDate="2011-03-01T15:47:24.607" Score="0" ViewCount="" Body="&lt;p&gt;No.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider any function $\phi:S^1\to\mathbb C$ which is smooth and which vanishes on a segment of the circle. Define $f:D\to\mathbb C$ so that $$f(z)=\int_{S^1}\frac{f(\zeta)}{\zeta-z}\,\mathrm d\zeta.$$ This will be holomorphic in the interior of $D$ and will coincide with $\phi$ on $S^1$.&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2011-03-01T15:47:24.607" CommentCount="2" />
  <row Id="24433" PostTypeId="2" ParentId="24409" CreationDate="2011-03-01T15:56:31.233" Score="3" ViewCount="" Body="&lt;p&gt;You are given that $f&amp;#39;&amp;#39;(x)&amp;gt;0$ everywhere. If this condition holds, then $f(x)$ is convex on its domain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any convex function will satisfy Jensen's Inequality. Using Jensen's inequality with two terms, you have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;f(2x)&amp;lt;\frac{f(x)+f(3x)}{2} &#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i.e.,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;f(2x) - f(x) &amp;lt; f(3x) -f(2x)&#xA;\end{equation}&lt;/p&gt;&#xA;" OwnerUserId="2641" LastEditorUserId="2641" LastEditorDisplayName="" LastEditDate="2011-03-01T16:24:57.620" LastActivityDate="2011-03-01T16:24:57.620" CommentCount="3" />
  <row Id="24434" PostTypeId="2" ParentId="24284" CreationDate="2011-03-01T15:56:48.103" Score="1" ViewCount="" Body="&lt;p&gt;I think that in analysis variables have a personality of their own which is not covered by the notions of &quot;free&quot; or &quot;bound&quot; from predicate calculus. We speak about independent or dependent variables and say, e.g., that the variables $x$, $y$ resp. $r$, $\phi$ are &quot;related&quot; by the equations $x=r \cos\phi$, $y=r\sin\phi$. A given variable $x$ even may give birth to an associated variable $dx$, and on and on. Concerning the formula $y=f(x)$ I would give it the following interpretation:  $x$ is a variable taking values in the domain of the function $f$,  and $y$ is a variable taking values in the range of $f$, and we assume that at any given time the values of these two variables are related by the equation $y=f(x)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It would be interesting to see this everyday use of variables being translated into official logolese . . .&lt;/p&gt;&#xA;" OwnerUserId="1303" LastActivityDate="2011-03-01T15:56:48.103" />
  <row Id="24435" PostTypeId="1" AcceptedAnswerId="24658" CreationDate="2011-03-01T16:07:22.280" Score="3" ViewCount="185" Body="&lt;p&gt;Let $X_i$ be a random variable, and the probability density function of X_i is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$P(X_i=k)=\frac{{k-1 \choose i-1}{m-k \choose n-i}}{m \choose n}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For more about the distribution, please refer to &lt;a href=&quot;http://www.math.uah.edu/stat/urn/OrderStatistics.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.math.uah.edu/stat/urn/OrderStatistics.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The mean of $X_i$ is $$E(X_i)=i\frac{m+1}{n+1}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to calculate the fourth central moment (kurtosis) of the $X_i$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it possible to use Maple to derive it ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is actually a 3-variable hyper geometric distribution.&lt;/p&gt;&#xA;" OwnerUserId="6904" LastEditorUserId="6904" LastEditorDisplayName="" LastEditDate="2011-03-01T17:58:51.433" LastActivityDate="2011-03-03T21:06:36.883" Title="How to use Maple derive the 4th central moment for this distribution?" Tags="&lt;probability&gt;&lt;maple&gt;" AnswerCount="1" CommentCount="6" FavoriteCount="2" />
  <row Id="24436" PostTypeId="2" ParentId="24429" CreationDate="2011-03-01T16:09:04.667" Score="10" ViewCount="" Body="&lt;p&gt;The only way the function can be zero on an arc of the circle is for the function to be identically zero. To see this, consider ${\displaystyle g(z) = \prod_{j = 0}^{n-1} f(e^{2\pi i j \over n}z)}$ for a large enough $n$ so that $g(z)$ is identically zero on the boundary of the disk. Then by the Cauchy integral formula, if $r &amp;lt; 1$ then for any $w$ in the interior one has&#xA;$$g(w) = {1 \over 2\pi i}\int_{|z| = r}{g(z) \over z - w}\,dz$$&#xA;Since $g(z)$ is continuous on the compact set $\bar{D}$, $g(z)$ is bounded, so one can apply the dominated convergence theorem and let $r$ go to $1$. One obtains&#xA;$$g(w) = {1 \over 2\pi i}\int_{|z| = 1}{g(z) \over z - w}\,dz$$&#xA;$$ = 0$$&#xA;Thus $g(z)$ is identically zero. Since the zeroes of a nonzero analytic function are isolated, we conclude that $f(z)$ is identically zero as well.&lt;/p&gt;&#xA;" OwnerUserId="3035" LastActivityDate="2011-03-01T16:09:04.667" CommentCount="6" />
  <row Id="24437" PostTypeId="1" AcceptedAnswerId="24500" CreationDate="2011-03-01T16:15:01.497" Score="2" ViewCount="129" Body="&lt;p&gt;I'm looking for a way to find the factor of a natural number, call it $x$, which is closest to $\sqrt{x}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If this question needs to be clarified, just let me know and I'll be happy to add more.&lt;/p&gt;&#xA;" OwnerUserId="5195" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2011-03-01T22:30:02.767" LastActivityDate="2011-03-01T22:30:02.767" Title="How would I find the factor of $x$ nearest to $\sqrt{x}$ ?" Tags="&lt;number-theory&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="24438" PostTypeId="1" AcceptedAnswerId="24441" CreationDate="2011-03-01T16:23:34.723" Score="1" ViewCount="115" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In numerical mathematics, we have to approximate the value of $\pi$ using three different methods. One of the three methods is the Monte-Carlo method, i.e. I let Matlab produce $n$ random numbers within the unit square, then I counted the numbers which were within the unit circle and looked at the quotient of those two numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Afterwards, an exercise states that we are to calculate the error of each of the three methods. As the other two methods were infinite sums, I assumed that I was to express the error in terms of $n$. However, I have no idea how I would calculate the formal error in terms of $n$ of the Monte-Carlo method.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it actually possible to do so, or at least to approximate that error? Or is it only statistics?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for any help.&lt;/p&gt;&#xA;" OwnerUserId="3787" LastActivityDate="2011-03-02T13:57:58.460" Title="Do random numbers have an &quot;error&quot;? If so, how can I calculate it?" Tags="&lt;numerical-methods&gt;&lt;pi&gt;&lt;monte-carlo&gt;" AnswerCount="1" />
  <row Id="24439" PostTypeId="1" CreationDate="2011-03-01T16:26:24.313" Score="0" ViewCount="67" Body="&lt;p&gt;I would like to create the throw effect like you see on the iPhone, which &lt;a href=&quot;http://www.azoffdesign.com/plugins/js/overscroll&quot; rel=&quot;nofollow&quot;&gt;this library&lt;/a&gt; implements (I asked this on StackOverflow regarding a javascript implementation a few months ago and no one really knew).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there an equation that I can basically plug in a viewport, a canvas, a bounding box, and a moving point (finger), and the equation will take the velocity of the point (averaged, or somehow tracked over time so it's a smooth throw), and move the canvas within the viewpoint, within the boundaries of some invisible bounding box, so it looks like it was thrown?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Something like an &lt;a href=&quot;https://github.com/viatropos/style/blob/master/src/style/utils/Modulate.as&quot; rel=&quot;nofollow&quot;&gt;easing equation like these methods&lt;/a&gt;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is, can the &quot;throw effect&quot; be boiled down to some f(x,y,z) equation?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for the help.&lt;/p&gt;&#xA;" OwnerUserId="5266" LastActivityDate="2011-03-01T16:26:24.313" Title="Is there a one-liner equation to calculate the “iPhone Throw Effect”?" Tags="&lt;calculus&gt;&lt;physics&gt;&lt;vector-calculus&gt;" CommentCount="4" />
  <row Id="24440" PostTypeId="2" ParentId="24424" CreationDate="2011-03-01T16:29:20.330" Score="1" ViewCount="" Body="&lt;p&gt;The way cryptographic signatures work in public key cryptosystems is as follows: Alice has a private key $d$ and a public key $e$. Everyone knows $e$, and presumably only Alice knows $d$. In order to &quot;sign&quot; a message, Alice writes, for instance, &quot;This is Alice, and I approve this message&quot; (plus some hash to ensure that the signature really belongs to that particular message) and then uses her &lt;strong&gt;private&lt;/strong&gt; key to encrypt this message. Everyone receives a cryptotext which they can decrypt with Alice's &lt;strong&gt;public&lt;/strong&gt; key to read the message. Since, presumably the only person who knows Alice's private key is Alice, and the only way to encrypt a message that will get decrypted with the public key is to use the private key, this assures the reader that it is indeed Alice who approves of the message. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a &lt;em&gt;proxy signature scheme&lt;/em&gt;, you have the same situation as in real life proxys. Alice doesn't want to take the trouble to sign a bunch of stuff, so she delegates this responsibilty to Bob. Bob will sign messages &lt;em&gt;in Alice's name.&lt;/em&gt; However, Alice does not want to tell Bob what her private key is so that Bob can sign the messages in Alice's name, so a set-up has to be provided whereby Alice gives Bob enough information so that Bob can sign Alices's name on the messages, without having the necessary information to go ahead an read all of Alice's mail with her private key.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is often done by the use of a &quot;proxy-server&quot; or an intermediary, who mediates between Bob and Alice; Bob signs his name, and the proxy &quot;translates&quot; Bob's signature into a valid Alice signature in some way (for example, the proxy and Bob are each given partial information about the private key, which is not enough by itself to break Alice's key, and each can generate a &quot;partial&quot; Alice-signature. They are then combined into a full signature.)  The proxy must be trusted, however, because if the proxy and Bob cooperate, then they can fake Alice's signature on any document, well beyond those that were originally delegated to him.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a proxy re-signature scheme, the idea is that the proxy is only semi-trusted (so protecting Alice's key and signature does not depend on the proxy not cooperating with Bob to usurp Alice's prerrogatives). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here, the proxy takes a valid signature from Alice in a document, and translates it into a valid signature from Bob in the same document or other documents, which are then further counter-signed by Bob for validity. The process is meant to be such that in order to do the translation, the proxy does not learn anything about Alice's signing key or Bob's; thus, there can be no cooperation between Bob and the proxy to fake Alice's signature. Moreover, the resignature is meant to be done without access to either key, so that Alice and Bob can keep their keys secure and off-line.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That said, I don't really know much about the mechanics of &lt;em&gt;how&lt;/em&gt; one sets these up, so I cannot answer that part of your question.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2011-03-01T16:29:20.330" />
  <row Id="24441" PostTypeId="2" ParentId="24438" CreationDate="2011-03-01T16:31:45.607" Score="3" ViewCount="" Body="&lt;p&gt;See my second (according to date) answer to &lt;a href=&quot;http://math.stackexchange.com/questions/19119/approximating-pi-using-monte-carlo-integration/19128#19128&quot;&gt;this question&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Elaborating&lt;/strong&gt; (based on results from the aforementioned thread, and using the same notation).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $\bar Z_n = \frac{{\sum\nolimits_{i = 1}^n {Z_i } }}{n}$ be the corresponding quotient approximating $\pi/4$ (with probability $1$, $\bar Z_n \to \pi/4$ as $n \to \infty$). It has expectation and variance given by&#xA;$$&#xD;&#xA;{\rm E}[\bar Z_n ] = \frac{\pi }{4}, \;\; {\rm Var}[\bar Z_n ] = \frac{\pi }{{4n}}\Big(1 - \frac{\pi }{4}\Big) &amp;lt; \frac{{10}}{{59n}},&#xD;&#xA;$$&#xA;leading to the following probabilistic error bound:&#xA;$$ &#xD;&#xA;{\rm P}\bigg[\bigg|\bar Z_n - \frac{\pi }{4} \bigg| \geq \varepsilon \bigg] &amp;lt; \frac{{10}}{{59n\varepsilon ^2 }}.&#xD;&#xA;$$&#xA;Thus, for example, &#xA;$$ &#xD;&#xA;{\rm P}\bigg[\bigg|\bar Z_{10^6} - \frac{\pi }{4} \bigg| \geq 10^{-2} \bigg] &amp;lt; \frac{{1}}{{590 }}.&#xD;&#xA;$$&#xA;Further, the central limit theorem leads to the following approximation (in distribution):&#xA;$$&#xD;&#xA;\bar Z_n - \frac{\pi }{4} \approx \frac{\sigma }{{\sqrt n }}\xi,&#xD;&#xA;$$&#xA;where &#xA;$$&#xD;&#xA;\sigma^2 = {\rm Var}[Z_1] = \frac{\pi }{4} \Big(1 - \frac{\pi }{4}\Big) &amp;lt; \frac{10}{59}&#xD;&#xA;$$&#xA;(hence $\sigma \approx 0.41$) and $\xi \sim {\rm Normal}(0,1)$.&#xA;For example, for $n=10^6$, this gives&#xA;$$&#xD;&#xA;\bar Z_{10^6} - \frac{\pi }{4} \approx 4.1 \times 10^{-4} \xi.&#xD;&#xA;$$&#xA;Now, you can use a &lt;a href=&quot;http://davidmlane.com/hyperstat/z_table.html&quot; rel=&quot;nofollow&quot;&gt;Normal Distribution Calculator&lt;/a&gt; to get a probabilistic estimate of the error. For example, ${\rm P}(|\xi| \leq 1) \approx 0.682689$, ${\rm P}(|\xi| \leq 2) \approx 0.9545$, and ${\rm P}(|\xi| \leq 3) \approx 0.9973$. &#xA;Hence, with high probability, the absolute error $|\bar Z_{10^6} - \pi/4|$ will be less than, say, $10^{-3}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally,  it is interesting to compare the above approximation with the following one (see my first answer to the related question). Let $\bar Y_n = \frac{{\sum\nolimits_{i = 1}^n {\sqrt {1 - U_i^2 } } }}{n}$, where $U_i$ are independent uniform$[0,1]$ variables (it too converges, with probability $1$, to $\pi / 4$ as $n \to \infty$). $\bar Y_n$ has expectation and variance given by&#xA;$$&#xD;&#xA;{\rm E}[\bar Y_n ] = \frac{\pi }{4}, \;\; {\rm Var}[\bar Y_n ] = \frac{1}{n} \bigg \{ \frac{2}{3} - \frac {{\pi ^2 }}{{16}}\bigg\}  &amp;lt; \frac{1}{{20n}},&#xD;&#xA;$$&#xA;leading to&#xA;$$&#xD;&#xA;{\rm P}\Big[\Big|\bar Y_n - \frac{\pi }{4} \Big| \geq \varepsilon \Big] &amp;lt; \frac{1}{{20n\varepsilon ^2 }}.&#xD;&#xA;$$&#xA;As before, the central limit theorem leads to the approximation &#xA;$$&#xD;&#xA;\bar Y_n - \frac{\pi }{4} \approx \frac{\sigma }{{\sqrt n }}\xi&#xD;&#xA;$$&#xA;(with $\xi \sim {\rm Normal}(0,1)$), but this time&#xA;$$&#xD;&#xA;\sigma^2 = {\rm Var}[Y_1] = \frac{2}{3} - \frac{{\pi ^2 }}{{16}} &amp;lt; \frac{1}{20}&#xD;&#xA;$$&#xA;(hence $\sigma \approx 0.223$). Note that the variance in this case is considerably smaller than the one in the previous case.&lt;/p&gt;&#xA;" OwnerUserId="2810" LastEditorUserId="2810" LastEditorDisplayName="" LastEditDate="2011-03-02T13:57:58.460" LastActivityDate="2011-03-02T13:57:58.460" CommentCount="3" />
  <row Id="24443" PostTypeId="2" ParentId="1748" CreationDate="2011-03-01T17:11:34.287" Score="3" ViewCount="" Body="&lt;p&gt;(answer moved from &lt;a href=&quot;http://math.stackexchange.com/questions/24408/are-there-any-good-algebraic-geometry-books-to-recommend/24411#24411&quot;&gt;closed duplicated question&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are interested in learning Algebraic Geometry I recommend the books of my Amazon list. They are in recommended order to learn from the beginning by yourself:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.amazon.com/lm/RHQS8Y3V7LJRQ/ref=cm_pdp_lm_title_1&quot; rel=&quot;nofollow&quot;&gt;http://www.amazon.com/lm/RHQS8Y3V7LJRQ/ref=cm_pdp_lm_title_1&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, from that list, a quick path to understand basic Algebraic Geometry would be to read &lt;strong&gt;Bertrametti et al.&lt;/strong&gt; &lt;em&gt;&quot;Lectures on Curves, Surfaces and Projective Varieties&quot;&lt;/em&gt;, &lt;strong&gt;Shafarevich&lt;/strong&gt;'s &lt;em&gt;&quot;Basic Algebraic Geometry&quot;&lt;/em&gt; vol. 1, 2 and &lt;strong&gt;Perrin&lt;/strong&gt;'s &lt;em&gt;&quot;Algebraic Geometry an Introduction&quot;&lt;/em&gt;. But then you are entering the world of abstract algebra. For that it is advisable to master the book of Miles &lt;strong&gt;Reid&lt;/strong&gt; &lt;em&gt;&quot;Undergraduate Commutative Algebra&quot;&lt;/em&gt; accompanied by &lt;strong&gt;Kemper&lt;/strong&gt;'s &lt;em&gt;&quot;A Course in Commutative Algebra&quot;&lt;/em&gt;. Both are very readable and basic, the former geometry-oriented and the latter has lots of solved problems. There is also a very nice and easy geometric introduction which may serve for undergraduates by &lt;strong&gt;Ueno&lt;/strong&gt; &lt;em&gt;&quot;Introduction to Algebraic Geometry&quot;&lt;/em&gt; but it should be used with other more standard titles since it is not a textbook.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My personal opinion is to avoid as main text books like &lt;em&gt;Harris&lt;/em&gt;' or &lt;em&gt;Cox et al&lt;/em&gt;., above all if you have limited time to learn. In my experience these kind of books will not get you very far within algebraic geometry by themselves, although they are good companions as sources for examples and computations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Focusing only at the undergraduate level the best books in order of sophistication are: &lt;strong&gt;Smith et al.&lt;/strong&gt; &lt;em&gt;&quot;An intitation to Algebraic Geometry&quot;&lt;/em&gt;, &lt;strong&gt;Reid&lt;/strong&gt; &lt;em&gt;&quot;Undergraduate Algebraic Geometry&quot;&lt;/em&gt;, &lt;strong&gt;Hulek&lt;/strong&gt; &lt;em&gt;&quot;Elementary Algebraic Geometry&quot;&lt;/em&gt;, &lt;strong&gt;Beltrametti et al.&lt;/strong&gt; *&quot;Lectures on Curves, Surfaces and Projective Varieties&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are several free online pdf courses. Concretely, the &lt;a href=&quot;http://www.math.lsa.umich.edu/~idolga/lecturenotes.html&quot; rel=&quot;nofollow&quot;&gt;introductory notes&lt;/a&gt; by &lt;strong&gt;Dolgachev&lt;/strong&gt; &lt;em&gt;&quot;Introduction to Algebraic Geometry&quot;&lt;/em&gt; (along with his &quot;Topics on Classical Algebraic Geometry&quot; to be published soon) are very algebraic and may be read without problems after or along with a book like &lt;strong&gt;Hulek&lt;/strong&gt;'s &lt;em&gt;&quot;Elementary Algebraic Geometry&quot;&lt;/em&gt;. For an introduction which covers very nicely even schemes and cohomology the very best is by far the &lt;a href=&quot;http://www.mathematik.uni-kl.de/~gathmann/alggeom.php&quot; rel=&quot;nofollow&quot;&gt;notes&lt;/a&gt; by &lt;strong&gt;Gathmann&lt;/strong&gt;'s &lt;em&gt;&quot;Algebraic Geometry&quot;&lt;/em&gt;. A nice &lt;a href=&quot;http://www.math.lsa.umich.edu/~wfulton/CurveBook.pdf&quot; rel=&quot;nofollow&quot;&gt;introductory course on algebraic curves&lt;/a&gt; is &lt;strong&gt;Fulton&lt;/strong&gt;'s &lt;em&gt;&quot;Algebraic Curves - An introduction to Algebraic Geometry&quot;&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When one dives into graduate level, it is very important to remember that Algebraic Geometry is a huge subject and there are different approaches to start. In my opinion, the word &quot;geometry&quot; is fundamental and foundational, so people should not forget that in the end there must be some geometric content or analogue. Therefore the bible book by &lt;strong&gt;Hartshorne&lt;/strong&gt; &lt;em&gt;&quot;Algebraic Geometry&quot;&lt;/em&gt; MUST be studied only after some mastering of the basics like Beltrametti et al., Shafarevich and Perrin and above all &lt;strong&gt;Mumford&lt;/strong&gt;'s &lt;em&gt;&quot;Algebraic Geometry: Complex Projective Varieties&quot;&lt;/em&gt; which is a masterpiece (you can continue to schemes by Mumford with the &lt;a href=&quot;http://www.math.upenn.edu/~chai/624_08/math624_08.html&quot; rel=&quot;nofollow&quot;&gt;unpublished notes here&lt;/a&gt;). In order to supplement Hartshorne's with another schematic point of view, the best books are &lt;strong&gt;Mumford&lt;/strong&gt;'s &lt;em&gt;&quot;The Red Book of Varieties and Schemes&quot;&lt;/em&gt; and the three volumes by &lt;strong&gt;Ueno&lt;/strong&gt; &lt;em&gt;&quot;Algebraic Geometry I. From Algebraic Varieties to Schemes&quot;&lt;/em&gt;, &lt;em&gt;&quot;Algebraic Geometry II. Sheaves and Cohomology&quot;&lt;/em&gt;, &lt;em&gt;&quot;Algebraic Geometry III. Further theory of Schemes&quot;&lt;/em&gt;. ONLY after all this materials one can understand the geometry behind the extremely algebraic but also good and interesting books like Liu Qing &quot;Algebraic Geometry and Arithmetic Curves&quot; (another approach into Arithmetic Geometry might be &lt;strong&gt;Hindry/Silverman&lt;/strong&gt; &lt;em&gt;&quot;Diophantine Geometry&quot;&lt;/em&gt; which starts with a review on algebraic geometry and proves Mordell and Faltings theorems without the use of schemes).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My personal learning path is this: Beltrametti et al.'s for classic and basic geometric foundations, Perrin's for a more algebraic introduction with a nice treatment of Riemann-Roch, Mumford's &quot;projective varieties&quot; for foundations along with a flavor on complex algebraic geometry, then study notes by Gathmann. After this, I would start approaching Mumford's red book and Ueno to supplement Hartshorne on schemes. With all this background mastering Hartshorne should not be a problem, BUT you must DO all the exercises you can since they are the most important stuff in Hartshorne's. Other sources on this regard is the new book by &lt;strong&gt;Görtz/Wedhorn&lt;/strong&gt; &lt;em&gt;&quot;Algebraic Geometry I, Schemes wiith Examples and Exercises&quot;&lt;/em&gt; and its future volume two. If one needs a different approach including some of the sheaf theory and homology needed along with Riemann surfaces, there is &lt;strong&gt;Harder&lt;/strong&gt;'s &lt;em&gt;&quot;Lectures on Algebraic Geometry I&quot;&lt;/em&gt; (and future vol. 2).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Further references about algebraic curves, surfaces, higher-dimensional varieties and other subjects can be found in my Amazon list.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(For those interested in the Complex Geometric side (Kähler, Hodge...) I recommend &lt;strong&gt;Moroianu&lt;/strong&gt;'s &lt;em&gt;&quot;Lectures on Kähler Geometry&quot;&lt;/em&gt;, &lt;strong&gt;Ballmann&lt;/strong&gt;'s &lt;em&gt;&quot;Lectures on Kähler Manifolds&quot;&lt;/em&gt; and &lt;strong&gt;Huybrechts&lt;/strong&gt;' &lt;em&gt;&quot;Complex Geometry&quot;&lt;/em&gt;. To connect this with Analysis of Several Complex Variables I recommend trying &lt;strong&gt;Fritzsche/Grauert&lt;/strong&gt; &quot;From Holomorphic Functions to Complex Manifolds&quot; and also &lt;strong&gt;Wells&lt;/strong&gt;' &lt;em&gt;&quot;Differential Analysis on Complex Manifolds&quot;&lt;/em&gt;. Or, to connect this with algebraic geometry, try, in this order, &lt;strong&gt;Miranda&lt;/strong&gt;'s &lt;em&gt;&quot;Algebraic Curves and Riemann Surfaces&quot;&lt;/em&gt;, &lt;strong&gt;Voisin&lt;/strong&gt;'s &lt;strong&gt;&quot;Hodge Theory and Complex Algebraic Geometry&quot;&lt;/strong&gt; vol. 1 and 2, and &lt;strong&gt;Griffiths/Harris&lt;/strong&gt; &lt;em&gt;&quot;Principles of Algebraic Geometry&quot;&lt;/em&gt;.)&lt;/p&gt;&#xA;" OwnerUserId="4058" LastEditorUserId="4058" LastEditorDisplayName="" LastEditDate="2011-03-01T18:20:45.303" LastActivityDate="2011-03-01T18:20:45.303" CommunityOwnedDate="2011-03-01T17:11:34.287" />
  <row Id="24444" PostTypeId="2" ParentId="24406" CreationDate="2011-03-01T17:12:07.820" Score="2" ViewCount="" Body="&lt;p&gt;You're right: $\overline{R}$ need not be a finitely generated $R$-module.  To the best of my knowledge, none of the counterexamples are &quot;easy&quot;.  Here is the one given by Kaplansky in his text &lt;em&gt;Commutative Rings&lt;/em&gt; (I quote verbatim):&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Theorem 100: Let $k$ be a field of characteristic $2$, and $T = k[[x]]$, the power series ring in an indeterminate.  With $u \in T$, let $K = k(x,u^2)$ and $L = k(x,u)$ &#xA;(field adjunction), $R = T \cap K$, $S = T \cap L$.  Then $[L:K] \leq 2$, $R$ and $S$ are discrete valuation rings with quotient fields $K$ and $L$, and $S$ is the integral closure of $R$ in $L$.  If $[L:K] = 2$, then $S$ is not a finitely generated $R$-module.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;He goes on to remark how the hypothesis $[L:K] = 2$ can be ensured, and then he gives the proof, which takes almost a page.  For what it's worth, I intend for this counterexample (or one like it) to be in my commutative algebra notes (in the section &quot;Normalization Theorems&quot;), but it is not there yet.&lt;/p&gt;&#xA;" OwnerUserId="299" LastActivityDate="2011-03-01T17:12:07.820" CommentCount="1" />
  <row Id="24445" PostTypeId="2" ParentId="20412" CreationDate="2011-03-01T17:13:51.920" Score="1" ViewCount="" Body="&lt;p&gt;Element-wise product of matrices is known as the Hadamard product, and can be notated as $A \circ B$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some &lt;a href=&quot;http://linear.ups.edu/jsmath/latest/fcla-jsmath-latestli101.html&quot; rel=&quot;nofollow&quot;&gt;basic properties of the Hadamard Product&lt;/a&gt; are described in this section from an open source linear algebra text. Wikipedia also mentions it in the article on Matrix Multiplication, with an alternate name as the Schur product.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the significance of element-wise multiplications (in signal processing), we encounter them frequently for time-windowing operations, as well as &lt;a href=&quot;https://ccrma.stanford.edu/~jos/mdft/Convolution_Theorem.html&quot; rel=&quot;nofollow&quot;&gt;pointwise multiplying in the DFT spectrum&lt;/a&gt; which is equivalent to convolution in time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wouldn't say this notation has completely caught on, in many cases $A \cdot B$ is used (like in the statement of the convolution theorem above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Searching for Hadamard Product on Math.SE will get you some other examples. (sorry, couldn't add as many links as I planned.)&lt;/p&gt;&#xA;" OwnerUserId="5610" LastActivityDate="2011-03-01T17:13:51.920" />
  <row Id="24446" PostTypeId="2" ParentId="24387" CreationDate="2011-03-01T17:26:32.947" Score="0" ViewCount="" Body="&lt;p&gt;Since the topic relates to ordered partitions of n, I would try to dovetail with existing nomenclature for ordered partitions. In particular, I suggest using 2 instead of D and 1 instead of S.&lt;/p&gt;&#xA;" OwnerUserId="1782" LastActivityDate="2011-03-01T17:26:32.947" CommentCount="1" />
  <row Id="24447" PostTypeId="2" ParentId="998" CreationDate="2011-03-01T17:43:28.890" Score="1" ViewCount="" Body="&lt;p&gt;I think Algebraic Geometry is too broad a subject to choose only one book. But &lt;a href=&quot;http://www.amazon.com/gp/richpub/listmania/fullview/RHQS8Y3V7LJRQ/ref=cm_sw_su_l&quot; rel=&quot;nofollow&quot;&gt;my personal choices for the BEST BOOKS are&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;UNDERGRADUATE: &#xA;&lt;strong&gt;Beltrametti et al.&lt;/strong&gt; &lt;em&gt;&quot;Lectures on Curves, Surfaces and Projective Varieties&quot;&lt;/em&gt; which starts from the very beginning with a classical geometric style. Very complete (proves Riemann-Roch for curves in an easy language) and concrete in classic constructions needed to understand the reasons about why things are done the way they are in advanced purely algebraic books. There are very few books like this and they should be a must to start learning the subject.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;HALF-WAY: &#xA;&lt;strong&gt;Shafarevich&lt;/strong&gt; - &lt;em&gt;&quot;Basic Algebraic Geometry&quot;&lt;/em&gt; vol. 1 and 2. They are the most complete on foundations and introductory into Schemes so they are very useful before more abstract studies. But the problems are almost impossible.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;GRADUATE FOR ALGEBRISTS AND NUMBER THEORISTS: &#xA;&lt;strong&gt;Liu Qing&lt;/strong&gt; - &lt;em&gt;&quot;Algebraic Geometry and Arithmetic Curves&quot;&lt;/em&gt;. It is a very complete book even introducing some needed commutative algebra and preparing the reader to learn arithmetic geometry like Mordell's conjecture, Faltings' or even Fermat-Wiles Theorem (If it were not out of stock, I would have said &lt;strong&gt;Iitaka&lt;/strong&gt;'s &lt;em&gt;&quot;Algebraic Geometry. Birational Geometry of Algebraic Varieties&quot;&lt;/em&gt; because it is a book very similar to Hartshorne but it proves results that the latter leaves as problems).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;GRADUATE FOR GEOMETERS: &#xA;&lt;strong&gt;Griffiths; Harris&lt;/strong&gt; - &lt;em&gt;&quot;Principles of Algebraic Geometry&quot;&lt;/em&gt;. By far the best for a complex-geometry-oriented mind. Also useful coming from studies on several complex variables.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;ONLINE NOTES:&#xA;&lt;strong&gt;Gathmann&lt;/strong&gt; - &lt;em&gt;&quot;Algebraic Geometry&quot;&lt;/em&gt; which can be found &lt;a href=&quot;http://www.mathematik.uni-kl.de/~gathmann/alggeom.php&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;. Just amazing notes; short but very complete, dealing even with schemes and cohomology and proving Riemann-Roch. It is the best free book you need to get enough algebraic geometry to understand the other titles.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;BEST ON SCHEMES:&#xA;&lt;strong&gt;Görtz; Wedhorn&lt;/strong&gt; - &lt;em&gt;Algebraic Geometry I, Schemes with Examples and Exercises&lt;/em&gt;. Tons of stuff on schemes; more complete than Mumford's Red Book. It does a great job complementing Hartshorne's treatment of schemes, above all because of the more solvable exercises.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;UNDERGRADUATE ON ALGEBRAIC CURVES:&#xA;&lt;strong&gt;Fulton&lt;/strong&gt; - &lt;em&gt;&quot;Algebraic Curves, an Introduction to Algebraic Geometry&quot;&lt;/em&gt; which can be found &lt;a href=&quot;http://www.math.lsa.umich.edu/~wfulton/CurveBook.pdf&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;. It is a classic and although the flavor is clearly of typed notes, it is by far the shortest and manageable book on curves, which serves as a very nice introduction to the whole subject. It does everything that is needed to prove Riemann-Roch for curves.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;GRADUATE ON ALGEBRAIC CURVES:&#xA;&lt;strong&gt;Arbarello; Cornalba; Griffiths; Harris&lt;/strong&gt; - &lt;em&gt;&quot;Geometry of Algebraic Curves&quot;&lt;/em&gt; vol 1 and 2. This one is focused on the reader, therefore many results are stated to be worked out. So some people find it the best way to really master the subject. Besides, the vol. 2 has finally appeared making the two huge volumes a complete reference on the subject.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;INTRODUCTORY ON ALGEBRAIC SURFACES:&#xA;&lt;strong&gt;Beauville&lt;/strong&gt; - &lt;em&gt;Complex Algebraic Surfaces&lt;/em&gt;. I have not found a quicker and simpler way to learn and clasify algebraic surfaces. The background already needed is minimum compared to other titles.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;ADVANCED ON ALGEBRAIC SURFACES:&#xA;&lt;strong&gt;Badescu&lt;/strong&gt; - &lt;em&gt;&quot;Algebraic Surfaces&quot;&lt;/em&gt;. For those needing a companion and expansion to Hartshorne's chapter. Done with more advanced tools than Beauville.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;ON MODULI SPACES AND DEFORMATIONS:&#xA;&lt;strong&gt;Hartshorne&lt;/strong&gt; - &lt;em&gt;&quot;Deformation Theory&quot;&lt;/em&gt;. Just the perfect complement to Hartshorne's main book, since it did not deal with these matters, and other books approach the subject from a different point of view (e.g. oriented for complex geometry or for physicists) than what a student of AG from Hartshorne's book may like to learn the subject.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;ON GEOMETRIC INVARIANT THEORY:&#xA;&lt;strong&gt;Mumford; Fogarty; Kirwan&lt;/strong&gt; - &lt;em&gt;&quot;Geometric Invariant Theory&quot;&lt;/em&gt;. Simply put, it is still the best and most complete. Besides, Mumford himself developed the subject. Alternatives are more introductory lectures by Dolgachev.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;ON HIGHER-DIMENSIONAL VARIETIES:&#xA;&lt;strong&gt;Debarre&lt;/strong&gt; - &lt;em&gt;&quot;Higher Dimensional Algebraic Geometry&quot;&lt;/em&gt;. The main alternative to this title is Kollar/Mori &quot;Birational Geometry of Algebraic Varieties&quot; but is regarded as much harder to understand by many students (and even professors).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="4058" LastEditorUserId="4058" LastEditorDisplayName="" LastEditDate="2011-03-01T18:07:15.047" LastActivityDate="2011-03-01T18:07:15.047" CommunityOwnedDate="2011-03-01T17:43:28.890" CommentCount="2" />
  <row Id="24448" PostTypeId="2" ParentId="24409" CreationDate="2011-03-01T17:44:16.213" Score="1" ViewCount="" Body="&lt;p&gt;Let $g(y) = f(2x+y)$. Then&#xA;$$g(y) = g(0) + y g&amp;#39;(0) + \frac{y^2}{2} g&amp;#39;&amp;#39;(\xi)$$&#xA;$$g(-y) = g(0) - y g&amp;#39;(0) + \frac{y^2}{2} g&amp;#39;&amp;#39;(\eta)$$&#xA;Note that $g&amp;#39;&amp;#39; &amp;gt; 0$. Hence, adding the above two we get $$g(y)+g(-y) = 2g(0) + \frac{y^2}{2} g&amp;#39;&amp;#39;(\xi) + \frac{y^2}{2} g&amp;#39;&amp;#39;(\eta) &amp;gt; 2g(0)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, we get $$f(2x+y) + f(2x-y) &amp;gt; 2f(2x)$$ Plug in $y=x$ and rearrange to get&#xA;$$f(2x)-f(x) &amp;lt; f(3x)-f(2x)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(PS: This is nothing but the derivation of Jensen's in the case when $\lambda = \frac{1}{2}$)&lt;/p&gt;&#xA;" OwnerUserId="3302" LastActivityDate="2011-03-01T17:44:16.213" />
  <row Id="24449" PostTypeId="1" AcceptedAnswerId="24450" CreationDate="2011-03-01T17:53:04.953" Score="1" ViewCount="67" Body="&lt;p&gt;Hey there world, I'm in year 11 right now and I just had a brief discussion with my maths teacher about function notation, specifically, how to write the result of a function squared, as apposed to the argument to a function squared.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Say we were using trigonomics functions&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a test, I wrote this, &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;sin(50)^2&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Assuming, as the programmer I am that it would be taken to mean sin(50)*sin(50)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was subsequently told that the correct notation is&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;sin^2 (50)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;or optionally&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;(sin (50))^2&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I'm going to get the marks back anyway as the calculator used the notation I thought correct, but I'm curious, is that actually 'proper' mathematical convention?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems a bit weird to bother using open close brackets to denote the argument to a function if your going to ignore them anyway. I suppose I should mention I'm Australian, just in case its some kind of English v American system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for the help!&lt;/p&gt;&#xA;" OwnerUserId="7655" LastActivityDate="2011-03-01T17:56:34.780" Title="Conventions for function notation" Tags="&lt;functions&gt;&lt;notation&gt;&lt;high-school&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="24450" PostTypeId="2" ParentId="24449" CreationDate="2011-03-01T17:56:34.780" Score="6" ViewCount="" Body="&lt;p&gt;This is a weird notational bug specific to trigonometric functions; chalk it up to historical inertia. We write $\sin^2 x$ for $(\sin x)^2$ but for a generic function $f$, more often than not $f^2(x)$ means $f(f(x))$ and does &lt;em&gt;not&lt;/em&gt; mean $(f(x))^2$ (or $f(x^2)$). On the other hand, $\sin^{-1} x$ means $\arcsin x$ rather than $\csc x$... &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is preferable to include extra parentheses when in doubt. Generally I would interpret $f(x)^2$ as $(f(x))^2$ but it is less clear whether $f(\log x)^2$ means $f((\log x)^2)$ or $(f(\log x))^2$. &lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2011-03-01T17:56:34.780" />
  <row Id="24452" PostTypeId="2" ParentId="24421" CreationDate="2011-03-01T18:22:34.057" Score="13" ViewCount="" Body="&lt;p&gt;Consider some random variables $X$ and $Y$ such that, for every $i$, $(X,Y)=(n_iy_i,n_i(1-y_i))$ with probability $w_i$. The OP asks a proof of an inequality equivalent to&#xA;$$&#xD;&#xA;E(g(X,Y))\le g(E(X),E(Y)),&#xD;&#xA;$$ &#xA;where, for every nonnegative $x$ and $y$,&#xA;$$&#xD;&#xA;g(x,y)=\sqrt{\frac{xy}{x+y+1}}.&#xD;&#xA;$$&#xA;The second partial derivatives $\partial^2_{xx}g$ and $\partial^2_{yy}g$ are negative and the determinant of the Hessian matrix of $g$ is $(xy+x+y)/(4xy(x+y+1)^3)$, which is positive. Hence both eigenvalues of the Hessian matrix are negative, the function $g$ is concave on its domain, and &lt;a href=&quot;http://en.wikipedia.org/wiki/Jensen%27s_inequality&quot; rel=&quot;nofollow&quot;&gt;Jensen's inequality&lt;/a&gt; yields the result.&lt;/p&gt;&#xA;" OwnerUserId="6179" LastEditorUserId="6179" LastEditorDisplayName="" LastEditDate="2011-03-01T20:46:56.530" LastActivityDate="2011-03-01T20:46:56.530" CommentCount="3" />
  <row Id="24454" PostTypeId="2" ParentId="1748" CreationDate="2011-03-01T18:41:26.537" Score="0" ViewCount="" Body="&lt;p&gt;karen smith wrote a little book &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0387989803&quot; rel=&quot;nofollow&quot;&gt;an invitation to algebraic geometry&lt;/a&gt; that has lots of pictures and is readable by an undergraduate.&lt;/p&gt;&#xA;" OwnerUserId="6925" LastActivityDate="2011-03-01T18:41:26.537" CommunityOwnedDate="2011-03-01T18:41:26.537" />
  <row Id="24455" PostTypeId="1" CreationDate="2011-03-01T18:46:47.013" Score="4" ViewCount="93" Body="&lt;p&gt;Does the sum &#xA;$$\sum_{z \in \mathbb{Z}^3\setminus \{(0,0,0)\}} \left( \frac{1}{|{\bf x} - {\bf z}|^2} - \frac{1}{|{\bf z}|^2} \right)$$&lt;br&gt;&#xA;converge pointwise or even uniformly for $\varepsilon &amp;lt; |{\bf x}| &amp;lt; 1-\varepsilon$?       &lt;/p&gt;&#xA;" OwnerUserId="7656" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-01T19:11:11.253" LastActivityDate="2011-03-01T21:18:47.190" Title="Convergence of Sum over Integer Lattice" Tags="&lt;sequences-and-series&gt;&lt;lattices&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" />
  <row Id="24456" PostTypeId="1" AcceptedAnswerId="24469" CreationDate="2011-03-01T18:51:09.077" Score="6" ViewCount="349" Body="&lt;p&gt;Hello,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have just watched the first half of the 3rd lecture of Gilbert Strang on the open course ware with link:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://ocw.mit.edu/courses/mathematics/18-06-linear-algebra-spring-2010/video-lectures/&quot; rel=&quot;nofollow&quot;&gt;http://ocw.mit.edu/courses/mathematics/18-06-linear-algebra-spring-2010/video-lectures/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems that with a matrix multiplication AB=C, that the entries as scalars, are formed from the dot product computations of the rows of A with the columns of B. Visual interpretations from mechanics of overlpaing forces come to mind immediately because that is the source for the dot product (inner product).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I see the rows of C as being the dot product of the rows of B, with the dot product of a particular row of A. Similar to the above and it is easy to see this from the individual entries in the matrix C as to which elements change to give which dot products.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For understanding matrix multiplication there is the geometrical interpretation, that the matrix multiplication is a change in the reference system since matrix B can be seen as a transormation operator for rotation, scalling, reflection and skew. It is easy to see this by constructing example B matrices with these effects on A. This decomposition is a strong argument and is strongly convincing of its generality. This interpreation is strong but not smooth because I would find smoother an explanation which would be an interpretation begining from the dot product of vectors and using this to explain the process and the interpretation of the results (one which is a bit easier to see without many examples of the putting numbers in and seeing what comes out which students go through).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can hope that sticking to dot products throughout the explanation and THEN seeing how these can be seen to produce scalings, rotations, and skewings would be better. But, after some simple graphical examples I saw this doesn't work as the order of the columns in matrix B are important and don't show in the graphical representation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The best explanation I can find is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://answers.yahoo.com/question/index?qid=20081012135509AA1xtKz&quot; rel=&quot;nofollow&quot;&gt;http://answers.yahoo.com/question/index?qid=20081012135509AA1xtKz&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;it is convincing but a bit dissapointing (explains why this approach preserves the &quot;composition of linear transformations&quot;; thanks@Arturo Magidin). So the question is: &lt;strong&gt;why does matrix multiplication happen as it does, and are there good practical examples to support it? Preferably not via rotations/scalings/skews&lt;/strong&gt; (thanks @lhf)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Best,&lt;/p&gt;&#xA;" OwnerUserId="1587" LastEditorUserId="1543" LastEditorDisplayName="" LastEditDate="2011-04-11T21:43:02.147" LastActivityDate="2011-04-11T21:43:02.147" Title="matrix multiplication: interpreting and understanding the multiplication process" Tags="&lt;linear-algebra&gt;&lt;matrices&gt;&lt;intuition&gt;&lt;faq&gt;" AnswerCount="2" CommentCount="7" FavoriteCount="2" />
  <row Id="24457" PostTypeId="2" ParentId="20096" CreationDate="2011-03-01T19:03:14.927" Score="1" ViewCount="" Body="&lt;p&gt;I heard somewhere about colouring graphs with a minimal number of colours. My son was four years old when we looked at this and he loved it! I am sure it is interesting for a three year old too.&lt;/p&gt;&#xA;" OwnerUserId="1154" LastActivityDate="2011-03-01T19:03:14.927" CommunityOwnedDate="2011-03-01T19:03:14.927" />
  <row Id="24458" PostTypeId="2" ParentId="24427" CreationDate="2011-03-01T19:22:11.757" Score="2" ViewCount="" Body="&lt;p&gt;Yes the answer is affirmative if I assume that no DVR defining $A$ contains the field of fraction of $A \cap F&amp;#39;$.&#xA;The collection of intersection of DVR's defining $A$ with the field $F&amp;#39;$ is the collection of DVR's defining $A\cap F&amp;#39;$. So let $(D_x,v_x)$ be the family of DVR's in in $F&amp;#39;$ defining the krull domain $A\cap F&amp;#39;$. I have to show that the field of fraction of $A\cap F&amp;#39;$ is $F&amp;#39;$. Choose any element $y$ in $F&amp;#39;$. Now by definition $\exists$ only finitely many valuations $v_x$ such that $v_x(y) &amp;lt; 0$ say $x_1,x_2, \ldots,x_n$. Note that for rest of the values of x the element $y$ is in $D_x$. Now consider the intersection of maximal ideals corresponding to $D_{x_1}, D_{x_2}, \ldots D_{x_n}$ with $A \cap F&amp;#39;$ which is non empty due to our assumption. Choose an element $z$ in the intersection and multiply $y$ by $z$ to push it inside $A \cap F&amp;#39;$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general the answer is NO.&#xA;Here is the counter example. You know $C[T]$ is a Dedekind domain and therefore is a krull domain with its field of fraction $C(T)$. Now consider the sub field to be $C(T + \frac{1}{T})$. Then the intersection $C(T + \frac{1}{T}) \cap C[T]$ is $C$.&lt;/p&gt;&#xA;" OwnerUserId="7565" LastEditorUserId="7565" LastEditorDisplayName="" LastEditDate="2011-03-01T20:48:22.510" LastActivityDate="2011-03-01T20:48:22.510" CommentCount="2" />
  <row Id="24459" PostTypeId="1" AcceptedAnswerId="24462" CreationDate="2011-03-01T19:30:25.683" Score="8" ViewCount="202" Body="&lt;p&gt;My roommates and I have an argument you guys can help to settle (peace is at stake, don't let us down!) In undergrad calculus courses, one usually explains what it means for a function to be differentiable at a point x, and then differentiable in a domain. Then the focus is entirely on this latter notion. My question is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Has the notion of differentiability at a point any interest?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That is, I'm looking for a theorem which is valid for a function regular at some point, but which needs significantly less regularity in a neighborhood of this point, or a good reason for which such a theorem doesn't exist. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, this question is very flexible, and any insight is welcome.&lt;/p&gt;&#xA;" OwnerUserId="7085" LastActivityDate="2011-03-01T20:13:35.847" Title="Differentiable at a point" Tags="&lt;calculus&gt;&lt;real-analysis&gt;&lt;soft-question&gt;&lt;definition&gt;" AnswerCount="3" FavoriteCount="1" />
  <row Id="24460" PostTypeId="1" CreationDate="2011-03-01T19:43:49.587" Score="1" ViewCount="62" Body="&lt;p&gt;Is there a straightforward algorithm for fitting data to an ellipse or other conic section? The data generally only approximately fits a portion of the ellipse. I am looking for something that doesn't involve a complicated iterative search, since this has to run at interactive speeds for data sets on the order of 100s of data points. I can downsample or cluster the data if it goes beyond that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found an article: &quot;Least-Squares Fitting of Circles and Ellipses&quot; (Gander et al 1994) and it does seem to address my needs, but it uses a lot of mathematical machinery that I either don't understand or have a library for. I'm sure I can grok it given time, but this isn't something I would like to allocate a week to doing.&lt;/p&gt;&#xA;" OwnerUserId="7657" LastActivityDate="2011-05-31T03:06:59.390" Title="Fitting data to a portion of an ellipse or conic section" Tags="&lt;geometry&gt;&lt;soft-question&gt;&lt;computational-geometry&gt;&lt;conic-sections&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="24461" PostTypeId="2" ParentId="24459" CreationDate="2011-03-01T19:53:28.267" Score="6" ViewCount="" Body="&lt;p&gt;Though somewhat trivial, there is the Local Extremum Theorem: If $f$ is differentiable at $c$ and has a local&#xA;extremum at $c$, then $f&amp;#39;(c) = 0$.&lt;/p&gt;&#xA;" OwnerUserId="2810" LastActivityDate="2011-03-01T19:53:28.267" />
  <row Id="24462" PostTypeId="2" ParentId="24459" CreationDate="2011-03-01T19:53:41.753" Score="3" ViewCount="" Body="&lt;p&gt;Differentiability at a point is useful when you want to formulate a theorem like the fundamental theorem of calculus in the Lebesgue setting. For instance a continuous nondecreasing $f:[a,b]\to\mathbb{R}$ that satisfies some nice properties (maps sets of measure 0 to sets of measure 0 and is absolutely continuous) will be differentiable almost everywhere and satisfy the familiar $f(b) - f(a) = \int_a^b f&amp;#39;(x)d\mu(x)$ where $\mu$ is the Lebesgue measure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here of course $f&amp;#39;(x)$ really means any function which is equal to $f&amp;#39;(x)$ whenever $f&amp;#39;$ exists.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Often functions will be differentiable almost everywhere or, you'll only need to consider functions differentiable almost everywhere, and in this case differentiability at a point is inherent in such a definition. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, sets of measure $0$ are small in some sense but such sets can still be dense so you can still easily have functions whose points of nondifferentiability are lurking in any interval about a point, although to construct such weird functions isn't trivial.&lt;/p&gt;&#xA;" OwnerUserId="2055" LastActivityDate="2011-03-01T19:53:41.753" />
  <row Id="24463" PostTypeId="2" ParentId="24455" CreationDate="2011-03-01T19:55:51.660" Score="1" ViewCount="" Body="&lt;p&gt;It seems to converge conditionally, with convergence if you sum over a sphere centered at the origin and then let the radius of the sphere increase without bound.  Let ${\bf x}=(0,0,d)$ and ${\bf z}=(x,y,z)$ (sorry for reusing $x$).  Then $$\begin{align}\sum_{z \in \mathbb{Z}^3\setminus \{(0,0,0)\}} \left( \frac{1}{|{\bf x} - {\bf z}|^2} - \frac{1}{|{\bf z}|^2} \right)&amp;amp;=\sum_{z \in \mathbb{Z}^3\setminus \{(0,0,0)\}} \left(\frac{1}{x^2+y^2+(z-d)^2}-\frac{1}{x^2+y^2+z^2}\right) \\ &amp;amp;=\sum_{z \in \mathbb{Z}^3\setminus \{(0,0,0)\}} \frac{(z-d)^2-z^2}{(x^2+y^2+(z-d)^2)(x^2+y^2+z^2)}\\ &amp;amp;=\sum_{z \in \mathbb{Z}^3\setminus \{(0,0,0)\}} \frac{-2dz+d^2}{(x^2+y^2+(z-d)^2)(x^2+y^2+z^2)}\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we add together the contributions from the points $(x,y,z)$ and $(x,y,-z)$ so the sum is over the upper half space  $$\begin{align}&amp;amp;=\sum \frac{-2dz+d^2}{(x^2+y^2+(z-d)^2)(x^2+y^2+z^2)}+\frac{2dz+d^2}{(x^2+y^2+(z+d)^2)(x^2+y^2+z^2)} \\ &amp;amp; =\sum \frac{-8d^2z^2+d^2(x^2+y^2)}{(x^2+y^2+(z-d)^2)(x^2+y^2+(z+d)^2)(x^2+y^2+z^2)}\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Far from the origin, where convergence is determined, this goes down at least as fast as $\frac{1}{r^4}$ and there are $r^2$ points in a spherical shell at a given radius, so we have a sum of $\frac{1}{r^2}$, which converges.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2011-03-01T21:18:47.190" LastActivityDate="2011-03-01T21:18:47.190" CommentCount="2" />
  <row Id="24464" PostTypeId="2" ParentId="20096" CreationDate="2011-03-01T19:59:44.853" Score="0" ViewCount="" Body="&lt;p&gt;Young children usually recognize shapes and geometric objects better than they understand the abstract concept of number. One possibility I've always wanted to try are arranging blocks. You can arrange blocks into two-dimensional grids to make rectangles. You can try to give a child the problem of making such a rectangle by just arranging the blocks in a grid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could try showing such a child that certain numbers cannot be arranged into rectangles without one of the sides having length 1, which corresponds to prime factorization. You don't have to explain prime factorization or factorization at all, but the experience might stick and later when the child grows up he or she might find the memory useful for understanding factorization.&lt;/p&gt;&#xA;" OwnerUserId="2055" LastActivityDate="2011-03-01T19:59:44.853" CommunityOwnedDate="2011-03-01T19:59:44.853" />
  <row Id="24465" PostTypeId="1" AcceptedAnswerId="24482" CreationDate="2011-03-01T20:00:25.883" Score="2" ViewCount="56" Body="&lt;p&gt;I'm having a disagreement with a client that I think is relatively simple mathematics, but I'd like a third party to confirm or dispute my assertions. Here's the situation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have a list of N &lt;em&gt;final prices&lt;/em&gt; p&lt;sub&gt;1&lt;/sub&gt;, ..., p&lt;sub&gt;N&lt;/sub&gt;. A &lt;em&gt;final price&lt;/em&gt; is tax-inclusive and includes two kinds of tax:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;em&gt;flat tax&lt;/em&gt;, which is the same for all items&lt;/li&gt;&#xA;&lt;li&gt;&lt;em&gt;sales tax&lt;/em&gt;, which is a percentage of the item's price&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;That means that you can define the &lt;em&gt;base price&lt;/em&gt; b&lt;sub&gt;k&lt;/sub&gt; of an item to be&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;p&lt;sub&gt;k&lt;/sub&gt; = (1 + t) &amp;middot; b&lt;sub&gt;k&lt;/sub&gt; + t&lt;sub&gt;0&lt;/sub&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;where t&lt;sub&gt;0&lt;/sub&gt; is the flat tax and t is the sales tax. Alternatively, you can define the price of an item to be &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;p&lt;sub&gt;k&lt;/sub&gt; = b&lt;sub&gt;k&lt;/sub&gt; + t&lt;sub&gt;k&lt;/sub&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;p&lt;sub&gt;k&lt;/sub&gt; = b&lt;sub&gt;k&lt;/sub&gt; + t &amp;middot; b&lt;sub&gt;k&lt;/sub&gt; + t&lt;sub&gt;0&lt;/sub&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;b&lt;sub&gt;k&lt;/sub&gt; = p&lt;sub&gt;k&lt;/sub&gt; - t&lt;sub&gt;k&lt;/sub&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;where t&lt;sub&gt;k&lt;/sub&gt; is the tax for that item.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the question is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Given only a list of prices (p&lt;sub&gt;1&lt;/sub&gt;, ..., p&lt;sub&gt;N&lt;/sub&gt;) and the &lt;em&gt;total tax&lt;/em&gt; T = t&lt;sub&gt;1&lt;/sub&gt; + ... + t&lt;sub&gt;k&lt;/sub&gt;, can you determine (b&lt;sub&gt;1&lt;/sub&gt;, ..., b&lt;sub&gt;n&lt;/sub&gt;)? If so, how would you express this?&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7658" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-01T21:24:09.530" LastActivityDate="2011-03-02T01:18:46.323" Title="Compute base price of each item in a list of items given the total sales tax" Tags="&lt;algebra&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="24467" PostTypeId="2" ParentId="24459" CreationDate="2011-03-01T20:13:35.847" Score="4" ViewCount="" Body="&lt;p&gt;There are functions that are only differentiable at one point, here is an example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the function $d:\mathbb{R}\to\mathbb{R}$ defined by $d(x)=0$ if $x$ is rational and $d(x)=1$ if $x$ is irrational. This function is not differentiable anywhere, since it is not continuous, however, it is surely bounded. Now look at $f(x)=x^2\cdot d(x)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that for a rational $x$ we have $f(x)=0$ and while for irrational $x$ we have $f(x)=x^2$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next since there are rational numbers arbitrary close to any irrational $x$ our function is not continuous at irrationals. The same argument holds for rational $x$'s whenever $x\ne0$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;At $x=0$ we have, for $h$ rational and $\ne0$,&#xA;$$\frac{f(x+h)-f(x)}{h}= \frac{f(0+h)-f(0)}{h}=\frac{f(h)-f(0)}{h}=\frac{0}{h}=0$$&#xA;while for irrational $h$ we get &#xA;$$\frac{f(x+h)-f(x)}{h} = $$\frac{f(h)-f(0)}{h}=$$\frac{h^2}{h}=h$$&#xA;which tends to $0$ as $h\to0$, in particular $f$ is differentiable at $0$ and $f&amp;#39;(0)=0$.&lt;/p&gt;&#xA;" OwnerUserId="1154" LastActivityDate="2011-03-01T20:13:35.847" CommentCount="1" />
  <row Id="24468" PostTypeId="1" AcceptedAnswerId="24473" CreationDate="2011-03-01T20:25:00.347" Score="0" ViewCount="43" Body="&lt;p&gt;Say I want to pay someone &lt;code&gt;$100.00&lt;/code&gt; but the service I use keeps 3% of the payment I make. Meaning they would only receive &lt;code&gt;$97.00&lt;/code&gt; since they would keep &lt;code&gt;$3&lt;/code&gt;. If I increase my payment to &lt;code&gt;$103&lt;/code&gt; they would keep $3.09 meaning they would only receive &lt;code&gt;$99.91&lt;/code&gt;. Is there a formula to calculate the exact extra amount I'd need to pay in order to offset the percentage fee?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I know this is tagged incorrectly but I can't create tags yet and I didn't see any that fit.)&lt;/p&gt;&#xA;" OwnerUserId="7660" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-01T21:26:58.810" LastActivityDate="2011-03-01T21:26:58.810" Title="Formula for offsetting some additional percentage like an added fee." Tags="&lt;arithmetic&gt;" AnswerCount="4" />
  <row Id="24469" PostTypeId="2" ParentId="24456" CreationDate="2011-03-01T20:32:14.803" Score="20" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Some comments first.&lt;/strong&gt; There are several serious confusions in what you write. For example, in the third paragraph, having seen that the entries of $AB$ are obtained by taking the dot product of the corresponding row of $A$ with column of $B$, you write that you view $AB$ as a dot product of &lt;strong&gt;rows&lt;/strong&gt; of $B$ and rows of $A$. It's not. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For another example, you talk about matrix multiplication &quot;happening&quot;. Matrices aren't running wild in the hidden jungles of the Amazon, where things &quot;happen&quot; without human beings. Matrix multiplication is &lt;strong&gt;defined&lt;/strong&gt; a certain way, and then the definition is why matrix multiplication is done the way it is done. You may very well ask &lt;em&gt;why&lt;/em&gt; matrix multiplication is defined the way it is defined, and whether there are other ways of defining a &quot;multiplication&quot;  on matrices (yes, there are; read further), but that's a completely separate question. &quot;Why does matrix multiplication happen the way it does?&quot; is pretty incoherent on its face.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another example of confusion is that not every matrix corresponds to a &quot;change in reference system&quot;. This is only true, viewed from the correct angle, for &lt;em&gt;invertible&lt;/em&gt; matrices.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Standard matrix multiplication.&lt;/strong&gt; Matrix multiplication is defined the way it is because it corresponds to composition of linear transformations. Though this is valid in extremely great generality, let's focus on linear transformations $T\colon \mathbb{R}^n\to\mathbb{R}^m$. Since linear transformations satisfy $T(\alpha\mathbf{x}+\beta\mathbf{y}) = \alpha T(\mathbf{x})+\beta T(\mathbf{y})$, if you know the value of $T$ at each of $\mathbf{e}_1,\ldots,\mathbf{e}_n$, where $\mathbf{e}^n_i$ is the (column) $n$-vector that has $0$s in each coordinate except the $i$th coordinate where it has a $1$, then you know the value of $T$ at every single vector of $\mathbb{R}^n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in order to describe the value of $T$, I just need to tell you what $T(\mathbf{e}_i)$ is. For example, we can take&#xA;$$T(\mathbf{e}_i) = \left(\begin{array}{c}a_{1i}\\a_{2i}\\ \vdots\\ a_{mi}\end{array}\right).$$&#xA;Then, since&#xA;$$\left(\begin{array}{c}k_1\\k_2\\ \vdots\\k_n\end{array}\right) = k_1\mathbf{e}_1 + \cdots +k_n\mathbf{e}_n,$$ we have&#xA;$$T\left(\begin{array}{c}k_1\\k_2\\ \vdots\\ k_n\end{array}\right) = k_1T(\mathbf{e}_1) + \cdots +k_nT(\mathbf{e}_n) = k_1\left(\begin{array}{c}a_{11}\\a_{21}\\ \vdots\\a_{m1}\end{array}\right) + \cdots + k_n\left(\begin{array}{c}a_{1n}\\a_{2n}\\ \vdots\\ a_{mn}\end{array}\right).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is very fruitful, then to keep track of the $a_{ij}$ in some way, and given the expression above, we keep track of them in a matrix, which is just a rectangular array of real numbers. We then think of $T$ as being &quot;given&quot; by the matrix&#xA;$$\left(\begin{array}{cccc}&#xD;&#xA;a_{11} &amp;amp; a_{12} &amp;amp; \cdots &amp;amp; a_{1n}\\&#xD;&#xA;a_{21} &amp;amp; a_{22} &amp;amp; \cdots &amp;amp; a_{2n}\\&#xD;&#xA;\vdots &amp;amp; \vdots &amp;amp; \ddots &amp;amp; \vdots\\&#xD;&#xA;a_{m1} &amp;amp; a_{m2} &amp;amp; \cdots &amp;amp; a_{mn}&#xD;&#xA;\end{array}\right).$$&#xA;If we want to keep track of $T$ this way, then for an arbitrary vector $\mathbf{x} = (x_1,\ldots,x_n)^t$ (the ${}^t$ means &quot;transpose&quot;; turn every rown into a column, every column into a row), then we have that $T(\mathbf{x})$ corresponds to:&#xA;$$\left(\begin{array}{cccc}&#xD;&#xA;a_{11} &amp;amp; a_{12} &amp;amp; \cdots &amp;amp; a_{1n}\\&#xD;&#xA;a_{21} &amp;amp; a_{22} &amp;amp; \cdots &amp;amp; a_{2n}\\&#xD;&#xA;\vdots &amp;amp; \vdots &amp;amp; \ddots &amp;amp; \vdots\\&#xD;&#xA;a_{m1} &amp;amp; a_{m2} &amp;amp; \cdots &amp;amp; a_{mn}&#xD;&#xA;\end{array}\right) \left(\begin{array}{c}&#xD;&#xA;x_1\\x_2\\ \vdots\\ x_n\end{array}\right) = \left(\begin{array}{c}&#xD;&#xA;a_{11}x_1 + a_{12}x_2 + \cdots + a_{1n}x_n\\&#xD;&#xA;a_{21}x_1 + a_{22}x_2 + \cdots + a_{2n}x_n\\&#xD;&#xA;\vdots\\&#xD;&#xA;a_{m1}x_1 + a_{m2}x_2 + \cdots + a_{mn}x_n&#xD;&#xA;\end{array}\right).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What happens when we have &lt;strong&gt;two&lt;/strong&gt; linear transformations, $T\colon \mathbb{R}^n\to\mathbb{R}^m$ and $S\colon\mathbb{R}^p\to\mathbb{R}^n$? If $T$ corresponds as above to a certain $m\times n$ matrix, then $S$ will likewise correspond to a certain $n\times p$ matrix, say&#xA;$$\left(\begin{array}{cccc}&#xD;&#xA;b_{11} &amp;amp; b_{12} &amp;amp; \cdots &amp;amp; b_{1p}\\&#xD;&#xA;b_{21} &amp;amp; b_{22} &amp;amp; \cdots &amp;amp; b_{2p}\\&#xD;&#xA;\vdots &amp;amp; \vdots &amp;amp; \ddots &amp;amp; \vdots\\&#xD;&#xA;b_{n1} &amp;amp; b_{n2} &amp;amp; \cdots &amp;amp; b_{np}&#xD;&#xA;\end{array}\right).$$&#xA;What is $T\circ S$? First, it is a linear transformation because composition of linear transformations yields a linear transformation. Second, it goes from $\mathbb{R}^p$ to $\mathbb{R}^m$, so it should correspond to an $m\times p$ matrix. Which matrix? If we let $\mathbf{f}_1,\ldots,\mathbf{f}_p$ be the (column) $p$-vectors given by letting $\mathbf{f}_j$ have $0$s everywhere and a $1$ in the $j$th entry, then the matrix above tells us that&#xA;$$S(\mathbf{f}_j) = \left(\begin{array}{c}b_{1j}\\b_{2j}\\ \vdots \\b_{nj}\end{array}\right) = b_{1j}\mathbf{e}_1+\cdots + b_{nj}\mathbf{e}_n.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, what is $T\circ S(\mathbf{f}_j)$? This is what goes in the $j$th column of the matrix that corresponds to $T\circ S$. Evaluating, we have:&#xA;\begin{align*}&#xA;T\circ S(\mathbf{f}_j) &amp;amp;= T\Bigl( S(\mathbf{f}_j)\Bigr)\\&#xA;&amp;amp;= T\Bigl( b_{1j}\mathbf{e}_1 + \cdots + b_{nj}\mathbf{e}_n\Bigr)\\&#xA;&amp;amp;= b_{1j} T(\mathbf{e}_1) + \cdots + b_{nj}T(\mathbf{e}_n)\\&#xA;&amp;amp;= b_{1j}\left(\begin{array}{c}&#xA;a_{11}\\ a_{21}\\ \vdots\\ a_{m1}\end{array}\right) + \cdots + b_{nj}\left(\begin{array}{c} a_{1n}\\a_{2n}\\ \vdots\\ a_{mn}\end{array}\right)\\&#xA;&amp;amp;= \left(\begin{array}{c}&#xA;a_{11}b_{1j} + a_{12}b_{2j} + \cdots + a_{1n}b_{nj}\\&#xA;a_{21}b_{1j} + a_{22}b_{2j} + \cdots + a_{2n}b_{nj}\\&#xA;\vdots\\&#xA;a_{m1}b_{1j} + a_{m2}b_{2j} + \cdots + a_{mn}b_{nj}&#xA;\end{array}\right).&#xA;\end{align*}&#xA;So if we want to write down the matrix that corresponds to $T\circ S$, then the $(i,j)$th entry will be&#xA;$$a_{i1}b_{1j} + a_{i2}b_{2j} + \cdots + a_{in}b_{nj}.$$&#xA;So we &lt;strong&gt;define&lt;/strong&gt; the &quot;composition&quot; or product of the matrix of $T$ with the matrix of $S$ to be precisely the matrix of $T\circ S$. We can make this definition without reference to the linear transformations that gave it birth: if the matrix of $T$ is $m\times n$ with entries $a_{ij}$ (let's call it $A$); and the matrix of $S$ is $n\times p$ with entries $b_{rs}$ (let's call it $B$), then the matrix of $T\circ S$ (let's call it $A\circ B$ or $AB$) is $m\times p$ and with entries $c_{k\ell}$, where&#xA;$$c_{k\ell} = a_{k1}b_{1\ell} + a_{k2}b_{2\ell} + \cdots + a_{kn}b_{n\ell}$$&#xA;&lt;strong&gt;by definition&lt;/strong&gt;. Why? Because then the matrix of the composition of two functions is &lt;em&gt;precisely&lt;/em&gt; the product of the matrices of the two functions. We can work with the matrices directly without having to think about the functions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In point of fact, there is nothing about the dot product which is at play in this definition. It is essentially by happenstance that the $(i,j)$ entry can be obtained as a dot product of &lt;em&gt;something&lt;/em&gt;. In fact, the $(i,j)$th entry is obtained as the &lt;strong&gt;matrix product&lt;/strong&gt; of the $1\times n$ matrix consisting of the $i$th row of $A$, with the $n\times 1$ matrix consisting of the $j$th column of $B$. Only if you transpose this column can you try to interpret this as a dot product. (In fact, the modern view is &lt;em&gt;the other way around&lt;/em&gt;: we &lt;em&gt;define&lt;/em&gt; the dot product of two vectors as a special case of a more general inner product, called the Frobenius inner product, which is defined in terms of matrix multiplication, $\langle\mathbf{x},\mathbf{y}\rangle =\mathrm{trace}(\overline{\mathbf{y}^t}\mathbf{x})$). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And because product of matrices corresponds to composition of linear transformations, all the nice properties that composition of linear functions has will &lt;em&gt;automatically&lt;/em&gt; also be true for product of matrices, &lt;em&gt;because products of matrices is nothing more than a book-keeping device for keeping track of the composition of linear transformations&lt;/em&gt;. So $(AB)C = A(BC)$, because composition of functions is associative. $A(B+C) = AB + AC$ because composition of linear transformations distributes over sums of linear transformations (sums of matrices are defined entry-by-entry because that agrees &lt;em&gt;precisely&lt;/em&gt; with the sum of linear transformations). $A(\alpha B) = \alpha(AB) = (\alpha A)B$, because composition of linear transformations behaves that way with scalar multiplication (products of matrices by scalar are defined the way they are &lt;em&gt;precisely&lt;/em&gt; so that they will correspond to the operation with linear transformations). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we &lt;strong&gt;define&lt;/strong&gt; product of matrices &lt;strong&gt;explicitly&lt;/strong&gt; so that it will match up composition of linear transformations. There really is no deeper hidden reason. It seems a bit incongruous, perhaps, that such a simple reason results in such a complicated formula, but such is life.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another reason why it is somewhat misguided to try to understand matrix product in terms of dot product is that the matrix product keeps track of &lt;em&gt;all&lt;/em&gt; the information lying around about the two compositions, but the dot product loses a lot of information about the two vectors in question. Knowing that $\mathbf{x}\cdot\mathbf{y}=0$ only tells you that $\mathbf{x}$ and $\mathbf{y}$ are perpendicular, it doesn't really tell you anything else. There is a lot of informational loss in the dot product, and trying to explain matrix product in terms of the dot product requires that we &quot;recover&quot; all of this lost information in some way. In practice, it means keeping track of all the original information, which makes trying to shoehorn the dot product into the explanation unnecessary, because you will &lt;em&gt;already&lt;/em&gt; have all the information to get the product directly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Examples that are not just &quot;changes in reference system&quot;.&lt;/strong&gt; Note that &lt;strong&gt;any&lt;/strong&gt; linear transformation corresponds to a matrix. But the only linear transformations that can be thought of as &quot;changes in perspective&quot; are the linear transformations that map $\mathbb{R}^n$ to itself, and which are one-to-one and onto. There are &lt;em&gt;lots&lt;/em&gt; of linear transfomrations that aren't like that. For example, the linear transformation $D$ from $\mathbb{R}^3$ to $\mathbb{R}^2$ defined by &#xA;$$D\left(\begin{array}{c}&#xD;&#xA;a\\b\\c\end{array}\right) = \left(\begin{array}{c}b\\2c\end{array}\right)$$&#xA;is not a &quot;change in reference system&quot; (because lots of nonzero vectors go to zero, but there is no way to just &quot;change your perspective&quot; and start seeing a nonzero vector as zero) but is a linear transformation nonetheless. The corresponding matrix is $2\times 3$, and is&#xA;$$\left(\begin{array}{cc}&#xD;&#xA;0 &amp;amp; 1 &amp;amp; 0\\&#xD;&#xA;0 &amp;amp; 0 &amp;amp; 2&#xD;&#xA;\end{array}\right).$$&#xA;Now consider the linear transformation $U\colon\mathbb{R}^2\to\mathbb{R}^2$ given by &#xA;$$U\left(\begin{array}{c}x\\y\end{array}\right) = \left(\begin{array}{c}3x+2y\\&#xD;&#xA;9x + 6y\end{array}\right).$$&#xA;Again, this is not a &quot;change in perspective&quot;, because the vector $\binom{2}{-3}$ is mapped to $\binom{0}{0}$. It has a matrix, $2\times 2$, which is&#xA;$$\left(\begin{array}{cc}&#xD;&#xA;3 &amp;amp; 2\\&#xD;&#xA;9 &amp;amp; 6&#xD;&#xA;\end{array}\right).$$&#xA;So the composition $U\circ T$ has matrix:&#xA;$$\left(\begin{array}{cc}&#xD;&#xA;3 &amp;amp; 2\\&#xD;&#xA;9 &amp;amp; 6&#xD;&#xA;\end{array}\right) \left(\begin{array}{ccc}&#xD;&#xA;0 &amp;amp; 1 &amp;amp; 0\\&#xD;&#xA;0 &amp;amp; 0 &amp;amp; 2&#xD;&#xA;\end{array}\right) = \left(\begin{array}{ccc}&#xD;&#xA;0 &amp;amp; 3 &amp;amp; 4\\&#xD;&#xA;0 &amp;amp; 9 &amp;amp; 12&#xD;&#xA;\end{array}\right),$$&#xA;which tells me that&#xA;$$U\circ T\left(\begin{array}{c}x\\y\\z\end{array}\right) = \left(\begin{array}{c} 3y + 4z\\ 9y+12z\end{array}\right).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Other matrix products.&lt;/strong&gt; Are there other ways to define the product of two matrices? Sure. There's the &lt;a href=&quot;http://en.wikipedia.org/wiki/Matrix_multiplication#Hadamard_product&quot; rel=&quot;nofollow&quot;&gt;Hadamard product&lt;/a&gt;, which is the &quot;obvious&quot; thing to try: you can multiply two matrices of the same size (and only of the same size), and you do it entry by entry, just the same way that you add two matrices. This has some nice properties, but it has nothing to do with linear transformations. There's the &lt;a href=&quot;http://en.wikipedia.org/wiki/Kronecker_product&quot; rel=&quot;nofollow&quot;&gt;Kronecker product&lt;/a&gt;, which takes an $m\times n$ matrix times a $p\times q$ matrix and gives an $mp\times nq$ matrix. This one is associated to the &lt;em&gt;tensor product&lt;/em&gt; of linear transformations. They are defined differently because they are meant to model other operations that one does with matrices or vectors. &lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2011-03-01T20:32:14.803" CommentCount="5" />
  <row Id="24470" PostTypeId="1" CreationDate="2011-03-01T20:35:42.100" Score="2" ViewCount="54" Body="&lt;p&gt;I was thinking about the fact that $\frac{1}{\zeta(n)}$ is the probability that $n$ randomly chosen integers are coprime, and wondered if there were &amp;ldquo;natural&amp;rdquo; ways in which $e^{-n}$ (for $n&gt;0$) and $\frac{1}{\Gamma(n)} $(provided that $0 &amp;lt; \frac{1}{\Gamma(n)} &amp;lt; 1$) might be interpreted as probabilities with respect to a property of some structure on $n$ objects?&lt;/p&gt;&#xA;" OwnerUserId="890" LastActivityDate="2011-03-01T21:00:04.340" Title="number-theoretic? probabilities associated with $e^{-n}$ and $1/\Gamma(n)$" Tags="&lt;number-theory&gt;&lt;real-analysis&gt;&lt;probability&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="24471" PostTypeId="2" ParentId="24468" CreationDate="2011-03-01T20:37:02.647" Score="1" ViewCount="" Body="&lt;p&gt;You want to pay &lt;code&gt;x&lt;/code&gt; to someone, but services take &lt;code&gt;r&lt;/code&gt;%, you have to pay them &lt;code&gt;y&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;x = y * (1 - (r/100))&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Then&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;y = x * / (1 - (r/100))&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="6258" LastEditorUserId="6258" LastEditorDisplayName="" LastEditDate="2011-03-01T21:10:07.733" LastActivityDate="2011-03-01T21:10:07.733" />
  <row Id="24472" PostTypeId="2" ParentId="24468" CreationDate="2011-03-01T20:37:49.080" Score="1" ViewCount="" Body="&lt;p&gt;If you are going to get 97% of x, and you want to get $100, then:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;100 = .97x &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(divide each side by .97)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;100/.97 = x&lt;/p&gt;&#xA;" OwnerUserId="1312" LastActivityDate="2011-03-01T20:37:49.080" />
  <row Id="24473" PostTypeId="2" ParentId="24468" CreationDate="2011-03-01T20:38:33.597" Score="2" ViewCount="" Body="&lt;p&gt;If $x$ is the quantity you pay, then they keep $0.97x$. Since you want $0.97x$ to equal $100$, you have&#xA;\begin{align*}&#xA;0.97 x &amp;amp;= 100\\&#xA;x &amp;amp; = &amp;amp; \frac{100}{0.97}\\&#xA;x &amp;amp; \approx &amp;amp; 103.0928&#xA;\end{align*}&#xA;so you would want to give them something between &lt;code&gt;$&lt;/code&gt;$103.09$ and &lt;code&gt;$&lt;/code&gt;$103.10$ (as it happens, 97% of 103.09 is 99.9973, which would be rounded up to 100 by most, and 97% of 103.10 is 100.007, which would be rounded up to 100.01 by most).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, if you want them to get $T$ and they only get $r$% of what you give them, then the formula is that you want to give them $x$, where&#xA;$$x = \frac{100T}{r}.$$&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2011-03-01T20:38:33.597" CommentCount="1" />
  <row Id="24474" PostTypeId="2" ParentId="24468" CreationDate="2011-03-01T20:39:44.820" Score="1" ViewCount="" Body="&lt;p&gt;In general suppose $FV$ is the future value of an asset and $PV$ is the present value of an asset. Then $FV = \frac{PV}{(1-d)^n}$ where $d = i/(i+1)$ (where $i$ is the annual effective interest rate) is the discount rate. In this example, $d = 0.03$, $n=1$ and $FV = 100$. &lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2011-03-01T20:39:44.820" />
  <row Id="24475" PostTypeId="1" CreationDate="2011-03-01T20:55:34.300" Score="1" ViewCount="152" Body="&lt;p&gt;Hi guys, I have the following exercise to do but don't know how to approach it:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $G$ be a graph with $n$ nodes ($n \ge 2$), and where every node has degree at least $3$. Show that $G$ has a cycle of length $\le 2\cdot \lceil\log n\rceil$. And it allows me to consider a breadth-first search tree.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First of all, what's the length of a cycle? Is it the number of arcs in the cycle? Secondly, how do I show it's $2\cdot\lceil\log n\rceil$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks, Sorin&lt;/p&gt;&#xA;" OwnerUserId="7662" LastEditorUserId="6622" LastEditorDisplayName="" LastEditDate="2011-03-01T21:14:45.560" LastActivityDate="2011-03-06T15:47:30.913" Title="Showing that a graph has a cycle length less than something" Tags="&lt;graph&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" />
  <row Id="24476" PostTypeId="2" ParentId="20096" CreationDate="2011-03-01T20:57:59.710" Score="2" ViewCount="" Body="&lt;p&gt;My suggestion is similar to that of oosterwal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My kids liked to hear &quot;stories&quot; of the following kind: If there are three trucks in the yard, and Thomas brings two more, how many are there now?  If Diesel comes and takes away one, then how many are there?   My experience was that for three and four year olds, simple arithmetic with numbers less than five was in reach,&#xA;and that visualizing the objects in terms of story settings and characters that they liked made it interesting and fun for them.  (Even just adding or subtracting one object is non-trivial at that age, and I think that practicing it helps develop the sense of numbers as cardinalities, and not just names.)&lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2011-03-01T20:57:59.710" CommunityOwnedDate="2011-03-01T20:57:59.710" />
  <row Id="24477" PostTypeId="1" AcceptedAnswerId="24498" CreationDate="2011-03-01T21:00:02.843" Score="1" ViewCount="142" Body="&lt;p&gt;I'm wondering if these matrices have a name? (I'm somehow tempted to call them subunitary but it seems to be reserved for &lt;a href=&quot;http://www.psi.toronto.edu/matrix/special.html&quot; rel=&quot;nofollow&quot;&gt;something else&lt;/a&gt;.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The matrix $M \in \mathbb{C}^{n\times n}$ is called ..., if  all the singular values $\sigma_1,\dots, \sigma_n$ are strictly smaller than 1. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that if all the singular values are 1 then $M$ is a unitary matrix. That is why I think it should be called subunitary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strike&gt;&#xA;The matrix $M \in \mathbb{C}^{n\times n}$ is called ..., if  it fulfills&#xA;$$M = \alpha U$$&#xA;with $|\alpha|&amp;lt;1, \alpha \in \mathbb{C}$ and $U \in U(n)$ ($U(n)$ is the group of unitary matrices). (maybe it has a name if one allows $|\alpha|=1$)&#xA;It seems like this space is closed under multiplication. But obviously it is not a group as the $0$ matrix is not invertible. &lt;/strike&gt;&lt;/p&gt;&#xA;" OwnerUserId="7266" LastEditorUserId="7266" LastEditorDisplayName="" LastEditDate="2011-05-17T07:13:00.990" LastActivityDate="2011-05-17T07:13:00.990" Title="Do these matrices have a name?" Tags="&lt;linear-algebra&gt;&lt;matrices&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="24478" PostTypeId="2" ParentId="24470" CreationDate="2011-03-01T21:00:04.340" Score="3" ViewCount="" Body="&lt;p&gt;This isn't exactly what you asked for, but with $H_n = \sum_{k=1}^n \frac{1}{k}$ the $n^{th}$ harmonic number, it turns out that $e^{-H_n}$ is asymptotically the probability that a permutation of a large set has no cycles of length at most $n$. This generalizes the well-known asymptotic for derangements when $n = 1$. I think you can prove it by inclusion-exclusion, but there is a much cleaner proof using the &lt;a href=&quot;http://qchu.wordpress.com/2009/06/24/gila-vi-the-cycle-index-polynomials-of-the-symmetric-groups/&quot; rel=&quot;nofollow&quot;&gt;exponential formula&lt;/a&gt;. &lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2011-03-01T21:00:04.340" />
  <row Id="24479" PostTypeId="2" ParentId="24460" CreationDate="2011-03-01T21:07:45.443" Score="0" ViewCount="" Body="&lt;p&gt;[Note: As lhf quite rightly pointed out in the comments, this answer is not very useful.]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think iterative methods should be able to yield results interactively for hundreds of data points, but if you want to optimize for speed and simplicity, it's important which objective function you choose. Two obvious candidates in the circular case are&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f(x_0,y_0,r) = \sum_i \left(\sqrt{(x_i-x_0)^2+(y_i-y_0)^2} - r\right)^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$g(x_0,y_0,r) = \sum_i \left((x_i-x_0)^2+(y_i-y_0)^2 - r^2\right)^2\;.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One might argue that the first alternative captures more accurately what you want to minimize, but the second one is computationally much simpler. There's a solution of the optimization conditions for it &lt;a href=&quot;http://www.dtcenter.org/met/users/docs/write_ups/circle_fit.pdf&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;. I believe that you can get quite a simple system of equations if you treat the elliptic case in the same vein:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$g(x_0,y_0,r,\mu,\nu) = \sum_i \left(\mu(x_i-x_0)^2+\nu(y_i-y_0)^2 - r^2\right)^2\;.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I haven't worked out all the details, but I believe the resulting equations are linear in $\mu$ and $\nu$ and you can solve them through a simple iteration. If you want to pursue this approach and need further assistance, let me know.&lt;/p&gt;&#xA;" OwnerUserId="6622" LastEditorUserId="6622" LastEditorDisplayName="" LastEditDate="2011-05-01T02:49:00.630" LastActivityDate="2011-05-01T02:49:00.630" CommentCount="5" />
  <row Id="24480" PostTypeId="1" CreationDate="2011-03-01T21:11:35.157" Score="1" ViewCount="91" Body="&lt;p&gt;I'm looking for solution to this non-homogeneus problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{\partial u}{\partial t}-\frac{\partial^2 u}{\partial x^2}=F(x,t)$           for     $0&amp;lt;x&amp;lt;\pi$,         $t&amp;gt;0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$u(x,0)=0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$u(0,t)=\frac{\partial u}{\partial x}(x=\pi)=0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anyone know how to proceed?&lt;/p&gt;&#xA;" OwnerUserId="7663" LastEditorUserId="7663" LastEditorDisplayName="" LastEditDate="2011-03-01T21:21:43.760" LastActivityDate="2011-03-01T22:11:48.800" Title="Resolution of a non-homogeneous heat equation" Tags="&lt;homework&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="24481" PostTypeId="2" ParentId="24456" CreationDate="2011-03-01T21:16:50.020" Score="2" ViewCount="" Body="&lt;p&gt;I think part of the problem people have with getting used to linear transformations vs. matrices is that they have probably never seen an example of a linear transformation defined without reference to a matrix or a basis. So here is such an example. Let $V$ be the vector space of real polynomials of degree at most $3$, and let $f : V \to V$ be the derivative.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$V$ does not come equipped with a natural choice of basis. You might argue that $\{ 1, x, x^2, x^3 \}$ is natural, but it's only convenient: there's no reason to privilege this basis over $\{ 1, (x+c), (x+c)^2, (x+c)^3 \}$ for any $c \in \mathbb{R}$ (and, depending on what my definitions are, it is literally impossible to do so). More generally, $\{ a_0(x), a_1(x), a_2(x), a_3(x) \}$ is a basis for any collection of polynomials $a_i$ of degree $i$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$V$ also does not come equipped with a natural choice of dot product, so there's no way to include those in the discussion without making an arbitrary choice. It really is just a vector space equipped with a linear transformation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since we want to talk about composition, let's write down a second linear transformation. $g : V \to V$ will send a polynomial $p(x)$ to the polynomial $p(x + 1)$. Note that, once again, I do not need to refer to a basis to define $g$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then the abstract composition $gf : V \to V$ is well-defined; it sends a polynomial $p(x)$ to the polynomial $p&amp;#39;(x + 1)$. I don't need to refer to a basis or multiply any matrices to see this; all I am doing is composing two functions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let's do everything in a particular basis to see that we get the same answer using the &lt;em&gt;correct and natural definition of matrix multiplication.&lt;/em&gt; We'll use the basis $ \{ 1, x, x^2, x^3 \}$. In this basis $f$ has matrix&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left[ \begin{array}{cccc} 0 &amp;amp; 1 &amp;amp; 0 &amp;amp; 0 \\\&#xD;&#xA; 0 &amp;amp; 0 &amp;amp; 2 &amp;amp; 0 \\\&#xD;&#xA; 0 &amp;amp; 0 &amp;amp; 0 &amp;amp; 3 \\\&#xD;&#xA; 0 &amp;amp; 0 &amp;amp; 0 &amp;amp; 0 \end{array} \right]$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and $g$ has matrix&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left[ \begin{array}{cccc} 1 &amp;amp; 1 &amp;amp; 1 &amp;amp; 1 \\\&#xD;&#xA; 0 &amp;amp; 1 &amp;amp; 2 &amp;amp; 3 \\\&#xD;&#xA; 0 &amp;amp; 0 &amp;amp; 1 &amp;amp; 3 \\\&#xD;&#xA; 0 &amp;amp; 0 &amp;amp; 0 &amp;amp; 1 \end{array} \right].$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I encourage you to go through all the generalities in Arturo's post in this example to verify that $gf$ has the matrix it is supposed to have. &lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2011-03-01T21:16:50.020" />
  <row Id="24482" PostTypeId="2" ParentId="24465" CreationDate="2011-03-01T21:17:44.047" Score="5" ViewCount="" Body="&lt;p&gt;From the comment, it is clear that you want to figure this out without knowing either the value of $t$ or $t_0$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;With a single price for all items you can: the base price is just the total minus the taxes, divided by the number of items. In fact, you can compute the base price of any item whose final price $p_i$ is exactly $\frac{1}{N}$th the total amount paid, $P=p_1+\cdots+p_N$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if you have at least two distinct prices among $p_1,\ldots,p_N$, then you cannot without knowing $t$ or $t_0$, or at least one $b_i$ that does not correspond to a $p_i$ that is exactly $\frac{1}{N}$th of $p_1+\cdots+p_N$. Below is the mathematical development of how to obtain examples, but of course to convince your client that this cannot be done, it will suffice to come up with an example in which the same set of data leads to two different possibilities. If that's all you need, you can skip to the paragraph that begins &lt;em&gt;&quot;So, to construct an example ...&quot;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Corrected.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the original prices are $(b_1,\ldots,b_N)$, with $B=b_1+\cdots+b_N$, then the total tax on them is $T = tB+Nt_0$. You also know $P=p_1+\cdots+p_N$, so you also know $(1+t)B + Nt_0$. Thus, $P-T = (1+t)B+Nt_0 - (tB+Nt_0) = B$.  So we know $B$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $T = tB + Nt_0$ and you know $T$, $B$, and $N$, you can parametrize $t_0$ in terms of $t$: solving for $Nt_0$ we get $Nt_0 = T-tB$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So each choice of $t$ for which $T-tB\gt 0$ gives a possible choice for $t_0$ that will maintain the same tax.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But you also know $p_i$ for each $i$. Since&#xA;$$p_i = (1+t)b_i + t_0 = (1+t)b_i + \frac{1}{N}\left(T-tB\right)$$&#xA;and since you know $B$, $T$, $N$, and have fixed a value of $t$, you get&#xA;$$b_i =  \frac{1}{1+t}\left(p_i + \frac{1}{N}(tB-T)\right) = \frac{Np_i + tB-T}{N(1+t)}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each selection of $t$ will give you a value for $b_i$ as above. If you add up the $b_i$ you get $B$; and if you take $(1+t)b_i + t_0$, you will get&#xA;$$p_i + \frac{1}{N}(tB-T) + t_0 = p_i$$&#xA;(since $t_0 = \frac{1}{N}(T-tB)$). Moreover, the tax paid in this case will be&#xA;\begin{align*}&#xA;\sum(tb_i) + T-tB &amp;amp;= \sum((1+t)b_i - b_i) + T-tB = \sum(p_i) - \sum b_i + T-tB \\&#xA;&amp;amp;= P-B+T-tB = P-(1+t)B+T = T&#xA;\end{align*}&#xA;again the right answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Will different choices of $t$ yield the same base prices? Not in general. Because if $b_i$ is the value obtained by using $t$ and $b_i&amp;#39;$ is the value obtain by using $t&amp;#39;$, with $t\neq t&amp;#39;$, then setting $b_i=b_i&amp;#39;$ we get&#xA;\begin{align*}&#xA;\frac{Np_i+tB-T}{N(1+t)} &amp;amp;=\frac{Np_i+t'B-T}{N(1+t')}\\&#xA;(1+t')(Np_i+tB-T) &amp;amp;= (1+t)(Np_i+t'B-T)\\&#xA;Np_i + tB - T + t'(Np_i+tB-T) &amp;amp;= Np_i + t'B - T + t(Np_i+t'B-T)\\&#xA;(t-t')B + (t'-t)(Np_i - T) &amp;amp;= 0\\&#xA;(t-t')(T + B - Np_i) &amp;amp;= 0&#xA;\end{align*}&#xA;which occurs if and only if $Np_i =T+ B = P$, if and only if $Np_i = P$, that is, $p_i$ is exactly $\frac{1}{N}$th the total price.  If that is the case, then all possible choices of $t$ will yield the same base price, namely $\frac{1}{N}(P-T)$. But for any $p_i$ which is not exactly $\frac{1}{N}$th of $P$ you will get different values of $b_i$ for each value of $t$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, you can get different base prices with different sales tax rates (provided they satisfy $tB\leq T$, with $tB=T$ being the case where the flat tax is $0$). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, to construct an example, consider the case where the final prices were $10$ and $20$, and the total tax was $3$. So $p_1 = 10$, $p_2 = 20$, $P=30$, $T=3$, $N=2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know that the total base prices were $B = P-T = 27$. This leads to the equation&#xA;$$2t_0 = 3 - 27t.$$&#xA;We want to select two distinct values of $t$ for which $3-27t\gt 0$. That is, $t\lt \frac{1}{9}$. Take $t=.1$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $t=.1$, then $2t_0 = 3-2.7 = 0.3$, so $t_0 = 0.15$. We have a 10% sales tax and a 15 cent flat tax. If this is the case, then&#xA;\begin{align*}&#xA;b_1 &amp;amp;= \frac{2p_1 + tB-T}{2(1+t)} = \frac{20+.1(27)-3}{2(1.1)} \approx 8.954\\&#xA;b_2 &amp;amp;= \frac{2p_2 + tB-t}{2(1+t)} = \frac{40+.1(27)-3}{2(1.1)} \approx 18.045.&#xA;\end{align*}&#xA;We can verify that this works: $\$ $8.95 plus a tax of 10% (rounded up) plus 15 cents gives $\$ $10.0. And $\$ $18.05 plus a tax of 10% (rounded down, since we rounded up the price) plus 15 cents gives &#xA;$\$ $20.00. The total tax paid was $.90 + .15 + 1.80 + .15 = 3.00$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now take $t=0.08$. Then $2t_0 = 3-.08(27) = 0.84$, so now we have a sales tax of 6% and a flat tax of 42 cents. In this case, we get&#xA;\begin{align*}&#xA;b_1 &amp;amp;= \frac{2p_1 + tB - T}{2(1+t)} = \frac{20 + .08(27)-3}{2(1.08)} \approx 8.87\\&#xA;b_2 &amp;amp;= \frac{2p_2 + tB - T}{2(1+t)} = \frac{40 + .08(27) - 3}{2(1.08)} \approx &#xA;18.13&#xA;\end{align*}&#xA;A base price of $\$ $8.87, plus an 8% sales tax (71 cents) plus 42 cents flat tax gives $\$ $10.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A base price of $\$ $18.13, plus an 8% sales tax ($\$ $1.45) plus 42 cents flat tax gives $\$ $20.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The total tax is $0.71 + 0.42 + 1.45 + 0.42 = 3.00$, again as expected.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we get two possibilities: a 10% sales tax with a 15 cent flat tax, with base prices of $\$ $8.95 and $\$ $18.05; or a sales tax of 8% and a flat tax of 42 cents, with base prices of $\$ $8.87 and $\$ $18.13. Different base prices, same individual final prices, same total tax paid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Knowing a single base price $b_i$ which does not correspond to a $p_i$ with $Np_i=P$ will also suffice, since from the equation&#xA;$$p_i = (1+t)b_i + \frac{1}{N}\left(T-tB\right)$$&#xA;you can solve for $t$, knowing $p_i$, $N$, $T$, and $B$, provided only that $tb_i\neq \frac{B}{N}$, which occurs if and only if $p_i = \frac{P}{N}$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in summary, without knowing at least one $b_i$ that does not correspond to a $p_i$ with $Np_i=P$, or knowing $t_0$, or knowing $t$, an at least one price $p_i$ not $\frac{1}{N}$th of $P$, you cannot figure out the base prices. But if you know at least one of $t$, $t_0$, or a $b_i$ with $Nb_i\neq B=P-T$, then you can (or if all $p_i$ are equal).&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-02T01:18:46.323" LastActivityDate="2011-03-02T01:18:46.323" CommentCount="1" />
  <row Id="24483" PostTypeId="2" ParentId="24480" CreationDate="2011-03-01T21:23:47.607" Score="1" ViewCount="" Body="&lt;p&gt;You should find the fundamental solution (Green's function)&#xA;$$(\partial_t - \partial_x^2) G(x,t) = \delta(x) \delta(t)$$&#xA;where $\delta$ is the Dirac delta function. This solution depends on the boundary conditions you impose on your solution. A particular solution of the inhomogeneous equation is then given by the convolution $G \star F$. ($F$ should also obey the boundary conditions)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For your concrete boundary conditions, one can get the fundamental solution. Fourier transforming from $x$ to $k$, the homogeneous equation reads&#xA;$$(\partial_t+ k^2) f=0.$$&#xA;The solution is given by $f=e^{-k^2 t}$. In order to satisfied the boundary conditions at $x=0$ and $x=\pi$, we have to add the solutions with $k$ and $-k$: $G(x,t) = \sum_k \sin(k x) e^{-k^2 t}$, where $k = \frac{1}{2} + n$ with $n\in \mathbb{N}_0$.&lt;/p&gt;&#xA;" OwnerUserId="7266" LastEditorUserId="7266" LastEditorDisplayName="" LastEditDate="2011-03-01T22:11:48.800" LastActivityDate="2011-03-01T22:11:48.800" CommentCount="1" />
  <row Id="24484" PostTypeId="2" ParentId="24284" CreationDate="2011-03-01T21:28:06.963" Score="0" ViewCount="" Body="&lt;p&gt;The common method to talk about a function in predicate logic is to say, for the desired $y = f(x)$,&#xA;$$&#xD;&#xA;\forall x \exists y \ f(x, y).&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This treats $f$ as a binary relation, with the quantifiers making the relation act more like a function, namely for every $x$ there is at least one $y$ that satisfies whatever conditions are in the (at this point) predicate $f$ (one could add guards that would restrict this further to say 'at most one $y$' or 'exactly one $y$', or to restrict the domain or codomain).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in that sense, in this traditional logical rewriting, both variables are bound (since they are mentioned in the quantifiers).&lt;/p&gt;&#xA;" OwnerUserId="1919" LastActivityDate="2011-03-01T21:28:06.963" />
  <row Id="24485" PostTypeId="1" CreationDate="2011-03-01T21:29:12.207" Score="2" ViewCount="98" Body="&lt;p&gt;I'm a bit rusty on my math, so please forgive me if my terminology is wrong or I'm overlooking  extending a simple formula to solve the problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a collection of points in 2D space (x, y coordinates). I want to find the &quot;average&quot; point within that collection. (Centroid, center of mass, barycenter might be better terms.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is the average point just that whose x coordinate is the average of the x's and y coordinate is y the average of the y's?&lt;/p&gt;&#xA;" OwnerUserId="7661" LastActivityDate="2011-03-01T21:35:48.283" Title="Find the average of a collection of points (in 2D space)" Tags="&lt;geometry&gt;&lt;average&gt;" AnswerCount="2" />
  <row Id="24486" PostTypeId="2" ParentId="24485" CreationDate="2011-03-01T21:32:20.897" Score="3" ViewCount="" Body="&lt;p&gt;Yes, you can compute the average for each coordinate separately because $$\frac{1}{n} \sum (x_i,y_i) = \frac{1}{n} (\sum x_i, \sum y_i) = (\frac{1}{n}\sum x_i, \frac{1}{n}\sum y_i)$$&lt;/p&gt;&#xA;" OwnerUserId="589" LastActivityDate="2011-03-01T21:32:20.897" CommentCount="3" />
  <row Id="24488" PostTypeId="2" ParentId="24485" CreationDate="2011-03-01T21:35:48.283" Score="3" ViewCount="" Body="&lt;p&gt;There are different types of averages. Only the average of numbers is unambigious. The average you are looking for depends on what you want to use it for.  If you take the avg. x and y coordinates separately, that will give you the center of mass.&lt;/p&gt;&#xA;" OwnerUserId="4929" LastActivityDate="2011-03-01T21:35:48.283" />
  <row Id="24489" PostTypeId="1" AcceptedAnswerId="24527" CreationDate="2011-03-01T21:51:28.070" Score="7" ViewCount="285" Body="&lt;p&gt;So I picked up a couple of good undergraduate-level books over the weekend and have been working through them...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In &lt;em&gt;Algebra: Chapter 0&lt;/em&gt;, the author of the text writes:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The prototype of the well-behaved relation is '=', which corresponds to 'the diagonal'&#xA;  $$\{ (a, b) \in S \times S \, | \, a = b \} = \{(a, a) \, | \, a \in S \} \subseteq S \times S $$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The problem I'm having is that the way I'm interpreting this expression makes it seem uninteresting, almost tautological, which I'm thinking can't be correct.  What I think this expression is saying is this:  Suppose I give you elements $a$ and $b$ in the set $S$ to compare for equality.  To do that, write $a = b$ as the ordered pair $(a,b)$, which is the same element as $(a, a)$, which by a previous definition we are told is true iff $\{b\} = \{a\}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But when I try to bring this to a concrete level, by letting $S$ be the set of all polynomials with integer coefficients, say, I become confused.  Since $(0, 6x + (-6)x)$ is not the same ordered pair as $(0, 0)$, aren't we left to conclude that $(6x + (-6)x) \neq 0$?  That doesn't seem right. The only thing I can see this definition being useful for is writing tautologies like $A = A$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And then I wondered: What does it mean to write an expression like $x^2 = 2x$?  Am I correct in saying that the &quot;$=$&quot; there is not the same &quot;$=$&quot; as in the definition given above?  How would one write the definition of this new &quot;$=$&quot; in set-theoretic terms, with universal quantifiers?&lt;/p&gt;&#xA;" OwnerUserId="6296" LastActivityDate="2011-03-02T12:30:39.653" Title="Naive set theory question on &quot;=&quot;" Tags="&lt;elementary-set-theory&gt;" AnswerCount="5" CommentCount="1" FavoriteCount="1" />
  <row Id="24490" PostTypeId="2" ParentId="24489" CreationDate="2011-03-01T21:56:43.630" Score="4" ViewCount="" Body="&lt;p&gt;Consider $S=\{1,2,3\}$. The diagonal is $\{ (1,1), (2,2), (3,3) \}$. There is nothing trivial or tautological about it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;BTW, $(0, 6x + (-6)x)$ &lt;em&gt;is&lt;/em&gt; the same ordered pair as $(0, 0)$ because $6x + (-6)x=0$ as polynomials. You need to recall the definition of equality of polynomials.&lt;/p&gt;&#xA;" OwnerUserId="589" LastEditorUserId="589" LastEditorDisplayName="" LastEditDate="2011-03-01T22:08:06.590" LastActivityDate="2011-03-01T22:08:06.590" CommentCount="3" />
  <row Id="24491" PostTypeId="2" ParentId="2949" CreationDate="2011-03-01T22:00:19.680" Score="2" ViewCount="" Body="&lt;p&gt;that properties of recursive function are not always provable. For example, existence of an algorithm which non-terminates and whose non-termination cannot be proved.&lt;/p&gt;&#xA;" OwnerUserId="4876" LastActivityDate="2011-03-01T22:00:19.680" CommunityOwnedDate="2011-03-01T22:00:19.680" />
  <row Id="24492" PostTypeId="2" ParentId="24480" CreationDate="2011-03-01T22:06:31.850" Score="1" ViewCount="" Body="&lt;p&gt;In case you don't know how to get the Green's function for the boundary conditions you have: You can find the solutions of the corresponding homogeneous equation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\partial u}{\partial t}-\frac{\partial^2 u}{\partial x^2}=0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;by separation of variables:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$u(x,t)=X(x)T(t)\;,$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{T&amp;#39;}{T}=\frac{X&amp;#39;&amp;#39;}{X}\;.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then if you have any solution of the inhomogeneous equation that doesn't satisfy the boundary conditions, you can expand the error in the solutions of the homogeneous equation and substract the result from your solution of the inhomogeneous equation.&lt;/p&gt;&#xA;" OwnerUserId="6622" LastActivityDate="2011-03-01T22:06:31.850" />
  <row Id="24493" PostTypeId="2" ParentId="24489" CreationDate="2011-03-01T22:12:37.523" Score="3" ViewCount="" Body="&lt;p&gt;The difference between the two is that on the left you start with a more general set(the set of all ordered pairs) and then reduce it to the one you want with an equivalence relation; while on the right you directly construct the set you want. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your polynomial example isn't a counterexample because you're confusing set equality with polynomial equality. If the object $(0,6x + (-6)x)$ is considered different than $(0,0)$, then, yes, $(6x + (-6)x)$ is considered a different object than $0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To give another example, if you look at the integers modulo $7$, the number $0$ is &lt;strong&gt;equivalent&lt;/strong&gt; to the number $7$ but as objects in the base system they are not &lt;strong&gt;equal&lt;/strong&gt;.&lt;/p&gt;&#xA;" OwnerUserId="5468" LastActivityDate="2011-03-01T22:12:37.523" />
  <row Id="24494" PostTypeId="1" AcceptedAnswerId="24523" CreationDate="2011-03-01T22:15:19.437" Score="2" ViewCount="66" Body="&lt;p&gt;For $\alpha, \beta \in \mathbb{Z}[\zeta_3]$ , is it true that $(\alpha \cdot \beta)&amp;#39;=\alpha&amp;#39;\cdot \beta&amp;#39;$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $\alpha, \beta \in \mathbb{Z}[\zeta_3]$ , is it true that $(\alpha + \beta)&amp;#39;=\alpha&amp;#39;+ \beta&amp;#39;$?&lt;/p&gt;&#xA;" OwnerUserId="2212" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-02T01:28:13.493" LastActivityDate="2011-03-02T04:25:16.357" Title="In $\mathbb{Z}[\zeta_3]$ is the product of conjugates equal to the conjugate of the product?" Tags="&lt;algebraic-number-theory&gt;" AnswerCount="2" CommentCount="5" />
  <row Id="24496" PostTypeId="2" ParentId="24475" CreationDate="2011-03-01T22:20:11.570" Score="2" ViewCount="" Body="&lt;p&gt;Use your hint. Start &quot;exploring&quot; the graph from some vertex. As long as the edges you explore don't close a cycle, you'll get some tree - here you get to use the assumption on the degrees (what does it tell you about the tree?). The tree cannot grow forever (why?), and this is how you get your cycle.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2011-03-01T22:20:11.570" CommentCount="4" />
  <row Id="24497" PostTypeId="1" AcceptedAnswerId="24502" CreationDate="2011-03-01T22:23:19.290" Score="2" ViewCount="80" Body="&lt;p&gt;In the well known matrix form of a least squares regression&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where I am trying to solve for &lt;strong&gt;B&lt;/strong&gt; in Y = B1X1 + B2X2 + B3&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I might be given X and Y sample data as something like &lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$X$ = &#xA;$\begin{bmatrix} &#xD;&#xA;1 &amp;amp; 2 \\&#xD;&#xA;2 &amp;amp; 3 \\ &#xD;&#xA;\end{bmatrix}$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$Y$ = &#xA;$\begin{bmatrix} &#xD;&#xA;7 \\&#xD;&#xA;8 \\ &#xD;&#xA;\end{bmatrix}$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In such an arrangement how do I include B3?  I would think I would want to add it in as always 1 in &lt;strong&gt;X&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;IE&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$X$ = &#xA;$\begin{bmatrix} &#xD;&#xA;1 &amp;amp; 2 \\&#xD;&#xA;2 &amp;amp; 3 \\ &#xD;&#xA;1 &amp;amp; 1 \\&#xD;&#xA;\end{bmatrix}$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sorry if the question is not clear.&lt;/p&gt;&#xA;" OwnerUserId="7605" LastEditorUserId="6335" LastEditorDisplayName="" LastEditDate="2011-03-01T23:01:53.950" LastActivityDate="2011-03-05T23:47:55.033" Title="How to handle constant term in Least Squares Regression?" Tags="&lt;linear-algebra&gt;&lt;regression&gt;" AnswerCount="1" />
  <row Id="24498" PostTypeId="2" ParentId="24477" CreationDate="2011-03-01T22:24:01.850" Score="1" ViewCount="" Body="&lt;p&gt;If you allow the eigenvalues to have different norms, all at most $1$, then it's a &lt;em&gt;contraction&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2011-03-01T22:24:01.850" CommentCount="4" />
  <row Id="24499" PostTypeId="2" ParentId="24287" CreationDate="2011-03-01T22:26:28.853" Score="1" ViewCount="" Body="&lt;p&gt;(Maybe it shouldn't be an answer, as I won't say anything no one said before.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A knot group is indeed a f.p. (obvious) group whose H1 is Z and H2 is 0 (by Alexander's duality). Moreover, as the meridian normally generates the group (that's a clear consequence of the Wirtinger's presentation), its weight is one. (The weight is the minimal number of &quot;normal&quot; generators. As far as I know, it's still a very mysterious group invariant.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of the early successes of surgery theory was to prove that these properties characterize knot groups of high dimension. (Michel Kervaire, &lt;em&gt;Les nœuds de dimension supérieure&lt;/em&gt; (1965), &lt;a href=&quot;http://archive.numdam.org/article/BSMF_1965__93__225_0.pdf&quot; rel=&quot;nofollow&quot;&gt;available here&lt;/a&gt;.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, as always, the dimension three is more restrictive. For example, a knot exterior is a &lt;a href=&quot;http://en.wikipedia.org/wiki/Haken_manifold&quot; rel=&quot;nofollow&quot;&gt;Haken manifold&lt;/a&gt; so Waldhausen's work imply a lot of restrictions. For a start, the word problem for this knot has to be solvable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm sure that all the things we know about geometric decomposition and 3-manifold groups give various restrictions (for example, Stallings's fibration theorem imply that if you have an epimorphism from a knot group to a f.g. group whose kernel if f.g., then it comes from a fibration on the circle; in particular, the kernel is a surface group and the image is Z).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I think that no definitive answer is known. Globally, 3-manifold groups are still mysterious (we still don't know if they are linear!) and they are just very particular quotients of link groups so I think our knowledge of them is still limited.&lt;/p&gt;&#xA;" OwnerUserId="7085" LastActivityDate="2011-03-01T22:26:28.853" />
  <row Id="24500" PostTypeId="2" ParentId="24437" CreationDate="2011-03-01T22:29:49.227" Score="3" ViewCount="" Body="&lt;p&gt;The strength of the RSA cryptosystem hinges on the same question - more specifically, given $n=pq$ where $p,q$ are primes of the same relative magnitude, is it easy to find $p,q$? If so, you should stop putting your money in the bank.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The best known algorithm for factoring $n$ (NFS) takes time $\exp(O((\log n)^{1/3} (\log\log n)^{2/3}))$. This is much better than &quot;trial division&quot; (dividing by all numbers less than $\sqrt{n}$), which takes time $O(\sqrt{n}) = O(2^{(\log n)/2})$.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2011-03-01T22:29:49.227" CommentCount="1" />
  <row Id="24501" PostTypeId="2" ParentId="23658" CreationDate="2011-03-01T22:37:29.077" Score="2" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Axioms used in Euclid's &lt;em&gt;elements&lt;/em&gt; as translated by J. L. Heiberg&lt;/strong&gt;&lt;br&gt;&#xA;Link: &lt;a href=&quot;http://farside.ph.utexas.edu/euclid/Elements.pdf&quot; rel=&quot;nofollow&quot;&gt;http://farside.ph.utexas.edu/euclid/Elements.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Definitions&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;A point is that of which there is no part. &lt;/li&gt;&#xA;&lt;li&gt;And a line is a length without breadth. &lt;/li&gt;&#xA;&lt;li&gt;And the extremities of a line are points. &lt;/li&gt;&#xA;&lt;li&gt;A straight-line is (any) one which lies evenly with points on itself. &lt;/li&gt;&#xA;&lt;li&gt;And a surface is that which has length and breadth only. &lt;/li&gt;&#xA;&lt;li&gt;And the extremities of a surface are lines. &lt;/li&gt;&#xA;&lt;li&gt;A plane surface is (any) one which lies evenly with the straight-lines on itself. &lt;/li&gt;&#xA;&lt;li&gt;And a plane angle is the inclination of the lines to one another, when two lines in a plane meet one another, and are not lying in a straight-line.&lt;/li&gt;&#xA;&lt;li&gt;And when the lines containing the angle are straight then the angle is called rectilinear.&lt;/li&gt;&#xA;&lt;li&gt;And when a straight-line stood upon (another) straight-line makes adjacent angles (which are) equal to one another, each of the equal angles is a right-angle, and the former straight-line is called a perpendicular to that upon which it stands.&lt;/li&gt;&#xA;&lt;li&gt;An obtuse angle is one greater than a right-angle. &lt;/li&gt;&#xA;&lt;li&gt;And an acute angle (is) one less than a right-angle. &lt;/li&gt;&#xA;&lt;li&gt;A boundary is that which is the extremity of something. &lt;/li&gt;&#xA;&lt;li&gt;A figure is that which is contained by some boundary or boundaries. &lt;/li&gt;&#xA;&lt;li&gt;A circle is a plane figure contained by a single line [which is called a circumference], (such that) all of the straight-lines radiating towards [the circumference] from one point amongst those lying inside the figure are equal to one another.&lt;/li&gt;&#xA;&lt;li&gt;And the point is called the center of the circle.&lt;/li&gt;&#xA;&lt;li&gt;And a diameter of the circle is any straight-line, being drawn through the center, and terminated in each direction by the circumference of the circle. (And) any such (straight-line) also cuts the circle in half.&lt;/li&gt;&#xA;&lt;li&gt;And a semi-circle is the figure contained by the diameter and the circumference cuts off by it. And the center of the semi-circle is the same (point) as (the center of) the circle.&lt;/li&gt;&#xA;&lt;li&gt;Rectilinear figures are those (figures) contained by straight-lines: trilateral figures being those contained by three straight-lines, quadrilateral by four, and multi- lateral by more than four.&lt;/li&gt;&#xA;&lt;li&gt;And of the trilateral figures: an equilateral trian- gle is that having three equal sides, an isosceles (triangle) that having only two equal sides, and a scalene (triangle) that having three unequal sides.&lt;/li&gt;&#xA;&lt;li&gt;And further of the trilateral figures: a right-angled triangle is that having a right-angle, an obtuse-angled (triangle) that having an obtuse angle, and an acute- angled (triangle) that having three acute angles.&lt;/li&gt;&#xA;&lt;li&gt;And of the quadrilateral figures: a square is that which is right-angled and equilateral, a rectangle that which is right-angled but not equilateral, a rhombus that which is equilateral but not right-angled, and a rhomboid that having opposite sides and angles equal to one an- other which is neither right-angled nor equilateral. And let quadrilateral figures besides these be called trapezia.&lt;/li&gt;&#xA;&lt;li&gt;Parallel lines are straight-lines which, being in the same plane, and being produced to infinity in each direc- tion, meet with one another in neither (of these direc- tions).&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Postulates&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Let it have been postulated† to draw a straight-line from any point to any point.&lt;/li&gt;&#xA;&lt;li&gt;And to produce a finite straight-line continuously in a straight-line.&lt;/li&gt;&#xA;&lt;li&gt;And to draw a circle with any center and radius. &lt;/li&gt;&#xA;&lt;li&gt;And that all right-angles are equal to one another. &lt;/li&gt;&#xA;&lt;li&gt;And that if a straight-line falling across two (other) straight-lines makes internal angles on the same side (of itself whose sum is) less than two right-angles, then the two (other) straight-lines, being produced to infinity, meet on that side (of the original straight-line) that the (sum of the internal angles) is less than two right-angles (and do not meet on the other side).&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Common Notions&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Things equal to the same thing are also equal to one another.&lt;/li&gt;&#xA;&lt;li&gt;And if equal things are added to equal things then the wholes are equal.&lt;/li&gt;&#xA;&lt;li&gt;And if equal things are subtracted from equal things then the remainders are equal.&lt;/li&gt;&#xA;&lt;li&gt;And things coinciding with one another are equal to one another.&lt;/li&gt;&#xA;&lt;li&gt;And the whole [is] greater than the part.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="7044" LastActivityDate="2011-03-01T22:37:29.077" />
  <row Id="24502" PostTypeId="2" ParentId="24497" CreationDate="2011-03-01T22:49:28.053" Score="3" ViewCount="" Body="&lt;p&gt;Usually the sample data is represented as a 'design matrix' $X$ in which each row represents one training vector. For your $X$ it would look like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\begin{bmatrix} &#xD;&#xA;1 &amp;amp; 2 &amp;amp; 1\\&#xD;&#xA;2 &amp;amp; 3 &amp;amp; 1\\ &#xD;&#xA;\end{bmatrix}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, you need to insert column of $1$'s as the first (or last) column in you design matrix.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The $X$ you proposed would be correct if each column, not row, represented a training example.&lt;/p&gt;&#xA;" OwnerUserId="7665" LastEditorUserId="7665" LastEditorDisplayName="" LastEditDate="2011-03-05T23:47:55.033" LastActivityDate="2011-03-05T23:47:55.033" CommentCount="1" />
  <row Id="24503" PostTypeId="2" ParentId="24489" CreationDate="2011-03-01T22:54:11.817" Score="4" ViewCount="" Body="&lt;p&gt;Writing a &lt;a href=&quot;http://en.wikipedia.org/wiki/Term_(first-order_logic)#Terms&quot; rel=&quot;nofollow&quot;&gt;term&lt;/a&gt;, like $(-6)+6$, usually means that the term should be evaluated. $(-6)+6=0$ makes sense because of evaluating $(-6)+6$. If we compare written terms as strings, then terms on the left and right sides of equality will be always identical, which is uninteresting, as you pointed out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for polynomials, a term describing a polynomial is evaluated (simplified) to the &lt;b&gt;canonical form&lt;/b&gt;. There are several, e.g. a partial function $f$ from natural numbers to $R$ (the set of coefficients in polynomials) which is everywhere $\neq 0$. $f(n)$ is a coefficient at n-th degree of the variable of the polynomial.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we do not have an evaluation function but have an &lt;a href=&quot;http://en.wikipedia.org/wiki/Equivalence_relation&quot; rel=&quot;nofollow&quot;&gt;equivalence relation&lt;/a&gt;, we can construct an evaluation function as a quotient map.&lt;/p&gt;&#xA;" OwnerUserId="7011" LastEditorUserId="7011" LastEditorDisplayName="" LastEditDate="2011-03-02T12:30:39.653" LastActivityDate="2011-03-02T12:30:39.653" />
  <row Id="24504" PostTypeId="1" AcceptedAnswerId="24506" CreationDate="2011-03-01T22:56:35.877" Score="5" ViewCount="90" Body="&lt;p&gt;What is the image of $\zeta_3$ under the non-identity embedding of $\mathbb{Q}(\zeta_3)$ in $\mathbb{C}$?&lt;/p&gt;&#xA;" OwnerUserId="2212" LastEditorUserId="900" LastEditorDisplayName="" LastEditDate="2011-03-02T09:00:50.850" LastActivityDate="2011-03-02T11:28:28.967" Title="What is the image of $\zeta_3$ under the non-identity embedding of $\mathbb{Q}(\zeta_3)$ in $\mathbb{C}$?" Tags="&lt;number-theory&gt;&lt;algebraic-number-theory&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="24505" PostTypeId="1" AcceptedAnswerId="24520" CreationDate="2011-03-01T23:02:36.950" Score="5" ViewCount="103" Body="&lt;p&gt;Wikipedia states that a Mahlo cardinal is hyper-inaccessible, hyper-hyper-inaccessible, etc. Is this a characterisation of Mahlo? If not what about &quot;alpha = hyper^alpha-inaccessible&quot; with the obvious transfinite recursive definition of hyper^alpha-inaccessible ?&lt;/p&gt;&#xA;" OwnerUserId="7666" LastEditorUserId="413" LastEditorDisplayName="" LastEditDate="2011-03-02T00:57:46.797" LastActivityDate="2011-03-02T01:02:47.573" Title="mahlo and hyper-inaccessible cardinals" Tags="&lt;set-theory&gt;&lt;cardinals&gt;&lt;large-cardinals&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="24506" PostTypeId="2" ParentId="24504" CreationDate="2011-03-01T23:03:54.447" Score="7" ViewCount="" Body="&lt;p&gt;If $\zeta_3$ is a primitive cubic root of $1$, then its image under the non-identity embedding is $\zeta_3^{-1}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Indeed, there is an endomorphism $\sigma:\mathbb Q(\zeta_3^{-1})\to\mathbb Q(\zeta_3^{-1})$ such that $\sigma(\zeta_3)=\zeta_3^{-1}$, which in fact generates the Galois group of the field over $\mathbb Q$. The non-trivial embedding is then the composition of $\sigma$ with the trivial embedding.&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2011-03-01T23:03:54.447" CommentCount="1" />
  <row Id="24507" PostTypeId="1" AcceptedAnswerId="24516" CreationDate="2011-03-01T23:08:04.137" Score="21" ViewCount="1186" Body="&lt;p&gt;Though I've understood the logic behind's Russell's paradox for long enough, I have to admit I've never really understood why mathematicians and mathematical historians thought it so important.  Most of them mark its formulation as an epochal moment in mathematics, it seems to me.  To my uneducated self, Russell's paradox seems almost like a trick question, like a child asking what the largest organism on Earth is, and getting back an answer talking about &lt;a href=&quot;http://en.wikipedia.org/wiki/Armillaria_solidipes&quot; rel=&quot;nofollow&quot;&gt;a giant fungus.&lt;/a&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;&lt;em&gt;Well, OK&lt;/em&gt;...&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The set of all sets admits all sorts of unmathematical objects right? Chickens, spoons, 2 $\times$ 4s, what have you...I can't imagine a situation where a mathematician solving a concrete problem would invoke a set like that.  The typical sets one sees in everyday mathematics are pretty well defined and limited in scope: $\mathbb{Z}$, $\mathbb{R}$, $\mathbb{C}$, $\mathbb{R}^n, \mathbb{Q}_p$, subsets of those things.  They are all built off each other in a nice way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But clearly important people think the result significant.  It prompted Gottlob Frege, a smart and accomplished man, who Wikipedia tells me extended the study of logic in non-trivial ways, to say that the &quot;foundations of [his] edifice were shaken.&quot;  So I suppose Russell's paradox is &quot;deep&quot; somehow?  Is there a practical example in mathematics that showcases this, where one might be led horribly astray if one doesn't place certain restrictions on sets, in say, dealing with PDEs, elliptic curves, functional equations?  Why isn't Russell's paradox just a &quot;gotcha&quot;?&lt;/p&gt;&#xA;" OwnerUserId="6296" LastEditorUserId="7011" LastEditorDisplayName="" LastEditDate="2011-03-04T16:18:10.317" LastActivityDate="2011-03-04T16:57:44.667" Title="Why did mathematicians take Russell's paradox seriously?" Tags="&lt;logic&gt;&lt;set-theory&gt;&lt;math-history&gt;" AnswerCount="11" CommentCount="14" FavoriteCount="8" />
  <row Id="24508" PostTypeId="1" CreationDate="2011-03-01T23:10:25.167" Score="0" ViewCount="45" Body="&lt;p&gt;I wonder how many different ways are there of writing the Baker-Hausdorff equation! This is a form which I recently encountered and haven't been able to figure out how it comes,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$e^ae^Xe^b = e^{[X+L_X.(b-a)+(L_X Coth L_X).(b+a)+..]}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $L_X$ is an operator which acts on another matrix say $f$ as,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$L_X.f = \frac{1}{2}[X,f]$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would be grateful if someone can help with the above.&lt;br&gt;&#xA;(and motivate it) &lt;/p&gt;&#xA;" OwnerUserId="965" LastActivityDate="2011-03-01T23:10:25.167" Title="A form of the Baker-Hausdorff equation" Tags="&lt;algebra&gt;&lt;matrices&gt;" CommentCount="2" />
  <row Id="24509" PostTypeId="1" AcceptedAnswerId="24512" CreationDate="2011-03-01T23:29:53.380" Score="2" ViewCount="189" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Possible Duplicate:&lt;/strong&gt;&lt;br&gt;&#xA;  &lt;a href=&quot;http://math.stackexchange.com/questions/287/how-come-32-5-31-5&quot;&gt;How come 32.5 = 31.5?&lt;/a&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/B5cCT.gif&quot; alt=&quot;Image&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="6335" LastEditorUserId="6296" LastEditorDisplayName="" LastEditDate="2011-03-07T11:25:21.417" LastActivityDate="2011-03-07T11:25:21.417" ClosedDate="2011-03-07T12:04:08.763" Title="Why does rearranging the pieces of this triangle illusion give a different area?" Tags="&lt;geometry&gt;&lt;recreational-mathematics&gt;" AnswerCount="4" CommentCount="3" />
  <row Id="24510" PostTypeId="2" ParentId="24509" CreationDate="2011-03-01T23:35:20.243" Score="1" ViewCount="" Body="&lt;p&gt;It isn't true. See the &lt;a href=&quot;http://en.wikipedia.org/wiki/Missing_square_puzzle&quot; rel=&quot;nofollow&quot;&gt;Wikipedia page&lt;/a&gt; about this puzzle.&lt;/p&gt;&#xA;" OwnerUserId="264" LastActivityDate="2011-03-01T23:35:20.243" />
  <row Id="24511" PostTypeId="2" ParentId="24509" CreationDate="2011-03-01T23:35:45.213" Score="2" ViewCount="" Body="&lt;p&gt;Eye trick! Look at the angles formed where the red and green triangle meet.&lt;/p&gt;&#xA;" OwnerUserId="5105" LastActivityDate="2011-03-01T23:35:45.213" CommentCount="2" />
  <row Id="24512" PostTypeId="2" ParentId="24509" CreationDate="2011-03-01T23:36:25.047" Score="8" ViewCount="" Body="&lt;p&gt;If you count carefully, you'll see that the base is meant to be $13$ units long, while the height is $5$ units long. That means that the triangle on the top on the top figure, which has a height of $2$, should have a base of length $b$, where&#xA;$$\frac{b}{2} = \frac{13}{5}$$&#xA;or $b = \frac{26}{5}$, longer than the $5$ units depicted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Likewise, the bottom red triangle, with a base of size $8$ should have a height of length $h$, with&#xA;$$\frac{h}{8} = \frac{5}{13}$$&#xA;or $h = \frac{40}{13}$, which is a little longer than the $3$ depicted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in fact, the &quot;missing square&quot; comes from misdrawing the pictures (or from having the individual figures drawn correctly, but the composed figures not being real triangles; the two inner triangles are not similar, though they &quot;should&quot; be). &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2011-03-01T23:57:19.423" LastActivityDate="2011-03-01T23:57:19.423" />
  <row Id="24513" PostTypeId="1" AcceptedAnswerId="25663" CreationDate="2011-03-01T23:46:32.497" Score="9" ViewCount="148" Body="&lt;p&gt;This is somewhat related to the question &lt;a href=&quot;http://math.stackexchange.com/questions/16203/why-doesnt-hom-commute-with-taking-stalks&quot;&gt;Why doesn&amp;#39;t Hom commute with taking stalks?&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is this: If $F$ and $G$ are locally free sheaves of $\mathcal{O}_X$ -modules on an arbitrary ringed space $(X,\mathcal{O}_X)$, then is the stalk of the Hom sheaf $\mathcal{H}om(F,G)$ at a point $p$ equal to $\text{Hom}_{\mathcal{O}_{X,p}}(F_p,G_p)$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I ask this because I feel I need it to solve Exercise 5.1(a) from Chapter II in Hartshorne. By proposition 6.8 of Chapter III, the answer to my question is affirmative IF $X$ is a Noetherian scheme, and it holds even if $F$ is only coherent and with no conditions on $G$, but his does not answer my question which is assuming less on $X$ and more on the sheaves. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In any case, is there a way to solve the exercise in Hartshorne without going to the stalks?&lt;/p&gt;&#xA;" OwnerUserId="7194" LastEditorUserId="7194" LastEditorDisplayName="" LastEditDate="2011-03-09T04:26:42.720" LastActivityDate="2011-03-09T04:26:42.720" Title="Does Hom commute with stalks for locally free sheaves?" Tags="&lt;algebraic-geometry&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="2" />
  <row Id="24514" PostTypeId="2" ParentId="2949" CreationDate="2011-03-01T23:53:45.877" Score="2" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Supersymmetry&lt;/strong&gt;. From a mathematical standpoint it implies that every bosonic (resp. fermionic) particle in the universe has a fermionic (resp. bosonic) superpartner, &lt;em&gt;i.e.&lt;/em&gt;, the existence of two &lt;em&gt;physical bijections&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2720" LastActivityDate="2011-03-01T23:53:45.877" CommunityOwnedDate="2011-03-01T23:53:45.877" />
  <row Id="24515" PostTypeId="2" ParentId="24494" CreationDate="2011-03-01T23:59:55.120" Score="1" ViewCount="" Body="&lt;p&gt;If conjugation means mapping $\zeta$ to $\zeta^2=1/\zeta$, then it coincides with complex conjugation, which is a ring homomorphism.&lt;/p&gt;&#xA;" OwnerUserId="589" LastActivityDate="2011-03-01T23:59:55.120" />
  <row Id="24516" PostTypeId="2" ParentId="24507" CreationDate="2011-03-02T00:07:37.183" Score="27" ViewCount="" Body="&lt;p&gt;Russell's paradox means that you can't just take some formula and talk about all the sets which satisfy the formula.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a big deal. It means that some collections cannot be sets, which in the universe of set theory means these collections are not elements of the universe. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a way, Russell's paradox is an early form of diagonilzation, namely we create some diagonal function in order to show some property. The most known proofs to feature diagonalization are the fact that the universe of Set Theory is not a set, and Cantor's theorem about the cardinality of the power set.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The point, eventually, is that some collections are not sets. This is a big deal when your universe is made out of sets, but I think that one really has to study some set theory in order to truly understand why this is a problem.&lt;/p&gt;&#xA;" OwnerUserId="622" LastActivityDate="2011-03-02T00:07:37.183" />
  <row Id="24517" PostTypeId="2" ParentId="24505" CreationDate="2011-03-02T00:10:36.510" Score="3" ViewCount="" Body="&lt;p&gt;$\kappa$ is Mahlo is it has stationary many inaccessibles below it, or alternatively every normal (continuous in terms of limit ordinals) has a fixed point within $\kappa$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the same because one can define a CLUB set as an image of a normal function, so having stationary many inaccessibles below $\kappa$ means there's always a fixed point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this way, a Mahlo cardinal is indeed inaccessible, as well as hyper-inaccessible and so on.&lt;/p&gt;&#xA;" OwnerUserId="622" LastActivityDate="2011-03-02T00:10:36.510" CommentCount="0" />
  <row Id="24518" PostTypeId="1" AcceptedAnswerId="24519" CreationDate="2011-03-02T00:29:06.837" Score="5" ViewCount="103" Body="&lt;p&gt;F is a linear functional in $V&amp;#39;$ a linear vector space which operates on $\phi\in V$. Show that there is a one-to one correspondence between F and $f\in V $ such that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$F(\phi)=(f,\phi)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $V$ is also a linear vector space.  $(,)$ represents inner product. Any hint will be helpful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Attempt: I don't even know if this is right but...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $F(\phi)=c$ , $c\in \mathbb{C}$&#xA;then let $(f_1, \phi)=c$ and $(f_2, \phi)=c$ $\rightarrow$ $(f_1-f_2, \phi)=0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But this only means orthogonality.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: so, from hint&lt;/p&gt;&#xA;&#xA;&lt;p&gt;let $\{e_i\}$ be a linearly independent basis for vectorspace $V$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$F(\Phi) = F (\sum_n \phi_n e_n) =\sum_n \phi_n F(e_n) $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(f,\Phi) = \sum_n f_n^* \phi_n$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore if $F(\phi)=c$ , $c\in \mathbb{C}$ and $(f,\Phi)=c$ then &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sum_n (f_n^* -F(e_n)) \phi_n =0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;implies (?) $f_n^* =F(e_n)$ $\forall n$? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;&#xA;Ok. So the above implies that if some $(e_i,\Phi)$ vanishes then we cannot have a unique $f$. So is this solved by specifying at the beginning that $\{e_i\}$ is a basis where $\phi_i=(e_i,\Phi)$ is non vanishing for all i?&lt;/p&gt;&#xA;" LastActivityDate="2011-03-02T01:21:59.193" Title="linear algebra basic proof " Tags="&lt;linear-algebra&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" />
  <row Id="24519" PostTypeId="2" ParentId="24518" CreationDate="2011-03-02T00:42:24.413" Score="0" ViewCount="" Body="&lt;p&gt;Hint: Like every linear transformation, a linear functional is determined by what it does on a basis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Make sure it's clear what you want to prove: For each linear functional $F$, there is a vector $f$ such that $F(\phi) = (f,\phi)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have computed $F(\phi)$ after expressing $\phi$ in a basis. Which vector $f$ do you need to take so that $F(\phi) = (f,\phi)$? Will choosing a special kind of basis help?&lt;/p&gt;&#xA;" OwnerUserId="589" LastEditorUserId="589" LastEditorDisplayName="" LastEditDate="2011-03-02T01:01:49.903" LastActivityDate="2011-03-02T01:01:49.903" CommentCount="5" />
  <row Id="24520" PostTypeId="2" ParentId="24505" CreationDate="2011-03-02T00:54:03.367" Score="2" ViewCount="" Body="&lt;p&gt;No amount of hyperinaccessibility or hyperhyperinaccessibility and so on can be provably equivalent to Mahloness (unless those notions are inconsistent). The reason is that if $\kappa$ is Mahlo, then all its hyperinaccessibility and hyperhyperinacessibility  properties and so on are expressible in the structure $\langle V_\kappa,\in\rangle$, once one knows that $\kappa$ is regular, since they have to do only with what is happening eventually below $\kappa$. But when $\kappa$ is Mahlo, then there are many inaccessible $\delta$ with $V_\delta\prec V_\kappa$, since the set of $\alpha$ with $V_\alpha\prec V_\kappa$ is club in $\kappa$ by a Lowenheim-Skolem argument. Any such $\delta$ will therefore have exactly the same hyperhyperinacessibility properties as $\kappa$, even though when $\kappa$ is the least Mahlo, then no such $\delta$ is Mahlo. So those properties do not imply Mahloness. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the property of $\kappa$ being Mahlo is naturally expressed in $V_{\kappa+1}$, since it makes reference to stationarity, which requires one to consider all subsets of $\kappa$. &lt;/p&gt;&#xA;" OwnerUserId="413" LastEditorUserId="413" LastEditorDisplayName="" LastEditDate="2011-03-02T01:02:47.573" LastActivityDate="2011-03-02T01:02:47.573" CommentCount="1" />
  <row Id="24521" PostTypeId="1" CreationDate="2011-03-02T00:57:43.493" Score="3" ViewCount="97" Body="&lt;p&gt;My algebra book has a quick practical example at the beginning of the chapter on polynomials and their functions. Unfortunately it just says &quot;this is why polynomial functions are important&quot; and moves on. &lt;strong&gt;&lt;em&gt;I'd like to know how to come up with the function&lt;/em&gt;&lt;/strong&gt; (the teacher considers this out of scope). Even suggesting google search terms to find out more information on this sort of thing would be helpful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The example&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider a stack of apples, in a pyramid shape. It starts with one apple at the top. The next layer is 2x2, then 3x3, and so on. How many apples are there, given x number of layers?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The polynomial function&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f(x) = \frac{2x^3 + 3x^2 + x}{6}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What I do and don't understand&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks to @DJC, I now know this is a standard function to generate &lt;a href=&quot;http://en.wikipedia.org/wiki/Square_pyramidal_number&quot; rel=&quot;nofollow&quot;&gt;Square Pyramidal Numbers&lt;/a&gt;, which is part of &lt;a href=&quot;http://en.wikipedia.org/wiki/Faulhaber%27s_formula&quot; rel=&quot;nofollow&quot;&gt;Faulhaber's formula&lt;/a&gt;. Faulhaber's formula appears to be about quickly adding sequential coefficients which all have the same exponent. Very cool. But how does one get from:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{k=1}^{n} k^p$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;to the spiffy function above? If I'm sounding stupid, how do I make the question better?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fwiw, I'm in intermediate algebra in the USA. The next course would be Trigonometry or Calculus. (to help people place my current knowledge level)&lt;/p&gt;&#xA;" OwnerUserId="7668" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-02T04:54:58.997" LastActivityDate="2011-03-02T11:27:30.517" Title="How do I come up with a function to count a pyramid of apples?" Tags="&lt;functions&gt;&lt;polynomials&gt;" AnswerCount="4" CommentCount="4" />
  <row Id="24522" PostTypeId="2" ParentId="24509" CreationDate="2011-03-02T01:25:52.950" Score="1" ViewCount="" Body="&lt;p&gt;Take your credit card, driver's license, or some other readily available straightedge and put it against the hypotenuse. You'll find the composite shape is not a triangle, but a cleverly disguised irregular quadrilateral. The better, more mathy answers made sense once I got my brain away from the idea that there were triangles involved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nothing but a cheap trick designed to confound.&lt;/p&gt;&#xA;" OwnerUserId="7668" LastActivityDate="2011-03-02T01:25:52.950" />
  <row Id="24523" PostTypeId="2" ParentId="24494" CreationDate="2011-03-02T01:26:45.303" Score="2" ViewCount="" Body="&lt;p&gt;If $F$ is a field of characteristic different from $2$, and $K$ is an extension with $[K:F]=2$, then the &lt;em&gt;conjugate of&lt;/em&gt; $\alpha\in K$ is the image of $\alpha$ under the unique nontidentity element $\sigma\in\mathrm{Gal}(K/F)$ (this agrees with the usual &quot;complex conjugation&quot;, since complex conjugation &lt;em&gt;is&lt;/em&gt; the unique nonidentity element of $\mathrm{Gal}(\mathbb{C}/\mathbb{F})$). Since $\sigma$ is a field homomorphism, you automatically get both&#xA;$$\sigma(\alpha+\beta) = \sigma(\alpha)+\sigma(\beta)\qquad\text{and}\qquad \sigma(\alpha\beta) = \sigma(\alpha)\sigma(\beta).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, this holds for the unique nonidentity automorphism of $\mathbb{Q}(\zeta_3)$ over $\mathbb{Q}$ (which, as it happens, is complex conjugation). In particular, it holds for elements of $\mathbb{Z}[\zeta_3]$ (which, as it happens, is closed under this automorphism) , so that the restriction of conjugation is in fact also an automorphism of the ring $\mathbb{Z}[\zeta_3]$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, you can try defining different kinds of &quot;conjugate&quot;. Instead of applying the automorphism of $\mathbb{Q}(\zeta_3)$, you could try defining &quot;conjugate&quot; by $a+b\zeta_3\mapsto a-b\zeta_3$ with $a,b\in\mathbb{Z}$. If you do that, then it's not a ring homomorphism, because&#xA;$$(a+b\zeta_3)(x+y\zeta_3) = ax+\zeta_3^2by + (ay+bx)\zeta_3 = ax-by + (ay+bx-by)\zeta_3$$&#xA;but &#xA;$$(a-b\zeta_3)(x-y\zeta_3) = ax+\zeta_3^2by - (ay+bx)\zeta_3 = ax-by - (ay+bx+by)\zeta_3$$&#xA;so the &quot;conjugate&quot; of the product is not the product of the &quot;conjugates&quot;. That's why one uses the field automorphism.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Added.&lt;/em&gt; I focused on extensions of degree $2$ to be able to refer to &lt;em&gt;the&lt;/em&gt; conjugate. One can of course consider extensions of larger degree or in charactersitic $2$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the general case, the &lt;strong&gt;conjugates&lt;/strong&gt; (plural in general) of $\alpha$ in $K$ are the images of $\alpha$ under automorphisms that lie in $\mathrm{Aut}(K/F)$. Under any particular automorphism $\sigma$ it is still true that $\sigma(\alpha\beta) = \sigma(\alpha)\sigma(\beta)$, but if you consider different automorphisms (different notions of &quot;conjugate&quot;) of course things go astray. One can also consider the &lt;em&gt;norm&lt;/em&gt;, $N_{K/F}$, which maps from $K$ to $F$ by $N_{K/F}(\alpha) = \prod_{\sigma\in G} \sigma(\alpha)$, where $G$ is the Galois group of the Galois closure of $K$ over $F$. The norm &lt;em&gt;is&lt;/em&gt; multiplicative: $N_{K/F}(\alpha\beta) = N_{K/F}(\alpha)N_{K/F}(\beta)$; and a very important map. But I doubt this is what you were refering to.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-02T04:25:16.357" LastActivityDate="2011-03-02T04:25:16.357" />
  <row Id="24524" PostTypeId="1" CreationDate="2011-03-02T01:31:59.687" Score="0" ViewCount="174" Body="&lt;p&gt;If we have an entire function $f(z)$, can we apply the Cauchy integration formula for derivative for $f(z)$ and integrate over $\Bbb R$ instead of simple closed curve, i.e.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is this formula is true:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f&amp;#39;(a_{0}) = \frac{1}{2\pi i}\int_{\Bbb R}\frac{f(t)}{(t-a_{0})^{2}}dt$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $f(t)$ is just $f(z)$ with $z=t\in \Bbb R$.&lt;/p&gt;&#xA;" OwnerUserId="7531" LastEditorUserId="7531" LastEditorDisplayName="" LastEditDate="2011-03-02T02:10:14.223" LastActivityDate="2011-05-01T10:13:20.107" Title="Cauchy Formula for derivatives" Tags="&lt;complex-analysis&gt;" AnswerCount="3" CommentCount="5" />
  <row Id="24525" PostTypeId="1" AcceptedAnswerId="24543" CreationDate="2011-03-02T01:36:01.917" Score="1" ViewCount="105" Body="&lt;p&gt;Let $X$ be a finite-dimensional inner product space, and $T$ a linear operator on $X$. Let $W$ be a subset of $X$ with the following property:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $T$ preserves norms on $W$, then $T$ is orthogonal on $X$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One might say that the set $W$ 'detects' orthogonality. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is: what is the smallest number of elements that $W$ may contain (as a function of dim$(X) = n$) ?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Quite simply, we can observe that $n + {n \choose 2}$ would form an upper bound for the minimal cardinality of $W$. However, I cannot think of a way to proof whether or not this upper bound is also a lower bound.&lt;/p&gt;&#xA;" OwnerUserId="647" LastEditorUserId="647" LastEditorDisplayName="" LastEditDate="2011-03-02T02:22:06.073" LastActivityDate="2011-03-03T04:33:42.057" Title="A Question about Orthogonal Operators" Tags="&lt;linear-algebra&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="24527" PostTypeId="2" ParentId="24489" CreationDate="2011-03-02T02:01:44.673" Score="6" ViewCount="" Body="&lt;p&gt;First of all the way you approach the definition is wrong and obfuscating which leads to your second question. Also, your confusion regarding the polynomials stems from a common problem a lot of people have when they start dealing with (a bit more) abstract mathematics: They mix up the object with the representation of the object.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What your definition says is this: Imagine I give you two objects $a,b\in S$ to compare for equality. These two objects are equal if and only if they are one and the same. Typically, the set theoretic approach to a binary relation on a set $S$, is an $R\subset S\times S$. This means that the equality relation is $E=\{(a,a)\in S\times S : a\in S\}$, which means that $a=b$ if and only if $(a,b)\in E$ which is true exactly when $a$ and $b$ are one and the same (or more concisely -which I purposely avoided- exactly when $a=b$). This may sound pointless and uninteresting and tautological. Actually Wittgenstein in &quot;Tractatus Logico-philosophicus&quot; points out this apparent pointlessness of &quot;$=$&quot;. But such a concept &lt;em&gt;is&lt;/em&gt; fruitful mathematically because the mathematical interest in the elements of $S$ lies in their properties: We want to know about the form, the structure of a mathematical universe and our only means of describing it is through a mathematical language. Saying $3=3$ is uninteresting but saying that for every $x,y$ we have $(x+y)^2=x^2+y^2+2xy$ tells you something and the fact that given any number $n\geq 3$ you can never find positive natural numbers $a,b,c$ such that $a^n+b^n=c^n$ is not trivial at all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now regarding your confusion, as others have pointed out you are mixing the expressions of polynomials with polynomials. $6x-6x$ is a &lt;em&gt;description&lt;/em&gt; of a polynomial and so is $0$. But these two descriptions talk about the same polynomial. This is essentially the information conveyed by writing $0=6x-6x$ (see the usefulness of the equality relation here). Using objects that are -in some sense- different to indicate the same thing is something we do all the time, even outside of mathematics. Without this we wouldn't be able to communicate, since words as &lt;em&gt;physical objects&lt;/em&gt; are different from one another, even though they may spell the same thing. Every time you see letters on a paper, you ignore their physical nature (their consisting of atoms that you have most probably never encountered before, and definitely not in that form) and only care about what they &lt;em&gt;symbolise&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, regarding you second question: Formally, no, it is the same definition. In a formal environment the standard definition of the meaning of the symbol &quot;$=$&quot; is always the diagonal. An expression such as $x^2=2x$ doesn't have any meaning whatsoever because $x$ is a free variable. We assign a truth value (true or false) in &lt;em&gt;sentences&lt;/em&gt; which are defined as formulas that contain no free variable. Intuitively, ask yourself the following: First of all what is this $x$? Secondly, a relation is defined as a subset of the Cartesian product of a set. What is this set? For &quot;$=$&quot; to be a relation in the formula $x^2=2x$ would mean that $x^2$ and $2x$ are elements of a set. So your question isn't well defined. What we mean when we say that some $a\in\mathbb{R}$ (for example) satisfy $x^2=2x$ is that $a^2=2a$, or that the $a^2$ is &lt;em&gt;the same number&lt;/em&gt; with $2a$, which is exactly the definition of the equality I (and the book you are studying) presented. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; When you write a mathematical sentence like &quot;$2+3=5$&quot; or &quot;$(\forall x)((2x)^2=4x^2)$&quot; you have to keep in mind that these are strings of symbols. &lt;a href=&quot;http://marcel-oehler.marcellosendos.ch/comics/ch/1992/09/19920901.gif&quot; rel=&quot;nofollow&quot;&gt;As Calvin points out&lt;/a&gt; these strings don't have a specific meaning. Meaning is something we &lt;em&gt;equip&lt;/em&gt; on a language. Tarski's definition of truth intuitively is pretty much what we do when we teach very young children the meaning of words. We show them the objects and say &quot;This is a CAR&quot;. Through repetition, children learn to equip the concept of an automobile to the sound &quot;car&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Tarski said roughly the same: Assume that we have a set $S$, a function $f:S\times S\to S$, a relation $E\subset S\times S$ and some elements $a,b,c\in S$. Now I come up with a language that has &lt;em&gt;symbols&lt;/em&gt; &quot;$+$&quot;, &quot;$=$&quot; and &quot;$0$&quot;, &quot;$1$&quot;, &quot;$2$&quot;. Then we can assign to each symbol  an actual objects, for example we will say that &quot;$+$&quot; will represent $f$, &quot;$=$&quot; will represent $E$ and each of &quot;$0$&quot;, &quot;$1$&quot;, &quot;$2$&quot; will represent $a,b,c$ respectively. Now we can say that the sentence &quot;$2+1=3$&quot; is true &lt;em&gt;in the world&lt;/em&gt; $(S,f,E,a,b,c)$ exactly when $(f(b,a),c)\in E$. In practice we try to use the same symbols for similar functions and relations and this is why we reserve the symbol &quot;$=$&quot; for the diagonal relation.&lt;/p&gt;&#xA;" OwnerUserId="1772" LastEditorUserId="1772" LastEditorDisplayName="" LastEditDate="2011-03-02T05:58:48.697" LastActivityDate="2011-03-02T05:58:48.697" CommentCount="13" />
  <row Id="24528" PostTypeId="2" ParentId="24524" CreationDate="2011-03-02T02:07:34.497" Score="0" ViewCount="" Body="&lt;p&gt;I know that the following analog to the Cauchy integral formula holds.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $z$ in the upper half-plane, and &#xA;$$&#xD;&#xA;h_z(\zeta) = \frac{1}{2 \pi i} \left( \frac{1}{\zeta - z} - \frac{1}{\zeta - \overline{z}}\right)&#xD;&#xA;$$&#xA;we have&#xA;$$&#xD;&#xA;f(z) = \int_{-\infty}^{\infty} f(t)h_z(t) dt&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To prove this, evaluate the integral via residue calculus.  I think another way to prove this is to start with the Cauchy integral formula for a circle and use a conformal mapping between the unit disc and the upper half-plane to get the above formula.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would think a similar analog of the Cauchy integral formula for derivatives holds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note the upper half-plane excludes the real line.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know this does not address your exact question, but I hope it is helpful in some way.&lt;/p&gt;&#xA;" OwnerUserId="6887" LastEditorUserId="6887" LastEditorDisplayName="" LastEditDate="2011-03-02T08:58:31.793" LastActivityDate="2011-03-02T08:58:31.793" CommentCount="1" />
  <row Id="24529" PostTypeId="1" CreationDate="2011-03-02T02:07:59.653" Score="4" ViewCount="297" Body="&lt;p&gt;Hello everyone. I'm very interested to know how I can factorise $a^{10} +a^5 +1$ in two factors with integer coefficients. I've tried a lot but I don't have any idea how do that.&lt;/p&gt;&#xA;" OwnerUserId="3128" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2011-03-23T17:42:30.817" LastActivityDate="2011-03-24T13:00:39.217" Title="Factoring $a^{10}+a^5+1$" Tags="&lt;polynomials&gt;" AnswerCount="4" CommentCount="3" FavoriteCount="1" />
  <row Id="24531" PostTypeId="1" CreationDate="2011-03-02T02:14:40.247" Score="0" ViewCount="43" Body="&lt;p&gt;I'm a programmer, but I'm no mathematician, so I need a little help understanding what is required to accomplish this task.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a short explanation of the data I have from the Coastal Data Information Program (CDIP) site: &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&#xA;//h-----------------------------------------------------------------------------&#xA;//h .root/data_access/MEM_2dspectra.cdip&#xA;//h&#xA;//h Creates a 2D (dir/freq) energy spectra from a standard CDIP spectral&#xA;//h file using the Maximum Entropy Method (MEM).&#xA;//h&#xA;//h NOTE: This is ONE WAY to estimate a 2d spectrum from the buoy data.  &#xA;//h CDIP uses it because it makes narrow directional spectra for swell in the &#xA;//h Pacific.&#xA;//h&#xA;//h By default the directions are grouped in 5-degree bins, and the&#xA;//h frequencies bands are the same as in the sp file. Values are given&#xA;//h in units of m*m/Hz/deg.&#xA;//h&#xA;//h Data is output as a matrix:                   +------Direction bands------&gt;&#xA;//h   - First column is centered at 5 degrees;    |        72 columns&#xA;//h   - Last column is centered at 360 degrees;   |&#xA;//h   - First row is the lowest freq (.025 Hz     |&#xA;//h       for directional buoys);              Frequency bands&#xA;//h   - Last row is the highest freq (.58 Hz      |  64 rows for &#xA;//h       for directional buoys);                 |  directional buoys&#xA;//h                                               |&#xA;//h                                               v&#xA;//h&#xA;//h Arguments: 1 - either a 3-digit station ID (or 5-digit MO ID) for the &#xA;//h                  latest data, or a 19-character sp (or mp) filename.&#xA;//h            2 - optional; set to 'even' to get evenly spaced freq bands,&#xA;//h                  all with a bandwidth of 0.01Hz (0.03Hz-0.58Hz)&#xA;//h &#xA;//h Usage:&#xA;//h   http://cdip.ucsd.edu/data_access/MEM_2dspectra.cdip?{args}&#xA;//h   where {args} examples are:&#xA;//h     1. h                          -&gt; returns this help&#xA;//h     2. 029                        -&gt; MEM spectra for most recent 029 sp&#xA;//h     3. sp10001200606232058        -&gt; MEM spectra for sp10001200606232058&#xA;//h     4. 106+even                   -&gt; Latest 106 spectra, 0.01Hz bands&#xA;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Here's a sample of the data for buoy 100 (Torrey Pines):&#xA;&lt;a href=&quot;http://cdip.ucsd.edu/data_access/MEM_2dspectra.cdip?100&quot; rel=&quot;nofollow&quot;&gt;http://cdip.ucsd.edu/data_access/MEM_2dspectra.cdip?100&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's the resulting heat or intensity map from CDIP:&#xA;&lt;a href=&quot;http://cdip.ucsd.edu/?nav=recent&amp;amp;sub=observed&amp;amp;units=metric&amp;amp;tz=UTC&amp;amp;pub=public&amp;amp;map_stati=1,2,3&amp;amp;stn=100&amp;amp;stream=p1&amp;amp;xitem=dir_spectrum&quot; rel=&quot;nofollow&quot;&gt;http://cdip.ucsd.edu/?nav=recent&amp;amp;sub=observed&amp;amp;units=metric&amp;amp;tz=UTC&amp;amp;pub=public&amp;amp;map_stati=1,2,3&amp;amp;stn=100&amp;amp;stream=p1&amp;amp;xitem=dir_spectrum&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;How do I take this data and create the heat or intensity map?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;P.S.&lt;/strong&gt;&#xA;&lt;em&gt;I'm looking at various image processing and graphics libraries, such as GD, ImageMagic and GnuPlot, as well as Ruby-specific solutions like RMagick or the ruby-gd gem, and also looked at Google Visualization options.  If you have any suggestions with respect to the programming side of things, by all means make a suggestion, but overall I need to understand the idea first before I can approach the programming aspects of this problem.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="7670" LastActivityDate="2011-03-02T02:14:40.247" Title="How do you create a spectral plot and then later a heat or intensity map from 2d data?" Tags="&lt;polar-coordinates&gt;" CommentCount="3" FavoriteCount="1" />
  <row Id="24532" PostTypeId="2" ParentId="24529" CreationDate="2011-03-02T02:16:30.277" Score="2" ViewCount="" Body="&lt;p&gt;Hint: It clearly does not have a linear factor since $\pm1$ is not a root. Try a quadratic factor $a^2+u a+v$, for some integers $u$ and $v$.&lt;/p&gt;&#xA;" OwnerUserId="589" LastActivityDate="2011-03-02T02:16:30.277" />
  <row Id="24533" PostTypeId="1" AcceptedAnswerId="24539" CreationDate="2011-03-02T02:20:34.947" Score="18" ViewCount="745" Body="&lt;p&gt;I read this quote attributed to &lt;a href=&quot;http://en.wikipedia.org/wiki/Vladimir_Arnold&quot; rel=&quot;nofollow&quot;&gt;VI Arnold&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;Who can't calculate the average value of the one hundredth power of the sine function within five minutes, doesn't understand mathematics - even if he studied supermanifolds, non-standard calculus or embedding theorems.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt; Source is &lt;a href=&quot;http://iopscience.iop.org/0036-0279/46/1/M14&quot; rel=&quot;nofollow&quot;&gt;&quot;A mathematical trivium&quot;&lt;/a&gt; A book of 100 problems that university students &quot;should be able to solve&quot;. The statement asks for calculation within 10% accuracy. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;So the average value over the entire domain should be the same as the average value over  $[0,\pi/2]$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$&amp;lt;\sin^{100} (x)&amp;gt;= \frac{\int_0^{\pi/2} \sin^{100}(x) dx}{\int_0^{\pi/2} dx}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So here's what I did:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, this graph would be a train of highly sharp peaks. The integrand would assume values close to zero a up till before it sharply rises to 1. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So up till some $\epsilon \in [0,\pi/2]$ we will have $\sin x \approx x$ and for the remaining $\pi/2 - \epsilon$ interval I could find the area of triangle with base $\pi/2 - \epsilon$ and height $1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$&amp;lt;\sin^{100} (x)&amp;gt; \approx \frac{2}{\pi} (\int_0^\epsilon x^{100} dx + .5 (\frac{\pi}{2}-\epsilon))$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I beleive in principal it should be possible to find an $\epsilon$ such that the above expression yields the exact answer. So I try to approximate it, no good. Then I try mathematica and it is looking like there is no $\epsilon$ for which the value I am expecting is even close to the actual value. I plot the original and find that my approximation is hopeless.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Not to mention that my 5 minutes were over. So I admit I do not understand mathematics and humbly ask if someone could:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Point out my mistake (Other than that $\epsilon$ is probably incomputable within 5 mins)&lt;/li&gt;&#xA;&lt;li&gt;How the hell is this done in 5 minutes?&lt;img src=&quot;http://i.imgur.com/wcDny.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The picture below has the $\sin^{100} x$ in blue (bottom) and my approximation of it plotted against $\epsilon$ (pink). ALthough there is no reason for them to be together, the upper graph has a minima quiet above the exact value of the integral. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;&#xA;Just realized&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $u=\cos x$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\int_0^{\pi/2} \sin^{100}(x) dx = \int_1^0 (1-u^2)^{49/2}du\approx \int_1^0 (1 - \frac{49}{2} u^2) du $&lt;/p&gt;&#xA;" LastActivityDate="2011-03-02T16:07:01.863" Title="Find the average of $\sin^{100} (x)$ in 5 minutes?" Tags="&lt;approximation&gt;" AnswerCount="5" CommentCount="8" FavoriteCount="4" />
  <row Id="24534" PostTypeId="2" ParentId="24525" CreationDate="2011-03-02T02:22:13.590" Score="2" ViewCount="" Body="&lt;p&gt;Edit: This answer gives a construction of $W$ which shows that it need not contain more than ${n+1\choose 2}$ elements (if $X$ is real), but does not answer the question posed, as Yuval's argument does.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The easiest way to construct a finite set $W\subset X$ which detects orthogonality is as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $\{\alpha_1,\dots,\alpha_n\}$ be a basis for $X$. Let &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$W=\{\alpha_j+c\alpha_k:j\leq k;c=0,1\text{ or }i\}.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then $|W|=n^2$. Suppose $T$ is a linear operator on $X$ such that $\|T\alpha\|=\|\alpha\|$ for every $\alpha\in W$. If $j\leq k$, then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\langle T\alpha_j,T\alpha_k\rangle=\frac{1}{2}\left(\|T\alpha_j+T\alpha_k\|^2+i\|T\alpha_j+iT\alpha_k\|^2\right)-\left(\|T\alpha_j\|^2+\|T\alpha_k\|^2\right)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$=\frac{1}{2}\left(\|T(\alpha_j+\alpha_k)\|^2+i\|T(\alpha_j+i\alpha_k)\|^2\right)-\left(\|T\alpha_j\|^2+\|T\alpha_k\|^2\right)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$=\frac{1}{2}\left(\|\alpha_j+\alpha_k\|^2+i\|\alpha_j+i\alpha_k\|^2\right)-\left(\|\alpha_j\|^2+\|\alpha_k\|^2\right)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$=\langle \alpha_j,\alpha_k\rangle.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $j&amp;gt;k$, the above calculation shows that $\langle T\alpha_k,T\alpha_j\rangle=\langle \alpha_k,\alpha_j\rangle$, hence&#xA;$\langle T\alpha_j,T\alpha_k\rangle=\overline{\langle T\alpha_k,T\alpha_j\rangle}=\overline{\langle \alpha_k,\alpha_j\rangle}=\langle\alpha_j,\alpha_k\rangle.$&#xA;Thus $T$ preserves inner products, and $T$ is unitary. Notice that if $X$ is real, $W$ need not contain more than ${n+1\choose2}$ elements; simply choose the set above, without the $n\choose2$ elements $\alpha_j+i\alpha_k$, $j\leq k$.&lt;/p&gt;&#xA;" OwnerUserId="7671" LastEditorUserId="7671" LastEditorDisplayName="" LastEditDate="2011-03-02T06:30:45.640" LastActivityDate="2011-03-02T06:30:45.640" CommentCount="5" />
  <row Id="24535" PostTypeId="1" AcceptedAnswerId="24540" CreationDate="2011-03-02T02:31:27.037" Score="13" ViewCount="290" Body="&lt;p&gt;If $\lambda_1,\dots,\lambda_n$ are distinct positive real numbers, then &#xA;$$\sum_{i=1}^n \prod_{j\neq i} {\lambda_j\over \lambda_j-\lambda_i}=1.$$&#xA;This identity follows from a probability calculation that you can find at the &#xA;top of page 311 in the 10th edition of &lt;em&gt;Introduction to Probability Models&lt;/em&gt; by Sheldon Ross. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a slick or obvious explanation for this identity? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This  question is sort of similar to &lt;a href=&quot;http://math.stackexchange.com/questions/4220/is-this-a-known-algebraic-identity&quot;&gt;my previous problem&lt;/a&gt;; clearly algebra is not my strong suit! &lt;/p&gt;&#xA;" OwnerUserId="940" LastEditorUserId="6296" LastEditorDisplayName="" LastEditDate="2011-03-04T05:59:45.047" LastActivityDate="2011-03-04T05:59:45.047" Title="Is this algebraic identity obvious?" Tags="&lt;algebra-precalculus&gt;&lt;sequences-and-series&gt;" AnswerCount="1" FavoriteCount="0" />
  <row Id="24536" PostTypeId="1" AcceptedAnswerId="26671" CreationDate="2011-03-02T02:33:45.793" Score="1" ViewCount="96" Body="&lt;p&gt;Is there a cis (CosX + iSinX) Button in CASIO graphic calculators ? if so where its ?&#xA;if no is there any turn around for it ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS. I got a CASIO fx-9860G&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="4987" LastActivityDate="2011-03-13T07:42:23.000" Title="is there a CIS button on graphic calculators (Casio/TI)" Tags="&lt;calculator&gt;" AnswerCount="2" />
  <row Id="24538" PostTypeId="2" ParentId="24533" CreationDate="2011-03-02T02:36:29.400" Score="9" ViewCount="" Body="&lt;p&gt;If you do an integration by parts, writing $\sin^{100}x$ as $\sin(x)\sin^{99}(x)$ you get:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_{-a}^a \sin^{100}x dx = \frac{99}{100} \int_{-a}^a \sin^{98}x dx$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I think that argument gives you an induction, and you get something like:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_{-\pi}^{\pi} \sin^{100}x dx = \frac{99\cdot 97 \cdots 1}{100 \cdot 98 \cdots 2} 2\pi$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;FYI: I've never studied supermanifolds. I &lt;em&gt;am&lt;/em&gt; a professional mathematician and I think the problem took me maybe 2 minutes, once I had finished reading your question. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;FYI number 2: A decent approximation to the average of $\sin^n(x)$ over $[-\pi,\pi]$ would be $\frac{1}{\sqrt{n}}$.  You get this from the above argument, using the approximation that $\ln(1+x) \simeq x$ for $x$ small, together with the approximation that the sum $1+1/2+1/3+\cdots+1/n \simeq \ln(n)$. &lt;/p&gt;&#xA;" OwnerUserId="642" LastEditorUserId="642" LastEditorDisplayName="" LastEditDate="2011-03-02T03:09:52.977" LastActivityDate="2011-03-02T03:09:52.977" CommentCount="3" />
  <row Id="24539" PostTypeId="2" ParentId="24533" CreationDate="2011-03-02T02:47:22.507" Score="32" ViewCount="" Body="&lt;p&gt;Since $$\sin x=\frac{e^{ix}-e^{-ix}}{2i} $$&#xA;and $$\int_0^{2\pi}e^{ikx}\,dx=\left\{\begin{array}{cl}0&amp;amp;k\ne0\\ 2\pi&amp;amp;k=0\end{array}\right.,$$&#xA;we have $$\int_0^{2\pi}\sin^{100}x\,dx=\frac1{2^{100}}\sum_{k=0}^{100}\binom{100}{k}\int_0^{2\pi}e^{ikx}(-1)^{100-k}e^{-i(100-k)x}\,dx=\frac{\binom{100}{50}}{2^{100}}2\pi,$$ and the average value is $$\frac{\binom{100}{50}}{2^{100}}.$$&lt;/p&gt;&#xA;" OwnerUserId="462" LastActivityDate="2011-03-02T02:47:22.507" CommentCount="2" />
  <row Id="24540" PostTypeId="2" ParentId="24535" CreationDate="2011-03-02T02:50:31.887" Score="22" ViewCount="" Body="&lt;p&gt;It's the Lagrange interpolation polynomial for the constant function $1$ evaluated at $0$:&#xA;$$\sum_{i=1}^n 1 \prod_{j\neq i} {{\lambda_j-0}\over \lambda_j-\lambda_i}=1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, you have that the polynomial below interpolates the data points $(\lambda_i,y_i$):&#xA;$$\sum_{i=1}^n y_i \prod_{j\neq i} {{\lambda_j-x}\over \lambda_j-\lambda_i}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See &lt;a href=&quot;http://en.wikipedia.org/wiki/Lagrange_polynomial&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Lagrange_polynomial&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="589" LastEditorUserId="589" LastEditorDisplayName="" LastEditDate="2011-03-02T03:04:40.920" LastActivityDate="2011-03-02T03:04:40.920" CommentCount="5" />
  <row Id="24541" PostTypeId="2" ParentId="24533" CreationDate="2011-03-02T02:55:10.103" Score="1" ViewCount="" Body="&lt;p&gt;According to a computer algebra package I'm using, which are pretty powerful these days, &#xA;$\int_{0}^{\pi/2} \frac{2\sin^{100}(x)}{\pi} dx = \frac{126114180XXXXXX24166851562157}{158456325028528675187087900672}$ (the point here being that even if one is not so great on methods of integration, computer algebra systems have come quite far in the past few decades). It's also worth pointing out that this answer is conceptually much less satisfying than either Andres or Ryan's answer. (six of the characters of the answer have been replaced by &quot;X&quot;s, because I don't want to just give you the answer).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moreover, there's the question of the intent of VI Arnold's quote, which is difficult to understand without his full biography. If you could perhaps provide a reference to the quote so the context is available, then it may be possible to explicate it, but that's kind of soft-questiony.&lt;/p&gt;&#xA;" OwnerUserId="890" LastActivityDate="2011-03-02T02:55:10.103" CommentCount="1" />
  <row Id="24543" PostTypeId="2" ParentId="24525" CreationDate="2011-03-02T03:09:37.503" Score="2" ViewCount="" Body="&lt;p&gt;Suppose a set $W$ is a 'detector'. This means that for any matrix $T$, if $w&amp;#39;T&amp;#39;Tw = w&amp;#39;w$ for all $w \in W$ then $T$ must be orthogonal. This is a set of $|W|$ linear equations on $T&amp;#39;T$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $|W| &amp;lt; \binom{n+1}{2}$, then there is a non-zero symmetric matrix $A$ such that $w&amp;#39;Aw = 0$ for all $w \in W$. Note that $A$ has some non-zero eigenvalue.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clearly $w&amp;#39;Iw = w&amp;#39;w$ for all $w \in W$, and so for all $\epsilon \in \mathbb{R}$, we have $w&amp;#39;(I+\epsilon A)w = w&amp;#39;w$ for all $w \in W$. Choosing $\epsilon &amp;gt; 0$ small enough, we get some matrix $P = I+\epsilon A$ which is positive semidefinite, and thus has a square root $T$ such that $P=T&amp;#39;T$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $\epsilon \neq 0$, $P$ has some positive eigenvalue different than $1$. Thus $T$ has some eigenvalue which is not of unit magnitude, and so cannot be orthogonal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Addendum: This argument shows that $W$ is a detector iff the outer products $\{ w \otimes w : w \in W\}$ generate the linear space of all symmetric matrices. An example is the set $$W = \{ e_i+e_j : 1 \leq i \leq j \leq n \}.$$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;What follows answers the related question where $T$ preserves all &lt;em&gt;inner products&lt;/em&gt; on $W$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can take any basis of $X$ as $W$: if $T$ preserves all inner products among a basis, it will preserve all inner products.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, if you take less than $n$ elements, then consider the projection into their linear span. On $W$ it acts as the identity, so it preserves inner products. But its not even regular.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same arguments show that $W$ detects orthogonality iff it spans $X$.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2011-03-03T04:33:42.057" LastActivityDate="2011-03-03T04:33:42.057" />
  <row Id="24544" PostTypeId="2" ParentId="24507" CreationDate="2011-03-02T03:20:21.817" Score="15" ViewCount="" Body="&lt;p&gt;The difficulty can be hard to see from a modern point of view, where set theory books have been written in a way that avoids the problem. So imagine that you had never learned any set theory, but you are familiar with the English word &quot;set&quot;. You probably think, as mathematicians did at the time, that any well-defined collection of objects forms a set, and any set is a particular well-defined collection of objects. This seems like a perfectly reasonable idea, but Russell's paradox shows that it's actually inconsistent.  Moreover, the &quot;set&quot; that we now associate with Russell's paradox -- the set of all sets that don't contain themselves -- seems like a perfectly well-defined set. After all, each particular set is either a member of itself or not, so the ones that are not seem to be a well-defined collection.  This casts doubt on the English language term &quot;set&quot;, which in turn seems to cast doubt on all mathematics done in English.  &lt;em&gt;That&lt;/em&gt; was why the paradox was so revolutionary -- because things people had been routinely using turned out to be inconsistent, and it seemed possible that more paradoxes might be found that would cast doubt on other mathematical objects.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The solution that mathematicians eventually adopted was to drop the normal English meaning of set (&quot;any well-defined collection of things&quot;) and replace it with a different concept, the &quot;iterative&quot; concept of sets.  But it took some time for that solution to even be proposed, much less accepted. Even now, a key argument that the new conception is consistent is that nobody has managed to find an inconsistency. If that seems less than completely certain, that's because it is.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Several other paradoxes were discovered around the same time. One of the more interesting ones in my opinion, which was discovered somewhat later, is &quot;Curry's paradox&quot;. It consists of this sentence: &quot;If this sentence is true then 0=1&quot;. If you use the normal natural-language proof technique for proving an if/then statement, you can actually prove that sentence is true (it's not hard - try it). But if that sentence is true, then 0=1 - and you can prove that too, in the normal mathematical way. The fact that the normal techniques we use in everyday mathematics can prove 0=1 is certainly paradoxical, since each of those techniques seems fine on its own.  Like Russell's paradox, Curry's paradox both casts doubt on our informal natural-language mathematics, and shows that certain formal theories that we might wish were consistent are actually inconsistent. &lt;/p&gt;&#xA;" OwnerUserId="630" LastActivityDate="2011-03-02T03:20:21.817" />
  <row Id="24545" PostTypeId="2" ParentId="24507" CreationDate="2011-03-02T03:29:13.453" Score="1" ViewCount="" Body="&lt;p&gt;Partially quoting Terence Tao, &quot;One problem with this axiom ('Universal specification' or 'axiom of comprehension') is that it creates sets which are far too 'large' - for instance one can talk of a set of all objects (a so called universal set). Since sets are themselves objects, this means sets are allowed to contain themselves, which is somewhat silly state of affairs.One way to informally resolve it to think of objects as being arranged in hierarchy. .... &quot; -- Analysis1, 2nd ed., Hindustan Book Agency, pp47&lt;/p&gt;&#xA;" OwnerUserId="1674" LastActivityDate="2011-03-02T03:29:13.453" />
  <row Id="24546" PostTypeId="2" ParentId="24507" CreationDate="2011-03-02T03:57:33.057" Score="-3" ViewCount="" Body="&lt;p&gt;Fortunately, set theory isn't that important to modern math.  But category theory is, and an analogue of Russell's paradox stops you from constructing what would otherwise be one of the most obvious things to construct, the category of categories.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In practice, I've seen a bunch of expositions that go &quot;Let $\mathbf{Cat}$ be the category of all (small) categories...&quot; with a bit of a wink and a nudge to get you on the right track.  So I don't think this is a terribly important issue -- but it does give a natural place for a working mathematician to get accidentally involved in set theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By Russell's paradox proper, there's also no set of groups, rings, topological spaces, and so on, but this can be remedied by introducing proper classes, and as far as I can tell, you'd never want to topologize the class of topological spaces or something like that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;: I realized your question was more historical.  Russell came at a time of tremendous vagueness in math, and his work was largely an effort to eliminate such vagueness.  It was surprising that such a project could possibly have nontrivial results.  Frege, specifically, had been working on a massive treatise on logic that was &lt;a href=&quot;http://en.wikipedia.org/wiki/Gottlob_Frege#Work_as_a_logician&quot; rel=&quot;nofollow&quot;&gt;totally destroyed&lt;/a&gt; by Russell's paradox, because it relied so heavily on sets whose elements could come from anywhere.  If you're interested, &lt;a href=&quot;http://www.logicomix.com/en/&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;Logicomix&lt;/em&gt;&lt;/a&gt; is a really good graphic novel about Russell's life and work -- addressed to non-mathematicians, but a good read and helps motivate issues like these.&lt;/p&gt;&#xA;" OwnerUserId="2215" LastEditorUserId="2215" LastEditorDisplayName="" LastEditDate="2011-03-02T04:06:58.573" LastActivityDate="2011-03-02T04:06:58.573" CommentCount="11" />
  <row Id="24547" PostTypeId="1" CreationDate="2011-03-02T03:59:05.303" Score="1" ViewCount="91" Body="&lt;p&gt;I have a graph with edge-set E and the following SDP (page 8 &lt;a href=&quot;http://etd.fcla.edu/UF/UFE0001011/butenko_s.pdf&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://yaroslavvb.com/upload/save/lovasz-sdp2.png&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm trying to use CVXOPT to solve this problem, which asks for SDP to be expressed in &lt;a href=&quot;http://abel.ee.ucla.edu/cvxopt/userguide/coneprog.html#semidefinite-programming&quot; rel=&quot;nofollow&quot;&gt;vectorized form&lt;/a&gt; as below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://yaroslavvb.com/upload/save/cvxopt-form.png&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do I go about turning my SDP into this form?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Update&lt;/b&gt;&#xA;From &lt;a href=&quot;http://www.or-exchange.com/questions/1256/free-tool-to-use-for-this-sdp/1258#1258&quot; rel=&quot;nofollow&quot;&gt;suggestion&lt;/a&gt; of Joachim Dahl it's easier to figure out values from the dual form. We need to set $G_1,h_1$ and $c$ as follows&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$G_1=\left(\begin{array}{c}&#xD;&#xA;\text{vec}(I)\\\\&#xD;&#xA;\text{vec}(U_{e1})\\\\&#xD;&#xA;\text{vec}(U_{e2})\\\\&#xD;&#xA;\cdots\\\\&#xD;&#xA;\text{vec}(U_{em})&#xD;&#xA;\end{array}\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;here $I$ is $n\times n$ identity matrix, $\text{vec}(A)$ is matrix $A$ taken as vector in row-major vector form, $U_{ek}$ is a $0,1$-valued matrix where entry $i,j$ is 1 iff $i,j$ is in the edge $ek$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$c = (-1,0,0,0,0,\ldots,0)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$h_1=- \left(\begin{array}{ccc}&#xD;&#xA;1&amp;amp;1&amp;amp;\ldots\\\\&#xD;&#xA;1&amp;amp;1&amp;amp;\ldots\\\\&#xD;&#xA;\ldots&amp;amp;\ldots&amp;amp;\ldots&#xD;&#xA;\end{array}&#xD;&#xA;\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that for a graph with $n$ nodes and $m$ edges, $G_1$ is an $m+1\times n^2$ matrix, $c$ is a vector of length $m+1$ and $h_1$ is an $n\times n$ matrix. Matrix $X$ from original formulation is the $z_1$ parameter, an $n^2$ vector representing the matrix in row-major form&lt;/p&gt;&#xA;" OwnerUserId="998" LastEditorUserId="998" LastEditorDisplayName="" LastEditDate="2011-03-03T19:01:46.260" LastActivityDate="2011-03-04T05:22:40.043" Title="Turning SDP into vectorized form" Tags="&lt;optimization&gt;&lt;math-software&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="24548" PostTypeId="1" CreationDate="2011-03-02T04:03:16.830" Score="4" ViewCount="102" Body="&lt;p&gt;Hi&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the classic &lt;a href=&quot;http://en.wikipedia.org/wiki/Coupon_collector%27s_problem&quot; rel=&quot;nofollow&quot;&gt;Coupon Collector's problem&lt;/a&gt;, it is well known that the time $T$ necessary to complete a set of $n$ randomly-picked coupons satisfies $E[T] \sim n \ln n $,$Var(T) \sim n^2$, and $\Pr(T &amp;gt; n \ln n + cn) &amp;lt; e^{-c}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This upper bound is better than the one given by the Chebyshev inequality, which would be roughly  $1/c^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is: is there a corresponding better-than-Chebyshev &lt;em&gt;lower bound&lt;/em&gt; for $T$? (e.g., something like $\Pr(T &amp;lt; n \ln n - cn) &amp;lt; e^{-c}$ ) ?&lt;/p&gt;&#xA;" OwnerUserId="7673" LastActivityDate="2011-03-02T04:03:16.830" Title="what is a tight lower bound on the coupon collector time?" Tags="&lt;probability&gt;&lt;combinatorics&gt;" CommentCount="2" FavoriteCount="3" />
  <row Id="24549" PostTypeId="1" AcceptedAnswerId="24583" CreationDate="2011-03-02T04:04:02.447" Score="3" ViewCount="77" Body="&lt;p&gt;I'm learning some group cohomology from the third section in Serre's &lt;em&gt;Local Fields&lt;/em&gt;, and I'm up to the section on change of group. If $f:G&amp;#39;\rightarrow G$ is a homomorphism of groups and $A$ is a $G$-module, there is an induced $G&amp;#39;$-module structure given by $s&amp;#39;\cdot a=f(s&amp;#39;)\cdot a$, for $s&amp;#39;\in G&amp;#39;$ and $a\in A$ (this is Serre's notation, I would reserve $G&amp;#39;$ for the commutator subgroup, but oh well). This induces both maps on cohomology $H^q(G,A)\rightarrow H^q(G&amp;#39;,A)$ and homology $H_q(G&amp;#39;,A)\rightarrow H_q(G,A)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is: If $A$ is a projective/injective/relatively projective/relatively injective $G$-module, must $A$-as-a-$G'$-module be such a $G&amp;#39;$-module? If not, are there any assumptions we could make about $f:G&amp;#39;\rightarrow G$ so that this would be true (e.g., $f$ surjective)?&lt;/p&gt;&#xA;" OwnerUserId="264" LastActivityDate="2011-03-02T06:25:09.343" Title="Is a projective (etc.) $G$-module also a projective $G'$-module?" Tags="&lt;group-cohomology&gt;" AnswerCount="1" />
  <row Id="24550" PostTypeId="1" CreationDate="2011-03-02T04:04:18.000" Score="0" ViewCount="41" Body="&lt;p&gt;I'm having an issue with a large amount of time stamped data for work. Each data point has a value and a time stamp.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We are going to be displaying this data on a graph, but we need a way to reduce the amount of points in the series due to the restriction on visible points on the graph.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Eg. A week time range will return approximately 5000 individual points, but the graphing component can only show up to 450 points in a series at a time (1 point per pixel).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So this means we need to simplify the amount of data in the series. We don't want to average points, but keep the raw values, just remove a certain number to get it underneath the 450 point maximum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These series tend to be sawtooth waveforms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We investigated the DouglasPeucker, but I wasn't sure how to apply it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there any other algorithms or suggestions on how this can be accomplished?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: This is an example of a typical sawtooth waveform.&#xA;&lt;img src=&quot;http://i.imgur.com/tBgiL.png&quot; alt=&quot;Sawtooth example&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="7672" LastEditorUserId="7672" LastEditorDisplayName="" LastEditDate="2011-03-02T04:38:19.640" LastActivityDate="2011-03-02T04:38:19.640" Title="Line simplification algorithm" Tags="&lt;sequences-and-series&gt;" CommentCount="5" FavoriteCount="1" />
  <row Id="24551" PostTypeId="2" ParentId="24524" CreationDate="2011-03-02T04:14:38.117" Score="0" ViewCount="" Body="&lt;p&gt;That's not True Sivaram; $f$ is entire so $f&amp;#39;$ is also entire function, but it doesn't say that $f$ (or $f'$) is real for real $z$!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I don't see the &quot;add comment icon on my screen&quot;)&lt;/p&gt;&#xA;" OwnerUserId="7630" LastEditorUserId="7630" LastEditorDisplayName="" LastEditDate="2011-03-02T04:19:40.530" LastActivityDate="2011-03-02T04:19:40.530" CommentCount="2" />
  <row Id="24552" PostTypeId="1" CreationDate="2011-03-02T04:25:30.163" Score="3" ViewCount="212" Body="&lt;p&gt;$N$ points are to be put on sides of a square. What's the maximum number of triangles which are formed by joining those points?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hints would be appreciated.&lt;/p&gt;&#xA;" OwnerUserId="696" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2011-03-02T15:17:42.787" LastActivityDate="2011-03-02T15:17:42.787" Title="Maximum number of triangles formed by points on a square?" Tags="&lt;combinatorics&gt;" AnswerCount="2" CommentCount="9" />
  <row Id="24553" PostTypeId="2" ParentId="18589" CreationDate="2011-03-02T04:29:33.000" Score="1" ViewCount="" Body="&lt;p&gt;Formally the Riemann zeta-function can be expressed as &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \zeta(z)=\prod_{k=0}^{\infty}\;\;\prod_{p\in \mathbb{P}}\bigg\{\left( 1 -\varphi^{-1}p^{-5^{k}z} +p^{-2\cdot5^{k}z}\right)\left( 1 +\varphi p^{-5^{k}z} +p^{-2\cdot5^{k}z}\right)\bigg\}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This follows from the fact that the zeta function can be expressed as &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \zeta(z)=\prod_{p\in \mathbb{P}}\;\;\sum_{k=0}^{\infty}\frac{1}{p^{k\;z}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and after some manipulations its easy to get the above representation.&lt;/p&gt;&#xA;" OwnerUserId="1747" LastActivityDate="2011-03-02T04:29:33.000" />
  <row Id="24554" PostTypeId="2" ParentId="24521" CreationDate="2011-03-02T04:33:05.870" Score="1" ViewCount="" Body="&lt;p&gt;You can prove it works by induction.  It works for $x=1$, as $f(x)=1$.  So assume it works for $x:  f(x) = \frac{x^3}{3} + \frac{x^2}{2} + \frac{x}{6}$ and we want to show it works for $x+1$:  $$\begin{align} f(x+1)&amp;amp;=f(x)+(x+1)^2 \\ &amp;amp;=\frac{x^3}{3} + \frac{x^2}{2} + \frac{x}{6} + (x+1)^2 \\ &amp;amp;=\frac{x^3}{3} + \frac{x^2}{2} + \frac{x}{6} +\frac{3x^2+3x+1}{3}+\frac{2x+1}{2}+\frac{1}{6} \\&amp;amp; =\frac{(x+1)^3}{3} + \frac{(x+1)^2}{2} + \frac{x+1}{6}\end{align}$$&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-03-02T04:33:05.870" />
  <row Id="24555" PostTypeId="2" ParentId="24521" CreationDate="2011-03-02T04:33:31.297" Score="4" ViewCount="" Body="&lt;p&gt;You can prove the formula works using induction.  I'm not sure what level you're at mathematically, so don't take this the wrong way if you have heard it... but if you haven't heard it, it's the following fact.  If $P(n)$ is some true/false statement based on a natural number $n$, and $P(1)$ is true, and whenever $P(k)$ is true then $P(k+1)$ is also true, then $P(n)$ is true for all natural numbers $n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this example, $P(1)$ is the statement that $1=\frac{1}{3}+\frac{1}{2}+\frac{1}{6}$, obviously true.  Now suppose that $P(k)$ is true, that is, that $f(k)$ is known to be $\frac{k^3}{3}+\frac{k^2}{2}+\frac{k}{6}$.  Then $f(k+1)$ is just the $k$th plus $(k+1)^2=k^2+2k+1$.  This is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle \frac{k^3}{3}+\frac{k^2}{2}+\frac{k}{6}+k^2+2k+1=\frac{2k^3+3k^2+k+6k^2+12k+6}{6}$&#xA;$\displaystyle =\frac{2k^3+9k^2+13k+6}{6}=\frac{(2k^3+6k^2+6k+2)+3k^2+6k+3)+(k+1)}{6}$&#xA;$\displaystyle =\frac{2(k+1)^3+3(k+1)^2+(k+1)}{6}=\frac{(k+1)^3}{3}+\frac{(k+1)^2}{2}+\frac{(k+1)}{6}.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How you actually come up with a formula like this in the first place is a different matter.  &quot;Watch for patterns&quot; is really what it boils down to, but there are a couple shortcuts.  The rule is simple enough that you can guess it's a polynomial.  The numbers you're adding to get from one stage to the next are squares, so you can guess that it's a cubic.  One way to check this is the &quot;method of common differences.&quot;  Your sequence is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle 0,1,5,14,30,55,\dotsc$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so the differences between successive terms are&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle 1,4,9,16,25,\dotsc$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and the differences between successive terms of that sequence are&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle 3,5,7,9,\dotsc$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and finally&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle 2,2,2,\dotsc$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The final sequence is constant, the one above that is linear, the one above that is quadratic, and the one above that is cubic.  So we have $f(n)=an^3+bn^2+cn+d$.  Then by substituting different values for $n$ and solving the system of equations, we can find out what $a,b,c,$ and $d$ are.  (It was obvious in this case what the first difference sequence was, but in other cases it may not be so obvious.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was going to conclude by noting that the $n$th triangular number is $\frac{(n)(n+1)}{1\cdot 2}$, and the $n$th pyramidal number is $\frac{(n)(n+1)(n+2)}{1\cdot 2\cdot 3}$, and guessing that things worked the same in higher dimensions.  But as it turns out, &lt;a href=&quot;http://mathworld.wolfram.com/FaulhabersFormula.html&quot; rel=&quot;nofollow&quot;&gt;they don't&lt;/a&gt;.  Take that how you will.&lt;/p&gt;&#xA;" OwnerUserId="2215" LastActivityDate="2011-03-02T04:33:31.297" CommentCount="2" />
  <row Id="24556" PostTypeId="2" ParentId="24547" CreationDate="2011-03-02T04:38:19.357" Score="2" ViewCount="" Body="&lt;p&gt;You can put $h_1 = 0$ and put $$ G_1 = -\frac{1}{2} \sum_{i \leq j} (e_{ij}+e_{ji}) x_{ij}. $$ This way you have that $x_{ij}$ are the entries of a PSD. You can add your constraints using $Ax=b$. Finally, notice that your objective function is linear in the entries of the matrix.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2011-03-02T04:38:19.357" CommentCount="2" />
  <row Id="24557" PostTypeId="2" ParentId="24521" CreationDate="2011-03-02T04:39:29.580" Score="4" ViewCount="" Body="&lt;p&gt;The first layer has $1^2$ elements; the second has $2^2$ elements; the next has $3^2$ elements, and so on. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So this is a matter of finding the formula for&#xA;$$1^2 + 2^2 + 3^2 + \cdots + n^2$$&#xA;in terms of $n$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For squares, this is a well known formula: $1^2+2^2+\cdots +n^2 = \frac{n(n+1)(2n+1)}{6}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that &#xA;$$\frac{n^3}{3} + \frac{n^2}{2} + \frac{n}{6}  = \frac{2n^3 + 3n^2 + n}{6} = \frac{n(2n^2+3n+1)}{6} = \frac{n(2n+1)(n+1)}{6}$$&#xA;same formula.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, you might be more interested in how to figure out the formulas in the first place. How does one express $1^k + 2^k + 3^k + \cdots + n^k$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is known as &lt;a href=&quot;http://en.wikipedia.org/wiki/Faulhaber%27s_formula&quot; rel=&quot;nofollow&quot;&gt;Faulhaber's formula&lt;/a&gt;. For the general case,&#xA;$$1^p + 2^p + \cdots + n^p = \frac{1}{p+1}\sum_{j=0}^p(-1)^jB_j\binom{p+1}{j} n^{p+1-j}$$&#xA;where $B_m$ is the $m$th &lt;a href=&quot;http://en.wikipedia.org/wiki/Bernoulli_number&quot; rel=&quot;nofollow&quot;&gt;Bernoulli number.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $p=2$, you get the spiffy formula with:&#xA;\begin{align*}&#xA;1^2+\cdots+n^2 &amp;amp;= \frac{1}{3}\left(\binom{3}{0}B_0n^3 - \binom{3}{1}B_1n^2 + \binom{3}{2}B_2n - \binom{3}{3}B_3\right)\\&#xA;&amp;amp;= \frac{1}{3}\left(B_0n^3 - 3B_1n^2 + 3B_2n - B_3\right)\\&#xA;&amp;amp;= \frac{1}{3}\left(n^3 - 3\left(-\frac{1}{2}\right)n^2 + 3\left(\frac{1}{6}\right)n - 0\right)\\&#xA;&amp;amp;= \frac{1}{3}n^3 +\frac{1}{2}n^2 + \frac{1}{6}n\\&#xA;&amp;amp;= \frac{n^3}{3} + \frac{n^2}{2} + \frac{n}{6},&#xA;\end{align*}&#xA;using the fact that $B_0 = 1$ $B_1 = -\frac{1}{2}$, $B_2 = \frac{1}{6}$, and $B_3=0$ (in fact, $B_n = 0$ for all odd $n\gt 1$). &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-02T04:50:04.943" LastActivityDate="2011-03-02T04:50:04.943" CommentCount="1" />
  <row Id="24558" PostTypeId="1" CreationDate="2011-03-02T04:47:03.547" Score="1" ViewCount="57" Body="&lt;p&gt;Hi, Everyone:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fundamental group Pi_1(G) of a connected graph G is defined to consist of all loops&#xA;  (i.e., closed paths)  based at  a given fixed basepoint/vertex g in G as elements,&#xA;  and concatenation as the group operation. (for G connected, we can show that the&#xA;  group is independent of the choice of basepoint; given g,g' we construct an isomorphism&#xA;  between the two groups by joining g,g' with a path.) Also given a connected graph G, there is is a result that G has a spanning tree T -- a tree T&amp;lt; G that uses all vertices in G ). &#xA;    Now, the main result (all else was a set-up until now) is that Pi_1(G) ~ Gp({E(T-G)}), i.e., Pi_1(G) is the free group generated by all edges in G that are not in T. My doubt is about this isomorphism; I do see how we need edges in T-G in order to generate loops based at a point, since T is acyclic (so that if G=T, the group is trivial). But I do not see how the group is isomorphic to the free group generated by edges in T-G. This is what I have so far: let fi denote a free edge, i.e., an edge in T-G. We can assume there is just one path joining any two edges in T (otherwise, T would contain a cycle)&#xA;Given g fixed in G (assume WOLG g in T), I think every closed loop based at g can be realized as the concatenation of a path within T, and an edge fi (aka, a pth in G-T), and then a(the) path from an endpoint of fi to g again. Then, for any edge fi, there is a (class) of loops based at g. Question: is the above the sketch of a proof for the fact that Pi_1(G) is the free group generated by all the edges in G-T?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sorry, I did not know how to make the question any more concise; I hope I don't get any Courics for writing too-long of a question.&#xA;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="7398" LastActivityDate="2011-03-02T04:47:03.547" Title="Why is the Fundamental Group of a Connected Graph G Free on elements in G-T; T spanning tree for G)" Tags="&lt;algebra&gt;&lt;topology&gt;&lt;graph-theory&gt;" CommentCount="4" />
  <row Id="24559" PostTypeId="2" ParentId="24536" CreationDate="2011-03-02T04:48:28.133" Score="1" ViewCount="" Body="&lt;p&gt;$cis(x)$ is equal to $e^{ix}$.&lt;/p&gt;&#xA;" OwnerUserId="173" LastActivityDate="2011-03-02T04:48:28.133" CommentCount="2" />
  <row Id="24561" PostTypeId="2" ParentId="24533" CreationDate="2011-03-02T04:58:57.297" Score="12" ViewCount="" Body="&lt;p&gt;Assuming Arnold means to find an approximate value -- I'd do this as follows: first, we may as well find the average of $\cos^{100} x$. I'll do this over a half-period, $-\pi/2 \le x \le \pi/2$. But $\cos x \approx 1-x^2/2$, so $\cos^{100} x \approx (1-x^2/2)^{100}$. If $x$ is small -- which it will have to be for $\cos x$ to be large (i. e. near 1) -- then $1-x^2/2 \approx e^{-x^2/2}$. So $\cos^{100} x \approx e^{-50x^2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the number we're looking for is about &#xA;$$ {1 \over \pi} \int_{-\pi/2}^{\pi/2} e^{-50x^2} \: dx. $$&#xA;But the integrand is so small far from zero that the limits of integration can be replaced with $-\infty$ and $\infty$ without changing much. That gives&#xA;$$ {1 \over \pi} \int_{-\infty}^\infty e^{-50x^2} \: dx. $$&#xA;Change variables, $u = x/\sqrt{50}$, to get&#xA;$$ {1 \over \pi \sqrt{50}} \int_{-\infty}^\infty e^{-u^2} \: du. $$&#xA;Finally, that integral is well-known to be $\sqrt{\pi}$; the approximate answer is $1/\sqrt{50\pi}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This method has the advantage that it works for high powers of &lt;i&gt;any&lt;/i&gt; function and isn't specialized to the trig functions. One source that explains this trick (and uses it to approximate the same integral) is Sanjoy Mahajan's book &lt;a href=&quot;http://mitpress.mit.edu/books/full_pdfs/Street-Fighting_Mathematics.pdf&quot; rel=&quot;nofollow&quot;&gt;Street-Fighting Mathematics&lt;/a&gt; (link goes to Creative-Commons downloadable version of the book).&lt;/p&gt;&#xA;" OwnerUserId="173" LastActivityDate="2011-03-02T04:58:57.297" CommentCount="2" />
  <row Id="24562" PostTypeId="2" ParentId="24529" CreationDate="2011-03-02T05:01:03.957" Score="11" ViewCount="" Body="&lt;p&gt;Another hint: can you factor $a^{15}-1$? In more than one way?&lt;/p&gt;&#xA;" OwnerUserId="173" LastActivityDate="2011-03-02T05:01:03.957" CommentCount="5" />
  <row Id="24563" PostTypeId="2" ParentId="24552" CreationDate="2011-03-02T05:03:09.180" Score="2" ViewCount="" Body="&lt;p&gt;Each triangle requires exactly three points, but not every set of three points defines a triangle. There are $n\choose 3$ sets of three points. You have to get rid of sets where all three points fall on the same side of the square. If $S_i$ points are on side $i$ of the square, $\sum_{i=1}^4 {{S_i}\choose 3}$ sets of three don't make triangles, and must be subtracted out. Since you're looking for the maximum number of triangles, minimize $\sum_{i=1}^4 {{S_i}\choose 3}$ subject to $S_1+S_2+S_3+S_4 = N$.  &lt;/p&gt;&#xA;" OwnerUserId="7675" LastActivityDate="2011-03-02T05:03:09.180" CommentCount="2" />
  <row Id="24564" PostTypeId="1" AcceptedAnswerId="24565" CreationDate="2011-03-02T05:06:06.650" Score="2" ViewCount="104" Body="&lt;p&gt;$\int_{0}^{2}\frac{1}{(x+2)(x-1)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Will I be able to use partial fractions on the indefinite integral and then evaluate my answer as f(2)-f(0)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any help is appreciated as always.  :)&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;EDIT:  I looked this up and apparently the integral doesn't converge so now I need help finding where I went wrong.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I had solved for  1 = $\frac{A}{x+2} + \frac{B}{x-1}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;sub in x = 1   and $B=\frac{1}{3}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;sub in x=-2 and $A=\frac{-1}{3}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Giving:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{-1}{3}\int\frac{dx}{x+2} + \frac{1}{3}\int\frac{dx}{x-1}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Simple u-substitution gives:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{-1}{3}ln|x+2|$  + $\frac{1}{3}ln|x-1|  +   C   $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which I was going to then evaluate f(2)-f(0)  but somewhere I went wrong.&lt;/p&gt;&#xA;" OwnerUserId="6423" LastEditorUserId="6423" LastEditorDisplayName="" LastEditDate="2011-03-02T05:25:19.613" LastActivityDate="2011-03-02T05:25:19.613" Title="Help integrating from $\int_{0}^{2}\frac{1}{(x+2)(x-1)}dx$" Tags="&lt;calculus&gt;&lt;integral&gt;" AnswerCount="1" CommentCount="6" />
  <row Id="24565" PostTypeId="2" ParentId="24564" CreationDate="2011-03-02T05:08:48.023" Score="5" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Hint 1:&lt;/strong&gt;&#xA;$$\frac{1}{(x+2)(x-1)}=\frac{1}{3}\left(\frac{1}{x-1}-\frac{1}{x+2}\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint 2:&lt;/strong&gt; Does this converge or diverge?  Think about $$\int_{-1}^{1}\frac{1}{x}dx.$$  Does the value of this integral even make sense?&lt;/p&gt;&#xA;" OwnerUserId="6075" LastActivityDate="2011-03-02T05:08:48.023" CommentCount="3" />
  <row Id="24566" PostTypeId="1" AcceptedAnswerId="24568" CreationDate="2011-03-02T05:12:56.737" Score="3" ViewCount="62" Body="&lt;p&gt;I just learned about modular numbers on wikipedia, such as $17 \equiv 3\pmod{7}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what is infinity $\pmod{n}$? It can't very well be all the numbers at once, so what happens?&lt;/p&gt;&#xA;" OwnerUserId="7676" LastEditorUserId="6075" LastEditorDisplayName="" LastEditDate="2011-03-02T05:15:57.677" LastActivityDate="2011-03-02T05:19:03.147" Title="Modular numbers" Tags="&lt;modular-arithmetic&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="24568" PostTypeId="2" ParentId="24566" CreationDate="2011-03-02T05:19:03.147" Score="3" ViewCount="" Body="&lt;p&gt;When we say $a (\bmod n)$, we need $a \in \mathbb{Z}$ and $n \in \mathbb{Z} \backslash \{0\}$. So your question &quot;$\infty (\bmod n)$&quot; doesn't make sense in the first place. It is like asking &quot;What is $\text{apple} (\bmod n)$?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What your probably mean and want to know is &quot;What is $\displaystyle \lim_{x \in \mathbb{Z}, x \rightarrow \infty} x (\bmod n)$?&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $n \neq \pm 1$, then the answer is &quot;It doesn't exist&quot;. If $n = \pm 1$, then the answer is $0$.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastActivityDate="2011-03-02T05:19:03.147" CommentCount="3" />
  <row Id="24569" PostTypeId="1" AcceptedAnswerId="24574" CreationDate="2011-03-02T05:23:59.350" Score="0" ViewCount="41" Body="&lt;p&gt;I solved a problem in Calculus 2 exercise lesson and I do not fully understand it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The task is to show that cross derivates differ in sign ( $f_{\text{xy}}(0;0)=-1$ and $f_{\text{yx}}(0;0)=1\, $ )&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The function is:&#xA;$ f(x,y)=\left\{&#xD;&#xA;\begin{array}{cc}&#xD;&#xA; \frac{x y \left(x^2-y^2\right)}{x^2+y^2} &amp;amp; x^2+y^2\neq 0 \\&#xD;&#xA; 0 &amp;amp; x^2+y^2=0&#xD;&#xA;\end{array}&#xD;&#xA;\right. $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Short solution is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/u9Jli.png&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/9Tl1m.png&quot; alt=&quot;Plot of function&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was told that lesson, what I should take away from this is, that cross derivates in this case&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;are not continuous&lt;/li&gt;&#xA;&lt;li&gt;do not equal&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Can someone explain:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;why it is not continuous&lt;/li&gt;&#xA;&lt;li&gt;why can't I just take $f_{xy}=\begin{cases}&#xD;&#xA; \frac{x^6+9 x^4 y^2-9 x^2 y^4-y^6}{\left(x^2+y^2\right)^3} &amp;amp; x^2+y^2\neq 0 \\&#xD;&#xA; 0 &amp;amp; x^2+y^2 = 0&#xD;&#xA;\end{cases} $ and where do I deduce, that I need to use limits at first place.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="2317" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2011-03-02T08:09:44.083" LastActivityDate="2011-03-02T08:09:44.083" Title="Second order cross derivate question" Tags="&lt;multivariable-calculus&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="24572" PostTypeId="2" ParentId="24529" CreationDate="2011-03-02T05:48:39.407" Score="5" ViewCount="" Body="&lt;p&gt;If you really have no other way: Substitute x for a^5, so you get x^2+x+1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then use some quadratic formula action, then use some factoring action on those binomials, then just try all possible products of the factors you get until something nice comes out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I could see this being not so nice a solution though.&lt;/p&gt;&#xA;" OwnerUserId="7676" LastActivityDate="2011-03-02T05:48:39.407" CommentCount="1" />
  <row Id="24573" PostTypeId="2" ParentId="24569" CreationDate="2011-03-02T05:48:49.720" Score="1" ViewCount="" Body="&lt;p&gt;The function $f_{xy}$ that you wrote down (presumably by using various standard derivative formulas) is not continuous at $0$.  As you asserted above, the limit as $(x,y) \to (0,0)$ depends on the path taken, so the limit does not exist.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps an analogous example from single-variable calculus will clear this up.  The function $g(x) = |x|$ defined for all reals has derivative&#xA;$ g&amp;#39;(x) = \dfrac {|x|} {x} = \left\{ \begin {array} {l l} 1 &amp;amp; x&amp;gt;0 \\ -1 &amp;amp; x&amp;lt;0 \end {array} \right. $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can extend the domain of definition to include $x = 0$ by defining the value however we like, but the function will &lt;strong&gt;not&lt;/strong&gt; be continuous.&lt;/p&gt;&#xA;" OwnerUserId="6509" LastActivityDate="2011-03-02T05:48:49.720" />
  <row Id="24574" PostTypeId="2" ParentId="24569" CreationDate="2011-03-02T05:53:58.437" Score="1" ViewCount="" Body="&lt;p&gt;The cross-derivative as you have worked out when $(x,y) \neq (0,0)$ is $\displaystyle \frac{x^6+9 x^4 y^2-9 x^2 y^4-y^6}{\left(x^2+y^2\right)^3}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So why is this function not continuous at the origin?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What does continuity at a point $(x_0,y_0)$ in $2D$ mean?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a function to be continuous at $(x_0,y_0)$, we need that irrespective of the way you approach the point $(x_0,y_0)$, the value of the limit should not change. (Note that in $2D$ we have infinite directions of approaching the point $(x_0,y_0)$. This is in contrast to $1D$ where we have only $2$ directions to approach a specific point.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the function $\displaystyle \frac{x^6+9 x^4 y^2-9 x^2 y^4-y^6}{\left(x^2+y^2\right)^3}$, let us see what is the value of the limit as we approach the origin $(0,0)$ along different lines passing through the origin $(0,0)$ i.e. along lines $y=mx$ for different values of $m$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\lim_{y=mx; x \rightarrow 0} \frac{x^6+9 x^4 y^2-9 x^2 y^4-y^6}{\left(x^2+y^2\right)^3} = \lim_{x \rightarrow 0} \frac{x^6+9 x^4 (mx)^2-9 x^2 (mx)^4-(mx)^6}{\left(x^2+(mx)^2\right)^3} = \frac{1+9 m^2-9 m^4-m^6}{\left(1+m^2\right)^3}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, the limit depends on $m$ i.e. the direction in which you approach.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Note that $\frac{\partial}{\partial x} \left(\frac{\partial f}{\partial y} \right)$ is obtained by plugging in $m=0$ and $\frac{\partial }{\partial y} \left(\frac{\partial f}{\partial x} \right) $ is obtained by plugging in $m= \infty$)&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2011-03-02T05:59:02.533" LastActivityDate="2011-03-02T05:59:02.533" />
  <row Id="24575" PostTypeId="1" CreationDate="2011-03-02T06:00:40.807" Score="2" ViewCount="112" Body="&lt;p&gt;This appeared on my professor's test review, and its taken me hours to, surprise surprise, get the wrong answer. Could someone help me with the method I should be using to solve this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$y^{\prime\prime}+y=\tan x$$&lt;/p&gt;&#xA;" OwnerUserId="6895" LastEditorUserId="6075" LastEditorDisplayName="" LastEditDate="2011-03-02T06:06:18.167" LastActivityDate="2011-03-02T08:27:40.303" Title="What is the quickest way to solve this 2nd Order Linear ODE?" Tags="&lt;differential-equations&gt;" AnswerCount="4" />
  <row Id="24576" PostTypeId="1" AcceptedAnswerId="24606" CreationDate="2011-03-02T06:05:22.230" Score="1" ViewCount="162" Body="&lt;p&gt;Pointwise convergence of Fourier series of a piecewise continuous (and Lipschitz continuous everywhere) function.I basically want to understand a proof for convergence of a Fourier series of $f(x)$ to $\frac{1}{2}(f(x^+)+f(x^-))$. I have seen a proof in Zygmund's book but I don't understand it as it apparently requires some number theory and also not self contained. Please suggest me a reference (book or even a web location) which gives a self contained proof without any reference to number theory. I guess the original proof by Dirichlet uses number theory.Thank you. Kindly note that I expect more than &quot;any introductory book on fourier analysis&quot; type of an answer.&lt;/p&gt;&#xA;" OwnerUserId="2987" LastEditorUserId="5363" LastEditorDisplayName="" LastEditDate="2011-03-02T10:13:17.383" LastActivityDate="2011-03-02T16:44:33.453" Title="Pointwise convergence of Fourier series of a piecewise continuous (and Lipschitz continuous everywhere) function - a reference request." Tags="&lt;reference-request&gt;&lt;fourier-analysis&gt;&lt;fourier-series&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="24577" PostTypeId="1" AcceptedAnswerId="24635" CreationDate="2011-03-02T06:09:20.973" Score="4" ViewCount="213" Body="&lt;p&gt;Let $\{ f_n \}$ be a sequence in a Hilbert space $L^2(\mathbb{R}^d)$. We say that this sequence converges weakly to an element $f \in L^2$ if $\langle f_n, g \rangle \to \langle f,g \rangle$ for every $g \in L^2$ (where $\langle \cdot,\cdot \rangle$ denotes the inner product on $L^2$). By definition, we are given that the weak limit $f$ is in $L^2$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, suppose we know that a sequence &quot;formally&quot; converges weakly to a limit $f$ (i.e. $\langle f_n, g \rangle \to \langle f,g \rangle$ for every $g \in L^2$ for some $f$ which we don't necessarily know yet to be in $L^2$) . &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does this, purely by the characteristics of weak convergence, directly imply that $f \in L^2$? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think you could also generalize this question to any Hilbert space, provided that taking the inner product of an element possibly not in the Hilbert space makes sense.&lt;/p&gt;&#xA;" OwnerUserId="1736" LastEditorUserId="6887" LastEditorDisplayName="" LastEditDate="2011-03-02T09:13:49.110" LastActivityDate="2011-03-02T14:27:13.407" Title="On the limits of weakly convergent subsequences" Tags="&lt;functional-analysis&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="24578" PostTypeId="1" CreationDate="2011-02-28T22:03:44.007" Score="0" ViewCount="108" Body="&lt;p&gt;I have the following homework, for which I'd like you to give me some hints. Thank you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are M food dishes and N bottles of different wines. There are 4 ways to pair food and wine: 1 (dish) to 1 (type of wine), 1 to many, many to 1, and many to many. A menu must use all M dishes and N wine bottles. How many possible menus can one make?&lt;/p&gt;&#xA;" OwnerUserId="6289" LastEditorUserId="3733" LastEditorDisplayName="" LastEditDate="2011-03-02T06:29:21.213" LastActivityDate="2011-05-31T07:07:08.173" Title="How many ways to pair wine and food?" Tags="&lt;homework&gt;&lt;combinatorics&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="24579" PostTypeId="2" ParentId="24575" CreationDate="2011-03-02T06:12:05.923" Score="5" ViewCount="" Body="&lt;p&gt;The quick rigorous method I can think of is by variation of parameters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The solution is given by $y = y_h + y_p$ where $y_h$ is the homogeneous part of the solution and $y_p$ is the particular solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The solution to the homogeneous part is $y_h(x) = c_1 \cos(x) + c_2 \sin(x)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$y_p$ is obtained by &lt;strong&gt;variation of parameters&lt;/strong&gt; as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason for writing it as a linear combination of $\cos(x)$ and $\sin(x)$ is that these two are the linearly independent solution to the homogeneous part.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$y_p = a(x) \cos(x) + b(x) \sin(x)$$ $y_p&amp;#39; = a&amp;#39;(x) \cos(x) + b&amp;#39;(x) \sin(x) - a(x) \sin(x) + b(x) \cos(x)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Set $a&amp;#39;(x) \cos(x) + b&amp;#39;(x) \sin(x) = 0$ and hence $y_p&amp;#39; = - a(x) \sin(x) + b(x) \cos(x)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$y_p&amp;#39;&amp;#39; = - a&amp;#39;(x) \sin(x) + b&amp;#39;(x) \cos(x) - a(x) \cos(x) -b(x) \sin(x) = b&amp;#39;(x) \cos(x) - a&amp;#39;(x) \sin(x) - y_p$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, we have $a&amp;#39;(x) \cos(x) + b&amp;#39;(x) \sin(x) = 0$ and $- a&amp;#39;(x) \sin(x) + b&amp;#39;(x) \cos(x) = \tan(x)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solve for $a(x)$ and $b(x)$ from the two equations to get $b&amp;#39;(x) = \sin(x)$ and $a&amp;#39;(x) = -\frac{\sin^2(x)}{\cos(x)}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From which we get $b(x) = -\cos(x)$ and $a(x) = \sin(x) + \log \left( \left| \frac{\cos(x/2) - \sin(x/2)}{\cos(x/2) + \sin(x/2)} \right| \right)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and plug it back in and simplify to get the particular solution as &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$y_p = \cos(x) \log \left( \left| \frac{\cos(x/2) - \sin(x/2)}{\cos(x/2) + \sin(x/2)} \right| \right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The final solution is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$y = c_1 \cos(x) + c_2 \sin(x) + \cos(x) \log \left( \left| \frac{\cos(x/2) - \sin(x/2)}{\cos(x/2) + \sin(x/2)} \right| \right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;: Taking a cue from Moron's post, the Green's function for this equation (which is nothing but a 1D Helmholtz equation with unit wavenumber) is $G(x) = -i \frac{e^{i|x|}}{2}$ and hence the particular solution is $\int G(x-y) \tan(y) dy$&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2011-03-02T07:42:36.710" LastActivityDate="2011-03-02T07:42:36.710" CommentCount="2" />
  <row Id="24580" PostTypeId="2" ParentId="24575" CreationDate="2011-03-02T06:13:46.173" Score="1" ViewCount="" Body="&lt;p&gt;If you feed d^2y/dx^2+y=tan(x) to Wolfram Alpha you get a solution.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-03-02T06:13:46.173" CommentCount="3" />
  <row Id="24581" PostTypeId="2" ParentId="24575" CreationDate="2011-03-02T06:17:07.367" Score="3" ViewCount="" Body="&lt;p&gt;Read this article:  &lt;a href=&quot;http://www.sosmath.com/diffeq/second/variation/variation.html&quot; rel=&quot;nofollow&quot;&gt;http://www.sosmath.com/diffeq/second/variation/variation.html&lt;/a&gt;&lt;br&gt;&#xA;It talks about &quot;Variation of Parameters&quot; which is what you need to use to solve it.  A solution ends up being:  $$y=-\cos(x)\ln\left(\sec(x)+\tan(x)\right).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All of this is covered in detail in the link.  Also see &lt;a href=&quot;http://en.wikipedia.org/wiki/Variation_of_parameters&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Variation_of_parameters&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="6075" LastEditorUserId="6075" LastEditorDisplayName="" LastEditDate="2011-03-02T08:27:40.303" LastActivityDate="2011-03-02T08:27:40.303" CommentCount="4" />
  <row Id="24583" PostTypeId="2" ParentId="24549" CreationDate="2011-03-02T06:25:09.343" Score="2" ViewCount="" Body="&lt;p&gt;If $f:G&amp;#39; \to G$ is injective, this is true in all cases: (relatively) projective/injective. Identify the $G$-modules with $\mathbb{Z}G$-modules and observe that $\mathbb{Z}G$ is a free (hence also projective and flat) $\mathbb{Z}G&amp;#39;$-module (it is free as a $\mathbb{Z}G&amp;#39;$-module because $\mathbb{Z}G \cong \bigoplus_{G/G&amp;#39;} \mathbb{Z}G&amp;#39;$ as a $\mathbb{Z}G&amp;#39;$-module). The functors $\mathbb{Z}G \otimes_{G&amp;#39;} {-}$ (induction) and $\operatorname{Hom}_{\mathbb{Z}G&amp;#39;}{(\mathbb{Z}G,{-})}$ (co-induction) are therefore exact and left/right adjoint to restriction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By checking the lifting/extension conditions, it is easy to see that a left/right adjoint to an exact functor preserves projectives/injectives. Here &quot;exact&quot; of course means &quot;sends $\mathbb{Z}$-split exact sequences to $\mathbb{Z}$-split exact sequences&quot; in the relative case. For instance, if $L$ has an exact right adjoint $R$ and $P$ is projective then $LP$ is projective because a lifting problem of a map $LP \to N$ over an epimorphism $M \twoheadrightarrow N$ corresponds to a lifting problem of a map $P \to RN$ over the morphism $RM \twoheadrightarrow RN$ which is epi because $R$ is exact.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if $f$ is not injective, this is wrong in general. Just consider the case where $G = \{e\}$ is the trivial group. It is certainly not the case that $\mathbb{Z}$ is (rel.) projective or $\mathbb{Q}$ is (rel.) injective as a $\mathbb{Z}G&amp;#39;$-module (otherwise group homology with $\mathbb{Z}$-coefficients or group cohomology with $\mathbb{Q}$-coefficients would always have to vanish).&lt;/p&gt;&#xA;" OwnerUserId="5363" LastActivityDate="2011-03-02T06:25:09.343" />
  <row Id="24584" PostTypeId="2" ParentId="24578" CreationDate="2011-03-02T06:36:23.937" Score="1" ViewCount="" Body="&lt;p&gt;I don't understand &quot;There are 4 ways to pair food and wine: 1 (dish) to 1 (type of wine), 1 to many, many to 1, and many to many&quot;  If you have a number of courses, you need that many dishes and  that many wines.  The naive answer is that you can pair n wines with the first dish, n-1 with the next, and so on, leading to n! choices.  But that doesn't respect the 4 ways to pair food and wine.  What am I missing?&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-03-02T06:36:23.937" />
  <row Id="24585" PostTypeId="1" CreationDate="2011-03-02T06:44:28.407" Score="3" ViewCount="138" Body="&lt;p&gt;Hi, everyone:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am trying to review my differential Geometry. In J.Lee's &lt;em&gt;Smooth Manifolds&lt;/em&gt;, there is an exercise in which one has to show that the product of smooth covering maps is a smooth covering map. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A smooth covering map is a smooth cover $\pi: M&amp;#39; \rightarrow M$, where&#xA;$M, \, M&amp;#39;$ are smooth manifolds, and in which every $p$ in $M&amp;#39;$ has a neighborhood $U_p$ in $M$ such that $p|_{p^{-1}} (U_p) \rightarrow U_p$ (i.e., the restriction of $p$ to the inverse image of $U_p$ is a diffeomorphism) . I think this exercise is straightforward, (basically we just use the fact that the product of diffeomorphisms is a diffeomorphism) but I am kind of rusty, and would appreciate your inputs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, we need to show that , given covers $p_1:M&amp;#39; \rightarrow M$ and $p_2:N&amp;#39; \rightarrow {N}$ that for any pair $(p,q)$ there is a neighborhood $W$ of $(p, q)$ such that $p_1 \times p_2 |_{p_1^{-1} \times p_2^{-1} (W)} : p_1^{-1} \times p_2^{-1} (W) \rightarrow W$ is a diffeomorphism. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But we know that for $p$ in $M$ there is a $U_p$ with $p_1|_{p_1^{-1} (U_p)} : p_1^{-1} (U_p) \rightarrow U_p$ and for any $q$ in $N$ there is a $U_q$ with $p_2|_{{p_2}^{-1} (U_q)} : p_2^{-1} (U_q) \rightarrow U_q$&#xA;such that both are diffeomorphisms. Then the product map is a diffeomorphism automatically, isn't it, i.e., isn't the map:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(p_1, p_2)|_{(p_1^{-1} \times p_2^{-1}) (U_q)} : (p_1^{-1} \times p_2^{-1}) (U_q) \rightarrow U_p \times U_q$$ a diffeomorphism?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know that if $f = (f_1(x_1,\ldots,x_n),f_2(x_1,\ldots,x_n))$ is differentiable with derivative $(f_1&amp;#39;,f_2&amp;#39;)$ and$f_1^{-1}$ and $f_2^{-1}$ are each differentiable (by the assumption of diffeomorphism), then  I think the differentiable inverse here is given by the identities fof&lt;sup&gt;-1&lt;/sup&gt;=Id.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I imagine there may be some issue in showing that the above is true for manifolds, and not just for subsets of $\mathbb{R}^n$. Other than that, is my setup correct?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically, I am curious as to whether this problem comes down to the fact that the&#xA;product of diffeomorphisms is a diffeomorphism.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="7679" LastEditorUserId="7687" LastEditorDisplayName="" LastEditDate="2011-03-02T10:33:43.460" LastActivityDate="2011-05-31T17:07:20.570" Title="Product of Smooth Covering Maps a Smooth Covering Map? ( J.Lee, 2-12)" Tags="&lt;differential-geometry&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="24586" PostTypeId="2" ParentId="24489" CreationDate="2011-03-02T06:58:03.027" Score="1" ViewCount="" Body="&lt;p&gt;I'd think about the answer to your question &quot;What does it mean to write an expression like $x^2 = 2x$?&quot; in terms of formal systems:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Formal_system&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Formal_system&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're given the additional &quot;axiom&quot; $x^2 = 2x$, you can then reason about this expression using the other axioms and rules of inference in the system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When asked to &quot;find $x$&quot;, you're being asked to derive a theorem stating that $x$ is equal to some other quantity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're interested in reading more about formal languages; I'd recommend Kleene's &quot;Introduction to Metamathematics&quot;, which is a classic.&lt;/p&gt;&#xA;" OwnerUserId="7438" LastActivityDate="2011-03-02T06:58:03.027" />
  <row Id="24587" PostTypeId="2" ParentId="24032" CreationDate="2011-03-02T07:05:32.630" Score="4" ViewCount="" Body="&lt;p&gt;Fairly random (and definitely unrelated) web surfing turned up &lt;a href=&quot;http://www.math.uga.edu/~pete/Cheng-Zheng99.pdf&quot; rel=&quot;nofollow&quot;&gt;this nice short paper of Cheng and Zheng&lt;/a&gt; which proves in a very constructive way that if $f: \mathbb{R} \rightarrow \mathbb{R}$ is any continuous function which is periodic with irrational period, then $f(\mathbb{N})$ is dense in $f(\mathbb{R})$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The ideas are first exhibited with respect to the function $f(x) = \sin x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your question is not literally a special case of this, since $\tan x$ is not continuous on all of $\mathbb{R}$.  Nevertheless I think it is close enough for the same methods to be carried over.  (For instance, $\tan x$ can be viewed as a continuous function with values in $\mathbb{R} \mathbb{P}^1 \cong S^1$.)  In any case the relation here is close enough so that I thought the paper would be of interest to readers of this question.  &lt;/p&gt;&#xA;" OwnerUserId="299" LastActivityDate="2011-03-02T07:05:32.630" />
  <row Id="24588" PostTypeId="1" CreationDate="2011-03-02T07:13:54.557" Score="2" ViewCount="99" Body="&lt;p&gt;Really, the curve in question is the polar plot $ r = cos( K * \theta) $, where $K$ is any irrational number (I use $\pi$), but the transformation to a parametric one on $x$ and $y$ with domain $t$ is an unsurprising one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It would appear that the curve is confined to the unit circle, and also that it never repeats -- that it is aperiodic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given these three things&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The domain of the function is any real number&lt;/li&gt;&#xA;&lt;li&gt;The range of the function is confined to a finite space&lt;/li&gt;&#xA;&lt;li&gt;The function is aperiodic&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Does it mean that the unit circle is completely filled for this parametric function from negative infinity to infinity?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can we say that for any given point in the unit circle, there is a number for $t$ where the curve intersects it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to say no.  I really do.  My intuition says so.  But why?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there any other curves with the three bullet pointed conditions above that can be shown to be more clearly non-space-filling?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the appropriate mathematical term for the way thi curve acts on the unit circle?&lt;/p&gt;&#xA;" OwnerUserId="51" LastEditorUserId="51" LastEditorDisplayName="" LastEditDate="2011-03-02T18:28:29.363" LastActivityDate="2011-03-02T18:28:29.363" Title="Is this parametric curve space-filling? Why or why not?" Tags="&lt;functions&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="24589" PostTypeId="1" AcceptedAnswerId="25218" CreationDate="2011-03-02T07:24:26.633" Score="6" ViewCount="123" Body="&lt;p&gt;The following problem has come up in my research and I don't have the tools (i.e., I don't know Algebraic Geometry, especially over $\mathbb{R}$) to solve it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider two subsets $X$ and $Y$ of $M_n(\mathbb{R})^k$ with $k\geq 3$.  The subset $X$ consists of all $k$ tuples of n x n matrices $(A_1,..., A_k)$ such that for any two $A_i$ and $A_j$, there is a vector $v_{ij}$ which is simultaneously an eigenvector for $A_i$ and $A_j$ (but perhaps with different eigenvalues).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The subset $Y$ consists of all those elements of $X$ such that $v_{ij}$ can be chosen independently from $i$ and $j$ - that is, if $(A_1,.., A_k)\in Y$ then all $k$ matrices have a common eigenvector.  (Of course, when $k=1$ or $k=2$, $X = Y$, hence the above restriction on $k$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, I have not been able to prove that $X$ is Zariski closed (though I have not tried that hard - it's not so important for my purposes), but I can prove that it's contained in a proper Zariski closed subset of $M_n(\mathbb{R})^k$ (thought of as $\mathbb{R}^{kn^2}$):  we have $f_{12} = det(A_1A_2 - A_2A_1) = 0$ since $A_1A_2 v_{12} = A_2A_1 v_{12}$.  Or, since we're worker over $\mathbb{R}$, we can put all the $f_{ij}$ into one big polynomial equation $$\sum_{1\leq i &amp;lt; j\leq k} f_{ij}^2 = 0.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I'd like to know is&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Is there a Zariski closed subset $F$ with $Y\subseteq F\subsetneq X$?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Said another way&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Is there a polynomial which is simultaneously satisfied by all k-tuples of matrices sharing a common eigenvector but for which there are elements in $X$ which do not solve it?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Finally, in case it helps, the case I'm most interested in is $n=3$ and $k = 5$, but I imagine the choice of $n$ won't affect the answer greatly and $k=3$ probably contains all the insight necessary to tackle the larger $k$ values.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you in advance for your help.&lt;/p&gt;&#xA;" OwnerUserId="331" LastEditorUserId="331" LastEditorDisplayName="" LastEditDate="2011-03-02T15:22:21.150" LastActivityDate="2011-03-05T23:10:29.957" Title="Given k real n x n matrices with a common eigenvector, is there some nontrivial polynomial equation the entries of the matrices satisfy?" Tags="&lt;linear-algebra&gt;&lt;algebraic-geometry&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="24590" PostTypeId="2" ParentId="24575" CreationDate="2011-03-02T07:25:34.027" Score="3" ViewCount="" Body="&lt;p&gt;The method below will solve equations of the form:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$$y&amp;#39;&amp;#39; + y = \frac{f&amp;#39;(x)}{\cos x}$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;First notice that $\displaystyle (h \cos x)&amp;#39;&amp;#39; = h&amp;#39;&amp;#39; \cos x - 2 h&amp;#39; \sin x - h \cos x$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus if $\displaystyle y = h \cos x$, then $\displaystyle y&amp;#39;&amp;#39; + y = h&amp;#39;&amp;#39; \cos x - 2h&amp;#39; \sin x$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus $\displaystyle (y&amp;#39;&amp;#39; + y&amp;#39;)\cos x = h&amp;#39;&amp;#39; \cos^2 x - 2h&amp;#39; \sin x \cos x = (h&amp;#39; \cos^2 x)&amp;#39;$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus we get $$h&amp;#39; \cos^2 x = f(x) + A$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And so &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$$y = \cos x  \int (f(x) + A)\sec^2 x \ \text{d}x$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In your case, $\displaystyle f(x) =  - \cos x$ and so&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$$y = \cos x \ \int (A - \cos x) \sec^2 x \ \text{d}x  = A\sin x -  \cos x \ \log (\sec x + \tan x) + B \cos x$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2011-03-02T07:25:34.027" CommentCount="2" />
  <row Id="24591" PostTypeId="1" CreationDate="2011-03-02T07:30:56.523" Score="0" ViewCount="87" Body="&lt;p&gt;For those who use this book to learn differential geometry, what do you think of this book?&lt;/p&gt;&#xA;" OwnerUserId="2177" LastEditorUserId="367" LastEditorDisplayName="" LastEditDate="2011-03-02T15:39:53.813" LastActivityDate="2011-03-02T15:39:53.813" Title="Comment and suggestion on the differential geometry of curves and surfaces by Manfredo P. do Carmo" Tags="&lt;soft-question&gt;" CommentCount="1" />
  <row Id="24592" PostTypeId="1" AcceptedAnswerId="24593" CreationDate="2011-03-02T07:32:00.267" Score="2" ViewCount="66" Body="&lt;p&gt;There's a problem I can't figure out in my homework. I can't really understand what it's asking. Maybe someone can help.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A meteor enters the Earth's atmosphere&#xA;  and burns up at a rate that, at each&#xA;  instant, is proportional to its&#xA;  surface area. Assuming that the meteor&#xA;  is always spherical, show that the&#xA;  radius decreases at a constant rate.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I think the problem is asking me either to show that $\frac{dr}{dt} = 0$ (which I don't know how to do) or that the relationship $\frac{dV}{dt} / \frac{dA}{dt}$ has no parameter $\frac{dr}{dt}$, which I've done, I think.. $\frac{dV}{dt} / \frac{dA}{dt} = \frac {r}{2}$&lt;/p&gt;&#xA;" OwnerUserId="7682" LastActivityDate="2011-03-02T07:42:50.347" Title="Rate of change question" Tags="&lt;calculus&gt;&lt;homework&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="24593" PostTypeId="2" ParentId="24592" CreationDate="2011-03-02T07:41:34.903" Score="2" ViewCount="" Body="&lt;p&gt;Use $\dfrac{dV}{dt}=kA$ to show that $\dfrac{dr}{dt}=l$, assuming that rate of burning is expressed in terms of volume.&lt;/p&gt;&#xA;" OwnerUserId="4594" LastActivityDate="2011-03-02T07:41:34.903" />
  <row Id="24595" PostTypeId="1" CreationDate="2011-03-02T07:48:54.190" Score="2" ViewCount="80" Body="&lt;p&gt;I am interested in somehow characterizing the zeros of the following function:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$P(z) = \sum_{i=1}^n \frac{\alpha_i}{z+\lambda_i}$, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;with $\lambda_1\geq \lambda_2 \geq \ldots \geq \lambda_n=0$ and $\sum_{i=1}^n \alpha_i = 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore, the values $\alpha_i$ are in general non-integer and may even be complex.  The poles $\lambda_i$, however, are all real.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I initially had hoped to use something like the Gauss-Lucas theorem for this, but the conditions on $\alpha_i$ seem to prevent this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While it would be great to make some strong statements about the zeros, I would be more then satisfied to find conditions guaranteeing that the zeros lie in the left-half of the complex plane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any thoughts/suggestions would be great.  Thanks!&lt;/p&gt;&#xA;" OwnerUserId="7683" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2011-05-01T12:54:23.697" LastActivityDate="2011-05-31T13:07:14.570" Title="Zeros of linear partial fractions" Tags="&lt;complex-analysis&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="24596" PostTypeId="1" AcceptedAnswerId="24613" CreationDate="2011-03-02T07:53:31.007" Score="1" ViewCount="94" Body="&lt;p&gt;Let $X_1,\dots,X_n$ be Gaussian random variables with mean $\mu$ and variance $\sigma^2$. What is known about the distribution of&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\sqrt{\frac1{n-1}\sum_{i=1}^n(X_i-\bar X)^2}}{\bar X} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with $\bar X = \frac 1n \sum_{i=1}^n X_i$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does this distribution have a name? I am especially interested in its cumulative distribution function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If this is too complicated, I'd settle for &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\sqrt{\frac1{n}\sum_{i=1}^n(X_i-\mu)^2}}{\mu} $$&lt;/p&gt;&#xA;" OwnerUserId="505" LastActivityDate="2011-03-02T12:57:11.697" Title="Distribution of the coefficient of variation of independent normal variables?" Tags="&lt;statistics&gt;&lt;probability-theory&gt;" AnswerCount="3" CommentCount="0" />
  <row Id="24597" PostTypeId="2" ParentId="24178" CreationDate="2011-03-02T08:00:25.750" Score="5" ViewCount="" Body="&lt;p&gt;There is a name for what you describe: it's called an &lt;a href=&quot;http://mathworld.wolfram.com/Orthomorphism.html&quot; rel=&quot;nofollow&quot;&gt;orthomorphism&lt;/a&gt;. If $S$ is the cyclic group of order $n$, then the number of orthomorphisms for $n=1,3,5,...$ is $1, 3, 15, 133, 2025, 37851,...$ (sequence &lt;a href=&quot;http://oeis.org/A006717&quot; rel=&quot;nofollow&quot;&gt;A006717&lt;/a&gt; in OEIS). (If $n$ is even, $S$ has no orthomorphisms.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can find lots of links for the case when $S$ is GF$(2^n)$ by googling &quot;orthomorphism of galois field&quot;. For instance, &lt;a href=&quot;http://ijns.femto.com.tw/contents/ijns-v10-n1/ijns-2010-v10-n1-p57-61.pdf&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; is a short paper which confirms my $244744192$ and goes on to discuss the construction of permutation polynomials.&lt;/p&gt;&#xA;" OwnerUserId="1508" LastActivityDate="2011-03-02T08:00:25.750" />
  <row Id="24598" PostTypeId="2" ParentId="24596" CreationDate="2011-03-02T08:06:14.243" Score="3" ViewCount="" Body="&lt;p&gt;This is a bit long to be a comment and hence posting this as an answer. Look up &lt;a href=&quot;http://en.wikipedia.org/wiki/F-distribution&quot; rel=&quot;nofollow&quot;&gt;Fisher distribution&lt;/a&gt; for the first one and &lt;a href=&quot;http://en.wikipedia.org/wiki/Chi-square_distribution&quot; rel=&quot;nofollow&quot;&gt;chi-squared distribution&lt;/a&gt; for the second one though both the random variables don't exactly fit in the distributions. (For the first one, you need to subtract of the mean from the denominator and square it and rescale it to be a Fisher distribution while in the second case, you need to square it and rescale it to be chi-squared distribution).&lt;/p&gt;&#xA;" OwnerUserId="3302" LastActivityDate="2011-03-02T08:06:14.243" />
  <row Id="24599" PostTypeId="2" ParentId="24577" CreationDate="2011-03-02T08:17:10.447" Score="0" ViewCount="" Body="&lt;p&gt;Well, there are answers to this on different levels. One answer is, that the norm in a Hilbert space is always weakly lower semicontinuous meaning that for a weakly converging sequence $(f_n)$ it holds that&#xA;$$&#xD;&#xA;\lim\inf \|f_n\| \geq \|w-\lim f_n\|.&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another answer is that due to the principle of uniform boundedness (or Banach-Steinhaus-Theorem) every weakly convergent sequence is bounded (see &lt;a href=&quot;http://en.wikipedia.org/wiki/Weak_convergence_%28Hilbert_space%29&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;).&lt;/p&gt;&#xA;" OwnerUserId="3148" LastActivityDate="2011-03-02T08:17:10.447" CommentCount="2" />
  <row Id="24600" PostTypeId="2" ParentId="24552" CreationDate="2011-03-02T08:25:12.363" Score="0" ViewCount="" Body="&lt;p&gt;Seems like you'd rather put the question here than have me a feast haha.... Expand the formula given above, then this problem is simply an integer inequality. A little trick is required :-). For N=4k this would be a bit easier.&lt;/p&gt;&#xA;" OwnerUserId="7684" LastActivityDate="2011-03-02T08:25:12.363" CommentCount="1" />
  <row Id="24601" PostTypeId="1" CreationDate="2011-03-02T08:35:14.147" Score="1" ViewCount="102" Body="&lt;p&gt;Hi, Everyone:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am trying to find all 1- or 2- dimensional Lie Algebras &quot;a&quot; up to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;isomorphism. This is what I have so far:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If a is 1-dimensional, then every vector (and therefore every tangent&lt;/p&gt;&#xA;&#xA;&lt;p&gt;vector field) is of the form cX. Then , by antisimmetry, and bilinearity:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[X,cX]=c[X,X]= -c[X,X]==0 . I think this forces a unique Lie algebra&lt;/p&gt;&#xA;&#xA;&lt;p&gt;because Lie algebra isomorphisms preserve the bracket .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also know Reals R are the only 1-dimensional Lie group, so its Lie algebra&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(R also) is also 1-dimensional. How can I show that every other 1-dimensional&lt;/p&gt;&#xA;&#xA;&lt;p&gt;algebra is isomorphic to this one? Do I use preservation of bracket?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For 2 dimensions, I am trying to use the fact that the dimension of&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the Lie algebra g of a group G is the same as the dimension of the ambient group/manifold&lt;/p&gt;&#xA;&#xA;&lt;p&gt;G. I know that all surfaces (i.e., groups of dimension 2) can be&lt;/p&gt;&#xA;&#xA;&lt;p&gt;classified as products of spheres and Tori, and I think the only 2-dimensional&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lie group is S&lt;sup&gt;1&lt;/sup&gt;xS&lt;sup&gt;1&lt;/sup&gt;, but I am not sure every Lie algebra&lt;/p&gt;&#xA;&#xA;&lt;p&gt;can be realized as the Lie algebra of a Lie group ( I think this is true in the&lt;/p&gt;&#xA;&#xA;&lt;p&gt;finite-dimensional case, but I am not sure).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know there is a result out there that I cannot yet prove that all 1- and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2-dimensional Lie algebras are isomorphic to Lie subalgebras of gl(2,R)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(using matrix multiplication, of course); would someone suggest how to show this&lt;/p&gt;&#xA;&#xA;&lt;p&gt;last? Thanks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any ideas?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="7687" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2011-03-02T11:16:39.770" LastActivityDate="2011-03-02T11:16:39.770" Title="Classsifying 1- and 2- dimensional Algebras, up to Isomorphism" Tags="&lt;differential-geometry&gt;&lt;lie-algebra&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="24604" PostTypeId="2" ParentId="24507" CreationDate="2011-03-02T08:45:38.640" Score="1" ViewCount="" Body="&lt;p&gt;Maybe this answer is just stating the obvious. If so I apologize in advance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Often when one proves things in mathematics, one uses a proof by contradiction. This means that we assume the opposite of what we want to prove, make some deductions, and eventually deduce something absurd like $0=1$ or &quot;$x\in X$ and $x \notin X$.&quot; The falsehood of this last statement then implies the truth of the original statement that we wanted to prove.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But you see now how this kind of logic breaks down if there is an inconsistency embedded in our everyday mathematics? Using proof by contradiction you could then &quot;prove&quot; any statement at all in the following way: &lt;em&gt;&quot;Assume by way of contradiction that A has a non-real eigenvalue. Let S be the set of all sets which do not contain themselves. [...] As this is a contradiction, A has only real eigenvalues.&quot;&lt;/em&gt; Even if it was in this case obvious that the contradiction had nothing to do with &lt;em&gt;A&lt;/em&gt; and everything to do with the tacked-on Russell's paradox, this might not be obvious in a more complicated situation.&lt;/p&gt;&#xA;" OwnerUserId="677" LastActivityDate="2011-03-02T08:45:38.640" />
  <row Id="24605" PostTypeId="2" ParentId="24595" CreationDate="2011-03-02T08:49:27.970" Score="0" ViewCount="" Body="&lt;p&gt;For given $\lambda_i$, you have $n$ non-zero, linearly independent functions and you form a linear combination of them with arbitrary complex coefficients $\alpha_i$ -- that means you can pick $n-1$ points in general position and make them zeros of $P$ by a suitable choice of the coefficients (e.g. by picking one of the functions and mimicking its function values at the $n-1$ chosen points by a suitable linear combintaion of the other $n-1$ functions). Conversely, since you can bring all the fractions to a common denominator and the numerator will then be of degree $n-1$, the function has exactly $n-1$ zeros in the general case. So generally you have $n-1$ zeros that can be almost anywhere you like (except for the poles, and sets of measure zero where the function values are linearly dependent), and the only thing that's left to characterize is what sort of special cases you can get.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I implicitly assumed that the $\lambda_i$ are different and the $\alpha_i$ are non-zero -- if either of these assumptions isn't fulfilled, you can reduce to a case with lower $n$ where they are.)&lt;/p&gt;&#xA;" OwnerUserId="6622" LastEditorUserId="6622" LastEditorDisplayName="" LastEditDate="2011-03-02T08:56:23.830" LastActivityDate="2011-03-02T08:56:23.830" CommentCount="3" />
  <row Id="24606" PostTypeId="2" ParentId="24576" CreationDate="2011-03-02T08:54:28.163" Score="0" ViewCount="" Body="&lt;p&gt;Try Chapter 11 of Apostol's &lt;em&gt;Mathematical Analysis, 2nd Ed&lt;/em&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Theorem 11.12 is the particular result you are after.&lt;/p&gt;&#xA;" OwnerUserId="6887" LastActivityDate="2011-03-02T08:54:28.163" CommentCount="1" />
  <row Id="24608" PostTypeId="2" ParentId="24504" CreationDate="2011-03-02T08:58:39.167" Score="1" ViewCount="" Body="&lt;p&gt;If you have a cyclotomic field $K = \mathbb{Q}(\zeta)$ where $\zeta := e^{2 \pi i /p}$ with $p$ an odd prime number, then it can be easily seen that its $p-1$ embeddings in $\mathbb{C}$ are given by $\sigma _i : K \hookrightarrow \mathbb{C}$ with $\sigma _i (\zeta) = \zeta ^{i}$ for $1 \leq i \leq p-1$. This is just a consequence of the fact that the minimum polynomial for $\zeta$ over $\mathbb{Q}$ is $$f(t) = \frac{t^p - 1}{t - 1} = t^{p-1} + t^{p-2} + \cdots + t + 1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and $f(t)$ factors as $$f(t) = (t - \zeta)(t - \zeta ^{2}) \cdots (t - \zeta ^{p-1})$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in general the image of $\zeta$ under the non-trivial embeddings of $K$ will be the powers $\zeta \mapsto \zeta^{i}$ for $2 \leq i \leq p-1$, which in your case for $p = 3$ will only correspond to $\zeta_{3} \mapsto \zeta_{3}^{2}$ but in this case $\zeta_{3}^{2} = \zeta_{3}^{-1}$ as in Mariano's answer. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And about the question you posed in your comment to Mariano's answer, yes, if $a + b \zeta_{3} \in \mathbb{Z}[\zeta_{3}]$ then $\sigma_{2}(a + b \zeta_{3}) = a + b \zeta_{3}^{2} = a + b \zeta_{3}^{-1}$.&lt;/p&gt;&#xA;" OwnerUserId="900" LastActivityDate="2011-03-02T08:58:39.167" />
  <row Id="24609" PostTypeId="1" AcceptedAnswerId="24610" CreationDate="2011-03-02T09:07:27.617" Score="0" ViewCount="94" Body="&lt;p&gt;I've been looking in vain (most books I came across give identities involving sums or recurrence relations, but do not give much attention to fixed values) for a reference to the following identity:&#xA;$$S(n,n-3)={n \choose 2}{n \choose 4},$$&#xA;where $S(n,k)$ is the unsigned &lt;a href=&quot;http://en.wikipedia.org/wiki/Stirling_numbers_of_the_first_kind&quot; rel=&quot;nofollow&quot;&gt;Stirling number of the first kind&lt;/a&gt;. Is this well-known, or too trivial to be mentioned anywhere?&lt;/p&gt;&#xA;" OwnerUserId="689" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2011-05-28T03:14:19.620" LastActivityDate="2011-05-28T03:14:19.620" Title="Is this identity involving Stirling numbers of the first kind well-known?" Tags="&lt;combinatorics&gt;&lt;reference-request&gt;&lt;stirling-numbers&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" />
  <row Id="24610" PostTypeId="2" ParentId="24609" CreationDate="2011-03-02T09:11:14.530" Score="4" ViewCount="" Body="&lt;p&gt;The identity and the proof for the identity are there in the &lt;a href=&quot;http://en.wikipedia.org/wiki/Stirling_numbers_of_the_first_kind#Simple_identities&quot; rel=&quot;nofollow&quot;&gt;wiki&lt;/a&gt; link you have sent.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastActivityDate="2011-03-02T09:11:14.530" CommentCount="2" />
  <row Id="24611" PostTypeId="2" ParentId="24588" CreationDate="2011-03-02T09:11:31.890" Score="6" ViewCount="" Body="&lt;p&gt;Consider the intersection of the curve with the ray $\theta = 0$. These are points at a distance $r = \cos(2\pi n K)$ from the origin, for all $n \in \mathbb Z$. Note that there are only countably many of them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The curve gets &lt;em&gt;arbitrarily close&lt;/em&gt; to all points in the unit disk, but does not &lt;em&gt;pass through&lt;/em&gt; all of them.&lt;/p&gt;&#xA;" OwnerUserId="856" LastActivityDate="2011-03-02T09:11:31.890" CommentCount="5" />
  <row Id="24612" PostTypeId="2" ParentId="24406" CreationDate="2011-03-02T09:20:53.683" Score="2" ViewCount="" Body="&lt;p&gt;Example (O.Zariski?):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the rational function field $k(x)$ over a field $k$ of characteristic $p&amp;gt;0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The field extension $k((x))/k(x)$ is transcendental; let $\alpha\in k[[x]]$ be transcendental &#xA; and let $y:=\alpha^p$. Then $L:=k(x,\alpha )$ is a purely inseparable&#xA;extension of $K:=k(x,y)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The discrete valuation ring $B:=k[[x]]\cap L$ is the integral&#xA;closure of the discrete valuation ring $A:=k[[x]]\cap K$ and $x$ is a prime element of $A$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For every $n\in\mathbb{N}$ the element $y$ can be written as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$&#xD;&#xA;y=f_n^p + x^{pn}y_n , f_n\in k[x], y_n\in k[[x]].&#xD;&#xA;$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(To see this consider the power series: $\alpha =c_0+c_1x+c_2x^2+c_3x^3+\ldots$ hence&#xA;$y =c_0^p+c_1^px^p+c_2^px^{2p}+c_3^px^{3p}+\ldots$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then $y_n\in A$ and one gets:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$&#xD;&#xA;y_n^{1/p} =x^{-n}(-f_n+\alpha )\;\; (*).&#xD;&#xA;$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now if $B$ were a finite $A$-module one could find $d\in A\setminus 0$ such that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$&#xD;&#xA;dB\subseteq A+A\alpha   +A\alpha^2 +\ldots +A\alpha^{p-1}=:R.&#xD;&#xA;$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular $d y_n^{1/p}\in R$, which using (*) yields that $d$ is divisible by $x^n$&#xA;for every $n$. (Note that $1, \alpha ,\ldots ,\alpha^{p-1}$ are linearly independent &#xA;over $K$.)&lt;/p&gt;&#xA;" OwnerUserId="2479" LastActivityDate="2011-03-02T09:20:53.683" CommentCount="1" />
  <row Id="24613" PostTypeId="2" ParentId="24596" CreationDate="2011-03-02T09:37:07.960" Score="3" ViewCount="" Body="&lt;p&gt;The second one, as Sivaram indicated, is straightforward. First, however, note that it is not defined for $\mu = 0$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume that $\mu &amp;gt; 0$ (the case $\mu &amp;lt; 0$ being analogous). Then, for any $x \geq 0$,$$&#xA;{\rm P}\bigg[\frac{{\sqrt {n^{ - 1} \sum\nolimits_{i = 1}^n {(X_i  - \mu )^2 } } }}{\mu } \le x \bigg] = {\rm P}\bigg[\sum\limits_{i = 1}^n {\bigg(\frac{{X_i  - \mu }}{\sigma }\bigg)^2 }  \le \frac{{n\mu ^2 x^2 }}{{\sigma ^2 }}\bigg].&#xA;$$&#xA;Now, $\frac{{X_i  - \mu }}{\sigma }$ are i.i.d. ${\rm N}(0,1)$, hence their sum of squares is chi-squared with $n$ degrees of freedom. So just substitute $\frac{{n\mu ^2 x^2 }}{{\sigma ^2 }}$ in the cdf of the latter (which is given by $&#xA;\frac{1}{{\Gamma (n/2)}}\gamma (n/2,x/2)$).&lt;/p&gt;&#xA;" OwnerUserId="2810" LastEditorUserId="2810" LastEditorDisplayName="" LastEditDate="2011-03-02T12:38:51.090" LastActivityDate="2011-03-02T12:38:51.090" CommentCount="2" />
  <row Id="24614" PostTypeId="1" CreationDate="2011-03-02T09:47:05.090" Score="0" ViewCount="66" Body="&lt;p&gt;Hello,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am looking for a clustering algorithm similar to DBSCAN but with a limitation on the intra-cluster distance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i.e. the input will be a set of points, a minimal distance, and a the minimal number of points in a cluster. The output will contain clusters of points such that the distance between each pair of points in a cluster will be smaller than the minimal distance given as an input parameter (as opposed to what happens in DBSCAN).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope this is understood. Any help would be appreciated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Katie&lt;/p&gt;&#xA;" OwnerUserId="7689" LastActivityDate="2011-03-02T09:47:05.090" Title="Clustering algorithm with intra-cluster distance limitation" Tags="&lt;algorithms&gt;&lt;machine-learning&gt;&lt;clustering&gt;" CommentCount="2" />
  <row Id="24616" PostTypeId="2" ParentId="1196" CreationDate="2011-03-02T10:46:51.317" Score="1" ViewCount="" Body="&lt;p&gt;$$&#xD;&#xA;\begin{align}&#xD;&#xA;n^3+2n &amp;amp;= n(n^2+2) \\&#xD;&#xA;       &amp;amp;= n(n^2−1+3)\\&#xD;&#xA;       &amp;amp;= n((n^2−1^2)+3)\\&#xD;&#xA;       &amp;amp;= n((n−1)(n+1)+3)\\&#xD;&#xA;       &amp;amp;= n(n−1)(n+1)+3n&#xD;&#xA;\end{align}&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The key is to identify the factor $(n^2-1^2)=(n-1)(n+1)$. Looking at the result you can tell that it must be divisable by 3 because one of $n$, $(n-1)$ or $(n+1)$ is a multiple of 3. &lt;a href=&quot;http://math.stackexchange.com/questions/1196/proof-that-n32n-is-divisible-by-3/1200#1200&quot;&gt;Ben Albert&lt;/a&gt; gives a similar solution but I believe $3n^2$ should be $3n$ instead.&lt;/p&gt;&#xA;" OwnerUserId="6678" LastEditorUserId="6678" LastEditorDisplayName="" LastEditDate="2011-03-04T20:33:13.167" LastActivityDate="2011-03-04T20:33:13.167" />
  <row Id="24617" PostTypeId="2" ParentId="24533" CreationDate="2011-03-02T10:49:46.923" Score="8" ViewCount="" Body="&lt;p&gt;I suspect Michael Lugo's answer was the intended one, but for what it's worth, Andres Caicedo's combinatorial answer has a combinatorial proof. The Riemann sum &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{1}{n} \sum_{k=0}^{n-1} \cos^{100} \frac{2 \pi k}{n}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;counts the probability that you return to where you started in a random walk of length $100$ on $\mathbb{Z}/n\mathbb{Z}$ (where adjacent residues are connected by an edge); this is because the adjacency matrix is $P + P^{-1}$ where $P$ is a permutation matrix describing an $n$-cycle, so the eigenvalues of the adjacency matrix are $e^{ \frac{2 \pi i k}{n} } + e^{- \frac{2 \pi i k}{n} } = 2 \cos \frac{2 \pi k}{n}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $n &amp;gt; 100$ this probability is clearly just $\frac{1}{2^{100}} {100 \choose 50}$, and taking the limit as $n \to \infty$ we obtain our result. And combinatorialists know that ${2n \choose n} \approx \frac{4^n}{\sqrt{\pi n}}$ by one of various methods (Stirling's formula, singularity analysis, the central limit theorem...). &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Actually (I had forgotten that I knew this) one can get the integral directly without looking at a Riemann sum. The key is that $\mathbb{Z}$ (the graph where adjacent integers are connected by an edge) is the &lt;a href=&quot;http://qchu.wordpress.com/2010/03/07/walks-on-graphs-and-tensor-products/&quot; rel=&quot;nofollow&quot;&gt;representation graph&lt;/a&gt; of $\text{SO}(2)$ acting on its standard representation $V$, so &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_0^{2\pi} (2 \cos x)^{100} \, dx$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is precisely the multiplicity of the trivial representation in $V^{\otimes 100}$, which is precisely the number of walks of length $100$ from the origin to itself on $\mathbb{Z}$. &lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2011-03-02T10:58:21.910" LastActivityDate="2011-03-02T10:58:21.910" CommentCount="1" />
  <row Id="24618" PostTypeId="2" ParentId="24601" CreationDate="2011-03-02T10:58:14.193" Score="1" ViewCount="" Body="&lt;p&gt;Although you've tagged your question &quot;differential-geometry&quot;, it's actually a pure algebra question.  While you're right that for Lie algebras over (e.g.) $\mathbb{R}$ there is a deep relationship with Lie groups which motivates their study and can be helpful for proving &quot;purely algebraic theorems&quot;, it seems to me that juggling between Lie groups and Lie algebras is distracting you from the issues at hand.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One more comment on your Lie-theoretic approach: there is indeed a bijection between connected, simply connected real Lie groups and finite-dimensional real Lie algebras, but the &lt;em&gt;group structure&lt;/em&gt; on the Lie group side cannot be ignored.  It is not enough just to classify manifolds which admit a Lie group structure, since the same manifold may admit a Lie group structure in multiple different ways.  An especially relevant example is that any nilpotent Lie group -- i.e., a Lie group with associated Lie algebra a nilpotent Lie algebra -- is as a manifold isomorphic to $\mathbb{R}^n$, but the group law need not be commutative.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Coming back to the classification of Lie algebras of small dimension:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have actually already done the one-dimensional case, as you have observed that the Lie bracket on any one-dimensional Lie algebra must be trivial.  Thus any two one-dimensional Lie algebras are isomorphic: any vector space isomorphism will do.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In two dimensions there is again the Lie algebra $L_1$ with trivial bracket, but there is also a noncommutative Lie algebra $L_2$.  Concretely, if we take a basis of $x,y$ of $\mathbb{R}^2$ and define $[x,x] = [y,y] = 0$ and $[x,y] = -[y,x] = y$ then this works to give a Lie algebra.  (Check this!)  Now the same construction can be done in many other ways, but they are all isomorphic to this one: start by assuming that &#xA;$[x,y] = ax + by$ with $a$ and $b$ not both zero and then find a new basis $X$, $Y$ &#xA;under which the bracket is again $[X,Y] = Y$.  Thus there are exactly two Lie algebras of dimension $2$ over $\mathbb{R}$.  In both cases, the corresponding Lie groups are isomorphic as manifolds to $\mathbb{R}^2$ (there are many ways to see this, magic words being &lt;strong&gt;exponential map&lt;/strong&gt; and &lt;strong&gt;Baker-Campbell-Hausdorff formula&lt;/strong&gt;; you'll probably learn them later on), but one of the group structures is the usual one on $\mathbb{R}^2$ and the other is a non-commutative group structure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note by the way that the situation is much different starting in dimension three: &#xA;there are then infinitely many isomorphism classes of Lie algebras, and indeed continuous families of Lie algebras.  See for instance Section 4 of &lt;a href=&quot;http://www.emis.de/journals/EM/expmath/volumes/14/14.1/deGraaf.pdf&quot; rel=&quot;nofollow&quot;&gt;this paper&lt;/a&gt;&#xA;which constructs, over an arbitrary field $F$, for each $a \in F$ a Lie algebra $L_a^3$ such that for $a,b \in F$, $L_a^3 \cong L_b^3 \iff a = b$.  Thus Lie algebras &quot;vary in moduli&quot; starting in dimension three.&lt;/p&gt;&#xA;" OwnerUserId="299" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2011-03-02T11:14:08.727" LastActivityDate="2011-03-02T11:14:08.727" />
  <row Id="24619" PostTypeId="1" AcceptedAnswerId="24621" CreationDate="2011-03-02T11:03:49.587" Score="0" ViewCount="59" Body="&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/Thkg3.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The curved line is a part of a circle with a center at the lower right corner.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What would be the name of the shaded region?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="7693" LastActivityDate="2011-03-02T11:09:48.843" Title="What's the name of a part of a circle that's formed by an arbitrary intersecting line?" Tags="&lt;geometry&gt;&lt;circle&gt;" AnswerCount="1" />
  <row Id="24620" PostTypeId="1" CreationDate="2011-03-02T11:05:05.810" Score="0" ViewCount="24" Body="&lt;p&gt;I've got 64 images of 96x192 pixels. I have to arrange them on a rectangle. I need the height and width of that rectangle to be powers of two. Given the dimensions of my images, they don't perfectly 'fit' on it (I can not rotate or cut the images into smaller ones); there is always some 'wasted space'.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;¿What are the dimensions of the rectangle that minimizes that wasted space?&lt;/li&gt;&#xA;&lt;li&gt;¿Can this problem be generalized to &lt;code&gt;n&lt;/code&gt; images of &lt;code&gt;w&lt;/code&gt; x &lt;code&gt;h&lt;/code&gt; pixels?&lt;/li&gt;&#xA;&lt;li&gt;¿Does this problem have a name?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Context: This is not a class exercise. I'm trying to arrange sprites into a image for using them on hobby videogames. The requirement of power-of-two comes from the fact that some old video-cards don't handle non-po2 images very well. I've been thinking about this for 24 hours and couldn't arrive to any satisfying solution.&lt;/p&gt;&#xA;" OwnerUserId="3893" LastEditorUserId="3893" LastEditorDisplayName="" LastEditDate="2011-03-02T11:46:10.190" LastActivityDate="2011-03-02T12:36:17.007" Title="minimal po2 surface for sprite arrangement" Tags="&lt;calculus&gt;&lt;discrete-mathematics&gt;&lt;minimal-surfaces&gt;&lt;packing-problem&gt;" AnswerCount="1" />
  <row Id="24621" PostTypeId="2" ParentId="24619" CreationDate="2011-03-02T11:09:48.843" Score="3" ViewCount="" Body="&lt;p&gt;It's called a &lt;a href=&quot;http://en.wikipedia.org/wiki/Circular_segment&quot; rel=&quot;nofollow&quot;&gt;circular segment&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="589" LastActivityDate="2011-03-02T11:09:48.843" />
  <row Id="24622" PostTypeId="2" ParentId="24521" CreationDate="2011-03-02T11:27:30.517" Score="1" ViewCount="" Body="&lt;p&gt;This formula is derived in many different ways in the book &lt;a href=&quot;http://en.wikipedia.org/wiki/Concrete_Mathematics&quot; rel=&quot;nofollow&quot;&gt;Concrete Mathematics&lt;/a&gt;. For example, one method which haven't been mentioned yet here is to use &quot;discrete calculus&quot;, which easily lets you compute sums of &quot;falling factorial powers&quot; $k^{\underline{m}} = k(k-1)(k-2) \dots (k-m+1)$ almost as when you integrate ordinary powers $x^m$. Upon writing $k^2 = k(k-1)+k = k^{\underline{2}} + k^{\underline{1}}$, your sum becomes&#xA;$$ \sum_{0 \le k &amp;lt; n+1} (k^{\underline{2}} + k^{\underline{1}})&#xD;&#xA;= \left[ \frac{k^{\underline{3}}}{3} + \frac{k^{\underline{2}}}{2} \right]_{k=0}^{n+1} = \frac{(n+1)n(n-1)}{3} + \frac{(n+1)n}{2}&#xD;&#xA;= \frac{(n+1)n(2n+1)}{6},&#xD;&#xA;$$&#xA;which is your polynomial $f(n)$. (See the book for explanation of why it this works.)&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2011-03-02T11:27:30.517" CommentCount="2" />
  <row Id="24623" PostTypeId="2" ParentId="24504" CreationDate="2011-03-02T11:28:28.967" Score="3" ViewCount="" Body="&lt;p&gt;If you define an algebraic extension of $\Bbb Q$ as $K={\Bbb Q}[x]/(f(x))$ where $f(x)$ is an irreducible polynomial of degree $d$, you obtain the $d$ embeddings of $K$ by sending $x\mapsto\alpha$ where $\alpha$ is a complex root of $f(x)$ (they always exist, thanks to Gauss).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of ${\Bbb Q}(\zeta_3)$, this is just ${\Bbb Q}[x]/(x^2-x+1)$ because $x^2-x+1$ is the quadratic irreducible factor of $x^3-1$. Thus you obtain two complex embeddings sending&#xA;$$&#xD;&#xA;x=\zeta_3\mapsto\alpha^\pm=\frac12(1\pm\sqrt{-3}).&#xD;&#xA;$$&#xA;If you decide to identify $\zeta_3=\alpha^{+}=(1+\sqrt{-3})/2$, a moment of thought (in the form of a short computation) will convince you that the &quot;other&quot; root, i.e. the other embedding, $\alpha^{-}$ is just $\zeta_3^2$. Mind that the two roots (and so the two embeddings) are switched by complex conjugation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The identities $\zeta_3^2=\overline{\zeta}_3=\zeta_3^{-1}$ are also clear observing that $1$, $\alpha^{+}$ and $\alpha^{-}$ are the three complex roots of $x^3-1$, so that their product must be $1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Adrian's answer gives the generalization to the $p$-th cyclotomic field&lt;/p&gt;&#xA;" OwnerUserId="688" LastActivityDate="2011-03-02T11:28:28.967" />
  <row Id="24624" PostTypeId="1" AcceptedAnswerId="24633" CreationDate="2011-03-02T11:28:41.930" Score="0" ViewCount="86" Body="&lt;p&gt;I'm confused about the different types of homology groups. For instance, on Wikipedia (&lt;a href=&quot;http://en.wikipedia.org/wiki/Orientability&quot; rel=&quot;nofollow&quot;&gt;Orientability&lt;/a&gt;) they write $H_n (M, \partial M; \mathbb{Z} )$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, where they explain the construction of homology groups, they write $H_n (X) := ker(\partial_n) / im(\partial_{n + 1})$. (&lt;a href=&quot;http://en.wikipedia.org/wiki/Homology_%28mathematics%29&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are these two the same?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason why I'm interested in this question is because I'm interested in an alternative definition of orientability, namely, that a compact $n$-manifold is orientable if and only if its top homology group is isomorphic to $\mathbb{Z}$. In fact, I'm only interested in surfaces, i.e. $n= 2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Seeing as every surface can be triangulated, I assume I can restrict myself to looking at simplicial &lt;a href=&quot;http://en.wikipedia.org/wiki/Simplicial_homology&quot; rel=&quot;nofollow&quot;&gt;homology groups&lt;/a&gt;. Are there any further restrictions that I can make to make it easier?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for any help.&lt;/p&gt;&#xA;" OwnerUserId="5798" LastActivityDate="2011-03-02T12:34:58.400" Title="A question regarding the notation/definition of homology group" Tags="&lt;algebraic-topology&gt;&lt;homology&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="24625" PostTypeId="1" CreationDate="2011-03-02T11:32:53.803" Score="0" ViewCount="61" Body="&lt;p&gt;We have to choose 4 types of soup from a supermarket's 10 varieties, such that at least 2 are identical. In how many ways can we do this ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have 10 varieties of soup like so {v1, .. v10}. We can partition our choices into those with 4 identical (e.g. {v1, v1, v1, v1}), those with 3 identical (e.g. {v1, v1, v1, v2}), and those with 2 identical (e.g. {v1, v1, v2, v3}). Our total number of choices is then:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;10 + 10.9 + 10.9C2 = 10 + 90 + 360 = 460&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The textbook answer is 505. Can anyone point out my error ?&lt;/p&gt;&#xA;" OwnerUserId="7597" LastActivityDate="2011-03-02T12:20:07.243" Title="Choose 4 types of soup from 10 varieties, with constraint ?" Tags="&lt;combinatorics&gt;" AnswerCount="1" CommentCount="0" />
  <row Id="24626" PostTypeId="2" ParentId="24620" CreationDate="2011-03-02T11:35:29.527" Score="1" ViewCount="" Body="&lt;p&gt;You need at least $96 \times 192 \times 64 = 1179648$ pixels.  This is slightly more than $2^{20} = 1048576$ but fewer than $2^{21} = 2097152$.  So you just need to check that you can fit them into $2^{21}$.  One of many possibilities would be a rectangle $2^{9} \times 2^{12} = 512 \times 4096$, you could have 4 images across (since $ 4 \times 96 = 384$ ) and 16 images down (since $16 \times 192 = 3072$ ) and so the 64 images would fit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a &lt;a href=&quot;http://en.wikipedia.org/wiki/Packing_problem&quot; rel=&quot;nofollow&quot;&gt;packing problem&lt;/a&gt;  &lt;/p&gt;&#xA;" OwnerUserId="6460" LastEditorUserId="6460" LastEditorDisplayName="" LastEditDate="2011-03-02T12:36:17.007" LastActivityDate="2011-03-02T12:36:17.007" CommentCount="6" />
  <row Id="24627" PostTypeId="1" AcceptedAnswerId="24640" CreationDate="2011-03-02T11:44:04.397" Score="2" ViewCount="50" Body="&lt;p&gt;Say I have three constants a, b and c (all &gt; 0) and three variables x, y and z (all &gt;= 0).  I want to find values for x, y, z which maximise the lowest of the following:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;ax - y - z&#xA;-x + by - z&#xA;-x -y + cz&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;How can I find these values?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;: just corrected expressions -- variables and constants were wrong way around&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;: x, y and z all must be &gt;=0&lt;/p&gt;&#xA;" OwnerUserId="7694" LastEditorUserId="7694" LastEditorDisplayName="" LastEditDate="2011-03-02T13:52:16.080" LastActivityDate="2011-03-02T13:52:16.080" Title="How do find the optimal values for these expressions?" Tags="&lt;optimization&gt;" AnswerCount="1" FavoriteCount="2" />
  <row Id="24628" PostTypeId="1" AcceptedAnswerId="24645" CreationDate="2011-03-02T11:52:28.040" Score="1" ViewCount="147" Body="&lt;p&gt;Hi! I am taking a course in linear algebra at the moment, and the book I have uses $1\times n$ matrices, $n\times 1$ matrices and $n$-tuples to represent vectors. In condition I have been taught that $1\times n$ and $n\times 1$ matrices &lt;em&gt;are&lt;/em&gt; vectors. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What, then, is the difference between an $n$-tuple and a $1\times n$ matrix? What do we need tuples for, that we can't use matrices for?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I see that the product between tuples are defined in another way than products between matrices. The product between two $1\times n$ matrices isn't even defined (if $n\neq 1$). But this we can easily solve with trasposing one of them. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope you can help to clarify this for me.&lt;/p&gt;&#xA;" OwnerUserId="7044" LastActivityDate="2011-03-02T15:01:26.240" Title="Relationship between tuples, vectors and column/row matrices." Tags="&lt;linear-algebra&gt;&lt;definition&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="24629" PostTypeId="1" AcceptedAnswerId="24657" CreationDate="2011-03-02T12:14:52.990" Score="0" ViewCount="167" Body="&lt;p&gt;Does there exist a reduced quadratic equation with solutions equal to both the $\sqrt{2}$ and $\sqrt{3}$?&lt;/p&gt;&#xA;" OwnerUserId="7695" LastEditorUserId="6296" LastEditorDisplayName="" LastEditDate="2011-03-03T12:48:52.250" LastActivityDate="2011-03-03T12:48:52.250" Title="What reduced quadratic equation has solutions $\sqrt{2}$ and $\sqrt{3}$" Tags="&lt;algebra&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="1" />
  <row Id="24630" PostTypeId="2" ParentId="24507" CreationDate="2011-03-02T12:18:30.983" Score="1" ViewCount="" Body="&lt;p&gt;I cannot say it for russels paradox itself, but for computer science analogons.&#xA;There are Questions, which can fundamentally not be answered. Not because of too few knowledge or power, but because of logical &quot;reason&quot;. With Undecidability, this seems to be &quot;obvious&quot; today. But:&#xA;Back then, the mathematicans thought, they just has to provide the right set, that they could solve every problem in the world. That there would be no theoretically unsolvable problems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a so huge fact, that it takes some time, to really acknowledge it without any trivialization..&lt;/p&gt;&#xA;" OwnerUserId="7696" LastActivityDate="2011-03-02T12:18:30.983" />
  <row Id="24631" PostTypeId="2" ParentId="24625" CreationDate="2011-03-02T12:20:07.243" Score="3" ViewCount="" Body="&lt;p&gt;You're missing a case!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a hint: you're 45 below the right answer, 45 = 10C2. What choice can you make that requires you to pick two varieties where the order is not important?&lt;/p&gt;&#xA;" OwnerUserId="360" LastActivityDate="2011-03-02T12:20:07.243" CommentCount="2" />
  <row Id="24632" PostTypeId="2" ParentId="24507" CreationDate="2011-03-02T12:29:16.653" Score="1" ViewCount="" Body="&lt;p&gt;I think the key point in Russell's paradox is the self-reference. It was important as well in subsequent developments like Gödel's incompleteness theorems. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The possibility of self-reference is what in a way opens up a theory. Something that makes it richer. But also something that from an axiomatic point of view allows for the possibility of paradoxes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Douglas_Hofstadter&quot; rel=&quot;nofollow&quot;&gt;Douglas Hofstadter&lt;/a&gt; has written a load of popular books on the subject.&lt;/p&gt;&#xA;" OwnerUserId="3567" LastActivityDate="2011-03-02T12:29:16.653" />
  <row Id="24633" PostTypeId="2" ParentId="24624" CreationDate="2011-03-02T12:34:58.400" Score="2" ViewCount="" Body="&lt;p&gt;If $S$ is your surface, you should consider $H_2(S,\partial S;Z)$, which means the second homology relative to the boundary, with $Z$ coefficients.  The notation is used for many different kinds of homology, but you are right, it is fine to use the simplicial version.  The default coefficients are (often) $Z$ so sometimes they are left out of the notation.  Thus if the surface is closed, you may just see $H_2(S)$ and if the surface has boundary you may see $H_2(S,\partial S)$ (without the $Z$'s).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The coefficients, while sometimes omitted, are important.  For instance if you use $Z/2Z$ coefficients then the second homology of a closed surface will be $Z/2Z$, whether it is orientable or not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, with $Z$ coefficients, if $S$ has no boundary then $H_2(S)$ will be $Z$ if $S$ is orientable and $0$ if $S$ is non-orientable.  If $S$ has boundary then $H_2(S,\partial S)$ will be $Z$ if $S$ is orientable and $0$ if $S$ is non-orientable.&lt;/p&gt;&#xA;" OwnerUserId="6071" LastActivityDate="2011-03-02T12:34:58.400" />
  <row Id="24634" PostTypeId="2" ParentId="24629" CreationDate="2011-03-02T12:46:49.727" Score="4" ViewCount="" Body="&lt;p&gt;A quadratic equation with roots $a$ and $b$ is of the form $c(x-a)(x-b)$. I assume that &quot;reduced&quot; here means that you take $c = 1$.&lt;/p&gt;&#xA;" OwnerUserId="5676" LastActivityDate="2011-03-02T12:46:49.727" CommentCount="4" />
  <row Id="24635" PostTypeId="2" ParentId="24577" CreationDate="2011-03-02T12:49:02.660" Score="2" ViewCount="" Body="&lt;p&gt;Let me elaborate on user3148's answer and comment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are two facts:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;A weak Cauchy sequence $(f_{n})$ is bounded.&lt;/li&gt;&#xA;&lt;li&gt;Every bounded sequence has a weakly convergent subsequence.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Combining these two facts it is easy to see that every weak Cauchy sequence converges. Recall that a weak Cauchy sequence is a sequence $(f_{n})$ such that $\langle f_{n}, g\rangle$ is Cauchy in $\mathbb{R}$ for all $g$. The condition you impose on the sequence $(f_{n})$ means in particular that it is a weak Cauchy sequence, so it necessarily converges to some $f \in L^2$.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Proof of 1. This follows immediately from the Banach-Steinhaus theorem applied to the operators $\langle f_{n}, \cdot \rangle: X^{\ast} \to \mathbb{R}$, see &lt;a href=&quot;http://arxiv.org/abs/1005.1585&quot; rel=&quot;nofollow&quot;&gt;Sokal's recent paper&lt;/a&gt; for a neat proof of that theorem (without Baire!).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proof of 2. This is immediate from the version of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Banach%2DAlaoglu_theorem&quot; rel=&quot;nofollow&quot;&gt;Banach-Alao&amp;#287;lu&lt;/a&gt; theorem saying that the unit ball in a separable &lt;a href=&quot;http://en.wikipedia.org/wiki/Reflexive_space&quot; rel=&quot;nofollow&quot;&gt;reflexive space&lt;/a&gt; is compact metrizable in the weak topology (= weak$^{\ast}$-topology by reflexivity).&lt;/p&gt;&#xA;" OwnerUserId="5363" LastEditorUserId="5363" LastEditorDisplayName="" LastEditDate="2011-03-02T14:27:13.407" LastActivityDate="2011-03-02T14:27:13.407" CommentCount="6" />
  <row Id="24636" PostTypeId="2" ParentId="24596" CreationDate="2011-03-02T12:57:11.697" Score="2" ViewCount="" Body="&lt;p&gt;For the first one, if you consider instead of $S/\bar{X}$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\bar{X}-\mu}{S/\sqrt{n}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;then, this is a &lt;a href=&quot;http://en.wikipedia.org/wiki/Student%27s_t-distribution&quot; rel=&quot;nofollow&quot;&gt;Student t-variable&lt;/a&gt; with $n-1$ degrees of freedom.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you don't want the $\mu$, you can use the &lt;a href=&quot;http://en.wikipedia.org/wiki/Noncentral_t-distribution&quot; rel=&quot;nofollow&quot;&gt;non-central t-distribution&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="3567" LastActivityDate="2011-03-02T12:57:11.697" />
  <row Id="24637" PostTypeId="1" CreationDate="2011-03-02T13:07:21.980" Score="2" ViewCount="132" Body="&lt;p&gt;Let $\alpha\in \mathbb{Z}_p$ be an $p$-adic integer and define for $n\in \mathbb{Z}_{\geq 0}$ &#xA;$${\alpha\choose n} := \frac{\alpha(\alpha-1)\cdot\ldots\cdot(\alpha-n+1)}{n!}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is again a $p$-adic integer, and we can define&#xA;$$\alpha_1:=\sum_{n=0}^\infty \overline{{\alpha \choose n}} p^n$$&#xA;and&#xA;$$\alpha_2:=\sum_{n=0}^\infty \overline{{\alpha \choose p^n}} p^n,$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\overline{(\cdot)}$ means reduction modulo $p$. &#xA;How are $\alpha,\alpha_1,\alpha_2$ related? Are there formulas expressing this relation?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks a lot!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;: Some thinking led me to the following conclusion: Using continuity of $x\mapsto {x\choose p^n}$ as a function $\mathbb{Z}_p\rightarrow \mathbb{Q}_p$ and &lt;a href=&quot;http://en.wikipedia.org/wiki/Lucas%27_theorem&quot; rel=&quot;nofollow&quot;&gt;Lucas' Theorem&lt;/a&gt; it follows that $\alpha=\alpha_2$. Does this seem correct?&lt;/p&gt;&#xA;" OwnerUserId="7698" LastEditorUserId="7698" LastEditorDisplayName="" LastEditDate="2011-03-03T13:06:39.160" LastActivityDate="2011-03-03T13:06:39.160" Title="p-adic numbers and binomial coefficients" Tags="&lt;number-theory&gt;&lt;binomial-coefficients&gt;&lt;p-adic-number-theory&gt;" CommentCount="3" />
  <row Id="24638" PostTypeId="1" AcceptedAnswerId="24643" CreationDate="2011-03-02T13:16:07.883" Score="3" ViewCount="127" Body="&lt;p&gt;Simplify $\sum\limits_{k = 0}^n {{2^k}\tan ({2^k}x)}$ which $k \in \{ 0,1,...,n + 1\} ,{2^k}x \notin \{ 0,\frac{\pi }{2}\}$&lt;/p&gt;&#xA;" OwnerUserId="7697" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2011-03-02T17:14:25.497" LastActivityDate="2011-03-02T17:14:25.497" Title="Simplifying $\sum 2^k \tan(2^k x)$" Tags="&lt;trigonometry&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" />
  <row Id="24639" PostTypeId="2" ParentId="1196" CreationDate="2011-03-02T13:18:45.690" Score="1" ViewCount="" Body="&lt;p&gt;$$n^3+2n=n(n^2+2)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $n$ is divisible by $3$, then obviously, so is $n^3+2n$ because you can factor out $n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $n$ is not divisible by $3$, it is sufficient to show that $n^2+2$ is divisible by 3. Now, if $n$ is not divisible by $3$, $n=3k+1$ or $n=3k+2$ for some integer $k$. Plug that into $n^2+2$ and you'll get $9k^2+6k+3$ and $9k^2+6k+6$ respectively. Both of which are divisible by 3.&lt;/p&gt;&#xA;" OwnerUserId="3567" LastActivityDate="2011-03-02T13:18:45.690" />
  <row Id="24640" PostTypeId="2" ParentId="24627" CreationDate="2011-03-02T13:27:26.337" Score="1" ViewCount="" Body="&lt;p&gt;If I understand, you are trying to solve a minimax (or rather, maximin) type problem.  Formally, your statement is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\max \min \{ax-y-z, -x+by-z, -x-y+cz\}$, and you have not listed any other constraints.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A standard trick is to reformulate this problem as a constrained maximization problem.  We can do this by introducing an auxiliary variable $\gamma$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\max_{\gamma, x,y,z} \gamma$ subject to $ \gamma \leq f_i(x,y,z)$ for $i=1,2,3$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $f_i(x,y,z)$ corresponds to each of the linear functions you stated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is nothing but a linear program now, which can be solved using your favorite method.&lt;/p&gt;&#xA;" OwnerUserId="7683" LastActivityDate="2011-03-02T13:27:26.337" CommentCount="4" />
  <row Id="24641" PostTypeId="2" ParentId="24507" CreationDate="2011-03-02T13:33:05.770" Score="3" ViewCount="" Body="&lt;p&gt;Much of what you say makes sense philosophically, and there are also ways to make it precise mathematically: For example, one result in &lt;a href=&quot;http://en.wikipedia.org/wiki/Reverse_mathematics&quot; rel=&quot;nofollow&quot;&gt;reverse mathematics&lt;/a&gt; is that mathematicians often need even less complex sets than they use.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I think you are actually asking several different questions.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Why did mathematicians take Russell's paradox seriously?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Did they? Those that were dealing with foundations certainly did, but I'd be surprised if that claim was true for the average mathematician.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The set of all sets admits all sorts of unmathematical objects right?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Well, that depends on how you define &quot;set.&quot; In naive set theory, the idea that a set can contain any given object is a common point of view; I think it appeals to some pre-existing linguistic intuition about the word &quot;set.&quot; But as you say:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I can't imagine a situation where a mathematician solving a concrete problem would invoke a set like that.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is correct for a lot of mathematics, but there also are a lot of areas where you can easily get into trouble. Cardinal and ordinal numbers are the first examples, historically speaking (BTW, the Burali-Forti paradox apparantly predates Russell's paradox). Category theory is a more complicated case. One isn't necessarily tempted to invoke a &quot;set of all sets,&quot; but similar things which are just as problematic (and arguably related).&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;But clearly important people think the result significant. It prompted Gottlob Frege, a smart and accomplished man, who Wikipedia tells me extended the study of logic in non-trivial ways, to say that the &quot;foundations of [his] edifice were shaken.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I think you are generalizing too much from this sentence. Russell first stated his paradox explicitly about Frege's system. This is because Frege was basically the first person who tried to reduce all of mathematics to a single formal system. And then it turned out that his system, which took years to work out, could in fact prove every statement and was thus useless, technically speaking. It's not hard to imagine he was a bit disappointed.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Is there a practical example in mathematics that showcases this, where one might be led horribly astray if one doesn't place certain restrictions on sets, in say, dealing with PDEs, elliptic curves, functional equations?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I would answer &quot;no&quot; because whenever set-theoretic difficulties come up, the problematic sets turn out to be &quot;tricky&quot; in exactly the way you describe. However, being &quot;tricky&quot; is not a particularly solid criterion, and one that different people might have different opinions about. So when in doubt, mathematicians prefer to delegate such questions to a specific formal system, usually ZFC. That is, the criterion that is actually used nowadays is: &quot;Can this set be constructed in ZFC?&quot; If it can't, that doesn't necessarily mean that a contraction can arise from talking about it; this is where things become more interesting.&lt;/p&gt;&#xA;" OwnerUserId="1386" LastActivityDate="2011-03-02T13:33:05.770" CommentCount="4" />
  <row Id="24642" PostTypeId="2" ParentId="24507" CreationDate="2011-03-02T13:48:53.410" Score="1" ViewCount="" Body="&lt;p&gt;Because it represented a contradiction in the current formulation of math. In math a contradiction will propagate and invalidate everything, you can use it to prove and disprove any statement. This paradox could not be solved by math, it required a reformulation of math.&lt;/p&gt;&#xA;" OwnerUserId="4929" LastActivityDate="2011-03-02T13:48:53.410" CommentCount="2" />
  <row Id="24643" PostTypeId="2" ParentId="24638" CreationDate="2011-03-02T13:54:50.290" Score="6" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Hint:&lt;/strong&gt; $\ln(\cos(x))&amp;#39;=\tan(x)$ and &lt;a href=&quot;http://math.stackexchange.com/questions/8439/prove-cos-a-cdot-cos-2a-cdot-cos-22a-cdot-cos-23a-cos-2n-1&quot;&gt;Recursion&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Alternative route:&lt;/strong&gt; Use the identity $\cot(x) - 2\cot(2x)=\tan(x)$ and telescoping.&lt;/p&gt;&#xA;" OwnerUserId="3567" LastEditorUserId="3567" LastEditorDisplayName="" LastEditDate="2011-03-02T14:38:23.013" LastActivityDate="2011-03-02T14:38:23.013" CommentCount="2" />
  <row Id="24644" PostTypeId="2" ParentId="24396" CreationDate="2011-03-02T14:43:17.533" Score="1" ViewCount="" Body="&lt;p&gt;&lt;em&gt;(As suggested by a thread in meta.stackexchange I moved my own solution (which is however still partial) into a separate answer)&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The following solves the problem of finding the Newton-iteration at least for the &quot;principal&quot; value.     &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using $c(b) = ln(ln(b))$ extends not only the range for numerical evaluation but also simplifies the whole problem. First I remove the use of the named variable &lt;em&gt;u&lt;/em&gt; by a function properly dependend on $\beta$ : $$u(\beta) = \frac{\beta}{ \sin(\beta) } *\exp( i * \beta) = \beta \cot(\beta) + i \beta $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then I consider the replacement of the binary search using $ v(\beta) = \ln(u(\beta))-u(\beta)  $ by the Newton/Raphson-iteration-scheme to approximate to a zero of $$ v(\beta,b) = v(\beta) - c(b) $$     &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Btw., here &lt;em&gt;v&lt;/em&gt; has a an interesting powerseries: its coefficients are rational scalings of bernoulli-numbers, resp. zeta at negative arguments:&#xA;$$v(x)=-1 + \sum_{k=2}^{\infty} (2 i)^k*(k+1)/k! \zeta (1-k) x^k $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That value &lt;em&gt;c(b)&lt;/em&gt; to which I want to approximate defines the constant deviation from &lt;em&gt;v(x)&lt;/em&gt; so I include that in the constant term:&#xA;$$v(x,b)=-1 - c(b) + \sum_{k=2}^{\infty} (2 i)^k*(k+1)/k! \zeta (1-k) x^k $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The derivative by termwise differentiation is&#xA;$$ dv(x) = v&amp;#39;(x) = \sum_{k=2}^{\infty} (2 i)^k*(k+1)*k/k! \zeta (1-k) x^{k-1} $$&#xA;and is the same as of $ v(x,b) $     &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the Newton-iteration for finding the root of $v(\beta,b) $ is&#xA;$$ \beta_0 = \frac{\pi}2 \text{  and  } \beta_{n+1} = \beta_n - \frac{v(\beta_n,b)}{v&amp;#39;(\beta_n)} $$&#xA;and with some n'th iteration I get $\beta_n$ which provides approximations to the parameters&#xA;$$ t \approx t_n = \exp(u(\beta_n)) $$&#xA;$$ B \approx b_n = \exp(\exp(v(\beta_n,B))) $$&#xA;where also $$ B = t^{\frac1t}  \approx t_n \ ^{\frac1{t_n}} $$ and &lt;strong&gt;&lt;em&gt;B&lt;/em&gt;&lt;/strong&gt; is the given real base for which I search the complex fixpoint &lt;em&gt;t&lt;/em&gt;.     &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Numerically this iteration converges well with few (say 64) terms; I need now only 4 to 8 iterations to find a well approximated &lt;em&gt;t&lt;/em&gt; for base &lt;em&gt;b=4&lt;/em&gt; where I needed hundreds of iterations using the binary search to get accuracy of, say, fourty digits.    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I've not yet tried to extend this to other complex solutions in &lt;em&gt;t&lt;/em&gt;, so this is still open)     &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&lt;strong&gt;[update2]&lt;/em&gt;&lt;/strong&gt; The selection of another branch (and meaning of another fixpoint) is as simple as possible: just adjust the search-interval for the iterative function to the appropriate interval of $k*\pi+\epsilon \ldots (k+1)*\pi-\epsilon$ Using the Pari/GP-builtin binary search with the new functions, I get the first seven fixpoints as follows    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \begin{array} {rrr}&#xD;&#xA;  k &amp;amp;  log(t) &amp;amp; t \ \ (=Fixpoint) &amp;amp; error: \ \ \  4 - t^{\frac1t} \\&#xD;&#xA;  0 &amp;amp; 0.0886671+1.51223*I &amp;amp; 0.0639598+1.09084*I &amp;amp; 2.04128E-202 \\&#xD;&#xA;  1 &amp;amp; 1.20116+4.44867*I &amp;amp; -0.866450-3.20904*I &amp;amp; 8.36925E-201 \\&#xD;&#xA;  2 &amp;amp; 1.73066+7.63096*I &amp;amp; 1.24841+5.50457*I &amp;amp; 1.49014E-200 \\&#xD;&#xA;  3 &amp;amp; 2.07153+10.8062*I &amp;amp; -1.49429-7.79501*I &amp;amp; 7.14449E-201 \\&#xD;&#xA;  4 &amp;amp; 2.32409+13.9723*I &amp;amp; 1.67648+10.0789*I &amp;amp; 1.16353E-200 \\&#xD;&#xA;  5 &amp;amp; 2.52508+17.1324*I &amp;amp; -1.82146-12.3584*I &amp;amp; 1.01043E-199 \\&#xD;&#xA;  6 &amp;amp; 2.69214+20.2884*I &amp;amp; 1.94197+14.6350*I &amp;amp; 1.85757E-200&#xD;&#xA; \end{array} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&lt;strong&gt;[update]&lt;/em&gt;&lt;/strong&gt; Hmm, the needed number of terms &lt;em&gt;m&lt;/em&gt; in the powerseries of &lt;em&gt;v(x)&lt;/em&gt; to get, for instance, accuracy to 40 decimal digits can be fairly well estimated by the following formula for bases &lt;em&gt;b where 2 &amp;lt;= b &amp;lt;= 2000  .&lt;/em&gt;&lt;br&gt;&#xA;$$ m \approx 62.448 * \ln(\ln(b)) + 100 \text{     (formula estimated using Excel)   } $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a fixed base &lt;em&gt;b&lt;/em&gt; one can determine the accuracy to number of digits depending on number of terms &lt;em&gt;n&lt;/em&gt; of the power series &lt;em&gt;v(ß)&lt;/em&gt;. Here is a plot for &lt;em&gt;b=2, b=4, b=16, b=256&lt;/em&gt;     &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, for base &lt;em&gt;B=4&lt;/em&gt; I get using 150 terms in the power series of v(x) the complex fixpoint &lt;em&gt;t&lt;/em&gt; to 40 digits using 6 iterations as&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;t =  0.06395981030134317249921085931098535528761 +     &#xA;       1.090843376539957576373598843800166909991*I&#xA;4 - t^(1/t) = 1.67294184828 E-46&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/7wica.png&quot; alt=&quot;Accuracy&quot;&gt;    &lt;/p&gt;&#xA;" OwnerUserId="1714" LastEditorUserId="1714" LastEditorDisplayName="" LastEditDate="2011-03-10T15:13:21.997" LastActivityDate="2011-03-10T15:13:21.997" />
  <row Id="24645" PostTypeId="2" ParentId="24628" CreationDate="2011-03-02T14:53:00.663" Score="3" ViewCount="" Body="&lt;p&gt;As with many aspects of linear algebra, this aspect is greatly cleared up by working in a coordinate-independent fashion. What an $n \times m$ matrix really is is a particular representation of a linear transformation $T$ from a vector space of dimension $m$ to a vector space of dimension $n$. To write down such a representation you need a basis of both the source and the target vector spaces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $V$ is a single $n$-dimensional vector space, then &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$n \times n$ matrices generally denote linear transformations $T : V \to V$ with respect to a basis $e_1, ... e_n$. Note that we are using the same basis for both source and target.&lt;/li&gt;&#xA;&lt;li&gt;$n \times 1$ matrices generally denote elements of $V$. Note that this is the same thing as a linear transformation $k \to V$ where $k$ is the base field (I assume $k = \mathbb{R}$ here).&lt;/li&gt;&#xA;&lt;li&gt;$1 \times n$ matrices generally denote linear transformations $V \to k$, otherwise known as elements of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Dual_space&quot; rel=&quot;nofollow&quot;&gt;dual space&lt;/a&gt; $V^{\ast}$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;When we pick a basis $e_1, ... e_n$ of $V$, it follows that every element $v \in V$ can be uniquely expressed in the form&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$v = \sum v_i e_i.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now (and this is somewhat confusing, but it is a lesson well worth learning) the coefficients $v_i$ actually define linear transformations $V \to k$; in other words, they define distinguished elements of $V^{\ast}$ called the &lt;strong&gt;dual basis&lt;/strong&gt; $e_i^{\ast}$ associated to $e_i$. Again, this is confusing, so I'll repeat: $e_i^{\ast}$ is the linear transformation $V \to k$ which sends a vector $v$ to the component $v_i$ of $e_i$ in the unique representation of $v$ in the basis $e_i$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem with working with matrices instead of linear transformations is that nobody ever tells you about the dual basis, and generally people treat the basis and the dual basis as if they were the same thing, which they're not; they transform differently under change of coordinates. When you take the transpose of a matrix, what you are actually doing is switching basis elements with dual basis elements. This operation is &lt;em&gt;coordinate-dependent&lt;/em&gt;, and nobody ever tells you this. (Another reason why you can go a long time without ever learning this lesson is that it doesn't matter if $V$ has an inner product on it with respect to which $e_i$ is an orthonormal basis, since then this operation &lt;em&gt;is&lt;/em&gt; coordinate-independent with respect to orthogonal change of coordinates.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can multiply a $n \times 1$ matrix with a $1 \times n$ matrix to get a $1 \times 1$ matrix; this is a basis-dependent way of talking about the dual pairing $V^{\ast} \times V \to k$ given by evaluating a linear transformation $V \to k$ at a given element of $V$. Note that the dual pairing is coordinate-independent, and it is determined by what it does to a basis of $V^{\ast}$ and a basis of $V$. Predictably it sends $(e_i^{\ast}, e_j)$ to $1$ if $i = j$ and $0$ otherwise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can also multiply a $n \times 1$ matrix with a $1 \times n$ matrix in the other direction to get an $n \times n$ matrix; this is a basis-dependent way of talking about the isomorphism between $\text{End}(V)$ (the space of endomorphisms of $V$, or linear transformations $V \to V$) and the &lt;a href=&quot;http://en.wikipedia.org/wiki/Tensor_product&quot; rel=&quot;nofollow&quot;&gt;tensor product&lt;/a&gt; $V^{\ast} \otimes V$. Explicitly, the isomorphism is as follows: if $T$ is a linear transformation with matrix $a_{ij}$, so that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$T(e_i) = \sum e_j a_{ji}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;then it gets sent to the element $\sum a_{ji} e_i^{\ast} \otimes e_j \in V^{\ast} \otimes V$. The dual pairing then gives a pairing $V \times (V^{\ast} \otimes V) \to V$ which is precisely the evaluation of a linear transformation at an element of $V$. Again, this is probably extremely confusing, but, again, it is a lesson well worth learning; it is the abstract way to talk about representing a linear transformation by a matrix. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;What the word &quot;vector&quot; means is also worth clearing up. A vector is just an element of a vector space. Here that means an element of $V$, but sometimes it will actually mean an element of $V^{\ast}$. This is a bad habit on the part of people who work with coordinates; they should actually call these &lt;strong&gt;dual vectors&lt;/strong&gt;, since vectors and dual vectors (column and row vectors) transform differently under change of coordinates. &lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2011-03-02T15:01:26.240" LastActivityDate="2011-03-02T15:01:26.240" CommentCount="3" />
  <row Id="24646" PostTypeId="1" AcceptedAnswerId="24647" CreationDate="2011-03-02T15:23:16.283" Score="2" ViewCount="94" Body="&lt;p&gt;The following is alpha &lt;strong&gt;of&lt;/strong&gt; t times x:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\alpha{}(t)x$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The following is alpha &lt;strong&gt;times&lt;/strong&gt; t times x:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\alpha{}(t)x$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My instructor had one interpretation, I used the other ;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there an easy or standard way to signify that we're talking about a funciton rather than a variable multiplication here?&lt;/p&gt;&#xA;" OwnerUserId="4276" LastActivityDate="2011-03-02T17:28:33.290" Title="Is there an easy way to signal to the reader the difference between a function and a multiplication?" Tags="&lt;notation&gt;" AnswerCount="3" />
  <row Id="24647" PostTypeId="2" ParentId="24646" CreationDate="2011-03-02T15:29:47.997" Score="5" ViewCount="" Body="&lt;p&gt;Just a suggestion: we can simply write $\alpha t x$ to mean the product of the three variables. Usually it is clear from the context what the meaning is: if $\alpha$ is a function, then $\alpha (t)$ would be the function evaluated at $t$.&lt;/p&gt;&#xA;" OwnerUserId="4594" LastActivityDate="2011-03-02T15:29:47.997" CommentCount="2" />
  <row Id="24648" PostTypeId="2" ParentId="24646" CreationDate="2011-03-02T15:31:44.773" Score="2" ViewCount="" Body="&lt;p&gt;For &quot;alpha times t times x&quot; I would suggest writing&#xA;$$&#xD;&#xA;\alpha\cdot t\cdot x&#xD;&#xA;$$&#xA;and adding brackets if appropriate, like $(\alpha\cdot t)\cdot x$. You could also leave the centered dots.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The bracket in $\alpha(t) x$ does not make much sense.&lt;/p&gt;&#xA;" OwnerUserId="367" LastActivityDate="2011-03-02T15:31:44.773" />
  <row Id="24649" PostTypeId="1" CreationDate="2011-03-02T16:03:33.590" Score="1" ViewCount="85" Body="&lt;p&gt;There is a square table and 2 persons are sitting on each side of it so there are 8 persons in total ... how many total number of permutation is possible?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does circular permutation rules applies here?&lt;/p&gt;&#xA;" OwnerUserId="3744" LastActivityDate="2011-03-02T19:35:29.080" Title="Yet another permutation problem" Tags="&lt;combinatorics&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="24650" PostTypeId="2" ParentId="24585" CreationDate="2011-03-02T16:15:19.187" Score="1" ViewCount="" Body="&lt;p&gt;You're right. To prove this you may use the fact that you have an atlas for $M_1\times M_2$ (it's the assumed differential structure of the exercise) whose elements are of the form $(U_1\times U_2,\varphi_1\times\varphi_2)$, with $(U_1,\varphi_1)$ being in an atlas for $M_1$ and $(U_2,\varphi_2)$ being in an atlas for $M_2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, given the diffeomorphisms $f_1:M_1\to M_1&amp;#39;$ and $f_2:M_2\to M_2&amp;#39;$ you have for every $(p_1,p_2)\in M_1\times M_2$ smooth coordinate charts $(U_1\times U_2,\varphi_1\times\varphi_2)$ and $(V_1\times V_2,\psi_1\times\psi_2)$, with $(p_1,p_2)\in U_1\times U_2$ and $(f_1(p_1),f_2(p_2))=(q_1,q_2)\in V_1\times V_2$. So, the smoothnes of&#xA;$$(\psi_1\times\psi_2)\circ(f_1\times f_2)\circ(\varphi_1\times\varphi_2)^{-1}= (\psi_1\circ f_1\circ\varphi_1^{-1})\times(\psi_2\circ f_2\circ\varphi_2^{-1})$$&#xA;and&#xA;$$(\varphi_1\times\varphi_2)\circ(f_1\times f_2)^{-1}\circ(\psi_1\times\psi_2)^{-1} = (\varphi_1\circ f_1^{-1}\circ\psi_1^{-1})\times(\varphi_2\circ f_2^{-1}\circ\psi_2^{-1})$$&#xA;come from the smoothness of&#xA;$$\psi_1\circ f_1\circ\varphi_1^{-1},$$&#xA;$$\psi_2\circ f_2\circ\varphi_2^{-1},$$&#xA;$$\varphi_1\circ f_1^{-1}\circ\psi_1^{-1},$$&#xA;and&#xA;$$\varphi_2\circ f_2^{-1}\circ\psi_2^{-1}$$&#xA;as in Euclidean spaces.&lt;/p&gt;&#xA;" OwnerUserId="370" LastActivityDate="2011-03-02T16:15:19.187" CommentCount="2" />
  <row Id="24652" PostTypeId="2" ParentId="24576" CreationDate="2011-03-02T16:44:33.453" Score="0" ViewCount="" Body="&lt;p&gt;Let me suggest the following (among many others):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Theorem 1 in chapter 2, section 4 of&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Harmonic Analysis:  A Gentle Introduction Carl L. DeVito, JONES AND BARTLETT PUBLISHERS, 2007&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Theorem 2.1 in&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Princeton Lectures in Analysis&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I Fourier Analysys&#xA;An Introduction&lt;/p&gt;&#xA;&#xA;&lt;p&gt;E. Stein, R. Shakarchi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PUP&lt;/p&gt;&#xA;" OwnerUserId="4791" LastActivityDate="2011-03-02T16:44:33.453" CommentCount="1" />
  <row Id="24653" PostTypeId="1" AcceptedAnswerId="24654" CreationDate="2011-03-02T16:46:21.017" Score="3" ViewCount="181" Body="&lt;p&gt;Suppose that $(X, \rho)$ is a metric space, $|X| &amp;gt; 1$. Is it possible to prove that $X$ is an open set without assuming the axiom of choice?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As I understand it, the challenge is to find a way to assign an open ball to each point of the space (assign a radius, basically). The way I found is to select some point $y \in X$ and assign to each point $x \in X\setminus\!\{y\}$ the ball $B_{2\rho(x,y)}(x)$. Then it is easy to show that the union of these balls is $X$, and so $X$ is open.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there another way to prove that $X$ is open without relying on the axiom of choice? Or am I wrong and the axiom of choice is not required for the above proof?&lt;/p&gt;&#xA;" OwnerUserId="3793" LastActivityDate="2011-03-03T00:14:55.447" Title="Is it possible to prove that the metric space is an open set without choice?" Tags="&lt;set-theory&gt;&lt;metric-spaces&gt;" AnswerCount="4" CommentCount="6" />
  <row Id="24654" PostTypeId="2" ParentId="24653" CreationDate="2011-03-02T16:50:12.980" Score="5" ViewCount="" Body="&lt;p&gt;You can pick the radius of the open ball centred at $x$ however you like, and the union of all the balls will be $X$. For example, you could just let all the radii be 1.&lt;/p&gt;&#xA;" OwnerUserId="5203" LastActivityDate="2011-03-02T16:50:12.980" CommentCount="7" />
  <row Id="24655" PostTypeId="2" ParentId="24653" CreationDate="2011-03-02T16:50:20.767" Score="1" ViewCount="" Body="&lt;p&gt;$X$ is an open set because every topology (=collection of open sets) on $X$ must contain $X$, see the &lt;a href=&quot;http://en.wikipedia.org/wiki/Topology#Mathematical_definition&quot; rel=&quot;nofollow&quot;&gt;definition&lt;/a&gt;. Nothing metric here.&lt;/p&gt;&#xA;" OwnerUserId="6179" LastActivityDate="2011-03-02T16:50:20.767" CommentCount="12" />
  <row Id="24657" PostTypeId="2" ParentId="24629" CreationDate="2011-03-02T17:14:50.163" Score="6" ViewCount="" Body="&lt;p&gt;A polynomial with coefficients in $\mathbb{Q}$ that has $\sqrt{2}$ as a root must also have $-\sqrt{2}$ as a root; likewise, if it has $\sqrt{3}$ as a root, then it must also have $-\sqrt{3}$ as a root. So any polynomial with coefficients in $\mathbb{Q}$ that has &lt;em&gt;both&lt;/em&gt; $\sqrt{2}$ and $\sqrt{3}$ as roots must be of degree at least $4$ or equal to $0$, and must be a multiple of $(x^2-2)(x^2-3)$. So there is no quadratic polynomial with coefficients in $\mathbb{Q}$ that will do it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To see this, suppose that $p(x)$ is a polynomial with coefficients in $\mathbb{Q}$ that has $\sqrt{2}$ as a root. Dividing with remainder by $x^2-2$ we get $p(x) = (x^2-2)q(x) + r(x)$, where $q(x)$ and $r(x)$ have rational coefficients, and $r(x)=0$ or else $r(x)$ is of degree 0 or 1. Plugging in $\sqrt{-2}$, we get $0 = 0q(\sqrt{2}) + r(\sqrt{2}) = r(\sqrt{2})$. But there are no polynomials of degree $0$ or $1$ with coefficients in $\mathbb{Q}$ that are $0$ at $\sqrt{2}$ (that would yield that $\sqrt{2}$ is rational), so $r(x)=0$. Therefore, $r(x) = 0$, so $p(x)$ is a multiple of $x^2-2$; and since $x^2-2$ is $0$ at $-\sqrt{2}$, then so is $p(x)$. The same argument shows that it is also a multiple of $x^2-3$, and so $p(x)$ will also have $-\sqrt{3}$ as a root.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $x^2-3$ and $x^2-2$ are relatively prime in $\mathbb{Q}[x]$, any polynomial that is a multiple of both is a multiple of their product. The smallest degree polynomial that does is $(x^2-2)(x^2-3) = x^4 - 5x^2 + 6$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you allow other coefficients, then the only quadratics that work are of the form $c(x-\sqrt{2})(x-\sqrt{3}) = cx^2 - c(\sqrt{2}+\sqrt{3})x + \sqrt{6}$ and its multiples.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-02T17:21:16.757" LastActivityDate="2011-03-02T17:21:16.757" CommentCount="2" />
  <row Id="24658" PostTypeId="2" ParentId="24435" CreationDate="2011-03-02T17:26:36.320" Score="5" ViewCount="" Body="&lt;p&gt;It seems the best way to describe the moments of $X_i$ is to use ascending factorials. Using Pochhammer's notation &#xA;$$&#xD;&#xA;(x)_p=x(x+1)\cdots(x+p-1)&#xD;&#xA;$$ &#xA;for every nonnegative integer $p$, one gets&#xA;$$&#xD;&#xA;E((X_i)_p)=(i)_p\frac{(m+1)_p}{(n+1)_p}.&#xD;&#xA;$$&#xA;Since $(x)_1=x$, this yields your formula for $E(X_i)$. Using the shorthand&#xA;$$&#xD;&#xA;a_p=(i+p-1)\frac{m+p}{n+p},&#xD;&#xA;$$&#xA;one gets $E((X_i)_p)=a_1a_2\cdots a_{p}$. This yields (I think) the value of the kurtosis of $X_i$ defined by&#xA;$$&#xD;&#xA;\beta_2(X_i)=\frac{E((X_i-E(X_i))^4)}{E((X_i-E(X_i))^2)^2}&#xD;&#xA;$$ &#xA;as the ratio&#xA;$$&#xD;&#xA;\beta_2(X_i)=\frac{a_2a_3a_4-6a_2a_3+7a_2-1-4a_1(a_2a_3-3a_2+1)+6a_1^2(a_2-1)-3a_1^3}{a_1(a_2-a_1-1)^2}.&#xD;&#xA;$$&#xA;And I am well aware that this solution does not use Maple...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt; To get the formula for $E((X_i)_p)$ written above, one can denote by $X_i^{m,n}$ the random variable considered by the OP and sum over every possible value of $k$ the algebraic identity&#xA;$$&#xD;&#xA;P(X_i^{m,n}=k)\cdot (k)_p=P(X_{i+p}^{m+p,n+p}=k+p)\cdot (i)_p\frac{(m+1)_p}{(n+1)_p}.&#xD;&#xA;$$&#xA;The LHS of the resulting equality is $E((X_i^{m,n})_p)$ and the RHS is the total mass of the distribution of $X_{i+p}^{m+p,n+p}$, which equals $1$, times the factor $(i)_p(m+1)_p/(n+1)_p$. This proves the formula.&lt;/p&gt;&#xA;" OwnerUserId="6179" LastEditorUserId="6179" LastEditorDisplayName="" LastEditDate="2011-03-03T21:06:36.883" LastActivityDate="2011-03-03T21:06:36.883" CommentCount="5" />
  <row Id="24659" PostTypeId="2" ParentId="24646" CreationDate="2011-03-02T17:28:33.290" Score="1" ViewCount="" Body="&lt;p&gt;I have noticed that misinterpretations of notation are becoming more common. For instance, many students have difficulty when differentiating something such as y = sin(tan(x)), mistakenly interpreting it as a product as opposed to a composition. Jasper makes a good point when he emphasizes that the context is usually a good guide as to how to interpret. In my example, the function sine has to act on something, which would be tan(x). As a rule, if you are unsure how to interpret, ask for clarification. If the situation does not allow that, or the instructor declines to comment, closely examine the context.&lt;/p&gt;&#xA;" OwnerUserId="2933" LastActivityDate="2011-03-02T17:28:33.290" CommentCount="1" />
  <row Id="24660" PostTypeId="1" AcceptedAnswerId="27788" CreationDate="2011-03-02T17:31:59.070" Score="2" ViewCount="78" Body="&lt;p&gt;Using the implicit function theorem one can prove the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $X,Y$ be Banach-spaces, $U\subset X$ open, $f\colon U\to \mathbf{R}$, $g\colon U\to Y$ continuously differentiable function. If $f|_{g^{-1}(0)}$ has a local extremum at $x$, and $g&amp;#39;(x)\in L(X;Y)$ has a right inverse in $L(Y;X)$, then there is a unique $\lambda\in Y^*$, such that $f&amp;#39;(x)=\lambda\circ g&amp;#39;(x)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One can also give some second order necessary and sufficient conditions. This generalizes the case where $Y$ is finite dimensional, and $g&amp;#39;(x)$ is surjective. However I have seen some texts that claimed that only surjectivity is sufficient even in the infinite dimensional case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is does that hold true? If it does what is the method of the proof, because I cannot figure how the implicit function theorem could be used. If the proof is complicated I would appreciate even just some good references.&lt;/p&gt;&#xA;" OwnerUserId="7598" LastActivityDate="2011-03-18T14:15:01.793" Title="Condition for existence of Lagrange-multiplier" Tags="&lt;calculus&gt;&lt;reference-request&gt;&lt;nonlinear-optimization&gt;" AnswerCount="1" />
  <row Id="24661" PostTypeId="2" ParentId="24653" CreationDate="2011-03-02T17:33:03.803" Score="0" ViewCount="" Body="&lt;p&gt;As an aside, note that you don't need Choice to validate the openness of an arbitrary open subset $U\subseteq X$.  That is, we can assign an open ball around each $u\in U$ without Choice: let $\delta_u = \frac12 \inf\{\rho(u,x) : x\notin U\}$&#xA;and set $F_u = B_{\delta_u}(u)$.&lt;/p&gt;&#xA;" OwnerUserId="3852" LastActivityDate="2011-03-02T17:33:03.803" />
  <row Id="24662" PostTypeId="1" AcceptedAnswerId="24692" CreationDate="2011-03-02T17:34:55.150" Score="19" ViewCount="450" Body="&lt;p&gt;Let $q\in (0,1)$. Is there a way of computing the series&#xA;$$&#xD;&#xA;\sum_{n=0}^\infty q^{n^2}&#xD;&#xA;$$&#xA;explicitly? Is there at least a nice accurate estimate?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All I could get is the estimate&#xA;$$\sqrt{\frac{\pi}{4\cdot\mathrm{ln}\frac{1}{q}}}\leq\sum_{n=0}^\infty q^{n^2}\leq 1+\sqrt{\frac{\pi}{4\cdot\mathrm{ln}\frac{1}{q}}}$$&#xA;via integration (quite possibly flawed).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $q=\frac{1}{2}$, Maple gives the values&#xA;$$&#xD;&#xA;1.064467020\leq 1.564468414\leq 2.064467020,&#xD;&#xA;$$&#xA;showing that my estimate is not very precise. (Of course, the sum of the two errors will always be $1$. Here both errors are coincidentally almost exactly $\frac{1}{2}$.)&lt;/p&gt;&#xA;" OwnerUserId="367" LastEditorUserId="367" LastEditorDisplayName="" LastEditDate="2011-04-10T17:58:05.843" LastActivityDate="2011-04-10T20:36:53.793" Title="How to compute the series $\sum_{n=0}^\infty q^{n^2}$?" Tags="&lt;sequences-and-series&gt;" AnswerCount="4" CommentCount="4" FavoriteCount="3" />
  <row Id="24663" PostTypeId="1" AcceptedAnswerId="24664" CreationDate="2011-03-02T18:01:53.927" Score="1" ViewCount="83" Body="&lt;p&gt;Hello,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;how can one show (hopefully in an elementary manner) that there exist irreducible polynomials of arbitrary degree and number of variables over arbitrary field?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;thank you&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S. induction?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: ehm, sorry, I meant for $n\geq2$, i.e. in at least two variables.&lt;/p&gt;&#xA;" OwnerUserId="7325" LastEditorUserId="7325" LastEditorDisplayName="" LastEditDate="2011-03-02T18:08:43.343" LastActivityDate="2011-03-02T18:15:19.053" Title="Does $\forall n,d\!\in\!\mathbb{N}$ $\forall$ field $\mathbb{F}$ exist an irreducible $f\!\in\!\mathbb{F}[x_1,\ldots,x_n]$ of degree $d$?" Tags="&lt;abstract-algebra&gt;&lt;commutative-algebra&gt;&lt;field-theory&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="24664" PostTypeId="2" ParentId="24663" CreationDate="2011-03-02T18:03:32.763" Score="4" ViewCount="" Body="&lt;p&gt;This is false. Any irreducible polynomial in one variable over an algebraically closed field (such as $\mathbb{C}$) has degree 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $n \ge 2$, the answer is yes: just take the irreducible polynomial $y-x^d$.&lt;/p&gt;&#xA;" OwnerUserId="5203" LastEditorUserId="5203" LastEditorDisplayName="" LastEditDate="2011-03-02T18:15:19.053" LastActivityDate="2011-03-02T18:15:19.053" CommentCount="0" />
  <row Id="24665" PostTypeId="1" AcceptedAnswerId="24666" CreationDate="2011-03-02T18:21:34.733" Score="3" ViewCount="62" Body="&lt;p&gt;I was recently reading about classifying the extrema of a continuous function of a single variable.  I came across the information that if the second derivative is zero then we can examine the higher derivatives such as $f^{\prime\prime\prime}$,$f^{(4)}$ and so on.  Suppose $f^{(n)}$ is the first non-zero derivative.  If n is odd, then the point is an inflection point and if n is even then a positive nth derivative means a minimum and a negative nth derivative means a maximum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, my question is, can anyone come up with an example of a function $f(x)$ where this information about the nth derivative can be applied?  I wanted to add one to my notes.  Further, can anyone come up with an example of both a inflection point and a maximum or minimum?&lt;/p&gt;&#xA;" OwnerUserId="1004" LastActivityDate="2011-03-02T18:27:24.993" Title="The maximum of a function of a single variable where all derivatives $f^{(n)}=0$ for $n\le 2$?" Tags="&lt;real-analysis&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="24666" PostTypeId="2" ParentId="24665" CreationDate="2011-03-02T18:27:24.993" Score="3" ViewCount="" Body="&lt;p&gt;Do the mononomials $x^n$ and $-x^n$ work? Any power series expansion which starts at the $n$-th term generalizes this idea. Another interesting function to study in this context is $x^k \sin(1/x)$, though it is a bit more subtle&lt;/p&gt;&#xA;" OwnerUserId="5882" LastActivityDate="2011-03-02T18:27:24.993" />
  <row Id="24667" PostTypeId="1" CreationDate="2011-03-02T18:27:48.530" Score="1" ViewCount="123" Body="&lt;p&gt;For natural numbers, I would like to prove that if $m &amp;lt; n$, then $k^m &amp;lt; k^n$ for $k$ not $0$ or $1$. Induction seems viable, but I don't know which variable to induct on. Any suggestions?&lt;/p&gt;&#xA;" OwnerUserId="7709" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2011-03-02T18:53:36.657" LastActivityDate="2011-03-02T18:53:36.657" Title="If $m &lt; n$, then $k^m &lt; k^n$?" Tags="&lt;induction&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="24668" PostTypeId="1" CreationDate="2011-03-02T18:29:39.487" Score="2" ViewCount="44" Body="&lt;p&gt;Again, I really just want a nudge in the right direction. Possibly a large nudge, but not the straight forward answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am trying to figure out how to solve &lt;a href=&quot;http://projecteuler.net/index.php?section=problems&amp;amp;id=191&quot; rel=&quot;nofollow&quot;&gt;Project Euler Problem 191&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe I have most of it figured out, but I cannot come up with an algorithm or pattern for the 3-consecutive-Absent days part. I can even find the number of permutations of the strings holding the total number of A-days constant (i.e. how many permutations with only 1 A-day, only 2 A-days, only 3 A-days, etc.), but I cannot figure out how many permutations have 3 consecutive A-days.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, is there a mathematical formula for this? Is there a way to break this up so that I have &quot;(some # permutations) +/- (another # of permutations) = (# of consecutive A-days)&quot;? I believe this is a combinatorial problem, but maybe it falls under a different category.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any help is appreciated. Thanks.&lt;/p&gt;&#xA;" OwnerUserId="7708" LastActivityDate="2011-03-02T19:00:16.103" Title="Need a nudge in the right direction - How do I find the total number of permutation with 3 consecutive characters?" Tags="&lt;combinatorics&gt;&lt;project-euler&gt;" AnswerCount="2" />
  <row Id="24670" PostTypeId="2" ParentId="24668" CreationDate="2011-03-02T18:39:35.070" Score="1" ViewCount="" Body="&lt;p&gt;For each string length $n$, you want to keep track of six quantities: the number of ``good'' (i. e. prize-winning) strings of length $n$ which end in $i$ $A$s ($i = 0, 1, 2$) and contain $j$ $L$s ($j = 0, 1$).  These six quantities for strings of length $n+1$ can be simply written in terms of those for length $n$.&lt;/p&gt;&#xA;" OwnerUserId="173" LastActivityDate="2011-03-02T18:39:35.070" CommentCount="1" />
  <row Id="24671" PostTypeId="1" CreationDate="2011-03-02T18:44:34.637" Score="1" ViewCount="121" Body="&lt;p&gt;Adding to the &lt;a href=&quot;http://math.stackexchange.com/search?q=%22for+dummies%22&quot;&gt;for dummies&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.paulsprojects.net/opengl/sh/technical.html&quot; rel=&quot;nofollow&quot;&gt;The real spherical harmonics are orthonormal basis functions on the surface of a sphere.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd like to fully understand that sentence and what it means.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Still grappling with &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Orthonormal basis functions (I believe this is like Fourier Transform's basis functions are sines and cosines, and sin is orthogonal to cos, and so the components can have a zero inner product..)&lt;/li&gt;&#xA;&lt;li&gt;&quot;.. are orthonormal basis functions ..&lt;strong&gt;on the surface of a sphere&lt;/strong&gt;&quot;.&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;What sphere&lt;/strong&gt;?  Where does the sphere come from?  Do you mean for each position on the sphere, we have a value?  Is the periodicity in space on the sphere exploited?  Is that how we get the higher order terms?&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1069" LastActivityDate="2011-03-02T19:31:34.003" Title="Spherical harmonics for dummies" Tags="&lt;spherical-harmonics&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="24672" PostTypeId="2" ParentId="24667" CreationDate="2011-03-02T18:48:53.953" Score="1" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Hint:&lt;/strong&gt; Consider $f(x)=a^{x}$ and prove that it's monotonically increasing.&lt;/p&gt;&#xA;" LastActivityDate="2011-03-02T18:48:53.953" />
  <row Id="24673" PostTypeId="1" AcceptedAnswerId="24695" CreationDate="2011-03-02T18:53:01.480" Score="2" ViewCount="99" Body="&lt;p&gt;Suppose $C$ is an algebraic curve (which has singular points) over an algebraically closed field $k$, and that $f$ is a rational function on $C$. How does one defines the Weil divisor of $f$? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is that the local rings of $C$ at singular points are not DVR's, so I do not have an obvious candidate for an order at a point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: Let me give an example, inspired by an answer from below. Suppose $C$ is curve $y^2=x^3$. What would be the order of the rational function $x/y$ at the origin?&lt;/p&gt;&#xA;" OwnerUserId="5491" LastEditorUserId="5491" LastEditorDisplayName="" LastEditDate="2011-03-03T08:56:19.563" LastActivityDate="2011-03-03T15:34:20.943" Title="Divisors of rational functions on curves at singular points" Tags="&lt;algebraic-geometry&gt;&lt;reference-request&gt;" AnswerCount="1" />
  <row Id="24674" PostTypeId="2" ParentId="24668" CreationDate="2011-03-02T19:00:16.103" Score="1" ViewCount="" Body="&lt;p&gt;Ignoring the L, you only need to count the number of strings in which the As appear doubly at the most.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To count this, first fix the number of Os and fix the number of As appearing singly (and so by implication the number of AA blocks).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now write the Os with dashes between them and to left, for instance with 3 Os.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;_ O _ O _ O _&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You now need to fill some of the dashes with the number of A and AA you chose.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Vary the number of Os and As and add up for the total.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope that helps.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2011-03-02T19:00:16.103" />
  <row Id="24675" PostTypeId="1" AcceptedAnswerId="24691" CreationDate="2011-03-02T19:05:10.170" Score="1" ViewCount="123" Body="&lt;p&gt;Suppose $H$ is embedded in $G$ and $H&amp;#39;$ is isomorphic to $H$ and embedded in $G&amp;#39;$. Then we can simultaneously embed $H$, $H&amp;#39;$, $G$ and $G&amp;#39;$ into a single object (the amalgamated free product) such that $H$ and $H&amp;#39;$ become identifiable. This seems similar to the Isomorphism Extension Theorem for fields which is important for developing Galois Theory. What are the practical uses of the amalgamated free product? &lt;/p&gt;&#xA;" OwnerUserId="2548" LastEditorUserId="5798" LastEditorDisplayName="" LastEditDate="2011-03-02T21:42:38.193" LastActivityDate="2011-03-03T04:44:42.447" Title="Amalgamated Free Product" Tags="&lt;group-theory&gt;&lt;intuition&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="24676" PostTypeId="1" AcceptedAnswerId="24685" CreationDate="2011-03-02T19:12:08.250" Score="2" ViewCount="246" Body="&lt;p&gt;How can I prove that a convex function ƒ defined on some open interval C is continuous on C?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you.&lt;/p&gt;&#xA;" OwnerUserId="6163" LastActivityDate="2011-03-02T23:37:36.843" Title="convex function in open interval is continuous" Tags="&lt;calculus&gt;" AnswerCount="3" />
  <row Id="24677" PostTypeId="1" AcceptedAnswerId="24679" CreationDate="2011-03-02T19:13:47.850" Score="2" ViewCount="99" Body="&lt;p&gt;I have the following exercise:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Show that if a measure-preserving system $(X, \mathcal B, \mu, T)$ has the property that for any $A,B \in \mathcal B$ there exists $N$ such that&#xA;$$\mu(A \cap T^{-n} B) = \mu(A)\mu(B)$$&#xA;for all $n \geq N$, then $\mu(A) = 0$ or $1$ for all $A \in \mathcal B$&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the back of the book states that I should fix $B$ with $0 &amp;lt; \mu(B) &amp;lt; 1$ and then find $A$ using the Baire Category Theorem. Edit: I'm now pretty sure that this &quot;$B$&quot; is what &quot;$A$&quot; is in the required result.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: This stopped being homework so I removed the tag. Any approach would be nice. I have some idea where I approximate $A$ with $T^{-n} B^C$ where the $n$ will be an increasing sequence and then taking the $\limsup$ of the sequence. I'm not sure if it is correct. I will add it later on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My attempt after Didier's comment: &quot;proof&quot;:&lt;/strong&gt;&#xA;First pick $B$ with $0 &amp;lt; \mu(B) &amp;lt; 1$. Then set $A_0 = B^C$ and&#xA;determine the smallest $N_0$ such that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mu(A_0 \cap T^{-N_0} B) = \mu(A_0) \mu(B)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Continue like this and set&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$A_k = T^{-N_{k - 1}} B^C$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we note that the $N_k$ are a strictly increasing sequence, since&#xA;suppose not, say $N_{k} \leq N_{k - 1}$ then&#xA;$$\mu \left ( T^{-N_{k - 1}} B^C \cap T^{-N_{k - 1}} B \right ) = 0 \neq&#xD;&#xA;    \mu(B^C) \mu(B) &amp;gt; 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Set $A = \limsup_n A_n$, then note that&#xA;\begin{align}&#xA;\sum_n \mu(A_n) = \sum_n \mu(B^C) = \infty&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So $\mu(A) = 1$, by the Borel-Cantelli lemma. Well, not yet, because&#xA;we are also required to show that the events are independent, so it is&#xA;sufficient to show that $\mu(A_{k + 1} \cap A_k) = \mu(A_{k + 1&#xD;&#xA;})\mu(A_k)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know that $\mu(T^{N_k} B^C \cap T^{N_{k + 1}} B) = \mu(B^C)\mu(B)$. So&#xA;does a similar result now hold if we replace $B$ with $B^C$ in the&#xA;second part?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note:&#xA;\begin{align}&#xA;\mu(A \cap T^{-M} B^C) &amp;amp;= \mu(A \setminus (T^{-M} B \cap A))\\&#xA;&amp;amp;= \mu(A) - \mu(A)\mu(B) \\&#xA;&amp;amp;= \mu(A) - \mu(A \cap T^{-M} B)\\&#xA;&amp;amp;= \mu(A)\mu(B^C)&#xA;\end{align}&#xA;which is what was required.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For this $A$ and $B$ we can find an $M$ and a $k$ such that $N_k \leq&#xD;&#xA;M &amp;lt; N_{k + 1}$. Now note that $\limsup_n A \cap T^{-M} B = \limsup_n&#xD;&#xA;(A \cap T^{-M} B)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Further,&#xA;$$\sum_n \mu(A_n \cap T^{-N_{k +1}}) = \mu(A_0 \cap T^{-N_{k + 1}}) +&#xD;&#xA;\ldots + \mu(A_{k + 1} \cap T^{N_{k + 1}}) &amp;lt; \infty$$&#xA;So again by the Borel-Cantelli Lemma we have &#xA;$\mu(\limsup_n A_n \cap T^{-M} B) = 0$.&#xA;Thus we get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mu(A) \mu(B) = \mu(B) = \mu(A \cap T^{-M} B) = 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is a contradiction since $\mu(B) &amp;gt; 0$. So, such $B$'s&#xA;violate the condition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added:&lt;/strong&gt; Actually the metric on the space of events $d(A,B) = \mu(A \Delta B)$ can work together with Baire's Category Theorem.&lt;/p&gt;&#xA;" OwnerUserId="1120" LastEditorUserId="1120" LastEditorDisplayName="" LastEditDate="2011-03-05T23:26:29.817" LastActivityDate="2011-03-05T23:26:29.817" Title="Stronger than strong-mixing." Tags="&lt;ergodic-theory&gt;" AnswerCount="1" />
  <row Id="24678" PostTypeId="2" ParentId="24649" CreationDate="2011-03-02T19:16:35.337" Score="2" ViewCount="" Body="&lt;p&gt;It depends on your definition of what sitting positions are to be considered different:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If chairs are considered all distict, the answer is obviously 8!=40320&lt;/li&gt;&#xA;&lt;li&gt;If only the sequence of neighbours matters (that is, if rotations of chairs does not matter), (but clock and counter clock wise are considered different) we have a circular permutation: 7!=5040&lt;/li&gt;&#xA;&lt;li&gt;If only 90 degrees rotations of the table induce equivalent arrangements (the most reasonable definition here, I'd say) one has 2 7! = 10080&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="312" LastEditorUserId="312" LastEditorDisplayName="" LastEditDate="2011-03-02T19:35:29.080" LastActivityDate="2011-03-02T19:35:29.080" />
  <row Id="24679" PostTypeId="2" ParentId="24677" CreationDate="2011-03-02T19:24:07.370" Score="1" ViewCount="" Body="&lt;p&gt;Hint: what happens if $A=T^{-N}B$?&lt;/p&gt;&#xA;" OwnerUserId="6179" LastActivityDate="2011-03-02T19:24:07.370" CommentCount="3" />
  <row Id="24680" PostTypeId="1" CreationDate="2011-03-02T19:24:07.527" Score="3" ViewCount="203" Body="&lt;p&gt;&lt;strong&gt;This is a problem from Hungerford's book:&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $E$ be an intermediate field of extension $K\subset F$ and assume that $E=K(u_1, \cdots ,u_r)$ where $u_i$ are (some of the) roots of $f\in K[x]$. Then $F$ is a splitting field of $f$ over $K$ if and only if $F$ is a splitting field of $f$ over E. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is my attempt.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $u_1, \cdots ,v_n$ be the roots of $f$. Then since $F$ is a splitting field of $f$ over $K$, $F=K(v_1, \cdots ,v_n)$ $\implies$ $F=E(v_1, \cdots ,v_n)$. So $F$ is a splitting field of $f$ over $E$ .  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conversely, suppose $v_1, \cdots ,v_n$ are the roots of $f$ in $F$.  Then $F=E(v_1, \cdots ,v_n)=K(u_1, \cdots ,u_r,v_1, \cdots, v_n)$. Thus $F$ is a splitting field of $f$ over $K$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My Question:&lt;/strong&gt; First I want to know if the proof above is right. The second part of the problem asks to extend the above to splitting fields of arbitrary  &lt;strong&gt;&lt;em&gt;set of polynomials&lt;/em&gt;.&lt;/strong&gt; This is where I'm lost. I'd like a little help on how to begin.&lt;br&gt;&#xA;Thanks.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update:&lt;/strong&gt;  I think I've got it now. I had some difficulty with the indexing. Thanks.  &lt;/p&gt;&#xA;" OwnerUserId="4426" LastEditorUserId="4426" LastEditorDisplayName="" LastEditDate="2011-03-03T17:22:41.977" LastActivityDate="2011-03-03T17:22:41.977" Title="Splitting Fields" Tags="&lt;galois-theory&gt;" CommentCount="4" />
  <row Id="24682" PostTypeId="2" ParentId="24662" CreationDate="2011-03-02T19:26:21.580" Score="16" ViewCount="" Body="&lt;p&gt;$\theta_3(q)=\sum_{n=-\infty}^\infty q^{n^2}$ where $q=e^{-\pi s}$ and since, for $Re(s)&amp;gt;0,$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sqrt{s} \theta_ 3(e^{-\pi s}) = \theta_3(e^{-\pi/s})$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{n=0}^\infty q^{n^2} = \sqrt{ \frac{\pi} {4 \log \left( \frac{1}{q} \right) } }&#xD;&#xA;\theta_3(e^{-\pi/s}) + \frac12.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if $s$ is small then $e^{-\pi/s}$ is large and so $\theta_3(e^{-\pi/s}) \approx 1,$&#xA;and hence you obtain the result you noticed:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{n=0}^\infty q^{n^2} \approx \sqrt{ \frac{\pi} {4 \log \left( \frac{1}{q} \right) } }&#xD;&#xA;+ \frac12.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For greater accuracy take more terms in the sum defining $\theta_3(e^{-\pi/s}).$&lt;/p&gt;&#xA;" OwnerUserId="1301" LastEditorUserId="1301" LastEditorDisplayName="" LastEditDate="2011-03-02T19:33:22.747" LastActivityDate="2011-03-02T19:33:22.747" CommentCount="5" />
  <row Id="24683" PostTypeId="1" AcceptedAnswerId="24687" CreationDate="2011-03-02T19:28:56.167" Score="5" ViewCount="89" Body="&lt;p&gt;We know from Calculus what a series is, and you might have seen infinite products as well. But the &lt;a href=&quot;http://en.wikipedia.org/wiki/Elementary_symmetric_polynomial&quot; rel=&quot;nofollow&quot;&gt;Elementary Symmetric Polynomials&lt;/a&gt; give an entire spectrum of operators between a sum and product over a finite set.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given a sequence $s_n$ and an $a \in \mathbb{N}$(representing which operator we're picking), define $S_n$ to be the set $\{s_k | 1 \leq k \leq n\}$. Then our &quot;generalized series&quot; is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$T_a(s) = \lim_{n \to \infty} e_a(S_n)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $a = 0$, you get $1$ no matter what the set is. If $a = 1$, then you get a standard Calculus series. If $a = 2$, then we get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$T_2(s) = \sum_{n = 1}^\infty s_n \left(\sum_{m = n+1}^\infty s_m\right)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $a=3$, then&#xA;$T_3(s) = \sum_{n=1}^\infty s_n \left(\sum_{m = n+1}^\infty s_m \left(\sum_{k = m+1}^\infty s_k \right) \right)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and so on. I can provide a Mathematica function I wrote to compute them, if you like. My questions are:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Is there a standard name or paper for these?&lt;/li&gt;&#xA;&lt;li&gt;I can argue to myself that if $a &amp;lt; b$ and $T_b(s)$ exists, then $T_a(s)$ must exist as well. Does there exist a sequence $s_n$ and a $b &amp;gt; 1$ such that $T_a(s)$ exists for every $a &amp;lt; b$, but $T_b(s)$ doesn't?&lt;/li&gt;&#xA;&lt;li&gt;You can't recover an infinite product with this $T$ function. Short of providing a new function $R$ that swaps $\sum$ for $\prod$ and multiplication for addition in the $T_2$ and $T_3$ expansions above, is there an easy way to recover them?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Thank you for your time,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-- Michael Burge&lt;/p&gt;&#xA;" OwnerUserId="5468" LastActivityDate="2011-03-02T19:43:05.113" Title="Has this Extension to a Series been Studied Before?" Tags="&lt;combinatorics&gt;&lt;analysis&gt;&lt;sequences-and-series&gt;&lt;polynomials&gt;&lt;generating-functions&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" />
  <row Id="24684" PostTypeId="2" ParentId="24671" CreationDate="2011-03-02T19:31:34.003" Score="1" ViewCount="" Body="&lt;p&gt;$\theta$ and $\phi$ the coordinates of a spherical surface.  They are similar to latitude ($\theta$) and longitude ($\phi$) except that $\theta$ goes from $0$ to $\pi$ and $\phi$ goes from $0$ to $2\pi$.  Each harmonic has a value at every point, for example $Y_1^{-1}(\theta,\phi)=\frac{1}{2}\sqrt{\frac{3}{2\pi}}\sin(\theta)e^{-i\phi}$.  Given the coordinates you can calculate the value.  The orthogonality is because if you integrate the product of any two different harmonics over the surface of the sphere, you get $0$.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-03-02T19:31:34.003" />
  <row Id="24685" PostTypeId="2" ParentId="24676" CreationDate="2011-03-02T19:31:59.677" Score="5" ViewCount="" Body="&lt;p&gt;In any closed interval subset of the open interval, the function satisfies $$\displaystyle \frac{|f(x) - f(y)|}{|x-y|} \leq C$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;($\displaystyle C$ depending on the interval).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Intuitively, the slope of the line joining $(x, f(x))$ to $(y,f(y))$ increases if you move $x$ or $y$ to the right.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2011-03-02T19:40:23.323" LastActivityDate="2011-03-02T19:40:23.323" />
  <row Id="24686" PostTypeId="1" CreationDate="2011-03-02T19:40:39.497" Score="-2" ViewCount="200" Body="&lt;p&gt;is $(a+1)^2 - a^2 + 2 + (a+1)^2 =(a+2)^2$ part of a bigger equation&lt;/p&gt;&#xA;" OwnerUserId="7712" LastEditorUserId="5105" LastEditorDisplayName="" LastEditDate="2011-03-03T01:32:08.030" LastActivityDate="2011-03-03T01:32:08.030" ClosedDate="2011-03-03T02:33:11.893" Title="consecutive squares" Tags="&lt;algebra-precalculus&gt;" AnswerCount="1" CommentCount="7" />
  <row Id="24687" PostTypeId="2" ParentId="24683" CreationDate="2011-03-02T19:43:05.113" Score="2" ViewCount="" Body="&lt;p&gt;There's a book on the subject: &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0198504500&quot; rel=&quot;nofollow&quot;&gt;Symmetric Functions and Hall Polynomial&lt;/a&gt;s, Macdonald. You can find all sorts of bases for the linear space of all symmetric functions, and they lead to the representation theory of $S_n$.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2011-03-02T19:43:05.113" CommentCount="1" />
  <row Id="24688" PostTypeId="2" ParentId="24686" CreationDate="2011-03-02T19:44:18.220" Score="6" ViewCount="" Body="&lt;p&gt;1) $(a+b)^2-a^2+2b^2+(a+b)^2=(a+2b)^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) $(a+3)^2=3(a+2)^2-3(a+1)^2+a^2$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From which the generalization to a formula for $(a+n)^2$ with $n\ge3$ as a linear combination of $(a+k)^2$ from $k=0$ to $k=n-1$ should be clear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) $(a+3)^2=3(a+1)^2-2a^2+6$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From which the generalization to a formula for $(a+n)^2$ with $n\ge3$ as an affine function of $(a+1)^2$ and $a^2$ should be clear.&lt;/p&gt;&#xA;" OwnerUserId="6179" LastEditorUserId="6179" LastEditorDisplayName="" LastEditDate="2011-03-02T20:01:54.480" LastActivityDate="2011-03-02T20:01:54.480" />
  <row Id="24689" PostTypeId="1" AcceptedAnswerId="24718" CreationDate="2011-03-02T20:04:32.680" Score="5" ViewCount="124" Body="&lt;p&gt;Let $k = \lceil \frac{3 \ln n}{\ln \ln n}\rceil$. How does one show that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \left(\frac{e}{k}\right)^k \frac{1}{1-\frac{e}{k}} \le n^{-2} ? $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is from p. 44 of Motwani and Raghavan, Randomized Algorithms, where they're talking about ball/bin probabilities. The $\left(\frac{e}{x}\right)^x$ is motivated but the $k$ just comes out of nowhere. Ignoring that for the moment, I don't see the derivation of the inequality; even assuming the the 3 is really an approximation for $e$, and throwing out the $\log\log n$... and the geometric series, I get (letting $k^{*}= e \log n$):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;\left(\frac{e}{k^{*}}\right)^{k^{*}}\approx n^{-\log\log n} \le n^{-2}&#xD;&#xA;$$&#xA;for $n \ge e^{e^2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I changed things quite a bit, and this is quite a large constant ($\approx 3^{27}$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So two questions:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;how does one derive the full inequality?&lt;/li&gt;&#xA;&lt;li&gt;why that particular $k$?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1919" LastActivityDate="2011-03-06T21:56:15.797" Title="Inequality on balls/bins with nested logs" Tags="&lt;probability&gt;&lt;inequality&gt;&lt;asymptotics&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" />
  <row Id="24690" PostTypeId="2" ParentId="24662" CreationDate="2011-03-02T20:21:39.470" Score="4" ViewCount="" Body="&lt;p&gt;The comment above, pointing out that the partial sums converge quickly, suggests you can improve your estimate nicely by using a bound for $R_n$, the remainder after the nth partial sum, given by the integral of $q^{x^2}$: &#xA;$$&#xD;&#xA;R_n = \int_n^\infty q^{x^2}\, dx = \sqrt{ \frac{\pi}{2\ln(1/q)}} ( 1- \mathrm{erf}( n \sqrt{\ln(1/q)}))&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even taking only $n=3$, for example, gives (for $q=1/2$)&#xA;$$&#xD;&#xA;0.56401 &amp;lt; \sum_{n=1}^{\infty} q^{x^2} &amp;lt; 0.56489&#xD;&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="4229" LastActivityDate="2011-03-02T20:21:39.470" />
  <row Id="24691" PostTypeId="2" ParentId="24675" CreationDate="2011-03-02T20:26:50.033" Score="4" ViewCount="" Body="&lt;p&gt;As Steve points out, amalgamated free product are key to the Seifert van-Kampen's theorem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other uses:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;HNN Extensions use amalgamated free products. HNN extensions are as follows: given a group $G$ and two subgroups $H$ and $K$ that are isomorphic as groups but not conjugate in $G$, one constructs and overgroup $\mathcal{G}$ of $G$ in which $H$ and $K$ are conjugate. Some consequences of HNN extensions: every countable group can be embedded in a 2-generated group; there exist infinite simple torsion-free groups; there are uncountably many isomorphism classes of finite generated groups; etc. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Embeddability of amalgams. Suppose that in your setting $G$ and $G&amp;#39;$ satisfy some further group-theoretic property (e.g., they are finite, torsion, nilpotent, $p$-groups, solvable, etc). Can you find a group $\mathcal{G}$ that contains both $G$ and $G&amp;#39;$ as subgroups, in which $G\cap G&amp;#39; = H=H&amp;#39;$ (or at least, $G\cap G'\supseteq H=H'$), and $\mathcal{G}$ satisfies the same group theoretic property as $G$ and $G&amp;#39;$? Here the answer depends on the group theoretic property; for example, if $G$ and $G&amp;#39;$ are $p$-groups, you may be unable to find a $p$-group into which you can embed them with the properties above. Finding conditions for embeddability is important in terms of separation properties of morphisms, and the concept of epimorphism in special categories of groups (especially varieties).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Completing &quot;partial groups&quot;. Often one has a structure in which only &lt;em&gt;some&lt;/em&gt; products are defined, and one wants to extend it to a group in which &lt;em&gt;all&lt;/em&gt; products are defined so that you get a group. The amalgamated free product is the key to this.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;They are the group-theoretic categorical pushout of an inclusion diagram. Important in dealing with some notions of category theory within groups.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added.&lt;/strong&gt; Let me add that I do not think your analogy to the Isomorphism Extension Theorem is very apt, at least not if by &quot;Isomorphism Extension Theorem&quot; you are thinking about the same theorem as I am. By the IET I understand the following theorem: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Theorem.&lt;/strong&gt; If $F$ is a field, $E$ is an algebraic extension of $F$, and $f\colon F\rightarrow K$ is a field embedding, then $f$ extends to an embedding $\overline{f}\colon E\to \overline{K}$ of $E$ into the algebraic closure of $K$. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I think you are thinking about the identification between $H$ and $H&amp;#39;$ as playing the role of the map $f$ above; but in the IET you already have a &lt;em&gt;fixed&lt;/em&gt; &quot;overfield&quot; into which $K$ is embedded &lt;em&gt;in a particular way&lt;/em&gt;, and into which you map $E$. Moreover, the extension $\overline{f}\colon E\to \overline{K}$ may &quot;overlap&quot; with $K$ in more than $f(F)$; that is, $\overline{f}(E)\cap K$ may be strictly larger than $f(F)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It would be more analogous to this if what you had was a group $G$ with a subgroup $H$, an embedding of $H$ into $G&amp;#39;$, and a group $\overline{G&amp;#39;}$, &lt;em&gt;independent of&lt;/em&gt; $G$, into which $G$ could then be embedded extending the the map $H\hookrightarrow G&amp;#39;$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think that if you want to find an analogous theorem in fields, it would be that given a field $F$ and two algebraic extensions $K$ and $L$, there exists an extension $M$ of $F$ into which you can embed $K$ and $L$ (in such a way that $K\cap L\supseteq F$), and which is &quot;universal&quot; in the sense that any field which contains an image of $L$ and of $K$ will contain an image of $M$ (this field, of course, is the compositum of $L$ and $K$ in the algebraic closure of $F$). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This makes a bit more sense in that both are instances of pushouts, even though in the field case you need not have a &lt;em&gt;strong&lt;/em&gt; embedding (one in which the intersection of the two objects is exactly their &quot;common&quot; object, and not something larger). &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-03T04:44:42.447" LastActivityDate="2011-03-03T04:44:42.447" />
  <row Id="24692" PostTypeId="2" ParentId="24662" CreationDate="2011-03-02T21:00:30.597" Score="15" ViewCount="" Body="&lt;p&gt;As Qiaochu says, the &lt;a href=&quot;http://en.wikipedia.org/wiki/Euler%E2%80%93Maclaurin_formula&quot; rel=&quot;nofollow&quot;&gt;Euler-Maclaurin formula&lt;/a&gt; is useful here.  It gives the error in the approximation &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{n=0}^{\infty} q^{n^2} \approx \sqrt{\frac{\pi}{4 \log (1/q)}} + \frac{1}{2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;to be exactly &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \lim_{a \to \infty}  \int_0^a 2xq^{x^2} (\log q) \left(x - \lfloor x \rfloor - \frac{1}{2}\right) dx.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $q^{x^2} \to 0$ quite rapidly, the value of this integral can be approximated closely by using small values of $a$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, if $q = 1/2$, the integral (i.e., the error in the approximation) evaluates to $1.39417 \times 10^{-6}$ (via Mathematica).&lt;/p&gt;&#xA;" OwnerUserId="2370" LastActivityDate="2011-03-02T21:00:30.597" />
  <row Id="24693" PostTypeId="1" CreationDate="2011-03-02T21:27:03.303" Score="2" ViewCount="122" Body="&lt;p&gt;I would like to learn some differential geometry: basically manifolds, differentiable manifolds, smooth manifolds, De Rham cohomology and everything else that is pretty much part of a course in differential geometry. I do however know some deal of category theory and algebraic geometry, and I would therefore like to learn differential geometry from a more &quot;abstract&quot; (categorical and algebraical) setting. Are there any good books for this? I was able to find a book called &quot;Sheaves on Manifolds&quot; but I don't know if it is a good book for learning the subject (AFAIK, the book might assume prior knowledge of differential geometry) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;/edit/ Or just lecture notes.&lt;/p&gt;&#xA;" OwnerUserId="7713" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2011-03-02T21:53:04.267" LastActivityDate="2011-03-02T21:53:04.267" Title="Want to learn differential geometry and want the sheaf perspective" Tags="&lt;differential-geometry&gt;" CommentCount="6" />
  <row Id="24694" PostTypeId="2" ParentId="24141" CreationDate="2011-03-02T21:39:52.630" Score="2" ViewCount="" Body="&lt;p&gt;You are asked to prove that&#xA;$$&#xD;&#xA;\int_{ - \infty }^\infty  {V_{n - 1} (s + x)dF(x)}  &amp;gt; s - c,&#xD;&#xA;$$&#xA;for all $n \geq 1$. For $n=1$, substituting from the definition of $V_0$, you need to show that&#xA;$$&#xD;&#xA;\int_{ - \infty }^\infty  {\max \lbrace s + x - c,0\rbrace dF(x)}  &amp;gt; s - c.&#xD;&#xA;$$&#xA;For this purpose, first note that &#xA;$$&#xD;&#xA;\max \lbrace s + x - c,0\rbrace  \ge s + x - c.&#xD;&#xA;$$&#xA;Then, by linearity of the integral, you can consider the sum&#xA;$$&#xD;&#xA;\int_{ - \infty }^\infty  {(s - c)dF(x)}  + \int_{ - \infty }^\infty  {xdF(x)} ,&#xD;&#xA;$$&#xA;from which the assertion for $n=1$ follows. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To complete the inductive proof, substituting from the definition of $V_n$, you need to show that&#xA;$$&#xD;&#xA;\int_{ - \infty }^\infty  {\max \bigg\lbrace s + x - c,\int_{ - \infty }^\infty  {V_{n - 1} (s + x + u)dF(u)} \bigg\rbrace dF(x)} &amp;gt; s - c,&#xD;&#xA;$$&#xA;under the induction hypothesis that, for any $s \in \mathbb{R}$,&#xA;$$&#xD;&#xA;\int_{ - \infty }^\infty  {V_{n - 1} (s + x)dF(x)}  &amp;gt; s - c.&#xD;&#xA;$$ &#xA;For this purpose, recall the end of the proof for $n=1$. &lt;/p&gt;&#xA;" OwnerUserId="2810" LastActivityDate="2011-03-02T21:39:52.630" />
  <row Id="24695" PostTypeId="2" ParentId="24673" CreationDate="2011-03-02T21:45:16.027" Score="2" ViewCount="" Body="&lt;p&gt;You can define the &quot;order&quot; of a regular function at a point $x$ to be the length of the quotient: $\mathcal O_{X,x}/(f)$ (which will be Artinian, so has finite length). You can probably find the details in Hartshorne's book or Fulton's &quot;Intersection theory&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Details added: here is a sketch that this is well-defined: Suppose your rational function can be represented by $f/g $ and $f&amp;#39;/g&amp;#39;$ at the stalk, which I will call $R$. $R$ is one-dimensional domain (you can assume less, but let's make it simple). We know $f/g =f&amp;#39;/g&amp;#39;$ and want to prove $l(R/fR) - l(R/gR) = l(R/f&amp;#39;R) - l(R/g&amp;#39;R)$ or &#xA;$$l(R/fR) + l(R/g&amp;#39;R) = l(R/f&amp;#39;R) + l(R/gR)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $fg&amp;#39; = f&amp;#39;g$, we are done by the following general fact: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$l(R/abR) = l(R/aR) + l(R/bR) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint: look at the sequence $0 \to aR/abR \to R/abR \to R/aR \to 0$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS: One can also give a definition by using the &lt;em&gt;normalization&lt;/em&gt; of $X$, but I think the above is more down-to-earth and computable, if less sexy.  &lt;/p&gt;&#xA;" OwnerUserId="223" LastEditorUserId="223" LastEditorDisplayName="" LastEditDate="2011-03-03T15:34:20.943" LastActivityDate="2011-03-03T15:34:20.943" CommentCount="3" />
  <row Id="24696" PostTypeId="1" CreationDate="2011-03-02T21:52:40.840" Score="3" ViewCount="141" Body="&lt;p&gt;A while back when I was taking a first year cs course, our professor had us write the algorithm for insertion sort in The Scheme programming language. There were also several other similar recursion type programs for us to write, and one of the question in the assignment was to prove that the programs are correct using Mathematical Induction. At first I tried to justify my program using &quot;Weak Induction&quot;, however my professor insisted on me using &quot;Strong Induction&quot; and I lost marks for not using it. I was not confident at the time to engage in an argument or ask for reasons why Strong over Weak and this has been bugging me ever since. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the question is this: Using weak induction in the ordinary sense such as, proving n = 0 for base case and proving the inductive step of the algorithm, was there anything wrong other than pedagogy? If so, could you enlighten me by explaining why there was a need to use strong induction to prove recursion type algorithms such as insertion sort?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance&lt;/p&gt;&#xA;" OwnerUserId="7667" LastEditorUserId="7667" LastEditorDisplayName="" LastEditDate="2011-03-02T21:58:14.817" LastActivityDate="2011-03-02T22:03:10.070" Title="Proving insertion sort using induction" Tags="&lt;computer-science&gt;&lt;proof-strategy&gt;&lt;induction&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="24697" PostTypeId="1" AcceptedAnswerId="25650" CreationDate="2011-03-02T22:00:55.440" Score="3" ViewCount="100" Body="&lt;p&gt;I am stuck proving the theorem that there exists $x$, $x^4 \equiv 2 \pmod p$ iff $p$ is of the form $A^2 + 64B^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So far I have got this (and I am not sure if it's correct)&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Let $p = a^2 + b^2$ be an odd prime,&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\left(\frac{a}{p}\right) = \left(\frac{p}{a}\right) = \left(\frac{a^2 + b^2}{a}\right) = \left(\frac{b^2}{a}\right) = 1$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;since $p \equiv 1 \pmod 4$&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\left(\frac{a+b}{p}\right) = \left(\frac{(a+b)^2-2ab}{a+b}\right) = \left(\frac{2}{a+b}\right) = (-1)^{((a+b)^2-1)/8}$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;using the Jacobi symbol and second supliment of quadratic reciprocity.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$(a+b)^{(p-1)/2} = (2ab)^{(p-1)/4}$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;since $(a+b)^2 \equiv 2ab \pmod p$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;and the last step which I'm stuck on now is for $p = a^2 + b^2$ let $x^2 \equiv -1 \pmod p$ then $2^{(p-1)/4} = x^{ab/2}$. And I don't see how to prove the theorem with this result.&lt;/p&gt;&#xA;" OwnerUserId="4290" LastActivityDate="2011-03-07T23:05:53.100" Title="Help complete a proof of Dirichlet on biquadratic character of 2?" Tags="&lt;number-theory&gt;&lt;quadratic-reciprocity&gt;" AnswerCount="1" />
  <row Id="24698" PostTypeId="2" ParentId="24696" CreationDate="2011-03-02T22:03:10.070" Score="1" ViewCount="" Body="&lt;p&gt;This is certainly an issue of pedagogy.  With Strong Induction, one doesn't have to mention a base case explicitly.  However, one has to be able to consider an empty set of inductive hypotheses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See:  &lt;a href=&quot;http://en.wikipedia.org/wiki/Strong_induction#Complete_induction&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Strong_induction#Complete_induction&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It would seem that a proof that uses Strong Induction as a tool is actually a weaker proof!&lt;/p&gt;&#xA;" OwnerUserId="6509" LastActivityDate="2011-03-02T22:03:10.070" />
  <row Id="24699" PostTypeId="1" AcceptedAnswerId="24740" CreationDate="2011-03-02T21:16:19.563" Score="4" ViewCount="319" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Darwinia_%28video_game%29&quot; rel=&quot;nofollow&quot;&gt;Darwinia&lt;/a&gt; features an intro which represents a modified version of &lt;a href=&quot;http://en.wikipedia.org/wiki/Conway%27s_Game_of_Life&quot; rel=&quot;nofollow&quot;&gt;Conway's Game of Life&lt;/a&gt;. &#xA;&lt;a href=&quot;http://www.youtube.com/watch?v=RIqPWw0sqOI&quot; rel=&quot;nofollow&quot;&gt;You can see it in action here.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The game developers added one more rule about the game: &lt;strong&gt;no cell may live longer than 50 generations.&lt;/strong&gt; This is supposed to kill all the cells eventually, which is what does happens in the game.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, when I tried to model this behaviour in my own implementation of the Game of Life, I've got different results. When four dots live together, they usually don't die. When one of them passes away, a new dot gets born in the next generations due to three other dots. Unless two or more dots disappear at the same time, the block will keep itself alive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the question is, why do four-dot blocks die in Darwinia and not die in my case?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update:&lt;/strong&gt;&#xA;To be clear, the starting pattern seems to be completely random. During the intro you may press 'g', and everything starts all over again with a random pattern. Also I've changed the link to the video, where you may see a longer version.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update 2:&lt;/strong&gt;&#xA;Tried to add the age of the cell into the birth mechanism, so that a cell past certain age can't give birth to new cells. It definitely works, with the lifespan of 50 generations, setting the last age when a cell can give birth to a new one to, say, 48 generations, works in most cases. Not sure whether the Introversion Software guys used rules like this, though, but it works.&lt;/p&gt;&#xA;" OwnerUserId="7732" LastEditorUserId="7732" LastEditorDisplayName="" LastEditDate="2011-03-06T21:46:01.357" LastActivityDate="2011-03-06T21:46:01.357" Title="How can I change the Conway's Game of Life so that, eventually, all cells die?" Tags="&lt;algorithms&gt;" AnswerCount="4" CommentCount="8" />
  <row Id="24700" PostTypeId="2" ParentId="24699" CreationDate="2011-03-02T22:34:19.587" Score="2" ViewCount="" Body="&lt;p&gt;If a four block decays depends on whether two of the cells were created at the same time or not. So it depends on the initial configuration, which were probably different in your case.&lt;/p&gt;&#xA;" OwnerUserId="4929" LastActivityDate="2011-03-02T22:34:19.587" CommentCount="1" />
  <row Id="24701" PostTypeId="1" CreationDate="2011-03-02T22:36:15.663" Score="8" ViewCount="188" Body="&lt;p&gt;I need an NPDA for the following language if it is context-free, and if it isn't I need a proof using the pumping lemma that it is not a CFL:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$L_1=\{w_1w_2 \in \{a,b\}^* : |w_1| = |w_2|,w_1\neq w_2\}$$&lt;/p&gt;&#xA;" OwnerUserId="7716" LastEditorUserId="6296" LastEditorDisplayName="" LastEditDate="2011-03-04T07:42:08.903" LastActivityDate="2011-05-13T19:57:17.917" Title="is this language context free?" Tags="&lt;computer-science&gt;&lt;formal-systems&gt;" AnswerCount="1" CommentCount="9" FavoriteCount="2" />
  <row Id="24702" PostTypeId="1" AcceptedAnswerId="24706" CreationDate="2011-03-02T22:46:42.110" Score="3" ViewCount="87" Body="&lt;p&gt;A submersion between smooth manifolds is an open map. Is the converse true? That is, is a smooth open map $f:M\to N$ between smooth manifolds a submersion? We can additionally assume that it is surjective, if necessary, because that is the only case I am interested in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can see how it is almost true, taking a chart $U$ about $m\in M$ and considering a chart $V$ about $f(m)\in N$ contained in the image of $U$ (which is open). Using the isomorphism between the charts and the tangent spaces, I'm fairly sure this gives us a submersion, but I feel there is a slight gap in my argument. So either, does my argument work, or is it true but my argument is incomplete, or is it false?&lt;/p&gt;&#xA;" OwnerUserId="3835" LastActivityDate="2011-03-02T23:12:47.980" Title="Given a smooth map which is open, is it a submersion?" Tags="&lt;differential-geometry&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="1" />
  <row Id="24703" PostTypeId="1" AcceptedAnswerId="24705" CreationDate="2011-03-02T23:04:42.673" Score="3" ViewCount="95" Body="&lt;p&gt;Given a published mathematics article, is there a way to find its tree of subsequent citations starting from the date of publication? Does such a (reverse bibliography) database exist? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I'm at a student in a relatively well-known university with access to virtually every mathematics journal imaginable, but I can't seem to find such a simple and useful tool.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="2720" LastEditorUserId="2720" LastEditorDisplayName="" LastEditDate="2011-03-02T23:29:11.390" LastActivityDate="2011-03-02T23:29:11.390" Title="A Reverse Citation Index" Tags="&lt;reference-request&gt;" AnswerCount="3" FavoriteCount="1" />
  <row Id="24704" PostTypeId="1" AcceptedAnswerId="24711" CreationDate="2011-03-02T23:09:42.987" Score="8" ViewCount="200" Body="&lt;p&gt;It seems that often in using counting arguments to show that a group of a given order cannot be simple, it is shown that the group must have at least $n_p(p^n-1)$ elements, where $n_p$ is the number of Sylow p-subgroups.  It is explained that the reason this is the case is because distinct Sylow p-subgroups intersect only at the identity, which somehow follows from Lagrange's Theorem.  I cannot see why this is true.  Can anyone quicker than I tell me why, I know it's probably very obvious.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: This isn't a homework question, so if the answer is obvious I'd really just appreciate knowing why.  Thanks!&lt;/p&gt;&#xA;" OwnerUserId="3595" LastEditorUserId="8575" LastEditorDisplayName="" LastEditDate="2011-05-19T09:10:00.580" LastActivityDate="2011-05-19T09:10:00.580" Title="Simple Group Theory Question Regarding Sylow Theorems" Tags="&lt;group-theory&gt;&lt;simple-groups&gt;" AnswerCount="4" CommentCount="2" FavoriteCount="2" />
  <row Id="24705" PostTypeId="2" ParentId="24703" CreationDate="2011-03-02T23:12:14.797" Score="4" ViewCount="" Body="&lt;p&gt;There are several, including the public &lt;a href=&quot;http://citeseer.ist.psu.edu/&quot; rel=&quot;nofollow&quot;&gt;CiteSeerX&lt;/a&gt; and &lt;a href=&quot;http://scholar.google.com/&quot; rel=&quot;nofollow&quot;&gt;Google Scholar&lt;/a&gt; as well as many commercial databases.&lt;/p&gt;&#xA;" OwnerUserId="6460" LastActivityDate="2011-03-02T23:12:14.797" />
  <row Id="24706" PostTypeId="2" ParentId="24702" CreationDate="2011-03-02T23:12:47.980" Score="3" ViewCount="" Body="&lt;p&gt;A nonconstant holomorphic function is an open map on $\mathbb{C}\cong\mathbb{R^2}$, but is a submersion only if its derivative is never zero.  So for example, $z\mapsto z^2$, a.k.a. $(x,y)\mapsto (x^2-y^2,2xy)$ is a counterexample.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2011-03-02T23:12:47.980" />
  <row Id="24707" PostTypeId="2" ParentId="24676" CreationDate="2011-03-02T23:14:25.007" Score="1" ViewCount="" Body="&lt;p&gt;elaborating on another answer: for $x\leq y\leq z$ we have&#xA;$$&#xD;&#xA;\frac{f(y)-f(x)}{y-x}\leq\frac{f(z)-f(x)}{z-x}&#xD;&#xA;$$&#xA;employing convex combination (i.e. $y=tx+(1-t)z, t\in[0,1]$)&#xA;$$&#xD;&#xA;y=\frac{z-y}{z-x}x+\frac{y-x}{z-x}z&#xD;&#xA;$$&#xA;and the definition of convexity&#xA;$$&#xD;&#xA;f(tx+(1-t)y)\leq tf(x)+(1-t)f(y).&#xD;&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="6925" LastEditorUserId="6925" LastEditorDisplayName="" LastEditDate="2011-03-02T23:37:36.843" LastActivityDate="2011-03-02T23:37:36.843" />
  <row Id="24708" PostTypeId="2" ParentId="24703" CreationDate="2011-03-02T23:14:41.777" Score="2" ViewCount="" Body="&lt;p&gt;MathSciNet has a &lt;em&gt;From References&lt;/em&gt; link that you can use to build the tree yourself, assuming your article is in MathSciNet.&lt;/p&gt;&#xA;" OwnerUserId="589" LastActivityDate="2011-03-02T23:14:41.777" />
  <row Id="24709" PostTypeId="2" ParentId="24676" CreationDate="2011-03-02T23:16:28.507" Score="1" ViewCount="" Body="&lt;p&gt;For each $x_0 \in C$, $f$ is both left- and right-differentiable at $x_0$, that is,&#xA;$$&#xD;&#xA;\mathop {\lim }\limits_{x \uparrow x_0 } \frac{{f(x) - f(x_0 )}}{{x - x_0 }} \in \mathbb{R} \;\; {\rm and} \;\; \mathop {\lim }\limits_{x \downarrow x_0 } \frac{{f(x) - f(x_0 )}}{{x - x_0 }} \in \mathbb{R},&#xD;&#xA;$$&#xA;respectively. For a proof, see Theorem 14.5(a) on p. 248 of &lt;a href=&quot;http://books.google.co.il/books?id=ErtmdgxSxA8C&amp;amp;printsec=frontcover&amp;amp;dq=%22lectures+on+real+analysis%22&amp;amp;source=bl&amp;amp;ots=2QKGe9G6wG&amp;amp;sig=b9_tyT5XsjYnEC1rycx3lTIRTxg&amp;amp;hl=iw#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;this book&lt;/a&gt; or Theorem 1(2) &lt;a href=&quot;http://elib.mi.sanu.ac.rs/files/journals/tm/20/tm1112.pdf&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;. It follows that $f$ is both left- and right-continuous at $x_0$, hence continuous there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remark: A convex function on a closed interval need not be continuous at the end points (for example, the function $f$ on $[0,1]$ defined by $f(x)=x^2$ for $|x|&amp;lt;1$, $f(x)=2$ for $x \in \lbrace -1,1 \rbrace)$.&lt;/p&gt;&#xA;" OwnerUserId="2810" LastActivityDate="2011-03-02T23:16:28.507" />
  <row Id="24710" PostTypeId="2" ParentId="24703" CreationDate="2011-03-02T23:17:10.730" Score="4" ViewCount="" Body="&lt;p&gt;If you go to the AMS MathSciNet database, from each paper you can access to the list of papers citing it by clicking on the &quot;From Reference&quot; link from a &quot;Citation&quot; box in the upper right of the page.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See for istance the 255 papers citing Pierre Deligne's first paper on the Weil Conjectures here: &#xA;&lt;a href=&quot;http://www.ams.org/mathscinet/search/publications.html?refcit=340258&amp;amp;loc=refcit&quot; rel=&quot;nofollow&quot;&gt;http://www.ams.org/mathscinet/search/publications.html?refcit=340258&amp;amp;loc=refcit&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="688" LastActivityDate="2011-03-02T23:17:10.730" />
  <row Id="24711" PostTypeId="2" ParentId="24704" CreationDate="2011-03-02T23:21:04.923" Score="8" ViewCount="" Body="&lt;p&gt;That's because it is not true in general. Look at $2$-Sylows in $S_5$: they have nontrivial intersection.&lt;/p&gt;&#xA;" OwnerUserId="2660" LastActivityDate="2011-03-02T23:21:04.923" />
  <row Id="24712" PostTypeId="2" ParentId="24704" CreationDate="2011-03-02T23:21:22.077" Score="3" ViewCount="" Body="&lt;p&gt;Suppose $P$ and $Q$ are Sylow p-subgroups of prime order p (so not just any power of p; as others remarked, then it is not true in general). Note that $P\cap Q$ is a subgroup of $P$ (and of $Q$). So by Lagrange, the order $|P\cap Q|$ divides p. As p is prime, it is 1 or p. But it cannot be p, as $P$ and $Q$ are distinct. So $|P\cap Q|=1$ and consequently the intersection is trivial.&lt;/p&gt;&#xA;" OwnerUserId="6490" LastActivityDate="2011-03-02T23:21:22.077" CommentCount="3" />
  <row Id="24713" PostTypeId="1" AcceptedAnswerId="24714" CreationDate="2011-03-02T23:24:26.237" Score="1" ViewCount="46" Body="&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/dXBu1.png&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(sorry, low level so I can't attach the image in the post)&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;a is given.&lt;/li&gt;&#xA;&lt;li&gt;b is given.&lt;/li&gt;&#xA;&lt;li&gt;c is given.&lt;/li&gt;&#xA;&lt;li&gt;For X, I don't have a clue where he&#xA;is.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I was thinking about rect / vector instersection, projection, and other things but still I couldn't come with a solution. If it's computer light, the better.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sorry for not making clear before: ab and bx are not always orthogonal.&lt;/p&gt;&#xA;" OwnerUserId="7717" LastEditorUserId="5363" LastEditorDisplayName="" LastEditDate="2011-03-02T23:37:28.660" LastActivityDate="2011-03-02T23:37:28.660" Title="Trying to find a point X" Tags="&lt;algebra&gt;&lt;vector-calculus&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="24714" PostTypeId="2" ParentId="24713" CreationDate="2011-03-02T23:28:41.047" Score="0" ViewCount="" Body="&lt;p&gt;From the drawing it seems that the segments $ab$ and $bx$ are orthogonal. So $x$ is the intersection of $ac$ and the line orthogonal to $ab$ that passes through $b$.&lt;/p&gt;&#xA;" OwnerUserId="589" LastActivityDate="2011-03-02T23:28:41.047" CommentCount="2" />
  <row Id="24715" PostTypeId="2" ParentId="24704" CreationDate="2011-03-02T23:31:53.750" Score="1" ViewCount="" Body="&lt;p&gt;It simply isn't true, Sylow p-subgroups can very well intersect non-trivially, Plop gave an example thereof.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, it seems like you actually cannot say the following, see comments. I'm just leaving it here as a mistake one shouldn't make, so I won't mind if a moderator deletes it since it's not an actual answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[wrong]You could say that the number of elements of order $p$ is at least $n_p(p^n-p^{n-1})+p^{n-1}$,  which is the case when all $p$-groups intersect maximally. [/wrong] Note that in this case however the intersection of all Sylow $p$-subgroups is a normal subgroup (even a  characteristic subgroup, it is called $\mathbf O_p(G)$), so this cannot occur in a simple group.&lt;/p&gt;&#xA;" OwnerUserId="5189" LastEditorUserId="5189" LastEditorDisplayName="" LastEditDate="2011-03-03T01:43:47.647" LastActivityDate="2011-03-03T01:43:47.647" CommentCount="8" />
  <row Id="24716" PostTypeId="2" ParentId="24713" CreationDate="2011-03-02T23:34:21.267" Score="0" ViewCount="" Body="&lt;p&gt;You have a triangle with vertices $a$, $b$ and $x$ which looks right-angled at $b$; you also have a point on the hypotenuse $c$.  So you need to draw the perpendicular line to $ab$ at $b$ and extend the line $ac$; where the two lines intersect must be $x$ (unless $ac$ is perpendicular to $ab$). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want a more detailed solution, than you could give us some information about how you are expressing the locations of $a$, $b$ and $c$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT in light of comment.  If you know nothing except the location of $a$, $b$ and $c$, not even the angle at $b$, then $x$ could be anywhere on the line $ac$ (possibly required to be beyond $c$)&lt;/p&gt;&#xA;" OwnerUserId="6460" LastActivityDate="2011-03-02T23:34:21.267" CommentCount="1" />
  <row Id="24717" PostTypeId="1" AcceptedAnswerId="24732" CreationDate="2011-03-02T23:40:10.227" Score="4" ViewCount="72" Body="&lt;p&gt;Suppose there is a second-order real-state stochastic process $X: \Omega \times T \rightarrow \mathbb{R}$ with $T= \mathbb{R}$ and probability space $(\Omega, \mathcal{F}, P)$. I was wondering if the following inequality holds for integrals over an interval $[a,b] \subset \mathbb{R}$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \Vert \int_a^b X_t dt \Vert_2 \leq \int_a^b \Vert X_t \Vert_2 dt \ ?$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the integral on LHS is a stochastic integral while the one on RHS is a deterministic one. $L_2$ norm is on $L_2$ space of squared-integrable random variables as measurable mappings from $(\Omega, \mathcal{F}, P)$ to $(\mathbb{R}, \mathcal{B})$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this &lt;a href=&quot;http://en.wikipedia.org/wiki/Jensen%27s_inequality#Measure-theoretic_and_probabilistic_form&quot; rel=&quot;nofollow&quot;&gt;Jensen's inequality&lt;/a&gt;? I don't think it is, because although the $L_2$ norm is convex, it is not a mapping from $\mathbb{R}$ to $\mathbb{R}$, but from $L_2(\Omega, \mathcal{F})$ to $\mathbb{R}$, and the types of integrals on LHS and on RHS are not the same. So I was wondering if this inequality is true and why?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can it be seen as a generalization of Jensen's inequality? If yes, is it only because of the similarity of their forms or the similarity of some deeper things?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks and regards!&lt;/p&gt;&#xA;" OwnerUserId="1281" LastEditorUserId="1281" LastEditorDisplayName="" LastEditDate="2011-03-03T01:02:15.273" LastActivityDate="2011-03-03T01:28:48.147" Title="Exchanging order of stochastic integral and $L_2$ norm" Tags="&lt;probability-theory&gt;&lt;stochastic-processes&gt;" AnswerCount="1" CommentCount="7" />
  <row Id="24718" PostTypeId="2" ParentId="24689" CreationDate="2011-03-02T23:44:23.080" Score="6" ViewCount="" Body="&lt;p&gt;The inequality does not hold. If you set $n=5.6 \times 10^{6}$ (close to $ e^{e^e}$) with $k=17$, you will find that the left hand side is larger than $3.4 \times 10^{-14}$ and the right hand side is smaller than $3.2 \times 10^{-14}$. To see why the inequality is wrong and why $k=\lceil \frac{4 \log n}{\log \log n} \rceil$ works, read on:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To get a &quot;good&quot; value for $k$, you want that the inequality becomes asymptotically $n\to\infty$ and equality. That is how you derive the particular form. Taking log of your inequality, we get (i switched the sign of the equation because I rather like positive numbers)&#xA;$$ L= k (\log k -1) +\log(1- e/k) \geq  2 \log n .$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most important term (for $n,k \to \infty$) is $k\log k$. So we would like to have&#xA;$$ k\sim \frac{2 \log n}{\log k}.$$&#xA;This can be solved iteratively. The first approximation is $k_0 = 2 \log n$. We set this approximation in the right hand side and obtain $k_1 = 2 \log n/ \log \log n$ (up to an irrelevant additive term $\log 2$ in the denominator). This is already almost the estimate, changing the 2 to 3 and the ceiling function are needed to have some space and make the inequality valid (and not only the expressions asymptotically equal). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, let us check the inequality and set $k= c \log n /\log \log n$ ($c$ we will determine in the end). We obtain&#xA;$$ L= \frac{c \log n}{\log \log n} ( \log c + \log \log n - \log \log \log n -1) + \log (1- e \log \log n /c \log n).$$ We will use that $(\log c -1) &amp;gt; 0$ (for $c&gt;e$), $(\log \log \log n / \log \log n) \leq 1/e$ (with the maximum attained for $n=e^{e^e}$), and $\log (1- e \log \log n /c \log n)&amp;gt;\log(c-1)-\log c$ (with the minimum at $n=e^e$). Thereby, we can show that $$L \geq c  (1- 1/e) \log n + \log (c-1) - \log c .$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The lowest value of $c$ for which we can hope that the inequality is always fulfilled is given by $c(1-1/e) =2$, i.e., $c= 2e/(e-1) \approx 3.2$. Assuming that $n&amp;gt;e$ such that $\log n &amp;gt;1$, we can further bound the expression&#xA;$$L \geq [c (1-1/e) + \log (c-1) - \log c ] \log n \geq 2 \log n$$&#xA;if $c\gtrsim 3.67$.&lt;/p&gt;&#xA;" OwnerUserId="7266" LastEditorUserId="7266" LastEditorDisplayName="" LastEditDate="2011-03-06T21:56:15.797" LastActivityDate="2011-03-06T21:56:15.797" CommentCount="5" />
  <row Id="24719" PostTypeId="2" ParentId="24675" CreationDate="2011-03-02T23:54:15.517" Score="2" ViewCount="" Body="&lt;p&gt;Another important use of amalgamated products arises from groups acting on trees. One easy way to interpret $G=\mbox{SL}_2(\mathbb{Z})$ as an amalgam is to observe how it naturally acts on a tree embedded in the upper half plane $H$ (the tree being simply the boundary of the tessellation of $H$ by fundamental domains of the action of $G$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See Serre's &quot;Trees&quot; (&quot;Arbres, Amalgames, SL_2&quot;), or &lt;a href=&quot;http://www.math.okstate.edu/~araghur/trees.pdf&quot; rel=&quot;nofollow&quot;&gt;these nice lecture notes&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="308" LastActivityDate="2011-03-02T23:54:15.517" />
  <row Id="24720" PostTypeId="2" ParentId="22859" CreationDate="2011-03-02T23:59:21.200" Score="3" ViewCount="" Body="&lt;p&gt;Actually, the continuity of the partials is not &lt;em&gt;needed&lt;/em&gt;, twice total differentiability &lt;em&gt;at the point&lt;/em&gt; is sufficient, so the one and the higher dimensional cases are totally analogous. &lt;em&gt;But&lt;/em&gt; the existence of the second order partials is &lt;em&gt;insufficient&lt;/em&gt; for twice total differentiability, so the Hessian is not necessarily the second total derivative just because the second partials exist, and you can compute the Hessian. But their continuity is a simple, sufficient condition for twice differentiability.&lt;/p&gt;&#xA;" OwnerUserId="7598" LastActivityDate="2011-03-02T23:59:21.200" />
  <row Id="24721" PostTypeId="2" ParentId="24653" CreationDate="2011-03-03T00:01:25.317" Score="0" ViewCount="" Body="&lt;p&gt;There is no choice issue here at all, and no need to pick a particular radius, since in the definition of open set, there is no obligation for us to assign to each point a unique radius. Rather, we can assign many radii to each point, and this will also verify opennes. We just need to know that every point has at least one ball around it contained in the set; perhaps there are many balls around that point contained in the set (and indeed, of course any smaller ball also works). So let us simply take all balls around any point of any possible radius to see that the whole space is open.&lt;/p&gt;&#xA;" OwnerUserId="413" LastEditorUserId="413" LastEditorDisplayName="" LastEditDate="2011-03-03T00:14:55.447" LastActivityDate="2011-03-03T00:14:55.447" />
  <row Id="24722" PostTypeId="1" AcceptedAnswerId="24724" CreationDate="2011-03-03T00:02:25.877" Score="7" ViewCount="164" Body="&lt;p&gt;In general, it is not too hard to find isomorphisms between two groups when their order is relatively low. However, as their orders grow, it becomes increasingly irritating to write down their entire Cayley tables and such. Is there a set of tricks that is generally useful when trying to prove that two groups are actually isomorphic? After all, it usually seems easier to prove that they aren't, as you just need to point out one property that doesn't correspond...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example: in Armstrong's Groups and Symmetry, it is asked to show that the dihedral group of order 8 and the subgroup of S4 generated by (1234) and (24) are isomorphic. It is easy to send D4's &quot;single-rotation&quot; element r to S4's (1234) and D4's &quot;flipping&quot; element s to S4's (24), as they are all part of the generating set and their orders coincide, but what is the way to go from here? Also, how far should one go in showing the isomorphism - might pointing out the correspondence in generating elements even be enough?&lt;/p&gt;&#xA;" OwnerUserId="1730" LastActivityDate="2011-03-03T00:39:54.613" Title="What are useful tricks for determining whether groups are isomorphic?" Tags="&lt;group-theory&gt;" AnswerCount="3" FavoriteCount="2" />
  <row Id="24723" PostTypeId="1" CreationDate="2011-03-03T00:11:01.400" Score="2" ViewCount="135" Body="&lt;p&gt;Hello, I am trying to prove the recurrence $2T(n-1) + 1$ has the bound $\theta(2^{n})$. $T(1) = \theta (1)$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My attempted solution:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align*}&#xA;T(n) &amp;amp;= 2T(n-1) + 1 \\&#xA;&amp;amp;= 2 \{ 2T(n-2) + 1 \} + 1 \\&#xA;&amp;amp;= 2 \{ 2 \{ 2T( n - 3 ) + 1 \} + 1 \} + 1 \\&#xA;&amp;amp; \ldots \\&#xA;&amp;amp;= 2^{i}T(n-i)+\sum_{i=0}^{k} 2^{i} + 1.&#xA;\end{align*}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I then proceed to substitute $n-1$ for $i$ which gives me&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align*}&#xA;&amp;amp; 2^{n-1}T(n-(n-1)) + \sum_{i=0}^{n-1} 2^{i} + 1 \\&#xA;&amp;amp;= 2^{n-1}T(n-n+1) + \sum_{i=0}^{n-1} 2^{i} + \sum_{i=0}^{n-1} 1 \\&#xA;&amp;amp;= 2^{n-1}T(1) + 2^{n-1+1} + n - 1 \\&#xA;&amp;amp;= 2^{n-1}\theta(1) + 2^{n} + n - 1\\&#xA;\end{align*}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At this point I can see that $2^{n}$ is the dominating factor, however in the solution hints of my book it states that I should use $i = n$. However when I do that I get $T(0)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suspect there is something wrong with my substitution/unraveling step, however I can't seem to figure it out. Any help would be appreciated&lt;/p&gt;&#xA;" OwnerUserId="7718" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2011-06-03T13:48:00.923" LastActivityDate="2011-06-03T13:48:00.923" Title="Proving the bound on a recurrence relation" Tags="&lt;algorithms&gt;&lt;asymptotics&gt;&lt;recurrence-relations&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="24724" PostTypeId="2" ParentId="24722" CreationDate="2011-03-03T00:17:32.970" Score="6" ViewCount="" Body="&lt;p&gt;Proving that two groups are isomorphic is a provably hard problem, in the sense that the &lt;a href=&quot;http://en.wikipedia.org/wiki/Group_isomorphism_problem&quot; rel=&quot;nofollow&quot;&gt;group isomorphism problem&lt;/a&gt; is undecidable. Thus there is literally no general algorithm for proving that two groups are isomorphic. To prove that two &lt;em&gt;finite&lt;/em&gt; groups are isomorphic one can of course run through all possible maps between the two, but that's not fun in general.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For your particular example, there's not much to say. It depends on what you know about $D_4$. If you know a presentation of it, then you can prove that what you've defined is actually a homomorphism which is moreover surjective. Since the two groups have the same order, you're done.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For recognizing a finite group via a presentation you can sometimes use the &lt;a href=&quot;http://en.wikipedia.org/wiki/Todd%E2%80%93Coxeter_algorithm&quot; rel=&quot;nofollow&quot;&gt;Todd-Coxeter algorithm&lt;/a&gt;. &lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2011-03-03T00:17:32.970" />
  <row Id="24725" PostTypeId="2" ParentId="24722" CreationDate="2011-03-03T00:18:07.713" Score="3" ViewCount="" Body="&lt;p&gt;There are a few ways to show that a homomorphism of groups is an isomorphism:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;A homomorphism which is a bijection of sets is an isomorphism.&lt;/li&gt;&#xA;&lt;li&gt;A homomorphism with a two-sided inverse is an isomorphism.&lt;/li&gt;&#xA;&lt;li&gt;A homomorphism which is surjective as a map of sets and with a trivial kernel is an isomorphism.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Of course, sometimes the hard part is showing that you have a homomorphism at all. For your example of $D_4$ and $S_4$, you need to show the map you specify extends to a homomorphism. It's automatically surjective because you've hit every element in the generating set. Then it suffices to show that it has a trivial kernel. However, the &lt;em&gt;easiest&lt;/em&gt; way to show that $D_4$ embeds as a subgroup of $S_4$ is to show that it acts faithfully on a set of 4 elements. This is easy: $D_4$ naturally acts on the vertices of a square, and it's certainly faithful because there's no non-trivial element of $D_8$ which acts trivially.&lt;/p&gt;&#xA;" OwnerUserId="5191" LastActivityDate="2011-03-03T00:18:07.713" />
  <row Id="24726" PostTypeId="2" ParentId="24722" CreationDate="2011-03-03T00:21:19.917" Score="4" ViewCount="" Body="&lt;p&gt;The usual way to show isomorphy between groups is indeed to simply construct the isomorphism, i.e. construct a function $\varphi$ that is a bijection and morphism. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However sometimes you can resort to neater 'tricks', by considering actions of groups and constructing equivalent permutation representations. For instance to show that $Aut(\mathbf C_2\times \mathbf C_2) \cong \mathbf S_3 \cong \mathbf D_6$ one could notice that all three groups act on three elements. The first one actually consists of all permutations of the non-trivial elements of  $C_2\times \mathbf C_2$, (1,0), (0,1) and (1,1), the second group is by definition the group of all permutations of $\{1,2,3\}$ and the third group acts on a triangle in the plane where once again it consists of all possible permutations of the three vertices of the triangle. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;All three groups act faithful (only the identity acts trivially) therefore any bijection between these sets (the non-trivial elements of $\mathbf C_2\times \mathbf C_2$, the set $\{1,2,3\}$, the angles of a triangle) will naturally induce an isomorphism between these groups. It's not really an alternative to constructing the isomorphism ofcourse, it just tells you where to start looking for one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another strategy that I've seen at work when dealing with groups defined by presentations is to first construct an epimorphism (which is rather easy, one only has to verify that all relations still hold) and then show by considerations about the group (e.g. simplicity) that its kernel must be trivial.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, it's often very hard to think of such things. Here's something I read: During the golden years of discovery of sporadic simple groups two groups of equal order were constructed and even though this was suspected and quite some non-trivial things about these groups had been proven, it remained unsure for quite another while whether or not they were actually equal. (It turned out they were. It would be interesting if someone had a reference of this fact, I forgot where I read it.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;--- added:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's how this may help to solve your particular problem. Let $\mathbf D_8$ act on a square with vertices numbered $1,2,3,4$. Note that indeed $a = (1\ 2\ 3\ 4)$ and $b = (2\ 4)$ correspond to valid isometries of the square, let's call them $R$ (for rotation) and $D$ (for reflection along the diagonal). This implies that the map &#xA;$$ \varphi \colon \langle R,D\rangle \to \langle a,b\rangle $$&#xA;is actually an isomorphism: each element of either groups uniquely corresponds to some permutation of $\{1,2,3,4\}$ resp. the vertices of the square. Therefore it is sufficient to show that $\mathbf D_8$ is generated by $R$ and $D$ (which is well-known) and you're done.&lt;/p&gt;&#xA;" OwnerUserId="5189" LastEditorUserId="5189" LastEditorDisplayName="" LastEditDate="2011-03-03T00:39:54.613" LastActivityDate="2011-03-03T00:39:54.613" CommentCount="1" />
  <row Id="24727" PostTypeId="2" ParentId="24723" CreationDate="2011-03-03T00:28:25.900" Score="1" ViewCount="" Body="&lt;p&gt;Maybe here?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align*}&#xD;&#xA;T(n) &amp;amp;= 2T(n-1) + 1 \\&#xD;&#xA;&amp;amp;= 2 \{ 2T(n-2) + 1 \} + 1 \\&#xD;&#xA;&amp;amp;= 2 \{ 2 \{ 2T( n - 3 ) + 1 \} + 1 \} + 1 \\&#xD;&#xA;&amp;amp; \ldots \\&#xD;&#xA;&amp;amp;= 2^{i}T(n-i)+\sum_{k=1}^{i-1} 2^{k} + 1.&#xD;&#xA;\end{align*}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also,&#xA;$$&#xD;&#xA;\begin{align*}&#xD;&#xA;&amp;amp; 2^{n-1}T(n-(n-1)) + \sum_{i=1}^{n-2} 2^{i} + 1 \\&#xD;&#xA;&amp;amp;= 2^{n-1}T(n-n+1) + \sum_{i=1}^{n-2} 2^{i} + 1 \\&#xD;&#xA;&amp;amp;= 2^{n-1}T(1) + 2^{n-1} -1 \\&#xD;&#xA;&amp;amp;= 2^{n-1}\theta(1) + 2^{n-1} - 1\\&#xD;&#xA;\end{align*}$$&lt;/p&gt;&#xA;" OwnerUserId="6016" LastActivityDate="2011-03-03T00:28:25.900" />
  <row Id="24728" PostTypeId="2" ParentId="24723" CreationDate="2011-03-03T00:30:46.273" Score="2" ViewCount="" Body="&lt;p&gt;Your fourth step should be $T(n) = 2^k T(n-k) + \displaystyle \sum_{i=1}^k 2^{i-1}$ and there are a few errors in your manipulation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A simpler way to go about would be to let $T(n) = f(n) - 1$. Plugging this in the recurrence, we get $$f(n)-1 = 2(f(n-1)-1) + 1$$ i.e. $$f(n) = 2f(n-1)$$ Solving this recurrence is child's play and we get $$f(n) = f(1) 2^{n-1}$$ and hence $$T(n) = f(1) 2^{n-1} - 1 = T(1) 2^{n-1} + 2^{n-1} - 1 $$ since $f(1) = T(1) + 1$&lt;/p&gt;&#xA;" OwnerUserId="3302" LastActivityDate="2011-03-03T00:30:46.273" />
  <row Id="24730" PostTypeId="2" ParentId="23958" CreationDate="2011-03-03T01:05:00.453" Score="2" ViewCount="" Body="&lt;p&gt;I've never see the following approach used, even though it seems obvious.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Find the stationary points of $F(x_{1}, x_{2}\ldots x_{n})$ &#xA;subject to the constrants $g_{i}(x_{1}, x_{2} \ldots x_{n}) = c_{i},\ i = 1, 2 \ldots l$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\nabla F=0$  defines the stationary condition for $F$, and so $\nabla g_{i} = 0$ must follow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because of the constraints, the individual $x_{1}\ldots x_{n}$ are not independent of one another and so $\frac{\partial F}{\partial x_{i}}$ cannot be set to zero for each i = 1,...n&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Eliminating an $x_{i}$ in $F$ using each constraint would mean the resulting $F&amp;#39;$ had $n-l$ remaining independent $x_{i}$s and each corresponding $\frac{\partial F&amp;#39;}{\partial x_i}$ could now be independently set to zero. In practice, this is difficult, if not impossible. Fortunately, we're only interested in differentials and so can use each constraint to each eliminate a $dx_{i}$ from $dF$. For the first contraint, for example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\nabla g_{1} = 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$dx_{i} = \frac {1}{\frac{\partial g_{1}}{\partial x_{i}}}(-\nabla g_{1} + \frac{\partial g_{1}}{\partial x_{i}}dx_i)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Substituting into $\nabla F = 0$ gives&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sum_{j=1}^{n-1} \frac {\partial F}{\partial x_j}dx_j+\frac {\frac {\partial F}{\partial x_{i}}}{\frac{\partial g_{1}}{\partial x_{i}}}(-\nabla g_{1} + \frac{\partial g_{1}}{\partial x_{i}}dx_i) dx_i = 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sum_{j=1}^{n-1} \left(\frac {\partial F}{\partial x_j}-\frac {\frac {\partial F}{\partial x_{i}}}{\frac{\partial g_{1}}{\partial x_{i}}}\frac {\partial g_{1}}{\partial x_{j}}\right)dx_{j} = 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sum_{j=1}^{n-1} \left(\frac {\partial F}{\partial x_j}-\lambda \frac {\partial g_{1}}{\partial x_{j}}\right)dx_{j} = 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The obvious limitations from the above are &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The constrained stationary points are&#xA;satisfied only along constrained&#xA;differentials and so the point is not&#xA;necessarily an unconstrained&#xA;stationary point&lt;/li&gt;&#xA;&lt;li&gt;$\frac{\partial g_{1}}{\partial&#xD;&#xA;   x_{i}}$ must not equal zero at a&#xA;stationary point.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Using all l constraints gives&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sum_{j=1}^{n-1} \left(\frac {\partial F}{\partial x_j}-\frac {\frac {\partial F}{\partial x_{a}}}{\frac{\partial g_{1}}{\partial x_{a}}}\frac {\partial g_{1}}{\partial x_{j}}-\frac {\frac {\partial F}{\partial x_{b}}}{\frac{\partial g_{2}}{\partial x_{b}}}\frac {\partial g_{2}}{\partial x_{i}}\cdots -\frac {\frac {\partial F}{\partial x_{l}}}{\frac{\partial g_{l}}{\partial x_{l}}}\frac {\partial g_{l}}{\partial x_{j}}\right)dx_{j} = 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sum_{j=1}^{n-1} \left(\frac {\partial F}{\partial x_j}-\lambda_1 \frac {\partial g_{1}}{\partial x_{j}}-\lambda_2\frac {\partial g_{2}}{\partial x_{j}}\cdots-\lambda_l \frac {\partial g_{l}}{\partial x_{j}}\right)dx_{j} = 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And $\frac{\partial g_{1}}{\partial x_{a}},\frac{\partial g_{2}}{\partial x_{b}}\ldots \frac{\partial g_{l}}{\partial x_{l}}$ must not equal zero.&lt;/p&gt;&#xA;" OwnerUserId="7544" LastEditorUserId="7544" LastEditorDisplayName="" LastEditDate="2011-03-03T02:32:54.523" LastActivityDate="2011-03-03T02:32:54.523" />
  <row Id="24731" PostTypeId="1" AcceptedAnswerId="25120" CreationDate="2011-03-03T01:14:07.330" Score="7" ViewCount="72" Body="&lt;p&gt;Let $\pi$ be an arbitrary permutation of the set $\lbrace 1,\ldots,n,n+1,\ldots,2n \rbrace$ for some $n \in \mathbb{N}$. &#xA;We call a swap &lt;em&gt;local&lt;/em&gt; if you swap two neighboring positions in $\pi$, i.e. if you change the positions $i$ and $i-1$ or $i$ and $i+1$ for some  $i$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A &lt;em&gt;$c$-separation&lt;/em&gt; of the pairs $(1,n+1),\ldots,(n,2n)$ is a partition of $\pi$ in $L := \lbrace \pi(1), \ldots, \pi(p) \rbrace$ and $R := \lbrace \pi(p+1), \ldots, \pi(2n) \rbrace$ such that for at least $c$ pairs $(k,k+n)$ hold $(k,k+n) \in L \times R$ or $(k,k+n) \in R \times L$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is a good upper bound on the number of local swaps I have to perform on $\pi$ to get a $c$-separation of the pairs $(1,n+1), \ldots, (n,2n)$?&lt;/p&gt;&#xA;" OwnerUserId="4857" LastEditorUserId="4857" LastEditorDisplayName="" LastEditDate="2011-03-05T01:14:02.140" LastActivityDate="2011-03-05T06:14:00.333" Title="Upper bound of the number of local swaps" Tags="&lt;combinatorics&gt;&lt;permutations&gt;&lt;counting&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="24732" PostTypeId="2" ParentId="24717" CreationDate="2011-03-03T01:22:09.320" Score="2" ViewCount="" Body="&lt;p&gt;Yes, this is true&lt;sup&gt;*&lt;/sup&gt;, and extends to the $L^p$ norm for all $1\le p\le\infty$,&#xA;$$&#xD;&#xA;\left\Vert\int_a^bX_s\,ds\right\Vert_p\le\int_a^b\left\Vert X_s\right\Vert_p\,ds.&#xD;&#xA;$$&#xA;Consider the case where $X$ is simple, so that $X_t=\sum_{k=1}^nU_k1_{\{t\in V_k\}}$ for random variables $U_k$ and disjoint measurable subsets $V_k\subseteq\mathbb{R}$. Using $\lambda$ to denote the Lebesgue measure.&#xA;$$&#xD;&#xA;\begin{align}&#xD;&#xA;\left\Vert\int_a^bX_s\,ds\right\Vert_p &amp;amp;= \left\Vert\sum_kU_k\lambda\left(V_k\cap(a,b)\right)\right\Vert_p\\&#xD;&#xA;&amp;amp;\le\sum_k\Vert U_k\Vert_p\lambda\left(V_k\cap(a,b)\right)\\&#xD;&#xA;&amp;amp;=\int_a^b\Vert X_s\Vert_p\,ds&#xD;&#xA;\end{align}&#xD;&#xA;$$&#xA;Then, approximate arbitrary measurable $X$ as a limit of simple functions. The same argument will hold for integrable functions $\mathbb{R}\to B$ for any Banach space $B$, and not just for $L^p$ spaces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;*&lt;/sup&gt; I'm assuming that $X$ is measurable as a map from $\Omega\times\mathbb{R}$ to $\mathbb{R}$. &lt;/p&gt;&#xA;" OwnerUserId="1321" LastEditorUserId="1321" LastEditorDisplayName="" LastEditDate="2011-03-03T01:28:48.147" LastActivityDate="2011-03-03T01:28:48.147" CommentCount="2" />
  <row Id="24734" PostTypeId="2" ParentId="24699" CreationDate="2011-03-03T01:33:59.337" Score="4" ViewCount="" Body="&lt;p&gt;Perhaps they're saying that no cell in the grid can have more than 50 generations of total life?  On a finite board, this obviously guarantees that every pattern dies out eventually while still offering potentially large lifespans for patterns as a whole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the implementation is such that a cell that would otherwise live dies only if it's been alive for 50 consecutive generations, then there are still infinite patterns that can arise; in particular, Phoenix patterns are oscillators in which all live cells die out every step (but new cells are born): see &lt;a href=&quot;http://www.conwaylife.com/wiki/index.php?title=Phoenix_1&quot; rel=&quot;nofollow&quot;&gt;http://www.conwaylife.com/wiki/index.php?title=Phoenix_1&lt;/a&gt; for a simple example.  Another, more practical easy example is the humble &lt;a href=&quot;http://en.wikipedia.org/wiki/Glider_%28Conway%27s_Life%29&quot; rel=&quot;nofollow&quot;&gt;Glider&lt;/a&gt;; as it's constantly in motion, none of its constituent cells is alive for more than a few frames at a time, and a pattern that evolves to a solitary glider roaming the grid would thus live forever.&lt;/p&gt;&#xA;" OwnerUserId="785" LastActivityDate="2011-03-03T01:33:59.337" CommentCount="1" />
  <row Id="24735" PostTypeId="1" AcceptedAnswerId="24848" CreationDate="2011-03-03T01:36:42.690" Score="14" ViewCount="339" Body="&lt;p&gt;On MO, Daniel Moskovich has &lt;a href=&quot;http://mathoverflow.net/questions/51531/theorems-that-are-obvious-but-hard-to-prove/51652#51652&quot; rel=&quot;nofollow&quot;&gt;this to say&lt;/a&gt; about the &lt;a href=&quot;http://en.wikipedia.org/wiki/Hauptvermutung&quot; rel=&quot;nofollow&quot;&gt;Hauptvermutung&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The Hauptvermutung is so obvious that it gets taken for granted everywhere, and most of us learn algebraic topology without ever noticing this huge gap in its foundations (of the text-book standard simplicial approach). It is implicit every time one states that a homotopy invariant of a simplicial complex, such as simplicial homology, is in fact a homotopy invariant of a polyhedron.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have to admit I find this statement mystifying. We recently set up the theory of simplicial homology in lecture and I do not see anywhere that the Hauptvermutung needs to be assumed to show that simplicial homology is a homotopy invariant. Doesn't this follow once you have simplicial approximation and you also know that simplicial maps which are homotopic induce chain-homotopic maps on simplicial chains? &lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2011-03-04T19:42:19.253" Title="Is there a gap in the standard treatment of simplicial homology?" Tags="&lt;algebraic-topology&gt;" AnswerCount="3" CommentCount="4" FavoriteCount="1" />
  <row Id="24736" PostTypeId="2" ParentId="19216" CreationDate="2011-03-03T01:37:05.467" Score="0" ViewCount="" Body="&lt;p&gt;if you know the E(X) then you can figure out how far your random variables lie from the mean(E(X)) by computing the Var(X). So that way you can figure out where all your x's are bundle up at.&#xA;And to figure that out just use the def. Var(X) or you can find you E(x^2) and take the difference one by one from you r.v. X&lt;/p&gt;&#xA;" OwnerUserId="7720" LastActivityDate="2011-03-03T01:37:05.467" />
  <row Id="24737" PostTypeId="2" ParentId="18198" CreationDate="2011-03-03T01:49:43.363" Score="0" ViewCount="" Body="&lt;p&gt;the sample space are the numbers that your random variable X can take. If you have a uniform distribution on the interval (0,1) then the value that your r.v. can take is any thing from O to 1. If your the r.v is outside this interval then your pdf is zero. So that means that the universal space are the all the real numbers but for this sample we are only consern with what happens from [0,1]&lt;/p&gt;&#xA;" OwnerUserId="7720" LastActivityDate="2011-03-03T01:49:43.363" />
  <row Id="24738" PostTypeId="2" ParentId="20501" CreationDate="2011-03-03T01:52:42.403" Score="1" ViewCount="" Body="&lt;p&gt;If you have measurements of $x$, $y$ and $z$ isolated, you can estimate the density functions (histograms, or whatever) of each one.. isolated. That is (assumming $x$, $y$ and $z$ are random variables that have a join density, i.e. that each ocurrence corresponds to a tuple $(x,y,z)$), you can estimate the marginal densities. And nothing more. The joint density is the product of the marginals if (and only if) the variables are independent. If you cannot assume that, and if you dont have more data, you can't estimate the joint density.&lt;/p&gt;&#xA;" OwnerUserId="312" LastActivityDate="2011-03-03T01:52:42.403" />
  <row Id="24739" PostTypeId="1" AcceptedAnswerId="24792" CreationDate="2011-03-03T02:26:29.517" Score="6" ViewCount="113" Body="&lt;p&gt;People in StackOverflow seems not so into this theme, so I thought I could have better luck in here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I had the idea of an spaceship game where the world is confined in the surface of an 4-D hypersphere (also called a 3-sphere). Thus, in seeing it from inside, it would look like a 3-D world, but by navigating in every direction, I would never leave the limited volume of the 3-sphere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To represent the 3-shpere as a &quot;flat&quot; 3-D space, I use a stereographic projection, which is very simple to implement, just need to divide the point in the 3-sphere by one minus its w coordinate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To represent the vertices of the objects I am using normalized 4d vectors, such that x²+y²+z²+w²=1, thus keeping them inside the 3-sphere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first problem to solve was rotation. But I soon figured out that ordinary 3d rotation matrices would suffice to rotate the world around the viewer in the 3d projection, since it does not mess up with the w coordinate (pretty much like rotating a sphere around the z-axis would also rotate its stereographic projection).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then I figured out that any rotation that included the w coordinate would be equivalent of translation inside the 3d projection (just not commutative, as ordinary 3d translations on &quot;flat&quot; spaces), then I could translate along the axis by using a simple around axis rotation matrix (x', y') = (x * cos a - y * sin a, x * sin a + y * cos a), but variating w along with another axis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is so far where I got, and I could not figure out how to navigate forward, based on the position the viewer is facing from the projection. I can apply the inverse transform to derive the normalized 4-D vector (called F) the viewer is facing in the hypersphere coordinates, but I don't know how to navigate in that direction by using a 4x4 matrix (what is optimal in OpenGL). I could think on a hackish solution: for every vertex V, do V' = normalize(d*F + V), where d is the distance moved forward (in some strange unit I can not exactly precise). This way only works for small values of d, there is no direct correlation between d and the angle variation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus the question is: how to move forward (using a 4x4 matrix transform) being in the surface of a 4-D hypersphere? In other words: if I am at (x, y, z, w) now and want to be at (x', y', z', w') next (both vectors of norm 1), how can I derive M such that $M \times (x, y, z, w) = (x&amp;#39;, y&amp;#39;, z&amp;#39;, w&amp;#39;)$ ?&lt;/p&gt;&#xA;" OwnerUserId="7721" LastActivityDate="2011-03-03T10:35:17.140" Title="Navigating though the surface of a hypersphere in a computer game" Tags="&lt;geometry&gt;&lt;analytic-geometry&gt;&lt;spherical-geometry&gt;" AnswerCount="1" CommentCount="0" />
  <row Id="24740" PostTypeId="2" ParentId="24699" CreationDate="2011-03-03T02:28:50.670" Score="2" ViewCount="" Body="&lt;p&gt;From what I see in the video, the cells are directly manipulated in order to zero them out. For example, if you see a square block of four alive cells (as happens in the video), and they continue alive in the next step, they will stay this way until some other cell &quot;crashes&quot; into them: since the alive/dead status of a cell is a function of the statuses of the eight neighbors, each of these four cells represents a fixed point of that function. That is, no set of rules from Conway's Life will mimic exactly the behavior in the video, and you will need some direct modification of the grid in order to achieve similar results.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, there are many sets of rules that are &quot;less prolific&quot;, or more hostile, than the usual one, and this in combination with some specific starting setup should guarantee that all cells die eventually.&lt;/p&gt;&#xA;" OwnerUserId="6427" LastEditorUserId="6427" LastEditorDisplayName="" LastEditDate="2011-03-03T02:48:35.607" LastActivityDate="2011-03-03T02:48:35.607" CommentCount="4" />
  <row Id="24741" PostTypeId="1" CreationDate="2011-03-03T02:36:58.987" Score="3" ViewCount="65" Body="&lt;p&gt;Suppose I have a graph like this&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://yaroslavvb.com/upload/save/petersen-graph.png&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and a  &lt;a href=&quot;http://mathbin.net/59552&quot; rel=&quot;nofollow&quot;&gt;list&lt;/a&gt; of its automorphisms. How do I go about getting a set of generators for this group?&lt;/p&gt;&#xA;" OwnerUserId="998" LastActivityDate="2011-03-03T04:08:27.660" Title="Getting generators of graphs automorphism group" Tags="&lt;group-theory&gt;&lt;graph-theory&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="2" />
  <row Id="24743" PostTypeId="1" CreationDate="2011-03-03T02:49:10.613" Score="19" ViewCount="308" Body="&lt;p&gt;Which of the following is more surprising?&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;In a group of 100 people, the tallest person is one inch taller than the second tallest person.&lt;/li&gt;&#xA;&lt;li&gt;In a group of one billion people, the tallest person is one inch taller than the second tallest person.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Put more precisely, suppose we have a normal distribution with given mean $\mu$ and standard deviation $\sigma$. If we sample from this distribution $N$ times, what is the expected difference between the largest and second largest values in our sample? In particular, does this expected difference go to zero as $N$ grows?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In &lt;a href=&quot;http://math.stackexchange.com/questions/11871/probability-distribution-of-the-maximum-of-random-variables&quot;&gt;another question&lt;/a&gt;, it is explained how to compute the distribution $MAX_N$ of the maximum, but I don't see how to extract an estimate for the expected value of the maximum from that answer. Though $E(MAX_N)-E(MAX_{N-1})$ isn't the number I'm looking for, it might be a good enough estimate to determine if the value goes to zero as $N$ gets large.&lt;/p&gt;&#xA;" OwnerUserId="60" LastActivityDate="2011-06-02T15:28:06.890" Title="Does exceptionalism persist as sample size gets large?" Tags="&lt;statistics&gt;" AnswerCount="6" CommentCount="3" FavoriteCount="4" />
  <row Id="24744" PostTypeId="1" AcceptedAnswerId="24747" CreationDate="2011-03-03T02:59:56.770" Score="5" ViewCount="122" Body="&lt;p&gt;What is the explicit form of the inverse of the function $f:\mathbb{Z}^+\times\mathbb{Z}^+\rightarrow\mathbb{Z}^+$  where $$f(i,j)=\frac{(i+j-2)(i+j-1)}{2}+i?$$ &lt;/p&gt;&#xA;" OwnerUserId="1004" LastActivityDate="2011-03-03T06:33:20.203" Title="The inverse of a certain tricky function" Tags="&lt;functions&gt;" AnswerCount="2" CommentCount="6" FavoriteCount="1" />
  <row Id="24745" PostTypeId="2" ParentId="24744" CreationDate="2011-03-03T03:16:21.427" Score="3" ViewCount="" Body="&lt;p&gt;Since your function seems to be Cantor's pairing function $p(x,y) = \frac{(x+y)(x+y+1)}{2} + y$ applied to $x= j-2, y = i$, and since the inverse of the pairing function is $p^{-1}(z) = (\frac{\lfloor \frac{\sqrt{8z+1}-1}{2} \rfloor^2 + 3\lfloor \frac{\sqrt{8z+1}-1}{2} \rfloor}{2}-z,z-\frac{\lfloor \frac{\sqrt{8z+1}-1}{2} \rfloor^2 + \lfloor \frac{\sqrt{8z+1}-1}{2} \rfloor}{2})$, the inverse of your function is: $f^{-1}(z)=(z-\frac{\lfloor \frac{\sqrt{8z+1}-1}{2} \rfloor^2 + \lfloor \frac{\sqrt{8z+1}-1}{2} \rfloor}{2},2+ \frac{\lfloor \frac{\sqrt{8z+1}-1}{2} \rfloor^2 + 3\lfloor \frac{\sqrt{8z+1}-1}{2} \rfloor}{2}-z)$, which can be a bit ugly. What is your motivation for inverting this function?&lt;/p&gt;&#xA;" OwnerUserId="6427" LastEditorUserId="6427" LastEditorDisplayName="" LastEditDate="2011-03-03T03:41:42.553" LastActivityDate="2011-03-03T03:41:42.553" CommentCount="2" />
  <row Id="24746" PostTypeId="1" AcceptedAnswerId="24749" CreationDate="2011-03-03T03:18:24.663" Score="2" ViewCount="133" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Two permutations are conjugate of each other if and&#xA;  only if they have the same cycle structure.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If I want to find the number of elemnts in $S_{13}$ that are both conjugate to (123)(45) and (12)(34), then is the answer 0? because (12)(34) and (123)(45) have not got the same cycle structure.&lt;/p&gt;&#xA;" OwnerUserId="7340" LastActivityDate="2011-03-03T03:43:04.783" Title="number of elements in $S_{13}$ conjugate to both (12)(34) and (123)(45)" Tags="&lt;group-theory&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" />
  <row Id="24747" PostTypeId="2" ParentId="24744" CreationDate="2011-03-03T03:22:50.337" Score="4" ViewCount="" Body="&lt;p&gt;Let $i+j-2 = n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have $f = 1 + 2 + 3 + \cdots + n + i$ with $1 \leq i \leq n+1$. Note that the constraint $1 \leq i \leq n+1$ forces $n$ to be the maximum possible $n$ such that the sum is strictly less than $f$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence given $f$, find the maximum $n_{max}$ such that $$1 + 2 + 3 + \cdots + n_{max} &amp;lt; f \leq 1 + 2 + 3 + \cdots + n_{max} + (n_{max} + 1)$$ and now set $i = f - \frac{n_{max}(n_{max}+1)}{2}$ and $j = n_{max} + 2 - i$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$n_{max}$ is given by $\left \lceil \frac{-1 + \sqrt{1 + 8f}}{2} - 1 \right \rceil$ which is obtained by solving $f = \frac{n(n+1)}{2}$ and taking the ceil of the positive root minus one. (since we want the sum to strictly smaller than $f$ as we need $i$ to be positive)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, &#xA;$$&#xD;&#xA;\begin{align}&#xD;&#xA;n_{max} &amp;amp; = &amp;amp; \left \lceil \frac{-3 + \sqrt{1 + 8f}}{2} \right \rceil\\\&#xD;&#xA;i &amp;amp; = &amp;amp; f - \frac{n_{max}(n_{max}+1)}{2}\\\&#xD;&#xA;j &amp;amp; = &amp;amp; n_{max} + 2 - i&#xD;&#xA;\end{align}&#xD;&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2011-03-03T06:33:20.203" LastActivityDate="2011-03-03T06:33:20.203" CommentCount="2" />
  <row Id="24748" PostTypeId="2" ParentId="24746" CreationDate="2011-03-03T03:23:35.793" Score="6" ViewCount="" Body="&lt;p&gt;Yes.  Even if you didn't know about cycle structure, conjugate elements have the same order, and those two don't.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2011-03-03T03:23:35.793" CommunityOwnedDate="2011-03-03T03:23:35.793" CommentCount="4" />
  <row Id="24749" PostTypeId="2" ParentId="24746" CreationDate="2011-03-03T03:24:48.117" Score="5" ViewCount="" Body="&lt;p&gt;That is correct.  For this example, also notice that the two elements you mention have different orders.  $(12)(34)$ has order 2 while $(123)(45)$ has order 6.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given a group $G$ suppose that $a,b\in G$ are conjugate.  Then there exists $c\in G$ such that $cac^{-1}=b$.  If $a$ has order $n$, that is if $a^n=e$, then $$b^n=\left(cac^{-1}\right)^n=cac^{-1}cac^{-1}\cdots cac^{-1}=ca^nc^{-1}=cc^{-1}=e$$ so that $\text{ord}(b)\leq \text{ord}(a)$.  By symmetry, we also have that $\text{ord}(a)\leq \text{ord}(b)$ and hence $\text{ord}(a)=\text{ord}(b)$.&lt;/p&gt;&#xA;" OwnerUserId="6075" LastEditorUserId="6075" LastEditorDisplayName="" LastEditDate="2011-03-03T03:43:04.783" LastActivityDate="2011-03-03T03:43:04.783" CommentCount="2" />
  <row Id="24750" PostTypeId="1" CreationDate="2011-03-03T03:42:48.180" Score="5" ViewCount="119" Body="&lt;p&gt;Assuming that you start out with a root node, and decide with 50% probability whether or not to add two children nodes. If they do, repeat this process for them. How can you find the average length of this random binary tree?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm thinking along the lines of &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle\lim_{n\to\infty}\sum\limits_{i=1}^n (\frac{n}{2})^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;because 1/2 * n represents 50% probability, and I'm squaring it because the tree gets exponentially larger. However, I feel like I've done something terribly wrong (probably because I have). Can anyone give me some help?&lt;/p&gt;&#xA;" OwnerUserId="7547" LastActivityDate="2011-03-03T14:11:31.163" Title="How do you calculate the average length of a random binary tree?" Tags="&lt;homework&gt;&lt;probability&gt;&lt;trees&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="1" />
  <row Id="24751" PostTypeId="2" ParentId="24750" CreationDate="2011-03-03T03:54:26.313" Score="2" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Hint&lt;/strong&gt;: These problems can be done by what is known in probability as the first step analysis. Assume that the expected length is $L$. Take one step and try to come up with a relation involving $L$ and solve for $L$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you find first step analysis difficult, think along the lines of what the probability that the length of the tree is $L$ and then use this to find expectation. For instance, the probability the length is $0$ is $\frac{1}{2}$, the probability the length is $1$ is $\frac{1}{2} \times \frac{1}{4}$ and so on. Try to find a pattern and sum it up.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2011-03-03T03:59:42.970" LastActivityDate="2011-03-03T03:59:42.970" CommentCount="4" />
  <row Id="24752" PostTypeId="2" ParentId="24743" CreationDate="2011-03-03T04:01:44.060" Score="5" ViewCount="" Body="&lt;p&gt;A quick heuristic attempt at this: first, standard results on &lt;a href=&quot;http://en.wikipedia.org/wiki/Order_statistic&quot; rel=&quot;nofollow&quot;&gt;order statistics&lt;/a&gt; tell us that if we take $n$ samples from any distribution, with CDF $F$, the $k$th shortest person out of $n$ will typically have height around $F^{-1}(k/(n+1))$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So fix $\mu = 0$ and $\sigma = 1$. Then we expect the height of the tallest out of $n-1$ people to be around $\Phi^{-1}(1-1/n)$, and the height of the second tallest to be around $\Phi^{-1}(1-2/n)$, where $\Phi$ is the standard normal CDF. The question, then, is what happens to $\Phi^{-1}(1-1/n)-\Phi^{-1}(1-2/n)$ as $n$ gets large. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, it's a standard estimate that for large $z$, $1-\Phi(z) \approx \phi(z)/z$, where $\phi(z) = e^{-z^2/2}/\sqrt{2\pi}$ is the standard normal PDF. So let $\epsilon =  1-\Phi(z)$; then we get $\epsilon \approx \phi(z)/z$. Inverting gives the approximation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Phi^{-1}(1-\epsilon) \approx W\left( {1 \over 2\epsilon^2 \pi} \right)^{1/2}$$, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $W$ is the Lambert $W$ function, the inverse of $x \rightarrow xe^x$. In particular, if $\epsilon = 1/n$, then we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Phi^{-1}(1-1/n) \approx W \left( {n^2 \over 2 \pi} \right)^{1/2}$$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So finally the question becomes, what happens to &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ W\left( {4n^2 \over 2 \pi} \right)^{1/2} - W\left( {n^2 \over 2\pi} \right)^{1/2} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;as $n$ gets large? It appears that this goes to zero as $n$ gets large; that is, smaller gaps are expected between the smallest and second smallest entries in larger samples from the normal distribution. So (2) is more surprising.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That being said, I've thrown out a lot here, but I'm guessing that this captures the correct asymptotics.&lt;/p&gt;&#xA;" OwnerUserId="173" LastActivityDate="2011-03-03T04:01:44.060" CommentCount="5" />
  <row Id="24753" PostTypeId="2" ParentId="24743" CreationDate="2011-03-03T04:02:44.277" Score="1" ViewCount="" Body="&lt;p&gt;I'd take this approach:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Call  $ p_d= P(x_1 &amp;gt; x_2 + d)$ &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The probability that sample $x_1$ is the largest value AND it exceeds the second largest in more than $d$ is just $p_d^{N-1}$ &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The probability that the largest value exceeds the second largest in more that $d$ is then $N p_d^{N-1}$ WRONG-fixed below&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;To compute $p_d$: that is the probability that the difference of two iid normals exceeds $d$. But the difference of two normals is a normal of media cero, and variance $2 \sigma^2$, so that probability is given by the integral of the normal cumulative distribution function. What matters to us is that it's a constant that don't depend on $N$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the probability in question (fixed $d$, $N \to \infty$) tends to zero.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;UPDATE: As correctly pointed out in the comments, the second step is wrong, the events are not independent. They are independent, though, if $x_1$ is fixed - that is, they are conditionally independent. So:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$P(x_1 &amp;gt; x_2 + d | x_1) = F_x(x_1 - d) $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$P(x_1 &amp;gt; x_i + d | x_1) = F_x^{N-1}(x_1 - d)$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And  $P(x_1 &amp;gt; x_i + d) = \int_{-\infty}^{\infty}F_x^{N-1}(x - d) f_x(x) dx$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(this tends to $1/N$ as $d \to 0^+$, as was to be expected)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, restricting to $d \ge 0$, the probability that the largest value exceeds the second largest in more that $d$ is (if I have not messed up anything else) :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ P(x_A - x_B &amp;gt; d)= N \int_{-\infty}^{\infty} F_x^{N-1}(x - d) f_x(x) dx $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $x_A$ is the largest value and $x_B$ the second one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is about the above formula going to zero or not for fixed $d$, and growing N. That seems to depends on the density.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a check: if $x$ is exponential with parameter $\lambda$, the integrals can be evaluated, and I get:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ P(x_A - x_B &amp;gt; d) = exp( - \lambda d) $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This (besides tending to 1 for $d \to 0^+$, as it should) does not depend on $N$. Hence, for an exponential distribution the two events of the original question are equally surprising.&lt;/p&gt;&#xA;" OwnerUserId="312" LastEditorUserId="312" LastEditorDisplayName="" LastEditDate="2011-03-03T18:14:01.747" LastActivityDate="2011-03-03T18:14:01.747" CommentCount="2" />
  <row Id="24754" PostTypeId="2" ParentId="24741" CreationDate="2011-03-03T04:08:27.660" Score="3" ViewCount="" Body="&lt;p&gt;Algorithmically, you take the elements one by one and use them to grow a subgroup.  You create strong generating sets so that the size of the subgroup is known and so that membership in the subgroup can be tested easily.  You only record a new generator when you get an element not in the current subgroup.  You can stop early if you notice the subgroup and the list have the same size (as long as you trust the list is correct).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;GAP provides a command &lt;code&gt;AsGroup&lt;/code&gt; to convert a list of group elements into a group (which can then be asked for small generating sets).  In fact, in this example it fairly automatically finds a good generating set.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the list is not too long, you can use:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;g := AsGroup( List( myListOfImages, PermList ) );&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This has some performance problems if the list had more like 10,000 elements, but for small groups like this it is instant.  It yields:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Group([ (3,6)(5,7)(8,10), (2,5)(3,4)(7,10)(8,9), (1,2)(3,5)(6,7)(8,10) ])&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;If you like the image list format for permutations:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;gap&amp;gt; PrintArray( List( GeneratorsOfGroup(g), ListPerm ) );&#xA;[ [   1,   2,   6,   4,   7,   3,   5,  10,   9,   8 ],&#xA;  [   1,   5,   4,   3,   2,   6,  10,   9,   8,   7 ],&#xA;  [   2,   1,   5,   4,   3,   7,   6,  10,   9,   8 ] ]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;You can also ask GAP (and grape) directly:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;LoadPackage(&quot;grape&quot;);&#xA;Petersen := Graph( SymmetricGroup(5), [[1,2]], OnSets, function(x,y) return Intersection(x,y)=[]; end );;&#xA;AutomorphismGroup(Petersen);&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The vertices are in a different order than yours, but otherwise everything is good.&lt;/p&gt;&#xA;" OwnerUserId="583" LastActivityDate="2011-03-03T04:08:27.660" />
  <row Id="24755" PostTypeId="2" ParentId="24743" CreationDate="2011-03-03T04:31:29.920" Score="3" ViewCount="" Body="&lt;p&gt;Revised answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A very accurate approximation for the case of the normal distribution can be found in &lt;a href=&quot;http://gupea.ub.gu.se/bitstream/2077/3092/1/correction.pdf&quot; rel=&quot;nofollow&quot;&gt;this paper&lt;/a&gt;.&#xA;Let $X_{1:n} \leq X_{2:n} \leq \cdots \leq X_{n:n}$ be the ordered statistics obtained from a random sample $X_1,X_2,\ldots,X_n$, where $X_i \sim {\rm Normal}(\mu,\sigma^2)$.&#xA;According to Eq. (2), for $i \geq n/2$ and as $n \to \infty$,&#xA;$$&#xD;&#xA;X_{i:n}  \approx \mu  + \sigma \bigg[\sqrt {2\ln n}  - \frac{{\ln (\ln n) + \ln (4\pi ) - 2W_{i:n} }}{{2\sqrt {2\ln n} }}\bigg],&#xD;&#xA;$$&#xA;where $W_{i:n}$ has the density&#xA;$$&#xD;&#xA;g_{i:n} (w) = \frac{1}{{(n - i)!}}\exp ( - (n - i + 1)w - \exp ( - w)), \;\; - \infty  &amp;lt; w &amp;lt; \infty .&#xD;&#xA;$$&#xA;Thus, for example,&#xA;$$&#xD;&#xA;g_{n:n} (w) = \exp ( - w - \exp ( - w)), \;\; - \infty  &amp;lt; w &amp;lt; \infty &#xD;&#xA;$$&#xA;and&#xA;$$&#xD;&#xA;g_{n-1:n} (w) = \exp ( - 2w - \exp ( - w)), \;\; - \infty  &amp;lt; w &amp;lt; \infty .&#xD;&#xA;$$&#xA;According to Eqs. (3) and (4) of that paper,&#xA;$$&#xD;&#xA;{\rm E}[X_{n:n} ] \approx \mu  + \sigma \bigg[\sqrt {2\ln n}  - \frac{{\ln (\ln n) + \ln (4\pi ) - 2 \cdot 0.5772}}{{2\sqrt {2\ln n} }}\bigg]&#xD;&#xA;$$&#xA;and&#xA;$$&#xD;&#xA;{\rm Var}[X_{n:n} ] \approx \frac{{\sigma ^2  \cdot 1.64493}}{{2\ln n}}.&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some general facts, which are somewhat useful in our context. If $X_{1:n} \leq X_{2:n} \leq \cdots \leq X_{n:n}$ are the ordered statistics obtained from a random sample $X_1,X_2,\ldots,X_n$, where the $X_i$ have cdf $F$ and pdf $f$, then &#xA;$$&#xD;&#xA;{\rm E}[X_{i:n}]  = \frac{{n!}}{{(i - 1)!(n - i)!}}\int_{ - \infty }^\infty  {x [ F(x)] ^{i - 1} [ 1 - F(x)] ^{n - i} f(x)\,dx}.&#xD;&#xA;$$&#xA;By an exercise in a book on order statistics,&#xA;$$&#xD;&#xA;{\rm E}[X_{r + 1:n}  - X_{r:n} ] = {n \choose r}\int_{ - \infty }^\infty  {[F(x)]^r [1 - F(x)]^{n - r}\, dx} ,\;\; r = 1, \ldots ,n - 1.&#xD;&#xA;$$&#xA;Letting $r=n-1$ thus gives&#xA;$$&#xD;&#xA;{\rm E}[X_{n:n}  - X_{n-1:n} ] = n \int_{ - \infty }^\infty  {[F(x)]^{n-1} [1 - F(x)]\, dx}.&#xD;&#xA;$$&#xA;Applying this formula to the case of exponential with mean $\theta$ gives a constant difference:&#xA;$$&#xD;&#xA;{\rm E}[X_{n:n}  - X_{n-1:n} ] = n\int_0^\infty  {(1 - e^{ - x/\theta } )^{n - 1} e^{ - x/\theta } \, dx}  = \theta.&#xD;&#xA;$$&#xA;Nevertheless, the corresponding pdf, $\theta ^{ - 1} e^{ - x/\theta } \mathbf{1}(x \geq 0)$, goes to zero much faster than, say, $1/x^2$ as $x \to \infty$. In fact, $X_{n:n}  - X_{n-1:n}$ is exponentially distributed with mean $\theta$ (see also leonbloy's answer). Indeed, substituting the exponential cdf $F(x)=(1-e^{-x/\theta})\mathbf{1}(x \geq 0)$ and pdf $f(x)=\theta^{-1} e^{-x/\theta}\mathbf{1}(x \geq 0)$ into the general formula&#xA;$$&#xD;&#xA;f_{X_{n:n}  - X_{n-1:n} } (w) = \frac{{n!}}{{(n - 2)!}}\int_{ - \infty }^\infty  {[F(x)]^{n - 2} f(x)f(x + w)\,dx},\;\; 0 &amp;lt; w &amp;lt; \infty&#xD;&#xA;$$&#xA;for the density of $X_{n:n}-X_{n-1:n}$ (which is a special case of the formula for $X_{j:n}-X_{i:n}$, $1 \leq i &amp;lt; j \leq n$), gives&#xA;$$&#xD;&#xA;f_{X_{n:n}  - X_{n-1:n} } (w) = \theta^{-1}e^{-w/\theta}, \;\; 0 &amp;lt; w &amp;lt; \infty, &#xD;&#xA;$$&#xA;that is, $X_{n:n}  - X_{n-1:n}$ is exponential with mean $\theta$.&lt;/p&gt;&#xA;" OwnerUserId="2810" LastEditorUserId="2810" LastEditorDisplayName="" LastEditDate="2011-03-06T13:07:20.157" LastActivityDate="2011-03-06T13:07:20.157" CommentCount="8" />
  <row Id="24756" PostTypeId="1" CreationDate="2011-03-03T05:06:52.060" Score="3" ViewCount="43" Body="&lt;p&gt;The solid angle spanned by a disc of unit radius, as observed from a point $(r,z)$ at a distance $z&amp;gt;0$ above a point in the disc plane with at distance $r&amp;gt;0$ to the center, can be expressed as  $\Omega_1+\Omega_2$ where&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Omega_1 =2\pi\Theta[-(r+1)(r-1)] +\frac{2 z}{\sqrt{(r+1)^2 &#xD;&#xA;+ z^2}}\frac{r-1}{r+1}\Pi\left[\alpha^2,k\right]$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Omega_2 = \frac{2z}{\sqrt{(r+1)^2 + z^2}} K(k).$$&#xA;Here&#xA;$$\alpha^2 = n = \frac{(r+1)^2-(r-1)^2}{(r+1)^2}, &#xD;&#xA;k^2=m=\frac{(r+1)^2-(r-1)^2}{(r+1)^2+z^2},$$&#xA;and $K$ and $\Pi$ are complete elliptic integrals (see &lt;a href=&quot;http://dlmf.nist.gov/19.2&quot; rel=&quot;nofollow&quot;&gt;DLMF Sec. 19.2&lt;/a&gt;) and $\Theta$ is Heaviside's theta function. Note that Mathematica defines the elliptic integrals in terms of $n$ and $m$. The formula is adapted from S. Tryka, see (&lt;a href=&quot;http://dx.doi.org/10.1063/1.1150012&quot; rel=&quot;nofollow&quot;&gt;Rev. Sci. Instrum., 70, 3915 (1999)&lt;/a&gt;). The paper quotes a special case for $r=1$ which is not included in the expression above, but the expression does have the correct limiting value from both sides.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This all well and fine, except that both terms of $\Omega_1$  are discontinuous at $r=1$ -- which is incidentally the region of interest to me. Is there any way to reformulate the expression to avoid explicit discontinuities?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update&lt;/strong&gt;: I suddenly realize that in the formula I quote, I have already made some transformations of the elliptic integrals compared to Tryka's paper. I'll try to dig out my notes to document these.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Note added by J.M.: the formulae in Tryka's paper referred to above were derived in &lt;a href=&quot;http://dx.doi.org/10.1016/S0030-4018%2896%2900789-4&quot; rel=&quot;nofollow&quot;&gt;a previous paper of the author&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2842" LastEditorUserId="2842" LastEditorDisplayName="" LastEditDate="2011-05-03T07:52:19.647" LastActivityDate="2011-05-03T07:52:19.647" Title="Solid angle spanned by disc/rewriting expression with elliptic integrals" Tags="&lt;special-functions&gt;" CommentCount="5" />
  <row Id="24757" PostTypeId="1" AcceptedAnswerId="24783" CreationDate="2011-03-03T05:07:59.123" Score="9" ViewCount="147" Body="&lt;p&gt;If you have a  group, (say you have group table or any other information), is there an algorithm to find the group presentation? What is the general way of finding presentation of a group?&lt;/p&gt;&#xA;" OwnerUserId="7340" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-03T06:10:20.830" LastActivityDate="2011-03-10T01:54:08.407" Title="Working out a Group Presentation" Tags="&lt;group-theory&gt;" AnswerCount="2" CommentCount="8" />
  <row Id="24758" PostTypeId="1" AcceptedAnswerId="24760" CreationDate="2011-03-03T05:29:47.917" Score="4" ViewCount="282" Body="&lt;p&gt;There are plenty of theorems out there as well as laws within Mathematics.  For example, in Boolean Algebra:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Theorems&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Indempotent&lt;/li&gt;&#xA;&lt;li&gt;Involution&lt;/li&gt;&#xA;&lt;li&gt;Theorem of Complementarity&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Laws&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Commutative&lt;/li&gt;&#xA;&lt;li&gt;Associative&lt;/li&gt;&#xA;&lt;li&gt;Distributive&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;There are countless other examples out there, but my real question is this:  What makes a Theorem a Theorem and a Law a Law?  &lt;/p&gt;&#xA;" OwnerUserId="1602" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-03T05:38:24.917" LastActivityDate="2011-05-28T15:32:07.447" Title="What's the difference between a theorem and a law?" Tags="&lt;definition&gt;" AnswerCount="1" CommentCount="8" FavoriteCount="1" />
  <row Id="24759" PostTypeId="1" CreationDate="2011-03-03T05:31:20.993" Score="3" ViewCount="258" Body="&lt;p&gt;I often get confused about&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;when a Markov chain has an equilibrium distribution;&lt;/li&gt;&#xA;&lt;li&gt;when this equilibrium distribution is unique; &lt;/li&gt;&#xA;&lt;li&gt;which starting states converge to the equilibrium distribution; and &lt;/li&gt;&#xA;&lt;li&gt;how finite and countably infinite Markov chains different with respect to the above.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;(Google isn't quite clearing up my confusion.) Is the following correct/am I missing anything?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An irreducible Markov chain (finite or countably infinite) has a unique equilibrium distribution if and only if all states are positive recurrent. (What about reducible Markov chains? A reducible Markov chain has a non-unique equilibrium distribution iff all states are positive recurrent?) However, not all starting states necessarily converge to the unique equilibrium, unless the Markov chain is also aperiodic; that is, an irreducible Markov chain converges to its unique equilibrium regardless of initial state, if and only if all states are positive recurrent and aperiodic.&lt;/p&gt;&#xA;" OwnerUserId="1546" LastActivityDate="2011-03-04T13:34:04.263" Title="Equilibrium distributions of Markov Chains" Tags="&lt;probability&gt;&lt;stochastic-processes&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="24760" PostTypeId="2" ParentId="24758" CreationDate="2011-03-03T05:34:59.803" Score="5" ViewCount="" Body="&lt;p&gt;(Edited answer according to all feedback given, thanks.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Theorems&lt;/em&gt; are results proven from &lt;em&gt;axioms&lt;/em&gt;, more specifically those of mathematical logic and the systems in question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Laws&lt;/em&gt; usually refer to axioms themselves, but can also refer to well-established and common formulas such as the &lt;em&gt;law of sines&lt;/em&gt; or the &lt;em&gt;law of cosines&lt;/em&gt;, which really are theorems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a particular context, &lt;em&gt;propositions&lt;/em&gt; are the more trivial theorems, &lt;em&gt;lemmas&lt;/em&gt; are intermediate results, while &lt;em&gt;corollaries&lt;/em&gt; are results deduced easily from others. However, lemmas and corollaries may be major results on their own.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that a system may be given axioms in more ways than one. For example, we can use the &lt;em&gt;least upper bound axiom&lt;/em&gt; to define the real numbers, or we can consider this axiom as a theorem if we were to construct the reals from the rationals using &lt;em&gt;Dedekind cuts&lt;/em&gt; and prove it instead. The difference here lies in which axioms we choose to start with.&lt;/p&gt;&#xA;" OwnerUserId="4594" LastEditorUserId="4594" LastEditorDisplayName="" LastEditDate="2011-05-28T15:32:07.447" LastActivityDate="2011-05-28T15:32:07.447" CommentCount="3" />
  <row Id="24761" PostTypeId="2" ParentId="24507" CreationDate="2011-03-03T05:37:39.427" Score="14" ViewCount="" Body="&lt;p&gt;My dad likes to tell of a quotation he once read in a book on philosophy of mathematics. He does not remember which book it was, and I have never tried to track it down; this is really hearsay to the fourth degree, so it may not even be true. But I think it is pretty apt. The quote describes a castle on a cliff where, after each a storm finally dies down, the spiders come out andrun around frantically rebuilding their spiderwebs, afraid that if they don't get them up quickly enough the castle will fall down. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The interesting thing about the quote was that it was attributed to a book on the logical foundations of mathematics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, note that you are looking at the problem from a perspective of someone who &quot;grew up&quot; with sets that were, in some way, &lt;em&gt;carefully&lt;/em&gt; &quot;built off each other in a nice way.&quot; This was not always the case. Mathematicians were not always very careful with their foundations: and when they started working with infinite sets/collections, they &lt;em&gt;were&lt;/em&gt; not being particularly careful. Dedekind does not start from the Axiom of Infinity to construct the naturals and eventually get to the reals; and moreover, when he gives his construction it is precisely to try to answer the question of just what &lt;em&gt;is&lt;/em&gt; a real number! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In some ways, Russell's paradox was a storm that sent the spiders running around to reconstruct the spiderwebs.  Mathematicians hadn't been working with infinite collections/sets for very long, at least not as &quot;completed infinities&quot;. The work of Dedekind on the reals, and even on algebraic number theory with the definitions of ideals and modules, was not without its critics. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Some&lt;/em&gt; mathematicians had become interested in the issues of foundations; one such mathematician was Hilbert, both through his work on the Dirichlet Principle (justifying the work of Riemann), and his work on Geometry (with the problems that had become so glaring in the &quot;unspoken assumptions&quot; of Euclid). Hilbert was such a towering figure at the time that his interest was itself interesting, of course, but there weren't that many mathematicians working on the foundations of mathematics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would think like Sebastian, that most &quot;working mathematicians&quot; didn't worry too much about Russell's paradox; much like they didn't worry too much about the fact that Calculus was not, originally, on solid logical foundation. Mathematics clearly &lt;em&gt;worked&lt;/em&gt;, and the occasional antinomy or paradox was likely not a matter of interest or concern.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, the 19th century had highlighted a &lt;em&gt;lot&lt;/em&gt; of issues with mathematics. During this century all sorts of tacit assumptions that mathematicians had been making had been exploded. Turns out, functions can be &lt;em&gt;very&lt;/em&gt; discontinuous, not just at a few isolated points; they can be continuous but not differentiable everywhere; you can have a curve that fills up a square; the Dirichlet Principle need not hold; there are geometries where there are no parallels, and geometries where there are an infinite number of parallels to a given line and through a point outside of it; etc. While it was clear that mathematics &lt;em&gt;worked&lt;/em&gt;, there was a general &quot;feeling&quot; that it would be a good idea to clear up these issues.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So some people began to study foundations specifically, and try to build a solid foundation (perhaps like Weierstrass had given a solid foundation to calculus). Frege was one such.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And to people who were very interested in logic and foundations, like Frege, Russell's paradox it &lt;em&gt;was&lt;/em&gt; a big deal because it pinpointed that one particular tool that was very widely used led to serious problems. This tool was unrestricted comprehension: any &quot;collection&quot; you can name was an object that could be manipulated and played with.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might say, &quot;well, but Russell's paradox arises in a very artificial context, it would never show up with a &quot;real&quot; mathematical collection.&quot; But then, one might say that functions that are continuous everywhere and nowhere differentiable are &quot;very artificial, and would never show up in a 'real' mathematical problem&quot;. True: but it means that certain results that had been taken for granted no longer can be taken for granted, and need to be restricted, checked, or justified anew, if you want to claim that the argument is valid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In context, Russell's paradox showed an entirely new thing: there can be collections that are not sets, that are not objects that can be dealt with mathematically. This is a very big deal if you don't even have that concept to begin with! Think about finding out that a &quot;function&quot; doesn't have to be &quot;essentially continuous&quot; and can be an utter mess: an entirely new concept or idea; and entirely new possibility that has to be taken into account when thinking about functions. So with Russell's, an entirely new idea that needs to be taken into account when thinking about collections and sets. All the work that had been done before which tacitly assumed that just because you could name a collection it was an object that could be mathematically manipulated was now, in some sense, &quot;up in the air&quot; (as much as those castle walls are &quot;up in the air&quot; until the spiders rebuild their webs, perhaps, or perhaps more so). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If nothing else, Russell's paradox creates an entirely new category of things that did not exist before: &lt;strong&gt;not-sets&lt;/strong&gt;. Now you think, &quot;oh, piffle; I could have told them that&quot;, but that's because you grew up in a mathematical world where the notion that there are such things as &quot;not-sets&quot; is taken for granted. At the time, it was the exact opposite that was taken for granted, and Russell's paradox essentially tells everyone that something they all thought was true just isn't true. Today we are so used to idea that it seems like an observation that is not worth that much, but that's because we grew up in a world that already knew it.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would say that Russell's paradox was a big deal and wasn't a big deal. It was a big deal for anyone who was concerned with foundations, because it said &quot;you need to go further back: you need to figure out what &lt;em&gt;is&lt;/em&gt; and what is &lt;em&gt;not&lt;/em&gt; a collection you can work with.&quot; It undermined all of Frege's attempt at setting up a foundation for mathematics (which is why Frege found it so important: he certainly had invested a lot of himself into efforts that were not only cast into doubt, but were essentially demolished before they got off the ground). It was such a big deal that it completely infuses our worldview today, when we simply &lt;em&gt;take for granted&lt;/em&gt; that some things are not sets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, it did not have a major impact on things like calculus of variations, differential equations, etc., because those fields do not really rely very heavily on their foundations, but only on the properties of the objects they are working with; just like most people don't care about the Kuratowski definition of an ordered pair; it's kind of nice to know it's there, but most will treat the ordered pair as a black box. I would expect most of them to think &quot;Oh, okay; get back to me when you sort that out.&quot; Perhaps like the servants living on the castle not worrying too much about whether the spiders are done building their webs or not. Also, much like the way that after Weierstrass introduced the notion of $\epsilon$-$\delta$ definitions and limits into calculus, and then re-established what everyone was using &lt;em&gt;anyway&lt;/em&gt; when they were using calculus, it had little impact in terms of applications of calculus.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That rambles a bit, perhaps. And I'm not a very learned student of history, so my impressions may be off anyway.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2011-03-03T05:37:39.427" CommunityOwnedDate="2011-03-03T05:37:39.427" CommentCount="10" />
  <row Id="24762" PostTypeId="2" ParentId="24743" CreationDate="2011-03-03T06:47:21.873" Score="3" ViewCount="" Body="&lt;p&gt;Let $\mu_{i:n}$ denote the mean of the $i$th largest observation from a sample of size $n$ from a distribution.  The question is about the behavior of $\mu_{n:n} - \mu_{n-1:n}$ for a normal distribution.  David and Nagaraja's &lt;em&gt;&lt;a href=&quot;http://books.google.com/books?id=3Ts1yDLWXmQC&amp;amp;printsec=frontcover&amp;amp;dq=david+and+nagaraja+order+statistics&amp;amp;source=bl&amp;amp;ots=UPvtKJlP9Y&amp;amp;sig=dBJ1P7zQhm1buOxkiplHyvglPNQ&amp;amp;hl=en&amp;amp;ei=o8pvTaXUNo6CsQP8zqC3Cw&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=1&amp;amp;ved=0CBQQ6AEwAA#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;Order Statistics&lt;/a&gt;&lt;/em&gt; says that&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;For an $N(0,1)$ distribution, $\mu_{n:n}$ has the asymptotically dominant term $\sqrt{2\log n}$ (pp. 302-303,  see also Shai Covo's answer).&lt;/li&gt;&#xA;&lt;li&gt;For any distribution, $\mu_{n-1:n} + (n-1)\mu_{n:n} = n \mu_{n-1,n-1}$ (p. 44).&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;From (2), we have $\mu_{n:n} - \mu_{n-1:n} = n (\mu_{n:n} - \mu_{n-1,n-1})$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Adding the asymptotic from (1), we see that $\mu_{n:n} - \mu_{n-1:n}$ has the asymptotically dominant term $n \left(\sqrt{2\log n} - \sqrt{2\log (n-1)} \right)$, which is dominated by $$\frac{n}{n \sqrt{2 \log n}} = \frac{1}{\sqrt{2 \log n}},$$&#xA;which tells us both that $\mu_{n:n} - \mu_{n-1:n} \to 0$ and the rate at which it does so.&lt;/p&gt;&#xA;" OwnerUserId="2370" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2011-03-03T17:49:53.437" LastActivityDate="2011-03-03T17:49:53.437" />
  <row Id="24763" PostTypeId="1" AcceptedAnswerId="24770" CreationDate="2011-03-03T06:53:58.727" Score="8" ViewCount="226" Body="&lt;p&gt;Hi, I am trying to prove that given a subspace $\mathbf{W}$ in $\mathbb{R^{n}}$, the subspace and its orthogonal complement 'cover' whole of $\mathbb{R^{n}}$ through '+' where we define $\mathbf{W}$+$\mathbf{W^{\perp}}$ as linear combinations of vectors both in the subspace and in its orthogonal complement. It seems intuitively right, and I can prove that the sum of their dimensions adds up to n, but I am not sure how to prove the question I am looking at. Thanks!&lt;/p&gt;&#xA;" OwnerUserId="5157" LastActivityDate="2011-03-07T17:34:25.940" Title="Proving that $\mathbf{W}$+$\mathbf{W^{\perp}}$=$\mathbb{R^{n}}$" Tags="&lt;linear-algebra&gt;" AnswerCount="3" FavoriteCount="1" />
  <row Id="24764" PostTypeId="2" ParentId="24429" CreationDate="2011-03-03T07:01:50.363" Score="2" ViewCount="" Body="&lt;p&gt;Let $E$ be a closed subset of the boundary of the disk.  Then there is a nonzero function $f$ continuous on the closed disk and analytic on the open disk such that $f\vert_E=0$ if and only if the Lebesgue measure of $E$ is zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fact that the Lebesgue measure has to be zero (which answers your question) follows from a more general theorem that if $f$ is in any of the Hardy spaces of the disk, then the log of the absolute value of the boundary function of $f$ is in $L^1$ of the circle with Lebesgue measure.  This is in Theorem 17.17 of Rudin's &lt;em&gt;Real and complex analysis&lt;/em&gt;, 3rd Edition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fact that every closed subset of the boundary with Lebesgue measure zero can consist of zeros of a nonzero function that is continuous on the closed disk and analytic on the interior is a theorem of Fatou.  This appears in the section &quot;Theorems of Fatou and Rudin&quot; in Chapter 6 of Hoffman's &lt;em&gt;Banach spaces of analytic functions&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See also Myke's &lt;a href=&quot;http://math.stackexchange.com/questions/6208/isolated-zeros-on-closure-of-a-domain&quot;&gt;related question&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2011-03-06T00:12:30.650" LastActivityDate="2011-03-06T00:12:30.650" />
  <row Id="24765" PostTypeId="2" ParentId="24763" CreationDate="2011-03-03T07:15:30.133" Score="1" ViewCount="" Body="&lt;p&gt;Can you produce a basis of the right size using vectors in $W$, $W^{\perp}$? There is a standard result that perpendicular vectors are&#xA; linearly-independent (assume not, then set a non-trivial combination to zero,then dot both sides by the &quot;right&quot; vector to get a contradiction). Start with&#xA; the fact that $W$, $W^{\perp}$ are both subspaces, then try using perpendicular-&gt; linearly-independent to produce a basis of the right size for $\mathbb{R}^n$. Overall, you have &#xA; $k$ linearly-independent vectors in a basis for $W$, $(n-k)$ linearly-indep in a basis for $W^{\perp}$, and you can show that their set union is a basis for $\mathbb{R}^n$.&lt;/p&gt;&#xA;" OwnerUserId="7679" LastEditorUserId="3731" LastEditorDisplayName="" LastEditDate="2011-03-07T17:34:25.940" LastActivityDate="2011-03-07T17:34:25.940" CommentCount="4" />
  <row Id="24766" PostTypeId="2" ParentId="24763" CreationDate="2011-03-03T07:29:30.670" Score="5" ViewCount="" Body="&lt;p&gt;If you know how to prove that the dimensions of $W$ and $W^\perp$ add up to $n$, then the same argument shows that the dimensions of $W+W^\perp$ and $(W+W^\perp)^\perp$ add up to $n$, and the latter is $0$.  (Any vector in $(W+W^\perp)^\perp$ is perpendicular to itself, because it is in both $W^\perp$ and $W^{\perp\perp}$.)&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2011-03-03T07:29:30.670" />
  <row Id="24767" PostTypeId="1" CreationDate="2011-03-03T07:35:56.327" Score="4" ViewCount="136" Body="&lt;p&gt;It is a well known fact that the geometric series &#xA;$$1+x+x^2+x^3+\ldots$$&#xA;has the following form&#xA;$$\frac{1}{1-x}$$&#xA;Another possible representation  is&#xA;$$\prod_{k=0}^{\infty}\left(1+x^{2^{k}}\right)$$&#xA;This comes from  the identity&#xA;$$1+x+x^2+x^3+\ldots+x^{2^{k}}=\frac{1-x^{2^{k}+1}}{1-x}$$&#xA;now taking the numerator of the rhs we have&#xA;$$1-x^{2^{k}+1}=\left(1-x^{2^{k}}\right)\left(1+x^{2^{k}}\right)=\left(1-x^{2^{k}-1}\right)\left(1+x^{2^{k}-1}\right)\left(1+x^{2^{k}}\right)$$&#xA;proceeding this way we eventually get&#xA;$$\left(1-x\right)\left(1+x\right)\left(1+x^{2}\right)\ldots\left(1+x^{2^{k}-2}\right)\left(1+x^{2^{k}-1}\right)\left(1+x^{2^{k}}\right)$$&#xA;Taking the limit for the geometric series &#xA;$$\sum_{k=0}^{\infty}x^{k}=\prod_{k=0}^{\infty}\left(1+x^{2^{k}}\right)$$&#xA;Now taking the zeta function&#xA;$$\zeta(z)=\prod_{p\in\mathbb{P}}\left(1+\frac{1}{p^{z}}+\frac{1}{p^{2z}}+\frac{1}{p^{3z}}+\ldots\right)$$&#xA;we can express it as &#xA;$$\zeta(z)=\prod_{k=0}^{\infty}\;\prod_{p\in\mathbb{P}}\left(1+\frac{1}{p^{z\;2^{k}}}\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now considere  for &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$G(z)=\prod_{k=1}^{\infty}\;\prod_{p\in\mathbb{P}}\left(1+\frac{1}{p^{z\;2^{k}}}\right)$$&#xA;note that now $k\geq 1$ and that $G(z)$ converges absolutely for $z&amp;gt;\frac{1}{2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Can we say that, after analytic continuation, that&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$H(z)=\sum_{k=0}^{^\infty}\frac{|\mu(k)|}{k^{z}}=\prod_{p\in\mathbb{P}}\left(1+\frac{1}{p^{z}}\right)$$ &#xA;&lt;strong&gt;has exactly the same zeros as $\zeta(z)$?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="1747" LastEditorUserId="1747" LastEditorDisplayName="" LastEditDate="2011-03-20T09:35:03.157" LastActivityDate="2011-03-20T09:35:03.157" Title="Square-free zeta function zeros." Tags="&lt;number-theory&gt;&lt;complex-analysis&gt;&lt;prime-numbers&gt;&lt;riemann-zeta&gt;&lt;zeta-function&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="1" />
  <row Id="24769" PostTypeId="2" ParentId="24759" CreationDate="2011-03-03T07:55:27.300" Score="1" ViewCount="" Body="&lt;p&gt;The answers you have given are true at least for finite Markov chains. (My courses did not cover any others, I am afraid. And all references I have are German, so of limited use to you =) ).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The part &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A reducible Markov chain has a non-unique equilibrium distribution iff all states are positive recurrent?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;is not true. Consider the Markov Chain on the states 0 and 1, which goes from 0 to 1 with probability 1 and then stays there. It has a unique equilibrium distribution ($\delta_1$), without the state 0 being positive recurrent.&lt;/p&gt;&#xA;" OwnerUserId="505" LastActivityDate="2011-03-03T07:55:27.300" />
  <row Id="24770" PostTypeId="2" ParentId="24763" CreationDate="2011-03-03T07:55:48.703" Score="3" ViewCount="" Body="&lt;p&gt;Suppose $W$ has dimension $k$, and start with a basis $v_1,\ldots,v_k$ of $W$.  This is a linearly independent subset of $\mathbb{R}^n$, so it may be extended to a basis $v_1,\ldots,v_n$ of $\mathbb{R}^n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Apply the &lt;a href=&quot;http://en.wikipedia.org/wiki/Gram%E2%80%93Schmidt_process&quot; rel=&quot;nofollow&quot;&gt;Gram-Schmidt process&lt;/a&gt; to $v_1,\ldots,v_n$ to obtain an orthonormal basis $w_1,\ldots,w_n$ for $\mathbb{R}^n$ such that $w_1,\ldots,w_k$ is an orthonormal basis for $W$.  Now take any vector $z = a_1 w_1 + \ldots + a_k w_k + a_{k+1} w_{k+1} + \ldots + a_n w_n$ of $\mathbb{R}^n$.  The conditions for $z \in W^{\perp}$ are &#xA;precisely   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\langle z,w_1 \rangle = \ldots = \langle z,w_k \rangle= 0$,  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;i.e., &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a_1 = \ldots = a_k = 0$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus $W^{\perp}$ is the span of $w_{k+1},\ldots,w_n$ and &#xA;$\mathbb{R}^n = W \oplus W^{\perp}$.  &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;In the above argument, the key is showing that $\dim W + \dim W^{\perp} = \dim \mathbb{R}^n$ (as Jonas Meyer says in his answer).  Another argument for this, valid for any nondegenerate symmetric bilinear form $(u,v) \mapsto B(u,v)$ on a vector space $V$ (not necessarily finite-dimensional) over a field $K$ is given in $\S 4$ of &lt;a href=&quot;http://math.uga.edu/~pete/quadraticforms.pdf&quot; rel=&quot;nofollow&quot;&gt;these notes on quadratic forms&lt;/a&gt;.  Note that a general nondegenerate bilinear form could be isotropic: that is, one may have nonzero vectors $v$ with $B(v,v) = 0$ and thus $Kv \cap (Kv)^{\perp} \neq 0$.  But by definition an inner product is anisotropic, which forces $W \cap W^{\perp} = 0$ for all subspaces $W$.  &lt;/p&gt;&#xA;" OwnerUserId="299" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2011-03-03T08:03:03.663" LastActivityDate="2011-03-03T08:03:03.663" />
  <row Id="24771" PostTypeId="2" ParentId="24767" CreationDate="2011-03-03T07:59:39.967" Score="3" ViewCount="" Body="&lt;p&gt;Hint: $$ \sum \frac{|\mu(k)|}{k^z} = \frac{\zeta(z)}{\zeta(2z)} $$ for $\Re(z) &amp;gt; 1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Check out &lt;a href=&quot;http://en.wikipedia.org/wiki/Dirichlet_series&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Dirichlet_series&lt;/a&gt; &#xA;It might help you out.&lt;/p&gt;&#xA;" OwnerUserId="2320" LastActivityDate="2011-03-03T07:59:39.967" CommentCount="1" />
  <row Id="24772" PostTypeId="1" CreationDate="2011-03-03T08:08:08.903" Score="1" ViewCount="37" Body="&lt;p&gt;I have the following generalized eigenvalue problem&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\det[P_i^TLP_j+zP_i^TP_j] = 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$L$ is a positive-semidefinite matrix with 1 eigenvalue at 0.  More precisely, it is the combinatorial Laplacian matrix for a connected graph.&#xA;$P_i$ ($P_j$) is the identity matrix with the i-th (j-th) column removed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What, if anything, can be said about the generalized eigenvalues $z$ of this problem?&lt;/p&gt;&#xA;" OwnerUserId="7683" LastActivityDate="2011-03-03T08:08:08.903" Title="Generalized Eigenvalues of $P_i^TLP_j+zP_i^TP_j$" Tags="&lt;linear-algebra&gt;&lt;eigenvalues&gt;" CommentCount="2" />
  <row Id="24773" PostTypeId="1" CreationDate="2011-03-03T08:13:00.943" Score="0" ViewCount="40" Body="&lt;p&gt;If one defines the loxodrome of a surface of revolution as the space curve that cuts the &lt;a href=&quot;http://mathworld.wolfram.com/Meridian.html&quot; rel=&quot;nofollow&quot;&gt;meridians&lt;/a&gt; of the surface of revolution at a constant angle $\alpha$, how does one derive the parametric equations of the loxodrome?&lt;/p&gt;&#xA;" OwnerUserId="7724" LastActivityDate="2011-03-03T08:13:00.943" Title="Loxodromes of a surface of revolution" Tags="&lt;differential-geometry&gt;" />
  <row Id="24774" PostTypeId="2" ParentId="24750" CreationDate="2011-03-03T08:27:20.493" Score="2" ViewCount="" Body="&lt;p&gt;Here are some empirical assertions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the probability that the tree is of height $n$ is $P_n$ then $P_0 = 1/2$ and &#xA;$$P_{n+1} = P_n \left( 1 - \sqrt{2 P_n} + \frac{P_n}{2} \right).$$&#xA;[This is related to &lt;a href=&quot;http://oeis.org/A076628&quot; rel=&quot;nofollow&quot;&gt;OEIS A076628&lt;/a&gt; where $P_n = 2 b(n+1)^2$] &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since for $n&amp;gt;10$,    $$\dfrac{1}{n^2} &amp;lt; P_n &amp;lt; \dfrac{2}{n^2}$$ [asymptotically $P_n$ is $2/n^2$], the expectation of the height of the tree is infinite.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If this is homework, there is probably an easier method.&lt;/p&gt;&#xA;" OwnerUserId="6460" LastEditorUserId="6460" LastEditorDisplayName="" LastEditDate="2011-03-03T14:11:31.163" LastActivityDate="2011-03-03T14:11:31.163" CommentCount="5" />
  <row Id="24775" PostTypeId="2" ParentId="24743" CreationDate="2011-03-03T08:31:11.083" Score="0" ViewCount="" Body="&lt;p&gt;I'm no statistician, but it would clearly depend on what type of distribution you're talking about.  The highest-rated answer is talking about PDFs and CDFs, and then he goes on to pick a CDF and do calculations off of that.  The problem with that is that not everything is a bell curve or other very common distribution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given a completely flat distribution, such as a computer's random number generator, outliers will evaporate very quickly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given a bell curve, such as your height example, outliers will be semi-persistent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a distribution function with infinite variance (i.e. +/- infinity,) then there is no reason for outliers to have close neighbors.&lt;/p&gt;&#xA;" OwnerUserId="7725" LastActivityDate="2011-03-03T08:31:11.083" />
  <row Id="24776" PostTypeId="1" AcceptedAnswerId="24781" CreationDate="2011-03-03T08:35:57.477" Score="1" ViewCount="119" Body="&lt;p&gt;Consider the inner product by $\langle f,g \rangle_{H^1} = \langle f, g \rangle_{L^2} + \sum_{|\alpha|=1} \langle D^\alpha f, D^\alpha g \rangle_{L^2}$ where $\alpha$ is a multi-index and $D$ denotes the weak derivative. Define $H^1(\Omega)$ as the space of functions that are finite under the norm induced from this inner product. It can be shown that $H^1(\Omega)$ is a Hilbert space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, suppose there exists a sequence of functions $\{ f_n \} \subset H^1(\Omega)$ that converges weakly in $H^1$ to some limit $f \in H^1(\Omega)$. Can I then say that this sequence converges weakly to the same limit under the $L^2$ inner product? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;By an application of the Banach-Alaoglu Theorem, I know that weak convergence of this sequence in $H^1$ will imply strong convergence of a subsequence in $H^1$. And then I think strong convergence of this subsequence in $H^1$ will imply strong convergence in $L^2$ as well. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I'm not sure if anything can be said about the entire sequence under the $L^2$ inner product and its weak/strong convergence properties.    &lt;/p&gt;&#xA;" OwnerUserId="1736" LastEditorUserId="1736" LastEditorDisplayName="" LastEditDate="2011-03-03T08:50:39.287" LastActivityDate="2011-03-03T09:28:02.673" Title="Weak convergence in Sobolev spaces" Tags="&lt;functional-analysis&gt;" AnswerCount="1" />
  <row Id="24777" PostTypeId="1" AcceptedAnswerId="24782" CreationDate="2011-03-03T08:41:24.133" Score="0" ViewCount="103" Body="&lt;p&gt;Please suggest a reference for a proof of Gibbs phenomenon at jump discontinuities of a function.&lt;/p&gt;&#xA;" OwnerUserId="2987" LastActivityDate="2011-04-10T15:17:07.243" Title="reference request for proof of Gibbs phenomenon at jump discontinuities" Tags="&lt;reference-request&gt;&lt;fourier-analysis&gt;&lt;fourier-series&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="24778" PostTypeId="1" AcceptedAnswerId="24957" CreationDate="2011-03-03T09:06:49.040" Score="0" ViewCount="89" Body="&lt;p&gt;I would like to rewrite this two-level program as a Non Linear Program using KKT conditions&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\min_{x \in X, \space y \in Y} \space f(x)$&lt;br&gt;&#xA;$\text{s.t. } \space g(x, y) \leq 0$&lt;br&gt;&#xA;$y \in \text{argmin}_{z \in Y} \space F(x,z)$&lt;br&gt;&#xA;$\space \space \space \space \space \space \text{s.t.} \space Ax + Bz - b \leq 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here $A$ and $B$ are matrices with dimension $m \times n_x$ and $m \times n_y$ respectively. The idea is to replace the $\text{argmin}$ constraint on the $y$ variables with KKT conditions for a point $(x,y)$. &lt;/p&gt;&#xA;" OwnerUserId="7145" LastEditorUserId="7145" LastEditorDisplayName="" LastEditDate="2011-03-04T01:24:57.383" LastActivityDate="2011-03-08T03:32:42.770" Title="Using KKT to Write a Non-Linear Program" Tags="&lt;optimization&gt;&lt;nonlinear-optimization&gt;" AnswerCount="1" />
  <row Id="24780" PostTypeId="2" ParentId="24704" CreationDate="2011-03-03T09:27:34.323" Score="2" ViewCount="" Body="&lt;p&gt;In some situations, to prove that groups of order $n$ cannot be simple, you can use the counting argument if all Sylow subgroups have trivial intersection, and a different argument otherwise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example let $G$ be a simple group of order $n=144 = 16 \times 9$. The number $n_3$ of Sylow 3-subgroups is 1, 4 or 16. If $n_3 = 1$ then there is normal Sylow subgroup and if $n_3= 4$ then $G$ maps nontrivially to $S_4$, so we must have $n_3 = 16$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If all pairs of Sylow 3-subgroups have trivial intersection, then they contain in total $16 \times 8$ non-identity elements, so the remaining 16 elements must form a unique and hence normal Sylow 2-subgroup of $G$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Otherwise two Sylow 3-subgroups intersect in a subgroup $T$ of order 3. Then the normalizer $N_G(T)$ of $T$ in $G$ contains both of these Sylow 3-subgroups, so by Sylow's theorem it has at least 4 Sylow 3-subgroups, and hence has order at least 36, so $|G:N_G(T)| \le 4$ and $G$ cannot be simple.&lt;/p&gt;&#xA;" OwnerUserId="2820" LastActivityDate="2011-03-03T09:27:34.323" />
  <row Id="24781" PostTypeId="2" ParentId="24776" CreationDate="2011-03-03T09:28:02.673" Score="1" ViewCount="" Body="&lt;p&gt;$L^2(\Omega)$ is a subspace of $H^{-1}(\Omega)$, the dual of $H^1(\Omega)$. Thus, a sequence converging weakly in $H^1$ converges weakly in $L^2$ to the same limit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, nothing can be said about strong convergence in $L^2$.&lt;/p&gt;&#xA;" OwnerUserId="4791" LastActivityDate="2011-03-03T09:28:02.673" />
  <row Id="24782" PostTypeId="2" ParentId="24777" CreationDate="2011-03-03T09:43:33.547" Score="1" ViewCount="" Body="&lt;p&gt;Check out Boggess and Narcowich's &lt;a href=&quot;http://books.google.com/books?id=paZVMSrlEw8C&amp;amp;printsec=frontcover&quot; rel=&quot;nofollow&quot;&gt;book&lt;/a&gt; on wavelets and signal processing.&lt;/p&gt;&#xA;" OwnerUserId="5523" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2011-04-10T15:17:07.243" LastActivityDate="2011-04-10T15:17:07.243" />
  <row Id="24783" PostTypeId="2" ParentId="24757" CreationDate="2011-03-03T09:49:03.423" Score="6" ViewCount="" Body="&lt;p&gt;Here is a very brief description of the method used by computer systems like GAP and Magma to find a presentation of a moderately small finite group (of order up to about $10^7$) on a given generating set $X$. This dates from about 1972 and is due originally to John Cannon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You start with an initial set of relators $R_0$, which could be empty, or you might include a few obvious short relators like $x^2$. Then run Todd-Coxeter coset enumeration for this presentation over the identity subgroup. It won't complete of course, and the standard tactic is to interrupt it after it has defined $c|G|$ cosets, for some constant $c&amp;gt;1$, such as $c=1.1$. You then look for the first equation between your defined cosets which is true in $G$ but not yet known in your incomplete coset table. This gives the shortest new relator of $G$ that is not an easy consequence of $R_0$, so you adjoin it to $R_0$ and resume the coset enumeration. Then just repeat this process of interrupting the coset enumeration and adding new relators until the coset enumeration completes with $|G|$ cosets. You then have a presentation of $G$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This tends to produce reasonably good presentations in terms of total relator length, particularly if you retrospectively go through all relators except the last one found, and try removing them to see if they are redundant. The computed presentations will not necessarily be &quot;nice&quot; in the sense that the relators consist of powers, commutators, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are lots of refinements for handling larger groups, particularly if you do not insist on having the presentation on the given set $X$. For example, for permutation groups with base and strong generating sets, a variant of this method can be used to find a presentation on a set of strong generators.&lt;/p&gt;&#xA;" OwnerUserId="2820" LastActivityDate="2011-03-03T09:49:03.423" />
  <row Id="24784" PostTypeId="1" AcceptedAnswerId="30408" CreationDate="2011-03-03T09:56:44.567" Score="10" ViewCount="210" Body="&lt;p&gt;This is one of those perhaps rare occasions when someone takes the advice of the FAQ and asks a question to which they already know the answer.  This puzzle took me a while, but I found it both simple and satisfying.  It's also great because the proof doesn't use anything fancy at all but it's still a very nice little result.&lt;/p&gt;&#xA;" OwnerUserId="2159" LastEditorUserId="2159" LastEditorDisplayName="" LastEditDate="2011-03-03T10:06:01.157" LastActivityDate="2011-04-01T22:08:51.767" Title="Why isn't $\mathbb{CP}^2$ a covering space for any other manifold?" Tags="&lt;algebraic-topology&gt;" AnswerCount="2" CommentCount="5" FavoriteCount="2" />
  <row Id="24785" PostTypeId="1" AcceptedAnswerId="24786" CreationDate="2011-03-03T09:57:31.797" Score="0" ViewCount="63" Body="&lt;p&gt;Define $D^n = \{ x \in \mathbb{R}^n : |x| \leq 1 \}$. By identifying all the points of $S^{n-1}$ we get a topological space which is intuitively homeomorphic to $S^n$. If n = 2, this can be visualised by pushing the centre of the disc $D^2$ down so you have a sack, then shrinking the boundry of the sack to a point which gives you a teardrop shaped object which is clearly homeomorphic to $S^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am new to algebraic topology. How do I prove that the quotient space is actually homeomorphic to $S^n$. I haven't been able to write down explicitly a continuous map between $D^n$ and $S^n$ which maps $S^{n-1}$ to a point on $S^n$, which at the moment is the only way I know how to begin showin that two spaces are homeomorphic. Is more machinery needed? If so I am interested to hear what is needed. If not, please tell me how stupid I am and give me a hint! &lt;/p&gt;&#xA;" OwnerUserId="1725" LastActivityDate="2011-03-03T11:00:11.933" Title="D^n quotient by its boundary" Tags="&lt;algebraic-topology&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="0" />
  <row Id="24786" PostTypeId="2" ParentId="24785" CreationDate="2011-03-03T10:11:33.947" Score="1" ViewCount="" Body="&lt;p&gt;Well, in the case $D^1 = [0, 1]$ and $S^1 = $ unit circle around $0$, can you write down a continuous map?&lt;/p&gt;&#xA;" OwnerUserId="5798" LastActivityDate="2011-03-03T10:11:33.947" />
  <row Id="24787" PostTypeId="1" CreationDate="2011-03-03T10:17:31.803" Score="0" ViewCount="38" Body="&lt;p&gt;If you form the Euler-Lagrange equation for some calculus of variations problem in x, f(x) and f'(x), and the resulting expression is always non-negative over the domain of x (because the expression is independent of f(x) and f'(x)), does that mean any perturbation that increases f(x) increases the functional? This seems natural given it should be some generalisation of &#xA;what one gets when the first derivative of a function is always non-negative... &lt;/p&gt;&#xA;" OwnerUserId="7728" LastActivityDate="2011-03-03T10:17:31.803" Title="Euler-Lagrange expression uniformly non-negative" Tags="&lt;functional-analysis&gt;" />
  <row Id="24788" PostTypeId="2" ParentId="24735" CreationDate="2011-03-03T10:20:14.360" Score="6" ViewCount="" Body="&lt;p&gt;One doesn't need to explicitly compare  with singular homology to get homotopy invariance (although that is certainly one way to do it), and one certainly doesn't need the Hauptvermutung (thankfully, since it is false in general).  Rather, as you say, one can simplicially approximate a continuous map between simplicial complexes, and thus make simplicial homology a functor on the category of simplicial complexes and continuous maps (which, as you observe, will then factor through the homotopy category).  Just thinking it over briefly, it seems to me that the hardest step in this approach will be to show that the induced map on homology really is independent of the choice of a  simplicial approximation. Presumably this was covered in the lectures you are attending.  (In fact, I guess this amounts to the fact that homotopic simplicial maps induce the same map on homology, which you say you have proved.)   &lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2011-03-03T10:20:14.360" CommentCount="1" />
  <row Id="24789" PostTypeId="1" AcceptedAnswerId="24816" CreationDate="2011-03-03T10:21:41.693" Score="4" ViewCount="167" Body="&lt;p&gt;Hi Math! I hope the simplicity of this problem doesn't offend anyone's intelligence, however i'm unsure where else I should post this, and fairly certain its due to some kind of math rule (or just my stupid mistake);&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm trying to figure out why the figures in the far right boxes** below are not equal.  How is the sum of the results of the division of individual items larger than when the division is applied to the already summed items?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;** &lt;a href=&quot;http://img535.imageshack.us/img535/9113/division.png&quot; rel=&quot;nofollow&quot;&gt;Here's a link to the model i've constructed&lt;/a&gt; (which did nothing but confuse, see below)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/H2B5z.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt; Unfortunately it looks like i've mislead with the 0.05 looking like an average (hence the /2 in comments &amp;amp; answers), when in fact it is not.  I've got my question into an equation now, so hopefully this is clearer (and yeah, my bad that this equation probably looks nothing like the original model, apologies)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Revised Question:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When $a_2/a_1 = b_2/b_1 , a_1 \neq b_1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{a_{1}}{a_{2}/a_{1}}+\dfrac{b_{1}}{b_{2}/b_{1}}=&#xD;&#xA;\frac{a_{1}+b_{1}}{1+\dfrac{(a_{2}+b_{2})-(a_{1}+b_{1})}{a_{1}+b_{1}}}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So why when $a_2/a_1 \neq b_2/b_1 , a_1 \neq b_1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{a_{1}}{a_{2}/a_{1}}+\dfrac{b_{1}}{b_{2}/b_{1}}\neq \frac{a_{1}+b_{1}}{1+\dfrac{(a_{2}+b_{2})-(a_{1}+b_{1})}{a_{1}+b_{1}}}?$$&lt;/p&gt;&#xA;" OwnerUserId="7727" LastEditorUserId="7727" LastEditorDisplayName="" LastEditDate="2011-03-08T02:24:31.063" LastActivityDate="2011-03-08T02:24:31.063" Title="Why does the sum of a division applied to individual items not equal the division applied to the sum of those items?" Tags="&lt;algebra-precalculus&gt;" AnswerCount="3" CommentCount="8" />
  <row Id="24790" PostTypeId="2" ParentId="24785" CreationDate="2011-03-03T10:28:53.697" Score="0" ViewCount="" Body="&lt;p&gt;Yes, you only need to find a suitable map between $D^n$ and $S^n$, and to do that, your visualisation should help you a lot. For this kind of problem, getting a visualisation is most of the work, and you did that already. The rest is just describing what you see.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's pick $n=2$. You said that if you push &quot;down&quot; the center of $D^2$ and &quot;close&quot; the sack, you would get $S^2$. This should tell you that you are looking at a map $f : D^2 \rightarrow S^2$ such that the $f(0,0) = (0,0,-1)$  and is $f(x,y) = (0,0,+1)$ for $(x,y) \in S^1$. Then you need to pick a way to choose for example the $z$ coordinate for intermediate points in a continuous way. Any continuous function should work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do you visualise your sack with a round shape all the time ? What should be the inverse image of the equator ? A circle of center $(0,0)$ in $D^1$ ? Then your map should be rotationnaly invariant around $(0,0)$ and you can settle to search for $f$ only on a diameter of $D^2$. What is the image of a diameter in your visualisation ? A circle of $S^2$ ? &#xA;So $f(x,y)$ should be $(kx,ky,z)$ for some functions $z$ and $k$ ? etc.&lt;/p&gt;&#xA;" OwnerUserId="6270" LastActivityDate="2011-03-03T10:28:53.697" />
  <row Id="24791" PostTypeId="1" CreationDate="2011-03-03T08:32:06.333" Score="2" ViewCount="118" Body="&lt;p&gt;symbolize the following in Lg using the the interruption function below. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;U: {o|o is a person}&#xA;L2: { , m loves n} &#xA;H1: { p|o, o is exaited}&#xA;C1: {o|o, o is a kid&#xA;b: Bob&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;if any loves Bob, then Bob loves everyone.&quot;   if theres a x (Lxb --&gt; Lcx)&#xA;right?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;happy people and kids love Bob.&quot; For any x[Hx ^ If therex a x Hbx) .... im not sure how to finish this one. does anyone know?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;there is some kid that everyone loves&quot;&#xA;if theres an X, if theres a y(Lyx)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;looks something like this... ]x]yLyx&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;there is a happy kid who loves all unhappy kids&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;at least two kids love Bob&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;]x]y(Lxb ^ Lyb)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;the kid who loves Bob loves everyone&quot;&#xA;]x(Lxb --&gt;VyLxy&lt;/p&gt;&#xA;&#xA;&lt;p&gt;any feedback. if i am wrong please correct me. &lt;/p&gt;&#xA;" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2011-03-03T16:08:24.577" LastActivityDate="2011-03-03T17:21:19.640" Title="quantificational translation" Tags="&lt;logic&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="24792" PostTypeId="2" ParentId="24739" CreationDate="2011-03-03T10:35:17.140" Score="2" ViewCount="" Body="&lt;p&gt;The $M$ you asked in your final question is in general non-unique, given the data you specified. The way to see this is as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given a point $(x,y,z,w)$ and another $(x&amp;#39;,y&amp;#39;,z&amp;#39;,w&amp;#39;)$, add to it the third point $(0,0,0,0)$ in four dimensional space. Assuming that the two original points do not coincide and that they are not antipodes, this three points are not collinear and so defines a plane in four dimensional space. So you can take a rotation in that plane that carries $(x,y,z,w)$ to $(x&amp;#39;,y&amp;#39;,z&amp;#39;,w&amp;#39;)$ while fixing the directions perpendicular to that plane fixed. Call this transformation $M_1$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, since your ambient space is four dimensional, the directions perpendicular to the given plane also is two dimensional (4-2 = 2). So you can equally take an arbitrary rotation in that plane which fixes all directions perpendicular to it. Call such a rotation $O$. Then you can check that $M_2 = OM_1$ also sends $(x,y,z,w)$ to $(x&amp;#39;,y&amp;#39;,z&amp;#39;,w&amp;#39;)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To be explicit. Assume your initial coordinate is $(1,0,0,0)$ and the final coordinate is $(0,1,0,0)$. Then we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ M_1 = \begin{pmatrix} 0 &amp;amp; -1 &amp;amp; 0 &amp;amp; 0\\ 1 &amp;amp; 0 &amp;amp; 0 &amp;amp; 0 \\ 0 &amp;amp; 0 &amp;amp; 1 &amp;amp; 0 \\ 0 &amp;amp; 0 &amp;amp; 0 &amp;amp; 1\end{pmatrix}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you take $O$, parametrized by $\theta$, to be&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ O(\theta) = \begin{pmatrix} 1 &amp;amp; 0 &amp;amp; 0 &amp;amp; 0 \\ 0 &amp;amp; 1 &amp;amp; 0 &amp;amp; 0\\ 0 &amp;amp; 0 &amp;amp; \cos\theta &amp;amp; \sin\theta \\ 0 &amp;amp; 0 &amp;amp; -\sin\theta &amp;amp; \cos\theta\end{pmatrix}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;then you can check that $O(\theta)M_1$ will send $(1,0,0,0)$ to $(0,1,0,0)$ for any $\theta$, but the matrices $O(\theta)M_1$ are all different for $\theta$ in the range $0 \leq \theta &amp;lt; 2\pi$. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;So what does this mean physically? Moving &quot;forward&quot; in Euclidean space is not as simple an issue as you think. The analogue of the different $O(\theta)$ in 3 dimension Euclidean space corresponds to rotating the whole space along the axis of travel! In other words, imagine you have a spaceship in Euclidean space and it spins (with the axis of spinning in the same direction of its travel) while it moves forward. This is what $O(\theta)$ captures, the spinning. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Without getting too much into the gory details, I will note that this ambiguity lies in the heart of differential/Riemannian geometry, and is closely connected with the notion of parallel transport. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In any case, using a bit of differential geometry, you can see that the matrix $M_1$ defined above is the &quot;correct&quot; notion of translation is you do not want &quot;spinning&quot;. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Okay, enough about that. How to actually implement this? Let's say that the viewer is sitting at coordinates $(x,y,z,w)$. And let's say the viewer is facing &lt;em&gt;in the direction of&lt;/em&gt; $(\delta x, \delta y, \delta z, \delta w)$ with $\delta x^2 + \delta y^2 + \delta z^2 + \delta w^2 = 1$. Notice that since the direction the viewer is facing is &lt;em&gt;tangential&lt;/em&gt; to the hypersphere, it must be perpendicular to the coordinates, that is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ x\cdot \delta x + y\cdot \delta y + z \cdot \delta z + w \cdot \delta w = 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given these two vectors, you can use some linear algebra to complete them to an orthonormal basis of four dimensional space (which can be obtained by solving some linear equations based on orthogonality to the two given vectors), call the two additional vectors $(a,b,c,d)$ and $(a&amp;#39;,b&amp;#39;,c&amp;#39;,d&amp;#39;)$. Then your translation matrix should be given by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ M(\phi) = \begin{pmatrix} x &amp;amp; \delta x &amp;amp; a &amp;amp; a&amp;#39;\\ y &amp;amp; \delta y &amp;amp; b &amp;amp; b&amp;#39; \\ z &amp;amp; \delta z &amp;amp; c &amp;amp; c&amp;#39; \\ w &amp;amp; \delta w &amp;amp; d &amp;amp; d&amp;#39;\end{pmatrix} \begin{pmatrix} \cos(\phi) &amp;amp; \sin(\phi) &amp;amp; 0 &amp;amp; 0\\ -\sin(\phi) &amp;amp; \cos(\phi) &amp;amp; 0 &amp;amp; 0 \\ 0 &amp;amp; 0 &amp;amp; 1 &amp;amp; 0 \\ 0 &amp;amp; 0 &amp;amp; 0 &amp;amp; 1\end{pmatrix} \begin{pmatrix} x &amp;amp; y &amp;amp; z &amp;amp; w\\ \delta x &amp;amp; \delta y &amp;amp; \delta z &amp;amp; \delta w \\ a &amp;amp; b &amp;amp; c &amp;amp; d \\ a&amp;#39; &amp;amp; b&amp;#39; &amp;amp; c&amp;#39; &amp;amp; d&amp;#39;\end{pmatrix} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A quick explanation: by fixing the orthonormal basis, we can use it to construct an orthogonal transformation to a coordinate system in which the viewer sits at $(1,0,0,0)$ and is facing in the $(0,1,0,0)$ direction. Then all we need to do is to conjugate back the transportation matrix for that situation. The $\phi$ parameter measures the (angular) distance you travelled on the hypersphere. &lt;/p&gt;&#xA;" OwnerUserId="1543" LastActivityDate="2011-03-03T10:35:17.140" CommentCount="2" />
  <row Id="24793" PostTypeId="1" AcceptedAnswerId="24829" CreationDate="2011-03-03T10:37:50.333" Score="0" ViewCount="124" Body="&lt;p&gt;The Fokker-Planck equation for several variables is :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{\partial W}{\partial t} = L_{FP}W$  (1)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$L_{FP} = -\frac{\partial}{\partial x_i}D_i(\{x\})+\frac{\partial^2}{\partial x_i \partial x_j}D_{ij}(\{x\}).$  (2)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The summation convention for Latin indices is used here. The drift vector $D_i$ and the diffusion tensor $D_{ij}$ generally depend on the N variables $x_1,...,x_N = \{x\}$. The Fokker-Planck equation is an equation for the distribution function $W(\{x\},t)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to [Risken 1989 ch6], If drift &amp;amp; diffusion coefficients do not depend on time &amp;amp; $D_{ij}$ is positive definite everywhere &amp;amp; if the drift coefficient has no singularities, a stationary solution $W_{st}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$L_{FP} W_{st} = 0$,  (3)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;may exist.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If one solves the above equation, a possible stationary solution can be&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$W_{st} =\frac{a}{D_{ij}}exp(\int^{x_j}_0 \frac{D_i}{D_{ij}} dt_j)$  (4)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where a is a normalization constant. Now I want to expand this probability distribution for i=1,2. If I use the Einstein summation convention, it becomes&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$W_{st} =\{\frac{a}{D_{11}}exp(\int^{x_1}_0 \frac{D_1}{D_{11}} dt_1)+\frac{a}{D_{12}}exp(\int^{x_2}_0 \frac{D_1}{D_{12}} dt_2)+\frac{a}{D_{21}}exp(\int^{x_1}_0 \frac{D_2}{D_{21}} dt_1)+\frac{a}{D_{22}}exp(\int^{x_2}_0 \frac{D_2}{D_{22}} dt_2)\}$. (5)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems very strange to me. Is it a really correct probability distribution or I made a mistake somewhere? And if it is correct how can I normalize it? Can anyone help?&lt;/p&gt;&#xA;" OwnerUserId="7553" LastEditorUserId="1543" LastEditorDisplayName="" LastEditDate="2011-05-03T09:48:05.007" LastActivityDate="2011-05-03T09:48:05.007" Title="expansion of an expression" Tags="&lt;matrices&gt;&lt;notation&gt;&lt;probability-distributions&gt;&lt;mathematical-physics&gt;" AnswerCount="1" CommentCount="7" FavoriteCount="1" />
  <row Id="24794" PostTypeId="1" AcceptedAnswerId="24944" CreationDate="2011-03-03T10:51:55.383" Score="10" ViewCount="373" Body="&lt;p&gt;I have solutions to Exercises 5.16&amp;ndash;5.19 in Atiyah&amp;ndash;Macdonald's Introduction to Commutative Algebra, but not in the order desired; I find myself needing later exercises to do earlier ones, and it's been frustrating me. Online solution sets (I count five, in various stages of completion) seem to either not notice this problem or treat it as something too obvious to merit consideration. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For context, the first part of 5.16 is Noether's normalization lemma, which is fine, and 5.18 is Zariski's Lemma, which is proved in the text two times, and is accessible to (at least) two  natural proofs at this point. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I can't do is to obtain the second part of 5.16 without using 5.17, or prove 5.17 without using 5.18. Moreover, 5.19 specifically asks one to prove 5.17 using 5.18 (which I can do), so the strong implication is that the original solutions should not require use of Zariski's Lemma. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$k$ is an infinite field, assumed algebraically closed in 5.17 but not 5.16.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first part of 5.17 is that if an affine variety $X$ in $k^n$ has ideal $I(X)$ a proper subset of $k[t_1,\ldots,t_n]$, then $X$ is not empty. This follows immediately from the second part of 5.16.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second part of 5.16 is that for any subvariety $X$ of $k^n$ there are an $r \leq n$ &#xA;and a linear map $k^n \to k^r$ taking $X$ onto all of $k^r$. The natural candidate map follows from Noether normalization, but my approach to surjectivity seems to require the second part of 5.17.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second part of 5.17 is that every maximal ideal of $k[t_1,\ldots,t_n]$ is of the form $(t_1 - a_1,\ldots,t_n - a_n)$ for some $a_i \in k$. From this one can show a similar result for the coordinate ring of a subvariety of $k^n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So this mess will be fixed if I can prove the second parts of 5.16 and 5.17 without reference to later material.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In more detail, my current approach to the second part of 5.16 is as follows. Noether normalization shows $A$ is integral over some polynomial ring $A&amp;#39; := k[y_1,\ldots,y_r]$ in $r \leq n$ indeterminates. The proof of Noether normalization, at least in case $k$ is infinite, gives the $y_i$ as $k$-linear combinations of the natural generators $x_1,\ldots,x_n$ of $A$, the coordinate functions on $X$. If we say $y_i = \sum a_{ij} x_j$, then the projection $k^n \to k^r$ should be given by $(z_1,\ldots,z_n) \mapsto&#xD;&#xA;\big(\sum_{j=1}^n a_{1j}z_j,\ldots,\sum_{j=1}^n a_{rj}z_j,\big)$. It's not immediately obvious to me that it is surjective. However, letting $\textrm{Max}(A) \subseteq \textrm{Spec}(A)$ denote the set of maximal ideals of $A$, results of Chapter 5 show that the map $\textrm{Max}(A) \to \textrm{Max}(A&amp;#39;)$ induced by $k^n \to k^r$ is surjective. If we know each maximal ideal of $A$ is of the form $(x_1-c_1,\ldots,x_n-c_n)$ for some $(c_1,\ldots,c_n) \in X$, then we can identify $X$ and $\textrm{Max}(A)$, and that will show the map $X \to k^r$ is surjective. That seems, however, not to be what they want, and also like too much work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My current approach to the second part of 5.17 is to use the weak Nullstellensatz (if $k$ is algebraically closed and $\mathfrak a$ is a proper ideal of $k[t_1,\ldots,t_n]$, then there is at least one point of $k^n$ at which $\mathfrak a$ vanishes). The weak Nullstellensatz implies the first part of 5.17, but the reverse implication is not obvious to me. This implication is certainly proved by, e.g., the strong Nullstellensatz, but that would be completely missing the point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update&lt;/strong&gt;: To clarify, the text of 5.16, excluding a long hint about the first part, is as follows.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Let $k$ be a field and let $A \neq 0$ be a finitely generated $k$-algebra.&#xA;  Then there exist elements $y_1,\ldots,y_r \in A$ which are algebraically &#xA;  independent over $k$ and such that $A$ is integral over &#xA;  $k[y_1,\ldots,y_r]$. We shall assume that $k$ is &lt;em&gt;infinite&lt;/em&gt;.&#xA;  (The result is still true if $k$ is finite, but a different proof is needed.)&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;...&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;From the proof  it follows that $y_1,\ldots, y_r$ &#xA;  may be chosen to be linear combinations of $x_1,\ldots,x_n$.&#xA;  This has the following geometrical interpretation:&#xA;  if $k$ is algebraically closed and $X\!$ is an affine algebraic variety in $k^n$&#xA;  with coordinate ring $A \neq 0$, then there exists a linear subspace $L$ &#xA;  of dimension $r$ in $k^n$ and a linear mapping of $k^n$ onto $L$ &#xA;  which maps $X\!$ &lt;em&gt;onto&lt;/em&gt; $L$ [Use Exercise 2].&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I had forgotten about that hint... Here is what Ex. 5.2 says.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Let $A$ be a subring of a ring $B$ such that $B$ is integral over $A$,&#xA;  and let $f\colon A \to \Omega$ be a homomorphism of $A$ into &#xA;  an algebraically closed field $\Omega$.&#xA;  Show that $f\!$ can be extended to a homomorphism of $B$ into $\Omega$. [Use (5.10).]&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7112" LastEditorUserId="7112" LastEditorDisplayName="" LastEditDate="2011-03-03T22:03:16.060" LastActivityDate="2011-03-05T06:47:28.197" Title="Atiyah-Macdonald Exercises 5.16-5.19" Tags="&lt;commutative-algebra&gt;" AnswerCount="2" CommentCount="10" FavoriteCount="1" />
  <row Id="24795" PostTypeId="1" AcceptedAnswerId="24831" CreationDate="2011-03-03T10:53:59.080" Score="2" ViewCount="47" Body="&lt;p&gt;I have a set of points $(x, y)$ where each one comes from either one of two linear functions:&#xA;\begin{align*}&#xA;    y &amp;amp;= m_1 x + b_1\\&#xA;    y &amp;amp;= m_2 x + b_2&#xA;\end{align*}&#xA;Is there a fitting method to find such functions, without knowing from which function each of the points come from?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS. can somebody add fit (or fitting) to the existing tags&lt;/p&gt;&#xA;" OwnerUserId="7731" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-03T14:56:12.463" LastActivityDate="2011-03-03T16:18:02.707" Title="How to fit fixed data from two linear functions" Tags="&lt;statistics&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="24796" PostTypeId="2" ParentId="24785" CreationDate="2011-03-03T10:54:36.930" Score="0" ViewCount="" Body="&lt;p&gt;You know that $S^n$ is a one point compactification of $R^n$. That is $S^n = R^n \cup \{\infty\}$ with appropriate topology. Now define a map $f\colon D^n \to S^n$ as follows if $x$ is in interior of $D^n$ then $f(x) = \frac{x}{1 -\mod{x}}$ but if $x$ is in boundary then $f(x) = \infty$. Now it is easy to see that $f$ is continuas in the interior and as well as in boundary. Sequential criterion may help you to conclude the statement. Now $D^n$ is compact so the map is a closed map and therefore a quotient map. Hence it induces a homeomorphism from $D^n$ to $S^n$.&lt;/p&gt;&#xA;" OwnerUserId="7565" LastEditorUserId="7565" LastEditorDisplayName="" LastEditDate="2011-03-03T11:00:11.933" LastActivityDate="2011-03-03T11:00:11.933" />
  <row Id="24797" PostTypeId="1" CreationDate="2011-03-03T11:04:01.077" Score="0" ViewCount="37" Body="&lt;p&gt;Take two weakly convex, weakly increasing non-negative functions, $g(x)$ and $h(x)$, domain of $x$ is $[a,d]$, $g(x)=0$ for $x \in [a,b]$, $h(x)=0$ for $x \in [a,c]$, and $g(d)=h(d)$. So $g$ and $h$ start at zero, stay at zero over some (possibly different) domain(s), and intersect again at $d$. Can we make a general statement on the relative convexities and slopes of $g$ and $h$ that achieves a weak ranking of $g$ and $h$ over all $x$ in $[a,d]$? i.e. $g(x)\geq h(x)$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clearly if $b&amp;lt;c$ and $g&amp;#39;(x)\leq h&amp;#39;(x)$ for $x \in [c,d]$ that will do it, but i'm more interested in situations where the first derivatives can't necessarily be ranked in this way (and, preferably, where we don't have to invoke a condition on $b$ or $c$). Intuitively it seems like there must be some condition on the scaled second derivative? like on $g&amp;#39;&amp;#39;(x)/g&amp;#39;(x)$ vs $h&amp;#39;&amp;#39;(x)/h&amp;#39;(x)$?&lt;/p&gt;&#xA;" OwnerUserId="7728" LastEditorUserId="1543" LastEditorDisplayName="" LastEditDate="2011-03-03T11:40:08.293" LastActivityDate="2011-03-03T11:40:08.293" Title="curvature and intersection of convex functions" Tags="&lt;calculus&gt;&lt;convex-analysis&gt;" CommentCount="4" />
  <row Id="24799" PostTypeId="2" ParentId="24789" CreationDate="2011-03-03T11:23:37.317" Score="2" ViewCount="" Body="&lt;p&gt;I'm not sure where the figures are coming from, but an interpretation is the inequality ($a,b,c &gt; 0$)&#xA;$$\frac{a}{a+b} + \frac{a}{a+c} \ge \frac{2a}{a+ (b+c)/2}, \quad (1)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which follows from&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$((a+b)-(a+c))^2 \ge 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and so&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(a+b)^2+(a+c)^2 \ge 2(a+b)(b+c).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, on adding $2(a+b)(b+c)$ to both sides,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$((a+b)+(a+c))^2 \ge 4(a+b)(b+c)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and dividing both sides by $((a+b)+(a+c))(a+b)(b+c)$ and multiplying by $a$ will give us $(1).$&lt;/p&gt;&#xA;" OwnerUserId="1301" LastEditorUserId="1301" LastEditorDisplayName="" LastEditDate="2011-03-03T12:06:07.873" LastActivityDate="2011-03-03T12:06:07.873" CommentCount="1" />
  <row Id="24800" PostTypeId="1" CreationDate="2011-03-03T11:46:38.430" Score="5" ViewCount="87" Body="&lt;p&gt;Suppose we have a $n\times m$ rectangular grid (namely: $nm$ points disposed as a matrix with $n$ rows and $m$ columns).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We randomly pick $h$ different points in the grid, where every point is equally likely.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If only horizontal or vertical movements between two points are allowed, what is the probability that the points define at least one closed path?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ps: we can suppose $m=n$ to simplify&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, let $n=m=4$ and $h=6$.&#xA;1 denotes a selected point, 0 a non-selected one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These $6$ points define a closed path:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1  0  0  1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;0  0  0  0&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1  1  0  0&lt;/p&gt;&#xA;&#xA;&lt;p&gt;0  1  0  1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;as these $6$ do (the $4$ in the bottom-right corner):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1  0  0  1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;0  0  0  0&lt;/p&gt;&#xA;&#xA;&lt;p&gt;0  1  0  1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;0  1  0  1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;while the following $6$ points do not:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1  0  0  0&lt;/p&gt;&#xA;&#xA;&lt;p&gt;0  0  0  1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1  1  0  0&lt;/p&gt;&#xA;&#xA;&lt;p&gt;0  1  0  1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Substantially, the $h$ points define a closed path if and only if there exist a subset of these $h$ points such that every point in the subset has one other point of the subset on the same row and one on the same column. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for your help.&lt;/p&gt;&#xA;" OwnerUserId="7733" LastActivityDate="2011-03-03T13:52:25.283" Title="Random points in a rectangular grid defining a closed path" Tags="&lt;probability&gt;&lt;combinatorics&gt;" AnswerCount="0" CommentCount="1" FavoriteCount="1" />
  <row Id="24801" PostTypeId="1" AcceptedAnswerId="24803" CreationDate="2011-03-03T12:08:51.640" Score="1" ViewCount="48" Body="&lt;p&gt;Hello,&#xA;Sorry for my English.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is the question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Graph (V,E).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Definition: Legal's up-path in a graph  from s to t is existent if and only if for every Vi, Vi+1 (for each i) fulfill w(Vi)&amp;lt;=w(Vi+1), when w(v) is a weight of the vertex.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know that there is an algorithm like BFS that solve the problem of finding the shorter path in a Graph. We need to find, by reduction, an algorithm that solve the problem of legal's up-path.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I try to split each vertex, but it didn’t work (for example: for Vi=4, I tried to create 4 Edge, and then activate the BFS, but it didn’t work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank u. &lt;/p&gt;&#xA;" OwnerUserId="7734" LastActivityDate="2011-03-03T12:58:19.870" Title="Graph - reduction" Tags="&lt;homework&gt;&lt;graph-theory&gt;" AnswerCount="1" />
  <row Id="24802" PostTypeId="1" CreationDate="2011-03-03T12:45:02.097" Score="0" ViewCount="95" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Possible Duplicate:&lt;/strong&gt;&lt;br&gt;&#xA;  &lt;a href=&quot;http://math.stackexchange.com/questions/4787/motivation-behind-standard-deviation&quot;&gt;Motivation behind standard deviation?&lt;/a&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;p&gt;This is a beginner question. The variance for a population is calculated using the squares of the deviations of the values from the mean and then calculating the average of these squares. In my understanding, the square is introduced as a way of ensuring that the values being added up are all positive. My question is - why not just use the absolute value of the difference between a value and the mean, instead of its square? Is there something else here in this formula that I am missing?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just as an exercise I tried calculating the variance both ways for a small set (1, 1, 2, 5, 4).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The 'squared' variance formula gives - 2.64. Square root of that is 1.624.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The absolute value formula gives - 1.52&lt;/p&gt;&#xA;" OwnerUserId="142" LastActivityDate="2011-03-03T12:45:02.097" ClosedDate="2011-03-03T12:51:05.643" Title="Why don't we calculate the variance with absolute values?" Tags="&lt;statistics&gt;" CommentCount="5" />
  <row Id="24803" PostTypeId="2" ParentId="24801" CreationDate="2011-03-03T12:58:19.870" Score="0" ViewCount="" Body="&lt;p&gt;Assuming the BFS is for directed graphs, and you start with an undirected graph.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint: Try to pick a direction (and in some cases two) for each edge in the undirected graph so that the resulting graph is directed, and on which you can activate BFS.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2011-03-03T12:58:19.870" CommentCount="3" />
  <row Id="24804" PostTypeId="2" ParentId="24759" CreationDate="2011-03-03T13:11:04.007" Score="5" ViewCount="" Body="&lt;p&gt;For a Markov chain with $N&amp;lt;\infty$ states, the set $I$ of invariant probability vectors is a non-empty simplex in ${\mathbb R}^N$ whose extreme points correspond to the recurrent classes of the chain. Thus, the vector is unique iff there is exactly one  recurrent class; the transient states (if any) play absolutely no role (as in Jens's example). The set $I$ is a point, line segment, triangle, etc. exactly when there are one, two, three, etc. recurrent classes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the invariant vector $\pi$ is unique, then there is only one recurrent class and the chain will eventually end up there. The vector $\pi$ necessarily puts zero mass on all transient states. Letting $\phi_n$ be the law of $X_n$, as you say, we have $\phi_n\to \pi$ only if the recurrent class is aperiodic. However, in general we have Ces&amp;agrave;ro convergence: &#xA;$${1\over n}\sum_{j=1}^n \phi_j\to\pi.$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;An infinite state space Markov chain need not have any recurrent states, and may have the zero measure as the only invariant measure, finite or infinite. Consider the chain on the positive integers which jumps to the right at every time step.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Generally, a Markov chain with countable state space  has invariant probabilities iff there are positive recurrent classes. If so, every invariant probability vector $\nu$ is a convex combination of &#xA;the unique invariant vector $m_j$ corresponding to each positive recurrent class $j\in J$, i.e., &#xA;$$\nu=\sum_{j\in J} c_j m_j,\qquad c_j\geq 0,\quad \sum_{j\in J}c_j=1.$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This result is Corollary 3.23 in Wolfgang Woess's &lt;em&gt;Denumerable Markov Chains&lt;/em&gt;. &lt;/p&gt;&#xA;" OwnerUserId="940" LastEditorUserId="940" LastEditorDisplayName="" LastEditDate="2011-03-04T13:34:04.263" LastActivityDate="2011-03-04T13:34:04.263" CommentCount="1" />
  <row Id="24805" PostTypeId="1" AcceptedAnswerId="24812" CreationDate="2011-03-03T13:24:40.643" Score="8" ViewCount="122" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of my professors wrote the following open question on the blackboard:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $M$ is a compact, connected smooth $4$-manifold such that $\pi_1(M) = 0$, $\pi_2(M) = 0$ (first two homotopy groups are trivial), does it follow that $M$ is diffeomorphic to the $4$-sphere?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and warned us, that if we managed to solve it, we would get an instant Ph.D. -- so, keen on getting a Ph.D. before my bachelor degree, I went to work immediately! ;-)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My first thought was the following: If one could endow $M$ with a Riemannian metric giving a Riemannian manifold with constant sectional curvature $1$, then by compactness $M$ would be a complete, connected, simply connected manifold of curvature $1$, which would imply the statement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I obviously didn't get much further than this (my dreams were shattered!).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyways, this leads to the question: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;When is it possible to endow a smooth manifold with a metric which has some desired properties (i.e. constant curvature or bounded curvature)?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Has there been much work on this? Are there any good books/papers I could take a look at (just to get some impression of how the experts approach this problem)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was also wondering whether the above is actually an approach to the problem taken by people working in the field? Or may it be completely hopeless to try and gain any control of the metric globally?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, as always I thank in advance for any comments, answers etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Best regards,&#xA;S.L.&lt;/p&gt;&#xA;" OwnerUserId="3208" LastActivityDate="2011-03-03T17:07:53.210" Title="Smooth Poincaré Conjecture" Tags="&lt;differential-geometry&gt;&lt;differential-topology&gt;&lt;riemannian-geometry&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="24806" PostTypeId="1" CreationDate="2011-03-03T13:41:54.317" Score="3" ViewCount="146" Body="&lt;p&gt;How do i prove the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For every irrational number q, given e&gt;0, there exists natural numbers N and M such that |Nq-M|&amp;lt; e&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you.&lt;/p&gt;&#xA;" OwnerUserId="7741" LastEditorUserId="6296" LastEditorDisplayName="" LastEditDate="2011-03-03T13:44:19.233" LastActivityDate="2011-03-03T18:08:09.980" Title="Multiples of a given irrational number can be arbitrarily close to a natural number" Tags="&lt;homework&gt;&lt;real-analysis&gt;" AnswerCount="3" FavoriteCount="1" />
  <row Id="24808" PostTypeId="2" ParentId="24806" CreationDate="2011-03-03T13:57:44.950" Score="5" ViewCount="" Body="&lt;p&gt;Hint: Look at the fractional parts of $q,2q, 3q, \dots $. There are an infinite of them squeezed into $(0,1)$.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2011-03-03T13:57:44.950" CommentCount="7" />
  <row Id="24809" PostTypeId="1" CreationDate="2011-03-03T14:03:40.610" Score="1" ViewCount="52" Body="&lt;p&gt;I've been presented with a function expansion which I'm told is correct but I can't figure out where the sign in the second term might be coming from.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ &#xD;&#xA;&#xD;&#xA;e^{i\alpha(x_\mu + \epsilon \, n_\mu)} = e^{i\alpha(x_\mu)} ( 1 - i\,\epsilon \,n^\nu\, \partial_\nu \alpha(x_\mu) ) &#xD;&#xA;&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x_\mu$, $n_\mu$ are four vectors, the metric signature is + - - - and $\epsilon$ is infinitesimal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Taylor expanding this myself about $x_\mu$ I get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ &#xD;&#xA;&#xD;&#xA;e^{i\alpha(x_\mu + \epsilon \, n_\mu)} = e^{i\alpha(x_\mu)} ( 1 + i\,\epsilon \,n^\nu \,\partial_\nu \alpha(x_\mu) ) &#xD;&#xA;&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Am I missing something or is the presented equation wrong? Perhaps to do with the metric, or maybe they are doing something other than a Taylor expansion ? :-/ &lt;/p&gt;&#xA;" OwnerUserId="7738" LastActivityDate="2011-03-03T14:03:40.610" Title="4 vector Taylor expansion, sign confusion" Tags="&lt;taylor-expansion&gt;" CommentCount="2" FavoriteCount="1" />
  <row Id="24810" PostTypeId="1" CreationDate="2011-03-03T14:16:17.710" Score="2" ViewCount="165" Body="&lt;p&gt;I've long dreamt of an occasion &lt;em&gt;where hundreds or thousands of mathematicians work on a single problem&lt;/em&gt; in a truly coordinated way. Well, now that I'm nearing my doctoral defense in mathematics, I've decided to consider making this &lt;em&gt;dream&lt;/em&gt; a &lt;em&gt;reality&lt;/em&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I can't do it alone....&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are some of your thoughts on how best to implement such a scheme? What are some practical pitfalls preventing such an effort taking place? How could an organizing body avoid them? Would it be feasible to try a test run on Math.SE of some coordinated work, perhaps a simple but unsolved problem in an area of mathematics with various connections to subject areas? What would be the most appropriate level so that most could contribute?&lt;/p&gt;&#xA;" OwnerUserId="2720" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2011-03-03T18:50:27.847" LastActivityDate="2011-03-03T20:01:00.893" CommunityOwnedDate="2011-03-03T16:39:37.680" Title="Large-scale Coordinated Research in Mathematics" Tags="&lt;reference-request&gt;&lt;open-problem&gt;&lt;advice&gt;&lt;problem-solving&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="3" />
  <row Id="24812" PostTypeId="2" ParentId="24805" CreationDate="2011-03-03T14:19:39.173" Score="6" ViewCount="" Body="&lt;p&gt;I imagine you'd get more than a Ph. D!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I get the impression that it's generally hopeless to gain control of the metric globally.  The problem is that if $g_1$ and $g_2$ are metrics, then so is $g_1 + g_2$, but the curvature of $g_1+g_2$ can be wildly different from $g_1$ and $g_2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Said another way, typically one uses a partition of unity argument to show that metrics exist on all smooth (paracompact) manifolds, but curvature is &lt;em&gt;not&lt;/em&gt; well behaved under partitions of unity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another very relevant point is the Hopf conjecture: that $S^2\times S^2$ does not carry a metric of positive sectional curvature.  If we were to adapt your method to this setting, we're just trying to prove positive curvature, which is a far cry from constant positive sectional curvature, but the efforts here have been fairly fruitless.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, Burkhard Wilking has shown $\mathbb{R}P^2\times\mathbb{R}P^3$ has a metric of &quot;almost positive curvature&quot; - it has an open dense set of points $U$ such that for any point in $U$, all sectional curvatures are positive.  It follows from the classical Synge Theorem, that this example cannot be deformed to positive curvature everywhere.  So, if you were to get your hands on a &quot;sort of&quot; nice metric on your hypothetical $S^4$, then there's still no reason to assume you can deform it to one even of positive curvature.  (Though, to be clear, if $M$ is simply connected with a metric of almost positive curvature, it is still completely open whether or not this can be deformed to positive curvature.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the positive side, Jeff Cheeger showed the connect sum of any pair of $\mathbb{R}P^n$, $\mathbb{C}P^n$, $\mathbb{H}P^n$, or $\mathbb{O}P^2$ (of appropriate dimensions) carries a metric of nonnegative sectional curvature.  To do this he looks at each of those manifolds minus a point, and shows that near the deleted point, one can deform the metric to be nonnegatively curved everywhere and locally isometric to the product $S^{k}\times \mathbb{R}$ near the removed point.  This guarantees he can glue the metrics together on each piece and get a smooth nonnegatively curved metric.&lt;/p&gt;&#xA;" OwnerUserId="331" LastEditorUserId="331" LastEditorDisplayName="" LastEditDate="2011-03-03T17:07:53.210" LastActivityDate="2011-03-03T17:07:53.210" CommentCount="1" />
  <row Id="24813" PostTypeId="2" ParentId="24789" CreationDate="2011-03-03T14:28:35.220" Score="1" ViewCount="" Body="&lt;p&gt;Your observation is correct!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $\displaystyle \frac{a_2}{a_1} = x$ and $\displaystyle \frac{b_2}{b_1} = y$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then suppose the expressions you have are equal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{a_1}{x} + \frac{b_1}{y} = \frac{(a_1 + b_1)^2}{xa_1 + yb_1}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This gives us&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(ya_1 + xb_1)(xa_1 + yb_1) = xy(a_1 + b_1)^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which gives us, after some algebra and cancelling $\displaystyle a_1 b_1$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x^2 + y^2 = 2xy$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i.e.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ (x-y)^2 = 0 $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and thus,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x = y$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if the two expressions you have are equal, then it is necessarily true that $\displaystyle \frac{a_2}{a_1} = \frac{b_2}{b_1}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact you will always have (for positive reals)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{a_1}{x} + \frac{b_1}{y} \ge \frac{(a_1 + b_1)^2}{xa_1 + yb_1}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the equality occurring iff $\displaystyle x = y$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another (possibly quicker than the above) way to see this inequality is to apply Cauchy Schwarz to $\displaystyle (\sqrt{\frac{a}{x}}, \sqrt{\frac{b}{y}})$ and $\displaystyle (\sqrt{ax}, \sqrt{by})$, equality occuring only when these are linearly dependent (implying $x=y$).&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2011-03-03T17:26:51.263" LastActivityDate="2011-03-03T17:26:51.263" CommentCount="4" />
  <row Id="24815" PostTypeId="2" ParentId="24810" CreationDate="2011-03-03T14:34:14.640" Score="7" ViewCount="" Body="&lt;p&gt;There is already a similar scheme the &lt;strong&gt;Polymath projects&lt;/strong&gt; which you can find at the &lt;a href=&quot;http://polymathprojects.org/&quot; rel=&quot;nofollow&quot;&gt;blog&lt;/a&gt; and &lt;a href=&quot;http://michaelnielsen.org/polymath1/index.php?title=Main_Page&quot; rel=&quot;nofollow&quot;&gt;wiki&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="6460" LastActivityDate="2011-03-03T14:34:14.640" CommunityOwnedDate="2011-03-03T16:39:37.680" CommentCount="6" />
  <row Id="24816" PostTypeId="2" ParentId="24789" CreationDate="2011-03-03T15:03:02.893" Score="1" ViewCount="" Body="&lt;p&gt;Because for $a_{1},a_{2},b_{1},b_{2}\neq 0$ we have the following equivalent inequalities:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{a_{1}}{\dfrac{a_{2}}{a_{1}}}+\dfrac{b_{1}}{\dfrac{b_{2}}{b_{1}}}\neq &#xD;&#xA;\frac{a_{1}+b_{1}}{1+\dfrac{(a_{2}+b_{2})-(a_{1}+b_{1})}{a_{1}+b_{1}}}\Leftrightarrow \frac{a_{2}}{a_{1}}\neq \frac{b_{2}}{b_{1}}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This can be shown as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{a_{1}}{\dfrac{a_{2}}{a_{1}}}+\dfrac{b_{1}}{\dfrac{b_{2}}{b_{1}}}\neq &#xD;&#xA;\frac{a_{1}+b_{1}}{1+\dfrac{(a_{2}+b_{2})-(a_{1}+b_{1})}{a_{1}+b_{1}}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Leftrightarrow \frac{a_{1}^{2}}{a_{2}}+\frac{b_{1}^{2}}{b_{2}}\neq \frac{%&#xD;&#xA;\left( a_{1}+b_{1}\right) ^{2}}{a_{1}+b_{1}+(a_{2}+b_{2})-(a_{1}+b_{1})}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Leftrightarrow \frac{a_{1}^{2}}{a_{2}}+\frac{b_{1}^{2}}{b_{2}}\neq \frac{%&#xD;&#xA;\left( a_{1}+b_{1}\right) ^{2}}{a_{2}+b_{2}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Leftrightarrow \frac{a_{1}^{2}b_{2}+a_{2}b_{1}^{2}}{a_{2}b_{2}}\neq \frac{%&#xD;&#xA;\left( a_{1}+b_{1}\right) ^{2}}{a_{2}+b_{2}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Leftrightarrow \left( a_{1}^{2}b_{2}+a_{2}b_{1}^{2}\right) \left(&#xD;&#xA;a_{2}+b_{2}\right) \neq \left( a_{1}+b_{1}\right) ^{2}a_{2}b_{2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Leftrightarrow&#xD;&#xA;a_{1}^{2}b_{2}^{2}+a_{2}^{2}b_{1}^{2}-2a_{2}b_{2}a_{1}b_{1}\neq 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Leftrightarrow \left( a_{1}b_{2}-a_{2}b_{1}\right) ^{2}\neq 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Leftrightarrow a_{1}b_{2}\neq a_{2}b_{1}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Leftrightarrow \frac{a_{2}}{a_{1}}\neq \frac{b_{2}}{b_{1}}.$$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Your numerical case seems to be for $a_{1}=b_{1}=a_{2}=1,b_{2}=1.1$ for&#xA;which we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{1}{1/1}+\frac{1}{1.1/1}\neq \frac{2}{1+\frac{(1+1.1)-\left( 1+1\right) }{2%&#xD;&#xA;}}=\frac{2}{1.05}\Leftrightarrow \frac{1}{1}\neq \frac{1.1}{1}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/z8wN7.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="752" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2011-03-03T17:22:01.603" LastActivityDate="2011-03-03T17:22:01.603" CommentCount="2" />
  <row Id="24817" PostTypeId="1" AcceptedAnswerId="24858" CreationDate="2011-03-03T15:07:20.533" Score="2" ViewCount="175" Body="&lt;p&gt;I checked most of the posts about Taylor expansion with scalar functions. Could anyone tell me what is the multivariate version of Taylor's Theorem, and how I can use it?&lt;/p&gt;&#xA;" OwnerUserId="7743" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-03T16:27:14.553" LastActivityDate="2011-03-03T20:47:27.423" Title="What is the vector form of Taylor's Theorem?" Tags="&lt;linear-algebra&gt;&lt;multivariable-calculus&gt;" AnswerCount="2" />
  <row Id="24818" PostTypeId="1" AcceptedAnswerId="24820" CreationDate="2011-03-03T15:09:38.270" Score="1" ViewCount="114" Body="&lt;p&gt;Given continuous and Periodic function,&#xA;How can I prove that it is Uniformly continuous?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you!&lt;/p&gt;&#xA;" OwnerUserId="6163" LastActivityDate="2011-03-03T21:26:03.417" Title="continuous and Periodic function is Uniformly continuous" Tags="&lt;calculus&gt;" AnswerCount="2" />
  <row Id="24819" PostTypeId="2" ParentId="24735" CreationDate="2011-03-03T15:18:09.453" Score="1" ViewCount="" Body="&lt;p&gt;I think maybe what he means is that people think that simplicial homology is a homotopy invariant of polyhedra, or at least take the fact for granted, not that simplicial homology is not a homotopy invariant of simplicial complexes. I guess his main point is that homotopy invariants of simplicial complexes are not necessarily homotopy invariants of polyhedra.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;does this help? or was this much obvious?&lt;/p&gt;&#xA;" OwnerUserId="627" LastActivityDate="2011-03-03T15:18:09.453" CommentCount="2" />
  <row Id="24820" PostTypeId="2" ParentId="24818" CreationDate="2011-03-03T15:24:08.683" Score="4" ViewCount="" Body="&lt;p&gt;Observe that it is uniformly continuous when restricted to a closed interval whose length is a period of your function (or twice that, to simplify things a little bit), and then use periodicity to extend to the whole of $\mathbb R$.&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2011-03-03T15:24:08.683" CommentCount="7" />
  <row Id="24821" PostTypeId="2" ParentId="24806" CreationDate="2011-03-03T15:28:39.147" Score="2" ViewCount="" Body="&lt;p&gt;Here's a small variant: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the quotient ${\Bbb R}/{\Bbb Z}$ which can be identified to the circle $S^1$. The irrationality of $q$ is equivalent to the injectivity of the quotient map $\pi:{\Bbb R}\rightarrow S^1$ restricted to ${\Bbb Z}q$. The assertion amounts to prove that for any open neighborhood $V\ni0$ in $S^1$ there is always a $n\bar{q}\in V$ with $n\neq0$, i.e. &#xA;$\pi({\Bbb Z}q)\cap(V\setminus(0))\neq\emptyset$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the translates $V_n=n\bar{q}+V\subset S^1$ for $n\in{\Bbb Z}$ (still open, of course). If you can find an $m\in{\Bbb Z}$ such that there exists $n\bar{q}\in V_m\setminus(m\bar{q})$ then $(n-m)\bar{q}\in V\setminus(0)$ and you are done.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If not, it means that for all $m\in{\Bbb Z}$, $\pi({\Bbb Z}q)\cap V_m=\{m\bar{q}\}$. But this says that $\pi({\Bbb Z}q)$ is discrete in $S^1$ and since $S^1$ is compact, this is in contradiction with its infinitess.&lt;/p&gt;&#xA;" OwnerUserId="688" LastEditorUserId="688" LastEditorDisplayName="" LastEditDate="2011-03-03T16:58:24.990" LastActivityDate="2011-03-03T16:58:24.990" />
  <row Id="24822" PostTypeId="2" ParentId="24795" CreationDate="2011-03-03T15:29:34.403" Score="0" ViewCount="" Body="&lt;p&gt;Your question reminds me a robust model fitting method: RANSAC, which is widely used in computer vision to remove image feature outliers. Suppose the data are roughly located on two lines. If traditional fitting methods are employed, you may get a linear function that is very different to the ones you expected. But if you use RANSAC, at least it can give you one correct estimated line, and the points on the other line will be regarded as outliers.&lt;/p&gt;&#xA;" OwnerUserId="7156" LastActivityDate="2011-03-03T15:29:34.403" />
  <row Id="24823" PostTypeId="1" CreationDate="2011-03-03T15:36:49.340" Score="0" ViewCount="29" Body="&lt;p&gt;Show that deleting at most (m-s)(n-t)/s edges from a K_{m,n} will never destroy all its K_{s,t} subgraphs. I'm completly stuck here and any help would be welcome!&lt;/p&gt;&#xA;" OwnerUserId="2394" LastActivityDate="2011-03-03T17:56:49.250" Title="Forcing bipartite graphs?" Tags="&lt;graph-theory&gt;" AnswerCount="1" />
  <row Id="24824" PostTypeId="1" CreationDate="2011-03-03T15:52:13.117" Score="-2" ViewCount="93" Body="&lt;p&gt;ok the question that i have is that the equation, (a+1)^2 - (a)^2 + 2 + (a+1)^2 = (a+2)^2 that i made whaen i was eleven. Part of a bigger equation, such as a missing part of a bigger calculus equation, or maybe even an equation such as pie. You can check the liability of this equation if you find any natural nummber (a&gt;0) that doesn't work please inform me immiedently. Other wise please if you know anything please inform me i think this is/could be something big. This was made by Kristopher Cotton (me), alone,first.&lt;/p&gt;&#xA;" OwnerUserId="7712" LastActivityDate="2011-03-03T16:09:07.440" ClosedDate="2011-03-04T14:26:29.700" Title="consequtive squares" Tags="&lt;algebra-precalculus&gt;" AnswerCount="1" CommentCount="9" />
  <row Id="24825" PostTypeId="2" ParentId="24817" CreationDate="2011-03-03T15:53:23.897" Score="2" ViewCount="" Body="&lt;p&gt;It is not entirely clear which of the following you are asking for: (a) Taylor expansion for a &lt;em&gt;vector-valued&lt;/em&gt; function or (b) Taylor expansion for a function defined on $\mathbb{R}^N$ with $N &amp;gt; 1$. In the former case, you just use the usual Taylor's theorem component by component. In the latter, look at &lt;a href=&quot;http://en.wikipedia.org/wiki/Taylor%27s_theorem#Taylor.27s_theorem_for_several_variables&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; for the statement and proof. &lt;/p&gt;&#xA;" OwnerUserId="1543" LastActivityDate="2011-03-03T15:53:23.897" CommentCount="1" />
  <row Id="24826" PostTypeId="2" ParentId="24824" CreationDate="2011-03-03T16:02:46.297" Score="5" ViewCount="" Body="&lt;p&gt;Written as $(a+2)^2 - 2(a+1)^2 + a^2 = 2$, this is just the $n=2$ case of the well-known fact that a degree $n$ polynomial has constant $n$th finite difference. (Known at least since the time of Newton!) See &lt;a href=&quot;http://mathworld.wolfram.com/FiniteDifference.html&quot; rel=&quot;nofollow&quot;&gt;http://mathworld.wolfram.com/FiniteDifference.html&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2011-03-03T16:02:46.297" CommentCount="3" />
  <row Id="24828" PostTypeId="1" AcceptedAnswerId="24835" CreationDate="2011-03-03T16:11:43.030" Score="1" ViewCount="43" Body="&lt;p&gt;Vector $\vec{a}$ can be broken down into its components $\vec{a}_\parallel$ and $\vec{a}_\perp$ relative to $\vec{e}$.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\vec{a}_\parallel = (\vec{a}\vec{e})\vec{e}$ &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;and &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\vec{a}_\perp = \vec{e} \times (\vec{a} \times \vec{e})$ &lt;strong&gt;(f1)&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The orthogonal part can be found via application of the triple product:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\vec{a}_\perp = \vec{a} - \vec{a}_\parallel = \vec{a}(\vec{e}\vec{e}) - \vec{e}(\vec{e}\vec{a}) = \vec{e} \times (\vec{a} \times \vec{e})$  &lt;strong&gt;(f2)&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This one causes me problems. I tried to use some values for the formulas and disaster strikes:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\vec{a} = \begin{pmatrix} 3 \\ 2 \\ 1 \end{pmatrix}$ and $\vec{e} = \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}$ makes $\vec{a}_\parallel = \begin{pmatrix} 6 \\ 6 \\ 6 \end{pmatrix}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I thought to follow the values &quot;all the way&quot; through the last formula &lt;strong&gt;f2&lt;/strong&gt;. So I calculate for $\vec{a}_{\perp subtraction} = \vec{a} - \vec{a}_\parallel = \begin{pmatrix} -3 \\ -4 \\ -5 \end{pmatrix}$ but I find for $\vec{a}_{\perp cross} = \vec{e} \times (\vec{a} \times \vec{e}) = \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} \times \begin{pmatrix} 1 \\ -2 \\ 1 \end{pmatrix} =\begin{pmatrix} 3 \\ 0 \\ -3 \end{pmatrix}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can someone point out my error? Must be some mis-calculation, as $\vec{a}_{\perp cross} + \vec{a}_\parallel \neq \vec{a}$ and I do not really find the two $\perp$-vectors parallel.&lt;/p&gt;&#xA;" OwnerUserId="5870" LastEditorUserId="6622" LastEditorDisplayName="" LastEditDate="2011-03-03T16:30:05.470" LastActivityDate="2011-03-03T16:45:56.363" Title="Cross product: problem with breakdown of vector to parallel + orthogonal components." Tags="&lt;vector-calculus&gt;&lt;learning&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="24829" PostTypeId="2" ParentId="24793" CreationDate="2011-03-03T16:15:24.010" Score="2" ViewCount="" Body="&lt;p&gt;(I will use the Einstein summation convention and $\partial_i = \frac{\partial}{\partial{x_i}}$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The stationary solutions fulfill $L_{FP} W_{st}=0$ as explained. This can be rewritten as $\partial_i J_i =0$ where we have introduced the current&#xA;$$J_i = D_i W_{st} - \partial_j  D_{ij} W_{st}.$$&#xA;Many FP equations allow for a stationary solution where the current vanishes $J_i=0$ (and not only the divergence is equal to 0). Assuming that, we obtain&#xA;$D_i W_{st} = \partial_j D_{ij} W_{st} = (\partial_j D_{ij}) W_{st} +  D_{ij} \partial_j W_{st} $ which is equivalent to &#xA;$$ \partial_j \log W_{st} = (D^{-1})_{ji} (D_i - \partial_k D_{ik}).$$&#xA;Integrating this equation, we obtain&#xA;$$ W_{st}(x) = c \exp\left[\int^x dx&amp;#39;_j  (D^{-1})_{ji} (D_i -\partial_k D_{ik}) \right]$$ with $c$ some normalization constant. So the last two equations of the questions are in fact wrong.&lt;/p&gt;&#xA;" OwnerUserId="7266" LastEditorUserId="7266" LastEditorDisplayName="" LastEditDate="2011-03-03T18:44:50.727" LastActivityDate="2011-03-03T18:44:50.727" CommentCount="11" />
  <row Id="24830" PostTypeId="1" AcceptedAnswerId="24832" CreationDate="2011-03-03T16:17:21.437" Score="0" ViewCount="70" Body="&lt;p&gt;Given two connected open sets $U,V \subset X$ such that $U \cap V$ is path connected and $U \cup V = X$, then $\pi_1(U) \ast_{\pi_1(U \cap V)} \pi_1(V) \cong \pi_1(X)$. This is of course the Seifert Van Kampen theorem, a question in Munkres asks if the homomorphism induced by the inclusion map of $i: V \rightarrow X$ is trivial what can you say about the homomorphism induced by $j: U \rightarrow X$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's clear to me that $j_\ast$ must be surjective and therefore $\pi_1(X) \cong \pi_1(U)/ker(j_\ast)$. My question is can you say anything more? Does the kernel of $j_\ast$ relate at all to $F$, as given in the usual &lt;a href=&quot;http://upload.wikimedia.org/wikipedia/commons/thumb/5/5e/VanKampen-01.png/750px-VanKampen-01.png&quot; rel=&quot;nofollow&quot;&gt;diagram&lt;/a&gt;. Would the kernel be at all related to the normal closure of the image of the induced homomorphism of the inclusion $k: U \cap V \rightarrow U$?&lt;/p&gt;&#xA;" OwnerUserId="1930" LastActivityDate="2011-03-03T17:47:10.503" Title="On certain cases of Seifert Van Kampen" Tags="&lt;algebraic-topology&gt;&lt;seifert-van-kampen&gt;" AnswerCount="2" />
  <row Id="24831" PostTypeId="2" ParentId="24795" CreationDate="2011-03-03T16:18:02.707" Score="1" ViewCount="" Body="&lt;p&gt;Imagine that your data is presented as a binary two dimensional image (assuming a unit value whenever the point coordinate (x,y) is present in your data). Then the problem is equivalent to straight line detection in binary images. The equivalent problem can be solved using the &lt;a href=&quot;http://en.wikipedia.org/wiki/Hough_transform&quot; rel=&quot;nofollow&quot;&gt;Hough transform&lt;/a&gt; for straight line detection.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The basic idea of the hough transform is as follows:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;It transforms an array from the geometrical coordinates (x,y) to the space of initial points and slopes.&lt;/li&gt;&#xA;&lt;li&gt;Every point contributes a unit to all sets of slopes and initial points of straight lines passing through this point.&lt;/li&gt;&#xA;&lt;li&gt;The Hough transform of the whole array is the linear superposition of the transform of it's individual points.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The detected lines appear as peaks in the hough transform image&lt;/p&gt;&#xA;" OwnerUserId="1430" LastActivityDate="2011-03-03T16:18:02.707" CommentCount="1" />
  <row Id="24832" PostTypeId="2" ParentId="24830" CreationDate="2011-03-03T16:21:42.213" Score="2" ViewCount="" Body="&lt;p&gt;Notice that the question is purely algebraic. If $G=\pi_1(U)$, $G&amp;#39;=\pi_1(V)$ and $H=\pi_1(U\cap V)$, you are asking what is the kernel of the obvious map $G\to G*_HG&amp;#39;$ (where the amalgamation happens along the images of $H$ in $G$ and $G'$) if the map $G&amp;#39;\to G*_HG&amp;#39;$ is trivial. &lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2011-03-03T16:21:42.213" CommentCount="2" />
  <row Id="24833" PostTypeId="1" AcceptedAnswerId="24851" CreationDate="2011-03-03T16:29:51.820" Score="9" ViewCount="569" Body="&lt;p&gt;Hello,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wonder is there an elementary proof for Fermat's last theorem. Why it's so difficult to prove this theorem by elementary method? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks,  &lt;/p&gt;&#xA;" OwnerUserId="5711" LastActivityDate="2011-03-04T05:24:59.777" Title="Is there an elementary proof for Fermat's last theorem?" Tags="&lt;elementary-number-theory&gt;" AnswerCount="3" CommentCount="13" />
  <row Id="24834" PostTypeId="2" ParentId="24828" CreationDate="2011-03-03T16:32:29.117" Score="2" ViewCount="" Body="&lt;p&gt;Your formulas are valid only if $\vec{e}$ is a unit vector, but the $\vec{e}$ you're using isn't. To use an arbitrary non-zero vector $\vec{e}$, you'd need to divide the right-hand side of both formulas by $\lvert \vec{e}\rvert^2$.&lt;/p&gt;&#xA;" OwnerUserId="6622" LastActivityDate="2011-03-03T16:32:29.117" />
  <row Id="24835" PostTypeId="2" ParentId="24828" CreationDate="2011-03-03T16:32:56.903" Score="1" ViewCount="" Body="&lt;p&gt;To get the projection along $\vec{e}$ i.e. $\vec{a_{||}}$, you need to project your vector $a$ along the unit vector in the direction of $\vec{e}$. Similarly, if you want the component of $\vec{a}$ perpendicular to $\vec{e}$, $\vec{a_{\perp}} = \frac{\vec{e}}{||\vec{e}||}_2 \times \left( \vec{a} \times \frac{\vec{e}}{||\vec{e}||}_2  \right)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, $\vec{a_{||}} = \left( \vec{a} \cdot \frac{\vec{e}}{||\vec{e}||}_2 \right) \frac{\vec{e}}{||\vec{e}||}_2$ and $\vec{a_{\perp}} = \vec{a} - \vec{a_{||}} = \frac{\vec{e}}{||\vec{e}||}_2 \times \left( \vec{a} \times \frac{\vec{e}}{||\vec{e}||}_2  \right)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your case, $\vec{a_{||}} = \frac{6}{\sqrt{3}} \frac1{\sqrt{3}} \left(1,1,1\right)^T = \left( 2,2,2 \right)$ and hence $\vec{a_{\perp}} = \left( 3,2,1 \right) - \left( 2,2,2 \right) = \left( 1,0,-1 \right) = \frac{\vec{e}}{||\vec{e}||}_2 \times \left( \vec{a} \times \frac{\vec{e}}{||\vec{e}||}_2  \right)$&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2011-03-03T16:45:56.363" LastActivityDate="2011-03-03T16:45:56.363" />
  <row Id="24836" PostTypeId="2" ParentId="20397" CreationDate="2011-03-03T16:41:45.887" Score="3" ViewCount="" Body="&lt;p&gt;Highbrow: Derivation of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Euler%E2%80%93Lagrange_equation&quot; rel=&quot;nofollow&quot;&gt;Euler-Lagrange equations&lt;/a&gt; describing how a physical system evolves through time from Hamilton's Least Action Principle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a very brief summary. Consider a very simple physical system consisting of a point mass moving under the force of gravity, and suppose you know the position $q$ of the point at two times $t_0$ and $t_f$. Possible trajectories of the particle as it moved from its starting to ending point correspond to curves $q(t)$ in $\mathbb{R}^3$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of these curves describes the physically-correct motion, wherein the particle moves in a parabolic arc from one point to the other. Many curves completely defy the laws of physics, e.g. the point zigs and zags like a UFO as it moves from one point to the other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hamilton's Principle gives a criteria for determining which curve is the physically correct trajectory; it is the curve $q(t)$ satisifying the variational principle&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\min_q \int_{t_0}^{t_f} L(q, \dot{q}) dt$$&#xA;subject to the constraints $q(t_0) = q_0, q(t_f) = q_f$. &#xA;Where $L$ is a scalar-valued function known as the Lagrangian that measures the difference between the kinetic and potential energy of the system at a given moment of time. (Pedantry alert: despite being historically called the &quot;least&quot; action principle, really instead of minimizing we should be extremizing; ie all critical points of the above functional are physical trajectories, even those that are maxima or saddle points.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It turns out that a curve $q$ satisfies the variational principle if and only if it is a solution to the ODE&#xA;$$ \frac{d}{dt} \frac{\partial L}{\partial \dot{q}} + \frac{\partial L}{\partial q} = 0,$$&#xA;roughly equivalent to the usual Newton's Second Law $ma-F=0$, and the key step in the proof of this equivalence is integration by parts. What is remarkable here is that we started with a boundary-value problem -- given two positions, how did we get from one to the other? -- and ended with an ODE, an initial-value problem -- given an initial position and velocity, how does the point move as we advance through time?&lt;/p&gt;&#xA;" OwnerUserId="7530" LastActivityDate="2011-03-03T16:41:45.887" CommunityOwnedDate="2011-03-03T16:41:45.887" />
  <row Id="24837" PostTypeId="2" ParentId="24791" CreationDate="2011-03-03T16:49:34.130" Score="4" ViewCount="" Body="&lt;p&gt;First of all some general remarks: We use the existential quantifier ($\exists$) when we want to talk about the existence of something. If the phrase is talking about a &lt;em&gt;specific&lt;/em&gt; individual then (most probably) the existential quantifier is what you need. On the other hand the universal quantifier ($\forall$) talks about everyone. If the phrase refers to everyone, or a group of people with a property or about people generally then chances are that you need to use the universal quantifier. For example if we say &quot;If there is a kid who is happy then everyone else is happy&quot; then the first part of the phrase refers to a &lt;em&gt;specific&lt;/em&gt; kid. Not to groups of kids or kids in general but about a specific individual. The second part of the phrase refers to everyone. So the first part will require an existential quantifier while the second a universal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When we say &quot;Anyone that has a property A also has property B&quot;, this is translated with quantifiers as $(\forall x)[A(x)\to B(x)]$. On the other hand when we want to say &quot;If there exists someone with property A then something else (unrelated to who that someone is)&quot; then we write $[(\exists x)A(x)]\to\ldots$. In this case observe that what goes after the &quot;$\to$&quot; isn't in the range of the existential quantifier. That means that it's independent of the $x$. Finally when we want to say that &quot;There exists someone with property A and property B&quot; then (as you may have guessed) we write $(\exists x)[A(x)\land B(x)]$. Returning now to the example I gave in the first paragraph: &quot;If there is a kid who is happy then everyone is happy&quot;. Observe that what goes after the mention of kid isn't related to the kid at all. So this will be written using quantifiers as $[(\exists x)(C(x)\land H(x))]\to(\forall x)H(x)$. If I say &quot;If someone is a kid then he is happy&quot; then observe that I am not referring to a specific kid but rather to kids in general. Thus I will use a universal quantifier: $(\forall x)(C(x)\to H(x))$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So when you try to turn English phrases into ones with quantifiers you have to first understand the meaning of the phrase and based on that, decide the type of the quantifier that you should use as well as its range. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now regarding your problems:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&quot;If any loves Bob, then Bob loves everyone.&quot; In this case that someone is specific. A certain person that loves Bob. After that it is a simple $p\to q$, so to figure this one out write using quantifiers the $p$ and $q$. &quot;someone loves Bob&quot; is $(\exists x)L(x,b)$ and &quot;Bob loves everyone&quot; is $(\forall y)L(b,y)$. So this one should be $(\exists x)L(x,b)\to(\forall y)L(b,y)$.&lt;/li&gt;&#xA;&lt;li&gt;&quot;Happy people and kids love Bob.&quot; There are two ways to see this. A first simpler approach is to write it as &quot;happy people love Bob and kids love Bob&quot;. This would be $[(\forall x)(U(x)\land H(x)\to L(x,b))]\land[(\forall y)(C(y)\to L(y,b))]$. But there is another way to approach this, namely: &quot;If someone is a happy person or he is a kid then he loves Bob&quot;. That would be $(\forall x)[(U(x)\land H(x))\lor C(x)\to L(x,b)]$.&lt;/li&gt;&#xA;&lt;li&gt;&quot;There is some kid that everyone loves&quot;. In this one your answer is not correct. What you wrote says &quot;there is a kid that is loved by someone&quot; but we want a kid that is loved by everyone. This would be $(\exists x)(\forall y)(L(y,x))$.&lt;/li&gt;&#xA;&lt;li&gt;&quot;There is a happy kid who loves all unhappy kids&quot;. Again this happy kid is specific. So this would be $(\exists x)[H(x)\land C(x)\land (\forall y)(\lnot H(y)\to L(x,y)]$. In this one since this kid is related the unhappy kids observe that the range of the existential quantifier is the whole phrase.&lt;/li&gt;&#xA;&lt;li&gt;&quot;At least two kids love Bob&quot;. Without use of the equality relation you cannot properly write this unless you assume something more to be true (for example that one person cannot love himself and that every kid loves another kid). Your phrase is almost correct except for the fact that it doesn't specify that these $y$ and $x$ are two different individuals. To do this simply add $x\neq y$. So this would be $\exists x\exists y(L(x,b)\land L(y,b)\land x\neq y)$.&lt;/li&gt;&#xA;&lt;li&gt;&quot;The kid who loves Bob loves everyone&quot;. This appears a bit ambiguous, since it makes it sound like there exists only one kid that loves Bob. But I will assume that the phrase means &quot;If a kid loves Bob then he loves everyone&quot;. This phrase doesn't refer to a specific kid but to any kid that loves Bob (if you have a hard time seeing this observe that the phrase as I wrote it means the same with the phrase &quot;Kids that love Bob love everyone&quot;). This would be $(\forall x)[(C(x)\land L(x,b))\to(\forall y)(L(x,y)]$. &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;P.S.: Regarding the last one if the phrase means &quot;There is a unique kid that loves Bob and that kid loves everyone&quot; then you have to convey the information that this kid is unique as well. This would be $(\exists x)[C(x)\land L(x,b)\land(\forall y)((C(y)\land L(y,b))\to y=x)]$. What this says is that if anyone is a kid and loves Bob then he is $x$. So the whole phrase would be: $(\exists x)[C(x)\land L(x,b)\land(\forall y)((C(y)\land L(y,b))\to y=x)\land(\forall z)L(x,z)]$.&lt;/p&gt;&#xA;" OwnerUserId="1772" LastEditorUserId="1772" LastEditorDisplayName="" LastEditDate="2011-03-03T17:21:19.640" LastActivityDate="2011-03-03T17:21:19.640" />
  <row Id="24838" PostTypeId="2" ParentId="24833" CreationDate="2011-03-03T17:28:33.127" Score="5" ViewCount="" Body="&lt;p&gt;As far as I know, the only proof there is, is Wiles's proof. And that is not an elementary proof. There is an article about his proof on &lt;a href=&quot;http://en.wikipedia.org/wiki/Wiles%27_proof_of_Fermat%27s_Last_Theorem&quot; rel=&quot;nofollow&quot;&gt;wikipedia&lt;/a&gt;, if you're interested. There is also this video &lt;a href=&quot;http://video.google.com/videoplay?docid=8269328330690408516#&quot; rel=&quot;nofollow&quot;&gt;documentary&lt;/a&gt; about it that I would highly recommend.&lt;/p&gt;&#xA;" OwnerUserId="7044" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2011-03-03T18:27:01.607" LastActivityDate="2011-03-03T18:27:01.607" CommentCount="4" />
  <row Id="24839" PostTypeId="1" CreationDate="2011-03-03T17:44:40.210" Score="2" ViewCount="117" Body="&lt;p&gt;There are few well-know styles of logical systems (LS): Hilbert-style, Sequent-style, Natural deducton style. And such lesser-know styles as Gottlob-Frege two-demensional notation, systems with graphical syntax etc. (&lt;a href=&quot;http://en.wikipedia.org/wiki/Proof_calculus&quot; rel=&quot;nofollow&quot;&gt;In wikipedia&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And I have some questions:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;For example, is the Resolution method (language of clauses and resolution rule) is another &lt;strong&gt;style&lt;/strong&gt; of LS? Or is it (resolution method) something other than style of LS?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Is the Lambda-calsulus is another &lt;strong&gt;style&lt;/strong&gt; of LS? Or one can consider it as kind of semantics? How Lambda-calculus relate to logical systems?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;And finally, how one can modify logical systems (classical LS)? For example, One can otherwise interpret the logical connectives (the other semantics of logical connectives) and as a result get a constructive logic; or one can introduce anoter logical operation, and as a result get a modal logic; or one can modify type of quantifiers and as a result get a high-order logics; etc. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="7529" LastActivityDate="2011-04-27T04:30:10.470" Title="Styles of logical systems" Tags="&lt;logic&gt;&lt;lambda-calculus&gt;" AnswerCount="1" />
  <row Id="24840" PostTypeId="1" AcceptedAnswerId="24864" CreationDate="2011-03-03T17:45:48.163" Score="8" ViewCount="146" Body="&lt;p&gt;Hi, I'm doing one of the exercises of Stewart and Tall's book on Algebraic Number Theory. The problem concerns finding an expression for the norm in the cyclotomic field $K = \mathbb{Q}(e^{2\pi i / 5})$. The exact problem is the following:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;blockquote&gt;&#xA;    &lt;p&gt;If $\zeta = e^{2 \pi i / 5}$, $K = \mathbb{Q}(e^{2\pi i / 5})$, prove that the norm of $\alpha \in \mathbb{Z}[\zeta]$ is of the form $\frac{1}{4}(A^2 -5B^2)$ where $A, B \in \mathbb{Z}$. &lt;/p&gt;&#xA;    &#xA;    &lt;p&gt;(&lt;strong&gt;Hint:&lt;/strong&gt; In calculating $\textbf{N}(\alpha)$, first calculate $\sigma_1 (\alpha) \sigma_4 (\alpha)$ where $\sigma_i (\zeta) := \zeta^{i}$. Show that this is of the form $q + r\theta + s\phi$ where $q, r, s \in \mathbb{Z}$, $\theta = \zeta + \zeta^{4}$ and $\phi = \zeta^{2} + \zeta^{3}$. In the same way establish $\sigma_2 (\alpha) \sigma_3 (\alpha) = q + s\theta + r\phi$  )&lt;/p&gt;&#xA;    &#xA;    &lt;p&gt;Using Exercise $3$ prove that $\mathbb{Z}[\zeta]$ has an infinite number of units.&lt;/p&gt;&#xA;  &lt;/blockquote&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now, I've already done what the hint says and arrived at the following. If we let $\alpha = a +b\zeta^{} + c\zeta^{2} + d\zeta^{3} \in \mathbb{Z}[\zeta]$ then after simplifying I get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\textbf{N}(\alpha) = \sigma_1 (\alpha) \sigma_4 (\alpha) \sigma_2(\alpha) \sigma_3(\alpha) = ( q + r\theta + s\phi ) ( q + s\theta + r\phi )$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ = q^2 + (qr + qs)(\theta + \phi) + rs(\theta^2 + \phi^2) + (r^2 + s^2)\theta \phi$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and then it is not that hard to see that $\theta + \phi = -1$, $\theta^2 + \phi^2 = 3$ and $\theta \phi = -1$ so that in the end&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\textbf{N}(\alpha) = q^2 - (qr + qs) + 3rs - (r^2 + s^2)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $q = a^2 + b^2 + c^2 + d^2$, $r = ab + bc + cd$ and $s = ac + ad + bd$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, here I got stuck because I just can't take the last expression for the norm into the form that the exercise wants. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The purpose is to get that nice form for the norm to find units by solving the diophantine equation $\textbf{N}(\alpha) = \pm 1$, which is what the Exercise $3$ mentioned in the statatement of the problem is about.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I already know how to prove the existence of infinitely many units in $\mathbb{Z}[\zeta]$ (without using Dirichlet's Unit Theorem of course), but the exercise also demands a proof that the norm is equal to $\frac{1}{4}(A^2 -5B^2)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I even asked my professor about this and we were not able to get the desired form for the norm. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;blockquote&gt;&#xA;    &lt;p&gt;So my question is if anybody knows how to prove that the norm has that form, and if so, how can I show that? Or if it could be that maybe the hint given in the exercise is not that helpful?&lt;/p&gt;&#xA;  &lt;/blockquote&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Thanks a lot in advance for any help with this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After looking at Derek Jennings' answer below, to get from the expression I had for the norm to the one in Derek's answer is just a matter of taking out a common factor of $1/4$ in the expression and then completing the square.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\textbf{N}(\alpha) = q^2 - (qr + qs) + 3rs - (r^2 + s^2) = q^2 - q(r+s) + rs - (r-s)^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ = \frac{1}{4}( 4q^2 - 4q(r+s) + 4rs - 4(r-s)^2  ) $$ $$=  \frac{1}{4} ( 4q^2 - 4q(r+s) +\overbrace{(r+s)^2} - \overbrace{(r+s)^2} + 4rs - 4(r-s)^2  )$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ = \frac{1}{4} ( (2q -(r+s))^2 -(r-s)^2 - 4(r-s)^2 )$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ = \frac{1}{4}( (2q - r - s)^2 - 5(r-s)^2 ) = \frac{1}{4}(A^2 - B^2)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;as desired. Of course it is easier if you already know what to get at =)&lt;/p&gt;&#xA;" OwnerUserId="900" LastEditorUserId="900" LastEditorDisplayName="" LastEditDate="2011-03-03T21:19:55.113" LastActivityDate="2011-03-03T21:19:55.113" Title="Finding the norm in the cyclotomic field $\mathbb{Q}(e^{2\pi i / 5})$" Tags="&lt;abstract-algebra&gt;&lt;algebraic-number-theory&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="24841" PostTypeId="1" CreationDate="2011-03-03T17:46:44.873" Score="0" ViewCount="19" Body="&lt;p&gt;I've been reading the following papers in access control.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://faculty.nps.edu/dedennin/publications/lattice76.pdf&quot; rel=&quot;nofollow&quot;&gt;http://faculty.nps.edu/dedennin/publications/lattice76.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.winlab.rutgers.edu/~trappe/Courses/AdvSec05/access_control_lattice.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.winlab.rutgers.edu/~trappe/Courses/AdvSec05/access_control_lattice.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The papers ultimately try to reduce the access control mechanisms to lattice based access control. But I do not understand why reducing to lattice model is so important.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone comment?&lt;/p&gt;&#xA;" OwnerUserId="7750" LastActivityDate="2011-03-03T17:46:44.873" Title="What is the importance of lattice based access control?" Tags="&lt;lattices&gt;" />
  <row Id="24842" PostTypeId="2" ParentId="24830" CreationDate="2011-03-03T17:47:10.503" Score="1" ViewCount="" Body="&lt;p&gt;Indeed, $j_*$ must be surjective.  There are certainly many possibilities for the kernel:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(a)  $j_*$ could be the zero map, for instance if $X$ is simply connected ($U$ need not be simply connected here);&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(b)  $j_*$ could be injective, for instance if $V$ is a point;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(c)  or it could be somewhere in between, for instance if $X$ is a torus, $U$ a punctured torus, $V$ a disk, and $U\cap V$ a circle.  (Here $j_*$ is the &quot;abelianization&quot; map from $F_2$ to $Z^2$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There must be many other simple examples.  I would guess that Munkres is looking for &quot;$j_*$ is surjective.&quot;&lt;/p&gt;&#xA;" OwnerUserId="6071" LastActivityDate="2011-03-03T17:47:10.503" CommentCount="1" />
  <row Id="24843" PostTypeId="1" AcceptedAnswerId="26182" CreationDate="2011-03-03T17:55:20.107" Score="2" ViewCount="110" Body="&lt;p&gt;Hi!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;NOTATION: Let $\mathcal{C}=\mathcal{V}(F)\subseteq\mathbb{P}^2$ be a curve of degree $3\!=\!deg(F)$ with no singularities and let $A_0\!\!\in\!\mathcal{C}$ be fixed. Let $Div(\mathcal{C})$ denote the group of divisors on $\mathcal{C}$, i.e. the set of all formal sums $$\{\sum\limits_{P\in\mathcal{C}} n_PP\,|\; n_P\!\in\!\mathbb{Z}, \text{only finitely many } n_P \text{ are not zero}\},$$&#xA;let $\mathbb{F}(\mathcal{C})$ be $\{\text{rational functions from }\mathcal{C}\text{ to }\mathbb{F}\}$, i.e. the field of fractions of $\mathbb{F}[x_0\!:\!x_1\!:\!x_2]/I(\mathcal{C})$. Let $\psi:\mathbb{F}(\mathcal{C})\setminus\{0\}\rightarrow Div(\mathcal{C})$ denote the mapping, that sends each rational function $f$ to the principal divisor $(f)=\sum_{P\in\mathcal{C}}\mu_P(f,F)P$ where $\mu_P(f,F)$ is the intersection multiplicity of curves $\mathcal{V}(f),\mathcal{V}(F)$ in $P$. Then $Cl(\mathcal{C})$ denotes the group of divisor classes on $\mathcal{C}$, i.e. $Div(\mathcal{C})/im(\psi)$. So any two divisors $D_1$ and $D_2$ are equivalent, $D_1\sim D_2$, iff $D_1-D_2=(f)$ for some $f\in\mathbb{F}(\mathcal{C})$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;QUESTION: Define $\varphi:\mathcal{C}\rightarrow Cl^0(\mathcal{C})\!=\!\{\text{divisor classes on }\mathcal{C}\text{ of degree }0\}$ as a mapping, that sends each $A$ to the divisor class of $A-A_0$. How can I &lt;strong&gt;prove that $\varphi$ is surjective&lt;/strong&gt;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;WHAT IS ALREADY KNOWN: on a smooth cubic curve $\mathcal{C}$ for $P,Q,R,S\in\mathcal{C}$:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$P\sim Q\Leftrightarrow P=Q$&lt;/li&gt;&#xA;&lt;li&gt;$P+Q\sim R+S \;\;\Longleftrightarrow\;\;$ the line through $P,Q$ intersects the line through $R,S$ on $\mathcal{C}$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;help&lt;/p&gt;&#xA;" OwnerUserId="7325" LastEditorUserId="7325" LastEditorDisplayName="" LastEditDate="2011-03-03T18:11:59.080" LastActivityDate="2011-03-10T10:31:00.460" Title="every divisor of degree $0$ on a smooth cubic curve $\mathcal{C}\subseteq\mathbb{P}^2$ is equivalent to $A-A_0$ for a fixed $A_0\in\!\mathcal{C}$" Tags="&lt;algebraic-geometry&gt;&lt;plane-curves&gt;&lt;algebraic-curves&gt;" AnswerCount="1" CommentCount="9" />
  <row Id="24844" PostTypeId="2" ParentId="24823" CreationDate="2011-03-03T17:56:49.250" Score="2" ViewCount="" Body="&lt;p&gt;Here is a hint:  If you can find some number $N$ of edge-disjoint $K_{s,t}$ subgraphs of $K_{m,n}$, then because they are edge-disjoint, you would have to delete at least $N$ edges from $K_{m,n}$ to be sure none of your subgraphs was intact.&lt;/p&gt;&#xA;" OwnerUserId="6071" LastActivityDate="2011-03-03T17:56:49.250" />
  <row Id="24845" PostTypeId="1" AcceptedAnswerId="24850" CreationDate="2011-03-03T17:57:45.567" Score="2" ViewCount="87" Body="&lt;p&gt;Let $f \in m(\Omega,\mathcal{F})$, i.e. $f \mapsto [-\infty,\infty]$ and let $A \in \mathcal{F}$ be an atom. Prove that $f$ is almost everywhere constant on A: there exists $k \in [-\infty,\infty]$ such that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\mu (\{\omega \in A : f(\omega) \neq k \} )=0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was thinking let $k=\frac{1}{\mu(A)}\int \limits_{A} f\,d\mu$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then let $B=\{\omega \in A :f(\omega) \neq k \}$.  Since A is an atom, and B is a subset of A then $\mu(B)=0$, in which case we're done, or $\mu(B)=\mu(A)$.  So I need to show that $\mu(B)&amp;lt;\mu(A)$.&lt;/p&gt;&#xA;" OwnerUserId="2505" LastActivityDate="2011-03-04T01:58:06.930" Title="A measurable function on an atom is almost everywhere constant." Tags="&lt;analysis&gt;&lt;measure-theory&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="24847" PostTypeId="1" AcceptedAnswerId="25762" CreationDate="2011-03-03T18:02:26.593" Score="8" ViewCount="125" Body="&lt;p&gt;I'm trying to prove that, given a principal fibration $\Omega B \rightarrow F \stackrel{p}{\rightarrow} E$ such that $p$ is a retraction of $F$ onto $E$, the total space $F$ is homotopy equivalent to the product $\Omega B \times E$. This is essentialy a problem from Hatcher (4.3, Q22).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, since $E$ is a retract of $F$, we have a map $p \colon E \rightarrow F$ such that $ps = 1$. So the long exact sequence in homotopy for the fibration breaks into split short exact sequences, and one approach I've been trying is to show that the isomorphism in the splitting is induced from some map $F \rightarrow \Omega B \times E$, so that Whitehead's theorem can finish the job (you can assume I'm always talking about CW complexes).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another approach I've tried - inspired by the proof of the similar statement about principal bundles and global cross-sections - is to try and define an action of the fibre $\Omega B$ on $F$. Up to homotopy, we can replace $F$ with the homotopy fibre $F_q$, where $E \stackrel{q}{\rightarrow} B$ is the rest of the fibration sequence in the principal fibration, and it's then very easy to define an action of $\Omega B$ on $F_q$ (again, see the exercises in Hatcher), so we have an action of $\Omega B$ on $F$ up to homotopy. So we can define a map $\Omega B \times E \rightarrow F$ as the composite $$\Omega B \times E \stackrel{1 \times s}{\rightarrow} \Omega B \times F \rightarrow F,$$ where the last map is the action, and I thought about trying to show that this induces isomorphisms in homotopy for, once again, a solution by Whitehead.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively, I could try and define a map $F \rightarrow \Omega B \times E$ and show that it's the homotopy inverse of the above map, but I've come a little unstuck in trying to do this. There's no obvious map $F \rightarrow\Omega B$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've been wrestling with this problem for the last few days, and I've reached the point where I'm just going around in circles and finding it hard to make much progress. Sorry for the hodgepodge nature of this post, but I wanted to explain the various ways I've been going about solving this. I'm going to leave it on the backburner for a few days, sleep on it a little more, and hopefully I'll come back to it and the answer will be obvious --- it does feel like I'm missing something important that's staring me in the face. In the meantime, though, I'd love it if someone could suggest a hint or two, or let me know if I'm tackling this in the right way! &lt;/p&gt;&#xA;" OwnerUserId="7748" LastEditorUserId="7748" LastEditorDisplayName="" LastEditDate="2011-03-03T19:54:03.837" LastActivityDate="2011-03-08T14:08:33.167" Title="Principal fibrations with a section are trivial" Tags="&lt;algebraic-topology&gt;" AnswerCount="1" FavoriteCount="2" />
  <row Id="24848" PostTypeId="2" ParentId="24735" CreationDate="2011-03-03T18:06:53.327" Score="7" ViewCount="" Body="&lt;p&gt;I didn't state it very well- what I meant is that standard algebraic topology textbooks take for granted (or cause the reader to take for granted) that topology of polyhedra is the same as topology of simplicial complexes. Sean Tilson's response is spot-on; I'll restate it in my own words.&lt;br&gt;&#xA;By simplicial approximation, continuous maps of polyhedra are homotopic to PL maps. However, homeomorphisms of polyhedra &lt;b&gt;might not be&lt;/b&gt; homotopic to PL homeomorphisms, merely to PL maps (the statement that they are is an equivalent formulation of the Hauptvermutung). So a (combinatorial) homotopy invariant of simplicial complexes might a-priori &lt;b&gt;fail&lt;/b&gt; to be a homotopy invariant of polyhedra, unless one proves also that it's independent of the choice of simplicial approximation (which Matt E. and Qiaochu both say). It isn't enough just to show that it's independent under combinatorial homotopy of simplicial complexes.&lt;br&gt;&#xA;I apologise for the confusion. My answer was essentially taken from Page 4 of &lt;a href=&quot;http://www.maths.ed.ac.uk/~aar/books/haupt.pdf&quot; rel=&quot;nofollow&quot;&gt;The Hauptvermutung Book&lt;/a&gt;- I apologise also for the lack of attribution. I interpret what it says there as the assertion that textbooks don't tend to check independence with respect to choice of simplicial approximation; and my own experience (or inattentiveness) leads me to suspect that this is indeed the case. I'll look in a library on Monday, and maybe edit this response again.&lt;/p&gt;&#xA;" OwnerUserId="7803" LastEditorUserId="7803" LastEditorDisplayName="" LastEditDate="2011-03-04T19:42:19.253" LastActivityDate="2011-03-04T19:42:19.253" CommentCount="5" />
  <row Id="24849" PostTypeId="2" ParentId="24806" CreationDate="2011-03-03T18:08:09.980" Score="0" ViewCount="" Body="&lt;p&gt;Not to toot my own horn or anything, but my answer to this question&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://math.stackexchange.com/questions/24032/can-we-make-tanx-arbitrarily-close-to-an-integer-when-x-in-mathbbz/24041#24041&quot;&gt;Can we make $\tan(x)$ arbitrarily close to an integer when $x\in \mathbb{Z}$?&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;includes an outline of how you would prove this (similar to Moron's hint).&lt;/p&gt;&#xA;" OwnerUserId="6071" LastActivityDate="2011-03-03T18:08:09.980" />
  <row Id="24850" PostTypeId="2" ParentId="24845" CreationDate="2011-03-03T18:18:30.503" Score="2" ViewCount="" Body="&lt;p&gt;Looks like you almost have it: Try splitting $B$ into two sets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More details:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Define $k$ as you did: $k\mu(A) = \int_{A} f \text{d}\mu$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $B_1 = A \cap \{x: f(x) \gt k\}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $\mu(B) \gt 0$, then since $A$ is atomic, we have $\mu(A) = \mu(B)$ and thus&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ k\mu(A) = \int_{A} f \text{d}\mu = \int_{B_1} f \text{d}\mu \gt \int_{B_1} k\ \text{d}\mu = \int_{A} k\ \text{d}\mu = k \mu(A)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus we must have that $\mu(B_1) \lt \mu(A)$ and so, $\mu(B_1) = 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly $\mu(B_2) = 0$ where $B_2 = A \cap \{x: f(x) \lt k\}$&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2011-03-04T01:58:06.930" LastActivityDate="2011-03-04T01:58:06.930" CommentCount="5" />
  <row Id="24851" PostTypeId="2" ParentId="24833" CreationDate="2011-03-03T18:20:07.357" Score="20" ViewCount="" Body="&lt;p&gt;It is a common false hunch that shortly-stated theorems should have short proofs. This hunch is easily falsified by employing basic results in logic. For any formal system that has nontrivial power (e.g. Peano arithmetic) there is no recursive algorithm that decides theoremhood.  Now if  there existed a recursive bound $\rm\ L(n)\ $ on the length of proofs of a statement of length $\rm\:n\:,\:$ then we could test for theoremhood simply be enumerating and testing all possible proofs of length $\rm\le L(n)\ $. Hence there can be no such recursive bound on the length of proofs. It follows that there exist short stated theorems with proofs so long that they are probably not amenable to human comprehension (these results date back to Goedel's 1936 paper on speedup theorems). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It remains to be seen whether or not there exists &lt;em&gt;mathematically interesting&lt;/em&gt; theorems like this. There may be examples in Collatz-like congruential iterations (similar to the difficult open $\rm\: 3\ x + 1\: $ problem) that were &lt;a href=&quot;http://groups.google.com/group/comp.lang.scheme/msg/b8c43aee2bc12241&quot; rel=&quot;nofollow&quot;&gt;discovered in the wild&lt;/a&gt; while analyzing Busy-beaver holdout machines (while attempting to find the smallest universal Turing machines). John Conway has shown that there exists such congruential iterations with undecidable halting problem. That such undecidable problems may be encoded so succinctly in programs for tiny Turing machines should not come as a surprise to anyone familiar with the above simple results from logic. They are a testament to the power of ingenuity - whether it be human (in powerful mathematical theories) or nature (the DNA-based programs designed by evolution).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a chess-theoretic analog see &lt;a href=&quot;http://groups.google.com/group/sci.math/msg/c9440917b82ca5fc&quot; rel=&quot;nofollow&quot;&gt;my post here,&lt;/a&gt; which discusses some massive brute-force computated chess endgame databases revealing optimal move sequences that are completely incomprehensible to human experts. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Returning to the specific topic at hand, it is known that Fermat's Last Theorem cannot be proved by certain types of descent proofs similar to the classical simple proofs known for small exponents. References to such work can probably be located by Googling &quot;Tate Shafarevich obstruction&quot;.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2011-03-03T20:19:32.367" LastActivityDate="2011-03-03T20:19:32.367" CommentCount="2" />
  <row Id="24852" PostTypeId="1" AcceptedAnswerId="24853" CreationDate="2011-03-03T18:43:33.030" Score="4" ViewCount="257" Body="&lt;p&gt;Hello ::- D. My first post here ::- &gt;. I got a rather simple question. But please, allow me to introduce myself a bit first. I think it's polite for a first post ::- D.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm a game developer (free Flash games) with a few &lt;em&gt;ahem&lt;/em&gt; big holes in my math (read: I haven't done any math since high school - thanks to the school, I was totally repulsed by it in University). Even so, I always liked math, but only out of school ::- D. Unfortunately, I haven't kept touch with it (it wasn't necessary so far).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, without further delay, I will get to my problem for a game I'm working on. Anybody helping me wins a free ticket to the game's credits &lt;em&gt;grin&lt;/em&gt;. And it's going to be a nice game, much nicer than my previous: &lt;a href=&quot;http://www.kongregate.com/games/Kyliathy/thunderbirdz&quot; rel=&quot;nofollow&quot;&gt;http://www.kongregate.com/games/Kyliathy/thunderbirdz&lt;/a&gt; (to see that I'm not a fraud, LOL).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I'm trying to do (and failing miserably) is to rotate a Rectangle around a Point, in Adobe Flash. The problem is that Flash rotates an object relative to it's X=0, Y=0 coordinate, which they call a registration point. And I want to rotate the Rectangle around ANOTHER POINT inside that Rectangle. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a sample Flash which illustrates my problem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.axonnsd.org/W/P002/MathSandBox.swf&quot; rel=&quot;nofollow&quot;&gt;http://www.axonnsd.org/W/P002/MathSandBox.swf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[LATER EDIT: all SWF samples we talked about in the comments below are now accessible at the same links, except that you have to add '/W/P002'after 'axonnsd.org' - see the above link.]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The BLUE circle should stay on top of the RED circle. Instead, the Rectangle rotates around its registration point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For people without Flash or who hate Flash, here is an image of the same problem which also show the registration point via a BLACK Square&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/NAGxQ.jpg&quot; alt=&quot;Rotation Problem&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now.......&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My solution to this problem would be to find an equation to MOVE the Rectangle to an appropriate X/Y so that the BLUE circle stays on top of the RED circle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Specifying the rotation in Flash is simple: object.rotation = X, where X can be any number. It will, however, always divide it by 360, of course.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But now I have to find SOME method by which I set my object.x and object.y so that the Rectangle appears to rotate around the BLUE/RED circles, NOT around its registration point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And, for the life of me, I don't even know where to START finding that mysterious equation ::- D. I bet it involves a bit of Pi and some drops of trigonometry. But I don't even know where to begin... Pointers, anybody?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you for reading my long first message! ::- D.&lt;/p&gt;&#xA;" OwnerUserId="7745" LastEditorUserId="7745" LastEditorDisplayName="" LastEditDate="2011-03-14T12:23:44.177" LastActivityDate="2011-03-14T12:23:44.177" Title="2D Rotation Around Point" Tags="&lt;geometry&gt;&lt;trigonometry&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="24853" PostTypeId="2" ParentId="24852" CreationDate="2011-03-03T18:52:06.690" Score="5" ViewCount="" Body="&lt;p&gt;The upper-left corner of the rectangle is the point we want to locate.  It travels along a circular path with radius $r=\sqrt{a^2+b^2}$ and center $(h,k)=$ $(x_r+a,y_r+b)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/GEMlu.png&quot; alt=&quot;orbit 1&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One possible parametric description of a circle with center $(h,k)$ and radius $r$ is $(x,y)=$ $(h+r\cos t,k+r\sin t)$, but this parameterization starts at $(h+r,k)$ when $t=0&amp;#176;$ and moves to $(h,k+r)$ when $t=90&amp;#176;$.  Swapping the sine and cosine gives a parameterization $(x,y)=$ $(h+r\sin t,k+r\cos t)$ that starts at $(h,k+r)$ when $t=0&amp;#176;$ and moves to $(h+r,k)$ when $t=90&amp;#176;$.  Reversing the sign of the cosine term in the $y$ coordinate gives a parameterization $(x,y)=$ $(h+r\sin t,k-r\cos t)$ that starts at $(h,k-r)$ when $t=0&amp;#176;$ and moves to $(h+r,k)$ when $t=90&amp;#176;$.  In the coordinate system you describe, this is the circle below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/Cg8qg.png&quot; alt=&quot;orbit 2&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we're fairly close, but $t$ is slightly off from $\theta$.  Letting $t=\theta-\arctan(\frac{a}{b})$ shifts the parameter to match the desired values.&#xA;$$\begin{align}&#xD;&#xA;x&amp;amp;=h+r\sin(\theta-\arctan\left(\frac{a}{b}\right))&#xD;&#xA;\\&#xD;&#xA;y&amp;amp;=k-r\cos(\theta-\arctan\left(\frac{a}{b}\right))&#xD;&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;&lt;em&gt;Some earlier failed attempts at thinking transformationally are in the revision history.&lt;/em&gt;&lt;/sup&gt;&lt;/p&gt;&#xA;" OwnerUserId="72" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2011-03-03T22:26:02.647" LastActivityDate="2011-03-03T22:26:02.647" CommentCount="40" />
  <row Id="24854" PostTypeId="1" CreationDate="2011-03-03T19:15:51.927" Score="20" ViewCount="502" Body="&lt;p&gt;This is spurred by the comments to my answer &lt;a href=&quot;http://math.stackexchange.com/questions/24507/why-did-mathematicians-take-russells-paradox-seriously/24546#24546&quot;&gt;here&lt;/a&gt;.  I'm unfamiliar with set theory beyond Cohen's proof of the independence of the continuum hypothesis from ZFC.  In particular, I haven't witnessed any real interaction between set-theoretic issues and the more conventional math I've studied, the sort of place where you realize &quot;in order to really understand this problem in homotopy theory, I need to read about large cardinals.&quot;  I've even gotten the feeling from several professional mathematicians I've talked to that set theory is no longer relevant, and that if someone were to find some set-theoretic flaw in their axioms (a non-standard model or somesuch), they would just ignore it and try again with different axioms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also don't personally care for the abstraction of set theory, but this is a bad reason to judge anything, especially at this early stage in my life, and I feel like I'd be more interested if I knew of some ways it interacted with the rest of the mathematical world.  So:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;What do set theorists today care about?&lt;/li&gt;&#xA;&lt;li&gt;How does set theory interact with the rest of mathematics?&lt;/li&gt;&#xA;&lt;li&gt;(more subjective but) Would mathematicians working outside of set theory benefit from thinking about large cardinals, non-standard models, or their ilk?&lt;/li&gt;&#xA;&lt;li&gt;Could you recommend any books or papers that might convince a non-set theorist that the subject as it's currently practiced is worth studying?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Thanks a lot!&lt;/p&gt;&#xA;" OwnerUserId="2215" LastActivityDate="2011-03-07T17:22:36.477" Title="What are the issues in modern set theory?" Tags="&lt;reference-request&gt;&lt;set-theory&gt;" AnswerCount="5" CommentCount="2" FavoriteCount="7" />
  <row Id="24855" PostTypeId="1" AcceptedAnswerId="24856" CreationDate="2011-03-03T19:17:36.683" Score="4" ViewCount="135" Body="&lt;p&gt;Hello, I want to show that there are no simple groups of order $p^{k}(p+1)$ where $k&amp;gt;0$ and $p$ is a prime number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So suppose there is such a group. Then if we let $n_{p}$ denote the number of $p$-Sylow subgroups of $G$ we have that $n_{p}=p+1$. Now by letting $G$ act on $Sylow_{P}(G)$ by conjugation we obtain a group homomorphism $G \rightarrow S_{p+1}$. Since $G$ is simple then either $ker(f)$ is trivial or all $G$. Now here's my question: assume $ker(f)=G$ this would imply then that $G$ has a unique $p$-Sylow subgroup no? but then such subgroup is normal which contradicts the fact that $G$ is simple. So the map in fact is injective but then $|G|$ divides $(p+1)!$ which cannot be.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically my question is if my argument is correct, namely thta if $ker(f)=g$ implies the existence of a unique $p$-Sylow subgroup which implies such subgroup is normal in $G$ which cannot be. In case this is wrong, how do you argue that $ker(f)$ cannot be all $G$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="6495" LastEditorUserId="8575" LastEditorDisplayName="" LastEditDate="2011-05-19T09:13:08.800" LastActivityDate="2011-05-19T09:13:08.800" Title="Showing groups of order $p^{k}(p+1)$ are not simple, p prime" Tags="&lt;group-theory&gt;&lt;simple-groups&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="24856" PostTypeId="2" ParentId="24855" CreationDate="2011-03-03T19:26:04.887" Score="4" ViewCount="" Body="&lt;p&gt;You are (mostly) correct.  If ker(f) = G, then the image of G in the symmetric group is just the identity, so G does not move any of its Sylow p-subgroups around.  However, G acts transitively on its Sylow p-subgroups (they are all conjugate) and the identity is not transitive unless p+1=1, which is silly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If ker(f) = 1, then G embeds in the symmetric group on p+1 points, so its order divides (p+1)!.  This is not a contradiction when k=1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Indeed, taking p=2, the non-abelian group of order six has order p(p+1) and has exactly p+1 Sylow p-subgroups, and the homomorphism f is injective.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course a group of order p(p+1) is also not simple, but probably for a different reason.&lt;/p&gt;&#xA;" OwnerUserId="583" LastActivityDate="2011-03-03T19:26:04.887" CommentCount="7" />
  <row Id="24857" PostTypeId="2" ParentId="24855" CreationDate="2011-03-03T19:26:11.000" Score="1" ViewCount="" Body="&lt;p&gt;&quot;Now here's my question: assume ker(f)=G this would imply then that G has a unique p-Sylow subgroup no? &quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;yes, the group action (conjugation on sylow subgroups) is transitive.&lt;/p&gt;&#xA;" OwnerUserId="6925" LastActivityDate="2011-03-03T19:26:11.000" />
  <row Id="24858" PostTypeId="2" ParentId="24817" CreationDate="2011-03-03T19:26:16.370" Score="1" ViewCount="" Body="&lt;p&gt;For a scalar-valued function $f$ of $n$ variables $x_k$ the multivariate form of Taylor's theorem can be brought into the following form:&#xA;$$f(p+X)=\sum_{r=0}^n {1\over r!} d^r f(p)(X)+ R_n.$$&#xA;Here $d^r f(p)(X)$ is a homogeneous polynomial of degree $r$ in the coordinates $X_1$, $\ldots$, $X_n$ of the increment vector $X$, namely&#xA;$$d^r f(p)(X)=\sum_{k_1,\ldots,k_r}f_{.k_1\ldots k_r}(p)X_{k_1}\ldots X_{k_r},$$&#xA;where the indices $k_i$ run from $1$ to $n$; so formally there are $n^r$ summands in total. The remainder term $R_n$ can assume various forms, e.g. $R_n(X)= o(|X|^n) \ (X\to 0)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Among the $n^r$ terms of $d^r f(p)(X)$ many are equal. If we have just two variables $x$,$y\ $ then the $2^r$ terms of $d^r f(p)(X,Y)$ (now the increment vector is $(X,Y)$ with scalar $X$, $Y$) can be arranged into&#xA;$$d^r f(p)(X,Y) =\sum_{k=0}^r {r \choose k}\ {\partial^r f\over \partial x^k\partial y^{r-k}}(p)\  X^k Y^{r-k}.$$&#xA;In the last formula the expressions ${\partial^r f\over \partial x^k\partial y^{r-k}}(p)$ are ${\it constants}$. In particular one has $$df(p)(X,Y)=f_x(p)X+f_y(p)Y, \qquad d^2f(p)(X,Y)=f_{xx}(p)X^2+2 f_{xy}(p) XY +f_{yy}(p)Y^2 .$$&lt;/p&gt;&#xA;" OwnerUserId="1303" LastEditorUserId="1303" LastEditorDisplayName="" LastEditDate="2011-03-03T20:47:27.423" LastActivityDate="2011-03-03T20:47:27.423" />
  <row Id="24859" PostTypeId="2" ParentId="24810" CreationDate="2011-03-03T19:28:57.773" Score="4" ViewCount="" Body="&lt;p&gt;I don't think that there is a way such that hundreds or thousands of people could collaborate on a problem without subdivision into smaller groups with certain synchronization points. Large software projects with thousands of developers are usually broken down to groups with 2-20 members. Imagine that you'd have to talk to each member in your group for 30 minutes per week, that would have to talking for 10 hours in the latter case, every week, just to synchronize with what the others are doing (this is sometimes called the combinatorial explosion of communication).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this sense I'd say that Wikipedia is very successful in collecting and explaining a lot of mathematical knowledge and pointing out open problems. If you supplement this wiki ansatz with links to more specialized wikis where subgroups document their work about certain open problems, you'd come close to a large scale collaboration. The &lt;a href=&quot;http://ncatlab.org/nlab/show/HomePage&quot; rel=&quot;nofollow&quot;&gt;nLab&lt;/a&gt; is an example of this, but it has not attracted a large crowd yet. The nLab has an accompanying forum, where people talk about their thoughts before there is an agreement about what should be written on the wiki. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can write about your work and where you are stuck on the wiki, explaining all the background and ask on the forum for help. Others can chime in either by supplementing what you wrote on the wiki, or by joining the discussion on the forum. I don't see why this could not work with a lot more participants, as long as they form subgroups and avoid a situation where every participant has to talk and listen to everyone else.&lt;/p&gt;&#xA;" OwnerUserId="7556" LastActivityDate="2011-03-03T19:28:57.773" CommunityOwnedDate="2011-03-03T19:28:57.773" />
  <row Id="24860" PostTypeId="1" AcceptedAnswerId="38304" CreationDate="2011-03-03T19:39:56.847" Score="0" ViewCount="78" Body="&lt;p&gt;Let $a ,$ $b ,$ $\alpha ,$ and $\beta$ be non-zero binary strings of length exactly $n$ where $$ \beta = a \alpha + b. $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, consider the following scenario. Fix $a$ and $b.$ I give you $\alpha$ and $\beta,$ but keep $a$ and $b$ as secrets from you. What is the probability that you can modify both strings (same length, just alter some bits), such that the altered strings $\alpha&amp;#39;$ and $\beta&amp;#39;$ still satisfy $$ \beta&amp;#39; = a \alpha&amp;#39; + b. $$ A cryptography paper which I am reading mentions it is $2^{-n}$ but I couldn't convince myself why.&lt;/p&gt;&#xA;" OwnerUserId="2468" LastActivityDate="2011-05-10T18:39:21.567" Title="Probability of modifying $\alpha$ and $\beta$ in $ \beta = a \alpha + b $ such that the equality still holds" Tags="&lt;probability&gt;" AnswerCount="1" CommentCount="7" />
  <row Id="24861" PostTypeId="2" ParentId="24854" CreationDate="2011-03-03T19:48:41.387" Score="7" ViewCount="" Body="&lt;p&gt;Maybe start with looking at the chapters for the Handbook of Set Theory &lt;a href=&quot;http://handbook.assafrinot.com/&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&#xA;As a topologist, I can say that set theory is still very useful in General Topology, as people run into many questions that are independent of set theory there, or that are helped by techniques from Set theory. &lt;a href=&quot;http://www.math.osu.edu/~friedman/&quot; rel=&quot;nofollow&quot;&gt;Harvey Friedmann&lt;/a&gt; has many (I think) interesting ideas about the role of large cardinals in &quot;normal&quot;, combinatoric questions. &lt;a href=&quot;http://shelah.logic.at/&quot; rel=&quot;nofollow&quot;&gt;Shelah&lt;/a&gt; is probably the most prolific author in Set Theory and has quite a breadth of subjects.&lt;/p&gt;&#xA;" OwnerUserId="4280" LastActivityDate="2011-03-03T19:48:41.387" />
  <row Id="24863" PostTypeId="2" ParentId="24810" CreationDate="2011-03-03T20:01:00.893" Score="3" ViewCount="" Body="&lt;p&gt;Maybe this response will seem oblique bordering on irrelevant, but...Your question reminds me of the classic Sandman story &lt;a href=&quot;http://thehoosh.tumblr.com/post/783668820/sandman-a-dream-of-a-thousand-cats&quot; rel=&quot;nofollow&quot;&gt;Dream of A Thousand Cats&lt;/a&gt;.  In it, a cat learns that reality could be completely restructured (or rather, re-restructured...) if only (any) one thousand cats would simultaneously believe and dream in unison of a different reality.  When asked what he thinks of this, one of the characters -- a cat -- says &quot;I would like to see anyone -- prophet, king or god -- persuade a thousand cats to do anything at the same time.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I feel like herding mathematicians is only a little easier than herding cats.  What grand project would convince a thousand of us to get involved in a coordinated way?&lt;/p&gt;&#xA;" OwnerUserId="299" LastActivityDate="2011-03-03T20:01:00.893" CommunityOwnedDate="2011-03-03T20:01:00.893" CommentCount="3" />
  <row Id="24864" PostTypeId="2" ParentId="24840" CreationDate="2011-03-03T20:12:42.447" Score="4" ViewCount="" Body="&lt;p&gt;You're almost there: set $A=2q-r-s$ and $B=r-s.$ Then your expression for $\textbf{N}(\alpha)$ reduces to the desired form. i.e. your&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\textbf{N}(\alpha) = \frac14 \left \lbrace (2q-r-s)^2 - 5(r-s)^2 \right \rbrace&#xD;&#xA;= \frac14(A^2-5B^2).$$&lt;/p&gt;&#xA;" OwnerUserId="1301" LastEditorUserId="1301" LastEditorDisplayName="" LastEditDate="2011-03-03T20:19:43.210" LastActivityDate="2011-03-03T20:19:43.210" CommentCount="2" />
  <row Id="24865" PostTypeId="2" ParentId="24840" CreationDate="2011-03-03T20:16:21.527" Score="5" ViewCount="" Body="&lt;p&gt;I agree completely with Qiaochu.  If you understand what he wrote, then you can use this to give a much shorter and easier proof than the hint is suggesting.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact I didn't even read the hint.  I saw the question, thought &quot;Oh, that looks like the norm from $\mathbb{Q}(\sqrt{5})$&quot;.  I also know that $\mathbb{Q}(\sqrt{5})$ is the unique quadratic subfield of $\mathbb{Q}(\zeta_5)$ (since $5 \equiv 1 \pmod 4$; there's a little number theory here), and that for any tower of finite field extensions $L/K/F$, we have &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$N_{L/F}(x) = N_{K/F}(N_{L/K}(x))$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Norms also carry algebraic integers to algebraic integers, so this shows that the &#xA;norm of any element of $\mathbb{Z}[\zeta_5]$ is also the norm of some algebraic integer of &#xA;$\mathbb{Q}(\sqrt{5})$, i.e., is of the form $(\frac{A+\sqrt{5}B}{2})(\frac{A-\sqrt{5}B}{2})$ for some $A,B \in \mathbb{Z}$.  I think we're done.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[I have never read Stewart and Tall, so it may be that they have not assumed or developed this much theory about norm maps at the point they give that exercise.  But if you know it, use it!]&lt;/p&gt;&#xA;" OwnerUserId="299" LastActivityDate="2011-03-03T20:16:21.527" CommentCount="7" />
  <row Id="24866" PostTypeId="1" CreationDate="2011-03-03T20:21:30.183" Score="4" ViewCount="147" Body="&lt;p&gt;I have run across the following multinomial series:&#xA;$$ 				\sum_{j = a}^{N}&#xD;&#xA;				\binom{N}{j} \binom{j}{a} d^{-j}&#xD;&#xA;$$&#xA;Here, $d&amp;gt;1$. This seems like a formula which has either a well-known identity, or which has no further closed form or simplification. Can anyone shed any light on this formula, or possibly point me to a standard reference?&lt;/p&gt;&#xA;" OwnerUserId="439" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-03T21:49:46.687" LastActivityDate="2011-03-03T21:49:46.687" Title="Identity for $\sum\limits_{j = a}^{N} 				\binom{N}{j} \binom{j}{a} d^{-j}$ ?" Tags="&lt;combinatorics&gt;&lt;sequences-and-series&gt;&lt;binomial-coefficients&gt;" AnswerCount="4" CommentCount="1" />
  <row Id="24867" PostTypeId="2" ParentId="24866" CreationDate="2011-03-03T20:29:35.930" Score="2" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Answer:&lt;/strong&gt; $$\binom{N}{a} d^{-a}(1+d^{-1})^{N-a} \; .$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll leave it up to you to check it.&lt;/p&gt;&#xA;" OwnerUserId="3567" LastActivityDate="2011-03-03T20:29:35.930" />
  <row Id="24868" PostTypeId="1" CreationDate="2011-03-03T20:30:26.637" Score="1" ViewCount="50" Body="&lt;p&gt;I mean if there is any construction that satisfies all the conditions for an vectorial space except it lacks an identity element? This questions was posed to me by a classmate last semester and I have been puzzling over it since then.&lt;/p&gt;&#xA;" OwnerUserId="7754" LastActivityDate="2011-03-03T22:29:21.957" Title="Is it possible to construct a quasi-vectorial space without an identity element?" Tags="&lt;vector-space&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="2" />
  <row Id="24869" PostTypeId="1" CreationDate="2011-03-03T20:31:38.990" Score="0" ViewCount="37" Body="&lt;p&gt;According to the Wikipedia article, the Simplex algorithm depends on constraining all the unknowns to be &gt;= 0.  I have a problem where one of my variables is highly likely to be negative in many cases.  How can I adapt the problem so I can use Simplex?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;: example: say I have x, y and z and I want to find values (both &gt;= 0) for x and y which minimize z, where I also have the constraints:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;1.4x -y +z &amp;gt;= 0&#xA;-x +2.2y +z &amp;gt;= 0&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="7755" LastEditorUserId="7755" LastEditorDisplayName="" LastEditDate="2011-03-03T20:44:02.277" LastActivityDate="2011-03-04T02:40:43.287" Title="How can I adapt my problem so that it is amenable to the simplex algorithm?" Tags="&lt;linear-algebra&gt;&lt;linear-programming&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="24870" PostTypeId="2" ParentId="24866" CreationDate="2011-03-03T20:34:12.213" Score="6" ViewCount="" Body="&lt;p&gt;$$\sum_{j = a}^{N} \binom{N}{j} \binom{j}{a} d^{-j} = \sum_{j = a}^{N} \binom{N}{a} \binom{N-a}{j-a} d^{-j} = \binom{N}{a} \sum_{j = 0}^{N-a} \binom{N-a}{j} d^{-j-a} $$&#xA;$$= d^{-a} \binom{N}{a} \sum_{j = 0}^{N-a} \binom{N-a}{j} d^{-j} =  d^{-a} \binom{N}{a} (1+d^{-1})^{N-a}.$$&#xA;The first step uses what's called &lt;em&gt;trinomial revision&lt;/em&gt;.  See, for example, &lt;em&gt;Concrete Mathematics&lt;/em&gt;, p. 174.  The last step uses the binomial theorem.&lt;/p&gt;&#xA;" OwnerUserId="2370" LastActivityDate="2011-03-03T20:34:12.213" />
  <row Id="24871" PostTypeId="2" ParentId="24866" CreationDate="2011-03-03T20:35:30.253" Score="4" ViewCount="" Body="&lt;p&gt;You can use $${N \choose j}{j \choose a} = \frac{ N!}{j! (N-j)!} \frac{j!}{a! (j-a)!} =\frac{N!}{a!(N-a)!} \frac{(N-a)!}{(j-a)! (N-j)!} = {N \choose a}{ N-a \choose j-a}$$ to write your expression as&#xA;$${N \choose a} \sum_{j=a}^N { N-a \choose j-a} d^{-j}= {N \choose a} d^{-a} \sum_{j=0}^{N-a} { N -a \choose j} d^{-j} = {N \choose a} d^{-a}  (1+1/d)^{N-a}.$$ (using binomial series)&lt;/p&gt;&#xA;" OwnerUserId="7266" LastActivityDate="2011-03-03T20:35:30.253" />
  <row Id="24873" PostTypeId="1" AcceptedAnswerId="24875" CreationDate="2011-03-03T20:38:04.887" Score="14" ViewCount="403" Body="&lt;p&gt;It is very elementary to show that $\mathbb{R}$ isn't homeomorphic to $\mathbb{R}^m$ for $m&amp;gt;1$: subtract a point and use the fact that connectedness is a homeomorphism invariant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Along similar lines, you can show that $\mathbb{R^2}$ isn't homeomorphic to $\mathbb{R}^m$ for $m&amp;gt;2$ by subtracting a point and checking if the resulting space is simply connected. Still straightforward, but a good deal less elementary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the general result that $\mathbb{R^n}$ isn't homeomorphic to $\mathbb{R^m}$ for $n\neq m$, though intuitively obvious, is usually proved using sophisticated results from algebraic topology, such as invariance of domain or extensions of the Jordan curve theorem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a more elementary proof of this fact? If not, is there intuition for why a proof is so difficult?&lt;/p&gt;&#xA;" OwnerUserId="7530" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-03T21:29:01.277" LastActivityDate="2011-03-03T23:21:01.833" Title="Elementary proof that $\mathbb{R}^n$ is not homeomorphic to $\mathbb{R}^m$" Tags="&lt;topology&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="1" />
  <row Id="24874" PostTypeId="2" ParentId="24873" CreationDate="2011-03-03T20:51:00.137" Score="6" ViewCount="" Body="&lt;p&gt;Well, I might recast your proofs of the first two cases as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose that $\mathbb{R}^n$ and $\mathbb{R}^m$ are homeomorphic.  Then for any $P \in \mathbb{R}^n$, there must exist a point $Q \in \mathbb{R}^m$ such that $\mathbb{R}^n \setminus \{P\}$ and $\mathbb{R}^m \setminus \{Q\}$ are homeomorphic.  (Since the homeomorphism group of $\mathbb{R}^m$ acts transitively, really any point $Q$ is okay, but I'm trying to be both simple and rigorous.)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now your proof when $n = 1$ is equivalent to the observation that $\pi_0(\mathbb{R}^1 \setminus \{P\})$ is nontrivial, while $\pi_0(\mathbb{R}^n \setminus \{Q\})$ is zero for all $n &amp;gt; 1$.  (Note that $\pi_0(X)$ is a free $\mathbb{Z}$-module on the set of path-components of $X$, so really we are using that $\mathbb{R}^1$ minus a point is path-connected and $\mathbb{R}^n$ minus a point is not, for $n &amp;gt; 1$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When $n =2$, your proof is literally that $\pi_1(\mathbb{R}^2 \setminus \{P\})$ is nonzero whereas $\pi_1(\mathbb{R}^n \setminus \{Q\})$ is nonzero for all $n &amp;gt; 2$.  It is a little questionable to me whether the fundamental group counts as &quot;elementary&quot; -- you certainly have to learn about homotopies and prove some basic results in order to get this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are okay with such things, then it seems to me like you might as well also admit the higher homotopy groups: the point here is that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;for all $m \in \mathbb{Z}^+$, $\pi_{m-1}(\mathbb{R}^m \setminus \{P\}) \cong \mathbb{Z}$ whereas for $n &amp;gt; m$, $\pi_{m-1}(\mathbb{R}^n \setminus \{Q\}) = 0$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I am remembering correctly, the higher homotopy groups are introduced very early on in J.P. May's &lt;em&gt;A Concise Course in Algebraic Topology&lt;/em&gt; and applied to essentially this problem, among others.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[By the way, if we are okay with homotopy, then we probably want to replace $\mathbb{R}^n \setminus \{P\}$ with its homotopy equivalent subspace $S^{n-1}$ throughout.  For some reason I decided to avoid mentioning this in the arguments above.  If I had, it would have saved me a fair amount of typing...]&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Added&lt;/b&gt;: Of course one could also use homology groups instead, as others have suggested in the comments.  One might argue that homotopy groups are easier to &lt;em&gt;define&lt;/em&gt; whereas homology groups are easier to &lt;em&gt;compute&lt;/em&gt;.  But this one computation of the &quot;lower&quot; homotopy groups of spheres is not very hard, and my guess is that if you want to start from scratch and prove everything, then for this problem homotopy groups will give the shorter approach.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As to why the problem is hard to solve in an elementary way: the point is that the two spaces $\mathbb{R}^m \setminus \{P\}$ and $\mathbb{R}^n \setminus \{Q\}$ look the same when viewed from the lens of general topology.  [An exception: one can develop &lt;strong&gt;topological dimension theory&lt;/strong&gt; to tell them apart.  For this I recommend the classic text of Hurewicz and Wallman.  Whether that's &quot;more elementary&quot;, I couldn't say.]  In order to distinguish them for $m,n$ not too small, it seems that you need to develop various notions of the &quot;higher connectivities&quot; of a space, which leads inevitably to homotopy groups and co/homology groups.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another alternative is to throw out homeomorphism and look instead at diffeomorphism.  This puts a reasonable array of tools from differentiable topology and manifold theory at your disposal rather quickly (see e.g. Milnor's book &lt;em&gt;Topology from a differentiable viewpoint&lt;/em&gt;).  It is not hard to show that the dimension of a manifold is a diffeomorphism invariant!  So maybe the subtlety comes from insisting on working in the topological category, which often turns out to be more difficult than working with topological spaces with nice additional structures.&lt;/p&gt;&#xA;" OwnerUserId="299" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2011-03-03T21:01:23.783" LastActivityDate="2011-03-03T21:01:23.783" CommentCount="1" />
  <row Id="24875" PostTypeId="2" ParentId="24873" CreationDate="2011-03-03T20:54:56.140" Score="14" ViewCount="" Body="&lt;p&gt;There are reasonably accessible proofs that are purely general topology. First one needs to show Brouwer's fixed point theorem (which has an elementary proof, using barycentric subdivion and Sperner's lemma), or some result of similar hardness. Then one defines a topological dimension function (there are 3 that all coincide for separable metric spaces, dim (covering dimension), ind (small inductive dimension), Ind (large inductive dimension)), say we use dim, and then we show (using Brouwer) that $\dim(\mathbb{R}^n) = n$ for all $n$. As homeomorphic spaces have the same dimension (which is quite clear from the definition), this gives the result. This is in essence the approach Brouwer himself took, but he used a now obsolete dimension function called Dimensionsgrad in his paper, which does coincide with dim etc. for locally compact, locally connected separable metric spaces. Lebesgue proposed the covering dimension, but had a false proof for $\dim(\mathbb{R}^n) = n$, which Brouwer corrected.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One can find such proofs in Engelking (general topology), Nagata (dimension theory), or nicely condensed in van Mill's books on infinite dimensional topology. These proofs do not use homology, homotopy etc., although one could say that the Brouwer proof of his fixed point theorem (via barycentric division etc.) was a precursor to such ideas.&lt;/p&gt;&#xA;" OwnerUserId="4280" LastEditorUserId="4280" LastEditorDisplayName="" LastEditDate="2011-03-03T21:40:37.540" LastActivityDate="2011-03-03T21:40:37.540" CommentCount="2" />
  <row Id="24876" PostTypeId="2" ParentId="24854" CreationDate="2011-03-03T21:00:29.683" Score="4" ViewCount="" Body="&lt;p&gt;One of the professors in my department was talking to me about it the other day, he joked that logic and set theory to &quot;conventional&quot; mathematics is like mathematics to physics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First of all, to me conventional mathematics is to take an idea and derive more ideas from it by logical inference - this relates to what my Algebra I professor said in the very first math lecture I attended to: mathematics is the science of deducing A from B and C.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Secondly, the aforementioned professor was joking but it was part true in a way. Set theory trickles into model theory and topology which in turn trickle into algebra and analysis. Those fields are conventional mathematics, I think.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lastly, I can't speak completely about set theorists but I can say what I see from the very narrow point of view I have right now - the dominance of ZF was established and now there is a search for &quot;measures of consistency&quot; how much more do you need to assume. The axiom of choice, its negation, existence of some large cardinals, and so forth and so on. This is my narrow point of view, as someone who's mostly studying forcing and large cardinals for the past few months and I might as well be talking trash.&lt;/p&gt;&#xA;" OwnerUserId="622" LastActivityDate="2011-03-03T21:00:29.683" CommentCount="2" />
  <row Id="24877" PostTypeId="1" CreationDate="2011-03-03T21:04:58.973" Score="0" ViewCount="150" Body="&lt;p&gt;In general, are papers published by mathematicians the same thing as PhD theses? In other words, if a mathematician submitted a paper for a PhD would it get accepted? Or are papers of higher quality than PhD theses?&lt;/p&gt;&#xA;" OwnerUserId="7756" LastActivityDate="2011-03-03T21:17:10.137" Title="Papers published by mathematicians" Tags="&lt;soft-question&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="24878" PostTypeId="2" ParentId="24866" CreationDate="2011-03-03T21:08:06.557" Score="3" ViewCount="" Body="&lt;p&gt;Since you mentioned the word &quot;multinomial&quot;, here is another way, using the Multinomial theorem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Expand &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(1 + x/d + x)^n = \sum_{p+q+r=n} {n \choose p,q,r} 1^p \cdot (x/d)^q \cdot x^r$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$= \sum_{a=0}^{n} \sum_{j=a}^{n} {n \choose a,j-a,n-j} 1^{a}  \cdot (x/d)^{j-a} \cdot x^{n-j}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ = \sum_{a=0}^{n} \left(\sum_{j=a}^{n} {n \choose j} {j \choose a} d^{a-j}\right) x^{n-a}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Expand $\displaystyle (1 + x(1+1/d))^n$ using binomial theorem and equate the coefficient of $\displaystyle  x^{n-a}$ on both sides to get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ {n \choose a} (1 + 1/d)^{n-a} = \sum_{j=a}^{n} {n \choose j} {j \choose a} d^{a-j}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and so&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ {n \choose a} d^{-a} (1 + 1/d)^{n-a} = \sum_{j=a}^{n} {n \choose j} {j \choose a} d^{-j}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are a couple more answers which use similar ideas:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://math.stackexchange.com/questions/4175/beautiful-identity-sum-k-mn-1k-m-binomkm-binomnk-delta/4187#4187&quot;&gt;Beautiful identity: $\sum_{k=m}^n (-1)^{k-m} \binom{k}{m} \binom{n}{k} = \delta_{mn}$&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://math.stackexchange.com/questions/10346/non-probabilistic-proofs-of-a-binomial-coefficient-identity-from-a-probability-qu/10605#10605&quot;&gt;Non-probabilistic proofs of a binomial coefficient identity from a probability question&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2011-03-03T21:08:06.557" />
  <row Id="24879" PostTypeId="1" AcceptedAnswerId="24886" CreationDate="2011-03-03T21:10:04.557" Score="1" ViewCount="132" Body="&lt;p&gt;The first few important algebraic number fields I have read about are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\mathbb{Q}$ The integers&lt;/li&gt;&#xA;&lt;li&gt;$\mathbb{Q}[\sqrt{d}]$ quadratic&lt;/li&gt;&#xA;&lt;li&gt;$\mathbb{Q}[e^{\frac{2 i \pi}{p}}]$ cyclotomic&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;What could be read about next?&lt;/p&gt;&#xA;" OwnerUserId="4290" LastActivityDate="2011-03-03T21:55:58.140" Title="What is the next important number field?" Tags="&lt;algebraic-number-theory&gt;" AnswerCount="1" CommentCount="8" />
  <row Id="24880" PostTypeId="2" ParentId="24868" CreationDate="2011-03-03T21:14:34.627" Score="8" ViewCount="" Body="&lt;p&gt;In an &lt;a href=&quot;http://www.math-atlas.org/99/lin_alg&quot; rel=&quot;nofollow&quot;&gt;old sci.math post by Dave Rusin&lt;/a&gt;, he discusses dropping sundry axioms form the usual set of axioms of vector spaces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, let's recall what axioms we are dealing with: normally, a vector space $\mathbf{V}$ over a field $\mathbb{F}$ is defined to be a set, together with operations $+\colon\mathbf{V}\times\mathbf{V}\to\mathbf{V}$ and $\cdot\colon \mathbb{F}\times\mathbf{V}\to\mathbf{V}$, written in their usual infix notation, which satisfies the following conditions:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;For all $x,y\in\mathbf{V}$, $x+y=y+x$.&lt;/li&gt;&#xA;&lt;li&gt;For all $x,y,z\in\mathbf{V}$, $(x+y)+z = x+(y+z)$.&lt;/li&gt;&#xA;&lt;li&gt;There exists a vector $\mathbf{0}\in\mathbf{V}$ such that for all $x\in\mathbf{V}$, $x+\mathbf{0}=x$.&lt;/li&gt;&#xA;&lt;li&gt;For each $x\in\mathbf{V}$ there exists $y\in\mathbf{V}$ such that $x+y=\mathbf{0}$.&lt;/li&gt;&#xA;&lt;li&gt;For all $x\in\mathbf{V}$, $\alpha,\beta\in \mathbb{F}$, $\alpha(\beta x) = (\alpha\beta)x$.&lt;/li&gt;&#xA;&lt;li&gt;For all $x\in\mathbf{V}$, $1x = x$.&lt;/li&gt;&#xA;&lt;li&gt;For all $x,y\in\mathbf{V}$, $\alpha\in\mathbb{F}$, $\alpha(x+y) = \alpha x + \alpha y$.&lt;/li&gt;&#xA;&lt;li&gt;For all $x\in\mathbf{V}$, $\alpha,\beta\in\mathbb{F}$, $(\alpha+\beta)x = \alpha x + \beta x$.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;You cannot just drop 3, because that would make 4 unintelligible. A way around it is to replace 4 with another statement which, in the presence of all &lt;em&gt;other&lt;/em&gt; axioms, is equivalent to 4; namely:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;4'. For all $x,y,w\in\mathbf{V}$, if $y + x = w + x$, then $y=w$. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That is, we have &lt;em&gt;right&lt;/em&gt; cancellation. (This is important; if you set up cancellation on the &quot;other side&quot; as the identity, you run into some difficulties below)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that if you have 1-8, then you get 4'. And if you have 1-3, 4', and 5-8, then you get 4: first, note that since $\mathbf{0}+0x = 0x = (0+0)x = 0x + 0x$, then 4' implies that $0x = \mathbf{0}$. Then given any $x\in\mathbf{V}$, we have $x + (-1)x = (1+(-1))x = 0x = \mathbf{0}$, so 4 holds in this case. That is, 1, 2, 3, 4', 5, 6, 7, and 8, are an alternative way of defining vector spaces, with the added advantage that now you can drop any of the eight and the remaining statements are still intelligible. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you take 1, 2, 3, 4', 5, 6, 7, and 8, then you can construct objects that are not vector spaces and satisfy any seven of these and not the eighth.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;All but 1: Take $\mathbf{V}=\mathbb{R}$, and define scalar multiplication by $\alpha x = x$, and $x+y = x$ for all $x$ and $y$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;All but 2: Take $\mathbf{V}=\mathbb{R}^2$, define scalar multiplication the usual way, and &#xA;$$x+y = \left\{\begin{array}{ll}&#xA;(0,0) &amp;amp;\mbox{if $x=(0,0)$ or $y=(0,0)$;}\\&#xA;|\cos(\theta)|(x+y) &amp;amp; \mbox{if $x\neq(0,0)\neq y$, and $\theta$ is the angle from $x$ to $y$.}&#xA;\end{array}\right.$$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;All but 3: Take $\mathbf{V}=\emptyset$, with the empty addition and multiplication!&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;All but 4 or 4': Take $\mathbf{V}=\mathbb{R}\cup\{\bigcirc\}$. Define addition so that it is the usual addition in $\mathbb{R}$, and $r+\bigcirc=\bigcirc+r = r$ for all $r\in\mathbb{R}$, and $\bigcirc+\bigcirc=\bigcirc$. Scalar multiplication is regular multiplication on $\mathbb{R}$, and $r\bigcirc = \bigcirc$ for all reals $r$. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;All but 5: Take $\mathbf{V}=\mathbb{R}$, and let $\sigma\colon\mathbb{R}\to\mathbb{Q}$ be any additive homomorphism of abelian groups. Define addition as usual, and scalar multiplication by $r x = \sigma(r)\cdot x$, where the multiplication on the right hand side is the usual real multiplication.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;All but 6: Take $\mathbf{V}=\mathbb{R}$ with the usual addition, but zero multiplication: $\alpha x = 0$ for all $x$ and all $\alpha$. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;All but 7: Take $\mathbf{V}=\mathbb{C}^2$, addition defined the usual way, and scalar multiplication given by:&#xA;$$\alpha(x,y) = \left\{\begin{array}{ll}&#xA;(\alpha x,\alpha y) &amp;amp; \mbox{if $x\neq 0$,}\\&#xA;(0, \overline{\alpha}y) &amp;amp; \mbox{if $x= 0$.}&#xA;\end{array}\right.$$&#xA;where $\overline{\alpha}$ is complex conjugation. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;All but 8: Take $\mathbf{V}=\mathbb{R}$ with usual addition, and scalar multiplication $r x = r^2\cdot x$, where the multiplication on the right hand side is the usual multiplication of real numbers.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Okay, but &quot;All but 3&quot; was almost cheating. What if we require that $\mathbf{V}$ be nonempty? Then you cannot have a structure that satisfies 1, 2, 4', 5, 6, 7, and 8, and does not satisfy 3:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose $\mathbf{V}$ satisfies 1, 2, 4', 5, 6, 7, and 8, and is nonempty. Let $x\in V$. Then for all $y\in V$, we have:&#xA;\begin{align*}&#xA;(0x + y) + 0x &amp;amp;= 0x + (0x+y) &amp;amp;\quad&amp;amp;\mbox{(by 1)}\\&#xA; &amp;amp;= (0x+0x) + y &amp;amp;\quad&amp;amp;\mbox{(by 2)}\\&#xA;&amp;amp;= (0+0)x + y &amp;amp;&amp;amp;\mbox{(by 8)}\\&#xA;&amp;amp;= 0x + y\\&#xA;&amp;amp;= y + 0x &amp;amp;&amp;amp;\mbox{(by 1)}&#xA;\end{align*}&#xA;By 4', this means that $0x+y = y$, so $\mathbf{0}=0x$ shows that 3 is satisfied.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-03T22:29:21.957" LastActivityDate="2011-03-03T22:29:21.957" CommentCount="2" />
  <row Id="24881" PostTypeId="2" ParentId="24877" CreationDate="2011-03-03T21:17:10.137" Score="1" ViewCount="" Body="&lt;p&gt;Papers published in the field of Mathematics are no different than any other field. In PhD thesis what you usually do is to usually put all your related work together and defend it against a committee. For example, if you have published three papers improving/related to your previous work then you combine your work together and defend it as your PhD thesis but if it is just one that you have published at all then you defend that alone as your PhD thesis. The question should be is the Conference/Journal you're publishing your work is of higher quality/ranked ? Higher the rank better the recognition. &lt;/p&gt;&#xA;" OwnerUserId="6335" LastActivityDate="2011-03-03T21:17:10.137" />
  <row Id="24882" PostTypeId="2" ParentId="24818" CreationDate="2011-03-03T21:26:03.417" Score="1" ViewCount="" Body="&lt;p&gt;a proof by contradiction &lt;a href=&quot;http://en.wikipedia.org/wiki/Heine%E2%80%93Cantor_theorem&quot; rel=&quot;nofollow&quot;&gt;but you should do it yourself!&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="6925" LastActivityDate="2011-03-03T21:26:03.417" />
  <row Id="24883" PostTypeId="1" AcceptedAnswerId="24887" CreationDate="2011-03-03T21:47:56.333" Score="2" ViewCount="230" Body="&lt;p&gt;Hello,&#xA;I have to compute this limit:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\lim_{x \to\infty} 3(\sqrt{\strut x}\sqrt{\strut x-3}-x+2)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;wolfram|alpha &lt;a href=&quot;http://www.wolframalpha.com/input/?i=Limit%5B3%20%28Sqrt%5Bx%5D%20Sqrt%5Bx%20-%203%5D%20-%20x%20%2B%202%29%2C%20x%20-%3E%20%5C%5BInfinity%5D%5D&quot; rel=&quot;nofollow&quot;&gt;says&lt;/a&gt; that answer is 3/2, but I can't get why. Does anyone know how to get this limit?&lt;/p&gt;&#xA;" OwnerUserId="7758" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2011-03-03T23:42:51.413" LastActivityDate="2011-05-17T07:00:38.020" Title="Limit of $\lim_{x \to\infty} 3\left(\sqrt{\strut x}\sqrt{\strut x-3}-x+2\right)$" Tags="&lt;limit&gt;" AnswerCount="5" CommentCount="1" FavoriteCount="0" />
  <row Id="24884" PostTypeId="1" AcceptedAnswerId="24891" CreationDate="2011-03-03T21:50:01.903" Score="1" ViewCount="59" Body="&lt;p&gt;$$\frac{(z x^{-1} y)^5 y^5}{x^{-4} z^{-4}}$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How would I evaluate this if $x = 10$, $y = -3$ and $z = 3$ ? I would like a step-by-step solution to help me fully understand it.&lt;/p&gt;&#xA;" OwnerUserId="7759" LastEditorUserId="6296" LastEditorDisplayName="" LastEditDate="2011-03-04T06:02:55.933" LastActivityDate="2011-03-04T06:02:55.933" Title="How to evaluate an expresssion" Tags="&lt;homework&gt;&lt;algebra-precalculus&gt;" AnswerCount="2" />
  <row Id="24885" PostTypeId="2" ParentId="24883" CreationDate="2011-03-03T21:55:19.200" Score="4" ViewCount="" Body="&lt;p&gt;Hint:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(\sqrt{x} \sqrt{x-3} - x)(\sqrt{x} \sqrt{x-3} + x) = -3x$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$3/2$ is correct.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2011-03-03T21:55:19.200" CommentCount="0" />
  <row Id="24886" PostTypeId="2" ParentId="24879" CreationDate="2011-03-03T21:55:58.140" Score="4" ViewCount="" Body="&lt;p&gt;Quadratic and cyclotomic fields are important because their structure is simple enough to allow the explicit determintion of some features. In other words, they are a remarkable source of examples.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More intrinsically, cyclotomic fields are important because, by the celebrated Kronecker-Weber theorem, every abelian extension of $\Bbb Q$ is a subextension of a cyclotomic field. An abelian extension is a Galois extension with abelian Galois group.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we fix a quadratic imaginary field $K$ (i.e. $K={\Bbb Q}(\sqrt{d})$ with $d\in{\Bbb Z}^{&amp;lt;0}$) the theory of complex multiplications tells us where to look for the abelian extensions of $K$. Namely, one considers the complex torus&#xA;$$&#xD;&#xA;T=\frac{\Bbb C}{{\Bbb Z}\oplus{\Bbb Z}\tau}&#xD;&#xA;$$&#xA;(where $K={\Bbb Q}(\tau)$) which embeds in the projective plane as a non-singular cubic $\cal C$. Then one knows that an abelian extension of $K$ is always a subextension of the field obtained adjoining to $K$ the $x$-coordinate of a point of $\cal C$ image of a point of $T$ of the form $a+b\tau$ with $a$, $b\in{\Bbb Q}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This may be a good candidate for the next important (class) of number field(s).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mind that these are the only established cases where we know explicitly the abelian extensions of a number field.&lt;/p&gt;&#xA;" OwnerUserId="688" LastActivityDate="2011-03-03T21:55:58.140" />
  <row Id="24887" PostTypeId="2" ParentId="24883" CreationDate="2011-03-03T21:56:33.623" Score="16" ViewCount="" Body="&lt;p&gt;The two standard techniques work: multiply and divide by the conjugate, then divide both numerator and denominator by the highest power of $x$.&#xA;\begin{align*}&#xA;\lim_{x\to\infty}3\left(\sqrt{x}\sqrt{x-3} - x+2\right) &amp;amp;= 3\lim_{x\to\infty}\left(\sqrt{x^2-3x} - (x-2)\right)\\&#xA;&amp;amp;= 3\lim_{x\to\infty}\frac{(\sqrt{x^2-3x}-(x-2))(\sqrt{x^2-3x}+(x-2))}{\sqrt{x^2-3x}+ (x-2)} \\&#xA;&amp;amp;= 3\lim_{x\to\infty}\frac{(x^2-3x) - (x-2)^2}{\sqrt{x^2-3x}+(x-2)}\\&#xA;&amp;amp;= 3\lim_{x\to\infty}\frac{x^2 - 3x - x^2 + 4x - 4}{\sqrt{x^2-3x} + (x-2)}\\&#xA;&amp;amp;= 3\lim_{x\to\infty}\frac{x - 4}{\sqrt{x^2-3x}+(x-2)}\\&#xA;&amp;amp;= 3\lim_{x\to\infty}\frac{\frac{1}{x}(x-4)}{\frac{1}{x}(\sqrt{x^2-3x}+(x-2))}\\&#xA;&amp;amp;= 3\lim_{x\to\infty}\frac {1 - \frac{4}{x}}{\sqrt{1 - \frac{3}{x}} + 1 - \frac{2}{x})}\\&#xA;&amp;amp;= 3\left(\frac{1}{\sqrt{1}+1}\right) = \frac{3}{2}.&#xA;\end{align*}&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2011-03-03T21:56:33.623" CommentCount="2" />
  <row Id="24888" PostTypeId="2" ParentId="24883" CreationDate="2011-03-03T21:56:33.973" Score="6" ViewCount="" Body="&lt;p&gt;$$ 3 (\sqrt{x}\sqrt{x-3} -x +2 ) =&#xD;&#xA;3 (\sqrt{x^2-3 x} -x +2 ) = 3 [x (\sqrt{1 -3/x} -1) +2 ]$$&#xA;$$=3 [x (1 -3/2x + O(x^{-2}) -1) +2 ] = 3/2 + O(x^{-1})$$&#xA;which implies that the limit is $3/2$.&lt;/p&gt;&#xA;" OwnerUserId="7266" LastEditorUserId="7266" LastEditorDisplayName="" LastEditDate="2011-05-17T07:00:38.020" LastActivityDate="2011-05-17T07:00:38.020" CommentCount="4" />
  <row Id="24889" PostTypeId="2" ParentId="24883" CreationDate="2011-03-03T22:00:36.233" Score="1" ViewCount="" Body="&lt;p&gt;See &lt;a href=&quot;http://math.stackexchange.com/questions/22928/limit-of-algebraic-function-avoiding-lhopitals-rule/22943#22943&quot;&gt;this&lt;/a&gt; MSE page, your question is similar.  Notice your limit can be rewritten as $$6+3\cdot \lim_{x\rightarrow\infty}\left(\sqrt{x^2-3x}-x\right)$$ but in general we have that &#xA; $$\lim_{x\rightarrow\infty}\sqrt[n]{x^n+a_{n-1}x^{n-1}+\cdots +a_1x+a_0}-x=\frac{a_n}{n}$$ which follows from &lt;a href=&quot;http://en.wikipedia.org/wiki/L%27H%C3%B4pital%27s_rule&quot; rel=&quot;nofollow&quot;&gt;l'hopitals rule&lt;/a&gt;.  Hence the answer to your problem is $$6-\frac{9}{2}=\frac{3}{2}.$$&lt;/p&gt;&#xA;" OwnerUserId="6075" LastActivityDate="2011-03-03T22:00:36.233" CommentCount="0" />
  <row Id="24890" PostTypeId="1" AcceptedAnswerId="24894" CreationDate="2011-03-03T22:02:18.333" Score="0" ViewCount="91" Body="&lt;p&gt;I have the following summation:&#xA;$$\sum_{i=0}^n\left(6in + n(8n+2)\right).$$&#xA;Evaluating I get&#xA;\begin{align*}&#xA;6n\left(\frac{n(n+1)}{2}\right) + n(8n+2) = 3n(n(n+1)) + n(8n+2)\\&#xA;&amp;amp;=3n^2(n+1) + n(8n+2)\\&#xA;&amp;amp;=3n^3 + 3n^2 + 8n^2 + 2n\\&#xA;&amp;amp;=3n^3 + 11n^2 + 2n.&#xA;\end{align*}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this correct? &lt;/p&gt;&#xA;" OwnerUserId="6331" LastEditorUserId="6296" LastEditorDisplayName="" LastEditDate="2011-03-04T01:37:19.347" LastActivityDate="2011-03-04T01:37:19.347" Title="Determining a closed-form solution for the following sum" Tags="&lt;algebra-precalculus&gt;&lt;sequences-and-series&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="24891" PostTypeId="2" ParentId="24884" CreationDate="2011-03-03T22:04:25.333" Score="5" ViewCount="" Body="&lt;p&gt;You can simply plug in and chug. &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Multiply $z=3$ by $x^{-1} = \frac{1}{10}$; get $A$.&lt;/li&gt;&#xA;&lt;li&gt;Multiply $A$ by $y=-3$ to get $B$. This is $(zx^{-1}y)$.&lt;/li&gt;&#xA;&lt;li&gt;Raise $B$ to the fifth power (multiply it by itself five times) to get $C$; that is, $C=(zx^{-1}y)^5$.&lt;/li&gt;&#xA;&lt;li&gt;Raise $y=-3$ to the fifth power to get $D$; this is $y^5$.&lt;/li&gt;&#xA;&lt;li&gt;Multiply $C$ and $D$ together to get $E$. This is the numerator of your expression.&lt;/li&gt;&#xA;&lt;li&gt;Take the reciprocal of $x^4$ (that is, $10^4$, then one over what you get). Call that $F$.&lt;/li&gt;&#xA;&lt;li&gt;Take the reciprocal of $z^4$; call that $G$.&lt;/li&gt;&#xA;&lt;li&gt;Multiply $F$ by $G$ and call it $H$; this is $x^{-4}z^{-4}$.&lt;/li&gt;&#xA;&lt;li&gt;Divide $E$ by $H$. That's the final answer.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Or you can do some algebra first: dividing by $x^{-4}$ is the same as &lt;em&gt;multiplying&lt;/em&gt; by $x^4$. Same with $z^{-4}$. So your expression is &lt;em&gt;exactly the same&lt;/em&gt; as&#xA;$$ x^4z^4(zx^{-1}y)^5y^5.$$&#xA;And since $(ab)^n = a^nb^n$, then $(zx^{-1}y)^5 = z^5 (x^{-1})^5 y^5 = z^5 x^{-5}y^5$. So your expression is the same as&#xA;$$x^4 z^4 z^5 x^{-5}y^5y^5.$$&#xA;Now simplify by putting all the $x$s, all the $y$s and all the $z$s together, plug in the values you have, and perform the operations that are left.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2011-03-03T22:04:25.333" CommentCount="1" />
  <row Id="24893" PostTypeId="2" ParentId="24884" CreationDate="2011-03-03T22:06:17.793" Score="1" ViewCount="" Body="&lt;p&gt;First, use the fact that $x^{-n}=\frac{1}{x^n}$ to find $$\frac{(z x^{-1} y)^5 y^5}{x^{-4} z^{-4}}=(zx^{-1}y)^5 \cdot \frac{y^5}{x^{-4}z^{-4}}=\left(\frac{zy}{x}\right)^5\cdot x^4z^4y^5 $$ Next, $$\left(\frac{zy}{x}\right)^5\cdot x^4z^4y^5=\frac{z^5y^5}{x^5}\cdot x^4z^4y^5=z^9y^{10}x$$  Can you finish from here?  &lt;/p&gt;&#xA;" OwnerUserId="6075" LastActivityDate="2011-03-03T22:06:17.793" />
  <row Id="24894" PostTypeId="2" ParentId="24890" CreationDate="2011-03-03T22:08:13.797" Score="3" ViewCount="" Body="&lt;p&gt;Your first step is wrong.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{i=0}^{n} (6in + n(8n+2)) = \sum_{i=0}^{n} (6in) + \sum_{i=0}^{n} n(8n+2)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You wrote&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{i=0}^{n} (6in) = 6n \sum_{i=0}^{n} i =  6n^2(n+1)/2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but missed this one:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{i=0}^{n} n(8n+2) = n(8n+2) \sum_{i=0}^n 1 =  n(n+1)(8n+2)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and wrote&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$n(8n+2)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;instead.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2011-03-03T22:18:23.463" LastActivityDate="2011-03-03T22:18:23.463" CommentCount="5" />
  <row Id="24895" PostTypeId="1" AcceptedAnswerId="24899" CreationDate="2011-03-03T22:08:48.760" Score="0" ViewCount="90" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;A curve has parametric equations &#xA;  $$\begin{array}{rcl}&#xD;&#xA;x &amp;amp; = &amp;amp; \cos(t)\\\&#xD;&#xA;y &amp;amp; = &amp;amp; \frac{1}{2}\sin(2t)&#xD;&#xA;\end{array}\qquad \text{ for }0\leq t \leq 2\pi.$$&#xA;  Show that the Cartesian equation of the curve is $y^2=(x^2)(1-x^2).$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What are the steps for showing this?&lt;/p&gt;&#xA;" OwnerUserId="7761" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-03T22:13:10.067" LastActivityDate="2011-03-03T22:16:09.597" Title="Converting parametric equation to Cartesian one" Tags="&lt;geometry&gt;&lt;analytic-geometry&gt;&lt;graph&gt;" AnswerCount="2" />
  <row Id="24896" PostTypeId="1" AcceptedAnswerId="24919" CreationDate="2011-03-03T22:11:37.340" Score="3" ViewCount="72" Body="&lt;p&gt;Let $R$ be a ring, $I$ an ideal. According to Atiyah-Macdonald, if $R$ is Noetherian, then, we have $\hat{I}=\hat{R}I$ where hat denotes $I$-adic completion of $R$ and (I presume) $\hat{I}$ denotes the induced completion on $I$. I don't understand how to arrive at this equality and why the Noetherian hypothesis is necessary. Essentially $\hat{I}$ consists of equivalence classes of Cauchy sequences with elements in $I$. Any element of $\hat{R}I$ is an equivalence class of Cauchy sequences consisting of elements of $I$. I don't see how every Cauchy sequence with elements in $I$ is equivalent to one which can be written as a sum of products of a Cauchy sequence and a constant sequence of an element of $I$.&lt;/p&gt;&#xA;" OwnerUserId="7763" LastEditorUserId="7763" LastEditorDisplayName="" LastEditDate="2011-03-04T00:31:09.527" LastActivityDate="2011-03-04T00:50:39.027" Title="I-adic completion of a ring" Tags="&lt;algebra&gt;&lt;commutative-algebra&gt;&lt;ring-theory&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="24897" PostTypeId="2" ParentId="24895" CreationDate="2011-03-03T22:12:22.423" Score="2" ViewCount="" Body="&lt;p&gt;Hint:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Use the trigonometric identities: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sin (2x) = 2 \sin x \cos x$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \sin^2 x + \cos^2 x = 1$$&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2011-03-03T22:12:22.423" CommentCount="1" />
  <row Id="24898" PostTypeId="1" CreationDate="2011-03-03T22:13:29.287" Score="0" ViewCount="133" Body="&lt;p&gt;How to show that $\sin(2 \pi nx)$ does not converge as n goes to infinity?  $x \in (0,1/2)$&lt;/p&gt;&#xA;" OwnerUserId="7762" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-03T22:17:06.990" LastActivityDate="2011-03-03T22:17:06.990" Title="$\sin(2\pi nx)$ does not converge for $x \in (0,1/2)$" Tags="&lt;calculus&gt;&lt;analysis&gt;" CommentCount="6" />
  <row Id="24899" PostTypeId="2" ParentId="24895" CreationDate="2011-03-03T22:16:09.597" Score="2" ViewCount="" Body="&lt;p&gt;First, evaluate $x^2(1-x^2)$ with $x=\cos(t)$ and verify that you get $y^2$. This will prove that every point in the parametric curve lies on the curve $y^2 = x^2(1-x^2)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But this is not enough; this only shows that the parametric curve is &lt;em&gt;contained&lt;/em&gt; inside the curve $y^2 = x^2(1-x^2)$. To show that the paramatric curve &lt;em&gt;is&lt;/em&gt; the curve $x^2(1-x^2)$, you also need to show that every point on the latter curve is &quot;covered&quot; by the parametric curve.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So show that every point on that curve can be obtained by picking an appropriate value of $t$ and plugging into the parametric equations. (For example, notice that if $(a,b)$ lies in the curve $y^2 = x^2(1-x^2)$, then you must have $-1\leq a\leq 1$ (otherwise, $1-x^2\lt 0$, and $y^2\lt 0$, which is impossible; that means that there exist a $t$ on $[0,2\pi]$ (in fact, two of them usually) for which $\cos(t) = a$. Then check to see if one of  those values of $t$ &quot;works&quot; for $y$).&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2011-03-03T22:16:09.597" />
  <row Id="24900" PostTypeId="2" ParentId="24873" CreationDate="2011-03-03T22:18:37.800" Score="5" ViewCount="" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;is there intuition for why a proof is so difficult?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sure: the topological category is horrible. A generic continuous function is bizarre and will &lt;a href=&quot;http://en.wikipedia.org/wiki/Space-filling_curve&quot; rel=&quot;nofollow&quot;&gt;violate your geometric intuitions&lt;/a&gt;. When we prove that $\mathbb{R}^n$ is not homeomorphic to $\mathbb{R}^m$ by proving that $S^n$ is not homotopy equivalent to $S^m$, much of the work goes into proving that, &lt;em&gt;up to homotopy&lt;/em&gt;, we can ignore how bizarre generic continuous functions are. That is, you think that &quot;homeomorphic&quot; is an intuitive condition, but it's not. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is why the corresponding question in the smooth category is much easier; generic smooth functions are much less bizarre in a way that is quantified by &lt;a href=&quot;http://en.wikipedia.org/wiki/Sard%27s_theorem&quot; rel=&quot;nofollow&quot;&gt;Sard's lemma&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Here is a specific example of what I mean. The reason we can distinguish $\mathbb{R}$ from $\mathbb{R}^m, m &amp;gt; 1$ by removing a point is because continuous functions send points to points. It is tempting to argue as follows: we can distinguish $\mathbb{R}^2$ from $\mathbb{R}^m, m &amp;gt; 2$ by removing a &lt;em&gt;line&lt;/em&gt;, since the result is not connected for $\mathbb{R}^2$ but is connected for $\mathbb{R}^m$. But of course this argument doesn't work because continuous functions need not send lines to lines; the image of a line can be much bigger, e.g. all of $\mathbb{R}^m$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is &lt;em&gt;weird.&lt;/em&gt; For $\mathbb{R}^2$, as you say, we can rescue this proof by removing a point because we know about simple connectedness and because, again, continuous functions send points to points. But for $\mathbb{R}^3$ we are stuck: removing a plane doesn't work, and removing a line doesn't even work, so if we want to stick to our &quot;removing a point&quot; strategy we had better figure out what the analogue of simple connectedness is in higher dimensions. This naturally leads to homotopy and homology, which happen to be strong enough tools to deal with the fact that continuous functions are bizarre, but they don't change the fact that continuous functions are bizarre. &lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2011-03-03T23:21:01.833" LastActivityDate="2011-03-03T23:21:01.833" CommentCount="4" />
  <row Id="24901" PostTypeId="1" AcceptedAnswerId="25036" CreationDate="2011-03-03T22:19:45.760" Score="10" ViewCount="222" Body="&lt;p&gt;Let $f$ be a strictly increasing positive continuous function defined on $[1,\infty)$ with limit $\infty$ as $x$ goes towards $\infty$. Then $\sum_{k=0}^{\infty} \frac{1}{f(k)}$ converges if and only if $\sum_{k=0}^{\infty} \frac{f^{-1}(k)}{k^2}$ converges, where $f^{-1}$ denotes the inverse of $f$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I saw this claim as an exercise in a analysis textbook, linked from this site. Can not remember which one unfortunately. It was listed as a challenging exercise and has proven too challenging for me.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My initial idea was to try to use the integral test. $\sum \frac{1}{f(k)}$ converges exactly when $\int \frac{1}{f(x)}$ converges. I thought I might do some smart change of variable to find that $\int \frac{f^{-1}(x)}{x^2}$ converges. I could not come up with one however and I also realized that I do not know that the sum $\sum \frac{f^{-1}(k)}{k^2}$ satisfies the conditions for using the integral test. Unfortunately I ran out of ideas at that point. &lt;/p&gt;&#xA;" OwnerUserId="4623" LastEditorUserId="4623" LastEditorDisplayName="" LastEditDate="2011-03-04T10:16:09.430" LastActivityDate="2011-03-04T16:43:40.043" Title="$\sum \frac{1}{f(k)}$ converges iff $\sum \frac{f^{-1}(k)}{k^2}$ converges" Tags="&lt;sequences-and-series&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="2" />
  <row Id="24902" PostTypeId="1" AcceptedAnswerId="24910" CreationDate="2011-03-03T22:20:12.450" Score="1" ViewCount="61" Body="&lt;p&gt;Hello!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have some LP problem and I'm willing to solve it &lt;em&gt;(this is an exercise from some optimization-related book).&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, Mathematica tells me that the problem is &lt;strong&gt;unbounded&lt;/strong&gt; and I want to make a generic proof of that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What would be the easiest way to prove that the selected linear optimization problem is &lt;strong&gt;unbounded&lt;/strong&gt; (and obviously, &lt;em&gt;doesn't have an exact solution?).&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;LinearProgramming[&#xA;&#xA;{1, 2, 1, -1, 0},&#xA;&#xA;{{0, 10, 1, 2, 3}, {-1, 5, 1, 1, 1}, {2, -1, 1, -3, 0}}, &#xA;&#xA;{{25, 0}, {10, 0}, {6, -1}}, &#xA;&#xA;{{-Infinity, Infinity}, {0, Infinity}, {0, Infinity}, &#xA; {-Infinity, Infinity}, {-Infinity, Infinity}}, &#xA;&#xA; Method -&amp;gt; &quot;RevisedSimplex&quot;]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="4971" LastActivityDate="2011-03-04T00:17:33.143" Title="Basic linear problem solving" Tags="&lt;optimization&gt;&lt;linear-programming&gt;" AnswerCount="1" />
  <row Id="24903" PostTypeId="2" ParentId="24883" CreationDate="2011-03-03T22:43:12.180" Score="5" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\ $ It is simply a first derivative: $\: $ changing variables $\rm\ x\to 1/x\ $ transforms it to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\rm\displaystyle\ 3\ \lim_{x\to\ 0}\ \frac{\sqrt{1-3x}+2\ x-1}{x}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But this $\rm\ lim\ $ has the form of a derivative, namely it is simply&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\rm\displaystyle\ \lim_{x\to\ 0}\ \frac{f(x)-f(0)}x\ =\ f\:&amp;#39;(0) \quad\ \ \ for\quad\ f(x)\ =\ \  \sqrt{1-3x}+2\ x-1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now it is easy to directly calculate that $\rm\ f\:&amp;#39;(0)\ =\ 1/2\ $ so the original limit $\to 3/2\:.\:$ Note that this solution employs only knowledge of the &lt;em&gt;definition&lt;/em&gt; of the derivative and basic &lt;em&gt;rules&lt;/em&gt; for calculating derivatives of polynomials and powers. It does &lt;em&gt;not&lt;/em&gt; employ related more advanced techniques such (binomial) power series or Taylor series, l'Hôpital's rule, the mean-value theorem, etc.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2011-03-04T02:00:48.063" LastActivityDate="2011-03-04T02:00:48.063" CommentCount="3" />
  <row Id="24904" PostTypeId="1" AcceptedAnswerId="24905" CreationDate="2011-03-03T22:46:01.370" Score="1" ViewCount="112" Body="&lt;p&gt;So I am having a lot of trouble proving this. It was for an assignment due tuesday, but the prof said I can have a bit of extra time with this question. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please do not give me the answer because I am sure my prof is on this website :P&lt;/p&gt;&#xA;&#xA;&lt;p&gt;okay so the question is as follows:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Let $R$ be a ring with identity. Show that the map $f: \mathbb{Z} \rightarrow R$ given by $f(k) = k1_R$ is a homomorphism.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So obviously I went about proving the axioms thinking the $k$ and $1_R$ are integer products. BUT as my prof pointed out, there not. Here is the explanation on his course webpage. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A couple of remarks about Assignment 8&#xA;  problems:&#xA;  In exercise 25, when H'ford writes&#xA;  something like  $k1_R$  , where  k is an integer, he of course is not&#xA;  speaking of a  PRODUCT of elements of&#xA;  the ring  R.  Given an element  a of&#xA;  R, the  inductive definition of  ka, &#xA;  k in ZZ, in connection with the &#xA;  definition of the characteristic of a&#xA;  ring, about which  you folks had an&#xA;  assignment problem, was this:&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;0a := 0_R,&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;and, for all  k  in  |N,  (k+1)a :=&#xA;  ka + a.  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Also, if  k  is a negative integer,&#xA;  then  ka := -((-k)a).&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;I believe we defined finite sums&#xA;  inductively a long time ago.  One&#xA;  defines  0a  as above, and we could&#xA;  then use our earlier inductive&#xA;  definition and just say that, for each&#xA;  k  in  ZZ^+,&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;               k &#xA;        ka := Sum a.&#xA;              j=1     &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;  &#xA;  &lt;p&gt;And then define  ka  for negative&#xA;  integers  k  as above.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;so how do I use his definition of what $k$ is and how do I apply it to my proof?&lt;/p&gt;&#xA;" OwnerUserId="287" LastEditorUserId="367" LastEditorDisplayName="" LastEditDate="2011-03-03T22:59:23.430" LastActivityDate="2011-03-04T08:00:34.210" Title="proving - ring theory and homomorphisms" Tags="&lt;abstract-algebra&gt;&lt;ring-theory&gt;" AnswerCount="2" />
  <row Id="24905" PostTypeId="2" ParentId="24904" CreationDate="2011-03-03T22:58:09.373" Score="1" ViewCount="" Body="&lt;p&gt;You have to show that $(m\cdot n)1_R=(m1_R)\cdot(n1_R)$ for all integers $m$ and $n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $m$ and $n$ are non-negative this follows from your definition $ka:=\sum_{j=1}^k a$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The general case is then an easy consequence of the axioms. You should try to work this out.&lt;/p&gt;&#xA;" OwnerUserId="367" LastEditorUserId="367" LastEditorDisplayName="" LastEditDate="2011-03-04T08:00:34.210" LastActivityDate="2011-03-04T08:00:34.210" />
  <row Id="24906" PostTypeId="2" ParentId="24901" CreationDate="2011-03-03T23:00:27.390" Score="11" ViewCount="" Body="&lt;p&gt;As the OP noted, by the integral test the sum converges iff $\int_0^\infty dx\,\frac{1}{f(x)}$ converges. Integrating by parts, we obtain&#xA;$$\int_0^\infty dx\,\frac{1}{f(x)} = - \int_0^\infty dx\, \frac{x f&amp;#39;(x)}{f(x)^2}.$$&#xA;Changing the integration variable from $x$ to $y=f(x)$ (which is admissible as $f(x)$ is monotonously increasing) yields&#xA;$$\int_0^\infty dx\,\frac{1}{f(x)}= - \int_{f(0)}^\infty dy\, \frac{f^{-1}(y)}{y^2}.$$ Using the integral test again, we have that the original series converges iff $$\sum \frac{f^{-1}(k)}{k^2}$$ converges.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: (trying to incorporate some comments)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) the OP did not specify that $f$ is differentiable. One can use a Spline interpolation to make $f \in C^1$. (thaks to user7530)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) there was a question about the fate of the boundary term $x f(x)$: one should prove the argument in two steps. First assume $\sum f(k)^{-1}$ converges, then the boundary term is zero as $f(k)$ falls off faster than $k^{-1}$ and you can prove that $\sum \frac{f^{-1}(k)}{k^2}$ converges. Then the other way round, assume $\sum \frac{f^{-1}(k)}{k^2}$ converges. The boundary term then reads $f^{-1}(y) y$ which you can prove to be finite by the same method and thereby showing that $\sum f(k)^{-1}$ converges.&lt;/p&gt;&#xA;" OwnerUserId="7266" LastEditorUserId="7266" LastEditorDisplayName="" LastEditDate="2011-03-04T09:19:55.900" LastActivityDate="2011-03-04T09:19:55.900" CommentCount="7" />
  <row Id="24907" PostTypeId="2" ParentId="24904" CreationDate="2011-03-03T23:01:23.397" Score="0" ViewCount="" Body="&lt;p&gt;So, I guess that for $m$ and $n$ positive you need to prove that&#xA;$$&#xD;&#xA;\left(\sum_{i=1}^m\,1_R\right)\left(\sum_{i=1}^n\,1_R\right)=&#xD;&#xA;\left(\sum_{i=1}^{mn}\,1_R\right).&#xD;&#xA;$$&#xA;How many addends you have when you apply the distributive property on the LHS ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then you have to come up with something for when $m$ and $n$ are not both positive.&lt;/p&gt;&#xA;" OwnerUserId="688" LastActivityDate="2011-03-03T23:01:23.397" />
  <row Id="24908" PostTypeId="1" CreationDate="2011-03-03T23:34:59.970" Score="1" ViewCount="188" Body="&lt;p&gt;I was curious (even though this is a very amateur question)... what would the multiplication equivalent of sigma (the summation symbol) be?&#xA;$$\sum$$&#xA;I want to do a series of multiplication of terms.... instead of addition...&lt;/p&gt;&#xA;" OwnerUserId="5970" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2011-03-04T02:00:23.510" LastActivityDate="2011-03-04T02:00:23.510" Title="multiplication equivalent of the summation symbol" Tags="&lt;notation&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="24909" PostTypeId="2" ParentId="24908" CreationDate="2011-03-03T23:36:29.043" Score="5" ViewCount="" Body="&lt;p&gt;It's a capital pi (for product): $\prod$. You can sometimes find useful symbols and guess what they mean from their &lt;a href=&quot;http://www.artofproblemsolving.com/Wiki/index.php/LaTeX:Symbols&quot; rel=&quot;nofollow&quot;&gt;$\LaTeX$ names&lt;/a&gt;, although I guess the most reliable way to find useful, well-known symbols is just to read a lot of math. &lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2011-03-03T23:36:29.043" CommentCount="2" />
  <row Id="24910" PostTypeId="2" ParentId="24902" CreationDate="2011-03-03T23:51:33.197" Score="2" ViewCount="" Body="&lt;p&gt;Use &lt;a href=&quot;http://en.wikipedia.org/wiki/Linear_programming#Duality&quot; rel=&quot;nofollow&quot;&gt;duality&lt;/a&gt;.  If you can prove that the dual problem is infeasible, then the original problem must either be infeasible or unbounded.  Then show that the original problem is feasible (i.e., by finding at least one point in the feasible region). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Detecting infeasibility for your dual should be straightforward.  You will have three variables, three equality constraints, and two inequality constraints in the dual.  Three equations with three dual variables will (in this case) produce a unique solution.  Then you just need to verify that this unique solution does not satisfy the two remaining inequality constraints. &lt;/p&gt;&#xA;" OwnerUserId="2370" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2011-03-04T00:17:33.143" LastActivityDate="2011-03-04T00:17:33.143" />
  <row Id="24911" PostTypeId="1" AcceptedAnswerId="24918" CreationDate="2011-03-03T23:55:23.053" Score="1" ViewCount="154" Body="&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt;  In Functional Analysis we can note things like: every closed subspace of a Banach space is Banach.  In this case, what does &quot;closed subspace&quot; mean?  Does this mean closed under the norm topology?  Or does this mean closed in the sense that multiplication of scalars and addition of vectors is closed?  Or does this mean closed with respect to limits?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm reviewing this material and I realized that even though I have this in my notes a number of times I am unsure of what this actually is.  I thought it was the second statement above, but the third statement makes the &quot;every closed subspace of a banach space is banach&quot; statement easy to prove.&lt;/p&gt;&#xA;" OwnerUserId="2959" LastActivityDate="2011-03-04T00:47:48.150" Title="Closed Subspaces of Vector Spaces." Tags="&lt;functional-analysis&gt;&lt;definition&gt;" AnswerCount="2" />
  <row Id="24912" PostTypeId="1" AcceptedAnswerId="24914" CreationDate="2011-03-04T00:08:44.340" Score="2" ViewCount="58" Body="&lt;p&gt;$$&#xD;&#xA;\begin{align*}&#xD;&#xA;&amp;amp;= (n+1)! − 1 + ( (n+1) &amp;#183; (n+1)! )\\&#xD;&#xA;&#xD;&#xA;&amp;amp;= (n+1)! (1+n+1) − 1\\&#xD;&#xA;&#xD;&#xA;&amp;amp;= (n+1)! (n+2) − 1\\&#xD;&#xA;&#xD;&#xA;&amp;amp;= (n+2)! − 1\\&#xD;&#xA;\end{align*}&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm confused at how the first step goes to the second step. Can someone please clarify this?&lt;/p&gt;&#xA;" OwnerUserId="7767" LastEditorUserId="6296" LastEditorDisplayName="" LastEditDate="2011-03-04T01:01:15.663" LastActivityDate="2011-03-04T01:01:15.663" Title="Use of algebra and factorials for a question related to proof by induction." Tags="&lt;algebra&gt;&lt;discrete-mathematics&gt;&lt;induction&gt;&lt;factorial&gt;" AnswerCount="1" />
  <row Id="24913" PostTypeId="2" ParentId="24911" CreationDate="2011-03-04T00:08:54.497" Score="1" ViewCount="" Body="&lt;p&gt;I think &quot;subspace&quot; usually refers to closure under scalar multiplication and vector addition, while &quot;closed&quot; refers to closure under the topology. Sometimes people like to say &quot;linear subspace&quot; instead of subspace to be more precise, but I've found that in a lot of functional analysis textbooks, the linear part is often implied.&lt;/p&gt;&#xA;" OwnerUserId="1736" LastActivityDate="2011-03-04T00:08:54.497" CommentCount="1" />
  <row Id="24914" PostTypeId="2" ParentId="24912" CreationDate="2011-03-04T00:13:22.337" Score="5" ViewCount="" Body="&lt;p&gt;It's just the &lt;em&gt;distributive law&lt;/em&gt; $\rm\ k!\ (1 + k)\ =\ k! + k\ k!\ $ for $\rm\ k = n+1.$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;TIP&lt;/strong&gt; $\ $ Often applications of simple laws are obfuscated in specific proofs. In such contexts, in order to abstract out the essential structure, it may help to substitute variables for expressions that play no role. Let's do this above. Substituting $\rm (n+1)!\ \to\ x,\ \ n+1\ \to\ y\ $ the first two steps become &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\rm x - 1 + y\ x\ = \ x\ (1 + y) - 1 $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which makes obvious the application of the distributive law.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2011-03-04T00:35:20.227" LastActivityDate="2011-03-04T00:35:20.227" />
  <row Id="24915" PostTypeId="2" ParentId="24908" CreationDate="2011-03-04T00:22:45.603" Score="1" ViewCount="" Body="&lt;p&gt;I google &quot;latex symbols&quot; when I need something I can't recall. That'll give you many lists and tips. For example, if you choose the first hit, the &lt;a href=&quot;http://www.artofproblemsolving.com/Wiki/index.php/LaTeX%3aSymbols&quot; rel=&quot;nofollow&quot;&gt;AoPS list&lt;/a&gt; and look for the sum symbol you'll find the product symbol right below it. This is but a simple example of a general technique of exploiting organization and classification on the web to discover information about similar items.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the inverse, when you know the symbol but not the name there is &lt;a href=&quot;http://detexify.kirelabs.org/classify.html&quot; rel=&quot;nofollow&quot;&gt;deTeXify&lt;/a&gt; which let's you draw the symbol and does OCR to find the name. For deeper questions there's the &lt;a href=&quot;http://tex.stackexchange.com/&quot;&gt;(La)Tex SE site.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2011-03-04T00:44:56.177" LastActivityDate="2011-03-04T00:44:56.177" />
  <row Id="24916" PostTypeId="1" CreationDate="2011-03-04T00:23:22.720" Score="1" ViewCount="46" Body="&lt;p&gt;Question:&#xA;The area covered by a newly-introduced species of invasive weed doubles every 4 years. Suppose the weed currently covers 23 km^2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a) Write an equation that gives the area, A, covered by the weed, in t years.&#xA;b) Estimate how many years it will take for the weed to cover 51 km^2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$A = 23(2)^{t/4}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I could keep plugging in values till I get the correct but how would I reverse the equation to get t when A = 51?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I really need to understand this stuff for a test tomorrow... I jumped into a course 3 weeks late and this is what I get hit with.&lt;/p&gt;&#xA;" OwnerUserId="7759" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-04T01:02:29.003" LastActivityDate="2011-03-04T01:03:38.317" Title="Solve for A (Exponents)" Tags="&lt;homework&gt;&lt;algebra-precalculus&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="24917" PostTypeId="1" CreationDate="2011-03-04T00:34:32.120" Score="3" ViewCount="100" Body="&lt;p&gt;I need to calculate the inverse Fourier transform of the following functions:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;$\displaystyle f(w) = e^{(-\pmb{i}5w)} * {\rm sinc}(2w) $&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$\displaystyle g(w) = \frac{\pmb{i}w}{(3+\pmb{i}w)(1+\pmb{i}w)} $&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I'm really stuck and I don't know where to start, any help would be appreciated!&lt;/p&gt;&#xA;" OwnerUserId="7766" LastEditorUserId="1543" LastEditorDisplayName="" LastEditDate="2011-04-17T20:31:09.930" LastActivityDate="2011-04-17T20:31:09.930" Title="Calculating the inverse Fourier transform of two given functions" Tags="&lt;fourier-analysis&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="24918" PostTypeId="2" ParentId="24911" CreationDate="2011-03-04T00:47:48.150" Score="4" ViewCount="" Body="&lt;p&gt;&quot;Does this mean closed under the norm topology?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yes, that's what it means.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Or does this mean closed in the sense that multiplication of scalars and addition of vectors is closed?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;No, that is what &quot;subspace&quot; means here.  (As user1736 says, sometimes people write &quot;linear subspace&quot; for emphasis, but in functional analysis it is generally safe to assume that &quot;subspace&quot; means &quot;linear subspace&quot;.)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Or does this mean closed with respect to limits?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again yes, because this is an equivalent characterization of closed subspaces of a topological space.  Since a Banach space is metrizable hence first countable, it is enough to take limits of sequences.  For a general topological space -- and even some non-Banach topological vector spaces -- in order to retain this equivalence, one must allow limits of &lt;a href=&quot;http://en.wikipedia.org/wiki/Net_%28topology%29&quot; rel=&quot;nofollow&quot;&gt;nets&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="299" LastActivityDate="2011-03-04T00:47:48.150" CommentCount="1" />
  <row Id="24919" PostTypeId="2" ParentId="24896" CreationDate="2011-03-04T00:50:39.027" Score="2" ViewCount="" Body="&lt;p&gt;If $R$ is Notherian, then $I$ must be finitely generated, say $I = \langle p_1,\ldots, p_n\rangle$. So if an element of $\hat{I}$ is represented by a sum $x = i_1 + i_2 + \cdots$, rewriting $i_m = p_1 i_{m1} + \ldots + p_n i_{mn}$ we can rewrite this as &#xA;$$x = p_1 \sum_{k_1}i_{1k} + \cdots + p_n\sum_{k_n}i_{nk} \in \hat{R}I$$&#xA;which is what you wanted.&lt;/p&gt;&#xA;" OwnerUserId="668" LastActivityDate="2011-03-04T00:50:39.027" CommentCount="2" />
  <row Id="24920" PostTypeId="2" ParentId="24916" CreationDate="2011-03-04T01:03:38.317" Score="3" ViewCount="" Body="&lt;p&gt;Logarithms are your friend. You have $A = 23(2)^{t/4}$, which means that &#xA;$$\ln(A) = \ln(23(2)^{t/4}) = \ln(23) + \ln(2^{t/4}) = \ln(23) + \frac{t}{4}\ln(2).$$&#xA;If you know $A$, you can use this to figure out $t$.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2011-03-04T01:03:38.317" />
  <row Id="24921" PostTypeId="1" AcceptedAnswerId="24923" CreationDate="2011-03-04T01:43:40.620" Score="1" ViewCount="110" Body="&lt;p&gt;While reading through my textbook it says &quot;the most important example of an inner-product space is $F^n$&quot;, where $F$ denotes $\mathbb{C}$ or $\mathbb{R}$ . &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Our definition of of an inner product on a vector space $V$ is as follows:&lt;br&gt;&#xA;1) Positive definite: $\langle v,v \rangle \ge 0$ with equality if and only if $v=0$&lt;br&gt;&#xA;2) Linearity in the first arguement: $\langle a_1v_1+a_2v_2,w \rangle = a_1 \langle v_1,w \rangle + a_2\langle v_2,w \rangle$&lt;br&gt;&#xA;3) Conjugate symmetric: $\langle u,v\rangle = \overline{\langle v,u\rangle}$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $$\displaystyle w=(w_1\ldots,w_n) , z=(z_1,\ldots,z_n)$$&lt;br&gt;&#xA;Then:&lt;br&gt;&#xA;$$\displaystyle \langle w,z\rangle =w_1\overline{z_1}+\cdots+w_n\overline{z_n}$$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm trying to verify that this is indeed true.  So first I want to check that $\langle w,z\rangle$ satisfies condition (1).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Say that $w,z\in \mathbb{C}$.&lt;br&gt;&#xA;Just looking at say $w_1=a+bi$ and $z_1=c+di$, how can we guarantee that $w_1\overline{z_1}\geq 0$?&lt;br&gt;&#xA;&lt;strong&gt;If we can observe this, it would need to hold true for the other coordinates as well.  So my question is, how do we know that $w_1\overline{z_1}\geq 0$?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="2533" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-04T06:50:00.837" LastActivityDate="2011-03-04T06:50:45.737" Title="Inner Product, definite positive?" Tags="&lt;linear-algebra&gt;&lt;inner-product-space&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="24922" PostTypeId="1" AcceptedAnswerId="24926" CreationDate="2011-03-04T01:53:25.967" Score="2" ViewCount="96" Body="&lt;p&gt;A two dimensional silo shaped figure is formed by placing a semi-circle of diameter 1 on top of a unit square, with the diameter coinciding with the top of the square. How do we find the radius of the smallest circle that contains this silo?&lt;/p&gt;&#xA;" OwnerUserId="5359" LastActivityDate="2011-03-04T02:25:07.917" Title="Find radius of the Smallest Circle that contains this figure" Tags="&lt;geometry&gt;" AnswerCount="1" FavoriteCount="0" />
  <row Id="24923" PostTypeId="2" ParentId="24921" CreationDate="2011-03-04T02:02:57.233" Score="4" ViewCount="" Body="&lt;p&gt;Item 1 in your definition of an inner product is incorrect.  A simple counter example from $\mathbb{R}^2$ is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(1,0).(-1,0) = -1.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It should read &#xA;$$\langle v, v \rangle \ge 0,$$&#xA;this guarantees that all vectors in your space have a non-negative length. &lt;/p&gt;&#xA;" OwnerUserId="2283" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-04T06:50:45.737" LastActivityDate="2011-03-04T06:50:45.737" />
  <row Id="24924" PostTypeId="1" AcceptedAnswerId="24925" CreationDate="2011-03-04T02:20:00.167" Score="1" ViewCount="129" Body="&lt;p&gt;Hi, I am given a square matrix $A$, and I need to prove that if c is its eigenvalue, then it is also an eigenvalue of its transpose. How should I approach this? Clearly $Av$=$cv$, but I am not sure how to bring transpose into the equation.&lt;/p&gt;&#xA;" OwnerUserId="5157" LastActivityDate="2011-03-04T03:58:09.060" Title="If c is an eigenvalue of $A$, is it an eigenvalue of $A^{\mathbf{T}}$?" Tags="&lt;linear-algebra&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="24925" PostTypeId="2" ParentId="24924" CreationDate="2011-03-04T02:23:27.573" Score="11" ViewCount="" Body="&lt;p&gt;A simple way would be to look at $\left | A^T-cI \right |$. $$\left | A^T-cI \right | = \left| (A-cI)^T \right| = \left| (A-cI) \right|$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\left| B \right|$ denotes the determinant of $B$. First note that for any matrix $B$, $\left| B \right| = \left| B^T \right|$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(This is true since you get the same determinant if you find the determinant along the row or column.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$c$ is an eigenvalue of $A$ iff $\left| A - cI\right| = 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that $\left| A^T - cI \right| = \left| A^T - cI^T \right| = \left| (A - cI)^T \right| = \left| A - cI\right| = 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, $c$ is an eigenvalue of $A^T$.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2011-03-04T02:42:35.733" LastActivityDate="2011-03-04T02:42:35.733" CommentCount="3" />
  <row Id="24926" PostTypeId="2" ParentId="24922" CreationDate="2011-03-04T02:25:07.917" Score="5" ViewCount="" Body="&lt;p&gt;Draw the line of length 1.5 that cuts both the square and the half circle into two identical pieces.  (Starting from the middle of the base of the square, go straight up)  Notice that the center of the larger circle must lie somewhere on this line.  Say that the center point is distance $x$ from the base of the square, and that this circle has radius $r$.  Than, the distance from the center point to the corner of the square should be $r$, so by Pythagoras we find $$r^2 = x^2 +(1/2)^2$$.  Now, the distance to the top of the half circle from this center point is $1.5-x$ and this should also be the radius.  Hence $$1.5-x=r.$$  Thus you have two equations and two unknowns, which you can solve from here.&lt;/p&gt;&#xA;" OwnerUserId="6075" LastActivityDate="2011-03-04T02:25:07.917" CommentCount="1" />
  <row Id="24927" PostTypeId="1" CreationDate="2011-03-04T02:27:31.770" Score="9" ViewCount="161" Body="&lt;p&gt;Does this mean that the first homotopy group in some sense contains more information than the higher homotopy groups? Is there another generalization of the fundamental group that can give rise to non-commutative groups in such a way that these groups contain more information than the higher homotopy groups? &lt;/p&gt;&#xA;" OwnerUserId="7769" LastActivityDate="2011-03-05T01:21:15.827" Title="What are the ramifications of the fact that the first homotopy group can be non-commutative, whilst the higher homotopy groups can't be?" Tags="&lt;algebraic-topology&gt;&lt;homotopy-theory&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="1" />
  <row Id="24928" PostTypeId="2" ParentId="24924" CreationDate="2011-03-04T02:38:12.540" Score="1" ViewCount="" Body="&lt;p&gt;Another way to look at it is to see that it's equivalent to the following statement: $A$ is regular iff $A^T$ is regular. That follows from the fact that the row rank is equal to the column rank. But I'm not sure how easy it is to show that directly. Anyone?&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2011-03-04T02:38:12.540" CommentCount="6" />
  <row Id="24929" PostTypeId="2" ParentId="24869" CreationDate="2011-03-04T02:40:43.287" Score="1" ViewCount="" Body="&lt;p&gt;The standard trick is to replace $z$ with $z^+ - z^-$ throughout.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2011-03-04T02:40:43.287" CommentCount="1" />
  <row Id="24932" PostTypeId="2" ParentId="24927" CreationDate="2011-03-04T03:13:20.230" Score="6" ViewCount="" Body="&lt;p&gt;Let $\mathcal{C}$ be a category with finite limits and a final object. In general, if $Y$ is an object in $\mathcal{C}$ such that $\hom(X, Y)$ is naturally a group for each $X \in \mathcal{C}$, then $Y$ is called a &quot;group object&quot; in $\mathcal{C}$; that is, there is a multiplication map $Y \times Y \to Y$ and an inversion $Y \to Y$ and an identity $\ast \to Y$ (for $\ast$ the final object) that satisfy a categorical version of the usual group axioms (stated arrow-theoretically). In the case of interest here, $\mathcal{C}$ is the homotopy category of pointed topological spaces, and the statement that the homotopy groups are groups is the statement that the spheres $S^n$ are group objects in the opposite category -- in other words, $S^n$ is a so-called &quot;H cogroup.&quot; When one writes out the arrows, one ends up with a &quot;comultiplication map&quot; $S^n \to S^n \vee S^n$ and a map $S^n \to \ast$ ($\ast$ the point)  that satisfy the dual of the usual group axioms, up to homotopy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason that the higher homotopy groups are abelian and $\pi_1$ is not is that $S^n$ is an abelian H cogroup for $n \geq 2$ and not  for $n=1$. This is basically a consequence of the Eckmann-Hilton argument (namely, there are two natural and mutually distributive ways of defining the H cogroup structure of $S^n$, depending on which coordinate one chooses; they must be equal and both commutative). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now to your more general question. So one can define covariant functors from the pointed homotopy category to the category of groups: just pick any H cogroup object and to consider maps from it into the given space. An easy way of getting these is to take the reduced suspension of any space $X$, $\Sigma X$, and to note that $\Sigma X$ can be made into an H cogroup (in kind of the same way as $S^n$ is---actually, the $S^n$ is a special case of this). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One may object that considering suspensions is not really anything new, because homotopy classes of $\Sigma X = S^1 \wedge X$ into a space $Y$ is the same as considering homotopy classes of maps $S^1 \to Y^X$ when $X$ is reasonable (say locally compact and Hausdorff), so we really have a variant of the fundamental group.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, there is the question of whether &lt;em&gt;all&lt;/em&gt; functors from the pointed homotopy category to the category of groups can be expressed in this way, that is, whether it is representable. On the pointed homotopy category of CW complexes, there are fairly &lt;a href=&quot;http://en.wikipedia.org/wiki/Brown_representability_theorem&quot; rel=&quot;nofollow&quot;&gt;weak conditions&lt;/a&gt; that will ensure representability.&lt;/p&gt;&#xA;" OwnerUserId="536" LastEditorUserId="536" LastEditorDisplayName="" LastEditDate="2011-03-05T01:21:15.827" LastActivityDate="2011-03-05T01:21:15.827" CommentCount="2" />
  <row Id="24934" PostTypeId="2" ParentId="24924" CreationDate="2011-03-04T03:58:09.060" Score="0" ViewCount="" Body="&lt;p&gt;Let $\lambda_1,v_1$ and $\lambda_2,v_2$ be eigenpairs of $A$ and $A^T$ respectively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$Av_1 = \lambda_1 v_1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$A^Tv_2 = \lambda_2 v_2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$v_2^TAv_1 = \lambda_2 v_2^T v_1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\lambda_1 v_2^Tv_1 = \lambda_2 v_2^T v_1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$v_2^T v_1 = 0$ and $\lambda_1 \neq \lambda_2$.&lt;/li&gt;&#xA;&lt;li&gt;$\lambda_1 = \lambda_2$&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="825" LastActivityDate="2011-03-04T03:58:09.060" CommentCount="1" />
  <row Id="24935" PostTypeId="1" AcceptedAnswerId="24937" CreationDate="2011-03-04T04:16:47.663" Score="3" ViewCount="75" Body="&lt;p&gt;The Nullstellensatz for $\mathbb{C}[x_1, \ldots, x_n]$ gives a dictionary between radical ideals and varieties, which makes the following assertion obvious: a radical ideal in $\mathbb{C}[x_1, \ldots, x_n]$ is the intersection of the maximal ideals containing it. Is there a simple purely algebraic way of seeing this? &lt;/p&gt;&#xA;" OwnerUserId="1491" LastActivityDate="2011-03-04T04:36:11.527" Title="Proving this algebra fact" Tags="&lt;commutative-algebra&gt;" AnswerCount="2" />
  <row Id="24936" PostTypeId="1" AcceptedAnswerId="24938" CreationDate="2011-03-04T04:20:09.897" Score="1" ViewCount="108" Body="&lt;p&gt;Smith Company produces desk lamps. The information for June indicated that the selling price was $\$25$ per unit, variable costs were  $\$15$ per unit, fixed costs totaled $\$6,000$, and the margin of safety in dollars  was $\$12,500$. Smith currently sells 1,100 lamps and earns $\$5,000$ of profit. How much is Smith’s margin of safety ratio?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A.) 45.5%&lt;/p&gt;&#xA;&#xA;&lt;p&gt;B.) 2.5%&lt;/p&gt;&#xA;&#xA;&lt;p&gt;C.) 2.2%&lt;/p&gt;&#xA;&#xA;&lt;p&gt;D.) 40.0%&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;em&gt;&lt;strong&gt;To compute Margin of Safety Ratio you take (Margin of Safety in&#xA;Dollars/Actual (expected) Sales)&lt;/em&gt;&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="7773" LastEditorUserId="7773" LastEditorDisplayName="" LastEditDate="2011-03-04T04:26:41.200" LastActivityDate="2011-03-04T04:31:06.063" Title="Accounting Question: Computing margin of Safety Ratio" Tags="&lt;homework&gt;&lt;analysis&gt;&lt;education&gt;&lt;finance&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="24937" PostTypeId="2" ParentId="24935" CreationDate="2011-03-04T04:25:18.450" Score="5" ViewCount="" Body="&lt;p&gt;Any radical ideal (in any ring) is an intersection of prime ideals. So the question becomes why a prime ideal is an intersection of maximal ideals. Such rings are called &lt;a href=&quot;http://en.wikipedia.org/wiki/Jacobson_ring&quot; rel=&quot;nofollow&quot;&gt;Jacobson rings&lt;/a&gt;. A field is clearly Jacobson. It is a general fact (a more general form of the Nullstellensatz) that a ring finitely generated over  a Jacobson ring is Jacobson. (This is in Eisenbud ch. 4.) So any affine coordinate ring is Jacobson.&lt;/p&gt;&#xA;" OwnerUserId="536" LastActivityDate="2011-03-04T04:25:18.450" CommentCount="1" />
  <row Id="24938" PostTypeId="2" ParentId="24936" CreationDate="2011-03-04T04:31:06.063" Score="0" ViewCount="" Body="&lt;p&gt;Hint:  If they expect to sell 1,100 at $25 each, what are the expected sales?  You already have the margin of safety, so it is just division from there.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-03-04T04:31:06.063" CommentCount="2" />
  <row Id="24939" PostTypeId="1" AcceptedAnswerId="24941" CreationDate="2011-03-04T04:33:42.160" Score="1" ViewCount="62" Body="&lt;p&gt;A univariate function $f$ is periodic with period $p_1,\ldots,p_k$ if &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f(z) = f(z + \sum_{i=1}^k n_i \cdot p_i)$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;for all complex $z$ and integers $n_i$. Elliptic function is an example of doubly periodic function. &lt;a href=&quot;http://mathworld.wolfram.com/TriplyPeriodicFunction.html&quot; rel=&quot;nofollow&quot;&gt;It is claimed&lt;/a&gt; that a triply periodic univariate function cannot exist, but why? Can't we follow the construction of elliptic functions that uses Schwarz-Christoffel formula to map upper-plane onto a triangle, and use the reflection principle to create a lattice which corresponds to three periods?&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Which step fails to hold when we mimic the construction of an elliptic function to create a triply period meromorphic univariate function?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sorry if the question is stupid. I do not have an access to the original proof that a triply periodic function cannot exist, either. So references are welcome!&lt;/p&gt;&#xA;" OwnerUserId="2767" LastEditorUserId="2767" LastEditorDisplayName="" LastEditDate="2011-03-04T04:39:57.973" LastActivityDate="2011-03-04T04:39:57.973" Title="Why does triply periodic univariate function not exist?" Tags="&lt;complex-analysis&gt;&lt;reference-request&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="24940" PostTypeId="2" ParentId="24935" CreationDate="2011-03-04T04:36:11.527" Score="2" ViewCount="" Body="&lt;p&gt;There is an elegant abstract approach to the Hilbert Nullstellensatz &#xA;discovered circa 1950 by Goldman, Krull and Zariski. For more &#xA;on this look up the terms: G-domain, Hilbert ring, Jacobson ring, e.g. see Kaplansky's textbook &lt;em&gt;Commutative Rings&lt;/em&gt;.&#xA;See also &lt;a href=&quot;http://cr.yp.to/zgk.html&quot; rel=&quot;nofollow&quot;&gt;D.J. Bernstein's ZGK page &lt;/a&gt; which has some online links, &#xA;including expositions of an &quot;elementary&quot; proof by R. Munshi (1999). &lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2011-03-04T04:36:11.527" />
  <row Id="24941" PostTypeId="2" ParentId="24939" CreationDate="2011-03-04T04:37:50.950" Score="6" ViewCount="" Body="&lt;p&gt;The problem is with the lattice of periods you want to consider in the construction...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One can easily show that if a subgroup $\Gamma$ of the additive group $\mathbb C$ contains three linearly independent elemtents over $\mathbb Z$, then it is not closed; so there is no sensible way to do all the reflections you are planning to do... To actually prove triply periodic meromorphic functions do not exist, using that fact about subgroups of $\mathbb C$ and properties of meromorphic functions should get you to where you want.&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2011-03-04T04:37:50.950" CommentCount="1" />
  <row Id="24942" PostTypeId="1" CreationDate="2011-03-04T04:41:13.467" Score="13" ViewCount="351" Body="&lt;p&gt;Take Boolean Algebra for instance, the underlying finite field/ring $0, 1, \{AND, OR\}$ is equivalent to $ 0, 1, \{NAND\} $ or $ 0, 1, \{ NOR \}$ where NAND and NOR are considered as universal gates. Is this property, that AND('multiplication') and OR ('addition') can be written in terms of a single universal binary relation (e.g. NAND or NOR) is true with every finite field(or finite rings)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT : I am interested in mathematical structures where boolean algebra holds (so that I can design a digital circuit.). Comments from JDK and jakiri points out that this is a valid question for finite rings at least and for finite fields in one case (i.e. $1, 0$ case). &lt;/p&gt;&#xA;" OwnerUserId="1674" LastEditorUserId="1674" LastEditorDisplayName="" LastEditDate="2011-03-04T10:38:39.487" LastActivityDate="2011-03-06T07:46:57.873" Title="Universal binary operation and finite fields (ring)" Tags="&lt;boolean-algebra&gt;&lt;finite-fields&gt;" CommentCount="11" FavoriteCount="4" />
  <row Id="24943" PostTypeId="2" ParentId="24284" CreationDate="2011-03-04T04:47:13.537" Score="0" ViewCount="" Body="&lt;p&gt;I think this is the extension (by definition) rule.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The extension rule is essentially introducing &lt;em&gt;new&lt;/em&gt; names (= variables) for terms/formulas. Formally, we take a &lt;em&gt;new&lt;/em&gt; variable, and add an axiom to our axioms that states the new variable satisfies the required property. In your case, $y$ is a new variable and it is value is $f(x)$, i.e. the axiom is $y=f(x)$ (it is more complicated that this since $y$ is an object variable not a function variable and therefore you have to treat $x$ as constant as long as you are using $y$). The situation is simpler for propositional logic since we don't have to deal with functional variables.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In short, you can think of the variables as free, but remember that now you have an extra axiom that states the relation between those free variables, and that you can &lt;em&gt;not&lt;/em&gt; quantify on $x$ as usual since now it is not an arbitrary object anymore but a specific object satisfying the definition axiom.&lt;/p&gt;&#xA;" OwnerUserId="468" LastEditorUserId="468" LastEditorDisplayName="" LastEditDate="2011-03-04T05:02:17.207" LastActivityDate="2011-03-04T05:02:17.207" />
  <row Id="24944" PostTypeId="2" ParentId="24794" CreationDate="2011-03-04T04:50:12.333" Score="5" ViewCount="" Body="&lt;p&gt;You haven't said what it means in this context for $X$ to be &quot;nonempty&quot;, but I presume it means that there is a $\overline{k}$-valued point of $X$.  (Certainly $X$ need not admit a $k$-valued point if $k$ is not algebraically closed, so considering $\overline{k}$-valued points seems to be the most sensible interpretation.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With this interpretation, the non-emptiness of $X$ follows directly from Noether normalization together with Exercise 5.2:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $A$ is the coordiante ring of $X$, assumed to be non-zero, then Noether normalization gives an integral injective map $k[y_1,\ldots,y_r] \hookrightarrow A,$ for some $0 \leq r \leq n$.  Exercise 5.2 then guarantees that any&#xA;map $k[y_1,\ldots,y_r] \to \overline{k}$ can be extended to a map&#xA;$A \to \overline{k}$; this is the desired surjectivity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[Added:] Having read the comments below, let me try again.  Rereading the second part of 5.16, I see that in fact $k$ is to be assumed algebraically closed, so that $k = \overline{k}$; this simplifies the above discussion, by removing the need to talk about $\overline{k}$-valued points.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we have $X \subset k^n$ cut out by some non-unit ideal $I \subset k[x_1,\ldots,x_n]$, an integral injection $k[y_1,\ldots,y_r] \hookrightarrow A := k[x_1,\ldots,x_n]/I,$ and our goal is to prove that the induced map&#xA;$X \to k^r$ is surjective.  The argument written above proves this: any point of $k^r$ corresponds to a map $k[y_1,\ldots,y_r] \to k.$  Exercise 5.2 extends this to a map $A \to k.$  This gives a point of $X$ mapping to the given point of $k^r$. QED.&lt;/p&gt;&#xA;" OwnerUserId="221" LastEditorUserId="221" LastEditorDisplayName="" LastEditDate="2011-03-04T13:08:04.257" LastActivityDate="2011-03-04T13:08:04.257" CommentCount="4" />
  <row Id="24945" PostTypeId="1" AcceptedAnswerId="25007" CreationDate="2011-03-04T04:54:19.553" Score="10" ViewCount="189" Body="&lt;p&gt;I was in a discussion today with a philosopher about the merit of the technique of &quot;proof by contradiction.&quot;  He mentioned the Law of Excluded Middle, wherein we (typically as mathematicians) assume that we either have P or Not P.  That is, if we show not &quot;Not P&quot;, then we must have P.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Further along in the conversation, he mentions Kolmogorov's tendencies toward Intuitionist Logic (where the Law of Excluded middle does not hold; i.e. we cannot infer p from not not p).  I located the source material of Kolmogorov, &quot;On the Principle of Excluded Middle,&quot; wherein he states:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;...it is illegitimate to use the principle of excluded middle in the domain of transfinite arguments.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Furthermore,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Only the finitary conclusions of mathematics can have significance in applications.  But the transfinite arguments are often used to provide a foundation for finitary conclusions.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Additionally,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;We shall prove that all the finitary conclusions obtained by means of a transfinite use of the principle of excluded middle are correct and can be proved even without its help.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My question:  What does Kolmogorov mean when he differentiates finitary conclusions from transfinite arguments?  Namely, what is an example of a finite conclusion, and what is an example of a corresponding transfinite argument?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Source material cited in Wikipedia: &lt;a href=&quot;http://en.wikipedia.org/wiki/Andrey_Kolmogorov#Bibliography&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Andrey_Kolmogorov#Bibliography&lt;/a&gt;)&lt;/p&gt;&#xA;" OwnerUserId="7051" LastActivityDate="2011-03-04T13:16:07.320" Title="What is an example of a Transfinite Argument?" Tags="&lt;logic&gt;" AnswerCount="2" />
  <row Id="24946" PostTypeId="1" CreationDate="2011-03-04T05:00:05.680" Score="1" ViewCount="50" Body="&lt;p&gt;How can I reversibly combing two very large numbers, such as 9,828,485,546,536,174,656,640,115,183 and 43,044,700,185? I need them to be combined into a single number, and the order of the numbers must be remained intact, so being required to flip them would be unfavorable, but I could just add a 0 or 1 at the front or end to specify whether or not the numbers had been flipped if I had too.&lt;/p&gt;&#xA;" OwnerUserId="7775" LastActivityDate="2011-04-03T05:39:12.300" Title="Reversibly Combining Very Large" Tags="&lt;algorithms&gt;" AnswerCount="1" />
  <row Id="24947" PostTypeId="2" ParentId="24927" CreationDate="2011-03-04T05:01:27.983" Score="1" ViewCount="" Body="&lt;p&gt;One geometrical fact from which the non-commutativity of the fundamental group is the following: two objects on a line can't switch relative position (i.e. left and right) through homotopy, as they are unable to &quot;pass&quot; each other.  Two objects in a higher dimensional space can, however; so intuitively, it seems that a homotopy theory based on mappings of $I^n$ will naturally be abelian for $n \ge 2$.&lt;/p&gt;&#xA;" OwnerUserId="1769" LastActivityDate="2011-03-04T05:01:27.983" CommentCount="1" />
  <row Id="24948" PostTypeId="2" ParentId="24946" CreationDate="2011-03-04T05:05:43.627" Score="1" ViewCount="" Body="&lt;p&gt;See this &lt;a href=&quot;http://math.stackexchange.com/questions/23503/create-unique-number-from-2-numbers/23526#23526&quot;&gt;prior question&lt;/a&gt; on &lt;a href=&quot;http://en.wikipedia.org/wiki/Pairing_function&quot; rel=&quot;nofollow&quot;&gt;pairing functions&lt;/a&gt;, e.g. the classical Cantor pairing or &lt;a href=&quot;http://szudzik.com/ElegantPairing.pdf&quot; rel=&quot;nofollow&quot;&gt;another &quot;elegant&quot; one&lt;/a&gt; useful for depth-first ordering.&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2011-03-04T05:05:43.627" />
  <row Id="24949" PostTypeId="1" AcceptedAnswerId="24951" CreationDate="2011-03-04T05:10:03.637" Score="3" ViewCount="62" Body="&lt;p&gt;I just read a paper in which an ODE was solved using a step I don't understand.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$${dI \over \kappa (x) dx} = I$$ &#xA;Let&#xA;$$\tau = \int_0^x \kappa(x)dx$$&lt;br&gt;&#xA;Rewrite equation as&#xA;$${dI \over d\tau} = I$$ &#xA;and solve.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How does the final expression follow from the definition of $\tau$?  &lt;/p&gt;&#xA;" OwnerUserId="7776" LastActivityDate="2011-03-04T05:14:42.963" Title="Mysterious step while solving ODE" Tags="&lt;calculus&gt;&lt;differential-equations&gt;" AnswerCount="2" />
  <row Id="24950" PostTypeId="2" ParentId="24949" CreationDate="2011-03-04T05:14:36.313" Score="2" ViewCount="" Body="&lt;p&gt;Chain rule:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{dI}{dx}=\frac{dI}{d\tau}\frac{d\tau}{dx}=\frac{dI}{d\tau}\kappa(x).$$&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2011-03-04T05:14:36.313" CommunityOwnedDate="2011-03-04T05:14:36.313" />
  <row Id="24951" PostTypeId="2" ParentId="24949" CreationDate="2011-03-04T05:14:42.963" Score="3" ViewCount="" Body="&lt;p&gt;If $\tau(x) = \displaystyle \int_{0}^x \kappa(y) dy$, then $d \tau = \kappa(x) dx$&lt;/p&gt;&#xA;" OwnerUserId="3302" LastActivityDate="2011-03-04T05:14:42.963" />
  <row Id="24952" PostTypeId="2" ParentId="24917" CreationDate="2011-03-04T05:16:19.290" Score="1" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Here are some hints:&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) If $e_t(x)= e^{itx}$,what is $\widehat{e_tf}$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) What is $\widehat{f&amp;#39;}$?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Also note that it is important to state the exact definition of the Fourier transform if you wish to compute explicit functions.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;" OwnerUserId="1154" LastActivityDate="2011-03-04T05:16:19.290" />
  <row Id="24953" PostTypeId="2" ParentId="24945" CreationDate="2011-03-04T05:20:22.227" Score="3" ViewCount="" Body="&lt;p&gt;I think you will find all your answers (and much more) in the very readable exposition by Thierry Coquand: &lt;em&gt;Kolmogorov's contribution to intuitionistic logic&lt;/em&gt;, in the book &lt;a href=&quot;http://experiment.iitalia.com/librarysplit2/Charpentier%20E.,%20Lesne%20A.,%20Nikolski%20N.%20%28eds%29%20Kolmogorov%27s%20heritage%20in%20mathematics%20%28Springer,%202007%29%28ISBN%203540363491%29%28326s%29.pdf&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;Kolmogorov's Heritage in Mathematics&lt;/em&gt;.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2011-03-04T05:20:22.227" CommentCount="3" />
  <row Id="24954" PostTypeId="2" ParentId="24547" CreationDate="2011-03-04T05:22:40.043" Score="1" ViewCount="" Body="&lt;p&gt;$x$= vec_c(X)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$c$= vec_c( $e e^T$ )&lt;/p&gt;&#xA;&#xA;&lt;p&gt;vec_c(A) here is a column-major vectorization.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Set of constraints can be expressed in the form Ax=b, where A is matrix $G_1$ in the question update, and vector b= (1, 0, ..., 0 ). First row takes care of trace constraint, and all the next take care of edge constraints.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;positive semi-definite constraint can be set via $G_1 x + s_1= h_1$, in particular $G_1$ is minus identity matrix of size $n^2 \times n^2$, and $h_1=0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is computationally more efficient to implement it in a functional form. There is corresponding interface in python, I do not know about other implementations. You also need to take care of matrix representation, CVXOPT assumes it is symmetric, and access only lower triangular part of it, see manual you linked for details.&lt;/p&gt;&#xA;" OwnerUserId="7774" LastActivityDate="2011-03-04T05:22:40.043" CommentCount="2" />
  <row Id="24956" PostTypeId="2" ParentId="24833" CreationDate="2011-03-04T05:24:59.777" Score="13" ViewCount="" Body="&lt;p&gt;Fermat's Last Theorem, although elementary to state, is a very subtle problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The general &lt;a href=&quot;http://en.wikipedia.org/wiki/Abc_conjecture&quot; rel=&quot;nofollow&quot;&gt;$ABC$ conjecture&lt;/a&gt; (still unproved) states (roughly) that if $C = A + B$ (with $A, B, C$ coprime integers), then it is not possible for $A$, $B$, and $C$ to be simultaneously divisible by high powers of integers. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fermat's equation considers the special case when $A$, $B$, and $C$ are all taken to be perfect $n$th powers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The $ABC$ conjecture in general, and Fermat in particular, are then subtle problems relating the additive and multiplicative nature of the integers, and so&#xA;there it is perhaps not too surprising that they are difficult to prove (or, in &#xA;the case of $ABC$, that it remains unproved!).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another famous conjecture relating the additive and multiplicative nature of the&#xA;integers is the Goldbach conjecture.  This is, if you like, the &quot;opposite&quot; of $ABC$; it states that for an even integer $N$, we may write $N = p_1 + p_2,$ where $p_1$ and $p_2$ are prime (which is a kind of &quot;opposite&quot; to being divisible by a high perfect power).  It also resists proof.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At a technical level, the tools that have been brought to bear on Fermat, and on $ABC$, are quite different from the tools that have been brought to bear on Goldbach, so perhaps one shouldn't take a comparison between them too seriously.&#xA;But they do share the common element of getting at something quite deep about the interrelationship between the additive and mutliplicative structure of the integers, and this is what makes them difficult (or so it seems to me). &lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2011-03-04T05:24:59.777" />
  <row Id="24957" PostTypeId="2" ParentId="24778" CreationDate="2011-03-04T06:06:00.817" Score="1" ViewCount="" Body="&lt;p&gt;You can write the bilevel program as follows: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\min_{x \in X, y \in Y} f(x)$&lt;br&gt;&#xA;s.t. $g(x,y) \leq 0$&lt;br&gt;&#xA;$y = z$&lt;br&gt;&#xA;$\nabla_{z} F(x,z) + \mu^\top\nabla_{z}(Ax+Bz-b) = 0$&lt;br&gt;&#xA;$Ax + Bz -b \leq 0$&lt;br&gt;&#xA;$0 \leq \mu \perp (Ax + Bz -b) \geq 0$&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The last line is a complementarity constraint that can also be written:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\mu_{i} \geq 0$&lt;br&gt;&#xA;$\mu_{i} [Ax + Bz - b]_{i} = 0$&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is an MPEC. I'm not sure if this formulation will give you acceptable solutions, but you can try it. There are rules for posing well-posed MPECs: see this paper for details: Baumrucker, B.T., J. G. Renfro and L.T. Biegler “MPEC Problem Formulations in Chemical Engineering&#xA;Applications,” Computers and Chemical Engineering, 32, pp. 2903-2913 (2008)&lt;/p&gt;&#xA;" OwnerUserId="1124" LastEditorUserId="1124" LastEditorDisplayName="" LastEditDate="2011-03-04T06:35:06.160" LastActivityDate="2011-03-04T06:35:06.160" CommentCount="2" />
  <row Id="24958" PostTypeId="2" ParentId="19485" CreationDate="2011-03-04T06:17:27.390" Score="1" ViewCount="" Body="&lt;p&gt;At the risk of seeming repetitive (see &quot;How to prove the mathematical induction is true?&quot;), the principle of mathematical induction is just one of Peano's axioms for the natural numbers (Axiom 5, below):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Axiom 1: 1 is a natural number. That is, our set is not empty; it contains an object called 1 (read &quot;one&quot;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Axiom 2: For each x there exists exactly one natural number, called the successor of x, which will be denoted by x'.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Axiom 3: We always have ~x'=1. That is, there exists no number whose successor is 1. That is, there exists no number whose successor is 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Axiom 4: If x'=y' then x=y. That is, for any given number there exists either no number or exactly one number whose successor is the given number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Axiom 5 (Axiom of Induction): Let there be given a set M of natural numbers, with the following properties: I. 1 belongs to M. II. If x belongs to M then so does x'. Then M contains all the natural numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Source: &lt;a href=&quot;http://www.ms.uky.edu/~lee/ma502/notes2/node7.html&quot; rel=&quot;nofollow&quot;&gt;http://www.ms.uky.edu/~lee/ma502/notes2/node7.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have trouble visualizing these relationships, I find it helps to think of the set of natural numbers as directed graph (a network of nodes connected by one-way arrows) as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) --&gt; (2) --&gt; (3) --&gt; ...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In words: There is a node (call it 1) that has no arrows going into it (Axioms 1 and 3, above). Every node has exactly one arrow leaving it (Axiom 2). Every node has at most one arrow going into it (Axiom 4). You can get to any node by starting at 1 and following the arrows; that is, there are no isolated nodes or portions of the graph off to the side (Axiom 5). &lt;/p&gt;&#xA;" OwnerUserId="3515" LastEditorUserId="3515" LastEditorDisplayName="" LastEditDate="2011-03-04T15:06:16.717" LastActivityDate="2011-03-04T15:06:16.717" />
  <row Id="24960" PostTypeId="2" ParentId="15479" CreationDate="2011-03-04T06:27:12.207" Score="4" ViewCount="" Body="&lt;p&gt;Filters and ultrafilters play a basic role in topology: they allow for a general theory of convergence in topological spaces when sequences do not suffice.  Moreover, they do so in an arguably somewhat more graceful and canonical way than the rival theory of nets: see &lt;a href=&quot;http://math.uga.edu/~pete/convergence.pdf&quot; rel=&quot;nofollow&quot;&gt;my notes on convergence&lt;/a&gt; where I develop nets, then filters, then compare the two.  But either way the theory of convergence in topological spaces is a sort of &quot;finished product&quot; -- it is there for you to use, and it is certainly useful, but the theory itself and how to apply it seem to be by now rather thoroughly understood: for instance, the theory of convergence stands at almost exactly the same point today as it did after the publication of Bourbaki's &lt;em&gt;Topologie Générale&lt;/em&gt; by 1950.  (Or so is my impression.  Please let me know if I am simply speaking from ignorance.)   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, the use of (especially ultra-)filters in mathematics generally is definitely on the rise, especially in recent years.  The fundamental construction here is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Ultraproduct&quot; rel=&quot;nofollow&quot;&gt;ultraproduct&lt;/a&gt; of structures.  Its importance in model theory is guaranteed by the &lt;a href=&quot;http://en.wikipedia.org/wiki/Los%27_theorem&quot; rel=&quot;nofollow&quot;&gt;theorem of Jerzy Łoś&lt;/a&gt;.  Very roughly, the ultraproduct of a family of structures inherits all the elementary properties shared by &quot;most&quot; of the structures in the family while at the same time being very conveniently &quot;large&quot; (in more model-theoretic terminology, ultraproducts enjoy good &lt;em&gt;saturation properties&lt;/em&gt;).  This leads swiftly to the idea of &lt;em&gt;nonstandard models&lt;/em&gt;, which were initially most popular in analytic contexts.  But nonstandard methods seem not to have really caught on with the majority of the analysis community.  It is not so in algebra, algebraic geometry and number theory, where &quot;nonstandard methods&quot; are becoming increasingly, um, standard.  My guess is that 10-20 years from now, ultrafilters will be a very familiar topic to graduate students, because of their model-theoretic applications.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To get just a little taste of the use of ultrafilters in model theory, you could read &lt;a href=&quot;http://www.math.uga.edu/~pete/modeltheory2010Chapter6.pdf&quot; rel=&quot;nofollow&quot;&gt;the final chapter&lt;/a&gt; of the lecture notes for the short summer course on model theory I taught in 2010.  (If you don't know any model theory already, you'll need to look at the earlier chapters as well to really understand it, but perhaps you can gain an impression of what's going on just by skimming this chapter.)  But that's barely the tip of the iceberg: there are whole books written on this subject, e.g. by (Bell and Slomson) and (Chang and Keisler).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, the study of saturation properties of ultraproducts leads quickly to some nontrivial set-theoretic considerations.  The fact that model theorists know and care much more about set theory than many other working mathematicians can probably be traced to this, at least in part.&lt;/p&gt;&#xA;" OwnerUserId="299" LastActivityDate="2011-03-04T06:27:12.207" />
  <row Id="24961" PostTypeId="1" AcceptedAnswerId="24962" CreationDate="2011-03-04T06:49:11.790" Score="3" ViewCount="103" Body="&lt;p&gt;Why does $\mu(k,n)=\mu(1,\frac{n}{k})$, where $\mu$ is the Mobius function and $k$ and $n$ are integers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm reading about the Mobius function on a poset of integers ordered by divisibility. The author uses this fact several times, but I don't see why it's so obvious. What exactly is this equality saying? Thanks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: The author computes $\mu(1,n)$ as $\mu(1,n)=1$ if $n=1$, $\mu(1,n)=(-1)^k$ if $n$ is the product of $k$ distinct primes, and $\mu(1,n)=0$ otherwise.&lt;/p&gt;&#xA;" OwnerUserId="6742" LastEditorUserId="6742" LastEditorDisplayName="" LastEditDate="2011-03-04T06:56:40.397" LastActivityDate="2011-03-04T19:39:07.477" Title="Why does $\mu(k,n)=\mu(1,\frac{n}{k})$?" Tags="&lt;number-theory&gt;&lt;combinatorics&gt;" AnswerCount="2" CommentCount="6" FavoriteCount="1" />
  <row Id="24962" PostTypeId="2" ParentId="24961" CreationDate="2011-03-04T07:07:55.243" Score="3" ViewCount="" Body="&lt;p&gt;&lt;em&gt;Note. This assumed that both $\mu$s were the Moebius function on a poset. Now the OP notes that $\mu(1,n)$ is being defined by $\mu(n)$, the usual Moebius function on the natural numbers. I have vague memories of having dealt with the poset Moebius function before, but I don't think I ever saw that for the natural numbers ordered by divisibility, the two agree; see below.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Moebius function on a poset is defined by:&#xA;$$\mu(x,y) = \left\{\begin{array}{ll}&#xA;1 &amp;amp; \mbox{if $x=y$,}\\&#xA;-\sum\limits_{x\leq z\lt y} \mu(x,z) &amp;amp; \mbox{if $x\lt y$,}\\&#xA;0 &amp;amp; \mbox{otherwise.}&#xA;\end{array}\right.$$&#xA;Here you are ordering the natural numbers by divisibility (you say integers, but they don't form a poset since divisibility is not antisymmetric, so either you are taking equivalence classes, or are workng on the naturals).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $k$ does not divide $n$, then both sides are $0$ and there is nothing to do. (I'm assuming that for non-integer rationals $q$, $\mu(1,q)$ is defined to be $0$ by fiat).  So assume that $k|n$. Then writing $n=kd$, the assertion is equivalent to showing that for all natural numbers $k$ and $d$,&#xA;$$\mu(k,kd) = \mu(1,d).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fix $k$. We prove this by strong induction on $d$. It holds if $d=1$, since $\mu(k,k) = 1 =\mu(1,1)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume the result holds for all $d$, $1\leq d \lt n$, and you want to show that it is true for $d=n$. We have:&#xA;\begin{align*}&#xA;\mu(k,kn) &amp;amp;= -\sum_{j=1}^{n-1}\mu(k,kj)\\&#xA;&amp;amp;= -\sum_{j=1}^{n-1}\mu(1,j) &amp;amp;\qquad&amp;amp;\mbox{(by the induction hypothesis)}\\&#xA;&amp;amp;= \mu(1,n).&#xA;\end{align*}&#xA;Thus, $\mu(k,kn) = \mu(1,n)$ for all positive integers $n$ (on the positive integers ordered by divisibility).&lt;/p&gt;&#xA;&#xA;&lt;hr/&gt;&#xA;&#xA;&lt;p&gt;Does the above definition agree with the usual Moebius function for integers, described in the original post? Yes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, consider the case in which $d$ is squarefree. $\mu(1,1)=1$ by the definition given. Assume that if $d$ is a product of fewer than $k$ distinct primes, then $\mu(1,d) = (-1)^r$, where $r$ is the number of distinct primes that divide $d$. If $d$ is the product of exactly $k$ distinct primes, then we have&#xA;\begin{align*}&#xA;\mu(1,d) &amp;amp;= -\sum_{m|d, m\neq d} \mu(1,m)\\&#xA;&amp;amp;= -\left(1 - \binom{k}{1} + \binom{k}{2} -\binom{k}{3}+\cdots+(-1)^{k-1}\binom{k}{k-1}\right)\\&#xA;&amp;amp;= -\left( (1-1)^k - \binom{k}{k}(-1)^k\right)&amp;amp;&amp;amp;\mbox{(binomial expansion)}\\&#xA;&amp;amp;= -(-1)^{k+1} = (-1)^{k+2} = (-1)^k.&#xA;\end{align*}&#xA;The second equality comes from the following consideration: the value of $\mu(1,m)$ on the proper divisors of $d$ only depends on the number of primes that divide them; there are $\binom{k}{1}$ divisors that are prime, $\binom{k}{2}$ that are products of two primes, etc. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So $\mu(1,d) = \mu(d)$ for the squarefree integers $d$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now assume that $n$ is not square free, not equal to $1$. If $d=p^2$ is the square of a prime, then &#xA;$$\mu(1,d) = -\mu(1,1) -\mu(1,p) = -1+1 = 0.$$&#xA;Now assume that for any $m\lt d$, $\mu(1,m)=\mu(m)$ (the usual arithmetic Moebius function), and that $d$ is divisible by the square of a prime $p$. If $m$ is a proper divisors of $\frac{d}{p^2}$, then $\mu(1,p^2m)=0$ by induction. So we only need to worry about the divisors of $\frac{d}{p}$.&#xA;And we have:&#xA;$$\mu(1,d) = -\sum_{m|\frac{d}{p}}\mu(1,m) = \left(-\sum_{m|\frac{d}{p},m\neq\frac{d}{p}} \mu(1,m)\right) - \mu\left(1,\frac{d}{p}\right) = \mu\left(1,\frac{d}{p}\right) - \mu\left(1,\frac{d}{p}\right) = 0.$$&#xA;So the two definitions agree on the natural numbers ordered by divisibility.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-04T19:39:07.477" LastActivityDate="2011-03-04T19:39:07.477" CommentCount="2" />
  <row Id="24963" PostTypeId="1" AcceptedAnswerId="24966" CreationDate="2011-03-04T07:13:21.347" Score="3" ViewCount="172" Body="&lt;p&gt;What are some ways to solve this recurrence relation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a(n+1)=2 a(n) - a(n-1) -1, \text{ with }a(0)=0, a(10)=0?$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried to first convert this inhomogeneous equation into a homogeneous one following &lt;a href=&quot;http://en.wikipedia.org/wiki/Recurrence_relation#General_methods&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$b_{n}=Ab_{n-1}+Bb_{n-2}+K \,$ can be&#xA;  converted into homogeneous form as&#xA;  follows: The steady state is found by&#xA;  setting $b_n = b_{n−1} = b_{n−2} = b^{*}$ to&#xA;  obtain $b^{*} = \frac{K}{1-A-B}. \, $.&#xA;  Then the non-homogeneous recurrence&#xA;  can be rewritten in homogeneous form&#xA;  as $[b_n -b^{*}]=A[b_{n-1}-b^{*}]+B[b_{n-2}-b^{*}], \, $&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But in the recurrence relation given at the beginning of this post, the denominator in $b^{*} = \frac{K}{1-A-B} \, $ is $1-2+1=0$. So how can I solve the recurrence relation? Thanks in advance! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS: Are there some books or websites with some summary of various ways to solve recurrence relation?&lt;/p&gt;&#xA;" OwnerUserId="4674" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2011-06-03T13:47:46.383" LastActivityDate="2011-06-03T13:47:46.383" Title="How to solve this recurrence relation" Tags="&lt;reference-request&gt;&lt;recurrence-relations&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="0" />
  <row Id="24964" PostTypeId="1" CreationDate="2011-03-04T07:23:29.590" Score="18" ViewCount="259" Body="&lt;p&gt;I was talking with friends about silly questions involving what numbers you can get using only a single digit &quot;3&quot; and unary operations. We eventually conjectured that using only factorials and square roots you can get arbitrarily close to any number greater than or equal to $1$. But we are having trouble proving or disproving the conjecture.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Precisely, start with the number $3$. Then take its factorial $m$ times, and then take the square root of its result $n$ times. Ie, $(3!!\ldots !)^{\frac{1}{2^n}}$ where there are $m$ factorials. Let the set of numbers achievable in this way be $X$. Is $X$ dense in $[1,\infty)$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only progress we have made is to show that any interval $[x,x^2]$ for $x&amp;gt;1$ there is a limit point $a \in [x,x^2)$ of $X$. This is true because for any $z &amp;gt; x^2$ we can square root it an appropriate number of times to get it in $[x,x^2)$. We can do this for infinitely many points of the sequence $3,3!,3!!,\ldots$. And all points we get through this process are distinct. Let $F(m)$ be $3$ with $m$ factorials. If $F(m)^{\frac{1}{2^n}} = F(a)^{\frac{1}{2^b}}$ then we can raise each side to a power and get an expression of the form $F(m) = F(a)^{\frac{1}{2^c}}$. But factorials are not squares. (To see this for $q!$, note that there is a prime in $[q/2,q]$ by bertrand's postulate. This prime appears only once in the factorization of $q!$). So all numbers we get are distinct. We have infinitely many distinct points in $X \cap [x,x^2)$ and therefore there is a limit point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other than that, we can't figure anything out. It feels like $X$ should be dense. Consider some interval $[x,x^2]$. Take lots of factorials of $3$. Then take square roots of that until it falls in $[x,x^2]$. It feels like the points we get will be somewhat uniformly distributed around $[x,x^2]$ and therefore dense.&lt;/p&gt;&#xA;" OwnerUserId="30" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-04T07:29:23.503" LastActivityDate="2011-03-04T14:55:48.160" Title="Repeated Factorials and Repeated Square Rooting" Tags="&lt;sequences-and-series&gt;&lt;limit&gt;&lt;factorial&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="6" />
  <row Id="24965" PostTypeId="1" CreationDate="2011-03-04T07:29:54.593" Score="1" ViewCount="208" Body="&lt;p&gt;How do I find an orthonormal basis for a plane defined as $x_1+x_2+x_3=0$? I don't know where to start -- not even given dimensions. Thank you.&lt;/p&gt;&#xA;" OwnerUserId="7781" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-04T07:45:53.447" LastActivityDate="2011-03-04T07:45:53.447" Title="Find an orthonormal basis of a plane" Tags="&lt;linear-algebra&gt;" AnswerCount="1" CommentCount="6" />
  <row Id="24966" PostTypeId="2" ParentId="24963" CreationDate="2011-03-04T07:34:21.427" Score="4" ViewCount="" Body="&lt;p&gt;Mary: You may consider your equation for $n+2$ instead of $n+1$; this gives us $$ a(n+2)=2a(n+1)-a(n)-1.$$ Subtract the first equation from this one. We obtain $$\begin{array}{rl}a(n+2)-a(n+1)&amp;amp;=(2a(n+1)-a(n)-1)-(2a(n)-a(n-1)-1)\\&amp;amp;=2a(n+1)-3a(n)+a(n-1),\end{array}$$ or $$a(n+2)=3a(n+1)-3a(n)+a(n-1).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, if your equation is &quot;almost homogeneous&quot; but for a constant, the same trick of subtracting consecutive instances of the recurrence gives us a homogeneous equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The associated characteristic equation is now $x^3-3x^2+3x-1=0$, or $(x-1)^3=0$. This means that $a(n)=A+Bn+Cn^2$ for some constants $A,B,C$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, if the equation has the form $(x-r)^k(x-b)^s\dots=0$, then the general solution will have the form $$a(n)=(A+Bn+\dots+C n^{k-1})r^n+(D+En+\dots+F n^{s-1})b^n+\dots$$ (in the most studied case, the characteristic equation has no repeated roots, but as in your example here, repeated roots may occur, so we need this more general version).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To find $A,B,C$ in our equation for $a(n)=A+Bn+Cn^2$, we use the given initial conditions. Are you sure the condition $a(10)=0$ is correct, rather than $a(1)=0$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With $a(1)=0$, we have $0=A+B+C$. Also, $a(0)=0$, so $A=0$. Finally, $a(2)=2a(1)-a(0)-1=-1$, using the original recurrence, so $A+2B+4C=-1$. This easily gives us $A=0$, $B=1/2$, and $C=-1/2$, or $$ a(n)=\frac{n-n^2}2. $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that we needed to compute $a(2)$ before we could find $A,B,C$. This is because the original equation was not homogeneous, so even though it is of second order, we needed an additional initial condition to the two given to us. (Note the homogeneous equation we obtained is of third order, needing 3 initial conditions.)&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, I do not know of a decent reference on recurrence relations at an elementary level (I know of one, in Spanish, a translation of a small Russian booklet published by Mir eds. years ago). There are several standard approaches which might not be as elementary as you may want, but you may enjoy looking at them anyway: Using linear algebra or generating functions. For the latter, a good reference is &quot;generatingfunctionology&quot;, &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://www.math.upenn.edu/~wilf/DownldGF.html&quot; rel=&quot;nofollow&quot;&gt;http://www.math.upenn.edu/~wilf/DownldGF.html&lt;/a&gt; &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and for the former a good reference is Volume II of Apostol's Calculus book.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Hmm... With $a(10)=0$, we can proceed as follows: We have $$a(n)=A+Bn+Cn^2$$ and so $$A=0$$ (since $a(0)=0$) and $$10B+100C=0$$ (since $a(10)=0$ and $A=0$). We also have $$a(n+2)=2a(n+1)-a(n)-1,$$ or&#xA;$$ B(n+2)+C(n+2)^2=2B(n+1)+2C(n+1)^2-Bn-Cn^2-1, $$ which simplifies to $2C=-1$, or $$C=-1/2.$$ Then $$B=5,$$ from the equation for $a(10)=0$, and $$ a(n)=5n-\frac{n^2}2. $$&lt;/p&gt;&#xA;" OwnerUserId="462" LastEditorUserId="462" LastEditorDisplayName="" LastEditDate="2011-03-04T08:03:01.293" LastActivityDate="2011-03-04T08:03:01.293" CommentCount="4" />
  <row Id="24967" PostTypeId="2" ParentId="24965" CreationDate="2011-03-04T07:43:53.713" Score="3" ViewCount="" Body="&lt;p&gt;The subspace in $\mathbb{R}^n$ defined by $x_1+x_2+x_3=0$ is of dimension $n-1$ and has a basis consisting of the vectors&#xA;\begin{align*}&#xA;&amp;amp;(1,0,-1,0,0,\ldots,0)\\&#xA;&amp;amp;(0,1,-1,0,0,\ldots,0)\\&#xA;&amp;amp;(0,0,0,1,0,\ldots,0)\\&#xA;&amp;amp;(0,0,0,0,1,\ldots,0)\\&#xA;&amp;amp;\qquad\qquad\vdots\\&#xA;&amp;amp;(0,0,0,0,0,\ldots,1)&#xA;\end{align*}&#xA;The last $n-3$ vectors are already normal, pairwise orthogonal, and orthogonal to the first two. So all you need to do is orthogonalize $(1,0,-1,0,\ldots,0)$ and $(0,1,-1,0,\ldots,0)$ by using Gram-Schmidt, and then normalize. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The dimension of the ambient space does not matter in the end, except in so far as padding the basis.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2011-03-04T07:43:53.713" />
  <row Id="24968" PostTypeId="1" AcceptedAnswerId="24969" CreationDate="2011-03-04T07:48:32.370" Score="1" ViewCount="80" Body="&lt;p&gt;Suppose phi is a bundle map from E to F, where E and F are two vector bundles over manifold M. Now for each x in M, phi restricted to Ex:Ex-&gt;Fx is a isomorphism, how to show that phi indeed a diffeomorphism?&lt;/p&gt;&#xA;" OwnerUserId="9016" LastEditorUserId="6296" LastEditorDisplayName="" LastEditDate="2011-03-04T07:51:58.710" LastActivityDate="2011-03-04T08:33:53.533" Title="If bundle map is a local isomorphism, then bundle map is diffeomorphic?" Tags="&lt;homework&gt;&lt;algebraic-topology&gt;&lt;vector-bundles&gt;" AnswerCount="2" />
  <row Id="24969" PostTypeId="2" ParentId="24968" CreationDate="2011-03-04T07:52:07.003" Score="1" ViewCount="" Body="&lt;p&gt;Using the fact that $\phi$ is bijective on the fibers, show that it is in fact bijective. To show it is a diffeo then, it is enough to do it locally on $M$. You can therefore assume that both $E$ and $F$ are trivial. Can you see how to proceed?&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2011-03-04T07:52:07.003" />
  <row Id="24970" PostTypeId="2" ParentId="24963" CreationDate="2011-03-04T07:55:00.657" Score="2" ViewCount="" Body="&lt;p&gt;You get all solutions of the inhomogeneous equation by finding any specific solution of the inhomogeneous equation and adding it to all solutions of the homogeneous equation. The steady-state ansatz is one way of finding a solution of the inhomogeneous equation, but if it doesn't work, as in this case, you can still pursue the general approach of first finding a specific solution to the inhomogeneous equation. Then the Wikipedia section you cited shows how to find all the homogeneous solutions that you need to add to it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the present case, there's also another solution: You can rewrite your recurrence relation as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$a(n+1)-2a(n)+a(n-1)=-1\;.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this form, you can see that the left-hand side is a double difference: Defining&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(\Delta b) (n) := b (n+1) - b(n)\;,$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;we can write the equation as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(\Delta\Delta a)(n-1) = -1\;.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This we can solve by twice using the fact that &quot;the inverse of $\Delta$ is $\Sigma$&quot;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(\Delta a)(n-1) = -n + c\;,$$&#xA;$$a(n-1)= -\frac{n(n+1)}{2}+cn+d$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with &quot;constants of summation&quot; $c$ and $d$ that you can determine to satisfy the initial conditions. Determining the constants will be simpler if you rewrite this as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$a(n)=-n^2/2+c&amp;#39;n+d&amp;#39;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with new constants $c&amp;#39;$, $d&amp;#39;$.&lt;/p&gt;&#xA;" OwnerUserId="6622" LastEditorUserId="6622" LastEditorDisplayName="" LastEditDate="2011-03-04T08:08:40.677" LastActivityDate="2011-03-04T08:08:40.677" CommentCount="1" />
  <row Id="24972" PostTypeId="1" CreationDate="2011-03-04T08:26:41.727" Score="2" ViewCount="76" Body="&lt;p&gt;how can i translate this sentence to a quantifier formula, when the universe - { o|o is a set }&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any master has as elements all and only sets which are not elements of themselves. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;i know it starts of with an existential quantifier since its dealing with a specific object &quot;master&quot; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;any feedback would be great. &lt;/p&gt;&#xA;" OwnerUserId="7779" LastActivityDate="2011-04-03T10:39:13.747" Title="quantifier translation" Tags="&lt;logic&gt;" AnswerCount="1" />
  <row Id="24973" PostTypeId="2" ParentId="24968" CreationDate="2011-03-04T08:33:53.533" Score="0" ViewCount="" Body="&lt;p&gt;So if $\phi$ is bijective on the fibers, isn't it already mean that $\phi$ is bijective on M? Now I have some trouble to show that $\phi$ is a local diffeomorphism(which is to show that it is a local homeomorphism). Any hint? And I don't quite understand what you mean by &quot;you can assume that both E and F are trivial&quot;&lt;/p&gt;&#xA;" OwnerUserId="7787" LastActivityDate="2011-03-04T08:33:53.533" CommentCount="3" />
  <row Id="24974" PostTypeId="1" AcceptedAnswerId="24977" CreationDate="2011-03-04T08:35:52.580" Score="1" ViewCount="54" Body="&lt;p&gt;Let V , W be finite dimensional vector spaces over R. Let A : V-&gt;W be a linear map.&#xA;Choose bases of V and W and the corresponding bases of $\Lambda^k$(V ) and of $\Lambda^k$(W). How to show that the entries of the matrix representing $\Lambda^k$(A) are polynomial in the entries of the matrix representing A.&lt;/p&gt;&#xA;" OwnerUserId="9016" LastEditorUserId="1242" LastEditorDisplayName="" LastEditDate="2011-03-04T08:43:03.943" LastActivityDate="2011-03-04T09:05:52.473" Title="Matrix representing $\Lambda^k$(A)" Tags="&lt;homework&gt;&lt;tensors&gt;" AnswerCount="1" />
  <row Id="24975" PostTypeId="1" AcceptedAnswerId="24979" CreationDate="2011-03-04T08:41:50.793" Score="3" ViewCount="37" Body="&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/aoRn2.png&quot; alt=&quot;1/2 of PI&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was wondering - how would I specify the interval (the amount that &lt;code&gt;n&lt;/code&gt; increases each time) between terms? Is that possible? What if I want it to increase by, say, 2, each time instead of one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In Python it would be &lt;code&gt;n += 2&lt;/code&gt; -&gt; I'm looking for an equivalent...&lt;/p&gt;&#xA;" OwnerUserId="5970" LastEditorUserId="30" LastEditorDisplayName="" LastEditDate="2011-03-04T09:08:21.460" LastActivityDate="2011-03-04T09:13:58.740" Title="interval for a product to infinity" Tags="&lt;calculus&gt;&lt;sequences-and-series&gt;&lt;notation&gt;&lt;infinity&gt;&lt;products&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="24976" PostTypeId="1" CreationDate="2011-03-04T08:45:18.070" Score="3" ViewCount="184" Body="&lt;p&gt;For a fixed positive integer n, if   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$D = \left|\begin{array}{ccc} &#xD;&#xA;n! &amp;amp; (n + 1)! &amp;amp; (n + 2)! \\&#xD;&#xA;(n + 1)! &amp;amp; (n + 2)! &amp;amp; (n + 3)! \\ &#xD;&#xA;(n + 2)! &amp;amp; (n + 3)! &amp;amp; (n + 4)!&#xD;&#xA;\end{array} \right|$    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;show that $\left(\dfrac{D}{(n!)^{3}} - 4 \right)$ is divisible by $n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any ideas on how to go about solving this??&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank You in advance.&lt;/p&gt;&#xA;" LastActivityDate="2011-03-19T04:56:35.173" Title="For a fixed positive integer n, show that the determinant below is divisible by n." Tags="&lt;integer-programming&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="24977" PostTypeId="2" ParentId="24974" CreationDate="2011-03-04T09:05:52.473" Score="2" ViewCount="" Body="&lt;p&gt;Apply $\Lambda^k(A)$ to any basis element $v_{i_1} \wedge \ldots \wedge v_{i_k}$, and you get $Av_{i_1} \wedge \ldots \wedge Av_{i_k}$. Write each $Av_{i_j}$ in terms of the basis for $W$. Expand, getting the coefficient of each term being a product of elements of $A$. Collecting terms to get an expression in the basis for $\Lambda^k(W)$. You get coefficients being sums of product of elements of $A$. Ie, polynomials in elements of $A$.&lt;/p&gt;&#xA;" OwnerUserId="7782" LastActivityDate="2011-03-04T09:05:52.473" />
  <row Id="24978" PostTypeId="1" AcceptedAnswerId="24989" CreationDate="2011-03-04T09:06:53.613" Score="8" ViewCount="155" Body="&lt;p&gt;Let $f: \mathbb{R} \to \mathbb{R}$ be a function. Usually when proving a theorem where $f$&amp;#160;is assumed to be continuous, differentiable, $C^1$ or smooth, it is enough to draw intuition by assuming that $f$ is piecewise smooth (something that one could perhaps draw on a paper without lifting your pencil). What I'm saying is that in all these cases my mental picture is about the same. This works most of the time, but sometimes it of course doesn't.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence I would like to ask for examples of continuous, differentiable and $C^1$&amp;#160;functions, which would highlight the differences between the different classes. I'm especially interested in how nasty differentiable functions can be compared to continuously differentiable ones. Also if it is the case that the one dimensional case happens to be uninteresting, feel free to expand your answer to functions $\mathbb{R}^n \to \mathbb{R}^m$. The optimal answer would also list some general minimal 'sanity-checks' for different classes of functions, which a proof of a theorem concerning a particular class would have to take into account.&lt;/p&gt;&#xA;" OwnerUserId="3776" LastActivityDate="2011-03-04T14:40:03.320" Title="Nasty examples for different classes of functions" Tags="&lt;real-analysis&gt;&lt;intuition&gt;&lt;examples&gt;&lt;counter-examples&gt;" AnswerCount="3" CommentCount="6" FavoriteCount="1" />
  <row Id="24979" PostTypeId="2" ParentId="24975" CreationDate="2011-03-04T09:13:15.763" Score="4" ViewCount="" Body="&lt;p&gt;You could do it two different ways. You could use a substack, like this&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \prod_{\substack{n=1 \\ 2 \mid n}}^\infty \frac{4n^2}{4n^2 -1} .$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or you could also let $n = 2m$ (or $n = 2m - 1$ if you want to start at $1$) and take the product as $m$ goes from $1$ to $\infty$, like this&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \prod_{m=1}^\infty \frac{4(2m)^2}{4(2m)^2-1} . $$&lt;/p&gt;&#xA;" OwnerUserId="7782" LastActivityDate="2011-03-04T09:13:15.763" />
  <row Id="24980" PostTypeId="2" ParentId="24975" CreationDate="2011-03-04T09:13:58.740" Score="1" ViewCount="" Body="&lt;p&gt;There are some different ways mathematicians deal with this. For simple cases like this one, the most popular way is to write the product as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\prod_{n=1}^\infty \frac{4(2n)^2}{4(2n)^2 - 1}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a more general case, one can use the notation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\prod_{n \in S} \frac{4n^2}{4n^2 - 1},$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $S$&amp;#160;is a set that $n$&amp;#160;takes values from, for example $S = \{2n \mid n \in \mathbb{N}\}$. This way is a little bit problematic, since the order of the summation or multiplication must be understood from the context.&lt;/p&gt;&#xA;" OwnerUserId="3776" LastActivityDate="2011-03-04T09:13:58.740" />
  <row Id="24981" PostTypeId="2" ParentId="24978" CreationDate="2011-03-04T09:29:39.030" Score="3" ViewCount="" Body="&lt;p&gt;Although this is maybe not a good example of very &quot;nasty&quot; functions, you could look at $f_i = x^i \sin(1/x)$ for $i=0,1,2,3$ in order to see the distinction between those classes of functions. If you set $f_i(0)=0$ for all $i$, then $f_0$ is not continuous in $0$, $f_1$ is continuous but not differentiable in $0$, $f_2$ is differentiable but not $C^1$ and $f_3$ is $C^1$. &lt;/p&gt;&#xA;" OwnerUserId="1766" LastActivityDate="2011-03-04T09:29:39.030" CommentCount="2" />
  <row Id="24982" PostTypeId="1" AcceptedAnswerId="25009" CreationDate="2011-03-04T09:32:57.967" Score="4" ViewCount="130" Body="&lt;p&gt;(1)In set theory, what is the purpose for defining the concept of cofinality?is it that important?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2)The concept of cofinality finally leads to 2 types of infinite cardinal, for which the first one is regular cardinal and another one is singular cardinal defined this way:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;An infinite cardinal $\aleph_{\alpha}$ is regular if $cf (\aleph_{\alpha})=\aleph_{\alpha}$ and singular if $cf (\aleph_{\alpha}) &amp;lt; \aleph_{\alpha}$ . I was wondering here...what is the meaning of this and its consequences(the fact that it has 2 types of cardinal)?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your explanation is very much appreciated. thanks ahead.  &lt;/p&gt;&#xA;" OwnerUserId="1683" LastEditorUserId="622" LastEditorDisplayName="" LastEditDate="2011-03-04T13:05:43.167" LastActivityDate="2011-03-04T13:24:00.247" Title="Cofinality and its Consequences" Tags="&lt;elementary-set-theory&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="24983" PostTypeId="1" AcceptedAnswerId="24988" CreationDate="2011-03-04T09:34:10.820" Score="2" ViewCount="62" Body="&lt;p&gt;I would like to show the following claim:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $l_1, l_2 \in \mathbb{R}^{n+1}$ be two distinct lines then $\exists$ a linear isomorphism $A : \mathbb{R}^{n+1} \rightarrow \mathbb{R}^{n+1}$ with &#xA;$Al_1 = (\lambda, 0, 0, \dots , 0)$ and $Al_2 = (0, \lambda, 0, \dots , 0)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I thought this $A$ needed to prove the claim is the basis transformation matrix from the standard basis of $\mathbb{R}^{n+1}$ into the basis consisting of $l_1$, $l_2$ and some linearly independent basis vectors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My &lt;strong&gt;question&lt;/strong&gt; now is: Are my thoughts correct? And if so, how can I construct the missing vectors to make $\{l_1, l_2 \}$ into a basis of dimension $n + 1$?&lt;/p&gt;&#xA;" OwnerUserId="5798" LastActivityDate="2011-03-04T10:19:50.237" Title="Constructing a basis using two vectors" Tags="&lt;linear-algebra&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="24984" PostTypeId="1" AcceptedAnswerId="24992" CreationDate="2011-03-04T09:40:36.200" Score="2" ViewCount="114" Body="&lt;p&gt;So the Tower of Hanoi numbers are given by the recurrence $h_n=2h_{n-1}+1$ and $h_1=1$. I let my generating function be &#xA;$$&#xD;&#xA;g(x)=\sum h_nx^n&#xD;&#xA;$$&#xA;Then &#xA;$$&#xD;&#xA;g(x)=\sum h_n x^n=\sum (2h_{n-1}+1)x^n=\sum 2h_{n-1}x^n+\sum x^n=2xg(x)+\frac{1}{1-x}.&#xD;&#xA;$$&#xA;Solving for $g(x)$ I find &#xA;$$&#xD;&#xA;g(x)=\frac{1}{(1-2x)(1-x)}=\frac{2}{1-2x}-\frac{1}{1-x}=2\sum (2x)^n-\sum x^n.&#xD;&#xA;$$&#xA;It seems then that the coefficient $h_n$ of $x^n$ is $2^{n+1}-1$, but &lt;a href=&quot;http://mathworld.wolfram.com/TowerofHanoi.html&quot; rel=&quot;nofollow&quot;&gt;wolfram mathworld&lt;/a&gt; says it should be $h_n=2^n-1$. What did I do wrong here? Thanks.&lt;/p&gt;&#xA;" OwnerUserId="6742" LastActivityDate="2011-03-04T11:19:20.430" Title="Using generating functions to find a formula for the Tower of Hanoi numbers." Tags="&lt;combinatorics&gt;&lt;generating-functions&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="24985" PostTypeId="2" ParentId="24972" CreationDate="2011-03-04T09:41:49.130" Score="1" ViewCount="" Body="&lt;p&gt;I think &quot;any master&quot; translates into $\forall$ masters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I had to answer this question, I think I would first try to translate &quot;all and only sets which are not elements of themselves&quot;. To do that, I might (as a partial answer) try to translate &quot;a set that is not an element of itself&quot;, &quot;all sets which are not elements of themselves&quot; and &quot;only sets which are not elements of themselves&quot;. Then I would try to merge these three partial answers into the answer to the original question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; I think &quot;all sets that are not an element of themselves&quot; looks like this: $$ \{ S \mid&amp;#160;  S \not\in S \} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope this helps!&lt;/p&gt;&#xA;" OwnerUserId="5798" LastEditorUserId="5798" LastEditorDisplayName="" LastEditDate="2011-03-04T10:15:13.593" LastActivityDate="2011-03-04T10:15:13.593" />
  <row Id="24986" PostTypeId="2" ParentId="24984" CreationDate="2011-03-04T10:02:59.063" Score="0" ViewCount="" Body="&lt;p&gt;I think you also need $h_0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ g(x) = \sum_{n \geq 0} h_n x^n = h_0 x_0 + \sum_{n \geq 1} (2 h_{n-1} + 1)x^n$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you write $\sum_{n \geq 0} 2 h_{n-1} x^n$, this is not defined because you don't know $h_{-1}$. Or is it given in the problem?&lt;/p&gt;&#xA;" OwnerUserId="5798" LastActivityDate="2011-03-04T10:02:59.063" CommentCount="4" />
  <row Id="24988" PostTypeId="2" ParentId="24983" CreationDate="2011-03-04T10:19:50.237" Score="3" ViewCount="" Body="&lt;p&gt;Your thinking is basically correct, except: $l_1$, $l_2$ are vectors, not lines; and you have to exclude the case that one of them is zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;About extending the basis, see &lt;a href=&quot;http://en.wikipedia.org/wiki/Basis_%28linear_algebra%29#Extending_to_a_basis&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/Gram-Schmidt_process&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="6622" LastActivityDate="2011-03-04T10:19:50.237" CommentCount="1" />
  <row Id="24989" PostTypeId="2" ParentId="24978" CreationDate="2011-03-04T10:45:29.447" Score="3" ViewCount="" Body="&lt;p&gt;The Wikipedia article &lt;a href=&quot;http://en.wikipedia.org/wiki/Pompeiu_derivative&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Pompeiu_derivative&lt;/a&gt; gives one example of how bad a non-continuous derivative can be. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One can show that any set whose complement is a dense intersection of countably many open sets is the point of discontinuities for some derivative. In particular a derivative can be discontinuous almost everywhere and on a dense set. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;See the book &quot;Differentiation of Real Functions&quot; by Andrew Bruckner for this and much more.  &lt;/p&gt;&#xA;" OwnerUserId="4623" LastActivityDate="2011-03-04T10:45:29.447" CommentCount="1" />
  <row Id="24990" PostTypeId="1" AcceptedAnswerId="24991" CreationDate="2011-03-04T10:54:57.117" Score="2" ViewCount="53" Body="&lt;p&gt;I'm doing some programming for a video game in which the player characters and monster enemies can move at different speeds on 2-dimensional, cell-based tile map. Examples of these speeds are as follows:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;1600 ms = 0.625 tiles/s&#xA;1000 ms = 1 tile/s&#xA;800 ms = 1.25 tiles/s&#xA;400 ms = 2.5 tiles/s&#xA;100 ms = 10 tiles/s&#xA;25 ms = 40 tiles/s&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The same mathematical formula is used to convert every one of the above equations (feed the formula the first value, out pops the second). Problem is my math skills are terrible and I can't figure out how to determine what that formula actually is. Help!?&lt;/p&gt;&#xA;" OwnerUserId="7791" LastActivityDate="2011-03-04T11:04:43.323" Title="How to find basic algebra conversion formula?" Tags="&lt;algebra&gt;" AnswerCount="1" />
  <row Id="24991" PostTypeId="2" ParentId="24990" CreationDate="2011-03-04T11:04:43.323" Score="1" ViewCount="" Body="&lt;p&gt;Second value = 1000/first value.&lt;/p&gt;&#xA;" OwnerUserId="1878" LastActivityDate="2011-03-04T11:04:43.323" CommentCount="2" />
  <row Id="24992" PostTypeId="2" ParentId="24984" CreationDate="2011-03-04T11:19:20.430" Score="5" ViewCount="" Body="&lt;p&gt;Setting $$g(x) = \sum_{n=0}^{\infty} h_{n+1} x^n,$$ we obtain using $h_n = 2 h_{n-1} +1$ and $h_1 =1$&#xA;$$g(x) = \sum_{n=0}^{\infty} h_{n+1} x^n = h_1 + \sum_{n=1}^{\infty} (2 h_{n} +1) x^n = 1 + \sum_{n=0}^{\infty} (2 h_{n+1} +1) x^{n+1} =1 +2 x g(x)  + \frac{x}{1-x}.$$&#xA;Solving for $g(x)$, we obtain $$g(x) = \frac{1}{(1-x) (1-2x)} = \frac{2}{1-2x} - \frac{1}{1-x} = \sum_{n=0}^{\infty} (2^{n+1} -1) x^n$$ such that $h_n = 2^n -1$.&lt;/p&gt;&#xA;" OwnerUserId="7266" LastActivityDate="2011-03-04T11:19:20.430" CommentCount="1" />
  <row Id="24993" PostTypeId="2" ParentId="24961" CreationDate="2011-03-04T11:19:50.617" Score="4" ViewCount="" Body="&lt;p&gt;I think the more conceptual way to see that this holds is that the Möbius function $\mu(x,y)$ only depends on the structure of the poset $[x,y] = \{ z \; : \; x \leq z \leq y\}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So it suffices to show that when the natural numbers are ordered by divisibility and &lt;em&gt;k&lt;/em&gt; divides &lt;em&gt;n&lt;/em&gt;, then the posets $[k,n]$ and $[1,\frac n k]$ are isomorphic. To see this, use the isomorphism $z \mapsto \frac z k$.  &lt;/p&gt;&#xA;" OwnerUserId="677" LastActivityDate="2011-03-04T11:19:50.617" />
  <row Id="24994" PostTypeId="1" AcceptedAnswerId="25155" CreationDate="2011-03-04T11:56:43.537" Score="2" ViewCount="181" Body="&lt;p&gt;I am trying to prove young's inequality for integrals&#xA;$$&#xD;&#xA;ab \leq \int_0^a \! f(x) \, \mathrm{d}x + \int_0^b \! f^{-1}(x) \, \mathrm{d}x.&#xD;&#xA;$$&#xA;Can you help me please?&lt;/p&gt;&#xA;" OwnerUserId="3755" LastEditorUserId="2937" LastEditorDisplayName="" LastEditDate="2011-03-05T14:52:43.393" LastActivityDate="2011-05-04T20:30:30.810" Title="Young inequality" Tags="&lt;calculus&gt;&lt;integral&gt;&lt;inequality&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="24995" PostTypeId="1" AcceptedAnswerId="25015" CreationDate="2011-03-04T12:09:39.173" Score="5" ViewCount="155" Body="&lt;p&gt;I know that all finite subgroups of $SL_2(\mathbb R)$ are cyclic by standard averaging argument. They are all conjugate to some finite subgroup of $SO _2(\mathbb R)$ and therefore cyclic. My question is how to classify all finite subgroups of $GL_2(\mathbb R)$. &#xA;Thanking you.&lt;/p&gt;&#xA;" OwnerUserId="7565" LastEditorUserId="7565" LastEditorDisplayName="" LastEditDate="2011-03-04T12:25:43.840" LastActivityDate="2011-03-04T22:14:26.183" Title="A question on $GL_2(\mathbb R)$" Tags="&lt;linear-algebra&gt;&lt;group-theory&gt;&lt;algebra&gt;" AnswerCount="2" CommentCount="0" />
  <row Id="24996" PostTypeId="1" CreationDate="2011-03-04T12:10:53.337" Score="3" ViewCount="235" Body="&lt;p&gt;&lt;strong&gt;Prove by induction&lt;/strong&gt;. Every &lt;strong&gt;partial order&lt;/strong&gt; on a &lt;strong&gt;nonempty finite&lt;/strong&gt; set has at least one &lt;strong&gt;minimal&lt;/strong&gt; element. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I solve that question ?&lt;/p&gt;&#xA;" OwnerUserId="7792" LastEditorUserId="367" LastEditorDisplayName="" LastEditDate="2011-03-04T13:03:15.600" LastActivityDate="2011-03-04T13:03:50.093" Title="Question related with partial order - finite set - minimal element" Tags="&lt;elementary-set-theory&gt;&lt;induction&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="24997" PostTypeId="1" AcceptedAnswerId="25041" CreationDate="2011-03-04T12:15:26.577" Score="2" ViewCount="155" Body="&lt;p&gt;In linear programming, an LP can have multiple optimal solutions if it contains degenerate vertices, i.e. where one of the base-variables is 0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can an LP also have multiple optimal solutions if it does not contain degenerate solutions? If so, what is an example LP?&lt;/p&gt;&#xA;" OwnerUserId="7793" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2011-03-04T17:24:42.987" LastActivityDate="2011-03-04T17:24:42.987" Title="Can a non-degenerate LP have multiple optimal solutions?" Tags="&lt;linear-programming&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="24998" PostTypeId="1" CreationDate="2011-03-04T12:15:36.080" Score="3" ViewCount="48" Body="&lt;p&gt;Let X an irreducible singular curve with a singular point $x$. Consider $A$, the normalization of $\mathcal{O}_{X,x}$, and $x_1,\dots,x_r$ the points over $x$ in the normalization of $X$. Why $A\cong \mathcal{O}_{x_1}\cap\dots\cap\mathcal{O}_{x_r}$?&#xA;Moreover assume that the curve is over a field $k$. Why do the dimension of $A/\mathcal{O}_{X,x}$ is r-1?&lt;/p&gt;&#xA;" OwnerUserId="6583" LastEditorUserId="6583" LastEditorDisplayName="" LastEditDate="2011-03-04T15:05:04.283" LastActivityDate="2011-03-04T15:05:04.283" Title="On local rings of a normalization" Tags="&lt;algebraic-geometry&gt;&lt;arithmetic-geometry&gt;" CommentCount="1" FavoriteCount="1" />
  <row Id="24999" PostTypeId="1" CreationDate="2011-03-04T12:18:10.667" Score="4" ViewCount="254" Body="&lt;p&gt;I could really use some step-by-step help on these two problems please. Thank You in advance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1.)   Let $V = \{{\bf{A|A}}$ is an $n \times n$ matrix, $n$ fixed, &#xA;det$({\bf{A}}) = 0$ }. Is $V$, with the usual addition and   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;scalar multiplication, a vector space? Give reason. If yes, find the &#xA;dimension and a basis for $V$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2.)   Let $V = \{f(x)|f(x) = (ax + b)e^{-x},\; a,b\; \in\; \mathbb{R}\}$.&#xA;Is $V$, with the usual addition and scalar  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;multiplication, a vector space? Give reason. If yes, find the dimension&#xA;and basis for $V$.&lt;/p&gt;&#xA;" LastActivityDate="2011-04-11T21:41:28.217" Title="Vector Spaces: Finding a basis and Dimension." Tags="&lt;linear-algebra&gt;&lt;vector-space&gt;" AnswerCount="3" />
  <row Id="25000" PostTypeId="1" AcceptedAnswerId="25001" CreationDate="2011-03-04T12:20:27.370" Score="5" ViewCount="113" Body="&lt;p&gt;Hi, I'm a Phd student who teaches part time at a high school and I noticed something when teaching sequences today.  I asked my students to find the nth term (the general term) for some sequences.  They observed:   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $a_n=n$ then the first differences will be $1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $a_n=n^2$ then the second differences will be $2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $a_n=n^3$ then the third differences will be $6$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $a_n=n^4$ then the fourth differences will be $27$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I can now construct a sequence: $1,2,6,27,120,720,...$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are these numbers?  Can I find $a_n$?  How?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="5981" LastEditorUserId="5981" LastEditorDisplayName="" LastEditDate="2011-03-04T12:32:38.263" LastActivityDate="2011-03-04T12:33:38.577" Title="nth term of sequences" Tags="&lt;sequences-and-series&gt;&lt;high-school&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="25001" PostTypeId="2" ParentId="25000" CreationDate="2011-03-04T12:28:28.890" Score="8" ViewCount="" Body="&lt;p&gt;Your $27$ should be $24$. Now it's easy to see the pattern!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edited to add:&lt;/strong&gt;&#xA;To see why, take (for example) $a_n=n^4$. The first difference is  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a_{n+1}-a_n = (n+1)^4 - n^4 = 4n^3 +$ lower powers of $n$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the fourth difference is four times the third difference of $n^3$ plus the third difference of lower powers of $n$. Lower powers have zero third difference, so this is $4 \times 6 = 24$.&lt;/p&gt;&#xA;" OwnerUserId="1508" LastEditorUserId="1508" LastEditorDisplayName="" LastEditDate="2011-03-04T12:33:38.577" LastActivityDate="2011-03-04T12:33:38.577" CommentCount="1" />
  <row Id="25002" PostTypeId="2" ParentId="24982" CreationDate="2011-03-04T12:47:39.860" Score="7" ViewCount="" Body="&lt;p&gt;Some interesting results regarding the cofinality of a cardinal are related to the exponential function of cardinals. For example we can prove that $\kappa^{cf(\kappa)}&amp;gt;\kappa$ or we can prove that $cf(2^\kappa)&amp;gt;\kappa$. Actually this last one (along with Cantor's $\kappa&amp;lt;2^\kappa$) happens to be the only restrain to the values of $2^\kappa$. For example we know that $2^{\aleph_0}$ cannot be any cardinal with cofinality $\aleph_0$, so it cannot be $\aleph_\omega$, $\aleph_{\aleph_\omega}$ etc. Also, it's easy to see that the cofinality of every ordinal is a regular cardinal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It can be proved in ZF that a cardinal $\kappa$ is singular exactly when there exists a cardinal $\lambda&amp;lt;\kappa$ and a family of sets $\{S_\xi : \xi&amp;lt;\lambda\}$ such that $|S_\xi|&amp;lt;\kappa$ and $\bigcup_{\xi&amp;lt;\lambda}S_\xi=\kappa$. This says that a singular cardinal $\kappa$ is a union of less than $\kappa$ subsets of $\kappa$, each with less than $\kappa$ elements. The size of every infinite cardinal depends solely on regular cardinals: Every infinite cardinal can be written as a regular sum of regular cardinals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Intuitively you can say that when we pass to a regular cardinal we perform a bigger leap in size. For example $\aleph_0$ is a regular cardinal. It cannot be written as a finite union of finite sets. In a sense moving from the finite to the infinite makes a huge difference. Regular cardinals have different properties than singular cardinals because of this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the axiom of choice one can show that every successor cardinal is regular but it is provable that we cannot prove in ZFC neither the existence of uncountable regular limit cardinals nor the consistency of such existence (interestingly enough we cannot prove in ZFC minus infinity neither the existence of infinite sets nor the consistency of such existence).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Limit regular cardinals (that are uncountable) are called weakly inaccessible. Strengthening this a bit, let's assume that an uncountable regular cardinal $\kappa$ is a strong limit (that is for every $\lambda&amp;lt;\kappa$ we have $2^\lambda&amp;lt;\kappa$). These cardinals are called (strongly) inaccessible cardinals. They are important metamathematically because at the height of such a cardinal the cumulative hierarchy is a model of ZFC. That is to say, these cardinals are natural bounds to the methods we have for creating new sets through the axioms, and thus assumptions of the existence of such cardinals create a &quot;bigger&quot; universe of set theory. We can further strengthen these results by assuming that a cardinal has other properties (combinatorial for example) and yield much stronger forms of cardinals. This gave rise to the theory of large cardinals which provides a measurement for the consistency strength of set theoretic statements.&lt;/p&gt;&#xA;" OwnerUserId="1772" LastActivityDate="2011-03-04T12:47:39.860" CommentCount="1" />
  <row Id="25003" PostTypeId="1" CreationDate="2011-03-04T12:48:50.107" Score="1" ViewCount="93" Body="&lt;p&gt;I have 2 polynomials $p_1(x_1,\ldots,x_n)$ and $p_2(x_1,\ldots,x_n)$, of which I have to compute the product, with a special property: The exponent of each variable is always either $0$ or $1$, where every exponent greater than $1$ gets cut down to $1$. E.g. $x_1 * x_1x_2 = x_1x_2$, and the result of $(2x_1x_2x_4 + 4x_2x_3) * (3x_2x_3 - x_1x_4)$ would be $2x_1x_2x_3x_4 -2x_1x_2x_4 +12x_2x_3$. Additionally, values for all but one variables are given. So the result would be some polynomial $p(x_k) = ax_k + b$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Currently I compute the product the usual way, multiplying each summand of $p_1$ with every summand in $p_2$, and then apply the values for all known $x_i$. But since both polynomials easily contain $10^5$ summands each, the algorithm iterates over $10^{10}$ elements, which simply takes too much time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is: Does there exists a more effective procedure to compute $p(x_k)$?&lt;/p&gt;&#xA;" OwnerUserId="6205" LastActivityDate="2011-06-02T16:09:14.407" Title="Efficient calculation of polynomial product" Tags="&lt;algorithms&gt;&lt;polynomials&gt;&lt;products&gt;" AnswerCount="1" />
  <row Id="25004" PostTypeId="2" ParentId="24999" CreationDate="2011-03-04T12:49:03.930" Score="6" ViewCount="" Body="&lt;p&gt;The first one  is not a vector space as sum of two  singular matrix may be nonsingular. For example &#xA;$$\begin{pmatrix} 1 &amp;amp; 0 \\\ 0 &amp;amp; 0 \end{pmatrix} + \begin{pmatrix} 0 &amp;amp; 0 \\\ 0 &amp;amp; 1 \end{pmatrix} = \begin{pmatrix} 1 &amp;amp; 0 \\\ 0 &amp;amp; 1 \end{pmatrix}$$&#xA;The second one is a vector space of dimension 2 as $xe^{-x}$ and $e^{-x}$ are linearly independent continuas functions. If $axe^{-x} + be^{-x} = 0$ for $a,b \in \mathbb{R}$, Then $ax +b = 0$ as a continuas function on $\mathbb{R}$. Putting $x = 0,1$ we have $b = 0$ and $a +b = 0$. Hence $a = b = 0$.&lt;/p&gt;&#xA;" OwnerUserId="7565" LastEditorUserId="7565" LastEditorDisplayName="" LastEditDate="2011-04-11T21:41:28.217" LastActivityDate="2011-04-11T21:41:28.217" CommentCount="4" />
  <row Id="25005" PostTypeId="2" ParentId="24976" CreationDate="2011-03-04T13:02:20.473" Score="7" ViewCount="" Body="&lt;p&gt;$$\begin{eqnarray*}&#xD;&#xA;D &amp;amp;=&amp;amp;%&#xD;&#xA;\begin{vmatrix}&#xD;&#xA;n! &amp;amp; (n+1)! &amp;amp; (n+2)! \\ &#xD;&#xA;(n+1)! &amp;amp; (n+2)! &amp;amp; (n+3)! \\ &#xD;&#xA;(n+2)! &amp;amp; (n+3)! &amp;amp; (n+4)!%&#xD;&#xA;\end{vmatrix}&#xD;&#xA;\\&#xD;&#xA;&amp;amp;=&amp;amp;%&#xD;&#xA;\begin{vmatrix}&#xD;&#xA;n! &amp;amp; (n+1)n! &amp;amp; (n+2)(n+1)n! \\ &#xD;&#xA;(n+1)n! &amp;amp; (n+2)(n+1)n! &amp;amp; (n+3)(n+2)(n+1)n! \\ &#xD;&#xA;(n+2)(n+1)n! &amp;amp; (n+3)(n+2)(n+1)n! &amp;amp; (n+4)(n+3)(n+2)(n+1)n!%&#xD;&#xA;\end{vmatrix}&#xD;&#xA;\\&#xD;&#xA;&amp;amp;=&amp;amp;n!^{3}(n+1)^{2}(n+2)\underset{\text{This determinant is 2 (see below)}}{\underbrace{%&#xD;&#xA;\begin{vmatrix}&#xD;&#xA;1 &amp;amp; (n+1) &amp;amp; (n+2)(n+1) \\ &#xD;&#xA;1 &amp;amp; (n+2) &amp;amp; (n+3)(n+2) \\ &#xD;&#xA;1 &amp;amp; (n+3) &amp;amp; (n+4)(n+3)%&#xD;&#xA;\end{vmatrix}%&#xD;&#xA;}} \\&#xD;&#xA;&amp;amp;=&amp;amp;n!^{3}\left( 2n^{3}+8n^{2}+10n+4\right) &#xD;&#xA;\end{eqnarray*}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have used the following property repeatedly: &quot;If $B$ results from $A$ by multiplying one row or column with a number $c$, then $\det(B) = c \cdot \det(A)$&quot;. (&lt;a href=&quot;http://en.wikipedia.org/wiki/Determinant&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;). $n!$ is a factor of rows 1, 2 and 3. $n+1$ is a factor of rows 2 and 3. $n+2$ is a factor of row 3.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{D}{n!^{3}}-4=2n^{3}+8n^{2}+10n=(2n^{2}+8n+10)n.$$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Evaluation of the last determinant &lt;a href=&quot;http://mathworld.wolfram.com/Determinant.html&quot; rel=&quot;nofollow&quot;&gt;expanding it by the minors&lt;/a&gt; of column 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$%&#xD;&#xA;\begin{vmatrix}&#xD;&#xA;1 &amp;amp; (n+1) &amp;amp; (n+2)(n+1) \\ &#xD;&#xA;1 &amp;amp; (n+2) &amp;amp; (n+3)(n+2) \\ &#xD;&#xA;1 &amp;amp; (n+3) &amp;amp; (n+4)(n+3)%&#xD;&#xA;\end{vmatrix}%&#xD;&#xA;$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$=%&#xD;&#xA;\begin{vmatrix}&#xD;&#xA;(n+2) &amp;amp; (n+3)(n+2) \\ &#xD;&#xA;(n+3) &amp;amp; (n+4)(n+3)%&#xD;&#xA;\end{vmatrix}%&#xD;&#xA;-%&#xD;&#xA;\begin{vmatrix}&#xD;&#xA;(n+1) &amp;amp; (n+2)(n+1) \\ &#xD;&#xA;(n+3) &amp;amp; (n+4)(n+3)%&#xD;&#xA;\end{vmatrix}%&#xD;&#xA;+%&#xD;&#xA;\begin{vmatrix}&#xD;&#xA;(n+1) &amp;amp; (n+2)(n+1) \\ &#xD;&#xA;(n+2) &amp;amp; (n+3)(n+2)%&#xD;&#xA;\end{vmatrix}%&#xD;&#xA;$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$=(n+2)(n+3)%&#xD;&#xA;\begin{vmatrix}&#xD;&#xA;1 &amp;amp; (n+3) \\ &#xD;&#xA;1 &amp;amp; (n+4)%&#xD;&#xA;\end{vmatrix}%&#xD;&#xA;-(n+1)(n+3)%&#xD;&#xA;\begin{vmatrix}&#xD;&#xA;1 &amp;amp; (n+2) \\ &#xD;&#xA;1 &amp;amp; (n+4)%&#xD;&#xA;\end{vmatrix}%&#xD;&#xA;$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$+(n+1)(n+2)%&#xD;&#xA;\begin{vmatrix}&#xD;&#xA;1 &amp;amp; (n+2) \\ &#xD;&#xA;1 &amp;amp; (n+3)%&#xD;&#xA;\end{vmatrix}%&#xD;&#xA;$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$=(n+2)(n+3)-2(n+1)(n+3)+(n+1)(n+2)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$=\left( n^{2}+5n+6\right) -\left( 2n^{2}+8n+6\right) +\left(&#xD;&#xA;n^{2}+3n+2\right) =2$&lt;/p&gt;&#xA;" OwnerUserId="752" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2011-03-04T18:28:19.393" LastActivityDate="2011-03-04T18:28:19.393" CommentCount="1" />
  <row Id="25006" PostTypeId="2" ParentId="24996" CreationDate="2011-03-04T13:03:50.093" Score="4" ViewCount="" Body="&lt;p&gt;One proof method (by induction) is on the size of your set.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The basis, a singleton. There is only one element so clearly it is minimal.&#xA;Assume the claim is true for sets of size $n$, now prove for $n+1$:&lt;br&gt;&#xA;Choose any element of $A$, denote it by $a$ now restrict your partial order to $A\setminus\{a\}$. By the induction hypothesis there is a minimal element there, $b$. Now look at the original order - if $a&amp;lt;b$ show that $a$ is minimal, otherwise $b$ is still minimal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Either way, you found a minimal element in $A$ and your partial order.&lt;/p&gt;&#xA;" OwnerUserId="622" LastActivityDate="2011-03-04T13:03:50.093" />
  <row Id="25007" PostTypeId="2" ParentId="24945" CreationDate="2011-03-04T13:10:01.990" Score="8" ViewCount="" Body="&lt;p&gt;The usage of &quot;transfinite&quot; there is not the same as what we now call &quot;transfinite induction&quot;. Kolmogorov essentially just means &quot;infinite&quot;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One example of the sort of thing that Kolmogorov calls &quot;excluded middle&quot; is now called the &quot;limited principle of omniscience&quot; (LPO).  LPO says that if you have any property $P(n)$ of a natural number $n$, so that each number either does or does not have the property, then either there is a number $m$ such that $P(m)$ holds, or, for every number $m$, $P(m)$ does not hold.  In other words, if $(\forall m)(P(m) \lor \lnot P(m))$ then $(\exists m)P(m) \lor (\forall m)\lnot P(m)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;LPO is trivially true in classical logic but it is not trivial in intuitionistic systems. The best way to understand this is to know that when an intuitionist says that a number exists with some property, he or she means that he or she already has an example of a specific number with that property. So LPO claims, on this reading, that there is some way to determine whether $(\exists m)P(m)$ holds, for any decidable property $P$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Classical mathematicians will accept that, while they believe that the sentence $(\exists m)P(m)$ is either true or false, they have no way in general to decide which is the case (in general). For intuitionists, the lack of ability to decide which is the case means that they cannot assert that the sentence is true, and the cannot assert it is false. Because of their redefinition of the word &quot;or&quot;, this means to them that they cannot assert the sentence is true or false. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The next place to go if you're new to this area is the Brouwer-Heyting-Kolmogorov (BHK) interpretation of intuitionistic logic. That provides a way for classically-trained mathematicians to understand what intuitionists mean when they say certain things, including the way that they redefine the terms &quot;or&quot; and &quot;there exists&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One example of what Kolmogorov means by &quot;finitary conclusion&quot; is an equation between numbers such as $2^2 = 4$ or $0=1$. He might even accept equations with free variables like $3x + 5x = 8x$ or $3^x \not = 4y$. Working in classical Peano arithmetic, it would be possible to make a proof which ends with that sort of statement, but which applies LPO along the way. Such a proof would not be acceptable in intuitionistic logic. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are &quot;conservation results&quot; that show that if sentences of a certain form are provable in Peano arithmetic, then they are actually provable (possibly with a different proof) in intuitionistic Heyting arithmetic. This is what Kolmogov is referring to when he says&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;We shall prove that all the finitary conclusions obtained by means of a transfinite use of the principle of excluded middle are correct and can be proved even without its help.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="630" LastEditorUserId="630" LastEditorDisplayName="" LastEditDate="2011-03-04T13:16:07.320" LastActivityDate="2011-03-04T13:16:07.320" />
  <row Id="25008" PostTypeId="1" CreationDate="2011-03-04T13:18:24.840" Score="1" ViewCount="78" Body="&lt;p&gt;I have two points $P_1$ and $P_2$ in the plane. For each of the points, I have two vectors $v_1$ and $v_2$. I want to find a quadratic Bézier curve from $P_1$ to $P_2$ of length $l$ leaving $P_1$ in the direction of $v_1$ and entering $P_2$ in the direction of $v_2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; It is not needed that I restrict myself to Bézier curves, it was just my first approach. Other ideas are more than welcome.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, as stated in the comments, this is not always possible, but I am only interested in the cases where it is possible.&lt;/p&gt;&#xA;" OwnerUserId="3317" LastEditorUserId="3317" LastEditorDisplayName="" LastEditDate="2011-03-04T13:40:49.470" LastActivityDate="2011-03-04T13:40:49.470" Title="Finding a quadratic Bézier curve of length $l$ between two points" Tags="&lt;geometry&gt;&lt;bezier-curve&gt;&lt;parametric&gt;" CommentCount="3" />
  <row Id="25009" PostTypeId="2" ParentId="24982" CreationDate="2011-03-04T13:24:00.247" Score="5" ViewCount="" Body="&lt;p&gt;The way I look at cofinality is:&#xA;$$cf(\kappa) = \min\{|A| \colon A\subseteq\kappa \wedge \forall \beta&amp;lt;\kappa\exists \alpha\in A (\beta\le\alpha)\}$$&#xA;That is, the smallest size you need in order to be unbounded.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clearly $cf(\kappa)\le\kappa$ for every cardinal, and for regular cardinals this is really an equality - you need to have $\kappa$ many elements in order to be unbounded, you also have that every set of cardinality $\kappa$ is unbounded.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, suppose $\kappa$ is singular, namely $cf(\kappa)&amp;lt;\kappa$, that means that you can take $&amp;lt;\kappa$ many steps and still be unbounded.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Example:&lt;/strong&gt;&lt;br&gt;&#xA;Consider $\kappa=\aleph_\omega$, that is the first cardinal number which is larger than $\aleph_n$ for every finite $n$. We have that $\langle\aleph_n\colon n\in\omega\rangle$ is a cofinal sequence, as every ordinal below $\kappa$ is smaller than some $\aleph_n$, but this is merely a countable set while $\kappa$ is very much uncountable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now suppose I want to prove some property about ordinals below $\aleph_\omega$ which has a somewhat inductive property, that is if it is true at one point it will be true below it. This process requires me only countably many steps and in fact it can be done with a simple induction over the natural numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, had I wanted to do the same on $\aleph_1$ (assuming the usual models of ZFC where it is regular) I would have to ensure the process to continue over $\aleph_1$ many elements. &lt;/p&gt;&#xA;" OwnerUserId="622" LastActivityDate="2011-03-04T13:24:00.247" CommentCount="2" />
  <row Id="25010" PostTypeId="2" ParentId="24999" CreationDate="2011-03-04T13:42:53.650" Score="1" ViewCount="" Body="&lt;p&gt;I think that Anjan really wanted to say that the sum of two singular matrices may be non-singular. For instance:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;\begin{pmatrix}&#xD;&#xA;1  &amp;amp;  0  \\\&#xD;&#xA;0  &amp;amp;  0&#xD;&#xA;\end{pmatrix}&#xD;&#xA;+&#xD;&#xA;\begin{pmatrix}&#xD;&#xA;0  &amp;amp;  0  \\\&#xD;&#xA;0  &amp;amp;  1&#xD;&#xA;\end{pmatrix}&#xD;&#xA;=&#xD;&#xA;\begin{pmatrix}&#xD;&#xA;1  &amp;amp;  0  \\\&#xD;&#xA;0  &amp;amp;  1&#xD;&#xA;\end{pmatrix}&#xD;&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="664" LastActivityDate="2011-03-04T13:42:53.650" CommentCount="1" />
  <row Id="25011" PostTypeId="1" AcceptedAnswerId="25065" CreationDate="2011-03-04T14:05:19.557" Score="4" ViewCount="135" Body="&lt;p&gt;Let $U$ is an ultrafilter on a set $X$, and $V$ an ultrafilter on a set $Y$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Rudin-Keisler_order#Ordering_on_ultrafilters&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt; says: Ultrafilters $U$ and $V$ are Rudin–Keisler equivalent, $U\equiv_{RK}V$, if there exist sets $A\in U$, $B\in V$, and a bijection $f: A → B$ which satisfies the condition above. (If $X$ and $Y$ have the same cardinality, the definition can be simplified by fixing $A = X$, $B = Y$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where &quot;the condition above&quot; is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$C\in V\iff f^{-1}[C]\in U.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How to prove that the special case of the same cardinality is equivalent to the case of arbitrary $X$ and $Y$?&lt;/p&gt;&#xA;" OwnerUserId="4876" LastEditorUserId="4876" LastEditorDisplayName="" LastEditDate="2011-03-04T17:41:40.490" LastActivityDate="2011-04-26T17:34:26.107" Title="Equivalence of two definitions of Rudin–Keisler equivalence" Tags="&lt;set-theory&gt;" AnswerCount="2" CommentCount="16" />
  <row Id="25012" PostTypeId="2" ParentId="24995" CreationDate="2011-03-04T14:17:41.717" Score="1" ViewCount="" Body="&lt;p&gt;Consider a finite subgroup $F$. It acts on $\mathbb R^2$. We can define a $F$-invariant positive definite inner product by Weyl's averaging trick. Then an isomomorphic image of $F$ sits inside $O(2,\mathbb R)$. &lt;/p&gt;&#xA;" OwnerUserId="7795" LastActivityDate="2011-03-04T14:17:41.717" />
  <row Id="25013" PostTypeId="2" ParentId="24964" CreationDate="2011-03-04T14:29:34.810" Score="2" ViewCount="" Body="&lt;p&gt;A related conjecture is posted at &lt;a href=&quot;http://www.cadaeic.net/conjectures.htm&quot; rel=&quot;nofollow&quot;&gt;this link&lt;/a&gt;-that with the floor function you can get all naturals.  I remember seeing a proof to that version with $\pi$ instead of $3$ a while ago, but can't find it now.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2011-03-04T14:55:48.160" LastActivityDate="2011-03-04T14:55:48.160" CommentCount="1" />
  <row Id="25014" PostTypeId="1" AcceptedAnswerId="25018" CreationDate="2011-03-04T14:33:48.470" Score="9" ViewCount="215" Body="&lt;p&gt;By definition of the Riemann Zeta Function, $$\zeta\left(\frac{1}{2}\right) = \sum_{n=1}^\infty \frac{1}{\sqrt{n}}.$$ Since $\forall n \geq 1 : \frac{1}{\sqrt{n}} \geq \frac{1}{n}$, we have that for all $N \geq 1$, $$\sum_{n=1}^N\frac{1}{\sqrt{n}} \geq \sum_{n=1}^N \frac{1}{n},$$ but it is well known that $$\lim_{N\rightarrow\infty}\sum_{n=1}^N\frac{1}{n}=\infty,$$ so $\zeta\left(\frac{1}{2}\right)$ diverges by the comparison test.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, $\zeta\left(\frac{1}{2}\right)$ should equal positive infinity, correct?  If so, why do Maple, Mathematica, and Matlab all return a value of around $-1.4604$ when asked to numerically approximate this value?  For example, &lt;a href=&quot;http://www.wolframalpha.com/input/?i=Zeta%5b1/2%5d&quot; rel=&quot;nofollow&quot;&gt;see here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="5022" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2011-04-19T03:06:26.547" LastActivityDate="2011-05-10T20:10:31.117" Title="Erroneous numerical approximations of $\zeta\left(\frac{1}{2}\right)$?" Tags="&lt;special-functions&gt;&lt;riemann-zeta&gt;&lt;computer-algebra-systems&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="3" />
  <row Id="25015" PostTypeId="2" ParentId="24995" CreationDate="2011-03-04T14:35:36.920" Score="4" ViewCount="" Body="&lt;p&gt;The same averaging argument you and rvk describe gives you that any finite subgroup $G$ of $\operatorname{GL}_2(\mathbb{R})$ is conjugate to a subgroup of $O_2(\mathbb{R})$.  (For that matter, this is also shown in $\S 1.3$ of &lt;a href=&quot;http://math.uga.edu/~pete/8410Chapter9.pdf&quot; rel=&quot;nofollow&quot;&gt;these notes&lt;/a&gt;.  The rest of the notes treat complex and $p$-adic analogues, with applications to classifying finite subgroups of $\operatorname{GL}_n(\mathbb{Q})$.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Without loss of generality, we may as well assume that $G \subset O_2(\mathbb{R})$.  Let $H = G \cap SO_2(\mathbb{R})$.  Since $[O_2(\mathbb{R}):SO_2(\mathbb{R})] = 2$, we have $[G:H] = 1$ or $2$.  If $G = H$ then $G$ itself is a finite subgroup of $SO_2(\mathbb{R})$, hence cyclic.  If $[G:H] = 2$, let $g \in G \setminus H$.  Then &#xA;$\det(g) = -1$, so by the &lt;a href=&quot;http://en.wikipedia.org/wiki/Cartan%E2%80%93Dieudonn%C3%A9_theorem&quot; rel=&quot;nofollow&quot;&gt;Cartan-Dieudonné Theorem&lt;/a&gt; $g$ is a linear reflection and &#xA;thus $G = \langle H, g \rangle$ is a dihedral group $D_n$.&lt;/p&gt;&#xA;" OwnerUserId="299" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2011-03-04T22:14:26.183" LastActivityDate="2011-03-04T22:14:26.183" CommentCount="8" />
  <row Id="25016" PostTypeId="1" AcceptedAnswerId="25021" CreationDate="2011-03-04T14:38:51.353" Score="2" ViewCount="301" Body="&lt;p&gt;How can I solve this cubic equation?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$H^{3} − 3\left[(1 + A\cos(T) )^{2} + \frac{2r \cdot A \sin(T)}{B}\right]H + 2(1 + A \cos(T))^{3} = 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solution in terms of H.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edited in order to give more insight to my problem:&#xA;It was an equation which comes as a part of a derivation in Computational Fluid Dynamics. My motivation is to get H in terms of A, B, r and T. And plot a graph between H and r keeping A and B and T as constants. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A general doubt: If a cubic equation consists of a imaginary root, then is it compulsory that the number of imaginary roots should always be 2?&lt;/p&gt;&#xA;" OwnerUserId="7488" LastEditorUserId="7488" LastEditorDisplayName="" LastEditDate="2011-03-08T11:09:10.200" LastActivityDate="2011-05-09T01:58:48.810" Title="How may I solve this Cubic Equation?" Tags="&lt;algebra&gt;" AnswerCount="1" CommentCount="7" FavoriteCount="1" />
  <row Id="25017" PostTypeId="2" ParentId="24978" CreationDate="2011-03-04T14:40:03.320" Score="3" ViewCount="" Body="&lt;p&gt;Another example of what can go wrong is &lt;a href=&quot;http://en.wikipedia.org/wiki/Volterra%27s_function&quot; rel=&quot;nofollow&quot;&gt;Volterra's function&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;It is differentiable everywhere.&lt;/li&gt;&#xA;&lt;li&gt;Its derivative is bounded everywhere.&lt;/li&gt;&#xA;&lt;li&gt;Its derivative is not Riemann-integrable.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="4791" LastActivityDate="2011-03-04T14:40:03.320" />
  <row Id="25018" PostTypeId="2" ParentId="25014" CreationDate="2011-03-04T14:41:50.737" Score="17" ViewCount="" Body="&lt;p&gt;The analytic continuation of $\zeta(s)$ for $Re(s)&amp;gt;0$ is given by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\zeta(s) = \frac{1}{1-2^{1-s} } \sum_{n=1}^\infty \frac{(-1)^{n-1}}{n^s}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So this is why WolframAlpha gives you&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\zeta \left( \frac12 \right) = -(1+\sqrt{2}) \sum_{n=1}^\infty \frac{(-1)^{n-1}}{\sqrt{n}}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See &lt;a href=&quot;http://mathworld.wolfram.com/RiemannZetaFunction.html&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; for more details.&lt;/p&gt;&#xA;" OwnerUserId="1301" LastEditorUserId="1301" LastEditorDisplayName="" LastEditDate="2011-03-04T14:48:18.253" LastActivityDate="2011-03-04T14:48:18.253" />
  <row Id="25019" PostTypeId="2" ParentId="25014" CreationDate="2011-03-04T14:46:02.377" Score="14" ViewCount="" Body="&lt;p&gt;To supplement Derek's answer, the Riemann zeta function $\zeta(s) = \sum_{n} n^{-s}$ was originally only considered for $s = \sigma + i t$, where $\sigma &amp;gt; 1$.  Note that when $\sigma &amp;gt; 1$, the zeta function always converges absolutely.  However, Riemann showed that the definition of the $\zeta$ function could be extended via &lt;a href=&quot;http://en.wikipedia.org/wiki/Analytic_continuation&quot; rel=&quot;nofollow&quot;&gt;analytic continuation&lt;/a&gt; to the whole complex plane (with a pole at $s = 1$).  There, it satisfies the functional equation  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;\Gamma(s/2) \pi^{-s/2} \zeta(s) = \Gamma\left(\frac{1-s}{2}\right) \pi^{- \frac{1-s}{2}}\zeta(1-s).&#xD;&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="6491" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2011-05-10T20:10:31.117" LastActivityDate="2011-05-10T20:10:31.117" />
  <row Id="25020" PostTypeId="2" ParentId="24927" CreationDate="2011-03-04T14:48:18.617" Score="6" ViewCount="" Body="&lt;p&gt;Thinking about the higher homotopy groups as just groups is in some sense missing the point. The higher homotopy groups are not just abelian groups: they are $\pi_1$-modules, for one thing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More loftily, from the &lt;a href=&quot;http://ncatlab.org/nlab/show/nPOV&quot; rel=&quot;nofollow&quot;&gt;n-categorical point of view&lt;/a&gt;, the homotopy groups are really just a convenient stand-in for a more fundamental structure, the &lt;a href=&quot;http://ncatlab.org/nlab/show/fundamental+infinity-groupoid&quot; rel=&quot;nofollow&quot;&gt;fundamental $\infty$-groupoid&lt;/a&gt; of a space. Roughly speaking, the fundamental $\infty$-groupoid is a gadget that incorporates information about the paths between points, homotopies between paths, homotopies between homotopies between paths, and so forth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is possible to truncate the fundamental $\infty$-groupoid into a collection of easier-to-understand objects, the fundamental $n$-groupoids $\Pi_n$:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The fundamental $0$-groupoid $\Pi_0$ is just the set of connected components. &lt;/li&gt;&#xA;&lt;li&gt;The fundamental $1$-groupoid is the &lt;a href=&quot;http://ncatlab.org/nlab/show/groupoid&quot; rel=&quot;nofollow&quot;&gt;groupoid&lt;/a&gt; of homotopy classes of paths between points; it is a generalization of the fundamental group that is independent of basepoint. If the space is connected, the fundamental $1$-groupoid is equivalent to the category with a single object whose morphisms are the elements of the fundamental group $\pi_1$.&lt;/li&gt;&#xA;&lt;li&gt;The fundamental $2$-groupoid is the &lt;a href=&quot;http://nlab.mathforge.org/nlab/show/2-groupoid&quot; rel=&quot;nofollow&quot;&gt;$2$-groupoid&lt;/a&gt; of paths and homotopy classes of homotopies between them; it is a generalization of the action of $\pi_1$ on $\pi_2$ that is independent of basepoint. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;And so forth: more generally the fundamental $n$-groupoid is a generalization of the relationship between the first $n$ homotopy groups. Unfortunately I can't think of a nice reference to these ideas off the top of my head; I've gleaned them from several sources. The references in Baez and Shulman's &lt;a href=&quot;http://arxiv.org/abs/math/0608420&quot; rel=&quot;nofollow&quot;&gt;Lectures on n-categories and cohomology&lt;/a&gt; might be a good start. &lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2011-03-04T14:48:18.617" CommentCount="2" />
  <row Id="25021" PostTypeId="2" ParentId="25016" CreationDate="2011-03-04T14:56:11.743" Score="2" ViewCount="" Body="&lt;ol&gt;&#xA;&lt;li&gt;Write it as $H^3 + qH + p = 0$;  &lt;/li&gt;&#xA;&lt;li&gt;use &lt;a href=&quot;http://www.wolframalpha.com/input/?i=solve+h%5E3%2Bq%2ah%2Bp%3D0+for+h&quot; rel=&quot;nofollow&quot;&gt;Wolfram Alpha&lt;/a&gt; or &lt;a href=&quot;http://en.wikipedia.org/wiki/Cubic_equation&quot; rel=&quot;nofollow&quot;&gt;formulas for roots&lt;/a&gt;;  &lt;/li&gt;&#xA;&lt;li&gt;plug in for &lt;em&gt;q&lt;/em&gt; and &lt;em&gt;p&lt;/em&gt;.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="4929" LastEditorUserId="10557" LastEditorDisplayName="" LastEditDate="2011-05-09T01:58:48.810" LastActivityDate="2011-05-09T01:58:48.810" CommentCount="3" />
  <row Id="25023" PostTypeId="2" ParentId="24976" CreationDate="2011-03-04T14:58:50.747" Score="7" ViewCount="" Body="&lt;p&gt;Define $\rm\ (n+k)_k\ :=\ (n+k)!/n!\: =\: (n+k)\ \cdots\ (n+1)\ \equiv\ k!\ \ (mod\ n)\:.\ $ Then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\rm \frac{D}{n!^3}\ \ =\ \ \begin{vmatrix}&#xD;&#xA;1 &amp;amp; n+1 &amp;amp; (n+2)_2 \\ &#xD;&#xA;n+1 &amp;amp; (n+2)_2 &amp;amp; (n+3)_3 \\ &#xD;&#xA;(n+2)_2 &amp;amp; (n+3)_3 &amp;amp; (n+4)_4&#xD;&#xA;\end{vmatrix}\ \ \equiv\ \ \begin{vmatrix}&#xD;&#xA;1 &amp;amp; 1 &amp;amp; 2! \\ &#xD;&#xA;1 &amp;amp; 2! &amp;amp; 3! \\ &#xD;&#xA;2! &amp;amp; 3! &amp;amp; 4!%&#xD;&#xA;\end{vmatrix}\ \ \equiv\ \ 4\ \ \ (mod\ n) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that $\rm\:det(a_{i\:j})\ \equiv\ det(a_{i\:j}\: mod\ n)\ \ (mod\ n)\ $ because $\rm\:det\:$ is a polynomial in the $\rm\:a_{i\:j}\:.$ Hence it's just a special case of $\rm\ f\:(g_1(n),\cdots,g_k(n))\ \equiv\ f\:(g_1(0),\cdots,g_k(0))\ \ (mod\ n)\ $ for polynomials $\rm\:f,\:g_i\:.$ Note how performing such modular reductions greatly simplifies the arithmetical calculations.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2011-03-06T15:46:16.117" LastActivityDate="2011-03-06T15:46:16.117" CommentCount="5" />
  <row Id="25024" PostTypeId="1" AcceptedAnswerId="25030" CreationDate="2011-03-04T15:07:40.160" Score="2" ViewCount="174" Body="&lt;p&gt;Given two polynomials $A = \sum_{0\le k&amp;lt;n} a_k x^k$ and $B =\sum_{0\le k&amp;lt;n} b_k x^k$ of the same degree $n$, which are equal for all $x$, is it always true that $\ a_k = b_k\ $ for all $0\le k&amp;lt;n?$. All Coefficients and $x$ are complex numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt;&#xA;Sorry, formulated the question wrong.&lt;/p&gt;&#xA;" OwnerUserId="5282" LastEditorUserId="5282" LastEditorDisplayName="" LastEditDate="2011-03-05T12:20:15.883" LastActivityDate="2011-03-05T12:20:15.883" Title="Equality of polynomials: formal vs. functional" Tags="&lt;abstract-algebra&gt;&lt;polynomials&gt;" AnswerCount="3" CommentCount="5" />
  <row Id="25025" PostTypeId="2" ParentId="25024" CreationDate="2011-03-04T15:13:43.867" Score="-1" ViewCount="" Body="&lt;p&gt;You just said A=B for all x. x is the only variable, so yes.&lt;/p&gt;&#xA;" OwnerUserId="4929" LastActivityDate="2011-03-04T15:13:43.867" CommentCount="1" />
  <row Id="25026" PostTypeId="2" ParentId="25024" CreationDate="2011-03-04T15:17:55.317" Score="7" ViewCount="" Body="&lt;p&gt;The answer is in general no. If the ground field is infinite,then it is true. In general it is not TRUE. In the polynomial algebra ${\mathbb{Z}/2\mathbb{Z}}[X]$ consider the polynomials $X^2$ and $X$. But they are different in ${\mathbb{Z}/2\mathbb{Z}}[X]$.&lt;/p&gt;&#xA;" OwnerUserId="7565" LastActivityDate="2011-03-04T15:17:55.317" CommentCount="7" />
  <row Id="25027" PostTypeId="1" CreationDate="2011-03-04T15:19:32.490" Score="1" ViewCount="128" Body="&lt;p&gt;Consider:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;if $\log u$ and $\log v$ are both subharmonic function on an open set $U$, then $\log(u+v)$ is also subharmonic on $U$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;How can I prove this?&lt;/p&gt;&#xA;" OwnerUserId="7428" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-04T16:11:36.373" LastActivityDate="2011-05-15T17:36:00.240" Title="Subharmonic functions" Tags="&lt;complex-analysis&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="25028" PostTypeId="2" ParentId="24901" CreationDate="2011-03-04T15:21:11.170" Score="2" ViewCount="" Body="&lt;p&gt;Use one further change of variable. Define $g(x) = 1/f(x)$. Then $g(x)$ is a strictly decreasing continuous function mapping $[1,\infty)$ to $(0,c]$. Without loss of generality we can take $c = 1$. Observe that $f^{-1}(x) = g^{-1}(1/x)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So it is enough to show that $\sum g(k)$ converges iff $\sum \frac{1}{k^2}g^{-1}(\frac{1}{k})$. (Note that here if you use the integral test you do not need to use $g$ is differentiable). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Observe that the partial sums and partial integrals have the following relation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \sum_{k = n+1}^{N+1} g(k) \leq \int_n^N g(x) dx \leq \sum_{k=n}^{N} g(k) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using monotonicity of $g$. Hence $\sum g(k)$ converges IFF the integral converges. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now look at the following (the leading 1 comes from the assumption that $g(1) = 1$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ 1 + \int_1^\infty g(x) dx = \int_0^1 g^{-1}(y) dy = \int_1^{\infty} \frac{g^{-1}(\frac1y)}{y^2} dy $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is the integral version of what we want to prove. But in any case, consider the partial Riemann sums&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \sum_{k = n}^N \left(\frac{1}{k} - \frac{1}{k+1}\right)g^{-1}(\frac{1}{k}) \leq \int_{N^{-1}}^{n^{-1}} g^{-1}(y) dy \leq \sum_{k = n}^N \left(\frac{1}{k} - \frac{1}{k+1}\right)g^{-1}(\frac{1}{k+1}) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;using now the monotonicity of $g^{-1}(y)$. Now notice that $\frac{1}k - \frac{1}{k+1} \sim \frac{1}{k^2}$ (in the sense that one can be controlled by the other by constant factors), you have that the partial sum&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \sum_{k = n}^N \frac{g^{-1}(\frac{1}{k})}{k^2} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is comparable to &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \int_{N^{-1}}^{n^{-1}} g^{-1}(y) dy $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and hence their convergences imply each other. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Chain everything together you have the desired result. &lt;/p&gt;&#xA;" OwnerUserId="1543" LastActivityDate="2011-03-04T15:21:11.170" />
  <row Id="25029" PostTypeId="2" ParentId="25003" CreationDate="2011-03-04T15:33:08.627" Score="1" ViewCount="" Body="&lt;p&gt;Well first of all, if you're making the substitution $x_i^2 = x_i$ for all $i$, then it had better be the case that the only values you are interested in (and so are substituting in) are $x_i = 0$ and $x_i=1$, since these are the only solutions to this equation.  So what you're saying is that you're interested in the value of the product $p_1p_2$ as a function of $x_k\in\{0,1\}$ when values in $\{0,1\}$ are substituted in for all the other variables.  So just compute $p_1$ and $p_2$ with the given values for indices $i\neq k$ and both values of $x_k$.  Multiply the resulting values for the two polynomials when $x_k = 0$ to get the value $p(0)$ and multiply the values for the two polynomials when $x_k =1$ to get the value $p(1)$.  Then $b = p(0)$ and $a = p(1) - p(0)$.  This will reduce your work from evaluating $10^{10}$ terms to evaluating $4\times 10^5$ terms.&lt;/p&gt;&#xA;" OwnerUserId="884" LastActivityDate="2011-03-04T15:33:08.627" CommentCount="3" />
  <row Id="25030" PostTypeId="2" ParentId="25024" CreationDate="2011-03-04T15:42:47.493" Score="7" ViewCount="" Body="&lt;p&gt;For $\rm\ f = A-B\in R[x]\:,\:$ it is equivalent to ask if $\rm\ f(r) = 0\ $ for all $\rm\: r\in R\ \Rightarrow\ f = 0\:,\: $ i.e. if $\rm\:f\ $ is zero as a function then is $\rm\:f\ $ zero as a formal polynomial, i.e. are all its coefficients zero? This is true if $\rm\:R\:$ is an integral domain of cardinality greater than the degree of $\rm\:f\:,\:$ e.g. if $\rm|R|$ is infinite, but it may fail otherwise, e.g. $\rm\ x^p = x\ $ for all $\rm\: x\in \mathbb Z/p\ $ by Fermat's little theorem, but $\rm\ x^p \ne x\ $ in $\rm\: \mathbb Z/p\:[x]\:.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Remark&lt;/strong&gt;  $\ $ In fact a ring $\rm\: D\:$ is a domain $\iff$ every nonzero polynomial $\rm\ f(x)\in D[x]\ $ has at most $\rm\ deg\ f\ $ roots in $\rm\:D\:.\:$ For the simple proof see my &lt;a href=&quot;http://math.stackexchange.com/questions/4597/simple-explanation-and-examples-of-the-miller-rabin-primality-test/4604#4604&quot;&gt;post here&lt;/a&gt;, where I illustrate it constructively in $\rm\: \mathbb Z/m\: $ by showing that, $\:$ given any $\rm\:f(x)\:$ with more roots than its degree,$\:$ we can quickly compute a nontrivial factor of $\rm\:m\:$ via a $\rm\:gcd\:$. The quadratic case of this result is at the heart of many integer factorization algorithms, which try to factor $\rm\:m\:$ by searching for a nontrivial square root in $\rm\: \mathbb Z/m\:,\:$ e.g. a square root of $1$ that is not $\:\pm 1$.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2011-03-04T15:51:13.447" LastActivityDate="2011-03-04T15:51:13.447" />
  <row Id="25031" PostTypeId="1" CreationDate="2011-03-04T15:44:13.313" Score="1" ViewCount="84" Body="&lt;p&gt;Let $U\subseteq \mathbb{C}$ be an open set and let $u(z)=-\log(\mathrm{dist}(z,\partial U))$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need to show that $u$ is subharmonic on $U$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\partial U$ it means the boundary of $U$.&lt;/p&gt;&#xA;" OwnerUserId="7428" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-04T16:24:44.880" LastActivityDate="2011-03-04T17:26:23.880" Title="subharmonic functions" Tags="&lt;analysis&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="25032" PostTypeId="1" CreationDate="2011-03-04T15:56:38.463" Score="5" ViewCount="386" Body="&lt;p&gt;Hi, I am currently a math major in college and my main problem is that it feels directionless. My college offers little in term of variety in undergraduate math so I moved on into taking graduate courses and I am actually loving it. But at this point I am just randomly taking courses that looks interesting and was wondering if there was specific directions to take within math. And even past that where will it lead.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;&lt;br&gt;&#xA;After reading the comments I decided to add extra info. On the courses that I like, I have taken Abstract Algebra and Linear Algebra and really enjoyed them and also Number theory. I have also taken the typical three semester Calculus courses, Differential Equations, Discrete Math, Probability, Numerical Analysis and Real Analysis. On these courses I like them though I am not as enthusiastic about them as I am for the other courses I have mentioned. Though I find their applications very interesting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And more specifically on the question it is on what paths are there to take in school and also what paths are there to take after graduation.&lt;/p&gt;&#xA;" OwnerUserId="7711" LastEditorUserId="7711" LastEditorDisplayName="" LastEditDate="2011-03-08T21:46:55.743" LastActivityDate="2011-03-09T00:56:54.727" Title="Where to go with Mathematics?" Tags="&lt;soft-question&gt;" AnswerCount="5" CommentCount="10" FavoriteCount="2" />
  <row Id="25033" PostTypeId="2" ParentId="24999" CreationDate="2011-03-04T16:23:16.223" Score="5" ViewCount="" Body="&lt;p&gt;Okay, this got a bit mangled. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) Is the set $\mathbf{V}=\{A\mid A\text{ is an }n\times n\text{ matrix and }\det(A)=0\}$ a vector space, under the usual addition and scalar multiplication of matrices?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the set of &lt;em&gt;all&lt;/em&gt; $n\times n$ matrices is a vector space, the question is really whether this is a subspace (all the axioms of a vector space will necessarily hold, except perhaps for the existence of a zero vector, the existence of inverses, and the &quot;hidden&quot; axioms that the set must be closed under vector addition and scalar multiplication: the sum of two vectors in $\mathbf{V}$ must lie in $\mathbf{V}$, and every scalar multiple of a vector in $\mathbf{V}$ lies in $\mathbf{V}). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Scalar multiplication is easy: if $\alpha\in\mathbb{R}$ and $A$ is any $n\times n$ matrix, then we know that $\det(\alpha A) = \alpha^n \det(A)$. So $\det(\alpha A)=0$ if and only if $\alpha = 0$ or $\det(A)=0$. So, if $A\in\mathbf{V}$, then $\alpha A\in\mathbf{V}$ for all scalars $\alpha$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What about vector addition? We would need to show that if $A$ and $B$ both have nonzero determinant, then so does $A+B$. But this is not the case, as Agustí Roig points out. It should be easy to come up with a similar example for any $n\gt 0$. So $\mathbf{V}$ is not closed under addition (exhibit an &lt;em&gt;explicit&lt;/em&gt; pair of matrices, both in $\mathbf{V}$, but whose sum is not in $\mathbf{V}$; &lt;em&gt;sometimes&lt;/em&gt; the sum of two matrices in $\mathbf{V}$ is in $\mathbf{V}$, the point is that it doesn't &lt;em&gt;always&lt;/em&gt; lie in $\mathbf{V}$, so give an example!).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2). Again, since all real valued functions form a vector space, the only issue is whether the set $\mathbf{V} = \{ f(x)\mid f(x) = (ax+b)e^{-x},\ a,b\in\mathbb{R}\}$ is closed under sums and scalar multiplication.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose $f(x),g(x)\in\mathbf{V}$. Will $f(x)+g(x)$ lie in $\mathbf{V}$ as well? Write&#xA;\begin{align*}&#xA;f(x) &amp;amp;= (ax+b)e^{-x}\\&#xA;g(x) &amp;amp;= (cx+d)e^{-x}\\&#xA;f(x)+g(x) &amp;amp;= \Bigl( (ax+b)e^{-x}\Bigr) + \Bigl( (cx+d)e^{-x}\Bigr)\\&#xA;&amp;amp;= \Bigl( (ax+b) + (cx+d)\Bigr) e^{-x}\\&#xA;&amp;amp;= \Bigl( (a+c)x + (b+d)\Bigr) e^{-x}.&#xA;\end{align*}&#xA;So setting $A=a+c$ and $B=b+d$, which are real numbers because all of $a,b,c,d$ are real numbers, then we see that we can write $f(x)+g(x)$ in the form $(Ax+B)e^{-x}$. So if each of $f(x)$ and $g(x)$ are in $\mathbf{V}$, then $f(x)+g(x)\in\mathbf{V}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now you need to show that if $f(x)\in\mathbf{V}$ and $\alpha\in\mathbb{R}$, then $\alpha f(x)\in\mathbf{V}$. I'll leave that to you to do.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What about the dimension of $\mathbf{V}$? If you want to describe an element of $\mathbf{V}$, you really only need to specify two things: the value of $a$ and the value of $b$. that suggests that the dimension will be $2$. Can you find a set of $2$ linearly independent functions, both in $\mathbf{V}$, that span $\mathbf{V}$?&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2011-03-04T16:23:16.223" />
  <row Id="25034" PostTypeId="1" AcceptedAnswerId="25035" CreationDate="2011-03-04T16:35:03.547" Score="2" ViewCount="133" Body="&lt;p&gt;I understand how to do differentiation when the elements are individual functions, but am having trouble applying the concepts with nested functions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, given $(\cos \pi x + \sin \pi y)^5= 24$&#xA;$$\frac{d}{dx}\left[\frac{d}{dx}(\cos \pi x)+\frac{d}{dx} (\sin \pi y)\right]^5=\frac{d}{dx}24$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Following the chain rule of ${f}&amp;#39;(g(x))\cdot {g}&amp;#39;(x), $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$5[(\cos(\pi x) +\sin (\pi y)]^4 \cdot [-\pi \sin(\pi x) + \pi \cos(\pi y)]=0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's where I get stuck.  First, as I see it, I have a few options for the next step:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;multiply ${g}&amp;#39;(x)$ by 5&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;multiply $g(x)$ by 4&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;What about after that?  How do I separate the contents inside the power block to move the $x$ functions to the other side of the equation?&lt;/p&gt;&#xA;" OwnerUserId="7006" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-04T16:52:20.037" LastActivityDate="2011-03-05T17:06:19.670" Title="Help with implicit differentation" Tags="&lt;calculus&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" />
  <row Id="25035" PostTypeId="2" ParentId="25034" CreationDate="2011-03-04T16:42:59.687" Score="1" ViewCount="" Body="&lt;p&gt;Jason.  You should not be solving for $y$, you should be solving for $\frac{dy}{dx}$.  There was an error in your calculation, however, when you applied the chain rule to find $\frac{d}{dx} (\sin (\pi y))$.  It should be:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;\frac{d}{dx} \left( \sin ( \pi y) \right) = \cos(\pi y) \frac{d}{dx} (\pi y) = \cos (\pi y) \pi \frac{dy}{dx},&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;since we are viewing $y$ as a function of $x$.  Try now to solve for $\frac{dy}{dx}$ with what you've been given here.&lt;/p&gt;&#xA;" OwnerUserId="6491" LastEditorUserId="6296" LastEditorDisplayName="" LastEditDate="2011-03-04T18:21:54.857" LastActivityDate="2011-03-04T18:21:54.857" CommentCount="1" />
  <row Id="25036" PostTypeId="2" ParentId="24901" CreationDate="2011-03-04T16:43:40.043" Score="3" ViewCount="" Body="&lt;p&gt;Define $A(n)$ to be $\{k: 2^n \leq f(k) &amp;lt; 2^{n+1}\}$, and $|A(n)|$ its cardinality. Let $S_n$ be the sum of the terms of the first sequence over $k$ in $A_n$. Then we have&#xA;$${1 \over 2^{n+1}}|A(n)| &amp;lt; S_n \leq {1 \over 2^n}|A(n)|$$&#xA;Thus the first sum $\sum_n S_n$ is within a factor of $2$ of ${\displaystyle \sum_n {|A(n)| \over 2^n}}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next, let $T_n$ be the sum of the terms of the second sequence over $2^n \leq k &amp;lt; 2^{n+1}$. In the sum $T_n$, since $f^{-1}$ is increasing, $f^{-1}(k)$ is less than $f^{-1}(2^{n+1}) = \sum_{m \leq n+1}|A(m)|$, and at least $f^{-1}(2^n) = \sum_{m \leq n}|A(m)|$. There are $2^n$ terms in the sum defining $T_n$, so we have&lt;br&gt;&#xA;$$2^{-2(n+1)}2^n\sum_{m \leq n}|A(m)| \leq T_n &amp;lt; 2^{-2n}2^n\sum_{m \leq n+1}|A(m)|$$&#xA;Summing the above in $n$, we have that $\sum_n T_n$ is within a constant factor of &#xA;$$\sum_n 2^{-n} \sum_{m \leq n} |A(m)|$$&#xA;Switching the order of summation, this becomes&#xA;$$\sum_m |A(m)| \sum_{n \geq m} 2^{-n}$$&#xA;$$= 2\sum_m{|A(m)| \over 2^m}$$&#xA;This is within a constant factor of $\sum_n {S_n}$ computed above, so one series converges iff the other one does.&lt;/p&gt;&#xA;" OwnerUserId="3035" LastActivityDate="2011-03-04T16:43:40.043" />
  <row Id="25038" PostTypeId="2" ParentId="24507" CreationDate="2011-03-04T16:57:44.667" Score="2" ViewCount="" Body="&lt;p&gt;The significance of Russel's paradox is not just philosophical. Russel's paradox implies a contradiction in the presence of the unbounded axiom of comprehension:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\exists s. \forall e. e\in s \leftrightarrow \phi$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\phi$ is the logical formula typically containing $e$. This axiom “creates” the set $s$ of all elements $e$ satisfying $\phi$. In other words, it turns properties into sets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $r$ be the set of elements $e$ satisfying $e\notin e$. Then $r$ satisfies: &#xA;$\forall e. e\in r \leftrightarrow e\notin e$. Instantiate $e$ to $r$: $r\in r \leftrightarrow r\notin r$. $p:=r\in r$ for clarity. Use $(p\leftrightarrow (\neg p))\rightarrow \bot$ valid in intuitionistic propositional logic. We've got a contradiction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you suspect there is a gap in my proof, I &lt;a href=&quot;http://beroal.livejournal.com/9260.html&quot; rel=&quot;nofollow&quot;&gt;formalized it in Coq&lt;/a&gt;, but the accompanying text is mostly in Russian. To prove a contradiction, I need  building blocks, which I introduced via “Variable” — the universe “all_sets”, the relation “in_set”, and our honey “comprehension”.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S. Contradictions in pure &lt;a href=&quot;http://en.wikipedia.org/wiki/Coq&quot; rel=&quot;nofollow&quot;&gt;Coq&lt;/a&gt; are not found. ;)&lt;/p&gt;&#xA;" OwnerUserId="7011" LastActivityDate="2011-03-04T16:57:44.667" CommentCount="1" />
  <row Id="25039" PostTypeId="2" ParentId="25032" CreationDate="2011-03-04T17:13:12.370" Score="0" ViewCount="" Body="&lt;p&gt;There are definitely particular directions you can take.  I can see several ways in which this could play out.  You might come across a mathematician whose style of mathematics is really interesting to you, you read all his papers and then you set yourself the goal of learning what you need to know to get into that area.  You might become impressed with a particular applied problem like modeling climate change or prediction of the stock market and then you try to learn all the mathematics you need to effectively work on that problem.  You could start by reading what people have written on the topic and learning enough to understand the approaches they are all using.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since you seem to have broad interests, try to get involved in some undergraduate research project that seems interesting.  If there is nothing in the mathematics department at your school, you can try other schools in your area or the Physics, Chemistry, Computer Science or Economics department.  It would be good for your CV and it might have the side benefit of helping you get a sense of what it feels like to be more focused.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A life in research is pretty boring in the abstract.  You pick some really small sub-problem that very few people are interested in.  You work at it and make marginal successes that might have applications 100 years after you are dead.  On the other hand, it's fun for some people and its a clean, safe, low-intensity way to make a living.&lt;/p&gt;&#xA;" OwnerUserId="1004" LastActivityDate="2011-03-04T17:13:12.370" />
  <row Id="25040" PostTypeId="2" ParentId="25031" CreationDate="2011-03-04T17:13:56.690" Score="1" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Hint:&lt;/strong&gt; For fixed $x\in\partial U$, the function $-\log(|z-x|)$ is harmonic on $U$.&lt;/p&gt;&#xA;" OwnerUserId="940" LastActivityDate="2011-03-04T17:13:56.690" CommentCount="2" />
  <row Id="25041" PostTypeId="2" ParentId="24997" CreationDate="2011-03-04T17:24:24.237" Score="1" ViewCount="" Body="&lt;p&gt;Yes.  A simple example is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\max x_1 + x_2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;subject to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ x_1 + x_2 \leq 1,$$&#xA;$$ x_1, x_2 \geq 0.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The level curves of the objective function are parallel to the boundary of the constraint $x_1 + x_2 \leq 1$, and so every point on the line segment $x_1 + x_2 = 1$ with $x_1, x_2 \geq 0$ is optimal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This example gives you the general case: You can have multiple, nondegenerate optimal solutions if the level surfaces of the objective function are parallel to one of the constraint inequalities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's an interesting connection with the dual problem, too.  If the original problem has multiple, nondegenerate optimal solutions, then the dual problem must have a unique, degenerate optimal solution.&lt;/p&gt;&#xA;" OwnerUserId="2370" LastActivityDate="2011-03-04T17:24:24.237" />
  <row Id="25042" PostTypeId="2" ParentId="25031" CreationDate="2011-03-04T17:26:23.880" Score="0" ViewCount="" Body="&lt;p&gt;Suppose $U$ is connected open. Fix a $z\in U.$ Since $U$ is open, there is an $s&amp;gt;0$ such that $B(z,s)\subset U.$ Now, $\frac{1}{|\partial B(z,s)|}\int_{\partial B(z,s)}dist(w,\partial U) \leq dist(w,\partial U).$ Now, use Jensen inequality to finish the proof. &#xA;Above argument shows that your function, say f(w) which is defined in the interior of U is subharmonic. You should also discuss, for instance, the riemann integrability of $dist(w,\partial U)$ on the boundary set $\partial B(z,r)$ if you want a complete solution.   &lt;/p&gt;&#xA;" OwnerUserId="7796" LastActivityDate="2011-03-04T17:26:23.880" CommentCount="1" />
  <row Id="25043" PostTypeId="1" CreationDate="2011-03-04T18:15:16.210" Score="0" ViewCount="80" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Possible Duplicate:&lt;/strong&gt;&lt;br&gt;&#xA;  &lt;a href=&quot;http://math.stackexchange.com/questions/20291/matrix-analysis-in-mathbbc&quot;&gt;Matrix Analysis in $\mathbb{C}$&lt;/a&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Let $A\in \mathrm{GL}_n(\mathbb{C})$. Show that $$\det(I+A)=1+tr(A)+\epsilon(A)$$ where Modulus of $\epsilon(A)$ by norm of $A=0$ as $A$ tends to $0$, for any matrix norm.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If I define $J(A)= \det(A)$ for $A$ is in $\mathrm{GL}_n(\mathbb{C})$, then $J$ is differentiable at all such $A$ and that, if $H$ is in $M_n(\mathbb{C})$,then $J&amp;#39;(A)(H)=\det(A)tr(A^*H)$. where $A^*$ is inverse of $A$.&lt;/p&gt;&#xA;" OwnerUserId="7799" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-04T19:07:00.590" LastActivityDate="2011-03-04T19:07:00.590" ClosedDate="2011-03-04T19:09:58.413" Title="matrix analysis in $\mathbb{C}$" Tags="&lt;homework&gt;" CommentCount="1" />
  <row Id="25045" PostTypeId="1" AcceptedAnswerId="25048" CreationDate="2011-03-04T18:20:06.783" Score="1" ViewCount="127" Body="&lt;p&gt;I have a question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could you please help me to solve this?&#xA;Thanks in advance&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $a$ be a finitely generated ideal of $A$, commutative ring with identity, such that $a^2 = a$. Show that $a \subseteq Ae$ with $e^2 = e$, and $e$ is not identity.&lt;/p&gt;&#xA;" OwnerUserId="1252" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-04T19:12:04.500" LastActivityDate="2011-03-28T16:06:02.797" Title="problem on finitely generated ideal " Tags="&lt;abstract-algebra&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" />
  <row Id="25046" PostTypeId="1" CreationDate="2011-03-04T18:42:03.623" Score="0" ViewCount="103" Body="&lt;p&gt;For the following argument, how can I produce an adequate translation, including interpretations function and formal sentence where universe $\mathcal{U} = \{\bigcirc\mid\bigcirc\text{ is a set}\}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And we can probably use:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;M1: and o is a master set&#xA;e: elements &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1.) Any master set has as elements all and only set which are not elements of themselves. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2.) There are no master sets&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and we can probabloy use:&#xA;M1: and o is a master set&#xA;e: elements &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Am I on the right track?&lt;/p&gt;&#xA;" OwnerUserId="7779" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-04T19:14:38.663" LastActivityDate="2011-03-04T19:53:23.660" Title="how to translate a sentence into a quantifier?" Tags="&lt;logic&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="25047" PostTypeId="1" AcceptedAnswerId="25062" CreationDate="2011-03-04T18:48:33.957" Score="4" ViewCount="271" Body="&lt;p&gt;Hello,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a programming problem that I'm trying to solve at TopCoder arena. However, I feel like it need Mathematics to prove its correctness.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Problem&lt;/strong&gt;&lt;br&gt;&#xA;Given a list $L$ of $n$ elements, and a number $k, k \leq n$, where $k$ defined as a number consecutive elements can be reversed. For example:&#xA;$L = \{ 4, 3, 2, 1 \}$ and $k = 3$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;reverse at $0$ yield a new list $L = \{ 2, 3, 4, 1 \}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So our goal is to sort the list in ascending order by reversing $k$ element each time if possible.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is, what condition must be hold in order to have an impossible case? And how to find the fewest moves in term of $n$ and $k$?&#xA;Here are few examples:  &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;{1,2,3}  &#xA;3  &#xA;Fewest move is 0&#xA;&#xA;{3,2,1}&#xA;3&#xA;Fewest move is 1&#xA;&#xA;{5,4,3,2,1}&#xA;2&#xA;Fewest move is 10&#xA;&#xA;{3,2,4,1,5}&#xA;4&#xA;This is impossible case&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Thanks,&lt;/p&gt;&#xA;" OwnerUserId="5711" LastEditorUserId="5711" LastEditorDisplayName="" LastEditDate="2011-03-07T07:30:10.537" LastActivityDate="2011-03-07T13:49:04.500" Title="Find the fewest moves to sort a list" Tags="&lt;elementary-number-theory&gt;" AnswerCount="2" CommentCount="8" />
  <row Id="25048" PostTypeId="2" ParentId="25045" CreationDate="2011-03-04T18:50:47.373" Score="2" ViewCount="" Body="&lt;p&gt;What follows is partially excerpted from my &lt;a href=&quot;http://mathforum.org/kb/thread.jspa?threadID=1111328&amp;amp;messageID=3648504&quot; rel=&quot;nofollow&quot;&gt;Historia Matematica post of 2005\01\04&lt;/a&gt; on this topic, in reply to questions by Colin McLarty and Martin Davis. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Below is a proof by Gilmer from a &lt;a href=&quot;http://www.jstor.org/pss/2316741&quot; rel=&quot;nofollow&quot;&gt;Monthly Classroom Note[1]&lt;/a&gt;.&#xA;Gilmer mentions it's also presented on p.58 of his &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0824712420&quot; rel=&quot;nofollow&quot;&gt;textbook[2]&lt;/a&gt;.$\:$&#xA;Remark that the use of Cramer's rule in Gilmer's proof is simply a special&#xA;case of the deduction of an equation of integral dependence over&#xA;an ideal (vs. ring), see Kaplansky, &lt;em&gt;Commutative Rings,&lt;/em&gt; p.11 Exer.1.&#xA;or see his later Theorem 75, viz.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;THEOREM&lt;/strong&gt; $\rm\ 75.\ \ $ Let $\rm\:R\:$ be a ring, $\rm\:J\:$ an ideal in $\rm\:R,\ B\:$ an $\rm\:R$-module generated by $\rm\:n\:$ elements,$\ \ $  and $\rm\:r\:$ an element of $\rm\:R\:$ satisfying $\rm\ r\:B \subset JB\:.\ $ Then $\rm\ (r^n - j)\ B = 0\ $ for some $\rm\ j \in J\:.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The desired proof now follows immediately, namely:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Specializing $\rm\ r=1,\ B=J\ $ yields $\rm\ (1-j)\ J = 0\ \Rightarrow\ J = (j),\ \ j^2 = j\:.\quad$ &lt;strong&gt;QED&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that this may be viewed as a generalization of the simpler &lt;a href=&quot;http://math.stackexchange.com/q/29454/242&quot;&gt;Dedekind domain&lt;/a&gt; case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The above proof doesn't work in the noncommutative case &#xA;because the &lt;em&gt;determinant trick&lt;/em&gt; no longer applies. However&#xA;one can prove Theorem 75 without using determinants by&#xA;instead appealing to Nakayama's Lemma. Namely, see Exercises 3.1, 3.2, p.43 in&#xA;Atiyah and Macdonald, &lt;em&gt;Introduction to Commutative Algebra&lt;/em&gt;. Below is said direct proof of Gilmer - using Cramer's rule.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;LEMMA&lt;/strong&gt;  $\ $ If $\rm\:B\:$ is a finitely generated idempotent ideal of a commutative&#xA;ring $\rm\:T\:$ then $\rm\:B\:$ is principal and is generated by an idempotent element.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof&lt;/strong&gt; $\ $ We first assume that $\rm\:T\:$ has an identity and we let $\rm\:\{b_{\:i}\:\}\:$ be a finite set of generators for $\rm\:B\:.\:$ Then $\rm\:B = B^2 = \sum\: B\ b_{\:i}\:$ so that we obtain a system of equations  $\rm\ b_k = \sum s_{\:k\:i}\ b_{\:i}\:,\ $ where $\rm\:s_{\:k\:i} \in B\:.\ $ This gives rise to a system of equations $\rm\ \sum\ (\delta_{\:k\:i}-s_{\:k\:i})\ b_{\:i} = 0\:,\:$ where $\delta =$  Kronecker delta.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By Cramer's rule $\rm\:d\ b_i = 0\:$ for all $\rm\:i\:,\:$ where $\rm\:d\:$ is the determinant of the matrix $\rm\ [\delta_{ki}-s_{ki}]\:.\ $ It is easy to see that $\rm\:d\:$ has the form $\rm\: 1-b\: $ for some $\rm\:b \in B\:.\ $ Since $\rm\ 0 = d\ b_i = b_i - b\ b_i\ $ for each $\rm\:i,\ B\ $ is the principal ideal generated by $\rm\:b\:.\ $ And since $\rm\:1-b\:$ kills $\rm\:B\:,\:$ we conclude $\rm\ (1-b)\ b = 0\:,\ $ or $\rm\ b = b^2\:.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $\rm\:T\:$ contains no identity element, we consider a commutative ring $\rm\:T&amp;#39;\:$ obtained by adjoining an identity element to $\rm\:T\:.\ $ Then $\rm\:B\:$ is a finitely&#xA;generated idempotent ideal of $\rm\:T&amp;#39;,\:$ and hence is principal as an ideal of $\rm\:T&amp;#39;,\:$ generated by an idempotent element $\rm\:v\:.\ $ Since $\rm\:T&amp;#39;$ is obtained by adjoining an identity element to $\rm\:T\:,\:$ it follows $\rm\:v\:$ also generates $\rm\:B\:$ as an ideal of $\rm\:T\:.\quad$ &lt;strong&gt;QED&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[1] Robert Gilmer, &lt;a href=&quot;http://www.jstor.org/pss/2316741&quot; rel=&quot;nofollow&quot;&gt;An Existence Theorem for Non-Noetherian Rings&#xA;(in Classroom Notes),&lt;/a&gt;&lt;br&gt;&#xA;The American Mathematical Monthly, Vol. 77, No. 6, pp. 621-623.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;[2] Robert Gilmer, &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0824712420&quot; rel=&quot;nofollow&quot;&gt;Multiplicative Ideal Theory,&lt;/a&gt;&lt;br&gt;&#xA;Queens University, Kingston, Ontario, 1968.  See p.58&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2011-03-28T16:06:02.797" LastActivityDate="2011-03-28T16:06:02.797" />
  <row Id="25049" PostTypeId="1" AcceptedAnswerId="25052" CreationDate="2011-03-04T18:53:17.153" Score="0" ViewCount="127" Body="&lt;p&gt;Alice released a new version of her software for removing red-eye from pictures. However, she wants to protect her secret algorithm from disassemblers and such while still letting her customers remove red-eye from pictures. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;She looks at standard encryption:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;A message is encrypted with a key and sent across the wire&lt;/li&gt;&#xA;&lt;li&gt;If some nefarious onlooker is dropping eaves on the network, all he gets is a garbled mess of bits.&lt;/li&gt;&#xA;&lt;li&gt;When it arrives, it's decrypted by someone. There are a variety of ways to distribute keys, etc. but the person you send your message to should be able to decrypt it somehow in all of them.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;And wonders if something like this might be possible:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The user wants to remove red-eye from his family photo. He 'encrypts' the image with a key of some sort.&lt;/li&gt;&#xA;&lt;li&gt;The software contains an 'encrypted algorithm' that operates only on 'encrypted data'. If you try to disassemble it and see what it's doing, it's like trying to read encrypted bits - it's garbled and you get no information from it.&lt;/li&gt;&#xA;&lt;li&gt;When the image is done, the client 'decrypts' his image. You might signal that the image is done by seeing if a bit is set somewhere, by trying to decrypt it every iteration and seeing if this is valid, or even having the last instruction in the 'encrypted' data-space decrypt it into regular data again.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;You can probably think of a few variations: Maybe the client has to request a public key from a website somewhere and then ask Alice's website to decrypt it with her private key. But the important properties are that&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;An onlooker can only determine the state of the image at 2 times: Before we request red-eye removal, and after the algorithm is done removing it. &lt;/li&gt;&#xA;&lt;li&gt;All the computation has to be done client-side. Sure, Alice could expose a web service somewhere that removes red-eye in images for you if you provide your product key, but that's not what I'm asking.&lt;/li&gt;&#xA;&lt;li&gt;Red-eye removal was just an example; you should be able to transform any sufficiently general type of algorithm into one of these special 'encrypted algorithms.'&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;My questions are: Does this concept already exist, and if so are there any standard references? Is there some technical flaw that I didn't consider? Surely if it worked like I described then it would be used for copy-protection in software(even though 'locking' an entire piece of software is very different from locking a theoretical algorithm).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you for your time,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-- Michael Burge&lt;/p&gt;&#xA;" OwnerUserId="5468" LastActivityDate="2011-03-05T06:17:55.763" Title="Running 'Encrypted Code' on a Computer" Tags="&lt;algorithms&gt;&lt;computer-science&gt;&lt;cryptography&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" />
  <row Id="25050" PostTypeId="2" ParentId="25049" CreationDate="2011-03-04T18:59:37.200" Score="5" ViewCount="" Body="&lt;p&gt;Look up &quot;Homomorphic encryption&quot;.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2011-03-04T18:59:37.200" />
  <row Id="25051" PostTypeId="1" CreationDate="2011-03-04T19:03:14.937" Score="1" ViewCount="105" Body="&lt;p&gt;Is there a name associated to rectangular matrices $M \times N$ that have exactly one entry equal to $1$ in each row and $0$ everywhere else?&lt;/p&gt;&#xA;" OwnerUserId="7801" LastActivityDate="2011-03-04T22:28:48.080" Title="Matrix with exactly one 1 in each row" Tags="&lt;linear-algebra&gt;&lt;matrices&gt;&lt;terminology&gt;" AnswerCount="3" CommentCount="4" />
  <row Id="25052" PostTypeId="2" ParentId="25049" CreationDate="2011-03-04T19:03:53.500" Score="3" ViewCount="" Body="&lt;p&gt;What you are looking for is code that works on encrypted data: see &lt;a href=&quot;http://cacm.acm.org/magazines/2010/3/76275-a-first-glimpse-of-cryptographys-holy-grail/fulltext&quot; rel=&quot;nofollow&quot;&gt;A First Glimpse of Cryptography's Holy Grail&lt;/a&gt; by &lt;a href=&quot;http://cseweb.ucsd.edu/~daniele/&quot; rel=&quot;nofollow&quot;&gt;Daniele Micciancio&lt;/a&gt; in &lt;em&gt;Communication of the ACM&lt;/em&gt; 3/2010. Processing arbitrary encrypted data is an open problem. The paper is behind a pay wall but if you are at an academic institution with a computer science department, your library most likely will have it available.&lt;/p&gt;&#xA;" OwnerUserId="6678" LastEditorUserId="6678" LastEditorDisplayName="" LastEditDate="2011-03-05T06:17:55.763" LastActivityDate="2011-03-05T06:17:55.763" CommentCount="3" />
  <row Id="25054" PostTypeId="1" CreationDate="2011-03-04T19:16:39.640" Score="4" ViewCount="143" Body="&lt;p&gt;We have a random generator that generates independent random bits with probability $P(x=1) = P$ and $P(x=0)=1-P$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given $N$ random independent bits, we estimate $P$ by $\hat{P} = N_1/(N_0+N_1)$.&#xA;where $N_0$ is the number of $0$'s and $N_1$ is number of $1$'s.&#xA;The expected value for $\hat{P}$ can be simply shown to be $P$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a. What is the expected value for the estimated entropy defined as following: &#xA; $\hat{H}=-[ \hat{P} \log(\hat{P}) + (1-\hat{P}) \log(1-\hat{P}) ]$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;b. If we take $M$ independent sets of $N$ random bits as above and each time estimate the entropy using the above equation, What is the expected value for the smallest estimated entropy among those $M$ sets ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;thanks, &#xA;MG&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S. If the solution to the integral for general P is too complicated, a solution to the special case $P=1/2$ would be appreciated as well. &lt;/p&gt;&#xA;" OwnerUserId="7138" LastEditorUserId="6179" LastEditorDisplayName="" LastEditDate="2011-05-21T11:15:42.650" LastActivityDate="2011-06-04T15:05:56.500" Title="expected value for min estimated entropy" Tags="&lt;probability&gt;&lt;statistics&gt;" AnswerCount="1" CommentCount="7" FavoriteCount="3" />
  <row Id="25057" PostTypeId="2" ParentId="25051" CreationDate="2011-03-04T19:36:50.083" Score="0" ViewCount="" Body="&lt;p&gt;&lt;strike&gt;For $M=N$, these are called &lt;a href=&quot;http://en.wikipedia.org/wiki/Permutation_matrix&quot; rel=&quot;nofollow&quot;&gt;permutation matrices&lt;/a&gt;.&lt;/strike&gt; (striked according to Moron's comment) Yours are a (admittedly very restricted) special case of &lt;a href=&quot;http://www.ic.unicamp.br/~meidanis/research/pqr/&quot; rel=&quot;nofollow&quot;&gt;matrices with the consecutive ones property&lt;/a&gt;, but I'm not sure how much that helps you.&lt;/p&gt;&#xA;" OwnerUserId="689" LastActivityDate="2011-03-04T19:36:50.083" CommentCount="2" />
  <row Id="25061" PostTypeId="2" ParentId="25046" CreationDate="2011-03-04T19:53:23.660" Score="1" ViewCount="" Body="&lt;p&gt;I think you are asking: &quot;How can I translate these sentences into formal ones using quantifiers?&quot; Using $x\in M$ to mean &quot;x is a master set&quot;, we'd get:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$\forall x\in \mathcal{U}:x\in M \implies \forall y\in x: y\not\in y$&lt;/li&gt;&#xA;&lt;li&gt;$\neg\exists x\in \mathcal{U}: x\in M$&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I'm not entirely sure what you mean by an interpretation function, but if you're just trying to generate an extension for these propositions, I think you can come up with some examples. Without knowing your universe though, I don't think we can do much to help.&lt;/p&gt;&#xA;" OwnerUserId="2549" LastActivityDate="2011-03-04T19:53:23.660" />
  <row Id="25062" PostTypeId="2" ParentId="25047" CreationDate="2011-03-04T19:53:58.267" Score="2" ViewCount="" Body="&lt;p&gt;If the length of each reversal you can use is exactly $k$, and not at most $k$, then as far as I know the computational complexity of the problem is open. However, it has been investigated by Chen and Skiena in the case of permutations, in their paper entitled &lt;a href=&quot;http://dx.doi.org/10.1016/S0166-218X%2896%2900069-8&quot; rel=&quot;nofollow&quot;&gt;&quot;Sorting with fixed-length reversals&quot;&lt;/a&gt;, where they give among other results conditions on the feasibility of the sorting problems with respect to $k$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update&lt;/strong&gt;: &lt;a href=&quot;http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.52.8588&amp;amp;rep=rep1&amp;amp;type=ps&quot; rel=&quot;nofollow&quot;&gt;here is a free version&lt;/a&gt; for those who do not have access to the above.&lt;/p&gt;&#xA;" OwnerUserId="689" LastEditorUserId="689" LastEditorDisplayName="" LastEditDate="2011-03-04T20:06:06.117" LastActivityDate="2011-03-04T20:06:06.117" CommentCount="4" />
  <row Id="25063" PostTypeId="2" ParentId="25011" CreationDate="2011-03-04T20:03:31.500" Score="0" ViewCount="" Body="&lt;p&gt;I hope the following works:  Let $U, V$ be ultrafilters on $X$ and $Y$ respectively. &#xA;One side is clear. If there is a bijection $f\colon X\rightarrow Y$ satisfying RK-condition then the restriction of $f$ to any $A\in U$ satisfies RK-condition and it is a bijection. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the other side: Assume that there is $A,B$ and a bijection $f\colon A\rightarrow B$ satisying RK-condition as given in the wikipedia page. Hence, we know that $f$ is defined at every point of $U,$ because $f$ satisfies RK-condition by assumption.  Hence, we have a function $f\colon X\rightarrow Y$ and we want to show that it is 1-1 and onto. &#xA;If $f$ is not onto, then $\exists C\subset Y$ such that $f(X)\cap C=\emptyset.$ Then trivially $f^{-1}(C)=\emptyset\in U$ by the RK-condition. This certainly not possible. Hence $f$ is onto. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, suppose that $f\colon X\rightarrow Y$ is not 1-1. This means there are $x_1,x_2\in X$ $x_1\neq x_2$ but $f(x_1)=f(x_2).$ Let $K:=\left\{x_1,x_2\right\}\subset X.$ If $K\cap A=\emptyset$ and $K\notin U$ then consider the set $A\cup K\in U$ since $U$ is an ultrafilter. The set $f(A\cup K)=f(A)\cup f(K)$ where $f(K)$ is a one point set. But since $V$ is an ultrafilter and $B\in B$ we necessarily have $f(K)\subset B$ which contradicts to $A\cap K=\emptyset.$ Hence, $A\cap K\neq \emptyset.$ $K$ contains two points. But these two points cannot be in $A$ simultaneously, because $f$ is a bijection on $A$. Therefore, WLOG say $x_1\in A$ and $x_2\notin A.$ But then consider $f(A)\in V.$ $f(A)$ contains $f(x_1).$ By RK-condition again inverse image $f^{-1}f(A)=A$ should contain $x_2$ too. This is again a contradiction. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, $f\colon X\rightarrow Y$ is 1-1 and onto.   &lt;/p&gt;&#xA;" OwnerUserId="7796" LastActivityDate="2011-03-04T20:03:31.500" CommentCount="2" />
  <row Id="25064" PostTypeId="1" AcceptedAnswerId="25095" CreationDate="2011-03-04T20:24:32.457" Score="4" ViewCount="141" Body="&lt;p&gt;Consider the following definition of &lt;strong&gt;closed maps&lt;/strong&gt;, defined in the book &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0471486000&quot; rel=&quot;nofollow&quot;&gt;Nonlinear Programming&lt;/a&gt; by Bazaraa &lt;em&gt;et al.&lt;/em&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Let $X$ and $Y$ be nonempty closed sets in $\mathbb{R}^p$ and $\mathbb{R}^q$, respectively. Let $\mathbf{A} \colon X \to Y$ be a point-to-set map. The map $\mathbf{A}$ is said to be &lt;strong&gt;closed&lt;/strong&gt; at $\mathbf{x} \in X$ if for any sequences $\{\mathbf{x}_k\}$ and &#xA;  $\{\mathbf{y}_k\}$ satisfying &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$\mathbf{x}_k \in X, \qquad\qquad\quad \mathbf{x}_k \to \mathbf{x}$&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$\mathbf{y}_k \in \mathbf{A}(\mathbf{x}_k), \qquad \mathbf{y}_k \to \mathbf{y}$&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;we have that $\mathbf{y} \in \mathbf{A}(\mathbf{x})$. The map $\mathbf{A}$ is said to be closed on $Z \subseteq X$ if it is closed at each point in $Z$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm trying to solve the following problem (which is problem 7.7 of the aforementioned book):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Let $\mathbf{A} \colon \mathbb{R}^n \times \mathbb{R}^n \to \mathbb{R}^n$ be the point-to-set map defined as follows. Given $\mathbf{x}, \mathbf{z} \in \mathbb{R}^n$, then &#xA;  $\mathbf{y} \in \mathbf{A}(\mathbf{x},\mathbf{z})$ means that &#xA;  $\mathbf{y} = c\mathbf{x} + (1-c)\mathbf{z}$ for some $c \in [0,1]$ and &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$\|\mathbf{y}\| \le \| \lambda \mathbf{x} + (1-\lambda)\mathbf{z} \|$ &#xA;  for all $\lambda \in [0,1]$.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Show that the map $\mathbf{A}$ is closed for each of the following cases:&lt;/p&gt;&#xA;  &#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;$\|\cdot\|$ denotes the Euclidean norm; that is, &#xA;  $\|\mathbf{g}\| = (\sum_{i=1}^n{{g_i}^2})^{1/2}$.&lt;/li&gt;&#xA;  &lt;li&gt;$\|\cdot\|$ denotes the $\ell_1$ norm; that is, &#xA;  $\|\mathbf{g}\| = \sum_{i=1}^n{|g_i|}$.&lt;/li&gt;&#xA;  &lt;li&gt;$\|\cdot\|$ denotes the sup norm; that is, &#xA;  $\|\mathbf{g}\| = \max_{1 \le i \le n}|g_i|$.&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I tried a lot of approaches, but I couldn't figure out what to do. Any hints and suggestions are welcome.&lt;/p&gt;&#xA;" OwnerUserId="7804" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-04T21:55:30.227" LastActivityDate="2011-03-05T02:44:45.693" Title="Why This Map is Closed?" Tags="&lt;topology&gt;&lt;topological-vector-spaces&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" />
  <row Id="25065" PostTypeId="2" ParentId="25011" CreationDate="2011-03-04T20:32:50.380" Score="1" ViewCount="" Body="&lt;p&gt;Assume that we have $|X|=|Y|$, let $\mathcal{U}$, $\mathcal{V}$ ultarfilters on $X$ and $Y$ respectively and let $A\in\mathcal{U}$, $B\in\mathcal{V}$ and a bijection $f:A\to B$ such that $C\in\mathcal{V}\iff f_{-1}[C]\in\mathcal{U}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Observe that if $|X-A|=|Y-B|$ you are done since both these sets are not in $\mathcal{U}$ and $\mathcal{V}$ (respectively) and you can extend $f$ arbitrarily at these points. Now assume that $|X-A|&amp;lt;|Y-B|$. Let $B&amp;#39;\supset B$ such that $|X-A|=|Y-B&amp;#39;|$ while $|B|=|B&amp;#39;|$ (such $B&amp;#39;$ exists; to see this observe that $|B|=|Y|$). We need to find a bijection $g:B\to B&amp;#39;$ that satisfies the condition of RK. Then we would be done, since we would be able to extend $g\circ f$ as I described above. Let $D=B&amp;#39;\setminus B$ and notice that $D\notin\mathcal{V}$. Let $E\subset B$ an infinite set of size greater than or equal to that of $D$ such that $E\notin\mathcal{V}$. Let $h:E\to D\cup E$ some arbitrary bijection and let $g$ be the identity for every $x\in B\setminus E$ and equal to $h$ for the elements of $E$. This is the $g$ we are looking for.&lt;/p&gt;&#xA;" OwnerUserId="1772" LastEditorUserId="1772" LastEditorDisplayName="" LastEditDate="2011-04-26T17:34:26.107" LastActivityDate="2011-04-26T17:34:26.107" CommentCount="3" />
  <row Id="25066" PostTypeId="1" AcceptedAnswerId="25071" CreationDate="2011-03-04T20:39:37.660" Score="1" ViewCount="63" Body="&lt;p&gt;This is an old exercise I wrote up, but I'm unhappy with my solution. I assume only the basic properties of addition and multiplication for natural numbers as sets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume that $m$ and $n$ are natural numbers with $m$ less than $n$. Show that there is some $p$ in $\omega$ for which $m+p^+=n$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;First see that $n\neq 0$, since $m\in n$. If $m=0$, then we can take $0+p^+=p^+=n$, and we know such a $p$ exists since $n$ is nonzero. Now suppose $k\in n$, and that for some $p$, $k+p^+=n$. Since $k\in n$, $k^+\ \underline{\in}\ n$. If $k^+=n$, then the conclusion holds trivially, since $k^+\not\in n$. If $k^+\in n$, then observe that&#xA;$$&#xD;&#xA;k+p^+=(k+p)^+=k^++p=n.&#xD;&#xA;$$&#xA;Since $k^+\neq n$, we have $p\neq 0$, and thus $p=q^+$ for some $q$. Hence $k^++q^+=n$, and thus conclusion holds for all $m\in n$. Essentially, since we know $0+n=n$, we can then find a $p$ such that $1+p^+=n$, and from this we can find a $q$ such that $2+q^+=n$, and so on for all $m\in n$. This must eventually terminate, as $n$ is finite.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But this feels more like an intuitive sketch than a real proof. I feel there should be a way to do this rigorously with induction, since pretty much all early set theory proofs dealing with naturals use induction. I don't know how to set it up though or what to induct on. Could someone show how to actually do this rigorously? Thanks.&lt;/p&gt;&#xA;" OwnerUserId="7805" LastActivityDate="2011-03-04T21:11:12.403" Title="Show that there is some $p$ in $\omega$ for which $m+p^+=n$. " Tags="&lt;set-theory&gt;&lt;natural-number&gt;" AnswerCount="1" />
  <row Id="25067" PostTypeId="1" AcceptedAnswerId="25070" CreationDate="2011-03-04T20:54:56.667" Score="4" ViewCount="181" Body="&lt;p&gt;Given two natural numbers I am supposed to reverse each of them and then sum them up and reverse the sum to get the final answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example if the numbers are $4358$ and $754$ then the answer should be $1998$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I always thought that there should be a mathematical trick avoid brute force (i.e reversing each numbers and summing them up and reverse the sum to get the final answer).I heard &quot;&lt;strong&gt;double reverse is no reverse&lt;/strong&gt;&quot; but never really get the trick myself.Could anybody help? &lt;/p&gt;&#xA;" OwnerUserId="2109" LastEditorUserId="6296" LastEditorDisplayName="" LastEditDate="2011-03-04T22:00:59.393" LastActivityDate="2011-03-04T22:00:59.393" Title="Is there any mathematical trick ?" Tags="&lt;algorithms&gt;&lt;recreational-mathematics&gt;" AnswerCount="1" />
  <row Id="25068" PostTypeId="1" AcceptedAnswerId="25072" CreationDate="2011-03-04T21:00:08.997" Score="2" ViewCount="91" Body="&lt;p&gt;Hi, Math SE.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've found in Wikipedia the formula for calculating an individual row in Pascal's Triangle:&#xA;$$v_c = v_{c-1}\left(\frac{r-c}{c}\right).$$&#xA;where $r = \mathrm{row}+1$, $c$ is the column starting from $0$ on left and $v_0 = 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, I've tried to do by hand and it works,  but I don't understand how to find this magic formula when I don't have access to Wikipedia...:)&lt;/p&gt;&#xA;" OwnerUserId="5046" LastEditorUserId="6296" LastEditorDisplayName="" LastEditDate="2011-03-05T01:13:20.147" LastActivityDate="2011-03-05T10:47:53.490" Title="How can I prove the formula for calculating successive entries in a given row of Pascal's triangle?" Tags="&lt;combinatorics&gt;&lt;algebra-precalculus&gt;&lt;binomial-coefficients&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="25069" PostTypeId="2" ParentId="25032" CreationDate="2011-03-04T21:04:23.577" Score="0" ViewCount="" Body="&lt;p&gt;Are you looking for direction in mathematics, or in the real world of life and career? Different questions, different answers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Within Math, in my experience, you need to be a generalist though a couple years in grad school because you will have a required body of knowledge that spans Analysis, Algebra, etc. In that time, you will/should develop an interest in some pretty narrow question for your research. The choice probably depends on a mix of what's interesting to you, and who is interesting to work with.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In life, Math is divided into the academic and the practical. You see academic mathematicians all the time. The practical ones are spread thin as actuaries in insurance companies, &quot;quantitative&quot; types on Wall Street, management scientists in advertising, operations analysis for the military, etc, etc, etc. My personal bias is that computer programming is the black hole that sucks in failed mathematicians (like me). Relative to other fields, it's easy to learn, easy to get work, easy to make decent money, but hard to get out of.&lt;/p&gt;&#xA;" OwnerUserId="1312" LastActivityDate="2011-03-04T21:04:23.577" />
  <row Id="25070" PostTypeId="2" ParentId="25067" CreationDate="2011-03-04T21:10:12.823" Score="11" ViewCount="" Body="&lt;p&gt;Line the numbers up flush left. When you carry a digit, carry to the right. Something like&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;4358&#xA;754  (+&#xA;------&#xA;11&#xA; 8&#xA;  9&#xA;   8&#xA;------&#xA;1998&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="1543" LastActivityDate="2011-03-04T21:10:12.823" CommentCount="1" />
  <row Id="25071" PostTypeId="2" ParentId="25066" CreationDate="2011-03-04T21:11:12.403" Score="2" ViewCount="" Body="&lt;p&gt;Your proof is essentially correct, but you mixed up the induction hypothesis a bit. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You should have written: Assume that if $k\in n$ there exists a $p$ such that $k+p^+=n$. &lt;em&gt;Now we want to prove it for $k^+$.&lt;/em&gt; So let $k^+\in n$. Because of this we have $n=m^+$ such that $k\in m$. By the induction hypothesis we have that there exists a $p$ such that $k+p^+=m$ or $(p^++k)^+=m^+$ (here I used the commutativity of addition) or $p^++k^+=n$ or $k^++p^+=n$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This proof though, uses commutativity and the fact $k^+\in m^+\Rightarrow k\in m$. To avoid these you can use induction on $n$ (it's a good idea to use induction on the right-most term because of how addition and multiplication are defined):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First of all if $n=0$ it is vacuously true.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume that it's true for $n$ and let $m&amp;lt;n^+$. This means that $m=n$ or $m&amp;lt;n$. If $m=n$ then $p=0$ (this is trivial). If $m&amp;lt;n$ then, because of the induction hypothesis there is a $p$ such that $m+p^+=n$. This means that $(m+p^+)^+=n^+$ or (by the definition of addition) $m+(p^+)^+=n^+$, so $p^+$ is the number we are looking for. &lt;/p&gt;&#xA;" OwnerUserId="1772" LastActivityDate="2011-03-04T21:11:12.403" CommentCount="1" />
  <row Id="25072" PostTypeId="2" ParentId="25068" CreationDate="2011-03-04T21:20:03.277" Score="5" ViewCount="" Body="&lt;p&gt;How much do you know about the entries in Pascal's Triangle? Do you know their relation to the coefficients of the binomial expansion?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The $n+1$st row of Pascal's Triangle gives you the coefficients of the binomial $(a+b)^n$. That is, if the row is&#xA;$$k_0\quad k_1\quad k_2\quad\cdots\quad k_n$$&#xA;that means that if you multiply out $(a+b)^n$, you will get&#xA;$$k_0a^n + k_1a^{n-1}b + k_2a^{n-2}b^2 + \cdots + k_nb^n.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The coefficients are given by the binomials: $k_i = \binom{n}{i}$, where&#xA;$$\binom{n}{i} = \frac{n!}{i!(n-i)!}$$&#xA;with $!$ symbolising the factorial, $r!=1\times 2\times\cdots \times r$. It is not hard to work out that&#xA;$$\binom{n}{m} = \frac {n(n-1)(n-2)\cdots(n-m+1)}{m(m-1)(m-2)\cdots 1}.$$&#xA;That is, in the numerator you multiply the integers from $n$ down to $n-m+1$ ($m$ factors), and in the denominator you multiply the integers from $m$ down to $1$ (also $m$ factors).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, suppose you already know the value $v_{c-1}$ in row $R$. This is &#xA;$$\binom{R-1}{c-1} = \frac{(R-1)(R-2)\cdots(R-c)}{(c-1)(c-2)\cdots 1}.$$&#xA;The next value, $v_{c}$, is $\binom{R-1}{c}$, which is:&#xA;\begin{align*}&#xA;v_c&amp;amp;=\binom{R-1}{c}\\&#xA;&amp;amp; = \frac{(R-1)(R-2)\cdots(R-c-1)}{c(c-1)(c-2)\cdots 1}\\&#xA; &amp;amp;= \frac{(R-1-c)\Bigl((R-1)(R-2)\cdots(R-c)\Bigr)}{c\Bigl((c-1)(c-2)\cdots 1\Bigr)}\\&#xA;&amp;amp;= \frac{R-1-c}{i}\left(v_{c-1}\right).&#xA;\end{align*}&#xA;Now, remember this was in row $R$. So letting $r = \mathrm{row}-1 = R-1$, we get&#xA;$$v_c = \binom{r}{c} = v_{c-1}\left(\frac{r-c}{c}\right),$$&#xA;exactly the formula given.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added.&lt;/strong&gt; In retrospect, this was very  silly of me. If you knew that the $i$th (starting with $0$) entry in the $R$th row is given by $\binom{R}{i}$, then why would you need to remember this &lt;em&gt;other&lt;/em&gt; formula?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, why is the entry given by $\binom{R}{i}$? First: $\binom{R}{i}$ counts the number of ways in which you can choose $i$ things from $R$ possibilities. So, for example, $\binom{4}{2}$ tells you in how many ways you can choose two things out of four possible ones: if the four options are 1, 2, 3, and 4, you could pick 1 and 2; or 1 and 3; or 1 and 4; or 2 and 3; or 2 and 4; or 3 and 4; that is, there are six possible ways of picking two out of four. So $\binom{4}{2}=6$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why does the formula I gave above give the right count? If you have $R$ things to choose from, you have $R$ ways to pick out the first choice; then $R-1$ ways to pick out the second; then $R-2$ to pick out the third, and so on. You end up with $R(R-1)(R-2)\cdots(R-i+1)$ ways of choosing $i$ things. However, each choice was counted many times. In the example I gave above, I don't count &quot;1 and 2&quot; as different from &quot;2 and 1&quot;. How many ways did I count each choice? Of the $i$ choices I made, I could have selected any of the $i$ things first; then any of the remaining $i-1$ second; then any of the remaining $i-2$ as my third choice, etc. That is, each choice of $i$ things can be made in $i(i-1)(i-2)\cdots (2)(1)$ ways. So, each choice was counted that many times; to get the &lt;em&gt;right&lt;/em&gt; count, we divide the original count by the factor by which we overcounted, so&#xA;$$\binom{R}{i}=\frac{R(R-1)(R-2)\cdots(R-i+1)}{i(i-1)\cdots 2\times 1}.$$&#xA;This is the formula I gave above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One small wrinkle: what happens when $i=0$? Well, in how many ways you make &lt;em&gt;no&lt;/em&gt; choices? Just one: don't do anything. So $\binom{R}{0}=1$ always. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And, why is this &quot;number of ways to pick&quot; related to the coefficients of $(a+b)^n$? Imagine that you need to do the product. Write them down one after the other, as if you were writing down a sum:&#xA;\begin{align*}&#xA;a &amp;amp;+b\\&#xA;a &amp;amp;+b\\&#xA;a &amp;amp;+b\\&#xA;&amp;amp;\vdots\\&#xA;a &amp;amp;+ b\\&#xA;a &amp;amp;+b&#xA;\end{align*}&#xA;You need to do all the possible products of each element with the elements of the other rows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How many times will you get $a^n$? Just once, because the only way to get $a^n$ is by multiplying all the $a$s together. How many times will you get $a^{n-1}b$? Well, you get $a^{n-1}b$ if you pick one $b$ in one row, and multiply it by all the $a$s in the other rows. You have $n$ different choices for the one $b$, so there are $\binom{n}{1}$ ways of getting $a^{n-1}b$. How many ways can you get $a^{n-2}b^2$? You have to pick two which rows will provide you with the two $b$s. There are $n$ possibilities, you need to pick $2$, so it's $\binom{n}{2}$ ways. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And so on. How many times will you get $a^{n-i}b^i$? You need to pick $i$ rows to provide you with the $b$s, and there are $\binom{n}{i}$ ways to pick them. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So when we write out $(a+b)^n$, the monomial $a^{n-i}b^i$ will show up $\binom{n}{i}$ times. So that's the coefficient.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means that the $i$th entry in the $(n+1)$st row of Pascal's triangle is $\binom{n}{i}$, which agrees with what we said above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And now that you know &lt;em&gt;this&lt;/em&gt;, this is really the easy formula to remember, not the &quot;magic formula&quot; form Wikipedia.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-05T01:01:24.360" LastActivityDate="2011-03-05T01:01:24.360" />
  <row Id="25073" PostTypeId="1" AcceptedAnswerId="25139" CreationDate="2011-03-04T21:28:18.297" Score="8" ViewCount="113" Body="&lt;p&gt;Suppose I chose n uniformly distributed random points in a 3D cube. What is the expected value for the percentage of points lying on the convex hull as a function of n?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just as a reference, I made the following experiment in Mathematica 8:  &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Needs[&quot;TetGenLink`&quot;]; Show[&#xA; DiscretePlot[&#xA;  1/k Mean[Length /@ Union /@ (Flatten /@ (TetGenConvexHull /@ &#xA;          RandomReal[{0, 1}, {500, k, 3}]))], {k, 4, 200, 3}, &#xA;  AxesOrigin -&amp;gt; {0, 0}, Joined -&amp;gt; True], Plot[.5, {x, 1, 200}]]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/KyC5T.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again as a reference, if we plot the mean number of points in the convex hull (not the percentage) as a function of the total number of points we get:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/TkZHK.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit 2&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second plot in Log and Log-Log forms:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/KqDad.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit 3&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As noted by @Raskolnikov in the comments below, and confirmed by the &quot;experimental&quot; result, the case n=5 can be though as the &lt;a href=&quot;http://mathworld.wolfram.com/CubeTetrahedronPicking.html&quot; rel=&quot;nofollow&quot;&gt;Cube Tetrahedron Picking Formula&lt;/a&gt; wich is basically the probability of the fifth point being inside the tetrahedron determined by the other four points. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As noted by @Steven Stadnicki  that is not completely obvious because you are choosing the four points beforehand and some permutation could have been left aside ... but the experiments confirm the @Raskolnicov reasoning.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit 4&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I fitted the data using &lt;a href=&quot;http://ccsl.mae.cornell.edu/eureqa&quot; rel=&quot;nofollow&quot;&gt;&lt;strong&gt;Eureqa&lt;/strong&gt;&lt;/a&gt;, a nice package from Cornell for guessing fitting functions, and got as a probable fit for the &lt;em&gt;number&lt;/em&gt; of points in the convex hull:  &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;f[x_] := 1.4723399 Log@x Log@(3.0543704 + x)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;which gives:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/f7z3V.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In line with Raskolnicov's answer about the asymptotic behavior. I wasn't able to read the cited paper, though (restricted access)&lt;/p&gt;&#xA;" OwnerUserId="2731" LastEditorUserId="2731" LastEditorDisplayName="" LastEditDate="2011-03-05T13:51:02.247" LastActivityDate="2011-03-05T13:51:02.247" Title="Expectancy value for the percentage of points lying in the Convex Hull (3D)" Tags="&lt;geometry&gt;&lt;statistics&gt;&lt;probability&gt;" AnswerCount="1" CommentCount="18" FavoriteCount="2" />
  <row Id="25074" PostTypeId="1" CreationDate="2011-03-04T21:49:40.823" Score="0" ViewCount="64" Body="&lt;p&gt;This is a subset of the problem of the A* algorithm. Basically I have a graph with a bunch of nodes (a node is just a point in space). Each node has an x and y coordinate in space, so I can calculate the distance between two nodes by using the regular Euclidean distance. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another data that I have is a cost to travel from node A to node B. This value is represented in feet. My main goal is to add this Euclidean distance with this associated cost in feet, how can I do this? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that the first step is to calculate a constant as a scaling number, say beta. The number is then divided by this beta and then we can add the two numbers together. In order to find beta we can use a sample data set and beta would be the same for the other... I am not sure if this is right though...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sample data:&#xA;Node, X-coor, Y-coor&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1,273078.139,353257.444&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2,273122.77,352868.571&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3,272963.687,353782.863&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4,273949.566,353370.127&lt;/p&gt;&#xA;&#xA;&lt;p&gt;5,274006.13,352910.551&#xA;............&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Node origin, Node destination, cost&#xA;1,3351,179,15,0.14&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2,3264,172,15,0.13&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2,3471,139,15,0.11&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3,3467,226,15,0.17&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4,3309,211,15,0.16&#xA;...........&lt;/p&gt;&#xA;" OwnerUserId="7806" LastEditorUserId="6296" LastEditorDisplayName="" LastEditDate="2011-03-05T03:12:21.677" LastActivityDate="2011-03-05T03:12:21.677" Title="scaling a Euclidean distance with distance in feet" Tags="&lt;linear-algebra&gt;&lt;algorithms&gt;&lt;graph-theory&gt;" CommentCount="1" />
  <row Id="25075" PostTypeId="1" CreationDate="2011-03-04T21:51:51.723" Score="0" ViewCount="77" Body="&lt;p&gt;What is the isomorphism between dual vector bundle $E^*$ and $\mathrm{Hom}(E,M\times \mathbb R)$? There is a natural isomorphism on bundle that is $\mathrm{Hom}(E,E&amp;#39;)=E^*\otimes E$, therefore I am wondering if I can use this isomorphism to get the result that $E^*$ is isomorphic to $E^*\otimes (M\times\mathbb R)$ and thus isomorphic to $\mathrm{Hom}(E, M\times \mathbb R)$?&lt;/p&gt;&#xA;" OwnerUserId="9016" LastEditorUserId="367" LastEditorDisplayName="" LastEditDate="2011-03-05T00:19:45.297" LastActivityDate="2011-05-04T00:49:12.773" Title="What is the isomorphism between E* and Hom(E, M$\times$R)" Tags="&lt;homework&gt;&lt;vector-bundles&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="25076" PostTypeId="2" ParentId="25051" CreationDate="2011-03-04T22:13:51.850" Score="2" ViewCount="" Body="&lt;p&gt;Such a matrix is precisely a matrix representation of an arbitrary function from a set of size $M$ to a set of size $N$, in the sense that multiplication by a row vector is a linearized version of evaluating the function. &lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2011-03-04T22:13:51.850" CommentCount="2" />
  <row Id="25077" PostTypeId="2" ParentId="20291" CreationDate="2011-03-04T22:17:58.257" Score="1" ViewCount="" Body="&lt;p&gt;You can use $\det(e^{M}) = \exp(\text{tr} M)$ valid for diagonalizable matrices $M$. For your case this implies $$\det(I+\lambda A) = \exp[ \text{tr} \log (I+ \lambda A)] = \exp[ \text{tr} ( \lambda A + \mathcal{O}(\lambda^2) )] = 1 + \lambda \,\text{tr} A + \mathcal{O}(\lambda^2).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Jacobian of $\det(A)$ can be obtained using this result&#xA;$$\det(A+ \lambda H) - \det(A) = \det(A) [ \det( I + \lambda A^{-1} H) -1]&#xD;&#xA; = \lambda \det(A) \,\text{tr} (A^{-1} H) + \mathcal{O}(\lambda^2)$$ proving the result stated in the question.&lt;/p&gt;&#xA;" OwnerUserId="7266" LastActivityDate="2011-03-04T22:17:58.257" />
  <row Id="25078" PostTypeId="2" ParentId="25068" CreationDate="2011-03-04T22:23:35.533" Score="2" ViewCount="" Body="&lt;p&gt;Sometimes memorizing a formula is easiest, sometimes it's easier to derive the formula as needed than memorize it (too many plus or minus 1's), but sometimes just remembering a concrete pattern works. For example, if you want to compute the row for &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$${6 \choose m}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$1,6,15,20,15,6,1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;start with 1 and multiply by $\frac{6}{1}$. Then to get the next one multiply by $\frac{5}{2}$, the next $\frac{4}{3}$... and so on, the &lt;strong&gt;numerator going down by one&lt;/strong&gt; each time, the &lt;strong&gt;denominator up by one&lt;/strong&gt; each time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$1 \ \ \frac{6}{1} \ \ 6 \ \ \frac{5}{2} \ \ 15 \ \ \frac{4}{3} \ \ 20 \ \ \frac{3}{4} \ \  15 \ \ \frac{2}{5} \ \ 6 \ \ \frac{1}{6} \ \ 1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the sequence is reversible because by the time you get to the middle of the sequence the fractions themselves are the inverse of their mirrored item.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Try it for 14:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$1 \ \ \frac{14}{1} \ \ 14 \ \ \frac{13}{2} \ \ 91 \ \ \frac{12}{3} \ \ 364 \ \ \frac{11}{4} \ \ 1001 \ \ \frac{10}{5} \ \ 2002 \ \ \frac{9}{6} \ \ 3003 \ldots$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The numbers always cancel, either by the fraction itself or with some factors of the value so far. This is actually a good way to compute binomial coefficients exactly by computer in keeping all the intermediate computations as low as possible (instead of using the factorials).&lt;/p&gt;&#xA;" OwnerUserId="1919" LastActivityDate="2011-03-04T22:23:35.533" CommentCount="1" />
  <row Id="25079" PostTypeId="2" ParentId="25051" CreationDate="2011-03-04T22:28:48.080" Score="0" ViewCount="" Body="&lt;p&gt;Wikipedia gives: &quot;A right stochastic matrix is a square matrix each of whose rows consists of nonnegative real numbers, with each row summing to 1.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This definition restricts you to square matrices, but in Henryk Minc's book &quot;Permanents&quot; he explicitly considers non-square matrices and is always careful to say &quot;$n$-square doubly stochastic&quot; when he means this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It fits in with Qiaochu Yuan's answer in that an arbitrary right stochastic matrix gives a 'function' where $M_{ij}$ is the probability that element $i$ in the domain is mapped to element $j$ in the co-domain.&lt;/p&gt;&#xA;" OwnerUserId="5468" LastActivityDate="2011-03-04T22:28:48.080" />
  <row Id="25080" PostTypeId="2" ParentId="25032" CreationDate="2011-03-04T22:57:43.213" Score="1" ViewCount="" Body="&lt;p&gt;unless you really truly see yourself getting a phd in math and eeking out some academic career, you should problem take a bunch of applied math, study engineering of some sort, or learn a bunch of programming so you can actually get a job.&lt;/p&gt;&#xA;" OwnerUserId="6925" LastActivityDate="2011-03-04T22:57:43.213" />
  <row Id="25081" PostTypeId="1" CreationDate="2011-03-04T23:12:57.487" Score="3" ViewCount="85" Body="&lt;p&gt;As I understand it, many of the ideas that were introduced into algebraic geometry in the mid 20th century by french mathematicians were done by transporting over ideas from the theory of manifolds (for example defining things locally). Instead of there being an analogy between these 2 areas, is it possible to do more and use a theory that generalizes both of them? (Perhaps like the analogy between function fields and number fields, and how they can be studied simultaneously with valuations as described in that paper by Artin and Whapples).&lt;/p&gt;&#xA;" OwnerUserId="7769" LastActivityDate="2011-03-04T23:12:57.487" Title="Is there a theory that generalizes both varieties and manifolds?" Tags="&lt;algebraic-geometry&gt;&lt;manifold&gt;" CommentCount="4" />
  <row Id="25082" PostTypeId="1" CreationDate="2011-03-04T23:39:15.753" Score="2" ViewCount="64" Body="&lt;p&gt;Can manifolds be uniformly approximated by varieties in the way that continuous functions can be uniformly approximated by polynomials?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I got the idea from reading the Princeton companion to mathematics when it gave: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Theorem (Nash) Let M be a manifold in $\mathbb{R}^n$. Fix any large number R. Then there is a polynomial f whose zero set is as close to M as we want, at least inside a ball of radius R around the origin.&lt;/p&gt;&#xA;" OwnerUserId="7769" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2011-03-05T00:15:32.917" LastActivityDate="2011-03-05T00:15:32.917" Title="Can manifolds be uniformly approximated by varieties?" Tags="&lt;algebraic-geometry&gt;&lt;manifold&gt;" CommentCount="3" FavoriteCount="1" />
  <row Id="25083" PostTypeId="2" ParentId="25075" CreationDate="2011-03-05T00:22:18.350" Score="1" ViewCount="" Body="&lt;p&gt;You are right, $E^*$ is isomorphic to $E^*\otimes (M\times\mathbb R)$. This follows from the fact that $V\otimes\mathbb R\cong V$ for every real vector space $V$.&lt;/p&gt;&#xA;" OwnerUserId="367" LastActivityDate="2011-03-05T00:22:18.350" />
  <row Id="25084" PostTypeId="1" CreationDate="2011-03-05T00:28:58.933" Score="0" ViewCount="109" Body="&lt;p&gt;The way that I first heard this was &quot;For every upside down A, there exists a backwards E, such that (something, something, something).  I'm having trouble remembering how it ended. Can anyone help me?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit 1&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is not a &quot;real&quot; math question.  It was more like a nursery rhyme to help us remember a few symbols and their meaning.&lt;/p&gt;&#xA;" OwnerUserId="4181" LastEditorUserId="4181" LastEditorDisplayName="" LastEditDate="2011-03-05T00:45:42.213" LastActivityDate="2011-03-05T00:45:42.213" ClosedDate="2011-03-05T02:02:14.307" Title="For every upside down A" Tags="&lt;soft-question&gt;" CommentCount="2" />
  <row Id="25085" PostTypeId="1" AcceptedAnswerId="25100" CreationDate="2011-03-05T01:17:53.117" Score="1" ViewCount="159" Body="&lt;p&gt;Given $(5x+5y~)^3= 125x^3+125y^3$, find the derivative.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the chain rule and power rule, I came up with&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$3(5x+5y)^2 \cdot (\frac{d}{dx}5x+\frac{dy}{dx}5y)= 3 \cdot 125x^2 +3 \cdot 125y^2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the derivative of $5x~$ is 5, but what about the derivative of $5y~$?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that $\frac{dy}{dx}5y~$ turns to $5(\frac{dy}{dx}(y~))$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What happens after that?  When I plugged the formula into Wolfram Alpha to double check my steps, it says that $\frac{dy}{dx}(y~)=0$  What is the reasoning behind that?&lt;/p&gt;&#xA;" OwnerUserId="7006" LastActivityDate="2011-03-06T13:25:02.743" Title="More help with implicit differentation" Tags="&lt;calculus&gt;&lt;homework&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="25087" PostTypeId="2" ParentId="25085" CreationDate="2011-03-05T01:24:17.837" Score="0" ViewCount="" Body="&lt;p&gt;You are applying $\frac{d}{dx}$ to both sides.  On the left side, you should have $\frac{d}{dx}5y$ instead of $\frac{dy}{dx}5y$, which is just $5\frac{dy}{dx}$.  And when you take $\frac{d}{dx}125y^3$ on the right, you need to use the chain rule to involve $\frac{dy}{dx}$ as well.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-03-05T01:24:17.837" CommentCount="3" />
  <row Id="25088" PostTypeId="2" ParentId="25027" CreationDate="2011-03-05T01:29:56.693" Score="2" ViewCount="" Body="&lt;p&gt;If $w$ is strictly positive and $C^2$ then a direct calculation shows that $\log w$ is subharmonic if and only if $w\Delta w+|\nabla w|^2 \ge 0$.  Taking $w=u+v$ we see that $\log(u+v)$ is subharmonic if and only if&#xA;$$&#xD;&#xA;(u\Delta u+|\nabla u|^2)+(v\Delta v+|\nabla v|^2) +\Delta(uv) \ge 0.&#xD;&#xA;$$&#xA;If $\log u$ and $\log v$ are subharmonic then  the first two terms on the left are non-negative. And so is the third because &#xA;$$&#xD;&#xA;\Delta(uv) = uv\cdot\left[\Delta(\log u+\log v) +|\nabla(\log uv|^2\right],&#xD;&#xA;$$&#xA;which is clearly non-negative.&lt;/p&gt;&#xA;" OwnerUserId="7562" LastActivityDate="2011-03-05T01:29:56.693" />
  <row Id="25090" PostTypeId="1" AcceptedAnswerId="25125" CreationDate="2011-03-05T01:47:12.103" Score="30" ViewCount="992" Body="&lt;p&gt;I've just sat through several lectures that proved most of the results in Tate's thesis: the self-duality of the adeles, the construction of &quot;zeta functions&quot; by integration, and the proof of the functional equation. However, while I was able to follow at least some of the arguments in the individual steps,  I understand almost nothing about the big picture. My impression so far is that Tate invented a new and fancier way of proving the functional equation that the Hecke analytic approach. But is there more to the story than &quot;this is a neat way of proving something already known&quot;?&#xA;I'm under the impression that Tate's thesis laid the foundations for the Langlands program, but I don't understand this properly yet. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can someone explain to me what's the real significance and meaning of Tate's thesis? &lt;/p&gt;&#xA;" OwnerUserId="536" LastActivityDate="2011-03-05T17:29:47.840" Title="What's the significance of Tate's thesis?" Tags="&lt;algebraic-number-theory&gt;" AnswerCount="1" CommentCount="6" FavoriteCount="10" />
  <row Id="25091" PostTypeId="1" AcceptedAnswerId="25094" CreationDate="2011-03-05T01:54:29.350" Score="2" ViewCount="133" Body="&lt;p&gt;Having some difficulty doing translations for complicated neither...nor sentences.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With these characters:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;~: Negation; $\vee$: Disjunction; &amp;amp;: Conjunction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm trying to translate and understand, for example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Neither John nor Mary are standing in front of either Jim or Cary&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have been told that a successful translation of &quot;Neither e nor a is to the right of c&quot; is translated as follows: ~(RightOf(e, c) $\vee$ RightOf(e, c))&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What about just doing a translation on: &quot;I like neither chocolate nor vanilla&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;~(Like(chocolate) $\vee$ Like(Vanilla))&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What confuses me the most is the sentence: &quot;I like neither chocolate nor vanilla&quot; is translated to ~((Like(chocolate) $\vee$ Like(vanilla)) and the sentence: &quot;Neither e nor a is to the right of c and to the left of b&quot; is translated to ~(RightOf(e, c) &amp;amp; LeftOf(e, b)) &amp;amp; ~(RightOf(a, c) &amp;amp; LeftOf(a, b)). Both sentences use neither...nor, however in the second sentence I see no disjunction, but in the first it exists.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any food for thought and help would be appreciated!&lt;/p&gt;&#xA;" OwnerUserId="7810" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-05T03:29:36.377" LastActivityDate="2011-03-05T03:29:36.377" Title="Translating &quot;neither...nor&quot; into a mathematical logical expression" Tags="&lt;logic&gt;&lt;propositional-calculus&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="1" />
  <row Id="25092" PostTypeId="1" CreationDate="2011-03-05T01:57:51.363" Score="1" ViewCount="79" Body="&lt;p&gt;suppose $v$ is a signed measure on $(X,M)$ and $E\in M$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;how do i go about showing that $|v|(E)=sup\{\sum_1^n |v(E_j)|: E_j\cap E_i=0 \forall i\neq j, \cup_1^n E_j=E\}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;sorry it took me awhile to fix the latex&lt;/p&gt;&#xA;" OwnerUserId="6556" LastActivityDate="2011-03-05T06:25:39.327" Title="question about total variation (measure theory)" Tags="&lt;measure-theory&gt;" AnswerCount="1" />
  <row Id="25094" PostTypeId="2" ParentId="25091" CreationDate="2011-03-05T02:08:41.010" Score="5" ViewCount="" Body="&lt;p&gt;Mathematically, there are two ways of &quot;translating&quot; &quot;I like neither chocolate nor vanilla&quot; (the two ways are logically equivalent, an instance of &lt;a href=&quot;http://en.wikipedia.org/wiki/De_Morgan%27s_laws&quot; rel=&quot;nofollow&quot;&gt;de Morgan's laws&lt;/a&gt;). You can write either:&#xA;$$\neg\bigl(\mathrm{like}(\mathrm{chocolate})\bigr)\ \&amp;amp;\ \neg\bigl(\mathrm{like}(\mathrm{vanilla})\bigr)$$&#xA;(that is, &quot;I don't like chocolate and I don't like vanilla&quot;) or as&#xA;$$\neg\bigl(\mathrm{like}(\mathrm{chocolate})\ \vee\ \mathrm{like}(\mathrm{vanilla})\bigr).$$&#xA;The two are equivalent, because $\neg(P\vee Q) \equiv (\neg P)\&amp;amp;(\neg Q)$ (this is one of De Morgan's Laws: for &quot;P or Q&quot; to be false, you need &lt;em&gt;both&lt;/em&gt; P to be false &lt;strong&gt;and&lt;/strong&gt; Q to be false).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second sentence is similar: &quot;Neither e nor a is to the right of c and to the left of b&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You mention you are confused about it; I think it's just a confusion of parsing (it is a bit awkwardly constructed). What is says is &quot;Neither e nor a satisfy &lt;code&gt;xxxx&lt;/code&gt;&quot;. That is: &quot;e does not satisfy &lt;code&gt;xxxx&lt;/code&gt;, and a does not satisfy &lt;code&gt;xxxx&lt;/code&gt;&quot;. What is this &lt;code&gt;xxxx&lt;/code&gt;? It is the condition &quot;is to the right of c and to the left of b&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the sentence is the same as &quot;neither &lt;code&gt;&lt;/code&gt;e is to the right of c and to the left of b&lt;code&gt;&lt;/code&gt;, nor &lt;code&gt;&lt;/code&gt;a is to the right of c and to the left of b&lt;code&gt;&lt;/code&gt;.&quot; (I added the grey spaces as parsing aides).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Note that &quot;e is not to the right of c and to the left of b&quot; means that either e is not to the right of c, or e is not to the left of b, or both; same with a).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you wanted to translate it in the same manner as the first translation above, with conjunction, you would have:&#xA;$$\Bigl(\neg\bigl(\mathrm{RightOf}(e,c)\&amp;amp;\mathrm{LeftOf}(e,b)\bigr) \Bigr) \&amp;amp; \Bigl(\neg\bigl(\mathrm{RightOf}(a,c)\&amp;amp;\mathrm{LeftOf}(e,b)\bigr)\Bigr).$$&#xA;But you can equally as well use the model of the second translation above, to get the equivalent statement:&#xA;$$\neg\Biggl( \bigl(\mathrm{RightOf}(e,c)\&amp;amp;\mathrm{LeftOf}(e,b)\bigr) \vee \bigl(\mathrm{Rightof}(a,c)\&amp;amp;\mathrm{LeftOf}(a,b)\bigr)\Biggr).$$&#xA;So now you see the disjunction in the second.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason why the first &quot;translation&quot; of this phrase might be better than the second is that the mix of conjunctions and disjunctions ( &quot;(a and b) or (x and y)&quot; ) is usually a little harder to parse than a sequence of conjunctions, even with negations ( &quot;not(a and b) and not(x and y)&quot;). So the latter form is slightly prefered for parsing reasons, but mathematically they are equivalent. The parsing problem does not show up in the vanilla/chocolate example, because there are no conjunctions hiding inside the clauses and complicating the parsing.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-05T03:26:12.293" LastActivityDate="2011-03-05T03:26:12.293" CommentCount="1" />
  <row Id="25095" PostTypeId="2" ParentId="25064" CreationDate="2011-03-05T02:26:03.510" Score="3" ViewCount="" Body="&lt;p&gt;For two points $a_1, a_2\in\mathbb{R}^n$, let $\mathcal{L}\left(a_1,a_2\right) =&#xD;&#xA;\left\{\lambda a_1 + \left(1-\lambda\right)a_2\: |\: \lambda \in \left[0,1\right] \right\}$, the closed line segment in $\mathbb{R}^n$ between the points $a_1$ and $a_2$. Your set map $A$ then takes a pair $\left(x,z\right)$ in $\mathbb{R}^n \times \mathbb{R}^n$ to the points in $\mathcal{L}\left(x,z\right)$ which are of minimal distance to the origin with respect to whatever norm your considering.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of the Euclidean norm, $A\left(x,z\right)$ will only consist of one point for all $\left(x,z\right)\in\mathbb{R}^n \times \mathbb{R}^n$. Suppose that&#xA;$\alpha_1$ and $\alpha_2$ are two distinct elements of $A\left(x,z\right)$, with $\left|\alpha_1\right| = \left|\alpha_2\right|=r$. Then $\alpha_1$ and $\alpha_2$ are two points on the boundary of an $n$-sphere of radius $r$ centered at the origin, and thus the interior points of $\mathcal{L}\left(\alpha_1,\alpha_2\right)$ are in the interior of this $n$-sphere. Since $\mathcal{L}\left(\alpha_1,\alpha_2\right)$ is contained in $\mathcal{L}\left(x,z\right)$, this contradicts our claim that $\alpha_1$ and $\alpha_2$ are of minimal distance to the origin within $\mathcal{L}\left(x,z\right)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's not so hard to show that the function $f:\mathbb{R}^n \times \mathbb{R}^n \rightarrow \mathbb{R}^n$, defined by letting $f\left(x,z\right)$ equal the one point in $A\left(x,z\right)$, is continuous. This will imply the result you want in the case of the Euclidean norm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the other norms, things are a little more complicated. The argument above won't go through because the unit ball in each of these cases will be an $n$ dimensional hyper cube. However, you can show that either $A\left(x,z\right)$ will consist of a single point or that $\mathcal{L}\left(x,z\right)$ will contain some segment that is contained in the boundary of some ball that is centered at the origin, in which case $A\left(x,z\right)$ will be this segment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $\mathcal{U}$ be the set of all $\left(x,z\right)$ in $\mathbb{R}^n \times \mathbb{R}^n$ for which $A\left(x,z\right)$ contains only one element and let $\mathcal{V}$ be the complement of $\mathcal{U}$. It's not too difficult to show that the function $f:\mathcal{U}\rightarrow \mathbb{R}^n$ defined by letting $f\left(x,z\right)$ be the unique point in $A\left(x,z\right)$ is continuous on $\mathcal{U}$, that $\mathcal{U}$ is open in $\mathbb{R}^n \times \mathbb{R}^n$, and that $\mathcal{V}$ has measure zero in $\mathbb{R}^n \times \mathbb{R}^n$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let $\left(x,z\right)$ be an element of $\mathbb{R}^n \times \mathbb{R}^n$, $\left(x_k,z_k\right)$ be a sequence in $\mathbb{R}^n \times \mathbb{R}^n$ that converges to $\left(x,z\right)$, and $y_k \in A\left(x_k,z_k\right)$ a sequence in $\mathbb{R}^n$ that converges to $y\in \mathbb{R}^n$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $\left(x_k,z_k\right)$ has a subsequence $\left(x_{n_k},z_{n_k}\right)$ such that $\left(x_{n_k},z_{n_k}\right) \in \mathcal{U}$, for each $n_k$, then $y_{n_k}$ will converge to $y$ and we can use the continuity of $f$ on $\mathcal{U}$ to get the result we want.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Otherwise $\left(x_k,z_k\right)$ will be in $\mathcal{V}$ for all sufficiently large $k$. This puts some restrictions on what $\left(x,z\right)$ can be, and I'm going to leave it to you to figure out how to handle this case.&lt;/p&gt;&#xA;" OwnerUserId="7323" LastEditorUserId="7323" LastEditorDisplayName="" LastEditDate="2011-03-05T02:44:45.693" LastActivityDate="2011-03-05T02:44:45.693" CommentCount="3" />
  <row Id="25096" PostTypeId="1" AcceptedAnswerId="25108" CreationDate="2011-03-05T02:26:50.980" Score="13" ViewCount="214" Body="&lt;p&gt;I have one proof of the following statement, and I would like to know if there is a simpler proof.  I am not sure if “simpler” is the right word or not but, for the purpose of this question, I prefer an elementary proof rather than a short proof depending on a powerful and difficult-to-prove theorem.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;For any integer &lt;i&gt;n&lt;/i&gt;≥171, there exists an integer &lt;i&gt;m&lt;/i&gt; coprime to &lt;i&gt;n&lt;/i&gt; satisfying &lt;i&gt;n&lt;/i&gt;&amp;lt;&lt;i&gt;m&lt;/i&gt;&lt;sup&gt;2&lt;/sup&gt;&amp;lt;2&lt;i&gt;n&lt;/i&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The proof which I have actually gives an additional guarantee that &lt;i&gt;m&lt;/i&gt; is a prime.  I do not need this guarantee and want a simpler proof if there is one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My proof uses the method used by Nagura [Nag52], which I suspect may be overkill for this purpose.  By the same argument as the one in [Nag52], we can prove that for every &lt;i&gt;x&lt;/i&gt;≥16769, there exists a prime between &lt;i&gt;x&lt;/i&gt; and 2&lt;sup&gt;1/4&lt;/sup&gt;&lt;i&gt;x&lt;/i&gt;, and we can verify that the same conclusion holds for &lt;i&gt;x&lt;/i&gt;≥32 by calculation.  This implies that for &lt;i&gt;n&lt;/i&gt;≥32&lt;sup&gt;2&lt;/sup&gt;=1024, there exist at least two primes between √&lt;i&gt;n&lt;/i&gt; and √(2&lt;i&gt;n&lt;/i&gt;).  Because both primes are greater than √&lt;i&gt;n&lt;/i&gt;, at least one of them must be coprime to &lt;i&gt;n&lt;/i&gt;.  The case of 171≤&lt;i&gt;n&lt;/i&gt;≤1023 can be verified by calculation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a simpler proof of the statement above?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[Nag52] Jitsuro Nagura.  On the interval containing at least one prime number.  &lt;em&gt;Proceedings of the Japan Academy&lt;/em&gt;, 28(4):177–181, 1952.  &lt;a href=&quot;http://projecteuclid.org/euclid.pja/1195570997&quot; rel=&quot;nofollow&quot;&gt;http://projecteuclid.org/euclid.pja/1195570997&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="926" LastEditorUserId="6075" LastEditorDisplayName="" LastEditDate="2011-03-26T17:15:53.503" LastActivityDate="2011-03-26T17:15:53.503" Title="On existence of an integer between √n and √(2n) coprime to n" Tags="&lt;number-theory&gt;&lt;analytic-number-theory&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" />
  <row Id="25097" PostTypeId="1" AcceptedAnswerId="25154" CreationDate="2011-03-05T02:27:32.477" Score="2" ViewCount="83" Body="&lt;p&gt;function $f$ on $R$ can be made periodic with period 1 by wrapping &#xA;$$g(x)=\sum_{k=-\infty}^{\infty}f(x+k)$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given $g$ with period 1. How to find continuous $f$ in general case? &lt;/p&gt;&#xA;" OwnerUserId="6052" LastActivityDate="2011-03-05T16:31:58.170" Title="Does inverse wrapping exist?" Tags="&lt;calculus&gt;" AnswerCount="1" CommentCount="7" />
  <row Id="25100" PostTypeId="2" ParentId="25085" CreationDate="2011-03-05T03:13:41.287" Score="1" ViewCount="" Body="&lt;p&gt;Others have pointed out that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{d}{dx}(5x+5y)^3 = 3 (5x + 5y)^2 \frac{d}{dx}(5x + 5y) = 3 (5x + 5y)^2 (5 + 5 \frac{dy}{dx})$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;because $\frac{d}{dx}(5x + 5y) = \frac{d}{dx}(5x) + \frac{d}{dx}(5y) = 5 + 5 \frac{dy}{dx}$.  In your calculation when you write $\frac{dy}{dx} y$, it means $y$ multiplied with $\frac{dy}{dx}$.  When you write $\frac{d}{dx}(y)$, it means $\frac{dy}{dx}$.  You can think of $\frac{d}{dx}$ as a function that requires another function as input.  Just like a function $f$ requires a number as input so that's why $f(x)$ doesn't mean $x$ multiplied with $f$; it means $x$ is the input to $f$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the right hand side, you forgot to apply the Chain Rule:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{d}{dx}(125x^3 + 125y^3) = \frac{d}{dx}(125x^3) + \frac{d}{dx}(125y^3) = 375x^2 + 375y^2 \frac{dy}{dx}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, after this you can solve for $\frac{dy}{dx}$ as usual by isolating $\frac{dy}{dx}$ to one side.&lt;/p&gt;&#xA;" OwnerUserId="7548" LastActivityDate="2011-03-05T03:13:41.287" />
  <row Id="25102" PostTypeId="1" AcceptedAnswerId="25106" CreationDate="2011-03-05T03:34:44.983" Score="7" ViewCount="173" Body="&lt;p&gt;In Leibniz notation, the 2nd derivative is written as&#xA;$$\frac{d^2y}{dx^2}\ ?$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is the location of the 2 in different places in the $dy/dx$ terms?&lt;/p&gt;&#xA;" OwnerUserId="1069" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-05T03:36:20.050" LastActivityDate="2011-03-05T04:13:06.680" Title="Why is the 2nd derivative written as $\frac{d^2y}{dx^2}$?" Tags="&lt;calculus&gt;&lt;notation&gt;" AnswerCount="3" CommentCount="7" />
  <row Id="25103" PostTypeId="2" ParentId="25102" CreationDate="2011-03-05T03:40:44.240" Score="2" ViewCount="" Body="&lt;p&gt;The $d$ is meant to represent the &quot;change in&quot;. And the Leibnitz notation is meant to remind you that you are computing the ratio between the change in $y$ and the change in $x$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you take the second derivative, you are computing how the &lt;em&gt;derivative&lt;/em&gt; is changing as $x$ changes; that is, you are trying to compute&#xA;$$\frac{d(y&amp;#39;)}{dx}.$$&#xA;Now, $y&amp;#39;$ is itself a rate of change: it is the rate at which $y$ changes. So the &quot;numerator&quot; of the differential notation is telling you that you are trying to consider the change in the &lt;em&gt;change&lt;/em&gt; in $y$, not the change in $y^2$ (which is what &quot;$dy^2$&quot; would represent).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you are trying to describe the change in &quot;the-change-in-$y$&quot;, relative to how $x$ is changing. $x$ is only changing &quot;once&quot;, so you should have a single $d$ in the &quot;denominator&quot; (remember, &lt;a href=&quot;http://math.stackexchange.com/questions/21199/dy-dx-is-not-a-ratio/21209#21209&quot;&gt;not &lt;em&gt;really&lt;/em&gt; a denominator&lt;/a&gt;). So why $x^2$? Because you are trying to figure out the change of &lt;code&gt;blah&lt;/code&gt; as $x$ changes, and &lt;code&gt;blah&lt;/code&gt; is a rate of change as $x$ changes as well. So you are taking $x$ &lt;strong&gt;twice&lt;/strong&gt;, but considering only one change. Hence, single $d$, but $x$ squared.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2011-03-05T03:40:44.240" />
  <row Id="25104" PostTypeId="2" ParentId="25102" CreationDate="2011-03-05T03:47:03.187" Score="8" ViewCount="" Body="&lt;p&gt;Purely symbolically, if we accept that $dy = f&amp;#39;(x)\,dx$, and treat $dx$ as a constant, then $$d^2y = d(dy) = d(f&amp;#39;(x)\,dx) = dx\,d(f&amp;#39;(x)) = dx\,f&amp;#39;&amp;#39;(x)\,dx = f&amp;#39;(x)\,(dx)^2,$$&#xA;so dividing yields:&#xA;$$\frac{d^2y}{(dx)^2} = \frac{d^2y}{dx^2} = f&amp;#39;&amp;#39;(x).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As to where this notation actually comes from, though: My guess is that it comes from a time when mathematicians primarily thought of $dx$ and $dy$ as &quot;infinitesimal quantities.&quot;  There are ways of doing so rigorously (via non-standard analysis), and perhaps there is a way of making this notation rigorous that way.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;However, we can still give rigorous meaning to these calculations without appealing to non-standard analysis by using the language of bilinear forms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $f$ is differentiable, we can define a map $$df\colon \mathbb{R} \to L(\mathbb{R}; \mathbb{R})$$ via $$df(x)(dx) = f&amp;#39;(x)\,dx.$$  Here, $L(\mathbb{R};\mathbb{R})$ denotes the set of linear maps from $\mathbb{R} \to \mathbb{R}$, and $dx$ is simply a real number.  Going one step further, we can consider the map $$d^2f = d(df)\colon \mathbb{R} \to L(\mathbb{R};L(\mathbb{R};\mathbb{R})).$$&#xA;By identifying $L(\mathbb{R}; L(\mathbb{R}; \mathbb{R}))$ with $B(\mathbb{R} \times \mathbb{R};\mathbb{R})$, we get a bilinear map $$d^2f(x)(dx^1, dx^2) = dx^1\, f&amp;#39;&amp;#39;(x) \,dx^2$$ whose associated quadratic form is $$d^2f(x)(dx) = f&amp;#39;&amp;#39;(x)\,(dx)^2.$$&#xA;It is now perfectly legal to divide on both sides by $(dx)^2$, obtaining $$\frac{d^2f}{dx^2} = f&amp;#39;&amp;#39;(x).$$&lt;/p&gt;&#xA;" OwnerUserId="640" LastEditorUserId="640" LastEditorDisplayName="" LastEditDate="2011-03-05T04:11:55.503" LastActivityDate="2011-03-05T04:11:55.503" />
  <row Id="25105" PostTypeId="1" CreationDate="2011-03-05T04:07:43.987" Score="1" ViewCount="55" Body="&lt;p&gt;Is it possible to get a MS in a biostatistics and then apply to a PhD program in Math?&lt;/p&gt;&#xA;" OwnerUserId="7813" LastActivityDate="2011-03-05T04:07:43.987" Title="Biostatistics and Math" Tags="&lt;soft-question&gt;" CommentCount="2" />
  <row Id="25106" PostTypeId="2" ParentId="25102" CreationDate="2011-03-05T04:13:06.680" Score="10" ViewCount="" Body="&lt;p&gt;Quite simply,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \frac{d}{dx}\left(\frac{d}{dx}(y)\right) = \frac{d}{dx}\left(\frac{dy}{dx}\right) = \frac{d\,dy}{dx\,dx} = \frac{d^2 y}{dx^2} $$&lt;/p&gt;&#xA;" OwnerUserId="45" LastActivityDate="2011-03-05T04:13:06.680" />
  <row Id="25107" PostTypeId="1" CreationDate="2011-03-05T04:14:07.117" Score="-1" ViewCount="83" Body="&lt;p&gt;Hello.&#xA;What is that puzzle game? an example::&#xA;&lt;img src=&quot;http://i.imgur.com/RlAoJ.gif&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And where can I found some games like that?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you ..&lt;/p&gt;&#xA;" OwnerUserId="2861" LastEditorUserId="3731" LastEditorDisplayName="" LastEditDate="2011-03-05T04:26:08.197" LastActivityDate="2011-03-05T05:35:16.863" Title="What is the name of that puzzle game? (slant)" Tags="&lt;reference-request&gt;&lt;puzzle&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" />
  <row Id="25108" PostTypeId="2" ParentId="25096" CreationDate="2011-03-05T04:15:49.603" Score="9" ViewCount="" Body="&lt;p&gt;I will show what you want for sufficiently large $n$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;First question:  How many squares are between $n$ and $2n$.  Well we have $$\#\text{of squares}=\lceil\sqrt{2n}-1\rceil-\lceil\sqrt{n}-1\rceil\geq\sqrt{2n}-\sqrt{n}-2=(\sqrt{2}-1)\sqrt{n}-2\geq \frac{\sqrt{n}}{3}$$ for sufficiently large $n$.  So lets look at the square roots of these squares, which form a sequence of more than $\sqrt{n}/3$ consecutive integers.  (This is fine since $\gcd(n,m)=1$ if and only if $\gcd(n^2,m)=1$)  The goal is then to show that &lt;em&gt;one&lt;/em&gt; of these integers in this sequence is relatively prime to $n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Recall the elementary &lt;a href=&quot;http://en.wikipedia.org/wiki/Legendre_sieve&quot; rel=&quot;nofollow&quot;&gt;Eratosthenes-Legendre&lt;/a&gt; sieve which says that:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Theorem:&lt;/strong&gt; For any real $x$ and any $y\geq 0$, we have $$S(x,y;n)=\frac{\phi(n)}{n}y+O\left(2^{\omega(n)}\right)$$ where $S(x,y;n)$ is the number of integers $m$ in the interval $x&amp;lt;m\leq x+y$ for which $(n,m)=1$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Using this theorem, combined with the fact that $2^{\omega(n)}\ll_{\epsilon} n^\epsilon$, and the fact that $$\frac{\phi(n)}{n}\geq \frac{C}{\log\log n},\ \ (C&amp;gt;0)$$ the result then follows for sufficiently large $n$.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice that this proof generalizes to show that for any $k$, there exists $N$ such that for any $n&amp;gt;N$ there exists $m\in \left(\sqrt[k]{n},\ \sqrt[k]{2n}\right)$ for which $\gcd(m,n)=1$.  In other words, we can show:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;For any $k$, taking $n$ large enough guarantees that there will be a $k^{th}$ power between $n$ and $2n$ that is also relatively prime to $n$.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hope that helps, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Remark:&lt;/strong&gt;  The key fact which makes this proof work is that $2^{\omega(n)}\ll_\epsilon n^\epsilon$ for every $\epsilon$.  The Eratosthenes-Legendre Sieve is really what you get when you try the simplest approach to evaluating $S(x,y;n)$.  It follows directly from the inclusion-exclusion principle.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added Proof:&lt;/strong&gt;&#xA;Why is it true that  $$\frac{\phi(n)}{n}\geq \frac{e^{-\gamma}}{\log\log n}+O\left(\frac{1}{(\log\log n)^2}\right)?$$  This follows from Mertens formulas along with Chebyshevs estimates.  The error term can be removed by making the constant smaller, which is what I wrote down above, but lets prove this form. &#xA;Look at the $n$ which minimize $\phi(n)/n$ for their size.  These will be of the form $$\prod_{p\leq y}p,$$ that is the product of the first few primes. (*Prove these numbers do indeed minimize) Taking logarithms introduces $\theta(y)$, so we can deduce $\log \log n = \log y +O(1)$ by Chebyshevs estimate.  Next $$\frac{n}{\phi(n)}=\prod_{p\leq y}=\left( 1-\frac{1}{p}\right)^{-1}=e^\gamma\log y+O(1)$$ is one of Mertens formulas. Taking reciprocals yields $$\frac{\phi(n)}{n} = \frac{e^{-\gamma}}{\log \log n} + O\left(\frac{1}{(\log \log n)^2}\right)$$ as desired.&lt;br&gt;&#xA;That result is actually stronger than we need.  All the above proof required was that $n^{-\epsilon}\ll_\epsilon\phi(n)/n$ for every $\epsilon$.&lt;/p&gt;&#xA;" OwnerUserId="6075" LastEditorUserId="6075" LastEditorDisplayName="" LastEditDate="2011-03-26T17:15:02.020" LastActivityDate="2011-03-26T17:15:02.020" CommentCount="4" />
  <row Id="25109" PostTypeId="1" AcceptedAnswerId="25114" CreationDate="2011-03-05T04:29:56.977" Score="1" ViewCount="144" Body="&lt;p&gt;I have been going through the RSA cipher and have been wondering if there is a way other than the Extended Euclid Method to find&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a^{-1} \mod n$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;where a,n $\in$ Z&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S : n is not necessarily prime&lt;/p&gt;&#xA;" OwnerUserId="149" LastEditorUserId="1016" LastEditorDisplayName="" LastEditDate="2011-03-05T05:18:46.723" LastActivityDate="2011-03-06T02:43:03.067" Title="Multiplicative Inverse and RSA" Tags="&lt;elementary-number-theory&gt;&lt;modular-arithmetic&gt;" AnswerCount="3" CommentCount="3" />
  <row Id="25110" PostTypeId="1" AcceptedAnswerId="25113" CreationDate="2011-03-05T04:30:34.877" Score="0" ViewCount="99" Body="&lt;p&gt;Hi everyone!&#xA;I have a problem in solving my number theory homework. My question is as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $p$ be an odd prime. Prove that $a$ is a quadratic residue mod $p$ if and only if the $I_{g}\(x\)$ (index with respect to any primitive root of $p$) is even.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please edit my writing. Thanks. Does anyone know where to start? Thank you very much for everything!&lt;/p&gt;&#xA;" OwnerUserId="6912" LastActivityDate="2011-03-05T05:57:05.677" Title="Quadratic residues and index w.r.t. primitive root of a odd prime p" Tags="&lt;homework&gt;&lt;elementary-number-theory&gt;" AnswerCount="1" />
  <row Id="25111" PostTypeId="1" AcceptedAnswerId="25134" CreationDate="2011-03-05T04:50:37.470" Score="1" ViewCount="95" Body="&lt;p&gt;I'm reading a tutorial on expectation maximization which gives an example of a coin flipping experiment (the description is at &lt;a href=&quot;http://www.nature.com/nbt/journal/v26/n8/full/nbt1406.html?pagewanted=all&quot; rel=&quot;nofollow&quot;&gt;http://www.nature.com/nbt/journal/v26/n8/full/nbt1406.html?pagewanted=all&lt;/a&gt;). Could you please help me understand where the probabilities in step 2 of the process (i.e. in the middle of part b in the below illustration) come from? Thank you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/mj0nb.gif&quot; alt=&quot;Expectation maximization&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;EM starts with an initial guess of the parameters. 2. In the E-step, a probability distribution over possible completions is computed using the current parameters. The counts shown in the table are the expected numbers of heads and tails according to this distribution. 3. In the M-step, new parameters are determined using the current completions. 4. After several repetitions of the E-step and M-step, the algorithm converges.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="6289" LastActivityDate="2011-03-05T09:26:43.713" Title="how does expectation maximization work?" Tags="&lt;statistics&gt;" AnswerCount="1" FavoriteCount="2" />
  <row Id="25112" PostTypeId="1" AcceptedAnswerId="25118" CreationDate="2011-03-05T05:00:22.947" Score="1" ViewCount="102" Body="&lt;p&gt;Hello everyone!&#xA;Can anybody help me out with this number theory question?&#xA;My question is as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $n$ is a positive integer and if an integer $x$ exists such that &#xA;$x^{n-1}\equiv 1 \ (\text{mod} \ n)$ and $x^{\frac{n-1}{q}} \neq 1 \ (\text{mod} \ n)$ for all prime divisors $q$ of $n-1$, then $n$ is prime. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please edit my writing because I cannot install MathJax in my computer.&#xA;Thank you for any help!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe we have to use some reasoning on the order of $x$, but I don't know where to start. Thanks!&lt;/p&gt;&#xA;" OwnerUserId="6912" LastEditorUserId="2000" LastEditorDisplayName="" LastEditDate="2011-03-05T05:58:47.140" LastActivityDate="2011-03-05T07:17:58.987" Title="Prime divisors of n-1, prove n is prime" Tags="&lt;homework&gt;&lt;elementary-number-theory&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="25113" PostTypeId="2" ParentId="25110" CreationDate="2011-03-05T05:08:18.837" Score="2" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\ $ Let $\rm\:c\:$ be a primitive root. If $\rm\ a \equiv b^2\: $ and $\rm\ b \equiv c^k\ $ then $\rm\ a \equiv (\cdots)^2\ $ so $\rm\:a\:$ has even index. Conversely suppose $\rm\:a\:$ has even index  $\rm\ I_c(a) = 2\ k\:.\:$ Then $\rm\ a \equiv (\cdots)^2\ $ so $\rm\:a\:$ is a quadratic residue.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Essentially the proof boils down to the identity $\rm\ (c^k)^2\ \equiv\ c^{\:2\: k}\:.\ $ It easily generalizes from $\rm\ 2\to n\:.$&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2011-03-05T05:57:05.677" LastActivityDate="2011-03-05T05:57:05.677" CommentCount="6" />
  <row Id="25114" PostTypeId="2" ParentId="25109" CreationDate="2011-03-05T05:18:21.980" Score="5" ViewCount="" Body="&lt;p&gt;Let $a \in \mathbb{Z}$ with $(a,n)=1$. Then by Euler's Theorem,&#xA;$$a^{\varphi(n)} \equiv 1 \pmod n$$&#xA;Then, it follows, since $a$ is invertible mod $n$, that&#xA;$$a^{-1} \equiv a^{\varphi(n)-1} \pmod n$$&#xA;As a note, in practice this is much harder to compute than by using the extended Euclidean algorithm, as computing $\varphi(n)$ requires factoring $n$.&lt;/p&gt;&#xA;" OwnerUserId="1016" LastActivityDate="2011-03-05T05:18:21.980" />
  <row Id="25115" PostTypeId="1" AcceptedAnswerId="25116" CreationDate="2011-03-05T05:26:47.357" Score="0" ViewCount="133" Body="&lt;p&gt;Hello!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $\{X(t), t&amp;gt;=0\}$ and $\{Y(t),t&amp;gt;=0\}$ be independent Poisson processes with parameters $\lambda_1$ and $\lambda_2$, respectively. Define $Z_1(t)=X(t)+Y(t)$, $Z_2(t)=X(t)-Y(t)$, $Z_3(t)=X(t)+k$, $k$ a positive integer. Determine which of the above processes are Poisson and find $\lambda$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any help is appreciated! This is not homework! Thanks.&lt;/p&gt;&#xA;" OwnerUserId="6912" LastEditorUserId="6912" LastEditorDisplayName="" LastEditDate="2011-03-07T16:02:29.610" LastActivityDate="2011-03-07T16:02:29.610" Title="Poisson process" Tags="&lt;homework&gt;&lt;probability-theory&gt;&lt;stochastic-processes&gt;" AnswerCount="1" />
  <row Id="25116" PostTypeId="2" ParentId="25115" CreationDate="2011-03-05T05:34:57.740" Score="3" ViewCount="" Body="&lt;p&gt;The last one is not Poisson since $Z_{3}(0) \neq 0$. Consider $t_1 \leq t_2 \leq t_3 \leq t_4$. Then look at $Z_{1}(t_2)-Z_{1}(t_1), Z_{1}(t_4)-Z_{1}(t_3)$ and see if they are independent. Do the same for $Z_{2}(t)$. Then you have to look at stationary increments.&lt;/p&gt;&#xA;" OwnerUserId="2548" LastEditorUserId="2548" LastEditorDisplayName="" LastEditDate="2011-03-05T05:40:15.567" LastActivityDate="2011-03-05T05:40:15.567" CommentCount="11" />
  <row Id="25117" PostTypeId="2" ParentId="25107" CreationDate="2011-03-05T05:35:16.863" Score="0" ViewCount="" Body="&lt;p&gt;It is called the snake game.  My favorite implementation is &lt;a href=&quot;http://krispixton.com/SlitherQuest/download.htm&quot; rel=&quot;nofollow&quot;&gt;SlitherQuest&lt;/a&gt;.  At the risk of offending some, my favorite logic puzzle game is PrismaPix, from the same author.  It is an expansion of minesweeper.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-03-05T05:35:16.863" CommentCount="0" />
  <row Id="25118" PostTypeId="2" ParentId="25112" CreationDate="2011-03-05T05:54:36.943" Score="3" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\ \ $ The conditions imply that the order $\rm\:k\:$ of $\rm\:x\:$ is a divisor of $\rm\:n-1\:$ but not a &lt;em&gt;proper&lt;/em&gt; divisor, therefore $\rm\: k = n-1\:.\ $  By Euler, $\rm\ k\ |\ \phi(n)\ $ so $\rm\ n-1\: \le\: \phi(n)\:.\ $ This implies that $\rm\:n\:$ is prime, since $\rm\ \phi(n) \:\le\: n-\color{red}{2}\ $ for composite $\rm\:n\:,\:$ since they have at least $\:\color{red}2\ $ smaller naturals non-coprime to $\rm\:n\:.$&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2011-03-05T07:17:58.987" LastActivityDate="2011-03-05T07:17:58.987" CommentCount="3" />
  <row Id="25119" PostTypeId="1" AcceptedAnswerId="25122" CreationDate="2011-03-05T06:09:18.747" Score="1" ViewCount="120" Body="&lt;p&gt;Let $x_1,\ldots,x_n$ be any numbers and $\bar{x} = (x_1+ \ldots + x_n)/n$. Then I need to prove that $$min_a \sum_{i=1}^n (x_i - a)^2 = \sum_{i=1}^n (x_i - \bar{x})^2$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I do not need the answer but I need an visual/intuitive understanding of the question. I do not understand what min$_a$ mean. Actually I'm kind of struggling to understand &quot;minimization concepts&quot; i.e. the value $a$ is not defined or anything so what does minimization of $a$ mean ? It would be great if someone can explain the concept in general too. Thanks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I got the answer perfectly for what I asked. I tried to solve the problem as well. I got stuck but here is my try.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{i=1}^n (x_i - a)^2 = \sum_{i=1}^n (x_i - \bar{x} + \bar{x} - a)^2$$&#xA;$$= \sum_{i=1}^n (x_i - \bar{x})^2 + 2 \sum_{i=1}^n (x_i - \bar{x})(x_i - a) + \sum_{i=1}^n(x_i - a)^2 $$ ...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone help me after this ? Thanks again.&lt;/p&gt;&#xA;" OwnerUserId="6335" LastEditorUserId="6335" LastEditorDisplayName="" LastEditDate="2011-03-05T08:02:29.017" LastActivityDate="2011-03-05T23:24:06.957" Title="Understanding some basics." Tags="&lt;statistics&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="25120" PostTypeId="2" ParentId="24731" CreationDate="2011-03-05T06:14:00.333" Score="0" ViewCount="" Body="&lt;p&gt;Fix $p=n$. For $m \in (n-c,n]$, $\pi(m)$ is either already separated (these form the set $S$), or not already separated (these form the set $N$). Let $X$ be the set of all &quot;partners&quot; of those $\pi(m)$ which are separated (i.e. $S$). Thus $X \subset \{\pi(n+1),\ldots,\pi(2n)\}$. We can use $O(c^2)$ right shifts so that none of $X$ lies in $\pi(n+1),\ldots,\pi(n+c)$. Now we can use $O(c^2)$ right shifts to move $N$ across the &quot;break&quot;, without ruining the separation of $S$.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2011-03-05T06:14:00.333" CommentCount="2" />
  <row Id="25121" PostTypeId="2" ParentId="25092" CreationDate="2011-03-05T06:25:39.327" Score="2" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Hint:&lt;/strong&gt; Consider the &lt;a href=&quot;http://en.wikipedia.org/wiki/Hahn_decomposition_theorem&quot; rel=&quot;nofollow&quot;&gt;Hahn Decomposition Theorem&lt;/a&gt;, which allows us to split $X$ into a positive and negative set for $\nu$.  That is, we may write $X=P\cup N$, $P\cap N=\emptyset$ where $P$ is a positive set for $\nu$ and $N$ is a negative set for $\nu$.&lt;/p&gt;&#xA;" OwnerUserId="6075" LastActivityDate="2011-03-05T06:25:39.327" CommentCount="1" />
  <row Id="25122" PostTypeId="2" ParentId="25119" CreationDate="2011-03-05T06:26:30.463" Score="5" ViewCount="" Body="&lt;p&gt;Given $x_1,\ldots,x_n$, you can define a function $$f(a) = \sum_{i=1}^n (x_i-a)^2.$$ This function is defined for all real $a$, and in some sense it measures the centrality of $a$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can try some simple example on Wolfram alpha, say &quot;plot((a-1)^2+(a-2)^2+(a-4)^2,a=0..5)&quot;. In this case the data points are $1,2,4$, and you get a nice, parabolic-like shape (this is no coincidence). The lowest point of the function is the minimum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The function $f(a)$ is bounded from below (since $f(a) \geq 0$), and so it cannot happen that $f(a)$ can be arbitrarily small. Therefore there must be a value below which $f(a)$ cannot reach, and moreover an optimal one - this is known as the infimum. The infimum is a value $L$ such that $f(a) \geq L$ always, and moreover for each $l &amp;gt; L$, there is some value of $a$ such that $f(a) &amp;lt; l$; that expresses the fact that $L$ is optimal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, it might be that $f(a)$ can get arbitrarily close to the infimum, but never reach it. In the case at hand, this doesn't happen, and the function $f(a)$ is actually minimized at some point, namely the average (or mean).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The equation you state can be used to justify the definition of average - the average is the value that minimizes the average squared distance from the datapoints. You could choose other criteria - for example, if you replace squared distance by absolute distance, then the optimal value of $a$ is the median.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason that people care about squared distance is that it's much easier to work with, and the resulting theory is very nice. For example, the central limit theorem states that in many cases, processes converge to a normal distribution whose parameters depend on the mean and variance of the original distribution - the variance is just $\min f(a)$ (normalized).&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2011-03-05T06:26:30.463" CommentCount="1" />
  <row Id="25124" PostTypeId="2" ParentId="24794" CreationDate="2011-03-05T06:47:28.197" Score="2" ViewCount="" Body="&lt;p&gt;This is just another wording of Matt E's very nice argument. That's why I'm making it a community wiki. (I voted for the question and for Matt E's answer.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll freely use Exercises 1.27 and 1.28. For any regular map $\varphi:V\to W$ between affine varieties, write $\varphi^*$ for the induced $k$-algebra morphism between coordinate rings (going into the reverse direction). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $V\subset k^n$ be an affine variety, let $A$ be its coordinate ring, let $B$ be the coordinate ring of $k^r$, and let $\varphi:V\to k^r$ be a regular map induced by a linear map from $k^n$ to $k^r$. Assume that $\varphi^*:B\to A$ is injective and that $A$  is integral over $\varphi^*(B)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $z$ be in $k^r$. We must find a $v$ in $V$ such that $\varphi(v)=z$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;View $z$ as a regular map from the zero dimensional affine space $\{0\}$ to $k^r$. By Exercise 5.2 there is a $k$-algebra morphism $v^*:A\to k$, coming from a $v$ in $V$ viewed as a regular map from $\{0\}$ to $V$, such that $v^*\circ\varphi^*=z^*:B\to k$, and we get $\varphi\circ v=z$. &lt;/p&gt;&#xA;" OwnerUserId="660" LastActivityDate="2011-03-05T06:47:28.197" CommunityOwnedDate="2011-03-05T06:47:28.197" CommentCount="8" />
  <row Id="25125" PostTypeId="2" ParentId="25090" CreationDate="2011-03-05T06:59:33.883" Score="38" ViewCount="" Body="&lt;p&gt;Tate's thesis introduces the concept, ubiquitous now, of doing analysis, and especially Fourier analysis, on the locally compact ring of adeles.    In this setting, the discrete subgroup $\mathbb Z \subset \mathbb R$ is replaced&#xA;by the discrete subgroup $\mathbb Q \subset \mathbb A$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This has a number of implications, some of which are:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;$\mathbb Q$ is a field, and $\mathbb A$ is essentially a product of fields.&#xA;It is technically almost always easier to work with fields rather than more general rings (such as $\mathbb Z$).  The adelic formalism allows one to have one's cake and eat it too (in some sense): one is working with the field $\mathbb Q$, not the ring $\mathbb Z$, but the primes are still present, in the factorization of $\mathbb A$ as a product.  (And this product structure of &#xA;$\mathbb A$, which is formally very simple, captures in some subtle way the&#xA;deeper sense of &quot;product&quot; in the statement of the fundamental theorem of arithmetic, i.e. that any natural number is a unique product of prime powers.)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Tate writes zeta-functions, or more generally, Hecke $L$-series, as integrals over $\mathbb A^{\times}$.  The Euler product structure of the $L$-series then becomes simply a factorization of this integral according to the product structure of $\mathbb A^{\times}$.  (This is a manifestation of the parenthetical remark at the end of point (1).)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The proof of the functional equation becomes (more-or-less) just an application of Poisson summation (in the adelic context).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is worth comparing this with the classical proof (which one can read in Lang's book, among other places, if memory serves).   Classically, one takes&#xA;the sum over ideals representation of the $L$-function, and decomposes it first&#xA;into a finite collection of sums, indexed by the ideal class group, each sum taking place over all the integral ideals in a given ideal class.  These individual series are then described as Mellin transforms of theta series, &#xA;and the functional equation is derived from the transformation properties of the&#xA;theta series, the latter being proved by an application of Poisson summation in the classical setting.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once one unpacks all the details, Tate's proof and Hecke's proof don't look so different; but the difference in packaging is enormous!  In Tate's approach there is no need to unpack everything (for example, the ideal class group is just lurking around in the background implicitly, and there is no need to bring it out explicitly), while in the classical arguments such unpacking is key to the whole thing.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As another example of the conceptual clarity and simplification that Tate's approach gives, you might consider the way he derives the formula for the residue at $s = 1$ of the zeta function of a number field (i.e. the general class number formula) and compare it with the classical derivation.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Working in the case of a function field over a finite field, Tate derives the Riemann--Roch formula (in the form $\dim H^0(C,\mathcal O(D)) - \dim&#xA;H^0(C,\mathcal O(K - D)) = 1 + \deg D - g$) as a straightforward consequence of Poisson summation.  Among other things, this provides a rather striking unification of (what we now call) Serre duality and Fourier duality.  (Although I don't know the precise history, this probably has antecedents in the literature: the original proof of the functional equation of the $\zeta$-function for a curve over a finite field, by  Schmidt, proceeded by applying Riemann--Roch; so Tate is essentially reversing this argument.)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Tate's derivation of the functional equation of $L$-series in terms of local functional equations shows that the global root number --- i.e. the constant that appears in the functional equation --- is a product of local numbers.  As far as I understand, this wasn't known (and perhaps not even suspected) prior to Tate's proof.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This may seem slightly esoteric, but experience shows that one should regard global root numbers, and their factorization into a product of local root numbers (or $\epsilon$-factors), to be of essentially equal importance to global $L$-series, and their (Euler product) factorization into local $L$-factors.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Summary/Conclusion: The aim of the above list is just to highlight some of the points to watch out for while studying Tate's thesis.  Let me now make some remarks at a more general level.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the classical theory of zeta and $L$-functions, there is a tension between the analytic tools, which are essentially additive Fourier theory (i.e. Poissson summation) and the multiplicative aspects of the theory (exemplified by the Euler product).  Tate's thesis resolves these tensions by moving to the adelic context.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the general theory of automorphic forms (say on a quotient $\Gamma&#xA;\backslash G(\mathbb R)$) for some congruence subgroup $\Gamma$ of the integral points $G(\mathbb Z)$ of a semi-simple or reductive Lie group $G(\mathbb R)$) there is the same tension between the harmonic analysis and Lie theory (which $\Gamma \backslash G(\mathbb R)$ is well set-up to accommodate) and the theory of Hecke operators (which pertain to the finite primes, which are not particular visible in this classical description), which is resolved by moving to the adelic picture $G(\mathbb Q)\backslash G(\mathbb A)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another thing to bear in mind is that the theories of $L$-series and of automorphic forms are quite technical in nature, and so conceptual and aesthetic simplifications (as in Tate's thesis) go hand in hand with technical simplifications.  (See e.g. points (1) and (3) above.)  One instance of this in the automorphic forms context is that conjugacy classes in $G(\mathbb Q)$ are much easier to understand than in a congruence subgroup $\Gamma$ of $G(\mathbb Z)$.  (Another instance of the technical superiority of fields over more general rings.)  One might also consider the &lt;a href=&quot;http://en.wikipedia.org/wiki/Weil_conjecture_on_Tamagawa_numbers&quot; rel=&quot;nofollow&quot;&gt;Tamagawa number one theorem&lt;/a&gt;, which gives an elegant reformulation and generalization of a myriad of classical results.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, to finish, Tate's thesis is significant because it improves the classical point of view in a number of ways, achieving conceptual, technical, and aesthetic simplifications.  At the same time, it suggests a way of unifying harmonic analytic and arithmetic considerations in the general context of automorphic forms, by working in the adelic context.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, I strongly suggest working through the details of Tate's thesis in the particular case of the Riemann zeta function, and seeing how his arguments and construction compare with the classical ones.  If you haven't already done this, it should be quite enlightening.  (In particular, it will illuminate points (1), (2), and (3) above.)&lt;/p&gt;&#xA;" OwnerUserId="221" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2011-03-05T17:29:47.840" LastActivityDate="2011-03-05T17:29:47.840" CommentCount="1" />
  <row Id="25126" PostTypeId="1" AcceptedAnswerId="25127" CreationDate="2011-03-05T07:05:03.567" Score="1" ViewCount="86" Body="&lt;p&gt;I'm trying to get insight into the eigenvalue spectrum of a square matrix (large N, symmetric,positive semi definite matrix) using Fourier transforms (I've tried transforming a bunch of thigngs: the diagonasl, the skew diagonal, the entrie matrix, each eigenvector, etc... Is there anything I should be looking at specifically?&lt;/p&gt;&#xA;" OwnerUserId="7815" LastActivityDate="2011-03-05T07:51:07.287" Title="Is it possible to link the eigenvalues of a matrix to the Fourier transform of the matrix?" Tags="&lt;matrices&gt;&lt;fourier-series&gt;&lt;eigenvalues&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="25127" PostTypeId="2" ParentId="25126" CreationDate="2011-03-05T07:34:11.110" Score="2" ViewCount="" Body="&lt;p&gt;I know of a couple of results for certain special types of matrices.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) If you have a circulant matrix, the eigenvalues of that matrix will be the discrete Fourier transform of the first row of the circulant matrix.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) For the second one I don't remember all the technical details. Suppose you have a symmetric Toeplitz matrix whose size goes to infinity, certain functions of the eigenvalues can be written in terms of the Fourier transform of the first row of the matrix. In fact, the expression is written as a Riemann sum, which becomes a Riemann integral in the Fourier domain. In other words, you can write an unwieldy expression involving the eigenvalues of this growing matrix in terms of an integral which is usually much easier to evaluate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to learn more, check out Robert Gray's notes at &lt;a href=&quot;http://ee.stanford.edu/~gray/toeplitz.pdf&quot; rel=&quot;nofollow&quot;&gt;http://ee.stanford.edu/~gray/toeplitz.pdf&lt;/a&gt; for an introduction to these kinds of results. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is also the book by Grenander and Szego which gives a lot more detail and proves some more general results here &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://books.google.com/books?id=CFhVdL78wGcC&amp;amp;printsec=frontcover&amp;amp;dq=toeplitz+forms&amp;amp;hl=en&amp;amp;ei=CudxTf_XFtL1gAfH9qRV&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=1&amp;amp;ved=0CDQQ6AEwAA#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;http://books.google.com/books?id=CFhVdL78wGcC&amp;amp;printsec=frontcover&amp;amp;dq=toeplitz+forms&amp;amp;hl=en&amp;amp;ei=CudxTf_XFtL1gAfH9qRV&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=1&amp;amp;ved=0CDQQ6AEwAA#v=onepage&amp;amp;q&amp;amp;f=false&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2641" LastEditorUserId="2641" LastEditorDisplayName="" LastEditDate="2011-03-05T07:51:07.287" LastActivityDate="2011-03-05T07:51:07.287" CommentCount="2" />
  <row Id="25128" PostTypeId="1" AcceptedAnswerId="25145" CreationDate="2011-03-05T07:42:26.823" Score="1" ViewCount="167" Body="&lt;p&gt;Hello,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Prove that $x^{2} \equiv 1 \pmod{2^k}$ has exactly four solution incongruent solutions.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My attempt:&lt;br&gt;&#xA;We have, &#xA;$x^2 - 1 = (x - 1) \times (x + 1)$, then &#xA;$(x - 1)(x + 1) \equiv 0 \pmod{2^k}$&#xA;which implies,&lt;br&gt;&#xA;$2^k|(x - 1)$ or $2^k|(x + 1) \implies x \equiv \pm 1 \pmod{2^k} (1)$&lt;br&gt;&#xA;Furthermore, $2^{k-1} \equiv 0 \pmod{2^k} \Leftrightarrow 2^{k-1} + 1 \equiv 1 \pmod{2^k}$.&lt;br&gt;&#xA;Multiply both sides by $-1$, we have another congruent namely $-(2^{k-1} + 1) \equiv -1 \pmod{2^k}$&#xA;Hence,  $x \equiv \pm(1 + 2^{k-1}) \pmod{2^k} (2)$&lt;br&gt;&#xA;From $(1)$ and $(2)$, we can conclude that $x^{2} \equiv 1 \pmod{2^k}$ have four incongruent solutions.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Am I in the right track?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks,&lt;/p&gt;&#xA;" OwnerUserId="5711" LastActivityDate="2011-03-05T20:22:24.520" Title="Prove that $x^{2} \equiv 1 \pmod{2^k}$ has exactly four solution incongruent solutions" Tags="&lt;elementary-number-theory&gt;" AnswerCount="3" CommentCount="4" />
  <row Id="25129" PostTypeId="1" CreationDate="2011-03-05T08:26:00.597" Score="1" ViewCount="50" Body="&lt;p&gt;I apologize in advance for how basic this question is...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $j:V\rightarrow V/U$ be the ultrapower map where U is an ultrafilter on a set S, and $j(x)=[c_x]$.  Now, let $f\in j(0)$.  Then $f$ is undefined almost everywhere on S, but we're assuming $f$ is a function defined on all of S, so we conclude no such $f$ exists, and so $j(0)=0$.  Okay, that I get.  Now look at $j(1)$.  By the same argument, $f\in j(1)$ iff $f=\emptyset$ a.e.  So now my issue; this isn't unique!  As long as it's 0 almost everywhere, I can give $f$ random values elsewhere.  And in any case, it's a function on S, not $\{\emptyset\}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you!&lt;/p&gt;&#xA;" OwnerUserId="7817" LastActivityDate="2011-03-05T08:44:09.560" Title="How do I explicitly see that the Ultrapower map is the identity below its critical point?" Tags="&lt;set-theory&gt;&lt;large-cardinals&gt;" AnswerCount="1" />
  <row Id="25130" PostTypeId="2" ParentId="25109" CreationDate="2011-03-05T08:33:15.030" Score="2" ViewCount="" Body="&lt;p&gt;Step 1: Compute the product $ax$ for all integers $1\leq x \leq n-1$. Stop when $ax\equiv 1 \pmod n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Step 2: The $x$ you get from the previous step is $a^{-1}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;DONE&lt;/p&gt;&#xA;" OwnerUserId="7676" LastActivityDate="2011-03-05T08:33:15.030" />
  <row Id="25131" PostTypeId="2" ParentId="25129" CreationDate="2011-03-05T08:44:09.560" Score="2" ViewCount="" Body="&lt;p&gt;Suppose $j$ is an elementary embedding, then it is 1-1, this is the definition of an embedding.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The critical point is the first ordinal such that $j(\kappa)\neq\kappa$. Suppose $\alpha=j(\kappa)&amp;lt;\kappa$ then what would $j(\alpha)$ be? It cannot be $\alpha$ because $j$ is 1-1. So you have to have $j(\kappa)&amp;gt;\kappa$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if $\kappa$ is the smallest one that moves, $j(\alpha)=\alpha$ for every ordinal below $\kappa$ and thus it is the identity map.&lt;/p&gt;&#xA;" OwnerUserId="622" LastActivityDate="2011-03-05T08:44:09.560" CommentCount="4" />
  <row Id="25132" PostTypeId="1" CreationDate="2011-03-05T09:04:27.190" Score="0" ViewCount="76" Body="&lt;p&gt;We have a forest of rooted trees. Two players makes alternating moves according to the following rule: one move is to cut vertex and all its children. Player which makes last move (no vertices remain) wins.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can we compute Grundy function for the positions in the game?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose we have a trees and we need to say whether current position is winning or losing?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you.&lt;/p&gt;&#xA;" OwnerUserId="7818" LastEditorUserId="6622" LastEditorDisplayName="" LastEditDate="2011-03-05T09:42:45.837" LastActivityDate="2011-03-05T16:31:15.610" Title="Game on the tree, cutting branch." Tags="&lt;game-theory&gt;" AnswerCount="2" />
  <row Id="25133" PostTypeId="1" CreationDate="2011-03-05T09:14:18.143" Score="5" ViewCount="123" Body="&lt;p&gt;I would like to determine the cardinality of the sets specified bellow. Nevertheless, I don't know how to approach or how to start such a proof. Any help will be appreciated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $X$ and $Y$ are well-ordered sets, then determine the cardinality of:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$\{f : f$ is a function from $X$ to $Y\}$&lt;/li&gt;&#xA;&lt;li&gt;$\{f : f$ is an order-preserving function from $X$ to $Y\}$&lt;/li&gt;&#xA;&lt;li&gt;$\{f : f$ is a surjective and order-preserving function from $X$ to $Y\}$&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="7819" LastEditorUserId="7819" LastEditorDisplayName="" LastEditDate="2011-03-05T17:42:50.383" LastActivityDate="2011-03-05T19:40:57.697" Title="Cardinality of sets of functions with well-ordered domain and codomain" Tags="&lt;set-theory&gt;&lt;axiom-of-choice&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="25134" PostTypeId="2" ParentId="25111" CreationDate="2011-03-05T09:26:43.713" Score="2" ViewCount="" Body="&lt;p&gt;These are the likelihoods of the corresponding set of $10$ coin tosses having been produced by the two coins (using the current estimate for their biases) normalized to add up to $1$. The estimated probability of $k$ out of $10$ tosses of coin $i$ ($i\in\{A,B\}$) yielding heads is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$p_i(k)=\left({10\atop k}\right) \theta_i^k (1-\theta_i)^{10-k}\;.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The binomial coefficient is the same for both coins, so it drops out in the normalization, and only the ratio of the remaining factors determines the result. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, in the second row, we have $9$ heads and $1$ tails. Given the current bias estimates $\theta_A=0.6$ and $\theta_B=0.5$, the factors are&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\theta_A^9 (1-\theta_A)^{10-9}\simeq0.004$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\theta_B^9 (1-\theta_B)^{10-9}\simeq0.001\;,$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;resulting in the numbers&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{0.004}{0.004+0.001}=0.8$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{0.001}{0.004+0.001}=0.2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;in the second row.&lt;/p&gt;&#xA;" OwnerUserId="6622" LastActivityDate="2011-03-05T09:26:43.713" CommentCount="1" />
  <row Id="25135" PostTypeId="2" ParentId="25119" CreationDate="2011-03-05T09:49:39.170" Score="4" ViewCount="" Body="&lt;p&gt;The question is about minimization with respect to a variable &lt;em&gt;a&lt;/em&gt;. For this we use the derivative: the minimum of the function $f(a)$ is where the first derivative is zero and the second is positive.&lt;br&gt;&#xA;Now write $$f(a) = \sum_{i=1}^n(x_i-a)^2 = \sum_{i=1}^n a^2-2ax_i+x_i^2 $$ &#xA;Then the derivative is $$f&amp;#39;(a) = \sum_{i=1}^n (2a - 2x_i) = 2(an - \sum_{i=1}^n x_i)  $$&#xA;We see: if $a*n$ equals the sum we have $f&amp;#39;(a)$ a zero. So we can compute $a_0$ by&#xA;$$ a_0 = \frac{\sum_{i=1}^n x_i }n $$ &#xA;Next, if the second derivative is positive, then indeed $f(a_0)$ is also a minimum:&#xA;$$ f&amp;#39;&amp;#39;(a) = 2n $$ and this is positive - independent of &lt;em&gt;a&lt;/em&gt; and/or the $x_i$ . But the formula for $a_0 $ is just the definition of the mean: $ x_{mean} =\frac{\sum_{i=1}^n x_i }n $ &lt;/p&gt;&#xA;" OwnerUserId="1714" LastEditorUserId="1714" LastEditorDisplayName="" LastEditDate="2011-03-05T23:24:06.957" LastActivityDate="2011-03-05T23:24:06.957" />
  <row Id="25136" PostTypeId="1" AcceptedAnswerId="25150" CreationDate="2011-03-05T10:02:13.077" Score="4" ViewCount="201" Body="&lt;p&gt;How do you prove $$\sum_{j=0}^n (x)^j (-1)^{n-j} \left\{{n \atop j}\right\} = x^n,$$&#xA;where $(x)^j=x(x+1)...x(x+j-1)$ and $\left\{{n \atop j}\right\}$ is a Stirling number of the second kind?&lt;/p&gt;&#xA;" OwnerUserId="6904" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2011-05-27T18:41:42.113" LastActivityDate="2011-05-27T18:41:42.113" Title="How to prove the equality $\sum_{j=0}^n (x)^j (-1)^{n-j} \left\{{n \atop j}\right\} = x^n$?" Tags="&lt;combinatorics&gt;&lt;stirling-numbers&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="1" />
  <row Id="25137" PostTypeId="1" CreationDate="2011-03-05T10:22:42.897" Score="2" ViewCount="115" Body="&lt;p&gt;Can someone help me to construct a linear functional in $\mathcal{C}[0,1]$ that does not attain its norm?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Actually, I want to prove that $\mathcal{C}[0,1]$ is not reflexive Banach space. Is it sufficient to construct that kind of functional?&lt;/p&gt;&#xA;" OwnerUserId="6990" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-05T20:53:51.703" LastActivityDate="2011-03-05T20:53:51.703" Title="construction of a linear functional in C[0,1]" Tags="&lt;functional-analysis&gt;" AnswerCount="2" />
  <row Id="25138" PostTypeId="2" ParentId="25137" CreationDate="2011-03-05T10:36:42.763" Score="4" ViewCount="" Body="&lt;p&gt;Try $\varphi(f) = \int_{0}^{1/2} f(x)\,dx - \int_{1/2}^{1} f(x)\,dx$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the second question, yes. For every $x \in X$ there is by Hahn-Banach a functional $\varphi \in X^{\ast}$ with $\|\varphi\| = 1$ such that $\varphi(x) = \|x\|$. Now apply this to $X^{\ast}$ and use reflexivity.&lt;/p&gt;&#xA;" OwnerUserId="5363" LastEditorUserId="5363" LastEditorDisplayName="" LastEditDate="2011-03-05T10:48:50.803" LastActivityDate="2011-03-05T10:48:50.803" />
  <row Id="25139" PostTypeId="2" ParentId="25073" CreationDate="2011-03-05T10:41:27.410" Score="2" ViewCount="" Body="&lt;p&gt;In the meantime, I have found &lt;a href=&quot;http://www.jstor.org/stable/3214289&quot; rel=&quot;nofollow&quot;&gt;this article&lt;/a&gt; which addresses an even more general issue, namely the same problem but for the interior of a $d$-dimensional polytope.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, for the 3D-cube, this implies $O((\log n)^2)$ for the number of points.&lt;/p&gt;&#xA;" OwnerUserId="3567" LastEditorUserId="3567" LastEditorDisplayName="" LastEditDate="2011-03-05T10:49:37.667" LastActivityDate="2011-03-05T10:49:37.667" CommentCount="2" />
  <row Id="25140" PostTypeId="2" ParentId="25068" CreationDate="2011-03-05T10:47:53.490" Score="2" ViewCount="" Body="&lt;p&gt;Just a very short derivation:&lt;br&gt;&#xA;If $\binom r i $ are the &lt;em&gt;i&lt;/em&gt;'th entries in the &lt;em&gt;r&lt;/em&gt;'th row, then this means $ \frac {r!}{i! (r-i)!} $ for the &lt;em&gt;i&lt;/em&gt;'th entry. Two consecutive entries are then connected by &#xA;$$ \frac {r!}{(i+1)! (r-(i+1))!} &#xD;&#xA;=  \frac {r!}{(i+1)*i! \frac{(r-i)!}{r-i}}&#xD;&#xA;=  \frac {r!}{i! (r-i)!}*\frac{r-i}{i+1} $$    &lt;/p&gt;&#xA;" OwnerUserId="1714" LastActivityDate="2011-03-05T10:47:53.490" />
  <row Id="25141" PostTypeId="1" AcceptedAnswerId="25143" CreationDate="2011-03-05T11:04:54.283" Score="0" ViewCount="43" Body="&lt;p&gt;Hi, I'm looking for some basic information about vector before going through a chapter in book. Can someone provide links to some good tutorials?&lt;/p&gt;&#xA;" OwnerUserId="7821" LastActivityDate="2011-03-05T11:27:50.227" Title="Vector Tutorials" Tags="&lt;vector-calculus&gt;&lt;vector-space&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="25142" PostTypeId="1" AcceptedAnswerId="25148" CreationDate="2011-03-05T11:06:05.200" Score="5" ViewCount="98" Body="&lt;p&gt;Question that seems pretty easy, but I can't formalize it:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let  $L \subset C$ be a lattice, and $f(z)$ be an elliptic function for $L$, that is a meromorphic function so that $f(z+w) = f(z)$ for all $\omega \in L$. Assume that $f$ is analytic except for double poles at each point of the lattice $L$. Show that $f = a\wp + b$ for some constants $a,b$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I tried:&#xA;$\displaystyle f(z) = \prod_{\omega \in L} {\frac{g(z)}{(z-\omega)^2}}$ , $g(z)$ is analytic and therefore constant in the fundamental domain. Now what is left to do, is to take the product apart to partial fractions, and then I get almost what needed, except it's not one constant $a$ and $b = \sum_{\omega \in L} -\frac1{\omega^2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Am I right? How do I proceed?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance.&lt;/p&gt;&#xA;" OwnerUserId="6707" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2011-05-04T05:16:52.503" LastActivityDate="2011-05-04T05:16:52.503" Title="Elliptic functions and Weierstrass $\wp$-function" Tags="&lt;complex-analysis&gt;&lt;special-functions&gt;" AnswerCount="1" />
  <row Id="25143" PostTypeId="2" ParentId="25141" CreationDate="2011-03-05T11:27:50.227" Score="0" ViewCount="" Body="&lt;p&gt;A search on Google gives this page as a first hit: &lt;a href=&quot;http://www.math.oregonstate.edu/home/programs/undergrad/CalculusQuestStudyGuides/vcalc/vcalc.html&quot; rel=&quot;nofollow&quot;&gt;http://www.math.oregonstate.edu/home/programs/undergrad/CalculusQuestStudyGuides/vcalc/vcalc.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there anything in specific you'd like some info on?&lt;/p&gt;&#xA;" OwnerUserId="7155" LastActivityDate="2011-03-05T11:27:50.227" CommentCount="2" />
  <row Id="25145" PostTypeId="2" ParentId="25128" CreationDate="2011-03-05T11:32:34.970" Score="5" ViewCount="" Body="&lt;p&gt;Note that $(x - 1)(x + 1) \equiv 0 \mod{2^k}$  will imply that $x$ must be an odd integer. So we may assume that $x = 2m + 1$. Putting value of $x$ in the equation we have $4m(m+1) \equiv 0 \mod{2^k}$. This means that $2^{k-2}$ divides $m(m+1)$ if I assume $k&amp;gt;2$. Note that if $k \leq 2$ then there is no condition on $m$. So all residue classes of odd integer satisfy the above equation. So now assume that $k &amp;gt; 2$. if $m$ is even then $m$ is divisible by $2^{k-2}$ and $x = 2^{k -1}t +1$ $t \in \mathbb{Z}$. But if $m$ is odd, then $m+1$ is divisible by $2^{k - 2}$ and in this case $x= 2(m +1)-1 = 2^{k-1}t - 1$ for $t \in \mathbb{Z}$. In the first we shall have only two non congruent solution namely $1$,$2^{k -1} +1$ whereas in the second case the incongruent solutions are $-1$ and $2^{k-1} - 1$.&lt;/p&gt;&#xA;" OwnerUserId="7565" LastActivityDate="2011-03-05T11:32:34.970" />
  <row Id="25146" PostTypeId="1" AcceptedAnswerId="25153" CreationDate="2011-03-05T11:43:45.570" Score="0" ViewCount="86" Body="&lt;p&gt;$$ y&amp;#39;= \log(ax+by+c)~ ;\qquad  a,b,c \in R$$ &lt;p&gt;So is this how I am supoosed to ask questions here?&lt;/p&gt;&lt;/p&gt;&#xA;" OwnerUserId="7822" LastEditorUserId="1543" LastEditorDisplayName="" LastEditDate="2011-03-05T12:56:47.783" LastActivityDate="2011-03-05T13:20:05.227" Title="differential equation" Tags="&lt;differential-equations&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="25147" PostTypeId="1" CreationDate="2011-03-05T11:53:38.457" Score="1" ViewCount="171" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm just getting started with Markov chains and have a few simple questions:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Is it possible to define a period for a reducible Markov chain? If so, how?&lt;/li&gt;&#xA;&lt;li&gt;Can we define balance equations and a stationary distribution for reducible Markov chains?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="7824" LastEditorUserId="822" LastEditorDisplayName="" LastEditDate="2011-03-05T14:07:30.703" LastActivityDate="2011-03-05T14:07:30.703" Title="Markov Chains - 2 clarification questions" Tags="&lt;probability&gt;&lt;discrete-mathematics&gt;&lt;stochastic-processes&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="25148" PostTypeId="2" ParentId="25142" CreationDate="2011-03-05T12:05:54.847" Score="3" ViewCount="" Body="&lt;p&gt;Here is a hint to help orient you: Suppose $f$ is doubly periodic on the lattice $L_{\tau} = \mathbb{Z} \oplus \tau \mathbb{Z}$, where $\tau \in \mathbb{H}$, and $f$ is assumed analytic everywhere save for double poles on the lattice points of $L_{\tau}$. What can you say about the function $f/\wp$ on $L_{\tau}$? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Compare the position of the poles of both $f$ and $\wp$. If they occur at the same place, then $f/\wp$ has no poles (provided that the order of the poles is the same), and so is doubly periodic as well as analytic, hence constant of $\Lambda_{\tau}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now generalize $f$ to an arbitrary lattice $\omega_1 \mathbb{Z} \oplus \omega_2 \mathbb{Z}$, as the one you post, and consider the function $g = (f - a)/\wp = f/\wp - a/\wp$, where $a$ is an arbitrary constant. What properties will $g$ have on the lattice?&lt;/p&gt;&#xA;" OwnerUserId="2720" LastEditorUserId="2720" LastEditorDisplayName="" LastEditDate="2011-03-05T13:47:22.763" LastActivityDate="2011-03-05T13:47:22.763" CommentCount="3" />
  <row Id="25150" PostTypeId="2" ParentId="25136" CreationDate="2011-03-05T12:49:11.193" Score="9" ViewCount="" Body="&lt;p&gt;To remove the $(-1)^{n-j}$ term we replace $x$ by $-x$ and prove&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x^n = \sum_{j=0}^n \left\{ { n \atop j } \right\} (x)_j , \quad (1)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $(x)_j = x(x-1)(x-2) \cdots (x-j+1)$ and $(x)_0 = 1.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We proceed by induction. Note that $(1)$ is true for $n=0$ and so we take $n&amp;gt;0$ and assume $(1)$ is true for $n-1.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align*}&#xD;&#xA;x^n &amp;amp;= x \cdot x^{n-1} \\&#xD;&#xA;&amp;amp;= x \sum_{j=0}^{n-1} \left\{ { n-1 \atop j } \right\} (x)_j \\&#xD;&#xA;&amp;amp;= \sum_{j=0}^{n-1} \left\{ { n-1 \atop j } \right\} (x)_j (x-j + j) \\&#xD;&#xA;&amp;amp;= \sum_{j=0}^{n-1} \left\{ { n-1 \atop j } \right\} (x)_{j+1} +&#xD;&#xA;\sum_{j=0}^{n-1}  j \left\{ { n-1 \atop j } \right\} (x)_j \\&#xD;&#xA;&amp;amp;= \sum_{j=1}^n \left\{ { n-1 \atop j-1 } \right\} (x)_j +&#xD;&#xA;\sum_{j=0}^{n-1}  j \left\{ { n-1 \atop j } \right\} (x)_j \\&#xD;&#xA;&amp;amp;= \sum_{j=0}^n \left\{ { n \atop j } \right\} (x)_j.&#xD;&#xA;\end{align*}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The last equality follows from the &lt;a href=&quot;http://en.wikipedia.org/wiki/Stirling_numbers_of_the_second_kind&quot; rel=&quot;nofollow&quot;&gt;recurrence relation&lt;/a&gt; for the Stirling numbers of the second kind:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \left\{ { n \atop j } \right\} = \left\{ { n-1 \atop j-1 } \right\} +&#xD;&#xA;j \left\{ { n-1 \atop j } \right\}.$$ &lt;/p&gt;&#xA;" OwnerUserId="1301" LastActivityDate="2011-03-05T12:49:11.193" CommentCount="2" />
  <row Id="25151" PostTypeId="1" CreationDate="2011-03-05T13:09:09.633" Score="0" ViewCount="87" Body="&lt;p&gt;I have a assignment for Minitab 15, which about varoius statistics concepts (basic and foundamental topics) but have a big problem with understanding of of questions! Also I do not know how it's possible to use Minitab® to answer this question, I would be glad some one get me clue. (- I don't need some one to solve the problem just help to how could do it in Minitab®)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the question: &lt;br/&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&amp;nbsp; Generate 15 000 samples of size&#xA;  n = 15 each, from the Poisson&#xA;  distribution with parameter λ=3&#xA;  Calculate the mean from each sample.&#xA;  You have now 15 000 means. &#xA;  &lt;br/&gt;  Draw a histogram of these 15&#xA;  000 means. What is the shape of the&#xA;  histogram? Calculate the mean of the&#xA;  15 000 means and the standard&#xA;  deviation. Use the Central Limit&#xA;  Theorem to explain why they should be&#xA;  close to 3 and 0.4472 respectively.&#xA;  Also explain the shape of the&#xA;  histogram.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Also as I'm new to statistics, can't understand how it's possible to have 1500 means?&lt;/p&gt;&#xA;" OwnerUserId="7826" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2011-05-05T16:06:48.267" LastActivityDate="2011-06-04T16:11:09.103" Title="Central Limit Theorem in Minitab 15" Tags="&lt;math-software&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="25153" PostTypeId="2" ParentId="25146" CreationDate="2011-03-05T13:20:05.227" Score="3" ViewCount="" Body="&lt;p&gt;You can &quot;solve&quot; this equation by &lt;a href=&quot;http://en.wikipedia.org/wiki/Separation_of_variables&quot; rel=&quot;nofollow&quot;&gt;separation of variables&lt;/a&gt;. First introduce a new function $z(x)= a x + b y(x) + c$. In terms of this function the differential equation reads&#xA;$$ z&amp;#39;(x) = b \log z(x) +a$$&#xA;and is separable (as it does not depend explicitly on $x$). An (implicit) solution reads&#xA;$$ \int_{z_0}^z dz&amp;#39; \frac{1}{b \log z&amp;#39; +a} = x- x_0$$&#xA;with the initial condition $z(x_0)  =z_0$ which corresponds to $y(x_0) = (z_0 -a x_0 -c)/b$ in terms of the old variables. The integral can be evaluated and gives (it is rather ugly looking)&#xA;$$ \frac{e^{-a/b} \text{Ei}\left( \frac{a}{b} + \log z \right)}{b} = x-x_0 + \frac{e^{-a/b} \text{Ei}\left( \frac{a}{b} + \log z_0 \right)}{b},$$&#xA;with Ei the &lt;a href=&quot;http://en.wikipedia.org/wiki/Exponential_integral&quot; rel=&quot;nofollow&quot;&gt;exponential integral&lt;/a&gt;. Being able to solve explicitly for $z(x)$ or $y(x)$ seems to be unlikely.&lt;/p&gt;&#xA;" OwnerUserId="7266" LastActivityDate="2011-03-05T13:20:05.227" />
  <row Id="25154" PostTypeId="2" ParentId="25097" CreationDate="2011-03-05T13:23:31.417" Score="7" ViewCount="" Body="&lt;p&gt;The solution to your problem is non-unique. But if you just want a candidate:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;assuming $g$ is continuous, let $c$ be the average value of $g$ over $[0,1]$. Then by intermediate value theorem, the function $g - c$ hits 0 at some point. By translating the axis we can assume that $g-c = 0$ at $0$. Define $f_1(x) = g-c$ if $x\in [0,1]$ and $0$ elsewhere. $f_1$ is continuous, has compact support, and its periodicization is $g-c$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now if suffices to find a continuous function $f_2$ such that its periodicization is $c$. For this we simply define &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ f_2 = \begin{cases}&#xD;&#xA;cx &amp;amp; x\in[0,1]\\&#xD;&#xA;c - c(x-1) &amp;amp; x\in(1,2]\\&#xD;&#xA;0 &amp;amp; x\notin [0,2]&#xD;&#xA;\end{cases} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You easily check that $f_2$ is continuous, has compact support, and has periodicization the constant value $c$. So $f = f_1 + f_2$ will &quot;wrap&quot; to $g$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(By suitable mollification, $f_2$ can clearly be made smooth. By considering derivatives, you can make $f_1$ as smooth as the function $g$.)&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Edit:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After thinking about this more, a slightly simpler construction is as follows. Observe that we assumed $g$ to be a continuous, periodic function. Extend $g$ to over $\mathbb{R}$ by periodicity. Then it suffices to construct a function $\psi$, with the following property:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\sum_{k = -\infty}^\infty \psi(x+k) = 1$&lt;/li&gt;&#xA;&lt;li&gt;$\psi$ is continuous (smooth)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This is because if you let $f(x) = g(x) \psi(x)$, using periodicity of $g$ and the above property of $\psi$, $f$ has the desired properties. Now let $\eta$ be a smooth transition function: $\eta = 0$ if $x &amp;lt; 0$, $\eta = 1$ if $x &amp;gt; 1$, and $\eta$ is smooth. Consider &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \phi(x) = \frac{1}c \eta(x) (1- \eta(x - k)) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $k$ is a positive integer and the constant $c$ is chosen such that the compactly supported function $\phi(x)$ verifies the condition above on its sum. (By construction the periodic sum of $\eta(x)(1- \eta(x-k))$ is some constant integer.)  &lt;/p&gt;&#xA;" OwnerUserId="1543" LastEditorUserId="1543" LastEditorDisplayName="" LastEditDate="2011-03-05T16:31:58.170" LastActivityDate="2011-03-05T16:31:58.170" CommentCount="1" />
  <row Id="25155" PostTypeId="2" ParentId="24994" CreationDate="2011-03-05T13:28:14.623" Score="2" ViewCount="" Body="&lt;p&gt;You can find a couple of short proofs in the &lt;a href=&quot;http://www.emis.de/journals/JIPAM/images/289_06_JIPAM/289_06_www.pdf&quot; rel=&quot;nofollow&quot;&gt;Journal of Inequalities in Pure and&#xA;Applied Mathematics&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1301" LastActivityDate="2011-03-05T13:28:14.623" />
  <row Id="25156" PostTypeId="1" CreationDate="2011-03-05T13:36:33.977" Score="0" ViewCount="63" Body="&lt;p&gt;What are applications of Rudin–Keisler order of ultrafilters?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the purpose we study it for?&lt;/p&gt;&#xA;" OwnerUserId="4876" LastActivityDate="2011-03-05T13:36:33.977" CommunityOwnedDate="2011-03-05T16:44:13.090" Title="What are applications of Rudin–Keisler order of ultrafilters?" Tags="&lt;set-theory&gt;" CommentCount="7" />
  <row Id="25157" PostTypeId="1" AcceptedAnswerId="25163" CreationDate="2011-03-05T13:42:07.993" Score="4" ViewCount="183" Body="&lt;p&gt;Hey there,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to prove the following theorem (no idea whether it has a name):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $V = \mathbb{R}^n$ or $\mathbb{C}^n$ and $||.||$ be a norm on $V$. Then, there exist $C_1, C_2 &amp;gt; 0$ such that for all $x \in V$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$C_1 ||x||_\infty \leq ||x|| \leq C_2 ||x||_\infty$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I first let $x \neq 0$ (otherwise it would be trivial). Then, I divided by $||x||_\infty$ and normalized the vector $x$ such that $||x||_\infty = 1$. That left me with&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$C_1 \leq ||x|| \leq C_2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but I don't see how this could help me or how I could possibly limit an unknown norm. How can I proceed? Or is this the wrong way anyway? Thanks for any answers.&lt;/p&gt;&#xA;" OwnerUserId="3787" LastEditorUserId="3787" LastEditorDisplayName="" LastEditDate="2011-03-19T20:15:16.230" LastActivityDate="2011-05-07T18:47:47.560" Title="How to prove $C_1 ||x||_\infty \leq ||x|| \leq C_2 ||x||_\infty$?" Tags="&lt;homework&gt;&lt;analysis&gt;&lt;norm&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="25158" PostTypeId="1" CreationDate="2011-03-05T13:46:53.293" Score="1" ViewCount="213" Body="&lt;p&gt;Dear all, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm trying to derive the LTE for CN applied to the linear heat equation; &#xA;$u_t = u_{xx}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is that I end up with terms of the form $\frac{{\Delta t}^k}{{\Delta x}^2}$&#xA;when using a two dimensional Taylor expansion around $(x,t)$ for the term:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;${\Delta x}^2 {\delta^2_x} = (u_{i+1}^{n+1} - 2 u_{i}^{n+1} + u_{i-1}^{n+1})$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What am I doing wrong?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(trying to prove LTE for the Crandall-Douglas scheme)&lt;/p&gt;&#xA;" OwnerUserId="7827" LastEditorUserId="7827" LastEditorDisplayName="" LastEditDate="2011-03-05T13:52:13.803" LastActivityDate="2011-06-03T20:10:43.797" Title="Crank-Nicolson Local Truncation Error" Tags="&lt;numerical-methods&gt;&lt;pde&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="25159" PostTypeId="2" ParentId="25128" CreationDate="2011-03-05T13:47:23.693" Score="3" ViewCount="" Body="&lt;p&gt;Existence is easy:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The solutions are $\{1,2^{k-1}-1,2^{k-1}+1,2^k-1\}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Squaring them gives $\{1, 2^{2k-2}-2^k+1, 2^{2k-2}+2^k+1, 2^{2k}-2^{k+1}+1\}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reducing $\pmod {2^k}$ gives 1 in each case (given that $2k-2 &amp;gt; k+1$, which forces $k \ge 3$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For uniqueness you could use the fact that the units group of $\mathbb{Z}/2^k \mathbb{Z}$ is $C_2 \times C_{2^{k-2}}$ on the other hand that also gives existence..&lt;/p&gt;&#xA;" OwnerUserId="4290" LastEditorUserId="4290" LastEditorDisplayName="" LastEditDate="2011-03-05T16:37:18.727" LastActivityDate="2011-03-05T16:37:18.727" CommentCount="4" />
  <row Id="25160" PostTypeId="2" ParentId="25147" CreationDate="2011-03-05T14:06:50.220" Score="2" ViewCount="" Body="&lt;p&gt;For (1), note that each state has its own period.  One can show that any two communicating states have the same period, so for an irreducible Markov chain, all states have the same period, and thus one may call this the period of the chain.  For a reducible chain, non-communicating states may have different periods, so the &quot;period of the chain&quot; is not well defined.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For (2), the notion of stationary distribution still makes sense for a non-reducible Markov chain, but in general there may be more than one of them.  Each of them may or may not satisfy the balance equations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The theory of reducible Markov chains effectively reduces to that for irreducible chains, because of the following fact:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Proposition&lt;/strong&gt;.  The state space $S$ of any Markov chain may be partitioned as $S = T \cup \bigcup_i C_i$, where the $C_i$ are closed and irreducible, and for every $x \in T$ there exists $y \in S$ such that $x \to y$ but $y \not\to x$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So each $C_i$ can be considered an irreducible Markov chain in its own right.  Each $C_i$ has a well-defined period $p_i$, but of course these periods may be different for different $i$.  Each $C_i$ has at most one stationary distribution; if $C_i$ has stationary distribution $\pi_i$, it extends to a stationary distribution for $S$ by setting $\pi_i(x)=0$ for $x \notin C_i$.  Thus the chain has (potentially) one stationary distribution for each $C_i$, and one can get more of them by taking convex combinations $\pi = \sum_i a_i \pi_i$ where $a_i \ge 0$, $\sum_i a_i = 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The states of $T$ are all transient, so they don't really enter into the limiting behavior of the chain.   For instance, any stationary distribution $\pi$ for $S$ must have $\pi(x) = 0$ for $x \in T$.&lt;/p&gt;&#xA;" OwnerUserId="822" LastActivityDate="2011-03-05T14:06:50.220" />
  <row Id="25161" PostTypeId="1" AcceptedAnswerId="25165" CreationDate="2011-03-05T14:32:37.317" Score="1" ViewCount="108" Body="&lt;p&gt;Hello! Suppose matrix $A\in R^{3 \times 3} $ and &#xA;$$A=&#xD;&#xA;\left[ \begin{array}{ccc} b_1 &amp;amp; b_2 &amp;amp; b_3 \end{array} \right]&#xD;&#xA;\left[ \begin{array}{ccc} c_1 &amp;amp; 0 &amp;amp; 0 \\\ 0 &amp;amp; c_2 &amp;amp; 0 \\\ 0 &amp;amp; 0 &amp;amp; c_3 \end{array} \right]&#xD;&#xA;\left[ \begin{array}{ccc} b_1^T \\\ b_2^T \\\ b_3^T \end{array} \right]&#xD;&#xA;$$&#xA;where $b_i\in R^3$ and $B=(b_1,b_2,b_3)$ is non-singular.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $Ab_i=\lambda_i b_i, i=1,2,3$, that is $b_i$ is the eigenvectors of A, can we say  $ b_1,b_2,b_3 $ are orthogonal eigenvectors of $A$ and $ c_1,c_2,c_3$ are eigenvalues of $A$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you very much.&#xA;Shiyu&lt;/p&gt;&#xA;" OwnerUserId="7156" LastActivityDate="2011-03-05T16:16:14.317" Title="A question on orthogonal eigenvectors of positive definite matrices" Tags="&lt;linear-algebra&gt;&lt;eigenvalues&gt;&lt;eigenvectors&gt;" AnswerCount="2" CommentCount="6" />
  <row Id="25162" PostTypeId="2" ParentId="4043" CreationDate="2011-03-05T14:47:00.903" Score="0" ViewCount="" Body="&lt;p&gt;There are lots of good books on symbolic logic, &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0132051524&quot; rel=&quot;nofollow&quot;&gt;Understanding Symbolic Logic&lt;/a&gt; being one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, there aren't many of them using a &quot;programming&quot; approach, &lt;a href=&quot;http://homepages.cwi.nl/~jve/HR/&quot; rel=&quot;nofollow&quot;&gt;Haskell Road to Logic, Maths, and programming&lt;/a&gt; being such a gem. &lt;/p&gt;&#xA;" OwnerUserId="7828" LastEditorUserId="7828" LastEditorDisplayName="" LastEditDate="2011-03-05T14:54:35.827" LastActivityDate="2011-03-05T14:54:35.827" CommunityOwnedDate="2011-03-05T14:47:00.903" />
  <row Id="25163" PostTypeId="2" ParentId="25157" CreationDate="2011-03-05T15:02:19.637" Score="2" ViewCount="" Body="&lt;p&gt;I try to make a self-contained answer to the question (more info can be found on &lt;a href=&quot;http://planetmath.org/encyclopedia/ProofThatAllNormsOnFiniteVectorSpaceAreEquivalent.html&quot; rel=&quot;nofollow&quot;&gt;this page&lt;/a&gt;). The short answer is: yes you are on the right track!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you are missing:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) $||x||$ is a continuous function with respect to the $||\cdot||_\infty$ norm. You can write $x= \sum_n x_n e_n$ with $e_n$ a basis with $||e_n||=1$ (similar $y= \sum_n y_n e_n$). We are interested in $$||x-y|| = || \sum_n (x_n - y_n) e_n || \leq \sum_n | (x_n - y_n)|  \leq || x - y||_\infty$$ using the triangle inequality. So for every $\epsilon&amp;gt;0$, we can chose a $\delta = \epsilon$ such that with $|| x - y||_\infty &amp;lt; \delta$ it follows that $||x-y|| &amp;lt; \epsilon$. Therefore, $||x||$ is $\infty$-continuous.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) The unit sphere $S= \{x :||x||_\infty = 1\}$ is a compact set thereby $||x||$ assumes a minimum $C_1$ and a maximum $C_2$ on it.&lt;/p&gt;&#xA;" OwnerUserId="7266" LastEditorUserId="7266" LastEditorDisplayName="" LastEditDate="2011-05-07T18:47:47.560" LastActivityDate="2011-05-07T18:47:47.560" CommentCount="2" />
  <row Id="25164" PostTypeId="2" ParentId="25136" CreationDate="2011-03-05T15:05:11.993" Score="7" ViewCount="" Body="&lt;p&gt;Use the definition! (Note we prove equation (1) which appears in Derek's answer).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Stirling numbers, $\displaystyle \left\{ { n \atop k } \right\}$ are the number of ways to partition set of $n$ items into $k$ non-empty subsets:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $\displaystyle x$ be any positive integer $\displaystyle \gt n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let us count the number of ways to choose an $\displaystyle n$-tuple $\displaystyle (a_1, a_2, \dots, a_n)$, by picking the $\displaystyle a_i$ from $\{1,2, \dots, x\}$ with replacement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The total possibilities is $\displaystyle x^n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now count the same, by counting tuples with exactly $\displaystyle j$ distinct colours, and adding up varying $\displaystyle j$ from $\displaystyle 0$ to $\displaystyle n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the identity is a polynomial in $\displaystyle x$, and is satisfied by infinitely many integers $\displaystyle x$, this is true even if $\displaystyle x$ is made complex.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2011-03-05T15:05:11.993" />
  <row Id="25165" PostTypeId="2" ParentId="25161" CreationDate="2011-03-05T15:18:57.930" Score="1" ViewCount="" Body="&lt;p&gt;We know that $A$ has a basis of eigenvectors, since $B$ is non-singular. Thus we can write:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$A = BDB^{-1}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $D$ is a diagonal matrix containing the eigenvalues. Since $A$ is positive definite, all eigenvalues are positive, so $D$ is invertible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$BDB^{-1} = BCB^T \Leftrightarrow DB^{-1} = CB^T \Leftrightarrow B^{-1} = D^{-1}CB^T$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So $B^{-1}B^{-T} = D^{-1}C$. Since $B^{-1}B^{-T}$ is invertible, $D^{-1}C$ has to be invertible (all $c_i \neq 0$), so we can write $B^TB = C^{-1}D$. Since $C^{-1}D$ is a diagonal matrix, this shows that the vectors $b_1, b_2, b_3$ are orthogonal (but not necessarily orthonormal).&lt;/p&gt;&#xA;" OwnerUserId="5966" LastEditorUserId="5966" LastEditorDisplayName="" LastEditDate="2011-03-05T15:43:28.680" LastActivityDate="2011-03-05T15:43:28.680" CommentCount="5" />
  <row Id="25166" PostTypeId="2" ParentId="25132" CreationDate="2011-03-05T15:51:54.580" Score="1" ViewCount="" Body="&lt;p&gt;The technique is well described in  &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/1568811306&quot; rel=&quot;nofollow&quot;&gt;Winning Ways&lt;/a&gt; or (in less detail) at &lt;a href=&quot;http://www.yucs.org/~gnivasch/cgames/spraguegrundy/index.html&quot; rel=&quot;nofollow&quot;&gt;Sprague-Gundy theory&lt;/a&gt;.  Each tree is a nim-heap.  To get its value, look at the value of all subtrees that you can move to and the value is the minimum value not in the set.  For example, consider a tree with a single stick from the root and three branches at the top.  You can move to zero or a similar tree with two branches at the top.  To get the value of the one with two branches, you can move to zero or two, so its value is *1.  Then the one with three can move to 0 or *1, so has value *2.  The value of the forest is the bitwise XOR of the binary values of all the trees.  The position is losing if the value is zero.  A single tree will never be zero as you can always chop it down at the root.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2011-03-05T16:21:57.723" LastActivityDate="2011-03-05T16:21:57.723" />
  <row Id="25167" PostTypeId="2" ParentId="25161" CreationDate="2011-03-05T16:07:03.393" Score="0" ViewCount="" Body="&lt;p&gt;If all the λs are different, then the eigenvectors of A are necessarily orthogonal and the cs are eigenvalues (assuming the $b$s are also normalized). If some of the λs are the same, this is not necessarily the case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an example what goes wrong when some $\lambda$ are equal, take the case $A= I_2$ (a $3\times 3$ example is also easy to construct). We have&#xA;$$ A= \begin{pmatrix} b_1 &amp;amp; b_2 \end{pmatrix}&#xD;&#xA;\begin{pmatrix} c_1 &amp;amp; 0  \\\ 0 &amp;amp; c_2   \end{pmatrix}&#xD;&#xA; \begin{pmatrix} b_1^T \\\ b_2^T \end{pmatrix} $$&#xA;with&#xA;$$&#xD;&#xA;\begin{align}&#xD;&#xA; c_1 &amp;amp;=1 &amp;amp; c_2 &amp;amp;=2 \\&#xD;&#xA; b_1 &amp;amp;= \frac{1}{\sqrt{2}}\begin{pmatrix} 1\\1\end{pmatrix} &amp;amp; &#xD;&#xA;b_2 &amp;amp; = \frac{1}{2} \begin{pmatrix} 1 \\ -1 \end{pmatrix}&#xD;&#xA;\end{align},&#xD;&#xA;$$&#xA;so the $b$s are orthogonal but the $c$s are not the eigenvalues. (Calle post shows that in this case the $b$s have to be orthogonal)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To find an example where the $b$s are not orthogonal, take $A=0_2$. We can write the matrix in the form given above with&#xA;$$&#xD;&#xA;\begin{align}&#xD;&#xA; c_1 &amp;amp;=1 &amp;amp; c_2 &amp;amp;=-1 \\&#xD;&#xA; b_1 &amp;amp;= \begin{pmatrix} 1\\1\end{pmatrix} &amp;amp; &#xD;&#xA;b_2 &amp;amp; = -b_1&#xD;&#xA;\end{align},&#xD;&#xA;$$&#xA;so neither are the $c$s the eigenvalues nor the $b$s orthogonal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the case when all the $\lambda$s are different, the $b$s are up to normalization uniquely determined. They are automatically orthogonal on each other. If they are not normalized the $c$s are not necessarily the eigenvalues. But if you assume them normalized then the $c$s correspond to the $\lambda$s.&lt;/p&gt;&#xA;" OwnerUserId="7266" LastEditorUserId="7266" LastEditorDisplayName="" LastEditDate="2011-03-05T16:16:14.317" LastActivityDate="2011-03-05T16:16:14.317" CommentCount="1" />
  <row Id="25168" PostTypeId="2" ParentId="25132" CreationDate="2011-03-05T16:31:15.610" Score="0" ViewCount="" Body="&lt;p&gt;It is known as &lt;a href=&quot;http://en.wikipedia.org/wiki/Hackenbush&quot; rel=&quot;nofollow&quot;&gt;hackenbush&lt;/a&gt;. It's theory is well developed and you may google a lot...&lt;/p&gt;&#xA;" OwnerUserId="2481" LastActivityDate="2011-03-05T16:31:15.610" CommentCount="1" />
  <row Id="25169" PostTypeId="1" AcceptedAnswerId="26879" CreationDate="2011-03-05T16:50:20.350" Score="13" ViewCount="451" Body="&lt;p&gt;Here is an interesting infinite series. It would be great to see a method to evaluate it, if possible. I know it converges to a little less than 11/40&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle\sum_{k=1}^{\infty}\frac{1}{4^{k}+k!}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I could not think of any good identities to start this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks a million to those who can show how to evaluate it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe even in general, $\displaystyle\sum_{k=1}^{\infty}\frac{1}{x^{k}+k!}$, where $x\geq 1$&lt;/p&gt;&#xA;" OwnerUserId="6358" LastEditorUserId="3731" LastEditorDisplayName="" LastEditDate="2011-03-05T17:09:03.177" LastActivityDate="2011-03-15T09:58:57.420" Title="Are there clever ways to evaluate this infinite series?" Tags="&lt;sequences-and-series&gt;" AnswerCount="1" CommentCount="10" FavoriteCount="4" />
  <row Id="25170" PostTypeId="1" CreationDate="2011-03-05T16:52:53.780" Score="10" ViewCount="136" Body="&lt;p&gt;I know that the complex projective line $\mathbb{C}P^1$ can be embedded in  the complex projective space $\mathbb{C}P^n$ (Veronese embedding). For example,   $\mathbb{C}P^1\rightarrow\mathbb{C}P^3$ is given explicitly by $(z,w)\mapsto(z^3,z^2w,zw^2,w^3)$ in homogenous coordinates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was wondering if the same could be done with quaternionic projective spaces, i.e. is there a 'veronese type' embedding: $\mathbb{H}P^1\rightarrow\mathbb{H}P^n$ ??&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I know that the non-comutativity of quaternions make the veronese map ill defined)&lt;/p&gt;&#xA;" OwnerUserId="7831" LastActivityDate="2011-04-14T20:51:25.960" Title="Quaternionic veronese Embedding" Tags="&lt;quaternions&gt;&lt;projective-geometry&gt;" AnswerCount="1" CommentCount="6" FavoriteCount="2" />
  <row Id="25171" PostTypeId="1" AcceptedAnswerId="25173" CreationDate="2011-03-05T16:53:18.790" Score="0" ViewCount="152" Body="&lt;p&gt;Hi ::- ). To find the Point where 2 lines intersect, is there any simpler method than this? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Line-line_intersection&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Line-line_intersection&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example could you use the equation of the line? (y = mx + b) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another question: what would happen if I use Determinants for 2 lines that never intersect? (parallel lines). I think I should mention that I intend to do this calculation in a programming language (AS3). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suppose that those determinants will return something along the lines of division by zero / +/- infinity for parallel lines. I didn't try it yet ::- D. Still trying to figure out the best way to get the intersection point. Determinants or... do you know another method?&lt;/p&gt;&#xA;" OwnerUserId="7745" LastActivityDate="2011-03-05T17:35:14.990" Title="Intersection Of Two Lines In 2D" Tags="&lt;geometry&gt;" AnswerCount="2" CommentCount="0" />
  <row Id="25172" PostTypeId="2" ParentId="25171" CreationDate="2011-03-05T17:06:03.133" Score="2" ViewCount="" Body="&lt;p&gt;What about a system of linear equations? Like&#xA;$$\begin{aligned}y&amp;amp;=m_1x+b_1\\y&amp;amp;=m_2x+b_2\end{aligned}$$&#xA;I actually thought you learn to do this kind of stuff at school... at least I did.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And determinants are just a result because you use an equation system. If there's no solution, you're right, the determinants get zero and if you have to divide it gets infinit at some point. But that could simply be checked with an if statement.&lt;/p&gt;&#xA;" OwnerUserId="7832" LastActivityDate="2011-03-05T17:06:03.133" CommentCount="3" />
  <row Id="25173" PostTypeId="2" ParentId="25171" CreationDate="2011-03-05T17:25:15.250" Score="2" ViewCount="" Body="&lt;p&gt;well you probably want to include vertical lines, so you should have a system of the form&#xA;$a_1x+b_1y=c_1, a_2x+b_2y=c_2$.  if $a_1b_2-a_2b_1=0$ you have parallel (or identical) lines.  else return the point of intersection, which is&#xA;$$&#xD;&#xA;(x,y)=\Big(\frac{c_1b_2-b_1c_2}{a_1b_2-b_1a_2},\frac{a_1c_2-c_1a_2}{a_1b_2-b_1a_2}\Big).&#xD;&#xA;$$&#xA;for instance using cramer's rule. in general, you should take linear algebra at some point if you haven't. you will need it.&lt;/p&gt;&#xA;" OwnerUserId="6925" LastEditorUserId="6925" LastEditorDisplayName="" LastEditDate="2011-03-05T17:35:14.990" LastActivityDate="2011-03-05T17:35:14.990" CommentCount="1" />
  <row Id="25174" PostTypeId="1" AcceptedAnswerId="25177" CreationDate="2011-03-05T17:58:18.097" Score="1" ViewCount="143" Body="&lt;p&gt;Hi ::- ). I'm working on a Flash game and I need to find when a shooting line intersects an enemy. The enemy's shape is rectangular. So far, I was thinking to test this like so:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Test if the shooting line intersects ANY of the lines composing the rectangle. &lt;strong&gt;BUT&lt;/strong&gt; while the shooting line can be endless, I need to only validate an intersection if the shooting line ACTUALLY TOUCHES THE INSIDE of any of the rectangle's lines, meaning that the X/Y of the intersection point has to be along one of those lines.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I'm wondering if there is an easier method. Like, for example, the one they used here for a Triangle:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://stackoverflow.com/questions/3590308/testing-if-a-line-has-a-point-within-a-triangle&quot;&gt;http://stackoverflow.com/questions/3590308/testing-if-a-line-has-a-point-within-a-triangle&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this applicable to a Rectangle as well? If so, could you please explain a bit of the reasoning? ::- ). It's been a long time since I did any math, but I'm faithfully coming back to it now, because I want to continue developing games. Any help is greatly appreciated and will not be forgotten ::- D. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S.: Also, my games are freeware! ::- )&lt;/p&gt;&#xA;" OwnerUserId="7745" LastActivityDate="2011-03-05T18:42:05.017" Title="Intersection Of Line and Rectangle in 2D" Tags="&lt;geometry&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="25176" PostTypeId="1" AcceptedAnswerId="25198" CreationDate="2011-03-05T18:19:58.687" Score="1" ViewCount="124" Body="&lt;p&gt;I'm unable to prove the following two statements: &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;$ \displaystyle x^n+x^{\frac{1}{n}}+n = 0$ has no real zeros for all $n\geqslant 2.$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$ \displaystyle x^n+x^{\frac{1}{n}}-n = 0$ has exactly one real zero for $n \geqslant 1$. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;(where $n$ is an integer and $x$ is a positive real number). &lt;/p&gt;&#xA;" OwnerUserId="4558" LastEditorUserId="4558" LastEditorDisplayName="" LastEditDate="2011-03-05T18:53:33.180" LastActivityDate="2011-03-05T21:34:17.580" Title="Show that $ \displaystyle x^n+x^{\frac{1}{n}}+n = 0$ has no real zeros for $n\geqslant 2$" Tags="&lt;calculus&gt;&lt;analysis&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="1" />
  <row Id="25177" PostTypeId="2" ParentId="25174" CreationDate="2011-03-05T18:34:47.627" Score="1" ViewCount="" Body="&lt;p&gt;Yes, that solution would work! And I think it is quite clever.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One observes that a line $ax + by + c = 0$ in the plane defines two semi-planes: the points $(\overline{x}, \overline{y})$such that $a\overline{x} + b\overline{y} + c &amp;gt; 0$ and those for which $a\overline{x} + b\overline{y} + c &amp;lt; 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The algorithm works because by substituting the coordinates of the vertices one finds in which semi-plane they lie. If some vertices lie in the upper semi-plane and some in the lower semi-plane then we must conclude that the line does intersect the polygon.&lt;/p&gt;&#xA;" OwnerUserId="4471" LastEditorUserId="4471" LastEditorDisplayName="" LastEditDate="2011-03-05T18:42:05.017" LastActivityDate="2011-03-05T18:42:05.017" CommentCount="1" />
  <row Id="25178" PostTypeId="2" ParentId="25176" CreationDate="2011-03-05T18:37:08.530" Score="2" ViewCount="" Body="&lt;p&gt;Here's a hint: observe that if $n &amp;lt; 0$ the function $f(x) = x^{n} + x^{1 \over n}$ is decreasing for $x \geq 0$, and if $n &amp;gt; 0$ then $f(x)$ is increasing for $x \geq 0$. &lt;/p&gt;&#xA;" OwnerUserId="3035" LastActivityDate="2011-03-05T18:37:08.530" />
  <row Id="25179" PostTypeId="2" ParentId="25176" CreationDate="2011-03-05T18:48:01.480" Score="1" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Hint:&lt;/strong&gt;  For 1. the left hand side is always positive, and for 2. notice the function is monotonic on $(0,\infty)$.  (You need to specify that $n\geq 1$ for the second one)&lt;/p&gt;&#xA;" OwnerUserId="6075" LastActivityDate="2011-03-05T18:48:01.480" />
  <row Id="25180" PostTypeId="1" CreationDate="2011-03-04T15:08:00.913" Score="1" ViewCount="37" Body="&lt;p&gt;I would like to know how the MSU - Video Quality measurement tool:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://compression.ru/video/quality_measure/info_en.html#ybluringmeasure&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;calculates the blur metric. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What could be the mathematical computation to calculate blur?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to do something similar in my Image pipeline.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT: If I have a blurry image, I upsize it / downsize it, and then compute the blur again for this resampled image, would it be same/more/less?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="1249" LastActivityDate="2011-03-05T18:49:15.247" Title="VQMT image blurring metric" Tags="&lt;image-processing&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="25181" PostTypeId="2" ParentId="25180" CreationDate="2011-03-04T17:45:15.833" Score="2" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;There are many possible blur metrics&lt;/strong&gt;.  For a list of some, see this &lt;a href=&quot;http://scien.stanford.edu/pages/labsite/2010/psych221/projects/2010/AdityaAnchuri/main.pdf&quot; rel=&quot;nofollow&quot;&gt;student paper&lt;/a&gt; (from Stanford U).  For many more, Google &quot;&lt;a href=&quot;http://www.google.com/search?q=blurring+metric+image&quot; rel=&quot;nofollow&quot;&gt;blurring metric image&lt;/a&gt;&quot;.  Most appear to measure some form of local contrast.  (This suggests downsized images would be measured as less blurry than the originals and upsized as more blurry, but as @mattdm comments, the blurriness of the upsized image depends on the resampling algorithm.)  Do not expect any simple mathematical formula.  For example, one method first runs an edge detector over the image and then estimates the edge widths (wider = blurrier of course).  For the details, and to answer the edited question, you would need to consult the MSU people, because they do not appear to have publicly documented their metric.&lt;/p&gt;&#xA;" OwnerUserId="1489" LastActivityDate="2011-03-04T17:45:15.833" />
  <row Id="25182" PostTypeId="1" AcceptedAnswerId="25194" CreationDate="2011-03-05T19:02:24.290" Score="3" ViewCount="112" Body="&lt;p&gt;Does anyone know how to derive a formula for the coefficients. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is if, &#xA;$f(x)=\sum _{n=0}^{\infty } a_nx^n$ and $g(x)=\sum _{n=0}^{\infty } b_nx^n$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;suppose the compostion is an analytic function, $h(x)=f(g(x))=\sum _{n=0}^{\infty } c_nx^n$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there an expression we can find for the coefficients $c_n$ in terms of $a_n$ and $b_n$? Can someone show me how its derived. I know we could substitute $g$ into $f$ and collect powers of $x$. But I believe a formula for general n may be written down. &lt;/p&gt;&#xA;" OwnerUserId="6386" LastActivityDate="2011-03-05T19:54:24.703" Title="composition of power series." Tags="&lt;real-analysis&gt;&lt;complex-analysis&gt;&lt;power-series&gt;&lt;analytic&gt;" AnswerCount="3" />
  <row Id="25183" PostTypeId="2" ParentId="25182" CreationDate="2011-03-05T19:05:25.503" Score="2" ViewCount="" Body="&lt;p&gt;Please refer this link:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://unapologetic.wordpress.com/2008/09/24/composition-of-power-series/&quot; rel=&quot;nofollow&quot;&gt;http://unapologetic.wordpress.com/2008/09/24/composition-of-power-series/&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" LastActivityDate="2011-03-05T19:05:25.503" CommentCount="1" />
  <row Id="25184" PostTypeId="2" ParentId="25137" CreationDate="2011-03-05T19:07:35.883" Score="1" ViewCount="" Body="&lt;p&gt;Knowing some deeper theorems will help to see what to do.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Riesz representation theorem says that the continuous linear functionals on $C([0,1])$ are precisely the signed Radon measures on $[0,1]$.  It's not hard to see that any bounded measurable function is a continuous linear functional on the space of signed measures, i.e. is an element of $C([0,1])^{**}$.  So the existence of bounded discontinuous functions on $[0,1]$ shows that $C([0,1])$ cannot be reflexive.&lt;/p&gt;&#xA;" OwnerUserId="822" LastActivityDate="2011-03-05T19:07:35.883" />
  <row Id="25186" PostTypeId="2" ParentId="25133" CreationDate="2011-03-05T19:17:05.267" Score="5" ViewCount="" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The cardinality of the set of functions from $X$ to $Y$ is the definition of the cardinal $Y^X$. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The number of order-preserving functions from $X$ to $Y$, given that well-orders of each set have been fixed, depends on the nature of those orders. For example, there are no such orders in the case that the order type of $X$ is longer than the order type of $Y$. If  $X$ and $Y$ are finite, then there is some interesting combinatorics involved to give the right answer. For example, if both are finite of the same size, there is only one order-preserving function. If $Y$ is one bigger, then there are $Y$ many (you can put the hole anywhere). And so on. If $Y$ is infinite, of size at least $X$, then you get $Y^X$ again, since you can code any function into the omitted part, by leaving gaps of a certain length. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;A surjective order-preserving map is an isomorphism, and for well-orders, this is unique if it exists at all, so the answer is either 0 or 1, depending on whether the orders are isomorphic or not.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="413" LastEditorUserId="413" LastEditorDisplayName="" LastEditDate="2011-03-05T19:40:57.697" LastActivityDate="2011-03-05T19:40:57.697" CommentCount="5" />
  <row Id="25187" PostTypeId="1" AcceptedAnswerId="25201" CreationDate="2011-03-05T19:19:47.403" Score="3" ViewCount="243" Body="&lt;p&gt;Does anybody have a simple proof this inequality &#xA;$$e^x\le x+e^{x^2}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="6016" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-05T20:26:24.233" LastActivityDate="2011-03-06T00:15:57.567" Title="proof of inequality $e^x\le x+e^{x^2}$" Tags="&lt;inequality&gt;" AnswerCount="4" CommentCount="0" />
  <row Id="25188" PostTypeId="2" ParentId="25187" CreationDate="2011-03-05T19:22:59.973" Score="2" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Hint:&lt;/strong&gt;  Consider the power series of $e^x$ which converges on all of $\mathbb{R}$.  Then $$e^x-x=1+\frac{x^2}{2}+\frac{x^3}{3!}+\frac{x^4}{4!}+\cdots$$ and $$e^{x^2}=1+x^2+\frac{x^4}{2}+\frac{x^6}{3!}+\cdots .$$ For small $x$ the $x^2$ dominates, and for $|x|\geq 1$ the second power series will dominate the first.  (The coefficient of $x^{2n}$ is large enough to cover both coefficients of $x^{2n-1}$ and $x^{2n}$ in the first expansion.)&lt;/p&gt;&#xA;" OwnerUserId="6075" LastActivityDate="2011-03-05T19:22:59.973" CommentCount="4" />
  <row Id="25189" PostTypeId="2" ParentId="25187" CreationDate="2011-03-05T19:25:45.993" Score="4" ViewCount="" Body="&lt;p&gt;Consider $f(x) = x+e^{x^2}-e^x$. Then $f&amp;#39;(x) = 1+ 2xe^{x^2}-e^x$. Find the critical points. So the minimum of $f(x)$ is $0$ which implies that $e^x \leq x+e^{x^2}$. &lt;/p&gt;&#xA;" OwnerUserId="2548" LastEditorUserId="2548" LastEditorDisplayName="" LastEditDate="2011-03-05T19:31:39.093" LastActivityDate="2011-03-05T19:31:39.093" CommentCount="1" />
  <row Id="25190" PostTypeId="1" AcceptedAnswerId="25195" CreationDate="2011-03-05T19:26:30.127" Score="1" ViewCount="65" Body="&lt;p&gt;Hello ::- ). I'm trying to check if a line hits a rectangle, and for that, I found this nice solution:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://math.stackexchange.com/questions/3534/line-triangle-intersection&quot;&gt;Line triangle intersection.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is that, having forgot almost all I ever knew about math, I don't know how to obtain that equation of a line which they use there: &lt;code&gt;4x − 3y + 2 = 0&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know a line's equation is &lt;code&gt;y = mx + b&lt;/code&gt;. But the above form and how can I obtain it is a mystery to me. Can somebody please bring a flashlight? ::- D.&lt;/p&gt;&#xA;" OwnerUserId="7745" LastActivityDate="2011-03-05T20:22:15.867" Title="How to obtain Line Equation of the form ax + bx + c = 0" Tags="&lt;geometry&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="0" />
  <row Id="25191" PostTypeId="2" ParentId="25182" CreationDate="2011-03-05T19:38:23.237" Score="1" ViewCount="" Body="&lt;p&gt;There is a closed form, but it is kind of complicated: &lt;a href=&quot;http://en.wikipedia.org/wiki/Fa%C3%A0_di_Bruno%27s_formula&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Fa%C3%A0_di_Bruno%27s_formula&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3035" LastActivityDate="2011-03-05T19:38:23.237" CommentCount="2" />
  <row Id="25192" PostTypeId="2" ParentId="25190" CreationDate="2011-03-05T19:39:28.270" Score="1" ViewCount="" Body="&lt;p&gt;You can transform the given equation to make it explicit in $y$ to obtain the form that you are familiar with:&#xA;$$&#xD;&#xA;\begin{align}&#xD;&#xA; 4x - 3y + 2 =&amp;amp; 0 \\&#xD;&#xA; 4 x + 2 =&amp;amp; 3y \\&#xD;&#xA; \frac{4}{3}x + \frac{2}{3} =&amp;amp; y \\&#xD;&#xA; y =&amp;amp; \frac{4}{3}x + \frac{2}{3}&#xD;&#xA;\end{align}&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From this you can tell that the line crosses the $y$-axis at $\frac{2}{3}$ and that the slope of the line is $\frac{4}{3}$: when $x$ increases by 1, $y$ increases by $\frac{4}{3}$.&lt;/p&gt;&#xA;" OwnerUserId="6678" LastActivityDate="2011-03-05T19:39:28.270" CommentCount="1" />
  <row Id="25193" PostTypeId="1" CreationDate="2011-03-05T19:52:57.940" Score="7" ViewCount="119" Body="&lt;p&gt;Suppose $p$, $k$ and $s$ are integers with $s,k \le p$. Consider the following polynomial in $x$ and $y$,&#xA;$$ \sum_{\ell=0}^k \binom{s}{\ell} \binom{p-s}{k-\ell} x^\ell &#xD;&#xA;y^{p-\ell}$$&#xA;Does this expression look familiar to anyone? Is there closed form?&lt;/p&gt;&#xA;" OwnerUserId="7202" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-05T20:30:59.687" LastActivityDate="2011-05-10T09:43:21.763" Title="A binomial coefficient identity?" Tags="&lt;combinatorics&gt;&lt;binomial-coefficients&gt;&lt;generating-functions&gt;" AnswerCount="2" />
  <row Id="25194" PostTypeId="2" ParentId="25182" CreationDate="2011-03-05T19:54:24.703" Score="1" ViewCount="" Body="&lt;p&gt;There are (rather unwieldly) &quot;closed-forms&quot; in terms of &lt;a href=&quot;http://eom.springer.de/B/b110250.htm&quot; rel=&quot;nofollow&quot;&gt;Bell polynomials&lt;/a&gt; and other closely related combinatorial objects. However, if you are really interested in efficiently &lt;em&gt;calculating&lt;/em&gt; compositions of power series then there are better algorithms, dating back at least to the work of &lt;a href=&quot;http://citeseer.ist.psu.edu/viewdoc/summary?doi=10.1.1.131.5030&quot; rel=&quot;nofollow&quot;&gt;Brent and Kung&lt;/a&gt;, from which you can find links to recent work in this area.&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2011-03-05T19:54:24.703" CommentCount="4" />
  <row Id="25195" PostTypeId="2" ParentId="25190" CreationDate="2011-03-05T20:05:17.643" Score="2" ViewCount="" Body="&lt;p&gt;If you know two points on the line, $(x_1,y_1)$ and $(x_2,y_2)$, then the vector $\langle x_2-x_1,y_2-y_1\rangle$ is in the direction of the line and the vector $\vec{n}=\langle y_1-y_2, x_2-x_1\rangle$ (swap the components, take the opposite of one component) is orthogonal (perpendicular) to the line.  For any point $(\pmb{x},\pmb{y})$ on the line, the vector $\langle \pmb{x}-x_1,\pmb{y}-y_1\rangle=\langle \pmb{x},\pmb{y}\rangle-\langle x_1,y_1\rangle$ is in the direction of the line, so it is also orthogonal to $\vec{n}$, so its dot product with $\vec{n}$ is $0$: $$\begin{align}&#xD;&#xA;\vec{n}\cdot\langle \pmb{x}-x_1,\pmb{y}-y_2\rangle&amp;amp;=0&#xD;&#xA;\\&#xD;&#xA;\vec{n}\cdot\left(\langle \pmb{x},\pmb{y}\rangle-\langle x_1,y_1\rangle\right)&amp;amp;=0&#xD;&#xA;\\&#xD;&#xA;\vec{n}\cdot\langle \pmb{x},\pmb{y}\rangle-\vec{n}\cdot\langle x_1,y_1\rangle&amp;amp;=0&#xD;&#xA;\\&#xD;&#xA;\langle y_1-y_2, x_2-x_1\rangle\cdot\langle \pmb{x},\pmb{y}\rangle-\langle y_1-y_2, x_2-x_1\rangle\cdot\langle x_1,y_1\rangle&amp;amp;=0&#xD;&#xA;\\&#xD;&#xA;(y_1-y_2)\pmb{x}+(x_2-x_1)\pmb{y}-\left((y_1-y_2)x_1+(x_2-x_1)y_1\right)&amp;amp;=0&#xD;&#xA;\\&#xD;&#xA;(y_1-y_2)\pmb{x}+(x_2-x_1)\pmb{y}+\left(-(y_1-y_2)x_1-(x_2-x_1)y_1\right)&amp;amp;=0&#xD;&#xA;\\&#xD;&#xA;(y_1-y_2)\pmb{x}+(x_2-x_1)\pmb{y}+\left((y_2-y_1)x_1-(x_2-x_1)y_1\right)&amp;amp;=0&#xD;&#xA;\end{align}$$&lt;/p&gt;&#xA;" OwnerUserId="72" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2011-03-05T20:22:15.867" LastActivityDate="2011-03-05T20:22:15.867" CommentCount="5" />
  <row Id="25196" PostTypeId="2" ParentId="25128" CreationDate="2011-03-05T20:22:24.520" Score="4" ViewCount="" Body="&lt;p&gt;For any positive integer $n$, let $\mathbb{Z}/n\mathbb{Z}$ be the usual ring of integers modulo $n$ and let $U(n) = (\mathbb{Z}/n \mathbb{Z})^{\times}$ be its &lt;strong&gt;unit group&lt;/strong&gt; -- i.e., the &quot;reduced residues&quot; modulo $n$ under multiplication.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your question can be viewed as asking about the structure of this unit group when &#xA;$n = 2^k$ for $k \geq 3$.  (Note that $U(2)$ is the trivial group and $U(4)$ has order $2$, so the structure of these groups is clear.)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact it is a standard result -- at the border of undergraduate number theory and undergraduate algebra -- to give an exact computation of $U(n)$ for all positive integers $n$.  See for instance Theorem 1 (and the discussion immediately preceding it, which reduces the general problem to the prime power case) &lt;a href=&quot;http://math.uga.edu/~pete/4400primitiveroots.pdf&quot; rel=&quot;nofollow&quot;&gt;of these notes&lt;/a&gt;.  Especially, for all $k \geq 3$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$U(2^k) \cong Z_2 \times Z_{2^{k-2}}$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where here $Z_a$ denotes a cyclic group of order $a$.  In a product $H_1 \times H_2$ &#xA;of finite (multiplicatively written) commutative groups, an element $h = (h_1,h_2)$ satisfies $h^2 = 1$ iff this holds separately for both coordinates $h_1^2 = h_2^2 = 1$.  Here $H_1$ and $H_2$ are both cyclic groups of even order, so each has exactly two elements which square to $1$, and thus $U(2^k)$ has $2 \times 2 = 4$ such elements.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course one can be much more explicit about what these elements are.  This takes place in the proof of the result as well as in the answers others have given to this question.  &lt;/p&gt;&#xA;" OwnerUserId="299" LastActivityDate="2011-03-05T20:22:24.520" />
  <row Id="25197" PostTypeId="2" ParentId="25187" CreationDate="2011-03-05T20:27:46.447" Score="1" ViewCount="" Body="&lt;p&gt;Consider $f(x) = x+\mathrm{e}^{x^2}-\mathrm{e}^x$. Then $f&amp;#39;(x) = 1+ 2x\mathrm{e}^{x^2}-\mathrm{e}^x$ and &#xA;$$f&amp;#39;&amp;#39;(x) = 2(1+2x^2)\mathrm{e}^{x^2}-\mathrm{e}^x\ge \mathrm{e}^xg(x),&#xD;&#xA;$$ &#xA;with $g(x)=2\mathrm{e}^{x^2-x}-1$. Since $x^2-x\ge\frac14$, $g(x)\ge2\mathrm{e}^{-1/4}-1&amp;gt;0$ for every $x$, hence $f&amp;#39;&amp;#39;$ is positive everywhere and $f&amp;#39;$ is increasing. Since $f&amp;#39;(0)=0$, $f&amp;#39;(x)&amp;lt;0$ if $x&amp;lt;0$ and $f&amp;#39;(x)&amp;gt;0$ if $x&amp;gt;0$. Thus, the function $x\mapsto f(x)$ is decreasing on $x\le0$ and increasing on $x\ge0$. Since $f(0)=0$, we are done.&lt;/p&gt;&#xA;" OwnerUserId="6179" LastActivityDate="2011-03-05T20:27:46.447" />
  <row Id="25198" PostTypeId="2" ParentId="25176" CreationDate="2011-03-05T20:37:11.457" Score="2" ViewCount="" Body="&lt;p&gt;Let $y=x^{1/n}$. Then $x^{n} + x^{1/n} = y^{n^2} + y$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, for any constant $k$, you have&#xA;$$\frac{d}{dy}\left(y^{n^2}+ y + k\right) = n^2y^{n^2-1} + 1.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For nonnegative $y$, this is always positive; so $x^n + x^{1/n}+a$ is strictly increasing on $[0,\infty)$. (Well, technically, $y^{n^2}+y+a$ is strictly increasing; but since $y$ is itself an increasing function of $x$ for $x\geq 0$ and $n\gt0$, then so is $x^n + x^{1/n}+a$). In particular, it has at most one zero on $[0,\infty)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $a\gt 0$, then the function is positive at $0$, hence has no zeros on $[0,\infty)$. If $a\leq 0$, then the function is nonpositive at $0$, hence has exactly one zero on $[0,\infty)$.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-05T21:34:17.580" LastActivityDate="2011-03-05T21:34:17.580" />
  <row Id="25199" PostTypeId="2" ParentId="25193" CreationDate="2011-03-05T20:44:08.047" Score="3" ViewCount="" Body="&lt;p&gt;It is the sum of $pth$ degree homogeneous terms of $y^{p-k}(1 +x)^s(1+y)^{p-s}$ &lt;/p&gt;&#xA;" OwnerUserId="7565" LastActivityDate="2011-03-05T20:44:08.047" CommentCount="2" />
  <row Id="25200" PostTypeId="1" CreationDate="2011-03-05T20:54:33.153" Score="-1" ViewCount="103" Body="&lt;p&gt;Let $a$ and $b$ be two nonzero elements of a PID $R$. Prove that direct sum $R/aR\oplus R/bR$ is isomorphic to the direct sum $R/cR\oplus R/dR$, where $c=\mathrm{lcm(a,b)}$, $d=\gcd(a,b)$.&lt;/p&gt;&#xA;" OwnerUserId="7837" LastEditorUserId="622" LastEditorDisplayName="" LastEditDate="2011-03-05T20:59:05.523" LastActivityDate="2011-03-05T21:01:58.230" Title="modules over PID" Tags="&lt;abstract-algebra&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="25201" PostTypeId="2" ParentId="25187" CreationDate="2011-03-05T21:01:28.483" Score="6" ViewCount="" Body="&lt;p&gt;For $x \geq 0$,&#xA;$$&#xD;&#xA;e^{x^2 - x} + x e^{-x} \geq 1 + (x^2 - x) + x (1-x) = 1 \&amp;gt; .&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $x &amp;lt; 0$, let $y = - x &amp;gt; 0$, whence,&#xA;$$&#xD;&#xA;e^{-x^2}(e^x - x) = e^{-y^2} (e^{-y} + y) \leq e^{-y^2} (1 - y + y^2/2 + y) \leq e^{-y^2}(1+y^2) \leq e^{-y^2} e^{y^2} = 1 \&amp;gt; .&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Notice that we've only used the basic facts that $e^x \geq 1 + x$ for all $x \in \mathbb{R}$ and that $e^{-x} \leq 1 - x + x^2/2$ for $x \geq 0$, both of which are trivial to derive by simple differentiation, similar to Didier's approach.&lt;/p&gt;&#xA;" OwnerUserId="7003" LastEditorUserId="7003" LastEditorDisplayName="" LastEditDate="2011-03-06T00:15:57.567" LastActivityDate="2011-03-06T00:15:57.567" CommentCount="1" />
  <row Id="25202" PostTypeId="2" ParentId="25200" CreationDate="2011-03-05T21:01:58.230" Score="2" ViewCount="" Body="&lt;p&gt;&lt;b&gt;Hint&lt;/b&gt;: The &lt;a href=&quot;http://en.wikipedia.org/wiki/Structure_theorem_for_finitely_generated_modules_over_a_principal_ideal_domain&quot; rel=&quot;nofollow&quot;&gt;structure theorem for modules over a PID&lt;/a&gt; gives a &lt;strong&gt;canonical form&lt;/strong&gt; for finitely generated torsion modules.  Here you have two finitely generated torsion modules, say $T_1$ and $T_2$, which you want to show are isomorphic, but neither one is (necessarily) in canonical form.  So the natural first thing to do is...&lt;/p&gt;&#xA;" OwnerUserId="299" LastActivityDate="2011-03-05T21:01:58.230" />
  <row Id="25203" PostTypeId="1" CreationDate="2011-03-05T21:09:10.310" Score="2" ViewCount="56" Body="&lt;p&gt;I am looking for all $k$-dimensional subspaces of $(\mathbb{Z}/2\mathbb{Z})^n$ up to permutational equivalence.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Is there a database of all $[n,k]$-codes up to equivalence for reasonable values of $(n,k)$?  &lt;sup&gt;For instance, $1 \leq k \leq 6$ and $6 \leq n \leq 12$ (giving around 35,000 codes).&lt;/sup&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When $k=1$, there are $n$ such subspaces, each uniquely determined by the Hamming weight of the nonzero element.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When $k=2$, there are 0, 1, 3, 6, 10, 16, 23, 32 … (&lt;a href=&quot;http://oeis.org/A034198&quot; rel=&quot;nofollow&quot;&gt;OEIS:A034198&lt;/a&gt;) such subspaces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When $k=3$, there are 0, 0, 1, 4, 10, 22, 43, 77, … (&lt;a href=&quot;http://oeis.org/A034357&quot; rel=&quot;nofollow&quot;&gt;OEIS:A034357&lt;/a&gt;) such subspaces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When $k=4$, there are 0, 0, 0, 1, 5, 16, 43, 106, &amp;hellip; (&lt;a href=&quot;http://oeis.org/A034358&quot; rel=&quot;nofollow&quot;&gt;OEIS:A034358&lt;/a&gt;) such subspaces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For each $k\geq 2$, I have trouble finding representatives of the codes for $n \geq 8$, but surely someone has recorded them somewhere.  For $n=8$, there are just 8, 32, 77, 106 such codes for $k=1,2,3,4$, so it seems reasonable to me that someone wrote them down.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have found several databases of &quot;good&quot; codes, but I actually want the bad codes too.&lt;/p&gt;&#xA;" OwnerUserId="583" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-05T21:19:16.367" LastActivityDate="2011-03-05T21:19:16.367" Title="Database of binary linear codes (including bad ones)" Tags="&lt;linear-algebra&gt;&lt;combinatorics&gt;&lt;coding-theory&gt;" />
  <row Id="25204" PostTypeId="2" ParentId="25136" CreationDate="2011-03-05T21:47:09.717" Score="2" ViewCount="" Body="&lt;p&gt;There's a simple but not very well-known formula on row sums for number triangles that can be applied here.  I believe it is due to Neuwirth.  Suppose you have a triangle of numbers $R(n,k)$ with $R(0,0) = 1$ and, for $n \geq 1$, satisfying $$R(n,k) = (\alpha (n-1) + \beta k + \gamma) R(n-1,k) + (\alpha&amp;#39; (n-1) + \beta&amp;#39; (k-1) + \gamma&amp;#39;) R(n-1,k-1).$$&#xA;There are several interesting number triangles that are special cases, such as the binomial coefficients and both kinds of Stirling numbers.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, the formula is that if $\beta + \beta&amp;#39; = 0$ then $$\sum_{k=0}^n R(n,k) = \prod_{i=0}^{n-1} \left((\alpha + \alpha&amp;#39;)i + \gamma + \gamma&amp;#39;\right).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is easy to verify that $R(n,k) = \left\{ { n \atop k } \right\} x^{\underline{k}}$ satisfies the above recurrence with $\beta = 1, \beta&amp;#39; = -1, \gamma&amp;#39; = x$ and all other parameters $0$.  Thus the row sum formula yields Derek's reformulation of the problem (Eq. (1)):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{k=0}^n \left\{ { n \atop k } \right\} x^{\underline{k}} = \prod_{i=0}^{n-1} x = x^n.$$&lt;/p&gt;&#xA;" OwnerUserId="2370" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2011-03-05T22:01:24.973" LastActivityDate="2011-03-05T22:01:24.973" />
  <row Id="25205" PostTypeId="1" AcceptedAnswerId="25211" CreationDate="2011-03-05T21:57:19.483" Score="17" ViewCount="399" Body="&lt;p&gt;My son was busily memorizing digits of $\pi$ when he asked if any power of $\pi$ was an integer.  I told him: $\pi$ is transcendental, so no non-zero integer power can be an integer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After tiring of memorizing $\pi$, he resolved to discover a new irrational whose expansion is easier to memorize.  He invented (probably re-invented) the number $J$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$J = 6.12345678910111213141516171819202122\ldots$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which clearly lets you name as many digits as you like pretty easily.  He asked me if $J$ is transcendental just like $\pi$, and I said it must be but I didn't know for sure. Is there an easy way to determine this?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can show that $\pi$ is transcendental (using Lindemann-Weierstrass) but it doesn't work for arbitrary numbers like $J$, I don't think.&lt;/p&gt;&#xA;" OwnerUserId="7162" LastEditorUserId="7162" LastEditorDisplayName="" LastEditDate="2011-03-05T22:12:07.393" LastActivityDate="2011-03-06T04:47:13.390" Title="Is this number transcendental?" Tags="&lt;irrational-numbers&gt;&lt;transcendental-numbers&gt;" AnswerCount="2" CommentCount="7" FavoriteCount="1" />
  <row Id="25206" PostTypeId="1" AcceptedAnswerId="25212" CreationDate="2011-03-05T22:00:54.583" Score="1" ViewCount="255" Body="&lt;p&gt;Hey.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I prove that $n^8 \lt 8^n-1$ for all $n$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm searching for  technical way to show that without using calculus theorems, using well known facts such as $n^2 \lt 2^n$ and so.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="6163" LastEditorUserId="6296" LastEditorDisplayName="" LastEditDate="2011-03-06T00:55:56.203" LastActivityDate="2011-03-06T19:39:29.460" Title="$n^8 \lt 8^n-1$" Tags="&lt;inequality&gt;" AnswerCount="4" CommentCount="7" />
  <row Id="25207" PostTypeId="2" ParentId="25205" CreationDate="2011-03-05T22:01:44.987" Score="12" ViewCount="" Body="&lt;p&gt;This is just the &lt;a href=&quot;http://en.wikipedia.org/wiki/Champernowne_constant&quot; rel=&quot;nofollow&quot;&gt;Champernowne constant&lt;/a&gt; plus $6$. Since the Champernowne constant is transcendental, so is this number (as $6$ is, of course, algebraic).&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2011-03-05T22:01:44.987" CommentCount="5" />
  <row Id="25208" PostTypeId="2" ParentId="25206" CreationDate="2011-03-05T22:04:17.293" Score="2" ViewCount="" Body="&lt;p&gt;&lt;em&gt;Written before the OP made it clear he wanted a proof without calculus, and when the problem asked how to prove that the inequality held for&lt;/em&gt; &lt;strong&gt;some&lt;/strong&gt; $n$ &lt;em&gt;instead of &quot;for all $n$&quot;.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that $\displaystyle \lim_{n\to\infty}\frac{n^8}{8^n-1} = 0$ (say, by applying L'Hopital's Rule enough times). That means that, eventually, the numerator has to be smaller than the denominator.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added.&lt;/strong&gt; The equality holds for $n=1$ and for $n\geq 9$. It fails for $n=2,\ldots,8$ (the fact that it fails for $n=8$ is particularly glaring, since you get $8^8$ and $8^8-1$). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can prove by simple induction that if $n\gt 8$ and the inequality holds for $n$, then it holds for $n+1$. &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-05T22:25:19.717" LastActivityDate="2011-03-05T22:25:19.717" CommentCount="3" />
  <row Id="25209" PostTypeId="1" AcceptedAnswerId="25216" CreationDate="2011-03-05T22:07:34.373" Score="6" ViewCount="219" Body="&lt;p&gt;Money is typically denominated in a way that allows for a greedy algorithm when computing a given amount $s$ as a sum of denominations $d_i$ of coins or bills:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;s = \sum_{i=1}^k n_i d_i\quad\text{with } n_i &amp;gt; 0, d_i &amp;gt; d_{i-1}&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We start with the highest denomination $d$ not exceeding $s$ and subtract it.  We repeat this process until $s$ is completely represented. As long as all denominations are available, this simple algorithm guarantees the minimum number of coins or bills used. (This is a little sketchy but I guess we are all familiar with it and it's not central to my question.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is: what property of $d_i$ makes this work? I once heard that the value of any coin must be bigger than the sum of all coins of lower denomination for the greedy algorithm to work. This holds for the typical denomination:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;1,2,5,10,20,50,100&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I wasn't able to come up with a counter example or even a proof. Surely this problem has been studied and I would be glad to know more about it. In fact, I'm not even sure about the right tags -- feel free to edit them. I assume the algorithm to find the sum with the minimum number of coins becomes tricky when we no longer can assume that all denominations are available.&lt;/p&gt;&#xA;" OwnerUserId="6678" LastEditorUserId="6678" LastEditorDisplayName="" LastEditDate="2011-03-06T11:42:46.917" LastActivityDate="2011-03-21T19:03:13.000" Title="Looking to understand the rationale for money denomination" Tags="&lt;combinatorics&gt;&lt;algorithms&gt;&lt;reference-request&gt;&lt;elementary-number-theory&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" />
  <row Id="25210" PostTypeId="2" ParentId="25206" CreationDate="2011-03-05T22:12:21.200" Score="1" ViewCount="" Body="&lt;p&gt;Let n=1, then you get 1&amp;lt;8-1. QED.&lt;/p&gt;&#xA;" OwnerUserId="4929" LastActivityDate="2011-03-05T22:12:21.200" />
  <row Id="25211" PostTypeId="2" ParentId="25205" CreationDate="2011-03-05T22:12:56.833" Score="20" ViewCount="" Body="&lt;p&gt;This is a transcendental number, in fact one of the best known ones, it is $6+$ &lt;a href=&quot;http://en.wikipedia.org/wiki/Champernowne_constant&quot; rel=&quot;nofollow&quot;&gt;Champernowne's number&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Kurt Mahler was first to show that the number is transcendental, a proof can be found on his &quot;Lectures on Diophantine approximations&quot;, available through &lt;a href=&quot;http://projecteuclid.org/DPubS?service=UI&amp;amp;version=1.0&amp;amp;verb=Display&amp;amp;page=toc&amp;amp;handle=euclid.ndml/1175197407&quot; rel=&quot;nofollow&quot;&gt;Project Euclid&lt;/a&gt;. The argument (as typical in this area) consists in analyzing the rate at which rational numbers can approximate the constant (see the section on &quot;Approximation by rational numbers: Liouville to Roth&quot; in the Wikipedia entry for &lt;a href=&quot;http://en.wikipedia.org/wiki/Transcendence_theory&quot; rel=&quot;nofollow&quot;&gt;Transcendence theory&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An excellent book to learn about proofs of transcendence is &quot;Making transcendence transparent: an intuitive approach to classical transcendental number theory&quot;, by Edward Burger and Robert Tubbs.&lt;/p&gt;&#xA;" OwnerUserId="462" LastEditorUserId="418" LastEditorDisplayName="" LastEditDate="2011-03-06T04:47:13.390" LastActivityDate="2011-03-06T04:47:13.390" />
  <row Id="25212" PostTypeId="2" ParentId="25206" CreationDate="2011-03-05T22:23:13.743" Score="7" ViewCount="" Body="&lt;p&gt;While this fails for some $n$, such as $n=8$, exponential functions with base larger than $1$ always grow faster than polynomials. Therefore you should expect the inequality to hold for sufficiently large $n$.  In fact, you can check that it holds for $n=9$, and if you're concerned only with integer solutions you can show by induction that it holds for all larger $n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $n^8&amp;lt;8^n-1$ for some $n\geq 9$, then &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\begin{align*}&#xD;&#xA;8^{n+1}-1&amp;amp;\gt 8(8^n-1)\gt 8n^8\\&#xD;&#xA; &amp;amp;=(n+1)^8\left(8\left(\frac{n}{n+1}\right)^8\right)\\&#xD;&#xA; &amp;amp;\geq (n+1)^8\left(8\left(\frac{9}{10}\right)^8\right)\\&#xD;&#xA;&amp;amp;\gt (n+1)^8 .&#xD;&#xA;\end{align*}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A way to think about why the right-hand side eventually dominates is to note that the ratio of successive terms on the left-hand side approaches $1$ (because $\frac{n+1}{n}=1+\frac{1}{n}$ approaches $1$), while the ratio of successive terms on the right-hand side is approximately $8$. This indicates that eventually the right-hand side grows much faster than the left-hand side.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2011-03-05T22:46:46.647" LastActivityDate="2011-03-05T22:46:46.647" />
  <row Id="25213" PostTypeId="1" AcceptedAnswerId="25223" CreationDate="2011-03-05T22:24:49.560" Score="9" ViewCount="79" Body="&lt;p&gt;&lt;strong&gt;In an additive category, why is finite products the same as finite coproducts?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is relatively easy to prove when the category is R-mod, but my intuition/creativity fails to see how the method can be extended to arbritrary additive categories&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Specifically, a category (in Weibel, &quot;An introduction to homological algebra&quot;) is called additive if the Hom-sets are abelian groups, composition of morphisms distribute over addition, and such that it has a distinguished zero object (that is, an object that is both initial and terminal).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After giving this definition, Weibel claims, without further explanation, that &quot;this structure is enough to make finite products the same as finite coproducts&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How is this?&lt;/p&gt;&#xA;" OwnerUserId="4284" LastActivityDate="2011-03-06T00:09:48.327" Title="In an additive category, why is finite products the same as finite coproducts?" Tags="&lt;category-theory&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" />
  <row Id="25214" PostTypeId="1" AcceptedAnswerId="25215" CreationDate="2011-03-05T22:28:57.150" Score="7" ViewCount="325" Body="&lt;p&gt;What does &lt;code&gt;:=&lt;/code&gt; mean?  &lt;p&gt; &lt;/p&gt; &lt;p&gt; &lt;/p&gt;&lt;/p&gt;&#xA;" OwnerUserId="4950" LastEditorUserId="1069" LastEditorDisplayName="" LastEditDate="2011-03-18T15:48:36.550" LastActivityDate="2011-03-18T23:48:45.033" Title="What does := mean?" Tags="&lt;notation&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="25215" PostTypeId="2" ParentId="25214" CreationDate="2011-03-05T22:33:05.587" Score="16" ViewCount="" Body="&lt;p&gt;It is borrowed from computer programming: it means that the item on the left hand side is being &lt;strong&gt;defined&lt;/strong&gt; to be what is on the right hand side. For example, &#xA;$$y := 7x+2$$&#xA;means that $y$ is &lt;em&gt;defined&lt;/em&gt; to be $7x+2$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is different from, say, writing&#xA;$$1 = \sin^2(\theta) + \cos^2(\theta)$$&#xA;where we are saying that the two sides are equal, but we are not &lt;em&gt;defining&lt;/em&gt; &quot;1&quot; to be the expression &quot;$\sin^2(\theta) + \cos^2(\theta)$&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically, some people think that there should be notational difference between saying &quot;I define &lt;code&gt;blah&lt;/code&gt; to be equal to &lt;code&gt;blankety&lt;/code&gt;&quot; and saying &quot;&lt;code&gt;blah&lt;/code&gt; is equal to &lt;code&gt;blankety&lt;/code&gt;&quot;. So they use &lt;code&gt;:=&lt;/code&gt; for the first and &lt;code&gt;=&lt;/code&gt; for the latter. Usually, it is clear from context which of the two uses of the equal sign is intended (often because of signal words like &quot;Let&quot;, &quot;We define&quot;, etc.)&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-06T05:01:44.853" LastActivityDate="2011-03-06T05:01:44.853" CommentCount="15" />
  <row Id="25216" PostTypeId="2" ParentId="25209" CreationDate="2011-03-05T23:00:47.037" Score="9" ViewCount="" Body="&lt;p&gt;The greedy solution is not optimal in general, even for the case when the value of any coin is bigger than the sum of all coins of lower denomination.  For example, suppose we have coins of denominations 25, 9, 4, and 1.  For 37 cents, the greedy solution uses five coins: one 25, one 9, and three 1s.  However, the optimal solution is four coins: one 25 and three 4s.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem of finding the minimum number of coins given a set of denominations is called the &lt;em&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Change-making_problem&quot; rel=&quot;nofollow&quot;&gt;change-making problem&lt;/a&gt;&lt;/em&gt;.  It's a variant of the famous knapsack problem and is known to be difficult in the general case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An integer-programming formulation for the change-making problem was given in the answer to this math.SE question:  &lt;a href=&quot;http://math.stackexchange.com/questions/21949/on-problems-of-coins-totaling-to-a-given-amount&quot;&gt;On problems of coins totaling to a given amount&lt;/a&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Added&lt;/em&gt;: While the Wikipedia article on the change-making problem claims that the greedy algorithm is optimal for U.S. coins, it provides no proof.  The article &quot;&lt;a href=&quot;http://arxiv.org/PS_cache/arxiv/pdf/0809/0809.0400v1.pdf&quot; rel=&quot;nofollow&quot;&gt;Canonical Coin Systems for Change-Making Problems&lt;/a&gt;&quot; discusses characterizations for when the greedy algorithm is optimal for coin systems of up to five denominations.  It also implies that for systems of larger than five denominations determining when the greedy algorithm is optimal is still an open problem.&lt;/p&gt;&#xA;" OwnerUserId="2370" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2011-03-06T14:29:55.997" LastActivityDate="2011-03-06T14:29:55.997" CommentCount="1" />
  <row Id="25217" PostTypeId="1" AcceptedAnswerId="25220" CreationDate="2011-03-05T23:05:58.463" Score="5" ViewCount="93" Body="&lt;p&gt;Attempt: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sum_{k=1}^n (p_k + \frac{1}{p_k})^2 = 2n + \sum_k p_k^2 + \sum_k \frac{1}{p_k^2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I used the Cauchy inequality to decompose 1 as $\sqrt{p_k}(\frac{1}{\sqrt{p_k}})$ and got &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$n^2\leq \sum_k \frac{1}{p_k}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I could, use the Cauchy inequality again on $\frac{1}{p_k}\cdot 1$ to get &#xA;$(\sum\frac{1}{p_k})^2 &amp;lt; n \sum\frac{1}{p_k^2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Rightarrow n^4&amp;lt; (\sum\frac{1}{p_k})^2 &amp;lt; n \sum\frac{1}{p_k^2} $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Rightarrow n^3\leq\sum\frac{1}{p_k^2} $ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;what about $\sum p_k^2$. There's a mental block here. Any help would be appreciated. &lt;/p&gt;&#xA;" LastActivityDate="2011-03-07T19:07:37.123" Title="Show $\sum_{k=1}^n (p_k + \frac{1}{p_k})^2\geq n^3 + 2n + \frac{1}{n}$ ; $p_k\geq 0 \forall k$ and $\sum_kp_k=1$" Tags="&lt;inequality&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="1" />
  <row Id="25218" PostTypeId="2" ParentId="24589" CreationDate="2011-03-05T23:10:29.957" Score="2" ViewCount="" Body="&lt;p&gt;Let us consider the problem over $\mathbb{C}$  first (because it is a question about joint spectrum and it is easier to tackle at first over an algebraically closed field). Then both $X$ and $Y$ are subsets of $\mathbb{A}^{k n^2} (\mathbb{C}) = MaxSpec(\mathbb{C}[x_{pqs}]_{p,q=1,\ldots,n,s=1,\ldots,k})$. Now a $k$-tuple of matrices, $(A_1,\ldots,A_k)$, has a joint eigenvector if and only if the long matrix $(\lambda_1 I -A_1 \lambda_2 I - A_2 \ldots \lambda_k I - A_K)$ has rank less then $n$ for some choice $(\lambda_1,\ldots,\lambda_k)$. Now you can consider the maximal minors of the long matrix as polynomials in $(n^2+1) k$ variables. The zeros of the ideal generated by those maximal minors is a Zariski closed subset of $\mathbb{A}^{k (n^2+1)} (\mathbb{C})=MaxSpec(\mathbb{C}[x_{pqs},\lambda_j])$. Let us call this ideal, generated by the maximal minors $\mathfrak{I}$. Then $\mathfrak{J} = \mathfrak{I} \cap \mathbb{C}[x_{pqs}]_{p,q=1,\ldots,n,s=1,\ldots,k}$ is the ideal of $Y$. Therefore $Y$ is closed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finding the polynomial you are looking for is a question of elimination using Groebner basis techniques. I'd advise Macaulay2 for this.&lt;/p&gt;&#xA;" OwnerUserId="7838" LastActivityDate="2011-03-05T23:10:29.957" CommentCount="4" />
  <row Id="25220" PostTypeId="2" ParentId="25217" CreationDate="2011-03-05T23:17:21.550" Score="4" ViewCount="" Body="&lt;p&gt;Let $$np&amp;#39; = \sum p_i$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider $$\sum (p_i - p&amp;#39;)^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;btw, I presume you meant $(\sum \frac{1}{p_k})^2$ instead of $\sum (\frac{1}{p_k})^2$&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2011-03-05T23:22:36.840" LastActivityDate="2011-03-05T23:22:36.840" CommentCount="5" />
  <row Id="25221" PostTypeId="1" AcceptedAnswerId="25945" CreationDate="2011-03-05T23:25:18.737" Score="10" ViewCount="334" Body="&lt;p&gt;Hello, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm finishing up reading about Hilbert's axioms, and this final problem from Hartshorne's Geometry: Euclid and Beyond is throwing me for a loop.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm working in a Hilbert plane, so the axioms of incidence, betweenness and congruence apply. I have the additional Archimedes' axiom, which states that for given line segments $AB$ and $CD$, there is a natural number $n$ such that $n$ copies of $AB$ added together will be greater than $CD$. However, I do not have use of the parallel postulate, nor the circle-circle intersection axiom.*&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I want to show that the exterior of a circle is a segment connected set, which means that if $B$ and $C$ are two points outside a circle $\Gamma$, then there exists a third point $D$ such that $BD$ and $DC$ are entirely outside $\Gamma$. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My intuitive argument is straightforward. We can draw line BC. If this is outside $\Gamma$, we are done. Otherwise we can find $l$ to be the perpendicular bisector of $BC$ and take points on $l$, further and further 'out' until we find some $D$ such that $BD$ and $DC$ are outside $\Gamma$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know how to formalize my idea, or why Archimedes' axiom is necessary for this to work. Hartshorne includes the warning that without it or the parallel postulate, the exterior of $\Gamma$ may not be segment connected. Also, I'm not quite sure how to show that it is indeed eventually outside $\Gamma$, since without the circle-circle intersection axiom, I can't find the center of $\Gamma$ nor can I find tangents to $\Gamma$ from $B$ or $C$. I'd be grateful for any explanation, thank you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;*This states that for two circles, if one contains a point inside the other and a point outside the other, then the two circles intersect in (necessarily) two points.&lt;/p&gt;&#xA;" OwnerUserId="1583" LastEditorUserId="1583" LastEditorDisplayName="" LastEditDate="2011-03-09T21:21:19.550" LastActivityDate="2011-03-10T19:18:10.477" Title="From baby Hartshorne: showing the exterior of a circle is segment connected." Tags="&lt;geometry&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="3" />
  <row Id="25222" PostTypeId="2" ParentId="25206" CreationDate="2011-03-05T23:25:47.977" Score="2" ViewCount="" Body="&lt;p&gt;It's easy to show  that powers grow faster than polynomials, i.e. $\rm\ c &amp;gt; 1\ \Rightarrow\ c^n &amp;gt; p(n)\ $ for $\rm\ n &amp;gt; n_0\:.\ $ It suffices to show eventually $\rm\ f(n) = c^n/p(n) &amp;gt; 1\:,\: $ or, equivalently, eventually $\rm\ f(n+1)/f(n) &amp;gt; 1 \ $ since, by &lt;strong&gt;multiplicative telescopy,&lt;/strong&gt; $\rm\:f(n)\:$ is a product of these adjacent term ratios, viz.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\rm f(0)\ \ \prod_{k\:=\:0}^{n-1}\ \frac{f(k+1)}{f(k)}\ \ = \ \ \ {\rlap{--}f(0)}\frac{\rlap{--}f(1)}{\rlap{--}f(0)}\frac{\rlap{--}f(2)}{\rlap{--}f(1)}\ \ \cdots\ \ \frac{f(n)}{\rlap{----}f(n-1)}\ =\ \ f(n) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But $\rm\ f(k+1)/f(k)\ =\ c\ p(k)/p(k+1)\ \to\ c &amp;gt; 1\ $ as $\rm\ k\to \infty\ $ since for equal degree  polynomials $\rm p(x),q(x)\:$ we have $\rm\ lim_{x\to\infty}\: p(x)/q(x) = p_0/q_0 = $ quotient of their leading coefficients, viz.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\rm \frac{p_0\ x^d + p_1\ x^{d-1}+\:\cdots\:+p_d}{q_0\ x^d + q_1\ x^{d-1}+\:\cdots\:+q_d}\ =\ \frac{p_0\  + p_1/x +\:\cdots\:+p_d/x^d}{q_0\  + q_1/x +\:\cdots\:+q_d/x^d}\ \to\ \frac{p_0}{q_0}\ \ \ as\ \ \  x\to\infty$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As I have mentioned here before in &lt;a href=&quot;http://math.stackexchange.com/search?q=user:242+telescopy&quot;&gt;many posts,&lt;/a&gt;  by means of cancelling complicated expressions, &lt;em&gt;telescopy&lt;/em&gt; often reduces induction problems to trivialities (here the fact that a product of terms eventually $&amp;gt; 1$ is itself eventually $&gt; 1$). For example, the inductive proof in the answer by Jonas is essentially a telescopic proof of this type. Difficult problems involving &lt;em&gt;hyperrational&lt;/em&gt; functions (i.e. $\rm\ f(n+1)/f(n) = $ rational function of $\rm\:n\:,\:$ such as powers and exponentials) are, after application of telescopy, greatly simplified to trivial problems about &lt;em&gt;rational&lt;/em&gt; functions - functions so simple that questions about such can be decided mechanically by algorithms.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2011-03-06T19:39:29.460" LastActivityDate="2011-03-06T19:39:29.460" />
  <row Id="25223" PostTypeId="2" ParentId="25213" CreationDate="2011-03-05T23:35:15.037" Score="5" ViewCount="" Body="&lt;p&gt;Note that a product $A \times B$ is the same as a pull-back diagram&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;A x B -&amp;gt; B&#xA;  |      |&#xA;  v      v&#xA;  A ---&amp;gt; 0&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;with maps $p:A \times B \to A$ and $q: A \times B \to B$. &#xA;In particular there is a map $i: A \to A \times B$ such that $pi = 1_{A}$ and $qi = 0$ as well as a map $j: B \to A \times B$ such that $qj = 1_{B}$ and $pj = 0$. Now note that $p(ip + jq) = pip + 0 = p$ and $q(ip + jq) = q$ so that $ip + jq = 1_{A \times B}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let us check that $i: A \to A \times B$ and $j: B \to A \times B$ define a coproduct. Given $f: A \to D$ and $g: B \to D$ we get a map $d: A \times B \to D$ by setting $d = fp + gq$. Since $di = (fp + gq)i = fpi +gqi = f$ and $dj = g$ it remains to prove uniqueness of $d$. But this is clear as any other such map will satisfy $(d-d&amp;#39;)1_{A \times B} = (d-d&amp;#39;)(ip+jq) = 0$. Summing up, we have proved that the product of two objects is also a coproduct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now if we want to say that finite products and coproducts exist and coincide, we need a zero object, since otherwise the empty product (the terminal object) and the empty coproduct (the initial object) would not coincide.&lt;/p&gt;&#xA;" OwnerUserId="5363" LastEditorUserId="5363" LastEditorDisplayName="" LastEditDate="2011-03-06T00:09:48.327" LastActivityDate="2011-03-06T00:09:48.327" CommentCount="2" />
  <row Id="25224" PostTypeId="2" ParentId="25217" CreationDate="2011-03-05T23:36:39.643" Score="2" ViewCount="" Body="&lt;p&gt;You have reduced the problem to showing that ${\displaystyle \sum_{k=1}^n p_k^2 + \sum_{k=1}^n {1 \over p_k^2} \geq {1 \over n} + n^3}$. Note that equality is achieved here when each $p_k = {1 \over n}$, when the first sum adds to ${1 \over n}$ and the second sum adds to $n^3$. Thus it makes sense to try to show that the first sum is at least ${1 \over n}$ and the second sum is at least $n^3$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the first sum, one can use the Cauchy-Schwartz inequality to conclude that $n \sum_{k=1}^n {p_k}^2 \geq (\sum_{k=1}^n {p_k})^2 = 1$, so that the first sum is in fact at least ${1 \over n}$. For the second sum, one similarly has &#xA;$n \sum_{k=1}^n {1 \over p_k^2} \geq (\sum_{k=1}^n {1 \over p_k})^2$ or equivalently, $\sum_{k=1}^n {1 \over p_k^2} \geq {1 \over n} (\sum_{k=1}^n {1 \over p_k})^2$. To analyze $\sum_{k=1}^n {1 \over p_k}$, one may use the arithmetic-harmonic inequality, which says that for positive real numbers $a_1,...,a_n$ one has&#xA;$${a_1 + ... + a_n \over n} \geq {n \over {1 \over a_1} + ... + {1 \over a_n}}$$&#xA;Applying this to $a_k = {1 \over p_k}$ gives that &#xA;$$\sum_{k=1}^n {1 \over p_k} \geq n^2{1 \over \sum_{k=1}^n p_k}$$&#xA;$$ = n^2$$&#xA;So the second sum is at least ${1 \over n}(n^2)^2 = n^3$ as needed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: Naturally, in the time it took to write this out, someone already solved it.. such is life :)&lt;/p&gt;&#xA;" OwnerUserId="3035" LastActivityDate="2011-03-05T23:36:39.643" />
  <row Id="25225" PostTypeId="1" CreationDate="2011-03-05T23:49:45.457" Score="1" ViewCount="129" Body="&lt;p&gt;Let be $n_1$, $n_2$ such natural numbers that $n_1\geq 3$ and  $n_2\geq 3$ and let be $G_{n1,n2}$ a graph that takes shape by taking $G_{n_1}$, the cycle of $n_1$ vertices, and $G_{n_2}$ the cycle of $n_2$ vertices, and linking all of $G_{n_1}$ vertices with all of $G_{n_2}$ vertices.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a) For which values of $n_1,n_2$ has the graph $G_{n1,n2}$ an Euler cycle ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;b) For which values of ${n_1,n_2}$ has the graph $G_{n1,n2}$ a Hamilton cycle ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;c) For which values of ${n_1,n_2}$ is the graph $G_{n1,n2}$ a 5-colorable but not 4-colorable?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Indication&lt;/strong&gt;: Show first that in each legal(valid) coloration of $G_{n1,n2}$ the sets of colours that are used for $G_{n_1}$ vertices and $G_{n_2}$ vertices they have to to be foreign from each other.&lt;/p&gt;&#xA;" OwnerUserId="7843" LastEditorUserId="7843" LastEditorDisplayName="" LastEditDate="2011-03-06T17:37:39.163" LastActivityDate="2011-05-05T21:49:44.587" Title="Graph Problems(Euler,Hamilton,Color)" Tags="&lt;homework&gt;" AnswerCount="1" CommentCount="8" />
  <row Id="25226" PostTypeId="1" AcceptedAnswerId="41684" CreationDate="2011-03-06T00:24:38.703" Score="2" ViewCount="88" Body="&lt;p&gt;Is there a connection between the Hardy Spaces on the unit disk and on $\mathbb R^n$?.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If so, can we use results from the Hardy Spaces on the unit disk to prove $(H^1)^* = \text{BMO}$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Further, what is the most fruitful way to define $H^1$ on $\mathbb R^n$ but avoiding (bounded) distributions? I was writing something for a project and this would be the only point where I would use them. Of course, I could take the completion of the $C_c^\infty$ functions in the $H^1$-norm, but is this workable enough?&lt;/p&gt;&#xA;" OwnerUserId="1120" LastActivityDate="2011-05-27T16:57:58.857" Title="Hardy Spaces on the unit disk and $\mathbb R^n$." Tags="&lt;soft-question&gt;&lt;functional-analysis&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="25227" PostTypeId="1" CreationDate="2011-03-06T01:10:01.350" Score="5" ViewCount="53" Body="&lt;p&gt;If $p:\tilde{X}\rightarrow X$ is a regular covering space of finite degree, why is it not obvious that if two curves $\gamma$ and $\delta$ are isotopic in $\tilde{X}$ their images are isotopic in $X$?  By my understanding, this is a nontrivial theorem (including the statement that the cover is possibly branched) in a paper by Birman and Hilden, and it further requires the condition that the group of deck transformations is solvable.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems to me that this statement should follow from the fact that induced map on fundamental group $p_*$ is well defined.  What am I missing?  Does anyone have a counterexample (preferably one involving surfaces, rather than anything higher dimensional)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If this is in fact nontrivial, will it be nontrivial with the added condition of the path lifting property?&lt;/p&gt;&#xA;" OwnerUserId="7845" LastActivityDate="2011-03-06T01:42:12.813" Title="Does isotopy in a covering space imply isotopy in a base space?" Tags="&lt;algebraic-topology&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="25228" PostTypeId="1" AcceptedAnswerId="25278" CreationDate="2011-03-06T01:22:53.040" Score="6" ViewCount="77" Body="&lt;p&gt;Let $1 &amp;lt; k \in \mathbb N$ and $M = \{(z_1, z_2) \in \mathbb C^2 : k|z_1|^2 + |z_2|^2 = 1\}$.  Let $S^1$ act on $M$ via $e^{i\theta}(z_1,z_2) = (e^{ik\theta} z_1, e^{i\theta} z_2)$.  Then I am told that $M/S^1$ is not a manifold (observe that the action is not free as $e^{2\pi i/k}$ stabilizes the points of the form $(z_1,0)$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am having trouble seeing why this fails to be a manifold (I'm guessing it fails to be locally $\mathbb R^2$ at the points $(z_1,0)$).  I've read that this space has a cone singularity and was wondering if someone could explain how to visually see this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="9" LastActivityDate="2011-03-06T12:31:49.647" Title="Visualising a specific orbifold." Tags="&lt;differential-topology&gt;&lt;visualization&gt;" AnswerCount="1" FavoriteCount="0" />
  <row Id="25229" PostTypeId="1" CreationDate="2011-03-06T01:26:02.060" Score="0" ViewCount="71" Body="&lt;p&gt;Let $a$, $b$ an element of $\mathbb{Z}$ with $a$ and $b$ not both zero and let $c$ be a nonzero integer. Prove that $$(ca,cb) = |c|(a,b)$$&lt;/p&gt;&#xA;" OwnerUserId="7846" LastEditorUserId="6296" LastEditorDisplayName="" LastEditDate="2011-03-06T01:35:51.747" LastActivityDate="2011-03-06T04:30:38.997" Title="Greatest common Divisors" Tags="&lt;elementary-number-theory&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="1" />
  <row Id="25230" PostTypeId="2" ParentId="25229" CreationDate="2011-03-06T01:32:17.060" Score="3" ViewCount="" Body="&lt;p&gt;Let $d = (ca,cb)$ and $d&amp;#39; = |c|(a,b)$. Show that $d|d&amp;#39;$ and $d&amp;#39;|d$. &lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2011-03-06T01:32:17.060" />
  <row Id="25231" PostTypeId="2" ParentId="25227" CreationDate="2011-03-06T01:42:12.813" Score="2" ViewCount="" Body="&lt;p&gt;If two paths are isotopic then they are homotopic via a homotopy $h_t$ that is injective for each $t$.  Projecting this to the base doesn't guarantee that $h_t$ will remain injective.&lt;/p&gt;&#xA;" OwnerUserId="9" LastActivityDate="2011-03-06T01:42:12.813" />
  <row Id="25232" PostTypeId="2" ParentId="25229" CreationDate="2011-03-06T01:45:02.737" Score="3" ViewCount="" Body="&lt;p&gt;Below is a proof of the &lt;em&gt;distributive law&lt;/em&gt; for GCDs that works in every domain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;THEOREM&lt;/strong&gt; $\rm\quad (a,b)\ =\ (ac,bc)/c\quad$  if $\rm\ (ac,bc)\ $ exists&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof&lt;/strong&gt; $\rm\quad d\ |\ a,b\ \iff\ dc\ |\ ac,bc\ \iff\ dc\ |\ (ac,bc)\ \iff\ d|(ac,bc)/c$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See my &lt;a href=&quot;http://math.stackexchange.com/questions/16042/about-greatest-common-divisor/16043#16043&quot;&gt;post here&lt;/a&gt; for further discussion of this property and its relationship with &lt;strong&gt;Euclid's Lemma&lt;/strong&gt;.&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2011-03-06T01:45:02.737" />
  <row Id="25233" PostTypeId="1" AcceptedAnswerId="25252" CreationDate="2011-03-06T01:46:02.300" Score="1" ViewCount="88" Body="&lt;p&gt;I'm reading Matsumura's proof on independence of valuation (p.87, &quot;Commutative Ring Theory&quot;), and I understand most of the proof, except the last bit about principal ideal ring, and if anyone could help me I would greatly appreciate it:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$K$ a field, $R_1$,...,$R_n\subset K$   DVRs of $K$ not containing one another, $A=\cap_{i}^{n}R_i$. Let $\mathfrak{m}_i\subset R_i$ be the maximal ideals and $\mathfrak{p}_i:=\mathfrak{m}_i\cap A$. I can see that $A$ is a semi-local ring with maximal ideals $\mathfrak{p}_1$,...,$\mathfrak{p}_n$, and that $A_{\mathfrak{p}_i}=R_i$. But I don't see why $A$ is a PID. Here is a passage from the book:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\mathfrak{m}_i\neq\mathfrak{m}_i^2$, and hence $\mathfrak{p}_i\neq\mathfrak{p}_i^{(2)}$, where $\mathfrak{p}^{(2)}:=\mathfrak{p}^2A_{\mathfrak{p}}\cap A$. Thus there exists $x_i\in\mathfrak{p}_i$ such that $x_i\notin\mathfrak{p}_i^{(2)}$, and $x_i\notin\mathfrak{p}_j$ for $i\neq j$ (I understand up to here); then $\mathfrak{p}_i=x_iA$ (I don't understand this). If $I$ is any ideal of $A$ and $IR_i=x^{\nu_i}_iR_i$ for $i=1$,...,$n$ then it is easy to see that $I=x_{1}^{\nu_1}...x_{n}^{\nu_n}A$ (I don't see this).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm sorry for this rather pedantic question but I have no one to help me...&lt;/p&gt;&#xA;" OwnerUserId="4341" LastEditorUserId="7243" LastEditorDisplayName="" LastEditDate="2011-03-06T04:18:51.957" LastActivityDate="2011-03-06T20:01:52.397" Title="Independence of valuations" Tags="&lt;commutative-algebra&gt;" AnswerCount="2" CommentCount="7" />
  <row Id="25234" PostTypeId="1" CreationDate="2011-03-06T01:54:57.970" Score="1" ViewCount="48" Body="&lt;p&gt;Let $a_1,\ldots,a_n$ be a set of $n$ positive numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there known lower and upper bounds on:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle\frac{\prod_{i} \Gamma(a_i)}{\Gamma(\sum_i a_i)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\Gamma$ is the Gamma function (a generalization of the factorial distribution)?&lt;/p&gt;&#xA;" OwnerUserId="7848" LastActivityDate="2011-06-04T04:10:56.463" Title="are there useful bounds on the &quot;gamma&quot; coeficients (generalization of multinomial coefficients)?" Tags="&lt;combinatorics&gt;" AnswerCount="1" />
  <row Id="25235" PostTypeId="2" ParentId="25234" CreationDate="2011-03-06T02:15:37.327" Score="1" ViewCount="" Body="&lt;p&gt;You can use Stirling's approximation. Considering the multinomial coefficient in $a_i$ instead of what you wrote, and putting $p_i = a_i/\sum_i a_i$, you should get that the logarithm of the multinomial coefficient is&#xA;$$(1+o(1))(\sum_i a_i) H(p_1,\ldots,p_n),$$&#xA;where $H(\cdots)$ is the entropy function:&#xA;$$H(p_1,\ldots,p_n) = -\sum_i p_i \log p_i.$$&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2011-03-06T02:15:37.327" />
  <row Id="25236" PostTypeId="2" ParentId="25109" CreationDate="2011-03-06T02:43:03.067" Score="0" ViewCount="" Body="&lt;p&gt;Do the standard Euclidean algorithm with n^2 and an+1.  The first remainder less than n that appears is the inverse of a.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Find the inverse of 16 mod 21:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Apply the Euclidean algorithm to 21^2 = 441 and 16*21+1 = 337 to get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;441 = 337 + 104&#xA;337 = 3*104 + 25&#xA;104 = 4*25 + 4&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since 4 is less than 21, 4 is the inverse of 16 mod 21.&lt;/p&gt;&#xA;" OwnerUserId="4253" LastActivityDate="2011-03-06T02:43:03.067" />
  <row Id="25237" PostTypeId="1" AcceptedAnswerId="25238" CreationDate="2011-03-06T02:44:41.037" Score="2" ViewCount="141" Body="&lt;p&gt;In the introduction of &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0387905189&quot; rel=&quot;nofollow&quot;&gt;Hungerford's Algebra&lt;/a&gt; (p. 2), he gives a rather trivial example of a class that is not a set, but what is the purpose of even having this term defined? Is it useful, other than to give a name to collections of objects that are not sets?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, how is this term related to equivalence and congruence classes? More specifically, are there equivalence or congruence classes that are not sets?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt; It turns out the &quot;rather trivial example&quot; is Russell's paradox and wasn't so trivial at the time of it's discovery.&lt;/p&gt;&#xA;" OwnerUserId="7840" LastEditorUserId="7840" LastEditorDisplayName="" LastEditDate="2011-03-07T18:53:33.647" LastActivityDate="2011-03-07T18:53:33.647" Title="What is the significance of &quot;classes&quot;?" Tags="&lt;abstract-algebra&gt;&lt;set-theory&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="25238" PostTypeId="2" ParentId="25237" CreationDate="2011-03-06T02:49:58.623" Score="2" ViewCount="" Body="&lt;p&gt;There is no connection to equivalence/congruence classes - that's just a coincidence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Classes are useful since sometimes we want to talk about them as sets. For example, instead of saying &quot;for every ordinal $\alpha$&quot;, you can say &quot;$\forall \alpha \in \mathrm{On}$&quot;. In other words, it's just for convenience.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2011-03-06T02:49:58.623" CommentCount="1" />
  <row Id="25239" PostTypeId="1" AcceptedAnswerId="25248" CreationDate="2011-03-06T02:56:32.410" Score="7" ViewCount="226" Body="&lt;p&gt;Can anyone give me a hint on showing (in a relatively elegant way, as I know the answer from WolframAlpha), that the complex valued function $z^2e^z-z$ has at most 2 roots with norm less than 2?  Obviously it has one root at $z=0$.  The other root is the number $W_0(1)$, where $W_0(1)$ is the principal branch of the product log function at 1 (the inverse of the function $ze^z$, and is approximately equal to .56.  I have seen that the next possible zero is outside of the ball of radius 2, but I cannot show it algebraically.  I am primarily trying to use Rouche's Theorem, and to find a suitable function $g(z)$ such that $z^2e^z-z-g(z)|&amp;lt;|g(z)|$, however I have not been able to find a good one yet.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any help would be much appreciated, especially ideas about what sort of function to consider in cases like this, rather than just &quot;here's the function...&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="3595" LastEditorUserId="3595" LastEditorDisplayName="" LastEditDate="2011-03-06T04:15:22.070" LastActivityDate="2011-04-10T22:10:15.720" Title="Complex Zeros of $z^2e^z-z$" Tags="&lt;complex-analysis&gt;&lt;approximation&gt;&lt;roots&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="1" />
  <row Id="25240" PostTypeId="2" ParentId="25237" CreationDate="2011-03-06T03:00:23.480" Score="1" ViewCount="" Body="&lt;p&gt;Classes were introduced so that we could have a &quot;collection&quot; of things which is not a set, for instance the set of all sets (does it contain itself...?).  A class that is not a set is called a proper class, e.g the &quot;class&quot; of all sets.  It's just sort of a sneaky way of avoiding paradoxes.  See &lt;a href=&quot;http://en.wikipedia.org/wiki/Proper_class&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Proper_class&lt;/a&gt; for more info.&lt;/p&gt;&#xA;" OwnerUserId="3595" LastActivityDate="2011-03-06T03:00:23.480" />
  <row Id="25241" PostTypeId="1" AcceptedAnswerId="25243" CreationDate="2011-03-06T03:02:57.197" Score="7" ViewCount="145" Body="&lt;p&gt;&lt;strong&gt;Problem statement:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let A be an uncountable set, B a countable subset of A, and C the complement of B in A.  Prove that there exists a one-to-one correspondence between A and C.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My thoughts:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's a bijection between A and A (the identity function).  There's a bijection between C and C (the identity function).  There's a bijection between B and $\mathbb{N}$.  That's all I know.&lt;/p&gt;&#xA;" OwnerUserId="6059" LastActivityDate="2011-03-06T03:16:01.013" Title="Proof there is a 1-1 correspondence between an uncountable set and itself minus a countable part of it" Tags="&lt;homework&gt;&lt;elementary-set-theory&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" />
  <row Id="25242" PostTypeId="1" AcceptedAnswerId="25246" CreationDate="2011-03-06T03:09:18.893" Score="3" ViewCount="78" Body="&lt;p&gt;I ha come across a book on groups as Galois groups, and in the introduction it mentions the paper by I.R. Shafarevich which says that every solvable group can be realized as Galois groups of some Galois extensions over $\mathbb{Q}$. It, however, does not tell us where to find that paper; therefore, I would like to know where I can find it.&lt;br&gt;&#xA;In any case, thanks very much.  &lt;/p&gt;&#xA;" OwnerUserId="6792" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-06T04:06:51.017" LastActivityDate="2011-03-06T19:58:31.607" CommunityOwnedDate="2011-03-06T07:17:23.123" Title="Where can I find the paper by Shafarevich on the result of the realization of solvable groups as Galois groups over $\mathbb{Q}$?" Tags="&lt;reference-request&gt;&lt;galois-theory&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" />
  <row Id="25243" PostTypeId="2" ParentId="25241" CreationDate="2011-03-06T03:10:19.490" Score="6" ViewCount="" Body="&lt;p&gt;Since $A\setminus B$ is uncountable, assuming countable choice it has a countably infinite subset $B&amp;#39;$.  Then $B&amp;#39;\cup B$ is countable, so there is a bijection $g:B&amp;#39;\cup B\to B&amp;#39;$.  Define $f:A\to A\setminus B$ by $f\vert_{B&amp;#39;\cup B}=g$ and $f(a)=a$ for $a\in A\setminus(B&amp;#39;\cup B)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically, you just take chunks off of $A$ and $A\setminus B$ that have equal size, respectively $B&amp;#39;\cup B$ and $B&amp;#39;$, leaving the remaining sets equal to $A\setminus(B&amp;#39;\cup B)$, then piece together the two bijections.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2011-03-06T03:16:01.013" LastActivityDate="2011-03-06T03:16:01.013" />
  <row Id="25244" PostTypeId="2" ParentId="25225" CreationDate="2011-03-06T03:13:55.187" Score="1" ViewCount="" Body="&lt;p&gt;For b), with high connectivity it is hard to avoid Hamiltonian cycles.  For c) have you proved what the hint asks?  How many colors are required for a cycle?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Added after the first two comments:  The description of Euler cycle is just a restatement of one of the comments to the question, and is correct.  For the Hamiltonian cycle, it is true there always is one.  To prove it, you should be able to describe one.  For example, if $n1=n2$, you could number the vertices in each cycle from 1 to n in order around the cycle.  The Hamiltonian cycle is from vertex $i$ in $G_{n1}$ to vertex $i$ in $G_{n2}$, then from vertex $i$ in $G_{n2}$ to vertex $i+1$ in $G_{n1}$ and finally from vertex $n$ in $G_{n2}$ to vertex $1$ in $G_{n1}$.  Can you find a construction that works if $n1 \ne n2$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Did you figure out how many colors it takes just for a single cycle as a function of the number of vertices?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Further addition, now that the due date has probably passed:  For b) there is always a Hamiltonian cycle.  Start at one vertex of $G_{n1}$ and follow the cycle to the last vertex before you close.  Then go to a vertex of $G_{n2}$, traverse the $G_{n2}$ cycle until the last vertex and go back to the vertex you started at.  This is a Hamiltonian cycle.  For c), coloring a cycle takes $2$ colors if the cycle is even and $3$ if it is odd.  As the colors in the two cycles must be distinct, the whole graph takes $4$ if $n1, n2$ are both even, $5$ if one is even and one is odd, $6$ if they are both odd.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2011-04-05T20:51:01.023" LastActivityDate="2011-04-05T20:51:01.023" CommentCount="2" />
  <row Id="25245" PostTypeId="1" CreationDate="2011-03-06T03:45:02.453" Score="6" ViewCount="99" Body="&lt;p&gt;This is an exercise from the book &lt;em&gt;Algebraic Number Theory&lt;/em&gt; by Jurgen Neukirch, on page 166. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And, after solving several previous exercises, I found this to be particularly difficult to solve. I am hoping only for hints, not a complete solution. And any hint is appreciated.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, I have made a conjecture that the number of all nonarchimedean valuations extending one of $Q$ is either 1 or $\phi(n)+1$, as to the archimedean ones, I think they are easy to find, where $\phi(n)$ is the Euler function of $n$.&lt;br&gt;&#xA;And, eventually, I stated my question once again to make it clear.&lt;br&gt;&#xA;Again, please provide me only with hints so I can work it out myself:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;How to determine all valuations of the field $\mathbb{K}$ where $\mathbb{K}$ is obtained by adjoining $n$th root of 2 to $\mathbb{Q}$?  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="6792" LastEditorUserId="6792" LastEditorDisplayName="" LastEditDate="2011-03-07T11:53:05.110" LastActivityDate="2011-03-07T11:53:05.110" Title="How to determine all valuations of the field $\mathbb{Q(\sqrt[n]{2})}$?" Tags="&lt;algebraic-number-theory&gt;&lt;absolute-value&gt;" CommentCount="6" FavoriteCount="2" />
  <row Id="25246" PostTypeId="2" ParentId="25242" CreationDate="2011-03-06T03:47:13.767" Score="2" ViewCount="" Body="&lt;p&gt;The original paper, in Russian, was:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;I.R. Shafarevich, &lt;em&gt;Construction of fields of algebraic numbers with given solvable Galois group&lt;/em&gt; (russian) Izv. Akad. Nauk SSSR, Ser. Mat. &lt;strong&gt;18&lt;/strong&gt; (1954), 525–578. English translation in Amer. Math. Soc. Transl. &lt;strong&gt;4&lt;/strong&gt; (1956), 185-237. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The original paper had an error in it, and there was a correction appended in 1989. A complete proof can be found in&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;J. Neukirch, A. Schmidt &amp;amp; K. Wingberg, &lt;em&gt;Cohomology of number fields,&lt;/em&gt; Grundlehren der Mathematischen Wissenschaften 323, Springer-Verlag, 2000.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;There is also a proof available &lt;a href=&quot;http://www.math.uiuc.edu/Algebraic-Number-Theory/0136/&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;, in a paper by Alexander Schmidt and Kay Wingberg. The paper is available in Postscript.  &lt;a href=&quot;http://www.ps2pdf.com&quot; rel=&quot;nofollow&quot;&gt;ps2pdf.com&lt;/a&gt; will create a PDF from an uploaded PS file, so it can be used to convert it if you cannot open PS documents.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-06T19:58:31.607" LastActivityDate="2011-03-06T19:58:31.607" CommunityOwnedDate="2011-03-06T07:17:23.123" CommentCount="8" />
  <row Id="25247" PostTypeId="1" AcceptedAnswerId="25250" CreationDate="2011-03-06T04:07:07.287" Score="3" ViewCount="162" Body="&lt;p&gt;I've been stumbling across a couple of questions in Ahlfors that fall along the lines of &quot;If $f$ is an analytic function with property $X$, show that $f$ reduces to a polynomial.&quot; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One concrete example would be:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Show that a function which is analytic in the whole plane and has a nonessential singularity at $\infty$ reduces to a polynomial. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was wondering what the general approach to these problems is. Is it to use some sort of factorization of the zeros/poles of the function and then to use Cauchy's estimate?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My thought process for this particular problem has been as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider $g(z) = \frac{1}{f(\frac{1}{z})}$. Then $h(z)=0$ for $z=0$. $h$ cannot vanish identically or else $f$ is identically $\infty$ and then, $f$ does not have a nonessential singularity at $\infty$. Thus, we know that $g(z) = z^h g_h(z)$ where $g_h$ is an analytic function with $g_h(0) \neq 0$. Then we get $\frac{1}{f(\frac{1}{z})} = z^h \frac{1}{f_h(\frac{1}{z})}$ (where $\frac{1}{f_h(\frac{1}{z})} = g_h(z)$), which implies that $f(\frac{1}{z}) = z^{-h} f_h(\frac{1}{z})$. This last expression can also be equivalently written as $f(z) = z^h f_h(z)$. Then, since $f_h(z)$ is bounded as $|z| \to \infty$, we can use Cauchy's estimate to show that the $(h+1)$-th derivative of $f$ is equal to $0$ for all $z$, which implies that the function must reduce to a polynomial.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd appreciate it if any of you could tell me if that's correct and/or if there is a better/different way to go about answering the question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an extra note, would it be wrong to assume that any approach toward these problems with polynomials is also going to extend in some sense toward problems about rational functions?&lt;/p&gt;&#xA;" OwnerUserId="1736" LastActivityDate="2011-03-06T05:05:02.697" Title="Relation between analytic functions and polynomials" Tags="&lt;complex-analysis&gt;" AnswerCount="1" CommentCount="6" FavoriteCount="1" />
  <row Id="25248" PostTypeId="2" ParentId="25239" CreationDate="2011-03-06T04:14:36.190" Score="7" ViewCount="" Body="&lt;p&gt;This is different from the approach you were thinking about, and perhaps not &quot;relatively elegant.&quot;  It uses no complex analysis.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a unique nonzero real solution, because if $x$ is negative then so is $xe^x$, while for positive $x$, $xe^x$ increases without bound.  Since $1e^1\gt1$, the real solution has modulus less than $1$.  It remains to be shown that there are no nonreal solutions of modulus less than 2.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $ze^z=1$, then the imaginary part of $ze^z$ must be zero, which means that if $z=x+iy$, then $x\sin(y)+y\cos(y)=0$.  Thus $x=-y\cot(y)$ if $\sin(y)\neq0$.  The real part of $ze^z$ must be $1$, which after substituting $-y\cot(y)$ for $x$ means $-\frac{y}{\sin(y)}e^{-y\cot(y)}=1$.  But for $0&amp;lt;|y|&amp;lt;\pi$ this is impossible because $-\frac{y}{\sin(y)}e^{-y\cot(y)}$ is negative.  Thus if $x+iy$ is a nonreal solution to the equation, then $|x+iy|\geq|y|\geq\pi$.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2011-03-06T04:32:31.393" LastActivityDate="2011-03-06T04:32:31.393" CommentCount="1" />
  <row Id="25249" PostTypeId="2" ParentId="25229" CreationDate="2011-03-06T04:30:38.997" Score="0" ViewCount="" Body="&lt;p&gt;If $(a,b)=d$, then the equation $ax+by=dz$ has a solution for all $z \in \mathbb{N}$, and this implies that $acx+bcy=(dc)z$ admits a solution for all $z \in \mathbb{N}$. And hence we can deduce the result which must appear in every elementary number theory book.&lt;br&gt;&#xA;Moreover, you have not offered your motivation which absolutely will make the post better.&lt;/p&gt;&#xA;" OwnerUserId="6792" LastActivityDate="2011-03-06T04:30:38.997" />
  <row Id="25250" PostTypeId="2" ParentId="25247" CreationDate="2011-03-06T04:34:12.423" Score="1" ViewCount="" Body="&lt;p&gt;When I look at this problem, &lt;a href=&quot;http://en.wikipedia.org/wiki/Cauchy%27s_integral_formula&quot; rel=&quot;nofollow&quot;&gt;Cauchy's Integral Formula&lt;/a&gt; definitely jumps out.  Other than that, I am not too sure of any &quot;general&quot; approaches.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Solution:&lt;/strong&gt; Recall the singularity at $\infty$ of $f(z)$ is the same as the singularity at $0$ of $f\left(\frac{1}{z}\right)$.  Say $f$ has a pole of order $m$ at $\infty$.  Then $$z^m\cdot f\left(\frac{1}{z}\right)$$ can be made is an analytic function at $z=0$.  Thus, $f(z)=O(z^m)$ as $z\rightarrow \infty$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now apply Cauchy's Integral Formula to a sufficiently high derivative of $f$.  Using the contour $C_R$, the circle of radius $R$, we can find that this derivative will be bounded on the complex plane.  (Show it is bounded in the circle of radius $r$ for every $r$ by taking $R$ large enough, and using $f(z)=O(z^m)$) Hence by &lt;a href=&quot;http://en.wikipedia.org/wiki/Liouville%27s_theorem_%28complex_analysis%29&quot; rel=&quot;nofollow&quot;&gt;Liouvilles Theorem&lt;/a&gt; it is a constant.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus we conclude $f$ is a polynomial.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Approaches:&lt;/strong&gt;  It looks like your solution is basically the same as what I wrote, so this answer may not be too much help.  (Although, I don't fully see the use of defining the function $g$?)  I am curious to see what other solutions people conceive.  In short, I think you are thinking about it in a fairly good way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Alternative:&lt;/strong&gt;  Suppose $f$ is analytic but not a polynomial.  Then in the Taylor expansion around $0$, there must be infinitely many terms.  Hence, the Taylor expansion of $f(1/z)$ at $0$ will have arbitrarily large negative powers, so that $0$ cannot be a pole of finite order.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Alternative 2:&lt;/strong&gt;  Recall that if $h(z)$ has a pole of order $n$ at $z=a$, then $h^&amp;#39;(z)$ has a pole of order $n+1$ at $z=a$ (n\geq 1).  Consequently, if $f(1/z)$ has a pole of order $m$ at 0, then $$f^&amp;#39;(1/z)\cdot \frac{-1}{z^2}$$ has a pole order $n+1$, and hence $f^&amp;#39;(1/z)$ has a pole order $n-1$.  Thus, $f^{(n)}(z)$ is analytic everywhere on the Riemann sphere, and hence a constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; I added two alternative solutions.  I view them as the same, since the key ideas employed in each are identical.&lt;/p&gt;&#xA;" OwnerUserId="6075" LastEditorUserId="6075" LastEditorDisplayName="" LastEditDate="2011-03-06T05:05:02.697" LastActivityDate="2011-03-06T05:05:02.697" CommentCount="2" />
  <row Id="25251" PostTypeId="1" AcceptedAnswerId="25467" CreationDate="2011-03-06T05:22:06.703" Score="1" ViewCount="70" Body="&lt;p&gt;Here the equivalent means:in the bundle ,B is bottom manifold(also called base space),S is section,we can also treat S as a bottom and B as a section in the same bundle.&#xA;Is it true in the vector bundle? and also in the fiber bundle?   &lt;/p&gt;&#xA;" OwnerUserId="4706" LastEditorUserId="4706" LastEditorDisplayName="" LastEditDate="2011-03-07T04:58:09.420" LastActivityDate="2011-03-07T06:22:31.703" Title="Does the section and the bottom manifold equivalent in the bundle?" Tags="&lt;vector-bundles&gt;&lt;fiber-bundles&gt;" AnswerCount="2" CommentCount="5" />
  <row Id="25252" PostTypeId="2" ParentId="25233" CreationDate="2011-03-06T05:57:57.013" Score="1" ViewCount="" Body="&lt;p&gt;If you have a map of $R$-modules $f: M \rightarrow N$, whether it is an injection / surjection / bijection can be checked locally: i.e., it is enough to check the corresponding property for $f_{\mathfrak{m}}: M_{\mathfrak{m}} \rightarrow N_{\mathfrak{m}}$ for all maximal ideals $\mathfrak{m}$ of $R$. (See for instance $\S 7.5$ of &lt;a href=&quot;http://www.math.uga.edu/~pete/integral.pdf&quot; rel=&quot;nofollow&quot;&gt;my commutative algebra notes&lt;/a&gt;.  I'm sure it's in Matsumura as well...)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case you have an inclusion of $R$-modules $\iota: x_i A \rightarrow \mathfrak{p}_i$ (moral: don't forget that &quot;ideal&quot; = &quot;$R$-submodule of $R$&quot;!). I think you'll find the hypotheses are rigged so that upon extending to each of the finitely many maximal ideals, $\iota$ is a bijection.  So $\iota$ is itself a bijection.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second statement you ask about can be proved in a similar way.&lt;/p&gt;&#xA;" OwnerUserId="299" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2011-03-06T20:01:52.397" LastActivityDate="2011-03-06T20:01:52.397" />
  <row Id="25253" PostTypeId="1" CreationDate="2011-03-06T06:11:15.393" Score="6" ViewCount="93" Body="&lt;p&gt;Let $R$ be a commutative ring with identity and $M$ be an $R$-module. I have trouble understanding the restriction map in the definition of the sheaf of $\mathcal{O}_X$ modules. Explicitly, let $f,g\in R$ s.t. $D(g)\subseteq D(f)$, then what is the map $M_f\to M_g$. My main difficulty is understanding what kind of module homomorphism to expect, since the left hand side module is over $R_f$ while the one on the right is over $R_g$. Most books I flipped through, said &quot;it is defined similarly&quot; referring to the definition of the restriction maps in the structure sheaf of Spec$R$&lt;/p&gt;&#xA;" OwnerUserId="7763" LastEditorUserId="7763" LastEditorDisplayName="" LastEditDate="2011-03-06T14:53:13.210" LastActivityDate="2011-03-06T14:53:13.210" Title="restriction map in a Sheaf of $\mathcal{O}_X$ modules" Tags="&lt;algebraic-geometry&gt;&lt;commutative-algebra&gt;&lt;category-theory&gt;" AnswerCount="1" />
  <row Id="25254" PostTypeId="1" CreationDate="2011-03-06T06:52:19.987" Score="5" ViewCount="160" Body="&lt;p&gt;Let $a_1, \ldots a_n$ be i.i.d random variables that take the value $1$ and $-1$ with equal probability. Fix a unit vector $(x_1,\ldots, x_n)$ (such that $\sum_j x_j^2 = 1 $). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider $v = \langle a, x\rangle$. It's easy to see that $E[v] = 0$. It's also easy to bound Pr($|v| &gt; t$) via standard Hoeffding bounds. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I'm looking for is a more accurate analysis of the distribution of $|v|$. Heuristically, it seems that $E[|v|]$ should be around $\sqrt{n}$, and what I'd like are concentration bounds &lt;strong&gt;around&lt;/strong&gt; $E[|v|]$ (specifically, bounds on lower and upper tails). My feeling is that this is either easy, well known, or both :), and I'm wondering if there's a quick reference or argument. &lt;/p&gt;&#xA;" OwnerUserId="865" LastActivityDate="2011-03-07T14:29:05.573" Title="Concentration bound for absolute value of sum of Bernoulli variables" Tags="&lt;probability-theory&gt;" AnswerCount="3" FavoriteCount="1" />
  <row Id="25255" PostTypeId="2" ParentId="25253" CreationDate="2011-03-06T07:03:13.577" Score="1" ViewCount="" Body="&lt;p&gt;If $D(g) \subset D(f)$, then some power of $g$ is a multiple of $f$. That is due to the fact that the radical of an ideal, is precisely the intersection of all of the prime ideals containing it. Then the restriction map is just the localization map $R_f \to R_{fg} = R_g$. Take a look at the first chapter of &quot;Geometry of Schemes&quot;, by Eisenbud and Harris.&lt;/p&gt;&#xA;" OwnerUserId="7838" LastActivityDate="2011-03-06T07:03:13.577" CommentCount="6" />
  <row Id="25256" PostTypeId="1" AcceptedAnswerId="25257" CreationDate="2011-03-06T07:15:28.460" Score="2" ViewCount="99" Body="&lt;p&gt;I am not sure whether or not this is a duplicate question. I'm wondering what is an efficient way to compute&#xA;$$x \equiv a^b \bmod{n}$$&#xA;where $a,b,n \in \mathbb{Z}$ and $a,b &amp;lt; n$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example say I need to compute $13^{59} \bmod{77}$, how do I do it manually?&lt;/p&gt;&#xA;" OwnerUserId="149" LastEditorUserId="6296" LastEditorDisplayName="" LastEditDate="2011-03-06T09:11:50.990" LastActivityDate="2011-03-06T09:57:43.290" Title="Efficient way to find $a^b \bmod {n}$" Tags="&lt;number-theory&gt;&lt;homework&gt;&lt;elementary-number-theory&gt;&lt;modular-arithmetic&gt;" AnswerCount="3" CommentCount="4" />
  <row Id="25257" PostTypeId="2" ParentId="25256" CreationDate="2011-03-06T07:31:50.740" Score="7" ViewCount="" Body="&lt;p&gt;The usual trick is &lt;em&gt;repeated squaring&lt;/em&gt;. If you want to raise a number $x$ to the 59th power, you follow this algorithm:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;x1 = x&#xA;x2 = x1 * x1&#xA;x4 = x2 * x2&#xA;x8 = x4 * x4&#xA;x16 = x8 * x8&#xA;x32 = x16 * x16&#xA;x59 = x32 * x16 * x8 * x2 * x1&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Usually the algorithm is given using the following recurrence equations:&#xA;$$ x^{2n} = (x^n)^2, x^{2n+1} = (x^n)^2\cdot x. $$&#xA;Using these equations, you can write a function that raises a number to a power.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since you're computing everything modulo $n$, don't forget to reduce modulo $n$ each time. This will keep the numbers small and so the effort manageable.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2011-03-06T07:31:50.740" CommentCount="1" />
  <row Id="25258" PostTypeId="2" ParentId="25256" CreationDate="2011-03-06T07:32:04.010" Score="8" ViewCount="" Body="&lt;p&gt;This case is rather simple, since $(13,77)=1$ and $\phi(77) = 60$. So by Euler's theorem $13^{60} \equiv 1 \pmod{77}$. Now $13^{59}\cdot 13 \equiv 1 \pmod{77}$, Hence you can find it using the generalized Euclidean algorithm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general if $(a,n)=1$, then I'd take $b \pmod{\phi(n)}$ and proceed with taking iterated squares.&lt;/p&gt;&#xA;" OwnerUserId="7838" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2011-03-06T07:36:48.397" LastActivityDate="2011-03-06T07:36:48.397" />
  <row Id="25259" PostTypeId="2" ParentId="25254" CreationDate="2011-03-06T07:43:36.613" Score="1" ViewCount="" Body="&lt;p&gt;You can try and see what these values should be under the Central Limit Theorem, and then use the Berry-Esseen bounds to prove that your guess is correct. You can't get large deviation bounds this way since the error is too big (and indeed, the normal approximation doesn't hold for the tails), but for your purposes it might be enough.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2011-03-06T07:43:36.613" />
  <row Id="25260" PostTypeId="2" ParentId="25239" CreationDate="2011-03-06T07:58:30.687" Score="4" ViewCount="" Body="&lt;p&gt;As I understand the question, you are also interested in how one finds functions for Rouché. That is why I give this answer, even though I cannot prove the last inequality (which is however true).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For Rouché you want a function which is close enough to $f(z) = z^2 e^{z}-z$ for $z=2e^{i\phi}$, but simple enough such that you know how many zeros there are in circle with $|z| \leq 2$. I cannot give you a general recipe. But here is how I proceed to find it:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first obvious choice $g(z) =z^2 e^{z}$ (for which we know the number of zeros) fails for $z$ around $-2$, simply because the $g(z)$ becomes exponential small whereas in $f(z)$ the $-z$ term dominates. Therefore, you want some other term in  $g(z)$ which takes around $-2$ whereas you still need to have an $e^{z}$ in order that the function approximates $f(z)$ around $2$ correctly ($f$ is quite large for $z=2$ and then quickly falls off before at some value the $z$ term takes over).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My second guess therefore is $g(z)= z^2 ( e^{z} + c)$ with some $c&amp;gt;0$. Playing around a bit, I see that the value $c=1/2$ seems to work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now what one has to proof: we have to show that $|f(z) - g(z)| &amp;lt; |g(z)|$ for $z=2 e^{i\phi}$. The left hand side is $\left|z(\frac{1}{2}+z)\right| =  4 \left| \cos (\phi/2)\right|$. The right hand side can be simplified to $$|g(z)|^2= 16 \left|e^{z} +\frac{1}{2}\right|^2 = 16 \left[\frac{1}{4} + e^{4 \cos \phi}  + e^{2 \cos \phi} \cos(2 \sin \phi)\right].$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Taking the difference $$ \frac{|g(z)|^2 -|f(z) - g(z)|^2}{16}  = \frac{1}{4} + e^{4 \cos \phi}  + e^{2 \cos \phi} \cos(2 \sin \phi) - \cos^2 (\phi/2) \geq 0.07 .$$ The last estimate was obtained by plotting the function. The minimum is attained around $\phi=1.9$. Too make the argument complete one should prove that the difference is larger than zero (which I couldn't do so far). Then we know that $f(z)$ and $g(z)$ have the same number of zeros with $|z| \leq 2$. As $g(z)$ has only two zeros (two times $z=0$), the same hold for $f(z)$. &lt;/p&gt;&#xA;" OwnerUserId="7266" LastEditorUserId="7266" LastEditorDisplayName="" LastEditDate="2011-04-10T22:10:15.720" LastActivityDate="2011-04-10T22:10:15.720" />
  <row Id="25261" PostTypeId="1" CreationDate="2011-03-06T08:46:50.730" Score="15" ViewCount="522" Body="&lt;p&gt;Back in college I tried to find characterizations of compact (metric) spaces that make use of Lebesgue's Covering Lemma. During the course of this, I defined the following (pretty arbitrarily named) property of subsets of topological spaces:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If $X=(X,\mathcal{O})$ is a topological space and $Y\subseteq X$, we say that $Y$ is &lt;em&gt;relatively finite&lt;/em&gt; (in $X$) if and only if for every neighborhood $N$ of the boundary $\partial Y$ the set $Y\setminus N$ is finite.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My professor found it quite interesting, but couldn't recall seeing something similar before.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My assumption is that this property is either not very meaningful, or that it has previously been used elsewhere under a different name. Can anyone enlighten me?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; Some context, in case it helps: As I said I was looking for a converse of the Covering Lemma &amp;ndash; i.e. what properties do I have to require of the metric space $X$ in addition to $X$ fulfilling the Lebesgue condition (every open cover has a Lebesgue number $\delta&amp;gt;0$ such that every $\delta$-ball is contained in some cover element) in order for $X$ to be compact?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A property I found was the &amp;ldquo;relative finiteness&amp;rdquo; of the set $X^\star$ of the isolated points of $X$. Since I did this out of mere curiosity and never actually used it for anything, the only additional properties of &amp;ldquo;relatively finite&amp;rdquo; sets I can give you are these small ones that I needed for the proof:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If, in addition to the above definition, $Y$ is open, then every sequence of points from $Y$ has a cluster point in $X$.&lt;/li&gt;&#xA;&lt;li&gt;If, in addition to the above definition, $X$ is first-countable, then the interior of $Y$ is at most countable.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="3" LastEditorUserId="3" LastEditorDisplayName="" LastEditDate="2011-03-12T19:53:26.977" LastActivityDate="2011-04-23T12:54:46.813" Title="A property of subsets of topological spaces" Tags="&lt;topology&gt;&lt;reference-request&gt;" AnswerCount="1" CommentCount="9" />
  <row Id="25262" PostTypeId="1" AcceptedAnswerId="25279" CreationDate="2011-03-06T08:55:41.463" Score="3" ViewCount="144" Body="&lt;p&gt;Hey guys,&#xA;I have this ex in Data structure course.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This ex is about big o notation, and as far I remember It means that $f_1$ and $f_2$ bound asymptotically $g_1, g_2,$ but i'm not quite sure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question:&#xA;$f_1(n)=O(g_1(n)),  f_2(n)=O(g_2(n))$&#xA;Two functions are from and to the natural numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need to prove:&#xA;$f_1(n)+f_2(n) = O(\max\{g_1(n),g_2(n)\})$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you for the help.&lt;/p&gt;&#xA;" OwnerUserId="6163" LastEditorUserId="7266" LastEditorDisplayName="" LastEditDate="2011-03-06T16:13:54.747" LastActivityDate="2011-03-06T16:13:54.747" Title="The big O notation" Tags="&lt;homework&gt;&lt;notation&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="25263" PostTypeId="1" CreationDate="2011-03-06T08:59:06.680" Score="7" ViewCount="102" Body="&lt;p&gt;The following statement is an exercise in point set topology: If $E \to X$ is a covering with nonempty finite fibers and $X$ is compact, then also $E$ is compact. Now Grothendieck generalized covering theory so that in particular separable field extensions may be regarded as coverings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; What is the corresponding statement in field theory?&lt;/p&gt;&#xA;" OwnerUserId="1650" LastActivityDate="2011-03-06T08:59:06.680" Title="algebraic version of &quot;finite covering of a compact space is compact&quot;" Tags="&lt;topology&gt;&lt;field-theory&gt;" CommentCount="5" FavoriteCount="2" />
  <row Id="25264" PostTypeId="1" AcceptedAnswerId="25275" CreationDate="2011-03-06T09:07:42.563" Score="4" ViewCount="64" Body="&lt;p&gt;I've seen this written several places without proof, so I assume it's not difficult, but I am not getting it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $\mathbb P$ be a $\kappa$-c.c. notion of forcing, and let $C\in M[G]$ be club in $\kappa$.  I want to show there exists $D\in M$ such that $D\subset C$ is club in $\kappa$.  Kunen suggests:  let $f\in M[G], f:\kappa\rightarrow\kappa$ such that $\forall\alpha&amp;lt;\kappa(\alpha&amp;lt;f(\alpha)\in C$.  Then because the forcing is $\kappa$-c.c., there exists $F\in M, F:\kappa\rightarrow P(\kappa)$ such that $\forall\alpha&amp;lt;\kappa(\alpha&amp;lt;f(\alpha)\in F(\alpha)\wedge(|F(\alpha)|&amp;lt;\kappa)^{M})$.  But now I'm stuck.  The union of the image of $F$ is a set cofinal in $\kappa$, which contains a cofinal subset of $C$ that is not necessarily closed.  Any ideas?  Thank you!&lt;/p&gt;&#xA;" OwnerUserId="7817" LastActivityDate="2011-03-06T11:25:46.193" Title="If P is k-c.c. and C is club in k in M[G] then C contains a club in M" Tags="&lt;set-theory&gt;&lt;forcing&gt;" AnswerCount="1" />
  <row Id="25266" PostTypeId="1" AcceptedAnswerId="25269" CreationDate="2011-03-06T09:41:44.680" Score="2" ViewCount="87" Body="&lt;p&gt;How does dot product of two vectors is A · B = |A| * |B| * cos(Θ) ?&lt;/p&gt;&#xA;" OwnerUserId="7821" LastActivityDate="2011-03-06T13:13:02.630" Title="Dot product of two vectors" Tags="&lt;vector-calculus&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="25267" PostTypeId="2" ParentId="25256" CreationDate="2011-03-06T09:57:43.290" Score="2" ViewCount="" Body="&lt;p&gt;$13^{59} \pmod 7 = (-1)^{59} \pmod 7 = 6$.&lt;br&gt;&#xA;$13^{59} \pmod {11} = 2^{59} \pmod {11} = 2^9 \pmod {11} = 6$.&lt;br&gt;&#xA;So the answer is $6$.&lt;/p&gt;&#xA;" OwnerUserId="1508" LastActivityDate="2011-03-06T09:57:43.290" />
  <row Id="25268" PostTypeId="2" ParentId="25266" CreationDate="2011-03-06T10:03:14.170" Score="1" ViewCount="" Body="&lt;p&gt;One way of showing this requires taking a Geometric look at stuff:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Think of $\vec{a},\vec{b}$ as the vertices of a triangle with one corner in the origin&#xA;and sides of length $|\vec a|,|\vec b|,|\vec{a-b}|$. Now, use the Law of cosines, and inner product properties (over $\mathbb R$) to calculate $|\vec{a-b}|$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The law of cosines gives you:&#xA;$$|\vec{a-b}|^2=|\vec{a}|^2+|\vec{b}|^2-2|\vec{a}||\vec{b}|\cos\theta(a,b)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Inner product gives:&#xA;$$|\vec{a-b}|^2=\langle\vec{a-b},\vec{a-b}\rangle=|\vec{a}|^2+|\vec{b}|^2-2\langle\vec{a},\vec{b}\rangle$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;from there on it's an easy proof (I left you the technical details).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope that helps&lt;/p&gt;&#xA;" OwnerUserId="7129" LastEditorUserId="7129" LastEditorDisplayName="" LastEditDate="2011-03-06T13:13:02.630" LastActivityDate="2011-03-06T13:13:02.630" CommentCount="1" />
  <row Id="25269" PostTypeId="2" ParentId="25266" CreationDate="2011-03-06T10:22:22.110" Score="3" ViewCount="" Body="&lt;p&gt;Think about a triangle with sidelengths $|\textbf{a}|,|\textbf{b}|,|\textbf{c}|$. Then we can use the law of cosines. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;\begin{align}&#xD;&#xA;|\textbf{c}|^2&amp;amp;=|\textbf{a}|^2+|\textbf{b}|^2-2|\textbf{a}||\textbf{b}| \cos \theta \\&#xD;&#xA;\implies 2|\textbf{a}||\textbf{b}| \cos \theta &amp;amp;= |\textbf{a}|^2+|\textbf{b}|^2-|\textbf{c}| = \textbf{a}\cdot \textbf{a} + \textbf{b} \cdot \textbf{b} - \textbf{c}\cdot \textbf{c} &#xD;&#xA;\end{align}&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the properties of the dot product and from the fact that $\textbf{c}=\textbf{b}-\textbf{a}$ we find that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;\begin{align}&#xD;&#xA;  \textbf{c}\cdot \textbf{c} &amp;amp;=(\textbf{b}-\textbf{a}) \cdot (\textbf{b}-\textbf{a}) \\&#xD;&#xA;  &amp;amp;=(\textbf{b}-\textbf{a})\cdot \textbf{b} - (\textbf{b}-\textbf{a}) \cdot \textbf{a} \\&#xD;&#xA;  &amp;amp;= \textbf{b}\cdot \textbf{b} - \textbf{a}\cdot \textbf{b} - \textbf{b}\cdot \textbf{a} + \textbf{a}\cdot \textbf{a}.&#xD;&#xA;\end{align}&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By substituting $\textbf{c}\cdot \textbf{c}$, we get&#xA;$$&#xD;&#xA;\begin{align}&#xD;&#xA;  2|\textbf{a}||\textbf{b}| \cos \theta &amp;amp;= \textbf{a}\cdot \textbf{a} + \textbf{b} \cdot \textbf{b} - (\textbf{b}\cdot \textbf{b} - \textbf{a}\cdot \textbf{b} - \textbf{b}\cdot \textbf{a} + \textbf{a}\cdot \textbf{a}) \\&#xD;&#xA;  &amp;amp;=2 \textbf{a}\cdot \textbf{b}.&#xD;&#xA;\end{align}&#xD;&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="7044" LastActivityDate="2011-03-06T10:22:22.110" />
  <row Id="25270" PostTypeId="1" AcceptedAnswerId="25297" CreationDate="2011-03-06T10:24:22.957" Score="19" ViewCount="285" Body="&lt;p&gt;I pick a random subset $S$ of $\{1,\ldots,N\}$, and you have to guess what it is. After each guess $G$, I tell you the number of elements in $G \cap S$. How many guesses do you need?&lt;/p&gt;&#xA;" OwnerUserId="857" LastActivityDate="2011-03-06T15:38:39.987" Title="Guessing a subset of {1,...,N}" Tags="&lt;combinatorics&gt;&lt;puzzle&gt;" AnswerCount="3" FavoriteCount="4" />
  <row Id="25271" PostTypeId="1" CreationDate="2011-03-06T10:35:24.037" Score="1" ViewCount="56" Body="&lt;p&gt;I have a question. Could you please help me to solve this problem?&#xA;Is R^2\A and R^2\B are homeomorphic eventhough A and B are non-homeomorphic closed subsets of R^2.&lt;/p&gt;&#xA;" OwnerUserId="7858" LastActivityDate="2011-03-06T11:14:31.443" Title="Non-Homeomorphic closed" Tags="&lt;general-topology&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="25272" PostTypeId="2" ParentId="25271" CreationDate="2011-03-06T10:41:16.647" Score="2" ViewCount="" Body="&lt;p&gt;A= {0}, B= closed unit ball is an example of this.&lt;/p&gt;&#xA;" OwnerUserId="6417" LastEditorUserId="6417" LastEditorDisplayName="" LastEditDate="2011-03-06T10:54:42.867" LastActivityDate="2011-03-06T10:54:42.867" CommentCount="0" />
  <row Id="25273" PostTypeId="1" AcceptedAnswerId="25292" CreationDate="2011-03-06T10:54:37.763" Score="4" ViewCount="165" Body="&lt;p&gt;Hello, is there any rule to differentiate the function $(a\,x)^{b\,x}$? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've got to find the derivative of $(x^2+1)^{\arctan x}$ and &lt;a href=&quot;http://www.wolframalpha.com/&quot; rel=&quot;nofollow&quot;&gt;Wolfram|Alpha&lt;/a&gt; says the answer is &#xA;$$\tan^{-1}(x) (x^2+1)^{\tan^{-1}(x)-1} \left(\frac{d}{dx}(x^2+1)\right)+\log(x^2+1) (x^2+1)^{\tan^{-1}(x)} \left(\frac{d}{dx}(\tan^{-1}(x))\right)$$&#xA;Is there any general rule to do that? &#xA;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="7758" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2011-03-06T17:57:01.773" LastActivityDate="2011-03-06T18:00:51.520" Title="Derivative of $(a\,x)^{b\,x}$" Tags="&lt;calculus&gt;" AnswerCount="4" />
  <row Id="25274" PostTypeId="2" ParentId="25273" CreationDate="2011-03-06T11:04:40.163" Score="5" ViewCount="" Body="&lt;p&gt;Assuming you mean $(ax)^{bx}$, I'd just write it as $(e^{\ln(ax)})^{bx}$ and use the chain rule (ie $e^{\ln(ax)bx} = e^{u(x)}$ and go from there).&lt;/p&gt;&#xA;" OwnerUserId="7155" LastActivityDate="2011-03-06T11:04:40.163" />
  <row Id="25275" PostTypeId="2" ParentId="25264" CreationDate="2011-03-06T11:14:17.997" Score="3" ViewCount="" Body="&lt;p&gt;This is how I think about it. Suppose that $\dot C$ is a $P$-name and some condition $p_0$ forces that $\dot C$ is club in $\kappa$. Consider the set $D$ of ordinal $\alpha\lt\kappa$ such that $p_0$ forces $\check\alpha\in\dot C$. This set is certainly closed, since if $p_0$ forces ordinals into $\dot C$, then by closure it will also force their supremum into $\dot C$. But $D$ is also unbounded. To see this, fix any ordinal $\alpha_0$. The club $C$ will have a next element after $\alpha_0$, and by the mixing lemma, we can find a name $\dot\beta_0$ such that $p_0$ forces that $\dot\beta_0$ is the next element of $\dot C$ after $\check\alpha$. (This is essentially the function $f$ you mention.) There is an antichain of possible values for $\dot\beta_0$---this is the function $F$ you mention---and so by the $\kappa$-c.c. of $P$, we can find an ordinal $\alpha_1$ such that $p_0$ forces $\dot\beta_0\lt\check\alpha_1$. Now continue with $\dot\beta_1$ and $\alpha_2$, $\dot\beta_2$ and so on in the same way. Thus, the condition $p_0$ forces that each $\dot\beta_n$ is in $\dot C$, so it forces that the supremum of these $\omega$ many ordinals is in $\dot C$. But this  supremum is the same as the supremum of the $\alpha_n$, and so $p_0$ forces that the ordinal $\alpha_\omega=\text{sup}\alpha_n$ is in $\dot C$. Thus, we have found an ordinal in $D$ above $\alpha_0$, and so $D$ is unbounded, as desired.&lt;/p&gt;&#xA;" OwnerUserId="413" LastEditorUserId="413" LastEditorDisplayName="" LastEditDate="2011-03-06T11:25:46.193" LastActivityDate="2011-03-06T11:25:46.193" CommentCount="7" />
  <row Id="25276" PostTypeId="2" ParentId="25271" CreationDate="2011-03-06T11:14:31.443" Score="0" ViewCount="" Body="&lt;p&gt;This sounds wrong. Maybe I can't see the picture, but if you take $A={0}$ and $B={0,1}$, then the first space will have two connected components and the second three.&lt;/p&gt;&#xA;" OwnerUserId="7838" LastActivityDate="2011-03-06T11:14:31.443" CommentCount="3" />
  <row Id="25277" PostTypeId="2" ParentId="25273" CreationDate="2011-03-06T11:22:51.107" Score="3" ViewCount="" Body="&lt;p&gt;Not answering the math part, since Stijn has &lt;a href=&quot;http://math.stackexchange.com/questions/25273/derivation-of-axbx/25274#25274&quot;&gt;already done that&lt;/a&gt;, but if you click on the &quot;show steps&quot; button, Wolfram|Alpha shows you one possible path for the derivation.  I've included the image for the derivation of &#xA;$\frac{\partial}{\partial x} ((a x)^{b x})$&#xA;&lt;img src=&quot;http://i.imgur.com/tymBK.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A similar set of steps is supplied for the other derivative that you want to take: &lt;code&gt;http://www.wolframalpha.com/input/?i=d/dx((x^2%2B1)^(tan^(-1)(x)))&lt;/code&gt;&lt;/p&gt;&#xA;" OwnerUserId="954" LastEditorUserId="954" LastEditorDisplayName="" LastEditDate="2011-03-06T11:41:26.263" LastActivityDate="2011-03-06T11:41:26.263" CommentCount="2" />
  <row Id="25278" PostTypeId="2" ParentId="25228" CreationDate="2011-03-06T11:28:07.143" Score="5" ViewCount="" Body="&lt;p&gt;You can parametrize the three-dimensional manifold $M$ by the magnitude of either $z_1$ or $z_2$ and the two arguments of $z_1$ and $z_2$. Let's call the one of $z_1$ and $z_2$ whose magnitude we use for the parametrization $z$ and the other $z&amp;#39;$. You can visualize this parametrization using the volume enclosed by a torus, with the magnitude of $z$ corresponding to the distance from the central ring, the argument of $z$ corresponding to the angle with respect to the central ring (the poloidal angle) and the argument of $z&amp;#39;$ corresponding to the angle with respect to the torus axis (the toroidal angle). That is, with the parameters used in &lt;a href=&quot;http://en.wikipedia.org/wiki/Torus#Geometry&quot; rel=&quot;nofollow&quot;&gt;the Wikipedia article on the torus&lt;/a&gt;, the magnitude of $z$ is $r$, the argument of $z$ is $v$, and the argument of $z&amp;#39;$ is $u$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To see what happens in the two interesting neighbourhoods around $z_1=0$ and $z_2=0$, we can use the appropriate torus where $z$ is the one of $z_1$ and $z_2$ that is close to zero. Dealing first with $z_1$ near $0$ (and thus $z=z_1$, $z'=z_2$), the orbits of the given action of $S^1$ are spirals that spiral around the central ring $k$ times before returning to the same point. We can pick any disk defined by some value of $u$ as the set of representatives of these orbits, and see that the space of these orbits has the manifold structure of an ordinary disk.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, if $z_2$ is near $0$ (and thus $z=z_2$, $z'=z_1$), the orbits are again spirals that spiral around the central ring, but this time, they only go around the ring $1/k$ times before returning to the same value of $u$, and only return to the same point after going around the torus axis $k$ times. If we now pick any disk defined by some value of $u$, it contains $k$ representatives of each orbit. Thus, the space of these orbits has the same structure as the quotient of the disk with respect to the cyclic group of rotations generated by a rotation around the origin through $2\pi/k$. The conical singularity arising from this is described &lt;a href=&quot;http://books.google.de/books?id=WgUkSTJWQacC&amp;amp;pg=PA360&amp;amp;lpg=PA360&amp;amp;dq=cone+singularity+orbifold&amp;amp;source=bl&amp;amp;ots=2V8E6snAtc&amp;amp;sig=VaLQsMN-ue8QRka-JLhzDF4nRuM&amp;amp;hl=de&amp;amp;ei=sWFzTeLKAYWVswaR5omEDg&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=7&amp;amp;ved=0CFIQ6AEwBg#v=onepage&amp;amp;q=cone%20singularity%20orbifold&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; for the case $k=2$. Note that it is only a singularity in the differentiable structure, not in the topological structure, since the cone is topologically a two-dimensional manifold at the origin.&lt;/p&gt;&#xA;" OwnerUserId="6622" LastEditorUserId="6622" LastEditorDisplayName="" LastEditDate="2011-03-06T12:31:49.647" LastActivityDate="2011-03-06T12:31:49.647" />
  <row Id="25279" PostTypeId="2" ParentId="25262" CreationDate="2011-03-06T11:35:25.380" Score="3" ViewCount="" Body="&lt;p&gt;$f_1(n)=O(g_1(n))$ (for $n\to\infty$) means that $|f_1(n)| \leq C_1 |g_1(n)|$ $\forall n&amp;gt; n_1$ with $C_1$ and $n_1$ finite positive numbers. Similarly for $f_2(n)=O(g_2(n))$ yields $|f_2(n)| \leq C_2 |g_2(n)|$ $\forall n&amp;gt; n_2$. (see &lt;a href=&quot;http://en.wikipedia.org/wiki/Big_O_notation&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Big_O_notation&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the &lt;a href=&quot;http://en.wikipedia.org/wiki/Triangle_inequality&quot; rel=&quot;nofollow&quot;&gt;triangle inequality&lt;/a&gt;, we get $$|f_1(n) + f_2(n)| \leq |f_1(n)| + |f_2(n)| \leq C_1 |g_1(n)| + C_2 |g_2(n)| \qquad \forall n&amp;gt; \max\{n_1,n_2\}.$$ &#xA;Using that both $|g_1|$ and $|g_2|$ are bounded from above by $\max\{|g_1|, |g_2|\}$, we obtain&#xA;$$|f_1(n) + f_2(n)| \leq (C_1 + C_2) \max \{ |g_1(n)| , |g_2(n)| \} \qquad n&amp;gt;\max\{n_1,n_2\},$$&#xA;or in terms of big-O notation&#xA;$$f_1(n) + f_2(n)  = O (\max \{ |g_1(n)| , |g_2(n)| \}).$$&lt;/p&gt;&#xA;" OwnerUserId="7266" LastEditorUserId="7266" LastEditorDisplayName="" LastEditDate="2011-03-06T12:04:19.043" LastActivityDate="2011-03-06T12:04:19.043" />
  <row Id="25280" PostTypeId="2" ParentId="23903" CreationDate="2011-03-06T11:38:12.040" Score="2" ViewCount="" Body="&lt;p&gt;There is an approach to quantum logic where you get topoi with quantum logic. An elementary topos is sometimes regarded as a &quot;place&quot; where you can do mathematics, but where classical logic doesn't necessarily apply. Thus you get a different sort of mathematics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know very little about these quantum topoi, so I cannot detail in what way their mathematics differ from the classical one. But I think the two articles referenced in the below PlanetMath articles may (or may not - I haven't read them) answer your question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://planetmath.org/encyclopedia/QuantumLogicsTopoi2.html&quot; rel=&quot;nofollow&quot;&gt;http://planetmath.org/encyclopedia/QuantumLogicsTopoi2.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://planetmath.org/encyclopedia/QuantumStateSpace.html&quot; rel=&quot;nofollow&quot;&gt;http://planetmath.org/encyclopedia/QuantumStateSpace.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5553" LastActivityDate="2011-03-06T11:38:12.040" />
  <row Id="25281" PostTypeId="2" ParentId="25233" CreationDate="2011-03-06T11:45:07.817" Score="1" ViewCount="" Body="&lt;p&gt;Let $v_i$ be the valuations corresponding to the DVR $R_i$. Note that according to your choice we have $v_i(x_j) = 0$ if $i \neq j$ but $v_i(x_j) =1$ if $i = j$. So $x_iA \subset \mathfrak{p}_i$. We have to show the opposite inclusion. Choose any any $y \in  \mathfrak{p}_i$. Then $y = x_it$ for some $t \in R_i$ since $\mathfrak{m}_i = x_i R_i$. Now it is easy to see that values of $t = y/x_i$ under all valuations $v_i$ are non negative. This means that $t \in A$. So we are done. in fact you can also show  that $\mathfrak{p}_i^{k} = \mathfrak{p}_i^{(k)}$ for any natural number $k$. Now note that the element $x_{1}^{\nu_1}...x_{n}^{\nu_n} \in I_{\mathfrak{p_i}}$ for all $\mathfrak{p_i}$. This means that $x_{1}^{\nu_1}...x_{n}^{\nu_n}A \in I$. Now the two ideals are equal locally. Therefore they are equal globally.&lt;/p&gt;&#xA;" OwnerUserId="7565" LastActivityDate="2011-03-06T11:45:07.817" />
  <row Id="25282" PostTypeId="2" ParentId="25151" CreationDate="2011-03-06T12:57:27.130" Score="0" ViewCount="" Body="&lt;p&gt;Actually non of them, and both of them, I have problem with Math in Minitab® ;)  I don't know how to generate Poisson random samples with 1500 sample of size n=15, I think of :&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&#xA; MTB &gt; Random 15000 C1-C15;&#xA;SUBC&gt;   Poisson 3.&#xA;MTB &gt;&#xA;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;But don't seems corect!!&lt;/p&gt;&#xA;" OwnerUserId="7826" LastActivityDate="2011-03-06T12:57:27.130" />
  <row Id="25283" PostTypeId="2" ParentId="25270" CreationDate="2011-03-06T13:00:09.920" Score="0" ViewCount="" Body="&lt;p&gt;$M$ denotes the solution set, $|M|$ denotes its cardinality (the number of elements).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that if I guess $\{1\dots N\}$, then you'll have to answer with $|M|$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I can just brute force your set by removing the elements of $\{1\dots N\}$ one-by-one. Every time your answer decreases I will have identified an element of the solution. By induction, your answer will have to reach zero eventually. By that time I will have identified all $x\in M$. It will take me one more guess to win the game.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the algorithm outlined above I would guess at most $N+1$ times.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am pretty sure in that you cannot do better than $O(n)$ in the worst case (I offer no proof though).&lt;/p&gt;&#xA;" OwnerUserId="1712" LastActivityDate="2011-03-06T13:00:09.920" CommentCount="1" />
  <row Id="25284" PostTypeId="1" CreationDate="2011-03-06T13:14:32.333" Score="3" ViewCount="83" Body="&lt;p&gt;I am faced with an approximation that replaces a probability density function with the indicator function and I am at a loss as to why this is valid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We want to model the lifetime $T$ of a website using the density function $Pr[X &amp;gt; t] \approx e^{-\rho t}$. Then, according to the math I am reading,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\begin{eqnarray}&#xD;&#xA;  P[X &amp;gt; k] &amp;amp; = &amp;amp; \int_0^t Pr[X &amp;gt; k\,|\,T = u] \frac{d Pr[T \leq u]}{du} du \\ &#xD;&#xA;           &amp;amp; \approx &amp;amp; \rho \int_0^t e^{-\rho u} Pr[X &amp;gt; k\,|\,T = u] du \\&#xD;&#xA;           &amp;amp; \approx &amp;amp; \rho \int_0^t e^{-\rho u} 1_{\{l(u) &amp;gt; k\}} du&#xD;&#xA;\end{eqnarray}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;if $l(u) \sim e^{\beta u}$ for some $\beta &amp;lt; \rho$ (Here, $\sim$ means that $l(u)$ grows as fast as $e^{\beta u}$ - some would write $l(u) = \Theta(e^{\beta u})$). $1_{\{l(u) &amp;gt; k\}}$ is the indicator function that is equal to $1$ if $l(u) &amp;gt; k$, otherwise it is $0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to understand the intermediate steps required to jump from the PDF to the indicator function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a bit of context, this expression comes from mathematics that tries to model the lifetime $t$ of a website. This is used to explain why the world-wide web network is (roughly) scale-free. &lt;/p&gt;&#xA;" OwnerUserId="3409" LastEditorUserId="3409" LastEditorDisplayName="" LastEditDate="2011-03-06T15:06:44.023" LastActivityDate="2011-03-06T15:06:44.023" Title="When is it valid to convert a function inside a probability integral to the indicator function?" Tags="&lt;probability&gt;&lt;integral&gt;&lt;approximation&gt;" CommentCount="11" />
  <row Id="25285" PostTypeId="2" ParentId="25270" CreationDate="2011-03-06T13:24:33.260" Score="-2" ViewCount="" Body="&lt;p&gt;The answer is $N$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $F(N)$ be the number of guesses needed for determining $S$ of $\{1,...,N\}$ when the size of $S$ is unknown and let $H(N)$ be the number of guesses needed for determining $S$ of $\{1,...,N\}$ when the size of S is known.  Obviously $H(0)=0, H(1) = 0, F(0)=0,F(1)=1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose we know the size $|S|.$  After one guess, the problem reduces to determining $G \cap S$ of G, where we know $|G \cap S|$ and $G^C \cap S$ of $G^C \cap \{1,...,N\}$, where we know $|G^C \cap S|$.  Thus&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$H(N) = 1+\min_{|G|} H(|G|) + H(N - |G|).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now suppose that $H(N) = N-1$ for $N=1,...,n$.  Then by our induction hypothesis, $H(|G|) + H(n+1-|G|)=|G|-1 + n-|G| = n-1$ when $\min(|G|,N-|G|) &amp;gt; 0$.  When $|G|=0$ or $|G|=n+1$, $H(|G|) + H(n+1-|G|)=H(n+1)$, yet it is impossible that $H(n+1) = 1 + H(n+1)$, so $|G|=0$ and $|G|=n+1$ cannot be the minimizing choices of $|G|$.  Therefore we conclude by induction that $H(N) = N-1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we will show that $F(N) = N$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose that $F(N) = N$ for $N=1,...,n$.  We have&#xA;$$F(N) = 1+\min_{|G|} H(|G|) + F(N - |G|).$$&#xA;and thus by a similar reasoning process as before, we conclude by induction that $F(n+1) = n+1$.&lt;/p&gt;&#xA;" OwnerUserId="7862" LastActivityDate="2011-03-06T13:24:33.260" CommentCount="4" />
  <row Id="25286" PostTypeId="1" AcceptedAnswerId="25288" CreationDate="2011-03-06T13:32:49.487" Score="1" ViewCount="135" Body="&lt;p&gt;&lt;strong&gt;LATER EDIT:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I managed to find the errors in my equations below. It was a sign mistake on one of the terms (- instead of +). Having corrected that, the method which I describe below WORKS, but DON'T USE IT ::- D. My method is convoluted and prone to mistakes (as I proved, haha). The 2 answers below offer much better choices. hardmath found a few of my mistakes and provided an alternative way of obtaining X2 and Y2 using the same system of formulas I used.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, joriki provided an alternate solution which is VERY efficient in terms of number of calculations which need to be done each frame. Eventually, I completely dropped my approach and will use his solution. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you hardmath and joriki, and, by all means, Math @ Stack ::- D. You people rule! I humbly thank you for helping a programmer in distress!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;ORIGINAL POST:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hi ::- D. I'm working on a (Flash) game and I have a weapon from which a projectile departs. The projectile travels along a line. I have the slope of the line as well as the origin (1 point). I want to calculate, depending on the Distance, where should I draw the projectile, frame by frame. The distance increases with a certain quantity each frame, as the projectile travels along the line.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have tried my best to solve this and this is what I did so far. I would not mind if you know a simpler way, but I would very much appreciate it if you would tell me why my method doesn't work (the coordinates I'm getting are very chaotic).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I used these equations to build a system:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.analyzemath.com/Calculators/TwoPointsCalculator.html&quot; rel=&quot;nofollow&quot;&gt;http://www.analyzemath.com/Calculators/TwoPointsCalculator.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;d = sqrt [ (x1 -x2)^ + (y1 - y2)^ ]&#xA;&#xA;m = (y2 - y1) / (x2 - x1)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I took out y2:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;y2 = m * x2 - m * x1 + y1&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;From that equation, I know everything except x2. &lt;code&gt;x1, y1 and m are all known&lt;/code&gt;. I'm out to find x2 and y2 which are the coordinates of my desired point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I removed the radical by squaring the first equation and also breaking (solving) the paranthesis. I should mention it's been a very long time since I did any math so please excuse the lack of correct terminology. Also, I'm unfamiliar with how to produce a fast MathJaX of the equations, but I think they're simple enough to be written normally. ^ - means square. I used some extra paranthesis to clear it up.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;d^ = x1^ - (2 * x2 * x1) + x2^ + y1^ - (2 * y2 * y1) + y2 ^&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Ok so I also know who &lt;code&gt;d&lt;/code&gt; is. Now, I took y2 and substituted it in the above equation. This is what resulted:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;d^ = x1^ - 2 * x2 * x1  + x2^ + y1^ - 2 * y1 * (m * x2 + K) + (m * x2 +K) ^&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Where K is that part which I can calculate from the above y2 equation: y2 = m* x2 &lt;code&gt;-m * x1 + y1&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So now I got an equation where I know everything except x2. I reduced this equation to a quadratic equation. The final form (I solved the squared parenthesis above as well) is:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;x2^ ( - 1 - m^) + x2 (2 * x1   + 2 *  y1 * m - 2 * m * K) + d^ - x1^ - y1^ + 2 * y1 * K - K ^ = 0&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Looks scary doesn't it? &lt;em&gt;laugh&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This seems to be correct so far. Because we're talking about a line, slope and distance, it's obvious that there can be &lt;strong&gt;TWO&lt;/strong&gt; points which are of distance D from my Origin (x1, y1). So I solve the quadratic equation in order to find x2-1 and x2-2 which I can then take to my first equation and get y2-1 and y2-2. So, I moved to solve the quadratic equation. First, I found a, b and c.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;a = -1 - m^&#xA;b = (2 * x1   + 2 *  y1 * m - 2 * m * K) &#xA;c = d^ -  x1^ - y1^  + 2 * y1 * K - K ^ &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;And now, I found the 2 roots:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;x2 = (-b + sqrt (b^ - 4ac)) / 2a&#xA;x2 = (-b - sqrt (b^ - 4ac)) / 2a&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Then, I found the two y2 points. And that's it ::- D.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It doesn't work &lt;em&gt;laugh&lt;/em&gt;......... I worked for hours this morning, flexing whatever pathetic math muscles I got left and I worked all this far to end up with:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(x=1509141.9643610462, y=20806970.245399687)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This, in a case where the origin is at 200, 250 and the distance is 1 and the slope is 2.2 or -0.7 or whatever.......&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sometimes, I even get NaN (that is, Not A Number, invalid value in ActionScript 3). So the values I'm getting are DEFINITELY not related to my origin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A point of interest however: for slope = 1.380657160570366, my projectile &lt;strong&gt;ACTUALLY SEEMS TO WORK&lt;/strong&gt;, as in, it appears a bit TOO far from the Origin (about 50 pixels), but it &lt;strong&gt;DOES MOVE&lt;/strong&gt; along the correct line, albeit, strangely, it ACCELERATES, even though I never increase its speed, I simply increase the distance:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;_ProjectileDistance += _ProjectileSpeed;&#xA;var newCoords: Point = GetPointAlongLine(weaponRootPointX, weaponRootPointY, _ProjectileDistance, _ProjectileSlope);&#xA;Projectile.x = newCoords.x;&#xA;Projectile.y = newCoords.y;&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;And for those who know programming as well, here is what all I said looks like in code:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;  var a: Number = - 1 - slope * slope;&#xA;  var b: Number = 2 * x1 + 2 * y1 * slope - 2 * slope * y2KnownPart;&#xA;  var c: Number = distance*distance - x1*x1 - y1*y1 - 2 * y1 * y2KnownPart - y2KnownPart*y2KnownPart;&#xA;&#xA;  var x2Root1: Number = (- b + Math.sqrt(b*b - 4 * a * c)) / (2 * a);&#xA;  var x2Root2: Number = (- b - Math.sqrt(b*b - 4 * a * c)) / (2 * a);&#xA;  var y2Root1: Number = slope * x2Root1 + y2KnownPart;&#xA;  var y2Root2: Number = slope * x2Root2 + y2KnownPart;&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Sorry for the long post ::- D. I hope somebody will have the patience to read until this line ::- D. If you did, know that you're doing it for a good cause ::- D. My games are free, so is my software, just look here ::- D &lt;a href=&quot;https://sourceforge.net/users/echysttas&quot; rel=&quot;nofollow&quot;&gt;https://sourceforge.net/users/echysttas&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7745" LastEditorUserId="7745" LastEditorDisplayName="" LastEditDate="2011-03-06T18:14:44.597" LastActivityDate="2011-03-06T18:14:44.597" Title="2D - Coordinates of a point along a line, based on d and m - Where am I messing up?" Tags="&lt;geometry&gt;" AnswerCount="2" />
  <row Id="25287" PostTypeId="1" AcceptedAnswerId="25300" CreationDate="2011-03-06T14:04:37.217" Score="1" ViewCount="89" Body="&lt;p&gt;Given $a,b&amp;gt;0$ such that $a\ne b$ but nothing else is given (e.g. $a,b$ are not known to be coprime), I want to understand the structure of the set of solutions to the following system of two linear equations:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a(x+y)+b(z+w)=0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a(z-w)+b(x-y)=0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I approach such a problem? Are additional assumptions on $a,b$ helpful in this case?&lt;/p&gt;&#xA;" OwnerUserId="1818" LastActivityDate="2011-03-07T04:16:18.950" Title="Solution in integer to a system of linear equations" Tags="&lt;linear-algebra&gt;&lt;elementary-number-theory&gt;" AnswerCount="2" />
  <row Id="25288" PostTypeId="2" ParentId="25286" CreationDate="2011-03-06T14:05:43.843" Score="3" ViewCount="" Body="&lt;p&gt;Your approach is way too complicated -- I won't take the time to find where you made a mistake, but instead show how it can be done much simpler:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The distance $d$ is the hypotenuse of a right triangle formed by the two axis-parallel segments of lengths $\Delta y := y_2-y_1$ and $\Delta x := x_2-x_1$, much like in this picture I found on the web:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://www.sternenwind.ch/mathproj/scans/stoffplaene/3g_lin_fkt_steigungsdreieck.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The slope $m$ is $\Delta y/\Delta x$, which is the tangent of the angle $\alpha$ between the line and the $x$-axis: $m=\tan\alpha$. The increments $\Delta x$ and $\Delta y$ are given by the cosine and sine, respectively, of that angle times the hypotenuse, so all you have to do is calculate $\alpha=\arctan m$ and then $\Delta x = d\cos \alpha$ and $\Delta y=d\sin\alpha$. If your calculations are time-critical, you can get by with a couple fewer transcendental operations, but this is the most straightforward way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: Actually, you'll want to use the &lt;code&gt;atan2&lt;/code&gt; function that a lot of programming environments have, with arguments $\Delta y$ and $\Delta x$, since you lose a sign when you form $m=\Delta y/\Delta x$, i.e. you no longer know which direction along the line the projectile is travelling.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: hardmath's answer is roughly what I had in mind when I said &quot;If your calculations are time-critical, you can get by with a couple fewer transcendental operations&quot;. Note, however, that in both answers the transcendental operations have to be performed only once (a square root in hardmath's case, three trigonometric functions in my case), and then calculating the projectile's position for each $d$ only involves multiplications.&lt;/p&gt;&#xA;" OwnerUserId="6622" LastEditorUserId="6622" LastEditorDisplayName="" LastEditDate="2011-03-06T14:13:27.033" LastActivityDate="2011-03-06T14:13:27.033" CommentCount="12" />
  <row Id="25289" PostTypeId="2" ParentId="25286" CreationDate="2011-03-06T14:08:01.000" Score="1" ViewCount="" Body="&lt;p&gt;I think you missed a chance to keep the problem simple.  Since &lt;code&gt;m, x1, y1, d&lt;/code&gt; are known, you are asking to solve two equations for the two unknowns &lt;code&gt;x2, y2&lt;/code&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ (x2 - x1)^2 + (y2 - y1)^2 = d^2 $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ (y2 - y1) / (x2 - x1) = m $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But this is simpler to state in terms of unknowns &lt;code&gt;u = x2 - x1&lt;/code&gt; and &lt;code&gt;v = y2 - y1&lt;/code&gt;.  Once you've found &lt;code&gt;u, v&lt;/code&gt; it is trivial to get &lt;code&gt;x2, y2&lt;/code&gt; by adding &lt;code&gt;x1, y1&lt;/code&gt; respectively.  So consider the two equations:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ u^2 + v^2 = d^2 $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ v/u = m $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now $v = m u$ from the 2nd eqn. can be substituted into the 1st eqn. giving us simply:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ (m^2 + 1)u^2 = d^2 $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;whose solution is obviously (assuming d &gt; 0):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ u = \pm d/ \sqrt{m^2 + 1} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and correspondingly:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ v = \pm m d/ \sqrt{m^2 + 1} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where the same choice of plus-or-minus sign must be made in both formulas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Out of an abundance of caution one should substitute these back into the equations for &lt;code&gt;u, v&lt;/code&gt; above, to make sure we haven't introduced artifact roots (we have not).  Geometrically the picture is a line through the origin intersecting the circle of radius &lt;code&gt;d&lt;/code&gt; at a point &lt;code&gt;(u,v)&lt;/code&gt; along a line of slope &lt;code&gt;m&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, add &lt;code&gt;(x1,y1)&lt;/code&gt; to &lt;code&gt;(u,v)&lt;/code&gt; to get &lt;code&gt;(x2,y2)&lt;/code&gt; and you are done.  Presumably there's some information about the direction you are traveling that tells which of the two solutions is the one you want...&lt;/p&gt;&#xA;" OwnerUserId="3111" LastActivityDate="2011-03-06T14:08:01.000" CommentCount="9" />
  <row Id="25290" PostTypeId="1" CreationDate="2011-03-06T14:15:02.503" Score="3" ViewCount="63" Body="&lt;p&gt;Given a theory (say $T$) which satisfies the hypotheses of the Godel- Rosser theorem. How to show whether a Rosser sentence $R$ for $T$ is true for the standard interpretation ($\Bbb N$) or not?.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I know it is true but how we can prove this claim!)&lt;/p&gt;&#xA;" OwnerUserId="7863" LastActivityDate="2011-03-06T19:22:26.783" Title="Rosser sentence" Tags="&lt;logic&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="25291" PostTypeId="1" CreationDate="2011-03-06T14:20:51.083" Score="1" ViewCount="72" Body="&lt;p&gt;Where a group can consist of 1 or more items, groups don't have to be equally sized and items can be duplicates. Example - Choose 3 groups:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Items: 1 2 2 3&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Groups: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) (2 2) (3)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1 2) (2) (3)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(3 1) (2) (2)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(3 2) (2) (1)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update&lt;/strong&gt;&#xA;Following Rasmus' comment, I see that this is more technically described as finding the possible partitions of a multiset where the number of cells is &lt;em&gt;x&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="7864" LastEditorUserId="7864" LastEditorDisplayName="" LastEditDate="2011-03-06T14:45:46.163" LastActivityDate="2011-06-04T21:11:38.897" Title="What is the number of ways to choose x groups from y items? (partitions with x cells of a multiset)" Tags="&lt;combinatorics&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="25292" PostTypeId="2" ParentId="25273" CreationDate="2011-03-06T14:32:26.340" Score="2" ViewCount="" Body="&lt;p&gt;Your derivative of $(x^2+1)^{\arctan x}$ is the particular case for $u(x)=x^2+1$ and $v(x)=\arctan x$ of  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{d}{dx}\left(\left[ u(x)\right] ^{v(x)}\right)=v(x)\left[ u(x)\right]&#xD;&#xA;^{v(x)-1}u^{\prime }(x)+\left( \ln u(x)\right) \left[ u(x)\right]&#xD;&#xA;^{v(x)}v^{\prime }(x),$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or omitting de variable $x$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left( u^{v}\right)^{\prime }=vu^{v-1}u^{\prime }+\left( \ln u\right) u^{v}v^{\prime }.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This can be derived observing that, since $u=e^{\ln u}$, we have $u^v=e^{v\ln u}$: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{eqnarray*}&#xD;&#xA;\frac{d}{dx}\left( u^{v}\right)  &amp;amp;=&amp;amp;\frac{d}{dx}\left( e^{v\ln u}\right)  \\&#xD;&#xA;&amp;amp;=&amp;amp;e^{v\ln u}\left( \ln u\frac{dv}{dx}+\frac{v}{u}\frac{du}{dx}\right)  \\&#xD;&#xA;&amp;amp;=&amp;amp;u^{v}\left( \ln u\frac{dv}{dx}+\frac{v}{u}\frac{du}{dx}\right)  \\&#xD;&#xA;&amp;amp;=&amp;amp;u^{v}\ln u\frac{dv}{dx}+u^{v}\frac{v}{u}\frac{du}{dx} \\&#xD;&#xA;&amp;amp;=&amp;amp;u^{v}(\ln u)v&amp;#39;+u^{v-1}vu&amp;#39;.&#xD;&#xA;\end{eqnarray*}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another possibility is to consider the variables $u$ and $v$ (both depending on $x$) in the function &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$z=f(u(x),v(x))=\left[ u(x)\right] ^{v(x)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and determine its &lt;a href=&quot;http://en.wikipedia.org/wiki/Total_derivative&quot; rel=&quot;nofollow&quot;&gt;total derivative&lt;/a&gt; with respect to $x$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{eqnarray*}&#xD;&#xA;\frac{dz}{dx} &amp;amp;=&amp;amp;\frac{d}{dx}\left( \left[ u(x)\right] ^{v(x)}\right) \\&#xD;&#xA;&amp;amp;=&amp;amp;\frac{%&#xD;&#xA;\partial z}{\partial u}\frac{du}{dx}+\frac{\partial z}{\partial v}\frac{dv}{%&#xD;&#xA;dx} \\&#xD;&#xA;&amp;amp;=&amp;amp;v(x)\left[ u(x)\right] ^{v(x)-1}u^{\prime }(x)+\left[ u(x)\right]&#xD;&#xA;^{v(x)}\left( \ln u(x)\right) v^{\prime }(x)&#xD;&#xA;\end{eqnarray*}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;because&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\partial z}{\partial u}=\frac{\partial }{\partial u}\left(&#xD;&#xA;u^{v}\right) =vu^{v-1}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\partial z}{\partial v}=\frac{\partial }{\partial v}\left(&#xD;&#xA;u^{v}\right) =\frac{\partial }{\partial v}\left( e^{\ln u\cdot v}\right)&#xD;&#xA;=e^{\ln u\cdot v}\ln u=u^{v}\ln u.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $u(x)=ax,v(x)=bx$, we get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{d}{dx}\left( \left( ax\right) ^{bx}\right) =bx\left( ax\right)&#xD;&#xA;^{bx-1}a+\left( ax\right) ^{bx}\left( \ln (ax)\right) b.$$&lt;/p&gt;&#xA;" OwnerUserId="752" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2011-03-06T18:00:51.520" LastActivityDate="2011-03-06T18:00:51.520" />
  <row Id="25293" PostTypeId="1" AcceptedAnswerId="25294" CreationDate="2011-03-06T14:32:35.987" Score="1" ViewCount="52" Body="&lt;p&gt;let f be defined on [0,2$\pi$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;denote $f_x(y)=f(y-x)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;how do i go about showing $||f_x||_{L_p}=||f||_{L_p}$ for all p&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i was trying to write out f in fourier series, and use the fact that the fourier coefficient of $f_x$ is $e^{inx}$ times the fourier coefficient of f. but i couldnt get anywhere.&lt;/p&gt;&#xA;" OwnerUserId="6556" LastActivityDate="2011-03-06T14:49:00.980" Title="show $||f_x||_{L_p}=||f||_{L_p}$ for all p" Tags="&lt;fourier-analysis&gt;" AnswerCount="1" />
  <row Id="25294" PostTypeId="2" ParentId="25293" CreationDate="2011-03-06T14:49:00.980" Score="0" ViewCount="" Body="&lt;p&gt;If I understand the question correctly, then $x \in [0,2 \pi]$ a fixed point. If so then it should be just a substitution in the integral defining the $L_p$ norm.&lt;/p&gt;&#xA;" OwnerUserId="7838" LastActivityDate="2011-03-06T14:49:00.980" />
  <row Id="25295" PostTypeId="1" AcceptedAnswerId="25325" CreationDate="2011-03-06T14:53:56.513" Score="3" ViewCount="263" Body="&lt;p&gt;A scheme to generate random variates distributed uniformly in  $S^2=\{x\in \mathbb{R}^n \mid \|x\|_2=1\}$ is well known: generate a standard normal variate in $R^n$ and normalize it to unit norm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a similarly simple and clever procedure to simulate uniformly distributed variates on the 1-ball $S^1=\{x \in \mathbb{R}^n \mid \|x\|_1=1\}$? &lt;/p&gt;&#xA;" OwnerUserId="7192" LastEditorUserId="2420" LastEditorDisplayName="" LastEditDate="2011-03-06T16:48:01.737" LastActivityDate="2011-03-06T17:59:49.197" Title="Simulating uniformly on $S^1=\{x \in \mathbb{R}^n \mid \|x\|_1=1\}$" Tags="&lt;geometry&gt;&lt;probability&gt;&lt;simulation&gt;" AnswerCount="3" CommentCount="10" />
  <row Id="25296" PostTypeId="2" ParentId="25291" CreationDate="2011-03-06T15:33:01.543" Score="0" ViewCount="" Body="&lt;p&gt;Note that this corresponds to the partition of $4$ into $3$ nonzero parts (or $2+1+1 = 4$).&lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2011-03-06T15:33:01.543" CommentCount="1" />
  <row Id="25297" PostTypeId="2" ParentId="25270" CreationDate="2011-03-06T15:38:39.987" Score="10" ViewCount="" Body="&lt;p&gt;An obvious upper bound is $N$ queries, since you can test each element individually.  On the other hand, it takes at least $\Omega(N/\log N)$ queries: $N$ bits of information are required to identify the target subset, and each query can yield at most $O(\log N)$ bits of information, since each query has only $O(N)$ possible answers.  To see that the upper bound is not sharp, consider the following strategy for $N=5$, which takes at most $4$ queries:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Guess $\{1,2,3,4,5\}$.  If the result is $0$ or $5$, we have the answer.  If the result is $1$ or $4$, bisection search (for the single member or the single missing element) gives the answer in three more queries.  Suppose the result is $2$ (the strategy for $3$ is the same by symmetry).&lt;/li&gt;&#xA;&lt;li&gt;Guess $\{1,2\}$.  If the result is $2$, we have the answer.  If the result is $0$, then bisection search on $\{3,4,5\}$ (for the single missing element) gives the answer in two more queries.  Suppose the result is $1$.  Then we know the answer is $\{a,b\}$ for some $a \in \{1,2\}$ and $b \in \{3,4,5\}$.&lt;/li&gt;&#xA;&lt;li&gt;Guess $\{1,3\}$.  If the result is $2$, we have the answer.  If the result is $0$, then the answer is $\{2,b\}$ for some $b\in\{4,5\}$, and one more query gives the answer.  Suppose the result is $1$.  Then we know the answer is $\{1,4\}$, $\{1,5\}$, or $\{2,3\}$.&lt;/li&gt;&#xA;&lt;li&gt;Guess $\{1,4\}$.  The answer is $\{1,4\}$ if the result is $2$, or $\{1,5\}$ if the result is $1$, or $\{2,3\}$ if the result is $0$.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;This example gives an improved upper bound asymptotic to $4N/5$.  It seems likely that the correct answer is strictly $o(N)$ (i.e., eventually less than $cN$ for any fixed $c$), but whether or not it's $\Theta(N/\log N)$, I can't say.&lt;/p&gt;&#xA;" OwnerUserId="5546" LastActivityDate="2011-03-06T15:38:39.987" CommentCount="2" />
  <row Id="25299" PostTypeId="2" ParentId="25273" CreationDate="2011-03-06T15:57:21.007" Score="4" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\rm\ \ (F^G)&amp;#39;\ =\ (e^{G\:ln\ F})&amp;#39;\: =\ F^G\ (GF&amp;#39;/F + G&amp;#39;\ ln\ F)$&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2011-03-06T16:32:07.293" LastActivityDate="2011-03-06T16:32:07.293" CommentCount="2" />
  <row Id="25300" PostTypeId="2" ParentId="25287" CreationDate="2011-03-06T15:58:05.733" Score="1" ViewCount="" Body="&lt;p&gt;You can look at this as finding the null space of the matrix&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\begin{pmatrix} x+y &amp;amp; z+w \\ z-w &amp;amp; x-y \end{pmatrix}$&#xA;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To have a non-trivial null space, the determinant of this matrix must be zero, which tells you something about x, y, w and z&lt;/p&gt;&#xA;" OwnerUserId="1470" LastActivityDate="2011-03-06T15:58:05.733" />
  <row Id="25301" PostTypeId="1" AcceptedAnswerId="25305" CreationDate="2011-03-06T16:03:33.287" Score="4" ViewCount="125" Body="&lt;p&gt;Hello guys!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a part from a homework. I solved all examples apart from this one. So the task is: We know that $3$ divides $a^2 + b^2$. Prove that $3$ divides $a$ and $3$ divides $b$. I cannot think of anything useful. I know that $a^2 + b^2 = (a + b)^2 - 2ab$, but I don't see how it can help me :(&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Best regards,&#xA;Petar&lt;/p&gt;&#xA;" OwnerUserId="6698" LastEditorUserId="2641" LastEditorDisplayName="" LastEditDate="2011-03-06T16:18:33.413" LastActivityDate="2011-03-06T17:05:37.973" Title="Division by 3 proof" Tags="&lt;homework&gt;&lt;elementary-number-theory&gt;" AnswerCount="4" CommentCount="1" FavoriteCount="1" />
  <row Id="25302" PostTypeId="2" ParentId="25295" CreationDate="2011-03-06T16:03:49.673" Score="5" ViewCount="" Body="&lt;p&gt;Does the following work? (read the comment of jokiri to convince yourself that the algorithm works --- thanks jokiri)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cut a line of length 1 into $n$ parts. Mathematically, throw $n-1$ random numbers between uniform between 0 and 1. Sort them to obtain $\mathbf{s}=(0,s_1, \dots, s_{n-1},1)$ with $s_i \leq s_j$; $\forall i &amp;lt;j$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take the point $\mathbf{x}= \left[\sigma_1(s_1- s_0), \sigma_2(s_2 - s_1), \dots, \sigma_n(s_n - s_{n-1})\right] \in S^1$. Here, $\sigma_i = \pm 1$ are  randomly choosen. It is quite clear that $\mathbf{x}$ is on the unit sphere. The question remains: is it uniform on the sphere?&lt;/p&gt;&#xA;" OwnerUserId="7266" LastEditorUserId="7266" LastEditorDisplayName="" LastEditDate="2011-03-06T17:54:52.417" LastActivityDate="2011-03-06T17:54:52.417" CommentCount="6" />
  <row Id="25303" PostTypeId="1" CreationDate="2011-03-06T16:13:25.597" Score="2" ViewCount="76" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Let f(x) and g(x) be two functions&#xA;defined on some subset of the real&#xA;numbers. There are two definitions&#xA;for $f \in o(g)\mbox{ as&#xD;&#xA;    }x\to\infty\,$, according to&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Big_O_notation#Little-o_notation&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$$\lim_{x \to    \infty}\frac{f(x)}{g(x)}=0.$$&lt;/li&gt;&#xA;&lt;li&gt;for every $M &amp;gt; 0$, there exists a&#xA;constant $x_0$, such that $$|f(x)| \le \; M |g(x)|\mbox{ for all }x&amp;gt;x_0. $$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I was wondering if the two&#xA;definitions are equivalent? Can it&#xA;be possible that the second one is&#xA;more general than the first one in&#xA;that the limit of the ratio may not&#xA;exist?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;Similar questions &lt;a href=&quot;http://en.wikipedia.org/wiki/Big_O_notation#Family_of_Bachmann.E2.80.93Landau_notations&quot; rel=&quot;nofollow&quot;&gt;for $f \in&#xD;&#xA;    \omega(g)$ and for $f \sim g$&lt;/a&gt;?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Thanks and regards!&lt;/p&gt;&#xA;" OwnerUserId="1281" LastEditorUserId="1281" LastEditorDisplayName="" LastEditDate="2011-03-06T17:10:02.997" LastActivityDate="2011-03-06T21:36:28.930" Title="$f \in o(g), f \in \omega(g), f \sim g$ and existence of $\lim_{x \rightarrow \infty} \frac{f(x)}{g(x)}$" Tags="&lt;real-analysis&gt;" AnswerCount="1" CommentCount="10" />
  <row Id="25304" PostTypeId="2" ParentId="25301" CreationDate="2011-03-06T16:14:27.853" Score="3" ViewCount="" Body="&lt;p&gt;We know that $a^2+b^2 = 3q$ for $q \in \mathbb{Z}$. Suppose for contradiction that $3$ does not divide $a$ or $3$ does not divide $b$. Then $a = 3l+1$ or $a=3l+2$ and $b = 3l&amp;#39;+1$ or $b=3l&amp;#39;+2$ for $l, l&amp;#39; \in \mathbb{Z}$.  &lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2011-03-06T16:14:27.853" />
  <row Id="25305" PostTypeId="2" ParentId="25301" CreationDate="2011-03-06T16:15:42.403" Score="7" ViewCount="" Body="&lt;p&gt;Well consider all of the squares modulo 3. $0^2 = 0$, $1^2 = 1$ and $2^2 = 1$. So now take the expression modulo 3, you know that $3 \mid a^2 + b^2$. So $a^2 + b^2 \equiv 0 \pmod 3$, but now if $3$ doesn't divide $a$ or $b$, then $a^2 \equiv 1 \pmod 3$ or $b^2 \equiv 1 \pmod 3$. But that contradicts the assumption that $a^2 + b^2 \equiv 0 \pmod 3$.&lt;/p&gt;&#xA;" OwnerUserId="7838" LastEditorUserId="1269" LastEditorDisplayName="" LastEditDate="2011-03-06T16:18:43.803" LastActivityDate="2011-03-06T16:18:43.803" CommentCount="3" />
  <row Id="25306" PostTypeId="2" ParentId="25301" CreationDate="2011-03-06T16:18:04.793" Score="3" ViewCount="" Body="&lt;p&gt;Any integer can be written as one of three forms $3k$, $3k+1$ or $3k+2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we take an integer of the form $3k+1$, then $(3k+1)^2 \equiv 1 (\mbox{mod }3)$. Similarly, if the integer is of the form $3k+2$, then $(3k+2)^2 \equiv 1 (\mbox{mod }3)$. Using these, you can prove your result.&lt;/p&gt;&#xA;" OwnerUserId="2641" LastActivityDate="2011-03-06T16:18:04.793" />
  <row Id="25308" PostTypeId="2" ParentId="25295" CreationDate="2011-03-06T16:23:45.043" Score="0" ViewCount="" Body="&lt;p&gt;Here's the method I proposed in the comment. Pick a random point $(x_1,\ldots,x_{n-1})$ from the $n-1$-dimensional hypercube. (This amounts to choosing $n-1$ real numbers uniformly in $[0,1]$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if $\sum_{i=1}^{n} x_i \geq 1$ throw the point out and start again, if not keep it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Transform the point using the following affine map:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f:\mathbb{R}^{n-1}\to\mathbb{R}^{n}: (x_1,\ldots,x_{n-1}) \mapsto (x_1,\ldots,x_{n-1},1-\sum_{i=1}^{n} x_i) \; .$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, similarly to what Fabian does, select $n$ signs $\sigma_i=\pm 1$ randomly and multiply each component of the vector you obtained by these signs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As pointed out already by joriki and Gerben, for high dimensions, this method is very wasteful since a fraction $\frac{(n-1)!-1}{(n-1)!}$ points will have to be thrown out.&lt;/p&gt;&#xA;" OwnerUserId="3567" LastEditorUserId="3567" LastEditorDisplayName="" LastEditDate="2011-03-06T16:29:18.083" LastActivityDate="2011-03-06T16:29:18.083" CommentCount="2" />
  <row Id="25309" PostTypeId="1" AcceptedAnswerId="25313" CreationDate="2011-03-06T16:29:33.977" Score="11" ViewCount="120" Body="&lt;p&gt;Can every function $\psi ~\colon \mathbb{F_{p}^n} \to  \mathbb{F_{p}}$ be regarded as a polynomial function for some polynomial in $\mathbb{F_{p}[x_1, \ldots,x_n]}$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe this is true, but am having trouble proving it.&lt;/p&gt;&#xA;" OwnerUserId="4806" LastActivityDate="2011-03-06T16:54:51.917" Title="Functions on vector spaces over finite fields" Tags="&lt;abstract-algebra&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="25310" PostTypeId="2" ParentId="25301" CreationDate="2011-03-06T16:45:22.840" Score="0" ViewCount="" Body="&lt;p&gt;$\rm\: mod\ 3:\ 0^2 \equiv 0,\ 1^2 \equiv 2^2\equiv 1\: $ so $\rm\: a^2 + b^2 \equiv 1\:$ or $2\ $ if $\rm\ a\ or\ b\not\equiv 0\:.\: $&lt;br&gt;&#xA;Therefore we may conclude that $\rm\ a^2 + b^2 \equiv 0\ \ \:\Rightarrow\:\ \ a\ and\ b\equiv 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More generally, if wlog $\rm\ b\not\equiv 0\ $ then $\rm\ a^2\equiv -b^2\ \Rightarrow\ (a/b)^2 \equiv -1\ $ contra $\rm\ x^{2\:}\: \not\equiv -1\ \ (mod\ 3)\:.$&lt;br&gt;&#xA;This proof works in &lt;em&gt;every&lt;/em&gt; domain where $-1$ is not a square, e.g. integers mod $\rm\ p = 4n+3\ $ prime, as per the 1st supplement to &lt;a href=&quot;http://en.wikipedia.org/wiki/Quadratic_reciprocity&quot; rel=&quot;nofollow&quot;&gt;the law of quadratic reciprocity.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2011-03-06T17:05:37.973" LastActivityDate="2011-03-06T17:05:37.973" />
  <row Id="25311" PostTypeId="2" ParentId="24699" CreationDate="2011-03-06T16:51:51.563" Score="0" ViewCount="" Body="&lt;p&gt;I think it may be more like what Steven Stadnicki suggested, except once the cell is cleared, the counter is reset. Check the central pixel in 3x1-&gt;1x3-&gt;3x1-&gt;... patterns.&lt;/p&gt;&#xA;" OwnerUserId="7869" LastActivityDate="2011-03-06T16:51:51.563" CommentCount="1" />
  <row Id="25312" PostTypeId="2" ParentId="25309" CreationDate="2011-03-06T16:54:12.557" Score="3" ViewCount="" Body="&lt;p&gt;The answer is yes. Suppose that the elements of $F_p$ are given by $\{{a_1,\dots,a_p\}}$. First, let us note that the function $f(x) =1$ if $x=a_i$, and $f(x)=0$ otherwise is a polynomial, given by (up to a constant) $f(x) = (x-a_1)\dots \widehat{(x-a_i)} \dots (x-a_p)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let us denote this function by $f_i$. For an $n$-tuple $(a_{i_1},\dots,a_{i_n})$ let $f_{i_1,\dots,i_n}(x_1,\dots,x_n) = f_{i_1}(x_1) \cdot f_{i_2}(x_2) \cdot \dots \cdot f_{i_n}(x_n)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, given a function $f:F_p^n \to F_p$, we may write it as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f(x_1,\dots,x_n) =  \sum_{ (a_{i_1},\dots,a_{i_n}) \in F_p^n } f_{i_1,\dots,i_n}(x) f(a_{i_1},\dots,a_{i_n})$ which is a finite sum of polynomials, and therefore, a polnomial.&lt;/p&gt;&#xA;" OwnerUserId="5491" LastActivityDate="2011-03-06T16:54:12.557" />
  <row Id="25313" PostTypeId="2" ParentId="25309" CreationDate="2011-03-06T16:54:51.917" Score="12" ViewCount="" Body="&lt;p&gt;For each point $y=(y_1,...,y_n)$ in $\mathbb{F}_p^n$, the polynomial&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\prod_{i=1}^n\prod_{z_i=0\atop z_i\neq y_i}^{p-1}(x_i-z_i)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is non-zero at $y$ and zero everywhere else, so these polynomials form a basis of the function space.&lt;/p&gt;&#xA;" OwnerUserId="6622" LastActivityDate="2011-03-06T16:54:51.917" CommentCount="1" />
  <row Id="25314" PostTypeId="1" AcceptedAnswerId="25318" CreationDate="2011-03-06T17:06:12.767" Score="-5" ViewCount="113" Body="&lt;p&gt;So afair general relativity states that in rotating reference frame value of π would grow due to lorentz contraction. Since it would no longer be equal to standard 3.1415, I assume it could be made any number, like 4, for example. Right? If so, what would be the most profound consequence of π rationality to those who lived in such a space?&lt;/p&gt;&#xA;" OwnerUserId="7869" LastActivityDate="2011-03-06T17:32:52.133" ClosedDate="2011-03-06T19:56:20.673" Title="consequences of rational π?" Tags="&lt;pi&gt;" AnswerCount="3" CommentCount="6" />
  <row Id="25315" PostTypeId="1" AcceptedAnswerId="25344" CreationDate="2011-03-06T17:09:08.193" Score="20" ViewCount="270" Body="&lt;p&gt;I've read more than once the analogy between simple groups and prime numbers, stating that any group is built up from simple groups, like any number is built from prime numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've recently started self-studying subgroup series, which is supposed to explain the analogy, but I'm not completely sure I understand how &quot;any group is made of simple groups&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given a group $G$ with composition series $$ \{e\}=G_0 \triangleleft G_1\triangleleft \dots \triangleleft G_{r-1} \triangleleft G_r=G$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;then $G$ has associated the simple factor groups $H_{i+1}=G_{i+1}/G_i$. But how is it &quot;built&quot; from them?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, if we have those simple groups $H_i$ then we can say the subnormal subgroups in the composition series can be recovered by taking certain extensions of $H_i$: $$ 1 \to K_i \to G_i \to H_i \to 1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $H_i = G_i/G_{i-1}$, $K_i\simeq G_{i-1}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then $G$ is built from some uniquely determined (Jordan-Hölder) simple groups $H_i$ by taking extensions of these groups.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this description accurate?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question now is: this description seems overly theoretical to me. I don't know how the extensions of $H_i$ look like, and I don't understand how $G$ puts these groups together. Can we describe more explicitly how a group $G$ is made of simple groups?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: I forgot a (not-so-tiny) detail. The previous explanation works for &lt;em&gt;finite&lt;/em&gt; groups, or more in general for groups with a composition series. But what about groups which don't admit a composition series? Is it correct to say that they are built from simple groups?&lt;/p&gt;&#xA;" OwnerUserId="2614" LastEditorUserId="2614" LastEditorDisplayName="" LastEditDate="2011-03-06T17:35:27.320" LastActivityDate="2011-03-07T14:30:47.207" Title="How is a group made up of simple groups?" Tags="&lt;abstract-algebra&gt;&lt;group-theory&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="2" />
  <row Id="25316" PostTypeId="2" ParentId="20393" CreationDate="2011-03-06T17:10:53.387" Score="1" ViewCount="" Body="&lt;p&gt;One way to define &quot;tail probability&quot; is by using the notion of &lt;em&gt;rarity&lt;/em&gt;. Suppose an outcome has probability $p$ under some distribution $\mathcal{D}$. Its &lt;em&gt;rarity&lt;/em&gt; is&#xA;$$\Pr[\Pr[\mathcal{D}] \leq p].$$&#xA;So the rarity is the probability that the outcome of $\mathcal{D}$ is as rare as your particular outcome. In the case of a $N(0,1)$, for example, the rarity of an outcome $x$ is equal to the total weight of both tails extending from $ \pm |x|$ outwards.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The nice thing is that the rarity of a random outcome of $\mathcal{D}$ (if it is continuous) is uniformly distributed (exercise). So it converts an arbitrary continuous random distribution to a uniform scale.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given a multivariate normal distribution $N(\mu,\Sigma)$, we see that the probability of $x$ depends only on $(x-\mu)&amp;#39;\Sigma^{-1}(x-\mu)$. In the case where $\mu=0$ and $\Sigma=I$, we simply get $x&amp;#39;x$, which is a sum of squares. In the general case, there is an affine transformation which yields a $N(0,I)$ variable. We are thus led to the &lt;a href=&quot;http://en.wikipedia.org/wiki/Chi-square_distribution&quot; rel=&quot;nofollow&quot;&gt;chi-square distribution&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Pearson%27s_chi-square_test&quot; rel=&quot;nofollow&quot;&gt;chi-square goodness-of-fit test&lt;/a&gt;, for example, starts with a multinomial distribution, approximates it using a multivariate distribution, and then outputs the rarity metric, which is chi-square distributed.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2011-03-06T17:10:53.387" />
  <row Id="25317" PostTypeId="2" ParentId="25315" CreationDate="2011-03-06T17:18:53.620" Score="1" ViewCount="" Body="&lt;p&gt;Have you already met the notion of a &lt;a href=&quot;http://en.wikipedia.org/wiki/Semidirect_product&quot; rel=&quot;nofollow&quot;&gt;semidirect product&lt;/a&gt;?&lt;/p&gt;&#xA;" OwnerUserId="6507" LastActivityDate="2011-03-06T17:18:53.620" CommentCount="4" />
  <row Id="25318" PostTypeId="2" ParentId="25314" CreationDate="2011-03-06T17:19:22.980" Score="3" ViewCount="" Body="&lt;p&gt;π is the ratio of the circumference of a circle to its diameter &lt;em&gt;in an ideal flat space&lt;/em&gt;. If a space is not flat, so that the ratio between the circumference and diameter of a circle is different, that doesn't mean that π is different in that space.&lt;/p&gt;&#xA;" OwnerUserId="5656" LastActivityDate="2011-03-06T17:19:22.980" />
  <row Id="25319" PostTypeId="2" ParentId="25314" CreationDate="2011-03-06T17:25:08.990" Score="3" ViewCount="" Body="&lt;p&gt;You seem to have a misunderstanding about the relation between mathematics and physics. $\pi$ is a mathematical constant, not a physical constant. Physical constants like the speed of light or the gravitational constant appear in several physical theories, and in more refined theories they might turn out to be non-constant. Mathematical constants can be defined entirely abstractly without appeal to measurements. Thus they are independent of any physical theories.&lt;/p&gt;&#xA;" OwnerUserId="1609" LastActivityDate="2011-03-06T17:25:08.990" />
  <row Id="25320" PostTypeId="1" AcceptedAnswerId="39443" CreationDate="2011-03-06T17:28:16.363" Score="19" ViewCount="499" Body="&lt;p&gt;I am trying to understand Apery's 1978 proof that $\zeta(3) = \displaystyle \sum_{n=1}^\infty \frac{1}{n^3}$ is irrational. The idea behind the proof is to find an 'accelerated' series for $\zeta(3)$ which converges too fast to $\zeta(3)$, thus proving that $\zeta(3)$ cannot be rational. In particular, a particular quantity is defined:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$e_{n,k} = \displaystyle \sum_{m=1}^k \frac{(-1)^{m-1} (m!)^2 (n-m)!}{2m^3 (n+m)!},\quad k \leq n.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The key is to show that $\displaystyle \lim_{n \rightarrow \infty} e_{n,k} = 0$ uniformly in $k$, and I have no idea why this sum converges to 0. Any ideas?&lt;/p&gt;&#xA;" OwnerUserId="7871" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2011-05-17T08:06:33.113" LastActivityDate="2011-06-02T18:50:31.553" Title="Apery's proof of the irrationality of $\zeta(3)$" Tags="&lt;number-theory&gt;&lt;sequences-and-series&gt;" AnswerCount="3" CommentCount="7" FavoriteCount="2" />
  <row Id="25321" PostTypeId="2" ParentId="25314" CreationDate="2011-03-06T17:32:52.133" Score="0" ViewCount="" Body="&lt;p&gt;When you say Pi will grow, you are not talking about the number Pi, which is defined by math and is contant in physics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lets say there is a functions that arise in physics that give Pi in some rest some restframe, and another number in different cirumstances, this does not mean the number Pi is changing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f(x)=x+Pi$&lt;br&gt;&#xA;for x=0, you get pi&lt;br&gt;&#xA;for x=1 you dont get pi, that doesnt mean pi changed.&lt;/p&gt;&#xA;" OwnerUserId="4929" LastActivityDate="2011-03-06T17:32:52.133" />
  <row Id="25322" PostTypeId="1" AcceptedAnswerId="25327" CreationDate="2011-03-06T17:33:14.443" Score="3" ViewCount="49" Body="&lt;p&gt;Hey there,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given the differential equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\dot{x} = y-x^2, \;\; \dot{y} = -x+y^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have to find the solutions of this differential equation which move/lie on a line. I am not quite sure how to handle this problem, I started by writing $y = mx + q$, so:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\dot{y} = m \dot{x} = m (y-x^2) = -x+y^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solving this equation, I eventually arrived at $y = -x-1$. Now, a friend of mine told me this already is the solution, but I think it is only the line on which the solutions of the differential equation move. If so, how can I proceed in order to determine the solutions?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for any answers in advance.&lt;/p&gt;&#xA;" OwnerUserId="3787" LastEditorUserId="3787" LastEditorDisplayName="" LastEditDate="2011-03-19T20:14:44.093" LastActivityDate="2011-03-19T20:14:44.093" Title="Determining the solutions of a differential equation which lie on a line" Tags="&lt;homework&gt;&lt;differential-equations&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="25323" PostTypeId="2" ParentId="25320" CreationDate="2011-03-06T17:39:24.523" Score="1" ViewCount="" Body="&lt;p&gt;If $k$ is fixed, all the terms are $O(1/n^2)$, and more over this is an alternating sum, so the entire sum is $O(1/n^2)$. For the uniformity I guess a more delicate argument is needed, but probably Stirling's approximation is your friend here.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2011-03-06T17:39:24.523" />
  <row Id="25324" PostTypeId="2" ParentId="23814" CreationDate="2011-03-06T17:53:37.330" Score="1" ViewCount="" Body="&lt;p&gt;If You want to see the connection between the term $\sqrt{\pi}$ and factorial, explained in the geometric terms, You can watch Knuth's lecture entitled &lt;a href=&quot;http://scpd.stanford.edu/knuth/index.jsp&quot; rel=&quot;nofollow&quot;&gt;Why Pi?&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4811" LastActivityDate="2011-03-06T17:53:37.330" />
  <row Id="25325" PostTypeId="2" ParentId="25295" CreationDate="2011-03-06T17:59:49.197" Score="4" ViewCount="" Body="&lt;p&gt;Flip $n$ fair coins to pick an orthant---that is, to pick the signs of the coordinates of the point you are choosing. Now &lt;a href=&quot;http://www.cs.cmu.edu/~nasmith/papers/smith+tromble.tr04.pdf&quot; rel=&quot;nofollow&quot;&gt;pick a point uniformly in a the standard simplex&lt;/a&gt;, and flip the signs of its coordinates according to what they coins told you.&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2011-03-06T17:59:49.197" CommentCount="1" />
  <row Id="25326" PostTypeId="1" AcceptedAnswerId="25328" CreationDate="2011-03-06T18:00:37.370" Score="11" ViewCount="133" Body="&lt;p&gt;Let $X$ be a topological space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Call $x,y\in X$ &lt;em&gt;swappable&lt;/em&gt; if there is a homeomorphism $\phi\colon X\to X$ with $\phi(x)=y$. This defines an equivalence relation on $X$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One might call $X$ &lt;em&gt;homogeneous&lt;/em&gt; if all pairs of points in $X$ are swappable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, for instance, topological groups are homogeneous, as well as discrete spaces. Also any open ball in $\mathbb R^n$ is homogeneous. On the other hand, I think, the closed ball in any dimension is not homogeneous.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I assume that these notions have already been defined elsewhere. Could you please point me to that?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there any interesting properties that follow for $X$ from homogeneity? I think for these spaces the group of homeomorphisms of $X$ will contain a lot of information about $X$. &lt;/p&gt;&#xA;" OwnerUserId="367" LastEditorUserId="367" LastEditorDisplayName="" LastEditDate="2011-05-30T23:57:04.020" LastActivityDate="2011-05-30T23:57:04.020" Title="Homogeneous topological spaces" Tags="&lt;topology&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="25327" PostTypeId="2" ParentId="25322" CreationDate="2011-03-06T18:05:21.460" Score="3" ViewCount="" Body="&lt;p&gt;Plug in $y = -x - 1$ to the second equation and get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \dot{y} = 1 +  y + y^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To solve this for $y(t)$, separate variables and get &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\int \frac{dy}{1 + y  + y^2} = \int dt = t + c$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the integral on the left is $\displaystyle \frac{2 \tan ^{-1}\left(\frac{2 y+1}{\sqrt{3}}\right)}{\sqrt{3}}$.  Now solve for $y$ and use this to get $x$.&lt;/p&gt;&#xA;" OwnerUserId="2515" LastActivityDate="2011-03-06T18:05:21.460" CommentCount="3" />
  <row Id="25328" PostTypeId="2" ParentId="25326" CreationDate="2011-03-06T18:29:30.687" Score="4" ViewCount="" Body="&lt;p&gt;Googling&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;topological space is homogeneous&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;brings up several articles that use the same terminology, for example &lt;a href=&quot;http://www.jstor.org/pss/2162148&quot; rel=&quot;nofollow&quot;&gt;this one&lt;/a&gt;.  It is also the terminology used in the question &lt;a href=&quot;http://math.stackexchange.com/questions/2194/why-is-the-hilbert-cube-homogeneous&quot;&gt;Why is the Hilbert cube homogeneous?&lt;/a&gt;.  The &lt;a href=&quot;http://en.wikipedia.org/wiki/Perfect_space#Connection_with_other_topological_properties&quot; rel=&quot;nofollow&quot;&gt;Wikipedia article on Perfect space mentions&lt;/a&gt; that a homogeneous space is either perfect or discrete.  The &lt;a href=&quot;http://en.wikipedia.org/wiki/Homogeneous_space&quot; rel=&quot;nofollow&quot;&gt;Wikipedia article on Homogeneous space&lt;/a&gt;, which uses a more general definition, may also help.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2011-03-06T18:29:30.687" CommunityOwnedDate="2011-03-06T18:29:30.687" CommentCount="2" />
  <row Id="25329" PostTypeId="1" AcceptedAnswerId="25334" CreationDate="2011-03-06T18:35:57.527" Score="10" ViewCount="458" Body="&lt;p&gt;This question is motivated by a step in the proof given &lt;a href=&quot;http://math.stackexchange.com/questions/25206/n8-lt-8n-1/25212#25212&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$\begin{align*}&#xD;&#xA;8^{n+1}-1&amp;amp;\gt 8(8^n-1)\gt 8n^8\\&#xD;&#xA; &amp;amp;=(n+1)^8\left(8\left(\frac{n}{n+1}\right)^8\right)\\&#xD;&#xA; &amp;amp;\geq (n+1)^8\left(8\left(\frac{9}{10}\right)^8\right)\\&#xD;&#xA;&amp;amp;\gt (n+1)^8 .&#xD;&#xA;\end{align*}$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I had no trouble following along with the proof until I hit the step that relied on&#xA;$$8\left(\frac{9}{10}\right)^8 &amp;gt; 1$$. So I whipped out a calculator and confirmed that this is indeed correct. And I could see, after some fooling around with pen and paper that any function in the form&#xA;\begin{align}&#xA;k \; \left(\frac{n}{n+1}\right)^k&#xA;\end{align}&#xA;where $n \in \mathbb{Z}$ and $k \rightarrow \infty$ is bound to fall below one and stay there. So it's not a given that any function in the above form will be greater than one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I'm actually curious about is &lt;strong&gt;whether there are nifty or simple little tricks or calculations you can do in your head or any handwavy type arguments that you can make&lt;/strong&gt; to confirm that $$8\left(\frac{9}{10}\right)^8 &amp;gt; 1$$ and even more generally, to confirm for certain natural numbers $k,n$ whether&#xA;\begin{align}&#xA;k \; \left(\frac{n}{n+1}\right)^k &gt; 1&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So are there? And if there are, what are they? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It can be geometrical. It can use arguments based on loose bounds of certain log values. It doesn't even have to be particularly simple as long as it is something you can do in your head and it is something you can explain reasonably clearly so that others can do also it (so if you're able to mentally calculate numbers like Euler, it's not useful for me).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can safely assume that I have difficulties multiplying anything greater two single digit integers in my head. But I do also know that $$\limsup_{k\rightarrow \infty} \log(k) - a\cdot k &amp;lt; 0$$ for any $a&amp;gt;0$ without having to go back to a textbook.&lt;/p&gt;&#xA;" OwnerUserId="6690" LastActivityDate="2011-03-07T06:52:17.510" Title="Simple proof that $8\left(\frac{9}{10}\right)^8 &gt; 1$" Tags="&lt;algebra&gt;" AnswerCount="5" CommentCount="2" FavoriteCount="1" />
  <row Id="25330" PostTypeId="2" ParentId="25329" CreationDate="2011-03-06T18:50:58.693" Score="5" ViewCount="" Body="&lt;p&gt;You can use the fact that $(1 + x/n)^n$ approaches $e^x$ for large $n$.  Then&#xA;$$&#xD;&#xA;\left(\frac{n-1}{n}\right)^{k} = \left(1 - \frac{1}{n}\right)^{k} \thicksim e^{-k/n}&#xD;&#xA;$$&#xA;as $n\rightarrow\infty$.  In your case, this would give $(9/10)^8 \approx e^{-4/5} &amp;gt; 1/e$, which is clearly greater than $1/8$.  (For $n=10$ and $x=-1$, the error in the exponential approximation is about 5%.)&lt;/p&gt;&#xA;" OwnerUserId="5546" LastActivityDate="2011-03-06T18:50:58.693" CommentCount="2" />
  <row Id="25331" PostTypeId="1" AcceptedAnswerId="25342" CreationDate="2011-03-06T18:51:56.943" Score="4" ViewCount="109" Body="&lt;ol&gt;&#xA;&lt;li&gt;Given the set of functions defined&#xA;on a subset in $\mathbb{R}$ and&#xA;taking values in $\mathbb{R}$, I was&#xA;wondering if $O$ as in $f \in O(g)$&#xA;is a total partial order? By total, I would like to know if every two functions can be compared by $O$? Is it&#xA;acting like $\geq$ on $\mathbb{R}$?&lt;/li&gt;&#xA;&lt;li&gt;How about $\Omega$, $\Theta$, $o$, $\omega$ and $\sim$? Are they acting &#xA;like $\leq$, $==$, $&amp;gt;$, $&amp;lt;$ and $==$&#xA;respectively?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;For definitions, see &lt;a href=&quot;http://en.wikipedia.org/wiki/Big_O_notation#Family_of_Bachmann.E2.80.93Landau_notations&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;.&#xA;Thanks and regards!&lt;/p&gt;&#xA;" OwnerUserId="1281" LastActivityDate="2011-04-26T21:10:54.650" Title="Is $O$ in $f \in O(g)$ a total partial order on the set of functions?" Tags="&lt;real-analysis&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="25332" PostTypeId="2" ParentId="25329" CreationDate="2011-03-06T18:53:56.977" Score="22" ViewCount="" Body="&lt;p&gt;For this particular example, it's pretty easy, since&#xA;$$&#xD;&#xA;(9/10)^8  \geq (8/10)^4 \geq (6/10)^2 \geq (3/10)&#xD;&#xA;$$&#xA;and so you're done. Notice, my bounds are quite crude in the last two cases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Repetitive squaring is the quickest way to get up to a large power. In fact, that's how a typical computer implementation will do it for integer powers.&lt;/p&gt;&#xA;" OwnerUserId="7003" LastActivityDate="2011-03-06T18:53:56.977" CommentCount="3" />
  <row Id="25333" PostTypeId="1" CreationDate="2011-03-06T18:57:59.023" Score="1" ViewCount="144" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Possible Duplicate:&lt;/strong&gt;&lt;br&gt;&#xA;  &lt;a href=&quot;http://math.stackexchange.com/questions/20969/prove-0-1-from-first-principles&quot;&gt;Prove $0! = 1$ from first principles&lt;/a&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;p&gt;Why does 0! = 1?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All I know of factorial is that x! is equal to the product of all the numbers that come before it. The product of 0 and anything is 0, and seems like it would be reasonable to assume that 0! = 0. I'm perplexed as to why I have to account for this condition in my factorial function (Trying to learn Haskell). Thanks. &lt;/p&gt;&#xA;" OwnerUserId="7707" LastActivityDate="2011-03-06T19:15:04.387" ClosedDate="2011-03-06T19:41:08.617" Title="Why does 0! = 1 ?" Tags="&lt;factorial&gt;" AnswerCount="5" CommentCount="3" />
  <row Id="25334" PostTypeId="2" ParentId="25329" CreationDate="2011-03-06T18:58:19.777" Score="11" ViewCount="" Body="&lt;p&gt;Note that $(1-.1)^8\geq 1-.8$ by &lt;a href=&quot;http://en.wikipedia.org/wiki/Bernoulli%27s_inequality&quot; rel=&quot;nofollow&quot;&gt;Bernoulli's inequality&lt;/a&gt;, as mentioned in a comment by lvb.  For other cases you can use $\left(1-\frac{1}{n+1}\right)^k\geq 1-\frac{k}{n+1}$, which makes it easier to find a sufficient condition on $n$ for $k\left(\frac{n}{n+1}\right)^k$ to be larger than $1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the motivating problem, the idea was that $k$ is fixed (at $8$), and since $\frac{n}{n+1}$ goes to $1$ as $n$ goes to infinity, $k\left(\frac{n}{n+1}\right)^k$ approaches $k$, and in particular is eventually bigger than $1$. One then finds out in a particular case what &quot;eventually&quot; means; here $n=9$ suffices.  In general, Bernoulli's inequality leads to the conclusion that $n&amp;gt;k$ suffices.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2011-03-06T19:33:16.880" LastActivityDate="2011-03-06T19:33:16.880" CommunityOwnedDate="2011-03-06T18:58:19.777" CommentCount="2" />
  <row Id="25335" PostTypeId="1" AcceptedAnswerId="25384" CreationDate="2011-03-06T18:59:48.387" Score="5" ViewCount="126" Body="&lt;p&gt;Show that all Eisenstein series $G_k$ can be expressed as polynomials in $G_4$ and $G_6$, e.g. express $G_8$ and $G_{10}$ in this way. &#xA;&#xA;Hint: Setting $a_n := (2n+1)G_{2n+2}$, show that &#xA; $2n(2n-1)a_n = 6(2a_n + \sum\limits_{k=1}^{n-2} a_{k}a_{n-1-k})$ , $n&amp;gt;2$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously, what I have trouble proving, is the Hint. It seems, like I should use the second derivative of Weierstrass function representation, using the Eisenstein series (namely $\wp = \displaystyle\frac{1}{z^2} + \sum\limits_{m=1}^{\infty} (2m+1)G_{2m+2}z^{2m}$), but I'm not sure where the sum on the right comes from.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any help will be appreciated.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Can someone help some more?&lt;/p&gt;&#xA;" OwnerUserId="6707" LastEditorUserId="6707" LastEditorDisplayName="" LastEditDate="2011-03-07T19:00:44.513" LastActivityDate="2011-03-07T19:36:04.280" Title="Eisenstein series" Tags="&lt;analytic-number-theory&gt;" AnswerCount="1" />
  <row Id="25336" PostTypeId="2" ParentId="25333" CreationDate="2011-03-06T19:02:38.633" Score="1" ViewCount="" Body="&lt;p&gt;Mostly it is based on convention, when one wants to define the quantity $\binom{n}{0} = \frac{n!}{n! 0!}$ for example. An intuitive way to look at it is $n!$ counts the number of ways to arrange $n$ distinct objects in a line, and there is only one way to arrange nothing.&lt;/p&gt;&#xA;" OwnerUserId="7871" LastActivityDate="2011-03-06T19:02:38.633" CommentCount="2" />
  <row Id="25337" PostTypeId="2" ParentId="25333" CreationDate="2011-03-06T19:03:02.853" Score="0" ViewCount="" Body="&lt;p&gt;We know that $\binom{n}{n}$ and $\binom{n}{0}= 1$. Thus $0! = 1$.&lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2011-03-06T19:03:02.853" CommentCount="1" />
  <row Id="25339" PostTypeId="2" ParentId="25333" CreationDate="2011-03-06T19:03:39.697" Score="1" ViewCount="" Body="&lt;p&gt;It has many reasons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, we can have a power series: $e^x = \sum_{n} x^n/n!$ and we would like the first term to be $1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, how many permutations are there of $0$ numbers? Well, one.&lt;/p&gt;&#xA;" OwnerUserId="1120" LastActivityDate="2011-03-06T19:03:39.697" />
  <row Id="25340" PostTypeId="2" ParentId="25333" CreationDate="2011-03-06T19:04:34.257" Score="0" ViewCount="" Body="&lt;p&gt;It's because $n! = \prod_{0&amp;lt;k\le n} k.$ ($n!$ is the product of all numbers $1, 2,\dots n$) For $n = 0$ there isn't any number greater then 0 and lesser or equal to $n$, so the product is empty; the empty product is defined  by convention as 1.&lt;/p&gt;&#xA;" OwnerUserId="5282" LastEditorUserId="5282" LastEditorDisplayName="" LastEditDate="2011-03-06T19:15:04.387" LastActivityDate="2011-03-06T19:15:04.387" />
  <row Id="25341" PostTypeId="2" ParentId="25333" CreationDate="2011-03-06T19:04:38.333" Score="1" ViewCount="" Body="&lt;p&gt;In a combinatorial sense, $n!$ refers to the number of ways of permuting $n$ objects.  There is exactly one way to permute 0 objects, that is doing nothing, so $0!=1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are plenty of resources that already answer this question.  Also see:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://mathforum.org/library/drmath/view/57905.html&quot; rel=&quot;nofollow&quot;&gt;http://mathforum.org/library/drmath/view/57905.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://wiki.answers.com/Q/Why_is_zero_factorial_equal_to_one&quot; rel=&quot;nofollow&quot;&gt;http://wiki.answers.com/Q/Why_is_zero_factorial_equal_to_one&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Factorial#Definition&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Factorial#Definition&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="6075" LastActivityDate="2011-03-06T19:04:38.333" CommentCount="2" />
  <row Id="25342" PostTypeId="2" ParentId="25331" CreationDate="2011-03-06T19:12:12.180" Score="13" ViewCount="" Body="&lt;p&gt;It is not a &lt;em&gt;partial&lt;/em&gt; order since you don't have anti-symmetry: $f\in O(g)$ and $g\in O(f)$ does not imply $f=g$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you identify functions which are equivalent in that sense, you end up with a partial order (reflexivity and transitivity are obvious), but this this will not be a &lt;em&gt;total&lt;/em&gt; order: for instance, $\sin(x)$ and $\cos(x)$ are not comparable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Summing up, $O$ only defines a &lt;em&gt;preorder&lt;/em&gt; on the set of functions from $\mathbb R$ to $\mathbb R$.&lt;/p&gt;&#xA;" OwnerUserId="367" LastEditorUserId="367" LastEditorDisplayName="" LastEditDate="2011-04-26T21:10:54.650" LastActivityDate="2011-04-26T21:10:54.650" />
  <row Id="25343" PostTypeId="2" ParentId="25290" CreationDate="2011-03-06T19:22:26.783" Score="1" ViewCount="" Body="&lt;p&gt;The way that you prove that something is true in the standard interpretation is, essentially, by just considering what the sentence means in the standard interpretation. In the standard model, when the Rosser sentence uses the &quot;Pvbl&quot; predicate, this actually refers to the &quot;is provable&quot; predicate of the metatheory.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now you reason in the metatheory as Apostolos did in a comment: if the sentence was provable in $T$, then its negation would also be provable in $T$, which is impossible because $T$ was assumed to be consistent. So the sentence is not provable. The sentence itself is an implication whose hypothesis is &quot;if this sentence is provable&quot;. Since it isn't provable, the hypothesis is false, which makes the entire implication vacuously true.  That's how you prove the Rosser sentence is true in the standard interpretation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One thing that can be confusing about this is that when we prove the sentence is true in the standard interpretation, we don't just work in $T$. We know the sentence is not provable in $T$, so we can't do that anyway: we have to add extra metatheoretic assumptions about a particular model if we want to show the sentence is true in that model.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The way we make those assumptions here is by working with a particular model of $T$ in which the formal provability predicate &quot;Pvbl&quot; corresponds to actual provability. That is the only special property of the standard model that we need in our proof. So the proof that the Rosser sentence is true in the standard model can be viewed equally well as a proof that the Rosser sentence is true in any model whose formal provability predicate agrees with the standard one for standard sentences. &lt;/p&gt;&#xA;" OwnerUserId="630" LastActivityDate="2011-03-06T19:22:26.783" />
  <row Id="25344" PostTypeId="2" ParentId="25315" CreationDate="2011-03-06T19:26:48.307" Score="13" ViewCount="" Body="&lt;p&gt;Everything you say is correct: the sense that a finite group is &quot;built&quot; from its simple Jordan-Hölder factors is by repeated extensions.  But this &quot;building&quot; process is much more complicated for groups than the analogous process of building integers from prime numbers because given a (multi-)set of building blocks -- i.e., a finite list $\mathcal{H} = \{ \{H_1,\ldots,H_n\} \}$ of finite simple groups -- there will be in general several (finitely many, obviously, but perhaps a large number) nonisomorphic groups $G$ with &#xA;composition factors $\mathcal{H}$.  The simplest example of this has already been given by Zhen Lin in a comment: if &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\mathcal{H} = \{ \{ C_2, C_2 \} \}$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;then the two groups with these composition factors are $C_4$ and $C_2 \times C_2$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems to be a working assumption of experts in the field that it is hopeless to expect a nice solution to the extension problem.  For instance, consider the special case $\mathcal{H} = \{ \{ C_p,\ldots,C_p \} \}$, in which every composition factor is cyclic of order $p$ -- i.e. a finite $p$-group.  It is known that the function $f(p,n)$ which counts the number of isomorphism classes of finite groups of order $p^n$ grows very rapidly as a function of $n$ for any fixed $p$.  For instance, &lt;a href=&quot;http://www.math.niu.edu/~rusin/known-math/99/p-groups&quot; rel=&quot;nofollow&quot;&gt;see here&lt;/a&gt; for a reference to the fact that $f(2,9) = 10494213$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nevertheless the group extension problem is an important and interesting one -- it is one of the historical sources for the field of &lt;strong&gt;group cohomology&lt;/strong&gt; and still plays a major role -- and in many special cases one can say something nice.  But the general &quot;program&quot; of classifying all finite groups by (i) classifying all simple groups and (ii) determining all finite groups with a given set $\mathcal{H}$ of composition factors does not seem realistic: step (i) was amazingly hard but in the end doable.  It looks very easy compared to step (ii)!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, you ask about infinite groups.  Here the Jordan-Hölder theory extends precisely to groups $G$ which admit at least one composition series, and a standard (necessary and sufficient) criterion for this is that there are no infinite sequences of subgroups &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$H_1 \subsetneq H_2 \subsetneq \ldots$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;with each $H_i$ normal in $H_{i+1}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$H_1 \supsetneq H_2 \supsetneq \ldots$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with each $H_{i+1}$ normal in $H_i$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So for instance an infinite cyclic group $\mathbb{Z}$ does not satisfy the descending chain condition on subgroups and there is no sense (known to me, at least) in which $\mathbb{Z}$ is built up out of simple groups.   &lt;/p&gt;&#xA;" OwnerUserId="299" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2011-03-07T14:30:47.207" LastActivityDate="2011-03-07T14:30:47.207" CommentCount="2" />
  <row Id="25345" PostTypeId="1" AcceptedAnswerId="25348" CreationDate="2011-03-06T19:31:23.137" Score="4" ViewCount="205" Body="&lt;p&gt;I've looked all over and I can't find a good proof of why the diagonals of a rhombus should intersect at right angles. I can intuitively see its true, just by drawing rhombuses, but I'm trying to prove that the slopes of the diagonals are negative reciprocals and its not working out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm defining my rhombus as follows: $[(0,0), (a, 0), (b, c), (a+b, c)]$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've managed to figure out that $c = \sqrt{a^2-b^2}$ and that the slopes of the diagonals are $\frac{\sqrt{a^2-b^2}}{a+b}$ and $\frac{-\sqrt{a^2-b^2}}{a-b}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I can't figure out is how they can be negative reciprocals of one another. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt; I mean to say that I could not find the algebraic proof. I've seen and understand the geometric proof, but I needed help translating it into coordinate form.&lt;/p&gt;&#xA;" OwnerUserId="7873" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2011-03-06T23:28:39.090" LastActivityDate="2011-03-06T23:28:39.090" Title="Why do rhombus diagonals intersect at right angles?" Tags="&lt;geometry&gt;&lt;algebra&gt;&lt;algebra-precalculus&gt;&lt;analytic-geometry&gt;" AnswerCount="3" CommentCount="5" />
  <row Id="25346" PostTypeId="1" AcceptedAnswerId="25358" CreationDate="2011-03-06T19:35:20.673" Score="1" ViewCount="81" Body="&lt;p&gt;Given the sequence $\{ f(i) \}_{0 \leq i \leq n}$ and the array $\{ g_{n}(i,j) \}_{0 \leq i, j \leq n}$ such that $f(i) = \sum_{j = 0}^{n} h_{j} \, g_{n}(i,j)$ for some integers $h_{0}, \dots, h_{n}$, is there a way to write $h_{j}$ as a sum over $f(i)$ and $g_{n}(i,j)$ using the calculus of finite differences?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Update:&lt;/em&gt; The first post below answers the question of writing $h_{j}$ as a function of $f$ and $g$, but I'd like to avoid matrix inversion if at all possible, hence &lt;em&gt;calculus of finite differences&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2720" LastEditorUserId="2720" LastEditorDisplayName="" LastEditDate="2011-03-06T21:37:04.987" LastActivityDate="2011-03-06T21:37:04.987" Title="Sequences and Finite Differences" Tags="&lt;combinatorics&gt;&lt;finite-differences&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" />
  <row Id="25347" PostTypeId="2" ParentId="25345" CreationDate="2011-03-06T19:37:07.607" Score="2" ViewCount="" Body="&lt;p&gt;Hint: Multiply the slopes together and simplify.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2011-03-06T19:37:07.607" />
  <row Id="25348" PostTypeId="2" ParentId="25345" CreationDate="2011-03-06T19:38:20.310" Score="5" ViewCount="" Body="&lt;p&gt;Another way to say that the slopes are opposite reciprocals is to say that their product is $-1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xD;&#xA;\frac{\sqrt{a^2-b^2}}{a+b}\cdot\frac{-\sqrt{a^2-b^2}}{a-b}&#xD;&#xA;&amp;amp;=\frac{-(\sqrt{a^2-b^2})^2}{(a+b)(a-b)}&#xD;&#xA;\\&#xD;&#xA;&amp;amp;=\frac{-(a^2-b^2)}{a^2-b^2}&#xD;&#xA;\\&#xD;&#xA;&amp;amp;=-1&#xD;&#xA;\end{align}$$&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2011-03-06T19:38:20.310" CommentCount="3" />
  <row Id="25349" PostTypeId="1" AcceptedAnswerId="25350" CreationDate="2011-03-06T19:40:53.570" Score="1" ViewCount="121" Body="&lt;p&gt;Let $k\in \mathbb Z$; is there a name for the function $f(k)$ below?&#xA;$$&#xD;&#xA;f(k) = \text{max}(k, 0)&#xD;&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="3009" LastActivityDate="2011-03-06T21:32:59.097" Title="What is the name of this function?" Tags="&lt;functions&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="25350" PostTypeId="2" ParentId="25349" CreationDate="2011-03-06T19:45:23.353" Score="6" ViewCount="" Body="&lt;p&gt;This could be called the &quot;&lt;a href=&quot;http://en.wikipedia.org/wiki/Positive_and_negative_parts&quot; rel=&quot;nofollow&quot;&gt;positive part&lt;/a&gt;&quot; of $k$ and denoted $f(k)=k^+$.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2011-03-06T19:45:23.353" />
  <row Id="25351" PostTypeId="2" ParentId="25349" CreationDate="2011-03-06T19:58:44.947" Score="0" ViewCount="" Body="&lt;p&gt;Write $|x| = |x|_{+} + |x|_{-}$, where $|x|_{\pm} = \pm x \theta(\pm x)$, where $\theta$ is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Heaviside_step_function&quot; rel=&quot;nofollow&quot;&gt;Heaviside step function&lt;/a&gt;. Then your function is the restriction of the function $|x|_{+}$, the &lt;em&gt;positive part of the absolute value function&lt;/em&gt;, that is, on $\mathbb{Z}$.&lt;/p&gt;&#xA;" OwnerUserId="2720" LastEditorUserId="2720" LastEditorDisplayName="" LastEditDate="2011-03-06T21:32:59.097" LastActivityDate="2011-03-06T21:32:59.097" />
  <row Id="25352" PostTypeId="1" AcceptedAnswerId="25377" CreationDate="2011-03-06T19:59:43.317" Score="8" ViewCount="178" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm working my way though a classical geometry book by Hartshorne right now, but this problem popped up in a section I'm reading. It is Problem 13.10 from Hartshorne's Geometry: Euclid and Beyond if you're curious.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, the problem states:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If $a,b\in\mathbb{Z}$, and if $a+b\sqrt{2}$ has a square root in $\mathbb{Q}(\sqrt{2})$, then the square root is actually in $\mathbb{Z}[\sqrt{2}]$. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm not super familiar with algebra, so I'm having trouble interpreting the question, but I would like to know how to solve it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I looked up $\mathbb{Q}(\sqrt{2})$ on wikipedia, and it seems that it is the set $\{a+b\sqrt{2}\ |\ a,b\in\mathbb{Q}\}$. I couldn't find  $\mathbb{Z}[\sqrt{2}]$, but I assume it is the set $\{a+b\sqrt{2}\ |\ a,b\in\mathbb{Z}\}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if $a+b\sqrt{2}$ has a square root in $\mathbb{Q}(\sqrt{2})$ means there exists some $c+d\sqrt{2}\in\mathbb{Q}(\sqrt{2})$ such that &#xA;$$&#xD;&#xA;(c+d\sqrt{2})^2=c^2+2d^2+2cd\sqrt{2}=a+b\sqrt{2}.&#xD;&#xA;$$&#xA;This implies (I think?) that $c^2+2d^2=a$ and $2cd=b$. If this is the correct path, is there then someway to conclude that $c$ and $d$ are in fact integers? Thanks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the way, is this exercise easily related to some aspect of classical geometry? It seems kind of out of the blue to me.&lt;/p&gt;&#xA;" OwnerUserId="1583" LastActivityDate="2011-03-06T23:19:23.673" Title="If $a,b\in\mathbb{Z}$, and if $a+b\sqrt{2}$ has a root in $\mathbb{Q}(\sqrt{2})$, then the root is actually in $\mathbb{Z}[\sqrt{2}]$." Tags="&lt;abstract-algebra&gt;&lt;field-theory&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="25353" PostTypeId="1" AcceptedAnswerId="25399" CreationDate="2011-03-06T20:01:20.993" Score="1" ViewCount="136" Body="&lt;p&gt;Shannon formally deﬁned the amount of information in a message as a function of the probability of occurrence of each possible message[1]. Given a universe of messages $\mathbf{M} = \{ m_1,m_2,..,m_n \}$ and a probability $\mathbf{p(m_i)}$&#xA; for the occurrence of each message, the information content of a &#xA;message in $\mathbf{M}$ is given by:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum\limits_{i=1}^{n} -p(m_{i})\log_{2}(p(m_{i})) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is this formula? How did Sannon add $log_2$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[1]C.E. Shannon. &lt;strong&gt;A mathematical theory of communication&lt;/strong&gt;. Bell Systems Technical Journal, 27:379–423, 623–656, 1948.&lt;/p&gt;&#xA;" OwnerUserId="7265" LastEditorUserId="3567" LastEditorDisplayName="" LastEditDate="2011-03-06T20:06:45.090" LastActivityDate="2011-03-07T00:13:59.367" Title="Shannon's formula" Tags="&lt;information-theory&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="25354" PostTypeId="1" AcceptedAnswerId="31296" CreationDate="2011-03-06T20:37:20.260" Score="5" ViewCount="156" Body="&lt;p&gt;Here is a rather tough integration. I think it looks like an Elliptic Integral of some sort.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_{1}^{\infty}\frac{1}{\sqrt{3x^{4}+6x^{2}-1}}dx$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since there are no odd terms in the quartic, I thought maybbe completing the square &lt;/p&gt;&#xA;&#xA;&lt;p&gt;would be OK. But I got nowhere. I even factored it into:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$3x^{4}+6x^2-1=((2\sqrt{3}-3)x^{2}+1)((2\sqrt{3}+3)x^{2}-1)$ and got nowhere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the solution will involve the Gamma function in some manner.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anyone have a good starting point for this...like a clever substitution?. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for any input. &lt;/p&gt;&#xA;" OwnerUserId="6358" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-06T21:12:02.997" LastActivityDate="2011-04-06T09:24:40.863" Title="Integrate a quartic in a radical in the denominator" Tags="&lt;integral&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="25355" PostTypeId="1" CreationDate="2011-03-06T20:42:55.247" Score="1" ViewCount="222" Body="&lt;p&gt;I need to integrate this expression:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int^{\pi}_0\frac{\gamma^{10} \theta^2 \sin\theta}{(\gamma^2 \theta^2 + 1)^5}  d\theta.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can use the fact that gamma is very large, which I think means I should rearrange the bottom line to allow an expansion, i.e.&#xA;$$\int^{\pi}_0\frac{\theta^2 \sin\theta}{(\theta^2 + \dfrac{1}{\gamma^2})^5}d\theta  ,$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but I'm still not sure how to continue from here. Any hints would be great, thanks.&lt;/p&gt;&#xA;" OwnerUserId="7877" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2011-03-06T21:08:52.790" LastActivityDate="2011-03-07T08:04:33.740" Title="Integration over solid angle" Tags="&lt;integral&gt;" AnswerCount="1" CommentCount="9" FavoriteCount="1" />
  <row Id="25357" PostTypeId="2" ParentId="25352" CreationDate="2011-03-06T20:48:22.340" Score="5" ViewCount="" Body="&lt;p&gt;First of all $\mathbb{Q}(\sqrt{2})$ is the smallest field containing $\mathbb{Q}$ and $\sqrt{2}$, where $\mathbb{Z}[\sqrt{2}]$ is the smallest ring containing $\mathbb{Z}$ and $\sqrt{2}$. (That is square brackets mean ring, parentheses mean field).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From what you have, I am confident you can conclude that $c$ and $d$ are integers.  Here's what i would do to find it directly:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $c=\frac{p}{q}, d=\frac{r}{s}$ with $gcd(p,q)=1$ and $gcd(r,s)=1$.  Since $cd=\frac{pr}{qs}$ is an integer, this implies that $qs$ divides $pr$ and because of the conditions $gcd(p,q)=1$ and $gcd(r,s)=1$, $qm=r$ and $sn=p$, with $n$ and $m$ integers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And we know $a=\frac{s^2n^2}{q^2}+2\frac{q^2m^2}{s^2}=\frac{s^4n^2+2q^4m^2}{q^2s^2}$ is an integer.  Hence $q^2s^2$ divides $s^4n^2+2q^4m^2$.  Since $q^2$ divides the second term and the whole thing, it must divide the first term, and the same with $s^2$.  But since $q^2$ and $s^2n^2$ are relatively prime, this implies $q^2$ is 1, and the same follows for $s$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm fairly certain there is a cleaner way to do this using the Gauss Lemma, but this is something you can work out directly, which is also nice.&lt;/p&gt;&#xA;" OwnerUserId="7845" LastActivityDate="2011-03-06T20:48:22.340" CommentCount="2" />
  <row Id="25358" PostTypeId="2" ParentId="25346" CreationDate="2011-03-06T21:09:23.477" Score="2" ViewCount="" Body="&lt;p&gt;Using the vector notations $f=(f(i))_i$ and $h=(h_i)_i$ and the matrix notation $g=(g_n(i,j))_{i,j}$, you impose that $f=g\cdot h$ and you ask for an expression of $h$ as a function of $f$ and $g$. Let $D=\det g$. If $g$ is not invertible, that is, if $D=0$, either this is impossible or the solution is not unique. If $g$ is invertible, that is, if $D\ne 0$, the unique solution is $h=g^{-1}\cdot f$ and you know that $D\cdot g^{-1}$ is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Invertible_matrix&quot; rel=&quot;nofollow&quot;&gt;comatrix&lt;/a&gt; $c$ of $g$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, in the invertible case $h_i=\displaystyle\frac1D\sum_{j=0}^nc(i,j)f(j)$ for every $0\le i\le n$.&lt;/p&gt;&#xA;" OwnerUserId="6179" LastActivityDate="2011-03-06T21:09:23.477" CommentCount="3" />
  <row Id="25359" PostTypeId="1" AcceptedAnswerId="25361" CreationDate="2011-03-06T21:29:03.303" Score="3" ViewCount="183" Body="&lt;p&gt;I'm trying to prove that&#xA;$$ \frac{\cos(A)}{1-\tan(A)} + \frac{\sin(A)}{1-\cot(A)} = \sin(A) + \cos(A)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can someone help me to get started? I've done other proofs but this one has me stumped! Just a start - I don't need the whole proof. Thanks.&lt;/p&gt;&#xA;" OwnerUserId="7879" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2011-03-06T21:33:06.177" LastActivityDate="2011-03-07T16:03:03.113" Title="Trig identity proof help" Tags="&lt;trigonometry&gt;" AnswerCount="3" FavoriteCount="1" />
  <row Id="25360" PostTypeId="2" ParentId="25359" CreationDate="2011-03-06T21:30:43.763" Score="3" ViewCount="" Body="&lt;p&gt;To start, I'd suggest rewriting the left side in terms of sine and cosine (the tangent and cotangent) and simplifying the complex fractions.&lt;/p&gt;&#xA;" OwnerUserId="72" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2011-03-06T21:37:41.773" LastActivityDate="2011-03-06T21:37:41.773" CommentCount="3" />
  <row Id="25361" PostTypeId="2" ParentId="25359" CreationDate="2011-03-06T21:31:41.297" Score="4" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Hint:&lt;/strong&gt; Write $\tan(A)$ as $\frac{\sin(A)}{\cos(A)}$ and $\cot(A)$ as $\frac{\cos(A)}{\sin(A)}$ do some algebra and use the fact that $\frac{x^2-y^2}{x-y} = x + y$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the equality makes sense only when $A \neq n \pi, n \pi + \frac{\pi}{4}, n \pi + \frac{\pi}{2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{\cos(A)}{1-\tan(A)} + \frac{\sin(A)}{1-\cot(A)} = \frac{\cos(A)}{1-\frac{\sin(A)}{\cos(A)}} + \frac{\sin(A)}{1-\frac{\cos(A)}{\sin(A)}} = \frac{\cos^2(A)}{\cos(A) - \sin(A)} - \frac{\sin^2(A)}{\cos(A) - \sin(A)} = \cos(A) + \sin(A)$&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2011-03-07T16:03:03.113" LastActivityDate="2011-03-07T16:03:03.113" CommentCount="2" />
  <row Id="25362" PostTypeId="1" CreationDate="2011-03-06T21:31:59.400" Score="0" ViewCount="158" Body="&lt;p&gt;In my Trig class we have begun working on graphing the trig functions and working with radians and I'm trying to wrap my head around them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the moment I'm having trouble understanding radian measures and how to find where certain points lie on the unit circle and how to know what quadrant they are in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, we are to find the reference angle of 5pi/6. My book says it terminates in QII.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This may be a dumb question, but how does one figure this out? What am I missing? How do you know that pi/2 &amp;lt; 5pi/6 &amp;lt; pi?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance...&lt;/p&gt;&#xA;" OwnerUserId="7880" LastActivityDate="2011-03-06T22:12:13.787" Title="Finding a point on the unit circle; more specifically, what quadrant it is in" Tags="&lt;trigonometry&gt;" AnswerCount="3" CommentCount="4" />
  <row Id="25363" PostTypeId="2" ParentId="25362" CreationDate="2011-03-06T21:35:36.893" Score="3" ViewCount="" Body="&lt;p&gt;Think of the fractions without π: $\frac{1}{2}&amp;lt;\frac{5}{6}&amp;lt;1$ (or, alternately, since we're dealing with sixths, 5 sixths is between 3 sixths ($\frac{1}{2}$) and 6 sixths ($1$).&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2011-03-06T21:35:36.893" />
  <row Id="25364" PostTypeId="2" ParentId="25303" CreationDate="2011-03-06T21:36:28.930" Score="2" ViewCount="" Body="&lt;p&gt;The two definitions of “&lt;i&gt;f&lt;/i&gt;(&lt;i&gt;x&lt;/i&gt;)∈o(&lt;i&gt;g&lt;/i&gt;(&lt;i&gt;x&lt;/i&gt;)) as &lt;i&gt;x&lt;/i&gt;→∞” you stated are equivalent as long as &lt;i&gt;g&lt;/i&gt;(&lt;i&gt;x&lt;/i&gt;)≠0 for every sufficiently large &lt;i&gt;x&lt;/i&gt;.  Usually we use the O-notation and its relatives only when this condition holds.  Under this condition, the second definition is just what we mean by the first definition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If this condition is not satisfied, then there are arbitrarily large points &lt;i&gt;x&lt;/i&gt; at which the function &lt;i&gt;f&lt;/i&gt;(&lt;i&gt;x&lt;/i&gt;)/&lt;i&gt;g&lt;/i&gt;(&lt;i&gt;x&lt;/i&gt;) is not defined.  In this case, it is not clear what we mean by lim&lt;sub&gt;&lt;i&gt;x&lt;/i&gt;→∞&lt;/sub&gt; &lt;i&gt;f&lt;/i&gt;(&lt;i&gt;x&lt;/i&gt;)/&lt;i&gt;g&lt;/i&gt;(&lt;i&gt;x&lt;/i&gt;) = 0.&lt;/p&gt;&#xA;" OwnerUserId="926" LastActivityDate="2011-03-06T21:36:28.930" />
  <row Id="25365" PostTypeId="2" ParentId="25345" CreationDate="2011-03-06T21:36:50.697" Score="3" ViewCount="" Body="&lt;p&gt;You don't have to work through square roots if you use the properties of the vector dot product and the parallelogram law to construct the rhombus.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I.e one of the rhombus's diagonals can be identified with $\mathbf{a + b}$ where $\mathbf{b}$ is a vector added head-to-tail to vector $\mathbf{a}$, according to the parallelogram law.  Similarly the other diagonal is given by $\mathbf{b - a}$.  The constraint for a rhombus is $\lVert \mathbf{a} \rVert^2 = \lVert \mathbf{b} \rVert^2$. Two vectors $\mathbf{u, v}$ are perpendicular iff $\mathbf{u} \cdot \mathbf{v} = 0$, and as $\mathbf{(a + b)} \, \mathbf{\cdot} \, \mathbf{(b - a)}$ = $ \lVert \mathbf{b} \rVert^2 - \lVert \mathbf{a} \rVert^2 = 0$, the two diagonals are perpendicular.&lt;/p&gt;&#xA;" OwnerUserId="6296" LastEditorUserId="6296" LastEditorDisplayName="" LastEditDate="2011-03-06T22:19:03.017" LastActivityDate="2011-03-06T22:19:03.017" />
  <row Id="25366" PostTypeId="2" ParentId="25362" CreationDate="2011-03-06T21:36:59.887" Score="0" ViewCount="" Body="&lt;p&gt;We know that $\pi = 6 \pi/6$ and $\pi/2  = 3 \pi/6$. &lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2011-03-06T21:36:59.887" />
  <row Id="25368" PostTypeId="1" AcceptedAnswerId="25388" CreationDate="2011-03-06T12:11:30.530" Score="0" ViewCount="129" Body="&lt;p&gt;Given two points $A =(x_1,y_1)$, $B = (x_2,y_2)$ and length $L$, how do I plot a parabolic segment of length $L$ that connects A and B? The vertex of the parabola $(p, \, q)$ should be such that $x_1 \leq p \leq x_2$ and $q \leq y_1, y_2$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, I need to draw a parabolic segment connecting $A$ and $B$ with length $L$ that looks like a 'U' shape, with the vertex being below $A$ and $B$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="7897" LastEditorUserId="6296" LastEditorDisplayName="" LastEditDate="2011-03-06T23:30:29.933" LastActivityDate="2011-03-07T13:29:45.287" Title="Plotting parabolic segment of fixed length between points A &amp; B" Tags="&lt;conic-sections&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" />
  <row Id="25369" PostTypeId="2" ParentId="25368" CreationDate="2011-03-06T19:22:32.510" Score="1" ViewCount="" Body="&lt;p&gt;Suppose we have a parabola defined by $y = a x^2$, with $a \gt 0$. Now we want to know its length between $x_1$ and $x_2$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\begin{align} L &amp;amp;= \int dl \\  &amp;amp;= \int \sqrt{dx^2 + dy^2} \end{align}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$y = a x^2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$dy = 2 a x dx$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\begin{align} L &amp;amp;= \int \sqrt{dx^2+ (2 a x dx)^2} \\&#xD;&#xA;   &amp;amp;= \int dx \sqrt{1+ (2 a x)^2} \end{align}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we look in a table of integral rules:&lt;br&gt;&#xA;$\int \sqrt{1 + (kx)^2} = \frac{1}{k} \sqrt{1 + k^2x^2} + \frac{1}{2} \ln\left(kx + \sqrt{1 + k^2x^2}\right) + C$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now comes the hard part: given x&lt;sub&gt;1&lt;/sub&gt;, x&lt;sub&gt;2&lt;/sub&gt; and L, solve for a. This looks like a trancendental equation and offhand I don't see any clever way to solve it analytically. So you'll have to solve it numerically. A simple loop of code should do it; L increases strictly with a.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Shifting to the desired x&lt;sub&gt;1&lt;/sub&gt;, x&lt;sub&gt;2&lt;/sub&gt;, y&lt;sub&gt;1&lt;/sub&gt; and y&lt;sub&gt;2&lt;/sub&gt; is trivial (but notice that the problem appears to be underconstrained).&lt;/p&gt;&#xA;" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2011-03-07T05:16:16.770" LastActivityDate="2011-03-07T05:16:16.770" CommentCount="3" />
  <row Id="25370" PostTypeId="2" ParentId="25359" CreationDate="2011-03-06T21:50:32.270" Score="1" ViewCount="" Body="&lt;p&gt;I would try multiplying the numerator and denominator both by (for the first term) $1+\tan{(A)}$ and for the second $1+\cot{(A)}$.  From there it should just be a little bit of playing with Pythagorean identities ($\sin^2{(A)}+\cos^2{(A)}=1$, $\tan^2{(A)}+1=\sec^2{(A)}$, and $1+\cot^2{(A)}=\csc^2{(A)}$) and writing $\tan{(A)}$ and $\cot{(A)}$ in terms of $\sin$ and $\cos$.&lt;/p&gt;&#xA;" OwnerUserId="7845" LastActivityDate="2011-03-06T21:50:32.270" />
  <row Id="25371" PostTypeId="1" AcceptedAnswerId="25373" CreationDate="2011-03-06T21:52:27.933" Score="2" ViewCount="243" Body="&lt;p&gt;What is the general way of finding the basis for intersection of two vector spaces?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose I'm given the bases of two vector spaces U and W:&#xA;$$ \mathrm{Base}(U)= \left\{ \left(1,1,0,-1\right), \left(0,1,3,1\right) \right\} $$&#xA;$$ \mathrm{Base}(W) =\left\{ \left(0,-1,-2,1\right), \left(1,2,2,-2\right) \right\} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I already calculated U+W, and the dimension is 3 meaning the dimension of $ U \cap W $ is 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer is supposedly obvious, one vector is the basis of $ U \bigcap W $ but how do I calculate it?&lt;/p&gt;&#xA;" OwnerUserId="7295" LastEditorUserId="7243" LastEditorDisplayName="" LastEditDate="2011-03-06T23:37:28.053" LastActivityDate="2011-04-18T22:18:15.730" Title="How to find basis for intersection of two vector spaces" Tags="&lt;linear-algebra&gt;&lt;vector-space&gt;&lt;intersection&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="25373" PostTypeId="2" ParentId="25371" CreationDate="2011-03-06T22:03:54.080" Score="5" ViewCount="" Body="&lt;p&gt;Assume $\textbf{v} \in U \cap W$. Then $\textbf{v} = a(1,1,0,-1)+b(0,1,3,1)$ and $\textbf{v} = x(0,-1,-2,1)+y(1,2,2,-2)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $\textbf{v}-\textbf{v}=0$, then $a(1,1,0,-1)+b(0,1,3,1)-x(0,-1,-2,1)-y(1,2,2,-2)=0$. If we solve for $a, b, x$ and $y$, we obtain the solution as $x=1$, $y=1$, $a=1$, $b=0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so $\textbf{v}=(1,1,0,-1)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can validate the result by simply adding $(0,-1,-2,1)$ and $(1,2,2,-2)$&lt;/p&gt;&#xA;" OwnerUserId="2548" LastEditorUserId="-1" LastEditorDisplayName="" LastEditDate="2011-04-18T22:18:15.730" LastActivityDate="2011-04-18T22:18:15.730" />
  <row Id="25374" PostTypeId="2" ParentId="25371" CreationDate="2011-03-06T22:06:35.060" Score="3" ViewCount="" Body="&lt;p&gt;Parameterize both vector spaces (using different variables!) and set them equal to each other.  Then you will get a system of 4 equations and 4 unknowns, which you can solve.  Your solutions will be in both vector spaces.&lt;/p&gt;&#xA;" OwnerUserId="7845" LastActivityDate="2011-03-06T22:06:35.060" />
  <row Id="25375" PostTypeId="1" AcceptedAnswerId="25381" CreationDate="2011-03-06T22:07:35.183" Score="0" ViewCount="38" Body="&lt;p&gt;The original function is $y^2 = kx^3$, and I'm being asked to find the orthogonal trajectory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ 2y\frac{dy}{dx} = 3kx^2 $$&#xA;$$ \frac{dy}{dx} = \frac{3kx^2}{2y}$$&#xA;$$ \text{New }\frac{dy}{dx} = \frac{2y}{3kx^2} $$&#xA;$$ \frac{1}{2y}\,dy = \frac{1}{3kx^2}\,dx$$&#xA;$$ \frac{\ln|y|}{2} = \frac{-1}{3kx} + c$$&#xA;$$ \ln|y| = \frac{-2}{3kx} + c$$&#xA;$$ y = e^{\frac{-2}{3kx} + c}$$&#xA;$$ y = e^ce^\frac{-2}{3kx}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm told that the final solution is $2x^2+3y^2 = d$, where d is a constant. However, I can't find a way to get there from where I currently am. I suppose you'd have to combine c and k in some way, but I don't see it. Any hints?&lt;/p&gt;&#xA;" OwnerUserId="7547" LastActivityDate="2011-03-06T22:39:07.017" Title="Isolating Constants in Orthogonal Trajectories" Tags="&lt;calculus&gt;&lt;homework&gt;" AnswerCount="1" CommentCount="0" />
  <row Id="25376" PostTypeId="2" ParentId="25371" CreationDate="2011-03-06T22:07:47.133" Score="2" ViewCount="" Body="&lt;p&gt;It is a one dimensional vector space, so find any non-zero vector which is in both spaces and it will be a basis.&lt;/p&gt;&#xA;" OwnerUserId="3595" LastEditorUserId="3595" LastEditorDisplayName="" LastEditDate="2011-03-07T20:43:50.157" LastActivityDate="2011-03-07T20:43:50.157" CommentCount="2" />
  <row Id="25377" PostTypeId="2" ParentId="25352" CreationDate="2011-03-06T22:11:36.597" Score="3" ViewCount="" Body="&lt;p&gt;First $\rm\ 2\: c^2\: $ times $\rm\ c^2+2\ d^2 =\: a\ $ yields $\rm\ 2\: c^4 + b^2 =\: 2\: a\  c^2\ $ hence $\rm\ 2\: c\in \mathbb Z\ $ by the &lt;a href=&quot;http://en.wikipedia.org/wiki/Rational_root_theorem&quot; rel=&quot;nofollow&quot;&gt;Rational Root Test.&lt;/a&gt;   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next $\:4\ $ times $\rm\ 2\ d^2 = \:a-c^2\: \ \to\ \ 8\ d^2 = \:4\ a\ - (2\:c)^2 \in \mathbb Z\ \:$ thus $\rm\: 2\: d\in \mathbb Z\ $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally $\rm\:  4\ a - 2\ (2\: d)^2 \:=\: (2\:c)^2\: \Rightarrow\ 2\:|\:(2\ c)^2\Rightarrow 2\: |\: 2\: c\ \Rightarrow\ c\in \mathbb Z\ \Rightarrow\ d\in \mathbb Z\quad\quad$ &lt;strong&gt;QED&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2011-03-06T23:19:23.673" LastActivityDate="2011-03-06T23:19:23.673" CommentCount="3" />
  <row Id="25378" PostTypeId="2" ParentId="25362" CreationDate="2011-03-06T22:12:13.787" Score="1" ViewCount="" Body="&lt;p&gt;Write whatever the angle is in radians as $\frac{p}{q}\pi$.  Since going around the top half of the circle once is equivalent to $pi$ radians, divide the top half up into $q$ pieces, and the bottom half into $q$ pieces (like pie slices).  So you're dividing your whole circle into $2q$ pieces.  Now, starting with the top of the first pie slice above the $x$-axis on the right side being 1, count out $p$ pie slices.  Where you end is $p$ &quot;copies&quot; of $\frac{\pi}{q}$ and is the quadrant you belong in.&lt;/p&gt;&#xA;" OwnerUserId="3595" LastActivityDate="2011-03-06T22:12:13.787" />
  <row Id="25379" PostTypeId="1" AcceptedAnswerId="25380" CreationDate="2011-03-06T22:17:55.637" Score="3" ViewCount="102" Body="&lt;p&gt;I am reading methods of solving recurrence relation on Wikipedia. There is &lt;a href=&quot;http://en.wikipedia.org/wiki/Recurrence_relation#General_linear_homogeneous_recurrence_relations&quot; rel=&quot;nofollow&quot;&gt;one method&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Many linear homogeneous recurrence&#xA;  relations may be solved by means of&#xA;  the generalized hypergeometric series.&#xA;  Special cases of these lead to&#xA;  recurrence relations for the&#xA;  orthogonal polynomials, and many&#xA;  special functions. For example, the&#xA;  solution to $$J_{n+1}=\frac{2n}{z}J_n-J_{n-1}$$&#xA;  is given by $$J_n=J_n(z), \,$$ the Bessel function.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There are no description regarding how to use the method of &quot;generalized hypergeometric series&quot;, nor can I find some on the article for &lt;a href=&quot;http://en.wikipedia.org/wiki/Generalized_hypergeometric_series&quot; rel=&quot;nofollow&quot;&gt;generalized hypergeometric series&lt;/a&gt; or on &lt;a href=&quot;http://en.wikipedia.org/wiki/Bessel_function&quot; rel=&quot;nofollow&quot;&gt;Bessel function&lt;/a&gt;. I was wondering if someone here can explain somehow or gives some references about that? Thanks and regards!&lt;/p&gt;&#xA;" OwnerUserId="1281" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2011-06-03T13:47:29.270" LastActivityDate="2011-06-03T13:47:29.270" Title="How to solve recurrence relations by the generalized hypergeometric series" Tags="&lt;sequences-and-series&gt;&lt;reference-request&gt;&lt;recurrence-relations&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="25380" PostTypeId="2" ParentId="25379" CreationDate="2011-03-06T22:24:43.093" Score="3" ViewCount="" Body="&lt;p&gt;See the (on-line, downloadable) book &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://www.math.upenn.edu/~wilf/AeqB.html&quot; rel=&quot;nofollow&quot;&gt;A = B, by Petkovsek, Wilf, and Zeilberger&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It gives all sorts of links between hypergeometric series and recurrence relations.&lt;/p&gt;&#xA;" OwnerUserId="1919" LastActivityDate="2011-03-06T22:24:43.093" />
  <row Id="25381" PostTypeId="2" ParentId="25375" CreationDate="2011-03-06T22:39:07.017" Score="2" ViewCount="" Body="&lt;p&gt;Note that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\dfrac{dy}{dx} = \dfrac{3kx^2}{2y} = \dfrac{3kx^3}{2xy} = \dfrac{3y^2}{2xy} = \dfrac{3y}{2x}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so the orthogonal curves satisfy&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\dfrac{dy}{dx} = -\dfrac{2x}{3y}.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rewrite this equation as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$2x\ dx + 3y\ dy = 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and integrate both sides.&lt;/p&gt;&#xA;" OwnerUserId="857" LastActivityDate="2011-03-06T22:39:07.017" />
  <row Id="25382" PostTypeId="1" AcceptedAnswerId="25386" CreationDate="2011-03-06T22:53:24.643" Score="3" ViewCount="64" Body="&lt;p&gt;In Linear equations, linear ODEs and linear recurrence relations, when solving homogeneous equations, there are a subspace of solutions, and when solving inhomogeneous equations, a particular inhomogeneous solution plus the subspace of homogeneous solutions will give the set of inhomogeneous solutions. I was wondering what the general common theory for Linear equations, linear ODEs and linear recurrence relations in this aspect? Thanks and regards!&lt;/p&gt;&#xA;" OwnerUserId="1281" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2011-06-03T13:46:45.557" LastActivityDate="2011-06-03T13:46:45.557" Title="Common theory for Linear equations, linear ODEs and linear recurrence relations" Tags="&lt;linear-algebra&gt;&lt;differential-equations&gt;&lt;recurrence-relations&gt;" AnswerCount="3" FavoriteCount="1" />
  <row Id="25384" PostTypeId="2" ParentId="25335" CreationDate="2011-03-06T22:57:47.837" Score="2" ViewCount="" Body="&lt;p&gt;Use the differential equaltion satisfied by $\wp$:&#xA;  $$ \wp&amp;#39;^2 = 4\wp^3 -g_2 \wp -g_3 $$ &#xA;from this you get easily (as $g_2 = 60 G_4$)&#xA;  $$ \wp&amp;#39;&amp;#39; = 6 \wp^2 -10 a_1 $$&#xA;which translates directly into the hint if you substitute $\wp$ with the series you give.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: TO clarify further, squaring &#xA;$$ \wp(z) = \frac{1}{z^2} + \sum_{m&amp;gt;=1} a_m z^{2m} $$ &#xA;gives &#xA;$$\begin{align*} \wp(z)^2 &amp;amp;= \frac{1}{z^4} + \frac{2}{z^2}\sum_{m\ge1} a_m z^{2m} + \left(\sum_{m&amp;gt;=1} a_m z^{2m}\right)^2 =\\&#xD;&#xA;&amp;amp;= \frac{1}{z^4} + 2\sum_{m\ge 1} a_m z^{2m-2} + \sum_{m\ge 1} z^{2m} \sum_{k,l \ge 1, k+l = m} a_ka_l\\&#xD;&#xA;&amp;amp;= \frac{1}{z^4} + 2\sum_{m\ge 1} a_m z^{2m-2} + \sum_{m\ge 2} z^{2m-2} \sum_{k=1}^{m-2} a_ka_{m-1-k}&#xD;&#xA;\end{align*} $$&#xA;I hope this helps.&lt;/p&gt;&#xA;" OwnerUserId="3274" LastEditorUserId="3274" LastEditorDisplayName="" LastEditDate="2011-03-07T19:36:04.280" LastActivityDate="2011-03-07T19:36:04.280" CommentCount="2" />
  <row Id="25385" PostTypeId="2" ParentId="25382" CreationDate="2011-03-06T22:59:30.970" Score="0" ViewCount="" Body="&lt;p&gt;Perhaps you are looking for the notion of an &lt;a href=&quot;http://en.wikipedia.org/wiki/Affine_space&quot; rel=&quot;nofollow&quot;&gt;affine space&lt;/a&gt;, or more generally, a &lt;a href=&quot;http://en.wikipedia.org/wiki/Torsor&quot; rel=&quot;nofollow&quot;&gt;torsor&lt;/a&gt;?&lt;/p&gt;&#xA;" OwnerUserId="884" LastActivityDate="2011-03-06T22:59:30.970" />
  <row Id="25386" PostTypeId="2" ParentId="25382" CreationDate="2011-03-06T23:03:28.767" Score="5" ViewCount="" Body="&lt;p&gt;This is simply a result of the fact that you are dealing with a linear operator, so you could say this is all subsumed in the study of linear transformations, which is one of the central concerns of Linear Algebra.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are dealing with a linear transformation $T$, so that $T(\alpha\mathbf{x}) = \alpha T(\mathbf{x})$ and $T(\mathbf{x}+\mathbf{y}) = T(\mathbf{x})+T(\mathbf{y})$, and you are considering the solution to an equation&#xA;$$T(\mathbf{x}) = \mathbf{a},$$&#xA;then the difference between any two solutions is a solution to $T(\mathbf{x})=\mathbf{0}$. This is simply the fact that if $T(\mathbf{x}_1) = \mathbf{a}= T(\mathbf{x}_2)$, then &#xA;$$T(\mathbf{x}_1-\mathbf{x}_2) = T(\mathbf{x}_1) - T(\mathbf{x}_2) = \mathbf{a}-\mathbf{a} = \mathbf{0}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means that if you can find &lt;em&gt;all&lt;/em&gt; solutions to $T(\mathbf{x}) = \mathbf{0}$, and a single solution to $T(\mathbf{x}) = \mathbf{a}$, then you can generate all solutions to the latter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This comes down to an application of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Isomorphism_theorem#First_isomorphism_theorem&quot; rel=&quot;nofollow&quot;&gt;First Isomorphism Theorem&lt;/a&gt;. The solutions to $T(\mathbf{x})=\mathbf{0}$ form the kernel or nullspace of the transformation, so the image of $T$ is isomorphic to $\mathbf{V}/\mathrm{ker}(T)$. The &quot;target&quot; $\mathbf{a}$ corresponds to a single element of this quotient, which is of the form $\mathbf{v}+\mathrm{ker}(T)$, where $\mathbf{v}$ is any particular solution.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2011-03-06T23:03:28.767" />
  <row Id="25387" PostTypeId="2" ParentId="25382" CreationDate="2011-03-06T23:04:22.537" Score="0" ViewCount="" Body="&lt;p&gt;You need look up Linear Algebra.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastActivityDate="2011-03-06T23:04:22.537" />
  <row Id="25388" PostTypeId="2" ParentId="25368" CreationDate="2011-03-06T23:06:06.063" Score="0" ViewCount="" Body="&lt;p&gt;To amplify Beta's answer, if you write the equation as $y=a(x-b)^2+c$ the fact that the two points need to be on the parabola gives two equations in $a,b,c$.  Your constraints on the vertex become constraints on $a, b$ and $c$.  The arclength from $x_1$ to $x_2$ is $L=\left.\frac{1}{2a}\sqrt{1+4a^2(x-b)^2}+\frac{1}{2}\ln\left(2a(x-b)+\sqrt{1+4a^2(x-b)^2}\right)\right|_{x_1}^{x_2}$ and this is a third equation in $a$ and $b$.  It looks like you are into a numeric solution, but you do have enough constraints.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2011-03-07T13:29:45.287" LastActivityDate="2011-03-07T13:29:45.287" CommentCount="5" />
  <row Id="25389" PostTypeId="2" ParentId="25331" CreationDate="2011-03-06T23:27:10.780" Score="4" ViewCount="" Body="&lt;p&gt;None of those is a total order.&lt;br&gt;&#xA;Take $f(n)=\begin{cases}&#xD;&#xA;n &amp;amp; \text{n odd}\\&#xD;&#xA;n^2 &amp;amp;\text{n even}\end{cases}$&#xA;&lt;br&gt;&#xA; $g(n)=\begin{cases}&#xD;&#xA;n^2 &amp;amp; \text{n odd}\\&#xD;&#xA;n &amp;amp;\text{n even}\end{cases}$&lt;/p&gt;&#xA;" OwnerUserId="7243" LastActivityDate="2011-03-06T23:27:10.780" />
  <row Id="25390" PostTypeId="1" AcceptedAnswerId="25392" CreationDate="2011-03-06T23:29:55.487" Score="1" ViewCount="210" Body="&lt;p&gt;Hello,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example:&#xA;$$7x \equiv 1 \pmod{31} $$&#xA;In this example, the inverse of $7$ is $9$. How can we find out that $9$? What are the steps that I need to do?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update&lt;/strong&gt;&lt;br&gt;&#xA;If I have a general modulo equation:&lt;br&gt;&#xA;$$5x + 1 \equiv 2 \pmod{6}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the fastest way to solve it? My initial thought was:&#xA;$$5x + 1 \equiv 2 \pmod{6}$$&#xA;$$\Leftrightarrow 5x + 1 - 1\equiv 2 - 1 \pmod{6}$$&#xA;$$\Leftrightarrow 5x \equiv 1 \pmod{6}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then solve for the inverse of $5$ modulo 6. Is it a right approach? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks,  &lt;/p&gt;&#xA;" OwnerUserId="5711" LastEditorUserId="5711" LastEditorDisplayName="" LastEditDate="2011-03-06T23:51:35.057" LastActivityDate="2011-03-07T00:27:16.697" Title="How to find the inverse modulo m?" Tags="&lt;elementary-number-theory&gt;&lt;modular-arithmetic&gt;" AnswerCount="5" />
  <row Id="25391" PostTypeId="2" ParentId="25390" CreationDate="2011-03-06T23:35:42.877" Score="3" ViewCount="" Body="&lt;p&gt;A hint: Try to use Fermats Little theorem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a^{p-1}=1 \mod p$ for $p$ prime, and all $a\in \mathbb{Z}$.  &lt;/p&gt;&#xA;" OwnerUserId="5882" LastActivityDate="2011-03-06T23:35:42.877" CommentCount="1" />
  <row Id="25392" PostTypeId="2" ParentId="25390" CreationDate="2011-03-06T23:36:25.933" Score="3" ViewCount="" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;One method is simply the Euclidean algorithm: &#xA;\begin{align*}&#xA;31 &amp;amp;= 4(7) + 3\\&#xA;7 &amp;amp;= 2(3) + 1.&#xA;\end{align*}&#xA;So $ 1 = 7 - 2(3) = 7 - 2(31 - 4(7)) = 9(7) - 2(31)$. Viewing the equation $1 = 9(7) -2(31)$ modulo $31$ gives $ 1 \equiv 9(7)\pmod{31}$, so the multiplicative inverse of $7$ modulo $31$ is $9$. This works in any situation where you want to find the multiplicative inverse of $a$ modulo $m$, provided of course that such a thing exists (i.e., $\gcd(a,m) = 1$). The Euclidean Algorithm gives you a constructive way of finding $r$ and $s$ such that $ar+ms = \gcd(a,m)$, but if you manage to find $r$ and $s$ some other way, that will do it too. As soon as you have $ar+ms=1$, that means that $r$ is the modular inverse of $a$ modulo $m$, since the equation immediately yields $ar\equiv 1 \pmod{m}$. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Another method is to play with fractions (Gauss's method):&#xA;$$\frac{1}{7} = \frac{1\times 5}{7\times 5} = \frac{5}{35} = \frac{5}{4} = \frac{5\times 8}{4\times 8} = \frac{40}{32} = \frac{9}{1}.$$&#xA;Here, you reduce modulo $31$ where appropriate, and the only thing to be careful of is that you should only multiply and divide by things relatively prime to the modulus. Here, since $31$ is prime, this is easy. At each step, I just multiplied by the smallest number that would yield a reduction; so first I multiplied by $5$ because that's the smallest multiple of $7$ that is larger than $32$, and later I multiplied by $8$ because it was the smallest multiple of $4$ that is larger than $32$.  &lt;em&gt;Added:&lt;/em&gt; As Bill notes, the method may fail for composite moduli. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Both of the above methods work for general modulus, not just for a prime modulus (though Method 2 may fail in that situation); of course, you can only find multiplicative inverses if the number is relatively prime to the modulus.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update.&lt;/strong&gt; Yes, your method for general linear congruences is the standard one.  We have a very straightforward method for solving congruences of the form $$ax \equiv b\pmod{m},$$&#xA;namely, it has solutions if and only if $\gcd(a,m)|b$, in which case it has exactly $\gcd(a,m)$ solutions modulo $m$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To solve such equations, you first consider the case with $\gcd(a,m)=1$, in which case $ax\equiv b\pmod{m}$ is solved either by finding the multiplicative inverse of $a$ modulo $m$, or as I did in method $2$ above looking at $\frac{b}{a}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you know how to solve them in the case where $\gcd(a,m)=1$, you can take the general case of $\gcd(a,m) = d$, and from&#xA;$$ax\equiv b\pmod{m}$$&#xA;go to&#xA;$$\frac{a}{d}x \equiv \frac{b}{d}\pmod{\frac{m}{d}},$$&#xA;to get the unique solution $\mathbf{x}_0$. Once you have that unique solution, you get all solutions to the original congruence by considering&#xA;$$\mathbf{x}_0,\quad \mathbf{x}_0 + \frac{m}{d},\quad \mathbf{x}_0 + \frac{2m}{d},\quad\ldots, \mathbf{x}_0 + \frac{(d-1)m}{d}.$$&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-06T23:57:11.273" LastActivityDate="2011-03-06T23:57:11.273" CommentCount="11" />
  <row Id="25393" PostTypeId="2" ParentId="25390" CreationDate="2011-03-06T23:37:19.377" Score="5" ViewCount="" Body="&lt;p&gt;We know that $(7,31) = 1$. So you can use the &lt;a href=&quot;http://en.wikipedia.org/wiki/Extended_Euclidean_algorithm&quot; rel=&quot;nofollow&quot;&gt;extended Euclidean  algorithm&lt;/a&gt;. In particular, you can find two integers $s,t$ such that $7s+31t = 1$. So $7s \equiv 1 \pmod {31}$.  &lt;/p&gt;&#xA;" OwnerUserId="2548" LastEditorUserId="2548" LastEditorDisplayName="" LastEditDate="2011-03-06T23:42:31.460" LastActivityDate="2011-03-06T23:42:31.460" CommentCount="1" />
  <row Id="25394" PostTypeId="1" AcceptedAnswerId="25397" CreationDate="2011-03-06T23:42:34.350" Score="4" ViewCount="73" Body="&lt;p&gt;I thought I had this question down, but while looking over my solution, I think I'm missing a step.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to show for $f_n$ the nth fibonacci number, that $f_n$ is divisible by $4$ if and only if $n$ is divisible by $6$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I approach this with strong induction. I was able to prove the backward implication by assuming the result holds for all $f_k$ up to $n$, and rewriting $f_{n+1}=8f_{n-4}+5f_{n-5}$. Also, I was able to prove that $f_n$ is divisible by $2$, if and only if $n$ is divisible by $3$ by rewriting $f_{n+1}=2f_{n-1}+f_{n-2}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I guess it remains to show that $f_{n+1}$ divisible by $4$ implies that $n$ is even, and thus $n$ will be divisible by $6$, but I just can't figure out how to show this last part. I've messed with proofs by contradiction, the contrapositive, but have come up with nothing. How would one do that?&lt;/p&gt;&#xA;" OwnerUserId="6742" LastActivityDate="2011-03-07T00:05:20.873" Title="$f_n$ is divisible by $4$ if and only if $n$ is divisible by $6$." Tags="&lt;elementary-number-theory&gt;&lt;induction&gt;" AnswerCount="2" FavoriteCount="0" />
  <row Id="25395" PostTypeId="2" ParentId="25390" CreationDate="2011-03-06T23:52:12.447" Score="1" ViewCount="" Body="&lt;p&gt;There are many methods, e.g. the &lt;a href=&quot;http://math.stackexchange.com/questions/20717/how-to-find-solutions-of-linear-diophantine-ax-by-c/20738#20738&quot;&gt;extended Euclidean algorithm&lt;/a&gt;, or &lt;a href=&quot;http://math.stackexchange.com/questions/14579/troubles-finding-inverse-modulus/14608#14608&quot;&gt;Gauss's special case for computing inverses modulo a prime&lt;/a&gt;, or, by Euler-Fermat $\rm\ (a,m)=1 \Rightarrow\ a^{-1} \equiv a^{\phi(m)-1}\ \ (mod\ m)\:.$&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2011-03-06T23:52:12.447" CommentCount="1" />
  <row Id="25396" PostTypeId="1" AcceptedAnswerId="25403" CreationDate="2011-03-06T23:58:40.007" Score="0" ViewCount="63" Body="&lt;p&gt;A signal which is a function of a set of periodic signals with periods $p_1,\dotsc, p_n$ will have a period $P$ which is the least common multiple of all $p_i$. (If $p_i$ are relatively prime, which is the only relevant case here, then $P = \prod_i p_i$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume that each $p_i$ must be an integer and greater than some constant $k$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is a straightforward algorithm which will list, for each possible $P$ in increasing order, the values of $n$ and $p_1,\dotsc, p_n$ which minimize $\sum_i (12 + 3\lceil (p_i - 1)/4 \rceil)$ (the cost of the signal generators)?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;For those curious about the application, or wondering if this is homework, I am constructing a virtual logic circuit in the game &lt;a href=&quot;http://www.minecraft.net/&quot; rel=&quot;nofollow&quot;&gt;Minecraft&lt;/a&gt;; I wish to &lt;a href=&quot;http://gaming.stackexchange.com/questions/14171/minecraft-inactivity-timeout-device&quot;&gt;produce a long (relative to the logic elements' propagation time) delay using a minimum of resources&lt;/a&gt;. I am using a group of clocks with differing periods and ORing the outputs, so that the output has a period which is the product of the individual periods as described. I have found a satisfactory result by trial and error, but I'd like to also find and document a systematic solution.&lt;/p&gt;&#xA;" OwnerUserId="1208" LastEditorUserId="1208" LastEditorDisplayName="" LastEditDate="2011-03-07T00:10:28.683" LastActivityDate="2011-03-07T00:35:26.917" Title="Minimize $\sum f(p_i)$ for given $\prod p_i$" Tags="&lt;algorithms&gt;&lt;elementary-number-theory&gt;&lt;optimization&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" />
  <row Id="25397" PostTypeId="2" ParentId="25394" CreationDate="2011-03-07T00:04:43.947" Score="4" ViewCount="" Body="&lt;p&gt;One way to look at it is to consider the sequence in $\mathbb{Z}/4\mathbb{Z}$.  Same definition: $f_1=f_2=1$ and $f_{n+1}=f_{n}+f_{n-1}$.  The sequence hence starts as follows: 1, 1, 2, 3, 1, 0, 1, 1, ...  Clearly the sequence is periodic of period 6.  You can probably find a way to write this hand-wavy argument more rigorously.&lt;/p&gt;&#xA;" OwnerUserId="7885" LastActivityDate="2011-03-07T00:04:43.947" CommentCount="1" />
  <row Id="25398" PostTypeId="2" ParentId="25394" CreationDate="2011-03-07T00:05:20.873" Score="1" ViewCount="" Body="&lt;p&gt;Let's compute the Fibonacci sequence modulo $4$:&#xA;$$0,1,1,2,3,1,0,1,\ldots$$&#xA;Here the sequence repeats, since $0,1$ is where we started. We get that&#xA;$$F_n \equiv F_{n \mod{6}} \pmod{4},$$&#xA;since $F_n \mod{4}$ is cyclic, with cycle of size $6$. Examining the first $6$ remainders, we see that only $F_0 \equiv 0 \pmod{4}$, hence your question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you really wanted, you could rephrase this argument as complete induction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's easy to generalize this argument to other moduli - in general the sequence will be eventually periodic, perhaps with an initial prefix which is not repeated.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2011-03-07T00:05:20.873" />
  <row Id="25399" PostTypeId="2" ParentId="25353" CreationDate="2011-03-07T00:13:59.367" Score="3" ViewCount="" Body="&lt;p&gt;Shannon proves in his paper that in order to code $n$ messages you need roughly $nH(M)$ bits. You can take that as the definition of entropy, and then derive Shannon's formula.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are also axiomatic derivations of $H(M)$ - you start with some expected properties of the entropy function, and you end up with exactly that formula. I think that's also in Shannon's paper.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The logarithm comes in basically since to represent $T$ different messages you need at least $\log_2 T$ bits.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2011-03-07T00:13:59.367" CommunityOwnedDate="2011-03-07T00:13:59.367" />
  <row Id="25400" PostTypeId="2" ParentId="25390" CreationDate="2011-03-07T00:27:16.697" Score="1" ViewCount="" Body="&lt;p&gt;We can visualize the solution to $7x \equiv 1 \pmod{31}$ as the intersection of the two &quot;lines&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ \ \ \ \ y \equiv 7x \pmod{31}$ and $y \equiv 1 \pmod{31}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first line is closed under addition and subtraction, since it passes through the origin. If $(x_1, y_1)$ and $(x_2, y_2)$ are points on the line $y \equiv 7x \pmod{31}$, then so are $(x_1+x_2, y_1+y_2)$ and $(x_1-x_2, y_1-y_2)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To find a point where the lines intersect, we start with the two points $(1, 7)$ and $(0,31)$ on the first line, and repeatedly subtract one point from another until the $y$-coordinate is 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ \ \ \ \ (0,31) - (1,7) = (-1, 24)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ \ \ \ \ (-1, 24) - (1,7) = (-2, 17)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ \ \ \ \ (-2, 17) - (1, 7) = (-3, 10)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ \ \ \ \ (-3, 10) - (1, 7) = (-4, 3)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ \ \ \ \ (1,7) - (-4, 3) = (5, 4)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ \ \ \ \ (5,4) - (-4, 3) = (9,1)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, $x \equiv 9 \pmod{31}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can speed up this procedure by subtracting a multiple of one point from another point instead. This leads to the Euclidean algorithm.&lt;/p&gt;&#xA;" OwnerUserId="857" LastActivityDate="2011-03-07T00:27:16.697" />
  <row Id="25401" PostTypeId="1" CreationDate="2011-03-07T00:34:23.233" Score="0" ViewCount="146" Body="&lt;p&gt;Does any one know of a particular textbook or reference  that proves existence and uniquence of the ODE $\displaystyle\frac{\mathrm{d}y}{\mathrm{d}x}=f(x,y)$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt;&#xA;Consider the initial value problem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{dy}{dx}=f(x,y)$, $y(x_0)=y_0$                                                (E)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume $f:D\to\mathbb{R}$ is a continuous where $D=\{(x,y):m\leq x\leq n, p\leq y\leq q\}$. Assume that $\phi(x_0)=y_0$, $y_0\in[p,q]$. Then $y=\phi(x)$ is a solution of (E) if and only if&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\phi(x)=y_0+\int_{x_0}^x f(t,\phi(t))dt$.&lt;/p&gt;&#xA;" OwnerUserId="7340" LastEditorUserId="7340" LastEditorDisplayName="" LastEditDate="2011-03-07T02:16:07.530" LastActivityDate="2011-03-07T05:12:43.903" Title="Existence and uniqueness theorem of ODE $\displaystyle\frac{\mathrm{d}y}{\mathrm{d}x}=f(x,y)$" Tags="&lt;real-analysis&gt;&lt;reference-request&gt;&lt;differential-equations&gt;" AnswerCount="3" CommentCount="6" />
  <row Id="25402" PostTypeId="2" ParentId="25329" CreationDate="2011-03-07T00:35:08.430" Score="3" ViewCount="" Body="&lt;p&gt;I'll start from the &quot;obvious&quot; fact that $(5/4)^3 &amp;lt; 2$. In fact, the cube root of 2 is around 1.26; of course you can explicitly compute $(5/4)^3 = 125/64 &amp;lt; 128/64 = 2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then cubing both sides of that inequality, you get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(5/4)^9 &amp;lt; 8$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But $(10/9)^8 &amp;lt; (10/9)^9 &amp;lt; (5/4)^9$, so $(10/9)^8 &amp;lt; 8$. Taking reciprocals, $(9/10)^8 &amp;gt; 1/8$; multiplying both sides by 8 gives your result.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Actually, my heuristic here is as follows: $(10/9)$ is roughly one whole tone (two semitones); so $(10/9)^8$ is around sixteen semitones, or an octave and a major third, or $2 \times (5/4) = 2.5$. So $(9/10)^8$ must be around $0.4$. See &lt;a href=&quot;http://web.mit.edu/6.055/handouts/singing-logarithms.pdf&quot; rel=&quot;nofollow&quot;&gt;Sanjoy Mahajan's handout on &quot;singing logarithms&quot;, originally due to I. J. Good&lt;/a&gt;. Of course this method is really only useful if you know a little music theory.&lt;/p&gt;&#xA;" OwnerUserId="173" LastActivityDate="2011-03-07T00:35:08.430" />
  <row Id="25403" PostTypeId="2" ParentId="25396" CreationDate="2011-03-07T00:35:26.917" Score="0" ViewCount="" Body="&lt;p&gt;If the $p_i$ are all prime, then $P$ must be squarefree and a given $P$ can only come from one set of $p_i$-its prime factorization.  In that sense there is no minimization to be done, unless you mean a $P$ which has a lower value of $\sum_i (12 + 3\lceil (p_i - 1)/4 \rceil)$ than lower $P$'s.  For a given prime $p$ its contribution to $\log P$ is $\log p$, while its contribution to the sum is as you see.  So the ratio of these gives a figure of merit for primes.  It is above 0.04 from about 5 to 33, so primes in there are what you want, and the $4n+1$ ones are somewhat better due to your ceiling function.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-03-07T00:35:26.917" CommentCount="4" />
  <row Id="25404" PostTypeId="1" AcceptedAnswerId="25408" CreationDate="2011-03-07T00:38:50.737" Score="2" ViewCount="131" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;If $p$ is a prime integer, then $\mathbb{Z}[i]/(p)$ is isomorphic to $\mathbb{Z}_p/(x^2 + 1)$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I suspect that there is a typo, and that the $\mathbb{Z}$ on the r.h.s. should be $\mathbb{Z}[x]$ (polynomials).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Either way, I need help getting started and understanding the notation.&#xA;$\mathbb{Z}[i]$ is the Gaussian integers, so $\mathbb{Z}[i]/p$ means taking modulus $p$... so any elements that differ by a multiple of $p$ are congruent? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What about the expression on the right? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks. &lt;/p&gt;&#xA;" OwnerUserId="7850" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-07T01:40:28.943" LastActivityDate="2011-03-07T01:40:28.943" Title="Isomorphism between two quotient rings" Tags="&lt;abstract-algebra&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="25405" PostTypeId="1" AcceptedAnswerId="25932" CreationDate="2011-03-07T00:42:17.987" Score="2" ViewCount="160" Body="&lt;p&gt;So the question asks, given that we have a undirected graph with unique edge weights, prove that the graph has a unique minimum spanning tree. &#xA;My Proof:&#xA;If the graph has unique edge weights, we can list the edges as &#xA;$$E_1,E_2,\ldots, E_N, \text{ s.t. } E_1 &amp;lt; E_2 &amp;lt; E_3 &amp;lt; \cdots &amp;lt; E_N.$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then applying Kruskal's Algorithm, we can say that the edges that will be in our minimum spanning tree will be all edges in the above list such that as we go through the list, the next element added does not create a cycle. Since the only time we are faced with multiple decision points is when there are two edges of the same weight to choose from, and since the edge weights are distinct, we are never confronted with a decision to choose between two edges, and hence our minimum spanning tree is unique. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I feel as though the part about decision points is flimsy. Any advice?   &lt;/p&gt;&#xA;" OwnerUserId="6758" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2011-03-07T01:31:11.063" LastActivityDate="2011-03-09T07:35:45.027" Title="How do I write this proof more formally?" Tags="&lt;algorithms&gt;&lt;graph-theory&gt;&lt;proof-strategy&gt;" AnswerCount="2" />
  <row Id="25406" PostTypeId="1" AcceptedAnswerId="25412" CreationDate="2011-03-07T00:45:52.070" Score="1" ViewCount="92" Body="&lt;p&gt;Using two forms of provability:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Identity Elimination/Transitivity&lt;/li&gt;&#xA;&lt;li&gt;AnaCon: Analytical Consequence&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Below, &quot;Larger(x,y)&quot; means &quot;x is larger than y&quot;, &quot;Smaller(x,y)&quot; means &quot;x is smaller than y&quot;, and &quot;SameSize(x,y)&quot; means &quot;x is the same size as y&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What would be the best way (makes more logical sense) to prove the following:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Larger(b, c)&lt;/li&gt;&#xA;&lt;li&gt;Smaller(b, d)&lt;/li&gt;&#xA;&lt;li&gt;SameSize(d, e)&lt;br/&gt;&#xA;--&gt; Larger(e, c)     &amp;lt;-- this is what we are trying to prove&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I have the following proofs:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proof # 1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4) Smaller(b, e)  AnaCon: 2, 3&lt;br/&gt;&#xA;5) Larger(e, b)   AnaCon: 4&lt;br/&gt;&#xA;6) Larger(e, c)   AnaCon: 5, 1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proof # 2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4) Smaller(b, e)  AnaCon: 2, 3&lt;br/&gt;&#xA;5) Larger(e, c)  AnaCon: 4, 1&lt;br/&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proof # 3&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4) Smaller(c, d)  AnaCon: 1, 2&lt;br/&gt;&#xA;5) Larger(e, c)   AnaCon: 4, 3&lt;/p&gt;&#xA;" OwnerUserId="7810" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-07T01:24:10.040" LastActivityDate="2011-03-07T01:39:51.680" Title="Which proof makes more logical sense?" Tags="&lt;logic&gt;&lt;proof-strategy&gt;" AnswerCount="1" CommentCount="10" FavoriteCount="1" />
  <row Id="25407" PostTypeId="2" ParentId="25404" CreationDate="2011-03-07T01:01:55.277" Score="1" ViewCount="" Body="&lt;p&gt;The expression on the right should be $\mathbb{Z}_p[x]/(x^2+1)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a number theory person, I prefer to reserve the notation $\mathbb{Z}_p$ for the $p$-adic integers, and stick to $\mathbb{Z}/p\mathbb{Z}$ (or $\mathbb{F}_p$, depending on the setting) for &quot;the integers mod $p$&quot;, but as Arturo points out $\mathbb{Z}_p$ is quite common outside of algebra / number theory circles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What's going on here is that both rings are (isomorphic to) the integer polynomials $\mathbb{Z}[x]$, mod the ideal $(p,x^2+1)$. This problem is asking you to show that it doesn't matter if you first &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;quotient $\mathbb{Z}[x]$ by the ideal $(x^2+1)$ (which produces $\mathbb{Z}[i]$), and then quotient the resulting ring $\mathbb{Z}[i]$ by the ideal generated by $p$, or&lt;/li&gt;&#xA;&lt;li&gt;quotient $\mathbb{Z}[x]$ by the ideal $(p)$ (which produces $\mathbb{Z}_p[x]$), and then quotient the resulting ring $\mathbb{Z}_p[x]$ by the ideal generated by $x^2+1$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Arturo's advice about how to approach proving that these rings are isomorphic is spot on. The above is more of an explanation for why you might expect these two rings to be isomorphic. When we &quot;mod out&quot;, we are making certain identifications of elements in our ring (specifically, when we mod out the ring $R$ by the ideal $I$, we identify $a\in R$ and $b\in R$ when $a-b\in I$), and this question is asking you to show that the order in which we identify elements of a ring doesn't matter.&lt;/p&gt;&#xA;" OwnerUserId="264" LastEditorUserId="264" LastEditorDisplayName="" LastEditDate="2011-03-07T01:28:46.743" LastActivityDate="2011-03-07T01:28:46.743" CommentCount="9" />
  <row Id="25408" PostTypeId="2" ParentId="25404" CreationDate="2011-03-07T01:09:26.377" Score="2" ViewCount="" Body="&lt;p&gt;The notation $\mathbb{Z}[i]/(p)$ means $\mathbb{Z}[i]$ modulo the ideal generated by $p$, i.e., the multiples of $p$ (in the Gaussian integers). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The expression on the right is supposed to be $\mathbb{Z}_p[x]/(x^2+1)$. So first, $\mathbb{Z}_p$ is the integers modulo $p$, $\mathbb{Z}/p\mathbb{Z}$. Then you take $\mathbb{Z}_p[x]$, which are the polynomials with coefficients in $\mathbb{Z}_p$. Finally, you take the quotient modulo $x^2+1$, which means two polynomials with coefficients in $\mathbb{Z}_p$ are equivalent if and only if their difference is a multiple of $x^2+1$ (multiple in $\mathbb{Z}_p[x]$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you know the Isomorphism Theorem for rings, then the simplest way to do this is to construct a ring homomorphism either from $\mathbb{Z}[i]$ to $\mathbb{Z}_p[x]/(x^2+1)$ that is onto, and has kernel equal to the multiples of $p$; or a ring homomorphism from $\mathbb{Z}_p[x]/(x^2+1)$ to $\mathbb{Z}[i]/(p)$ that is onto and whose kernel are exactly the multiples of $x^2+1$. The latter is probably simpler.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you don't know the Isomorphism Theorem for rings, you can try to define a homomorphism directly. You want to decide how to map an equivalence class of the form $a+bi + (p)$ ($a,b\in\mathbb{Z}$) to an equivalence class of the form $\overline{r}+\overline{s}x + (x^2+1)$, where $\overline{r}$ and $\overline{s}$ are integers modulo $p$ (you should first prove that every element in $\mathbb{Z}_p[x]$ is equivalent modulo $x^2+1$ to a polynomial of degree at most $1$; try using the division algorithm). There is an obvious choice: map $a+bi + (p)$ to $\overline{a}+\overline{b}x + (x^2+1)$ (the reason you would want to do this is that $x+(x^2+1)$ satisfies $t^2 + 1 = 0$ in $\mathbb{Z}_p[x]/(x^2+1)$; that is, it is a &quot;square root of $-1$&quot;, just like $i$). You would need to show this is well-defined, one-to-one (on classes), onto, and a ring homomorphism. &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-07T01:21:06.287" LastActivityDate="2011-03-07T01:21:06.287" CommentCount="1" />
  <row Id="25409" PostTypeId="2" ParentId="25401" CreationDate="2011-03-07T01:11:54.317" Score="0" ViewCount="" Body="&lt;p&gt;I presume you mean existence and uniqueness of the &lt;em&gt;solution&lt;/em&gt; of this ODE? It seems you're referring to the Picard–Lindelöf theorem? That's proved in &lt;a href=&quot;http://en.wikipedia.org/wiki/Picard%E2%80%93Lindel%C3%B6f_theorem&quot; rel=&quot;nofollow&quot;&gt;the corresponding Wikipedia article&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="6622" LastActivityDate="2011-03-07T01:11:54.317" CommentCount="1" />
  <row Id="25410" PostTypeId="2" ParentId="25405" CreationDate="2011-03-07T01:30:15.510" Score="0" ViewCount="" Body="&lt;p&gt;In proving the optimality of Kruskal's algorithm, you show that the spanning tree that the algorithm generates is at least as good as any other spanning tree. The same argument shows that the tree is better than any other spanning tree if the weights are unique.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2011-03-07T01:30:15.510" CommentCount="2" />
  <row Id="25411" PostTypeId="2" ParentId="25401" CreationDate="2011-03-07T01:30:21.873" Score="0" ViewCount="" Body="&lt;p&gt;The usually way they do this is via some kind of contract mapping theorem. This theorem is called the fundamental theorem of ODE, it can be extended to $n$ variable situation via similar arguments. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The book I learned this in high school is probably this one:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0387906096&quot; rel=&quot;nofollow&quot;&gt;http://www.amazon.com/Partial-Differential-Equations-Mathematical-Sciences/dp/0387906096/ref=sr_1_1?s=books&amp;amp;ie=UTF8&amp;amp;qid=1299461183&amp;amp;sr=1-1&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think you can find similar ones in any standard ODE books. In particular Arnold's book has an incomplete proof based on some assumption in the very start. &lt;/p&gt;&#xA;" OwnerUserId="7887" LastActivityDate="2011-03-07T01:30:21.873" />
  <row Id="25412" PostTypeId="2" ParentId="25406" CreationDate="2011-03-07T01:39:51.680" Score="1" ViewCount="" Body="&lt;p&gt;As edited they now all look equivalent.  Proof #1 is slightly more explicit when using the fact that &lt;em&gt;Smaller(y,x)&lt;/em&gt; implies &lt;em&gt;Larger(x,y)&lt;/em&gt; , which seems to be implicit in the others. You could also have another 3 step version &lt;/p&gt;&#xA;&#xA;&lt;p&gt;4) Larger(d, b) from 2     - reverse &lt;/p&gt;&#xA;&#xA;&lt;p&gt;5) Larger(d, c) from 4, 1  - transitive &lt;/p&gt;&#xA;&#xA;&lt;p&gt;6) Larger(e, c) from 5, 3  - substitution &lt;/p&gt;&#xA;&#xA;&lt;p&gt;and I suspect there are other variants&lt;/p&gt;&#xA;" OwnerUserId="6460" LastActivityDate="2011-03-07T01:39:51.680" />
  <row Id="25413" PostTypeId="1" CreationDate="2011-03-07T01:50:48.993" Score="1" ViewCount="88" Body="&lt;p&gt;In &lt;a href=&quot;http://en.wikipedia.org/wiki/Recurrence_relation#Theorem&quot; rel=&quot;nofollow&quot;&gt;Wikipedia about difference equations&lt;/a&gt;, there is some description about correspondence between ODE and difference equation:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If you consider the Taylor series of&#xA;  the solution to a linear differential&#xA;  equation:$$\sum_{n=0}^\infty \frac{f^{(n)}(a)}{n!} (x-a)^n$$ you&#xA;  see that the coefficients of the&#xA;  series are given by the nth derivative&#xA;  of $f(x)$ evaluated at the point $a$. The&#xA;  differential equation provides a&#xA;  linear difference equation relating&#xA;  these coefficients.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The rule of thumb (for equations in&#xA;  which the polynomial multiplying the&#xA;  first term is non-zero at zero) is&#xA;  that:$$ y^{[k]} \to f[n+k]$$and more&#xA;  generally $$x^m*y^{[k]} \to n(n-1)(n-m+1)f[n+k-m]$$&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Example: The recurrence relationship&#xA;  for the Taylor series coefficients of&#xA;  the equation: $$(x^2 + 3x -4)y^{[3]} -(3x+1)y^{[2]} + 2y = 0\,$$ is given by$$n(n-1)f[n+1] + 3nf[n+2] -4f[n+3] -3nf[n+1] -f[n+2]+ 2f[n] = 0\, $$ or $$-4f[n+3] +2nf[n+2] + n(n-4)f[n+1]  +2f[n] = 0.\,$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My questions are:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;I was wondering what the rationale&#xA;behind this conversion from an ODE to&#xA;a difference equation is? Although&#xA;having tried to read it several&#xA;times, I was not able to understand&#xA;it.&lt;/li&gt;&#xA;&lt;li&gt;In reverse direction, can a&#xA;difference equation be converted to&#xA;an ODE using this correspondence?&#xA;How to?&lt;/li&gt;&#xA;&lt;li&gt;Is the conversion of an ODE into a&#xA;difference equation in numerical&#xA;methods for solving an ODE related&#xA;to the correspondence between the&#xA;two mentioned above?&lt;/li&gt;&#xA;&lt;li&gt;&lt;blockquote&gt;&#xA;  &lt;p&gt;This equivalence can be used to&#xA;  quickly solve for the recurrence&#xA;  relationship for the coefficients in&#xA;  the power series solution of a linear&#xA;  differential equation.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Problems generally solved using the&#xA;  power series solution method taught in&#xA;  normal differential equation classes&#xA;  can be solved in a much easier way.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I was wondering how exactly the&#xA;correspondence can make solving an&#xA;ODE or a difference equation easier?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Thanks and regards!&lt;/p&gt;&#xA;" OwnerUserId="1281" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2011-06-03T13:46:30.737" LastActivityDate="2011-06-03T13:46:30.737" Title="Correspondence between ODE and difference equation" Tags="&lt;differential-equations&gt;&lt;recurrence-relations&gt;&lt;difference-equations&gt;" AnswerCount="1" />
  <row Id="25414" PostTypeId="1" AcceptedAnswerId="25418" CreationDate="2011-03-07T01:50:57.243" Score="3" ViewCount="124" Body="&lt;p&gt;I typed $\pi$ into Wolfram Alpha and in the short list of definitions there appeared&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \pi = -i \log(-1)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which really bothered me. Multiplying on both sides by $2i$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ 2\pi i = 2 \log(-1) = \log(-1)^2 = \log 1= 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is clearly false.  I guess my error is $\log 1 = 0$ when $\log$ is complex-valued.  I need to use $1 = e^{2 \pi i}$ instead.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So my question: is it correct for WA to say $\pi = -i \log(-1)$?  Or should be they specifying &quot;which&quot; $-1$ they mean?  Clearly $ -1 = e^{i \pi}$ is the &quot;correct&quot; value of $-1$ here.&lt;/p&gt;&#xA;" OwnerUserId="7162" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-07T01:55:55.587" LastActivityDate="2011-03-07T02:01:24.407" Title="Is it standard to say $-i \log(-1)$ is $\pi$?" Tags="&lt;complex-numbers&gt;&lt;logarithms&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="25415" PostTypeId="1" CreationDate="2011-03-07T01:51:16.637" Score="0" ViewCount="60" Body="&lt;p&gt;Suppose that $A=\mathbb{R}[x]$ is the polynomial ring of one variable over the real numbers and $B$ is the real vector space which is the set $B = \{nx\mid n\text{ is a real number}\}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Show that $A\otimes_{\mathbb{R}} B \otimes_{\mathbb{R}} A$ is a free $A \otimes_{\mathbb{R}} A$-module.&lt;/p&gt;&#xA;" OwnerUserId="7889" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-07T01:55:01.127" LastActivityDate="2011-03-07T02:27:26.453" Title="free modules in algebra" Tags="&lt;abstract-algebra&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="25417" PostTypeId="2" ParentId="25414" CreationDate="2011-03-07T02:00:26.593" Score="4" ViewCount="" Body="&lt;p&gt;The (principal value) of the complex logarithm is defined as $\log z = \ln |z| + i Arg(z)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\log(-1) = \ln|-1| + i Arg(-1) = 0 + i \pi.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and then, one simply gets&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;-i \log(-1) = -i (i \pi) = \pi.&#xD;&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="6491" LastActivityDate="2011-03-07T02:00:26.593" />
  <row Id="25418" PostTypeId="2" ParentId="25414" CreationDate="2011-03-07T02:01:24.407" Score="7" ViewCount="" Body="&lt;p&gt;It is not which $-1$ but which $\log$ which is the issue. Wolfram Alpha has used some kind of &lt;em&gt;principal value&lt;/em&gt; of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Complex_logarithm&quot; rel=&quot;nofollow&quot;&gt;complex logarithm&lt;/a&gt;. But using this, $a \log (b) = \log (b^a)$ is no longer always true.&lt;/p&gt;&#xA;" OwnerUserId="6460" LastActivityDate="2011-03-07T02:01:24.407" />
  <row Id="25419" PostTypeId="2" ParentId="25415" CreationDate="2011-03-07T02:20:07.583" Score="3" ViewCount="" Body="&lt;p&gt;Since $B\cong \mathbb{R}$ as a vector space/module, what does that tell you about $A\otimes_{\mathbb{R}} B$? And then about $(A\otimes_{\mathbb{R}}B)\otimes_{\mathbb{R}} A$?&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-07T02:27:26.453" LastActivityDate="2011-03-07T02:27:26.453" />
  <row Id="25420" PostTypeId="1" CreationDate="2011-03-07T02:31:08.827" Score="0" ViewCount="51" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Calls to a customer service center&#xA;  last on average 2.7 minutes with a&#xA;  standard deviation of 1.4 minutes. An&#xA;  operator in the call center is&#xA;  required to answer 74 calls each day.&#xA;  Assume the call times are independent.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;em&gt;What is the standard deviation of the&#xA;  total amount of time in minutes the&#xA;  operator will spend on the calls each&#xA;  day? Give your answer to four decimal&#xA;  places.&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I initially attempted to do: &lt;code&gt;74*1.4=103.6&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, this doesn't seem to work. I'm not exactly sure how to go about solving this problem.&lt;/p&gt;&#xA;" OwnerUserId="7890" LastActivityDate="2011-03-07T02:53:38.057" Title="Probability Problem - Standard Deviation" Tags="&lt;probability&gt;&lt;probability-distributions&gt;" AnswerCount="1" />
  <row Id="25422" PostTypeId="1" AcceptedAnswerId="25424" CreationDate="2011-03-07T02:36:53.587" Score="2" ViewCount="87" Body="&lt;p&gt;Say I have a complex polynomial:&lt;br&gt;&#xA;$$a_0+a_1x+\cdots+a_nx^n,$$&lt;br&gt;&#xA;where $a_0,\ldots,a_n$ are complex numbers.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the conjugate of this polynomial?  How is it defined?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, if we have an inner product on the vector space $V$ defined by (for complex polynomials):&lt;br&gt;&#xA;$$\int_0^1 p(x)\overline{q(x)}dx$$&lt;br&gt;&#xA;Then how do we know that:&#xA;$$\langle v,v \rangle = \int_0^1 (a_0+a_1x+\cdots+a_nx^n)(\overline{a_0}+\overline{a_1}x+\cdots+\overline{a_n}x^n)dx$$&lt;br&gt;&#xA;is positive?  Since there could be negative numbers in there, do we just know the positive ones outweigh the negative?&lt;/p&gt;&#xA;" OwnerUserId="2533" LastEditorUserId="2533" LastEditorDisplayName="" LastEditDate="2011-03-07T03:32:33.127" LastActivityDate="2011-03-07T04:11:08.227" Title="Conjugate of complex polynomial?" Tags="&lt;complex-conjugate&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="25423" PostTypeId="2" ParentId="13124" CreationDate="2011-03-07T02:44:08.523" Score="1" ViewCount="" Body="&lt;p&gt;Extremely late answer, but Eisenbud discusses this in his appendix on homological algebra in &quot;Commutative algebra with a view towards algebraic geometry&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the Yoneda interpretation of Ext, and gives rise to a multiplicative structure on Ext-modules.&lt;/p&gt;&#xA;" OwnerUserId="4284" LastActivityDate="2011-03-07T02:44:08.523" />
  <row Id="25424" PostTypeId="2" ParentId="25422" CreationDate="2011-03-07T02:47:30.967" Score="4" ViewCount="" Body="&lt;p&gt;It is simply the polynomial you get by replacing each $a_i$ by its complex conjugate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You should think of this as the map induced on $\mathbb{C}[x]$ by the map $\mathbb{C}\to\mathbb{C}$ given by complex conjugation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added:&lt;/strong&gt; How do we know that $\int_0^1 p(x)\overline{p(x)}\,dx$ is positive?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Write each $a_j = \alpha_j + i\beta_j$, with $\alpha_j,\beta_j\in\mathbb{R}$. Then note that $v = q(x)+ir(x)$, where &#xA;\begin{align*}&#xA;q(x) &amp;amp;= \alpha_0 + \alpha_1 x + \cdots + \alpha_nx^n\\&#xA;r(x) &amp;amp;= \beta_0 + \beta_1x + \cdots + \beta_nx^n. &#xA;\end{align*}&#xA;So you have:&#xA;\begin{align*}&#xA;\langle v,v\rangle &amp;amp;= \int_0^1 (a_0+a_1x + \cdots + a_nx^n)(\overline{a_0}+\overline{a_1}x + \cdots \overline{a_n}x^n)\,dx\\&#xA;&amp;amp;= \int_0^1(q(x)+ir(x))(q(x)-ir(x))\,dx\\&#xA;&amp;amp;= \int_0^1\Bigl( \left(q(x)\right)^2 - i^2\left(r(x)\right)^2\Bigr)\,dx\\&#xA;&amp;amp;= \int_0^1\Bigl( \left(q(x)\right)^2 + \left(r(x)\right)^2\Bigr)\,dx&#xA;\end{align*}&#xA;and since this is the integral of two nonnegative &lt;em&gt;real&lt;/em&gt; valued functions, it is nonnegative (and equal to $0$ if and only if both $q(x)$ and $r(x)$ are zero, i.e., if and only if $v=\mathbf{0}$). &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-07T04:11:08.227" LastActivityDate="2011-03-07T04:11:08.227" CommentCount="6" />
  <row Id="25426" PostTypeId="2" ParentId="25420" CreationDate="2011-03-07T02:53:38.057" Score="1" ViewCount="" Body="&lt;p&gt;Hint:  the variance, not the standard deviation, increases linearly with the number of calls (under the assumptions that you are expected to work under, especially independent variation from one call to another.)&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-03-07T02:53:38.057" />
  <row Id="25427" PostTypeId="1" AcceptedAnswerId="25497" CreationDate="2011-03-07T02:57:47.900" Score="0" ViewCount="86" Body="&lt;p&gt;Let p be prime.  For $n|p-1$, let $f_n(x)=(x-a_1)(x-a_2)\cdots(x-a_s)$ where $a_1$ through $a_s$ belong to the exponent $n \pmod{p}$ and $s=\phi(n)$.  For sufficiently large x, $f_n(x)$ is positive.  Prove that&#xA;$$f_n(x)=\prod_{m|n} (x^{n/m} - 1)^{\mu(m)} \pmod{p}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I did prove (using logs) that if we let $f(n)$ be a positive function over the natural numbers and let $g(n) = \prod_{m|n} f(m)$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;then&#xA;$$f(n)=\prod_{m|n} g(n/m)^{\mu(m)}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, if I can prove $x^n-1=\prod_{m|n} f_m(x) \pmod{p}$, then I can use what I have already proven to finish the proof.  I'm not sure where to go from here, though.&lt;/p&gt;&#xA;" OwnerUserId="7578" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-07T04:47:42.440" LastActivityDate="2011-03-07T10:15:40.787" Title="Prove $x^n-1=\prod_{m|n} f_m(x) \pmod{p}$, where $f_n(x)=(x-a_1)(x-a_2)...(x-a_s)$ and $a_1$ through $a_s$ belong to the exponent $n \pmod{p}$" Tags="&lt;number-theory&gt;&lt;homework&gt;" AnswerCount="2" />
  <row Id="25428" PostTypeId="2" ParentId="25427" CreationDate="2011-03-07T03:22:44.543" Score="0" ViewCount="" Body="&lt;p&gt;Check Serre's book &quot;A course in arithemetic&quot;. This is like in the second page. You will learn a lot by reading that book. &lt;/p&gt;&#xA;" OwnerUserId="7887" LastActivityDate="2011-03-07T03:22:44.543" />
  <row Id="25429" PostTypeId="1" CreationDate="2011-03-07T03:26:41.293" Score="0" ViewCount="48" Body="&lt;p&gt;Let the random variable X be equal to the number of days that it takes a high-risk driver to have an accident. Assume that X has a exponential distribution. If P(X &amp;lt; 50) = 0.25, compute P(X &gt; 100 | X &gt; 50). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, P(X &gt; 50) = 1 - P(X &amp;lt; 50) = 1 - .25 = .75 &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, P(X &gt; 100 | X &gt; 50) = .75?&lt;/p&gt;&#xA;" OwnerUserId="7893" LastActivityDate="2011-03-07T03:26:41.293" Title="Memory Lapse Property of Exponential Distributions...." Tags="&lt;probability&gt;&lt;probability-distributions&gt;" CommentCount="1" />
  <row Id="25430" PostTypeId="1" AcceptedAnswerId="25516" CreationDate="2011-03-07T03:32:48.663" Score="6" ViewCount="284" Body="&lt;p&gt;I was under the mistaken impression that if one could find the generating function for a sequence of numbers, you could just plug in a natural number $n$ to find the nth term of the sequence. I realize now that I was confusing this with a closed form formula. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if that is not the case, then what is the point of generating functions? How do they make understanding counting sequences easier? For example, suppose  I had a problem where I wanted to count how many ways I could buy $n$ pieces of apples, oranges, and pears given that I want an even number of apples, an odd number of oranges, and at most 3 pears. This would be the number of nonnegative integer solutions to $a+b+c=n$ with $a$ even, $b$ odd, and $0\leq c\leq 3$. This is the same as the coefficient of $x^n$ in the product&#xA;$$&#xD;&#xA;(1+x^2+x^4+\cdots)(x+x^3+x^5+\cdots)(1+x+x^2+x^3) = \frac{1}{1-x^2}\cdot\frac{x}{1-x^2}\cdot\frac{1-x^4}{1-x}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But what good is that? I don't see how this is much better. Also, with use of exponential generating functions, it seems the choose of monomials we use as place holders for the terms of the sequence can be arbitrary. Then is sequence is just the coefficient of the nth monomial that you've chosen to build the generating function with. What is the real advantage of doing things like this? Many problems I see tend to ask me find the generating function, but then I'm rarely asked to do anything with it.&lt;/p&gt;&#xA;" OwnerUserId="6742" LastActivityDate="2011-03-07T17:58:05.247" Title="Why are generating functions useful?" Tags="&lt;combinatorics&gt;&lt;generating-functions&gt;" AnswerCount="6" CommentCount="6" FavoriteCount="2" />
  <row Id="25431" PostTypeId="1" AcceptedAnswerId="25432" CreationDate="2011-03-07T03:38:17.660" Score="1" ViewCount="115" Body="&lt;p&gt;&lt;a href=&quot;http://crazyproject.wordpress.com/2010/12/20/the-ring-of-polynomials-over-a-field-with-no-linear-term-is-not-a-ufd/&quot; rel=&quot;nofollow&quot;&gt;My question pertains to this link&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(We are not required to show that this is indeed a subring, only that it is not a unique factorization domain.)&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand (more or less) everything stated, through &quot;Now $x^6 = x^3x^3 = x^2x^2x^2$&quot;.&#xA;&lt;br&gt;&#xA;Does this conclude the proof because we have found an element in R with two distinct factorizations? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How did we use the requirement that the linear terms be zero? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks. &lt;/p&gt;&#xA;" OwnerUserId="7850" LastActivityDate="2011-03-07T16:47:15.217" Title="Subring of F[x], all with linear term = 0, is not a UFD" Tags="&lt;abstract-algebra&gt;" AnswerCount="4" />
  <row Id="25432" PostTypeId="2" ParentId="25431" CreationDate="2011-03-07T03:46:27.150" Score="4" ViewCount="" Body="&lt;p&gt;It's not enough to just find two distinct factorizations (after all, in $\mathbb{Z}$, you have $8 = 2\times\times 2$ &lt;strong&gt;and&lt;/strong&gt; $8=2\times 4$). What is important is that you find two distinct factorization &lt;em&gt;into irreducibles&lt;/em&gt; (that are not related via multiplication by units; so again, $4=2\times 2 = (-2)\times(-2)$ does not disprove that $\mathbb{Z}$ is a UFD).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here, you need to show that both $x^2$ and $x^3$ are irreducibles in this ring, before you can conclude from $x^6 = x^3x^3 = x^2x^2x^2$ that you have two (essentially) distinct factorizations of $x^6$ into irreducibles. And in order to prove that both $x^2$ and $x^3$ are irreducible in this ring you &lt;em&gt;will&lt;/em&gt; have to use that the linear terms of the elements of this ring are zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The link you provide proves this before exhibiting the factorization; when it says &quot;since neither of these is degree $1$&quot;, &lt;strong&gt;that's&lt;/strong&gt; where they are using there is no linear term.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2011-03-07T03:46:27.150" CommentCount="1" />
  <row Id="25433" PostTypeId="2" ParentId="25431" CreationDate="2011-03-07T03:47:30.673" Score="1" ViewCount="" Body="&lt;p&gt;Because no element of $R$ has degree $1$ which shows that $x^2$ and $x^3$ are irreducible in $R$.&lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2011-03-07T03:47:30.673" CommentCount="1" />
  <row Id="25434" PostTypeId="2" ParentId="25430" CreationDate="2011-03-07T03:50:30.683" Score="7" ViewCount="" Body="&lt;p&gt;Have you read Generatingfunctionology by Herbert Wilf? The book is loaded with methods and techniques based on generating functions for solving a number of different problems. I guess the book can answer your question better than I could.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I feel that generating functions are powerful because they allow you to use tools from calculus and analysis to solve problems in areas such as discrete mathematics and combinatorics, where such tools don't seem readily applicable. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: I just wanted to add one more point. Suppose you have translated the problem from the sequence to the generating function and you don't see any simple closed form solutions, you could use Wolfram Alpha to make absolutely sure. Wolfram alpha can easily expand out expressions involving large polynomial terms and will often give you a series expansion if it exists in closed form. &lt;/p&gt;&#xA;" OwnerUserId="2641" LastEditorUserId="2641" LastEditorDisplayName="" LastEditDate="2011-03-07T17:34:11.900" LastActivityDate="2011-03-07T17:34:11.900" />
  <row Id="25435" PostTypeId="1" AcceptedAnswerId="25438" CreationDate="2011-03-07T03:56:35.000" Score="3" ViewCount="100" Body="&lt;p&gt;Show that if $A$ is a retract of $X$ then for all $n \ge 0$ $$H_n(X) \simeq H_n(A) \oplus H_n(X,A)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we have a retraction $r:X \to A$, which is surjective.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the long exact sequence&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\cdots \to H_n(A) \to H_n(X) \to H_n(X,A) \to H_{n-1}(A) \cdots$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As $r$ is surjective we have that $H_n(X) \to H_n(X,A)$ is surjective, and hence $H_n(A) \to H_n(X)$ is injective. Thus we have a short exact sequence&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$0 \to H_n(A) \to H_n(X) \to H_n(X,A) \to 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am unsure how to go from the fact this is exact, to the result (assuming the above is correct!)&lt;/p&gt;&#xA;" OwnerUserId="2219" LastEditorUserId="2219" LastEditorDisplayName="" LastEditDate="2011-03-07T04:43:45.970" LastActivityDate="2011-03-07T04:43:45.970" Title="Relative homology of a retract" Tags="&lt;algebraic-topology&gt;&lt;homological-algebra&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="25436" PostTypeId="2" ParentId="25430" CreationDate="2011-03-07T04:00:36.253" Score="1" ViewCount="" Body="&lt;p&gt;When trying to understand a counting situation, the best thing to have is a closed form formula, a function that is easily calculable by well known methods ('closed-form' is a bit slippery in meaning, but usually means, without any recursion, but there's room for disagreement over whether summations or products are allowed (though factorial is often allowed). From such functions, it is usually straightforward to determine approximations and asymptotics and to combine with other other functions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The next best thing is a recurrence. It allows calculation often in a very straightforward manner with no approximation needed. But recurrences are a bit inscrutable; it's very difficult to simplify or read off properties directly from the recurrence (whereas closed form is very easy to understand).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Somewhere between these two, generating functions allow ease in combinatorial manipulation (multiplying two ogfs means that one is getting an ordered pair of the objects the gfs describe). Also there are analytical techniques that allow extraction of asymptotics from gfs. And ostensibly one can get values for $f(n)$ from the gf by differentiating it $n$ times and evaluating at 0. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wilf's generatingfunctionology, as mentioned in the comments, is the place to go for getting these techniques for gfs.&lt;/p&gt;&#xA;" OwnerUserId="1919" LastActivityDate="2011-03-07T04:00:36.253" />
  <row Id="25437" PostTypeId="1" CreationDate="2011-03-07T04:02:34.880" Score="0" ViewCount="120" Body="&lt;p&gt;The Lake Shore Inn is trying to determine its break-even point. The inn has 50 rooms that are rented at $60 per night. Operating costs are as follows:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Salaries      $7,200  per month&#xA;Utilities     $1,500  per month&#xA;Depreciation  $1,200  per month&#xA;Maintenance   $300    per month&#xA;Maid Service  $8      per room&#xA;Other Costs   $28     per room&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Determine the inn's break-even point in (1) number of rented rooms per month and (2) dollars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Break-even number of rooms rented per month:&lt;strong&gt;&lt;em&gt;_&lt;/em&gt;__&lt;/strong&gt;&lt;br&gt;&#xA;Break-even sales dollars:&lt;strong&gt;&lt;em&gt;_&lt;/em&gt;_&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Formulas used:&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fixed Costs / Contribution Margin per Unit (Unit selling price - Unit Variable Costs) = Break-even point in Units.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fixed Costs / Contribution Margin Ratio (Contribution Margin per Unit / Unit selling price) = Break-even point in Dollars.&lt;/p&gt;&#xA;" OwnerUserId="7773" LastEditorUserId="1736" LastEditorDisplayName="" LastEditDate="2011-03-07T07:07:44.140" LastActivityDate="2011-03-07T07:07:44.140" Title="Accounting Question: Computing Break-even #'s" Tags="&lt;homework&gt;&lt;education&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="25438" PostTypeId="2" ParentId="25435" CreationDate="2011-03-07T04:08:04.360" Score="1" ViewCount="" Body="&lt;p&gt;If you can find a homomorphism $H_n(X,A) \to H_n(X)$ which 'splits' the quotient map $H_n(X) \to H_n(X,A)$ then you can use the &lt;a href=&quot;http://en.wikipedia.org/wiki/Splitting_lemma&quot; rel=&quot;nofollow&quot;&gt;splitting lemma&lt;/a&gt; to give you your direct sum.&lt;/p&gt;&#xA;" OwnerUserId="668" LastActivityDate="2011-03-07T04:08:04.360" CommentCount="1" />
  <row Id="25439" PostTypeId="1" AcceptedAnswerId="25445" CreationDate="2011-03-07T04:12:12.563" Score="3" ViewCount="105" Body="&lt;p&gt;I understand how to apply the chain rule to two functions, but trip up when it comes to three or more.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given $f~(x~)= x~\sqrt{1-x^{2}}$, I can see there are three values:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a: 1-x^{2}$&lt;br&gt;&#xA;$b: \sqrt{a}$&lt;br&gt;&#xA;$c: x~\cdot b$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, $$f(b(a))= \frac{1}{2~\sqrt{1-x^{2}}}~\cdot~2x= x \sqrt{1-x^{2}},$$ but how do I factor in the last x value?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Am I on the right track here?&lt;/p&gt;&#xA;" OwnerUserId="7006" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-07T04:20:27.357" LastActivityDate="2011-03-07T04:43:43.223" Title="Applying Chain Rule to more than two functions" Tags="&lt;calculus&gt;&lt;homework&gt;" AnswerCount="3" />
  <row Id="25440" PostTypeId="2" ParentId="25430" CreationDate="2011-03-07T04:12:56.963" Score="3" ViewCount="" Body="&lt;p&gt;To answer your question for the example you state:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \frac{1}{1-x^2} \cdot \frac{x}{1-x^2} \cdot \frac{1-x^4}{1-x} = (x+x^2 +x^3 + x^4)(1-x^2)^{-2} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By binomial theorem (yes, it is valid for negative exponents too) we get that this is equal to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ (x + x^2 + x^3 + x^4) \left(\sum_{j=0}^{\infty} (-1)^j {-2 \choose j} x^{2j}\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you now figure out a formula for the coefficient of $\displaystyle x^n$, thus solving your problem?&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2011-03-07T04:12:56.963" CommentCount="4" />
  <row Id="25441" PostTypeId="2" ParentId="25437" CreationDate="2011-03-07T04:14:13.853" Score="1" ViewCount="" Body="&lt;p&gt;Hint: Unit Variable costs are the costs that increase depending on the number of units that are rented, while the costs per month are fixed.  The basic idea is that you spend a certain number of dollars to stay open (taxes, mortgage, desk staff) and there are costs that increase per room rented (cleaning rooms). So what would the Contribution Margin per Unit be?  It is defined in the first formula.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-03-07T04:14:13.853" />
  <row Id="25442" PostTypeId="2" ParentId="25431" CreationDate="2011-03-07T04:14:31.107" Score="1" ViewCount="" Body="&lt;p&gt;$\rm (x^2)^3 = (x^3)^2\ $ is a nonunique factorization into irreducibles, because the irreducibles $\rm\:x^2\:,\ x^3$ are nonassociate, i.e.  $\rm\ x^3\nmid x^2\ $ (by degrees) and $\rm\: x^2\nmid x^3\: $ since $\rm\ x\not\in R\:.\: $ It is a bit simpler to prove that $\rm\:R\:$ is not a $\rm UFD$ by simply noting that $\rm\:x^2\:$ is irreducible but not prime, since $\rm\ x^2\: | \:(x^3)^2\ $ but $\rm\ x^2\nmid x^3\:.$ Even simpler: $\rm UFD \Rightarrow$ integrally closed, but $\rm\:R\:$ isn't: $\rm\ y^2 - x^2\ $ has the root $\rm\ y = x^3/x^2 = x\not\in R\:.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That $\rm\:R\:$ is a subring is easy, $\rm\ mod\ x^2\: F[x]:\ f\equiv f_0,\ g\equiv g_0\: \Rightarrow\ f+g\equiv f_0+g_0,\ f\ g\equiv f_0\ g_0$&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2011-03-07T06:25:40.137" LastActivityDate="2011-03-07T06:25:40.137" />
  <row Id="25443" PostTypeId="2" ParentId="25287" CreationDate="2011-03-07T04:16:18.950" Score="0" ViewCount="" Body="&lt;p&gt;Elaborating a bit on Chris' post, the determinant condition is $x^{2} + w^{2} = y^{2} + z^{2}$. Adding and subtracting the original system of equations gives &#xA;\begin{eqnarray}&#xA;a(x + y + z - w) + b(x - y + z + w) = 0 \quad \text{and} \quad a(x + y - z + w) + b(w - x + y + z) = 0.&#xA;\end{eqnarray}&#xA;which places more constraints on the possible solutions $(x,y,z,w)$ if $a$ and $b$ are non-zero.&lt;/p&gt;&#xA;" OwnerUserId="2720" LastActivityDate="2011-03-07T04:16:18.950" />
  <row Id="25444" PostTypeId="1" CreationDate="2011-03-07T04:18:17.763" Score="4" ViewCount="118" Body="&lt;p&gt;On a uni-directional circular road where all vehicles travel in the same direction with varying speeds and overtaking is allowed on the entire road, what is the expected number of times &#xA;(a) a particular vehicle will overtake others?&#xA;(b) a particular vehicle will be overtaken by others?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, what I want to find out: Is (a) = (b)?&#xA;Thanks,&lt;/p&gt;&#xA;" OwnerUserId="7896" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2011-03-07T04:59:27.730" LastActivityDate="2011-03-10T05:43:43.067" Title="Overtaking Probability" Tags="&lt;probability&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" />
  <row Id="25445" PostTypeId="2" ParentId="25439" CreationDate="2011-03-07T04:22:53.110" Score="6" ViewCount="" Body="&lt;p&gt;The way you're trying to break up your function isn't quite right.  The step where $x$ is multiplied with $\sqrt{1-x^2}$ should be seen as a product, not a composition.  So you could look at this as $f(x)=g(x)\cdot h(x)$, where $h(x)=b(a(x))$ with $a$ and $b$ are as you defined them and $g(x)=x$.  You can apply the product rule first, and only apply the chain rule for the part where you differentiate $h(x)$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f&amp;#39;(x)=g&amp;#39;(x)h(x)+g(x)h&amp;#39;(x)=h(x)+xb&amp;#39;(a(x))a&amp;#39;(x)=\sqrt{1-x^2}+x\frac{1}{2\sqrt{1-x^2}}(-2x),$$&#xA;etc.  (You had a sign error in the derivative of $a$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other problem is the notation where you wrote &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f(b(a))= \frac{1}{2~\sqrt{1-x^{2}}}~\cdot~2x= x \sqrt{1-x^{2}}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This doesn't really make sense.  $f(b(a))$ is something different entirely from what you want to consider.  The middle term is an attempted derivative of $b(a(x))$.  And the right-hand term is $f(x)$.  Thus, none of these are equal, and &quot;$=$&quot; is best reserved for equality, both to avoid errors and to allow meaningful communication to others (such as to whomever grades your assignment if this is for a class).&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2011-03-07T04:35:42.563" LastActivityDate="2011-03-07T04:35:42.563" />
  <row Id="25447" PostTypeId="2" ParentId="25439" CreationDate="2011-03-07T04:23:58.240" Score="2" ViewCount="" Body="&lt;p&gt;If you know how to apply the chain rule for two functions, then the simplest thing to do to avoid getting tripped up is to the work one step at a time. For instance,&#xA;\begin{eqnarray*}&#xA;\frac{d}{dx}c\Biggl(b\Bigl(a(x)\Bigr)\Biggr) &amp;amp;= c'\Biggl(b\Bigl(a(x)\Bigr)\Biggr)\Biggl(b\Bigl(a(x)\Bigr)\Biggr)'\\&#xA;&amp;amp;= c'\Biggl(b\Bigl(a(x)\Bigr)\Biggr)\Biggl( b'\Bigl(a(x)\Bigr)\Bigl(a(x)\Bigr)'\Biggr)\\&#xA;&amp;amp;= c'\Biggl(b\Bigl(a(x)\Bigr)\Biggr)b'\Bigl(a(x)\Bigr)a'(x).&#xA;\end{eqnarray*}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That said, in your example, you should &lt;strong&gt;not&lt;/strong&gt; think of $f(x)$ as the composition of three functions, but rather first and foremost as a &lt;strong&gt;product&lt;/strong&gt;, so the first thing you should do is apply the product rule, $(gh)&amp;#39; = g&amp;#39;h + gh&amp;#39;$, with $g(x) = x$ and $h(x) = \sqrt{1-x^2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(If you want to think about $c$ as a function of the form $c(u) = xu$, then what would the derivative of $c$ with respect to $x$ be? First you would use the product rule: $\frac{d}{dx}c(u) = \frac{d}{dx}(xu) = u + x\frac{du}{dx}$, and then you would find the derivative of $u$.)&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-07T04:43:43.223" LastActivityDate="2011-03-07T04:43:43.223" CommentCount="4" />
  <row Id="25448" PostTypeId="2" ParentId="25439" CreationDate="2011-03-07T04:24:41.237" Score="1" ViewCount="" Body="&lt;p&gt;You want to work you're way from the outside in to take derivatives.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, apply the product rule to $x$ and $\sqrt{1-x^2}$. That is, $f&amp;#39;(x)g(x)+f(x)g&amp;#39;(x)$&lt;br&gt;&#xA;In this case, you can let $f(x)=x$ and $g(x)=\sqrt{1-x^2}$.&lt;br&gt;&#xA;$$1*\sqrt{1-x^2} + x(\sqrt{1-x^2})&amp;#39;$$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now you have only 2 functions to work with.  The outer square root function and the inner square function.&lt;br&gt;&#xA;Now you get:&#xA;$$\sqrt{1-x^2} + x(\frac{1}{2}(1-x^2)^{-1/2}(-x^2)&amp;#39;)$$&#xA;And finally:&lt;br&gt;&#xA;$$\sqrt{1-x^2} + x(\frac{1}{2}(1-x^2)^{-1/2}(-2x))$$&lt;br&gt;&#xA;Which you can simplify to:&lt;br&gt;&#xA;$$\sqrt{1-x^2} + \frac{-x^2}{\sqrt{1-x^2}}$$&lt;br&gt;&#xA;And further to:&lt;br&gt;&#xA;$$\frac{-2x^2+1}{\sqrt{1-x^2}}$$  &lt;/p&gt;&#xA;" OwnerUserId="2533" LastEditorUserId="2533" LastEditorDisplayName="" LastEditDate="2011-03-07T04:33:11.240" LastActivityDate="2011-03-07T04:33:11.240" CommentCount="2" />
  <row Id="25449" PostTypeId="1" CreationDate="2011-03-07T04:24:54.517" Score="7" ViewCount="128" Body="&lt;p&gt;Consider the homogeneous space $GL(3)/GL(2) = GL(3,\mathbb{R})/GL(2,\mathbb{R})$ where $GL(2)$ fixes the first coordinate axis (so can be identified with the subgroup of $2\times 2$ blocks sitting in the 'bottom right' corner of matrices in $GL(3)$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any 'explicitly known' other homogeneous space which is isomorphic to this one? Say by using orthogonal groups? Is it some sort of Grasmannian/Stiefel manifold?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Am I correct in guessing that $GL(3)/GL(2)$ has dimension 5?&lt;/p&gt;&#xA;" OwnerUserId="3835" LastActivityDate="2011-03-11T02:25:56.913" Title="Do we know this homogeneous space by another name?" Tags="&lt;differential-geometry&gt;&lt;lie-groups&gt;&lt;manifold&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="25450" PostTypeId="2" ParentId="25444" CreationDate="2011-03-07T04:35:43.357" Score="1" ViewCount="" Body="&lt;p&gt;Certainly not unless the speed of the vehicle is the mean speed of the traffic.  I think in that case you can make a symmetry argument-go to a coordinate system that rotates at the mean speed.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-03-07T04:35:43.357" CommentCount="1" />
  <row Id="25451" PostTypeId="1" CreationDate="2011-03-07T05:02:13.200" Score="4" ViewCount="172" Body="&lt;p&gt;In page 76 of Atiyah's K-theory, he discussed the possibilities of extending Bott periodicity to the case when $\mathbb{S}^{2}\times X$ is replaced by a fibration over $X$ with structure group $U_{1}$. He quote an example as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;..In particular let $$E=L\oplus 1$$ be the sum of a line-bundle $L$ and the trivial-bundle $L$. Then $\mathbb{P}(E)$ has fibre the complex projective line on $\mathbb{S}^{2}$. Moreover on $\mathbb{P}(E)$ we have the natural tautologous line-bundle $H^{*}$. Consider of determinants then shows that, on $\mathbb{P}(E)$, we have an exact sequence $$0\rightarrow H^{*}\rightarrow \mathbb{P}^{*}(E)\rightarrow H\otimes p^{*}L\rightarrow 0$$ where $p: \mathbb{P}(E)\rightarrow X$ denotes the projection. This exact sequence shows that, in the $K(X)$ module $K(\mathbb{P}(E))$, the element $[H]$ satisfies the relation $$[E]=[H][L]+[H]^{*}&amp;quot;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This &quot;particular case&quot; was later expanded by Atiyah into a theorem claiming $K(\mathbb{P}(L\oplus 1))$ is generated by $[H]$ and is subject to the single relation $([H]-[1])([H][L]-[1])=0$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My questions are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose $H$ is the canonical line bundle over $\mathbb{S}^{2}\cong \mathbb{C}\mathbb{P}_{1}$, how do we get this exact sequence? I know the question is kind of dumb but I do not see how $P(E)$ (I understand as the projectification of the fibre in E) has fibre the complex projective line on $\mathbb{S}^{2}$, since $\mathbb{P}(E)$ is a vector bundle over $X$ and not over $\mathbb{S}^{2}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also I do not understand what Atiyah meant that &quot;consideration of determinants&quot;. Someone suggested me that this may mean by applying $\wedge $ operation on the sequence to see they are actually the same; unfortunately the fact I stuck in the former question made me unable to proceed to exterior powers. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am suspecting that Atiyah actually meant $H$ to be something different, since he used $H^{*}$ for &quot;the tautologous line bundle&quot;, it maybe the pull back of some $H$ defined on $X$. But this looks quite unlikely so I decided to ask. Originally I thought this is simply technical but later Atiyah used similar constructions to get incredible results like:&#xA; $$K(X)[t]/\prod^{n}_{i=1}(t-[L^{*}_{i}])\cong K(\mathbb{P}(L_{1}\otimes...\otimes L_{n})$$ with $L_{i}$s being line bundles over $X$ and $H$ be &quot;the standard bundle over $\mathbb{P}(L_{1}\otimes...\otimes L_{n})$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And $$K(\mathbb{P}(\mathbb{C}^{n}))\cong \mathbb{Z}[t]/(t-1)^{n}$$ under $t\rightarrow H$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A proof in detail (as I am an undergraduate) would be mostly welcome.  &lt;/p&gt;&#xA;" OwnerUserId="7887" LastEditorUserId="7887" LastEditorDisplayName="" LastEditDate="2011-03-07T06:16:55.590" LastActivityDate="2011-03-11T19:46:41.507" Title="Asking an exact sequence of $K(X)$ modules" Tags="&lt;topological-k-theory&gt;" CommentCount="11" FavoriteCount="1" />
  <row Id="25452" PostTypeId="1" CreationDate="2011-03-07T05:10:32.043" Score="3" ViewCount="135" Body="&lt;p&gt;Suppose that $p$ is a prime. Prove that the sum of the primitive roots modulo $p$ is congruent to&#xA;$\mu(p − 1) \pmod{p}$.&lt;/p&gt;&#xA;" OwnerUserId="7578" LastEditorUserId="6491" LastEditorDisplayName="" LastEditDate="2011-03-07T05:18:33.507" LastActivityDate="2011-06-03T00:12:17.240" Title="Prove sum of primitive roots congruent to $\mu(p-1) \pmod{p}$" Tags="&lt;number-theory&gt;&lt;homework&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="25453" PostTypeId="2" ParentId="25401" CreationDate="2011-03-07T05:12:43.903" Score="1" ViewCount="" Body="&lt;p&gt;The intro ODE text by Boyce and DiPrima gives a fairly complete outline of this proof.  Some of the details are left as exercises for the reader, but since the text is aimed at an introductory audience, anyone who knows enough to ask this question would presumably have little trouble filling in the gaps.&lt;/p&gt;&#xA;" OwnerUserId="4229" LastActivityDate="2011-03-07T05:12:43.903" />
  <row Id="25454" PostTypeId="1" AcceptedAnswerId="25459" CreationDate="2011-03-07T05:16:14.613" Score="1" ViewCount="109" Body="&lt;p&gt;Given $f(x)= \sin(\pi x)^{2}$, find the derivative.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the chain rule my work is as follows:  $(\sin(\pi x)^2)&amp;#39;$ becomes&#xA;$$2 \sin(\pi x) \cdot \frac{d}{dx}(\sin(\pi x)$$&#xA;The derivative of sin is cos, thus&#xA;$$2 \sin(\pi x) \cdot \cos(\pi x) \cdot \frac{d}{dx}(\pi x)$$ &#xA;The derivative of $\pi x$ is $\pi$, and the equation stretches to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$2 \sin(\pi x) \cos(\pi x) \pi == 2 \pi \sin(\pi x) \cos(\pi x)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the book states the answer as $$2 \pi^{2}~ x ~\cos(\pi x)^{2}$$ and that definitely doesn't match my result.  Where did I go wrong? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks to Arturo Madigan, Jonas Meyer, et al for their help.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I re-did the problem based on having $(\pi x)^{2}$, having the exponent rather than the sin function, and it seems I have a missing exponent as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Differentiating the terms of the function via the chain rule, I get $$(\pi x)^{2} [\frac{d}{dx}sin] \cdot \frac{d}{dx}(\pi x)^{2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ cos(\pi x)^{2} \cdot 2\pi x= 2~\pi~ x cos~(\pi x)^{2}$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to the book answer, $2~\pi x$ should actually be $2~ \pi^{2} x$&lt;/p&gt;&#xA;" OwnerUserId="7006" LastEditorUserId="7006" LastEditorDisplayName="" LastEditDate="2011-03-07T06:14:08.043" LastActivityDate="2011-03-07T06:14:08.043" Title="Chain Rule applied to Trig Functions" Tags="&lt;calculus&gt;&lt;homework&gt;&lt;trigonometry&gt;" AnswerCount="3" CommentCount="9" />
  <row Id="25455" PostTypeId="1" AcceptedAnswerId="25515" CreationDate="2011-03-07T05:16:57.083" Score="10" ViewCount="365" Body="&lt;p&gt;I am comfortable with the definition of adjoint functors. I have done a few exercises proving that certain pairs of functors are adjoint (tensor and hom, sheafification and forgetful, direct image and inverse image of sheaves, spec and global sections ect) but I am missing the bigger picture.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why should I care if a functor has a left adjoint? What does it tell me about the functor? &lt;/p&gt;&#xA;" OwnerUserId="1725" LastActivityDate="2011-03-07T22:16:47.677" Title="Why should I care about adjoint functors" Tags="&lt;algebraic-geometry&gt;&lt;commutative-algebra&gt;&lt;category-theory&gt;" AnswerCount="4" FavoriteCount="6" />
  <row Id="25456" PostTypeId="2" ParentId="25251" CreationDate="2011-03-07T05:19:26.363" Score="0" ViewCount="" Body="&lt;p&gt;I think what you meant is some duality between $\Gamma(E)$ and $B$. I have never seen bottom manifold, etc defined in literature. I think there is no such relationship. Maybe you can ask some Chinese for help in English so others can understand you. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The best way to treat sections is to think them as an equivalence between the category of vector bundles over $X$ and the categrory of finitely generated projective modules over $\mathbb{C}(X)$ (complex value functions on $X$). &lt;/p&gt;&#xA;" OwnerUserId="7887" LastActivityDate="2011-03-07T05:19:26.363" />
  <row Id="25457" PostTypeId="2" ParentId="25454" CreationDate="2011-03-07T05:19:35.033" Score="0" ViewCount="" Body="&lt;p&gt;The derivative is $\pi \cos(\pi x)$.&lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2011-03-07T05:19:35.033" CommentCount="2" />
  <row Id="25458" PostTypeId="2" ParentId="25454" CreationDate="2011-03-07T05:28:42.483" Score="0" ViewCount="" Body="&lt;p&gt;Think of the function $\sin(\pi x)$ like this: $x \to \pi x \to \sin(\pi x)$. In which case it is intuitively clear that the rate of change of the function should be the multiplication of the rate of changes of $x \to \pi x$ and $y \to \sin (y)$, the latter being evaluated at $y=x$. Hence PEV's answer. &lt;/p&gt;&#xA;" OwnerUserId="4806" LastActivityDate="2011-03-07T05:28:42.483" />
  <row Id="25459" PostTypeId="2" ParentId="25454" CreationDate="2011-03-07T05:33:05.533" Score="3" ViewCount="" Body="&lt;p&gt;Given the answer, the question was for the derivative of $\sin\Bigl( (\pi x)^2\Bigr)$; instead, you computed the derivative of $\Bigl(\sin(\pi x)\Bigr)^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you were computing the derivative of the latter, then your computations are correct; the derivative is $2\pi\sin(\pi x)\cos(\pi x)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if you were asked for the derivative of $\sin\Bigl((\pi x)^2\Bigr)$, then of course you were looking at the wrong function, and that's why the answers don't match. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's possible you had &quot;$\sin(\pi x)^2$&quot; and interpreted this as $(\sin(\pi x))^2$; usually, $\sin^2(\pi x)$ is used for the latter, so &quot;$\sin(\pi x)^2$&quot; would be interpreted as $\sin\Bigl((\pi x)^2\Bigr)$.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-07T05:40:46.460" LastActivityDate="2011-03-07T05:40:46.460" CommentCount="5" />
  <row Id="25460" PostTypeId="2" ParentId="25455" CreationDate="2011-03-07T05:38:41.300" Score="13" ViewCount="" Body="&lt;p&gt;For one thing, it tells you that the functor respects colimits. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, the &quot;free group&quot; functor is the left adjoint of the &quot;underlying set&quot; functor from $\mathcal{G}roup$ to $\mathcal{S}et$. The fact that it is a left adjoint tells you that it respects colimits, so the free group of a coproduct is the coproduct of the free groups. The &quot;coproduct&quot; in $\mathcal{S}et$ is the disjoint union, and the coproduct in $\mathcal{G}roup$ is the free product: so the free group on a disjoint union $X\cup Y$, $F(X\cup Y)$, is (isomorphic to) the free product of the free groups on $X$ and $Y$, $F(X)*F(Y)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dually, right adjoints respect limits; so in the case above, the underlying set of a product of groups is the product of the underlying sets of the groups.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Added:&lt;/em&gt; Ever wondered why the underlying set of a product of topological spaces is the product of the underlying sets, &lt;strong&gt;and&lt;/strong&gt; the underlying set of a coproduct of topological spaces is &lt;em&gt;also&lt;/em&gt; the coproduct/disjoint union of the underlying sets of the topological spaces? Why the constructions in topological spaces &lt;em&gt;always&lt;/em&gt; seem to start by doing the corresponding thing to underlying sets, but in other categories like $\mathcal{G}roup$, $R-\mathcal{M}od$, only &lt;em&gt;some&lt;/em&gt; of the constructions do that? (I know &lt;em&gt;I&lt;/em&gt; did) It's because while in $\mathcal{G}roup$ the underlying set functor has a left adjoint but not a right adjoint, in $\mathcal{T}op$, the underlying set functor has &lt;em&gt;both&lt;/em&gt; a left &lt;em&gt;and&lt;/em&gt; a right adjoint (given by endowing the set with the discrete and indiscrete topologies).&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-07T05:49:55.607" LastActivityDate="2011-03-07T05:49:55.607" CommentCount="2" />
  <row Id="25461" PostTypeId="1" AcceptedAnswerId="25593" CreationDate="2011-03-07T05:40:03.230" Score="11" ViewCount="322" Body="&lt;p&gt;Hello,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $R$ be a ring and for $x,y\!\in\!R$, define the &lt;em&gt;commutator&lt;/em&gt; as $[x,y]:=xy-yx$. An $r\!\in\!R$ is &lt;em&gt;idempotent&lt;/em&gt; iff $r^2=r$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can one prove, that if every commutator is idempotent, then the whole ring is commutative, i.e. all commutators are zero?&lt;/p&gt;&#xA;" OwnerUserId="7325" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-07T05:56:59.060" LastActivityDate="2011-03-10T10:03:35.733" Title="if every commutator is idempotent, then the ring is commutative" Tags="&lt;abstract-algebra&gt;&lt;ring-theory&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="1" />
  <row Id="25462" PostTypeId="2" ParentId="25455" CreationDate="2011-03-07T05:41:32.217" Score="2" ViewCount="" Body="&lt;p&gt;According to MacLane, the slogan is, &quot;Adjoint functors arise everywhere.&quot; So there is a good reason to study them!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Having an adjoint tells you that the functor commutes with (either) limits or colimits. If a functor has a left adjoint, then it commutes with colimits, while if it has a right adjoint, it commutes with limits. For nice categories, one can sometimes conclude the &lt;a href=&quot;http://ncatlab.org/nlab/show/adjoint+functor+theorem&quot; rel=&quot;nofollow&quot;&gt;converse&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One example of this is in an abelian category. In the case of $R$-modules, for instance, the adjunction between Hom and the tensor product shows that the tensor product is right-exact (a.k.a. commutes with finite colimits). This is a somewhat more conceptual argument than the usual one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another example is the following. Let $X$ be a locally compact Hausdorff space. Then the functor $Z \mapsto Z \times X$ has an adjoint (namely, the functor $Y \mapsto Y^X$). It follows that taking products with $X$ preserves push-out diagrams, and more generally all colimits. This is useful sometimes in algebraic topology. For instance, if you have a push-out $A \cup_B C$ and homotopies $A \times I \to Z$ and $C \times I \to Z$ that agree on $B \times I$, you get a homotopy $(A \cup_B C) \times I \to Z$, the continuity of which might not be immediately obvious otherwise.&lt;/p&gt;&#xA;" OwnerUserId="536" LastActivityDate="2011-03-07T05:41:32.217" />
  <row Id="25463" PostTypeId="2" ParentId="24854" CreationDate="2011-03-07T05:44:42.407" Score="3" ViewCount="" Body="&lt;p&gt;One place where you need to care about set-theoretical issues is in category theory. Even once you get past the obvious 'problems', like the existence of the category of sheaves on a non-small category (one of the usual solutions is to assume at least one &lt;em&gt;universe&lt;/em&gt; - a set that acts like $V_\kappa$ for some inaccessible $\kappa$, and work just with sets of size bounded by the size of elements of $V_\kappa$, or perhaps enough such sets such that every set in contained in some universe), there are interesting questions that are affected by set-theory like Vopěnka's principle - see the &lt;a href=&quot;http://ncatlab.org/nlab/show/Vop%C4%9Bnka%27s+principle&quot; rel=&quot;nofollow&quot;&gt;nLab page&lt;/a&gt; for a brief discussion.&lt;/p&gt;&#xA;" OwnerUserId="3835" LastActivityDate="2011-03-07T05:44:42.407" CommentCount="1" />
  <row Id="25464" PostTypeId="1" AcceptedAnswerId="25471" CreationDate="2011-03-07T06:16:16.480" Score="2" ViewCount="73" Body="&lt;p&gt;A follow up to &lt;a href=&quot;http://math.stackexchange.com/questions/25435/relative-homology-of-a-retract&quot;&gt;a previous question of mine&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I thought it was true, but according to my book $G \simeq H \oplus K$, with $G \simeq H$ does not imply $K=0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a simple counter example? In what situation is this true?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Motivation is, from the previous question, to show that $H_n(X,A)=0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; The groups are Abelian&lt;/p&gt;&#xA;" OwnerUserId="2219" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-07T06:30:09.520" LastActivityDate="2011-03-07T06:30:09.520" Title="Direct sum with isomorphic groups" Tags="&lt;group-theory&gt;&lt;abelian-groups&gt;" AnswerCount="1" />
  <row Id="25465" PostTypeId="2" ParentId="25461" CreationDate="2011-03-07T06:18:25.353" Score="11" ViewCount="" Body="&lt;p&gt;This does not answer the general question, but a simple special case, which I find cute:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose $R$ is a finite dimensional algebra over a field of characteristic zero. Pick $x$, $y\in R$. Since $[x,y]$ is a commutator, its action on every finite dimensional $R$-module is of trace zero; on the other hand, since $[x,y]$ is by hypothesis idempotent, its trace on a module is the dimension of its image. It follows that $[x,y]$ acts by zero on every module so, because faithful modules do exist, it must be zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Later.&lt;/strong&gt; Consider the non-commutative polynomial $f(x,y)=[x,y]^2-[x,y]$, and define new polynomial $h$ by $$h(x,y)=f(x,y)-f(y,x)$$ A simple computation shows that $h(x,y)=2[x,y]$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we plug elements of $R$ into $f$ we get zero, so the same is true of $h$. The above observation means then that $2[x,y]=0$ for all $x$, $y\in R$. If now $2$ is invertible in $R$, or at least not a divisor of zero, then we see that $R$ is commutative.&lt;/p&gt;&#xA;" OwnerUserId="274" LastEditorUserId="274" LastEditorDisplayName="" LastEditDate="2011-03-07T07:49:03.107" LastActivityDate="2011-03-07T07:49:03.107" CommentCount="3" />
  <row Id="25466" PostTypeId="1" AcceptedAnswerId="25469" CreationDate="2011-03-07T06:21:48.687" Score="5" ViewCount="105" Body="&lt;p&gt;I have the following problem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $ \cot{C} = \frac{\sqrt{3}}{7} $, find $ \csc{C} $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From my trig identities, I know that $ \cot{\theta} = \frac{1}{\tan{\theta}} $, and $ \csc{\theta} = \frac{1}{\sin{\theta}} $, and also $ \cot{\theta} = \frac{\cos{\theta}}{\sin{\theta}} $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I can't seem to see how to connect the dots to get from cotangent to cosecant. I figure I might be able to use the last identity if I can somehow make $ \cos{C} = 1 $, but I don't really see how to do that, either.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is homework, so please provide me with some pointers rather than complete solutions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="3384" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-07T06:22:27.157" LastActivityDate="2011-03-07T06:30:09.907" Title="Finding $ \csc \theta $ given $ \cot \theta $" Tags="&lt;homework&gt;&lt;trigonometry&gt;" AnswerCount="3" />
  <row Id="25467" PostTypeId="2" ParentId="25251" CreationDate="2011-03-07T06:22:31.703" Score="3" ViewCount="" Body="&lt;p&gt;First, a minor point: As Mariano notes in a comment, in English one says &lt;em&gt;base manifold&lt;/em&gt;, not &lt;em&gt;bottom manifold&lt;/em&gt;.  In any case, I will let $E$ denote the total space of the fibre bundle and $B$ denote the base.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $\pi: E \to B$ be a fibre bundle, and let $\sigma:B \to E$ be a section, with image $S = \sigma(B) \subset E$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your question is, as far as I understand, the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;can one find a projection $\pi&amp;#39;: E \to S$ which makes $E$ a bundle over $S$,&#xA;so that furthermore $B$ is now realized as a section.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In some sense the answer is &quot;Yes&quot;, but for trivial reasons:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One can always find such a projection $\pi&amp;#39;$, because $\pi$ induces a diffeomorphism $S \to B$, so that abstractly $B$ and $S$ are the same space. &#xA;More concretely, define $\pi&amp;#39;: E \to S$ via $\pi&amp;#39; = \sigma\pi.$  This realizes&#xA;$E$ as a fibre bundle over $S$.  However, nothing very exciting has happened; we have just replaced the base $B$ by the diffeomorphic manifold $S$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We may now define the section $\sigma&amp;#39;: S \to E$ via $\sigma&amp;#39;(s) = s.$&#xA;The image of $\sigma&amp;#39;$ is just $S$ again, thought of as a submanifold of $E$,&#xA;but if you want we can identify this with $B$ via the projection $\pi$.&#xA;Again, this is not very exciting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My sense is that you are thinking of something a little deeper than this, but I'm not sure what.  One problem in general is that the base $B$ and a section&#xA;$S$ have a different nature: $S$ maps into $E$, while $E$ maps onto $B$ (so $B$ is not really &quot;in&quot; the fibre bundle at all).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps the vector bundle case is the one you want to focus on: then&#xA;$B$ can be made into a section via the zero section.  But I still don't see&#xA;what you are going to do with this.  (One thing you can do is that you can intersect $B$ and $S$, and this is how one produces characteristic classes geometrically; is that of any interest to you at all?)&lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2011-03-07T06:22:31.703" CommentCount="1" />
  <row Id="25468" PostTypeId="2" ParentId="25466" CreationDate="2011-03-07T06:23:17.377" Score="3" ViewCount="" Body="&lt;p&gt;From&#xA;$\sin^2\theta + \cos^2\theta = 1$, &#xA;divide through by $\sin^2\theta$ to get a relation between $\cot^2\theta$ and $\csc^2\theta$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S. The information given is not enough, though, to determine the value of $\csc\theta$ unless you happen to know which quadrant you are working in; you know that you are in either quadrant I or III, since the cotangent is positive; but that does not tell you whether the cosecant is positive or negative; you'll have two possible answers. This is pretty much the same situation as how, if you know that $\sin\theta=\frac{1}{2}$, this only determines $\cos\theta$ up to sign.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2011-03-07T06:23:17.377" CommentCount="2" />
  <row Id="25469" PostTypeId="2" ParentId="25466" CreationDate="2011-03-07T06:24:17.523" Score="4" ViewCount="" Body="&lt;p&gt;In &lt;a href=&quot;http://math.stackexchange.com/questions/15514/how-can-i-find-the-derivative-of-y-sin-arctan-x-tan-arcsin-x/15517#15517&quot;&gt;my answer here&lt;/a&gt;, I describe some general ideas for how to use one trig function of an angle to determine another trig function of the same angle.  The basic idea is to draw triangle(s) on a coordinate system that correspond to the given trig function and use those to compute the other trig function.&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2011-03-07T06:24:17.523" CommentCount="2" />
  <row Id="25471" PostTypeId="2" ParentId="25464" CreationDate="2011-03-07T06:28:00.853" Score="6" ViewCount="" Body="&lt;p&gt;This will be true if the groups are finitely generated, or if they are finite dimensional vector spaces over a field (with the isomorphism being a vector space isomorphism), or in other contexts in which one can appeal to some kind of finite invariant like dimension, free rank, size of the torsion subgroup, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a counterexample, consider $G = \mathbb Z \oplus \mathbb Z \oplus \cdots $ (countably infinite direct sum).  (Exercise: find a suitable choice of $K$.)&lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2011-03-07T06:28:00.853" CommentCount="3" />
  <row Id="25472" PostTypeId="2" ParentId="25466" CreationDate="2011-03-07T06:30:09.907" Score="3" ViewCount="" Body="&lt;p&gt;Here's a simple way I usually think about it. Suppose you have a right triangle in the first quadrant. Since $\cot C=\frac{\sqrt{3}}{7}$, you know the ratio of the leg adjacent to $C$ to the leg opposite of $C$ is $\frac{\sqrt{3}}{7}$. So let's just say the opposite leg has length $7$ and the adjacent leg has length $\sqrt{3}$. Then by the Pythagorean theorem, the hypotenuse has length $\sqrt{52}=2\sqrt{13}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now $\csc C$ is just the ratio of the hypotenuse to the opposite leg, essentially the inverse of $\sin$, which is the ratio of the opposite leg to the hypotenuse. &lt;/p&gt;&#xA;" OwnerUserId="1583" LastActivityDate="2011-03-07T06:30:09.907" />
  <row Id="25473" PostTypeId="2" ParentId="25455" CreationDate="2011-03-07T06:38:08.760" Score="1" ViewCount="" Body="&lt;p&gt;The perspective on adjoint functors that I have found helpful (though one that I don't know how useful it will be in the long run) is that adjoint functors correspond to certain universal constructions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a concrete example, then for any set $S$ we can characterize the free $R$-module $FS$ on the set $S$ (together with inclusion map $i\colon S\to FS$) up to isomorphism by the obvious universal property: if $f\colon S\to N$ is any set-function from the set $S$ into the module $N$, then there exists a unique module homomorphism $\phi\colon FS\to N$ such that $\phi\circ i=f$, i.e. a unique module homomorphism determined by the image of the basis elements $S$ of $FS$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, this is sloppy since what we're doing is mixing categories. To untangle the mess, we introduce the forgetful functor $G:R$-$Mod\to Sets$ which associates to every module over $R$ its underlying set, and to every module homomorphism its underlying set-function). Then we can express the above universal property more cleanly by saying that for a module $FS$ together with an inclusion set-map $i\colon S\to GFS$ (that lives in $Sets$) is the free module on the set $S$ if for every set-map $f\colon S\to GN$ where $N$ is an $R$-module, there exists a unique module homomorphism $\phi\colon FS\to N$ such that $G\phi\circ i=f$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively, this can be expressed by defining for every set $S$ the comma category $(S\downarrow G)$ whose objects are pairs $(f,N)$ where $N$ is an $R$-module and $f\colon S\to GN$ is a set-function, and whose morphisms between $(f_1,M)$ and $f(f_2,N)$ are given by morphisms $\phi\colon M\to N$ such that $f_2=G\phi\circ f_1$, i.e. morphisms that make the appropriate triangle commute. Then we can say that $(i,FS)$ is the free module on the set $S$ if it is the initial object in the comma category $(S\downarrow G)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It turns out that if there exists an initial object in $(S\downarrow G)$ for every $S$, then any choice of initial objects $(\epsilon_S, FS)$ in each such category $(S\downarrow G)$ allows us to extend the association $S\to FS$ to a functor, and that functor will be (up to natural equivalence) the left adjoint to $G$ (this makes $\epsilon\colon I\to GF$ the unit). In other words, that $G$ has a left adjoint means that it encodes a(n initial) universal property that holds for every object in the codomain category of $G$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Showing the equivalence of the three definitions of adjoint functors on the wikipedia article (http://en.wikipedia.org/wiki/Adjoint_functors#Formal_definitions) I've found to be a tedious but highly rewarding exercise (much moreso than that of checking whether particular pairs of functors are adjoint or not).&lt;/p&gt;&#xA;" OwnerUserId="400" LastActivityDate="2011-03-07T06:38:08.760" />
  <row Id="25474" PostTypeId="1" CreationDate="2011-03-07T06:42:53.950" Score="3" ViewCount="109" Body="&lt;p&gt;Hi, can anyone help me to inform the name of any book where I can get the following theorem, or give some detailed hint to solve this one:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $X$ and $Y$ be two random variables, if the moment generating functions of $X$ and $Y$ are equal then the probability distributions of $X$ and $Y$ will be same. In other words, if $E(e^{tX})=E(e^{tY})$ for all $t$ real, then $X$ and $Y$ have the same distribution.&lt;/p&gt;&#xA;" OwnerUserId="6990" LastEditorUserId="1543" LastEditorDisplayName="" LastEditDate="2011-03-07T09:35:01.477" LastActivityDate="2011-03-07T09:45:12.337" Title="moment generating function" Tags="&lt;probability&gt;&lt;reference-request&gt;" AnswerCount="4" CommentCount="1" />
  <row Id="25475" PostTypeId="1" AcceptedAnswerId="25481" CreationDate="2011-03-07T06:43:57.010" Score="4" ViewCount="112" Body="&lt;p&gt;Any natural number can be expressed as the sum of three triangular numbers, or as four square numbers.  The natural analog for expressing numbers as the sum of two others would apparently be the sum of two &quot;linear&quot; numbers, but all natural numbers are &quot;linear&quot;, so this is rather unsatisfying.  Is there a well-known sparser set of integers (or, half-integers, for that matter) that has this property?&lt;/p&gt;&#xA;" OwnerUserId="7882" LastActivityDate="2011-03-07T16:12:47.900" Title="Natural set to express any natural number as sum of two in the set." Tags="&lt;combinatorics&gt;&lt;elementary-number-theory&gt;&lt;recreational-mathematics&gt;&lt;partitions&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="25476" PostTypeId="2" ParentId="24854" CreationDate="2011-03-07T06:45:58.300" Score="2" ViewCount="" Body="&lt;p&gt;I'll take a stab at this part:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;How does set theory interact with the rest of mathematics?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Henno Brandsma mentioned that set theory is useful in topology. Another area, which has quite a few analogies to topology, is measure theory, and by extension, probability theory. We are very interested in, for example, infinite product spaces, because many useful probabilistic models have infinitely many random variables. But we can't just summon them into existence without some care. To do so in first-order logic requires the axiom of choice (or something like it). But then non-measurable sets exist, so every set of outcomes must be accompanied by a $\sigma$-algebra of sets for which measurement (probability or volume) makes sense...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Further, much like in topology, our basic definitions rely quite heavily on &quot;set-construction&quot; axioms like comprehension, union and powerset. For example, the common operation &quot;take the coarsest $\sigma$-algebra $\mathcal{A}$ such that $\phi(\mathcal{A})$&quot; appeals directly to those axioms: powerset twice, then comprehension, and then union (with one indirection, from an intersection of uncountably many $\sigma$-algebras).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So probability theory is intricately tied with the axiom of choice, in such a way that anybody doing serious work in it is always aware of it. And we use the axioms directly, even if we often forget that that's what we're doing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;I've even gotten the feeling from several professional mathematicians I've talked to that set theory is no longer relevant, and that if someone were to find some set-theoretic flaw in their axioms (a non-standard model or somesuch), they would just ignore it and try again with different axioms.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What else would we do but try again? :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The real question is: how easy would it be? Much of analysis would be fine; the reals seem to have the same properties no matter what theory you build them in. Measure theory and probability theory, however, which are so tied to set theory, would (I think) come crashing down - everything would be suspect, and would have to be rebuilt almost from scratch. What a terrifying and exciting idea!&lt;/p&gt;&#xA;" OwnerUserId="2122" LastActivityDate="2011-03-07T06:45:58.300" CommentCount="2" />
  <row Id="25477" PostTypeId="2" ParentId="25329" CreationDate="2011-03-07T06:52:17.510" Score="1" ViewCount="" Body="&lt;p&gt;Just another mental approach. We write&#xA; $$8\left(\frac9{10}\right)^8 &amp;gt;1 \rightarrow \frac9{10}&amp;gt;\frac18^{\frac18}=\left(1-\frac78\right)^{\frac18} $$&#xA;Then if one remembers that $ (1-x)^{\frac1a} = 1 - \frac1a x - O\left(\frac{x^2}a\right) $ then we get&#xA;$$\frac9{10}&amp;gt; 1-\frac7{64}-\epsilon \rightarrow 9&amp;gt;10-\frac{70}{64}-10\epsilon  $$ which is obviously true. It just depends on whether you remember the beginning of the binomial-formula for fractional exponents.&lt;/p&gt;&#xA;" OwnerUserId="1714" LastActivityDate="2011-03-07T06:52:17.510" />
  <row Id="25478" PostTypeId="1" CreationDate="2011-03-07T06:57:08.757" Score="3" ViewCount="81" Body="&lt;p&gt;Customers arrive at a casino as a Poisson process of rate 100 customers per hour. Upon&#xA;arriving, each customer must flip a coin, and only those customers who flips heads actually enter the casino. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So how can I find&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;the PMF of N? Let's say for the&#xA;number of customers who arrive&#xA;between 5PM to 7PM. &lt;/li&gt;&#xA;&lt;li&gt;the pdf of T? The    time between&#xA;successive customers    arriving.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="7900" LastActivityDate="2011-03-07T09:59:10.620" Title="Poisson Distritubtion" Tags="&lt;calculus&gt;&lt;stochastic-processes&gt;&lt;probability-distributions&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" />
  <row Id="25479" PostTypeId="2" ParentId="25478" CreationDate="2011-03-07T07:08:27.230" Score="0" ViewCount="" Body="&lt;p&gt;Note that the pdf of $T$ is an exponential distribution. &lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2011-03-07T07:08:27.230" />
  <row Id="25480" PostTypeId="2" ParentId="25474" CreationDate="2011-03-07T07:19:27.310" Score="2" ViewCount="" Body="&lt;p&gt;One can look at the contrapositive: $$F_{X}(x) \neq F_{Y}(y) \implies M_{X}(t) \neq M_{Y}(t)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think it would be easier to look at characteristic functions (i.e. $\varphi_{X}(t) = E[e^{it}X]$).&lt;/p&gt;&#xA;" OwnerUserId="2548" LastEditorUserId="2548" LastEditorDisplayName="" LastEditDate="2011-03-07T07:35:35.447" LastActivityDate="2011-03-07T07:35:35.447" CommentCount="2" />
  <row Id="25481" PostTypeId="2" ParentId="25475" CreationDate="2011-03-07T07:40:21.180" Score="3" ViewCount="" Body="&lt;p&gt;Assuming Goldbach's conjecture, you can take the set of all primes and successors of primes (plus some small numbers). These have density $2/\log n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Not assuming the conjecture, you can take primes, almost-primes and successors of primes (plus some small numbers) - this is the famous Chen's theorem. The resulting density is $\log\log n/\log n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another suggestion is to take the set of all numbers whose binary expansion is &quot;supported&quot; on only odd powers of $2$ or only even powers of $2$. The resulting density is roughly $2/\sqrt{n}$, so this is close to optimal (you need at least $1/\sqrt{n}$).&lt;/p&gt;&#xA;" OwnerUserId="1277" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2011-03-07T16:12:47.900" LastActivityDate="2011-03-07T16:12:47.900" CommentCount="2" />
  <row Id="25482" PostTypeId="2" ParentId="25475" CreationDate="2011-03-07T07:42:55.797" Score="0" ViewCount="" Body="&lt;p&gt;This math overflow post may be of some interest: &lt;a href=&quot;http://mathoverflow.net/questions/9731/polynomial-representing-all-nonnegative-integers&quot; rel=&quot;nofollow&quot;&gt;http://mathoverflow.net/questions/9731/polynomial-representing-all-nonnegative-integers&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="6075" LastEditorUserId="6075" LastEditorDisplayName="" LastEditDate="2011-03-07T07:50:02.853" LastActivityDate="2011-03-07T07:50:02.853" CommunityOwnedDate="2011-03-07T07:50:18.190" />
  <row Id="25483" PostTypeId="2" ParentId="25478" CreationDate="2011-03-07T07:43:37.747" Score="1" ViewCount="" Body="&lt;p&gt;The key observation is that the new process, denote it by $N = \lbrace N(t) : t \geq 0 \rbrace$, has rate $\lambda$ which is half the original, that is $\lambda=50$.&#xA;Then the answers to your questions follow immediately. Denote the original process by $N_0$, and set $\lambda_0 = 100$.&#xA;Then, the distribution of $N(1)$ (that is, the number of customers who actually enter the casino in one hour) is obtained (using the law of total probability) as follows.&#xA;$$&#xD;&#xA;{\rm P}(N(1) = k) = \sum\limits_{j = k}^\infty  {{\rm P}(N(1) = k|N_0 (1) = j) {\rm P}(N_0 (1) = j)}.&#xD;&#xA;$$&#xA;Given $N_0 (1) = j$, $N(1)$ is binomial$(j,1/2)$. Hence,&#xA;$$&#xD;&#xA;{\rm P}(N (1) = k) = \sum\limits_{j = k}^\infty  {\frac{{e^{ - \lambda _0 } \lambda _0^j }}{{j!}}{j \choose k}\frac{1}{{2^j }}} = \cdots = \frac{{e^{ - \lambda _0 /2} (\lambda _0 /2)^k }}{{k!}}.&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;: For a proof that the new process $N$ indeed has stationary independent increments, see, for example, p. 211 &lt;a href=&quot;http://www.mast.queensu.ca/~stat455/lecturenotes/set4.pdf&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; (where $\lambda$ and $p$ correspond to $100$ and $1/2$, respectively).&lt;/p&gt;&#xA;" OwnerUserId="2810" LastEditorUserId="2810" LastEditorDisplayName="" LastEditDate="2011-03-07T09:59:10.620" LastActivityDate="2011-03-07T09:59:10.620" CommentCount="1" />
  <row Id="25484" PostTypeId="2" ParentId="25474" CreationDate="2011-03-07T07:48:55.057" Score="2" ViewCount="" Body="&lt;p&gt;An even stronger result is &lt;a href=&quot;http://en.wikipedia.org/wiki/L%C3%A9vy%27s_continuity_theorem&quot; rel=&quot;nofollow&quot;&gt;Levy's continuity theorem&lt;/a&gt;. Your question is easier, look it up &lt;a href=&quot;http://en.wikipedia.org/wiki/Characteristic_function_%28probability_theory%29#Inversion_formulas&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;. If your variable is continuous, then the characteristic function is the Fourier transform of the probability density function, so all you need to do is run the inverse Fourier transform. For the general case, see the reference.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2011-03-07T07:48:55.057" />
  <row Id="25485" PostTypeId="1" AcceptedAnswerId="25486" CreationDate="2011-03-07T07:50:32.557" Score="7" ViewCount="168" Body="&lt;p&gt;In elementary number theory, I find the following notations being used interchangeably, which leads me to have many ambiguous assumptions:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$\mathbb{Z}_p^\times$;  &lt;/li&gt;&#xA;&lt;li&gt;$\mathbb{Z}_p$,  &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;where &lt;em&gt;p&lt;/em&gt; is any integer. What's the difference between them?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One more question to add to the kitty: what do they mean when they say &quot;a subgroup of $\mathbb{Z}_p$, where &lt;em&gt;p&lt;/em&gt; is prime&quot;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It will be nice if someone can enlighten me on this elementary notation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="7488" LastEditorUserId="10557" LastEditorDisplayName="" LastEditDate="2011-05-09T01:43:25.807" LastActivityDate="2011-05-09T01:43:25.807" Title="Number Theory: confusion with notation" Tags="&lt;elementary-number-theory&gt;&lt;notation&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="25486" PostTypeId="2" ParentId="25485" CreationDate="2011-03-07T07:57:57.173" Score="6" ViewCount="" Body="&lt;p&gt;Usually the notation $\mathbb{Z}_p$ or $\mathbb{Z}/p\mathbb{Z}$ mean the integers modulo $p$, that is $\{0,\ldots,p-1\}$ where you add and multiply as usually and then take the reminder modulo $p$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The notation of $\mathbb{Z}_p^\times$ is for those numbers which have a multiplicative inverse modulo $p$, namely all $n$ such that exists $m$ such that $n\cdot m$ is $1$ modulo $p$. These are the numbers coprime to $p$, the greatest common divisor of them and $p$ is $1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A subgroup of $\mathbb{Z}_p^\times$ is a subset of these numbers which is closed under multiplication (but not necessarily addition), and every number in this subset also has its multiplicative inverse in there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\mathbb{Z}_5 = \{0,1,2,3,4\}$&lt;br&gt;&#xA;$\mathbb{Z}_5^\times = \{1,2,3,4\}$&lt;br&gt;&#xA;$\{1,4\}$ is a subgroup of $\mathbb{Z}_5^\times$. Can you see why?&lt;/p&gt;&#xA;" OwnerUserId="622" LastEditorUserId="622" LastEditorDisplayName="" LastEditDate="2011-03-07T08:26:54.230" LastActivityDate="2011-03-07T08:26:54.230" CommentCount="5" />
  <row Id="25488" PostTypeId="2" ParentId="4422" CreationDate="2011-03-07T08:03:22.480" Score="0" ViewCount="" Body="&lt;p&gt;Hm, I would have suggested Stein+Shakarchi as well, but alright.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You say you want a book that's to the point, and you've already gone through Baby Rudin... Did you like Baby Rudin?  If so, you might consider Big Rudin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The relevant chapters are Chapters 4, 5, 9, which discuss Hilbert spaces (hence Fourier series), Banach spaces, and the Fourier transform, respectively.  Admittedly, Chapters 1-3 focus on the Lebesgue integral and $L^p$ spaces, but I'm not sure how much of that you'll actually need.  (Well, alright, you might need the most basic properties of $L^p$, but that shouldn't be a problem.)  My only warning would be that Big Rudin is, of course, somewhat terse and rather abstract.&lt;/p&gt;&#xA;" OwnerUserId="640" LastActivityDate="2011-03-07T08:03:22.480" CommunityOwnedDate="2011-03-07T08:03:22.480" />
  <row Id="25489" PostTypeId="2" ParentId="25355" CreationDate="2011-03-07T08:04:33.740" Score="4" ViewCount="" Body="&lt;p&gt;As it is not clear whether to have the larger or small $\gamma$ expansion of the integral $$I(\gamma)= \int^{\pi}_0\frac{\gamma^{10} \theta^2 \sin\theta}{(\gamma^2 \theta^2 + 1)^5}  d\theta,$$ I will provide both.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For large $\gamma$, we observe that the integral is dominated for $\theta \approx \gamma^{-1}$. That is why can hope that we can expand the $\sin$ function. We will show later that this is really admissible when we see that successive terms are smaller when $\gamma\to\infty$ (the expansion will be an &lt;a href=&quot;http://en.wikipedia.org/wiki/Asymptotic_expansion&quot; rel=&quot;nofollow&quot;&gt;asymptotic expansion&lt;/a&gt;). The substitution $x= \gamma\theta$ yields $$I(\gamma) = \gamma^{7} \int_0^{\pi \gamma}  \frac{x^2 \sin(x/\gamma)}{(1+x^2)^5} dx \sim \gamma^{6} \int_0^\infty \frac{x^3}{(1+x^2)^5} dx + O(\gamma^5).$$&#xA;Here, we have used that every term in the expansion of $\sin$ gives an additional $\gamma^{-1}$ and that the integral is dominated for $x\approx 1$ such that we could put the integration boundary to $\infty$. The remaining integral is a number which can be evaluated to $1/24$. In total we have $$I(\gamma) \sim \frac{\gamma^6}{24}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For small $\gamma$, we can obtain a converging series by expanding the integrand in a  Taylor series. The first term in the expansion reads&#xA;$$I(\gamma) = \gamma^{10} \int_0^\pi \theta^2 \sin \theta d\theta + O(\gamma^{11}) = (\pi^2 -4) \gamma^{10} + O(\gamma^{11}).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Knowing the two asymptotics, we (more or less) can draw a plot of the function. It starts off as $(\pi^2 -4) \gamma^{10}$ and then turns over into $\gamma^6/24$. The crossover region is for $\gamma \approx 1$. In this spirit, I expect $$I(\gamma) \approx \frac{\gamma^{10}}{(\pi^2-4)^{-1} + 24 \gamma^4}$$ to be a reasonable good approximation to the initial integral.&lt;/p&gt;&#xA;" OwnerUserId="7266" LastActivityDate="2011-03-07T08:04:33.740" />
  <row Id="25490" PostTypeId="2" ParentId="25474" CreationDate="2011-03-07T08:11:45.147" Score="2" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://www.dartmouth.edu/~chance/teaching_aids/books_articles/probability_book/Chapter10.pdf&quot; rel=&quot;nofollow&quot;&gt;This chapter&lt;/a&gt; (which I found immediately by googling for 'moment generating function proof') contains a proof in the discrete case and a sketch of a proof in the continuous case. It also says that the theorem holds only under the hypothesis that the random variables have finite range.&lt;/p&gt;&#xA;" OwnerUserId="6622" LastActivityDate="2011-03-07T08:11:45.147" CommentCount="1" />
  <row Id="25491" PostTypeId="1" CreationDate="2011-03-07T08:34:48.947" Score="2" ViewCount="66" Body="&lt;p&gt;Hello all,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a computational problem that I am solving that ultimately relies on either the areas (for $\mathbb{R}^2$) or the volumes (for $\mathbb{R}^3$ and above) for a pair of cropped objects.  We can think of these cropped objects as closed, concave (but sometimes convex), orientable polytopes with finite surface area and no self-intersections that may have holes in their interior.  In the case where the polytopes are in $\mathbb{R}^2$ or $\mathbb{R}^3$, one can use Green's theorem and Gauss' divergence theorem, respectively, to arrive at a closed-form solution to the contour integral.  For $\mathbb{R}^d$, it is my guess that Stokes' theorem could be applied.  Is this correct?  For instance, when $d = 4$ and assuming no holes, would I write $$\int\int\int\int_V (\nabla \cdot {\mathbf F})dV = \oint\oint\oint_{\partial V} ({\mathbf F} \cdot {\mathbf n})dS$$&#xA;where $\mathbf{F}$ is a suitably chosen, continuously differentiable vector field defined on $V$ and $\mathbf{n}$ is the normal for the closed surface boundary $\partial V$?&lt;/p&gt;&#xA;" OwnerUserId="7901" LastEditorUserId="7901" LastEditorDisplayName="" LastEditDate="2011-03-07T19:58:36.260" LastActivityDate="2011-04-11T00:39:59.580" Title="(Closed, Concave, Orientable) Polytope Volume" Tags="&lt;calculus&gt;&lt;geometry&gt;" AnswerCount="1" />
  <row Id="25492" PostTypeId="1" CreationDate="2011-03-07T09:23:21.257" Score="1" ViewCount="86" Body="&lt;p&gt;How can I compute of the $2\pi$-periodic function whose restriction to the range $[-\pi,\pi]$ is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \begin{cases} 1 &amp;amp;&amp;amp; 0 &amp;lt; x &amp;lt; \pi \\ 0 &amp;amp;&amp;amp; -\pi &amp;lt; x &amp;lt; 0 \end{cases} $$&lt;/p&gt;&#xA;" OwnerUserId="7663" LastEditorUserId="1543" LastEditorDisplayName="" LastEditDate="2011-04-06T12:44:16.960" LastActivityDate="2011-04-06T12:44:16.960" Title="Compute a Fourier Transform" Tags="&lt;homework&gt;&lt;fourier-analysis&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" />
  <row Id="25493" PostTypeId="1" AcceptedAnswerId="25778" CreationDate="2011-03-07T09:23:43.597" Score="6" ViewCount="82" Body="&lt;p&gt;Greetings,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've been reading Maclane's &quot;Homology&quot; and ran into the following question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $(R,S)$ be a resolvent pair of ring, i.e $R$ is an $S$-algebra and we have a functor $\Psi \colon \operatorname{R-Mod} \to \operatorname{S-Mod}$, that is additive exact and faithful. We also have an left adjoint functor of $\Psi$, namely $F \colon \operatorname{S-Mod} \to \operatorname{R-Mod}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One defines then relative $Ext_{(R,S)}$ functor, using the bar-resolution. Can you please help me find a concrete example of a case when $Ext^1_{(R,S)} \neq Ext^1_R$ (where $Ext^1_R$ is the regular $Ext$ functor).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My thought on the matter. One can identify $Ext^1_{(R,S)}(A,B)$ with the set of extensions of $B$ by $A$, that are $S$-split. One must then find an extension in $R$ that is not $S$-split to do the trick. But this argument feels a bit like cheating.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any help will be appreciated. &lt;/p&gt;&#xA;" OwnerUserId="7838" LastEditorUserId="7838" LastEditorDisplayName="" LastEditDate="2011-03-08T14:50:18.013" LastActivityDate="2011-03-08T16:38:31.197" Title="Example of relative Ext functor" Tags="&lt;category-theory&gt;&lt;homological-algebra&gt;&lt;abelian-categories&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="25494" PostTypeId="2" ParentId="25474" CreationDate="2011-03-07T09:24:01.523" Score="2" ViewCount="" Body="&lt;p&gt;The idea is to reduce the problem to the well-known case of characteristic functions (cf. Yuval's or PEV's answer). For a complete proof (for moment-generating functions), see Theorem 4.9 &lt;a href=&quot;http://books.google.co.il/books?id=-l5nOfQf_CAC&amp;amp;pg=PA108&amp;amp;lpg=PA108&amp;amp;dq=%22moment+generating+function+and+characteristic+functions%22&amp;amp;source=bl&amp;amp;ots=BHUgN7AAck&amp;amp;sig=gBpZB31nbP6Qj_pofDP3eUP4hiY&amp;amp;hl=iw#v=onepage&amp;amp;q=%22moment%20generating%20function%20and%20characteristic%20functions%22&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;. Further, see Theorem 8.1 &lt;a href=&quot;http://books.google.co.il/books?id=CnBxaLaOcBAC&amp;amp;pg=PA189&amp;amp;lpg=PA189&amp;amp;dq=uniqueness+theorem+moment-generating+functions&amp;amp;source=bl&amp;amp;ots=bEN3LZzQaU&amp;amp;sig=X4YlLtmZFRQYNmpDvWHSYiHGqbw&amp;amp;hl=iw#v=onepage&amp;amp;q=uniqueness%20theorem%20moment-generating%20functions&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;, and property (M1) &lt;a href=&quot;http://books.google.co.il/books?id=_mayRBczVRwC&amp;amp;pg=PA260&amp;amp;lpg=PA260&amp;amp;dq=uniqueness+theorem+moment+generating+functions+characteristic+functions+m1&amp;amp;source=bl&amp;amp;ots=J8xrl-pELv&amp;amp;sig=LmoIGUHjkMuu-_CYWTWrNyGyxTY&amp;amp;hl=iw#v=onepage&amp;amp;q=uniqueness%20theorem%20moment%20generating%20functions%20characteristic%20functions%20m1&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2810" LastEditorUserId="2810" LastEditorDisplayName="" LastEditDate="2011-03-07T09:45:12.337" LastActivityDate="2011-03-07T09:45:12.337" />
  <row Id="25495" PostTypeId="1" AcceptedAnswerId="25498" CreationDate="2011-03-07T09:42:51.233" Score="6" ViewCount="117" Body="&lt;p&gt;I read a little proposition that any quantity obtainable from rational numbers by a finite number of operations $+$, $-$, $\cdot$, $\div$, $\sqrt{{}}$ can be put in a standard form $\frac{p}{q}\cdot A$, where $\frac{p}{q}\in\mathbb{Q}$ and $A$ is an expression of only integers and $+$, $-$, $\cdot$, and $\sqrt{{}}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't think it's difficult to believe this is the case, but I don't know of a nice, clear way to prove it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried to simplify the argument first by saying that any operation $+$, $-$, $\cdot$, $\div$ can be rewritten entirely in terms of $+$, since $-$ is the addition of a the additive inverse, $\cdot$ is repeated addition, and $\div$ is multiplication of the reciprocal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also decided that given a radicand consisting of rationals and operations $+$, $-$, $\cdot$, $\div$, you could then rewrite the radicand as a sum of rationals, and multiply the radical by $\frac{\sqrt{\Delta^2}}{\Delta}$, where $\Delta$ is some multiple of the denominators. This would put the radicand in standard form, with a fraction $\frac{1}{\Delta}$ outside the radical. For example,&#xA;$$&#xD;&#xA;\sqrt{\frac{a}{b}+\frac{c}{d}}=\sqrt{\frac{ad+cb}{bd}}=\frac{\sqrt{(bd)^2}}{bd}\sqrt{\frac{ad+cb}{bd}}=\frac{1}{bd}\sqrt{(bd)(ad+cb)}.&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suppose for nested radicals, the process could be repeated. Then any sequence of these operations can be put into a sum of terms with rational denominators by rationalizing any denominator, and then one could do the same procedure of pulling a common multiple of all the terms out in front to put the expression in standard form. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I wanted to formalize this idea, but it seems like there are so many different things to consider, that I don't know how to go about it. Is there a way to make basic induction work, or perhaps some other method? Thanks.&lt;/p&gt;&#xA;" OwnerUserId="1583" LastEditorUserId="1583" LastEditorDisplayName="" LastEditDate="2011-03-07T09:52:27.273" LastActivityDate="2011-03-08T06:03:54.360" Title="Is there a nice way to show that any finite sequence of arithmetic operations on the rationals can be put in 'standard form'?" Tags="&lt;arithmetic&gt;&lt;induction&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="25496" PostTypeId="2" ParentId="4422" CreationDate="2011-03-07T09:46:12.397" Score="4" ViewCount="" Body="&lt;p&gt;You mentioned graduate level. So you really should &lt;em&gt;first&lt;/em&gt; learn Lebesgue integration. (Stein and Shakarchi volume 3 is not bad, as are many of the usual suspects -- Big Rudin and Royden's book on measure theory, just to name a couple.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then I would recommend any/all of the following:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Stein and Weiss, &lt;em&gt;Introduction to Fourier Analysis on Euclidean Spaces&lt;/em&gt; (after that you may also be interested in Stein's &lt;em&gt;Singular Integrals and Differentiability Properties of Functions&lt;/em&gt; and &lt;em&gt;Harmonic Analysis&lt;/em&gt;)&lt;/li&gt;&#xA;&lt;li&gt;Grafakos, &lt;em&gt;Classical and Modern Fourier Analysis&lt;/em&gt; (which has been republished in the GTM series as two separate books; you should start with the &lt;em&gt;Classical Fourier Analysis&lt;/em&gt; volume).&lt;/li&gt;&#xA;&lt;li&gt;Sogge, &lt;em&gt;Fourier Integrals in Classical Analysis&lt;/em&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;For one aspect of how Fourier analysis fits into the broader framework of analysis, I also recommend studying some distribution theory, and theory of partial/pseudo/para-differential operators. Some interesting texts in that regard include:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Friendlander and Joschi, &lt;em&gt;Introduction to the theory of distributions&lt;/em&gt;&lt;/li&gt;&#xA;&lt;li&gt;Hörmander, &lt;em&gt;Analysis of Linear Partial Differential Operators&lt;/em&gt;, volumes 1-4 (the first volume includes a quick &quot;review&quot; of the parts of Fourier analysis used; I put the word in quotes because, well, it is Hörmander...)&lt;/li&gt;&#xA;&lt;li&gt;Alinhac and Gérard, &lt;em&gt;Pseudo-differential Operators and the Nash-Moser Theorem&lt;/em&gt; (and if you read French, you should consider looking at the French original)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1543" LastEditorUserId="1543" LastEditorDisplayName="" LastEditDate="2011-03-07T09:54:41.060" LastActivityDate="2011-03-07T09:54:41.060" CommunityOwnedDate="2011-03-07T09:46:12.397" CommentCount="3" />
  <row Id="25497" PostTypeId="2" ParentId="25427" CreationDate="2011-03-07T10:15:40.787" Score="1" ViewCount="" Body="&lt;p&gt;The set of solutions of $x^n-1$ are all elements of order $d$ with $d|n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $f_d(x)$ is just the elements of order $d$ then $f_n(x) = \frac{x^n-1}{\prod_{d|n, d \not = n}f_d(x)}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rearranging to get the formula which you can apply Mobius inversion to. By the way, you don't have to use logarithms because this is just the normal mobius inversion written in multiplicative notation instead of additive.&lt;/p&gt;&#xA;" OwnerUserId="4290" LastActivityDate="2011-03-07T10:15:40.787" />
  <row Id="25498" PostTypeId="2" ParentId="25495" CreationDate="2011-03-07T10:32:08.117" Score="8" ViewCount="" Body="&lt;p&gt;This sort of thing is proved by induction over the number of operations by treating the &quot;outermost&quot; operation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Case $+$: $\left(\frac{p}{q}A\right)+\left(\frac{r}{s}B\right)=\frac{1}{qs}(psA+qrB)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Case $-$: $\left(\frac{p}{q}A\right)-\left(\frac{r}{s}B\right)=\frac{1}{qs}(psA-qrB)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Case $\cdot$: $\left(\frac{p}{q}A\right)\cdot\left(\frac{r}{s}B\right)=\frac{pr}{qs}(AB)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Case $\div$: $\left(\frac{p}{q}A\right)\div\left(\frac{r}{s}B\right)=\frac{ps}{qr}\frac{A}{B}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Case $\sqrt{}$: $\sqrt{\frac{p}{q}A}=\frac{1}{q}\left(\sqrt{pqA}\right)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the first version of this answer, I claimed: &quot;Now all that remains to be shown is that you can deal with the $B$ in the denominator in the $\div$ case, and this you can show with the same kind of induction by cases over the number of operations in $B$.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It turns out this was overly optimistic, and actually the more subtle part of the proof lies in dealing with that case. So here's a proper proof:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The proof is by induction over $(n,m)$, where $m$ is the number of operations in $B$ and $n$ is the number of different radicands that appear in $B$. For given $(n,m)$, assume that we can move all roots to the numerator if the denominator has $(n&amp;#39;,m&amp;#39;) &amp;lt; (n,m)$ in lexicographical order, i.e. either $n&amp;#39; &amp;lt; n$, or $n&amp;#39;=n$ and $m&amp;#39;&amp;lt;m$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To simplify things, let's replace each &quot;$-$&quot; by &quot;$+(-1)\cdot$&quot;, so only the operations $+$, $\cdot$ and $\sqrt{}$ remain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Case $\sqrt{}$: This is easy:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{1}{B}=\frac{1}{\sqrt{C}}=\frac{\sqrt{C}}{C}\;,$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and we've reduced the denominator's $n$ by $1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Case $\cdot$: This is also easy:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{1}{B}=\frac{1}{C\cdot D}=\frac{1}{C}\frac{1}{D}\;,$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and we can apply the induction hypothesis separately to $C$ and $D$ (both of which have lower $m$ than $B$ and at most equal $n$) to move all roots in both fractions to the numerator.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Case $+$: This is the harder part. We multiply out any products of sums that are not inside a root. This may increase $m$ but leaves $n$ unchanged, and it yields a sum of products of one or more factors, each of which is an integer or a root. If the same root appears twice in a product, we replace the two occurrences by the radicand (which doesn't increase $n$) and then again multiply out all products of sums, iterating until there are no more roots occurring twice in the same term of the outer sum. (This process must terminate since each radicand inserted when resolving a duplicate root contains fewer operations than the root, and only roots in this radicand can lead to new duplicates.) Now we choose a root among the roots in the products that contains the greatest number of operations (picking one arbitrarily if there are several with the same number of operations), and we factor it out of any terms in which it occurs. Since it occurs at most once in each term (since we resolved double occurrences) we get an expression of the form $C + D \cdot \sqrt{E}$, where $\sqrt{E}$ occurs nowhere else in the expression. (It cannot occur outside of roots since we factored it out, and it cannot occur inside roots since it is the root containing the greatest number of operations.) Now we expand the fraction with $C - D\sqrt{E}$, yielding&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{1}{B}=\frac{1}{C+D\cdot\sqrt{E}}=\frac{C-D\cdot\sqrt{E}}{C\cdot C - D\cdot D \cdot E}\;.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We may have greatly increased $m$, but we have decreased $n$, since $E$ no longer occurs as a radicand in the denominator and no new radicands have been introduced. That completes the induction.&lt;/p&gt;&#xA;" OwnerUserId="6622" LastEditorUserId="6622" LastEditorDisplayName="" LastEditDate="2011-03-08T06:03:54.360" LastActivityDate="2011-03-08T06:03:54.360" CommentCount="11" />
  <row Id="25499" PostTypeId="1" CreationDate="2011-03-07T10:52:34.743" Score="1" ViewCount="93" Body="&lt;p&gt;Hello!&#xA;I am trying to export a matrix in a file using Mathematica.&#xA;I get the matrix and then use&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Export[&quot;myfile.dat&quot;, MyMatrix, &quot;???Format???&quot;]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;There are many formats. I want the one that prints the matrix in this way:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;{{0,1,0,....},{0,0,1,...},...}&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;That is the classic Mathematica format....&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I do this?&#xA;Which one is the correct format?&lt;/p&gt;&#xA;" OwnerUserId="4175" LastActivityDate="2011-03-07T11:57:44.220" Title="Exporting a matrix in Mathematica in curly brackets format" Tags="&lt;matrices&gt;&lt;math-software&gt;&lt;mathematica&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="25500" PostTypeId="1" CreationDate="2011-03-07T10:52:41.063" Score="6" ViewCount="88" Body="&lt;p&gt;This is Exercise III.2.8 of Bourbaki's &lt;em&gt;Theory of Sets&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An ordered set $E$ is said to be ramified if, for each pair of elements $x,y$ of $E$ such that $x&amp;lt;y$, there exists $z&amp;gt;x$ such that $y$ and $z$ are not comparable. $E$ is said to be completely ramified if it is ramified and has no maximal elements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(a) Let $E$ be an ordered set and let $a$ be an element of $E$. Let $\mathfrak{R}_a$ denote the set of ramified subsets of $E$ which have $a$ as least element. Show that $\mathfrak{R}_a$, ordered by inclusion, has a maximal element.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(b) If $E$ is branched, show that every maximal element of $\mathfrak{R}_a$ is completely ramified. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(An ordered set $E$ is said to be branched if for each $x\in E$ there exist $y,z$ in $E$ such that $x\leq y$, $x\leq z$ and the intervals $\left[y,\rightarrow\right[$ and $\left[z,\rightarrow\right[$ do not intersect.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The proof of (a) is a straightforward application of Zorn's Lemma. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For (b), I would like to argue in the following way. Suppose $F$ is a maximal element of $\mathfrak{R}_a$ and $x$ is a maximal element of $F$. Then there is $y,z$ in $E$ such that $x\leq y$, $x\leq z$ and $\left[y,\rightarrow\right[\cap\left[z,\rightarrow\right[=\emptyset$. Then $F\cup\{y,z\}$ is ramified.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I haven't been able to show that $F\cup\{y,z\}$ is ramified, and I'm not sure that it's true, since if $b\in F$ and $b&amp;lt;y$, I don't see a way to conclude that $b\leq x$ or $b&amp;lt;z$. On the other hand, there is no natural way of adding further elements to $F\cup\{y,z\}$ such that it becomes a ramified set.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Any hints at the solution or general thoughts about the problem are greatly appreciated.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Bourbaki seems to be the only mathematician to use the words &quot;branched&quot; and &quot;ramified&quot; for those properties. Are there other more common words, or are these properties without interest?   &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="1878" LastActivityDate="2011-03-07T13:48:23.093" Title="Bourbaki exercise about &quot;ramified&quot; ordered sets" Tags="&lt;order-theory&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" />
  <row Id="25501" PostTypeId="2" ParentId="25499" CreationDate="2011-03-07T11:09:15.197" Score="2" ViewCount="" Body="&lt;p&gt;Export[&quot;file.txt&quot;,MyMatrix,&quot;String&quot;] works in Mathematica 6&lt;/p&gt;&#xA;" OwnerUserId="7838" LastActivityDate="2011-03-07T11:09:15.197" />
  <row Id="25502" PostTypeId="1" CreationDate="2011-03-07T11:17:56.870" Score="1" ViewCount="31" Body="&lt;p&gt;I am trying to find a way to calculate the probability that a user will pass a test at a certain certification authority, based on the user's history with answering questions using our service. Looking at a single question at a time (I know how to combine the probabilities), I have the following data:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;code&gt;pass&lt;/code&gt; - The percentage of users who passed the test at the certification authority on their first attempt who answered the  question correctly.&lt;/li&gt;&#xA;&lt;li&gt;&lt;code&gt;fail&lt;/code&gt; - he percentage of users who failed the test at the certification authority on their first attempt who answered the question correctly.&lt;/li&gt;&#xA;&lt;li&gt;&lt;code&gt;correct&lt;/code&gt; - if the current user answered the question correctly.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Given an example question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$pass = 0.96$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$fail = 0.42$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the user answered the question correctly, I believe the probability that he will pass (given that question as the only evidence), is something like:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$P(pass) = \frac{0.96}{(0.96 + 0.42)} = 0.697$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Am I correct? What if the user answered the question wrongly the first time?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On average, 82% of our candidates pass the certification on the first attempt. Of everyone who tries to take the certification, only 50% passes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;: What about a second question, where 70% of candidates who pass the test on their first attempt answer correctly, but 80% of candidates who fail the test on their first attempt answer correctly? (meaning $pass = 0.7$ and $fail = 0.8$) Answering this question incorrectly will contribute to passing the test, even though it logically shouldn't.&lt;/p&gt;&#xA;" OwnerUserId="7902" LastEditorUserId="7902" LastEditorDisplayName="" LastEditDate="2011-03-07T13:28:07.257" LastActivityDate="2011-03-07T13:28:07.257" Title="Bayesian classifier to predict user behavior" Tags="&lt;probability&gt;" AnswerCount="1" />
  <row Id="25503" PostTypeId="2" ParentId="25492" CreationDate="2011-03-07T11:21:52.777" Score="1" ViewCount="" Body="&lt;p&gt;By working out the integral $c_n=\frac{1}{2\pi}\int_{-\pi}^\pi f(x)e^{-i n x}dx$, which in this case simplifies to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$c_n=\frac{1}{2\pi}\int_0^{\pi}e^{-i n x}dx$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From here on it shouldn't be too hard...&lt;/p&gt;&#xA;" OwnerUserId="5882" LastActivityDate="2011-03-07T11:21:52.777" />
  <row Id="25504" PostTypeId="1" AcceptedAnswerId="25817" CreationDate="2011-03-07T11:22:50.297" Score="5" ViewCount="144" Body="&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; Let’s call a real number $x$ &lt;em&gt;approachable&lt;/em&gt; if there exists a Turing machine $M$ such that $M(1), M(2),\dots$ is a sequence than converge to $x$. If we can prove (&lt;strong&gt;edit&lt;/strong&gt;: In ZFC) that the sequence converge, we say that $x$ is &lt;em&gt;provably approachable&lt;/em&gt;. Is every approachable number also provably approachable?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Clearification:&lt;/strong&gt; Another way to ask the same question is: Is it true that: &quot;If there exists a Turing machine M such that x is the limit of the output values, then there exists a Turing machine M' that converge to the same number, and we can prove that it converge&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Comments:&lt;/strong&gt;  I don't know these concepts have names, I just invented the terminology: If they have names, please make a comment, so I can change it (&lt;strong&gt;edit&lt;/strong&gt;: Approachable is called &quot;computable in the limit&quot;, but I don't know if there is a name for provably appoachable. &quot;Approachable&quot; have been used a lot in the answers, so I have decided not the change it in the question). All computable numbers are provably approachable but not all provably approachable numbers are computable (Chatins constant is a counterexample). I don't know what the relation is between approachable and definable (I don't even know if there is an accepted definition of definable number?)&lt;/p&gt;&#xA;" OwnerUserId="3078" LastEditorUserId="3078" LastEditorDisplayName="" LastEditDate="2011-03-09T10:28:29.387" LastActivityDate="2011-03-09T10:28:29.387" Title="Approachable = provably approachable?" Tags="&lt;computability&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="25505" PostTypeId="1" AcceptedAnswerId="26032" CreationDate="2011-03-07T11:29:41.147" Score="4" ViewCount="160" Body="&lt;p&gt;What is the value of $$\int_{[0,1]^2} \!\!\!dx\,dy\,\mathcal{P} \frac{\log(x)}{x-y},$$&#xA;where $\mathcal{P}$ denotes Cauchy's principal value. I solved this once for a homework, but I can neither remember the answer nor reproduce it right now... (bad memory)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any help or comment is highly welcome.&lt;/p&gt;&#xA;" OwnerUserId="7266" LastEditorUserId="7266" LastEditorDisplayName="" LastEditDate="2011-03-09T16:30:50.877" LastActivityDate="2011-03-09T18:17:44.173" Title="What is $\int_{[0,1]^2}\!dx\,dy\,\mathcal{P} \frac{\log(x)}{x-y}$?" Tags="&lt;calculus&gt;&lt;homework&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="25506" PostTypeId="1" CreationDate="2011-03-07T11:34:05.270" Score="2" ViewCount="224" Body="&lt;p&gt;What textbook is good for introductory group theory?&lt;/p&gt;&#xA;" OwnerUserId="7340" LastActivityDate="2011-03-07T15:06:56.337" Title="Introductory Group theory textbook" Tags="&lt;group-theory&gt;&lt;reference-request&gt;" AnswerCount="5" CommentCount="3" FavoriteCount="1" />
  <row Id="25507" PostTypeId="2" ParentId="25254" CreationDate="2011-03-07T11:41:50.253" Score="1" ViewCount="" Body="&lt;p&gt;(I'm just a student so I'm sorry if I offend anyone with obvious comments)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If one takes the intuitive case $x_i = 1/\sqrt{n}$, the exact distribution is to linear transformations that of the Binomial distribution. Naturally, one should not expect to understand the tail bounds of your case better than this well-studied one. Additionally, if one is interested in &quot;the worst case&quot;, then I phantom it should be precisely this. If not, perhaps the more successful approaches carry over.&lt;/p&gt;&#xA;" OwnerUserId="7904" LastActivityDate="2011-03-07T11:41:50.253" />
  <row Id="25508" PostTypeId="2" ParentId="25504" CreationDate="2011-03-07T11:49:48.890" Score="1" ViewCount="" Body="&lt;p&gt;Your question is vague, which prevents an answer. This is too long for a comment, so I am writing it here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First of all, you haven't said what the axioms of your proof system are. If we have a very weak formal system, of course, then you will get a negative answer, since we won't be able to prove very much. But as the system gets stronger, we will be able to prove more. (In an inconsistent system, of course, we can prove anything.) So perhaps you are really asking whether, say, ZFC or PA are already strong enough to prove the instances you are interested in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second, however, what statement is it that should be provable, exactly? You seem to want to suppose that a real $x$ is approachable, and ask of this particular real whether it is provable that it is approachable. But of course ZFC doesn't know which particular $x$ you are talking about, unless you provide a description of it. So what do you mean? Since we have supposed that $x$ is approachable by some Turing machine $M$, then the only sensible description of $x$ we have is simply &quot;the limit of the values output by $M$&quot;. But in this case, the question becomes the triviality of proving the statement: &quot;The real $x$ that is the limit of the output values of Turing machine $M$ is the limit of the output values of some Turing machine.&quot; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So it seems that the question wants clarification.&lt;/p&gt;&#xA;" OwnerUserId="413" LastActivityDate="2011-03-07T11:49:48.890" CommentCount="3" />
  <row Id="25509" PostTypeId="2" ParentId="25502" CreationDate="2011-03-07T11:49:54.053" Score="1" ViewCount="" Body="&lt;p&gt;You are missing some information, for example what proportion of candidates pass the test on the first attempt (though it might be something else like the probability that this question is answered correctly).  You also have some notation issues: &lt;em&gt;pass&lt;/em&gt; seems to have two different meanings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For illustration, let's suppose the proportion of candidates who pass the test on the first attempt is 0.8. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then if the user answered the question correctly (on a first attempt), the probability that he will pass overall (on a first attempt) is &#xA;$$\frac{0.8 \times 0.96}{0.8 \times 0.96 + (1-0.8) \times 0.42} \approx 0.901$$&#xA;while if the user answered the question incorrectly (on a first attempt), the probability that he will pass overall (on a first attempt) is &#xA;$$\frac{0.8 \times (1-0.96)}{0.8 \times (1-0.96) + (1-0.8) \times (1-0.42)} \approx 0.216$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answers are sensitive to the 0.8 illustration.  If instead the value had been 0.2 then the two conditional probabilities would have been about 0.364 and 0.017.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your value of 0.697 (or 0.696) corresponds to the case where the overall pass rate for the whole test is 0.5; in this case the second conditional probability would be about $\frac{1-0.96}{(1-0.96) + (1-0.42)} \approx 0.0.065$.  In the same situation, your 0.304 is the probability that the candidate will fail overall given a correct answer to this particular question. But you should make the 0.5 explicit and include it in the fractions.&lt;/p&gt;&#xA;" OwnerUserId="6460" LastActivityDate="2011-03-07T11:49:54.053" CommentCount="2" />
  <row Id="25510" PostTypeId="2" ParentId="25499" CreationDate="2011-03-07T11:57:44.220" Score="2" ViewCount="" Body="&lt;p&gt;In really you can do this just with &lt;code&gt;Put&lt;/code&gt;:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;a = Table[RandomInteger[], {10}, {10}];&#xA;a &amp;gt;&amp;gt; &quot;C:\\test.txt&quot;&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="7617" LastActivityDate="2011-03-07T11:57:44.220" />
  <row Id="25512" PostTypeId="2" ParentId="25485" CreationDate="2011-03-07T12:05:37.030" Score="4" ViewCount="" Body="&lt;p&gt;$\mathbb{Z}_p$ is, in my opinion, problematic notion. In slightly less elementary number theory it refers to the &lt;a href=&quot;http://en.wikipedia.org/wiki/P-adic_number&quot; rel=&quot;nofollow&quot;&gt;p-adic numbers&lt;/a&gt;, which are very different from the integers $\bmod p$. An unambiguous, if slightly more cumbersome, notation would be $\mathbb{Z}/(p)$ or $\mathbb{Z}/p\mathbb{Z}$. &lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2011-03-07T12:05:37.030" CommentCount="2" />
  <row Id="25513" PostTypeId="2" ParentId="25506" CreationDate="2011-03-07T12:15:56.987" Score="4" ViewCount="" Body="&lt;p&gt;What about the book &lt;em&gt;An introduction to group theory&lt;/em&gt; by &lt;strong&gt;&lt;em&gt;Joseph Rotman&lt;/em&gt;&lt;/strong&gt;?&lt;br&gt;&#xA;It is in my opinion a classic.&lt;/p&gt;&#xA;" OwnerUserId="6792" LastActivityDate="2011-03-07T12:15:56.987" />
  <row Id="25514" PostTypeId="2" ParentId="25431" CreationDate="2011-03-07T12:24:56.990" Score="1" ViewCount="" Body="&lt;p&gt;$F[x^2,x^3]$ is an integral domain with field of fraction $F(x)$. $x \in F(x)$ is integral over  $F[x^2,x^3]$ but not in $F[x^2, x^3]$. So it is not a UFD as any UFD is integrally closed domain.&lt;/p&gt;&#xA;" OwnerUserId="7565" LastEditorUserId="7565" LastEditorDisplayName="" LastEditDate="2011-03-07T16:47:15.217" LastActivityDate="2011-03-07T16:47:15.217" />
  <row Id="25515" PostTypeId="2" ParentId="25455" CreationDate="2011-03-07T12:30:03.910" Score="10" ViewCount="" Body="&lt;p&gt;The answer to this question is the same as the answer to every question of this genre (&quot;why should I care about groups,&quot; &quot;why should I care about rings&quot;): they show up everywhere and are an extremely useful organizing principle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a meta-principle that any time you're trying to understand something about categories, it's a good idea to restrict to the special case of posets first, regarded as categories where $a \le b$ means there is a single arrow $a \to b$. For example:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Products are the same thing as infima. In particular, the terminal object is the empty infimum, which is the maximum element. (This may seem backwards, but it really is what you get out of the definitions.)&lt;/li&gt;&#xA;&lt;li&gt;Coproducts are the same thing as suprema. In particular, the initial object is the empty supremum, which is the minimum element.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So what's a pair of adjoint functors in this context? Well, if $P, Q$ are two posets, then a functor $f : P \to Q$ is just an order-preserving function. So a pair of adjoint functors is first of all a pair $f : P \to Q, g : Q \to P$ of order-preserving functions. The definition I think is most convenient when studying posets is that $f, g$ must satisfy&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\text{Hom}_Q(fa, b) \cong \text{Hom}_P(a, gb)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for all $a \in P, b \in Q$. But this is the same thing as requiring that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$fa \le b \Leftrightarrow a \le gb.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This relationship is called a &lt;a href=&quot;http://en.wikipedia.org/wiki/Galois_connection&quot; rel=&quot;nofollow&quot;&gt;Galois connection&lt;/a&gt;. Important examples of Galois connections include:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$K \to L$ is a finite Galois extension, $P$ is the poset of subgroups of $\text{Gal}(L/K)$, $Q$ is the poset of subfields $K \to M \to L$, $f$ sends a subgroup to its fixed field, $g$ sends a subfield to $\text{Gal}(L/M)$. (One has to reverse the order of one of these posets for this to work.) &lt;/li&gt;&#xA;&lt;li&gt;$P$ is the poset of ideals of $\mathbb{C}[x_1, ... x_n]$, $Q$ is the poset of subsets of $\mathbb{C}^n$, $f$ sends an ideal to the set of points defined by its elements vanishing, $g$ sends a set of points to the ideal of functions vanishing on it. (Again, one has to reverse the order of one of these posets.)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Galois connections exist in extreme generality and are, by themselves, already an important organizing principle in mathematics. So adjoint functors are &lt;em&gt;even more important than that!&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; It's probably worth explaining what's going on in the above examples abstractly. Let $A, B$ be two sets, and let $r : A \times B \to \{ 0, 1 \}$ be a relation. Then $r$ induces an order-reversing Galois connection (a pair of contravariant adjoint functors) between the poset $\mathcal{P}(A)$ of subsets of $A$ and the poset $\mathcal{P}(B)$ of subsets of $B$ as follows: if $S \subset A$ then $f(S) = \{ b \in B : r(a, b) = 1 \forall a \in S \}$ and if $S \subset B$ then $g(S) = \{ a \in A : r(a, b) = 1 \forall b \in S \}$. I'll leave it as an exercise to figure out what $A, B, r$ are in the above examples. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note also that the fact that left adjoints preserve colimits and right adjoints preserve limits continues to hold for Galois connections, and shows that some of the properties of the Galois connections above are purely formal (in the sense that they follow from this &quot;abstract nonsense&quot;). Unfortunately it is generally not emphasized which properties those are. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Adjoint_functors#Motivation&quot; rel=&quot;nofollow&quot;&gt;Wikipedia article&lt;/a&gt; does a nice job of explaining some broad general motivation (and has a lot of good discussion on this question besides): very roughly, an adjoint is the best substitute for an inverse that exists in a lot of cases that we care about. You can sort of see how this works in the above examples.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An important property of adjoint pairs is that they restrict to equivalences on subcategories, and this is what we get in the Galois theory and algebraic geometry examples above: the first adjoint pair is an equivalence by the fundamental theorem of Galois theory, and the second adjoint pair restricts to an equivalence between reduced ideals and varieties by the Nullstellensatz. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Since your question is tagged [algebraic-geometry], here is an important &lt;em&gt;non-example&lt;/em&gt; related to the second half of Arturo's answer. There is a functor $F : \text{Aff} \to \text{Set}$ sending an affine scheme to its set of points (the prime ideals of the corresponding ring), and it does &lt;em&gt;not&lt;/em&gt; have a left adjoint: there is no &quot;free affine scheme&quot; on a set. The reason is that $F$ does not preserve limits. (Note that a functor &lt;em&gt;has&lt;/em&gt; a left adjoint if and only if it &lt;em&gt;is&lt;/em&gt; a right adjoint.) In fact, it does not even preserve products. The product of two affine schemes $\text{Spec } R, \text{Spec } S$ is $\text{Spec } R \otimes_{\mathbb{Z}} S$, and it is a basic property of schemes that this is not the set-theoretic product. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It follows that the functor $F : \text{Aff} \to \text{Top}$ sending an affine scheme to its set of points in the Zariski topology also does not have a left adjoint. If you've ever wondered why the Zariski topology on $\mathbb{A}^2$ isn't the product topology on $\mathbb{A}^1 \times \mathbb{A}^1$, now you know. &lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2011-03-07T22:16:47.677" LastActivityDate="2011-03-07T22:16:47.677" CommentCount="2" />
  <row Id="25516" PostTypeId="2" ParentId="25430" CreationDate="2011-03-07T12:47:58.380" Score="9" ViewCount="" Body="&lt;p&gt;Closed form formulas are overrated. When they exist, generating function techniques can often help you find them; when they don't, the generating function is the next best thing, and it turns out to be much more powerful than it looks at first glance. For example, most generating functions are actually &lt;em&gt;meromorphic functions&lt;/em&gt;, and this means that one can deduce asymptotic information about a sequence from the locations of the poles of its generating function. This is, for example, how one deduces the asymptotic of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Partition_(number_theory)#Asymptotic_behaviour&quot; rel=&quot;nofollow&quot;&gt;partition numbers&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your particular example, the generating function is rational, so it has a finite number of poles. That means you can use partial fraction decomposition on it to immediately get a closed form.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might be interested in reading &lt;a href=&quot;http://web.mit.edu/~qchu/Public/TopicsInGF.pdf&quot; rel=&quot;nofollow&quot;&gt;my notes on generating functions&lt;/a&gt;, which have several examples and which I hope will be enlightening. The first basic thing to grasp is that manipulating generating functions is much easier than manipulating sequences, but the power of generating functions goes much deeper than this. For a really thorough discussion I highly recommend Flajolet and Sedgewick's &lt;a href=&quot;http://algo.inria.fr/flajolet/Publications/books.html&quot; rel=&quot;nofollow&quot;&gt;Analytic Combinatorics&lt;/a&gt;, which is available for free online. &lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2011-03-07T12:47:58.380" CommentCount="1" />
  <row Id="25517" PostTypeId="2" ParentId="25413" CreationDate="2011-03-07T12:55:08.980" Score="3" ViewCount="" Body="&lt;p&gt;All that's going on here is that when you differentiate a Taylor series expansion&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f(x) = f^{(0)}(a) + \frac{f^{(1)}(a)}{1!} (x - a)^1 + \frac{f^{(2)}(a)}{2!} (x - a)^2 + ...$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;you get the Taylor series expansion&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f&amp;#39;(x) = f^{(1)}(a) + \frac{f^{(2)}(a)}{1!} (x - a)^1 + \frac{f^{(3)}(a)}{2!} (x - a)^2 + ....$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, what you've done is precisely shifted the Taylor coefficients one to the left. That's all the Wikipedia article is saying. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I disagree that this makes ODEs any easier to solve, though. ODEs were &lt;em&gt;already&lt;/em&gt; this easy to solve, you just weren't taught the right language for seeing this.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2011-03-07T12:55:08.980" CommentCount="2" />
  <row Id="25518" PostTypeId="1" CreationDate="2011-03-07T13:15:34.137" Score="2" ViewCount="58" Body="&lt;p&gt;Let $W$ be a real Hilbert space of dimension $n$ and $V$ a Hilbert subspace of dimension $m$. Assume that $f_1,\cdots,f_k$ are points in $W$ such that the following holds:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;there exists $\sigma&amp;gt;0$ such that $dist(f_i,V)\geq\sigma$ for $i=1,2,\cdots,k$,&lt;/li&gt;&#xA;&lt;li&gt;there exists $0&amp;lt;\delta\leq\sigma$ such that the balls $B(f_i,\delta)$ are disjoint for $i=1,2,\cdots,k$,&lt;/li&gt;&#xA;&lt;li&gt;there exists $L&amp;gt;0$ such that $||f_i||_2\leq L$ for $i=1,2,\cdots,k$.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Can you find a good bound on $k$ (the number of the points $f_i$) involving $n,m,\sigma,\delta,L$?&lt;/p&gt;&#xA;" OwnerUserId="1754" LastActivityDate="2011-03-07T13:15:34.137" Title="Counting balls in Hilbert spaces" Tags="&lt;real-analysis&gt;&lt;hilbert-spaces&gt;" />
  <row Id="25519" PostTypeId="1" CreationDate="2011-03-07T13:21:51.257" Score="4" ViewCount="169" Body="&lt;p&gt;Is there an easy way to construct, on the same filtered probability space, a Brownian motion $W$ and a Poisson process $N$, such that $W$ and $N$ are not independent ?  &lt;/p&gt;&#xA;" OwnerUserId="7406" LastActivityDate="2011-03-07T20:51:28.203" Title="Correlated Brownian motion and Poisson process" Tags="&lt;probability&gt;&lt;stochastic-processes&gt;" AnswerCount="1" />
  <row Id="25520" PostTypeId="2" ParentId="25047" CreationDate="2011-03-07T13:23:24.017" Score="1" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; This is only a helpful hint if you consider the minimum amount of swaps needed to sort a sequence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just two hints:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Permutations are uniquely given by their cyclic form, e.g. $2143$ is $(1,2)(3,4)$.&lt;/li&gt;&#xA;&lt;li&gt;If you have a cycle of $k$ elements, you need $k-1$ swaps to bring the elements to their positions in the sorted sequence&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Work for you: proof optimality of this and implement a function that finds all circles in a given sequence.&lt;/p&gt;&#xA;" OwnerUserId="3330" LastEditorUserId="3330" LastEditorDisplayName="" LastEditDate="2011-03-07T13:49:04.500" LastActivityDate="2011-03-07T13:49:04.500" CommentCount="2" />
  <row Id="25521" PostTypeId="2" ParentId="25506" CreationDate="2011-03-07T13:37:43.093" Score="3" ViewCount="" Body="&lt;p&gt;I think the group theory part (= first 6 chapters) of &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0471433349&quot; rel=&quot;nofollow&quot;&gt;Abstract Algebra&lt;/a&gt; by Dummit and Foote is quite good. Personally, I dislike Armstrong's book &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/144193085X&quot; rel=&quot;nofollow&quot;&gt;Groups and Symmetry&lt;/a&gt;; his style is too informal to my taste, and definitions are hidden in the text.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A concise, clear one is Humprhey's &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0198534531&quot; rel=&quot;nofollow&quot;&gt;A Course in Group Theory&lt;/a&gt;, it gets you quickly to the core of the subject.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a 'second' course I like the Universitext &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0387405100&quot; rel=&quot;nofollow&quot;&gt;The Theory of Finite Groups: An Introduction&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="6490" LastEditorUserId="6490" LastEditorDisplayName="" LastEditDate="2011-03-07T15:04:13.810" LastActivityDate="2011-03-07T15:04:13.810" />
  <row Id="25522" PostTypeId="2" ParentId="25500" CreationDate="2011-03-07T13:48:23.093" Score="1" ViewCount="" Body="&lt;p&gt;I'm not sure if this counts as a rigourous proof, but what if you think of $F$ as a directed graph.  Then $F&amp;#39; = F \cup \{y,z\}$ is formed from $F$ by adding $y$ and $z$ as vertices, and just two edges, one from $x$ to $y$ and one from $x$ to $z$. Now two elements satisfy $u&amp;lt;v$ if and only if there is a directed path from $u$ to $v$. Then it's clear that if $b&amp;lt;y$ then also $b\leqslant x$ since you can just take a directed path from $b$ to $y$ and remove the last arrow, and you get a directed path from $b$ to $x$; the same logic shows that $b&amp;lt;z$, and since $y$ and $z$ are by construction incomparible, this shows that $F&amp;#39;$ is a ramified set, contradicting the maximality of $F$.&lt;/p&gt;&#xA;" OwnerUserId="668" LastActivityDate="2011-03-07T13:48:23.093" CommentCount="2" />
  <row Id="25523" PostTypeId="2" ParentId="25506" CreationDate="2011-03-07T13:48:25.217" Score="0" ViewCount="" Body="&lt;p&gt;If you mean an intro to abstract algebra, &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0486474178&quot; rel=&quot;nofollow&quot;&gt;A book of abstract algebra&lt;/a&gt; by Charles C. Pinter is great. See the reviews on Amazon.&lt;/p&gt;&#xA;" OwnerUserId="6763" LastActivityDate="2011-03-07T13:48:25.217" />
  <row Id="25524" PostTypeId="2" ParentId="25506" CreationDate="2011-03-07T13:54:36.973" Score="1" ViewCount="" Body="&lt;p&gt;A very intuitive one is this: &quot;Visual Group Theory&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Preview here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://books.google.de/books?id=T_o0CnMZecMC&amp;amp;lpg=PP1&amp;amp;ots=VHFMDQx8yi&amp;amp;dq=visual%20group%20theory&amp;amp;hl=en&amp;amp;pg=PP1#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;http://books.google.de/books?id=T_o0CnMZecMC&amp;amp;lpg=PP1&amp;amp;ots=VHFMDQx8yi&amp;amp;dq=visual%20group%20theory&amp;amp;hl=en&amp;amp;pg=PP1#v=onepage&amp;amp;q&amp;amp;f=false&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="346" LastActivityDate="2011-03-07T13:54:36.973" />
  <row Id="25525" PostTypeId="1" CreationDate="2011-03-07T14:04:12.247" Score="1" ViewCount="92" Body="&lt;p&gt;hi, can any one say how to relate stone weirstrass approximation theorem in this problem? given that $\mu$ and $\nu$ are two probability measure on Real line, if for all bounded continous function $f$, $\int f d \mu = \int f d \nu$, then $\mu=\nu$?&lt;/p&gt;&#xA;" OwnerUserId="6990" LastEditorUserId="1766" LastEditorDisplayName="" LastEditDate="2011-03-07T14:55:04.650" LastActivityDate="2011-03-07T15:11:28.920" Title="probability measure" Tags="&lt;homework&gt;&lt;probability&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="25526" PostTypeId="1" CreationDate="2011-03-07T14:15:36.580" Score="2" ViewCount="99" Body="&lt;p&gt;Hi I really need help in how to determine sub fields $L$ of $\mathbb{C}$ over $\mathbb{Q}$ and finding $[L:\mathbb{Q}]$.  The question is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Find subfields $L$ of $\mathbb{C}$ that are splitting fields over $\mathbb{Q}$, find $[L:\mathbb{Q}]$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{array}{ll}&#xD;&#xA;\mathrm{(i)} &amp;amp; t^8 + 1 \\&#xD;&#xA;\mathrm{(ii)} &amp;amp; t^4  -4t^2  +1\\&#xD;&#xA;\mathrm{(iii)} &amp;amp; t^3  -5\\&#xD;&#xA;\mathrm{(iv)}  &amp;amp; t^{17}  - 1\\&#xD;&#xA;\end{array}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any help would be great!&lt;/p&gt;&#xA;" OwnerUserId="7907" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-07T18:42:18.827" LastActivityDate="2011-04-06T19:39:35.380" Title="Sub fields of $\mathbb{C}$ that are splitting fields over $\mathbb{Q}$" Tags="&lt;abstract-algebra&gt;&lt;group-theory&gt;" AnswerCount="1" />
  <row Id="25527" PostTypeId="1" CreationDate="2011-03-07T14:16:04.073" Score="5" ViewCount="73" Body="&lt;p&gt;Hi all,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know how to find a point on a circle given a radius and an angle, but my knowledge of trigonometry doesn't extend much further than that. My question is probably best explained diagrammatically:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/EsPnr.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given a centre point, radius, and two points on the circle, how do I find the 'middle' point between those two points? I'm also interested in knowing the distance between the internal line formed between the two points and this new middle point. (Referring to the diagram, I'm trying to determine the distance, q, and the unknown point, x?, y?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Stewart&lt;/p&gt;&#xA;" OwnerUserId="7908" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2011-03-07T14:37:01.227" LastActivityDate="2011-03-07T15:36:18.603" Title="Finding point on a circle" Tags="&lt;euclidean-geometry&gt;&lt;circle&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="25528" PostTypeId="1" AcceptedAnswerId="25535" CreationDate="2011-03-07T14:24:08.977" Score="2" ViewCount="103" Body="&lt;p&gt;excuse my lack of knowledge and expertise in math;)&#xA;but to me it would came naturally that the cubic root of -8 would be -2 since -2 ^ 3 = -8.&#xA;but when i check wolfram alpha for cbrt(-8), real it tells me it doesnt exist:&#xA;&lt;a href=&quot;http://www.wolframalpha.com/input/?i=cbrt%28-8%29+real&quot; rel=&quot;nofollow&quot;&gt;http://www.wolframalpha.com/input/?i=cbrt%28-8%29+real&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i came to trust woflram alpha so i thought i'd ask you guys, to explain the sense of that to me.&lt;/p&gt;&#xA;" OwnerUserId="5043" LastActivityDate="2011-03-07T14:42:02.393" Title="cubic root of negative numbers" Tags="&lt;arithmetic&gt;" AnswerCount="4" />
  <row Id="25530" PostTypeId="2" ParentId="25254" CreationDate="2011-03-07T14:29:05.573" Score="1" ViewCount="" Body="&lt;p&gt;This is another question instead of an answer, but I was thinking about a similar problem and I was wondering if one could use the cumulants to derive bounds, since they are relatively simple in the binomial case.  Sorry if this is inappropriate.&lt;/p&gt;&#xA;" OwnerUserId="7862" LastActivityDate="2011-03-07T14:29:05.573" />
  <row Id="25531" PostTypeId="2" ParentId="25526" CreationDate="2011-03-07T14:32:14.913" Score="1" ViewCount="" Body="&lt;p&gt;For example take $t^3 - 5$ the roots of this polynomial are: $\sqrt{5}, \xi \sqrt{5}, \xi^2 \sqrt{5}$. Here $\xi$ is the primitive root of unity of order $3$. Now you note that you need to add both $\sqrt{5}$ and $\xi$ to $\mathbb{Q}$, for the polynomial to split. Hence $L = \mathbb{Q}(\xi,\sqrt{5})$. Now if you take $F = \mathbb{Q}(\xi)$, then $[L:\mathbb{Q}] = [L:F][F:\mathbb{Q}]$. The extension $L/F$ is Kummer and $F/\mathbb{Q}$ is cyclotomic, hence $[L:F] = 2$ and $[F:\mathbb{Q}] = 2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rest are very similar and most rely on cyclotomic extensions. Remember that if you add a primitive root of unity of order $n$, you'll get a field extension of degree $\phi(n)$, where $\phi$ is the Euler totient function.&lt;/p&gt;&#xA;" OwnerUserId="7838" LastActivityDate="2011-03-07T14:32:14.913" />
  <row Id="25532" PostTypeId="2" ParentId="25527" CreationDate="2011-03-07T14:34:20.260" Score="2" ViewCount="" Body="&lt;p&gt;First, you have two points $(x_1,y_1)$ in your diagram.  I will use that for the one on the circle.  You can calculate $d$ from the coordinates of $(x_1,y_1)$ and $(x_2,y_2)$ as $d=\frac{1}{2}\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$.  If you draw the line segment from  the center to $(x_1,y_1)$ you have a right triangle with sides $d, (r-q)$, and  $r$, so $r^2=d^2+(r-q)^2$ which you can solve for $q$ by the quadratic equation.  Then the coordinates of the intersection of the radius and the chord are $(\frac{x_2+x_1}{2},\frac{y_2+y_1}{2})$ and you can extend the line segment from the center through that until it has length $r$&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-03-07T14:34:20.260" />
  <row Id="25533" PostTypeId="2" ParentId="25528" CreationDate="2011-03-07T14:36:37.667" Score="2" ViewCount="" Body="&lt;p&gt;When non-computers calculate the cube root of (-8), we can think of it as $(-1*8)^{1/3}$&#xA;Then we have $-1*8^{1/3} = -1*2 = -2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wolfram is using the polar complex form of -8 = 8cis(π)&#xA;Then the cube root of this is 2cis(π/3), which is 1 + i√3 (an alternate form on Wolfram)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Incidentally, if you take $(1 + i\sqrt3)^3$, you will get -8!&lt;/p&gt;&#xA;" OwnerUserId="7850" LastActivityDate="2011-03-07T14:36:37.667" />
  <row Id="25534" PostTypeId="2" ParentId="25528" CreationDate="2011-03-07T14:36:49.027" Score="4" ViewCount="" Body="&lt;p&gt;See &lt;a href=&quot;http://mathworld.wolfram.com/CubeRoot.html&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt;. In particular, the prinicipal cube root has nonzero imaginary part.&lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2011-03-07T14:36:49.027" CommentCount="1" />
  <row Id="25535" PostTypeId="2" ParentId="25528" CreationDate="2011-03-07T14:37:08.747" Score="5" ViewCount="" Body="&lt;p&gt;-8 has three cube roots: $ -2 $, $1 + i \sqrt{ 3 } $ and $1 - i \sqrt{ 3 } $. So you can't answer the question &quot;Is $ \sqrt[3]{-8} $ real&quot; without specifying which of them you're talking about.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For some reason, WolframAlpha is only giving $1 + i \sqrt{ 3 } $ as an answer -- that looks like a bug in WolframAlpha to me. &lt;/p&gt;&#xA;" OwnerUserId="5656" LastActivityDate="2011-03-07T14:37:08.747" CommentCount="1" />
  <row Id="25536" PostTypeId="2" ParentId="25528" CreationDate="2011-03-07T14:42:02.393" Score="1" ViewCount="" Body="&lt;p&gt;Of course, you're absolutely right about $-2$ being &lt;em&gt;a&lt;/em&gt; cubic root of $-8$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The  point might be that there are actually, &lt;em&gt;three&lt;/em&gt; different cubic roots of $-8$, namely the roots of the polynomial $x^3+8$. One of this roots is real ($-2$), the other two are complex and conjugate of each other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you &lt;em&gt;ask&lt;/em&gt; Wolfram for the cubic root of $-8$ you get one of these two non real roots, namely $1+(1.732050807568877293527446341505872366942805253810380628055...)i$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I gather that Wolfram is instucted to choose one of the roots by some criteria that in this case leads to the exclusion of the real root. Maybe browsing the Wolfram site may help understanding what these criteria are. (My guess is that it outputs the root $\alpha=re^{i\theta}$ with smaller $\theta$ in the range $[0,2\pi)$.) &lt;/p&gt;&#xA;" OwnerUserId="688" LastActivityDate="2011-03-07T14:42:02.393" />
  <row Id="25537" PostTypeId="2" ParentId="25430" CreationDate="2011-03-07T14:53:21.937" Score="0" ViewCount="" Body="&lt;p&gt;They are quite useful in problems 137 and 140 at &lt;a href=&quot;http://projecteuler.net/index.php?section=problems&quot; rel=&quot;nofollow&quot;&gt;Project Euler&lt;/a&gt;, though you have to take them seriously as convergent series.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2011-03-07T15:05:23.467" LastActivityDate="2011-03-07T15:05:23.467" />
  <row Id="25538" PostTypeId="2" ParentId="25506" CreationDate="2011-03-07T15:06:56.337" Score="0" ViewCount="" Body="&lt;p&gt;Here are some good books:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Algebra, Abstract and concrete by Fred Goodman, (link :http://www.math.uiowa.edu/~goodman/algebrabook.dir/algebrabook.html) it's available for download.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Abstract Algebra by Dummit and Foote.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;A first course in Abstract Algebra by John. B Fraleigh.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" LastActivityDate="2011-03-07T15:06:56.337" />
  <row Id="25539" PostTypeId="2" ParentId="25525" CreationDate="2011-03-07T15:11:28.920" Score="1" ViewCount="" Body="&lt;p&gt;I believe the intended solution is to approximate the indicator function of a measurable set (which is not continuous!) by bounded continuous functions. A measure is determined by the measure it assigns to measurable sets, so if you can establish $\mu(A) = \nu(A)$ for every interval $A \subseteq \mathbb{R}$ (or indeed, for any other $\pi$-system generating the $\sigma$-algebra in question), that proves that $\mu = \nu$.&lt;/p&gt;&#xA;" OwnerUserId="5191" LastActivityDate="2011-03-07T15:11:28.920" />
  <row Id="25540" PostTypeId="2" ParentId="25320" CreationDate="2011-03-07T15:13:31.443" Score="5" ViewCount="" Body="&lt;p&gt;&lt;em&gt;Update&lt;/em&gt;: edited text and formatting. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can write the sum $e_{n,k}$ in the form (see section 4 of &lt;a href=&quot;http://www.ift.uni.wroc.pl/~mwolf/Poorten_MI_195_0.pdf&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; article by Alf van der Poorten)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$e_{n,k}=\sum_{m=1}^{k}\frac{(-1)^{m-1}}{2m^{3}\dbinom{n}{m}\dbinom{n+m}{m}},&#xD;&#xA;\quad (1\leq k\leq n).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $u_{n,m}=m^{3}\binom{n}{m}\binom{n+m}{m}$. If $1=m\leq n$, then $u_{n,m}=n\left( n+1\right)$. To show that  $u_{n,m}&amp;gt;n(n+1)$ for $1&amp;lt;m\leq n$ we consider the following two cases:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;if $1&amp;lt;m=n$, then $u_{n,m}=n^{3}\binom{2n}{n}&amp;gt;n(n+1)$; &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;if $1&amp;lt;m\leq n-1$, then $m^{3}\binom{n}{m}\geq m^{3}\binom{n}{1}=m^{3}n&amp;gt;n$ and $\binom{n+m}{m}\geq \binom{n+m}{1}=n+m&amp;gt;n+1$. Hence $u_{n,m}&amp;gt;n(n+1)$. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Hence, for $1&amp;lt;k\leq n$, we get:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{eqnarray*}&#xD;&#xA;\left\vert e_{n,k}\right\vert  &amp;amp;=&amp;amp;\left\vert \sum_{m=1}^{k}\frac{(-1)^{m-1}}{%&#xD;&#xA;2m^{3}\binom{n}{m}\binom{n+m}{m}}\right\vert \leq \sum_{m=1}^{k}\left\vert &#xD;&#xA;\frac{(-1)^{m-1}}{2m^{3}\binom{n}{m}\binom{n+m}{m}}\right\vert\leq \sum_{m=1}^{n}\frac{1}{2m^{3}\binom{n}{m}\binom{n+m}{m}}\\&amp;amp;&amp;lt;&amp;amp;\sum_{m=1}^{n}\frac{1}{2n(n+1)}=\frac{n}{2n(n+1)}&amp;lt;\frac{1}{n}.&#xD;&#xA;\end{eqnarray*}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $k=1$, we get $\left\vert e_{n,1}\right\vert =\frac{1}{2n\left( n+1\right) }%&#xD;&#xA;\leq \frac{1}{2(n+1)}&amp;lt;\frac{1}{n}$. Thus for every integer $1\leq k\leq n$, we proved that $\left\vert e_{n,k}\right\vert &amp;lt;\frac{1}{n}$, which implies that $e_{n,k}$ converges uniformly in $k$ to $0$.&lt;/p&gt;&#xA;" OwnerUserId="752" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2011-05-18T12:12:09.267" LastActivityDate="2011-05-18T12:12:09.267" />
  <row Id="25541" PostTypeId="1" AcceptedAnswerId="25542" CreationDate="2011-03-07T15:16:21.533" Score="1" ViewCount="77" Body="&lt;p&gt;Is it possible to reduce this expression further ? (2x+y)(x-y)/(2x-y)(x+y)&lt;/p&gt;&#xA;" OwnerUserId="7833" LastActivityDate="2011-03-07T15:31:12.950" Title="Is it possible to reduce this algebra expression - (2x+y)(x-y)/(2x-y)(x+y) ?" Tags="&lt;algebra&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="25542" PostTypeId="2" ParentId="25541" CreationDate="2011-03-07T15:25:39.990" Score="1" ViewCount="" Body="&lt;p&gt;If you mean $\frac{(2x+y)(x-y)}{(2x-y)(x+y)}$-parentheses, please- you can write it as $\frac{2x^2-xy-y^2}{2x^2+xy-y^2}=1-\frac{2xy}{2x^2+xy-y^2}$  Is either one a reduction?  It is in the eye of the beholder.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-03-07T15:25:39.990" CommentCount="1" />
  <row Id="25543" PostTypeId="2" ParentId="25541" CreationDate="2011-03-07T15:31:12.950" Score="0" ViewCount="" Body="&lt;p&gt;No, it can't be further reduced. However it can also be written as (2x^2 -xy - y^2)/(2x^2 + xy - y^2)&lt;/p&gt;&#xA;" OwnerUserId="7909" LastActivityDate="2011-03-07T15:31:12.950" />
  <row Id="25544" PostTypeId="2" ParentId="25527" CreationDate="2011-03-07T15:36:18.603" Score="0" ViewCount="" Body="&lt;p&gt;Write $(x_1,y_1)=(Rcos\theta_1,Rsin\theta_1)$ and $(x_2,y_2)=(Rcos\theta_2,Rsin\theta_2)$, where $R = \sqrt{x_1^2+y_1^2}$ is the radius of the circle, and solve for $\theta_1$ and $\theta_2$; then clearly $(x_?,y_?)=(Rcos(\theta_1+\delta),Rsin(\theta_1+\delta)$, where $\delta = \frac{\theta_1-\theta_2}{2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now that we have all the $(x,y)$ coordinates for all the relevant points, it's easy to set up the linear system of 2 equations with two unknown's, and to solve for the point of intersection of the two lines.  The two equations are of course the equations for two lines drawn in the figure.&lt;/p&gt;&#xA;" OwnerUserId="1417" LastActivityDate="2011-03-07T15:36:18.603" />
  <row Id="25545" PostTypeId="1" AcceptedAnswerId="25555" CreationDate="2011-03-07T15:40:46.293" Score="0" ViewCount="159" Body="&lt;p&gt;I need help solving &lt;strong&gt;(b)&lt;/strong&gt; in this math problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/Dczrt.png&quot; alt=&quot;Determine the matrix of f in the standard basis (Linear Algebra)&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the relationship between the matrix &lt;em&gt;f&lt;/em&gt; and the transformation matrix in the standard basis?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for any suggestions!&lt;/p&gt;&#xA;" OwnerUserId="713" LastActivityDate="2011-03-07T17:03:48.870" Title="Determine the matrix of the linear transformation &quot;f&quot; in the standard basis (Linear Algebra)" Tags="&lt;linear-algebra&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="25546" PostTypeId="1" CreationDate="2011-03-07T15:49:50.167" Score="2" ViewCount="51" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to estimate the quantile of some data. The data are so huge that they can not be accommodated in the memory. And data are not static, new data keep coming. Does anyone know any algorithm to monitor the quantile(s) of the data observed so far with very limited memory and computation? I find &lt;a href=&quot;http://citeseerx.ist.psu.edu/viewdoc/download;jsessionid=742445C867F4FC0A1658981316B79CC3?doi=10.1.1.74.2672&amp;amp;rep=rep1&amp;amp;type=pdf&quot; rel=&quot;nofollow&quot;&gt;P2 algorithm&lt;/a&gt; useful. But it does not work very well for my data, which are extremely heavy-tailed distributed.&lt;/p&gt;&#xA;" OwnerUserId="7910" LastActivityDate="2011-03-08T05:00:38.170" Title="algorithm to dynamically monitor quantile(s)" Tags="&lt;algorithms&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="25547" PostTypeId="2" ParentId="136" CreationDate="2011-03-07T15:52:20.233" Score="0" ViewCount="" Body="&lt;p&gt;o.k. you see the difference between the squares, but you dont understand why. I also have noticed this differnce, but i took it a step further when i came up with equation &#xA;(a+1)^2- (a)^2 + 2 + (a+1)^2 = (a+2)^2. i asked my question which you can see in the section&#xA;tagged &quot;consecutive squares&quot;. which was, &quot;if it was part of a bigger equation but found out it wasn't&quot;. none the less it took the differences of a certain square and corelated it to the next consecutive square (such as the fact that 1^2=1, and 2^2=4 and the difference between them, as shown in the chart above, is 3 sure, great! but what of it ?), well if you add 2 to that difference than add it back to the (2^2, or 4) the larger consecutive square. Then you get 9 which is the next cosecutive square, (3^2). That is the concept of the equation above, technically i didn`t explain that specific aspect of the difference of consecutive squares, but i did explain the general correlation between consecutive squares.    &lt;/p&gt;&#xA;" OwnerUserId="7911" LastActivityDate="2011-03-07T15:52:20.233" CommentCount="1" />
  <row Id="25548" PostTypeId="2" ParentId="25504" CreationDate="2011-03-07T16:03:51.863" Score="4" ViewCount="" Body="&lt;p&gt;This post is marked community wiki because it is just an elaboration on how I understand the question. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To make these question precise, I would make this definition:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A &lt;em&gt;provably approachable real number&lt;/em&gt; in a theory $T$ is a Turing machine $A$ such that $T$ proves that $A$ enumerates a convergent sequence of rational numbers. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;On the other hand, a real number $z$ is &lt;em&gt;approachable&lt;/em&gt; if there is a Turing machine $M$ such that $M$ enumerates a sequence of rationals converging to $z$.  The statement ``every approachable real number is provably approachable'' then means that, for every approachable real number $z$, there is a machine $M$ such that $M$ enumerates a sequence of rationals converging to $z$ and $T$ proves that $M$ enumerates a convergent sequence of rationals. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The approachable reals are usually called &quot;limit computable&quot; in the literature, because of the following result. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Theorem.&lt;/strong&gt; A real number is approachable if and only if its binary expansion is computable from the halting problem, which happens if and only if the binary expansion is a $\Delta^0_2$ set when viewed as a subset of $\mathbb{N}\times\{0,1\}$. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here is the question as I read it. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Question 1.&lt;/strong&gt; Is every approachable real number provably approachable in ZFC? (ZFC can be replaced with any other effective theory.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The issue is not: given a Turing machine which we know to be total, can we prove in ZFC that the machine is total - the answer to that is certainly &quot;no&quot;. Instead, the question is, essentially: if we have a Turing machine $A$ that happens to be total in the standard model, is there some other Turing machine $M$ that computes the same function on standard inputs, but which is also provably total? (Note that in general the function computed in an arbitrary model is an extension of the function computed in the standard model because now there are also nonstandard inputs.) I don't know the answer to that yet, but I think it's a very interesting question. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I first wrote this comment, I had the idea of &quot;provably computable&quot; real number in my head instead of &quot;provably approachable&quot;. The same question can be asked for that:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Question 2.&lt;/strong&gt; Is every computable real number provably computable in ZFC, where computable means that the binary expansion is c.e. and provably computable means there is some machine that enumerates the binary expansion and which ZFC proves to be total? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is essentially equivalent to the first question, because the first question is just the relativization of the second to $0\prime$. &lt;/p&gt;&#xA;" OwnerUserId="630" LastEditorUserId="630" LastEditorDisplayName="" LastEditDate="2011-03-07T17:07:22.797" LastActivityDate="2011-03-07T17:07:22.797" CommunityOwnedDate="2011-03-07T16:03:51.863" />
  <row Id="25549" PostTypeId="1" AcceptedAnswerId="25558" CreationDate="2011-03-07T16:10:12.340" Score="1" ViewCount="69" Body="&lt;p&gt;Why is my overall delta percentage of 0.9% (bottom right) less than all 5 of the values that make it up (1.7%, 3.2%, ...)?  I know with the difference in magnitude of each row and column that it could skew the percentages, and I know that I'm subtracting percentages, but I'd still expect at least 1 of the values in the right-most column to be lower than 0.9%!&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;      A        B    B/A            C        D    D/C       D/C - B/A&#xA; 77,629   74,645   96.2%       7,986    7,814   97.8%        1.7%&#xA; 59,496   55,606   93.5%       9,779    9,448   96.6%        3.2%&#xA; 48,764   40,484   83.0%      11,413   10,320   90.4%        7.4%&#xA; 25,685   18,777   73.1%       9,901    7,882   79.6%        6.5%&#xA;  2,915    1,956   67.1%       1,165      822   70.6%        3.5%&#xA;&#xA;214,489  191,468   89.3%      40,244   36,286   90.2%        0.9%  Totals&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="7912" LastActivityDate="2011-03-07T17:07:08.583" Title="Percentages don't add up" Tags="&lt;algebra-precalculus&gt;&lt;arithmetic&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" />
  <row Id="25550" PostTypeId="1" AcceptedAnswerId="25608" CreationDate="2011-03-07T16:12:10.480" Score="0" ViewCount="141" Body="&lt;p&gt;This should be something relatively simple. I know there's a trick to this, I just can't remember it.&#xA;I have an equation&#xA;$$\frac{3x}{x-3}\geq 4.$$&#xA;I remember being shown at some point in my life that you could could multiply the entire equation by $(x-3)^2$ &#xA;in order to get rid of the divisor. However, &#xA;$$3x(x-3)\geq 4(x-3)^2.$$ &#xA;Really doesn't seem to go anywhere.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyone have any clue how to do this, that could perhaps show me/ point me in the right direction?&lt;/p&gt;&#xA;" OwnerUserId="7913" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-07T23:22:41.033" LastActivityDate="2011-03-09T15:59:26.193" Title="Squaring across an inequality with an unknown number" Tags="&lt;algebra&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="25551" PostTypeId="1" AcceptedAnswerId="25553" CreationDate="2011-03-07T16:12:49.193" Score="3" ViewCount="99" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Let $L=\mathbb{C}(t)$ (where $t$ is an indeterminate). Let $\phi\colon L\to L$ be the $\mathbb{C}$-automorphism of $L$ given by &#xA;  $$\phi(t) = \frac{3t-2}{4t-3},$$&#xA;  and let $G$ be the group generated by $\phi$. Find the fixed field of $G$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Could really use some help with this. Thanks!&lt;/p&gt;&#xA;" OwnerUserId="7907" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-07T18:39:06.927" LastActivityDate="2011-03-07T19:43:05.107" Title="Question regarding galois theory : finding a fixed field of a group generated by a C-automorphism" Tags="&lt;abstract-algebra&gt;&lt;group-theory&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="25552" PostTypeId="2" ParentId="25550" CreationDate="2011-03-07T16:24:43.407" Score="3" ViewCount="" Body="&lt;p&gt;$$\begin{align}&#xD;&#xA;3x(x-3)&amp;amp;\geq 4(x-3)^2 \\&#xD;&#xA;3x^2-9x &amp;amp;\geq 4(x^2-6x+9) \\&#xD;&#xA;3x^2-9x &amp;amp;\geq 4x^2-24x+36&#xD;&#xA;\end{align}$$&#xA;By rearranging we get&#xA;$$\begin{align}&#xD;&#xA;x^2-15x+36 &amp;amp;\leq 0 \\&#xD;&#xA;(x-3)(x-12) &amp;amp;\leq 0&#xD;&#xA;\end{align}&#xD;&#xA;$$&#xA;So $(x-3)&amp;gt; 0 \land (x-12)\leq0$ or $(x-3)&amp;lt; 0 \land (x-12)\geq 0$. ($x-3\neq0$)&lt;/p&gt;&#xA;" OwnerUserId="7044" LastEditorUserId="7044" LastEditorDisplayName="" LastEditDate="2011-03-08T08:29:32.730" LastActivityDate="2011-03-08T08:29:32.730" CommentCount="2" />
  <row Id="25553" PostTypeId="2" ParentId="25551" CreationDate="2011-03-07T16:39:45.587" Score="4" ViewCount="" Body="&lt;p&gt;Note that $\phi^2$ = identity on $L$. Let $K$ be the fixed field of $G = \langle\phi\rangle$. Then $[L : K] = 2$. Also $t$ satisfies the polynomial $(X - t)(X - \phi(t)) \in C(t+\phi(t), t\phi(t))[X] $. So the fixed field is $K = C(t+\phi(t), t\phi(t))$&lt;/p&gt;&#xA;" OwnerUserId="7565" LastEditorUserId="7565" LastEditorDisplayName="" LastEditDate="2011-03-07T19:43:05.107" LastActivityDate="2011-03-07T19:43:05.107" CommentCount="2" />
  <row Id="25555" PostTypeId="2" ParentId="25545" CreationDate="2011-03-07T16:43:28.553" Score="1" ViewCount="" Body="&lt;p&gt;Say $\gamma=[g_1,g_2]$, $\beta$ is the standard ordered basis, and $[T]_{\gamma}$ is the matrix from part $(a)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can find the change-of-basis matrix from $\gamma$ to $\beta$ by writing $g_1$ and $g_2$ as linear combinations of ${1 \choose 0}, {0 \choose 1}$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;${5 \choose 4}=5{1 \choose 0}+ 4{0 \choose 1}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;${-4 \choose 5}=-4{1 \choose 0}+ 5{0 \choose 1}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These coefficients form the columns of the change-of-basis-matrix from $\gamma$ to $\beta$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$[Q]_{\gamma}^\beta=&#xD;&#xA; \begin{pmatrix}&#xD;&#xA;  5 &amp;amp; -4 \\&#xD;&#xA;  4 &amp;amp; 5&#xD;&#xA; \end{pmatrix}&#xD;&#xA;$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can also find the change-of-basis-matrix from $\beta$ to $\gamma$ by reversing the process above (writing ${1 \choose 0}$ and ${0 \choose 1}$ as linear combinations of $g_1$ and $g_2$) or by taking the inverse of $[Q]_{\gamma}^\beta$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In either case, $[Q]_{\beta}^\gamma=\begin{pmatrix}&#xD;&#xA;  \frac{5}{41} &amp;amp; \frac{4}{41} \\&#xD;&#xA;  \frac{-4}{41} &amp;amp; \frac{5}{41}&#xD;&#xA; \end{pmatrix}&#xD;&#xA;$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then we multiply the matrices in the proper order. (The matrix $[T]_{\beta}$ is the matrix representation of $f$ with respect to the standard ordered basis.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$[T]_{\beta}=[Q]_{\gamma}^\beta[T]_{\gamma}[Q]_{\beta}^\gamma= \begin{pmatrix}&#xD;&#xA;  5 &amp;amp; -4 \\&#xD;&#xA;  4 &amp;amp; 5&#xD;&#xA; \end{pmatrix}&#xD;&#xA;\begin{pmatrix}&#xD;&#xA;  1 &amp;amp; 0 \\&#xD;&#xA;  0 &amp;amp; -1&#xD;&#xA; \end{pmatrix}&#xD;&#xA;\begin{pmatrix}&#xD;&#xA;  \frac{5}{41} &amp;amp; \frac{4}{41} \\&#xD;&#xA;  \frac{-4}{41} &amp;amp; \frac{5}{41}&#xD;&#xA; \end{pmatrix}=&#xD;&#xA;\begin{pmatrix}&#xD;&#xA;  \frac{9}{41} &amp;amp; \frac{40}{41} \\&#xD;&#xA;  \frac{40}{41} &amp;amp; \frac{-9}{41}&#xD;&#xA; \end{pmatrix}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope that helps!&lt;/p&gt;&#xA;" OwnerUserId="7840" LastEditorUserId="7840" LastEditorDisplayName="" LastEditDate="2011-03-07T17:03:48.870" LastActivityDate="2011-03-07T17:03:48.870" CommentCount="1" />
  <row Id="25556" PostTypeId="1" CreationDate="2011-03-07T16:51:53.137" Score="3" ViewCount="94" Body="&lt;p&gt;Why is the zero set in $\mathbb{C}$ of a polynomial f(x,y) in two complex variables always non-discrete (no zero of f is isolated)?&lt;/p&gt;&#xA;" OwnerUserId="7475" LastActivityDate="2011-03-08T00:29:40.683" Title="Why are there no discrete zero sets of a polynomial in two complex variables?" Tags="&lt;polynomials&gt;&lt;complex-variables&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="25557" PostTypeId="1" AcceptedAnswerId="25559" CreationDate="2011-03-07T16:54:38.290" Score="0" ViewCount="54" Body="&lt;p&gt;I have equations for two lines, one of which is linear and the other is logarithmic, ie:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;y = m1 * x + c1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;y = m2 * ln(x) + c2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;..and I need to find out where (if at all) these lines intersect.  I realise that I need to solve:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;m1 * x + c1 = m2 * ln(x) + c2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;..for x, but apart from shuffling the constants around I'm not sure how to do this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a general solution to this problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="7914" LastActivityDate="2011-03-07T18:59:27.197" Title="Find intersection of linear and logarithmic lines" Tags="&lt;geometry&gt;&lt;algebra&gt;&lt;graph&gt;&lt;intersection&gt;" AnswerCount="2" />
  <row Id="25558" PostTypeId="2" ParentId="25549" CreationDate="2011-03-07T16:55:25.553" Score="3" ViewCount="" Body="&lt;p&gt;Here's a simpler example that makes it clear how this can happen:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;     A      B    B/A          C      D   D/C     D/C-B/A      &#xA; 10000   9000    90%        100     95   95%        5%        &#xA;   100     86    86%      10000   9100   91%        5%        &#xA;&#xA; 10100   9086    90%      10100   9195   91%        1%  Totals&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Your &quot;Totals&quot; row is not a fair average of the other rows at all; when you just look at the percentages, you lose all information about which rows have big numbers and which rows have small numbers.  In your example, the upper rows for A and B are comparatively more significant, which skews the total B/A percentage up, while the lower rows for C and D are comparatively more significant, skewing the total D/C percentage down.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;If you really want to measure some kind of average of the difference of the percentages among all five rows, you have several options-- you could just take the average of your rightmost column, or you could take a weighted average (weighted by A or C, maybe).  The best choice depends on what the data means and on precisely what you want to measure.&lt;/p&gt;&#xA;" OwnerUserId="1461" LastEditorUserId="1461" LastEditorDisplayName="" LastEditDate="2011-03-07T17:07:08.583" LastActivityDate="2011-03-07T17:07:08.583" />
  <row Id="25559" PostTypeId="2" ParentId="25557" CreationDate="2011-03-07T17:00:09.210" Score="2" ViewCount="" Body="&lt;p&gt;There is no algebraic solution.  You can solve this numerically.  As log changes so slowly, iterative methods converge quickly.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-03-07T17:00:09.210" CommentCount="1" />
  <row Id="25561" PostTypeId="1" AcceptedAnswerId="25582" CreationDate="2011-03-07T17:17:33.333" Score="4" ViewCount="72" Body="&lt;p&gt;I am not that familiar with arguments using ordinals so this may be pretty simple. Let $T$ be a topology on a set $X$. For each ordinal $\alpha$ let $T_{\alpha}$ be a topology on $X$ satisfying $T\subseteq T_{\alpha+1}\subsetneq T_{\alpha}$ (notice the proper inclusion) and $T_{\alpha}=\bigcap_{\beta&amp;lt;\alpha}T_{\beta}$ when $\alpha$ is a limit ordinal. Apparently, this should imply that there is some ordinal $\gamma$ such that $T=T_{\gamma}$. Why is this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: I see there is a problem with the above question. I will try to right this here. Henno is right that I am referring to a certain $f$ that makes a specific topology containing $T$ coarser but so that the new topology still contains $T$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;New question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We start with a topology $T_0$ on $X$ containing $T$ and have $T\subseteq T_{\alpha+1}=f(T_{\alpha})$ for successor ordinals. We have $T_{\alpha}=\bigcap_{\beta&amp;lt;\alpha}T_{\beta}$ for limit ordinals. We also have that $T_{\alpha+1}=T$ whenever $T_{\alpha+1}=T_{\alpha}$. Apparently there should be a $\gamma$ such that $T=T_{\gamma}$. Why is this?&lt;/p&gt;&#xA;" OwnerUserId="2423" LastEditorUserId="2423" LastEditorDisplayName="" LastEditDate="2011-03-07T18:39:32.733" LastActivityDate="2011-03-07T18:53:32.260" Title="Transfinite sequences of topologies" Tags="&lt;topology&gt;&lt;ordinals&gt;" AnswerCount="2" />
  <row Id="25562" PostTypeId="1" CreationDate="2011-03-07T17:21:47.063" Score="0" ViewCount="70" Body="&lt;p&gt;Consider two groups $(\mathbb{Z},+)$ and $(\{1,-1,i,-i\},\cdot)$ where $i^2=-1$, show that the mapping defined by $f(n)=i^n$ for $n$ belonging to $\mathbb{Z}$ is a homomorphism from $(\mathbb{Z},+)$ onto $(\{1,-1,i,-i\},\cdot)$, and determine its kernel.&lt;/p&gt;&#xA;" OwnerUserId="7916" LastEditorUserId="1583" LastEditorDisplayName="" LastEditDate="2011-03-07T17:50:08.243" LastActivityDate="2011-03-07T17:50:08.243" Title="abstract algebra homomorphism" Tags="&lt;abstract-algebra&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="25563" PostTypeId="2" ParentId="24854" CreationDate="2011-03-07T17:22:36.477" Score="22" ViewCount="" Body="&lt;p&gt;Set theory today is a vibrant, active research area,&#xA;characterized by intense fundamental work both on set&#xA;theory's own questions, arising from a deep historical&#xA;wellspring of ideas, and also on the interaction of those&#xA;ideas with other mathematical subjects. It is fascinating and I would encourage anyone to learn more about it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the field is simply too vast to summarize easily,&#xA;allow me merely to describe a few of the major topics that&#xA;are actively studied in set theory today.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Large_cardinal&quot; rel=&quot;nofollow&quot;&gt;Large&#xA;cardinals&lt;/a&gt;.&lt;/b&gt;&#xA;These are the strong axioms of infinity, first studied by&#xA;Cantor, which often generalize properties true of $\omega$&#xA;to a larger context, while providing a robust hierarchy of&#xA;axioms increasing in consistency strength. Large cardinal&#xA;axioms often express combinatorial aspects of infinity,&#xA;which have powerful consequences, even low down. To give&#xA;one deep example, if there are sufficiently many Woodin&#xA;cardinals, then all projective sets of reals are Lebesgue&#xA;measurable, a shocking but very welcome situation. You may&#xA;recognize some of the various large cardinal&#xA;concepts---inaccessible, Mahlo, weakly compact,&#xA;indescribable, totally indescribable, unfoldable, Ramsey,&#xA;measurable, tall, strong, strongly compact, supercompact,&#xA;almost huge, huge and so on---and new large cardinal&#xA;concepts are often introduced for a particular purpose.&#xA;(For example, in recent work Thomas Johnstone and I proved&#xA;that a certain forcing axiom was exactly equiconsistent&#xA;with what we called the uplifting cardinals.) I encourage&#xA;you to follow the Wikipedia link for more information.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Forcing_%28set_theory%29&quot; rel=&quot;nofollow&quot;&gt;Forcing&lt;/a&gt;.&lt;/b&gt;&#xA;The subject of set theory came to maturity with the&#xA;development of forcing, an extremely flexible technique for&#xA;constructing new models of set theory from existing models.&#xA;If one has a model of set theory $M$, one can construct a&#xA;forcing extension $M[G]$ by adding a new ideal element $G$,&#xA;which will be an $M$-generic filter for a forcing notion&#xA;$\mathbb{P}$ in $M$, akin to a field extension in the sense&#xA;that every object in $M[G]$ is constructible algebraically&#xA;from $G$ and objects in $M$. The interaction of a model of&#xA;set theory with its forcing extensions provides an&#xA;extremely rich, intensely studied mathematical context.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Independence Phenomenon.&lt;/b&gt; The initial uses of forcing&#xA;were focused on proving diverse independence results, which&#xA;show that a statement of set theory is neither provable nor&#xA;refutable from the basic ZFC axioms. For example, the&#xA;Continuum Hypothesis is famously independent of ZFC, but we&#xA;now have thousands of examples. Although it is now the norm&#xA;for statements of infinite combinatorics to be independent,&#xA;the phenomenon is particularly interesting when it is shown&#xA;that a statement from outside set theory is independent,&#xA;and there are many prominent examples.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Forcing Axioms.&lt;/b&gt; The first forcing axioms were often&#xA;viewed as unifying combinatorial assertions that could be&#xA;proved consistent by forcing and then applied by&#xA;researchers with less knowledge of forcing. Thus, they&#xA;tended to unify much of the power of forcing in a way that&#xA;was easily employed outside the field. For example, one&#xA;sees applications of &lt;a href=&quot;http://en.wikipedia.org/wiki/Martin%27s_axiom&quot; rel=&quot;nofollow&quot;&gt;Martin's&#xA;Axiom&lt;/a&gt;&#xA;undertaken by topologists or algebraists. Within set&#xA;theory, however, these axioms are a focal point, viewed as&#xA;expressing particularly robust collections of consequences,&#xA;and there is intense work on various axioms and finding&#xA;their large cardinal strength.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Inner_model_theory&quot; rel=&quot;nofollow&quot;&gt;Inner model&#xA;theory&lt;/a&gt;.&lt;/b&gt;&#xA;This is a huge on-going effort to construct and understand&#xA;the canonical fine-structural inner models that may exist&#xA;for large cardinals, the analogues of G&amp;ouml;del's&#xA;constructible universe $L$, but which may accommodate large&#xA;cardinals. Understanding these inner models amounts in a&#xA;sense to the ability to take the large cardinal concept&#xA;completely apart and then fit it together again. These&#xA;models have often provided a powerful tool for showing that&#xA;other mathematical statements have large cardinal strength.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;&lt;a href=&quot;http://www.math.lsa.umich.edu/~ablass/hbk.pdf&quot; rel=&quot;nofollow&quot;&gt;Cardinal characteristics of the&#xA;continuum&lt;/a&gt;.&lt;/b&gt;&#xA;This subject is concerned with the diverse cardinal&#xA;characteristics of the continuum, such as the size of the&#xA;smallest non-Lebesgue measurable set, the additivity of the&#xA;null ideal or the cofinality of the order $\omega^\omega$&#xA;under eventual domination, and many others. These cardinals&#xA;are all equal to the continuum under CH, but separate into&#xA;a rich hierarchy of distinct notions when CH fails.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Descriptive_set_theory&quot; rel=&quot;nofollow&quot;&gt;Descriptive set&#xA;theory&lt;/a&gt;.&lt;/b&gt;&#xA;This is the study of various complexity hierarchies at the&#xA;level of the reals and sets of reals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;&lt;a href=&quot;http://www.math.ucla.edu/~greg/223b.1.06s/hjorth.dvi&quot; rel=&quot;nofollow&quot;&gt;Borel equivalence relation&#xA;theory&lt;/a&gt;.&lt;/b&gt;&#xA;Arising from descriptive set theory, this subject is an&#xA;exciting comparatively recent development in set theory,&#xA;which provides a precise way to understand what otherwise&#xA;might be a merely informal understanding of the comparative&#xA;difficulty of classification problems in mathematics. The&#xA;idea is that many classification problems arising in&#xA;algebra, analysis or topology turn out naturally to&#xA;correspond to equivalence relations on a standard Borel&#xA;space. These relations fit into a natural hierarchy under&#xA;the notion of Borel reducibility, and this notion provides&#xA;us with a way to say that one classification problem in&#xA;mathematics is at least as hard as or strictly harder than&#xA;another. Researchers in this area are deeply knowledgable&#xA;both about set theory and also about the subject area in&#xA;which their equivalence relations arise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Philosophy of set theory.&lt;/b&gt; Lastly, let me also&#xA;mention the emerging subject known as the philosophy of set&#xA;theory, which is concerned with some of the philosophical&#xA;issues arising in set theoretic research, particularly in&#xA;the context of large cardinals, such as: How can we decide&#xA;when or whether to adopt new mathematical axioms? What does&#xA;it mean to say that a mathematical statement is true? In&#xA;what sense is there an intended model of the axioms of set&#xA;theory? Much of the discussion in this area weaves together&#xA;profoundly philosophical concerns with extremely technical&#xA;mathematics concerning deep features of forcing, large&#xA;cardinals and inner model theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Remark.&lt;/b&gt; I see in your answer to the linked question&#xA;you mentioned that you may not have been exposed to much&#xA;set theory at Harvard, and I find this a pity. I would&#xA;encourage you to look beyond any limiting perspectives you&#xA;may have encountered, and you will discover the rich,&#xA;fascinating subject of set theory. The standard&#xA;introductory level graduate texts would be Jech's book Set&#xA;Theory and Kanamori's book The Higher Infinite, on large&#xA;cardinals, and both of these are outstanding.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I apologize for this too-long answer...&lt;/p&gt;&#xA;" OwnerUserId="413" LastActivityDate="2011-03-07T17:22:36.477" CommentCount="3" />
  <row Id="25564" PostTypeId="2" ParentId="25561" CreationDate="2011-03-07T17:26:34.793" Score="2" ViewCount="" Body="&lt;p&gt;For every $\alpha$ we have $T_\alpha\subseteq P(P(X))$, if you have a decreasing chain of anything longer than $2^{2^X}$ then it has to stabilize somewhere by a simple argument of cardinality.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since all the topologies include $T$ it has to be that at some point we reached it.&lt;/p&gt;&#xA;" OwnerUserId="622" LastActivityDate="2011-03-07T17:26:34.793" CommentCount="3" />
  <row Id="25565" PostTypeId="2" ParentId="25562" CreationDate="2011-03-07T17:38:12.243" Score="3" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Hint.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;For proving $f: (G,\circ) \to (G&amp;#39;,\cdot)$ is a group homomorphism, show that $f$ satisfies $f( x \circ y) = f(x) \cdot f(y)$. Here your binary operations are $+$ and $\cdot$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" LastActivityDate="2011-03-07T17:38:12.243" />
  <row Id="25566" PostTypeId="2" ParentId="2194" CreationDate="2011-03-07T17:39:48.530" Score="3" ViewCount="" Body="&lt;p&gt;In Jan van Mill's books (infinite dimensional topology, a prerequisite; the infinite-dimensional topology of function spaces) there are complete self-contained proofs (starting from basic stuff about separable metric spaces) [in essence the same proof in both, but the second book is more recent so might be more readily available]. It is based on Anderson's proof of this, and uses the Inductive Convergence criterion, which states in essence that when (for compact metrisable $X$) $h_n$ is a sequence of homeomorphisms of $X$, there is a sequence of $\epsilon_n &amp;gt; 0$ such that if $d(h_n, 1_X) &amp;lt; \epsilon_n$, where $1_X$ is the identity on $X$, and $d$ is a complete distance on the space of homeomorphisms of $X$, that then the limit of $h_n \circ h_{n-1} \circ \ldots h_1$ as $n$ tends to infinity exists and is a homeomorphism of $X$. Also, each $\epsilon_n$ depends only on the $h_i$ with $i &amp;lt; n$. This allows one to construct homeomorphisms of compact metrisable spaces in a very controlled way. Note that a metric on the Hilbert cube is such that a movement in a high coordinate only contributes a very small amount to the distance, so we can push problems to higher and higher coordinates and &quot;solve&quot; them there by a sufficiently small movement. With this criterion the proof is just 2 pages or so.&lt;/p&gt;&#xA;" OwnerUserId="4280" LastActivityDate="2011-03-07T17:39:48.530" />
  <row Id="25567" PostTypeId="1" CreationDate="2011-03-07T17:40:49.320" Score="0" ViewCount="80" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Let $K$ be any field and $s$ an indeterminate. Then $K(s)$ is a field extension of $K (s^n )$. Prove that $[K (s):K (s^n )]=n$. Hence   show    that    the minimum polynomial of $s$ over $K(s^n)$ is $t^n − s^n$.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;[Hint: first show that $s$ satisfies a poly of degree $n$ over $K(s^n)$; this gives $\leq$. Then show that $\{1, s, \ldots, s^{n−1}\}$ is Linearly Independent over $K(s^n)$; this gives $\geq$.]&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I would really like some help with this please!&lt;/p&gt;&#xA;" OwnerUserId="7907" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-07T18:28:41.917" LastActivityDate="2011-04-06T18:39:35.267" Title="Galois theory help needed for field extensions" Tags="&lt;abstract-algebra&gt;&lt;group-theory&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="25568" PostTypeId="1" CreationDate="2011-03-07T17:44:33.700" Score="13" ViewCount="188" Body="&lt;p&gt;A pair of dice is rolled repeatedly until each outcome (2 through 12) has occurred at least once. What is the expected number of rolls necessary for this to occur?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notes: This is not very deep conceptually, but because of the unequal probabilities for the outcomes, it seems that the calculations involved are terribly messy.  It must have been done already (dice have been studied for centuries!) but I can't find a discussion in any book, or on line. Can anybody give a reference?&lt;/p&gt;&#xA;" OwnerUserId="7917" LastActivityDate="2011-03-07T18:15:14.517" Title="A Question About Dice" Tags="&lt;probability&gt;&lt;combinatorics&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" />
  <row Id="25569" PostTypeId="1" CreationDate="2011-03-07T17:48:03.353" Score="4" ViewCount="57" Body="&lt;p&gt;Where could I find a formula that produces integers $n$ and $m$ such that $\binom{n}{2}/\binom{m}{2}=1/2$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, this questions can be reformulated as: How to find all the integer values of n and m such that $$2n(n-1) = m(m-1)\quad ?$$&lt;/p&gt;&#xA;" OwnerUserId="7918" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2011-03-07T18:01:26.630" LastActivityDate="2011-03-07T18:25:01.897" Title="Formula for $\binom{n}{2}/\binom{m}{2}=1/2$ or $2n(n-1) = m(m-1)$?" Tags="&lt;combinatorics&gt;&lt;elementary-number-theory&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="25570" PostTypeId="2" ParentId="25569" CreationDate="2011-03-07T17:56:22.457" Score="7" ViewCount="" Body="&lt;p&gt;Multiply by $4$ to get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ 2(2n-1)^2 - 1 = (2m-1)^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a form of Pell's equation ($\displaystyle 2x^2 - y^2 = 1$) and has infinite solutions in which both $x$ and $y$ are odd. These can be read off from the continued fraction of $\sqrt{2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The solutions to $\displaystyle 2x^2 - y^2 = 1$ with $\displaystyle x$ and $\displaystyle y$ odd, can be read off from the recurrence&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle x_{n+1} = 3x_n + 2y_n$&lt;br/&gt;&#xA;$\displaystyle y_{n+1} = 4x_n + 3y_n$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;starting with $\displaystyle x_1 = 1,\ y_1 = 1$.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2011-03-07T18:10:59.667" LastActivityDate="2011-03-07T18:10:59.667" />
  <row Id="25571" PostTypeId="2" ParentId="25430" CreationDate="2011-03-07T17:58:05.247" Score="2" ViewCount="" Body="&lt;p&gt;I like the book &lt;a href=&quot;http://algo.inria.fr/flajolet/Publications/book.pdf&quot; rel=&quot;nofollow&quot;&gt;&quot;Analytic Combinatorics&quot;&lt;/a&gt; by Philippe Flajolet and Robert Sedgewick  Part A. It doesn't only show a way how you translate combinatorical problems in terms functional relations of  generating functions (the show that unlabelled structures relate to ordinary generating functions,  and labelled structures relate to exponential generating functions), but it also provides ways (asymptotic expansion, ...) such that you can actually calculate something.&lt;/p&gt;&#xA;" OwnerUserId="7266" LastActivityDate="2011-03-07T17:58:05.247" />
  <row Id="25572" PostTypeId="2" ParentId="25567" CreationDate="2011-03-07T18:03:06.307" Score="2" ViewCount="" Body="&lt;p&gt;Shwoing that $[K(s):K(s^n)] \leq n$ is easy - in fact you have already written down a polynomial of $s$ over $K(s^n)$, which is $t^n - s^n$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To show the other direction of the inequality, it suffices to show that $\{1,s, \cdots, s^{n-1}\}$ is linearly independent over $K(s^n)$. Write down a linear dependence relation. Clearing denominators, we can assume that there is a relation $\sum_{i=0}^{n-1} a_i s^i = 0$, with $a_i \in K[s^n]$. Now read off the degree of each term. Can you deduce that $a_i = 0$ for all $i$?&lt;/p&gt;&#xA;" OwnerUserId="325" LastActivityDate="2011-03-07T18:03:06.307" />
  <row Id="25573" PostTypeId="2" ParentId="25568" CreationDate="2011-03-07T18:03:41.947" Score="3" ViewCount="" Body="&lt;p&gt;Suppose the probabilities are $p_i$. For each set $S$, the probability that in the first $n$ terms we have only seen throws from $S$ is&#xA;$$ p_S^n, \quad p_S \triangleq \sum_{i \in S} p_i. $$&#xA;The probability that we have &lt;em&gt;not&lt;/em&gt; seen all outcomes by the $n$th throw is&#xA;$$ r_n = \sum_{S \neq \emptyset} (-1)^{|S|+1} p_{\overline{S}}^n. $$&#xA;The required expectation is&#xA;$$ E = \sum_{n \geq 1} r_n. $$&#xA;Changing the order of summation, the summand corresponding to $S$ contributes to the sum&#xA;$$ (-1)^{|S|+1} \sum_{n \geq 1} p_{\overline{S}}^n = \frac{(-1)^{|S|+1} p_{\overline{S}}}{p_S}. $$&#xA;So we get the formula&#xA;$$ E = \sum_{S \neq \emptyset} (-1)^{|S|+1} \left(\frac{1}{p_S} - 1 \right) = \sum_{S \neq \emptyset} \frac{(-1)^{|S|+1}}{p_S} - 1. $$&#xA;You can now in principle plug in the values of the dice and get the result. Since there are $11$ outcomes, you need to sum $2^{11} = 2048$ reciprocals (including the final $-1$).&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2011-03-07T18:03:41.947" CommentCount="1" />
  <row Id="25574" PostTypeId="2" ParentId="25569" CreationDate="2011-03-07T18:08:05.383" Score="4" ViewCount="" Body="&lt;p&gt;In &lt;a href=&quot;http://math.stackexchange.com/questions/9269/how-to-solve-inhomogeneous-quadratic-forms-in-integers/9288#9288&quot;&gt;my answer here&lt;/a&gt; you will find Lagrange's method for reducing the general bivariate quadratic Diophantine equation to a Pell equation, along with some interesting links, including a Java applet that will show you the complete solution, including complete step-by-step details of the solution.&#xA;According to &lt;a href=&quot;http://math.stackexchange.com/questions/9269/how-to-solve-inhomogeneous-quadratic-forms-in-integers/9288#9288&quot;&gt;said applet&lt;/a&gt; the solutions are given by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\rm N_{i+1}\ =\ 3\ N_i + 2\ M_i -2 $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\rm M_{i+1}\ =\ 4\ N_i + 3\ M_i -3 $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with $\rm\ N_0,M_0\ =\ 0,1\:;\ \ 1,1\:.\ $  To see complete details of the solution, including the relationship with the continued fraction of $\rm\:\sqrt{2}\:,\:$ choose the &quot;step by step&quot; solution method.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2011-03-07T18:25:01.897" LastActivityDate="2011-03-07T18:25:01.897" />
  <row Id="25575" PostTypeId="1" AcceptedAnswerId="25579" CreationDate="2011-03-07T18:11:47.053" Score="3" ViewCount="142" Body="&lt;p&gt;How do I prove&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(2^r-1)(1-x)x^{2^{r}-2}+x^{2^{r}-1}&amp;gt;x^{2^{r}-r-1}$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;for $\frac{1}{2}&amp;lt;x&amp;lt;1$ and $r \in \mathbb N$ $(r\neq 1)$?&lt;/p&gt;&#xA;" OwnerUserId="5877" LastEditorUserId="5877" LastEditorDisplayName="" LastEditDate="2011-03-07T19:00:25.743" LastActivityDate="2011-03-07T20:38:18.133" Title="Proof by induction for an exponential inequality" Tags="&lt;inequality&gt;&lt;induction&gt;" AnswerCount="3" CommentCount="5" />
  <row Id="25576" PostTypeId="2" ParentId="25568" CreationDate="2011-03-07T18:15:14.517" Score="11" ViewCount="" Body="&lt;p&gt;This is the coupon collectors problem with unequal probabilities. There is a treatment of this problem in &lt;strong&gt;Example 5.17&lt;/strong&gt; of the 10th edition of &lt;em&gt;Introduction to Probability Models&lt;/em&gt; by Sheldon Ross (page 322). He solves the problem by embedding it into a Poisson process. Anyway, the answer is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ E[T]=\int_0^\infty \left(1-\prod_{j=1}^m (1-e^{-p_j t})\right) dt, $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;when there are $m$ events with probability $p_j, j=1, \dots, m$ of occurring.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your particular problem with two fair dice, my calculations give&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$E[T] = {769767316159\over 12574325400} = 61.2173.$$&lt;/p&gt;&#xA;" OwnerUserId="940" LastActivityDate="2011-03-07T18:15:14.517" CommentCount="2" />
  <row Id="25579" PostTypeId="2" ParentId="25575" CreationDate="2011-03-07T18:37:21.690" Score="4" ViewCount="" Body="&lt;p&gt;Assuming $1/2 \lt x \lt 1$ and $r \in \mathbb{N}$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Set $y = 1/x$ and multiply by $y^{2^{r} - 1}$ to get (which is equivalent to the original for $y \gt 0$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(2^r - 1)(y-1)  + 1 \gt y^r$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which is true if (as $y \gt 1$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(2^r - 1) \gt \frac{y^r-1}{y-1}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which is same as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ 1 + 2 + 2^2 + \dots + 2^{r-1} \gt 1 + y + y^2 + \dots + y^{r-1}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which is true if $ 1 \lt y \lt 2$ (i.e $1/2 \lt x \lt 1$) and $r \neq 1$.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2011-03-07T20:38:18.133" LastActivityDate="2011-03-07T20:38:18.133" CommentCount="5" />
  <row Id="25580" PostTypeId="1" CreationDate="2011-03-07T18:45:19.727" Score="1" ViewCount="46" Body="&lt;p&gt;Given is a set of rigid 3D transformations, $(R_1, t_1) \ldots (R_n, t_n)$ (i.e. several poses = rotation $R$ + translation $t$). Each of these poses represents a transformation from local to world coordinates of an object that was rotated about a fixed axis. How can the &quot;best fit&quot; rotation axis in local coordinates be computed?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is my approach (I am not sure how valid it is):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $x$ is a point on the local rotation axis, its world position $R_ix + t_i$ will be identical for all given poses (because points on the rotation axis are fixed points under rotation about this axis):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$R_1 x + t_1 = R_2 x + t_2 = R_3 x + t_3$ and so on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some rearranging yields a system of equations:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(R_1 - R_2) x = (t_2 - t_1)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(R_2 - R_3) x = (t_3 - t_2)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(R_3 - R_1) x = (t_1 - t_3)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &quot;chain equation&quot; given above translates to $n (n-1)/2$ equations (by enumerating all possible pairs of transformations, which are transitively equal).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This equation system is an instance of $Ax = b$. Now, I am at a loss of how to solve this... on the one hand, this equation system is clearly overdetermined (because there are many more equations than unknowns), but on the other hand there are infinitely many solutions (because all points on the rotation axis solve it), so the usual least-squares approach will not work here (because $A^TA$ will be singular).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not even sure if the approach above makes any sense... any hints at how to solve this problem in a &quot;best fit&quot; sense are appreciated.&lt;/p&gt;&#xA;" OwnerUserId="7921" LastEditorUserId="6622" LastEditorDisplayName="" LastEditDate="2011-03-07T19:33:46.017" LastActivityDate="2011-03-07T19:33:46.017" Title="Determine axis of rotation from pose measurements" Tags="&lt;linear-algebra&gt;" />
  <row Id="25581" PostTypeId="2" ParentId="25556" CreationDate="2011-03-07T18:46:03.593" Score="4" ViewCount="" Body="&lt;p&gt;Write $f(x,y)$ as a polynomial in $y$ with coefficients in $\mathbb C[x]$, so that $f(x,y)=\sum_{i=0}^nf_i(x)y^i$. If $x\in\mathbb C$ is such that $f_n(x)\neq0$, then there exist $n$ values of $y$ such that $f(x,y)=0$. It follows that if $X\subseteq \mathbb C\times\mathbb C$ is the zero locus of $f$, then the first projection $\pi_x:X\to\mathbb C$ has an image with finite complement. This implies, in particular, that $X$ is uncountable, and no uncountable subset of $\mathbb C\times\mathbb C$ is discrete.&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2011-03-07T18:46:03.593" CommentCount="2" />
  <row Id="25582" PostTypeId="2" ParentId="25561" CreationDate="2011-03-07T18:53:32.260" Score="4" ViewCount="" Body="&lt;p&gt;The reason is that as you proceed with your recursion, you are gradually throwing out objects from your set. Since you only have a set of objects to begin with, you must eventually run out of things to throw out. The only way this can happen, under the assumptions you have set up, is if you have hit the bottom topology $T$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One can turn this idea into a proof by noting that if the recursion never hit $T$, then you would have an injection from the class of ordinals into the power set of $T_0$, by the map $\alpha\mapsto (T_\alpha-T_{\alpha+1})$. (Or you could also select an element from this set and get a map directly into $T_0$.) This would contradict Hartog's theorem among others, that for every set there is an ordinal that does not inject into it. &lt;/p&gt;&#xA;" OwnerUserId="413" LastActivityDate="2011-03-07T18:53:32.260" CommentCount="1" />
  <row Id="25583" PostTypeId="1" AcceptedAnswerId="25587" CreationDate="2011-03-07T18:54:26.057" Score="1" ViewCount="147" Body="&lt;p&gt;$y = \sin(\pi x), 0 \le x \le 1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$y&amp;#39; = \pi \cos(\pi x)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(y&amp;#39;)^2 = \pi^2\cos^2(\pi x)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ds = \sqrt{1 + \pi^2 \cos^2(\pi x)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$r = y = \sin(\pi x)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$S = \displaystyle\int_0^1 2 \pi\sin(\pi x)\sqrt{1 + \pi^2 \cos^2(\pi x)} dx$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm a little lost as to what to do next. Should I have simplified ds more or do I need to do a substitution with $u = \pi^2\cos^2(\pi x)$?&lt;/p&gt;&#xA;" OwnerUserId="6331" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2011-03-08T12:18:20.477" LastActivityDate="2011-03-08T12:18:20.477" Title="Find the area of the surface obtained by rotating the curve about the x axis" Tags="&lt;calculus&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" />
  <row Id="25584" PostTypeId="2" ParentId="25557" CreationDate="2011-03-07T18:59:27.197" Score="0" ViewCount="" Body="&lt;p&gt;The general solution involves the &lt;a href=&quot;http://en.wikipedia.org/wiki/Lambert_W_function&quot; rel=&quot;nofollow&quot;&gt;Lambert W function&lt;/a&gt;, whose defining equation is $z = W(z)e^{W(z)}$ for complex numbers $z$.  If either $m_1$ or $m_2$ is zero in the given problem, then the solution is elementary, so suppose $m_1,m_2 \neq 0$.  Then&#xA;$$&#xD;&#xA;m_1 x + c_1 = m_2 \ln x + c_2&#xD;&#xA;$$&#xA;can be rewritten as&#xA;$$&#xD;&#xA;-\frac{m_1}{m_2}e^{(c_1-c_2)/m_2} = -\frac{m_1 x}{m_2} e^{-m_1 x / m_2},&#xD;&#xA;$$&#xA;which has the solution&#xA;$$&#xD;&#xA;-\frac{m_1 x}{m_2} = W\left(-\frac{m_1}{m_2}e^{(c_1-c_2)/m_2}\right),&#xD;&#xA;$$&#xA;or&#xA;$$&#xD;&#xA;x = -\frac{m_2}{m_1} W\left(-\frac{m_1}{m_2}e^{(c_1-c_2)/m_2}\right).&#xD;&#xA;$$&#xA;For instance, if $m_1=1$, $m_2=-1$, and $c_1=c_2=0$, this gives $x=W(1)=\Omega=0.56714329...$ (the &lt;a href=&quot;http://en.wikipedia.org/wiki/Omega_constant&quot; rel=&quot;nofollow&quot;&gt;Omega constant&lt;/a&gt;), which is correct, since $\Omega = -\ln\Omega$.&lt;/p&gt;&#xA;" OwnerUserId="5546" LastActivityDate="2011-03-07T18:59:27.197" />
  <row Id="25585" PostTypeId="2" ParentId="25217" CreationDate="2011-03-07T19:07:37.123" Score="1" ViewCount="" Body="&lt;p&gt;For reference, this problem is exercise 1.6 (page 14) of &lt;em&gt;The Cauchy-Schwarz Master Class&lt;/em&gt; by J. Michael Steele. His solution is similar to Zarrax's, but for the lower bound on $\sum_{k=1}^n {1\over p_k}$  he uses the Cauchy-Schwarz inequality twice; first on the splitting $1=\sqrt{p_k}\cdot {1\over \sqrt{p_k}}$ and then on the splitting ${1\over p_k}=1\cdot {1\over p_k}$. &lt;strong&gt;Edit:&lt;/strong&gt; I forgot to mention that the OP did this in the same way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In Exercise 13.6 Steele gives a generalization, that for $\alpha&amp;gt;0$, we have &#xA;$$\sum_{k=1}^n\left(p_k+{1\over p_k}\right)^\alpha \geq {(n^2+1)^\alpha\over n^{\alpha-1}}.$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you like inequalities, I strongly suggest getting this book. It is a gem!&lt;/p&gt;&#xA;" OwnerUserId="940" LastActivityDate="2011-03-07T19:07:37.123" />
  <row Id="25586" PostTypeId="1" AcceptedAnswerId="25591" CreationDate="2011-03-07T19:08:39.510" Score="3" ViewCount="181" Body="&lt;p&gt;We know that $$\int_0^\infty \left(\frac{\sin x}{x}\right)^2 dx=\int_0^\infty \frac{\sin x}{x} dx=\frac{\pi}{2}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do I show that $$\int_0^\infty \left\vert\frac{\sin x}{x}\right\vert dx$$ converges? &lt;/p&gt;&#xA;" OwnerUserId="7815" LastEditorUserId="1543" LastEditorDisplayName="" LastEditDate="2011-05-03T09:47:36.800" LastActivityDate="2011-05-03T09:47:36.800" Title="How to prove absolute summability of sinc function?" Tags="&lt;calculus&gt;&lt;integral&gt;&lt;definite-integral&gt;" AnswerCount="3" />
  <row Id="25587" PostTypeId="2" ParentId="25583" CreationDate="2011-03-07T19:08:49.970" Score="1" ViewCount="" Body="&lt;p&gt;Hint:  I think a substitution of $u=\pi \cos(\pi x)$, will help as you have $du$ available.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-03-07T19:08:49.970" CommentCount="3" />
  <row Id="25588" PostTypeId="1" CreationDate="2011-03-07T19:11:12.750" Score="4" ViewCount="169" Body="&lt;p&gt;Hi, I have read (I think) all of the previous threads on this website (and many others) &#xA;on this topic &amp;amp; unfortunately have not found an answer to my question. Due to the fact that I am only beginning to study the subjects of tensors, differential forms etc... &#xA;I appeal to your good nature not to respond with anything too advanced as it just goes over my head. The structure of my question is as follows: first a look at the cross product; then a look at a wedge product calculation &amp;amp; it's similarities (that I think &#xA;are far more explicit if you interpret it in the way I've explained below) to the cross product &amp;amp; finally 4 questions (in bold) that are motivated by the wedge product calculation. &#xA;The 4 questions are all I'm hoping for answers to as I barely understand anything beyond &#xA;what I've written about this subject &amp;amp; just don't understand what a lot of the online &#xA;fora threads on this topic are saying in the responses to this topic. I think these &#xA;questions are my way in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The cross product is a strange animal, it really has very little justification as it is &#xA;taught in elementary linear algebra books. It took me a long time to learn that the &#xA;cross product is really no more than the dot product in disguise. It is actually quite &#xA;easy to derive the result that a cross product gives, through clever algebra, as is done in the cross product pdf's &lt;a href=&quot;http://www.owlnet.rice.edu/~fjones/&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; &amp;amp; here: synechism.org/drupal/cfsv/ (sorry about the text-link!). By doing &#xA;your own algebra you can &#xA;justify the anti-symmetric property of the cross product,&#xA;$\overline{u} \ x \ \overline{v} \ = \ - \ \overline{v} \ x \ \overline{u}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So understanding the cross product in this way is quite satisfying to me as we can &#xA;easily justify why $\ \overline{u} \ x \ \overline{u} \ = \ 0$ without relying on &#xA;these properties as definitions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My questions are based on the fact that these properties can be justified in such an &#xA;elementary way. If you've never seen the cross product explained they way it is in the .pdf's then I urge you to read them &amp;amp; think seriously about it. I'm sure these are &#xA;justified in more advanced works in other ways but if an explanation can be given &#xA;at this level I see no reason not to take it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So lets look at an example &amp;amp; the steps taken that I think have explanations analogous &#xA;to those of the cross product above:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ \overline{v}$ = v₁$\ \hat{e_1}$ + v₂$\ \hat{e_2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ \overline{w}$ = w₁$\ \hat{e_1}$ + w₂$\ \hat{e_2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(where $\ \hat{e_1}$ = (1,0) &amp;amp; $\ \hat{e_2}$ = (0,1)).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;v ⋀ w = (v₁$\ \hat{e_1}$ + v₂$\ \hat{e_2}$) ⋀ (w₁$\ \hat{e_1}$ + w₂$\ \hat{e_2}$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;v ⋀ w = v₁w₁$\ \hat{e_1}$⋀$\ \hat{e_1}$ + v₁w₂$\ \hat{e_1}$⋀$\ \hat{e_2}$ + v₂w₁$\ \hat{e_2}$⋀$\ \hat{e_1}$ + v₂w₂$\ \hat{e_2}$⋀$\ \hat{e_2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;v ⋀ w = v₁w₂$\ \hat{e_1}$⋀$\ \hat{e_2}$ + v₂w₁$\ \hat{e_2}$⋀$\ \hat{e_1}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;v ⋀ w= (v₁w₂ - v₂w₁)$\ \hat{e_1}$⋀$\ \hat{e_2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is interpreted as the area contained in the parallelogram formed by v &amp;amp; w.&#xA;No doubt you noticed that all of the manipulations with the $\ \hat{e_i}$'s have &#xA;the exact same form as the cross product. Notice also the fact that this two &#xA;dimensional calculation comes out with the exact same result as the cross product of&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ \overline{v&amp;#39;}$ = v₁$\ \hat{e_1}$ +_v₂$\ \hat{e_2}$_+ 0$\ \hat{e_3}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ \overline{w&amp;#39;}$ = w₁$\ \hat{e_1}$ + w₂$\ \hat{e_2}$ + 0$\ \hat{e_3}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;in ℝ³.Also the general &lt;/p&gt;&#xA;&#xA;&lt;p&gt;x ⋀ y = (x₁$\ \hat{e_1}$ + x₂$\ \hat{e_2}$ + x₃$\ \hat{e_3}$) ⋀ (y₁$\ \hat{e_1}$ + y₂$\ \hat{e_2}$ + y₃$\ \hat{e_3}$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;comes out with the exact same result as the cross product. The important thing is that &#xA;the cross product of the two vector results in a vector orthogonal to $\ \overline{v}$ &amp;amp; $\ \overline{w}$ and that the result is the same as the wedge product calculation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;1: Can $\ \hat{e_1}$⋀$\ \hat{e_2}$ be interpreted as $\ \hat{e_3}$ in my above &#xA;calculation?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I mean is that can $\ \hat{e_1}$⋀$\ \hat{e_2}$ be interpreted as a (unit) vector &#xA;orthogonal to the two vectors involved in the calculation that is scaled up by some &#xA;factor β, i.e. β$\ \hat{e_1}$⋀$\ \hat{e_2}$ where β is the scalar representing the &#xA;area of the parallelogram.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;2: Just as we can algebraically validate why $\overline{u} \ x \ \overline{v} \ = \ - \ \overline{v} \ x \ \overline{u}$ why doesn't the exact same logic validate &#xA;$\ \hat{e_1}$⋀$\ \hat{e_2}$ = - $\ \hat{e_2}$⋀$\ \hat{e_1}$?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we think along these lines I think we can justify why $\ \hat{e_1}$⋀$\ \hat{e_1}$ = 0,&#xA;just as it occurs analogously in the cross product. They seems far too similar for it to be coincidence but I can't find anyone explaining this relationship.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;3: In general, if you are taking the wedge product of (n - 1) vectors in n-space &#xA;will you always end up with a new vector orthogonal to all of the others?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are taking the wedge product of (n - 1) vectors then will you end up with &#xA;λ($\ \hat{e_1}$⋀$\ \hat{e_2}$⋀...⋀$\ \hat{e_n}$)&#xA;where the term ($\ \hat{e_1}$⋀$\ \hat{e_2}$⋀...⋀$\ \hat{e_n}$) is orthogonal to all &#xA;the vectors involved in the calculation &amp;amp; the term λ represents the area/volume/hypervolume (etc...) contained in the (n - 1) vectors?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;4: I have seen it explained that we can interpret the wedge product of &#xA;$\ \hat{e_1}$⋀$\ \hat{e_2}$ &#xA;as in the picture &lt;a href=&quot;http://upload.wikimedia.org/wikipedia/commons/thumb/9/92/Wedge_product.JPG/170px-Wedge_product.JPG&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;, as a kind of two-dimensional &#xA;vector. Still, the result given is no different to that of the 3-D cross product so is &#xA;it not justifiable to think of $\ \hat{e_1}$⋀$\ \hat{e_2}$ as if it were just an &#xA;orthogonal vector in the same way you would the cross product if you think along the &#xA;lines I have been tracing out in this post? When you go on to take the wedge product &#xA;of (n - 1) vectors in n-space can I not think in the same (higher dimensional) way?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's it, thanks a lot for taking the time to read this I have tried to be as clear &#xA;as possible, any contradictions/errors are as a result of my poor knowledge of all of &#xA;this! :D&lt;/p&gt;&#xA;" OwnerUserId="7923" LastActivityDate="2011-03-07T20:39:04.513" Title="Wedge Product, A Novel Interpretation or Just Plain Wrong?" Tags="&lt;exterior-algebra&gt;&lt;cross-product&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" />
  <row Id="25589" PostTypeId="1" AcceptedAnswerId="25590" CreationDate="2011-03-07T19:12:41.660" Score="2" ViewCount="92" Body="&lt;p&gt;With a perfect use of the condom you have a 2% chance to get pregnant each year. These figures seem surprisingly high to me. So I was wondering &lt;strong&gt;what the chance is to have a baby after a lifelong use&lt;/strong&gt; (let's assume 25-35 years).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What if you use the pill? And with both the pill &amp;amp; condom? What with the &quot;typical use&quot; scenario?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Comparison_of_birth_control_methods#Effectiveness_of_various_methods&quot; rel=&quot;nofollow&quot;&gt;Pregnancy chances after one year of use&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Pill | Perfect use: 0.3% &lt;/li&gt;&#xA;&lt;li&gt;Pill | Normal use: 8% &lt;/li&gt;&#xA;&lt;li&gt;Condom | Perfect use: 2% &lt;/li&gt;&#xA;&lt;li&gt;Condom | Normal use: 15%&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="5867" LastActivityDate="2011-03-07T19:18:35.357" Title="Probability - Contraception" Tags="&lt;probability&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="25590" PostTypeId="2" ParentId="25589" CreationDate="2011-03-07T19:18:35.357" Score="5" ViewCount="" Body="&lt;p&gt;If you have a probability of failure for one year of $p$, you have a probability of no failure for one year of $1-p$.  Then the chance of no failure for $n$ years is $(1-p)^n$ if you assume the chances are independent.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-03-07T19:18:35.357" CommentCount="3" />
  <row Id="25591" PostTypeId="2" ParentId="25586" CreationDate="2011-03-07T19:25:46.073" Score="12" ViewCount="" Body="&lt;p&gt;It doesn't. Using the convexity of $1/x$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_0^\infty \left\vert\frac{\sin x}{x}\right\vert \,\mathrm{d}x=\sum_{k=0}^\infty\int_{k\pi}^{(k+1)\pi}\left\vert\frac{\sin x}{x}\right\vert \,\mathrm{d}x&amp;gt;\sum_{k=0}^\infty\int_{k\pi}^{(k+1)\pi}\frac{\left\vert\sin x\right\vert}{(k+1/2)\pi} \,\mathrm{d}x=\frac{2}{\pi}\sum_{k=0}^\infty\frac{1}{k+1/2}\;,$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which diverges since the harmonic series diverges.&lt;/p&gt;&#xA;" OwnerUserId="6622" LastActivityDate="2011-03-07T19:25:46.073" CommentCount="4" />
  <row Id="25592" PostTypeId="2" ParentId="25586" CreationDate="2011-03-07T19:26:00.667" Score="9" ViewCount="" Body="&lt;p&gt;It doesn't.  This function is a sequence of bumps of decreasing size.  The $n^{\text{th}}$ &quot;bump&quot; bounds area on the order of $\frac{1}{n}$ (there are a number of ways to show this), but $\sum_{n=1}^\infty \frac{1}{n} = \infty$.&lt;/p&gt;&#xA;" OwnerUserId="884" LastActivityDate="2011-03-07T19:26:00.667" />
  <row Id="25593" PostTypeId="2" ParentId="25461" CreationDate="2011-03-07T19:33:25.417" Score="8" ViewCount="" Body="&lt;p&gt;Proof from &lt;strong&gt;&quot;Two elementary generalisations of boolean rings&quot; AMM Feb.1986&lt;/strong&gt;:&#xA;Let $Z(R):=\{x\!\in\!R;\: \forall y\!\in\!R\!: xy=yx\}$ denote the center of the ring. We notice that &#xA;    \begin{equation*}&#xA;        \label{eq:7.1}&#xA;        xy-yx=(xy-yx)^2=(yx-xy)^2=yx-xy.&#xA;        \tag*{(1)}&#xA;    \end{equation*}&#xA;First we show a general lemma, that if $xy\!=\!0$ implies $yx\!=\!0$ $(\ast)$, then every idempotent $e$ is central. For arbitrary $r\!\in\!R$ we have $(e^2\!-\!e)r=0=e(er\!-\!r)$ and by $(\ast)$ we have $0=(er\!-\!r)e=ere\!-\!re$. Similarly, from $r(e^2\!-\!e)=0=(re\!-\!r)e$ by $(\ast)$ we get $0=e(re\!-\!r)=ere\!-\!er$. Therefore $re=ere=er$ which proves $e\!\in\!Z(R)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next we prove, that all commutators are central. This follows from the previous paragraph, because the condition $(\ast)$ holds: if $xy=0$, then $yx$ $=yx-xy$ $=(yx\!-\!xy)^2$ $=(yx)^2$ $=y(xy)x$ $=0$. We have proved, that &#xA;    \begin{equation*}&#xA;        \label{eq:7.2}&#xA;        \forall x,y\in R:\; xy-yx\in Z(R).&#xA;        \tag*{(2)}&#xA;    \end{equation*}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next we prove that all squares are central. This we get from the equality  $x(xy-yx)$ $\overset{(1)}{=}x(yx-xy)$ $\overset{(2)}{=}(yx-xy)x$, which tells us that $x^2y=yx^2$ , i.e.&#xA;\begin{equation*}&#xA;    \label{eq:7.3}&#xA;    \forall x\in R:\; x^2\in Z(R).&#xA;    \tag*{(3)}&#xA;\end{equation*}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next we prove, that $(xy)^2=(yx)^2$. This we achieve via $(3)$ by writing $yxy$ as a sum of squares:&#xA;$$(xy)^2 =x(yxy)=x\big((yx)^2+y^2-(yx\!-\!y)^2-y^2x\big)$$&#xA;\begin{equation*}&#xA;    \label{eq:7.4}&#xA;    \overset{(3)} {=}\big((yx)^2+y^2-(yx-y)^2-y^2x\big)x=(yxy)x=(yx)^2&#xA;    \tag*{(4)}&#xA;\end{equation*}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, we show that all elements are central:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$xy\!-\!yx=(xy\!-\!yx)^2=xy(xy\!-\!yx)\!-\!yx(xy\!-\!yx) \overset{(1)}{=} xy(xy\!-\!yx)\!-\!yx(yx\!-\!xy)$$&#xA;$$= (xy)^2\!-\!xy^2x\!-\!(yx)^2\!+\!yx^2y\overset{(4)}{=}-\!xy^2x\!+\!yx^2y \overset{(3)}{=} -\!x^2y^2\!+\!x^2y^2=0$$&#xA;This completes the proof.&lt;/p&gt;&#xA;" OwnerUserId="7325" LastEditorUserId="7325" LastEditorDisplayName="" LastEditDate="2011-03-08T02:27:03.977" LastActivityDate="2011-03-08T02:27:03.977" CommentCount="1" />
  <row Id="25594" PostTypeId="2" ParentId="25586" CreationDate="2011-03-07T19:45:48.527" Score="8" ViewCount="" Body="&lt;p&gt;$\int_0^\infty \frac{\sin x}{x} dx$ is in fact my most favorite example of a function which is Riemann integrable but not Lesbegue integrable. (just to tease people which think that Lesbegue is &lt;strong&gt;so&lt;/strong&gt; much better than Riemann; in fact this integral might appear more frequent in applications than any integral which is Lesbegue but not Riemann integrable)&lt;/p&gt;&#xA;" OwnerUserId="7266" LastActivityDate="2011-03-07T19:45:48.527" CommentCount="3" />
  <row Id="25595" PostTypeId="1" CreationDate="2011-03-07T19:50:32.773" Score="2" ViewCount="115" Body="&lt;p&gt;Consider the equation&#xA;$$\mathbf{v} = \mathbf{v}_0 + A\mathbf{x}$$&#xA;where $A$ is an $m\times n$ matrix ($m\gt n$) with entries equal to $-1$, $0$, or $1$ only. Additionally, $1$ and $-1$ only appear once in each row $A$ so the sum of a row is always zero. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathbf{x}_0 = \mathbf{0}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is to find an $\mathbf{x}$ that minimizes the L-infinity norm of $\mathbf{v}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What method should I use? &#xA;Thanks. &lt;/p&gt;&#xA;" OwnerUserId="7926" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-07T23:44:48.527" LastActivityDate="2011-03-08T00:36:03.080" Title="Minimize vector's L-infinity norm" Tags="&lt;optimization&gt;" AnswerCount="1" CommentCount="0" />
  <row Id="25596" PostTypeId="1" CreationDate="2011-03-07T19:52:34.167" Score="1" ViewCount="183" Body="&lt;p&gt;I originally posted this over at stackoverflow and they suggested asking it over here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;link to original: &lt;a href=&quot;http://stackoverflow.com/questions/5222781/calculating-perpendicular-and-angular-distance-between-line-segments-in-3d&quot;&gt;http://stackoverflow.com/questions/5222781/calculating-perpendicular-and-angular-distance-between-line-segments-in-3d&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Text of original:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am working on implementing a clustering algorithm in C++. Specifically, this algorithm: &lt;a href=&quot;http://www.cs.uiuc.edu/~hanj/pdf/sigmod07_jglee.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.cs.uiuc.edu/~hanj/pdf/sigmod07_jglee.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At one point in the algorithm (sec 3.2 p4-5), I am to calculate perpendicular and angular distance (d┴ and dθ) between two line segments: p1 to p2, p1 to p3.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It has been a while since I had a math class, I am kinda shaky on what these actually are conceptually and how to calculate them. Can anyone help?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;It looks like I can calculate the perpendicular distance by the following (d1^3 + d2^3) / (d1^2 + d2^2), where d1 is the euclidean distance between the starting points and d2 is the euclidean distance between the ending points. How would I calculate euclidean distances though?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The angular distance looks to be calculated by the dot product of the vectors. It has been about 10 years since linear algebra so I am very rusty on this but I think I can get a handle on this one hopefully. Just a bit unsure of how to conceptually transfer line segments to vectors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, any help you all could offer would be greatly appreciated.&lt;/p&gt;&#xA;" OwnerUserId="7925" LastActivityDate="2011-03-17T23:16:10.070" Title="calculating perpendicular and angular distance between line segments in 3d" Tags="&lt;geometry&gt;&lt;3d&gt;" AnswerCount="2" />
  <row Id="25597" PostTypeId="2" ParentId="25596" CreationDate="2011-03-07T20:08:47.650" Score="2" ViewCount="" Body="&lt;p&gt;How are you defining the distance between two segments?  It appears the two segments share an endpoint, so the distance would normally be zero.  Then if you take the two vectors $\vec {v_1}= \overrightarrow{p_2-p_1}$ and $\vec {v_2}= \overrightarrow{p_3-p_1}$, you have $\cos(\theta)=\frac{\vec {v_1}\cdot \vec {v_2}}{|\vec {v_1}||\vec {v_2}|}$ where $\theta$ is the angle between them.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-03-07T20:08:47.650" CommentCount="11" />
  <row Id="25598" PostTypeId="1" AcceptedAnswerId="25603" CreationDate="2011-03-07T20:14:11.933" Score="1" ViewCount="109" Body="&lt;p&gt;I am having trouble proving this theorem:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;For a field $k$, let $f \in k[x]$ be&#xA;  irreducible, and $g \in k[x]$ be a&#xA;  polynomial. Let $h\in k[x]$ be an&#xA;  irreducible factor of $f\circ g$. Show&#xA;  that $\deg f \mid \deg h$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My attempt so far was to prove this:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Let $\mathrm{Sq}(n,k)$ be the set of&#xA;  square $n\times n$ matrices with&#xA;  entries in $k$. Let $A\in&#xD;&#xA;&amp;gt; \mathrm{Sq}(n,k)$ which has an&#xA;  irreducible characteristic polynomial&#xA;  $f\in k[x]$. Let $g \in k[x]$. Then&#xA;  there exists a $B\in \mathrm{Sq}(n,k)$&#xA;  satisfying $A=g(B)$ iff $f\circ g$ has&#xA;  a factor in $k[x]$ of degree $n$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And here's my proof.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Proof.&lt;/strong&gt; ($\Rightarrow$) Suppose $B$ exists and let $p$ be the minimal polynomial&#xA;  of $B$. Since $k[B]$ contains $k[A]$,&#xA;  $\deg p = n$. Since $f(g(B)) = f(A) =&#xD;&#xA;&amp;gt; 0$, we have $p \mid f\circ g$.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;($\Leftarrow$) Let $h\mid f\circ g$ and&#xA;  $\deg h = n$. Let $C$ be the companion&#xA;  matrix of $h$. Then $f(g(C)) = 0$, and&#xA;  since $f$ is irreducible with $\deg f = n$,&#xA;  $g(C)$ is similar to the companion matrix of $f$. Therefore&#xA;  $g(C)$ is similar to $A$. As such, $A = T^{-1} g(C) T$&#xA;  for some $T\in\mathrm{GL}(n,k)$. Then&#xA;  $g(T^{-1}CT)=A$, so $B=T^{-1}CT$. $\blacksquare$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm not sure where to go from here. Is this even relevant? Is there a better way to go about this? Hints would be helpful, a proof even more appreciated.&lt;/p&gt;&#xA;" OwnerUserId="2666" LastActivityDate="2011-03-13T11:21:13.030" Title="A question on a theorem about polynomials" Tags="&lt;abstract-algebra&gt;&lt;polynomials&gt;" AnswerCount="1" />
  <row Id="25599" PostTypeId="2" ParentId="25575" CreationDate="2011-03-07T20:16:35.550" Score="1" ViewCount="" Body="&lt;p&gt;The inequality is equivalent to $(2^r-1)(1-x)x^{r-1}+x^r&amp;gt;1$, that is $(2^r-1)x^{r-1}-(2^r-2)x^r&amp;gt;1$.&#xA;To prove this one you can differentiate the LHS with respect to $x$, to find $x^{r-2} \left((2^r-1)(r-1)-(2^r-2)r x \right)$.&#xA;From that you get that the expression $(2^r-1)(1-x)x^{r-1}+x^r$ is either (as a function of $x$) increasing on $[1/2,1]$, increasing on $[1/2,x_0]$ and decreasing on $[x_0,1]$, or decreasing on $[1/2,1]$ (actually it is the second one since $x_0 = \frac{(2^r-1)(r-1)}{(2^r-2)r}$ is between $1/2$ and $1$, but it doesn't matter).&#xA;So we only need to check that the values at $x=1/2$ and $x=1$ are $\geq 1$, and in fact they are $1$, so that the inequality is sharp.&lt;/p&gt;&#xA;" OwnerUserId="2660" LastActivityDate="2011-03-07T20:16:35.550" />
  <row Id="25600" PostTypeId="2" ParentId="25575" CreationDate="2011-03-07T20:17:10.057" Score="3" ViewCount="" Body="&lt;p&gt;The assertion holds for every $x$ in $(1/2,1)$ and every real number $r&amp;gt;1$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To see this, one can begin like Moron and try to show that $f(y)&amp;lt;f(2)$ with $y=1/x$ and&#xA;$$&#xD;&#xA;f(y)=(y^r-1)/(y-1).&#xD;&#xA;$$&#xA;Since $y&amp;lt;2$, it is enough to prove that $f$ is nondecreasing. The sign of $f&amp;#39;(y)$ is the sign of $g(y)$ with&#xA;$$&#xD;&#xA;g(y)=(r-1)y^r-ry^{r-1}+1.&#xD;&#xA;$$&#xA;Now, $g&amp;#39;(y)=r(r-1)y^{r-2}(y-1)$ is positive on $y&amp;gt;1$ because $r&amp;gt;1$ hence $y\mapsto g(y)$ is increasing on $y&amp;gt;1$, and $g(1)=0$ hence $g(y)&amp;gt;0$ for $y&amp;gt;1$. Finally, $y\mapsto f(y)$ is increasing on $y&amp;gt;1$ hence we are done.&lt;/p&gt;&#xA;" OwnerUserId="6179" LastActivityDate="2011-03-07T20:17:10.057" />
  <row Id="25601" PostTypeId="1" AcceptedAnswerId="25617" CreationDate="2011-03-07T20:18:59.307" Score="1" ViewCount="190" Body="&lt;p&gt;Lets say we are given a primal linear programming problem:&lt;br&gt;&#xA;$\begin{array}{ccc}&#xD;&#xA;\text{minimize } &amp;amp; c^{T}x &amp;amp; &amp;amp;\\&#xD;&#xA;\text{subject to: } &amp;amp; Ax &amp;amp; \ge &amp;amp; b \\&#xD;&#xA;&amp;amp; x &amp;amp; \ge &amp;amp; 0&#xD;&#xA;\end{array}$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The dual problem is defined as:&lt;br&gt;&#xA;$\begin{array}{ccc}&#xD;&#xA;\text{maximize } &amp;amp; b^{T}y &amp;amp; &amp;amp;\\&#xD;&#xA;\text{subject to: } &amp;amp; A^{T}y &amp;amp; \le &amp;amp; c \\&#xD;&#xA;&amp;amp; y &amp;amp; \ge &amp;amp; 0&#xD;&#xA;\end{array}$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to the duality theorem&#xA;$c^{T}x \ge b^{T}y$ for every feasible solution $x$ and $y$, and in addition when $x$ and $y$ are optimal solutions to the primal and the dual task then $c^{T}x=b^{T}y$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if we define linear programming task with following constraints:&lt;br&gt;&#xA;$\begin{array}{ccc}&#xD;&#xA;Ax &amp;amp; \ge &amp;amp; b \\&#xD;&#xA;x &amp;amp; \ge &amp;amp; 0 \\&#xD;&#xA;A^{T}y &amp;amp; \le &amp;amp; c \\&#xD;&#xA;y &amp;amp; \ge &amp;amp; 0 \\&#xD;&#xA;b^{T}y &amp;amp; \ge &amp;amp; c^{T}x&#xD;&#xA;\end{array}$&lt;br&gt;&#xA;Then any feasible solution to this task should be an optimal solution to the primal and the dual task, because the last constraint might be satisfied only if $x$ and $y$ are optimal.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is why this approach is not used?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I see three potential reasons:&lt;br&gt;&#xA;1) I've made somewhere mistake and it doesn't make any sense.&lt;br&gt;&#xA;2) It is often the case when primal or dual problem is infeasible. I've seen such examples, but in all of them the optimal solution was unbounded, is it the only case when exactly one of the primal and dual problem is infeasible?&lt;br&gt;&#xA;3) Finding any feasible solution might be hard. The so called Phase 1 of simplex method can be used to find a feasible solution. I couldn't find the complexity of this phase, is it exponential just like the simplex algorithm? The other question is what is the fastest method to determine whether there exist any feasible solution? This solution doesn't have to be found.&lt;/p&gt;&#xA;" OwnerUserId="4811" LastActivityDate="2011-03-07T22:14:36.387" Title="Primal and dual solution to linear programming" Tags="&lt;linear-programming&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="25602" PostTypeId="1" AcceptedAnswerId="25610" CreationDate="2011-03-07T20:18:59.563" Score="0" ViewCount="42" Body="&lt;p&gt;Given $\alpha(s)$ a smooth arc-length parametrized curve, how can you write the equation for the surface $f(s,t) = \alpha(s) + t \alpha&amp;#39;(s)$ component-wise?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is, I want to write $f = (f_x, f_y, f_z)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can write $\alpha(s) = (x(s), y(s), z(s))$, but that doesn't give much.  For instance, I don't think $f_x(s,t) = x(s) + t x&amp;#39;(s)$.  Is that right?&lt;/p&gt;&#xA;" OwnerUserId="7029" LastActivityDate="2011-03-07T20:50:24.203" Title="Surface given by an arc-length parametrized curve" Tags="&lt;calculus&gt;&lt;differential-geometry&gt;" AnswerCount="1" CommentCount="0" />
  <row Id="25603" PostTypeId="2" ParentId="25598" CreationDate="2011-03-07T20:28:23.730" Score="4" ViewCount="" Body="&lt;p&gt;Suppose $h$ is an irreducible factor of $f \circ g$. Let $L = k(\alpha)$ for some root $\alpha$ of $h$ then $\dim_k (L)=\deg(h)$. Notice that $h(x) \mid f\circ g(x)$ so $\alpha$ is also a root of $f\circ g$ and so $g(\alpha)$ is a root of $f$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $\beta=g(\alpha) \in k(\alpha)$ then $F=k(\beta)\subseteq L$. So to finish the proof $\deg(h)=[L:k]=[L:F][F:k]=[L:F] \deg(f)$.&lt;/p&gt;&#xA;" OwnerUserId="2125" LastEditorUserId="2125" LastEditorDisplayName="" LastEditDate="2011-03-13T11:21:13.030" LastActivityDate="2011-03-13T11:21:13.030" CommentCount="3" />
  <row Id="25605" PostTypeId="1" CreationDate="2011-03-07T20:29:53.973" Score="1" ViewCount="93" Body="&lt;p&gt;Let $F$ be an arbitrary field of characteristic $0$, $K$ its algebraic closure. Define $M=\{ (x,y)\in M_n(F)&amp;#215;M_n(F) \mid [x,y]=0\}$ and let $N$ be the Zariski closure of $M$ in $K^{2n^2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can one show that $N$ contains the set $\{(axa^{-1},aya^{-1}) \mid (x,y)\in N, a\in \mathrm{GL}(n,K)\}$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you.&lt;/p&gt;&#xA;" OwnerUserId="7927" LastActivityDate="2011-06-05T23:12:38.617" Title="algebraic closure of commuting pairs of matrices" Tags="&lt;algebraic-geometry&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="25606" PostTypeId="2" ParentId="25588" CreationDate="2011-03-07T20:39:04.513" Score="5" ViewCount="" Body="&lt;p&gt;The best introduction I know of to the exterior product is Sergei Winitzki's &lt;em&gt;free&lt;/em&gt; book &lt;a href=&quot;http://www.lulu.com/product/file-download/linear-algebra-via-exterior-products/6214035?productTrackingContext=search_results/search_shelf/center/1&quot; rel=&quot;nofollow&quot;&gt;Linear Algebra via Exterior Products&lt;/a&gt;. Chapter $2$ in particular I think addresses all of your questions (it is unclear how much of Chapter $1$ you need to read in order to read Chapter $2$, I guess that depends on how much linear algebra you've had).&lt;/p&gt;&#xA;" OwnerUserId="400" LastActivityDate="2011-03-07T20:39:04.513" CommentCount="1" />
  <row Id="25607" PostTypeId="1" AcceptedAnswerId="25611" CreationDate="2011-03-07T20:44:26.000" Score="1" ViewCount="52" Body="&lt;p&gt;PROBLEM: Let $R$ be a ring with $1$ and $M$ be a unital $R$-module (i.e. $1x=x$). Let there for each submodule $M_1\neq M$ exist a submodule $M_2\neq M$, such that $M_1\cap M_2=\{0\}$. How can I &lt;strong&gt;prove, that $M$ is semisimple&lt;/strong&gt;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;DEFINITIONS: A module $M$ is &lt;em&gt;semisimple&lt;/em&gt; iff $\exists$ simple submodules $M_i\leq M$, such that $M=\bigoplus_{i\in I}M_i$. A module $M_i$ is &lt;em&gt;simple&lt;/em&gt; iff it has no submodules (other than $\{0\}$ and $M$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;KNOWN FACTS: $M$ is semisimple $\Leftrightarrow$ $\exists$ simple submodules $M_i\leq M$, such that $M=\sum_{i\in I}M_i$ (the sum need not be direct) $\Leftrightarrow\forall\!M_1\!\leq\!M\;\exists M_2\!\leq\!M$ such that $M_1\oplus M_2=M$ (i.e. every submodule is a direct sumand).&lt;/p&gt;&#xA;" OwnerUserId="7325" LastEditorUserId="7325" LastEditorDisplayName="" LastEditDate="2011-03-07T20:53:35.957" LastActivityDate="2011-03-07T20:53:35.957" Title="In a unital $R$-module $M$, if $\forall M_1\!\lneq\!M\;\;\exists M_2\!\lneq\!M$, such that $M_1\!\cap\!M_2\!=\!\{0\}$, then $M$ is semisimple." Tags="&lt;abstract-algebra&gt;&lt;modules&gt;" AnswerCount="1" CommentCount="8" />
  <row Id="25608" PostTypeId="2" ParentId="25550" CreationDate="2011-03-07T20:44:54.970" Score="4" ViewCount="" Body="&lt;p&gt;That method will work, but there's actually a simpler more general way. But first let's finish that method. $\:$ After multiplying through by $\rm\: (x-3)^2\: $ (&lt;em&gt;squared&lt;/em&gt; to preserve the sense of the inequality) we obtain $\rm\ 3\:x\:(x-3) \ge\: 4\:(x-3)^2\:.\:$ Putting all terms on one side and factoring out $\rm\ x-3\ $ we obtain $\rm\: p(x) = (x-3)\:(a\:x-b) \ge 0\ $ for some $\rm\:a\:,\:b\:.\:$ The (at most) two roots partition the real line into (at most) three intervals where $\rm\ p(x)\ $ has constant sign. So the answer follows by simply testing these few intervals to determine where $\rm\:p(x)\:$ has the desired sign.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But multiplying through by $\rm\: (x-3)^2\: $ creates more work than necessary. Instead we can simply multiply through by $\rm\: x-3\: $ and worry about the signs later, since we really only need to know the other root (besides $\rm\ x = 3\:$) of the quadratic $\rm\: p(x)\:.\:$ Indeed, if we multiply by $\rm\ x-3\ $ we get either $\rm\ 3x \ge 4\:x-12\ \Rightarrow\ 12 \ge x\ $ or we get the reverse $\rm\ 12 \le x\:,\:$ depending on the sign of $\rm\ x-3\:.\: $ Either way we get the same root $\rm\: x = 12\:,\: $ so we obtain the same partition of the real line into intervals where the function has constant sign. This method is simpler since one works with lower degree polynomials, here linear vs. quadratic in the first method above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similar remarks hold true for arbitrary rational functions. Suppose that $\rm\: A,B,C,D\: $ are polynomials in $\rm\:x\:$ and we seek to determine where $\rm\: A/B\ &amp;gt;\ C/D\:.\: $ The first method converts to polynomials by multiplying it by $\rm\: (BD)^2\: $ resulting in $\rm\: ABD^2 &amp;gt;\: CDB^2\ $ or $\rm\ BD\ (AD-BC) &amp;gt; 0\:.\:$ For the second method, we partition the real line by the roots of the denominators $\rm\: B,D\ $ and also by the roots of the polynomial obtained by multiplying through by $\rm\:BD\:,\:$ i.e. the roots of $\rm\: AD-BC\:.\:$ However the union of these root sets is precisely the same as the roots of $\rm\: BD\ (AD-BC)\: $ from the first method. Since both methods partition the real line into the same constant-sign intervals, they're equivalent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you study rational (and algebraic) functions and their Riemann surfaces you'll see that this is just a special case of how their roots (including poles = roots at $\infty$) serve to characterize them. From another viewpoint, the above method can be viewed as a very special case of &lt;a href=&quot;http://math.stackexchange.com/questions/23068/are-there-any-sets-other-than-the-usual-in-which-we-can-apply-sturms-axioms/23110#23110&quot;&gt;Tarski's quantifier elimination algorithm&lt;/a&gt; for deciding arbitrary real polynomial inequalities - which, like above, works by decomposing $\rm\: \mathbb R^n\: $ into a &lt;em&gt;finite&lt;/em&gt; number constant-sign (semi-)algebraic &quot;cylinders&quot;.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2011-03-09T15:59:26.193" LastActivityDate="2011-03-09T15:59:26.193" CommentCount="3" />
  <row Id="25610" PostTypeId="2" ParentId="25602" CreationDate="2011-03-07T20:50:24.203" Score="2" ViewCount="" Body="&lt;p&gt;$f(s,t)=(x(s)+tx&amp;#39;(s),y(s)+ty&amp;#39;(s),z(s)+tz&amp;#39;(s))$ just as you're saying.&lt;/p&gt;&#xA;" OwnerUserId="6925" LastActivityDate="2011-03-07T20:50:24.203" />
  <row Id="25611" PostTypeId="2" ParentId="25607" CreationDate="2011-03-07T20:50:45.297" Score="3" ViewCount="" Body="&lt;p&gt;Let $M$ be your module, and let $M_1$ be a submodule. Consider the set $\mathcal S$ of all submodules $N$ of $M$ such that $M_1\cap N=0$, and order $\mathcal S$ by inclusion. It is easy to see that $\mathcal S$ satisfies the hypothesis of Zorn's Lemma, so there exists an element $M_2\in\mathcal S$ which is maximal. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have $M_1\cap M_2=0$ and we want to show that $M_1+M_2=M$. If that were not the case, your hypothesis would provide a submodule $P\subseteq M$ such that $(M_1+M_2)\cap P=0$. Can you see how to reach a contradiction now?&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2011-03-07T20:50:45.297" CommentCount="6" />
  <row Id="25612" PostTypeId="2" ParentId="25519" CreationDate="2011-03-07T20:51:28.203" Score="1" ViewCount="" Body="&lt;p&gt;Consider the following. &#xA;Let $Y$ be a Brownian motion, and $N$ an independent Poisson process with rate $\lambda$. Let $t$ satisfy $e^{-\lambda t} = 1/2$. Define $W$ as follows. If $N_t = 0$, then $W = {\rm sgn}(Y_1) Y$; if $N_t &amp;gt; 0$, then $W = -{\rm sgn}(Y_1) Y$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyone is welcome to give opinion on this suggestion.&lt;/p&gt;&#xA;" OwnerUserId="2810" LastActivityDate="2011-03-07T20:51:28.203" CommentCount="5" />
  <row Id="25614" PostTypeId="1" AcceptedAnswerId="25615" CreationDate="2011-03-07T21:05:39.330" Score="3" ViewCount="55" Body="&lt;p&gt;I know I am making a mistake somewhere but consider the following truth table:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;p0   p1   p2  |&#xA;-------------------&#xA;0    0    0   | 1  &#xA;0    0    1   | 0  &#xA;0    1    0   | 0  &#xA;0    1    1   | 1  &#xA;1    0    0   | 1  &#xA;1    0    1   | 1  &#xA;1    1    0   | 1  &#xA;1    1    1   | 1&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This coresponds to the boolean formula in DNF:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;p0 v (~p1 ^ ~p2) v (p1 ^ p2)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;or in CNF:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;(p0 v ~p1 v p2) ^ (p0 v p1 v ~p2)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Using the last two clauses and applying the resolution inference rule on p1:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;(p0 v ~p1 v p2) ^ (p0 v p1 v ~p2) |- (p0 v p2 v ~p2) = (p0 v T) = p0&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;So from the two clauses I can infer p0. This does not seem right since p0 does not always have to be true according to the truth table.&lt;/p&gt;&#xA;" OwnerUserId="7929" LastActivityDate="2011-03-07T21:39:23.237" Title="A case of theorem proving through resolution" Tags="&lt;logic&gt;" AnswerCount="1" />
  <row Id="25615" PostTypeId="2" ParentId="25614" CreationDate="2011-03-07T21:13:38.090" Score="6" ViewCount="" Body="&lt;p&gt;Your mistake is that $(p_0 \vee T) = T$ and not $p_0$ as you wrote. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact it is provable from the empty set that $(p_0 \vee p_2\vee\neg p_2)$.&lt;/p&gt;&#xA;" OwnerUserId="622" LastActivityDate="2011-03-07T21:13:38.090" CommentCount="1" />
  <row Id="25617" PostTypeId="2" ParentId="25601" CreationDate="2011-03-07T21:24:31.533" Score="4" ViewCount="" Body="&lt;p&gt;There are three things that have to be satisfied in order for a solution to a linear programming problem to be optimal:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The primal solution must be feasible.&lt;/li&gt;&#xA;&lt;li&gt;The dual solution must be feasible.&lt;/li&gt;&#xA;&lt;li&gt;Complementary slackness must be satisfied.  (Remember that primal variables are paired with dual slack variables and dual variables are paired with primal slack variables.  Complementary slackness is the requirement that, for each of these pairs, at least one variable must be zero.)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The primal simplex method (after Phase I) keeps (1) and (3) always true and searches for a solution that satisfies (2).  The dual simplex method (again, after Phase I), keeps (2) and (3) always true and searches for a solution that satisfies (1).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The approach you are describing (minus the $b^Ty \geq c^T x$ constraint) &lt;em&gt;is&lt;/em&gt; used.  It's the other option, in which (1) and (2) are always kept true while the algorithm searches for a solution that satisfies (3).  As Yuval Filmus indicates, this is called a &lt;em&gt;primal-dual method&lt;/em&gt; or the &lt;em&gt;parametric self-dual simplex method&lt;/em&gt;.  See, for example, Rader's &lt;em&gt;Deterministic Operations Research&lt;/em&gt;, pp. 432-440, or Vanderbei's &lt;em&gt;&lt;a href=&quot;http://books.google.com/books?id=T-BW1g69wbYC&amp;amp;pg=PA119&amp;amp;lpg=PA119&amp;amp;dq=in+the+previous+section,+we+illustrated+vanderbei+linear+programming&amp;amp;source=bl&amp;amp;ots=OfP7GQRETC&amp;amp;sig=q-g53N-MPHoYB0yB0DmWJf03nlQ&amp;amp;hl=en&amp;amp;ei=Kkt1TfaaPI6-sAPcpNTGBA&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=1&amp;amp;ved=0CBQQ6AEwAA#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;Linear Programming: Foundations and Extensions&lt;/a&gt;&lt;/em&gt;, pp 119-121.  (See also Vanderbei's text for how to find an initial feasible solution to both problems; i.e., Phase I.)  The idea dates back at least to George Dantzig, the inventor of the simplex method. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a side comment, Vanderbei indicates that the parametric self-dual simplex method is more amenable to probabilistic analysis than the other versions of the simplex method.&lt;/p&gt;&#xA;" OwnerUserId="2370" LastActivityDate="2011-03-07T21:24:31.533" />
  <row Id="25618" PostTypeId="1" AcceptedAnswerId="25620" CreationDate="2011-03-07T21:24:47.457" Score="0" ViewCount="65" Body="&lt;p&gt;I'm being asked to find the arc length of $y=sin x$ for [0, $\frac{pi}{2}$] using $M_8$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've determined that $y\prime^2=cos^2x$.  So, using the formula for arc length, I get $\sqrt{1+cos^2x}$ as my function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, they want me to evaluate this using $M_8$, so I end up with 8 midpoints from $\frac{pi}{32}$ to $\frac{15pi}{32}$ moving up in increments of  $\frac{pi}{16}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, I know that $\Delta{x} = \frac{pi}{16}$.  So, i end up with this estimated integral:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{pi}{16}(\sqrt{1+cos^2(\frac{pi}{32})} + \sqrt{1+cos^2(\frac{3pi}{32})} + ... \sqrt{1+cos^2(\frac{15pi}{32})}) = 1.91009889$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please feel free to correct any errors here.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question, however, has to do with simplifying this answer using trig.  I would assume that there's a fancy way to do this, rather than just handing in the answer 1.91009889 (though the prof will accept this).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd like to sharpen my trig skills, so I figured this was a great change.&lt;/p&gt;&#xA;" OwnerUserId="5248" LastActivityDate="2011-03-07T21:32:32.553" Title="Simplifying $\sqrt{1+cos^2(x)}$ for various values of x" Tags="&lt;calculus&gt;&lt;homework&gt;&lt;trigonometry&gt;" AnswerCount="1" />
  <row Id="25620" PostTypeId="2" ParentId="25618" CreationDate="2011-03-07T21:32:32.553" Score="1" ViewCount="" Body="&lt;p&gt;Offhand, not much comes to mind with regard to $\sqrt{1+\cos^2x}$, but $\cos\frac{k\pi}{2^n}$ can be evaluated exactly using the half-angle identity $\cos\frac{x}{2}=\pm\sqrt{\frac{1+\cos x}{2}}$ (choose + or – based on the value of $\frac{x}{2}$) repeatedly.&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2011-03-07T21:32:32.553" CommentCount="4" />
  <row Id="25621" PostTypeId="2" ParentId="19153" CreationDate="2011-03-07T21:46:28.217" Score="1" ViewCount="" Body="&lt;p&gt;$S_{n}=\sum^{n}_{k=0}\cos(t\sqrt{k})=\sum^{n}_{k=0}\frac{e^{it\sqrt{k}}+e^{-it\sqrt{k}}}{2}$. Any closed form of this must somehow evaluate $\sum^{n}_{k=0}e^{it\sqrt{k}}$. Consider $e^{x}=\sum^{\infty}_{i=0}x^{i}/i!$, $e^{it\sqrt{k}}$ would be $\sum^{\infty}_{j=0}(it\sqrt{k})^{j}/j!$. Hence the series would be:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$+_{n}(t)=\sum^{\infty}_{k=0}\sum^{\infty}_{j=0}(it\sqrt{k})^{j}/j!$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This can be decomposed into even and odd $j$ terms. Since $j$ evaluate from $0$ we name the even one to be $S_{0}$ and the odd one $S_{1}$. The even terms $j=2s$ sums up to:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$+_{0}(t)=\sum^{\infty}_{k=0}\sum^{\infty}_{s=0}-(t^{2s}k^{s})/(2s)!$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And the odd one $j=2s+1$ sums up to:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$+_{1}(t)=-i\sum^{\infty}_{k=0}\sum^{\infty}_{s=0}(t\sqrt{k})^{2s+1}/(2s+1)!$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, what is the closed form of the original  series? It is the same as $+_{0}(t)$. Switch $k$ and $s$ we should have $$\sum^{\infty}_{s=0}\sum^{\infty}_{k=0}-[(t^{2s})/(2s)!]k^{s}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think there is NO such closed form to evaluate this double series. &lt;/p&gt;&#xA;" OwnerUserId="7887" LastEditorUserId="7887" LastEditorDisplayName="" LastEditDate="2011-03-08T02:14:33.163" LastActivityDate="2011-03-08T02:14:33.163" CommentCount="1" />
  <row Id="25622" PostTypeId="1" AcceptedAnswerId="25628" CreationDate="2011-03-07T21:46:35.927" Score="1" ViewCount="58" Body="&lt;p&gt;Let $\{x_k\}$ be an infinite series. It is known that $\sum_{k=-\infty}^\infty \vert x_k\vert^2&amp;lt;\infty$ and $\sum_{k=-\infty}^\infty \vert x_k\vert$ does not converge. How do I prove that the Fourier transform of $\{x_k\}$, $\mathcal{F}(\omega)$ exists (or what other conditions need to be satisfied for it to exist)? I can solve this if $x_k$ is absolutely summable, but given that this is not satisfied, I don't know how.&lt;/p&gt;&#xA;" OwnerUserId="7815" LastActivityDate="2011-03-07T22:20:05.963" Title="Existence of Fourier transform" Tags="&lt;calculus&gt;&lt;fourier-analysis&gt;&lt;fourier-series&gt;" AnswerCount="1" />
  <row Id="25623" PostTypeId="1" AcceptedAnswerId="25627" CreationDate="2011-03-07T21:49:08.997" Score="3" ViewCount="167" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Prove $(2n+1) + (2n+3) + (2n+5) +&#xD;&#xA;&amp;gt; \cdots + (4n-1) = 3n^{2}$ for all&#xA;  positive integers $n$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So the provided solution avoids induction and makes use of the fact that $1 + 3 + 5 + \cdots + (2n-1) = n^{2}$ however I cannot understand the first step: $(2n+1) + (2n+3) + (2n+5) + \cdots + (4n-1) = (1 + 3 + 5 + \cdots + (4n-1)) -(1 + 3 + 5 + \cdots + (2n-1))$. Once that has been established I can follow the rest, but I was hoping someone could help me understand why $(1 + 3 + 5 + \cdots + (4n-1))$ which already looks like less than the LHS can be made equal to the LHS by subtracting a positive number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Additionally, I wanted to prove the equality using induction, but had trouble with that as well. I think I am thrown off by the last term $4n-1$. Even the initial case of $n=1$ is not totally clear to me: does it hold because $2(1)+1 = 3(1)^{2}$ or is it because $4(1)-1 = 3(1)^{2}$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Either way, my approach was to replace every $n$ on the LHS with $n+1$ which resulted in: $$(2(n+1)+1) + (2(n+1)+3)+ \cdots + (4(n+1)-1)$$ and I am not sure what the second to last term in the sequence would be... I tried simplifying anyway $$(2n+3) + (2n+5) + \cdots + (4n+3)$$ and I this point I thought I could subtract $(2n+1)$ from both sides of the induction assumption resulting in $(2n+3) + (2n+5) + \cdots ? = 3n^{2} - (2n+1) - (4n-1)$ substituting this yields: $$3n^{2} - 2n - 1 - 4n + 1 + 4n + 3 = 3n^{2} + 6n + 3 - 8n + 1 = 3(n+1)^{2} - 8n + 1$$ but I guess I don't want the $-8n + 1$... I also tried substituting $3n^{2} - (2n+1)$ without that last term being subtracted, but that did not work out either. If anyone can help me understand how to do this properly I would really appreciate it. Thanks!&lt;/p&gt;&#xA;" OwnerUserId="3711" LastActivityDate="2011-03-07T22:15:23.297" Title="Proving $(2n+1) + (2n+3) + \cdots + (4n-1) = 3n^{2}$ for all positive integers $n$" Tags="&lt;induction&gt;" AnswerCount="5" />
  <row Id="25624" PostTypeId="1" CreationDate="2011-03-07T21:53:17.753" Score="0" ViewCount="102" Body="&lt;p&gt;I saw some place where someone wrote&#xA;$$\frac{1}{1-f(x)} = \sum_{n=0}^{\infty} f(x)^n, \forall x \in \mathbb{C}, $$&#xA;where $f$ is some function s.t. $\vert f(x)\vert &amp;lt; 1, \forall x \in \mathbb{C}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then he proved that the RHS $\sum_{n=0}^{\infty} f(x)^n$ absolutely converges $, \forall x \in \mathbb{C}$, in order that the RHS is well-defined and finite. I was wondering why we have to have the absolute convergence? Isn't that $\vert f(x)\vert &amp;lt; 1, \forall x \in \mathbb{C}$ can guarantee $\frac{1}{1-f(x)} = \sum_{n=0}^{\infty} f(x)^n, \forall x \in \mathbb{C}$? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I admit I am not familiar with complex analysis, and I am not sure if there is also a similar situation in Real analysis? I would appreciate if someone could point out what kinds of materials (such as Wikipedia articles or other internet links) will help me in this regard, besides explanation. Thanks for clarification!&lt;/p&gt;&#xA;" OwnerUserId="1281" LastEditorUserId="1281" LastEditorDisplayName="" LastEditDate="2011-03-08T02:52:38.437" LastActivityDate="2011-03-08T02:52:38.437" Title="Why to prove a series absolutely converges when expanding a function into the series?" Tags="&lt;sequences-and-series&gt;&lt;complex-analysis&gt;" AnswerCount="1" CommentCount="6" />
  <row Id="25625" PostTypeId="2" ParentId="25623" CreationDate="2011-03-07T21:54:14.353" Score="1" ViewCount="" Body="&lt;p&gt;Gauss gives this solution:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(4n-1+2n+1)*\frac{(\frac{(4n-1-(2n+1))}{2}+1)}{2}$$&lt;/p&gt;&#xA;" OwnerUserId="7887" LastActivityDate="2011-03-07T21:54:14.353" />
  <row Id="25626" PostTypeId="2" ParentId="9999" CreationDate="2011-03-07T21:59:27.403" Score="2" ViewCount="" Body="&lt;p&gt;As for ways to check is a number is a Fibonacci number, I'll speak to efficient testing since the other aspects seem covered.  First, you should check if the residue mod some fixed number is possible for a Fibonacci number.  Sloane's &lt;a href=&quot;http://oeis.org/A189761&quot; rel=&quot;nofollow&quot;&gt;A189761&lt;/a&gt; gives the sequence of 'good' moduli to use: for example, there are only 54 distinct residues mod 39088169 that contain Fibonacci numbers.  A binary search on these lets you quickly remove 99.9998% of possible numbers.  (Depending on how large you want to go, a smaller or larger number may be advisable.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course for very small numbers you may choose to do a binary search directly before checking residues to the chosen modulus.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For numbers that pass the test, the 5n&lt;sup&gt;2&lt;/sup&gt; ± 4 test is best, and it's probably sensible to do this directly&amp;mdash;the residue test above ensures that the number is not of a prohibited congruence class to your modulus.  That is, compute the square root of the number and see if it is close to an integer.  If so, square it and test if the result is the original number.  The methods of Bernstein's paper &quot;Detecting perfect powers in essentially linear time&quot; (section 8) can be used here if desired.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that if n &amp;lt; 0 you need only test 5n&lt;sup&gt;2&lt;/sup&gt; + 4.&lt;/p&gt;&#xA;" OwnerUserId="1778" LastEditorUserId="1778" LastEditorDisplayName="" LastEditDate="2011-06-03T01:49:55.897" LastActivityDate="2011-06-03T01:49:55.897" />
  <row Id="25627" PostTypeId="2" ParentId="25623" CreationDate="2011-03-07T22:02:12.487" Score="4" ViewCount="" Body="&lt;p&gt;The first item, $(2n+1) + (2n+3) + (2n+5) + \cdots + (4n-1)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ = (1 + 3 + 5 + \cdots + (4n-1)) -(1 + 3 + 5 + \cdots + (2n-1))$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;comes because you are just adding and subtracting the same set of terms on the RHS.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your base case for the induction has just one term on the left side, which I would write $2*1+1=3*1^2$  For the induction, assume $(2n+1) + (2n+3) + (2n+5) + \cdots + (4n-1)=3n^2$ then extend it to $n+1$: $(2n+1) + (2n+3) + (2n+5) + \cdots + (4n-1) + (4n+1)+(4n+3)-(2n+1)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$=3n^2+4n+1+4n+3-(2n+1)=3n^2+6n+1=3(n+1)^2$&lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2011-03-07T22:08:26.317" LastActivityDate="2011-03-07T22:08:26.317" CommentCount="5" />
  <row Id="25628" PostTypeId="2" ParentId="25622" CreationDate="2011-03-07T22:08:03.250" Score="1" ViewCount="" Body="&lt;p&gt;Whether it exists depends on which space you allow $\mathcal{F}$ to be. Using &lt;a href=&quot;http://en.wikipedia.org/wiki/Parseval%27s_theorem&quot; rel=&quot;nofollow&quot;&gt;Parseval&lt;/a&gt;, you can immediately show from $\sum_k |x_k|^2$ that in your case $\mathcal{F}(\omega) \in L^2([0,2\pi])$. So if you allow for functions in $L^2$ the Fourier transform exist without further assumtions.&lt;/p&gt;&#xA;" OwnerUserId="7266" LastEditorUserId="7266" LastEditorDisplayName="" LastEditDate="2011-03-07T22:20:05.963" LastActivityDate="2011-03-07T22:20:05.963" CommentCount="3" />
  <row Id="25629" PostTypeId="2" ParentId="25623" CreationDate="2011-03-07T22:08:13.887" Score="2" ViewCount="" Body="&lt;p&gt;Your initial problem seems to be seeing&#xA;$$1 + 3 + 5 + \cdots + (4n-1) = 1 + 3 + 5 + \cdots + (2n-1) + (2n+1) + (2n+3) + \cdots + (4n-1)$$&#xA;so the LHS is $(2n)^2 = 4n^2$ and the left half of the RHS is $n^2$, leaving $3n^2$ for the right half of the RHS.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For induction, you need to start with one term where $(2 \times 1 +1) = 3 \times 1^2$, and then given  $(2n+1) + (2n+3) + \cdots + (4n-1) = 3n^2$ that you can prove $(2n+3) + \cdots + (4n-1) + (4n+1) + (4n+3) = 3(n+1)^2$, i.e. that $3n^2 - (2n+1) +  (4n+1) + (4n+3) = 3(n+1)^2$, which is not difficult. &lt;/p&gt;&#xA;" OwnerUserId="6460" LastEditorUserId="6460" LastEditorDisplayName="" LastEditDate="2011-03-07T22:14:23.460" LastActivityDate="2011-03-07T22:14:23.460" CommentCount="1" />
  <row Id="25630" PostTypeId="2" ParentId="25623" CreationDate="2011-03-07T22:08:27.890" Score="1" ViewCount="" Body="&lt;p&gt;The simple solution using $1+3+\ldots+(2n-1)=n^2$ is as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(2n+1)+(2n+3)+\ldots+(4n-1)=2n\cdot n+ (1+3+\ldots+(2n-1))=3n^2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fact that $1+3+\ldots+(2n-1)=n^2$ is true is a simple induction: Assume it's true for $n$. Then for $n+1$ we get $1+3+\ldots+(2n-1)+(2(n+1)-1)=n^2+2n+1=(n+1)^2$&lt;/p&gt;&#xA;" OwnerUserId="1772" LastEditorUserId="1772" LastEditorDisplayName="" LastEditDate="2011-03-07T22:15:23.297" LastActivityDate="2011-03-07T22:15:23.297" />
  <row Id="25631" PostTypeId="2" ParentId="25623" CreationDate="2011-03-07T22:08:28.470" Score="2" ViewCount="" Body="&lt;p&gt;$\rm\ S(n) := \sum_{k=0}^{n-1}\ 2k+1 = 1 + 3 +\:\cdots\:+ 2n-1\ = n^2\ $ so that it's clear that your sum is simply $\rm\ S(2\:n) - S(n) = 4n^2 - n^2 = 3n^2\:.\:$ Therefore you need only prove that the first sum holds true. Hint: the inductive step is equivalent to the fact that $\rm\ S(n+1) - S(n) = (n+1)^2 - n^2 = 2n+1\:.\:$&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2011-03-07T22:08:28.470" CommentCount="1" />
  <row Id="25632" PostTypeId="2" ParentId="25601" CreationDate="2011-03-07T22:14:36.387" Score="1" ViewCount="" Body="&lt;p&gt;I think that you're missing some additional nitpicky conditions here and there to guarantee optimality complementary slackness, but the idea is essentially correct. You should read more about &lt;a href=&quot;http://en.wikipedia.org/wiki/Karush%E2%80%93Kuhn%E2%80%93Tucker_conditions&quot; rel=&quot;nofollow&quot;&gt;Karush Kuhn Tucker (KKT) conditions&lt;/a&gt; in order to see how these equations apply not only to linear programming, but constrained optimization in general. &lt;/p&gt;&#xA;" OwnerUserId="7145" LastActivityDate="2011-03-07T22:14:36.387" />
  <row Id="25633" PostTypeId="1" AcceptedAnswerId="25636" CreationDate="2011-03-07T22:22:43.507" Score="0" ViewCount="14" Body="&lt;p&gt;Let's say I have a resource, call it E.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are four important values&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Em - the maximum amount of E we can have&lt;/li&gt;&#xA;&lt;li&gt;Ec - the current amount of E&lt;/li&gt;&#xA;&lt;li&gt;Er - the rate at which E replenishes (this is in % of Em)&lt;/li&gt;&#xA;&lt;li&gt;Ea - the cost amount in E of an action&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Let us assume that we start at max capacity.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;at t=0 : Ec = Em&#xA;at t=i : Ec = (Ec at i-1) - Ea     + Er*Em&#xA; aka   curr = previous    - action + replenish&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;How can I make a formula that models this behavior to determine at what t will I run out of resources.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can make some estimates when I use values, let's say Em is 1000, Er is 0.01, and Ea is 50.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1000 / 50 = 20, so at t = 20, I will only have what I gained from my resource replenishment. 1000 * 0.01 = 10 * 20 = 200 more resources to spend.&#xA;200 / 50 = 4, so in 4 more t's, I would have acquired 40. I can estimate that between t=24 and t=25 I will run out of resources.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Estimating to within a t is acceptable, but doing it iteratively is a pain on my real datasets. I really don't remember how to turn an infinite series into a formula, so I turn to you guys.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="2333" LastActivityDate="2011-03-07T22:28:43.153" Title="Determine total expendature with constant return" Tags="&lt;convergence&gt;" AnswerCount="1" />
  <row Id="25634" PostTypeId="1" AcceptedAnswerId="25659" CreationDate="2011-03-07T22:24:02.013" Score="3" ViewCount="85" Body="&lt;p&gt;Why is the area between these two functions positive?&#xA;\begin{align*}&#xA;f(x) &amp;amp;= (x-1)^3\\&#xA;g(x) &amp;amp;= (x-1).&#xA;\end{align*}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The area between $f(x)$ and $g(x)$ is in two parts. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first part is in the 4th quadrant with lower limit $0$ and upper limit $1$, $f(x)$ is the upper function. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second part is in the 1st quadrant with lower limit 1 and upper limit 2, $g(x)$ is the upper function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I compute the areas individually I get two equal positive numbers the sum of which is 0.5&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Shouldn't the area in the fourth quadrant be negative? Or am I missing something?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've been trying to understand why area in the fourth quadrant is positive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you.&lt;/p&gt;&#xA;" OwnerUserId="7931" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-07T23:25:52.387" LastActivityDate="2011-03-07T23:31:31.087" Title="Why is the area between these two graphs positive?" Tags="&lt;calculus&gt;&lt;integral&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="1" />
  <row Id="25635" PostTypeId="2" ParentId="25634" CreationDate="2011-03-07T22:27:34.390" Score="4" ViewCount="" Body="&lt;p&gt;Here is a graph of the two functions with f in pink and g in blue.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/IBOMX.png&quot; alt=&quot;plot of f and g&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By symmetry, the area of each of the two enclosed regions is the same—area, here, meaning&#xA;the geometric concept which is nonnegative (or perhaps positive), as opposed to signed area.  In terms of signed area, one might think of the two pieces as having opposite sign because f is above g for oen piece and g is above f for the other piece.  However, typically, a problem asking for the area between curves or the area enclosed by one or more curves is asking for geometric area.  The typical method of solution in that instance is to consider each piece separately, integrating (top function) – (bottom function) for each piece, to guarantee a positive (nonnegative) result.&lt;/p&gt;&#xA;" OwnerUserId="72" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2011-03-07T22:37:50.710" LastActivityDate="2011-03-07T22:37:50.710" CommentCount="2" />
  <row Id="25636" PostTypeId="2" ParentId="25633" CreationDate="2011-03-07T22:28:43.153" Score="1" ViewCount="" Body="&lt;p&gt;As you have written the equation, you have a fixed change of $-Ea + ErEm$ per period, so it will last $\frac{Em}{Ea-ErEm}$ periods.  Is this what you meant?&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-03-07T22:28:43.153" CommentCount="1" />
  <row Id="25637" PostTypeId="2" ParentId="25634" CreationDate="2011-03-07T22:28:55.390" Score="2" ViewCount="" Body="&lt;p&gt;it depends on how you're being asked to measure it, i.e. whether you consider&#xA;$$&#xD;&#xA;\int_0^2(f-g), \int_0^2(g-f), \text{ or } \int_0^2|f-g|&#xD;&#xA;$$&#xA;or if you want the area totally enclosed (as in another answer) then you only integrate from 1 to 2 as in the graph pictured (as in another answer). &lt;/p&gt;&#xA;" OwnerUserId="6925" LastActivityDate="2011-03-07T22:28:55.390" />
  <row Id="25638" PostTypeId="2" ParentId="25605" CreationDate="2011-03-07T22:29:01.193" Score="0" ViewCount="" Body="&lt;p&gt;I believe that $N=M$, since if we consider two generic matrices $X$ and $Y$ in variables $x_{ij}$ and $y_{ij}$, respectively, then the equation $[X,Y] = 0$ translates into a set of polynomial equations in the aforementioned $2 n^2$ variables. Let $I$ be the ideal generated by the above polynomials in the ring of polynomials in $2 n^2$ variables. Then $M = Z(I)$. Now the second statement follows easily from the fact that for every $A \in GL(n,K)$ we have $[AXA^{-1},AYA^{-1}] = A[X,Y]A^{-1}$.&lt;/p&gt;&#xA;" OwnerUserId="7838" LastActivityDate="2011-03-07T22:29:01.193" />
  <row Id="25639" PostTypeId="1" AcceptedAnswerId="25672" CreationDate="2011-03-07T22:31:28.377" Score="3" ViewCount="164" Body="&lt;p&gt;This is something that confuses me.&#xA;I have read a few mathematical texts and they often seem to use Lemma/Proposition/Theorem if they have a particular statement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, which one to use? A lemma can be something you need to prove a more important theorem, but then what about Fatou's Lemma?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When to pick Proposition or Theorem?&lt;/p&gt;&#xA;" OwnerUserId="1120" LastActivityDate="2011-03-08T01:29:59.683" Title="Lemma/Proposition/Theorem, which one should we pick?" Tags="&lt;soft-question&gt;" AnswerCount="3" CommentCount="6" FavoriteCount="1" />
  <row Id="25640" PostTypeId="1" AcceptedAnswerId="25644" CreationDate="2011-03-07T22:31:52.750" Score="2" ViewCount="118" Body="&lt;p&gt;Is there a general result about the existence of (non-trivial) solutions of the diophantine equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$Ax^2 + By^2 = Cz^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for A,B,C known positive integers, pair-wise relatively prime?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What if we know C is 1 or 3?&lt;/p&gt;&#xA;" OwnerUserId="7933" LastEditorUserId="7933" LastEditorDisplayName="" LastEditDate="2011-03-08T17:18:04.890" LastActivityDate="2011-03-08T17:18:04.890" Title="Existence of solutions to diophantine quadratic form" Tags="&lt;number-theory&gt;&lt;diophantine-equations&gt;&lt;quadratic-forms&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="1" />
  <row Id="25641" PostTypeId="1" AcceptedAnswerId="25647" CreationDate="2011-03-07T22:33:47.663" Score="0" ViewCount="71" Body="&lt;p&gt;Question: if f is analytical and $ |f(z)| &amp;lt; M $ for $ |z| =&amp;lt; R $ find an upper bound for $ |f(&amp;#39;n)(z)|$ in $ |z| =&amp;lt; \rho &amp;lt; R $  (where $f(&amp;#39;n)$ means the nth derivative of $f$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I cited Cauchy's Inequality, to find the bound $\frac{n!M_R} {R^n}$ , but since we have the additional limit of $\rho$ I think we should be able to find a better bound. The geometric interpretation is that the the circle of radius $\rho$ limits the radius of $|z|$. Is there further geometric intuition that I should see to solve this problem? &lt;/p&gt;&#xA;" OwnerUserId="6758" LastActivityDate="2011-03-07T23:02:57.703" Title="Can I get a better bound on this function?" Tags="&lt;complex-analysis&gt;" AnswerCount="1" />
  <row Id="25642" PostTypeId="2" ParentId="25639" CreationDate="2011-03-07T22:37:24.127" Score="3" ViewCount="" Body="&lt;p&gt;I don't know if there are any hard and fast rules, but here is a rough start for others to nitpick:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;A Theorem is a major result that you care about (e.g. &quot;the goal of this paper is to prove the following theorem&quot;).&lt;/li&gt;&#xA;&lt;li&gt;A Lemma is a useful result that needs to be invoked repeatedly to prove some Theorem or other. Note that sometimes Lemmas can become much more useful than the Theorems they were originally written down to prove. &lt;/li&gt;&#xA;&lt;li&gt;A Proposition is a technical result that does not need to be invoked as often as a Lemma.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2011-03-07T23:19:57.520" LastActivityDate="2011-03-07T23:19:57.520" CommentCount="7" />
  <row Id="25643" PostTypeId="2" ParentId="25624" CreationDate="2011-03-07T22:39:41.590" Score="1" ViewCount="" Body="&lt;p&gt;Yeah, it's not clear to me why he would need absolute convergence, given that he knowns |f(x)|&amp;lt;1.  He might have wanted to show the RHS is continuous, but that seems obvious since it is equal to the LHS, which is continuous.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is possible that, for some reason, the author originally needed this argument, but, due to a later simplification, he failed to drop this line of reasoning, not realizing it had become redundant.&lt;/p&gt;&#xA;" OwnerUserId="7933" LastActivityDate="2011-03-07T22:39:41.590" />
  <row Id="25644" PostTypeId="2" ParentId="25640" CreationDate="2011-03-07T22:45:01.247" Score="3" ViewCount="" Body="&lt;p&gt;Yes; actually, this is one of the only classes of Diophantine equations for which such a result exists! First we will make some simplifying observations. Observe that it is pretty easy to tell what happens when $z = 0$, so suppose $z \neq 0$. Next observe that finding integer solutions is equivalent to finding rational solutions, and since we can scale all three variables by the same constant we can assume $z = 1$, so we are solving $Ax^2 + By^2 = C$ for rationals $x, y$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's a classical result that if there is one solution, there is a straightforward way to describe all of the other solutions: if $(x_0, y_0)$ is a solution, then any line of the form $(x_0 + at, y_0 + bt)$ where $a, b$ are fixed rationals intersects the curve $Ax^2 + By^2 = C$ in exactly one other point, and this intersection must be rational; conversely, every other rational solution arises in this way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So it suffices to find a single solution. To do this the key is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Hasse%E2%80%93Minkowski_theorem&quot; rel=&quot;nofollow&quot;&gt;Hasse-Minkowski theorem&lt;/a&gt;, which tells you that solutions exist over $\mathbb{Q}$ if and only if they exist over $\mathbb{R}$ and over the &lt;a href=&quot;http://en.wikipedia.org/wiki/P-adic_number&quot; rel=&quot;nofollow&quot;&gt;p-adic numbers&lt;/a&gt; $\mathbb{Q}_p$ for all primes $p$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is very easy to check if a solution exists over $\mathbb{R}$, so it suffices to check if solutions exist over $\mathbb{Q}_p$ for all $p$. If $p \nmid 2ABC$, then the &lt;a href=&quot;http://en.wikipedia.org/wiki/Chevalley%E2%80%93Warning_theorem&quot; rel=&quot;nofollow&quot;&gt;Chevalley-Warning theorem&lt;/a&gt; shows that the equation has a solution in $\mathbb{Z}/p\mathbb{Z}$, and by &lt;a href=&quot;http://en.wikipedia.org/wiki/Hensel%27s_lemma&quot; rel=&quot;nofollow&quot;&gt;Hensel's lemma&lt;/a&gt; these solutions can be upgraded to solutions in $\mathbb{Z}_p \subset \mathbb{Q}_p$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we are reduced to checking the finitely many primes dividing $2ABC$. But for any particular such prime, this is more or less an application of &lt;a href=&quot;http://en.wikipedia.org/wiki/Quadratic_reciprocity&quot; rel=&quot;nofollow&quot;&gt;quadratic reciprocity&lt;/a&gt; together with Hensel's lemma again.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is classical material; I think you can find a more thorough exposition in the beginning of Cassels' &lt;em&gt;Lectures on Elliptic Curves.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2011-03-07T22:45:01.247" CommentCount="4" />
  <row Id="25645" PostTypeId="1" CreationDate="2011-03-07T23:00:14.057" Score="0" ViewCount="45" Body="&lt;p&gt;To motivate this question, let me begin with a picture:&#xA;&lt;img src=&quot;http://owen.maresh.info/julia.png&quot; /&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each letter labels a &quot;blob&quot; of this quartic julia set. (is there a technical term for these parts?). Because of resolution limitations I haven't been able to color and label every blob. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only answer to the the MO question &lt;a href=&quot;http://mathoverflow.net/questions/38236/symmetries-of-the-julia-sets-for-z2c&quot; rel=&quot;nofollow&quot;&gt;Symmetries of Julia sets for $z^2 - c$&lt;/a&gt; mentions that &amp;ldquo;This means that there is a quasi-conformal map (thus of bounded distortion) which maps &lt;i&gt;parts of the Julia set to the whole&lt;/i&gt;.&amp;rdquo; The transformation that this labelling seems to suggest maps the entire Julia set to itself. &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Does the set of quasiconformal transformations $\Xi$ (as motivated by the one above) which map the Julia set to itself form a group? I am aware that the quasiconformal transformation is not a quasiconformal conjugacy -- while it fixes two points in the plane it also fixes the entire &lt;b&gt;shape&lt;/b&gt; of the Julia set)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If it exists, Is $\Xi$ the discrete subgroup of a continuous group? (I want to make a movie of such a transformation as this)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;What relation, if any, exists between $\Xi$ and Teichmüller space?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The reason that automorphism is in quotes in the question title is because the phrase &quot;quasiconformal automorphism groups&quot; was used in &lt;a href=&quot;http://www.math.harvard.edu/~ctm/papers/home/text/papers/qciii/qciii.pdf&quot; rel=&quot;nofollow&quot;&gt;this paper&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="890" LastActivityDate="2011-05-30T14:14:50.903" Title="quasiconformal &quot;automorphism&quot; groups of julia sets" Tags="&lt;lie-groups&gt;&lt;fractals&gt;&lt;quasiconformal&gt;&lt;mappings&gt;" AnswerCount="1" />
  <row Id="25646" PostTypeId="2" ParentId="25640" CreationDate="2011-03-07T23:01:21.443" Score="2" ViewCount="" Body="&lt;p&gt;As Qiaochu mentioned, this is a classical problem that is a prototypical example of an equation that is amenable to the &lt;em&gt;local-global&lt;/em&gt; approach. Legendre obtained a complete solution by descent circa 1785. Later, as $p$-adic methods emerged, it was realized that Legendre's solution could be elegantly reformulated via such local-global techniques. You can find a nice readable four-page introduction on &lt;a href=&quot;http://books.google.com/books?id=yMGeElJ8M0wC&amp;amp;pg=PA238&quot; rel=&quot;nofollow&quot;&gt;pp. 238-242 of Harvey Cohn: Advanced Number Theory.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2011-03-07T23:01:21.443" />
  <row Id="25647" PostTypeId="2" ParentId="25641" CreationDate="2011-03-07T23:02:57.703" Score="2" ViewCount="" Body="&lt;p&gt;No, there is no better bound. Consider the function $f(z)=M \frac{z^n}{R^n}$ which fulfills the requirement (ok, that $|f(z)|$ is really smaller than $M$ one should replace $M$ by $M-\epsilon$ and in the end let $\epsilon\to0$). It is easy to check that $$f^{(n)}(z) = \frac{n! M}{R^n},$$ i.e., it exhausts the Cauchy bound. Note however that the $n$-th derivative is independent on $z$. So there is no way you get a better bound by restricting $|z|&amp;lt;\rho$.&lt;/p&gt;&#xA;" OwnerUserId="7266" LastActivityDate="2011-03-07T23:02:57.703" />
  <row Id="25648" PostTypeId="1" AcceptedAnswerId="25658" CreationDate="2011-03-07T23:03:59.577" Score="1" ViewCount="98" Body="&lt;p&gt;My problem is how to somehow relate the the gcd and congruence. I know that (a,b) = ax + by. I also know that $a \equiv b$ (mod n) means n|a-b. Any hints?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="7934" LastActivityDate="2011-03-08T02:19:20.820" Title="Show that if $a \equiv b$ (mod n), gcd(a,n)=gcd(b,n)" Tags="&lt;homework&gt;&lt;elementary-number-theory&gt;" AnswerCount="3" />
  <row Id="25649" PostTypeId="1" CreationDate="2011-03-07T23:04:32.910" Score="1" ViewCount="235" Body="&lt;p&gt;$T(1) = 1 , T(n) = 2T(n/2) + n^3$? Divide and conquer, need help, I dont know how to solve it?&lt;/p&gt;&#xA;" OwnerUserId="7935" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2011-06-03T13:46:06.570" LastActivityDate="2011-06-03T13:46:06.570" Title="$T(1) = 1 , T(n) = 2T(n/2) + n^3$? Divide and conquer" Tags="&lt;algorithms&gt;&lt;asymptotics&gt;&lt;recurrence-relations&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="1" />
  <row Id="25650" PostTypeId="2" ParentId="24697" CreationDate="2011-03-07T23:05:53.100" Score="1" ViewCount="" Body="&lt;p&gt;We can asume that 2 is a quadratic residue mod $p$ and so that $p \equiv 1 \pmod 8$ and this implies that if we pick $a$ odd and $b$ even then $b$ is a multiple of 4. We have to prove that $b$ is a multiple of 8. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;First observe thatas $x^2 \equiv -1 \pmod{p}$ and $a^2 + b^2 = p$ we have &#xA;    $$ \left(\frac{a+b}{p}\right) \equiv (-1)^{((a+b)^2-1)/8} \equiv x^{(p+2ab-1)/4} \pmod{p} $$&#xA;note that the exponent of $x$ is even because $b$ is multiple of 4, so we can chose the sign of the base as we wish.&lt;br&gt;&#xA;In adition we also have &#xA;    $$ a^2 \equiv -b^2 \pmod{p} $$&#xA;so $a^2 \equiv (xb)^2 \pmod{p}$ and &#xA;   $$ax \equiv \pm b \pmod{p}$$&#xA;and picking the sign of $x$ we can asume that $ax \equiv b \pmod{p}$. So  $ab \equiv a^2 x$ and &#xA;  $$ (ab)^{(p-1)/4} = a^{(p-1)/2} x^{(p-1)/4} \equiv x^{(p-1)/4} \pmod{p} $$&#xA;because $a$ is a quadratic residue mod $p$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The identity you obtained:&#xA;    $$\left(\frac{a+b}{p}\right) = (a+b)^{(p-1)/2} \equiv (2ab)^{(p-1)/4} \pmod p $$&#xA;now becomes&#xA;    $$x^{(p+2ab-1)/4} \equiv 2^{(p-1)/4} x^{(p-1)/4} \pmod p $$&#xA;and in consequence&#xA;    $$2^{(p-1)/4} \equiv x^{ab/2} \pmod p $$&#xA;As $b$ is multiple of 4 say $b = 4b&amp;#39;$, we have&#xA;    $$2^{(p-1)/4} \equiv (-1)^{ab&amp;#39;} \pmod p $$&#xA;so $2$ is a biquadratic residue iif $b&amp;#39;$ is even or what is the same thing iif $b$ is multiple of 8 and we are done. &lt;/p&gt;&#xA;" OwnerUserId="3274" LastActivityDate="2011-03-07T23:05:53.100" CommentCount="2" />
  <row Id="25651" PostTypeId="2" ParentId="25649" CreationDate="2011-03-07T23:08:56.243" Score="2" ViewCount="" Body="&lt;p&gt;Use &lt;a href=&quot;http://en.wikipedia.org/wiki/Akra%E2%80%93Bazzi_method&quot; rel=&quot;nofollow&quot;&gt;Akra-Bazzi&lt;/a&gt; which is more useful than the &lt;a href=&quot;http://en.wikipedia.org/wiki/Master_theorem&quot; rel=&quot;nofollow&quot;&gt;Master Theorem&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using Akra-Bazzi, I believe you get $$T(x) = \theta(x^3)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can also use the Case 3 of Master theorem in the wiki link above. (Note: That also gives $\theta(x^3)$.)&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2011-03-07T23:08:56.243" CommentCount="9" />
  <row Id="25653" PostTypeId="2" ParentId="25648" CreationDate="2011-03-07T23:10:57.533" Score="2" ViewCount="" Body="&lt;p&gt;We know that $a-b = nx$. So $a-nx = b$ and $b+nx = a$. &lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2011-03-07T23:10:57.533" CommentCount="1" />
  <row Id="25654" PostTypeId="2" ParentId="25640" CreationDate="2011-03-07T23:11:57.677" Score="0" ViewCount="" Body="&lt;p&gt;I would happily add Serre's book. &lt;/p&gt;&#xA;" OwnerUserId="7887" LastActivityDate="2011-03-07T23:11:57.677" CommentCount="2" />
  <row Id="25655" PostTypeId="2" ParentId="25639" CreationDate="2011-03-07T23:13:24.737" Score="4" ViewCount="" Body="&lt;p&gt;While generally  the terms are used as suggested by Qiaochu, there are some authors who are bothered by these nebulous subjective terms. For example, Kaplansky wrote in the preface of his classic textbook &lt;em&gt;Commutative Rings&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In the style of Landau, or Hardy and Wright, I have presented the material as an unbroken series of theorems. I prefer this to the n-place decimal system favored by some authors, and I have also grown tired of seeing a barrage of lemmas, propositions, corollaries, and scholia (whatever they are). I admit that this way the lowliest lemma gets elevated to the same eminence as the most awesome theorem. Also, the number of theorems becomes impressive,  so impressive that I felt the need to add an index of theorems. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="242" LastActivityDate="2011-03-07T23:13:24.737" />
  <row Id="25656" PostTypeId="2" ParentId="25648" CreationDate="2011-03-07T23:18:11.227" Score="5" ViewCount="" Body="&lt;p&gt;Another way to understand $a \equiv b \pmod n$ is that there is some $k$ such that $a = b + kn$. This is the view in terms of Ideals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now you can conclude by proving that $$(b + kn,n) = (b,n).$$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Two useful facts about gcd for every $a,b$:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\exists x,y,\,\,(a,b)=ax+by.$&lt;/li&gt;&#xA;&lt;li&gt;$\forall x,y,\,\,(a,b)|ax+by.$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Now the idea is to prove that both $(b+kn,n)|(b,n)$ and $(b,n)|(b+kn,n)$ which implies that $(b+kn,n) = (b,n).$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You've already shown the first one $(b+kn,n)=(b+kn)x+ny=bx+knx+ny=bx+n(kx+y)$ hence $(b,n)|(b+kn,n).$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can use the same algebra methods to prove the second part and conclude. (HINT: u = u + v - v)&lt;/p&gt;&#xA;" OwnerUserId="4290" LastEditorUserId="4290" LastEditorDisplayName="" LastEditDate="2011-03-08T00:27:20.657" LastActivityDate="2011-03-08T00:27:20.657" CommentCount="2" />
  <row Id="25657" PostTypeId="2" ParentId="25649" CreationDate="2011-03-07T23:20:51.707" Score="7" ViewCount="" Body="&lt;p&gt;The homogeneous part of the equation $T(n) =2 T(n/2)$ has the general solution $$T_0(n) = C n.$$ So all what we have to do is find a particular solution of the inhomogeneous equation $$T(n) = 2 T(n/2) + n^{3} .$$ Quite often it is good to try an ansatz which has the same form as the inhomogeneous term. Therefore, we try $T_p(n) = c n^3$ which yields $$c n^3 = \frac{c}{4} n^3 + n^3.$$ Solving for $c$, we get the particular solution $T_p(n)= \frac{4}{3} n^3$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The general solution therefore has the form $T(n) =T_0(n) + T_p(n)$. With the initial condition $T(1)=1$, we obtain $C= - \frac{1}{3}$. So the solution is given by &#xA;$$ T(n) = \frac{n}{3} (4 n^2 -1). $$&lt;/p&gt;&#xA;" OwnerUserId="7266" LastEditorUserId="7266" LastEditorDisplayName="" LastEditDate="2011-03-07T23:30:45.953" LastActivityDate="2011-03-07T23:30:45.953" CommentCount="5" />
  <row Id="25658" PostTypeId="2" ParentId="25648" CreationDate="2011-03-07T23:22:05.990" Score="3" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\: $ If $\rm\ d\ |\ n\ |\ a-b\  $ then $\rm\ d\ |\ a\ \iff\ d\ |\ b\:, \ $ i.e. $\: $ if $\rm\ a\equiv b\ $ then $\rm\ a\equiv 0\ \iff\ b\equiv 0\ \ \ (mod\ d)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus $\rm\ \{a,n\}\:,\ \{b,n\}\ $ have the same common divisors $\rm\:d\:,\:$ so the same &lt;em&gt;greatest&lt;/em&gt; common divisor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;REMARK&lt;/strong&gt;  $\ $ Note how much simpler the problem becomes after eliminating the obfuscated divisibility &lt;em&gt;relations&lt;/em&gt; in favor of &lt;em&gt;equations&lt;/em&gt; (here congruences). $\ $ After converting it to manipulation of equations it is completely trivial, viz. if $\rm\ a = b\ $ then $\rm\ a = n = 0\ \iff\ b = n = 0\:.\:$ That is the tremendous power of congruences - they allow us to transform diverse problem to &lt;em&gt;equational form&lt;/em&gt; - allowing us to reuse our well-honed intuition on manipulating (integer) equations. To succeed in number theory and algebra it is &lt;em&gt;absolutely essential&lt;/em&gt; to master such techniques. They lead to powerful methods of &quot;modular reduction&quot; - an algebraic way of &quot;dividing and conquering&quot; - reducing problems to simpler problems that hopefully combine to yield the complete solution. &lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2011-03-08T02:19:20.820" LastActivityDate="2011-03-08T02:19:20.820" CommentCount="3" />
  <row Id="25659" PostTypeId="2" ParentId="25634" CreationDate="2011-03-07T23:31:31.087" Score="4" ViewCount="" Body="&lt;p&gt;&quot;Area between two graphs&quot; is, by definition, positive regardless of where in the plane it lies. This is different from the concept of &quot;net signed area between the graph and the $x$-axis&quot;, which is what the regular Riemann integral measures.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is much like we say that the area of the unit circle is $2\pi$, regardless of where in the plane we &quot;draw&quot; it. (If you were considering the &lt;em&gt;net signed area&lt;/em&gt; inside the unit circle centered at the origin, then this would be $0$, since we count the area below the $X$-axis as &quot;negative&quot; and the area over the $X$-axis as &quot;positive&quot;, but the &lt;em&gt;net signed area&lt;/em&gt; is not the same as the &lt;em&gt;area&lt;/em&gt;.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So here, you are being asked to measure the &lt;em&gt;area&lt;/em&gt;, in the usual sense of the word, of the region enclosed by the two graphs. This means integrating the absolute value of their difference (which automatically takes care of converting &quot;net signed area&quot; into regular area), which in turn is done by, as you do, by breaking it up to figure out where $|f-g|$ equals $f-g$ and where it equals $g-f$. But in any case, the &quot;area enclosed&quot; refers to the actual area enclosed, not the &quot;net signed area enclosed&quot;.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2011-03-07T23:31:31.087" />
  <row Id="25660" PostTypeId="1" AcceptedAnswerId="25669" CreationDate="2011-03-07T23:36:23.583" Score="14" ViewCount="189" Body="&lt;p&gt;I recently saw a conjecture on a blog ( &lt;a href=&quot;http://blog.tanyakhovanova.com/?p=311&quot; rel=&quot;nofollow&quot;&gt;http://blog.tanyakhovanova.com/?p=311&lt;/a&gt; ) which the author refers to as the 86 conjecture. The conjecture claims that all powers of 2 greater than $2^{86}$ have a zero in their base 10 representation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;At first glance this seems like a numerical curiosity, and I wasn't sure that this was an interesting or deep mathematical question. I wanted to ask the community if they knew of any existing work that deals with either this type of question or something similar. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried searching for 86 conjecture on Google and didn't get anything useful.&lt;/p&gt;&#xA;" OwnerUserId="2641" LastEditorUserId="3731" LastEditorDisplayName="" LastEditDate="2011-03-08T00:26:57.160" LastActivityDate="2011-03-08T01:02:57.763" Title="Status of a conjecture about powers of 2" Tags="&lt;number-theory&gt;&lt;open-problem&gt;&lt;conjectures&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="25661" PostTypeId="2" ParentId="25556" CreationDate="2011-03-07T23:50:58.050" Score="2" ViewCount="" Body="&lt;p&gt;This actually holds for any analytic function of two complex variables on an open set. Suppose $f(x_0,y_0) = 0$. Then viewed as a function of $y$, $f(x_0,y)$ must have an isolated zero of some order $k$ at $y = y_0$. By complex analysis (the residue theorem will work for example), for some small $r$ one has &#xA;$${1 \over 2\pi i} \int_{|z - y_0| = r} {f&amp;#39;(x_0, z) \over f(x_0,z)}\,dz = k$$&#xA;But the integrand above is a continuous function of $x$ near $x = x_0$ and takes integer values. Hence for $x$ close enough to $x_0$ one also has&#xA;$${1 \over 2\pi i} \int_{|z - y_0| = r} {f&amp;#39;(x, z) \over f(x,z)}\,dz = k$$&#xA;This in turn implies that whenever $x$ is close enough to $x_0$, $f(x,y)$ (viewed as a function of $y$) has $k$ zeroes inside $|z - y_0| = r$. Since this holds for arbitrarily small $r$, we conclude the zero of $f(x,y)$ at $(x_0,y_0)$ is not isolated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is actually a watered-down version of the proof of the famous Weierstrass Preparation Theorem (which incidentally also implies this fact in pretty short order).&lt;/p&gt;&#xA;" OwnerUserId="3035" LastEditorUserId="3035" LastEditorDisplayName="" LastEditDate="2011-03-08T00:29:40.683" LastActivityDate="2011-03-08T00:29:40.683" />
  <row Id="25662" PostTypeId="2" ParentId="24513" CreationDate="2011-03-07T23:53:04.757" Score="3" ViewCount="" Body="&lt;p&gt;As $F$ and $G$ are locally free, and as the stalk at $p$ depends only of what happens in an open neighborhood of $p$, you can assume that $F$ and $G$ are in fact &lt;em&gt;free&lt;/em&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the context of that exercise, the sheaves are actually of finite rank, so we end up with $F$ and $G$ free of finite rank. Since everything in sight is an additive functor, additivity allows us to reduce to the case where $F$ and $G$ are in fact free of rank $1$. Everything is then obvious :)&lt;/p&gt;&#xA;" OwnerUserId="274" LastEditorUserId="274" LastEditorDisplayName="" LastEditDate="2011-03-08T00:13:21.097" LastActivityDate="2011-03-08T00:13:21.097" CommentCount="2" />
  <row Id="25663" PostTypeId="2" ParentId="24513" CreationDate="2011-03-08T00:16:41.600" Score="6" ViewCount="" Body="&lt;p&gt;Presumably the solution you have in mind is this. There is a natural homomorphism $\newcommand{\E}{\mathcal{E}}\newcommand{\O}{\mathcal{O}}\newcommand{\Hom}{\mathcal{H}om}&#xD;&#xA;\E\to(\E^\vee)^\vee$. It suffices to check that this homomorphism is an isomorphism at every stalk.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's the right approach, but there's no need to go all the way to stalks. It suffices to show that on any open set $U\subseteq X$ where $\E$ is actually free, the natural map is an isomorphism. Since $\E|_U\cong \O_U^n$, it suffices to check that the natural map $\O^n\to \Hom(\Hom(\O^n,\O),\O)$ is an isomorphism. This is easy to check, but requires you to unravel the map.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Remark:&lt;/strong&gt; Note that to show two sheaves on $X$ are isomorphic, it &lt;em&gt;is not enough&lt;/em&gt; to find an open cover $X$ and isomorphisms between the two sheaves on each open set. The reason is that the isomorphisms may not agree the intersections of the open sets in the cover. Naturality of the map plays a very important role here: it ensures that the isomorphisms on the open cover will glue. To put it another way, we first constructed the morphism $\E\to (\E^\vee)^\vee$, and &lt;em&gt;then&lt;/em&gt; checked that it is an isomorphism on an open cover. I want to stress that it is &lt;em&gt;not enough&lt;/em&gt; to just check that $\O^n$ and $\Hom(\Hom(\O^n,\O),\O)$ are isomorphic, you must check that the specific map $s\mapsto (\phi\mapsto \phi(s))$ is an isomorphism.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This same remark holds if you want to use the stalks approach. You must first construct the global map, and then verify that it induces isomorphisms on stalks. Just showing that the stalks are isomorphic is not enough. If it were, any two locally free sheaves of the same rank would be isomorphic.&lt;/p&gt;&#xA;" OwnerUserId="60" LastActivityDate="2011-03-08T00:16:41.600" CommentCount="1" />
  <row Id="25664" PostTypeId="1" AcceptedAnswerId="25666" CreationDate="2011-03-08T00:22:55.313" Score="2" ViewCount="72" Body="&lt;p&gt;Can anyone point me to any reference on the following inequality:&#xA;Let $c&amp;gt;0, a,b \in \Bbb{R}$&#xA;$$&#xD;&#xA;\|a+b\|^2 \leq (1+c)\|a\|^2+(1+{\textstyle\frac{1}{c}})\|b\|^2&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not even sure if &quot;Bohr's Inequality&quot; is this inequality's name, since I can't find it anywhere in the internet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm having a bad day and I got stuck in this one, apparently. Any suggestions or hints on how to prove it will be appreciated.&lt;/p&gt;&#xA;" OwnerUserId="7928" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-08T00:29:27.283" LastActivityDate="2011-03-08T00:33:41.430" Title="Bohr's Inequality" Tags="&lt;homework&gt;&lt;inequality&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="25665" PostTypeId="2" ParentId="25664" CreationDate="2011-03-08T00:31:49.713" Score="2" ViewCount="" Body="&lt;p&gt;Do you mean&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(a+b)^2 \le (1+c)a^2 + (1 + 1/c)b^2$$&#xA;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This follows from $\text{AM} \ge \text{GM}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ca^2 + b^2/c \ge 2 ab$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;No add $a^2 + b^2$ to both sides.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you meant $\mathbb{R}^n$ and the Euclidean Norm, you can prove it by applying the above $n$ times.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2011-03-08T00:31:49.713" CommentCount="1" />
  <row Id="25666" PostTypeId="2" ParentId="25664" CreationDate="2011-03-08T00:33:41.430" Score="1" ViewCount="" Body="&lt;p&gt;Here's a hint. $a,b$ and $c$ are real, and the inequality boils down to showing that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;c|a|^2 + \frac{1}{c}|b|^2 -2ab \geq 0&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the function&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;f(x) = x|a|^2 + \frac{1}{x}|b|^2 -2ab&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the minimum value of this function?&lt;/p&gt;&#xA;" OwnerUserId="2641" LastActivityDate="2011-03-08T00:33:41.430" CommentCount="2" />
  <row Id="25667" PostTypeId="2" ParentId="25595" CreationDate="2011-03-08T00:36:03.080" Score="1" ViewCount="" Body="&lt;p&gt;The problem is easily solvable using linear programming.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively, you can &quot;roll out&quot; the simplex algorithm. Start with the assignment $\mathbf{x} \equiv 0$. At each step, select an extremal entry $v_i$ and one of the two corresponding variables $x_j$. Try either increasing or decreasing $x_j$, depending on its sign, until the situation &quot;balances&quot; at a smaller $L_\infty$ norm (this is also worth doing if the norm doesn't change, though then you need to ensure somehow that you don't &quot;cycle&quot;). Whenever this method offers no improvement for all extremal entries and all corresponding variables, we are at an optimum.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2011-03-08T00:36:03.080" />
  <row Id="25668" PostTypeId="2" ParentId="25649" CreationDate="2011-03-08T01:00:03.877" Score="2" ViewCount="" Body="&lt;p&gt;Hmm, possibly another way of heuristics is instructive.&lt;br&gt;&#xA;First write the undisputable elements of the sequence:&lt;br&gt;&#xA;$$\begin{array}  &amp;amp;&#xD;&#xA;  a(1) &amp;amp;=&amp;amp;a(2^0) &amp;amp; = &amp;amp; 1 \\\&#xD;&#xA;   a(2)&amp;amp;=&amp;amp;a(2^1) &amp;amp;=&amp;amp;10 &amp;amp;= &amp;amp; 2^3 + 2*1  &amp;amp;=&amp;amp; 2*(4^1+1) \\\&#xD;&#xA;   a(4)&amp;amp;=&amp;amp;a(2^2) &amp;amp;=&amp;amp;84 &amp;amp;=&amp;amp; 4^3 + 2*10  &amp;amp;=&amp;amp; 4*(4^2+4^1+1)  \\\&#xD;&#xA;   a(8)&amp;amp;=&amp;amp;a(2^3) &amp;amp;=&amp;amp;680 &amp;amp;=&amp;amp; 8^3 + 2*84 &amp;amp;=&amp;amp; 8^3+2*4^3+4*2^3+8*1^3\\\&#xD;&#xA;       &amp;amp; &amp;amp;       &amp;amp; &amp;amp;                   &amp;amp;=&amp;amp; 8*(4^3+4^2+4^1+1)  \\\  &#xD;&#xA;   \ldots &amp;amp;=&amp;amp;a(2^k)&amp;amp;=&amp;amp; \ldots &#xD;&#xA;   \end{array} $$&#xA;It is obvious how this can be continued, because at the exponent &lt;em&gt;k&lt;/em&gt; we get always $8^k$ plus two times the previous, thus the weighted sum of all powers of &lt;em&gt;8&lt;/em&gt; which can be expressed as consecutive powers of &lt;em&gt;4&lt;/em&gt;:&#xA;$$ a(2^k) = 2^k*(4^k+4^{k-1} \ldots +4^0)= 2^k*\frac{4^{k+1}-1}{4-1} $$&#xA;Now the step &quot;divide&quot; can be taken: the above gives also a meaningful possibility for interpolation of the non-explicitely defined elements. If we allow base-&lt;em&gt;2&lt;/em&gt; logarithms for &lt;em&gt;k&lt;/em&gt; we get for&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{array} &amp;amp;&#xD;&#xA; &amp;amp; a(2^k) &amp;amp;= &amp;amp; 2^k*\frac{4^{k+1}-1}{4-1} \\\&#xD;&#xA;  &amp;amp;      &amp;amp;= &amp;amp; 2^k*\frac{4*(2^{k})^2-1}{3} \\\&#xD;&#xA;      \text{assuming }&amp;amp; k&amp;amp;=&amp;amp; \frac{\log(n)}{\log(2)} \\\&#xD;&#xA; &amp;amp; a(n)  &amp;amp;=&amp;amp; n*\frac{4*n^2-1}{3} \\\&#xD;&#xA;  &amp;amp;      &amp;amp;=&amp;amp; n^3 + \frac{(n-1)n(n+1)}{3!} \\\&#xD;&#xA;   &amp;amp;     &amp;amp;=&amp;amp; n^3 + 2*\binom{n+1}{3} \\\&#xD;&#xA; \end{array} $$&#xA;where the expression in the fourth line is the same as Fabian's result.&lt;/p&gt;&#xA;" OwnerUserId="1714" LastEditorUserId="1714" LastEditorDisplayName="" LastEditDate="2011-03-08T01:12:19.180" LastActivityDate="2011-03-08T01:12:19.180" />
  <row Id="25669" PostTypeId="2" ParentId="25660" CreationDate="2011-03-08T01:02:57.763" Score="9" ViewCount="" Body="&lt;p&gt;Richard Guy's &lt;a href=&quot;http://books.google.com/books?id=1AP2CEGxTkgC&amp;amp;pg=PA398&amp;amp;lpg=PA398&amp;amp;dq=decimal+representation+of+powers+of+two+unsolved+problems+in+number+theory&amp;amp;source=bl&amp;amp;ots=TiqCeYDJpx&amp;amp;sig=fPyNdKdS4-LirADqTw4UA2RDoJI&amp;amp;hl=en&amp;amp;ei=83x1TaayC4mqsAOUutm8BA&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=1&amp;amp;ved=0CBgQ6AEwAA#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;Unsolved Problems in Number Theory&lt;/em&gt;, Problem F24&lt;/a&gt;, mentions only that Dan Hoey has verified this conjecture for $2^n$ up to $n = 2,500,000,000$.  Guy tends to be fairly complete in his references.  Since he doesn't give any others I doubt there's much more out there. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Several other related questions on decimal representations of powers of integers appear in F24 as well, most of which also appear to be open problems.  There are some sequences in the &lt;a href=&quot;http://oeis.org/&quot; rel=&quot;nofollow&quot;&gt;On-line Encyclopedia of Integer Sequences&lt;/a&gt; mentioned, too; those might be worth chasing down.&lt;/p&gt;&#xA;" OwnerUserId="2370" LastActivityDate="2011-03-08T01:02:57.763" CommentCount="2" />
  <row Id="25670" PostTypeId="1" CreationDate="2011-03-08T01:08:13.287" Score="3" ViewCount="107" Body="&lt;p&gt;I've seen one proof of $\mathrm{psl}_2(\mathbf Z) \cong C_2 * C_3$ based on the &lt;a href=&quot;http://en.wikipedia.org/wiki/Ping-pong_lemma&quot; rel=&quot;nofollow&quot;&gt;Ping-pong lemma&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But what I'm looking for is how this result was historically obtained first.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've checked Coxeter &amp;amp; Moser's &quot;Generators and relations for discrete groups&quot; and they (§7.2) do things that I find weird, they let the matrices act on circles in the half-plane $y&amp;gt;0$ which in turn they regard as lines of a hyperbolic plane, it then follows this group can be interpreted as triangle group acting on a hyperbolic triangle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this indeed the 'classic' way to arrive at this result? And if so, I'd be happy with a bit of an explanation of what's going on, or a reference to a text where this is explained a bit more in detail, because I'm really not that comfortable with hyperbolic triangles.&lt;/p&gt;&#xA;" OwnerUserId="5189" LastActivityDate="2011-03-10T01:48:23.780" Title="Well known proofs that $\mathrm{psl}_2(\mathbf Z) \cong C_2 * C_3$" Tags="&lt;group-theory&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="3" />
  <row Id="25671" PostTypeId="1" CreationDate="2011-03-08T01:19:36.443" Score="1" ViewCount="228" Body="&lt;p&gt;I have seen a picture for $dV$ so that $dV = r^{2} \sin(\theta) dr d\theta d\phi$. But how can I deduce things like dA and dV? In a simpler coordinate (not sure about the name), $dA = r dr d\phi$, again no idea &lt;em&gt;how to derive it&lt;/em&gt; (just given to me to memorize it). So explain what is dA in polar coordinates?&lt;/p&gt;&#xA;" OwnerUserId="7628" LastActivityDate="2011-03-08T21:54:18.590" Title="dA in polar coordinates?" Tags="&lt;polar-coordinates&gt;" AnswerCount="2" CommentCount="9" FavoriteCount="2" />
  <row Id="25672" PostTypeId="2" ParentId="25639" CreationDate="2011-03-08T01:29:59.683" Score="5" ViewCount="" Body="&lt;p&gt;There seem to be two issues here.  One is why certain well-known results are called Lemmas, such as Zorn's, Yoneda's, Nakayama's, and so on.  I don't know the answer to this; presumably it is a mixture of what was written in some original source and the results of the transmission of that original source through the mathematical tradition.  (As one interesting example of how labels can be changed in the course of transmission, there is a result in the theory of automorphic forms and Galois representations, very well known to experts, universally referred to as &quot;Ribet's Lemma&quot;; however, in the original paper it is labelled as a proposition!)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second issue is how contemporary writers label the results in their papers.  My experience is that typically the major results of the paper are called theorems, the lesser results are called propositions (these are typically ingredients in the proofs of the theorems which are also stand-alone statements that may be of independent interest), and the small technical results are called lemmas.   This probably varies quite a bit from writer to writer (and perhaps also from field to field?).  &lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2011-03-08T01:29:59.683" CommentCount="2" />
  <row Id="25673" PostTypeId="1" AcceptedAnswerId="25687" CreationDate="2011-03-08T01:38:07.193" Score="0" ViewCount="72" Body="&lt;p&gt;I came across the following sentence in an article:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$f$ is injective on $L^1(\mathbb{R})$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;where $f$ is some function of $g$. Now I have a vague idea that injective= one-to-one, and can visualize it in $L^1$. Does it mean anything to say injective on $L^n$? Can someone illustrate it with an example in $L^2$?&lt;/p&gt;&#xA;" OwnerUserId="7815" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-08T02:25:42.417" LastActivityDate="2011-03-08T02:34:11.733" Title="Injective on $L^n$: what does it mean?" Tags="&lt;linear-algebra&gt;&lt;functions&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="25674" PostTypeId="1" AcceptedAnswerId="25678" CreationDate="2011-03-08T01:40:37.070" Score="5" ViewCount="106" Body="&lt;p&gt;Suppose you have an exponential generating function&#xA;$$&#xD;&#xA;F(x)=F_0+F_1x+F_2\frac{x^2}{2!}+\cdots+F_n\frac{x^n}{n!}+\cdots&#xD;&#xA;$$&#xA;and you want to get only the even terms&#xA;$$&#xD;&#xA;F_e(x)=F_0+F_2\frac{x^2}{2!}+F_4\frac{x^4}{4!}+\cdots&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it possible to write $F_e(x)$ in terms of $F(x)$? I noticed that taking the derivative of $F(x)$ seems to shift the coefficients down a term, but couldn't get much further than that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm also interested in the odd case, but I guess that follows easily since $F(x)-F_e(x)=F_o(x)$. &lt;/p&gt;&#xA;" OwnerUserId="7938" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2011-03-08T02:38:33.710" LastActivityDate="2011-03-08T02:38:33.710" Title="Given an exponential generating function, is it possible to isolate only the even terms?" Tags="&lt;generating-functions&gt;" AnswerCount="3" FavoriteCount="2" />
  <row Id="25675" PostTypeId="2" ParentId="25671" CreationDate="2011-03-08T01:41:17.903" Score="4" ViewCount="" Body="&lt;p&gt;The way to calculate this without appeal to a geometric construction is to use the &lt;a href=&quot;http://en.wikipedia.org/wiki/Jacobian_matrix_and_determinant&quot; rel=&quot;nofollow&quot;&gt;Jacobian determinant&lt;/a&gt;.  The Jacobian matrix is constructed by taking the partial derivative of the new coordinates ($\epsilon_i$) with respect to the old coordinates ($\delta_j$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$J_{ij} = \frac{\partial \epsilon_i}{\partial \delta_j}.$$&lt;/p&gt;&#xA;" OwnerUserId="2283" LastEditorUserId="2283" LastEditorDisplayName="" LastEditDate="2011-03-08T01:48:18.767" LastActivityDate="2011-03-08T01:48:18.767" />
  <row Id="25676" PostTypeId="2" ParentId="25674" CreationDate="2011-03-08T01:44:57.003" Score="3" ViewCount="" Body="&lt;p&gt;$F_e(x)=\frac{1}{2}(F(x)+F(-x))$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$F_o(x)=\frac{1}{2}(F(x)-F(-x))$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See also &lt;a href=&quot;http://math.stackexchange.com/questions/5274/how-do-i-divide-a-function-into-even-and-odd-sections&quot;&gt;How do I divide a function into even and odd sections?&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2011-03-08T01:44:57.003" CommunityOwnedDate="2011-03-08T01:44:57.003" CommentCount="2" />
  <row Id="25677" PostTypeId="1" CreationDate="2011-03-08T01:46:47.207" Score="1" ViewCount="46" Body="&lt;p&gt;Suppose you have a classifier $C^n$ which continuously outputs a stream of classification labels $K^n_i$ and corresponding timestamps $T^n_i$. Also, we know the prior probability $P(K^n) \forall n$. Now, suppose we have several classifiers obtaining classification labels from different sources of data but all attempting to recognize the same object.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do I match the information from these streams? I could store a set of $k$ observations from all classifiers ({$K^n_i,T^n_i$} $\forall n$ and $i\in [1,k]$) and assume that observations correspond to each other based on the distance between their timestamps. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;E.g. if we have 2 classifiers and $1\le a,b \le k$, {$K^1_a,T^1_a$} , {$K^2_b,T^2_b$} match if $|T^1_a-T^2_b|&amp;lt;\epsilon$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a more &quot;intelligent&quot; way to do this?&lt;/p&gt;&#xA;" OwnerUserId="825" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2011-03-08T06:15:12.130" LastActivityDate="2011-05-07T06:50:12.677" Title="Asynchronous data stream matching" Tags="&lt;statistics&gt;&lt;machine-learning&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="25678" PostTypeId="2" ParentId="25674" CreationDate="2011-03-08T01:47:07.210" Score="10" ViewCount="" Body="&lt;p&gt;They're called series &lt;em&gt;bisections&lt;/em&gt;, e.g. bisecting into even and odd parts the power series for $\:\rm e^{\:i\:x} \:,\;$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xD;&#xA;\rm f(x) \ \ &amp;amp;=\ \ \rm\frac{f(x)+f(-x)}{2} \;+\; \frac{f(x)-f(-x)}{2} \\\\ &#xD;&#xA;\\\\&#xD;&#xA;\Rightarrow\quad\quad \rm e^{\:i\:x} \ \ &amp;amp;=\ \ \rm\cos(x) \ +\ i \ \sin(x)  \end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly one can perform &lt;em&gt;multisections&lt;/em&gt; into $\rm\:n\:$ parts using $\rm\:n\:$'th roots of unity - see &lt;a href=&quot;http://math.stackexchange.com/questions/3510/how-to-prove-eulers-formula-expi-t-costi-sint/3514#3514&quot;&gt;my post here&lt;/a&gt; for some examples and see Riordan's classic &#xA;textbook &lt;em&gt;Combinatorial Identities&lt;/em&gt; for many applications. Briefly, &#xA;with  $\rm\:\zeta\ $ a primitive $\rm\:n$'th root of unity,  the $\rm\:m$'th $\rm\:n$-section selects the linear progression of $\rm\: m+k\:n\:$ indexed terms from a series $\rm\ f(x)\ =\ a_0 + a_1\ x + a_2\ x^2 +\:\cdots\ $ as follows&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\quad\quad\quad  a_m\ x^m\ +\ a_{m+n}\ x^{m+n}\ +\ a_{m+2\:n}\ x^{m+2\:n}\ +\:\cdots $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\quad =\ \frac{1}{n} \big(f(x)\ +\ f(x\zeta)\ \zeta^{-m}\ +\ f(x\zeta^{\:2})\ \zeta^{-2m}\ +\:\cdots\: +\ f(x\zeta^{\ n-1})\ \zeta^{\ (1-n)\:m}\big)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EXERCISE&lt;/strong&gt; $\;$ Use multisections to give elegant proofs of the following&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\quad\quad\rm\displaystyle sin(x)/e^{x} \quad\:$ has every $\rm 4\ k\;$'th term zero in its power series&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\quad\quad\rm\displaystyle cos(x)/e^{x} \quad$ has every $\rm 4\ k+2\;$'th term zero in its power series&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See the posts in &lt;a href=&quot;http://groups.google.com/group/sci.math/msg/642159e4f5db4e44&quot; rel=&quot;nofollow&quot;&gt;this thread&lt;/a&gt; for various solutions and more on multisections. When you later study representation theory of groups you will learn that this is a special case of much more general results, with relations to Fourier and other transforms. It's also closely related to various Galois-theoretic results on modules, e.g. see my remark about Hilbert's Theorem 90 in the linked thread.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2011-03-08T02:12:43.120" LastActivityDate="2011-03-08T02:12:43.120" CommentCount="1" />
  <row Id="25679" PostTypeId="1" CreationDate="2011-03-08T01:58:33.917" Score="3" ViewCount="80" Body="&lt;p&gt;This is a homework exercise, so please don't post full solutions to the question below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Grothendieck (I believe) introduced several axioms an abelian category A voluntarily could satisfy. In particular, we have the two following:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;AB5&lt;/strong&gt; A is cocomplete and filtered colimits of exact sequences are exact.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2 &lt;strong&gt;AB5*&lt;/strong&gt; A is complete and filtered inverse limits of exact sequences are exact.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Then my problem is the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;A.4.7 (Weibel)&lt;/strong&gt; Show that if $A\neq 0$, then A cannot satisfy both axiom AB5 and AB5*. Hint: consider $\oplus A_i \to \prod A_i$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;_&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have stared at this problem for a long time but no ideas have (nor much intuition for filtered limits) been born. If anyone has any helpful explanations or wonderful hints, I'd be really grateful. Thanks.&lt;/p&gt;&#xA;" OwnerUserId="4284" LastActivityDate="2011-03-08T01:58:33.917" Title="Abelian categories, axioms AB5 and AB5* and incompatability" Tags="&lt;homework&gt;&lt;category-theory&gt;&lt;abelian-categories&gt;" CommentCount="4" />
  <row Id="25680" PostTypeId="2" ParentId="25674" CreationDate="2011-03-08T02:08:42.533" Score="4" ViewCount="" Body="&lt;p&gt;The general technique of isolating the terms which are congruent to $a \bmod n$ is known in the US high school competition circuit as the &quot;roots of unity filter&quot; and known to most others as the &lt;a href=&quot;http://en.wikipedia.org/wiki/Discrete_Fourier_transform&quot; rel=&quot;nofollow&quot;&gt;discrete Fourier transform&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The abstract idea is that the &lt;a href=&quot;http://en.wikipedia.org/wiki/Cyclic_group&quot; rel=&quot;nofollow&quot;&gt;cyclic group&lt;/a&gt; $\mathbb{Z}/n\mathbb{Z}$ acts on the space of, say, formal power series in one variable $z$ over $\mathbb{C}$ via $z \mapsto e^{\frac{2 \pi i}{n} } z$. This &lt;a href=&quot;http://en.wikipedia.org/wiki/Group_representation&quot; rel=&quot;nofollow&quot;&gt;group representation&lt;/a&gt; has $n$ isotypic components (subspaces where the group acts the same way) corresponding to the different residues $\bmod n$, and what we are interested in computing is the projections to these components, which can be done quite explicitly by averaging appropriately over the group action. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, to isolate the terms divisible by $3$ you use $\frac{F(z) + F(\omega z) + F(\omega^2 z)}{3}$ where $\omega = e^{ \frac{2 \pi i}{3} }$. The fact that all the terms cancel out nicely when you do this is a special case of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Character_theory#Orthogonality_relations&quot; rel=&quot;nofollow&quot;&gt;orthogonality relations&lt;/a&gt; in representation theory. &lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2011-03-08T02:08:42.533" />
  <row Id="25681" PostTypeId="2" ParentId="25671" CreationDate="2011-03-08T02:10:19.690" Score="5" ViewCount="" Body="&lt;p&gt;Completing rcollyer's answer&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\int \int_S f(x,y) dx dy = \int \int_S f(x(u,v),y(u,v)) Det[J(u,v)] du dv$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for 2-D polar coordinates $u=r$ and $v=\theta$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Find out $Det[J(r,\theta)] =r$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore $dA= r dr d\theta$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in general your area element is $dA=Det[J(r,\theta)] du dv$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Jacobians are used extensively in Statistical mechanics, but usually most physics problems come in flavors of one in three coordinate systems(cart, spherical, cylindrical) so we do not bother with the Jacobian when geometric intuition gets us the right answer for these three cases readily, which I have written in the comment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Jacobian is just a product of the scale factors when you change from one (orthogonal) coordinate system to another. Please note, that all this is a part of nonstandard analysis which exploits the notion of an infinitesimal. Start working from 2-D. Your area element would be a region bounded by the lines $x,x+\Delta x,y, y+\Delta y$ and your area element is $\Delta x \Delta y$. Finally take the limit in $\Delta x ,\Delta y\rightarrow 0$ and this is understood when you write $\int \int dx dy$. Repeat with polar coordinates. To locate a curve start with the constant coordinate $r$ which is a circle of radius $r$. Then draw a curve of constant $\theta$. This is a ray which goes from the origin to infinity making an angle $\theta$ with the positive x axis. Now draw another constant curve $r+\Delta r$ and $\theta + \Delta \theta$. Look at the intersection of these two. When you take the limits, this can be approximated as a rectangle, so, area is $\Delta r (r \Delta \theta)$. The factor $r$ which converts the infinitesimal change in $\theta$ ti the corresponding displacement $rd\theta$ is called the scale factor $h_\theta$. For an orthogonal system the jacobian is the product of scale factors. For spherical coordinate system it is easy to show  that:&#xA;$h_r=1, h_\theta=r, h_\phi=r \sin \theta $ so $J(r,\theta,\phi)=r^2\sin\theta$&#xA;For cylindrical $h_\rho=1,h_\phi=\rho,h_z=1$ etc.. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now regarding your question about the integral, consider a transformation of a cartesian coordinate $x$ to some new coordinate $\xi$, you know that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\int f(x) dx = \int f(x(\xi)) \frac{dx}{d\xi}d\xi$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The jacobian is $J(\xi)=\frac{dx}{d\xi}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For surface integrals, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\int \int_S f(x_1,x_2) dx_1 dx_2 = \int \int_S f(x_1(\xi_1,\xi_2),x_2(\xi_1,\xi_2)) Det[J(\xi_1,\xi_2)] d\xi_1 d\xi_2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For volume integrals&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\int \int \int f(x_1,x_2,x_3) dx_1 dx_2 dx_3 = \int \int \int f(x_1(\xi_1,\xi_2,\xi_3),x_2(\xi_1,\xi_2,\xi_3),x_3(\xi_1,\xi_2,\xi_3)) |J(\xi_1,\xi_2,\xi_3)| d\xi_1 d\xi_2 d\xi_3$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT2&lt;/strong&gt;&#xA;As I mentioned in the earlier comment, when you have to find the surface element in a 3D coordinate system, you have to first know what your surface is. This would depend on what actually your finction $f(x,y)$ is. Mostly in physical problems, the surface is traced out by keeping one coordinate system fixed and varying the other two (examples- sphere's surface is fixed radii and varying azimuthal and polar angles, a plane's surface is a fixed polar angle and varying azimuthal angle and radial coordinate, etc...) &lt;/p&gt;&#xA;" LastActivityDate="2011-03-08T21:54:18.590" CommentCount="2" />
  <row Id="25682" PostTypeId="1" AcceptedAnswerId="25737" CreationDate="2011-03-08T02:18:03.787" Score="2" ViewCount="84" Body="&lt;p&gt;I've been trying to understand de la Vallee Poussin's &quot;Demonstration Simplifiee du Theorem de Dirichlet sur la Progression Arithmetique&quot; and I've got stuck at the following step on pg 18 where Poussin takes the logarithmic derivative of:&#xA;$$\Sigma_{n}\frac{\chi(n)}{n^{s}} = \prod_{q}(1-\frac{\chi(q)}{q^{s}})^{-1}$$ &#xA;in order to obtain:&#xA;$$-D\log\sum_{n}\frac{\chi(n)}{n^{s}} = \Sigma_{q}\frac{\chi(q)\log(q)}{q^{s}-\chi(q)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Specifically, when I try to work through the calculation, I don't see where the -1 on the left hand side of the equation comes from.  When I tried to take logs and then differentiate the first equation, I ended up with the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$D\log\sum_{n}\frac{\chi(n)}{n^{s}} = \Sigma_{q}\frac{\chi(q)\log(q)}{q^{s}-\chi(q)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could someone help me understand where I'm going wrong please?&lt;/p&gt;&#xA;" OwnerUserId="1179" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2011-03-08T06:16:05.220" LastActivityDate="2011-03-08T07:23:23.180" Title="Question on de la Vallee Poussin's simplified proof of Dirichlet's theorem on primes in arithmetic progressions" Tags="&lt;analytic-number-theory&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="25683" PostTypeId="1" CreationDate="2011-03-08T02:23:10.220" Score="0" ViewCount="55" Body="&lt;p&gt;What property of a cryptographic function is described below?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;B = F(A)&#xA;A = F(B)&lt;/p&gt;&#xA;" OwnerUserId="7939" LastActivityDate="2011-03-08T03:23:01.860" Title="Property of a cryptographic function" Tags="&lt;cryptography&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="25684" PostTypeId="1" AcceptedAnswerId="25690" CreationDate="2011-03-08T02:23:20.850" Score="2" ViewCount="148" Body="&lt;p&gt;Can someone help me understand the idea behind constructing a formula for the following:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;For $n\in\mathbb{N}$, $n\geq 2$, find and prove a formula for:&#xA;  $$\prod_{i=2}^n \left(1 - \frac{1}{i^2}\right).$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Please, Please, I need to know HOW to do this not just how to solve this question. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Confused about:&lt;br /&gt;&#xA;- Why does I start at 2 and not 1 like regular series, is it because it makes the solution zero?&lt;br /&gt;&#xA;- Why does $n \geq 2$ have to be true?&lt;br /&gt;&#xA;- Do you start with $i = 2$ for the first element?&lt;br /&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks any help is appreciated!&lt;/p&gt;&#xA;" OwnerUserId="5505" LastEditorUserId="1543" LastEditorDisplayName="" LastEditDate="2011-05-03T10:03:51.107" LastActivityDate="2011-05-03T10:03:51.107" Title="Formula for Geometric Progression" Tags="&lt;sequences-and-series&gt;&lt;products&gt;" AnswerCount="2" CommentCount="8" />
  <row Id="25685" PostTypeId="1" AcceptedAnswerId="25686" CreationDate="2011-03-08T02:27:23.333" Score="3" ViewCount="72" Body="&lt;p&gt;Is this a valid proof for (2n+1,3n+1)=1?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\exists \ d \in \mathbb{Z}$&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$d | 2n+1$ means $2n+1 \equiv 0$ (mod $d$)&lt;/li&gt;&#xA;&lt;li&gt;$d | 3n+1$ means $3n+1 \equiv 0$ (mod $d$)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Multiply 1. by 3 and 2. by 2:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$6n+3 \equiv 0$ (mod $d$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$6n+2 \equiv 0$ (mod $d$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Subtract the second from the first to get:&#xA;$ 1 \equiv 0$ (mod $d$), which means d|1, which means d=1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="7934" LastActivityDate="2011-03-08T02:43:35.623" Title="Valid proof for (2n+1,3n+1)=1?" Tags="&lt;homework&gt;&lt;elementary-number-theory&gt;" AnswerCount="1" />
  <row Id="25686" PostTypeId="2" ParentId="25685" CreationDate="2011-03-08T02:33:35.857" Score="4" ViewCount="" Body="&lt;p&gt;Yes, nice job! That's a very nice application of the method that I described in your &lt;a href=&quot;http://math.stackexchange.com/questions/25648/show-that-if-a-equiv-b-mod-n-gcda-n-gcdb-n/25658#25658&quot;&gt;prior question&lt;/a&gt;. Notice again how simple the problem becomes once it is reduced to equational (congruence) form.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2011-03-08T02:43:35.623" LastActivityDate="2011-03-08T02:43:35.623" CommentCount="2" />
  <row Id="25687" PostTypeId="2" ParentId="25673" CreationDate="2011-03-08T02:34:11.733" Score="1" ViewCount="" Body="&lt;p&gt;Presumably, it means that $f$ takes distinct functions in $L^1(\mathbb{R})$ to distinct functions (you say $f$ is &quot;some function of $g$&quot;, but you don't say who $g$ is).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even though $L^1(\mathbb{R})$ is all sorts of things (a measure space, an algebra, a ring of functions, a vector space over $\mathbb{R}$, etc), it is still a set of &lt;em&gt;things&lt;/em&gt;. And, being a set of things, you can talk about functions whose domain is $L^1(\mathbb{R})$, and you can talk about such functions being one-to-one, surjective, bijective, etc. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Same is true for $L^n$, which is likewise a set.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance: take a finite interval $[a,b]$, and let $\chi_{[a,b]}$ be the characteristic function of $[a,b]$; that is,&#xA;$$\chi_{[a,b]}(x) = \left\{\begin{array}{ll}&#xA;0 &amp;amp; \mbox{if $x\notin [a,b]$,}\\&#xA;1 &amp;amp; \mbox{if $x\in [a,b]$.}&#xA;\end{array}\right.$$&#xA;Now, I can define a function $\mathcal{F}\colon L^1(\mathbb{R})\to L^1(\mathbb{R})$ by the following rule: if $g\in L^1(\mathbb{R})$, then $\mathcal{F}(g) = g\chi_{[a,b]}$ (product of functions).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $g\in L^1(\mathbb{R})$, then so is $g\chi_{[a,b]}$, so $\mathcal{F}$ is indeed a function from $L^1(\mathbb{R})$ to $L^1(\mathbb{R})$. (In fact, it is a linear transformation, which you should check for yourself).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, $\mathcal{F}$ is not injective on $L^1(\mathbb{R})$: if you take an interval $[c,d]$ with $[c,d]\cap[a,b]=\emptyset$, then $\mathcal{F}(\chi_{[c,d]}) = \mathcal{F}(\mathbf{0})$ (where $\mathbf{0}$ is the constant function with value $0$). Since $\chi_{[c,d]}\neq\mathbf{0}$, then $\mathcal{F}$ is not 1-to-1 on $L^1(\mathbb{R})$. &lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2011-03-08T02:34:11.733" />
  <row Id="25688" PostTypeId="1" AcceptedAnswerId="25692" CreationDate="2011-03-08T02:49:50.107" Score="6" ViewCount="161" Body="&lt;p&gt;I'd like to solve this one similarly to my previous question: &lt;a href=&quot;http://math.stackexchange.com/questions/25685/valid-proof-for-2n1-3n1-1&quot;&gt;Valid proof for (2n+1,3n+1)=1?&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I did find a somewhat related post that uses a different method: &lt;a href=&quot;http://math.stackexchange.com/questions/6382/let-n-be-a-positive-integer-n-denotes-the-factorial-of-n-let-d-gcdn-1&quot;&gt;Let n be a positive integer, n! denotes the factorial of n. Let d = gcd(n! + 1, (n + 1)! + 1). Show that d divides n.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So how would I go about this? I can write the above like:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\exists \ d \ \in \mathbb{Z}$&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$n!+1 \equiv 0$ (mod $d$)&lt;/li&gt;&#xA;&lt;li&gt;$(n+1)!+1 \equiv 0$ (mod $d$)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Not sure what to do next. Any hints?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="7934" LastEditorUserId="7934" LastEditorDisplayName="" LastEditDate="2011-03-08T02:58:44.993" LastActivityDate="2011-03-08T16:25:53.473" Title="Proof that (n!+1,(n+1)!+1)=1" Tags="&lt;homework&gt;&lt;elementary-number-theory&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="25689" PostTypeId="2" ParentId="25684" CreationDate="2011-03-08T02:56:30.040" Score="2" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT:&lt;/strong&gt; $$\prod_{i=2}^n \left( 1 - \frac{1}{i^2}\right) = \prod_{i=2}^n \left(\frac{i^2 - 1}{i^2} \right)=\prod_{i=2}^n \frac{(i-1)(i+1)}{i^2} = \frac{\displaystyle\prod_{i=2}^n (i-1) \prod_{i=2}^n (i+1)}{\left(\displaystyle\prod_{i=2}^n i\right)^2}.$$&lt;/p&gt;&#xA;" OwnerUserId="6491" LastActivityDate="2011-03-08T02:56:30.040" CommentCount="3" />
  <row Id="25690" PostTypeId="2" ParentId="25684" CreationDate="2011-03-08T02:56:58.100" Score="8" ViewCount="" Body="&lt;p&gt;You can start a summation or a product &lt;em&gt;anywhere&lt;/em&gt;, not just at $1$. There is no law mandating that they must start at $1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case, if you were to allow $i=1$, then the first factor of the product would be&#xA;$$\left(1 - \frac{1}{1^2}\right) = 0$$&#xA;and so the entire product would be zero, which would make everything rather silly. So you start at $2$ instead (you could instead &lt;em&gt;reindex&lt;/em&gt; and consider&#xA;$$\prod_{j=1}^{n-1}\left(1 - \frac{1}{(j+1)^2}\right)$$&#xA;but that just makes the expression look more complicated; better to change the indices instead).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why do you have $n\geq 2$? Because if $n\lt 2$, then the product has no factors (since no value of $i$ can be at least $2$ and also less than or equal to some $n\lt 2$). The empty product is equal to $1$ by definition, but most people have trouble with that assertion, so why introduce it? Better to stick to $n\geq 2$ so that it is clear there are always factors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yes: you start by plugging in $i=2$, then $i=3$, then $i=4$, and so on until you get to $i=n$. For example, with $n=4$, you have&#xA;\begin{align*}&#xA;\prod_{i=2}^4\left(1 - \frac{1}{i^2}\right) &amp;amp;= \left(1 - \frac{1}{2^2}\right)\left(1 - \frac{1}{3^2}\right)\left(1 - \frac{1}{4^2}\right)\\&#xA;&amp;amp;= \left(1 - \frac{1}{4}\right)\left(1 - \frac{1}{9}\right)\left(1 - \frac{1}{16}\right)\\&#xA;&amp;amp;= \left(\frac{3}{4}\right)\left(\frac{8}{9}\right)\left(\frac{15}{16}\right)\\&#xA;&amp;amp;= \frac{360}{576} = \frac{5}{8}.&#xA;\end{align*}&#xA;As to finding a formula, well, you might try a few values and see if you spot a pattern; then you can try using induction to prove it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Added.&lt;/em&gt; Okay, you say you are stuck, you don't see a pattern. This is just &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the factors you are multiplying: $1 - \frac{1}{i^2}$. Writing them as a single fraction, you have&#xA;$$1 - \frac{1}{i^2} = \frac{i^2 - 1}{i^2} = \frac{(i-1)(i+1)}{i^2}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, what are you doing at each step? Let's write them out explicitly:&#xA;$$\left(\frac{(1)(3)}{2^2}\right)\left(\frac{(2)(4)}{3^2}\right)\left(\frac{(3)(5)}{4^2}\right)\cdots\left(\frac{(n-1)(n+2)}{n^2}\right).$$&#xA;The denominator is easy: you can just rearrange it, throw in a couple of extra factors of $1$ (who cares if you multiply by $1$? Doesn't do anything) and write:&#xA;\begin{align*}&#xA;&amp;amp;(2)(2)(3)(3)(4)(4)\cdots(n)(n)\\&#xA;&amp;amp;=(2)(3)(4)\cdots(n)(2)(3)(4)\cdots(n)= \Bigl((1)(2)(3)(4)\cdots(n)\Bigr)\Bigl((1)(2)(3)(4)\cdots(n)\Bigr).&#xA;\end{align*}&#xA;Now, what do you call the product of all the positive integers from $1$ through $n$? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Great! We now have a formula for the denominator.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What about the numerator? Again, reorder them putting first all the &lt;em&gt;first&lt;/em&gt; factors, and next all the &lt;em&gt;second&lt;/em&gt; factors; we have:&#xA;\begin{align*}&#xA;&amp;amp;(1)(3)(2)(4)(3)(5)(4)(6)\cdots (n-1)(n+1)\\&#xA;&amp;amp;= \Bigl( (1)(2)(3)(4)\cdots(n-1)\Bigr)\Bigl((3)(4)(5)\cdots(n+1)\Bigr).&#xA;\end{align*}&#xA;Hmmm... Looks like another set of those &quot;products of all positive integers from $1$ up to &lt;code&gt;something&lt;/code&gt;.&quot; Now, granted, the second factor isn't that, but just throw in a $(1)(2)$, and then divide by $2$ to keep it honest:&#xA;$$\Bigl((1)(2)(3)(4)\cdots(n-1)\Bigr)\Bigl((1)(2)(3)(4)(5)\cdots(n+1)\Bigr)\left(\frac{1}{2}\right).$$&#xA;Now, express it using the same kind of symbol as the denominator.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, step back and look at what you get. Is there some simplification/cancellation that you can make to get something simple? What?&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-08T03:55:32.927" LastActivityDate="2011-03-08T03:55:32.927" CommentCount="9" />
  <row Id="25691" PostTypeId="1" AcceptedAnswerId="25695" CreationDate="2011-03-08T02:58:51.847" Score="2" ViewCount="147" Body="&lt;p&gt;I have the following parametric equations:&#xA;$$x = \sin\theta,\qquad y = \cos\theta,\qquad    0 \leq \theta \leq \pi.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The corresponding Cartesian equation is $x^2 + y^2 = \sin^2\theta + \cos^2\theta = 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So it would be $x^2 + y^2 = 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How would I develop the curve for the equation? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Should I plot points? I know that this is a circle centered at 1, but with the limits on $\theta$, I'm not sure how to determine the final graph. &lt;/p&gt;&#xA;" OwnerUserId="6331" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-08T03:03:04.147" LastActivityDate="2011-03-08T05:38:17.650" Title="How to sketch the curve of parametric equations" Tags="&lt;calculus&gt;" AnswerCount="1" />
  <row Id="25692" PostTypeId="2" ParentId="25688" CreationDate="2011-03-08T03:00:21.250" Score="4" ViewCount="" Body="&lt;p&gt;Here's the purely equational proof. Simply put $\rm\ k = (n-1)!\ $ in &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;THEOREM&lt;/strong&gt; $\rm\quad ((n+1)\ n\ k+1,\ n\ k+1)\ =\ 1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof&lt;/strong&gt; $\ \ $ Working modulo the gcd $\rm\: := d\:$ we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(1)\rm\quad\quad (n+1)\ n\ k\ \equiv\: -1\quad\quad$ by $\rm\ d\ |\ (n+1)\ n\ k+1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(2)\rm\quad\quad\phantom{(n+1)\ } n\ k\ \equiv\: -1\quad\quad$ by $\rm\ d\ |\ n\ k+1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(3)\rm\quad\quad\phantom{(n+1)\ n\ } n\ \equiv\ \ 0\quad\quad\ $ by substituting $\:(2)\:$ in $\:(1)\:$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(4)\rm\quad\quad\phantom{(n+1)\ n\ } 0\ \equiv\: -1\quad\quad$ by substituting $\:(3)\:$ in $(2)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we conclude $\rm\: 0\ \equiv\ 1\ $ i.e. $\rm\ d\ |\ 1\quad$ &lt;strong&gt;QED&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unwinding the linear relations used in the above proof (or, equivalently, using the extended Euclidean algorithm) yields the Bezout relation that I gave in the question that you linked to, viz.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \rm 1\ =\ (n-1)!\ ((n+1)!+1)\ +\ (1-(n+1)!/n)\ (n!+1)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice how what seems like magic viewed in terms of divisibility relations is reduced to a purely &lt;em&gt;mechanical&lt;/em&gt; elimination process in equational form. In higher number theory you'll learn more precisely how linear algebra methods such as Gaussian elimination extend from fields to certain rings, e.g. google Hermite and Smith normal forms.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2011-03-08T07:20:14.407" LastActivityDate="2011-03-08T07:20:14.407" CommentCount="3" />
  <row Id="25695" PostTypeId="2" ParentId="25691" CreationDate="2011-03-08T03:04:49.230" Score="8" ViewCount="" Body="&lt;p&gt;So now you know that every point in the parametric curve lies on the unit circle, since every point satisfies the equation $x^2+y^2=1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some questions to ask yourself:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Where do you start? That is, when $\theta=0$, &lt;em&gt;where&lt;/em&gt; on the unit circle are you?&lt;/li&gt;&#xA;&lt;li&gt;As $\theta$ increases from $0$ towards $\pi$, what happens to the point $(x,y) = (\sin\theta,\cos\theta)$?&lt;/li&gt;&#xA;&lt;li&gt;Once $\theta$ reaches $\pi$, where are you?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Now you know &lt;em&gt;which&lt;/em&gt; parts of the unit circle are given by the parametric curve.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added.&lt;/strong&gt; Like many students early on, you seem to want to plot a few points and then interpolate; this is not a good idea, because it relies on you just happening to hit the correct points to get an accurate picture of what is going on. To give you an example, if you were trying to plot a graph for the function $y = \sin(\pi x)$, and tried a few points, say, $x=0$, $x=1$, $x=2$, $x=3$, etc., you might think that your function is the constant function $0$, because your selection of points happens to miss all the important stuff that is going on with $y$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You &lt;em&gt;don't&lt;/em&gt; want to do that. Instead, you want to think about what those functions are doing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of the ways in which I find most fruitful to think about parametric equations is to think of the parameter as giving &lt;em&gt;time&lt;/em&gt;, and the equations as describing the movement of a point; imagine an animation with a glowing point moving along the plane, and leaving a &quot;trail&quot; of light behind it. That trail is the parametric curve, the point is the position at the &quot;present $t$&quot;. You want to think about what that point is doing as your parameter ranges from its initial value to its final value (that is, as the animation goes from beginning to end).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, start with $\theta=0$, the first frame of your animation. Your glowing point will be at $x(0) = \sin(0) = 0$, and $y(0)=\cos(0) = 1$. So you start at the point $(0,1)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, press the &lt;code&gt;PLAY&lt;/code&gt; button. What happens as $\theta$ starts advancing from $0$ towards $\pi$? The $x$ coordinate will follow the graph of $y(\theta)=\sin(\theta)$, so first it will rise from $0$ to $1$ (at $\theta=\frac{\pi}{2}$), and then drop again from $1$ to $0$ (at $\theta=\pi$). It will do so without jumps or breaks. So if you were looking only at the &quot;shadow&quot; of our glowing point on the $x$-axis, it starts at $x=0$, then moves towards the right in a smooth way (no jerks, no jumps, no skips) until it hits $1$ at the midpoint of the movie, and then moves back towards $0$ until it returns to $0$ at the end of the movie.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What about $y$? It starts at $1$; it will behave like the graph of $\cos\theta$. As you press &lt;code&gt;PLAY&lt;/code&gt;, it will start at $1$, and then drop towards $0$, again in a smooth way, without jumps, jerks, or skips, until it hits $0$ halfway through the movie (at $\theta=\frac{\pi}{2}$). Then, it will &lt;em&gt;keep going&lt;/em&gt; in the same direction, from $0$ down to $-1$, and reach $-1$ at the end of the &quot;movie&quot; (when $\theta=\pi$). So if you look at the &quot;shadow&quot; of the glowing point on the $y$-axis, it will start at $1$, then drop down to $0$, and keep going down to $-1$, all in a generally smooth manner, with no jumps, jerks, hesitations, backtracks, etc. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now put those two motions together: you start at $(0,1)$, the top of the unit circle. Then as $\theta$ moves from $0$ to $\pi/2$, the glowing point starts moving both right and down, always along the unit circle, until $\theta=\pi/2$, when it is at point &lt;code&gt;&amp;lt;fill in the blank&amp;gt;&lt;/code&gt;. Press &lt;code&gt;PAUSE&lt;/code&gt; on the video, and take stock. Where are we? What part of the unit circle have we traced? How many times? Any backtrack? Any long period of time spent without moving? Was the motion generally &quot;fluid&quot;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Okay, ready to continue? Press &lt;code&gt;PLAY&lt;/code&gt; again, and our $\theta$ starts increasing from $\pi/2$ towards $\pi$. That glowing point representing $(x(\theta),y(\theta))$ moves, but now down and to the left, always along the unit circle, until finally when $\theta=\pi$, the &quot;end of the movie&quot;, it reaches its final destination at point &lt;code&gt;&amp;lt;fill in the other blank&amp;gt;&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All along, the movement was without jumps, hesitations, or backtracks, because the functions $x=\sin\theta$ and $y=\cos\theta$ have those motions: no jumps, no skips, no jerks, no hesitations, no backtracks, just a smooth motion (imagine your hand drawing their graphs). That glowing point has now traced &lt;em&gt;part&lt;/em&gt; of the unit circle, exactly once, without backtracking. Which part?&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-08T05:38:17.650" LastActivityDate="2011-03-08T05:38:17.650" CommentCount="12" />
  <row Id="25696" PostTypeId="1" CreationDate="2011-03-08T03:11:01.970" Score="1" ViewCount="83" Body="&lt;p&gt;find the value of R so that the line through (3, 13) and (R, -5) has a slope of 8&lt;/p&gt;&#xA;" OwnerUserId="7940" LastEditorUserId="1016" LastEditorDisplayName="" LastEditDate="2011-03-08T04:01:58.100" LastActivityDate="2011-03-08T04:01:58.100" Title="Help Me Figure Out What R Is." Tags="&lt;homework&gt;&lt;algebra-precalculus&gt;" AnswerCount="2" />
  <row Id="25697" PostTypeId="2" ParentId="25688" CreationDate="2011-03-08T03:11:19.047" Score="1" ViewCount="" Body="&lt;p&gt;Suppose $m$ is a positive integer dividing both $n! + 1$ and $(n+1)! + 1$. The goal is to show that $m = 1$. Note that $m$ also divides $(n+1)*(n! + 1) - ((n+1)! + 1)$, which is equal to $(n+1)! + (n+1) - (n+1)! - 1 = n$. Since $m$ divides $n$, it also divides $n!$. Since it divides $n! + 1$ as well, it divides $n! + 1 - n!$ or just $1$. Hence $m = 1$ as needed.&lt;/p&gt;&#xA;" OwnerUserId="3035" LastEditorUserId="3035" LastEditorDisplayName="" LastEditDate="2011-03-08T03:35:14.843" LastActivityDate="2011-03-08T03:35:14.843" CommentCount="3" />
  <row Id="25698" PostTypeId="2" ParentId="25696" CreationDate="2011-03-08T03:12:12.780" Score="3" ViewCount="" Body="&lt;p&gt;We have $$m = \frac{-5-13}{R-3} = 8$$ and solve for $R$. &lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2011-03-08T03:12:12.780" />
  <row Id="25699" PostTypeId="2" ParentId="25696" CreationDate="2011-03-08T03:15:10.560" Score="4" ViewCount="" Body="&lt;p&gt;You know that the formula for slope is given by: $$ m = \frac{y_{2}-y_{1}}{x_{2}-x_{1}}$$ and given that the value for $m=8$. Use this to find the value of $R$.&lt;/p&gt;&#xA;" LastActivityDate="2011-03-08T03:15:10.560" />
  <row Id="25700" PostTypeId="2" ParentId="25683" CreationDate="2011-03-08T03:23:01.860" Score="2" ViewCount="" Body="&lt;p&gt;$F$ is an involution. You don't want this property for a block cipher since the function doesn't look &quot;random&quot;. On the other hand, a stream cipher works exactly that way, but then it's a bit misleading to think of it as a function $F$: the function is missing an important parameter, namely time.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2011-03-08T03:23:01.860" CommentCount="2" />
  <row Id="25701" PostTypeId="1" AcceptedAnswerId="25706" CreationDate="2011-03-08T03:28:16.300" Score="4" ViewCount="395" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Show that $13$ divides $2^{70} + 3^{70}$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My main problem here is trying to figure out how to separate the two elements in the sum, and then use Fermat's little theorem. So how can I separate the two?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="7934" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-08T03:38:09.937" LastActivityDate="2011-05-08T01:42:46.667" Title="Show that 13 divides $2^{70}+3^{70}$." Tags="&lt;homework&gt;&lt;elementary-number-theory&gt;" AnswerCount="6" CommentCount="3" />
  <row Id="25704" PostTypeId="2" ParentId="25221" CreationDate="2011-03-08T03:34:58.460" Score="0" ViewCount="" Body="&lt;p&gt;I know it is not an answer and maybe discouraging, but from my modern geometry class I remember the Hilbert axioms are seriously limited, ambiguous, with some conflating definitions. Maybe you should turn to another book. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For your approach, I feel what you meant by &quot;further and further 'out'..&quot; etc is essentially magnifying the line segment. Without the Archimedes' axiom you are not guaranteed you can pass over the circle. I know it is kind of silly but you need it to work. Similarly you may run into other problems with similar approaches without appropriate axioms. For the fifth postulate maybe you can try the Poincare model or Klein model. I am quite sure without it you cannot find $D$. &lt;/p&gt;&#xA;" OwnerUserId="7887" LastActivityDate="2011-03-08T03:34:58.460" />
  <row Id="25705" PostTypeId="2" ParentId="25701" CreationDate="2011-03-08T03:39:12.077" Score="4" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\rm\: $ Little Fermat shows $\rm\ mod\ 13:\ \ 2^{70} + 3^{70}\ \equiv\ 2^{-2} + 3^{-2}\ \equiv\ 1/4+1/9\ \equiv 13/36\ \equiv\ 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;NOTE&lt;/strong&gt; $\ $ For a slight generalization &lt;a href=&quot;http://math.stackexchange.com/questions/37556/can-this-number-theory-mcq-be-solved-in-4-minutes/37654#37654&quot;&gt;see here.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2011-05-07T18:44:35.110" LastActivityDate="2011-05-07T18:44:35.110" />
  <row Id="25706" PostTypeId="2" ParentId="25701" CreationDate="2011-03-08T03:39:25.007" Score="10" ViewCount="" Body="&lt;p&gt;Compute $2^{70}$ and $3^{70}$ modulo $13$ &lt;em&gt;separately&lt;/em&gt; (e.g., using Fermat's Little Theorem). If $2^{70}\equiv a\pmod{13}$ and $3^{70}\equiv b\pmod{13}$, then what is $2^{70}+3^{70}$ congruent to modulo 13?&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2011-03-08T03:39:25.007" CommentCount="1" />
  <row Id="25708" PostTypeId="2" ParentId="25701" CreationDate="2011-03-08T03:41:52.170" Score="1" ViewCount="" Body="&lt;p&gt;So by FlT, you know that $2^{12}\equiv 1\pmod{13}$ and $3^{12}\equiv 1\pmod{13}$. So &#xA;$$&#xD;&#xA;2^{70}+3^{70}\equiv (2^{12})^5\cdot 2^{10}+(3^{12})^5\cdot 3^{10}\equiv 2^{10}+3^{10}\pmod{13}.&#xD;&#xA;$$&#xA;However, again by FlT, &#xA;$$&#xD;&#xA;2^{12}\equiv 2^{10}\cdot 2^{2}\equiv 1\pmod{13}\implies 2^{10}\equiv 2^{-2}\pmod{13}.&#xD;&#xA;$$&#xA;That is to say, $2^{10}$ is congruent to the multiplicative inverse of $4$ modulo $13$. You can do the same for $3^{10}$, except you will need to find the multiplicative inverse to $9$ modulo $13$. You'll see that the sum of these inverses is $0$ modulo $13$, to get your answer.&lt;/p&gt;&#xA;" OwnerUserId="1583" LastActivityDate="2011-03-08T03:41:52.170" />
  <row Id="25709" PostTypeId="2" ParentId="25701" CreationDate="2011-03-08T03:42:26.677" Score="5" ViewCount="" Body="&lt;p&gt;$2^{12} \equiv 1 \pmod{13}$ and $3^{12} \equiv 1 \pmod{13}$ by Fermat's Little Theorem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, $2^{72} \equiv 1 \pmod{13}$ and $3^{72} \equiv 1 \pmod{13}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$2^{72} \equiv 1 \pmod{13} \Rightarrow 2^{72} \equiv 40 \pmod{13} \Rightarrow 2^{70} \equiv 10 \pmod{13}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$3^{72} \equiv 1 \pmod{13} \Rightarrow 3^{72} \equiv 27 \pmod{13} \Rightarrow 3^{70} \equiv 3 \pmod{13}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, you get the result.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2011-03-08T04:16:47.460" LastActivityDate="2011-03-08T04:16:47.460" CommentCount="6" />
  <row Id="25710" PostTypeId="1" AcceptedAnswerId="25714" CreationDate="2011-03-08T03:42:40.760" Score="5" ViewCount="76" Body="&lt;p&gt;In S. Wagon's &quot;The Banach-Tarski Paradox,&quot; amenable groups are defined on p. 12 as follows:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;[amenable] groups bear a left-invariant, finitely additive measure of total measure one that is defined on all subsets.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;He defines $SO_2$ to be the group of rotations of the unit circle, which he has used to show that $S_1$, the unit circle, is $SO_2$-paradoxical (as an analogue to the usual non-measurable set defined in the interval $[0,1)$ ).  I am taking measure theory this term, but am not sure how to assign a measure to subsets of $SO_2$.  Thus, I am not really sure where to start in showing whether or not $SO_2$ is an amenable group.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I look at the Wikipedia entry about amenable groups, I'm unable to make much more sense of the definition in the context of the material.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is $SO_2$, the group of rotations of the unit circle, an amenable group?  If not, why (so that I may build an intuition for these objects)?&lt;/p&gt;&#xA;" OwnerUserId="7051" LastActivityDate="2011-03-08T04:03:54.527" Title="Is $SO_2$ an amenable group?" Tags="&lt;group-theory&gt;&lt;measure-theory&gt;&lt;paradox&gt;" AnswerCount="2" CommentCount="0" />
  <row Id="25711" PostTypeId="2" ParentId="25701" CreationDate="2011-03-08T03:45:10.677" Score="3" ViewCount="" Body="&lt;p&gt;The &quot;quick&quot; way to do this is as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$2^{6}=64=-1$(mod 13), $3^{3}=1$ (mod 13)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence you have $2^{70}=(2^{6})^{11}*2^{4}=(-1)*(3)=-3$ (mod 13)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And $3^{70}=(3^{3})^{13}*3=3$ (mod 13)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore the result is $-3+3=0$ (mod 13). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fermat's little theorem implies $a^{p-1}\cong 1$ (mod p), but this may be more slick, I am not sure. Mind that Fermat's little theorem is not optimal in many cases. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &quot;ultra quick&quot; way may be $2^{70}+3^{70}=3^{70}*(1+(2/3)^{70})$. Now notice $2/3=2*9=5$. And $5^2=-1$, hence $(1+(5^{2})^{25})=1+-1=0$. Therefore $2^{70}+3^{70}=0$. &lt;/p&gt;&#xA;" OwnerUserId="7887" LastEditorUserId="7887" LastEditorDisplayName="" LastEditDate="2011-03-12T01:38:28.290" LastActivityDate="2011-03-12T01:38:28.290" CommentCount="2" />
  <row Id="25712" PostTypeId="1" CreationDate="2011-03-08T03:45:51.760" Score="3" ViewCount="111" Body="&lt;p&gt;Working my way through a combinatorics text and I'm hung up on a couple of questions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1.) Let $p=p_1 p_2\cdots p_n$ be a permutation.  An inversion of $p$ is a pair of entries $(p_i,p_j)$ so that $i&amp;lt;j$ but $p_i&amp;gt;p_j$.  Let us a call a permutation even (resp. odd) if it has an even (resp. odd) number of inversion.  Prove that the permuation consisting of the one cycle $(a_1a_2\cdots a_k)$ is even if $k$ is odd, and is odd if $k$ is even.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2.)  Let $I(n,k)$ be the number of $n$-permutations that have $k$ inversions.  Prove that $I(n,k)=I(n,\binom{n}{2}-k)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3.)  Find an explicit formula for $I(n,3)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These are from a combinatorics text, but I vaguely remember this topic popping up in undergrad abstract algebra and possibly in an algorithm design course as well.&lt;/p&gt;&#xA;" OwnerUserId="7463" LastActivityDate="2011-03-08T14:36:22.613" Title="Permutation Inversion Questions (3)" Tags="&lt;combinatorics&gt;&lt;abstract-algebra&gt;&lt;permutations&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="25713" PostTypeId="2" ParentId="25701" CreationDate="2011-03-08T03:47:55.627" Score="17" ViewCount="" Body="&lt;p&gt;Okay, I'm on a little different wavelength so I'll turn my comment into an answer.. if $n$ is odd the polynomial $x + y$ divides $x^n + y^n$. So letting $x = 2^2, y = 3^2,$ and $n = 35$ you get that $13 = 2^2 + 3^2$ divides $2^{70} + 3^{70}$. &lt;/p&gt;&#xA;" OwnerUserId="3035" LastActivityDate="2011-03-08T03:47:55.627" CommentCount="2" />
  <row Id="25714" PostTypeId="2" ParentId="25710" CreationDate="2011-03-08T04:00:49.083" Score="6" ViewCount="" Body="&lt;p&gt;For groups with a topology, like $SO_2$, one usually uses a slightly different definition. You will find it in Wikipedia.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, it is easy to see that a compact group is amenable, and $SO_2$ is compact.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you insist on ignoring the topology of $SO_2$, though, you are still left with an abelian group. And all discrete abelian groups are amenable---but this is harder to see.&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2011-03-08T04:00:49.083" CommentCount="0" />
  <row Id="25715" PostTypeId="2" ParentId="25710" CreationDate="2011-03-08T04:03:54.527" Score="3" ViewCount="" Body="&lt;p&gt;According to Terry Tao's notes &lt;a href=&quot;http://terrytao.wordpress.com/2009/01/08/245b-notes-2-amenability-the-ping-pong-lemma-and-the-banach-tarski-paradox-optional/&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; and &lt;a href=&quot;http://terrytao.wordpress.com/2009/04/14/some-notes-on-amenability/&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; it is amenable.&lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2011-03-08T04:03:54.527" CommentCount="1" />
  <row Id="25716" PostTypeId="1" AcceptedAnswerId="25721" CreationDate="2011-03-08T04:22:26.200" Score="4" ViewCount="164" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to show that $a^n | b^n$ implies $a | b$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I thought I could convert it to congruences and work backwards, but as far as I remember, $a \equiv b \pmod{m}$ implies $a^n \equiv b^n \pmod{m}$, not the opposite, unless m is prime. Is that right?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I am not sure how to approach this one. Any ideas?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="7934" LastActivityDate="2011-03-09T16:48:50.107" Title="Show that a^n | b^n implies a | b" Tags="&lt;homework&gt;&lt;elementary-number-theory&gt;" AnswerCount="3" CommentCount="5" />
  <row Id="25717" PostTypeId="2" ParentId="25716" CreationDate="2011-03-08T04:28:13.730" Score="2" ViewCount="" Body="&lt;p&gt;Any prime $p$ that divides $a$ $k$ times, is such that $p^n$ divides $a^n$ $nk$ times and thus $b^n$ as well. But if a prime divides $b^n$ $nk$ times, it divides $b$ $k$ times as well. So every prime divisor of $a$ is one of $b$, with at least the same multiplicity. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or use that $(b-a)$ divides $b^n - a^n$ for all $n \ge 1$.  &lt;/p&gt;&#xA;" OwnerUserId="4280" LastActivityDate="2011-03-08T04:28:13.730" CommentCount="2" />
  <row Id="25719" PostTypeId="2" ParentId="25716" CreationDate="2011-03-08T04:28:50.663" Score="4" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Hint:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Write $b^n = q_1^{n \alpha_1} \dots q_k^{n \alpha_k}$, where $\alpha_i &amp;gt; 0$.  If $a^n | b^n$, then $a^n = q_1^{n \beta_1} \dots q_k^{n \beta_k}$, where $0 \leq \beta_k \leq \alpha_k$.&lt;/p&gt;&#xA;" OwnerUserId="6491" LastActivityDate="2011-03-08T04:28:50.663" CommentCount="1" />
  <row Id="25720" PostTypeId="2" ParentId="25712" CreationDate="2011-03-08T04:28:55.063" Score="5" ViewCount="" Body="&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;if the permutation consists of one cycle, (this doesn't seem to have anything to do with inversions), how many repetitions of the cycle will it take to return to identity if $k$ is even? if $k$ is odd?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;for $I(n,k)$, how many total possible inversions are there? How many inversions are there of the reverse of a given permutation?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;for $I(n,3)$, what is $I(n,1)$? $I(n,2)$? Look at some values, guess a formula, and use induction to quickly prove. Or think combinatorially: you have 3 inversions; if they don't interact then it's straightforward; if they do interact then a little more thought on counting them.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1919" LastEditorUserId="1919" LastEditorDisplayName="" LastEditDate="2011-03-08T14:36:22.613" LastActivityDate="2011-03-08T14:36:22.613" />
  <row Id="25721" PostTypeId="2" ParentId="25716" CreationDate="2011-03-08T04:29:06.350" Score="6" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\: $ For $\rm\ c = b/a\ $ it is $\rm\ c^n = d\in \mathbb Z\ \Rightarrow\ c\in \mathbb Z\:.\:$ Now apply the &lt;a href=&quot;http://en.wikipedia.org/wiki/Rational_root_theorem&quot; rel=&quot;nofollow&quot;&gt;Rational Root Test&lt;/a&gt; to $\rm\ x^n-d\:.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice again, just as in your prior few questions, converting it from divisibility relational form into equational form helps us to solve it efficiently. Here it allows us to notice that the problem is simply the special &lt;em&gt;monic&lt;/em&gt; case of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Rational_root_theorem&quot; rel=&quot;nofollow&quot;&gt;Rational Root Test.&lt;/a&gt; Recall that this says that the leading coefficient of a polynomial $\rm\ f(x)\in \mathbb Z[x]\ $ serves as a denominator for all its rational roots. In particular, when the leading coefficient is one this implies that every rational root can be written with denominator $= 1$, i.e. is an integer. This is a &lt;em&gt;fundamental&lt;/em&gt; property in number theory. Indeed it is the property that is used to generalize the notion of integer from rationals to algebraic number fields. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For some generalizations see &lt;a href=&quot;http://math.stackexchange.com/questions/5801/for-any-positive-integers-a-b-one-has-a4b3-implies-ab/5806#5806&quot;&gt;my post here.&lt;/a&gt; Here's a pertinent excerpt:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;THEOREM&lt;/strong&gt; $\:$ TFAE for $\rm\: a,b\: $ in domain $\rm\:Z\:,\ \: r \in Q \:=\:$ fraction field of $\rm\: Z\:,\ n\in \mathbb N$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(1)\ \ \rm\ \ r = \sqrt[n]a  \ \Rightarrow\ r \in Z$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(2)\ \ \rm\ \ r^n \in \:Z \ \ \Rightarrow\  r \in Z$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(3)\ \ \rm\ \ \ a^n\:|\:b^n \ \ \Rightarrow\:\  a\:|\:b$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(4)\ \ \rm\ \ (a^n) =  (b^n,\: c^n) \ \Rightarrow\  (a) = (b,c)\ $  as ideals in $\rm\:Z$&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2011-03-09T16:27:43.917" LastActivityDate="2011-03-09T16:27:43.917" CommentCount="1" />
  <row Id="25722" PostTypeId="2" ParentId="25546" CreationDate="2011-03-08T05:00:38.170" Score="1" ViewCount="" Body="&lt;p&gt;The state of the art in this area has advanced some ways since the mid-1980s. The keywords you should be using in your search are &quot;stream&quot; and &quot;online&quot;, both terms used to denote the situation when the size of the input data set (or stream) is far too large to store in memory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Specifically, I would suggest starting with &lt;a href=&quot;http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.90.8495&amp;amp;rep=rep1&amp;amp;type=pdf&quot; rel=&quot;nofollow&quot;&gt;&quot;Space-Efficient Online Computation of Quantile Summaries&quot;&lt;/a&gt; by Greenwald &amp;amp; Khanna and &lt;a href=&quot;http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.75.4895&amp;amp;rep=rep1&amp;amp;type=pdf&quot; rel=&quot;nofollow&quot;&gt;&quot;How to Summarize the Universe&quot;&lt;/a&gt; by Gilbert, et. al.&lt;/p&gt;&#xA;" OwnerUserId="2643" LastActivityDate="2011-03-08T05:00:38.170" CommentCount="1" />
  <row Id="25723" PostTypeId="1" AcceptedAnswerId="25728" CreationDate="2011-03-08T05:08:24.710" Score="4" ViewCount="98" Body="&lt;p&gt;Let $X,Y$ denote (real) vector spaces. The vector space of $n$-linear maps $X^n \to Y$ will be denoted by $L^n(X,Y)$. Unless I'm much mistaken&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$L(X,L(X,Y)) \ \ \  L^2(X,Y) \ \ \ L(X \otimes X,Y)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;are pairwise isomorphic. One could even view the three as pairwise naturally isomorphic functors from vector spaces to vector spaces (contravariant in $X$ and covariant in $Y$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But what if we want to work with Banach spaces and bounded linear maps instead?Operator norms allow us to make $B(X,B(X,Y))$ into a Banach space without any headache. A 2-linear map $f:X^2 \to Y$ is continuous in the product topology if and only if the norm $\|f\| = \sup_{\|x\| = \|x&amp;#39;\| = 1} \|f(x,x&amp;#39;)\|$ is finite and this makes $B^2(X,Y)$ the set of continuous $2$-linear maps into a Banach space which is isometrically isomorphic to $B(X,B(X,Y))$. But what about $B(X \otimes X, Y)$? According to &lt;a href=&quot;http://en.wikipedia.org/wiki/Topological_tensor_product#Cross_norms_and_tensor_products_of_Banach_spaces&quot; rel=&quot;nofollow&quot;&gt;wikipedia&lt;/a&gt; there are various ways of putting a norm on $X \otimes X$. Which is the right one if we want the same sort of equivalence in this setting?&lt;/p&gt;&#xA;" OwnerUserId="6608" LastActivityDate="2011-03-08T05:50:40.797" Title="What is the ''right&quot; norm for the Banach space tensor product in this situation?" Tags="&lt;functional-analysis&gt;&lt;banach-spaces&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="25724" PostTypeId="1" CreationDate="2011-03-08T05:09:56.180" Score="1" ViewCount="51" Body="&lt;p&gt;how to add two integers in their cantor expansion?&lt;/p&gt;&#xA;" OwnerUserId="7944" LastActivityDate="2011-03-08T05:26:26.013" Title="addition in cantor expansion" Tags="&lt;addition&gt;" AnswerCount="2" />
  <row Id="25725" PostTypeId="2" ParentId="25724" CreationDate="2011-03-08T05:26:14.357" Score="0" ViewCount="" Body="&lt;p&gt;The same way you usually do it. Addition with carry from the least significant digit to the most significant digit. That works for any base where digit $k$ is worth $\prod_{i=1}^k a_i$ for arbitrary integers $a_i \geq 2$.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2011-03-08T05:26:14.357" />
  <row Id="25726" PostTypeId="2" ParentId="25724" CreationDate="2011-03-08T05:26:26.013" Score="2" ViewCount="" Body="&lt;p&gt;You could define a Cantor Expansion so we don't have to search for it, please.  The Cantor Expansion is defined as $x=\sum_1^n a_i i!$ where $0 \le a_i \le i$ So if $y=\sum_1^n b_i i!$, $x+y=\sum_1^n (a_i+b_i) i!$.  The only problem comes if $a_i+b+i \gt i$ but then a carry is in order.  Coefficient $i$ becomes $a_i+b_i-i-1$ and coefficient $i+1$ is increased by 1.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-03-08T05:26:26.013" />
  <row Id="25727" PostTypeId="2" ParentId="25677" CreationDate="2011-03-08T05:29:23.077" Score="0" ViewCount="" Body="&lt;p&gt;Check out &lt;a href=&quot;http://en.wikipedia.org/wiki/Boosting&quot; rel=&quot;nofollow&quot;&gt;Boosting&lt;/a&gt;, also known as &quot;multiplicative weights&quot; or &quot;mixture of experts&quot;.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2011-03-08T05:29:23.077" CommentCount="1" />
  <row Id="25728" PostTypeId="2" ParentId="25723" CreationDate="2011-03-08T05:34:40.763" Score="4" ViewCount="" Body="&lt;p&gt;You're looking for the &lt;em&gt;projective tensor norm&lt;/em&gt;. Explicitly, for $\omega$ in the algebraic tensor product $X \otimes_{\mathbb{R}} Y$ it is given by&#xA;\[&#xA;\Vert \omega\Vert_{\pi} = \inf\left\{\sum \|x_{i}\|\,\|y_{i}\|\,:\,\omega = \sum_{i=1}^{n} x_{i} \otimes y_{i}\right\}&#xA;\]&#xA;and the associated tensor product $X \hat{\otimes}_{\pi} Y$ is the completion of the algebraic tensor product with respect to that norm (the above expression obviously defines a semi-norm, in order to show that it is a norm, you have to think a little bit).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $X \otimes_{\mathbb{R}} Y$ sits inside $X \hat{\otimes}_{\pi} Y$ you get a bilinear map $m:X \times Y \to X \hat{\otimes}_{\pi} Y$ by sending $(x,y)$ to $x \otimes y$ and by definition $\Vert x \otimes y\Vert_{\pi} \leq \|x\|\,\|y\|$ so that $\|m\| \leq 1$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The basic thing you have to check is that the above inequality is in fact an equality $\Vert x \otimes y\Vert_{\pi} = \|x\|\,\|y\|$ and using this it is easy to prove that a continuous bilinear map $\Phi: X \times Y \to Z$ yields a unique linear map $\varphi: X \hat{\otimes}_{\pi} Y \to Z$ of the same norm (by the property of the algebraic tensor product $\Phi$ gives a linear map on $X \otimes_{\mathbb{R}} Y$ and if you've proven that this linear map is continuous, it is continuous on a dense subspace, hence extends uniquely to $X \hat{\otimes}_{\pi} Y$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words the map $m: X \times Y \to X \hat{\otimes}_{\pi} Y$ yields an isometric isomorphism&#xA;$$\displaystyle&#xD;&#xA;B(X \hat{\otimes}_{\pi} Y, Z) \to B^2(X,Y;Z) &#xD;&#xA;$$&#xA;by pre-composition $\varphi \mapsto \varphi \circ m = \Phi$. As you already observed $B^2(X,Y;Z) \cong B(X,B(Y,Z))$ which yields the isometric isomorphisms&#xA;\[&#xA;B(X \hat{\otimes}_{\pi} Y, Z) \cong B(X,B(Y,Z)) \cong B^2(X,Y;Z)&#xA;\]&#xA;as in the case of vector spaces. In fact, you can convince yourself that you have to define the projective tensor norm as above if you want the first isomorphism to hold, already for $Z = \mathbb{R}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can find all this and much more in great detail in R.A. Ryan's book &lt;em&gt;Introduction to tensor products of Banach spaces&lt;/em&gt;, Springer 2001.&lt;/p&gt;&#xA;" OwnerUserId="5363" LastEditorUserId="5363" LastEditorDisplayName="" LastEditDate="2011-03-08T05:50:40.797" LastActivityDate="2011-03-08T05:50:40.797" CommentCount="1" />
  <row Id="25729" PostTypeId="2" ParentId="25688" CreationDate="2011-03-08T05:55:43.307" Score="3" ViewCount="" Body="&lt;p&gt;By the Euclidean Algorithm  $$\left(n!+1,\ (n+1)!+1\right)=\left(n!+1,\ (n+1)!+1-(n!+1)\right)$$  $$=\left(n!+1,n\cdot n!\right).$$  It is clear the last two must be relatively prime, since if $p|\ (n!\cdot n)$ then $p| n!$ (and there is an extra +1 hanging around).  &lt;/p&gt;&#xA;" OwnerUserId="6075" LastEditorUserId="6075" LastEditorDisplayName="" LastEditDate="2011-03-08T16:25:53.473" LastActivityDate="2011-03-08T16:25:53.473" />
  <row Id="25730" PostTypeId="1" AcceptedAnswerId="25731" CreationDate="2011-03-08T05:59:59.133" Score="5" ViewCount="133" Body="&lt;p&gt;Given the following problem:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;integrate $\cos^3(2x)$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I was given the solution &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;\begin{align*}&#xA;  \int\cos^3(2x)\, \mathrm dx &amp;amp;= \int\cos^2(2x) \cos 2x\, \mathrm dx = \int(1-\sin^2 2x)\cos 2x\, \mathrm dx\\&#xA;  &amp;amp;= \frac12\int (1-u^2)\, \mathrm du = \cdots&#xA;  \end{align*}&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;but the problem is I am stuck on the 1/2. Where did it come from?&lt;/p&gt;&#xA;" OwnerUserId="7946" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2011-04-17T16:06:11.860" LastActivityDate="2011-04-17T16:06:11.860" Title="The integral $\int \cos^3(2x) \, \mathrm dx$" Tags="&lt;calculus&gt;&lt;integral&gt;" AnswerCount="2" CommentCount="5" />
  <row Id="25731" PostTypeId="2" ParentId="25730" CreationDate="2011-03-08T06:02:27.560" Score="6" ViewCount="" Body="&lt;p&gt;When the substitution is made, $u=\sin 2x$, so $du=2\cos 2x\;dx$, or $\frac{1}{2}du=\cos 2x\;dx$.&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2011-03-08T06:02:27.560" />
  <row Id="25732" PostTypeId="1" AcceptedAnswerId="25733" CreationDate="2011-03-08T06:25:20.057" Score="2" ViewCount="81" Body="&lt;p&gt;I need to simplify $\sin^3(x)+\cos^2(x)\sin(x)$:&lt;br /&gt;&#xA;First thing I noticed was the pythagorean identity.&lt;br /&gt;&#xA;$$\sin(x)\sin^2(x)+\cos^2(x)\sin(x) \rightarrow \sin(x)(1)\sin(x)\rightarrow\sin^2(x),$$ &#xA;but this doesn't work obviously.&lt;br /&gt;&#xA;This is where I am stuck, I just can't see any way to simplify. &lt;/p&gt;&#xA;" OwnerUserId="7947" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-08T06:30:37.790" LastActivityDate="2011-03-08T06:30:37.790" Title="Simplify Trig Expressions" Tags="&lt;trigonometry&gt;" AnswerCount="1" />
  <row Id="25733" PostTypeId="2" ParentId="25732" CreationDate="2011-03-08T06:28:42.303" Score="7" ViewCount="" Body="&lt;p&gt;I think you're on the right track, but your algebra is going awry.  You rewrote $\sin^3 x+\cos^2 x\sin x$ as $\sin x\sin^2 x+\cos^2 x\sin x$, which is a good start.  Before you can use the Pythagorean identity, you should &lt;em&gt;factor out&lt;/em&gt; the common $\sin x$: $$\sin x\sin^2 x+\cos^2 x\sin x=\sin x(\sin^2 x+\cos^2 x)$$&#xA;Now, apply the Pythagorean identity.&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2011-03-08T06:28:42.303" />
  <row Id="25734" PostTypeId="1" AcceptedAnswerId="25758" CreationDate="2011-03-08T06:33:12.310" Score="4" ViewCount="196" Body="&lt;p&gt;Hello,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've been working my way through Enderton's Elements of Set Theory for a while, and I feel I have a decent grasp on some of the basics of elementary set theory. My question is, where should I look to next in set theory? What is a good book for set theory that may be considered 'the next step up'?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it helps any, my background knowledge consists of some basic abstract algebra, general topology, linear algebra, etc., but I'm not sure how often they are used in real set theory. Thanks.&lt;/p&gt;&#xA;" OwnerUserId="1583" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2011-03-08T06:36:57.020" LastActivityDate="2011-03-08T10:34:47.083" Title="What is a good text in intermediate set theory?" Tags="&lt;reference-request&gt;&lt;set-theory&gt;&lt;books&gt;" AnswerCount="4" />
  <row Id="25735" PostTypeId="1" AcceptedAnswerId="25736" CreationDate="2011-03-08T06:40:13.053" Score="5" ViewCount="108" Body="&lt;p&gt;Question:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Consider a square matrix of order $m$.&#xA;  At each step you can move one step to&#xA;  the right or one step to the top. How&#xA;  many possibilities are to reach $(m,m)$&#xA;  from $(0,0)$?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I think it is just counting the &lt;a href=&quot;http://oeis.org/A000984&quot; rel=&quot;nofollow&quot;&gt;Central binomial coefficients&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Am I right? If not what is be the correct answer and why?&lt;/p&gt;&#xA;" OwnerUserId="2109" LastActivityDate="2011-03-08T06:57:14.777" Title="Counting paths in a square matrix" Tags="&lt;combinatorics&gt;&lt;matrices&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="25736" PostTypeId="2" ParentId="25735" CreationDate="2011-03-08T06:44:09.097" Score="11" ViewCount="" Body="&lt;p&gt;You are correct. The reason is to get to $(m,m)$ you need to take a total of $m+m$ steps. However, you need to choose $m$ of those steps to be steps up, so the total number of paths is $\binom{m+m}{m}=\binom{2m}{m}$, since the central binomial coefficient picks which of the $2m$ steps will be up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To add to this, in general, the number of paths from $(0,0)$ to $(m,n)$ is then $\binom{m+n}{m}=\binom{m+n}{n}$ for the same reasoning.&lt;/p&gt;&#xA;" OwnerUserId="1583" LastEditorUserId="1583" LastEditorDisplayName="" LastEditDate="2011-03-08T06:57:14.777" LastActivityDate="2011-03-08T06:57:14.777" />
  <row Id="25737" PostTypeId="2" ParentId="25682" CreationDate="2011-03-08T07:05:12.027" Score="1" ViewCount="" Body="&lt;p&gt;In what follows, D denotes differentiation with respect to $s$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Sigma_{n}\frac{\chi(n)}{n^{s}} = \prod_{q}(1-\frac{\chi(q)}{q^{s}})^{-1}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We take the log of each side and then proceed to take the derivative.  Thus we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$Dlog\Sigma_{n}\frac{\chi(n)}{n^{s}} = Dlog \prod_{q}(1-\frac{\chi(q)}{q^{s}})^{-1}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the right hand side of the equation is equal to:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$D\Sigma_{q}log(1-\frac{\chi(q)}{q^{s}})^{-1} = -D\Sigma_{q}log(1-\frac{\chi(q)}{q^{s}})$ using properties of log.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we differentiate term by term.  For an arbitrary term in the sum, we have:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$Dlog(1-\frac{\chi(q)}{q^{s}})=-(1-\frac{\chi(q)}{q^{s}})^{-1}\chi(q)Dq^{-s}$ by the chain rule.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But we know that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$Dq^{-s}=-log(q)q^{-s}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus we have &#xA;$Dlog(1-\frac{\chi(q)}{q^{s}})=(1-\frac{\chi(q)}{q^{s}})^{-1}\chi(q)log(q)q^{-s}$ because the negative signs cancel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But then&#xA;$-D\Sigma_{q}log(1-\frac{\chi(q)}{q^{s}}) = \Sigma(1-\frac{\chi(q)}{q^{s}})^{-1}\chi(q)log(q)q^{-s}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And after re-arranging, this gives us $-D\Sigma_{q}log(1-\frac{\chi(q)}{q^{s}})=-\Sigma_{q}\frac{\chi(q)log(q)}{q^{s}-\chi(q)}$.  Thus we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$Dlog\Sigma_{n}\frac{\chi(n)}{n^{s}}= -\Sigma_{q}\frac{\chi(q)log(q)}{q^{s}-\chi(q)}$ and so swapping the negative sign from the right hand side to the left hand side gives us the result.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(P.S. Please forgive me if there are typos in the above-I'm not very good with latex code!)&lt;/p&gt;&#xA;" OwnerUserId="1179" LastEditorUserId="1179" LastEditorDisplayName="" LastEditDate="2011-03-08T07:23:23.180" LastActivityDate="2011-03-08T07:23:23.180" CommentCount="2" />
  <row Id="25738" PostTypeId="2" ParentId="25734" CreationDate="2011-03-08T07:20:23.970" Score="1" ViewCount="" Body="&lt;p&gt;hmm, well I'm no expert, but you could try Jech's &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/3540440852&quot; rel=&quot;nofollow&quot;&gt;&quot;Set Theory&quot;&lt;/a&gt;. I haven't read it, but I have glanced through it quite a lot, and it is a huge book, 700+ pages. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It covers basic as well as advanced (majority) set theory (+selected topics) and it starts from zero, although it doesn't teach logic. It is in no way a a foundation of mathematics (like Principia Mathematica), because it is reader friendly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is also (illegally) available as an ebook on the net.&lt;/p&gt;&#xA;" OwnerUserId="7325" LastEditorUserId="7325" LastEditorDisplayName="" LastEditDate="2011-03-08T07:32:07.497" LastActivityDate="2011-03-08T07:32:07.497" />
  <row Id="25739" PostTypeId="1" AcceptedAnswerId="25742" CreationDate="2011-03-08T07:32:51.923" Score="4" ViewCount="154" Body="&lt;p&gt;Let $R$ be a commutative ring with identity. I've been trying to prove the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $M \subset R$, i.e. $M \neq R$, is a maximal multiplicative set, then $R \setminus M$ is a prime ideal of $R$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have spent some time in it, but nothing is coming to my mind. Any idea/solution will be appreciated.&lt;/p&gt;&#xA;" OwnerUserId="7819" LastActivityDate="2011-03-08T17:42:51.497" Title="Maximal multiplicative set implies complement is a prime ideal" Tags="&lt;ring-theory&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="1" />
  <row Id="25740" PostTypeId="2" ParentId="25739" CreationDate="2011-03-08T07:40:39.653" Score="0" ViewCount="" Body="&lt;p&gt;We must assume $\rm 0\not\in M $ since $\rm\: 0\in M\: \Rightarrow\: 0 \not\in M&amp;#39; := R\setminus M\:,\: $ so $\rm\:M&amp;#39;\:$ is not an ideal. A simple Zorn lemma argument (see below) shows that, since $\rm\:M\:$ is multiplicatively closed, the ideal $\rm\{0\}\subset M&amp;#39;\:$ can be enlarged to an ideal $\rm\:P\:$ maximal w.r.t. to exclusion of $\rm\:M\:,\:$ and such an ideal must be prime. Hence $\rm\: P = M&amp;#39;\:,\:$ else we could enlarge $\rm\:M\:$ to the monoid $\rm\:P&amp;#39;\:,\:$ contra maximality of $\rm\:M\:.\quad$ &lt;strong&gt;QED&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Note&lt;/strong&gt; $\: $ The prime $\rm\:P\:$ above may be alternatively constructed  as the contraction of a maximal ideal $\rm\:Q\:$ of the localization $\rm\: R_M = M^{-1} R\:.\: $ $\rm\:Q\:$ exists since $\rm\ 0\not\in M\ \Rightarrow\ R_M \ne \{0\}\:.\:$ Generally, there is a bijective order-preserving correspondence between all prime-ideals in $\rm\:R_M\:$ and all prime ideals in $\rm R$ disjoint from $\rm\:M\:,\:$ see Theorem 34 in Kaplansky's &lt;em&gt;Commutative Rings&lt;/em&gt;. His Theorem 1, pp. 1-2, is the above-invoked form of this result employing no localization theory. I've appended it below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/OtewU.jpg&quot; alt=&quot;enter image description here&quot;&gt;&#xA;&lt;img src=&quot;http://i.imgur.com/cM5Ei.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2011-03-08T17:42:51.497" LastActivityDate="2011-03-08T17:42:51.497" />
  <row Id="25742" PostTypeId="2" ParentId="25739" CreationDate="2011-03-08T08:00:36.427" Score="5" ViewCount="" Body="&lt;p&gt;I assume that all multiplicative subsets discussed here are disjoint from $\{0\}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me write $S$ for the maximal multiplicative subset of $R$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Step 1: I claim that $S^{-1} R$ is a local ring.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Indeed, if not, there would be at least two maximal ideals in the localization, i.e., &#xA;two prime ideals $\mathfrak{p}_1 \neq \mathfrak{p}_2$, each maximal with respect to &#xA;being disjoint from $S$.  Thus $S$ is contained in both $R \setminus \mathfrak{p}_1$ and $R \setminus \mathfrak{p}_2$, but these are two different sets so at least one of the containments must be proper, contradicting the maximality of $S$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Step 2: Therefore there is a unique maximal ideal in the localization, which corresponds to a prime ideal $\mathfrak{p}$ of $R$ which is disjoint from $S$.  Arguing as above, maximality of $S$ implies $S = R \setminus \mathfrak{p}$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Remark 1&lt;/b&gt;: When $R$ is a domain, the unique maximal multiplicative subset is obviously $R \setminus \{0\}$.  In this case it is tempting to construe &quot;maximal multiplicative subset&quot; to mean &quot;multiplicative subset maximal with respect to being properly contained in $R \setminus \{0\}$.&quot; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Remark 2&lt;/b&gt;: Conversely, the primes $\mathfrak{p}$ such that $R \setminus \mathfrak{p}$ is maximal are precisely the primes which are &lt;em&gt;minimal&lt;/em&gt; with respect to containing $\{0\}$.  (When $R$ is a domain and the question is reconstrued as above, we want $\mathfrak{p}$ to be minimal with respect to &lt;em&gt;properly&lt;/em&gt; containing $\{0\}$.)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mariano's answer makes this clear as well.&lt;/p&gt;&#xA;" OwnerUserId="299" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2011-03-08T08:42:10.417" LastActivityDate="2011-03-08T08:42:10.417" CommentCount="1" />
  <row Id="25743" PostTypeId="2" ParentId="25739" CreationDate="2011-03-08T08:15:32.337" Score="2" ViewCount="" Body="&lt;p&gt;Another way to proceed is to show&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;First, that every multiplicative subset $S$ of a ring is contained in a &lt;em&gt;saturated&lt;/em&gt; multiplicative subset $S&amp;#39;$, that is, one such that $$ab\in S&amp;#39;\implies a\in S&amp;#39;\wedge b\in S&amp;#39;.$$ In fact, there is a smallest saturated multiplicative subset containing $S$, called the &lt;em&gt;saturation&lt;/em&gt;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Second, that a saturated multiplicative subset $S$ of a ring (which does not contain $0$) is the intersection of the complements of the prime ideals which are disjoint from it. (The hard part here is to show that such primes actually exist: you can construct them as the maximal elements in the family of ideals disjoint from $S$)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Once you have these two facts, your claim follows easily.&lt;/p&gt;&#xA;" OwnerUserId="274" LastEditorUserId="274" LastEditorDisplayName="" LastEditDate="2011-03-08T08:23:33.250" LastActivityDate="2011-03-08T08:23:33.250" CommentCount="1" />
  <row Id="25744" PostTypeId="2" ParentId="25734" CreationDate="2011-03-08T08:18:25.100" Score="3" ViewCount="" Body="&lt;p&gt;While Jech's book (Leon's recommendation) certainly is an outstanding book, it may be a bit advanced. In my daily work, I find what's covered in Ciesielski's book &lt;em&gt;Set theory for the working mathematician&lt;/em&gt;, LMS student texts &lt;strong&gt;32&lt;/strong&gt;,  amply enough for my needs. It makes for an easy read and prepares the reader gently towards forcing. I don't know Enderton's book, but I imagine there's quite a bit of overlap in the beginning. Another book that's often recommended is Kunen's &lt;em&gt;Set theory, an introduction to independence proofs&lt;/em&gt;, but I find it a bit hard on the casual reader. Concerning the philosophical background, I found the beginning of Fraenkel, Bar-Hillel, Levy, &lt;em&gt;Foundations of set theory&lt;/em&gt; an exciting read (Fraenkel is the F in ZF: Zermelo-Fraenkel).&lt;/p&gt;&#xA;" OwnerUserId="5363" LastActivityDate="2011-03-08T08:18:25.100" CommentCount="1" />
  <row Id="25745" PostTypeId="1" CreationDate="2011-03-08T08:31:29.680" Score="2" ViewCount="79" Body="&lt;p&gt;Minutes ago I read that if $F$ is the field of algebraic numbers over $\mathbb{Q}$, then every polynomial in $F[x]$ splits over $F$. That's awesome! Nevertheless, I don't fully understand why it is true. Can you throw some ideas about why this is true?&lt;/p&gt;&#xA;" OwnerUserId="7819" LastActivityDate="2011-03-08T09:17:52.520" Title="Interesting property of a polynomial ring with coefficients algebraic over $\mathbb{Q}$" Tags="&lt;field-theory&gt;" AnswerCount="2" CommentCount="5" FavoriteCount="1" />
  <row Id="25746" PostTypeId="2" ParentId="25745" CreationDate="2011-03-08T08:58:03.737" Score="5" ViewCount="" Body="&lt;p&gt;The fact you mention is that the field $\overline{\mathbb{Q}}$ of algebraic numbers is &lt;strong&gt;algebraically closed&lt;/strong&gt;.  This is true because it is the &lt;strong&gt;algebraic closure&lt;/strong&gt; of the field $\mathbb{Q}$ inside the larger algebraically closed field $\mathbb{C}$, i.e., the set of all complex numbers which satisfy some nonzero polynomial equation with $\mathbb{Q}$-coefficients. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notwithstanding the terminology, that the algebraic closure of a field in an algebraically closed field is algebraically closed requires (nontrivial) proof!  See for instance $\S 4$ of &lt;a href=&quot;http://math.uga.edu/~pete/FieldTheory.pdf&quot; rel=&quot;nofollow&quot;&gt;my notes on field theory&lt;/a&gt;, especially Proposition 17 and Corollary 18.&lt;/p&gt;&#xA;" OwnerUserId="299" LastActivityDate="2011-03-08T08:58:03.737" />
  <row Id="25747" PostTypeId="1" AcceptedAnswerId="25752" CreationDate="2011-03-08T09:04:03.747" Score="1" ViewCount="56" Body="&lt;p&gt;Hello,&#xA;I have differential equation $x\cdot\sqrt{4+y^2}\,dx + y\cdot\sqrt{1+x^2}\,dy = 0$, which is simple to solve on paper, but I have problem solving it with Maple (I use 10th, tried on 14th also). When applying dsolve it complains that it is not ODE.&#xA;Any suggestions?&#xA;Thank you in advance&lt;/p&gt;&#xA;" OwnerUserId="7949" LastActivityDate="2011-03-08T09:41:19.983" Title="Differential equation problem in Maple" Tags="&lt;differential-equations&gt;&lt;maple&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="25748" PostTypeId="1" AcceptedAnswerId="25751" CreationDate="2011-03-08T09:07:59.013" Score="1" ViewCount="52" Body="&lt;p&gt;Follow up from: &lt;a href=&quot;http://stackoverflow.com/questions/5225061/exponential-decay-surrounding-bounding-box&quot;&gt;http://stackoverflow.com/questions/5225061/exponential-decay-surrounding-bounding-box&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I manipulate a simple logistic function like this one:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/edGT1.png&quot; alt=&quot;logistic graph&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So that the lower bound is at 0, and the (-3, 3) range is roughly doubled and depends on some value (the lower edge of the bounding box as described in the original question)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And then how would I flip the function horizontally, so that it can work for the upper edge of the box?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maths n00b, sorry&lt;/p&gt;&#xA;" OwnerUserId="5539" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2011-03-08T09:21:17.813" LastActivityDate="2011-03-08T09:21:17.813" Title="Stretch Logistic Function" Tags="&lt;algebra-precalculus&gt;&lt;functions&gt;" AnswerCount="1" />
  <row Id="25749" PostTypeId="2" ParentId="25745" CreationDate="2011-03-08T09:10:51.950" Score="2" ViewCount="" Body="&lt;p&gt;Consider some polynomial $$x^n = \sum_{i=0}^{n-1} c_i x^i,$$ where the $c_i$ are algebraic numbers. Thus for each $i$ we have a similar identity $$c_i^{n_i} = \sum_{j=0}^{n_i-1} d_{i,j} c_i^j,$$ where this time the $d_{i,j}$ are rationals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose that $\alpha$ is a root of the original polynomial. By using the above identities, every power of $\alpha$ can be expressed as a linear combination &lt;em&gt;with rational coefficients&lt;/em&gt; of terms of the form $$\alpha^m c_0^{m_0} \cdots c_{n-1}^{m_{n-1}},$$ where $$0 \leq m &amp;lt; n \text{ and } 0 \leq m_i &amp;lt; n_i.$$ Putting all these $N = nn_0\cdots n_{n-1}$ elements into a vector $v$, we get that there are &lt;em&gt;rational&lt;/em&gt; vectors $u_k$ such that $$\alpha^k = \langle u_k,v \rangle.$$ Among the first $N+1$ vectors $u_k$ there must be some non-trivial &lt;em&gt;rational&lt;/em&gt; linear combination that vanishes: $$\sum_{k=0}^N t_k u_k = 0.$$ Therefore $$\sum_{k=0}^N t_k \alpha^k = 0,$$ and so $\alpha$ is also algebraic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This proof is taken from &lt;a href=&quot;http://people.reed.edu/~jerry/361/lectures/lec09.pdf&quot; rel=&quot;nofollow&quot;&gt;these lecture notes&lt;/a&gt;, but it's pretty standard.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2011-03-08T09:10:51.950" />
  <row Id="25751" PostTypeId="2" ParentId="25748" CreationDate="2011-03-08T09:20:49.577" Score="4" ViewCount="" Body="&lt;p&gt;If you take $y=\text{erf}(x)+1$, that will shift the graph up by 1, so the lower limit is 0 and the upper limit is 2.  If you want to keep the upper limit at 1, taking half ($y=\frac{1}{2}(\text{erf}(x)+1)$) will return the upper limit to 1 and keep the lower limit at 0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to double the input from $(-3,3)$ to $(-6,6)$, halving $x$ will do that, $y=\frac{1}{2}(\text{erf}(\frac{1}{2}x)+1)$.  If you then want to move it so that what was at $-6$ is now at $a$, use $y=\frac{1}{2}(\text{erf}(\frac{1}{2}(x-a-6))+1)$.  To flip it horizontally, take the opposite of $x$, $y=\frac{1}{2}(\text{erf}(-\frac{1}{2}x)+1)$.&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2011-03-08T09:20:49.577" CommentCount="3" />
  <row Id="25752" PostTypeId="2" ParentId="25747" CreationDate="2011-03-08T09:30:53.400" Score="2" ViewCount="" Body="&lt;p&gt;The command you are looking for is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;dsolve(x*sqrt(4+y(x)^2)+y(x)*sqrt(1+x^2)*diff(y(x),x), y(x))&lt;/code&gt;.&lt;/p&gt;&#xA;" OwnerUserId="7266" LastEditorUserId="7266" LastEditorDisplayName="" LastEditDate="2011-03-08T09:41:19.983" LastActivityDate="2011-03-08T09:41:19.983" CommentCount="2" />
  <row Id="25753" PostTypeId="1" CreationDate="2011-03-08T09:34:39.417" Score="3" ViewCount="113" Body="&lt;p&gt;In a standard deck, there are 13 cards in each suit,such as hearts. For simplicity we'll number them from 1 to 13. This question is about these 13 cards. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Start with the cards in the usual order. This time you are allowed to divide the deck into three piles, consisting of the top n cards, the next m cards, and the remaining 13-n-m cards. Keeping the order within these piles unchanged, you place one pile at the top, one in the middle, and one on the bottom. You may repeat this as many times as you like, changing m and n if you wish. How many arrangements can you get?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My answer:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we can number them from 1 to 13 in the usual order:&#xA;1  2  3  4  5  6  7  8  9  10  11  12  13&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So as one case I can have firs pile=4 cards, second pile=4 cards and last pile=5 cards. If I do the described permutation, I get:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1  5  9  2  6  10  3  7  11  4  8  12  13&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this correct? If this is correct, then this question is asking me to consider different values of m and n and consider different arrangements based on these values? &lt;/p&gt;&#xA;" OwnerUserId="7340" LastActivityDate="2011-03-09T10:17:13.717" Title="number of arrangements" Tags="&lt;homework&gt;&lt;group-theory&gt;" AnswerCount="3" CommentCount="8" />
  <row Id="25754" PostTypeId="2" ParentId="25734" CreationDate="2011-03-08T09:42:45.397" Score="0" ViewCount="" Body="&lt;p&gt;Paul Halmos &lt;em&gt;Naive Set Theory&lt;/em&gt; is an Excellent text&lt;/p&gt;&#xA;" LastActivityDate="2011-03-08T09:42:45.397" CommentCount="2" />
  <row Id="25755" PostTypeId="2" ParentId="25753" CreationDate="2011-03-08T09:43:29.283" Score="3" ViewCount="" Body="&lt;p&gt;Try to think of a generating set of $S_{13}$ that you can implement using this operation.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2011-03-08T09:43:29.283" CommentCount="3" />
  <row Id="25756" PostTypeId="1" AcceptedAnswerId="25757" CreationDate="2011-03-08T09:46:26.057" Score="1" ViewCount="41" Body="&lt;p&gt;Very simple question but I keep getting this wrong!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have two points e.g. A$(13, 6$) &amp;amp; B$(11, 12)$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the gradient formula $m = (y_2 - y_1)/(x_2 - x_1)$ how do you know which of A or B corresponds to $(x_1, y_1)$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can work this out by drawing a diagram of the line. If the slope is negative, the higher of the two points is $(x_1, y_1)$. Otherwise the lower is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But this takes too long in an exam. Is there a quicker way?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="6409" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2011-03-08T09:50:54.373" LastActivityDate="2011-03-08T09:50:54.373" Title="Calculating gradient of a line: how do you know which way to order the points?" Tags="&lt;calculus&gt;&lt;geometry&gt;&lt;algebra-precalculus&gt;&lt;analytic-geometry&gt;" AnswerCount="1" />
  <row Id="25757" PostTypeId="2" ParentId="25756" CreationDate="2011-03-08T09:49:17.087" Score="4" ViewCount="" Body="&lt;p&gt;It doesn't matter which point is which, since $$\frac{y_2-y_1}{x_2-x_1}\cdot\frac{-1}{-1}=\frac{-y_2+y_1}{-x_2+x_1}=\frac{y_1-y_2}{x_1-x_2}.$$  In your specific example, $$\frac{6-12}{13-11}=\frac{-6}{2}=-3$$ and $$\frac{12-6}{11-13}=\frac{6}{-2}=-3.$$&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2011-03-08T09:49:17.087" CommentCount="0" />
  <row Id="25758" PostTypeId="2" ParentId="25734" CreationDate="2011-03-08T10:34:47.083" Score="3" ViewCount="" Body="&lt;p&gt;I have not read it myself, however I got a good recommendation from one of my teachers - &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Azriel Levy's &lt;em&gt;Basic Set Theory&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Jech's &lt;em&gt;Set Theory&lt;/em&gt; is a great book but I think it is indeed slightly too advanced, he writes that the first part contains full proofs (I only read chapters from the second parts, in which proofs are many times sketched out and the details are left for the reader). Once you've got the basic theorems down, one might also check &lt;em&gt;The Handbook Of Set Theory&lt;/em&gt; written by an ensemble of competent writers, for more specific topics.&lt;/p&gt;&#xA;" OwnerUserId="622" LastActivityDate="2011-03-08T10:34:47.083" CommentCount="2" />
  <row Id="25759" PostTypeId="2" ParentId="25753" CreationDate="2011-03-08T10:48:57.713" Score="3" ViewCount="" Body="&lt;p&gt;To put the card in position $n$ at the top without changing the order of the other cards, split the deck into $(n-1)$, $1$ and $13-n$ piles and restack as $1$, $(n-1)$ and $(13-n)$.  You don't need to do this if $n=1$, while if $n=13$ then split into something like $10$, $2$ and $1$ and restack as $1$, $10$ and $2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do this up to 13 times and you can have any order.     &lt;/p&gt;&#xA;" OwnerUserId="6460" LastActivityDate="2011-03-08T10:48:57.713" CommentCount="6" />
  <row Id="25761" PostTypeId="1" AcceptedAnswerId="25777" CreationDate="2011-03-08T12:34:06.417" Score="3" ViewCount="131" Body="&lt;p&gt;I just opened vol.1 of the Bourbaki treatise to take a look at how they define mathematical structure. I was amazed by its sheer complexity. Can you recommend an introductory text that wouldn't require as much effort to understand?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, a few related soft questions:&lt;br&gt;&#xA;1) Is category theory more general than this theory?&lt;br&gt;&#xA;2) In Bourbaki approach, &lt;em&gt;theory&lt;/em&gt; corresponds to &lt;em&gt;category&lt;/em&gt;, and &lt;em&gt;structure&lt;/em&gt; corresponds to &lt;em&gt;object&lt;/em&gt;, am I right?&lt;br&gt;&#xA;3) I recently started reading MacLane's &quot;Categories for a working mathematician&quot;, and category theory seems much simpler to understand. Is it an illusion due to less formal exposition, or is it really the case? If it's the latter, was this the reason why everyone adopted category theory in place of Bourbaki approach?&lt;/p&gt;&#xA;" OwnerUserId="3793" LastActivityDate="2011-03-08T17:00:57.700" Title="Introduction to Bourbaki structures, and their relation to category theory" Tags="&lt;soft-question&gt;&lt;reference-request&gt;&lt;category-theory&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="2" />
  <row Id="25762" PostTypeId="2" ParentId="24847" CreationDate="2011-03-08T12:40:46.830" Score="3" ViewCount="" Body="&lt;p&gt;My supervisor pointed me in the right direction for the solution, which I'm recording here for posterity. It (unsurpringly) involves the action of $\Omega B$ on the homotopy fibre $F$, just like in the bundle case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Recall that I had defined a map $\Omega B \times E \rightarrow F$ by composition of $1 \times s$ with the action $*$. Now, remembering that the product of two fibrations is a fibration, we have the following commutative diagram of long exact sequences of homotopy groups and maps between them:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{array}{ccccc}\pi_{n+1} E &amp;amp; = &amp;amp; \pi_{n+1} E &amp;amp; = &amp;amp; \pi_{n+1} E \\&#xD;&#xA;&#xD;&#xA;\downarrow &amp;amp; &amp;amp; \downarrow &amp;amp; &amp;amp; \downarrow \\&#xD;&#xA;&#xD;&#xA;\pi_{n} \Omega B &amp;amp; \longrightarrow &amp;amp; \pi_{n} \Omega B \times \pi_{n} \Omega B &amp;amp; &#xD;&#xA;\longrightarrow &amp;amp; \pi_n \Omega B\\&#xD;&#xA;&#xD;&#xA;\downarrow &amp;amp; &amp;amp; \downarrow &amp;amp; &amp;amp; \downarrow \\&#xD;&#xA;&#xD;&#xA;\pi_{n} \Omega B \times \pi_{n} E &amp;amp; \longrightarrow &amp;amp; \pi_{n} \Omega B \times \pi_n F &amp;amp; \longrightarrow &amp;amp; \pi_{n} F&#xD;&#xA;&#xD;&#xA;\\ \downarrow &amp;amp; &amp;amp; \downarrow &amp;amp; &amp;amp; \downarrow \\&#xD;&#xA;&#xD;&#xA;\pi_{n} E &amp;amp; = &amp;amp; \pi_{n} E &amp;amp; = &amp;amp; \pi_{n} E \\ &#xD;&#xA;&#xD;&#xA;\downarrow &amp;amp; &amp;amp; \downarrow &amp;amp; &amp;amp; \downarrow \\&#xD;&#xA;&#xD;&#xA;\pi_{n-1} \Omega B &amp;amp; \longrightarrow &amp;amp; \pi_{n-1} \Omega B \times \pi_{n-1} \Omega B &amp;amp; \longrightarrow &amp;amp; \pi_{n-1} \Omega B&#xD;&#xA;\end{array}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where the second and fifth rows come from (left) inclusion and $H$-group multiplication, and the middle row is the homomorphism induced by the map $\ast \circ (1 \times s)$ defined above. The result then follows from the 5-lemma and Whitehead's theorem. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Sorry about the shoddiness of the diagram, I wasn't sure if this textbox is xy enabled or not, so I stuck with an array.)&lt;/p&gt;&#xA;" OwnerUserId="7748" LastEditorUserId="7748" LastEditorDisplayName="" LastEditDate="2011-03-08T14:08:33.167" LastActivityDate="2011-03-08T14:08:33.167" />
  <row Id="25763" PostTypeId="1" CreationDate="2011-03-08T12:54:57.910" Score="2" ViewCount="99" Body="&lt;p&gt;I have recently come across this problem trying to visualise a certain economic model and I'm finding the solution is just beyond my reach. As far as I can tell there is a simplification of the problem which is easier and would still be good to have answered.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are two moving point particles A and B on a Cartesian plane. Particle A is trying to reach B as quickly as possible, it can do so by applying acceleration in any direction. Its acceleration and speed have constant upper limits. Particle A knows the position and velocity of itself and of B, and can continuously[1] adjust its acceleration (which doesn't have to be continuous). What function of the particles' position and velocity should A use to reach and stop at B in as little time as possible?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the simplest version B is moving with a constant velocity, which I think will produce a better result for the complete version too, in which B can have acceleration. For many nodes targeting each other my current approach of &quot;accelerate as fast as possible to B's current position&quot; quickly resembles chaos for a few particles targeting each other in a chain. I know why that's a poor approach, I just don't know how to make a better one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I haven't been clear enough, I'd be happy to provide a visualisation of the problem and my not-working solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Pretty please work this all the way through. I've shown a lot of people this problem and the pattern has been for them solve it for a single dimension then tell me that it should be easy to just do it for both dimensions. It really isn't, there's almost certainly a derivation and optimisation equation in there somewhere since there are three unknowns: acceleration on each axis and time but only two equations to solve them with (position of both particles with respect to time). Another way of looking at it is that the dimensions can't be considered separately assuming they can move with maximum acceleration, since the magnitude of the acceleration is limited there is a trade-off there. As is probably clear from my fumbling explanations, I'm not math savvy enough to translate this idea into number and figures.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[1] Not actually continuous, since it runs on a computer in discrete (but tiny) steps. &lt;/p&gt;&#xA;" OwnerUserId="7953" LastEditorUserId="7953" LastEditorDisplayName="" LastEditDate="2011-03-09T12:32:24.147" LastActivityDate="2011-05-08T12:50:37.580" Title="Optimising closest approach with third order motion" Tags="&lt;vector-calculus&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="25765" PostTypeId="1" AcceptedAnswerId="25767" CreationDate="2011-03-08T13:29:42.650" Score="11" ViewCount="364" Body="&lt;p&gt;How many circles (radius – r) are needed to cover circle which radius is 2 times bigger (radius – 2r).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think we need to use area which is $S=\pi R^2$ but I don't really know what to do&lt;/p&gt;&#xA;" OwnerUserId="7149" LastActivityDate="2011-03-08T14:23:15.623" Title="How many circles to cover 2 times bigger circle?" Tags="&lt;geometry&gt;&lt;circle&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="25766" PostTypeId="1" AcceptedAnswerId="25776" CreationDate="2011-03-08T13:46:28.403" Score="3" ViewCount="91" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've got a question regarding the constructible universe and I'm a bit confused about the Condensation Lemma for the universe constructible from some set $A$. Help will be greatly appreciated:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's assume that we have $(M,E)$ a model of ZF (or maybe of ZFC) but not necessarily standard or transitive. Since it's a model of ZF we can create its constructible universe $L^{(M,E)}$. What can we say about this model? I think it is a model of ZF both inside $(M,E)$ and outside of it. Is this correct? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What about the continuum hypothesis and the axiom of choice? To show that the actual $L$ satisfies GCH we use the Condensation Lemma which is based on the fact that $V=L$ is satisfied in every $L_\gamma$. Furthermore to show the consistency of the axiom of constructibility and of its consequences we need to show $(V=L)^L$ which is done through absoluteness. Can we do something similar in the case of the arbitrary model? For example can we say that $L^{(M,E)}$ is an actual model of ZFC+GCH? Or it is so only inside $(M,E)$?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Regarding relative constructibility: I can see how $L[A]$ is a model of ZFC and I can see how for an inner model $M$ of ZF that has $A\cap M\in M$ we have that $L[A]\subset M$ through the Gödel operations. Furthermore it appears to me that since satisfiability even if we add a set as a predicate is absolute, we can through this show the absoluteness of &quot;$x$ is relative constructible&quot; and thus show both that $L[A]$ is the least inner model (under the restriction mentioned above)  and $(\exists X\quad V=L[X])^{L[A]}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My problem is with the generalized continuum hypothesis and the condensation lemma. In Jech's book it states that GCH is true in $L[A]$ &lt;em&gt;above&lt;/em&gt; some ordinal $\alpha$. But then the Condensation Lemma is stated as:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If $\mathcal{M}\prec(L_\delta[A],\in,A\cap L_\delta[A])$ where $\delta$ is a limit ordinal, then the transitive collapse of $\mathcal{M}$ is $L_\gamma[A]$ for some $\gamma\leq\delta$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If this is indeed how the condensation lemma generalizes then why can't we prove the GCH much like we do in the case of $L$? For every $\alpha$, taking $X\in\mathcal{P}^{L[A]}(\omega_\alpha)$ there is of course some $\delta$ such that $X\in L_\delta[A]$. Then taking the Skolem Hull of $\omega_\alpha\cup\{X\}$ we get a model $\mathcal{M}\prec(L_\delta[A],\in,A\cap L_\delta[A])$ with $|\mathcal{M}|=|\omega_\alpha|$. Its transitive collapse would be some $L_\gamma[A]$ and since $|L_\gamma[A]|=|\gamma|$  we would have that $\gamma&amp;lt;\omega_{\alpha+1}$. I can't find any gap in this syllogism. What am I missing?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance,&lt;/p&gt;&#xA;" OwnerUserId="1772" LastEditorUserId="1772" LastEditorDisplayName="" LastEditDate="2011-03-08T15:58:34.463" LastActivityDate="2011-03-08T16:23:15.833" Title="Some questions regarding (relative) constructibility and the condensation lemma" Tags="&lt;set-theory&gt;" AnswerCount="1" />
  <row Id="25767" PostTypeId="2" ParentId="25765" CreationDate="2011-03-08T14:23:15.623" Score="15" ViewCount="" Body="&lt;p&gt;This is a &lt;a href=&quot;http://en.wikipedia.org/wiki/Disk_covering_problem&quot; rel=&quot;nofollow&quot;&gt;disk covering problem&lt;/a&gt;.  As you have stated it, it is not quite as difficult as some others. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first task is to find the minimum number of small circles which cover the circumference of the bigger circle rather than the whole area.  If this is $m$ then it will be impossible to have a regular $m$-gon with edges of length $2r$ fitting strictly inside the circle of radius $2r$.  This implies $m \ge 6$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It turns out to be just possible to cover the circumference of the bigger circle with 6 smaller circles, and ignoring symmetries there is only one way to do it (you get a regular hexagon whose vertices lie on the bigger circle and whose edges are diameters of the smaller circles).  But this leaves an uncovered area in the middle, which needs at least one (and in fact exactly one) more.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the answer is seven smaller circles.   &lt;/p&gt;&#xA;" OwnerUserId="6460" LastActivityDate="2011-03-08T14:23:15.623" />
  <row Id="25769" PostTypeId="1" AcceptedAnswerId="26056" CreationDate="2011-03-08T14:34:58.700" Score="1" ViewCount="85" Body="&lt;p&gt;So, I had a quiz last night in my discrete structures class and this question came up:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;There are 3 processors in a computer. The instructions to a computer are labeled 1,2,3,4,5,6,7,8,9,10 each taking 1 second to execute. The relation between a pair of instructions (x,y) is given by R:'x' is executed at the same time or before 'y'. Using the properties of relations, design an execution process including the sequence of execution that takes the minimum time. Relation between the instructions is R={(1,1)(2,2)(3,3)(4,4)(5,5)(6,6)(7,7)(8,8)(9,9)(10,10)(1,7)(5,8)(2,3)(8,10)(5,10)}&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How would I go about solving this? I want to understand how to do it....but I don't know where to start...&lt;/p&gt;&#xA;" OwnerUserId="7800" LastActivityDate="2011-03-09T21:29:44.907" Title="Execution process with properties of relations" Tags="&lt;discrete-mathematics&gt;&lt;relations&gt;" AnswerCount="2" />
  <row Id="25770" PostTypeId="1" AcceptedAnswerId="25773" CreationDate="2011-03-08T14:48:33.987" Score="4" ViewCount="132" Body="&lt;p&gt;Hello everyone!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $x_i (i=1,...,n, n&amp;gt;d)$ be a unit vector in $R^d$. $c_i&amp;gt;0$ is a positive real scalar. How to prove the following fact? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fact: There exist some vectors $x_i$ such that $\sum_{i=1}^n c_i x_i=0$ if and only if $c_i\le\sum_{j\neq i}c_j, \forall i$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The necessity is easy to prove. (My proof of the necessity: If $\sum_{i=1}^n c_i x_i=0$, then $c_i x_i = -\sum_{j\neq i}c_j x_j$. So  $\| c_i x_i \|=c_i\le\sum_{j\neq i}\| c_j x_j\|=\sum_{j\neq i}c_j)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But how to prove the sufficiency? That is: if $c_i\le\sum_{j\neq i}c_j, \forall i$, can we always find some $x_i$ such that $\sum_{i=1}^n c_i x_i=0$? It seems a very basic result, but not easy to prove for me. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you for your help.&#xA;Shiyu&lt;/p&gt;&#xA;" OwnerUserId="7156" LastEditorUserId="6622" LastEditorDisplayName="" LastEditDate="2011-03-08T15:14:55.750" LastActivityDate="2011-03-08T15:52:12.000" Title="Prove a basic fact on a linear combination of vectors" Tags="&lt;linear-algebra&gt;&lt;vector-space&gt;&lt;norm&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="25771" PostTypeId="1" AcceptedAnswerId="25772" CreationDate="2011-03-08T15:01:00.283" Score="4" ViewCount="182" Body="&lt;p&gt;Measure theory was established on naive set theory(Not totally sure). But after Russell discovered the paradox named by him, set theory was reconstructed in the sense of axiomatization.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is in the first chapter of many measure theory textbooks, there is a set theory introduction, most likely describing the naive set theory. How can I be convinced that measure theory is rigorous? Or I can just take as granted that it is, and the approach by introducing naive set theory is only because it is easier to understand?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And could we encounter such a example of the paradox when using measure theory, e.g to analyze integrability.&lt;/p&gt;&#xA;" OwnerUserId="7920" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-08T19:08:01.727" LastActivityDate="2011-03-08T19:08:01.727" Title="Russell's paradox and the foundations of measure theory" Tags="&lt;measure-theory&gt;" AnswerCount="1" />
  <row Id="25772" PostTypeId="2" ParentId="25771" CreationDate="2011-03-08T15:12:32.867" Score="11" ViewCount="" Body="&lt;p&gt;The only practical difference (as long as one doesn't work near to the foundations of mathematics at least) between naive set theory and axiomatic set theory is that the axioms say that there are certain things you just can't do - such as unrestricted comprehension and sets have to be wellfounded, and neither crops up in measure theory since the most complicated set theoretic &quot;stuff&quot; you do starts with a set $X$, considers $\sigma$-algebras on $X$, i.e. certain subsets of $\mathcal{P}(X)$, the power set of $X$, so the collection of all $\sigma$-algebras on $X$ is a subset of $\mathcal{P}(\mathcal{P}(X))$. In particular you never run into unrestricted comprehension because you're always guaranteed to have a set to do comprehension in, and these &quot;nested structures&quot; that lead to non-wellfoundedness don't happen either for essentially the same reason.&lt;/p&gt;&#xA;" OwnerUserId="1269" LastActivityDate="2011-03-08T15:12:32.867" CommentCount="1" />
  <row Id="25773" PostTypeId="2" ParentId="25770" CreationDate="2011-03-08T15:52:12.000" Score="1" ViewCount="" Body="&lt;p&gt;Hint:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume $c_1 \le c_2 \le \dots \le c_n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you try to find a $k$ such that $c_1 + \dots + c_{k-1}$, $c_k$ and $c_{k+1} + \dots + c_n$ form the sides of a triangle?&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2011-03-08T15:52:12.000" CommentCount="6" />
  <row Id="25774" PostTypeId="1" AcceptedAnswerId="25775" CreationDate="2011-03-08T16:09:49.517" Score="1" ViewCount="73" Body="&lt;p&gt;Hey guys, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have this question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $f$ be a Polynomial so that $\mathbb{Q}[x] \ni f=ax+b$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With the initial information: $f(\sqrt{2})=0$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can someone prove that $f=0$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="6163" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2011-03-08T17:26:17.173" LastActivityDate="2011-03-08T17:26:17.173" Title="Polynomial ring" Tags="&lt;linear-algebra&gt;" AnswerCount="1" />
  <row Id="25775" PostTypeId="2" ParentId="25774" CreationDate="2011-03-08T16:18:10.243" Score="7" ViewCount="" Body="&lt;p&gt;Suppose $f(\sqrt{2})=a\sqrt{2}+b=0$, where $a$ and $b$ are rational numbers. If $a\neq 0$, then after rearranging, we have $\sqrt{2}=\frac{-b}{a}$; but $\frac{-b}{a}$ is rational (since $a$ and $b$ are) and $\sqrt{2}$ is not, which is a contradiction. Thus, $a=0$, and we have $0\cdot\sqrt{2}+b=b=0$, so $b=0$. Thus $f=0\cdot x+0=0$.&lt;/p&gt;&#xA;" OwnerUserId="264" LastActivityDate="2011-03-08T16:18:10.243" CommentCount="4" />
  <row Id="25776" PostTypeId="2" ParentId="25766" CreationDate="2011-03-08T16:23:15.833" Score="3" ViewCount="" Body="&lt;p&gt;Apostolos: About the first question, yes. The point is $(*)$: If $(M,\dot\in)$ is a model of set theory, $\phi$ is a sentence, and $(M,\dot\in)\models$&quot;$(N,E)$ is a model of $\phi$&quot;, then $(N,E)\models\phi$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;More precisely, $(N,E)^*\models\phi$, where $(N,E)^*$ is the model that $M$ thinks $(N,E)$ is. Here, $(N,E)^*$ has universe $\{a\in M\mid (M,\dot\in)\models a\dot\in N\}$ and its relation is $\{(a,b)\in M\times M\mid (M,\dot\in)\models a,b\dot\in N\land aE b\}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(*)$ is easily established by an induction on the complexity of formulas. Now, if $\phi$ is an axiom of ZFC, then $(M,\dot\in)$ thinks that $\phi$ is an axiom of ZFC (more precisely, if $n$ is a Gödel number for $\phi$, then in $M$, the numeral corresponding to $n$ is a Gödel number for a formula, and that formula is precisely $\phi$), and therefore, $(N,E)^*$ satisfies all ZFC axioms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that $M$ may fail to be an $\omega$-model, in which case there are &quot;natural numbers&quot; in $M$ that code what $M$ believes are ZFC axioms, and $M$ may believe that $(N,E)$ satisfies them. We do not care about these &quot;fake formulas&quot;. Similarly, there may be an $(N,E)$ in $M$ that $M$ thinks is not a model of ZFC but $(N,E)^*$ is, in fact, a ZFC model. The reason is similar: $M$ may think that $(N,E)$ does not satisfy one of the fake axioms, but this does not matter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the above, it doesn't matter whether $(N,E)$ is a proper class or a set in the sense of $M$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The argument by induction in the complexity of formulas is perfectly general, so in fact $((L,\in)^M)^*$ is a model of every $\phi$ that follows from ZFC$+V=L$. In particular, GCH, diamonds, squares, etc, hold in $(L^M)^*$. Of course, there are also additional properties this model has that are not provable from ZFC$+V=L$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;About the second question: Suppose that CH fails, and let $A$ be a subset of $\omega_2$ that codes an injective $\omega_2$-sequence of reals, say the $\alpha$-th real is coded in $A\cap(\omega\cdot\alpha,\omega\cdot(\alpha+1))$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then $L[A]$ cannot possibly be a model of GCH, because $A\in L[A]$ so $L[A]$ sees at last $\omega_2^V\ge\omega_2^{L[A]}$ many different reals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you see, the usual proof of CH breaks down because we cannot ensure that for every real there is a countable $\alpha$ such that the real belongs to $L_\alpha[A]$. In fact, if we are careful we could even have a situation where $\omega_1=\omega_1^L$, ${\mathfrak c}=\omega_2$, $A$ codes all the reals, and $L_{\omega_1}[A]=L_{\omega_1}$, i.e., none of the &quot;new&quot; reals are visible in $L_{\omega_1}[A]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The issue is that $A$ may not collapse correctly, meaning, what you do is take $\delta$ large and a countable elementary $X\prec L_\delta[A]$ that contains $r$. Then you collapse $X$. Its collapse may not be $L_\alpha[A]$ for any $\alpha$, because the collapse of $A$ needs not be $A$. Of course, an initial segment of $A$ coincides with an initial segment of its collapse, but its collapse may &quot;code information faster&quot;, in particular, it will code $r$ by a countable stage. &lt;/p&gt;&#xA;" OwnerUserId="462" LastActivityDate="2011-03-08T16:23:15.833" CommentCount="4" />
  <row Id="25777" PostTypeId="2" ParentId="25761" CreationDate="2011-03-08T16:31:16.697" Score="5" ViewCount="" Body="&lt;p&gt;I have seen a bit of Bourbaki's structure definition. The Bourbaki group defines structure roughly as a collection of sets with functions and relations on them. They take as an example a topological space which is a set together with some subset of its power set. Bourbaki's structures are anything which can be defined in this manner, like groups, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Category theory studies classes of objects and their morphisms. It tries to classify constructions based on abstract properties of morphisms and diagrams. For instance, in category we have a description of a product without ever using &quot;elements&quot;, and this definition of a product applies to all categories; whether or not a product actually exists is another issue. Sometimes it does, and sometimes it doesn't.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Category theory does not give us an obvious way of constructing familiar structures like groups in the first place, although there is interesting mathematics behind how far one can go just with the concept of a category.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suggest forgetting Bourbaki altogether unless (a) you need a specific result or (b) you are interested in historical treatments. Since Bourbaki covered much, what you should choose instead depends on what you want to learn. If you are interested in Category Theory, keep looking at Mac Lane or try Awodey's book. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, if you are interested in analyzing mathematics from its models and structure, try model theory. It's a branch of logic that studies the types of models that can occur for a given set of axioms and how those models differ, both from a first-order perspective (i.e. what can you distinguish just by first-order sentences) and from an outside perspective (are they isomorphic?). David Marker's book on model theory is a nice introduction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are just interested in specific structures like groups, rings, etc. read some abstract algebra. You should have a good familiarity with abstract algebra, which will help you understand why category theory is so important. Depending on how much algebra you know, picking up an introductory book like Rotman's on Homological Algebra might interest you. You can also look at Weibel's book on Homological Algebra, which is my favourite, although it is more advanced.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most graduate algebra textbooks introduce some really basic category theory and use it in algebra. Jacobson's Basic Algebra 2 is pretty nice and includes various diagram definitions.&lt;/p&gt;&#xA;" OwnerUserId="2055" LastActivityDate="2011-03-08T16:31:16.697" />
  <row Id="25778" PostTypeId="2" ParentId="25493" CreationDate="2011-03-08T16:38:31.197" Score="1" ViewCount="" Body="&lt;p&gt;Well, if $R=S$, then an $S$-split extension is obviously also $R$-split, so $\operatorname{Ext}^1_{(R,S)}$ is always zero. On the other hand, $\operatorname{Ext}^1_{R}$ can be pretty much anything you like.&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2011-03-08T16:38:31.197" />
  <row Id="25779" PostTypeId="1" AcceptedAnswerId="25819" CreationDate="2011-03-08T16:44:06.950" Score="2" ViewCount="182" Body="&lt;p&gt;How can we prove that all change-of-basis matrices are invertible? The trivial case when it's a change of basis for $\mathbb{R^{n}}$ is easily demonstratable using, for example, determinants. But I am struggling to rigorously show this for all bases, for example for a two-dimensional subspace of $\mathbb{R^{4}}$. I am sure that there are many ways to go about this proof, and I would be very appreciative for as many ways of demonstration as possible, to back up my intuition!&lt;/p&gt;&#xA;" OwnerUserId="7961" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-08T19:18:43.180" LastActivityDate="2011-03-08T20:39:02.210" Title="Rigorously proving that a change-of-basis matrix is always invertible" Tags="&lt;linear-algebra&gt;" AnswerCount="1" CommentCount="14" FavoriteCount="1" />
  <row Id="25780" PostTypeId="1" AcceptedAnswerId="25789" CreationDate="2011-03-08T16:52:06.780" Score="1" ViewCount="78" Body="&lt;p&gt;I need to find integer x with which function's y gets lowest integer values&#xA;$$f(x)=\frac{x^2-x-17}{x-2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried to find derivative, but it never equals 0.&#xA;Other steps was to change expression&#xA;$$f(x)=\frac{x^2-x-17}{x-2}=1+\frac{(x-5)(x+3)}{x-2}=x-2+\frac{3(x-7)}{x-2}$$&#xA;But didn't notice any solutions&lt;/p&gt;&#xA;" OwnerUserId="7149" LastActivityDate="2011-03-09T22:26:17.840" Title="Find lowest integer value with integer x of given function" Tags="&lt;functions&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="25781" PostTypeId="2" ParentId="25761" CreationDate="2011-03-08T17:00:57.700" Score="0" ViewCount="" Body="&lt;p&gt;My take on this is that category theory is a well-defined mathematical theory with a small set of definitions and axioms specifically for describing and proving things for categories, with many instances of these theorems applicable throughout mathematics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, the Bourbaki school really provides an -approach- to all mathematics that is very formal. It's more of an approach, a general way of presentation than anything else. There's no Bourbaki 'theory' (collection of theorems). The Bourbaki approach is an attempt to present traditional mathematics in a particular manner.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So with the Bourbaki approach, you'll get a kind of culture of doing mathematics, but with category theory you'll get category specific theorems (which have their application to other specific fields of mathematics).&lt;/p&gt;&#xA;" OwnerUserId="1919" LastActivityDate="2011-03-08T17:00:57.700" />
  <row Id="25782" PostTypeId="1" AcceptedAnswerId="25790" CreationDate="2011-03-08T17:00:59.633" Score="2" ViewCount="86" Body="&lt;p&gt;hello im having trouble showing the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;let $u$ be a positive measure. if $\int_E f\, du= \int_E g\, du$ for all measurable $E$ then $f=g$ a.e.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i was trying to argue by contradiction: if $f\neq g$ a.e. then there must exist some set $E=\{x: f(x)\neq g(x)\}$ such that $u(E) \gt 0$. Then let $E^+=\{x: f(x)\gt g(x)\}$ and $E^-=\{x: f(x)\lt g(x)\}$. Now, if $E^+$ or $E^-$ is measurable and have positive measure then $\int_{E^+} f\, du \gt \int_{E^+} g\, du$ or $\int_{E^-} f\, du \lt \int_{E^-} g\, du$, contradiction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you can see, the argument hinges on $E^+$ or $E^-$ being measurable. This is the part im having trouble with.&lt;/p&gt;&#xA;" OwnerUserId="6556" LastActivityDate="2011-03-08T17:33:24.357" Title="measure theory question" Tags="&lt;measure-theory&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" />
  <row Id="25784" PostTypeId="2" ParentId="25763" CreationDate="2011-03-08T17:03:54.447" Score="0" ViewCount="" Body="&lt;p&gt;If B has constant velocity, then A can predict B's location at any point in time.  A can then calculate the maximum distance it can cover in any amount of time, including the need to slow and stop.  Let $a$ be the maximum acceleration of A and $v$ be the maximum velocity.  So if A starts with zero velocity, it can accelerate at $a$ for a time span of $\frac{v}{a}$, covering  distance $\frac{v^2}{2a}$ in the process.  Then it can go along at $v$ as long as it wants, then decelerate and stop.  So in time $t$ it can go $\frac{v^2}{a} + v(t-\frac{2v}{a})$ for time greater than $\frac{2v}{a}$.  A can find a point where B will be at that time and go there.  Some correction is needed for short times, but they should be easy to figure out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Added in response to comment:  Working in 2D, The position of B is $(B_x+v_{Bx}t,B_y+v_{By}t)$.  Assume A starts at the origin.  Then the distance from where A is now to where B will be is $\sqrt{(B_x+v_{Bx}t)^2+(B_y+v_{By}t)^2}$.  So we solve for $t$ in $\sqrt{(B_x+v_{Bx}t)^2+(B_y+v_{By}t)^2}=\frac{v^2}{a} + v(t-\frac{2v}{a})$ and meet B there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If A is moving at $t=0$, the same philosophy will work, but things are much messier.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2011-03-08T18:08:10.807" LastActivityDate="2011-03-08T18:08:10.807" CommentCount="5" />
  <row Id="25785" PostTypeId="2" ParentId="25782" CreationDate="2011-03-08T17:11:22.247" Score="4" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;hint&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The difference of two measurable functions is measurable&lt;/li&gt;&#xA;&lt;li&gt;$(0,\infty)\subset\mathbb{R}$ is Borel, so for a measurable function $F$, the set on which it takes positive values is a measurable set. &lt;/li&gt;&#xA;&lt;li&gt;The collection of measurable sets form a $\sigma$ algebra, and in particular intersection of two measurable sets is measurable. &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="1543" LastActivityDate="2011-03-08T17:11:22.247" CommentCount="6" />
  <row Id="25786" PostTypeId="2" ParentId="25444" CreationDate="2011-03-08T17:15:57.563" Score="1" ViewCount="" Body="&lt;p&gt;Let the speeds be $s_1 \le s_2 \le \cdots \le s_n$ (in laps per unit time).  Relative to vehicle $k$, vehicle $j$ moves at a rate of $s_j - s_k$.  Assuming the vehicles are presently uniformly and independently distributed around the road, the expected number of overtakings in a period of time $t$ therefore is $t|s_j - s_k|$.  Vehicles $1, 2, \ldots, k-1$ will be overtaken and vehicles $k+1, k+2, \ldots, n$ will overtake vehicle $k$.  Therefore:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(a) expected number of times vehicle $k$ overtakes others = $t\sum_{j=1}^{k-1}(s_k-s_j)$;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(b) expected number of times vehicle $k$ is overtaken = $t\sum_{j=k+1}^{n}(s_j-s_k)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Empty sums equal zero, of course.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When, as Ross Millikan suggests, vehicle $k$ moves at the mean speed of traffic, the sums (a) and (b) must be equal, because $s_k = \frac{1}{n}\sum_{j=1}^{n}s_j$ is algebraically equivalent to (b) - (a) equalling zero.&#xA;Clearly the implication works in the other direction, too.&lt;/p&gt;&#xA;" OwnerUserId="1489" LastActivityDate="2011-03-08T17:15:57.563" />
  <row Id="25788" PostTypeId="2" ParentId="25780" CreationDate="2011-03-08T17:21:26.900" Score="0" ViewCount="" Body="&lt;p&gt;For $n &amp;lt; -13$ we have $n+1 &amp;lt; f(n) &amp;lt; n+2$ and so  $f(n)$ is not an integer for integer $n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $n &amp;gt; 17$ we have $n &amp;lt; f(n) &amp;lt; n+1$ amd so $f(n)$ is not an integer for integer $n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you just have to check the 31 integers remaining.  The integer values are $f(-13)=-11$, $f(-3)=1$, $f(-1)=5$, $f(1)=17$, $f(3)=-11$, $f(5)=1$, $f(7)=5$, $f(17)=17$.  So the integer minima are $-11$ when $n=-13$ or $n=3$.&lt;/p&gt;&#xA;" OwnerUserId="6460" LastActivityDate="2011-03-08T17:21:26.900" />
  <row Id="25789" PostTypeId="2" ParentId="25780" CreationDate="2011-03-08T17:23:26.450" Score="4" ViewCount="" Body="&lt;p&gt;If you want $x$ and $f(x)$ to both be integers, you have a Diophantine equation.  Expressing $f(x)=x+1-\frac{15}{x-2}$ is the key, as you need the fraction to be integral.  The only ways to have that is for $x-2$ to equal $\pm 1,3,5$, or $15$ and you can just try all eight.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2011-03-08T17:50:32.027" LastActivityDate="2011-03-08T17:50:32.027" />
  <row Id="25790" PostTypeId="2" ParentId="25782" CreationDate="2011-03-08T17:27:13.367" Score="2" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Hint&lt;/strong&gt; We have that $f$ and $g$ map into $\mathbb{R}$ from some unknown measure space, say $(X,\mathcal{M})$.  Let $h(x)=f(x)-g(x)$.  Then $E^+\subset h^{-1}(0,\infty)$ and $E^- \subset h^{-1}(-\infty,0)$.  Then recall that the sum of two measurable functions is measurable (addition is continuous), and $h$ being $\mathcal{M}$-measurable is equivalent to $f^{-1}(a,\infty)\in\mathcal{M}$ for every $a\in \mathbb{R}$ and it is also equivalent to $f^{-1}(-\infty, a)\in\mathcal{M}$ for every $a\in \mathbb{R}$. (Since these sets generate the Borel sigma algebra over $\mathbb{R}$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then conclude $h^{-1}(0,\infty)$ and $h^{-1}(-\infty,0)$ are measurable, and integrate over them.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope that helps,&lt;/p&gt;&#xA;" OwnerUserId="6075" LastEditorUserId="6075" LastEditorDisplayName="" LastEditDate="2011-03-08T17:33:24.357" LastActivityDate="2011-03-08T17:33:24.357" CommentCount="3" />
  <row Id="25791" PostTypeId="1" AcceptedAnswerId="25804" CreationDate="2011-03-08T17:27:16.050" Score="5" ViewCount="208" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;The ordered pair $(a,b)$ is defined to be the set $\{\{a\},\{a,b\}\}$.&quot; ~ Hungerford's &lt;em&gt;Algebra&lt;/em&gt; (p.6)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I think this is the first time that i've seen this definition. I've read the &lt;a href=&quot;http://en.wikipedia.org/wiki/Ordered_pair&quot; rel=&quot;nofollow&quot;&gt;wiki page&lt;/a&gt;. Is it defined this way, as opposed to a definition relating to functions as in a Cartesian product, because this definition is considered more elementary (or foundational) being that it is related directly to sets?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, the definition of an ordered $n$-tuple, according to the &lt;a href=&quot;http://en.wikipedia.org/wiki/N-tuple&quot; rel=&quot;nofollow&quot;&gt;wiki page&lt;/a&gt; seems vague (perhaps i'm misunderstanding it). For an ordered triple it gives the example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(1,2,3) = \{\{(1,2)\},\{(1,2),3\}\}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but how do we know this is &lt;strong&gt;not&lt;/strong&gt; the ordered pair $((1,2),3)$? Or is the difference between $(1,2,3)$ and $((1,2),3)$ considered trivial?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thirdly, and perhaps unrelated, what does it mean for &lt;a href=&quot;http://math.stackexchange.com/questions/14828/set-theoretic-definition-of-numbers&quot;&gt;a natural number to be defined&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$2_{\mathbb{N}} = \{\emptyset,\{\emptyset\} \},$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and is this also done so that we can define $\mathbb{N}$ in terms of sets?&lt;/p&gt;&#xA;" OwnerUserId="7840" LastEditorUserId="7840" LastEditorDisplayName="" LastEditDate="2011-03-08T17:44:09.757" LastActivityDate="2011-03-08T20:13:32.980" Title="Definition of an Ordered Pair" Tags="&lt;elementary-set-theory&gt;&lt;definition&gt;" AnswerCount="5" CommentCount="6" />
  <row Id="25792" PostTypeId="1" CreationDate="2011-03-08T17:35:45.697" Score="2" ViewCount="90" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am looking for some good text/reference on complex Fourier series resp. Fourier analysis for complex (in particular holomoprhic) functions (of one variable). The more it contains on this particular subject, the better.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Background: For my diploma thesis, I need in particular to understand asymptotics of the Fourier coefficients for certain entire functions, so I need to study it fast, that is, more straightforward, well-structured theory without much &quot;bla-bla&quot;, and less exercises... Nevertheless, I would like to learn the more general theory of &lt;strong&gt;Fourier analysis for complex/holomorphic functions&lt;/strong&gt; as it has a great deal of applications in Analytic Number Theory, which is one of the subjects of interest to me. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance!&lt;/p&gt;&#xA;" OwnerUserId="5994" LastEditorUserId="5994" LastEditorDisplayName="" LastEditDate="2011-03-08T18:59:42.143" LastActivityDate="2011-03-08T18:59:42.143" CommunityOwnedDate="2011-03-08T17:46:13.467" Title="Please recommend good text on complex Fourier series/analysis" Tags="&lt;complex-analysis&gt;&lt;fourier-analysis&gt;" CommentCount="7" FavoriteCount="1" />
  <row Id="25793" PostTypeId="2" ParentId="25791" CreationDate="2011-03-08T17:35:49.773" Score="3" ViewCount="" Body="&lt;p&gt;We want to define the most with as little as possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That way we only define what sets are, and by that we define ordered pairs, and so on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The usual way is to define an ordered pair $\langle a,b\rangle = \{\{a\},\{a,b\}\}$. This is just because it's easy to work with.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can define an ordered pair as the image of a function from the domain which is the power set of the power set of the empty set, the first element is the image of $\emptyset$ and the second is the image of $\{\emptyset\}$. (Yes, functions are usually defined as collections of ordered pairs. I'm talking about existence of a formula with two free variables.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again, these are just conventions and we work with that we find comfortable and as clear as possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the second issue, we only define pairs, but there is a natural identification between $\langle a,\langle b,c\rangle\rangle$ and $\langle a,b,c\rangle$ and of course $\langle\langle a,b\rangle, c\rangle$. So once again we only define as little as possible and somewhat abuse our own notation because we know that the formal backbone exists and is strong.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And lastly, as I said before, we want to define the most with as little as possible. In the world of set theory it's nice to have only sets. So we define $0=\emptyset$, and inductively we can define the natural numbers in terms of sets, so $n=\{0,\ldots,n-1\}$.&lt;/p&gt;&#xA;" OwnerUserId="622" LastEditorUserId="7840" LastEditorDisplayName="" LastEditDate="2011-03-08T18:56:25.270" LastActivityDate="2011-03-08T18:56:25.270" />
  <row Id="25794" PostTypeId="1" AcceptedAnswerId="25799" CreationDate="2011-03-08T17:39:10.330" Score="0" ViewCount="147" Body="&lt;p&gt;If I am correct in stating that a factorial of a number ( of entities ) is the number of ways in which those entities can be arranged, then my question is as simple as asking - how do you conceive the idea of &lt;em&gt;arranging&lt;/em&gt; &lt;strong&gt;nothing&lt;/strong&gt; ? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Its easy to conceive of a null element in the context of arrays, for example - so you say that there is only one way to present a null element. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, in layman terms - if there are three humans h1, h2, h3 that need to be arranged to sit on three chairs c1, c2, c3 - then how do you conceive of a) a null human, and b) to arrange those ( that ? ) null humans ( human ? ) on the three chairs ? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please note that referral to humans is just for easy explanation - not trying to be pseudo-philosophical. Three balls to be arranged on three corners of a triangle works just fine. So basically, how do you conceive of an object that doesn't exist, and &lt;strong&gt;then&lt;/strong&gt; conceive of &lt;strong&gt;arranging&lt;/strong&gt; that object ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, in essence ... is 0! = 1, a convenience for mathematicians ? Not that its the only convenience, but just asking. Of course, there are many. &#xA;If yes, then its a pity that I can't find it stated like so anywhere. &#xA;If not, then can anybody suggest resources to read actual, good proofs ? &lt;/p&gt;&#xA;" OwnerUserId="7963" LastActivityDate="2011-03-08T18:02:33.030" Title="Factorial of 0 - a convenience ?" Tags="&lt;permutations&gt;&lt;factorial&gt;" AnswerCount="4" CommentCount="6" />
  <row Id="25795" PostTypeId="2" ParentId="25791" CreationDate="2011-03-08T17:41:11.310" Score="0" ViewCount="" Body="&lt;p&gt;For your first question, you are correct. An ordered pair defined as &quot;syntactic sugar&quot; for the set construction is more fundamental.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For your second question, when we think about a $k$-tuple, the fundamental property is that we have &quot;projections&quot; from the $k$-tuple to a particular element of the tuple. In other words, all that we essentially require is that we can somehow discern one element from another, and there is an ordering.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With the definition given, $((1,2),3)$ is indeed equivalent to $(1,2,3)$. This is &lt;em&gt;almost&lt;/em&gt; like Lisp where &lt;code&gt;(1 2 3)&lt;/code&gt; is equivalent to &lt;code&gt;(1 . (2 . (3 . ())))&lt;/code&gt;. The principle is the same; we are just making a sort of binary tree. As you might imagine, $(1,2,3,4)$ would be equivalent to $(((1,2),3),4)$ by the definition given.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is really an unfortunate problem with notation, and not so much the definition of tuples themselves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For your last question, basically we can give a formal &lt;strong&gt;construction&lt;/strong&gt; to the natural numbers by using more primitive pieces of mathematics. You can think of it as just a particular encoding for natural numbers --- or even a &lt;em&gt;canonical&lt;/em&gt; encoding if you want. Another encoding might be a natural number defined as a composition of &quot;successor&quot; functions, etc. But of course they're all isomorphic.&lt;/p&gt;&#xA;" OwnerUserId="2666" LastActivityDate="2011-03-08T17:41:11.310" />
  <row Id="25796" PostTypeId="2" ParentId="25791" CreationDate="2011-03-08T17:42:12.367" Score="0" ViewCount="" Body="&lt;p&gt;As regards your first question, yes, your intuition that this way of defining things is more set-theory-like and foundational is correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since in a set no ordering is defined (i.e., the set $\{a,b\}$ is the same as the set $\{ b, a \}$ ), if you want to express everything just in terms of sets you have to fix a convention to do so. So, in a sense, writing $\{\{a\},\{a,b\}\}$ is a bit like saying, &quot;we are interested in the set $\{a, b\}$, but with $a$ in a distinguished role&quot;.&lt;/p&gt;&#xA;" OwnerUserId="6507" LastActivityDate="2011-03-08T17:42:12.367" />
  <row Id="25797" PostTypeId="2" ParentId="25794" CreationDate="2011-03-08T17:52:30.330" Score="1" ViewCount="" Body="&lt;p&gt;The definition can be motivated by the $\Gamma$-function, but let's ignore that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can think of $0!$ as an empty product if we wish, the notion of &quot;nothing&quot; times &quot;nothing&quot; is $1$. But how do we arrive at this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instead of $1\times 2\times 3\times 4$, let's write $\times(1,2,3)$. I suppose you agree that we could break this up into arbitrary sub-products: $\times(1,2,3) = \times(\times(1,2),\times(3))$. Naturally, we can interpret $\times(3)$ to just be $3$. But suppose we broke this up even more: $\times(\times(3),\times())$. What would the most natural choice of $\times()$ be? Of course, $1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of factorial, we can interpret $1!$ to be $\times(1, \times()) = 1\cdot(0!)=1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For other binary operations, we can determine the &quot;empty&quot;-operation by a simple rule-of-thumb: if we have the commutative operation $\oplus$ (just a generic operation), we can define the &quot;empty-$\oplus$&quot; as the value of $\iota$ such that $\forall x.\, \iota \oplus x = x$ --- that is, the $\oplus$-identity element.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Examples: $\max() = -\infty$, $\min() = \infty$, $+() = 0$, $\gcd() = 0$, etc.&lt;/p&gt;&#xA;" OwnerUserId="2666" LastEditorUserId="2666" LastEditorDisplayName="" LastEditDate="2011-03-08T17:58:07.000" LastActivityDate="2011-03-08T17:58:07.000" CommentCount="4" />
  <row Id="25798" PostTypeId="1" AcceptedAnswerId="25806" CreationDate="2011-03-08T17:53:33.393" Score="1" ViewCount="118" Body="&lt;p&gt;can someone give an insight on the following problem? I'm not sure how to start the problem. It's a practice problem for &quot;mean value theorem&quot; and &quot;Taylor's Theorem&quot; so I'm assuming they might be necessary for the proof. Thanks!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let$f: \mathbb{R} \to \mathbb{R}$ be a function. Suppose that $f$ is differentiable, that $f(0)=1$, and that $|f&amp;#39;(x)| \leq 1$ for all $x \in \mathbb{R}$. Prove that $|f(x)| \leq |x|+1$ for all $x \in \mathbb{R}$.&lt;/p&gt;&#xA;" OwnerUserId="7429" LastActivityDate="2011-03-08T18:03:20.547" Title="practice problem from Mean Value Theorem in Real Analysis" Tags="&lt;real-analysis&gt;&lt;analysis&gt;" AnswerCount="3" />
  <row Id="25799" PostTypeId="2" ParentId="25794" CreationDate="2011-03-08T17:57:05.810" Score="3" ViewCount="" Body="&lt;p&gt;I don't see any difficulty with the idea of &quot;arranging a set of zero things&quot;...&#xA;There is no need to &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;conceive of an object that doesn't exist, and then conceive of arranging that object&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;mostly, because that does not make much sense at all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I give you 2 apples and I tell you to arrange them in a line above the table, there are 2 possible outcomes. If I give you zero apples, how many possible possible things can I find on the table after you are done? Exactly one: no apples on the table.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&amp;nbsp;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the way, &lt;em&gt;every&lt;/em&gt; definition in math is done out of convenience. &lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2011-03-08T17:57:05.810" CommentCount="6" />
  <row Id="25800" PostTypeId="2" ParentId="25794" CreationDate="2011-03-08T17:58:20.493" Score="3" ViewCount="" Body="&lt;p&gt;There is exactly one way to arrange nothing: the null arrangement. You've misused your chair analogy: when you arrange null humans, you do it on &lt;em&gt;null&lt;/em&gt; chairs, and there is exactly one way to do this. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps the following alternate definition will make things clearer. Suppose I have $n$ cards labeled $1, 2, ... n$ in order, and I shuffle them. Then $\frac{1}{n!}$ is the probability that they will stay in the original order. If $n = 0$, this probability is $1$ since $0$ cards can only be arranged in one possible order. &lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2011-03-08T17:58:20.493" />
  <row Id="25801" PostTypeId="2" ParentId="25798" CreationDate="2011-03-08T17:58:31.160" Score="3" ViewCount="" Body="&lt;p&gt;As an example, I will show how to prove something different using the mean value theorem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$|\cos y - \cos x|  \leq |y - x|$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The mean value theorem says that for $x \lt y$, there is some $c \in (x,y)$ such that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f(y) - f(x) = (x-y) f&amp;#39;(c)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, when when $f(x) = \cos x$ we have that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\cos y - \cos x = (y-x) (-\sin c)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now if we take the modulus,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$|\cos y - \cos x| = |y - x| |\sin c| \leq |y - x|$, (as $|\sin c| \leq 1$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus using the mean value theorem, we have proven that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$|\cos y - \cos x|  \leq |y - x|$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do you see how to solve your problem now?&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2011-03-08T17:58:31.160" />
  <row Id="25802" PostTypeId="1" AcceptedAnswerId="25808" CreationDate="2011-03-08T18:00:25.403" Score="1" ViewCount="113" Body="&lt;p&gt;What is difference between &quot;recognizable&quot; and &quot;decidable&quot; in context of Turing machines?&lt;/p&gt;&#xA;" OwnerUserId="7965" LastActivityDate="2011-03-08T18:19:16.673" Title="Recognizable vs Decidable" Tags="&lt;computer-science&gt;&lt;computability&gt;" AnswerCount="1" />
  <row Id="25803" PostTypeId="2" ParentId="25798" CreationDate="2011-03-08T18:00:34.277" Score="0" ViewCount="" Body="&lt;p&gt;Try using the mean value theorem: for any $x$ we know there is some $a$ between $0$ and $x$ such that $$\frac{f(x)-1}{x} = f&amp;#39;(a)$$.  Rearranging gives $f(x) = 1 + xf&amp;#39;(a)$.  Now take absolute values and use the triangle inequality.&lt;/p&gt;&#xA;" OwnerUserId="9" LastActivityDate="2011-03-08T18:00:34.277" />
  <row Id="25804" PostTypeId="2" ParentId="25791" CreationDate="2011-03-08T18:01:09.897" Score="4" ViewCount="" Body="&lt;p&gt;You may be interested in reading Kuratowski's &quot;Set Theory&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's what I remember from it:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;First one defines pairs $\langle a,b\rangle = \{\{a\},\{a,b\}\}$, with this definition one can define $A\times B$ as the set of all pairs $(a,b)$ with $a\in A$ and $b\in B$. However, that's not a good way to proceed, because of the problems you note.&lt;/li&gt;&#xA;&lt;li&gt;With this definition one defines $\prod_{i\in I} X_i$ as the set of functions $f\colon I \to X_i$ such that $f(i)\in X_i$, here $\{X_i \mid i\in I\}$ is a collection of sets (in other words a function $I\to \mathcal P(\cup X_i)$).&lt;/li&gt;&#xA;&lt;li&gt;In particular $A^2$ is the set of functions $2\to A$, where $2 = \{\varnothing, \{\varnothing\}\}$, and $A\times B$ is the set of functions $f\colon 2 \to A\cup B$ where $f(\varnothing)\in A$ and $f(\{\varnothing\})\in B$. &lt;/li&gt;&#xA;&lt;li&gt;Now we forget about that first definition, and proceed with the latter. (Even though we use the former definition to state the latter!) The practical advantage is that now in fact $A\times B\times C$ is actually well-defined, just like any other product, no matter how large the index-set $I$. &lt;/li&gt;&#xA;&lt;li&gt;It is still not true that $(A\times B)\times C = A\times (B\times C)$, and in fact both are still different from $A\times B\times C$. However, there are bijections between these three sets that are so obvious that for all practical purposes one may consider them to be equal.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="5189" LastActivityDate="2011-03-08T18:01:09.897" CommentCount="3" />
  <row Id="25805" PostTypeId="2" ParentId="25794" CreationDate="2011-03-08T18:02:33.030" Score="1" ViewCount="" Body="&lt;p&gt;Building on Qiaochu's answer, set theory defines functions from $X$ to $Y$ as particular subsets of $X \times Y$ : a subset $f \subset X \times Y$ is a function if $\forall x \in X \exists ! y \in Y, (x,y) \in f$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you use this definition, then there is exactly one function from $\emptyset$ to $\emptyset$, which is $\emptyset$ (and it's also a bijection !), so there is exactly one permutation of the empty set.&lt;/p&gt;&#xA;" OwnerUserId="6270" LastActivityDate="2011-03-08T18:02:33.030" />
  <row Id="25806" PostTypeId="2" ParentId="25798" CreationDate="2011-03-08T18:03:20.547" Score="1" ViewCount="" Body="&lt;p&gt;First i'll look at the positive reals (I leave the negative ones for you)&#xA;Lets work by contradiction.  Suppose there is a point, $x_0$ such that $f(x_0)&amp;gt;1+x_0$. (Note: Here $|x|=x$ since we are dealing with positive reals)  Then we have $$\frac{f(x_0)-f(0)}{x_0}&amp;gt;\frac{x_0+1-1}{x_0}=1$$ and hence by the mean value theorem, there is a point $c\in (0,x_0)$ such that $f^&amp;#39;(c)&amp;gt;1$ which is a contradiction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope that helps, &lt;/p&gt;&#xA;" OwnerUserId="6075" LastActivityDate="2011-03-08T18:03:20.547" CommentCount="1" />
  <row Id="25808" PostTypeId="2" ParentId="25802" CreationDate="2011-03-08T18:12:57.193" Score="4" ViewCount="" Body="&lt;p&gt;A language is Recognizable iff there is a Turing Machine which will halt and accept &lt;em&gt;only&lt;/em&gt; the strings in that language and for strings not in the language, the TM either rejects, or does not halt at all. Note: there is no requirement that the Turing Machine should halt for strings not in the language. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A language is Decidable iff there is a Turing Machine which will accept strings in the language and reject strings not in the language.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps this link will be helpful: &lt;a href=&quot;http://kilby.stanford.edu/~rvg/154/handouts/decidability.html&quot; rel=&quot;nofollow&quot;&gt;http://kilby.stanford.edu/~rvg/154/handouts/decidability.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2011-03-08T18:19:16.673" LastActivityDate="2011-03-08T18:19:16.673" CommentCount="2" />
  <row Id="25809" PostTypeId="1" CreationDate="2011-03-08T18:15:11.780" Score="0" ViewCount="60" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Possible Duplicate:&lt;/strong&gt;&lt;br&gt;&#xA;  &lt;a href=&quot;http://math.stackexchange.com/questions/13129/automorphism-of-the-field-of-rational-functions&quot;&gt;Automorphism of the Field of rational functions.&lt;/a&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do I show: $Aut_K(K(t))$ is isomorphic to $PGL_2(K)$, here $K$ is any field and $t$ indeterminate. Actually how do I show all K-automorphisms take the form $t \mapsto \frac{\alpha_1+t\alpha_2}{\alpha_2+t\alpha_4}$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many thanks.&lt;/p&gt;&#xA;" OwnerUserId="7966" LastActivityDate="2011-03-08T18:15:11.780" ClosedDate="2011-03-08T19:19:48.063" Title="Isomorphisms of Aut(K(t))" Tags="&lt;field-theory&gt;" CommentCount="1" />
  <row Id="25811" PostTypeId="1" CreationDate="2011-03-08T18:24:19.217" Score="1" ViewCount="93" Body="&lt;p&gt;I am currently woking on a program where I simulate the following problem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine you have an empty tank with a volume of $100$ litres and every $p$ seconds you add $m$ litres of water to it, beginning from the time of $s$ seconds (we may assume $s \leq p$). Now other people also add water to the tank with corresponding values $\hat p,\hat m,\hat s$ and so on. The task is to find out at which time the tank will be full (it can happen that its volume is actually exceeded).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried to formulate the problem mathematically and noticed that the following function can be used to model the added water of one person:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f(x,m,p,s)=m \cdot \lfloor \frac{p - s + x}{p} \rfloor$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have for example $m=20,p=3,s=1$ you will get the function&#xA;$f(x,20,3,1)=20 \left\lfloor \frac{x+2}{3}\right\rfloor$&#xA;we will also consider the function&#xA;$g(x,20,3,1)=20 \frac{x+2}{3}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/LpnwJ.png&quot; alt=&quot;Function graph&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The red function is $f(x)$, the blue function is $g(x)$. It can be seen that the linear function $g(x)$ perfectly solves our problem to see that the tank will be full after 13 seconds. However, lets add an additional person who helps filling the tank like $\hat f(x)=f(x, 10, 3, 2)$ and $\hat g(x)$ set analog to above. Then we get:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/pTZeH.png&quot; alt=&quot;Plot of the new situation&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where the red graph is $f(x)+\hat f(x)$ and the blue line is $g(x)+\hat g(x)$. Now we cannot simply use the linear functions to solve the problem and the step functions are very nasty to solve, it is funny that Mathematica already fails at the situation where only one person adds water ( NSolve[20 Floor[(2 + x)/3] &gt;= 100, x] returns &quot;This system cannot be solved with the methods available to NSolve.&quot; ).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I hope you can help me to find a way for finding the time where the tank is full given an amount of people who add water to the tank and their vales $p,m,s$. I don't want a closed formula, a way to algorithmically solve the problem would already be enough. Because currently I simulate the problem step by step and when there is a person who adds very tiny amounts of water very frequently you have to run a very high amount of iterations to get the exact result.&lt;/p&gt;&#xA;" OwnerUserId="3123" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2011-03-08T18:29:34.803" LastActivityDate="2011-03-08T18:29:34.803" Title="Sum of periodic step functions" Tags="&lt;algorithms&gt;" CommentCount="11" FavoriteCount="2" />
  <row Id="25812" PostTypeId="1" AcceptedAnswerId="25813" CreationDate="2011-03-08T18:29:45.960" Score="4" ViewCount="141" Body="&lt;p&gt;Say $z \in \mathbb{C}$ and $\bar{z}$ the complex conjugate (i.e. with $\bar{z} z = \left|z \right|^2$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can a function of $z$ and $\bar{z}$ be analytical?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example: $f(z,\bar{z}) = Az^3 + B \bar{z} z$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I thought no, because the partial derivatives will depend on the direction in the complex plane (i.e. the phase of the line along which you take the derivative limit).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="939" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2011-03-08T18:38:33.733" LastActivityDate="2011-03-08T21:44:37.040" Title="Analyticity of a function depending on $z$ and $\bar{z}$" Tags="&lt;complex-analysis&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="1" />
  <row Id="25813" PostTypeId="2" ParentId="25812" CreationDate="2011-03-08T18:32:07.393" Score="8" ViewCount="" Body="&lt;p&gt;One of the many equivalent definitions for a function to be holomorphic is $\displaystyle \frac{\partial f}{\partial \bar{z}} = 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle \frac{\partial f}{\partial \bar{z}} = 0$ is equivalent to &lt;a href=&quot;http://en.wikipedia.org/wiki/Cauchy%E2%80%93Riemann_equations&quot; rel=&quot;nofollow&quot;&gt;Cauchy Riemann&lt;/a&gt; equations as shown below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x = \frac{z+\bar{z}}{2} \text{ and } y = \frac{z-\bar{z}}{2i}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\partial f}{\partial \bar{z}} = \frac{\partial f}{\partial x} \frac{\partial x}{\partial \bar{z}} + \frac{\partial f}{\partial y} \frac{\partial y}{\partial \bar{z}} = \frac{1}{2} \left( \frac{\partial f}{\partial x} + i \frac{\partial f}{\partial y} \right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if $f = u(x,y) + i v(x,y)$, where $u,v: \mathbb{R} \times \mathbb{R} \rightarrow \mathbb{R}$, then&#xA;$$\frac{\partial f}{\partial x} = \frac{\partial u}{\partial x} + i \frac{\partial v}{\partial x}$$&#xA;$$\frac{\partial f}{\partial y} = \frac{\partial u}{\partial y} + i \frac{\partial v}{\partial y}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence,$$\frac{\partial f}{\partial \bar{z}} = \frac{1}{2} \left( \frac{\partial u}{\partial x} - \frac{\partial v}{\partial y} \right) + \frac{i}{2} \left( \frac{\partial u}{\partial y} + \frac{\partial v}{\partial x} \right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, you find that $$\left( \frac{\partial f}{\partial \bar{z}} = 0 \right) \iff \left(\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y} \text{ and } \frac{\partial u}{\partial y} = - \frac{\partial v}{\partial x} \right)$$&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2011-03-08T21:44:37.040" LastActivityDate="2011-03-08T21:44:37.040" CommentCount="6" />
  <row Id="25814" PostTypeId="1" CreationDate="2011-03-08T18:34:01.037" Score="1" ViewCount="104" Body="&lt;p&gt;Is there any theorem that apllies to non simple cycles  and chromatic number?&#xA;For example we know that x(G)=2 if G does not contains odd cycles.&#xA;What about a non-simple cycle that contains  odd cycles.&#xA;E.g A square and a triangle having a common vertex is (n=6,3-chromatic)&#xA;    A petagon and a triangle having a common vertex is (n=7, 3-chromatic)&#xA;Is any proof for saying that at least 3 colors are needed to use for non simple cycles that contain an odd cycle?And is that related to the number of its vertices?&#xA;PS it is not a homework but basic to use it for one&#xA;Thnx in advance&lt;/p&gt;&#xA;" OwnerUserId="7843" LastActivityDate="2011-03-08T23:50:23.347" Title="Chromatic number and non-simple cycles" Tags="&lt;graph-theory&gt;" AnswerCount="1" />
  <row Id="25816" PostTypeId="2" ParentId="25814" CreationDate="2011-03-08T18:37:17.757" Score="2" ViewCount="" Body="&lt;p&gt;It is a well known theorem that a graph is bipartite (2 colourable) if and only if all its cycles are of even length.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if you have a graph which has an odd cycle, it will require at least 3 colours.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance you can find a proof here: &lt;a href=&quot;http://books.google.com/books?id=aR2TMYQr2CMC&amp;amp;pg=PA18&quot; rel=&quot;nofollow&quot;&gt;http://books.google.com/books?id=aR2TMYQr2CMC&amp;amp;pg=PA18&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In any case, for a simpler proof of: To a cycle of odd length, to each vertex you cannot assign one of two colours so that adjacent vertices get different colours&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Say you try and walk around the cycle. Each step you take, you flip the colour. So to get back to the original vertex, you need to take an even number of steps. That is not possible with an odd cycle. So you need at least 3 colours.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that if the graph you have is exactly two cycles which share exactly one vertex (like in your examples), then that graph is 3 colourable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is because a cycle is 3 colourable, even if you pre-assign a vertex with a given colour. You colour one cycle first, then based on the colour of the common vertex, you colour the other cycle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, though, determining the chromatic number of a graph is a hard problem. In fact, it is NP-Complete.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2011-03-08T23:50:23.347" LastActivityDate="2011-03-08T23:50:23.347" CommentCount="5" />
  <row Id="25817" PostTypeId="2" ParentId="25504" CreationDate="2011-03-08T19:04:39.547" Score="4" ViewCount="" Body="&lt;p&gt;After a discussion with Russell Miller, I've got an answer to the question. There is in fact an approachable real that is not provably approachable. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let us adopt PA as the base theory, although the construction can easily adapt to other theories. I will describe a Turing machine $M$ that produces a convergent sequence of rational numbers, but any Turing machine $M&amp;#39;$ that PA proves to produce a convergent sequence of rational numbers converges to a different number than $M$ does. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The idea is to diagonalize against any such proofs that may be discovered. Fix an enumeration of the Turing machines $M_n$. Consider the following Turing machine $M$. Our machine will at stage $k$ produce a rational number $r_k$ by giving finitely many trinary  digits, but using only the digits 0 and 1 and not 2, to avoid non-unique readibility issues. As the construction proceeds, we systematically enumerate all possible proofs from the theory. We may find at some stage $k$ that there is a proof that Turing machine $M_n$ produces a convergent sequence of rational numbers. In this case, we run $M_n$ for $k$ steps, getting the current rational $q_{n,k}$ approximation to the real to which $M_n$ is converging. In this case, if $r_k$ is different from $q_{n,k}$ by digit $n$, then we use $r_{k+1}=r_k$; otherwise, we produce $r_{k+1}$ by flipping the $n$-th digit of $r_k$ from $0$ to $1$ or vice versa to ensure that $r_{k+1}$ is different from $q_{n,k}$. (Note, we are flipping the $n$-th digit, not the $k$-th digit, so each machine $M_n$ is tied to the $n$-th digit of our limit real.) And simply proceed with this plan, considering the new proofs as they appear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that each machine $M_n$ that provably produces a convergent sequence will be considered infinitely many times, for increasingly large $k$, since there are many proofs that it does so. So the relevant $k$ for $M_n$ will become arbitrarily large. Since $M_n$ was proved to produce a convergent sequence, it follows that the values of $q_{n,k}$ really do converge, and thus eventually stabilize in their first $n$ bits (if those bits are all $0$s and $1$s). Thus, we will flip the $n$-th bit of our rational $r_k$ at most finitely many times. It follows that our sequence $r_k$ is a convergent sequence of rational numbers. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But also, it follows that whenever there is a proof that some machine $M_n$ produces a convergent sequence of rational numbers, then the limit of our real will differ from the limit of that machine by digit $n$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, we have an approachable real that is not provably approachable, as desired.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me remark on the confusing subtle point here about in which theory we have proved our machine to produce a convergent sequence of rationals. The answer is that we have done so in any theory that knows that any statement that is provable in the base theory is in fact true, since we needed to know that when we found a proof that $M_n$ produces a converging sequence of rationals, that those rationals really did converge. For example, ZFC has this relation to PA, since ZFC proves that if $\varphi$ is provable in PA, then $\varphi$ is true. But more generally, for any $\omega$-consistent theory $T$, we may extend to a stronger theory $T^+$ that knows this implication.&lt;/p&gt;&#xA;" OwnerUserId="413" LastEditorUserId="413" LastEditorDisplayName="" LastEditDate="2011-03-08T19:58:12.200" LastActivityDate="2011-03-08T19:58:12.200" CommentCount="2" />
  <row Id="25818" PostTypeId="1" AcceptedAnswerId="25832" CreationDate="2011-03-08T19:15:37.320" Score="0" ViewCount="78" Body="&lt;p&gt;Let $\{a_k\}$ be an infinite sequence, with $\sum_{k=\infty}^\infty \vert a_k\vert^2&amp;lt;\infty$. Let $f(\omega)=\sum_{k=-\infty}^\infty a_k e^{ik\omega}$ be its Fourier series. By Plancheral's theorem, $f$ is bounded. How do I get a reasonable bound for $|f|$, i.e., if $|f|\leq g$, how do I find $g$?&lt;/p&gt;&#xA;" OwnerUserId="7815" LastActivityDate="2011-03-08T20:49:18.487" Title="Bound for Fourier series" Tags="&lt;fourier-series&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="25819" PostTypeId="2" ParentId="25779" CreationDate="2011-03-08T19:35:04.470" Score="5" ViewCount="" Body="&lt;p&gt;What &lt;em&gt;is&lt;/em&gt; a change-of-basis matrix? You have a vector space $\mathbf{V}$ (and it doesn't matter if $\mathbf{V}$ is all of $\mathbb{R}^n$, or some subspace thereof, or even something entirely different), and two different ordered bases for $\mathbf{V}$, $\beta_1$ and $\beta_2$ (necessarily of the same size, since two bases of the same vector space always have the same size):&#xA;\begin{align*}&#xA;\beta_1 &amp;amp;= \Bigl[ \mathbf{v}_1,\mathbf{v}_2,\ldots,\mathbf{v}_n\Bigr]\\&#xA;\beta_2 &amp;amp;= \Bigl[ \mathbf{w}_1,\mathbf{w}_2,\ldots,\mathbf{w}_n\Bigr].&#xA;\end{align*}&#xA;A &quot;change of basis&quot; matrix is a matrix that translates from $\beta_1$ coordinates to $\beta_2$ coordinates. That is, $A$ is a change-of-basis matrix (from $\beta_1$ to $\beta_2$) if, given the coordinate vector $[\mathbf{x}]_{\beta_1}$ of a vector $\mathbf{x}$ relative to $\beta_1$, then $A[\mathbf{x}]_{\beta_1}=[\mathbf{x}]_{\beta_2}$ gives the coordinate vector of $\mathbf{x}$ relative to $\beta_2$, for all $\mathbf{x}$ in $\mathbf{V}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do we get a change-of-basis matrix? We write each vector of $\beta_1$ in terms of $\beta_2$, and these are the columns of $A$:&#xA;\begin{align*}&#xA;\mathbf{v}_1 &amp;amp;= a_{11}\mathbf{w}_1 + a_{21}\mathbf{w}_2+\cdots+a_{n1}\mathbf{w}_n\\&#xA;\mathbf{v}_2 &amp;amp;= a_{12}\mathbf{w}_1 + a_{22}\mathbf{w}_2 +\cdots + a_{n2}\mathbf{w}_n\\&#xA;&amp;amp;\vdots\\&#xA;\mathbf{v}_n &amp;amp;= a_{1n}\mathbf{w}_1 + a_{2n}\mathbf{w}_2 + \cdots + a_{nn}\mathbf{w}_n.&#xA;\end{align*}&#xA;We know we can do this because $\beta_2$ is a basis, so we can express any vector (in particular, the vectors in $\beta_1$) as linear combinations of the vectors in $\beta_2$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then the change-of-basis matrix translating from $\beta_1$ to $\beta_2$ is&#xA;$$ A = \left(\begin{array}{cccc}&#xD;&#xA;a_{11} &amp;amp; a_{12} &amp;amp; \cdots &amp;amp; a_{1n}\\&#xD;&#xA;a_{21} &amp;amp; a_{22} &amp;amp; \cdots &amp;amp; a_{2n}\\&#xD;&#xA;\vdots &amp;amp; \vdots &amp;amp; \ddots &amp;amp; \vdots \\&#xD;&#xA;a_{n1} &amp;amp; a_{n2} &amp;amp; \cdots &amp;amp; a_{nn}&#xD;&#xA;\end{array}\right).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is $A$ always invertible? Because just like there is a change-of-basis from $\beta_1$ to $\beta_2$, there is also a change-of-basis from $\beta_2$ to $\beta_1$. Since $\beta_1$ is a basis, we can express every vector in $\beta_2$ using the vectors in $\beta_1$:&#xA;\begin{align*}&#xA;\mathbf{w}_1 &amp;amp;= b_{11}\mathbf{v}_1 + b_{21}\mathbf{v}_2 + \cdots + b_{n1}\mathbf{v}_n\\&#xA;\mathbf{w}_2 &amp;amp;= b_{12}\mathbf{v}_2 + b_{22}\mathbf{v}_2 + \cdots + b_{n2}\mathbf{v}_n\\&#xA;&amp;amp;\vdots\\&#xA;\mathbf{w}_n &amp;amp;= b_{1n}\mathbf{v}_n + b_{2n}\mathbf{v}_n + \cdots + b_{nn}\mathbf{v}_n.&#xA;\end{align*}&#xA;So the matrix $B$, with&#xA;$$B = \left(\begin{array}{cccc}&#xD;&#xA;b_{11} &amp;amp; b_{12} &amp;amp; \cdots &amp;amp; b_{1n}\\&#xD;&#xA;b_{21} &amp;amp; b_{22} &amp;amp; \cdots &amp;amp; b_{2n}\\&#xD;&#xA;\vdots &amp;amp; \vdots &amp;amp; \ddots &amp;amp; \vdots\\&#xD;&#xA;b_{n1} &amp;amp; b_{nn} &amp;amp; \cdots &amp;amp; b_{nn}&#xD;&#xA;\end{array}\right),$$&#xA;has the property that given any vector $\mathbf{x}$, if $[\mathbf{x}]_{\beta_2}$ is the coordinate vector of $\mathbf{x}$ relative to $\beta_2$, then $B[\mathbf{x}]_{\beta_2}=[\mathbf{x}]_{\beta_1}$ is the coordinate vector of $\mathbf{x}$ relative to $\beta_1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But now, consider what the matrix $BA$ does to the standard basis of $\mathbb{R}^n$ (or $\mathbf{F}^n$, in the general case): what is $BA\mathbf{e}_i$, where $\mathbf{e}_i$ is the vector that has a $1$ in the $i$th coordinate and zeros elsewhere? It's a matter of interpreting this correctly: $\mathbf{e}_i$ is the coordinate vector relative to $\beta_1$ of $\mathbf{v}_i$, because $[\mathbf{v}_i]={\beta_1}=\mathbf{e}_i$. Therefore, since $A[\mathbf{x}]_{\beta_1} = [\mathbf{x}]_{\beta_2}$ and $B[\mathbf{x}]_{\beta_2}=[\mathbf{x}]_{\beta_1}$ for every $\mathbf{x}$, we have:&#xA;$$BA\mathbf{e}_i = B(A\mathbf{e}_i) = B(A[\mathbf{v}_i]_{\beta_1}) = B[\mathbf{v}_i]_{\beta_2} = [\mathbf{v}_i]_{\beta_1} = \mathbf{e}_i.$$&#xA;That is, $BA$ maps $\mathbf{e}_i$ to $\mathbf{e}_i$ for $i=1,\ldots,n$. The only way for this to happen is if $BA=I_n$ is the identity. The same argument, now interpreting $\mathbf{e}_i$ as $[\mathbf{w}_i]_{\beta_2}$, shows that $AB$ is also the identity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So $A$ and $B$ are both invertible. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So every change-of-basis matrix is necessarily invertible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It doesn't really matter if you are considering a subspace of $\mathbb{R}^N$, a vector space of polynomials or functions, or any other vector space. So long as it is finite dimensional (so that you can define the &quot;change-of-basis&quot; matrix), change-of-basis matrices are always invertible. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Added.&lt;/em&gt; I just saw the comment where you give the definition you have of change-of-basis matrix: &lt;em&gt;a matrix $C$ which, when multiplied by a matrix $B$ whose columns form a basis of a certain subspace, produces another matrix $A$ whose columns form a basis for the same subspace.&lt;/em&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This matrix $C$ is just the matrix that expresses the columns of $A$ in terms of the columns of $B$. That is, it's the change-of-basis matrix from &quot;columns-of-A&quot; coordinates to &quot;columns-of-B&quot; coordinates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, take the subspace of $\mathbb{R}^4$ given by $x=z$ and $y=w$, with basis &#xA;$$\left(\begin{array}{c}1\\0\\1\\0\end{array}\right),\quad\left(\begin{array}{c}0\\1\\0\\1\end{array}\right),$$&#xA;and now consider the same space, but with basis&#xA;$$\left(\begin{array}{c}1\\1\\1\\1\end{array}\right),\quad \left(\begin{array}{r}1\\-2\\1\\-2\end{array}\right).$$&#xA;The matrix $C$ such that&#xA;$$ \left(\begin{array}{rr}&#xD;&#xA;1 &amp;amp; 1\\&#xD;&#xA;1 &amp;amp; -2\\&#xD;&#xA;1 &amp;amp; 1\\&#xD;&#xA;1 &amp;amp; -2&#xD;&#xA;\end{array}\right) = C\left(\begin{array}{cc}&#xD;&#xA;1 &amp;amp; 0\\&#xD;&#xA;0 &amp;amp; 1\\&#xD;&#xA;1 &amp;amp; 0\\&#xD;&#xA;0 &amp;amp; 1&#xD;&#xA;\end{array}\right)$$&#xA;is obtained by writing each vector in the columns of $A$ in terms of the columns of $B$:&#xA;\begin{align*}&#xA;\left(\begin{array}{r}&#xA;1\\1\\1\\1\end{array}\right) &amp;amp;= 1\left(\begin{array}{c}1\\ 0\\ 1\\ 0\end{array}\right) + 1\left(\begin{array}{c}0 \\ 1 \\ 0 \\ 1\end{array}\right),\\&#xA;\left(\begin{array}{r}&#xA;1\\ -2\\ 1\\ -2\end{array}\right) &amp;amp;= 1\left(\begin{array}{c}1\\0\\1\\0\end{array}\right) -2\left(\begin{array}{c}0\\1\\0\\1\end{array}\right).&#xA;\end{align*}&#xA;And so, the matrix $C$ is &#xA;$$C = \left(\begin{array}{rr}&#xD;&#xA;1 &amp;amp; 1\\&#xD;&#xA;1 &amp;amp; -2&#xD;&#xA;\end{array}\right).$$&#xA;Expressing the columns of $B$ in terms of the columns of $A$ give the inverse:&#xA;\begin{align*}&#xA;\left(\begin{array}{c}1\\ 0\\ 1\\ 0\end{array}\right) &amp;amp;= &#xA;\frac{2}{3}\left(\begin{array}{c}1 \\ 1\\ 1\\ 1\end{array}\right) + \frac{1}{3}\left(\begin{array}{r}1 \\ -2\\ 1\\ -2\end{array}\right)\\&#xA;\left(\begin{array}{c}0\\ 1\\ 0\\ 1\end{array}\right) &amp;amp;= \frac{1}{3}\left(\begin{array}{c} 1\\ 1\\ 1\\ 1\end{array}\right) -\frac{1}{3}\left(\begin{array}{r}1\\ -2\\ 1\\ -2\end{array}\right),&#xA;\end{align*}&#xA;so the inverse of $C$ is:&#xA;$$C^{-1} = \left(\begin{array}{rr}&#xD;&#xA;\frac{2}{3} &amp;amp; \frac{1}{3}\\&#xD;&#xA;\frac{1}{3} &amp;amp; -\frac{1}{3}&#xD;&#xA;\end{array}\right),$$&#xA;which you can verify by multiplying by $C$. &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="6622" LastEditorDisplayName="" LastEditDate="2011-03-08T20:39:02.210" LastActivityDate="2011-03-08T20:39:02.210" CommentCount="11" />
  <row Id="25820" PostTypeId="1" AcceptedAnswerId="25823" CreationDate="2011-03-08T19:39:14.170" Score="2" ViewCount="95" Body="&lt;p&gt;(Someone may please change the title if they can think of a better one)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have a Hilbert Space $\mathcal{H}$ that consists of all functions $\psi(x)$ such that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\int_{-\infty}^{\infty} |\psi(x)|^2 dx \lt \infty $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I have to show that there are functions in $\mathcal{H}$  such that $Q\phi(x)=x\psi(x)$ is not in $\mathcal{H}$ . Does it suffice if I think of a counterexample.. but even then, the only two square integrables i can think of are the Gaussian and the dirac-delta &quot;function&quot;. For both the cases, the corresponding $x\psi(x)$ is also L2. Could someone give me more examples of functions that are  square integrable over the entire real axis?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore, if I now consider the function space $\Omega$ with the infinite set of conditions, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\int_{-\infty}^{\infty} |\psi(x)|^2 (1+|x|^n)dx \lt \infty$ for $n=0,1,2,\cdots$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have to show now, that for any $\psi(x)$ in $\Omega$ the function $Q\phi(x)=x\psi(x)$ is also in $\Omega$. I can think of a weak argument like &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\int_{-\infty}^{\infty} |Q\psi(x)|^2 (1+|x|^n)dx = \int_{-\infty}^{\infty} |\psi(x)|^2 (x^2+|x|^{n+2})dx \approx \int_{-\infty}^{\infty} |\psi(x)|^2 (1+|x|^n)dx \lt \infty $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The last step because we would be concerned only with the behaviour of $x$ when it is very large, as the function would have to go to zero at large $x$ for it to be in $\Omega$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not so sure about this, and face a basic conceptual difficulty. Can we conclude if  $\int_{-\infty}^{\infty} |\psi(x)|^2 dx \lt \infty $ then $\psi(x)$ should fall off as $1/x^2$? &lt;/p&gt;&#xA;" LastActivityDate="2011-03-08T20:11:15.893" Title="Hilbert spaces, square integrability etc..." Tags="&lt;functional-analysis&gt;&lt;hilbert-spaces&gt;" AnswerCount="2" CommentCount="6" FavoriteCount="1" />
  <row Id="25821" PostTypeId="1" AcceptedAnswerId="25826" CreationDate="2011-03-08T19:49:18.587" Score="0" ViewCount="44" Body="&lt;p&gt;For an assignment, I've been asked to find the $\frac{dy}{dx}$ for the formula $$2 \cos(4x) \sin(9y)=7$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The major problem I'm having with understanding implicit differentiation is understanding what distinguishes a typical $\frac{d}{dx}$ from $\frac{dy}{dx}$  An example I came across gave the process of $$\frac{d}{dx}[xy^{2}]= x\frac{d}{dx}[y^{2}]+y^{2}\frac{d}{dx}[x]= x(2y\frac{dy}{dx}+y^{2}(1)= 2xy\frac{dy}{dx}+y^{2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Applying that approach to the above formula, I came up with&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$2 \cos(4x) \sin (9y)=7 \rightarrow \frac{d}{dx}[2 \cos(4x) \sin(9y)]= \frac{d}{dx}(7)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Applying the product rule to the elements:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$2[\cos(4x) \frac{d}{dx}\sin(9y)]+[\sin(9y)\frac{d}{dx} \cos(4x)]=0$$&#xA;$$2[\cos(4x)9 \cos(9y)\frac{dy}{dx}]+[\sin(9y)(-4 \sin(4x))\frac{dy}{dx}]=0$$&#xA;$$18 \cos(4x)\cos(9y)\frac{dy}{dx} + \sin(9y)-4 \sin(4x)\frac{dy}{dx}=0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, am I on the right track as far as assigning the correct $\frac{dy}{dx}$ terms?&#xA;Assuming I am, how do I separate the $\frac{dy}{dx}$ terms from their respective functions?  Because if I divide $18 \cos(4x)$ to isolate $\sin(9y)$, dividing against zero will produce zero.&lt;/p&gt;&#xA;" OwnerUserId="7006" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-08T20:11:10.157" LastActivityDate="2011-03-08T20:33:56.277" Title="Help with Implicit Differentation and Trigonometry" Tags="&lt;calculus&gt;&lt;homework&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="25822" PostTypeId="1" AcceptedAnswerId="25828" CreationDate="2011-03-08T19:49:24.497" Score="0" ViewCount="245" Body="&lt;p&gt;How can I prove, that a polynomial function $$f(x) = \sum_{0\le k \le n}a_k x^k\qquad n\in\mathbb N,\ a_k\in\mathbb C$$ is zero for at most $n$ different values of $x$? (Except $n=0$ where $f(x)$ is $0$ in all cases)&lt;/p&gt;&#xA;" OwnerUserId="5282" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2011-03-08T20:25:45.063" LastActivityDate="2011-03-08T20:33:32.813" Title="How to prove that a polynomial of degree $n$ has at most $n$ roots?" Tags="&lt;polynomials&gt;" AnswerCount="2" CommentCount="12" />
  <row Id="25823" PostTypeId="2" ParentId="25820" CreationDate="2011-03-08T19:56:12.980" Score="3" ViewCount="" Body="&lt;p&gt;Hint: Think along lines of Cauchy distribution function. Say $\psi(x) = \frac{1}{\sqrt{1+x^2}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the second part of your question, you could define your function space $\Omega$ to be functions such that $\displaystyle \int_{-\infty}^{\infty} \left| \psi(x) \right|^2 \left|x \right|^n dx &amp;lt; \infty$, for $n \in \mathbb{N}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Your function space and the function space defined in the above line are the same. Since if $\displaystyle \int_{-\infty}^{\infty} \left| \psi(x) \right|^2 \left|x \right|^n dx &amp;lt; \infty$, for $n \in \mathbb{N}$, then $\displaystyle \int_{-\infty}^{\infty} \left| \psi(x) \right|^2 dx &amp;lt; \infty$ by taking $n=0$ and adding the two gives your definition and also the function satisfying your definition also fits in the definition given in this answer since $\displaystyle \int_{-\infty}^{\infty} \left| \psi(x) \right|^2 \left|x \right|^n dx \leq \displaystyle \int_{-\infty}^{\infty} \left| \psi(x) \right|^2 (1 + \left|x \right|^n) dx &amp;lt; \infty$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The conclusion now is trivial since $2 \in \mathbb{N}$ and if $n \in \mathbb{N}$, then so does $n+2$&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2011-03-08T20:11:15.893" LastActivityDate="2011-03-08T20:11:15.893" CommentCount="2" />
  <row Id="25824" PostTypeId="2" ParentId="25820" CreationDate="2011-03-08T19:57:03.263" Score="1" ViewCount="" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Also I know I can construct piecewise functions, but is there something better than that?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I don't know what's wrong with piecewise functions, but how about $f(x)=1/(1+|x|)$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As Akhil mentioned, the Dirac delta function is not in $L^2$.  A densely defined linear functional on a space is typically a very different object from an element of the space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To show that $Q(\Omega)\subseteq\Omega$, note that $x^2+|x|^{n+2}\leq 1+|x|^{n+3}$ outside of a bounded interval, and $x^2+|x|^{n+2}$ is bounded on bounded intervals.  &lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2011-03-08T19:57:03.263" CommentCount="6" />
  <row Id="25825" PostTypeId="2" ParentId="20969" CreationDate="2011-03-08T19:58:20.197" Score="1" ViewCount="" Body="&lt;p&gt;One can prove this by convention (that is based on what 'works', based on base cases for product of a list).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But for a proof from 'nothing' as it were...define what factorial is supposed to mean. I think of it as the number of one-to-one, onto (=bijective) functions from a set of size $n$ to itself. if$n$ is 0 then the set is the empty set. How many bijective functions are there from $\emptyset$ to $\emptyset$? (a function is a set of ordered pairs (with some restrictions)? There are no legal pairs for these restrictions, and so there are no legal subsets of these pairs...other than the empty set. So $\emptyset$ -is- a function (it -is- a set of pairs (empty of course), all of whose pairs satisfy the function criteria). No other functions will work so $\emptyset$ is the only function that works. So there is only 1 bijective function on a set of size 0. So $0! = 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yes, this is weird, but it works. Negative numbers, complex numbers, they're all weird even when you just manipulate their properties. But you'll get over it.&lt;/p&gt;&#xA;" OwnerUserId="1919" LastEditorUserId="1919" LastEditorDisplayName="" LastEditDate="2011-03-08T22:14:48.677" LastActivityDate="2011-03-08T22:14:48.677" />
  <row Id="25826" PostTypeId="2" ParentId="25821" CreationDate="2011-03-08T20:02:43.020" Score="1" ViewCount="" Body="&lt;p&gt;Going from $\frac{d}{dx}[2 \cos(4x) \sin(9y)]=0$ to $2[\cos(4x) \frac{d}{dx}\sin(9y)]+[\sin(9y)\frac{d}{dx} \cos(4x)]=0$ you lost a factor $2$ on the second term.  Then, considering only the second term $[\sin(9y)\frac{d}{dx} \cos(4x)]=[\sin(9y)(-4\sin(4x)]$ there should not be a $\frac{dy}{dx}$ as the argument of the $\cos$ does not depend on $y$, just on $x$.  From your next to last line to the last line you lost some parentheses that subtracted the terms instead of multiplying them.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-03-08T20:02:43.020" CommentCount="3" />
  <row Id="25828" PostTypeId="2" ParentId="25822" CreationDate="2011-03-08T20:12:14.947" Score="3" ViewCount="" Body="&lt;p&gt;Using Linear Algebra,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the $n+1$ distinct roots are $\alpha_i$, then we have that $x = [a_0, a_1, \dots, a_n]^{T}$ is a solution of $Ax = 0$ where $A$ is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Vandermonde_matrix&quot; rel=&quot;nofollow&quot;&gt;Vandermonde matrix&lt;/a&gt; using the $\alpha_i$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the Vandermonde matrix is invertible for distinct $\alpha_i$, it follows that $x = [0, 0, \dots, 0]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus if $a_j \neq 0$ for some $j$, then your polynomial can have at most $n$ different roots.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: This is basically saying that given a field $K$, any polynomial of degree $n$ in $K[x]$ has at most $n$ distinct roots.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fundamental Theorem of Algebra is an assertion of the fact that $\mathbb{C}$ is algebraically closed, and the $K$ above need not be algebraically closed.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2011-03-08T20:25:45.053" LastActivityDate="2011-03-08T20:25:45.053" />
  <row Id="25829" PostTypeId="2" ParentId="25791" CreationDate="2011-03-08T20:13:32.980" Score="0" ViewCount="" Body="&lt;p&gt;The need for an explicit definition of ordered pair comes from the simplicity of he concept of set: there is no order defined in the set axioms - an object is either a member of a set or not. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But it is a convenient mechanism to be able to have the first of a series (and the second also). And the definition of 'ordered pair' allows that. Of course you have to show that, whatever your definition of 'ordered pair' is, you can show that &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;if $(a, b) = (c, d)$ then $a = c$ and $b=d$, &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;which you can do with the given definition (the notation $\{b,a,d,c\}$ is the same set as $\{a,b,c,d\}$, but  $(b,a,d,c)$ is different from $(a,b,c,d)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To get n-tuples, it should be straightforward that (though not equivalent in underlying set notation) one can say $(a,b,c)$ can be implemented using ordered pairs as $((a,b),c)$ or $(a, (b,c))$ where you can extract the first element easily, but also to get the second element of the triple you get the first element of the $(b,c)$ which is the second element of the pair $(a,(b,c))$.&lt;/p&gt;&#xA;" OwnerUserId="1919" LastActivityDate="2011-03-08T20:13:32.980" />
  <row Id="25830" PostTypeId="2" ParentId="25821" CreationDate="2011-03-08T20:23:29.783" Score="1" ViewCount="" Body="&lt;p&gt;Two mistakes: you lost a factor of $2$, and you introduced a spurious $\frac{dy}{dx}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{dy}{dx}$ only comes in when you are differentiating a function of $y$ (or $y$ itself), because $y$ is itself a function of $x$, and so the Chain Rule applies; you don't know what the derivative of $y$ is, so you leave it indicated with $\frac{dy}{dx}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It might be less confusing if you use $y&amp;#39;$ instead of $y$ in this instance. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, we have:&#xA;\begin{align*}&#xA;2\cos(4x)\sin(9y) &amp;amp;= 7\\&#xA;\frac{d}{dx}\Bigl(2\cos(4x)\sin(9y)\Bigr) &amp;amp;= \frac{d}{dx}7\\&#xA;2\frac{d}{dx}\Bigl(\cos(4x)\sin(9y)\Bigr) &amp;amp;= 0\quad\mbox{(factor out a $2$)}\\&#xA;2\Biggl( \cos(4x)\Bigl(\frac{d}{dx}\sin(9y)\Bigr) + \sin(9y)\Bigl(\frac{d}{dx}\cos(4x)\Bigr)\Biggr) &amp;amp;= 0\quad\mbox{(product rule)}\\&#xA;2\Biggl( \cos(4x)\Bigl( \cos(9y)\frac{d}{dx}(9y)\Bigr) + \sin(9y)\Bigl(-\sin(4x)\frac{d}{dx}(4x)\Bigr)\Biggr) &amp;amp;=0\quad\mbox{(chain rule)}\\&#xA;2\Biggl( \cos(4x)\Bigl(\cos(9y)9\frac{d}{dx}(y)\Bigr) + \sin(9y)\Bigl(-\sin(4x)4\Bigr)\Biggr) &amp;amp;= 0 \quad\mbox{(chain rule)}\\&#xA;2\Biggl(\cos(4x)\Bigl(9\cos(9y)y'\Bigr) + \sin(9y)\Bigl(-4\sin(4x)\Bigr)\Biggr) &amp;amp;= 0 \quad\mbox{(because $\frac{d}{dx}y = y'$)}\\&#xA; 18\cos(4x)\cos(9y)y' - 8\sin(9y)\sin(4x) &amp;amp;= 0\quad\mbox{(algebra)}\\&#xA;18\cos(4x)\cos(9y)y' &amp;amp;= 8\sin(9y)\sin(4x)\\&#xA;y' &amp;amp;= \frac{8\sin(9y)\sin(4x)}{18\cos(4x)\cos(9y)}\\&#xA;\frac{dy}{dx} &amp;amp;= \frac{4}{9}\left(\frac{\sin(9y)}{\cos(9y)}\right)\left(\frac{\sin(4x)}{\cos(4x)}\right)\\&#xA;\frac{dy}{dx} &amp;amp;= \frac{4}{9}\tan(9y)\tan(4x).&#xA;\end{align*}&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-08T20:33:56.277" LastActivityDate="2011-03-08T20:33:56.277" />
  <row Id="25831" PostTypeId="2" ParentId="25822" CreationDate="2011-03-08T20:33:32.813" Score="5" ViewCount="" Body="&lt;p&gt;You don't need the fundamental theorem of algebra or the Vandermonde determinant, only the &lt;strong&gt;factor theorem&lt;/strong&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proposition:&lt;/strong&gt; A polynomial of degree at most &lt;em&gt;n&lt;/em&gt; with more than &lt;em&gt;n&lt;/em&gt; roots vanishes identically.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Proof:&lt;/em&gt; By induction. The base case is $n=0$, which is obvious. Now take a polynomial &lt;em&gt;f&lt;/em&gt; of degree at most &lt;em&gt;n&lt;/em&gt;, and let $x_1,\ldots,x_{n+1}$ be distinct roots of &lt;em&gt;f&lt;/em&gt;. By the factor theorem, we can write $$f(x) = (x-x_{n+1})g(x)$$ where &lt;em&gt;g&lt;/em&gt; plainly has degree at most $n-1$. Now substitute $x = x_i$ for $i=1,\ldots,n$. For all these values of &lt;em&gt;x&lt;/em&gt; the left hand side vanishes and the factor $(x_i-x_{n+1})$ is nonzero. Hence all these $x_i$ must be roots of &lt;em&gt;g&lt;/em&gt; and by induction &lt;em&gt;g&lt;/em&gt; is identically zero. &lt;strong&gt;QED&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This same proof works over any field (or even integral domain).&lt;/p&gt;&#xA;" OwnerUserId="677" LastActivityDate="2011-03-08T20:33:32.813" CommentCount="1" />
  <row Id="25832" PostTypeId="2" ParentId="25818" CreationDate="2011-03-08T20:49:18.487" Score="1" ViewCount="" Body="&lt;p&gt;The fact that $\sum_{k \in Z} |a_k|^2$ is finite just means that $f(\omega)$ is an $L^2$ function, and Plancherel's theorem says that any $L^2$ function arises as $f(\omega)$ for some such sequence $\{a_k\}_{k \in Z}$. If you want $f(\omega)$ to be bounded, you have to assume further conditions on $\{a_k\}_{k \in Z}$. For example, if you know that $\sum_k |a_k|$ is finite as well, then by summing the absolute values of the terms you get that $|f(\omega)| \leq \sum_k |a_k|$.&lt;/p&gt;&#xA;" OwnerUserId="3035" LastActivityDate="2011-03-08T20:49:18.487" CommentCount="1" />
  <row Id="25833" PostTypeId="1" CreationDate="2011-03-08T21:06:21.937" Score="1" ViewCount="92" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;If $w(X)\leq n$ ($n$ is finite), and if $B_1$ is a base of $X$ such that $|B_1|\leq n$, then for any base $B$ of $X$ we have $B_1$ is contained in $B$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Can you help me in proving this fact?&lt;/p&gt;&#xA;" OwnerUserId="7970" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-08T23:37:43.940" LastActivityDate="2011-03-08T23:48:22.587" Title="topology question" Tags="&lt;general-topology&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="25835" PostTypeId="1" CreationDate="2011-03-08T21:13:43.287" Score="0" ViewCount="36" Body="&lt;p&gt;As a user is producing a series of point samples (by drawing on a graphics tablet), I want to generate a curve or series of curves that approximates (and smoothes) the samples. However, because he sees the results in real-time, I'd like the algorithm to be incremental - as I receive point N, I'd like the curve to be extended towards point N.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I currently have an implementation which fits a series of bezier curves to the entire set of points and although it produces a good-looking curve, it jiggles about too much as you're drawing. I feel an incremental update would be better.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What's a good algorithm or approach to take here?  &lt;/p&gt;&#xA;" OwnerUserId="7969" LastActivityDate="2011-03-08T21:13:43.287" Title="What's the best approach to incrementally extend an interpolated curve" Tags="&lt;algorithms&gt;&lt;bezier-curve&gt;" CommentCount="1" />
  <row Id="25836" PostTypeId="1" AcceptedAnswerId="25863" CreationDate="2011-03-08T21:17:33.647" Score="6" ViewCount="165" Body="&lt;p&gt;I am wondering if there are some studies about the number of reverses of direction to return to the starting point in random walk (either symmetric or non-symmetric), for example, its distribution and expectation etc..&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you.&lt;/p&gt;&#xA;" OwnerUserId="6016" LastActivityDate="2011-03-09T13:42:23.730" Title="number of reverses of direction to return in random walk" Tags="&lt;probability&gt;&lt;random&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="2" />
  <row Id="25839" PostTypeId="2" ParentId="25032" CreationDate="2011-03-08T21:59:13.433" Score="1" ViewCount="" Body="&lt;p&gt;I would say math is more a tool that you have to apply to the real world to make any practical use of it. For example, learn some physics, computer science, or engineering, then you will find yourself applying Mathematics in many real world examples.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A great example: if you master the priniples of physics (which requires a mastery of mathematics), then you can create the most realistic physics inside video games.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Math is only a framework from which to describe the real world. Staying strictly in mathematics is something I'd recommend not doing unless you want to teach mathematics. Find the area in the sciences that you like, and apply mathematics to it.&lt;/p&gt;&#xA;" OwnerUserId="7971" LastActivityDate="2011-03-08T21:59:13.433" />
  <row Id="25840" PostTypeId="1" AcceptedAnswerId="25854" CreationDate="2011-03-08T22:00:07.613" Score="2" ViewCount="65" Body="&lt;p&gt;Given the formula $x^{2}y^{2}-8x=3$, find the second derivative.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I calculated the first derivative as $$-\frac{xy^{2}+4}{x^{2}y}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Working from that, I calculated the second derivative starting with&#xA;$$\frac{([x^{2}y)\frac{d}{dx}(-xy^{2}+4)]-[(-xy^{2}+4)\frac{d}{dx}(x^{2}y)]}{(x^{2}y)^{2}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The left $\frac{d}{dx}$ was calculated by&#xA;$$\frac{d}{dx}(-xy^{2}+4)= [\frac{d}{dx}(-xy^{2})]+[\frac{d}{dx}(4)]= -x\frac{d}{dx}(y^{2})+ y^{2}\frac{d}{dx}(-x)= -2xy\frac{dy}{dx}- y^{2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The right $\frac{d}{dx}$ was calculated by&#xA;$$\frac{d}{dx}(x^{2}y)= x^{2}\frac{d}{dx}(y)+ y\frac{d}{dx}(x^{2})= x^{2}\frac{dy}{dx}+2xy$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Plugging everything into the formula resulted in&#xA;$$\frac{[(x^{2}y)(-2xy\frac{d}{dx}-y^{2}] - [(-xy^{2}+4)(x^2\frac{dy}{dx}+2xy)]}{(x^{2}y)^2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{[(-2x^{2}y^{2}\frac{dy}{dx}- 2x^{2}y^{3}]- [(x^{3}y^{2}\frac{dy}{dx}+2x^{2}y^{3}- 4x^2\frac{dy}{dx}-8xy]}{x^{4}y{2}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Combining like terms resulted in&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{-x^{3}y^{2}\frac{dy}{dx}-4x^{2}\frac{dy}{dx}-8xy}{x^{4}y^{3}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is where I stall out.  All my previous examples haven't been in quotient form, and how do I produce a $\frac{dy}{dx}$ from a quotient?&lt;/p&gt;&#xA;" OwnerUserId="7006" LastActivityDate="2011-03-08T23:34:50.263" Title="Implicitly find the second derivative" Tags="&lt;calculus&gt;&lt;homework&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="25841" PostTypeId="1" AcceptedAnswerId="25862" CreationDate="2011-03-08T22:05:51.290" Score="3" ViewCount="131" Body="&lt;p&gt;I have two boolean matrices:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;A = |1 1 0|&#xA;    |0 1 0|&#xA;    |0 0 1|&#xA;&#xA;and&#xA;&#xA;B = |1 0 0|&#xA;    |1 1 1|&#xA;    |0 0 1|&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;What is the result of &lt;strong&gt;A x B&lt;/strong&gt; and what are the steps needed to attain the result?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Note:&lt;/strong&gt; My textbook says that the answer to the above is:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;A x B = |1 1 1|&#xA;        |1 1 1|&#xA;        |0 0 1|&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;and that &lt;strong&gt;&lt;code&gt;A * B&lt;/code&gt;&lt;/strong&gt; is not equal to &lt;strong&gt;&lt;code&gt;A x B&lt;/code&gt;&lt;/strong&gt;. Unfortunately, it does not give the steps needed to find the solution.&lt;/p&gt;&#xA;" OwnerUserId="7971" LastEditorUserId="1543" LastEditorDisplayName="" LastEditDate="2011-03-10T14:21:44.767" LastActivityDate="2011-03-10T14:21:44.767" Title="How do you do a cross product of two $3 \times 3$ boolean matrices?" Tags="&lt;matrices&gt;&lt;computer-science&gt;" AnswerCount="1" CommentCount="9" />
  <row Id="25842" PostTypeId="2" ParentId="25840" CreationDate="2011-03-08T22:13:11.690" Score="1" ViewCount="" Body="&lt;p&gt;So all your work has found that&#xA;$$&#xD;&#xA;\frac{d^2y}{dx^2}=\frac{-x^{3}y^{2}\frac{dy}{dx}-4x^{2}\frac{dy}{dx}-8xy}{x^{4}y^{3}}&#xD;&#xA;$$&#xA;but you have that &#xA;$$&#xD;&#xA;\frac{dy}{dx}=-\frac{xy^2+4}{x^2y}&#xD;&#xA;$$&#xA;so you can substitute that in to find the second derivative in terms of $x$ and $y$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I think the first derivative is slightly different from what you have. &#xA;$$&#xD;&#xA;\frac{d}{dx}[x^2]y^2+x^2\frac{d}{dx}[y^2]-\frac{d}{dx}[8x]=\frac{d}{dx}[3]&#xD;&#xA;$$&#xA;gives&#xA;$$&#xD;&#xA;2xy^2+x^2\cdot 2y\frac{dy}{dx}-8=0\implies \frac{dy}{dx}=\frac{xy^2-4}{-x^2y}&#xD;&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="1583" LastEditorUserId="1583" LastEditorDisplayName="" LastEditDate="2011-03-08T22:26:33.643" LastActivityDate="2011-03-08T22:26:33.643" CommentCount="4" />
  <row Id="25843" PostTypeId="1" AcceptedAnswerId="25846" CreationDate="2011-03-08T22:30:46.647" Score="0" ViewCount="71" Body="&lt;p&gt;First off, is this question appropriate for this Stack Exchange? If not, I'll remove it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now to my question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there an easy way to apply the trend line formula from a chart to any given X value in Excel?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, I want to get the Y value for a given X = $2,006.00. I've already taken the formula and retyped it out be:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;=-0.000000000008*X^3 - 0.00000001*X^2 + 0.0003*X - 0.0029&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am continually making adjustments to the trend line by adding more data, and don't want to retype out the formula every time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/tgq8c.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="7972" LastActivityDate="2011-03-08T22:55:16.273" Title="Use a Trendline Formula to Get Values for Any Given X with Excel" Tags="&lt;statistics&gt;" AnswerCount="1" />
  <row Id="25844" PostTypeId="2" ParentId="25836" CreationDate="2011-03-08T22:38:14.163" Score="5" ViewCount="" Body="&lt;p&gt;For the symmetric random walk with $X_0=0$ and conditioned on $X_{2n}=0$, the average number of reversals should be $n$. Each outcome is a random ordering of $n$ plus signs and $n$ minus signs. A reversal takes place at any of the $2n-1$ time points from $1$ to $2n-1$, when the neighboring signs are opposite. Given the sign of one neighbor, the chance that the other neighbor has an opposite sign is ${n\over 2n-1}$. Adding up over the time points shows that the average number of sign changes is $n$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The original question however is about the average number of reversals up to the &lt;strong&gt;first&lt;/strong&gt; return $T$ to the origin. The argument above can be modified to show that, for $n&amp;gt;1$, &#xA;$$E(\mbox{reversals}\ |\ T=2n) = n - 1  = T/2-1,$$  so that the expected number of reversals is infinite.   &lt;/p&gt;&#xA;" OwnerUserId="940" LastActivityDate="2011-03-08T22:38:14.163" CommentCount="4" />
  <row Id="25845" PostTypeId="1" AcceptedAnswerId="25847" CreationDate="2011-03-08T22:50:04.860" Score="2" ViewCount="82" Body="&lt;p&gt;A problem in Lee's Introduction to Topological Manifolds got me thinking about this question. I can easily construct a locally euclidean space that is not second countable, by taking a disjoint union of $\mathbb R^n$ over an uncountable index set, but this is not connected of course. Similarly one can easily find connected first countable spaces that are not second countable, I presume an infinite dimensional Hilbert Space would suffice (no?).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The difficulty that I find in constructing such a space, is a lack of &quot;tools&quot;. Essentially I can either take a disjoint union or a connected sum. It's appealing then to attempt to take an uncountable connected sum of n-manifolds, but it's not apparent to me if this is well-defined. For instance we can't enumerate an uncountable index, so we can't just attach one after another. Then it seems we would have to attach all of them at the same time, but because an $n$-manifold is second countable there can't be an enough open balls to do this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a clever way to construct such a space with connected sums? The only thing I can think of is to assume the negation of the axiom of choice and take the limit of an infinite chain of connected sums of $n$-manifolds, then this doesn't have to be countable. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it isn't true how do you prove it? I can't see any clear reason why it would be false. &lt;/p&gt;&#xA;" OwnerUserId="1930" LastActivityDate="2011-03-09T01:21:15.413" Title="Does there exist a Connected Locally Euclidean Space that is not second countable?" Tags="&lt;general-topology&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="1" />
  <row Id="25846" PostTypeId="2" ParentId="25843" CreationDate="2011-03-08T22:55:16.273" Score="1" ViewCount="" Body="&lt;p&gt;I think that excel has built-in formulas for any type of trendline that you might use (i.e. you can use &quot;linest()&quot; if you had your chart options set for a linear trendline). Knowing this, you might just want to use the corresponding the function to get your Y values. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another method might be figuring out the VBA code used to create charts and trendlines and then extracting what you need. The following might be a good reference: &lt;a href=&quot;http://groups.google.com/group/microsoft.public.excel.programming/browse_frm/thread/c28678590192d449?hl=en&amp;amp;rnum=2&amp;amp;prev=/groups%3Fhl%3Den%26q%3Dtrendline%2Bcoefficients%26meta%3Dgroup%253Dmicrosoft.public.excel.programming&quot; rel=&quot;nofollow&quot;&gt;http://groups.google.com/group/microsoft.public.excel.programming/browse_frm/thread/c28678590192d449?hl=en&amp;amp;rnum=2&amp;amp;prev=/groups%3Fhl%3Den%26q%3Dtrendline%2Bcoefficients%26meta%3Dgroup%253Dmicrosoft.public.excel.programming&lt;/a&gt;  &lt;/p&gt;&#xA;" OwnerUserId="1736" LastActivityDate="2011-03-08T22:55:16.273" />
  <row Id="25847" PostTypeId="2" ParentId="25845" CreationDate="2011-03-08T22:58:09.030" Score="7" ViewCount="" Body="&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Long_line_%28topology%29&quot; rel=&quot;nofollow&quot;&gt;long line&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="3835" LastActivityDate="2011-03-08T22:58:09.030" CommentCount="2" />
  <row Id="25849" PostTypeId="1" CreationDate="2011-03-08T23:17:34.747" Score="5" ViewCount="166" Body="&lt;p&gt;The group of units of the rings $\text{GF}(p^r)$ and $\mathbb{Z}/p^r\mathbb{Z}$ are both cyclic (except for the exception of prime powers are not cyclic when $p=2$ and $r\ge 3$). This is a strong result which I have been using a lot but I don't understand it properly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to know more alternative proofs of these results. If there are any generalizations (e.g. can both cases be treated together?) and specific cases (e.g. is there something special about the intersection $\text{GF}(p) = \mathbb{Z}/p\mathbb{Z}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope that question doesn't seem vague, it's more like a couple of questions together - I will greatly appreciate anything on this topic so thanks very much!&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Inspired by &lt;a href=&quot;http://math.stackexchange.com/questions/28913/consider-the-product-b-1-b-2-cdots-b-phim-when-is-this-congruent-to&quot;&gt;lhf's answer to a question about Wilson's theorem&lt;/a&gt; is there any way to prove that if the product of all elements of a group are $-1$ then the units group is cyclic?&lt;/p&gt;&#xA;" OwnerUserId="4290" LastEditorUserId="4290" LastEditorDisplayName="" LastEditDate="2011-03-25T05:49:48.543" LastActivityDate="2011-03-25T05:49:48.543" Title="Other ways to deduce Cyclicity of the Units of certain groups?" Tags="&lt;number-theory&gt;&lt;group-theory&gt;&lt;finite-fields&gt;" AnswerCount="3" CommentCount="10" FavoriteCount="1" />
  <row Id="25850" PostTypeId="2" ParentId="25769" CreationDate="2011-03-08T23:21:45.663" Score="2" ViewCount="" Body="&lt;p&gt;The interesting part of $R$ is&#xA;$$ \begin{align*} &amp;amp;1 \leq 7 \\ &amp;amp;2 \leq 3 \\ &amp;amp;5 \leq 8 \leq 10 \end{align*} $$&#xA;Missing are $4,6,9$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does that help?&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2011-03-08T23:21:45.663" CommentCount="6" />
  <row Id="25851" PostTypeId="1" CreationDate="2011-03-08T23:26:26.670" Score="4" ViewCount="117" Body="&lt;p&gt;I would like to know the value of the following improper integral:&#xA;$$\int \limits_{-\infty}^{\infty}\frac{2x}{1+x^2}dx$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;as the function $f(x)=\displaystyle\frac{2x}{1+x^2}$ satisfies $f(-x)=-f(x)$ can I immediately  conclude that the integral is equal to zero?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you in advance for any suggestion.&lt;/p&gt;&#xA;" OwnerUserId="7979" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-08T23:58:47.457" LastActivityDate="2011-03-08T23:58:47.457" Title="Improper integral" Tags="&lt;improper-integrals&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="25852" PostTypeId="2" ParentId="25851" CreationDate="2011-03-08T23:28:01.513" Score="4" ViewCount="" Body="&lt;p&gt;This improper integral does not exist. It falls off like $1/x$ and that is not integrable. What you note works if you can show that it converges.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I have taken the definition as computing the integral from $a$ to $b$ and send $a$ to negative infinity and then $b$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you send them to infinity at the same time, then the integral is zero by what you have noted. You can also note that the indefinite integral of your integrand is equal to $\ln(1 + x^2)$.&lt;/p&gt;&#xA;" OwnerUserId="1120" LastActivityDate="2011-03-08T23:28:01.513" />
  <row Id="25853" PostTypeId="2" ParentId="25851" CreationDate="2011-03-08T23:32:40.717" Score="7" ViewCount="" Body="&lt;p&gt;You can't conclude that the integral is zero. What you can conclude though is that the Cauchy principal value is zero. The integral is defined as below. $$\int_{-\infty}^{\infty} \frac{2x}{1+x^2} dx = \lim_{R_1 \rightarrow - \infty, R_2 \rightarrow \infty} \int_{R_1}^{R_2} \frac{2x}{1+x^2} dx = \lim_{R_1 \rightarrow - \infty} \int_{R_1}^c \frac{2x}{1+x^2} dx + \lim_{R_2 \rightarrow \infty} \int_{c}^{R_2} \frac{2x}{1+x^2} dx $$&#xA;where $c \in \mathbb{R}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unless both the integrals are finite, you cannot make sense of the integral since you will get something &quot;like $(-\infty) + (+ \infty$)&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However note that if $R_1 \rightarrow -\infty$ and $R_2 \rightarrow \infty$ at the same pace i.e. $R_2 = -R_1 = R$, then it is called the Cauchy principal value and is zero.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2011-03-08T23:37:59.307" LastActivityDate="2011-03-08T23:37:59.307" />
  <row Id="25854" PostTypeId="2" ParentId="25840" CreationDate="2011-03-08T23:34:50.263" Score="2" ViewCount="" Body="&lt;p&gt;It's actually a bit simpler to work directly with the original (since then you don't have to worry about the quotient rule). Simply take derivatives twice, and then solve for $y&amp;#39;&amp;#39;$ in terms of $x$, $y$, and $y&amp;#39;$; only then plug in $y&amp;#39;$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Start with&#xA;$$x^2y^2 - 8x=3.$$&#xA;Taking derivatives once, we get&#xA;\begin{align*}&#xA;\frac{d}{dx}\Bigl(x^2y^2 - 8x\Bigr) &amp;amp;= \frac{d}{dx}3\\&#xA;x^2\frac{d}{dx}y^2 + y^2\frac{d}{dx}x^2 - 8 &amp;amp;= 0\\&#xA;x^2\bigl(2yy'\bigr) + y^2\bigl(2x\bigr) -8 &amp;amp;= 0\\&#xA;2x^2yy' + 2xy^2 - 8&amp;amp;= 0\\&#xA;x^2yy' + xy^2 - 4&amp;amp;= 0.&#xA;\end{align*}&#xA;Now take derivatives again, and solve for $y&amp;#39;&amp;#39;$:&#xA;\begin{align*}&#xA;\frac{d}{dx}\Bigl(x^2yy' + xy^2-4\Bigr) &amp;amp;= 0\\&#xA;x^2y\left(\frac{d}{dx}y'\right) + x^2y'\left(\frac{d}{dx}y\right) + yy'\left(\frac{d}{dx}x^2\right)\\&#xA;\quad+x\left(\frac{d}{dx}y^2\right) + y^2\left(\frac{d}{dx}x\right) &amp;amp;=0\\&#xA;x^2yy'' + x^2(y')^2 + 2xyy' + 2xyy' + y^2 &amp;amp;=0\\&#xA;x^2yy'' +x^2(y')^2 + 4xyy' + y^2 &amp;amp;=0\\&#xA;x^2yy'' &amp;amp;= -\Bigl( x^2(y')^2 + 4xyy' + y^2\Bigr)\\&#xA;y''&amp;amp;= -\frac{x^2(y')^2 + 4xyy' + y^2}{x^2y}\\&#xA;y''&amp;amp;= -\frac{(y')^2}{y} - \frac{4y'}{x} - \frac{y}{x^2}.&#xA;\end{align*}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to get the value entirely in terms of $x$ and $y$, you can go back to the formula we had with the first derivative,&#xA;$$x^2yy&amp;#39; + xy^2 - 4 = 0,$$&#xA;we can solve for $y&amp;#39;$ to get&#xA;\begin{align*}&#xA;x^2yy' &amp;amp;= 4 - xy^2\\&#xA;y' &amp;amp;= \frac{4-xy^2}{x^2y}\\&#xA;y' &amp;amp;= \frac{4}{x^2y} - \frac{y}{x}.&#xA;\end{align*}&#xA;So we can plug in this value of $y&amp;#39;$ into the formula for $y&amp;#39;&amp;#39;$:&#xA;\begin{align*}&#xA;y'' &amp;amp;= -\frac{(y')^2}{y} - \frac{4y'}{x} - \frac{y}{x^2}\\&#xA;y'' &amp;amp;= -\frac{1}{y}\left(\frac{4}{x^2y} - \frac{y}{x}\right)^2 - \frac{4}{x}\left(\frac{4}{x^2y}-\frac{y}{x}\right) - \frac{y}{x^2}\\&#xA;y'' &amp;amp;= -\frac{1}{y}\left(\frac{16}{x^4y^2} - \frac{8y}{x^3y} + \frac{y^2}{x^2}\right)-\frac{16}{x^3y} + \frac{4y}{x^2} - \frac{y}{x^2}\\&#xA;y'' &amp;amp;= -\frac{16}{x^4y^3} + \frac{8}{x^3y} - \frac{y}{x^2} - \frac{16}{x^3y} + \frac{3y}{x^2}\\&#xA;y'' &amp;amp;= -\frac{16}{x^4y^3} - \frac{8}{x^3y}+ \frac{2y}{x^2}.&#xA;\end{align*}&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2011-03-08T23:34:50.263" />
  <row Id="25855" PostTypeId="2" ParentId="25851" CreationDate="2011-03-08T23:35:11.880" Score="8" ViewCount="" Body="&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Cauchy_principal_value&quot; rel=&quot;nofollow&quot;&gt;Cauchy principal value&lt;/a&gt; of your integral is defined to be&#xA;$$ \lim_{a \to \infty} \int_{-a}^a {2x \over 1+x^2} \: dx $$&#xA;and this is zero by symmetry. So there is a sense in which the answer is $0$. However, if we let the limits go to infinity at different rates then we can make this integral &quot;equal&quot; whatever we want; for example, for any positive real constant $k$,&#xA;$$ \lim_{a \to \infty} \int_{-a}^{ka} {2x \over 1+x^2} \: dx = \lim_{a \to \infty} \log \left( {1+k^2 a^2 \over 1+a^2} \right) = 2 \log k. $$&#xA;Basically, since there is infinitely much area between this curve and the $x$-axis both in the first and third quadrants, we have to be careful about how we arrange things in order to make them &quot;cancel out&quot;; this is why we just throw up our hands and say that the integral doesn't converge.&lt;/p&gt;&#xA;" OwnerUserId="173" LastActivityDate="2011-03-08T23:35:11.880" />
  <row Id="25856" PostTypeId="1" AcceptedAnswerId="25864" CreationDate="2011-03-08T23:39:52.300" Score="3" ViewCount="118" Body="&lt;p&gt;Can someone show me the proof that the convolution of a compactly supported real valued function on $\mathbb{R}$ with a locally integrable function is also continuous?  I feel that this is a standard analysis result, but cannot remember how to prove it.  Thanks! &lt;/p&gt;&#xA;" OwnerUserId="3595" LastEditorUserId="3595" LastEditorDisplayName="" LastEditDate="2011-03-09T00:03:01.347" LastActivityDate="2011-03-09T00:21:15.147" Title="Basic Convolution Result" Tags="&lt;calculus&gt;&lt;analysis&gt;&lt;functional-analysis&gt;&lt;convolution&gt;&lt;integral-transforms&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="25857" PostTypeId="2" ParentId="25833" CreationDate="2011-03-08T23:46:42.380" Score="0" ViewCount="" Body="&lt;p&gt;Let $B_1$ be a base for $X$. Consider another base $B$ of $X$. Suppose $w(X) \leq n$ and $|B_1| \leq n$. Suppose for contradiction that $B \subset B_1$. You can probably use the following facts to come up with a contradiction:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The base elements cover $X$.&lt;/li&gt;&#xA;&lt;li&gt;Let $B$ be a base. Suppose $x_1, x_2 \in B$. Let $l = x_1 \cap x_2$. Then for each $x \in l$, there is a base element $x_3$ containing $x$ and contained in $l$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2011-03-08T23:46:42.380" CommentCount="1" />
  <row Id="25858" PostTypeId="2" ParentId="25833" CreationDate="2011-03-08T23:48:22.587" Score="2" ViewCount="" Body="&lt;p&gt;Assuming $w(X)$ is the minimun cardinality of a basis...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If a space $X$ has a finite basis, then it has finitely many open sets, and then &lt;em&gt;every&lt;/em&gt; intersection of open sets is an open set. It follows that from this that for each $x\in X$ there is a minimal open set $U_x\subseteq X$ such that $x\in U$,&#xA;and it is easy to see that every basis must contain the set $\mathcal B=\{U_x:x\in X\}$. As one can check that $\mathcal B$ is actually a basis, it is then the &lt;em&gt;unique&lt;/em&gt; basis of cardinal $w(B)$ and it is contained in every other basis.&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2011-03-08T23:48:22.587" CommentCount="2" />
  <row Id="25859" PostTypeId="1" AcceptedAnswerId="25860" CreationDate="2011-03-08T23:50:09.853" Score="3" ViewCount="47" Body="&lt;p&gt;I am going crazy showing something that is clearly wrong, but I can't see the error in my logic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am calculating $H_1(D^2,S^1)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the disc is contractible so $H_n(D^2)=0$ for all $n$. Also $H_n(S^1)=\mathbb{Z}$ for $n=0,1$ and is $0$ otherwise. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we have a long exact sequence of homology groups&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\cdots H_1(D^2)\to H_1(D^2,S^1) \to H_0(S^1) \to H_0(D^2) \cdots$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the above results we have the sequence&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$0 \to H_1(D^2,S^1) \to \mathbb{Z} \to 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and hence $H_1(D^2,S^1) \simeq \mathbb{Z}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is clearly not true as this would imply that $H_1(S^2)=\mathbb{Z}$, but I can't identify the gap in my logic (note that I am not working in reduced homology)&lt;/p&gt;&#xA;" OwnerUserId="2219" LastActivityDate="2011-03-08T23:54:26.260" Title="Relative homology of $D^2$ and $S^1$" Tags="&lt;algebraic-topology&gt;" AnswerCount="1" CommentCount="0" />
  <row Id="25860" PostTypeId="2" ParentId="25859" CreationDate="2011-03-08T23:54:26.260" Score="5" ViewCount="" Body="&lt;p&gt;$H_n(D^2)=0$ for all &lt;strong&gt;positive&lt;/strong&gt; $n$...&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2011-03-08T23:54:26.260" CommentCount="1" />
  <row Id="25862" PostTypeId="2" ParentId="25841" CreationDate="2011-03-08T23:55:52.077" Score="9" ViewCount="" Body="&lt;p&gt;I think it is the same as conventional matrix multiplication just that the multiplication is replaced by the &quot;and&quot; operation while the addition is replaced by the &quot;or&quot; operation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, $$A \times B = &#xD;&#xA;\begin{bmatrix} 1 &amp;amp; 1 &amp;amp; 0 \\\ 0 &amp;amp; 1 &amp;amp; 0 \\\ 0 &amp;amp; 0 &amp;amp; 1 \end{bmatrix} \times \begin{bmatrix} 1 &amp;amp; 0 &amp;amp; 0 \\\ 1 &amp;amp; 1 &amp;amp; 1 \\\ 0 &amp;amp; 0 &amp;amp; 1 \end{bmatrix}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$A \times B = \begin{bmatrix} (1 \&amp;amp; 1) || (1 \&amp;amp; 1) || (0 \&amp;amp; 0) &amp;amp; (1 \&amp;amp; 0) || (1 \&amp;amp; 1) || (0 \&amp;amp; 0) &amp;amp; (1 \&amp;amp; 0) || (1 \&amp;amp; 1) || (0 \&amp;amp; 1) \\\ (0 \&amp;amp; 1) || (1 \&amp;amp; 1) || (0 \&amp;amp; 0) &amp;amp; (0 \&amp;amp; 0) || (1 \&amp;amp; 1) || (0 \&amp;amp; 0) &amp;amp; (0 \&amp;amp; 0) || (1 \&amp;amp; 1) || (0 \&amp;amp; 1) \\\ (0 \&amp;amp; 1) || (0 \&amp;amp; 1) || (1 \&amp;amp; 0) &amp;amp; (0 \&amp;amp; 0) || (0 \&amp;amp; 1) || (1 \&amp;amp; 0) &amp;amp; (0 \&amp;amp; 0) || (0 \&amp;amp; 1) || (1 \&amp;amp; 1) \end{bmatrix}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$A \times B = \begin{bmatrix} 1 &amp;amp; 1 &amp;amp; 1 \\\ 1 &amp;amp; 1 &amp;amp; 1 \\\ 0 &amp;amp; 0 &amp;amp; 1 \end{bmatrix}$$&lt;/p&gt;&#xA;" OwnerUserId="3302" LastActivityDate="2011-03-08T23:55:52.077" />
  <row Id="25863" PostTypeId="2" ParentId="25836" CreationDate="2011-03-09T00:00:30.600" Score="4" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;In the symmetric case,&lt;/strong&gt; one assumes that $X_0=0$ and that $X_n=Y_1+\cdots+Y_n$ for $n\ge1$, where $(Y_n)_{n\ge1}$ is i.i.d. Bernoulli and centered. For $n\ge1$, let $R_n$ denote the number of reversals of $(X_k)_{0\le k\le n}$. Then $R_1=0$ and $R_n=U_2+\cdots+U_n$ where $U_k=[Y_kY_{k-1}=-1]$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The conditional expectation of $R_{n+1}$ with respect to the $\sigma$-algebra $F_n=\sigma(X_k;0\le k\le n)$ is $R_n+P(Y_nY_{n+1}=-1|F_n)=R_n+\frac12$, hence $M_n=2R_n-n$ defines a martingale $(M_n)_{n\ge1}$ starting from $M_1=-1$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, for every uniformly integrable stopping time such as the first hitting time $T_h$ of the set $\{0,h\}$ with $h\ge1$ by $(X_n)_n$, $E(M_{T_h})=-1$, hence &#xA;$$&#xD;&#xA;2E(R_{T_h})=E(T_h)-1.&#xD;&#xA;$$ &#xA;When $h\to+\infty$, $T_h$ converges to the first return time $T$ to $0$ and $T$ is not integrable hence $R_T$ is not integrable.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;In the asymmetric case,&lt;/strong&gt; assume that $P(Y_n=+1)=p$ and $P(Y_n=-1)=1-p$ for a given $p$ in $(0,1)$. If $p\ne\frac12$, $(X_n)_n$ has a positive probability to never hit $0$ again, in which case the total number of reversals of its path is almost surely infinite, hence not integrable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One way to save the day is to assume that $p&amp;lt;\frac12$ (for example) and to condition on the event $[X_1=1]$. Write $P^+$ for this conditioned probability measure and $E^+$ for the expectation with respect to $P^+$. Then the first return time $T$ to $0$ is (at last!) integrable for $P^+$ and $R_T\le T-1$ hence $R_T$ is integrable for $P^+$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To compute the value of $E^+(R_T)$, one can mimick the argument given in the symmetric case to show that the formula&#xA;$$&#xD;&#xA;M_n=2R_n-n-(1-2p)X_{n-1}&#xD;&#xA;$$ &#xA;defines a martingale $(M_n)_{n\ge1}$ with respect to $P^+$, starting from $M_1=-1$. Since $X_{T-1}=+1$ almost surely for $P^+$, this yields &#xA;$$&#xD;&#xA;2E^+(R_{T})=E^+(T)-2p,&#xD;&#xA;$$ &#xA;and it remains to compute $E^+(T)$. This can be done by the usual first-step decomposition: on $[Y_2=-1]$, $T=2$, and on $[Y_2=+1]$, $T=T&amp;#39;+T&amp;#39;&amp;#39;$ for two independent copies of $T$. Hence $E^+(T)=2(1-p)+2E^+(T)p$, which yields the value of $E^+(T)$. Finally the mean number of reversals is&#xA;$$&#xD;&#xA;E^+(R_T)=\frac{1-2p(1-p)}{1-2p}.&#xD;&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="6179" LastEditorUserId="6179" LastEditorDisplayName="" LastEditDate="2011-03-09T13:42:23.730" LastActivityDate="2011-03-09T13:42:23.730" CommentCount="3" />
  <row Id="25864" PostTypeId="2" ParentId="25856" CreationDate="2011-03-09T00:21:15.147" Score="1" ViewCount="" Body="&lt;p&gt;On your comment for $L^2$, we can do it as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Recall that for $1 \leq p &amp;lt; \infty$ translation is continuous in $L^p$, that is if $f \in L^p(\mathbb R^n)$ and $z \in \mathbb R^n$ then $\lim_{y \to 0} \|\tau_{y + z} f - \tau_z f\|_p = 0$. Here $\tau_y$ is the shift over $y$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Okay, fine. By Young's inequality for convolutions, the convolution actually exists. So&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\|\tau_y (f \ast g) -  (f \ast g)\|_\infty = \|(\tau_y f - f) \ast g)\|_\infty \leq \|\tau f - f\|_2 \|g\|_2 \to 0 \text{ as $y \to 0$.}$$&#xA;So actually we get uniform continuity. Note that this also works for $f$ in $L^p$ and $g$ in $L^q$.&lt;/p&gt;&#xA;" OwnerUserId="1120" LastActivityDate="2011-03-09T00:21:15.147" CommentCount="3" />
  <row Id="25865" PostTypeId="2" ParentId="25849" CreationDate="2011-03-09T00:51:16.153" Score="6" ViewCount="" Body="&lt;p&gt;It seems like you aren't aware of the result yoyo cites in the comments, so I think it's worth pointing out as a start.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Theorem:&lt;/strong&gt; A finite subgroup of the multiplicative group of any field is cyclic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Proof.&lt;/em&gt; Let $G$ be a finite subgroup of $F^{\ast}$, $F$ a field, and $n = |G|$. Then the elements of $G$ must be precisely the roots of the polynomial $x^n - 1$. We have the factorization&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x^n - 1 = \prod_{d | n} \Phi_d(x)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\Phi_d$ is the $d^{th}$ &lt;a href=&quot;http://en.wikipedia.org/wiki/Cyclotomic_polynomial&quot; rel=&quot;nofollow&quot;&gt;cyclotomic polynomial&lt;/a&gt;. It is not hard to see that the roots of $\Phi_d$ over $F$ must in fact be precisely the set of elements of order exactly $d$ in $G$; in particular, there exists an element of order exactly $n$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've been told that the result for $\mathbb{Z}/p^n\mathbb{Z}$ is clearer if one first proves the corresponding result for the &lt;a href=&quot;http://en.wikipedia.org/wiki/P-adic_number&quot; rel=&quot;nofollow&quot;&gt;$p$-adic integers&lt;/a&gt;, but I'm not sure if this actually saves you any work. &lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2011-03-09T00:51:16.153" />
  <row Id="25866" PostTypeId="2" ParentId="25032" CreationDate="2011-03-09T00:56:54.727" Score="1" ViewCount="" Body="&lt;p&gt;My advice to you is to take whatever kind of mathematics you find interesting. The subject is vast and profound, filled with diverse ideas and methods, each of which can be pursued further than one might imagine. For any course you take, there is a further course, and another and perhaps another and then published research work carrying those same ideas further and further into distant abstract lands. So just follow the path where it leads you. As long as you have talent and find it interesting, you will do well. The fact that you have enjoyed abstract algebra, linear algebra and number theory suggests that you might enjoy more pure  mathematics courses, and surely there are plenty. The various subject areas of mathematics have a structure that will become apparant to you as you learn more, and there is little need at first to grasp the whole structure. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the elementary level, such as in high school, mathematics can often seem to follow a linear progression---one might have elementary algebra, then geometry, then trigonometry, then pre-calculus, then calculus, etc.---but of course at higher levels, mathematics does not follow a linear progression. Rather, it branches out into diverse areas, each of which develop the structure of its own ideas. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I was an undergraduate student, I had taken a sophomore course (Math 6 at Caltech) that had included a semester on mathematical logic, which I thoroughly enjoyed. Later, when looking at the catalogue to register for the next semester, I realized that there was an entire subject of mathematical logic, with many courses and an entire research program. I was enthralled!&lt;/p&gt;&#xA;" OwnerUserId="413" LastActivityDate="2011-03-09T00:56:54.727" />
  <row Id="25867" PostTypeId="1" CreationDate="2011-03-09T00:58:45.650" Score="3" ViewCount="98" Body="&lt;p&gt;From the wiki page &lt;a href=&quot;http://en.wikipedia.org/wiki/Catalan_number&quot; rel=&quot;nofollow&quot;&gt;Catalan number&lt;/a&gt;, we know the number of Young tableaux whose diagram is a 2-by-n rectangle given $2n$ distinct numbers is $C_n$. In general, given $m\times n$ distinct numbers, how many Young tableaux whose diagram is a $m\times n$  rectangle are there? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, what if these numbers can be repeated?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many thanks.&lt;/p&gt;&#xA;" OwnerUserId="6016" LastActivityDate="2011-03-09T08:11:37.467" Title="the number of Young tableaux in general" Tags="&lt;combinatorics&gt;" AnswerCount="2" CommentCount="0" />
  <row Id="25868" PostTypeId="1" AcceptedAnswerId="25870" CreationDate="2011-03-09T01:02:51.247" Score="2" ViewCount="86" Body="&lt;p&gt;I feel this must be an old question, but so far I have never seen it mentioned in any of the books I know. Most of modern treatment of advanced linear algebra is centered on tensor products and eigendecomposition, etc. So here is the question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know ordinary matrices provided a way of measuring a list of vectors $a_{i},i\in I_{n}$ such that for two different list of vectors we have the addition and multiplication. Further multiplication is associative, generates another list of vectors. Regardless of the specifics in row and column operations, I want to ask:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) Is it possible to define this for an array of vectors $a_{ij},i,j\in I_{n}$? I wish it to fulfill at least associativity, has an identity element, and makes some sense(such that respect to the individual $a_{ik}$ with $k$ fixed it resemble matrix multiplication?). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) Is it possible to define an exterior operation on it that resembling ordinary determinant?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(3) If the answer for above questions are both positive, is there any way to define a canonical decomposition of such $3$-dimensional matrices? I wish there be something resembling various decomposition theorems in ordinary linear algebra we have. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I had been thinking about this since I was a freshmen but I never got much progress (maybe I am too lazy). I will be happy to see if such an operation exists (or the reason why it should not exist). &lt;/p&gt;&#xA;" OwnerUserId="7887" LastActivityDate="2011-03-09T01:08:05.437" Title="Is there any thing like matrices for three-dimensional linear algebra?" Tags="&lt;linear-algebra&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="0" />
  <row Id="25869" PostTypeId="1" CreationDate="2011-03-09T01:07:39.177" Score="2" ViewCount="69" Body="&lt;p&gt;What are the exact meanings of the terms &quot;operation&quot;, &quot;function&quot; and &quot;map&quot; (are they even exactly defined)? I have always believed that an &quot;operation&quot; was a map&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$S \times S \times [...] \times S \rightarrow S.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, according to the always-so-reliable Wikipedia, an operation can be any map on the form&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$A_1 \times A_2 \times [...] \times A_3 \rightarrow S.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fine, but Wolfram's MathWorld disagrees with Wikipedia and agrees with my original definition. Which is correct?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, above I've used the word &quot;map&quot; to describe operations. Is this correct use of the term? Could I have replaced &quot;map&quot; with &quot;function&quot; without changing any of the meaning, i.e., do &quot;map&quot; and &quot;function&quot; mean exactly the same thing, or is there a slight difference in formal meaning and/or customary meaning?&lt;/p&gt;&#xA;" OwnerUserId="4576" LastActivityDate="2011-03-09T01:07:39.177" Title="Meaning of the terms operation, function and map" Tags="&lt;terminology&gt;" CommentCount="2" />
  <row Id="25870" PostTypeId="2" ParentId="25868" CreationDate="2011-03-09T01:08:05.437" Score="3" ViewCount="" Body="&lt;p&gt;There was a good discussion of this &lt;a href=&quot;http://mathoverflow.net/questions/48045/why-are-matrices-ubiquitous-but-hypermatrices-rare&quot; rel=&quot;nofollow&quot;&gt;on mathoverflow&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="884" LastActivityDate="2011-03-09T01:08:05.437" CommentCount="1" />
  <row Id="25871" PostTypeId="1" CreationDate="2011-03-09T01:12:00.963" Score="1" ViewCount="75" Body="&lt;p&gt;I have seen the phrase &quot;minimal number of generators of an ideal&quot; (in a Noetherian local ring) several times. I am unable to see how this is a well defined. Explicitly, how do we show, if $x_1,...,x_m$ and $y_1,...,y_n$ are minimal generating sets of an ideal in a Noetherian local ring, then $m=n$.&lt;/p&gt;&#xA;" OwnerUserId="7763" LastEditorUserId="7763" LastEditorDisplayName="" LastEditDate="2011-03-09T01:24:29.093" LastActivityDate="2011-03-09T02:21:31.330" Title="Minimal number of generators of an ideal in a Noetherian local ring" Tags="&lt;commutative-algebra&gt;" AnswerCount="1" />
  <row Id="25873" PostTypeId="2" ParentId="25845" CreationDate="2011-03-09T01:21:15.413" Score="1" ViewCount="" Body="&lt;p&gt;Problem 3-8 in Lee constructs the line with two origins (maybe this is the problem that got you thinking about this question).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, what about a line with uncountably many origins? &lt;/p&gt;&#xA;" OwnerUserId="566" LastActivityDate="2011-03-09T01:21:15.413" CommentCount="2" />
  <row Id="25874" PostTypeId="2" ParentId="25867" CreationDate="2011-03-09T01:32:14.947" Score="7" ViewCount="" Body="&lt;p&gt;For the answer to your main question, you need to use the &lt;a href=&quot;http://en.wikipedia.org/wiki/Hook-length_formula#Dimension_of_a_representation&quot; rel=&quot;nofollow&quot;&gt;hook-length formula&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://oeis.org/A060854&quot; rel=&quot;nofollow&quot;&gt;OEIS A060854&lt;/a&gt; gives the result &#xA;$$(mn)! \prod_{i=0}^{n-1} \frac{i!}{(m+i)!}  \textrm{ or equivalently } (mn)! \prod_{j=0}^{m-1} \frac{j!}{(n+j)!}  $$&#xA;and some more information.&lt;/p&gt;&#xA;" OwnerUserId="6460" LastActivityDate="2011-03-09T01:32:14.947" CommentCount="3" />
  <row Id="25875" PostTypeId="1" AcceptedAnswerId="25877" CreationDate="2011-03-09T01:33:08.053" Score="1" ViewCount="47" Body="&lt;p&gt;Let $Q(x,y)$  be the statement $y = 2x +1$ what are the values of the following. The universe is $Z^+$ {1,2,3,....}.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(a) $\forall x\exists y Q(x,y)$ This is true&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(b) $\exists x \forall y Q(x,y)$ This is false.&lt;/p&gt;&#xA;" OwnerUserId="1040" LastActivityDate="2011-03-09T01:38:28.517" Title="Is this quantification correct?" Tags="&lt;logic&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="25876" PostTypeId="1" AcceptedAnswerId="25878" CreationDate="2011-03-09T01:36:15.727" Score="14" ViewCount="609" Body="&lt;p&gt;I have been looking at the birthday problem (http://en.wikipedia.org/wiki/Birthday_problem) and I am trying to figure out what the probability of 3 people sharing a birthday in a room of 30 people is. (Instead of 2).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I thought I understood the problem but I guess not since I have no idea how to do it with 3.&lt;/p&gt;&#xA;" OwnerUserId="7742" LastEditorUserId="1543" LastEditorDisplayName="" LastEditDate="2011-04-16T18:32:52.030" LastActivityDate="2011-04-16T18:32:52.030" Title="Probability of 3 people in a room of 30 having the same birthday" Tags="&lt;probability&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="2" />
  <row Id="25877" PostTypeId="2" ParentId="25875" CreationDate="2011-03-09T01:38:28.517" Score="1" ViewCount="" Body="&lt;p&gt;The first is true.  You are correct.  For the second, truth would imply that 2x+1=1 has a positive integer as a solution, which is false, so you are correct again.&lt;/p&gt;&#xA;" OwnerUserId="3595" LastActivityDate="2011-03-09T01:38:28.517" CommentCount="1" />
  <row Id="25878" PostTypeId="2" ParentId="25876" CreationDate="2011-03-09T01:52:39.623" Score="17" ViewCount="" Body="&lt;p&gt;The birthday problem with 2 people is quite easy because finding the probability of the complementary event &quot;all birthdays distinct&quot; is straightforward. For 3 people, the complementary event includes &quot;all birthdays distinct&quot;, &quot;one pair and the rest distinct&quot;, &quot;two pairs and the rest distinct&quot;, etc. To find the exact value is pretty complicated. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Poisson approximation is pretty good, though. Imagine checking every triple and calling it a &quot;success&quot; if all three have the same birthdays. The total number of successes is approximately Poisson with mean value ${30 \choose 3}/365^2$. Here $30\choose 3$ is the number of triples, and $1/365^2$ is the chance that any particular triple is a success.&#xA;The probability of getting at least one success is obtained from the Poisson distribution:&#xA;$$ P(\mbox{ at least one triple birthday with 30 people})\approx 1-\exp(-{30 \choose 3}/365^2)=.0300. $$     &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can modify this formula for other values, changing either 30 or 3. For instance, &#xA;$$ P(\mbox{ at least one triple birthday with 100 people})\approx 1-\exp(-{100 \choose 3}/365^2)=.7029,$$&#xA;$$ P(\mbox{ at least one double birthday with 25 people })\approx 1-\exp(-{25 \choose 2}/365)=.5604.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Poisson approximation is very useful in probability, not only for birthday problems!    &lt;/p&gt;&#xA;" OwnerUserId="940" LastActivityDate="2011-03-09T01:52:39.623" CommentCount="3" />
  <row Id="25879" PostTypeId="1" CreationDate="2011-03-09T01:58:27.020" Score="0" ViewCount="111" Body="&lt;p&gt;Pretty easy when its only a function of the form $y = ax + b$, however I'm trying to find if this function:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f(x,y) = 2x + y$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is onto and one-to-one. Is there a procedure to follow to check this?&#xA;Thanks!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: My reasoning.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;onto:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$z = 2x + y$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x = (z-y)/ 2$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$y = z - 2x$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Seeing as both $x$ and $y$ can be represented in &quot;safe&quot; ways (no dividing by zero), I'm assuming that the function is onto.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One-to-One:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$2x_1 + y_1 = 2x_2 + y_2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$2(x_1-x_2) = (y_2-y_1)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore if $y_2 = y_1$, $x_2 = x_1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I guess my reasoning is wrong :/&lt;/p&gt;&#xA;" OwnerUserId="7983" LastEditorUserId="7840" LastEditorDisplayName="" LastEditDate="2011-03-09T03:16:59.577" LastActivityDate="2011-03-09T03:42:19.220" Title="Onto and One-to-One of a 2 variable function" Tags="&lt;functions&gt;" AnswerCount="2" CommentCount="7" />
  <row Id="25880" PostTypeId="2" ParentId="25876" CreationDate="2011-03-09T02:06:42.253" Score="12" ViewCount="" Body="&lt;p&gt;An exact formula  can be found in Anirban DasGupta, &lt;a href=&quot;http://www.math.ucdavis.edu/~tracy/courses/math135A/UsefullCourseMaterial/birthday.pdf&quot; rel=&quot;nofollow&quot;&gt;&#xA;The matching, birthday and the strong birthday problem: a contemporary review&lt;/a&gt;, Journal of Statistical Planning and Inference 130 (2005), 377-389. This paper claims that if $W$ is the number of triplets of people having the same birthday, $m$ is the number of days in the year, and $n$ is the number of people, then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ P(W \ge 1) = 1 - \sum_{i=0}^{\lfloor n/2 \rfloor} {m! n! \over i! (n-2i)! (m-n+i)! 2^i m^n} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;No derivation or source is given; I think the idea is that the term corresponding to $i$ is the probability that there are $i$ birthdays shared by 2 people each and $n-2i$ birthdays with one person each.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, if $m = 365, n = 30$ this formula gives $0.0285$, not far from Byron's approximation.&lt;/p&gt;&#xA;" OwnerUserId="173" LastActivityDate="2011-03-09T02:06:42.253" />
  <row Id="25881" PostTypeId="2" ParentId="25871" CreationDate="2011-03-09T02:21:31.330" Score="1" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\ $ Over a local ring there is a nicely behaved notion of minimal set of generators by way of Nakayama's lemma. See section 20.1, The uniqueness of free resolutions, in Eisenbud: &lt;em&gt;Commutative Algebra: with a View Toward Algebraic Geometry&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2011-03-09T02:21:31.330" CommentCount="1" />
  <row Id="25883" PostTypeId="1" AcceptedAnswerId="25894" CreationDate="2011-03-09T02:28:05.893" Score="6" ViewCount="164" Body="&lt;p&gt;How to show that Rolle's theorem, the Mean Value Theorem are equivalent to the least upper bound property?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm thinking of starting like this: Let F be an ordered field that does not satisfy the least upper bound property, and then deduce that F does not satisfy either Rolle's or MVT. But I'm not sure how to continue, any help please? thanks!&lt;/p&gt;&#xA;" OwnerUserId="7429" LastEditorUserId="6296" LastEditorDisplayName="" LastEditDate="2011-03-09T03:05:54.570" LastActivityDate="2011-03-09T05:59:53.573" Title="Equivalence of Rolle's theorem, the mean value theorem, and the least upper bound property?" Tags="&lt;real-analysis&gt;" AnswerCount="3" CommentCount="6" />
  <row Id="25884" PostTypeId="2" ParentId="25879" CreationDate="2011-03-09T02:45:13.363" Score="2" ViewCount="" Body="&lt;p&gt;(I lack the reputation to &lt;em&gt;just&lt;/em&gt; comment, so here goes...)&#xA;Your reasoning about 1-1 is wrong because, while&#xA;&quot;Therefore if y2 = y1, x2 = x1.&quot; may be true, there is no guarantee that y2 = y1!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice that f(2, 0) = f(1, 2) = f(0, 4)&lt;/p&gt;&#xA;" OwnerUserId="7850" LastActivityDate="2011-03-09T02:45:13.363" CommentCount="2" />
  <row Id="25885" PostTypeId="1" CreationDate="2011-03-09T02:47:09.603" Score="4" ViewCount="202" Body="&lt;p&gt;Let $l^2$ be the set of sequences $x = (x_n)_{n\in\mathbb{N}}$ ($x_n \in \mathbb{C}$) such that $\sum_{k\in\mathbb{N}} \left|x_k\right|^2 &amp;lt; \infty$, how can I prove that $l^2$ is a Hilbert space (with dot-product $\left(x,y\right) = \sum_{k\in\mathbb{N}} x_k\overline{y_k}$). This is a standard textbook exercise: apparently this is easy and, even to me, it seems self-evident. However, I don't know what to do with the infinite sum.&lt;/p&gt;&#xA;" OwnerUserId="7985" LastActivityDate="2011-03-09T09:36:20.040" Title="How to prove that square-summable sequences form a Hilbert space?" Tags="&lt;sequences-and-series&gt;&lt;hilbert-spaces&gt;" AnswerCount="3" CommentCount="5" FavoriteCount="2" />
  <row Id="25886" PostTypeId="1" AcceptedAnswerId="25891" CreationDate="2011-03-09T02:49:01.157" Score="1" ViewCount="85" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Let $n$ be an integer.&#xA;  If $3$ does not divide $n$, then $3$ divides $n^2 - 1$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm trying to prove this using a &quot;proof by cases&quot;. However, I'm lost as to how to start. I thought proof by cases started by making a logical expression, but I can't seem to find the right one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="7983" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-09T03:01:11.923" LastActivityDate="2011-03-09T03:43:43.713" Title="Using proof by cases -- stuck" Tags="&lt;logic&gt;&lt;discrete-mathematics&gt;" AnswerCount="4" CommentCount="0" />
  <row Id="25887" PostTypeId="1" CreationDate="2011-03-09T02:52:44.693" Score="3" ViewCount="262" Body="&lt;p&gt;How to show that the Peano Arithmetic theory is not scapegoat?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Peano Arithmetic is a consistent theory.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;A theory T is scapegoat if for every formula $A$ with only one free variable there exist a closed term $s$ such that $T$ proves:&#xA;        $$(\exists x \; (\neg A(x)) )\Rightarrow \neg A(s)$$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;(&quot;$\neg$&quot; means &quot;not&quot;).&#xA;I think we can start by assuming that the theory is scapegoat then we get a contradiction, but how!?&lt;/p&gt;&#xA;" OwnerUserId="7986" LastEditorUserId="630" LastEditorDisplayName="" LastEditDate="2011-03-09T13:03:49.363" LastActivityDate="2011-03-09T13:14:49.490" Title="Scapegoat Theory!" Tags="&lt;logic&gt;&lt;proof-theory&gt;" AnswerCount="4" CommentCount="6" />
  <row Id="25888" PostTypeId="2" ParentId="25886" CreationDate="2011-03-09T02:53:10.277" Score="2" ViewCount="" Body="&lt;p&gt;Look at the possible remainders when $3$ divides an integer. The set of possible remainders when a number is divided by $3$ is $\{0,1,2\}$. You are given that $3$ does not divide $n$. Hence, the possible remainders when $3$ divides $n$ are $1$ and $2$. Hence $n=3k+1$ or $n=3k+2$. Now look at the square of the two cases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $n=3k+1$, then $n^2 = 9k^2 + 6k + 1 = 3(3k^2+2k) + 1 \Rightarrow (n^2-1) = 3(3k^2+2k)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $n=3k+2$, then $n^2 = 9k^2 + 12k + 4 = 3(3k^2+4k+1) + 1 \Rightarrow (n^2-1) = 3(3k^2+4k+1)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence in both the cases $3|(n^2-1)$.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2011-03-09T03:00:20.143" LastActivityDate="2011-03-09T03:00:20.143" />
  <row Id="25889" PostTypeId="2" ParentId="25886" CreationDate="2011-03-09T02:53:29.360" Score="1" ViewCount="" Body="&lt;p&gt;Case: $n = 1 \pmod{3}$.&#xA;Then $n^2 = 1 \pmod{3}$ &lt;br&gt;&#xA;i.e. $n^2 - 1 = 0 \pmod{3}$&lt;br&gt;&#xA;i.e. $3$ divides $n^2 - 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Case: $n = 2 \pmod{3}$.&#xA;(The next lines are identical to the above, since 2^2 = 4 = 1 (mod 3))&lt;/p&gt;&#xA;" OwnerUserId="7850" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-09T02:54:49.047" LastActivityDate="2011-03-09T02:54:49.047" />
  <row Id="25891" PostTypeId="2" ParentId="25886" CreationDate="2011-03-09T03:00:23.313" Score="6" ViewCount="" Body="&lt;p&gt;If you want to do a proof by cases, the first step is to identify a complete list of possible cases (in principle, they need not be mutually exclusive, but in practice they usually are).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you have your completely list of possible cases, then you examine each case in turn. If in all cases you get the desired conclusion, then the conclusion holds in all cases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, here you want to show that if $3$ does not divide $n$, then $3$ divides $n^2-1$, and you want to do it &quot;by cases&quot;. So, what is the complete list of &quot;cases&quot; that covers all possibilities? &quot;Cases&quot; for what or whom?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the only thing that can change from one instance of this problem to the other is $n$, then the cases will refer to $n$. There are many ways to break up all possible $n$ into cases. The simplest here, because you are dealing with divisibility by $3$, is to consider the possible remainders when you divide $n$ by $3$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what are all the possible situations when you divide $n$ by $3$? You can get a remainder of $0$, a remainder of $1$, or a remainder of $2$. So we go through each case in turn.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, suppose that $3$ does not divide $n$; we want to conclude that $3$ divides $n^2-1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;CASE 1: The remainder is $0$. This case is impossible, because in this case $3$ divides $n$, and we are assuming that $3$ does not divide $n$. So we can discard this case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;CASE 2: The remainder is $1$. Then $n$ can be written as $n=3k+1$; square both sides, subtract $1$, see what happens.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;CASE 3: The remainder is $2$. Then $n$ can be written as $n=3k+2$; &lt;code&gt;&amp;lt;continue here&amp;gt;&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since, in all cases we get the conclusion we want, the conclusion holds for all $n, as desired.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-09T03:43:43.713" LastActivityDate="2011-03-09T03:43:43.713" />
  <row Id="25892" PostTypeId="2" ParentId="25885" CreationDate="2011-03-09T03:09:13.807" Score="2" ViewCount="" Body="&lt;p&gt;Show that your definition $\left(x,y\right) \equiv \sum_{k\in\mathbb{N}} x_k\overline{y_k}$  Satisfies all the inner product axioms. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint: axioms are &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;conjugate symmetric&lt;/li&gt;&#xA;&lt;li&gt;linearity in one of the arguments&lt;/li&gt;&#xA;&lt;li&gt;Positive definite: $(u,u)\geq 0$ equality iff u=0&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Now show that $F(x) = (x,z)$ defines a bounded linear functional $F$ on your inner product space $X$ of norm $||z||$. If the mapping $z\rightarrow F$ (  $X\rightarrow X&amp;#39;$ where X' is space of continuous linear functionals) is surjective, it must be a hilbert space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt; OK, since you're worried about infinites  let me share something I discovered recently in a book called the Cauchy Schwarz masterclass, the author calls it the principle of qualitative inference.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given: $\sum_{k\in\mathbb{N}} a_k^2 \lt \infty$  and $\sum_{k\in\mathbb{N}} b_k^2 \lt \infty$ then it can be shown that $\sum_{k\in\mathbb{N}} a_k  b_k \lt \infty$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For large k &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a_k b_k \leq \frac{1}{2} (a_k^2+b_k^2)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Rightarrow \sum_{k\in\mathbb{N}} a_k b_k \leq \sum_{k\in\mathbb{N}}  \frac{1}{2} (a_k^2+b_k^2)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This already gives you a glimpse at the truth of the assertion. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(On a sidenote, you can go on to introduce $\hat{a}_k=\frac{a_k}{\sqrt{\sum_{k\in\mathbb{N}} a_k^2}}$ &#xA;and derive the Cauchy inequality!)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now do for conjugate symmetric what was done for real symmetric. &lt;/p&gt;&#xA;" LastActivityDate="2011-03-09T03:49:26.750" CommentCount="5" />
  <row Id="25893" PostTypeId="2" ParentId="25887" CreationDate="2011-03-09T03:12:49.160" Score="1" ViewCount="" Body="&lt;p&gt;We know that Peano Arithmetic is a consistent theory. So it has a consistent extension $K&amp;#39;$ such that $K&amp;#39;$ is a scapegoat theory and contains denumerably many closed terms.&lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2011-03-09T03:12:49.160" CommentCount="1" />
  <row Id="25894" PostTypeId="2" ParentId="25883" CreationDate="2011-03-09T03:15:46.523" Score="8" ViewCount="" Body="&lt;p&gt;Suppose $F$ is an ordered field, and $A \subset F$ is bounded above but has no least upper bound. Define a function $f \colon F \to F$ by $f(x)=1$ if $x$ is an upper bound of $A$, $f(x)=0$ otherwise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I claim that for every $x$, there is an open interval around $x$ on which $f$ is constant. If $x$ is an upper bound of $A$, then since $x$ is not a least-upper-bound there exists an upper bound $y$ of $A$ with $y &amp;lt; x$, and so $(y,x+1)$ works. If $x$ is not an upper bound of $A$, then there exists an element $a$ of $A$ such that $a&amp;gt;x$, and so $(x-1,a)$ works.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The existence of these intervals shows that $f$ is everywhere differentiable with derivative $0$. Since $f$ is not constant, this contradicts the mean value theorem. Thus the MVT implies the least-upper-bound property.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The fact that the least-upper-bound property implies Rolle's theorem, and that Rolle's theorem implies the MVT, is proved in every first course of analysis.&lt;/p&gt;&#xA;" OwnerUserId="5203" LastEditorUserId="5203" LastEditorDisplayName="" LastEditDate="2011-03-09T03:21:04.413" LastActivityDate="2011-03-09T03:21:04.413" CommentCount="2" />
  <row Id="25895" PostTypeId="2" ParentId="25885" CreationDate="2011-03-09T03:19:44.553" Score="1" ViewCount="" Body="&lt;p&gt;You need to show that $\ell^2$ is a vector space, that the dot product satisfies the properties of inner product, and that the space is complete relative to the induced metric&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The tricky bit in showing it is a vector space is showing that it is closed under sums; this is a consequence of the &lt;a href=&quot;http://en.wikipedia.org/wiki/H%C3%B6lder_inequality&quot; rel=&quot;nofollow&quot;&gt;Holder inequality&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Showing that the dot product is an inner product is straightforward: you may need to use the Holder inequality or &lt;a href=&quot;http://en.wikipedia.org/wiki/Cauchy%E2%80%93Schwarz_inequality&quot; rel=&quot;nofollow&quot;&gt;Cauchy-Schwartz inequality&lt;/a&gt; to show that $\sum x_k\overline{y_k}$ converges when both $(x_n)$ and $(y_n)$ are square summ&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And then you have the issue of showing that $\ell^2$ is actually complete relative to the induced norm. This is pretty standard (in fact, one shows that the spaces $\ell_p$ are Banach spaces for all $p\geq 1$, hence complete, so $\ell^2$ is complete. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added.&lt;/strong&gt; From the comments, it seems your main difficulty is operating with the infinite sums and making sure they have the properties you want, similar to finite sums. (Let me commend you for your caution: it is indeed important not to simply assume that properties that hold for finite sums will transfer to infinite series). We use the fact that, by definition,&#xA;$$\sum_{k=1}^{\infty} a_k = \lim_{n\to\infty}\sum_{k=1}^n a_n.$$&#xA;In the sense that the series on the left converges to the limit, if the limit exists. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, for example, you have:&#xA;\begin{align*} \overline{\langle (x_n),(y_n)\rangle} &amp;amp;= \overline{\sum_{k=1}^{\infty}x_k\overline{y_k}}\\&#xA;&amp;amp;=\overline{\lim_{n\to\infty}\sum_{k=1}^n x_k\overline{y_k}}\\&#xA;&amp;amp;= \lim_{n\to\infty}\overline{\sum_{k=1}^n x_k\overline{y_k}}&amp;amp;&amp;amp;\mbox{(since conjugation is continuous)}\\&#xA;&amp;amp;= \lim_{n\to\infty}\sum_{k=1}^n\overline{x_k\overline{y_k}} &amp;amp;&amp;amp;\mbox{(since the sum is finite)}\\&#xA;&amp;amp;= \lim_{n\to\infty}\sum_{k=1}^n\overline{x_k}\overline{\overline{y_k}}\\&#xA;&amp;amp;= \lim_{n\to\infty}\sum_{k=1}^n\overline{x_k}y_k\\&#xA;&amp;amp;= \lim_{n\to\infty}\sum_{k=1}^ny_k\overline{x_k}\\&#xA;&amp;amp;= \sum_{k=1}^{\infty} y_k\overline{x_k}\\&#xA;&amp;amp;= \langle (y_k),(x_k)\rangle.&#xA;\end{align*}&#xA;Etc.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-09T03:31:51.543" LastActivityDate="2011-03-09T03:31:51.543" CommentCount="1" />
  <row Id="25896" PostTypeId="2" ParentId="25886" CreationDate="2011-03-09T03:21:05.843" Score="1" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\ $ If not, and $\rm\ n-1,\ n,\ n+1 \ $ is the least consecutive triple of naturals all not divisible by $3$ then $\rm\ n-2,\ n-1,\ n\ $ would be a smaller triple (since $\rm\ 3\ |\ n-2\iff 3\ |\ n+1)\:,\:$ a contradiction. The case analysis arises when showing that the leftshifted sequence remains nonnegative (and also we use sign case analysis since we exploit $\rm\ n\to -n\ $ symmetry to reduce to the nonnegative case). &lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2011-03-09T03:21:05.843" />
  <row Id="25898" PostTypeId="2" ParentId="25887" CreationDate="2011-03-09T03:31:39.047" Score="3" ViewCount="" Body="&lt;p&gt;Let $P(x)$ be the formula &quot;if PA is inconsistent, then $x$ is the Gödel number of a proof of $0=1$&quot;. Then PA proves $\exists x P(x)$, but (assuming PA is consistent) it does not prove $P(s)$ for any closed term $s$.&lt;/p&gt;&#xA;" OwnerUserId="5203" LastActivityDate="2011-03-09T03:31:39.047" CommentCount="1" />
  <row Id="25899" PostTypeId="2" ParentId="25879" CreationDate="2011-03-09T03:42:19.220" Score="3" ViewCount="" Body="&lt;p&gt;I'm assuming your function has domain all of $\mathbb{R}^2$. For the codomain, some subset of $\mathbb{R}$. But you should state them explicitly, &lt;em&gt;especially&lt;/em&gt; when you are trying to figure out if the function is one-to-one and/or onto; those properties depend on the domain the codomain, not just on the formula. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To show the function is onto, given any $z$ you just need to produce a single pair $(x,y)$ that maps to $z$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your &quot;argument&quot; does not really work. You are defining $x$ in terms of both $z$ and $y$, and then defining $y$ in terms of both $z$ and $x$; you have a circular definition. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, say I ask you to use your formula to produce a pair $(x,y)$ that maps to $z=1$. You say: &quot;take $x = (z-y)/2 = (1-y)/2$, and then $y=z-2x = 1-2x$. So $x=(1-y)/2$, and $y=1-2x$.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ehr... okay. But what &lt;em&gt;are&lt;/em&gt; $x$ and $y$? If you plug in the value of $x$ into the formula for $y$, you get $y=y$; if you plug in the value of $y$ into the formula for $x$, you get $x=x$. You aren't giving any pair that maps to $1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, instead, given $z$, produce a &lt;em&gt;specific&lt;/em&gt; value of $x$ and of $y$ (that depends on $z$, of course, but not on $x$ or $y$) that will map to $z$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For one-to-one, your reasoning breaks down. Once you get to $2(x_1-x_2) = (y_2-y_1)$, you have no warrant for the assertion that $y_1=y_2$ and $x_1=x_2$. &lt;em&gt;If&lt;/em&gt; one of them holds, then the other holds; but what if &lt;em&gt;neither&lt;/em&gt; holds? For instance, if $x_1=2$ and $x_2=1$, can you find $y_1$ and $y_2$ that will make the equality true anyway? If so, then you can simply &lt;em&gt;exhibit&lt;/em&gt; the pairs $(2,y_1)$ and $(1,y_2)$ to show that the function is not one-to-one, since you can find two distinct pairs that map to the same thing.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2011-03-09T03:42:19.220" />
  <row Id="25900" PostTypeId="2" ParentId="25883" CreationDate="2011-03-09T03:51:48.023" Score="3" ViewCount="" Body="&lt;p&gt;First lets prove equivalence of the Mean Value Theorem (MVT) and Rolles theorem.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;$\text{MVT}\iff \text{Rolles}$:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clearly MVT implies Rolles.  For the converse, consider a differentiable function $f$ on the interval $[a,b]$ with $$\frac{f(b)-f(a)}{b-a}=c.$$  Then set $g(x)=f(x)-c(x-a)$.  Then $g(a)=f(a)$ and $g(b)=f(b)-(f(b)-f(a))=f(a)$ so we can apply rolles theorem to find a point such that $g&amp;#39;(x)=0$.  Since $$g^&amp;#39;(x)=f^&amp;#39;(x)-\frac{f(b)-f(a)}{b-a}$$ we have proven the MVT.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Chris Eagle showed why the MVT implies the least upper bound property.  Lets do the other direction:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;$\text{Least Upper Bound}\Rightarrow \text{Rolles}$:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose $f$ is differentiable on the interval $(a,b)$ with $f(a)=f(b)$, and continuous on $[a,b]$.  Since it will be uniformly continuous, it attains a maximum and minimum in $[a,b]$.  These cannot both be endpoints (otherwise it is constant), so assume WLOG the maximum occurs in $(a,b)$ at the point $\gamma$.  Then look at the difference quotients.  In some neighborhood of $\gamma$, if $x&amp;lt;\gamma$ then $$\frac{f(x)-f(\gamma)}{x-\gamma}\geq 0$$ and if $x&amp;gt;\gamma$ then  $$\frac{f(x)-f(\gamma)}{x-\gamma}\leq 0$$ since $f(\gamma)$ is a maximum.  Since $f$ is differentiable, the right and left hand limits must be equal.  Hence $f&amp;#39;(\gamma)=0$ as desired.&lt;/p&gt;&#xA;" OwnerUserId="6075" LastActivityDate="2011-03-09T03:51:48.023" />
  <row Id="25901" PostTypeId="1" CreationDate="2011-03-09T04:04:59.157" Score="1" ViewCount="57" Body="&lt;p&gt;Let $f_0 = 1$, and $f_1 = 1$, and $f_n = f_{n-1}+f_{n-2}$ when $n \gt 1$ (the Fibonacci sequence) &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Prove using induction that $f_n\gt 2n$, when $n \geq 6$. (note the $f_6 = 13$, $f_7=21$)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I want to rewrite $f_n \gt 2n$ as $f_n\gt f_{2n}$ is this legal?&lt;/p&gt;&#xA;" OwnerUserId="1040" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-09T04:06:10.707" LastActivityDate="2011-03-09T05:13:55.240" Title="Can I rewrite part of this induction?" Tags="&lt;induction&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="25902" PostTypeId="1" CreationDate="2011-03-09T04:26:11.447" Score="2" ViewCount="205" Body="&lt;p&gt;A machine works for an exponentially distributed time with rate μ and then fails. A repair crew checks the machine at times distributed according to a Poisson process with rate λ; if the machine is found to have failed then it is immediately replaced. Find the expected time between replacements of machines.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;What I have so far: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe that if M represents the failure time of the machine and T represents the time when the machine is checked, then the probability of the machine being replaced is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P(M &amp;lt; T) = μ/(μ+λ)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since both M and T are exponentially distributed. So, that got me into thinking that this as Exponential(μ+λ) distributed, and so the expected time for the machine to be fixed is just (μ+λ)/μ.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another thought was to find the expected time of the failure of the machine (1/μ) and then the expected time of checking the machine conditioned on the machine having failed, which I think is just 1/λ by the no memory property.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I feel like I do not understand what I am doing and am just aimlessly trying to solve the question. I can't seem to find a similar problem or example in my textbook (I'm on Ch 5 of &quot;Introduction to Probability Models&quot; by Ross, 10ed), and if anyone can help me with the problem or refer me to some good resources, I would be extremely grateful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Quick edit:&#xA;It looks like the question asks for the expected time &lt;em&gt;between&lt;/em&gt; replacements, so now I feel like I did things completely wrong :(&lt;/p&gt;&#xA;" OwnerUserId="7990" LastActivityDate="2011-05-28T18:53:04.110" Title="Expected value for a Poisson Process" Tags="&lt;stochastic-processes&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="25903" PostTypeId="1" AcceptedAnswerId="25939" CreationDate="2011-03-09T04:34:07.957" Score="5" ViewCount="427" Body="&lt;p&gt;We know that the Fourier transform of the sinc function is the rectangular function (or top hat). However, I'm at a loss as to how to prove it. Most textbooks and online sources start with the rectangular function, show that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_{-\infty}^\infty \text{rect}(x)e^{i\omega x}dx=\int_{-1/2}^{1/2}e^{i\omega x}dx=\left.\frac{e^{i\omega x}}{i\omega}\right\vert_{-1/2}^{1/2}=\text{sinc}(\omega/2)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and then just invoke duality and claim that the Fourier transform of the sinc function is the rectangular function. Is there any way of deriving this directly? i.e., starting with the sinc function? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've tried, but I'm not sure as to how to proceed. I know that the sinc is not Lebesgue integrable and only improper Riemann integrable. Some vague recollection of these being important in the Fourier transform hinders my thought process. Can someone clear things up for me?&lt;/p&gt;&#xA;" OwnerUserId="7815" LastEditorUserId="7815" LastEditorDisplayName="" LastEditDate="2011-03-09T07:10:47.230" LastActivityDate="2011-03-18T22:36:16.820" Title="Derive Fourier transform of sinc function" Tags="&lt;integral&gt;&lt;fourier-series&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="25904" PostTypeId="2" ParentId="25901" CreationDate="2011-03-09T04:42:12.887" Score="3" ViewCount="" Body="&lt;p&gt;It's not correct, as $f_{n} = f_{n-1} + f_{n-2}$, so, if you recursively write $f_{2n}$ in terms of $f_{n}$ and $f_{n-1}$, you'll, see that $f_{2n}\gt f_{n}$ if $n\geq 1$.&lt;/p&gt;&#xA;" OwnerUserId="7928" LastEditorUserId="7928" LastEditorDisplayName="" LastEditDate="2011-03-09T05:13:55.240" LastActivityDate="2011-03-09T05:13:55.240" CommentCount="1" />
  <row Id="25905" PostTypeId="2" ParentId="25902" CreationDate="2011-03-09T04:44:54.537" Score="1" ViewCount="" Body="&lt;p&gt;The time in between replacements of machines will be exponentially distributed. &lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2011-03-09T04:44:54.537" CommentCount="1" />
  <row Id="25906" PostTypeId="1" CreationDate="2011-03-09T04:49:24.420" Score="1" ViewCount="66" Body="&lt;p&gt;I am aware of the formula.. but can someone give me a clear definition of what the total area enclosed by a polar curve represents.&#xA;Thanks&lt;/p&gt;&#xA;" OwnerUserId="7984" LastActivityDate="2011-03-09T04:49:24.420" Title="What is the total area enclosed by a polar curve and the x-axis?" Tags="&lt;polar-coordinates&gt;" CommentCount="3" />
  <row Id="25907" PostTypeId="1" CreationDate="2011-03-09T04:56:40.390" Score="1" ViewCount="109" Body="&lt;p&gt;Homework problem I recently got:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Find the length of the curve of $y=\sqrt{a^2-x^2}$&#xA;  from $x=-2$ to $x=1$. Also find the area of the surface of revolution about the $x$-axis.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Having trouble understanding this.  I'm not even sure what it would look like.  I assume a is supposed to be a constant.  Is it just a semi-circle with radius $a$.  $X$-intercepts of $(-a,0)$ and $(a,0)$ and a max at $(0,a)$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm a little iffy about how to first find the length.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I do know that I should start with $\int_{-2}^{1} 2\pi y ds$ since I am revolving about the x-axis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for anyone who can help me out.&lt;/p&gt;&#xA;" OwnerUserId="6423" LastEditorUserId="6423" LastEditorDisplayName="" LastEditDate="2011-03-09T06:06:33.813" LastActivityDate="2011-03-09T19:32:30.337" Title="Find the length of the curve from x=-2 to x=1 of $y=\sqrt{a^2-x^2}$" Tags="&lt;calculus&gt;&lt;homework&gt;" AnswerCount="1" CommentCount="0" />
  <row Id="25908" PostTypeId="2" ParentId="25901" CreationDate="2011-03-09T04:57:20.623" Score="2" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\ $ Prove by induction that any increasing sequence, once positive, remains positive, then apply that to $\rm\ g_{\:n} =\ f_n - 2\ n\:,\: $ which is increasing since $\rm\ g_{\:n+1}-g_{\:n}\ =\ f_{n-1} - 2\ &amp;gt;\ 0\ $ for $\rm\ n \ge 6\:.$ &lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2011-03-09T04:57:20.623" />
  <row Id="25909" PostTypeId="1" AcceptedAnswerId="25912" CreationDate="2011-03-09T04:59:43.427" Score="2" ViewCount="63" Body="&lt;p&gt;I'm wondering if there is a function $f(x,y)$ such that&lt;br&gt;&#xA;$$\frac{\partial f}{\partial x}(x, y)=2x-3y$$&lt;br&gt;&#xA;and&lt;br&gt;&#xA;$$\frac{\partial f}{\partial y}(x, y)=3x+4y$$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Integrating the first equation with respect to x and the second with respect to y, we get:&lt;br&gt;&#xA;$$\int (2x-3y)dx = x^2-3xy+c_1$$ where $c_1$ can depend on y and:&lt;br&gt;&#xA;$$\int (3x+4y)dy = 2y^2+3xy+c_2$$ where $c_2$ can depend on x.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;At this point my argument is that in order to get a $-3y$ from a partial with respect to $x$, we need $-3xy$ in the function.  But to get $3x$ from the partial with respect to $y$, we need $-3xy$.  Therefore, in order to combine these 2 equations above, we would need $c_1$ to depend on $x$, which is not allowed, and $c_2$ to depend on $y$.  How can I formalize this argument?  Am I missing something?  Do I have the right idea?&lt;/p&gt;&#xA;" OwnerUserId="2533" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-09T05:06:43.140" LastActivityDate="2011-03-09T07:20:18.857" Title="Finding a function $f(x,y)$ with specified partials" Tags="&lt;calculus&gt;&lt;homework&gt;" AnswerCount="3" CommentCount="0" />
  <row Id="25910" PostTypeId="2" ParentId="25907" CreationDate="2011-03-09T05:05:57.027" Score="4" ViewCount="" Body="&lt;p&gt;The graph, as you surmised, of $y=\sqrt{a^2-x^2}$ is the upper half of the semicircle with center at $(0,0)$ and radius $a$. (Square both sides to get $y^2 = a^2-x^2$, or $x^2+y^2=a^2$, the equation of the circle described; since $y\geq 0$, it's the upper half). (Of course, you need $|a|\geq 2$ for your limits to make sense).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The formula for the length of the graph of $y=f(x)$ from $x=r$ to $x=s$ is&#xA;$$\mathrm{Arc\ Length} = \int_r^s \sqrt{1 + \left(\frac{dy}{dx}\right)^2}&#xD;&#xA;\,dx.$$&#xA;So here, you have $y=\sqrt{a^2-x^2}$, so take the derivative, square, plug it into the formula, that will give the arc length.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the area of the surface, I don't know which formula you know. The one I know is that the area is given by&#xA;$$\mathrm{Area} = \int_r^s 2\pi f(x)\sqrt{1 + \left(\frac{dy}{dx}\right)^2}\,dx.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added.&lt;/strong&gt; So, since you've essentially worked it out: if $|a|\gt 2$ (so we don't have to worry about the derivative and the curve being defined over the entire interval $[-2,1]$), we have&#xA;$$\frac{dy}{dx} = -\frac{x}{\sqrt{a^2-x^2}}.$$&#xA;So&#xA;$$ 1 + \left(\frac{dy}{dx}\right)^2 = 1 + \frac{x^2}{a^2-x^2} = \frac{a^2}{a^2-x^2}.$$&#xA;So for arc length, we have:&#xA;\begin{align*}&#xA;\mathrm{Arc\ Length} &amp;amp;= \int_{-2}^1 \sqrt{1 + \left(\frac{dy}{dx}\right)^2}\,dx \\&#xA;&amp;amp;= \int_{-2}^1 \sqrt{\frac{a^2}{a^2-x^2}}\,dx\\&#xA;&amp;amp;= \int_{-2}^1 \frac{a}{\sqrt{a^2-x^2}}\,dx\\&#xA;&amp;amp;= a\int_{-2}^1\frac{dx}{\sqrt{a^2-x^2}}\\&#xA;&amp;amp;= a\arcsin\left(\frac{x}{a}\right)\Biggm|_{-2}^1\\&#xA;&amp;amp;= a\arcsin\left(\frac{1}{a}\right) - a\arcsin\left(\frac{-2}{a}\right)\\&#xA;&amp;amp;= a\arcsin\left(\frac{1}{a}\right) + a\arcsin\left(\frac{2}{a}\right).&#xA;\end{align*}&#xA;For the surface area, you have:&#xA;\begin{align*}&#xA;\mathrm{Surface\ Area} &amp;amp;= \int_{-2}^1 2\pi f(x)\sqrt{1+\left(\frac{dy}{dx}\right)^2}\,dx\\&#xA;&amp;amp;= 2\pi \int_{-2}^1 \sqrt{a^2-x^2}\left(\frac{a}{\sqrt{a^2-x^2}}\right)\,dx\\&#xA;&amp;amp;= 2\pi a \int_{-2}^1 \,dx = 2\pi a(1-(-2)) = 6\pi a.&#xA;\end{align*}&#xA;If $a=2$, then the two integrals are improper and you need to use limits, but you'll end up with the same answer in both cases. If $0\lt a\lt 2$, the integrals are not well-defined (but then neither is the problem).&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-09T19:32:30.337" LastActivityDate="2011-03-09T19:32:30.337" CommentCount="17" />
  <row Id="25911" PostTypeId="2" ParentId="25909" CreationDate="2011-03-09T05:07:08.930" Score="8" ViewCount="" Body="&lt;p&gt;From the first equation ${\displaystyle {\partial \over \partial y} {\partial f \over \partial x} = -3}$, while from the second equation ${\displaystyle {\partial \over \partial x} {\partial f \over \partial y} = 3}$. Since the mixed partials have to be the same for your purposes, the equations can't be solved simultaneously.&lt;/p&gt;&#xA;" OwnerUserId="3035" LastActivityDate="2011-03-09T05:07:08.930" CommentCount="2" />
  <row Id="25912" PostTypeId="2" ParentId="25909" CreationDate="2011-03-09T05:11:29.890" Score="2" ViewCount="" Body="&lt;p&gt;Zarrax's solution is the straightforward way to do it. If you want to push through your own, it's best to write $c_1$ and $c_2$ explicitly as functions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You would need a function $f(x,y)$ with&#xA;$$f(x,y) = x^2 - 3xy + c_1(y) = 2y^2 + 3xy + c_2(x).$$&#xA;For equality to hold, you would need &#xA;$$c_1(y) = 2y^2 - x^2 + 6xy + c_2(x) = 2y^2 + 6xy + \Bigl(c_2(x)-x^2\Bigr).$$&#xA;Since $c_1(y)$ can only depend on $y$, this tells you that the part that depends on $x$ alone, $c_2(x)-x^2$, must be equal to $0$; but that still leaves the $6xy$ term, which cannot cancel out. So no such function $c_1(y)$ exists that depends only on $y$, so there are no solutions.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2011-03-09T05:11:29.890" CommentCount="1" />
  <row Id="25913" PostTypeId="1" AcceptedAnswerId="25915" CreationDate="2011-03-09T05:20:10.083" Score="5" ViewCount="74" Body="&lt;p&gt;Hello,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $p - a \equiv -a \pmod{p}$ then what is $\frac{p-1}{2} \equiv ? \pmod{p}$? Where $p$ is an odd prime.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I read in the book, they claimed:&#xA;$$p - 1 \equiv -1 \pmod{p}$$&#xA;$$p - 2 \equiv -2 \pmod{p}$$&#xA;$$p - 3 \equiv -3 \pmod{p}$$&#xA;$$ ... $$&#xA;$$\frac{p - 1}{2} \equiv -(\frac{p - 1}{2}) \pmod{p}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I realized that the last statement was wrong because if it's true then:&#xA;$$\frac{p - 1}{2} + \frac{p - 1}{2}\equiv 0 \pmod{p}$$&#xA;$$p - 1 \equiv 0 \pmod{p}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/mtF0X.png&quot; alt=&quot;This is my scan version from the book 6th edition&quot;&gt;&#xA;Any idea?   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks, &lt;/p&gt;&#xA;" OwnerUserId="5711" LastEditorUserId="5711" LastEditorDisplayName="" LastEditDate="2011-03-09T06:12:23.777" LastActivityDate="2011-03-09T06:31:28.823" Title="If $p - a \equiv -a \pmod{p}$ then what is $\frac{p-1}{2} \equiv ? \pmod{p}$?" Tags="&lt;elementary-number-theory&gt;" AnswerCount="2" CommentCount="11" FavoriteCount="1" />
  <row Id="25914" PostTypeId="1" AcceptedAnswerId="25919" CreationDate="2011-03-09T05:27:05.450" Score="1" ViewCount="148" Body="&lt;p&gt;Can someone guide me in the right direction on this question?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Prove that every $n$ in $\mathbb{N}$ can be written as a product of an odd integer and a non-negative integer power of $2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance: $36 = 2^2(9)$ , $80 = 2^4(5)$ , $17 = 2^0(17)$ , $64 = 2^6(1)$ , etc...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any hints in the right direction is appreciated (please explain for a beginner).&#xA;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="5505" LastEditorUserId="7840" LastEditorDisplayName="" LastEditDate="2011-03-09T05:35:52.090" LastActivityDate="2011-03-09T15:38:31.857" Title="Proof by Strong Induction" Tags="&lt;induction&gt;" AnswerCount="3" />
  <row Id="25915" PostTypeId="2" ParentId="25913" CreationDate="2011-03-09T05:28:26.297" Score="4" ViewCount="" Body="&lt;p&gt;The sequence as written doesn't make sense, since the last term does not follow the pattern of the previous ones. However, you &lt;em&gt;can&lt;/em&gt; argue that&#xA;$$p - \left(\frac{p-1}{2}\right) \equiv -\frac{p-1}{2}\pmod{p}.$$&#xA;This &lt;em&gt;does&lt;/em&gt; follow the same pattern as the rest of the terms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And since&#xA;$$p - \frac{p-1}{2} = \frac{2p-p+1}{2} = \frac{p+1}{2}$$&#xA;then you have that&#xA;$$\frac{p+1}{2}\equiv -\frac{p-1}{2}\pmod{p},$$&#xA;hence $\frac{p-1}{2}\equiv -\frac{p+1}{2}\pmod{p}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or peharps, it went like this:&#xA;\begin{align*}&#xA;p-1 &amp;amp;\equiv -1\pmod{p}\\&#xA;p-2 &amp;amp;\equiv -2 \pmod{p}\\&#xA;&amp;amp;\vdots\\&#xA;\frac{p-1}{2} = p-\left(\frac{p+1}{2}\right) &amp;amp;\equiv -\frac{p+1}{2}\pmod{p}&#xA;\end{align*}&#xA;and you have the wrong sign in the numerator on the right hand side?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added.&lt;/strong&gt; So, the rest of the proof. According to the scans Sivaram has posted, this was part of the computation of $\left(\frac{p-1}{2}\right)!^2$. The idea then is that we can get $\left(\frac{p-1}{2}\right)!$ by multiplying the terms on the right hand side, and then get it again by multiplying the terms on the left hand side. That is, we look at&#xA;\begin{align*}&#xA;p-1 &amp;amp;\equiv -1 \pmod{p}\\&#xA;p-2 &amp;amp;\equiv -2 \pmod{p}\\&#xA;&amp;amp;\vdots\\&#xA;p-\left(\frac{p-1}{2}\right) &amp;amp;\equiv -\frac{p-1}{2}\pmod{p}&#xA;\end{align*}&#xA;and note that if $p\equiv 3\pmod{2}$, then we have an odd number of negative signs on the right hand side. So we have that:&#xA;$$(-1)\left(\frac{p-1}{2}\right)! = (-1)\Bigl(1\times 2\times\cdots\times \frac{p-1}{2}\Bigr) \equiv (p-1)(p-2)\cdots\left(p - \frac{p-1}{2}\right)\pmod{p}.$$&#xA;Now, $p - \frac{p-1}{2} = \frac{p+1}{2} = \frac{p-1}{2}+1$. So therefore we have that:&#xA;\begin{align*}&#xA;\left(\frac{p-1}{2}\right)!^2 &amp;amp;= \left(\frac{p-1}{2}\right)!\left(\frac{p-1}{2}\right)!\\&#xA;&amp;amp;\equiv  \left(\frac{p-1}{2}\right)!\Biggl( -(p-1)(p-2)\cdots\left(\frac{p-1}{2}+1\right)\Biggr)\pmod{p}\\&#xA;&amp;amp;\equiv -\left(1\times 2\times 3\times\cdots\times\left(\frac{p-1}{2}\right)\times\left(\frac{p-1}{2}+1\right)\times\cdots\times (p-1)\right)\pmod{p}\\&#xA;&amp;amp;\equiv -(p-1)!\pmod{p}.&#xA;\end{align*}&#xA;But we know that $(p-1)!\equiv -1\pmod{p}$ by Wilson's Theorem, so&#xA;$$\left(\frac{p-1}{2}\right)!^2 \equiv -(p-1)!\equiv -(-1) = 1\pmod{p},$$&#xA;so if we let $x = \left(\frac{p-1}{2}\right)!$, then $x^2 \equiv 1 \pmod{p}$. This means that $p|x^2-1 = (x-1)(x+1)$, so either $p|x-1$ or $p|x+1$. that is, eithe $x\equiv 1 \pmod{p}$ or $x\equiv-1\pmod{p}$; giving: &#xA;$$\text{either }\left(\frac{p-1}{2}\right)!\equiv -1\pmod{p}\quad\text{or}\quad \left(\frac{p-1}{2}\right)!\equiv 1\pmod{p}.$$&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-09T06:31:28.823" LastActivityDate="2011-03-09T06:31:28.823" CommentCount="3" />
  <row Id="25916" PostTypeId="2" ParentId="25914" CreationDate="2011-03-09T05:31:37.870" Score="0" ViewCount="" Body="&lt;p&gt;Proof:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the fundamental theorem of algebra, every integer $N$ can be uniquely factored as $\prod^{n}_{i=1}p_{i}^{a_{i}}$. Now, mark $2=p_{1}$, note $a_{i}$ can take value of $0$. You got the theorem. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the &quot;inductive&quot; proof, suppose for $n&amp;lt;k$ this is true. For $n+1$ its factors must be in previous $n$ numbers. Hence $n+1=\prod n_{i}$. Decompose $n_{i}$ by induction hypothesis you get the statement. &lt;/p&gt;&#xA;" OwnerUserId="7887" LastActivityDate="2011-03-09T05:31:37.870" CommentCount="9" />
  <row Id="25917" PostTypeId="1" CreationDate="2011-03-09T05:35:15.473" Score="4" ViewCount="91" Body="&lt;p&gt;$\sqrt{2+\sqrt{2+\sqrt{2+\dots}}}$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\dots\sqrt{2+\sqrt{2+\sqrt{2}}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why they are different?&lt;/p&gt;&#xA;" OwnerUserId="7992" LastActivityDate="2011-03-09T05:48:13.073" Title="How to compute the following formulas?" Tags="&lt;arithmetic&gt;" AnswerCount="2" CommentCount="6" />
  <row Id="25918" PostTypeId="2" ParentId="25917" CreationDate="2011-03-09T05:39:39.903" Score="4" ViewCount="" Body="&lt;p&gt;Define a sequence $(x_n)_{n\geq1}$ so that $x_1=\sqrt2$ and $x_{n+1}=\sqrt{2+x_n}$. Then the second formula you give can be said to colorfully describe the limit $\lim_{n\to\infty}x_n$. If we suppose the limit $L$ does exist (and it is not difficult to show it does exist!), then since for all $n$ we have $x_{n+1}^2=2+x_n$, passing to the limit we see that $L^2=2+L$. The polynomial $x^2-x-2$ has two roots, $-1$ and $2$: since all the $x_n$ are positive, the only possible value for $L$ is $2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you do the other one?&lt;/p&gt;&#xA;" OwnerUserId="274" LastEditorUserId="274" LastEditorDisplayName="" LastEditDate="2011-03-09T05:48:13.073" LastActivityDate="2011-03-09T05:48:13.073" CommentCount="3" />
  <row Id="25919" PostTypeId="2" ParentId="25914" CreationDate="2011-03-09T05:42:08.917" Score="7" ViewCount="" Body="&lt;p&gt;To prove something by strong induction, you have to prove that&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If all natural numbers strictly less than $N$ have the property, then $N$ has the property.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;is true for all $N$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So: our induction hypothesis is going to be:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Every natural number $k$ that is strictly less than $n$ can be written as a product of a power of $2$ and an odd number.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And we want to prove that from this hypothesis, we can conclude that $n$ itself can be written as the product of a power of $2$ and an odd number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, we have two cases: either $n$ is odd, or $n$ is even. If we can prove the result holds in both cases, we'll be done.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Case 1:&lt;/strong&gt; $n$ is odd. Then we can write $n=2^0\times n$, and we are done. So in Case 1, the result holds for $n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Case 2:&lt;/strong&gt; If $n$ is even, then we can write $n=2k$ for some natural number $k$. But then $k\lt n$, so we can apply the induction hypothesis to $k$. We conclude that &lt;code&gt;...and you should finish this part...&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we conclude that the result holds for all natural numbers by strong induction.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2011-03-09T05:42:08.917" CommentCount="5" />
  <row Id="25920" PostTypeId="2" ParentId="25917" CreationDate="2011-03-09T05:42:18.477" Score="4" ViewCount="" Body="&lt;p&gt;Suppose that the first converges to some value $x$.  Because the whole expression is identical to the first inner radical, $\sqrt{2+\sqrt{2+\sqrt{2+\cdots}}}=x=\sqrt{2+x}$ and solving for $x$ gives $x=2$.  Of course, I haven't justified that it converges to some value.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second can be thought of as starting with $\sqrt{2}$ and repeatedly applying the function $f(x)=\sqrt{2+x}$.  Trying this numerically suggests that the values converge to 2.  Solveing $f(x)=x$ shows that $2$ is a fixed point of that function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Looking at the second expression is actually how I'd justify (though it is perhaps not a rigorous proof) that the first expression converges.&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2011-03-09T05:42:18.477" />
  <row Id="25921" PostTypeId="2" ParentId="25883" CreationDate="2011-03-09T05:45:27.260" Score="5" ViewCount="" Body="&lt;p&gt;The purpose of this answer is to provide some supplementary insight into Chris Eagle's answer and to show in particular that his proof of MVT $\implies$ LUB is a very natural one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In $\S 4$ of &lt;a href=&quot;http://math.uga.edu/~pete/realinduction.pdf&quot; rel=&quot;nofollow&quot;&gt;this note&lt;/a&gt;, I discuss the notions of completeness and Dedekind completeness in linearly ordered sets in terms of the induced order topology.  (Recall an ordered set is Dedekind complete if it satisfies LUB -- every nonempty subset which is bounded above has a least upper bound -- and is complete if it is Dedekind complete and has a minimal element and a maximal element.)  In particular, I show that a linear order is &lt;strong&gt;dense&lt;/strong&gt; and Dedekind complete iff it is connected in the order topology.  Every ordering on a field (compatible with the field structure, of course) is dense, so we see that an ordered field satisfies LUB iff it is connected in the order topology.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(In fact Chris Eagle's answer gives a snappy proof of &quot;connected $\implies$ LUB&quot;: if LUB does not hold, take a nonempty set $A$ which is bounded above but has no least upper bound.  Then the set of upper bounds of $A$ is nonempty, proper, open and closed.  The other implication is the one which is more familiar from real analysis, and is not quite so easy.  In the above note, I take the perspective that a very natural proof is given using the technique of &lt;strong&gt;real induction&lt;/strong&gt;.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if we want to show that MVT $\implies$ LUB (as Chris Eagle points out, LUB $\implies$ Rolle's Theorem $\implies$ MVT is part of the standard honors calculus / elementary real analysis curriculum), it is natural to go by contrapositive: if an ordered field $(F,+,\cdot,&amp;lt;)$ does not satisfy LUB, then it is not connected, i.e., there exists a nonconstant continuous function $f: F \rightarrow \{0,1\}$.  It follows immediately from the definition of the order topology that for each $x \in F$, there exists an open interval $I$ containing $x$ such that $f|_I$ is constant.  Thus $f$ is a nonconstant function which is everywhere differentiable with derivative identically zero, contradicting MVT.&lt;/p&gt;&#xA;" OwnerUserId="299" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2011-03-09T05:59:53.573" LastActivityDate="2011-03-09T05:59:53.573" CommentCount="1" />
  <row Id="25922" PostTypeId="2" ParentId="25913" CreationDate="2011-03-09T05:59:32.137" Score="1" ViewCount="" Body="&lt;p&gt;In the 5th edition, they have got the signs right, at least thats what the scanned version of the ebook says.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/YxGZv.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have $p-k \equiv -k \pmod{p}$ and hence $$ \prod_{k=1}^{\frac{p-1}{2}} (p-k) \equiv \prod_{k=1}^{\frac{p-1}{2}} (-k) \pmod{p}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that $$(p-1)! = \left(\prod_{k=1}^{\frac{p-1}{2}} k \right) \left(\prod_{k=1}^{\frac{p-1}{2}} (p-k) \right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, $$(p-1)! \equiv \left(\prod_{k=1}^{\frac{p-1}{2}} k \right) \left(\prod_{k=1}^{\frac{p-1}{2}} (p-k) \right) \pmod{p} \equiv \left(\prod_{k=1}^{\frac{p-1}{2}} k \right) \left(\prod_{k=1}^{\frac{p-1}{2}} (-k) \right) \pmod{p}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, $$(p-1)! \equiv (-1)^{\frac{p-1}{2}} \left(\prod_{k=1}^{\frac{p-1}{2}} k \right)^2 \pmod{p}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left(\prod_{k=1}^{\frac{p-1}{2}} k \right) = \left(\frac{p-1}{2}\right)!$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From Wilson's theorem, $$(p-1)! \equiv -1 \pmod{p}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, $$-1 \equiv (-1)^{\frac{p-1}{2}} \left[ \left(\frac{p-1}{2}\right)! \right]^2 \pmod{p}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, $$\left[ \left(\frac{p-1}{2}\right)! \right]^2 \equiv (-1)^{\frac{p+1}{2}}  \pmod{p}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, if $p = 4k+3$, then $$(-1)^{\frac{p+1}{2}} = 1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, if $p = 4k+3$, $$\left[ \left(\frac{p-1}{2}\right)! \right]^2 \equiv 1  \pmod{p}$$&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2011-03-09T06:27:13.163" LastActivityDate="2011-03-09T06:27:13.163" CommentCount="6" />
  <row Id="25923" PostTypeId="2" ParentId="25849" CreationDate="2011-03-09T06:25:44.150" Score="6" ViewCount="" Body="&lt;p&gt;For what it's worth, &lt;a href=&quot;http://math.uga.edu/~pete/4400primitiveroots.pdf&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; is my writeup of the theorem on non/existence of primitive roots modulo $N$, used for a course in number theory at the advanced undergraduate level.  (It does not use Hensel's Lemma.  As others have pointed out, although there is definitely something Henselesque going on here, it is not so straightforward to use HL to give a proof.)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The basic idea for showing that $U(p^n) = (\mathbb{Z}/p^n \mathbb{Z})^{\times}$ is cyclic for any odd prime $p$ was communicated to me by David Savitt: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As usual, we use the fact that $(\mathbb{Z}/p\mathbb{Z})^{\times}$ is cyclic, which I prove &lt;a href=&quot;http://math.uga.edu/~pete/4400primitiveroots.pdf&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; in the usual way (i.e., in essentially the same way as in Qiaochu's answer to this question).  So we know there exists an element of order $p-1$ (any lift of a generator mod $p$ will have order divisible by $p-1$, so take a suitable power of that element).  Since &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\# U(p^n) = \varphi(p^n) = p^{n-1}(p-1)$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;and &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\operatorname{gcd}(p^{n-1},p-1) = 1$, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;it is enough to find an element of order $p^{n-1}$.  In fact the element $1+p \pmod{p^n}$ has order $p^{n-1}$, which is proved using elementary and relatively painless binomial coefficient considerations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This proof is about 50% shorter than most others I have seen.  It does use a little familiarity with finite commutative groups, so would not be appropriate for the most general possible number-theoretic audience.&lt;/p&gt;&#xA;" OwnerUserId="299" LastActivityDate="2011-03-09T06:25:44.150" CommentCount="2" />
  <row Id="25924" PostTypeId="1" AcceptedAnswerId="25933" CreationDate="2011-03-09T06:41:32.037" Score="3" ViewCount="129" Body="&lt;p&gt;I'm having a little difficulty in proving what are probably simple induction proofs. Here is the question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Define function $f(n)$ as follows. $f(1) = 2$ and $f(n) = n*f(n-1)$ when $n &amp;gt; 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Use induction to prove that $f(n) &amp;gt; 2^n$ for all $n&amp;gt;2$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that $f(1) =2$, $f(2)=4$. I understand the basis step, so I'm not going to write that out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we have. $f(k) = k*f(k-1)&amp;gt; 2^k$ we want to show the following.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f(k) + f(k+1) = (k+1)*f(k)&amp;gt; 2^{k+1}$ This is where I'm getting stuck. Can someone give me a hint as to how to proceed?&lt;/p&gt;&#xA;" OwnerUserId="1040" LastActivityDate="2011-03-16T19:21:49.173" Title="Proof by induction" Tags="&lt;induction&gt;" AnswerCount="3" CommentCount="4" />
  <row Id="25925" PostTypeId="1" AcceptedAnswerId="26011" CreationDate="2011-03-09T06:46:04.210" Score="3" ViewCount="126" Body="&lt;p&gt;Let $S$ be the set of functions $f:\mathbb{R}\to \mathbb{R}$ such that $\sqrt{f(1)+\sqrt{f(2)+\sqrt{f(3)+\dots}}}$ converges.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A function $q(x)$ dominates $p(x)$ if there exist an m such that $q(x)\gt p(x)$ for all $x\gt m$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take all functions f(x) from S and put $O(f(x))$ in S2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which function $g(x)$ in $S2$ dominates all others?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there asymptotic lower and upper bounds on g(x)?&lt;/p&gt;&#xA;" OwnerUserId="4929" LastEditorUserId="4929" LastEditorDisplayName="" LastEditDate="2011-03-09T15:10:24.720" LastActivityDate="2011-03-09T16:56:23.687" Title="Nested radicals" Tags="&lt;sequences-and-series&gt;&lt;functions&gt;&lt;convergence&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="25927" PostTypeId="1" AcceptedAnswerId="25928" CreationDate="2011-03-09T07:10:53.207" Score="6" ViewCount="113" Body="&lt;p&gt;I have the following problem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose that $A$ is a symmetric matrix, with $A$ = $A^{T}$ . Suppose $\vec{v}$ and $\vec{w}$ are eigenvectors of $A$ associated&#xA;with distinct eigenvalues. Show that $\vec{v}$ and $\vec{w}$ must be orthogonal.&#xA;(Hint: Show that a$\vec{v}$ $\cdot$ $\vec{w}$ = $\vec{v}$ $\cdot$ b$\vec{w}$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am unsure how to approach this, even with the hint taken into account. I tried to use the fact that orthogonal complement of Im(A) is in Ker of A transpose, and since they are equal it is also in ker A, but that didn't get me anywhere (I am probably thinking in the wrong direction). Thanks in advance for your hints!&lt;/p&gt;&#xA;" OwnerUserId="7997" LastEditorUserId="7997" LastEditorDisplayName="" LastEditDate="2011-03-09T07:17:42.670" LastActivityDate="2011-03-09T13:57:32.103" Title="How to show that the following eigenvectors have to be orthogonal?" Tags="&lt;homework&gt;&lt;linear-algebra&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="25928" PostTypeId="2" ParentId="25927" CreationDate="2011-03-09T07:16:57.233" Score="5" ViewCount="" Body="&lt;p&gt;Let $\vec{v}$ be the eigenvector corresponding to $\lambda$ and $\vec{w}$ be the eigenvector corresponding to $\mu$, then we have $Av = \lambda v$ and $Aw = \mu w$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$v^T(Aw) = (Aw)^Tv$ since it is a scalar. And hence $v^TAw = w^TA^Tv$. Further since $A=A^T$, we get $v^TAw = w^TAv$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now use the fact that $Aw = \mu w$ and $Av = \lambda v$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And use the fact that $\lambda \neq \mu$, to get that the eigenvectors are orthogonal.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2011-03-09T07:28:46.313" LastActivityDate="2011-03-09T07:28:46.313" />
  <row Id="25929" PostTypeId="2" ParentId="25909" CreationDate="2011-03-09T07:20:18.857" Score="1" ViewCount="" Body="&lt;p&gt;Instead of integrating &lt;em&gt;both&lt;/em&gt; equations and then trying to match $c_1$ and $c_2$ to each other, it's usually simpler (especially for larger systems) to integrate just &lt;em&gt;one&lt;/em&gt; equation, say the first one. This gives&#xA;$$f(x,y) = x^2 - 3xy + c_1(y)$$&#xA;for some (still unknown) function $c_1(y)$.&#xA;Plugging that into the second differential equation, one finds that it holds iff&#xA;$$-3x + \frac{dc_1}{dy}(y) = 3x + 4y.$$&#xA;But it is impossible to satisfy this, since there is no function of $y$ alone such that its derivative equals $6x+4y$. Consequently, the system has no solution $f(x,y)$.&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2011-03-09T07:20:18.857" />
  <row Id="25930" PostTypeId="2" ParentId="25914" CreationDate="2011-03-09T07:27:34.913" Score="1" ViewCount="" Body="&lt;p&gt;This may be deduced from a more general result that's both simpler to prove and more insightful. Namely, the result follows immediately this frequently applicable &lt;em&gt;multiplicative&lt;/em&gt; form of induction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;LEMMA&lt;/strong&gt; $\ $ (Multiplicative Induction) $\ $ If a set $\rm\:N\subset\mathbb N\:$ contains $1$ and $\rm\:N\:$ contains every prime and, further, $\rm\:N\:$ is closed under multiplication, then $\rm\: N = \mathbb N\:.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof&lt;/strong&gt; $\ $ (by strong induction) $\ $ Suppose that $\rm\:N\:$ contains all naturals $\rm &amp;lt; n\:.\:$ If $\rm\:n\:$ is $1$ or prime then by hypothesis $\rm\:n\in N\:.\:$ Otherwise $\rm\:n\:$ is composite so $\rm\ n = j\ k\ $ for $\rm\: j,k &amp;lt; n\:.\:$ By the induction hypothesis $\rm\ j,k\in N\ $ hence $\rm\ n \ =\ j\ k\in N\ $ since $\rm\:N\:$ is closed under multiplication by hypothesis. $\quad$ &lt;strong&gt;QED&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the desired result follows easily by the lemma. Let $\rm\:N\:$ be the naturals that have the form $\rm\:2^i\ n\:$ for odd $\rm\:n\in \mathbb N\:.\ $ Clearly $1$ and all primes are in $\rm\:N\:.\ $ Furthermore $\rm\:N\:$ is closed under multiplication since $\rm\ (2^i\ m)\ (2^j\ n)\ =\ 2^{i+j}\ m\: n\ $ and $\rm\:m\:n\:$ is odd since $\rm\:m,n\:$ are odd. $\: $ Thus $\rm\ N = \mathbb N\ $ by the Lemma.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2011-03-09T15:38:31.857" LastActivityDate="2011-03-09T15:38:31.857" />
  <row Id="25931" PostTypeId="2" ParentId="25903" CreationDate="2011-03-09T07:27:53.590" Score="0" ViewCount="" Body="&lt;h2&gt;This computation has serious unknown flaws&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;I think for function $f(x)$ the fourier transform is defined by $$f(t)=\int f(x)e^{-2\pi ixt}dx$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the original function is $$\frac{sin(\pi x)}{\pi x}=\frac{e^{i\pi x}-e^{-i\pi x}}{2i\pi x}$$ hence the transformed one become $$(2i\pi)^{-1}\int \frac{(e^{i\pi(1-2t) x}-e^{-i\pi(1+2t) x})}{ x}dx$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence we need to evaluate $$\int \frac{e^{kix}}{x}dx, k=\pi(\pm 1-2t)$$ and we need to consider function $$g(z)=\frac{e^{ikz}}{z}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the indented semi circle on $x$-axis. with $x\rightarrow \infty$ the upper part just evaporated. Hence evaluating the inner circle we have $$\int \frac{e^{kix}}{x}dx=\int^{0}_{\pi}ie^{ikre^{i\phi}}d\phi=?$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and $$\int \frac{e^{-kix}}{x}dx=\int^{0}_{\pi}ie^{-ikre^{i\phi}}d\phi=?$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case $|t|=1/2$, either left or the right side becomes $1$. Now note $$\frac{1}{2i}\int \frac{e^{ix}-1}{x}=\frac{\pi}{2}$$ hence $$\frac{1}{2i\pi}\int \frac{e^{ikx}-1}{x}=\frac{1}{2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Otherwise the result should be ideally 0. I examined my computation several times, could not find the problem when $|t|&amp;lt;1/2$. If it does transform to the rectangular function then when $|t|&amp;lt;1/2$ it should be 1. &lt;/p&gt;&#xA;" OwnerUserId="7887" LastEditorUserId="7887" LastEditorDisplayName="" LastEditDate="2011-03-09T08:59:49.497" LastActivityDate="2011-03-09T08:59:49.497" CommentCount="8" />
  <row Id="25932" PostTypeId="2" ParentId="25405" CreationDate="2011-03-09T07:35:45.027" Score="2" ViewCount="" Body="&lt;p&gt;Suppose the graph has two MSTs. We list the first MST's edges increasingly according to the weights. Then we list the second MST's edges by the same way. Then we check the two lists. Two lists are denoted by $L_1$ and $L_2$. $L_{ij}$ represents the $j$-th edge in $L_i$. Clearly, there exists $k$ such that the weight of $L_{1k}$ is less than that of $L_{2k}$ and $L_{1j}$ is the same as $L_{2j}$ for $j&amp;lt;k$. Then we add $L_{1k}$ to the second MST, and some cycle appears. According to the Kruskal's algorithm, we could delete $L_{2j}$ for some $j&amp;gt;k$ to make $L_2$ the MST again. Clearly the weight of the new $L_2$ is lighter than that of the original $L_2$. Contradiction!&lt;/p&gt;&#xA;" OwnerUserId="7992" LastActivityDate="2011-03-09T07:35:45.027" />
  <row Id="25933" PostTypeId="2" ParentId="25924" CreationDate="2011-03-09T07:40:45.207" Score="0" ViewCount="" Body="&lt;p&gt;$$\begin{align}&#xD;&#xA;P(3):&amp;amp;\quad f(3)=12 &amp;gt; 8 = 2^3 \\&#xD;&#xA;P(k):&amp;amp;\quad f(k)&amp;gt;2^k \\&#xD;&#xA;P(k+1):&amp;amp;\quad f(k+1) = (k+1)f(k) &amp;gt; (k+1) 2^k &amp;gt;2 \cdot 2^k = 2^{k+1}&#xD;&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what we do is first to show that $f(n)&amp;gt;2^n$ for $n=3$. Then we assume that it is also true for $n=k$. At last we prove that it is true for $k+1$ if it is true for $k$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I did in the last step is this:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;I wrote down the expression for $f(k+1)$.&lt;/li&gt;&#xA;&lt;li&gt;Because of the step $P(k)$ we know (assume) that $f(k)&amp;gt;2^k$ which implies that $(k+1)f(k) &amp;gt; (k+1) 2^k$ since $k+1&amp;gt;0$.&lt;/li&gt;&#xA;&lt;li&gt;We also know that $k+1 &amp;gt; 2$ which implies that $(k+1) 2^k &amp;gt;2 \cdot 2^k$.&lt;/li&gt;&#xA;&lt;li&gt;$2 \cdot 2^k = 2^{k+1}$. So it follows that $f(k+1)&amp;gt;2^{k+1}$.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="7044" LastActivityDate="2011-03-09T07:40:45.207" />
  <row Id="25934" PostTypeId="2" ParentId="25924" CreationDate="2011-03-09T07:42:33.897" Score="0" ViewCount="" Body="&lt;p&gt;It might be clearer if you wrote $f(3)=12&amp;gt;8$ as well, to start you off.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now with the inductive hypothesis $f(n)&amp;gt;2^n$ and $n&amp;gt;2$ you look at $f(n+1)$ and this implies&#xA;$$f(n+1) = n \times f(n) &amp;gt; n \times 2^n &amp;gt; 2 \times 2^n = 2^{n+1} \textrm{ and } n+1&amp;gt;n&amp;gt;2$$ (essentially what Asaf Karagila  wrote backwards), i.e. &#xA;$$f(n+1) &amp;gt; 2^{n+1} \textrm{ and } n+1&amp;gt;2$$&#xA;which is all you need.&lt;/p&gt;&#xA;" OwnerUserId="6460" LastActivityDate="2011-03-09T07:42:33.897" />
  <row Id="25935" PostTypeId="1" AcceptedAnswerId="25946" CreationDate="2011-03-09T07:46:00.430" Score="8" ViewCount="193" Body="&lt;p&gt;Can anyone think of sequences $\{a_n\}$, $\{b_n\}$ such that $\sum a_n$ diverges, ${b_n}\to\infty$, but $\sum a_nb_n $ converges?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that $\{a_n\}$ must have infinitely many positive terms and infinitely many negative terms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: I get the feeling that only Qiaochu Yuan could answer this one... ;)&lt;/p&gt;&#xA;" OwnerUserId="7999" LastEditorUserId="7999" LastEditorDisplayName="" LastEditDate="2011-03-09T08:12:07.397" LastActivityDate="2011-03-09T08:34:51.917" Title="Example of Convergent Series" Tags="&lt;sequences-and-series&gt;" AnswerCount="1" CommentCount="10" FavoriteCount="2" />
  <row Id="25938" PostTypeId="1" CreationDate="2011-03-09T07:55:09.843" Score="0" ViewCount="85" Body="&lt;p&gt;i always had an intuition that formulas of coordinate formulas can be derived by use of cones (such as general equations of parabola etc.) Can anybody please help me.&#xA;And another Question : is there any relation between semivertical angle of a cone and the angle between coordinate axes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;some more observations of mine - a cone can be cut into a plane extending to infinity in all directions. also shifting the vertex of cone like shifting of origin can help explain presence of imaginary lines passing through any point on the coordinate plane &lt;/p&gt;&#xA;" OwnerUserId="8000" LastEditorUserId="8000" LastEditorDisplayName="" LastEditDate="2011-03-11T04:39:29.483" LastActivityDate="2011-03-11T04:39:29.483" Title="Relation of cone with formulas of coordinate geometry" Tags="&lt;conic-sections&gt;" CommentCount="3" />
  <row Id="25939" PostTypeId="2" ParentId="25903" CreationDate="2011-03-09T08:01:03.997" Score="2" ViewCount="" Body="&lt;p&gt;Since sinc is an entire function and decays with $1/\omega$, we can slightly shift the contour of integration in the inverse transform, and since there's no longer a singularity then, we can split the integral in two:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{eqnarray}\int_{-\infty}^\infty e^{i\omega x}\frac{e^{i\omega/2}-e^{-i\omega/2}}{i\omega}\mathrm{d}\omega&#xD;&#xA;&amp;amp;=&amp;amp;&#xD;&#xA;\int_{-\infty+\epsilon i}^{\infty+\epsilon i} e^{i\omega x}\frac{e^{i\omega/2}-e^{-i\omega/2}}{i\omega}\mathrm{d}\omega\\\&#xD;&#xA;&amp;amp;=&amp;amp;&#xD;&#xA;\int_{-\infty+\epsilon i}^{\infty+\epsilon i} e^{i\omega x}\frac{e^{i\omega/2}}{i\omega}\mathrm{d}\omega - &#xD;&#xA;\int_{-\infty+\epsilon i}^{\infty+\epsilon i} e^{i\omega x}\frac{e^{-i\omega/2}}{i\omega}\mathrm{d}\omega\;.&#xD;&#xA;\end{eqnarray}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we can apply different substitutions to the two parts, $\omega&amp;#39;=\omega(x+\frac{1}{2})$ in the first part and $\omega&amp;#39;=\omega(x-\frac{1}{2})$ in the second part. That transforms the integrand into $e^{i\omega&amp;#39;}/(i\omega&amp;#39;)$ in both cases. Now if $x$ lies outside the rectangle, the signs of the factors in the substitutions are the same, so the two integrals stay on the same side of the origin and go in the same direction, and hence yield the same value and cancel to $0$. But if $x$ lies inside the rectangle, then there's a sign change due to $x-\frac{1}{2}$ but not due to $x+\frac{1}{2}$, so we get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_{-\infty+\epsilon i}^{\infty+\epsilon i} \frac{e^{i\omega&amp;#39;}}{i\omega&amp;#39;}\mathrm{d}\omega&amp;#39; + &#xD;&#xA;\int_{\infty-\epsilon i}^{-\infty-\epsilon i} \frac{e^{i\omega&amp;#39;}}{i\omega&amp;#39;}\mathrm{d}\omega&amp;#39;\;,$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is (again using the sufficient decay at infinity)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\oint \frac{e^{i\omega&amp;#39;}}{i\omega&amp;#39;}\mathrm{d}\omega&amp;#39;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;on a contour that encloses the pole at the origin, and hence the value is $2\pi$.&lt;/p&gt;&#xA;" OwnerUserId="6622" LastEditorUserId="6622" LastEditorDisplayName="" LastEditDate="2011-03-09T08:19:24.410" LastActivityDate="2011-03-09T08:19:24.410" CommentCount="7" />
  <row Id="25941" PostTypeId="2" ParentId="25924" CreationDate="2011-03-09T08:03:19.860" Score="1" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\rm\quad\displaystyle \frac{f(n)}{2^n}\ =\ \frac{2}2\ \frac{2}{2}\ \frac{3}2\ \frac{4}2\ \cdots\ \frac{n}2\ &amp;gt; 1\ $ for $\rm\ n &amp;gt; 2\ $ since each factor is $&amp;gt;1$ after the 2nd factor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Generally that works to show that factorials grow faster than powers, i.e. $\rm\  f(n) &amp;gt; c^n\ $ for $\rm\ n &amp;gt; n_0\:.\ $ It suffices to show: eventually $\rm\ g(n) = f(n)/c^n &amp;gt; 1\:,\: $ or, equivalently, eventually $\rm\: g(n+1)/g(n) &amp;gt; 1 \ $ since, by &lt;strong&gt;multiplicative telescopy,&lt;/strong&gt; $\rm\:g(n)\:$ is a product of these adjacent term ratios, namely &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\rm g(0)\ \ \prod_{k\:=\:0}^{n-1}\ \frac{g(k+1)}{g(k)}\ =  \ \ {\rlap{--}g(0)}\frac{\rlap{--}g(1)}{\rlap{--}g(0)}\frac{\rlap{--}g(2)}{\rlap{--}g(1)}\ \ \cdots\ \ \frac{g(n)}{\rlap{----}g(n-1)}\ =\ \ g(n) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yours has $\rm\ g(k+1)/g(k)\ =\ (k+1)/2\ &amp;gt;\ 1\ $ for $\rm\ k &amp;gt; 1\ $ so $\rm\:g(n) &amp;gt; 1\:,\:$ as a product of terms  $&amp;gt; 1\:.$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As I have emphasized here before in &lt;a href=&quot;http://math.stackexchange.com/search?q=user:242+telescopy&quot;&gt;many posts,&lt;/a&gt;  by means of cancelling complicated expressions, &lt;em&gt;telescopy&lt;/em&gt; often reduces induction problems to trivialities (e.g. a product of terms $&amp;gt; 1$ is itself $&gt; 1$). Difficult problems involving &lt;em&gt;hyperrational&lt;/em&gt; functions (i.e. $\rm\ f(n+1)/f(n) = $ rational function of $\rm\:n\:,\:$ such as powers and exponentials) are, after application of telescopy, greatly simplified to trivial problems about &lt;em&gt;rational&lt;/em&gt; functions - functions so simple that questions about such can be decided mechanically by algorithms.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2011-03-16T19:21:49.173" LastActivityDate="2011-03-16T19:21:49.173" />
  <row Id="25943" PostTypeId="2" ParentId="25867" CreationDate="2011-03-09T08:11:37.467" Score="1" ViewCount="" Body="&lt;p&gt;It not quite clear what you mean by allowing repeated numbers, but what one usually considers in that case is so-called &lt;em&gt;semi-standard&lt;/em&gt; Young tableaux, i.e., tableaux which are increasing (strict inequality) down each column, but only nondecreasing (equality allowed) along each row. The number of such arrangements on a given Young diagram, where the numbers $1,2,\dots,N$ are allowed, is counted as follows: define the &quot;content&quot; of box $(i,j)$ in the diagram to be $i-j$. Here's an illustration:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; 0  1  2  3  4&#xA;-1  0  1  2&#xA;-2 -1  0&#xA;-3 -2&#xA;-4 -3&#xA;-5&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Hook lengths are defined as for the usual hook-length formula for counting &lt;em&gt;standard&lt;/em&gt; Young tableaux:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;10  8  5  3  1&#xA; 8  6  3  1&#xA; 6  4  1&#xA; 4  2&#xA; 3  1&#xA; 1&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;To get the answer, take the product over all boxes of (($N$ plus the content of that box) divided by (the hook length for that box)).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(This is a special case of something called Stanley's hook-content formula.)&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2011-03-09T08:11:37.467" />
  <row Id="25945" PostTypeId="2" ParentId="25221" CreationDate="2011-03-09T08:28:52.093" Score="6" ViewCount="" Body="&lt;p&gt;The claim made &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;... that the exterior of a circle is a segment connected set, which means that if $B$ and $C$ are two points outside a circle $\Gamma$, then there exists a third point $D$ such that $\overline{BD}$ and $\overline{DC}$ are entirely outside $\Gamma$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;is &lt;em&gt;not&lt;/em&gt; true in &lt;a href=&quot;http://en.wikipedia.org/wiki/Absolute_geometry&quot; rel=&quot;nofollow&quot;&gt;neutral plane geometry&lt;/a&gt; (a.k.a. the Hilbert plane). More precisely, it is not provable since it is true in Euclidean geometry and false in hyperbolic geometry, both of which are models of neutral geometry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, using the definition of a segment-connected set given in &lt;a href=&quot;http://books.google.com/books?id=EJCSL9S6la0C&amp;amp;pg=PA97&amp;amp;lpg=PA97&amp;amp;dq=%22Hilbert+plane%22&amp;amp;source=bl&amp;amp;ots=uyYV0QtZvf&amp;amp;sig=RgL8UPLVJ635maNMtwrcPmIBC20&amp;amp;hl=en&amp;amp;ei=0oh2TbSDLor6swOz4_S5BA&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=8&amp;amp;ved=0CE8Q6AEwBw#v=onepage&amp;amp;q=%22segment-connected%22&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;Hartshorne's &lt;em&gt;Geometry&lt;/em&gt; page 80&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A subset $W$ of the plane is&#xA;  segment-connected if given any two&#xA;  points $A, B \in W$, there is a finite&#xA;  sequence of points $A = A_1, A_2, \dots , A_n = B$&#xA;  such that for each $i = 1, 2,..., n-1$,&#xA;  the segment $\overline {A_iA_{i+1}}$ is entirely contained&#xA;  within $W$,&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;it can be shown that, in neutral geometry, the exterior of a circle &lt;em&gt;is&lt;/em&gt; segment-connected.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Proof that the claim is true in Euclidean geometry&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Let circle $\Gamma$ have radius $r$ and center $O$. If the line $\ell$ connecting $B$ and $C$ does not pass through $\Gamma$, we are done. Otherwise, let us construct a line $b$ passing through $B$ perpendicular to $\overleftrightarrow{OB}$ and a line $c$ passing through $C$ perpendicular to $\overleftrightarrow{OC}$ If line $\ell$ does not pass through $O$, $B$ and $C$ are on the same side of a line $m$ parallel to $\ell$ and passing through $O$, and the measure of angle $\angle BOC$ is less than $180$. If we consider $\ell$ a transversal cutting lines $b$ and $c$ and consecutive interior angles $\beta$ and $\gamma$ sum to less than $180$. Thus by Euclid's fifth postulate lines $b$ and $c$ intersect at some point $D$ and because all points on $b$ are at least $OB$ from $O$ and all points on $c$ are at least $OC$ from $O$ the segments $\overline{BD}$ and $\overline{DC}$ are entirely outside $\Gamma$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/cbjrH.png&quot; alt=&quot;Connecting two points with two segments in Euclidean geometry&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $\ell$ does pass through $O$, then without loss of generality let us assume $OB \leq OC$. Set $\epsilon = \frac{OB-r}{2}$ and construct a line $n$ perpendicular to $\ell$ at $O$. Find a point $O&amp;#39;$ on $n$ a distance $\epsilon$ from $O$.  Let us now consider a second circle $\Gamma&amp;#39;$ with center $O&amp;#39;$ and radius $r+\epsilon$. The circle $\Gamma&amp;#39;$ fully contains $\Gamma$ since, if point $X$ is in or on $\Gamma$, $OX \leq r$ and, by the triangle inequality, $O&amp;#39;X \leq OX + \epsilon \leq r+ \epsilon$. So $X$ is in or on $\Gamma&amp;#39;$. Point $B$ is not in $\Gamma&amp;#39;$ since $m$ is perpendicular to $\ell$ at $O$ and $O&amp;#39;B &amp;gt; OB = r + 2\epsilon &amp;gt; r+ \epsilon$. We get a similar result for point $C$. Hence, we can apply the first part of this proof to $B$, $C$ and the new circle $\Gamma&amp;#39;$ to address this case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/ESqJF.png&quot; alt=&quot;The bigger circle contains the smaller, but no the points&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Proof that the claim is false in hyperbolic geometry&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Consider any circle $\Gamma$ with center $O$ and radius $r$ and with diameter $\overline{QQ&amp;#39;}$. Let $\ell$ be the line perpendicular to line $\overleftrightarrow{QQ&amp;#39;}$ at $O$, and let $P$ and $P&amp;#39;$ be the points of $\ell$ which intersect the circle. Let $m$ be the line passing through $P$ that is asymptotic to line $\overleftrightarrow{QQ&amp;#39;}$ on the $Q$ side of $O$. Drop a perpendicular from $O$ to the line $m$, and let $F$ be the foot of that perpendicular. The point $F$ will be the closest approach that $m$ comes to $O$ and, since $m$ is not perpendicular to $\ell$ at $P$, $OF &amp;lt; r$. Let $R$ be a point of $\overrightarrow{OF}$ between $F$ and the circle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $n$ be the line through $R$ that is asymptotic to $m$ in the other direction. since $m$ and $n$ are asymptotic, any line passing through one, in the direction of the asymptote, passes through the other. So the line $\ell$ passes through both. Consider any one of the points of $\ell$ between $m$ and $n$ — call it $B$. Any line through $B$ that avoids the circle must pass through $n$ and must also pass through $m$ but on the $Q&amp;#39;$ side of $\ell$. It consequently cannot cross $m$ on the $Q$ side and so cannot cross the ray $\overrightarrow{OQ}$. We can make the same argument on the other side of $\ell$ to find points, $B$, near $P$ just outside the circle such that if a line passes through $B$ but not the circle, it cannot intersect ray $\overrightarrow{OQ&amp;#39;}$. There is, apparently, no line through these points that can avoid the circle and go across line $\overleftrightarrow{QQ&amp;#39;}$. Of course, we can make a similar argument about points, $C$, near $P&amp;#39;$ just outside the circle. However, if we are to find a $D$ for a pair of these $B$'s and $C$'s such that $\overline{BD}$ and $\overline{DC}$ are entirely outside $\Gamma$, one or the other of $\overline{BD}$ or $\overline{DC}$ must cross $\overleftrightarrow{QQ&amp;#39;}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/8kqJG.png&quot; alt=&quot;Showing segment-connection may require more than two segments in hyperbolic geometry&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have shown that in hyperbolic geometry there are points in the exterior of any circle that require more than two segments to connect them and stay in the exterior. Fooling around with the Poincaré disk, reveals that the larger the circle, the more segments may be needed.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Proof that, in neutral geometry, the exterior of a circle is segment-connected&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Consider any circle $\Gamma$ with center $O$ and radius $r$ and any points $B$ and $C$ in the circle's exterior. Construct a perpendicular line $m_1$ to line $\overleftrightarrow{OB}$ at point $B$. Note that all of line $m_1$ lies outside the circle. Find a point $A_1$ on line $m_1$ to one side of $B$ at a distance of $2r$ from $B$ and let angle $\theta$ be the measure of angle $\angle BOA_1$. If $C$ falls on or in the interior of angle $\angle BOA_1$, then the ray $\overrightarrow{OC}$ will intersect $m_1$ at a point $E$. Connect $B$ to $E$ with a segment that lies on $m_1$ and so lies entirely outside $\Gamma$, and then finish by connecting $E$ to $C$ with a segment which will also lie outside since $E$ ad $C$ both lie outside $\Gamma$. If $C$ does not fall in the interior of angle $\angle BOA_1$, connect B to $A_1$, and then find the next point of our segment connection on ray $\overrightarrow{OA_1}$ at distance $OB$ from $O$ and call the point $B_1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now repeat the process with $B_1$. Construct a perpendicular line $m_2$ to line $\overleftrightarrow{OB_1}$ at point $B_1$, and find a point $A_2$ on line $m_2$ to the side of $B_1$ opposite $B$ at a distance of $2r$ from $B_1$. By SAS, the triangle $\triangle OBA_1$ is congruent to $\triangle OB_1A_2$ and so the measure of the angle $\angle B_1OA_2$ is also $\theta$. Again, if $C$ falls on or in the interior of angle $\angle B_1OA_2$, then we can find, as before, a point $E$ on $m_2$  and connect $B_1$ to $E$ to $C$ with segments. If not, connect to $A_2$ and find the next point of our segment connection on ray $\overrightarrow{OA_2}$ at distance $OB$ from $O$ and call the point $B_2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can repeat the process until, by Archimedian property of angle measures, $C$ lies within $\angle B_iOA_{i+1}$ by which time we will have constructed a segment connection between $B$ and $C$ entirely in the exterior of $\Gamma$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/kZkf9.png&quot; alt=&quot;How to segment-connect outside a circle in neutral geometry&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="6272" LastEditorUserId="6272" LastEditorDisplayName="" LastEditDate="2011-03-10T19:18:10.477" LastActivityDate="2011-03-10T19:18:10.477" CommentCount="5" />
  <row Id="25946" PostTypeId="2" ParentId="25935" CreationDate="2011-03-09T08:31:08.270" Score="9" ViewCount="" Body="&lt;p&gt;Let $c_n = (-1)^n/\sqrt{n}$. Then $\sum c_n$ converges by the alternating series test. Let $b_n=\sqrt{n}$ or $n$ depending on whether $n$ is even or odd. Then $b_n \to \infty$. Let $a_n = c_n / b_n$. Then $a_n = 1/n$ if $n$ is even and $a_n = -1/n\sqrt{n}$ if $n$ is odd, so the negative $a_n$'s have a finite sum, which the positive $a_n$'s don't. Hence $\sum a_n$ diverges.&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2011-03-09T08:31:08.270" CommentCount="1" />
  <row Id="25948" PostTypeId="1" AcceptedAnswerId="25950" CreationDate="2011-03-09T09:14:11.947" Score="3" ViewCount="153" Body="&lt;p&gt;If I write (1 3 4 6)(2 3 4)(4 6 1) as a product of disjoint cycles, I get (164)(23), is it true that order of (1 3 4 6)(2 3 4)(4 6 1) is lcm(2,3)=6 (just the order of its disjoint cycles)?&lt;/p&gt;&#xA;" OwnerUserId="7340" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-10T01:44:43.347" LastActivityDate="2011-03-10T01:44:43.347" Title="order of a permutation" Tags="&lt;group-theory&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="25949" PostTypeId="1" AcceptedAnswerId="42390" CreationDate="2011-03-09T09:16:08.263" Score="1" ViewCount="76" Body="&lt;p&gt;Hi, I need to build up the matrix of matrices. Say, for all $1\leq i,j\leq n$ I have to define a matrix $p_{ij}$ which will be a matrix $n\times n$. What I can do - it is to build up a matrix $P$ which is $n^2\times n^2$ - but for $n=20$ there is an error about memory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could you please tell me how to solve this problem?&lt;/p&gt;&#xA;" OwnerUserId="5887" LastActivityDate="2011-05-31T17:04:06.607" Title="Matrix of matrices in matlab" Tags="&lt;matlab&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="0" />
  <row Id="25950" PostTypeId="2" ParentId="25948" CreationDate="2011-03-09T09:33:40.433" Score="1" ViewCount="" Body="&lt;p&gt;Yes, it's true that the order of a product of disjoint cycles is the least common multiple of the orders of the disjoint cycles in the product.&lt;/p&gt;&#xA;" OwnerUserId="5798" LastActivityDate="2011-03-09T09:33:40.433" />
  <row Id="25951" PostTypeId="2" ParentId="25885" CreationDate="2011-03-09T09:36:20.040" Score="-1" ViewCount="" Body="&lt;p&gt;This is more of an addendum for later, when you dig deeper into functional analysis: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Complex analysis tells us, that every holomorphic function can be represented by its Taylor series, locally. Actually, the space of square summable complex numbers is, as a Hilbert space, isomporph to all holomorphic functions on the unit disc $D := \{ z \in \mathbb{C}, |z| \le 1 \}$, that are square integrable:&#xA;$$&#xD;&#xA;\int_D |f(z)|^2 dz \le \infty&#xD;&#xA;$$ &#xA;with the obvious scalar product&#xA;$$&#xD;&#xA;(f, g) = \int_D \bar f (z) g(z) dz &#xD;&#xA;$$&#xA;via the mapping&#xA;$$&#xD;&#xA;f(z) \to \sum_{i = 0}^{\infty} a_i z^i&#xD;&#xA;$$&#xA;and vice versa.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It may help to compare the proof of the Hilbert space axioms for the space of sequences with the proof of the axioms for the function space (the latter needs some &quot;advanced&quot; knowledge of complex calculus). Some may seem to be easier, some more involved. For example it is easy to see that the scalar product actually is a scalar product, without the need to handle infinite sums.&lt;/p&gt;&#xA;" OwnerUserId="7556" LastActivityDate="2011-03-09T09:36:20.040" CommentCount="3" />
  <row Id="25952" PostTypeId="1" CreationDate="2011-03-09T09:52:50.540" Score="0" ViewCount="112" Body="&lt;p&gt;Hello,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm sure this question has a simple answer, but I'm a very beginner to calculus...&lt;br&gt;&#xA;This is the problem I have: Given a vertex point and two x-axis cutting points, how do I find a formula for parabola-like diagram?&lt;br&gt;&#xA;Important part: the vertex point is not in the middle of the parabola.    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I know, there are infinite solutions - how can find one specific?    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you.&lt;/p&gt;&#xA;" OwnerUserId="8005" LastActivityDate="2011-03-09T10:39:56.630" Title="Draw a parabola-like graph according to 3 points" Tags="&lt;conic-sections&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="25953" PostTypeId="2" ParentId="25952" CreationDate="2011-03-09T10:02:34.423" Score="1" ViewCount="" Body="&lt;p&gt;EDIT in light of myself's comment.  By &quot;parabola-like graph&quot; I assume you mean quadratic in $x$.  Three points determines a quadratic in $x$ uniquely.  See &lt;a href=&quot;http://en.wikipedia.org/wiki/Polynomial_interpolation&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Polynomial_interpolation&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="7862" LastEditorUserId="7862" LastEditorDisplayName="" LastEditDate="2011-03-09T10:38:01.150" LastActivityDate="2011-03-09T10:38:01.150" CommentCount="2" />
  <row Id="25954" PostTypeId="2" ParentId="25753" CreationDate="2011-03-09T10:17:13.717" Score="0" ViewCount="" Body="&lt;p&gt;Every permutation of a finite set can be expressed as the product of transpositions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously, under the system I have described, I can transpose the top two cards. but I do not know how to devise a method of moving any two cards to the top without disturbing the other cards?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is unclear whether m or n are allowed to be zero. Can my method deal with the situation if they are not?&lt;/p&gt;&#xA;" OwnerUserId="7340" LastActivityDate="2011-03-09T10:17:13.717" />
  <row Id="25955" PostTypeId="2" ParentId="25952" CreationDate="2011-03-09T10:17:39.863" Score="3" ViewCount="" Body="&lt;p&gt;In general, one of the many ways to start, write &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$Ax^2+Bx +C=y$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Write three of these for your three pairs, i.e translate the information given to: $(x_1,y_1)$,$(x_2,y_2)$,$(x_3,y_3)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You've got three equations in three unknowns. Solve it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or, if $a,b$ are your x-axis cutting points, then parabola will be of the form&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$y= K (x-a) (x-b)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Determine K by using dy/dx =0 at vertex coordinates.&lt;/p&gt;&#xA;" LastActivityDate="2011-03-09T10:39:56.630" />
  <row Id="25956" PostTypeId="1" CreationDate="2011-03-09T08:53:07.290" Score="0" ViewCount="148" Body="&lt;p&gt;Hello, can anyone help me?,i have a line Ax+By+C=0(let's call it Line1); i want to draw a perpicidular line(Line2) from the center of line1 with a length of 10..&#xA; what i did is getting the slope and intercept of line1 as well as the it's middle point(point m).&#xA;and i used the length equation L=sqrt((line2.x-m.x)^2+(line2.y-m.y)^2);but i dont get the right result.&#xA;Thanks in advance&lt;/p&gt;&#xA;" LastActivityDate="2011-03-09T11:24:33.217" Title="draw line which is perpindicular to another line with a wanted length" Tags="&lt;geometry&gt;" AnswerCount="5" CommentCount="2" />
  <row Id="25957" PostTypeId="2" ParentId="25956" CreationDate="2011-03-09T08:59:03.687" Score="0" ViewCount="" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;i have a line &lt;strong&gt;Ax+By+C=0&lt;/strong&gt;(let's call&#xA;  it Line1); i want to draw a&#xA;  perpicidular line(Line2) &lt;strong&gt;from the&#xA;  center of line1&lt;/strong&gt; with a length of 10&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am curious: how do you define the center of an infinite line? Is your line in fact a segment defined by two points? If so, do you want to draw a perpendicular line from the middle point of those segment ends and perpendicular to that?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In that case, you can use the vector that is perpendicular to the original vector. If those 2 point are named A and B. The vector being AB has components (x, y). A perpendicular vector to AB has the components (y, x). Then you simply need to draw a line following that vector from the point in the middle of [A, B] (simply add coordinates of the points and divide by 2).&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;If you have already have the middle point and the equation, then you can have a look at &lt;a href=&quot;http://www.purplemath.com/modules/strtlneq3.htm&quot; rel=&quot;nofollow&quot;&gt;this tutorial to help you solve the equation&lt;/a&gt;. (find the slope of the line then invert and negate to get the slope of the perpendicular)&lt;/p&gt;&#xA;" LastActivityDate="2011-03-09T09:04:09.130" CommentCount="1" />
  <row Id="25958" PostTypeId="2" ParentId="25956" CreationDate="2011-03-09T08:59:32.080" Score="0" ViewCount="" Body="&lt;p&gt;Defining the line Ax + By + C = 0 as a function f(x), you can use linear interpolation between the two endpoints where f(a) = p(0) [beginning] and f(b) = p(t) [end]. The value t = 0.5 will then be at the center of the line and the perpendicular slope is the reciprocal of f(x).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Useful resource for performing this task:&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Linear_interpolation&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Linear_interpolation&lt;/a&gt;&lt;/p&gt;&#xA;" LastActivityDate="2011-03-09T08:59:32.080" />
  <row Id="25959" PostTypeId="2" ParentId="25956" CreationDate="2011-03-09T09:05:53.703" Score="0" ViewCount="" Body="&lt;p&gt;The line &lt;strong&gt;l&lt;/strong&gt; = (A,B,C) where &lt;strong&gt;lx&lt;/strong&gt; = 0 has the normal (A,B). There are two perpendicular vectors to this normal namly (A, -B) and (-A, B).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Two lines are perpendicular if their normals &lt;a href=&quot;http://en.wikipedia.org/wiki/Cross_product&quot; rel=&quot;nofollow&quot;&gt;cross proudct&lt;/a&gt; equals zero. We can easily see that (A,B) x (A,-B) = A*-B - B*A&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The C value of the perpendicular line to be created does not matter and can be set to zero or C for example.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we have line Ax + Bx + C = 0 have a perpendicular line Ax - By + C = 0&lt;/p&gt;&#xA;" OwnerUserId="8011" LastActivityDate="2011-03-09T09:05:53.703" CommentCount="2" />
  <row Id="25960" PostTypeId="2" ParentId="25956" CreationDate="2011-03-09T09:19:24.350" Score="0" ViewCount="" Body="&lt;p&gt;Your code calculates the length of half the existing line. There are many ways to solve your problem. Get start and end coords, sx, sy, ex, ey. Calculate the slope angle with angle=atan(dy/dx), where dx=ex-sx, dy=ey-sy. Add or subtract pi/2 to angle. Calculate midpoint with mx=(ex-sx)/2 and my=(ey-sy)/2. This is the start point of the perpendicular line. Get the end point with mex=mx+10*cos(angle), mey=my+10*sin(angle). Easy as pi/2. :)&lt;/p&gt;&#xA;" LastActivityDate="2011-03-09T09:19:24.350" />
  <row Id="25961" PostTypeId="2" ParentId="25956" CreationDate="2011-03-09T09:56:55.440" Score="1" ViewCount="" Body="&lt;p&gt;Thanks guys for your support. here's what i did ( and yaay it works)&#xA;1-As i have the line equation (or points) i got dy and dx.&#xA;2- i got the normal vector to the line(-dy,dx) and (dy,-dx) let's call it NormalVec1 and NormalVec2.&#xA;3- i got the magnitude of NormalVec&#xA;4- i got the UNIT normalvector by dividing NormalVec by its Magnitude.&#xA;5- assume that i want a line with length L.&#xA;   so it gonna be L/2 on both sides.&#xA;6- assume that i want to draw this perpindicular line crosing the original line at its center point M.&#xA;7-so the perpindicular line's points are:&#xA;endpoint=(M.X+unitNormalVec1*L/2 ,M.X+unitNormalVec1*L/2) ; for normalVec1 is (-dy,dx)&#xA;startpoint=(M.X+unitNormalVec2*L/2 ,M.X+unitNormalVec2*L/2) ; for normalVec1 is (dy,-dx)&#xA;then you can draw a line segment from these points.&#xA;Thanks, &lt;/p&gt;&#xA;" LastActivityDate="2011-03-09T09:56:55.440" />
  <row Id="25962" PostTypeId="2" ParentId="6262" CreationDate="2011-03-09T11:30:18.687" Score="0" ViewCount="" Body="&lt;p&gt;I suspect that the Wikipedia article's number of $41,472 = 24^4 / 8$ arrangements of the four cubes comes from the idea that each of the four cubes can have 24 orientations.  However, the whole stack can be rotated into any of four positions, removing a factor of 4 without altering the four visible side face patterns.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In addition, the whole stack can be turned upside down (or equivalently, since order does not matter, each cube can be turned upside down) without altering the four visible face patterns, just whether you encounter them clockwise or anti-clockwise. That removes a further factor of 2.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So Wikipedia uses your $3 \times 24^{n-1}$ answer.&lt;/p&gt;&#xA;" OwnerUserId="6460" LastActivityDate="2011-03-09T11:30:18.687" />
  <row Id="25963" PostTypeId="1" AcceptedAnswerId="25984" CreationDate="2011-03-09T11:51:58.597" Score="1" ViewCount="56" Body="&lt;p&gt;I am trying to figure out if I made a mistake in the following proof - in particular I have been trying to verify the inequality (*) below.  I have attached my proof but I have been getting some error messages in Mathematica that make me think something is wrong when I try to compare the left hand side of the inequality to the right hand side.  The background of this problem is related to the study of moments of self similar processes and I have no idea if the estimates are valid besides from some crude numerical calculations for $N= 2,3,4$ and all $H_i = 1/2$.  I have attached the proof for the case N =2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Did I make a mistake in one of the estimates or applying fubini somewhere?  In particular I am starting to worry about line (**) is breaking the integral up like this and then using $x_1 &amp;lt; x_2$ and $x_2 &amp;lt; x_1$ in the respective regions of integration to arrive at the estimate on the next line valid?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $H_1 \in (0,1), c$ a positive constant and let $\Phi(t) = \int_{t}^{\infty} e^{\frac{-x^2}{2}}dx$.  Then we have the following inequality &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(*) $\int_{1}^{\infty} \int_{1}^{\infty}  e^{-\frac{ x_{1}^{2}}{2} -\frac{ x_{2}^{2}}{2} + \frac{1}{c} |x_1|  |x_2|^{H_1} } dx_1 dx_2 \leq 2 \Phi(1) \int_{1}^{\infty}  e^{ -\frac{ x_{2}^{2}}{2} + \frac{1}{c} |x_2| ^{1+H_1} }dx_2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proof:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\int_{1}^{\infty} \int_{1}^{\infty}  e^{-\frac{ x_{1}^{2}}{2} -\frac{ x_{2}^{2}}{2} + \frac{1}{c} |x_1|  |x_2|^{H_1} } dx_1 dx_2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(**) $ = \int_{1}^{\infty} \int_{1}^{x_2}  e^{-\frac{ x_{1}^{2}}{2} -\frac{ x_{2}^{2}}{2} + \frac{1}{c} |x_1|  |x_2|^{H_1} } dx_1 dx_2 &#xD;&#xA;+ \int_{1}^{\infty} \int_{x_2}^{\infty}  e^{-\frac{ x_{1}^{2}}{2} -\frac{ x_{2}^{2}}{2} + \frac{1}{c} |x_1|  |x_2|^{H_1} } dx_1 dx_2 $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ \leq \int_{1}^{\infty} \int_{1}^{x_2}  e^{-\frac{ x_{1}^{2}}{2} -\frac{ x_{2}^{2}}{2} + \frac{1}{c} |x_2| ^{1+H_1} } dx_1 dx_2 + \int_{1}^{\infty} \int_{x_2}^{\infty}  e^{-\frac{ x_{1}^{2}}{2} -\frac{ x_{2}^{2}}{2} + \frac{1}{c}  |x_1|^{1+H_1} } dx_1 dx_2 $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ =  \int_{1}^{\infty}  e^{ -\frac{ x_{2}^{2}}{2} + \frac{1}{c} |x_2| ^{1+H_1} } (\int_{1}^{x_2} e^{-\frac{ x_{1}^{2}}{2}}dx_1)dx_2 + \int_{1}^{\infty}  e^{ -\frac{ x_{1}^{2}}{2} + \frac{1}{c} |x_1| ^{1+H_1} } (\int_{x_2}^{\infty} e^{-\frac{ x_{2}^{2}}{2}}dx_2)dx_1 $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ \leq 2  \int_{1}^{\infty}  e^{ -\frac{ x_{2}^{2}}{2} + \frac{1}{c} |x_2| ^{1+H_1} } (\int_{1}^{\infty} e^{-\frac{ x_{1}^{2}}{2}}dx_1)dx_2 &#xD;&#xA; = 2 \Phi(1) \int_{1}^{\infty}  e^{ -\frac{ x_{2}^{2}}{2} + \frac{1}{c} |x_2| ^{1+H_1} }dx_2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2)I have tried comparing the LHS to the right hand side for the case $N = 2, N= 4$ and all H_i = 1/2 but Mathematica gives me the following errors attached below.  I am thinking I will just need to test the above inequality's for  $N &amp;gt; 3$ in matlab but I am not sure what the easiest method would be.  Would monte carlo integration apply to these type of integrals even though they are over an unbounded region?  Any numerical suggestions/references are greatly appreciated&lt;/p&gt;&#xA;&#xA;&lt;p&gt;NIntegrate[ &#xA; 2*(1/Sqrt[2*Pi])^2*Exp [1/2 (-x ^2 - y^2)] + 1/10*x*y^(1/2), {x, 0, &#xA;  Infinity}, {y, 0, Infinity}]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;NIntegrate::slwcon: Numerical integration converging too slowly; suspect one of the following: singularity, value of the integration is 0, highly oscillatory integrand, or WorkingPrecision too small. &gt;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;NIntegrate::ncvb: NIntegrate failed to converge to prescribed accuracy after 18 recursive bisections in x near {x,y} = {4.557658617658054*10^1348705,0.184661}. NIntegrate obtained 1.541751650430305&lt;code&gt;15.954589770191005*^88741243 and 3.653228868630915&lt;/code&gt;15.954589770191005*^88741243 for the integral and error estimates. &gt;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;0.*10^88741243&lt;/p&gt;&#xA;&#xA;&lt;p&gt;NIntegrate[&#xA; 2*(1/Sqrt[2*Pi])^4 Exp [&#xA;   1/2 (-x^2 - y^2 - z^2 - w^2) + &#xA;    1/10 x &lt;em&gt;Sqrt[y]&lt;/em&gt; z^(1/4)* w^(1/8)], {x, 0, [Infinity]}, {y, &#xA;  0, [Infinity]}, {z, 0, [Infinity]}, {w, 0, [Infinity]}]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;NIntegrate::slwcon: Numerical integration converging too slowly; suspect one of the following: singularity, value of the integration is 0, highly oscillatory integrand, or WorkingPrecision too small. &gt;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;NIntegrate::eincr: The global error of the strategy GlobalAdaptive has increased more than 2000 times. The global error is expected to decrease monotonically after a number of integrand evaluations. Suspect one of the following: the working precision is insufficient for the specified precision goal; the integrand is highly oscillatory or it is not a (piecewise) smooth function; or the true value of the integral is 0. Increasing the value of the GlobalAdaptive option MaxErrorIncreases might lead to a convergent numerical integration. NIntegrate obtained 0.1321276729737014&lt;code&gt;and 4.5493840913708495&lt;/code&gt;*^-7 for the integral and error estimates. &gt;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;0.132128&lt;/p&gt;&#xA;" OwnerUserId="7980" LastActivityDate="2011-03-09T14:43:38.627" Title="Help with the proof of a Gaussian type inequality and some numerical results" Tags="&lt;calculus&gt;&lt;real-analysis&gt;&lt;analysis&gt;&lt;numerical-methods&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="25964" PostTypeId="1" AcceptedAnswerId="25965" CreationDate="2011-03-09T11:54:30.273" Score="2" ViewCount="109" Body="&lt;p&gt;I'm reading Hatcher and on page 12 he writes:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;... It is less trivial to show that there are path-connected spaces that do not deformation retract onto a point....&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The definition of deformation retract is given as:&#xA;&quot;A deformation retraction of a space $X$ onto a subspace $A$ is a family of maps $f_t :X→X$, $t \in I$, such that $f_0 = 1 $ (the identity map), $f_1(X) = A$, and $f_t |A = 1$ $\forall t$. The family $f_t$ should be continuous in the sense that the associated map $X&amp;#215;I→X, (x,t) \rightarrow f_t(x)$ is continuous.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now my &lt;strong&gt;question&lt;/strong&gt; is: how is it possible for a space $X$ to be path-connected but not retractable to a single point? Can one not just pick $x \in X$ and then retract any $y \in X$ along the path between $x$ and $y$? &lt;/p&gt;&#xA;" OwnerUserId="5798" LastActivityDate="2011-03-09T16:35:42.820" Title="Deformation retract on path-connected space" Tags="&lt;algebraic-topology&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="25965" PostTypeId="2" ParentId="25964" CreationDate="2011-03-09T12:08:01.693" Score="5" ViewCount="" Body="&lt;p&gt;Suppose $X$ is path-connected, then for any $y \in X$, and for any $x$, there is a path $\phi_x : I \to X$ such that $\phi_x(0)=x$ and $\phi_x(1) = y$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With those maps, you can form a map $\phi : X \times I \to X$, with $\phi(x,t) = \phi_x(t)$.&#xA;This map has all the properties you would want of a deformatino retract from $X$ onto $y$, except that it has no reason at all to be continuous.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An example of a non retractable path connected space is the circle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's say you pick $y = (1,0)$, and decide to connect any $x$ to $y$ in a clockwise motion, via the top of the circle.&#xA;Then the map you $\phi$ you will obtain will not be continuous at $(y,t)$ for any $t \in ]0;1[$, since the neighboor points below $y$ will travel all around the circle, while $y$ stays where he is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And as Hatcher says, it is not trivial to show that the circle is not retractable onto a point.&lt;/p&gt;&#xA;" OwnerUserId="6270" LastActivityDate="2011-03-09T12:08:01.693" />
  <row Id="25966" PostTypeId="1" AcceptedAnswerId="25970" CreationDate="2011-03-09T12:15:10.623" Score="2" ViewCount="173" Body="&lt;p&gt;$f \left (\frac{x}{x+1} \right) = x^2$ then $f(x)=?$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I encountered this exercice, and and don't know how to solve it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;in what category of math does this belong?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;in what book/website I can study/exercise myself?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;thank you very much.&lt;/p&gt;&#xA;" OwnerUserId="8009" LastEditorUserId="3793" LastEditorDisplayName="" LastEditDate="2011-03-09T13:19:54.740" LastActivityDate="2011-03-09T13:19:54.740" Title="If f(x/(x+1))=x² then f(x)=?" Tags="&lt;calculus&gt;&lt;functions&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="2" />
  <row Id="25967" PostTypeId="2" ParentId="25887" CreationDate="2011-03-09T12:23:56.913" Score="4" ViewCount="" Body="&lt;p&gt;Seeing this semantically: Gödel's incompleteness theorems say that there is a model $\mathcal{M}$ of PA that &lt;em&gt;thinks&lt;/em&gt; that PA is inconsistent. If $A(x)$ is the formula that says &quot;$x$ &lt;em&gt;is a proof of a contradiction in PA&lt;/em&gt;&quot; then that model satisfies $\exists x A(x)$. Now, the closed terms of the language represent the standard natural numbers. Thus, since it's assumed that PA is consistent there doesn't exist a closed term $s$ of the language of PA such that $A(s)$ is true in $\mathcal{M}$. This means that $(\exists x A(x))\to A(s)$ is also false in $\mathcal{M}$ for every term. Since it's false, PA  doesn't prove it (due to soundness). Thus PA is not a scapegoat theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Seeing this syntactically: Assume that $A(x)$ is defined as above. Then if for any closed term $s$ the sentence $(\exists x A(x))\to A(s)$ was provable in PA, PA would prove $A(s)\lor(\lnot(\exists x A(x))$. At the same time since PA is assumed consistent it of course proves $\lnot A(s)$ (this is because finding out if a sequence of formulas constitutes a proof is a decidable problem). The consequence is that PA proves $\lnot(\exists x A(x))$ which states the consistency of PA, which is impossible because of the second incompleteness theorem.&lt;/p&gt;&#xA;" OwnerUserId="1772" LastActivityDate="2011-03-09T12:23:56.913" />
  <row Id="25968" PostTypeId="1" AcceptedAnswerId="25974" CreationDate="2011-03-09T12:27:51.043" Score="0" ViewCount="53" Body="&lt;p&gt;In the presentation here about Level set methods&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.cs.au.dk/~bang/smokeandwater2006/Lecture9_IntroToWaterAndLS.ppt&quot; rel=&quot;nofollow&quot;&gt;http://www.cs.au.dk/~bang/smokeandwater2006/Lecture9_IntroToWaterAndLS.ppt&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the author constructs a linear PDE &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{\partial \phi}{\partial t} + \triangledown \phi \cdot V = 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to solve this numerically, first the author uses forward Euler time step for discretization. Then, he &quot;discretizes the gradient&quot; and mentions various choices for this, one of which being&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{\phi_i - \phi^n_{i-1}}{\Delta t} = 1$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other choices he mentions are discretizations equaling 0, or 0.5. I do not understand how these values, 1, 0, 0.5 are chosen. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know gradient is a vector calculated using partial derivatives, at a specific point x_0, y_0, .. which can also result in a scalar if we want to calculate them for a certain direction, then we take a dot product. It does not make any sense to me discretization of a gradient can automatically be made equal to a specific value.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any ideas? &lt;/p&gt;&#xA;" OwnerUserId="6786" LastActivityDate="2011-03-09T13:21:43.570" Title="Discretizing gradient in upwind differencing, hamilton-jacobi equation" Tags="&lt;calculus&gt;&lt;numerical-methods&gt;&lt;vector-calculus&gt;&lt;finite-differences&gt;" AnswerCount="1" />
  <row Id="25970" PostTypeId="2" ParentId="25966" CreationDate="2011-03-09T12:44:41.367" Score="2" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Hint:&lt;/strong&gt; Write $f(y/(y+1))= y^2$, now solve $y/(y+1)=x$ for $y$.&lt;/p&gt;&#xA;" OwnerUserId="3567" LastActivityDate="2011-03-09T12:44:41.367" CommentCount="4" />
  <row Id="25972" PostTypeId="2" ParentId="25887" CreationDate="2011-03-09T13:01:33.643" Score="5" ViewCount="" Body="&lt;p&gt;A theory $T$ is said to have the &lt;em&gt;witness property&lt;/em&gt; if for every formula $A(x)$, if $T$ proves $(\exists x) A(x)$ then there is a term $t$ such that $T$ proves $A(t)$. The witness property is one of the criteria that are used to tell, qualitatively, whether a theory is constructive. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The definition of being a scapegoat theory is stronger than this: it requires the implication $(\exists x)A(x) \Rightarrow A(t)$ to be provable in the theory.  In particular, every scapegoat theory has the witness property. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Fact: PA does not have the witness property. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To prove this, use the same sentence $A(x)$ that Apostolos gave, which says &quot;if there is any coded proof of 0=1 then $x$ is a code for such a proof&quot;.  PA proves $(\exists x)A(x)$. However, there is no single term $t$ such that PA can prove &quot;if there is any coded proof of 0=1 then $t$ is such a code&quot;. In models where there is such a code, it will never be represented by a term. &lt;/p&gt;&#xA;" OwnerUserId="630" LastEditorUserId="630" LastEditorDisplayName="" LastEditDate="2011-03-09T13:14:49.490" LastActivityDate="2011-03-09T13:14:49.490" />
  <row Id="25973" PostTypeId="2" ParentId="25966" CreationDate="2011-03-09T13:03:52.627" Score="5" ViewCount="" Body="&lt;p&gt;This is an exercise about &quot;change of variable&quot; or &quot;variable substitution&quot;. This is an important technique in elementary calculus and topics that use calculus. You'll encounter it in integration theory first, and again with problems involving differential equations etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &quot;trick&quot; is to introduce an old variable x (already there in the problem statement) and a new variable, let's call it z, then we already know z as a function of x,&#xA;$$&#xD;&#xA;z := \frac{x}{x+1}&#xD;&#xA;$$&#xA;Now you have to calculate the inverse, x as a function of z:&#xA;$$&#xD;&#xA;x(z) = ...? &#xD;&#xA;$$&#xA;When you know that, you can calculate $x(z)^2$ and finally the right hand side of&#xA;$$&#xD;&#xA;f(z) = f(\frac{x}{x+1}) = ...&#xD;&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="7556" LastActivityDate="2011-03-09T13:03:52.627" />
  <row Id="25974" PostTypeId="2" ParentId="25968" CreationDate="2011-03-09T13:21:43.570" Score="1" ViewCount="" Body="&lt;p&gt;It will probably be helpful to have a look at the &lt;a href=&quot;http://en.wikipedia.org/wiki/Finite_difference_method&quot; rel=&quot;nofollow&quot;&gt;finite difference method&lt;/a&gt; on Wikipedia, and here at the example of the heat equation in one spatial dimension.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On slide 21, the author picks a concrete example with concrete values of the potential function at certain grid points, the &quot;magical numbers&quot; on slide 22 come from there (the values -0.5 and 0.5 are the prescribed values of the potential function at the nodes). The three standard ways of discretization correspond to the forward, backward and central difference approximation to the differential. For every difference approximation, you can calculate the number for the potential function $\phi$ at node $j$ from the prescribed values from slide 21.&lt;/p&gt;&#xA;" OwnerUserId="7556" LastActivityDate="2011-03-09T13:21:43.570" CommentCount="0" />
  <row Id="25975" PostTypeId="1" CreationDate="2011-03-09T13:26:13.860" Score="1" ViewCount="127" Body="&lt;p&gt;For finite group $G$, Mobius function is defined on the lattice of subgroups of $G$ as: $\sum_{H\geq K} {\mu (H)}=\delta_{K,G}$. From this, $\mu (H)$ can be obtained by induction argument.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;1.&lt;/strong&gt; Can we define $\mu(H)$  explicitly (i.e. is there formula) for any $H \leq G$? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;2.&lt;/strong&gt; For cyclic groups of order $n$, does this Mobius function match with our number-theoretic Mobius function? How? &lt;/p&gt;&#xA;" OwnerUserId="8186" LastActivityDate="2011-03-09T22:40:26.653" Title="Mobius function for finite groups" Tags="&lt;group-theory&gt;" AnswerCount="0" CommentCount="0" FavoriteCount="1" />
  <row Id="25976" PostTypeId="1" AcceptedAnswerId="26051" CreationDate="2011-03-09T13:30:25.323" Score="1" ViewCount="43" Body="&lt;p&gt;There's a homework exercise I have to do for my scheduling course about which I have some questions. The exercise is this:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;blockquote&gt;&#xA;    &lt;p&gt;Consider the $1|\text{ }|\sum w_jT_j$ problem. (This means that we look at one machine which has to process a number of jobs and where the objective function is the total weighted tardiness of a schedule.) &lt;/p&gt;&#xA;    &#xA;    &lt;p&gt;Prove that if for jobs $j$ and $k$ we have that $\frac{w_j}{p_j} &amp;gt; \frac{w_k}{p_k}$, $d_j &amp;lt; d_k$ and $p_j &amp;lt; p_k$ then there exists an optimal schedule where job $j$ is scheduled before $k$. &lt;/p&gt;&#xA;  &lt;/blockquote&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The tardiness of job $i$ is defined as $T_i := \max\{0,C_i-d_i\}$ where $C_i$ is the completion time of job $i$ and each tardiness has an associated weight $w_i$. Furthermore, $p_i$ and $d_i$ denote the processing time and due date of job $i$, respectively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have been able to prove a lemma in the course notes, which is a weaker version of the statement in the exercise. The lemma states that&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If there are jobs $j,k$ such that $d_j \leq d_k$, $p_j \leq p_k$ and $w_j \geq w_k$ then there exists an optimal schedule for $1|\text{ } | \sum w_jT_j$ in which job $j$ is scheduled before job $k$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I've proved this lemma (correctly, I hope) by contradiction: I assumed an optimal schedule has job $k$ scheduled before job $j$ and showed that the weighted tardiness of this schedule is bigger than the schedule with job $j$ scheduled before job $k$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is: is there an easy way to use the lemma to prove the statement in the exercise? Does the condition $\frac{w_j}{p_j} &amp;gt; \frac{w_k}{p_k}$ imply anything useful that we can use? Thanks in advance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The thing about this lemma is that, while we have stronger conditions on $d_j,d_k$ and $p_j,p_k$ we only have that $w_j &amp;gt; \frac{p_j}{p_k} w_k$. Since $\frac{p_j}{p_k} &amp;lt; 1$ this does not imply that $w_j &amp;gt; w_k$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS: there was no &quot;scheduling&quot; tag so I put this question under &quot;optimization&quot;. I hope that's ok.&lt;/p&gt;&#xA;" OwnerUserId="7155" LastEditorUserId="7155" LastEditorDisplayName="" LastEditDate="2011-03-09T19:42:48.410" LastActivityDate="2011-03-09T20:12:39.047" Title="About the existence of a certain optimal schedule for the $1|\text{ } | \sum w_j T_j$ problem" Tags="&lt;homework&gt;&lt;optimization&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="25977" PostTypeId="1" AcceptedAnswerId="25980" CreationDate="2011-03-09T13:52:30.097" Score="2" ViewCount="75" Body="&lt;p&gt;Hi all,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the past two weeks, I've tried to prove two different results that hold the same structure: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose a property that holds for a dense subset of a metric space.&#xA;Prove that it holds for the entire metric space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case, both questions were related to uniform convergence:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i) Show that if $(f_{n})$ is a sequence of continuous functions on M, and if $\sum_{n=1}^{\infty}f_{n}$ is converges uniformly on a dense subset A of M, then the series converges uniformly on all of M.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ii) Let A be a dense subset of M. If $(f_{n})$ is a sequence of continuous functions on M, and if the sequence converges uniformly on A, prove that $(f_{n})$ converges uniformly on M.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My questions are the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the framework I should have in my mind when trying to prove these kinds of properties? What is the basic frame of mind you have when proving such results?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because I feel that I'm missing something, that is, my intuition about how a dense subset works is not strong enough.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the way, I'm sorry for the wall of text :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance!&lt;/p&gt;&#xA;" OwnerUserId="7053" LastActivityDate="2011-03-09T15:20:07.363" Title="Prove a property holds on a metric space if it holds on a dense subset." Tags="&lt;real-analysis&gt;&lt;analysis&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" />
  <row Id="25978" PostTypeId="1" AcceptedAnswerId="25981" CreationDate="2011-03-09T13:55:45.750" Score="5" ViewCount="84" Body="&lt;p&gt;Let $(X, d)$ be a compact metric space and let $T:X \to X$ be a continuous mapping. Now, does there exist a probability measure $\nu$ such that $T_* \nu = \nu$ (the first thing is the image measure)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I want to do this using a fixed-point theorem. So I start like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Define the operator $\psi_T$ from $C(X)$ into itself by $\psi_T f = f \circ T$. Now we also know that $C(X)^* = P(X)$ where $P(X)$ are the Borel probability measures on $X$. So we have a mapping $\psi_T^* : P(X) \to P(X)$. Further we have that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\langle f, \underbrace{\psi_T^* \mu}_{= \nu} \rangle = \langle \psi_T f, \mu \rangle = \int f \circ T \, d\mu.$$&#xA;And $\nu$ is completely determined by&#xA;$$\int f \, d\nu = \int f \circ T \, d\mu$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Further, since $X$ is compact I know that $P(X)$ is compact (hence complete) with respect to the Bounded Lipschitz metric. This one is given by&#xA;$$d_\text{BL}(\mu, \nu) := \sup \left \{ \left | \int f \, d\mu - \int f \, d\nu \right | : f \in \text{BL}(X,d), \|f\|_\text{BL} \leq 1 \right \}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So everything is fine for the Banach fixed point theorem except one thing: is $d_\text{BL}$ a contraction? I don't think so. Does someone have an idea if I'm on the right track or does someone have a suggestion how to continue?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the way, this is homework.&lt;/p&gt;&#xA;" OwnerUserId="1120" LastEditorUserId="1120" LastEditorDisplayName="" LastEditDate="2011-03-09T17:59:22.687" LastActivityDate="2011-03-09T17:59:22.687" Title="Existence of a measure that preserves a given mapping." Tags="&lt;homework&gt;&lt;measure-theory&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="25979" PostTypeId="2" ParentId="25927" CreationDate="2011-03-09T13:57:32.103" Score="1" ViewCount="" Body="&lt;p&gt;Just writing Sivaram's solution in another way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For any matrix $A \in \mathfrak{gl}(\mathbb K,n)$, considering the inner product $\langle v,w\rangle = \sum_{i=1}^nv_iw_i$, we have that&#xA;$$\langle Av,w\rangle = \langle v,A^Tw\rangle,$$&#xA;for all $v$ and $w \in \mathbb K^n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, if $Av=\lambda v$ and $Aw=\mu w$ and $A=A^T$, then&#xA;$$\lambda\langle v,w\rangle = \langle Av,w\rangle = \langle v,Aw\rangle = \mu\langle v,w\rangle.$$&#xA;From where we conlude that $\lambda = \mu$ or $\langle v,w\rangle=0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Observe that, for $\mathbb K = \mathbb C$ and the inner product given by $\langle v,w\rangle = \sum_{i=1}^nv_i\overline{w_i}$, we would need the condition $A=\overline{A}^T$ in order to get&#xA;$$\langle Av,w\rangle = \langle v,Aw\rangle,$$&#xA;for all $v$ and $w \in \mathbb K^n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, for any matrix $A \in \mathfrak{gl}(\mathbb K,n)$ and inner product $\langle,\rangle$ there is a matrix $A^* \in \mathfrak{gl}(\mathbb K,n)$ such that&#xA;$$\langle Av,w\rangle = \langle v,A^*w\rangle,$$&#xA;for all $v$ and $w \in \mathbb K^n$. This matrix $A^*$ is called the dual (or adjoint) matrix of $A$. For more details see &quot;Linear Algebra&quot; by Greub.&lt;/p&gt;&#xA;" OwnerUserId="370" LastActivityDate="2011-03-09T13:57:32.103" />
  <row Id="25980" PostTypeId="2" ParentId="25977" CreationDate="2011-03-09T14:06:36.567" Score="1" ViewCount="" Body="&lt;p&gt;Perhaps the most useful characterization of dense subsets is this: if $A \subset X$ is dense, then for any $x \in X$ there exists a sequence $\{x_n\} \subset A$ such that $x_n \to x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using this, it is easy to show the following: if $g : X \to \mathbb{R}$ is continuous, and $g \le a$ on $A$ for some $a \in \mathbb{R}$, then $g \le a$ everywhere on $x$.  Taking $g$ to be something like $f_n - f_m$ will help address your second question, which is easily equivalent to the first one.&lt;/p&gt;&#xA;" OwnerUserId="822" LastActivityDate="2011-03-09T14:06:36.567" />
  <row Id="25981" PostTypeId="2" ParentId="25978" CreationDate="2011-03-09T14:15:54.420" Score="4" ViewCount="" Body="&lt;p&gt;If I remember correctly, the map in question is &lt;em&gt;not&lt;/em&gt; a contraction; instead, you have to use the compactness of the unit ball of measures in the weak* topology and a fixed point result. Namely, you consider the space of measures of total mass one (which is compact), which is acted on by $T_*$ (in a linear manner). Now if you are in a Banach space and have  a compact convex subset which is invariant under a linear operator, it has a fixed point. (Hint: consider successive averages $a_n$ of a given element, show that $Ta_n - a_n$ is small for $n$ large, and choose a limit point of the $\{a_n\}$.) So this gives the result.&lt;/p&gt;&#xA;" OwnerUserId="536" LastActivityDate="2011-03-09T14:15:54.420" CommentCount="4" />
  <row Id="25982" PostTypeId="1" CreationDate="2011-03-09T14:30:27.820" Score="6" ViewCount="91" Body="&lt;p&gt;Let $a$ be a positive irrational number. Let $p_k/q_k, p_{k+1}/q_{k+1}$ be two consecutive&#xA;convergents of its simple continued fraction, where $k\ge 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it possible that both $$|a-(p_k/q_k)|&amp;lt;1/(2q_k^2)$$ and $$|a-(p_{k+1}/q_{k+1})|&amp;lt;1/(2q_{k+1}^2)$$&#xA;are true?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can only prove that at least one of these inequalities is true.&lt;/p&gt;&#xA;" OwnerUserId="3249" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2011-03-09T15:51:39.597" LastActivityDate="2011-03-09T15:51:39.597" Title="A question on continued fraction" Tags="&lt;number-theory&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" />
  <row Id="25983" PostTypeId="1" CreationDate="2011-03-09T14:33:09.810" Score="2" ViewCount="140" Body="&lt;p&gt;I know that vector space of all real valued continuous functions on compact Hausdorff space is [&quot;can be&quot;? -- PLC] infinite dimensional.  My question is when it will be finite dimensional?  And how I will identify that vector space with $\mathbb{R}^n$ for some $n$?&lt;/p&gt;&#xA;" OwnerUserId="8018" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2011-03-09T15:46:03.380" LastActivityDate="2011-03-09T18:22:02.907" Title="vector space of continuous functions on compact Hausdorff space" Tags="&lt;topology&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="25984" PostTypeId="2" ParentId="25963" CreationDate="2011-03-09T14:43:38.627" Score="0" ViewCount="" Body="&lt;p&gt;There is an error in your computation in the change of order of integration in the second integral of the RHS. It should be&#xA;$$&#xD;&#xA;\int_1^\infty\Bigl(\int_{x_2}^\infty \exp(-\frac{x_1^2}{2}-\frac{x_2^2}{2}+\frac{x_1^{1+H_1}}{c})dx_1\Bigr)dx_2=&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;\int_1^\infty\exp(-\frac{x_1^2}{2}+\frac{x_1^{1+H_1}}{c})\Bigl(\int_{1}^{x_1}\exp(-\frac{x_2^2}{2})dx_2\Bigr)dx_1.&#xD;&#xA;$$&#xA;As for the Mathematica calculation, there is also an error. The term&lt;/p&gt;&#xA;&#xA;&lt;p&gt;+1/10*x*y^(1/2)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;should go inside the Exp .&lt;/p&gt;&#xA;" OwnerUserId="4791" LastActivityDate="2011-03-09T14:43:38.627" CommentCount="1" />
  <row Id="25986" PostTypeId="1" AcceptedAnswerId="25996" CreationDate="2011-03-09T14:53:24.920" Score="3" ViewCount="326" Body="&lt;p&gt;Barring very elementary arithmetic, which skills from elementary school are essential for understanding the world better? &lt;/p&gt;&#xA;" OwnerUserId="8019" LastEditorUserId="2720" LastEditorDisplayName="" LastEditDate="2011-03-09T15:16:01.787" LastActivityDate="2011-03-10T00:37:51.217" CommunityOwnedDate="2011-03-09T15:55:13.277" Title="Most important elementary math skills" Tags="&lt;soft-question&gt;&lt;teaching&gt;&lt;advice&gt;" AnswerCount="6" CommentCount="5" FavoriteCount="1" />
  <row Id="25987" PostTypeId="1" CreationDate="2011-03-09T14:56:24.433" Score="1" ViewCount="118" Body="&lt;p&gt;If we have three random variables (not independent in general) $X$, $Y$, $Z$ can we conclude that $(X+Y, X+Z)\approx (Y,Z)$ about their joint distributions? If not, what is the minimum requirement on $X$, $Y$, $Z$ to hold this? &#xA;$(X+Y, X+Z)$ is a two dimensional random variable and so $(Y,Z)$. &lt;/p&gt;&#xA;" OwnerUserId="8020" LastEditorUserId="8020" LastEditorDisplayName="" LastEditDate="2011-03-09T15:04:34.247" LastActivityDate="2011-05-08T15:50:42.417" Title="Joint distributions of random variables" Tags="&lt;probability&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="25988" PostTypeId="2" ParentId="25986" CreationDate="2011-03-09T15:06:13.457" Score="3" ViewCount="" Body="&lt;p&gt;Wow, can i say everything? But let me just mention three that might be the most important:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Any money related mathematics will be very useful for obvious reasons. It amazes me how many of my students cannot even intuitively tell whether their answer to a problem involving tax makes sense. As consumers, we should be able to see a price tag and know about how much we will have to pay including taxes. And of course, there are many other applications involving money.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Spatial awareness can be very important. Geometry (composite shapes, three dimensional shapes, etc.) helps teach the brain to literally view the physical world.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) Logic and algebra teach us everyday problem solving. Word problems, although they are often a dreaded topic, are SO beneficial! It's being able to comprehend a problem, assess what needs to be done, and find a way to get the desired result, that will likely be the most significant take-way from elementary mathematics for a person in a non-math-related (or really in any) field.&lt;/p&gt;&#xA;" OwnerUserId="7840" LastEditorUserId="7840" LastEditorDisplayName="" LastEditDate="2011-03-09T19:35:27.857" LastActivityDate="2011-03-09T19:35:27.857" CommunityOwnedDate="2011-03-09T15:55:13.277" />
  <row Id="25989" PostTypeId="2" ParentId="25987" CreationDate="2011-03-09T15:10:24.150" Score="0" ViewCount="" Body="&lt;p&gt;You would have to show that $M_{X+Y,X+Z}(s,t) = M_{Y,Z}(s,t)$ (i.e. the joint moment generating functions are the same).&lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2011-03-09T15:10:24.150" CommentCount="4" />
  <row Id="25990" PostTypeId="2" ParentId="25986" CreationDate="2011-03-09T15:11:00.063" Score="8" ViewCount="" Body="&lt;p&gt;Not everyone will agree, but I'd say the most important adaptive traits that a child can learn in elementary school are &lt;a href=&quot;http://en.wikipedia.org/wiki/Deductive_reasoning&quot; rel=&quot;nofollow&quot;&gt;Deductive Reasoning&lt;/a&gt;, &lt;a href=&quot;http://en.wikipedia.org/wiki/Inductive_reasoning&quot; rel=&quot;nofollow&quot;&gt;Inductive Reasoning&lt;/a&gt;, &lt;a href=&quot;http://en.wikipedia.org/wiki/Abductive_reasoning&quot; rel=&quot;nofollow&quot;&gt;Abductive Reasoning&lt;/a&gt; and Healthy &lt;a href=&quot;http://en.wikipedia.org/wiki/Social_skill&quot; rel=&quot;nofollow&quot;&gt;Social Skill&lt;/a&gt;. &lt;/p&gt;&#xA;" OwnerUserId="2720" LastEditorUserId="2720" LastEditorDisplayName="" LastEditDate="2011-03-09T22:35:26.850" LastActivityDate="2011-03-09T22:35:26.850" CommunityOwnedDate="2011-03-09T15:55:13.277" CommentCount="2" />
  <row Id="25991" PostTypeId="2" ParentId="25983" CreationDate="2011-03-09T15:18:27.107" Score="8" ViewCount="" Body="&lt;p&gt;For a topological space $X$, let me write $C(X,\mathbb{R})$ for the set of continuous functions $f: X \rightarrow \mathbb{R}$.  Note that $C(X,\mathbb{R})$ forms a ring under pointwise addition and multiplication which contains $\mathbb{R}$ as the subring of constant functions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) If $X = \{x_1,\ldots,x_n\}$ is finite, then since it is Hausdorff it has the discrete topolgy so $C(X,\mathbb{R})$ is the set of all functions from $\{x_1,\ldots,x_n\}$ to $\mathbb{R}$.  In this case a natural basis is given by &quot;$\delta$ functions'': i.e., for $1 \leq i \leq n$, let $e_i: X \rightarrow \mathbb{R}$ be such that $e_i(x_j) = \delta_{i,j}$ (i.e., $1$ if $i = j$, $0$ otherwise).  Note that in this case $e_1,\ldots,e_n$ is also a family of idempotent elements giving rise to a direct product decomposition &lt;em&gt;of rings&lt;/em&gt; $C(X,\mathbb{R}) \cong \mathbb{R}^n$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) If $X$ is infinite, then for every positive integer $n$ there is a subspace &#xA;$X_n$ consisting of exactly $n$ points.  By Step 1, the dimension of the space &#xA;$C(X_n,\mathbb{R})$ is $n$.  Moreover, since $X$ is Hausdorff, $X_n$ is closed in $X$, &#xA;and, since $X$ is compact Haudorff, the &lt;a href=&quot;http://en.wikipedia.org/wiki/Tietze_extension_theorem&quot; rel=&quot;nofollow&quot;&gt;Tietze Extension Theorem&lt;/a&gt; applies to show that every continuous function on $X_n$ extends to a continuous function on $X$.  In other words, the natural restriction map $r_n: C(X,\mathbb{R}) \rightarrow C(X_n,\mathbb{R})$ is surjective.  Since $r_n$ is an $\mathbb{R}$-linear map (indeed a homomorphism of $\mathbb{R}$-algebras), it follows that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\operatorname{dim} C(X,\mathbb{R}) \geq \operatorname{dim} C(X_n,\mathbb{R}) = n.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $n$ was arbitrary, we conclude that $\operatorname{dim} C(X,\mathbb{R})$ is infinite.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Added&lt;/b&gt;: As Qiaochu points out, it is enough to require that $X$ be what I call &quot;C-separated&quot;: i.e., for the continuous $\mathbb{R}$-valued functions to separate points of $X$.  And for this it is enough that $X$ be Tychonoff.  (Recall compact Hausdorff $\implies$ normal $\implies$ Tychonoff.)  In fact these considerations come up in $\S 5.2$ of &lt;a href=&quot;http://math.uga.edu/~pete/integral.pdf&quot; rel=&quot;nofollow&quot;&gt;my commutative algebra notes&lt;/a&gt;.  See especially the second exercise in that section, which asks for a justification of the claim I made in the first sentence of this paragraph.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Indeed, since &quot;C-separated&quot; implies Hausdorff, looking back at my answer it shows that one can replace &quot;compact Hausdorff&quot; with &quot;C-separated&quot; and the result still holds.  However, one cannot replace &quot;C-separated&quot; with &quot;regular&quot;.  As I have mentioned elsewhere, there are infinite regular spaces $X$ in which the only continuous functions are the constant functions, so $C(X,\mathbb{R}) = \mathbb{R}$ is a one-dimensional space.&lt;/p&gt;&#xA;" OwnerUserId="299" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2011-03-09T18:22:02.907" LastActivityDate="2011-03-09T18:22:02.907" CommentCount="5" />
  <row Id="25992" PostTypeId="2" ParentId="25977" CreationDate="2011-03-09T15:20:07.363" Score="2" ViewCount="" Body="&lt;p&gt;A simple property of dense sets $D$: if a closed set $C$ contains $D$, then $C$ equals the whole space $X$: $D \subset C \rightarrow X = \overline{D} \subset \overline{C} = C \subset X$ which gives $C = X$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The example Nate gave: $g \le a$ on $D$ then this also holds on $X$, follows right away: $D \subset C = g^{-1}[(-\infty, a] ]$, which is closed, and hence equals $X$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose that $(f_n)$ converges uniformly to $f$ on $D$, which is dense. This is saying the same as $(f_n)$ is Cauchy in the uniform norm when restricted to $D$. So for all $\epsilon &amp;gt; $ we have that there is some $N$ in $\mathbb{N}$ such that $ \bigcap_{n,m \ge N} \{ x : |f_n(x) - f_m(x) | \le \epsilon \}$ contains $D$, but the latter set is closed and hence equals $X$ (using continuity of the $f_n$ and the norm). It follows that the $(f_n)$ form a Cauchy sequence on all of $X$ and by completeness of the sup norm it converges uniformly on $X$.&#xA;One could say that being Cauchy is a &quot;closed condition&quot; and thus if a dense set has it, all points have it...&lt;/p&gt;&#xA;" OwnerUserId="4280" LastActivityDate="2011-03-09T15:20:07.363" />
  <row Id="25994" PostTypeId="1" AcceptedAnswerId="26001" CreationDate="2011-03-09T15:24:46.363" Score="1" ViewCount="113" Body="&lt;p&gt;Suppose $(X,\mathcal {M})$ is a measurable space. For function $f(r,x): \mathbb R\times X\to\mathbb R$, suppose each r-section of it is $\mathcal M$-measurable. For constant $R\in\mathbb R$, do we have $\limsup \limits_{r\to R}f(r,x)$(this is a function of $x\in X$) always $\mathcal M$-measurable?&#xA;The definition of $\limsup \limits_{r\to R} f(r)$ is $\inf\limits_{\delta&amp;gt;0}\{\sup\limits_{0&amp;lt;|r-R|&amp;lt;\delta}f(r)\}$. Thanks!&lt;/p&gt;&#xA;" OwnerUserId="7321" LastActivityDate="2011-03-09T22:58:13.737" Title="measurability of limit superior for functions" Tags="&lt;real-analysis&gt;&lt;measure-theory&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="1" />
  <row Id="25995" PostTypeId="2" ParentId="25982" CreationDate="2011-03-09T15:31:47.583" Score="4" ViewCount="" Body="&lt;p&gt;At least it can happen that not both are true. Example: 333/106 is a convergent to $\pi$, but $(\pi-(333/106)) \cdot 2 \cdot 106^2 \approx 1.87 &amp;gt; 1$.&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2011-03-09T15:31:47.583" CommentCount="4" />
  <row Id="25996" PostTypeId="2" ParentId="25986" CreationDate="2011-03-09T15:36:42.783" Score="2" ViewCount="" Body="&lt;p&gt;Let me point that some ideas in the direction of what you are looking for can be found, for example, in Polya's work on heuristics (How to solve it), or in Davis and Hersh (Mathematical experience). Maybe after some reflection and elaboration, but there they are!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe I should say something more about this. Mathematics exist because problems exist, particularly problems that admit solutions by a certain kind of means we could call mathematical (I know this sounds circular, but what other name could we give them?). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would say that the two most prominent skills are idealization and abstraction. Idealization because in order to find a solution one needs to understand the problem and that involves being able to separate its relevant features from the irrelevant ones. Abstraction because the techniques developed for a particular problem become of general applicability in many other, possibly unsuspected, situations.&lt;/p&gt;&#xA;" OwnerUserId="5405" LastEditorUserId="5405" LastEditorDisplayName="" LastEditDate="2011-03-09T16:19:13.250" LastActivityDate="2011-03-09T16:19:13.250" CommunityOwnedDate="2011-03-09T15:55:13.277" />
  <row Id="25997" PostTypeId="1" AcceptedAnswerId="26000" CreationDate="2011-03-09T15:48:57.990" Score="1" ViewCount="68" Body="&lt;p&gt;I know that the functions $f$ and $g$ are Riemann-integrable on some closed set. &#xA;Also, $f$ has a min ($m&gt;0$) on the closed set.&lt;br&gt;&#xA;Let $h(x) = f(x)^{g(x)}$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to prove that $h$ is Riemann-integrable.&lt;br&gt;&#xA;I've been trying to use the Lebesgue's Theorem  but I'm not sure how to prove the set of discontinuities of $h$ is measure zero without assuming it is Riemann-integrable.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've tried breaking up the integral of $f$ into it's continuous sections, then the discontinuities of $f^g$ for each section is equal to the discontinuities of $g$. Hence, we have a countable union of measure zero sets, which is measure zero. But, am I allowed to assume the integral of $f^g$ exists for each section?&lt;/p&gt;&#xA;" OwnerUserId="7175" LastEditorUserId="822" LastEditorDisplayName="" LastEditDate="2011-03-09T16:53:44.700" LastActivityDate="2011-03-09T16:53:44.700" Title="Set of discontinuities" Tags="&lt;real-analysis&gt;" AnswerCount="1" />
  <row Id="25998" PostTypeId="2" ParentId="26691" CreationDate="2011-03-09T15:54:38.937" Score="5" ViewCount="" Body="&lt;p&gt;you can define moebius functions for posets (&lt;a href=&quot;http://en.wikipedia.org/wiki/Incidence_algebra&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Incidence_algebra&lt;/a&gt;).  As for the cyclic group, the answer is yes since the lattice of subgroups (under inclusion) is isomorphic to the lattice of divisors of the order of the group.  there are also some closed formulas for some groups (for example $p$-groups as in the wikipedia article)&lt;/p&gt;&#xA;" OwnerUserId="6925" LastEditorUserId="274" LastEditorDisplayName="" LastEditDate="2011-03-09T22:40:26.653" LastActivityDate="2011-03-09T22:40:26.653" CommentCount="4" />
  <row Id="25999" PostTypeId="1" CreationDate="2011-03-09T16:07:03.213" Score="2" ViewCount="134" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Possible Duplicate:&lt;/strong&gt;&lt;br&gt;&#xA;  &lt;a href=&quot;http://math.stackexchange.com/questions/23729/what-is-the-condition-for-a-field-to-make-the-degree-of-its-algebraic-closure-ove&quot;&gt;What is the condition for a field to make the degree of its algebraic closure over it infinite?&lt;/a&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;p&gt;More generally, what is known about subfields $\mathbb{F}\subset \mathbb{C}$ such that the degree $[\mathbb{C}:\mathbb{F}]$ is finite?  We all know, for example, that $[\mathbb{C}:\mathbb{C}]=1$ and $[\mathbb{C}:\mathbb{R}]=2$ (or we &lt;em&gt;should&lt;/em&gt; all know this).  However, $\mathbb{C}$ has &lt;em&gt;many&lt;/em&gt; subfields.  Can you construct/identify one with $[\mathbb{C}:\mathbb{F}]=3$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a question which I think about whenever I teach abstract algebra.&lt;/p&gt;&#xA;" OwnerUserId="6766" LastEditorUserId="153" LastEditorDisplayName="" LastEditDate="2011-04-12T02:09:45.703" LastActivityDate="2011-04-12T02:09:45.703" ClosedDate="2011-03-09T16:53:29.660" Title="Is there a subfield F of the complex field with [C:F]=3?" Tags="&lt;abstract-algebra&gt;&lt;field-theory&gt;" CommentCount="2" />
  <row Id="26000" PostTypeId="2" ParentId="25997" CreationDate="2011-03-09T16:09:51.550" Score="2" ViewCount="" Body="&lt;p&gt;I assume when you say &quot;$f$ has a min ($m&gt;0$)&quot; we mean $f(x)&amp;gt;0$ on this set. Let $E_g$ and $E_f$ be the set of all points of discontinuity for $g$, $f$ respectively. (which have Lebesgue measure 0)  Then let $$U=E_f^c\cap E_g^c.$$  Both $f,g$ must be continuous on $U$.  Since $x^y$ is a continuous function from $(0,\infty)\times\mathbb{R}\rightarrow \mathbb{R}$ we see $f(x)^{g(x)}$ will be continuous on $U$ as the composition of continuous functions is continuous.  (Note:  For the range/domains to work out, we need $f(x)&gt;0$).  Now, as $m (U^c)=0$, that is the Lebesgue measure of $U^c$ is 0, we conclude $f(x)^{g(x)}$ is Riemannn integrable as desired.  (The set of discontinuities is a subset of $U^c$)  &lt;/p&gt;&#xA;" OwnerUserId="6075" LastEditorUserId="6075" LastEditorDisplayName="" LastEditDate="2011-03-09T16:15:47.587" LastActivityDate="2011-03-09T16:15:47.587" />
  <row Id="26001" PostTypeId="2" ParentId="25994" CreationDate="2011-03-09T16:11:33.653" Score="3" ViewCount="" Body="&lt;p&gt;The principle is disprovable in ZFC. (I had a previous answer where I showed that it was disprovable from ZFC+CH, which I've edited with an improved solution). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, fix some map $g$ from $(0,1)$ to itself such that for every $s \in (0,1)$ there are $r$ arbitrarily close to $0$ such that $g(r) = s$. The map $g$ can be taken to be $(1+\sin(1/r))/2$ but the exact definition isn't important.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $A$ be a nonmeasurable subset of $(0,1)$. For $r \in (0,1)$ we define the function $f_r: (0,1) \to \{0,1\}$ as the characteristic function of $A \cap \{g(r)\}$. So $f_r(g(r))$ is $1$ if $g(r) \in A$, and $f_r(g(r)) = 0$ otherwise, and $f_r(x) = 0$ for all $x \not = g(r)$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $f: (0,1) \to \{0,1\}$ be the limit function $f(s) = \lim_{r \to 0^+} \sup_{\{t: 0 &amp;lt; t &amp;lt; r\}} f_t(s)$. Then $f$ is the characteristic function of $A$, and so $f$ is not measurable. &lt;/p&gt;&#xA;" OwnerUserId="630" LastEditorUserId="630" LastEditorDisplayName="" LastEditDate="2011-03-09T22:58:13.737" LastActivityDate="2011-03-09T22:58:13.737" CommentCount="2" />
  <row Id="26002" PostTypeId="2" ParentId="25986" CreationDate="2011-03-09T16:21:27.343" Score="2" ViewCount="" Body="&lt;p&gt;Definitely the &lt;a href=&quot;http://en.wikipedia.org/wiki/Intercept_theorem&quot; rel=&quot;nofollow&quot;&gt;Intercept theorem&lt;/a&gt; or the related concept of similarity. Knowing that when two objects are similar, all length have the same ratio $\lambda$, all areas have the same ratio $\lambda^2$, ... . This leads directly to the concept of dimensional analysis. Knowing that you can estimate almost everything in life (and also in mathematics, see, e.g., &lt;a href=&quot;http://mitpress.mit.edu/books/full_pdfs/Street-Fighting_Mathematics.pdf&quot; rel=&quot;nofollow&quot;&gt;Street-Fighting Mathematics&lt;/a&gt;).&lt;/p&gt;&#xA;" OwnerUserId="7266" LastActivityDate="2011-03-09T16:21:27.343" CommunityOwnedDate="2011-03-09T16:21:27.343" />
  <row Id="26003" PostTypeId="2" ParentId="25964" CreationDate="2011-03-09T16:35:42.820" Score="1" ViewCount="" Body="&lt;p&gt;It's probably worth pointing out that you should be careful with the term retraction versus &lt;em&gt;deformation&lt;/em&gt; retraction. Every space $X$ retracts onto a point, via the map $r \colon X \rightarrow \ast$ that sends everything to a single point, but as Hatcher says, not every space deformation retracts to a point. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What Hatcher's definition of deformation retract is really saying is that a deformation retract is a retract $r$ along with a homotopy $r \simeq 1$, where $1$ is the identity. So if a space deformation retracts to another space, those two spaces have the same homotopy type, whereas spaces can retract onto spaces that that are not of their homotopy type (as you'll soon see if you keep reading the book).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, if every space deformation retracted to a point, every space would be contractible, and algebraic topology would be very boring indeed!&lt;/p&gt;&#xA;" OwnerUserId="7748" LastActivityDate="2011-03-09T16:35:42.820" CommentCount="2" />
  <row Id="26004" PostTypeId="1" AcceptedAnswerId="26013" CreationDate="2011-03-09T16:36:42.627" Score="1" ViewCount="128" Body="&lt;p&gt;There's a question over on stack overflow about adding negative numbers to a negative number. The question surrounds why 10 - -5 is equal to 5.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm very happy that the result is 5, but why is it 5 and not 15. It is philosophical or is there a concrete reason for it? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The original question was asked here:&#xA;&lt;a href=&quot;http://stackoverflow.com/questions/5248734/odd-result-using-multiple-operator-in-c&quot;&gt;http://stackoverflow.com/questions/5248734/odd-result-using-multiple-operator-in-c&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;UPDATE&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So apart from the obvious transcription error, left in place for posterity, I think the original question stands. 10 - (-5) equals 15, fine, but at what point in mathematical history was this decided? In the same way the 0 wasn't always used there came a point in history when it was in common use. In the same way, when did a double negative become a positive in common mathematical parlance and more importantly why?&lt;/p&gt;&#xA;" OwnerUserId="8024" LastEditorUserId="8024" LastEditorDisplayName="" LastEditDate="2011-03-09T16:56:03.120" LastActivityDate="2011-03-09T17:46:23.430" Title="Why does adding a negative number to a positive number reduce the original positive number?" Tags="&lt;logic&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="1" />
  <row Id="26005" PostTypeId="2" ParentId="26004" CreationDate="2011-03-09T16:40:54.603" Score="1" ViewCount="" Body="&lt;p&gt;You are not adding a negative number, you are subtracting it!&lt;br&gt;&#xA;10-(-5)=15&lt;br&gt;&#xA;10+(-5)=5&lt;br&gt;&#xA;10+-(-5)=15 &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for pholsophical reasons:&lt;br&gt;&#xA;I am not unhappy = I am happy&lt;/p&gt;&#xA;" OwnerUserId="4929" LastEditorUserId="4929" LastEditorDisplayName="" LastEditDate="2011-03-09T17:08:02.897" LastActivityDate="2011-03-09T17:08:02.897" />
  <row Id="26006" PostTypeId="2" ParentId="26004" CreationDate="2011-03-09T16:45:09.093" Score="2" ViewCount="" Body="&lt;p&gt;I think you've misinterpreted the SO thread. The poster was unhappy that the computer gave 15 for 10 - - 5 and he was (incorrectly) expecting the answer to be 5. It's not peculiar to C# either, all programming languages I know of work that way and that's the way basic arithmetic is taught. I don't see what the problem is.&lt;/p&gt;&#xA;" OwnerUserId="7953" LastActivityDate="2011-03-09T16:45:09.093" />
  <row Id="26007" PostTypeId="1" AcceptedAnswerId="26015" CreationDate="2011-03-09T16:46:52.523" Score="2" ViewCount="156" Body="&lt;p&gt;More important than knowing inequalities by hearts is knowing when and how to apply them and what they express.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regarding &lt;a href=&quot;http://en.wikipedia.org/wiki/Chebyshev%27s_inequality&quot; rel=&quot;nofollow&quot;&gt;Chebyshev's&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/Markov%27s_inequality&quot; rel=&quot;nofollow&quot;&gt;Markov's&lt;/a&gt; inequality. What is the relation (if any) between them? Which one is more strict (and in which situation)? Is there an easy way to understand what they express (kind of like drawing a triangle for the triangle inequality)? What is a typical application in probability theory? What are applications outside of probability theory?&lt;/p&gt;&#xA;" OwnerUserId="7266" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2011-03-09T16:53:58.943" LastActivityDate="2011-03-09T17:10:59.403" Title="Chebyshev's versus Markov's inequality" Tags="&lt;probability&gt;&lt;probability-theory&gt;&lt;inequality&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="1" />
  <row Id="26008" PostTypeId="1" AcceptedAnswerId="26033" CreationDate="2011-03-09T16:47:45.540" Score="2" ViewCount="141" Body="&lt;p&gt;Consider $\displaystyle\sum_{1}^{p-1}\frac{1}{x}$, where $p$ is an odd prime. I want to find a general formula for this sum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can't believe that I am having trouble figuring this out, but I can't figure out the summation in the question. I have to find a general formula $A_p/B_p$ for the summation from 1 to $p-1$ of $\frac{1}{x}$. Then I have to prove that this is correct and make a conjecture about $A_p \mod p^2$. Somehow, I am having trouble on the first step. Is there a general formula for the summation of $\frac{1}{x}$ alone? That would probably give me a hint where to look. Thanks.&lt;/p&gt;&#xA;" OwnerUserId="7435" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-09T17:55:22.330" LastActivityDate="2011-03-09T18:28:02.453" Title="General formula for $\sum_1^{p-1} \frac{1}{x}$, where $p$ is an odd prime" Tags="&lt;homework&gt;&lt;elementary-number-theory&gt;" AnswerCount="1" CommentCount="10" FavoriteCount="2" />
  <row Id="26010" PostTypeId="2" ParentId="26007" CreationDate="2011-03-09T16:55:21.770" Score="1" ViewCount="" Body="&lt;p&gt;As you can see in the wikipedia links, Markov's inequality can be used to prove Chebyshev's Inequality. &lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2011-03-09T16:55:21.770" />
  <row Id="26011" PostTypeId="2" ParentId="25925" CreationDate="2011-03-09T16:56:23.687" Score="1" ViewCount="" Body="&lt;p&gt;Herschfeld &lt;a href=&quot;http://mathworld.wolfram.com/HerschfeldsConvergenceTheorem.html&quot; rel=&quot;nofollow&quot;&gt;proved&lt;/a&gt; that the nested radical converges iff $f_n^{2^-n}$ is bounded. In view of this, you consider bounded sequences under the same domination relation. It is clear that the increasing sequence of all integral constant sequences is cofinal (bounds every sequence). Coming back to the original problem, the following sequence is increasing and cofinal:&#xA;$$&#xD;&#xA;\begin{align*}&#xD;&#xA;&amp;amp;1,1,1,1,\ldots \\&#xD;&#xA;&amp;amp;2,4,16,256,\ldots \\&#xD;&#xA;&amp;amp;3,9,81,6561,\ldots \\&#xD;&#xA;&amp;amp;4,16,256,65536,\ldots \\&#xD;&#xA;&amp;amp;5,25,625,390625,\ldots \\&#xD;&#xA;&amp;amp;\ldots&#xD;&#xA;\end{align*}&#xD;&#xA;$$&#xA;The $k$th sequence starts with $k$ and continues by repeated squaring.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2011-03-09T16:56:23.687" />
  <row Id="26012" PostTypeId="1" AcceptedAnswerId="26029" CreationDate="2011-03-09T17:03:53.083" Score="5" ViewCount="111" Body="&lt;p&gt;Given two sets of finite measure in $\mathbb{R}$ say, $E$ and $F$, and their characteristic functions $\chi_E$ and $\chi_F$, can somebody show that $\chi_E\ast\chi_F(x)$ (the convolution) is a continuous function of $x$?  This is a qual problem from an old qual that I'm studying, and I cannot figure it out.  If we were dealing with continuous functions or mollifiers or something it would be straightforward, but what if the sets $E$ and $F$ are somehow pathological, like the Cantor set, or something like that?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="3595" LastActivityDate="2011-03-09T19:28:09.183" Title="Measure Theory and Integrals of Characteristic Functions" Tags="&lt;measure-theory&gt;&lt;functional-analysis&gt;" AnswerCount="1" CommentCount="7" FavoriteCount="1" />
  <row Id="26013" PostTypeId="2" ParentId="26004" CreationDate="2011-03-09T17:05:04.080" Score="4" ViewCount="" Body="&lt;p&gt;In fact $\rm\ 10\ -\ -5\ =\ 15\:,\ $ is correct, see the proof below. $\:$ See also the &lt;a href=&quot;http://math.stackexchange.com/questions/9933/why-negative-times-negative-positive/9940#9940&quot;&gt;Law of Signs.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\quad\quad\quad\quad\quad\quad\quad\quad\ \ \ \: n\ =\ 10\ \ -\ -5$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\iff\ \ n\ +\ -5 \ =\ 10\quad\quad\quad\quad\ $ by adding $-5\:$ to both sides of above&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\iff\quad\quad\quad\quad\quad\ \ n \ =\ 10\ +\ 5\quad$ by adding $\ \ 5\ \ \:$ to both sides of above&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively $\rm\ - (-x)\ =\: - (-x) +(-x + x) \ =\ (-(-x) + -x) + x\ =\ x$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i.e. both $\rm\: x\: $ and $\rm\:-(-x)\:$ are inverses of $\rm-x\:$ so they are equal by &lt;strong&gt;uniqueness of inverses&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The proof uses only the basic laws of arithmetic, that addition is associative and commutative, and every integer $\rm\:n\:$ has an additive inverse $\rm -n\:,\ $ i.e. the integers comprise an additive abelian &lt;a href=&quot;http://en.wikipedia.org/wiki/Group_%28mathematics%29&quot; rel=&quot;nofollow&quot;&gt;group.&lt;/a&gt; The laws for negative integers weren't &quot;decided&quot; - they are forced upon us by &quot;persistence of form&quot;, i.e. by requiring the the enlargement from naturals to integers obeys the same laws as the original structure. In particular, if you follow the above link you will see how the law of signs follows from the ring axioms - most notably the &lt;em&gt;distributive&lt;/em&gt; law. The notion of a ring axiomatizes these familiar arithmetic properties of the integers, rational, polynomials, etc. The ring abstraction allows us to generalize properties of integers to diverse algebraic structures that share essential algebraic &quot;integer-like&quot; properties - structures which, in turn, may allow us to more simply deduce properties of the integers, e.g. solving Diophantine equations by passing to Gaussian integers or algebraic number fields. That we can pull back results from the general to more specific structures is only because of said persistence of form - the laws are the same whether the number is positive or negative, rational or irrational, real or imaginary. Thus whatever ring theorems we deduce about integers will specialize to true theorems about naturals, and whatever ring-theoretic properties we deduce about Gaussian integers $\rm\ m + n\ i\ $ will specialize to true properties of integers, simply because we invoked only ring laws that hold true &lt;em&gt;universally&lt;/em&gt;, i.e. in every ring.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2011-03-09T17:46:23.430" LastActivityDate="2011-03-09T17:46:23.430" />
  <row Id="26014" PostTypeId="1" AcceptedAnswerId="26031" CreationDate="2011-03-09T17:10:20.523" Score="2" ViewCount="106" Body="&lt;p&gt;I'm reading Hatcher and I'm doing exercise 9 on page 19. Can you tell me if my answer is correct?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Exercise: Show that a retract of a contractible space is contractible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proof:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $X$ be a contractible space, i.e. $\exists f :X \rightarrow \{ * \}$,$g: { * } \rightarrow X$ continuous such that $fg \cong id_{\{ *\}}$ and $gf \cong id_X$ and let $r:X \rightarrow X$ be a retraction of $X$ i.e. $r$ continuous and $r(X) = A$ and $r\mid_A = id_A$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(&lt;strong&gt;Edited&lt;/strong&gt; using Anton's answer)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Define $f^\prime := f\mid_A$ and $g^\prime := r \circ g$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we need to show $f^\prime \circ g^\prime \cong id_{ \{ * \} }$  and $g^\prime \circ f^\prime \cong id_A$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f^\prime \circ g^\prime \cong id_{ \{ * \} }$ follows from $f^\prime \circ g^\prime = id_{ \{ * \} }$ (because there is only one function ${ * } \rightarrow { * }$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From $gf \cong id_X$ we have a homotopy $H: I \times X \rightarrow X$ such that $H(0,x) = g \circ f (x)$ and $H(1,x) = id_X$. From this we build a homotopy $H^\prime : I \times A \rightarrow A$ by defining $H^\prime := r \circ H \mid_{I \times A}$. Then $H^\prime (0,x) = g^\prime \circ f\mid_A (x)$ and $H^\prime (1,x) = id_A$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm particularly &lt;s&gt;dis&lt;/s&gt;satisfied with the amount of detail in my reasoning &lt;s&gt;but I can't seem to produce what I'm looking for.&lt;/s&gt; Many thanks for your help!&lt;/p&gt;&#xA;" OwnerUserId="5798" LastEditorUserId="5798" LastEditorDisplayName="" LastEditDate="2011-03-09T22:22:26.953" LastActivityDate="2011-03-09T22:22:26.953" Title="Proof that retract of contractible space is contractible" Tags="&lt;algebraic-topology&gt;" AnswerCount="1" />
  <row Id="26015" PostTypeId="2" ParentId="26007" CreationDate="2011-03-09T17:10:35.520" Score="5" ViewCount="" Body="&lt;p&gt;Markov's inequality is a &quot;large deviation bound&quot;. It states that the probability that a non-negative random variable gets values much larger than its expectation is small.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Chebyshev's inequality is a &quot;concentration bound&quot;. It states that a random variable with finite variance is concentrated around its expectation. The smaller the variance, the stronger the concentration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both inequalities are used to claim that most of the time, random variables don't get &quot;unexpected&quot; values.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A typical application of concentration bounds is polling - if you poll 200 people, then the standard deviation is so-and-so, and so the probability that the results are off by some amount $x$ is only $p$ which is very small. You can bound $p$ using Chebyshev's inequality, although there are better bounds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another application is in the statistics of patients in a hospital. Suppose they conduct a survey and find that the average number of patients per day is 10. Suppose they want to be able all patients 90% of the time. Then they need to handle at most 100 patients.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The list of probability results goes on. The law of large numbers states that the average of identical random variables tends (almost certainly) to the expectation. The central limit theorem describes the behavior of the noise, but only &quot;close&quot; to the expectation. Chernoff bounds are large deviation bounds useful for understanding the behavior of the noise away from the expectation.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2011-03-09T17:10:35.520" CommentCount="2" />
  <row Id="26016" PostTypeId="2" ParentId="26007" CreationDate="2011-03-09T17:10:59.403" Score="0" ViewCount="" Body="&lt;p&gt;$t\mu\{x\in X : |f(x)|\geq t\}\leq\int_Xf$ is simple to remember.  take a value $t$ use it for a lower bound for $f$ on $\{x\in X : |f(x)|\geq t\}$.  This is definitely less or equal $\int_X f$.  take the graph of a nice function $f$, draw a horizontal line at height $t$, and take the area underneath the line and under the graph.&lt;/p&gt;&#xA;" OwnerUserId="6925" LastActivityDate="2011-03-09T17:10:59.403" />
  <row Id="26017" PostTypeId="1" AcceptedAnswerId="26020" CreationDate="2011-03-09T17:11:33.503" Score="1" ViewCount="68" Body="&lt;p&gt;Not sure if many of you speak binary, but the question is to prove that 2n bits are sufficient to store the product of two unsigned n-bit numbers (i.e., there will be no overflow).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've thought of a simple method by contradiction, but its way too simple and I feel theres something wrong with it. So here it is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose not, that is, suppose that 2n bits are &lt;strong&gt;not&lt;/strong&gt; sufficient to store the product of two unsigned n-bit numbers. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then consider the product of two binary numbers 1111 x 1111. The result is 11100001, which is 8 bits=2n where n=4. So 2n was indeed sufficient to hold the product. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Have we proved anything here? Is this legit. If not, what are some ways I could prove this?&lt;/p&gt;&#xA;" OwnerUserId="2985" LastActivityDate="2011-03-09T17:18:43.487" Title="Proving a statement by contradiction" Tags="&lt;boolean-algebra&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="26018" PostTypeId="2" ParentId="26017" CreationDate="2011-03-09T17:16:25.547" Score="4" ViewCount="" Body="&lt;p&gt;Your proof isn't valid - maybe there are some other numbers that do require more than $2n$ digits? You only provide a specific example which doesn't. This way you can prove that the product requires only $1$ digit - consider $0 \times 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your idea actually works, but there's no need for proof by contradiction. For the case $n=4$, each of the $n$-bit numbers is at most $15$, and so the product is at most $225$, which requires only $2n=8$ bits. This generalizes.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2011-03-09T17:16:25.547" CommentCount="4" />
  <row Id="26019" PostTypeId="2" ParentId="25849" CreationDate="2011-03-09T17:17:48.510" Score="5" ViewCount="" Body="&lt;p&gt;Once you show that $\mathbb{Z}/p\mathbb{Z}$ has a primitive root, my favorite proof of the &quot;ramping up&quot; that lets you demonstrate the existence of primitive roots in $\mathbb{Z}/p^n \mathbb{Z}$ (for $p$ odd) is in LeVeque &quot;Fundamentals of Number Theory&quot; Theorem 4.4 and the comment following it.  This proof also makes clear that if an integer $g$ is a primitive root in $\mathbb{Z}/p\mathbb{Z}$, then either $g$ or $g+p$ is a primitive root in $\mathbb{Z}/p^n \mathbb{Z}$ for all $n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;He proves this as a consequence of the following (his Theorem 4.4):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose $p$ is a prime relatively prime to $a$. Let $t_n$ be the order of $a \mod p^n$ and assume that $p^z$ exactly divides $p^{t_1}-1$.  Then if $p&amp;gt;2$ or $z&amp;gt;1$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ t_n = \begin{cases} t_1, \quad &amp;amp;\text{for $n \leq z$}\\ t_1 p^{n-z}, \quad &amp;amp;\text{for $n &amp;gt; z$.}\end{cases}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I like this because I imagine modifying it to give the following alternative proof of the existence of primitive roots.  First, find a primitive root $g$ in $\mathbb{Z}/p \mathbb{Z}$.  Employ Dirichlet's Theorem on primes in arithmetic progressions to find a prime $a$ in the same class as $g$.  In the cyclotomic $\mathbb{Z}_p$ extension of $\mathbb{Q}$, $a$ must have intertial degree $p$ between the $z$th and $z+1$st layers (sitting inside the fields $\mathbb{Q} (\zeta_{p^z})$ and $\mathbb{Q} (\zeta_{p^{z+1}})$.  Thus, $a$ has intertial degree $p$ between all of the higher layers of the cyclotomic $\mathbb{Z}_p$ extension, giving LeVeque's formula for the orders $t_n$.  The proof of the existence of primitive roots is very short once this is established.&lt;/p&gt;&#xA;" OwnerUserId="4253" LastActivityDate="2011-03-09T17:17:48.510" />
  <row Id="26020" PostTypeId="2" ParentId="26017" CreationDate="2011-03-09T17:18:43.487" Score="4" ViewCount="" Body="&lt;p&gt;Your mistake is negating (for every $n$ bit numbers $a,b$ the product $ab$ is a $2n$ bit number or less) to (for every $n$ bit numbers $a,b$ the product $ab$ takes more than $2n$ bits) (which you did contradict with your example) but the actual negation is (for some $n$ bit numbers $a,b$ the product $ab$ takes more than $2n$ bits). You can't contradict a &quot;for some&quot; just by giving an example though.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for proving the theorem, you could prove the lemma if $x \le y$ then $x$ takes less than or equal to the number of bits that $y$ takes. Then 1111x1111 = 11100001 does prove the theorem in the case of n=4, but you will have to do this multiplication in a general way to prove it for every n.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another useful thing to make proving this easier is that 2^n-1 = 1111111...111 for n 1's.&lt;/p&gt;&#xA;" OwnerUserId="4290" LastActivityDate="2011-03-09T17:18:43.487" CommentCount="2" />
  <row Id="26021" PostTypeId="1" CreationDate="2011-03-09T17:20:39.803" Score="3" ViewCount="195" Body="&lt;p&gt;&lt;em&gt;[Update 3]&lt;/em&gt;: I realized, that the series $s(h)$ below is simply a &quot;(general) Dirichlet-series&quot; (1), so after I know the principle how to find the Taylor-series for some function (which may be given by its Dirichlet-series) I can now make my question more precise as:     &lt;/p&gt;&#xA;&#xA;&lt;p&gt;how to find the Dirichlet-series-representation of a function, given its Taylor-series?     &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;(1) - the attribute &quot;general&quot; means a series $ \sum_{k=1}^{\infty} a_k*e^{- \lambda _k s}$ where the $\lambda _k$ form a strictly monotone sequence of real numbers increasing to $+\infty$ (K. Knopp, &quot;Infinite series...&quot;, german ed.)&lt;/em&gt;&lt;br&gt;&#xA;&lt;em&gt;(I did not change the text below) [end update 3]&lt;/em&gt;     &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;In the study of functional iteration and the Schröder function I'm concerned with series of the type:&#xA;$$ s(h)= a_0* (u^0)^h + a_1* (u^1)^h + a_2 * (u^2)^h + ... $$ &#xA;where &lt;em&gt;h&lt;/em&gt; is the (iteration-) &quot;height&quot;-parameter.&lt;br&gt;&#xA;Usually, if this converges, in the numerical evaluation I change order of computation:&#xA;$$ s(h)= a_0*(u^h)^0 + a_1*(u^h)^1 + a_2*(u^h)^2 + ... $$ &#xA;which means in many cases I handle this as a power series in $u^h$ . But when I considered further analysis, for instance the derivative with respect to &lt;em&gt;h&lt;/em&gt;, I differentiate using $h$ as highest exponent. &lt;em&gt;(BTW, what is the name of that type of series?)&lt;/em&gt;           &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now by accident I asked Pari/GP for the coefficients of $s(x)$ and what I got back was a power series in &lt;em&gt;x&lt;/em&gt;. I never thought deeper about such a conversion, except that I remember the case of the conversion of the zeta-series into a power series using the Stieltjes constants.&lt;br&gt;&#xA;Example:   &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;s(h) = 2*1^h -1.09662*1/4^h + 0.100215*1/16^h  -0.00366327*1/64^h &#xA;     + 0.0000717362*1/256^h  -0.000000874084*1/1024^h + ... - ... &#xA;ps(x) = 1.00000 + 1.25723*x - 0.699161*x^2 + 0.172874*x^3 + 0.0350749*x^4    &#xA;     - 0.0550780*x^5 + 0.0288564*x^6 - 0.00942667*x^7 + O(x^8)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Well, I think, what Pari/GP internally does is to compute the derivatives of $s(h)$ and construct the Taylor series. But now I'm curious (and that's my question): How could I convert a power series &lt;em&gt;ps(x)&lt;/em&gt; into a series of the &lt;em&gt;s(h)&lt;/em&gt;-type?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&lt;strong&gt;[update]&lt;/em&gt;&lt;/strong&gt; &lt;em&gt;I think the comment to Mitch would be a good additional background and explanation for the question, so I copy it to here and extend it a bit&lt;/em&gt;   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm originally interested in the &lt;strong&gt;&lt;em&gt;method&lt;/em&gt;&lt;/strong&gt;: &lt;em&gt;how&lt;/em&gt; one would do such a conversion of $ps(x)$ to $s(h)$ ?&lt;br&gt;&#xA;Note that some values for $s(h)$ are $s(1) \approx 1.73205$ , $s(2) \approx 1.93185$.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My example comes from iteration of the function $$f(x) = \sqrt{2+x}$$ $$f(f(x))=\sqrt{2+\sqrt{2+x}} $$  $$f(f(f(x))) =\sqrt{2+\sqrt{2+\sqrt{2+x}}} $$ and so on.     &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now $f(x)$ has a power series in $x$, as well as $f(f(x))$ and for each iterate there is another power series. The method of Schröder functions applied to a recentered version of $f(x)$, which has no constant term (also called &quot;regular iteration&quot;), allows to find a power series $F(x,h)$ in two variables where $h$ means the iteration-&quot;height&quot;. In such a function $F(x,h)$  the coefficients at powers of $x$ are polynomials in $\lambda^{ h}$ (where $\lambda$ is an eigenvalue of the function which I usually denote as $u$ for ASCII-readability).     &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I set $x=1$ and reorder summation collecting like powers of $\lambda^h $ I get a series in $\lambda^h$ only which I called $S(h)$ This function gives exactly the value of the &lt;em&gt;h&lt;/em&gt;'th iterate of $f(x)$ beginning at &lt;em&gt;x=1&lt;/em&gt; simply meaning $  s(h) = F(1,h) $    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But $s(h)$ has not the usual form of a power series in $h$ as shown above. When I asked Pari/GP for an evaluation at some &lt;em&gt;h&lt;/em&gt; I accidentally typed $s(x)$ and Pari/GP gave back a power series in &lt;em&gt;x&lt;/em&gt;, I called it above $ps(x)$. $s()$ and $ps()$ have completely different forms but give the same result: $s(h) = ps(h)$ where both functions converge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The $s(h)$ as well as the $ps(x)$-form allow also continuous (i.e fractional) iteration giving the same values. Interestingly the behave of the convergence of the two forms is completely different. The limit for $h\to\infty$ is immediately visible in $s(h)$ but for $ps(h)$ that would likely be a divergent series, while for $h=0$ we see immediately that $ps(h)=1$ but would not recognize it by $s(h)$ &lt;/p&gt;&#xA;" OwnerUserId="1714" LastEditorUserId="1714" LastEditorDisplayName="" LastEditDate="2011-05-04T15:19:43.300" LastActivityDate="2011-05-04T15:19:43.300" Title="How to convert a powerseries into a Dirichlet series?" Tags="&lt;sequences-and-series&gt;&lt;functional-analysis&gt;&lt;power-series&gt;" CommentCount="7" FavoriteCount="1" />
  <row Id="26023" PostTypeId="1" AcceptedAnswerId="26030" CreationDate="2011-03-09T17:34:33.333" Score="2" ViewCount="105" Body="&lt;p&gt;Hello,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have that:&lt;br&gt;&#xA;If $a \equiv 1 \pmod{n}$, then does $a^n \equiv 1 \pmod{n}$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I don't know if this is true:&lt;br&gt;&#xA;If $a \not\equiv 1 \pmod{n}$, then does $a^n \not\equiv 1 \pmod{n}$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried to find a counter example, but it always turned out to be true. Any idea?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks,&lt;/p&gt;&#xA;" OwnerUserId="5711" LastActivityDate="2011-03-09T20:15:13.923" Title="If $a \not\equiv 1 \pmod{n}$, then does $a^n \not\equiv 1 \pmod{n}$?" Tags="&lt;elementary-number-theory&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="26024" PostTypeId="1" AcceptedAnswerId="26027" CreationDate="2011-03-09T17:37:17.310" Score="2" ViewCount="103" Body="&lt;p&gt;Hello,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm really confused about these two. For example if $n = 6$, then:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Divisors: $2, 3$&lt;br&gt;&#xA;Proper Divisors: $1, 2, 3, 6$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it right?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update&lt;/strong&gt;&lt;br&gt;&#xA;From Elementary Number Theory and Its Application by Kenneth H. Rosen 6th edition, page 256:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Because of certain mystical beliefs, the ancient Greeks were interested in those integers that are equal to the sum of all their proper positive divisors. Such integers are called perfect numbers.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Example:&lt;br&gt;&#xA;  $\sigma(6) = 1 + 2 + 3 + 6 = 12$, we see that $6$ is perfect. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Thanks,  &lt;/p&gt;&#xA;" OwnerUserId="5711" LastEditorUserId="5711" LastEditorDisplayName="" LastEditDate="2011-03-10T06:41:52.633" LastActivityDate="2011-03-10T06:41:52.633" Title="What is the difference between divisors and proper divisors?" Tags="&lt;elementary-number-theory&gt;" AnswerCount="2" CommentCount="9" />
  <row Id="26025" PostTypeId="2" ParentId="26023" CreationDate="2011-03-09T17:38:58.733" Score="2" ViewCount="" Body="&lt;p&gt;Counterexample: $a = 9$, $n = 10$...&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2011-03-09T17:38:58.733" CommunityOwnedDate="2011-03-09T17:42:24.130" CommentCount="1" />
  <row Id="26026" PostTypeId="1" AcceptedAnswerId="26035" CreationDate="2011-03-09T17:40:39.963" Score="4" ViewCount="63" Body="&lt;p&gt;Let $f \in C^1_c(\Omega)$ where $\Omega \subset \mathbb{R}^d$ is a bounded domain. Let $\phi \in C^1_c(\mathbb{R}^d)$ be an approximation of the identity (i.e. $\int_{\mathbb{R}^d} \phi=1$, $\phi \geq 0$, $\phi_\epsilon := \frac{1}{\epsilon^d} \phi(\frac{x}{\epsilon})$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How would you prove that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_\Omega |f(x) - f \ast \phi_\epsilon(x)| dx \leq \epsilon \int_\Omega |\nabla f| dx?$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm trying to show that the family of $C^1_c$ functions convolved with a mollifier is uniformly close to the function in $L^1$ (which would be true after having this result if we assume something like the family of functions being bounded in $W^{1,1}(\Omega)$). &lt;/p&gt;&#xA;" OwnerUserId="1736" LastActivityDate="2011-03-09T22:42:51.613" Title="Bounding the integral of a $C^1$ function using its gradient" Tags="&lt;real-analysis&gt;&lt;functional-analysis&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" />
  <row Id="26027" PostTypeId="2" ParentId="26024" CreationDate="2011-03-09T17:43:12.430" Score="8" ViewCount="" Body="&lt;p&gt;I think most of the time the convention would be: divisors = $\{1,2,3,6\}$ and proper divisors= $ \{1,2,3\} $.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance 6 is perfect because $\sigma(n)=2n$ where σ is the sum of its &lt;em&gt;divisors&lt;/em&gt; or $s(n)=n$ where $s$ is the sum of its &lt;em&gt;proper divisors&lt;/em&gt; . (Note that the notation $s$ may not be completely standard.) &lt;/p&gt;&#xA;" OwnerUserId="5189" LastEditorUserId="5189" LastEditorDisplayName="" LastEditDate="2011-03-09T19:48:28.500" LastActivityDate="2011-03-09T19:48:28.500" CommentCount="2" />
  <row Id="26028" PostTypeId="2" ParentId="3815" CreationDate="2011-03-09T17:46:49.250" Score="0" ViewCount="" Body="&lt;p&gt;Your presupposition is incorrect; there are a number of set theories (some of them provably equiconsistent with ZF) in which the set of all singletons exists; see &lt;a href=&quot;http://math.stackexchange.com/questions/10086/limiting-set-theory-using-symmetry/22026#22026&quot;&gt;Limiting set theory using symmetry&lt;/a&gt;, or Forster’s &lt;a href=&quot;http://www.oup.com/us/catalog/general/subject/Mathematics/Logic/?view=usa&amp;amp;ci=9780198514770&quot; rel=&quot;nofollow&quot;&gt;Oxford Logic Guide&lt;/a&gt; on the subject.  (Disclaimer:  Forster discusses my work in the book, so I’m hardly unbiased, but it is the standard work.)&lt;/p&gt;&#xA;" OwnerUserId="7075" LastEditorUserId="7075" LastEditorDisplayName="" LastEditDate="2011-03-30T16:09:44.193" LastActivityDate="2011-03-30T16:09:44.193" CommentCount="2" />
  <row Id="26029" PostTypeId="2" ParentId="26012" CreationDate="2011-03-09T17:50:05.800" Score="0" ViewCount="" Body="&lt;p&gt;let $f(x)=\int\chi_E(y)\chi_F(x-y)dy$.  Then $|f(s)-f(t)|=|\int\chi_E(y)(\chi_F(s-y)-\chi_F(t-y))dy|\leq\mu(E)|\int(\chi_F(s-y)-\chi_F(t-y))dy|$.  approximate $F$ with finitely many open intervals (call this union of intervals $U$) with $\mu(F\Delta U)&amp;lt;\epsilon$.  if $|s-t|$ is less than the distance between any two of the intervals making up $U$, then we have $\int(\chi_U(s-y)-\chi_U(t-y))dy=0$ (hopefully this is obvious).  also note that $|\int(\chi_F(s-y)-\chi_F(t-y))dy-\int(\chi_U(s-y)-\chi_U(t-y))dy|\leq2\epsilon$.  Hence for $|s-t|$ small with respect to the chosen $U$ we have $|f(s)-f(t)|\leq2\epsilon\mu(E)$.&lt;/p&gt;&#xA;" OwnerUserId="6925" LastEditorUserId="6925" LastEditorDisplayName="" LastEditDate="2011-03-09T19:28:09.183" LastActivityDate="2011-03-09T19:28:09.183" CommentCount="2" />
  <row Id="26030" PostTypeId="2" ParentId="26023" CreationDate="2011-03-09T17:50:46.687" Score="6" ViewCount="" Body="&lt;p&gt;Think about equalities: it is certainly true that if $x=1$, then $x^n=1$; but it is &lt;em&gt;not&lt;/em&gt; true that if $x\neq 1$ then $x^n\neq 1$ (e.g., $x=-1$, $n$ even).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same is true with congruences: &lt;em&gt;congruences modulo&lt;/em&gt; $n$ may be added and multiplied, just like equalities: if $a\equiv b\pmod{n}$ and $c\equiv d\pmod{n}$, then $a+c\equiv b+d\pmod{n}$ (adding the two congruences) and $ac\equiv bd\pmod{n}$ (multiplying the two congruences). Same as with equalities. This immediately gives you that if $a\equiv 1 \pmod{n}$, then multiplying the congruence by itself $m$ times you will get $a^m \equiv 1^m=1\pmod{n}$, and in particular $a^n\equiv 1\pmod{n}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, we cannot just add and multiple &lt;em&gt;non-equalities&lt;/em&gt;: that is, if $a\neq b$ and $c\neq d$, then it does &lt;em&gt;not&lt;/em&gt; follow that $a+c\neq b+d$ or that $ac\neq bd$ (I'll leave it to you to find easy counterexamples). &lt;strong&gt;Note well:&lt;/strong&gt; these are not inequalities of the form $a\lt b$, but merely of the form &quot;not equal to&quot;. Likewise with congruences. Just because $a\not\equiv 1 \pmod{n}$, it does not follow that $a^2\not\equiv 1^2\pmod{n}$, or $a^n\not\equiv 1\pmod{n}$ in general.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It &lt;em&gt;does&lt;/em&gt; follow for $n$ prime, but that's not because of the properties of congruences, but rather because of Fermat's Little Theorem: because Fermat's Little Theorem tells you that $a^p\equiv a\pmod{p}$ when $p$ is a prime, so you can conclude that if $a\not\equiv 1\pmod{p}$, then $a^p \equiv a\not\equiv 1\pmod{p}$. But this is a particular property of primes, not of congruences.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added.&lt;/strong&gt; Going from the comments, some situations where it is true and some where it is false:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;If $\gcd(a,n)\gt 1$, then of course $a\not\equiv 1\pmod{n}$, and likewise $a^n\not\equiv 1\pmod{n}$; this is the &quot;trivial case&quot;, one might say.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If $\gcd(a,n)=1$, then of course we have $a^{\varphi(n)}\equiv 1 \pmod{n}$ by Euler's Theorem; thus, $a^n\equiv a^{\varphi(n)}a^{n-\varphi(n)}\equiv a^{n-\varphi(n)}\pmod{n}$. If $\gcd(\varphi(n),n-\varphi(n))=\gcd(\varphi(n),n)=1$, then from $a^{n-\varphi(n)}\equiv 1\pmod{n}$ you can conclude that $a\equiv 1\pmod{n}$, since no element, other than $1$, of the multiplicative group modulo $n$ has order dividing $n-\varphi(n)$ (since they all have order dividing $\varphi(n)$). &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;However, if $\gcd(\varphi(n),)\gt 1$, then we can always pick a prime $p$ that divides $\gcd(\varphi(n),n)$, and find an element $a$ of the multiplicative group modulo $n$ that has order exactly $p$. That will mean that $a\not\equiv 1\pmod{n}$, but $a^p\equiv 1\pmod{n}$. Then we will have $a^{n} = a^{\varphi(n)}a^{n-\varphi(n)} = (a^{p})^k\equiv 1\pmod{n}$, where $kp = n-\varphi(n)$, so in this case there will always be a counterexample.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In short, the implication &#xA;$$a\not\equiv 1\pmod{n}\Longrightarrow a^n\not\equiv 1\pmod{n}$$ &#xA;holds if and only if $\gcd(n,a)\gt 1$ or $\gcd(a,n)=\gcd(\varphi(n),n)=1$, where $\varphi(n)$ is Euler's phi function.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-09T20:15:13.923" LastActivityDate="2011-03-09T20:15:13.923" CommentCount="8" />
  <row Id="26031" PostTypeId="2" ParentId="26014" CreationDate="2011-03-09T18:10:25.587" Score="4" ViewCount="" Body="&lt;p&gt;You're glossing over the most important things. You are given the information that $X$ is contractible. This means that there exists &lt;em&gt;a specific homotopy&lt;/em&gt; $H:I\times X\to X$ which contracts the identity map. Your goal is to construct &lt;em&gt;a specific homotopy&lt;/em&gt; $H&amp;#39;:I\times A\to A$ which contracts the identity map. Your proof never explicitly makes use of $H$ and never explicitly (maybe not even implicitly) defines $H&amp;#39;$. This makes it feel like a bunch of symbol-pushing. (This isn't meant to be insulting; I'm just explaining why the proof &lt;em&gt;feels&lt;/em&gt; unsatisfying.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Two preliminary points:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;$\newcommand{\pt}{{\{\ast\}}}f&amp;#39;=f\circ r$ isn't a map between the right things; it should be a map $A\to\pt$, but it's a map $X\to \pt$. This isn't too important since there's only one map from anything to a point, but the algebraic manipulation will be clearer if you set $f&amp;#39;:=f|_A$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;You don't need to check $f&amp;#39;\circ g&amp;#39;\cong id_\pt$ since there's only one map $\pt\to\pt$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So you only need to find a homotopy from $id_A$ to $g&amp;#39;\circ f&amp;#39;$. That is, you need to make a movie (homotopy) which continuously crushes $A$ to a point. But you already have a movie $H:I\times X\to X$ which crushes $X$ to a point (i.e. $H(0,-)=id_X$ and $H(1,-)=g\circ f$), and you have a way to retract &lt;em&gt;anything&lt;/em&gt; in $X$ to something in $A$, so you can just retract your movie, setting $H&amp;#39; = r\circ H|_{I\times A}$. Then $H&amp;#39;(0,-) = r\circ H(0,-)|_A = r\circ id_A = id_A$ and $H&amp;#39;(1,-)=r\circ H(1,-)|_A = r\circ (g\circ f)|_A = r\circ g\circ f|_A = g&amp;#39;\circ f&amp;#39;$.&lt;/p&gt;&#xA;" OwnerUserId="60" LastActivityDate="2011-03-09T18:10:25.587" CommentCount="8" />
  <row Id="26032" PostTypeId="2" ParentId="25505" CreationDate="2011-03-09T18:17:44.173" Score="2" ViewCount="" Body="&lt;p&gt;Let $f(x,y)=\frac{\log x}{x-y}$.&#xA;Since the principal value does not seem well-defined here, let's integrate $f(x,y)+f(y,x)= \frac{\log (x/y)}{x-y}$ (which is positive) on the triangle defined by $0&amp;lt;y&amp;lt;x&amp;lt;1$.&#xA;Now $\int_0^x \frac{\log (x/y)}{x-y} dy = -\int_0^1 \frac{\log u}{1-u} du$ (change of variable $y=xu$), which does not depend on $x$, and so this is equal to the first integral (integrating for $x$ between $0$ and $1$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now $-\int_0^{1-\epsilon} \frac{\log u}{1-u} du = \int_0^{1-\epsilon} \sum_{ n \geq 1} \frac{u^{n-1}}{n} du = \sum_{n \geq 1} \frac{(1-\epsilon)^n}{n^2}$ and letting $\epsilon$ go to $0$ gives $\zeta(2)=\frac{\pi^2}{6}$.&lt;/p&gt;&#xA;" OwnerUserId="2660" LastActivityDate="2011-03-09T18:17:44.173" CommentCount="2" />
  <row Id="26033" PostTypeId="2" ParentId="26008" CreationDate="2011-03-09T18:28:02.453" Score="6" ViewCount="" Body="&lt;p&gt;Here's a bigger hint.  Try pairing terms in the sum like so:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{1}{1} + \frac{1}{2} + \cdots + \frac{1}{p-1} = \left(\frac{1}{1} + \frac{1}{p-1}\right) + \left(\frac{1}{2} + \frac{1}{p-2}\right) + \cdots + \left(\frac{1}{\frac{p-1}{2}} + \frac{1}{p - \frac{p-1}{2}}\right).$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, get a common denominator for each &lt;em&gt;pair&lt;/em&gt;.  That will tell you something about $A_p \mod p$.  From there maybe you can make some conjectures about $A_p \mod p^2$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I guess I'm also a little confused as to what you're being asked to do with finding a formula for $\frac{A_p}{B_p}$.  This is called the $p$th &lt;em&gt;harmonic number&lt;/em&gt; $H_p$, and there is no known nice formula for $H_p$ or for $A_p$ when the fraction is in reduced form.  I'm not even sure there is a nice formula for $B_p$ when the fraction is in reduced form.  So I think Yuval Filmus's comment is right: You're not going to be able to do much better than the &quot;gook&quot; you are describing in your comment above.&lt;/p&gt;&#xA;" OwnerUserId="2370" LastActivityDate="2011-03-09T18:28:02.453" CommentCount="1" />
  <row Id="26034" PostTypeId="1" CreationDate="2011-03-09T18:33:01.310" Score="0" ViewCount="132" Body="&lt;p&gt;The given curve is rotated about the y-axis. Find the area of the resulting surface&#xA;y= (1/4 x^2) - (1/2 lnx) x is in between 1 and 2 (including 1 and 2)If anyone could please point me in the right direction of how to solve this I would be very grateful :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To  Arturo Magidin, I'm very sorry, this is my first time using this or hearing about this site so I didn't really know how it worked.  I'll implement your suggestions in the future&lt;/p&gt;&#xA;" OwnerUserId="8032" LastEditorUserId="8032" LastEditorDisplayName="" LastEditDate="2011-03-10T14:13:03.303" LastActivityDate="2011-05-09T14:55:56.373" Title="Area of a Surface of Revolution?" Tags="&lt;calculus&gt;&lt;homework&gt;" AnswerCount="1" CommentCount="6" />
  <row Id="26035" PostTypeId="2" ParentId="26026" CreationDate="2011-03-09T18:40:18.277" Score="6" ViewCount="" Body="&lt;p&gt;WLOG assume $\phi(x)$ supported in the unit ball (it has compact support, so it is supported in &lt;em&gt;some&lt;/em&gt; ball). First look at&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ |f(x) - f*\phi_\epsilon(x)| \leq \int_{|z|\leq 1} \phi(z) |f(x) - f(x-\epsilon z)| dz $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;replace the integrand&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ f(x) - f(x-\epsilon z) = - \int_0^{\epsilon|z|} D_rf(x - r\omega) dr $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with $\omega = z / |z|$ and using the fundamental theorem of calculus. So&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Integrating the whole thing over $x$, and changing the order of integration, you have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \int_{\Omega}|f(x) - f*\phi_\epsilon(x)|dx \leq \int_{|z|\leq 1} \phi(z) \int_0^{\epsilon|z|} \int_{\Omega} |D_rf(x - r\omega)| dx~ dr~ dz $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The inside most integral for fixed $r\omega$ gives you $\int_\Omega |\nabla f| dx$. The integral over $r$ gives you the factor of $\epsilon$. And integrating $\phi(z)$ over the ball of radius 1 gives you 1. &lt;/p&gt;&#xA;" OwnerUserId="1543" LastEditorUserId="1543" LastEditorDisplayName="" LastEditDate="2011-03-09T22:42:51.613" LastActivityDate="2011-03-09T22:42:51.613" CommentCount="7" />
  <row Id="26036" PostTypeId="2" ParentId="26034" CreationDate="2011-03-09T18:55:12.643" Score="2" ViewCount="" Body="&lt;p&gt;The formula for the area of a surface of revolution about the $y$-axis formed by $(x(t),y(t))$ on $a\le t\le b$ is $$2\pi \int_a^b x(t)\sqrt{(x&amp;#39;(t))^2+(y&amp;#39;(t))^2} dt.$$  In your case, let $x(t)=t$ so that $y(t)=\frac{1}{4}t^2-\frac{1}{2}t\ln t$.&lt;/p&gt;&#xA;" OwnerUserId="72" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2011-03-09T19:16:59.777" LastActivityDate="2011-03-09T19:16:59.777" CommentCount="5" />
  <row Id="26037" PostTypeId="1" CreationDate="2011-03-09T19:14:07.473" Score="5" ViewCount="244" Body="Potentially related-questions, shown before posting, didn't have anything like this, so I apologize in advance if this is a duplicate.&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I know there are many ways of calculating (or should I say &quot;ending up at&quot;) the constant &lt;em&gt;e&lt;/em&gt;.  How would you explain &lt;em&gt;e&lt;/em&gt; concisely?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's a rather beautiful number, but when friends have asked me &quot;what &lt;strong&gt;is&lt;/strong&gt; &lt;em&gt;e&lt;/em&gt;?&quot; I'm usually at a loss for words -- I always figured the math explains it, but I would really like to know how others conceptualize it, especially in common-language (say, &quot;English&quot;).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;hr&gt;&#xA;related but not the same: &lt;a href=&quot;http://math.stackexchange.com/questions/3006/could-you-explain-why-dex-dx-ex-intuitively&quot;&gt;Could you explain why $de^x/dx = e^x$ &amp;quot;intuitively&amp;quot;?&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1774" LastActivityDate="2011-03-09T19:48:48.217" Title="Intuitive Understanding of the constant &quot;e&quot;" Tags="&lt;intuition&gt;" AnswerCount="5" CommentCount="6" FavoriteCount="2" />
  <row Id="26038" PostTypeId="2" ParentId="26037" CreationDate="2011-03-09T19:19:49.953" Score="3" ViewCount="" Body="&lt;p&gt;&lt;em&gt;One&lt;/em&gt; way of understanding what is $e$, is to see it as a rate of growth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://betterexplained.com/articles/an-intuitive-guide-to-exponential-functions-e/&quot; rel=&quot;nofollow&quot;&gt;This article&lt;/a&gt; explains it very well.&lt;/p&gt;&#xA;" OwnerUserId="7476" LastActivityDate="2011-03-09T19:19:49.953" CommentCount="2" />
  <row Id="26039" PostTypeId="2" ParentId="26037" CreationDate="2011-03-09T19:23:11.517" Score="3" ViewCount="" Body="&lt;p&gt;For the somewhat-calculus-literate, your &quot;related but not the same&quot; question is what I'd go for: $e$ is the number for which the exponential function with that base is its own derivative.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Without calculus, I'd go for the notion of compound interest: With a rate $r$ per period, compounded $n$ times per period, $A$ grows to $A(1+\frac{r}{n})^n$ after 1 period; as $n\to\infty$, $A(1+\frac{r}{n})^n\to Ae^r$.&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2011-03-09T19:23:11.517" CommentCount="1" />
  <row Id="26040" PostTypeId="2" ParentId="26037" CreationDate="2011-03-09T19:33:42.023" Score="2" ViewCount="" Body="&lt;p&gt;It is my opinion there is no &quot;intuitive understanding of the number $e$&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Presumably, what you want to explain to your friends is not some mythical intuitive content of the number but some actual, concrete property it has which makes you appreciate it. Explain &lt;em&gt;that&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="274" LastEditorUserId="274" LastEditorDisplayName="" LastEditDate="2011-03-09T19:41:51.500" LastActivityDate="2011-03-09T19:41:51.500" />
  <row Id="26041" PostTypeId="2" ParentId="25596" CreationDate="2011-03-09T19:36:32.350" Score="1" ViewCount="" Body="&lt;p&gt;I finally clicked through to the paper and found the definitions of distances you are referring to.  The hard part is finding the points $p_s$ and $p_e$ in figure 5.  If you parameterize segment 1 as $s_i+t(e_i-s_i)$ we can find the distance from $s_j$ to a point and minimize over $t$.  The distance squared is $d^2=\sum(s_j-s_i-t(e_i-s_i))^2$ &lt;strong&gt;note: sign before $t$ changed to -, does not affect the following.&lt;/strong&gt; where the sum is over the various components.  If we take the derivative with respect to $t$ and set to zero we find $t=\frac{\sum(s_j-s_i)(e_i-s_i)}{\sum(e_i-s_i)^2}$.  Plugging this $t$ into $s_i+t(e_i-s_i)$ gives the point $p_s$. A similar expression using $e_j$ instead of $s_j$ gives the value of $t$ for $p_e$.  Now you can calculate all the distances using the usual sum of squares technique.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Added:  so imagine $s_i=(1,2), e_i=(5,4), s_j=(2,1)$  If you plot them, you will see that $p_s=(1.4,2.2)$.  The calculation would be $t=\frac{(2-1)(5-1)+(4-2)(1-2)}{(5-1)^2+(4-2)^2}=0.1$, so $p_s$ lies 0.1 of the way from $s_i$ to $e_i$ and is at $(1.4,2.2)$.  Similarly, if $e_j$ is at $(2,5)$, $t=\frac{(2-1)(5-1)+(5-2)(1-2)}{(5-1)^2+(4-2)^2}=0.5$ and $e_j=(3,3)$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then we can say $l_{||1}=\sqrt{(1.4-1)^2+(2.2-2)^2}=.447$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$l_{||2}=\sqrt{(3-1)^2+(3-2)^2}=2.236$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$l_{\bot1}=\sqrt{(2-1.4)^2+(1-2.2)^2}=1.34$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$l_{\bot2}=\sqrt{(3-5)^2+(3-4)^2}=2.236$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$L_j=4, d_\theta=l_{\bot2}-l_{\bot1}=0.896$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have three coordinates all these sums of two things become three.  From here you should be able to follow the equations in the article.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2011-03-17T23:16:10.070" LastActivityDate="2011-03-17T23:16:10.070" CommentCount="13" />
  <row Id="26042" PostTypeId="1" CreationDate="2011-03-09T19:37:37.447" Score="3" ViewCount="144" Body="&lt;p&gt;I am reading the Wikipedia article on &lt;a href=&quot;http://en.wikipedia.org/wiki/Conditional_independence&quot; rel=&quot;nofollow&quot;&gt;conditional independence&lt;/a&gt;. There seems to be Two definitions for conditional independence of Two random variables $X$ and $Y$ given another one $Z$:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Two random variables $X$ and $Y$ are&#xA;conditionally independent given a&#xA;third random variable $Z$ if and&#xA;only if they are independent in&#xA;their conditional probability&#xA;distribution given $Z$. That is, $X$&#xA;and $Y$ are conditionally&#xA;independent given $Z$ if and only&#xA;if, given any value of $Z$, the&#xA;probability distribution of $X$ is&#xA;the same for all values of $Y$ and&#xA;the probability distribution of $Y$&#xA;is the same for all values of $X$.&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Two random variables $X$ and $Y$ are&#xA;conditionally independent given a&#xA;random variable $Z$ if they are&#xA;independent given $\sigma(Z)$: the&#xA;$\sigma$-algebra generated by $Z$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Two events $R$ and $B$ are&#xA;conditionally independent given a&#xA;$\sigma$-algebra $\Sigma$ if&#xA;    $$\Pr(R \cap B \mid \Sigma) = \Pr(R \mid \Sigma)\Pr(B \mid&#xD;&#xA;    \Sigma)\ a.s.$$ where $\Pr(A \mid&#xD;&#xA;    \Sigma)$ denotes the conditional&#xA;expectation of the indicator&#xA;function of the event $A$, given the&#xA;sigma algebra $\Sigma$. That is,&lt;br&gt;&#xA;$$ \Pr(A \mid \Sigma) :=&#xD;&#xA;    \operatorname{E}[\chi_A\mid\Sigma].$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Two random variables X and Y are&#xA;conditionally independent given a&#xA;$\sigma$-algebra $\Sigma$ if the&#xA;above equation holds for all $R$ in&#xA;$\sigma(X)$ and $B$ in $\sigma(Y)$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I can understand the second definition, but my questions are:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;What does the first definitions mean&#xA;actually? I have tried several times&#xA;on reading it, but fail to get what&#xA;it means? Can someone rephrase it&#xA;using rigorous and clean language,&#xA;for example, by writing the definition in terms of some formulae?&lt;/li&gt;&#xA;&lt;li&gt;Do the two definitions agree with&#xA;each other? Why?&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;ADDED:&lt;/strong&gt; I was wondering if the following is the correct way to understand the first definition. Notice that $P(\chi_A \mid Z)$ is defined as $E(\chi_A \mid Z)$ and therefore is a random variable. When the conditional probability $P(\cdot \mid Z)$ is  &quot;regular&quot;, i.e. when $P(\cdot \mid Z)(\omega)$ is a probability measure for each point $\omega$ in the underlying sample space $(\Omega, \mathcal{F}, P)$, does conditional independence between $X$ and $Y$ given $Z$ mean that $X$ and $Y$ are independent w.r.t. every probability measure defined by $P(\cdot \mid Z)(\omega), \forall \omega \in \Omega$? If yes, is the conditional probability $P(\cdot \mid Z)$ always  guaranteed to be &quot;regular&quot;? So that there is no need to explicitly write this &quot;regular&quot; assumption?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Thanks and regards!&lt;/p&gt;&#xA;" OwnerUserId="1281" LastEditorUserId="1281" LastEditorDisplayName="" LastEditDate="2011-03-10T01:18:26.490" LastActivityDate="2011-03-10T01:18:26.490" Title="Understanding conditional independence of two random variables given a third one" Tags="&lt;probability-theory&gt;&lt;measure-theory&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="26043" PostTypeId="2" ParentId="26037" CreationDate="2011-03-09T19:38:03.290" Score="1" ViewCount="" Body="&lt;p&gt;Suppose you have a particle with the following property: starting from rest at time zero (in seconds), the magnitude of its velocity $v$ (in m/s) at time $t$ is exactly equal to the distance $d$ from its starting place (at the same time). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is, at 1 meter its speed is 1 m/s, at 2 meters, 2 m/s and so on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the distance function $d = d(t)$ of this particle as a function of time, you might ask? Since $v(t) = d^{\prime}(t) = d(t)$, then $d(t) = e^{t}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The constant $e$ is the magnitude of this particle's distance or velocity at time $t = 1$.&lt;/p&gt;&#xA;" OwnerUserId="2720" LastEditorUserId="2720" LastEditorDisplayName="" LastEditDate="2011-03-09T19:48:48.217" LastActivityDate="2011-03-09T19:48:48.217" />
  <row Id="26044" PostTypeId="2" ParentId="25986" CreationDate="2011-03-09T19:40:39.380" Score="1" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Innumeracy_(book)&quot; rel=&quot;nofollow&quot;&gt;Numeracy&lt;/a&gt;. &lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2011-03-09T19:40:39.380" CommunityOwnedDate="2011-03-09T19:40:39.380" CommentCount="1" />
  <row Id="26045" PostTypeId="2" ParentId="26023" CreationDate="2011-03-09T19:44:32.943" Score="1" ViewCount="" Body="&lt;p&gt;This can be completely analyzed by basic group thory. Mod $\rm\:n\:$, the elements coprime to $\rm\:n\:$ comprise the &lt;em&gt;unit group&lt;/em&gt; of order $\rm\:\phi = \phi(n)\:.\ $ Thus $\rm\ x^n = x^{\phi} = 1\ \Rightarrow\ x^{(n,\ \phi)} = 1\:.\:$ So if they are coprime, i.e. $\rm\:(n,\phi)=1\:,\:$ then we deduce that $\rm\:x=1\:$ and the result is true. Otherwise it fails since then we may choose a prime $\rm\:p\ |\ (n,\phi)\ $ properly, and a counterxample arises by choosing an element of order $\rm\:p\:.$&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2011-03-09T20:07:50.840" LastActivityDate="2011-03-09T20:07:50.840" />
  <row Id="26046" PostTypeId="2" ParentId="26037" CreationDate="2011-03-09T19:47:30.463" Score="0" ViewCount="" Body="&lt;p&gt;I also believe this to be equivalent to the cited question about e^x, but will answer the explicit question for the sake of argument. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To be concise, I remark to friends and students that e is the most important number in calculus, just as pi is the most important constant in geometry. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Certainly these claims are arguable, but my friends/students don't argue!&lt;/p&gt;&#xA;" OwnerUserId="7850" LastActivityDate="2011-03-09T19:47:30.463" CommentCount="4" />
  <row Id="26048" PostTypeId="2" ParentId="26024" CreationDate="2011-03-09T20:01:59.300" Score="2" ViewCount="" Body="&lt;p&gt;Generally for order relations, the adjective &quot;proper&quot; is usually used to denote a strict ordering, i.e. $\rm\ a \preceq b\ $ &lt;em&gt;properly&lt;/em&gt; means $\rm\ a \preceq b\ $ but not $\rm\ b \preceq a\:.\:$ Thus we have proper divisors, proper subsets, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence $\rm\:a\:$ is a proper divisor of $\rm\:b\:,\:$ or $\rm\ a\ |\ b\ $ properly, $\:$ simply means that $\rm\ a\ |\ b\ $ but not $\rm\ b\ |\ a\:.$&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2011-03-09T20:20:41.283" LastActivityDate="2011-03-09T20:20:41.283" />
  <row Id="26049" PostTypeId="1" CreationDate="2011-03-09T20:07:53.587" Score="2" ViewCount="135" Body="&lt;p&gt;Given that $(\psi_i)_0^{|\chi|-1}$ is an orthonormal basis in $l^2(\pi)$ of real eigenfunctions of $K$ where $K$ is a Markov operator, then why is $\sum_0^{|\chi|-1} \psi_i^2(x)\pi(x) = 1$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we were summing over $x$, then it would make sense since $\psi_i$ is of unit norm.  This time though, we're summing over $i$ so I don't see why the above still holds.  I think it's false, but I want to make sure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And the source of this is here: &lt;a href=&quot;http://projecteuclid.org/DPubS/Repository/1.0/Disseminate?view=body&amp;amp;id=pdf_1&amp;amp;handle=euclid.aoap/1034968224&quot; rel=&quot;nofollow&quot;&gt;http://projecteuclid.org/DPubS/Repository/1.0/Disseminate?view=body&amp;amp;id=pdf_1&amp;amp;handle=euclid.aoap/1034968224&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(At the end of page 702.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm also stuck trying to prove corollary 2.2 in the paper, which is $$\|h_t^x-1\|_2^2 \leq \frac{1}{\pi(x)}e^{-t}+\|k_x^{[t/2]}-1\|_2^2.$$  Help would be much appreciated!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;UPDATE&lt;/em&gt; I GOT IT.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I would still appreciate someone explaining to me why $\left\|H_t\right\|_{2\to \infty}=\sup_x h_{2t}(x,x)&amp;gt;1$ for all finite $t&amp;gt;0$, yet the limit as $t\to \infty$ is $1.$  That's unclear to me.&#xA;Also why is $\left\|H_t\right\|_{2\to q}$ decreasing with respect to $t.$  I'd especially appreciate an intuitive explanation.&lt;/p&gt;&#xA;" OwnerUserId="7329" LastEditorUserId="7329" LastEditorDisplayName="" LastEditDate="2011-03-18T04:25:54.750" LastActivityDate="2011-03-18T04:25:54.750" Title="eigenfunction summation" Tags="&lt;probability&gt;" CommentCount="5" />
  <row Id="26050" PostTypeId="2" ParentId="25986" CreationDate="2011-03-09T20:08:54.447" Score="1" ViewCount="" Body="&lt;p&gt;A short answer would be &lt;strong&gt;Identifying and Explaining Patterns&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="3302" LastActivityDate="2011-03-09T20:08:54.447" CommunityOwnedDate="2011-03-09T20:08:54.447" />
  <row Id="26051" PostTypeId="2" ParentId="25976" CreationDate="2011-03-09T20:12:39.047" Score="1" ViewCount="" Body="&lt;p&gt;I don't know that the lemma, per se, can be used to prove the stronger result, but I suspect that the ideas you used in proving it will also apply.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That said, your question seems essentially to be about the effect of the condition $\frac{w_j}{p_j} &amp;gt; \frac{w_k}{p_k}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose you have some optimal schedule with task $k$ occurring before task $j$.  Then consider the schedule in which you swap $j$ and $k$ in the order in which you do the tasks.  Ignoring what happens with the tasks and times in between $j$ and $k$, the effect of this swap is that task $k$ is completed $p_j$ time units later and task $j$ is completed $p_k$ time units earlier.  If both tasks are tardy this means that the overall tardiness of this new schedule is changed by $w_k p_j - w_j p_k$.  Since $\frac{w_j}{p_j} &amp;gt; \frac{w_k}{p_k}$, this quantity will be negative.  So the new schedule will have smaller overall tardiness, which is a contradiction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This isn't the full argument, since you do have to worry about the intervening tasks and about the cases in which $j$ and $k$ are not tardy, but this is the essence of the argument, and it explains the effect of the condition $\frac{w_j}{p_j} &amp;gt; \frac{w_k}{p_k}$.&lt;/p&gt;&#xA;" OwnerUserId="2370" LastActivityDate="2011-03-09T20:12:39.047" />
  <row Id="26052" PostTypeId="1" AcceptedAnswerId="26370" CreationDate="2011-03-09T20:55:41.690" Score="7" ViewCount="184" Body="&lt;p&gt;In grade school we learn that a square can be &lt;em&gt;equidecomposed&lt;/em&gt; into two congruent right isosceles triangles. Does the following three dimensional generalization hold?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider a &lt;em&gt;trirectangular tetrahedron&lt;/em&gt; with vertices at $(0,0,0)$, $(1,0,0)$, $(0,1,0)$ and $(0,0,1)$ and volume $\frac{1}{6}$. (It has three orthogonal right triangular faces and one equilateral triangular face opposite the (tri-right-angled) common vertex.) The inscribing cube of the tetrahedron has unit side length and unit volume. Is there a dissection of this cube into 6 tetrahedra of the above form? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Provided that such a dissection exists and this is the way it is to be done, it's easy to see how 4 tetrahedra can be glued together to give the boundary of the cube, but I don't quite see how the other two should fit inside.&lt;/em&gt; &lt;/p&gt;&#xA;" OwnerUserId="2720" LastEditorUserId="2720" LastEditorDisplayName="" LastEditDate="2011-03-24T23:36:56.633" LastActivityDate="2011-03-24T23:38:30.970" Title="Equidecomposability of a Cube into 6 Trirectangular Tetrahedra" Tags="&lt;geometry&gt;" AnswerCount="3" CommentCount="4" FavoriteCount="1" />
  <row Id="26053" PostTypeId="1" AcceptedAnswerId="26574" CreationDate="2011-03-09T20:56:42.103" Score="2" ViewCount="180" Body="&lt;p&gt;Suppose I have a random variable X that is known to follow the binomial distribution B, but whose parameter $p$ is unknown.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have observed 100 samples of X, and they all came out true.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I calculate a 95% confidence interval for the value of $p$ (which stands for P(true)) of the underlying distribution B?&lt;/p&gt;&#xA;" OwnerUserId="5910" LastEditorUserId="5910" LastEditorDisplayName="" LastEditDate="2011-03-12T17:52:49.133" LastActivityDate="2011-03-12T21:52:16.737" Title="How to calculate a confidence interval for $p$, the parameter of the binomial distribution?" Tags="&lt;probability&gt;&lt;statistics&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="2" />
  <row Id="26055" PostTypeId="2" ParentId="26042" CreationDate="2011-03-09T21:08:55.300" Score="1" ViewCount="" Body="&lt;p&gt;The first definition is the informal one, but at the same time seems rather convoluted to me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd prefer: X and Y are conditionally independent with respect to a given Z iff&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$P(X \; Y | Z) = P(X | Z ) P(Y | Z)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Recall that conditioning one (or several) variables on the value of another, is (informally) the same as restricting the whole universe to a part of it.&#xA;Then, if you are given the value of $Z$, you can think as if you are defining new variables that are the same as the unconditioned but that are restricted to our new (smaller universe) $X&amp;#39; \equiv X | Z$ $Y&amp;#39; \equiv Y | Z$&#xA;The above formula simply states that $X&amp;#39;$ and $Y&amp;#39;$ are independent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first definition says the same, but applying (in words) the property that two variables are independent iff their conditioned probabilities are the same as the unconditioned : $A$ indep $B$ iff $P(A | B ) = P (A)$&lt;/p&gt;&#xA;" OwnerUserId="312" LastActivityDate="2011-03-09T21:08:55.300" CommentCount="3" />
  <row Id="26056" PostTypeId="2" ParentId="25769" CreationDate="2011-03-09T21:29:44.907" Score="0" ViewCount="" Body="&lt;p&gt;I dont think so. youve got the first instruction called more than once. ive got this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P1: 1 | 4 | 7 | 10&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P2: 2 | 5 | 8 &lt;/p&gt;&#xA;&#xA;&lt;p&gt;P3: 3 | 6 | 9&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Its simple. It only takes four seconds, and all the relationships are satisfied. It is partial order (transitive, reflexive, and antisymmetric) so there is more than one right answer. Also, the 10 could be executed by any of the processors. &lt;/p&gt;&#xA;" OwnerUserId="7988" LastActivityDate="2011-03-09T21:29:44.907" CommentCount="1" />
  <row Id="26057" PostTypeId="1" CreationDate="2011-03-09T21:30:41.007" Score="0" ViewCount="99" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Possible Duplicate:&lt;/strong&gt;&lt;br&gt;&#xA;  &lt;a href=&quot;http://math.stackexchange.com/questions/9188/is-mathbbq-sqrt2-cong-mathbbq-sqrt3&quot;&gt;Is $\mathbb{Q}(\sqrt{2}) \cong \mathbb{Q}(\sqrt{3})$?&lt;/a&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;p&gt;Is it correct that $\mathbb{Q}(\sqrt{2})$ and $\mathbb{Q}(\sqrt{3})$ are isomorphic as fields?&lt;/p&gt;&#xA;" OwnerUserId="8039" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2011-03-09T22:43:42.533" LastActivityDate="2011-03-09T22:43:42.533" ClosedDate="2011-03-09T21:47:28.183" Title="field isomorphism" Tags="&lt;abstract-algebra&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="26058" PostTypeId="1" AcceptedAnswerId="26073" CreationDate="2011-03-09T21:37:39.720" Score="4" ViewCount="104" Body="&lt;p&gt;If $f$ is an $L^2$ function and $\hat{f}$, its Fourier transform, also in $L^2$, can the Fourier transform and its inverse be written as &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\hat{f}(\omega)=\int_{-\infty}^\infty f(x) e^{i\omega x}dx$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f(x)=\int_{-\infty}^\infty \hat{f}(\omega) e^{-i\omega x}d\omega$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;respectively, disregarding the constants? When can a Fourier transform and its inverse exist, but the above definitions not hold?&lt;/p&gt;&#xA;" OwnerUserId="7815" LastActivityDate="2011-03-09T23:25:42.980" Title="Inverse Fourier transform relation for $L^2$ function" Tags="&lt;fourier-analysis&gt;&lt;fourier-series&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="26059" PostTypeId="1" AcceptedAnswerId="26067" CreationDate="2011-03-09T21:39:12.107" Score="3" ViewCount="77" Body="&lt;p&gt;Question: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Find the coordinates of those points on the curve given by the equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ x^2-0.25xy+y^2=16 $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;at which the tangent line has slope 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Implicitly derived:$\ \frac{(-2x+0.25y)}{(2y-0.25x)} $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The book talked something about solving for y (from the original equation): &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ \pm \sqrt{-x^2+0.25xy+16}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;then substituting it back into the implicitly derived equation and solving, but I am not sure that is going to work or else it is going to be very cumbersome. Anyone know how to solve this, perhaps in a simpler (possibly more correct) fashion? Thanks!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S. I spent 30mins figuring out the mathTex stuff so it looks pretty, it was the least I could do I suppose? Only for all you picky math geniuses who are going to help me! :)&lt;/p&gt;&#xA;" OwnerUserId="7144" LastEditorUserId="1543" LastEditorDisplayName="" LastEditDate="2011-04-18T12:05:37.300" LastActivityDate="2011-04-18T12:05:37.300" Title="Implicit Differentiation: Multiple Solutions?" Tags="&lt;calculus&gt;&lt;homework&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="26060" PostTypeId="2" ParentId="26057" CreationDate="2011-03-09T21:39:46.677" Score="2" ViewCount="" Body="&lt;p&gt;No. If $f$ were an isomorphism between $\mathbb{Q}(\sqrt{2})$ and $\mathbb{Q}(\sqrt{3})$ then $f$ is the identity on $\mathbb{Q}$ (it maps 1 to 1, so $\mathbb{Z}$ to itself, etc.). But what is $a = f(\sqrt(2))$? $\sqrt{2}$ satisfies the equation $x^2 - 2 = 0$ in the domain, so applying $f$ to this equation applies that $a$ also satisfies that same equation (same, as the coefficients are fixed under $f$). Now try to show that this cannot be, i.e. no $y$ of the form $c + d\sqrt{3}$ can satisfy this equation.&lt;/p&gt;&#xA;" OwnerUserId="4280" LastActivityDate="2011-03-09T21:39:46.677" />
  <row Id="26061" PostTypeId="1" AcceptedAnswerId="26257" CreationDate="2011-03-09T21:50:54.030" Score="1" ViewCount="136" Body="&lt;p&gt;I am trying to understand the concept of strong Markov property quoted from &lt;a href=&quot;http://en.wikipedia.org/wiki/Markov_property#Strong_Markov_Property&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Suppose that $X=(X_t:t\geq 0)$ is a&#xA;  stochastic process on a probability&#xA;  space&#xA;  $(\Omega,\mathcal{F},\mathbb{P})$ with&#xA;  natural filtration&#xA;  $\{\mathcal{F}\}_{t\geq 0}$. Then $X$&#xA;  is said to have the strong Markov&#xA;  property if, for each stopping time&#xA;  $\tau$, conditioned on the event&#xA;  $\{\tau &amp;lt; \infty\}$, the process&#xA;  $X_{\tau + \cdot}$ (which maybe needs&#xA;  to be defined) is independent from&#xA;  $\mathcal{F}_{\tau}:=\{A \in  \mathcal{F}: \tau \cap A \in  \mathcal{F}_t ,\, \ t \geq 0\}$ and&#xA;  $X_{\tau + t} − X_{\tau}$ has the same&#xA;  distribution as $X_t$ for each $t \geq  0$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here are some questions that make me stuck:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;In $\mathcal{F}_{\tau}:=\{A \in&#xD;&#xA;    \mathcal{F}: \tau \cap A \in&#xD;&#xA;    \mathcal{F}_t ,\, \ t \geq 0\} $,&#xA;what does $\tau \cap A $ mean?&#xA;$\tau$ is a stopping time and&#xA;therefore a random variable and $A$&#xA;is a $\mathcal{F}$-measurable&#xA;subset, but what does $\tau \cap A$&#xA;mean?&lt;/li&gt;&#xA;&lt;li&gt;How is the process $X_{\tau + \cdot}$ defined from the process $X_{\cdot}$ ? Is it the translated&#xA;version of the latter by $\tau$?&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;How is the conditional independence&#xA;between a process, such as $X_{\tau&#xD;&#xA;    + \cdot}$,  and the sigma algebra, such as $\mathcal{F}_{\tau}$, given&#xA;an event, such as $\{\tau &amp;lt;&#xD;&#xA;    \infty\}$, defined? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Related question, is independence&#xA;between a random variable and a&#xA;sigma algebra defined as&#xA;independence between the sigma&#xA;algebra of the random variable and&#xA;the sigma algebra?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;Is &quot;$X_{\tau+ t}  − X_{\tau}$ has the&#xA;same distribution as $X_t$ for each&#xA;$t \geq 0$&quot; also conditional on the&#xA;event $\{\tau &amp;lt; \infty\}$?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Thanks and regards!&lt;/p&gt;&#xA;" OwnerUserId="1281" LastEditorUserId="1281" LastEditorDisplayName="" LastEditDate="2011-03-09T21:57:25.493" LastActivityDate="2011-03-10T20:46:55.550" Title="Questions about the concept of strong Markov property" Tags="&lt;probability-theory&gt;&lt;stochastic-processes&gt;" AnswerCount="2" CommentCount="8" />
  <row Id="26062" PostTypeId="1" AcceptedAnswerId="26087" CreationDate="2011-03-09T21:54:10.247" Score="5" ViewCount="95" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm new to all this large cardinal thing and I have trouble in proving the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $\kappa$ is a $\gamma$-strong cardinal, for some large enough $\gamma$, then $\kappa$ is $\Sigma_2$-reflecting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any ideas or hints will be welcome.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance&lt;/p&gt;&#xA;" OwnerUserId="2926" LastActivityDate="2011-03-10T00:38:02.793" Title="Strong cardinals and reflection" Tags="&lt;set-theory&gt;&lt;large-cardinals&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="26063" PostTypeId="2" ParentId="26052" CreationDate="2011-03-09T22:00:12.823" Score="1" ViewCount="" Body="&lt;p&gt;I'm pretty sure there isn't.  The most straightforward way of proving this is probably to (a) show that any embedding of the trirectangular tetrahedron in the cube is 'extremal' (that is, that its right-triangle faces are of necessity aligned with the faces of the cube) and then (b) show that any possible extremal embedding of a trirectangular tetrahedron misses some point in the cube.  (a) is a little bit tricky, but once you've got that, (b) is easy (what point is most likely to be missed?).&lt;/p&gt;&#xA;" OwnerUserId="785" LastActivityDate="2011-03-09T22:00:12.823" />
  <row Id="26064" PostTypeId="2" ParentId="26052" CreationDate="2011-03-09T22:02:38.610" Score="0" ViewCount="" Body="&lt;p&gt;Here is what I've discovered so far. Four trirectangular tetrahedra glued into a cubical shape leaves a regular tetrahedral void (a 3-simplex actually) within the cube. There is no way to get two more trirectangular tetrahedra into that space without further dissection (a proof follows by considering angles).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This, however, doesn't settle the question because I've looked at only one gluing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The slickest proof of the above fact is that the &lt;em&gt;Dehn invariant&lt;/em&gt; of the trirectangular tetrahedron is non-zero, while that of the cube is zero. Since the Dehn invariant is additive in equidecomposition, the result follows.&lt;/p&gt;&#xA;" OwnerUserId="2720" LastEditorUserId="2720" LastEditorDisplayName="" LastEditDate="2011-03-24T23:38:30.970" LastActivityDate="2011-03-24T23:38:30.970" CommentCount="2" />
  <row Id="26065" PostTypeId="2" ParentId="26059" CreationDate="2011-03-09T22:14:35.880" Score="2" ViewCount="" Body="&lt;p&gt;You have already implicitly derived the expression for $\frac{dy}{dx}$. You can set that equal to 1 and simplify to get $x = -y$. Substitute that in the original equation and you have a quadratic that can be easily solved. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I got the solutions $x = \pm 8/3$ and $y = \mp 8/3$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The book seems to be making a big deal out of a very simple question.&lt;/p&gt;&#xA;" OwnerUserId="2641" LastActivityDate="2011-03-09T22:14:35.880" />
  <row Id="26066" PostTypeId="2" ParentId="25780" CreationDate="2011-03-09T22:20:41.667" Score="0" ViewCount="" Body="&lt;p&gt;Suppose that $\rm\:p(x)\in \mathbb Z[x]\:.\ $ By the root-factor theorem we have $\rm\ p(x)\ \equiv\ p(2)\ \ (mod\ x-2)\ $ Therefore $\rm\ x-2\ |\ p(x)\ \Rightarrow\ x-2\ |\ p(2)\ \Rightarrow\ |x-2| \le |p(2)|\:.\ $ This inequality constrains the possible solutions to a finite set which can be explicitly tested. Now put $\rm\ p(x) = x^2 -x - 17\:.$&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2011-03-09T22:26:17.840" LastActivityDate="2011-03-09T22:26:17.840" />
  <row Id="26067" PostTypeId="2" ParentId="26059" CreationDate="2011-03-09T22:24:59.600" Score="2" ViewCount="" Body="&lt;p&gt;Since it's an equation of degree 2, the curve in question is a &lt;a href=&quot;http://en.wikipedia.org/wiki/Conic_section&quot; rel=&quot;nofollow&quot;&gt;conic section&lt;/a&gt;. Looking at the coefficients, it's probably an ellipse. (The criterion for that is that the quadratic form should always be positive.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The equation is symmetric in $x$ and $y$. In other words, a reflection at the line $y=x$ will map the ellipse to itself. This also means that the line $y=x$ is also a main axis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/41Klg.png&quot; alt=&quot;Ellipse has main axis $y=x$&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Coincidentally&lt;/em&gt;, the question also asks for a tangent with &lt;em&gt;exactly the same slope&lt;/em&gt; as the main axis. This can only happen at the points where the other main axis intersects the ellipse. In our case, the other axis is the line $y=-x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, the points in question have the form $(x,-x)$ and can be obtained by solving the equation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ x^2 + 1/4&amp;#183;x^2 + x^2 = 16 $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i.e. $x = \pm\sqrt{64/9} = \pm 8/3$.&lt;/p&gt;&#xA;" OwnerUserId="434" LastActivityDate="2011-03-09T22:24:59.600" CommentCount="2" />
  <row Id="26068" PostTypeId="2" ParentId="25948" CreationDate="2011-03-09T22:26:59.180" Score="2" ViewCount="" Body="&lt;p&gt;Yes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $\pi \in S$ be a permutation; then write $\pi = S_1 \cdots S_k$ where $S_1, \ldots, S_k \in S$ are disjoint cycles. Define $n_i = \#S_i$ for $i \in \{1,\ldots,k\}$. Now $\pi^{n_i}$ fixes the elements of $S_i$. So the smallest number which fixes all elements of $\pi$ is $\#\pi = \mathcal{lcm} \{n_1,\ldots,n_k\}$.&lt;/p&gt;&#xA;" OwnerUserId="8042" LastEditorUserId="8042" LastEditorDisplayName="" LastEditDate="2011-03-09T22:46:42.673" LastActivityDate="2011-03-09T22:46:42.673" CommentCount="2" />
  <row Id="26069" PostTypeId="2" ParentId="26058" CreationDate="2011-03-09T22:57:05.813" Score="1" ViewCount="" Body="&lt;p&gt;There are many cases when the Fourier transform can be defined and the function is not $L^2$. Take  a look at &lt;a href=&quot;http://en.wikipedia.org/wiki/Distribution_%28mathematics%29#Tempered_distributions_and_Fourier_transform&quot; rel=&quot;nofollow&quot;&gt;tempered distributions&lt;/a&gt; for a class of &quot;functions&quot; for which the Fourier transform and its inverse can be defined. The most prominent example is the delta function (distribution). &lt;/p&gt;&#xA;" OwnerUserId="7266" LastEditorUserId="7266" LastEditorDisplayName="" LastEditDate="2011-03-09T23:08:30.313" LastActivityDate="2011-03-09T23:08:30.313" />
  <row Id="26070" PostTypeId="2" ParentId="8978" CreationDate="2011-03-09T22:58:27.417" Score="1" ViewCount="" Body="&lt;p&gt;'Combinatorial' usually refers to finite or discrete objects so using an integral really is going quite the other direction to a analytic are, so maybe one would want just an (unqualified) interpretation of what it means when one takes a gf (of an already combinatorial situation) and replaces the summation symbol with an integral to get an analytic situation (which accords more with an analytic continuation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the question really might be 'What is the -analytic' significance of the integral?' or possibly 'What is an analytic analog of a generating function?'&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Neither is obvious, and the integral you give seems to be unknown (even if $a_n = 1$). so I'll try to say something about gfs that might help.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A gf is a way to capture a function on the naturals by other means; the natural number exponents of the gf mark the values of the original function. So by trying to use an integral, you're trying to interpret the exponent as a real (rather than an integer). The furthest I've seen gfs generalized is to Puiseux series which allow rational exponents (but with some restrictions). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only appropriate 'conversion' I can think of is something like a Fourier or Laplace transform; those will convert a function to another 'domain' where manipulation can occur, and then convert back to the functional domain.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Edited out (the following is not what the OP is looking for):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The coefficient of $x^n/n!$  in an egf is usually combinatorially interpreted as the number of labeled sets of size $n$ (with a given structure).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A more likely integral related to an egf is to integrate (formally, forget the constant) the entire egf with respect to the egf variable, which has an immediate calculation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \int E(z) \ dz = \sum_{n=0}^{\infty} \int\frac{a_{n}z^{n}}{n!}\ dz = &#xD;&#xA;\sum_{n=1}^{\infty} \frac{a_{n}z^{n+1}}{(n+1)!} = \sum_{n=1}^{\infty} \frac{a_{n-1}z^n}{n!}$$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All this does is in a sense shift the function: $\int E(z)$ is the egf of $a_{n-1}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, the derivative of the egf shifts the other direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For ordinary generating functions (ogf), the derivative usually means combinatorially that you are 'pointing to' (distinguishing, choosing) one particular element out of a structure of size $n$ (and the integral 'undoes' the pointing). For egfs, the labeling provides a sort of 'pointing' already for every object, so the $n$ multiplier is irrelevant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I realize I have answered a different question than you have asked, in the hopes that it is what you are really after. You are asking, given an egf for a function over naturals, to presume the existence of an analytic continuation of the original function, and then integrate over that analytic continuation (along with $z^n/n!$ also interpreted over reals). For the moment, I really can't say, there than it looks like some fractional derivative.&lt;/p&gt;&#xA;" OwnerUserId="1919" LastEditorUserId="1919" LastEditorDisplayName="" LastEditDate="2011-03-10T21:34:06.163" LastActivityDate="2011-03-10T21:34:06.163" CommentCount="1" />
  <row Id="26071" PostTypeId="1" AcceptedAnswerId="26082" CreationDate="2011-03-09T22:59:44.113" Score="2" ViewCount="147" Body="&lt;p&gt;Let $X$, $Y$ be Banach spaces. Let $T:X\rightarrow Y$ be a bounded linear operator.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Under what circumstances is the image of $T$ closed in $Y$ (except finite-dimensional image).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, I wonder, under which assumptions $T:X\rightarrow T(X)$ is a bounded linear bijection between Banach spaces, so it is at least an isomorphism onto its image by bounded inverse theorem.&lt;/p&gt;&#xA;" OwnerUserId="3557" LastActivityDate="2011-03-10T00:24:13.217" Title="When is the image of a linear operator closed?" Tags="&lt;functional-analysis&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="26072" PostTypeId="1" AcceptedAnswerId="26075" CreationDate="2011-03-09T23:24:08.593" Score="1" ViewCount="56" Body="&lt;p&gt;Let $k&amp;#39;$ be an field extension of an infinite field $k$. If $\{v_i\}$ is a set of vectors over $k$, and they satisfy $\sum_{i=1}^{n} c_iv_i =0$ for some nontrivial $\{c_i\} \subset k&amp;#39;$, then is the same set of vectors $\{v_i\}$ linearly dependent over $k$? In other words, can the $c_i$'s be chosen in $k$ only? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I see that if $k&amp;#39;$ is a finite extension of $k$, then I can quickly prove why this fact is true. But for infinite extensions I'm less sure. Maybe I'm missing something very obvious? Thanks in advance.&lt;/p&gt;&#xA;" OwnerUserId="6369" LastActivityDate="2011-03-09T23:52:44.047" Title="When can linear independence of vectors over a field be 'carried over' to linear independence its subfield?" Tags="&lt;linear-algebra&gt;&lt;abstract-algebra&gt;" AnswerCount="1" />
  <row Id="26073" PostTypeId="2" ParentId="26058" CreationDate="2011-03-09T23:25:42.980" Score="3" ViewCount="" Body="&lt;p&gt;Jonas T already hit on the important point that for the first integral to be defined, $f$ must be in $L^1$. Similarly, for the second integral to be defined, $\hat{f}$ must be in $L^1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fabian has mentioned that the Fourier tranform &lt;a href=&quot;http://eom.springer.de/f/f041170.htm&quot; rel=&quot;nofollow&quot;&gt;can be generalized&lt;/a&gt; even beyond $L^2$.  But if you're only concerned with $L^2$, then one way to define the Fourier transform is to first show that it defines an isometric (with respect to the $L^2$ norm) isomorphism on the &lt;a href=&quot;http://en.wikipedia.org/wiki/Schwartz_space&quot; rel=&quot;nofollow&quot;&gt;Schwartz space&lt;/a&gt;, which is dense in $L^2$, and then take the unique extension to all of $L^2$.  The fact that the Fourier transform defines an isometry with dense range on a dense subspace of $L^2$, and thus has a unique extension to a unitary operator on $L^2$, is known as &lt;a href=&quot;http://en.wikipedia.org/wiki/Plancherel%27s_theorem&quot; rel=&quot;nofollow&quot;&gt;Plancherel's theorem&lt;/a&gt;.  I have only mentioned one approach, which is the one I learned from Chapter 10 of J.B. Conway's &lt;em&gt;&lt;a href=&quot;http://books.google.com/books?id=ix4P1e6AkeIC&amp;amp;dq=a+course+in+functional+conway&amp;amp;source=gbs_navlinks_s&quot; rel=&quot;nofollow&quot;&gt;A course in functional analysis&lt;/a&gt;&lt;/em&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the answer to your final question, at least restricted to $L^2$, is that every element of $L^2\setminus L^1$ provides an example, strictly speaking.  However, you can show that $\frac{1}{\sqrt{2\pi}}\int_{-R}^R f(x)e^{-i\omega x}dx$ converges to $\hat{f}(\omega)$ in $L^2$ norm as $R\to\infty$, and similarly $\frac{1}{\sqrt{2\pi}}\int_{-R}^R \hat{f}(\omega)e^{i\omega x}dx$ converges to $f(x)$ in $L^2$ norm as $R\to\infty$.  I don't have a reference to a proof handy, but this and more is summarized in the Springer online encyclopedia's &lt;a href=&quot;http://eom.springer.de/f/f041150.htm&quot; rel=&quot;nofollow&quot;&gt;article on the Fourier transform&lt;/a&gt;, which includes helpful references and links.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2011-03-09T23:25:42.980" CommentCount="4" />
  <row Id="26074" PostTypeId="1" AcceptedAnswerId="26079" CreationDate="2011-03-09T23:26:07.563" Score="0" ViewCount="102" Body="&lt;p&gt;Okay, so this was labeled as a &quot;fun problem&quot;, but I'm having trouble knowing how to approach it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm given: $\lim\limits_{x \to 0^+} f(x) = A$ and $\lim\limits_{x \to 0^-} f(x) = B$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need to find (or at least know where to start):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a) $\lim\limits_{x \to 0^+} f(x^3 - x)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;b) $\lim\limits_{x \to 0^-} f(x^3 - x)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any insight on how to approach this (or even a solution) would be greatly appreciated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="8044" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-10T02:03:44.350" LastActivityDate="2011-03-10T02:04:13.083" Title="Calculus Limit Problem" Tags="&lt;calculus&gt;" AnswerCount="3" CommentCount="3" />
  <row Id="26075" PostTypeId="2" ParentId="26072" CreationDate="2011-03-09T23:27:02.043" Score="5" ViewCount="" Body="&lt;p&gt;Any non trivial linear combination with coefficients from $k&amp;#39;$ will involve only finitely many elements of $k&amp;#39;$. Consider the subfield $k&amp;#39;&amp;#39;$ of $k&amp;#39;$ generated by $k$ and by these coefficients. Then $k&amp;#39;&amp;#39;$ is a finite extension of $k$, the $v_i$'s satisfy a linear relation with coefficients in $k&amp;#39;&amp;#39;$, and you claim to be able to do what you want in that case. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;N.B.:&lt;/strong&gt; In any case, your question, as stated, does not make a lot of sense really: your $v_i$'s are presumably elements of a $k$-vector space, so what sense does it make to compute a linear combination with them using coefficients from $k&amp;#39;$? What I wrote above answers a version of what you meant: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;suppose that $V$ is a $k$-vector space, that $(v_i)_{i\in I}$ is a family of elements of $V$, and that the family  $(1\otimes v_i)_i$ of elements of the $k&amp;#39;$-vector space $k&amp;#39;\otimes_k V$ is linearly dependent over $k&amp;#39;$: then $(v_i)_{i\in I}$ is linearly dependent in $V$ over $k$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There is no need, though, to proceed by reducing first to a finite subextension to do this. An alternative is the following: suppose that we have a non-trivial zero linear combination $\sum_i a_i\otimes v_i=0$ of the family $(1\otimes v_i)_{i\in I}$ with coefficients $(a_i)_{i\in I}$ in $k&amp;#39;$. Let $(z_j)_{j\in J}$ be a basis of $k&amp;#39;$ as a $k$-vector space, and suppose for all $i\in I$ we have $a_i=\sum_{j\in J}a_{i,j}z_j$, with all the $a_{i,j}\in k$. Then $$0 = \sum_{i\in I} a_i\otimes v_i = \sum_{j\in J}z_j\otimes\Bigl(\sum_{i\in I}a_{i,j}v_i\Bigr).$$ As $(z_j)_{j\in J}$ is linearly independent, this means that for all $j\in J$ we have $$\sum_{i\in I}a_{i,j}v_i=0,$$ and since we have assumed one of the $a_i$s is not zero, one of these linear combinations, which have coefficients in $k$, is itself non-trivial.&lt;/p&gt;&#xA;" OwnerUserId="274" LastEditorUserId="274" LastEditorDisplayName="" LastEditDate="2011-03-09T23:52:44.047" LastActivityDate="2011-03-09T23:52:44.047" CommentCount="1" />
  <row Id="26076" PostTypeId="2" ParentId="26074" CreationDate="2011-03-09T23:33:14.693" Score="0" ViewCount="" Body="&lt;p&gt;You know what $f$ does when its input is negative and going to zero and when its input is positive and going to zero.  When $x$ is negative and going to zero, then $x^3-x$ is also going to zero, but because $x^2&amp;lt;1$ when $|x|&amp;lt;1$, $x^3-x=x(x^2-1)$ will be positive.  Similarly, when $x$ is positive and going to zero, $x^3-x$ will be negative and going to zero.  This should allow you to determine the one-sided limits from the information given.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2011-03-09T23:33:14.693" />
  <row Id="26077" PostTypeId="1" CreationDate="2011-03-09T23:33:21.597" Score="0" ViewCount="100" Body="&lt;p&gt;I'm looking at a textbook problem, which has the answer as well, but don't understand what's going on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem: A part with lifetime X has a normal distribution with mean of 7 years and standard deviation of 1.2 years.&#xA;If the warranty is for 5 years, what % will fail inside the warranty?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer is: P(X&amp;lt;=5) = P(Z&amp;lt;=(5-7)/1.2) = P(Z&amp;lt;=-1.67)&#xA;= .5-.4525 = 0.0475&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I follow everything until the .5-.4525 part. Where do these numbers come from?&lt;/p&gt;&#xA;" OwnerUserId="8045" LastActivityDate="2011-04-28T07:43:02.197" Title="Normal Distribution" Tags="&lt;homework&gt;&lt;probability&gt;&lt;probability-distributions&gt;&lt;normal-distribution&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="26078" PostTypeId="2" ParentId="26074" CreationDate="2011-03-09T23:34:26.957" Score="0" ViewCount="" Body="&lt;p&gt;When $x$ goes to zero, $x^3-x$ also goes to zero. But we can be more precise: in fact,&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;when $x$ goes to zero from the left, $x^3-x$ goes to zero from the right, and&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;when $x$ goe to zero from the right, $x^3-x$ goes to zero from the left.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Can you see how to use this to solve the problem? &lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2011-03-09T23:34:26.957" />
  <row Id="26079" PostTypeId="2" ParentId="26074" CreationDate="2011-03-09T23:36:36.020" Score="1" ViewCount="" Body="&lt;p&gt;Let $z(x)=x^3-x$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have&#xA;$$\lim_{x\rightarrow 0^- }f(z(x)) =\lim_{z\rightarrow 0^+ }f(z) $$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;i.e $z$ approaches from $0+\delta$ when x approaches from $0-\delta$ for $\delta&amp;gt;0$ and vice versa for the other limit. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So whatever limits you have for x in will be inverted in z, i.e, answer for a) is B and b) is A &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To get a visual hint, &lt;a href=&quot;http://www.wolframalpha.com/input/?i=x%5E3+-x+plot&quot; rel=&quot;nofollow&quot;&gt;plot&lt;/a&gt; $x^3-x$&lt;/p&gt;&#xA;" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-10T02:04:13.083" LastActivityDate="2011-03-10T02:04:13.083" />
  <row Id="26080" PostTypeId="1" CreationDate="2011-03-09T23:41:19.500" Score="7" ViewCount="265" Body="&lt;p&gt;Why do all school algebra texts define simplest form for expressions with radicals to not allow a radical in the denominator. For the classic example, $1/\sqrt{3}$ needs to be &quot;simplified&quot; to $\sqrt{3}/3$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a mathematical or other reason?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And does the same apply to exponential notation -- are students expected to &quot;simplify&quot; $3^{-1/2}$ to $3^{1/2}/3$ ?&lt;/p&gt;&#xA;" OwnerUserId="271" LastActivityDate="2011-03-10T01:18:46.457" Title="No radical in the denominator -- why?" Tags="&lt;algebra&gt;&lt;roots&gt;" AnswerCount="4" CommentCount="4" FavoriteCount="1" />
  <row Id="26081" PostTypeId="2" ParentId="26061" CreationDate="2011-03-09T23:44:26.597" Score="2" ViewCount="" Body="&lt;p&gt;For the first one the notation is wrong it should I think $A\cap\{\tau\le t\}\in\mathcal{F}_t$ instead of $A\cap\tau$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And for the fourth point, look at George Lowther comments below, that fully address the problematic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regard&lt;/p&gt;&#xA;" OwnerUserId="4437" LastEditorUserId="4437" LastEditorDisplayName="" LastEditDate="2011-03-10T10:55:09.077" LastActivityDate="2011-03-10T10:55:09.077" CommentCount="4" />
  <row Id="26082" PostTypeId="2" ParentId="26071" CreationDate="2011-03-09T23:45:33.657" Score="2" ViewCount="" Body="&lt;p&gt;An answer to your last question is that a bounded linear map $T$ between Banach spaces is injective with closed range if and only if it is bounded below, meaning that there is a constant $c&amp;gt;0$ such that for all $x$ in the domain, $\|Tx\|\geq c\|x\|$.  You can read more about this in Chapter 2 of &lt;em&gt;&lt;a href=&quot;http://books.google.com/books?id=5h8bEWNalwMC&amp;amp;printsec=frontcover&amp;amp;source=gbs_atb#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;An invitation to operator theory&lt;/a&gt;&lt;/em&gt; by Abramovich and Aliprantis.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2011-03-10T00:24:13.217" LastActivityDate="2011-03-10T00:24:13.217" />
  <row Id="26083" PostTypeId="1" AcceptedAnswerId="26113" CreationDate="2011-03-09T23:53:27.907" Score="1" ViewCount="92" Body="&lt;p&gt;I recently learned about spectral sequences (which I am still trying to understand).  In class we defined a (co)homologically graded exact couple (of $A_p^n$s and $C_p^n$'s with vertical arrows in the $A$ columns) to be a certain grid of exact sequences, which given some convergence assumptions, allows us to calculate the direct limit over $p$ of the $A_p^n$ in terms of $E_\infty (C_p^n)$ (this is a pretty rough sketch).  I know inverse limits in general behave worse than direct limits, but in some nice cases, is there something like spectral sequences to compute inverse limits instead?  There are lots of interesting things in number theory that are inverse limits is why I am wondering about this.&lt;/p&gt;&#xA;" OwnerUserId="5265" LastActivityDate="2011-03-10T02:55:21.257" Title="Spectral Sequences and Inverse Limits" Tags="&lt;algebraic-topology&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="26084" PostTypeId="2" ParentId="26080" CreationDate="2011-03-09T23:57:23.537" Score="6" ViewCount="" Body="&lt;p&gt;I heard from a high school math teacher once that this was done back in the day (before calculators) so that people could look up the values of these expressions in tables.  If this is indeed the case, there is no good reason for this practice now.&lt;/p&gt;&#xA;" OwnerUserId="5265" LastActivityDate="2011-03-09T23:57:23.537" CommentCount="2" />
  <row Id="26085" PostTypeId="2" ParentId="26080" CreationDate="2011-03-10T00:06:25.183" Score="6" ViewCount="" Body="&lt;p&gt;I have had students seem surprised when I told them that I have no preference for $\tan\left(\frac{\pi}{6}\right)$ to be &quot;simplified&quot; to $\frac{\sqrt 3}{3}$, for example.  A high school teacher whom I told this was even more surprised, but no reason was given in defense of requiring the practice.  Nonetheless, to some extent I am glad that high school teachers often tell students that they have to rationalize denominators, because it means that many of them will at least be familiar with the idea when the time comes that &quot;rationalizing&quot; (both denominators and numerators) &lt;em&gt;does&lt;/em&gt; serve a purpose in simplifying algebraic expressions, in particular when finding limits in calculus.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So maybe $\frac{1}{4+\sqrt 3}$ is no less simple than $\frac{4-\sqrt 3}{13}$, but $\frac{h}{\sqrt{x+h}-\sqrt{x}}$ is harder to deal with than $\sqrt{x+h}+\sqrt{x}$ when $h$ goes to $0$, and this comes more easily if you already know how to rationalize denominators with numbers.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2011-03-10T00:12:27.033" LastActivityDate="2011-03-10T00:12:27.033" CommentCount="2" />
  <row Id="26086" PostTypeId="2" ParentId="26080" CreationDate="2011-03-10T00:07:35.360" Score="5" ViewCount="" Body="&lt;p&gt;The usual reason I've heard is that dividing by integers is computationally easier -- it's easier to find, say, $(5\sqrt{3})/3$ by computing $5 \times \sqrt{3} \approx 8.66 $ and then dividing by $3$ to get $2.89$ then to find $5/\sqrt{3}$ by dividing $5/1.73$ directly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The slightly cynical teacher in me wants to say that the reason for demanding no radicals in the denominator is so there is only one right answer to each question, which simplifies grading.&lt;/p&gt;&#xA;" OwnerUserId="173" LastActivityDate="2011-03-10T00:07:35.360" CommentCount="3" />
  <row Id="26087" PostTypeId="2" ParentId="26062" CreationDate="2011-03-10T00:08:57.057" Score="5" ViewCount="" Body="&lt;p&gt;A cardinal $\kappa$ is $\Sigma_2$ reflecting if whenever a sentence $\varphi$ is true in some $V_\alpha$, then there is an $\alpha\lt\kappa$ such that $V_\alpha\models\varphi$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(This can be seen to be equivalent to the &quot;reflecting&quot; version of $\Sigma_2$-reflecting, since every $\Sigma_2$ statement $\psi$ is equivalent to a statement of the form &quot;$\exists\alpha V_\alpha\models\varphi$&quot;, where $\varphi$ is some sentence (with no bound on the complexity of $\varphi$). But let us leave this aside.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if $\kappa$ is a strong cardinal, and there is some $\alpha$ above $\kappa$ such that $V_\alpha\models\varphi$ for some statement $\varphi$, then fix an $\alpha$-strongness embedding $j:V\to M$ with critical point $\kappa$. Since $V_\alpha^M=V_\alpha$, the model $M$ thinks there is $\alpha\lt j(\kappa)$ for which $V_\alpha\models\varphi$. Thus, by elementarity, there is $\alpha\lt\kappa$ in $V$ for which $V_\alpha\models \varphi$, thereby verifying that $\kappa$ is $\Sigma_2$-reflecting, as desired.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, to answer your question, pick any $\gamma\gt\kappa$ such that all $\Sigma_2$ statements reflect from $V$ down to $V_\gamma$. Such a cardinal $\gamma$ exists by the Reflection theorem. Now, I claim that any $\gamma$-strong cardinal $\kappa$ is $\Sigma_2$ reflecting, since any statement $\varphi$ true in some $V_\alpha$ will be true in some $V_\alpha$ with $\alpha\lt\gamma$ by the choice of $\gamma$, and thus $\varphi$ will be true in $V_\alpha$ for some $\alpha\lt\kappa$ by the argument of the previous paragraph.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, it is easy to see that if $\kappa$ is $\gamma$-strong for this $\gamma$, then in fact it is fully strong, since any violation of strongness of $\kappa$ above $\gamma$ would be reflected below $\gamma$.&lt;/p&gt;&#xA;" OwnerUserId="413" LastEditorUserId="413" LastEditorDisplayName="" LastEditDate="2011-03-10T00:38:02.793" LastActivityDate="2011-03-10T00:38:02.793" CommentCount="4" />
  <row Id="26088" PostTypeId="1" AcceptedAnswerId="26093" CreationDate="2011-03-10T00:10:33.077" Score="0" ViewCount="34" Body="&lt;p&gt;A set $\Gamma$ of probability measures on $X$ is said to be &lt;em&gt;tight&lt;/em&gt; if for every $\epsilon &amp;gt; 0$ there exists a compact set $K_\epsilon$ of $X$ such that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mu(K) \geq 1 - \epsilon \text{ for all $\mu \in \Gamma$}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fine, now let $\mathcal M$ be the set of measurable mappings on the probability space $(\Omega, \mathcal F, P)$ if there exists $M &amp;gt; 0$ such that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_\Omega |f(\omega)| \, dP(\omega) \leq M \text{ for all $f \in \mathcal M$}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do I now show that the set $\{f_{\#}P : f \in \mathcal M\}$ is tight?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please only give me a hint as this is &quot;homework&quot;. What I tried to do was assume it is not tight and then try to construct a compact set that gives me a contradiction, but that seems to lead nowhere.&lt;/p&gt;&#xA;" OwnerUserId="1120" LastActivityDate="2011-03-10T00:55:48.787" Title="Sufficient condition for a set of measures to be tight." Tags="&lt;homework&gt;&lt;measure-theory&gt;" AnswerCount="1" />
  <row Id="26089" PostTypeId="1" AcceptedAnswerId="26101" CreationDate="2011-03-10T00:27:47.340" Score="8" ViewCount="302" Body="&lt;p&gt;I have two right triangles One is a 6-8-10 and inside is a 3-4-5 and the space between the two triangles is a uniform amount.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I made a really awkward and weird pic of the diagram and I need to solve for X&#xA;&lt;img src=&quot;http://i.imgur.com/wHO71.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How would I go about solving this? I couldn't figure out any good approaches.&lt;/p&gt;&#xA;" OwnerUserId="5063" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-10T05:39:24.963" LastActivityDate="2011-03-10T05:39:24.963" Title="Finding the distance between two triangles, one inside the other" Tags="&lt;geometry&gt;" AnswerCount="4" CommentCount="2" />
  <row Id="26091" PostTypeId="2" ParentId="26089" CreationDate="2011-03-10T00:38:50.297" Score="1" ViewCount="" Body="&lt;p&gt;One way to solve this would be to set up a coordinate system with the bottom left corner of the outer triangle at $(0,0)$. The bottom left corner of the inner triangle would then be at $(t,t)$ for some positive real $t$. ($t$ is what you called $x$, but that's confusing in the coordinate context.) Then you just need to set $t$ so that the distance between the two hypotenuses are equal; see the &lt;a href=&quot;http://en.wikipedia.org/wiki/Parallel_(geometry)&quot; rel=&quot;nofollow&quot;&gt;Wikipedia article on parallel lines&lt;/a&gt; for the relevant formulas. (They're not too hard to derive on your own.)&lt;/p&gt;&#xA;" OwnerUserId="173" LastEditorUserId="173" LastEditorDisplayName="" LastEditDate="2011-03-10T01:43:18.557" LastActivityDate="2011-03-10T01:43:18.557" CommentCount="3" />
  <row Id="26092" PostTypeId="1" AcceptedAnswerId="26099" CreationDate="2011-03-10T00:48:23.163" Score="1" ViewCount="51" Body="&lt;p&gt;Suppose you have a triangle with vertices $(1,0),(0,1), (0,0)$, and let $P$ be the number of all integer coordinates in the triangle. So $P=3$. Then the Minkowski sum $P+P$ is the set of points found by adding all possible pairs of integer coordinates in the triangle. So $P+P$ consists of points in the original triangle, as well as points $(2,0), (0,2), (1,1)$, and thus $P+P=6$. It was mentioned to me that the generating function for this sequence of sums is given by &#xA;$$&#xD;&#xA;p(x)=\frac{1}{(1-x)^3}&#xD;&#xA;$$&#xA;by I don't see why. What is the derivation of this? Thanks.&lt;/p&gt;&#xA;" OwnerUserId="6742" LastActivityDate="2011-03-10T01:33:51.237" Title="Generating function for Minkowski Sum." Tags="&lt;generating-functions&gt;" AnswerCount="2" />
  <row Id="26093" PostTypeId="2" ParentId="26088" CreationDate="2011-03-10T00:55:48.787" Score="2" ViewCount="" Body="&lt;p&gt;Try to use Markov's inequality; $P(|X|\geq a)\leq E(|X|)/a.$ &lt;/p&gt;&#xA;" OwnerUserId="940" LastActivityDate="2011-03-10T00:55:48.787" CommentCount="1" />
  <row Id="26095" PostTypeId="1" AcceptedAnswerId="26103" CreationDate="2011-03-10T01:08:10.070" Score="1" ViewCount="125" Body="&lt;p&gt;Let me start by saying I graduated High School in 2002 and have worked as a technician, a systems administrator, and a part time web developer up till now.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have started taking courses at community college for my general requirements, and I am planning which major I should pursue.  I have always had a passion for mathematics since a young age, but never pursued it, and currently I would say my level of math is at a algebra level.  My understanding for computers and programming is much greater than mathematics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Should I consider taking Math as my major or go with Computer Science?  I do not want to do a double major, since I have no real plans for staying in College for that long.  My goal is to receive a degree, gain educational experience, and take my career to the next level.  I plan on working as a programmer professionally.&lt;/p&gt;&#xA;" OwnerUserId="7810" LastActivityDate="2011-03-10T01:31:09.287" CommunityOwnedDate="2011-03-10T01:51:31.610" Title="Considering math or computer science" Tags="&lt;soft-question&gt;&lt;computer-science&gt;&lt;education&gt;" AnswerCount="1" CommentCount="0" />
  <row Id="26097" PostTypeId="2" ParentId="26080" CreationDate="2011-03-10T01:18:46.457" Score="5" ViewCount="" Body="&lt;p&gt;This transformation is known as $\:$ &lt;em&gt;rationalizing the denominator&lt;/em&gt;.$\ \:$ As the terminology suggests, $\ $it serves to simplify by transforming an irrational divisor into a rational divisor. This can lead to all sorts of simplifications. Below are a couple examples.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this &lt;a href=&quot;http://math.stackexchange.com/questions/23909/help-solving-a-limit/23917#23917&quot;&gt;prior question&lt;/a&gt; is an example where RTD transforms a limit of indeterminate form into a simple determinate limit by way of cancelling apparent singularity at $\rm\ x = a\ $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\rm \frac{x^2-a\sqrt{ax}}{\sqrt{ax}-a}\ =\ \frac{x^2-a\sqrt{ax}}{\sqrt{ax}-a} \ \frac{\sqrt{ax}+a}{\sqrt{ax}+a}\ =\ \frac{ax\:(x-a)+\sqrt{ax}\ (x^2-a^2) }{a\:(x-a) }\ =\ x+(x+a)\sqrt{\frac{x}{a}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's another example from number theory showing how RTD serves to reduce divisibility of algebraic integers to rational integers. Consider the Gaussian integers $\rm\ \mathbb I = \{ m + n\ i\ : \ m,n\in \mathbb Z \}\ $. As in any ring we define divisibilty by $\rm\ a\ |\ b\ in\ \mathbb I \iff b/a \in \mathbb I\:.\ $ Suppose we wish to know if $\rm\ 2+3\ i\ |\ 91\ in\ \mathbb I\:,\:$ i.e. is $\rm\ w = 91/(2+3\ i)\in \mathbb I\ ?\ $ Now in fact $\rm\:\mathbb I\:$ happens to have a division algorithm which we could apply. But it is simpler to RTD yielding $\rm\ w = 91\ (2-3\ i)/(2^2+3^2) = 7\ (2-3\ i)\ $ so, indeed, $\rm\: w\in \mathbb I\:.\ $ More generally we can often reduce problems about algebraic numbers to problems about rational numbers by taking norms, traces, etc. In fact this is how Kronecker constructed his divisor theory for algebraic integers, see e.g. Harold Edwards: &lt;em&gt;Divisor Theory&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2011-03-10T01:18:46.457" CommentCount="4" />
  <row Id="26098" PostTypeId="2" ParentId="26089" CreationDate="2011-03-10T01:20:56.223" Score="1" ViewCount="" Body="&lt;p&gt;Find the lengths of the segments labeled A, B, C, and D.  These should add up to 6.  To find the lengths of C and D, use the fact that the triangles I have drawn in are similar to the inside (or outside) triangle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/VmGCw.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="5265" LastActivityDate="2011-03-10T01:20:56.223" />
  <row Id="26099" PostTypeId="2" ParentId="26092" CreationDate="2011-03-10T01:23:45.127" Score="2" ViewCount="" Body="&lt;p&gt;The third stage $P+P+P$ has the existing points plus (3,0), (2,1), (1,2), (0,3) to give a total of 10 points and it should be obvious that the $n^{\text{th}}$ stage adds $n+1$ additional points thus producing the (offset) triangle numbers at each stage.  You should also be able to see that the starting place or $0^{\text{th}}$ stage is just $(0,0)$ to which you add $P$ to get the first stage, so has 1 point.  So the sequence of numbers of points is $1,3,6,10,\ldots$.  You want this as a generating function, so you want an expression for &#xA;$$g(x) = 1 + 3x + 6x^2 + 10x^3 + \cdots + \frac{(n+1)(n+2)}{2} x^n +  \cdots$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some of us can spot such things from familiarity but one way to derive it is to see that&lt;br&gt;&#xA;$$ \int \int g(x) \, dx \, dx = A + Bx + x^2 /2 + x^3 /2 + x^4 /2 + x^5/2 + \cdots + x^{n+2} /2 +  \cdots $$&#xA;This is essentially a simple geometric series with sum $A+Bx +\frac{x^2}{2(1-x)}$ and then reversing the double integration we find&lt;br&gt;&#xA;$$g(x) = \frac{d}{dx}  \frac{d}{dx} \left( A+Bx +\frac{x^2}{2(1-x)} \right) = \frac{1}{(1-x)^3}$$&lt;/p&gt;&#xA;" OwnerUserId="6460" LastActivityDate="2011-03-10T01:23:45.127" CommentCount="3" />
  <row Id="26100" PostTypeId="2" ParentId="26089" CreationDate="2011-03-10T01:24:14.193" Score="1" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Algebraic Solution:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Extend the lines making up the side length $3$ of the inner triangle until it hits the walls of the larger one, and consider the triangle formed.  What is the length of this line particular line?  Well it is $$x+3+(?)$$ where $(?)=\frac{5}{4}x$ by similar triangles.  What about the base line?  It has length $8-x$.  And the hypotenuse?  It has length $10-x\frac{5}{4}$ by similar triangles again.  Then by the Pythagorean Theorem, $$\left(10-x\frac{5}{4}\right)^{2}=\left(8-x\right)^{2}+\left(3+\frac{9}{4}x\right)^{2}$$ and solving this quadratic yields $x=1$ as a solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Quick Solution:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice the side lengths of the larger triangle sum to equal its area, 24. (That is why we get 1)  Draw straight tubes, a $x\times 6$ tube, a $x\times 8$ tube and a $x\times 10$ tube along the sides on the triangle, on the inside so that all of the area of the triangle except the missing middle piece is covered. (The middle piece is the smaller triangle)  Then the overlapping parts, and the parts hanging outside the triangle can be moved around to make up the smaller inside triangle which is missing.  Hence the areas are equal so $8x+6x+10x=\frac{6\cdot 8}{2}=24$ so $x=1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Remark:&lt;/strong&gt;  I also hope this is understandable without a picture.&lt;/p&gt;&#xA;" OwnerUserId="6075" LastEditorUserId="6075" LastEditorDisplayName="" LastEditDate="2011-03-10T01:33:23.683" LastActivityDate="2011-03-10T01:33:23.683" CommentCount="4" />
  <row Id="26101" PostTypeId="2" ParentId="26089" CreationDate="2011-03-10T01:27:07.853" Score="14" ViewCount="" Body="&lt;p&gt;Draw lines connecting the &quot;respective vertices&quot;. Add up the areas to solve for $x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/saED3.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Area of a right triangle with the non-hypotenuse sides being $a$ and $b$ is $\frac{1}{2}(a \times b)$ while the area of a trapezium with parallel sides being $a$ and $b$ and the height being $h$ is $\frac{1}{2}h(a+b)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{1}{2}x(5+10) + \frac{1}{2}x(3+6) + \frac{1}{2}x(4+8) + \frac{1}{2}(3 \times 4) = \frac{1}{2} (6 \times 8)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$36x + 12 = 48 \Rightarrow x = 1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let us try to look at a slightly general case. Take a triangle and scale it to another similar triangle with the scale factor being $t$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/DKAWk.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let the sides of the inner triangle be $a$, $b$ and $c$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The perimeter and area of the inner triangle is $P$ and $A$ respectively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sides of the outer triangle are $ta$, $tb$ and $tc$ while the perimeter and area are $tP$ and $t^2A$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As before join the &quot;respective&quot; vertices and summing the areas give us,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$A + \frac{1}{2}x(ta + a) + \frac{1}{2}x(tb + b) + \frac{1}{2}x(tc + c) = t^2A$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$A + \frac{1}{2}x(t+1)P = t^2A \Rightarrow \frac{1}{2}x(t+1)P = (t^2-1)A$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x = \frac{2A}{P}(t-1)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the problem asked, $t=2$ with $P = 2A = 12$ and hence we get $x=1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, $t=1$ gives $x=0$ as expected.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This also gives a &lt;em&gt;nice proof&lt;/em&gt; that the &lt;strong&gt;radius of the incircle&lt;/strong&gt; of a triangle is $$r_{in} = \frac{2A}{P}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is got by plugging in $t=0$ and realizing that $\left| x \right|$ is nothing but the radius of the incircle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, $$x = (t-1)r_{in}$$&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2011-03-10T05:35:53.760" LastActivityDate="2011-03-10T05:35:53.760" CommentCount="8" />
  <row Id="26103" PostTypeId="2" ParentId="26095" CreationDate="2011-03-10T01:31:09.287" Score="4" ViewCount="" Body="&lt;p&gt;To be honest, this is a personal question that likely can not and should not be answered by a group of strangers. With that said, programming jobs are often filled by both CS and math students, though CS students generally start at a higher position due to experience. Those with strong math backgrounds typically have great intuition on how to approach problems, but a lack of experience in things like formal software engineering practices can limit your options. From my experience as a double major in math and CS, the job opportunities for CS graduates are everywhere. My math background has helped me within the job, but is seemingly neglected in the application/interview process. Ultimately the decision should be made based on your existing background and exactly what you want to accomplish in school (gain educational experience is a little too vague).&lt;/p&gt;&#xA;" OwnerUserId="1016" LastActivityDate="2011-03-10T01:31:09.287" CommunityOwnedDate="2011-03-10T01:51:31.610" CommentCount="1" />
  <row Id="26104" PostTypeId="2" ParentId="26092" CreationDate="2011-03-10T01:33:51.237" Score="1" ViewCount="" Body="&lt;p&gt;The number of points after adding $n$ of the $P$s is the number of solutions of non-negative solutions of $a + b \leq n$. Imagine the points $0$ to $n$ on a line. Pick two points $p_1&amp;lt;p_2$, and let $a = p_1$, $b = p_2 - p_1 - 1$. This process is reversible, and so the number of solutions is $\binom{n+2}{2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A useful identity is&#xA;$$ \sum_{m=0}^n \binom{m + k}{k} = \binom{n+k+1}{k+1}. $$&#xA;This is true since each $k+1$ subset of $n+k+1$ with maximal point $M \geq k+1$ is a $k$ subset of $M - 1 \geq k$, and this process is reversible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given a generating series, multiplying it by $1/(1-x) = 1 + x + x^2 + \ldots$ is like computing &quot;running sums&quot;. So starting with&#xA;$$ 1,0,0,0,\ldots, $$&#xA;multiplying by $1/(1-x)$ we get&#xA;$$ 1,1,1,1,\ldots, $$&#xA;multiplying again we get&#xA;$$ 1,2,3,4,\ldots, $$&#xA;and once again&#xA;$$ 1,3,6,10,\ldots. $$&#xA;The general term for $1/(1-x)^d$ is $\binom{n+d-1}{d-1} x^n$, as can be easily proven by induction using the identity above.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2011-03-10T01:33:51.237" CommentCount="0" />
  <row Id="26105" PostTypeId="1" AcceptedAnswerId="26112" CreationDate="2011-03-10T01:40:11.290" Score="0" ViewCount="70" Body="&lt;p&gt;This post is a continuation of &lt;a href=&quot;http://math.stackexchange.com/questions/26042/understanding-conditional-independence-of-two-random-variables-given-a-third-one&quot;&gt;my previous one&lt;/a&gt;, but can be read as being stand-alone. I am trying to understand the concept of conditional independence of two random variables given a third one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Following is the definition that I think is the right one:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Two random variables $X$ and $Y$ are&#xA;  conditionally independent given a&#xA;  random variable $Z$ if they are&#xA;  independent given $\sigma(Z)$: the&#xA;  $\sigma$-algebra generated by $Z$.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Two events $R$ and $B$ are&#xA;  conditionally independent given a&#xA;  $\sigma$-algebra $\Sigma$ if&#xA;      $$\Pr(R \cap B \mid \Sigma) = \Pr(R \mid \Sigma)\Pr(B \mid \Sigma)\ a.s.$$ where $\Pr(A \mid \Sigma)$&#xA;  denotes the conditional expectation of&#xA;  the indicator function of the event&#xA;  $A$, given the sigma algebra $\Sigma$.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Two random variables $X$ and $Y$ are&#xA;  conditionally independent given a&#xA;  $\sigma$-algebra $\Sigma$ if the above&#xA;  equation holds for all $R$ in&#xA;  $\sigma(X)$ and $B$ in $\sigma(Y)$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My questions are&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Is the definition equivalent to that&#xA;$X$ and $Y$ are conditional&#xA;independent given $Z$ iff $E(X&#xD;&#xA;    \times Y \mid Z) = E(X\mid Z) \times&#xD;&#xA;    E(Y \mid Z)$? If yes, can this be&#xA;proved from the definition, by that&#xA;any measurable function (here random&#xA;variable) can be pointwise&#xA;approximated by a sequence of simple&#xA;functions (a simple function is the finite linear combination of indicator functions), just as used in&#xA;definition of Lebesgue integral, and&#xA;that conditional expectation and taking pointwise limit can be exchangeable?&lt;/li&gt;&#xA;&lt;li&gt;Since both $E(X\mid Z)$ and $E(Y&#xD;&#xA;    \mid Z)$ are random variables, is&#xA;conditional independence between $X$&#xA;and $Y$ given $Z$ equivalent to that&#xA;$E(X\mid Z)$ and $E(Y \mid Z)$ are&#xA;independent random variables w.r.t.&#xA;the underlying probability measure&#xA;$P$ of the underlying sample space&#xA;$(\Omega, \mathcal{F}, P)$? Or are the two related somehow?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Thanks and regards!&lt;/p&gt;&#xA;" OwnerUserId="1281" LastEditorUserId="6296" LastEditorDisplayName="" LastEditDate="2011-03-10T09:15:08.853" LastActivityDate="2011-03-10T09:15:08.853" Title="Further questions about conditional independence of two random variables given a third one" Tags="&lt;probability-theory&gt;&lt;measure-theory&gt;" AnswerCount="1" />
  <row Id="26106" PostTypeId="1" CreationDate="2011-03-10T01:46:34.187" Score="1" ViewCount="183" Body="&lt;p&gt;I can not figure out how to integrate &#xA;$$\int\frac{3x^2+x+4}{x^3+x} \, dx.$$&#xA;I got as far as factoring the denominator which gives us x(x^2+1). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So from there I got&#xA;$$\frac{3x^2+x+4}{x(x^2+1)} = \frac{A}{x} + \frac{Bx+C}{x^2+1}$$&#xA;which would give us: $(3x^2 + x + 4) = A(x^2 + 1) + (Bx+C)x$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is where Im stuck because I can not figure out the values for B and C. Please help.&lt;/p&gt;&#xA;" OwnerUserId="8051" LastEditorUserId="2937" LastEditorDisplayName="" LastEditDate="2011-03-10T03:33:39.360" LastActivityDate="2011-04-15T16:16:35.857" Title="integration by partial fraction" Tags="&lt;calculus&gt;&lt;integral&gt;" AnswerCount="4" CommentCount="3" />
  <row Id="26107" PostTypeId="2" ParentId="25670" CreationDate="2011-03-10T01:48:23.780" Score="1" ViewCount="" Body="&lt;p&gt;I don't know about the history, but if you want to understand 'what's going on' the keyword is 'orbifold'.  I recommend Peter Scott's article 'The geometries of 3-manifolds', available as a pdf on &lt;a href=&quot;http://www.math.lsa.umich.edu/~pscott/&quot; rel=&quot;nofollow&quot;&gt;his website&lt;/a&gt;, for an introduction to 2-dimensional orbifolds.&lt;/p&gt;&#xA;" OwnerUserId="2287" LastActivityDate="2011-03-10T01:48:23.780" />
  <row Id="26108" PostTypeId="2" ParentId="24757" CreationDate="2011-03-10T01:54:08.407" Score="5" ViewCount="" Body="&lt;p&gt;No, in general there's not an algorithm.  In &lt;a href=&quot;http://arxiv.org/abs/1003.5117&quot; rel=&quot;nofollow&quot;&gt;this paper&lt;/a&gt;, Bridson and I prove that there are finite sets of integer matrices that generate finitely presented groups, but for which there's no algorithm to compute a presentation.&lt;/p&gt;&#xA;" OwnerUserId="2287" LastActivityDate="2011-03-10T01:54:08.407" />
  <row Id="26109" PostTypeId="2" ParentId="26106" CreationDate="2011-03-10T01:56:23.777" Score="1" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Hint:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the equation $(3x^2 + x + 4) = A(x^2 + 1) + (Bx+C)x$, by matching coefficients of the quadratic term $x^2$ we find:&#xA;$$A+B=3,$$&#xA;by matching coefficients of the linear term $x$ we find:&#xA;$$C=1,$$&#xA;and lastly, by matching the constant term we find:&#xA;$$A=4$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you solve it from here?&lt;/p&gt;&#xA;" OwnerUserId="6075" LastActivityDate="2011-03-10T01:56:23.777" CommentCount="3" />
  <row Id="26110" PostTypeId="2" ParentId="26106" CreationDate="2011-03-10T01:58:31.920" Score="6" ViewCount="" Body="&lt;p&gt;Okay, first a bunch of nitpicks:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;You are not trying to integrate the integral, you are trying to integrate the function itself.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The integral does not equal the partial fraction decomposition, only the rational function itself does.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;You are missing the $dx$ in all the integrals.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So. You are trying to integrate $\displaystyle \frac{3x^2+x+4}{x^3+x}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, you make sure the numerator has smaller degree than the denominator, doing long division if necessary to get it into that form (done).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, you factor the denominator completely (done).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then you set up the partial fraction decomposition problem (done):&#xA;$$\frac{3x^2+x+4}{x(x^2+1)} = \frac{A}{x} + \frac{Bx+C}{x^2+1}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, you perform the operation on the right hand side to the expression you want (done):&#xA;$$\frac{3x^2+x+4}{x(x^2+1)} = \frac{A(x^2+1) + (Bx+C)x}{x(x^2+1)}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, you know the numerators are equal (done):&#xA;$$3x^2 + x+4 = A(x^2+1)+ (Bx+c)x.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, you figure out the values of $A$, $B$, and $C$. There are two strategies:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Do the operations on the right and write it as a polynomial; since two polynomials are equal if and only if they have the same coefficients, the coefficient of $x^2$ on the right equals $3$, etc. This will set up a system of linear equations for $A$, $B$, and $C$, which you can solve:&#xA;$$3x^2 + x + 4 = Ax^2 + A + Bx^2 + Cx = (A+B)x^2 + Cx + A$$&#xA;so $A+B=3$, $C=1$, and $A=4$. From this, you get $A=4$, $B=-1$, and $C=1$, so&#xA;$$\frac{3x^2+x+4}{x(x+1)} = \frac{4}{x} + \frac{-x+1}{x^2+1}.$$&#xA;Now you just need to do the simpler integrals&#xA;$$\int\frac{3x^2+x+4}{x^3+x}\,dx = \int\frac{4}{x}\,dx - \int\frac{x}{x^2+1}\,dx + \int\frac{1}{x^2+1}\,dx.$$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Plug in some values of $x$ to get information about $A$, $B$, and $C$. Specifically, pick values that make some of the terms equal to $0$ (the roots of the original polynomial), and start simplifying. For example, from&#xA;$$3x^2 + x + 4 = A(x^2+1) + (Bx+C)x,$$&#xA;plugging in $x=0$ you get $4 = A(1) + 0$, so $A=4$; now we have&#xA;$$3x^2 + x + 4 = 4x^2 + 4 + (Bx+C)x.$$&#xA;Moving all the known factors to the left, we have&#xA;$$-x^2 + x = (Bx+C)x$$&#xA;and factoring out $x$, we get&#xA;$x(-x+1) = (Bx+C)x$, from which you can cancel $x$ to get&#xA;$$-x+1 = Bx+C$$&#xA;which immediately gives $B=-1$ and $C=1$, as before. Now that we know $A$, $B$, and $C$, proceed as in 1.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-10T02:58:36.377" LastActivityDate="2011-03-10T02:58:36.377" CommentCount="3" />
  <row Id="26111" PostTypeId="1" CreationDate="2011-03-10T01:59:57.187" Score="2" ViewCount="46" Body="&lt;p&gt;There are many proofs that there is a unique division quaternion algebra over a locally compact field that is not $\mathbb{C}$. For instance this set of notes/book by John Voight contains two proofs: &lt;a href=&quot;http://www.cems.uvm.edu/~voight/crmquat/book/quat-modforms-041310.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.cems.uvm.edu/~voight/crmquat/book/quat-modforms-041310.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That said, every proof that I've seen first splits the proof into the archimedean and non-archimedean cases (and then sometimes into the residue characteristic 2 and $\ne$ 2 cases).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even Weil's Basic Number Theory doesn't seem to have a proof that treats all places equally (maybe I missed it?). Does anyone know of such a proof?&lt;/p&gt;&#xA;" OwnerUserId="2580" LastActivityDate="2011-03-10T01:59:57.187" Title="Proof that there's a unique division quaternion algebra over a locally compact field?" Tags="&lt;algebra&gt;&lt;absolute-value&gt;" CommentCount="3" FavoriteCount="1" />
  <row Id="26112" PostTypeId="2" ParentId="26105" CreationDate="2011-03-10T02:09:12.250" Score="4" ViewCount="" Body="&lt;p&gt;For (1), the forward direction is true and can be proved as you suggest, provided $X,Y$ are integrable.  The converse is false even for ordinary independence (where $\Sigma$ is the trivial $\sigma$-field $\{\Omega, \emptyset\}$ or $Z = c$); we have $E[XY] = E[X] E[Y]$ iff $X,Y$ are uncorrelated, which is weaker than independence.  (Wikipedia gives as an example $X \sim U(-1,1), Y=X^2$).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is true is that $X,Y$ are conditionally independent given $\Sigma$ (where $\Sigma = \sigma(Z)$ is a special case) iff $$E[f(X) g(Y) | \Sigma] = E[f(X) | \Sigma] E[g(Y)|\Sigma]$$ for all $f,g$ in some suitably large class of measurable functions from $\mathbb{R}$ to $\mathbb{R}$.  For instance, one could take: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;all bounded measurable functions&#xA;&lt;li&gt;indicator functions of measurable sets&#xA;&lt;li&gt;bounded continuous functions&#xA;&lt;li&gt;smooth and compactly supported functions&#xA;&lt;li&gt;indicators of open/closed/half-open intervals or half-lines.&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;For (2), this is false.  You can get counterexamples to both directions by taking $X,Y$ to be not (unconditionally) independent and considering the extreme cases where $\Sigma$ is trivial or $\Sigma = \mathcal{F}$.&lt;/p&gt;&#xA;" OwnerUserId="822" LastEditorUserId="822" LastEditorDisplayName="" LastEditDate="2011-03-10T04:56:05.993" LastActivityDate="2011-03-10T04:56:05.993" />
  <row Id="26113" PostTypeId="2" ParentId="26083" CreationDate="2011-03-10T02:22:22.753" Score="1" ViewCount="" Body="&lt;p&gt;I think that the best thing to do when you want to start learning about spectral sequences is to do computations, see this answer for an outline: &lt;a href=&quot;http://math.stackexchange.com/questions/7966/reference-for-spectral-sequences/12726#12726&quot;&gt;Reference for spectral sequences&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I really believe what I said there, it is not a plug.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was going to recommend a paper, but I looked at your profile and that drasticly changes my advice. Ask Mike Boardman! Or at least try to trace through the first part of his conditionally convergent spectral sequences! That paper is really flipping cool. It is not easy, but it isn't really hard. Also, you can probably skip the alternative descriptions (the weird big formulas in the first section) of the $E_r$ terms for now.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Seriously, you want to read that paper.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: I feel bad for not explicitly answering your question, but all I would do is look up something in that paper and mess it up trying to copy it down.&lt;/p&gt;&#xA;" OwnerUserId="627" LastEditorUserId="627" LastEditorDisplayName="" LastEditDate="2011-03-10T02:55:21.257" LastActivityDate="2011-03-10T02:55:21.257" CommentCount="10" />
  <row Id="26114" PostTypeId="2" ParentId="22065" CreationDate="2011-03-10T02:37:45.017" Score="0" ViewCount="" Body="&lt;p&gt;There are two ways to answer your question:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;what does foundation mean? The two areas address 'foundation' from different perspectives. Set theory (and logic) are attempts at foundations of mathematical knowledge about how we can know mathematical truths - how do you know that a theorem is 'true' (and what does true mean) ? (set theory does this by allowing a translation from your favorite branch of mathematics to sets, and then formalizing the inference operations on just the set  objects; everything can be converted to a set) &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Category theory is a foundations for -what- we know about mathematics - what similarities/commonalities are among disparate areas of mathematics. Take your favorite brach of mathematics, define some objects and some suitable functions and then you get all sorts of theorems for free (everything can be viewed through the 'category' lens. everything is a category)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course you can use one of these for the other (give a -logical- foundation of category theory, e.g. the objects and morphisms are implemented with sets; give a -categorical- foundation for set theory/logic, e.g. one can have a category of 'set' with 'functions' as the morphisms).&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;as to your title, how can there be more than one foundation at all, even of the same kind?There can certainly be more than one truth foundation (intuitionistic, constructivist, allowing different set axioms, different rules of inference, ZFC, NBG, etc.) each allowing desired features and disallowing unwanted features. And there have been many unification foundations (set theory can be viewed as one (just sets as the things that can be manipulated), group theory, algebra as a whole (symbolic manipulation), just category theory is the latest and best unification.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1919" LastActivityDate="2011-03-10T02:37:45.017" />
  <row Id="26115" PostTypeId="1" AcceptedAnswerId="26206" CreationDate="2011-03-10T02:39:29.747" Score="1" ViewCount="64" Body="&lt;p&gt;Let $v(x)\in L^1\cap L^2$ be a function, and $w(y)\in L^2$ be its Fourier transform. Let $\{w_k\}$ be an infinite sequence of samples from $w(y)$, sampled $T$ apart. Now,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$v(x)=\sum_{k=-\infty}^{\infty}w_k e^{ikx}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;in the $L^2$ norm sense. Is it possible to write $w_k=\displaystyle\int_0^{2\pi}v(x)e^{-ikx}dx$, since $v(x)$ is Riemann integrable, or should I use the inverse discrete transform? &lt;/p&gt;&#xA;" OwnerUserId="7815" LastEditorUserId="7815" LastEditorDisplayName="" LastEditDate="2011-03-10T05:43:13.743" LastActivityDate="2011-03-10T15:50:28.150" Title="Reconstruct elements of sequence from inverse transform" Tags="&lt;fourier-analysis&gt;&lt;fourier-series&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="26116" PostTypeId="1" AcceptedAnswerId="26119" CreationDate="2011-03-10T03:14:46.973" Score="4" ViewCount="62" Body="&lt;p&gt;I'm trying to find a local minimum for $f(x,y) = x^4 + 3xy^2 + y^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After finding that a critical point is at $f(x,y)$ is $(0,0)$, I am wondering if this is proper formal justification to prove that this is indeed the case?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $\epsilon &amp;gt;0$ be given.  Let $|y|=-x$&lt;br&gt;&#xA;(Is this proper? I just want to keep x and y to have the same absolute value basically)&lt;br&gt;&#xA;Let $x = \max\{\frac{-1}{3}, \frac{-\epsilon}{2}\}$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then $\sqrt{x^2+y^2}=\sqrt{2\frac{\epsilon}{2}^2} = \frac{\epsilon}{2} &amp;lt; \epsilon$&lt;br&gt;&#xA;(Is this proper to show that I can make x, y as close to $\epsilon$ as I want?)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $\frac{-1}{3}\leq x\lt 0$ then $3x \geq -1$ and $x^4+3xy^2+y^2 \geq 0$&lt;br&gt;&#xA;$\therefore (0,0)$ is a local minimum of $f(x,y)$&lt;/p&gt;&#xA;" OwnerUserId="8054" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-10T03:36:21.353" LastActivityDate="2011-03-10T03:36:21.353" Title="Is this argument that $(0,0)$ is a local minimum for $f(x,y)=x^4+3xy^2+y^2$ correct?" Tags="&lt;calculus&gt;&lt;homework&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="26117" PostTypeId="2" ParentId="26106" CreationDate="2011-03-10T03:16:34.250" Score="1" ViewCount="" Body="&lt;p&gt;Observe that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;\begin{align}\frac{3x^2+x+4}{x^3+x}=\frac{3(x^2+1)-2x+1}{x(x^2+1)}&amp;amp;=\frac{3}{x}-\frac{2}{x^2+1}+\frac{1}{x}-\frac{x}{x^2+1}\\&#xD;&#xA;&amp;amp;=\frac{4}{x}-\frac{2+x}{x^2+1}&#xD;&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and the rest should be easy.&lt;/p&gt;&#xA;" OwnerUserId="7340" LastEditorUserId="7340" LastEditorDisplayName="" LastEditDate="2011-03-12T04:47:21.873" LastActivityDate="2011-03-12T04:47:21.873" />
  <row Id="26118" PostTypeId="1" AcceptedAnswerId="26139" CreationDate="2011-03-10T03:20:24.347" Score="12" ViewCount="257" Body="&lt;p&gt;&lt;em&gt;Motivation&lt;/em&gt;: I'm writing a proof, and within it, I need to prove:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;em&gt;Conjecture&lt;/em&gt;: Let $p$ be an odd prime (i.e. $p \neq 2$).  Let $c \geq 2$, $d \geq 1$ and $r \geq 1$.  If $p^r$ divides $cd$, then $d(c-1) \geq r+1$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This conjecture, should it prove to be true (and I suspect it will [it seems to be like &quot;prove elephants are bigger than mosquitos&quot;], unless I've missed a few small cases), would be a rather uninteresting result and no real insight into the problem I'm studying.  This leads me to the question:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Question&lt;/strong&gt;: Does there exist a software package that would allow me to automatically prove (or disprove) this conjecture (or results like this)?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="139" LastEditorUserId="139" LastEditorDisplayName="" LastEditDate="2011-03-10T03:29:05.073" LastActivityDate="2011-03-10T19:53:26.970" Title="Is there an automated way to prove really boring elementary number theoretic results?" Tags="&lt;elementary-number-theory&gt;&lt;automated-theorem-proving&gt;" AnswerCount="1" CommentCount="9" />
  <row Id="26119" PostTypeId="2" ParentId="26116" CreationDate="2011-03-10T03:23:25.513" Score="8" ViewCount="" Body="&lt;p&gt;In order to show that $f(x,y)$ has a local minimum at $(0,0)$, you need to show that there is an $\epsilon\gt 0$ such that for all $(x,y)$, if $\sqrt{(x-0)^2+(y-0)^2}\lt\epsilon$ (that is, if $(x,y)$ is $\epsilon$-close to $(0,0)$), then $f(0,0)\leq f(x,y)$. In other words: there is an open ball around $(0,0)$ with the property that $f(0,0)$ is the smallest value that $f$ takes in that ball.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you are not &quot;given an $\epsilon\gt 0$&quot;, as you are with limits. Instead, &lt;em&gt;you&lt;/em&gt; get to pick the $\epsilon$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, you don't get to restrict the values of $x$ and $y$: &lt;em&gt;any&lt;/em&gt; value of $x$ and $y$ that falls within $\epsilon$ of $(0,0)$ has to be accounted for. So you don't get to say &quot;Let $|y|=-x$.&quot; And you don't get to pick $x$, either. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, I'm afraid that this is not a valid way to go about it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instead, let's think about $f(x,y)$. Both the $x^4$ and the $y^2$ will contribute positive factors; so really we only have a potential problem (i.e., a value of $f$ smaller than $0$) when $x\lt 0$; if $y=0$ the factor $3xy^2$, the only possible source of problems, vanishes, so we just need to worry about the case $y\neq 0$. Then we have $f(x,y) = x^4 + 3xy^2 + y^2 = x^4 + y^2(3x+1)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means that if $3x+1\geq 0$, then $f(x,y)\geq 0 = f(0,0)$, even if $x$ is negative and has absolute value much larger than $y$. And this will be achieved if $x\geq - \frac{1}{3}$. So all you need to do is make sure that your $x$ is larger than $-\frac{1}{3}$ (as you probably noticed, given the value you wanted to &lt;em&gt;assign&lt;/em&gt; to $x$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So: all you need to do is find some $\epsilon$ with the property that if $\sqrt{x^2+y^2}\lt \epsilon$, then you'll know that $x\geq -\frac{1}{3}$. Then you can argue that for &lt;em&gt;that&lt;/em&gt; $\epsilon$, you have the property you want (every point $(x,y)$ within $\epsilon$ of $(0,0)$ will have $f(x,y)\geq f(0,0)$). How do you find the $\epsilon$? Draw a picture: $\epsilon$ is the radius of a circle around $(0,0)$; you want every point inside the circle to have $x$-coordinate greater than or equal to $-\frac{1}{3}$. What value(s) can you pick for $\epsilon$? (There is no single answer, since if you have a value that works, any smaller value will also work).&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-10T03:31:18.690" LastActivityDate="2011-03-10T03:31:18.690" CommentCount="2" />
  <row Id="26120" PostTypeId="2" ParentId="20925" CreationDate="2011-03-10T03:38:33.267" Score="2" ViewCount="" Body="&lt;p&gt;Let $\displaystyle S_x=\sum_{x=1}^{14}4^x=4+4^2+\ldots+4^{14}$ and then multiply  both sides of $S_x$ by 4 to get $\displaystyle 4S_x=4^2+\ldots+4^{15}$. Now $4S_x-S_x=4^{15}-4$ so you can say that $\displaystyle S_x=\frac{4}{3}(4^{14}-1)$&lt;/p&gt;&#xA;" OwnerUserId="7340" LastActivityDate="2011-03-10T03:38:33.267" CommentCount="1" />
  <row Id="26121" PostTypeId="1" AcceptedAnswerId="26125" CreationDate="2011-03-10T03:44:53.777" Score="3" ViewCount="92" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Let n be a positive natural number , $n\ge 0$, then. $\displaystyle\sum_{i=0}^n i*i!= (n+1)!-1$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here is my attempt.I'm not going to write the base case because I understand that part.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming  $\displaystyle\sum_{i=0}^n i*i!= (n+1)!-1$ is true. We wish to show.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle \sum_{i=0}^{n+1} i*i!= (n+2)!-1$. Thus.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle\sum_{i=0}^{n+1} i*i!= (\sum_{i=0}^n i*i!) $ This is where I get stuck.&lt;/p&gt;&#xA;" OwnerUserId="1040" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-10T03:46:04.793" LastActivityDate="2011-03-10T04:46:57.950" Title="Proof by induction" Tags="&lt;induction&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="26122" PostTypeId="2" ParentId="4594" CreationDate="2011-03-10T03:46:12.637" Score="3" ViewCount="" Body="&lt;p&gt;You ask: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;suppose it could be proven for every&#xA;  algorithm that either:&lt;/p&gt;&#xA;  &#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;the algorithm is correct and uses an internal representation such&#xA;  that no two distinct input cliques&#xA;  lead to the same internal&#xA;  representation, or&lt;/li&gt;&#xA;  &lt;li&gt;the algorithm is incorrect.&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;  &#xA;  &lt;p&gt;Would this be a worthwhile result? How&#xA;  important would it be? Or is it wrong?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It's wrong. You can always take a correct algorithm for clique and add a preprocessing step that removes edges which are clearly not in any maximal clique. This will still be a correct algorithm for clique, and obviously does not satisfy either (1) or (2). &lt;/p&gt;&#xA;" OwnerUserId="2501" LastActivityDate="2011-03-10T03:46:12.637" CommentCount="2" />
  <row Id="26124" PostTypeId="2" ParentId="26121" CreationDate="2011-03-10T03:48:51.063" Score="1" ViewCount="" Body="&lt;p&gt;Well, your last equality isn't true! The left hand side has $n+2$ summands, and you dropped the biggest one on the right. What happened to $(n+1)*(n+1)!$? No wonder you got stuck...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Everything is correct until &quot;Thus&quot;. Now, write the $n+1$ case as &quot;the $n$ case and a little extra&quot;, so you can apply the induction hypothesis to the $n$ case. Then see if you can manipulate what you obtain into the formula you want.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here you have&#xA;$$\sum_{i=0}^{n+1}(i*i!) = \left(\sum_{i=0}^n(i*i!)\right) + \Biggl((n+1)*(n+1)!\Biggr).$$&#xA;Now apply the induction hypothesis to the first summand, and see what you can get.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-10T03:54:38.793" LastActivityDate="2011-03-10T03:54:38.793" CommentCount="8" />
  <row Id="26125" PostTypeId="2" ParentId="26121" CreationDate="2011-03-10T03:51:22.727" Score="0" ViewCount="" Body="&lt;p&gt;Your last step should read $$\sum_{i=0}^{k+1} \left( i \times i! \right)= \sum_{i=0}^k \left(i \times i!\right) + (k+1)(k+1)! $$ Now use your induction assumption to get $$\sum_{i=0}^{k+1} \left(i \times i! \right)= (k+1)! - 1 + (k+1)(k+1)! = (k+1)! (k+2) -1 = (k+2)! - 1$$&lt;/p&gt;&#xA;" OwnerUserId="3302" LastActivityDate="2011-03-10T03:51:22.727" />
  <row Id="26126" PostTypeId="1" AcceptedAnswerId="26127" CreationDate="2011-03-10T04:05:47.377" Score="1" ViewCount="77" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Find the length of the given curve $\displaystyle x=3y^{\frac{4}{3}}-\frac{3y^{\frac{2}{3}}}{32} $&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;where the bounds are given by $-8 \leq y \leq 343$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can solve for the positive part (0 to 343), but I am unsure of how I would go about integrating&#xA;$$\int_{-8}^{0} \sqrt{1+\left(\frac{dx}{dy}\right)^2}dy$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks again for any help!&lt;/p&gt;&#xA;" OwnerUserId="6423" LastEditorUserId="6996" LastEditorDisplayName="" LastEditDate="2011-05-24T06:38:46.637" LastActivityDate="2011-05-24T06:38:46.637" Title="Find the length of the given curve $x=3y^\frac{4}{3}-\frac{3y^{\frac{2}{3}}}{32} $" Tags="&lt;calculus&gt;&lt;homework&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="26127" PostTypeId="2" ParentId="26126" CreationDate="2011-03-10T04:16:38.213" Score="2" ViewCount="" Body="&lt;p&gt;$f(y)=3y^{4/3}-\frac{3}{32}y^{2/3}$ is an even function of $y$, i.e., $f(-y)=f(y)$, and therefore $f&amp;#39;(-y)=-f&amp;#39;(y)$, meaning that $f&amp;#39;(y)$ is an odd function.  This implies that $f&amp;#39;(-y)^2=f&amp;#39;(y)^2$.  So if you know how to deal with positive $y$, then you also can deal with negative $y$, e.g., $$\int_{-8}^0\sqrt{1+\left(\frac{dx}{dy}\right)^2}dy=\int_0^8\sqrt{1+\left(\frac{dx}{dy}\right)^2}dy.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even without noticing this, I don't see how you would be able to do the positive part if the negative part causes trouble.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2011-03-10T04:16:38.213" CommentCount="2" />
  <row Id="26128" PostTypeId="2" ParentId="26126" CreationDate="2011-03-10T04:19:45.823" Score="0" ViewCount="" Body="&lt;p&gt;Hi, the length of a curve with $r(t)=(x(t),y(t))$ can be found by $\int^{t_{1}}_{t_{0}}|r&amp;#39;(t)|dt$. In your problem you confused $\frac{dy}{dx}$ with $\frac{dx}{dy}$. That's why you cannot integrate. &lt;/p&gt;&#xA;" OwnerUserId="7887" LastActivityDate="2011-03-10T04:19:45.823" CommentCount="1" />
  <row Id="26129" PostTypeId="2" ParentId="21769" CreationDate="2011-03-10T04:24:07.663" Score="1" ViewCount="" Body="&lt;p&gt;You variation seems to have 2 parameters, $n$ and $m$, but no matter - for any $n$, {\em all} primes, except the (finite) set of prime factors, are coprime to any number $n$ (except itself), so the set of those primes $\ge m$ will still be infinite.&lt;/p&gt;&#xA;" OwnerUserId="1919" LastActivityDate="2011-03-10T04:24:07.663" CommentCount="1" />
  <row Id="26130" PostTypeId="2" ParentId="26121" CreationDate="2011-03-10T04:26:00.090" Score="1" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\: $ First trivially inductively prove the &lt;strong&gt;Fundamental Theorem of Difference Calculus&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\rm\ F(n)\ =\ \sum_{i=0}^n\ f(i)\ \  \iff\ \ \ F(n) - F(n-1)\ =\ f(n),\ \ \ F(0) = 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your result follows by $\rm\ F(n)-F(n-1)\ =\ (n+1)! - n!\ =\ (n+1 -1)\ n!\ =\ n\ n!\ =\ f(n)\:.$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that by employing the Fundamental Theorem we have reduced the proof to the trivial verification of an equation. No ingenuity is required. In fact for hyperrational functions like factorials it is so trivial that there are algorithms that can mechanically verify such equalities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the proof of the Fundamental Theorem is much more obvious than that for your special case because the &lt;em&gt;telescopic cancellation&lt;/em&gt; is obvious at this level of generality, whereas it is usually obfuscated in most specific instances. For further discussion see my &lt;a href=&quot;http://math.stackexchange.com/search?q=user:242+telescopy&quot;&gt;many posts on telescopy.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2011-03-10T04:46:57.950" LastActivityDate="2011-03-10T04:46:57.950" />
  <row Id="26131" PostTypeId="1" AcceptedAnswerId="26202" CreationDate="2011-03-10T04:44:07.597" Score="5" ViewCount="149" Body="&lt;p&gt;Hello. &#xA;I vaguely remember seeing something like a &quot;picture&quot; of various groups a while back. It was as if the elements of the group were each associated with a point and many points had segments connecting them, but not all were connected. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anyone know what I am talking about? If so, would you care to take the time to explain the basics (or point me to a resource, if you think google won't help)? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks. &lt;/p&gt;&#xA;" OwnerUserId="7850" LastActivityDate="2011-03-10T14:51:56.617" Title="Visual representation of groups." Tags="&lt;abstract-algebra&gt;&lt;soft-question&gt;" AnswerCount="3" FavoriteCount="1" />
  <row Id="26132" PostTypeId="1" AcceptedAnswerId="26164" CreationDate="2011-03-10T04:44:45.393" Score="2" ViewCount="101" Body="&lt;p&gt;Reframed question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Team A and Team B will be playing each other in a sporting event where there is one winner and one loser; a tie is impossible.  A man, wanting to make money from placing a bet on the winner, consults with three omniscient seers.  These seers know the outcome of the game already, being omniscient.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first seer informs the man (truthfully) that he randomly lies 20% of the time when telling the outcome of a sporting event, telling the truth the rest of the time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second seer says (truthfully) that he is just like the first, except he lies 40% of the time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The third seer says (truthfully) that he is just like the first and second seers, except he lies 70% of the time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first seer tells the man that team A will win.&#xA;The second seer tells the man that team A will lose.&#xA;The third seer tells the man that team A will win.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What is the probability that team A will win?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each seers' determination of whether to lie is independent of the others.  You can imagine that each seer rolls a ten sided die to make their decision of whether to lie.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After the above question is answered, I am curious if someone can come up with a general formula for any number of seers with any probability of lying, with any set of predictions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Follow-up Question&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Varying sample sizes&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now suppose that we do not know with what frequency the seers are lying, but we do have the following information:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first seer lied 200 times out of 1000 in the past.  That is, out of 1000 total trials, he lied 200 times.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second seer lied 40 times out of 100 in the past.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The third seer lied 7 times out of 10 in the past.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The average seer in the world (there are many) lies half the time.  The frequency with which all seers lie follows a normal distribution.  The three seers do not make up a significant percentage of total seers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The seers make the same predictions as in the original question.  What is the probability that team A will win?&lt;/p&gt;&#xA;" OwnerUserId="8055" LastEditorUserId="8055" LastEditorDisplayName="" LastEditDate="2011-03-11T06:22:28.503" LastActivityDate="2011-03-11T06:22:28.503" Title="Probability from a collection of independent predictions" Tags="&lt;probability&gt;&lt;statistics&gt;" AnswerCount="1" CommentCount="8" FavoriteCount="2" />
  <row Id="26133" PostTypeId="2" ParentId="26131" CreationDate="2011-03-10T04:50:01.560" Score="7" ViewCount="" Body="&lt;p&gt;See the  &lt;a href=&quot;http://en.wikipedia.org/wiki/Cayley_graph&quot; rel=&quot;nofollow&quot;&gt;Cayley graph&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/Schreier_coset_graph&quot; rel=&quot;nofollow&quot;&gt;Schreier  coset graphs.&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I had much enjoyment playing with such graphs when I was an MIT undergrad. In fact I recall that I submitted a very interesting one to prove a hairy identity on a problem set in a knot theory class taught by George Whitehead. He found it quite interesting (and amusing). Apparently no one had done if before so simply and vividly using a Cayley graph. It was not an easy task to impress Prof. Whitehead. Such is the beauty that some of these geometric techniques bring to the austere algebra. They are well worth your study.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2011-03-10T05:12:25.357" LastActivityDate="2011-03-10T05:12:25.357" CommentCount="5" />
  <row Id="26134" PostTypeId="2" ParentId="26131" CreationDate="2011-03-10T04:50:55.303" Score="4" ViewCount="" Body="&lt;p&gt;I believe you are referring to the &lt;a href=&quot;http://en.wikipedia.org/wiki/Cayley_graph&quot; rel=&quot;nofollow&quot;&gt;Cayley Graph&lt;/a&gt; of a group.&lt;/p&gt;&#xA;" OwnerUserId="2720" LastActivityDate="2011-03-10T04:50:55.303" />
  <row Id="26135" PostTypeId="1" AcceptedAnswerId="26136" CreationDate="2011-03-10T04:55:52.753" Score="10" ViewCount="351" Body="&lt;p&gt;I've read that mathematicians believe that factorization of integers is hard (in some precise computational sense.) &lt;strong&gt;Is it known whether polynomial factorization is computationally easy or hard?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One thing I originally thought was that if we could factor polynomials easily, then we could factor $n$ by finding a polynomial $p(x)$ with $p(m)=n$ for some $m$, then &quot;easily factorizing&quot; $p$ to get $p(x)=q(x)\cdot r(x)$. Then $q(m)\cdot r(m)$ would be a factorization of $n$. But we wouldn't get any information if one of these happened to be 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anyone have a solution or a reference for this problem? (I searched online, but all I could find was how to do factorization like in high school problems.)&lt;/p&gt;&#xA;" OwnerUserId="7676" LastEditorUserId="7676" LastEditorDisplayName="" LastEditDate="2011-03-10T21:14:41.093" LastActivityDate="2011-03-10T21:14:41.093" Title="Is factoring polynomials as hard as factoring numbers?" Tags="&lt;polynomials&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="2" />
  <row Id="26136" PostTypeId="2" ParentId="26135" CreationDate="2011-03-10T05:13:32.737" Score="20" ViewCount="" Body="&lt;p&gt;There is a polynomial time algorithm for factoring polynomials with rational coefficients (the &lt;a href=&quot;http://en.wikipedia.org/wiki/Polynomial_factorization#LLL_algorithm&quot; rel=&quot;nofollow&quot;&gt;LLL algorithm&lt;/a&gt; of Lenstra, Lenstra, and Lovasz), so factoring polynomials over the rationals is known to be &quot;easy&quot; (polynomial time is considered to make the problem &quot;computationally easy&quot;, even if in practice it does not perform well). Of the known methods for factoring general integers, on the other hand, the best is the &lt;a href=&quot;http://en.wikipedia.org/wiki/General_number_field_sieve&quot; rel=&quot;nofollow&quot;&gt;General number field sieve&lt;/a&gt;, which is believed to be super-polynomial and sub-exponential (the complexity of the algorithm is estimated heuristically), which is worse.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, this kind of problem tends to be easier for polynomials than for integers, thanks to the existence of the derivative. Thus, there is a fairly easy proof of &quot;FLT for polynomials&quot; (about one page long); it is trivial to detect if a polynomial is squarefree (and no such easy algorithm is known for integers), and so on. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your idea for factoring integers with polynomials is very natural, but unfortunately fails in both directions: not every factorization of $n$ would be detected by a factorization of $p(x)$, and not every factorization of $p(x)$ would give a factorization of $n$. For example, $p(x) = x^2+x+1$ is irreducible over $\mathbb{Q}$ (even over $\mathbb{R}$), but $p(10) = 111 = 3\times 37$, so this factorization cannot be &quot;detected&quot; by factoring $p(x)$. On the other hand, $q(x) = x^4 - 1 = (x-1)(x^3+x^2+x+1)$ would not detect factors of $q(2)=15$. If you have to start rummaging around for polynomials, trying several to see if you can detect a nontrivial factorization, even if you eventually &quot;hit&quot; on one that gives the answer the ease of factoring the polynomial may be overshadowed by the difficulty in finding a &quot;good&quot; polynomial in the first place, so that the entire thing may be a wash.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-10T05:34:25.280" LastActivityDate="2011-03-10T05:34:25.280" CommentCount="3" />
  <row Id="26137" PostTypeId="2" ParentId="26135" CreationDate="2011-03-10T05:20:08.750" Score="7" ViewCount="" Body="&lt;p&gt;As was mentioned there are (theoretical) polynomial time algorithms for factoring polynomials such as LLL. However, they do not perform well in practice, so various other algorithms are typically employed (e.g. Berlekamp, Zippel's sparse modular, etc). For some references see &lt;a href=&quot;http://math.stackexchange.com/questions/868/what-is-the-best-way-to-factor-arbitrary-polynomials/904#904&quot;&gt;my answer&lt;/a&gt; to a prior question here on the best ways to factor polynomials.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The method you propose seems related to an old method due to Bernoulli, Schubert, Kronecker, Hausman, et al. You can find a nice presentation of this algorithm with worked examples in Sec.16.1 pp. 347ff in: Mora. &lt;em&gt;Solving polynomial equation systems, I. The Kronecker-Duval Philosophy&lt;/em&gt;. This may be accessible by Google books. To find said section quickly simply search for &quot;Schubert&quot;. See also Mignotte's interesting paper [2] on the history (abstract appended below).&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;ABSTRACT.&lt;/strong&gt;  &lt;em&gt;The First General Method of Factorization of Polynomials.&lt;br&gt;&#xA;  On a Memoir of F.T. Schubert.&lt;/em&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;We analyse two little known papers of N. Bernoulli (1708) and F.T. Schubert (1794) on the factorization of integer polynomials as well as the work of L. Kronecker and B.A. Hausmann on the same topic. The factorization method of Bernoulli-Schubert uses the calculus and the interpolation of finite differences. It was rediscovered by Kronecker (1882), who used Lagrange interpolation. Both procedures allow the effective factorization of polynomials having small degrees and coefficients. An algorithm combining the results of Bernoulli-Schubert and Kronecker was obtained by B.A. Hausmann. His method is particularly useful for the factorization of stable polynomials. The three methods are briefly compared with modern factorization algorithms.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;[1] Edwards, H. Kronecker's Algorithmic Mathematics&lt;br&gt;&#xA;(Presented at &quot;Computability in Europe 2008,&quot; Athens, June 19, 2008.)&lt;br&gt;&#xA;&lt;a href=&quot;http://www.math.nyu.edu/faculty/edwardsd/athens.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.math.nyu.edu/faculty/edwardsd/athens.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[2] Maurice Mignotte; Doru Stefanescu&lt;br&gt;&#xA;La première méthode générale de factorisation des polynômes.&lt;br&gt;&#xA;Autour d'un mémoire de F.T. Schubert&lt;br&gt;&#xA;Revue d'histoire des mathématiques 7, fascicule 1 (2001), 67-89&lt;br&gt;&#xA;&lt;a href=&quot;http://smf4.emath.fr/Publications/RevueHistoireMath/7/html/smf_rhm_7_67-89.html&quot; rel=&quot;nofollow&quot;&gt;http://smf4.emath.fr/Publications/RevueHistoireMath/7/html/smf_rhm_7_67-89.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2011-03-10T06:29:47.507" LastActivityDate="2011-03-10T06:29:47.507" />
  <row Id="26139" PostTypeId="2" ParentId="26118" CreationDate="2011-03-10T05:58:44.883" Score="3" ViewCount="" Body="&lt;p&gt;You may want to look into the proof assistant &lt;a href=&quot;http://isabelle.in.tum.de/index.html&quot; rel=&quot;nofollow&quot;&gt;Isabelle&lt;/a&gt;. It's freely available under a BSD license, and is a start towards the sort of thing your looking for. It's an interactive theorem proving environment that offers a language, Isar, for expressing formal proofs, some built in automated tools, and a sledgehammer tool that will run several external automated theorem provers in parallel. However, I have only just begun to use this system this week, so I am unable to verify for myself if it can handle the simple proof from your question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This &lt;a href=&quot;http://arxiv.org/abs/cs.AI/0509025&quot; rel=&quot;nofollow&quot;&gt;paper&lt;/a&gt; outlines the work that went into producing a formally verified proof of the prime number theorem using Isabelle. The authors discuss some of the difficulties they had in proving small lemmas, but they seem confident that the available automated tools will eventually be strong enough to handle these sorts of things. Since this paper was written in 2005, it's conceivable that sufficiently strong tools are available today, but I am unaware of the extent of Isabelle's current capabilities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: With some coaxing, I was able to produce a partially automated proof of your result using Isabelle. I found it more challenging to get Isabelle to prove this result than to simply prove it myself though. At the moment this probably isn't a practical means of saving labor. Perhaps someone expert in this system with access to a wider library of relevant theorems and lemmas could&#xA;automate the proofs of a wide class of routine results, but I don't think the functionality you want is quite there yet. Nevertheless, you might find it more fun to try to get a computer to check an apparently routine conjecture than to try to grind through a proof yourself. &lt;/p&gt;&#xA;" OwnerUserId="7323" LastEditorUserId="7323" LastEditorDisplayName="" LastEditDate="2011-03-10T19:53:26.970" LastActivityDate="2011-03-10T19:53:26.970" />
  <row Id="26141" PostTypeId="1" CreationDate="2011-03-10T06:06:19.583" Score="0" ViewCount="34" Body="&lt;p&gt;Attach a (Laurent) monomial weight $x_1^{i_1} \cdots x_n^{i_n}$ to each point $(i_1, \dots, i_n)$ of $\mathbb{Z}^{n}$ and call it $\mathbb{Z}^{n}[x_1, x_{1}^{-1}, \dots, x_n, x_n^{-1}]$. Does this formal space have a name (other than the title) or any special properties?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand that &lt;a href=&quot;http://en.wikipedia.org/wiki/Newton_polygon&quot; rel=&quot;nofollow&quot;&gt;Newton polygons&lt;/a&gt; are studied using this space. How else is it useful?&lt;/p&gt;&#xA;" OwnerUserId="2720" LastEditorUserId="2720" LastEditorDisplayName="" LastEditDate="2011-03-10T06:25:56.993" LastActivityDate="2011-03-10T06:29:46.230" Title="Applications of the Formal Laurent Lattice" Tags="&lt;combinatorics&gt;&lt;algebraic-geometry&gt;&lt;commutative-algebra&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="26142" PostTypeId="1" AcceptedAnswerId="26162" CreationDate="2011-03-10T06:10:20.720" Score="2" ViewCount="84" Body="&lt;p&gt;Let $X_1,X_2,...$ be a sequence of integer-valued random variables that converge &lt;em&gt;in distribution&lt;/em&gt; to some random variable $X$. Am I right in thinking that we can always pick $X$ to be integer valued?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I thought like this: Let $F_n$ be the distribution function of $X_n$ and $F$ be the distribution function of $X$. Then $X$ cannot be integer valued only if there are two points $x_1$ and $x_2$ such that&#xA;$$F(x_1) \neq F(x_2)$$ &#xA;and&#xA;$$|x_1-x_2|&amp;lt;1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now since a distribution function cannot have more than a countable number of points of discontinuity, we can pick $x_1$ and $x_2$ such that they are points of continuity of $F$. But then for large enough $n$&#xA;$$F_n(x_1) \neq F_n(x_2)$$&#xA;But this contradicts the fact that $X_n$ is integer-valued.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Am I right?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[This is motivated by exercise 5.12 of Wasserman's &lt;em&gt;All of Statistics&lt;/em&gt; where he assumes that $X$ is integer-valued]&lt;/p&gt;&#xA;" OwnerUserId="1195" LastActivityDate="2011-03-10T08:16:03.817" Title="Integer-valued random variables must converge in distribution to a integer-valued random variable?" Tags="&lt;probability-theory&gt;" AnswerCount="3" FavoriteCount="0" />
  <row Id="26143" PostTypeId="1" AcceptedAnswerId="26168" CreationDate="2011-03-10T06:29:43.870" Score="3" ViewCount="137" Body="&lt;p&gt;I have $n$ points $(x_1, y_1), (x_2, y_2), \ldots, (x_n, y_n)$ all located in the unit square $[0,1] \times [0,1]$. I am trying to compute the largest distance from a point in the unit square to the closest $(x_i,y_i)$. Is there some sort of formula for this quantity in terms of the $x_i,y_i$? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks. &lt;/p&gt;&#xA;" OwnerUserId="8059" LastActivityDate="2011-03-10T08:52:15.940" Title="Formula for the largest distance to a set of points" Tags="&lt;elementary-geometry&gt;" AnswerCount="1" />
  <row Id="26144" PostTypeId="2" ParentId="26141" CreationDate="2011-03-10T06:29:46.230" Score="1" ViewCount="" Body="&lt;p&gt;It is the group algebra of the group $\mathbb Z^n$, so it shows up in lots of places. One place where you see it a lot is in &lt;em&gt;toric geometry&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;What exactly do you want to know?&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2011-03-10T06:29:46.230" CommentCount="1" />
  <row Id="26145" PostTypeId="1" AcceptedAnswerId="26148" CreationDate="2011-03-10T06:35:18.300" Score="2" ViewCount="107" Body="&lt;p&gt;Hello,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Problem:&lt;br&gt;&#xA;  If $x^{n-1} \equiv 1 \pmod{n}$, and for all divisors $q$ of $n - 1$, $a^{q} \not\equiv 1 \pmod{n}$, then $n$ is prime. $(n &amp;gt; 1)$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I read the proof in the book and it was very straightforward; however, I wonder is there a way to prove it by just using congruence property?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And another related question about power residue:&lt;br&gt;&#xA;If we have $a^{n - 1} \equiv 1 \pmod{n}$. Is there any relation between $n - 1$ and $\phi(n)$?  Because I thought of $a^{\phi(n)} \equiv 1 \pmod{n}$, when $(a, n) = 1$. I try to think of away to connect these two ideas, but I still cannot see it.&lt;br&gt;&#xA;Any idea?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks,  &lt;/p&gt;&#xA;" OwnerUserId="5711" LastEditorUserId="5711" LastEditorDisplayName="" LastEditDate="2011-03-10T07:24:10.697" LastActivityDate="2011-03-11T06:18:45.730" Title="How to prove Lucas's Converse of Fermat's Little Theorem without using primitive root?" Tags="&lt;elementary-number-theory&gt;&lt;modular-arithmetic&gt;" AnswerCount="2" />
  <row Id="26146" PostTypeId="2" ParentId="26142" CreationDate="2011-03-10T06:37:15.440" Score="1" ViewCount="" Body="&lt;p&gt;The CDFs $F_i$ all have the property that $F_i$ is constant on $(n,n+1)$ for every integer $n$. This property is conserved by the limiting function, and is in fact equivalent to the random variable being supported on the integers.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2011-03-10T06:37:15.440" CommentCount="4" />
  <row Id="26147" PostTypeId="2" ParentId="26142" CreationDate="2011-03-10T06:40:37.167" Score="0" ViewCount="" Body="&lt;p&gt;$X$ has to be an integer valued random variable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$X_n$ converges to $X$ in distribution if $F_n$ is the cdf of $X_n$ and if $F$ is the cdf of $X$, then $$\lim_{n \rightarrow \infty} F_n(x) = F(x)$$ Note that if $X_n$ is an integer valued random variable then $F_n(x) = F_n(\lfloor x \rfloor)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider $|F(x) - F(\lfloor x \rfloor)|$ and argue that this can be made smaller than $\epsilon$, $\forall$ $\epsilon &amp;gt;0$, by adding and subtracting $F_n(x)$ and making use of the fact that $F_n(x) = F_n(\lfloor x \rfloor)$ and then use triangle inequality and choose $N(\epsilon)$ such that both the $\frac{\epsilon}{2}$ are satisfied.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastActivityDate="2011-03-10T06:40:37.167" CommentCount="1" />
  <row Id="26148" PostTypeId="2" ParentId="26145" CreationDate="2011-03-10T06:42:22.267" Score="1" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\ \ $ The conditions imply that the order $\rm\:k\:$ of $\rm\:x\:$ is a divisor of $\rm\:n-1\:$ but not a &lt;em&gt;proper&lt;/em&gt; divisor, therefore $\rm\: k = n-1\:.\ $  By Euler, $\rm\ k\ |\ \phi(n)\ $ so $\rm\ n-1\: \le\: \phi(n)\:.\ $ This implies that $\rm\:n\:$ is prime, since $\rm\ \phi(n) \:\le\: n-\color{red}{2}\ $ for composite $\rm\:n\:,\:$ since they have at least $\:\color{red}2\ $ smaller naturals non-coprime to $\rm\:n\:.$&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2011-03-10T06:42:22.267" CommentCount="5" />
  <row Id="26149" PostTypeId="1" CreationDate="2011-03-10T06:49:37.580" Score="4" ViewCount="60" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Proofs_of_quadratic_reciprocity&quot; rel=&quot;nofollow&quot;&gt;Eisenstein's proof&lt;/a&gt; of the Quadratic Reciprocity (QR) (and its Jacobi symbol generalization) both rely on counting lattice points in two congruent triangles. If we take $t$-dilates of these triangles, does an extension of Gauss' Lemma still hold with some dependence on $t$? That is, can we assign a Legendre symbol $(p|q)$, or some simple extension thereof depending on $t$, to $(-1)^{M(t)}$, where $M(t)$ is the number of lattice points in the &lt;em&gt;$t$-dilate&lt;/em&gt; of the same triangle used in the standard proof and derive from it a &quot;$t$-dilated&quot; version of QR?&lt;/p&gt;&#xA;" OwnerUserId="2720" LastEditorUserId="2720" LastEditorDisplayName="" LastEditDate="2011-03-11T17:06:13.767" LastActivityDate="2011-03-11T17:06:13.767" Title="Generalizing Quadratic Reciprocity Law with Dilates" Tags="&lt;number-theory&gt;&lt;combinatorics&gt;" CommentCount="2" FavoriteCount="2" />
  <row Id="26150" PostTypeId="1" AcceptedAnswerId="27370" CreationDate="2011-03-10T07:02:07.820" Score="5" ViewCount="125" Body="&lt;p&gt;Given a bipartite graph G with two partition sets $U$ and $V$ of the same size $n$. In each set, there are d vertices of degree (n-d+1), and n-d vertices of degree (n-d). Can we find a perfect matching in this graph?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So basically, the two sets are the same in terms of degree sequence, although one has no idea how vertices are adjacent to each other. I tried to use Hall's theorem here, but i got stuck. Take any subset S of U that contain some vertices of degree n-d+1 and some of degree n-d, then I am not be able to show $|N(S)| \geq |S|$.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;EDIT: The first question is solved and now a more general and perhaps research-oriented question is: What if in each set $U$ and $V$, there are $d$ vertices of degree AT LEAST $(n-d+1)$, and $n-d$ vertices of degree AT LEAST $(n-d)$. Can we still find a perfect matching?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The method for the previous question fails here.&lt;/p&gt;&#xA;" OwnerUserId="8057" LastEditorUserId="8057" LastEditorDisplayName="" LastEditDate="2011-04-06T05:45:34.273" LastActivityDate="2011-04-06T05:45:34.273" Title="Perfect Matching in a bipartite graph with a constrained degree sequence" Tags="&lt;graph-theory&gt;" AnswerCount="1" />
  <row Id="26151" PostTypeId="1" AcceptedAnswerId="26158" CreationDate="2011-03-10T07:04:02.440" Score="6" ViewCount="96" Body="&lt;p&gt;I'm working through some of the exercises in &lt;em&gt;generatingfunctionology&lt;/em&gt;. One of the questions is to find the generating function where the $n$th term $a_n=P(n)$ for $P$ a polynomial. The answer is $P(xD)(1/(1-x))$. I thought it meant something like if $P(n)=3n+n^2$, then $P(xD)=3xD+x^2D^2$, and then you would apply that to $1/(1-x)$. However, I know the generating function for $a_n=n^2$ is $$\frac{x(x+1)}{(1-x)^3}$$&#xA;which is not the same as $x^2D^2(1/(1-x))$. So what does it mean? Thanks.&lt;/p&gt;&#xA;" OwnerUserId="6742" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2011-03-10T19:11:50.793" LastActivityDate="2011-03-10T19:11:50.793" Title="In generatingfunctionology, for a polynomial $P$ and a differential operator $D$, what does $P(xD)$ mean?" Tags="&lt;notation&gt;&lt;generating-functions&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" />
  <row Id="26152" PostTypeId="1" AcceptedAnswerId="26154" CreationDate="2011-03-10T07:05:53.080" Score="1" ViewCount="61" Body="&lt;p&gt;Hello,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In my textbook, they said:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$2x^{3} + 7x - 4 \equiv 0 \pmod{5}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The solution of this equation are the integers with $x \equiv 1 \pmod{5}$, as can be seen by testing $x = 0, 1, 2, 3,$ and $4.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And I have no clue how do they had $x \equiv 1 \pmod{5}$. I tested as they suggested:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $y = 2x^{3} + 7x - 4$, we have:&lt;br&gt;&#xA;$$x = 0: y = -4, \implies y \equiv 1 \pmod{5}$$&#xA;$$x = 1: y = 5, \implies y \equiv 0 \pmod{5}$$&#xA;$$x = 2: y = 26, \implies y \equiv 1 \pmod{5}$$&#xA;$$x = 3: y = 71, \implies y \equiv 1 \pmod{5}$$&#xA;$$x = 4: y = 152, \implies y \equiv 2 \pmod{5}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I did not understand is how these five modulo equations become $x \equiv 1 \pmod{5}$? Can anyone help me explain this?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks, &lt;/p&gt;&#xA;" OwnerUserId="5711" LastActivityDate="2011-03-10T07:56:14.603" Title="Question regarding solving polynomial of congruence?" Tags="&lt;elementary-number-theory&gt;&lt;modular-arithmetic&gt;" AnswerCount="4" />
  <row Id="26153" PostTypeId="1" AcceptedAnswerId="26208" CreationDate="2011-03-10T07:08:31.627" Score="0" ViewCount="53" Body="&lt;p&gt;I want to find an $n$-vector $\hat{\theta}$ that maximizes a function $f(\theta)$ subject to the $p$-norm constraint $||\theta||_p = c$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a general parametrization of $p$-norm hyperspheres that makes this easy?&lt;/p&gt;&#xA;" OwnerUserId="8058" LastActivityDate="2011-03-10T16:01:02.157" Title="parametrize hypersphere" Tags="&lt;optimization&gt;&lt;vector-space&gt;&lt;norm&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="26154" PostTypeId="2" ParentId="26152" CreationDate="2011-03-10T07:09:22.297" Score="2" ViewCount="" Body="&lt;p&gt;Any integer is going to be congruent to one of $0,1,2,3,4$ modulo $5$. As the testing shows, only those integers which are congruent to $1$ modulo $5$ are solutions, and any integer congruent to $1$ mod $5$ is a solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Those tests are using an arbitrary $x$ each time, so you shouldn't think of them as a system of congruences of a single variable $x$ but instead they imply the result described above.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Here's another thing to add which may help. Don't just check $0,1,2,3,4$. Take $x$ to be any arbitrary integer, not necessarily one of those just listed. So like I said before, this arbitrary $x$ is congruent to one of $0,1,2,3,4$ since they form a complete residue system. If $x\equiv 0\pmod{5}$, then&#xA;$$&#xD;&#xA;2x^3+7x-4\equiv 2(0)^3+7(0)-4\equiv -4\equiv 1\pmod{5}&#xD;&#xA;$$&#xA;so $x$ is not a solution. Likewise if $x\equiv 2,3,4\pmod{5}$. So the only choice which works is when $x\equiv 1\pmod{5}$. Hopefully that's clearer?&lt;/p&gt;&#xA;" OwnerUserId="1583" LastEditorUserId="1583" LastEditorDisplayName="" LastEditDate="2011-03-10T07:17:24.520" LastActivityDate="2011-03-10T07:17:24.520" CommentCount="5" />
  <row Id="26155" PostTypeId="2" ParentId="26152" CreationDate="2011-03-10T07:10:11.090" Score="2" ViewCount="" Body="&lt;p&gt;What your textbook means is $$\left(2x^{3} + 7x - 4 \equiv 0 \pmod{5} \right) \iff \left(x \equiv 1 \pmod{5} \right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is what you checked by plugging in the different cases for $x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By taking $x \equiv 1 \pmod{5}$, you proved that $2x^{3} + 7x - 4 \equiv 0 \pmod{5}$ and hence $$ \left(x \equiv 1 \pmod{5} \right) \Rightarrow \left(2x^{3} + 7x - 4 \equiv 0 \pmod{5} \right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now by taking $x \neq 1 \pmod{5}$, you proved that $2x^{3} + 7x - 4 \neq 0 \pmod{5}$ and hence $$\left(2x^{3} + 7x - 4 \equiv 0 \pmod{5} \right) \Rightarrow \left(x \equiv 1 \pmod{5} \right)$$&lt;/p&gt;&#xA;" OwnerUserId="3302" LastActivityDate="2011-03-10T07:10:11.090" CommentCount="1" />
  <row Id="26156" PostTypeId="2" ParentId="26151" CreationDate="2011-03-10T07:14:15.287" Score="6" ViewCount="" Body="&lt;p&gt;Without checking out the book, this would be my guess:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let for instance $P(x) = x^2 + 2x$, then $P(xD) = (xD)^2 + 2xD$, which is an operator.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let this operator work on your function $f(x) = \frac{1}{1-x}$. It's obvious that $(2xD)(f) = 2xf&amp;#39;$. (Here $f&amp;#39;$ is a bit an abuse of notation for $f&amp;#39;(x)$.) The other term is probably the confusing one. But if you write it as $xDxDf$ and work everything out from the right to the left you'll get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ xD(xf&amp;#39;) = x(1f&amp;#39;+xf&amp;#39;&amp;#39;) = xf&amp;#39; + x^2 f&amp;#39;&amp;#39; .$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And similarly for higher order terms, so with my example polynomial $P$ we have that $P(xD)(f) = 2xf&amp;#39; + xf&amp;#39; + x^2 f&amp;#39;&amp;#39;$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in summary your problem might be that $(xD)^2 = xDxD = xD(xD) = xD + x^2D^2 \neq x^2 D^2$, in other words $x$ and $D$ don't commute.&lt;/p&gt;&#xA;" OwnerUserId="5189" LastActivityDate="2011-03-10T07:14:15.287" />
  <row Id="26157" PostTypeId="2" ParentId="26152" CreationDate="2011-03-10T07:15:42.630" Score="2" ViewCount="" Body="&lt;p&gt;$$\begin{aligned}&#xD;&#xA;2x^3+7x-4 \equiv 0 ( mod 5)\\&#xD;&#xA;2x^3+2x-4 \equiv 0 ( mod 5)\\&#xD;&#xA;2(x-1)(x^2+x+2) \equiv 0 ( mod 5) &#xD;&#xA;\end{aligned}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now you have two solutions $x-1 \equiv 0 (mod5)$ or $x^2+x+2 \equiv 0 (mod5)$.&#xA;You can continue and verify if $x \equiv 1 ( mod 5)$ is the only solution.&lt;/p&gt;&#xA;" OwnerUserId="8057" LastActivityDate="2011-03-10T07:15:42.630" CommentCount="2" />
  <row Id="26158" PostTypeId="2" ParentId="26151" CreationDate="2011-03-10T07:42:25.427" Score="11" ViewCount="" Body="&lt;p&gt;The problem is that you've mistakenly assumed commutativity. Here $\rm\:x\:$ and $\rm\:D\:$ do not commute. Indeed $\rm\ D\ x = x\ D + 1\ $ since $\rm\ (D\ x)\ f\ =\ D\ (x\ f\:)\ =\ x\ f\:&amp;#39;+f\ =\ (x\ D + 1)\ f\:.\ $ So it's not true that $\rm\ (x\ D)^2\: =\ x^2\ D^2\:.\:$ Instead $\rm\ x\ D\ x\ D\ =\ x\ (x\ D + 1)\ D\ =\ x^2\ D^2 + x\ D\:.\ $ Generally&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\rm (x\ D)^n\ =\ \sum_{k\ =\ 0}^n\ S(n,k)\ x^k\ D^k $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\rm\ S(n,k)\ $ are the Stirling numbers of the second kind.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Beware that because this operator algebra is noncommutative, many familiar polynomial identities do not hold true. For example, consider the proof of the difference of squares identity&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\rm\ (x-y)\ (x+y)\ =\ x^2 + x\ y - y\ x - y^2\ =\ x^2 - y^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice how commutativity is assumed to cancel the middle terms. Thus the identity needn't remain true if we substitute elements from some noncommutative ring, e.g. we cannot substitute $\rm\:D\:$ for $\rm\:y\:.\:$ Said slightly more technically, the polynomial evaluation map is a ring homomorphism iff coefficients and indeterminates commute. See &lt;a href=&quot;http://math.stackexchange.com/questions/4270/is-a-field-ring-an-algebra-over-itself/4300#4300&quot;&gt;my post here&lt;/a&gt; for further remarks on this topic.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2011-03-10T08:32:41.343" LastActivityDate="2011-03-10T08:32:41.343" />
  <row Id="26159" PostTypeId="2" ParentId="4621" CreationDate="2011-03-10T07:48:53.223" Score="0" ViewCount="" Body="&lt;p&gt;I was wrestling with some of these same questions myself earlier and never really found completely solid answers for what makes a good curve.  I can, however, share my functions which I wound up using.  They're all constant-power ones, but with different levels of fade/cut.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To understand what makes a curve constant-power, you have to understand that the signal is a sound-pressure level signal and that power goes as sound pressure squared.  So if we have input signal $w_1$ and $w_2$ and we're attenuating the signal by multiplying $w_1 \cdot f(x)$ and $w_2 \cdot f(1-x)$ then $f$ is constant-power if $f^2(x)+f^2(1-x) = 1$.  So, in this case, the easiest way would be to make $f(x) = cos(\frac{\pi}{2}x)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, we can generalize this and say that any function $g$ with range [0,1] for domain [0,1] can be used to produce a constant-power crossfade function $f(x)=cos(\frac{\pi}{2}g(x))$.  However, it'll make the most sense if $g(0) = 0$, $g(0.5)=0.5$, $g(1)=1$, and $g$ is monotonic.  With this in mind, I more looked at functions $h$ where $h(-1) = -1$, $h(0)=0$, and $h(1)=1$ and then just did a simple linear transform between [-1,1] and [0,1].  So, the first thing I tried was $h(x)=x^{2n+1}$ for non-negative integers n.  This turned out to work quite well.  $n=0$ gives the constant power curve you show above and then as I go to $n=1,2,3$ I get curves a lot like slow fade, slow cut, and fast cut (although not identical since these are all constant power).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So my final function is $f(x)=cos(\frac{\pi}{4}((2x-1)^{2n+1}+1))$.  I don't have the graphs handy, but if you graph them, you'll see graphs similar to slow fade, slow cut, and fast cut, except that the plateau in the middle is at about 0.7 rather than at 1 and the track which doesn't fall off rises to 1 at the outside edge.  I've also tried them out in a software fader in Pure Data, and they sound pretty good to my ears.  I was also happy because it meant that I could switch between the four different curves with only a single parameter so that keeps the logic simple.&lt;/p&gt;&#xA;" OwnerUserId="8060" LastActivityDate="2011-03-10T07:48:53.223" />
  <row Id="26160" PostTypeId="1" CreationDate="2011-03-10T07:55:47.227" Score="11" ViewCount="351" Body="&lt;p&gt;Is there some proven statement that tells us &quot;not all X satisfy Y&quot;, but there are currently no examples of X which do not satisfy Y?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it necessarily true that there are always explicit counterexamples? Or are there statements that only guarantee the existence of such counterexamples, whose construction is impossible in ZFC.&lt;/p&gt;&#xA;" OwnerUserId="8062" LastActivityDate="2011-03-16T20:36:19.890" Title="Statements with no counterexample" Tags="&lt;logic&gt;" AnswerCount="4" FavoriteCount="3" />
  <row Id="26161" PostTypeId="2" ParentId="26152" CreationDate="2011-03-10T07:56:14.603" Score="1" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\rm\ \ x\ p(x)\ =\ 2\ x^4 + 7\ x^2 - 4\ x\ \equiv\ 2\ (x-1)^2\ \ (mod\ 5)\ $ for $\rm\ x\not\equiv 0\ $ since then $\rm\ x^4\equiv 1\ $&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2011-03-10T07:56:14.603" />
  <row Id="26162" PostTypeId="2" ParentId="26142" CreationDate="2011-03-10T07:58:48.717" Score="1" ViewCount="" Body="&lt;p&gt;There are problems with your proof.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I thought like this: Let $F_n$ be the&#xA;  distribution function of $X_n$ and $F$&#xA;  be the distribution function of $X$.&#xA;  Then $X$ cannot be integer valued only&#xA;  if there are two points $x_1$ and&#xA;  $x_2$ such that $$F(x_1) \neq F(x_2)$$&#xA;  and $$|x_1-x_2|&amp;lt;1$$ (**)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Not necessarily. What if the range of $X$ are non-integers with spaces between them bigger or equal to 1? For example , all the points $n+1/2$, $n\in \mathbb{Z}$ &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Now since a distribution function&#xA;  cannot have more than a countable&#xA;  number of points of discontinuity, we&#xA;  can pick $x_1$ and $x_2$ such that&#xA;  they are points of continuity of $F$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You pick such points of continuity, but at the same time you pick them with the property (**) from the first part. You have to justify that.&lt;/p&gt;&#xA;" OwnerUserId="7243" LastEditorUserId="7243" LastEditorDisplayName="" LastEditDate="2011-03-10T08:16:03.817" LastActivityDate="2011-03-10T08:16:03.817" CommentCount="3" />
  <row Id="26163" PostTypeId="2" ParentId="26145" CreationDate="2011-03-10T08:00:48.457" Score="1" ViewCount="" Body="&lt;p&gt;Do you understand the definition of $\phi(n)$?  It's the number of natural numbers less than $n$ which are relatively prime to $n$. If $n$ is prime, then $\phi(n)$ is necessarily $n-1$ (since only 1 is relatively prime with $n$).  Likewise, if $\phi(n) = n-1$ then only one number less than $n$ is relatively prime to it, which must be the number 1, so $n$ is prime.  So that would be the relationship between them.&lt;/p&gt;&#xA;" OwnerUserId="8060" LastEditorUserId="8060" LastEditorDisplayName="" LastEditDate="2011-03-11T06:18:45.730" LastActivityDate="2011-03-11T06:18:45.730" CommentCount="2" />
  <row Id="26164" PostTypeId="2" ParentId="26132" CreationDate="2011-03-10T08:09:15.540" Score="2" ViewCount="" Body="&lt;p&gt;There's not enough information to give a probability for team A winning. You can see this by imagining that you know &lt;em&gt;a priori&lt;/em&gt; that the probability for team A to win is $0$ or $1$ -- then this won't change if you listen to the seers. The information from the seers can only modify your &lt;em&gt;a priori&lt;/em&gt; knowledge, insofar as it's incomplete, but it can't give you a probability for team A to win if you don't already have some assumption about that probability.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Having said that, on the assumption that &lt;em&gt;a priori&lt;/em&gt; the teams are equally likely to win (a &quot;flat prior&quot;), it's straightforward to calculate the conditional probability of team A winning given the information from the seers:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For team A to win, the seers would have had to roll dice with probabilities 80%, 40% and 30%, respectively. For team A to lose, they would have had to roll dice with probabilities 20%, 60% and 70%, respectively. So the conditional probability for team A to win given the information from the seers is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{0.8\cdot0.4\cdot0.3}{0.8\cdot0.4\cdot0.3 + 0.2\cdot0.6\cdot0.7}\approx 53\%\;.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Update in response to the added last sentence in the question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The generalization to any number of seers with any probability of lying with any set of predictions is straightforward: Given $n$ seers with probability $p_i$ of lying and predictions $b_i\in\{0,1\}$ of team A winning, again assuming a 50/50 prior, the probability of team A winning is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\prod_i \left(b_i(1-p_i)+(1-b_i)p_i\right)}{\prod_i \left(b_i(1-p_i)+(1-b_i)p_i\right)+\prod_i \left(b_ip_i+(1-b_i)(1-p_i)\right)}\;.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Further update:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can also generalize this to an arbitrary prior by taking into account the prior probabilities. Assuming that &lt;em&gt;a priori&lt;/em&gt; the probability for team A to win is $p$, the conditional probability given the information from the seers is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{p\prod_i \left(b_i(1-p_i)+(1-b_i)p_i\right)}{p\prod_i \left(b_i(1-p_i)+(1-b_i)p_i\right)+(1-p)\prod_i \left(b_ip_i+(1-b_i)(1-p_i)\right)}\;,$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and you recover the above result by setting $p=1/2$. You can also get the two cases where your &lt;em&gt;a priori&lt;/em&gt; knowledge is complete and hence isn't modified by the new information from the seers by choosing $p=0$ or $p=1$, in which case the conditional probability also comes out as $0$ and $1$, respectively.&lt;/p&gt;&#xA;" OwnerUserId="6622" LastEditorUserId="6622" LastEditorDisplayName="" LastEditDate="2011-03-10T08:31:47.737" LastActivityDate="2011-03-10T08:31:47.737" CommentCount="1" />
  <row Id="26165" PostTypeId="2" ParentId="26160" CreationDate="2011-03-10T08:22:50.503" Score="8" ViewCount="" Body="&lt;p&gt;The statement &quot;not all X satisfy Y&quot; ($\lnot \forall X\,Y(X)$) is equivalent to &quot;there is an X such that Y is not true of X&quot; ($\exists X\,\lnot Y(X)$), which is equivalent to &quot;there is an X such that 'not Y' is true of X&quot; ($\exists X\,(\lnot Y)(X)$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is, for any proposition $Y$, asking for an explicit counterexample for $Y$ is the same as asking for an explicit example of the negation of $Y$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are many existence results for which no explicit example is known. In particular, for any such result relying strongly on the Axiom of Choice, we cannot explicitly construct an example. And any such existence result can be rephrased as an example of what you are asking for simply by taking the negation of the property in question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, &quot;There exists a nonmeasurable subset of $\mathbb{R}$&quot; can be rephrased as &quot;Not all subsets of $\mathbb{R}$ are measurable.&quot;&lt;/p&gt;&#xA;" OwnerUserId="7062" LastActivityDate="2011-03-10T08:22:50.503" CommentCount="2" />
  <row Id="26166" PostTypeId="1" CreationDate="2011-03-10T08:26:11.517" Score="-1" ViewCount="221" Body="&lt;p&gt;My friend got some job offers:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Go to a tier 3 college for teaching;&lt;/li&gt;&#xA;&lt;li&gt;Go to a company, apply for academic jobs again next year;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Can anyone that have some experience comments that, from career development point of view, which choice is better?His career goal is in academic instituition.&#xA;Thank you!&lt;/p&gt;&#xA;" OwnerUserId="1944" LastActivityDate="2011-05-23T05:53:19.567" ClosedDate="2011-03-10T23:25:14.387" Title="which choice is better(career)" Tags="&lt;soft-question&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="26167" PostTypeId="1" CreationDate="2011-03-10T08:49:27.320" Score="6" ViewCount="285" Body="&lt;p&gt;Given $n$ independent geometric random variables $X_n$, each with probability parameter $p$ (and thus expectation $E\left(X_n\right) = \frac{1}{p}$), what is&#xA;$$E_n = E\left(\max_{i \in 1 .. n}X_n\right)$$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;If we instead look at a continuous-time analogue, e.g. exponential random variables $Y_n$ with rate parameter $\lambda$, this is simple:&#xA;$$E\left(\max_{i \in 1 .. n}Y_n\right) = \sum_{i=n}^1\frac{1}{i\lambda}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I think this is right... that's the time for the first plus the time for the second plus ... plus the time for the last.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I can't find something similarly nice for the discrete-time case.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;What I &lt;em&gt;have&lt;/em&gt; done is to contruct a Markov chain modelling the number of the $X_n$ that haven't yet &quot;hit&quot;. (i.e. at each time interval, perform a binomial trial on the number of $X_n$ remaining to see which &quot;hit&quot;, and then move to the number that didn't &quot;hit&quot;.) This gives&#xA;$$E_n = 1 + \sum_{i=0}^n \left(\begin{matrix}n\\i\end{matrix}\right)p^{n-i}(1-p)^iE_i$$&#xA;which gives the correct answer, but is a nightmare of recursion to calculate. I'm hoping for something in a shorter form.&lt;/p&gt;&#xA;" OwnerUserId="5001" LastEditorUserId="5001" LastEditorDisplayName="" LastEditDate="2011-03-10T10:02:52.723" LastActivityDate="2011-03-11T05:47:03.777" Title="Expectation of the maximum of IID geometric random variables" Tags="&lt;probability&gt;&lt;geometric-distribution&gt;" AnswerCount="3" CommentCount="4" FavoriteCount="1" />
  <row Id="26168" PostTypeId="2" ParentId="26143" CreationDate="2011-03-10T08:52:15.940" Score="1" ViewCount="" Body="&lt;p&gt;There is certainly a formula for this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\max_{(x,y)\in[0,1]\times[0,1]}\min_i\; d((x,y),(x_i,y_i))\;.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I suspect that what you're actually interested in is an algorithm for computing the value of that formula :-). I think your best bet will be to compute the &lt;a href=&quot;http://en.wikipedia.org/wiki/Delaunay_triangulation&quot; rel=&quot;nofollow&quot;&gt;Delaunay triangulation&lt;/a&gt;, take the maximum of the radii of the circumcircles, and compare that with the maximum on the boundary of the square, which I think you can get by intersecting the bisectors of the exterior Delaunay edges with the boundary and considering the corners of the square separately.&lt;/p&gt;&#xA;" OwnerUserId="6622" LastActivityDate="2011-03-10T08:52:15.940" CommentCount="2" />
  <row Id="26169" PostTypeId="1" CreationDate="2011-03-10T08:56:46.923" Score="0" ViewCount="69" Body="&lt;p&gt;Hi, &#xA;I have values from two sensors stored in two vectors A and B. They both represent values of the sensors at times TA and TB which is stored in two other vectors(since it is not uniform sampling) Both A and B represent the same data but A is shifted a bit to the right because of the delay in starting the sensors. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is, how do I calculate this delay and more importantly, how do I shift A to match B or vice versa, such that I can do a one-to-one correspondence of the data? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Right now, I am finding the peak of the cross-correlation function of A and B to find the offset (in number of samples, not time) and padding the arrays with zero upto that number, but I am 100% sure that is wrong. Problem is, I don't know how else.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PLEASE help!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="7101" LastActivityDate="2011-03-10T08:56:46.923" Title="Cross correlation in MATLAB HELP" Tags="&lt;matlab&gt;&lt;correlation&gt;" CommentCount="1" />
  <row Id="26170" PostTypeId="1" AcceptedAnswerId="26200" CreationDate="2011-03-10T09:03:30.003" Score="5" ViewCount="176" Body="&lt;p&gt;Hello All, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was working on this problem here below, but seem to not know a precise or clean way to show the proof to this question below. I had about a few ways of doing it, but the statements/operations were pretty loosely used. The problem is as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Show that ${\bf A}^{-1}$ exists if and only if the eigenvalues $ \lambda _i$ ,  $1 \leq i \leq n$ of $\bf{A}$ are all non-zero, and then ${\bf A}^{-1}$ has the eigenvalues given by $ \frac{1}{\lambda _i}$, $1 \leq i \leq n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" LastActivityDate="2011-03-10T17:35:36.460" Title="Matrix Inverses and Eigenvalues" Tags="&lt;linear-algebra&gt;&lt;matrices&gt;&lt;eigenvalues&gt;&lt;inverse&gt;" AnswerCount="4" FavoriteCount="1" />
  <row Id="26171" PostTypeId="1" AcceptedAnswerId="26184" CreationDate="2011-03-10T09:14:11.963" Score="6" ViewCount="93" Body="&lt;p&gt;I'm reading Hatcher and I did exercise 10 on page 19. Can you tell me if my answer is correct? Many thanks for your help!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Claim: $X$ contractible $\Leftrightarrow \forall$ arbitrary maps $f:X \rightarrow Y$, $Y$ arbitrary, $f \cong const.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proof:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\implies$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given the homotopy $H: I \times X \rightarrow X$, $H(0,x) = id_X$, $H(1,x) = id_{ \{ \ast \}}$ and an arbitrary map $f: X &amp;#160;\rightarrow Y$ construct a homotopy $H^\prime: I \times X \rightarrow Y$, $H^\prime(0,x) = f(x)$, $H^\prime(1,x)=const_{y_0 \in Y}$ as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$H^\prime (t,x) := f(H(t,x))$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that even though not stated in the exercise, $f$ is assumed to be continuous. (At least I think that's the case)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\impliedby$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given $\forall f:X \rightarrow Y$: $f \cong const.$ we pick $Y := X$ then $f:X \rightarrow X$ $\cong const_{x_0} \forall f$. Now pick $f := id_X$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second part of question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Claim: $X$ contractible $\iff \forall f: Y \rightarrow X$: $f \cong const_X$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proof:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\implies$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Define $H^\prime(t,x) := H(t, f(x))$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\impliedby$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Choose $Y:=X$ and $f:=id_X$.&lt;/p&gt;&#xA;" OwnerUserId="5798" LastEditorUserId="5798" LastEditorDisplayName="" LastEditDate="2011-03-10T09:33:13.373" LastActivityDate="2011-03-10T10:41:13.343" Title="Proof that $X$ contractible $\iff \forall f:X \rightarrow Y$ : $f \cong const.$" Tags="&lt;algebraic-topology&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="26172" PostTypeId="1" CreationDate="2011-03-10T09:16:20.193" Score="5" ViewCount="148" Body="&lt;p&gt;After my &lt;a href=&quot;http://math.stackexchange.com/questions/25763/optimising-closest-approach-with-third-order-motion&quot;&gt;last question&lt;/a&gt; I have worked through the math quite a bit and now I'm stuck again. This time my question is less wordy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have two equations for $t$, one with respect to each $a_{x}$ and $a_{y}$:&#xA;\begin{align*}&#xA;t&amp;amp;=\frac{-2(u_{x}^{A}-v_{x}^{B})\pm\sqrt{4(u_{x}^{A}-v_{x}^{B})^{2}-8a_{x}(c_{x}^{A}-c_{x}^{B})}}{2a_{x}}\\&#xA;t&amp;amp;=\frac{-2(u_{y}^{A}-v_{y}^{B})\pm\sqrt{4(u_{y}^{A}-v_{y}^{B})^{2}-8a_{y}(c_{y}^{A}-c_{y}^{B})}}{2a_{y}}&#xA;\end{align*}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Additionally $a_{x}$ can be specified as an inequality in terms of $a_{y}$:&#xA;$a_{x}\leq\sqrt{a_{max}^{2}-a_{y}^{2}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All the variables except $t$, $a_{x}$ and $a_{y}$ are constants. How can I find a function to yield the values for $a_{x}$ and $a_{y}$ that give the lowest value for $t$? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: and this is how I arrived there:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$s_{x}^{A}, s_{y}^{A}, s_{x}^{B}, s_{y}^{B}$ : Functions for position of particles A and B on the x and y axis&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$t$ : Time elapsed&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a_{x}^{A}, a_{y}^{A}$ : Acceleration of A on each axis&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a_{max}$ : Maximum magnitude of acceleration of A&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$c_{x}^{A}, c_{y}^{A}, c_{x}^{B}, c_{y}^{B}$: Starting positions of A and B (constant)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$u_{x}^{A}, u_{y}^{A}$ - Starting velocity of A (constant)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$v_{x}^{B}, v_{y}^{B}$ - Velocity of B (constant)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$s^{A}=u^{A}t + \frac{a^{A}t^{2}}{2} + c^{A}$ : Function for position of A in terms of time and acceleration on one axis&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$s^{B}=v^{B}t + c^{B}$ : Formula for position of B in terms of time on one axis&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a_{y}^{A}\le \sqrt{a_{max}^{2} - {a_{x}^{A}}^{2}}$ : Inequality yielding maximum acceleration on one axis in terms of acceleration on another&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Optimize for t where $s^{A}=s^{B}$ on both axis:&#xA;\begin{align*}&#xA;s^{A}&amp;amp;=s^{B}\\&#xA;u^{A}t + a^{A}t2/2 + c^{A} &amp;amp;= v^{B}t + c^{B}\\&#xA;u^{A}t + a^{A}t2/2 - v^{B}t &amp;amp;=  c^{B} - c^{A}\\&#xA;2u^{A}t + a^{A}t2 - 2v^{B}t &amp;amp;=  2c^{B} - 2c^{A}\\&#xA;a^{A}t2 + 2t(u^{A}-v^{B}) + (2c^{A}-2c^{B}) &amp;amp;= 0&#xA;\end{align*}&#xA;\begin{align*}&#xA;t&amp;amp;=\frac{-2(u^{A}-v^{B})\pm\sqrt{4(u^{A}-v^{B})^{2}-8a(c^{A}-c^{B})}}{2a}&amp;amp;&amp;amp;\mbox{(quadratic formula)}\\&#xA;t&amp;amp;=\frac{-2(u_{x}^{A}-v_{x}^{B})\pm\sqrt{4(u_{x}^{A}-v_{x}^{B})^{2}-8a_{x}(c_{x}^{A}-c_{x}^{B})}}{2a_{x}}&#xA;&amp;amp;&amp;amp;\mbox{For X}\\&#xA;t&amp;amp;=\frac{-2(u_{y}^{A}-v_{y}^{B})\pm\sqrt{4(u_{y}^{A}-v_{y}^{B})^{2}-8a_{y}(c_{y}^{A}-c_{y}^{B})}}{2a_{y}}&amp;amp;&amp;amp;\mbox{For Y}&#xA;\end{align*}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By my questionable reasoning these functions for $t$ should give how long it will take to reach the target for a given acceleration. The two values for acceleration could be plotted on the X and Z axis in a 3D space, with $t$ on the Y axis. The constraints to maximum acceleration would reduce the function to a circular disc. The points where the acceleration would never reach the target will be undefined or infinite or something. The lowest positive point on the 'disc' on the Y axis is a point where there is a valid combination of accelerations in each direction to reach the target, and should be the fastest. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Have I gone wrong somewhere? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT 2: &lt;strong&gt;It would be really super great if someone could point me in the right direction to learn about optimisation in 3D&lt;/strong&gt;. Even if you think I'm approaching the problem presented in my other question wrong, I'd still like to know how to do it, which is why I phrased this question the way I did.&lt;/p&gt;&#xA;" OwnerUserId="7953" LastEditorUserId="7953" LastEditorDisplayName="" LastEditDate="2011-03-11T05:23:14.890" LastActivityDate="2011-05-10T05:56:15.970" Title="Find minimum in a constrained three-variable equation" Tags="&lt;calculus&gt;&lt;optimization&gt;&lt;3d&gt;" AnswerCount="1" CommentCount="9" />
  <row Id="26173" PostTypeId="2" ParentId="26170" CreationDate="2011-03-10T09:17:58.647" Score="5" ViewCount="" Body="&lt;p&gt;Here is a short proof using the fact that the eigenvalues of ${\bf A}$ are precisely the solutions in $\lambda$ to the equation $\det ({\bf A}-\lambda {\bf I})=0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose one of the eigenvalues is zero, say $\lambda_k=0$. Then $\det ({\bf A}-\lambda_k {\bf I})=\det ({\bf A})=0$, so ${\bf A}$ is not invertible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, suppose all eigenvalues are nonzero. Then zero is &lt;em&gt;not&lt;/em&gt; a solution to the equation $\det ({\bf A}-\lambda {\bf I})=0$, from which we conclude that $\det({\bf A})$ cannot be zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll leave the second question to you.&lt;/p&gt;&#xA;" OwnerUserId="7676" LastEditorUserId="7676" LastEditorDisplayName="" LastEditDate="2011-03-10T09:24:50.763" LastActivityDate="2011-03-10T09:24:50.763" CommentCount="1" />
  <row Id="26174" PostTypeId="2" ParentId="26170" CreationDate="2011-03-10T09:22:39.237" Score="4" ViewCount="" Body="&lt;p&gt;a different approach from Joseph's (which also shows what the form of $A^{-1}$ is)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let us assume for simplicity that $A$ is diagonalizable (otherwise one can most probably extend the proof using the Jordan normal form). The matrix $A$ can be brought on the form $$A= T D T^{-1}$$ with $D = \text{diag}(\lambda_i)$ a diagonal matrix containing the eigenvalues, and $T$ and invertible matrix. The inverse of $A$ therefore reads&#xA;$$A^{-1} = (T D T^{-1})^{-1} = T D^{-1} T^{-1}.$$ This inverse exists if $D^{-1}$ exist. But $D^{-1}$ we can calculate easily. It is given by $D^{-1} =\text{diag}(\lambda_i^{-1})$ which exists as long as all $\lambda_i$ are nonzero.&lt;/p&gt;&#xA;" OwnerUserId="7266" LastActivityDate="2011-03-10T09:22:39.237" CommentCount="4" />
  <row Id="26175" PostTypeId="2" ParentId="26167" CreationDate="2011-03-10T09:25:17.830" Score="2" ViewCount="" Body="&lt;p&gt;$P(\max Y_i=k)=P(\max Y_i\leq k)-P(\max Y_i&amp;lt;k)=F(k)^n-(F(k)-f(k))^n$.&lt;br&gt;&#xA;Thus $E(\max Y_i)=\sum_{k=0}^{\infty} k*[F(k)^n-(F(k)-f(k))^n]=\sum_{k=1}^{\infty}k*[(1-(1-p)^k)^n-(1-(1-p)^{k-1})^n]$. Not a closed form though.&lt;br&gt;&#xA;See also &#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Order_statistic&quot; rel=&quot;nofollow&quot;&gt;Order statistic&lt;/a&gt; for both continuous and discrete case.&#xA;The formula for the continuous case appears in Shai Covo's post &lt;a href=&quot;http://math.stackexchange.com/questions/24743/does-exceptionalism-persist-as-sample-size-gets-large/24755#24755&quot;&gt;here&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7243" LastEditorUserId="7243" LastEditorDisplayName="" LastEditDate="2011-03-10T09:36:12.680" LastActivityDate="2011-03-10T09:36:12.680" CommentCount="1" />
  <row Id="26176" PostTypeId="2" ParentId="26166" CreationDate="2011-03-10T09:28:56.923" Score="8" ViewCount="" Body="&lt;p&gt;Just to give you some basis for judging this answer: I've been an academic mathematician for a little more than ten years, and my direct experience is with tier I research universities.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now for my advice (for what it's worth):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Although there are probably counterexamples, my feeling is that once you leave academia, it is rather difficult to reenter.  For example, it is difficult to keep up with current research if you are not in an environment where there are seminars and colloquia, and where you have access to MathSciNet, journals, and so on.  Also, building a teaching resume is also an important part of pursuing an academic career, and you can't do this (at least in any conventional way) outside of academia.  (There are exceptions to this --- in the old days there&#xA;was Bell, later ATT, labs, but I'm not sure it exists anymore; the modern counterpart is Microsoft Research, and perhaps Google's research division --- but I presume that is not the kind of company job your friend is considering.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Realistically, it is also probably not easy to move from a teaching position at a tier 3 college to a more research-oriented position at a research university.  But it does seem realistic (at least to me) to use this position as a basis for establishing a teaching-oriented academic career, and such a career does offer opportunities for pursuing research as well (although obviously not in the same focussed way that a career at a more research-oriented institution would).   So your friend's decision should probably be based to a large extent on whether they would be happy to have a career in academia whose primary focus is teaching.&lt;/p&gt;&#xA;" OwnerUserId="221" LastEditorUserId="221" LastEditorDisplayName="" LastEditDate="2011-05-23T05:53:19.567" LastActivityDate="2011-05-23T05:53:19.567" />
  <row Id="26177" PostTypeId="1" AcceptedAnswerId="26178" CreationDate="2011-03-10T09:46:53.600" Score="0" ViewCount="74" Body="&lt;p&gt;Hi there,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My notes for determining the Maximum Likelihood decision rule for classifying two fish given two features (lightness and width) have a different decision rule if the features are correlated to when they're uncorrelated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ML Decision rule for 2 independant features:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;if p(lightness | sea bass)p(width | sea bass) &gt;= p(lightness | salmon)p(width | salmon)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;no problem with that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For 2 correlated features, the ML Decision Rule is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;if p(lightness, width | sea bass) &gt;= p(lightness, width | salmon)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The notation is a bit confusing for me. what does p(width, lightness) actually mean. I've seen things like p(x , $\omega$) before but what does that mean? I know that p(x | $\omega$) means the probability of x given $\omega$, but not what p(x , $\omega$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, my notes say that when two features are correlated, their covariance matrices will be different. Why is this? What does having different covariance matrices mean in terms of understanding the data?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many thanks in advance&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Joe&lt;/p&gt;&#xA;" OwnerUserId="1094" LastActivityDate="2011-03-10T09:56:33.697" Title="Maximum Likelihood notation" Tags="&lt;statistics&gt;" AnswerCount="1" />
  <row Id="26178" PostTypeId="2" ParentId="26177" CreationDate="2011-03-10T09:56:33.697" Score="1" ViewCount="" Body="&lt;p&gt;You seem to use $p(a,b|c)$ to denote $p(d|c)$ where $d$ denotes &quot;$a$ and $b$&quot; or, equivalently $d=a\cap b$.&lt;/p&gt;&#xA;" OwnerUserId="6179" LastActivityDate="2011-03-10T09:56:33.697" CommentCount="1" />
  <row Id="26179" PostTypeId="1" AcceptedAnswerId="26183" CreationDate="2011-03-10T10:01:22.490" Score="1" ViewCount="79" Body="&lt;p&gt;Hatcher uses the words &quot;path-component&quot; and &quot;component&quot; without giving their definition. The first occurrence of &quot;path-component&quot; is on p. 19, exercise 12. The first occurrence of &quot;component&quot; is on p. 6, example 0.2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think &quot;path-component&quot; means &quot;path-connected component&quot;. But what is a &quot;component&quot; of a topological space? He seems to use it to mean a $k$-cell, for any $k$. Is this the definition?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you for your help.&lt;/p&gt;&#xA;" OwnerUserId="5798" LastActivityDate="2011-03-10T10:36:44.660" Title="Hatcher's terminology" Tags="&lt;algebraic-topology&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="26180" PostTypeId="2" ParentId="25461" CreationDate="2011-03-10T10:03:35.733" Score="0" ViewCount="" Body="&lt;p&gt;Another nice little thing, also from &quot;Two elementary generalisations of boolean rings&quot; AMM Feb.1986: &lt;strong&gt;if every product is idempotent then the ring is commutative&lt;/strong&gt;, i.e.&#xA;$$\forall x,y\in R: (xy)^2=xy\Longrightarrow R\text{ is commutative}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proof: we use the lemma, mentioned in the previous proof: if $xy=0$ implies $yx=0$, then every idempotent is central. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In our case, if $xy=0$, then $yx=(yx)^2=y(xy)x=0$, so by the lemma, every idempotent is central. That means that $xy$ is central, i.e &lt;em&gt;every product is central&lt;/em&gt;. Hence we get &#xA;$$xy=(xy)^2=x(yxy)=(yxy)x=(yx)^2=yx,$$&#xA;so commutativity indeed holds.&lt;/p&gt;&#xA;" OwnerUserId="7325" LastActivityDate="2011-03-10T10:03:35.733" />
  <row Id="26181" PostTypeId="2" ParentId="26167" CreationDate="2011-03-10T10:06:26.263" Score="6" ViewCount="" Body="&lt;p&gt;First principle:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;To deal with maxima $M$ of independent random variables, use as much as possible events of the form $[M\le x]$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Second principle:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;To compute the expectation of a nonnegative random variable $Z$, use as much as possible the complementary cumulative distribution function $P(Z\ge z)$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In the discrete case, $E(M)=\displaystyle\sum_{k\ge0}P(M&amp;gt;k)$, the event $[M&amp;gt;k]$ is the complement of $[M\le k]$, and the event $[M\le k]$ is the intersection of the independent events $[X_i\le k]$, each of probability $F_X(k)$. Hence,&#xA;$$&#xD;&#xA;E(M)=\sum_{k\ge0}(1-P(M\le k))=\sum_{k\ge0}(1-P(X\le k)^n)=\sum_{k\ge0}(1-F_X(k)^n).&#xD;&#xA;$$&#xA;The continuous case is even simpler. For i.i.d. nonnegative $X_1$, $X_2$, ..., $X_n$,&#xA;$$&#xD;&#xA;E(M)=\int_0^{+\infty}(1-F_X(t)^n)\mathrm{d}t.&#xD;&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="6179" LastEditorUserId="6179" LastEditorDisplayName="" LastEditDate="2011-03-10T20:58:50.310" LastActivityDate="2011-03-10T20:58:50.310" CommentCount="2" />
  <row Id="26182" PostTypeId="2" ParentId="24843" CreationDate="2011-03-10T10:31:00.460" Score="3" ViewCount="" Body="&lt;p&gt;Given what is already known in the original question, you don't need Riemann--Roch to prove surjectivity.  (The content of Riemann--Roch is already encoded in the given facts.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rather, given two points $P$ and $Q$, draw a line through them, which meets $\mathcal C$ in a third point $R$.  Now draw a line through $R$ and $A_0$, which meets $\mathcal C$ in a third point $S$.   From the given facts, we find that&#xA;$P + Q \sim A_0 + S$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now if $D$ is a divisor of degree zero, write $D = D_+ - D_-$, with all&#xA;the coefficients of $D_+$ and $D_-$ being positive.  Repeatedly applying the&#xA;procedure of the preceding paragraph, we may write $D_+ \sim A_+ + n A_0$&#xA;for some point $A_+$, and $D_- \sim A_- + n A_0$ for some point $A_-$.&#xA;(We get the same number $n$ in both cases because $D$ has degree zero by assumption.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus $D \sim A_+ - A_-.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now an evident variation on the preceding construction shows that if we have&#xA;points $P$ and $S$, we may find a point $Q$ such that&#xA;$P + Q \sim A_0 + S.$   (Draw the line through $A_0$ and $S$, which meets&#xA;$\mathcal C$ in a third point $R$.  Now draw the line through $R$ and $P$, which meets $\mathcal C$ in a third point $Q$, which is the desired point.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, we may find a point $A$ so that&#xA;$A_- + A \sim A_0 + A_+$.  Thus $D \sim A - A_0$, as required.&lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2011-03-10T10:31:00.460" />
  <row Id="26183" PostTypeId="2" ParentId="26179" CreationDate="2011-03-10T10:36:44.660" Score="1" ViewCount="" Body="&lt;p&gt;Hatcher just means connected components. It is assumed that if you are reading Hatcher you have some background in general topology. &lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2011-03-10T10:36:44.660" />
  <row Id="26184" PostTypeId="2" ParentId="26171" CreationDate="2011-03-10T10:41:13.343" Score="3" ViewCount="" Body="&lt;p&gt;This looks basically right, although to nitpick I might argue that you should either write that $H(0,-)=id_X$ and $H(1,-)=const._*$ (both as maps $X\rightarrow X$) or you should write that $H(0,x)=x$ and $H(1,x)=*$ for all $x\in X$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, I think I can confidently say that you can assume every map in any topology book is assumed to be continuous.  Similarly, any function of groups can generally be assumed to be a homomorphism, etc.  The fancy way of saying this here is that we're working in the category of topological spaces, whose morphisms are continuous maps.&lt;/p&gt;&#xA;" OwnerUserId="2159" LastActivityDate="2011-03-10T10:41:13.343" />
  <row Id="26185" PostTypeId="1" CreationDate="2011-03-10T10:51:55.500" Score="0" ViewCount="63" Body="&lt;p&gt;Hey you guys, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;i have derived the following expressions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(\sum _{j=1}^i \prod _{z=1}^n \left(1-x_z\right))/i $ and &#xA;$\prod _{z=1}^n \left(1-\frac{\sum _{j=1}^i x_z}{i}\right)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;they are identical and give : &#xA;$\prod _{z=1}^n \left(1-x_z\right)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is now: Are they still identical if $x_z$ is modified by a Gaussian deviation? e.g. $x_z$ = ($x_z$ + Gaussian) ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit:&#xA;corrected expression 1&#xA;What i want to have in the end is the average of $i$ runs over the multiplication of all elements in n. Essentially what i want is the result of the expression:&#xA;$\prod _{z=1}^n \left(1-(x_z+gaussian\right))$ with the $gaussian$ averaged out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;edit 2: &#xA;I made a mistake. The elements are of course dependent on the variable of the sum, since i want to average over $i$ elements. But now i am not sure anymore whether i can compute the average before and then do the multiplication, or if i will have to do the multiplication with each $x_z$ and average over the results.&lt;/p&gt;&#xA;" OwnerUserId="2362" LastEditorUserId="2362" LastEditorDisplayName="" LastEditDate="2011-03-10T11:59:38.063" LastActivityDate="2011-03-10T11:59:38.063" Title="Averages and Gaussians" Tags="&lt;average&gt;" CommentCount="4" />
  <row Id="26186" PostTypeId="1" AcceptedAnswerId="26188" CreationDate="2011-03-10T10:56:11.540" Score="5" ViewCount="241" Body="&lt;p&gt;I'm doing &lt;s&gt;stuck with the first half of&lt;/s&gt; exercise 12 on page 19 in Hatcher:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Exercise:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Show that a homotopy equivalence $f : X \rightarrow Y$ induces a bijection between the set of path-components of $X$ and the set of path-components of $Y$. &lt;s&gt;and that $f$ restricts to a homotopy equivalence from each path-component of $X$ to the corresponding path-component of $Y$&lt;/s&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;s&gt;By the definition of homotopy equivalence, there exists $g$ such that $gf \simeq id_X$ and $fg \simeq id_Y$. So $g$ is a two-sided inverse of $f$, therefore $f$ is a bijection.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm stuck now because I don't know what there is to prove to show that &quot;$f$ restricts to a homotopy equivalence from each path-component of $X$ to the corresponding path-component of $Y$&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there anything to prove? If $f$ is a homotopy equivalence, is it not obvious that any restriction is also a homotopy equivalence? Many thanks for helping me, I really appreciate it.&lt;/s&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So here is an attempt of proof that $f^\ast:\pi_0(X) \rightarrow \pi_0(Y)$ is a bijection:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Define $f^\ast$ as $f^\ast (A) := B$ where $B$ is the path-(connected)-component that contains $f(A)$ and $A \in \pi_0 (X)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Claim&lt;/strong&gt;: $f^\ast$ is &lt;strong&gt;bijective&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;(i) injective:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $f(A) = f(B)$ where $A, B \in \pi_0(X)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then $f(A) \cup f(B)$ is a path-(connected)-component. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then $f(A \cup B)$ is also a path-component. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the definition of homotopy equivalence, there exists $g$ (continuous) such that $gf \simeq id_X$ and $fg \simeq id_Y$. So $g(f(A \cup B)) = A \cup B$ is path-connected. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\implies A = B$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the proof above I used the following claim:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Claim: $A$ path-connected, $f$ continuous $\implies f(A)$ path-connected.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proof: Assume $f(A)$ not path-connected, then $f(A)$ not connected, then $g(f(A))=A$ not connected $\implies$ contradiction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;(ii) surjective:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $B \in \pi_0(Y)$. Then for $A := g(B)$, $f(A) = B$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you tell me if my proof is correct? I really appreciate your help!&lt;/p&gt;&#xA;" OwnerUserId="5798" LastEditorUserId="5798" LastEditorDisplayName="" LastEditDate="2011-03-23T16:18:36.063" LastActivityDate="2011-03-23T16:18:36.063" Title="Proof of another Hatcher exercise: homotopy equivalence induces bijection" Tags="&lt;algebraic-topology&gt;" AnswerCount="2" CommentCount="5" FavoriteCount="4" />
  <row Id="26187" PostTypeId="2" ParentId="26186" CreationDate="2011-03-10T11:02:14.687" Score="2" ViewCount="" Body="&lt;p&gt;If $D$ is a path component of $X$, then $f$ restricts to a map $D \to Y$ which lands in some path component $E$ of $Y$. If you want to claim that $f$ automatically restricts to a homotopy equivalence from $D$ to $E$, you have to show first that $g(E) \subseteq D$ and second that the induced homotopy $H$ between $gf$ and $\text{id}_D$ never has image outside of $D$ (and the corresponding statement for $E$). This is stronger than what you're being asked to prove; all you need to show is that $f$ and $g$ induce inverse maps on path components. &lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2011-03-10T11:15:32.037" LastActivityDate="2011-03-10T11:15:32.037" CommentCount="3" />
  <row Id="26188" PostTypeId="2" ParentId="26186" CreationDate="2011-03-10T11:19:03.793" Score="6" ViewCount="" Body="&lt;p&gt;Be careful: A homotopy equivalence is by no means a bijection in general: Let $X = \mathbb{R}$ and $Y = \{\ast\}$. Also, a homotopy equivalence need not restrict to a homotopy equivalence of subspaces. For instance, the orthogonal projection of $\mathbb{R}^2$ onto the $x$ axis is a homotopy equivalence. However, it maps the unit circle to the interval $[-1,1]$ and these two spaces certainly aren't homotopy equivalent (but this isn't completely obvious). Even simpler, take the subspace $([-1,1] \times \{0\}) \cup ([-1,1] \times \{1\})$ of $\mathbb{R}^2$ and its image $[-1,1]$ under the projection: you'll get a contradiction to your exercise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The expressions $gf \simeq \operatorname{id}_{X}$ and $fg \simeq \operatorname{id}_{Y}$ mean that $gf$ and $fg$ are &lt;em&gt;homotopic&lt;/em&gt; to the respective identities. (btw: it is more customary to use $\simeq$ &lt;code&gt;\simeq&lt;/code&gt; instead of $\cong$ for &quot;homotopic&quot;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A continuous map sends path components &lt;em&gt;into&lt;/em&gt; path components (because it sends paths to paths). The path component of a point $x$ is sent into the path component of its image $f(x)$ and the path component of $f(x)$ is sent into the path component of $gf(x)$. But as $gf \simeq \operatorname{id}_{X}$, a homotopy $H: X \times [0,1] \to X$ such that $H(\cdot,0) = gf$ and $H(\cdot,1) = \operatorname{id}_{X}$ gives us the path $\gamma(t) = H(x,t)$ connecting $gf(x)$ with $x$, so the path component of $gf(x)$ is the same as the path component of $x$. I let you finish up the argument yourself.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added&lt;/strong&gt; in response to the edited question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $[x]$ be the path component of $x$. Let me write $f_{\ast}: \pi_{0}(X) \to \pi_{0}(Y)$ instead of $f^\ast$. By definition $f_{\ast}[x] = [f(x)]$ (check that this is well-defined!). In my last paragraph above I argued that $[gf(x)] = [x]$ so $g_\ast f_\ast [x] = [x]$ (check that $(gf)_\ast = g_\ast f_\ast$!). In other words $g_\ast f_\ast = (\operatorname{id}_{X})_{\ast} = \operatorname{id}_{\pi_{0}(X)}$ (check that $(\operatorname{id}_X)_\ast = \operatorname{id}_{\pi_{0}(X)}$!). In particular, $f_{\ast}$ is injective and $g_\ast$ is surjective. By symmetry we have $f_\ast g_\ast = \operatorname{id}_{\pi_{0}(Y)}$, so $f_{\ast}$ and $g_\ast$ are mutually inverse bijections.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;A bit later&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe it's better to start from scratch.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Define an equivalence relation on $X$ by $x \sim x&amp;#39;$ if and only if there is a path connecting $x$ and $x&amp;#39;$. The equivalence classes of $\sim$ are precisely the path components of $X$. In other words, $\pi_{0}(X) = X/\!\!\sim$. Write $[x]$ for the $\sim$-equivalence class of $x$ (so $[x]$ is the path component of $x$).&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Fact 1:&lt;/strong&gt; A continuous map $f: X \to Y$ sends path components into path components. In other words, if $x \sim x$ then $f(x) \sim f(x&amp;#39;)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Indeed, if $\gamma: [0,1] \to X$ is a path with $\gamma(0) = x$ and $\gamma(1) = x&amp;#39;$ then $f \circ \gamma: [0,1] \to Y$ is a path and $f(\gamma(0)) = f(x)$ and $f(\gamma(1)) = f(x&amp;#39;)$, so $f(x) \sim f(x&amp;#39;)$.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Again in other words Fact 1 tells us that $f$ yields a map $f_{\ast}: \pi_{0}(X) \to \pi_{0}(Y)$. Explicitly,&#xA;\[&#xA;f_{\ast}([x]) = [f(x)].&#xA;\]&#xA;This is well-defined because for $x \sim x&amp;#39;$ we have $f(x) \sim f(x&amp;#39;)$, so $[f(x)] = [f(x&amp;#39;)]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From this description we see that $(\operatorname{id}_{X})_{\ast}([x]) = [\operatorname{id}_{X}(x)] = [x] = \operatorname{id}_{\pi_{0}(X)}([x])$, so $(\operatorname{id}_{X})_{\ast} = \operatorname{id}_{\pi_{0}(X)}$. Also $(gf)_{\ast}([x]) = [gf(x)] = g_{\ast}([f(x)]) = g_{\ast}f_{\ast}([x])$, so $(gf)_{\ast} = g_{\ast}f_{\ast}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the two identities I've just proven are so important, let me state them again for emphasis:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Fact 2:&lt;/strong&gt; For the identity $\operatorname{id}_{X}: X \to X$ and any two maps $f:X \to Y$ and $g: Y \to Z$ &#xA;we have $$\displaystyle&#xD;&#xA;(\operatorname{id}_{X})_{\ast} = \operatorname{id}_{\pi_{0}(X)}:&#xD;&#xA;\pi_{0}(X) \to \pi_{0}(X) \qquad\text{and}\qquad (gf)_{\ast} = g_\ast f_\ast : \pi_{0}(X) \to \pi_{0}(Z)&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Fact 3:&lt;/strong&gt; If $f, f&amp;#39;: X \to Y$ are homotopic, $f \simeq f&amp;#39;$, then $f_{\ast} = f&amp;#39;_{\ast}: \pi_{0}(X) \to \pi_{0}(Y)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Indeed, pick a homotopy $H: X \times [0,1] \to Y$ such that $f = H(\cdot,0)$ and $f&amp;#39; = H(\cdot,1)$. Since $t \mapsto H(x,t)$ is a path connecting $f(x)$ to $f&amp;#39;(x)$ we see that $f(x) \sim f&amp;#39;(x)$, so $f_{\ast}[x] = [f(x)] = [f&amp;#39;(x)] = f&amp;#39;_\ast[x]$.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Now we are finally in shape to solve the exercise. Let $f: X \to Y$ and $g: Y \to X$ be such that $gf \simeq \operatorname{id}_{X}$ and $fg \simeq \operatorname{id}_{Y}$. Then combining facts 2 and 3 we have that&#xA;$$\displaystyle&#xD;&#xA;(\operatorname{id}_{X})_\ast = g_{\ast} f_{\ast}&#xD;&#xA;\qquad \text{and}\qquad&#xD;&#xA;(\operatorname{id}_{Y})_\ast = f_{\ast} g_{\ast}&#xD;&#xA;$$&#xA;and as $(\operatorname{id}_X)_\ast = \operatorname{id}_{\pi_{0}(X)}$ we see that $f_{\ast}$ and $g_{\ast}$ are mutually inverse bijections.&lt;/p&gt;&#xA;" OwnerUserId="5363" LastEditorUserId="5363" LastEditorDisplayName="" LastEditDate="2011-03-22T14:39:48.707" LastActivityDate="2011-03-22T14:39:48.707" CommentCount="2" />
  <row Id="26189" PostTypeId="2" ParentId="22164" CreationDate="2011-03-10T11:43:34.567" Score="1" ViewCount="" Body="&lt;p&gt;@Imelza If the data is not uniformly sampled in time, xcorr will not work correctly (and shifting it by samples will not give the right correspondence.) You could use &lt;a href=&quot;http://www.mathworks.com/help/techdoc/ref/interp1.html&quot; rel=&quot;nofollow&quot;&gt;interp1&lt;/a&gt; to resample it uniformly first.&lt;/p&gt;&#xA;" OwnerUserId="5610" LastActivityDate="2011-03-10T11:43:34.567" CommentCount="2" />
  <row Id="26190" PostTypeId="1" CreationDate="2011-03-10T11:51:28.330" Score="0" ViewCount="168" Body="&lt;p&gt;I'm trying to do exercise 5 on page 18 in Hatcher:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Show that if a space $X$ deformation retracts to a point $x \in X$, then for each neighborhood $U$ of $x$ in $X$ there exists a neighborhood $V \subset U$ of $x$ such that the inclusion map $V \rightarrow U$ is nullhomotopic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does the existence part of the neighbourhood $V$ follow from taking $V=U$, i.e. there is at least one neighbourhood in $U$($U$ itself)? Or does it have to be a proper neighbourhood? If it has to be a proper neighbourhood: is the idea to retract $U$ &quot;a bit&quot;, just enough to get a new neighbourhood $V \subset U$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for any hints, I appreciate your help!&lt;/p&gt;&#xA;" OwnerUserId="5798" LastActivityDate="2011-03-10T14:04:25.033" Title="Request for hint for exercise in Hatcher's Algebraic Topology" Tags="&lt;algebraic-topology&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="26192" PostTypeId="1" AcceptedAnswerId="26198" CreationDate="2011-03-10T12:16:14.790" Score="2" ViewCount="78" Body="&lt;p&gt;Hi all,&lt;br/&gt;&#xA;Are there any good online resources to learn multivariate statistics? ( With topics such as Multiple Linear Regression, Principal Component Analysis, Factor Analysis, Cluster Analysis and Discriminant Analysis)&lt;/p&gt;&#xA;" OwnerUserId="8067" LastActivityDate="2011-03-10T14:06:06.550" Title="Good Online Resources to learn Multivariate Statistics?" Tags="&lt;statistics&gt;&lt;online-resources&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="26193" PostTypeId="2" ParentId="26160" CreationDate="2011-03-10T12:28:36.883" Score="12" ViewCount="" Body="&lt;p&gt;There is a standard way to construct these things in classical logic.  Your question is the same as &quot;if we know that at least one $x$ satisfies $Z$, can we find an example of an $x$ that satisfies $Z$&quot; and I'll work with the positive version. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, choose some well-defined, unambiguous statement whose truth is not known at the moment. For example, P=NP or the Hodge conjecture. Now we can prove there is a natural number $x$ with the following property: &quot;If P=NP then $x=1$ and otherwise $x=0$&quot;.  But we can't give an explicit example of an $x$ with that property. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can actually make the property even more elementary by changing it to &quot;If ZFC proves P=NP then $x=1$ and otherwise $x=0$&quot;. That has only one quantifier, over ZFC-proofs, which can be regarded as one quantifier over natural numbers if proofs are represented by numbers.  As before, &quot;P=NP&quot; can be replaced with any unsolved, unambiguous conjecture, e.g. the Hodge conjecture, and the final sentence is still a sentence just about natural numbers. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The point of this example is that it avoids any difficulties with uncountable sets, the axiom of choice, etc. The existence of examples like this shows that systems like PA and ZFC do not have the witness property [1]. That is, just because you can prove a formula of the form $(\exists x)\phi(x)$ does not mean that there is a specific $t$ such that you can prove $\phi(t)$. One of the benefits of constructive systems is that they do have the witness property (meaning you always can find such a $t$) and this is one reason that it's interesting to ask what sorts of mathematics can be proved in such systems. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1: &lt;a href=&quot;http://en.wikipedia.org/wiki/Disjunction_and_existence_properties&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Disjunction_and_existence_properties&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="630" LastActivityDate="2011-03-10T12:28:36.883" CommentCount="13" />
  <row Id="26194" PostTypeId="1" CreationDate="2011-03-10T13:21:27.863" Score="0" ViewCount="97" Body="&lt;p&gt;Hello,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sorry for my English.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is the question:&#xA;$G=(v,e)$, undirected graph, $V=\{v_1,v_2,\ldots,v_n\}$. the vertices are organized in sequence from the smaller one to the biggest $v_1,\ldots,v_n$.&#xA;We will define a spread distribution $(C_1,\ldots,C_k)$ as follows:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;For each $i,j$, $C_i \cap C_j = \emptyset$.&lt;/li&gt;&#xA;&lt;li&gt;$\bigcup_{i=1}^k C_i = V$.&lt;/li&gt;&#xA;&lt;li&gt;Every $C_i$ consists of consecutive vertices, i.e. $v_{j+1},\ldots,v_{j+d}$.&lt;/li&gt;&#xA;&lt;li&gt;The vertices in $C_i$ form an independent set.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I need to find a Greedy Algorithm for this problem in $O(|V|+|E|)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried many ways, but the main problem for me is to handle the situation where $(v_3,v_4) \notin E$ and $(v_4,v_1) \in E$ (I check in my algorithm only &quot;neighboring&quot; vertices, $v_1-v_2,v_2-v_3,\ldots,v_{n-1}-v_n$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any ideas? Thank you!&lt;/p&gt;&#xA;" OwnerUserId="8069" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2011-03-10T17:28:11.620" LastActivityDate="2011-03-10T17:28:11.620" Title="Partition of graph into independent sets of consecutive vertices" Tags="&lt;homework&gt;&lt;algorithms&gt;&lt;graph-theory&gt;" CommentCount="4" />
  <row Id="26195" PostTypeId="1" AcceptedAnswerId="26196" CreationDate="2011-03-10T13:46:06.400" Score="3" ViewCount="73" Body="&lt;p&gt;Define the $q$-analog $(a;q)_n = \prod_{k=0}^n \left(1 - aq^k\right)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to prove the identity $\frac{(q^2;q^2)_\infty}{(q;q)_\infty}=\frac{1}{(q;q^2)_\infty}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I viewed the LHS this way:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{1 - q^2}{1 - q} \frac{1 - q^4}{1 - q^2} \frac{1 - q^6}{1 - q^3} \frac{1 - q^8}{1 - q^4} \frac{1 - q^{10}}{1 - q^5} \frac{1 - q^{11}}{1 - q^6} \cdots$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and you can imagine getting the RHS by canceling the $k$th term on the top with the $2k+1$th term on the bottom.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can we really cancel that way though? Since the gap between $k$ and $2k+1$ keeps getting bigger it might not be valid.&lt;/p&gt;&#xA;" OwnerUserId="4290" LastEditorUserId="4290" LastEditorDisplayName="" LastEditDate="2011-04-28T22:07:51.127" LastActivityDate="2011-04-28T22:07:51.127" Title="Validity of a q-series theorem" Tags="&lt;special-functions&gt;&lt;partitions&gt;&lt;q-series&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="26196" PostTypeId="2" ParentId="26195" CreationDate="2011-03-10T13:53:19.043" Score="6" ViewCount="" Body="&lt;p&gt;Take the logarithm of the quotient of infinite products to get a difference of infinite sums. Both sums converge absolutely, and hence you can rearrange their terms at will and let them cancel.&lt;/p&gt;&#xA;" OwnerUserId="6622" LastActivityDate="2011-03-10T13:53:19.043" />
  <row Id="26197" PostTypeId="2" ParentId="26190" CreationDate="2011-03-10T14:04:25.033" Score="2" ViewCount="" Body="&lt;p&gt;I don't want to give a complete answer here, and I second Theo Buehler's advice in the comments above.  But I will give a hint: the definition of deformation retract involves continuous maps.  And in a question in topology involving continuous maps, if you have to produce new open sets from old, this is typically done by invoking the definition of continuity in terms of open sets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are not naturally visual in your thinking, it will take time to learn how to think topologically/geometrically, and to learn how to translate visual intuitions into the formal language of spaces and continuous maps --- but it is well worth the effort and time!&lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2011-03-10T14:04:25.033" />
  <row Id="26198" PostTypeId="2" ParentId="26192" CreationDate="2011-03-10T14:06:06.550" Score="1" ViewCount="" Body="&lt;p&gt;Statsoft has a pretty good comprehensive online guide which I can recommend: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.statsoft.com/textbook/&quot; rel=&quot;nofollow&quot;&gt;http://www.statsoft.com/textbook/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-Ralph Winters&lt;/p&gt;&#xA;" OwnerUserId="8071" LastActivityDate="2011-03-10T14:06:06.550" CommentCount="1" />
  <row Id="26199" PostTypeId="2" ParentId="26172" CreationDate="2011-03-10T14:07:01.973" Score="0" ViewCount="" Body="&lt;p&gt;It is very simple to convince yourself (at least intuitively, but also a proof can be given if necessary) that in order to have the time $t$ as small as possible, it is best to have the largest acceleration possible. Thereby, $a_x^2+a_y^2=a_\text{max}^2$. The equations which should be fulfilled are given by&#xA;$$\frac{a_x t^2}{2} + v^A_x t+ c^A_x =  v^B_x t+ c^B_x \qquad   \frac{a_y t^2}{2} + v^A_y t+ c^A_y = t v^B_y + c^B_y.$$&#xA;Replacing $a_y$ by $a_y = \sqrt{a_\text{max}^2 - a_x^2}$,  we obtain&#xA;$$\frac{\sqrt{a_\text{max}^2 - a_x^2} t^2}{2} + v^A_y t+ c^A_y = t v^B_y + c^B_y.$$&#xA;Solving the first equation for $a_x$, and plugging everything the result into the second equation yields&#xA;$$\frac{\sqrt{a_\text{max}^2 - \left[\frac{2 (c^A_x- c^B_x +t (v^A_x -v^B_x)}{t} \right]^2} t^2}{2} + v^A_y t+ c^A_y =  v^B_y t + c^B_y.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After some manipulations the last equation can be brought into the form&#xA;$$\begin{multline}&#xD;&#xA;\frac{a_\text{max}^2}{4} t^4 + [(v^A_y- v^B_y)^2 - (v^A_x- v^B_x)^2] t^2  - 2[(c^A_x - c^B_x)(v^A_x - v^B_x)+ (c^A_y - c^B_y)(v^A_y - v^B_y)] t\\ -(c^A_x -c^B_x)^2 - (c^A_y -c^B_y)^2 =0.&#xD;&#xA;\end{multline}$$&#xA;Solving this equation for $t$ gives the optimal (minimal) time required. Note however that this is a quartic equation such that the solution becomes quite messy and that is why I will not write it down here.&lt;/p&gt;&#xA;" OwnerUserId="7266" LastActivityDate="2011-03-10T14:07:01.973" CommentCount="4" />
  <row Id="26200" PostTypeId="2" ParentId="26170" CreationDate="2011-03-10T14:38:27.900" Score="7" ViewCount="" Body="&lt;p&gt;(Assuming $\mathbf{A}$ is a square matrix, of course). Here's a solution that does not invoke determinants or diagonalizability, but only the definition of eigenvalue/eigenvector, and the characterization of invertibility in terms of the nullspace. (&lt;em&gt;Added for clarity:&lt;/em&gt; $\mathbf{N}(\mathbf{A}) = \mathrm{ker}(\mathbf{A}) = \{\mathbf{x}\mid \mathbf{A}\mathbf{x}=\mathbf{0}\}$, the nullspace/kernel of $\mathbf{A}$.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align*}&#xA;\mbox{$\mathbf{A}$ is not invertible} &amp;amp;\Longleftrightarrow \mathbf{N}(\mathbf{A})\neq{\mathbf{0}}\\&#xA;&amp;amp;\Longleftrightarrow \mbox{there exists $\mathbf{x}\neq\mathbf{0}$ such that $\mathbf{A}\mathbf{x}=\mathbf{0}$}\\&#xA;&amp;amp;\Longleftrightarrow \mbox{there exists $\mathbf{x}\neq\mathbf{0}$ such that $\mathbf{A}\mathbf{x}=0\mathbf{x}$}\\&#xA;&amp;amp;\Longleftrightarrow \mbox{there exists an eigenvector of $\mathbf{A}$ with eigenvalue $\lambda=0$}\\&#xA;&amp;amp;\Longleftrightarrow \mbox{$\lambda=0$ is an eigenvalue of $\mathbf{A}$.}&#xA;\end{align*}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that this argument holds even in the case where $\mathbf{A}$ has no eigenvalues (when working over a non-algebraically closed field, of course), where the condition &quot;the eigenvalues of $\mathbf{A}$ are all nonzero&quot; is true by vacuity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $\mathbf{A}$ invertible:&#xA;\begin{align*}&#xA;\mbox{$\lambda\neq 0$ is an eigenvalue of $\mathbf{A}$} &amp;amp;\Longleftrightarrow \mbox{$\lambda\neq 0$ and there exists $\mathbf{x}\neq \mathbf{0}$ such that $\mathbf{A}\mathbf{x}=\lambda\mathbf{x}$}\\&#xA;&amp;amp;\Longleftrightarrow\mbox{there exists $\mathbf{x}\neq\mathbf{0}$ such that $\mathbf{A}({\textstyle\frac{1}{\lambda}}\mathbf{x}) = \mathbf{x}$}\\&#xA;&amp;amp;\Longleftrightarrow\mbox{there exists $\mathbf{x}\neq \mathbf{0}$ such that $\mathbf{A}^{-1}\mathbf{A}({\textstyle\frac{1}{\lambda}}\mathbf{x}) = \mathbf{A}^{-1}\mathbf{x}$}\\&#xA;&amp;amp;\Longleftrightarrow\mbox{there exists $\mathbf{x}\neq \mathbf{0}$ such that $\frac{1}{\lambda}\mathbf{x} = \mathbf{A}^{-1}\mathbf{x}$}\\&#xA;&amp;amp;\Longleftrightarrow\mbox{$\frac{1}{\lambda}$ is an eigenvalue of $A^{-1}$.}&#xA;\end{align*}&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-10T16:14:59.910" LastActivityDate="2011-03-10T16:14:59.910" CommentCount="4" />
  <row Id="26202" PostTypeId="2" ParentId="26131" CreationDate="2011-03-10T14:51:56.617" Score="5" ViewCount="" Body="&lt;p&gt;Another visualization of a group on a graph is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Cycle_graph_(algebra)&quot; rel=&quot;nofollow&quot;&gt;cycle graph&lt;/a&gt;.  These are much less regular than a Cayley graph, so might better match your &quot;not all points were connected.&quot;  In fact, Cayley, Schreier coset, and cycle graphs are all connected, but they get less regular down the list.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The cycle graph has vertices the elements of the group, and for each maximal cyclic subgroup, a generator x is chosen, and edges are drawn between x&lt;sup&gt;i&lt;/sup&gt; and x&lt;sup&gt;i+1&lt;/sup&gt;.  It is not entirely clear to me if every cycle graph of a group is isomorphic, but they are for small groups at least.  Here are a few pictures from wikipedia: the elementary abelian group of order 16, the dihedral group of order 12, and the modular (but not Dedekind) group of order 16.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://upload.wikimedia.org/wikipedia/commons/0/0c/GroupDiagramMiniC2x4.png&quot;/&gt;&#xA;&lt;img src=&quot;http://upload.wikimedia.org/wikipedia/commons/b/bc/GroupDiagramMiniD12.png&quot;/&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://upload.wikimedia.org/wikipedia/commons/3/31/GroupDiagramMOD16.png&quot;/&gt;&lt;/p&gt;&#xA;" OwnerUserId="583" LastActivityDate="2011-03-10T14:51:56.617" CommentCount="1" />
  <row Id="26203" PostTypeId="1" AcceptedAnswerId="26207" CreationDate="2011-03-10T15:10:53.403" Score="3" ViewCount="108" Body="&lt;p&gt;Please help me prove by induction that &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$\displaystyle\forall n\in {{\mathbb{N}}^{*}}$, $\displaystyle\forall {{a}_{1}},\ldots ,{{a}_{n}}\in {\mathbb{R}}^{*}_{+}$, $\displaystyle \ln \left( \prod\limits_{j=1}^{n}{{{a}_{j}}} \right)=\sum\limits_{j=1}^{n}{\ln \left( {{a}_{j}} \right)}$. &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Deduce that $\displaystyle \forall n\in \mathbb{Z},\forall a\in {\mathbb{R}}^{*}_{+}$, $\displaystyle \ln \left( {{a}^{n}} \right)=n\ln a$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7022" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-10T16:24:07.473" LastActivityDate="2011-03-10T18:18:57.547" Title="Induction problem" Tags="&lt;algebra&gt;&lt;logarithms&gt;&lt;induction&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="26204" PostTypeId="1" AcceptedAnswerId="26218" CreationDate="2011-03-10T15:11:25.273" Score="2" ViewCount="83" Body="&lt;p&gt;Take for example $f(z)=e^z$, so the inverse is $z(f) = \ln(f) + n\pi i$ for an arbitrarily chosen (but fixed) branch $n\in\mathbb N$. Now if $f$ is restricted to e.g. $0&amp;lt;|1-f|&amp;lt;1$ such that the essential singularity at $f=0$ is not part of the definition region, is $z(f)$ then analytical in that region?&lt;/p&gt;&#xA;" OwnerUserId="163" LastActivityDate="2011-03-10T17:02:04.310" Title="Is the local inverse of an analytical function locally analytical as well?" Tags="&lt;functional-analysis&gt;&lt;inversion&gt;" AnswerCount="1" CommentCount="6" />
  <row Id="26205" PostTypeId="1" AcceptedAnswerId="26209" CreationDate="2011-03-10T15:46:40.250" Score="2" ViewCount="190" Body="&lt;p&gt;Please give me an intuitive explanation of 'implicit function theorem'. I read some bits and pieces of information from some textbook, but they look too confusing, especially I do not understand why they use Jacobian matrix to illustrate this theorem. &lt;/p&gt;&#xA;" OwnerUserId="7743" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-10T16:21:26.320" LastActivityDate="2011-03-10T18:49:05.773" Title="What is the 'implicit function theorem'?" Tags="&lt;real-analysis&gt;&lt;intuition&gt;&lt;multivariable-calculus&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="1" />
  <row Id="26206" PostTypeId="2" ParentId="26115" CreationDate="2011-03-10T15:50:28.150" Score="1" ViewCount="" Body="&lt;p&gt;The distance between the sampling points $T$ should be 1 for your equations to hold. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming everything to be nice the discrete back transform goes over to a Riemann integral as $n\to\infty$. So you can use your formula to calculate the back transform... &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As always: functions where everything goes well lie dense and the resulting $L^2$ norms are bounded by Plancherel such that this argument can be extended to arbitrary functions in $L^2$.&lt;/p&gt;&#xA;" OwnerUserId="7266" LastActivityDate="2011-03-10T15:50:28.150" CommentCount="3" />
  <row Id="26207" PostTypeId="2" ParentId="26203" CreationDate="2011-03-10T15:51:50.700" Score="3" ViewCount="" Body="&lt;p&gt;In certain &quot;transitions&quot; classes I have taught -- i.e., for undergraduate math majors getting used to formal proof and abstraction -- I assign problems like these as (rather easy) exercises in induction, the point being that the usual &quot;binary&quot; form of the identity is assumed, so here&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\log(xy) = \log x + \log y$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I am also guessing that the English translation of &quot;par récurrence&quot; is &quot;by induction&quot; and not, for instance, &quot;by recurrence&quot;.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If this is the case, see e.g. $\S 5$ &lt;a href=&quot;http://www.math.uga.edu/~pete/3200induction.pdf&quot; rel=&quot;nofollow&quot;&gt;of this handout on induction&lt;/a&gt; for some similar examples of such induction proofs.  The main idea here is that you have a product like $x_1 \cdots x_n x_{n+1}$ and you &quot;cleverly&quot; regroup it as $(x_1 \cdots x_n) \cdot x_{n+1}$ -- i.e., first a product of $n$ terms, to which your induction hypothesis applies, and then a binary product, to which your basic identity applies.&lt;/p&gt;&#xA;" OwnerUserId="299" LastActivityDate="2011-03-10T15:51:50.700" CommentCount="1" />
  <row Id="26208" PostTypeId="2" ParentId="26153" CreationDate="2011-03-10T16:01:02.157" Score="2" ViewCount="" Body="&lt;p&gt;The standard 2-norm hypersphere is parametrized by &lt;a href=&quot;http://en.wikipedia.org/wiki/N-sphere#Hyperspherical_coordinates&quot; rel=&quot;nofollow&quot;&gt;hyperspherical coordinates&lt;/a&gt;, which you can easily turn into a parametrization of the $p$-norm hypersphere by transforming each coordinate as $x_i \mapsto \operatorname{sgn}(x_i) \lvert x_i \rvert^{2/p}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I would not recommend using this for solving an optimization problem, because the parametrization is singular at the poles, and at many other points when $p \le 1$ or $p = \infty$. You should look into &lt;a href=&quot;http://en.wikipedia.org/wiki/Lagrange_multiplier&quot; rel=&quot;nofollow&quot;&gt;Lagrange multipliers&lt;/a&gt; instead.&lt;/p&gt;&#xA;" OwnerUserId="856" LastActivityDate="2011-03-10T16:01:02.157" />
  <row Id="26209" PostTypeId="2" ParentId="26205" CreationDate="2011-03-10T16:01:18.717" Score="5" ViewCount="" Body="&lt;p&gt;Let's use a simple example with only two variables. Assume there is some relation $f(x,y)=0$ between these variables (which is a general curve in 2D). An example would be $f(x,y) =x^2+y^2-1$ which is the unit circle in $\mathbb{R}^2$. Now you are interested to figure out the slope of the tangent to this curve at some point $x_0,y_0$ on the curve [with $f(x_0,y_0)=0$].&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you can do is to change $x$ a little bit $x = x_0 + \Delta x$. You are interested then how $y$ changes ($y= y_0 + \Delta y$); remember that we are interested in points on the curve with $f(x,y)=0$. Using Taylor expansion on $f(x,y)=0$ yields (up to lowest order in $\Delta x$ and $\Delta y$)&#xA;$$f(x,y)= \partial_x f(x_0,y_0) \Delta x + \partial_y f(x_0,y_0) \Delta y =0.$$&#xA;The slope is thereby given by $$&#xD;&#xA;\frac{\Delta y}{\Delta x} = - \frac{ \partial_x f(x_0,y_0)}{\partial_y f(x_0,y_0)}.$$ As $\Delta x \to 0$ higher order terms in the Taylor expansion (which we neglected) vanish and $\frac{\Delta y}{\Delta x}$ becomes the slope of the curve implicitly defined via $f(x,y)=0$ at $(x_0,y_0)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More variables and higher dimensional spaces can be treated similarly (using Taylor series in several variables). But the example above should provide you with enough intuition and insight to understand the 'implicit function theorem'.&lt;/p&gt;&#xA;" OwnerUserId="7266" LastActivityDate="2011-03-10T16:01:18.717" />
  <row Id="26210" PostTypeId="2" ParentId="26205" CreationDate="2011-03-10T16:01:24.387" Score="1" ViewCount="" Body="&lt;p&gt;Here is a nice explanation given by one of our Professor at an Undergraduate training camp. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Link: &lt;a href=&quot;http://www.mtts.org.in/expository-articles&quot; rel=&quot;nofollow&quot;&gt;http://www.mtts.org.in/expository-articles&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Please see under the section geometry for the &lt;em&gt;Implicit function theorem&lt;/em&gt;.&lt;/p&gt;&#xA;" LastActivityDate="2011-03-10T16:17:05.513" />
  <row Id="26211" PostTypeId="2" ParentId="10554" CreationDate="2011-03-10T16:07:41.323" Score="0" ViewCount="" Body="&lt;p&gt;I wish I had this reference handy, but Atiyah said something to the effect that algebra is about time and geometry is about space. The &quot;processes&quot; in mathematics are, in the broadest sense, the things concerned with time. (This doesn't take into account the idea of &quot;reification&quot;, the transformation of a process into an object, which is central to mathematical practice.)&lt;/p&gt;&#xA;" OwnerUserId="941" LastActivityDate="2011-03-10T16:07:41.323" />
  <row Id="26212" PostTypeId="1" AcceptedAnswerId="26250" CreationDate="2011-03-10T16:35:06.223" Score="2" ViewCount="113" Body="&lt;p&gt;Hi guys!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a question about Legendre symbol. &#xA;Let $a$, $b$ be integers. Let $p$ be a prime not dividing $a$. Show that the Legendre symbol verifies:&#xA;$$\sum_{m=0}^{p-1} \left(\frac{am+b}{p}\right)=0.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that $\displaystyle\sum_{m=0}^{p-1} \left(\frac{m}{p}\right)=0$, but how do I connect this with the previous formula? &#xA;Any help is appreciated.&lt;/p&gt;&#xA;" OwnerUserId="6912" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-10T16:37:56.963" LastActivityDate="2011-03-10T19:51:24.190" Title="Legendre Symbol" Tags="&lt;homework&gt;&lt;elementary-number-theory&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="26213" PostTypeId="2" ParentId="4808" CreationDate="2011-03-10T16:39:48.803" Score="2" ViewCount="" Body="&lt;p&gt;Here is a physicist's answer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine a rope enclosing a two-dimensional gas, with vacuum outside the rope.  The gas will expand, pushing the rope to enclose a maximal area at equilibrium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the system is at equilibrium, the tension in the rope must be constant, because if there were a tension gradient at some point, there would be a non-zero net force at that point in the direction of the rope, but at equilibrium the net force must be zero in all directions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The gas exerts a force outward on the rope, so tension must cancel this force.  Take a small section of rope, so that it can be thought of as a part of some circle, called the osculating circle.  The force on this rope segment due to pressure is $P l$, with $P$ pressure and $l$ the length.  The net force due to tension is $2 T \sin(l/2R)$, with $T$ tension and $R$ the radius of the osculating circle. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because the pressure is the same everywhere, and the force from pressure must be canceled by the force from tension, the net tension force must be the same for any rope segment of the same length.  That means the radius of the osculating circle is the same everywhere, so the rope must be a circle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: I'm not sure what the general attitude is here towards these sorts of physics-based intuitive answers; this is my first answer on math.SE.  Please let me know if this sort of reasoning is not what this site looks for. &lt;/p&gt;&#xA;" OwnerUserId="3305" LastActivityDate="2011-03-10T16:39:48.803" />
  <row Id="26214" PostTypeId="2" ParentId="26167" CreationDate="2011-03-10T16:50:49.790" Score="2" ViewCount="" Body="&lt;p&gt;There is no nice, closed-form expression for the expected maximum of IID geometric random variables.  However, the expected maximum of the corresponding IID exponential random variables turns out to be a very good approximation.  More specifically, we have the hard bounds&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{1}{\lambda} H_n \leq E_n \leq 1 + \frac{1}{\lambda} H_n,$$&#xA;and the close approximation&#xA;$$E_n \approx \frac{1}{2} + \frac{1}{\lambda} H_n,$$&#xA;where $H_n$ is the $n$th harmonic number $H_n = \sum_{k=1}^n \frac{1}{k}$, and $\lambda = -\log (1-p)$, the parameter for the corresponding exponential distribution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's the derivation.  Let $q = 1-p$.  Use Didier Piau's expression with the fact that if $X$ is geometric with parameter $p$ then $P(X \leq k) = 1-q^k$ to get &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$E_n = \sum_{k=0}^{\infty} (1 - (1-q^k)^n).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By viewing this infinite sum as right- and left-hand Riemann sum approximations of the corresponding integral we obtain &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_0^{\infty} (1 - (1 - q^x)^n) dx \leq E_n \leq 1 + \int_0^{\infty} (1 - (1 - q^x)^n) dx.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The analysis now comes down to understanding the behavior of the integral.  With the variable switch $u = 1 - q^x$ we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_0^{\infty} (1 - (1 - q^x)^n) dx = -\frac{1}{\log q} \int_0^1 \frac{1 - u^n}{1-u} du = -\frac{1}{\log q} \int_0^1 \left(1 + u + \cdots + u^{n-1}\right) du $$&#xA;$$= -\frac{1}{\log q}  \left(1 + \frac{1}{2} + \cdots + \frac{1}{n}\right) = -\frac{1}{\log q} H_n,$$&#xA;which is exactly the expression the OP has above for the expected maximum of $n$ corresponding IID exponential random variables, with $\lambda = - \log q$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This proves the hard bounds, but what about the more precise approximation?  The easiest way to see that is probably to use the &lt;a href=&quot;http://en.wikipedia.org/wiki/Euler%E2%80%93Maclaurin_formula&quot; rel=&quot;nofollow&quot;&gt;Euler-Maclaurin summation formula&lt;/a&gt; for approximating a sum by an integral.  Up to a first-order error term, it says exactly that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$E_n = \sum_{k=0}^{\infty} (1 - (1-q^k)^n) \approx \int_0^{\infty} (1 - (1 - q^x)^n) dx + \frac{1}{2},$$&#xA;yielding the approximation&#xA;$$E_n \approx -\frac{1}{\log q} H_n + \frac{1}{2},$$&#xA;with error term given by &#xA;$$\int_0^{\infty} n (\log q) q^x (1 - q^x)^{n-1} \left(x - \lfloor x \rfloor - \frac{1}{2}\right) dx.$$&#xA;One can verify that this is quite small unless $n$ is also small or $q$ is extreme.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All of these results, including a more rigorous justification of the approximation, the OP's recursive formula, and the additional expression &#xA;$$E_n = \sum_{i=1}^n \binom{n}{i} (-1)^{i+1} \frac{1}{1-q^i},$$&#xA;are in Bennett Eisenberg's paper &quot;On the expectation of the maximum of IID geometric random variables&quot;  (&lt;em&gt;Statistics and Probability Letters&lt;/em&gt; 78 (2008) 135-143).  &lt;/p&gt;&#xA;" OwnerUserId="2370" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2011-03-11T05:47:03.777" LastActivityDate="2011-03-11T05:47:03.777" />
  <row Id="26215" PostTypeId="1" AcceptedAnswerId="26220" CreationDate="2011-03-10T16:52:25.530" Score="5" ViewCount="193" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Let $f$ be a non-negative measurable function in $\mathbb{R}$. Suppose that $$\iint_{\mathbb{R}^2}{f(4x)f(x-3y)\,dxdy}=2\,.$$ Calculate $\int_{-\infty}^{\infty}{f(x)\,dx}$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My first thought was to change the order of integration so that I integrate in $y$ first, since there's only a single $y$ in the integrand... but I'm not sure how/if that even helps me. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then the more I thought about it, the less clear it was to me that Fubini's theorem even applies as it's written. Somehow I need a function of two variables. So should I set $g(x,y) = f(4x)f(x-3y)$ and do something with that? At least Fubini's theorem applies for $g(x,y)$, since we know it's integrable on $\mathbb{R}^2$.    .... Maybe?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm just pretty lost on this, so any help you could offer would be great. Thanks!&lt;/p&gt;&#xA;" OwnerUserId="2313" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-10T22:06:47.523" LastActivityDate="2011-03-10T23:14:38.110" Title="Fubini's theorem problem" Tags="&lt;homework&gt;&lt;real-analysis&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="3" />
  <row Id="26216" PostTypeId="1" AcceptedAnswerId="26217" CreationDate="2011-03-10T12:22:44.893" Score="0" ViewCount="155" Body="&lt;p&gt;For example, two Green functions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;G_1(\tau_1 - \tau_2, x) = \alpha_1(\tau_1 - \tau_2)\beta_1(x) H(\tau_1 - \tau_2) + \mu_1(\tau_1 - \tau_2) \nu_1 (x)  H(\tau_2 - \tau_1)&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;G_2(\tau_2 - \tau_3, x') = \alpha_2(\tau_2 - \tau_3)\beta_2(x') H(\tau_2 - \tau_3) + \mu_2(\tau_2 - \tau_3) \nu_2 (x')  H(\tau_3 - \tau_2)&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where the H is a step function, $H(\tau_a - \tau_b)=1$ when $\tau_a &amp;gt; \tau_b$ and zero otherwise. Can a product of these two Green functions be integrated? -&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;\int_0^y d\tau_2 G_1(\tau_1 - \tau_2, x) G_2(\tau_2 - \tau_3, x')&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If they can be, is there a general method to do so?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; For an easier comment below, I'll use&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;G_1(\tau_1 - \tau_2, x) = c H(\tau_1 - \tau_2) + d  H(\tau_2 - \tau_1)&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;G_2(\tau_2 - \tau_3, x') = a H(\tau_2 - \tau_3) + b H(\tau_3 - \tau_2)&#xA;\end{equation}&lt;/p&gt;&#xA;" OwnerUserId="10090" LastEditorUserId="1543" LastEditorDisplayName="" LastEditDate="2011-03-10T17:02:52.810" LastActivityDate="2011-04-09T17:39:57.287" Title="Can a product of 2 step functions be integrated?" Tags="&lt;integral&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="26217" PostTypeId="2" ParentId="26216" CreationDate="2011-03-10T12:56:41.247" Score="0" ViewCount="" Body="&lt;p&gt;Yes I think so. Expand the integrand into the 4 components like:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;[\alpha_1(\tau_1 - \tau_2)\beta_1(x) H(\tau_1 - \tau_2)][ \alpha_2(\tau_2 - \tau_3) \beta_2 (x)  H(\tau_2 - \tau_3)] + ....&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;giving&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;[\alpha_1(\tau_1 - \tau_2)\beta_1(x) \alpha_2(\tau_2 - \tau_3) \beta_2 (x) H(\tau_1 - \tau_2) H(\tau_2 - \tau_3)] + ....&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;to proceed we need to determine the relation between $\tau_1$ and  $\tau_3$, say $\tau_1&amp;gt; \tau_3$. Now the conditions from the 4 double H terms will expand onto conditions between the three $\tau$s. The fourth case is $\tau$ inconsistent so we get three such cases. Then proceed with each $\alpha$ $\tau_2$ integral. You may need to determine the relation between the $\tau$s and 0 and y as well.&lt;/p&gt;&#xA;" LastActivityDate="2011-03-10T12:56:41.247" CommentCount="2" />
  <row Id="26218" PostTypeId="2" ParentId="26204" CreationDate="2011-03-10T17:02:04.310" Score="2" ViewCount="" Body="&lt;p&gt;See the &lt;a href=&quot;http://en.wikipedia.org/wiki/Lagrange_inversion_theorem&quot; rel=&quot;nofollow&quot;&gt;Lagrange inversion theorem&lt;/a&gt;: &quot;$g$ is analytic at the point $b=f(a)$&quot;.&lt;/p&gt;&#xA;" OwnerUserId="6622" LastActivityDate="2011-03-10T17:02:04.310" />
  <row Id="26219" PostTypeId="2" ParentId="26160" CreationDate="2011-03-10T17:04:35.557" Score="0" ViewCount="" Body="&lt;p&gt;Cheat: both the universal and existential quantifiers require a &quot;universal set&quot; of discussion, although this is often omitted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you choose your universal set as the empty set, I the seemingly contradictory conditions are both vacuously true.&lt;/p&gt;&#xA;" OwnerUserId="2469" LastActivityDate="2011-03-10T17:04:35.557" />
  <row Id="26220" PostTypeId="2" ParentId="26215" CreationDate="2011-03-10T17:20:28.470" Score="4" ViewCount="" Body="&lt;p&gt;Your thought is a good one.  If you integrate in $y$ first you can change variable to $u=x-3y$.  You are considering $x$ a constant, so you get a decoupled product of integrals of $f$ over the real line.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2011-03-10T18:14:53.680" LastActivityDate="2011-03-10T18:14:53.680" CommentCount="1" />
  <row Id="26222" PostTypeId="1" AcceptedAnswerId="26232" CreationDate="2011-03-10T17:35:28.397" Score="0" ViewCount="279" Body="&lt;p&gt;What are the general uses of the hat and star symbol in math? Or could you please point me to a page that discusses this? Thanks&lt;/p&gt;&#xA;" OwnerUserId="6289" LastActivityDate="2011-03-10T21:46:51.913" CommunityOwnedDate="2011-03-10T17:55:08.863" Title="What does a hat or star means in math?" Tags="&lt;notation&gt;" AnswerCount="7" CommentCount="3" />
  <row Id="26223" PostTypeId="2" ParentId="26170" CreationDate="2011-03-10T17:35:36.460" Score="0" ViewCount="" Body="&lt;p&gt;I'd like to add a few things to complement &lt;a href=&quot;http://math.stackexchange.com/questions/26170/matrix-inverses-and-eigenvalues/26200#26200&quot;&gt;Arturo's&lt;/a&gt; and &lt;a href=&quot;http://math.stackexchange.com/questions/26170/matrix-inverses-and-eigenvalues/26174#26174&quot;&gt;Fabian's&lt;/a&gt; answers.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you take the outer product of a unit vector, $\hat{e}$ ($\lvert e \rangle$, &lt;a href=&quot;http://en.wikipedia.org/wiki/Dirac_notation&quot; rel=&quot;nofollow&quot;&gt;Dirac notation&lt;/a&gt;), with its &lt;a href=&quot;http://en.wikipedia.org/wiki/Dual_vector#Dual_vectors_and_bilinear_forms&quot; rel=&quot;nofollow&quot;&gt;dual&lt;/a&gt;, $\hat{e}^\ast$ ($\langle e \rvert$), you get a matrix that projects vectors onto the space defined by the unit vector, i.e.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{aligned}&#xD;&#xA;&#xD;&#xA;\mathbf{P}_e &amp;amp;= \hat{e} \otimes \hat{e}^\ast \\&#xD;&#xA;&amp;amp;= \lvert e \rangle \langle e \rvert \\&#xD;&#xA;&amp;amp;= \left( \begin{array}{ccc} \lVert e_{1} \rVert^2 &amp;amp; e_{1} e_2^\ast &amp;amp; \ldots \\ &#xD;&#xA;e_2 e_1^\ast &amp;amp; \lVert e_{2} \rVert^2 &amp;amp; \ldots \\ &#xD;&#xA;\vdots &amp;amp; \vdots &amp;amp; \ddots \end{array} \right)&#xD;&#xA;&#xD;&#xA;\end{aligned}&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \mathbf{P}_e \vec{v} = (\vec{v} \cdot \hat{e}) \hat{e}. $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, $\mathbf{P}_e$ is a projection operator. Using this, you can rewrite a matrix, $\mathbf{A}$, in terms of its eigenvalues, $\lambda_i$, and eigenvectors, $\lvert e_i \rangle$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \mathbf{A} = \sum_i \lambda_i \lvert e_i \rangle \langle e_i \rvert$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is called the &lt;a href=&quot;http://en.wikipedia.org/wiki/Spectral_decomposition&quot; rel=&quot;nofollow&quot;&gt;spectral decomposition&lt;/a&gt;.  From this, it is plain to see that any eigenvector, $\hat{e}_i$, with a zero eigenvalue does not contribute to the matrix, and for any vector component in one of those spaces, $\mathbf{P}_{e_i}\vec{v} = \vec{v}_i$, $$\mathbf{A} \vec{v}_i = 0.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This implies that the dimensionality of the space that $\mathbf{A}$ operates on is smaller than the dimension of the space used to describe it.  In other words, $\mathbf{A}$ does not posses &lt;a href=&quot;http://en.wikipedia.org/wiki/Matrix_rank&quot; rel=&quot;nofollow&quot;&gt;full rank&lt;/a&gt;, and is not invertible.&lt;/p&gt;&#xA;" OwnerUserId="2283" LastActivityDate="2011-03-10T17:35:36.460" />
  <row Id="26224" PostTypeId="2" ParentId="26222" CreationDate="2011-03-10T17:37:21.300" Score="0" ViewCount="" Body="&lt;p&gt;The &lt;strong&gt;Hat&lt;/strong&gt; symbol can be used to denote a vector. And the &lt;strong&gt;Star&lt;/strong&gt; symbol may possibly used to denote a binary operation. For example a non empty set $G$ with a binary operation $\star$ is said to be  a Group if.....&lt;/p&gt;&#xA;" LastActivityDate="2011-03-10T17:37:21.300" CommunityOwnedDate="2011-03-10T17:55:08.863" CommentCount="5" />
  <row Id="26225" PostTypeId="2" ParentId="26203" CreationDate="2011-03-10T17:37:31.343" Score="1" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\rm\displaystyle\ \ \ \ f(n)\ =\ \prod_{j\ =\ 1}^n\ a_j $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad \iff\ \ \ \:f\:(n)\ =\:\ a_n\ \: * \ \ f\:(n-1),\:\ \ \ f\:(0)\: = 1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad \iff\ \ F(n)\ =\ A_n + F(n-1),\ \ F(0) = 0\:,\ $ with $\rm\ \ F(n) = \ln\: f(n)\:,\ \ A_n = \ln\: a_n$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\displaystyle\quad \iff\ \ F(n)\ =\ \sum_{j\ =\ 1}^n\ A_n$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first and last equivalences are the recursive definitions of $\rm\:\Pi\:$ and $\rm\:\Sigma\:.$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The middle equivalence follows from $\rm\ \ln\ (x\ *\ y)\ =\ \ln\ x\ +\ \ln\ y\:.$&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2011-03-10T18:18:57.547" LastActivityDate="2011-03-10T18:18:57.547" CommentCount="1" />
  <row Id="26226" PostTypeId="2" ParentId="26222" CreationDate="2011-03-10T17:41:05.007" Score="1" ViewCount="" Body="&lt;p&gt;$\hat{}$ can also be used to denote the Fourier transform $\hat{f}$ of an integrable function $f$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ast$ can be used to denote the convolution product $f \ast g$ of two functions $f$ and $g$.&lt;/p&gt;&#xA;" OwnerUserId="5798" LastActivityDate="2011-03-10T17:41:05.007" CommunityOwnedDate="2011-03-10T17:55:08.863" CommentCount="0" />
  <row Id="26227" PostTypeId="2" ParentId="26222" CreationDate="2011-03-10T17:43:18.077" Score="2" ViewCount="" Body="&lt;p&gt;I have seen the star used for multiplication, hermitian conjugate of a matrix, special values of a variable (given a function $f(x), x^*$ might be the value of $x$ that minimizes $f$), among others.  In Conway's theory of games, * is the game that wins for the first player.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2011-03-10T18:36:25.147" LastActivityDate="2011-03-10T18:36:25.147" CommunityOwnedDate="2011-03-10T17:55:08.863" />
  <row Id="26228" PostTypeId="1" AcceptedAnswerId="26230" CreationDate="2011-03-10T17:47:10.900" Score="1" ViewCount="49" Body="&lt;p&gt;How can I find the volume of the solid generated when the region enclosed by $y=0, x=0, x=1$ and $(1+e^-2x)^0.5$ is rotated through 360 degrees about the x axis? &lt;/p&gt;&#xA;" OwnerUserId="8078" LastEditorUserId="8078" LastEditorDisplayName="" LastEditDate="2011-03-10T17:56:08.460" LastActivityDate="2011-03-10T18:06:34.843" Title="Volume of a rotated region?" Tags="&lt;calculus&gt;&lt;geometry&gt;&lt;integral&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="26229" PostTypeId="2" ParentId="26228" CreationDate="2011-03-10T17:53:28.943" Score="2" ViewCount="" Body="&lt;p&gt;The volume of an object can be found by integrating the cross-sectional area through all cross-sections.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A cross-section of this solid, taken perpendicularly to the $x$-axis at a distance $x$ from the $y$-axis, is a disk with radius $(1+e^{-2x})^{0.5}$.  Express the area of this disk as a function of $x$, and then integrate this function from $x=0$ to $x=1$.&lt;/p&gt;&#xA;" OwnerUserId="2937" LastEditorUserId="2937" LastEditorDisplayName="" LastEditDate="2011-03-10T18:06:34.843" LastActivityDate="2011-03-10T18:06:34.843" />
  <row Id="26230" PostTypeId="2" ParentId="26228" CreationDate="2011-03-10T17:53:42.140" Score="1" ViewCount="" Body="&lt;p&gt;I'm not completely sure that I understand what you mean, but try this:&#xA;$$\pi\int_0^1 f(x)^2 \text{ d}x$$&#xA;where $f(x)$ in this case is your function $(1+e^{-2x})^{\frac{1}{2}}$.&lt;/p&gt;&#xA;" OwnerUserId="7044" LastActivityDate="2011-03-10T17:53:42.140" CommentCount="4" />
  <row Id="26231" PostTypeId="1" AcceptedAnswerId="26241" CreationDate="2011-03-10T17:57:10.513" Score="3" ViewCount="177" Body="&lt;p&gt;Someone asked this question about how many ways there are to prove $0.999\dots = 1$ and I posted this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$  0.99999 \dots = \sum_{k = 1}^\infty \frac{9}{10^k} = 9  \sum_{k = 1}^\infty \frac{1}{10^k} = 9 \Big ( \frac{1}{1 - \frac{1}{10}} - 1\Big ) = \frac{9}{9} = 1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question was a duplicate so in the end it was closed but before that someone  wrote in a comment to the question: &quot;Guys, please stop posting pseudo-proofs on an exact duplicate!&quot; and I got down votes, so I assume this proof is wrong.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I would like to know, of course, why this proof is wrong. I have thought about it but somehow I can't seem to find the mistake. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many thanks for your help. The original can be found &lt;a href=&quot;http://math.stackexchange.com/questions/21674/proof-of-o-999-1-using-different-methods-approaches&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="5798" LastEditorUserId="5798" LastEditorDisplayName="" LastEditDate="2011-03-10T19:41:29.363" LastActivityDate="2011-03-10T19:41:29.363" Title="Regarding a proof involving geometric series" Tags="&lt;real-analysis&gt;&lt;alternative-proof&gt;" AnswerCount="2" CommentCount="11" />
  <row Id="26232" PostTypeId="2" ParentId="26222" CreationDate="2011-03-10T18:16:09.037" Score="3" ViewCount="" Body="&lt;p&gt;Different branches of mathematics may have varying conventional usages of these kind of &quot;decorations&quot;.  Typically they denote a transformed version of the base variable (e.g. $\hat{f}$ denoting the Fourier transform of $f$ as mentioned in another answer).  Or, they may denote a special or specific value of a variable ($x^*$ giving the value of $x$ minimizing $f$ from another answer.)  The $*$ symbol is often used for arbitrary associative binary operations.  Etc. etc.&lt;/p&gt;&#xA;" OwnerUserId="3852" LastActivityDate="2011-03-10T18:16:09.037" CommunityOwnedDate="2011-03-10T18:16:09.037" CommentCount="2" />
  <row Id="26233" PostTypeId="1" CreationDate="2011-03-10T18:18:17.440" Score="0" ViewCount="80" Body="&lt;p&gt;Suppose $H$,$K$ are normal subgroups of a group $G$. We don't assume $G$ is finite. Furthermore assume $[G:H]$ and $[G:K]$ are both finite. Then it follows that $[G: H \cap K]$ is divisible by $[G:H]$. Why is this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's what I thought. Suppose we write $G = \bigcup_{i \in I} a_{i}(H \cap K)$ as a disjoint&#xA;union of left cosets in $H \cap K$, here $I$ is some index set.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now $a_{i}(H \cap K) \subset a_{i}H$ hence $G \subset \bigcup_{i \in I} a_{i}H$. So we get that $I$ is at least as big as $[G:H]$, thus $[G: H \cap K]$ divides $[G: H]$ no. ? Why is this wrong? why is the other way around? &lt;/p&gt;&#xA;" OwnerUserId="8079" LastActivityDate="2011-04-09T19:39:57.293" Title="Question about indices of subgroups" Tags="&lt;group-theory&gt;" AnswerCount="2" CommentCount="5" />
  <row Id="26234" PostTypeId="2" ParentId="26222" CreationDate="2011-03-10T18:34:13.947" Score="5" ViewCount="" Body="&lt;p&gt;There is a nice list for $*$ in this &lt;a href=&quot;http://en.wikipedia.org/wiki/Asterisk&quot; rel=&quot;nofollow&quot;&gt;article&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I guess another (more general) term for &quot;hat&quot; is &lt;a href=&quot;http://en.wikipedia.org/wiki/Circumflex&quot; rel=&quot;nofollow&quot;&gt;Circumflex&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4266" LastActivityDate="2011-03-10T18:34:13.947" CommunityOwnedDate="2011-03-10T18:34:13.947" />
  <row Id="26235" PostTypeId="2" ParentId="26231" CreationDate="2011-03-10T18:35:40.823" Score="1" ViewCount="" Body="&lt;p&gt;I got this as an assignment in my first university-level calculus course: To prove $1 = 0.999...$. My proof was essentially the same as yours, and it was good enough for them. The only &quot;mistake&quot; I can think of must be the one Qiaochu is pointing out, that you assume to much about convergence of series. &lt;/p&gt;&#xA;" OwnerUserId="7044" LastActivityDate="2011-03-10T18:35:40.823" />
  <row Id="26236" PostTypeId="2" ParentId="26233" CreationDate="2011-03-10T18:40:21.023" Score="0" ViewCount="" Body="&lt;p&gt;I assume that you are familiar with this result: if M,N,G are groups and N &amp;lt; M &amp;lt; G (where &amp;lt; means &quot;to be a subgroup of&quot;), then [G:N]=[G:M][M:N]. In your case, take N=H\cap K, and M=H and apply the result I mentioned.&lt;/p&gt;&#xA;" OwnerUserId="8050" LastActivityDate="2011-03-10T18:40:21.023" />
  <row Id="26237" PostTypeId="2" ParentId="26233" CreationDate="2011-03-10T18:41:50.870" Score="1" ViewCount="" Body="&lt;p&gt;Your argument goes wrong here: &quot;So we get that $I$ is at least as big as $[G:H]$, thus $[G: H \cap K]$ divides $[G: H]$&quot;. Since $|I|=[G: H \cap K]$, &quot;$I$ is at least as big as $[G:H]$&quot; means $[G: H \cap K] \ge [G: H]$, which doesn't go well with, let alone implies, &quot;$[G: H \cap K]$ divides $[G: H]$&quot;.&lt;/p&gt;&#xA;" OwnerUserId="6622" LastActivityDate="2011-03-10T18:41:50.870" />
  <row Id="26238" PostTypeId="2" ParentId="26222" CreationDate="2011-03-10T18:42:48.793" Score="2" ViewCount="" Body="&lt;p&gt;how about omitted terms&#xA;$$&#xD;&#xA;\partial\langle x_0,...,x_n\rangle=\sum_{i=0}^n(-1)^i\langle x_0,...,\hat{x_i},...x_n\rangle&#xD;&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="6925" LastEditorUserId="6925" LastEditorDisplayName="" LastEditDate="2011-03-10T21:46:51.913" LastActivityDate="2011-03-10T21:46:51.913" CommunityOwnedDate="2011-03-10T18:42:48.793" />
  <row Id="26239" PostTypeId="2" ParentId="26222" CreationDate="2011-03-10T18:47:49.533" Score="0" ViewCount="" Body="&lt;p&gt;According to &lt;a href=&quot;http://en.wikipedia.org/wiki/ISO_31-11&quot; rel=&quot;nofollow&quot;&gt;ISO 31-11&lt;/a&gt;:&#xA;$$\begin{align}&#xD;&#xA;\mathbb{N}^* = \mathbb{N}-\{0\} \\&#xD;&#xA;\mathbb{Z}^* = \mathbb{Z}-\{0\}&#xD;&#xA;\end{align}$$&#xA;The same goes for $\mathbb{Q}, \mathbb{R}, \mathbb{C}$.&#xA;$$\begin{align}&#xD;&#xA;z^* = \text{complex conjugate of } z.&#xD;&#xA;\end{align}$$&lt;/p&gt;&#xA;" OwnerUserId="7044" LastActivityDate="2011-03-10T18:47:49.533" CommunityOwnedDate="2011-03-10T18:47:49.533" />
  <row Id="26240" PostTypeId="2" ParentId="26205" CreationDate="2011-03-10T18:49:05.773" Score="1" ViewCount="" Body="&lt;p&gt;The implicit function theorem really just boils down to this: if I can write down $m$ (sufficiently nice!) equations in $n + m$ variables, then, near any sufficiently nice solution point, there is a function of $n$ variables which give me the remaining $m$ coordinates of nearby solution points. In other words, I can, in principle, solve those equations and get the last $m$ variables in terms of the first $n$ variables. But (!) in general this function is only valid on some small set and won't give you all the solutions either. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a concrete example. Consider the equation $x^2 + y^2 = 1$. This is a single equation in two variables, and for a fixed $x_0 \ne 1, y_0 \ne$ satisfying the equation, there is a function $f$ of $x$ such that $x^2 + f(x)^2 = 1$ for $y$ near $x_0$, and $f(x_0) = y_0$. (Explicitly, for $y_0 &amp;gt; 0$, $f(x) = \sqrt{1 - x^2}$, and for $y_0 &amp;lt; 0$, $f(x) = -\sqrt{1 - x^2}$.) Notice that the function doesn't give you all the solution points — but this isn't surprising, since the solution locus of this equation is a circle, which isn't the graph of any function. Nonetheless, I have basically solved the equation and written $y$ in terms of $x$.&lt;/p&gt;&#xA;" OwnerUserId="5191" LastActivityDate="2011-03-10T18:49:05.773" />
  <row Id="26241" PostTypeId="2" ParentId="26231" CreationDate="2011-03-10T18:59:31.880" Score="4" ViewCount="" Body="&lt;p&gt;The problem is that you are assuming 1) that multiplication by constants distributes over infinite sums, and 2) the validity of the geometric series formula. Most of the content of the result is in 2), so it doesn't make much sense to me to assume it in order to prove the result. Instead you should prove 2), and if you really want to be precise you should also prove 1). &lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2011-03-10T18:59:31.880" CommentCount="6" />
  <row Id="26242" PostTypeId="2" ParentId="10554" CreationDate="2011-03-10T19:05:42.250" Score="0" ViewCount="" Body="&lt;p&gt;If you consider TCS to be maths, then formal semantics, in particular small-step or big-step semantics, might be of interest. Those map input data and a series of operations to a sequence of universes/states, on per time step.&lt;/p&gt;&#xA;" OwnerUserId="3330" LastActivityDate="2011-03-10T19:05:42.250" />
  <row Id="26243" PostTypeId="1" AcceptedAnswerId="26248" CreationDate="2011-03-10T19:08:09.127" Score="14" ViewCount="235" Body="&lt;p&gt;I'm reading &lt;a href=&quot;http://www.math.uconn.edu/~kconrad/blurbs/grouptheory/subgpseries1.pdf&quot; rel=&quot;nofollow&quot;&gt;this article&lt;/a&gt; by Keith Conrad, on subgroup series. I'm having trouble with a statement he does at page 6:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Any subgroup of $G$ which contains $[G,G]$ is normal in $G$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;He says this as evidence that commutator and center play dual roles, since any subgroup of $G$ contained in $Z(G)$ is normal in $G$. Now, this I'm sure I understand, but I don't see how the quoted line holds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I have read is that $[G,G]$ is the least normal subgroup of $G$ such that the quotient is abelian, which seems related.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, while we're at it: are the center and the commutator &quot;really&quot; (as in, categorically) dual constructions? I'm quite a novice in category theory, so please excuse me if this question is trivial.&lt;/p&gt;&#xA;" OwnerUserId="2614" LastActivityDate="2011-03-11T14:37:56.613" Title="Center-commutator duality" Tags="&lt;abstract-algebra&gt;&lt;group-theory&gt;" AnswerCount="3" FavoriteCount="8" />
  <row Id="26244" PostTypeId="2" ParentId="26243" CreationDate="2011-03-10T19:20:59.197" Score="2" ViewCount="" Body="&lt;p&gt;Let $H$ be a subgroup of $G$ containing all commutators.  To show $H$ is normal it would suffice to take any $h\in H$ and $g\in G$ and show that $g^{-1}hg\in H$.  But this element is the product of $h$ with a commutator (namely $h^{-1}g^{-1}hg$) so by assumption it is in $H$.&lt;/p&gt;&#xA;" OwnerUserId="6071" LastActivityDate="2011-03-10T19:20:59.197" CommentCount="1" />
  <row Id="26246" PostTypeId="2" ParentId="26243" CreationDate="2011-03-10T19:31:31.540" Score="3" ViewCount="" Body="&lt;p&gt;&quot;Dual&quot; means many things, and &quot;categorically dual&quot; is only one of them. To me the center and the commutator are very different constructions. The commutator is nothing more than the kernel of the abelianization. Abelianization $G \mapsto G/[G, G]$ defines a functor $\text{Grp} \to \text{Ab}$ which is left adjoint to the forgetful functor $\text{Ab} \to \text{Grp}$, realizing $\text{Ab}$ as a &lt;a href=&quot;http://en.wikipedia.org/wiki/Reflective_subcategory&quot; rel=&quot;nofollow&quot;&gt;reflective subcategory&lt;/a&gt; of $\text{Grp}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The center, however, is &lt;em&gt;not&lt;/em&gt; a functorial construction. Rather it should be thought of as a kind of &quot;higher automorphism group&quot; of $G$. More precisely, every group $G$ defines a category $C_G$ with one object whose morphisms are the elements of $G$ (composing like the elements of $G$ do), and it turns out that $Z(G)$ is precisely the automorphism group of the identity functor $C_G \to C_G$ (that is, it's the group of natural equivalences from this functor to itself). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; in particular the center is not universal for homomorphisms from an abelian group into $G$. More precisely, the forgetful functor $\text{Ab} \to \text{Grp}$ does not have a &lt;em&gt;right&lt;/em&gt; adjoint, and this is because it does not preserve coproducts. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course a more typical way to think of $Z(G)$ is as the kernel of the map $G \to \text{Aut}(G)$ given by conjugation. Again, this isn't functorial, but it is natural in a certain sense. &lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2011-03-10T19:39:20.940" LastActivityDate="2011-03-10T19:39:20.940" CommentCount="1" />
  <row Id="26248" PostTypeId="2" ParentId="26243" CreationDate="2011-03-10T19:49:37.203" Score="20" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Rewritten to ease the &quot;start-up&quot; cost.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;hr/&gt;&#xA;First, your first question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fact that any subgroup of $G$ that contains $[G,G]$ is normal in $G$ follows by the isomorphism theorems: if $H$ contains $N\triangleleft G$, then $H/N$ is a subgroup of $G/N$, and $H\triangleleft G$ if and only if $(H/N)\triangleleft (G/N)$. In the instant case, $N=[G,G]$ and $G/N$ is abelian, hence $H/[G,G]$ is necessarily normal in $G/[G,G]$ (since &lt;em&gt;every&lt;/em&gt; subgroup of an abelian group is abelian), so $H$ is normal in $G$, as claimed.&lt;/p&gt;&#xA;&#xA;&lt;hr/&gt;&#xA;&#xA;&lt;p&gt;Your second question is actually much more interesting (well, to me). The short answer is that the center and the commutator subgroup are not dual constructions in the &lt;em&gt;categorical sense&lt;/em&gt;, but they are dual constructions in a different sense (which &lt;em&gt;is&lt;/em&gt; related to a special kind of category that turns up a lot in General Algebra). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, let me give &quot;universal&quot; definitions of both subgroups, so we can see how the two are directly connected to one another.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;The commutator and the center.&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;As you note, the commutator subgroup of $G$ is the smallest normal subgroup $N$ of $G$ such that $G/N$ is abelian. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One way to characterize abelian groups is via &lt;em&gt;identities&lt;/em&gt;: equations in the language of group theory that may or may not be satisfied by a group. For example, the identity &quot;$xy=yx$&quot; is not satisfied by every group (meaning, given a group $G$, it need not be true that for all $x,y\in G$ you have $xy=yx$, though it may be true  for &lt;em&gt;some&lt;/em&gt; choices of $x,y\in G$). However, those groups that &lt;em&gt;do&lt;/em&gt; satisfy this identity are &lt;em&gt;precisely&lt;/em&gt; the abelian groups.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any identity in group theory can be rewritten in the form $\mathbf{w}=1$, where $\mathbf{w}$ is a word in the language of group theory (a product of variables and their inverses); just move everything to the same side. For abelian groups, we have $(xy)(yx)^{-1}=1$, or $xyx^{-1}y^{-1}=1$. You may recognize the left hand side as none other than the commutator $[x,y]$ of $x$ and $y$.  Call it $\mathbf{w}(x,y)$; that is, $\mathbf{w}(x,y) = [x,y] = xyx^{-1}y^{-1}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So another way of defining the commutator subgroup of $G$ is to use the word $\mathbf{w}(x,y)$ that characterizes the class of all abelian groups (as &quot;the groups for which every possible evaluation of that word equals the identity&quot;): the commutator subgroup of $G$ is &lt;em&gt;precisely&lt;/em&gt; the subgroup generated by all values that $\mathbf{w}(x,y)$ takes as $x$ and $y$ range over the elements of the group. Because we define the subgroup as &quot;the subgroup generated by all values that the word $\mathbf{w}(x,y)$ takes&quot;, we say that this is a &lt;strong&gt;verbal subgroup&lt;/strong&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is an interesting connection between the center and the commutator subgroup: you can think of the word $\mathbf{w}(x,y)$ as giving a function (&lt;em&gt;not&lt;/em&gt; a homomorphism) $G\times G\to G$, that maps $(x,y)$ to $\mathbf{w}(x,y)$. Even though it is not a homomorphism, we may ask: does it factor through something? Is there a normal subgroup $N$ of $G$ such that the map $G\times G\to G$ factors through $(G/N)\times (G/N)$? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words: what is the set of all elements $z$ of $G$ such that, for all $x,y\in G$, $[xz,y] = [zx,y] = [x,zy] = [x,yz] = [x,y]$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll spill the beans: it's the &lt;em&gt;center&lt;/em&gt;. To see this, note that if $z$ has this property, then $[z,y] = [1,y] = 1$ for all $y\in G$, so necessarily $z\in Z(G)$. Conversely, if $z\in Z(G)$, then &#xA;$$[xz,y] = xzy(xz)^{-1}y^{-1} = xzyz^{-1}x^{-1}y^{-1} = xzz^{-1}yx^{-1}y^{-1}=xyx^{-1}y^{-1}=[x,y]$$&#xA;and similarly for $[zx,y]$, $[x,zy]$, and $[x,yz]$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this context, the center is refered to as the &lt;strong&gt;marginal subgroup&lt;/strong&gt; of $G$ relative to the word $\mathbf{w}(x,y)$.  The term was coined by Phillip Hall; the idea is that these are elements that can be &quot;pushed to the margins&quot; when computing the values of $\mathbf{w}(x,y)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So here's a sense in which the commutator subgroup and the center are &quot;dual&quot;: the commutator is the subgroup generated by all &lt;em&gt;values&lt;/em&gt; of $\mathbf{w}(x,y)$, and the center is the subgroup of all elements that &lt;em&gt;don't affect&lt;/em&gt; the values of $\mathbf{w}(x,y)$ when you introduce them as factors. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, there are several ways in which the two constructions are &lt;em&gt;not&lt;/em&gt; truly &quot;dual&quot; in a categorical sense:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;The commutator subgroup is associated to the left adjoint of the forgetful functor from $\mathcal{A}b\mathcal{G}roup$ to $\mathcal{G}roup$. The left adjoint maps a group $G$ to $G/[G,G]$; that is, there is a natural bijection between the homomorphism from $G$ to an abelian group $A$, and the (abelian group) homomorphisms between $G/[G,G]$ and $A$. The group $G/[G,G]$ is left universal among all abelian groups into which $G$ maps, etc. All the things that are related to the fact that we have a left adjoint and that $\mathcal{A}b\mathcal{G}roup$ is a reflective subcategory of $\mathcal{G}roup$, as Qiaochu mentioned. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the center is not associated to a right adjoint.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The commutator subgroup is the equalizer of all maps from $G$ into abelian groups; the center, however, is not the coequalizer of all map from abelian groups into $G$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;While $[G,G]$ is always the smallest normal subgroup of $G$ such that the quotient is abelian, it is not true in general that $Z(G)$ is the largest normal subgroup of $G$ that is abelian.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The set of all marginal elements is already, directly, a subgroup; we don't need to talk about the &quot;subgroup generated by all marginal elements of $G$&quot;. But the set of all values of $\mathbf{w}(x,y)$ is &lt;em&gt;not necessarily&lt;/em&gt; a subgroup, we need to talk about the &quot;subgroup they generate&quot; to get the corresponding verbal subgroup.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Nonetheless, the construction of verbal and marginal subgroups does establish &lt;em&gt;some&lt;/em&gt; &quot;dual relation&quot; between the two. In fact, that relation fits into a much larger picture of verbal and marginal subgroups (which also explains the &quot;duality&quot; between the upper and lower central series of a group, which is what Keith Conrad's notes point out).&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Verbal and marginal subgroups in general&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;The commutator subgroup of $G$ is an example of a &lt;strong&gt;verbal subgroup&lt;/strong&gt; of $G$, and the center is the corresponding &lt;strong&gt;marginal&lt;/strong&gt; subgroup.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Varieties.&lt;/strong&gt; The key to these notions is the idea of a &lt;em&gt;variety of groups.&lt;/em&gt; A &lt;em&gt;variety of groups&lt;/em&gt; is any collection of groups which is closed under subgroups, homomorphic images, and arbitrary direct products. That is, if $\mathfrak{V}$ is the class, then you require that if $H\lt G$ and $G\in\mathfrak{V}$, then $H\in\mathfrak{V}$. That if $\varphi\colon G\to K$ is an onto group homomorphism and $G\in\mathfrak{V}$, then $K\in\mathfrak{V}$; and that if $\{G_i\}_{i\in I}$ is an arbitrary family of groups with $G_i\in\mathfrak{V}$ for all $i\in I$, then $\prod_{i\in I}G_i\in\mathfrak{V}$.  (Note that since the empty product is the trivial group, a variety of groups is always nonempty.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Examples of varieties are: the class of all groups; the class of all abelian groups; the class of all groups of exponent $n$ for a given $n$; the class of all nilpotent groups of class at most $c$ (for a given $c$); the class of all solvable groups of solvability length at most $s$ (for a given, fixed $s$). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Examples of collections that are &lt;em&gt;not&lt;/em&gt; varieties are: divisible groups (subgroup of divisible need not be divisible); finite groups (arbitrary products of finite groups need not be finite); solvable groups (arbitrary products of solvable groups need not be solvable); torsionfree groups (homomorphic image of torsionfree groups need not be torsionfree).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Verbal subgroups.&lt;/strong&gt; Given any variety of groups $\mathfrak{V}$, and a group $G$, there is a smallest normal subgroup of $G$, called $\mathfrak{V}(G)$, with the property that $G/\mathfrak{V}(G)\in\mathfrak{V}$. To see this, note that the family of all such normal subgroups is nonempty, since it contains $G$ itself; and given any family $\{N_i\}_{i\in I}$ of normal subgroups of $G$ such that $G/N_i\in\mathfrak{V}$ for all $i\in I$, then mapping $G$ to $\prod G/N_i \in\mathfrak{V}$ in the obvious way shows that $G/\cap N_i$ is a subgroup of a group in $\mathfrak{V}$, hence is itself in $\mathfrak{V}$. So the class of such subgroups has a smallest element (the intersection of all subgroups in the class). This subgroup is called the $\mathfrak{V}$-&lt;strong&gt;verbal subgroup of&lt;/strong&gt; $G$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(In the case where $\mathfrak{V}=\mathfrak{A}$, the variety of all abelian groups, the $\mathfrak{A}$-verbal subgroup is precisely the commutator subgroup).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason it's called a &quot;verbal subgroup&quot; is that to every variety there corresponds a set of &quot;identities&quot; that every group in $\mathfrak{V}$ satisfies, and such that every group that satisfies the identities is in $\mathfrak{V}$; these identities can be coded as words in the free group of countable rank (they are none other than the &lt;em&gt;elements&lt;/em&gt; of $\mathfrak{V}(\mathbf{F}_{\infty})$, where $\mathbf{F}_{\infty}$ is the free group of countable rank), and the $\mathfrak{V}$-verbal subgroup corresponds to the subgroup of $G$ given by all the possible evaluations of these words in $G$ (equivalently, it is the subgroup of $G$ generated by all the images of $\mathfrak{V}(\mathbf{F}_{\infty})$ under all possible homomorphism $\mathbf{F}_{\infty}\to G$). &#xA;In general you don't need &lt;em&gt;all&lt;/em&gt; the elements of $\mathfrak{V}(\mathbf{F}_{\infty})$, you only a set of elements that generates this subgroup as a fully invariant subgroup of $\mathbf{F}_{\infty}$ (see below for the definition of &quot;fully invariant subgroup&quot;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of abelian groups, for instance, the identity that defines abelian groups is $xy=yx$ (every group in which every possible evaluation makes $xy=yx$ true is abelian, and in an abelian group every possible evaluation  makes $xy=yx$ true), which can be &quot;coded&quot; as the single word $xyx^{-1}y^{-1}$. This word completely defines the variety of abelian groups in the sense that a group is abelian if and only if every evaluation of $xyx^{-1}y^{-1}$ in the group is trivial. (Though abelian groups satisfy many other identities that are not true for arbitrary groups, they can be shown to be consequences of the fact that $[x,y]=1$ for all $x$ and $y$). The verbal subgroup corresponding to the class of all abelian groups is then precisely the subgroup generated by all the possible evaluations of this word, namely the commutator subgroup.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Verbal subgroups are very, very nice: they sit at the top of a hierarchy of subgroups, that is ordered in terms of how well they behave with respect to homomorphisms:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;At the bottom we have the plain, &quot;vanilla&quot; subgroups. &lt;/li&gt;&#xA;&lt;li&gt;Then we have the normal subgroup: a subgroup $H$ of $G$ is normal if and only if for every $\varphi\in\mathrm{Inn}(G)$, $\varphi(H)\subseteq H$. ($\mathrm{Inn}(G)$ are the inner automorphisms of $G$, the automorphisms of the form $x\mapsto g^{-1}xg$, for some fixed $g\in G$).&lt;/li&gt;&#xA;&lt;li&gt;Then we have the characteristic subgroups: a subgroup $H$ of $G$ is characteristic if and only if for every $\varphi\in\mathrm{Aut}(G)$, we have $\varphi(H)\subseteq H$.&lt;/li&gt;&#xA;&lt;li&gt;Then we have the fully invariant subgroups: a subgroup $H$ of $G$ is fully invariant if and only if for every $\varphi\in\mathrm{End}(G)$, we have $\varphi(H)\subseteq H$.&lt;/li&gt;&#xA;&lt;li&gt;Verbal subgroups, however, have an even stronger property: for every group homomorphism $\varphi\colon G\to K$, between any two groupos, $\varphi(\mathfrak{V}(G))\subseteq\mathfrak{V}(K)$. That is, the image of a verbal subgroup under &lt;em&gt;any&lt;/em&gt; homomorphism is contained in the corresponding verbal subgroup of the image. In particular, verbal subgroups are fully invariant (hence characteristic, hence normal). &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;(It so happens that a subgroup of a free group is fully invariant if and only if it is verbal, which is why when we are looking for a set of generators for $\mathfrak{V}(\mathbf{F}_{\infty})$ it is enough to find one that generates it as a fully invariant subgroup). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So verbal subgroups are absolutely the bee's knees of subgroup (at least, relative to homomorphisms). They are also related to the subcategory $\mathfrak{V}$ of $\mathcal{G}roup$ in exactly the same way as the commutator subgroup is related to $\mathcal{A}b\mathcal{G}roup$: the map $G\to G/\mathfrak{V}(G)$ is a functor from $\mathcal{G}roup$ to $\mathfrak{V}$, and it is the left adjoint of the forgetful functor from $\mathfrak{V}$ to $\mathcal{G}roup$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Marginal subgroups&lt;/strong&gt; are related to verbal subgroups. Given any element $\mathbf{w}\in\mathbf{F}_{\infty}$, $\mathbf{w}$ is a word in finitely many letters, say $\mathbf{w}=\mathbf{w}(\mathbf{x}_1,\ldots,\mathbf{x}_n)$. Given a group $G$, you can look at the subgroup generated by all the values of the word $\mathbf{w}$,&#xA;$$\mathbf{w}(G) = \Bigl\langle \mathbf{w}(g_1,\ldots,g_n)\Bigm| g_1,\ldots,g_n\in G\Bigr\rangle.$$&#xA;This is just the $\mathbf{w}$-verbal subgroup of $G$, of course. But you can consider the set of all elements of $G$ that &quot;don't matter&quot; relative to $\mathbf{w}$: that is, &#xA;$$\mathbf{w}^*(G) = \left\langle x\in G\left| \begin{array}{l}&#xD;&#xA;\text{for all }g_1,\ldots,g_n\in G,\quad 1\leq i\leq n,\\&#xD;&#xA;\mathbf{w}(g_1,\ldots,g_{i-1},xg_i,g_{i+1},\ldots,g_n) =\mathbf{w}(g_1,\ldots,g_n)\\\&#xD;&#xA;\mathbf{w}(g_1,\ldots,g_{i-1},g_ix,g_{i+1},\ldots,g_n) = \mathbf{w}(g_1,\ldots,g_n).\end{array}\right.\right\rangle.$$&#xA;It's easy to check that $\mathbf{w}^*(G)$ is a subgroup of $G$, called the &quot;marginal subgroup of $G$ relative to $\mathbf{w}$.&quot;  In fact, it is always characteristic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(In the case of $\mathbf{w}(x,y) = [x,y]$, which yields the commutator subgroup as a verbal subgroup, has marginal subgroup equal to the center, as we saw above.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given a variety $\mathfrak{V}$, we can consider the marginal subgroup relative to $\mathfrak{V}$, $\mathfrak{V}^*(G)$, by taking the intersection of all $\mathbf{w}^*(G)$ for $\mathbf{w}\in\mathfrak{V}(\mathbf{F}_{\infty})$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Marginal subgroups are not quite as nice as verbal subgroup: they are always characteristic, but seldom fully invariant, and they don't have nice functorial properties like the verbal subgroups do. But of course they are closely related to the verbal subgroups. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other examples of marginal subgroups: if you consider the word $\mathbf{w}_c(x_1,\ldots,x_{c}) = [x_1,x_2,\ldots,x_c]$ (where $[a,b,c]=[[a,b],c]$), then the verbal subgroup $\mathbf{w}_c(G)$ is precisely the $c$th term of the lower central series of $G$; the corresponding marginal subgroup $\mathbf{w}_c^*(G)$ is the $c-1$st center of $G$, $Z_{c-1}(G)$, the $c-1$st term of the &lt;em&gt;upper&lt;/em&gt; central series of $G$; and so, a group $G$ is nilpotent of class at most $c$ if and only if $\mathbf{w}_{c+1}(G) = \{1\}$, if and only if $\mathbf{w}_{c+1}^*(G) = G$; that is, if and only if $G_{c+1}=\{1\}$, if and only if $Z_c(G) = G$.  The word $\mathbf{w}_{c+1}$ defines to the variety of nilpotent groups of class at most $c$.  The fact that the upper central series occurs precisely as the marginal subgroups of the lower central series (which is defined as a series of verbal subgroups) is behind many of the nice &quot;dualities&quot; between the two.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As another example, if you take $\mathbf{w}(x) = x^n$, this word corresponds to the Burnside variety $\mathfrak{B}_n$ of all groups of exponent $n$ (groups in which every element is of exponent $n$; some call them &quot;groups of exponent &lt;em&gt;dividing&lt;/em&gt; $n$&quot;), so named due to the Burnside and restricted Burnside problems; the Burnside problem for exponent $n$ is true if and only if $\mathbf{F}_k/\mathfrak{B}_n(\mathbf{F}_k)$ is finite for every $k$, where $\mathbf{F}_k$ is the free group of rank $k$. The marginal subgroup corresponding to $\mathbf{w}$ is the set of all central element of exponent $n$; that is, $\mathbf{w}^*(G) = \{x\in Z(G)\mid x^n = 1\}$. &lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Some duality&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;So there &lt;em&gt;is&lt;/em&gt; duality between verbal subgroups and the corresponding marginal subgroups, and it is exploited when dealing with things like the Schur multiplier, Baer invariants, and isologism, among other topics. You also see it at play when you look at nilpotent groups, and consider the upper and lower central series of $G$, which play kind of &quot;dual roles&quot; (one going up, the other going down; if $G$ is nilpotent of class exactly $c$, then $Z_{i+1}(G)\subseteq G_{c-i}$, where $G_{c-i}$ is the $(c-i)$th term of the lower central series of $G$, etc.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For another example of a duality between the verbal and marginal subgroups, consider the following; given a variety $\mathfrak{V}$, the following are equivalent:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$G\in\mathfrak{V}$.&lt;/li&gt;&#xA;&lt;li&gt;$\mathfrak{V}(G) = \{1\}$.&lt;/li&gt;&#xA;&lt;li&gt;$\mathfrak{V}^*(G) = G$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The first two equivalences are just from the definition of $\mathfrak{V}(G)$ as the smallest normal subgroup of $G$ for which $G/\mathfrak{V}(G)\in\mathfrak{V}$. To see the final equivalence, note that if $\mathbf{w}(x_1,\ldots,x_n)$ is an identity of $\mathfrak{V}$ (an identity that every group in $\mathfrak{V}$ must satisfy), then for all $g_1,\ldots,g_n\in\mathfrak{V}^*(G)$ we have&#xA;$$\mathbf{w}(g_1,\ldots,g_n) = \mathbf{w}(1g_1,\ldots,1g_n) = \mathbf{w}(1,\ldots,1) = 1.$$&#xA;Therefore, if $\mathfrak{V}^*(G)=G$, then $\mathfrak{V}(G)=\{1\}$. Converserly, if $\mathfrak{V}(G)=1$, then for all words $\mathbf{w}$ corresponding to $\mathfrak{V}$, all $x_1,\ldots,x_n\in G$, all $i$, $1\leq i\leq n$, and all $g\in G$ we have &#xA;\begin{align*}&#xA;1 &amp;amp;= \mathbf{w}(x_1,\ldots,x_{i-1},x_i,x_{i+1},\ldots,x_n) = \mathbf{w}(x_1,\ldots,x_{i-1},gx_{i},x_{i+1},\ldots,x_n)\\&#xA;&amp;amp;= \mathbf{w}(x_1,\ldots,x_{i-1},x_ig,x_{i+1},\ldots,x_n),&#xA;\end{align*}&#xA;so $g\in\mathfrak{V}^*(G)$ for all $g\in G$.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;References&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;A great place to learn about varieties of groups is Hanna Neumann's classic book, &lt;strong&gt;Varieties of Groups&lt;/strong&gt; Ergebnisse der Mathematik und ihrer Grenzgebiete Band 37, Springer-Verlag, 1967. It is of course a bit dated (many of the problems posed were solved in the 70s, and there is work of Kleiman in the late 70s and early 80s that essentially shows that things can get pretty nasty when dealing with arbitrary varieties; I'm sad to say that the varietal point of view has all fallen by the way side in today's group theory). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Marginal subgroups were introduced in a very short note by Phillip Hall: &lt;em&gt;Verbal and marginal subgroups&lt;/em&gt;, Phillip Hall, J. Reine Angew. Math. &lt;strong&gt;182&lt;/strong&gt; (1940), pp. 156-157, MR 2, 125i.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to see verbal and marginal subgroups play a major role and their interplay with homology, there is the tour de force by C.R. Leedham-Green and S. McKay, &lt;em&gt;Baer invariants, isologisms, varietal laws, and homology&lt;/em&gt;, Acta Math. &lt;strong&gt;136&lt;/strong&gt; (1976) no. 1-2, pp. 99-150, MR 0435250 (55 #8210). This one is &lt;em&gt;very&lt;/em&gt; hard going, though, so since you are only beginning, you might want to set it aside for a while before trying to dig in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From a slightly more general point of view, varieties of groups (and marginal subgroups) are the intersection of Group Theory and General Algebra (also called &quot;Universal Algebra&quot;); the idea of &quot;identities&quot; giving important subcategories of the collection of all your objects is central to General Algebra, and a lot of the universal and functorial properties of verbal subgroups are true in much more general settings. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can get a pretty good introduction to the &lt;em&gt;viewpoint&lt;/em&gt; of varieties of groups (if not to the specific properties they have by virtue of they being varieties &lt;em&gt;of groups&lt;/em&gt;) with the oft-recommended (by me, anyway) &lt;a href=&quot;http://math.berkeley.edu/~gbergman/245/&quot; rel=&quot;nofollow&quot;&gt;General Algebra book by George Bergman&lt;/a&gt;. The files are in postscript, but you can get them converted to PDF with freely available software/websites like ps2pdf. &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-11T14:37:56.613" LastActivityDate="2011-03-11T14:37:56.613" CommentCount="12" />
  <row Id="26249" PostTypeId="1" CreationDate="2011-03-10T19:50:02.650" Score="2" ViewCount="193" Body="&lt;p&gt;&lt;strong&gt;Definition&lt;/strong&gt;. A &lt;em&gt;&lt;strong&gt;subspace&lt;/strong&gt;&lt;/em&gt;  &lt;em&gt;of a vector space V is a subset H of V that has three properties&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(a) &lt;em&gt;The zero vector of V is in H&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(b) $\forall$ &lt;em&gt;&lt;strong&gt;u&lt;/strong&gt;&lt;/em&gt;, &lt;em&gt;&lt;strong&gt;v&lt;/strong&gt;&lt;/em&gt; $\in H$, &lt;em&gt;we have&lt;/em&gt;  &lt;em&gt;&lt;strong&gt;u&lt;/strong&gt;&lt;/em&gt; + &lt;em&gt;&lt;strong&gt;v&lt;/strong&gt;&lt;/em&gt; $\in H$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(c) $\forall$ &lt;em&gt;&lt;strong&gt;u&lt;/strong&gt;&lt;/em&gt; $\in H$, &lt;em&gt;c scalar, we have c&lt;/em&gt;  &lt;em&gt;&lt;strong&gt;u&lt;/strong&gt;&lt;/em&gt; $\in H$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $n \geq$ 0, denote $\mathbb{P}_n$ to be the set of all polynomials of degree at most $n$.  Which of the following are subspaces of $\mathbb{P}_n$ for an appropriate value of $n$? Support your answer with reason.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(a) All polynomials of the form $p(t) = a + t^2$, where $a \in \mathbb{R}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(b) All polynomials of degree at most 3, with integers as coefficients.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Solution&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both the statements (a) and (b) above are not subspaces of $\mathbb{P}_n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reason:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For part (a), All polynomials of the form $p(t) = a + t^2$, where $a \in \mathbb{R}$, is not a subspace because it failed to meet the standards of the first property stated above, that is, the coefficient of the $t^2$ term is always a 1, therefore the zero vector is not in this subspace of polynomials.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For part (b), All polynomials of degree at most 3, with integers as coefficents is also not a subspace becuase it as well failed to meet the standards of the third property, that is, integer coefficents, when multiplied by a real scalar, will not necessarily be of integer type leading to integer coefficents. Thusly, the resulting polynomial will not be in the subset of polynomials $\mathbb{P}_n$. $\Box$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can someone please look at this and see if I have the right idea and approach to this question. Thank You&lt;/p&gt;&#xA;" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-19T21:10:53.950" LastActivityDate="2011-03-19T21:10:53.950" Title="Which of the following polynomials are subspaces of $\mathbb{P}_n$ for an appropriate value of n?" Tags="&lt;linear-algebra&gt;&lt;polynomials&gt;&lt;vector-space&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="26250" PostTypeId="2" ParentId="26212" CreationDate="2011-03-10T19:51:24.190" Score="1" ViewCount="" Body="&lt;p&gt;To allow the question to be marked as answered, then:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Show that as $m$ ranges from $0$ to $p−1$, $am$ ranges over all residue classes modulo $p$, and hence $am+b$ ranges over all residue classes modulo $p$.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2011-03-10T19:51:24.190" CommunityOwnedDate="2011-03-10T19:51:24.190" />
  <row Id="26251" PostTypeId="1" AcceptedAnswerId="26254" CreationDate="2011-03-10T19:52:22.117" Score="2" ViewCount="154" Body="&lt;p&gt;So, I'm given: &#xA;$$f(x) = \left\{\begin{array}{ll}&#xA;x^2\sin\left(\frac{1}{x}\right) &amp;amp; \mbox{if $x\neq 0$,}\\&#xA;0 &amp;amp;\mbox{if $x=0$.}&#xA;\end{array}\right.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm asked to show that this piecewise function is both continuous and differentiable on the real number line. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure how to state what I know, i.e. that since $x^2$ and $\sin(1/x)$ are both continuous and differentiable on their domains, the product is also continuous and differentiable (on their domains where obviously $\sin(1/x)$ can't have $x = 0$.) But that case is covered since $f(x) = 0$ when $x = 0$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I guess I just to know how I would state all that in a mathematically accurate fashion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any assistance is greatly appreciated.&lt;/p&gt;&#xA;" OwnerUserId="8044" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2011-03-10T20:57:26.903" LastActivityDate="2011-03-10T20:57:26.903" Title="Showing a Function is Both Continuous and Differentiable" Tags="&lt;calculus&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="26252" PostTypeId="2" ParentId="26249" CreationDate="2011-03-10T19:53:10.710" Score="1" ViewCount="" Body="&lt;p&gt;Yes, you are right in both cases.&lt;/p&gt;&#xA;" OwnerUserId="8050" LastActivityDate="2011-03-10T19:53:10.710" />
  <row Id="26253" PostTypeId="2" ParentId="26251" CreationDate="2011-03-10T20:00:27.540" Score="2" ViewCount="" Body="&lt;p&gt;use the definition of differentiability, say at 0:&#xA;$$&#xD;&#xA;\lim_{x\to0}\frac{f(x)-f(0)}{x-0}=\lim_{x\to0}\frac{x^2\sin(1/x)}{x}=\lim_{x\to0}\ x\sin(1/x)=0&#xD;&#xA;$$&#xA;where the last step can be justified various ways.  for instance $|x\sin(1/x)|\leq|x|$.&lt;/p&gt;&#xA;" OwnerUserId="6925" LastActivityDate="2011-03-10T20:00:27.540" />
  <row Id="26254" PostTypeId="2" ParentId="26251" CreationDate="2011-03-10T20:00:43.937" Score="8" ViewCount="" Body="&lt;p&gt;Differentiability at a point implies continuity at that point.  Since $\sin$ is differentiable, $1/x$ is differentiable, $x$ is differentiable, compositions of differentiable functions are differentiable, and products of differentiable functions are differentiable, it follows that $f$ is differentiable at each nonzero point.  To show that $f$ is differentiable at $0$, compute $f&amp;#39;(0)$ using the definition of the derivative. Once you know this exists, you'll know that $f$ is differentiable, and hence continuous, on all of the real number line.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2011-03-10T20:22:11.487" LastActivityDate="2011-03-10T20:22:11.487" />
  <row Id="26255" PostTypeId="1" CreationDate="2011-03-10T20:14:35.393" Score="0" ViewCount="62" Body="&lt;p&gt;I don't understand this: a sequence of at most 5 paths. It is a sequence of paths which each is at most 5, or the sequence can only be as long as 5? Really confusing about this sentence. &lt;/p&gt;&#xA;" OwnerUserId="2419" LastActivityDate="2011-03-10T20:14:35.393" Title="what does &quot;a sequence of at most 5 paths&quot; mean?" Tags="&lt;topology&gt;" CommentCount="2" />
  <row Id="26256" PostTypeId="1" CreationDate="2011-03-10T20:33:35.887" Score="1" ViewCount="78" Body="&lt;p&gt;Hello all,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How would I go about showing that the third coefficient of the 2nd order Newton's interpolating polynomial is :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$a3=D^2y1= \frac{Dy2-Dy1}{x3-x1}=&#xD;&#xA;\frac{\frac{y_3-y_2}{x_3-x_2}-\frac{y_2-y_1}{x_2-x_1}}{x_3-x_1}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I dont know how to use math equations here, but I hope putting this in word and pressing the INSERT EQUATION you can see what I'm trying to prove ...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whenever I try doing it the normal way (placing $x=x_0$, $x=x_1$, $x=x_2$ etc..), I get :&#xA;$$\frac{\frac{y_3-y_1}{x_3-x_1}-\frac{y_2-y_1}{x_2-x_1}}{x_3-x_2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So how can I prove that both of these equations are the same !!?&lt;/p&gt;&#xA;" OwnerUserId="8083" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-10T21:48:55.417" LastActivityDate="2011-03-10T21:48:55.417" Title="Newton’s interpolating Polynomial - PROOF -" Tags="&lt;polynomials&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="26257" PostTypeId="2" ParentId="26061" CreationDate="2011-03-10T20:46:55.550" Score="1" ViewCount="" Body="&lt;p&gt;Here is a less garbled version of the Wikipedia definition. (Use TheBridge's correction for the definition of ${\cal F}_\tau$.) &#xA;  The post-$\tau$ process $X_{\tau+\cdot}$ is defined on the event $\{\tau&amp;lt;\infty\}$ by&#xA;$$&#xD;&#xA;X_{\tau+t}(\omega) = X_{\tau(\omega)+t}(\omega),\qquad t\ge 0,&#xD;&#xA;$$&#xA;for  $\omega\in\{\tau&amp;lt;\infty\}$. One way to state the strong Markov property is this: The conditional distribution of $X_{\tau+\cdot}$ given ${\cal F}_\tau$ is (a.s.) equal to the conditional distribution of &#xA;$X_{\tau+\cdot}$ given $\sigma\{X_\tau\}$, on the event $\{\tau&amp;lt;\infty\}$. More precisely,&#xA;$$&#xD;&#xA;P[ X_{\tau+t}\in B|{\cal F}_\tau] = P[ X_{\tau+t}\in B|X_\tau],\qquad \hbox{almost surely on }\{\tau&amp;lt;\infty\},&#xD;&#xA;$$&#xA;for all $t\ge 0$, and all measurable subsets $B$ of the state space of $X$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is equivalent to the statement that $X_{\tau+\cdot}$ and ${\cal F}_\tau$ are conditionally independent, given $X_\tau$:&#xA;$$&#xD;&#xA;P[ F\cap \{X_{\tau+t}\in B\}|X_\tau] = P[ F|X_\tau]\cdot P[X_{\tau+t}\in B|X_\tau],\qquad \hbox{almost surely on }\{\tau&amp;lt;\infty\},&#xD;&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="8084" LastActivityDate="2011-03-10T20:46:55.550" CommentCount="3" />
  <row Id="26258" PostTypeId="1" CreationDate="2011-03-10T21:00:41.480" Score="5" ViewCount="208" Body="&lt;p&gt;Let $B_t$ be the standard Brownian motion process, $a &amp;gt; 0$, and let $H_a = \inf \{ t : B_t &amp;gt; a \}$ be a stopping time. I want to show that the Laplace transform of $H_a$ is&#xA;$$\mathbb{E}[\exp(-\lambda H_a)] = \exp (-\sqrt{2\lambda} H_a)$$&#xA;by considering the martingale $$M_t = \exp \left(\theta B_t -\frac{1}{2}\theta^2 t\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's an obvious argument to follow here: &lt;em&gt;assuming&lt;/em&gt; the optional stopping theorem applies, we have&#xA;$$1 = \mathbb{E}[M_{H_a}] = \mathbb{E} \left[ \exp \left(\theta a - \frac{1}{2}\theta^2 H_a\right) \right] = \exp(\sqrt{2\lambda} a) \mathbb{E} \left[ \exp(-\lambda H_a) \right]$$&#xA;where $\theta = \sqrt{2\lambda}$. This is exactly what we wished to show. However, as far as I can tell, the hypotheses of the optional stopping theorem are &lt;em&gt;not&lt;/em&gt; satisfied here. Here is the statement I have:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If $(X_n)$ is a martingale and $T$ is an a.s. bounded stopping time, then $\mathbb{E}[X_T] = \mathbb{E}[X_0]$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I think not all is lost yet. $M_t &amp;gt; 0$ for all $t$, so the martingale convergence theorem applies, and $M_t \to M_\infty$ a.s. for some integrable random variable $M_\infty$. For each $t$, $H_a \wedge t = \min \{ H_a, t \}$ is a bounded stopping time, so certainly $\mathbb{E}[M_{H_a \wedge t}] = \mathbb{E}[M_0]$. But,&#xA;$$\mathbb{E}[M_{H_a \wedge t}] = \mathbb{E}[M_{H_a} \mathbf{1}_{\{H_a \le t\}}] + \mathbb{E}[M_t \mathbf{1}_{\{H_a &amp;gt; t\}}]$$&#xA;and clearly what one wants to do is to take $t \to \infty$ on both sides. But here's where I get stuck: I'm sure I need a convergence theorem here in order to conclude that the equation remains valid in the limit. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, $0 &amp;lt; M_{H_a} = \exp \left(\theta a - \frac{1}{2} \theta^2 H_a \right) \le \exp(\theta a)$, so the dominated convergence theorem applies, and so&#xA;$$\lim_{t \to \infty} \mathbb{E}[M_{H_a} \mathbf{1}_{\{H_a \le t\}}] = \mathbb{E}[M_{H_a} \mathbf{1}_{\{H_a &amp;lt; \infty\}}]$$&#xA;and I believe Fatou's lemma gives me that&#xA;$$\liminf_{t \to \infty} \mathbb{E}[M_t \mathbf{1}_{\{H_a &amp;gt; t\}}] \ge \mathbb{E}[M_{\infty} \mathbf{1}_{\{H_a = \infty\}}]$$&#xA;but I think what I need is the &lt;em&gt;equality&lt;/em&gt;&#xA;$$\lim_{t \to \infty} \mathbb{E}[M_t \mathbf{1}_{\{H_a &amp;gt; t\}}] = \mathbb{E}[M_\infty \mathbf{1}_{\{H_a = \infty\}}]$$&#xA;and as far as I can tell neither the monotone convergence theorem nor the dominated convergence theorem applies here. Is there anything I can do to rescue this line of thought?&lt;/p&gt;&#xA;" OwnerUserId="5191" LastEditorUserId="9901" LastEditorDisplayName="" LastEditDate="2011-05-10T19:07:31.923" LastActivityDate="2011-05-10T19:07:31.923" Title="The Laplace transform of the first hitting time of Brownian motion" Tags="&lt;probability-theory&gt;&lt;measure-theory&gt;&lt;stochastic-processes&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="26259" PostTypeId="1" AcceptedAnswerId="26261" CreationDate="2011-03-10T21:08:41.253" Score="5" ViewCount="155" Body="&lt;p&gt;I've been reading &lt;em&gt;generationgfunctionology&lt;/em&gt; by Herbert S. Wilf (you can find a copy of the second edition on the author's page &lt;a href=&quot;http://www.math.upenn.edu/~wilf/DownldGF.html&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On page 33 he does something I find weird. He wants to shuffle the index forward and does so like this:&#xA;\begin{align*}&#xA;(f_{n+1})_{n\in N_0} &amp;amp;= \frac{(f(X)-f(0))}{X}\\&#xA;(f_{n})_{n\in N_0} &amp;amp;= \sum_{n\in N_0} f_nx^n&#xA;\end{align*}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is this allowed? Namely $X$ has a noninvertible (0) constant term, so how is this division (multiplication with the reciprocal) defined within the ring of formal power series?&lt;/p&gt;&#xA;" OwnerUserId="7602" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2011-03-11T03:50:09.040" LastActivityDate="2011-03-11T14:37:36.923" Title="Formal power series - a question" Tags="&lt;algebra&gt;&lt;generating-functions&gt;&lt;power-series&gt;" AnswerCount="5" CommentCount="1" />
  <row Id="26260" PostTypeId="2" ParentId="26256" CreationDate="2011-03-10T21:13:19.893" Score="1" ViewCount="" Body="&lt;p&gt;I think if, for each of the two expressions, you get a common denominator with the two fractions in the numerator of the overall fraction, you'll be able to rewrite each expression into an identical form.  That is, rewrite each of the two expressions as a simplified polynomial expression divided by $(x_3-x_1)(x_3-x_2)(x_2-x_1)$.&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2011-03-10T21:13:19.893" />
  <row Id="26261" PostTypeId="2" ParentId="26259" CreationDate="2011-03-10T21:15:00.990" Score="7" ViewCount="" Body="&lt;p&gt;If a power series $g(X)$ has zero constant term, then it is true that it is not invertible, so you cannot divide by it in general.  However, it may happen that you can divide &lt;em&gt;some&lt;/em&gt; power series by $g(X)$, namely those that have $g(X)$ as a factor.  If $f(X)$ is a power series and $g(X)$ is a nonzero power series, then we can divide $h(X)=g(X)f(X)$ by $g(X)$ by defining $\frac{h(X)}{g(X)}$ to be $f(X)$.  This is unambiguous because, as Qiaochu noted, the power series ring is an integral domain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is what's going on here. Because $f(X)-f(0)$ has zero constant term, you can write it as $Xg(X)$ for some power series $g(X)$, and $\frac{f(X)-f(0)}{X}$ is another name for $g(X)$.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2011-03-10T21:35:14.840" LastActivityDate="2011-03-10T21:35:14.840" CommentCount="1" />
  <row Id="26262" PostTypeId="2" ParentId="26259" CreationDate="2011-03-10T21:17:49.157" Score="5" ViewCount="" Body="&lt;p&gt;The ring of formal power series is an integral domain, so you can cancel factors from two sides of an equation. (This is equivalent to saying that it embeds into a field, the ring of formal Laurent series.) This is the same thing you do when you say that $2 = \frac{6}{3}$ even though $3$ isn't invertible in $\mathbb{Z}$. &lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2011-03-10T21:17:49.157" />
  <row Id="26263" PostTypeId="2" ParentId="26259" CreationDate="2011-03-10T21:19:32.373" Score="1" ViewCount="" Body="&lt;p&gt;Are you allowed to write the following?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align*}&#xA;X =  \frac{X^2}{X}\&#xA;\end{align*}&lt;/p&gt;&#xA;" OwnerUserId="7367" LastActivityDate="2011-03-10T21:19:32.373" />
  <row Id="26265" PostTypeId="2" ParentId="26256" CreationDate="2011-03-10T21:30:22.187" Score="1" ViewCount="" Body="&lt;p&gt;$$\begin{align} \frac{\frac{y_3-y_2}{x_3-x_2}-\frac{y_2-y_1}{x_2-x_1}}{x_3-x_1} &#xD;&#xA;&amp;amp; =\frac{y_3-y_2}{(x_3-x_2)(x_3-x_1)}-\frac{y_2-y_1}{(x_2-x_1)(x_3-x_1)} \\ &#xD;&#xA;&amp;amp; =\frac{y_3-y_1}{(x_3-x_2)(x_3-x_1)}-\frac{y_2-y_1}{(x_3-x_2)(x_3-x_1)}-\frac{y_2-y_1}{(x_2-x_1)(x_3-x_1)} \\ &#xD;&#xA;&amp;amp; = \frac{y_3-y_1}{(x_3-x_2)(x_3-x_1)}-\frac{y_2-y_1}{(x_3-x_1)}\left(\frac{1}{x_3-x_2}+\frac{1}{x_2-x_1}\right) \\ &#xD;&#xA;&amp;amp; = \frac{y_3-y_1}{(x_3-x_2)(x_3-x_1)}-\frac{y_2-y_1}{(x_3-x_1)}\frac{x_3-x_1}{(x_3-x_2)(x_2-x_1)} \\ &#xD;&#xA;&amp;amp; = \frac{y_3-y_1}{(x_3-x_2)(x_3-x_1)}-\frac{y_2-y_1}{(x_3-x_2)(x_2-x_1)} \\&#xD;&#xA;&amp;amp;= \frac{\frac{y_3-y_1}{x_3-x_1}-\frac{y_2-y_1}{x_2-x_1}}{x_3-x_2}\end{align}$$&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-03-10T21:30:22.187" />
  <row Id="26266" PostTypeId="1" AcceptedAnswerId="26285" CreationDate="2011-03-10T21:41:18.860" Score="2" ViewCount="129" Body="&lt;p&gt;What is the 2D kernel, k, that when convoluted with a 2D signal, f, that is convoluted again with a gaussian 2D kernel, g, produces a result that is closest to the original signal, f'.&#xA;Something like this, I think:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For&lt;/p&gt;&#xA;&#xA;&lt;p&gt;f' = f $\star$ k $\star$ g (where $\star$ denotes convolution)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Find k, such that |f - f'| is minimal.&lt;/p&gt;&#xA;" OwnerUserId="8086" LastEditorUserId="8086" LastEditorDisplayName="" LastEditDate="2011-03-10T23:04:12.423" LastActivityDate="2011-03-11T07:03:34.693" Title="&quot;anti-gaussian&quot; 2D convolution kernel" Tags="&lt;convolution&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="26267" PostTypeId="2" ParentId="26215" CreationDate="2011-03-10T21:44:00.653" Score="1" ViewCount="" Body="&lt;p&gt;I think both Vitali and Matt are right. As soon as $G(x,y)$ is integrable on $\mathbb{R}^2$, $$\iint_{\mathbb{R}^2}{G(x,y)\,dxdy}=\int_{\mathbb{R}}dx\,\int_{\mathbb{R}}{G(x,y)\,dy}=\int_{\mathbb{R}}dy\,\int_{\mathbb{R}}{G(x,y)\,dx} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you can substitute: $$ u=4x $$ $$ v=x-3y $$ $$ x=\frac{u}{4} $$ $$ y=\frac{u}{12}-\frac{v}{3} $$&#xA;Jacobian $$ J=\det D(x(u,v),y(u,v))=\begin{vmatrix} \frac{\partial x}{\partial u} &amp;amp; \frac{\partial x}{\partial v} \\\\ \frac{\partial y}{\partial u} &amp;amp; \frac{\partial y}{\partial v} \end{vmatrix}=\begin{vmatrix}\frac{1}{4} &amp;amp; 0\\\\\frac{1}{12} &amp;amp; -\frac{1}{3}\end{vmatrix}=-\frac{1}{12} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then $$\iint_{\mathbb{R}^2}{f(4x)\,f(x-3y)\,dxdy}=-\frac{1}{12}\int_{-\infty}^{\infty}\,du\,\int_{\infty}^{-\infty}{f(u)\,f(v)\,dv}=\frac{1}{12}(\int_{-\infty}^{\infty}{f(z)\,dz})^2=2$$&#xA;Thus giving us&#xA;$$\int_{-\infty}^{\infty}{f(z)\,dz}=2\sqrt 6$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please tell me if I $\mathbb{F}\bigoplus$ something up.&lt;/p&gt;&#xA;" OwnerUserId="6554" LastEditorUserId="6554" LastEditorDisplayName="" LastEditDate="2011-03-10T23:14:38.110" LastActivityDate="2011-03-10T23:14:38.110" CommentCount="14" />
  <row Id="26268" PostTypeId="1" AcceptedAnswerId="26313" CreationDate="2011-03-10T22:07:04.043" Score="4" ViewCount="92" Body="&lt;p&gt;I'm trying to rewrite Taylor's formula on $\mathbb{R}^n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $p\in \mathbb{N}$, $V \subset \mathbb{R}^n$ open and convex, $\mathbf{x, a} \in V$, and $f:V\to \mathbb{R}$ where $f \in C^p(V)$. Then&#xA;$$f(\mathbf{x}) = \sum_{k=0}^{p-1} \frac{1}{k!} D^{(k)} f(\mathbf{a}; \mathbf{x-a}) + \frac{1}{(p-1)!} \int_0^1 (1-t)^{p-1} D^{(p)} f(\mathbf{a} + t(\mathbf{x-a}); \mathbf{x-a}) dt.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Surely, for $p=1$ the property evidently holds. I can then use an induction argument and integration by parts to prove this, is that correct? It was listed as a difficult exercise so I'm a bit cautious to waving my arms in the air too soon.&lt;/p&gt;&#xA;" OwnerUserId="8042" LastActivityDate="2011-03-11T03:53:48.010" Title="Integral form of Taylor's formula" Tags="&lt;homework&gt;&lt;real-analysis&gt;" AnswerCount="1" />
  <row Id="26269" PostTypeId="1" AcceptedAnswerId="26272" CreationDate="2011-03-10T22:15:59.127" Score="4" ViewCount="59" Body="&lt;p&gt;Let $\mathscr{C}$ be a strict monoidal category. I will denote the product of $\mathscr{C}$ by $\otimes$. The Drinfeld center $\mathscr{Z(C)}$ of $\mathscr{C}$ is the category with object $(X,\phi)$ where $X$ is an object of $\mathscr{C}$ and $\phi$ is a natural isomorphism from $X \otimes -$ to $ - \otimes X$. Morhphisms from $(X,\phi)$ to $(Y,\psi)$ in $\mathscr{Z(C)}$ are elements all $f \in \mathrm{hom_\mathscr{C}(X,Y)}$ such that $(\mathrm{id}_W \otimes f) \circ \phi_W = \psi_W \circ (f \otimes \mathrm{id}_W)$ for all $W \in \mathscr{C}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is the following. There must be a problem in the follwoing reasoning, but I cannot find it. I wonder if anybody can point out the mistake. &#xA;Fix any field $k$ of characteristic $0$. If $G$ is a finite group, then $\mathrm{Vec}_G$ the category of $G$-graded vector spaces is strict monoidal and moreover it is semisimple. Its simple objects are one dimensional vector spaces $V_g$ with grading given by an element $g \in G$. Morphisms are grading-preserving linear maps. In particular $V_g$ and $V_h$ are isomorphic if and only if $g = h$. The monoidal structure of $\mathrm{Vec}_G$ is given on simple object by multiplication of elements in $G$: $V_g \otimes V_h = V_{gh}$. Now assume that $G$ has a trivial center. Then for any $g \in G$ there is no natural isomorphism from $V_g \otimes - $ to $ - \otimes V_g$, since $V_{gh}$ and $V_{hg}$ are not isomorphic for some $h \in G$. Hence, its Drinfeld center is trivial.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the conclusion cannot be true, since $\mathscr{Z}(\mathrm{Vec}_G)$ is the representation category of the Drinfeld double $k[G] \ltimes \mathrm{Fun}(G)$ of $\mathrm{Fun}(G)$, where $k[G]$ is the group ring of $G$, $\mathrm{Fun}(G)$ are the $k$-valued functions on $G$ and $\ltimes$ denotes the crossed product with respect to the natural action.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you very much for your help.&lt;/p&gt;&#xA;" OwnerUserId="8085" LastActivityDate="2011-03-10T22:34:04.510" Title="Drinfeld Center" Tags="&lt;representation-theory&gt;&lt;monoidal-categories&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="26270" PostTypeId="1" AcceptedAnswerId="26271" CreationDate="2011-03-10T22:18:25.553" Score="1" ViewCount="73" Body="&lt;p&gt;Why is the following true?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{j=0}^{n-1}w(2\pi j/n)\left[\frac{1}{n}\sum_{k=0}^{n-1}e^{2\pi ik(j-m)/n}\right]=w(2\pi m/n)$$&lt;/p&gt;&#xA;" OwnerUserId="7815" LastEditorUserId="1543" LastEditorDisplayName="" LastEditDate="2011-05-03T09:47:17.020" LastActivityDate="2011-05-03T09:47:17.020" Title="Why is the following sum true?" Tags="&lt;fourier-series&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="26271" PostTypeId="2" ParentId="26270" CreationDate="2011-03-10T22:23:29.480" Score="4" ViewCount="" Body="&lt;p&gt;$$\sum_{k=0}^{n-1}e^{2\pi ik(j-m)/n} = 0 $$ unless $j = m+ ln$ for some $l \in \mathbb{Z}$. This can be seen from a simple geometric progression argument.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence when you perform the outer sum only the non-zero term namely at $j=m$ remains and hence &#xA;$$\sum_{j=0}^{n-1}w(2\pi j/n)\left[\frac{1}{n}\sum_{k=0}^{n-1}e^{2\pi ik(j-m)/n}\right]=w(2\pi m/n)$$&lt;/p&gt;&#xA;" OwnerUserId="3302" LastActivityDate="2011-03-10T22:23:29.480" />
  <row Id="26272" PostTypeId="2" ParentId="26269" CreationDate="2011-03-10T22:34:04.510" Score="3" ViewCount="" Body="&lt;p&gt;Your observation only means that there is no object $(X,\phi)$ in the Drinfeld center which has $X$ simple and non-isomorphic to $V_e$ ($e$ being the identity element of $G$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But let $C\subseteq G$ be a conjugacy class, and let $V_C=\bigoplus_{g\in C}V_g$. Then you should be able to find a natural isomorphism $\phi_C:V_C\otimes(\mathord-)\to(\mathord-)\otimes V_C$, so that $(V_C,\phi_C)$ is a non-trivial element in $\mathcal{Z}(\mathrm{Vec}_G)$.&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2011-03-10T22:34:04.510" CommentCount="2" />
  <row Id="26273" PostTypeId="1" AcceptedAnswerId="26414" CreationDate="2011-03-10T22:42:21.983" Score="10" ViewCount="215" Body="&lt;p&gt;I was assigned a few problems in my Honors Calculus II class, and one of them was kind of interesting to do:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Suppose that $f_{n}$ are nonnegative bounded functions on $A$ and let $M_{n} =  \sup f_{n}$. If $\displaystyle\sum\limits_{n=1}^\infty f_{n}$ converges uniformly on $A$, does it follow that $\displaystyle\sum\limits_{n=1}^\infty M_{n}$ converges (a converse to the Weirstrass $M$-test)?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I know that this question has been &lt;a href=&quot;http://math.stackexchange.com/questions/20519/question-regarding-weierstrass-m-test&quot;&gt;asked before&lt;/a&gt;, but I'm trying not to just copy an answer off the internet and instead to come up with an example of my own to see if I can actually understand the theorems that I'm learning.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To provide a counterexample, I tried to create a function which has a diverging $\sup$, but I'm not too confident that my proof is valid. Here it goes:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;\text Let \ f_{n}(x) = \begin{cases}&#xD;&#xA;  \begin{cases}&#xD;&#xA;    \frac{1 + x}{2}, &amp;amp; \text{if }x \in (-1,0]\\&#xD;&#xA;    \frac{1 - x}{2}, &amp;amp; \text{if }x \in (0,1)\\&#xD;&#xA;    0, &amp;amp; \text{elsewhere}&#xD;&#xA;  \end{cases}, &amp;amp; \text{if }n \text{ even}\\&#xD;&#xA;  \begin{cases}&#xD;&#xA;    x, &amp;amp; \text{if }x \in (0,1]\\&#xD;&#xA;    2 - x, &amp;amp; \text{if }x \in (1,2)\\&#xD;&#xA;    0, &amp;amp; \text{elsewhere}&#xD;&#xA;  \end{cases}, &amp;amp; \text{if }n \text{ odd}&#xD;&#xA;\end{cases}&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, $\text Let f(x) = 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From this definition, I can conclude that&#xA;$$&#xD;&#xA;\sup\{f_{n}\} = \begin{cases}&#xD;&#xA;  \frac{1}{2}, &amp;amp; \text{if }n \text{ even}\\&#xD;&#xA;  1, &amp;amp; \text{if }n \text{ odd}&#xD;&#xA;\end{cases}&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, to show that ${f_{n}(x)}$ is uniformly convergent, the definition of uniform convergence is used:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;\forall_{\epsilon &amp;gt; 0}\ \exists_{N}\ \text s.t.\ \forall_{x}\ \text if\ n &amp;gt; N, |f(x) - f_{n}(x)| &amp;lt; \epsilon&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $f_{n}(x)$ is strictly nonnegative and $f(x) = 0$, $|f(x) - f_{n}(x)| = f_{n}(x)$. By definition, $\epsilon &amp;gt; 0$, and since $f_{n}(x) = 0$ for $x \geq 2, f(x) - f_{n}(x) = 0 &amp;lt; \epsilon$ for $x \geq 2$. Therefore there exists a $N$ (namely, $N = 1$) which proves that the sequence is uniformly convergent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $\lim_{n\to\infty} f_{n} \neq 0$, by the Limit Test, the infinite sum  $\displaystyle\sum\limits_{n=1}^\infty \sup{f_{n}}$ diverges, which disproves the converse of the Weierstrass $M$-Test. $\blacksquare$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the first time I've actually used LaTeX, so I'm sorry for the way it looks. Is there anything that I can do to make this proof better (or even valid, if it's wrong), or is it fine the way it is?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This might be a bit of a long question...&lt;/p&gt;&#xA;" OwnerUserId="4450" LastEditorUserId="4450" LastEditorDisplayName="" LastEditDate="2011-03-11T17:51:55.443" LastActivityDate="2011-03-12T18:23:00.750" Title="Converse of the Weierstrass $M$-Test?" Tags="&lt;calculus&gt;&lt;homework&gt;&lt;real-analysis&gt;" AnswerCount="1" CommentCount="12" />
  <row Id="26274" PostTypeId="1" AcceptedAnswerId="26278" CreationDate="2011-03-10T23:03:02.497" Score="1" ViewCount="112" Body="&lt;p&gt;I had the following on an exam today.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let the universe = $Z^+$, $Z = (1,2,3,....100)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$|A| = 10$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$|B| = 10$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$|A\cap B| = 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(a) What is $|A^c \cup B^c|$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I thought about it like this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let set A be dogs&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let set B be cats. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or let A be even numbers and B be odd numbers. Sticking with dogs and cats though.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A-compliment is cats&lt;/p&gt;&#xA;&#xA;&lt;p&gt;B-compliment is dogs&lt;/p&gt;&#xA;&#xA;&lt;p&gt;their union is the whole Z so  $|A^c \cup B^c|= 100$ is this correct?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(b) What is $|A^c \cap B^c|$ I said this was 0 using the same logic above&lt;/p&gt;&#xA;" OwnerUserId="1040" LastEditorUserId="1040" LastEditorDisplayName="" LastEditDate="2011-03-10T23:30:05.610" LastActivityDate="2011-03-10T23:44:14.653" Title="Is the logic correct?" Tags="&lt;logic&gt;" AnswerCount="2" CommentCount="6" />
  <row Id="26275" PostTypeId="2" ParentId="20886" CreationDate="2011-03-10T23:08:04.257" Score="0" ViewCount="" Body="&lt;p&gt;The planar 3-connected graphs, which are the vertex-edge graphs of 3-dimensional convex polyhedra have essentially only one way to be embedded in the plane, by a theorem of Hassler Whitney. Though such embeddings may look different because the number of sides of the infinite face in different drawings may differ, the different embeddings are isomorphic. What can be accomplished using two vertices joined by paths of different lengths when the graph might be embedded in the plane in different ways?&lt;/p&gt;&#xA;" OwnerUserId="1369" LastActivityDate="2011-03-10T23:08:04.257" />
  <row Id="26276" PostTypeId="2" ParentId="26274" CreationDate="2011-03-10T23:13:10.803" Score="2" ViewCount="" Body="&lt;p&gt;&lt;s&gt;The reasoning is more or less correct.&lt;/s&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your analogies to other $A$ and $B$ (cats/dogs/even/odd) are misleading, because they have $A$ and $B$ as complementary sets themselves.  Because $|A|=10$ and $|B|=10$ and there are 100 elements in the universe, there are at least 80 other elements in the universe that are not in $A$ and $B$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As to part (a), I tend to think about it this way: Since $A$ and $B$ have no common elements, all of $B$ is contained in the complement of $A$, so the union of a set that contains all of $B$ with the complement of $B$ has to be the whole universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Beyond that, &lt;a href=&quot;http://math.stackexchange.com/questions/26274/is-the-logic-correct/26278#26278&quot;&gt;Calle's answer&lt;/a&gt; does a good job of answering the question.&lt;/p&gt;&#xA;" OwnerUserId="72" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2011-03-10T23:38:52.513" LastActivityDate="2011-03-10T23:38:52.513" CommentCount="4" />
  <row Id="26278" PostTypeId="2" ParentId="26274" CreationDate="2011-03-10T23:19:18.823" Score="5" ViewCount="" Body="&lt;p&gt;Let $A^c$ and $B^c$ be the complements of $A$ and $B$ respectively. We know that $|A \cap B| = 0$, that is, $A$ and $B$ have no elements in common, which implies that $A \subset B^c$ and $B \subset A^c$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So $A^c$ contains all elements, except $A$. $B^c$ contains all elements except $B$. But $A^c$ contains $B$ and $B^c$ contains $A$, so $A^c \cup B^c = Z$, thus $|A^c \cup B^c| = 100$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your reasoning is not correct, however. If you mean that $A$ is all dogs in $Z$ and $B$ is all cats in $Z$, then there are 80 other things in $Z$ which are neither cats or dogs, say elephants. Thus $A^c$ is the set of all cats and elephants and $B^c$ is the set of all dogs and elephants.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you mean that $A$ consist of only dogs, but not necessarily all dogs in $Z$, $B$ consists only of cats, but not necessarily all cats in $Z$, and $Z$ contains only dogs and cats, the situation is a bit different. Then three things are possible: $A^c$ contains dogs, $B^c$ contains cats, or both.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the b-part. We know that $|A^c| = 90$ and $|B^c| = 90$. As per above, we know that $A \subset B^c$ and $B \subset A^c$, but $A$ is not part of $A^c$ and $B$ is not part of $B^c$. However, &quot;the rest&quot; of $A^c$ and $B^c$ is the same, so $|A^c \cap B^c| = 80$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the cats, dogs and elephants: $A^c$ is all cats and elephants, $B^c$ is all dogs and elephants. The elements in common are the elephants.&lt;/p&gt;&#xA;" OwnerUserId="5966" LastEditorUserId="5966" LastEditorDisplayName="" LastEditDate="2011-03-10T23:44:14.653" LastActivityDate="2011-03-10T23:44:14.653" CommentCount="1" />
  <row Id="26279" PostTypeId="1" AcceptedAnswerId="26282" CreationDate="2011-03-10T23:20:08.043" Score="3" ViewCount="82" Body="&lt;p&gt;please help me establish(etablir): &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\forall n\in \mathbb{N}-\left\{ 0,\left. 1 \right\} \right.$ , $x\in \mathbb{R}-\left\{ \pi \mathbb{Z} \right\}$ , $\left| \sin \left( nx \right) \right|&amp;lt;n\left| \sin x \right|$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;thx in advance... &lt;/p&gt;&#xA;" OwnerUserId="7022" LastActivityDate="2011-03-11T02:09:19.027" Title="trigonometric inequality" Tags="&lt;algebra&gt;&lt;trigonometry&gt;&lt;inequality&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="26280" PostTypeId="1" AcceptedAnswerId="26283" CreationDate="2011-03-10T23:22:23.407" Score="1" ViewCount="79" Body="&lt;p&gt;Let $H^s(\mathbb{R}^d):= \{ u \in \mathcal{S}&amp;#39; : (1+|\xi|^2)^{s/2}\hat{u}(\xi) \in L^2(\mathbb{R}^d)\}$. It can be shown that this space is a Hilbert space and that $H^s \subset H^t$ if $t \leq s$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now suppose we have $t &amp;gt; s&amp;gt;0$ such that $H^s \subsetneq H^t$. Then we know that $(H^s)^* \supsetneq (H^t)^*$, where $^*$ denotes the dual. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the same time, since $H^s$ is a Hilbert space, the Riesz Representation theorem tells us that $H^s \cong (H^s)^*$ and similarly, $H^t \cong (H^t)^*$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But then doesn't $(H^s)^* \supsetneq (H^t)^*$ imply that $H^s \supsetneq H^t$, which contradicts our earlier statement that $H^s \subsetneq H^t$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What part of the argument here falls apart? &lt;/p&gt;&#xA;" OwnerUserId="1736" LastActivityDate="2011-03-11T00:21:24.827" Title="On the properties of the Sobolev Spaces $H^s$" Tags="&lt;functional-analysis&gt;" AnswerCount="2" CommentCount="0" />
  <row Id="26282" PostTypeId="2" ParentId="26279" CreationDate="2011-03-10T23:27:59.550" Score="3" ViewCount="" Body="&lt;p&gt;Hint: Have you heard of induction?&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2011-03-10T23:27:59.550" CommentCount="1" />
  <row Id="26283" PostTypeId="2" ParentId="26280" CreationDate="2011-03-10T23:28:18.043" Score="2" ViewCount="" Body="&lt;p&gt;When you say that $H^s \subsetneq H^t$, what you're saying is that there is a distinguished compatible family of strict inclusions $f_{s,t} : H^s \subsetneq H^t$. Dualizing these inclusions gives maps $f_{s,t}^{\ast} : (H^t)^{\ast} \to (H^s)^{\ast}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;These maps don't appear to be inclusions to me. At least, one can't conclude this from abstract properties of Banach spaces; judging by the finite-dimensional case I would guess that the dual of an inclusion tends to be a &lt;em&gt;quotient.&lt;/em&gt; So, if you like, the argument fails here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I think there is a second point you're missing, so I'll press on. All Hilbert spaces are canonically &lt;em&gt;anti-isomorphic&lt;/em&gt; to their duals $d : H \to H^{\ast}$ (anti-isomorphic means that scalar multiplication is sent to its conjugate), hence in particular there are anti-isomorphisms $d : H^s \to (H^s)^{\ast}$ and $H^t \to (H^t)^{\ast}$.  (You can upgrade these to isomorphisms by composing with the canonical anti-isomorphism between a Hilbert space and its &quot;opposite&quot; Hilbert space, but I don't think you get anything from doing this, especially since all it tells you is that the &quot;opposite&quot; and dual are naturally isomorphic.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's no possible way to get a contradiction from an argument like this because the maps $f_{s,t}$ are completely different from the maps $d$. (This is what happens when mathematical notation obscures what's really important, which is the morphisms, not the objects.) Hilbert spaces which strictly include each other can still be abstractly isomorphic; remember that all separable Hilbert spaces are isomorphic to $\ell^2$. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Willie brings up a point in the comments that is an &lt;em&gt;extra&lt;/em&gt; source of confusion here: the Sobolev spaces all have different inner products, but in the first part of the argument I guess you were implicitly thinking of all of them as subspaces of $L^2$ with the induced inner product? Unfortunately the notation here also hides the choice of inner product. It really is quite awful. &lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2011-03-10T23:42:42.870" LastActivityDate="2011-03-10T23:42:42.870" CommentCount="5" />
  <row Id="26284" PostTypeId="1" AcceptedAnswerId="26292" CreationDate="2011-03-10T23:45:37.887" Score="2" ViewCount="139" Body="&lt;p&gt;context: i am currently learning about the Radon Nicodym Theorem. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was wondering what the notation $du$ represent, where $u$ is a measure. I mean just $du$ by itself, not within an integral or anything.&lt;/p&gt;&#xA;" OwnerUserId="6556" LastActivityDate="2011-03-11T02:09:01.757" Title="what does $du$ mean by itself?" Tags="&lt;measure-theory&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="26285" PostTypeId="2" ParentId="26266" CreationDate="2011-03-10T23:48:56.567" Score="1" ViewCount="" Body="&lt;p&gt;For concreteness, let us define the Gaussian kernel $g$ as &#xA;$$g(x,y) = \frac{1}{2\pi \sigma_x \sigma_y} \exp\left[-\frac{x^2}{2 \sigma_x^2}-&#xD;&#xA;\frac{y^2}{2\sigma_y^2}\right].$$ Its Fourier transform is again a Gaussian&#xA;$$\hat g(k_x,k_y) = \int\!dx\, dy\, e^{i kx x + i ky y}\, k(x,y)=&#xD;&#xA; \exp\left[-\frac{\sigma_x^2 k_x^2}{2}-&#xD;&#xA;\frac{\sigma_y^2 k_y^2}{2}\right]$$&#xA;is again a Gaussian. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Under Fourier transform the convolution becomes a product, i.e.,&#xA;$\widehat{k\star g} = \hat{k} \times \hat{g}$. &#xA;The difference is given by $$\begin{align}\int\!dx\,dy\,|f(x,y)-f&amp;#39;(x,y)|^2&#xD;&#xA;&amp;amp;=\int\frac{dk_x}{2\pi}\frac{dk_y}{2\pi}\,|\hat{f}(k_x,k_y)-\hat{f&amp;#39;}(k_x,k_y)|^2\\&#xD;&#xA;&amp;amp;=\int\frac{dk_x}{2\pi}\frac{dk_y}{2\pi}\,|\hat{f}|^2|1-\hat{k} \hat{g}|^2&#xD;&#xA;\end{align}.$$ In order to minimize the difference, $\hat{k} \hat{g}$ should be as close as possible to 1 (and of course $k$ still normalized).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, we have to minimize&#xA;$$\int\frac{dk_x}{2\pi}\frac{dk_y}{2\pi}\,[|1-\hat{k} \hat{g}|^2+\lambda|\hat{k}|^2],$$ where $\lambda$ is a Lagrange multiplier to keep the norm of $\hat k$ at 1. We can variate this equation, and obtain&#xA;$$\int\frac{dk_x}{2\pi}\frac{dk_y}{2\pi}\,[2 \hat{g}(1- \hat{k} \hat{g})+2 \lambda\hat{k}] \delta \hat k,$$&#xA;with the solution&#xA;$$\hat{k} = \frac{\hat{g}}{\hat{g}^2+\lambda}=&#xD;&#xA;\frac{\exp\left[-\frac{\sigma_x^2 k_x^2}{2}-&#xD;&#xA;\frac{\sigma_y^2 k_y^2}{2}\right]}{\exp\left[-\sigma_x^2 k_x^2-&#xD;&#xA;\sigma_y^2 k_y^2\right]+\lambda}.$$ $\lambda$ is determined from the normalization condition.&lt;/p&gt;&#xA;" OwnerUserId="7266" LastEditorUserId="7266" LastEditorDisplayName="" LastEditDate="2011-03-11T07:03:34.693" LastActivityDate="2011-03-11T07:03:34.693" CommentCount="9" />
  <row Id="26286" PostTypeId="2" ParentId="26258" CreationDate="2011-03-11T00:02:17.720" Score="0" ViewCount="" Body="&lt;p&gt;I m not sure that I understand properly the question.&lt;br&gt;&#xA;We have $M_{H_a\wedge t}\to M_{H_a}$ almost surely, and&lt;br&gt;&#xA;$M_{H_a\wedge t}&amp;lt;e^{\theta.a}$&#xA;The right hand side is constant so integrable, so doesn't the dominated convergence readily applies for $t \to +\infty$ ? &#xA;Am i missing something here ?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regards&lt;/p&gt;&#xA;" OwnerUserId="4437" LastEditorUserId="4437" LastEditorDisplayName="" LastEditDate="2011-03-11T16:54:19.727" LastActivityDate="2011-03-11T16:54:19.727" CommentCount="4" />
  <row Id="26287" PostTypeId="1" AcceptedAnswerId="26352" CreationDate="2011-03-11T00:08:08.453" Score="5" ViewCount="74" Body="&lt;p&gt;In the introduction to André Weil's &lt;em&gt;Number of solutions of equations in finite fields&lt;/em&gt; he mentions article 358 of Gauss' &lt;em&gt;Disquisitiones&lt;/em&gt;. Can someone please show me the connection here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How does the Gaussian sum of order $3$, for primes of the form $p = 3n+1$ determine the number of solutions for all congruence $ax^3 - by^3 = 1 \pmod p$?&lt;/p&gt;&#xA;" OwnerUserId="4290" LastActivityDate="2011-03-11T18:10:17.430" Title="Counting solutions of a cubic congruence using Gauss sums" Tags="&lt;number-theory&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="26288" PostTypeId="1" CreationDate="2011-03-11T00:09:26.883" Score="2" ViewCount="58" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Question: If $f\colon G\to H$ is a group homomorphism, and $|G:\mathrm{Ker}(f)|$ and $|H|$ are coprime, then show that $f(G)=1$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Has anyone experienced the &quot;:&quot; notation before? Does it have something to do with cosets? any hints would be greatly appreciated. Thanks for reading.&lt;/p&gt;&#xA;" OwnerUserId="8091" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-11T00:12:29.730" LastActivityDate="2011-03-11T00:14:22.817" Title="Modern Alg. : Subgroups and Cosets" Tags="&lt;abstract-algebra&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="26289" PostTypeId="2" ParentId="26259" CreationDate="2011-03-11T00:10:07.463" Score="1" ViewCount="" Body="&lt;p&gt;The ring of formal power series over a domain is itself a domain, hence nonzero elements are cancellable, quotients are unique, etc. But you don't need these results for the whole ring - only for the element $\rm\:X\:,\: $ and this case can be proved quite simply. It is obvious that $\rm\:X\:$ is not a zero-divisor, i.e. $\rm\ X\ f\:(X) = 0\ \Rightarrow\ f\:(X)=0\:,\: $ from which closely related results follow immediately, viz.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;THEOREM&lt;/strong&gt; $\ \ $ TFAE for $\rm\ a \in R\ $ any ring&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(1)\quad\ \rm a\: b\ =\ a\: c\ \ \Rightarrow\ \ b\ =\ c\quad\quad\quad\quad\quad\quad\ $ ($\rm\:a\:$ is cancellable)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(2)\quad\rm\ a\:b\ =\ \ 0\ \ \ \Rightarrow\ \ b\ =\ 0\quad\quad\quad\quad\quad\quad\ $ ($\rm\:a\:$ is not a zero-divisor)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(3)\quad $ roots of $\rm\ a\ x\ =\ b\ $ are unique $\quad\ \ $ (uniqueness of division by $\rm\:a\:$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof&lt;/strong&gt; $\rm\ \ \ (1\Rightarrow 2)\ \ $ Specialize $\rm\: c = 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm (2\Rightarrow 3)\ \ \ a\ x = b = a\ x&amp;#39;\ \Rightarrow\ \ a\ (x-x&amp;#39;) = 0\ \ \Rightarrow\ \ x = x&amp;#39;$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm (3 \Rightarrow 1)\ \ \ a\ x = a\: c\ $ has roots $\rm\ b,c\ $ so $\rm\ b = c$ &lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2011-03-11T00:19:19.140" LastActivityDate="2011-03-11T00:19:19.140" CommentCount="2" />
  <row Id="26290" PostTypeId="2" ParentId="26288" CreationDate="2011-03-11T00:14:22.817" Score="3" ViewCount="" Body="&lt;p&gt;$|G\colon\mathrm{ker}(f)|$ is the index of $\mathrm{ker}(f)$ in $G$; that is, the cardinality of the set of cosets of $\mathrm{ker}(f)$ in $G$. You may have seen it denoted by $[G:\mathrm{ker}(f)]$, or $[G:K]$ where $K$ is an arbitrary subgroup of $G$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since you are talking about finite groups here, $|G\colon \mathrm{ker}(f)|$ is just the number of left (equivalently, right) cosets of $\mathrm{ker}(f)$ in $G$.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2011-03-11T00:14:22.817" />
  <row Id="26291" PostTypeId="2" ParentId="26280" CreationDate="2011-03-11T00:21:24.827" Score="2" ViewCount="" Body="&lt;p&gt;All the Sobolev spaces $H^s$ are isomorphic as Hilbert spaces simply because they have the same dimension.. the isomorphism from $H^s$ to $H^t$ is given by multiplying the Fourier transform by $(1 + |\xi|^2)^{s - t \over 2}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When one says that $H^s \subset H^t$ for $s &amp;gt; t$ one is looking at the finiteness of $||f||_{H_s} = ||(1 + |\xi|^2)^{s \over 2}\hat{f}(\xi)||_{L^2}$ and similarly for $H_t$. If you want a natural dual spaces that reverse the inclusion you have to view them as subspaces of a fixed Hilbert space with a single inner product, say that of $H^t$. In this case one can view the dual space of $H^s$ as $H^{-s}$ and the dual space of $H^t$ as $H^{-t}$ by taking $&amp;lt;f, g&amp;gt; = \int \hat{f}(\xi)\bar{\hat{g}(\xi)}\,d\xi$. In this situation the reversal of inclusions is clear. &lt;/p&gt;&#xA;" OwnerUserId="3035" LastActivityDate="2011-03-11T00:21:24.827" />
  <row Id="26292" PostTypeId="2" ParentId="26284" CreationDate="2011-03-11T00:40:42.970" Score="0" ViewCount="" Body="&lt;p&gt;It represents the measure $u$. If (a function) $f$ is the Radon-Nikodym derivative (or density) of $u$ w.r.t. $v$, it is denoted by $du/dv$. Now, if $f=du/dv$, then $du=f dv$. For example, suppose that $u$ is an absolutely continuous probability distribution with density function $f$. Then, this can be denoted by $du(x)=f(x)dx$. That is, the right-hand side determines the measure $u$ by $u((a,b]) = \int_a^b {f(x)dx}$ (for all $-\infty &amp;lt; a &amp;lt; b &amp;lt; \infty$). Here, $dx$ stands for the Lebesgue measure on $\mathbb{R}$. Remark: an alternative notation for $du(x)$ is $u(dx)$.&lt;/p&gt;&#xA;" OwnerUserId="2810" LastEditorUserId="2810" LastEditorDisplayName="" LastEditDate="2011-03-11T02:09:01.757" LastActivityDate="2011-03-11T02:09:01.757" CommentCount="1" />
  <row Id="26293" PostTypeId="2" ParentId="26284" CreationDate="2011-03-11T00:43:35.333" Score="0" ViewCount="" Body="&lt;p&gt;Rudin for instance uses $d\mu$ in the following sense: $d\mu = fd\lambda$ just means the measure $\mu$ is defined by $\mu(E) = \int_E fd\lambda$ where for instance $f\in L^1(\mathbb{R})$. That's all it is: just a shorthand for this.&lt;/p&gt;&#xA;" OwnerUserId="2055" LastActivityDate="2011-03-11T00:43:35.333" />
  <row Id="26294" PostTypeId="2" ParentId="25449" CreationDate="2011-03-11T01:21:23.467" Score="1" ViewCount="" Body="&lt;p&gt;More of a comment than an answer (I know you tagged this as &quot;differential-geometry&quot; and not &quot;homotopy-theory&quot;):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By using simultaneous continuous Gram-Schmidt orthonormalization, this deformation retracts onto $O(3)/(GL(2)\cap O(3))=O(3)/O(2)$.  Note that $O(3)/(O(2)\times O(1))=:Gr_2(\mathbb{R}^3)\cong Gr_1(\mathbb{R}^3)=\mathbb{RP}^2$, so taking an $O(1)$-quotient yields $\mathbb{RP}^2$.  As $O(3)/O(2)$ is connected (the coset of any $A\in O(3)$ contains an element of $SO(3)$), it must be that $O(3)/O(2)\cong S^2$.  (In fact, the fiber bundle $O(2)\rightarrow O(3)\rightarrow S^2$ is just the circle bundle $S(TS^2)$, i.e. the sphere bundle of $TS^2$.)  So, your space is homotopy equivalent to $S^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fact that you're quotienting by an embedded Lie subgroup means that the dimensions do indeed subtract: $\dim(GL(3)/GL(2))=3^2-2^2=5$.  So this isn't just $TS^2$ or anything like that.  I'd nevertheless be willing to bet that it's some sort of fiber bundle over $S^2$, but I couldn't prove it...&lt;/p&gt;&#xA;" OwnerUserId="2159" LastActivityDate="2011-03-11T01:21:23.467" />
  <row Id="26295" PostTypeId="1" AcceptedAnswerId="26349" CreationDate="2011-03-11T01:40:52.787" Score="0" ViewCount="161" Body="&lt;p&gt;What is the most general object that defines a solution to an ordinary differential equation?  (I don't know enough to know if this question is specific enough.  I am hoping the answer will be something like &quot;a function&quot;, &quot;a continuous function&quot;, &quot;a piecewise continuous function&quot; ... or something like this.)&lt;/p&gt;&#xA;" OwnerUserId="1004" LastActivityDate="2011-03-12T17:46:35.073" Title="What is the most general mathematical object that defines a solution to an ordinary differential equation?" Tags="&lt;differential-equations&gt;" AnswerCount="3" CommentCount="7" />
  <row Id="26296" PostTypeId="1" AcceptedAnswerId="26304" CreationDate="2011-03-11T01:48:23.730" Score="0" ViewCount="75" Body="&lt;p&gt;A steepest tangent vector to the explicit surface z=f[x,y] at the fixed point (x,y,f[x,y]) is given by the formula:&#xA;steepTan =   (?)&#xA;             (?)&#xA;             (?)&#xA;(You can replace the ?s with formulas like ([PartialD]f/[PartialD]x)[x,y]&#xA;One of the hints is that the steepest tan is the (dx,dy,?). The hardest part is figuring out the third variable. Which can be figured out by dz = mdx +ndy.&lt;/p&gt;&#xA;" OwnerUserId="8093" LastActivityDate="2011-03-11T02:53:32.740" Title="Steepest Tangent vector to an explicit surface" Tags="&lt;calculus&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="26297" PostTypeId="1" AcceptedAnswerId="26308" CreationDate="2011-03-11T01:49:47.257" Score="1" ViewCount="91" Body="&lt;p&gt;I have the following question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $I,J$ are ideals in $R$, then denote $I+J = \{ r \in R | r = x+y, x \in I, y \in J  \}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is not hard to show that $I+J$ is again an ideal in $R$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we then assume that $I+J = R$ we have the identity $R/(I \cap J) \simeq R/I \times R/J$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I managed to show this by constructing a function $f: R \to R/I \times R/J$ and then using the first isomorphism theorem. The proof then basically follows how one would derive the second isomorphism theorem from the first, which leads me the question: is it possible to approach this problem using the 2nd/3rd isomorphism theories directly?&lt;/p&gt;&#xA;" OwnerUserId="2219" LastActivityDate="2011-03-11T03:19:18.457" Title="Sum of ideals in a ring" Tags="&lt;ring-theory&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="26298" PostTypeId="2" ParentId="25449" CreationDate="2011-03-11T01:57:01.463" Score="3" ViewCount="" Body="&lt;p&gt;As a bundle it fibers over $\mathbb RP^2$ with fiber $\mathbb R^3 \setminus \mathbb R^2$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the most natural way to describe this space would be to consider two bundles over $\mathbb RP^2$: (1) $\mathbb RP^2 \times \mathbb R^3$ and (2) $\{(L,v) : L \in \mathbb RP^2, v \in \mathbb R^3, v \perp L \}$ for this purpose $\mathbb RP^2$ is considered the space of lines in $\mathbb R^3$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the total space of bundle (2) is a subspace of the total space of bundle (1).  The bundle you're interested in, is the complement of (2) in (1). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So another way to say what this space is, is it's all pairs $(L, v)$ such that $L$ is in $\mathbb RP^2$ and $v$ projects to a non-zero vector in $L$ (using orthogonal projection). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So another way to describe it would be the tangent bundle of $\mathbb RP^2$ fiber-product with a certain bundle over $\mathbb RP^2$ -- as a bundle over $\mathbb RP^2$ this &quot;certain bundle&quot; is the map $\mathbb R \times S^2 \to \mathbb RP^2$ where $(p,q) \longmapsto \pi(q)$ where $\pi : S^2 \to \mathbb RP^2$ is the 2:1 covering map. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even simpler, as a space it's diffeomorphic to $S^2 \times \mathbb R^3$.  You can make this equivarient with respect to the left $GL(3)$ action as well -- $GL(3)$ has its standard (projectivized) action on $S^2$, similarly it acts on $\mathbb R^3$.  &lt;/p&gt;&#xA;" OwnerUserId="642" LastEditorUserId="642" LastEditorDisplayName="" LastEditDate="2011-03-11T02:25:56.913" LastActivityDate="2011-03-11T02:25:56.913" CommentCount="1" />
  <row Id="26299" PostTypeId="2" ParentId="26259" CreationDate="2011-03-11T02:05:57.030" Score="0" ViewCount="" Body="&lt;p&gt;The concept of ring formalizes all the allowed operations, but I think this can be presented without all reference to the machinery.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The original function is $f_n$. The generating function is $f(x)$ where $x$ is the variable ($X$ in the OP's question):  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f(x) = \sum_{n\ge0} f_n x^n$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then we can operate on this as follows using naive power series operations: $f(0) = f_0$, so &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f(x) - f(0) = \sum_{n\ge1} f_n x^n$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(whatever the value of $f_0$; it can be invertible or not - we're not trying to invert the gf here at all). So then &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{f(x) - f(0)}{x} = \frac{1}{x}\sum_{n\ge1} f_n x^n = \sum_{n\ge1} f_n x^{n-1}  = \sum_{n\ge0} f_{n+1} x^n$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is what is desired, the gf for $f_{n+1}$.&lt;/p&gt;&#xA;" OwnerUserId="1919" LastEditorUserId="1919" LastEditorDisplayName="" LastEditDate="2011-03-11T14:37:36.923" LastActivityDate="2011-03-11T14:37:36.923" />
  <row Id="26300" PostTypeId="2" ParentId="26279" CreationDate="2011-03-11T02:09:19.027" Score="1" ViewCount="" Body="&lt;p&gt;What is the range of $\sin(z)$ What is the range of $\sin(n z)$? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;All the restrictions on $n$ and $x$ just remove annoying counterexamples.&lt;/p&gt;&#xA;" OwnerUserId="1919" LastActivityDate="2011-03-11T02:09:19.027" />
  <row Id="26301" PostTypeId="1" AcceptedAnswerId="26302" CreationDate="2011-03-11T02:27:02.700" Score="3" ViewCount="47" Body="&lt;p&gt;Does anybody know of a good reference (preferably online) where I can find a good, rigorous description of $Hom_R(C_\bullet,D_\bullet)_\bullet$, which is a cochain complex where the module is the nth degree is given as $\Pi_{i-j=n}Hom_R(C_i,D_j)$, where $C_\bullet$ and $D_\bullet$ are chain complexes?  I would give what the coboundary maps are, but that is the main piece of information I am trying to find a definition of, as well as some other things about it.  Thanks!&lt;/p&gt;&#xA;" OwnerUserId="3595" LastEditorUserId="3595" LastEditorDisplayName="" LastEditDate="2011-03-11T02:54:15.937" LastActivityDate="2011-03-11T02:54:15.937" Title="Hom cochain complex of two chain complexes" Tags="&lt;homological-algebra&gt;" AnswerCount="1" />
  <row Id="26302" PostTypeId="2" ParentId="26301" CreationDate="2011-03-11T02:30:31.650" Score="2" ViewCount="" Body="&lt;p&gt;The boundary in the complex $\hom(C_\bullet, D_\bullet)$ is simply the map &#xA;$$\delta:\prod_{i-j=n}Hom_R(C_j,D_i) \to \prod_{i-j=n-1}Hom_R(C_j,D_i)$$ such that for each $f\in Hom_R(C_i,D_j)$ we have $$\delta(f)=f\circ d_C-d_D\circ f.$$&#xA;(You do have to figure out in which factor of the codomain of $\delta$ each of these two summands is; notice also that $0$-cycles are precisely the morphisms of complexes, and you should try to see what exactly does it mean for two such $0$-cocycles to be homologous)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This should be explained in pretty much &lt;em&gt;every&lt;/em&gt; textbook on homological algebra. For example, this is one of the examples in section V.1 of P. J. Hilton, U. Stammbach's always useful &lt;em&gt;Course on homological algebra&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="274" LastEditorUserId="274" LastEditorDisplayName="" LastEditDate="2011-03-11T02:36:10.447" LastActivityDate="2011-03-11T02:36:10.447" CommentCount="3" />
  <row Id="26303" PostTypeId="1" AcceptedAnswerId="26579" CreationDate="2011-03-11T02:42:13.233" Score="3" ViewCount="203" Body="&lt;p&gt;Let $c,d&amp;gt;0$. I do not know how to integrate the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_{-d}^d  \frac{z}{2\sqrt{z+c}\ \cdot \sqrt{d^2-z^2}}\ \text{d}z.$$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Sorry~~ it should be $z^2$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think it can be simplified to the following: for $c\ge d&amp;gt;0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\sqrt{d}}{2} \int_{-\pi/2}^{\pi/2} \frac{\sin y}{\sqrt{\sin&#xD;&#xA;y+c/d}}\ \text{d}y.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks again!&lt;/p&gt;&#xA;" OwnerUserId="6145" LastEditorUserId="6145" LastEditorDisplayName="" LastEditDate="2011-03-13T04:10:29.977" LastActivityDate="2011-04-18T03:39:44.750" Title="How to find: $\int_{-d}^d  \frac{z}{2\sqrt{z+c}\ \cdot \sqrt{d^2-z^2}} dz$" Tags="&lt;calculus&gt;&lt;integral&gt;" AnswerCount="3" CommentCount="4" FavoriteCount="1" />
  <row Id="26304" PostTypeId="2" ParentId="26296" CreationDate="2011-03-11T02:53:32.740" Score="1" ViewCount="" Body="&lt;p&gt;The surface tangent vector you're looking for is the tangent vector to the curve on the surface that you get by travelling in the direction of the gradient. That curve is given by   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left(x + \lambda\frac{\partial f}{\partial x},y + \lambda\frac{\partial f}{\partial y},z\left(x + \lambda\frac{\partial f}{\partial x},y + \lambda\frac{\partial f}{\partial y}\right)\right)\;,$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so the tangent vector, the derivative with respect to $\lambda$, is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left(\frac{\partial f}{\partial x},\frac{\partial f}{\partial y},\frac{\partial f}{\partial x}\cdot\frac{\partial f}{\partial x} + \frac{\partial f}{\partial y}\cdot \frac{\partial f}{\partial y}\right)=\left(\frac{\partial f}{\partial x},\frac{\partial f}{\partial y},\lvert\nabla f\rvert^2\right)\;.$$&lt;/p&gt;&#xA;" OwnerUserId="6622" LastActivityDate="2011-03-11T02:53:32.740" />
  <row Id="26305" PostTypeId="1" AcceptedAnswerId="26356" CreationDate="2011-03-11T03:09:59.620" Score="0" ViewCount="108" Body="&lt;p&gt;What math subjects are used in quantitative trading firms?&lt;/p&gt;&#xA;" OwnerUserId="7810" LastActivityDate="2011-03-11T09:07:17.847" Title="Quantitative trading" Tags="&lt;soft-question&gt;&lt;math-software&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="1" />
  <row Id="26306" PostTypeId="1" CreationDate="2011-03-11T03:10:45.970" Score="2" ViewCount="122" Body="&lt;p&gt;Let us assume that directional derivative of a function $f$ exists at a point $p$ (i.e.,$ D_v(f)$) for all vectors $v \in \mathbb{R}^{n}$. Does it imply that the function is differentiable?&lt;/p&gt;&#xA;" OwnerUserId="1252" LastActivityDate="2011-03-12T09:41:34.550" Title="directional derivative in a manifold." Tags="&lt;calculus&gt;&lt;analysis&gt;&lt;differential-geometry&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="26307" PostTypeId="1" AcceptedAnswerId="26310" CreationDate="2011-03-11T03:19:02.967" Score="2" ViewCount="82" Body="&lt;p&gt;Hello,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $a^{n-1} \equiv 1 \pmod{n}$, can we say that $(a,n) = 1$? If not, what conditions do we need to make this argument true? Any idea?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks,  &lt;/p&gt;&#xA;" OwnerUserId="5711" LastActivityDate="2011-03-11T04:21:19.980" Title="If $a^{n-1} \equiv 1 \pmod{n}$, can we say that $(a,n) = 1$?" Tags="&lt;elementary-number-theory&gt;" AnswerCount="2" />
  <row Id="26308" PostTypeId="2" ParentId="26297" CreationDate="2011-03-11T03:19:18.457" Score="3" ViewCount="" Body="&lt;p&gt;In general, you always have that $R/(I\cap J)$ embeds into $R/I \times R/J$. This simply follows because if $f$ and $g$ are &lt;em&gt;any&lt;/em&gt; two ring homomorphism $f\colon R\to S$ and $g\colon R\to T$, then letting $f\times g\colon R\to S\times T$ by the map $(f\times g)(r) = (f(r),g(r))$, you always have $\mathrm{ker}(f\times g) = \mathrm{ker}(f)\cap\mathrm{ker}(g)$. Thus, $R/(I\cap J)$ is isomorphic to its image in $R/I\times R/J$ by the First Isomorphism Theorem. This &lt;em&gt;always&lt;/em&gt; holds, without any conditions on $I$ or on $J$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only issue here is showing that this map is onto, and for this you use the fact that $I+J=R$: given $x,y\in R$ and $(x+I,y+J)\in R/I\times R/J$, there exists $a_1,a_2\in I$, $b_1,b_2\in J$ such that $x=a_1+b_1$, $y=a_2+b_2$. Then $(b_1+a_2)+I = b_1+I = x+I$ and $(b_1+a_2)+J = a_2+J = y+J$, so $b_1+a_2$ maps to $(x+I,y+J)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as I can tell, this only uses the first isomorphism theorem, and&#xA;I don't see any use or reconstruction of the second isomorphism theorem (usually, that there is an inclusion preserving bijection between the subrings of $R/I$ and the subrings of $R$ that contains $I$, and the bijection maps ideals to ideals) nor the third (usually, that for a subring $R&amp;#39;$ and an ideal $I$, $(R'+I)/I \cong R'/(R'\cap I)$). Perhaps you have a different numbering, but I still don't see the application of anything other than the first isomorphism theorem and some basic properties about products in the proof above.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2011-03-11T03:19:18.457" CommentCount="4" />
  <row Id="26309" PostTypeId="2" ParentId="26307" CreationDate="2011-03-11T03:21:34.947" Score="1" ViewCount="" Body="&lt;p&gt;Yes. Suppose $n&amp;gt;1$. Since $a^{n-1}$ is relatively prime to $n$, $a$ and $n$ can share no common prime factors. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, you may note that the invertible elements in the ring of integers mod $n$ are precisely the integers (or residue classes thereof) prime to $n$. So since $a$ has an inverse in this ring (namely, $a^{n-2}$), $a$ must be relatively prime to $n$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(If $n =1$, the statement is true automatically.)&lt;/p&gt;&#xA;" OwnerUserId="536" LastEditorUserId="536" LastEditorDisplayName="" LastEditDate="2011-03-11T04:21:19.980" LastActivityDate="2011-03-11T04:21:19.980" CommentCount="3" />
  <row Id="26310" PostTypeId="2" ParentId="26307" CreationDate="2011-03-11T03:22:19.173" Score="4" ViewCount="" Body="&lt;p&gt;We have the following two elementary facts that you may be overlooking:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;If $a\equiv b\pmod{n}$, then $\gcd(n,a) = \gcd(n,b)$ (since $\gcd(n,a) = \gcd(n,a+kn)$ for any $k$, and $a\equiv b\pmod{n}$ implies that $b=a+kn$ for some $k$). &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If $a|b$, then $\gcd(n,a)|\gcd(n,b)$. Simply note that any common divisor of $n$ and $a$ must be a common divisor of $n$ and $b$, since $a|b$ holds.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So, if $m\geq 1$ and $a^m\equiv 1\pmod{n}$, then since $a|a^m$ we have&#xA;$$\gcd(n,a) \;|\; \gcd(n,a^m) = \gcd(n,1) = 1,$$&#xA;hence $\gcd(n,a)=1$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, for $a^{n-1}\equiv 1 \pmod{n}$: if $n\gt 1$, then apply the observation above with $m=n-1$. If $n=1$ then you cannot apply the observation, since then you have $n-1 = 0$, but if $n=1$, then $\gcd(n,a) = \gcd(1,a) = 1$ as well.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-11T03:32:45.247" LastActivityDate="2011-03-11T03:32:45.247" CommentCount="5" />
  <row Id="26311" PostTypeId="1" AcceptedAnswerId="26314" CreationDate="2011-03-11T03:30:36.670" Score="4" ViewCount="103" Body="&lt;p&gt;I am trying to prove that the set of irrational numbers $\mathbb{R} \setminus \mathbb{Q}$ is not an $F_{\sigma}$ set. Here's my attempt:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume that indeed $\mathbb{R} \setminus \mathbb{Q}$ is an $F_{\sigma}$ set. Then we may write it as a countable union of closed subsets $C_i$:&#xA;$$&#xD;&#xA;\mathbb{R} \setminus \mathbb{Q} = \bigcup_{i=1}^{\infty}  \ C_i&#xD;&#xA;$$&#xA;But $\text{int} ( \mathbb{R} \setminus \mathbb{Q}) = \emptyset$, so in fact each $C_i$ has empty interior as well. But then each $C_i$ is nowhere dense, hence $&#xD;&#xA;\mathbb{R} \setminus \mathbb{Q} = \bigcup_{i=1}^{\infty}  \ C_i$ is thin. But we know $\mathbb{R} \setminus \mathbb{Q}$ is thick, a contradiction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This seems a bit too simple. I looked this up online, and although I haven't found the solution anywhere, many times there is a hint: Use Baire's Theorem. Have I skipped an important step I should explain further or is Baire's Theorem used implicitly in my proof? Or is my proof wrong? Thanks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Thin and thick might not be the most standard terms so:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thin = meager = 1st Baire category&lt;/p&gt;&#xA;" OwnerUserId="5105" LastEditorUserId="5105" LastEditorDisplayName="" LastEditDate="2011-03-11T03:37:44.417" LastActivityDate="2011-03-11T04:10:42.207" Title="Proof That $\mathbb{R} \setminus \mathbb{Q}$ Is Not an $F_{\sigma}$ Set" Tags="&lt;homework&gt;&lt;real-analysis&gt;" AnswerCount="2" CommentCount="0" />
  <row Id="26312" PostTypeId="2" ParentId="26311" CreationDate="2011-03-11T03:37:28.117" Score="2" ViewCount="" Body="&lt;p&gt;Suppose $\mathbb{R} \setminus \mathbb{Q}$ is an $F_{\sigma}$ set. Then $\mathbb{Q}$ is a $G_{\delta}$ set which is a contradiction. In other words, suppose $$&#xD;&#xA;\mathbb{R} \setminus \mathbb{Q} = \bigcup_{i=1}^{\infty}  \ C_i&#xD;&#xA;$$ Then by Demorgan's Laws $$&#xD;&#xA; \mathbb{Q} = \bigcap_{i=1}^{\infty}  \ C_i^{c}&#xD;&#xA;$$ which is a contradiction since $\mathbb{R} \setminus \mathbb{Q}$ is a countable intersection of open sets and we know that the intersection between the rational is irrationals is $\emptyset$. So $\emptyset$ is a countable intersection of open dense subsets which contradicts Baire's category theorem. &lt;/p&gt;&#xA;" OwnerUserId="2548" LastEditorUserId="2548" LastEditorDisplayName="" LastEditDate="2011-03-11T04:03:31.493" LastActivityDate="2011-03-11T04:03:31.493" CommentCount="1" />
  <row Id="26313" PostTypeId="2" ParentId="26268" CreationDate="2011-03-11T03:53:48.010" Score="3" ViewCount="" Body="&lt;p&gt;Use Taylor's theorem in one dimension with remainder in integral form on the function $g(t) = f(a + t(x- a))$. If you write the Taylor expansion of $g(t)$ about $0$ and plug in $t = 1$ it translates into what you're trying to prove.  &lt;/p&gt;&#xA;" OwnerUserId="3035" LastActivityDate="2011-03-11T03:53:48.010" CommentCount="1" />
  <row Id="26314" PostTypeId="2" ParentId="26311" CreationDate="2011-03-11T04:10:42.207" Score="4" ViewCount="" Body="&lt;p&gt;Your solution is correct. You could also argue that $\mathbb{R} = \bigcup_{i =1}^{\infty} C_{i} \cup \bigcup_{q \in \mathbb{Q}} \{q\}$, so by Baire one of the $C_{i}$ must have non-empty interior, contradicting the fact that $\mathbb{R} \smallsetminus \mathbb{Q}$ has empty interior.&lt;/p&gt;&#xA;" OwnerUserId="5363" LastActivityDate="2011-03-11T04:10:42.207" CommentCount="1" />
  <row Id="26315" PostTypeId="1" CreationDate="2011-03-11T04:11:54.823" Score="0" ViewCount="54" Body="&lt;p&gt;Being a lit. student who doesn't even really know what all the math signs are, I have a question.  It is probably super easy for anyone with q math iq higher than my room-temperature one. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The scores of students on ACT college entrance exams in recent years have had a mean of $\mu = 18.6$ and a standard deviation of $5.9$. Assuming that admitted students from a college have the same mean and standard deviation, find the probability that $50$ randomly chosen students would have a sample mean higher than $20$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance!!&lt;/p&gt;&#xA;" OwnerUserId="8095" LastEditorUserId="8054" LastEditorDisplayName="" LastEditDate="2011-03-11T04:36:05.273" LastActivityDate="2011-03-11T04:36:05.273" Title="Stats help please!" Tags="&lt;statistics&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="26316" PostTypeId="1" AcceptedAnswerId="26319" CreationDate="2011-03-11T04:12:53.077" Score="2" ViewCount="67" Body="&lt;p&gt;I am wondering how to prove that the intersection of an infinite number of convex sets is convex. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can prove that the intersection of two convex sets is convex, and I believe that I can simply do an induction on this result, but I've heard that it would be wrong to do this since I am working with infinity. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I guess another way to think of this question is whether or not I have to take something special into consideration since the word &quot;infinite&quot; is involved. &lt;/p&gt;&#xA;" OwnerUserId="7145" LastEditorUserId="622" LastEditorDisplayName="" LastEditDate="2011-03-11T04:20:22.713" LastActivityDate="2011-03-11T04:25:06.497" Title="What is the right way to prove that the intersection of an infinite number of convex sets is convex?" Tags="&lt;convex-sets&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="26317" PostTypeId="2" ParentId="26316" CreationDate="2011-03-11T04:17:30.423" Score="0" ViewCount="" Body="&lt;p&gt;Suppose $x$, $y$ is in $F=\bigcap^{\infty}_{i=1}F_{i}$. We argue that the segment $xy$ is in $F$ because it is in all $F_{i}$. Hence by definition $F$ is convex. There is no problem with $\infty$ in here. &lt;/p&gt;&#xA;" OwnerUserId="7887" LastActivityDate="2011-03-11T04:17:30.423" CommentCount="2" />
  <row Id="26318" PostTypeId="1" AcceptedAnswerId="26332" CreationDate="2011-03-11T04:23:22.367" Score="4" ViewCount="105" Body="&lt;p&gt;I'm reading &lt;em&gt;Fractals Everywhere&lt;/em&gt; by Michael Barnsley. On pp. 6-8 [1] he defines a &lt;em&gt;linear space&lt;/em&gt; which, he says, &quot;is also called a vector space.&quot; However, his definition of a linear space only requires closure under vector addition and scalar multiplication. There is no mention, for example, of additive inverses. (I realize other properties are required.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;He also defines the Riemann Sphere (see section 1.5), and later, in an exercise, asks the reader to show that the Riemann Sphere is not a vector space. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can see that the Riemann Sphere is not a vector space in the traditional sense because $\infty$ has no additive inverse.  But it seems that using Barnsley's definition, it &lt;em&gt;would&lt;/em&gt; be a vector space, if we adopt the convention that $\infty + x = \infty$ for all $x \in \mathbb{C}$ and that $a \cdot \infty = \infty$ for all $a \in \mathbb{R}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Am I missing something?  Thanks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[1] &lt;a href=&quot;http://goo.gl/d0NzN&quot; rel=&quot;nofollow&quot;&gt;http://goo.gl/d0NzN&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="6449" LastActivityDate="2011-03-11T06:09:48.717" Title="Riemann Sphere and a Strange Vector Space Definition" Tags="&lt;vector-space&gt;" AnswerCount="2" />
  <row Id="26319" PostTypeId="2" ParentId="26316" CreationDate="2011-03-11T04:25:06.497" Score="2" ViewCount="" Body="&lt;p&gt;Let $S_{\alpha}$, $\alpha \in \Gamma$ be an infinite collection of convex sets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $S = \displaystyle \cap_{\alpha \in \Gamma} S_{\alpha}$ be the intersection of these sets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider $x,y \in S$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x,y \in S \Rightarrow x,y \in S_{\alpha}$, $\forall \alpha \in \Gamma$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $S_{\alpha}$ is convex $\forall \alpha \in \Gamma$, any convex combination of $x,y$ i.e. $\lambda x + (1 - \lambda) y \in S_{\alpha}$, $\forall \alpha \in \Gamma$, $\forall \lambda \in [0,1]$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, every convex combination is in $S_{\alpha}$, $\forall \alpha \in \Gamma$ and hence it is also in $\displaystyle \cap_{\alpha \in \Gamma} S_{\alpha}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, $S = \displaystyle \cap_{\alpha \in \Gamma} S_{\alpha}$ is convex.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastActivityDate="2011-03-11T04:25:06.497" CommentCount="2" />
  <row Id="26320" PostTypeId="2" ParentId="26315" CreationDate="2011-03-11T04:29:18.367" Score="1" ViewCount="" Body="&lt;p&gt;Hint:  the expected mean of the sample is the same as the mean of the parent distribution.  The variance of the mean goes down as 1/n.  So how many standard deviations high would the sample have to be?  See Estimation of Parameters in &lt;a href=&quot;http://en.wikipedia.org/wiki/Normal_distribution&quot; rel=&quot;nofollow&quot;&gt;the Wikipedia page&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-03-11T04:29:18.367" />
  <row Id="26321" PostTypeId="1" CreationDate="2011-03-11T04:41:43.503" Score="0" ViewCount="37" Body="&lt;p&gt;I have discrete time series containing signal with smoothly varied frequency over time (called a &quot;sweep&quot;). How can I design a discrete filter (low-pass in my case) of a finite length with linearly varying cut-frequency over time and constant cut-slope?&lt;/p&gt;&#xA;" OwnerUserId="3806" LastActivityDate="2011-03-11T04:41:43.503" Title="How to design a continuously time-variant digital filter?" Tags="&lt;signal-processing&gt;" FavoriteCount="1" />
  <row Id="26322" PostTypeId="2" ParentId="26318" CreationDate="2011-03-11T04:44:34.160" Score="1" ViewCount="" Body="&lt;p&gt;I don't feel I really contributed anything. I think the usually we treat $C^{*}$ is to treat it as $\mathbb{C}\mathbb{P}_{1}$, in other words the complex &quot;lines&quot; passing the complex &quot;plane&quot;. Hence for two elements in the plane we have $(c_{1},c_{2})\approx (\frac{c_{1}}{c_{2}},1)$ for $c_{2}\not=0$ and vector addition just follows. At $\infty$ we have $c_{2}=0$ and the whole line $(c_{1},0)$ was identified as $\infty$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now if you add two points in $C^{*}$, we are essentially adding two lines, and we may define $(a,b)+(c,d)=(ad+bc,bd)=(\frac{a}{b}+\frac{c}{d},1)$ if $b,d$ are not both zero. In case $d$ ended up being $0$ we should have $(a,b)+(c,d)=(ad,0)$, which is identified to the infinity point. This kind of justified $a+\infty=\infty$ in the Riemann Sphere. &lt;/p&gt;&#xA;" OwnerUserId="7887" LastActivityDate="2011-03-11T04:44:34.160" />
  <row Id="26323" PostTypeId="1" AcceptedAnswerId="26326" CreationDate="2011-03-11T05:15:20.353" Score="5" ViewCount="129" Body="&lt;p&gt;Today we were discussing how for an nxn orthogonal projection matrix from $\mathbb{R^{n}}$ onto a subspace W, Ker($A$)=$(Im$A$)^{\perp}$=$W^{\perp}$ and that Ker($A^{T}$) is also $W^{\perp}$. This prompted the question of what conditions are necessary for a matrix so that the kernels of it and its transpose are equal. It looks like it always works when it's an orthogonal projection, but we struggled to find an example of a square matrix for which the aforementioned identities would not hold, and consequently for which Ker($A$)$\neq$Ker($A^{T}$). It looks like we need to find a transformation whose Kernel would not consist only of things perpendicular to its image, but we were wondering if that was possible or if our reasoning was correct at all. Could you please clarify this issue (I know it's convoluted) and try to point out where we were right and wrong, and when do those identities hold? &lt;/p&gt;&#xA;" OwnerUserId="5157" LastEditorUserId="5157" LastEditorDisplayName="" LastEditDate="2011-03-11T05:31:26.800" LastActivityDate="2011-03-11T05:36:29.290" Title="A confusion about Ker($A$) and Ker($A^{T}$)" Tags="&lt;linear-algebra&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="26324" PostTypeId="2" ParentId="26323" CreationDate="2011-03-11T05:19:39.090" Score="0" ViewCount="" Body="&lt;p&gt;Hi, $A$ and $A^{T}$ are similar. So the answer should be the same. You can try standard Jordan decomposition and see how the Jordan block works. A few cases in $2\times 2$ would be very helpful. &lt;/p&gt;&#xA;" OwnerUserId="7887" LastActivityDate="2011-03-11T05:19:39.090" CommentCount="8" />
  <row Id="26325" PostTypeId="1" CreationDate="2011-03-11T05:20:12.800" Score="17" ViewCount="394" Body="&lt;p&gt;In some integration by parts problems, such as evaluating the integral of $e^x \cos x$ or $\sec^ 3 x$, one performs integration by parts (possibly more than once, and possibly together with algebraic manipulations) and eventually the original integral appears again.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To beginning students, this may superficially appear to be &quot;circular reasoning&quot; that doesn't solve the problem.  But it does, because if we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\int f(x) dx = g(x) + K \int f(x) dx$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $K \ne 1$, then rearranging gives&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\int f(x) dx = \frac{1}{1-K} g(x)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Does this technique have a commonly used name?&lt;/strong&gt; I once saw it called &quot;integration by parts with deja vu&quot; in some supplemental study materials for a calculus course.  I don't know who thought of that name but I've taken to using it with my students.&lt;/p&gt;&#xA;" OwnerUserId="7635" LastActivityDate="2011-04-12T21:31:35.620" Title="Does integration by parts with &quot;deja vu&quot; have a name?" Tags="&lt;calculus&gt;&lt;integral&gt;&lt;terminology&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="26326" PostTypeId="2" ParentId="26323" CreationDate="2011-03-11T05:26:25.713" Score="2" ViewCount="" Body="&lt;p&gt;Things always work nicely for projections, because unlike other matrices we have $\text{Range}\oplus\text{kernel}=\text{whole space}$  Let $V=\mathbb{R}^2$  Consider $$A=\left[\begin{array}{cc}&#xD;&#xA;0 &amp;amp; 1\\&#xD;&#xA;0 &amp;amp; 0\end{array}\right]$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then $$\text{ker}(A)=\left\{ \left[\begin{array}{c}&#xD;&#xA;a\\&#xD;&#xA;0\end{array}\right]\ :\ a\in\mathbb{R}\right\},$$ but $$A^{t}=\left[\begin{array}{cc}&#xD;&#xA;0 &amp;amp; 0\\&#xD;&#xA;1 &amp;amp; 0\end{array}\right]$$ so that $$\text{ker}(A^{t})=\left\{ \left[\begin{array}{c}&#xD;&#xA;0\\&#xD;&#xA;a\end{array}\right]\ :\ a\in\mathbb{R}\right\}.$$ These are definitely different.  (In contrast to the previous example, their direct sum is the whole space $V$)  &lt;/p&gt;&#xA;" OwnerUserId="6075" LastEditorUserId="6075" LastEditorDisplayName="" LastEditDate="2011-03-11T05:36:29.290" LastActivityDate="2011-03-11T05:36:29.290" CommentCount="1" />
  <row Id="26327" PostTypeId="1" AcceptedAnswerId="26336" CreationDate="2011-03-11T05:41:05.357" Score="7" ViewCount="121" Body="&lt;p&gt;Here is a problem from &lt;em&gt;generatingfunctionology&lt;/em&gt; that I'm stuck on:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/z30tr.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm trying to get started on part (a). I broke the string up like this. If the last digit is $0$, the number of possible strings is then $f(n-1,m,k)$. If the last digit is $1$, there are two subcases. If the $n-1^{th}$ digit is $0$, then we can cut them both off and the number of strings is $f(n-2,m-1,k)$. However, if the $n-1^{th}$ digit is $1$, then I don't know what to say, since even if I cut both last $1$s off, I can't have the last $k-2$ numbers of my $n-2$ long string be $1$s, but it's entirely possible that I could have $k-2$ $1$s earlier in the string. So I have something like:&#xA;$$&#xD;&#xA;f(n,m,k)=f(n-1,m,k)+f(n-2,m-1,k)+???&#xD;&#xA;$$&#xA;and I don't know what third term to put there. What is the third term? Thanks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it's not trouble, I may ask questions on parts (b) and (c) when I get there. &lt;/p&gt;&#xA;" OwnerUserId="6742" LastEditorUserId="6296" LastEditorDisplayName="" LastEditDate="2011-03-11T05:43:20.003" LastActivityDate="2011-04-14T23:52:25.947" Title="How many strings of length $n$ with $m$ $1$s have no more than $k$ consecutive $1$s? - from generatingfunctionology" Tags="&lt;combinatorics&gt;&lt;generating-functions&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="26328" PostTypeId="1" AcceptedAnswerId="26450" CreationDate="2011-03-11T05:47:27.923" Score="2" ViewCount="100" Body="&lt;p&gt;The question is to calculate the homology groups of the chain complex:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$0 \to A \stackrel{n}{\to} A \to 0$, where $A$ is an Abelian group and $n \in \mathbb{N}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't see a nice way to get anything out of this other than $H_1 = \mathrm{ker} \ \partial_1$ and $H_0 = A / \mathrm{Im} \ \partial_1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some quick calculations on small groups like $\mathbb{Z}/ 4\mathbb{Z}$ seem to confirm no 'nice' pattern, but I am probably missing something&lt;/p&gt;&#xA;" OwnerUserId="2219" LastActivityDate="2011-03-11T22:33:06.547" Title="Homology of a simple chain complex" Tags="&lt;homological-algebra&gt;" AnswerCount="2" CommentCount="9" />
  <row Id="26329" PostTypeId="1" AcceptedAnswerId="26330" CreationDate="2011-03-11T05:58:12.143" Score="0" ViewCount="72" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to move one object (dot) in circular path.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By using x and y position of that object.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="5404" LastActivityDate="2011-03-11T06:22:37.240" Title="Formula to Move the object in Circular Path" Tags="&lt;analytic-geometry&gt;&lt;circle&gt;" AnswerCount="3" FavoriteCount="1" />
  <row Id="26330" PostTypeId="2" ParentId="26329" CreationDate="2011-03-11T06:00:44.780" Score="2" ViewCount="" Body="&lt;p&gt;x and y should satisfy $x^2+y^2=r^2$, where r is radius of the circle.&lt;/p&gt;&#xA;" OwnerUserId="4929" LastActivityDate="2011-03-11T06:00:44.780" CommentCount="3" />
  <row Id="26332" PostTypeId="2" ParentId="26318" CreationDate="2011-03-11T06:03:31.513" Score="5" ViewCount="" Body="&lt;p&gt;You might call this thing (vector addition without inverses and scalar multiplication) a semi-module (&lt;a href=&quot;http://en.wikipedia.org/wiki/Semimodule#Generalizations&quot; rel=&quot;nofollow&quot;&gt;wikipedia&lt;/a&gt;) over a field. The definition is as follows (I assume that you have an additive identity, even though you don't specify one):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A &lt;em&gt;semimodule over a field&lt;/em&gt; $k$ is a set $M$ together with a function $+:M\times M \to M$ (vector addition) which defines an associative and commutative operation, an element $0\in M$ which is an identity element for $+$, and an action $k\times M \to M$ (scalar multiplication) which distributes over $+$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is, once you have scalar multiplication distributing over vector addition, you have $v + (-1)v = (1-1)v = 0$, thus $(-1)v$ is an additive inverse! Thus a semimodule over a field is the same as a module over a field which is just another name for a vector space. Thus the definition of a 'linear space' implies the remaining vector space axioms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can get around this by forgetting about some of the structure on the field, namely the additive inverse, but most often people drop both the additive and the multiplicative inverse to get a semiring = &lt;em&gt;rig&lt;/em&gt; (&lt;a href=&quot;http://en.wikipedia.org/wiki/Semiring&quot; rel=&quot;nofollow&quot;&gt;wikipedia&lt;/a&gt;). Then semimodules over a rig are very interesting objects, and feature in an area called &lt;a href=&quot;http://en.wikipedia.org/wiki/Tropical_geometry&quot; rel=&quot;nofollow&quot;&gt;tropical geometry&lt;/a&gt;. A basic example in tropical geometry is the rig $\mathbb{R}\cup \{\infty\}$ (similar to your example), but where 'addition' is $x\oplus y = min\{x,y\}$ and 'multiplication' is $x\otimes y = x+y$.&lt;/p&gt;&#xA;" OwnerUserId="3835" LastEditorUserId="3835" LastEditorDisplayName="" LastEditDate="2011-03-11T06:09:48.717" LastActivityDate="2011-03-11T06:09:48.717" CommentCount="3" />
  <row Id="26333" PostTypeId="2" ParentId="26329" CreationDate="2011-03-11T06:05:58.063" Score="1" ViewCount="" Body="&lt;p&gt;Is this part of a piece of software? Clarification on what you need would be very helpful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the object each frame has a velocity vector which it follows, just rotate the velocity vector through some fixed angle each frame. The easiest way would be to build a rotation matrix and multiply by the matrix. In a very naive implementation, you would then add the velocity to the position vector each frame. In most applications, this is insufficient, and it is preferable to use &lt;a href=&quot;http://en.wikipedia.org/wiki/Verlet_integration&quot; rel=&quot;nofollow&quot;&gt;Verlet integration&lt;/a&gt;, mainly in the presence of acceleration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or you can parameterize the points on the circle as a function of time using the relations $x = c_x + r\cos(t), y = c_y + r\sin(t)$ for $(c_x,c_y)$ the circle's center, $r$ its radius, and $t$ the time.&lt;/p&gt;&#xA;" OwnerUserId="251" LastActivityDate="2011-03-11T06:05:58.063" CommentCount="1" />
  <row Id="26334" PostTypeId="2" ParentId="26329" CreationDate="2011-03-11T06:07:54.743" Score="1" ViewCount="" Body="&lt;p&gt;There are a few ways to choose from, but a nice one that doesn't require per-step trig functions (so can be calculated by a computer very quickly) is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Midpoint_circle_algorithm&quot; rel=&quot;nofollow&quot;&gt;midpoint circle algorithm&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Otherwise, you can use x=cos(theta)*radius, y=sin(theta)*radius for 0 &amp;lt; theta &amp;lt; 360.&lt;/p&gt;&#xA;" OwnerUserId="7953" LastActivityDate="2011-03-11T06:07:54.743" CommentCount="1" />
  <row Id="26336" PostTypeId="2" ParentId="26327" CreationDate="2011-03-11T06:33:06.043" Score="7" ViewCount="" Body="&lt;p&gt;Assume $m$ and $n$ are large, say larger than $k$. If you continue your what-is-the-end-of-the-string argument, you will get $f(n,m,k)$ as the sum of $f(n-i,m-i+1,k)$ from $i=1$ to $i=k$, each of these terms counting the strings ending by $01^{i-1}$. Now, this sum has $k$ terms and you want only three. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Compare the $k$ terms sums for $f(n,m,k)$ and for $f(n-1,m-1,k)$. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The same terms appear in both sums with two exceptions: the first term of $f(n,m,k)$ is $f(n-1,m,k)$, which is not in $f(n-1,m-1,k)$, and the last term of $f(n-1,m-1,k)$ is $f(n-1-k,m-k,k)$, which is not in $f(n,m,k)$. Of course this reflects a combinatorial fact, which I let you explicit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, at least for large enough $n$ and $m$,&#xA;$$&#xD;&#xA;f(n,m,k)=f(n-1,m-1,k)+f(n-1,m,k)-f(n-1-k,m-k,k).&#xD;&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="6179" LastEditorUserId="6179" LastEditorDisplayName="" LastEditDate="2011-03-14T06:25:28.027" LastActivityDate="2011-03-14T06:25:28.027" CommentCount="4" />
  <row Id="26337" PostTypeId="1" CreationDate="2011-03-11T06:34:21.880" Score="1" ViewCount="125" Body="&lt;p&gt;Hi, I have this question that I took a shot at but I am not very familiar with Hermite or Laguerre, this my first time running across these type of polynomials and need some help please.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(a) The first four Hermite polynomials are $1, 2t,-2+4t^2,$ and $-12t+8t^3$. These polynomials arise naturally in the study of certain important differential equations in mathematical physics. Show that the first four Hermite polynomials form a basis of $\mathbb{P}_3$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(b) The first four Laguerre polynomials are $1, 1-t, 2-4t + t^2,$ and $6-18t + 9t^2- t^3$. Show&#xA;that these four Laguerre polynomials form a basis of $\mathbb{P}_3$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Results:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(a) The first four Hermite polynomials will be shown to form a basis of $\mathbb{P}_3$ by showing that they are linearly independent and that the number of polynomials equals the dimension of $\mathbb{P}_3$.&#xA;Consider the following linear combination of the four Hermite polynomials:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x(1) + y(2t) + z(-2+4t^2) + w(-12t + 8t^3) = at^3 + bt^2 + ct + d$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first four Hermite polynomials will be shown to be linearly independent by showing that the only linear combination of them that produces the zero polynomial ($0t^3+0t^2+0t+0$) is the trivial combination of zero times each polynomial.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is all I could come about thus far with it. Can anyone refine or correct this if this is the wrong approach to the problem.&lt;/p&gt;&#xA;" LastActivityDate="2011-03-11T12:29:59.840" Title="Show that the Hermite polynomials form a basis of $\mathbb{P}_3$" Tags="&lt;differential-equations&gt;&lt;polynomials&gt;" AnswerCount="2" />
  <row Id="26338" PostTypeId="2" ParentId="26337" CreationDate="2011-03-11T06:44:05.223" Score="1" ViewCount="" Body="&lt;p&gt;Forget all this Hermite and Laguerre stuff. The fact is that &lt;em&gt;any&lt;/em&gt; family of polynomials with all different degrees is linearly free. Hence &lt;em&gt;any&lt;/em&gt; family of polynomials with degrees $0$, $1$, $\ldots$, $n$ is a basis of the vector space of polynomials of degree at most $n$ (the space you denote by $\mathbb{P}_n$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A relatively more sophisticated way of saying the same thing is that any triangular matrix with no zero on its diagonal is invertible.&lt;/p&gt;&#xA;" OwnerUserId="6179" LastEditorUserId="6179" LastEditorDisplayName="" LastEditDate="2011-03-11T08:51:42.150" LastActivityDate="2011-03-11T08:51:42.150" />
  <row Id="26339" PostTypeId="1" AcceptedAnswerId="26488" CreationDate="2011-03-11T06:44:44.380" Score="4" ViewCount="221" Body="&lt;p&gt;Hello,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My attempt was:&#xA;$x^3 + 2x + 2 \equiv 0 \pmod{25}$&#xA;By inspection, we see that $x \equiv 1 \pmod{5}$. is a solution of $x^3 + 2x + 2 \equiv 0 \pmod{5}$. Let $x = 1 + 5k$, then we have:&#xA;$$(1 + 5k)^3 + 2(1 + 5k) + 2 \equiv 0 \pmod{25}$$&#xA;$$\Leftrightarrow 125k^3 + 75k^2 + 25k + 5 \equiv 0 \pmod{25}$$ &#xA;$$\Leftrightarrow 5 \equiv 0 \pmod{25}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And I was stuck here :( because k was completely cancelled out, how can we find solution for this equation?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks,&lt;/p&gt;&#xA;" OwnerUserId="5711" LastActivityDate="2011-03-12T00:50:15.653" Title="How to solve $x^3 + 2x + 2 \equiv 0 \pmod{25}$?" Tags="&lt;elementary-number-theory&gt;" AnswerCount="5" CommentCount="5" FavoriteCount="1" />
  <row Id="26341" PostTypeId="2" ParentId="26339" CreationDate="2011-03-11T07:15:58.410" Score="1" ViewCount="" Body="&lt;p&gt;This shows, that there is no modular zero for $x^3+2x+2\equiv 0  \; (\text{mod}\; 25)$ of the form $x = 1 + 5k$, since clearly $ 5 \not\equiv 0 \; (\text{mod}\; 25)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now go ahead and check $x = 3 + 5k$, because $x \equiv 3 \; (\text{mod}\; 5)$ is the other solution to $x^3+2x+2\equiv 0 \; (\text{mod}\; 5)$.&lt;/p&gt;&#xA;" OwnerUserId="8087" LastActivityDate="2011-03-11T07:15:58.410" />
  <row Id="26342" PostTypeId="2" ParentId="26339" CreationDate="2011-03-11T07:20:05.950" Score="2" ViewCount="" Body="&lt;p&gt;Hint by others, we use Hensel's lemma. By Hensel's lemma we solve $x^{3}+2x+2=0$ (mod 5). This has solution $1$ and $3$. But $1$ satisfies $f&amp;#39;(1)=0$, by Chan's previous work it should be cast aside. Hensel's lemma suggests solutions of the form $3+5k$ with $3(3+5k)^{2}+2\not=0$(mod 5). Hensel's lemma give $k$ to be:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$k(29+90 k+75 k^2)=k^{3}+2k+2$ (mod 5)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solve this we should get $k=2$. I do not know if this is the best way to solve it. &lt;/p&gt;&#xA;" OwnerUserId="7887" LastEditorUserId="7887" LastEditorDisplayName="" LastEditDate="2011-03-11T07:33:06.140" LastActivityDate="2011-03-11T07:33:06.140" CommentCount="3" />
  <row Id="26343" PostTypeId="2" ParentId="26303" CreationDate="2011-03-11T07:22:03.397" Score="1" ViewCount="" Body="&lt;p&gt;I wouldn't know how to do it manually, but contrarily to the case of yunone, Wolfram does yield an answer to me:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$-\frac{\pi  d^2 \left(2\frac{d^2}{c^2}\right)}{8 c^{3/2}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Under the conditions that $$\Re(c)&amp;gt;0\land d&amp;gt;0\land (c\notin \mathbb{R}\lor d\leq \Re(c))$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;p.s. I'm new to this forum and wouldn't know how to convert this statement to an nice format, could a moderator do this plz?&lt;/p&gt;&#xA;" OwnerUserId="8064" LastEditorUserId="1583" LastEditorDisplayName="" LastEditDate="2011-03-11T07:29:55.247" LastActivityDate="2011-03-11T07:29:55.247" CommentCount="3" />
  <row Id="26344" PostTypeId="2" ParentId="26339" CreationDate="2011-03-11T07:40:59.543" Score="1" ViewCount="" Body="&lt;p&gt;For this, I think the quickest way is simply to work out the 25 possibilities.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For 0,1,2,...24 you have $x^3+2x+2$ giving 2, 5, 14, 35, 74, 137, 230, 359, 530, 749, 1022, 1355, 1754, 2225, 2774, 3407, 4130, 4949, 5870, 6899, 8042, 9305, 10694, 12215, 13874 equivalent modulo 25 to 2, 5, 14, 10, 24, 12, 5, 9, 5, 24, 22, 5, 4, 0, 24, 7, 5, 24, 20, 24, 17, 5, 19, 15, 24. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the solution is $x \equiv 13$&lt;/p&gt;&#xA;" OwnerUserId="6460" LastActivityDate="2011-03-11T07:40:59.543" />
  <row Id="26345" PostTypeId="1" CreationDate="2011-03-11T07:44:11.310" Score="4" ViewCount="100" Body="&lt;p&gt;Hi! Let $G$ be a finite group such that for any two subgroups $H_{1}$ and $H_{2}$ of $G$ we have $H_{1} \subseteq H_{2}$ or $H_{2} \subseteq H_{1}$. Why this implies $G$ is a cyclic group?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ah. I think this works: Suppose $g \in G$ then $G = \langle g \rangle$. Suppose otherwise, then we can find $z$ such that $z \not \in \langle g  \rangle$. Now let $H_{1}=\langle z \rangle$ and $H_{2}=\langle g \rangle$. By assumption we have $H_{1} \subseteq H_{2}$ or $H_{2} \subseteq H_{1}$. In both cases we get the contradiction $z \in \langle g \rangle$.&lt;/p&gt;&#xA;" OwnerUserId="8079" LastEditorUserId="8079" LastEditorDisplayName="" LastEditDate="2011-03-11T08:25:59.263" LastActivityDate="2011-03-12T07:02:07.327" Title="Inclusion of subgroups implies the group is cyclic" Tags="&lt;group-theory&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="26346" PostTypeId="2" ParentId="26345" CreationDate="2011-03-11T07:45:45.723" Score="1" ViewCount="" Body="&lt;p&gt;Hi! Let $a$, $b$ be two random elements. Consider the subgroup $\langle a \rangle$ and $\langle b \rangle$, I think you know how to do the rest. &lt;/p&gt;&#xA;" OwnerUserId="7887" LastActivityDate="2011-03-11T07:45:45.723" CommentCount="2" />
  <row Id="26347" PostTypeId="1" AcceptedAnswerId="26348" CreationDate="2011-03-11T07:49:32.360" Score="0" ViewCount="63" Body="&lt;p&gt;Hello,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have equation &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/TIpOG.png&quot; alt=&quot;bezier cubic curve&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As much I know, I have four points lets say A(x1,y1), B(x2,y2), C(x3,y3), D(x4, y4). How these 'P' are related to these set of points. And what might a Cubic Bézier curves be for the this (A, B, C, D) given set of points.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[UPDATE]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Having problem uploading image, but image somewhat looks like this&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://stackoverflow.com/questions/5142556/algorithm-to-add-color-in-bezier-curves&quot;&gt;http://stackoverflow.com/questions/5142556/algorithm-to-add-color-in-bezier-curves&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regards,&lt;/p&gt;&#xA;" OwnerUserId="2199" LastEditorUserId="2199" LastEditorDisplayName="" LastEditDate="2011-03-11T08:35:26.767" LastActivityDate="2011-03-11T08:35:26.767" Title="what 'P' s mean in bezier curve equation" Tags="&lt;geometry&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="26348" PostTypeId="2" ParentId="26347" CreationDate="2011-03-11T07:55:21.513" Score="1" ViewCount="" Body="&lt;p&gt;That depends on what relationship you want your points $A$ to $D$ to have to the spline. It may be that you want them to have exactly the relationship to the splines that the points $P_0$ to $P_3$ have, which is explained &lt;a href=&quot;http://en.wikipedia.org/wiki/B%C3%A9zier_curve#Cubic_B.C3.A9zier_curves&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;. Strangely, the equation in your question looks as if you copied it from there, but in that case you'd have to explain what part of the explanation there of what $P_0$ to $P_3$ mean you don't understand, and what you want your points $A$ to $D$ to mean instead.&lt;/p&gt;&#xA;" OwnerUserId="6622" LastActivityDate="2011-03-11T07:55:21.513" CommentCount="3" />
  <row Id="26349" PostTypeId="2" ParentId="26295" CreationDate="2011-03-11T08:02:58.493" Score="3" ViewCount="" Body="&lt;p&gt;(All links go to Wikipedia unless stated otherwise.)&#xA;Elaborating on joriki's answer: The most general spaces where it does make sense to talk about differential equations are certain classes of &lt;a href=&quot;http://en.wikipedia.org/wiki/Topological_vector_space&quot; rel=&quot;nofollow&quot;&gt; topological vector spaces&lt;/a&gt;, it is for example rather straight forward to formulate the concept of a differential equation in Banach spaces. (Here the esolution to an equation is a mapping of topological vector spaces)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For differential equations in $\mathbb{R}^n$, the solutions themselves are elements of certain topological vector spaces.The most general topological vector spaces that are considered are AFAIK &lt;a href=&quot;http://en.wikipedia.org/wiki/Sobolev_space&quot; rel=&quot;nofollow&quot;&gt;Sobolev spaces&lt;/a&gt;, these are function spaces such that each individual function is normable with respect to a prescribed $L^p$ norm, has generalized derivatives in the sense of distributions to the order that is necessary to formulate the &lt;a href=&quot;http://en.wikipedia.org/wiki/Weak_formulation&quot; rel=&quot;nofollow&quot;&gt;weak formulation&lt;/a&gt; (see also the &lt;a href=&quot;http://www.azimuthproject.org/azimuth/show/Partial+differential+equation&quot; rel=&quot;nofollow&quot;&gt;Azimuth wiki&lt;/a&gt;) of the equation one would like to solve, such that the generalized derivatives are again normable with respect to the $L^p$ norm. Some Sobolev spaces have similar characterizations like &quot;all piecwise continuous function that...&quot; which are a little bit more complicated than that, and differ from space to space, so I'd rather like to refer to the extensive literature instead of repeating that here :-(&lt;/p&gt;&#xA;" OwnerUserId="7556" LastActivityDate="2011-03-11T08:02:58.493" CommentCount="1" />
  <row Id="26350" PostTypeId="1" CreationDate="2011-03-11T08:03:27.323" Score="2" ViewCount="102" Body="&lt;p&gt;Determine if the set $Z$ of all matricies form&#xA;$ \left[ \begin{array}{cc}&#xD;&#xA;a &amp;amp; b \\&#xD;&#xA;0 &amp;amp; d &#xD;&#xA;\end{array} \right] $&#xA;is a subspace of $M_{2 \times 2}$ (the set of all $2 \times 2$ matrices). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;% This is something I came up with. Can someone look at it and let me know any useful corrections/suggestions to the question please.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Answer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Without specification as to the nature of $a,b$ and $d$, it is assumed that $a,b,d \in \mathbb{R}$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, $H$ is determined to be a subspace of $M_{2 \times 2}$ because it is closed under scalar addition and scalar multiplication and contains the zero vector when $a=b=d=0$.&lt;/p&gt;&#xA;" LastEditorUserId="664" LastEditorDisplayName="" LastEditDate="2011-03-19T08:30:02.153" LastActivityDate="2011-03-19T08:30:02.153" Title="Determine whether the set $H$ of all matrices form a subspace of $M_{2 \times 2}$." Tags="&lt;linear-algebra&gt;&lt;matrices&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="26351" PostTypeId="2" ParentId="26295" CreationDate="2011-03-11T08:07:32.570" Score="-2" ViewCount="" Body="&lt;p&gt;Hi!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The object you are looking for is called a phase curve. For ODE $\frac{dy}{dx}=f(x)$ with $x$, we are looking for a solution of the form:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$y=y(t), x=x(t)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The result is a graph on the $xy$ plane that pass the initial condition $y=y_{0}$, $x=x_{0}$. All of this are quite standard. The best literature on this is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Differential Geometry: Manifolds, Curves, and Surfaces &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0387966269&quot; rel=&quot;nofollow&quot;&gt;http://www.amazon.com/Differential-Geometry-Manifolds-Surfaces-Mathematics/dp/0387966269/ref=sr_1_8?ie=UTF8&amp;amp;qid=1299830752&amp;amp;sr=8-8&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And &lt;em&gt;Ordinary Differential Equations, by V.I.Arnold.&lt;/em&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/3540345639&quot; rel=&quot;nofollow&quot;&gt;http://www.amazon.com/Ordinary-Differential-Equations-Universitext-Vladimir/dp/3540345639/ref=sr_1_4?s=books&amp;amp;ie=UTF8&amp;amp;qid=1299830826&amp;amp;sr=1-4&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7887" LastActivityDate="2011-03-11T08:07:32.570" />
  <row Id="26352" PostTypeId="2" ParentId="26287" CreationDate="2011-03-11T08:20:30.833" Score="4" ViewCount="" Body="&lt;p&gt;By ignoring a small subcase ($c = 0$ or $d = 0$), we can count the solutions for $cd \neq 0$ by the sum&#xA;$\displaystyle \sum_{\stackrel{c+d = 1}{c,d \neq 0}} \#\{x:ax^3 = c\} \times \#\{y: -by^3 = d\} ... (*)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Expressing the term in this sum, using multiplicative characters:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we look at $\{x: x^3 = u\}$ for some $u$, and we want to calculate its size. This can be interpreted as a sum of cubic characters: $\chi(u) + \chi^2(u) + \chi^3(u)$, where $\chi$ is a cubic character on $\mathbb{Z}/p\mathbb{Z}$, i.e. $\chi \neq 1$, but $\chi^3 = 1$. Now where does this interpretation come from?&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;If $x^3 = u$ has one solution, it has 3 because $p \equiv 1 \pmod{3}$ tells us that 1 has 3 cube roots in $\mathbb{Z}/p\mathbb{Z}$. On the character side, if $u$ is a cube of something, using $\chi^3 = 1$ w see that $\chi(u) + \chi^2(u) + \chi^3(u) = 3$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If $x^3 = u$ has no solution, then notice that $\chi(u) \neq 1$. Then if we write $S = \chi(u) + \chi^2(u) + \chi^3(u)$, note that $\chi(u)S = \chi(u)$, which means $S = 0$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;So how does this interpretation in terms of characters help? Plug in the character sums in (*). We are going to have a lot of cross terms. These cross terms can be splitted into sums of &lt;a href=&quot;http://en.wikipedia.org/wiki/Jacobi_sum&quot; rel=&quot;nofollow&quot;&gt;Jacobi sums&lt;/a&gt;, which are closely related to Gauss sums. (See &lt;a href=&quot;http://books.google.com/books?id=99v9XcOjhO4C&amp;amp;lpg=PA58&amp;amp;dq=Koblitz%27%20Introduction%20to%20Elliptic%20Curves%20and%20Modular%20forms%20Section%202.2&amp;amp;pg=PA56#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;Koblitz' Introduction to Elliptic Curves and Modular forms Section 2.2&lt;/a&gt;, or &lt;a href=&quot;http://books.google.com/books?id=jhAXHuP2y04C&amp;amp;lpg=PP1&amp;amp;dq=ireland%20and%20rosen&amp;amp;pg=PA92#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;Ireland and Rosen's A Classical Introduction to Modern Number Theory&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The situation here:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In our case,&#xA;$(*) = \sum_{c+d = 1, c,d \neq 0} \left(1 + \chi(c/a) + \chi^2(c/a)\right)\left(1 + \chi(-d/b) + \chi^2(-d/b)\right)$. &#xA;There are three types of sum in the expansion. I would do one in each type to show you how it feels like.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;$\displaystyle \sum_{c+d = 1, c,d \neq 0} 1 = p-2$, since $c$ can't be 0 or 1 and can be everything else.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$\displaystyle \sum_{c+d = 1, c,d \neq 0} \chi(c/a) = -\chi(1/a)$. To show this, note that for a nontrivial character $\chi$, $\displaystyle \sum_{u \in \mathbb{Z}/p\mathbb{Z}} \chi(u) = 0$, by an argument similar to what we showed in step 2 in the last block.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$\displaystyle \sum_{c+d = 1, c,d \neq 0} \chi(c/a) \psi (-d/b)$, where $\psi = \chi$ or $\chi^2$. This is the one we need Jacobi sums $\displaystyle J(\chi, \psi) = \sum_{x \in \mathbb{Z}/p\mathbb{Z}} \chi(x)\psi(1-x)$. Note that &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle \sum_{c+d = 1, c,d \neq 0} \chi(c/a) \psi (-d/b) = \chi(1/a)\psi(-1/b) J(\chi, \psi)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and that if you check out the books I mentioned before, $J$ can be expressed by Gauss sums of $\chi$ or $\chi^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Summary:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So after all these work, we can find $\displaystyle \sum_{\stackrel{c+d = 1}{c,d \neq 0}} \#\{x:ax^3 = c\} \times \#\{y: -by^3 = d\} $ in terms of Gauss sums. There remains the case $c = 0$ or $d = 0$, which is easy to do directly. Sum them up, and you get the answer. The two books I mentioned both have examples of these types, so you may want to read them.&lt;/p&gt;&#xA;" OwnerUserId="325" LastEditorUserId="325" LastEditorDisplayName="" LastEditDate="2011-03-11T18:10:17.430" LastActivityDate="2011-03-11T18:10:17.430" />
  <row Id="26353" PostTypeId="2" ParentId="26350" CreationDate="2011-03-11T08:27:03.333" Score="-1" ViewCount="" Body="&lt;p&gt;Let $A=&#xD;&#xA;\left[ &#xD;&#xA;\begin{array}{cc}&#xD;&#xA;a &amp;amp; b \\&#xD;&#xA;0 &amp;amp; d &#xD;&#xA;\end{array} &#xD;&#xA;\right]$ and &#xA;$B=&#xD;&#xA;\left[ \begin{array}{cc}&#xD;&#xA;x &amp;amp; y \\&#xD;&#xA;0 &amp;amp; z &#xD;&#xA;\end{array} \right]$&#xA;Then $A*B[2,1]=d*z$ and not 0. It doesn't seem to be closed under multiplication on its own set.&#xA;{{{{I don't really know how to use Math thingy, language, here}}}}&lt;/p&gt;&#xA;" OwnerUserId="6388" LastEditorUserId="6460" LastEditorDisplayName="" LastEditDate="2011-03-11T08:54:55.033" LastActivityDate="2011-03-11T08:54:55.033" CommentCount="6" />
  <row Id="26354" PostTypeId="2" ParentId="26350" CreationDate="2011-03-11T08:27:50.657" Score="1" ViewCount="" Body="&lt;p&gt;What you have done looks reasonable except for one tiny point.  When you say &quot;because it is closed under scalar addition and scalar multiplication&quot;, you should delete the first (but not the second) &lt;em&gt;scalar&lt;/em&gt;, as you are trying to show that adding two elements give a third member of the subspace.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you should end with &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Hence, $H$ is determined to be a subspace of $M_{2 \times 2}$ because it is closed under  addition and scalar multiplication and contains the zero vector when $a=b=d=0$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="6460" LastActivityDate="2011-03-11T08:27:50.657" CommentCount="1" />
  <row Id="26355" PostTypeId="1" AcceptedAnswerId="26357" CreationDate="2011-03-11T08:28:13.293" Score="2" ViewCount="46" Body="&lt;p&gt;Given a function $f:z\mapsto f(z)$ for a discrete set of points in the real interval $z\in[a,b]$ and the knowledge that $f$ is analytical along the real axis and that its Fourier transform is real valued along $\mathbb R$ (and thus $f(-z)=f^*(z)$), is it possible to reconstruct $f(z)$ for all $z\in\mathbb C$? If not, what approximations are possible?&lt;/p&gt;&#xA;" OwnerUserId="163" LastEditorUserId="163" LastEditorDisplayName="" LastEditDate="2011-03-11T08:44:09.600" LastActivityDate="2011-03-11T09:17:49.383" Title="Extrapolation of an analytical function" Tags="&lt;functional-analysis&gt;&lt;analytic&gt;&lt;extrapolation&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="26356" PostTypeId="2" ParentId="26305" CreationDate="2011-03-11T09:07:17.847" Score="4" ViewCount="" Body="&lt;p&gt;You could start with topics such as &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Probability&lt;/li&gt;&#xA;&lt;li&gt;Statistics&lt;/li&gt;&#xA;&lt;li&gt;Analysis and calculus&lt;/li&gt;&#xA;&lt;li&gt;Stochastic processes and stochastic calculus&lt;/li&gt;&#xA;&lt;li&gt;Numerical methods&lt;/li&gt;&#xA;&lt;li&gt;Computational methods&lt;/li&gt;&#xA;&lt;li&gt;Modelling&lt;/li&gt;&#xA;&lt;li&gt;Optimisation&lt;/li&gt;&#xA;&lt;li&gt;Decision theory&lt;/li&gt;&#xA;&lt;li&gt;Data mining&lt;/li&gt;&#xA;&lt;li&gt;Time series&lt;/li&gt;&#xA;&lt;li&gt;Operational research&lt;/li&gt;&#xA;&lt;li&gt;Mathematical finance&lt;/li&gt;&#xA;&lt;li&gt;Mathematical economics&lt;/li&gt;&#xA;&lt;li&gt;Mathematical psychology&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;and the list could go on&lt;/p&gt;&#xA;" OwnerUserId="6460" LastActivityDate="2011-03-11T09:07:17.847" CommunityOwnedDate="2011-03-11T09:07:17.847" CommentCount="1" />
  <row Id="26357" PostTypeId="2" ParentId="26355" CreationDate="2011-03-11T09:12:43.210" Score="4" ViewCount="" Body="&lt;p&gt;No, it's not possible to reconstruct $f(z)$. You can augment your set of given function values by the corresponding points $-z$ to fulfill the symmetry condition, then pick an arbitrary sufficiently decaying even analytic function $g$, e.g. any Gaussian centered at $z=0$, and divide all the function values by the value of $g$ at that point. Then you can choose arbitrary polynomials that interpolate between the resulting values, and each such polynomial multiplied by $g$ will have the required properties. You have a lot of freedom of choice in $g$ and in the polynomials, so $f$ is very much not determined by these conditions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Concerning the question what approximations are possible: For the polynomials, there's a natural choice, the unique polynomial of lowest degree that interpolates the function values divided by $g$. But there's no natural choice for $g$; different functions $g$ will lead to different interpolations for $f$, all with the required properties.&lt;/p&gt;&#xA;" OwnerUserId="6622" LastEditorUserId="6622" LastEditorDisplayName="" LastEditDate="2011-03-11T09:17:49.383" LastActivityDate="2011-03-11T09:17:49.383" CommentCount="2" />
  <row Id="26358" PostTypeId="1" AcceptedAnswerId="26359" CreationDate="2011-03-11T09:15:16.607" Score="5" ViewCount="277" Body="&lt;p&gt;How would you be solve sin x = 1 - x, without drawing the graph and manually measuring the point of intersection?&lt;/p&gt;&#xA;" OwnerUserId="6520" LastActivityDate="2011-03-11T11:42:04.940" Title="Solve sin x = 1 - x" Tags="&lt;trigonometry&gt;" AnswerCount="5" FavoriteCount="1" />
  <row Id="26359" PostTypeId="2" ParentId="26358" CreationDate="2011-03-11T09:25:30.810" Score="6" ViewCount="" Body="&lt;p&gt;There's no analytical solution for this kind of transcendental equation. You can either solve it graphically (as you said) or numerically. There are various methods for solving it numerically, see &lt;a href=&quot;http://en.wikipedia.org/wiki/Root-finding_algorithm#Specific_algorithms&quot; rel=&quot;nofollow&quot;&gt;this Wikipedia article&lt;/a&gt; for an overview.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're not looking for a particularly efficient solution, the simplest thing to do in this case might be to write the equation as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x=1-\sin x$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and iterate the map&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x_{n+1}=1-\sin x_n$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;until you achieve satisfactory convergence. This is very inefficient, though; it only makes sense if you do it on a computer and don't mind if it takes a lot of operations; you'll need a lot of patience to get anywhere with it with a calculator. :-)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A far more efficient method that yields $x$ to double-precision machine accuracy within a couple of iterations would be &lt;a href=&quot;http://en.wikipedia.org/wiki/Newton%27s_method&quot; rel=&quot;nofollow&quot;&gt;Newton's method&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f(x) = x + \sin x - 1 \stackrel{!}{=} 0\;,$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x_{n+1}=x_n-\frac{f(x_n)}{f&amp;#39;(x_n)}=x_n - \frac{x+\sin x - 1}{1 + \cos x}\;.$$&lt;/p&gt;&#xA;" OwnerUserId="6622" LastEditorUserId="6622" LastEditorDisplayName="" LastEditDate="2011-03-11T09:40:10.797" LastActivityDate="2011-03-11T09:40:10.797" />
  <row Id="26360" PostTypeId="2" ParentId="26358" CreationDate="2011-03-11T09:34:54.820" Score="3" ViewCount="" Body="&lt;p&gt;For such cases, I would use &lt;a href=&quot;http://wims.unice.fr/wims/en_tool~analysis~function.en.html&quot; rel=&quot;nofollow&quot;&gt;Wims Function Calculator&lt;/a&gt;. Type the function $f(x)=\sin(x)-(1-x)$, and check the last box to find the root of the equation $\sin(x)-(1-x) = 0$. You can obtain the value of the root even up to $200$ digits. In your example, the root is approximately $0.51097342938856910952001397114508063204535889262375023988987953800752819385525987164073419474333797073844705378201625612601584766524335342369803019732463083576632594596747582089441042803074463769962092$.&lt;/p&gt;&#xA;" OwnerUserId="2810" LastActivityDate="2011-03-11T09:34:54.820" />
  <row Id="26361" PostTypeId="2" ParentId="26358" CreationDate="2011-03-11T09:35:33.213" Score="1" ViewCount="" Body="&lt;p&gt;It depends on what you mean by &quot;solving&quot;. If you want to show that a solution exists, use the intermediate value theorem. Expressing the solution by &quot;elementary functions&quot; (also called an analytical solution or closed-form solution, but note that these terms don't have universal definitions) is not possible. If you want a numerical solution, then use Newton's method.&lt;/p&gt;&#xA;" OwnerUserId="1609" LastActivityDate="2011-03-11T09:35:33.213" CommentCount="1" />
  <row Id="26362" PostTypeId="2" ParentId="26358" CreationDate="2011-03-11T09:41:54.330" Score="1" ViewCount="" Body="&lt;p&gt;As joriki mentions, simple-minded fixed-point iteration works if you're not too concerned about efficiency in obtaining the numerical solution (One usually has to be very lucky to obtain a closed form solution for a transcendental equation; this is not one of those lucky situations.). I'll only note that making use of the fact that $\sin\;x\approx x$, you can then say that there is a solution very near $x=.5$. Feed that to a fixed-point iteration, Newton, or Halley, and you can then obtain a good numerical approximation (which Shai has already given for you).&lt;/p&gt;&#xA;" OwnerUserId="8101" LastActivityDate="2011-03-11T09:41:54.330" />
  <row Id="26363" PostTypeId="1" AcceptedAnswerId="26369" CreationDate="2011-03-11T09:43:07.333" Score="3" ViewCount="120" Body="&lt;p&gt;It is perhaps a bit embarrassing that while doing higher-level math, I have forgot some more fundamental concepts. I would like to ask whether the square root of a number includes both the positive and the negative square roots.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that for an equation $x^2 = 9$, the solution is $x = \pm 3$. But if simply given $\sqrt{9}$, does one assume that to mean only the positive root? And when simply talking about the square root of a number in general, would one be referring to both roots or just the positive one, when neither is specified?&lt;/p&gt;&#xA;" OwnerUserId="3143" LastActivityDate="2011-03-11T16:18:03.073" Title="Square roots -- positive and negative" Tags="&lt;roots&gt;" AnswerCount="3" />
  <row Id="26364" PostTypeId="2" ParentId="26363" CreationDate="2011-03-11T09:49:59.737" Score="0" ViewCount="" Body="&lt;p&gt;I would say &lt;a href=&quot;http://math.stackexchange.com/questions/13801&quot;&gt;this&lt;/a&gt; can answer your question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See also &lt;a href=&quot;http://math.stackexchange.com/questions/13094/significance-of-displaystyle-sqrtnan/13095#13095&quot;&gt;this&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="8101" LastEditorUserId="1543" LastEditorDisplayName="" LastEditDate="2011-03-11T10:21:28.657" LastActivityDate="2011-03-11T10:21:28.657" CommunityOwnedDate="2011-03-11T10:20:51.340" />
  <row Id="26365" PostTypeId="1" CreationDate="2011-03-11T09:55:57.057" Score="6" ViewCount="170" Body="&lt;p&gt;What are the theorems/results/widely applicable results in graph theory that everyone should know about?&lt;/p&gt;&#xA;" OwnerUserId="705" LastActivityDate="2011-03-11T11:50:31.253" CommunityOwnedDate="2011-03-11T10:54:44.810" Title="What are the most important results in graph theory?" Tags="&lt;graph-theory&gt;&lt;big-list&gt;" AnswerCount="3" CommentCount="7" FavoriteCount="1" />
  <row Id="26366" PostTypeId="1" CreationDate="2011-03-11T10:32:58.693" Score="0" ViewCount="36" Body="&lt;p&gt;Hi all,&lt;br/&gt;&#xA;Do any of you know an online resource where I can find exercises on multivariate statistics?&lt;br/&gt; ( With topics such as Multiple Linear Regression, Principal Component Analysis, Factor Analysis, Cluster Analysis and Discriminant Analysis)&lt;/p&gt;&#xA;" OwnerUserId="8067" LastActivityDate="2011-03-11T10:32:58.693" Title="Online exercises on multivariate statistics?" Tags="&lt;statistics&gt;&lt;online-resources&gt;" />
  <row Id="26367" PostTypeId="2" ParentId="26306" CreationDate="2011-03-11T10:41:47.237" Score="6" ViewCount="" Body="&lt;p&gt;You need to recall the definition of differentiability for functions on $\mathbb{R}^n$. The standard definition is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Definition&lt;/strong&gt; A function $f$ defined on $\mathbb{R}^n$ taking values in $\mathbb{R}$ (or $\mathbb{C}$) is said to be differentiable at $p\in\mathbb{R}^n$ if there exists a &lt;em&gt;linear map&lt;/em&gt; $J_p:\mathbb{R}^n\to\mathbb{R}$ (or $\mathbb{C}$) such that for all vectors $v$ in $\mathbb{R}^n$ you have that&#xA;$$ \lim_{s\to 0^+} \frac{ f(p + sv) - f(p) - J_p(sv)}{s} = 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The key thing to note is that the mapping induced by the &quot;directional derivatives&quot; that sends a vector $v$ to the derivative $D_v f$ is required to be linear map! (This reflects our intuitive understanding that a derivative gives a linear approximation to the function near a point.) So in your situation, if you can prove that the map $v \mapsto D_v f$ is linear, that is, given any $v,w\in\mathbb{R}^n$ and $a,b\in\mathbb{R}$ you have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ av + bw \mapsto a D_v f + bD_w f $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;then you have that the function is differentiable. The example linked to by Akhil in his comment shows an example where this linearity fails, and hence the function, depsite having all directional derivatives, is not differentiable. &lt;/p&gt;&#xA;" OwnerUserId="1543" LastActivityDate="2011-03-11T10:41:47.237" CommentCount="1" />
  <row Id="26368" PostTypeId="1" AcceptedAnswerId="26388" CreationDate="2011-03-11T10:43:22.667" Score="6" ViewCount="193" Body="&lt;p&gt;For any module $P$, we define $\mathrm{tr}(P)=\sum \mathrm{im}(f)$, where $f$ ranges over all elements of $\mathrm{Hom}(P,R)$， and call it trace.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why does it have such a name? Does it have any relation to the trace of matrix？&lt;/p&gt;&#xA;" OwnerUserId="4706" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-11T17:55:59.010" LastActivityDate="2011-03-11T17:55:59.010" Title="Why do we call it trace?" Tags="&lt;algebra&gt;&lt;modules&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="0" />
  <row Id="26369" PostTypeId="2" ParentId="26363" CreationDate="2011-03-11T10:46:35.473" Score="7" ViewCount="" Body="&lt;p&gt;If you want your square-root function $\sqrt x$ to be a &lt;em&gt;function&lt;/em&gt;, then it needs to have the properties of a function, in particular that for each element of the domain the function gives a single value from the codomain.  If you take a function to be a set of ordered pairs, then each of the initial values of the pairs must appear exactly once.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So to be a function, square-root needs to be single valued; the multi-valued version is really a &lt;em&gt;relation&lt;/em&gt;, at which point you might get into issues of principal values.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For convenience, the square root of non-negative real numbers is usually taken to be the non-negative real value, but there is nothing other than practicality to stop you from taking some other pattern.  Such arbitrary choices can raise significant issues when considering, for example, cube-root functions defined on the real and complex numbers. &lt;/p&gt;&#xA;" OwnerUserId="6460" LastEditorUserId="6460" LastEditorDisplayName="" LastEditDate="2011-03-11T15:48:41.310" LastActivityDate="2011-03-11T15:48:41.310" />
  <row Id="26370" PostTypeId="2" ParentId="26052" CreationDate="2011-03-11T10:51:25.570" Score="6" ViewCount="" Body="&lt;p&gt;That such a decomposition doesn't exist can be shown by looking at the interior solid angles at the vertices.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The cube has eight interior solid angles of $\pi/2$ each.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each of the six tetrahedra has one interior solid angle of $\pi/2$ and three that can be computed as shown &lt;a href=&quot;http://en.wikipedia.org/wiki/Solid_angle#Tetrahedron&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Omega=\phi_{ab}+\phi_{bc}+\phi_{ac}-\pi=\pi/2 + \arccos\frac{1}{\sqrt{3}}+ \arccos\frac{1}{\sqrt{3}}-\pi=2\arccos\frac{1}{\sqrt{3}}-\pi/2\;.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, the sum of the interior solid angles of the cube is $4\pi$, and the sum of the interior solid angles of the tetrahedra is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$6\left(\pi/2+3\left(2\arccos\frac{1}{\sqrt{3}}-\pi/2\right)\right)=6\left(6\arccos\frac{1}{\sqrt{3}}-\pi\right)\;.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is not equal to $4\pi$, since $\arccos\frac{1}{\sqrt{3}}$ is not a rational multiple of $\pi$. (Here's &lt;a href=&quot;http://www.uni-math.gwdg.de/jahnel/Preprints/cos.pdf&quot; rel=&quot;nofollow&quot;&gt;an elementary proof&lt;/a&gt; of that fact.) In fact it is almost $5\pi$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[Edit: I just realized in the shower that all the following analysis of potential inner vertices is actually not necessary. No matter what happens at the inner vertices, we have to fill the eight vertices of the cube with vertices of the tetrahedra, and the six vertices with interior solid angle $\pi/2$ are not enough to do it, so we have to use at least one of the others, but then we can't make the sum come out to $4\pi$, a rational multiple of $\pi$.]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This establishes that you can't fill the cube with the six tetrahedra by letting all their vertices coincide with the cube's vertices. It seems geometrically obvious that you can't do it with any of the vertices inside the cube, either, but this, too, can be proved rigorously using the solid angles. If there were a vertex inside the cube, the entire solid angle of $4\pi$ around that inner vertex would have to be filled. The candidates for filling it are a face of a tetrahedron, which subtends a solid angle of $2\pi$, an edge of a tetrahedron, which subtends a solid angle of twice the dihedral angle of the intersecting planes, i.e. either $\pi$ or $2\arccos\frac{1}{\sqrt{3}}$, or an interior solid angle of a vertex of a tetrahedron. Thus, adding up all the solid angles at the eight outer vertices and at $v$ inner vertices, the following equation would have to have solutions with non-negative integer values of $j$, $k$, $l$, $m$, $n$ and $v$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$6\left(6\arccos\frac{1}{\sqrt{3}}-\pi\right)+j(2\pi)+k(\pi)+l\left(2\arccos\frac{1}{\sqrt{3}}\right)+m(\pi/2)+n\left(\arccos\frac{1}{\sqrt{3}}-\pi/2\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$=4\pi+v(4\pi)\;.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is impossible, since the coefficient in front of $\arccos\frac{1}{\sqrt{3}}$ is non-zero, and thus the equation would imply that this is a rational multiple of $\pi$, which it isn't (see above).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S.: I just realized I forgot to mention an essential part of the proof:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\cos 2\phi = 2 \cos^2 \phi - 1\;,$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\arccos x = \frac{1}{2}\arccos(2x^2-1)\;,$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\arccos \frac{1}{\sqrt{3}}=\frac{1}{2}\arccos(-\frac{1}{3})\;,$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so $\arccos \frac{1}{\sqrt{3}}$ is rational iff $\arccos(-\frac{1}{3})$ is rational; then we can apply the theorem I linked to above.&lt;/p&gt;&#xA;" OwnerUserId="6622" LastEditorUserId="6622" LastEditorDisplayName="" LastEditDate="2011-03-11T13:44:27.203" LastActivityDate="2011-03-11T13:44:27.203" CommentCount="4" />
  <row Id="26371" PostTypeId="1" AcceptedAnswerId="26422" CreationDate="2011-03-11T10:52:09.230" Score="4" ViewCount="274" Body="&lt;p&gt;The definition of one-to-one was pretty straight forward. If the vectors are lin.indep the transformation would be one-to-one. But could it also be onto? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The definition of onto was a little more abstract. Would a zero-row in reduced echelon form have any effect on this? I just assumed that because it has a couple of free variables it would be onto, but that zero-row set me off a bit. Say if the matrix is 4 by 5. With two free variables. And the zero-row in echelon form. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Sorry for not posting the given matrix, but that is to specific. And I don't want to get a ban from uni for asking online. The task is determine the onto/one-to-one of to matrices)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll check back after class and update the question if more information is desirable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Update&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The definitions in the book is this; &#xA;Onto: &#xA;$T: \mathbb R^n \to \mathbb R^m $ is said to be onto $\mathbb R^m  $ if each b in &#xA;$\mathbb R^m  $ is the image of at least one x in $\mathbb R^n  $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One-to-one:&#xA;$T: \mathbb R^n \to \mathbb R^m $ is said to be one-to-one $\mathbb R^m  $ if each b in &#xA;$\mathbb R^m  $ is the image of at most one x in $\mathbb R^n  $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And then, there is another theorem that states that a linear transformation is one-to-one iff the equation T(x) = 0 has only the trivial solution. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That doesn't say anything about onto. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then there is this bit that confused be about onto: &#xA;Let $T: \mathbb R^n \to \mathbb R^m $ be a linear transformation and let A be the standard matrix for T. Then: &#xA;T maps $T: \mathbb R^n$ onto $\mathbb R^m $ iff the columns of A span $\mathbb R^m $.&lt;/p&gt;&#xA;" OwnerUserId="2230" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-11T16:19:38.987" LastActivityDate="2011-03-11T19:32:33.143" Title="Is a linear tranformation onto or one-to-one?" Tags="&lt;homework&gt;&lt;linear-algebra&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" />
  <row Id="26372" PostTypeId="2" ParentId="26365" CreationDate="2011-03-11T10:58:45.133" Score="4" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Hall%27s_marriage_theorem&quot; rel=&quot;nofollow&quot;&gt;Hall's marriage theorem&lt;/a&gt; is widely applicable. Remarkably it happens to be equivalent to other theorems in graph theory and combinatorics which are also widely applicable:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/K%C3%B6nig%27s_theorem_(graph_theory)&quot; rel=&quot;nofollow&quot;&gt;Konig's theorem&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Menger%27s_theorem&quot; rel=&quot;nofollow&quot;&gt;Menger's theorem&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Max-flow_min-cut_theorem&quot; rel=&quot;nofollow&quot;&gt;Max-flow min-cut&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Doubly_stochastic_matrix&quot; rel=&quot;nofollow&quot;&gt;Birkhoff-von Neumann theorem&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Dilworth%27s_theorem&quot; rel=&quot;nofollow&quot;&gt;Dilworth's theorem&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="232" LastActivityDate="2011-03-11T10:58:45.133" CommunityOwnedDate="2011-03-11T10:58:45.133" />
  <row Id="26373" PostTypeId="1" AcceptedAnswerId="26374" CreationDate="2011-03-11T11:01:17.897" Score="1" ViewCount="150" Body="&lt;p&gt;Hi everybody!!!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am trying to calculate the center of gravity of a polygon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My problem is that I need to be able to calculate the center of gravity for both regular and irregular polygons and even self intersecting polygons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is that possible?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've also read that: &lt;a href=&quot;http://paulbourke.net/geometry/polyarea/&quot; rel=&quot;nofollow&quot;&gt;http://paulbourke.net/geometry/polyarea/&lt;/a&gt; But this is restricted to non self intersecting polygons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I do this? Can you point me to the right direction?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sub-Question: Will it matter if the nodes are not in order? if for example you have a square shape and you name the top right point (X1Y1) and then the bottom right point (X3Y3)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words if your shape is like 4-1-2-3 (naming the nodes from left to right top to bottom)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: Might be a stupid question but I'm not a maths student or anything!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="8104" LastActivityDate="2011-03-21T19:45:31.013" Title="Center of gravity of a self intersecting irregular polygon" Tags="&lt;geometry&gt;&lt;algorithms&gt;" AnswerCount="3" CommentCount="3" />
  <row Id="26374" PostTypeId="2" ParentId="26373" CreationDate="2011-03-11T11:28:21.580" Score="1" ViewCount="" Body="&lt;p&gt;I think your best bet will be to convert the self-intersecting polygon into a set of non-self-intersecting polygons and apply the algorithm that you linked to to each of them. I don't think it's possible to solve your problem without finding the intersections, and if you have to find the intersections anyway, the additional effort of using them as new vertices in a rearranged vertex list is small compared to the effort of finding them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To answer your subquestion: Yes, the order of the nodes does matter, especially if the polygon is allowed to be self-intersecting since in that case the order is an essential part of the specification of the polygon and different orders specify different polygons -- for instance, the &quot;square&quot; with the ordering you describe would be the polygon on the right-hand side of the two examples of self-intersecting polygons that the page you linked to gives (rotated by $\pi/2$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S.: I just realized that different orders can also specify different non-self-intersecting  (but not convex) polygons, so the only case where you could specify a polygon by its vertices alone is if you know it's convex. But even then you have to use the vertices in the right order in the algorithm you linked to.&lt;/p&gt;&#xA;" OwnerUserId="6622" LastActivityDate="2011-03-11T11:28:21.580" CommentCount="1" />
  <row Id="26375" PostTypeId="2" ParentId="26365" CreationDate="2011-03-11T11:32:10.083" Score="2" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Euler_characteristic&quot; rel=&quot;nofollow&quot;&gt;Euler's formula&lt;/a&gt; $V - E + F = 1$ for planar graphs is extremely important; in some sense it motivated much of modern topology. (An excellent introduction to this thesis is Richeson's book &lt;a href=&quot;http://www.eulersgem.com/&quot; rel=&quot;nofollow&quot;&gt;Euler's Gem&lt;/a&gt;.) It also leads to a reasonably short proof of the classification of the Platonic solids, so even before generalization, it's quite important. The generalization to graphs on surfaces is used to prove the easy direction of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Heawood_conjecture&quot; rel=&quot;nofollow&quot;&gt;Heawood conjecture&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course nowadays it is recognized that Euler's formula is really a statement about &lt;a href=&quot;http://en.wikipedia.org/wiki/Triangulation_(topology)&quot; rel=&quot;nofollow&quot;&gt;triangulations&lt;/a&gt; of the sphere, and the generalization to triangulations of arbitrary manifolds and other characterizations of Euler characteristic is fundamental. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Geometry Junkyard has a nice list of &lt;a href=&quot;http://www.ics.uci.edu/~eppstein/junkyard/euler/&quot; rel=&quot;nofollow&quot;&gt;nineteen proofs of Euler's formula&lt;/a&gt;. One of them uses another basic and useful fact about graphs, which is that they always have &lt;a href=&quot;http://en.wikipedia.org/wiki/Spanning_tree&quot; rel=&quot;nofollow&quot;&gt;spanning trees&lt;/a&gt;. This is used, for example, in topology to prove that the geometric realization of a graph is homotopy equivalent to a wedge of circles, which shows that subgroups of free groups are free. &lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2011-03-11T11:32:10.083" CommunityOwnedDate="2011-03-11T11:32:10.083" CommentCount="1" />
  <row Id="26376" PostTypeId="2" ParentId="26358" CreationDate="2011-03-11T11:37:04.117" Score="1" ViewCount="" Body="&lt;p&gt;Below is a calculus-type approach. Remark that to first order $\sin x \simeq x,$ so your result will be somewhere around $x = \frac{1}{2}.$ That suggests a Taylor expansion of sine around $\frac{\pi}{6} \simeq \frac{1}{2}.$ You can introduce $\epsilon = x - \frac{\pi}{6},$ such that $$\sin(x) = \frac{1}{2} + \frac{\sqrt{3}}{2}\epsilon - \frac{1}{4}\epsilon^2 + \mathcal{O}(\epsilon^3).$$&#xA;Identifying $1 - x = 1 - (\frac{\pi}{6} + \epsilon),$ you can equate both sides of your equation and solve a linear, quadratic, ..., polynomial equation in $\epsilon$, which will however involve knowing a reasonable approximation to $\pi$ and/or $\frac{\pi}{6}.$&lt;/p&gt;&#xA;" OwnerUserId="6004" LastEditorUserId="6004" LastEditorDisplayName="" LastEditDate="2011-03-11T11:42:04.940" LastActivityDate="2011-03-11T11:42:04.940" />
  <row Id="26377" PostTypeId="2" ParentId="26365" CreationDate="2011-03-11T11:40:58.140" Score="2" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Kuratowski%27s_theorem&quot; rel=&quot;nofollow&quot;&gt;Kuratowski's and Wagner's theorems&lt;/a&gt; which give necessary and sufficient conditions for a finite graph to be planar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I should add that it's certainly possible to argue that these are actually theorems of topology rather than graph theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a simple proof of the non-planarity of $K_{3,3}$ and $K_5$ one can consult Munkres's &lt;em&gt;Topology&lt;/em&gt; 2nd ed. §64 and at the end of the section he notes &quot;It is a remarkable theorem, due to Kuratowski, that the converse is also true! The proof is not easy.&quot;&lt;/p&gt;&#xA;" OwnerUserId="1269" LastEditorUserId="1269" LastEditorDisplayName="" LastEditDate="2011-03-11T11:50:31.253" LastActivityDate="2011-03-11T11:50:31.253" CommunityOwnedDate="2011-03-11T11:40:58.140" CommentCount="1" />
  <row Id="26378" PostTypeId="2" ParentId="26371" CreationDate="2011-03-11T11:43:43.093" Score="1" ViewCount="" Body="&lt;p&gt;A function is onto if its codomain is equal to its image, that is, $f: X \to Y$ is onto if $f(X) = Y$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming you are working with real matrices, a matrix $A$ is usually seen as mapping vectors from $\mathbb R^n$ to $\mathbb R^m$. You want to show that for all $\vec y \in \mathbb R^m$ there exists a $\vec x \in \mathbb R^n$ such that $A \vec x = \vec y$. In some cases this is not possible depending on $m$ and $n$. How does $m$ and $n$ relate to the size of the matrix?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One way to check that a matrix is onto is to check if you can find vectors $\vec v_1, \vec v_2, \dots, \vec v_m$ such that $A\vec v_i = \vec e_i$ for $i = 1, 2, \dots, m$, where $\vec e_1, \vec e_2, \dots, \vec e_m$ is the standard basis in $\mathbb R^m$ (why?). Can you use the echelon form for this? Do you actually need to get the standard basis or is some other set of vectors ok?&lt;/p&gt;&#xA;" OwnerUserId="5966" LastEditorUserId="5966" LastEditorDisplayName="" LastEditDate="2011-03-11T12:01:06.077" LastActivityDate="2011-03-11T12:01:06.077" CommentCount="3" />
  <row Id="26379" PostTypeId="1" CreationDate="2011-03-11T11:44:19.133" Score="3" ViewCount="146" Body="&lt;p&gt;How should you break $\displaystyle \frac{x^n}{(1-x)(1-2x)(1-3x)\ldots(1-nx)}$ into partial fractions?&lt;/p&gt;&#xA;" OwnerUserId="7340" LastActivityDate="2011-03-28T19:25:33.497" Title="Partial fraction of $\displaystyle \frac{x^n}{(1-x)(1-2x)(1-3x)\ldots(1-nx)}$" Tags="&lt;algebra&gt;&lt;polynomials&gt;" AnswerCount="3" />
  <row Id="26380" PostTypeId="2" ParentId="26379" CreationDate="2011-03-11T11:56:18.887" Score="4" ViewCount="" Body="&lt;p&gt;First you have to take out a multiple of the denominator to make the degree of the numerator smaller than that of the denominator:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{x^n}{(1-x)(1-2x)(1-3x)\ldots(1-nx)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$= \frac{(-1)^n}{n!} + \frac{x^n-\frac{(-1)^n}{n!}(1-x)(1-2x)(1-3x)\ldots(1-nx)}{(1-x)(1-2x)(1-3x)\ldots(1-nx)}\;.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now you can deal with the remaining fraction. You already have the denominator in factorized form, so you know the partial fraction expansion will have the form&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{a_1}{1-x}+\frac{a_2}{1-2x}+\frac{a_3}{1-3x}+\ldots+\frac{a_n}{1-nx}\;.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To get the coefficients $a_i$, you can use the fact that the residues of the poles must be equal to the corresponding residues in the fraction.&lt;/p&gt;&#xA;" OwnerUserId="6622" LastEditorUserId="6622" LastEditorDisplayName="" LastEditDate="2011-03-11T12:06:24.790" LastActivityDate="2011-03-11T12:06:24.790" CommentCount="3" />
  <row Id="26381" PostTypeId="2" ParentId="26379" CreationDate="2011-03-11T12:08:45.370" Score="4" ViewCount="" Body="&lt;p&gt;To expand on joriki's comment, I claim that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\lim_{x \to \frac{1}{k}} \frac{x^n (1 - kx)}{(1 - x)...(1 - nx)} = a_k.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is not hard to see by multiplying both sides by $1 - kx$. On the other hand, this limit is quite easy to compute since $1 - kx$ cancels, giving&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$a_k = \frac{1}{k(k-1)...(k-n)} = \frac{(-1)^{n-k}}{k! (n-k)!}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that this technique works to give you the partial fraction decomposition of any rational function, at least when the denominator has simple roots: instead of canceling a factor, you can just use l'Hopital's rule.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note also that the function you're looking at happens to be the ordinary generating function of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Stirling_numbers_of_the_second_kind&quot; rel=&quot;nofollow&quot;&gt;Stirling numbers of the second kind&lt;/a&gt;, and here you can cheat because you can also find their &lt;em&gt;exponential&lt;/em&gt; generating function.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2011-03-11T12:08:45.370" />
  <row Id="26382" PostTypeId="2" ParentId="26328" CreationDate="2011-03-11T12:13:50.747" Score="1" ViewCount="" Body="&lt;p&gt;You actually picked out the right class of counterexamples, since for your average abelian group (Z or R), you won't find anything interesting. $H_0$ always equals $A/nA$, but $H_1$ 'defines' something that is called torsion, i.e. the vanishing of finite-order elements in a group. In some literature (for example, Hatcher's AT), you'll see the definition&#xA;$$n-{\rm Torsion}(G) = {\rm Ker}(g \mapsto n g | g \in G),$$ which coincides with your $H_1$. I hope that'll give you something to look up in your favorite algebra reference.&lt;/p&gt;&#xA;" OwnerUserId="6004" LastActivityDate="2011-03-11T12:13:50.747" CommentCount="1" />
  <row Id="26383" PostTypeId="2" ParentId="26337" CreationDate="2011-03-11T12:29:59.840" Score="0" ViewCount="" Body="&lt;p&gt;The only other thing to note, that Hermite $H_i$ and Laguerre $L_i$ polynomials are orthogonal.  In a sense that,&#xA;$\int H_i(x) H_j(x) w(x) dx = 0$ when $j \neq i$, where $w(x) = e^{x^2}$,&#xA;$\int L_i(x) L_j(x) v(x) dx = 0$ when $j \neq i$, where $v(x) = e^{-x}.$&lt;/p&gt;&#xA;" OwnerUserId="8109" LastActivityDate="2011-03-11T12:29:59.840" />
  <row Id="26384" PostTypeId="1" AcceptedAnswerId="26385" CreationDate="2011-03-11T12:32:09.343" Score="1" ViewCount="83" Body="&lt;p&gt;Hi! I'm trying to learn number theory on my own, and here's a proof I'm not quite sure I got right. It feels too simple(?), I'm thinking maybe I'm missing something. So the question is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Prove that if $n$ is a square, then each exponent in its prime-power decomposition is even.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My proof:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $n=m^2$, with $m$ having prime factors $p_i$ with exponents $e_i$ so that  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$m= p^{e_1}_1 p^{e_2}_2 \ldots p^{e_n}_n.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When squared, this gives&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$m^2 = (p^{e_1}_1 p^{e_2}_2 \ldots p^{e_n}_n)^2 = p^{2e_1}_1 p^{2e_2}_2 \ldots p^{2e_n}_n,$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where all the exponents are even.&lt;/p&gt;&#xA;" OwnerUserId="8110" LastActivityDate="2011-03-11T12:48:21.090" Title="Prime-power decomposition of a square" Tags="&lt;elementary-number-theory&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" />
  <row Id="26385" PostTypeId="2" ParentId="26384" CreationDate="2011-03-11T12:48:21.090" Score="2" ViewCount="" Body="&lt;p&gt;That is fine.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can even use it in the other direction to prove that if each exponent in the prime-power decomposition of $n$ is even, then $n$ is a square by saying&lt;br&gt;&#xA;$$n = p^{2e_1}_1 p^{2e_2}_2 \ldots p^{2e_n}_n = (p^{e_1}_1 p^{e_2}_2 \ldots p^{e_n}_n)^2$$ so $n = m^2$ where $m= p^{e_1}_1 p^{e_2}_2 \ldots p^{e_n}_n.$&lt;/p&gt;&#xA;" OwnerUserId="6460" LastActivityDate="2011-03-11T12:48:21.090" />
  <row Id="26386" PostTypeId="1" AcceptedAnswerId="26412" CreationDate="2011-03-11T12:59:56.523" Score="3" ViewCount="170" Body="&lt;p&gt;Hello, I have some trouble understanding the record value for a sequence of i.i.d. random variables of geometric distribution. Following quotation is from &lt;a href=&quot;http://books.google.com/books?id=JchiadWLnykC&amp;amp;pg=PA449&amp;amp;lpg=PA449&amp;amp;dq=geometric+distribution+%22record+value%22&amp;amp;source=bl&amp;amp;ots=pVHK8azotw&amp;amp;sig=KizbfcNqzC8Ri9q2sDVNa0om9CM&amp;amp;hl=en&amp;amp;ei=6Ip5TevPDIi30QG-t5DSAw&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=5&amp;amp;ved=0CDgQ6AEwBA#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;Univariate discrete distributions By Norman Lloyd Johnson, Adrienne W. Kemp, Samuel Kotz&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The lack-of-memory property of the&#xA;  geometric distribution gives it a role&#xA;  comparable to that of the exponential&#xA;  distribution. There are a number of&#xA;  characterizations of the geometric&#xA;  distribution based on record values.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;For the  record time $T_n$, &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If $X_j$ is observed at time $j$ ,&#xA;  then the record time sequence $\{T_n ,n \geq 0\}$ is defined as&#xA;  $T_0 = 1$&#xA;  with probability $1$ and $T_n = \min\{j : X_j &amp;gt; X_{t_{n−1}} \}$ for&#xA;  $n\geq 1$. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Is there a typo in $T_n = \min \{j : X_j &amp;gt;&#xD;&#xA;    X_{t_{n−1}} \}$? Should it be instead $T_n&#xD;&#xA;    = \min\{j : X_j &amp;gt; X_{T_{n−1}} \}$?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;For the  record value $R_n$, &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The record value sequence&#xA;  $\{R_n \}$ is defined as $R_n =X_{T_n} , n = 0, 1, 2, ...$. Suppose&#xA;  that the $X_j$ ’s are iid geometric&#xA;  variables with pmf $$p_x = p(1 − p)^{x−1}, x = 1, 2, ...$$ Then $R_n&#xD;&#xA;    =X_{T_n} = \sum_{j=0}^{n} X_j$ is&#xA;  distributed as the sum of $n + 1$ iid&#xA;  geometric variables.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;why does the second equality in&#xA;&quot;$R_n =X_{T_n} = \sum_{j=0}^{n} X_j$&#xA;&quot; hold?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;For the process of the record values&#xA;$\{ R_n, n \in \mathbb{N}\}$, &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Each of the following properties&#xA;  characterizes the geometric&#xA;  distribution: &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;(i) Independence: The rv’s $R_0 , R_1  - R_0 , ... , R_{n+1} - R_n ,...$ are independent. &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;(ii) Same Distribution: $R_{n+1} - R_n$ has the same distribution as&#xA;  $R_0$ . &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;(iii) Constant Regression: $E[R_{n+1}- R_n |R_n ]$ is constant.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;How to show that the three&#xA;properties hold? Are they derived&#xA;from the memoryless property of&#xA;geometric distribution?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What else can we say about the&#xA;process based on the memoryless&#xA;property of geometric distribution?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Thanks for your advice!&lt;/p&gt;&#xA;" OwnerUserId="3616" LastEditorUserId="5001" LastEditorDisplayName="" LastEditDate="2011-03-11T14:18:26.490" LastActivityDate="2011-03-11T21:39:21.957" Title="Questions about geometric distribution" Tags="&lt;probability&gt;&lt;statistics&gt;&lt;stochastic-processes&gt;&lt;geometric-distribution&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" />
  <row Id="26387" PostTypeId="1" CreationDate="2011-03-11T13:01:02.157" Score="0" ViewCount="114" Body="&lt;p&gt;How do I express the matlab function abs(M), on a matrix M, in mathematical terms?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I thought about norms or just |M|, but these return scalars, not another matrix of the same size as M.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sorry for the rough explanation, english isn't my primary language.&lt;/p&gt;&#xA;" OwnerUserId="8111" LastActivityDate="2011-03-11T13:22:52.447" Title="Absolute value of all values in a matrix" Tags="&lt;matrices&gt;&lt;absolute-value&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="26388" PostTypeId="2" ParentId="26368" CreationDate="2011-03-11T13:19:16.740" Score="6" ViewCount="" Body="&lt;p&gt;I have always viewed the term &lt;em&gt;trace&lt;/em&gt; as used to name $$\displaystyle \operatorname{tr}(P)=\sum_{f:P\to R}f(P)$$ as a reference to the fact that that ideal is the part of $R$ which you can reach from $P$... whatever that may mean :) One can check, for example, that $P$ is a generator of the category of modules iff $\operatorname{tr}(P)=R$, and that makes sense (to me!)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A more serious connection is the following. Suppose $V$ is a finite dimensional vector space over a field $k$, and let $\hom_k(V,k)$ be its dual space. Then there is a natural isomorphism $$\phi:V\otimes\hom_k(V,k)\cong\hom(V,V).$$ On the other hand, we have the usual trace map $\operatorname{tr}:\hom(V,V)\to k$, so we can consider the composition $$\operatorname{Tr}=\operatorname{tr}\circ\phi:V\otimes\hom_k(V,k)\to k,$$ which is also sometimes called a trace map. If you now replace $k$ by ring, $V$ by a left $R$-module, then what you wrote $\operatorname{tr}(P)$ is the image of my $\operatorname{Tr}$: it follows that the trace ideal is the image of the trace map.&lt;/p&gt;&#xA;" OwnerUserId="274" LastEditorUserId="274" LastEditorDisplayName="" LastEditDate="2011-03-11T13:24:58.967" LastActivityDate="2011-03-11T13:24:58.967" CommentCount="1" />
  <row Id="26389" PostTypeId="1" AcceptedAnswerId="26395" CreationDate="2011-03-11T13:20:06.757" Score="7" ViewCount="175" Body="&lt;p&gt;In most Descriptive Set Theory books, the rationale for working with the Baire space ($\mathbb{N}^{\mathbb{N}}$) as opposed to the real line ($\mathbb{R}$) is that the connectedness of the latter causes 'technical difficulties'. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is, what are these technical difficulties, and why does Descriptive Set Theory (normally?) stick to zero-dimensional Polish spaces?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance.&lt;/p&gt;&#xA;" OwnerUserId="7835" LastEditorUserId="622" LastEditorDisplayName="" LastEditDate="2011-05-06T17:11:48.093" LastActivityDate="2011-05-06T17:11:48.093" Title="Why is the real line not used in Descriptive Set Theory?" Tags="&lt;topology&gt;&lt;set-theory&gt;&lt;descriptive-set-theory&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="26390" PostTypeId="2" ParentId="26387" CreationDate="2011-03-11T13:20:51.237" Score="1" ViewCount="" Body="&lt;p&gt;I think you can also denote the matrix M by giving its generic (i, j)-entry with parenthesis around:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;M = (m&lt;sub&gt;ij&lt;/sub&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So Matlab's &quot;abs(M)&quot; would simply be: (|m&lt;sub&gt;ij&lt;/sub&gt;|).&lt;/p&gt;&#xA;" OwnerUserId="8112" LastActivityDate="2011-03-11T13:20:51.237" />
  <row Id="26391" PostTypeId="2" ParentId="26387" CreationDate="2011-03-11T13:22:52.447" Score="1" ViewCount="" Body="&lt;p&gt;'help abs' returns ABS(X) is the absolute value of the elements of X.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $ M = [a _{ij}] $ where $a _{ij} \in R$, $i=1...m$ , $j=1..n$ Then $ABS(M) =  [ABS(a _{ij})]$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: it's not a must for $a _{ij} \in R$ and can be also complex. &lt;/p&gt;&#xA;" OwnerUserId="5312" LastActivityDate="2011-03-11T13:22:52.447" />
  <row Id="26392" PostTypeId="1" AcceptedAnswerId="26401" CreationDate="2011-03-11T13:28:47.890" Score="0" ViewCount="30" Body="&lt;p&gt;Let $m &amp;gt; 0$ and $ a:[0,1] \rightarrow \mathbb R$ be a function with $a(\epsilon) \rightarrow 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then $ \epsilon^m a(\epsilon) \rightarrow 0 $ for $ \epsilon \rightarrow 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But is $ \epsilon^m a(\epsilon) \in o(\epsilon^m) $, with O-Notation used? Put into other words, if we multiply a zero sequence which converges by order $m$ with another zero sequence, is the decay strictly faster than by order $m$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You may assume $a$ to be uniformly continuous if that helps.&lt;/p&gt;&#xA;" OwnerUserId="3557" LastActivityDate="2011-03-11T14:33:03.153" Title="Accelerated Order of Convergence" Tags="&lt;sequences-and-series&gt;&lt;convergence&gt;&lt;classical-analysis&gt;" AnswerCount="1" />
  <row Id="26393" PostTypeId="1" AcceptedAnswerId="26398" CreationDate="2011-03-11T13:42:23.043" Score="1" ViewCount="85" Body="&lt;p&gt;Hello,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a graph with a cycle ($v_1,\ldots,v_k, v_1=v_k$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Claim: If there is a cycle with 2 edges of the same weight, and they are the heaviest edges in this cycle, then there is more than one Minimum spanning tree.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am trying to think about contradiction, but I can't find one. All my examples create more than one Minimum Spanning Tree.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks a lot.&lt;/p&gt;&#xA;" OwnerUserId="8069" LastEditorUserId="3317" LastEditorDisplayName="" LastEditDate="2011-03-11T14:14:41.343" LastActivityDate="2011-03-11T14:36:44.583" Title="Graph - Minimum spanning tree" Tags="&lt;homework&gt;&lt;graph&gt;" AnswerCount="2" />
  <row Id="26394" PostTypeId="2" ParentId="26339" CreationDate="2011-03-11T13:46:24.817" Score="1" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\ $ Find the roots $\rm\ x\equiv a\ (mod\ 5)\:.\:$ Substitute $\rm\ x = a + 5\ y\ $ into the equation yielding $\rm\ 10\ (1 - a^2)\ y + a^3+ 2\ a+ 2\equiv 0\ (mod\ 25)\:.\: $ Now check which roots $\rm\:a\:$ yield solutions for $\rm\:y\:.\:$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;REMARK&lt;/strong&gt; $\ $ This is a special case of a general method due to Hensel for lifting solutions  of congruences $\rm\:(mod\ p)\:$ to $\rm\:(mod\ p^n)\:$. However you don't need to know anything about this method to solve the problem using the above hint.&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2011-03-11T13:46:24.817" />
  <row Id="26395" PostTypeId="2" ParentId="26389" CreationDate="2011-03-11T13:48:22.317" Score="13" ViewCount="" Body="&lt;p&gt;There are several useful properties of $\mathbb{N}^\mathbb{N}$:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\mathbb{N}^\mathbb{N}$ is homeomorphic to $\mathbb{N}^\mathbb{N}\times\mathbb{N}^\mathbb{N}$.  So we can view each element of $\mathbb{N}^\mathbb{N}$ as a code for a pair of elements, and the decoding maps are continuous. This is not the case for $\mathbb{R}$.  &lt;/li&gt;&#xA;&lt;li&gt;$\mathbb{N}^\mathbb{N}$ is not connected. Because $\mathbb{R}$ is connected, there are no nonconstant continuous functions from $\mathbb{R}$ to $\{0,1\}$; you have to move up a few levels in the Borel hierarchy to get such functions.  This doesn't matter if you just care about the functions being Borel, but when you're looking at the actual levels of the Borel hierarchy it's more convenient to start with continuous functions rather than starting a little higher up. &lt;/li&gt;&#xA;&lt;li&gt;It's easy to construct an element of $\mathbb{N}^\mathbb{N}$: you just have to construct a sequence of natural numbers. On the other hand, the representations of real numbers (Cauchy sequences, Dedekind cuts) are not as straightforward to work with. That makes proofs technically more difficult without making them more interesting. For example, compare the diagonalization proof that $\mathbb{N}^\mathbb{N}$ is uncountable with the diagonalization proof that $\mathbb{R}$ is uncountable.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;There are also a few reasons that the use of $\mathbb{N}^\mathbb{N}$ does not result in a loss of generality:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;For any uncountable complete separable metric space (c.s.m.s.) $X$, there is a bijection between $X$ and $\mathbb{N}^\mathbb{N}$ that is both Borel measurable and has a Borel measurable inverse.  So if the property we are studying is preserved by Borel isomorphisms, we can just replace an uncountable c.s.m.s. $X$ with $\mathbb{N}^\mathbb{N}$. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Every c.s.m.s. is a continuous image of $\mathbb{N}^\mathbb{N}$. In fact, for any c.s.m.s. $X$ there is a closed subset $C$ of $\mathbb{N}^\mathbb{N}$ and a continuous bijection from $C$ to $X$. So if we are studying a property preserved by continuous maps, we can work with $\mathbb{N}^\mathbb{N}$ or with its closed subspaces without losing generality. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Those types of reasons are why it's safe to stick with $\mathbb{N}^\mathbb{N}$ most of the time: the goal is to study an arbitary c.s.m.s. (including $\mathbb{R}$) but for most purposes there's no loss of generality in studying $\mathbb{N}^\mathbb{N}$. &lt;/p&gt;&#xA;" OwnerUserId="630" LastActivityDate="2011-03-11T13:48:22.317" CommentCount="7" />
  <row Id="26396" PostTypeId="1" CreationDate="2011-03-11T13:51:12.720" Score="3" ViewCount="67" Body="&lt;p&gt;As we all know, the theory of class fields began at the paper by Takagi on the abelian extensions of the field of rational numbers. Then one naturally has the following  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Where can one find the English version of that paper, or had this paper been translated into English yet?  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="6792" LastActivityDate="2011-03-11T13:51:12.720" Title="Where can the original paper by Takagi in English be found?" Tags="&lt;reference-request&gt;&lt;algebraic-number-theory&gt;" CommentCount="2" FavoriteCount="2" />
  <row Id="26397" PostTypeId="2" ParentId="26393" CreationDate="2011-03-11T13:58:24.150" Score="0" ViewCount="" Body="&lt;p&gt;Of course, usually there is a group of MST(minimum spanning tree) and what really represents the MST is it's weight and not the exact tree, and your example it's exactly where more the one MST come from.&#xA;claim: if the graph has a cycle with two equal weighted edges then it has at least TWO MST, prof: you have MST containing edge a1 if you close the cycle with a2 and remove a1 you have another MST!&lt;/p&gt;&#xA;" OwnerUserId="8113" LastEditorUserId="8113" LastEditorDisplayName="" LastEditDate="2011-03-11T14:36:44.583" LastActivityDate="2011-03-11T14:36:44.583" />
  <row Id="26398" PostTypeId="2" ParentId="26393" CreationDate="2011-03-11T14:06:53.110" Score="1" ViewCount="" Body="&lt;p&gt;The claim is false.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider a graph $G$ with points $a,b,c,d,e$ and edges $(a,b),(b,c),(c,d),(d,e),(e,a),(a,d)$ where all edges have weight 1, except $(a,b)$ and $(a,e)$ which have weight 100. Then there is only one minimum spanning tree, namely $(b,c),(c,d),(d,e),(a,e)$.&#xA;&lt;img src=&quot;http://i.imgur.com/MVkYp.jpg&quot; alt=&quot;The Graph&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If instead the claim was the following: &quot;If a graph $G$ is a cycle, and two of the edges $e_1$ and $e_2$ have weight $w$ which is the maximum weight in $G$, then there are at least two different minimum spanning trees in $G$&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then the claim is true. Consider some spanning tree $T$ in $G$. Since $T$ is spanning, one of the edges, lets say $e_1$ with weight $w$ must be in $T$. Now consider a new tree $T&amp;#39; = T - e_1 + e_2$. Now one can easily show that $T&amp;#39;$ is also spanning and has the same weight as $T$.&lt;/p&gt;&#xA;" OwnerUserId="3317" LastEditorUserId="3317" LastEditorDisplayName="" LastEditDate="2011-03-11T14:34:12.573" LastActivityDate="2011-03-11T14:34:12.573" CommentCount="6" />
  <row Id="26399" PostTypeId="1" CreationDate="2011-03-11T14:27:30.580" Score="2" ViewCount="60" Body="&lt;p&gt;Let $f(m,n,w)$ be the probability density function of F variable with m numerator df and n denominator df, i.e.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f(m,n,w)=\frac{\Gamma\left(\frac{m+n}{2}\right)(m/n)^{m/2}}{\Gamma(m/2)\Gamma(n/2)}w^{(m/2)-1}\left(1+\frac{mw}{n}\right)^{-(m+n)/2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am interested in the infimum of $\int_1^\infty f(m,n,w) dw$ over all $m,n$. From my&#xA;exploration, this seems to be&#xA; $$1-erf(1/\sqrt{2})\approx 0.3173$$ where $erf(x)=\frac{2}{\sqrt{\pi}}\int_0^x e^{-t^2} dt$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone prove this or point me to a reference?&lt;/p&gt;&#xA;" OwnerUserId="3249" LastEditorUserId="3249" LastEditorDisplayName="" LastEditDate="2011-03-11T14:53:12.150" LastActivityDate="2011-03-11T14:53:12.150" Title="A question about F-distribution" Tags="&lt;statistics&gt;" CommentCount="5" FavoriteCount="1" />
  <row Id="26400" PostTypeId="1" CreationDate="2011-03-11T14:28:07.320" Score="1" ViewCount="88" Body="&lt;p&gt;I'm trying to find the equation of a line that passes through (1, 2, -1) and creates equivalent angles with the three positive axes. I'm at a loss, though I'd like just a small hint, not the full answer. Thanks!&lt;/p&gt;&#xA;" OwnerUserId="1523" LastActivityDate="2011-05-10T15:56:20.220" Title="Finding the implicit equation for a line" Tags="&lt;linear-algebra&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="26401" PostTypeId="2" ParentId="26392" CreationDate="2011-03-11T14:33:03.153" Score="1" ViewCount="" Body="&lt;p&gt;Yes. If you have two functions $a,b : [0,1]\to\mathbb{R} $ satisfying $a(\epsilon)\in o(f(\epsilon))$ and $b(\epsilon)\in O(g(\epsilon))$ for $\epsilon \to 0$ then $a(\epsilon)b(\epsilon) \in o(f(\epsilon)g(\epsilon))$ wich specializes to $\epsilon^ma(\epsilon)\in o(\epsilon^m)$ in your case, since clearly $a(\epsilon)\in o(1)$.&lt;/p&gt;&#xA;" OwnerUserId="8087" LastActivityDate="2011-03-11T14:33:03.153" />
  <row Id="26402" PostTypeId="1" CreationDate="2011-03-11T14:33:58.007" Score="0" ViewCount="37" Body="&lt;p&gt;I was trying to understand the Ergodic theory recently, but I don't really have any knowledge about linear operators, Markov semigroups etc. so I didn't even fully understand the definition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could someone recommend a good book, notes or similar resource to learn about Linear operators and Markov semigroups?&lt;/p&gt;&#xA;" OwnerUserId="8114" LastActivityDate="2011-03-11T14:37:17.363" Title="Linear operators and Markov semigroups" Tags="&lt;books&gt;&lt;semigroup&gt;" AnswerCount="1" FavoriteCount="2" />
  <row Id="26403" PostTypeId="2" ParentId="26402" CreationDate="2011-03-11T14:37:17.363" Score="1" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://www.impan.pl/~rams/r48-ladek.pdf&quot; rel=&quot;nofollow&quot;&gt;This&lt;/a&gt; seems to be thorough and well written.&lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2011-03-11T14:37:17.363" />
  <row Id="26404" PostTypeId="2" ParentId="26400" CreationDate="2011-03-11T14:46:13.830" Score="1" ViewCount="" Body="&lt;p&gt;Let $a$ be the angle between  $\textbf{v}$ and $\textbf{x}$, $b$ be the angle between $\textbf{v}$ and $\textbf{y}$ and $c$ be the angle between $\textbf{v}$ and $\textbf{z}$. Then you are looking at the &lt;a href=&quot;http://mathworld.wolfram.com/DirectionCosine.html&quot; rel=&quot;nofollow&quot;&gt;direction cosines&lt;/a&gt;. In other words, you are looking at the following: $$ \alpha := \cos a := \frac{\textbf{v} \cdot \hat{\textbf{x}}}{|\textbf{v}|}$$ $$\beta := \cos b := \frac{\textbf{v} \cdot \hat{\textbf{y}}}{|\textbf{v}|}$$ and $$\gamma := \cos c := \frac{\textbf{v} \cdot \hat{\textbf{z}}}{|\textbf{v}|}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then you can solve for $a$, $b$, and $c$. &lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2011-03-11T14:46:13.830" />
  <row Id="26405" PostTypeId="2" ParentId="26400" CreationDate="2011-03-11T14:48:51.280" Score="0" ViewCount="" Body="&lt;p&gt;Some hints:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have two points $x,y$ on a line, you can parameterize the line. How?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The vector $x-y$ is parallel to the line. How do you compute the angle between two vectors?&lt;/p&gt;&#xA;" OwnerUserId="5882" LastActivityDate="2011-03-11T14:48:51.280" />
  <row Id="26406" PostTypeId="2" ParentId="536" CreationDate="2011-03-11T15:08:24.347" Score="0" ViewCount="" Body="&lt;p&gt;P = probability that if you go to one, you will return to 0&#xA;Q = probability that if you go to 2 you will return to 0&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming you start at 0, you will either go to 1 or to -1, so finding the probability P for 1 is the same as the probability if you go to -1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From P, there are 2 possibilities.  1) you can go immediately back to 0, or you can go to 2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, P = ½ + ½ * Q&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is Q:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From Q, you have 2 possibilities.  You can go even further out, or you can return to 0.  What it the probability you will return to 0?  Well, the probability that you will return to 1 is P because returning to 1 from 2 is the same probability as returning to 0 from 1.  And, you will have to return to 1 before you can return to 0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore,  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, P = ½ + ½ P2&#xA;2P = 1+P2&#xA;P2 – 2P + 1 = 0&#xA;(P – 1) (P – 1) = 0&#xA;P = 1&lt;/p&gt;&#xA;" OwnerUserId="8115" LastActivityDate="2011-03-11T15:08:24.347" CommentCount="2" />
  <row Id="26407" PostTypeId="1" AcceptedAnswerId="26411" CreationDate="2011-03-11T15:18:07.547" Score="0" ViewCount="139" Body="&lt;p&gt;Is it possible to define a bijection $f:\mathbb{R}\setminus\mathbb{R}^{-}\rightarrow[0,1)$ such that $f$ is continuously differentiable on its entire domain?&lt;/p&gt;&#xA;" OwnerUserId="8116" LastActivityDate="2011-03-11T19:03:59.553" Title="Does my function exist?" Tags="&lt;analysis&gt;" AnswerCount="3" CommentCount="10" />
  <row Id="26409" PostTypeId="1" AcceptedAnswerId="26426" CreationDate="2011-03-11T15:27:44.040" Score="0" ViewCount="90" Body="&lt;p&gt;The Fourier transform of a sinc function is the top hat function. So, if $\{y_k\};\ k\in\{0,1,...,n-1\}$ are samples of the sinc function, sampled $T$ apart, the discrete Fourier transform is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$Y_m=\sum_{k=0}^{n-1}y_k e^{i 2\pi mk/n}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if I choose $T$ to be greater than 1, it doesn't satisfy the sampling criterion, and so should be aliased. Is there a way to determine how the aliasing occurs mathematically? To illustrate my point, here's a matlab code that calculates two sinc functions: one properly sampled and the other inadequately.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;x=-1000;1000;&#xA;y=sinc(x/2);   %#properly sampled&#xA;z=sinc(1.2*x); %#insufficiently sampled&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Plot of the magnitude of the FFT of &lt;code&gt;y&lt;/code&gt; looks like this &#xA;&lt;img src=&quot;http://i.imgur.com/b1Yt6.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and that of &lt;code&gt;z&lt;/code&gt; looks like this (I apologize, the y axis reads |Y|, when it should be |Z|)&#xA;&lt;img src=&quot;http://i.imgur.com/sNXjS.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Try as I might, I can't get the 'flipped' top-hat as in the above figure out of the expression for the DFT. Also, the top-hat no longer has zeros outside the interval, but some value. I can intuitively understand aliasing, and I've always thought that only the frequencies just above the sampling frequency get aliased into a particular lower frequency... however this seems to alias in everywhere&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can someone help me with this?&lt;/p&gt;&#xA;" OwnerUserId="7815" LastEditorUserId="2641" LastEditorDisplayName="" LastEditDate="2011-03-11T16:08:30.540" LastActivityDate="2011-03-11T18:03:03.160" Title="Aliasing in DFT: mathematical expression" Tags="&lt;fourier-analysis&gt;&lt;fourier-series&gt;&lt;signal-processing&gt;" AnswerCount="1" FavoriteCount="2" />
  <row Id="26410" PostTypeId="2" ParentId="26407" CreationDate="2011-03-11T15:30:24.393" Score="5" ViewCount="" Body="&lt;p&gt;If your domain is the reals greater than or equal to zero, $1-\exp(-x)$ seems to fill the bill.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-03-11T15:30:24.393" CommentCount="0" />
  <row Id="26411" PostTypeId="2" ParentId="26407" CreationDate="2011-03-11T15:30:28.830" Score="5" ViewCount="" Body="&lt;p&gt;Yes. For example, $f(x)=1-\frac{1}{1+x}$.&lt;/p&gt;&#xA;" OwnerUserId="5203" LastActivityDate="2011-03-11T15:30:28.830" />
  <row Id="26412" PostTypeId="2" ParentId="26386" CreationDate="2011-03-11T15:41:46.387" Score="2" ViewCount="" Body="&lt;p&gt;Your second point looks wrong in the book.  I think&#xA;$$R_n = X_{T_n} = X_{T_0} + \sum_{i=1}^n \left(X_{T_i} - X_{T_{i-1}}\right)$$ &#xA;or &#xA;$$R_n = X_{T_n} = R_0 + \sum_{i=1}^n (R_i - R_{i-1})$$&#xA;might be better (and obvious).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $T_0 = 1$, $R_0 = X_{T_0} = X_1$, identically distributed as each of the $X_j$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is where the memorylessness of the geometric distribution applies, because it has the property that $R_n - R_{n-1} = X_{T_n} - X_{T_{n-1}}$ has the same distribution as $X_j$ because if $X_{T_{n-1}} = k$ then&#xA;$$Pr(X_{T_n} - X_{T_{n-1}} = m) = Pr(X_{T_n} = m+k | X_{T_n} &amp;gt; k) = \frac{p(1 − p)^{m+k−1}}{\sum_{j \ge 1} p(1 − p)^{j+k−1} } = p(1 − p)^{m−1}$$&lt;br&gt;&#xA;and is independent of k and of all the $X_j$ for $j &amp;lt; T_{n-1}$ and $j&amp;gt;T_n$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So $R_j - R_{j-1}$ has the same distribution as $X_j$ and combined with independence leads to $R_n$ having the same distribution as $\sum_{j=0}^n X_j$, i.e. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;as the sum of $n + 1$ iid geometric variables.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;as the book says.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the book is wrong with its equals sign: two (sums of) random variables can have the same distribution without being the same.  If $A$ is uniformly distributed on $[0,1]$ and $B = 1-A$, then $A$ and $B$ have the same distribution, but they are unlikely to be equal to each other. &lt;/p&gt;&#xA;" OwnerUserId="6460" LastEditorUserId="6460" LastEditorDisplayName="" LastEditDate="2011-03-11T21:39:21.957" LastActivityDate="2011-03-11T21:39:21.957" CommentCount="4" />
  <row Id="26413" PostTypeId="1" CreationDate="2011-03-11T16:03:15.527" Score="-2" ViewCount="69" Body="&lt;p&gt;When one talks about the category $V_K$ of vector spaces over a field $K$ and considers the dual functor $D$ which maps a vector space $V$ to its dual $V^{*}$ I believe to have in mind something like a &lt;strong&gt;labeled&lt;/strong&gt; category, the labels letting me know which object is the dual of another object. (Or can I see this by carefully looking at the morphisms?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I want to know: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Is there - analogously to graphs - a&#xA;  distinction between labelled and&#xA;  unlabeled categories? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Side remark:&lt;/em&gt; I see something like a predominance of unlabeled graphs over labeled ones, the former being the more &quot;genuine&quot; graphs (as abstract structures). What's the situation in category theory?&lt;/p&gt;&#xA;" OwnerUserId="1792" LastEditorUserId="1792" LastEditorDisplayName="" LastEditDate="2011-03-11T16:12:14.490" LastActivityDate="2011-03-11T16:36:29.297" Title="Labeled and unlabeled categories" Tags="&lt;graph-theory&gt;&lt;category-theory&gt;" AnswerCount="1" CommentCount="11" />
  <row Id="26414" PostTypeId="2" ParentId="26273" CreationDate="2011-03-11T16:07:45.867" Score="1" ViewCount="" Body="&lt;p&gt;let $f_n(x):\mathbb{R}\to\mathbb{R}$ be $\frac{1}{n}\chi_{(n-1,n)}$.  then $\sum f_n$ converges uniformly but $\sum M_n=\sum 1/n$ diverges.  something like this would work (not trying to give an answer away, if this is valid, but trying to show along what lines you should be thinking)&#xA;&lt;hr&gt;&#xA;(for a set $A\subseteq\mathbb{R}$, $\chi_A(x)=1$ if $x\in A$, $\chi_A(x)=0$ if $x\not\in A$.  sometimes called the &lt;em&gt;indicator function&lt;/em&gt; of the set $A$, also denoted by $1_A$)&lt;/p&gt;&#xA;" OwnerUserId="6925" LastEditorUserId="6925" LastEditorDisplayName="" LastEditDate="2011-03-12T18:23:00.750" LastActivityDate="2011-03-12T18:23:00.750" CommentCount="1" />
  <row Id="26415" PostTypeId="2" ParentId="26363" CreationDate="2011-03-11T16:18:03.073" Score="2" ViewCount="" Body="&lt;p&gt;For positive real $x$, $\sqrt x$ denotes the positive square root of $x$, by definition. &lt;a href=&quot;http://en.wikipedia.org/wiki/Square_root&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt; agrees with me on this.&lt;/p&gt;&#xA;" OwnerUserId="1508" LastActivityDate="2011-03-11T16:18:03.073" />
  <row Id="26416" PostTypeId="1" AcceptedAnswerId="26421" CreationDate="2011-03-11T16:22:07.697" Score="0" ViewCount="53" Body="&lt;p&gt;Given the prior probability of 2 distributions  N(x,y) and N(a,b), where N(&amp;#x03BC;,&amp;#x03C3;^2): &lt;/p&gt;&#xA;&#xA;&lt;p&gt;how do you make a decision rule to minimize the probability of error, if the prior probabilities are equal?  Can you give an example? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What if the prior probabilities are different, such as&lt;br&gt;&#xA; P(Distribution 1) = 0.70&lt;br&gt;&#xA; P(Distribution 2) = 0.30 ?&lt;/p&gt;&#xA;" OwnerUserId="7265" LastEditorUserId="6460" LastEditorDisplayName="" LastEditDate="2011-03-11T16:57:23.540" LastActivityDate="2011-03-11T17:45:25.493" Title="Bayes classifier of 2 classes" Tags="&lt;probability&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="26417" PostTypeId="2" ParentId="26413" CreationDate="2011-03-11T16:36:29.297" Score="4" ViewCount="" Body="&lt;p&gt;$\text{Vect}_k$ is &lt;a href=&quot;http://en.wikipedia.org/wiki/Enriched_category&quot; rel=&quot;nofollow&quot;&gt;enriched&lt;/a&gt; over itself: for any two vector spaces $V, W$, the set $\text{Hom}_k(V, W)$ naturally acquires the structure of a vector space (since linear combinations of linear operators are linear operators), hence defines a functor $\text{Vect}_k \times \text{Vect}_k \to \text{Vect}_k$ contravariant in the first argument.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, for every $V$ the set $\text{Hom}_k(V, k) \equiv V^{\ast}$ naturally acquires the structure of a vector space, and this is the abstract origin of the dual vector space functor $\text{Vect}_k \to \text{Vect}_k$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;By introducing enough extra data, there is no need for &quot;labelings&quot; (and I don't understand what this means anyway). In particular, if you aren't comfortable with singling out $k$ as a $1$-dimensional vector space, you can introduce the &lt;a href=&quot;http://en.wikipedia.org/wiki/Monoidal_category&quot; rel=&quot;nofollow&quot;&gt;monoidal&lt;/a&gt; structure on $\text{Vect}_k$, in which the identity object $k$ is singled out as part of the extra data. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Side remark:&lt;/em&gt; I do not understand your side remark. There are many categories of graphs, and people are interested in both labeled and unlabeled graphs. What does &quot;genuine&quot; mean? &lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2011-03-11T16:36:29.297" CommentCount="3" />
  <row Id="26418" PostTypeId="2" ParentId="26345" CreationDate="2011-03-11T16:46:54.890" Score="3" ViewCount="" Body="&lt;p&gt;It might be worth noting that you do need $G$ to be finite; otherwise, the claim is not true. For example, the &lt;a href=&quot;http://en.wikipedia.org/wiki/Pr%C3%BCfer_group&quot; rel=&quot;nofollow&quot;&gt;Prufer $p$-group&lt;/a&gt; has the property that given any two elements $a$ and $b$, either $\langle a\rangle \subseteq \langle b\rangle $ or $\langle b\rangle \subseteq \langle a\rangle$, but the group itself is not cyclic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, we want to prove:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Suppose $G$ is a &lt;strong&gt;finite&lt;/strong&gt; group such that for all $a,b\in G$, either $\langle a\rangle\subseteq \langle b\rangle$ or else $\langle b\rangle \subseteq \langle a\rangle$. Then $G$ is cyclic.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Your &quot;I think this works&quot; is very badly phrased. You cannot begin by assuming that $G=\langle g\rangle$, because this is tantamount to assuming that $G$ is cyclic; but this is what you want to prove! You don't get to assume it to begin with. At best, you should say: &quot;&lt;strong&gt;If&lt;/strong&gt; $G=\langle g\rangle$, then we're done; otherwise, let...&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The argument works reasonably well: suppose $G=\{g_1,\ldots,g_n\}$ (remember, $G$ has to be &lt;em&gt;finite&lt;/em&gt;). Let $a_1=g_1$. Now, consider $g_2$; either $g_2\in\langle a\rangle$ or $\langle a\rangle\subseteq \langle g_2\rangle$. In the former case, let $a_2=g_1$; in the latter, let $a_2=g_2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume you have already defined $a_1,\ldots,a_{\ell}$, with $1\leq \ell\lt n$. Consider $g_{\ell+1}$. If $g_{\ell+1}\in\langle a_{\ell}\rangle$, let $a_{\ell+1}=a_{\ell}$. Otherwise, we have $\langle a_{\ell}\rangle \subseteq \langle g_{\ell+1}\rangle$; then set $a_{\ell+1}=g_{\ell+1}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, I claim that $G=\langle a_{n}\rangle$. Indeed, note that by construction, $g_k\in\langle a_{k}\rangle\subseteq \langle a_{k+1}\rangle$. Therefore, $g_1,\ldots,g_k\in\langle a_{k}\rangle$. In particular, $G=\{g_1,\ldots,g_n\}\subseteq \langle a_n\rangle \subseteq G$, giving equality. Thus, $G$ is cyclic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(There are of course easier ways of doing this; but this follows the intuitive idea of your argument, only written up more formally).&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-11T19:34:46.023" LastActivityDate="2011-03-11T19:34:46.023" CommentCount="3" />
  <row Id="26419" PostTypeId="1" AcceptedAnswerId="26442" CreationDate="2011-03-11T17:17:14.817" Score="3" ViewCount="163" Body="&lt;p&gt;Hello,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While playing around with programming, a problem suddenly came across my mind. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Given $n$ let's say $2^{64}$ (64-bit unsigned integer). Find a positive integer $x$, $1 \leq x \leq n$, such that $x$ has maximum divisors.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My attempt was:&lt;br&gt;&#xA;Let $n = p_1^{a_1} \times p_2^{a_2} \times ... \times p_k^{a_k}.$&lt;br&gt;&#xA;The number of divisors is given by:&#xA;$$(a_1 + 1) \times (a_2 + 1) \times ... \times (a_k + 1)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We need to find the maximum of these products. Since $(a_1 + 1)_{max} \implies {a_1}_{max}$, the product becomes: $a_1 \times a_2 \times .... \times a_k$&lt;br&gt;&#xA;Let $P = a_1 \times a_2 \times .... \times a_k$.&lt;br&gt;&#xA;The maximum of $P$ means $a_i = a_j$ where $1 \leq i, j \leq k$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I guess my question is how can we find a number that is less than $n$ which has this property?   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks, &lt;/p&gt;&#xA;" OwnerUserId="5711" LastEditorUserId="2937" LastEditorDisplayName="" LastEditDate="2011-03-11T19:09:28.760" LastActivityDate="2011-03-11T21:38:39.760" Title="Find maximum divisors of a number in range" Tags="&lt;number-theory&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="26420" PostTypeId="1" CreationDate="2011-03-11T17:28:26.117" Score="2" ViewCount="111" Body="&lt;p&gt;So the cup product is not well defined over co-chain groups, but all the books claim it is well defined over co-homology groups. The only thing I am not clear on is invariance under ordering/re-ordering of simplices when we go to the co-homology level. Every book seems to gloss over this, and after doing a few examples, I can't seem to figure out how to get this to work out right. Can someone fill me in?&lt;/p&gt;&#xA;" OwnerUserId="8122" LastActivityDate="2011-03-14T00:27:54.093" Title="cup product well definedness" Tags="&lt;cohomology&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="26421" PostTypeId="2" ParentId="26416" CreationDate="2011-03-11T17:45:25.493" Score="0" ViewCount="" Body="&lt;p&gt;I think what you are looking at is the Bayes optimal classifier. This is the classifier that minimizes the Bayes error. In other words, we have:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\text{classify} \ \textbf{x} \ \text{as} \ \begin{cases} C_{0} \ \ \ \ \text{if} \ P(C_0| \textbf{x}) &amp;gt; P(C_1| \textbf{x}) \\ C_1 \ \ \ \ \text{otherwise} \end{cases}$$&lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2011-03-11T17:45:25.493" CommentCount="1" />
  <row Id="26422" PostTypeId="2" ParentId="26371" CreationDate="2011-03-11T17:46:49.050" Score="8" ViewCount="" Body="&lt;p&gt;&quot;One-to-one&quot; and &quot;onto&quot; are properties of functions in general, not just linear transformations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Definition.&lt;/strong&gt; Let $f\colon X\to Y$ be a function.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$f$ is &lt;strong&gt;one-to-one&lt;/strong&gt; if and only if for every $y\in Y$ there is &lt;em&gt;at most&lt;/em&gt; one $x\in X$ such that $f(x)=y$; equivalently, if and only if $f(x_1)=f(x_2)$ implies $x_1=x_2$.&lt;/li&gt;&#xA;&lt;li&gt;$f$ is &lt;strong&gt;onto&lt;/strong&gt; (or &lt;strong&gt;onto&lt;/strong&gt; $Y$, if the codomain is not clear from context) if and only if for every $y\in Y$ there &lt;em&gt;at least&lt;/em&gt; one $x\in X$ such that $f(x)=y$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This definition applies to linear transformations as well, and in particular for linear transformations $T\colon \mathbb{R}^n\to\mathbb{R}^m$, and by extension to matrices, since an $m\times n$ matrix $A$ can be identified with the linear transformation $L_A\colon\mathbb{R}^n\to\mathbb{R}^m$ given by $L_A(\mathbf{x}) = A\mathbf{x}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the definitions are for &lt;em&gt;any&lt;/em&gt; functions. But when our sets $X$ and $Y$ have more structure to them, and the functions are not arbitrary, but special kinds of functions, we can often obtain &lt;em&gt;other&lt;/em&gt; ways of characterizing a function as one-to-one or onto which is easier/better/more useful/more conceptual/has interesting applications. This is indeed the case when we have such a rich structure as linear transformations and vector spaces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One-to-one is probably the easiest; this is because whether a function is one-to-one depends &lt;em&gt;only&lt;/em&gt; on its domain, and not on its codomain. By contrast, whether a function is onto depends on &lt;em&gt;both&lt;/em&gt; on the domain and the codomain (so, for instance, $f(x)=x^2$ is onto if we think of it as a function $f\colon\mathbb{R}\to[0,\infty)$, but not if we think of it as a function $f\colon\mathbb{R}\to\mathbb{R}$, or $f\colon[2,\infty)\to[0,\infty)$). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Theorem.&lt;/strong&gt; &lt;em&gt;Let&lt;/em&gt; $T\colon\mathbb{R}^n\to\mathbb{R}^m$ &lt;em&gt;be a linear transformation. The following are equivalent:&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$T$ &lt;em&gt;is one-to-one&lt;/em&gt;.&lt;/li&gt;&#xA;&lt;li&gt;$T(\mathbf{x})=\mathbf{0}$ &lt;em&gt;has only the trivial solution&lt;/em&gt; $\mathbf{x}=\mathbf{0}$.&lt;/li&gt;&#xA;&lt;li&gt;&lt;em&gt;If&lt;/em&gt; $A$ &lt;em&gt;is the standard matrix of&lt;/em&gt; $T$, &lt;em&gt;then the columns of&lt;/em&gt; $A$ &lt;em&gt;are linearly independent.&lt;/em&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Proof.&lt;/em&gt; The equivalence of (1) and (2) is basic in linear algebra, so let's deal with that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1)$\Rightarrow$(2): If $T$ is one-to-one, then for all $\mathbf{x}$, since $T(\mathbf{0})=\mathbf{0}$ (being linear), then $T(\mathbf{x})=\mathbf{0}=T(\mathbf{0})$ implies $\mathbf{x}=\mathbf{0}$; this proves (2).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2)$\Rightarrow$(1): Suppose $T(\mathbf{x}_1)=T(\mathbf{x}_2)$. Then&#xA;$$\mathbf{0} = T(\mathbf{x}_1) - T(\mathbf{x}_2) = T(\mathbf{x}_1-\mathbf{x}_2),$$&#xA;since $T$ is linear; because we are assuming (2), $T(\mathbf{x}_1-\mathbf{x}_2)=\mathbf{0}$ implies that $\mathbf{x}_1-\mathbf{x}_2 = \mathbf{0}$, so $\mathbf{x}_1=\mathbf{x}_2$, which proves that $T$ is indeed one-to-one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The key to the connection with (3) (and eventually to your confusion) is that multiplying a matrix by a vector can be seen as an operation on columns. If&#xA;$$A=\left(\begin{array}{cccc}&#xD;&#xA;a_{11} &amp;amp; a_{12} &amp;amp; \ldots &amp;amp; a_{1n}\\&#xD;&#xA;a_{21} &amp;amp; a_{22} &amp;amp; \ldots &amp;amp; a_{2n}\\&#xD;&#xA;\vdots &amp;amp; \vdots &amp;amp; \ddots &amp;amp; \vdots\\&#xD;&#xA;a_{m1} &amp;amp; a_{m2} &amp;amp; \ldots &amp;amp; a_{mn}&#xD;&#xA;\end{array}\right),$$&#xA;then let the &lt;em&gt;columns of&lt;/em&gt; $A$, $A_1$, $A_2,\ldots,A_n$ be:&#xA;$$A_1 = \left(\begin{array}{c}a_{11}\\a_{21}\\ \vdots\\a_{m1}\end{array}\right),\quad A_2 = \left(\begin{array}{c}a_{12}\\a_{22}\\ \vdots\\ a_{m2}\end{array}\right),\quad\ldots,A_n = \left(\begin{array}{c}a_{1n}\\a_{2n}\\ \vdots\\ a_{mn}\end{array}\right).$$&#xA;Then we have the following:&#xA;$$A\left(\begin{array}{c}x_1\\x_2\\ \vdots\\x_n\end{array}\right) = x_1 A_1 + a_2 A_2 + \cdots + a_nA_n.$$&#xA;That is, multiplying $A$ by $\mathbf{x}$ gives a linear combination of the columns of $A$. This gives the direct connection we need between conditions (1) and (2), and condition (3).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2)$\Rightarrow$(3): To show that the columns of $A$ are linearly independent, we need to show that if $\alpha_1A_1 + \cdots + \alpha_nA_n = \mathbf{0}$, then $\alpha_1=\cdots=\alpha_n=0$. So suppose $\alpha_1A_1+\cdots+\alpha_nA_n = \mathbf{0}$. Then&#xA;$$T(\mathbf{\alpha}) = A\mathbf{\alpha} = \alpha_1A_1+\cdots+\alpha_nA_n = \mathbf{0},\qquad\text{where }\mathbf{\alpha}=\left(\begin{array}{c}\alpha_1\\ \alpha_2\\ \vdots\\ \alpha_n\end{array}\right).$$&#xA;Because we are assuming (2), that means that from $T(\mathbf{\alpha}) = \mathbf{0}$ we can conclude taht $\alpha=\mathbf{0}$; therefore, $\alpha_1=\cdots=\alpha_n=0$. This proves that $A_1,\ldots,A_n$ are linearly independent. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(3)$\Rightarrow$(2): Suppose the columns of $A$ are linearly independent, and &#xA;$$\mathbf{0} = T(\mathbf{x}) = A\mathbf{x}\quad\text{where }\mathbf{x}=\left(\begin{array}{c}a_1\\a_2\\ \vdots\\ a_n\end{array}\right).$$&#xA;This means that $a_1A_1 + \cdots a_nA_n = \mathbf{0}$; since the columns of $A$ are assumed to be linearly independent, we conclude that $a_1=\cdots=a_n=0$, so $\mathbf{x}=\mathbf{0}$, proving (2). &lt;strong&gt;QED&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What about onto? There are two things here. One is a theorem similar to the one above; the other is the Rank-Nullity Theorem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Theorem.&lt;/strong&gt; &lt;em&gt;Let&lt;/em&gt; $T\colon\mathbb{R}^n\to\mathbb{R}^m$ &lt;em&gt;be a linear transformation. The following are equivalent:&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$T$ &lt;em&gt;is onto&lt;/em&gt;.&lt;/li&gt;&#xA;&lt;li&gt;&lt;em&gt;The equation&lt;/em&gt; $T(\mathbf{x})=\mathbf{b}$ &lt;em&gt;has solutions for every&lt;/em&gt; $\mathbf{b}\in\mathbb{R}^m$. &lt;/li&gt;&#xA;&lt;li&gt;&lt;em&gt;If&lt;/em&gt; $A$ &lt;em&gt;is the standard matrix of&lt;/em&gt; $T$, &lt;em&gt;then the columns of&lt;/em&gt; $A$ &lt;em&gt;span&lt;/em&gt; $\mathbb{R}^m$.  That is: every $\mathbf{b}\in\mathbb{R}^m$ is a linear combination of the columns of $A$.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Proof.&lt;/em&gt; (1)$\Leftrightarrow$(2) is essentially the definition, only cast in terms of equations for the sake of similarity to the previous theorem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2)$\Rightarrow$(3) Let $\mathbf{b}\in\mathbb{R}^m$. Then by (2) there exists an $\mathbf{a}\in\mathbb{R}^n$ such that $T(\mathbf{a})=\mathbf{b}$. We have:&#xA;$$\mathbf{b} = T(\mathbf{a}) = A\mathbf{a} = A\left(\begin{array}{c}a_1\\a_2\\ \vdots\\a_n\end{array}\right) = a_1A_1 + a_2A_2 + \cdots + a_nA_n.$$&#xA;That is, we can express $\mathbf{b}$ as a linear combination of the columns of $A$. Since $\mathbf{b}$ is arbitrary, every vector in $\mathbb{R}^m$ can be expressed as a linear combination of the columns of $A$, so the columns of $A$ span $\mathbb{R}^m$; this proves (3).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(3)$\Rightarrow$(2) Suppose the columns of $A$ span $\mathbb{R}^m$ and let $\mathbf{b}\in\mathbb{R}^m$. We want to show that $T(\mathbf{x})=\mathbf{b}$ has at least one solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the columns of $A$ span $\mathbb{R}^m$, there exist scalars $\alpha_1,\ldots,\alpha_n$ such that&#xA;$$\mathbf{b} = \alpha_1 A_1 + \cdots + \alpha_n A_n = A\left(\begin{array}{c}\alpha_1\\ \alpha_2\\ \vdots\\ \alpha_n\end{array}\right) = T(\mathbf{\alpha}).$$&#xA;So $\mathbf{\alpha})$, where&#xA;$$\mathbf{\alpha} = \left(\begin{array}{c}\alpha_1\\ \alpha_2\\ \vdots\\ \alpha_n\end{array}\right),$$&#xA;is a solution to $T(\mathbf{x})=\mathbf{b}$. This establishes (2). &lt;strong&gt;QED&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So: &quot;one-to-one&quot;-ness is related to linear independence; &quot;onto&quot;-ness is related to spanning properties. Note that linear independence is an &lt;em&gt;intrinsic&lt;/em&gt; property (it depends only on the set of vectors), whereas spanning is an &lt;em&gt;extrinsic&lt;/em&gt; properties (it depends also on the space we are considering; it is &lt;em&gt;contextual&lt;/em&gt;). This matches the fact that whether a function is one-to-one or not depends &lt;em&gt;only&lt;/em&gt; on the domain, but whether it is onto depends on &lt;em&gt;both&lt;/em&gt; the domain and the codomain of the function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But there is a deep connection between the two. Remember the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Definition.&lt;/strong&gt; Let $A$ be an $m\times n$ matrix. The &lt;em&gt;nullity of&lt;/em&gt; $A$, $\mathrm{nullity}(A)$, is the dimension of the kernel of $A$, that is, of the subspace of $\mathbb{R}^n$ given by&#xA;$$\mathrm{ker}(A) = \Bigl\{ \mathbf{x}\in\mathbb{R}^n\Bigm| A\mathbf{x}=\mathbf{0}\Bigr\}.$$&#xA;The &lt;em&gt;rank of&lt;/em&gt; $A$, $\mathrm{rank}(A)$ is the dimension of the image of $A$; that is, of the subspace of $\mathbb{R}^m$ given by&#xA;\begin{align*}&#xA;\mathrm{Im}(A) &amp;amp;= \Bigl\{ \mathbf{b}\in\mathbb{R}^m\Bigm| A\mathbf{x}=\mathbf{b}\text{ has at least one solution}\Bigr\}\\&#xA;&amp;amp;= \Bigl\{ A(\mathbf{x})\Bigm|\mathbf{x}\in\mathbb{R}^n\Bigr\}.&#xA;\end{align*}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The deep connection between them is given by the Rank-Nullity Theorem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Rank-Nullity Theorem.&lt;/strong&gt; Let $A$ be an $m\times n$ matrix. Then&#xA;$$\mathrm{rank}(A) + \mathrm{nullity}(A) = n.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we get two more equivalences for one-to-one and onto:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Theorem.&lt;/strong&gt; &lt;em&gt;Let&lt;/em&gt; $T\colon\mathbb{R}^n\to\mathbb{R^m}$ &lt;em&gt;be a linear transformation. The following are equivalent:&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$T$ &lt;em&gt;is one-to-one&lt;/em&gt;.&lt;/li&gt;&#xA;&lt;li&gt;*The equation $T(\mathbf{x})=\mathbf{0}$ &lt;em&gt;has only the trivial solution&lt;/em&gt; $\mathbf{x}=\mathbf{0}$.&lt;/li&gt;&#xA;&lt;li&gt;&lt;em&gt;If&lt;/em&gt; $A$ &lt;em&gt;is the standard matrix of&lt;/em&gt; $T$, &lt;em&gt;then the columns of&lt;/em&gt; $A$ &lt;em&gt;are linearly independent.&lt;/em&gt;&lt;/li&gt;&#xA;&lt;li&gt;$\mathrm{ker}(A) = \{\mathbf{0}\}$.&lt;/li&gt;&#xA;&lt;li&gt;$\mathrm{nullity}(A) = 0$.&lt;/li&gt;&#xA;&lt;li&gt;$\mathrm{rank}(A) = n$.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Proof.&lt;/em&gt; The equivalence of (4) and (5) follows because only the trivial subspace has dimension $0$; the equivalence of (4) and (2) follows by definition of the kernel. The equivalence of (5) and (6) follows from the Rank-Nullity Theorem, since $n = \mathrm{nullity}(A)+\mathrm{rank}(A)$, so $\mathrm{nullity}(A) = 0$ if and only if $\mathrm{rank}(A) = n$. Since we already know (1), (2), and (3) are equivalent, the result follows. &lt;strong&gt;QED&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Theorem.&lt;/strong&gt; &lt;em&gt;Let&lt;/em&gt; $T\colon\mathbb{R}^n\to\mathbb{R}^m$ &lt;em&gt;be a linear transformation. The following are equivalent:&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$T$ &lt;em&gt;is onto&lt;/em&gt;.&lt;/li&gt;&#xA;&lt;li&gt;&lt;em&gt;The equation&lt;/em&gt; $T(\mathbf{x})=\mathbf{b}$ &lt;em&gt;has solutions for every&lt;/em&gt; $\mathbf{b}\in\mathbb{R}^m$. &lt;/li&gt;&#xA;&lt;li&gt;&lt;em&gt;If&lt;/em&gt; $A$ &lt;em&gt;is the standard matrix of&lt;/em&gt; $T$, &lt;em&gt;then the columns of&lt;/em&gt; $A$ &lt;em&gt;span&lt;/em&gt; $\mathbb{R}^m$.  That is: every $\mathbf{b}\in\mathbb{R}^m$ is a linear combination of the columns of $A$.&lt;/li&gt;&#xA;&lt;li&gt;$\mathrm{Im}(A) = \mathbb{R}^m$.&lt;/li&gt;&#xA;&lt;li&gt;$\mathrm{rank}(A) = m$.&lt;/li&gt;&#xA;&lt;li&gt;$\mathrm{nullity}(A) = n-m$.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Proof.&lt;/em&gt; We already know that (1), (2), and (3) are equivalent. The equivalence of (4) and (2) follows by the definition of the image. The equivalence of (4) and (5) follows because the only subspace of $\mathbb{R}^m$ that has dimension $m$ is the whole space. Finally, the equivalence of (5) and (6) follows from the rank nullity theorem: since $n = \mathrm{rank}(A)+\mathrm{nullity}(A)$, then $\mathrm{nullity}(A) = n - \mathrm{rank}(A)$. So the rank equals $m$ if and only if the nullity equals $n-m$. &lt;strong&gt;QED&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So now you have a whole bunch of ways of checking if a matrix is one-to-one, and of checking if a matrix is onto. None of them is &quot;better&quot; than the others: for some matrices, one will be easier to check, for other matrices, it may be a different one which is easy to check.  Also, the rank of a matrix is closely related to its row-echelon form, so that might help as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note a few things: generally, &quot;onto&quot; and &quot;one-to-one&quot; are independent of one another. You can have a matrix be onto but not one-to-one; or be one-to-one but not onto; or be both; or be neither. The Rank-Nullity Theorem does place some restrictions: if $A$ is $m\times n$ and $m\lt n$, then the matrix cannot be onto (because $1\leq\mathrm{rank}(A)\leq m$, so if $\mathrm{rank}(A)+\mathrm{nullity}(A) = n$, we must have $\mathrm{nullity}(A)\gt 0$); dually, if $m\gt n$ then $A$ cannot be onto. In particular, the only matrices that can be &lt;em&gt;both&lt;/em&gt; one-to-one and onto are square matrices. On the other hand, you can have an $m\times n$ matrix with $m\lt n$ that is onto, or one that is &lt;em&gt;not&lt;/em&gt; onto. And you can have $m\times n$ matrices with $m\gt n$ that are one-to-one, and matrices that are &lt;em&gt;not&lt;/em&gt; one-to-one. &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-11T19:32:33.143" LastActivityDate="2011-03-11T19:32:33.143" CommentCount="5" />
  <row Id="26423" PostTypeId="1" AcceptedAnswerId="26435" CreationDate="2011-03-11T17:48:34.687" Score="1" ViewCount="135" Body="&lt;p&gt;The answer is probably gonna be very straightforward but I'm missing it. I want to prove the following claim:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;blockquote&gt;&#xA;    &lt;p&gt;Suppose we play a game where we win with probability $p$ and lose with probability $q = 1 - p$, where $p \neq \frac{1}{2}$. Denote by $X_i$ the outcome of game $i$ ($X_i = 1$ if we win and $X_i = -1$ if we lose). Prove that $$M_n = (\frac{p}{q})^{S_n}$$ is a martingale, where $S_n = \sum_{i=1}^{n}X_i$ and $M_0 = 1$.&lt;/p&gt;&#xA;  &lt;/blockquote&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;My attempt:&lt;/b&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We want to show that $E(M_n|\mathcal{F}_{n-1}) = M_{n-1}$. So:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$E(M_n|\mathcal{F}_{n-1}) = \prod_{i=1}^{n-1}(\frac{p}{q})^{X_i} E((\frac{p}{q})^{X_n})$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I omitted the $\mathcal{F}_{n-1}$ since it has no influence on the outcome of the $n$-th game.We are done if we can show that $E((\frac{p}{q})^{X_n}) = 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, $$E((\frac{p}{q})^{X_n}) = (\frac{p}{q})^{1}P(X_n = 1) + (\frac{p}{q})^{-1}P(X_n = -1) = \frac{p^2}{q} + \frac{q^2}{p} = \frac{p^3 + q^3}{pq}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At this point I'm having this feeling I am making things more difficult than it is. Can anyone give an easier way of showing this?&lt;/p&gt;&#xA;" OwnerUserId="7155" LastEditorUserId="7155" LastEditorDisplayName="" LastEditDate="2011-03-11T18:01:51.773" LastActivityDate="2011-03-17T09:20:41.207" Title="Proving that an unfair game is martingale" Tags="&lt;stochastic-processes&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="26424" PostTypeId="1" AcceptedAnswerId="26535" CreationDate="2011-03-11T17:51:24.933" Score="11" ViewCount="232" Body="&lt;p&gt;Suppose a samurai wants to try out his new sword and cuts a piece of bamboo twice, randomly, so now there are $3$ lenghts of bamboo. What is the probability of these 3 pieces being able to form a triangle?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have never came across a continuous probability problem before, but I tried doing it anyway and got a result of 0.25 probability.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My solution: Let $L$ be the original lenght of the bamboo, $x$ be the place of the first cut and $y$ be the place of the second cut. Writing out all the 3 triangle inequalities, we come to the conclusion that no piece of bamboo can have more than $L/2$ lenght, then the probability we're looking for is:&#xA;$$&#xD;&#xA;\frac{\int_{x=0}^{L/2}(\int_{y=L/2}^{x+L/2}(1)dy)dx}{\int_{x=0}^{L}(\int_{y=x}^L(1)dy)dx}=0.25&#xD;&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="7928" LastActivityDate="2011-03-12T08:10:07.630" Title="A samurai cuts a piece of bamboo" Tags="&lt;probability&gt;" AnswerCount="2" CommentCount="6" FavoriteCount="3" />
  <row Id="26425" PostTypeId="1" AcceptedAnswerId="26434" CreationDate="2011-03-11T17:55:36.177" Score="10" ViewCount="183" Body="&lt;p&gt;&lt;em&gt;Note:This is more a math-recreational question&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the series $\exp_p(x)=\sum_{k=0}^{\infty}\frac{x^k}{(k!)^p}$ which is some systematic modification of the exponential function. It's $\exp_1(x)=\exp(x)  $ if &lt;em&gt;p=1&lt;/em&gt;. I'm interested in the behaviour for $x=0\to -\infty$ depending on change of the parameter $p$.    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;[edit] For the plots, and also for the discussion of the oscillation I removed the constant 1 from the function. Then I edited that question for correction. But after I got some answer I better roll back the edits of the title and the formula and state that removal for the plots and the discussion of the oscillation only here explicitely instead. Please excuse that inconvenience [edit 2]&lt;/em&gt;    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;First oberservation is, that if $p&amp;gt;1$ the function begins to oscillate. If $p=2$ the oscillation diminuishes but for $p&amp;gt;2$ in general the oscillation seems to increase and even for a couple of values $p=2 + \epsilon $ which I checked manually where the (possibly) eventual increase occurs after an initial decrease.     &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So my questions are:    &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;is it true, that at p=2 and x from $ 0\to -\infty$ the amplitude of the oscillation diminueshes to zero?    &lt;/li&gt;&#xA;&lt;li&gt;is it true, that at $p=2+\epsilon, \epsilon&amp;gt;0$ and x from $ 0\to -\infty$ the amplitude of the oscillation eventually increases without bound?&lt;/li&gt;&#xA;&lt;li&gt;How could I approach that question, for instance by considering the form of the power series, the analysis of the powers of the factorials. Are there any &quot;keystones&quot; which might be helpful?    &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;This is a plot for $p=2$ so $f(x)=\exp_2(-(x^2)) -1 $. I used the squaring of the &lt;em&gt;x&lt;/em&gt; to see more of the oscillation. The &lt;em&gt;-1&lt;/em&gt; is to locate the oscillation around the x-axis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/8Lq83.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a plot for $p=2.02$ so $f(x)=\exp_{2.02}(-x^2) -1 $. Again I used the squaring of the &lt;em&gt;x&lt;/em&gt; to see more of the oscillation. Bigger epsilons let the oscillation increase at an &lt;em&gt;x&lt;/em&gt; nearer to the vertical $x=0$-line.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/i5fXy.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1714" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2011-04-28T09:39:18.187" LastActivityDate="2011-04-28T09:39:18.187" Title="Behaviour of the series $\exp_p(x)=\sum_{k=0}^{\infty}\frac{x^k}{(k!)^p}$ depending on $p\approx 2$?" Tags="&lt;sequences-and-series&gt;&lt;special-functions&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="26426" PostTypeId="2" ParentId="26409" CreationDate="2011-03-11T18:03:03.160" Score="1" ViewCount="" Body="&lt;p&gt;The transform is doing exactly what one would theoretically expect it to do. To get a better understanding of this, you might want to look at the result with some intermediate values of $T$ between $0.5$ and $1.2$. At $0.5$, you're getting back the proper top-hat function. As you increase $T$, you will still get a top-hat function, but the edge will move closer to the right-hand edge of the graph as you increase $T$, since the right-hand edge represents the value $1/T$. As $T$ reaches $1$, the edge of the top hat will reach the edge of the graph, and the entire graph will just have a constant value $1$. If you now further increase $T$, you're moving part of the top hat over the edge, into the part of the spectrum that can't be represented by the sampled values, and this part will get aliased. Aliasing works such that it &quot;folds&quot; the spectrum back onto itself, so instead of moving beyond the edge of the graph, that part of the spectrum gets reflected at the edge and added to the right-hand part of the graph. For instance, at $T=1.2$, the part from $1000$ to $1200$ can't be properly represented, is reflected at the $1000$ line and ends up between $800$ and $1000$ instead (with the component at $800$ corresponding to what &quot;should&quot; have been a component at $1200$, the one at $900$ corresponding to $1100$, and so on). That's why you're getting twice the correct value between $800$ and $1000$ -- half of this is the correct spectrum from $800$ to $1000$ and the other half is the aliased spectrum from $1000$ to $1200$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To see why aliasing works in this way, look at what happens to a cosine just past the edge of the spectrum, say, $\cos ((\pi + \epsilon)x)$ sampled at intervals of $1$, i.e. $x\in\mathbb{N}$. Then we have &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;\begin{eqnarray}&#xD;&#xA;\cos ((\pi + \epsilon)x)&#xD;&#xA;&amp;amp;=&amp;amp;&#xD;&#xA;\frac{1}{2}\left(\mathrm{e}^{\mathrm{i}(\pi+\epsilon)x}+\mathrm{e}^{-\mathrm{i}(\pi+\epsilon)x}\right)&#xD;&#xA;\\\&#xD;&#xA;&amp;amp;=&amp;amp;&#xD;&#xA;\frac{1}{2}\left(\mathrm{e}^{-\mathrm{i}(\pi-\epsilon)x}+\mathrm{e}^{\mathrm{i}(\pi-\epsilon)x}\right)&#xD;&#xA;\\\&#xD;&#xA;&amp;amp;=&amp;amp;&#xD;&#xA;\cos ((\pi - \epsilon)x)\;,&#xD;&#xA;\end{eqnarray}&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so the frequency $\pi+\epsilon$ has been reflected at $\pi$, the highest frequency properly representable with this sampling, and folded back to $\pi-\epsilon$. The same would happen with a sine, except that we'd pick up an additional minus sign, but since sinc is an even function, we only have cosines in this case, so the aliased parts of the spectrum are added constructively to the &quot;proper&quot; spectrum.&lt;/p&gt;&#xA;" OwnerUserId="6622" LastActivityDate="2011-03-11T18:03:03.160" />
  <row Id="26427" PostTypeId="1" AcceptedAnswerId="26430" CreationDate="2011-03-11T18:03:38.317" Score="2" ViewCount="71" Body="&lt;p&gt;I have proved the following problem, can you help me check if there is any loopholes in my proof?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let I be an open interval in R, let $c \in I$, and let $f, g\colon I\to \mathbb{R}$ be functions. Suppose that $f(c) = g(c)$, and that $f(x) \leq g(x)$ for all $x \in I$. Prove that if f and g are differentiable at c, then f'(c) = g'(c).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the following is my proof: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $x \in I$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;if $x \lt c$, then $\lim\limits_{x\rightarrow c-}\frac{f(x) &#xD;&#xA;        - f(c)}{x-c}$ = $\lim\limits_{x\rightarrow c-}\frac{f(c) &#xD;&#xA;        - f(x)}{c-x} \geq \lim\limits_{x\rightarrow c-}\frac{g(c) &#xD;&#xA;        - g(x)}{c-x} = \lim\limits_{x\rightarrow c-}\frac{g(x) &#xD;&#xA;        - g(c)}{x-c}$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;if $x \gt c$, then $\lim\limits_{x\rightarrow c+}\frac{f(x) &#xD;&#xA;        - f(c)}{x-c} \leq \lim\limits_{x\rightarrow c+}\frac{g(x) &#xD;&#xA;        - g(c)}{x-c}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $f$ and $g$ are differentiable at $c$, we know that $\lim\limits_{x\rightarrow c-}\frac{f(x) &#xD;&#xA;        - f(c)}{x-c} = \lim\limits_{x\rightarrow c+}\frac{f(x) &#xD;&#xA;        - f(c)}{x-c}=f&amp;#39;(c)$ and $\lim\limits_{x\rightarrow c-}\frac{g(x) &#xD;&#xA;        - g(c)}{x-c} = \lim\limits_{x\rightarrow c+}\frac{g(x) &#xD;&#xA;        - g(c)}{x-c} = g&amp;#39;(c)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus $\lim\limits_{x\rightarrow c-}\frac{f(x) &#xD;&#xA;        - f(c)}{x-c} = \lim\limits_{x\rightarrow c+}\frac{f(x) &#xD;&#xA;        - f(c)}{x-c} = \lim\limits_{x\rightarrow c-}\frac{g(x) &#xD;&#xA;        - g(c)}{x-c} = \lim\limits_{x\rightarrow c+}\frac{g(x) &#xD;&#xA;        - g(c)}{x-c}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore $f&amp;#39;(c) = g&amp;#39;(c)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can't find any problem in my proof, but for some reason I'm not feeling comfortable with it. However, if it's totally correct, just tell me :) Thanks!!&lt;/p&gt;&#xA;" OwnerUserId="7429" LastEditorUserId="2548" LastEditorDisplayName="" LastEditDate="2011-03-11T18:16:31.943" LastActivityDate="2011-03-11T18:17:15.967" Title="proving derivative in real analysis" Tags="&lt;real-analysis&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="26428" PostTypeId="1" AcceptedAnswerId="26498" CreationDate="2011-03-11T18:09:21.917" Score="2" ViewCount="96" Body="&lt;p&gt;Hello,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;G=(v,e) , with weight on the edges than can be only a or b (when $a&#xA;&#xA;&lt;p&gt;I need to find MST of the graph in O(v+e).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think to put all the edges in array, and than scanning the array. first only check about a, and after about b. The algorithm is like Kruskal's: check about evey edge if it doesnt form a cycle. but I'm not sure that this is taking O(v+e). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank u!&lt;/p&gt;&#xA;" OwnerUserId="8069" LastEditorUserId="8069" LastEditorDisplayName="" LastEditDate="2011-03-11T19:20:24.023" LastActivityDate="2011-03-12T02:28:17.283" Title="Graph - MST in O(v+e)" Tags="&lt;homework&gt;&lt;graph-theory&gt;" AnswerCount="1" />
  <row Id="26430" PostTypeId="2" ParentId="26427" CreationDate="2011-03-11T18:17:15.967" Score="1" ViewCount="" Body="&lt;p&gt;The final equality needs better justification; otherwise, it looks fine (modulo a couple of typos). To give a better justification for the final equality, do something like this:&#xA;\begin{align*}&#xA;g'(c) &amp;amp;= \lim_{x\to c}\frac{g(x)-g(c)}{x-c} = \lim_{x\to c-}\frac{g(x)-g(c)}{x-c}\\&#xA;&amp;amp;\leq \lim_{x\to c-}\frac{f(x)-f(c)}{x-c} = \lim_{x\to c}\frac{f(x)-f(c)}{x-c} = f'(c)\\&#xA;&amp;amp;= \lim_{x\to c+}\frac{f(x)-f(c)}{x-c}\\&#xA;&amp;amp;\leq \lim_{x\to c+}\frac{g(x)-g(c)}{x-c} = \lim_{x\to c}\frac{g(x)-g(c)}{x-c} = g'(c).&#xA;\end{align*}&#xA;Thus, we have $g&amp;#39;(c)\leq f&amp;#39;(c)\leq g&amp;#39;(c)$, so equality holds. &lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2011-03-11T18:17:15.967" CommentCount="1" />
  <row Id="26431" PostTypeId="2" ParentId="26419" CreationDate="2011-03-11T18:33:32.427" Score="3" ViewCount="" Body="&lt;p&gt;I don't know the exact answer to the question, but here is a way to get an approximate solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We are maximizing $a_1a_2a_3\ldots$ subject to a constraint $p_1^{a_1}p_2^{a_2}p_3^{a_3}\ldots \leq n$.  Let's treat the $a_i$ as continuous variables that we can differentiate.  In that case we might as well set the constraint $=n$ rather than $\leq n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take logarithms of both products.  Now we are maximizing  $\ln a_1 + \ln a_2 + \ln a_3 \ldots$ subject to $a_1 \ln p_1 + a_2 \ln p_2 + a_3 \ln p_3 \ldots = \ln n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Introduce a Langrange multiplier and set the gradients of the two sums to be proportional to each other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \frac{1}{a_1} \hat{a_1} + \frac{1}{a_2} \hat{a_2} + \frac{1}{a_3} \hat{a_3 }+ \ldots = \lambda (\ln p_1 \hat{a_1} + \ln p_2 \hat{a_2} + \ln p_3 \hat{a_3} + \ldots) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dotting both sides with $\hat{a_i}$ gives&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{1}{a_i} = \lambda \ln p_i$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$a_i = \frac{1}{\lambda \ln p_i}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\lambda$ is then chosen to satisfy the original constraint.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This should specify the maximum $a_i$ treating $a_i$ as continuous.  To get the $a_i$ to be integers you would round some up and some down.  Since you're writing code, you could simply brute-force search this much smaller space of possible integer solutions.&lt;/p&gt;&#xA;" OwnerUserId="3305" LastActivityDate="2011-03-11T18:33:32.427" CommentCount="2" />
  <row Id="26432" PostTypeId="1" AcceptedAnswerId="26458" CreationDate="2011-03-11T18:43:42.870" Score="3" ViewCount="295" Body="&lt;p&gt;Hi all,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While studying the various algorithm implementations available on-line of the Fast Fourier Transform algorithm, I've come to a question related to the way the DFT works in theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose you have a sequence of $N$ points $x_0, ..., x_{N-1}$. For $ k = 0, ..., N-1 $, let $ X_k = \sum_{n=0}^{N-1} x_n e^{-2ik\pi \frac{n}{N}} $.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've noticed that many algorithms are easier to implement, or faster, when the size of the input can be expressed as a power of 2. To pad the signal, I've seen two approaches. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Pad the signal with $0$s, settings $x_N, ..., x_{2^p-1} = 0$, and $X_k = \sum_{n=0}^{N-1} x_n e^{-2ik\pi \frac{n}{2^p}}$&lt;/li&gt;&#xA;&lt;li&gt;Interpolate the original values, by setting $\tau=N/2^p$ the new spacing between consecutive points and then guessing the values at $0, \tau, 2\tau, ..., (2^p-1)\tau$ through linear interpolation.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I've heard people saying different things: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some people oppose the first approach very strongly (I recently had a discussion with a physics teacher about this). They say that padding the signal with extra zeros give you the Fourier coefficients of a different function, which will bear no relation to those of the original signal. On the other hand, they say that interpolation works great. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, most libraries, if not all, that I have reviewed use the second solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All the references that I could find on the internet were pretty vague on this topic. Some say that the best band-limited interpolation that you can do in frequency domain is obtained through time-domain padding, but I couldn't find any proof of such statement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could you please help me figure out which advantages and drawbacks both approaches have? Ideally, I'm searching for something with a mathematical background, not only visual examples =)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="2964" LastActivityDate="2011-03-11T22:20:50.240" Title="Discrete Fourier Transform: Effects of zero-padding compared to time-domain interpolation" Tags="&lt;fourier-analysis&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="2" />
  <row Id="26433" PostTypeId="2" ParentId="26420" CreationDate="2011-03-11T18:43:51.020" Score="3" ViewCount="" Body="&lt;p&gt;Strictly speaking, the cup product is not commutative, though it is commutative up to sign on the level of cohomology.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is an abstract way of seeing this: namely, we can use the method of acyclic models.  Consider the following two functors from the category of spaces to the category of chain complexes. The first is $X \mapsto C_*(X \times X)$; the second is  $X \mapsto C_*(X)\otimes C_*(X)$. Since these are free and acyclic functors on the subset of standard simplices (this means that a) both can be represented as a sum of free abelian groups on sets which are representable functors in $X$, represented by simplices and b) evaluated on a simplex, they lead to acyclic complexes), there is a natural chain equivalence between two&#xA;$$ C_*(X \times X) \simeq C_*(X) \otimes C_*(X)$$&#xA;which itself is unique up to chain homotopy. This is the acyclic model theorem (as in Spanier, for instance).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the category of chain complexes over a commutative ring is not just an abelian category; it is a monoidal category. We can tensor two chain complexes and get a new chain complex. Moreover, it is a &lt;em&gt;symmetric&lt;/em&gt; monoidal category because there is an isomorphism $A_* \otimes B_* \simeq B_* \otimes A_*$ for chain complexes $A, B$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus if we are given one chain equivalence (fixed throughout the following) $C_*(X \times X) \simeq C_*(X) \otimes C_*(X)$, we get another by composing it with the swap map on the latter. These are both natural in $X$ and so, by the &lt;em&gt;uniqueness&lt;/em&gt; (up to chain homotopy) in the acyclic model theorem, we find that the two maps &#xA;$$C_*(X \times X) \rightrightarrows C_*(X) \otimes C_*(X)$$ are naturally chain homotopic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the dual of this means that the two maps&#xA;$$C^*(X) \otimes C^*(X) \rightrightarrows C^*(X \times X)$$ are naturally chain homotopic. On the level of cohomology, the two maps $H^*(X) \otimes H^*(X) \rightrightarrows H^*(X \times X)$$ are thus equal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the map $C_*(X \times X) \to C_*(X) \otimes C_*(X)$ is precisely the homology cross product, and its dual is the cohomology cross product. So we have seen that if we consider the cross-product map $H^*(X) \otimes H^*(X) \to H^*(X \times X)$, it is invariant under switching the two factors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might now object that I said that the cross-product (and thus the cup-product, which is obtained from the cross product by pulling back by the diagonal) is skew-commutative, not commutative. This comes from a feature of how the tensor product of complexes is actually defined: as a result, when you define the swap morphism, you have to introduce a sign (for it to be a chain map).&lt;/p&gt;&#xA;" OwnerUserId="536" LastActivityDate="2011-03-11T18:43:51.020" CommentCount="2" />
  <row Id="26434" PostTypeId="2" ParentId="26425" CreationDate="2011-03-11T18:48:32.933" Score="16" ViewCount="" Body="&lt;p&gt;For p = 2,  $exp_2(x) = J_0(2 \sqrt{x})$, where $J_0$ is a Bessel function.  The asymptotics of this are known: as $x \to -\infty$, $exp_2(x) = \frac {\sin \left( 2\,\sqrt {x}+1/4\,\pi  \right) }{\sqrt {\pi }\, &#xD;&#xA;x^{1/4}} + O(x^{-3/4})$.&lt;/p&gt;&#xA;" OwnerUserId="8508" LastActivityDate="2011-03-11T18:48:32.933" CommentCount="7" />
  <row Id="26435" PostTypeId="2" ParentId="26423" CreationDate="2011-03-11T18:50:36.490" Score="1" ViewCount="" Body="&lt;p&gt;This is wrong.&#xA;$$p^3+q^3 = (p+q) \times (p^2-p q+q^2)$$&#xA;Since $p+q=1$,&#xA;$$p^3+q^3 = p^2-p q+q^2 = (p-q)^2 + p q = (1-2 q)^2 + p q $$&#xA;And what you would expect to be equal to 1 is not unless $p=q=\frac{1}{2}$&#xA;$$\frac{p^3+q^3}{p q} = 1 + \frac{(1-2 q)^2}{p q}$$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Edit to clarify a point: If this is taken from an exercise book, there is a mistake in the claim: see Byron Schmuland's comment for the correction, and Eelvex's comment for the name of the martingale (after correcting the claim).&lt;/p&gt;&#xA;" OwnerUserId="2380" LastEditorUserId="2380" LastEditorDisplayName="" LastEditDate="2011-03-17T09:20:41.207" LastActivityDate="2011-03-17T09:20:41.207" CommentCount="3" />
  <row Id="26436" PostTypeId="1" CreationDate="2011-03-11T18:51:42.277" Score="1" ViewCount="55" Body="&lt;p&gt;Hi there,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've seen questions like this on a lot of old qualifying exams in algebraic topology. The type of question is usually of the form:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $X$, $Y$ be some topological spaces. Prove that $X$ can't be embedded in $Y$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem I have is that I don't see how homology or homotopy or really anything I know helps. For example, in the general case taking e.g. $S^1$ we can embed it into the disk $D^2$. However, the latter is contractible, so all homotopy and homology groups are zero. Thus, knowing the homotopy/homology groups for the larger space does not give us any information about whether or not we can embed something into it. This would suggest that homotopy and homology doesn't provide any general methods as the spaces in this particular example are pretty much as nice as anything can get.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is a somewhat soft question, but I'm curious about which tools in algebraic topology are usually used to tackle questions like this and what are some of the standard tricks involved?&lt;/p&gt;&#xA;" OwnerUserId="8125" LastActivityDate="2011-03-11T18:51:42.277" Title="Using algebraic topology to disprove existence of embeddings" Tags="&lt;algebraic-topology&gt;" CommentCount="8" FavoriteCount="1" />
  <row Id="26437" PostTypeId="1" AcceptedAnswerId="26453" CreationDate="2011-03-11T18:53:15.240" Score="3" ViewCount="96" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Let $u=p_1^{n_1}...p_t^{n_t}$ for $p_i$ prime and $n_i \in N$. Let $m_i = \frac{u}{p_i^{n_i}}$. Then there exist $c_i \in Z$ such that $c_1m_1 + ... + c_tm_t=1$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is given as a property we are allowed to use in one of our homework questions, but we're not sure why we are allowed to use it. We know it's true for $m_i$ that are relatively prime, but these $m_i$ are specifically &lt;em&gt;not&lt;/em&gt;. Can you explain why it still holds?&lt;/p&gt;&#xA;" OwnerUserId="7840" LastActivityDate="2011-03-11T20:26:38.430" Title="Integer Multiples of Integers Sum to 1" Tags="&lt;homework&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="26438" PostTypeId="1" CreationDate="2011-03-11T18:55:22.033" Score="12" ViewCount="300" Body="&lt;p&gt;EDIT: I found a brief discussion of this in Husemoller's Fibre Bundles, chapter 16 section 12.  Here to compute $\tilde K(\mathbb R P^{2n+1})$ he says to consider the map&#xA;$$&#xD;&#xA;\mathbb R P^{2n+1} = S^{2n+1}/\pm 1\to \mathbb C P^n = S^{2n+1}/U(1).&#xD;&#xA;$$&#xA;Under this map the canonical line bundle over $\mathbb C P^n$ pulls back to the complexification of the canonical line bundle over $\mathbb R P^{2n+1}$.  Then he says from looking at the (Atiyah-Hirzebruch) spectral sequence we get that $\tilde K(\mathbb R P^{2n+1}) = \mathbb Z/2^n$.  I don't see how looking at the spectral sequence helps (and what we learn from the map from $\tilde K(\mathbb C P^{n}) \to \tilde K(\mathbb R P^{2n+1})$).  All I can see is that $\tilde K(\mathbb R P^{2n+1})$ is pure torsion (by the Chern character isomorphism) and that it has order $2^n$ (from the spectral sequence).  I can't see why, for example, it isn't just a direct sum of $\mathbb Z/2$'s.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I found a homework assignment online from an old K-theory course and one of the problems says to compute $K(\mathbb R P^n)$ by using a suitable comparison map with $\mathbb C P^k$ and knowledge of $K(\mathbb C P^k)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have attempted this but have not been able to get anywhere.  The only map $\mathbb R P^n \to \mathbb C P^n$ I can think of is the one sending the equivalence class of $(x_0,\ldots,x_n) \in \mathbb R P^n$ to its equivalence class in $\mathbb C P^n$.  Under this (I think) the tautological line bundle over $\mathbb C P^n$ (which generates $K(\mathbb C P^n)$) gets sent to the complexification of the tautological line bundle over $\mathbb R P^n$.  But I really don't see where to go from here; if I had a map going the other way maybe I'd be able to say something but the map I have is neither injective nor surjective.  I also can't see how torsion is going to come out of this: $K(\mathbb C P^n)$ is torsionfree but $K(\mathbb R P^n)$ isn't.&lt;/p&gt;&#xA;" OwnerUserId="9" LastEditorUserId="1543" LastEditorDisplayName="" LastEditDate="2011-04-23T11:45:01.757" LastActivityDate="2011-04-23T11:45:01.757" Title="$K(\mathbb R P^n)$ from $K(\mathbb C P^k)$" Tags="&lt;algebraic-topology&gt;&lt;topological-k-theory&gt;" AnswerCount="3" CommentCount="6" FavoriteCount="1" />
  <row Id="26440" PostTypeId="2" ParentId="26437" CreationDate="2011-03-11T18:59:06.647" Score="1" ViewCount="" Body="&lt;p&gt;Hi!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you need to prove is if $gcd(a_{1},...,a_{i})=1$, then there exists $c_{i}$ such that $\sum c_{i}a_{i}=1$. You may do it via induction. &lt;/p&gt;&#xA;" OwnerUserId="7887" LastActivityDate="2011-03-11T18:59:06.647" />
  <row Id="26441" PostTypeId="2" ParentId="26419" CreationDate="2011-03-11T19:00:11.177" Score="2" ViewCount="" Body="&lt;p&gt;This answer might be useful if you are loking for an actual algorithm to solve the problem (and not a thoretical approach). &#xA;If $x$ is an integer in the range $[1,n]$ with maximum number of divisors write &#xA;  $$ x = q_1^{a_1} q_2^{a_2} \dots q_k^{a_k} $$&#xA;with the $q_i$ distinct primes and &#xA;  $$ a_1 \ge a_2 \ge \cdots \ge a_k. \quad(*)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then we see that the integer&#xA;  $$ x&amp;#39; = 2^{a_1} 3^{a_2} \dots p_k^{a_k} \quad (**)$$&#xA;has the same number of divisors and is $\le x$. So we can limit our search to the integers of the form $(**)$ which verify $(*)$. This makes the search really easy, and fast in small ranges: start with $(a_1,\dots,a_k) = (t,0,\dots,0)$ and $k$ such that $2\cdot3\cdot\dots\cdot p_{k+1} &amp;gt; n$, and $2^t \le n &amp;lt; 2^{t+1}$ and go through all the $k$-tuples verifying $(*)$ and $(**)$ in decreasing lexicographic order, keeping their product maximal and $&amp;lt;n$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can find a C++ program that performs this task in &lt;a href=&quot;http://forums.topcoder.com/?module=Thread&amp;amp;threadID=594206&amp;amp;start=0&amp;amp;mc=25#892980&quot; rel=&quot;nofollow&quot;&gt;this thread&lt;/a&gt; at TopCoder, though you need to adapt it to arrive up to $2^{64}$, since it is limited to $n\le 2^{64}/47$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: I have written a PARI-GP version which can arrive much further, for $n=2^{64}$ the probably non-unique integer with maximum number of divisors is:&#xA;$$ 18401055938125660800 = 2^7\cdot 3^4\cdot 5^2\cdot 7^2\cdot 11 \cdot 13 \cdot 17 \cdot 19 \cdot 23 \cdot 29 \cdot 31 \cdot 37 \cdot 41 $$ with 184320 divisors.&lt;/p&gt;&#xA;" OwnerUserId="3274" LastEditorUserId="3274" LastEditorDisplayName="" LastEditDate="2011-03-11T19:52:04.867" LastActivityDate="2011-03-11T19:52:04.867" CommentCount="2" />
  <row Id="26442" PostTypeId="2" ParentId="26419" CreationDate="2011-03-11T19:01:57.590" Score="5" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://oeis.org/A002182&quot; rel=&quot;nofollow&quot;&gt;Sequence A002182&lt;/a&gt; gives the highly composite numbers, where the number of divisors sets a record.  You want the largest under 2^64.  Unfortunately the table doesn't go that high, but the references do.  Somewhat in the spirit of Mark Eicenlaub's answer, if you think of $n = p_1^{a_1} \times p_2^{a_2} \times ... \times p_k^{a_k}.$ and $d(n)=(a_1 + 1) \times (a_2 + 1) \times ... \times (a_k + 1)$ and think of adding one to the exponent of $p_i$, $\log (n)$ increases by $\log (p_i)$ and $\log (d(n))$ increases by $\log(a_i+2)-\log(a_i+1)$.  So the figure of merit for an increase is $\frac{\log(a_i+2)-\log(a_i+1)}{\log (p_i)}$ and you can just look through the primes to find which you should add in.  This will miss some, where taking out one and adding another gets you a step up.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2011-03-11T21:38:39.760" LastActivityDate="2011-03-11T21:38:39.760" CommentCount="7" />
  <row Id="26443" PostTypeId="2" ParentId="26407" CreationDate="2011-03-11T19:03:59.553" Score="1" ViewCount="" Body="&lt;p&gt;Let us consider $f\,:\,\left[0;+\infty\right[\rightarrow\left[0;1\right[$ defined as:&#xA;$$f = \frac{2}{\pi} \arctan$$&lt;/p&gt;&#xA;" OwnerUserId="2380" LastActivityDate="2011-03-11T19:03:59.553" />
  <row Id="26444" PostTypeId="1" AcceptedAnswerId="26448" CreationDate="2011-03-11T19:25:16.443" Score="4" ViewCount="164" Body="&lt;p&gt;I have been trying this &lt;em&gt;notoriously difficult problem&lt;/em&gt; for quite some time but i haven't made any progress. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $\mathscr{I}(V)$ denote the set of all homomorphisms $f : V \to V$. That is $\mathscr{I}(V) = \text{Hom}(V,V)$.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Suppose $V$ is a vector space over a field $K$ and $\text{dim}_{K}(V)&amp;gt;1$, then prove that $\mathscr{I}(V)$ has no &lt;em&gt;proper two sided ideals&lt;/em&gt;. This means we have to show that $\mathscr{I}(V)$ has no two sided ideals other than $(0)$ and $\mathscr{I}(V)$. Next, does the conclusion of the above problem remain true if $V$ is &lt;em&gt;infinite dimensional&lt;/em&gt;.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Moreover, since Field's are the ubiquitous algebraic structures to have trivial ideals i tried thinking of proving $\mathscr{I}(V)$ to be a field. But i haven't yet taken a single step forward in terms of this problem.&lt;/p&gt;&#xA;" LastActivityDate="2011-03-13T13:46:52.633" Title="Ideals of $\mathscr{I}(V) = \text{Hom}(V,V)$." Tags="&lt;linear-algebra&gt;&lt;algebra&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="26445" PostTypeId="1" AcceptedAnswerId="26452" CreationDate="2011-03-11T19:36:26.310" Score="4" ViewCount="219" Body="&lt;p&gt;I thought it was elementary to me, but I started to do some exercises and came up some definitions I have sort of difficulty to distinguish. In parentheses  are my questions.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;$x/0$ is Impossible (  If it's impossible it can't have neither infinite solutions or even one. Nevertheless, both 1. and 2. are divided by zero, but only 2. has infinite solutions so as 1. has none solution, how and why ?)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$0/0$ is Undefined and has infinite solutions. (How come one be Undefined and yet has infinite solutions ?)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$0/x$ it's okay for me, no problem, but if someone else wants to add something about it, feel free to do it.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance.&lt;/p&gt;&#xA;" OwnerUserId="8127" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-11T19:54:35.900" LastActivityDate="2011-03-11T21:18:00.523" Title="Division by $0$" Tags="&lt;algebra&gt;&lt;algebra-precalculus&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="26446" PostTypeId="1" AcceptedAnswerId="26447" CreationDate="2011-03-11T19:37:22.783" Score="0" ViewCount="56" Body="&lt;p&gt;I have a bipartite graph and I want to partition each of the two sets of nodes (I don't know the technical term for these sets I mean each part of the bipartite graph) separately.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each node hase a multidimensional location (x1, x2, x3, xn) as well I want to consider the edges on this graph in the clustering (if two nodes have the same distance in X from a third node the one that shares more connections to the nods connecting to third node is nearer to the third node).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a method that is most appropriate to this situation. I have done lots of different clustering methods on points located in a continuous space but I can't see how I might apply similar methods to this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any comments appreciated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance.&lt;/p&gt;&#xA;" OwnerUserId="5875" LastActivityDate="2011-03-11T19:41:59.930" Title="Bipartite graph partitioning" Tags="&lt;graph&gt;&lt;clustering&gt;" AnswerCount="1" />
  <row Id="26447" PostTypeId="2" ParentId="26446" CreationDate="2011-03-11T19:41:59.930" Score="1" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://www.osti.gov/bridge/servlets/purl/816202-TkwqEq/native/&quot; rel=&quot;nofollow&quot;&gt;This&lt;/a&gt; should be helpful (article about how to partition a bipartite graph into partite sets) will help. I think you want to find the partite sets.&lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2011-03-11T19:41:59.930" CommentCount="1" />
  <row Id="26448" PostTypeId="2" ParentId="26444" CreationDate="2011-03-11T19:45:00.850" Score="4" ViewCount="" Body="&lt;p&gt;The first part, for finite dimensional spaces, amounts to showing that the ring of $n\times n$ matrices with coefficients in $K$ is simple. &lt;a href=&quot;http://math.stackexchange.com/questions/22629/why-is-the-ring-of-matrices-over-a-field-simple/22631#22631&quot;&gt;This is covered in this previous question and answer.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For infinite dimensional spaces, I have a two word hint: &lt;strong&gt;finite support.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are incorrect on your final paragraph: $\mathscr{I}(V)$ cannot be a field for $n\gt 1$, since it always nontrivial idempotents (projections onto nontrivial proper subspaces), nonzero nilpotent elements (the left-shift operator on a basis), and zero divisors (any non-invertible map is a zero divisor: compose with an appropriate projection onto a complement of the image). &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-11T21:18:53.703" LastActivityDate="2011-03-11T21:18:53.703" />
  <row Id="26449" PostTypeId="2" ParentId="26445" CreationDate="2011-03-11T19:49:43.867" Score="5" ViewCount="" Body="&lt;p&gt;The key is to realize what a fraction $\frac ab$ really represents: $\frac ab$ is the number with the property that $\frac ab \cdot b = a$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, in the first case if $x \ne 0$, there is no number $\frac x0$ with the property that $\frac x0 \cdot 0 = x$ since anything times zero is zero. So $\frac x0$ is undefined. In the second, any number $y$ has the property that $y \cdot 0 = 0$, so $\frac 00$ could represent any number $y$ according to the above characterization of a fraction, so $\frac 00$ is said to be indeterminate. Finally, if $x \ne 0$ then $0$ has the property that $0 \cdot x = 0$, so $\frac 0x$ is the number $0$.&lt;/p&gt;&#xA;" OwnerUserId="1914" LastActivityDate="2011-03-11T19:49:43.867" CommentCount="1" />
  <row Id="26450" PostTypeId="2" ParentId="26328" CreationDate="2011-03-11T19:50:50.760" Score="2" ViewCount="" Body="&lt;p&gt;We are given a chain complex: $ \ldots \to C_n \stackrel{\partial_n}{\to} \ldots \to C_1 \stackrel{\partial_1}{\to} C_0 \stackrel{\partial_0}{\to} \ldots$ where $C_0 = C_1 = A$, $\partial_0$ is the zero map and $\partial_1$ is multiplication by $n$. Also $C_n = 0$ for $n \neq 0,1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $C_n = 0$ for $n \neq 0,1$, we must have $\mathrm{Ker}\partial_n = 0$ for $n \neq 0,1$ (because $\mathrm{Ker}\partial_n\leq C_n = 0$). So $H_n = 0$ for $n \neq 0,1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next observe that, because $C_2 = 0$, we must have $\mathrm{Im}\partial_2 = 0$. So $H_1 = \mathrm{Ker}\partial_1/\mathrm{Im}\partial_2 \cong \mathrm{Ker}\partial_1 = \{a \in A | na = 0\}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, $H_0 = \mathrm{Ker}\partial_0/\mathrm{Im}\partial_1$. Since $\partial_0: A \to A$ is the zero map, $\mathrm{Ker}\partial_0 = A$. Also since $\partial_1: A \to A$ is multiplication by $n$, $\mathrm{Im}\partial_1 = nA$. So $H_0 = A/nA$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We conclude that for the given Chain complex:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$H_0 = A/nA$, $H_1 = \{a \in A | na =0\}$ and $H_n = 0$ otherwise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The issue you are describing (with $\mathbb{Z}/4\mathbb{Z}$, and $n=3$) is not really a problem. The above calculation is for an &lt;em&gt;arbitrary&lt;/em&gt; abelian group $A$ and integer $n$. You should think of the final answer as a formula of sorts: once you input a choice of abelian group $A$ and integer $n$, you are able to quickly determine the homology without having to go through the lengthy argument every time. However, just as is the case with functions, if you change your input (in this case the pair $(A,n)$) then the value of the function (in this case the Homology of the Chain complex) will also change.&lt;/p&gt;&#xA;" OwnerUserId="7493" LastEditorUserId="7493" LastEditorDisplayName="" LastEditDate="2011-03-11T22:33:06.547" LastActivityDate="2011-03-11T22:33:06.547" CommentCount="1" />
  <row Id="26451" PostTypeId="2" ParentId="23153" CreationDate="2011-03-11T19:52:52.693" Score="0" ViewCount="" Body="&lt;p&gt;Although I don't understand the specifics of your problem, let me try to &#xA;help you with a more generic argument in line with the title of your question. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In contrast to the idea of studying ODE's taking the &#xA;time-one flow (&quot;discrete-izing&quot;), the idea of studying the &quot;continuous time &#xA;analog&quot; is, &quot;usually&quot;, not very fruitful despite how attractive it is. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To sort of exemplify my comment, there is the&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Collatz_conjecture&quot; rel=&quot;nofollow&quot;&gt; Collatz conjecture &lt;/a&gt;.&#xA;I'm aware of some tentatives of considering the &quot;continuous time &#xA;analog&quot;, and some have radically different dynamics (but are relevant &#xA;since provide other ways and equivalences to study the conjecture). &#xA;Let's not go off topic, though: we're talking homeomorphisms. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bottom line is that the dynamics of homeomorphisms is much richer &#xA;that the dynamics of flows: the key feature (perhaps) is that orbits of &#xA;(autonomous) flows cannot cross. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, this feels more like your problem:&#xA;Perhaps the orbits minimizing costs in infinite steps were &#xA;&quot;crossing&quot;, in some sense, but you destroyed these solutions &#xA;by making your problem smooth. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again: take my intuition with a grain of salt, but I sincerely hope that this helped somehow.&lt;/p&gt;&#xA;" OwnerUserId="8126" LastActivityDate="2011-03-11T19:52:52.693" />
  <row Id="26452" PostTypeId="2" ParentId="26445" CreationDate="2011-03-11T19:54:17.240" Score="20" ViewCount="" Body="&lt;p&gt;The first question you need to ask is: What does &quot;$a/b$&quot; mean?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer is: &quot;$a/b$ is the &lt;strong&gt;unique&lt;/strong&gt; solution to the equation $bz = a$.&quot; (I'm using $z$ as the unknown, since you are using $x$ for other things).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given that answer, let's discuss your points out of order:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(3) is perfectly fine: $0/x$, with $x\neq 0$, is the solution to $xz = 0$; the unique solution is $z=0$, so $0/x = z$. The reason it's &lt;em&gt;unique&lt;/em&gt; is because $x\neq 0$, so the only way for the product to be $0$ is if $z$ is $0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In (1), by &quot;impossible&quot; we mean that the equation that defines it has no solutions: for something to be equal to $x/0$, with $x\neq 0$, we would need $0z = x$. But $0z=0$ for any $z$, so there are &lt;em&gt;no&lt;/em&gt; solutions to the equation. Since there are no solutions to the equation, there is no such thing as &quot;$x/0$&quot;. So $x/0$ does &lt;em&gt;not&lt;/em&gt; represent any number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In (2), the situation is a bit trickier; in terms of the defining equation, the problem here is that the equation $0z=0$ has &lt;em&gt;any&lt;/em&gt; value of $z$ as a solution (that's what the &quot;infinite solutions&quot; means). Since the expression $a/b$ means &quot;the &lt;strong&gt;unique&lt;/strong&gt; solution to $bx= a$, then when $a=b=0$, you don't have a unique answer, so there is no &quot;unique solution&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Generally speaking, we simply do not define &quot;division by $0$&quot;. The issue is that, once you get to calculus, you are going to find situations where you have two &lt;em&gt;variable&lt;/em&gt; quantities, $a$ and $b$, and you are considering $a/b$; and as $a$ and $b$ changes, you want to know what happens to $a/b$. In those situations, if $a$ is approaching $x$ and $b$ is approaching $y\neq 0$, then $a/b$ will approach $x/y$, no problem. If $a$ approaches $x\neq 0$, and $b$ approaches $0$, then $a/b$ does not approach &lt;em&gt;anything&lt;/em&gt; (the &quot;limits does not exist&quot;). But if &lt;em&gt;both&lt;/em&gt; $a$ and $b$ approach $0$, then you don't know what happens to $a/b$; it can exist, not exist, or approach pretty much any number. We say this kind of limit is &quot;indeterminate&quot;. So there is a reason for separating out cases (1) and (2): very soon you will see an important qualitative difference between the first kind of &quot;does not exist&quot; and the second kind.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-11T21:18:00.523" LastActivityDate="2011-03-11T21:18:00.523" />
  <row Id="26453" PostTypeId="2" ParentId="26437" CreationDate="2011-03-11T20:05:30.353" Score="2" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\: \ \rm\ \{ c_1\ m_1 +\:\cdots\:+c_t\ m_t\ :\ c_i \in \mathbb Z\}\ $ is closed under subtraction hence, by the lemma below, its least positive element $\rm\:d\:$ is a common divisor of all the $\rm\:m_i\:.\:$ Necessarily $\rm\:d = 1\:,\:$ since $\rm\ d\:|\:m_i\ $ and $\rm\:m_i\:$ is coprime to $\rm\:p_i\:,\ $ so $\rm\:d\:$ is a divisor of $\rm\ p_1^{n_{\ 1}}\:\cdots\:p_t^{n_{\ t}}\ $ coprime to all its prime factors,$\:$ thus $\rm\ d = 1\:$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;NOTE&lt;/strong&gt; $\ $ Furthermore, $\rm\ d\ = gcd(m_1,\:\cdots\:,m_t)\:$ since, being an integer linear combination of the $\rm\:m_i\:$ it is divisible by every common divisor $\rm\:c\:$ of all of the $\rm\ m_i\:,\ $  $\rm\ c\:|\:m_i\ \Rightarrow\ c\:\ |\ c_1\ m +\:\cdots\:+c_t\ m_t\  =\ d\:.\ $ So $\rm\:d\:$ is the &lt;em&gt;greatest&lt;/em&gt; common divisor, being a common divisor divisible by every common divisor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;LEMMA&lt;/strong&gt; $\: $ Suppose $\;\rm D\subset\mathbb Z \;$ is closed under subtraction and that $\rm D$ contains a nonzero element.&lt;br&gt;&#xA;Then $\rm\: D \: $ has a positive element and the least positive element of $\rm\: D\: $ divides every element of $\rm\: D\:$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof&lt;/strong&gt; $\ $ First $\rm D$ has a positive element since $\rm\: \ 0 \ne d\in D \ \Rightarrow\ d-d = 0\in D\ \Rightarrow\ 0-d = -d\in D\:.\ $ Let $\rm d$ be the least positive element in $\rm D$. Since $\rm d|n \iff d\;|-n, \;$ if $\rm c\in D$ is not divisible by $\rm d$ then we &#xA;may assume that $\rm c$ is positive, and the least such element. But $\rm c-d$ is a positive element of $\rm D$ not divisible by $\rm d$ &#xA;and smaller than $\rm c$, contra leastness of $\rm c$. So $\rm d$ divides every element of $\rm D\:.\ $ &lt;strong&gt;QED&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;REMARK&lt;/strong&gt; $\ $ Alternatively, if you already have available the $\rm\:CRT\:$ (Chinese Remainder Theorem), then simply apply it to solve the system $\rm\ x \equiv 1\ \ (mod\ p_i^{n_{\ i}})\:.\ $ It gives $\rm\ c_i \equiv\ m_{\:i}^{-1}\ \ (mod\ p_i^{n_{\ i}})\:.$&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2011-03-11T20:26:38.430" LastActivityDate="2011-03-11T20:26:38.430" />
  <row Id="26454" PostTypeId="2" ParentId="26432" CreationDate="2011-03-11T20:58:28.393" Score="1" ViewCount="" Body="&lt;p&gt;I'll try and give an intuitive answer that can be made mathematically precise with some careful analysis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, imagine someone gives you the DFT of a function, and it turns out to be a constant everywhere (i.e., for $k = 0,1,2, \ldots N-1$). Which function in the time domain has such a DFT? The answer is a &quot;delta function&quot;, i.e., $f(n) = C$ for $n=0$ and $f(n)= 0$ for $n \neq 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These arguments work when you reverse domains, so if you have a constant function in the time domain, you will have a &quot;delta&quot; function in the frequency domain. If you decide to pad you time domain sequence with zeros and then take DFT, you will get a sinc function instead of a delta function. Therefore, strictly speaking, by padding with zeros, you are distorting the DFT of the function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, there is another way of looking at this. Your padding by zeroes operation is essentially the same as taking an extended set of points of the original time series and multiplying it by a rectangular function (whose non zero region is limited to the points $x_0,\ldots x_{N-1}$). Therefore, in the frequency domain your original DFT is convolved by a sinc function. Ideally, you want the DFT of the original signal to be convolved with something that is close to a delta function (as this doesn't distort anything). In our case, this means that we should at least ensure that the sinc has a very sharp main lobe and side lobes that decay very fast (to approximate a delta function).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If your sinc function has a very spread out main lobe, your DFT will be highly distorted whereas if the main lobe is sharply peaked, the distortion is very small. In other words, if you tried to pad your original time series by a large number of zeros, the effect of the convolution is much more pronounced because the main lobe of the sinc will be very spread out. If you pad it with a small number of zeros, you are very close to convolving the DFT with a delta function, and distortion is small.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regarding interpolation, I guess that method could also work without giving much distortion, but the correct approach would be to use an interpolation filter instead of doing simple linear interpolation (as explained in &lt;a href=&quot;http://en.wikipedia.org/wiki/Upsampling&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Upsampling&lt;/a&gt; ). In practice, I'm not sure there will be a significant difference when dealing with well behaved signals that are properly sampled.&lt;/p&gt;&#xA;" OwnerUserId="2641" LastEditorUserId="2641" LastEditorDisplayName="" LastEditDate="2011-03-11T22:20:50.240" LastActivityDate="2011-03-11T22:20:50.240" CommentCount="1" />
  <row Id="26455" PostTypeId="1" AcceptedAnswerId="26457" CreationDate="2011-03-11T20:59:27.933" Score="2" ViewCount="108" Body="&lt;p&gt;I have a cyclotomic field $\mathbb{Q}(\zeta_8)$, and want to know how I can find a minimal polynomial of this element $\zeta_8-2*\zeta_8^3$. Can we generalize this to any number field?&lt;/p&gt;&#xA;" OwnerUserId="8129" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-11T21:01:08.540" LastActivityDate="2011-03-11T22:53:48.367" Title="Minimal polynomial" Tags="&lt;algebraic-number-theory&gt;" AnswerCount="2" CommentCount="5" />
  <row Id="26456" PostTypeId="1" AcceptedAnswerId="26472" CreationDate="2011-03-11T21:11:15.917" Score="2" ViewCount="120" Body="&lt;p&gt;Hi all,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Question 2.23 out of Boyd and Vanderberghe:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Give an example of two closed convex sets that are disjoint but cannot be strictly separated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The obvious idea is to take something like unbounded sets which are disjoint but approach each other in the limit. For example, f(x) = 1/x and g(x) = -1/x. But isn't x=0 a strictly separating hyperplane here?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="8130" LastActivityDate="2011-03-12T14:41:37.557" Title="Easy convex sets question" Tags="&lt;convex-analysis&gt;&lt;convex-optimization&gt;&lt;convex-sets&gt;" AnswerCount="3" FavoriteCount="0" />
  <row Id="26457" PostTypeId="2" ParentId="26455" CreationDate="2011-03-11T21:16:50.960" Score="3" ViewCount="" Body="&lt;p&gt;If you know the automorphisms of $\mathbb{Q}(\zeta_8)$, then the simplest way is to use them: if $\sigma_1,\ldots,\sigma_4$ are the four automorphisms of $\mathbb{Q}(\zeta_8)$ over $\mathbb{Q}$ (the degree of the $n$th cyclotomic field is $\varphi(n)$, and $\varphi(8) = 4$), then $\zeta_8 - 2\zeta_8^3$ certainly satisfies&#xA;$$\prod_{i=1}^4\left( x - \sigma_i(\zeta_8 - 2\zeta_8^3)\right) = \prod_{i=1}^4\left(x - \sigma_i(\zeta_8) - 2\sigma_i(\zeta_8)^3\right).$$&#xA;So the minimal polynomial will necessarily divide this degree $4$ polynomial; it is not linear, so it is either &lt;em&gt;this&lt;/em&gt; polynomial, or is an irreducible quadratic factor (this easily generalizes). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The four automorphisms are the maps $\zeta_8\mapsto\zeta_8$, $\zeta_8\mapsto\zeta_8^3$, $\zeta_8\mapsto\zeta_8^5$, and $\zeta_8\mapsto\zeta_8^7$. You can then compute the polynomial explicitly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively, since $\zeta_8$ is an algebraic integer, you can use the &quot;determinant method&quot;, outlined in Daniel Marcus's &lt;strong&gt;Number Fields&lt;/strong&gt; (Theorem 2, page 15):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that both $\mathbb{Z}[\zeta_8-2\zeta_8]$ is contained in a finitely generated $\mathbb{Z}$-module. Let $a_1,\ldots,a_n$ generate this module.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let $\alpha$ be the algebraic integer you want: $\alpha=\zeta_8 - 2\zeta_8^3$; for each $i$, $1\leq i\leq n$, express $\alpha a_i$ as a linear combination of the generators $a_i$. You obtain a matrix equation of the form&#xA;$$\left(\begin{array}{c}&#xD;&#xA;\alpha a_1\\&#xD;&#xA;\alpha a_2\\&#xD;&#xA;\vdots\\&#xD;&#xA;\alpha a_n\end{array}\right) = M\left(\begin{array}{c}&#xD;&#xA;a_1\\&#xD;&#xA;a_2\\&#xD;&#xA;\vdots\\&#xD;&#xA;a_n&#xD;&#xA;\end{array}\right)$$&#xA;where $M$ is an $n\times n$ matrix with coefficients in $\mathbb{Z}$. This means that $\det(\alpha I - M) = 0$, since the equation $(\alpha I - M)\mathbf{x}=\mathbf{0}$ has nontrivial solutions. Expressing this determinant in terms of the $n^2$ coefficients of $\alpha I - M$ you get an expression of the form $\alpha^n + \text{(lower degree terms)} = 0$, giving a monic polynomial with integer coefficients which has $\alpha$ as a root. Factoring it will yield the minimal polynomial. (This last step may be, of course, somewhat difficult in practice).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added.&lt;/strong&gt; I gave the above answers because you implied you want to know general methods. In this case, though, we can proceed explicitly. (Of course, since $\zeta_8 = \frac{\sqrt{2}}{2} + i\frac{\sqrt{2}}{2}$, we know that $\zeta_8 - 2\zeta_8^3=\frac{3\sqrt{2}}{2} - i\frac{\sqrt{2}}{2}$ has degree $4$ over $\mathbb{Q}$, so it is necessarily the monic polynomial of degree $4$ that you found above; we know it's degree $4$ because if it were degree $2$ it would lie in $\mathbb{Q}(\sqrt{d})$ for some square free integer $d$, and you would need both $\sqrt{2}$ and $\sqrt{-2}$ to be in this quadratic field, which is impossible).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $\mathbb{Q}(\zeta_8) = \mathbb{Q}(\sqrt{2},\sqrt{-2}) = \mathbb{Q}(\sqrt{2},i)$, the four automorphisms are the identity, complex conjugation, the map that sends $\sqrt{2}$ to $-\sqrt{2}$ and $i$ to $i$; and the map that sends $\sqrt{2}$ to $-\sqrt{2}$ and $i$ to $-i$. So the minimal polynomial is:&#xA;\begin{align*}&#xA;p(x) &amp;amp;= \left( x- \left(\frac{3\sqrt{2}}{2}-i\frac{\sqrt{2}}{2}\right)\right)&#xA;\left( x- \left(\frac{3\sqrt{2}}{2}+i\frac{\sqrt{2}}{2}\right)\right)\\&#xA;&amp;amp;\qquad\qquad \times &#xA;\left( x- \left(-\frac{3\sqrt{2}}{2}-i\frac{\sqrt{2}}{2}\right)\right)&#xA;\left( x- \left(-\frac{3\sqrt{2}}{2}+i\frac{\sqrt{2}}{2}\right)\right)\\&#xA;&amp;amp;= \left( \left(x-\frac{3\sqrt{2}}{2}\right)^2 + \left(\frac{\sqrt{2}}{2}\right)^2\right)\left( \left(x+\frac{3\sqrt{2}}{2}\right)^2 + \left(\frac{\sqrt{2}}{2}\right)^2\right)\\&#xA;&amp;amp;= \left( x^2 -3\sqrt{2}x + 5\right)(x^2+3\sqrt{2}x+5)\\&#xA;&amp;amp;= (x^2+5)^2 - \left(3\sqrt{2}x\right)^2\\&#xA;&amp;amp;= x^4 + 10x^2 + 25 -18x^2\\&#xA;&amp;amp;= x^4 - 8x^2 + 25.&#xA;\end{align*}&#xA;(Unless I made some silly arithmetical error, anyway).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To see the second method in action, we can take $1,\zeta_8, \zeta_8^2,\zeta_8^3$ as our generating set. Then we want to express the products of $\alpha$ with these four elements in terms of these four elements (where $\alpha = \zeta_8 - 2\zeta_8^3$). Noting that $\zeta_8$ is a root of $x^4+1$, we have&#xA;\begin{align*}&#xA;\alpha 1 &amp;amp;= \zeta_8 - 2\zeta_8^3\\&#xA;\alpha \zeta_8 &amp;amp;= 2 + \zeta_8^2\\&#xA;\alpha \zeta_8^2 &amp;amp;= 2\zeta_8 + \zeta_8^3\\&#xA;\alpha\zeta_8^3 &amp;amp;= -1+2\zeta_8^2.&#xA;\end{align*}&#xA;So the matrix $M$ in question is:&#xA;$$M = \left(\begin{array}{rrrr}&#xD;&#xA;0 &amp;amp; 1 &amp;amp; 0 &amp;amp; -2\\&#xD;&#xA;2 &amp;amp; 0 &amp;amp; 1 &amp;amp; 0\\&#xD;&#xA;0 &amp;amp; 2 &amp;amp; 0 &amp;amp; 1\\&#xD;&#xA;-1 &amp;amp; 0 &amp;amp; 2 &amp;amp; 0&#xD;&#xA;\end{array}\right).$$&#xA;Then we compute the determinant of $\alpha I - M$. We have:&#xA;\begin{align*}&#xA;\det(\alpha I - M) &amp;amp;= \left|\begin{array}{rrrr}&#xA;\alpha &amp;amp; -1 &amp;amp; 0 &amp;amp; 2\\&#xA;-2 &amp;amp; \alpha &amp;amp; -1 &amp;amp; 0\\&#xA;0 &amp;amp; -2 &amp;amp; \alpha &amp;amp; -1\\&#xA;1 &amp;amp; 0 &amp;amp; -2 &amp;amp; \alpha&#xA;\end{array}\right|\\&#xA;&amp;amp;= \alpha\left|\begin{array}{rrr}&#xA;\alpha &amp;amp; -1 &amp;amp; 0\\&#xA;-2 &amp;amp; \alpha &amp;amp; -1\\&#xA;0 &amp;amp; -2 &amp;amp; \alpha&#xA;\end{array}\right| + 2 \left|\begin{array}{rrr}&#xA;-1 &amp;amp; 0 &amp;amp; 2\\&#xA;-2 &amp;amp; \alpha &amp;amp; -1\\&#xA;0 &amp;amp; -2 &amp;amp; \alpha&#xA;\end{array}\right| - \left|\begin{array}{rrr}&#xA;-1 &amp;amp; 0 &amp;amp; 2\\&#xA;\alpha &amp;amp; -1 &amp;amp; 0\\&#xA;-2 &amp;amp; \alpha &amp;amp; -1&#xA;\end{array}\right|\\&#xA;&amp;amp;= \alpha(\alpha^3 - 2\alpha - 2\alpha) +2(-\alpha^2 + 8 + 2) - (-1 + 2\alpha^2 - 4)\\&#xA;&amp;amp;= \alpha^4 - 4\alpha^2 - 2\alpha^2 + 20 + 5 - 2\alpha^2\\&#xA;&amp;amp;= \alpha^4 - 8\alpha^2 + 25.&#xA;\end{align*}&#xA;So $\alpha$ satisfies $x^4 - 8x^2 + 25$, as before.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-11T21:56:23.970" LastActivityDate="2011-03-11T21:56:23.970" CommentCount="2" />
  <row Id="26458" PostTypeId="2" ParentId="26432" CreationDate="2011-03-11T21:37:01.287" Score="2" ViewCount="" Body="&lt;p&gt;Zero-padding in the time domain &lt;a href=&quot;https://ccrma.stanford.edu/~jos/dft/Zero_Padding_Theorem_Spectral.html&quot; rel=&quot;nofollow&quot;&gt;corresponds to interpolation in the Fourier domain&lt;/a&gt;. It is frequently used in audio, for example for picking peaks in sinusoidal analysis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While it doesn't increase the resolution, which really has to do with the &lt;em&gt;window shape&lt;/em&gt; and length. As mentioned by @svenkatr, taking the transform of a signal that's not periodic in the DFT size is like multiplying with a &lt;a href=&quot;https://ccrma.stanford.edu/~jos/sasp/Rectangular_Window.html&quot; rel=&quot;nofollow&quot;&gt;rectangular window&lt;/a&gt;, equivalent in turn to convolving it's spectrum with the transform of the rectangle function (a sinc), which has high energy in sidelobes (off-center frequencies), making the true sinusoidal peaks harder to find. This is known as spectral leakage.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I disagree with @svenkatr that zero-padding is causing the rectangular windowing, they are separate issues. If you multiply your non-periodic signal by a suitable window (like the Hann or Hamming) that has appropriate length to have the frequency resolution that you need and then zero-pad for interpolation in frequency, things should work out just fine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the way, zero-padding is not the only interpolation method that can be used. For example, in estimating parameters of sinusoidal peaks (amplitude, phase, frequency) in the DFT, local quadratic interpolation (take 3 points around a peak and fit a parabola) can be  used because it is more computationally efficient than padding to the exact frequency resolution that you want (would mean a much larger DFT size).&lt;/p&gt;&#xA;" OwnerUserId="5610" LastEditorUserId="5610" LastEditorDisplayName="" LastEditDate="2011-03-11T21:58:56.383" LastActivityDate="2011-03-11T21:58:56.383" CommentCount="3" />
  <row Id="26459" PostTypeId="1" AcceptedAnswerId="26462" CreationDate="2011-03-11T21:49:45.520" Score="4" ViewCount="157" Body="&lt;p&gt;This is a question from Spivak's Calculus. Question statement:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $\lim\limits_{x\to a}f(x)$ exists, and $\lim\limits_{x\to a}[f(x) + g(x)]$ exists, does it follow that $\lim\limits_{x\to a} g(x)$ exists?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I have not been able to find a counterexample, so it seems that it exists. Here is my attempt at proof, and I'm mostly interested in validity of my proof.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Following is given:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\forall \epsilon, \exists \delta_1: 0 &amp;lt; |x-a| &amp;lt; \delta_1 \Rightarrow |f(x) - L_1| &amp;lt; \epsilon$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\forall \epsilon, \exists \delta_2: 0 &amp;lt; |x-a| &amp;lt; \delta_2 \Rightarrow |f(x) + g(x) - L_2| &amp;lt; \epsilon$&lt;br&gt;&#xA;Let $\delta_2 \le \delta_1$. If greater $\delta_2$ works, then smaller one will of course work, and if smaller one works than that inequality follows from that.&#xA;Let $L_2 - L_1 = L_3$, that is $L_2 = L_3 + L_1$. Then we substitute:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$|f(x) + g(x) - L_1 - L_3| &amp;lt; \epsilon$&lt;br&gt;&#xA;$|(g(x) - L_3) - (L_1 - f(x))| &amp;lt; \epsilon$&lt;br&gt;&#xA;And by the inequality $|a-b| \ge |a| - |b|$:&lt;br&gt;&#xA;$|g(x) - L_3| - |L_1 - f(x)| \le  |(g(x) - L_3) - (L_1 - f(x))| &amp;lt; \epsilon$&lt;br&gt;&#xA;$|g(x) - L_3| - |f(x) - L_1| &amp;lt; \epsilon$&lt;br&gt;&#xA;$|g(x) - L_3| &amp;lt; \epsilon + |f(x) - L_1|$&#xA;Since $\delta_2 \le \delta_1$, we can always choose epsilon greater than $|f(x) - L_1|$ so we can make substitution:&lt;br&gt;&#xA;$|g(x) - L_3| &amp;lt; 2\epsilon$&lt;br&gt;&#xA;And we can make $2\epsilon$ as small as we wish. This completes the proof.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Thank you for any help!&lt;/p&gt;&#xA;" OwnerUserId="5501" LastEditorUserId="6622" LastEditorDisplayName="" LastEditDate="2011-03-11T22:05:22.947" LastActivityDate="2011-03-11T22:25:34.257" Title="If $\lim\limits_{x\to a}f(x)$ exists, and $\lim\limits_{x\to a}[f(x) + g(x)]$ exists, does it follow that $\lim\limits_{x\to a}g(x)$ exists?" Tags="&lt;calculus&gt;&lt;limit&gt;" AnswerCount="2" CommentCount="9" />
  <row Id="26460" PostTypeId="2" ParentId="26455" CreationDate="2011-03-11T21:58:25.963" Score="1" ViewCount="" Body="&lt;p&gt;Computing the min polynomial of your number $\rm\:w\:$ is easy since $\rm\ v\: :=\: w^2 $ is a &lt;em&gt;quadratic&lt;/em&gt; number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Namely $\rm\ w = \zeta_8-2\ \zeta_8^3\ \Rightarrow\ w^2  = 4 - 3\ i = v\:,\ $ root of$\rm\ \ p(x)= (x-v)\:(x-v') = x^2 - 8\ x +25\:.$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, since $\rm\: p(w^2) = p(v) = 0\ $ we conclude that $\rm\ q(w) = 0\ $ for $\rm\ q(x) = p(x^2) = x^4 - 8\ x^2 + 25\:.$&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2011-03-11T22:53:48.367" LastActivityDate="2011-03-11T22:53:48.367" />
  <row Id="26462" PostTypeId="2" ParentId="26459" CreationDate="2011-03-11T22:11:42.317" Score="5" ViewCount="" Body="&lt;p&gt;So it can be marked off as answered...&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Do you already know that if $\lim\limits_{x\to c}F(x)$ exists and $\lim\limits_{x\to c} G(x)$ exists, then $\lim\limits_{x\to c}\bigl(F(x)-G(x)\bigr)$ exists? If so, set $F(x) = f(x)+g(x)$ and $G(x) = f(x)$ to get the desired result.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The proof is (essentially) valid; it can be streamlined a bit if you simply start from $|g(x)-L_3|$ and use the triangle inequality:&#xA;\begin{align*}|g(x)-L_3| &amp;amp;= |g(x)-L_3 + f(x)-f(x)+L_2-L_2|\\ &#xA;&amp;amp;= \left|\Bigl(g(x)+f(x)-L_2\Bigr)-\Bigl(f(x)-(L_2-L_3)\Bigr)\right|\\&#xA;&amp;amp;\leq \left|\Bigl(g(x)+f(x)\bigr)-L_2\right|+\Bigl|f(x)-L_1\Bigr|\\&#xA;&amp;amp;\lt 2\epsilon.&#xA;\end{align*}&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-11T22:25:34.257" LastActivityDate="2011-03-11T22:25:34.257" CommunityOwnedDate="2011-03-11T22:11:42.317" />
  <row Id="26463" PostTypeId="2" ParentId="26459" CreationDate="2011-03-11T22:12:21.477" Score="4" ViewCount="" Body="&lt;p&gt;I think your demonstration is correct. Maybe it is &quot;more elegant&quot; to take $\epsilon/2$ in the following sentences:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;∀ϵ,∃δ1:0&amp;lt;|x−a|&amp;lt;δ1⇒$|f(x)−L_1|$&amp;lt;ϵ/2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;∀ϵ,∃δ2:0&amp;lt;|x−a|&amp;lt;δ2⇒$|f(x)+g(x)−L_2|$&amp;lt;ϵ/2,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;in a way that $|g(x)-L_3|&amp;lt;\epsilon$, given that $|x-a| &amp;lt; \delta_2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that this is the same as the following: If $\lim_{x\to a} f(x) = L$ and $\lim_{x\to a} h(x) = M$, then $\lim_{x\to a} [f(x)+h(x)] = L+M$.&lt;/p&gt;&#xA;" OwnerUserId="1178" LastActivityDate="2011-03-11T22:12:21.477" CommentCount="4" />
  <row Id="26464" PostTypeId="2" ParentId="26373" CreationDate="2011-03-11T22:22:27.993" Score="0" ViewCount="" Body="&lt;p&gt;If you are given just a list of points $z_i=(x_i, y_i)$ $\&amp;gt;(1\leq i\leq n)$ then it is not immediate how these points should determine a certain polygon. Maybe you want the convex hull $C$ of these points. There are algorithms that accept your list as input and produce a second list $(w_i)_{1\leq i\leq m}$ (a subset of the $z_i$) containing the corners of $C$ in counter-clockwise order. Using this second list you can compute the area and centroid of $C$ by means of the formulas given in the quoted source.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These formulas come from an application of Green's theorem to $C$ and its boundary $\partial C$. It reads as follows:&#xA;$${\rm area}(C)={1\over2}\int_{\partial C}(x\,dy- y\,dx).$$&#xA;If you apply this formula to an arbitrary closed curve, as, e.g., the piecewise linear curve $\gamma$ determined by the original $z_i$ you get very strange results: Every part enclosed by $\gamma$ is counted as many times as it is surrounded counterclockwise by $\gamma$.&lt;/p&gt;&#xA;" OwnerUserId="1303" LastActivityDate="2011-03-11T22:22:27.993" CommentCount="3" />
  <row Id="26465" PostTypeId="1" AcceptedAnswerId="26512" CreationDate="2011-03-11T22:24:56.830" Score="2" ViewCount="105" Body="&lt;p&gt;An on-going effort to answer &lt;a href=&quot;http://math.stackexchange.com/questions/23247/fourier-series-representations-of-symmetric-and-non-symmetric-polygons&quot;&gt;this question&lt;/a&gt; has brought up various &lt;a href=&quot;http://math.stackexchange.com/questions/22850/linear-algebra-of-symmetric-sums&quot;&gt;other&lt;/a&gt; &lt;a href=&quot;http://math.stackexchange.com/questions/23762/a-symmetric-sum-question-that-may-use-complex-analysis&quot;&gt;questions&lt;/a&gt;. Most recently, I have been wondering: given a set of $n$ distinct non-zero complex numbers, $z_1, \dots, z_j, \dots , z_n$, and given the values of the first $n+1$ the power sums, $\mathcal{Z}_0, \mathcal{Z}_1, \dots, \mathcal{Z}_n$: &#xA;$$\begin{bmatrix}&#xD;&#xA;  1 &amp;amp; 1 &amp;amp; \cdots &amp;amp; 1 \\&#xD;&#xA;  z_{1}  &amp;amp; z_{2} &amp;amp; \cdots &amp;amp; z_{n} \\&#xD;&#xA;  z_{1}^{2} &amp;amp; z_{2}^2 &amp;amp;  \cdots &amp;amp; z_{n}^{2} \\&#xD;&#xA;  \vdots  &amp;amp; \vdots &amp;amp; \ddots &amp;amp; \vdots  \\&#xD;&#xA;  z_{1}^n  &amp;amp; z_{2}^n &amp;amp;  \cdots &amp;amp; z_{n}^n&#xD;&#xA; \end{bmatrix}&#xD;&#xA;\begin{bmatrix}&#xD;&#xA;  1 \\&#xD;&#xA;  1 \\&#xD;&#xA;  1 \\&#xD;&#xA;  \vdots  \\&#xD;&#xA;  1&#xD;&#xA; \end{bmatrix}&#xD;&#xA;=&#xD;&#xA;\begin{bmatrix}&#xD;&#xA;  \mathcal{Z}_0  \\&#xD;&#xA;  \mathcal{Z}_1 \\&#xD;&#xA;  \mathcal{Z}_2  \\&#xD;&#xA;  \vdots  \\&#xD;&#xA;  \mathcal{Z}_n&#xD;&#xA; \end{bmatrix}&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;from the &lt;em&gt;&lt;a href=&quot;http://math.stackexchange.com/questions/22850/linear-algebra-of-symmetric-sums&quot;&gt;Linear Algebra of Symmetric Sums&lt;/a&gt;&lt;/em&gt; question, we know that these $n$ equations (if you don't include $\mathcal{Z}_0 = n$) determine the $z_j$'s. What if we consider the converse? How does requiring the $z_j$'s to be non-zero and distinct restrict the possible values for the power sums, the $\mathcal{Z}_k$'s? Clearly, they cannot all be zero, but there must be a large number of other choices for $\mathcal{Z}_1, \dots, \mathcal{Z}_n$ that produce non-distinct or zero values for some of the $z_j$'s.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&lt;strong&gt;Note&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This may appear to ask two different questions: &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;What is the effect of requiring the&#xA;$z_j$'s to be non-zero? &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;What is the&#xA;effect of requiring them to be&#xA;distinct?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;But a $z_j$ that is zero or two, say $z_j$ and $z_l$, that are the same have a very similar effect on the power sum equations: they shrink the problem from a problem with $n$ independent equations to one with $n-1$ (or fewer) independent equations and an extra dependent equation or two. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, considering this question in this fashion, I may have a sort of answer, albeit a horrendously ugly answer. Consider each of the myriad, but finite number of, ways the $z_j$'s can be degenerate: one $z_j$ zero, two of them zero, one zero and two not distinct, etc., and with that as an assumption, using the technique of the second half of &lt;a href=&quot;http://math.stackexchange.com/questions/23762/a-symmetric-sum-question-that-may-use-complex-analysis/24930#24930&quot;&gt;this answer&lt;/a&gt;, attempt to calculate the dependent $\mathcal{Z}_k$ equations from the others. If it works, the present set of $\mathcal{Z}_k$'s is degenerate. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, I am looking for a more tractable answer which might allow me to look at some sets of values for the $\mathcal{Z}_k$'s (especially sets with lots of zeros) and recognize them as not possible.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&lt;strong&gt;A Reformulation of the Question&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given the discussion in the note, or given Arturo's determinant calculation in his comment, we can reformulate the question equivalently as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given a set of &lt;em&gt;any&lt;/em&gt; $n$ complex numbers, $z_1, \dots , z_n$, and given the values of the first $n+1$ the power sums, $\mathcal{Z}_0, \mathcal{Z}_1, \dots, \mathcal{Z}_n$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \mathcal{Z}_0 =n $$&#xA;$$\begin{bmatrix}&#xD;&#xA;  z_{1}  &amp;amp; z_{2} &amp;amp; \cdots &amp;amp; z_{n} \\&#xD;&#xA;  z_{1}^{2} &amp;amp; z_{2}^2 &amp;amp;  \cdots &amp;amp; z_{n}^{2} \\&#xD;&#xA;  \vdots  &amp;amp; \vdots &amp;amp; \ddots &amp;amp; \vdots  \\&#xD;&#xA;  z_{1}^n  &amp;amp; z_{2}^n &amp;amp;  \cdots &amp;amp; z_{n}^n&#xD;&#xA; \end{bmatrix}&#xD;&#xA;\begin{bmatrix}&#xD;&#xA;  1 \\&#xD;&#xA;  1 \\&#xD;&#xA;  \vdots  \\&#xD;&#xA;  1&#xD;&#xA; \end{bmatrix}&#xD;&#xA;=&#xD;&#xA;\begin{bmatrix}&#xD;&#xA;  \mathcal{Z}_1 \\&#xD;&#xA;  \mathcal{Z}_2  \\&#xD;&#xA;  \vdots  \\&#xD;&#xA;  \mathcal{Z}_n&#xD;&#xA; \end{bmatrix}&#xD;&#xA;,$$&#xA;can you use the values of the power sums to determine whether the matrix is singular or not?&lt;/p&gt;&#xA;" OwnerUserId="6272" LastEditorUserId="6272" LastEditorDisplayName="" LastEditDate="2011-03-12T15:27:20.253" LastActivityDate="2011-03-12T20:55:57.000" Title="Linear Algebra of Symmetric Sums: a converse question" Tags="&lt;algebra&gt;&lt;complex-numbers&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" />
  <row Id="26466" PostTypeId="5" CreationDate="2011-03-11T22:40:03.883" Score="0" ViewCount="" Body="&lt;p&gt;Statistics is the science of the collection, organization, and interpretation of data. It deals with all aspects of this, including the planning of data collection in terms of the design of surveys and experiments. (From &lt;a href=&quot;http://en.wikipedia.org/wiki/Statistics&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More specifically, mathematical statistics is the study of statistics from a mathematical standpoint, using probability theory as well as other branches of mathematics such as linear algebra and analysis. (From &lt;a href=&quot;http://en.wikipedia.org/wiki/Mathematical_statistics&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For questions which are more generally about collecting and treating data, it might be advisable to consider posting questions on &lt;a href=&quot;http://stats.stackexchange.com/&quot;&gt;Cross Validated&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="-1" LastEditorUserId="3567" LastEditorDisplayName="" LastEditDate="2011-03-11T22:47:10.260" LastActivityDate="2011-03-11T22:47:10.260" />
  <row Id="26467" PostTypeId="4" CreationDate="2011-03-11T22:40:03.883" Score="0" ViewCount="" Body="Mathematical statistics is the study of statistics from a mathematical standpoint, using probability theory as well as other branches of mathematics such as linear algebra and analysis." OwnerUserId="-1" LastEditorUserId="3567" LastEditorDisplayName="" LastEditDate="2011-03-11T22:47:19.400" LastActivityDate="2011-03-11T22:47:19.400" />
  <row Id="26468" PostTypeId="1" AcceptedAnswerId="26473" CreationDate="2011-03-11T22:47:55.970" Score="1" ViewCount="79" Body="&lt;p&gt;I have Borel measures $\mu$, $\nu$ on $\mathbb R^2$ with densities $g, h$ respectively with respect to the Lebesgue measure. Now I assume that $\mu(A \times \mathbb R) = \nu(A \times \mathbb R)$ for every Borel set $A$. Now I want to find a Borel measure $\gamma$ on $\mathbb R^3$ such that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mu(A \times B) = \gamma(A \times B \times \mathbb R)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\nu(A \times B) = \gamma(A \times \mathbb R \times B)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fine. I try to write the measure $\gamma$ as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\gamma(Z) = \int 1_Z(x,y,z) \, d \nu(x,y) \, dz$$ and the same with $\mu$ instead of $\nu$, then I use the density wrt Lebesgue measure but this quickly becomes really messy. Is there an elegant way to do this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Further, what would be the point of this exercise? Why would someone want this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is homework, so only hints.&lt;/p&gt;&#xA;" OwnerUserId="1120" LastEditorUserId="1120" LastEditorDisplayName="" LastEditDate="2011-03-11T23:32:42.843" LastActivityDate="2011-03-11T23:32:42.843" Title="Disintegration with densities." Tags="&lt;measure-theory&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="26469" PostTypeId="1" AcceptedAnswerId="26484" CreationDate="2011-03-11T22:52:38.920" Score="2" ViewCount="136" Body="&lt;p&gt;I am looking to build a repertoire of olympiad type problems which have non-intuitive elegant solutions, If possible instead of a resource, I think problems would be the best. (i.e. select the best problem to post here). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem I like best that falls into this category is to prove that if a bigger rectangle has some smaller rectangles completely space filling inside it, and the small rectangles have at least one side of integer length, then we need to show that the big rectangle has at least one integer length. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The non-intuitive(to me) solution is to place each smaller rectangle on a checkerboard with side length of the pattern = 1/2. Then to note that each smaller rectangle must have an equal area of black and white. Then to prove that for any checkerboard to have an equal area of black and white, it must have one of the lengths of integer length. &lt;/p&gt;&#xA;" OwnerUserId="4728" LastActivityDate="2011-03-12T00:19:10.340" Title="Solutions to interesting problems with elegant and unintuitive methods" Tags="&lt;intuition&gt;&lt;contest-math&gt;" AnswerCount="3" CommentCount="10" />
  <row Id="26470" PostTypeId="1" AcceptedAnswerId="26471" CreationDate="2011-03-11T22:55:49.193" Score="2" ViewCount="131" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Let $f$ be a homomorphism mapping $G$ to $J$ groups. Assume that $J$ is abelian. Prove that if $H$ is a subgroup of $G$ and if $\mathrm{ker}(f)$ is a subset of $H$, then $H$ is normal in $G$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I honestly have no idea how to start this proof. The fact that $J$ is abelian gives me that for all $a$ and $b$ in $G$, $f(ab) = f(ba)$; but this tells me nothing since $f$ is not injective. Furthermore, what exactly can I get from the fact that $\mathrm{ker}(f)$ is a subset of $H$? I understand the given information but I am having issues putting it all together as I see no possible relation!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My plan was to piece the information all together to somehow set things up for the normal subgroup test, however since I  see no connection with the given information my approach seems to be rather futile.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another thought I just got was to prove that $f$ is injective then use the fact that $J$ is abelian to show that $ah = ha$ for all $a$ in $G$ and $h$ in $H$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="6523" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-11T23:06:31.240" LastActivityDate="2011-03-11T23:09:31.737" Title="Homomorphisms and normal subgroups proof" Tags="&lt;homework&gt;&lt;group-theory&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" />
  <row Id="26471" PostTypeId="2" ParentId="26470" CreationDate="2011-03-11T23:01:10.590" Score="2" ViewCount="" Body="&lt;p&gt;You won't be able to prove that $f$ is injective, since there is absolutely no reason why it would be; and you can probably come up with examples of groups which satisfy the set-up but where $f$ is not injective. If you cannot, here's one: take $G=Q_8$, the quaternion group of order $8$. Let $J = C_2\times C_2$, the Klein $4$-group. Let $f\colon G\to J$ be the map that sends $1$ and $-1$ to $(0,0)$, $i$ and $-i$ to $(1,0)$; $j$ and $-j$ to $(0,1)$; and $k$ and $-k$ to $(1,1)$. Now take $H = \{1, -1, i, -i\}$. (Of course, $H$ is normal in $G$; but the point is that this set-up satisfies all the hypothesis, but $f$ is not injective; so you won't be able to prove, in general, that $f$ is injective).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you already know the isomorphism theorems, you should remember that one of them says:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Theorem.&lt;/strong&gt; Let $f\colon G\to K$ be a group homomorphism. The subgroups of $G$ that contain $\mathrm{ker}(f)$ are in one-to-one, inclusion preserving, normality preserving correspondence with the subgroups of $f(G)$. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and that will solve your problem directly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you &lt;em&gt;don't&lt;/em&gt; know the isomorphism theorems, start from what you need to start: let $h\in H$ and $g\in G$; you need to show that $ghg^{-1}\in H$, right?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, $f(ghg^{-1}) = f(g)f(h)f(g)^{-1} = f(g)f(g)^{-1}f(h) = f(h)$, since $J$ is abelian.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What do you know about two elements that map to the same thing under a group homomorphism (in terms of the kernel of course)? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(&lt;em&gt;Hint:&lt;/em&gt; $f(a) = f(b)$ if and only if $f(a)f(b)^{-1}=1$, if and only if $f(ab^{-1})=1$, if and only if ....)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What do you know about the kernel of $f$ in this case? Can you use these facts to conclude that $ghg^{-1}\in H$?&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-11T23:09:31.737" LastActivityDate="2011-03-11T23:09:31.737" CommentCount="14" />
  <row Id="26472" PostTypeId="2" ParentId="26456" CreationDate="2011-03-11T23:11:46.577" Score="1" ViewCount="" Body="&lt;p&gt;To say that two convex sets $C$ and $D$ in $\mathbb{R}^n$ can be separated by a hyperplane $\mathcal{H}$ is to say that there is a linear functional $f$ on $\mathbb{R}^n$ such that $\mbox{sup} \{f\left(x\right)\: : \: x\in C\} \leq \mbox{inf} \{f\left(x\right)\: : \: x \in D\}$. $\mathcal{H}$ will then be the&#xA;hyperplane $\{x \in \mathbb{R}^n \: : \: f\left(x\right) = c\}$, which is a coset&#xA;of the kernel of $f$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We say that $C$ and $D$ can be strictly separated if a linear functional $f$ can be chosen so that the above inequality is strict. Your intuition about  what sorts of closed convex sets can be separated but not strictly separated is correct. This can happen when $C$ and $D$ are not compact and approach arbitrarily close to each other without meeting. So if you let $C$ be the points in $\mathbb{R}^2$ such that $y \geq \frac{1}{x}$ with $x &amp;gt; 0$ and $D$ be the points such that $y \leq -\frac{1}{x}$ with $x &amp;gt; 0$, then $C$ and $D$ are separated by the functional that projects onto the $y$ axis, but they are not strictly separated by this functional.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Noah's answer gives another good example. joriki's answer clears up a lot of things, and the paper he links to is a good introduction to this topic.&lt;/p&gt;&#xA;" OwnerUserId="7323" LastEditorUserId="7323" LastEditorDisplayName="" LastEditDate="2011-03-12T14:41:37.557" LastActivityDate="2011-03-12T14:41:37.557" CommunityOwnedDate="2011-03-12T06:34:22.690" />
  <row Id="26473" PostTypeId="2" ParentId="26468" CreationDate="2011-03-11T23:13:18.210" Score="1" ViewCount="" Body="&lt;p&gt;You are trying to find a random $(X,Y,Z)$ whose marginal distributions $(X,Y)$ and $(X,Z)$ are given. The simplest possible solution is to let $Y$ and $Z$ be conditionally independent given $X$. The existence of densities will help you to make this explicit.   &lt;/p&gt;&#xA;" OwnerUserId="940" LastActivityDate="2011-03-11T23:13:18.210" CommentCount="4" />
  <row Id="26474" PostTypeId="2" ParentId="26469" CreationDate="2011-03-11T23:13:44.120" Score="1" ViewCount="" Body="&lt;p&gt;Here's what I thought was the cannonical example, but five minutes of internet search can't find a reference, so I must be misremembering some piece of it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose you have two mountain climbers. Each climber can move up or down the mountain as they choose (i.e. their height is not necessarily monotonically increasing or decreasing), but their height as a function of time is continuous and differentiable (i.e. they can't magically hop from one place to another).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now suppose A descends from the summit at 6 AM and reaches the base at 6 PM. The next day, B starts ascending at 6 AM and reaches the summit at 6 PM. Will there be some time at which they were both at the same height? For example, if they both moved at the same rate, then at 12 PM they would both be halfway up. But they don't necessarily need to go at the same speed, nor are they even required to always move in the same direction, so we can't guarantee they'll meet at 12.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &quot;trick&quot; is to plot them as two lines on a graph. Since they're continuous and differentiable, they must intersect somewhere, which implies that there must be some time that they're at the same height.&lt;/p&gt;&#xA;" OwnerUserId="2549" LastActivityDate="2011-03-11T23:13:44.120" CommentCount="2" />
  <row Id="26475" PostTypeId="1" CreationDate="2011-03-11T23:15:02.767" Score="6" ViewCount="86" Body="&lt;p&gt;Plotting the Bessel functions of the first kind $J_n(z)$ versus $n$ for some fixed $z\gg1$, it appears that there is a sharp cutoff just before $n=z$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Three questions:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;What is a reference describing this sharp cutoff?&lt;/li&gt;&#xA;&lt;li&gt;What is an expression for the location of the maximum of $J_n(z)$ with respect to $n$, for fixed (large) $z$?&lt;/li&gt;&#xA;&lt;li&gt;What is a nice expression for the &lt;em&gt;envelope&lt;/em&gt; of the function $J_n(z)$ with respect to $z$? I.e, what is a function that (approximately) goes through all the maxima of the following plot, and then dies off appropriately for $n&amp;gt;z$?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/Oh62v.png&quot; alt=&quot;Plot of the Bessel function J_n(z) for z=100, n from 0 to 200.&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="2191" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2011-04-17T13:24:39.510" LastActivityDate="2011-04-17T13:24:39.510" Title="On the Bessel function $J_n(z)$ for high $z$, with respect to $n$..." Tags="&lt;special-functions&gt;" AnswerCount="1" FavoriteCount="2" />
  <row Id="26476" PostTypeId="2" ParentId="26468" CreationDate="2011-03-11T23:15:45.217" Score="2" ViewCount="" Body="&lt;p&gt;In the parlance of probability, when $\mu$ and $\nu$ have mass $1$, you look for the distribution $\gamma$ of a triple $(X,Y,Z)$ of random variables such that $\mu$ is the distribution of $(X,Y)$ and $\nu$ is the distribution of $(X,Z)$. Now we understand the condition that the first marginals of $\mu$ and $\nu$ coincide. And, since we have some latitude on the coupling of $(X,Y,Z)$, we could choose $\gamma$ such that, for example, $Y$ and $Z$ are independent conditionally on $X$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since you asked for a hint and not a solution, I stop here.&lt;/p&gt;&#xA;" OwnerUserId="6179" LastActivityDate="2011-03-11T23:15:45.217" />
  <row Id="26477" PostTypeId="1" AcceptedAnswerId="26493" CreationDate="2011-03-11T23:15:47.990" Score="5" ViewCount="219" Body="&lt;p&gt;Show that over any field $F$, the polynomial $x^3-3x+1$ is either irreducible or splits into linear factors.&lt;/p&gt;&#xA;&#xA;&lt;hr/&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edited:&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;This is my attempt:&lt;/strong&gt; Let $f(x)=x^3-3x+1$. Let $a_1,a_2,a_3$ be the roots of $f$. Suppose char $F\neq 2,3$. Suppose also that $f$  is neither irreducible nor splits in $F$. Then $f$ is reducible which implies that $a_1 \in F$. i.e. $f(x)=(x-a_1)g(x)$, where $g(x)\in F[x]$ is irreducible with deg $g=2$. Let $K$ be the splitting field of $g$. The $K$ is Galois over $F$.&#xA;So if $\sigma \in $ Aut($K/F$), then $\sigma (a_1)=a_1$ since $\sigma $ fixes $F$ and $a_1 \in F$. Since $\sigma$ permutes the roots of $f$, WLOG suppose $\sigma (a_2)=a_3$. Then&lt;br&gt;&#xA;$\sigma(\triangle) = \sigma((a_1-a_2)(a_1-a_3)(a_2-a_3))=-\triangle$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But $\triangle^2=D(f)=81$, so $\triangle = \pm 9 \in F$, so $\triangle \in F$. Therefore,&#xA;$9=-9$ $\implies 1=-1 \implies$ char $F =2$ which is a contradiction. So $f$ is either irreducible or splits in $F$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next suppose char $F=2$. Well, I'm not exactly sure what I can say about $f$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to know if my approach is correct and also what to do in the second case. Thanks.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;ADDED:&lt;/strong&gt;&lt;br&gt;&#xA;If char $F=2$, then $f=x^3+x+1$. Suppose $b$ is a root of $f$. Then $b^2 $ is also a root, since $f(b^2)=(b^2)^3+b^2+1=(b+1)^2+b^2+1=2b^2+2=0$. Is it enough to conclude that $f$ splits?  &lt;/p&gt;&#xA;" OwnerUserId="4426" LastEditorUserId="4426" LastEditorDisplayName="" LastEditDate="2011-03-13T20:30:20.690" LastActivityDate="2011-03-13T20:37:06.680" Title="Irreducibility and Splitting Fields" Tags="&lt;homework&gt;&lt;field-theory&gt;" AnswerCount="3" CommentCount="6" FavoriteCount="1" />
  <row Id="26478" PostTypeId="2" ParentId="26456" CreationDate="2011-03-11T23:20:11.313" Score="0" ViewCount="" Body="&lt;p&gt;Take $X = \{(x,y) \mid xy\geq 1, x,y&amp;gt;0\}$ and $Y = \{(x,y) \mid x\leq 0\}$.&lt;/p&gt;&#xA;" OwnerUserId="884" LastActivityDate="2011-03-11T23:20:11.313" />
  <row Id="26479" PostTypeId="2" ParentId="26456" CreationDate="2011-03-11T23:20:59.357" Score="0" ViewCount="" Body="&lt;p&gt;Some comments on the question and Albert Steppi's answer (written as an answer because they're too long for a comment):&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;You both seem to have ignored the &quot;convex&quot; part of the question. What you need is not the graphs of these functions but the sets they bound. Also you need to select one of the two branches of each of $1/x$ and $-1/x$ in order to get a convex set.&lt;/li&gt;&#xA;&lt;li&gt;BoB, I think where your otherwise correct ideas went wrong is that you considered the inequality between the values of the functional instead of the inequality between their infimum and supremum.&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.johndcook.com/SeparationOfConvexSets.pdf&quot; rel=&quot;nofollow&quot;&gt;This paper&lt;/a&gt; has more on the topic.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="6622" LastActivityDate="2011-03-11T23:20:59.357" CommentCount="2" />
  <row Id="26480" PostTypeId="1" AcceptedAnswerId="26485" CreationDate="2011-03-11T23:46:58.227" Score="0" ViewCount="26" Body="&lt;p&gt;I have this problem about account record pagination. I have to write the function that describes the total pages needed to contain those account records.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the problem:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;# Account record quantity to store per page&#xA;per_page € R &#xA;# Account records to insert in pages&#xA;tot_acc € R &#xA;&#xA;# The total pages needed to store the `tot_account` quantity&#xA;f(per_page,tot_acc) = tot_page € R = ? &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I need to find &lt;code&gt;f(per_page,tot_acc)&lt;/code&gt;.&lt;/p&gt;&#xA;" OwnerUserId="8136" LastEditorUserId="8136" LastEditorDisplayName="" LastEditDate="2011-03-12T00:18:26.413" LastActivityDate="2011-03-12T00:23:55.037" Title="Trouble on writing the function for an account record pagination problem" Tags="&lt;calculus&gt;&lt;functions&gt;&lt;problem-solving&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="26481" PostTypeId="2" ParentId="26469" CreationDate="2011-03-11T23:48:52.380" Score="1" ViewCount="" Body="&lt;p&gt;All good riddles satisfy your requirements. &lt;a href=&quot;http://mathoverflow.net/questions/7493/a-riddle-about-zeros-ones-and-minus-ones/32425#32425&quot; rel=&quot;nofollow&quot;&gt;This&lt;/a&gt; riddle is excellent since the statement and the solution are both easy, but the solution is quite non-obvious.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2011-03-11T23:48:52.380" CommentCount="1" />
  <row Id="26482" PostTypeId="2" ParentId="26424" CreationDate="2011-03-12T00:06:39.370" Score="1" ViewCount="" Body="&lt;p&gt;That works, assuming &lt;em&gt;randomly&lt;/em&gt; means each point is uniformly distributed on $[0,L]$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A similar approach is to note that the density of $x$ and $y$ is each $\frac{1}{L}$ so the probability of a triangle is &#xA;$$\int_{x=0}^{L/2}\int_{y=L/2}^{x+L/2}\frac{1}{L^2} dy\,dx + \int_{x=L/2}^{1}\int_{y=x-L/2}^{L/2}\frac{1}{L^2}dy\,dx = \int_{x=0}^{L/2}\frac{x}{L^2}dx + \int_{x=L/2}^{1}\frac{L-x}{L^2}dx = \frac{1}{4}$$&lt;/p&gt;&#xA;" OwnerUserId="6460" LastActivityDate="2011-03-12T00:06:39.370" />
  <row Id="26483" PostTypeId="2" ParentId="25491" CreationDate="2011-03-12T00:18:47.723" Score="2" ViewCount="" Body="&lt;p&gt;Your intuition is correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is best phrased in the language of differential forms (see &lt;a href=&quot;http://en.wikipedia.org/wiki/Differential_form&quot; rel=&quot;nofollow&quot;&gt;the Wikipedia page&lt;/a&gt;), where we have an $(n-1)$-form $\omega$ defined on a smooth $n$-manifold with boundary ($n=4$ in you example).  There are two useful things to do with this form:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Restrict it to the boundary:  $\omega|_{\partial M}$&lt;/li&gt;&#xA;&lt;li&gt;Take the exterior derivative to make an $n$-form:  $d\omega$&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Then, Stokes' Theorem says&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$\displaystyle \int_M d\omega = \int_{\partial M} \omega$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;All forms of Green, Gauss, Stokes theorems are higher-dimensional analogs of the Fundamental Theorem of Calculus.&lt;/p&gt;&#xA;" OwnerUserId="6509" LastActivityDate="2011-03-12T00:18:47.723" CommentCount="2" />
  <row Id="26484" PostTypeId="2" ParentId="26469" CreationDate="2011-03-12T00:19:10.340" Score="2" ViewCount="" Body="&lt;p&gt;The checkerboard and domino puzzle ( &lt;a href=&quot;http://www.jimloy.com/puzz/dominos.htm&quot; rel=&quot;nofollow&quot;&gt;http://www.jimloy.com/puzz/dominos.htm&lt;/a&gt;) has an interesting, simple and elegant solution. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, consider the generalization of this problem, where instead of removing squares at diagonal corners you remove one white square and one black square from anywhere on the board. When is it possible to tile such a reduced checkerboard?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer is that it is always possible to tile such a checkerboard, but the proof relies on a very elegant but non-obvious construction. I remember seeing the solution in Ross Honsberger's Mathematical Gems and it blew me away.&lt;/p&gt;&#xA;" OwnerUserId="2641" LastActivityDate="2011-03-12T00:19:10.340" CommentCount="2" />
  <row Id="26485" PostTypeId="2" ParentId="26480" CreationDate="2011-03-12T00:23:55.037" Score="1" ViewCount="" Body="&lt;p&gt;Are you asking something like:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With all variables as integers, given the first page number of a set of accounts $i$, the number of accounts that can fit on a page $d$, the total number of accounts $n$, what is the number of the last page?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If so, then you need to use $\left\lceil \frac{n}{d}\right\rceil$ pages for the accounts where the brackets mean &lt;em&gt;ceiling&lt;/em&gt; i.e. round up to the next integer. You need to add that to the page before the starting point $i-1$ to give the final page of the accounts&#xA;$$f(i,d,n) = \left\lceil \frac{n}{d}\right\rceil + i -1$$&#xA;or if you prefer&#xA;$$\text{tot_page} = f(\text{page, per_page, tot_acc})  = \left\lceil \frac{\text{tot_acc}}{\text{per_page}}\right\rceil + \text{page} -1.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps you were asking something else.&lt;/p&gt;&#xA;" OwnerUserId="6460" LastActivityDate="2011-03-12T00:23:55.037" />
  <row Id="26486" PostTypeId="1" AcceptedAnswerId="26491" CreationDate="2011-03-12T00:23:59.823" Score="2" ViewCount="271" Body="&lt;p&gt;How to prove:&#xA;if $f: (a,b) \rightarrow R$ is uniformly continuous, then $f$ can be extended to a continuous function $F: [a,b] \rightarrow R$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's suffice to show that $f$ can  be extended to a continuous function $G: [a,b) \rightarrow R$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I saw similar theorem (Tietze extension theorem) in topology books, but I'm not very familiar with lots of topology concepts. Also I've read one proof for this, it uses Cauchy sequence and Cauchy Completeness Theorem, which I'm also not that comfortable with. Can you think of any other proof for this using methods from uniform continuity, limits, etc. in real analysis? Thanks! :)&lt;/p&gt;&#xA;" OwnerUserId="7429" LastEditorUserId="536" LastEditorDisplayName="" LastEditDate="2011-03-15T22:16:20.883" LastActivityDate="2011-03-15T22:16:20.883" Title="Continuous Extension Theorem in real analysis" Tags="&lt;real-analysis&gt;" AnswerCount="2" CommentCount="5" FavoriteCount="2" />
  <row Id="26487" PostTypeId="1" AcceptedAnswerId="26490" CreationDate="2011-03-12T00:32:50.063" Score="1" ViewCount="66" Body="&lt;p&gt;Dear all, I am studying symmetry method for ODEs.&#xA;The book written by George W.bluman (can be found in google books) discusses the procedure of determining ODE admitting a given group in details. However, as to the converse way, it only gives some theories, but no examples.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am trying to solve the exercise 3.2 - 9, P120:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;given ODE&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(y-\frac{3}{2}x-3)y&amp;#39;+y=0$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;find a nontrivial Lie group of transformations admitted by the ODE;&#xA;and find the general solution of the ODE.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please help me out here.&#xA;Thank you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cheers,&#xA;Jan&lt;/p&gt;&#xA;" OwnerUserId="7920" LastActivityDate="2011-03-12T01:45:04.480" Title="How to find a nontrivial Lie group transformation?" Tags="&lt;differential-equations&gt;&lt;lie-groups&gt;" AnswerCount="1" />
  <row Id="26488" PostTypeId="2" ParentId="26339" CreationDate="2011-03-12T00:50:15.653" Score="3" ViewCount="" Body="&lt;p&gt;I wanted to post some excerpts from Niven's book that gives a nice recurrence relation to lifting nonsingular roots. I always found it helpful to use as a kind of plug and chug way to go. I also included an example of it in action.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/kBaE7.jpg&quot; alt=&quot;enter image description here&quot;&gt;&#xA;&lt;img src=&quot;http://i.imgur.com/mRBf7.jpg&quot; alt=&quot;enter image description here&quot;&gt;&#xA;&lt;img src=&quot;http://i.imgur.com/kC2Q1.jpg&quot; alt=&quot;enter image description here&quot;&gt;&#xA;&lt;img src=&quot;http://i.imgur.com/nHJwV.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To address your specific problem, you see that $x\equiv 1,3$ are roots mod $5$. However, &#xA;$$&#xD;&#xA;f&amp;#39;(1)=3(1)^2+2\equiv 0\pmod{5}\quad\text{and}\quad f&amp;#39;(3)=3(3)^2+2\equiv 4\pmod{5}&#xD;&#xA;$$&#xA;so $x\equiv 3$ is nonsingular, and may be lifted to a solution $a_2$ mod $25$ by Hensel's Lemma. Notice that $\overline{f&amp;#39;(3)}=4$. By the recurrence (2.6) of the excerpt above, you have&#xA;$$&#xD;&#xA;a_2=3-35(4)\equiv 3+(15)(4)\equiv 3+10\equiv 13\pmod{25}&#xD;&#xA;$$&#xA;which is the solution found by others. I hope it's useful when the numbers you're working with may not be as nice.&lt;/p&gt;&#xA;" OwnerUserId="1583" LastActivityDate="2011-03-12T00:50:15.653" CommentCount="1" />
  <row Id="26489" PostTypeId="1" AcceptedAnswerId="26513" CreationDate="2011-03-12T01:30:47.953" Score="2" ViewCount="43" Body="&lt;p&gt;Hello,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am reading Boyd's Convex Optimization book and I am stuck on the reasoning behind one of the statements. Specifically, I am looking at page 45, line 3 from the bottom. The statement is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$A \preceq B \iff \mathcal{E}_A \subseteq \mathcal{E}_B$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that I interpret $A \preceq B$ to mean $B-A \subset \mathbf{S}^n_+$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is my attempt at going from left to right:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$B-A \subset \mathbf{S}^n_+ \Rightarrow x^T (B - A) x \geq 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x^T (B - A) x \geq 0 \Rightarrow x^T B x - x^T A x \geq 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x^T B x - x^T A x \geq 0 \Rightarrow x^T B B^{-1} B x - x^T A A^{-1} A x \geq 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x^T B B^{-1} B x - x^T A A^{-1} A x \geq 0 \Rightarrow z^T B^{-1} z - y^T A^{-1} y \geq 0$ (eq: 1)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So here I get stuck because I'm not exactly sure how this implies ellipsoid A is a subset of ellipsoid B.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My attempt from going right to left got stuck because &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\mathcal{E}_A \subseteq \mathcal{E}_B \iff (x \in \{y \mid y^T A^{-1} y \leq 1\} \Rightarrow x \in \{y \mid y^T B^{-1} y \leq 1\})$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I made the assumption the following was true (for circles this makes some sense):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(x \in \{y \mid y^T A^{-1} y \leq 1 \} \Rightarrow x \in \{y \mid y^T B^{-1} y \leq 1\}) \Rightarrow x^T A^{-1} x \leq x^T B^{-1} x$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that after some algebraic manipulations, I reach point (eq: 1).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any help on how to represent ellipsoid A is a subset of ellipsoid B in matrix form? And also how do I work past (eq: 1)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="8140" LastEditorUserId="8140" LastEditorDisplayName="" LastEditDate="2011-03-12T02:33:07.320" LastActivityDate="2011-03-12T04:33:20.840" Title="Proving ellipsoid A is a subset of ellipsoid B iff (B-A) is positive semi-definite" Tags="&lt;convex-optimization&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="26490" PostTypeId="2" ParentId="26487" CreationDate="2011-03-12T01:32:29.577" Score="3" ViewCount="" Body="&lt;p&gt;Before I say anything, let me say that I very much think you should try to get a hold of Peter Olver's &lt;em&gt;Applications of Lie Groups to Differential Equations&lt;/em&gt;, published by Springer in its GTM series. It is literally packed with examples.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for your equation: notice that you can rewrite it as $$\frac{\mathrm d y}{\mathrm d x}=\frac{-y}{y-\tfrac32x-3}.$$ There is a point $(x_0,y_0)$ such that if we change variables to $X=x-x_0$ and $Y=y-y_0$, the equation becomes $$\frac{\mathrm d Y}{\mathrm d X}=\frac{-Y}{Y-\tfrac32X}.$$ Now this equation is homogeneous, &lt;em&gt;i.e.&lt;/em&gt;, the right hand side depends only on $Y/X$. This observation can be used to solve it explicitly, by changing variables once more so that the new dependent variable is $T(X)=Y(X)/X$. But it also tells us that for each $\lambda\in\mathbb R\setminus\{0\}$ the transformation &lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;X &amp;amp;\leadsto \lambda X \\&#xA;Y &amp;amp;\leadsto \lambda Y&#xA;\end{align} &lt;/p&gt;&#xA;&#xA;&lt;p&gt;leaves the equation invariant, so we have a $1$-dimensional Lie group of symmetries for the equation for $X$ and $Y$. Undoing the change of variables that got us there will give us a group of symmetries for the original equation.&lt;/p&gt;&#xA;" OwnerUserId="274" LastEditorUserId="274" LastEditorDisplayName="" LastEditDate="2011-03-12T01:45:04.480" LastActivityDate="2011-03-12T01:45:04.480" CommentCount="1" />
  <row Id="26491" PostTypeId="2" ParentId="26486" CreationDate="2011-03-12T01:41:03.033" Score="5" ViewCount="" Body="&lt;p&gt;There are two parts to this statement. The first half does not require completeness. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Claim&lt;/strong&gt; If $f:(a,b)\to\mathbb{R}$ is uniformly continuous, $\limsup_{x\to a^+} f \leq \liminf_{x\to a^+} f &amp;lt; \infty$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Sketch of proof&lt;/em&gt; First we show that $\limsup f$ must be bounded. Suppose it were not. Let $\delta$ be some small number such that $\delta &amp;lt; b-a$. Consider $x_0 = a + \delta$, and $y_0 = f(x_0)$. Construct a sequence $(x_i,y_i)$ in the following manner. Pick $x_i &amp;lt; a + \frac12 (x_{i-1} - a)$ such that $y_i = f(x_i) &amp;gt; 2 y_{i-1}$. (Notice that if this sequence terminates after finitely many terms and cannot be continued, you have will have that $f|_{(a,\frac12(x_N + a))}$ is bounded.) However, we have that $|x_{i+1} - x_{i}| &amp;lt; \frac{\delta}{2^i}$ while $|y_{i+1} - y_{i}| &amp;gt; 2^i y_0$. So it is easy to see that by taking $i\nearrow \infty$ you get a contradiction to uniform continuity. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly $f$ must be bounded below. Now suppose the $\limsup f &amp;gt; \liminf f$ are not equal. Then by a similar argument, you can construct an alternating sequence $x_1, x&amp;#39;_1, x_2, x&amp;#39;_2$ such that $2 |x&amp;#39;_i - a| &amp;lt; |x_i - a| &amp;lt; \frac12 |x&amp;#39;_{i-1} - a|$ and $f(x_i) \to \limsup f$ while $f(x&amp;#39;_i) \to\liminf f$. Then you have that for $0 &amp;lt; \epsilon &amp;lt; \limsup f -\liminf f$, for any $\delta$, there exists some $x_i$ such with a point $x&amp;#39;_i$ within $\delta$ of it taking value more than $\epsilon$ away under $f$, contradicting uniform continuity. Q.E.D. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Like Nate said in his comments, now you are required to use the completeness of $\mathbb{R}$. By using the completeness you can say that since $\limsup f \leq \liminf f$, the limit $\lim_{x\to a^+}f$ exists, so setting $f(a) = \lim_{x\to a^+} f$ you get a continuous function. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For illustration, imagine instead of you have the function $f:(0,1)\to \mathbb{R}^2\setminus \{ 0\}$ given by &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ f(s) = (s \cos s, s\sin s) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This function is uniformly continuous on $(0,1)$, but does not have continuous extension to $[0,1)$ with codomain $\mathbb{R}^2\setminus \{ 0\}$. &lt;/p&gt;&#xA;" OwnerUserId="1543" LastActivityDate="2011-03-12T01:41:03.033" CommentCount="3" />
  <row Id="26492" PostTypeId="1" AcceptedAnswerId="26509" CreationDate="2011-03-12T01:57:11.597" Score="4" ViewCount="131" Body="&lt;p&gt;Could I please get some help with the following true/sometimes true/false questions?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) If $A$ is 3-by-4, then for every $\vec{z}$ $\in$ $\mathbb{R^{3}}$ there are infinitely many solutions $\vec{x}^{*}$ to $A^{T}A\vec{x}^{*}$=$A^{T}\vec{z}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is related to least-squares, I can see how it could be sometimes true, for example if A is the zero matrix. But I don't know how to provide a more general answer, can you explain the reasoning in this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not sure if this is what I am looking for, but I think we said that $\mathbb{R^{n}}$ can always be decomposed into a subspace and its perp. So could I argue that since I am decomposing z, and its in $\mathbb{R^{n}}$, it will always be a combination of basis vectors of Im A and its perp?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) $A$ is n-by-n and magnitude of $A\vec{u}$=1 for all unit vectors $\vec{u}$. $A$ is orthogonal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Frin what is given to me I know that $A\vec{u}$ dot $A\vec{u}$=1 so $u^{T}A^{T}Au$=1. Not sure how to go from here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) If $S$ is 2-by-2 and invertible, and $T$: $\mathbb{R^{2x2}}\to\mathbb{R^{2x2}}$ is represented as $T(M)=S^{-1}MS$ then $T$ is an isomorphism.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can prove that M=0 if T(M)=0 by taking $S^{-1}$ from the left of $SMS^{-1}$=0 and then $S$ from the right to get M=0. That's how any matrix N can be constructed from M, too =&gt; $M=S^{-1}NS$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strike&gt;4) A product of a matrix times its transpose is always symmetric. &lt;/strike&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Figured this one out -- always true.&lt;/p&gt;&#xA;" OwnerUserId="8141" LastEditorUserId="8141" LastEditorDisplayName="" LastEditDate="2011-03-12T04:45:37.133" LastActivityDate="2011-03-12T06:05:48.300" Title="Several true/false questions" Tags="&lt;linear-algebra&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="26493" PostTypeId="2" ParentId="26477" CreationDate="2011-03-12T02:00:11.767" Score="5" ViewCount="" Body="&lt;p&gt;There's a slight problem with your argument: from $9=-9$ you can only conclude $1=-1$ if $3$ is invertible. What if $\mathrm{char}(F) = 3$? So you may want to start by positing that $\mathrm{char}(F)\neq 2,3$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(There is a typo in the fourth sentence: &quot;Then $f$ irreducible&quot; should be &quot;Then $f$ reducible...&quot;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, you need to deal with $\mathrm{char}(F) = 2$ and with $\mathrm{char}(F) = 3$. If $\mathrm{char}(F) = 3$, then $f(x) = x^3+1 = (x+1)^3$, so there is nothing to do in that case. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What about $\mathrm{char}(F)=2$? Your polynomial is just $f(x)=x^3+x+1$. Suppose $\alpha$ is a root. Try to see if you can construct some other root of the polynomial using $\alpha$. For instance, $\alpha+1$ doesn't work, because you have &#xA;$$(\alpha+1)^3 + (\alpha+1)+1 = \alpha^3+\alpha^2+\alpha+1+\alpha +1 = \alpha^2+\alpha+1 = \alpha^3+\alpha^2 = \alpha^2(\alpha+1)=\alpha^5\neq 0$$&#xA;but maybe some other expression involving $\alpha$ will do?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Added.&lt;/em&gt; So, now you have that if $\mathrm{char}(F)=2$, and $b$ is a root of $f(x)$, then so is $b^2$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can $b=b^2$? That is, can you simply get the same root again? If $b=b^2$, then $b^2-b=0$, so $b(b-1)=0$. Since you are in a field, either $b=0$ or $b=1$. But neither $b=0$ nor $b=1$ are roots of $f(x)$. So that means that if $b$ is a root of $f$, then $b\neq b^2$, and b^2$ is also a root of $f(x)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if $f(x)$ has at least one root in $F$, then it has at least &lt;strong&gt;two&lt;/strong&gt; roots in $F$ (namely, $b$ and $b^2$). Which means that...&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-13T20:37:06.680" LastActivityDate="2011-03-13T20:37:06.680" CommentCount="3" />
  <row Id="26494" PostTypeId="2" ParentId="26492" CreationDate="2011-03-12T02:14:51.957" Score="0" ViewCount="" Body="&lt;p&gt;For question 4, write down the matrix product in summation form. It can be found here: &lt;a href=&quot;http://en.wikipedia.org/wiki/Matrix_multiplication&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Matrix_multiplication&lt;/a&gt;. Then, set one matrix to the transpose of the other. It can be found here: &lt;a href=&quot;http://en.wikipedia.org/wiki/Transpose&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Transpose&lt;/a&gt;. Then, you will notice in the summation that a*b = b*a, thus you will find $B = A*A^T \iff B_{ij} = B_{ji} \iff B = B^T$&lt;/p&gt;&#xA;" OwnerUserId="8140" LastActivityDate="2011-03-12T02:14:51.957" CommentCount="1" />
  <row Id="26495" PostTypeId="2" ParentId="26477" CreationDate="2011-03-12T02:18:01.890" Score="5" ViewCount="" Body="&lt;p&gt;I like your approach, it's not one I would have thought of.  Here's an idea that seems to work except when characteristic of F is 3.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $x$ satisfies $x^3 - 3x + 1 = 0$, then so does $1 - \frac{1}{x}$.  Clearly zero is not a root of the given polynomial, so this makes sense.  Also if $x = 1 - \frac{1}{x}$, then $x$ would satisfy the equation $x^2 - x + 1 = 0$, and taken in combination with the cubic polynomial above, we would infer $3 = 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So except for characteristic 3, the formula $1 - \frac{1}{x}$ turns out to cyclically permute the three roots of the cubic polynomial.  Thus the polynomial either splits in a field F or is irreducible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added:&lt;/strong&gt; Arturo has nailed the characteristic 3 case.&lt;/p&gt;&#xA;" OwnerUserId="3111" LastEditorUserId="3111" LastEditorDisplayName="" LastEditDate="2011-03-12T02:59:33.087" LastActivityDate="2011-03-12T02:59:33.087" CommentCount="1" />
  <row Id="26496" PostTypeId="2" ParentId="26477" CreationDate="2011-03-12T02:23:01.203" Score="1" ViewCount="" Body="&lt;p&gt;Here is one way to do it using almost no theory, just playing with algebra.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose the polynomial has a root $a$ in $F$.  If you divide $x^3 - 3x + 1$ by $x-a$ the quotient is the polynomial&#xA;$$&#xD;&#xA;x^2 + ax + a^2 - 3&#xD;&#xA;$$&#xA;in $F(a)[x]$.  From the quadratic formula (assuming characteristic $\neq 2$ here, for the moment, I guess) you can see that this will have its roots in $F(a)$ if and only if $12-3a^2$ is a square in $F(a)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You may know that any element in $F(a)$ can be written in the form $pa^2 + qa + r$ for some $p,q,r$ in $F$.  The idea is to square this symbolic expression, rewrite it as a polynomial in $a$ of degree at most $2$, and then see if values $p, q, r$ in $F$ can be found making this equal to $-3a^2 + 0a + 12$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the identity $a^3 = 3a - 1$ (from the fact that $a$ is a root of the given polynomial) and $a^4 = 3a^2 - a$ (from multiplying the previous identity by $a$) one gets&#xA;$$&#xD;&#xA;(pa^2 + qa + r)^2 = (3p^2 + 2pr + q^2) a^2 + (-p^2 + 2qr + 6pq) a + (r^2 - 2pq).&#xD;&#xA;$$&#xA;So the goal is to find $p,q,r$ satisfying $3p^2 + 2pr + q^2 = -3$, $-p^2 + 2qr + 6pq = 0$, and $r^2 - 2pq = 12$.  Since we don't know anything about $F$, we might optimistically look for these $p, q, r$ in $\mathbb{Z}$.  This system of $3$ polynomial equations in $3$ integer unknowns can be fed to software, and one finds that e.g. $p = 2$, $q = 1$, and $r = -4$ give a solution (no matter what $F$ is!).  So the roots of $x^3 - 3x + 1$ are all in $F(a)$, and we can actually write formulas for them: $a$, $\frac{-a + (2a^2 + a -4)}{2}$, and $\frac{-a - (2a^2 + a - 4)}{2}$, which simplify to $a$, $a^2 - 2$, $-a^2 - a + 2$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Although we assumed characteristic $\neq 2$ to use the quadratic formula, we can immediately check that the single identity $a^3 - 3a + 1 = 0$ is indeed enough to ensure that&#xA;$$&#xD;&#xA;(x-a)(x - (a^2 - 2)) (x - (-a^2 - a + 2)) = x^3 - 3x + 1.&#xD;&#xA;$$&#xA;[In detail: expanding, the coefficient of $x^3$ is $1$ on the nose, the coefficient of $x^2$ is $0$ on the nose, and the coefficient of $x$ is a polynomial in $a$ that, when divided by $a^3 - 3a + 1$, has a remainder of $-3$; similarly the constant term is a polynomial in $a$ that, when divided by $a^3 - 3a + 1$, has remainder of $1$.]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if $x^3 -3x + 1$ has one root in $F$, it splits in $F$ for the reason that we can explicitly write the other two roots as polynomials in the one that we already have.&lt;/p&gt;&#xA;" OwnerUserId="3840" LastActivityDate="2011-03-12T02:23:01.203" />
  <row Id="26497" PostTypeId="1" AcceptedAnswerId="26550" CreationDate="2011-03-12T02:25:19.177" Score="4" ViewCount="193" Body="&lt;p&gt;Hi I'm having trouble with this proof.  I'm not sure if I did the first part right and how I should use the second part properly.  I am following the hint in the book to use the Axiom of Replacement to prove step three below and follow through with Replacement and Union to finalize the proof.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Prove that the Cartesian Product of two sets is a set&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ A \times B = \left \{ \left \langle u, v \right \rangle \mid u \in A \land v \in B \right \} $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Taking the Axiom of Replacement&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ \forall u \forall v \forall w \left ( \psi \left ( u, v \right ) \land \psi \left ( u, w \right ) \rightarrow v = w  \right ) \rightarrow \forall z \exists y \forall v \left ( v \in y \leftrightarrow \left ( \exists u \in z \right ) \psi \left ( u, v \right ) \right ) $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) First prove that the following is a set &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ \forall x \left ( \lbrace \langle x, y \rangle \mid y \in t \rbrace \right ) $ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;4) Taking (1) from Section 6&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ \langle x, y \rangle = \langle u, v \rangle \iff x = u \land y = v $    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;5) Use then this u and v for the Axiom of Replacement reduced to the following where v is unique&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ \forall z \exists y \forall v \left ( v \in y \leftrightarrow \left ( \exists u \in z \right ) \psi \left ( u, v \right ) \right ) $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;6) $ \forall t \left ( t \in y \right ) $ and $ z = \langle x, y \rangle $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ \forall \langle x, y \rangle \exists t \forall v \left ( v \in t \leftrightarrow \left ( \exists u \in \langle x, y \rangle \right ) \langle u, v \rangle \right ) $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence $ y \in t $ proving that $ \forall x \left ( \lbrace \langle x, y \rangle \mid y \in t \rbrace \right ) $ is a set by the Axiom of Replacement&lt;/p&gt;&#xA;&#xA;&lt;p&gt;7) Now taking the Axiom of Union&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ A \cup B = \lbrace x \mid x \in A \lor x \in B \rbrace $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;8) And the Axiom of Replacement&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ \forall u \forall v \forall w \left ( \psi \left ( u, v \right ) \land \psi \left ( u, w \right ) \rightarrow v = w  \right ) \rightarrow \forall z \exists y \forall v \left ( v \in y \leftrightarrow \left ( \exists u \in z \right ) \psi \left ( u, v \right ) \right ) $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;9) Prove that the Cartesian Product of two sets is a set&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ A \times B = \left \{ \left \langle u, v \right \rangle \mid u \in A \land v \in B \right \} $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there is something I can make more clear that would allow others to more easily aid me let me know. :|&lt;/p&gt;&#xA;" OwnerUserId="7816" LastEditorUserId="7816" LastEditorDisplayName="" LastEditDate="2011-03-12T07:30:57.850" LastActivityDate="2011-03-12T13:06:20.317" Title="Proof: Cartesian Product of Two Sets is a Set ZF" Tags="&lt;set-theory&gt;" AnswerCount="3" CommentCount="4" FavoriteCount="1" />
  <row Id="26498" PostTypeId="2" ParentId="26428" CreationDate="2011-03-12T02:28:17.283" Score="0" ViewCount="" Body="&lt;p&gt;The running time of the algorithm of Kruskal is dominated by the sorting time of the edges according to their weights. In your case you can do this in linear time. &#xA;The rest of Kruskal's algorithm also runs in linear time. So you get linear running time after all.&lt;/p&gt;&#xA;" OwnerUserId="4857" LastActivityDate="2011-03-12T02:28:17.283" CommentCount="4" />
  <row Id="26499" PostTypeId="1" AcceptedAnswerId="26511" CreationDate="2011-03-12T02:34:15.620" Score="6" ViewCount="156" Body="&lt;p&gt;Prove or disprove the following statment: For all integers b, c,and d, if x is a rational number such that $x^2+bx+c=d$, then x is an integer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a homework question from the book Discrete Mathematics for Computer Scientists by Stein, Drysdale and Bogart.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I since x is rational I thought I could start off with:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;${(\frac{m}{n})}^2+b\frac{m}{n}=d-c$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I don't know where to go from here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or I could try using the quadratic formula&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x=\frac{1}{2}\left(\pm\sqrt{b^{2}-4c+4d}-b\right)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but I am very weak with elementary number theory that I don't know where to go.  I am thinking that regardless of if $\sqrt{b^{2}-4c+4d}$ is an integer or not, the fact that I have &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x=\frac{1}{2}*\pm$ SomeNumber&lt;/p&gt;&#xA;&#xA;&lt;p&gt;means that x is not an integer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am new to writing proofs, and unfortunately, I don't really know how to prove this.  Any hints would be appreciated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit:  By plugging in simple numbers, for example x=1, b=1, c=1 and d=3 I can see that x is probably an integer, for all integers b,c,and d - so that means my thinking about the quadratic formula is not correct.  I will still work on this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2nd Edit:  Now I plug in more numbers and don't get integers. For example $x^2+2x+3=4$.  I am also new to this site, so I am not sure if I should continue to edit the post or write in the comments sections anytime I think of something new.  Please advise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3rd Edit: I think I know what to do.  The last section of the book covered universal quantifiers.  I believe the authors are are trying to get me to realize that they are saying $\forall b, c, d \in Z$ and I only need to give one one example for which the assertion is untrue.  And in my previous edit, b=2, c=3, and d=4 did not result in x being an integer.&lt;/p&gt;&#xA;" OwnerUserId="6391" LastEditorUserId="6391" LastEditorDisplayName="" LastEditDate="2011-03-12T03:27:41.783" LastActivityDate="2011-03-13T18:48:11.380" Title="For all integers b, c, and d, if x is rational such that x^2+bx+c=d, then x is an integer" Tags="&lt;homework&gt;&lt;elementary-number-theory&gt;&lt;discrete-mathematics&gt;" AnswerCount="2" CommentCount="5" />
  <row Id="26500" PostTypeId="1" AcceptedAnswerId="26552" CreationDate="2011-03-12T02:43:02.263" Score="5" ViewCount="81" Body="&lt;p&gt;I was reading &lt;a href=&quot;http://theorymine.co.uk/?go=faq#trivial&quot; rel=&quot;nofollow&quot;&gt;something&lt;/a&gt; which contained the following statement:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;It is a well-established mathematical result that theories consisting only of recursive definitions... are inherently consistent.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Does this result have a name? Could anyone point me to literature about this?&lt;/p&gt;&#xA;" OwnerUserId="2549" LastActivityDate="2011-03-12T13:37:15.627" Title="Recursively defined systems are always consistent?" Tags="&lt;logic&gt;&lt;proof-theory&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="2" />
  <row Id="26502" PostTypeId="2" ParentId="26497" CreationDate="2011-03-12T03:12:10.197" Score="8" ViewCount="" Body="&lt;p&gt;How does this proof suit you? I'm afraid it makes no use of the axiom of Replacement though, but I think it may be a little clearer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider any ordered pair $\langle u, v\rangle$ such that $u\in A$ and $v\in B$. By definition, $\langle u, v\rangle=\{\{u\},\{u,v\}\}$. Now $\{u\}\subseteq A$ so $\{u\}\subseteq A\cup B$, hence $\{u\}\in\mathscr{P}(A\cup B)$. Similarly, $\{u,v\}\subseteq A\cup B$, so $\{u,v\}\in\mathscr{P}(A\cup B)$. It follows that $\{\{u\},\{u,v\}\}\subseteq\mathscr{P}(A\cup B)$, so $\{\{u\},\{u,v\}\}\in\mathscr{PP}(A\cup B)$, and all of these sets are known to exist by the Union Axiom and the Power Set Axiom. Then by the axiom schema of specification, you know there exists a set &#xA;$$&#xD;&#xA;E=\{t\in\mathscr{PP}(A\cup B)\ |\ \exists u\exists v(u\in A\land v\in B\land t=\langle u,v\rangle\}.&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This set is unique by the Axiom of Extensionality, and is the set of all ordered pairs $\langle u, v\rangle$ with $u\in A$ and $v\in B$. &lt;/p&gt;&#xA;" OwnerUserId="1583" LastActivityDate="2011-03-12T03:12:10.197" CommentCount="1" />
  <row Id="26503" PostTypeId="1" CreationDate="2011-03-12T03:23:20.150" Score="3" ViewCount="105" Body="&lt;p&gt;&lt;strong&gt;Problem&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Given $g(x)$, solve the equation $f&amp;#39;(f^{-1}(x)) = \frac{1}{g(x)}$ for an invertible and differentiable function $f(x)$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So far I have tried setting $y = f^{-1}(x) \Leftrightarrow x = f(y)$, obtaining the differential equation&#xA;$$&#xD;&#xA;f&amp;#39;(y) = \frac{dx}{dy} = \frac{1}{g(x)}&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which we then solve to obtain $y$ in terms of $x$, i.e. to obtain the inverse function $f^{-1}(x)$. If this is easily inverted then we can find $f(x)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I am interested in is if anyone knows a better way to solve this, desirably one which allows us to determine $f(x)$ directly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm new to this kind of equation so please correct me if there's a better term for it than &quot;differential-functional&quot; :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt;&lt;br&gt;&#xA;I probably should say that in the particular context I am considering, $g(x)$ is the norm of a non-zero vector $\vec{r}(x)$ and is hence always positive.&lt;/p&gt;&#xA;" OwnerUserId="7090" LastEditorUserId="7090" LastEditorDisplayName="" LastEditDate="2011-03-13T11:33:56.063" LastActivityDate="2011-03-13T13:32:20.487" Title="A differential-functional equation: $f'(f^{-1}(x)) = 1/g(x)$" Tags="&lt;differential-equations&gt;&lt;functional-equations&gt;" AnswerCount="1" FavoriteCount="0" />
  <row Id="26506" PostTypeId="1" AcceptedAnswerId="26523" CreationDate="2011-03-12T03:40:38.013" Score="5" ViewCount="112" Body="&lt;p&gt;We all agree that the theory of class fields plays an eminent role in modern number theory.&lt;br&gt;&#xA;Nevertheless, what was our main concern is that how to solve various Diophantine equations to which the clues do not appear clear in the theory of class fields. And my question is:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;With respect to the beauty and perfection of the theory of class fields, can we use it to solve some mysterious Diophantine equations? Or can we apply it to other fields in Mathematics, such as Complex Analysis?  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The last question arises as one of my friends used to tell me that there is a mathematician who developes the theory of complex analysis in terms of class fields in one of his books. Nonetheless, when I looked up to his books, I didn't find any hint to this. Moreover, when I asked him about this again, he didn't say much. This stimulates somehow my imagination such that I am wondering if there indeed is such a theory.&lt;br&gt;&#xA;Any response is appreciated.   &lt;/p&gt;&#xA;" OwnerUserId="6792" LastActivityDate="2011-03-12T15:53:44.317" Title="Any resource of the applications of the theory of class fields." Tags="&lt;reference-request&gt;&lt;algebraic-number-theory&gt;&lt;applications&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="4" />
  <row Id="26507" PostTypeId="2" ParentId="26499" CreationDate="2011-03-12T03:43:22.430" Score="2" ViewCount="" Body="&lt;p&gt;Hint:  First, you can combine $c$ and $d$ as you only care about their difference.  There is the theorem that the square root of a positive integer is either integer or irrational.  You are right that one example disproves the assertion &quot;for all b,c, and d&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Added:  If you look at the quadratic formula, the only threat to x being integral (if it is rational) is the division by 2.  But if the square root is integral, it must have the same parity as b.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2011-03-12T15:00:41.170" LastActivityDate="2011-03-12T15:00:41.170" />
  <row Id="26508" PostTypeId="1" CreationDate="2011-03-12T03:48:00.800" Score="0" ViewCount="81" Body="&lt;p&gt;Suppose &lt;em&gt;M&lt;/em&gt; is a 20 by 3 complex matrix, and I'd like to SVD. For example, in Matlab, I can do easily with:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;[U, S, V] = svd(M);&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where &lt;em&gt;U&lt;/em&gt;, &lt;em&gt;S&lt;/em&gt;, and &lt;em&gt;V&lt;/em&gt; are complex matrices of20x20, 20x3, and 3x3, respectively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I'd like to make &lt;em&gt;U&lt;/em&gt;, a 20x20 matrix, to be either &lt;em&gt;real&lt;/em&gt; matrix or &lt;em&gt;near-real&lt;/em&gt; matrix (i.e., imaginary parts in the matrix are very small). &lt;em&gt;S&lt;/em&gt; and &lt;em&gt;V&lt;/em&gt; are okay with complex matrices. Having a real matrix is desirable for me in that it can minimize the overhead of computation in my program code.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have no idea such approximation or decomposition exists. Is it possible?&lt;/p&gt;&#xA;" OwnerUserId="4400" LastActivityDate="2011-05-11T10:56:34.923" Title="SVD for a complex matrix and approximation to a real matrix" Tags="&lt;matrices&gt;&lt;approximation&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" />
  <row Id="26509" PostTypeId="2" ParentId="26492" CreationDate="2011-03-12T04:03:07.730" Score="2" ViewCount="" Body="&lt;p&gt;So, (1) is at least &quot;sometimes true&quot;, which you can establish by simply producing an example (which you have). The only question is whether it is &lt;em&gt;always&lt;/em&gt; true.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, from studying least squares you probably saw that $\mathrm{rank}(A^TA) = \mathrm{rank}(A)$. Since $A$ is $3\times 4$ and $A^TA$ is $4\times 4$, then $A^TA$ is not invertible. That means that &lt;em&gt;if&lt;/em&gt; you can find a solution to $A^TA\mathbf{x} = A^T\mathbf{z}$ for some specific $\mathbf{z}$, then that equation has infinitely many solutions (take any specific solution $\mathbf{s}$, and for every solution $\mathbf{n}$ to $A^TA\mathbf{x}=\mathbf{0}$, of which there are infinitely many because $A^TA$ is not invertible, you have that $\mathbf{s}+\mathbf{n}$ is a solution. So it really comes down to asking &lt;em&gt;whether&lt;/em&gt; $A^TA\mathbf{x} = A^T\mathbf{z}$ always has at least &lt;em&gt;one&lt;/em&gt; solution for all $\mathbf{z}$, or whether there is at least one $\mathbf{z}$ for which there are no solutions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;At this point, thinking about least squares, you should either figure out how to argue that there &lt;em&gt;has&lt;/em&gt; to be at least one solution (and therefore infinitely many) for all $\mathbf{z}$; or else find an example where it does not have at least one solution. If you can find an example, then the example where it fails and the example where it works shows that it is &quot;sometimes true&quot;; if you can show it always has solutions, then you have your proof that it is &quot;true.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Added.&lt;/em&gt; There are &lt;em&gt;two&lt;/em&gt; parts to this argument as I sketched it: first you need to show that for every choice of $\vec{z}$, the system $A^TA\vec{x}=A^T\vec{z}$ has at least &lt;em&gt;one&lt;/em&gt; solution. For that, what you know about least squares solutions should be useful, so you are on the right track when you realized that this is related to least squares.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remember: if $\vec{z}$ is any vector, will $A\vec{x}=\vec{z}$ &lt;em&gt;always&lt;/em&gt; have a least squares solutions (yes; you already explained why: it's just the vector in the range of $A$ that is closed to $\vec{z}$); and, will a least squares solution &lt;em&gt;always&lt;/em&gt; be a solution to $A^TA\vec{x}=A^T\vec{z}$? (You probably &lt;em&gt;proved&lt;/em&gt; or &lt;em&gt;saw&lt;/em&gt; a theorem about this when you were figuring out how to do least squares; if so, you can just invoke that result).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After proving that for every $\vec{z}$ the system $A^TA\vec{x}=A^T\vec{z}$ always has at least one solution, you need to show that it always has infinitely many solutions. This is equivalent to asking whether $A^TA$ is invertible or not (you should explain why this is so; that is, that if it has more than one solution then $A^TA$ is not invertible, if $A^TA$ is not invertible, then it has more than one solution). Then explain why $A^TA$ cannot be invertible in this case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) Again: you know that this is at least &quot;sometimes true&quot; (for example, $\mathbf{A}=\mathbf{I}$). The question is whether it is &lt;em&gt;always&lt;/em&gt; true or not. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What characterizations of orthogonal matrices do you know? I assume it's that $A$ is orthogonal if and only if $A^TA = AA^T = I$? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Added.&lt;/em&gt; You are on the right track. You know that $A\vec{u}\cdot A\vec{u}=1$, and therefore that $\vec{u}\cdot(A^TA\vec{u}) = 1$ for all unit vectors $\vec{u}$, and therefore that $\vec{u} \cdot (\vec{u}-A^TA\vec{u}) = 0$ for all unit vectors $\vec{u}$. &#xA;Now notice that $I - A^TA$ is symmetric; so you can find an orthonormal basis of eigenvectors of $I - A^TA$. If $\vec{u}$ is a unit eigenvector of $I-A^TA$, what does $\vec{u}\cdot(I- A^TA)\vec{u} = 0$ tell you? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(3): Again, you know this is at least &quot;sometimes true&quot;. So, you need to determine: Is $T$ linear? Is $T$ one-to-one? Is it onto? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Linearity I'll leave to you. For one-to-one: suppose $T(M)=0$; does it follow that $M=0$? Start playing with $SMS^{-1}=0$ and see what you can conclude.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For surjectivity: given a matrix $N$, can you find an $M$ such that $T(M)=N$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Added&lt;/em&gt;: You've solved (3); it's always true.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-12T06:05:48.300" LastActivityDate="2011-03-12T06:05:48.300" CommentCount="24" />
  <row Id="26510" PostTypeId="2" ParentId="20328" CreationDate="2011-03-12T04:17:36.253" Score="0" ViewCount="" Body="&lt;p&gt;gary, I think your question is too general. For instance, there is something called 'cofibration' in topology, which deals with this type of problem under a strong condition: Namely, $f:A\rightarrow Y$ is continuous and has an extension iff every $g\colon A\rightarrow Y$ satisfies this property where $g$ is homotopic to f. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If a space $(X,A)$ has homotopy extension property with respect to $Y$ then it is easier to check if a map $f\colon A\rightarrow X$ has an extension, because , now, there are billions of maps that should simultaneously have extensions. However, even under this condition, there is no general theorem that I know.  &lt;/p&gt;&#xA;" OwnerUserId="7796" LastActivityDate="2011-03-12T04:17:36.253" />
  <row Id="26511" PostTypeId="2" ParentId="26499" CreationDate="2011-03-12T04:19:10.200" Score="4" ViewCount="" Body="&lt;p&gt;This is simply the monic quadratic case of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Rational_root_theorem&quot; rel=&quot;nofollow&quot;&gt;Rational Root Test&lt;/a&gt;. You could specialize that proof, or else proceed similarly to various irrationality proofs for square-roots. $\: $ E.g. $\:$  below is a &lt;a href=&quot;http://math.stackexchange.com/questions/4467/a1-2-is-either-an-integer-or-an-irrational-number/4585#4585&quot;&gt;proof that I discovered in high-school.&lt;/a&gt; First I present the proof for square-roots - where the idea is clearer. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;THEOREM&lt;/strong&gt; $\ $ For $\rm\: c\in \mathbb Z\:,\:$ any rational root $\rm\:r\:$ of $\rm\ x^2 = c\ $ is integral.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof&lt;/strong&gt; $\  $ Put $\rm\ r = m/n\ $ with $\rm\:(m,n) = 1\:.\:$ Then $\rm\ j\ m-k\ n \;=\; 1\;$  for some $\:\rm j,k \in \mathbb{Z}\;\;$ by Bezout.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence $\;\;\:\rm 0\: =\: (m-n\:r)\:(k+j\:r) \:=\: mk-njc + (j\ m-k\ n)\ r  \ \Rightarrow\ r = -mk+njc \ \in\ \mathbb{Z}\quad$ &lt;strong&gt;QED&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This proof easily extends to the root of a general &lt;em&gt;monic&lt;/em&gt; quadratic as follows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;THEOREM&lt;/strong&gt; $\ $ For $\rm\:b,c\in\mathbb Z\:,\:$ any rational root $\rm\:r\:$ of $\rm\ x^2 = b\ x + c\ $ is integral.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof&lt;/strong&gt; $\ $ Put $\rm\ r = m/n,\ (m,n)=1\:.\ $ Then $\rm\:(m-nb,n)=1\ $ so $\rm\ \exists\ j,k\in \mathbb Z\: :\ 1 = j\ (m-nb)-k\ n $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence $\rm\ 0 = (m-n\:r)\:(k+j\:r)\ =mk + (jm-kn)\:r-nj\:(br+c) = mk-njc + (j\ (m-nb)-k\ n)\ r$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the coefficient of $\rm\:r\:$ in the prior term is $1\:,\ $ so we deduce $\rm\ r \:=\: -mk+njc \ \in\ \mathbb Z\quad$ &lt;strong&gt;QED&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you learn about &lt;a href=&quot;http://math.stackexchange.com/questions/4713/if-the-order-divides-a-prime-p-then-the-order-is-p-or-1/4715#4715&quot;&gt;denominator ideals&lt;/a&gt; then you'll see that the above proof simply says that the denominator ideal of $\rm\:r\:$ contains $\rm\:n\:$ and $\rm\:n\:r = m\:,\:$ so it contains their gcd $\rm\:(n,m) = 1\:,\ $ so $\rm\ r\in \mathbb Z\:.$ Using Dedekind's notion of &lt;a href=&quot;http://mathoverflow.net/questions/30220/abstract-thought-vs-calculation/30313#30313&quot; rel=&quot;nofollow&quot;&gt;conductor ideal,&lt;/a&gt; the proof easily generalizes to higher degree monic polynomials, yielding that PIDs are integrally closed. &lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2011-03-13T18:48:11.380" LastActivityDate="2011-03-13T18:48:11.380" CommentCount="5" />
  <row Id="26512" PostTypeId="2" ParentId="26465" CreationDate="2011-03-12T04:29:27.567" Score="2" ViewCount="" Body="&lt;p&gt;In a comment you say that you want to know whether from knowing the values of the power sum symmetric polynomials on $z_1,\ldots,z_n$ you can determine whether the $z_i$ are pairwise distinct and nonzero, or not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The power sum symmetric polynomials&#xA;\begin{align*}&#xA;p_1(z_1,\ldots,z_n) &amp;amp;= z_1+\cdots + z_n\\&#xA;p_2(z_1,\ldots,z_n) &amp;amp;= z_1^2 + \cdots + z_n^2\\&#xA;&amp;amp;\vdots\\&#xA;p_n(z_1,\ldots,z_n) &amp;amp;= z_1^n + \cdots + z_n^n&#xA;\end{align*}&#xA;generate the ring of symmetric polynomials over $\mathbb{Q}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, you can express the elementary symmetric polynomials in $z_1,\ldots,z_n$ as sums of rational multiples of products of the power sum symmtric polynomials. So you can express&#xA;\begin{align*}&#xA;e_1(z_1,\ldots,z_n) &amp;amp;= z_1+\cdots + z_n\\&#xA;e_2(z_1,\ldots,z_n) &amp;amp;= z_1z_2 + z_1z_3+\cdots+z_1z_n+z_2z_3+\cdots+z_{n-1}z_n\\&#xA;&amp;amp;\vdots\\&#xA;e_n(z_1,\ldots,z_n) &amp;amp;= z_1\cdots z_n&#xA;\end{align*}&#xA;in terms of $p_1,\ldots,p_n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Plugging in the values $\mathcal{Z}_1,\ldots,\mathcal{Z}_n$ that you know for $p_1,\ldots,p_n$, that would determine the values of $e_1,\ldots,e_n$. With the values of $e_1,\ldots,e_n$, you get the coefficients of&#xA;$$(x-z_1)(x-z_2)\cdots(x-z_n)$$&#xA;because the coefficients are precisely the elementary symmetric polynomials in $z_1,\ldots,z_n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But it is a trivial matter to determine if this polynomial has zero as a root, or if it has repeated roots (for the latter, check the gcd with the derivative). Thus, from the values of $p_1,\ldots,p_n$ you can determine if the $z_i$ are pairwise distinct and nonzero, or not, which in turn tells you if $M$ is nonsingular or not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In summary: yes, you can tell if $M$ is singular or nonsingular, by computing the polynomial $(x-z_1)\cdots(x-z_n)$ via symmetric polynomials.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How does this relate to the question in the first paragraph, though? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;First: if you write the symmetric polynomial $z_1\cdots z_n$ in terms of the power symmetric polynomials, then setting this equal to $0$ gives you a polynomial with rational coefficients in $\mathcal{Z}_1,\ldots,\mathcal{Z}_n$. The $z_i$ are nonzero if and only if the $\mathcal{Z}_i$ do &lt;em&gt;not&lt;/em&gt; satisfy this polynomial. This gives you precisely the locus of vectors $(\mathcal{Z}_1,\ldots,\mathcal{Z}_n)^T$ that are excluded by the condition &quot;all $z_i$ are nonzero.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example: take the case $n=2$. Then we express $z_1z_2$ in terms of $z_1+z_2$ and $z_1^2 + z_2^2$:&#xA;$$ z_1z_2 = \frac{1}{2}\left( (z_1+z_2)^2 - (z_1^2+z_2^2)\right) = \frac{1}{2}(\mathcal{Z}_1^2 - \mathcal{Z}_2).$$&#xA;Therefore, one of $z_1$ and $z_2$ is zero if and only if $\mathcal{Z}_1^2 - \mathcal{Z}_2 = 0$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, the vectors $(\mathcal{Z}_1,\mathcal{Z}_2)$ that are excluded by the condition &quot;$z_1$ and $z_2$ are both nozero&quot; are exactly the points that lie on the grap of $y=x^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is fairly straightforward to do in any value of $n$, once you express $z_1\cdots z_n$ in terms of the power sum symmetric polynomials. So certainly, it is easy to write down a polynomial expression that determines exactly the locus of all &quot;bad $(\mathcal{Z}_1,\ldots,\mathcal{Z}_n)$&quot; that lead to at least one $z_i$ equal to $0$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added, Edited.&lt;/strong&gt; (&lt;em&gt;This is what I get for posting at 1am. I forget things like &quot;discriminant&quot; and &quot;resultant.&quot;&lt;/em&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To determine if the polynomial has repeated roots, you can just look at the &lt;a href=&quot;http://en.wikipedia.org/wiki/Discriminant&quot; rel=&quot;nofollow&quot;&gt;discriminant&lt;/a&gt;. The discriminant is a function of the coefficients of $(x-z_1)\cdots(x-z)n)$, and since we can compute the coefficients using the $\mathcal{Z}_i$, then we can compute the discriminant. It will be  a polynomial in the coefficient, hence a polynomial in the $\mathcal{Z}_i$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The key is that the discriminant is also equal to the product $\displaystyle\prod_{1\leq i\lt j\leq n}(z_i - z_j)^2$, and therefore the discriminant is equal to $0$ if and only if the polynomial has repeated roots. (In the $n=2$ case, what I was looking at, $\frac{1}{2}\mathcal{Z}_1^2 =\mathcal{Z}_2$, was just the discriminant of the quadratic polynomial, but with my algebra error and the hour, I failed to notice).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So: setting the discriminant equal to $0$ gives you the condition for repeated roots, and setting the constant term equal to $0$ gives you the condition for a root to be zero. The two together determine two varieties in affine $n$-space, and the locus of &quot;bad tuples&quot; is precisely their union. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In short: yes, you can determine exactly which $(\mathcal{Z}_1,\ldots,\mathcal{Z}_n)$ correspond to singular matrices and which to nonsingular matrices; it is described as the union of two affine varieties in $\mathbb{C}^n$, which can be written as two polynomials (with rational coefficients) in the $\mathcal{Z}_i$. &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-12T20:55:57.000" LastActivityDate="2011-03-12T20:55:57.000" CommentCount="5" />
  <row Id="26513" PostTypeId="2" ParentId="26489" CreationDate="2011-03-12T04:33:20.840" Score="2" ViewCount="" Body="&lt;p&gt;I'm going to assume that $0 \preceq A$, that is, that $A$ is itself positive semidefinite, as the result fails for more general $A$ (and the thing defined as the &quot;ellipsoid&quot; of $A$ is not a geometrical ellipsoid; e.g. if $A$ is the diagonal $2 \times 2$ matrix with diagonal $(-1, 0)$ and $B$ is the $2 \times 2$ identity matrix, we do have $A \preceq B$ in the sense that $B - A$ is positive semidefinite, but the &quot;ellipsoid&quot; of $A$ is an unbounded set whose boundary is the hyperbola $\{(x,y) \in \mathbb{R}^2: y^2 - x^2 = 1\}$ and this is not contained in the ellipsoid of $B$, aka the unit disc in $\mathbb{R}^2$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once we assume that $A$ is positive semidefinite it follows from $A \preceq B$ that $B = (B-A) + A$ is also positive semidefinite (since a sum of PSD matrices is PSD).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the definition of ellipsoid involves the inverse we should also assume that $A$ is invertible (and this implies that $B$ is--- e.g. because it implies the existence of a positive number $c$ with $cI \preceq A$ and transitivity of $\preceq$ implies then that $cI \preceq B$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now it is well known (but nontrivial) that $0 \preceq A \preceq B$ and $A$ invertible together imply that $0 \preceq B^{-1} \preceq A^{-1}$.  [The easiest way to prove this depends sensitively on how &quot;positive semidefinite&quot; is defined, and there are various approaches to this.  A functional analysis book, for example, would define the concept in a way that does not reference eigenvalues, since the notion makes perfectly good sense for operators that don't have any.]  Look it up your favorite reference or find a proof that you like online.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyhow: once you know the given hypotheses imply that $B^{-1} \preceq A^{-1}$ it is simple.  For any vector $x$ you have from this operator relation that $x^t (A^{-1} - B^{-1}) x \geq 0$, which implies (by expanding the matrix product and moving one term over to the other side) that for any $x$&#xA;$$&#xD;&#xA;x^t A^{-1} x \geq x^t B^{-1} x;&#xD;&#xA;$$&#xA;thus if the left hand side is $\leq 1$, so is the right hand side, thus the ellipsoid of $A$ is contained in that of $B$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the converse: if it is not the case that $A \preceq B$ (under the standing hypotheses that $A, B$ are both PSD and invertible) then it cannot be the case that $B^{-1} \preceq A^{-1}$ (or we could deduce from this that $A \preceq B$ by the well known but nontrivial fact I mentioned earlier, with the roles of $A$ and $B$ now played by $B^{-1}$ and $A^{-1}$), so there must be a vector $x_0$ with $x_0^t(A^{-1} - B^{-1}) x_0 &amp;lt; 0$, and for this vector we have&#xA;$$&#xD;&#xA;x_0^t B^{-1} x_0 &amp;gt; x_0^t A^{-1} x_0.&#xD;&#xA;$$&#xA;Since $A^{-1}$ is invertible, the right hand side must be positive, so by scaling $x_0$ as necessary we can make the right hand side equal to $1$.  The inequality then shows us that the vector $x_0$ is in the ellipsoid of $A$, but not in that of $B$.&lt;/p&gt;&#xA;" OwnerUserId="3840" LastActivityDate="2011-03-12T04:33:20.840" />
  <row Id="26514" PostTypeId="1" AcceptedAnswerId="26518" CreationDate="2011-03-12T04:34:20.527" Score="2" ViewCount="110" Body="&lt;p&gt;How to prove: Continuous function maps compact set to compact set using real analysis?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i.e. if $f: [a,b] \rightarrow \mathbb{R}$ is continuous, then $f([a,b])$ is closed and bounded. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have proved the bounded part. So now I need some insight on how to prove $f([a,b])$ is closed, i.e. $f([a,b])=[c,d]$. From Extreme Value Theorem, we know that $c$ and $d$ can be achieved, but how to prove that if $c &amp;lt; x &amp;lt; d$, then $x \in f([a,b])$ ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="7429" LastEditorUserId="2548" LastEditorDisplayName="" LastEditDate="2011-03-12T05:01:42.537" LastActivityDate="2011-03-12T05:03:38.773" Title="Proving continuous image of compact sets are compact" Tags="&lt;real-analysis&gt;" AnswerCount="1" CommentCount="7" />
  <row Id="26515" PostTypeId="2" ParentId="17482" CreationDate="2011-03-12T04:35:56.317" Score="1" ViewCount="" Body="&lt;p&gt;Johan, isn't $\displaystyle \int_C Fdz=2i\iint_{\Omega}\frac{\partial F}{\partial \bar{z}}dxdy$ true where F=u+iv? I think every example that you mention in your first sentence, is infact, an application of Green's theorem (I didn't check my last sentence.)&lt;/p&gt;&#xA;" OwnerUserId="7796" LastActivityDate="2011-03-12T04:35:56.317" />
  <row Id="26516" PostTypeId="1" AcceptedAnswerId="26517" CreationDate="2011-03-12T04:36:03.003" Score="1" ViewCount="68" Body="&lt;p&gt;I don't know how can I  to prove the following (this isn't homework, i'm just flipping through a book):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;if   a &amp;isin; &lt;strong&gt;F&lt;/strong&gt;, v &amp;isin; &lt;strong&gt;V&lt;/strong&gt;, and av=0&lt;/p&gt;&#xA;&#xA;&lt;p&gt;then a=0 or v=0&lt;/p&gt;&#xA;" OwnerUserId="5171" LastEditorUserId="5171" LastEditorDisplayName="" LastEditDate="2011-03-12T04:45:10.407" LastActivityDate="2011-03-12T04:51:13.730" Title="property of vector-space" Tags="&lt;linear-algebra&gt;&lt;vector-space&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="26517" PostTypeId="2" ParentId="26516" CreationDate="2011-03-12T04:51:13.730" Score="1" ViewCount="" Body="&lt;p&gt;If it hasn't been proven already, try proving the preliminary facts that $0_Fv=0_V$ and $a0_V=0_V$ for $a\in F$ and $v\in V$. This follows by using that fact that $0=0+0$ (for both the zero element of $F$ and of $V$), and using distributivity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $a=0$, you are done. So suppose $a\neq 0$, and hence has a multiplicative inverse. Try multiplying both sides of your equation by the inverse of $a$, and recall the axiom that $1v=v$ where $1$ is the multiplicative identity of $F$, to get your conclusion.&lt;/p&gt;&#xA;" OwnerUserId="1583" LastActivityDate="2011-03-12T04:51:13.730" CommentCount="2" />
  <row Id="26518" PostTypeId="2" ParentId="26514" CreationDate="2011-03-12T05:03:38.773" Score="1" ViewCount="" Body="&lt;p&gt;Lindsay, what you need is the intermediate value theorem, its proof is given in wikipedia.&lt;/p&gt;&#xA;" OwnerUserId="7796" LastActivityDate="2011-03-12T05:03:38.773" />
  <row Id="26519" PostTypeId="1" AcceptedAnswerId="26581" CreationDate="2011-03-12T05:13:30.390" Score="2" ViewCount="90" Body="&lt;p&gt;I'm asked to find the optimal $k$ such that $\cos{x}-1+x^2/2=O(x^k)\quad (x\to 0)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By Taylor's theorem I know that $\cos{x}-1+x^2/2=(1/4!)\xi^4$ for some $\xi\in (0,x)$. So $0&amp;lt;\cos{x}-1+x^2/2&amp;lt;(1/4!)x^4$, that is $\cos{x}-1+x^2/2=O(x^4)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My reasoning is as follows: if $0&amp;lt;x&amp;lt;1$ (which can be assumed since we're interested in $x\to 0$), we have $x^4&amp;lt;x^{4-\delta}$, so $x^{4-\delta}$ is not a 'better' order.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next I try to see whether $\cos{x}-1+x^2/2=O(x^{4+\delta})$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\cos{x}-1+x^2/2}{x^{4+\delta}}=\frac{1}{x^\delta}\frac{\cos{x}-1+x^2/2}{x^4}.$$ &#xA;But by repeated application of L'Hôpital's, &#xA;$$\lim_{x\to 0}\frac{\cos{x}-1+x^2/2}{x^4}=1/24,$$ &#xA;so $\displaystyle\lim_{x\to 0} \frac{\cos{x}-1+x^2/2}{x^{4+\delta}}=\infty$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This seems to show that $x^4$ is indeed the optimal order.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So is all that correct? Also, $1/24=1/4!$, so surely that's not a coincidence, but I don't see how to arrive at that limit in a different way (I don't know how that $\xi$ above depends on $x$).&lt;/p&gt;&#xA;" OwnerUserId="1313" LastEditorUserId="1313" LastEditorDisplayName="" LastEditDate="2011-03-12T11:30:25.760" LastActivityDate="2011-03-12T17:56:00.603" Title="&quot;optimal&quot; big-O order for $\cos{x}-1+x^2 / 2$" Tags="&lt;calculus&gt;&lt;numerical-methods&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="26520" PostTypeId="2" ParentId="26503" CreationDate="2011-03-12T05:19:00.663" Score="2" ViewCount="" Body="&lt;p&gt;Chen, let $h=f^{-1}.$ Multiply both sides by $h&amp;#39;$ to get $(f&amp;#39;h)h&amp;#39;=h&amp;#39;/g.$ But observe that $f&amp;#39;(f^{-1}(x))(f^{-1})&amp;#39;(x)=(f\circ f^{-1})&amp;#39;(x)=1.$ Hence, we have $(f^{-1})&amp;#39;(x)=g(x).$  So, $f^{-1}(x)=\int_0^x g(t) dt$.. &lt;/p&gt;&#xA;" OwnerUserId="7796" LastEditorUserId="1543" LastEditorDisplayName="" LastEditDate="2011-03-13T13:32:20.487" LastActivityDate="2011-03-13T13:32:20.487" CommentCount="6" />
  <row Id="26521" PostTypeId="1" AcceptedAnswerId="26562" CreationDate="2011-03-12T05:31:05.210" Score="1" ViewCount="152" Body="&lt;p&gt;How can I write $\frac{1}{a+x}$, $a$ a non-zero positive constant, in exponential terms in the form of $y = Ce^{-kx}$? I've tried to use to Taylor series but they only seem to work for $x &amp;lt; 1$.&lt;/p&gt;&#xA;" OwnerUserId="8143" LastEditorUserId="6296" LastEditorDisplayName="" LastEditDate="2011-03-12T20:15:45.677" LastActivityDate="2011-04-07T08:35:38.907" Title="How can I write $\frac{1}{(a+x)}$ as an exponential function $y = Ce^{-kx}$?" Tags="&lt;functional-analysis&gt;&lt;taylor-expansion&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="1" />
  <row Id="26522" PostTypeId="2" ParentId="26521" CreationDate="2011-03-12T05:46:51.717" Score="0" ViewCount="" Body="&lt;p&gt;You will not find a way to make $\frac{1}{a+x}=C\exp(-kx)$ as they are functions with very different shapes.  They both go to 0 as $x \to \infty$, but $C\exp(-kx)$ does it &lt;em&gt;much&lt;/em&gt; faster.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-03-12T05:46:51.717" CommentCount="2" />
  <row Id="26523" PostTypeId="2" ParentId="26506" CreationDate="2011-03-12T06:33:04.417" Score="13" ViewCount="" Body="&lt;p&gt;Let me address the question as to whether class field theory can be used to solve some Diophantine equations.  The answer is certainly &lt;em&gt;yes&lt;/em&gt;.  One historical example is given by Kummer's work on Fermat's Last Theorem; this relied on algebraic number theoretic results to do with cyclotomic fields, much of which can be reinterpreted as special cases of class field theory.  (One place to see this discussed is the nice historical chapter on Kummer's work by Michael Rosen&#xA;in the book &lt;em&gt;Modular forms and Fermat's Last Theorem&lt;/em&gt; (Cornell, Silverman, Stevens eds.).)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At a more basic level, consider the question of solving the following three equations:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x^2 + y^2 = p$$&#xA;$$ x^2 + 5 y^2 = p$$&#xA;$$x^2 + 23 y^2 = p$$&#xA;where $p$ is a prime and $x$ and $y$ are integers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Essentially (i.e. excluding some small primes, namely $p = 2$ in the first,&#xA;$p = 2$ or $5$ in the second, and $p = 2$ or $23$ in the third) each of&#xA;these questions can be rephrased as asking whether the prime $p$ splits into a product of &lt;em&gt;pricipal&lt;/em&gt; ideals in the extension $K:= \mathbb Q(\sqrt{-D})$,&#xA;where $D = 4,$ $20$, and $23$ respectively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the question of whether $p$ splits is easily answered; it is just a &#xA;question of whether the Jacobi symbol $\bigl( \frac{p}{D} \bigr)$ is equal&#xA;to $1$ or not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the question of whether $p$ splits into a product of principal ideals is more subtle; it is a question of whether $p$ splits completely in the Hilbert class field&#xA;$H$ of $K.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now in the case $D = 4$, we know that $K$ has class number one, and so $K = H$.&#xA;Thus we can solve $x^2 + y^2 = p$ if and only if $p \equiv 1 \bmod 4$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $D = 20$, then $K$ has class number two, and in fact $H = K(\sqrt{-1})$&#xA;is the compositum of the field $\mathbb Q(\sqrt{-4})$ and $\mathbb Q(\sqrt{-20})$.  Thus $p$ splits completely in $H$ if and only if $\bigl(\frac{p}{4}\bigr) = 1$&#xA;and $\bigl(\frac{p}{20}) = 1$, i.e. if and only if $\bigl(\frac{p}{4}\bigr) = 1$&#xA;and $\bigl(\frac{p}{5}) = 1$, i.e. if and only if $p \equiv 1 \bmod 4$ and $p \equiv \pm 1 \bmod 5$, i.e. if and only if $p \equiv 1 \text{ or } 9&#xD;&#xA;\bmod 20$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, if $D = 23$ then $K$ has class number three, and so $H$ is &lt;em&gt;not&lt;/em&gt; an abelian extension of $\mathbb Q$; rather it is an $S_3$-extension.  The equivalence of class fields and abelian extensions says that $H$ is &lt;em&gt;not&lt;/em&gt; a class field of $\mathbb Q$, and so there is &lt;em&gt;no&lt;/em&gt; congruence condition on $p$ that determines whether or not $p$ splits completely in $H$.  Thus there is &lt;em&gt;no&lt;/em&gt; congruence condition on $p$ that determines whether or not we can solve&#xA;$x^2 + 23y^2 = p$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instead, one has the following result from non-abelian class field theory&#xA;(due, in this case, to Hecke, but best understood from a modern viewpoint of being a special case of Langlands's general program for non-abelian class field theory):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can solve $x^2 + 23y^2 = p$ if and only if the coefficient of $q^p$ in the product &#xA;$$q\prod_{n=1}^{\infty}(1-q^n)(1-q^{23 n}),$$&#xA;which is &lt;em&gt;a priori&lt;/em&gt; equal to $-1,0,$ or $2$, is in fact equal to $2$.&#xA;Computing the product, one finds e.g. that the first such prime is $p = 59 =&#xD;&#xA;6^2 + 23\cdot 1^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(This infinite product is a certain &lt;em&gt;modular form&lt;/em&gt;, which is a particular kind of non-abelian analogue of a Jacobi symbol.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whether class field theory has applications to complex analysis, I don't know,&#xA;but the appearance of modular forms at the end of the above discussion shows that complex analysis has applications to class field theory.&lt;/p&gt;&#xA;" OwnerUserId="221" LastEditorUserId="221" LastEditorDisplayName="" LastEditDate="2011-03-12T15:53:44.317" LastActivityDate="2011-03-12T15:53:44.317" CommentCount="8" />
  <row Id="26524" PostTypeId="1" CreationDate="2011-03-12T06:41:25.557" Score="0" ViewCount="73" Body="&lt;p&gt;I have a set of N elements. I want to select few elements from the set. Now each element i of the set has a probability P(i) of getting selected. Then how can I figure out the probability of selecting atleast K elements from the set?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;K elements are required to be different. Some of the probabilities may be zero, but we are assured that there are atleast K elements whose probability is non zero.&lt;/p&gt;&#xA;" OwnerUserId="8144" LastEditorUserId="8144" LastEditorDisplayName="" LastEditDate="2011-03-13T15:33:42.770" LastActivityDate="2011-03-13T18:59:23.957" Title="Probability of selecting K elements of different probability from a set of N elemets" Tags="&lt;probability&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="26525" PostTypeId="2" ParentId="26444" CreationDate="2011-03-12T06:51:34.840" Score="4" ViewCount="" Body="&lt;p&gt;Imagine that you have a non-zero map $\phi:V \to V$, say with $V$ finite dimensional.  Now ask yourself what other maps can you get by pre- and post-composing with other endomorphisms, and then taking linear combinations.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you understand the finite dimensional case well, you will be ready for the infinite dimensional case.  (Arturo's hint will play an important role!)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: In order to prepare yourself for the infinite dimensional case, you will be better off trying to prove simplicity in the finite dimensional case using arguments with endomorphisms, rather than shifting to the language of matrices and relying on matrix manipulations.  (At least, if you do the latter, try to translate what you have done back into the language of endomorphisms and reinterpret the argument in that language.)&lt;/p&gt;&#xA;" OwnerUserId="221" LastEditorUserId="221" LastEditorDisplayName="" LastEditDate="2011-03-13T13:46:52.633" LastActivityDate="2011-03-13T13:46:52.633" />
  <row Id="26526" PostTypeId="2" ParentId="26345" CreationDate="2011-03-12T07:02:07.327" Score="1" ViewCount="" Body="&lt;p&gt;Another (marginally different) proof: Consider a maximal cyclic subgroup $H$ of $G$ (one exists since $G$ is finite).  If $g \in G$ then the assumption implies either that $\langle g \rangle \subset H$ or else that $H \subset \langle g \rangle$.  In the latter case, we have equality, by maximality of $H$, and so in all cases $g \in H$.  Thus $H = G$,&#xA;and so $G$ is cyclic.&lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2011-03-12T07:02:07.327" />
  <row Id="26527" PostTypeId="1" AcceptedAnswerId="26534" CreationDate="2011-03-12T07:22:50.650" Score="2" ViewCount="114" Body="&lt;p&gt;If I know the formal power series, I know how to find the closed form:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle F = \sum_{n=0}^{\infty} {X^n} = 1 + X^1 + X^2 + X^3 + ...$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle F \cdot X = X \cdot \sum_{n=0}^{\infty} {X^n} = X^1 + X^2 + X^3 + X^4 + ...$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle F - F \cdot X = 1 $$&#xA;$$\displaystyle F = \frac 1 {1 - X} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if I only know the closed form $\frac 1 {1 - X}$, how do I turn it back into the series $1 + X^1 + ...$? In other words, how do I do extract the coefficients if I only know the closed form and I do not know that $\frac 1 {1 - X}$ corresponds to $1 + X^1 + ...$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My textbook and everywhere I looked at seems to avoid talking about this, and somehow magically transform things back and forth with a set of known formulas. Is there a better way to do this, or is formula matching the best we can do?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; This is the type of questions I need to solve:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Find the coefficient of $X^8$ in the formal power series $(1 - 3X^4)^{-6}$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2166" LastEditorUserId="2166" LastEditorDisplayName="" LastEditDate="2011-03-12T07:35:23.573" LastActivityDate="2011-03-12T19:08:39.033" Title="Turning a closed-form generating function back to ordinary power series" Tags="&lt;discrete-mathematics&gt;&lt;generating-functions&gt;&lt;power-series&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="26528" PostTypeId="1" AcceptedAnswerId="26558" CreationDate="2011-03-12T07:31:28.770" Score="6" ViewCount="243" Body="&lt;p&gt;I encountered this problem in my probability class, and we weren't able to solve it, so I would like to know the answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have $m$ balls and $n$ boxes, with $n &amp;lt; m$, and you insert the balls into the boxes randomly, what is the probability that all the boxes have at least one ball in it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem doesn't specify if the balls are distinguishable or not, so you may assume either, so another question would be, if you assume they are distinguishable will you get the same answer as assuming they are not distinguishable? (This would be great because I think the non-distinguishable case is easier).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I appreciate any insight on the problem.&lt;/p&gt;&#xA;" OwnerUserId="1318" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2011-03-12T07:36:39.317" LastActivityDate="2011-03-14T16:26:36.947" Title="$m$ balls $n$ boxes probability problem" Tags="&lt;probability&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="3" />
  <row Id="26529" PostTypeId="2" ParentId="26528" CreationDate="2011-03-12T07:34:21.983" Score="3" ViewCount="" Body="&lt;p&gt;I have provided the solution to a similar problem &lt;a href=&quot;http://math.stackexchange.com/questions/11468/if-there-are-200-students-in-the-library-how-many-ways-are-there-for-them-to-be/11475#11475&quot;&gt;here&lt;/a&gt;. The solution has the  general case worked out as well.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastActivityDate="2011-03-12T07:34:21.983" />
  <row Id="26530" PostTypeId="1" CreationDate="2011-03-12T07:42:04.567" Score="3" ViewCount="65" Body="&lt;p&gt;If we are given two sides say $a$,$b$ and an angle $X$,How can we determine whether this angle $X$ is opposite angle to $a$ or $b$ (i.e $A$,$B$) or the third included angle $C$ ?&lt;/p&gt;&#xA;" OwnerUserId="2109" LastActivityDate="2011-03-12T18:40:24.937" Title="Determining Angle" Tags="&lt;geometry&gt;&lt;trigonometry&gt;" AnswerCount="2" />
  <row Id="26531" PostTypeId="2" ParentId="26530" CreationDate="2011-03-12T07:44:40.147" Score="1" ViewCount="" Body="&lt;p&gt;Generally, you can't.  It is quite possible that the given angle measure and side lengths allow the angle to be the included angle C (determining 1 triangle that way), or either A or B (possibly determining 0, 1, or 2 triangles in each of these two configurations).&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2011-03-12T07:44:40.147" />
  <row Id="26532" PostTypeId="1" CreationDate="2011-03-12T07:58:51.240" Score="0" ViewCount="71" Body="&lt;p&gt;Rules: hardly any, except:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;No additional figure;&lt;/li&gt;&#xA;&lt;li&gt;Functions are well known;&lt;/li&gt;&#xA;&lt;li&gt;It is correct;&lt;/li&gt;&#xA;&lt;li&gt;Plus, please make it &lt;em&gt;entertaining&lt;/em&gt;!&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Here are a few possible solutions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\DeclareMathOperator{\Ceil}{Ceil}&#xD;&#xA;\DeclareMathOperator{\rank}{rank}&#xD;&#xA;\DeclareMathOperator{\zeros}{zeros}&#xD;&#xA;\begin{eqnarray*}&#xD;&#xA;24&amp;amp;=&amp;amp;2\times(10-3)+10\\&#xD;&#xA;&amp;amp;=&amp;amp;2^{10}-10^3\\&#xD;&#xA;&amp;amp;=&amp;amp;(2+3-10\div10)!\\&#xD;&#xA;&amp;amp;=&amp;amp;(\sqrt{10\times10}-2)\times3\\&#xD;&#xA;&amp;amp;=&amp;amp;2\times10+3+\lg10\\&#xD;&#xA;&amp;amp;=&amp;amp;(10\div2-\lg10)\times3!\\&#xD;&#xA;&amp;amp;=&amp;amp;3!\times2\times(\lg10+\lg10)\\&#xD;&#xA;&amp;amp;=&amp;amp;\dbinom{10}{3}\times2\div10\\&#xD;&#xA;&amp;amp;=&amp;amp;(10\div2-10\bmod3)!\\&#xD;&#xA;&amp;amp;=&amp;amp;\Ceil(10\div3)+10\times2\\&#xD;&#xA;&amp;amp;=&amp;amp;(10-2)\times\min(3,10)\\&#xD;&#xA;&amp;amp;=&amp;amp;(10+2)\div\sin(3*10^\circ)\\&#xD;&#xA;&amp;amp;=&amp;amp;3\ll(2+10\div10)\\&#xD;&#xA;&amp;amp;=&amp;amp;3\times(10-2+\rank(\zeros(10)))\hspace{30 pt}\text{//In MatLab}\\&#xD;&#xA;&amp;amp;=&amp;amp;(2&amp;#39;!+3&amp;#39;!+10&amp;#39;!+10&amp;#39;!)!\hspace{30 pt}\text{//The Ultimate Universal Formulae}\\&#xD;&#xA;&amp;amp;=&amp;amp;(\cos(2&amp;#39;)+\cos(3&amp;#39;)+\cos(10&amp;#39;)+\cos(10&amp;#39;))!\hspace{30 pt}\text{//Another Universal Formulae}&#xD;&#xA;\end{eqnarray*}&#xD;&#xA;$&lt;/p&gt;&#xA;" OwnerUserId="8106" LastEditorUserId="8106" LastEditorDisplayName="" LastEditDate="2011-03-12T08:07:35.897" LastActivityDate="2011-03-12T08:07:35.897" Title="Solve 2, 3, 10, 10 for 24 game, other than these ..." Tags="&lt;symbolic-computation&gt;" CommentCount="6" />
  <row Id="26533" PostTypeId="1" CreationDate="2011-03-12T07:59:22.907" Score="0" ViewCount="79" Body="&lt;p&gt;Is any Lie Group a homogeneous space?&lt;/p&gt;&#xA;" OwnerUserId="5312" LastActivityDate="2011-03-12T07:59:22.907" Title="Are Lie Groups Homogeneous Spaces?" Tags="&lt;differential-geometry&gt;&lt;lie-groups&gt;&lt;riemannian-geometry&gt;" CommentCount="4" />
  <row Id="26534" PostTypeId="2" ParentId="26527" CreationDate="2011-03-12T08:09:40.423" Score="5" ViewCount="" Body="&lt;p&gt;The quickest way to transform your generating function to a power series is to have a table of formulae handy. For a given rational function, you would use partial fraction expansion if necessary and switch back to the power series by looking at the appropriate entry in the table for each term. If you look at Herbert Wilf's book &lt;a href=&quot;http://www.math.upenn.edu/~wilf/gfologyLinked2.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.math.upenn.edu/~wilf/gfologyLinked2.pdf&lt;/a&gt; , in section 2.5, he has a list of such formulae.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;\frac{1}{(1-x)^{k+1}} = \sum_n \binom{n+k}{n} x^n&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, for your case, we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;\frac{1}{(1-3x^4)^6} = \sum_n \binom{n+5}{n} 3^n x^{4n}&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The coefficient of $x^8$ in this expansion is $3^2 \times \binom{7}{2} = 189$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is actually not difficult to derive the identity. You start with&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;\frac{1}{1-x} = 1+x+x^2 +\ldots&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take derivative on both sides k times and you will get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;\frac{k!}{(1-x)^{k+1}} = \sum_n [(n+k)(n+k-1) \ldots n]x^n&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;this simplifies to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;\frac{1}{(1-x)^{k+1}} = \sum_n \binom{n+k}{n} x^n&#xA;\end{equation}&lt;/p&gt;&#xA;" OwnerUserId="2641" LastActivityDate="2011-03-12T08:09:40.423" />
  <row Id="26535" PostTypeId="2" ParentId="26424" CreationDate="2011-03-12T08:10:07.630" Score="4" ViewCount="" Body="&lt;p&gt;There is also a nice geometric-probability solution to the problem.  For simplicity, let $L=1$, with $x$ and $y$ as you describe.  The space of all possible values of $x$ and $y$ is the unit square $[0,1]\times[0,1]$, with each point being equally likely (as $x$ and $y$ are uniformly distributed).  In order for the three pieces to form a triangle, each piece must have length less than $\frac{1}{2}$, so:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;if $x&amp;lt;y$: $x&amp;lt;\frac{1}{2}$, $y-x&amp;lt;\frac{1}{2}$, and $1-y&amp;lt;\frac{1}{2}$;&lt;/li&gt;&#xA;&lt;li&gt;or if $x&amp;gt;y$: $y&amp;lt;\frac{1}{2}$, $x-y&amp;lt;\frac{1}{2}$, and $1-x&amp;lt;\frac{1}{2}$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Graphing these in the unit square gives the shaded region shown below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/u7yNE.png&quot; alt=&quot;graph of inequalities&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This region is $\frac{1}{4}$ of the total area of the square, so the probability is $\frac{1}{4}$.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;It's worth noting &lt;a href=&quot;http://math.stackexchange.com/questions/676/probability-that-a-stick-randomly-broken-in-two-places-can-form-a-triangle&quot;&gt;this similar but slightly different question&lt;/a&gt;, which arose from a mis-written Monte Carlo simulation of this problem.&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2011-03-12T08:10:07.630" CommentCount="2" />
  <row Id="26536" PostTypeId="2" ParentId="26528" CreationDate="2011-03-12T09:06:01.357" Score="5" ViewCount="" Body="&lt;p&gt;Let S be the number of ways to distribute m balls into n boxes with no restriction and T be number of ways to distribute m balls into n boxes with each boxes having at least one balls.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we can calculate S and T, then the probability that all the boxes have at least one ball in it is simply $\frac T S$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, we count the number of ways to distribute m balls into n boxes with no restriction. In other words, the number of solutions to the equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x_1 + x_2 + x_3 + ... + x_n = m$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can calculate this with: $S = {m+n-1 \choose n-1}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second, we count the number of ways to &lt;em&gt;distribute&lt;/em&gt; m balls into n boxes, with the restriction that all boxes have at least 1 balls. In other words, the number of solutions to the equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x_1 + x_2 + x_3 + ... + x_n = m \textrm{ where } x_i &amp;gt;= 1 \textrm { for all } i = 1..n$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's reformulate this second question. Let's put a single ball into each boxes ($x_i = y_i + 1$) and then distribute the other $m-n$ balls:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(y_1 + 1) + (y_2 + 1) + (y_3 + 1) + ... + (y_n + 1) = m$$&#xA;$$y_1 + y_2 + y_3 + ... + y_n = m-n$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So now, we can calculate $T = {(m-n)+n-1 \choose n-1} = {m-1 \choose n-1}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, the probability of each box having at least one ball is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;\frac T S&#xD;&#xA;=&#xD;&#xA;\frac {m-1 \choose n-1} {m+n-1 \choose n-1} &#xD;&#xA;=&#xD;&#xA;\frac {(m-1)!m!} {(m-n)!(m+n-1)!}&#xD;&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="2166" LastEditorUserId="2166" LastEditorDisplayName="" LastEditDate="2011-03-12T13:38:58.290" LastActivityDate="2011-03-12T13:38:58.290" CommentCount="8" />
  <row Id="26537" PostTypeId="2" ParentId="26306" CreationDate="2011-03-12T09:41:34.550" Score="0" ViewCount="" Body="&lt;p&gt;Subramani: Let $f\colon U\rightarrow V$ be defined where $U\subset M$ and $V\subset N,$ and $M,N$ are two manifolds. Every manifold comes with tangent spaces. If $x\in M$ then the tangent space at $x$ is the space of &quot;directional derivative&quot; operators. Indeed, locally for any nice curve passing though $x$ one can differentiate $f$ at $x$ in the direction of that curve, which in return, gives a quantity which describes the rate of change of $f$ on this curve at $x.$ The space of maps $D_{\gamma}$ ,whose input is the function and whose output is the directional derivative of that function,  is the tangent space. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Differentiability is a different thing: Namely, $f$ is differentiable at $x\in U$ if  (1) There is a linear map between the tangent space at $x$ and the tangent space of $f(x)$ (2) This maps approximates $f$ near the point $x.$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When (1) and (2) both holds one can compute $f$ locally near $x$ via the linear map. But the domain and range of this linear map are linear spaces, not the points of manifolds. To approximate $f,$ one projects the tangent space onto the manifold.&#xA;In summary:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(*) a function can have directional derivative along only one curve, but to define the derivative it should have directional derivatives in all directions+ they should form a linear map&lt;/p&gt;&#xA;" OwnerUserId="7796" LastActivityDate="2011-03-12T09:41:34.550" />
  <row Id="26538" PostTypeId="2" ParentId="17482" CreationDate="2011-03-12T09:59:39.597" Score="3" ViewCount="" Body="&lt;p&gt;It is easy to compute $\int_{-\infty}^\infty e^{-x^2/2}dx$ by an elementary argument, but maybe you want the integral $\int_{-\infty}^\infty e^{-x^2/2}\cos(\lambda x)dx$ without having complex analysis (Cauchy's theorem) at your disposal. So you can apply Green's theorem to the vector field &#xA;$$(P,Q):=e^{(y^2-x^2)/2}\bigl(\cos(x y),\sin(x y)\bigr)$$&#xA;and an elongated rectangle $[-a,a]\times[0,\lambda]$; then let $a\to\infty$.&lt;/p&gt;&#xA;" OwnerUserId="1303" LastActivityDate="2011-03-12T09:59:39.597" />
  <row Id="26539" PostTypeId="2" ParentId="26508" CreationDate="2011-03-12T10:13:36.777" Score="2" ViewCount="" Body="&lt;p&gt;You are asking for something impossible. Remember that though a Hermitian matrix has all its eigenvalues real, its eigenvectors are usually complex. Since the singular value decomposition of a matrix $\mathbf A$ is essentially equivalent to the eigendecomposition of the matrix $\mathbf A^T \mathbf A$ (or $\mathbf A\mathbf A^T$), which is Hermitian positive definite, $\mathbf U$ is very likely unitary (as opposed to orthogonal).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, it seems that the way you are using MATLAB, you are having the &lt;code&gt;svd()&lt;/code&gt; routine output the so-called &quot;full SVD&quot; instead of the &quot;economy SVD&quot; version. For the economy version (which should help you a great deal in minimizing computational overhead), $\mathbf U$ is 20×3, and $\mathbf S$ and $\mathbf V$ are 3×3. Fortunately, it is a simple matter to have MATLAB output the economy version: &lt;code&gt;[U, S, V] = svd(M, 0);&lt;/code&gt; For what it's worth, the related QR decomposition also supports the output of an economy-sized version as well. Have a look into MATLAB documentation for more details.&lt;/p&gt;&#xA;" OwnerUserId="8147" LastActivityDate="2011-03-12T10:13:36.777" />
  <row Id="26540" PostTypeId="2" ParentId="26486" CreationDate="2011-03-12T10:14:54.717" Score="3" ViewCount="" Body="&lt;p&gt;If you have Cauchy's criterion for functions at your disposal there is a simpler proof: Given an $\epsilon&amp;gt;0$, by uniform continuity there is a $\delta&amp;gt;0$ with $|f(x)-f(x&amp;#39;)|&amp;lt;\epsilon$ for all $x, x&amp;#39;\in\ ]a,a+\delta[\&amp;gt;$. Since $\epsilon$ was arbitrary it follows by Cauchy's criterion that $\lim_{x\to a+} f(x)=:f(a)$ exists, and similarly at $b$.&lt;/p&gt;&#xA;" OwnerUserId="1303" LastActivityDate="2011-03-12T10:14:54.717" CommentCount="1" />
  <row Id="26541" PostTypeId="1" AcceptedAnswerId="26658" CreationDate="2011-03-12T10:27:26.117" Score="2" ViewCount="54" Body="&lt;p&gt;For Linear complementarity problems (LCP) like&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\mathbf{Mz}+\mathbf{q} \ge \mathbf{0}$&lt;/li&gt;&#xA;&lt;li&gt;$\mathbf{z} \ge \mathbf{0}$&lt;/li&gt;&#xA;&lt;li&gt;$\mathbf{z}^{\mathrm{T}}(\mathbf{Mz}+\mathbf{q}) = 0$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;there exists a vast amount of material and algorithms. But what about this kind of problem:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\mathbf{Mz}+\mathbf{q} \ge \mathbf{0}$&lt;/li&gt;&#xA;&lt;li&gt;$\mathbf{Az} +\mathbf{b}\ge \mathbf{0}$&lt;/li&gt;&#xA;&lt;li&gt;$(\mathbf{Az} +\mathbf{b})^{\mathrm{T}}(\mathbf{Mz}+\mathbf{q}) = 0$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Is it also called LCP? Can it be transformed into the form above?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The real question is, how to solve it without the need to invert one of the matrices explicitly.&lt;/p&gt;&#xA;" OwnerUserId="4514" LastEditorUserId="6622" LastEditorDisplayName="" LastEditDate="2011-03-12T10:29:49.987" LastActivityDate="2011-03-13T05:45:25.943" Title="Linear complementarity problem - Classification" Tags="&lt;optimization&gt;&lt;inequality&gt;" AnswerCount="1" FavoriteCount="0" />
  <row Id="26542" PostTypeId="1" AcceptedAnswerId="26572" CreationDate="2011-03-12T10:41:27.680" Score="5" ViewCount="142" Body="&lt;p&gt;They are similar.All contain a map and can define the section,the fiber of the fiber bundle is just like the stalk of the sheaf.But what are the differences between them,maybe sheaf is more abstract and can break down, the fibre bundle is more geometric and must keep itself continue. Any other differences?  &lt;/p&gt;&#xA;" OwnerUserId="4706" LastActivityDate="2011-03-12T17:11:04.863" Title="What are the differences between fiber bundle and sheaf?" Tags="&lt;sheaf-theory&gt;&lt;fiber-bundles&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="1" />
  <row Id="26543" PostTypeId="1" AcceptedAnswerId="26560" CreationDate="2011-03-12T10:45:57.990" Score="0" ViewCount="108" Body="&lt;p&gt;I have been doing multiply and divide but never really tried to understand it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;E.g.  &lt;code&gt;(4 x 12 / 4)&lt;/code&gt; could be written as: (&lt;code&gt;12&lt;/code&gt;, &lt;code&gt;4 x 3&lt;/code&gt;, &lt;code&gt;48 / 4&lt;/code&gt;) but yet the result is always same, can anyone prove or explain why?&lt;/p&gt;&#xA;" OwnerUserId="8149" LastEditorUserId="8149" LastEditorDisplayName="" LastEditDate="2011-03-12T11:10:21.417" LastActivityDate="2011-03-12T15:21:32.633" Title="Multiplication and Division" Tags="&lt;number&gt;" AnswerCount="2" CommentCount="6" FavoriteCount="1" />
  <row Id="26544" PostTypeId="2" ParentId="26475" CreationDate="2011-03-12T10:53:40.847" Score="4" ViewCount="" Body="&lt;p&gt;1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The so-called &quot;transition region&quot; for $J_n(z)$ is well known; among its many applications, the usual three-term recursion relation for Bessel functions is numerically sound in the direction of increasing $n$ for as long as $z&amp;gt;n$. As a guide of sorts,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$J_n(z)\approx\frac{x^n}{2^n\Gamma(n+1)}\quad\mathrm{if}\quad 0 &amp;lt; z \ll n$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$J_n(z)\approx\sqrt{\frac{2}{\pi z}}\cos\left(z-\frac{\pi}{4}-\frac{n\pi}{2}\right)\quad\mathrm{if}\quad n \ll z$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The expressions for $\frac{\mathrm d}{\mathrm d\nu}J_\nu(z)$ are &lt;a href=&quot;http://dlmf.nist.gov/10.15&quot; rel=&quot;nofollow&quot;&gt;rather complicated&lt;/a&gt;; there's no reason to expect a simple expression for the solution of $\frac{\mathrm d}{\mathrm d\nu}J_\nu(z)=0$, as is usual for any transcendental equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are a lot of asymptotic results for Bessel functions with varying order in the DLMF; you might want to look into them. Offhand, I recall results for envelopes of $J_n(z)$ for fixed $n$ and varying $z$, but not for your situation; I'll edit this answer when I come across results of relevance to you.&lt;/p&gt;&#xA;" OwnerUserId="8147" LastActivityDate="2011-03-12T10:53:40.847" />
  <row Id="26545" PostTypeId="1" AcceptedAnswerId="26546" CreationDate="2011-03-12T11:43:50.793" Score="4" ViewCount="217" Body="&lt;p&gt;I am taking a topology course and we are now learning open and closed set. I am a bit confused to how to prove that a set is closed or opened, how should I approach these kind of problems. For example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question 1:&lt;/strong&gt; Let $(\mathcal{X},d)$ be an arbitrary metric space. Prove that any set which contains a finite number of points $\{x_1,x_2,\ldots,x_n\}$ is closed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Solution:&lt;/strong&gt; If we take the point $x_i$ where $1\leq i\leq n$ and no matter how small we make $r$, in the ball some points are outside of our set. Hence the set is closed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question 2:&lt;/strong&gt; Let $(\mathcal{X},d)$ be an arbitrary metric space. Prove that $B(\mathcal{X},r)=\{y\in\mathcal{X}:d(x,y)&amp;lt;r\}$ is open.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Solution:&lt;/strong&gt; Take a point $y_0$ in the ball, and make a ball with radius $\frac{1}{2}(r-d(x,y_0))$, the all the points in this ball are in the actual ball in the question. Hence the ball in the question is open.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are these proves right? I myself do not feel that I am proving anything. Could you please teach me the correct proof if these are not correct?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="7340" LastActivityDate="2011-03-12T12:19:44.530" Title="The general argument to prove a set is closed/open" Tags="&lt;topology&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="26546" PostTypeId="2" ParentId="26545" CreationDate="2011-03-12T11:59:18.013" Score="4" ViewCount="" Body="&lt;p&gt;For the first question: let's first prove that a one-point set is closed. Fix the point $x$, for the set $\{x\}$. Now consider any other point $y$. If $d(x,y)=\epsilon$, take $B(y,\epsilon/2)$. This open ball does not intersect $\{x\}$ (why?). This means that the complement of $\{x\}$ is open, which implies the set is closed. Now you know that any finite union of closed sets is closed (straight from the definition of topology by application of De Morgan's laws), and the result follows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It isn't true that any ball around $x_i$, no matter how small, will contain points other than itself. To see this, consider the set of integers with $d(m,n)=|m-n|$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the second question, your approach is the right one, just don't forget (unless you didn't) to show that the ball you constructed is indeed contained in the original ball, by using properties of $d$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So one way of determining whether a set is closed, is by looking at whether its complement is open, which in some cases could be easier.&lt;/p&gt;&#xA;" OwnerUserId="1313" LastEditorUserId="1313" LastEditorDisplayName="" LastEditDate="2011-03-12T12:19:44.530" LastActivityDate="2011-03-12T12:19:44.530" />
  <row Id="26547" PostTypeId="2" ParentId="26543" CreationDate="2011-03-12T12:16:07.880" Score="1" ViewCount="" Body="&lt;p&gt;You want to interpret the expression&#xA;$$a\times b\,\,\, /\,\,\, c$$&#xA;We could first multiply $a$ and $b$, and then divide the result by $c$, i.e.&#xA;$$(a\times b)\,\,\, /\,\,\, c\hskip0.5in (1)$$&#xA;or we could multiply $a$ by the quantity $b/c$, i.e.&#xA;$$a\times(b/c)\hskip0.5in (2)$$&#xA;Here is the reason expressions (1) and (2) are equal: dividing by a number $c$ is the same thing as multiplying by its reciprocal $1/c$. This is because, for any number $x$, the number $x/c$ is the unique solution to the equation&#xA;$$\_\_\_\times c = x\hskip0.5in (\ast)$$&#xA;Anything that can go in the spot $\_\_\_$ is equal to $x/c$. &#xA;But we also have $$\left(x\times (1/c)\right)\times c = x\times\left((1/c)\times c\right)=x\times 1 =x$$&#xA;where the first equality is justified by the associative law of multiplication. This means that $x\times (1/c)$ is also the solution to the equation $(\ast)$, and hence must be equal to $x/c$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, how does this help? We can rewrite (1) and (2) as&#xA;$$(a\times b)\times (1/c)\hskip0.5in (1)$$&#xA;and&#xA;$$a\times (b\times (1/c))\hskip0.5in (2)$$&#xA;respectively, and the associative law of multiplication tells us that these two quantities must be equal.&lt;/p&gt;&#xA;" OwnerUserId="264" LastActivityDate="2011-03-12T12:16:07.880" CommentCount="1" />
  <row Id="26548" PostTypeId="1" CreationDate="2011-03-12T12:31:57.627" Score="9" ViewCount="201" Body="&lt;p&gt;Fermat solved the Diophantine equation $(x^2)^2 + (y^2)^2 = z^2$ using descent, the key step was using the Pythagorean triples:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$x^2 = u^2 - v^2$&lt;/li&gt;&#xA;&lt;li&gt;$y^2 = 2 u v$&lt;/li&gt;&#xA;&lt;li&gt;$z = u^2 + v^2$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;but then it is seen that $x,v,u$ is another Pythagorean triple,&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$x = m^2 - n^2$&lt;/li&gt;&#xA;&lt;li&gt;$v = 2 m n$&lt;/li&gt;&#xA;&lt;li&gt;$u = m^2 + n^2$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;then the observation that $y^2 = 2^2 u m n$ (along with some coprimality conditions) implies that $u$,$m$ and $n$ are squares - which gives us the descent step needed.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I wanted to use this technique to solve $(x^2)^2 + (y^2)^2 = 2 z^2$ but it does not seem to work out. Parameterizing the curve through $(1,1,1)$ gives:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$x^2 = -m^2 - 2 m n + n^2$&lt;/li&gt;&#xA;&lt;li&gt;$y^2 = m^2 - 2 m n - n^2$&lt;/li&gt;&#xA;&lt;li&gt;$z = m^2 + n^2$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Nothing here seems to match up so well with the original problem (as it did in Fermat's case) and parameterizing through different points doesn't help either.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Is there any way to make descent work here? Was Fermat just lucky or is there a reason why descent can't work out here?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;update: I found the parametrization&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$x^2 = 2m^2 - n^2$&lt;/li&gt;&#xA;&lt;li&gt;$y^2 = 2m^2 - 4mn + n^2$&lt;/li&gt;&#xA;&lt;li&gt;$z = 2m^2 - 2mn + n^2$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;which looks like a better candidate for descent but I haven't been able to actually perform it. Since I cannot see how to show that $n$ and $x$ are both squares.&lt;/p&gt;&#xA;" OwnerUserId="4290" LastEditorUserId="4290" LastEditorDisplayName="" LastEditDate="2011-03-12T22:49:42.143" LastActivityDate="2011-05-04T06:35:04.780" Title="How to get Fermat descent working on other conics?" Tags="&lt;number-theory&gt;&lt;descent&gt;" AnswerCount="1" CommentCount="7" FavoriteCount="2" />
  <row Id="26549" PostTypeId="2" ParentId="26497" CreationDate="2011-03-12T13:03:18.607" Score="2" ViewCount="" Body="&lt;p&gt;I don't really follow what your book is saying, but maybe this is the proof they wanted:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;First, given $x$ and $y$, $\{x,y\}$ exists by the axiom of pairing, and $\{x\} = \{x,x\}$ exists by the axiom of pairing, so $\{\{x\},\{x,y\}\} = \langle x,y\rangle $ exists by a third application of pairing.  Now, viewing $x$ as a parameter, we have proved that for each $y \in Y$, the set $\langle x,y\rangle$ exists, so by the axiom of replacement, for each $x$ the set $S_x = \{ \langle x,y\rangle : y \in Y\}$ exists. This last step may also require the axiom of comprehension depending on exactly how your book phrases the axiom of replacement. Now, since $X$ is a set, apply the axiom of replacement again to get that $\{\langle x,y\rangle : x \in X, y \in Y\}$ exists.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'll leave it up to you to formalize that proof to verify what formulas are used in the applications of replacement. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a strange way to prove that cartesian products exist, however, because the theorem can be proved as yunone indicated using the axiom of powerset but not the axiom of replacement. That fact is relevant in set theory because we sometimes work with models that do satisfy the axiom of powerset but not the axiom of replacement (e.g. $V_{\omega + \omega}$). &lt;/p&gt;&#xA;" OwnerUserId="630" LastActivityDate="2011-03-12T13:03:18.607" CommentCount="3" />
  <row Id="26550" PostTypeId="2" ParentId="26497" CreationDate="2011-03-12T13:06:20.317" Score="2" ViewCount="" Body="&lt;p&gt;Regarding the use of Replacement and Union without Powerset, the proof can be as follows: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I will assume that $(x,y)=\{\{x\},\{x,y\}\}$. First you show that for every $x\in A$ the set $\{(x,y)\ : y\in B\}$. To do this, fix an $x\in A$. Then you define a class-function that sends every element of $Y$ to $(x,y)$. The formula used in replacement would be $\phi(y,z):= z=(x,y)$ (keep in mind that $x$ is fixed). Or more typically $z=\{\{x\},\{x,y\}\}$ or even more typically: $$(\forall w)(w\in z\iff[((\forall u)(u\in w\iff u=x))\lor((\forall u)(u\in w\iff u=x\lor u=y))])$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To see that this satisfies the uniqueness of $z$ that is required for replacement simply observe that the sentence says that $z$ is a set that contains exactly $\{x\}$ and $\{x,y\}$ and through the axiom of extensionality this $z$ is unique. So for every set $x\in X$ we have defined a set $B_x=\{(x,y)\ :y\in B\}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again using the axiom of replacement we obtain a set $C=\{B_x\ :x\in A\}$. Intuitively we create a class-function that sends every $x\in A$ to $B_x$. The formula that will be used for replacement will be $\psi(x,v):= v=B_x$. Typically this will be written: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(\forall z)(z\in v\iff$&#xA;$$(\forall w)(w\in z\iff[(\exists y\in B)((\forall u)(u\in w\iff u=x))\lor((\forall u)(u\in w\iff u=x\lor u=y))]))$$&#xA;The proof that this $v$ is a set is done in the previous paragraphs and the proof of the uniqueness is again due to extensionality. So we have proved that $C=\{B_x\ :x\in A\}$ is indeed a set. From the axiom of union $\bigcup C$ is a set. Now I claim that $\bigcup C$ is the cartesian product $A\times B$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To see this first take $a\in\bigcup C$. This means that there exists an $x\in A$ such that $a\in B_x$. Which means that there is an $y\in B$ such that $a=(x,y)$. Now take $x\in A$ and $y\in B$. First observe that $(x,y)\in B_x$. Then since $B_x\in C$ we have $(x,y)\in \bigcup C$.&lt;/p&gt;&#xA;" OwnerUserId="1772" LastActivityDate="2011-03-12T13:06:20.317" />
  <row Id="26551" PostTypeId="1" CreationDate="2011-03-12T13:20:51.147" Score="12" ViewCount="376" Body="&lt;p&gt;If we can use Whitney embedding to smoothly embed every manifold into Euclidean space, then why do we bother studying abstract manifolds, instead of their embeddings in $\mathbb{R}^n$? A vague explanation I have heard is that from this abstract viewpoint, we gain understanding into the intrinsic behavior of the manifold, without knowing anything about the ambient space in which it can be embedded. Can anyone give some examples of this, or any other reason why abstraction is necessary?&lt;/p&gt;&#xA;" OwnerUserId="8152" LastActivityDate="2011-03-12T16:29:43.867" Title="Why abstract manifolds?" Tags="&lt;differential-topology&gt;" AnswerCount="7" CommentCount="1" FavoriteCount="3" />
  <row Id="26552" PostTypeId="2" ParentId="26500" CreationDate="2011-03-12T13:37:15.627" Score="7" ViewCount="" Body="&lt;p&gt;It is not true in the most general sense but it is true in a certain sense.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One way that it is true in the sense of the Knaster–Tarski theorem:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If $L$ is a complete lattice and $\phi\colon L \to L$ is an order-preserving function, then the set of fixed points of $\phi$ is a complete sublattice of $L$, and in particular the set of fixed points in nonempty.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Many recursive constructions correspond to order-preserving maps on a complete lattice, and so by this theorem the constructions do have fixed points. Those fixed points correspond to the thing that the recursive definition is trying to construct. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, to construct the Borel sets on a topological space $X$, consider the lattice of all sets of subsets of $X$. This is a powerset lattice and so it is complete. Let $\phi$ be the following operator: given a family $F$, $\phi(F)$ contains every set that can be obtained as a countable union of sets from $F \cup F^c \cup O$ where $F^c$ is the set of complements of sets in $F$ and $O$ is the set of open sets in $X$. Then $\phi$ is a order-preserving operator, and its least fixed point is the family of Borel sets of $X$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now here is a sense in which recursive definitions do not always define something. It's related to the field of &quot;general recursive functions&quot; which was an early way that people studied computability. Suppose that we we write down a set of recursive definitions for a function:&#xA;$$&#xD;&#xA;f(0) = 5 \qquad f(x+1) = 2f(x)+x&#xD;&#xA;$$&#xA;That particular definition does define a total function $\mathbb{N} \to \mathbb{N}$. But in general there is no reason to expect an arbitrary set of rules will work. For example,&#xA;$$&#xD;&#xA;f(0) = 1 \qquad f(3x) = x + 2 \qquad f(x+1) = 2&#xD;&#xA;$$&#xA;does not define a function. The point here is we are considering arbitrary definitions, not just definitions in the form of the first example. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is an example based on the Collatz conjecture; does it give a total function? &#xA;$$ f(0) = 0 \qquad f(1) = 1 \qquad f(2x) = f(x) \qquad f(2x+1) = f(6x+4)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you can see, the problem of telling which sets of rules actually define a total function is far from simple, and it only gets worse if you have a system of several mutually recursively defined functions.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a third aspect of the problem, relating to the foundational importance of inductive definitions. For example, in unformalized mathematics, the natural numbers are the smallest set of objects which contains 0 and is closed under successor. This sort of concrete inductive definition is one reason that people often feel that the natural numbers are more directly accessible to intuition than, say, arbitrary uncountable sets. &lt;/p&gt;&#xA;" OwnerUserId="630" LastActivityDate="2011-03-12T13:37:15.627" CommentCount="2" />
  <row Id="26553" PostTypeId="2" ParentId="26551" CreationDate="2011-03-12T13:42:58.677" Score="19" ViewCount="" Body="&lt;p&gt;The same reason we study abstract groups instead of their embeddings into symmetry groups. If we know some properties of a manifold $M$ which are intrinsic and don't depend on embeddings, and we later encounter that manifold $M$ somewhere else in a different embedding than the one we're used to, then we automatically know a list of things we can say about this new copy of $M$. The point is that the study of things like groups and manifolds naturally separates into the study of their abstract structure and the study of their concrete representations, and failing to make this separation is unnecessarily confusing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, topological and smooth manifolds don't have a notion of length of paths or volume: these things are dependent on an embedding into $\mathbb{R}^n$, so if I meet a manifold in a different embedding I can't assume that things like length and volume are the same, and I will only get confused if I do assume this. I can assume, however, that things like the number of connected components, the homotopy groups, the cohomology, etc. are the same, so whenever I meet a manifold whose cohomology groups I know I can always apply that information. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wouldn't say that the abstraction is &quot;necessary.&quot; Nothing in mathematics is &lt;em&gt;necessary.&lt;/em&gt; We make the definitions we make so we can more easily understand certain mathematical phenomena, and the phenomena we are interested in with regard to manifolds are easier to understand if we define abstract manifolds first and then worry about their embeddings later. &lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2011-03-12T13:51:07.283" LastActivityDate="2011-03-12T13:51:07.283" CommentCount="1" />
  <row Id="26554" PostTypeId="2" ParentId="26551" CreationDate="2011-03-12T13:47:22.963" Score="11" ViewCount="" Body="&lt;p&gt;The canonical example of the value of the intrinsic viewpoint is probably General Relativity. How would one make physical sense of a 4-dimensional curved space-time if one had to embed it in a higher-dimensional Euclidean space? What would you embed the universe in? It's (by definition) &lt;em&gt;everything&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1269" LastActivityDate="2011-03-12T13:47:22.963" />
  <row Id="26555" PostTypeId="2" ParentId="26551" CreationDate="2011-03-12T14:04:56.180" Score="10" ViewCount="" Body="&lt;p&gt;A practical point to complement the answers already given: A lot of things can relatively easily be described as manifolds, e.g. as quotient spaces of other manifolds, but their embedding in a higher-dimensional Euclidean space is more difficult to think about or imagine or even to come up with. For instance, the Klein bottle is easily described as a square with certain edge identifications, and projective space is easily described as the set of all directions, and it's rather immediate that these are manifolds, and yet I wouldn't know off the top of my head how to embed either of them in a higher-dimensional Euclidean space. If the theory of manifolds were built on embeddings, you'd first have to figure out how to embed these things before you could start investigating them.&lt;/p&gt;&#xA;" OwnerUserId="6622" LastActivityDate="2011-03-12T14:04:56.180" />
  <row Id="26556" PostTypeId="2" ParentId="26551" CreationDate="2011-03-12T14:05:13.947" Score="3" ViewCount="" Body="&lt;p&gt;Many manifolds don't have a nice, &quot;natural&quot; embedding into a Euclidean space.  For example, the usual definition of projective spaces is not as a submanifold of $\mathbb R^n$.  Using the abstract definition, it is easy to see that a projective space is a manifold.  However, it is very non-trivial to find an embedding of it into a Euclidean space and I have never heard of such an embedding being useful.&lt;/p&gt;&#xA;" OwnerUserId="9" LastActivityDate="2011-03-12T14:05:13.947" />
  <row Id="26557" PostTypeId="2" ParentId="26551" CreationDate="2011-03-12T14:09:14.537" Score="3" ViewCount="" Body="&lt;p&gt;As noted by Yuan, I think, the main reason is intrinsic properties of manifolds. A very nice result related to this is Theorema Egregium of Gauss. There are quantities which can be measured by the people of the manifold without passing to another realms. When you embed the manifold into another space, you also add extra structure to your manifold. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A second factor, I think, is the definition of manifold itself. A manifold is an equivalence classes of atlases. We then think it as a set of points in our head. More precisely, what makes $S^1$ $S^1$ is not the set of points of $S^1$ but the atlas you abstractly consider. Then we say that the point set $S^1$ is a manifold with that atlas since $S^1$ admits that atlas. Since we have an abstract theory that work for our atlas, results are independent of the uniformization.   &lt;/p&gt;&#xA;" OwnerUserId="7796" LastActivityDate="2011-03-12T14:09:14.537" />
  <row Id="26558" PostTypeId="2" ParentId="26528" CreationDate="2011-03-12T14:58:53.510" Score="6" ViewCount="" Body="&lt;p&gt;This answer treats the balls and boxes as distinguishable so each pattern is of equal probability.  I doubt there is a practical experiment to test this which does not also have them distinguishable.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The number of ways of putting $m$ balls in $n$ boxes is $n^m$ since each ball can go in any of the boxes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The calculation for the number of ways of putting $m$ balls in $n$ boxes with each box having at least one ball is more complicated.  Let's call it $A(m,n)$.  If, when deciding what to do with the last ball, all the other boxes are full, then it can go in any of the $n$ boxes; if however one of the $n$ boxes is empty and the others full then it must go in the empty one.  So &#xA;$$A(m,n) = n A(m-1,n) + n A(m-1,n-1)$$&#xA;and clearly $A(m,1) = 1$ (there is only one way with one box) and  $A(m,n) = 0$ for $m&amp;lt;n$.  It turns out that &#xA;$$A(m,n) = n! \, S_2 (m,n)$$ &#xA;where $S_2$ means &lt;a href=&quot;http://en.wikipedia.org/wiki/Stirling_numbers_of_the_second_kind&quot; rel=&quot;nofollow&quot;&gt;Stirling numbers of the second kind&lt;/a&gt;  (&lt;a href=&quot;http://oeis.org/A019538&quot; rel=&quot;nofollow&quot;&gt;OEIS A019538&lt;/a&gt;). $A(n,n)=n!$ as one would expect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the probability that all the boxes have at least one ball is &#xA;$$ \frac{n! \, S_2 (m,n)}{n^m}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, with 3 balls and 2 boxes, this gives $\frac{2 \times 3}{8} = 3/4$.&lt;/p&gt;&#xA;" OwnerUserId="6460" LastEditorUserId="6460" LastEditorDisplayName="" LastEditDate="2011-03-14T16:26:36.947" LastActivityDate="2011-03-14T16:26:36.947" CommentCount="2" />
  <row Id="26559" PostTypeId="2" ParentId="26551" CreationDate="2011-03-12T15:07:03.583" Score="3" ViewCount="" Body="&lt;p&gt;Semi-serious: You don't spend time with proving the Whitney embedding theorem although it is a largely useless for any of your purposes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Serious: You don't know whether such an embedding exists if you impose further structure onto the manifold (such as a metric tensor) and demand a structure-preserving embedding into $\mathbb R^n$. Vice versa, in case you generalize the concept of manifold (say, singular manifolds), it is again unclear whether such an embedding exists or has any meaning, whereas the instrinsic point of view appears much more natural and allows for convenient mathematics.&lt;/p&gt;&#xA;" OwnerUserId="3557" LastEditorUserId="367" LastEditorDisplayName="" LastEditDate="2011-03-12T15:31:56.873" LastActivityDate="2011-03-12T15:31:56.873" CommentCount="1" />
  <row Id="26560" PostTypeId="2" ParentId="26543" CreationDate="2011-03-12T15:16:06.477" Score="1" ViewCount="" Body="&lt;p&gt;It may help to think of the numbers as products of smaller numbers. It may make it easier to understand what's going on.&#xA;$$4\times \frac{12}{4} = 2\times 2 \times \frac{2\times 2\times 3}{2\times 2}=2\times 2\times 3 = 4 \times 3 = 4\times 3 \times \frac{4}{4}=12$$&#xA;Notice how you can always multiply a number by a fraction $\frac{a}{a}$ for any $a \neq 0$ and get the same number (because $\frac{a}{a}=1$):&#xA;$$ b = 1 \times b= \frac{a}{a}\times b = \frac{a\times b}{a}=a\times \frac{b}{a}$$&#xA;We can for instance use it on your example:&#xA;$$ 12 = 12 \times \frac{4}{4} = \frac{12\times 4}{4} = \frac{48}{4}$$&lt;/p&gt;&#xA;" OwnerUserId="7044" LastEditorUserId="7044" LastEditorDisplayName="" LastEditDate="2011-03-12T15:21:32.633" LastActivityDate="2011-03-12T15:21:32.633" CommentCount="1" />
  <row Id="26562" PostTypeId="2" ParentId="26521" CreationDate="2011-03-12T16:02:41.927" Score="1" ViewCount="" Body="&lt;p&gt;So as has been pointed out, this can't be done.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, let's say that you want to do a good job near $x = 0$. In particular let $f(x) = 1/(a+x)$ where $a$ is fixed in advance; let $g(x) = Ce^{-kx}$ where $C$ and $k$ are what we're trying to find. Then we'll find $C, k$ such that $f(0) = g(0)$ and $f^\prime(0) = g^\prime(0)$. (Why do I find two equations here? Because we have two unknowns.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f(0) = g(0)$ and $f^\prime(0) = g^\prime(0)$ are, doing the computation, $1/a = C$ and $-1/a^2 = -Ck$. Plug $1/a = C$ into the second equation and get $k = 1/a$. So the function $g(x) = (1/a)e^{-x/a}$ is, in some sense, a good approximation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You say you used Taylor series. I don't know what exactly you tried. But one thing you might do in such a context would be to write the Taylor series &#xA;$$1/(a+x) = 1/a - x/a^2 + x^2/a^3 + x^3/a^4 + \cdots$$&#xA;and &#xA;$$Ce^{-kx} = C - Ckx + {C k^2 x^2 \over 2} - {C k^3 x^3 \over 6} + \cdots $$&#xA;and set the first and second terms of these equal; that of course gives the solution I already gave.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Edited to add&lt;/b&gt;: Let's look for approximations of the form $g(x) = \sum_{i=1}^n C_i e^{-k_i x}$ for $n = 2, 3$. For convenience set $a = 1$. When $n = 2$ we have four parameters, so we'll find $C_1, C_2, k_1, k_2$ so that $f(0) = g(0)$ and the first three derivatives are equal at zero. This gives, after some computation (Maple-assisted), approximately&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$g(x) = 0.854 e^{-0.586x} + 0.146 e^{-3.414x}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or exactly&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$g(x) = {2 + \sqrt{2} \over 4} e^{-(2-\sqrt{2})x} + {2 - \sqrt{2} \over 4} e^{-(2+\sqrt{2})x}. $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With $n = 3$ the corresponding approximation (setting the first six derivatives equal) is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$g(x) = 0.711 e^{-0.416x} + 0.279 e^{-2.294x} + 0.010 e^{-6.290x} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now unfortunately this is not an area I know very well.  I'm actually finding these by this procedure: set $f(0) = g(0)$ and asking Maple to solve for one variable in terms of the other $2n-1$; make this substitution, then set $f^\prime(0) = g^\prime(0)$ and solve for one of the remaining $2n-1$ variables in terms of the remaining $2n-2$; and repeating. Unfortunately the expressions involved get very complicated and Maple is having trouble do the $n = 4$ case symoblically. I'm mentioning this in case anyone recognizes this as some standard procedure in numerical analysis.&lt;/p&gt;&#xA;" OwnerUserId="173" LastEditorUserId="173" LastEditorDisplayName="" LastEditDate="2011-03-12T17:40:47.723" LastActivityDate="2011-03-12T17:40:47.723" />
  <row Id="26563" PostTypeId="1" CreationDate="2011-03-12T16:12:26.520" Score="4" ViewCount="179" Body="&lt;p&gt;Suppose a source produces an indefinite sequence of positive integers. How can you check whether the numbers are generated truly randomly?&lt;/p&gt;&#xA;" OwnerUserId="6978" LastEditorUserId="2937" LastEditorDisplayName="" LastEditDate="2011-03-13T01:15:15.313" LastActivityDate="2011-03-13T19:46:54.053" Title="How do you check if a sequence of numbers is truly random?" Tags="&lt;algorithms&gt;&lt;stochastic-processes&gt;" AnswerCount="2" CommentCount="7" />
  <row Id="26564" PostTypeId="1" AcceptedAnswerId="26582" CreationDate="2011-03-12T16:23:14.287" Score="1" ViewCount="114" Body="&lt;p&gt;Hey, !&#xA;In my pre-calculus class the teacher showed the solution of the following example:&#xA;\begin{align}&#xA;\vert x-3 \vert   \lt \vert  x - 4 \vert  + x&#xA;\end{align}&#xA;He started by stated the domains needed to be checked:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{aligned}&#xA;\lbrack 4, +\infty ) \newline&#xA;\lbrack 3, 4 ) \newline&#xA;( -\infty, 3)&#xA;\end{aligned}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which I don't have a based lead on how he come to these domains and then deducted the following inequalities ( for each domain respectively):&#xA;\begin{align}&#xA;x-3 \lt x - 4 + x \newline&#xA;x-3 \lt -(x-4) + x \newline&#xA;-(x-3) \lt -(x-4) + x&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The final solution was \begin{aligned} (-1, +\infty ) \end{aligned}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I cannot understand how he deducted the domains and the right inequalities(there is one missing possibility):&#xA;\begin{align} -(x-3) \lt x-4 + x \end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It may be apparent but I still cannot wrap my mind around it.&#xA;&lt;strong&gt;Thanks!&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="7295" LastActivityDate="2011-03-12T17:59:39.283" Title="Solving inequality with two absolute values" Tags="&lt;algebra-precalculus&gt;&lt;inequality&gt;&lt;absolute-value&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="26565" PostTypeId="2" ParentId="26563" CreationDate="2011-03-12T16:24:33.807" Score="4" ViewCount="" Body="&lt;p&gt;There are two answers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In classical probability theory, the question doesn't even make sense. From the usual perspective of probability theory, if I roll a fair die, I get a &quot;random number&quot; from 1 to 6, but none of those numbers is &quot;random&quot; on its own.  &quot;Randomness&quot; here corresponds to the process of obtaining a measurement; it's a property of a random variable, not the property of a particular value I measure from the random variable.  So I roll the die over and over and get &quot;1,1,1,1,1,...&quot;, that's still the outcome of a random variable, and in this sense that sequence was still &quot;generated randomly&quot;.  Individual measurements are not random on their own, and so any sequence of numbers from 1-6 could be generated randomly by rolling a fair die. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a separate theory, called &quot;Kolmogorov complexity&quot; or &quot;algorithmic randomness&quot;, which can be used to measure &quot;how random&quot; certain objects are, but the meaning of &quot;random&quot; here is not the same. Instead, a sequence of numbers is called &quot;algorithmically random&quot; if it satisfies a large collection of randomness properties (more formally, the sequence is random if it does not lie in any effective $G_\delta$ set of measure 0). This area is also well studied, and a lot is known about the sequences that are random in this sense. But a &quot;random process&quot;, such as repeatedly rolling a fair die, is perfectly capable of generating a sequence that is not algorithmically random. &lt;/p&gt;&#xA;" OwnerUserId="630" LastActivityDate="2011-03-12T16:24:33.807" />
  <row Id="26568" PostTypeId="2" ParentId="26551" CreationDate="2011-03-12T16:29:43.867" Score="4" ViewCount="" Body="&lt;p&gt;If you restricted the definition of manifolds to be &quot;submanifolds of Euclidean space&quot;, you'd run into the problem that sometimes you want to do calculus on an object -- natural objects like homogeneous spaces, Grassmann manifolds, cut-and-paste objects like the &quot;pac man&quot; torus and such.  But the formal restrictions on usage of manifolds would mean you would have to embed the manifold in Euclidean space before you could apply the machinery of calculus to it.  In that sense the submanifold restiction is like a type of formal red-tape that gets in the way of understanding.  Abstract manifolds dispense with that. &lt;/p&gt;&#xA;" OwnerUserId="642" LastActivityDate="2011-03-12T16:29:43.867" />
  <row Id="26569" PostTypeId="1" AcceptedAnswerId="26577" CreationDate="2011-03-12T16:40:56.633" Score="3" ViewCount="141" Body="&lt;p&gt;Why is the generalized quaternion group $Q_n$ not a semidirect product?&lt;/p&gt;&#xA;" OwnerUserId="8581" LastEditorUserId="583" LastEditorDisplayName="" LastEditDate="2011-03-12T17:25:38.943" LastActivityDate="2011-03-16T08:13:40.623" Title="Why is the generalized quaternion group not a semi-direct product?" Tags="&lt;homework&gt;&lt;abstract-algebra&gt;&lt;group-theory&gt;" AnswerCount="2" />
  <row Id="26570" PostTypeId="1" AcceptedAnswerId="26578" CreationDate="2011-03-12T17:01:58.257" Score="1" ViewCount="79" Body="&lt;p&gt;Let $r_{2,3}(n)$ and $r_{t,3}(n)$ denote the number of ways to write $n$ as a sum of three positive squares (&lt;a href=&quot;http://oeis.org/search?q=0,+0,+1,+0,+0,+3,+0,+0,+3,+0,+3,+1,+0,+6,+0,+0,+3,+3,+3,+0&amp;amp;sort=&amp;amp;language=english&amp;amp;go=Search&quot; rel=&quot;nofollow&quot;&gt;A063691&lt;/a&gt;) and as a sum of three non-negative triangular numbers (&lt;a href=&quot;http://oeis.org/search?q=1,+3,+3,+4,+6,+3,+6,+9,+3,+7,+9,+6,+9&amp;amp;sort=&amp;amp;language=english&amp;amp;go=Search&quot; rel=&quot;nofollow&quot;&gt;A008443&lt;/a&gt;), respectively. I have noticed that $r_{2,3}(8k+3) = r_{t,3}(k)$ for $k \geq 1$. For example, $r_{2,3}(11) = 3$ because $11 = 3^2 + 1^2 + 1^2 = 1^2 + 3^2 + 1^2 = 1^2 + 1^2 + 3^2$ and $r_{t,3}(1) = 3$ because $1 = 1 + 0 + 0 = 0 + 1 + 0 = 0 + 0 + 1$, where $0$ and $1$ are triangular numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this identity well-known? If so, where can I find its proof?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A proof should follow from showing that the coefficient of the $q^{8k + 3}$ of the $q$-series of $(\sum_{n \geq 1} q^{n^{2}})^{3}$ is equal to the corresponding coefficient of the $q$-series of $\frac{1}{8} \theta^{3}_{2}(q^{4})$, where $\theta_2(q) = \theta_2(0,q)$ is a &lt;a href=&quot;http://mathworld.wolfram.com/JacobiThetaFunctions.html&quot; rel=&quot;nofollow&quot;&gt;Jacobi theta function&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2720" LastEditorUserId="2720" LastEditorDisplayName="" LastEditDate="2011-03-12T21:21:09.837" LastActivityDate="2011-03-12T21:29:34.710" Title="Representability as a Sum of Three Positive Squares or Non-negative Triangular Numbers" Tags="&lt;number-theory&gt;&lt;combinatorics&gt;&lt;complex-analysis&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="2" />
  <row Id="26571" PostTypeId="1" CreationDate="2011-03-12T17:08:42.193" Score="3" ViewCount="106" Body="&lt;p&gt;About a year ago I took a Linear Algebra class that was required for my degree.  Unfortunately that class had an unidentified pre-requisite and started at a much higher level then I really needed.  Going in I had no prior experience with linear algebra.  I can definitely see how understanding linear algebra would be a very good thing to have in my field so I've been trying to piece it together ever since and have felt like I'm close but I just don't quite get it.  I understand that linear algebra is a way to solve a lot of equations rapidly... In my mind this seems like that means finding values for the variables... but it didn't seem like we ever did... Instead we were doing things like multiplication of matrices and that made no sense.  Or we would apply advance algorithms to get the matrix into certain forms which the reason for never made sense to me. So what does it mean to solve a system of equations?  What are some real world examples that might make understanding linear algebra easier?  Why are orthogonal and other types of matrices so special?  Any insights, examples, suggestions are greatly appreciated!&lt;/p&gt;&#xA;" OwnerUserId="8160" LastActivityDate="2011-03-12T19:25:35.223" Title="What is happening in a linear algebra computation?" Tags="&lt;linear-algebra&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="26572" PostTypeId="2" ParentId="26542" CreationDate="2011-03-12T17:11:04.863" Score="0" ViewCount="" Body="&lt;p&gt;If $(X,\mathcal{O}_X)$ is a ringed topological space, you can look at locally free sheaves of $\mathcal{O}_X$-modules on $X$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $\mathcal{O}_X$ is the sheaf of continuous functions on a topological manifold (=Hausdorff and locally homeomorphic to $\mathbb{R}^n$), or the sheaf of smooth functions on a smooth manifold, you get fiber bundles (the sheaf associated to a fiber bundle is the sheaf of &quot;regular&quot; (=continuous or smooth here) sections).&lt;/p&gt;&#xA;" OwnerUserId="2660" LastActivityDate="2011-03-12T17:11:04.863" CommentCount="1" />
  <row Id="26573" PostTypeId="1" AcceptedAnswerId="26576" CreationDate="2011-03-12T17:11:58.577" Score="3" ViewCount="64" Body="&lt;p&gt;Is there a system $\{s_1, \cdots, s_m\}$ of symmetric polynomials of $z_1, \cdots, z_n \in \mathbb{C}$ such that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$s_1(z_1, \cdots, z_n) = c_1$$&#xA;$$s_2(z_1, \cdots, z_n) = c_2$$&#xA;$$\cdots$$&#xA;$$s_m(z_1, \cdots, z_n) = c_m$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;has at most one solution $(z_1, \cdots, z_n)$ up to permutation, for all choices $c_1, \cdots, c_m \in \mathbb{C}$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If so, what is the minimum value of $m$?  And what are all such systems of polynomials $\{s_1, \cdots, s_m\}$ with this property?&lt;/p&gt;&#xA;" OwnerUserId="7862" LastEditorUserId="7862" LastEditorDisplayName="" LastEditDate="2011-03-12T17:17:47.387" LastActivityDate="2011-03-12T18:03:19.250" Title="What is the least number of symmetric polynomials needed to determine a unique solution z_1,...,z_n?" Tags="&lt;algebraic-geometry&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" />
  <row Id="26574" PostTypeId="2" ParentId="26053" CreationDate="2011-03-12T17:17:54.780" Score="5" ViewCount="" Body="&lt;p&gt;First of all, the usual use of the notation $\hat{p}$ refers to the (point) &lt;em&gt;estimate&lt;/em&gt; and &lt;strong&gt;not&lt;/strong&gt; the true underlying value of the parameter. Many answers to your question are available in standard texts. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Binomial_proportion_confidence_interval#Clopper-Pearson_interval&quot; rel=&quot;nofollow&quot;&gt;Clopper--Pearson&lt;/a&gt; interval is the most popular. It is &lt;em&gt;guaranteed&lt;/em&gt; to give at least 95% coverage for &lt;em&gt;all&lt;/em&gt; values of the underlying parameter. However, as a consequence it tends to be quite conservative. That said, the typical &quot;approximate intervals&quot; (e.g., Wald and score) both fail miserably if the true proportion is close to 0 or 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another method which tends to yield a slightly narrower interval than Clopper--Pearson can be found in&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;H. Blaker, &lt;a href=&quot;http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.34.9682&amp;amp;rep=rep1&amp;amp;type=pdf&quot; rel=&quot;nofollow&quot;&gt;Confidence curves and improved exact confidence intervals for discrete distributions&lt;/a&gt;, &lt;em&gt;Canadian J. Stat.&lt;/em&gt;, vol 28, 2000, 783--798.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Agresti also covers this in Section 1.4.4 (pp. 18-21) of &lt;em&gt;Categorical Data Analysis&lt;/em&gt;, 2nd. edition. (This is the larger green book, not the smaller one of almost the same name.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For this particular example, the Clopper--Pearson confidence interval is $(0.9638, 1.0)$.&lt;/p&gt;&#xA;" OwnerUserId="7003" LastEditorUserId="7003" LastEditorDisplayName="" LastEditDate="2011-03-12T21:52:16.737" LastActivityDate="2011-03-12T21:52:16.737" CommentCount="2" />
  <row Id="26575" PostTypeId="1" AcceptedAnswerId="26585" CreationDate="2011-03-12T17:25:49.503" Score="1" ViewCount="92" Body="&lt;p&gt;If two sequences $\{a_k\}$ and $\{b_k\}$ are such that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\lim_{n\to\infty}\frac{1}{n}\sum_{k=0}^{n-1}(a_k-b_k)=0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\lim_{n\to\infty}\frac{1}{n}\sum_{k=0}^{n-1}(a_k^2-b_k^2)=0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;does it mean that the two sequences are asymptotically equally distributed? i.e., you can't distinguish one sequence from the other?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume that two sequences are such that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\lim_{n\to\infty}\frac{1}{n}\sum_{k=0}^{n-1}(a_k^q-b_k^q)=0;\ \forall\ q\in\mathbb{N}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm asking this because a sentence in a paper claims that for two asymptotically equally distributed sequences, the above holds. I'm wondering if the converse is true, i.e., if the above hold for any sequence, are they asymptotically equally distributed?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT 2&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Definition:&lt;/strong&gt; &lt;em&gt;Asymptotically equally distributed sets&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Two sets $\{a_k\}$ and $\{b_k\}$ are said to be asymptotically equally distributed if &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$-\infty&amp;lt;\alpha_1\leq \{a_k,b_k\}\leq\alpha_2&amp;lt;\infty$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\lim_{n\to\infty}\frac{1}{n}\sum_{k=0}^{n-1}\left[f(a_k)-f(b_k)\right]=0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $f(x)$ is any function continuous on $[\alpha_1,\alpha_2]$. This is called Weyl's theorem or Weyl's definition, but I don't remember the original paper this was discussed.&lt;/p&gt;&#xA;" OwnerUserId="7815" LastEditorUserId="7815" LastEditorDisplayName="" LastEditDate="2011-03-12T18:18:42.033" LastActivityDate="2011-03-12T19:31:27.620" Title="Do equal mean and equal moment imply equal distribution?" Tags="&lt;asymptotics&gt;&lt;probability-distributions&gt;" AnswerCount="1" CommentCount="8" />
  <row Id="26576" PostTypeId="2" ParentId="26573" CreationDate="2011-03-12T17:28:21.620" Score="5" ViewCount="" Body="&lt;p&gt;This is true whenever the $s_i$ generate the ring of symmetric functions since if you know the &lt;a href=&quot;http://en.wikipedia.org/wiki/Elementary_symmetric_polynomial&quot; rel=&quot;nofollow&quot;&gt;elementary symmetric polynomials&lt;/a&gt; $e_1, ... e_n$ in the $z_i$ you can write down a polynomial they satisfy. This is true, in particular, if the $s_i$ are&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;the elementary symmetric polynomials,&lt;/li&gt;&#xA;&lt;li&gt;the first $n$ &lt;a href=&quot;http://en.wikipedia.org/wiki/Power_sum_symmetric_polynomial&quot; rel=&quot;nofollow&quot;&gt;power symmetric polynomials&lt;/a&gt;, or&lt;/li&gt;&#xA;&lt;li&gt;the first $n$ &lt;a href=&quot;http://en.wikipedia.org/wiki/Complete_homogeneous_symmetric_polynomial&quot; rel=&quot;nofollow&quot;&gt;complete homogeneous symmetric polynomials&lt;/a&gt;.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I know that the second can be replaced with &quot;any $n$ consecutive power symmetric polynomials,&quot; and I would expect the third can as well. I don't know about more general characterizations; that seems like a hard problem. I think the minimum $m$ should be $n$.  &lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2011-03-12T18:03:19.250" LastActivityDate="2011-03-12T18:03:19.250" CommentCount="4" />
  <row Id="26577" PostTypeId="2" ParentId="26569" CreationDate="2011-03-12T17:28:28.833" Score="5" ViewCount="" Body="&lt;p&gt;How many elements of order 2 does a generalized quaternion 2-group have?  How many elements of order 2 must each factor in the semi-direct product have?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that dicyclic groups (generalized quaternion groups that are not 2-groups) can be semi-direct products.  The dicyclic group of order 24 is a semi-direct product of a group of a quaternion group of order 8 acting on a cyclic group of order 3.&lt;/p&gt;&#xA;" OwnerUserId="583" LastActivityDate="2011-03-12T17:28:28.833" CommentCount="2" />
  <row Id="26578" PostTypeId="2" ParentId="26570" CreationDate="2011-03-12T17:37:30.410" Score="2" ViewCount="" Body="&lt;p&gt;It is easy enough to prove something like this for squares and triangular numbers since modulo 4, squares are 0 or 1, so any three squares adding to $8k+3$ must each be odd, and the equation&#xA;$$k = \frac{a(a+1)}{2} + \frac{b(b+1)}{2} + \frac{c(c+1)}{2}$$ &#xA;implies and is implied by&lt;br&gt;&#xA;$$8k+3 = (2a+1)^2 + (2b+1)^2 + (2c+1)^2 .$$ &lt;/p&gt;&#xA;" OwnerUserId="6460" LastEditorUserId="6460" LastEditorDisplayName="" LastEditDate="2011-03-12T21:29:34.710" LastActivityDate="2011-03-12T21:29:34.710" CommentCount="2" />
  <row Id="26579" PostTypeId="2" ParentId="26303" CreationDate="2011-03-12T17:41:52.097" Score="3" ViewCount="" Body="&lt;p&gt;$$\frac12\int_{-d}^d\frac{z}{\sqrt{(d^2-z^2)(c+z)}}\mathrm{d}z$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is equivalent to (by depressing the cubic within the square root):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$=-\int_{-d-c/3}^{d-c/3}\frac{z+c/3}{\sqrt{\frac{8cd^2}{3}-\frac{8c^3}{27}-\left(\frac{4c^2}{3}+4d^2\right)z+4z^3}}\mathrm{d}z$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the substitution&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$z=\wp\left(u;\frac{4c^2}{3}+4d^2,\frac{8c^3}{27}-\frac{8cd^2}{3}\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\wp(u;g_2,g_3)$ is a Weierstrass elliptic function, and making use of the differential relation for the Weierstrass function:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;${\wp^{\prime}}^2=4\wp^3-g_2\wp-g_3$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$-\int_{\wp^{(-1)}(-d-c/3)}^{\wp^{(-1)}(d-c/3)}\left(\wp\left(u;\frac{4c^2}{3}+4d^2,\frac{8c^3}{27}-\frac{8cd^2}{3}\right)+\frac{c}{3}\right)\mathrm du$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;whose &lt;a href=&quot;http://functions.wolfram.com/09.13.21.0001.01&quot; rel=&quot;nofollow&quot;&gt;integration&lt;/a&gt; I'll leave to the interested reader.&lt;/p&gt;&#xA;" OwnerUserId="8156" LastActivityDate="2011-03-12T17:41:52.097" />
  <row Id="26580" PostTypeId="2" ParentId="26295" CreationDate="2011-03-12T17:46:35.073" Score="5" ViewCount="" Body="&lt;p&gt;It's rather meaningless to ask for a &quot;most general mathematical object&quot;, since it carries an implicit prediction that no progress will be made in the future.  Also, it is usually pretty easy to make useless generalizations of existing notions.  Anyway, I'll offer a construction from Beilinson and Drinfeld's book &lt;em&gt;Chiral algebras&lt;/em&gt; that may be sufficiently encompassing for your current needs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take an ODE $F(x, y, y&amp;#39;, \dots, y^{(n)}) = 0$, where $F$ is some function (did you want it to be smooth?) in $n+2$ real variables.  Ordinarily, a solution would be a function $y(x)$ that has $n$ derivatives (defined on some open interval $I$ of reals), such that at each point $x \in I$, the values of $y$ and its $n$ derivatives satisfy the equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If your equation lacks solutions in the above sense, you can work with a very formal notion of solution, that arises from manipulating algebras of differential operators.  First, choose your favorite category of functions closed under multiplication and differentiation (e.g., polynomial, analytic, smooth...), and define three basic sheaves:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$\mathscr{O}$ is the sheaf of functions on $\mathbb{R}$ in your category.  To each open set, it assigns the ring of (polynomial, analytic, smooth...) functions on that set, and to each inclusion of open sets, it assigns the restriction homomorphism.&lt;/li&gt;&#xA;&lt;li&gt;$\mathscr{D}$ is the sheaf of linear differential operators.  It is the unital $\mathscr{O}$-algebra generated by $\frac{d}{dx}$, with the usual relation satisfied by derivatives.  Commutative $\mathscr{D}$-algebras are commutative $\mathscr{O}$-algebras with integrable connection, and $\mathscr{O}$ has a natural commutative $\mathscr{D}$-algebra structure.&lt;/li&gt;&#xA;&lt;li&gt;$\mathscr{P}$ is the sheaf of all finite order ordinary differential operators.  It assigns to each open set $I \subset \mathbb{R}$ the commutative algebra of functions $G(x, y, y&amp;#39;,\dots)$ for $x \in I$ that depend on only finitely many derivatives of $y$,.  Global sections of $\mathscr{P}$ form a space of functions on an infinite dimensional vector space that has coordinates labeled by $x$ and the derivatives of $y$ (which is viewed as a formal variable).&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The operator $F$ defines an ideal in the commutative $\mathscr{D}$-algebra $\mathscr{P}$, and the quotient $A = \mathscr{P}/(F)$ is a commutative $\mathscr{D}$-algebra.  Given a commutative $\mathscr{D}$-algebra $R$, an $R$-solution to your ODE is a $\mathscr{D}$-algebra map from $A$ to $R$.  If $R = \mathscr{O}$ and $\mathscr{O}$ is the sheaf of smooth functions, then $R$-solutions are ordinary solutions.  Even if the set of ordinary solutions is empty, you'll get a nonempty set of $A$-solutions.&lt;/p&gt;&#xA;" OwnerUserId="3189" LastActivityDate="2011-03-12T17:46:35.073" />
  <row Id="26581" PostTypeId="2" ParentId="26519" CreationDate="2011-03-12T17:56:00.603" Score="2" ViewCount="" Body="&lt;p&gt;Yes, that is all correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You're correct that $24=4!$ is not a coincidence here.  From Taylor's theorem you can derive a power series expansion for cosine,&#xA;$$\cos(x)=1-\frac{x^2}{2}+\frac{x^4}{4!}-\frac{x^6}{6!}+\cdots,$$ so that&#xA;$$\cos(x)-1+\frac{x^2}{2}=\frac{x^4}{4!}-\frac{x^6}{6!}+\cdots,$$ and&#xA;$$\frac{\cos(x)-1+\frac{x^2}{2}}{x^4}=\frac{1}{4!}-\frac{x^2}{6!}+\frac{x^4}{8!}-\cdots.$$  The series on the right-hand side of the the last equation is defined and continuous everywhere, and goes to $\frac{1}{4!}$ at $0$.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2011-03-12T17:56:00.603" />
  <row Id="26582" PostTypeId="2" ParentId="26564" CreationDate="2011-03-12T17:59:39.283" Score="2" ViewCount="" Body="&lt;p&gt;In &lt;a href=&quot;http://math.stackexchange.com/questions/3153/what-is-the-best-way-to-solve-an-equation-involving-multiple-absolute-values/3154#3154&quot;&gt;my answer here&lt;/a&gt;, I describe a method for solving multiple-absolute-value equations that can similarly be applied to inequalities.  It is the same idea as what your teacher showed and what's discussed in the comments, but the particular organization of the work by dividing up a number line and working under each section may be helpful.&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2011-03-12T17:59:39.283" />
  <row Id="26583" PostTypeId="2" ParentId="26053" CreationDate="2011-03-12T18:33:03.537" Score="2" ViewCount="" Body="&lt;p&gt;If they all come out true, then this means that $x=n$ with $n=100$ the sample size.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The probability of this happening for a binomial distribution with parameter $p$ is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathbb{P}(X=n)=p^n \; .$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you require a $1-\alpha$ confidence level, then this implies that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathbb{P}(X=n)=p^n \geq 1-\alpha$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(1-\alpha)^{1/n}\leq p \leq 1 \; .$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $n=100$ and $\alpha=0.05$ this gives a very narrow interval of&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$0.9995 \leq p \leq 1 \; .$$&lt;/p&gt;&#xA;" OwnerUserId="3567" LastActivityDate="2011-03-12T18:33:03.537" CommentCount="2" />
  <row Id="26585" PostTypeId="2" ParentId="26575" CreationDate="2011-03-12T18:38:40.287" Score="2" ViewCount="" Body="&lt;p&gt;The answer to the question as it stands is no. Let $b_n = 0$ identically and let $a_n$ be a sparse enough sequence such that all of its asymptotic moments are zero but such that it is unbounded (e.g. let $a_n = k$ if $n = 2^k$ and $0$ otherwise). Then the first condition is not satisfied.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you add the boundedness condition, then the answer is yes. Like yoyo says in the comments, by Stone-Weierstrass we can uniformly approximate any continuous function $f$ on an interval by polynomials, and it is not hard to see that the desired expression in $f$ respects uniform limits for bounded sequences.&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2011-03-12T19:31:27.620" LastActivityDate="2011-03-12T19:31:27.620" />
  <row Id="26586" PostTypeId="2" ParentId="26530" CreationDate="2011-03-12T18:40:24.937" Score="1" ViewCount="" Body="&lt;p&gt;$X$ can always be the included angle $C$.  Just draw an angle $X$, mark off lengths $a$ and $b$ along the sides, and connect the endpoints.  Your are following the side-angle-side method of proving triangles congruent.  If $X \ge \pi/2$, it must be opposite the longest side of the triangle, so can only be opposite the longer of $a$ and $b$.  If $X \lt \pi/2$, and it is opposite $a$, you can use the law of sines to find $B$, which may have one or two legal values, then use $\pi-A-B$ to find $C$.  Then you can try making $X$ opposite $b$.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-03-12T18:40:24.937" CommentCount="2" />
  <row Id="26587" PostTypeId="1" AcceptedAnswerId="26615" CreationDate="2011-03-12T19:00:28.353" Score="4" ViewCount="76" Body="&lt;p&gt;I have got a question on so-called time-dependet Sobolev spaces - in particular as introduced in Evans book on PDE for the treatment of parabolic and hyperbolic PDE.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let us take a look at a linear hyperbolic PDE in $n$ spatial dimension and $1$ time dimension.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$u_{,tt} - Lu = 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$Lu = \sum_{i,j} a^{ij} u_{,i} u_{,j} + \sum_{i} b^i u_{,i} + c u$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore, we impose zero-boundary conditions on a open, bounded, smooth-boundary set $U \subset \mathbb R^n$, and regard a finite time interval $R := [0,T]$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First of all, it is natural to grant the time variable not special treatment, and regard this PDE as a PDE on $R \times U$. We then apply the test function machinery and obtain the notion of weak solution in a natural way. We might want the solution to be two-times weakly differentiable in time and space directions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We then assume our function lies within $H^2(R) \otimes H^2_0{U}$ with the topological tensor product. I am not too well-versed with this construction, as it does not belong to my university's canon, but for $u \in H^2(R) \otimes H^2_0{U}$ you would expect $u_{,tt} \in H^0(R) \otimes H^2_0{U}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, we can regard our hyperbolic equation as an second-order ODE on a Hilbert space. Then we might want our solution to be $u \in H^2( R, H^2_0(U) )$. In that case $u_{,tt} \in H^2( R, H^2_0(U) )$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In most of the above, we can assume weaker regularity, i.e. replace $H^2$ by $H^1$. This works, as in the weak formulation, the second distributional derivatives may be handed over to a test function. Then we the second-derivatives in any direction can be found in $H^{-1}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This makes sense, at least to me. However, the book of Evans treats weak derivatives in the above setting in in a different way, and I do not understand the transition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, he defines the solution of the above hyperbolic equation, assuming zero boundary conditions, to have the properties [1]&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$u \in L^2( \mathbb R, H_0^1(U) )$&lt;/li&gt;&#xA;&lt;li&gt;$u_{,t} \in L^2( \mathbb R, L^2(U) )$&lt;/li&gt;&#xA;&lt;li&gt;$u_{,tt} \in L^2( \mathbb R, H^{-1}(U) )$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This is clearly the ODE-on-VS approach, but it appears the time-derivative is taken over to the spatial derivatives. Of course, he may do so, as this setting is more general than what proposed as a weak solution in the above paragraphs. But then we might use, say, distributions as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whence I wonder why he does so - whether this is just for the reader convenience for whatever reason, whether it really points to what the solution really behaves like most regularly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can somebody explain this to me?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[1] L.C.Evans, Partial Differential Equations, 2nd Edition, p.400.&lt;/p&gt;&#xA;" OwnerUserId="3557" LastEditorUserId="1543" LastEditorDisplayName="" LastEditDate="2011-04-21T22:09:27.883" LastActivityDate="2011-04-21T22:09:27.883" Title="(Question) on Time-dependent Sobolev spaces for evolution equations" Tags="&lt;differential-equations&gt;&lt;pde&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="2" />
  <row Id="26588" PostTypeId="2" ParentId="26527" CreationDate="2011-03-12T19:08:39.033" Score="1" ViewCount="" Body="&lt;p&gt;Differentiation $n$ times, setting the variable to 0, multiplying by $n!$ gets you the $n$th coefficient of the ogf of a function $f$. That is, if $F(x)$ is the ogf of $f(n)$ then  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f(n) = [(D^nf)(0)]\cdot n!$$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But you want the closed form of $f$ from the ogf. For the most part, that is an ad hoc process (svenkatr's answer gives a particular method for $\frac{1}{(1-x)^k}$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A general method is to use the above differentiation trick (essentially computing the Taylor series), but finding a pattern from it. For a simple example, if $F(x)= \log \frac{1}{1-x}$, differentiating numerous times, you'll see the pattern 1, 1/2, 1/3, 1/4,..., so you'll notice and be able to prove) that $f(n) = 1/n$.&lt;/p&gt;&#xA;" OwnerUserId="1919" LastActivityDate="2011-03-12T19:08:39.033" />
  <row Id="26589" PostTypeId="2" ParentId="26571" CreationDate="2011-03-12T19:15:50.087" Score="2" ViewCount="" Body="&lt;p&gt;Linear algebra is not necessarily about solving linear equations rapidly. Indeed, most algebraist don't care for speed except for mathematicians in numerics and computer algebra.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Linear Algebra is taught because of two dual reasons (i) it is best understood (ii) it is omnipresent within any higher mathematics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, constructs from linear de facto appear in any science that does anything beyond simple pen-and-paper-computation. Most prominent examples where linear algebra is used 'immediatly' compromise computer graphics and modeling biological/chemical/social systems, not to forget linear optimization or numerical linear algebra.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In further answer would (probably) end up in a huge, long list.&lt;/p&gt;&#xA;" OwnerUserId="3557" LastActivityDate="2011-03-12T19:15:50.087" />
  <row Id="26590" PostTypeId="2" ParentId="26571" CreationDate="2011-03-12T19:25:35.223" Score="7" ViewCount="" Body="&lt;p&gt;I'm not going to try and answer all of your questions because it's really a very broad question. I will at least give you a start and then the best thing you can do is either take another course or open a book and learn some things, coming back to ask more specific questions along the way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Surely you are familiar with equations like $5x = 6$ where we want to find all solutions $x$ where $x$ is in some field, say in this case the real numbers. In this case every nonzero element has an inverse so you can multiply by $1/5$ to get $x = 5/6$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Although linear algebra isn't really just about solving equations, that's where it starts. It's called linear because we only want to solve equations that are linear in the unknown variable. The simplest case would be something like&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;x + y = 4,&#xD;&#xA;2x - y = -1&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Actually we can just write this in matrix form. If you remember how to multiply a matrix, then we can write this system as $Ax = b$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;\begin{pmatrix}&#xD;&#xA;1 &amp;amp; 1\\&#xD;&#xA;2 &amp;amp; -1&#xD;&#xA;\end{pmatrix}&#xD;&#xA;\begin{pmatrix}&#xD;&#xA;x \\&#xD;&#xA;y&#xD;&#xA;\end{pmatrix}&#xD;&#xA;=&#xD;&#xA;\begin{pmatrix}&#xD;&#xA;4\\&#xD;&#xA;-1&#xD;&#xA;\end{pmatrix}&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason why we multiply matrices is because we want to solve $Ax = b$ by multiplying by the inverse of $A$ to get $x = A^{-1}b$. Of course, not all matrices have inverses. So the set of all $n\times n$ matrices, with addition and multiplication is a ring but not a field. A ring is sort of like a field but now we remove the requirement where inverses exist for all nonzero elements. Also matrix multiplication is not commutative: $AB$ is not necessarily equal to $BA$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the above case $A$ does have an inverse, and you can multiply on the left by $A^{-1}$ (see if you can find it) to get the solution to this system of equations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus we have found: &lt;strong&gt;multiplication of matrices helps us solve equations.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, we are only beginning because finding the inverse of a matrix is tricky, so we study the different ways to represent matrices and calculate with matrices in order to more efficiently move them around. This is a bit vague but intentionally so since there is so much mathematics going on in the background which you need to learn.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Linear algebra is really about vector spaces. To appreciate the idea of a vector space you should first get some experience with abstraction by doing hundreds of problems. A vector space is just a set of elements together with addition and scalar multiplication that satisfy certain axioms. It turns out that matrices correspond to maps between vector spaces in a chosen basis of that space. This may not make too much sense to you now, but the important point is that putting matrices in different forms corresponds to changing the basis of the vector space in different ways.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason why we like to use vector spaces is because then we can concentrate on the algebraic properties of vector spaces without having to worry about specific numbers or equations, which then can be applied to all sorts of problems which have little do with solving equations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The best thing you can do to understand linear algebra is to take a course/read a book and just start solving problems. It is impossible to really understand what it is about first and then practice doing it. The understanding comes with the practice.&lt;/p&gt;&#xA;" OwnerUserId="2055" LastActivityDate="2011-03-12T19:25:35.223" />
  <row Id="26591" PostTypeId="1" AcceptedAnswerId="26593" CreationDate="2011-03-12T19:26:19.930" Score="4" ViewCount="120" Body="&lt;p&gt;I would like to know any way of solving the diophantine equation $x^4+y^4=2z^2$. Or ideas that seem worth trying out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By solving I mean find all solutions and prove there are no more.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Keith Conrad showed how to reduce this equation to a different one which was solved by Fermat in his notes about Fermat descent, other than I have no ideas how to solve it. I tried to do descent on it directly but that seems completely impossible so I am interested in other techniques. Thanks very much.&lt;/p&gt;&#xA;" OwnerUserId="4290" LastActivityDate="2011-03-12T20:38:13.613" Title="How would you solve the diophantine $x^4+y^4=2z^2$" Tags="&lt;diophantine-equations&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="26592" PostTypeId="1" AcceptedAnswerId="26598" CreationDate="2011-03-12T19:39:52.167" Score="2" ViewCount="131" Body="&lt;p&gt;Hi, all. I am reading a paper and I could not understand the dimension of the vector space $Hom_{\Gamma}(\rho_i\otimes \rho, \rho_j)$. The paper is &lt;a href=&quot;http://pages.uoregon.edu/njp/Lusztig92.pdf&quot; rel=&quot;nofollow&quot;&gt;http://pages.uoregon.edu/njp/Lusztig92.pdf&lt;/a&gt;, section 1.1, page 113. Here $\rho$ is a 2-dimensional C-vector space. $\Gamma$ is a finite subgroup of the special linear group $SL(\rho)$. Then $\rho$ is a $\Gamma$-module. Let $I$ be the set of isomorphism classes of simple $\Gamma$-modules. For each $i\in I$, $\rho_i$ is a simple $\Gamma$-module in class $i$. It is said that $dim Hom_{\Gamma}(\rho_i\otimes \rho, \rho_j)$ is 0, 1, or 2. I don't know why. Could you explain this? Thank you very much.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have another question about this paper. On page 114, equation (c), it is said that $(M, M&amp;#39;)=|\Gamma|^{-1}\sum_{\gamma\in \Gamma}tr(\gamma, M)tr(\gamma^{-1}, M&amp;#39;)(2-tr(\gamma, \rho))$. Here $(M, M&amp;#39;)$ is defined on page 113, section 1.2. $(M, M&amp;#39;)=(M\otimes C^2: M&amp;#39;)-(M\otimes \rho: M&amp;#39;)$, $(M, M&amp;#39;)=dim Hom_{\Gamma}(M, M&amp;#39;)$. Why the equation (c) is true? What does $tr(\gamma, \rho)$ mean? We could define trace of a linear map of a matrix. But here $(\gamma, \rho)$ is not a matrix or a map. Thank you very much.&lt;/p&gt;&#xA;" LastActivityDate="2011-03-13T19:29:16.020" Title="Dimension of Hom(U, V)" Tags="&lt;linear-algebra&gt;&lt;representation-theory&gt;" AnswerCount="2" />
  <row Id="26593" PostTypeId="2" ParentId="26591" CreationDate="2011-03-12T19:57:03.837" Score="4" ViewCount="" Body="&lt;p&gt;A quick spot of googling found &lt;a href=&quot;http://www.mathreference.com/num-zext,442.html&quot; rel=&quot;nofollow&quot;&gt;Two fourths and a square.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Look at pattern six: the unique coprime solution is $1^4+1^4=2 \cdot 1^2 .$&lt;/p&gt;&#xA;" OwnerUserId="1301" LastEditorUserId="1301" LastEditorDisplayName="" LastEditDate="2011-03-12T20:38:13.613" LastActivityDate="2011-03-12T20:38:13.613" CommentCount="2" />
  <row Id="26594" PostTypeId="1" AcceptedAnswerId="26599" CreationDate="2011-03-12T20:02:24.647" Score="2" ViewCount="133" Body="&lt;p&gt;Hello all, I'm a physics grad student working, for my research, in a matrix vector space over the complex numbers. That is, both operators and the vectors they act upon are $n\times n$ complex matrices living in the same space. When it comes to the eigenvalue equation I've been using&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$AB-BA=\lambda B$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $A$ and $B$ are $n\times n$ complex matrices. I can &quot;justify&quot; this choice based on the Heisenberg picture of Quantum Mechanics, but I would like to give it a more solid mathematical background. So in this regard I have several questions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is the choice above sensible? what other choices could I try? and last what is the relevant literature dealing with matrix vector spaces?&#xA;I mean, every semi-decent book mentions the space of matrices as a vector space, but that's just about it. I've found some valuable pieces of information scattered througout the problems in the Hoffman linear algebra text, but again, that's just about it. I have yet to find a text that treats matrix vector spaces in any depth. Any help would be really appreciated.&lt;/p&gt;&#xA;" OwnerUserId="8163" LastEditorUserId="8163" LastEditorDisplayName="" LastEditDate="2011-03-12T21:26:45.713" LastActivityDate="2011-03-13T06:03:53.460" Title="Regarding eigenvalue equation in a matrix vector space" Tags="&lt;linear-algebra&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="1" />
  <row Id="26595" PostTypeId="2" ParentId="26592" CreationDate="2011-03-12T20:11:52.950" Score="1" ViewCount="" Body="&lt;p&gt;&lt;s&gt;I don't think the first question is a general feature of groups&lt;/s&gt;, so it is probably something special about finite subgroups of SL(2,C).  They are basically cyclic or dihedral, and so ρi and ρj have dimension at most 2, and the tensor product has dimension at most 4, so you immediately get Tij is at most 4, and I guess you just rule out having too many linear constituents (and the few exceptions).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Actually this is a general feature of finite groups and their irreducible representations.  Suppose χ and θ are irreducible characters and ρ is a character with complex conjugate ρ′.  Then [ χρ, θ ] = [ χ, ρθ ].  Whenever φ is a character, [ χ, φ ] ≤ φ(1)/χ(1), and equality holds iff φ = (φ(1)/χ(1))⋅χ.  In our case we get [ χρ, θ ] ≤ χ(1)/θ(1) ⋅ ρ(1) but also [ χ, ρθ ] ≤ θ(1)/χ(1) ⋅ ρ′(1) = θ(1)/χ(1) ⋅ ρ(1).  Since at most one of { θ(1)/χ(1), χ(1)/θ(1) } is bigger than 1, we get that [ χρ, θ ] ≤ ρ(1).  In the paper, ρ(1) = 2.  (This does not require faithful or self-dual characters.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For your third question, I believe tr(g,V) means the trace of the group element g acting on the G-module V.  Sometimes he uses M to mean the module and sometimes he uses ρ to mean the module.&lt;/p&gt;&#xA;" OwnerUserId="583" LastEditorUserId="583" LastEditorDisplayName="" LastEditDate="2011-03-13T19:29:16.020" LastActivityDate="2011-03-13T19:29:16.020" CommentCount="6" />
  <row Id="26596" PostTypeId="2" ParentId="26594" CreationDate="2011-03-12T20:19:25.740" Score="-2" ViewCount="" Body="&lt;p&gt;Here's something that might help.  Using Krnoecker products, we can write&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$AX-XA = 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;as &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$[(I \otimes A) - (A^T \otimes I)] \vec X = 0.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is taken directly from Horn and Johnson's &lt;em&gt;Topics in Matrix Analysis&lt;/em&gt;, pg. 257. &amp;nbsp;&lt;/p&gt;&#xA;" OwnerUserId="7862" LastEditorUserId="6296" LastEditorDisplayName="" LastEditDate="2011-03-12T20:39:31.323" LastActivityDate="2011-03-12T20:39:31.323" />
  <row Id="26597" PostTypeId="1" AcceptedAnswerId="26600" CreationDate="2011-03-12T20:21:16.053" Score="1" ViewCount="44" Body="&lt;p&gt;So I am trying to solve this practice problem, and so far I managed to get that the kernel of transformation is zero. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Show that $L$($\vec{x}$)=$A\vec{x}$ from $Im(A^{T})$ to $Im(A)$ is an isomorphism.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let L(x)=0. Then L(x)=$A$$A^{T}y$ for some y. We know that Ker(A)=$Im(A^{T})^{\perp}$. So if $A$$A^{T}y$=0, $A^{T}y$ is in $Im(A^{T})^{\perp}$, which is possible only if $A^{T}y$=0, since it is both in the $Im(A^{T})^{\perp}$ and  $Im(A^{T})$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't want use the fact that dim of these two subspaces (Im($A$) and Im($A^{T}$)) are equal, as I am trying to use this problem as another proof of that. So I need to show that Im($L$)=Im($A$) to complete the proof of the isomorphism. What would be a way of showing that everything in Im($A$) gets hit given this transformation (seems intuitive, but can't show it)? Thanks&lt;/p&gt;&#xA;" OwnerUserId="5157" LastActivityDate="2011-03-12T20:33:25.833" Title="Show that $L$($\vec{x}$)=$A\vec{x}$ from $Im(A^{T})$ to $Im(A)$ is an isomorphism." Tags="&lt;linear-algebra&gt;" AnswerCount="1" />
  <row Id="26598" PostTypeId="2" ParentId="26592" CreationDate="2011-03-12T20:21:50.153" Score="2" ViewCount="" Body="&lt;p&gt;This follows from the lemmas proven in &lt;a href=&quot;http://qchu.wordpress.com/2010/04/27/the-mckay-correspondence-i/&quot; rel=&quot;nofollow&quot;&gt;this blog post&lt;/a&gt; about the McKay correspondence (see the paragraph under the $\tilde{D}_n$ diagrams). The more general result is also true and also follows by the lemmas in the post: if $\rho$ is a faithful &lt;em&gt;self-dual&lt;/em&gt; representation of a finite group $G$ and $\rho_i, \rho_j$ are irreducible, then $\dim \text{Hom}_G(\rho_i \otimes \rho, \rho_j) \le \dim \rho$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lemma 6 implies something stronger: the inequality is strict unless $G$ has at most two irreducible representations! Unfortunately I don't know a more direct proof. One is tempted to write down an argument using characters and the triangle inequality, but it doesn't seem to work and I don't know if it can be fixed. &lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2011-03-12T20:55:37.207" LastActivityDate="2011-03-12T20:55:37.207" CommentCount="3" />
  <row Id="26599" PostTypeId="2" ParentId="26594" CreationDate="2011-03-12T20:26:49.180" Score="4" ViewCount="" Body="&lt;p&gt;The linear operator $\text{ad}_A : B \mapsto AB - BA = [A, B]$ defines the &lt;a href=&quot;http://en.wikipedia.org/wiki/Adjoint_endomorphism&quot; rel=&quot;nofollow&quot;&gt;adjoint representation&lt;/a&gt; of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Lie_algebra&quot; rel=&quot;nofollow&quot;&gt;Lie algebra&lt;/a&gt; $\mathfrak{gl}_n$, and what you are studying are its eigenvalues. You don't want a linear algebra book; you want a book on Lie theory. The classic reference (although it is not particularly directed towards physicists) is Fulton and Harris's &lt;em&gt;Representation Theory: a First Course.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The important structure here is not the vector space but the action on it, which is why books on linear algebra won't say anything specific about &quot;matrix vector spaces&quot; because they are, after all, still vector spaces. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What other choices you have depends, of course, on what it is you're doing. &lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2011-03-12T20:33:12.873" LastActivityDate="2011-03-12T20:33:12.873" CommentCount="5" />
  <row Id="26600" PostTypeId="2" ParentId="26597" CreationDate="2011-03-12T20:33:25.833" Score="2" ViewCount="" Body="&lt;p&gt;What you have shown is that $L$ is an injective linear map from $Im(A^T)$ to $Im(A)$. In particular, $\dim Im(A^T)\leq\dim Im(A)$. If this were an equality, then $L$ would be necessarily be an isomorphism since it's injective. But now you can use the same argument with $A$ and $A^T$ interchanged to show that $\dim Im(A)\leq\dim Im(A^T)$ (since $(A^T)^T=A$).&lt;/p&gt;&#xA;" OwnerUserId="5116" LastActivityDate="2011-03-12T20:33:25.833" CommentCount="3" />
  <row Id="26602" PostTypeId="1" CreationDate="2011-03-12T20:44:18.650" Score="0" ViewCount="119" Body="&lt;p&gt;Let $(X_{n})$ be a sequence of nonnegative uniformly integrable random variables. Is it true that $\limsup X_{n} \in L_{1}$? Thanks!&lt;/p&gt;&#xA;" OwnerUserId="7762" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2011-03-12T21:58:47.943" LastActivityDate="2011-03-12T21:58:47.943" Title="limit superior of uniformly integrable random variables is integrable?" Tags="&lt;probability-theory&gt;" AnswerCount="0" CommentCount="10" FavoriteCount="1" />
  <row Id="26604" PostTypeId="1" CreationDate="2011-03-12T21:16:20.103" Score="0" ViewCount="67" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Possible Duplicate:&lt;/strong&gt;&lt;br&gt;&#xA;  &lt;a href=&quot;http://math.stackexchange.com/questions/14721/three-variable-third-degree-diophantine-equation&quot;&gt;Three variable, third degree Diophantine equation&lt;/a&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;p&gt;Show the diophantine equation x^3 +y^3 =3z^3 has no nonzer0 integral solutions.&lt;/p&gt;&#xA;" OwnerUserId="8166" LastActivityDate="2011-03-12T21:16:20.103" ClosedDate="2011-03-12T21:40:47.150" Title="fermat like equation" Tags="&lt;number-theory&gt;" CommentCount="2" />
  <row Id="26605" PostTypeId="1" AcceptedAnswerId="26612" CreationDate="2011-03-12T21:17:56.097" Score="7" ViewCount="179" Body="&lt;p&gt;Solve $\displaystyle \cos(x-\alpha)\cos(x-\beta) = \cos{\alpha}\cos{\beta}+\sin^2{x}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My attempt:&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle \cos(x-\alpha)\cos(x-\beta) = \cos{\alpha}\cos{\beta}+\sin^2{x} \Rightarrow \cos(x-\alpha)\cos(x-\beta)-\cos{\alpha}\cos{\beta} = \sin^2{x},$ and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{aligned}LHS &amp;amp; = \frac{1}{2}\cos\left(2x-\alpha-\beta\right)+\frac{1}{2}\cos\left(\alpha-\beta\right)-\frac{1}{2}\cos\left(\alpha-\beta\right)-\frac{1}{2}\cos\left(\alpha+\beta\right)\\&amp;amp; = \frac{1}{2}\cos\left(2x-\alpha-\beta\right)-\frac{1}{2}\cos\left(\alpha+\beta\right) \\&amp;amp; = -\sin\left(\frac{2x-\alpha-\beta+\alpha+\beta}{2}\right)\sin\left(\frac{2x-\alpha-\beta-\alpha-\beta}{2}\right) \\&amp;amp; = -\sin{x}\sin\left(x-\alpha-\beta\right)\end{aligned}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, we have: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{aligned} &amp;amp; -\sin{x}\sin\left(x-\alpha -\beta\right) = \sin^2{x} \\&amp;amp; \Rightarrow \sin^2{x}+\sin{x}\sin\left(x-\alpha-\beta\right) = 0 \\&amp;amp; \Rightarrow \sin{x}\left(\sin{x}+\sin\left(x-\alpha-\beta\right)\right) = 0\end{aligned} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So either $\sin{x} = 0$ or $\sin{x} = \sin\left(\alpha+\beta-x\right)$. If $\sin{x} = 0$, then we have $\sin{x} = 0 \Rightarrow \sin{x}$ $= \sin\left(0\right)$ $\Rightarrow x = n\pi$ or $(2n+1)\pi$ for $n\in\mathbb{Z}$ -- we can write this as $k\pi$, where $k\in\mathbb{Z}$. If, on the other hand, $\sin{x} = \sin\left(\alpha+\beta-x\right)$ then $x = 2n\pi+\alpha+\beta-x \Rightarrow x = n\pi+\frac{1}{2}\left(\alpha+\beta\right)$, or $ x = 2n\pi+\alpha+\beta-x $ (EDIT: this was meant to be $x = (2n+1)\pi-\alpha-\beta+x$), which contains no solutions (is that the right way to put it?). Thus the solutions for the equation are $n\pi+\frac{1}{2}\left(\alpha+\beta\right)$ and $k\pi$, for any integers $k$ and $n$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; The answer in the book has the condition $\alpha+\beta \ne (2m+1)\pi$ -- why is that?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Request:&lt;/strong&gt; If you have the time, please critique the way my solution is presented (reasoning, use of notation, flow etc  - I've an admission test in which this plays big part soon). Is my use of $\Rightarrow$ ok?  &lt;/p&gt;&#xA;" OwnerUserId="8164" LastEditorUserId="8164" LastEditorDisplayName="" LastEditDate="2011-03-12T22:16:25.797" LastActivityDate="2011-03-12T22:40:30.497" Title="Solving $\displaystyle \cos(x-\alpha)\cos(x-\beta) = \cos{\alpha}\cos{\beta}+\sin^2{x}$" Tags="&lt;trigonometry&gt;&lt;high-school&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="2" />
  <row Id="26607" PostTypeId="1" CreationDate="2011-03-12T21:20:47.327" Score="4" ViewCount="119" Body="&lt;p&gt;Hello,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was wondering if $\mathbb{Z} \wr S_n$, where $\mathbb{Z}$ is the usual group of integers, $S_n$ the symmetric group on n elements and $\wr$ the wreath product of two groups, contains the braid group $B_n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was also wondering if $n+1$-dimensional matrices of the form :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{bmatrix} 1&amp;amp;0&amp;amp;0 \\\\ 1&amp;amp;0&amp;amp;1 \\\\ 1&amp;amp;1&amp;amp;0 \end{bmatrix}$$ for B2 &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{bmatrix} 1&amp;amp;0&amp;amp;0&amp;amp;0 \\\\ 1&amp;amp;0&amp;amp;1&amp;amp;0 \\\\ 1&amp;amp;1&amp;amp;0&amp;amp;0 \\\\ 0&amp;amp;0&amp;amp;0&amp;amp;1 \end{bmatrix}$$&#xA;and&#xA;$$\begin{bmatrix} 1&amp;amp;0&amp;amp;0&amp;amp;0 \\\\ 0&amp;amp;1&amp;amp;0&amp;amp;0 \\\\ 1&amp;amp;0&amp;amp;0&amp;amp;1 \\\\ 1&amp;amp;0&amp;amp;1&amp;amp;0 \end{bmatrix}$$&#xA; for B3 &lt;/p&gt;&#xA;&#xA;&lt;p&gt;and so on... form a representation of the braid group $B_n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you for your help&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A. Popoff&lt;/p&gt;&#xA;" OwnerUserId="8167" LastEditorUserId="8167" LastEditorDisplayName="" LastEditDate="2011-03-12T23:02:29.527" LastActivityDate="2011-03-13T01:13:07.483" Title="Braid groups and representations" Tags="&lt;group-theory&gt;&lt;algebra&gt;" AnswerCount="1" CommentCount="13" />
  <row Id="26608" PostTypeId="1" AcceptedAnswerId="26610" CreationDate="2011-03-12T21:40:07.413" Score="2" ViewCount="128" Body="&lt;p&gt;Hello,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Problem&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Solve $f(x) = 6x^3 + 27x^2 + 17x + 20 \equiv 0 \pmod{30}$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My attempt was:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $30 = 2.3.5$, we then have:&lt;br&gt;&#xA;$$&#xD;&#xA;\begin{cases}&#xD;&#xA;f(x) \equiv 0 \pmod{2}\\&#xD;&#xA;f(x) \equiv 0 \pmod{3}\\&#xD;&#xA;f(x) \equiv 0 \pmod{5}\\&#xD;&#xA;\end{cases}&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By inspection, we see that:&#xA;$$&#xD;&#xA;\begin{cases}&#xD;&#xA;x \equiv 0 \pmod{2}\\&#xD;&#xA;x \equiv 1 \pmod{2}\\&#xD;&#xA;x \equiv 2 \pmod{3}\\&#xD;&#xA;x \equiv 0 \pmod{5}\\&#xD;&#xA;x \equiv 1 \pmod{5}\\&#xD;&#xA;\end{cases}&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, there will be four cases:&lt;br&gt;&#xA;Case 1&#xA;$$&#xD;&#xA;\begin{cases}&#xD;&#xA;x \equiv 0 \pmod{2}\\&#xD;&#xA;x \equiv 2 \pmod{3}\\&#xD;&#xA;x \equiv 1 \pmod{5}\\&#xD;&#xA;\end{cases}&#xD;&#xA;$$&#xA;Case 2&lt;br&gt;&#xA;$$&#xD;&#xA;\begin{cases}&#xD;&#xA;x \equiv 0 \pmod{2}\\&#xD;&#xA;x \equiv 2 \pmod{3}\\&#xD;&#xA;x \equiv 0 \pmod{5}\\&#xD;&#xA;\end{cases}&#xD;&#xA;$$&#xA;Case 3 &#xA;$$&#xD;&#xA;\begin{cases}&#xD;&#xA;x \equiv 1 \pmod{2}\\&#xD;&#xA;x \equiv 2 \pmod{3}\\&#xD;&#xA;x \equiv 1 \pmod{5}\\&#xD;&#xA;\end{cases}&#xD;&#xA;$$&#xA;Case 4&#xA;$$&#xD;&#xA;\begin{cases}&#xD;&#xA;x \equiv 1 \pmod{2}\\&#xD;&#xA;x \equiv 2 \pmod{3}\\&#xD;&#xA;x \equiv 0 \pmod{5}\\&#xD;&#xA;\end{cases}&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Apply Chinese Remainder Theorem for the four system of equation, where $M = 30$:&lt;br&gt;&#xA;$$&#xD;&#xA;\begin{cases}&#xD;&#xA;M_1 = \frac{30}{2} = 15\\&#xD;&#xA;M_2 = \frac{30}{3} = 10\\&#xD;&#xA;M_3 = \frac{30}{5} = 6\\&#xD;&#xA;\end{cases}&#xD;&#xA;$$&#xA;And, &#xA;$$&#xD;&#xA;\begin{cases}&#xD;&#xA;15y_1 \equiv 1 \pmod{2} \implies y_1 = 1\\ &#xD;&#xA;10y_2 \equiv 1 \pmod{3} \implies y_2 = 1\\ &#xD;&#xA;6y_3 \equiv 1 \pmod{5} \implies y_3 = 1\\ &#xD;&#xA;\end{cases}&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, the four solutions are:&#xA;$$&#xD;&#xA;\begin{cases}&#xD;&#xA;x_1 = 1.15.0 + 1.10.2 + 1.6.1 = 26\\&#xD;&#xA;x_2 = 1.15.0 + 1.10.2 + 1.6.0 = 20\\&#xD;&#xA;x_3 = 1.15.1 + 1.10.2 + 1.6.1 = 41\\&#xD;&#xA;x_4 = 1.15.1 + 1.10.2 + 1.6.0 = 35\\&#xD;&#xA;\end{cases}&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;&lt;br&gt;&#xA;Add missing cases suggested by &lt;em&gt;yunone&lt;/em&gt;&lt;br&gt;&#xA;Case 5 &#xA;$$&#xD;&#xA;\begin{cases}&#xD;&#xA;x \equiv 0 \pmod{2}\\&#xD;&#xA;x \equiv 2 \pmod{3}\\&#xD;&#xA;x \equiv 2 \pmod{5}\\&#xD;&#xA;\end{cases}&#xD;&#xA;$$&#xA;Case 6&#xA;$$&#xD;&#xA;\begin{cases}&#xD;&#xA;x \equiv 1 \pmod{2}\\&#xD;&#xA;x \equiv 2 \pmod{3}\\&#xD;&#xA;x \equiv 2 \pmod{5}\\&#xD;&#xA;\end{cases}&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And the last two solutions are:&#xA;$$&#xD;&#xA;\begin{cases}&#xD;&#xA;x_5 = 1.15.0 + 1.10.2 + 1.6.2 = 32 \equiv 2 \pmod{30}\\&#xD;&#xA;x_6 = 1.15.1 + 1.10.2 + 1.6.2 = 47 \equiv 17 \pmod{30}\\&#xD;&#xA;\end{cases}&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Am I in the right track? Any idea?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks,&lt;/p&gt;&#xA;" OwnerUserId="5711" LastEditorUserId="5711" LastEditorDisplayName="" LastEditDate="2011-03-12T21:59:04.603" LastActivityDate="2011-03-13T01:14:12.973" Title="Solve $f(x) = 6x^3 + 27x^2 + 17x + 20 \equiv 0 \pmod{30}$" Tags="&lt;elementary-number-theory&gt;&lt;modular-arithmetic&gt;" AnswerCount="3" />
  <row Id="26609" PostTypeId="1" CreationDate="2011-03-12T21:40:39.870" Score="6" ViewCount="120" Body="&lt;p&gt;Consider a particle undergoing geometric brownian motion with drift $\mu$ and volatility $\sigma$ e.g. as in &lt;a href=&quot;http://en.wikipedia.org/wiki/Black%E2%80%93Scholes&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;. Let $W_t$ denote this geometric brownian motion with drift at time $t$. I am looking for a formula to calculate:&lt;br&gt;&#xA;$$&#xD;&#xA;\mathbb{P}\big(\max_{0 \leq t \leq n} W_t - \min_{0\leq t \leq n} W_t &amp;gt; z\big)  &#xD;&#xA;$$&#xA;The inputs to the formula will be $\mu$, $\sigma$, $z$, and $n$.&lt;/p&gt;&#xA;" OwnerUserId="8168" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2011-03-12T22:03:19.453" LastActivityDate="2011-03-12T22:03:19.453" Title="Given particle undergoing Geometric Brownian Motion, want to find formula for probability that max-min &gt; z after n days" Tags="&lt;probability&gt;&lt;stochastic-processes&gt;" CommentCount="6" FavoriteCount="1" />
  <row Id="26610" PostTypeId="2" ParentId="26608" CreationDate="2011-03-12T21:48:05.973" Score="3" ViewCount="" Body="&lt;p&gt;So far this looks like you're on the right track. You may want to reduce the solutions $41$ and $35$ modulo $30$ to $11$ and $5$, respectively. Also, note that $x\equiv 2$ is a solution to $f(x)\equiv 0 \pmod{5}$, so you should address those two extra cases to find a total of $6$ solutions modulo $30$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a side note, you can use the $\LaTeX$ code &lt;code&gt;\cdot&lt;/code&gt; to produce the multiplication symbol. For example, &lt;code&gt;$30=2\cdot 3\cdot 5$&lt;/code&gt; produces $30=2\cdot 3\cdot 5$.&lt;/p&gt;&#xA;" OwnerUserId="1583" LastEditorUserId="1583" LastEditorDisplayName="" LastEditDate="2011-03-12T21:53:30.807" LastActivityDate="2011-03-12T21:53:30.807" CommentCount="2" />
  <row Id="26611" PostTypeId="1" CreationDate="2011-03-12T21:50:05.007" Score="1" ViewCount="62" Body="&lt;p&gt;Suppose there is a function $f: X \rightarrow \mathbb{R}$, where $X \subseteq \mathbb{R}^n$.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;If $x^*$ is a local minimizer of $f$&#xA;over $X$, must $x^*$ be either of the&#xA;two cases:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;if $f$ is differentiable at $x^*$, then $x^*$ must be a stationary point&#xA;i.e.  $\nabla{f}(x^*)=0$&lt;/li&gt;&#xA;&lt;li&gt;$f$ is non-differentiable at $x^*$?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In other words, are there other&#xA;possible cases for a local&#xA;minimizer besides the two? I seem to have seen&#xA;this as a conclusion from somewhere that I now&#xA;cannot recall. However the following&#xA;proposition seems to challenge this.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;From p194 on &lt;a href=&quot;http://books.google.com/books?ei=UvV7TYHMAu-80QGBnOTQAw&amp;amp;ct=result&amp;amp;id=QeweAQAAIAAJ&amp;amp;dq=Bertsekas+nonlinear+programming&amp;amp;q=Proposition+2.1.2#search_anchor&quot; rel=&quot;nofollow&quot;&gt;Nonlinear&#xA;Programming&lt;/a&gt; by Dimitri P. Bertsekas:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $X$ is a convex set.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Proposition 2.1.2: (Optimality&#xA;  Condition)&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;(a) If $x^*$ is a local minimum of $f$&#xA;  over $X$. then $$ \nabla{f}(x^*)&amp;#39;(x - x^*) &amp;gt; 0, \forall x \in X. $$&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;(b) If $f$ is convex over $X$, then&#xA;  the necessary condition of part (a) is&#xA;  also sufficient for $x^*$ to minimize&#xA;  $f$ over $X$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If the conclusion in Part 1 is&#xA;true, then if $f$ is&#xA;differentiable at $x^*$, we will have&#xA;$ \nabla{f}(x^*)=0$. Under this logic, I was confused&#xA;why we have the proposition (a) as&#xA;above in the book? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How to interpret the proposition geometrically? &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Thanks and regards!&lt;/p&gt;&#xA;" OwnerUserId="1281" LastEditorUserId="1281" LastEditorDisplayName="" LastEditDate="2011-03-12T22:38:52.187" LastActivityDate="2011-03-12T22:38:52.187" Title="Local min and differentiability of a function" Tags="&lt;real-analysis&gt;&lt;optimization&gt;&lt;convex-optimization&gt;" CommentCount="7" />
  <row Id="26612" PostTypeId="2" ParentId="26605" CreationDate="2011-03-12T22:02:32.807" Score="4" ViewCount="" Body="&lt;p&gt;One point: why are you going back from $x = n\pi+\frac{1}{2}(\alpha+\beta)$ to $x=2n\pi + \alpha+\beta-x$? And why do you say there are no solutions?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That said: you are incorrect in claiming that $\sin(x) = \sin(\alpha+\beta-x)$ implies that $x = 2n\pi + \alpha+\beta-x$. That is not the only ways in which two values of sine may be equal. After all, sine takes the same value at $0$ and at $\pi$, even though $0$ and $\pi$ don't differ by a multiple of $2\pi$. And it takes the value $\frac{1}{2}$ at both $\frac{\pi}{6}$ and at $\frac{5\pi}{6}$, even though the difference is not even an integer multiple of $\pi$. So that part of the derivation is incorrect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added/Edit.&lt;/strong&gt; From the comments, it seems you meant to write the correct conditions: either $x$ and $\alpha+\beta-x$ differ by a multiple of $2n\pi$ (this is the condition you checked), or else they are &quot;symmetric around $\pi/2$ or $3\pi/2$ up to a multiple of $2\pi$&quot;, which gives the condition $x = (2m+1)\pi - (\alpha+\beta-x)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your error was discarding/completely ignoring the second condition. The second condition does not give an equation for $x$, but it &lt;strong&gt;does&lt;/strong&gt; give an equation for $\alpha+\beta$! It says that you must have $0 = (2m+1)\pi - (\alpha+\beta)$, or that $\alpha+\beta=(2m+1)\pi$ must hold. In other word: if $\alpha+\beta = (2m+1)\pi$, then you will &lt;em&gt;always&lt;/em&gt; have $\sin(x) = \sin(\alpha+\beta-x)$, &lt;em&gt;regardless&lt;/em&gt; of what the value of $x$ is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, in summary: if $\alpha+\beta = (2m+1)\pi$ for some integer $m$, then &lt;em&gt;every&lt;/em&gt; $x$ is a solution. And if $\alpha+\beta\neq (2m+1)\pi$, then your derivation is correct (once you fix the analysis of this second case).&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-12T22:40:30.497" LastActivityDate="2011-03-12T22:40:30.497" CommentCount="8" />
  <row Id="26613" PostTypeId="2" ParentId="26438" CreationDate="2011-03-12T22:32:42.393" Score="0" ViewCount="" Body="&lt;p&gt;I could not comment on this now for unknown reason, so I will venture a proof, which means myself is not very sure. All the $K$ groups below are reduced $K$ groups as $\overline{K}$ is quite complicated to type. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $n=2k+1$ we have $$K^{0}(S_{n},X)\rightarrow K^{0}(S_{n})\rightarrow K^{0}(X)\rightarrow K^{-1}(S^{n},X)\rightarrow K^{-1}(S_{n})\rightarrow K^{-1}(X)\rightarrow K^{0}(S_{n},X)..$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From Bott periodicity we have $K^{0}(S^{n})=0$, $K^{1}(S^{n})=\mathbb{Z}$. Note $S_{n}/\mathbb{R}\mathbb{P}_{n}\cong S_{n}$. Hence we have:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$0 \rightarrow 0\rightarrow K^{0}(X)\rightarrow \mathbb{Z} \rightarrow \mathbb{Z} \rightarrow K^{-1}(X)\rightarrow 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The middle map from is $f:z\rightarrow 2z$. The exactness implies $K^{0}(X)=0$ and $K^{1}(X)=\mathbb{Z}/2\mathbb{Z}$. Now proceed from reduced K-group to normal K-group we get:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$K^{0}(X)=\mathbb{Z}$, $K^{1}(X)=\mathbb{Z}\oplus \mathbb{Z}/2\mathbb{Z}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The $n=2k$ case should be smiliar. I am not sure if $K(\mathbb{S}\wedge \mathbb{R}\mathbb{P}^{n})=K(\mathbb{R}\mathbb{P}^{n+1})$ holds. If yes then it should be much easier to calculate. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The above computation may have unknown problems as $RP_{1}\cong S_{1}$ but the result $K$-groups are different. &lt;/p&gt;&#xA;" OwnerUserId="7887" LastEditorUserId="7887" LastEditorDisplayName="" LastEditDate="2011-03-14T04:58:57.150" LastActivityDate="2011-03-14T04:58:57.150" CommentCount="17" />
  <row Id="26614" PostTypeId="1" CreationDate="2011-03-12T22:39:40.920" Score="0" ViewCount="127" Body="&lt;p&gt;Let π:E→B be a smooth vector bundle. Prove π is a submersion.&lt;/p&gt;&#xA;" OwnerUserId="8169" LastEditorUserId="1543" LastEditorDisplayName="" LastEditDate="2011-04-21T22:09:08.523" LastActivityDate="2011-04-21T22:09:08.523" Title="Smooth Vector Bundle is a Submersion" Tags="&lt;differential-geometry&gt;&lt;riemannian-geometry&gt;&lt;vector-bundles&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="26615" PostTypeId="2" ParentId="26587" CreationDate="2011-03-12T22:47:03.457" Score="2" ViewCount="" Body="&lt;p&gt;Firstly, you shouldn't think of the weak solutions as ODE on VS. For ODE on VS (using something like the Hille-Yosida theorem for semigroups of linear operator), the correct regularity should be strong continuity: that $u \in C^1(R,\mathcal{H})$ where $\mathcal{H}$ is the Hilbert space. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Secondly, an intuitive explanation of why $u_{tt} \in L^2(R, H^{-1}(U))$. Just use the hyperbolic equation: $u_{tt} = L u$. So $u_{tt}$ should belong to the same space as $Lu$. With two (spatial) derivatives acting on $u\in L^2(R,H^1(U))$, it is natural that $Lu \in L^2(R,H^{-1}(U))$, and hence also $u_{tt}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thirdly: I don't think a naive application of the weak solution idea should give you what you claimed ($H^2(R)\otimes H^2_0(U)$). Compare, say, to the elliptic case of a function in a box. Demanding that a function admits 2 weak derivatives in each direction separately is rather stronger than demanding the function admits 2 weak derivative overall. The former is not isotropic. The latter is. (In particular, the former says that $\partial_x\partial_x\partial_y\partial_y\partial_z\partial_z u \in L^2$, which is much stronger than just $u\in H^2$.) In particular, if you have a function $u \in H^1(R\times U)\cap C^\infty(R\times U)$, you'd see that the $H^1(R\times U)$ norm is comparable to the sum of the $u\in L^2(R, H^1(U))$ and $u_t \in L^2(R, L^2(U)) = L^2(R\times U)$ norms. &lt;/p&gt;&#xA;" OwnerUserId="1543" LastActivityDate="2011-03-12T22:47:03.457" CommentCount="2" />
  <row Id="26616" PostTypeId="2" ParentId="26614" CreationDate="2011-03-12T22:49:15.220" Score="1" ViewCount="" Body="&lt;p&gt;We need to show locally $U$ of $E$ submerged to $V$ of $B$. But note $U\cong V\times F$, this should be automatic. &lt;/p&gt;&#xA;" OwnerUserId="7887" LastActivityDate="2011-03-12T22:49:15.220" CommentCount="5" />
  <row Id="26617" PostTypeId="2" ParentId="26608" CreationDate="2011-03-12T22:54:28.243" Score="1" ViewCount="" Body="&lt;p&gt;With yunone's help you now have the six solutions modulo 30.  But with questions like this it is often quicker, though perhaps less educational, to use brute force at least as a check.&lt;br&gt;&#xA;So for $x$ in 0, 1, 2, 3, ..., 24 you find $6x^3 + 27x^2 + 17x + 20$ is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;20, 70, 210, 476, 904, 1530, 2390, 3520, 4956, 6734, 8890, 11460, 14480, 17986, 22014, 26600, 31780, 37590, 44066, 51244, 59160, 67850, 77350, 87696, 98924, 111070, 124170, 138260, 153376, 169554 &lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is equivalent modulo 30 to &lt;/p&gt;&#xA;&#xA;&lt;p&gt;20, 10, 0, 26, 4, 0, 20, 10, 6, 14, 10, 0, 20, 16, 24, 20, 10, 0, 26, 4, 0, 20, 10, 6, 14, 10, 0, 20, 16, 24&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and so the solutions are 2, 5, 11, 17, 20, and 26 modulo 30, more or less as you have.&lt;/p&gt;&#xA;" OwnerUserId="6460" LastActivityDate="2011-03-12T22:54:28.243" CommentCount="2" />
  <row Id="26618" PostTypeId="1" AcceptedAnswerId="26712" CreationDate="2011-03-12T22:58:45.880" Score="1" ViewCount="117" Body="&lt;p&gt;Is there a parameterization by integers of the modular group SL(2,Z)? What I mean is, some expression for a matrix in terms of several variables (a,b,c,...) such that for each n-tuple of integers (a,b,c,...), the expression produces a (preferably distinct) unimodular matrix.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an example of what I mean, think of Pythagorean triples. It's easy to say &quot;Pythagorean triples are countably infinite, therefore a bijection Z &amp;lt;-&gt; Pythagorean triples exists, QED&quot;, but that doesn't actually help you, e.g. write a computer program to generate them, does it? But an explicit parameterization does exist, viz. (a(b^2-c^2), 2abc, a(b^2+c^2)) for integers a, b, and c.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does a similar explicit parameterization exist for unimodular matrices of integers?&lt;/p&gt;&#xA;" OwnerUserId="4519" LastEditorUserId="4519" LastEditorDisplayName="" LastEditDate="2011-03-13T08:41:36.343" LastActivityDate="2011-03-13T19:12:44.123" Title="Parameterization of modular group" Tags="&lt;group-theory&gt;&lt;matrices&gt;" AnswerCount="2" CommentCount="12" />
  <row Id="26619" PostTypeId="1" CreationDate="2011-03-12T08:10:52.830" Score="2" ViewCount="44" Body="&lt;p&gt;Given the number of combinations w/o repetition for a set of size n from which you choose k is given by:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;n! / k! * (n - k)!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How does one calculate the number of these combinations that include a given element.  That amount should be the same for all elements of the original set.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, given the set {A, B, C, D}, there are 6 different ways to pick 2 elements: AB, AC, AD, BC, BD, CD.  However there are only 3 of these subsets with 'A' in it (AB, AC, AD).  I am stumped on getting to this 3 beyond brute forcing it.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I assume I am missing some formula?&lt;/p&gt;&#xA;" LastActivityDate="2011-03-12T23:04:52.030" Title="Number of combinations w/o repetition that include a certain element" Tags="&lt;combinatorics&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="26620" PostTypeId="2" ParentId="26619" CreationDate="2011-03-12T08:13:17.843" Score="2" ViewCount="" Body="&lt;p&gt;You must choose 1 element, then you can choose any k-1 elements from the remaining n-1. It's the same formula but with n and k replaced with n-1 and k-1 respectively.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;(n-1)! / (k-1)! * ((n-1) - (k-1))!&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="2027" LastActivityDate="2011-03-12T08:13:17.843" CommentCount="1" />
  <row Id="26621" PostTypeId="2" ParentId="26619" CreationDate="2011-03-12T08:14:12.890" Score="0" ViewCount="" Body="&lt;p&gt;Since you already have 1 element picked for you, you need to pick k-1 element from n-1 remaining, that gives the following formula:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;(n-1)!/(k-1)!(n-k)!&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="1017" LastActivityDate="2011-03-12T08:14:12.890" />
  <row Id="26622" PostTypeId="2" ParentId="26619" CreationDate="2011-03-12T08:15:10.040" Score="0" ViewCount="" Body="&lt;p&gt;In the {A, B, C, D} example, picking A is already decided, to it reduces to picking the other one out of the three {B, C, D}, which is 3!/1!*(3-1)! = 3.&lt;/p&gt;&#xA;" LastActivityDate="2011-03-12T08:15:10.040" />
  <row Id="26623" PostTypeId="1" CreationDate="2011-03-12T23:07:11.243" Score="1" ViewCount="158" Body="&lt;p&gt;So, I'm a little confused about one such statement made in class today : if M is a smooth manifold without boundary such that the tangent bundle of M is trivial, then M is orientable. Is this always true ? &lt;/p&gt;&#xA;" OwnerUserId="8169" LastActivityDate="2011-03-12T23:19:13.853" Title="Smooth Manifold with trivial tangent bundle" Tags="&lt;topology&gt;&lt;differential-geometry&gt;&lt;manifold&gt;" AnswerCount="1" />
  <row Id="26624" PostTypeId="1" CreationDate="2011-03-12T23:09:34.757" Score="2" ViewCount="71" Body="&lt;p&gt;Given $a, d, n, x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suggest me a suitable algorithm to compute the  number of times the digit $x$ appearing in the  arithmetic sequence $a, a + d, a + 2 \times d, \cdots, a + n \times d$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;For example :&lt;/strong&gt; If $ a= 10, d=5,n= 2,x = 0$ the sequence is ($10, 15,20$) so the answer is $2$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Domain specification:&lt;/strong&gt; $- 1 \le a, n \le 10^9. \text{ and } - 1 \le d \le 20,000. $&lt;/p&gt;&#xA;" OwnerUserId="2109" LastEditorUserId="2109" LastEditorDisplayName="" LastEditDate="2011-03-12T23:23:05.553" LastActivityDate="2011-05-12T01:56:47.180" Title="Counting digits in an arithmetic sequence" Tags="&lt;sequences-and-series&gt;&lt;algorithms&gt;&lt;contest-math&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="26625" PostTypeId="2" ParentId="26623" CreationDate="2011-03-12T23:11:06.230" Score="7" ViewCount="" Body="&lt;p&gt;If you think of an orientation of a manifold as an orientation of the tangent space at each point which varies continuously, then if your tangent bundle is of the form $M\times \mathbb R^n$, you can use a fixed orientation on $\mathbb R^n$ to orient each $T_pM$. So yes, this is true in general. :)&lt;/p&gt;&#xA;" OwnerUserId="2941" LastActivityDate="2011-03-12T23:11:06.230" CommentCount="4" />
  <row Id="26626" PostTypeId="1" AcceptedAnswerId="26695" CreationDate="2011-03-12T23:11:43.347" Score="14" ViewCount="270" Body="&lt;p&gt;Fix a complex number $z=x+iy$ where $x,y\in \mathbf{Z}$ Consider the sequence generated by the powers $$z^0, z^1, z^2, z^3,z^4 \ldots$$ The question is whether it is possible to capture &lt;strong&gt;any &lt;em&gt;positive&lt;/em&gt; integer&lt;/strong&gt; as either the real part or the imaginary part of this sequence of the powers of a fixed complex number. From some calculation and reasoning, &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;it is very easy to see that in order to generate all positive integers, $x$ and $y$  have to be relatively prime;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;it appears that the sequence of real parts and the sequence of imaginary parts blows up very early and hence it would not be unwise to conjecture that such a complex number does not exist.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any way to get hold of this problem? Thanks in advance. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;My curiosity partly grew out of observations on various polynomial bijection questions &lt;a href=&quot;http://mathoverflow.net/questions/21003/polynomial-bijection-from-qxq-to-q&quot; rel=&quot;nofollow&quot;&gt;e.g.&lt;/a&gt;, I saw on &lt;a href=&quot;http://www.mathoverflow.com&quot; rel=&quot;nofollow&quot;&gt;MathOverflow&lt;/a&gt;. I hope that at least the simple statement of the problem appeals to one's thoughts.  &lt;/p&gt;&#xA;" OwnerUserId="5595" LastEditorUserId="5595" LastEditorDisplayName="" LastEditDate="2011-03-13T09:50:40.163" LastActivityDate="2011-03-13T10:18:15.970" Title="Sequence of powers of Gaussian integers capturing all positive integers?" Tags="&lt;number-theory&gt;&lt;complex-analysis&gt;&lt;polynomials&gt;" AnswerCount="2" CommentCount="12" />
  <row Id="26628" PostTypeId="1" AcceptedAnswerId="26630" CreationDate="2011-03-12T23:22:27.130" Score="3" ViewCount="143" Body="&lt;p&gt;I already know that the function&#xA;$$&#xD;&#xA;f(x) = &#xD;&#xA;\begin{cases}&#xD;&#xA;\exp(- \frac{1}{x^2}), \quad  x &amp;gt; 0 \\&#xD;&#xA;0 , \quad x \leq 0&#xD;&#xA;&#xD;&#xA;\end{cases}&#xD;&#xA;&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is infinitely differentiable throughout $\mathbb R$. The only real problem, of course, lies in showing that $f^{(k)} (0) = 0$ for any positive integer $k$. What I have not been able to deduce is that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;\phi(x) = &#xD;&#xA;\begin{cases}&#xD;&#xA;\exp(- \frac{1}{1 - x^2}), \quad  |x| &amp;lt; 1 \\&#xD;&#xA;0 , \quad |x| \geq 1&#xD;&#xA;&#xD;&#xA;\end{cases}&#xD;&#xA;&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is &lt;em&gt;also&lt;/em&gt; infinitely differentiable throughout $\mathbb R$, using the previous function. The problem now is finding out what happens at $x = 1,-1$. Does the substitution $\zeta ^2 = 1 - x^2$ work, or is there another way to prove this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you all!&lt;/p&gt;&#xA;" OwnerUserId="6209" LastActivityDate="2011-03-13T00:12:22.870" Title="Infinitely differentiable function with compact support" Tags="&lt;calculus&gt;" AnswerCount="2" />
  <row Id="26629" PostTypeId="1" CreationDate="2011-03-12T23:36:21.450" Score="5" ViewCount="139" Body="&lt;p&gt;I haven't used my algebra skills much for years and they seem to have atrophied significantly!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm having real trouble working out how to re-arrange a formula I've come across to get $x$ by itself on the left hand side. It looks like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{x}{\sqrt{A^{2}-x^{2}}}=\frac{B+\sqrt{C+Dx}}{E+\sqrt{F+G\sqrt{A^{2}-x^{2}}}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've tried every method I can remember but I can't get rid of those pesky square roots!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any ideas?&lt;/p&gt;&#xA;" OwnerUserId="8170" LastActivityDate="2011-03-13T02:37:34.790" Title="How can I re-arrange this equation?" Tags="&lt;algebra&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="26630" PostTypeId="2" ParentId="26628" CreationDate="2011-03-12T23:46:29.477" Score="3" ViewCount="" Body="&lt;p&gt;This is much easier if you first show that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$g(x) = \begin{cases}&#xD;&#xA;  e^{-\frac{1}{x}} &amp;amp; \text{for } x \gt 0 \\\\&#xD;&#xA;  0                &amp;amp; \text{for } x \leq 0&#xD;&#xA;\end{cases}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is smooth. Then it's easy to see that $f(x) = g(x^2)$ and $\phi(x) = g(1-x^2)$, i.e. $f$ and $\phi$ are compositions of smooth functions.&lt;/p&gt;&#xA;" OwnerUserId="1269" LastEditorUserId="1178" LastEditorDisplayName="" LastEditDate="2011-03-13T00:12:22.870" LastActivityDate="2011-03-13T00:12:22.870" CommentCount="1" />
  <row Id="26631" PostTypeId="2" ParentId="26628" CreationDate="2011-03-12T23:47:29.913" Score="1" ViewCount="" Body="&lt;p&gt;Hint 1: For every positive $c$ and every real $a$, the function $f_{a,c}$ defined by $f_{a,c}(x)=\exp(-c/(x-a))$ if $x&amp;gt;a$ and $f_{a,c}(x)=0$ if $x\le a$ is infinitely differentiable on the real line.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint 2: use $f_{a,c}$ to build a function infinitely differentiable on the real line, zero for $x\ge a$ and positive for $x&amp;lt;a$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint 3 : use the decomposition of $1/(1-x^2)$ as a linear combination of $1/(1\pm x)$ to write $\phi$ as a product of two infinitely differentiable functions.&lt;/p&gt;&#xA;" OwnerUserId="6179" LastActivityDate="2011-03-12T23:47:29.913" CommentCount="2" />
  <row Id="26632" PostTypeId="1" CreationDate="2011-03-12T23:51:54.163" Score="8" ViewCount="141" Body="&lt;p&gt;I have the following problem related to a statistics question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Prove that the function defined for $x\ge 1, y\ge 1$,&#xA;$$f(x,y)=\frac{\Gamma\left(\frac{x+y}{2}\right)(x/y)^{x/2}}{\Gamma(x/2)\Gamma(y/2)}\int_1^\infty w^{(x/2)-1}\left(1+\frac{xw}{y}\right)^{-(x+y)/2} dw$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is increasing in $x$ for each $y\ge 1$ and decreasing in $y$ for each $x\ge 1$. (Here $\Gamma$ is the gamma function.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Trying to prove by using derivatives seems difficult.&lt;/p&gt;&#xA;" OwnerUserId="3249" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2011-03-14T05:40:29.727" LastActivityDate="2011-03-14T05:40:29.727" Title="Monotonic behavior of a function" Tags="&lt;analysis&gt;&lt;statistics&gt;&lt;special-functions&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="1" />
  <row Id="26634" PostTypeId="2" ParentId="26521" CreationDate="2011-03-12T23:58:22.570" Score="1" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Stone%E2%80%93Weierstrass_theorem&quot; rel=&quot;nofollow&quot;&gt;Stone-Weierstrass theorem&lt;/a&gt; is your friend here. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To wit, the function $f$ defined on $[0,1]$ by $f(0)=0$ and $f(u)=\displaystyle\frac1{a-\log(u)}$ if $u\in(0,1]$ is continuous, hence for every positive $\varepsilon$ there exists a polynomial function $P_\varepsilon$ such that $|f(u)-P_\varepsilon(u)|\le\varepsilon$ for every $u$ in $[0,1]$. Writing $P_\varepsilon$ as&#xA;$\displaystyle P_\varepsilon(u)=\sum_{k=0}^{n_\varepsilon}C_k^\varepsilon u^k$ and using this for $u=\mathrm{e}^{-x}$, one gets that, for every nonnegative $x$,&#xA;$$&#xD;&#xA;\left|\frac1{a+x}-\sum_{k=0}^{n_\varepsilon}C_k^\varepsilon\mathrm{e}^{-kx}\right|\le\varepsilon.&#xD;&#xA;$$&#xA;If one wishes to get rid of the $k=0$ term, one proceeds as follows: the limit when $x\to+\infty$ of the inequality above indicates that $|C_0^\varepsilon|\le\varepsilon$, hence, for every nonnegative $x$,&#xA;$$&#xD;&#xA;\left|\frac1{a+x}-\sum_{k=1}^{n_\varepsilon}C_k^\varepsilon\mathrm{e}^{-kx}\right|\le2\varepsilon.&#xD;&#xA;$$&#xA;Caveat: nothing says that the coefficients $C_k^\varepsilon$ are nonnegative.&lt;/p&gt;&#xA;" OwnerUserId="6179" LastEditorUserId="6179" LastEditorDisplayName="" LastEditDate="2011-04-07T08:35:38.907" LastActivityDate="2011-04-07T08:35:38.907" />
  <row Id="26635" PostTypeId="2" ParentId="26629" CreationDate="2011-03-12T23:58:37.013" Score="1" ViewCount="" Body="&lt;p&gt;I would start by multiplying the numerator and denomenator on the right by $E-\sqrt{F+G\sqrt{A^2-x^2}}$:&#xA;$$\frac{x}{\sqrt{A^2 - x^2}} = \frac{\left(B + \sqrt{C + Dx}\right)\left(E - \sqrt{F + G\sqrt{A^2 - x^2}}\right)}{E^2-F+G\sqrt{A^2 - x^2}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It may also help the manipulation to set $y = \sqrt{A^2 - x^2}$ for a while.&lt;/p&gt;&#xA;" OwnerUserId="8137" LastEditorUserId="8137" LastEditorDisplayName="" LastEditDate="2011-03-13T02:37:34.790" LastActivityDate="2011-03-13T02:37:34.790" CommentCount="3" />
  <row Id="26637" PostTypeId="1" CreationDate="2011-03-13T01:00:14.847" Score="0" ViewCount="54" Body="&lt;p&gt;Let $p(t,x_1,...,x_n)$ be a continuous function which, evaluated at a real number $t$, gives a polynomial in $n+1$ variables over an algebraically closed field $k$. Let $V(t)$ be the vanishing set in affine $n$ -space of the ideal $(p(t,x_1,...,x_n))$. Is there any way to determine a map from $V(t_1)$  to $V_(t_2)$  by looking at $p(t,x_1,...,x_n)$? Furthermore, if we look at $f: V(t) \rightarrow V(t + dt)$, and define a functional $L[f]$ as the integral of $||f(x_1,...,x_n) - (x_1,...,x_n)||$  over a finite region $D$ , can we find the minimal such $f$ at each $t$ and from these construct the function $g: V(t_1) \rightarrow V(t_2)$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, if the above are possible for an algebraically closed field, what would happen if this field where replaced with a non-commutative division ring (which is still algebraically closed), such as the quaternions?&lt;/p&gt;&#xA;" OwnerUserId="8173" LastActivityDate="2011-03-13T01:00:14.847" Title="Maps between algebraic varieties induced by maps between corresponding ideals" Tags="&lt;abstract-algebra&gt;&lt;algebraic-geometry&gt;" CommentCount="2" />
  <row Id="26638" PostTypeId="2" ParentId="26607" CreationDate="2011-03-13T01:13:07.483" Score="4" ViewCount="" Body="&lt;p&gt;It is certainly not true that $\mathbb{Z}\wr S_n$ contains $B_n$, for any $n&amp;gt;2$.  Indeed, by definition, $\mathbb{Z}\wr S_n$ has a free abelian subgroup of finite (indeed, $n!$)  index.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, the kernel of the natural map of pure braid groups $PB_n\to PB_{n-1}$ obtained by forgetting a strand is the free group of rank $n-1$ (this follows from the Birman Exact Sequence, I suppose), so if $n&amp;gt;2$, $PB_n$ (and hence $B_n$) contains a non-abelian free group.  If it were contained in $\mathbb{Z}\wr S_n$ then this non-abelian free group would have an abelian subgroup of finite index, which is absurd.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[Note: this answer is essentially the same as Steve D's comment.]&lt;/p&gt;&#xA;" OwnerUserId="2287" LastActivityDate="2011-03-13T01:13:07.483" CommentCount="4" />
  <row Id="26639" PostTypeId="2" ParentId="26608" CreationDate="2011-03-13T01:14:12.973" Score="2" ViewCount="" Body="&lt;p&gt;$\rm\ mod\ 2\::\ \ 0\ =\ x^2 + x\ =\ x\ (x + 1)\ \ \Rightarrow\ \ x\ =\ 0,\:1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\ mod\ 3\::\ \ 0\ =\: -x + 2\ \ \Rightarrow\ \ x\ =\ 2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\ mod\ 5\::\ \ 0\ =\ x\ (x^2 - 3\ x + 2)\ =\ x\ (x-1)\ (x-2)\ \ \Rightarrow\ \ x\ =\ 0,\:1,\:2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\ x = 2\ (mod\ 3)\ \Rightarrow\ x = 2,5,8,11,14\ (mod\ 15)\ $ of which only $\rm\ S = \{2,5,11\}\ $ are $\rm\ 0,1,2\ (mod\ 5)\:.\:$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So $\rm\ x \ \in\ S\: \cup\: (S+15)\ =\ \{2,5,11,17,20,26\}\ \ (mod\ 30)\ $  since all values mod $2$ are solutions.&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2011-03-13T01:14:12.973" />
  <row Id="26640" PostTypeId="1" AcceptedAnswerId="41541" CreationDate="2011-03-13T01:29:13.617" Score="16" ViewCount="388" Body="&lt;p&gt;The story in nearly every introductory Calculus book is well known by everybody: you don't have the &quot;right&quot; to raise a number to an irrational power, so forget exponents for now and let's take a look at $y=x^{-1}$. How odd, the innocent formula for a power function's antiderivative breaks down but gee, it &lt;em&gt;must&lt;/em&gt; have an antiderivative, it's smooth! Let's examine its properties...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;...and in the end, Rosebud was his sled, no, wait, the mysterious antiderivative turns out to have an inverse that corresponds exactly to the elementary school concept of exponents, only it works for irrational exponents too! The hero wins! The End.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But what if we start from the opposite end? Start with the innocent, only-defined-for-rationals (so far) exponential function $y=k^x$, $k&amp;gt;0$, and if $x_0$ is irrational, prove that, as $x$ (while staying rational) approaches $x_0$, $k^x$ approaches some specific real number. Define that such number is $k^{x_0}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And from there, prove that our New! Improved! $k^x$ is continuous, has a derivative that's also an exponential, that there is some $k=e$ for which the exponential is its own derivative, that the inverse of $e^x$ has $x^{-1}$ as its derivative etc etc...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do you know of any Calculus text that takes that approach?&lt;/p&gt;&#xA;" OwnerUserId="2612" LastActivityDate="2011-05-26T18:15:32.267" Title="Do you know of any Calculus text defining the exponential and logarithm functions in an alternative way?" Tags="&lt;calculus&gt;&lt;logarithms&gt;&lt;definition&gt;" AnswerCount="3" CommentCount="8" FavoriteCount="2" />
  <row Id="26641" PostTypeId="1" CreationDate="2011-03-13T01:24:13.930" Score="1" ViewCount="88" Body="&lt;p&gt;I am having trouble understanding this step in the paper: we can apply the numerical inversion (16)-(17) to ^ faB (s) and ^ fbA(s) to derive faB (t) and fbA(t). How do you apply the numerical inversion to derive the function/what is faB (t) and fbA(t)? This is from this paper which has more information abou &lt;a href=&quot;https://docs.google.com/viewer?a=v&amp;amp;pid=explorer&amp;amp;chrome=true&amp;amp;srcid=0B2dDCV15EnaiNWNiOTJlYjEtZmFkYS00YTVmLWExZGYtYTA2YTQ0YTlkMWNj&amp;amp;hl=en&amp;amp;authkey=CNumrtMI&quot; rel=&quot;nofollow&quot;&gt;https://docs.google.com/viewer?a=v&amp;amp;pid=explorer&amp;amp;chrome=true&amp;amp;srcid=0B2dDCV15EnaiNWNiOTJlYjEtZmFkYS00YTVmLWExZGYtYTA2YTQ0YTlkMWNj&amp;amp;hl=en&amp;amp;authkey=CNumrtMI&lt;/a&gt; the screen shots are from pages 15-17 &#xA;-thanks&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The laplace transform of the function:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;img src=&quot;http://i.imgur.com/Qrtvx.png&quot; alt=&quot;the laplace transform of the function&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The numerical inversion processes:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;img src=&quot;http://i.imgur.com/sNAjO.png&quot; alt=&quot;the numerical inversion processes&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" LastEditorUserId="8174" LastEditorDisplayName="" LastEditDate="2011-03-14T01:47:36.753" LastActivityDate="2011-03-14T01:47:36.753" Title="What is the function of this laplace transform using the numerical inversion reference included." Tags="&lt;probability-theory&gt;" FavoriteCount="2" />
  <row Id="26642" PostTypeId="2" ParentId="26624" CreationDate="2011-03-13T01:33:45.380" Score="0" ViewCount="" Body="&lt;p&gt;I cannot tell you how best to go about this problem, but here is the approach I would use: &#xA;The number of digits that occur is the sum of the number of digits that occur in the ones place, the tens place, etc.  For the ones place, treat all these operations mod 10, and this becomes easy because n cycles from 0 to 9.  For other places, I would suggest finding the mth from last digit by looking at the sequence of terms (nth term mod 10^m - nth term mod 10^(m-1))/10^m.&lt;/p&gt;&#xA;" OwnerUserId="8173" LastActivityDate="2011-03-13T01:33:45.380" />
  <row Id="26644" PostTypeId="1" AcceptedAnswerId="26646" CreationDate="2011-03-13T01:43:30.030" Score="3" ViewCount="85" Body="&lt;p&gt;This is a concept that seems very intuitive to me, but I feel my proof is messy. Could someone more experienced than I perhaps offer some criticism/tell me if I'm even correct?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $A\subset\omega$. I want to show that $A$ is infinite if and only if&#xA;$$&#xD;&#xA;\forall_m(m\in\omega\implies\exists_n(n\in A\land (m\lt n))).&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here $a&amp;lt;b\iff a\in b$ for $a,b\in\omega$. Also, $A\sim B$ means two sets are in bijection.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think my proof is too roundabout. I prove $\implies$ by the contrapositive. Suppose the above property is not true. So there exists an $m$ such that there is no $n\in A$ such that $m&amp;lt;n$. This means for all $n\in A$, $n\leq m$, which implies $n\lt m^{+}$. So $A\subseteq m^+$. If $A=m^+$, then obviously $A\sim m^+$, so by definition $A$ is finite.&#xA;If $A\subsetneq m^+$, then we know $A\sim p$ for some $p&amp;lt;m^+$, so $A$ is finite by definition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $\impliedby$, I feel things get even worse. I try to prove the contrapositive by contradiction. I suppose $A$ is finite, but the property holds. Now if $A$ is empty, the property is false, a contradiction, so the contrapositive holds as needed. &lt;strong&gt;So assume $A$ is nonempty. Since $A$ is finite, $A$ must have a greatest element.&lt;/strong&gt; (I try to justify this by applying the well-ordering principle on the reverse ordering. This is probably my biggest concern, is it acceptable?) Denote this element $k$. Consider $k^+$. Then $k^+\in\omega$, and by the property, we have that there is a $p\in A$ such that $k^+&amp;lt;p$, but $k&amp;lt;p$, a contradiction. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Apologies for this messy proof. How can I clean it up/correct it? Many thanks.&lt;/p&gt;&#xA;" OwnerUserId="6742" LastEditorUserId="1583" LastEditorDisplayName="" LastEditDate="2011-03-13T01:46:58.077" LastActivityDate="2011-03-13T02:36:02.433" Title="Showing a subset $A$ of $\omega$ is infinite." Tags="&lt;elementary-set-theory&gt;" AnswerCount="1" />
  <row Id="26645" PostTypeId="1" AcceptedAnswerId="26675" CreationDate="2011-03-13T01:56:34.893" Score="2" ViewCount="82" Body="&lt;p&gt;I want to determine if a singular point is a local min, max or saddle point.&lt;br&gt;&#xA;We are dealing with singular points so we cannot use the hessian matrix.&lt;br&gt;&#xA;What I have written, and I think I must of missed something is :  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Say we have a function $f(x,y,z)$.  To show that $(2,2,2)$ is a saddle point, we want to show :&lt;br&gt;&#xA;$\forall \epsilon &amp;gt;0$, we want to find $k,w,h$ such that if $k^2+w^2+h^2&amp;lt;\epsilon^2$ then&lt;br&gt;&#xA;And here is where what I have writen is incomplete:&lt;br&gt;&#xA;$f(2+k, 2+w, 2+h) &amp;gt; f(2,2,2)$&lt;br&gt;&#xA;...................................$&amp;lt; f(2,2,2)$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this correct as $f(2-k, 2-w, 2-h)&amp;lt;f(2,2,2)$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: I'm not sure how standard these terms are, so to be clear a singular point here is a point at which the partial of the function does not exist.&lt;/p&gt;&#xA;" OwnerUserId="2533" LastEditorUserId="2533" LastEditorDisplayName="" LastEditDate="2011-03-13T02:03:54.480" LastActivityDate="2011-03-13T08:33:40.727" Title="Classifying singular points as local min, max or saddle points." Tags="&lt;optimization&gt;" AnswerCount="1" />
  <row Id="26646" PostTypeId="2" ParentId="26644" CreationDate="2011-03-13T02:36:02.433" Score="2" ViewCount="" Body="&lt;p&gt;It seems the crux of what you want to prove is that every finite ordered set has a largest element. To show this is true, I would use induction. I suppose you could also argue that all finite ordered sets are well-ordered as you suggest but that seems to be equally complicated. So to use induction, you suppose that every size $n$ ordered set has a maximal element. Then consider a size $n+1$ ordered set $S$, and remove an element $a$. Then $S\setminus\{a\}$ has a largest element $b$. Now the larger of $a$ and $b$ is a largest element for $S$.&lt;/p&gt;&#xA;" OwnerUserId="2941" LastActivityDate="2011-03-13T02:36:02.433" CommentCount="6" />
  <row Id="26647" PostTypeId="1" AcceptedAnswerId="26661" CreationDate="2011-03-13T03:06:29.433" Score="1" ViewCount="131" Body="&lt;p&gt;This is related to my previous asked question &lt;a href=&quot;http://math.stackexchange.com/questions/26303/how-to-find-int-dd-fracz2-sqrtzc-cdot-sqrtd2-z2-dz&quot;&gt;here&lt;/a&gt;. The integration can be simplified as (As pointed out by  Sivaram, the integral diverges to $-\infty$ as $c\to1+$):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_{-\pi/2}^{\pi/2} \frac{\sin y}{\sqrt{\sin y+c}}\ \text{d}y,\ \  c&amp;gt;1.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am appreciated if there is a specific answer (it would be great if there are some details).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="6145" LastEditorUserId="6145" LastEditorDisplayName="" LastEditDate="2011-03-13T04:48:50.523" LastActivityDate="2011-03-13T06:05:52.277" Title="How to find $\int_{-\pi/2}^{\pi/2} \frac{\sin y}{\sqrt{\sin y+c}}\ \text{d}y$?" Tags="&lt;calculus&gt;&lt;integral&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="26648" PostTypeId="1" CreationDate="2011-03-13T03:20:43.863" Score="2" ViewCount="45" Body="&lt;p&gt;I'm a bit stumped by the exponential family representation of a multi-variate Gaussian distribution. Basically, the exponential form is a generic form for a large class of probability distributions. The standard form is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f_X(x) = \exp[\theta&amp;#39; T(x) + F(\theta)]$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\theta$ is a set of parameters (based on $\mu$ and $\Sigma$), $T(x)$ is a vector of sufficient statistics, and $F$ is a function of the parameters that ensures the distribution is a pdf, i.e., sums to one. For more information on this form, see &lt;a href=&quot;http://www.cs.columbia.edu/~jebara/4771/tutorials/lecture12.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.cs.columbia.edu/~jebara/4771/tutorials/lecture12.pdf&lt;/a&gt;, &lt;a href=&quot;http://en.wikipedia.org/wiki/Exponential_family&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Exponential_family&lt;/a&gt;, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &quot;conversion&quot; for a multi-variate Gaussian distribution to exponential family form is listed as &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\theta = [\Sigma^{-1}\mu, -\frac{1}{2}\Sigma^{-1}]&amp;#39;$$&#xA;$$T(x) = [x, x x&amp;#39;]&amp;#39;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but this is confusing because the outer product $x x&amp;#39;$ is a matrix and $-\frac{1}{2}\Sigma^{-1}$ is also a matrix. Thus, it seems the product between $\theta$ and $T(x)$ should result in a scalar &quot;entry&quot; and a matrix &quot;entry&quot;. Obviously, this expression needs to evaluate to a scalar. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The inner product works fine in the scalar case, and I understand this conversion is computed by manipulating to the quadratic form $x&amp;#39;Ax + b&amp;#39;x$.&#xA;Still, it seems that I am completely missing something here.  Thanks for your help.&lt;/p&gt;&#xA;" OwnerUserId="5799" LastActivityDate="2011-03-13T03:20:43.863" Title="Exponential family representation of multi-variate Gaussians" Tags="&lt;probability-distributions&gt;" CommentCount="4" />
  <row Id="26649" PostTypeId="2" ParentId="26632" CreationDate="2011-03-13T03:25:22.347" Score="6" ViewCount="" Body="&lt;p&gt;Let $W \sim F(x, y)$ where $F(x,y)$ stands for an $F$ distribution with degrees of freedom $x$ and $y$. Then, the quantity&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;\mathbb{P}(W \geq 1 ) = f(x,y)=\frac{\Gamma\left(\frac{x+y}{2}\right)(x/y)^{x/2}}{\Gamma(x/2)\Gamma(y/2)}\int_1^\infty w^{(x/2)-1}\left(1+\frac{xw}{y}\right)^{-(x+y)/2} \mathrm{d}w \&amp;gt; .&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From this, I think you can find the answer in the reference below.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;B. K. Ghosh, &lt;a href=&quot;http://www.jstor.org/pss/2985113&quot; rel=&quot;nofollow&quot;&gt;Some monotonicity theorems for $\chi^2$, $F$ and $t$ distributions with applications&lt;/a&gt;, &lt;em&gt;J. Royal Stat. Soc. B&lt;/em&gt;, vol. 35, no. 3 (1973), pp. 480-492. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Incidentally, note that $W = \frac{y}{x} \frac{U_{xy}}{1-U_{xy}}$ where $U_{xy} \sim \mathrm{Beta}(x/2, y/2)$ and so $\mathbb{P}(W \geq 1) = \mathbb{P}(U_{xy} \geq (1+y/x)^{-1})$.&lt;/p&gt;&#xA;" OwnerUserId="7003" LastEditorUserId="7003" LastEditorDisplayName="" LastEditDate="2011-03-13T05:03:49.907" LastActivityDate="2011-03-13T05:03:49.907" CommentCount="6" />
  <row Id="26650" PostTypeId="1" CreationDate="2011-03-13T04:06:38.780" Score="1" ViewCount="106" Body="&lt;p&gt;Here is a silly mistake I am making: where exactly is the mistake?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that torus cannot hold a metric of constant curvature -1 ( hyperbolic metric ). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But what if I do this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The upper half-plane and $\mathbb{C}$ are diffeomorphic by a diffeomorphism, say $\phi$; pull the hyperbolic metric from the upper half plane to the complex plane $\mathbb{C}$, and then quotient it by $ Z\oplus Z$ to get a hyperbolic metric on the torus. Impossible! But what am I missing?&lt;/p&gt;&#xA;" OwnerUserId="6920" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-13T04:23:17.077" LastActivityDate="2011-03-13T12:47:12.647" Title="Hyperbolic metric on the torus ?" Tags="&lt;hyperbolic-geometry&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="26651" PostTypeId="1" CreationDate="2011-03-13T04:20:56.547" Score="0" ViewCount="148" Body="&lt;p&gt;Is there any way I can calculate derivatives using a Casio Graphing Calculator?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a program available for the calculator which can calculate them?&lt;/p&gt;&#xA;" OwnerUserId="8176" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2011-03-13T04:22:01.083" LastActivityDate="2011-03-13T07:25:05.060" Title="Calculate derivatives on Casio Graphics Calculator?" Tags="&lt;calculus&gt;&lt;calculator&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="26652" PostTypeId="2" ParentId="26626" CreationDate="2011-03-13T04:25:26.773" Score="3" ViewCount="" Body="&lt;p&gt;Update: This proof no longer works as is for the new question about &lt;em&gt;positive&lt;/em&gt; integers -- I suspect it could be fixed, but I won't bother since the other one is simpler and not affected by the change.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To generate $1$, we need to have $z^k=a+\mathrm{i}b$, and hence $(\lvert z \rvert^2)^k=a^2+b^2$, with $a=1$ or $b=1$. Since $\lvert z \rvert^2$ is an integer, $k$ cannot be even, for otherwise we would have two squares that differ by $1$, which is only possible if  $\lvert z \rvert=1$, which is obviously not a solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So $k$ must be odd. Now consider $(x + \mathrm{i}y)^k$ for $k$ odd (and still $a=1$ or $b=1$). Then all terms in the real part contain a factor $x$ and all terms in the imaginary part contain a factor $y$. Since one of these parts is $1$, it follows that one of $x$ and $y$ is $\pm 1$. They cannot both be $\pm 1$, since that is obviously not a solution. Denote the one that isn't $\pm1$ by $w$. (Also clearly $w\neq0$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now consider $(x + \mathrm{i}y)^k$ for arbitrary $k$. All terms in the real and imaginary parts are either $\pm 1$ or multiples of $w$. Thus the only residues modulo $w$ that occur are $\pm1$ and $0$. It follows that $\lvert w \rvert \le 3$, i.e. $w=\pm3$ or $w=\pm2$. Both cases are easily excluded: With $w=\pm3$, both parts of $z^2$ are even, and hence we cannot generate any odd integers other than $\pm1$ and $\pm3$. With $w=\pm2$, the residues modulo 5 of the real and imaginary parts are periodic in $k$ with period $4$ (period $2$ if we drop their sign) and none of the residues in the period is $0$, and hence we cannot generate any multiples of $5$. $\Box$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S.: I just realized that I killed this proof when I proposed to save the question by including $z^0$ :-). Fortunately the proof can be saved just as easily; just replace $1$ by $-1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.P.S.: This avenue has now also been closed by the restriction to positive integers.&lt;/p&gt;&#xA;" OwnerUserId="6622" LastEditorUserId="6622" LastEditorDisplayName="" LastEditDate="2011-03-13T10:18:15.970" LastActivityDate="2011-03-13T10:18:15.970" CommentCount="2" />
  <row Id="26653" PostTypeId="1" CreationDate="2011-03-13T05:04:46.103" Score="0" ViewCount="54" Body="&lt;p&gt;I was working on a scheme in cryptography and came up with the following scenario.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To put it in proper words.&lt;br&gt;&#xA;1. We have an element 1/(x+m).&lt;br&gt;&#xA;2. The 2 elements x and x1 are known.&lt;br&gt;&#xA;3. We want to transform 1/(x+m) to 1/(x1+m).&#xA;i.e, We need a k such that k/(x+m) = 1/(x1+m).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can consider x,x1,m to be elements of Z&lt;sub&gt;p&lt;/sub&gt;&lt;sup&gt;*&lt;/sup&gt;&lt;br&gt;&#xA;You can introduce any extra dummy variables if you want for the conversion. It'd be of great of help if you can give me an idea with this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="7488" LastEditorUserId="7488" LastEditorDisplayName="" LastEditDate="2011-03-13T05:30:14.950" LastActivityDate="2011-03-13T05:51:33.163" Title="Intuition for a simple conversion" Tags="&lt;number-theory&gt;&lt;algebra&gt;&lt;cryptography&gt;&lt;problem-solving&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="26654" PostTypeId="1" AcceptedAnswerId="26700" CreationDate="2011-03-13T05:09:11.467" Score="2" ViewCount="116" Body="&lt;p&gt;This is related to a homework question for a class that I am TAing for. I'm using Sipser terminology here (recognizable for computably enumerable, decidable for computable). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given that $w^r$ is the reverse of $w$, and that&#xA;$$T = \{ \langle M\rangle \mid \mbox{if $M$ accepts $w$ then $M$ accepts $w^r$}\}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now here are the questions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clearly $T$ is undecidable. Is $T$ unrecognizable? My intuition tells me yes it's unrecongnizable. Since we have to confirm that $w^r$ is accepted for all $w$, this seems to be a good sign that it is unrecognizable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is $\overline{T}$ recognizable? This also seems unrecognizable as we would have no way to tell if the machine is looping on a particular input.&lt;/p&gt;&#xA;" OwnerUserId="6299" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-13T05:11:47.503" LastActivityDate="2011-03-14T01:08:40.827" Title="Recognizable or not?" Tags="&lt;computability&gt;" AnswerCount="2" />
  <row Id="26655" PostTypeId="1" AcceptedAnswerId="26659" CreationDate="2011-03-13T05:36:25.797" Score="6" ViewCount="80" Body="&lt;p&gt;Hello,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $p$ be prime, and $a$ be integer. When does $(p - 1)! + 1 = p^a$ for some $a$ hold?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example:&#xA;$$p = 5 \implies (5 - 1)! + 1 = 25 = 5^2$$&#xA;$$p = 7 \implies (7 - 1)! + 1 = 721 = 7 \cdot 103$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any idea? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks,&lt;/p&gt;&#xA;" OwnerUserId="5711" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-13T05:42:07.617" LastActivityDate="2011-03-13T06:02:05.133" Title="Question regarding Wilson's Theorem" Tags="&lt;elementary-number-theory&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="26656" PostTypeId="1" AcceptedAnswerId="26673" CreationDate="2011-03-13T05:37:07.917" Score="3" ViewCount="76" Body="&lt;p&gt;I guess I understand some reasons why we should care for complex structure on manifolds, but what is the reason why product structure is studied? Does it arise naturally somewhere?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The product structure on a smooth manifold $M$ is given by a (1,1)-tensor $E$ such that $E \neq \pm1$ and $E^2 = 1$ and the following integrability condition holds: $$E[X,Y] = [EX,Y] + [X,EY] - E[EX,EY]$$&#xA;Compare it to a complex structure $J$ such that $J^2 = -1$ and $$J[X,Y] = [JX,Y] + [X,JY] + J[JX,JY]$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm still studying basic literature, one thing I do know is that on a Lie group it $E$ induces a double algebra structure on a correspoinding Lie algebra, and thus two foliations on the group, if I remember correctly; I guess it is generalized for an arbitrary manifold, so this 'splitting' of the tangent bundle and how it's compatible with other structures (complex, Hermitian, Kähler etc., e.g. a complex product structure is defined by the two respective structures such that $JE = -EJ$) is probably one of the reasons to study it, but I'm not sure this reason is good enough on its own.&lt;/p&gt;&#xA;" OwnerUserId="3793" LastEditorUserId="3793" LastEditorDisplayName="" LastEditDate="2011-03-13T07:39:48.590" LastActivityDate="2011-03-13T08:08:30.110" Title="Motivation for product structure" Tags="&lt;soft-question&gt;&lt;differential-geometry&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="26657" PostTypeId="2" ParentId="26653" CreationDate="2011-03-13T05:40:53.653" Score="1" ViewCount="" Body="&lt;p&gt;You are given $y=\frac{1}{x+m}$. Then $\frac{1}{y}= x+m$, and $x_1+m = \frac{1}{y}-x+x_1$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So $$\frac{1}{x_1+m} = \frac{y}{1-y(x-x_1)}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, simply set &#xA;$$k = \frac{1}{1-y(x-x_1)},\quad \text{where}\quad y = \frac{1}{1+m},$$&#xA;which can be done since you know $x$, $x_1$, and $y$.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-13T05:51:33.163" LastActivityDate="2011-03-13T05:51:33.163" CommentCount="2" />
  <row Id="26658" PostTypeId="2" ParentId="26541" CreationDate="2011-03-13T05:45:25.943" Score="0" ViewCount="" Body="&lt;p&gt;To answer your last question, you can formulate your problem as a &lt;a href=&quot;http://en.wikipedia.org/wiki/Quadratic_programming&quot; rel=&quot;nofollow&quot;&gt;quadratic program&lt;/a&gt; and then solve it using one of the standard methods for solving quadratic programs.  (And, yes, your problem can be transformed into a linear complementarity problem.  See the end of my answer.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rewrite the equality constraint as $$0 = {\bf z}^T {\bf A}^T {\bf Mz} + {\bf z}^T {\bf A}^T {\bf q} + {\bf b}^T {\bf Mz} + {\bf b}^T {\bf q} = {\bf z}^T {\bf A}^T {\bf Mz} + {\bf z}^T ({\bf A}^T {\bf q} + {\bf M}^T {\bf b}) + {\bf b}^T {\bf q}.$$  The second equality is valid because ${\bf b}^T {\bf Mz}$ is a scalar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, ${\bf A}^T {\bf M}$ is not necessarily symmetric, but by doing the matrix multiplication you can find some symmetric matrix ${\bf Q}$ so that ${\bf z}^T {\bf A}^T {\bf Mz} = \frac{1}{2} {\bf z}^T {\bf Qz}$.  Then your problem is equivalent to solving the quadratic program &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\min \frac{1}{2}{\bf z}^T {\bf Q z} + {\bf z}^T {\bf c} + {\bf b}^T {\bf q}$$&#xA;subject to&#xA;$${\bf Mz} + {\bf q} \geq {\bf 0},$$&#xA;$${\bf Az} + {\bf b} \geq {\bf 0},$$&#xA;where &#xA;${\bf c} = {\bf A}^T {\bf q} + {\bf M}^T {\bf b}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is this equivalent to your problem?  Let ${\bf z}^*$ be the optimal solution to the quadratic program.  The value of ${\bf z}^*$ cannot be smaller than $0$ because the problem constraints force $({\bf Az} + {\bf b})^T({\bf Mz} + {\bf q}) \geq 0$.  If the value of ${\bf z}^*$ is $0$, then ${\bf z}^*$ is a solution to your second problem.  If the value of ${\bf z}^*$ is strictly greater than $0$, then you cannot force $({\bf Az} + {\bf b})^T({\bf Mz} + {\bf q})$ to $0$ while satisfying the first two constraints, and so there is no solution to your second problem.    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;HR&gt;&#xA;If you want, you could also transform the quadratic program into a linear complementarity problem.  The Wikipedia page on &lt;a href=&quot;http://en.wikipedia.org/wiki/Linear_complementarity_problem#Relation_to_quadratic_programming&quot; rel=&quot;nofollow&quot;&gt;linear complementarity problems&lt;/a&gt; explains how to do that, so I'll leave it to you to work that out if you want.  You will have to introduce a nonnegativity constraint on the variables in the quadratic program first, though.  The standard way to do that is to replace ${\bf z}$ with ${\bf z}^+ - {\bf z}^-$ throughout the quadratic program and add the nonnegativity constraints ${\bf z}^+ \geq 0, {\bf z}^- \geq 0$.&lt;/p&gt;&#xA;" OwnerUserId="2370" LastActivityDate="2011-03-13T05:45:25.943" />
  <row Id="26659" PostTypeId="2" ParentId="26655" CreationDate="2011-03-13T06:02:05.133" Score="3" ViewCount="" Body="&lt;p&gt;A &lt;a href=&quot;http://en.wikipedia.org/wiki/Wilson_prime&quot; rel=&quot;nofollow&quot;&gt;Wilson prime&lt;/a&gt; is a prime number $p$ such that $p^2$ divides $(p-1)!+1$. The only known Wilson primes are 5, 13, and 563.&lt;/p&gt;&#xA;" OwnerUserId="857" LastActivityDate="2011-03-13T06:02:05.133" CommentCount="1" />
  <row Id="26660" PostTypeId="2" ParentId="26594" CreationDate="2011-03-13T06:03:53.460" Score="-1" ViewCount="" Body="&lt;p&gt;Hi Ricardo,  it is good to see you here. First let me start why your problem is not correctly stated. This also explains why my MO answer was correct. One can see an equation by using different glasses, but no one considers $3x=\lambda x$ as an eigenvalue value problem or a PDE! To have an eigenvalue problem you should have  a correctly stated linear operator $T$  whose action on a vector space is a priori known. Solving an eigenvalue problem means determining the eigenvalues and the corresponding eigenspaces of that operator. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The main reason why not every matrix equation containing a $\lambda$ is an eigenvalue equation is that not every matrix corresponds to a well defined operator. (remember that a matrix is nothing more than a rectangular representaion of a linear map)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, answers to your questions: First of all, I still don't see a correctly stated problem. Is $\lambda$ given and are you looking for $A+B?$ Or, is $A$ given and are you looking for $\lambda$ or $B?$ And most importantly, what do you want to obtain at the end of the day?         &lt;/p&gt;&#xA;" OwnerUserId="7796" LastActivityDate="2011-03-13T06:03:53.460" CommentCount="4" />
  <row Id="26661" PostTypeId="2" ParentId="26647" CreationDate="2011-03-13T06:05:52.277" Score="0" ViewCount="" Body="&lt;p&gt;Using formula 288.03 in &lt;a href=&quot;http://books.google.com/books?id=W6GV17SP0kQC&quot; rel=&quot;nofollow&quot;&gt;Byrd and Friedman's handbook&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_{\varphi}^{\pi/2}\frac{\sin^m u}{\sqrt{a+b\sin\;u}}\mathrm du=\frac2{\sqrt{a+b}}\int_0^{F\left(\arcsin\left(\sqrt{\frac{1-\sin\;\varphi}{2}}\right)\mid\frac{2b}{a+b}\right)}\left(1-2\mathrm{sn}^2\left(u\mid\frac{2b}{a+b}\right)\right)^m\mathrm du$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for $a &amp;gt; b &amp;gt; 0$, where $F(\phi|m)$ is the incomplete elliptic integral of the first kind and $\mathrm{sn}(u|m)$ is a Jacobian elliptic function, we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_{-\pi/2}^{\pi/2} \frac{\sin y}{\sqrt{\sin y+c}}\mathrm{d}y=2\left(\sqrt{1+c}\;E\left(\frac2{1+c}\right)-\frac{c}{\sqrt{1+c}}K\left(\frac2{1+c}\right)\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $K(m)$ is the complete elliptic integral of the first kind, and $E(m)$ is the complete elliptic integral of the second kind.&lt;/p&gt;&#xA;" OwnerUserId="8178" LastActivityDate="2011-03-13T06:05:52.277" />
  <row Id="26662" PostTypeId="1" CreationDate="2011-03-13T06:13:30.647" Score="2" ViewCount="238" Body="&lt;p&gt;How can we physically interpret an eigenvalue or an eigenvector in linear algebra?&lt;/p&gt;&#xA;" OwnerUserId="8180" LastEditorUserId="6296" LastEditorDisplayName="" LastEditDate="2011-03-13T06:48:47.897" LastActivityDate="2011-03-13T15:45:11.840" Title="Eigenvalues and eigenvectors in physics?" Tags="&lt;linear-algebra&gt;&lt;eigenvalues&gt;" AnswerCount="7" FavoriteCount="1" />
  <row Id="26663" PostTypeId="2" ParentId="26662" CreationDate="2011-03-13T06:32:16.023" Score="0" ViewCount="" Body="&lt;p&gt;in quantum mechanics, you can only measure the &lt;em&gt;stable&lt;/em&gt; state of a system, which is defined to be the eigen state of some operator. For example, $p=-i\frac{h}{2\pi}\nabla$. Hence eigen vectors and eigen values play a fundamental role in quantum mechanics. You may read Chapter 1-2 in Dirac's book &lt;em&gt;Principle of Quantum Mechanics&lt;/em&gt; to get a better understanding. I am also an undergraduate, I expect a much detailed answer from others as eigenvectors just appeared in everywhere. A not bad place for start is here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Eigenvalue&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Eigenvalue&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7887" LastActivityDate="2011-03-13T06:32:16.023" />
  <row Id="26664" PostTypeId="2" ParentId="26662" CreationDate="2011-03-13T06:37:16.630" Score="6" ViewCount="" Body="&lt;p&gt;In classical mechanics, the eigenvectors of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Inertia_tensor&quot; rel=&quot;nofollow&quot;&gt;inertia tensor&lt;/a&gt; correspond to the &lt;a href=&quot;http://en.wikipedia.org/wiki/Principal_axis_%28mechanics%29#Principal_axes_of_inertia&quot; rel=&quot;nofollow&quot;&gt;principal axes&lt;/a&gt; of the object; these are directions about which the angular momentum will be point in the same direction as the direction of rotation, so that  the object can rotate without precession in this direction (without any torque).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In quantum mechanics, to say that a state is an eigenfunction of an observable (i.e. Hermitian operator) equates to saying that measurement of that observable on that state is certain to return a definite value. In the special case of a (time-independent) Hamiltonian, the &lt;a href=&quot;http://en.wikipedia.org/wiki/Schrodinger_equation&quot; rel=&quot;nofollow&quot;&gt;Schrodinger equation&lt;/a&gt; implies that the eigenstate in question is a &lt;em&gt;stationary&lt;/em&gt; one---it evolves in time by multiplication by an exponential (of frequency proportional to the energy), and the expectation value of any measurement on the state remains constant in time.&lt;/p&gt;&#xA;" OwnerUserId="536" LastEditorUserId="536" LastEditorDisplayName="" LastEditDate="2011-03-13T14:13:59.527" LastActivityDate="2011-03-13T14:13:59.527" CommentCount="2" />
  <row Id="26665" PostTypeId="1" AcceptedAnswerId="26667" CreationDate="2011-03-13T06:38:16.877" Score="2" ViewCount="75" Body="&lt;p&gt;Please help me count.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have an alphabet with 26 English letters. I can reduce it to 25 letters by representing two of the letters (e.g. O and X) with a new letter (e.g. $\otimes$).&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;25&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;To get down to a 25 letter alphabet I have $26 \choose 2$ possibilities. So 325 possible reductions to alphabet of size 25.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;2&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;Getting down to an alphabet of size 2 is also pretty easy - I just need to count all possible ways of dividing the alphabet into 2 partitions. That would be $26 \choose 1$ + $26 \choose 2$ + ... + $26 \choose 13$. &#xA;In the same way I can reduce my alphabet to any size from 1 to 25. That's 38,754,731 (computed in &lt;a href=&quot;http://en.wikipedia.org/wiki/R_%28programming_language%29&quot; rel=&quot;nofollow&quot;&gt;R&lt;/a&gt; like this &lt;code&gt;sum(choose(26,1:13))&lt;/code&gt;).&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;The Rest&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;How can do the counting for size 3 or 24?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the total number of such alphabet reductions?&lt;/p&gt;&#xA;" OwnerUserId="8179" LastEditorUserId="8179" LastEditorDisplayName="" LastEditDate="2011-03-13T18:46:32.023" LastActivityDate="2011-03-13T18:46:32.023" Title="Given an alphabet with 26 letters, how many reduced alphabets are possible?" Tags="&lt;counting&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="26666" PostTypeId="2" ParentId="26662" CreationDate="2011-03-13T06:38:40.407" Score="1" ViewCount="" Body="&lt;p&gt;In terms of transformations of $E^{n}$ ($n$-dimensional Euclidean space), the eigenvectors of a transformation are axes along which the transformation does not rotate the space (but may or may not stretch or compress it). For each eigenvector there is a corresponding eigenvalue, which measures how much stretching or compression occurs along the axis defined by the eigenvector.&lt;/p&gt;&#xA;" OwnerUserId="8173" LastActivityDate="2011-03-13T06:38:40.407" CommentCount="2" />
  <row Id="26667" PostTypeId="2" ParentId="26665" CreationDate="2011-03-13T07:09:04.873" Score="3" ViewCount="" Body="&lt;p&gt;You're probably looking for the &lt;a href=&quot;http://en.wikipedia.org/wiki/Bell_number&quot; rel=&quot;nofollow&quot;&gt;Bell numbers&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2011-03-13T07:09:04.873" CommentCount="2" />
  <row Id="26668" PostTypeId="1" CreationDate="2011-03-13T07:22:07.997" Score="0" ViewCount="52" Body="&lt;p&gt;This is a follow up of the question: &lt;a href=&quot;http://math.stackexchange.com/questions/26653/intuition-for-a-simple-conversion&quot;&gt;Tranformation &lt;/a&gt;   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We are given&#xA;$$g^{1/(x+m)},$$&#xA;(it is not possible to find $\frac{1}{x+m}$ due to the Discrete log problem), can we find a $k$ such that&#xA;$$g^{k/(x+m)} = g^{1/(x_1+m)}\quad\text{?}$$&lt;/p&gt;&#xA;" OwnerUserId="8182" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-04-12T15:33:58.823" LastActivityDate="2011-05-12T15:56:57.493" Title="Feasability of the transformation" Tags="&lt;number-theory&gt;&lt;algebra&gt;&lt;cryptography&gt;&lt;problem-solving&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="26669" PostTypeId="2" ParentId="26654" CreationDate="2011-03-13T07:24:41.060" Score="4" ViewCount="" Body="&lt;p&gt;You can reduce both the halting problem and its complement to $T$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The easier part is reducing the halting problem. Given a machine $M$, construct a new machine $M&amp;#39;$ that usually just accepts, but on input $01$, say, runs $M$, and then accepts. Now $M$ halts iff $M&amp;#39; \in T$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let's reduce the converse. Given a machine $M$, construct a new machine $M&amp;#39;$ that usually just accepts, but on input $0^n1$ runs $M$ and accepts iff $M$ has &lt;em&gt;not&lt;/em&gt; halted within the first $n$ steps. Now $M$ halts iff $M&amp;#39; \notin T$.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2011-03-13T07:24:41.060" CommentCount="1" />
  <row Id="26670" PostTypeId="2" ParentId="26651" CreationDate="2011-03-13T07:25:05.060" Score="2" ViewCount="" Body="&lt;p&gt;I don't have that model at hand to check for sure, but I think you can get a numerical approximation of the derivative at a point by graphing the function and using the trace feature with derivative turned on in the settings.  Here is &lt;a href=&quot;http://support.casio.com/pdf/004/fx_plus_chapter08.pdf&quot; rel=&quot;nofollow&quot;&gt;the relevant section of the manual, from chapter 8&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/OSvKr.png&quot; alt=&quot;excerpt from section 8-6 of the manual&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2011-03-13T07:25:05.060" CommentCount="2" />
  <row Id="26671" PostTypeId="2" ParentId="24536" CreationDate="2011-03-13T07:42:23.000" Score="3" ViewCount="" Body="&lt;p&gt;From &lt;a href=&quot;http://support.casio.com/pdf/004/fx9860GSD_9860G_E.pdf&quot; rel=&quot;nofollow&quot;&gt;the instruction manual&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/Pg5zc.png&quot; alt=&quot;product of two polar-form complex numbers&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, entering &lt;code&gt;2∠30&lt;/code&gt; is equivalent to entering $2\text{ cis }30&amp;#176;$.&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2011-03-13T07:42:23.000" CommentCount="1" />
  <row Id="26672" PostTypeId="1" AcceptedAnswerId="26674" CreationDate="2011-03-13T07:52:24.773" Score="6" ViewCount="112" Body="&lt;p&gt;$$\lim_{(x,y)\to(0,0)} \frac{xy}{\sqrt{x^2+y^2}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Approaching (0,0) along x or along y both result in the limit approaching 0, so you want to make sure that the limit exists by doing more tests.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My solutions manual uses x = y^2 (or y = x^2). Why either of those? Why not y=x or x=y? Why a parabola?&lt;/p&gt;&#xA;" OwnerUserId="3514" LastEditorUserId="6296" LastEditorDisplayName="" LastEditDate="2011-03-13T09:21:35.723" LastActivityDate="2011-03-13T09:21:35.723" Title="Finding limit of function with more than one variable" Tags="&lt;calculus&gt;&lt;homework&gt;&lt;multivariable-calculus&gt;" AnswerCount="1" />
  <row Id="26673" PostTypeId="2" ParentId="26656" CreationDate="2011-03-13T08:08:30.110" Score="1" ViewCount="" Body="&lt;p&gt;I don't know a lot about this but in the context of a complex manifold, $E$ is called a real structure.  Searching the literature for that term should turn up a lot of material.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The conditions $JE = -EJ$ and $E^2 = 1$ make $E$ analogous to a conjugation operation.  This allows you to split up your tangent space into real and complex parts corresponding to the +1 and -1 eigenvalues of $E$, respectively (note that $JE = -EJ$ means that $J$ is an isomorphism between the +1 and -1 eigenspaces).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not exactly sure what the use of such a structure is, but it seems natural to want to generalize the notion of conjugation so we can divide our tangent space into real and imaginary parts.&lt;/p&gt;&#xA;" OwnerUserId="9" LastActivityDate="2011-03-13T08:08:30.110" CommentCount="3" />
  <row Id="26674" PostTypeId="2" ParentId="26672" CreationDate="2011-03-13T08:10:16.563" Score="6" ViewCount="" Body="&lt;p&gt;None of those choices suffices to prove that the limit is $0$, so I don't know what the solution writer meant.  No finite number of ways to approach $(0,0)$ can be enough to show that the limit exists.  On the other hand, in a case where the limit &lt;em&gt;doesn't&lt;/em&gt; exist at a point, one way to show it in some cases is to show that the function approaches different limits as you approach the point along two different curves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case, $(|x|-|y|)^2\geq 0$ implies $|xy|\leq\frac{1}{2}(x^2+y^2)$, so that $$\left|\frac{xy}{\sqrt{x^2+y^2}}\right|\leq\frac{1}{2}\sqrt{x^2+y^2}.$$  This makes it easy to see that the limit is $0$.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2011-03-13T08:10:16.563" CommentCount="4" />
  <row Id="26675" PostTypeId="2" ParentId="26645" CreationDate="2011-03-13T08:28:32.163" Score="1" ViewCount="" Body="&lt;p&gt;A saddle point is neither a local max nor a min.  For $(2,2,2)$ to be a saddle point we need the following.  For every $\epsilon$$&gt;0$ there exists $(x_1,y_1,z_1)$ and $(x_2,y_2,z_2)$ such that $d=\sqrt{(2-x_k)^2+(2-y_k)^2+(2-z_k)^2}$$&amp;lt;\epsilon$ for $k=1,2$ and &#xA;                 $f(x_1,y_1,z_1)&amp;lt;f(2,2,2)&amp;lt;f(x_2,y_2,z_2)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ie, no mater how small of an interval we put around the domain point $(2,2,2)$ there are points in that interval producing a larger and smaller value of $f$.&lt;/p&gt;&#xA;" OwnerUserId="3971" LastEditorUserId="3971" LastEditorDisplayName="" LastEditDate="2011-03-13T08:33:40.727" LastActivityDate="2011-03-13T08:33:40.727" />
  <row Id="26676" PostTypeId="1" AcceptedAnswerId="26681" CreationDate="2011-03-13T08:29:29.933" Score="4" ViewCount="149" Body="&lt;p&gt;we know that there are sets which are not lebesgue measurable ,we also know that outer measure is defined for any subset of R . &lt;/p&gt;&#xA;" OwnerUserId="7447" LastActivityDate="2011-03-13T09:07:06.367" Title="What is the difference between outer measure and lebesgue measure" Tags="&lt;measure-theory&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" />
  <row Id="26677" PostTypeId="1" AcceptedAnswerId="26686" CreationDate="2011-03-13T08:33:51.330" Score="1" ViewCount="77" Body="&lt;p&gt;How many symmetric polynomials of n degree with all their coefficients $\ =1 $ are there?Is there a type that computes their number?  &lt;/p&gt;&#xA;" OwnerUserId="5422" LastActivityDate="2011-03-13T19:22:57.710" Title="The number of symmetric polynomials of n degree" Tags="&lt;combinatorics&gt;&lt;polynomials&gt;" AnswerCount="1" />
  <row Id="26678" PostTypeId="1" AcceptedAnswerId="26679" CreationDate="2011-03-13T08:33:57.133" Score="1" ViewCount="159" Body="&lt;p&gt;I am trying to understand the value of $\bar{x_{1} x_{2}}+E(x_{1} x_{2})$. For all $i$, $E(x_{i})$ and $\sigma_{i}$ are given. Wikipedia gives the joint probability density function:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$E(XY) = \int \int x y j(x,y) dx dy$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;then I can find out from wikipedia that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$E(XY) = Cov(X,Y) + E(X)E(Y)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and by Cauchy-Swartz:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$| Cov(X,Y) | \leq \sigma(X) \sigma(Y)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but I cannot find a precise formula to find the value of $E(XY)$, only an upper bound with finite variances. A crux point to find the matrix $\rho$ so that I can calculate the $\sigma$ matrix. So how can I calculate the $E(XY)$ when only $E(x_{i})$ and $\sigma_{i}$ for all $i$ are given?&lt;/p&gt;&#xA;" OwnerUserId="7628" LastActivityDate="2011-03-13T09:34:44.233" Title="How to calculate the expected value $E(XY)$ with known $E(X)$, $E(Y)$ and $\sigma_{i}$?" Tags="&lt;statistics&gt;&lt;standard-deviation&gt;" AnswerCount="1" />
  <row Id="26679" PostTypeId="2" ParentId="26678" CreationDate="2011-03-13T08:43:21.750" Score="4" ViewCount="" Body="&lt;p&gt;Such a formula &lt;em&gt;cannot exist&lt;/em&gt;, since it would compute $E(XY)$ only using (characteristics associated to) the separate distributions of $X$ and $Y$. $E(XY)$ depends, as you noted, on the joint distribution of $X$ and $Y$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The missing figure you need to get a formula as you would like is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Correlation_and_dependence&quot; rel=&quot;nofollow&quot;&gt;correlation coefficient&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="367" LastEditorUserId="367" LastEditorDisplayName="" LastEditDate="2011-03-13T09:34:44.233" LastActivityDate="2011-03-13T09:34:44.233" CommentCount="1" />
  <row Id="26680" PostTypeId="1" CreationDate="2011-03-13T08:44:36.713" Score="7" ViewCount="160" Body="&lt;p&gt;It is a theorem that any function $f$ defined for positive real numbers satisfying&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$f(1)=1$&lt;/li&gt;&#xA;&lt;li&gt;$f(x+1)=x\cdot f(x)$&lt;/li&gt;&#xA;&lt;li&gt;$f$ is log convex&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;is identically equal to the gamma function. (Condition 2 means that this function interpolates a shifted factorial function.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, a beginner (such as myself) might ask: &lt;strong&gt;What if we weaken condition 2 by instead requiring $f$ to be merely convex, not log convex?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would imagine that such functions would look not too different, since intuitively, I can't wildly deviate the graph of the gamma function if I want to maintain condition 2 and stay convex.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just a follow-up musing---What if instead of condition 3, we require convexity and infinite differentiability? Do we still uniquely determine the gamma function?&lt;/p&gt;&#xA;" OwnerUserId="7676" LastActivityDate="2011-04-12T10:40:13.463" Title="The uniqueness of the Gamma Function" Tags="&lt;real-analysis&gt;&lt;special-functions&gt;&lt;gamma-function&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="26681" PostTypeId="2" ParentId="26676" CreationDate="2011-03-13T08:49:42.357" Score="8" ViewCount="" Body="&lt;p&gt;The Lebesgue measure and Lebesgue outer measure coincide on Lebesgue measurable sets, which can be defined in several equivalent ways. Let $m$ and $m^*$ denote the Lebesgue measure and the Lebesgue outer measure respectively. These are some possible definitions of $A\subset\mathbb{R}^n$ being measurable:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;For all $B\subset\mathbb{R}^n$&#xA;$$&#xD;&#xA;m^*(A)=m^*(A\cap B)+m^*(A\setminus B)&#xD;&#xA;$$&lt;/li&gt;&#xA;&lt;li&gt;For all $\epsilon&amp;gt;0$ there exist an&#xA;open set $G$ and a closed set $F$&#xA;such that $F\subset A\subset G$ and&#xA;$m^*(G\setminus F)&amp;lt;\epsilon$. (Note&#xA;that since $G\setminus F$ is open,&#xA;it is measurable, so that&#xA;$m^*(G\setminus F)=m(G\setminus F)$.)&lt;/li&gt;&#xA;&lt;li&gt;$A=F\cup N$, where $F$ is an&#xA;$F_\sigma$ (i.e. a countable union&#xA;of closed sets) and $m(N)=0$.&lt;/li&gt;&#xA;&lt;li&gt;$A=G\setminus N$, where $G$ is a&#xA;$G_\delta$ (i.e. a countable&#xA;intersection of open sets) and&#xA;$m(N)=0$.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The reason for the need of two different concepts is that neither of them is &quot;perfect&quot;:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$m$ is a measure, but is not defined for all subsets of $\mathbb{R}^n$&lt;/li&gt;&#xA;&lt;li&gt;$m^*$ is defined for all subsets of $\mathbb{R}^n$, but is not additive: here exist disjoint sets $A$ and $B$ such that $m^*(A\cup B)\ne m^*(A)+m^*(B)$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="4791" LastActivityDate="2011-03-13T08:49:42.357" />
  <row Id="26682" PostTypeId="1" AcceptedAnswerId="26690" CreationDate="2011-03-13T09:06:54.973" Score="3" ViewCount="121" Body="&lt;p&gt;I am looking for a good approximation for the $W_0$ branch of the Lambert $W$ function. I am looking for values $0 &amp;lt; x &amp;lt; e$ only, so I expect something simpler than the general Taylor expansion. Thanks.&lt;/p&gt;&#xA;" OwnerUserId="8184" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2011-03-16T05:22:57.473" LastActivityDate="2011-03-16T05:22:57.473" Title="Approximation for Lambert W function near zero" Tags="&lt;approximation&gt;&lt;special-functions&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="26683" PostTypeId="1" AcceptedAnswerId="26715" CreationDate="2011-03-13T09:07:03.677" Score="2" ViewCount="160" Body="&lt;p&gt;This problem comes from page 99 in Folland's &quot;real analysis: modern techniques and their applications&quot;, 2nd edition, as the image below shows.&#xA;&lt;img src=&quot;http://i.imgur.com/dnlUp.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried to prove condition (ii) for &lt;em&gt;n&lt;/em&gt;=1 under condition (i), The author says it follows from Theorems 1.16 and 1.18, but I can not see any relation between condition (ii) and the two Theorems mentioned. What $\inf\{...\}$ defines is an outer measure. But, since the collection of all open set, i.e., the topology of $\mathbb R^n$, is not a ring, I can not prove that the outer measure restricted to $\mathbb R^n$ is a measure. In consequence, I can not use subtractivity property of measure to prove condition (ii) using the method similar to that of Theorem 1.14. Could you please help me with this problem? Thanks!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS: Since this is a specific problem in Folland's book, in addition to the good will to answer, I have to assume the answerer have at hand a copy of Folland's book and, as a must have for owners of this book, its two versions of erratas.&lt;/p&gt;&#xA;" OwnerUserId="7321" LastEditorUserId="7321" LastEditorDisplayName="" LastEditDate="2011-03-13T11:00:24.233" LastActivityDate="2011-03-26T18:53:34.713" Title="A problem regarding regular measure" Tags="&lt;real-analysis&gt;&lt;measure-theory&gt;" AnswerCount="2" CommentCount="5" FavoriteCount="1" />
  <row Id="26684" PostTypeId="2" ParentId="26676" CreationDate="2011-03-13T09:07:06.367" Score="7" ViewCount="" Body="&lt;p&gt;As a supplement to Julián Aguirre's answer, note that Lebesgue originally wanted a measure $m$ to satisfy certain properties:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$m(S)&amp;gt;0$ for any set $S$&lt;/li&gt;&#xA;&lt;li&gt;$m$ is identical to length when considering intervals&lt;/li&gt;&#xA;&lt;li&gt;$m$ is translation invariant, i.e. if you slide your set up or down the real line, its measure should be unchanged&lt;/li&gt;&#xA;&lt;li&gt;$m$ should be (countably) additive.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Now, Lebesgue originally introduced both inner and outer measure, which were (respectively) under and over estimates of a set's true &quot;size&quot;, but these fail to be countably additive. Instead of trying to find some new measure which satisfies all 4 properties, he restricted to a smaller collection of sets (as in Julián's answer) called &quot;measurable sets&quot; for which outer measure &lt;em&gt;does&lt;/em&gt; satisfy 1-4.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This was a smart move, since it turns out that there is no nontrivial function satisfying 1-4 for every subset of $\mathbb{R}$.&lt;/p&gt;&#xA;" OwnerUserId="7676" LastActivityDate="2011-03-13T09:07:06.367" />
  <row Id="26685" PostTypeId="1" AcceptedAnswerId="26688" CreationDate="2011-03-13T09:15:54.843" Score="2" ViewCount="93" Body="&lt;p&gt;Hello,  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hensel's Lemma&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Suppose that f(x) is a polynomial with integer coefficients k is an integers with $k \geq  2$, and p is prime.&#xA;  Suppose further that $r$ is a solution of the congruence $f(x) \equiv 0 \pmod{p^{k-1}}$. Then,&lt;br&gt;&#xA;  If $f&amp;#39;(r) \not\equiv 0 \pmod{p}$, then there is a unique integer t, $0 \leq t &amp;lt; p$, such that &#xA;  $$f(r + tp^{k-1}) \equiv 0 \pmod{p^k}$$ given by&#xA;  $$t \equiv \overline{-f&amp;#39;(r)}\frac{f(r)}{p^{k-1}} \pmod{p}$$ where $\overline{-f&amp;#39;(r)}$ is an inverse of f'(r) modulo p.&#xA;  If $f&amp;#39;(r) \equiv 0 \pmod{p}$ and $f(r) \equiv 0 \pmod{p^k}$, then $f(r+tp^{k-1}) \equiv 0 \pmod{p^k}$ $\forall$ integers t.&#xA;  If $f&amp;#39;(r) \equiv 0 \pmod{p}$ and $f(r) \not\equiv 0 \pmod{p^k}$, then $f(x) \equiv 0 \pmod{p^k}$ has no solutions with $x \equiv r \pmod{p^{k-1}}$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm practicing solving congruence equation using Hensel's Lemma, however, I was a little confused with the last two cases.&lt;br&gt;&#xA;To be clear, I use this congruence equation as an example $f(x) = x^4 + 4x^3 + 2x^2 + 2x + 12 \equiv 0 \pmod{625}$&lt;br&gt;&#xA;My attempt was:&lt;br&gt;&#xA;By inspecting all remainders of $5$, i.e $0, 1, 2, 3, 4$ \&#xA;We can see that&#xA;$ x \equiv 3 \pmod{5}$ is the solution to $f(x) \equiv 0 \pmod{5}$ \&#xA;Apply Hensel's Lemma for $5^2 = 25$, we have:&#xA;$$f&amp;#39;(x) = 4x^3 + 12x^2 + 4x + 2$$&#xA;And,&#xA;$$f&amp;#39;(3) = 4.3^3 + 12.3^2 + 4.3 + 2 = 230 \equiv 0 \pmod{5}$$&#xA;$$f(3) = 3^4 + 4.3^3 + 2.3^2 + 2.3 + 12 = 225 \equiv 0 \pmod{5^2}$$&#xA;Hence,&#xA;$$x \equiv 3 \pmod{5^2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Apply Hensel's Lemma for $5^3 = 125$, we have:&#xA;$$f(3) = 225 \not\equiv 0 \pmod{5^3}$$&#xA;So $f(x) \equiv 0 \pmod{5^3}$ has no solutions with $x \equiv 3 \pmod{5^2}$. \&#xA;Therefore, there are no solutions to $f(x) = x^4 + 4x^3 + 2x^2 + 2x + 12 \equiv 0 \pmod{625}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I understood about Hensel's Lemma is, it let us lift up the solution from $p^k$ to $p^{k + 1}$ each time we found a solution of a current $k$. But, in the case 2, when it said for all integers t, I was confused. Does it mean I can use the previous solution with $p^k$. By that I mean, if I have $x \equiv 3 \pmod{5}$, then if case 2 satisfies, then $x \equiv 3 \pmod{25}$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the question is vague, please let me know. I will try my best to rewrite it again. Sorry for my poor English writing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks,&lt;/p&gt;&#xA;" OwnerUserId="5711" LastActivityDate="2011-03-13T09:27:11.490" Title="Question regarding Hensel's Lemma" Tags="&lt;elementary-number-theory&gt;" AnswerCount="1" />
  <row Id="26686" PostTypeId="2" ParentId="26677" CreationDate="2011-03-13T09:20:21.670" Score="5" ViewCount="" Body="&lt;p&gt;The number of different symmetric &lt;em&gt;monomials&lt;/em&gt; of degree $n$ in $k$ variables with all the coefficients 1 is equal to the number $p(n,k)$ of partitions of $n$ in up to $k$ parts, for every such partition say $a_1+a_2+\dots + a_k = n$ with $a_1 \ge a_2 \ge \dots \ge a_k \ge 0$ we have the symmetric &quot;&lt;em&gt;monomial&lt;/em&gt;&quot;:&#xA;  $$ x_1^{a_1} x_2^{a_2} \dots x_k^{a_k} + \dots $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The number of different symmetric &lt;em&gt;homogeneous&lt;/em&gt; polynomials of degree $n$ in $k$ variables &#xA;with all the coefficients 1 is as a consequence  $ 2^{p(n,k)}-1 $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The number of different symmetric polynomials of degree (exactly) $n$ in $k$ variables &#xA;with all the coefficients 1 is then&#xA;  $$ (2^{p(n,k)}-1)2^{p(n-1,k)+p(n-2,k)+ \dots+ p(1,k)+p(0,k)} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if you write &#xA;  $$P(n,k) = p(n,k) + p(n-1,k)+ \dots +p(0,k) $$&#xA;the number of partitions of numbers up to $n$ in at most $k$ parts then the number you ask for (if I have understood the question and I haven't made a silly mistake is)&#xA;  $$ 2^{P(n,k)} -2^{P(n-1,k)} $$&#xA;EDIT: This was wrong. Sorry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT 2: The answer above is answering the question if we consider the &lt;em&gt;total degree&lt;/em&gt; of the symmetric functions, in view of the comments we are looking instead for polynomials with largest degree $n$ in any of the variables then we have the following: &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The number of different symmetric monomials with degree up to $n$ in all the variables is $n^k$.&lt;/li&gt;&#xA;&lt;li&gt;The total number of symmetric polynomials composed with this monomials is $2^{n^k}$. &lt;/li&gt;&#xA;&lt;li&gt;So the number of symmetric polynomials having at least one term of degree $n$ in one of the variables is&#xA;$$ 2^{n^k}-2^{(n-1)^k} $$&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="3274" LastEditorUserId="3274" LastEditorDisplayName="" LastEditDate="2011-03-13T19:22:57.710" LastActivityDate="2011-03-13T19:22:57.710" CommentCount="4" />
  <row Id="26687" PostTypeId="1" CreationDate="2011-03-13T09:20:41.797" Score="4" ViewCount="197" Body="&lt;p&gt;How do you solve this equation for A:&#xA;$~~\sin(5A) + \cos(5A)\sin(A) - \cos(3A) = 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've tried expanding it many times, but I can't seem to be able to reduce it to a format I can work with. Is there a simpler method of solution than repeated expansion?&lt;/p&gt;&#xA;" OwnerUserId="6520" LastEditorUserId="6520" LastEditorDisplayName="" LastEditDate="2011-03-13T09:23:39.000" LastActivityDate="2011-05-12T09:56:53.587" Title="Solve sin(5A) + cos(5A)sin(A) - cos(3A) = 0" Tags="&lt;algebra&gt;&lt;trigonometry&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" />
  <row Id="26688" PostTypeId="2" ParentId="26685" CreationDate="2011-03-13T09:27:11.490" Score="1" ViewCount="" Body="&lt;p&gt;You go off track at the word &quot;Hence&quot;. If $f&amp;#39;(3)\equiv 0\pmod 5$ and $f(3)\equiv 0 \pmod{25}$ (I assume you've done this correctly; I didn't check), that means that $x\equiv 3,8,13,18,23 \pmod{25}$ are &lt;em&gt;all&lt;/em&gt; solutions modulo 25. You only verified that there are no solutions modulo 125 which are 3 modulo 25. There may still be solutions which are 8, 13, 18, or 23 modulo 25.&lt;/p&gt;&#xA;" OwnerUserId="60" LastActivityDate="2011-03-13T09:27:11.490" CommentCount="2" />
  <row Id="26689" PostTypeId="2" ParentId="26687" CreationDate="2011-03-13T09:27:27.297" Score="1" ViewCount="" Body="&lt;p&gt;Try using $\cos{A} = \sin\frac{\pi}{2}-A$ and the $\sin{A} + \sin{B}$ or $\cos{A} + \cos{B}$ formulas. You can also write $2 \cos{A}\sin{B} = \sin(A+B) - \sin(A-B)$, which actually will reduce $\cos{5A}\cdot\sin(A)$.&lt;/p&gt;&#xA;" LastActivityDate="2011-03-13T09:27:27.297" />
  <row Id="26690" PostTypeId="2" ParentId="26682" CreationDate="2011-03-13T09:40:42.527" Score="1" ViewCount="" Body="&lt;p&gt;I don't know how simple you need it, and since you never said anything on how accurate you want your approximant to be (i.e., to how many correct decimal places should the approximant match the Lambert function?),&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$W_0(z)\approx\ln(1+z)\frac{1+\frac{123}{40}z+\frac{21}{10}z^2}{1+\frac{143}{40}z+\frac{713}{240}z^2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;should be good enough, which has a maximum error of around $1.6\times 10^{-4}$ for $z\in[0,e]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rational portion here is a Padé approximant; probably one might do better with a minimax rational approximation, but I don't have the patience and inclination to derive it since your question's rather vague to begin with.&lt;/p&gt;&#xA;" OwnerUserId="8185" LastActivityDate="2011-03-13T09:40:42.527" CommentCount="6" />
  <row Id="26691" PostTypeId="1" AcceptedAnswerId="26853" CreationDate="2011-03-13T09:42:46.017" Score="1" ViewCount="173" Body="&lt;p&gt;I was reading a paper, related to group theory. I came across two doubts (may be simple, but puzzling me): &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Let $G$ be a finite group and consider the lattice of subgroups of $G$. On this lattice, we define a Mobius function $\mu$ by $\sum_{H\geq K} {\mu(H)=\delta_{K,G}}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want two verifications: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;1)&lt;/strong&gt; If $G=C_n$ is cyclic group of order $n$, and $H=C_m$ is any subgroup of order $m$, then $\mu(C_m)=\mu(n/m)$, the Mobius function in number theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;2)&lt;/strong&gt; If $G=D_n$, the dihedral group of order $2n$, then for each $m$, $m|n$, consider the subgroups $C_m$ (cyclic group of order $m$), and $D_m$, the dihedral group of order $m$. The valules of $\mu(H)$ for these subgroups are given as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\mu(C_m)=-\frac{n}{m}\mu(\frac{n}{m})$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\mu(D_m)=\mu(n/m)$ (for both cases of $m$ -odd or even)&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;(Here the $\mu$ on R.H.S. denotes the Mobius function of number theory).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried to verify these, but confused too. Please help. (This question can not be posted on MO, so posted lastly here).&lt;/p&gt;&#xA;" OwnerUserId="8186" LastActivityDate="2011-03-14T20:45:26.010" Title="Mobius function" Tags="&lt;group-theory&gt;" AnswerCount="2" CommentCount="9" FavoriteCount="4" />
  <row Id="26692" PostTypeId="1" AcceptedAnswerId="26696" CreationDate="2011-03-13T09:44:01.573" Score="2" ViewCount="212" Body="&lt;p&gt;Can we say that an isomorphism $V \cong V^*$ is not natural? It seems so intuitively, but formally the notion of a natural transformation between a functor and a cofunctor is not defined (or is it?).&lt;/p&gt;&#xA;" OwnerUserId="3793" LastActivityDate="2011-03-13T14:30:59.267" Title="Can we say that $V \cong V^*$ is not natural?" Tags="&lt;category-theory&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="26694" PostTypeId="2" ParentId="26680" CreationDate="2011-03-13T09:47:14.207" Score="1" ViewCount="" Body="&lt;p&gt;(This should be either a comment or CW.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Peter Luschny &lt;a href=&quot;http://www.luschny.de/math/factorial/hadamard/HadamardsGammaFunctionMJ.html&quot; rel=&quot;nofollow&quot;&gt;studied&lt;/a&gt; a number of gamma-like functions that do not have the log-convex imposition; you might want to look into them for inspiration.&lt;/p&gt;&#xA;" OwnerUserId="8185" LastActivityDate="2011-03-13T09:47:14.207" CommentCount="1" />
  <row Id="26695" PostTypeId="2" ParentId="26626" CreationDate="2011-03-13T10:16:41.010" Score="17" ViewCount="" Body="&lt;p&gt;Here's another, simpler proof:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$z^5=(x^5-10x^3y^2+5xy^4) + \mathrm{i}(y^5-10y^3x^2+5yx^4)&#xD;&#xA;\stackrel{\mathrm{A}}{=}&#xD;&#xA;x^5+\mathrm{i}y^5&#xD;&#xA;\stackrel{\mathrm{B}}{=}&#xD;&#xA;x+\mathrm{i}y=z\mod 10\;.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A is because either $x$ or $y$ must be even, else $z^2$ has only even parts. B is because as a ring $\mathbb{Z}/10\mathbb{Z}$ is isomorphic to $(\mathbb{Z}/5\mathbb{Z})\oplus(\mathbb{Z}/2\mathbb{Z})$ and taking the fifth power is the identity in each component.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus the residues modulo $10$ have period $4$, so there are at most $8$ different ones and we can't produce numbers with the remaining two.&lt;/p&gt;&#xA;" OwnerUserId="6622" LastActivityDate="2011-03-13T10:16:41.010" CommentCount="2" />
  <row Id="26696" PostTypeId="2" ParentId="26692" CreationDate="2011-03-13T10:25:59.040" Score="9" ViewCount="" Body="&lt;p&gt;Indeed, a natural transformation is defined for two parallel functors, that is, functors with the same domain and codomain. The identity functor I and the single dual functor * are not parallel:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$I:\sf{Vect_K}\to \sf{Vect_K} $&#xA;$*:\sf{Vect_K}^{op}\to \sf{Vect_K} $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words: I is covariant and * is contravariant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But let's ignore that. Suppose we have a collection of linear isomorphisms &#xA;$\{a_V:V\to V^*\}_{V\in\sf{Vect_K}}.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The naturality square would read&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$&#xD;&#xA;\begin{array}{ccccc}&#xD;&#xA;&amp;amp; V &amp;amp; \longrightarrow^{a_V} &amp;amp; V^* &amp;amp; \\&#xD;&#xA;L &amp;amp; \downarrow &amp;amp; &amp;amp; \uparrow &amp;amp; L^*\\&#xD;&#xA;&amp;amp; W &amp;amp; \longrightarrow^{a_W} &amp;amp; W^* &amp;amp;&#xD;&#xA;&#xD;&#xA;\end{array}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where L* is the usual dual (precomposition) linear map defined by &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$L^*f := [v\mapsto f(Lv)]=f\circ L,\ \ \ \ \ f\in W^*$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now this diagram commutes iff &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\alpha_V = L^*\alpha_W L$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for &lt;em&gt;all&lt;/em&gt; linear maps $L:V\to W$. But this cannot be since $\alpha_V$ is an isomorphism, and L not necessarily, e.g. take L=0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[PS: I copied this answer from my answer &lt;a href=&quot;http://www.physicsforums.com/showpost.php?p=2846497&amp;amp;postcount=9&quot; rel=&quot;nofollow&quot;&gt;at physicsforums&lt;/a&gt; ]&lt;/p&gt;&#xA;" OwnerUserId="6490" LastActivityDate="2011-03-13T10:25:59.040" />
  <row Id="26697" PostTypeId="1" AcceptedAnswerId="26714" CreationDate="2011-03-13T10:26:25.747" Score="5" ViewCount="206" Body="&lt;p&gt;Here is an interesting limit of an integral I do not know how to begin. Any help is greatly appreciated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\lim_{n\to \infty}n\int_{0}^{\frac{\pi}{2}}(1-\sqrt[n]{sin(x)})dx$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know it converges to $\frac{\pi ln(2)}{2}$, but how?.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks much&lt;/p&gt;&#xA;" OwnerUserId="6358" LastActivityDate="2011-03-13T16:33:59.460" Title="limit of integral involving root of sin(x)" Tags="&lt;integral&gt;" AnswerCount="4" />
  <row Id="26698" PostTypeId="2" ParentId="26697" CreationDate="2011-03-13T10:32:39.550" Score="2" ViewCount="" Body="&lt;p&gt;Suggestion: substitute $u=\sin x$, and consider the &lt;a href=&quot;http://en.wikipedia.org/wiki/Beta_function&quot; rel=&quot;nofollow&quot;&gt;Beta function&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: A better suggestion. To arrive to&#xA;$$&#xD;&#xA; - \int_0^{\pi /2} {\ln (\sin x)dx} \, \bigg( = \frac{{\pi \ln 2}}{2}\bigg),&#xD;&#xA;$$&#xA;consider&#xA;$$&#xD;&#xA;\mathop {\lim }\limits_{r \to \infty } \frac{{\int_0^{\pi /2} {[1 - (\sin x)^{1/r} ]dx} }}{{1/r}} = \mathop {\lim }\limits_{r \to \infty } \frac{{\int_0^{\pi /2} {[1 - e^{\ln (\sin x)/r} ]dx} }}{{1/r}},&#xD;&#xA;$$&#xA;followed by L'Hospital's rule.&lt;/p&gt;&#xA;" OwnerUserId="2810" LastEditorUserId="2810" LastEditorDisplayName="" LastEditDate="2011-03-13T16:33:59.460" LastActivityDate="2011-03-13T16:33:59.460" CommentCount="2" />
  <row Id="26699" PostTypeId="2" ParentId="26697" CreationDate="2011-03-13T11:10:54.087" Score="0" ViewCount="" Body="&lt;p&gt;$$n\int_0^{\frac{\pi}{2}}(1-\sqrt[n]{\sin(x)})\mathrm dx=\pi\left(\frac{n}{2}-n^2 2^{1-\frac1{n}}\frac{\Gamma\left(\frac1{n}\right)}{\Gamma\left(\frac1{2n}\right)^2}\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\approx\pi\left(\ln\sqrt{2}-\left(\left(\ln\sqrt{2}\right)^2+\frac{\pi^2}{48}\right)\frac1{n}+O\left(\frac1{n^2}\right)\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, take the limit...&lt;/p&gt;&#xA;" OwnerUserId="8188" LastActivityDate="2011-03-13T11:10:54.087" CommentCount="3" />
  <row Id="26700" PostTypeId="2" ParentId="26654" CreationDate="2011-03-13T11:59:18.227" Score="6" ViewCount="" Body="&lt;p&gt;The first thing to do when you get a problem like this is to try what Rogers calls the &quot;Tarski-Kuratowsi algorithm&quot; to estimate the complexity. Your problem &#xA;$$T = \{ \langle M\rangle \mid \mbox{if $M$ accepts $w$ then $M$ accepts $w^r$}\}.$$&#xA;can be written equivalently as&#xA;$$T = \{ \langle M\rangle \mid (\forall w)( (\exists t)(\mbox{$M$ accepts $w$ in $t$ steps}) \Rightarrow (\exists s)(\mbox{$M$ accepts $w^r$ in $s$ steps}))\}.$$&#xA;Because &quot;$M$ accepts $w$ in $t$ steps&quot; can be written with only bounded quantifiers, the quantifier pattern for the formula defining $T$ is $(\forall)((\exists)\to(\exists))$, so the prenex normal form has quantifier pattern $(\forall)(\exists)$. Hence the problem is definable by a $\Pi^0_2$ formula in the arithmetical hierarchy. This means that &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The problem is computable from $0&amp;#39;&amp;#39;$, the jump of the halting problem.  In fact it is many-one reducible to any $\Pi^0_2$ hard problem, such as the complement of the set $0&amp;#39;&amp;#39;$. &lt;/li&gt;&#xA;&lt;li&gt;Heuristically, the problem should compute $0&amp;#39;&amp;#39;$.  There is a phenomenon that most problems are at exactly the level of the arithmetical hierarchy that the Tarski-Kuratowski algorithm assigns.  This heuristic suggests you should be able to many-one reduce some $\Pi^0_2$ complete problem to your problem.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;One example of a $\Pi^0_2$ complete problem is &#xA;$$\{e : \mbox{$e$ halts on every input}\}$$&#xA;In this case, there &lt;strong&gt;is&lt;/strong&gt; a many-one reduction of that problem to yours, which I will let you work out, since it's a nice exercise in computability. The answer of Yuval Filmus gives a hint. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In summary: the precise answer in terms of the arithmetical hierarchy is that your problem is $\Pi^0_2$ complete. &lt;/p&gt;&#xA;" OwnerUserId="630" LastEditorUserId="630" LastEditorDisplayName="" LastEditDate="2011-03-14T01:08:40.827" LastActivityDate="2011-03-14T01:08:40.827" CommentCount="3" />
  <row Id="26701" PostTypeId="2" ParentId="26650" CreationDate="2011-03-13T12:47:12.647" Score="6" ViewCount="" Body="&lt;p&gt;A bit more detail for the answers by Jason and Ryan:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have some action of $\mathbb{Z}^2$ on the plane.  Conjugate that action by $\phi$ to find an action of $\mathbb{Z}^2$ on the upper-half plane.  The question now is: how can $\mathbb{Z}^2$ act on the upper-half plane via isometries? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $\alpha$ and $\beta$ be the two generators of $\mathbb{Z}^2$ and suppose that neither acts via the identity.  The classification of isometries tells us that $\alpha$ and $\beta$ are either elliptic, parabolic, or hyperbolic.  Since they commute with each other an analysis of fixed points shows that they must have the same type.  Thus their action on the plane can be reduced to an isometric action on a circle (elliptic case) or on a line.  After some $1$-dimensional work we find that the reduced, and hence the original, action is either unfaithful or not discrete.&lt;/p&gt;&#xA;" OwnerUserId="1307" LastActivityDate="2011-03-13T12:47:12.647" />
  <row Id="26703" PostTypeId="2" ParentId="20996" CreationDate="2011-03-13T13:00:49.513" Score="0" ViewCount="" Body="&lt;p&gt;Dear All&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I only found the following more technical explanation what is &#xA;going on in Gödels proof (from (*) p.160):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;(b) Gödel's First Incompleteness&#xA;  Theorem and Related Topics Fixed&#xA;  points theorems are particularly&#xA;  fruitful when applied  to formulas&#xA;  related to the set of all T-proofs or&#xA;  the set of  all T-provable formulas&#xA;  (where T is a theory containing Q). &#xA;  If T is rich enough to make&#xA;  arithmetization possible (IΣ_1 subset &#xA;  T suffices) then it is natural to work&#xA;  with the formalized  proof predicate&#xA;  Proof* (see Chap. I, Sect. 4); but for&#xA;  Gödel's  first incompleteness theorem&#xA;  it suffices that T contain Q and  that&#xA;  the set of all proofs is Δ_1 and hence&#xA;  has a Σ_1 binumeration.  (Gödel's&#xA;  second incompleteness theorem is more&#xA;  delicate and  to prove it we have to&#xA;  guarantee some provability conditions,&#xA;  see below.) The theorems are proved in&#xA;  a constructive way, i.e.  we give&#xA;  examples of sentences that are&#xA;  independent (neither  provable nor&#xA;  refutable). Since these examples are&#xA;  both  famous and useful we give their&#xA;  names (Gödel's formula,  Rosser's&#xA;  formula).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So there is a difference between what means are knowingly needed &#xA;for the first and second incompleteness theorem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lets only look at the first incompletness theorem. I guess&#xA;working with the weak requirement T contains Q only works when&#xA;we switch to the meta-meta level. So lets not do this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now is the result finitistically obtainable? Which means can&#xA;we proof it in PRA? I guess no, since PRA allows only induction&#xA;over quantifier free formulas, but we need existential quantified&#xA;bounded formulas (IΣ_1).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bye&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(*)&#xA;Metamathematics of First-Order Arithmetic&#xA;Petr Hájek, and Pavel Pudlák&#xA;Perspectives in Mathematical Logic, Volume 3&#xA;Berlin: Springer-Verlag, 1998. 2nd printing&#xA;&lt;a href=&quot;http://projecteuclid.org/DPubS?service=UI&amp;amp;version=1.0&amp;amp;verb=Display&amp;amp;handle=euclid.pl/1235421926&quot; rel=&quot;nofollow&quot;&gt;http://projecteuclid.org/DPubS?service=UI&amp;amp;version=1.0&amp;amp;verb=Display&amp;amp;handle=euclid.pl/1235421926&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4414" LastActivityDate="2011-03-13T13:00:49.513" />
  <row Id="26704" PostTypeId="2" ParentId="26692" CreationDate="2011-03-13T13:21:17.063" Score="14" ViewCount="" Body="&lt;p&gt;This is more of a long comment on wildildildlife's answer. A natural question after reading that proof is: what happens if we restrict to the &lt;em&gt;groupoid&lt;/em&gt; of finite dimensional vector spaces and their isomorphisms? (It may seem that picking $L=0$ is a bit of a &quot;cheat&quot;.) The fact is that there is still no equivalence between the identity and the dual functor. The point is that the collection of isomorphisms $\{\alpha_V \colon V \to V^\ast \}$ is &lt;em&gt;only&lt;/em&gt; allowed to depend on $V$, i.e. there must be a specific morphism for every $V$ that makes every diagram you can cook up using your category commutative. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular we can look at diagrams of the form&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$&#xD;&#xA;\begin{array}{ccc}&#xD;&#xA; V &amp;amp; \longrightarrow^{\alpha_V} &amp;amp; V^*  \\&#xD;&#xA; \downarrow &amp;amp; &amp;amp; \uparrow  \\&#xD;&#xA; V &amp;amp; \longrightarrow^{\alpha_V} &amp;amp; V^*. &#xD;&#xA;\end{array}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we get $\alpha_V = A^\ast \alpha_V A$ for all $A \in \mathrm{GL}(V)$. Equivalently, if we use $\alpha_V$ to identify two bases of $V$ and $V^\ast$, we must have $\mathrm{id}_V = A^T A$ for all $A \in \mathrm{GL}(V)$, which is absurd.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This point of view also leads directly to what condition we need to impose on the category for the existence of an equivalence like this: we need to find a category of vector spaces where the automorphisms $A \colon V \to V$ are the orthogonal matrices (wrt a basis). The most natural thing to consider is then finite dimensional spaces equipped with an inner product&lt;sup&gt;1&lt;/sup&gt; and morphisms preserving the inner product. Hence we have rediscovered the principle that you can identify a vector space with its dual precisely when it has an inner product.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;1&lt;/sup&gt; Remark: By an inner product I really mean an arbitrary nondegenerate bilinear form, so over $\mathbf R$ or $\mathbf C$ this is non-standard usage. Note that this category is automatically a groupoid, since any map of vector spaces preserving the inner product is an isomorphism.&lt;/p&gt;&#xA;" OwnerUserId="677" LastEditorUserId="677" LastEditorDisplayName="" LastEditDate="2011-03-13T14:30:59.267" LastActivityDate="2011-03-13T14:30:59.267" CommentCount="7" />
  <row Id="26705" PostTypeId="1" CreationDate="2011-03-13T13:22:54.940" Score="1" ViewCount="221" Body="&lt;p&gt;Follow up to &lt;a href=&quot;http://math.stackexchange.com/questions/15556/is-zero-odd-or-even&quot;&gt;this&lt;/a&gt; question. Is &lt;code&gt;0&lt;/code&gt; a positive number?&lt;/p&gt;&#xA;" OwnerUserId="8190" LastActivityDate="2011-03-13T20:40:46.643" Title="Is zero positive or negative?" Tags="&lt;number-theory&gt;" AnswerCount="4" CommentCount="1" />
  <row Id="26706" PostTypeId="2" ParentId="26705" CreationDate="2011-03-13T13:26:48.977" Score="3" ViewCount="" Body="&lt;p&gt;$0$ is &lt;a href=&quot;http://en.wikipedia.org/wiki/0_%28number%29&quot; rel=&quot;nofollow&quot;&gt;neither positive nor negative&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="149" LastActivityDate="2011-03-13T13:26:48.977" />
  <row Id="26707" PostTypeId="1" AcceptedAnswerId="26710" CreationDate="2011-03-13T13:50:28.923" Score="2" ViewCount="78" Body="&lt;p&gt;$$A:= \{ x^2 | x \in (2,3\sqrt{2}] , x \in \mathbb{R} \textrm{ and } x^2 \in \mathbb{Z}\}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$B:= \{ x^3 | x \in (-2\sqrt[3]{3},3] , x \in \mathbb{R} \textrm{ and } x^3 \in \mathbb{Z}\}$$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;s(A) + s(B) = ?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;s(A) is number of elements of A set. &#xA;For example if $\ A = \{a,b,c\} $&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;s(A) = 3&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="6097" LastEditorUserId="677" LastEditorDisplayName="" LastEditDate="2011-03-13T14:24:43.703" LastActivityDate="2011-03-13T14:24:43.703" Title="s(A) + s(B) = ? Set Theory" Tags="&lt;elementary-set-theory&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="26708" PostTypeId="2" ParentId="26705" CreationDate="2011-03-13T13:53:00.500" Score="11" ViewCount="" Body="&lt;p&gt;It really depends on context. In common use in English language, zero is unsigned, that is, it is neither positive nor negative. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In typical French mathematical usage, zero is &lt;em&gt;both&lt;/em&gt; positive and negative. Or rather, in mathematical French &quot;$x$ est positif&quot; (literally &quot;$x$ is positive&quot;) allows the case $x = 0$, while &quot;$x$ est positif strictement&quot; (literally &quot;$x$ is strictly positive&quot;) does not. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sometimes for computational purposes, it may be necessary to consider &lt;a href=&quot;http://en.wikipedia.org/wiki/Signed_zero&quot; rel=&quot;nofollow&quot;&gt;signed zeros&lt;/a&gt;, that is, treating $+0$ and $-0$ as two different numbers. One may think of this a capturing the different divergent behaviour of $1/x$ as $x\to 0$ from the left and from the right. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are interested in mathematical analysis, and especially &lt;a href=&quot;http://en.wikipedia.org/wiki/Semi-continuity&quot; rel=&quot;nofollow&quot;&gt;semi-continuous functions&lt;/a&gt;, then it sometimes makes more sense to consider intervals that are closed on one end and open on the other. Then depending on which situation are in it may be more natural to group 0 with the positive or negative numbers. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are certainly much more subtleties, but unless you clarify why exactly you are asking and in what context you are thinking about this, it is impossible to give an answer most suited to your applications. &lt;/p&gt;&#xA;" OwnerUserId="1543" LastActivityDate="2011-03-13T13:53:00.500" CommentCount="1" />
  <row Id="26709" PostTypeId="2" ParentId="26668" CreationDate="2011-03-13T14:05:32.813" Score="0" ViewCount="" Body="&lt;p&gt;If you were working with real numbers, then if you already found a $k$ such that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{k}{x+m} = \frac{1}{x_1+m}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;then just making each side an exponent on $g$, we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$g^{\frac{k}{x+m}} = g^{\frac{1}{x_1+m}}$$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could therefore use the same $k$ that Arturo gave in the previous question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, since you reference the discrete log problem, it is clear you are not working with rational numbers.  You should state clearly the arithmetic system within which you are working.  For simplicity, I will assume that you are working modulo some prime number $p$ and that $g$ is a primitive root modulo $p$.  When working with exponents on $g$, the exponents &quot;behave modulo p-1&quot;, i.e., you can replace an exponent with an integer equal to it modulo &quot;p-1&quot; without changing the value of the exponential expression.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order for your fractional exponents to make sense, if $g$ is indeed a primitive root, then you need $x+m$ and $x_1+m$ to be relatively prime to $p-1$. Since you don't know $m$, I don't see how it would be possible to know this.  But if so, then Arturo's expression still works, but you must interpret it as an integer modulo p-1. Recall that Arturo's answer was&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$k = \frac{1}{1-\frac{1}{x+m}(x-x_1)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can invert $x+m$ modulo $p-1$ (say using the Extended Euclidean Algorithm) since $x+m$ was assumed relatively prime to $p-1$.  Take the result, multiply by $x-x_1$, and subtract from $1$.  This result will still be relatively prime to $p-1$ since we assumed that $x_1+m$ was, so you can invert it modulo p-1 resulting in the $k$ that you desired.&lt;/p&gt;&#xA;" OwnerUserId="4253" LastActivityDate="2011-03-13T14:05:32.813" CommentCount="1" />
  <row Id="26710" PostTypeId="2" ParentId="26707" CreationDate="2011-03-13T14:18:58.560" Score="4" ViewCount="" Body="&lt;p&gt;Given $x \in (2, 3\sqrt{2}]$, what can $x^2$ be? It must be&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ 2^2 = 4 &amp;lt; x^2 \leq 18 = (3\sqrt{2})^2 $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So $A$ comprises of all integers ($x^2 \in \mathbb{Z}$) bigger than 4 and less-than-or-equal-to 18. There are 14 of them. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Given $x \in (-2\sqrt[3]{3}, 3]$, you similarly find that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ -24 &amp;lt; x^3 \leq 27 $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which tells you that there are $27 - (-24) = 51$ elements in $B$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Add them together you get $s(A) + s(B) = 65$. &lt;/p&gt;&#xA;" OwnerUserId="1543" LastActivityDate="2011-03-13T14:18:58.560" CommentCount="7" />
  <row Id="26711" PostTypeId="2" ParentId="26662" CreationDate="2011-03-13T14:38:25.357" Score="4" ViewCount="" Body="&lt;p&gt;The clearest interpretation I've found of the eigenvalues and eigenvectors of an operator in physics is in the &lt;a href=&quot;http://en.wikipedia.org/wiki/Wave_equation&quot; rel=&quot;nofollow&quot;&gt;wave equation&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\partial^2 u}{\partial t^2} = \Delta u$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\Delta$ denotes the Laplacian. The eigenvectors of $\Delta$ with negative eigenvalues satisfy $\Delta v = -\lambda^2 v$, and from any such eigenvector one can construct a solution&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$u = A \cos (\lambda t + B) v$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;of the wave equation. The solutions one gets in this way are precisely the  &lt;a href=&quot;http://en.wikipedia.org/wiki/Standing_wave&quot; rel=&quot;nofollow&quot;&gt;standing waves&lt;/a&gt; of frequency $\frac{\lambda}{2\pi}$, and under nice conditions every solution is a linear combination of standing waves. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In common physical situations, the Hamiltonian one uses in the &lt;a href=&quot;http://en.wikipedia.org/wiki/Schr%C3%B6dinger_equation&quot; rel=&quot;nofollow&quot;&gt;Schrödinger equation&lt;/a&gt; in quantum mechanics is closely related to the Laplacian, which is one reason solutions to the Schrödinger equation are referred to as wave functions. &lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2011-03-13T14:38:25.357" CommentCount="1" />
  <row Id="26712" PostTypeId="2" ParentId="26618" CreationDate="2011-03-13T14:38:44.513" Score="4" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://annals.math.princeton.edu/2010/171-2/p07&quot; rel=&quot;nofollow&quot;&gt;This paper&lt;/a&gt; (free access) answers your question theoretically, but the result doesn't look useful in practice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you Google for &quot;parametrization of the modular group&quot;, you get &lt;a href=&quot;http://www.deepdyve.com/lp/american-physical-society-aps/modular-invariance-of-the-thermo-partition-function-and-global-phase-60J0VINMq3&quot; rel=&quot;nofollow&quot;&gt;this paper&lt;/a&gt;, which you can rent for a day for $0.99; the Google context says &quot;Therefore, we have a complete parametrization of the modular group&quot;. (I didn't rent it to look inside.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://adsabs.harvard.edu/abs/2003AcAri.110..331Z&quot; rel=&quot;nofollow&quot;&gt;Here&lt;/a&gt;'s another paper that appears to address the question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.google.de/url?sa=t&amp;amp;source=web&amp;amp;cd=1&amp;amp;ved=0CBwQFjAA&amp;amp;url=https://www.uni-hohenheim.de/~gzim/Publications/habil.pdf&amp;amp;rct=j&amp;amp;q=%22Coherent%20States%20from%20Nonunitary%20Representations%22&amp;amp;ei=tdN8TZq4JoPzsgaomtzYAQ&amp;amp;usg=AFQjCNEY7SuSK-PT1a2qR9FP8g-0zTntww&amp;amp;sig2=AiwGoEY8WThU65HFecdUSQ&amp;amp;cad=rja&quot; rel=&quot;nofollow&quot;&gt;This dissertation&lt;/a&gt; says &quot;there is no polynomial parametrization in three integer variables of $SL_2(\mathbb{Z})$&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since you mention the parametrization of the Pythagorean triples: &lt;a href=&quot;http://books.google.de/books?id=08hPoGgSQFIC&amp;amp;lpg=PA352&amp;amp;ots=ugtsbYzwyu&amp;amp;dq=parametrization%20sl2&amp;amp;pg=PA351#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;this parametrization of $SL_2(\mathbb{R})$&lt;/a&gt; uses matrices of trigonometric and hyperbolic functions that can be restricted to rational values using Pythagorean triples:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left(\begin{array}{cc}\cosh\theta&amp;amp;\sinh\theta\\\sinh\theta&amp;amp;\cosh\theta\end{array}\right)=&#xD;&#xA;\left(\begin{array}{cc}\frac{c}{b} &amp;amp; \pm \frac{a}{b}\\\pm \frac{a}{b} &amp;amp; \frac{c}{b}\end{array}\right)\;,&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left(\begin{array}{cc}\cos\theta&amp;amp;-\sin\theta\\\sin\theta&amp;amp;\cos\theta\end{array}\right)=&#xD;&#xA;\left(\begin{array}{cc}\pm\frac{a}{c} &amp;amp; \mp \frac{b}{c}\\\pm \frac{b}{c} &amp;amp; \pm\frac{a}{c}\end{array}\right)\;.&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conversely, $\cos\theta=a/r$ and $\sin\theta=b/s$ in reduced form yields&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left(\frac{a}{r}\right)^2+\left(\frac{b}{s}\right)^2=1\;,$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is only possible if $r=s=c$ and thus $a$, $b$ and $c$ form a Pythagorean triple; and similarly for the hyperbolic case. This yields a parametrization of $SL_2(\mathbb{Q})$, but I'm afraid it doesn't help you much with $SL_2(\mathbb{Z})$, since I don't see how to get rid of the denominator or how to deal with the diagonal matrices $d_1$ and $d_2$ in the integer case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S.: A more pedestrian way of enumerating the matrices in $SL_2(\mathbb{Z})$ would be to enumerate the pairs of coprime integers $a$, $b$, solve $ad-bc=1$ for $c$ and $d$, and use all tuples of the form $(a,b,c+na,d+nb)$.&lt;/p&gt;&#xA;" OwnerUserId="6622" LastEditorUserId="6622" LastEditorDisplayName="" LastEditDate="2011-03-13T19:12:44.123" LastActivityDate="2011-03-13T19:12:44.123" CommentCount="3" />
  <row Id="26713" PostTypeId="2" ParentId="26618" CreationDate="2011-03-13T14:47:50.123" Score="6" ViewCount="" Body="&lt;p&gt;In the comments Theo refers to the classical result that the matrices $S = \left[ \begin{array}{cc} 0 &amp;amp; -1 \\\ 1 &amp;amp; 0 \end{array} \right], T = \left[ \begin{array}{cc} 1 &amp;amp; 1 \\\ 0 &amp;amp; 1 \end{array} \right]$ generate $\text{SL}_2(\mathbb{Z})$, and in fact we know exactly what the relations are: they are $S^2 = (ST)^3 = -1$ (where $-1$ is a central element of order $2$). This is essentially an application of the Euclidean algorithm, although generally the first place you will read about it is a book on modular forms, where it is more interesting to give a geometric proof in terms of the action on the upper half plane. See, for example, the last section of Serre's &lt;em&gt;A Course in Arithmetic.&lt;/em&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whether this counts as a parameterization of the modular group is up to you. Until you are more precise I don't know if this is the kind of answer you're looking for.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In practice, if all you want to do is generate a lot of elements of $\text{SL}_2(\mathbb{Z})$, it seems easier to me to pick two random integers and apply the &lt;a href=&quot;http://en.wikipedia.org/wiki/Extended_Euclidean_algorithm&quot; rel=&quot;nofollow&quot;&gt;extended Euclidean algorithm&lt;/a&gt; to them. Of course, the above result also gives an explicit algorithm that will generate all elements of $\text{SL}_2(\mathbb{Z})$ since it is easy to recursively generate words in two letters. &lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2011-03-13T14:56:24.640" LastActivityDate="2011-03-13T14:56:24.640" />
  <row Id="26714" PostTypeId="2" ParentId="26697" CreationDate="2011-03-13T15:11:17.333" Score="2" ViewCount="" Body="&lt;p&gt;You can integrate by parts in your integral, differentiating $1 - (\sin(x))^{1 \over n}$ and integrating 1. You get&#xA;$$n\int_0^{\pi \over 2}(1 -  \sin(x)^{1 \over n})\,dx = \int_0^{\pi \over 2}x \sin(x)^{{1 \over n} - 1}\cos(x)\,dx$$&#xA;(Both of the boundary terms evaluate to zero). In the right-hand integral above, the integrand increases monotonically to $x(\sin(x))^{-1}\cos(x) = x\cot(x)$ as &#xA;$n$ goes to infinity. So your limit is exactly&#xA;$$\int_0^{\pi \over 2}x\cot(x)\,dx$$&#xA;This is a pretty famous integral in itself, but you can integrate by parts again in the reverse direction (making sure the endpoints go to zero in the improper integral) to convert it into the integral&#xA;$$-\int_0^{\pi \over 2}\ln(\sin(x))\,dx$$&#xA;As you indicated this is well known and evaluates to ${\displaystyle{\pi \ln2 \over 2}}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Effectively what is going on here is that you are using that the limit as $\epsilon \rightarrow 0$ of ${\displaystyle{1 - u^{\epsilon} \over \epsilon}}$ is negative the derivative of $u^x$ at $x = 0$, or $-\ln(u)$. Letting $\epsilon = {1 \over n}$ and $u = \sin(x)$ gives that your integrand (including the $n$ factor) converges to $-\ln(\sin(x))$. So your answer should be the integral of that. But making this rigorous seems a little tricky (to me) so I integrated by parts and then took limits, then integrated by parts back again.&lt;/p&gt;&#xA;" OwnerUserId="3035" LastActivityDate="2011-03-13T15:11:17.333" CommentCount="1" />
  <row Id="26715" PostTypeId="2" ParentId="26683" CreationDate="2011-03-13T15:13:37.760" Score="1" ViewCount="" Body="&lt;p&gt;Do the finite case first (by restricting to a compact): $\nu$ is a finite Borel measure on a metric space $X$.&#xA;Look at the set $\mathcal{C}$ of $E \in \mathcal{B}$ such that $\nu(E) = \inf_{E \subset U\ \mathrm{open}} \nu(U) = \sup_{K \subset E\ \mathrm{closed}} \nu(K)$.&#xA;$\mathcal{C}$ is obviously non-empty and closed under complementation.&#xA;Since any open set can be written as the countable union of closed sets (use the metric here), open sets are in $\mathcal{C}$.&#xA;Finally, $\mathcal{C}$ is closed under countable union.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Going from the finite case to the &quot;regular&quot; case is achieved by writing $\mathbb{R}^n$ as the union of the closed balls of radii $n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Let us prove that $\mathcal{C}$ is closed under countable union. Let $E_n$ ($n \geq 1$) be in $\mathcal{C}$, and $E = \bigcup_n E_n$. Let $\epsilon &amp;gt; 0$. For every $n \geq 1$, there is an open set $U_n$ containing $E_n$ s.t. $\nu(U_n \setminus E_n)&amp;lt;\epsilon 2^{-n}$. Let $U = \bigcup_n U_n$. Then $U \setminus E \subset \bigcup_n \left(U_n \setminus E_n \right)$ so $\nu(U \setminus E) &amp;lt; \sum_{n \geq 1} \epsilon 2^{-n} = \epsilon$.&#xA;The sup equality is proved in a very similar way, except that you have to choose $m$ such that $\nu\left(E \setminus \bigcup_{n \leq m} \right) &amp;lt; \epsilon$ first (you can only take finite unions of closed sets to get closed sets).&lt;/p&gt;&#xA;" OwnerUserId="2660" LastEditorUserId="2660" LastEditorDisplayName="" LastEditDate="2011-03-13T18:57:29.207" LastActivityDate="2011-03-13T18:57:29.207" CommentCount="18" />
  <row Id="26716" PostTypeId="2" ParentId="26662" CreationDate="2011-03-13T15:17:44.017" Score="3" ViewCount="" Body="&lt;p&gt;Mathematically an eigenvector of a linear transformation is just a vector whose direction is fixed under the transformation, while its eigenvalue is the amount its magnitude changes by. To understand the physical meaning of it requires first understanding the physical meaning of a linear transformation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Akhil already gave a very good answer detailing two common examples. Let me here give a slightly more esoteric one. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider electromagnetism as a relativistic theory (Maxwell's theory is inherently compatible with special relativity). We can unite the electric field and magnetic field, which are each represented as spatial vectors (3 components), into a space-time quantity, a 4-by-4 matrix known as the &lt;a href=&quot;http://en.wikipedia.org/wiki/Electromagnetic_tensor&quot; rel=&quot;nofollow&quot;&gt;Faraday tensor&lt;/a&gt; denoted by $F_{ab}$. It is an antisymmetric matrix $F_{ab} = - F_{ba}$, and transforms covariantly under coordinate transformations of Minkowski space. (In other words, geometrically it is a two-form.) Expressed in terms of the components of the electric and magnetic fields, you have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ F = \begin{pmatrix}&#xD;&#xA;0 &amp;amp; -E_x &amp;amp; -E_y &amp;amp; - E_z \\&#xD;&#xA;E_x &amp;amp; 0 &amp;amp; -B_z &amp;amp; B_y \\&#xD;&#xA;E_y &amp;amp; B_z &amp;amp; 0 &amp;amp; -B_x \\&#xD;&#xA;E_z &amp;amp; -B_y &amp;amp; B_x &amp;amp; 0 \end{pmatrix} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Written as such we cannot consider the eigenvectors of $F_{ab}$. Geometrically it sends vectors into co-vectors, and eigenvectors are only meaningfully defined if the domain and codomain represent the same vector space. However, as we are working geometrically, we have a canonical map of co-vectors into vectors using the &quot;raising operation&quot; associated to the Minkowski metric. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, you multiply $F_{ab}$ on the right by the matrix $\eta^{ca}$ which has $(-1,1,1,1)$ along its diagonal, and 0 elsewhere. If you don't follow the whole shpiel above, don't worry too much. We can summarise it in the following: associated to an electromagnetic field on space-time, there is a natural linear transformation $\tilde{F}^a{}_b$ which can be expressed as $\sum_c \eta^{ac}F_{cb}$ and as a matrix&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \tilde{F} = \begin{pmatrix}&#xD;&#xA;0 &amp;amp; E_x &amp;amp; E_y &amp;amp;  E_z \\&#xD;&#xA;E_x &amp;amp; 0 &amp;amp; -B_z &amp;amp; B_y \\&#xD;&#xA;E_y &amp;amp; B_z &amp;amp; 0 &amp;amp; -B_x \\&#xD;&#xA;E_z &amp;amp; -B_y &amp;amp; B_x &amp;amp; 0 \end{pmatrix} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let us consider its eigenvectors! Suppose $v$ is a 4-vector such that $\tilde{F}v = \lambda v$. Now we take the Minkowski inner product against $v$ itself. You have that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \lambda \eta(v,v) = \eta( \tilde{F} v,v) = \eta( \eta\circ F v,v) = F(v,v) = 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The third equality uses the fact that $\eta = \eta^{-1}$, and the last inequality uses that $F$ is antisymmetric. (Notice that the above is also true with respect to the Euclidean inner product if $\tilde{F}$ were defined that way. And in particular it says that relative to a Euclidean inner product there are no real eigenvectors with non-zero eigenvalue...) So we conclude that $v$ either lies in the null-space of $\tilde{F}$ (from $\lambda = 0$; and hence in the null-space of $F$), or, $\eta(v,v) = 0$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The latter condition says that $v$ is a &lt;em&gt;light-like vector&lt;/em&gt; in Minkowski space. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A somewhat involved linear algebra computation using the symmetries of the problem gives us the following theorem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Theorem&lt;/strong&gt; &lt;em&gt;(Classification of electromagnetic fields)&lt;/em&gt; Given an arbitrary electromagnetic field, one of the following three must hold:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$\tilde{F}$ is the zero linear transformation: it sends all vectors to zero.&lt;/li&gt;&#xA;&lt;li&gt;$\tilde{F}$ has only one real eigenvector. It is a repeated eigenvector of multiplicity two, and its eigenvalue is 0. &lt;/li&gt;&#xA;&lt;li&gt;$\tilde{F}$ has exactly two real eigenvectors. They each have multiplicity one, and their eigenvalues agree in magnitude but differs in sign.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;(There's also a bit more you can say about the 3rd case, and I'll refer you to the &lt;a href=&quot;http://en.wikipedia.org/wiki/Classification_of_electromagnetic_fields&quot; rel=&quot;nofollow&quot;&gt;wikipedia article&lt;/a&gt;.  )&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case where $F$ is non-zero, the eigenvectors are called the &lt;em&gt;principal null directions&lt;/em&gt; of the electromagnetic field. The precise physical interpretation of the object is tricky in general, but in the second case, where $\tilde{F}$ has only one real eigenvector, the interpretation is clear: the unique principal null direction represents the direction of travel for the electromagnetic wave (recall that light always travels at the speed of light and on a light-like curve). (Recall that the eigenvalue is zero here, so there isn't much to interpret in this case.) (And you can think of the more complicated case 3 as situation where at the point the electromagnetic field is not described by the propagation of just one wave alone, but the superposition of many.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a last side remark: the same procedure can be used to describe, analogously, the propagation of gravitational disturbances, and is extremely useful in general relativity. &lt;/p&gt;&#xA;" OwnerUserId="1543" LastActivityDate="2011-03-13T15:17:44.017" />
  <row Id="26717" PostTypeId="2" ParentId="26662" CreationDate="2011-03-13T15:35:01.953" Score="6" ViewCount="" Body="&lt;p&gt;To add yet another answer: I'm surprised that normal modes haven't been mentioned yet. Whenever you expand the behaviour of a mechanical system around a point of equilibrium, you get coupled equations of motion of the form&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ddot{\vec{x}}=A\vec{x}\;,$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\vec{x}$ is a vector containing all coordinates of the system and $A$ is a square matrix. Such a system is solved by diagonalizing $A$, i.e. transforming to a basis of eigenvectors of $A$ in which $A$ becomes diagonal; this decouples the equations of motion, which are transformed into a set of simple harmonic oscillator equations that can be solved individually. The eigenvalues of $A$ correspond to the eigenfrequencies of the system (they are their negative squares), and the eigenvectors are the so-called normal modes -- patterns of displacement that allow the system to oscillate at a single frequency.&lt;/p&gt;&#xA;" OwnerUserId="6622" LastActivityDate="2011-03-13T15:35:01.953" />
  <row Id="26718" PostTypeId="2" ParentId="26662" CreationDate="2011-03-13T15:35:31.067" Score="3" ViewCount="" Body="&lt;p&gt;They are characteristic of oscillating systems. There can be certain &quot;modes&quot; which always oscillate at the same frequencies, these are the eigenvalues and the frequencies the eigenvectos. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Have a look at the &lt;a href=&quot;http://musicthing.blogspot.com/2006/05/incredible-spooky-video-about.html&quot; rel=&quot;nofollow&quot;&gt;video&lt;/a&gt; here. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;At resonance frequencies, the rice forms a pattern of gaps , which maps out the nodes of the standing wave pattern. The pattern you see is an eigenstate of that shape; the frequency squared is its eigenvalue.&lt;/p&gt;&#xA;" LastActivityDate="2011-03-13T15:45:11.840" />
  <row Id="26719" PostTypeId="2" ParentId="26697" CreationDate="2011-03-13T15:58:10.207" Score="6" ViewCount="" Body="&lt;p&gt;A trick (and it's only a trick) is to set $$f(x) = n(1-\sqrt[n]{\sin(x)})$$ so&#xA;$$ \sin(x) = \left(1-\frac{f(x)}{n}\right)^n$$ which in the limit $n\to\infty$ is&#xA;$$ \sin(x) = e^{-f_{\infty}(x)}$$ or $$f_{\infty}(x) = -\ln (\sin(x))$$ which you integrate to the result: $$-\int_0^{\frac{\pi}{2}}\ln(\sin(x))dx = \frac{\pi\ln 2}{2}$$&lt;/p&gt;&#xA;" OwnerUserId="7476" LastActivityDate="2011-03-13T15:58:10.207" CommentCount="3" />
  <row Id="26720" PostTypeId="1" AcceptedAnswerId="26723" CreationDate="2011-03-13T16:01:50.080" Score="1" ViewCount="83" Body="&lt;p&gt;Hey there,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to find out for which $p, q$ exists $C &amp;gt; 0$ with $||f||_p \leq C ||f||_q$ for all $f \in C([0,1]), p \in \mathbb{R}_{\geq 1} \cup \{\infty\}$. I first let $p,q \geq 1$ and I looked at the case where $q &amp;gt; p$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I defined $\tilde{f} := |f|^p$ and $\tilde{p} := \frac{q}{p}$, $f \in C([0,1])$ of course. Also, I let $g(x) = 1$ and applied Hölder's inequality:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$||\tilde{f} \cdot g||_1 \leq ||\tilde{f}||_\tilde{p} \cdot ||g||_\tilde{q}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where of course $||g||_\tilde{q} = 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\implies \int_0^1 |f|^p \mathrm dx \leq \left ( \int_0^1 |f|^q \mathrm dx \right )^\frac{p}{q} \iff ||f||_p \leq ||f||_q \cdot C \;\; \forall C \geq 1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So my solution is, that for $q &amp;gt; p \geq 1$, there always exists $C &amp;gt; 0$ such that the inequality holds good. Now, before I look at $p &amp;gt; q$, I wanted to know whether my solution is correct. It doesn't seem to be complete to me, as there might also exist $C \in (0,1)$ such that the inequality holds good. Am I wrong? If not, how can I find out for which $p$ and $q$ there exists a $C \in (0,1)$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance for any answers...&lt;/p&gt;&#xA;" OwnerUserId="3787" LastEditorUserId="3787" LastEditorDisplayName="" LastEditDate="2011-03-19T20:14:33.780" LastActivityDate="2011-03-19T20:14:33.780" Title="For which $p, q$ exists $C &gt; 0$ with $||f||_p \leq C ||f||_q$ for all $f \in C([0,1])$?" Tags="&lt;homework&gt;&lt;analysis&gt;" AnswerCount="2" />
  <row Id="26721" PostTypeId="1" AcceptedAnswerId="26732" CreationDate="2011-03-13T16:26:54.650" Score="1" ViewCount="33" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Let $f: A \rightarrow B$ be given. Prove:&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;(a) For each subset $X \subset A, X \subset f^{-1}(f(X)).$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;At first I didn't understand why $X$ wouldn't just be equal to $f^{-1}(f(X))$, but I guess that is the difference between an inverse &lt;em&gt;image&lt;/em&gt; and an inverse &lt;em&gt;function&lt;/em&gt;, right? For instance if I had $A = \{1,2,3\}, B = \{1,2\}, X = \{1,2\}, W = \{2,3\}$, and $f(x) = 1$ then $X, W \subset A$, but $f^{-1}(f(X)) = \{1,2,3\} \neq X$ and of course $X \subset f^{-1}(f(X))$, right?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So am I understanding at least that much correctly? My problem is that I don't know what I am actually supposed to write down to prove this in general, could someone please help me with that?&lt;/p&gt;&#xA;" OwnerUserId="3711" LastActivityDate="2011-03-13T16:48:10.707" Title="Proving elementary inverse image consequence" Tags="&lt;functions&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="26722" PostTypeId="1" CreationDate="2011-03-13T16:27:10.597" Score="2" ViewCount="125" Body="&lt;p&gt;What is the fastest way (general method) to calculate the quantity $a^b \!\mod c$? For example $a=2205$, $b=23$, $c=4891$.&lt;/p&gt;&#xA;" OwnerUserId="5010" LastEditorUserId="7003" LastEditorDisplayName="" LastEditDate="2011-03-13T16:39:17.297" LastActivityDate="2011-03-13T18:02:36.953" Title="calculating $a^b \!\mod c$" Tags="&lt;number-theory&gt;" AnswerCount="5" />
  <row Id="26723" PostTypeId="2" ParentId="26720" CreationDate="2011-03-13T16:31:24.743" Score="3" ViewCount="" Body="&lt;p&gt;You have proved what you set out to: if $q &amp;gt; p \ge 1$ there exists $C &amp;gt; 0$ (namely, $C=1$) such that $||f||_p \le C ||f||_q$ for all $f \in C([0,1])$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note you are &lt;em&gt;not&lt;/em&gt; trying to prove that $||f||_p \le C ||f||_q$ for all $C &amp;gt; 0$.  This is easily seen to be false.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might ask whether this result is sharp; whether there exists a number $0 &amp;lt; C &amp;lt; 1$ which could work.  To find out, consider $f=1$.&lt;/p&gt;&#xA;" OwnerUserId="822" LastEditorUserId="822" LastEditorDisplayName="" LastEditDate="2011-03-13T16:40:21.133" LastActivityDate="2011-03-13T16:40:21.133" CommentCount="1" />
  <row Id="26724" PostTypeId="1" AcceptedAnswerId="26747" CreationDate="2011-03-13T16:34:42.003" Score="3" ViewCount="125" Body="&lt;p&gt;Let $\mathcal{A}$ denote the collection of all subsets A of an uncountable set $\Omega$ for which either A or $A^c$ are countable. Let $\mu(A)$ denote the cardinality of A. Deﬁne $\phi(A)$ equal to $\emptyset$ or $\infty$ according as $A$ is countable or uncountable. Show that $\phi ≪ \mu$. Then show that the Radon-Nikodym theorem fails.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is an assignment. I found that this problem is not rigorous unless in this case we treat $+\infty$ as a number; and this equation, $+\infty+\infty=+\infty$ holds where we do not distinguish the cardinality of all the rational number, $\aleph_0$, and the cardinality of real number, $\aleph_1$, and so on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we treat $\infty$ as a number, it is easy to show that $\mathcal{A}$ is a $\sigma$-field, $\mu$ and $\phi$ a measure and $\phi ≪ \mu.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is: Does this make sense $\aleph_0+\aleph_1+\aleph_2$? (Set theory is not my suit.)&lt;/p&gt;&#xA;" OwnerUserId="8183" LastActivityDate="2011-03-13T17:58:58.000" Title="Does this make sense $\aleph_0+\aleph_1+\aleph_2$?" Tags="&lt;set-theory&gt;&lt;measure-theory&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="26725" PostTypeId="2" ParentId="26720" CreationDate="2011-03-13T16:36:52.670" Score="2" ViewCount="" Body="&lt;p&gt;Your argument seems to be correct. It shows that indeed $\Vert f\Vert_{p} \leq \Vert f\Vert_{q}$ for $1 \leq p \leq q \lt \infty$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instead of using H&amp;ouml;lder's inequality, you could also argue with &lt;a href=&quot;http://en.wikipedia.org/wiki/Jensen%27s_inequality&quot; rel=&quot;nofollow&quot;&gt;Jensen's inequality&lt;/a&gt; (which is often used in the proofs of H&amp;ouml;lder's inequality) as follows: For $p \leq q$ the function $\varphi(x) = x^{q/p}$ is convex, hence $\left(\int |f|^p \right)^{q/p} = \varphi\left(\int |f|^p\right) \leq \int \varphi(|f|^p) = \int |f|^{q}$ and extracting $q$th roots on both sides gives the desired $\Vert f \Vert_{p} \leq \Vert f \Vert_{q}$ for $p \leq q \lt \infty$. The case $q = \infty$ needs to be dealt with separately in this argument and I leave this to you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To see that $C = 1$ is the best possible constant simply take $f$ to be the constant function $f = 1$, which has $\Vert f \Vert_{p} = 1$ for all $p$, so there can't be a constant $C \lt 1$.&lt;/p&gt;&#xA;" OwnerUserId="5363" LastEditorUserId="5363" LastEditorDisplayName="" LastEditDate="2011-03-13T16:58:32.177" LastActivityDate="2011-03-13T16:58:32.177" CommentCount="2" />
  <row Id="26726" PostTypeId="1" CreationDate="2011-03-13T16:40:20.273" Score="0" ViewCount="120" Body="&lt;p&gt;What are the left and right ideals of matrix ring? How about the two sided ideals?&lt;/p&gt;&#xA;" OwnerUserId="8195" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2011-03-14T01:45:20.453" LastActivityDate="2011-03-14T01:45:20.453" Title="What are the left and right ideals of matrix ring? How about the two sided ideals?" Tags="&lt;algebra&gt;&lt;ring-theory&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="1" />
  <row Id="26727" PostTypeId="2" ParentId="26722" CreationDate="2011-03-13T16:41:57.060" Score="1" ViewCount="" Body="&lt;p&gt;One common way of fast exponentiation, either in modular arithmetic or in general, is &lt;a href=&quot;http://en.wikipedia.org/wiki/Exponentiation_by_squaring#Further_application&quot; rel=&quot;nofollow&quot;&gt;exponentiation by squaring&lt;/a&gt;. Here is &lt;a href=&quot;http://en.wikipedia.org/wiki/Exponentiation_by_squaring#Further_applications&quot; rel=&quot;nofollow&quot;&gt;the section&lt;/a&gt; where they talk specifically about $a^b\bmod c$.&lt;/p&gt;&#xA;" OwnerUserId="264" LastEditorUserId="264" LastEditorDisplayName="" LastEditDate="2011-03-13T16:47:34.387" LastActivityDate="2011-03-13T16:47:34.387" />
  <row Id="26728" PostTypeId="2" ParentId="26722" CreationDate="2011-03-13T16:42:15.233" Score="2" ViewCount="" Body="&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Exponentiation_by_squaring&quot; rel=&quot;nofollow&quot;&gt;square-and-multiply algorithm&lt;/a&gt; is generally the fastest way to do modular exponentiation.&lt;/p&gt;&#xA;" OwnerUserId="1016" LastActivityDate="2011-03-13T16:42:15.233" />
  <row Id="26729" PostTypeId="2" ParentId="26722" CreationDate="2011-03-13T16:42:16.657" Score="1" ViewCount="" Body="&lt;p&gt;The &lt;a href=&quot;http://mathforum.org/dr.math/faq/faq.peasant.html&quot; rel=&quot;nofollow&quot;&gt;Russian peasant's method&lt;/a&gt; is pretty straightforward for computing $a^b\bmod m$:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;c=a;d=b;r=1;&#xA;while d≠0&#xA;if d is odd then r=(cr) mod m;&#xA;d=d div 2; \\ integer division; one usually right-shifts bits in practice&#xA;c=c^2 mod m;&#xA;endwhile&#xA;r&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="8196" LastActivityDate="2011-03-13T16:42:16.657" CommentCount="1" />
  <row Id="26730" PostTypeId="1" AcceptedAnswerId="26734" CreationDate="2011-03-13T16:42:56.440" Score="1" ViewCount="76" Body="&lt;p&gt;Hey there,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have another short question. Let $U \subseteq \mathbb{R}^2$ be open and $f: U \rightarrow \mathbb{R}$ be continuously differentiable. Also, $\partial_y f(x,y) = 0$ for all $(x,y) \in U$. I want to find a sufficient condition for $U$ such that $f$ only depends on $x$. Of course, the condition shouldn't be too restrictive. Is it sufficient for $U$ to be connected?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks a lot for any help.&lt;/p&gt;&#xA;" OwnerUserId="3787" LastEditorUserId="3787" LastEditorDisplayName="" LastEditDate="2011-03-19T20:14:07.947" LastActivityDate="2011-03-19T20:14:07.947" Title="A sufficient condition for $U \subseteq \mathbb{R}^2$ such that $f(x,y) = f(x)$" Tags="&lt;homework&gt;&lt;analysis&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="26731" PostTypeId="2" ParentId="26726" CreationDate="2011-03-13T16:47:56.370" Score="1" ViewCount="" Body="&lt;p&gt;the right/left ideals (over a division ring) consist of matrices that are zero except for a given row (or column) and sums of these. (try taking the simplest matrix you can, say 1 in the $(i,j)$-position and zero elsewhere.  multiply by other junk on the left or right and see what you get.)  there are no non-trivial two-sided ideals (over a division ring).&lt;/p&gt;&#xA;" OwnerUserId="6925" LastEditorUserId="6925" LastEditorDisplayName="" LastEditDate="2011-03-13T17:06:47.337" LastActivityDate="2011-03-13T17:06:47.337" CommentCount="2" />
  <row Id="26732" PostTypeId="2" ParentId="26721" CreationDate="2011-03-13T16:48:10.707" Score="1" ViewCount="" Body="&lt;p&gt;Let $Y\subset B$. To say that $x\in f^{-1}(Y)$ is the same as saying that $f(x)\in Y$. Now, if $x\in X$, then $f(x)\in f(X)$, and hence $x\in f^{-1}(f(X))$.This proves that $X\subset f^{-1}(f(X))$. The inclusion may be strict if $f$ is not injective. For instance, if $A=B=\mathbb{R}$ the real numbers, $f(x)=x^2$ and $X=[0,\infty)$, then $f^{-1}(f(X))=\mathbb{R}$. This happens because $f$ is not injective: for each $y&amp;gt;0$ there are two values of $x$ such thet $f(x)=x^2=y$: namely, $\pm\sqrt{y}$. Of these two values only the positive one is in $X$.&lt;/p&gt;&#xA;" OwnerUserId="4791" LastActivityDate="2011-03-13T16:48:10.707" CommentCount="1" />
  <row Id="26733" PostTypeId="1" AcceptedAnswerId="26737" CreationDate="2011-03-13T16:54:01.603" Score="2" ViewCount="85" Body="&lt;p&gt;Suppose $(\Omega, \mathcal{A}, P)$ is the sample space and $X: (\Omega, \mathcal{A}) \rightarrow (\mathbb{R}, \mathcal{B})$ is a random variable. &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Using language of measure theory,&#xA;$P(A \mid X)$, the conditional&#xA;probability of an event given a&#xA;random variable, is defined from&#xA;conditional expectation as $E(I_A&#xD;&#xA;    \mid X)$. So $P(\cdot \mid X)$ is in&#xA;fact a mapping $\mathcal{A} \times&#xD;&#xA;    \Omega \rightarrow [0,1]$.&lt;/li&gt;&#xA;&lt;li&gt;In elementary probability, we&#xA;learned that $$P(A \mid X \in B): =&#xD;&#xA;    \frac{P(A \cap \{X \in B\})}{P(X \in&#xD;&#xA;    B)}.$$ If I understand correctly,&#xA;this requires and implies $P(X \in B) \neq 0$. So &#xA;$P(\cdot \mid X \in \cdot)$ is in&#xA;fact a mapping $\mathcal{A} \times&#xD;&#xA;    \mathcal{B} \rightarrow [0,1]$.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My questions&lt;/strong&gt; are: &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;When will $P(\cdot \mid X)$ in the&#xA;first definition and $P(\cdot \mid X&#xD;&#xA;    \in \cdot)$ in the second coincide/become consistent&#xA;with each other and how?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Is there some case when they can&#xA;both apply but do not agree with&#xA;other? Is the first definition a more&#xA;general one that include the second&#xA;as a special case?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Similar questions for conditional&#xA;expectation. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;In elementary probability, $E(Y \mid    X \in B)$ is defined as&#xA;expectation    of $Y$ w.r.t. the&#xA;p.m. $P(\cdot \mid    X \in B)$. So&#xA;$E(Y \mid X \in \cdot)$    is a&#xA;mapping $\mathcal{B} \rightarrow   &#xD;&#xA;    \mathbb{R}$.&lt;/li&gt;&#xA;&lt;li&gt;In measure theory,  $E(Y \mid X )$ is    a random variable $\Omega&#xD;&#xA;    \rightarrow    \mathbb{R}$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I was also wondering how $E(Y \mid X&#xD;&#xA;    \in \cdot)$ in elementary&#xA;probability and $E(Y \mid X )$ in&#xA;measure theory can coincide/become&#xA;consistent? Is the latter a general&#xA;definition which includs the former&#xA;as a special case?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Thanks and regards!&lt;/p&gt;&#xA;" OwnerUserId="1281" LastEditorUserId="7003" LastEditorDisplayName="" LastEditDate="2011-03-13T17:59:39.643" LastActivityDate="2011-03-13T17:59:39.643" Title="When do the measure-theoretic and elementary definitions of conditional probability/expectation coincide?" Tags="&lt;probability-theory&gt;" AnswerCount="1" />
  <row Id="26734" PostTypeId="2" ParentId="26730" CreationDate="2011-03-13T17:01:00.307" Score="3" ViewCount="" Body="&lt;p&gt;It's enough for $U$ to have the property that, whenever $(x,y_1)$ and $(x,y_2)$ are in $U$, so is the line segment between them. This can be proved by applying the mean value theorem to the function $g(y)=f(x,y)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is &lt;em&gt;not&lt;/em&gt; enough for $U$ to be connected. For example, take $U=\mathbb{R} \setminus \{(x,0)|x \le 0 \}$. Let $f(x,y)=0$ for $x \ge 0$, $f(x,y)=x^2$ for $x&amp;lt;0, y&amp;lt;0$ and $f(x,y)=-x^2$ for $x&amp;lt;0, y&amp;gt;0$.&lt;/p&gt;&#xA;" OwnerUserId="5203" LastActivityDate="2011-03-13T17:01:00.307" CommentCount="6" />
  <row Id="26735" PostTypeId="1" CreationDate="2011-03-13T17:11:26.153" Score="3" ViewCount="156" Body="&lt;p&gt;Let $W_t$ be a standard Brownian motion with $W_0 = 0$ and let $Z_t$ solve the stochastic differential equation $dZ_t = 2 Z_t W_t \mathrm{d}W_t$. This has solution&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;Z_t=\exp\Big\{W_t^2-\int_0^t{(2W_s^2+1)ds}\Big\} \&amp;gt; .&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is easy to show that $Z_t$ is a &lt;em&gt;local&lt;/em&gt; martingale since $P(\int_0^T{(Z_sW_s)^2ds}&amp;lt;\infty)=1.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could we show that $E[\int_0^T{(Z_sW_s)^2ds}]&amp;lt;\infty$, which implies $Z_t$ is a martingale in the interval $[0,T]?$ &lt;/p&gt;&#xA;" OwnerUserId="8183" LastEditorUserId="8183" LastEditorDisplayName="" LastEditDate="2011-03-13T22:24:32.607" LastActivityDate="2011-03-19T23:43:01.243" Title="Is this a martingale?" Tags="&lt;stochastic-processes&gt;&lt;stochastic-integrals&gt;" AnswerCount="1" CommentCount="10" FavoriteCount="1" />
  <row Id="26736" PostTypeId="2" ParentId="26730" CreationDate="2011-03-13T17:14:48.627" Score="1" ViewCount="" Body="&lt;p&gt;This is from a problem in Rudin, if I recall correctly, in the section on functions of several variables. The important point to note is that this problem relies on the mean value theorem, which is a statement about a function from $\mathbf{R}$ to $\mathbf{R}$. The trick in this situation is to restrict your domain $E$ to a single straight line, in which $x$ is fixed and $y$ is variable. By the mean value theorem, the distance between two points along this line will depend only on x, and so all points on this line will map to the same point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a notion in analysis of convexity. If $(x_1,y_1)$ and $(x_2,y_2)$ are in $E$, so is the line in between them. Our proof above required this to use the mean value theorem. However, we didn't require convexity in every direction. Since we only took partials in the $y$-direction, we just required convexity in the $y$-direction. That is to say, that if two points have the same $y$ value, then the line connecting them would remain in $E$. &lt;/p&gt;&#xA;" OwnerUserId="647" LastActivityDate="2011-03-13T17:14:48.627" />
  <row Id="26737" PostTypeId="2" ParentId="26733" CreationDate="2011-03-13T17:21:27.920" Score="2" ViewCount="" Body="&lt;p&gt;$$\int_C P(A|X)(\omega)dP(\omega)=\int_C E(I_A|X)(\omega)dP(\omega)=\int_C I_A(\omega)dP(\omega)=P(A\cap C)$$ for $C$ in the sigma algebra generated by $X$. So, for $C=\{X\in B\}$,&#xA;$$\frac{\int_{\{X\in B\}} P(A|X)(\omega)dP(\omega)}{P(X\in B)}=P(A\mid X\in B).$$&lt;/p&gt;&#xA;" OwnerUserId="367" LastEditorUserId="367" LastEditorDisplayName="" LastEditDate="2011-03-13T17:43:18.747" LastActivityDate="2011-03-13T17:43:18.747" CommentCount="9" />
  <row Id="26738" PostTypeId="2" ParentId="26722" CreationDate="2011-03-13T17:36:14.470" Score="1" ViewCount="" Body="&lt;p&gt;Generally &lt;a href=&quot;http://en.wikipedia.org/wiki/Exponentiation_by_squaring&quot; rel=&quot;nofollow&quot;&gt;repeated squaring&lt;/a&gt; works well. It deserves to be better known that this arises simply from writing the exponent in binary radix in Horner polynomial form, i.e. $\rm\ d_0 + x\ (d_1 + x\ (d_2\ +\:\cdots))\:.\ $ Below is an example of computing $\rm\ a^{101}\ $ by repeated squaring. Note that the repeated square form arises simply from performing various substitutions into the binary polynomial Horner form  namely $\rm\ 1\to a,\ \ 0\to 1,\ \ (x)\:2\to (x)^2\ $ into $101_{10} = 1100101_2\ $ expanded into Horner form, viz.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;     1   1   0   0   1   0   1  =   101_10 in binary radix &#xA;&#xA;  =  1)2+1)2+0)2+0)2+1)2+0)2+1  =   101 as binary Horner poly &#xA;&#xA;       2   2   2   2   2   2        101 &#xA; -&amp;gt;  a)  a)   )   )  a)   )  a  =  a    via repeated squaring &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="242" LastActivityDate="2011-03-13T17:36:14.470" />
  <row Id="26739" PostTypeId="1" AcceptedAnswerId="26742" CreationDate="2011-03-13T17:37:18.510" Score="2" ViewCount="78" Body="&lt;p&gt;The difference is $a^2 - b^2 = (a - b).(a + b)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But what about when I have $a^{25} + 1$ ? According to wolfram alpha, the alternate form is:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$(a+1) (a^4 -a^3 + a^2 -a + 1)( a^{20} - a^{15} + a^{10} -a^5 +1)$&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;However, the square root of 25 is a rational number 5.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But If I had 50, where the square root of 50 is irrational ?&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$(a^2 + 1) (a^8 -a^6 +a^4 -a^2 +1) (a^{40} -a^{30} +a^{20} -a^{10} +1 )$&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;In fact, I'm just wondering and trying to find out patterns, I'm very curious about that, since I haven't found anything related, only the alternate forms generated by wolfram. My question is how and what mathematical algorithm they used to find $k$ forms for $a^n + 1$ ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance.&lt;/p&gt;&#xA;" OwnerUserId="8127" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2011-03-13T17:41:10.160" LastActivityDate="2011-03-13T17:54:19.587" Title="Factorization of sum of two square" Tags="&lt;algebra&gt;&lt;algebra-precalculus&gt;" AnswerCount="4" />
  <row Id="26740" PostTypeId="2" ParentId="26705" CreationDate="2011-03-13T17:38:06.583" Score="6" ViewCount="" Body="&lt;p&gt;No. $\textbf{}   \textbf{}   \textbf{}    $&lt;/p&gt;&#xA;" OwnerUserId="299" LastActivityDate="2011-03-13T17:38:06.583" />
  <row Id="26741" PostTypeId="2" ParentId="26739" CreationDate="2011-03-13T17:43:31.640" Score="2" ViewCount="" Body="&lt;p&gt;There is a &lt;a href=&quot;http://en.wikipedia.org/wiki/Factorization_of_polynomials&quot; rel=&quot;nofollow&quot;&gt;Wikipedia article&lt;/a&gt; about algorithms for factoring polynomials. For the special case of polynomials of the form $a^n + 1$ the factors will be &lt;a href=&quot;http://en.wikipedia.org/wiki/Cyclotomic_polynomial&quot; rel=&quot;nofollow&quot;&gt;cyclotomic polynomials&lt;/a&gt;: more precisely,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$a^n + 1 = \prod_{d | 2n, d \nmid n} \Phi_d(a).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $n = 25$ the corresponding values of $d$ are $d = 2, 10, 50$, and for $n = 50$ the corresponding values of $d$ are $d = 4, 20, 100$. &lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2011-03-13T17:43:31.640" />
  <row Id="26742" PostTypeId="2" ParentId="26739" CreationDate="2011-03-13T17:44:21.233" Score="5" ViewCount="" Body="&lt;p&gt;Such factorizations are special cases of general formulas for factorizations of cylotomic polynomials, e.g. see &lt;a href=&quot;http://en.wikipedia.org/wiki/Cyclotomic_polynomial&quot; rel=&quot;nofollow&quot;&gt;wikipedia&lt;/a&gt;. These formulas follow by &lt;a href=&quot;http://en.wikipedia.org/wiki/M%C3%B6bius_inversion_formula&quot; rel=&quot;nofollow&quot;&gt;Mobius inversion&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Such polynomial factorizations come in handy for integer factorization. For example, &lt;a href=&quot;http://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.114.548&quot; rel=&quot;nofollow&quot;&gt;Aurifeuille, Le Lasseur and Lucas&lt;/a&gt; discovered so-called &lt;a href=&quot;http://www.ams.org/journals/mcom/2006-75-253/S0025-5718-05-01766-7/S0025-5718-05-01766-7.pdf&quot; rel=&quot;nofollow&quot;&gt;&lt;strong&gt;Aurifeuillian factorizations&lt;/strong&gt;&lt;/a&gt; of cyclotomic polynomials $\rm\;\Phi_n(x) = C_n(x)^2 - n\ x\  D_n(x)^2\;$. These play a role in factoring numbers of the form $\rm\; b^n \pm 1\:$, cf. the &lt;a href=&quot;http://homes.cerias.purdue.edu/~ssw/cun/&quot; rel=&quot;nofollow&quot;&gt;Cunningham Project&lt;/a&gt;. Below are some simple examples of such factorizations:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{array}{rl}&#xD;&#xA;x^4 + 2^2 \quad=&amp;amp;  (x^2 + 2x + 2)\;(x^2 - 2x + 2) \\\\&#xD;&#xA;\frac{x^6 + 3^2}{x^2 + 3} \quad=&amp;amp;  (x^2 + 3x + 3)\;(x^2 - 3x + 3) \\\\&#xD;&#xA;\frac{x^{10} - 5^5}{x^2 - 5} \quad=&amp;amp;  (x^4 + 5x^3 + 15x^2 + 25x + 25)\;(x^4 - 5x^3 + 15x^2 - 25x + 25) \\\\&#xD;&#xA;\frac{x^{12} + 6^6}{x^4 + 36} \quad=&amp;amp;  (x^4 + 6x^3 + 18x^2 + 36x + 36)\;(x^4 - 6x^3 + 18x^2 - 36x + 36) \\\\&#xD;&#xA;\end{array}$$&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2011-03-13T17:44:21.233" />
  <row Id="26744" PostTypeId="2" ParentId="26739" CreationDate="2011-03-13T17:48:28.907" Score="1" ViewCount="" Body="&lt;p&gt;Let $z=\mathrm{e}^{\mathrm{i}\pi/n}$, hence $z^n=-1$ and $z^{2k}$ from $k=1$ to $k=n$ are the $n$th roots of $1$. Then $a^n+1=a^n-z^n$ is the product of $a-z^{2k+1}$ from $k=1$ to $k=n$. One recovers a factorization on the real numbers by multiplying the $k$ term with the $n-k-1$ term for every $k\ne(n-1)/2$, since $(a-z^{2k+1})(a-z^{2n-2k-1})=a^2-2\cos((2k+1)\pi/n)a+1$.&lt;/p&gt;&#xA;" OwnerUserId="6179" LastActivityDate="2011-03-13T17:48:28.907" />
  <row Id="26745" PostTypeId="2" ParentId="26739" CreationDate="2011-03-13T17:54:19.587" Score="5" ViewCount="" Body="&lt;p&gt;If $n$ is odd then the polynomial $p(x)=x^n+1$ has a zero at $x=-1$, $p(-1)=(-1)^n+1=-1+1=0$.  By the factor theorem, $x+1$ is therefore a factor of $p(x)$.  Using polynomial long division (or synthetic division) you can show that $x^n+1=(x+1)(x^{n-1}-x^{n-2}+x^{n-3}-\cdots-x+1)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, consider $p(x)=x^5+1$.  By the above, since $5$ is odd, $p(x)=(x+1)(x^4-x^3+x^2-x+1)$.  Next notice that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\begin{align*}x^{25}+1&amp;amp;=(x^5)^5+1=p(x^5)\\&#xD;&#xA;&amp;amp;=(x^5+1)((x^5)^4-(x^5)^3+(x^5)^2-x^5+1)\\&#xD;&#xA;&amp;amp;=(x+1)(x^4-x^3+x^2-x+1)(x^{20}-x^{15}+x^{10}-x^5+1).\end{align*}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, notice that $x^{50}+1=(x^2)^{25}+1=p((x^2)^5)$, so you can just substitute $x^2$ for $x$ in the factorization of $x^{25}+1$.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2011-03-13T17:54:19.587" />
  <row Id="26746" PostTypeId="1" AcceptedAnswerId="26749" CreationDate="2011-03-13T17:55:52.293" Score="7" ViewCount="219" Body="&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/EKmG6.png&quot; alt=&quot;Inscribed kissing circles in an equilateral triangle&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Triangle is equilateral (AB=BC=CA), I need to find AB and R.&#xA;Any hints? &#xA;I was trying to make another triangle by connecting centers of small circles but didn't found anything&lt;/p&gt;&#xA;" OwnerUserId="7149" LastEditorUserId="1543" LastEditorDisplayName="" LastEditDate="2011-03-15T01:11:59.113" LastActivityDate="2011-03-15T01:11:59.113" Title="Inscribed kissing circles in an equilateral triangle" Tags="&lt;homework&gt;&lt;geometry&gt;&lt;circle&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="2" />
  <row Id="26747" PostTypeId="2" ParentId="26724" CreationDate="2011-03-13T17:58:58.000" Score="6" ViewCount="" Body="&lt;p&gt;A measure (in your case) takes values in $[-\infty, +\infty]$ or even $[0,\infty]$ probably.&#xA;The extended reals do have the arithmetric &quot;rule&quot; or convention that $+\infty + x = \infty$ for $x$ a normal real or $\infty$, while $+\infty + (-\infty)$ is undefined. &#xA;This means that the description of the measure $\mu$, the so-called counting measure, &lt;strong&gt;must&lt;/strong&gt; mean that any infinite set $A$, either countable or uncountable, gets $\mu(A) = +\infty$; this is what is meant by the fact that the measure is the cardinality: we only distinguish the finite ones, and &lt;strong&gt;all&lt;/strong&gt; infinite ones get value $\infty$. The addition convention ensures that this will work as a measure, as you said. The values thus are &lt;strong&gt;not&lt;/strong&gt; cardinals, but extended reals. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;With this information I hope you can figure out the question.&lt;/p&gt;&#xA;" OwnerUserId="4280" LastActivityDate="2011-03-13T17:58:58.000" CommentCount="1" />
  <row Id="26748" PostTypeId="2" ParentId="26722" CreationDate="2011-03-13T18:02:36.953" Score="2" ViewCount="" Body="&lt;p&gt;Let's assume that &lt;code&gt;a,b,c&lt;/code&gt; referred to here are positive integers, as in your example.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a specific exponent &lt;code&gt;b&lt;/code&gt;, there may be a faster (shorter) method of computing &lt;code&gt;a^b&lt;/code&gt; than binary exponentiation.  Knuth has a discussion of the phenomenon in Art of Computer Programming Vol. 2 (Semi-numerical Algorithms), Sec. 4.6.3 and the index term &quot;addition chains&quot;.  He cites &lt;code&gt;b=15&lt;/code&gt; as the smallest case where binary exponentiation is not optimal, in that it requires six multiplication but &lt;code&gt;a^3&lt;/code&gt; can be computed in two multiplications, and then &lt;code&gt;(a^3)^5&lt;/code&gt; in three more for a total of five multiplications.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the specific exponent &lt;code&gt;b=23&lt;/code&gt; the parsimonious addition chain involves the exponents (above &lt;code&gt;1&lt;/code&gt;) &lt;code&gt;2,3,5,10,13&lt;/code&gt;, at which point &lt;code&gt;a^23 = (a^10)*(a^13)&lt;/code&gt;, for a total of six multiplications.  Binary exponentiation for &lt;code&gt;b=23&lt;/code&gt; requires seven multiplications.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another approach that can produce faster results when &lt;code&gt;b&lt;/code&gt; is large (not in your example) depends on knowing something about the base &lt;code&gt;a&lt;/code&gt; and modulus &lt;code&gt;c&lt;/code&gt;.  Recall from Euler's generalization of Fermat's Little Thm. that if &lt;code&gt;a,c&lt;/code&gt; are coprime, then &lt;code&gt;a^d = 1 mod c&lt;/code&gt; for &lt;code&gt;d&lt;/code&gt; the Euler &lt;code&gt;phi&lt;/code&gt; function of &lt;code&gt;c&lt;/code&gt; (the number of positive integers less than &lt;code&gt;c&lt;/code&gt; and coprime to it).  In particular if &lt;code&gt;c&lt;/code&gt; is a prime, then by Fermat's Little Thm. either &lt;code&gt;c&lt;/code&gt; divides &lt;code&gt;a&lt;/code&gt; and &lt;code&gt;a^b = 0 mod c&lt;/code&gt; or else &lt;code&gt;a^b = a^e mod c&lt;/code&gt; where &lt;code&gt;e = b mod (c-1)&lt;/code&gt; since &lt;code&gt;phi(c) = c-1&lt;/code&gt; for a prime &lt;code&gt;c&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the base &lt;code&gt;a&lt;/code&gt; and modulus &lt;code&gt;c&lt;/code&gt; are not coprime, then it might be advantageous to factor &lt;code&gt;a&lt;/code&gt; into its greatest common factor with &lt;code&gt;c&lt;/code&gt; and its largest factor that is coprime to &lt;code&gt;c&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also it might be advantageous if &lt;code&gt;c&lt;/code&gt; is not prime to factor it into prime powers and do separate exponentiations for each such factor, piecing them back together via the Chinese Remainder Thm.  In your example &lt;code&gt;c = 4891 = 67*73&lt;/code&gt;, so you might compute &lt;code&gt;a^b mod 67&lt;/code&gt; and &lt;code&gt;a^b mod 73&lt;/code&gt; and combine those results to get &lt;code&gt;a^b mod c&lt;/code&gt;.  This is especially helpful if you are limited in the precision of integer arithmetic you can do.&lt;/p&gt;&#xA;" OwnerUserId="3111" LastActivityDate="2011-03-13T18:02:36.953" />
  <row Id="26749" PostTypeId="2" ParentId="26746" CreationDate="2011-03-13T18:03:58.143" Score="7" ViewCount="" Body="&lt;p&gt;Two hints:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Draw the line segments from $A$ and $B$ through the center of the large circle and use the 30°-60°-90° triangles and similar triangles to get a relationship between $AB$ and $R$.&lt;/li&gt;&#xA;&lt;li&gt;Draw the line parallel to $\overline{BC}$ through the point of tangency of the large circle and the small circle closest to $A$ to get a small equilateral triangle, and use what you learned about the relationship between $AB$ and $R$ on the small triangle.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;edit&lt;/em&gt; (related to comments below):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/qSvMt.png&quot; alt=&quot;small triangle at the top&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Above is the small triangle formed at the top of your diagram.  $DF=DE=4$, since both are radii of the small circle.  You can use the 30°-60°-90° triangles to find $AD$ and $AE$.  In particular, what do you find is $\frac{AD}{AF}$?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;edit 2&lt;/em&gt;: Since the homework problem is now done, here's how I would actually have done the problem myself, though it is not the solution I would expect from a geometry student:  In an equilateral triangle, the median, altitude, angle bisector, perpendicular bisector, etc. are all the same segment.  The point of concurrency of the angle bisectors is the center of the inscribed circle and the point of concurrency of the medians divides the medians in the ratio $2:1$, so the height of the smaller triangle shown above is $R=3\cdot 4=12$ and the height of the large triangle is $3R=36$.  That height is the $\sqrt{3}$ side in a $1:\sqrt{3}:2$ right triangle (30°-60°-90°), where $AB$ is the $2$ side, so $AB=\frac{2}{\sqrt{3}}\cdot 36=\frac{72}{\sqrt{3}}=24\sqrt{3}$.&lt;/p&gt;&#xA;" OwnerUserId="72" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2011-03-13T19:57:22.773" LastActivityDate="2011-03-13T19:57:22.773" CommentCount="11" />
  <row Id="26750" PostTypeId="1" AcceptedAnswerId="26752" CreationDate="2011-03-13T18:29:15.150" Score="0" ViewCount="164" Body="&lt;p&gt;Hi all,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm facing the problem of solving&#xA;    $$\sin(x)+x \cdot \cos(x)=0$$&#xA;using&#xA;    $$\tan(x)=\sin(x)/\cos(x)$$&#xA;I end at&#xA;    $$x+\tan(x)=0$$&#xA;on the other hand, I also tried&#xA;    $\cos(x)= \pm \sqrt{1-\sin^2(x)}$&#xA;resulting in $x^2-x^2\cdot \sin^2(x)-\sin^2(x)=0$&#xA;OR using $\sin(x)= \pm \sqrt{1-\cos^2(x)}$ I get&#xA;$1-\cos^2(x)-x^2\cdot \cos^2(x)=0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In all these cases I have no idea on how to get to a solution like $x=...$.&#xA;As far is I can see, I should try getting $x$ into the trigonometric functions, but I don't see how I could?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;tanks for all your help&lt;/p&gt;&#xA;" OwnerUserId="8198" LastEditorUserId="3567" LastEditorDisplayName="" LastEditDate="2011-03-13T18:35:51.283" LastActivityDate="2011-03-13T18:59:29.227" Title="How to solve trigonometric equation $\sin(x)+x\cdot \cos(x)=0$ ?" Tags="&lt;functions&gt;&lt;trigonometry&gt;" AnswerCount="2" />
  <row Id="26751" PostTypeId="2" ParentId="26746" CreationDate="2011-03-13T18:34:21.497" Score="2" ViewCount="" Body="&lt;p&gt;Let $a$ be the side of the triangle. If $A$ denotes the area and $P$ denotes the perimeter, then the radius of the incircle is given by $R = \frac{2A}{P} = \frac{2\sqrt{3} a^2/4}{3a} = \frac{\sqrt{3} a}{6}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $x$ be the distance of the center of the smaller circle to the nearest vertex.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The altitude is $x + 8 + 2R = \frac{\sqrt{3}a}{2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From similar triangle, we also have $\frac{x+4}{4} = \frac{x+8+R}{R} \Rightarrow \frac{x}{4} = \frac{x+8}{R}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have three equations in $x$,$R$ and $a$ solve them to get your $R$ and $a$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x+8 = \frac{\sqrt{3}a}{2} - 2R = \frac{\sqrt{3}a}{2} - \frac{\sqrt{3}a}{3} = \frac{\sqrt{3}a}{6} = R$. Hence, $\frac{x}{4} = 1 \Rightarrow x = 4$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$R = x + 8 = 12 \Rightarrow a = 2 \sqrt{3} R = 24 \sqrt{3}$.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2011-03-13T18:47:53.030" LastActivityDate="2011-03-13T18:47:53.030" CommentCount="1" />
  <row Id="26752" PostTypeId="2" ParentId="26750" CreationDate="2011-03-13T18:47:59.150" Score="3" ViewCount="" Body="&lt;p&gt;Generally, trig functions don't talk to polynomials, so you're not going to get an algebraic solution to this one.  Your favorite numeric solver will have no problem.  If you draw the graphs of $y=\tan x$ and $y=-x$ you'll have an easy time finding where to look for roots.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-03-13T18:47:59.150" />
  <row Id="26754" PostTypeId="1" AcceptedAnswerId="26758" CreationDate="2011-03-13T18:56:44.587" Score="2" ViewCount="75" Body="&lt;p&gt;My question is as titled. Here, the radical of a Lie algebra means its maximal solvable ideal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This question occurs in the proof of the theorem which states that &quot;A Lie algebra is semsimple if and only if its killing form is nondegenerate.&quot; In the proof, it is written that &quot;to prove that $L$ is semisimple, it will suffice to prove that every abelian ideal of $L$ lies in the radical of the killing form.&quot; So I am wondering what if $L$ contains no abelian ideal whiling being semisimple.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it possible?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many thanks.&lt;/p&gt;&#xA;" OwnerUserId="4736" LastActivityDate="2011-03-13T21:18:33.217" Title="Is there any example of a Lie Algebra, who has nontrivial radical but contains no abelian ideal?" Tags="&lt;lie-algebra&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="26755" PostTypeId="2" ParentId="26524" CreationDate="2011-03-13T18:59:23.957" Score="1" ViewCount="" Body="&lt;p&gt;There's no easy solution, but if you want a slick expression you can try the generating function approach. Write $Q_i = 1-P_i + P_it$ and extract the coefficient of $t^{K-1}$ in $$\frac{\prod_{i=1}^n Q_i}{1-t}.$$&#xA;This is the probability that less than $K$ elements were selected.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2011-03-13T18:59:23.957" />
  <row Id="26756" PostTypeId="2" ParentId="26750" CreationDate="2011-03-13T18:59:29.227" Score="1" ViewCount="" Body="&lt;p&gt;$$\sin{x} + x\cos{x} = 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\cos{x} (\tan{x} + x) =0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This would have been true if $\cos{x}=0$ but note that in this case the $\tan{x}$ term blows up and the equation is $0\times\infty$. The other possibility is when $\tan{x}=-x$ which has an infinite number of solutions. That $x=0$ is a solution is obvious, the others can be found &lt;a href=&quot;http://www.wolframalpha.com/input/?i=plot+tanx+=+-x&quot; rel=&quot;nofollow&quot;&gt;graphically&lt;/a&gt;/numerically.&lt;/p&gt;&#xA;" LastActivityDate="2011-03-13T18:59:29.227" />
  <row Id="26757" PostTypeId="1" AcceptedAnswerId="26867" CreationDate="2011-03-13T19:08:20.233" Score="4" ViewCount="238" Body="&lt;p&gt;Hello &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to prove that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\lim_{h\rightarrow\infty}\left(\int_{0}^{\infty}\left(\cos ht-1\right)\underset{t}{\triangle}\left[\frac{\phi(t)\exp\left(-itx\right)}{it}\right]dt\right)=-\int_{0}^{\infty}\underset{t}{\triangle}\left[\frac{\phi(t)\exp\left(-itx\right)}{it}\right]dt$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where &#xA;$\underset{t}{\triangle}\eta(t)=\eta(t)+\eta(-t)$ and $\phi$ is an integrable function (in the lebesgue sense), to be precise it is the fourier transform of an integrable density function and thus continuous. Also $\phi$ is differentiable at $0$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to the authors of this &lt;a href=&quot;http://ora.ouls.ox.ac.uk/objects/uuid%3Aa4c3ad11-74fe-458c-8d58-6f74511a476c/datastreams/ATTACHMENT01&quot; rel=&quot;nofollow&quot;&gt;paper&lt;/a&gt; (see proof of theorem 3), this can be achieved by showing $\underset{t}{\triangle}\left[\frac{\phi(t)\exp\left(-itx\right)}{it}\right]$ is integrable and the result will follow from the Riemann Lebesgue lemma. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;They do this by showing that $\underset{t}{\triangle}\left[\frac{\phi(t)\exp\left(-itx\right)}{it}\right]$ is uniformly bounded. And this is the part of the proof I am stuck on. Can anyone show me how to prove $\underset{t}{\triangle}\left[\frac{\phi(t)\exp\left(-itx\right)}{it}\right]$ is uniformly bounded and integrable? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="6386" LastEditorUserId="6386" LastEditorDisplayName="" LastEditDate="2011-03-14T17:09:45.033" LastActivityDate="2011-03-14T17:09:45.033" Title="How can I compute this limit" Tags="&lt;calculus&gt;&lt;analysis&gt;&lt;complex-analysis&gt;&lt;fourier-analysis&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="1" />
  <row Id="26758" PostTypeId="2" ParentId="26754" CreationDate="2011-03-13T19:16:16.063" Score="6" ViewCount="" Body="&lt;p&gt;If the radical $\mathfrak r$ of a Lie algebra $\mathfrak g$, then $\mathfrak r$ is a solvable Lie algebra. It follows that either $[\mathfrak r, \mathfrak r]$ is zero, so that $\mathfrak r$ is abelian, or $[\mathfrak r, \mathfrak r]$ is a non-trivial nilpotent ideal in $\mathfrak r$. In the last case, then the center of $[\mathfrak r,\mathfrak r]$, which is not trivial because of nilpotency, is an abelian ideal of $\mathfrak g$.&lt;/p&gt;&#xA;" OwnerUserId="274" LastEditorUserId="274" LastEditorDisplayName="" LastEditDate="2011-03-13T19:37:43.220" LastActivityDate="2011-03-13T19:37:43.220" CommentCount="2" />
  <row Id="26759" PostTypeId="2" ParentId="26754" CreationDate="2011-03-13T19:17:26.333" Score="2" ViewCount="" Body="&lt;p&gt;Let $\mathfrak{g}$ be a Lie algebra, $\mathfrak{r}$ its radical, and $\mathfrak{a}$ the center of $\mathfrak{r}$ (the set of $x \in \mathfrak{r}$ s.t. for all $y \in \mathfrak{r}$, $[x,y]=0$).&#xA;By the Jacobi identity, and since $\mathfrak{r}$ is an ideal of $\mathfrak{g}$, $\mathfrak{a}$ is an ideal of $\mathfrak{g}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moreover a nontrivial solvable Lie algebra has nontrivial center.&#xA;EDIT: This last sentence is indeed false (thank you Matt E). So it's easier to formulate it this way: by Jacobi, whenever $\mathfrak{a}$ and $\mathfrak{b}$ are ideals of a Lie algebra, $[\mathfrak{a},\mathfrak{b}]$ is also an ideal, so the last non-trivial $\mathcal{D}^n \mathfrak{r}$ is an abelian ideal of $\mathfrak{g}$.&lt;/p&gt;&#xA;" OwnerUserId="2660" LastEditorUserId="2660" LastEditorDisplayName="" LastEditDate="2011-03-13T21:18:33.217" LastActivityDate="2011-03-13T21:18:33.217" CommentCount="5" />
  <row Id="26760" PostTypeId="1" AcceptedAnswerId="26763" CreationDate="2011-03-13T19:24:17.840" Score="1" ViewCount="68" Body="&lt;p&gt;This problem is as titled.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The textbook states that the order of the Weyl group of type $E_8$, $F_4$ are $2^{14}3^55^27$ and 1152 respectively, but I am wondering how are these groups like, namely, how can they be decomposed into simpler groups, or what kind of subgroup or ideals do they have.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for any attention and help~&lt;/p&gt;&#xA;" OwnerUserId="4736" LastActivityDate="2011-03-13T19:47:16.837" Title="What are the Weyl group of type $E_8$, $F_4$,$G_2$ ?" Tags="&lt;lie-algebra&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="26761" PostTypeId="1" CreationDate="2011-03-13T19:27:48.237" Score="2" ViewCount="81" Body="&lt;p&gt;What is the sum of all positive integers $n$ for which $2^n + 65$ is a perfect square?&lt;/p&gt;&#xA;" OwnerUserId="3128" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2011-03-13T19:33:16.163" LastActivityDate="2011-03-13T19:56:08.070" Title="Perfect square. Factorising problem." Tags="&lt;homework&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="26762" PostTypeId="1" AcceptedAnswerId="26774" CreationDate="2011-03-13T19:30:12.843" Score="3" ViewCount="142" Body="&lt;p&gt;Hello. I would like to compute $\sum_{k=0}^n S(n,k) k$, where $S(n,k)$ is a Stirling number of the second kind. Any ideas? It is like I convolute the Stirling numbers of the second kind with the positive integers. Thank you very much!&lt;/p&gt;&#xA;" OwnerUserId="8201" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2011-05-28T15:10:14.327" LastActivityDate="2011-05-28T15:10:14.327" Title="Formula for $\sum_{k=0}^n S(n,k) k$, where $S(n,k)$ is a Stirling number of the second kind?" Tags="&lt;combinatorics&gt;&lt;stirling-numbers&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" />
  <row Id="26763" PostTypeId="2" ParentId="26760" CreationDate="2011-03-13T19:33:22.647" Score="2" ViewCount="" Body="&lt;p&gt;According to GAP, $F_4$ is&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;(((((C2 x D8) : C2) : C3) : C3) : C2) : C2&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Here &lt;code&gt;C2&lt;/code&gt; and so on denote cyclic groups, &lt;code&gt;D8&lt;/code&gt; denotes the dihedral group of order $8$, &lt;code&gt;x&lt;/code&gt; denotes direct product, and &lt;code&gt;:&lt;/code&gt; denotes semidirect product.&#xA;That is not the most enlightening description of the group, but shows that it is made up from rather simple pieces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You should download GAP and play with it a lit to explore the subgroups and what not in the groups. It's fun.&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2011-03-13T19:33:22.647" />
  <row Id="26764" PostTypeId="2" ParentId="26563" CreationDate="2011-03-13T19:46:54.053" Score="0" ViewCount="" Body="&lt;p&gt;Given some time dependent data-set, autocorrelation can be used to detect randomness and lack thereof. Take a look at for example &lt;a href=&quot;http://en.wikipedia.org/wiki/Correlogram&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Correlogram&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3317" LastActivityDate="2011-03-13T19:46:54.053" />
  <row Id="26765" PostTypeId="2" ParentId="26760" CreationDate="2011-03-13T19:47:16.837" Score="3" ViewCount="" Body="&lt;p&gt;Every Weyl group is a &lt;a href=&quot;http://en.wikipedia.org/wiki/Coxeter_group&quot; rel=&quot;nofollow&quot;&gt;Coxeter group&lt;/a&gt; with Coxeter diagram given more or less by the Dynkin diagram of the corresponding Lie algebra (up to some minor alterations). A good reference is Humphreys' &lt;em&gt;Reflection groups and Coxeter groups.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2011-03-13T19:47:16.837" />
  <row Id="26766" PostTypeId="2" ParentId="26761" CreationDate="2011-03-13T19:54:38.800" Score="3" ViewCount="" Body="&lt;p&gt;First show that $n$ cannot be odd.  This can be done by looking at $2^n + 65$ modulo 4.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So $n$ must be even, say $n=2m$.  If $2^{2m} +65=x^2$, then $65=(x-2^m)(x+2^m)$.  Now look at all the factorizations of 65.&lt;/p&gt;&#xA;" OwnerUserId="6312" LastActivityDate="2011-03-13T19:54:38.800" CommentCount="1" />
  <row Id="26767" PostTypeId="2" ParentId="26761" CreationDate="2011-03-13T19:54:59.647" Score="3" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First note that for $2^n + 65$ to be square, $n$ must be even. You can check this since a square $\pmod{10}$ can be $0,1,4,5,6,9$. Hence, $n=2k$. Hence the problem now boils down to finding $k$ such that $4^k + 65$ is a square.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$4^k + 65 = x^2$ i.e. $x^2- (2^k)^2 = 65$ i.e. $(x+2^k)(x-2^k) = 65$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now look at the different possible cases of factorizing $65$ over $\mathbb{Z}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Analyze the different cases and get the different values of $k$ and add them up. I am unable to think of direct way to evaluate the sum instead of computing the individual $k$'s.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastActivityDate="2011-03-13T19:54:59.647" />
  <row Id="26769" PostTypeId="1" CreationDate="2011-03-13T19:59:49.243" Score="0" ViewCount="37" Body="&lt;p&gt;As is known, $B_2$ can be realized as linear Lie algebra of elements of the form $x=&#xD;&#xA;\begin{pmatrix}&#xD;&#xA; 0 &amp;amp; b_1 &amp;amp; b_2 \\ &#xD;&#xA; c_1 &amp;amp; m &amp;amp; n \\ &#xD;&#xA; c_2 &amp;amp; p &amp;amp; q&#xD;&#xA;\end{pmatrix}$, where $c_1=-b_2^t$, $c_2=-b_1^t$, $q=-m^t$, $n^t=-n$, $p^t=-p$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My problem is, how can I find a Chevalley basis in this algebra?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think $H$ may be the subalgebra whose elements have nontrivial trace entries. It's not difficult for me to find a base for $H$. But the Chevalley constraints turn out very hard to follow. So I think maybe there are some methods for finding Chevalley basis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am told that in any semisimple algebra, when a root system $\phi$ and its base $\Delta$ are given, $H$ is determined, so does its base relative to each simple root in $\Delta$. I am eager to know how does this determination goes, for example, in the linear Lie algebra of type $B_2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not sure whether there are mistakes in my expression as I am not from an English speaking country, but I really appreciate any attention and help. Many thanks.&lt;/p&gt;&#xA;" OwnerUserId="4736" LastActivityDate="2011-03-13T21:06:37.900" Title="How can I find a Chevalley basis of $B_2$ in the matrix realization of this group?" Tags="&lt;lie-algebra&gt;" AnswerCount="1" />
  <row Id="26770" PostTypeId="1" CreationDate="2011-03-13T20:01:10.110" Score="0" ViewCount="58" Body="&lt;p&gt;I'm using Godunov operator-splitting method to solve a system of non-linear equations. I'm wondered about the stability of the method. I use finite difference scheme to solve the PDE part but I cannot understand the relation between the time step $dt$ and spatial step $dx$.&lt;/p&gt;&#xA;" OwnerUserId="8204" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2011-03-14T01:02:39.413" LastActivityDate="2011-03-14T01:02:39.413" Title="Stability of Godunov method" Tags="&lt;differential-equations&gt;&lt;numerical-methods&gt;" />
  <row Id="26771" PostTypeId="1" CreationDate="2011-03-13T20:12:26.777" Score="26" ViewCount="496" Body="&lt;p&gt;We have $n$ boxes. And initially there are $x_1, x_2, x_3, \ldots, x_n$ marbles in each box. We randomly (with equal probabilities) select one of the boxes. We take one marble from it and we put it into another (different from the origin) box chosen randomly (with equal probabilities). We continue this process until one of the boxes become empty. How many operations we do on average?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is not a homework. I don't know whether a closed form solution exists. My current results are: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}{}&#xA;x_1 x_2 &amp;amp; \text{ for } n=2\\&#xA;\frac{3x_1 x_2 x_3}{x_1 + x_2 + x_3} &amp;amp; \text{ for } n=3&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have crossposted in &lt;a href=&quot;http://www.artofproblemsolving.com/Forum/viewtopic.php?f=498&amp;amp;t=396424&quot; rel=&quot;nofollow&quot;&gt;artofproblemsolving&lt;/a&gt;. This &lt;a href=&quot;http://mathoverflow.net/questions/41939/a-balls-and-colours-problem/53108&quot; rel=&quot;nofollow&quot;&gt;problem&lt;/a&gt; is related and maybe (or not) useful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Update2:&#xA;As i learned: this problem has been studied before. As usual :)&#xA;It seems very hard even for $n=4$. No explicit solution is known, only asymptotics for the case $f(x,x,x,x)$. Nevertheless the solution is much much more easier if we change slightly the problem. For &lt;a href=&quot;http://www.jstor.org/stable/27642669&quot; rel=&quot;nofollow&quot;&gt;example&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Big thanks to Viktor for pointing the reference!&lt;/p&gt;&#xA;" OwnerUserId="2481" LastEditorUserId="2481" LastEditorDisplayName="" LastEditDate="2011-03-16T11:54:38.420" LastActivityDate="2011-03-16T13:03:53.077" Title="What is the expected number of steps in the following process?" Tags="&lt;probability&gt;" AnswerCount="1" CommentCount="14" FavoriteCount="9" />
  <row Id="26772" PostTypeId="1" AcceptedAnswerId="26777" CreationDate="2011-03-13T20:13:14.333" Score="8" ViewCount="216" Body="&lt;p&gt;I heard an interesting question recently:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What is the minimum number of people required to make it more likely than not that all 365 possible birthdays are covered?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Monte Carlo simulation suggests 2287 ($\pm 1$, I think).  More generally, with $p$ people, what is the probability that for each of the 365 days of the year, there is at least one person in the group with that birthday?  (Yes, ignoring the leap-day.)&lt;/p&gt;&#xA;" OwnerUserId="72" LastEditorUserId="1543" LastEditorDisplayName="" LastEditDate="2011-04-16T18:32:32.513" LastActivityDate="2011-04-16T18:32:32.513" Title="Birthday-coverage problem" Tags="&lt;probability&gt;" AnswerCount="3" FavoriteCount="1" />
  <row Id="26773" PostTypeId="2" ParentId="26772" CreationDate="2011-03-13T20:19:01.490" Score="4" ViewCount="" Body="&lt;p&gt;This is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Coupon_collector%27s_problem&quot; rel=&quot;nofollow&quot;&gt;Coupon collector problem&lt;/a&gt;  The expected value to cover them all (not quite what you asked) is $365 \sum_{i=1}^{365}\frac{1}{i}\approx 365n \ln 365 + 365\gamma + \frac{1}{2}$&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-03-13T20:19:01.490" CommentCount="2" />
  <row Id="26774" PostTypeId="2" ParentId="26762" CreationDate="2011-03-13T20:19:08.503" Score="10" ViewCount="" Body="&lt;p&gt;$$\sum_{k=0}^n \left\{ n \atop k\right\} k = \varpi(n+1) - \varpi(n),$$&#xA;where $\varpi(n)$ is the $n$th &lt;a href=&quot;http://en.wikipedia.org/wiki/Bell_number&quot; rel=&quot;nofollow&quot;&gt;Bell number&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using generating functions, I prove this and the generalizations&#xA;$$\sum_{k=0}^n \left\{ n \atop k\right\} k^m = \sum_{i=0}^m \binom{m}{i} R(m-i) \varpi(n+i),$$&#xA;$$\sum_{k=0}^n \left\{ n \atop k\right\} (-1)^k k^m = \sum_{i=0}^m \binom{m}{i} \varpi(m-i) R(n+i),$$&#xA;where $R(n)$ is the $n$th &lt;a href=&quot;http://oeis.org/A000587&quot; rel=&quot;nofollow&quot;&gt;Rao-Uppuluri-Carpenter number&lt;/a&gt;, in the paper &quot;&lt;a href=&quot;http://math.pugetsound.edu/~mspivey/Recurrences.pdf&quot; rel=&quot;nofollow&quot;&gt;On Solutions to a General Combinatorial Recurrence&lt;/a&gt;.&quot;  See Identities 12 and 13, which are at the very end.  This paper has been submitted for publication, and I am still waiting on referee reports.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know whether these results are new, but I had not seen them before.  (They are not the main point of the paper.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;HR&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added&lt;/strong&gt;: Here are some additional derivations for $\sum_{k=0}^n \left\{ n \atop k\right\} k = \varpi(n+1) - \varpi(n)$.  They are shorter than the one needed in the paper for the more general result.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;First&lt;/strong&gt;: Use the recurrence for the Stirling numbers of the second kind.  (Due to OP - see comments.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{k=0}^n \left\{ n \atop k\right\} k = \sum_{k=0}^n \left\{ n+1 \atop k\right\} - \sum_{k=0}^n \left\{ n \atop k-1\right\} = \sum_{k=0}^{n+1} \left\{ n+1 \atop k\right\} - \sum_{k=0}^n \left\{ n \atop k\right\}  =\varpi(n+1) - \varpi(n).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Second&lt;/strong&gt;: Use &lt;a href=&quot;http://mathworld.wolfram.com/BellPolynomial.html&quot; rel=&quot;nofollow&quot;&gt;Bell polynomials&lt;/a&gt; $B_n(x)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is known that $\sum_{k=0}^n \left\{ n \atop k \right\} x^k = B_n(x)$ (Eq. 14 on the linked page), $\frac{d}{dx} B_n(x) = \frac{B_{n+1}(x)}{x} - B_n(x)$ (Eq. 16), and $B_n(1) = \varpi(n)$ (Eq. 1).  Thus&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{k=0}^n \left\{ n \atop k\right\} k = \frac{d}{dx} \left.\sum_{k=0}^n \left\{ n \atop k\right\} x^k \right|_{x=1} = \frac{d}{dx} \left. B_n(x) \right|_{x=1} = \frac{B_{n+1}(1)}{1} - B_n(1) = \varpi(n+1) - \varpi(n).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Third&lt;/strong&gt;: Use the double generating function for the Stirling numbers of the second kind (see, for example, &lt;em&gt;Concrete Mathematics&lt;/em&gt;, 2nd edition, p. 351)&#xA;$$\sum_{n,k \geq 0} \left\{ n \atop k\right\} w^k \frac{z^n}{n!} = e^{w(e^z-1)}.$$&#xA;(The right-hand side is actually the exponential generating function for the Bell polynomials, so this derivation is a variation on the second one.)&#xA;Differentiating both sides with respect to $w$ and then letting $w = 1$ yields the exponential generating function (egf) for the sum in question:&#xA;$$\sum_{n \geq 0} \left(\sum_{k=0}^n \left\{ n \atop k\right\} k \right) \frac{z^n}{n!} = e^{e^z-1} e^z - e^{e^z-1}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is known that $e^{e^z-1}$ is the egf for the Bell numbers.  Since $e^z$ is the egf of the infinite sequence of $1$'s, and multiplication of exponential generating functions corresponds to &lt;a href=&quot;http://math.stackexchange.com/questions/8713/how-this-operation-is-called/8714#8714&quot;&gt;binomial convolutions&lt;/a&gt; of the sequences in question, this means    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{k=0}^n \left\{ n \atop k\right\} k = \sum_{k=0}^n \binom{n}{k} \varpi(k) - \varpi(n).$$&#xA;Finally, $\sum_{k=0}^n \binom{n}{k} \varpi(k) = \varpi(n+1)$ is a &lt;a href=&quot;http://en.wikipedia.org/wiki/Bell_number#Properties_of_Bell_numbers&quot; rel=&quot;nofollow&quot;&gt;well-known identity for the Bell numbers&lt;/a&gt;, so we have&#xA;$$\sum_{k=0}^n \left\{ n \atop k\right\} k = \varpi(n+1) - \varpi(n).$$&lt;/p&gt;&#xA;" OwnerUserId="2370" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2011-04-13T22:29:49.360" LastActivityDate="2011-04-13T22:29:49.360" CommentCount="6" />
  <row Id="26775" PostTypeId="2" ParentId="26772" CreationDate="2011-03-13T20:31:32.930" Score="10" ViewCount="" Body="&lt;p&gt;As Ross also states, this can be framed as a coupon collector problem where birthdays correspond to coupons and individuals correspond to trials. Then, good bounds and a strong asymptotic result are both known for the time (number of trials) needed to collect at least one coupon of each unique type.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $T$ be the number of trials at which all coupons have been collected. Let there be $n$ coupons (so $n = 365$ in our case). For each $n$, and any $c &amp;gt; 0$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;\mathbb{P}(T &amp;gt; n \log n + c n) &amp;lt; e^{-c}&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and, also,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;\mathbb{P}(T &amp;lt; n \log n - c n) &amp;lt; e^{-c} \&amp;gt; .&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the first inequality we get that no more than &lt;strong&gt;2407&lt;/strong&gt; trials are needed so that there is greater than 50% chance all coupons are collected, and from the second we know that &lt;em&gt;at least&lt;/em&gt; &lt;strong&gt;1900&lt;/strong&gt; are needed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For more details, see the recent question and answer &lt;strong&gt;&lt;a href=&quot;http://stats.stackexchange.com/questions/7774/what-is-a-tight-lower-bound-on-the-coupon-collector-time&quot;&gt;here&lt;/a&gt;&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The following classical asymptotic result holds for $c \in \mathbb{R}$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;\mathbb{P}(T &amp;gt; n \log n + c n ) \to 1 - e^{-e^{-c}} .&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See, e.g., Motwani and Raghavan, &lt;em&gt;Randomized Algorithms&lt;/em&gt;, pp. 60--63 for a proof.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that if you plug in $c = -\log \log 2$ here, one gets &lt;strong&gt;2287.240&lt;/strong&gt; which matches very closely to Byron's exact answer and the Monte Carlo estimate reported in the question.&lt;/p&gt;&#xA;" OwnerUserId="7003" LastEditorUserId="7003" LastEditorDisplayName="" LastEditDate="2011-03-14T12:22:03.230" LastActivityDate="2011-03-14T12:22:03.230" CommentCount="1" />
  <row Id="26776" PostTypeId="2" ParentId="26705" CreationDate="2011-03-13T20:35:25.557" Score="1" ViewCount="" Body="&lt;p&gt;0 is the result of the addition of an element (x) in a set with its negation (-x). Hence, it is not necessary to conceive 0 as having a negative element since it would produce itself. Therefore , by occam's razor (i.e. simplicity clause) it is not necessary for 0 to have a negative element. However, by definition, the given set must have a negative element for all the positive elements. Therefore, it makes no sense to conceive it as a positive number. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, 0 is neither positive nor negative. That is intuitive since 0 is null, defines nullity which is the absence of some abstract object.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if one does not agree with the simplicity clause, he can admit it as being both a positive and a negative number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, as many things it is a matter of definition. &lt;/p&gt;&#xA;" OwnerUserId="8207" LastEditorUserId="8207" LastEditorDisplayName="" LastEditDate="2011-03-13T20:40:46.643" LastActivityDate="2011-03-13T20:40:46.643" />
  <row Id="26777" PostTypeId="2" ParentId="26772" CreationDate="2011-03-13T20:35:25.557" Score="8" ViewCount="" Body="&lt;p&gt;For the coupon collector's problem with $n$ objects, let $T$ be the number of trials needed to get a complete set. Then we have the formula $$P(T\leq k)=n^{-k}\ n!\ \left\lbrace {k\atop n}\right\rbrace.$$ Here the braces indicate Stirling numbers of the second kind.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With $n=365$, Maple gives me $P(T\leq 2286)=.4994$ while $P(T\leq 2287)=.5003$, so that $2287$ &lt;em&gt;is&lt;/em&gt; the smallest number to give a 50% chance to get all 365 birthdays.&lt;/p&gt;&#xA;" OwnerUserId="940" LastActivityDate="2011-03-13T20:35:25.557" CommentCount="10" />
  <row Id="26778" PostTypeId="1" AcceptedAnswerId="26783" CreationDate="2011-03-13T20:46:03.963" Score="1" ViewCount="61" Body="&lt;p&gt;Let $F\subset K$ be an algebraic field extension. Is the set of all elements of $K$ that are purely inseparable over $F$ necessarily a subfield of $K$?&lt;/p&gt;&#xA;" OwnerUserId="4341" LastEditorUserId="4341" LastEditorDisplayName="" LastEditDate="2011-03-13T20:51:19.497" LastActivityDate="2011-03-13T21:08:25.270" Title="Purely inseparable extension" Tags="&lt;abstract-algebra&gt;&lt;field-theory&gt;" AnswerCount="2" />
  <row Id="26779" PostTypeId="1" AcceptedAnswerId="26786" CreationDate="2011-03-13T20:51:20.643" Score="4" ViewCount="102" Body="&lt;p&gt;I'm confronted with this question:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Let $V$ be an inner product space and $B=\{u_{1}, ..., u_{n}\}$ a basis of $V$.  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Suppose there exists $\lambda_{1},...,\lambda_{n} \in F (=R \text{ or } C)$ such that: $$||\sum_{k=1}^{n}\lambda_{k}u_{k}||^2=\sum_{k=1}^{n}|\lambda_{k}|^2$$&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Prove or disprove: $B$ is an orthonormal basis.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Not sure where to begin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried finding a counter example in the case of $V=R^2$ using the identity: $$||u+v||^2=||u||^2+2Re\langle u, v\rangle+||v||^2$$ which gave me this result: $$||\lambda_{1}u_{1}+\lambda_{2}u_{2}||^2=|\lambda_{1}|||u_{1}||^2+2Re\langle\lambda_{1}u_{1},\lambda_{2}u_{2}\rangle+|\lambda_{2}|||u_{2}||^2=|\lambda_{1}|^2+|\lambda_{2}|^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I can't see how this helps me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;edit:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a little follow up question if anyone is interested: suppose the equality above is true for all $\lambda_{1},...,\lambda_{n}$, does that imply the basis is orthonormal?&lt;/p&gt;&#xA;" OwnerUserId="3237" LastEditorUserId="3237" LastEditorDisplayName="" LastEditDate="2011-03-14T12:45:16.000" LastActivityDate="2011-03-14T12:45:16.000" Title="Regarding orthonormal basis" Tags="&lt;linear-algebra&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="26780" PostTypeId="2" ParentId="26769" CreationDate="2011-03-13T21:06:37.900" Score="1" ViewCount="" Body="&lt;p&gt;Probably what you mean is that a possible Cartan subalgebra is the subspace $\mathfrak h$ of diagonal matrices of that form. That is true.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For each $h\in\mathfrak h$ you can find the map $\operatorname{a}(h):\mathfrak g\to\mathfrak g$. The root spaces and the roots can be found by looking for simultaneous eigenvectors for the maps $\operatorname{ad}(h)$, $h\in\mathfrak h$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you get this far?&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2011-03-13T21:06:37.900" CommentCount="1" />
  <row Id="26781" PostTypeId="1" AcceptedAnswerId="26784" CreationDate="2011-03-13T21:07:05.670" Score="4" ViewCount="188" Body="&lt;p&gt;Hey there!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Q: If we consider $\mathbb{R}$ and $\mathbb{C}$ as $\mathbb{Q}$-vector spaces, then how can we show they are isomorphic?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that if two vector spaces have bases with the same cardinality, then they are isomorphic.  Also, Zorn's lemma tells us that every vector space has a basis. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case, answering my question amounts to showing that any bases of  $\mathbb{R}$ and $\mathbb{C}$ over $\mathbb{Q}$ have the same cardinality. In other words, I need to show dim$ \mathbb{R} =$ dim $\mathbb{C}$ over $\mathbb{Q}$, that is, they have bases with the same cardinality.  Can anyone help?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you!!&lt;/p&gt;&#xA;" OwnerUserId="8209" LastEditorUserId="6296" LastEditorDisplayName="" LastEditDate="2011-03-13T22:18:36.217" LastActivityDate="2011-03-14T02:45:04.400" Title="$\mathbb{R}$ and $\mathbb{C}$ as $\mathbb{Q}$ vector spaces" Tags="&lt;linear-algebra&gt;&lt;abstract-algebra&gt;&lt;algebra&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="2" />
  <row Id="26782" PostTypeId="2" ParentId="26778" CreationDate="2011-03-13T21:07:47.843" Score="4" ViewCount="" Body="&lt;p&gt;Yes. This is clearer if you use the following definition of pure inseparability: an element $\alpha$ of an algebraic extension $F/K$ is purely inseparable if $|\text{Hom}_K(K(\alpha), \bar{K})| = 1$. Now it is obvious that if $\alpha, \beta$ are purely inseparable then any $K$-homomorphism $K(\alpha, \beta) \to \bar{K}$ is determined by what it does to $\alpha$ and $\beta$, so is unique. &lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2011-03-13T21:07:47.843" CommentCount="1" />
  <row Id="26783" PostTypeId="2" ParentId="26778" CreationDate="2011-03-13T21:08:25.270" Score="5" ViewCount="" Body="&lt;p&gt;Yes. An element $a\in K$ is purely inseparable iff $a^{p^n}\in F$ for some $n\geq 0$ (see &lt;a href=&quot;http://en.wikipedia.org/wiki/Separable_extension#Purely_inseparable_extensions&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;). Let $E$ be the subset of purely inseparable elements of $K$. For any $a,b\in E$, we have $a^{p^n}\in F$ and $b^{p^m}\in F$ for some $n,m\geq0$, so&#xA;$$(-a)^{p^n}\in F\hskip 0.5in (a^{-1})^{p^n}=1/(a^{p^n})\in F$$&#xA;$$(a+b)^{p^{n+m}}=a^{p^{n+m}}+b^{p^{n+m}}=(a^{p^n})^{p^m}+(b^{p^m})^{p^n}\in F$$&#xA;and&#xA;$$(ab)^{p^{n+m}}=a^{p^{n+m}}b^{p^{n+m}}=(a^{p^n})^{p^m}(b^{p^m})^{p^n}\in F$$&#xA;Thus, $E$ is a subfield.  &lt;/p&gt;&#xA;" OwnerUserId="264" LastActivityDate="2011-03-13T21:08:25.270" />
  <row Id="26784" PostTypeId="2" ParentId="26781" CreationDate="2011-03-13T21:11:04.963" Score="6" ViewCount="" Body="&lt;p&gt;If $k$ is a field and $V$ is an infinite dimensional $k$-vector space, then $V\cong V\oplus V$. Using this, what you want to show follows from the fact that  as $\mathbb Q$-vector spaces, $\mathbb C\cong \mathbb R\oplus \mathbb R$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you see how to prove those two claims?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Later.&lt;/strong&gt; Ok, apparently not. Let's do it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;(1)&lt;/strong&gt; If $B$ is a basis for $V$, then the set $$B&amp;#39;=\{(b,0):b\in B\}\cup \{(0,b):b\in B\}$$ is a basis for $V\oplus V$. There is an obvious bijection $B&amp;#39;\cong \{1,2\}\times B$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if $X$ is an infinite set, then $X$ and $\{1,2\}\times X$ are in bijection. It follows from this that there is a bijection between the basis $B$ of $V$ and the basis $B&amp;#39;$ of $V\oplus V$. As you know, this implies that there is a linear isomorphism between $V$ and $V\oplus V$. This proves my first claim above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;(2)&lt;/strong&gt; Consider the map $$\phi:a+bi\in\mathbb C\mapsto (a,b)\in\mathbb R\oplus\mathbb R.$$ It is very easy to show that it is an isomorphism of $\mathbb Q$-vector spaces, so that $\mathbb C\cong\mathbb R\oplus\mathbb R$, as my second claim states.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;(3)&lt;/strong&gt; Finally, let's prove your claim that if $\mathbb R$ and $\mathbb C$ are isomorphic $\mathbb Q$-vector spaces: since $\mathbb R$ is a $\mathbb Q$-vector space of infinite dimensions, my first claim tells us that $\mathbb R\cong\mathbb R\oplus\mathbb R$ as $\mathbb Q$-vector spaces. On the other hand, my second claim tells us that $\mathbb R\oplus\mathbb R\cong\mathbb C$ as $\mathbb Q$-vector spaces. Transitivity, then, allows us to conclude that $\mathbb R\cong\mathbb C$ as $\mathbb Q$-vector spaces.&lt;/p&gt;&#xA;" OwnerUserId="274" LastEditorUserId="274" LastEditorDisplayName="" LastEditDate="2011-03-14T02:45:04.400" LastActivityDate="2011-03-14T02:45:04.400" CommentCount="6" />
  <row Id="26785" PostTypeId="1" AcceptedAnswerId="26787" CreationDate="2011-03-13T21:15:51.613" Score="2" ViewCount="82" Body="&lt;p&gt;I was working through Pugh's &lt;em&gt;Real Mathematical Analysis&lt;/em&gt; when I came across this proof:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Theorem&lt;/strong&gt; $\mathbb{N} \times \mathbb{N}$ is denumerable.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;Proof&lt;/strong&gt; Think of $\mathbb{N} \times \mathbb{N}$ as an $\infty \times \infty$ matrix and walk along the successive counter-diagonals.  This gives a list&#xA;  $$(1, 1), (2, 1), (1, 2), (3, 1), (2, 2), (1, 3), (4, 1) (3, 2) (2, 3) (1, 4) (5, 1)...$$&#xA;  of $\mathbb{N} \times \mathbb{N}$ and proves that $\mathbb{N} \times \mathbb{N}$ is denumerable. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The proof is accompanied by a visual diagram showing successive diagonals crossed out where for each $a_{ij} \in \mathbb{N} \times \mathbb{N}$, $i + j = d$, $d$ fixed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now from the diagram, it is visually easy to see that the graph of the function given is surjective.  And I understand &quot;intuitively&quot; why the graph is injective; we can just list the ordered pairs out as &quot;1, 2, 3...&quot;, where each one is different.  However, it was quite frustrating to explicitly come up with a function and check it was injective to my satisfaction.  The book gives hints in the exercises so you can eventually come up with the function $2^{i}(2j + 1)$, which I verified was injective.   Since primes seem to be &quot;fundamental&quot; when discussing the natural numbers, I thought all Cantor pairing functions might be of the form $p^{i}(pj + k)$ where $0 &amp;lt; k &amp;lt; p$, and $p$ is prime.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Looking it up, however, Wikipedia informs me that this is not so as,&#xA;$$n = f(i, j) = \dfrac{(i + j)(i + j + 1)}{2} + j$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is a bijection from $\mathbb{N} \times \mathbb{N} \rightarrow \mathbb{N}$, which I again verified.&#xA;So my question is, what do these Cantor pairing functions generally look like?  Is there a sufficiency criterion?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;: Zev's answer below convinces me that the general question I asked above is very uninteresting.  Can something more interesting be said if one requires the function to be continuous on $\mathbb{R^2}$, and still send $\mathbb{N} \times \mathbb{N} \rightarrow \mathbb{N}$?   I think that's was what was intuitively coming to my mind when I was thinking about &quot;an explicit form.&quot;&lt;/p&gt;&#xA;" OwnerUserId="6296" LastEditorUserId="6296" LastEditorDisplayName="" LastEditDate="2011-03-13T22:05:38.117" LastActivityDate="2011-03-13T22:05:38.117" Title="Is there a sufficiency criterion for Cantor pairing functions?" Tags="&lt;real-analysis&gt;&lt;elementary-set-theory&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="26786" PostTypeId="2" ParentId="26779" CreationDate="2011-03-13T21:21:00.090" Score="1" ViewCount="" Body="&lt;p&gt;It is false. A simple example would be to take $\lambda_i = 0$, $\forall i$ or for $i$'s which are not orthogonal to the rest.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you assume $\lambda_i \neq 0$, $\forall i$, then here is a counter example in $2D$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $u_1 = 2e_1 + e_2$ and $u_2 = e_1 + 2e_2$, where $e_1, e_2$ form the conventional orthonormal basis. Clearly this forms a basis. Let $\lambda_i \in \mathbb{R}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then $||\lambda_1 u_1 + \lambda_2 u_2 ||_2^2 = |\lambda_1|^2 + |\lambda_2|^2$ implies $(2 \lambda_1 + \lambda_2)^2 + (\lambda_1 + 2 \lambda_2)^2 = \lambda_1^2 + \lambda_2^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, $(4 \lambda_1^2 + 4 \lambda_1 \lambda_2 + \lambda_2^2) + (4 \lambda_2^2 + 4 \lambda_1 \lambda_2 + \lambda_1^2) = \lambda_1^2 + \lambda_2^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, $(4 \lambda_1^2 + 4 \lambda_1 \lambda_2) + (4 \lambda_2^2 + 4 \lambda_1 \lambda_2) = 0 \Rightarrow (\lambda_1 + \lambda_2)^2 = 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, if we choose $\lambda_1 = - \lambda_2$ we are done.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, if $u_1 = 2e_1 + e_2$ and $u_2 = e_1 + 2e_2$ and if $\lambda_1 = - \lambda_2 \in \mathbb{R}$, then we have&#xA;$$||\lambda_1 u_1 + \lambda_2 u_2||_2^2 = \lambda_1^2 + \lambda_2^2$$&lt;/p&gt;&#xA;" OwnerUserId="3302" LastActivityDate="2011-03-13T21:21:00.090" CommentCount="0" />
  <row Id="26787" PostTypeId="2" ParentId="26785" CreationDate="2011-03-13T21:23:05.927" Score="5" ViewCount="" Body="&lt;p&gt;I don't see how there could possibly any sort of &quot;characterization&quot; - the bijections from $\mathbb{N}\times\mathbb{N}$ to $\mathbb{N}$ are just the bijections from $\mathbb{N}\times\mathbb{N}$ to $\mathbb{N}$, whether or not there's a &quot;formula&quot; for them is irrelevant. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or, if you like, take any bijection $f:\mathbb{N}\times\mathbb{N}\rightarrow\mathbb{N}$ that is given by a &quot;formula&quot;, and compose it with any permutation $\sigma:\mathbb{N}\rightarrow\mathbb{N}$. That's a new bijection from $\mathbb{N}\times\mathbb{N}$ to $\mathbb{N}$ (in fact they can all be gotten this way), and there's no reason why $\sigma$ should preserve the form of the &quot;formula&quot; for $f$, it could do anything.&lt;/p&gt;&#xA;" OwnerUserId="264" LastEditorUserId="264" LastEditorDisplayName="" LastEditDate="2011-03-13T21:28:19.980" LastActivityDate="2011-03-13T21:28:19.980" CommentCount="1" />
  <row Id="26788" PostTypeId="2" ParentId="26779" CreationDate="2011-03-13T21:23:54.493" Score="1" ViewCount="" Body="&lt;p&gt;First of all, $\lambda_1$ and $\lambda_2$ should be squared in the middle expression of the last equality in your question.&#xA;Next, for the question to be nontrivial, an asumption like $\lambda_k\ne0$ for all $k$ must be added. Otherwise, a counterexample is given by taking all $\lambda_k=0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer to the question is that the system need not be an orthonormal basis. In $\mathbb{R}^2$ take $u_1=(1,0)$, $\lambda_1=1$. Then, for any $u_2\in\mathbb{R}^2$ with $\langle u_1,u_2\rangle\ne0$, let $\lambda_2=2\langle u_1,u_2\rangle/(1-\|u_2\|^2)$. Then $\|\lambda_1u_1+\lambda_2u_2\|^2=|\lambda_1|^2+|\lambda_2|^2$ but $u_1$ and $u_2$ are not orthogonal. They may be even linearly dependent. The example may be generalized to $\mathbb{R}^n$.&lt;/p&gt;&#xA;" OwnerUserId="4791" LastActivityDate="2011-03-13T21:23:54.493" />
  <row Id="26789" PostTypeId="1" CreationDate="2011-03-13T21:35:02.840" Score="2" ViewCount="86" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Prove the following statement: if $\frac{p}{q} (p \in \mathbb{N}, q \in \mathbb{N})$ is a rational number that corresponds to an infinite periodic decimal fraction $\alpha$, then the rational number $10^{k} \frac{p}{q} (k \in \mathbb{N})$ corresponds to an infinite periodic decimal fraction obtained from the decimal fraction $\alpha$ by moving the decimal separator (period) by $k$ places to the right.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here is my proof (not sure about its correctness):&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;As the problem states, $\frac{p}{q} = \alpha = a_1a_2\ldots a_n.a_{n+1}a_{n+2}\ldots a_{n+k}a_{n+k+1}\ldots$ (i.e. $a_1a_2\ldots a_n$ - integer part, $a_{n+1}a_{n+2}\ldots a_{n+k}\ldots$ - fractional part), $a_i$ - decimal digit for $\forall i \in \mathbb{N}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\begin{align*}&#xD;&#xA;10^{k} \frac{p}{q} &#xD;&#xA;&amp;amp;= 10^k\alpha = 10^k(a_1a_2\ldots a_n + \frac{a_{n+1}}{10} + \frac{a_{n+2}}{10^2} + \ldots + \frac{a_{n+k}}{10^k} + \frac{a_{n+k+1}}{10^{k+1}} + \ldots) \\&#xD;&#xA;&amp;amp;= a_1a_2\ldots a_n\underbrace{00\ldots 0}_k + a_{n+1}\underbrace{00\ldots 0}_{k-1} + a_{n+2}\underbrace{00\ldots 0}_{k-2} + \ldots \\&#xD;&#xA;&amp;amp;\ + a_{n+k} + \frac{a_{n+k+1}}{10} + \frac{a_{n+k+2}}{10^2} + \ldots \\&#xD;&#xA;&amp;amp;= a_1a_2\ldots a_{n+k-1}a_{n+k}.a_{n+k+1}a_{n+k+2}\ldots&#xD;&#xA;\end{align*}$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The question is: Is it correct to represent an infinite decimal fraction as a sum of infinite number of rational numbers (i.e. is it always true that, for example, $0.a_1a_2\ldots a_na_{n+1}\ldots = \frac{a_1}{10} + \frac{a_2}{10^2} + \ldots + \frac{a_n}{10^n} + \frac{a_{n+1}}{10^{n+1}} + \ldots$?)&lt;/p&gt;&#xA;" OwnerUserId="6286" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-13T21:45:17.133" LastActivityDate="2011-03-13T22:01:23.017" Title="Is this a correct proof of a fact about rational numbers?" Tags="&lt;homework&gt;&lt;fractions&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="26790" PostTypeId="2" ParentId="26781" CreationDate="2011-03-13T21:44:06.403" Score="3" ViewCount="" Body="&lt;p&gt;Two ways: one using only the fact that $|\mathbb{R}|=|\mathbb{C}|=\mathfrak{c}$; the other using the fact that we can identify the additive structure of $\mathbb{C}$ with the plane.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;If $\mathbf{F}$ is a field and $\mathbf{V}$ is a vector space over $\mathbf{V}$, what is the cardinality of $\mathbf{V}$? If $\beta$ is a basis for $\mathbf{V}$, then every vector of $\mathbf{V}$ can be written uniquely as an $\mathbf{F}$-linear combination of vectors in $\beta$. Therefore, there is a bijection between the elements of $\mathbf{V}$ and the set&#xA;$$\bigl\{ f\colon\beta\to \mathbf{F}\bigm| f(\mathbf{b})=\mathbf{0}\text{ for almost all }\mathbf{b}\in\beta\bigr\}.$$&#xA;That is, the set of functions of finite support from $\beta$ to $\mathbf{F}$, $\mathbf{F}^{(\beta)}$. If $\mathbf{F}$ is infinite, then this cardinality is equal to $|\mathbf{F}||\beta|=\max\{|\mathbf{F}|,|\beta|\}$. Here, $\mathbf{F}=\mathbb{Q}$. So if $\beta$ is a basis for $\mathbb{R}$ over $\mathbb{Q}$, we need $|\mathbb{R}| = \aleph_0|\beta|$. What is $|\beta|$? For $\mathbb{C}$, we need $|\mathbb{C}| = \aleph_0|\gamma|$, where $\gamma$ is a basis for $\mathbb{C}$ over $\mathbb{Q}$. What is $|\gamma|$?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Since the additive structure of $\mathbb{C}$ is just the same as the additive structure of $\mathbb{R}\oplus\mathbb{R}$, then they are isomorphic as $\mathbb{Q}$-vector spaces. So if $\{\mathbf{v}_{b}\}_{b\in\beta}$ is a basis for $\mathbb{R}$ over $\mathbb{Q}$, then &#xA;$$\bigl\{\mathbf{v}_{b}\bigr\}_{b\in\beta}\cup \bigl\{i\mathbf{v}_{b}\bigr\}_{b\in\beta}$$&#xA;is a basis for $\mathbb{C}$ over $\mathbb{Q}$ (prove it). So $\dim_{\mathbb{Q}}(\mathbb{C}) = 2\dim_{\mathbb{Q}}(\mathbb{R})$. If the dimension of $\mathbb{R}$ over $\mathbb{Q}$ were finite, this would show they are &lt;em&gt;not&lt;/em&gt; isomorphic. But is the dimension finite or infinite? And what does that tell you about the dimensions?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-13T22:01:48.947" LastActivityDate="2011-03-13T22:01:48.947" CommentCount="7" />
  <row Id="26791" PostTypeId="1" AcceptedAnswerId="26794" CreationDate="2011-03-13T21:45:08.097" Score="2" ViewCount="41" Body="&lt;p&gt;Lets assume that we have a element, which can have value from 1 till n. (let's set it on 20 to make it easier)&#xA;And we have the Set, that consists of object, which consists of three elements $\langle e_1, e_2, e_3 \rangle$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have also one rule regarding to objects in the set: $e_1 \geq e_2 \geq e_3$&#xA;- example of good objects: $\langle n, n, n\rangle$, $\langle n, n-1, n-1\rangle$, $\langle 20, 19, 18\rangle$, $\langle 3, 2, 1\rangle$, $\langle 3, 3, 3\rangle$, $\langle 3, 2, 2\rangle$.&lt;br&gt;&#xA;- example of bad objects: $\langle n, n+1, n\rangle$, $\langle 2, 3, 2\rangle$, $\langle 3, 2, 4\rangle$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the question:&#xA;How to count the amount of all good objects, which pass to this Set (they don't violate the rule ) ?&#xA;Can you give me any hints?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can solve this with brute force method. But probably there is a short way.&lt;/p&gt;&#xA;" OwnerUserId="8211" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-13T22:03:48.273" LastActivityDate="2011-03-13T22:10:12.183" Title="Is it Variation? Counting elements" Tags="&lt;combinatorics&gt;" AnswerCount="3" />
  <row Id="26792" PostTypeId="1" AcceptedAnswerId="26823" CreationDate="2011-03-13T21:56:47.993" Score="3" ViewCount="82" Body="&lt;p&gt;(not sure if &quot;decreasing-difference&quot; is the right way to put it - please edit away if you know the proper technical term)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From my analysis homework:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Does there exist a bounded sequence&#xA;  $\{x_n\}_{n \in \mathbb{N}}$ such that&#xA;  $x_{n+1}-x_n \to 0$, but the sequence&#xA;  $(\sum_{i=1}^n x_i)/n$ does not have a&#xA;  limit? Give an example or show none&#xA;  exist.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In a previous part, I showed that there exists a bounded sequence $\{x_n\}_{n \in \mathbb{N}}$ such that $\left| x_{n+1}-x_n \right| \to 0$, yet $x_n$ does not converge, using a rearrangement of the alternating harmonic series. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Intuitively, I believe none exist. I've tried showing that the sequence is Cauchy, but that does not seem to help:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left| \frac{ \sum_{i=1}^m x_i}{m} - \frac{ \sum_{i=1}^n x_i}{n} \right| = \left| \frac{ (n-m) \sum_{i=1}^m x_i - m \sum_{i=m+1}^n x_i}{mn} \right| $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I feel like I can't use my normal series tools because I am looking at an average, and the series does not necessarily converge.&lt;/p&gt;&#xA;" OwnerUserId="541" LastActivityDate="2011-03-14T01:28:08.377" Title="The average of a bounded, decreasing-difference sequence" Tags="&lt;homework&gt;&lt;real-analysis&gt;&lt;analysis&gt;&lt;sequences-and-series&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" />
  <row Id="26794" PostTypeId="2" ParentId="26791" CreationDate="2011-03-13T22:00:44.890" Score="1" ViewCount="" Body="&lt;p&gt;If the first number is $k$, and the second number is $j$, where $j \leq k$ then the last number has $j$ choices.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the number of favorable cases is $$\sum_{k=1}^n \sum_{j=1}^k  j = \sum_{k=1}^n \frac{k(k+1)}{2} = \frac{n(n+1)(n+2)}{6}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, if you have elements from $1$ to $n$ and want to choose an $m$ element set with the ordering you want the answer is $$\binom{n+m-1}{m}$$ which can be seen by induction on $m$ or by a combinatorial argument as proved &lt;a href=&quot;http://math.stackexchange.com/questions/20475/proving-n-choose-r-n-1-choose-r-1n-1-choose-r-when-1-leq-r/20478#20478&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2011-03-13T22:06:44.993" LastActivityDate="2011-03-13T22:06:44.993" />
  <row Id="26795" PostTypeId="2" ParentId="26789" CreationDate="2011-03-13T22:01:23.017" Score="1" ViewCount="" Body="&lt;p&gt;This really depends on what level we are talking about.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Formally, the &lt;em&gt;definition&lt;/em&gt; of the decimal expansion is that it is equal to the limit of the infinite sum/series. That is, &lt;em&gt;by definition&lt;/em&gt;, when we write&#xA;$$0.a_1a_2a_3\cdots a_n\cdots$$&#xA;we &lt;em&gt;really&lt;/em&gt; mean&#xA;$$\sum_{i=1}^{\infty}\frac{a_i}{10^i},$$&#xA;by which we &lt;em&gt;really&lt;/em&gt; mean&#xA;$$\lim_{n\to\infty}\left(\sum_{i=1}^n\frac{a_i}{10^i}\right).$$&#xA;So not only is it correct, it's what it &lt;em&gt;is&lt;/em&gt;. But this requires you to know what limits and series are.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you do, that gives you a way of writing out what you did a bit clearer. If you have&#xA;$$\frac{p}{q} = \sum_{i=1}^{\infty}\frac{a_i}{10^i},$$&#xA;then&#xA;\begin{align*}&#xA;10^k\frac{p}{q} &amp;amp;= 10^k\sum_{i=1}^{\infty}\frac{a_i}{10^i}\\&#xA;&amp;amp;= 10^k\lim_{n\to\infty}\left(\sum_{i=1}^{n}\frac{a_i}{10^i}\right)\\&#xA;&amp;amp;= \lim_{n\to\infty}\left(\sum_{i=1}^n\frac{10^ka_i}{10^i}\right)\\&#xA;&amp;amp;= \lim_{n\to\infty}\left(10^{k-1}a_1 + 10^{k-2}a_2+\cdots+a_k + \sum_{i=k+1}^{n}\frac{a_i}{10^{i-k-1}}\right)\\&#xA;&amp;amp;= 10^{k-1}a_1 + 10^{k-2}a_2 + \cdots + a_k + \lim_{n\to\infty}\sum_{i=k+1}^n\frac{a_i}{10^{i-k-1}}\\&#xA;&amp;amp;= 10^{k-1}a_1 + 10^{k-2}a_2+\cdots + a_k + \lim_{n\to\infty}\sum_{j=1}^{n-k-1}\frac{a_{j+k}}{10^{j}}\\&#xA;&amp;amp;= 10^{k-1}a_1+10^{k-2}a_2+\cdots+a_k + \sum_{j=1}^{\infty}\frac{a_{j+k}}{10^{j}}.&#xA;\end{align*}&#xA;which when translated back into decimal notation is&#xA;$$a_1a_2\cdots a_k.a_{k+1}a_{k+2}\cdots$$&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2011-03-13T22:01:23.017" CommentCount="3" />
  <row Id="26796" PostTypeId="2" ParentId="26791" CreationDate="2011-03-13T22:01:41.083" Score="1" ViewCount="" Body="&lt;p&gt;There are many ways.  One is first to ask how many 2-element things there are.  If $e_1=i$, there are $n+1-i$ of them.  So the total number of 2-element things is $$\sum_{i=1}^n(n+1-i)=n^2+n-\frac{n(n+1)}{2}=\frac{n(n+1)}{2}$$&lt;br&gt;&#xA;Then the number of 3-element things is $$\sum_{i=1}^n\frac{i(i+1)}{2}$$&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-03-13T22:01:41.083" />
  <row Id="26797" PostTypeId="2" ParentId="26791" CreationDate="2011-03-13T22:10:12.183" Score="1" ViewCount="" Body="&lt;p&gt;The simplest way is to simply pick three objects from your set allowing repetitions. Then you'll just &lt;em&gt;list them&lt;/em&gt; in nonincreasing order. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is just a count of combinations with repetitions. You have $n$ possibilities, and you want to pick $r$ elements, then the number of ways of doing it is $\binom{n+r-1}{r}$. See for example &lt;a href=&quot;http://en.wikipedia.org/wiki/Stars_and_bars_%28probability%29&quot; rel=&quot;nofollow&quot;&gt;Wikipedia's page on the stars and bars problem&lt;/a&gt; for a proof.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your example, $n=20$, $r=3$, so you would want&#xA;$$\binom{20+3-1}{3} = \binom{22}{3} = 1540.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This agrees with Ross's answer:&#xA;$$\sum_{i=1}^{20}\frac{i(i+1)}{2} = 1540.$$&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2011-03-13T22:10:12.183" />
  <row Id="26798" PostTypeId="2" ParentId="26735" CreationDate="2011-03-13T22:15:26.360" Score="1" ViewCount="" Body="&lt;p&gt;Yes, $Z$ is a proper martingale. However, $\int_0^T(Z_sW_s)^2\,ds$ is &lt;em&gt;not&lt;/em&gt; integrable for large $T$. As the quadratic variation of $Z$ is $[Z]_t=4\int_0^t(Z_sW_s)^2\,ds$, Ito's isometry says that this is integrable if and only if $Z$ is a square-integrable martingale, and you can show that $Z$ is not square integrable at large times (see below).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, it is &lt;em&gt;conditionally&lt;/em&gt; square integrable over small time intervals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;\begin{align}&#xD;&#xA;\mathbb{E}\left[Z_t^2W_t^2\;\Big\vert\;\mathcal{F}_s\right]&amp;amp;\le\mathbb{E}\left[W_t^2\exp(W_t^2)\;\Big\vert\;\mathcal{F}_s\right]\\&#xD;&#xA;&amp;amp;=\frac{1}{\sqrt{2\pi(t-s)}}\int x^2\exp\left(x^2-\frac{(x-W_s)^2}{2(t-s)}\right)\,dx&#xD;&#xA;\end{align}&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's a bit messy, but you can evaluate this integral and check that it is finite for $s \le t &amp;lt; s+\frac12$. In fact, integrating over the range $[s,s+h]$ (any $h &amp;lt; 1/2$) with respect to $t$ is finite. So, conditional on $W_s$, you can say that $Z$ is a square integrable martingale over $[s,s+h]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is enough to conclude that $Z$ is a proper martingale. We have $\mathbb{E}[Z_t\vert\mathcal{F}_s]=Z_s$ (almost surely) for any $s \le t &amp;lt; s+\frac12$. By induction, using the tower rule for conditional expectations, this extends to all $s &amp;lt; t$. Then, $\mathbb{E}[Z_t]=\mathbb{E}[Z_0] &amp;lt; \infty$, so $Z$ is integrable and the martingale conditions are met.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I mentioned above that the suggested method in the question cannot work because $Z$ is not square integrable. I'll elaborate on that now. If you write out the expected value of an expression of the form $\exp(aX^2+bX+c)$ (for $X$ normal) as an integral, it can be seen that it becomes infinite exactly when $a{\rm Var}(X)\ge1/2$ (because the integrand is bounded away from zero at either plus or minus infinity). Let's apply this to the given expession for $Z$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The expression for $Z$ can be made more manageable by breaking the exponent into independent normals. Fixing a positive time $t$, then $B_s=\frac{s}{t}W_t-W_s$ is a Brownian bridge independent of $W_t$. Rearrange the expression for $Z$&#xA;$$&#xD;&#xA;\begin{align}&#xD;&#xA;Z_t&amp;amp;=\exp\left(W_t^2-\int_0^t(2(\frac{s}{t}W_t+B_s)^2+1)\,ds\right)\\&#xD;&#xA;&amp;amp;=\exp\left(W_t^2-2\int_0^t\frac{s^2}{t^2}W_t\,ds+\cdots\right)\\&#xD;&#xA;&amp;amp;=\exp\left((1-2t/3)W_t^2+\cdots\right)&#xD;&#xA;\end{align}&#xD;&#xA;$$&#xA;where '$\cdots$' refers to terms which are at most linear in $W_t$. Then, for any $p &amp;gt; 0$,&#xA;$$&#xD;&#xA;Z_t^p=\exp\left(p(1-2t/3)W_t^2+\cdots\right).&#xD;&#xA;$$&#xA;The expectation $\mathbb{E}[Z_t^p\mid B]$ of $Z_t^p$ conditional on $B$ is infinite whenever&#xA;$$&#xD;&#xA;p(1-2t/3){\rm Var}(W_t)=p(1-2t/3)t \ge \frac12.&#xD;&#xA;$$&#xA;The left hand side of this inequality is maximized at $t=\frac34$, where it takes the value $3p/8$. So, $\mathbb{E}[Z_{3/4}^p\mid B]=\infty$ for all $p\ge\frac43$. The expected value of this must then be infinite, so $\mathbb{E}[Z^p_{3/4}]=\infty$. It is a standard application of Jensen's inequality that $\mathbb{E}[\vert Z_t\vert^p]$ is increasing in time for any $p\ge1$ and martingale $Z$. So, $\mathbb{E}[Z_t^p]=\infty$ for all $p\ge 4/3$ and $t\ge3/4$. In particular, taking $p=2$ shows that $Z$ is not square integrable.&lt;/p&gt;&#xA;" OwnerUserId="1321" LastEditorUserId="1321" LastEditorDisplayName="" LastEditDate="2011-03-19T23:43:01.243" LastActivityDate="2011-03-19T23:43:01.243" CommentCount="2" />
  <row Id="26800" PostTypeId="2" ParentId="23776" CreationDate="2011-03-13T22:19:19.260" Score="2" ViewCount="" Body="&lt;p&gt;Gambling is a good start-point for probability. We can treat $\sigma$-field is a structure of event as we need to define the addition and multiplication for numbers. The completeness of real number is suitable for our interesting calculations; and $\sigma$-filed plays the same roles. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope the following gambling example would help you to understand the filtration and conditional expectation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming that two person, say player A and player B, bet on the results of two consequences of coin tossing. &#xA;H: head  T: tail&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the time $0$, A and B do not know anything about the result except one of the event in $\Omega=\{HH,HT,TH,HH\}$ would happen. Hence the information at time $0$ that they both know is $\mathcal{F}_0=\{\emptyset,\Omega\}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the time $1$, the coin had been tossed only once; and they know that the events in the $\sigma$-field $\mathcal{F}_1=\{\emptyset, \Omega, \{HH,HT\},\{TH,TT\}\}\supset \mathcal{F}_0 $ could happen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the time $2$, the coin had been tossed twice; and they know that the events in the $\sigma$-field $\mathcal{F}_2=\{\emptyset, \Omega,\{HH\},\{HT\},\{TH\},\{TT\}\}\supset \mathcal{F}_1$ could happen which means they know everything about the gambling results.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please notice the evolution of information characterized by filtration $\mathcal{F}_0,\mathcal{F}_1,\mathcal{F}_2.$ With time passing, the unknown world $\Omega$ is divided much finer.  It is something like water flows through a pipe. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming that they bet on the following results and the coin is fair.&#xA;$$X(\omega)=\left\{ \begin{array}{l}&#xD;&#xA;2, \omega=HH,\mbox{means the first tossing is H and the second tossing is H}\\&#xD;&#xA;1, \omega=HT,\mbox{means the first tossing is H and the second tossing is T}\\&#xD;&#xA;1, \omega=TH,\mbox{means the first tossing is T and the second tossing is H} \\&#xD;&#xA;0, \omega=TT,\mbox{means the first tossing is T and the second tossing is T}\\&#xD;&#xA;\end{array} \right.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$E[X|\mathcal{F}_0]=1 $$&#xA;$$E[X|\mathcal{F_2}]=X(\omega) $$&#xA;$$E[X|\{HH,HT\}]=2P(HH|\{HH,HT\})+1P(HT|\{HH,HT\})$$&#xA;$$+1P(TH|\{HH,HT\})+0P(TT|\{HH,HT\})=\frac{3}{2}$$&#xA;$$E[X|\{TH,TT\}]=2P(HH|\{TH,TT\})+1P(HT|\{TH,TT\})$$&#xA;$$+1P(TH|\{TH,TT\})+0P(TT|\{TH,TT\})=\frac{1}{2} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$E[X|\mathcal{F_1}]=\left\{ \begin{array}{l} &#xD;&#xA;\frac{3}{2}, \omega\in \{HH,HT\}\\&#xD;&#xA;\frac{1}{2}, \omega \in \{TH,TT\}&#xD;&#xA;\end{array} \right.&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope those would be helpful.&lt;/p&gt;&#xA;" OwnerUserId="8183" LastActivityDate="2011-03-13T22:19:19.260" />
  <row Id="26801" PostTypeId="1" AcceptedAnswerId="26813" CreationDate="2011-03-13T22:20:11.317" Score="2" ViewCount="100" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm trying to prove that Cauchy distribution is stable, ie. if $X_{1}, X_{1}, ...$ are independent and have the Cauchy distributions, then $\frac{1}{n}(X_{1}+...+X_{n})$ has the same distribution as $X_{1}$ for $n \geq 2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suspect the proof has something to do with characteristic functions, but haven't been able to write it out. Anyone have any hints on how to approach this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="2076" LastActivityDate="2011-03-13T22:53:38.543" Title="Prove that Cauchy distribution is stable?" Tags="&lt;probability&gt;&lt;probability-theory&gt;&lt;probability-distributions&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="0" />
  <row Id="26802" PostTypeId="2" ParentId="26726" CreationDate="2011-03-13T22:22:33.983" Score="1" ViewCount="" Body="&lt;p&gt;For two-sided ideals, see &lt;a href=&quot;http://math.stackexchange.com/questions/22629/why-is-the-ring-of-matrices-over-a-field-simple&quot;&gt;the previous question which deals with fields&lt;/a&gt;. In the comments I state the description of ideals over a commutative ring with $1$: they are precisely the subrings that consist of matrices with coefficients in $\mathcal{J}$, where $\mathcal{J}$ is an ideal of $R$. The result for not-necessarily commutative rings is the obvious one. For rings without an identity, it gets a bit more complicated, but essentially the same kind of argument works.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For one-sided ideals, think about some of the argument given in the 2-sided case. You can do &lt;em&gt;some&lt;/em&gt; of the stuff described, though perhaps not all. For instance, for left ideals, you can isolate rows. So perhaps you can modify the answer yoyo gives for the division ring case to a general case, along the same lines as the result for two-sided ideals generalizes from fields to the general case.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2011-03-13T22:22:33.983" />
  <row Id="26803" PostTypeId="1" CreationDate="2011-03-13T22:23:16.277" Score="3" ViewCount="202" Body="&lt;p&gt;Now, I am working through a particular case in the book on smooth manifolds by John.M.Lee used in my graduate math class, let's say we have a smooth manifold X which has positive dimension. He then claims that the vector space $C^\infty$ of all smooth functions from $X \rightarrow \mathbb{R} $ will have infinite dimension over $\mathbb{R}$. While I do have a fair idea of the  claim, I'm a little bit lost to show that the vector space has infinite dimension.&lt;/p&gt;&#xA;" OwnerUserId="8169" LastActivityDate="2011-03-13T23:57:36.037" Title="vector space of all smooth functions has infinite dimension" Tags="&lt;differential-geometry&gt;&lt;manifold&gt;" AnswerCount="3" CommentCount="0" />
  <row Id="26804" PostTypeId="1" AcceptedAnswerId="26808" CreationDate="2011-03-13T22:26:02.543" Score="3" ViewCount="80" Body="&lt;p&gt;So when first developing the natural numbers in basic set theory, most properties are proven by induction. This is very convenient since $\omega$ is the smallest inductive set. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I saw a proof of the following proposition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For all $n\in\omega$ and $n\neq 0$, $n\times\omega\sim\omega$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The induction goes as let &#xA;$$&#xD;&#xA;K=\{n\in\omega\ |\ (n=0)\lor (n\times\omega\sim\omega)\}.&#xD;&#xA;$$&#xA;Base case, $0\in K$. Suppose $n\in K$. Then &#xA;$$&#xD;&#xA;\sigma(n)\times \omega=(n\cup\{n\})\times\omega=n\times\omega\cup\{n\}\times\omega\sim\omega&#xD;&#xA;$$&#xA;since $n\times\omega\sim\omega$ and $\{n\}\times\omega\sim\omega$, and the union of two countable sets is countable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What puts me off is that since the proposition is not true for $n=0$, we just throw it into the set to conclude that $K=\omega$, not $K=\omega\setminus\{0\}$. Shouldn't we start the base case at $n=1$ instead? Why is it still acceptable to begin at $0$ if the proposition is not actually true for $0$? &lt;/p&gt;&#xA;" OwnerUserId="6742" LastEditorUserId="6742" LastEditorDisplayName="" LastEditDate="2011-03-13T22:33:26.627" LastActivityDate="2011-03-13T22:46:20.920" Title="Question about base cases in induction." Tags="&lt;elementary-set-theory&gt;&lt;induction&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="26805" PostTypeId="2" ParentId="26803" CreationDate="2011-03-13T22:29:19.373" Score="0" ViewCount="" Body="&lt;p&gt;I'm not sure of the details for smooth manifolds, but certainly the polynomials have infinite dimension and are contained in $C^\infty$.  &lt;/p&gt;&#xA;" OwnerUserId="4266" LastActivityDate="2011-03-13T22:29:19.373" CommentCount="3" />
  <row Id="26807" PostTypeId="1" CreationDate="2011-03-13T22:34:51.520" Score="3" ViewCount="126" Body="&lt;p&gt;I'm trying to prove that every simple graph $G$ of girth $g(G)=5$ (length of smallest cycle), and minimum degree $\delta$, has at least $\delta^2 + 1$ vertices. &#xA;I tried using induction on $\delta$ without any results, and also tried to apply the pigeonhole principle, to no avail.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Help? &amp;nbsp;&lt;/p&gt;&#xA;" OwnerUserId="8214" LastEditorUserId="6296" LastEditorDisplayName="" LastEditDate="2011-03-14T00:25:41.863" LastActivityDate="2011-03-14T00:25:41.863" Title="If given the girth and the minimum degree of a simple graph $G$, can we give a lower bound on the number of vertices it has?" Tags="&lt;graph-theory&gt;" AnswerCount="2" />
  <row Id="26808" PostTypeId="2" ParentId="26804" CreationDate="2011-03-13T22:37:33.907" Score="3" ViewCount="" Body="&lt;p&gt;The proof is not correct as stated, precisely for the reason you point out: if $n=0$, then it is false that $n\times \omega\sim\omega$. So it's a good thing you are put off.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rather, from $n\times\omega \cup \{n\}\times\omega$, the argument should then say something like: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$\{n\}\times\omega\sim\omega$ holds; and since $n\in K$, either $n=0$ and $n\times\omega=\emptyset$, or else $n\times\omega\sim\omega$. In either case, we have $\sigma(n)\times\omega\sim\omega$, so $\sigma(n)\in K$. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(or else the inductive step could try to handle the $\sigma(n)=1$ case as a &quot;special case&quot;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's okay to begin at $0$ because the set $K$ &quot;takes care&quot; of the $n=0$ case explicitly. You can do this in all sorts of cases; for example, if you wanted to prove that every natural number greater than $99$ needs three or more digits to write in base $10$, you could let $K$ be&#xA;$$K = \{n\in\omega\mid (n\leq 99)\lor (n\text{ requires three or more digits to write in base }10)\}$$&#xA;and use induction to show that $K=\omega$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, when one does this, it is very often the case that the first case in which the original statement is true has to be dealt with independently or directly, which is what happens in the argument you quote. In the one I just made up, you would have to consider the case $\sigma(n)=100$ separately from all others: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;For $n=0$ or $\sigma(n)\lt 99$ you have $\sigma(n)\in K$ by definition; &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;for $\sigma(n)\gt 100$ you can use the fact that $\sigma(n)$ requires either the same number of digits of $n$ or one more, and since $n\in K$, and $\neg(n\leq 99)$, then $n$ requires at least $3$ digits, hence so does $\sigma(n)$; so $\sigma(n)\in K$. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;But if $\sigma(n)=100$, then the fact that $n\in K$ doesn't help you: you actually have to stare at $100$ and see that it takes three digits to conclude $\sigma(n)\in K$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Essentially, you get all the &quot;false cases&quot; for free, so that your &quot;true&quot; base, the first value for which the statement holds, usually needs to be taken care of directly. &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-13T22:46:20.920" LastActivityDate="2011-03-13T22:46:20.920" CommentCount="3" />
  <row Id="26809" PostTypeId="2" ParentId="26807" CreationDate="2011-03-13T22:40:53.140" Score="0" ViewCount="" Body="&lt;p&gt;Suppose for a simple graph $G$, $g(G) = 5$ and $\delta(G)$ is the minimum degree. Suppose for contradiction that $G$ has $\delta^2$ vertices. You can probably use Menger's Theorem to come up with a contradiction.&lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2011-03-13T22:40:53.140" />
  <row Id="26810" PostTypeId="2" ParentId="26807" CreationDate="2011-03-13T22:50:06.157" Score="4" ViewCount="" Body="&lt;p&gt;If you start from any vertex, it is connected to at least $\delta$ vertices.  Each of those is connected to at least $\delta-1$ others.  These are all distinct or you would have a 4 (or less)-cycle.  $1+\delta+\delta(\delta-1)=\delta^2+1$.  The pentagon and Petersen graph follow exactly this, then connect appropriate ones of the second tier of vertices.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2011-03-13T23:57:50.530" LastActivityDate="2011-03-13T23:57:50.530" CommentCount="10" />
  <row Id="26811" PostTypeId="2" ParentId="26803" CreationDate="2011-03-13T22:50:33.460" Score="7" ViewCount="" Body="&lt;p&gt;This follows from the existence of &lt;a href=&quot;http://en.wikipedia.org/wiki/Bump_function&quot; rel=&quot;nofollow&quot;&gt;bump functions&lt;/a&gt;.  Take an open set $U$ in your manifold with a chart $\phi:U\to \mathbb{R}^n$.  Let $k$ be a positive integer, and let $x_1,x_2,\ldots,x_k$ be distinct points in $\phi(U)$.  For each $j$ from $1$ to $k$, let $f_j:\mathbb{R}^n\to\mathbb{R}$ be a smooth function whose support is a compact subset of $\phi(U)$ such that $f_j(x_j)=1$ and $f_j(x_m)=0$ if $m\neq j$.  Define $g_j:X\to\mathbb{R}$ by $g_j(x)=0$ if $x$ is in $X\setminus U$, and $g_j(x)=f_j(\phi(x))$ if $x$ is in $U$.  Then $\{g_1,\ldots,g_k\}$ is a linearly independent set of smooth functions on $X$.  Since $k$ was arbitrary, the space of smooth functions is infinite dimensional.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2011-03-13T22:50:33.460" />
  <row Id="26812" PostTypeId="1" CreationDate="2011-03-13T22:53:15.500" Score="1" ViewCount="89" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Possible Duplicate:&lt;/strong&gt;&lt;br&gt;&#xA;  &lt;a href=&quot;http://math.stackexchange.com/questions/15055/in-a-family-with-two-children-what-are-the-chances-if-one-of-the-children-is-a&quot;&gt;In a family with two children, what are the chances, if one of the children is a girl, that both children are girls?&lt;/a&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;p&gt;I was introduce to the Two Child Problem a while back, which is stated like this:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Mr. Smith has two children. At least one of them is a boy. What is the probability that both children are boys?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There are two ways that I have seen to figure out the probability, and they both seem to me valid logic; but they cannot both be, because they yield different solutions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;First way&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given two children, there are four equally possible combinations:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;older boy, younger boy&#xA;older boy, younger girl&#xA;older girl, younger boy&#xA;older girl, younger girl&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Knowing that at least one is a boy, only one possibility can be eliminated, namely, both children being girls. Out of the three equally likely possibilities left, only one involves both children being boys. Thus, the probability of both children being boys is $1/3$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Second Way&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given two children where at least one is a boy, then the older child is a boy and/or the younger child is a boy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the older child is a boy, then the probability of the younger child being a boy is $1/2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the younger child is a boy, then the probability of the older child being a boy is $1/2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, for all possible situations with two children where at least one child is a boy, the probability of both children being boys is $1/2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which way is the correct way of evaluating the probability? Why is the other way wrong?&lt;/p&gt;&#xA;" OwnerUserId="7994" LastActivityDate="2011-03-13T22:53:15.500" ClosedDate="2011-03-13T23:08:54.177" Title="The Two Child Problem" Tags="&lt;probability&gt;" CommentCount="0" />
  <row Id="26813" PostTypeId="2" ParentId="26801" CreationDate="2011-03-13T22:53:38.543" Score="2" ViewCount="" Body="&lt;p&gt;The characteristic function for a Cauchy r.v centered around zero and with scale $\gamma$ is $\exp(- \gamma |t|)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $X_i$ are r.v. with characteristic function $\psi_i(t)$, then $aX_i$ has characteristic function $\psi_i(at)$ and characteristic function of $\sum X_i$ where $X_i$ are independent is $\prod \psi_i(t)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Use the above information to conclude what you want.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastActivityDate="2011-03-13T22:53:38.543" CommentCount="0" />
  <row Id="26814" PostTypeId="5" CreationDate="2011-03-13T23:10:38.823" Score="0" ViewCount="" Body="" OwnerUserId="-1" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-13T23:10:38.823" LastActivityDate="2011-03-13T23:10:38.823" />
  <row Id="26815" PostTypeId="4" CreationDate="2011-03-13T23:10:38.823" Score="0" ViewCount="" Body="For basic questions about limits, derivatives, integrals, and applications, mainly of one-variable functions." OwnerUserId="-1" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-13T23:10:38.823" LastActivityDate="2011-03-13T23:10:38.823" />
  <row Id="26816" PostTypeId="2" ParentId="26771" CreationDate="2011-03-13T23:20:02.180" Score="1" ViewCount="" Body="&lt;p&gt;Here are some results for very small numbers, when there are $n$ variables:&#xA;$$&#xD;&#xA;\begin{align*}&#xD;&#xA;f(1,1,1,\ldots,1) &amp;amp;= 1, \\&#xD;&#xA;f(2,1,1,\ldots,1) &amp;amp;= \frac{n}{n-1}, \\&#xD;&#xA;f(3,1,1,\ldots,1) &amp;amp;= \frac{n^3-2n^2+3n}{n^3-3n^2+4n-2} = \frac{n}{n-1} \cdot \frac{n^2-2n+3}{n^2-2n+2}, \\&#xD;&#xA;f(2,2,1,\ldots,1) &amp;amp;= \frac{n^3-n^2+2n}{n^3-3n^2+4n-2} = \frac{n}{n-1} \cdot \frac{n^2-n+2}{n^2-2n+2}.&#xD;&#xA;\end{align*}&#xD;&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2011-03-13T23:20:02.180" />
  <row Id="26817" PostTypeId="2" ParentId="26803" CreationDate="2011-03-13T23:51:01.193" Score="4" ViewCount="" Body="&lt;p&gt;I originally posted the following as a comment but I think it is a particularly simple argument, so worth upgrading to a answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Choose any $\theta\in C^\infty(M)$ which is not constant on a connected component of $M$. Then its image is an infinite subset of $\mathbb{R}$ (it includes a nontrivial interval). As the space $\mathbb{R}[X]$ of real polynomials is infinite dimensional, and a polynomial is fully determined by its values on an infinite set, you can conclude that&#xA;$$&#xD;&#xA;\left\{f\circ\theta\colon f\in\mathbb{R}[X]\right\}&#xD;&#xA;$$&#xA;is an infinite dimensional subspace of $C^\infty(M)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that this argument is applicable much more generally. E.g., if a Riemann surface has at least one non-constant meromorphic function then the space of meromorphic functions is infinite dimensional.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you like, you can modify this method to show that the dimension of $C^\infty(M)$ is (at least) the cardinality of the continuum, by proving the same is true for smooth functions on any nontrivial interval in $\mathbb{R}$. (Hint: the exponentials $x\mapsto e^{ax}$ are linearly independent).&lt;/p&gt;&#xA;" OwnerUserId="1321" LastEditorUserId="1321" LastEditorDisplayName="" LastEditDate="2011-03-13T23:57:36.037" LastActivityDate="2011-03-13T23:57:36.037" CommentCount="2" />
  <row Id="26818" PostTypeId="2" ParentId="26420" CreationDate="2011-03-14T00:27:54.093" Score="0" ViewCount="" Body="&lt;p&gt;right, ok. I asked about if cup product is well defined, here is why I am confused.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You wrote:  &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Strictly speaking, the cup product is not commutative, though it is commutative up to sign on the level of cohomology.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is an abstract way of seeing this...&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The cup product being commutative should have &lt;strong&gt;nothing&lt;/strong&gt; to do with it being well defined. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Commutativity of the the product should be talking about if $a \smile b = b \smile a$. This seems irrelevant for a conversation about the cup product being well defined. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So again, what I want to know if why is the cup product well defined. What I am asking is: is it the case that $(\alpha \smile \beta)(a)$ is the same as $(\alpha \smile \beta)(b)$ whenever $a$ is equivalent to $b$. Where equivalent is as as co-homology classes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now in your second comment, you have said that that it is not the case that that the cup product is well defined, and it's again unclear if you mean over co-chain groups or co-homology groups. I&lt;/p&gt;&#xA;&#xA;&lt;p&gt;n either case, what I want to know is:&#xA;When is the cup product well defined? Does your definition of well defined differ from the classical one? I do not at this point want to underst&lt;/p&gt;&#xA;" OwnerUserId="8122" LastActivityDate="2011-03-14T00:27:54.093" CommentCount="2" />
  <row Id="26819" PostTypeId="1" AcceptedAnswerId="26832" CreationDate="2011-03-14T00:39:32.443" Score="3" ViewCount="148" Body="&lt;p&gt;I'm studying for my Intro to Algebra class. I've reached the point where I want to understand the first Noether isomorphism theorem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is the definition I was given in class:&#xA;Let $f :R \rightarrow S$ be a surjective ring homomorphism. Then we have a &quot;commutative diagram&quot; of ring homomorphisms. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is the scanned diagram I received: Could someone please explain this diagram to me in basic terms? In addition, I'd deeply appreciate examples or proof of examples.&#xA;&lt;img src=&quot;http://i.imgur.com/7xgEE.png&quot; alt=&quot;scanned&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ultimately, can someone answer what it means to say something is equivalent to another thing &quot;up to isomorphism?&quot;&lt;/p&gt;&#xA;" OwnerUserId="287" LastEditorUserId="6296" LastEditorDisplayName="" LastEditDate="2011-03-14T01:12:05.567" LastActivityDate="2011-03-14T02:03:01.390" Title="The first Noether isomorphism theorem" Tags="&lt;abstract-algebra&gt;" AnswerCount="4" CommentCount="2" FavoriteCount="2" />
  <row Id="26820" PostTypeId="1" CreationDate="2011-03-14T00:41:37.087" Score="0" ViewCount="59" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Possible Duplicate:&lt;/strong&gt;&lt;br&gt;&#xA;  &lt;a href=&quot;http://math.stackexchange.com/questions/14721/three-variable-third-degree-diophantine-equation&quot;&gt;Three variable, third degree Diophantine equation&lt;/a&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;p&gt;How can I show that $x^3+y^3=3z^3$ has no integer solutions except when $z=0$?&lt;/p&gt;&#xA;" OwnerUserId="2505" LastActivityDate="2011-03-14T00:48:04.030" ClosedDate="2011-03-14T00:49:31.297" Title="Cubic Diophantine equation" Tags="&lt;number-theory&gt;&lt;diophantine-equations&gt;" AnswerCount="0" CommentCount="3" />
  <row Id="26821" PostTypeId="2" ParentId="26819" CreationDate="2011-03-14T00:45:27.090" Score="3" ViewCount="" Body="&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Rank%E2%80%93nullity_theorem&quot; rel=&quot;nofollow&quot;&gt;rank nullity theorem&lt;/a&gt; in linear algebra is an example of this theorem. In other words, two vector spaces are isomorphic if and only if they have the same dimension. So we have $$\text{dim}(R/ \text{ker} \ f) = \text{dim}(R) - \text{dim}(\text{ker} \ f) = \text{dim}(\text{Im} \ f)$$ or $$\text{dim} \ R = \text{dim}(\text{ker} \ f)+ \text{dim}(\text{Im} \ f)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The phrase &quot;up to isomorphism&quot; just means that we consider isomorphic structures to be equivalent. So $\mathbb{Z}_3$ and $\mathbb{Z}/3 \mathbb{Z}$ are equivalent up to isomorphism. &lt;/p&gt;&#xA;" OwnerUserId="2548" LastEditorUserId="2548" LastEditorDisplayName="" LastEditDate="2011-03-14T00:50:59.283" LastActivityDate="2011-03-14T00:50:59.283" />
  <row Id="26823" PostTypeId="2" ParentId="26792" CreationDate="2011-03-14T00:57:18.950" Score="2" ViewCount="" Body="&lt;p&gt;Edit 2: for your actual question, the answer is yes, you can find counterexamples. Consider the sequence $(1/n)$, and build your sequence as follows. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Start at $x_1 = 1$, and let $\alpha_1$ be a running counter which we start at 2. We then alternate between a decreasing phase and an increasing phase. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Decreasing phase: let $x_i = x_{i-1} - 1/\alpha_{i}$ if $x_{i-1} &amp;gt; 1/\alpha_i$. And increase $\alpha_{i+1} = \alpha_i + 1$. If $x_{i-1} &amp;lt; 1/\alpha_i$, set &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ x_i = x_{i+1} = \cdots = x_{100i} = x_{i-1}\qquad \qquad \alpha_{100i + 1} = \alpha_{i} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and then enter the increasing phase starting from the $100i+1$'th term. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Increasing phase: let $x_i = x_{i-1} + 1/\alpha_i$ if $x_{i-1} + 1/\alpha_i &amp;lt; 1$. And increase $\alpha_{i+1} = \alpha_{i} + 1$. Else set &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ x_i = x_{i+1} = \cdots = x_{100i} = x_{i-1} \qquad \qquad \alpha_{100i + 1} = \alpha_{i} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and enter the decreasing phase starting from the $100i+1$'th term. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Due to the long constant phases, you see that $\limsup_{n\to\infty} \frac{1}{n} \sum_1^n x_i = 1$, and the $\liminf = 0$. (The length of the constant phases grow exponentially, so that each constant phase completely dominates all previous terms in the average.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sequence looks like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1, 1/2, 1/6, {300 terms of 1/6}, 5/12, 37/60, 47/60, 389/420, {repeat 30700 times}, 673/840 ...&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Edit: oh wait, the below doesn't actually address your question.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;As a side note, the limit $\lim_{n\to\infty} \frac1n \sum_1^n x_i $ is known as the &lt;a href=&quot;http://en.wikipedia.org/wiki/Ces%C3%A0ro_mean&quot; rel=&quot;nofollow&quot;&gt;Cesàro mean&lt;/a&gt; of the sequence $x_i$. It is a fact that for any converging sequence $x_i\to \bar{x}$, the sequence $c_n = \frac{1}{n} \sum_1^n x_i$ also converges to $\bar{x}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To do this, let $\delta &amp;gt; 0$. It suffices to find $N$ large enough that for all $m &amp;gt; N$, $|c_m - \bar{x}| &amp;lt; \delta$. Because $x_i\to \bar{x}$, we can find $I$ large such that for all $j &amp;gt; I$, $|x_i - \bar{x}| &amp;lt; \delta / 3$. Now choose $N &amp;gt; I$ large enough such that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \sum_1^I x_i &amp;lt; \frac{\delta}{3}N  \qquad \textrm{and} \qquad  \frac{I}{N} &amp;lt; \frac{\delta}{3|\bar{x}|} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then we have for all $m &amp;gt;  N$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ m c_m = \sum_1^I x_i + \sum_{I+1}^m x_i $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ |c_m - \bar{x}| \leq | \frac{1}{m} \sum_1^I x_i | + | \frac{1}{m} \sum_{I+1}^m x_i - \bar{x} | $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first term on the right hand side is bounded by $\delta/3$. The second term we estimate&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ |\frac{1}m \sum_{I+1}^m x_i - \bar{x}| \leq |\frac{1}{m} \sum_{I+1}^m (x_i - \bar{x})| + |\frac{I}{m} \bar{x}| $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and by construction both of the terms on the RHS are bounded by $\delta / 3$. So we have the desired inequality. &lt;/p&gt;&#xA;" OwnerUserId="1543" LastEditorUserId="1543" LastEditorDisplayName="" LastEditDate="2011-03-14T01:28:08.377" LastActivityDate="2011-03-14T01:28:08.377" CommentCount="2" />
  <row Id="26824" PostTypeId="1" AcceptedAnswerId="26836" CreationDate="2011-03-14T01:16:59.573" Score="1" ViewCount="129" Body="&lt;p&gt;When calculating determinants it can be nice to multiply a row by a number or to add one row to another (your basic row operations).  Each has an easy to understand effect on the determinant.  Today I ran across a type of row operation whose effect on the determinant is a lot less clear to me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take an n×n matrix M and an n×n matrix A.  While calculating the determinant of M, I might mess around with the rows of M.   Every time I mess with the first row of M, I would like to be able to change my mind and mess with the second row of M instead.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I multiply the first row of M by 2 to get M′, or if I multiply the second row of M by 2 to get M″, I get the same determinants, det(M′) = det(M″).  This does not depend on the matrix M; I always have that 2 times the first row is equivalent to 2 times the second row.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, if M = [ a, b ; c, d ] then M′ = [ 2a, 2b ; c, d ] and M″ = [ a, b ; 2c, 2d ] and det(M′) = det(M″) = 2det(M) = 2ad−2bc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can I do something similar with &quot;A&quot; instead of &quot;2&quot;?&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;If I multiply the first row of M by A to get M′, then what matrix B do I need to multiply the second row of M by to get M″ so that det(M′)=det(M″) have the same determinant?&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I would like the answer to be depend only on A, not on M.  I calculated an answer for 2×2 matrices, but I'm not happy with it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, if A = [ 2, 0 ; 0, 1 ], and M = [ a, b ; c, d ] then M′ = [ 2a, b ; c, d ] and det(M′) = 2ad−bc.  I can choose B =  [ 1, 0 ; 0, 2 ] so that M″ = [ a, b ; c, 2d ] and det(M″) = 2ad−bc = det(M′).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I'd like to understand more clearly the relationship between A and B.  Perhaps there is a fair amount of freedom in choosing B and I have chosen a bad one in my work.  What is a natural choice of B (for general A)?&lt;/p&gt;&#xA;" OwnerUserId="583" LastActivityDate="2011-03-14T06:07:09.353" Title="Determinants: multiply a row by a scalar, all is good; multiply a row by a square matrix?" Tags="&lt;linear-algebra&gt;&lt;matrices&gt;" AnswerCount="2" CommentCount="5" />
  <row Id="26825" PostTypeId="1" CreationDate="2011-03-14T01:19:34.657" Score="0" ViewCount="45" Body="&lt;p&gt;Hello,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am reading Boyd's Convex Optimization text, and I am looking at page 55, line 7-5 from the bottom. The equation reads:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(\lambda \neq 0 \wedge \lambda^T(x-y) \leq \mu)  \forall (y \in K) \Rightarrow \lambda \succeq_{K^*} 0$ (eq: 1)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some context: K is a proper cone, K* is the dual cone of K, defined as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$K^* = \{y | y^Tx \geq 0 \forall (x \in K)\}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The part before the right arrow is &quot;the first inequality&quot;, and the part after the right arrow is &quot;we conclude $\lambda \succeq_{K^*} 0$&quot;. The exact Boyd text is &quot;From the first inequality we conclude $\lambda \succeq_{K^*} 0$&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a little more complicated than the earlier examples. Before, if a linear function was bounded above or below, the derivative was equal to zero (which I understood, page 49). There was also one equation before (eq: 1) which might be used here,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\lambda^Ty \geq \mu \forall (y \in K) \Rightarrow (\lambda \in K^*) \wedge (\mu \leq 0)$ (eq: Page 54, line 8 of Boyd).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From (eq:1), we get:&#xA;$(\lambda \neq 0 \wedge \lambda^T(x-y) \leq \mu)  \forall (y \in K) \Rightarrow (\lambda^T(y-x) \geq -\mu)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But (y-x) might not be in K? Also if $(\lambda^T(y-x) \geq -\mu)$, wouldn't this imply $-\mu \leq 0$, then $\mu \geq 0$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, on line 5 from the bottom, Boyd states: $x \in S \wedge x \in (x - K) \Rightarrow \lambda^T x = \mu$. The first $x \in S$ means $\lambda^T x \geq \mu$. I don't see how $x \in (x - K)$ will cause a $\lambda^T x \leq \mu$ to appear?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks, and I have also attached an image of the text.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/BeA2d.png&quot; alt=&quot;Boyd Page 55&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="8140" LastEditorUserId="8140" LastEditorDisplayName="" LastEditDate="2011-03-14T01:59:35.137" LastActivityDate="2011-03-14T01:59:35.137" Title="Convex Sets, Minimal Elements, and Minimizers" Tags="&lt;convex-optimization&gt;" />
  <row Id="26826" PostTypeId="1" CreationDate="2011-03-14T01:27:30.080" Score="1" ViewCount="94" Body="&lt;p&gt;Suppose $a$, $b$, $c$, and $d$ are positive numbers and each is not equal to $1$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If  $\log_a(d)$, $\log_b(d)$, and $\log_c(d)$  are an arithmetic progression in this order, then what is $(ac)^{\log_a(b)}$ equal to?&lt;/p&gt;&#xA;" OwnerUserId="3128" LastEditorUserId="6296" LastEditorDisplayName="" LastEditDate="2011-03-14T02:32:53.573" LastActivityDate="2011-03-14T03:34:16.403" Title="An arithmetic sequence of positive logarithms" Tags="&lt;homework&gt;&lt;algebra-precalculus&gt;" AnswerCount="2" CommentCount="7" />
  <row Id="26827" PostTypeId="2" ParentId="6244" CreationDate="2011-03-14T01:33:14.463" Score="0" ViewCount="" Body="&lt;p&gt;Let $\{a_1,\ldots,a_n\}$ be a finite set of real numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, since $\mathbb{Q}$ is countable, the set $\left\{\sum_{i\,=\,1}^n{q_ia_i}:q_i\in\mathbb{Q}\right\}$ is countable, whereas $\mathbb{R}$ is uncountable. Hence, no finite subset of $\mathbb{R}$ can span $\mathbb{R}$ with rational scalars, so $\mathbb{R}$ cannot have a finite basis over $\mathbb{Q}$.&lt;/p&gt;&#xA;" OwnerUserId="8209" LastEditorUserId="308" LastEditorDisplayName="" LastEditDate="2011-04-21T16:52:40.080" LastActivityDate="2011-04-21T16:52:40.080" CommentCount="2" />
  <row Id="26828" PostTypeId="2" ParentId="26819" CreationDate="2011-03-14T01:35:14.237" Score="0" ViewCount="" Body="&lt;p&gt;The first isomorphism theorem relates two rings $A$ and $B$ which have a surjective homomorphism from $A$ to $B$, meaning that every element of $B$ is the image of an element of $A$. In order to understand the theorem, it is crucial to understand what an isomorphism is. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;An isomorphism is a surjective homomorphism between two rings which can be inverted to produce another surjective homomorphism. What is important about an isomorphism is that, for any algebraic purposes, if the rings $A$ and $B$ have an isomorphism $f$ from $A$ to $B$, $A$ and $B$ differ only in notation. This is because if something is true for $A$, then the same thing can be verified for $B$ by taking $f^{-1}(b)$ for each element of $B$, performing computations in $A$, and then returning to ring $B$ by taking $f(f^{-1}(b)) = b$. An example of two isomorphic rings are $\mathbb{Z}_2 = {0,1}$, the integers mod 2 with addition and multiplication defined by $0+0 = 0, 0+1 = 1+0 = 1, 1+1=0, 0*0 = 1*0 = 0*1 = 0$ and $1*1 = 1$, and $R = {a,b}$ with addition and multiplication defined by $a+a = a, a+b=b+a=b, b+b=a, a*a = b*a = a*b = a$ and $b*b = b$. Clearly, these rings are essentially the same, we've just replaced &quot;0&quot; with &quot;a&quot; and &quot;1&quot; with &quot;b&quot;, and thus there is an isomorphism $f$ from $\mathbb{Z}_2$ to $R$ given by $f(0) = a$ and $f(1) = b$ (you should verify for yourself that this is an isomorphism. It is because isomorphisms preserve the important properties of a ring that we say rings are &quot;equivalent up to isomorphism&quot; to mean that they are algebraically identical.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another concept we need is the quotient ring $R/ \text{ker}f$. This ring is formed by taking the ring $R$ and considering the elements $a$ and $b$ &quot;equivalent&quot; if $f(a) = f(b)$. For example, if the function $f$ is a homomorphism from the ring of integers to $\mathbb{Z}_2$ defined by $f(x) = 0$ if $x$ is even and $f(x) = 1$ if $x$ is odd, then it is easy to see that the ring $R/ \text{ker}f$ consists of the set of even integers (all of which are equivalent to zero) and the set of odd integers (all of which are equivalent to one) and thus this is a ring with two elements (which is isomorphic to $\mathbb{Z}_2$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hopefully I did a good enough job explaining these concepts. The theorem says that $S$ and $R/ \text{ker}f$ are isomorphic - that is, essentially the same in the sense I described above.&lt;/p&gt;&#xA;" OwnerUserId="8173" LastActivityDate="2011-03-14T01:35:14.237" />
  <row Id="26829" PostTypeId="2" ParentId="26819" CreationDate="2011-03-14T01:40:33.253" Score="1" ViewCount="" Body="&lt;p&gt;To say that a diagram of objects and morphisms is commutative (or that the diagram commutes) is to say that for any two objects of the diagram, the morphism obtained by following a path between the two objects in the direction of the arrows is independent of the choice of the path.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your diagram, we have to paths from $R$ to $S$: 1) the arrow (ring morphism) $f$, and 2) the path given by the composition of the arrow (ring morphism) from $R$ to $R/\ker f$ (let's call it $h$) and the arrow (ring morphism) from $R/\ker f$ to $S$ (let's call it $g$). In this case, to say that your diagram is commutative means that $f=gh$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically, the theorem says that any ring epimorphism $f:r\rightarrow S$ can be factored as $gh$, where $h:R\rightarrow R/\ker f$ is epi and $g:R/\ker f\rightarrow S$ is iso (in many textbooks this last part is refereed to as the first isomorphism theorem).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's take a look at (a sketch of) the proof: since $f:R\rightarrow S$ is an (epi)morphism, its kernel $\ker f$ is a two-sided ideal of $R$, so we immediately have the quotient ring $R/\ker f$ and the canonical epimorphism from a ring to one of its quotients (in the general case, if $I$ is a two-sided ideal of a ring $R$ then the canonical epimorphism $\pi:R\rightarrow R/I$ is naturally defined by $\pi(r):=r+I$, for all $t\in R$). In our specific case, we have that $h:R\rightarrow R/\ker f$ is defined by $h(r)=r+\ker f$, for all $r\in R$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we must construct an isomorphism $g:R/\ker f\rightarrow S$. For this, we simply define (again, this definition is natural) $g(r+\ker f):=f(r)$, for all $r+\ker f\in R/\ker f$. It's easy to show that $g$, as we defined it, is in fact a ring isomorphism.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, from our definitions, it follows that for all $r\in R$ we have $gh(r)=g\big(h(r)\big)=g(r+\ker f)=f(r)$ so $f=gh$.&lt;/p&gt;&#xA;" OwnerUserId="8050" LastEditorUserId="8050" LastEditorDisplayName="" LastEditDate="2011-03-14T01:52:35.750" LastActivityDate="2011-03-14T01:52:35.750" />
  <row Id="26830" PostTypeId="2" ParentId="26726" CreationDate="2011-03-14T01:41:59.257" Score="1" ViewCount="" Body="&lt;p&gt;The left-ideals of the n×n matrix ring S = M&lt;sub&gt;n&lt;/sub&gt;(R) over a (unital, associative) ring R are in 1–1 correspondence with the left submodules of a free left R-module R&lt;sup&gt;n&lt;/sup&gt; of rank n.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given any left-R-submodule K ≤ R&lt;sup&gt;n&lt;/sup&gt; define the ideal I(K) as those matrices whose rows are elements of K.  If M is such a matrix and A is an arbitrary matrix (n×n over R), then AM is a matrix whose rows are R-linear combinations of the rows of M, and so also contained in K.  Similarly, by multiplying by elementary matrices (matrix units even), you get that every left ideal I determines a submodule K(I) consisting of the first rows of elements of the ideal.  You then check that I(K(I))=I and K(I(K))=K.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is called &lt;a href=&quot;http://en.wikipedia.org/wiki/Morita_equivalence&quot; rel=&quot;nofollow&quot;&gt;Morita equivalence&lt;/a&gt;.  The left ideal P = I(R⊕0⊕…⊕0) = I(R⊕o&lt;sup&gt;(n−1)&lt;/sup&gt;) is a direct summand of the left S-module S, and so is called a projective module.  It is faithful, since S ≅ P&lt;sup&gt;n&lt;/sup&gt; as left S-modules.  In other words, it is a progenerator. I(K) = &lt;sub&gt;S&lt;/sub&gt;P ⊗&lt;sub&gt;R&lt;/sub&gt; K and K(I) = &lt;sub&gt;R&lt;/sub&gt;Hom&lt;sub&gt;S&lt;/sub&gt;(P,I).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The short version is just that modules over S are just big versions of modules over R.  In particular, left ideals of S are just left submodules of R&lt;sup&gt;n&lt;/sup&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Over some rings R, the submodules of R&lt;sup&gt;n&lt;/sup&gt; can be significantly more complicated.  Even for fields, the submodule lattice of R=&lt;strong&gt;R&lt;/strong&gt; is very simple: just 0 ≤ R.  For R&lt;sup&gt;n&lt;/sup&gt; one has the &lt;a href=&quot;http://en.wikipedia.org/wiki/Grassmannian&quot; rel=&quot;nofollow&quot;&gt;Grassmannian&lt;/a&gt;, which to my mind is quite a bit more complex than R.  Other rings can be even worse, so one should be careful of leaning too heavily on the intuition for fields when dealing with matrix rings, though any Morita property will transfer between R and S without trouble using the &quot;I()&quot; and &quot;K()&quot; constructions.&lt;/p&gt;&#xA;" OwnerUserId="583" LastActivityDate="2011-03-14T01:41:59.257" />
  <row Id="26831" PostTypeId="2" ParentId="26826" CreationDate="2011-03-14T01:46:47.987" Score="3" ViewCount="" Body="&lt;p&gt;It is equal to a lot of things, but I suspect that $c^2$ is one of them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can show this by knowing that $\log_a(d) = \dfrac{\log_e(d)}{\log_e(a)}$ and similarly with other letter combinations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then the arithmetic progression implies a &lt;a href=&quot;http://en.wikipedia.org/wiki/Harmonic_progression_%28mathematics%29&quot; rel=&quot;nofollow&quot;&gt;harmonic progression&lt;/a&gt; of the logarithms of $a$, $b$ and $c$.  That will then tell you about $\log_e(ac)$, which you can then multiply by $\log_e(b) / \log_e(a)$.  If you then tidy up the right hand side, you should get something like $2 \log_e(c)$.  And then it is one simple step to the solution.&lt;/p&gt;&#xA;" OwnerUserId="6460" LastEditorUserId="6460" LastEditorDisplayName="" LastEditDate="2011-03-14T01:57:02.097" LastActivityDate="2011-03-14T01:57:02.097" CommentCount="1" />
  <row Id="26832" PostTypeId="2" ParentId="26819" CreationDate="2011-03-14T01:46:59.330" Score="5" ViewCount="" Body="&lt;p&gt;The first thing in the diagram are the solid arrows: those are maps that you &quot;know.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You &quot;know&quot; the map $f\colon R\to S$, because you are assuming that you are &lt;em&gt;given&lt;/em&gt; this map.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You also &quot;know&quot; the map $\pi\colon R\to R/\mathrm{ker}(f)$. This is simply the &lt;strong&gt;canonical projection onto the quotient.&lt;/strong&gt; Remember that whenever you have a ring $R$ and an ideal $\mathcal{I}$ of $R$, you can form the quotient ring $R/\mathcal{I}$; and there is a canonical projection map $\pi\colon R\to R/\mathcal{I}$ given by $\pi(r) = r+\mathcal{I}$. In this case, since $f$ is a homomorphism, you know that $\mathrm{ker}(f)$ is an ideal, so you automatically get the quotient ring $R/\mathrm{ker}(f)$ and the quotient map $\pi\colon R\to R/\mathrm{ker}(f)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second thing is that the &quot;dashed arrow&quot; represents a homomorphism which the theorem asserts exists; so one thing the theorem asserts is that there is a ring homomorphism $R/\mathrm{ker}(f)\to S$, and which is an isomorphism (assuming $f$ is onto).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, saying the diagram &quot;is commutative&quot; means the following: whenever you have two paths from one ring in the diagram to another (in this case, there are two ways to get from $R$ to $S$: either through the arrow $f$, or by first going to $R/\mathrm{ker}(f)$ via the map I called $\pi$, and the map $R/\mathrm{ker}(f)\to S$ that we are asserting exists), then the results you get are the same no matter which path you take (in this case: if $r\in R$, then $f(r)$ will be the same as first taking $\pi(r)$, and then applying the map $R/\mathrm{ker}(f)\to S$ to $\pi(r)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For an example, take $R=\mathbb{Z}\times\mathbb{Z}$, $S=(\mathbb{Z}/10\mathbb{Z})\times(\mathbb{Z}/3\mathbb{Z})$, and $f(a,b) = (a\bmod 10, b\bmod 3)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a ring homomorphism:&#xA;\begin{align*}&#xA;f\bigl((a,b) + (c,d)\bigr) &amp;amp;= f(a+c,b+d) = \bigl( a+c\bmod 10, b+d\bmod 3\bigr)\\&#xA;&amp;amp;= (a\bmod 10,b\bmod 3) + (c\bmod 10, d\bmod 3) = f(a,b) + f(c,d),\\&#xA;f\bigl((a,b)(c,d)\bigr) &amp;amp;= f(ac,bd) = \bigl(ac\bmod 10, bd\bmod 3\bigr)\\&#xA;&amp;amp;= (a\bmod 10, b\bmod 3)(c\bmod 10,d\bmod 3) = f(a,b)f(c,d).&#xA;\end{align*}&#xA;It is also onto: given $(x\bmod 10,y\bmod 3)\in S$, we can always find an element $(a,b)\in R$ such that $f(a,b)=(x\bmod 10, y\bmod 3)$. For instance, we can take $a=x$, $b=y$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;what is the kernel of this map? $f(a,b) = (0\bmod 10,0\bmod 3)$ if and only if $a\equiv 0 \pmod{10}$ and $b\equiv 0 \pmod{3}$. So the kernel is $10\mathbb{Z}\times 3\mathbb{Z}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, we have a map from $R$ to $R/\mathrm{ker}(f)$; namely,&#xA;\begin{align*}&#xA;\pi\colon \mathbb{Z}\times\mathbb{Z} &amp;amp;\longmapsto \frac{\mathbb{Z}\times\mathbb{Z}}{10\mathbb{Z}\times 3\mathbb{Z}}\\&#xA;(a,b) &amp;amp;\longmapsto (a,b) + 10\mathbb{Z}\times 3\mathbb{Z}&#xA;\end{align*}&#xA;What the theorem states is that there is an isomorphism&#xA;$$\varphi\colon \frac{\mathbb{Z}\times\mathbb{Z}}{10\mathbb{Z}\times3\mathbb{Z}} \to \left(\frac{\mathbb{Z}}{10\mathbb{Z}}\right)\times\left(\frac{\mathbb{Z}}{3\mathbb{Z}}\right)$$&#xA;which makes &quot;the diagram commute&quot;. That is, for every $(a,b)\in R$, we will have&#xA;$$f(a,b) = \phi(\pi(a,b)).$$&#xA;What is $\phi$? &#xA;$$\phi\bigl((a,b) + 10\mathbb{Z}\times 3\mathbb{Z}\bigr) = (a\bmod 10,b\bmod 3).$$&#xA;Does this work? First, $\phi$ is well defined: if $(a,b)+10\mathbb{Z}\times 3\mathbb{Z} = (c,d)+10\mathbb{Z}\times3\mathbb{Z}$, then $(a,b)-(c,d)\in 10\mathbb{Z}\times 3\mathbb{Z}$; this means that $a-c\in 10\mathbb{Z}$ and $b-d\in 3\mathbb{Z}$, so $a\bmod 10 = c\bmod 10$ and $b\bmod 3$ = $d\bmod 3$. Thererfore, $(a\bmod 10,b\bmod 3) = (c\bmod 10, d\bmod 3)$. Thus,&#xA;$$\phi\bigl( (a,b)+10\mathbb{Z}\times 3\mathbb{Z}\bigr) = \phi\bigl( (c,d)+10\mathbb{Z}\times 3\mathbb{Z}\bigr),$$&#xA;which shows that $\phi$ is well defined.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is also straightforward to check that $\phi$ is a ring homomorphism, and that $f(a,b) = \phi(\pi(a,b))$ for all $(a,b)\in R$. To show that $\phi$ is an isomorphism, note that it must be a surjection (because $\phi\circ\pi = f$ is a surjection, and if a composition is surjective, then the second map is surjective), and if &#xA;$$(0\bmod 10,0\bmod 3) = \phi\bigl( (a,b) + 10\mathbb{Z}\times 3\mathbb{Z}\bigr) = (a\bmod 10, b\bmod 3),$$&#xA;then $a\equiv 0\pmod{10}$ and $b\equiv 0 \pmod{3}$, so $(a,b)\in 10\mathbb{Z}\times 3\mathbb{Z}$; this means that&#xA;$$(a,b)+ 10\mathbb{Z}\times 3\mathbb{Z} = (0,0)+10\mathbb{Z}\times3\mathbb{Z},$$&#xA;so the map is one-to-one. Thus, $\phi$ is an isomorphism, as desired.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, $\phi$ is the &lt;em&gt;only&lt;/em&gt; homomorphism that &quot;fits&quot; into the diagram: because if $\psi$ also &quot;fits&quot;, then given &#xA;$(a,b)+10\mathbb{Z}\times 3\mathbb{Z}$, we can write this as $\pi(a,b)$, so&#xA;$$\psi\bigl((a,b)+10\mathbb{Z}\times 3\mathbb{Z}\bigr) = \psi(\pi(a,b)) = f(a,b) = \phi(\pi(a,b)) = \phi\bigl( (a,b)+10\mathbb{Z}\times 3\mathbb{Z}\bigr),$$&#xA;so $\psi=\phi$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&quot;Up to isomorphism.&quot;&lt;/strong&gt; What about your final question? Consider the integers &quot;in English&quot; and the integers &quot;in Spanish&quot; (&quot;los enteros&quot;). The names of the elements are different (&quot;one&quot;, &quot;two&quot;, etc., vs. &quot;uno&quot;, &quot;dos&quot;, etc.) But &quot;the integers&quot; and &quot;los enteros&quot; are &lt;strong&gt;essentially&lt;/strong&gt; the same ring: it's just a question of what we call the elements, not how we add them or multiply them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Explicitly, we have a &quot;translation&quot; betwen &quot;the integers&quot; and &quot;los enteros.&quot; This translation is such that if you take two integers, add them, and then translate the answer, you get the same thing as if you first translate the integer and then add them in Spanish. And the same thing with multiplication. Even though &quot;the integers&quot; and &quot;los enteros&quot; are &lt;em&gt;technically&lt;/em&gt; different things (as sets), if all we are concerned about is the ring-theoretic properties of these rings, then they are essentially &quot;the same ring.&quot; We don't really care whether we call the unit element &quot;one&quot; or &quot;uno&quot;, the important point is that it is a unit, that if you add it to itself you will never get the zero element (&quot;zero&quot; or &quot;cero&quot;, depending on which ring you are in) etc. The translation from English to Spanish and the translation from Spanish to English lets us go back and forth between the two at will, and none of the ring-theoretic properties will depend on which language we are using, but only on the ring-theoretic properties of the elements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, for all &lt;em&gt;ring theoretic mathematical purposes&lt;/em&gt;, &quot;the integers&quot; are &lt;em&gt;the same&lt;/em&gt; as &quot;los enteros&quot;: because we have a perfect translation between the two rings, namely, an isomorphism. So, the two are really &quot;equivalent&quot; (for any ring-theoretic investigation you might want to undertake).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means that there is really no point in distinguishing between the integers and los enteros. They are &quot;equivalent rings&quot;, because they is an isomorphism between them. We say this by saying that they are &quot;equivalent up to isomorphism.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(More formally: if you imagine the collection of all rings (okay, there are foundational problems with that, but ignore them; they are easy to get around), then the relation $R\cong S$ given by &quot;there is a ring isomorphism between $R$ and $S$&quot; is an equivalence relation: it is reflexive (the identity map shows $R\cong R$); symmetric (if $R\cong S$, then there is an isomorphism $f\colon R\to S$, so $f^{-1}\colon S\to R$ is an isomorphism from $S$ to $R$, proving $S\cong R$); and transitive: if $R\cong S$ and $S\cong T$, then there exist $f\colon R\to S$ and $g\colon S\to T$ that are isomorphisms, and $g\circ f\colon R\to T$ is an isomorphism, showing $R\cong T$. So we have an equivalence relation between rings, which partitions the collection of all rings into equivalence classes. When two rings are in the same equivalence class, we say they are &quot;equivalent up to isomorphism&quot;, because the equivalence relation is &quot;there is an isomorphism between&quot;.)&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-14T02:03:01.390" LastActivityDate="2011-03-14T02:03:01.390" />
  <row Id="26833" PostTypeId="1" CreationDate="2011-03-14T01:57:44.687" Score="2" ViewCount="123" Body="&lt;p&gt;I know the equation, $y = mx + b$ where $m$ is &lt;em&gt;slope&lt;/em&gt; and $b$ is &lt;em&gt;$y$-intercept&lt;/em&gt;, is a straight line. But I know also that $Ax + By = C$ is a straight line equation, but how does it represent a straight line?&lt;/p&gt;&#xA;" OwnerUserId="8218" LastEditorUserId="6296" LastEditorDisplayName="" LastEditDate="2011-03-14T03:14:42.170" LastActivityDate="2011-03-14T03:14:42.170" Title="How is  $Ax + By = C$ the equation of a straight line?" Tags="&lt;algebra-precalculus&gt;" AnswerCount="3" />
  <row Id="26834" PostTypeId="2" ParentId="26833" CreationDate="2011-03-14T01:59:32.297" Score="4" ViewCount="" Body="&lt;p&gt;If you solve for $y$ in $Ax+By = C$ ($B \neq 0$) you get slope-intercept form. In other words, $$By = C-Ax$$ $$y = \frac{C}{B}-\frac{Ax}{B}$$ so that $m = -A/B$ and $b = C/B$ in $y = mx+b$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As Gonzalo points out, if $B = 0$ then we get $Ax = C$ or $x = C/A$ which is a line parallel to the $y$-axis through the point $(C/A,0)$.&lt;/p&gt;&#xA;" OwnerUserId="2548" LastEditorUserId="2548" LastEditorDisplayName="" LastEditDate="2011-03-14T02:06:52.103" LastActivityDate="2011-03-14T02:06:52.103" CommentCount="1" />
  <row Id="26835" PostTypeId="2" ParentId="26833" CreationDate="2011-03-14T02:04:03.347" Score="1" ViewCount="" Body="&lt;p&gt;If $B=0$ and $A\neq 0$, then the equation is simply $x=C/A$, which is a straight line parallel to the $y$-axis passing through the point $(C/A,0)$; if $B\neq 0$, then you can rewrite $Ax+By=C$ as $y=\dfrac{-A}{B}x+\dfrac{C}{B}$ which is a straight line with slope $\dfrac{-A}{B}$ and $y$-intercept $\dfrac{C}{B}$.&lt;/p&gt;&#xA;" OwnerUserId="8050" LastActivityDate="2011-03-14T02:04:03.347" />
  <row Id="26836" PostTypeId="2" ParentId="26824" CreationDate="2011-03-14T02:05:09.970" Score="6" ViewCount="" Body="&lt;p&gt;Okay, I think I see what you want, and I claim that it is impossible to find an answer in the general case. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you consider the $n\times n$ matrix $M$ as a collection of $n$ row vectors, $m_1\ldots m_n$ in $\mathbb{R}^n$, computing the determinant is equivalent to computing the exterior product &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ m_1 \wedge m_2 \wedge \ldots \wedge m_n $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your operation seems to be trying to compare the above to &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ (Am_1) \wedge m_2 \wedge \ldots \wedge m_n $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and you are asking whether there exists a linear transformation $B$ such that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ (Am_1) \wedge \ldots \wedge m_n = m_1\wedge (Bm_2) \wedge m_3 \wedge \ldots \wedge m_n $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for all sets of vectors $\{m_1,\ldots,m_n\}$. I claim that this is not possible. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Fix a set of $\{ m_1, m_2, \ldots, m_n\}$ linearly independent. Let $m&amp;#39;_1 = m_1 + m_3$. (Assume $n \geq 3$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider an arbitrary linear transformation $A$ which gives $Am_1 = m_2$, and $Am&amp;#39;_1 = m_1$. This set of linear equations can be solved with a possibly non-unique solution. So such $A$ exists.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, acting on the matrix $M$ corresponding to $\{m_1, \ldots, m_n\}$, the transformation $Am_1 = m_2$ implies that multiplying $A$ to $m_1$ makes the transformed matrix have determinant zero. Whereas the original matrix, being full rank, has non-zero determinant. So if $B$ were a linear transformation to act on $m_2$ such that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ 0 = (Am_1) \wedge m_2 \ldots m_n = m_1\wedge (Bm_2) \wedge m_3 \ldots m_n $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;one must have that $Bm_2$ is in the span of $m_1, m_3, \ldots m_n$. By construction, this implies that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ m&amp;#39;_1 \wedge (Bm_2) \wedge m_3 \ldots m_n = 0 $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, we have that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ 0 \neq m_1 \wedge m_2 \wedge m_3\ldots m_n = (Am&amp;#39;_1) \wedge m_2 \wedge m_3\ldots m_n $$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And so we have shown that for dimension $n \geq 3$, what you want to do is impossible (there are too many degrees of freedom). &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;For dimension 2, on the other hand, things are just right. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\det M = M_{11} M_{22} - M_{12} M_{21} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Acting on the first row by $A$, you have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\det M&amp;#39; = M_{11}A_{11}M_{22} + M_{12} A_{21} M_{22} - M_{12}A_{22} M_{21} - M_{11} A_{12} M_{21}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which we regroup as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \det M&amp;#39; = M_{11} ( A_{11} M_{22} - A_{12} M_{21} ) - M_{12} (A_{22} M_{21} - A_{21} M_{22} ) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which you can solve to get that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ B = \begin{pmatrix} A_{22} &amp;amp; - A_{21} \\ - A_{12} &amp;amp; A_{11} \end{pmatrix} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or that $B$ is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Adjugate_matrix&quot; rel=&quot;nofollow&quot;&gt;adjugate matrix&lt;/a&gt; of $A$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Here we actually geometrically used the fact that the &lt;a href=&quot;http://en.wikipedia.org/wiki/Hodge_star&quot; rel=&quot;nofollow&quot;&gt;Hodge star&lt;/a&gt; operator $*$ sends vectors to vectors in 2 dimensions, and so $B = *A^T*$ is what we want.)&lt;/p&gt;&#xA;" OwnerUserId="1543" LastEditorUserId="1543" LastEditorDisplayName="" LastEditDate="2011-03-14T02:27:48.657" LastActivityDate="2011-03-14T02:27:48.657" CommentCount="11" />
  <row Id="26837" PostTypeId="1" AcceptedAnswerId="26844" CreationDate="2011-03-14T02:08:52.670" Score="15" ViewCount="1039" Body="&lt;p&gt;Is the concept of an odd imaginary number defined/well-defined/used in mathematics? I searched around but couldn't find anything. Thanks!&lt;/p&gt;&#xA;" OwnerUserId="3731" LastEditorUserId="3731" LastEditorDisplayName="" LastEditDate="2011-05-22T17:10:51.130" LastActivityDate="2011-05-31T22:40:32.123" Title="Do odd imaginary numbers exist?" Tags="&lt;algebra-precalculus&gt;&lt;complex-numbers&gt;&lt;parity&gt;" AnswerCount="6" CommentCount="2" FavoriteCount="2" />
  <row Id="26838" PostTypeId="2" ParentId="26837" CreationDate="2011-03-14T02:10:23.490" Score="2" ViewCount="" Body="&lt;p&gt;A complex number is of the form $z = x+iy$. The real part is $x$ and the imaginary part is $y$. So the imaginary part can be odd.&lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2011-03-14T02:10:23.490" />
  <row Id="26839" PostTypeId="2" ParentId="26826" CreationDate="2011-03-14T02:12:09.853" Score="0" ViewCount="" Body="&lt;p&gt;I take this to mean that $\log_c(d) - \log_b(d) = \log_b(d) - \log_a(d) = x$ for some $x$. In this case, the key is to use the properties of logarithms (including addition and multiplication of logarithms and changing the base) to manipulate these into a useful expression. My approach would be to start with $(ac)^{\log_{a}b}$, which I simplified to $b*c^{\log_{a}b}$, and using the difference relation derive $\log_{a}b = b/(1/(\log_{b}c)-1+b)$. Assuming that I made no mistakes in manipulation for the last derivation, this should easily give you an answer.&lt;/p&gt;&#xA;" OwnerUserId="8173" LastEditorUserId="8173" LastEditorDisplayName="" LastEditDate="2011-03-14T03:34:16.403" LastActivityDate="2011-03-14T03:34:16.403" CommentCount="1" />
  <row Id="26840" PostTypeId="2" ParentId="26837" CreationDate="2011-03-14T02:12:23.877" Score="3" ViewCount="" Body="&lt;p&gt;Sure.  If you define an even number as any number $k$ expressible as $k=2z$ where $z$ is a Gaussian integer, then an odd number would be any number $k$ for which no such $z$ can be found.  This would be any number whose real or imaginary part (or both) is odd.&lt;/p&gt;&#xA;" OwnerUserId="7862" LastActivityDate="2011-03-14T02:12:23.877" />
  <row Id="26841" PostTypeId="2" ParentId="26824" CreationDate="2011-03-14T02:13:44.397" Score="1" ViewCount="" Body="&lt;p&gt;The case of only two rows is special.  I don't think there is an answer independent of M with more than two rows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the first row of M happens to be zero, it doesn't matter what we do; the modified M will still have determinant zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Multiplying the first row of M by A (from the right) amounts to an arbitrary substitution of one row for another.  If we assume A nonsingular, which is essential even to your two row solution, then we are only restricting ourselves to replacing one nonzero row by any other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you are asking is therefore to replace the first row of M (assumed nonzero) by another nonzero row, and compensate for the effect on $det$ M by doing something to the second row of M, without benefit of knowing anything about the rest of matrix M.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All that matrix A gives us is the replacement for the first row of M.  All that matrix B defines for us is the replacement for the second row of M.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the 2x2 case we can take B to be A scaled by the reciprocal of $det$ A.  If M has three or more rows, there is no choice of B that would work for all matrices M.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An easy way to show this is to construct nonsingular A and M such that replacing the first row of M by its product with A produces a singular result regardless of what is done to the second row of M.  For example, take A to be any nonsingular matrix other than a multiple of the identity matrix, so that for some row vector $u$, $u$ and $u$A will be linearly independent.  Then take $v$ to be any row vector linearly independent of $u$ and $u$A, and define M to be the three rows $u$, $v$, and $u$A.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then if you perform your operation on the first two rows of M, regardless of choice of B, the result will be a singular matrix M&quot;, since the first and third rows of M&quot; are equal.  Hence $det$ M&quot; cannot be equal to $det$ M, since we constructed a nonsingular M.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added:&lt;/strong&gt; In the 2x2 case with A nonsingular, we know that multiplying both rows of M amounts to getting matrix product MA, whose determinant is ($det$ A)($det$ M).  So if we use B = (1/$det$ A) A, it amounts to multiplying M on the right by A and then factoring out $det$ A from the second row of M, which gets us back to a 2x2 matrix with the same determinant as before.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Correcting my misunderstanding:&lt;/strong&gt;  Jack wants to require, given matrix A, a matrix B such that for any matrix M = $ \left( \begin{array} {c} u \\ v \end{array} \right) $, for:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ M&amp;#39; = \left( \begin{array} {c} uA \\ v \end{array} \right) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ M&amp;#39;&amp;#39; = \left( \begin{array} {c} u \\ vB \end{array} \right) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;it is true that $det M&amp;#39; = det M&amp;#39;&amp;#39;$.  When my analysis is applied &quot;backwards&quot; to go from M' to M to M&quot;, the result is the same as Willie Wong's, namely B is the adjugate of A.  For if we apply my recipe, changing A to $A^{-1}$ for going from M' to M, then:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ B = (1/det A^{-1}) A^{-1} = adj(A) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which as Willie pointed out is defined even for singular A.&lt;/p&gt;&#xA;" OwnerUserId="3111" LastEditorUserId="3111" LastEditorDisplayName="" LastEditDate="2011-03-14T06:07:09.353" LastActivityDate="2011-03-14T06:07:09.353" CommentCount="8" />
  <row Id="26842" PostTypeId="2" ParentId="26837" CreationDate="2011-03-14T02:21:29.530" Score="2" ViewCount="" Body="&lt;p&gt;The concepts of &quot;even&quot; or &quot;odd&quot; come from the ideal structure of the ring of integers (I am not certain if this will mean anything to you). In the integers, numbers are classified as &quot;even&quot; if their image in $\mathbb{Z}/(2)$ is $0$ and &quot;odd&quot; if it is 1. The more useful concept of &quot;even&quot; or &quot;odd&quot; in the Gaussian integers would be to classify Gaussian integers by their image in $(\mathbb{Z}+\mathbb{Z}i)/(2)$. This would lead to four possibilities: numbers with both parts even, numbers with even real part and odd imaginary part, numbers with odd real part and even imaginary part, and numbers with both parts odd.&lt;/p&gt;&#xA;" OwnerUserId="8173" LastActivityDate="2011-03-14T02:21:29.530" />
  <row Id="26843" PostTypeId="2" ParentId="26837" CreationDate="2011-03-14T02:25:38.423" Score="10" ViewCount="" Body="&lt;p&gt;As I mentioned in an &lt;a href=&quot;http://math.stackexchange.com/questions/16011/what-makes-9-special/16015#16015&quot;&gt;earlier post&lt;/a&gt; on casting out nines, one can apply &lt;strong&gt;parity&lt;/strong&gt; arguments in any ring that has $\ \mathbb Z/2\ $ as an image, e.g. all rationals with odd denominator, or the Gaussian integers $\rm\:\mathbb Z[i]\:$, where the image $\rm\ \mathbb Z[i]/(2,i-1) \cong \mathbb Z/2\ $ yields the natural parity definition that $\rm\ a+b\:i\ $ is even iff $\rm\ a\equiv b\ \ (mod\ 2)\ $, i.e. if $\rm\ a+b\:i\ $ maps to $\:0\:$ via the above isomorphism, which maps $\rm\ 2\to 0,\ i\to 1\:$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Generally it's easy to show that if $\rm\:2\nmid f(x)\not\in\mathbb Z\:$ then the number of ways to define parity in the ring $\rm\ \mathbb Z[w] \cong \mathbb Z[x]/(f(x))\ $ is given by the number of roots of $\rm\: f(x)\: $ modulo $2\:.\ $ For suppose that there exists a homomorphism $\rm\ h\: :\: \mathbb Z[w]\to \mathbb Z/2\:.\:$ Then $\rm\:w\:$ must map to a root of $\rm\:f(x)\:$ in $\rm\ \mathbb Z/2\:.\ $ Thus if $\rm\ f(0)\equiv 0\ (mod\ 2)\ $ then $\rm\: \mathbb Z[w]/(2,w) \cong \mathbb Z[x]/(2,x,f(x)) \cong \mathbb Z/2\ $ since $\rm\: x\: |\: f(x)\ (mod\ 2)\:.\: $ Similarly if $\rm\ f(1)\equiv 0\ (mod\ 2)\ $ then $\rm\: \mathbb Z[w]/(2,w-1) \cong \mathbb Z[x]/(2,x-1,f(x)) \cong \mathbb Z/2\ $ since $\rm\: x-1\: |\: f(x)\ (mod\ 2)\:. $ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's consider some simple examples. Since $\rm\ x^2+1\ $ has the unique root $\rm\ x\equiv 1\ (mod\ 2)\:,\:$ the Gaussian integers $\rm\ \mathbb Z[i]\cong \mathbb Z[x]/(x^2+1)\ $ have a unique definition of parity - with $\rm\:i\:$ being odd. Since $\rm\ x^2+x+1\ $ has no roots modulo $\: 2\:,\: $ there is no way to define parity for the Eisenstein integers $\rm\ \mathbb Z[w] \cong \mathbb Z[x]/(x^2+x+1)\:.\ $ Indeed since $\rm\ w^3 = 1\ $ we infer that $\rm\: w \equiv 1\ (mod\ 2)\ $ contra  $\rm\ w^2+w+1 = 0\:.\ $ On the other hand $\rm\ \mathbb Z[w] \cong \mathbb Z[x]/(x^2+x+2)\ $ has two parity structures since both $\:0\:$ and $\rm\:1\:$ are roots of $\rm\ x^2 + x + 2\ $ modul0 $\rm\:2\:,\:$ so we can define $\rm\:w\:$ to be either even or odd.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any parity argument that's truly ring-theoretic will generalize to any ring $\rm\:R\:$ with parity, e.g. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;PARITY ROOT TEST&lt;/strong&gt;  $\ $ Suppose $\rm\:f(x)\:$ is a polynomial with coefficients in a ring $\rm\:R\:$ with parity. Then $\rm\:f(x)\:$ has no roots in $\rm\:R\:$ if $\rm\:f(x)\:$ has constant coefficient and coefficient sum both being odd. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof&lt;/strong&gt; $\ $ The test simply verifies  $\rm\ f(0)\: =\: f(1)\: =\: 1\:\ (mod\ 2)\:,\ $ i.e. $\rm\: f(x)\:$ has no roots $\rm\: (mod\ 2)\:.\ $ Therefore $\rm\:f(x)\:$ has no roots in $\rm\:R\:.\quad$  &lt;strong&gt;QED&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a corollary, we deduce that no ring $\rm\:R\:$ with parity contains a primitive cube root of one. Indeed, if so, $\rm\:R\:$ contains a root of $\rm\ f(x) = x^2+x+1\:,\ $ contra $\rm\:f(x)\:$ has no roots in $\rm\:R\:$ by the parity root test. Hence, for example, the Gaussian integers do not contain a primitive cube root of one.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2011-05-31T22:40:32.123" LastActivityDate="2011-05-31T22:40:32.123" CommentCount="7" />
  <row Id="26844" PostTypeId="2" ParentId="26837" CreationDate="2011-03-14T02:26:29.663" Score="39" ViewCount="" Body="&lt;p&gt;&quot;Odd&quot; has several meanings in mathematics: you have odd integers (those which are not multiples of $2$); you have odd functions (those that satisfy $f(-x) = -f(x)$ for all $x$); and possibly others.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to stick to the first meaning, then there's two things to keep in mind: even for just &lt;em&gt;real&lt;/em&gt; numbers, &quot;odd&quot; in the sense of &quot;not a multiple of $2$&quot; doesn't really work very well, because &lt;em&gt;every&lt;/em&gt; real number is a multiple of $2$: given $r\in\mathbb{R}$, $r = 2\left(\frac{r}{2}\right)$, and $\frac{r}{2}$ is a real number. The same is true for complex numbers: if $a+bi\in\mathbb{C}$, then $a+bi = 2\left(\frac{a}{2} + \frac{b}{2}i\right)$, so every complex number would be &quot;a multiple of $2$&quot;, so no complex number would be odd. So this concept does not really do much for complex numbers as a whole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, you can restrict to those complex numbers which have &lt;em&gt;integer&lt;/em&gt; real and complex part: $a+bi$ with $a,b\in\mathbb{Z}$ (instead of $a,b\in\mathbb{R}$, like in $\mathbb{C}$). These are called the &lt;strong&gt;Gaussian integers&lt;/strong&gt; because they were first studied by Gauss. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For these numbers, you can talk about &quot;multiples of $2$&quot;: a Gaussian integer $a+bi$ is a multiple of $2$ if and only if both $a$ and $b$ are even: because if $a+bi = 2(x+yi)$ with $a,b,x,y\in\mathbb{Z}$, then $a=2x$ is even and $b=2y$ is also even. So the &quot;odd Gaussian integers&quot; would be all Gaussian integers that are &lt;em&gt;not&lt;/em&gt; multiples of $2$, namely the $a+bi$ that have &lt;strong&gt;either&lt;/strong&gt; $a$ or $b$ odd. Note that $1$ would be an &quot;odd Gaussian integer&quot; (which looks good, because $1$ is an odd integer), but then again so would $1+2i$ (which may not look so good). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is also a slight wrinkle: in the integers, if you add two integers of the same parity, you will always get something that is &quot;even&quot;, and if you add two integers of different parity you will get something that is &quot;odd.&quot; This does not happen with the above notion of &quot;even&quot; in the Gaussian integers. For instance, $1+2i$ is &quot;odd&quot;, and so is $2+i$; if we add them, we get $3+3i$ which is &lt;strong&gt;also&lt;/strong&gt; &quot;odd.&quot; In fact, we have four different kinds of Gaussian integers: the &quot;even&quot; ones (both real and imaginary parts are even); the &quot;even-odd&quot; ones (real part even, imaginary part odd); the &quot;odd-even&quot; ones (real part odd, imaginary part even); and the &quot;odd&quot; ones (both real and imaginary part odd). It is only if you add two of the same &lt;em&gt;kind&lt;/em&gt; that you will get an &quot;even&quot; Gaussian, and if you add two different kinds you will get an &quot;odd&quot; Gaussian. So this concept of &quot;even&quot; and &quot;odd&quot; does not seem to behave like it does in the integers. &lt;em&gt;Added.&lt;/em&gt; What is worse, as Bill points out in comments, even this does not work well with multiplication, since for example the product of an &quot;even-odd&quot; by an &quot;even-odd&quot; gives an &quot;odd-even&quot;, not an &quot;even-odd&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We might, then, want to look at another possibility. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another possibility is to notice that $2i = (1+i)^2$, and $i$ is invertible in the Gaussian integers. So instead of looking at the multiples of $2$, you can try looking at the multiples of $1+i$ (just like you don't define &quot;odd&quot; in terms of multiples of $4$ in the integers; I bring up $4$ because $4=2^2$). When is a Gaussian integer a multiple of $1+i$? &#xA;$$(x+yi)(1+i) = (x-y) + (x+y)i.$$&#xA;Can we recognize such numbers? I claim they are precisely those Gaussian integers $a+bi$ with $a+b$ even. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Indeed, if $a+bi$ is a multiple of $1+i$, then as above we have $a=x-y$ and $b=x+y$ for some integers $x$ and $y$, so $a+b = (x-y)+(x+y) = 2x$ is even. Conversely, suppose that $a+bi$ has $a+b$ even, $a+b = 2k$. Then $a-b$ is also even (since $a-b = (a+b)-2b$), so we can write $a-b = 2\ell$. Then&#xA;\begin{align*}&#xA;(k -\ell i)(1+i) &amp;amp;= (k+\ell) + (k-\ell)i\\&#xA;&amp;amp;= \left( \frac{a+b}{2} + \frac{a-b}{2}\right) + \left(\frac{a+b}{2} - \frac{a-b}{2}\right)i\\&#xA;&amp;amp;= a + bi,&#xA;\end{align*}&#xA;so $a+bi$ is a multiple of $1+i$. So if you define &quot;odd&quot; in terms of &quot;multiple of $1+i$, then it corresponds precisely to whether or not $a\equiv b\pmod{2}$: if $a$ and $b$ have the same parity, then $a+bi$ is &quot;even&quot;; if $a$ and $b$ have different parity then $a+bi$ is &quot;odd&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It also has the advantage of mirroring a bit better what happens with parity in the integers: if you add two &quot;even&quot; or two &quot;odd&quot; Gaussian integers (under this definition), then the sum is &quot;even&quot;; and if you add an &quot;even&quot; and an &quot;odd&quot; Gaussian integer you get an &quot;odd&quot; Gaussian integer. Also, if you multiply an &quot;even&quot; Gaussian by any Gaussian you get an &quot;even&quot; Gaussian: for if $a$ and $b$ have the same parity, then&#xA;$$(a+bi) (x+yi) = (ax-by) + (ay+bx)i.$$&#xA;If both $a$ and $b$ are even, then so are $ax-by$ and $ay+bx$, so the result is even. If both $a$ and $b$ are odd, then either $x$ and $y$ have the same parity, in which case both $ax-by$ and $ay+bx$ are even; or else $x$ and $y$ have different parity, so that both $ax-by$ and $ay+bx$ are odd. Either way, the product is &quot;even.&quot; Similarly, if you multiply two &quot;odd&quot; Gaussians, the result will be &quot;odd.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I think the latter concept is a bit more intuitive, but that may be just me. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Post data.&lt;/em&gt; There is in fact a lot of very interesting stuff in the background of the above; considering $1+i$ instead of $2$ in the Gaussian integers comes from &lt;em&gt;Algebraic Number Theory&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-14T16:06:44.243" LastActivityDate="2011-03-14T16:06:44.243" CommentCount="10" />
  <row Id="26845" PostTypeId="2" ParentId="26833" CreationDate="2011-03-14T02:41:15.537" Score="3" ViewCount="" Body="&lt;p&gt;It is possible to think about $Ax+By=C$ in a few ways.  As in other answers, you can rewrite it either in the form $x=k$ or in the form $y=mx+b$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you know about vectors, you can think of the vector  $\langle A,B\rangle$ as being some vector orthogonal (perpendicular) to the line and $C=\langle A,B\rangle\cdot\langle x_0,y_0\rangle$ for any given point $(x_0,y_0)$ on the line (see &lt;a href=&quot;http://math.stackexchange.com/questions/25190/how-to-obtain-line-equation-of-the-form-ax-bx-c-0/25195#25195&quot;&gt;my answer here&lt;/a&gt; for more details on this idea).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $A=0$ or $B=0$, then the equation describes a vertical or horizontal line, since it says that $y$ or $x$ is a constant.  Given that, from here on, let's assume that $A,B\neq 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As $x$ decreases by $\frac{1}{A}$, $y$ has to increase by $\frac{1}{B}$ to keep the sum $Ax+By$ constantly equal to $C$ (and vice versa if $x$ increases).  Since the change in $y$ is constant for a constant change in $x$, the relationship between $x$ and $y$ is linear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In terms of graphing from $Ax+By=C$ form, let $x=0$ so that $By=C$ or $y=\frac{C}{B}$, to get that $(0,\frac{C}{B})$ is the $y$-intercept.  Similarly, $(\frac{C}{A},0)$ is the $x$-intercept.  Plotting these two points and drawing the unique line through them gives the graph of $Ax+By=C$.&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2011-03-14T02:41:15.537" />
  <row Id="26846" PostTypeId="1" CreationDate="2011-03-14T02:43:54.600" Score="2" ViewCount="194" Body="&lt;p&gt;&lt;strong&gt;(See edits at the bottom)&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm trying to use Bézier curves as an animation tool. Here's an image of what I'm talking about:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/iLxst.jpg&quot; alt=&quot;Example of a Bézier curve&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically, the value axis can represent anything that can be animated (position, scaling, color, basically any numerical value). The Bézier curve is used to control the speed at which the value is changing as well as it start and ending value and time. In this graphic, the animated value would slowly accelerate to a constant speed, then decelerate and stop.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is, that Bézier curve is defined with parametric equations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f_x(t):=(1-t)^3p_{1x} + 3t(1-t)^2p_{2x} + 3t^2(1-t)p_{3x} + t^3p_{4x}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f_y(t):=(1-t)^3p_{1y} + 3t(1-t)^2p_{2y} + 3t^2(1-t)p_{3y} + t^3p_{4y}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I need is a representation of that same Bézier curve, but defined as &lt;code&gt;value = g(time)&lt;/code&gt;, that is, &lt;code&gt;y = g(x)&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've tried solving for &lt;code&gt;t&lt;/code&gt; in the &lt;code&gt;x&lt;/code&gt; equation and substituting it in the &lt;code&gt;y&lt;/code&gt; equation, but that 3rd degree is giving me some difficulty.&#xA;I also tried integrating the derivative of the Bézier curve (&lt;code&gt;dy/dx&lt;/code&gt;) with respect to &lt;code&gt;t&lt;/code&gt;, but no luck.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any ideas?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Note&lt;/strong&gt; : &quot;Undefined&quot; situations are avoided by preventing the tangent control points from going outside the hull horizontally, preventing any overlap in the time axis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt; :&#xA;I have found two possible solutions. One uses Decasteljau's algorithm to approximate the $s$ parameter from the $t$ parameter, $s$ being the parameter of the parametric curve and $t$ being the time parameter. &lt;a href=&quot;https://collada.org/public_forum/viewtopic.php?p=4016&amp;amp;sid=ae21755569bd1c1df6cff58061cd6ddb#p4016&quot; rel=&quot;nofollow&quot;&gt;Here (at the bottom)&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other, from what I can understand of it, recreates a third degree polynomial equation matching the curve by solving a system of linear equation. &lt;a href=&quot;http://www.gamedev.net/topic/545021-hermite-splines/page__view__findpost__p__4514299&quot; rel=&quot;nofollow&quot;&gt;Here&lt;/a&gt;. I understand the idea, but I'm not sure of the implementation. Any help?&lt;/p&gt;&#xA;" OwnerUserId="8222" LastEditorUserId="8222" LastEditorDisplayName="" LastEditDate="2011-03-14T19:46:55.787" LastActivityDate="2011-03-15T23:54:33.780" Title="Is there an explicit form for cubic Bézier curves?" Tags="&lt;bezier-curve&gt;&lt;parametric&gt;" AnswerCount="4" CommentCount="7" FavoriteCount="1" />
  <row Id="26847" PostTypeId="1" AcceptedAnswerId="26851" CreationDate="2011-03-14T03:14:00.673" Score="3" ViewCount="172" Body="&lt;p&gt;I have heard much about the numerous appearances of the ratio found in nature: 1.6180339887.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there any actual mathematical uses that have been found of this number? What are its advantages? Just curiosity, really. &lt;/p&gt;&#xA;" OwnerUserId="8171" LastEditorUserId="8171" LastEditorDisplayName="" LastEditDate="2011-03-14T04:27:56.350" LastActivityDate="2011-03-14T16:23:04.097" Title="Uses of the 'Golden Ratio'" Tags="&lt;soft-question&gt;&lt;math-history&gt;" AnswerCount="3" CommentCount="3" />
  <row Id="26849" PostTypeId="2" ParentId="26847" CreationDate="2011-03-14T03:18:35.450" Score="3" ViewCount="" Body="&lt;p&gt;This is one realm in which Wikipedia can be amazingly helpful. &lt;a href=&quot;http://en.wikipedia.org/wiki/Golden_ratio&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Golden_ratio&lt;/a&gt; lists all sorts of applications, with citations if you're picky (which is commendable).&lt;/p&gt;&#xA;" OwnerUserId="8173" LastActivityDate="2011-03-14T03:18:35.450" CommentCount="2" />
  <row Id="26850" PostTypeId="2" ParentId="26846" CreationDate="2011-03-14T03:25:17.720" Score="2" ViewCount="" Body="&lt;p&gt;Assuming the intended meaning of $p1x$, etc. was as Ross believed, you should be able to solve for t explicitly in either equation using the formula at &lt;a href=&quot;http://en.wikipedia.org/wiki/Cubic_equation#General_formula_of_roots&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Cubic_equation#General_formula_of_roots&lt;/a&gt;. Of course, this is a bulky solution and you would need to identify which solution is real (though there should be only 1 assuming that the curve can be represented as $y = g(x)$).&lt;/p&gt;&#xA;" OwnerUserId="8173" LastActivityDate="2011-03-14T03:25:17.720" CommentCount="2" />
  <row Id="26851" PostTypeId="2" ParentId="26847" CreationDate="2011-03-14T04:18:27.213" Score="4" ViewCount="" Body="&lt;p&gt;I'm not sure if you mean &quot;man-made&quot; as in real world applications or as in mathematical applications, but one of my favorite things about the golden ratio is that it is somehow the hardest number to approximate by rationals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For one thing, it has the slowest converging simple continued fraction expansion. This is reflected in the fact that if you apply Euclid's algorithm to two successive Fibonacci numbers, you get a quotient of 1 at every step.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is also Hurwitz's theorem, which says that for any rational number $\xi$, there are infinitely rationals $\frac{n}{m}$ such that$$\left| \xi - \frac{n}{m} \right| &amp;lt; \frac{1}{\sqrt{5}m^2}.$$ The constant on the RHS can't be improved since, if we replace $\sqrt{5}$ with some larger number, then the statement of the theorem doesn't work when we let $\xi$ be the golden ratio. (Informally, we can't get too close, in a number theoretic way, to the golden ratio.)&lt;/p&gt;&#xA;" OwnerUserId="7676" LastActivityDate="2011-03-14T04:18:27.213" CommentCount="2" />
  <row Id="26852" PostTypeId="2" ParentId="26846" CreationDate="2011-03-14T04:28:54.890" Score="2" ViewCount="" Body="&lt;p&gt;Notice that if you happen to have $p_{1y}=0$, $p_{2y}=1/3$, $p_{3y}=2/3$ and $p_{4y}=1$, so that $f_y(t)=t$, then the graph you are after is actually of the inverse function of $f_x(t)$, which is going to be a mess and inevitable as complicated as the usual formula for the roots of a cubic polynomial.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That shows that the general formula necessarily has to be complicated. I doubt there is anything useful to be done...&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2011-03-14T04:28:54.890" />
  <row Id="26853" PostTypeId="2" ParentId="26691" CreationDate="2011-03-14T04:40:18.937" Score="1" ViewCount="" Body="&lt;p&gt;Be careful that μ(H) can mean different things in the literature.  I'll use the more specific μ(H,G) to indicate the Mobius function for the interval between H and G in the subgroup lattice of G.   Philip Hall uses μ(H) for μ(H,G) in his paper:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Hall, P. &quot;The Eulerian functions of a group.&quot;&#xA;  Quart. J. Math., Oxf. Ser. &lt;strong&gt;7&lt;/strong&gt;, 134-151 (1936).&#xA;  JFM:&lt;a href=&quot;http://www.zentralblatt-math.org/zmath/en/search/?q=an:62.0082.02&quot; rel=&quot;nofollow&quot;&gt;62.0082.02&lt;/a&gt;&#xA;  DOI:&lt;a href=&quot;http://dx.doi.org/10.1093/qmath/os-7.1.134&quot; rel=&quot;nofollow&quot;&gt;10.1093/qmath/os-7.1.134&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In order to calculate μ(H,G), I first calculate μ(1,G).  This will allow me to use some nice reductions.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;My copy of Hall is at the office, but I remembered the paper I was reading and it appears to be perfect for getting started with Möbius functions on groups (especially with regards to getting into the later rep theory):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Hawkes, T.; Isaacs, I. M.; Özaydin, M.&#xA;  &quot;On the Möbius function of a finite group.&quot;&#xA;  Rocky Mountain J. Math. &lt;strong&gt;19&lt;/strong&gt; (1989), no. 4, 1003–1034.&#xA;  MR&lt;a href=&quot;http://www.ams.org/mathscinet-getitem?mr=1039540&quot; rel=&quot;nofollow&quot;&gt;1039540&lt;/a&gt;&#xA;  DOI:&lt;a href=&quot;http://dx.doi.org/10.1216/RMJ-1989-19-4-1003&quot; rel=&quot;nofollow&quot;&gt;10.1216/RMJ-1989-19-4-1003&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This gives a remarkably useful reduction formula (Cor 3.3) for:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;b&gt;μ(1,G) = −k⋅μ(1,G/A)&lt;/b&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;where A is any abelian minimal normal subgroup of G with exactly k complements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This formula (in its Eulerian form) is due to:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Gaschütz, Wolfgang.&#xA;  &quot;Die Eulersche Funktion endlicher auflösbarer Gruppen.&quot;&#xA;  Illinois J. Math. &lt;strong&gt;3&lt;/strong&gt; (1959) 469–476.&#xA;  MR&lt;a href=&quot;http://www.ams.org/mathscinet-getitem?mr=107670&quot; rel=&quot;nofollow&quot;&gt;107670&lt;/a&gt;&#xA;  URL:&lt;a href=&quot;http://projecteuclid.org/getRecord?id=euclid.ijm/1255455454&quot; rel=&quot;nofollow&quot;&gt;euclid.ijm/1255455454&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Using this reduction formula trivializes the cyclic and dihedral groups, along with the following observation on p. 1010 of Hawkes, et al.  The Möbius function μ(1,H) does not depend on the embedding of H in G, it is an invariant of the isomorphism type of the group H.  In particular, we can forget about the surrounding group G and just think about the cyclic group and the dihedral group (rather than cyclic subgroups or dihedral subgroups).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If G is a cyclic group of non-square-free order, then it has a cyclic Sylow p-subgroup P that is not order p.  The only minimal normal subgroup of G contained in P is Ω(P), the subgroup of G consisting of the elements of order dividing p.  This subgroup has no complements at all (just like 2Z/4Z is not complemented in Z/4Z, because Z/4Z has only a single involution).  Hence k = 0 and μ(1,G) = −0⋅μ(1,G/Ω(P)) = 0.  The normal Möbius function of an integer is 0 if the integer is not square-free.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If G is cyclic of square-free order, then each of its Sylow p-subgroups is order p, and so a minimal normal subgroup P.  Each such subgroup P has exactly one complement (the Hall P′-subgroup of G consisting of those elements of order coprime to p).  Hence k = 1 and we have μ(1,G) = −μ(1,G/P).  Since G/P is a cyclic group of square-free order with one less prime factor we get the formula: μ(1,G) = (−1)&lt;sup&gt;j&lt;/sup&gt; where |G|=n is a square-free number with exactly j prime factors.  This is exactly the formula for μ(n).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now if G is dihedral of order 2n, then μ(1,G) = −n⋅μ(n), again by considering a minimal normal subgroup of order p, which has p complements, unless p&lt;sup&gt;2&lt;/sup&gt; divides n.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Now we handle the case of μ(H,G).  If N is a normal subgroup of G contained within H, then μ(H,G) = μ(H/N,G/N) by Noether's lattice isomorphism theorem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This handles the cyclic groups immediately, since H=N is normal, we just get μ(H,G) = μ(1,G/H) = μ([G:H]).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This also handles the cyclic subgroups (of rotations) in a dihedral group: if G is dihedral and H ≤ [G,G] then H is cyclic and normal in G, with a dihedral quotient (possibly of order 2 or 4).  Hence μ(H,G) = μ(1,G/H) = −kμ(k), where [G:H]=2k.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If H is a dihedral subgroup of a dihedral group G, then N=H∩[G,G] is cyclic, [H:N]=2 and N is normal in G (N is the group of rotations in H).  μ(H,G) = μ(H/N,G/N), and so we are left with the case that G is dihedral and H is a reflection (order 2 subgroup not contained within the group of rotations).  In this case, H is contained only in dihedral subgroups, one for each divisor of n.  In other words, the subgroup lattice from H to G is isomorphic to the divisor lattice of [G:H].  Hence μ(H,G) = μ([G:H]).&lt;/p&gt;&#xA;" OwnerUserId="583" LastEditorUserId="583" LastEditorDisplayName="" LastEditDate="2011-03-14T20:45:26.010" LastActivityDate="2011-03-14T20:45:26.010" CommentCount="2" />
  <row Id="26854" PostTypeId="1" CreationDate="2011-03-14T04:42:51.253" Score="16" ViewCount="562" Body="&lt;p&gt;I am aware of an intuitive explanation  for curl grad F = 0 (a block placed on a mountainous frictionless surface will slide to lower ground without spinning), and was wondering if there were a similar explanation for div curl F = 0.&lt;/p&gt;&#xA;" OwnerUserId="8223" LastActivityDate="2011-03-15T21:04:09.533" Title="What is an intuitive explanation for div curl F = 0?" Tags="&lt;vector-calculus&gt;" AnswerCount="5" CommentCount="1" FavoriteCount="4" />
  <row Id="26855" PostTypeId="2" ParentId="26854" CreationDate="2011-03-14T04:53:13.260" Score="4" ViewCount="" Body="&lt;p&gt;The curl of a vector field $F$ at a point on an infinitesimal surface $S$ can be thought of as how the vector field &quot;spins&quot; $S$. The divergence, on the other hand, describes how much the vector field &quot;passes through&quot; $S$. If you take the a vector field which spins a surface, how much does it pass through the surface? None.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, this is a &lt;strong&gt;highly&lt;/strong&gt; non-technical explanation.&lt;/p&gt;&#xA;" OwnerUserId="8173" LastEditorUserId="8173" LastEditorDisplayName="" LastEditDate="2011-03-15T01:34:53.787" LastActivityDate="2011-03-15T01:34:53.787" CommentCount="2" />
  <row Id="26856" PostTypeId="2" ParentId="26757" CreationDate="2011-03-14T05:14:28.330" Score="4" ViewCount="" Body="&lt;p&gt;The proof of Theorem 3 is discussing $\int_0^\infty exp(-itx) \phi(t) \frac{\cos(ht)-1}{t}\, dt$.  That -1 makes a big difference.&lt;/p&gt;&#xA;" OwnerUserId="8508" LastActivityDate="2011-03-14T05:14:28.330" CommentCount="4" />
  <row Id="26857" PostTypeId="1" CreationDate="2011-03-14T05:20:26.030" Score="2" ViewCount="55" Body="&lt;p&gt;Could someone tell me the time complexity of a convex quadratically constrained quadratic program (QCQP) problem? And any references?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you very much.&lt;/p&gt;&#xA;" OwnerUserId="8224" LastEditorUserId="8224" LastEditorDisplayName="" LastEditDate="2011-03-14T05:40:29.913" LastActivityDate="2011-03-14T05:40:29.913" Title="time complexity of a convex quadratically constrained quadratic program (QCQP) problem" Tags="&lt;convex-optimization&gt;" CommentCount="2" FavoriteCount="1" />
  <row Id="26859" PostTypeId="2" ParentId="26846" CreationDate="2011-03-14T05:40:40.610" Score="2" ViewCount="" Body="&lt;p&gt;Any implicit Cartesian equation you'll attempt to derive from the Bézier curve is necessarily complicated because the explicit solution for the cubic equation is complicated, as explained by previous posters; if you desperately need a $y=f(x)$ for manipulations, you're probably better off constructing a (piecewise) Hermite interpolant, which assumes you have an appropriate number of derivatives available in addition to function values.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the cubic case, one only needs two triples $(x_i,y_i,y_i^\prime)$ and $(x_{i+1},y_{i+1},y_{i+1}^\prime)$ available; then, there is a unique cubic $p_3(x)$ that satisfies the conditions $p_3(x_i)=y_i$ and $p_3^\prime(x_i)=y_i^\prime$ (and similarly for the other triple). If, say, you have $y_i$ as position values and $y_i^\prime$ as velocity values, you're all set. If you additionally have second derivative values, you can construct a quintic, but since that sort is rarely needed/wanted, a cubic usually suffices.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that cubic splines are but a special case of cubic Hermite interpolation; the additional thing is that continuity conditions (the agreement of first and second derivative values at certain points) are implemented.&lt;/p&gt;&#xA;" OwnerUserId="8226" LastActivityDate="2011-03-14T05:40:40.610" />
  <row Id="26860" PostTypeId="1" AcceptedAnswerId="26861" CreationDate="2011-03-14T05:59:45.803" Score="3" ViewCount="150" Body="&lt;p&gt;I get myself confused sometimes when to use &lt;code&gt;=&amp;gt;&lt;/code&gt; or &lt;code&gt;=&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;= x + y = x + 5 = x = -5&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;or&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;=&amp;gt; x + y =&amp;gt; x + 5 =&amp;gt; x = -5&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="8218" LastEditorUserId="6491" LastEditorDisplayName="" LastEditDate="2011-03-14T16:06:07.307" LastActivityDate="2011-03-14T16:06:07.307" Title="Implies sign in math" Tags="&lt;algebra-precalculus&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="26861" PostTypeId="2" ParentId="26860" CreationDate="2011-03-14T06:12:09.027" Score="12" ViewCount="" Body="&lt;p&gt;Neither of those are right.  It should be:&#xA;$$x + y = x + 5 \implies x = -5.$$&#xA;(Though I have to say, I'm not exactly sure how $x = -5$ directly follows from $x + y = x + 5$ to begin with.)&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Let's first distinguish between &lt;em&gt;expressions&lt;/em&gt; and &lt;em&gt;equations&lt;/em&gt;.  Expressions are things like $x + 2$ or $37$ or $x^2 - 2x$.  On the other hand, an equation is an equality between two expressions, such as $x + 2 = 3$ or $x = 1$ or $8 + 9 = 17$ and so on.  So, an equality says that two expressions are the same, and we use the equals sign $=$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that an equality is making a logical claim (true or false), whereas an expression is not.  For example, $6 = 6$ is a true equality, whereas $1 = 2$ is a false equality, whereas $17$ is just an expression (not true or false).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the implies sign $\implies$ is used to show that one equation can be logically deduced from another.  So, the implication&#xA;$$x + 2 = 3 \implies x = 1.$$&#xA;means that from the &lt;em&gt;equation&lt;/em&gt; $x + 2 = 3$, we can conclude that $x = 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that it does &lt;em&gt;not&lt;/em&gt; make sense to use the implication arrow between &lt;em&gt;expressions&lt;/em&gt;.  For example $x + y \implies x = -5$ does not make any sense, because the $x + y$ is an expression.  After all, what is the truth value of $x+ y$ anyway?  There isn't one.&lt;/p&gt;&#xA;" OwnerUserId="640" LastActivityDate="2011-03-14T06:12:09.027" />
  <row Id="26862" PostTypeId="2" ParentId="26860" CreationDate="2011-03-14T06:14:58.487" Score="7" ViewCount="" Body="&lt;p&gt;The equal sign $&amp;quot;=&amp;quot;$ should be used when you have two quantities which are equal.  For example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x^2 + 3x - 4 = (x+4)(x-1).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The $&amp;quot;\Rightarrow&amp;quot;$ should be used when relating two statements.  For example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \text{Today is Tuesday} \Rightarrow \text{Tomorrow is Wednesday}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is okay to mix them, say for instance you are solving an equation like the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ 3x + 4 = 10 \Rightarrow 3x = 6 \Rightarrow x = 2.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The quantities $3x + 4$ and $10$ are equal.  However, the statement $3x + 4 = 10$ &lt;em&gt;implies&lt;/em&gt; that $3x = 6$.&lt;/p&gt;&#xA;" OwnerUserId="6491" LastActivityDate="2011-03-14T06:14:58.487" />
  <row Id="26863" PostTypeId="2" ParentId="26854" CreationDate="2011-03-14T06:56:07.117" Score="2" ViewCount="" Body="&lt;p&gt;Think of the surface of water, and a two dimensional vector field $\mathbf{F}(x,y)$ which gives its velocity everywhere. Think that you put a needle that floats somewhere on its surface. If the needle is rotating, then it is in a region of non-zer curl, and if the needle (&lt;strong&gt;EDIT&lt;/strong&gt;: a bunch of them) appears to be translating inwards or outwards towards or away from some point, then it is a region of non zero divergence. Finite divergence just means there's like a hose pipe somewhere which is sucking in or spewing out fluid radially.  This is how you can think of the divergence and the curl individually.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For your question, think of a mixed region with non zero divergence and non-zero curl. The direction of the curl would be along the axis of rotation and hence perpendicular to the surface. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathbf{\nabla} \times \mathbf{F}(x,y) = g(x,y) \hat{z}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\nabla \cdot \mathbf{\nabla} \times \mathbf{F}(x,y) =  \frac{\partial g(x,y) }{\partial z} = 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or intuitively, the curl is perpendicular to the divergence, which is an abuse of notation because divergence is really an operator, so this is as far as the analogy goes.&lt;/p&gt;&#xA;" LastActivityDate="2011-03-14T07:39:39.810" CommentCount="2" />
  <row Id="26864" PostTypeId="1" AcceptedAnswerId="26866" CreationDate="2011-03-14T06:57:39.983" Score="1" ViewCount="49" Body="&lt;p&gt;I have read &lt;a href=&quot;http://ncatlab.org/nlab/show/Eckmann-Hilton+argument&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; and in some other places that if $$(a \circ b) \star (c \circ d) = (a \star c) \circ (b \star d)$$ for all $a,b,c,d$ in a set with two binary operations $\circ$ and $\star$, then each one is a monoid homomorphism.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My problem is that I don't quite understand in what sense these are monoid homomorphisms, since neither is a function $M \to M&amp;#39;$, but instead $M^2 \to M$, which confuses me somewhat. So, what exactly are these homomorphisms? Is $M^2$ supposed to be a monoid? If so, how?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="6427" LastActivityDate="2011-03-14T07:05:45.207" Title="Monoid homomorphism" Tags="&lt;abstract-algebra&gt;&lt;algebra&gt;&lt;monoid&gt;" AnswerCount="1" />
  <row Id="26865" PostTypeId="1" CreationDate="2011-03-14T07:00:58.180" Score="1" ViewCount="130" Body="&lt;p&gt;( &lt;em&gt;Leontief input-output model&lt;/em&gt; ) Suppose that three industries are interrelated so that their outputs are used as inputs by themselves, according to the $3 \times 3$ &lt;strong&gt;consumption matrix&lt;/strong&gt;   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;A&lt;/strong&gt; = [$a_{jk}$] = &#xA;$ \left[ \begin{array}{ccc}&#xD;&#xA;0.1&amp;amp;0.5&amp;amp;0\\&#xD;&#xA;0.8&amp;amp;0&amp;amp;0.4\\&#xD;&#xA;0.1&amp;amp;0.5&amp;amp;0.6&#xD;&#xA;  \end{array} \right] $  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $a_{jk}$ is the fraction of the output of industry $k$ consumed (purchased) by industry $j$. Let $p_{j}$ be the price charged by industry $j$ for its total output. A problem is to find prices so that for each industry, total expenditures equal total income. Determine that there is a price vector such that $~~~~$ &lt;strong&gt;p&lt;/strong&gt; = $[~~p_{1} ~~~ p_{2} ~~~ p_{3}~~]^{T}$ $~$ for this scenario.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any ideas on how to go about solving this??&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank You in advance.&lt;/p&gt;&#xA;" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2011-03-18T22:22:02.730" LastActivityDate="2011-03-18T22:22:02.730" Title="Find a price vector p for various prices of industries. " Tags="&lt;linear-algebra&gt;&lt;matrices&gt;&lt;applications&gt;&lt;economics&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="26866" PostTypeId="2" ParentId="26864" CreationDate="2011-03-14T07:05:45.207" Score="5" ViewCount="" Body="&lt;p&gt;If $M$ is a monoid, $M^2=M\times M$ is a monoid with coordinate-wise multiplication: that is, if $(a,b)$, $(a&amp;#39;,b&amp;#39;)\in M^2$, then the product is such that $$(a,b)\cdot(a&amp;#39;,b&amp;#39;)=(aa&amp;#39;,bb&amp;#39;).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now suppose $M$ is a monoid, and that $M^2$ is endowed with this monoid structure. Suppose additionally that we have a function $\phi:M^2\to M$. It then makes sense to say that $\phi$ is a morphism of monoids. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, it may well happen that the function $\phi$ defines itself a monoid structure on $M$. That is the situation of the lemma you have in mind, which is called the &lt;em&gt;Eckmann-Hilton lemma&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2011-03-14T07:05:45.207" />
  <row Id="26867" PostTypeId="2" ParentId="26757" CreationDate="2011-03-14T07:21:01.510" Score="1" ViewCount="" Body="&lt;p&gt;I hope I'm not misunderstanding the question, but here's what I think (I apologize ahead of time as it is late!)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Write $\eta(t) = \displaystyle\mathop{\Delta}_{t} \left[\frac{\phi(t) e^{-itx}}{it} \right] = \frac{\phi(t) e^{-itx} - \phi(-t) e^{itx}}{it}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;\begin{array}&#xD;&#xA;\\&#xD;&#xA;\int_0^{\infty} (cos(ht) - 1) \eta(t) dt &amp;amp; = \left.\left( \frac{1}{h} \sin(ht) - t\right) \eta(t)\right|_0^{\infty} + \int_0^{\infty} \left(1 - \frac{\sin(ht)}{ht}\right)t \eta&amp;#39;(t) dt \\&#xD;&#xA;&amp;amp;= \int_0^{\infty} \left( 1 - \frac{\sin(ht)}{ht} \right) t \eta&amp;#39;(t) dt&#xD;&#xA;\end{array}&#xD;&#xA;$$&#xA; by parts (which is justified as the author assumes $\phi$ is differentiable at the origin).  The Dominated Convergence Theorem implies that&#xA;$$&#xD;&#xA;\lim_{h \to \infty} \int_0^{\infty} \left( 1 - \frac{\sin(ht)}{ht} \right) t \eta&amp;#39;(t) dt = \int_0^{\infty} t \eta&amp;#39;(t) dt.&#xD;&#xA;$$&#xA;Again, by parts, we have&#xA;$$&#xD;&#xA;\int_0^{\infty} t \eta&amp;#39;(t) dt = \left. t \eta(t)\right|_0^{\infty} - \int_0^{\infty} \eta(t) dt = - \int_0^{\infty} \eta(t) dt.&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Putting it altogether gives:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;\int_0^{\infty} \left( \cos(ht) - 1\right) \eta(t) dt = - \int_0^{\infty} \eta(t) dt.&#xD;&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="6491" LastEditorUserId="6491" LastEditorDisplayName="" LastEditDate="2011-03-14T07:26:23.577" LastActivityDate="2011-03-14T07:26:23.577" CommentCount="4" />
  <row Id="26869" PostTypeId="1" AcceptedAnswerId="26878" CreationDate="2011-03-14T07:37:09.340" Score="3" ViewCount="132" Body="&lt;p&gt;After finishing a first calculus course, I know how to integrate by parts, for example, $\int x \ln x dx$, letting $u = \ln x$, $dv = x dx$: $$\int x \ln x dx = \frac{x^2}{2} \ln x - \int \frac{x^2}{2x} dx.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, what I could not figure out is why we assume from $dv = x dx$ that $v = \frac{x^2}{2}$, when it could be $v = \frac{x^2}{2} + C$ for any constant $C$. The second integral would be quite different, and not only by a constant, so I would like to understand why we &quot;forget&quot; this constant of integration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="6427" LastActivityDate="2011-03-14T16:44:12.033" Title="Constants of integration in integration by parts" Tags="&lt;calculus&gt;&lt;integral&gt;" AnswerCount="5" CommentCount="1" />
  <row Id="26870" PostTypeId="2" ParentId="26869" CreationDate="2011-03-14T07:41:51.967" Score="3" ViewCount="" Body="&lt;p&gt;Your observation that $dv=xdx$ does not imlpy $v=x^2/2$ is correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your confusion resolves when you say it this way: we set $v=x^2/2$ and this implies $dv=xdx$. &lt;/p&gt;&#xA;" OwnerUserId="367" LastActivityDate="2011-03-14T07:41:51.967" />
  <row Id="26871" PostTypeId="2" ParentId="26869" CreationDate="2011-03-14T07:45:07.783" Score="2" ViewCount="" Body="&lt;p&gt;Your observation is correct&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int x \ln x dx = \frac{x^2}{2} \ln x - \int \frac{x^2}{2x} dx.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could always write $$v = \frac{x^2}{2} + C$$ but that won't matter much because the final result would also involve a constant (Say $K$ which would be equal to $C+k$ )&lt;/p&gt;&#xA;" OwnerUserId="149" LastEditorUserId="149" LastEditorDisplayName="" LastEditDate="2011-03-14T16:44:12.033" LastActivityDate="2011-03-14T16:44:12.033" CommentCount="3" />
  <row Id="26872" PostTypeId="1" CreationDate="2011-03-14T08:05:30.817" Score="4" ViewCount="351" Body="&lt;p&gt;Can you give me few examples of binary operation that it is not associative, not commutative but has an identity element? &lt;/p&gt;&#xA;" OwnerUserId="7340" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-14T19:55:46.153" LastActivityDate="2011-03-19T01:55:28.063" Title="Non-associative, non-commutative binary operation with a identity" Tags="&lt;abstract-algebra&gt;" AnswerCount="5" CommentCount="10" FavoriteCount="1" />
  <row Id="26874" PostTypeId="2" ParentId="26872" CreationDate="2011-03-14T08:13:16.460" Score="8" ViewCount="" Body="&lt;p&gt;Pick any non-associative, non-commutative operation $\star$ on a set $X$. Now pick a new element $1$ which is &lt;em&gt;not&lt;/em&gt; in $X$, let $Y=X\cup\{1\}$, and define a new operation $\bullet$ on $Y$ extending $\star$ and such that $$1\bullet x=x\bullet1=x$$ for all $x\in X$, and $1\bullet1=1$.&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2011-03-14T08:13:16.460" CommentCount="5" />
  <row Id="26875" PostTypeId="1" AcceptedAnswerId="26889" CreationDate="2011-03-14T08:17:26.110" Score="2" ViewCount="121" Body="&lt;p&gt;So I want to prove that for natural numbers $m,n,k$, if $k$ is no $0$ or $1$, and if $m&amp;lt;n$, then $k^m&amp;lt;k^n$. I jotted down the following proof, but I'm not sure how acceptable it is. I only have use of basic set theoretic properties of the naturals with addition, multiplication, and powers. I also have use of cancellation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let&#xA;$$&#xD;&#xA;K=\{k\in\omega\ |\ k^m&amp;lt;k^n\}.&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $m&amp;lt;n$ there exists a $p^+\in\omega$ such that $m+p^+=n$. Consider $2$. I claim $2^{p^+}&amp;gt;1$. Now $p^+\neq 0$, since $0$ is not in the range of the successor. If $p^+=1$, then from the work above, $2^1=2&amp;gt;1$. Suppose then that $2^{p^+}&amp;gt;1$. Then $2^{p^{++}}=2^{p^+}\cdot 2&amp;gt;1$. This follows by transitivity since $1&amp;lt;2^{p^+}$, and $1&amp;lt;2$ implies $2^{p^+}\cdot 1&amp;lt;2^{p^+}\cdot 2$, that is $2^{p^+}&amp;lt;2^{p^{++}}$. Then from $1&amp;lt;2^{p^+}$ we have $2^m=2^m\cdot 1&amp;lt;2^m\cdot 2^{p^+}=2^{m+p^+}=2^n$, so $2\in K$. By the same reasoning we can show that any $k\neq 0,1$ is in $K$. If $k\neq 0,1$, then $1&amp;lt;k$. Again, I claim $k^{p^+}&amp;gt;1$. If $p^+=1$, we are done. So suppose $k^{p^+}&amp;gt;1$. Then since $1&amp;lt;k$, we have $k^{p^+}=k^{p^+}\cdot 1&amp;lt;k^{p^+}\cdot k=k^{p^{++}}$, so $1&amp;lt;k^{p^{++}}$ by transitivity, and so $k^{p^+}&amp;gt;1$ for all $p$, since $p^+\neq 0$. Then from $1&amp;lt;k^{p^+}$ we have $k^m\cdot 1&amp;lt;k^m\cdot k^{p^+}=k^{m+p^+}=k^n$, so $k\in K$, and the result follows. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;All in all, this approach seems very ugly. Is there a way to clean it up? I wanted to use the time tested method of induction, but assuming $k^m&amp;lt;k^n$ seems to have little to do with showing $(k^+)^m&amp;lt;(k^+)^n$, since I don't know how to distribute powers. Thanks for any constructive criticism.&lt;/p&gt;&#xA;" OwnerUserId="8228" LastActivityDate="2011-03-14T09:35:39.457" Title="If $m&lt;n$, then $k^m&lt;k^n$?" Tags="&lt;elementary-set-theory&gt;" AnswerCount="3" />
  <row Id="26876" PostTypeId="2" ParentId="26875" CreationDate="2011-03-14T08:28:32.823" Score="0" ViewCount="" Body="&lt;p&gt;I think for natural numbers since $k\ge 1$, $k^{n-m}$ should be greater than 1. Then it should be automatic. &lt;/p&gt;&#xA;" OwnerUserId="7887" LastActivityDate="2011-03-14T08:28:32.823" CommentCount="2" />
  <row Id="26877" PostTypeId="2" ParentId="26872" CreationDate="2011-03-14T08:30:03.010" Score="10" ViewCount="" Body="&lt;p&gt;Here are a couple easy examples, I'll leave it to you to verify their properties.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The natural numbers where $n\ast m=n^m$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The integers where $n\ast m=n-m$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The real numbers without zero where $x \ast y = \frac{x}{y}$.&lt;/p&gt;&#xA;" OwnerUserId="1930" LastActivityDate="2011-03-14T08:30:03.010" CommentCount="5" />
  <row Id="26878" PostTypeId="2" ParentId="26869" CreationDate="2011-03-14T08:30:15.040" Score="6" ViewCount="" Body="&lt;p&gt;The second integral would change, but also the first term... Have you actually checked to see what happens if you change the constant?&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2011-03-14T08:30:15.040" />
  <row Id="26879" PostTypeId="2" ParentId="25169" CreationDate="2011-03-14T08:30:26.257" Score="2" ViewCount="" Body="&lt;p&gt;The only approach I can think of is to narrow down the difference in the remainder term between the exact value of the series and the value of some partial sum that we use as  estimate and judge how good or close our estimate is. Take the remainder term $R_n=\sum_{k=n+1}^\infty \frac{1}{4^k+k!}$ and $T_n=\sum_{k=n+1}^\infty \frac{1}{4^k}$ we know that $R_n\lt T_n\lt\int_{n+1}^\infty  \frac{1}{4^k}dk $. The higher you choose your $n$ to be the lower the remainder or the error term. For example, $n=8$ partial sum is $s_7=0.2745411421$ and $R_7\lt 0.000011006$. So the partial sum is correct to atleast three or four decimal places. &#xA;(This was too long for a comment.)&lt;/p&gt;&#xA;" LastActivityDate="2011-03-14T08:30:26.257" CommentCount="2" />
  <row Id="26880" PostTypeId="2" ParentId="26875" CreationDate="2011-03-14T08:48:20.403" Score="1" ViewCount="" Body="&lt;p&gt;I assume you have something like $k^0 = 1$, $ k^a \cdot k = k^{a^+}$, if $0&amp;lt;a$ and $1&amp;lt;b$ then $a &amp;lt; a \cdot b$, and the typical properties of order such as if $a&amp;lt;b$ and $b&amp;lt;c$ then $a&amp;lt;c$ and if $a&amp;lt;b$ and $b=c$ then $a&amp;lt;c$.  In this case you also have $0 &amp;lt; 1 &amp;lt; k$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First you want to show that $0 &amp;lt; k^m$.  You have the initial step $0 &amp;lt; 1 = k^0$ (adjust this if instead you have $k^1 = k$) and the inductive step $0 &amp;lt; k^m &amp;lt; k^m \cdot k = k^{m^+}$.      &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then for the main hypothesis you have the initial step $k^m &amp;lt;  k^m \cdot k = k^{m^+}$ and the inductive step $k^m &amp;lt; k^n &amp;lt;  k^n \cdot k = k^{n^+}$ for $m&amp;lt;n$. &lt;/p&gt;&#xA;" OwnerUserId="6460" LastActivityDate="2011-03-14T08:48:20.403" CommentCount="2" />
  <row Id="26881" PostTypeId="1" CreationDate="2011-03-14T08:58:10.607" Score="0" ViewCount="361" Body="&lt;p&gt;Does anyone know where I can find (if they exist) full solutions to the exercises of Alan Hatcher's &lt;em&gt;Algebraic Topology&lt;/em&gt;? Thanks.&lt;/p&gt;&#xA;" OwnerUserId="5422" LastEditorUserId="6296" LastEditorDisplayName="" LastEditDate="2011-03-14T12:16:23.217" LastActivityDate="2011-03-14T12:16:23.217" Title="Solutions to Alan Hatcher's &quot;Algebraic Topology&quot;" Tags="&lt;education&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="26882" PostTypeId="1" AcceptedAnswerId="26905" CreationDate="2011-03-14T09:06:05.850" Score="0" ViewCount="113" Body="&lt;p&gt;Hello,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How to prove that $Aut(PSL(2,7))=PGL(2,7)$? Is this result extendible to $PSL(n,q)$ where $q=p^n$?&lt;/p&gt;&#xA;" OwnerUserId="8186" LastActivityDate="2011-04-23T15:27:17.900" Title="Automorphisms of projective special linear group" Tags="&lt;group-theory&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="3" />
  <row Id="26883" PostTypeId="1" AcceptedAnswerId="26901" CreationDate="2011-03-14T09:15:19.397" Score="2" ViewCount="55" Body="&lt;p&gt;Let $\Omega \in R^n$ be open, bounded and with smooth boundary. Can you prove the existence of a system of vectors that simultaneously forms an orthogonal basis both in $L^2(\Omega)$ and $H^1_0(\Omega)$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you generalize this construction to obtain a simultaneous orthogonal basis for $H^0(\Omega),\cdots,H^k_0(\Omega)$?&lt;/p&gt;&#xA;" OwnerUserId="3557" LastEditorUserId="5363" LastEditorDisplayName="" LastEditDate="2011-03-14T14:53:58.993" LastActivityDate="2011-03-14T14:53:58.993" Title="Simultaneous orthogonal basis for $L^2$, $H^1_0$, ... $H^k_0$" Tags="&lt;functional-analysis&gt;" AnswerCount="1" />
  <row Id="26884" PostTypeId="1" AcceptedAnswerId="26936" CreationDate="2011-03-14T09:29:37.107" Score="2" ViewCount="68" Body="&lt;p&gt;The first equation is $m_1x_1 + m_2x_2 + \cdots + m_{n}x_{n} = K$, and then&lt;br&gt;&#xA;$x_1 \geq x_2 \geq x_3 \geq \cdots \geq x_{n-1} \geq x_{n}$. Rather than solving is there any method which can tell me if this is solvable or not.   All the numbers numbers are positive intgers or zero.&lt;/p&gt;&#xA;" OwnerUserId="5403" LastEditorUserId="5403" LastEditorDisplayName="" LastEditDate="2011-03-14T14:21:50.413" LastActivityDate="2011-03-14T15:05:47.070" Title="how can I solve the following linear inequalities" Tags="&lt;linear-algebra&gt;&lt;inequality&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" />
  <row Id="26885" PostTypeId="1" CreationDate="2011-03-14T09:30:25.047" Score="3" ViewCount="98" Body="&lt;p&gt;I am currently working on source material for a math-related software project with my mother, who has a PhD in Elementary Education and specializes in math education. While she has quite a strong fundamental understanding of the current techniques and topics taught to children at an early grade school level, I am relatively unfamiliar with the subject area. Lately, I've been looking for sources of inspiration and information, such as &lt;a href=&quot;http://www.ted.com/talks/lang/eng/conrad_wolfram_teaching_kids_real_math_with_computers.html&quot; rel=&quot;nofollow&quot;&gt;Conrad Wolfram's TED talk &quot;Teaching kids real math with computers&quot;&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remembering that the Stack Exchange family now has this excellent math site, I figured I would ask this question here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What are some interesting concepts in math that can be taught to elementary school children, that aren't traditionally taught?&lt;/strong&gt; This can be because the topics aren't considered suitable, or perhaps because no one's thought to do it yet. For example, at the end of his talk, Wolfram describes a technique to visualize calculus using limits and shapes inscribed in a circle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(While I realize this question is slightly unusual compared to other questions I see on the front page -- in other words, university-level math questions -- I think it is appropriate given what the FAQ mentions as appropriate topics for this site.)&lt;/p&gt;&#xA;" OwnerUserId="4972" LastActivityDate="2011-03-14T09:44:39.950" Title="Non-traditional math concepts for early education" Tags="&lt;teaching&gt;" AnswerCount="1" />
  <row Id="26886" PostTypeId="2" ParentId="26881" CreationDate="2011-03-14T09:31:33.720" Score="4" ViewCount="" Body="&lt;p&gt;This should probably be a comment, but I felt was too long. I'm sure searching &quot;allen hatcher solutions&quot; is about the best you can do with google. But look at this quote from Hatcher's personal website:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I have not written up solutions to the exercises. The main reason for this is that the book is used as a textbook at a number of universities where the problems sets count for part of a student's grade (that is how I teach the course for example). However, individuals who are studying the book on their own and would like hints for specific problems should feel free to email me and I will try to respond.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://gershwin.ens.fr/vdaniel/Doc-Locale/Cours-Mirrored/Maths-Stuff/algtop/index.html&quot; rel=&quot;nofollow&quot;&gt;His homepage&lt;/a&gt; lists his email address, so if you're interested in working through his book, I have a feeling he'd be glad to answer your questions. &lt;/p&gt;&#xA;" OwnerUserId="1583" LastActivityDate="2011-03-14T09:31:33.720" />
  <row Id="26887" PostTypeId="2" ParentId="26854" CreationDate="2011-03-14T09:32:12.270" Score="7" ViewCount="" Body="&lt;p&gt;&lt;b&gt;Formal Intuition&lt;/b&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Formal intuition&quot; is a bit of a contradiction, but look at &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathbf{A} \cdot (\mathbf{A} \times \mathbf{B})$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cross products are perpendicular to the things being crossed, so $\mathbf{A \times B}$ is perpendicular to $\mathbf{A}$. The dot product of perpendicular vectors is zero, so&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathbf{A} \cdot (\mathbf{A} \times \mathbf{B}) = 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Replacing $\mathbf{A}$ with $\nabla$ would give the identity, but I'm not sure if this counts as intuitive.  Note: This isn't supposed to be a proof.  It's only supposed to suggest the identity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Geometric Intuition&lt;/b&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here, I'll use Gauss's theorem and Stoke's theorem, which themselves have intuitive explanations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Make a coordinate system like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/5UzQs.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider a cubical box.  Make it centered at the origin with sides of length $1$, so the corners are $(.5,.5,.5)$, $(.5,.5,-.5)$, $(.5,-.5,.5)$, etc.  This is for convenience - the area of each side is $1$ and the volume is $1$, so any time I need to divide or multiply by the area of a side or by the volume, nothing changes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We are looking at $\nabla \cdot (\nabla \times \mathbf{F})$.  Ignoring the stuff inside the parentheses, we're looking at a divergence of something.  Let's call that thing $\mathbf{C}$ since it's a curl.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gauss' theorem says that the integral of the divergence over the volume of the box is equal to the net flux coming out.  This means that we can find the average divergence over the entire box by thinking only about the faces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the top face.  To find the flux coming out of the top face, first find the average of $\mathbf{C}$ over that face.  Then project that vector onto a unit normal vector, $\hat{z}$.  This gives a vector $\mathbf{C}_z$ that either points into or out of the box.  If it points out, it contributes positive flux, and if it points in, it contributes negative flux.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/SqZV9.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we can show that all six such vectors (one for each face) contribute zero net flux when combined, we know that the average divergence over our box is zero.  How do we get information about the average of $\mathbf{C}$, projected onto a normal vector, over some face?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is where the curl comes in.  Stokes' theorem says that the average of $\mathbf{C}_z = (\nabla \times \mathbf{F})\cdot\hat{z}$ over the top is equal to the line integral over the edges of the top face of $\mathbf{F}\cdot \hat{r}$, with $\hat{r}$ a unit tangent vector.  In other words, each of the four edges of the top gives some contribution to the average of $\mathbf{C}_z$ on the top.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Grab one particular edge - the one from $(.5, -.5, .5)$ to $(.5,.5,.5)$ will do.  Suppose we traverse it from left to right, and get a positive number, say $4.2$, out of the line integral of $\mathbf{F} \cdot \hat{r}$ along that edge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/Jk9BB.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The purple vector is the original vector field $\mathbf{F}$.  It gets projected onto $\hat{r}$ to make $F_r$, the purple segment.  This gets integrated along the edge, and contributes to the black vector $\mathbf{C_z}$, which itself contributes to the flux.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The important thing to note is that this $4.2$ value of the line integral along the edge gets used twice in finding the flux because the edge borders on two different faces.  For the top face, going left-to-right on our edge means circling clockwise as viewed from above.  The right-hand rule tells us that the postive $4.2$ of the line integral along our edge contributes positively to a vector $\mathbf{C_z}$ point out of the top.  Since it points up out of the top, it contributes negative flux.  All told, traversing our edge, thinking of it as part of the top of the cube, contributes $-4.2$ to the total flux.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When we look at the front of the cube, going left-to-right on our edge means going clockwise around the front, as viewed from in front of the cube.  The right hand rule means that a positive clockwise line integral around the front creates a vector $\mathbf{C}_x$ pointing into the cube - positive flux.  So this time our edge contributes $+4.2$ to the total flux.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/5XyaL.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The edge's two contributions to the flux cancel each other because one point outwards and the other points inwards.  There is nothing special about this edge, so all edges will be canceled by their two contributions to the flux.  The total flux for this cube is zero.  That means the average of the divergence over the cube is zero.  There was nothing special about this cube, so the average divergence over any cube is zero, and the divergence itself must simply be zero everywhere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Physical Intuition&lt;/b&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This theorem about the divergence of the curl is responsible for the conservation of charge.  Begin with Maxwell's equation for the curl of the magnetic field&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\nabla \times \mathbf{B} = \frac{1}{c}\frac{\partial \mathbf{E}}{\partial t} + \frac{4 \pi}{c} \mathbf{J}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take the divergence of both sides.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\nabla \cdot (\nabla \times \mathbf{B}) = \frac{1}{c} \frac{\partial}{\partial t} \nabla \cdot \mathbf{E} + \frac{4\pi}{c}\nabla\cdot\mathbf{J}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now use Maxwell's equation for the divergence of the electric field&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\nabla \cdot \mathbf{E} = 4\pi \rho$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Insert this into the above to get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\nabla \cdot (\nabla \times \mathbf{B}) = \frac{4 \pi}{c} \left(\frac{\partial \rho}{\partial t} + \nabla\cdot\mathbf{J}\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We expect that electric charge will be conserved.  That means the the flux of charge current coming out of some region must be compensated by a loss of charge density there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\nabla \cdot \mathbf{J} = -\frac{\partial \rho}{\partial t}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Plugging this into our previous expression gives&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\nabla \cdot (\nabla \times \mathbf{B}) = \frac{4 \pi}{c} \left(\frac{\partial \rho}{\partial t} -\frac{\partial \rho}{\partial t}\right) = 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means that if electromagnetic fields obey Maxwell's equations and if electric charge is conserved, then the divergence of the curl of any magnetic field must be zero.  However, I'm not sure how intuitive this is, and also the magnetic field cannot be chosen arbitrarily because it must be divergenceless.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We could repeat the procedure for Maxwell's equation for the curl of the electric field.  There, we'd get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\nabla \cdot (\nabla \times \mathbf{E}) = \frac{\partial}{\partial t}\left(\nabla \cdot \mathbf{B}\right) = 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;because the magnetic field is divergenceless.  So the absence of magnetic charge implies that the divergence of the curl of all electric fields is zero.  Electric fields, unlike magnetic fields, can be essentially arbitrary as far as I know, and so viewed in a certain light Maxwell's equations imply the sought vector identity. &lt;/p&gt;&#xA;" OwnerUserId="3305" LastEditorUserId="3305" LastEditorDisplayName="" LastEditDate="2011-03-14T19:17:49.430" LastActivityDate="2011-03-14T19:17:49.430" CommentCount="8" />
  <row Id="26888" PostTypeId="1" AcceptedAnswerId="27219" CreationDate="2011-03-14T09:32:45.810" Score="9" ViewCount="438" Body="&lt;p&gt;The union of a sequence of $\sigma$-algebras need not be a $\sigma$-algebra, but how do I prove the stronger statement below?&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Let $\mathcal{F}_n$ be a sequence of&#xA;  $\sigma$-algebras. If the inclusion&#xA;  $\mathcal{F}_n \subsetneqq \mathcal{F}_{n+1} $ is strict, then&#xA;  the union&#xA;  $\bigcup_{n=1}^{\infty}\mathcal{F}_n$&#xA;  is &lt;strong&gt;not&lt;/strong&gt; a $\sigma$-algebra.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5593" LastEditorUserId="5593" LastEditorDisplayName="" LastEditDate="2011-03-15T12:49:17.753" LastActivityDate="2011-04-03T18:37:02.500" Title="The union of a strictly increasing sequence of $\sigma$-algebras is not a $\sigma$-algebra" Tags="&lt;homework&gt;&lt;measure-theory&gt;" AnswerCount="2" CommentCount="15" FavoriteCount="5" />
  <row Id="26889" PostTypeId="2" ParentId="26875" CreationDate="2011-03-14T09:35:39.457" Score="1" ViewCount="" Body="&lt;p&gt;The idea is to show that for $k\geq2$ we have $k^{m^+}&amp;gt;k^m$. For that you need to show that $m\cdot k&amp;gt;m$ for $k\geq2$ and $m\geq1$. And to show that you need to show that $m+n&amp;gt;m$ for $n\geq1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's start with the last one, $m+n&amp;gt;m$ for $n\geq1$: We work with induction on $n$. For $n=1$ $m+n=m+1=m^+$. By the definition of the natural numbers you have $m\in m^+$ so it's true. Now assume that it's true for $n$. For $n^+$ we have $m+n^+=(m+n)^+$. Now $(m+n)&amp;lt;(m+n)^+$. By the induction hypothesis we have $m&amp;lt;m+n&amp;lt;(m+n)^+$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let's show that $m\cdot k&amp;gt;m$ for $k\geq2$ and $m\geq1$ : We work with induction on $k$. For $k=2$ we have $m\cdot2=m\cdot1+m=m+m$. From the previous result (and since $m\geq1$) we have $m+m&amp;gt;m$ and thus $m\cdot2&amp;gt;m$. Assume that it's true for $k$. We have $m\cdot k^+=m\cdot k+m$. By the previous result $m\cdot k+m&amp;gt;m\cdot k$, By the induction hypothesis we have $m\cdot k&amp;gt;m$. Therefore $m\cdot k^+=m\cdot k+m&amp;gt;m\cdot k&amp;gt;m$ and we are done.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can now show that $k\geq2$ we have $k^{m^+}&amp;gt;k^m$: We have $k^{m^+}=k^m\cdot k$. Since $k\geq2$ from the previous result we get $k^m\cdot k&amp;gt;k^m$ which is what we want.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finalizing the proof let's show what you want by induction on $n$. If $n=0$ then the sentence is vacuously true. Assume that it's true for $n$. If $m&amp;lt;n^+$ then we have $m&amp;lt;n$ or $m=n$. Previously I showed that $k^n&amp;lt;k^{n^+}$. In case $m=n$ that suffices. In case $m&amp;lt;n$ by the induction hypothesis we have $k^m&amp;lt;k^n$. But $k^n&amp;lt;k^{n^+}$ and we are done.&lt;/p&gt;&#xA;" OwnerUserId="1772" LastActivityDate="2011-03-14T09:35:39.457" CommentCount="2" />
  <row Id="26890" PostTypeId="2" ParentId="26885" CreationDate="2011-03-14T09:44:39.950" Score="2" ViewCount="" Body="&lt;p&gt;I've had success teaching an elementary school student about binary numbers.  I listed powers of two up to 512 and claimed I could make any number up to 1000 by adding the powers of two, using each one at most once.  I had the  student give me numbers and demonstrated how to make them out of the powers of two. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then I had the student do it herself for some small numbers using guess-and-check.  Then we made a table with powers of two as columns and sequential integers, from 0, as rows, and put an X in any entry if the power of two for that column was used to make the number in that row.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We looked for patterns in the X's, and were introduced to binary numbers that way.  By the end of two hours, she could convert any number between binary and base ten, and could add and multiply in binary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I haven't tried to do it and don't know how, but I suspect you can introduce basic group theory to students of this age, too.&lt;/p&gt;&#xA;" OwnerUserId="3305" LastActivityDate="2011-03-14T09:44:39.950" />
  <row Id="26892" PostTypeId="1" CreationDate="2011-03-14T09:55:09.160" Score="0" ViewCount="54" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Possible Duplicate:&lt;/strong&gt;&lt;br&gt;&#xA;  &lt;a href=&quot;http://math.stackexchange.com/questions/8241/generating-integers-from-a-linear-combination-of-integers&quot;&gt;Generating integers from a linear combination of integers&lt;/a&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;p&gt;Consider the linear equation $a_{1}x_{1} + a_{2}x_{2}+ \cdots +a_{n}x_{n}=b$ such that $a_i,b\geq 0$, $gcd(a_1,\cdots a_n)|b$. For $n=2$, it is easy to show that for large value of $b&amp;gt;0$, the equation has a non-negative integer solution and the soultions are in arithmatic-progression. Is it true for $n&amp;gt;2$? i.e. equations with more that 2 variables?&lt;/p&gt;&#xA;" OwnerUserId="8186" LastEditorUserId="1778" LastEditorDisplayName="" LastEditDate="2011-03-14T13:01:45.833" LastActivityDate="2011-03-14T13:01:45.833" ClosedDate="2011-03-14T10:12:35.540" Title="Solutions of linear equations over Z" Tags="&lt;number-theory&gt;&lt;diophantine-equations&gt;" CommentCount="5" FavoriteCount="1" />
  <row Id="26893" PostTypeId="1" AcceptedAnswerId="26899" CreationDate="2011-03-14T10:03:06.083" Score="3" ViewCount="94" Body="&lt;p&gt;Does there exists a function $f \in C^2[0,\infty]$ (that is, $f$ is $C^2$ and has finite limits at $0$ and $\infty$) with $f&amp;#39;&amp;#39;(0) = 1$, such that for any $g \in L^p(0,T)$ (where $T &amp;gt; 0$ and $1 \leq p &amp;lt; \infty$ may be chosen freely) we get&#xA;$$&#xD;&#xA;\int_0^T \int_0^\infty \frac{u^2-s}{s^{5/2}} \exp{\left( -\frac{u^2}{2s} \right)} f(u) g(s) du ds = 0?&#xD;&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="8229" LastEditorUserId="8229" LastEditorDisplayName="" LastEditDate="2011-03-14T10:10:45.703" LastActivityDate="2011-03-14T14:53:05.463" Title="Existence of a function such that given integral vanishes" Tags="&lt;calculus&gt;&lt;real-analysis&gt;&lt;integral&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="26894" PostTypeId="1" AcceptedAnswerId="27001" CreationDate="2011-03-14T10:09:42.750" Score="4" ViewCount="130" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;If the complex numbers 0, $z_1$, $z_2$ and $z_3$ are concyclic, prove that $\frac{1}{z_1}$,$\frac{1}{z_2}$,$\frac{1}{z_3}$ are collinear.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I really can't seem to get anywhere on this problem, but all I've deduced is that there might be some relationship between circle geometry properties and the arguments of the complex numbers.&lt;/p&gt;&#xA;" OwnerUserId="6520" LastActivityDate="2011-03-14T19:09:20.577" Title="Complex number property" Tags="&lt;algebra&gt;&lt;complex-numbers&gt;&lt;circle&gt;&lt;complex-geometry&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="26896" PostTypeId="1" AcceptedAnswerId="26910" CreationDate="2011-03-14T11:01:09.370" Score="3" ViewCount="99" Body="&lt;p&gt;Dear All&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am currently studying &lt;a href=&quot;http://en.wikipedia.org/wiki/Herbrand%27s_theorem&quot; rel=&quot;nofollow&quot;&gt;Herbrand's Theorem&lt;/a&gt; and wonder in which form it would hold for intuitionistic logic. I guess in intuitionistic logic we will have only one witness.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To be practical I am experimenting with the following derivation, which is classically valid:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\lnot p(a) \rightarrow p(b) \vdash \exists x p(x)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I don't find an intuitionistic derivation. Isn't there any? Could eventually an intuitionistic version of Herbrands theorem be used to establish that there isn't a derivation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Best Regards  &lt;/p&gt;&#xA;" OwnerUserId="4414" LastActivityDate="2011-03-14T12:48:39.963" Title="Is there an intuitionistic proof of $\lnot p(a) \rightarrow p(b) \vdash \exists x p(x)$, what would Herbrand's Theorem say?" Tags="&lt;logic&gt;&lt;proof-theory&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="26897" PostTypeId="1" CreationDate="2011-03-14T11:16:43.957" Score="2" ViewCount="67" Body="&lt;p&gt;I've doing a bit of self study on the Fourier Transform and I keep coming across &quot;basis function&quot;, I have attempted to understand the theory behind it, but I'm finding a lot of explanations go into more maths concepts that I'm not fully understanding.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm wondering if someone can in simple english give me an explanation of &quot;Basis Function&quot; that I can understand?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any help is appreciated&lt;/p&gt;&#xA;" OwnerUserId="807" LastActivityDate="2011-03-14T12:37:55.740" Title="Basis functions in Fourier Series" Tags="&lt;functions&gt;&lt;fourier-analysis&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="26898" PostTypeId="1" AcceptedAnswerId="26908" CreationDate="2011-03-14T11:22:20.930" Score="0" ViewCount="46" Body="&lt;p&gt;Does the slope depend on how the &lt;code&gt;y&lt;/code&gt; of a straight line changes or on both x and y?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, A straight line &lt;code&gt;2x - y = 0&lt;/code&gt; passing through origin has a slope &lt;code&gt;m = 2&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, when&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;x = 1  |  y = 2&#xA;x = 2  |  y = 4&#xA;x = 3  |  y = 6&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;It looks like the value of &lt;code&gt;y&lt;/code&gt; is increasing by 2, which is confusing me because &lt;code&gt;m = 2&lt;/code&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In &lt;strong&gt;case 2&lt;/strong&gt;, A straight line &lt;code&gt;2x + 3y = 18&lt;/code&gt; has slope m = &lt;code&gt;-2/3&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and, when&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;x = 1 | y = 5.33&#xA;x = 2 | y = 4.66&#xA;x = 3 | y = 4&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;In this case it looks like the &lt;code&gt;y&lt;/code&gt; is decreasing by &lt;code&gt;-2/3&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so, does the slopes of a straight line only effect &lt;code&gt;y&lt;/code&gt;? &lt;/p&gt;&#xA;" OwnerUserId="8218" LastEditorUserId="8218" LastEditorDisplayName="" LastEditDate="2011-03-14T11:46:56.830" LastActivityDate="2011-03-14T19:20:20.077" Title="Slope of a line in 2D Co-ordinates" Tags="&lt;linear-algebra&gt;" AnswerCount="2" />
  <row Id="26899" PostTypeId="2" ParentId="26893" CreationDate="2011-03-14T11:38:30.537" Score="6" ViewCount="" Body="&lt;p&gt;I'm pretty sure the answer is no, there exists no such $f$. Here I first give a physical argument. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;First define the function $K(u,s) = \frac{1}{\sqrt{s}} \exp (-u^2 / 2s )$. Up to some constant normalisation factors, this is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Heat_equation#Fundamental_solutions&quot; rel=&quot;nofollow&quot;&gt;Heat Kernel&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Observe that your integral can be written as, by an explicit calculuation, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \int_0^T \int_0^\infty (\partial_{uu}^2K)(u,s)f(u)g(s) ~ du ~ ds $$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Now, notice that the condition is linear. If $f$ and $\tilde{f}$ are solutions, then $af + b\tilde{f}$ is also a solution. Next, observe that using the fundamental theorem of calculus, $f(u) = \textrm{constant}$ is a solution: since $\partial_uK$ vanishes both at 0 and infinity. Therefore we can, without loss of generality, assume that the solution we are looking for has $f(0) = 0$. So we can extend $f$ continuously to the negative real line by setting $f(x) = 0$ whenever $x &amp;lt; 0$. Then using that $K$ is a even function:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \int_0^{\infty} \partial_{uu}^2K(u,s)f(u) du = \int_{-\infty}^\infty \partial_{uu}^2K(0-u,s)f(u) du = \partial_{uu}^2 (K_s*f)(0) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $K_s*f$ is the convolution of the heat kernel $K(u,s)$ against $f(u)$ extended to the whole real line. In other words, that is the evaluation of the second derivative of $K_s*f$ at the origin. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, $K_s*f$ is a solution to the heat equation with initial data at $s=0$ being $f$. So in particular, up to a constant factor, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \partial_{uu}^2(K_s*f) = \partial_t(K_s*f)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So your desired integral condition, since you allow $g$ to be arbitrary, tells you that $K_s*f(0) = 0$ for all $s$. (Which is, in fact, basically what joriki wrote in his comment.) &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The condition that $f&amp;#39;&amp;#39;(0) = 1$ tells you that the initial temperature fluctuation is non-zero near the origin. As heat is diffusive, to the left of the origin you have no heat content, to the right you start with some non-zero temperature arbitrarily close to the origin. So in arbitrary short time you should feel some heat at the origin. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The mathematical argument follows: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, using joriki's comment, the problem reduces to considering on constant $s$ slices. Integrating by parts twice in $u$, we have that your condition implies&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \int_0^\infty K(u,s) f&amp;#39;&amp;#39;(u) du + \frac{1}{\sqrt{s}} f&amp;#39;(0) = 0 $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Taking $s\to 0$, the first term converges to some finite value which is non-zero by the assumption that $f&amp;#39;&amp;#39;(0) = 1\neq 0$ and $f&amp;#39;&amp;#39;$ is continuous. This gives a contradiction as, if $f&amp;#39;(0) = 0$ then the above equation would require $f&amp;#39;&amp;#39;(0) = 0$. And if $f&amp;#39;(0) \neq 0$, the above equation shows that the integral $\int_0^\infty K(u,s) f&amp;#39;&amp;#39;(u) du \nearrow \infty$ as $s\searrow 0$. &lt;/p&gt;&#xA;" OwnerUserId="1543" LastActivityDate="2011-03-14T11:38:30.537" CommentCount="1" />
  <row Id="26900" PostTypeId="2" ParentId="26898" CreationDate="2011-03-14T11:43:11.150" Score="0" ViewCount="" Body="&lt;p&gt;The slope of a straight line here is the change in $y$ divided by the change in $x$.  Your equation can be written as $y = 2x$ and, as your numbers show, each increase of $x$ by 1 leads to an increase in $y$ by 2.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Draw the lines $y = 2x$ and $y = x$ on co-ordinate axes with the $x$-axis horizontal and $y$-axis vertical, and you will see that the line with a larger slope is steeper.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added later:&lt;/strong&gt; On your second added question, you have correctly worked out the slope as close to $-2/3$. But this is not just the change in $y$.  If you had only the data&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;x = 1 | y = 5.33&#xA;x = 3 | y = 4    &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;then the change in $y$ would be $-1.33$, but the change in $x$ would be $2$.  You have to divide one by the other. So the slope would be $-1.33/2 \approx -2/3$, as before.&lt;/p&gt;&#xA;" OwnerUserId="6460" LastEditorUserId="6460" LastEditorDisplayName="" LastEditDate="2011-03-14T19:20:20.077" LastActivityDate="2011-03-14T19:20:20.077" />
  <row Id="26901" PostTypeId="2" ParentId="26883" CreationDate="2011-03-14T11:47:41.057" Score="1" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;hint&lt;/strong&gt; Look up the spectral theorem for self-adjoint operators. Apply it to the Laplacian on your domain $\Omega$. (The spectral theory of the Laplacian is rather standard, and should be done in most text books on PDEs or functional analysis; see, e.g. Evans, &lt;em&gt;Partial Differential Equations&lt;/em&gt;). Now notice that on $H^1_0(\Omega)$ you have the following integration by parts identity:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \int_{\Omega} \nabla u \cdot \nabla v ~dx = - \int_{\Omega} u \triangle v ~dx $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\triangle$ is the Laplace operator. &lt;/p&gt;&#xA;" OwnerUserId="1543" LastActivityDate="2011-03-14T11:47:41.057" />
  <row Id="26902" PostTypeId="2" ParentId="26896" CreationDate="2011-03-14T11:47:50.670" Score="2" ViewCount="" Body="&lt;p&gt;I assume $a$ and $b$ are constants of the language, and you want to prove&#xA;$$ \vdash \exists x (\lnot p(a) \rightarrow p(b)) \rightarrow p(x) $$&#xA;which is provable in classical logic by doing a contraction on the right, then instantiating $x$ and $x$ with $a$ and $b$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have a cut-free intuitionistic proof of that statement, then it must start with giving a witness term $t$ for $x$, and then proving $\vdash (\lnot p(a) \rightarrow p(b)) \rightarrow p(t)$. But there is no term $t$ for which this is provable. It is not long to look for all possible cut-free proofs in this case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The cut-elimination theorem here is really doing all the work for you because proofs that are cut-free are much easier to look for (especially since there is no contraction on the right in intuitionistic logic).&lt;/p&gt;&#xA;" OwnerUserId="6270" LastActivityDate="2011-03-14T11:47:50.670" CommentCount="2" />
  <row Id="26903" PostTypeId="1" CreationDate="2011-03-14T11:48:00.103" Score="0" ViewCount="103" Body="&lt;p&gt;What is a good graphing software ? The one that has the ability to accept an equation (linear, quadratic, etc.) from users and output a graph for that equation (software equivalnet of T1-84 graphing calculator).&lt;/p&gt;&#xA;" OwnerUserId="8235" LastActivityDate="2011-04-03T09:21:29.727" Title="What is a good graphing software ?" Tags="&lt;math-software&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="26904" PostTypeId="1" AcceptedAnswerId="27004" CreationDate="2011-03-14T12:00:41.020" Score="2" ViewCount="190" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can calculate the derivative of a function using the product rule, chain rule or quotient rule.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I find the resulting derivative function however, I have no way to check if my answer is correct!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I check if the calculated derivative equation is correct? (ie I haven't made a mistake factorising, or with one of the rules).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a graphics calculator.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="8176" LastActivityDate="2011-03-14T20:13:21.643" Title="How to check if derivative equation is correct?" Tags="&lt;calculus&gt;" AnswerCount="5" CommentCount="4" />
  <row Id="26905" PostTypeId="2" ParentId="26882" CreationDate="2011-03-14T12:35:59.363" Score="8" ViewCount="" Body="&lt;p&gt;OK, here is a quick direct argument to show that Aut(PSL$(2,7)$) = PGL$(2,7)$. Let $\alpha \in {\rm Aut}({\rm PSL}(2,7)$. Now PSL$(2,7)$ acts 2-transitively by conjugation on its 8 Sylow $7-$subgroups. Since these must also be permuted under the action of $\alpha$, $\alpha$ is induced by conjugation by an element $a \in S_8$. By multiplying $a$ by an inner automorphism, we may assume that it fixes a specific Sylow $7-$subgroup $S$ of PSL$(2,7)$. Since the full automorphism group of $S$ (i.e. the cyclic group of order 6) is induced on $S$ within PGL$(2,7)$, by&#xA;multiplying $a$ be an element of PGL$(2,7)$, we may assume that $a$ centralizes $S$. Then, by multiplying $a$ by an element of $S$, we can assume that $a$ fixes some other point as an element of $S_8$, but then the fact that it centralizes $S$ forces $a=1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same argument works for ${\rm Aut}({\rm PSL}(2,p)$ for any odd prime $p$. For ${\rm PSL}(n,p^k)$ with $k&amp;gt;1$ there are also field automorphisms, and for $n \ge 3$, there is also the graph automorphism (induced by inverse-transpose on  ${\rm SL}(n,p^k)$), but you would probably need to learn some more general theory of classical groups or groups of Lie type to understand that.&lt;/p&gt;&#xA;" OwnerUserId="2820" LastActivityDate="2011-03-14T12:35:59.363" />
  <row Id="26906" PostTypeId="2" ParentId="26904" CreationDate="2011-03-14T12:36:14.930" Score="0" ViewCount="" Body="&lt;p&gt;You can check certain values, like the saddle points, extremal points and local minima/maxima by setting the first derivative equal to zero/deriving further and checking these derivatives too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you found them right, putting the values into the original function plus/minus some $\Delta x$ should make things clear.&lt;/p&gt;&#xA;" OwnerUserId="5870" LastEditorUserId="5870" LastEditorDisplayName="" LastEditDate="2011-03-14T12:41:25.627" LastActivityDate="2011-03-14T12:41:25.627" CommentCount="1" />
  <row Id="26907" PostTypeId="2" ParentId="26897" CreationDate="2011-03-14T12:37:55.740" Score="1" ViewCount="" Body="&lt;p&gt;As Jonas points out, a basis function is simply a function in the basis of a function space.  Most likely, you are (whether it explicitly says so or not) working in the function space $L^2[0,1]$ (or something like it), which is an infinite-dimensional Hilbert space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The functions $\{\exp(2 \pi inx) : n \in \mathbb{Z}\}$ form a infinite (but countable) basis (sometimes called a Schauder basis) for all functions $f \in L^2[0,1]$.  That is, every function (in $L^2$) can be expressed as a linear combination of complex exponentials.  See, for example, &lt;a href=&quot;http://en.wikipedia.org/wiki/Hilbert_space#Fourier_analysis&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.  A basis function in this sense is simply one of those complex exponentials $f(x) = e^{2 \pi i n x}$.&lt;/p&gt;&#xA;" OwnerUserId="6491" LastActivityDate="2011-03-14T12:37:55.740" />
  <row Id="26908" PostTypeId="2" ParentId="26898" CreationDate="2011-03-14T12:41:17.953" Score="0" ViewCount="" Body="&lt;p&gt;The slope of the straight line is only affecting y in your case because you are deciding to measure y and not x, i.e. you are setting the values of x and calculating the values of y from there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you were to draw y = 2x while measuring y (that is, you are setting the values of y and calculating the values of x) you would find that x increases by 1/2 (i.e. 1/m). &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;y = 1 | x = 0.5&#xA;y = 2 | x = 1.0&#xA;y = 3 | x = 1.5&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;So in actuality m effects whichever value you are calculating, as you are setting the other value yourself (you're affecting the other value)&lt;/p&gt;&#xA;" OwnerUserId="8238" LastActivityDate="2011-03-14T12:41:17.953" />
  <row Id="26909" PostTypeId="1" AcceptedAnswerId="31358" CreationDate="2011-03-14T12:42:24.337" Score="6" ViewCount="142" Body="&lt;p&gt;Let $G$ be a compact (Hausdorff) group and $V$ a faithful (complex, continuous, finite-dimensional) representation of it. (Hence $G$ is a Lie group.) Is it true that every irreducible representation of $G$ occurs as a summand of $V^{\otimes n} \otimes (V^{\ast})^{\otimes m}$ for some $m, n$? (The original question only asked for summands of $V^{\otimes n}$ and, as anon mentions below, $\text{U}(1)$ is an easy counterexample.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that the corresponding result is true for finite groups, but the proof I know doesn't seem to easily generalize. It seems we ought to be able to apply Stone-Weierstrass: the characters you get from summands of $V^{\otimes n} (V^{\ast})^{\otimes m}$ form an algebra of class functions closed under addition, multiplication, and complex conjugate, so if we know that they separate points, they ought to be dense in the space of class functions. But &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) I'm not sure if we can show that these functions separate points, and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) I'm not sure if the space of conjugacy classes (with the quotient topology from $G$) is even Hausdorff.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Motivation:&lt;/strong&gt; I was looking for cheap ways to set up the representation theory of $\text{SU}(2)$. In this rather special case the character of the defining representation $V$, which is self-dual, already separates conjugacy classes, and I &lt;em&gt;think&lt;/em&gt; the above argument works. Then Clebsch-Gordan allows me to quickly classify the irreducible representations of $\text{SU}(2)$ without using Lie algebras.&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2011-03-14T17:26:45.330" LastActivityDate="2011-04-06T16:29:52.350" Title="Does every irreducible representation of a compact group occur in tensor products of a faithful representation (and its dual)?" Tags="&lt;representation-theory&gt;&lt;lie-groups&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="26910" PostTypeId="2" ParentId="26896" CreationDate="2011-03-14T12:48:39.963" Score="2" ViewCount="" Body="&lt;p&gt;Herbrand's theorem also holds for intuitionistic first order logic, actually you can argue directly but simply saying if something is provable in intuitionistic logic, then it is also provable in classical logic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general you can not expect to have only one witness to get a stronger version of the Herbrand's theorem because classical logic is an extension of intuitionistic logic and therefore you can prove theorems in a classical theory and if you have a stronger version it would apply to the classical logic. On the other hand, for certain theories like $\mathbf{HA}$ it is possible to prove stronger versions like the &lt;em&gt;existential property&lt;/em&gt; which essentially means that if you prove that a number satisfying a formula exists, then there is a natural number that witnesses it provably.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To prove that the formula you are discussing is not provable you can either use cut-elimination theorem or just come up with a model (e.g. Kripke model) which will make the statement false. It is not difficult to come up with such a model.&lt;/p&gt;&#xA;" OwnerUserId="468" LastActivityDate="2011-03-14T12:48:39.963" CommentCount="8" />
  <row Id="26911" PostTypeId="2" ParentId="10085" CreationDate="2011-03-14T13:02:11.073" Score="0" ViewCount="" Body="&lt;p&gt;Because $\omega$ is exactly &lt;em&gt;defined&lt;/em&gt; to be the smallest infinite ordinal number. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You may ask why is there a smallest ordinal? The answer is: because ordinals are defined in a way that they will be well-ordered w.r.t. membership relation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since it is the smallest infinite ordinal, all smaller ordinals are finite and (the cardinality of) the powerset of any finite ordinal is a finite ordinal, so it cannot be the powerset of any finite ordinal. And by Cantor's diagonal argument, the cardinality of any set is strictly smaller than the cardinality of its powerset, so it cannot be the powerset of itself (or any other ordinal since any infinite ordinal contains it). So it cannot be (the cardinality of) the powerset of any ordinal.&lt;/p&gt;&#xA;" OwnerUserId="468" LastEditorUserId="468" LastEditorDisplayName="" LastEditDate="2011-03-14T13:08:28.293" LastActivityDate="2011-03-14T13:08:28.293" />
  <row Id="26912" PostTypeId="1" CreationDate="2011-03-14T13:04:43.007" Score="0" ViewCount="82" Body="&lt;p&gt;Which of the following are true?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1:for every positive integer &lt;B&gt;m&lt;/B&gt;, there is a positive integer &lt;B&gt;n&lt;/B&gt; such that &lt;B&gt;m+n+1&lt;/B&gt; is a perfect square.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2:for every positive integer &lt;B&gt;m&lt;/B&gt;, there is a positive integer &lt;B&gt;n&lt;/B&gt; such that &lt;B&gt;mn+1&lt;/B&gt; is a perfect cube&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3:for every positive integer &lt;B&gt;m&lt;/B&gt;, there is a positive integer &lt;B&gt;n&lt;/B&gt; such that &lt;B&gt;m+n+1&lt;/B&gt; is a perfect cube&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4:for every positive integer &lt;B&gt;m&lt;/B&gt;, there is a positive integer &lt;B&gt;n&lt;/B&gt; such that &lt;B&gt;mn+1&lt;/B&gt; is a perfect square&lt;/p&gt;&#xA;" OwnerUserId="8000" LastEditorUserId="7850" LastEditorDisplayName="" LastEditDate="2011-03-14T15:35:00.497" LastActivityDate="2011-03-14T15:35:00.497" Title="Questions on perfect squares and perfect cubes" Tags="&lt;algebra-precalculus&gt;&lt;elementary-number-theory&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="26913" PostTypeId="2" ParentId="26912" CreationDate="2011-03-14T13:22:04.253" Score="1" ViewCount="" Body="&lt;p&gt;I'll do the first one, which should help on (3)...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given some &lt;strong&gt;m&lt;/strong&gt; $\geq 1$ &lt;br&gt;&#xA;Consider $(m + 1)^2 = m^2 + 2m + 1$ &lt;br&gt;&#xA;The left hand side $(n + m + 1)$ will equal a perfect square (namely, $m^2 + 2m + 1$) if we let $n = m^2 + m &gt; 0$&lt;br&gt;&lt;/p&gt;&#xA;" OwnerUserId="7850" LastEditorUserId="7850" LastEditorDisplayName="" LastEditDate="2011-03-14T13:42:45.623" LastActivityDate="2011-03-14T13:42:45.623" CommentCount="2" />
  <row Id="26914" PostTypeId="2" ParentId="26869" CreationDate="2011-03-14T13:22:38.827" Score="1" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\rm\ \  C&amp;#39;=0\ \ \Rightarrow\ \ (UV)&amp;#39;-U&amp;#39;\:V\ =\ UV&amp;#39;\: =\ U(V+C)&amp;#39;\: =\ (U(V+C))&amp;#39;-U&amp;#39;\:(V+C) $ &lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2011-03-14T13:22:38.827" />
  <row Id="26915" PostTypeId="2" ParentId="2949" CreationDate="2011-03-14T13:30:21.277" Score="2" ViewCount="" Body="&lt;p&gt;One of the most surprising results I have ever seen is the &lt;a href=&quot;http://mathworld.wolfram.com/VoroninUniversalityTheorem.html&quot; rel=&quot;nofollow&quot;&gt;Universality Theorem of Voronin&lt;/a&gt; which states that any nonvanishing analytic function can be well -approximated by $\zeta(s)$ somewhere in the critical strip for $0 &amp;lt; Re(s) &amp;lt; 1$.&lt;/p&gt;&#xA;" OwnerUserId="6491" LastActivityDate="2011-03-14T13:30:21.277" CommunityOwnedDate="2011-03-14T13:30:21.277" />
  <row Id="26916" PostTypeId="2" ParentId="2969" CreationDate="2011-03-14T13:39:18.073" Score="1" ViewCount="" Body="&lt;p&gt;I was inspired by the problem you presented, and wrote a program for finding the solutions. &#xA;I hope you don't mind me saying, after seeing the program solve this cube with less than 2 million placements and removals per solution on average (with no clever heuristics), the figure of 50 million seems to indicate some kind of problem in the implementation. I'll be glad to share if this could do good to you or anyone.&lt;/p&gt;&#xA;" OwnerUserId="8154" LastActivityDate="2011-03-14T13:39:18.073" />
  <row Id="26917" PostTypeId="2" ParentId="2949" CreationDate="2011-03-14T13:55:44.660" Score="2" ViewCount="" Body="&lt;p&gt;When young, that there exist &lt;a href=&quot;http://en.wikipedia.org/wiki/Dense_set&quot; rel=&quot;nofollow&quot;&gt;dense sets&lt;/a&gt; with zero measure and smaller cardinality, and later that there exist &lt;a href=&quot;http://en.wikipedia.org/wiki/Nowhere_dense_set#Nowhere_dense_sets_with_positive_measure&quot; rel=&quot;nofollow&quot;&gt;nowhere dense sets with positive measure&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="6460" LastActivityDate="2011-03-14T13:55:44.660" CommunityOwnedDate="2011-03-14T13:55:44.660" />
  <row Id="26918" PostTypeId="1" AcceptedAnswerId="30272" CreationDate="2011-03-14T13:55:56.190" Score="3" ViewCount="361" Body="&lt;p&gt;Inspired by a paper (from 2001) entitled &lt;a href=&quot;http://www.math.utah.edu/~palais/pi.pdf&quot; rel=&quot;nofollow&quot;&gt;Pi is Wrong&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is $\pi$ = 3.14... instead of 6.28... ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Setting $\pi$ = 6.28 would seem to simplify many equations and constants in math and physics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there an intuitive reason we relate the circumference of a circle to its diameter instead of its radius, or was it an arbitrary choice that's left us with multiplicative baggage?&lt;/p&gt;&#xA;" OwnerUserId="3372" LastActivityDate="2011-04-10T21:40:28.160" Title="Why is $\pi$ = 3.14... instead of 6.28...?" Tags="&lt;pi&gt;" AnswerCount="1" CommentCount="10" FavoriteCount="2" />
  <row Id="26919" PostTypeId="1" CreationDate="2011-03-14T14:01:16.530" Score="0" ViewCount="63" Body="&lt;p&gt;For which integral values of n can a set (1,2,3,...,4n) be split into n disjoint 4-element subsets (a,b,c,d) such that in each of these sets a=(b+c+d)/3 ?&lt;/p&gt;&#xA;" OwnerUserId="8000" LastActivityDate="2011-03-14T16:45:11.467" Title="Splitting of set of natural numbers" Tags="&lt;set-theory&gt;&lt;elementary-set-theory&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="26920" PostTypeId="2" ParentId="26912" CreationDate="2011-03-14T14:07:32.860" Score="1" ViewCount="" Body="&lt;p&gt;Second is easy too. Let's assume that $mn+1$ is a perfect cube and recall that $(a + 1)^3 = a^3 + 3a^2 + 3a+ 1$. If both of them are equal we get $(mn + 1) = a^3 + 3a^2 + 3a+ 1$ which is the same as $mn = a(a^2 + 3a + 3)$. This means that for every &lt;em&gt;m&lt;/em&gt; if you take $n = m^2 + 3m + 3$ then $mn+1$ will be a perfect cube.&lt;/p&gt;&#xA;" OwnerUserId="4215" LastEditorUserId="4215" LastEditorDisplayName="" LastEditDate="2011-03-14T14:47:41.700" LastActivityDate="2011-03-14T14:47:41.700" CommentCount="3" />
  <row Id="26921" PostTypeId="2" ParentId="26919" CreationDate="2011-03-14T14:07:44.350" Score="0" ViewCount="" Body="&lt;p&gt;What have you tried so far?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hints:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;since they all have to be integers, how many ways can $b+c+d$ be divisible by 3 and less than 40/3?&lt;/li&gt;&#xA;&lt;li&gt;since there are no other restrictions on $b,c,d$, how many ways can these three be permuted and still give the same sum?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1919" LastActivityDate="2011-03-14T14:07:44.350" />
  <row Id="26922" PostTypeId="1" CreationDate="2011-03-14T14:14:30.310" Score="0" ViewCount="78" Body="&lt;p&gt;This isn't really a question...My brother and his high school math teacher put together this puzzle quest all about Pi for 3/14. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's 3 years running and a great way to get people excited about math. I thought the community here would really appreciate it, which is why I'm posting. If you think it's spammy, please feel free to close the topic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Good luck!&#xA;&lt;a href=&quot;http://www.pidaychallenge.com/&quot; rel=&quot;nofollow&quot;&gt;http://www.pidaychallenge.com/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="8242" LastActivityDate="2011-03-14T14:14:30.310" ClosedDate="2011-03-14T14:49:39.857" Title="Can you Complete the Pi Day Challenge?" Tags="&lt;pi&gt;" CommentCount="2" />
  <row Id="26923" PostTypeId="2" ParentId="26904" CreationDate="2011-03-14T14:25:24.573" Score="4" ViewCount="" Body="&lt;p&gt;You can confirm it numerically by approximating its value at some particular points. &#xA;By definition: $$f\,&amp;#39;(x)=\lim_{\Delta x \to 0}{\frac{f(x + \Delta x) - f(x)}{\Delta x}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example:&#xA;$$\sin(x)&amp;#39;=\cos(x)$$&#xA;Let $\Delta x=0.000001$ and $x=1$&#xA;$$\sin(1)&amp;#39;\approx\frac{\sin(1+0.000001)-\sin(1)}{0.000001}\approx\frac{0.841471525 - 0.841470985}{0.000001}\approx0.54030189$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now your derivative:&#xA;$$\cos(1)\approx0.54030231$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which is very close. The smaller $\Delta x$, the closer the result. Pick a number of points on x-axis and check them like that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can probably define a custom function using your calculator and build its graph: $$ \mathrm{error}(x)=\left|\frac{f(x+0.0000001)-f(x-0.0000001)}{0.0000002}-f\,&amp;#39;(x)\right|$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take into account that the approximation error needs to be compared to the value of your derivative at that point.&lt;/p&gt;&#xA;" OwnerUserId="6554" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-14T20:13:21.643" LastActivityDate="2011-03-14T20:13:21.643" />
  <row Id="26924" PostTypeId="1" CreationDate="2011-03-14T14:25:35.120" Score="1" ViewCount="107" Body="&lt;p&gt;I'm finding it hard to find a derivation of the DFT from books, was wondering if someone can point me to some resources or attempt to explain it here.&lt;/p&gt;&#xA;" OwnerUserId="4789" LastActivityDate="2011-03-14T17:21:36.883" Title="Derivation of the discrete Fourier transform - for dummies" Tags="&lt;fourier-analysis&gt;" AnswerCount="0" CommentCount="2" />
  <row Id="26925" PostTypeId="1" AcceptedAnswerId="27189" CreationDate="2011-03-14T14:30:58.393" Score="7" ViewCount="426" Body="&lt;p&gt;Prove...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\int^{\infty}_{0}{x^{2n}e^{-x^{2}}}dx=\frac{(2n)!}{2^{2n}n!}.\frac{\sqrt{\pi}}{2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="8245" LastEditorUserId="8245" LastEditorDisplayName="" LastEditDate="2011-03-15T09:49:40.070" LastActivityDate="2011-03-16T03:09:03.227" Title="Can someone please show that $\int^{\infty}_{0}{x^{2n}e^{-x^{2}}}dx=\frac{(2n)!}{2^{2n}n!}.\frac{\sqrt{\pi}}{2}$  without using gamma function? " Tags="&lt;analysis&gt;" AnswerCount="4" CommentCount="1" FavoriteCount="5" />
  <row Id="26926" PostTypeId="1" CreationDate="2011-03-14T14:34:20.790" Score="2" ViewCount="246" Body="&lt;p&gt;I'm making some ListPlot[]'s of the binomial distribution, and encountering some strange Mathematica 7 behavior.  If I execute the following statements &lt;em&gt;together&lt;/em&gt;, the last test returns true (by together, I mean highlighting the three cells and Evaluating).&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;In[109]:= n = 15;&#xA;          p = 1/4;&#xA;          awef = Table[{k, Binomial[n, k] p^k (1 - p)^(n - k)}, {k, 0, n}];&#xA;&#xA;In[112]:= awef == Table[{k, Binomial[n, k] p^k (1 - p)^(n - k)}, {k, 0, n}]&#xA;&#xA;Out[112]= True&#xA;&#xA;In[114]:= awef == Table[{k, Binomial[n, k] p^k (1 - p)^(n - k)}, {k, 0, n}]&#xA;&#xA;Out[114]= False&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;However, if I then rerun the last test, I get false, as shown.  What's going on?&lt;/p&gt;&#xA;" OwnerUserId="3009" LastActivityDate="2011-03-31T07:40:31.640" Title="What is Mathematica doing?" Tags="&lt;mathematica&gt;" AnswerCount="3" FavoriteCount="1" />
  <row Id="26927" PostTypeId="2" ParentId="26893" CreationDate="2011-03-14T14:36:39.087" Score="3" ViewCount="" Body="&lt;p&gt;As Joriki pointed out in his comment, this is equivalent to finding an $f(u)$ such that for all $0 \leq s \leq T$ one has&#xA;$$\int_0^{\infty}(u^2 - s)\exp{(-{u^2 \over 2s})}f(u) \,du = 0$$ &#xA;Write  $\int_0^{\infty}u^2\exp{-({u^2 \over 2s})}f(u)\,du$ as&#xA;$-\int_0^{\infty}-{u \over s}\exp{(-{u^2 \over 2s})}suf(u)\,du$ and integrate by parts, integrating $-{u \over s}\exp{(-{u^2 \over 2s})}$ to $\exp{(-{u^2 \over 2s})}$, and differentiating the rest. The result is the expression&#xA;$$\int_0^{\infty}\bigg(s\exp{(-{u^2 \over 2s})}f(u) + suf&amp;#39;(u)\exp{(-{u^2 \over 2s})}\bigg)\,du$$&#xA;The first term of this cancels out the second term of your original expression, so what you need is a function $f(u)$ such that for all $0 \leq s \leq T$ one has&#xA;$$s\int_0^{\infty}uf&amp;#39;(u)\exp{(-{u^2 \over 2s})}\,du = 0$$&#xA;You can cancel out the $s$ factor in front, then change variables from $u$ to $\sqrt{u}$ to get&#xA;$$\int_0^{\infty}f&amp;#39;(\sqrt{u})\exp{(-{u \over 2s})}\,du = 0$$&#xA;Lastly, one can replace $s$ by ${1 \over 2s}$ to get that for all $s \geq {1 \over 2T}$ you need&#xA;$$\int_0^{\infty}f&amp;#39;(\sqrt{u})\exp{(-{su})}\,du = 0$$&#xA;This can't happen; the above defines analytic function of $s$ which can't be zero on a segment without being identically zero. So the Laplace transform of $f&amp;#39;(\sqrt{u})$ is identically zero, which for reasonable $f&amp;#39;$ will not happen unless $f&amp;#39;$ is identically zero.&lt;/p&gt;&#xA;" OwnerUserId="3035" LastEditorUserId="3035" LastEditorDisplayName="" LastEditDate="2011-03-14T14:53:05.463" LastActivityDate="2011-03-14T14:53:05.463" CommentCount="1" />
  <row Id="26929" PostTypeId="2" ParentId="2949" CreationDate="2011-03-14T14:47:14.350" Score="2" ViewCount="" Body="&lt;p&gt;Henry Ernest Dudeney's &lt;a href=&quot;http://www.btinternet.com/~se16/js/cuboid.htm#SpiderFly&quot; rel=&quot;nofollow&quot;&gt;Spider and Fly Problem&lt;/a&gt;: &lt;em&gt;With a cuboid 30x12x12, what is the minimum surface distance from a point which is on a 12x12 face and in 1 from the mid-point of an edge to the opposite point across the centre of the cuboid?&lt;/em&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The surprise is that the minimum distance requires a route using five of the six faces of the cuboid. &#xA;&lt;img src=&quot;http://i.imgur.com/GApE9.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="6460" LastEditorUserId="6460" LastEditorDisplayName="" LastEditDate="2011-03-14T19:08:40.963" LastActivityDate="2011-03-14T19:08:40.963" CommunityOwnedDate="2011-03-14T14:47:14.350" />
  <row Id="26930" PostTypeId="2" ParentId="26925" CreationDate="2011-03-14T14:47:46.550" Score="9" ViewCount="" Body="&lt;p&gt;Making a change of variable $u=x^2$ gives&#xA;$$&#xD;&#xA;\int_0^\infty  {x^{2n} e^{ - x^2 } dx}  = \frac{1}{2}\int_0^\infty  {u^{n - 1/2} e^{ - u} du}  = \frac{1}{2}\Gamma (n + 1/2).&#xD;&#xA;$$&#xA;Then from the well-known formula for the gamma function&#xA;$$&#xD;&#xA;\Gamma (n + 1/2) = \frac{{(2n)!}}{{4^n n!}}\sqrt \pi  &#xD;&#xA;$$&#xA;we get&#xA;$$&#xD;&#xA;\int_0^\infty  {x^{2n} e^{ - x^2 } dx}  = \frac{{(2n)!}}{{2^{2n} n!}}\frac{{\sqrt \pi  }}{2}.&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Second approach.&lt;/strong&gt; Writing&#xA;$$&#xD;&#xA;\int_0^\infty  {x^{2n} e^{ - x^2 } dx}  = \frac{1}{2} \frac{{\sqrt {2\pi (1/2)} }}{{\sqrt {2\pi (1/2)} }} \int_{ - \infty }^\infty  {x^{2n} \exp \bigg( - \frac{{x^2 }}{{2(1/2)}}\bigg)dx} &#xD;&#xA;$$&#xA;shows that&#xA;$$&#xD;&#xA;\int_0^\infty  {x^{2n} e^{ - x^2 } dx}  = \frac{{\sqrt \pi  }}{2}{\rm E}[X^{2n} ],&#xD;&#xA;$$&#xA;where ${\rm E}[X^{2n} ]$ is the $2n$-th moment of the Normal$(0,1/2)$ distribution.&#xA;Now you can see how others, &lt;a href=&quot;http://science.niuz.biz/higher-t19846.html&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; for example, find ${\rm E}[X^{2n} ]$ (for a Normal$(0,\sigma^2)$ distribution).&#xA;The simplest approach may be to use integration by parts. I'll leave it to you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt; (in light of the OP's edit): Integration by parts gives the result, using induction, as follows:&#xA;$$&#xD;&#xA;\int_0^\infty  {x^{2(n + 1)} e^{ - x^2 } dx}  = \frac{{2n + 1}}{2}\int_0^\infty  {x^{2n} e^{ - x^2 } dx}  = \frac{{[2(n + 1)]!}}{{2^{2(n + 1)} (n + 1)!}}\frac{{\sqrt \pi  }}{2}.&#xD;&#xA;$$&#xA;For the base case $n=0$, note that $\int_0^\infty  {e^{ - x^2 } dx}  = \frac{{\sqrt \pi  }}{2}$.&lt;/p&gt;&#xA;" OwnerUserId="2810" LastEditorUserId="2810" LastEditorDisplayName="" LastEditDate="2011-03-15T13:52:44.213" LastActivityDate="2011-03-15T13:52:44.213" />
  <row Id="26931" PostTypeId="2" ParentId="26872" CreationDate="2011-03-14T14:50:15.970" Score="8" ViewCount="" Body="&lt;p&gt;the octonions!  more generally (multiplication in) further algebras coming from the &lt;a href=&quot;http://en.wikipedia.org/wiki/Cayley%E2%80%93Dickson_construction&quot; rel=&quot;nofollow&quot;&gt;cayley-dickson&lt;/a&gt; construction.&lt;/p&gt;&#xA;" OwnerUserId="6925" LastEditorUserId="6925" LastEditorDisplayName="" LastEditDate="2011-03-14T19:32:40.453" LastActivityDate="2011-03-14T19:32:40.453" />
  <row Id="26932" PostTypeId="2" ParentId="26904" CreationDate="2011-03-14T14:53:16.023" Score="3" ViewCount="" Body="&lt;p&gt;Many derivative problems can be done more than one way.  One way to check your work is to try them both ways and see if you get the same thing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example:  $y=\frac{1}{x^2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can write this as $y=x^{-2}$ and use the power rule to get $y&amp;#39;=-2x^{-3}=\frac{-2}{x^3}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or you can use the quotient rule $y&amp;#39;=\frac{0*x^2-2*x}{{(x^2)}^2}=\frac{-2x}{x^4}=\frac{-2}{x^3}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is probably too simple of an example, but I hope you get the idea.&lt;/p&gt;&#xA;" OwnerUserId="4266" LastActivityDate="2011-03-14T14:53:16.023" />
  <row Id="26933" PostTypeId="2" ParentId="26925" CreationDate="2011-03-14T14:53:35.243" Score="21" ViewCount="" Body="&lt;p&gt;Alternatively, set $$I(\alpha) = \int_0^\infty e^{-\alpha x^2}\mathrm{d}x,$$ differentiate $n$ times with respect to $\alpha$ and evaluate at $\alpha = 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT:&#xA;To spell things a little more out, this technique is known as &lt;a href=&quot;http://en.wikipedia.org/wiki/Differentiation_under_the_integral_sign&quot; rel=&quot;nofollow&quot; title=&quot;Differentiation under the integral sign&quot;&gt;Differentiation under the integral sign&lt;/a&gt;.&#xA;Using the fact that $I(\alpha) =\frac12\sqrt{\frac{\pi}{\alpha}}$ and differentiating to obtain $$\frac{\mathrm{d}^n}{\mathrm{d}\alpha^n} I(\alpha) = (-1)^n\int_0^\infty x^{2n} e^{-\alpha x^2}\mathrm{d}x, $$&#xA;some algebraic manipulation and evaluating at $\alpha = 1$ will yield the wanted identity.&lt;/p&gt;&#xA;" OwnerUserId="8246" LastEditorUserId="8246" LastEditorDisplayName="" LastEditDate="2011-03-15T11:26:02.633" LastActivityDate="2011-03-15T11:26:02.633" CommentCount="2" />
  <row Id="26934" PostTypeId="1" AcceptedAnswerId="26965" CreationDate="2011-03-14T14:56:04.227" Score="4" ViewCount="42" Body="&lt;p&gt;Let $f:R \rightarrow R&amp;#39;$ be a ring homomorphism that is epic in the category of rings. Let $M,N$ be $R&amp;#39;$-modules. Why is it that a homorphism $h:M \rightarrow N$ is $R&amp;#39;$-linear if and only if it is $R$-linear? I have used this result in specific cases in the past (and know why it's true),e.g., if $f$ is the usual map of $R$ to $S^{-1}R$,$S$ a multiplicatively closed subset of $R$, but never in the generality I stated. But now I need the general result and would like to know why it works. A reference or a sketch for a proof would be appreciated. &lt;/p&gt;&#xA;" OwnerUserId="2933" LastActivityDate="2011-03-14T17:21:16.927" Title="restriction of scalars, reference or suggestion for proof" Tags="&lt;modules&gt;" AnswerCount="1" />
  <row Id="26935" PostTypeId="1" AcceptedAnswerId="26942" CreationDate="2011-03-14T15:01:50.843" Score="1" ViewCount="17" Body="&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/XvEht.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have sample data like in above format along X, Y axis. Now what i would like to do is to devide  it in &quot;n&quot; number of slabs having fixed values. Now how do i achive this in mathematics(statistics).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please guide me to good and simple tutorial so that i can understand it. I would like to use this in computer algorithm to solve one problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I learned it in school, college and now i am having hard time to remember, understand and interpret it. I wish i could have done it more seriously.&lt;/p&gt;&#xA;" OwnerUserId="8243" LastActivityDate="2011-03-14T15:35:38.540" Title="dividing samples in equal slabs" Tags="&lt;analysis&gt;&lt;statistics&gt;&lt;analytic-geometry&gt;&lt;graph&gt;" AnswerCount="2" />
  <row Id="26936" PostTypeId="2" ParentId="26884" CreationDate="2011-03-14T15:05:47.070" Score="1" ViewCount="" Body="&lt;p&gt;Response to comment-does that mean integers is the answer to my question:  But CRT will tell you when it cannot be done.  If all the $m_i$ share a common factor that $K$ does not, there is no solution.  If $m_1$ and $K$ have no common factor, you can pick $x_2$ through $x_n$ arbitrarily (subject to your inequalities), and solve the equation mod $m_1$.  If $x_1&amp;lt;x_2$, decrease $x_2$ through $x_n$ by $m_1$ and increase $x_1$ appropriately and you will have an answer.  If $m_1$ and $K$ do have a common factor, pick all the $x_i, i \gt 1$ to share that factor and it still works.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-03-14T15:05:47.070" CommentCount="2" />
  <row Id="26937" PostTypeId="2" ParentId="26935" CreationDate="2011-03-14T15:08:54.663" Score="0" ViewCount="" Body="&lt;p&gt;It looks like you want your slabs to all (except the first?) have the same thickness.  So slab $i$ would be at $a+di$ where $d$ is the thickness.  In your example, $a=4, d=2$.  I don't see what use you are making of the points, the line, or the curve through the points.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-03-14T15:08:54.663" CommentCount="2" />
  <row Id="26938" PostTypeId="2" ParentId="26869" CreationDate="2011-03-14T15:13:01.133" Score="4" ViewCount="" Body="&lt;p&gt;Take your example, $$\int x\ln x\,dx.$$&#xA;Note $x\gt 0$ must be assumed (so the integrand makes sense). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we let $u = \ln x$ and $dv= x\,dx$, then we can take $v$ to be any function with $dv = x\,dx$. So the &quot;generic&quot; $v$ will be, as you note, $v = \frac{1}{2}x^2 + C$. What happens then if we use this &quot;generic&quot; $v$?&#xA;\begin{align*}&#xA;\int x\ln x\,dx &amp;amp;= \ln x\left(\frac{1}{2}x^2 + C\right) - \int \left(\frac{1}{2}x^2+C\right)\frac{1}{x}\,dx\\&#xA;&amp;amp;= \frac{1}{2}x^2\ln x + C\ln x - \int\left(\frac{1}{2}x + \frac{C}{x}\right)\,dx\\&#xA;&amp;amp;= \frac{1}{2}x^2\ln x + C\ln x - \frac{1}{4}x^2 - C\ln x + D\\&#xA;&amp;amp;= \frac{1}{2}x^2\ln x - \frac{1}{4}x^2 + D,&#xA;\end{align*}&#xA;so in the end, we get the same result no matter what value of $C$ we take for $v$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This says that we can take &lt;em&gt;any&lt;/em&gt; value of $C$ and still get the same answer. Since we can take any value of $C$, why not take the simplest one, the one that does not require us to carry around an extra term that is going to cancel out anyway? Say..., $C=0$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This works in general. If you replace $v$ with $v+C$ in the integration by parts formula, you have&#xA;\begin{align*}&#xA;\int u\,dv &amp;amp;= u(v+C) - \int(v+C)\,du = uv + Cu - \int v\,du - \int C\,du\\&#xA; &amp;amp;= uv+Cu - \int v\,du - Cu = uv-\int v\,du.&#xA;\end{align*}&#xA;So the answer is the same regardless of the value of $C$, and so we take $C=0$ because that makes our life simpler. &lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2011-03-14T15:13:01.133" />
  <row Id="26939" PostTypeId="2" ParentId="26904" CreationDate="2011-03-14T15:16:35.263" Score="5" ViewCount="" Body="&lt;p&gt;For any specific derivative, you can ask a computer to check your result, as several other answers suggest.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if you want to be self-sufficient in taking derivatives (for an exam or other work), I recommend lots of focused practice.  Most calculus textbooks include answers to the odd-numbered problems in the back of the book, and if you search for &quot;derivative worksheet&quot; you'll find lots of problem lists online.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Work through a list of at least 20 problems, and check your answers-- if you get less than 80% or 90% right, you know you need more practice.  Here's the most important part: &lt;strong&gt;Track down your mistakes.&lt;/strong&gt;  Watch out for them in the future, and be sure you understand the right way to go.  Pay attention to simplifying your answers, too, because a lot of people make algebra mistakes after getting the calculus right.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rules you have are the best way to take these derivatives, you just have to be able to use them accurately.&lt;/p&gt;&#xA;" OwnerUserId="1461" LastActivityDate="2011-03-14T15:16:35.263" />
  <row Id="26940" PostTypeId="2" ParentId="26926" CreationDate="2011-03-14T15:19:59.500" Score="2" ViewCount="" Body="&lt;p&gt;Works fine in 8.0.1&#xA;BTW, what did your In[113] do?&lt;/p&gt;&#xA;" OwnerUserId="2710" LastActivityDate="2011-03-14T15:19:59.500" />
  <row Id="26941" PostTypeId="1" CreationDate="2011-03-14T15:34:45.060" Score="1" ViewCount="70" Body="&lt;p&gt;The question is:&#xA;When the piggy bank was opened, it yielded $5.22 in nickels and pennies. If there were 162 nickels and pennies altogether, how many of each were in the bank?&lt;/p&gt;&#xA;" OwnerUserId="8247" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-14T16:36:55.327" LastActivityDate="2011-03-14T17:50:36.573" Title="Word Problem (Algebra 2)" Tags="&lt;algebra&gt;" AnswerCount="3" />
  <row Id="26942" PostTypeId="2" ParentId="26935" CreationDate="2011-03-14T15:35:38.540" Score="0" ViewCount="" Body="&lt;p&gt;Are you interested in binning the data? If so, there are a few algorithms that you can use to group the data. But first you need to decide what method to use to bin the data. The most popular methods are &quot;equal width&quot; and &quot;equal frequency&quot;.  Sound like you need &quot;equal width&quot;. Here are a couple of links.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://cs-people.bu.edu/dgs/courses/cs105/lectures/data_mining_preparation.pdf&quot; rel=&quot;nofollow&quot;&gt;http://cs-people.bu.edu/dgs/courses/cs105/lectures/data_mining_preparation.pdf&lt;/a&gt;&#xA;http://en.wikipedia.org/wiki/Histogram&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-Ralph Winters&lt;/p&gt;&#xA;" OwnerUserId="8071" LastActivityDate="2011-03-14T15:35:38.540" CommentCount="1" />
  <row Id="26943" PostTypeId="2" ParentId="26941" CreationDate="2011-03-14T15:36:46.733" Score="3" ViewCount="" Body="&lt;p&gt;What have you tried?  Hint:  If you have $n$ nickels and $p$ pennies, how much money do you have?&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-03-14T15:36:46.733" CommentCount="4" />
  <row Id="26944" PostTypeId="1" AcceptedAnswerId="26971" CreationDate="2011-03-14T15:39:13.237" Score="3" ViewCount="118" Body="&lt;p&gt;Hey there,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have to solve a problem which states that $(m,n) = 1$, but I have no idea what this means. Maybe, the problem itself will help:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose $m, n \in \mathbb{Z}$, with $n &amp;gt; 0$ and $(m,n) = 1$. Show that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left \{ e^{2 \pi i m k/n}: 0 \leq k &amp;lt; n \right \} = \left \{ e^{2 \pi i j/n}: 0 \leq j &amp;lt; n\right \}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks a lot for any help.&lt;/p&gt;&#xA;" OwnerUserId="3787" LastActivityDate="2011-03-14T17:33:28.350" Title="What does $(m, n) = 1$ mean?" Tags="&lt;notation&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="26946" PostTypeId="2" ParentId="26944" CreationDate="2011-03-14T15:41:57.737" Score="4" ViewCount="" Body="&lt;p&gt;$(m,n)=1$ simply means that the &lt;strong&gt;G.C.D&lt;/strong&gt; of $m,n$ is $1$. In other words they are relatively prime.&lt;/p&gt;&#xA;" LastActivityDate="2011-03-14T15:41:57.737" />
  <row Id="26947" PostTypeId="2" ParentId="26941" CreationDate="2011-03-14T15:42:47.283" Score="2" ViewCount="" Body="&lt;p&gt;&quot;How many of each...&quot; actually gives a &lt;em&gt;hint&lt;/em&gt; about how to go about solving this. We have two things that we don't know: P, the number of pennies, and N, the number of nickels. &lt;br&gt;&#xA;You are probably learning systems of equations, so I'll take that approach. &lt;br&gt;&#xA;You need two equations. One will simply be the number of total coins:&lt;br&gt;&#xA;N + P = 162. (They gave you this!)&lt;br&gt;&#xA;&lt;br&gt;&#xA;Then there will be an equation about the &lt;em&gt;value&lt;/em&gt; of the coins. I'm going to write this in cents.&lt;br&gt;&lt;br&gt;&#xA;The total amount of cents is 522 (given), which comes from 5 cents per nickel and 1 cent per penny. In other words, &lt;br&gt;&#xA;5N + 1P = 522.&lt;br&gt;&#xA;I suggest elimination. Reply if you still need help...&lt;/p&gt;&#xA;" OwnerUserId="7850" LastActivityDate="2011-03-14T15:42:47.283" CommentCount="2" />
  <row Id="26948" PostTypeId="1" CreationDate="2011-03-14T15:55:52.213" Score="1" ViewCount="73" Body="&lt;p&gt;In localization of a ring $R$ or a module $M$ over $R$ at a multiplicative subset $S$ of $R$, we define an equivalence relation on $R\times S$ or $M\times S$ and define addition and multiplication on equivalence classes. I think, if you don't define an equivalence relation, and define addition and multiplication in the same way, you would still get a ring (respectively an $R$-module). Although, I don't expect this structure to have the nice properties of localization, I was wondering if such a structure has been studied.&lt;/p&gt;&#xA;" OwnerUserId="7763" LastActivityDate="2011-03-14T15:55:52.213" Title="A question about localization" Tags="&lt;abstract-algebra&gt;&lt;commutative-algebra&gt;" CommentCount="4" />
  <row Id="26949" PostTypeId="1" AcceptedAnswerId="26951" CreationDate="2011-03-14T16:01:09.737" Score="0" ViewCount="112" Body="&lt;p&gt;Find four consecutive odd integers such that 5 times the sum of the first and the third is 15 more than 5 times the fourth.&lt;/p&gt;&#xA;" OwnerUserId="8247" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2011-03-14T23:00:38.007" LastActivityDate="2011-03-15T01:35:30.740" Title="Finding Consecutive Integers" Tags="&lt;algebra&gt;" AnswerCount="4" CommentCount="4" />
  <row Id="26951" PostTypeId="2" ParentId="26949" CreationDate="2011-03-14T16:08:02.363" Score="4" ViewCount="" Body="&lt;p&gt;We have to find 4 odd numbers so let us take &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$2n-1$ , $2n+1$ , $2n+3$ , $2n+5$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we have $5 ((2n-1) + (2n+3)) - 15 = 5 (2n + 5)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can now solve this .....&lt;/p&gt;&#xA;" OwnerUserId="8250" LastEditorUserId="8137" LastEditorDisplayName="" LastEditDate="2011-03-14T19:15:51.687" LastActivityDate="2011-03-14T19:15:51.687" CommentCount="2" />
  <row Id="26952" PostTypeId="1" AcceptedAnswerId="27019" CreationDate="2011-03-14T16:10:49.083" Score="2" ViewCount="174" Body="&lt;p&gt;I have to solve find the value of&#xA;$$\sum_{k=1}^{n/2} k\log k$$&#xA;as a part of question. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How should I proceed on this ?&lt;/p&gt;&#xA;" OwnerUserId="8250" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-14T16:19:58.383" LastActivityDate="2011-03-14T20:47:21.647" Title="sum of logarithms" Tags="&lt;sequences-and-series&gt;" AnswerCount="2" CommentCount="8" FavoriteCount="1" />
  <row Id="26953" PostTypeId="2" ParentId="26949" CreationDate="2011-03-14T16:11:28.397" Score="0" ViewCount="" Body="&lt;p&gt;Let the numbers be $2m+1,2m+3,2m+5,2m+7$ then solve for $$ 5 \cdot \Bigl[ 2m+1 + 2m+5\Bigr] = 5(2m+7) + 15$$&lt;/p&gt;&#xA;" LastActivityDate="2011-03-14T16:11:28.397" />
  <row Id="26954" PostTypeId="1" CreationDate="2011-03-14T16:12:52.740" Score="12" ViewCount="202" Body="&lt;p&gt;For which $m$, $n$ (if any) is the following true: if $M$ and $M&amp;#39;$ are smooth manifolds of dimension $n$, and $\Phi$ is a bijection from $M$ to $M&amp;#39;$ such that for any subset $S$ of $M$, $\Phi(S)$ is an embedded submanifold of $M&amp;#39;$ of dimension $m$ iff $S$ is an embedded submanifold of $M$ of dimension $m$, then $\Phi$ is a diffeomorphism?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is clearly false when $m=0$ and when $m=n$.  I thought it looked plausible when, for example, $m=1$ and $n\geq 2$; but I can't see how to prove it.  &lt;/p&gt;&#xA;" OwnerUserId="4909" LastActivityDate="2011-03-25T08:46:59.473" Title="When is an $n$-dimensional manifold characterized by its $m$-dimensional submanifolds?" Tags="&lt;differential-geometry&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="26955" PostTypeId="2" ParentId="26847" CreationDate="2011-03-14T16:23:04.097" Score="1" ViewCount="" Body="&lt;p&gt;One part that I can think of mathematically would be the use of golden ratio in &lt;code&gt;fibonacci numbers.&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The golden ratio is the part of series of well known &lt;code&gt;fibonacci number series&lt;/code&gt; .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is the link since I am unable to write the actual series for the fibonacci here......&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Fibonacci_number&quot; rel=&quot;nofollow&quot;&gt;fibonacci&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="8250" LastActivityDate="2011-03-14T16:23:04.097" />
  <row Id="26956" PostTypeId="1" CreationDate="2011-03-14T16:25:17.643" Score="1" ViewCount="117" Body="&lt;p&gt;How would I solve this:&#xA;Multiply:  $(4 + x)(x^2 + 2x +3)$&lt;/p&gt;&#xA;" OwnerUserId="8247" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-14T16:30:11.883" LastActivityDate="2011-03-14T18:20:48.590" Title="Multiply: $(4 + x)(x^2 + 2x +3)$" Tags="&lt;algebra-precalculus&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="26957" PostTypeId="2" ParentId="26956" CreationDate="2011-03-14T16:33:00.203" Score="4" ViewCount="" Body="&lt;p&gt;This question, as you probably know, requires the use of the distributive property. To use DJC's suggestion $$(a + b)*c = a*c + b*c$$ &lt;br&gt;&#xA;Let &quot;a + b&quot; be your $4 + x$ and let &quot;c&quot; be your $x^2 + 2x + 3$ &lt;br&gt;&lt;br&gt;&#xA;Then we need to multiply a*c, or $4*(x^2 + 2x + 3)$ and add it to &lt;br&gt;&lt;br&gt;&#xA;b*c, which is $x*(x^2 + 2x + 3)$ &lt;br&gt;&lt;br&gt;&#xA;After taking these steps, combine like terms and write the result in order of decreasing exponents (for convention's sake)&lt;/p&gt;&#xA;" OwnerUserId="7850" LastActivityDate="2011-03-14T16:33:00.203" />
  <row Id="26958" PostTypeId="2" ParentId="26919" CreationDate="2011-03-14T16:45:11.467" Score="2" ViewCount="" Body="&lt;p&gt;If $a=\frac{b+c+d}{3}$, then $a+b+c+d=4a$ is a multiple of $4$. If $\{1,\dots,4n\}$ can be split in quadruples $\{a,b,c,d\}$ with $a=\frac{b+c+d}{3}$, then $1+2+\dots+4n$ must also be a multiple of $4$. I leave you to check that this is so if and only if $n$ is even.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $n=2$ we have the solution $\{1,\dots,8\}=\{4,1,3,8\}\cup\{5,2,6,7\}$. From here it is easy to construct a solution for any even $n$.&lt;/p&gt;&#xA;" OwnerUserId="4791" LastActivityDate="2011-03-14T16:45:11.467" CommentCount="1" />
  <row Id="26960" PostTypeId="1" CreationDate="2011-03-14T16:52:37.293" Score="-2" ViewCount="58" Body="&lt;p&gt;If f(x) = x^2 + 2x + 2, find f(m - 1)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can someone please solve this for me, with the steps.&lt;/p&gt;&#xA;" OwnerUserId="8247" LastActivityDate="2011-03-14T17:00:27.793" Title="Algebra Function Problem" Tags="&lt;algebra&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="26961" PostTypeId="2" ParentId="26952" CreationDate="2011-03-14T16:55:59.570" Score="3" ViewCount="" Body="&lt;p&gt;Here is an asymptotic expression using &lt;a href=&quot;http://en.wikipedia.org/wiki/Euler%E2%80%93Maclaurin_formula&quot; rel=&quot;nofollow&quot;&gt;EulerMcLaurin Summation&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \sum _{k=1}^{n} k \log k = \int_{1}^{n} x \log x\ \text{d}x + (n\log n)/2 + C&amp;#39; + (\log n + 1)/12+ \mathcal{O}(1/n^2)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ = n^2(2 \log n - 1)/4 + (n\log n)/2 + (\log n)/12 + C + \mathcal{O}(1/n^2)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for some constant $C$. &lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2011-03-14T17:41:10.840" LastActivityDate="2011-03-14T17:41:10.840" CommentCount="9" />
  <row Id="26962" PostTypeId="2" ParentId="26960" CreationDate="2011-03-14T16:59:43.673" Score="3" ViewCount="" Body="&lt;p&gt;Evaluating functions with inputs that &lt;em&gt;aren't just &quot;x&quot;&lt;/em&gt; is an important skill.&lt;br&gt;&lt;br&gt;&#xA;Let me walk you through some examples, using this function:&lt;br&gt;&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f(x) = x^2 + 2x + 2$ We will just plug in some numbers and letters, using parenthesis to show where we substituted...&lt;br&gt;&lt;br&gt;&#xA;$f(1) = (1)^2 + 2*(1) + 2 = 1 + 2 + 2 = 5$&lt;br&gt;&#xA;$f(2) = (2)^2 + 2*(2) + 2 = 4 + 4 + 2 = 10$&lt;br&gt;&#xA;$f(3a) = (3a)^2 + 2*(3a) + 2 = 9a^2 + 6a + 2$ Notice that our answer isn't &lt;em&gt;just&lt;/em&gt; a number; it is a variable expression.&lt;br&gt;&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now $f(m - 1) = (m - 1)^2 + 2*(m - 1) + 2$. &lt;br&gt;&lt;br&gt;I'll leave it to you to expand/distribute and combine like terms. &lt;/p&gt;&#xA;" OwnerUserId="7850" LastActivityDate="2011-03-14T16:59:43.673" CommentCount="3" />
  <row Id="26964" PostTypeId="2" ParentId="26960" CreationDate="2011-03-14T17:00:27.793" Score="1" ViewCount="" Body="&lt;p&gt;$$f(m-1) = (m-1)^{2} + 2(m-1) + 2 = m^{2} -2m + 1 + 2m-2 +2 = m^{2}+1$$&lt;/p&gt;&#xA;" LastActivityDate="2011-03-14T17:00:27.793" />
  <row Id="26965" PostTypeId="2" ParentId="26934" CreationDate="2011-03-14T17:08:08.997" Score="1" ViewCount="" Body="&lt;p&gt;Recall the characterization of epimorphisms in the category of rings (see for example &lt;a href=&quot;http://mathoverflow.net/questions/109/what-do-epimorphisms-of-commutative-rings-look-like&quot; rel=&quot;nofollow&quot;&gt;this question on MathOverflow&lt;/a&gt;):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Silver-Mazet-Isbell Zigzag Lemma for rings.&lt;/strong&gt; &lt;em&gt;Let&lt;/em&gt; $f\colon R\to S$ &lt;em&gt;be a ring homomorphism. Then&lt;/em&gt; $f$ &lt;em&gt;is an epimorphism in the category of rings if and only if for every&lt;/em&gt; $s\in S$ &lt;em&gt;there exist matrices&lt;/em&gt; $C$, $D$, and $E$, &lt;em&gt;where&lt;/em&gt; $C$ &lt;em&gt;is&lt;/em&gt; $1\times n$, $D$ &lt;em&gt;is&lt;/em&gt; $n\times n$, $E$ &lt;em&gt;is&lt;/em&gt; $n\times 1$; $C$ &lt;em&gt;and&lt;/em&gt; $E$ &lt;em&gt;have coefficients in&lt;/em&gt; $S$; $CD$, $D$, &lt;em&gt;and&lt;/em&gt; $DE$ &lt;em&gt;have coefficients in&lt;/em&gt; $f(R)$; &lt;em&gt;and&lt;/em&gt; $s=CDE$. &lt;em&gt;This is called a &quot;zigzag in&lt;/em&gt; $S$ &lt;em&gt;over&lt;/em&gt; $f(R)$ &lt;em&gt;with value&lt;/em&gt; $s$.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For simplicity, suppose that $R\subseteq R&amp;#39;$ and that $f$ is the canonical inclusion. Now I'm going to be a bit sloppy below, by identifying $M$ with its image in obvious matrix rings, but I think the idea carries through:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose $h\colon M\to N$ is $R$-linear, and let $m\in M$, $b\in R&amp;#39;$. We want to show that $h(bm) = bh(m)$. Since $R\hookrightarrow R&amp;#39;$ is an epimorphism, there is a zigzag in $R&amp;#39;$ over $R$ with value $b$, $b=CDE$, where $D$, $CD$, and $CE$ are matrices with coefficients in $R$. Since $h$ is $R$ linear, we have:&#xA;\begin{align*}&#xA;h(bm) &amp;amp;= h(CDEm)\\&#xA;&amp;amp;= CDh(Em) &amp;amp;\qquad&amp;amp;\mbox{(since $CD$ has coefficients in $R$)}\\&#xA;&amp;amp;= Ch(DEm) &amp;amp;&amp;amp;\mbox{(since $D$ has coefficients in $R$)}\\&#xA;&amp;amp;= C(DEh(m)) &amp;amp;&amp;amp;\mbox{(since $DE$ has coefficients in $R$)}\\&#xA;&amp;amp;= (CDE)h(m)\\ &#xA;&amp;amp;= bh(m).&#xA;\end{align*}&#xA;As I said, that's a bit sloppy, since in the middle I'm actually dealing with first with $\mathcal{M}_{1\times n}(M)$ as an $\mathcal{M}_{1\times n}(R&amp;#39;)$ module, with the induced map $h$; and later I'm dealing with $\mathcal{M}_{n\times n}(M)$ and $\mathcal{M}_{n\times 1}(M)$ over suitable matrix rings. But essentially that is what is going on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(You can also probably see why it's called a &quot;zigzag&quot;: you zigzag between $CD$ and $DE$ to pull the entire scalar out.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Isbell's Zigzag Lemma (characterizing elements that are in the dominion of a subalgebra in a large class of subalgebras) can be found in &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Isbell, John R.  &lt;em&gt;Epimorphisms and dominions.&lt;/em&gt;&#xA;1966  Proc. Conf. Categorical Algebra (La Jolla, Calif., 1965) &#xA;pp. 232-246 Springer, New York, MR0209202 (35 #105a)  &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The statement of the Zigzag Lemma for rings is incorrect in that paper. A correct version appears in:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Isbell, John R.  &lt;a href=&quot;http://jlms.oxfordjournals.org/cgi/pdf_extract/s2-1/1/265&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;Epimorphisms and dominions. IV.&lt;/em&gt;&lt;/a&gt;&#xA;J. London Math. Soc. (2)  &lt;strong&gt;1&lt;/strong&gt;  1969, pp. 265-273. MR0257120 (41 #1774)  &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;More generally, the collection of all $s\in S$ for which there is a Zigzag in $S$ over $f(R)$ with value $s$ is called the &lt;em&gt;dominion of&lt;/em&gt; $f(R)$ &lt;em&gt;in&lt;/em&gt; $S$. It is precisely the subring of $S$ given by:&#xA;$$\mathrm{dom}_S(f(R)) = \Bigl\{ s\in S\Bigm| \forall T,\ \forall g,h\colon S\to T( g|_{f(R)} = h|_{f(R)}\Rightarrow g(s)=h(s))\Bigr\}.$$&#xA;The argument above shows that any module map that is $R$ linear will necessarily be $\mathrm{dom}_{R&amp;#39;}(f(R))$-linear; in the special case when $\mathrm{dom}_{R&amp;#39;}(f(R)) = R&amp;#39;$, which is precisely the case where $f$ is an epimorphism, the conclusion you want follows.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-14T17:21:16.927" LastActivityDate="2011-03-14T17:21:16.927" CommentCount="1" />
  <row Id="26966" PostTypeId="1" CreationDate="2011-03-14T17:12:53.750" Score="-2" ViewCount="142" Body="&lt;p&gt;$2x + 2y + 2 = 0$ and $7x + 7y + 7 = 0$. Are these equations consistent, inconsistent, or dependent?&lt;/p&gt;&#xA;" OwnerUserId="8247" LastEditorUserId="1583" LastEditorDisplayName="" LastEditDate="2011-03-14T17:22:45.100" LastActivityDate="2011-03-20T07:44:52.227" Title="Are the equations $2x + 2y + 2 = 0$ and $7x + 7y + 7 = 0$ consistent, inconsistent, or dependent?" Tags="&lt;homework&gt;&lt;algebra-precalculus&gt;" AnswerCount="4" CommentCount="5" />
  <row Id="26967" PostTypeId="1" AcceptedAnswerId="26979" CreationDate="2011-03-14T17:15:19.753" Score="6" ViewCount="135" Body="&lt;p&gt;I'm looking for reasonably real, non-abstract applications modeled by quadratic equations where both solutions make sense. I'd like them to be accessible to high school algebra students.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One I come up with is firing a bullet through a sphere -- the two solutions correspond to the entry and exit points of the bullet, that is the intersections of the bullet's line and the sphere's surface. To keep it simple, we could do this instead with a line and circle in the xy plane, but the idea is the same. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Applications outside of physics would be particularly welcome.&lt;/p&gt;&#xA;" OwnerUserId="271" LastActivityDate="2011-03-14T18:45:22.787" CommunityOwnedDate="2011-03-14T18:39:45.237" Title="Both solutions to a quadratic make sense -- looking for applications" Tags="&lt;algebra&gt;&lt;applications&gt;&lt;quadratics&gt;" AnswerCount="4" CommentCount="1" />
  <row Id="26969" PostTypeId="2" ParentId="26909" CreationDate="2011-03-14T17:21:38.633" Score="1" ViewCount="" Body="&lt;p&gt;It doesn't work for $U(1)$: take $V$ to be the standard representation, you only get the &quot;nonnegative&quot; representations.&#xA;Indeed to use Stone Weierstrass in the complex case, the algebra has to be stable under complex conjugation.&#xA;Maybe consider the $V^n \otimes (V^*)^m$ instead.&lt;/p&gt;&#xA;" OwnerUserId="8251" LastActivityDate="2011-03-14T17:21:38.633" CommentCount="1" />
  <row Id="26970" PostTypeId="2" ParentId="26966" CreationDate="2011-03-14T17:30:41.213" Score="1" ViewCount="" Body="&lt;p&gt;(I'll try to spoon-feed a little less...)&lt;br&gt;&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, we try to decide if these two equations have &lt;em&gt;any&lt;/em&gt; solutions. In other words, is there (at least) one ordered pair (x, y) that is a solution (i.e. yields a true equality) for both equations? &lt;br&gt;&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suggest that you use the &quot;elimination&quot; method for solving systems of equations. &#xA;&lt;br&gt;&lt;br&gt;&#xA;Here's a hint: if you ever get an equation with no x's or y's that is &lt;em&gt;always false&lt;/em&gt;, like &#xA;3 = 0, then the system is inconsistent. &lt;br&gt;&lt;br&gt;&#xA;Likewise, if you come to an equation that is &lt;em&gt;always true&lt;/em&gt;, like 2 = 2 or 0 = 0, then the system is consistent and dependent. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Otherwise, you'll have one solution, which is called &quot;independent and consistent&quot;.&lt;/p&gt;&#xA;" OwnerUserId="7850" LastActivityDate="2011-03-14T17:30:41.213" />
  <row Id="26971" PostTypeId="2" ParentId="26944" CreationDate="2011-03-14T17:33:28.350" Score="6" ViewCount="" Body="&lt;p&gt;The notation $\rm\ (a,b)\ $ can denote either $\rm\ gcd(a,b)\ $ or the ideal $\rm\ a\ \mathbb Z + b\ \mathbb Z\:.\ $ In $\rm\:\mathbb Z\:$ (or in any Bezout domain $\rm\:Z\:$) these denote essentially the same object since $\rm\ a\ Z + b\ Z\ =\ c\ Z\ \ \iff\ \ c = gcd(a,b)\:.$ The advantage of the notation is that it allows one to simultaneously prove results for both gcds and ideals by using only those laws that hold true for both. For example, see my &lt;a href=&quot;http://math.stackexchange.com/questions/10400/comaximal-ideals-in-a-commutative-ring/10416#10416&quot;&gt;post on the Freshman's dream&lt;/a&gt; $\rm\ (a,b)^n = (a^n,b^n)\ $ for gcds and invertible ideals. Such analogies are exploited to the hilt when one studies &lt;em&gt;divisor theory&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2011-03-14T17:33:28.350" />
  <row Id="26972" PostTypeId="1" CreationDate="2011-03-14T17:34:24.777" Score="6" ViewCount="257" Body="&lt;p&gt;I have difficulties in showing the series $f(x)=\sum_{n=1}^\infty \frac{\cos(\sqrt{n}x)}{\sqrt{n}}$ is divergent at every real numbers $x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However I cannot find any elementary methods to do this. Can anyone help me on this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="8252" LastEditorUserId="8252" LastEditorDisplayName="" LastEditDate="2011-03-15T05:06:58.893" LastActivityDate="2011-03-21T07:24:49.510" Title="Divergence of $\sum\frac{\cos(\sqrt{n}x)}{\sqrt{n}}$" Tags="&lt;real-analysis&gt;&lt;sequences-and-series&gt;&lt;trigonometry&gt;&lt;convergence&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" />
  <row Id="26973" PostTypeId="2" ParentId="26967" CreationDate="2011-03-14T17:35:29.987" Score="3" ViewCount="" Body="&lt;p&gt;The simplest question I can think of is to characterize rectangles with side lengths $x, y$ and fixed perimeter $2x + 2y = p$ and area $xy = A$. The two roots of the corresponding quadratic equation (say in $x$), when they exist, correspond to the &quot;tall&quot; solution and the &quot;wide&quot; solution. But note that the statement of the problem is symmetric in $x$ and $y$, so if you know that a solution exists with $x \neq y$ then from one solution you automatically deduce the other. I think this is a nice example for students to see. &lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2011-03-14T17:35:29.987" CommunityOwnedDate="2011-03-14T18:39:45.237" CommentCount="1" />
  <row Id="26974" PostTypeId="1" CreationDate="2011-03-14T17:39:42.790" Score="3" ViewCount="107" Body="&lt;p&gt;how can i determine wether a circle can be circumscribed about a quadrilateral?&lt;/p&gt;&#xA;" OwnerUserId="8253" LastActivityDate="2011-03-14T17:54:23.323" Title="rules for circle circumscribing" Tags="&lt;circle&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="26975" PostTypeId="2" ParentId="26941" CreationDate="2011-03-14T17:50:36.573" Score="2" ViewCount="" Body="&lt;p&gt;You can either use explicit elimination as suggested, or implicitly as follows. If the bank contained $\:162\:$ pennies and $\:0\:$ nickels then the total amount would be $\:\$1.62\:$, which is $\:\$3.60\:$ short of $\:\$5.22\:.\:$ Exchanging a penny for a nickel adds $\:\$0.04\:,\:$ so we need to make $\ 3.6/0.04 = 360/4\ $ exchanges.&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2011-03-14T17:50:36.573" />
  <row Id="26976" PostTypeId="2" ParentId="26974" CreationDate="2011-03-14T17:54:23.323" Score="4" ViewCount="" Body="&lt;p&gt;If you're given a convex quadrilateral, a circle can be circumscribed about it if and only the quadrilateral is cyclic. A nice fact about cyclic quadrilaterals is that their opposite angles are supplementary. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://aleph0.clarku.edu/~djoyce/java/elements/bookIII/propIII22.html&quot; rel=&quot;nofollow&quot;&gt;Proposition III.22&lt;/a&gt; of Euclid's Elements gives a proof that the opposite angles of cyclic quadrilaterals are equal to two right angles. The converse is also true, that if the opposite angles of a quadrilateral are supplementary, then the quadrilateral is cyclic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another way to identify if a quadrilateral is cyclic is given in Hartshorne's book on classical geometry. A nice proof can be find in Hartshorne's Euclid: Geometry and Beyond, which I will include here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/ykoAf.jpg&quot; alt=&quot;enter image description here&quot;&gt;&#xA;&lt;img src=&quot;http://i.imgur.com/XKW4K.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you can determine either of these facts hold about your quadrilateral, then you know there exists a possible circle circumscribed about it.&lt;/p&gt;&#xA;" OwnerUserId="1583" LastActivityDate="2011-03-14T17:54:23.323" />
  <row Id="26977" PostTypeId="2" ParentId="26956" CreationDate="2011-03-14T17:56:13.400" Score="0" ViewCount="" Body="&lt;p&gt;First, distribute the 4. Add the products of 4 and $x^2$, 4 and 2x, 3 and 4. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, distribute the x. Add the products of x and $x^2$, x and 2x, 3 and x. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then all you have to do is add all of the items you have left. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;After the distributions, you should have &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$4x^2 + 8x + 12 + x^3 + x^2 + 3x =$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x^3 + 5x^2  +11x + 12$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Yay cool text!) &lt;/p&gt;&#xA;" OwnerUserId="8171" LastEditorUserId="8171" LastEditorDisplayName="" LastEditDate="2011-03-14T18:20:48.590" LastActivityDate="2011-03-14T18:20:48.590" CommentCount="1" />
  <row Id="26978" PostTypeId="2" ParentId="26865" CreationDate="2011-03-14T17:59:56.087" Score="3" ViewCount="" Body="&lt;p&gt;$Income = Expenditures$ condition would give you $A\,\vec p=\vec p$ or $(I-A)\vec p=0$. &#xA;Such homogeneous SLE will have infinitely many solutions (a parametric family): $$\vec p=c\left[\begin{array}{c} p_1 \\ p_2 \\ p_3 \end{array}\right]$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please see: Elementary Linear Algebra: Applications Version, Howard Anton, Chris Rorres. - 10.8 Leontief Economic Models&lt;/p&gt;&#xA;" OwnerUserId="6554" LastEditorUserId="6554" LastEditorDisplayName="" LastEditDate="2011-03-15T11:34:37.467" LastActivityDate="2011-03-15T11:34:37.467" CommentCount="6" />
  <row Id="26979" PostTypeId="2" ParentId="26967" CreationDate="2011-03-14T18:02:07.287" Score="6" ViewCount="" Body="&lt;p&gt;Throw a ball upwards, you wanna know how long until it reaches some height?&lt;br&gt;&#xA;Solve a quadratic, the two solutions correspond to when its on its way up and when its coming down again.&lt;/p&gt;&#xA;" OwnerUserId="4929" LastActivityDate="2011-03-14T18:02:07.287" CommunityOwnedDate="2011-03-14T18:39:45.237" />
  <row Id="26980" PostTypeId="2" ParentId="26949" CreationDate="2011-03-14T18:05:06.000" Score="2" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\rm\  \ 5\ (n_1 + n_3)\ =\ 5\ n_4 + 15\: \ \iff\: \ n_1 + n_3 - n_4\ =\ 3\:.\: $ So  $\rm\ n_3 - n_4 = -2\ \Rightarrow\ n_1 = \ \cdots$   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The hypothesis that the $\rm\:n_i\:$ are odd is stronger than needed. To solve for the $\rm\:n_i\:$ from the given equation it suffices to know only the values $\rm\ n_{i+1} - n_i\:.$&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2011-03-15T00:18:48.890" LastActivityDate="2011-03-15T00:18:48.890" />
  <row Id="26981" PostTypeId="1" AcceptedAnswerId="26989" CreationDate="2011-03-14T18:08:04.503" Score="2" ViewCount="63" Body="&lt;p&gt;Suppose that the population of cats in a has a rate of growth proportional to the population itself. Write down a differential equation for the population $P (t)$ at time $t$ ?&lt;/p&gt;&#xA;" OwnerUserId="7984" LastEditorUserId="7984" LastEditorDisplayName="" LastEditDate="2011-03-14T18:34:53.267" LastActivityDate="2011-03-14T19:09:44.117" Title="What is the differential equation expression for this question?" Tags="&lt;homework&gt;&lt;differential-equations&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="26982" PostTypeId="2" ParentId="26837" CreationDate="2011-03-14T18:11:29.383" Score="0" ViewCount="" Body="&lt;p&gt;I thought it might be interesting to consider when the &lt;em&gt;norm&lt;/em&gt; $\bar{z} = a^2+b^2$ of a gaussian integer $z = a+bi$ $a,b \in \mathbb{Z}$ is even or odd. In particular, this agrees with Artuo Magidin's definition as it being even if $a+b$ is, since squaring doesn't change the parity of an integer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the norm is multiplicative, we immediately recover that odd times odd is even, etc. Furthermore, since the norm of $z+x$, where x is another gaussian integer, $(a+c) + (b+d)i$, is $a^2 + c^2 + 2ac + b^2 + d^2 + 2bd = \bar{z} + \bar{x} + 2(ac+bd)$, and so also respects parity additively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strike&gt;However, now we have a homomorphism to the integers. If we quotient the gaussian integers by its kernel (namely, 1 and i), the resulting ring is, of course, (isomorphic to) the integers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if we take the parity of the norm as the definition of parity, then in some sense this matches up with the parity of the original integers because when we go back to the integers, the parity is the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope this makes sense to someone besides me.&lt;/strike&gt; As pointed out in the comments, this is incorrect.&lt;/p&gt;&#xA;" OwnerUserId="2954" LastEditorUserId="2954" LastEditorDisplayName="" LastEditDate="2011-03-14T19:45:53.813" LastActivityDate="2011-03-14T19:45:53.813" CommentCount="3" />
  <row Id="26983" PostTypeId="2" ParentId="26981" CreationDate="2011-03-14T18:11:54.107" Score="1" ViewCount="" Body="&lt;p&gt;Hint:  how would you express the rate of growth in terms of $P(t)$?&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-03-14T18:11:54.107" CommentCount="2" />
  <row Id="26984" PostTypeId="1" CreationDate="2011-03-14T18:16:52.307" Score="8" ViewCount="230" Body="&lt;p&gt;Consider the $n$-sphere $S^n$ and the real projective space $\mathbb{RP}^n$. There is a universal covering map $p: S^n \to \mathbb{RP}^n$, and it's clear that it's the coequaliser of $\mathrm{id}: S^n \to S^n$ and the antipodal map $a: S^n \to S^n$. At first I thought I might use this fact and invoke abstract nonsense to compute the homology groups of $\mathbb{RP}^n$ but then I realised that it didn't give the correct answer. The rubric of the question suggests the following approach:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Cut $S^n$ along two parallels on opposite sides of the equator to obtain two subspaces $L$, homeomorphic to two disjoint copies of the closed ball $B^n$, and $M$, homeomorphic to the closed cylinder $S^{n-1} \times B^1$.&lt;/li&gt;&#xA;&lt;li&gt;The antipodal map sends $L$ to $L$ and $M$ to $M$, and so they are both double covers of their images $Q$ and $R$ (resp.) in $\mathbb{RP}^n$. Clearly $Q$ is homemorphic to a single copy of $B^n$, and with some thought it's seen that $R$ is a $n$-dimensional generalisation of the Möbius strip. &lt;/li&gt;&#xA;&lt;li&gt;Use the fact that the $a$ and $p$ act nicely on $L$ and $M$ to obtain chain maps between the Mayer—Vietoris sequences for $S^n = L \cup M$ and $\mathbb{RP}^n = Q \cup R$.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;: Following Jim Conant's comment about deforming $R$ into $\mathbb{RP}^{n-1}$, I managed to compute the homology of $\mathbb{RP}^n$ by induction (on $n$, of course). However, I'm not convinced this was the intended solution, as the question explicitly refers to the fact that any nice simplicial map $f: K \to P$ sending subcomplexes $L, M$ into $Q, R$ (resp.), where $K = L \cup M$ and $P = Q \cup R$, induces a chain map $f_*$ between the Mayer—Vietoris sequences&#xA;$$\cdots \to H_{r+1} (K) \to H_r (L \cap M) \to H_r(L) \oplus H_r(M) \to H_r(K) \to H_{r-1} (L \cap M) \to \cdots$$&#xA;and&#xA;$$\cdots \to H_{r+1} (P) \to H_r (Q \cap R) \to H_r(Q) \oplus H_r(R) \to H_r(P) \to H_{r-1} (Q \cap R) \to \cdots$$&#xA;but I have not used this fact anywhere in my solution. Have I made a mistake somewhere?&lt;/p&gt;&#xA;" OwnerUserId="5191" LastEditorUserId="5191" LastEditorDisplayName="" LastEditDate="2011-03-14T21:18:40.747" LastActivityDate="2011-03-14T21:18:40.747" Title="Simplicial homology of real projective space by Mayer-Vietoris" Tags="&lt;homework&gt;&lt;algebraic-topology&gt;&lt;homology&gt;" CommentCount="3" />
  <row Id="26985" PostTypeId="1" AcceptedAnswerId="27244" CreationDate="2011-03-14T18:22:40.110" Score="3" ViewCount="106" Body="&lt;p&gt;There are two proofs of Nielsen-Schreier that I know of. The theorem states that every subgroup of a free group is free. The first proof uses topology and covering space theory and is rather elegant. The second uses combinatorial techniques on a free group of words with no relations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a more algebraic proof which somehow just uses the universal property of free groups and maybe other properties of groups that are proved more &quot;algebraically&quot;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm interested because groups are defined purely algebraically by equations, and some proofs that a subgroup of a free abelian group is free abelian have a far more algebraic flavour. So perhaps there is some proof of Nielsen-Schreier that also has a more algebraic flavour?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ideally I would like a proof that does not involve combinatorial properties of a group of words on generators; in other words preferably no facts from combinatorial group theory.&lt;/p&gt;&#xA;" OwnerUserId="2055" LastEditorUserId="5363" LastEditorDisplayName="" LastEditDate="2011-03-14T18:29:11.617" LastActivityDate="2011-03-15T19:06:11.500" Title="Alternative &quot;functorial&quot; proof of Nielsen-Schreier?" Tags="&lt;group-theory&gt;" AnswerCount="1" CommentCount="8" FavoriteCount="1" />
  <row Id="26986" PostTypeId="2" ParentId="26967" CreationDate="2011-03-14T18:39:25.937" Score="1" ViewCount="" Body="&lt;p&gt;An application from number theory is elementary recurrence relations, with the Fibonacci numbers providing the easy example: the equation $F_{n+2} = F_{n+1}+F_n$ ties neatly into the quadratic equation $x^2-x-1=0$, and the Fibonacci numbers themselves turn out to be a linear combination of powers of the two solutions of this equation, $\phi$ and $-\phi^{-1}$.  (The fact that the second solution is 'small' means that its contribution tends to $0$, but this isn't always the case; for instance, the equation $a_{n+2} = 5a_{n+1}-6a_{n}$ has as its 'core' solutions $2^n$ and $3^n$ and the general solution is a linear combination of these).&lt;/p&gt;&#xA;" OwnerUserId="785" LastActivityDate="2011-03-14T18:39:25.937" CommunityOwnedDate="2011-03-14T18:39:45.237" CommentCount="2" />
  <row Id="26987" PostTypeId="2" ParentId="26981" CreationDate="2011-03-14T18:40:25.377" Score="1" ViewCount="" Body="&lt;p&gt;To the OP: the equation you stated in a comment to Ross is crucial (and in contradiction with your post, I should say).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I understand that the question is to solve $P&amp;#39;(t)=rP(t)(1-P(t)/K)$, presumably with an initial condition $P(0)$ in $(0,K)$ (and if this is not what you ask for, please say so and I shall delete this answer).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint: The differential equation $P&amp;#39;(t)=rP(t)(1-P(t)/K)$ can be rewritten as $f(P(t))P&amp;#39;(t)=r$ for a function $f$ you should be able to write down. Let $g$ be any differentiable function and $G(t)=g(P(t))$. Compute $G&amp;#39;(t)$. Choose $g$ such that $G&amp;#39;(t)=P&amp;#39;(t)f(P(t))$. The equation $G&amp;#39;(t)=r$ is equivalent to $G(t)=rt+c$ for a given constant $c$. Use all this to solve the original equation.&lt;/p&gt;&#xA;" OwnerUserId="6179" LastActivityDate="2011-03-14T18:40:25.377" />
  <row Id="26988" PostTypeId="2" ParentId="26967" CreationDate="2011-03-14T18:45:22.787" Score="3" ViewCount="" Body="&lt;p&gt;Knowing two sides $a$ and $c$ of a triangle and an angle $\alpha$ not between them—i.e. the SSA condition—does not determine the length of the third side $b$. See &lt;a href=&quot;http://mathandmultimedia.com/tag/ssa-triangle-congruence/&quot; rel=&quot;nofollow&quot;&gt;Figure 3 on this page&lt;/a&gt;. (Unfortunately, the convention at the length of the side &lt;em&gt;opposite&lt;/em&gt; the vertex $A$ is denoted $a$ is going to make the notation a little confusing below.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Set the origin at the vertex $A$ with the $x$-axis along $AC$. Then the vertex $B$ is at $(c \cos \alpha, c \sin \alpha)$ and $C$ is at $(b,0)$, and we have $$(c \cos \alpha - b)^2 + (c \sin \alpha - 0)^2 = a^{2},$$ which is a quadratic in $b$ both of whose solutions evidently make sense.&lt;/p&gt;&#xA;" OwnerUserId="856" LastActivityDate="2011-03-14T18:45:22.787" CommunityOwnedDate="2011-03-14T18:45:22.787" CommentCount="1" />
  <row Id="26989" PostTypeId="2" ParentId="26981" CreationDate="2011-03-14T18:49:05.663" Score="2" ViewCount="" Body="&lt;p&gt;Well, we know that $k*P(t)=dP/dt$. This simply states that $P(t)$ is proportional to $dP/dt$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You may be able to stop here, based on the wording of your question. It simply asks for a differential equation involving P(t) and t, which this classifies as. You can move on if you have to solve for k at some point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, we rearrange this to be $dP/P = kdt$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we integrate both sides, &#xA;$\int dP/P = \int kdt$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This leaves us with $ln(P) = kt + C$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, we do the opposite operation of the natural log (putting everything to the power of e) and we are left with:&#xA;$P = e^{kt+c}$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then you solve for the initial y and the final form of the equation is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$P=P_{0}e^{kt}$&lt;/p&gt;&#xA;" OwnerUserId="8171" LastEditorUserId="8171" LastEditorDisplayName="" LastEditDate="2011-03-14T19:09:44.117" LastActivityDate="2011-03-14T19:09:44.117" CommentCount="1" />
  <row Id="26990" PostTypeId="2" ParentId="26966" CreationDate="2011-03-14T18:51:32.980" Score="1" ViewCount="" Body="&lt;p&gt;Since &lt;a href=&quot;http://math.stackexchange.com/questions/26966/are-the-equations-2x-2y-2-0-and-7x-7y-7-0-consistent-inconsisten/26970#26970&quot;&gt;The Chaz has given hints in the direction of symbolic manipulation&lt;/a&gt;, I'll hint at a graphical way of thinking:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;What are the slopes of the two lines?  What does that tell you about the lines?&lt;/li&gt;&#xA;&lt;li&gt;Can you find the $y$-intercepts of the lines?  Or the coordinates of other point(s) on the lines?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="72" LastActivityDate="2011-03-14T18:51:32.980" CommentCount="3" />
  <row Id="26991" PostTypeId="1" AcceptedAnswerId="26994" CreationDate="2011-03-14T18:55:55.800" Score="1" ViewCount="51" Body="&lt;p&gt;I need to brush up my knowledge about quadric surfaces and conics. I find myself understanding things better when I work with problems. Do you know any good textbook or website with problems about these topics? Thanks!&lt;/p&gt;&#xA;" OwnerUserId="1523" LastActivityDate="2011-03-14T19:04:14.580" Title="Where can I find good resources to practice quadric surfaces and conics?" Tags="&lt;analytic-geometry&gt;" AnswerCount="1" />
  <row Id="26992" PostTypeId="1" CreationDate="2011-03-14T18:58:41.730" Score="1" ViewCount="46" Body="&lt;p&gt;Heres the &lt;a href=&quot;https://docs.google.com/viewer?a=v&amp;amp;pid=explorer&amp;amp;chrome=true&amp;amp;srcid=0ByWInOLgnYZUZGJkOTJkOTMtMjUyMi00MmNmLWFkOTAtMWQ4MTUyYzdkYTRi&amp;amp;hl=en&quot; rel=&quot;nofollow&quot;&gt;&quot;Link&quot;&lt;/a&gt;, my issue circled in red. Can anyone tell me how they get this this step?; How does $6s_k = 6(5^k-1)$?  What rule are they using here?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$s_{k+1} =  6s_k - 5s_{k-1} = 6(5^k -1)-5(5^{k-1}-1)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks everyone! -Kyle&lt;/p&gt;&#xA;" OwnerUserId="8258" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-14T19:03:15.073" LastActivityDate="2011-03-14T19:06:01.920" Title="Induction problem." Tags="&lt;induction&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="26993" PostTypeId="2" ParentId="26903" CreationDate="2011-03-14T19:03:01.393" Score="2" ViewCount="" Body="&lt;p&gt;Grapher.app on any Mac from the past few years; &lt;a href=&quot;http://www.geogebra.org/&quot; rel=&quot;nofollow&quot;&gt;Geogebra&lt;/a&gt;, &lt;a href=&quot;http://www.peda.com/grafeq/&quot; rel=&quot;nofollow&quot;&gt;GrafEq&lt;/a&gt;, WinPlot, AutoGraph, &lt;a href=&quot;http://www.dynamicgeometry.com/&quot; rel=&quot;nofollow&quot;&gt;Geometer's Sketchpad&lt;/a&gt;, &lt;a href=&quot;http://education.ti.com/educationportal/sites/US/productDetail/us_smartview.html&quot; rel=&quot;nofollow&quot;&gt;TI SmartView&lt;/a&gt; (official TI-84 emulator), ...&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2011-03-14T19:03:01.393" />
  <row Id="26994" PostTypeId="2" ParentId="26991" CreationDate="2011-03-14T19:04:14.580" Score="1" ViewCount="" Body="&lt;p&gt;There is &lt;a href=&quot;http://tutorial.math.lamar.edu/Classes/CalcIII/QuadricSurfaces.aspx&quot; rel=&quot;nofollow&quot;&gt;Paul's Online Math Notes&lt;/a&gt;. Here is a set of related &lt;a href=&quot;http://dtc.pima.edu/~hacker/vectorcalc/homework-sets/vector-calc-hw3.pdf&quot; rel=&quot;nofollow&quot;&gt;problems&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2011-03-14T19:04:14.580" />
  <row Id="26995" PostTypeId="2" ParentId="26972" CreationDate="2011-03-14T19:04:22.767" Score="6" ViewCount="" Body="&lt;p&gt;Since you asked for elementary methods, here is one. Intuitively the idea for your first series is that $\sqrt{n}x$ increases more and more slowly and that one can find larger and larger intervals of integers $n$ such that, on each of these intervals the cosines are uniformly larger than a positive constant, a fact which brings you back to the evaluation of finite sums of $1/\sqrt{n}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More precisely, once you solved the case $x=0$, assume without loss of generality that $x$ is positive and choose your favorite angle whose cosine is positive, for example $\cos(\pi/3)=1/2$. Hence $\cos(\sqrt{n}x)\ge1/2$ for every $n$ such that there exists an integer $k$ such that &#xA;$$&#xD;&#xA;2k\pi-\pi/3\le\sqrt{n}x\le2k\pi+\pi/3.&#xD;&#xA;$$&#xA;This condition translates as $n\in I_k$ where $I_k$ is an integer interval of width of order $(wk)$ around an integer of order $(ck)^2$, where $c=2\pi/x$ and $w=8\pi^2/(3x^2)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sum of $\cos(\sqrt{n}x)/\sqrt{n}$ over $I_k$ is at least $1/2$ times the sum over $I_k$ of $1/\sqrt{n}$. This last sum is of order $1/(ck)$ times the number of terms in $I_k$, which is of order $(wk)$. Thus the sum of $\cos(\sqrt{n}x)/\sqrt{n}$ over $I_k$ is at least $w/(2c)+o(1)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When a series converges, each of its partial sums for $n$ in an interval $[n_0,n_1]$ is as small as one wants provided $n_0$ is large enough (this in fact characterizes convergent series). We disproved this property, hence the series diverges.&lt;/p&gt;&#xA;" OwnerUserId="6179" LastEditorUserId="6179" LastEditorDisplayName="" LastEditDate="2011-03-21T07:24:49.510" LastActivityDate="2011-03-21T07:24:49.510" CommentCount="3" />
  <row Id="26996" PostTypeId="2" ParentId="26992" CreationDate="2011-03-14T19:05:14.300" Score="1" ViewCount="" Body="&lt;p&gt;Just compute. You have $6S_{k}-5S_{k-1}= 6(5^{k}-1) - 5(5^{k-1}-1)=6\cdot 5^{k}-5^{k}-6+5=5^{k+1}-1=S_{k+1}$&lt;/p&gt;&#xA;" LastActivityDate="2011-03-14T19:05:14.300" />
  <row Id="26997" PostTypeId="2" ParentId="26992" CreationDate="2011-03-14T19:05:43.177" Score="2" ViewCount="" Body="&lt;p&gt;$s_i = 5^i - 1$ by inductive hypothesis (i.e. you are assuming this now), so &lt;br&gt;&#xA;$s_k = 5^k - 1$&lt;br&gt;&lt;br&gt;&#xA;That's it!&lt;/p&gt;&#xA;" OwnerUserId="7850" LastActivityDate="2011-03-14T19:05:43.177" />
  <row Id="26998" PostTypeId="2" ParentId="26992" CreationDate="2011-03-14T19:06:01.920" Score="2" ViewCount="" Body="&lt;p&gt;They use the property $P(n)$ they defined at the top of that page.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is known as the &quot;induction hypothesis.&quot;  The form of argument that induction takes is to prove a very simple &quot;base&quot; case ($P(0)$ and $P(1)$ here), and then make one more generalized argument that, if $P(n)$ holds (or, for strong induction as done here, $P(k)$ holds for all $k\le n$), then $P(n+1)$ holds.  This means that, when proving $P(n+1)$, &lt;em&gt;you can use $P(n)$&lt;/em&gt; (which, in this case, is the fact that $s_k = 5^k - 1$).  Finally, you say that because you can just &quot;keep doing this&quot; to get to any $n$, the hypothesis holds for all $n$.  To go any deeper, you need a bit of set theory.&lt;/p&gt;&#xA;" OwnerUserId="8137" LastActivityDate="2011-03-14T19:06:01.920" />
  <row Id="26999" PostTypeId="1" AcceptedAnswerId="27005" CreationDate="2011-03-14T19:06:24.320" Score="4" ViewCount="122" Body="&lt;p&gt;Given a matrix $A\in D^{n\times (n+1)}$ where $D$ is an integral domain, the system $Ax=0$, where $x\in D^{n+1}$, has a nontrivial solution (which can be proved looking at the field of fractions of $D$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would this result still be true if instead of $D$ we had a general ring $R$ with identity? Does $R$ need to be commutative?&#xA;(Obviously the above proof fails in this case, however I don't seem to see a reason why the result wouldn't be true in general.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="766" LastActivityDate="2011-03-14T20:18:22.013" Title="Nontrivial solution to a system of equations over a general ring" Tags="&lt;linear-algebra&gt;&lt;abstract-algebra&gt;&lt;ring-theory&gt;" AnswerCount="1" />
  <row Id="27000" PostTypeId="2" ParentId="26966" CreationDate="2011-03-14T19:06:58.427" Score="0" ViewCount="" Body="&lt;p&gt;Both of your equations can be simplified to $y=-1x-1$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What does this equation look like to you? You can graph it if you have a calculator or compare it to basic functions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You should also know what properties the three words you provided hold, and The Chaz has explained them very clearly. &lt;/p&gt;&#xA;" OwnerUserId="8171" LastEditorUserId="8171" LastEditorDisplayName="" LastEditDate="2011-03-14T19:14:07.483" LastActivityDate="2011-03-14T19:14:07.483" />
  <row Id="27001" PostTypeId="2" ParentId="26894" CreationDate="2011-03-14T19:09:20.577" Score="1" ViewCount="" Body="&lt;p&gt;The transformation $z\to\frac{1}{\bar{z}}$ is an inversion across the unit circle—the conjugation keeps the image at the same argument as the preimage.  Inversion across the unit circle maps lines and circles to lines and circles, with only lines and circles that pass through $0$ (the center of the unit circle) mapping to lines.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$z\to\frac{1}{z}$ can be thought of as the inversion across the unit circle composed with a reflection over the real axis (complex conjugation $z\to\bar{z}$).&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2011-03-14T19:09:20.577" />
  <row Id="27003" PostTypeId="2" ParentId="24701" CreationDate="2011-03-14T19:23:08.420" Score="0" ViewCount="" Body="&lt;p&gt;$$ S \to aSa | bSb | aXb | bXa $$&#xA;$$ X \to aXa | bXb | aXb | bXa | \epsilon $$&lt;/p&gt;&#xA;" OwnerUserId="8260" LastActivityDate="2011-03-14T19:23:08.420" CommentCount="4" />
  <row Id="27004" PostTypeId="2" ParentId="26904" CreationDate="2011-03-14T19:25:40.563" Score="0" ViewCount="" Body="&lt;p&gt;I have a TI-84, and it has a Numeric Derivative function. It can be found by pressing the MATH button, and it is the 8th function in the list. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;nDeriv( will appear on the screen. You then can enter in the original function, comma, the variable in the original function (probably $x$ unless you have something else stored), comma, and the value for x you wish for the calculator to plug in to the derivative. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, if I am finding the value of the derivative of $x^3-5x+9$ at $x=10$, my screen would show this before pressing enter: &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;nDeriv(X^3-5X+9,X,10) &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;You then plug in the same x value to your supposed derivative which you are checking and manually solve it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the values match, your derivative is correct! &lt;/p&gt;&#xA;" OwnerUserId="8171" LastActivityDate="2011-03-14T19:25:40.563" />
  <row Id="27005" PostTypeId="2" ParentId="26999" CreationDate="2011-03-14T19:34:18.387" Score="7" ViewCount="" Body="&lt;p&gt;It may fail for noncommutative rings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are noncommutative rings (with identity) $R$ such that $R^2\cong R$. For example, let $k$ be a field, let $\mathbf{V}$ be an infinite dimensional vector space over $k$, and let $R$ be the ring of endomorphisms of $\mathbf{V}$. The fact that $\mathbf{V}\cong\mathbf{V}\oplus\mathbf{V}$ can be translated into a proof that $R^2\cong R$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $R^2\cong R$, there is a $1\times 2$ matrix that corresponds to this isomorphism. For such a matrix $A$, $A\mathbf{x} = 0$ if and only if $\mathbf{x}=0$, since $A$ is an isomorphism. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For commutative rings, though, it works. Every commutative ring satisfies the &lt;em&gt;strong rank condition&lt;/em&gt;, which is that for any $n\lt \infty$, any set of linearly independent elements in the right module $R^n$ has cardinality at most $n$. This is equivalent to saying that any homogeneous system of $n$ linear equations with $m$ unknowns and $m\gt n$ has a nontrivial solution. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fact that commutative rings satisfy the strong rank condition is proven in T-Y Lam's &lt;strong&gt;Lectures on Rings and Modules&lt;/strong&gt;, pp 14-15. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Lemma.&lt;/strong&gt; Let $A$ and $B$ be right modules over $R$, where $B\neq 0$. If $A\oplus B$ can be embedded in $A$, then $A$ is not a noetherian module.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Proof.&lt;/em&gt; $A$ has a submodule $A_1\oplus B_1$ with $A_1\cong A$ and $B_1\cong B$. So $A\oplus B$ can be embedded in $A_1$, which therefore contans a submodule $A_2\cong B_2$ with $A_2\cong A$ and $B_2\cong B$. Continuing this way we get an infinite direct sum $B_1\oplus B_2\oplus\cdots$ in $A$, each $B_i\neq 0$, which shows that $A$ is not a noetherian module. &lt;strong&gt;QED&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Theorem.&lt;/strong&gt; Any right noetherian ring $R\neq 0$ satisfies the strong rank condition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Proof.&lt;/em&gt; Let $R\neq 0$ be right noetherian. Then for any $n$, the free right module $A=R^n$ is noetherian, so if $B\neq 0$, then $A\oplus B$ cannot be embedded in $A$; in particular, for any $m\gt n$, $R^m=A\oplus R^{m-n}$ cannot be embedded in $A=R^n$. &lt;strong&gt;QED&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Corollary.&lt;/strong&gt; Any commutative ring $R\neq 0$ satisfies the strong rank condition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Proof.&lt;/em&gt; Consider a system of $n$ linear equations in $m\gt n$ unknowns, with coefficients in $R$. Let $R_0$ be the subring of $R$ generated over $\mathbb{Z}\cdot 1$ by the coefficients $a_{ij}$. This is a nonzero noetherian ring by the Hilbert Basis Theorem, so the system of equations has a nontrivial solution over $R_0$, hence over $R$. Thus, $R$ has the strong rank condition. &lt;strong&gt;QED&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another nontrivial instance of rings with strong rank condition: if $R=A\times B$, then $R$ has the strong rank condition if and only if either $A$ or $B$ have the strong rank condition.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-14T20:18:22.013" LastActivityDate="2011-03-14T20:18:22.013" CommentCount="1" />
  <row Id="27006" PostTypeId="1" CreationDate="2011-03-14T19:47:50.620" Score="0" ViewCount="80" Body="&lt;p&gt;Hi everyone,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How does one averages a function over a solid angle? I'm used to do this with conventional angles but it's the first time I come across this problem with a solid angle. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In my specific problem, the function is only dependent on the azimuthal angle, which should simplify the problem somewhat, and the average is over the 4 $\pi$ steradians, but a generic solution would be greatly appreciated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance&lt;/p&gt;&#xA;" OwnerUserId="8261" LastActivityDate="2011-03-14T19:56:30.483" Title="Averaging over a solid angle" Tags="&lt;geometry&gt;" AnswerCount="1" />
  <row Id="27007" PostTypeId="2" ParentId="27006" CreationDate="2011-03-14T19:56:30.483" Score="1" ViewCount="" Body="&lt;p&gt;Similarly to a 2D angle, you integrate the function over the solid angle of interest and divide by the solid angle.  If it depended upon both $\theta$ and $\phi$ and you were interested in the whole sphere, you would do $\overline{f}=\frac{\int f(\theta,\phi)\cos(\theta)d\theta d\phi}{4\pi}$.  If your function does not depend upon $\theta$, that integral is 2 ($\cos\theta$ from $0$ to $\pi$) and now you have a usual 2D style average.  So $\overline{f}=\frac{\int f(\phi) d\phi}{2\pi}$&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-03-14T19:56:30.483" CommentCount="1" />
  <row Id="27008" PostTypeId="1" CreationDate="2011-03-14T19:57:41.363" Score="1" ViewCount="99" Body="&lt;p&gt;Why do we have dy/dx with the regular d, and 'del y/del x' with the 'funny' d?  I can easily find definitions for each expresion, but the definitions appear to be logically equivalent.  However, they are informal enough that it is possible that I am not understanding the definitions properly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Specifically, can anyone show me specific inputs for which the d/dx and del/delx operators return different outputs?  And do they return functions, or values?&lt;/p&gt;&#xA;" OwnerUserId="7546" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-14T20:00:53.323" LastActivityDate="2011-03-14T20:08:38.813" Title="Why are there two different Leibniz notations?" Tags="&lt;calculus&gt;&lt;notation&gt;&lt;definition&gt;" AnswerCount="0" CommentCount="6" />
  <row Id="27010" PostTypeId="2" ParentId="26854" CreationDate="2011-03-14T20:08:16.927" Score="10" ViewCount="" Body="&lt;p&gt;&lt;em&gt;Update:&lt;/em&gt; I felt that in my original answer, the intuition got lost in all of the formalism, so I've rewritten it to make it clearer. The core idea is still much the same as &lt;a href=&quot;http://mathoverflow.net/questions/21881/how-should-one-present-curl-and-divergence-in-an-undergraduate-multivariable-calc/21887#21887&quot; rel=&quot;nofollow&quot;&gt;Qiaochu's answer&lt;/a&gt; on the linked MO question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First off, I don't think you're going to get an explanation for $\nabla\cdot (\nabla\times \mathbf F) = 0$ that's on quite the same level as your sliding block example, because while the gravitational force is well-known to be the gradient of the negative potential, there aren't any tangible vector fields that are the &lt;em&gt;curl&lt;/em&gt; of something. The curl, like the cross product, is a little &quot;unnatural&quot; as it depends on the choice of handedness; classical physics, on the other hand, is independent of handedness, so you can never observe a curl directly, only hidden under an integral or cross product to make the handedness goes away.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nevertheless, the most mathematically natural argument that $\nabla\cdot (\nabla\times \mathbf F) = 0$ is quite easy to grasp. For a warm-up, let's look at another explanation for why $\nabla\times (\nabla f) = 0$. The curl of any vector field $\mathbf F$ describes how much $\mathbf F$ &quot;spins around&quot;, right? So take any closed loop in space; the net curl inside the loop is nothing but how much $\mathbf F$ &quot;circulates&quot; around the loop, i.e. $\oint \mathbf F \cdot d\mathbf x$. But if $\mathbf F$ is a gradient of something, say $f$, then $\mathbf F \cdot d\mathbf x$ is actually how much $f$ is going up or down as you walk along $d\mathbf x$. So if you walk along a closed loop and come back to where you started, the net change in $f$ has to be zero! Or as &lt;a href=&quot;http://mathoverflow.net/questions/21881/how-should-one-present-curl-and-divergence-in-an-undergraduate-multivariable-calc/21887#21887&quot; rel=&quot;nofollow&quot;&gt;wzzx's comment&lt;/a&gt; states: you can't walk from home to school and back and have gone uphill both ways.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we can do the same thing for $\nabla\cdot(\nabla\times \mathbf F)$. Intuitively, the divergence of a vector field $\mathbf G$ measures how much $\mathbf G$ is &quot;spreading out&quot; or &quot;pulling in&quot;. In other words, pick any region of space; what does the total divergence of $\mathbf G$ inside it tell you? It tells you exactly how much $\mathbf G$ is flowing out of the surface of the region, i.e. $\oint \mathbf G \cdot d\mathbf A$. But if $\mathbf G$ is the curl of another vector field $\mathbf F$, then $\int \mathbf G \cdot d\mathbf A$ on a surface just measures how much $\mathbf F$ circulates around the boundary of that surface. What's the boundary of a closed surface? Imagine taking a portion of the entire surface and growing it to cover the whole. Its boundary eventually gets smaller and smaller, and then disappears. So there is nothing for $\mathbf F$ to circulate around, and the circulation must be zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All I've shown here is that the integrals of $\nabla\times(\nabla f)$ and $\nabla\cdot(\nabla\times\mathbf F)$ are zero on any arbitrary region, but that should be enough to see that (at least for smooth $f$ and $\mathbf F$) their values must be zero everywhere.&lt;/p&gt;&#xA;" OwnerUserId="856" LastEditorUserId="856" LastEditorDisplayName="" LastEditDate="2011-03-15T21:04:09.533" LastActivityDate="2011-03-15T21:04:09.533" CommentCount="1" />
  <row Id="27011" PostTypeId="1" CreationDate="2011-03-14T20:08:18.817" Score="10" ViewCount="96" Body="&lt;p&gt;If one has a quasiprojective complex variety X, there is a natural map from the algebraic Picard group to the analytic Picard group.  Is this map either injective or surjective?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I assume the latter is not true, but couldn't think of an example (nor a definitive opinion for the former question)  If you take a normal crossings projective compactification Xbar, then both notions coincide, but while an algebraic line bundle on Xbar that is trivial on X must be associated to the boundary divisor, I couldn't see if the same holds in the analytic setting.  &lt;/p&gt;&#xA;" OwnerUserId="8259" LastActivityDate="2011-05-24T21:18:07.457" Title="algebraic versus analytic line bundles" Tags="&lt;algebraic-geometry&gt;&lt;analytic-geometry&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="3" />
  <row Id="27013" PostTypeId="2" ParentId="26854" CreationDate="2011-03-14T20:10:15.907" Score="4" ViewCount="" Body="&lt;p&gt;Here's an explanation I was given while I was still in high school (but not in school, of course!). I didn't fully understand it at the time, but it might be illuminating if you're willing to accept the &lt;em&gt;generalised Stokes theorem&lt;/em&gt;. Essentially, it's the claim that&#xA;$$\int_{M} d\omega = \int_{\partial M} \omega$$&#xA;where $\omega$ is a differential $(n - 1)$-form, $d\omega$ is its exterior derivative, $M$ is an $n$-dimensional smooth manifold with boundary, and $\partial M$ its boundary. (There are some technical conditions which I'm glossing over.) I should probably explain what these things are though: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;A &lt;em&gt;smooth $n$-manifold with boundary&lt;/em&gt; is an $n$-dimensional smooth geometric object where every interior point has a neighbourhood which looks like a piece of our familiar Euclidean $n$-space $\mathbb{R}^n$, and every boundary point has a neighbourhood which looks like a piece of $\mathbb{R}^{n-1} \times [0, \infty)$, which is just $\mathbb{R}^n$ chopped in half. For example, the interval $[0, 1]$ is a smooth 1-manifold with boundary, and its boundary is the set $\{ 0, 1 \}$; and the closed unit disc is a 2-manifold with boundary the unit circle.&lt;/li&gt;&#xA;&lt;li&gt;Note that the unit circle is a 1-manifold &lt;em&gt;without&lt;/em&gt; boundary; this is an instance of the general fact that the boundary of a smooth $n$-manifold with boundary is a smooth $(n-1)$-manifold &lt;em&gt;without&lt;/em&gt; boundary.&lt;/li&gt;&#xA;&lt;li&gt;A &lt;em&gt;differential $n$-form&lt;/em&gt; is essentially what you integrate when doing an $n$-fold integral. For example, in $\displaystyle \int_{-\infty}^{\infty} \exp(-x^2) \, dx$, you are integrating the differential 1-form $\exp(-x^2) \, dx$ over the 1-manifold $\mathbb{R}$, and in $\displaystyle \int_{0}^{2\pi} \int_0^1 r \, dr \, d\theta$, you are integrating the differential 2-form $r \, dr \wedge d\theta$ over the closed unit disc.&lt;/li&gt;&#xA;&lt;li&gt;The &lt;em&gt;exterior derivative&lt;/em&gt; of a differential $(n-1)$-form is a differential $n$-form. It's simultaneously a generalisation of the gradient, curl, and divergence operators. (More on this later.)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Now, I suppose I should explain what this has to do with div, grad, and curl. Let's start with a function $f$ of 3 real variables $(x, y, z)$. This can be regarded as a differential 0-form, and so it has an exterior derivative $df$. This is essentially the same thing as $\nabla f$; indeed, we can expand $df$ using the chain rule to get&#xA;$$df = \frac{\partial f}{\partial x} \, dx + \frac{\partial f}{\partial y} \, dy + \frac{\partial f}{\partial z} \, dz$$&#xA;which, neglecting certain subtleties, is the same as $\nabla f$ expanded out in the standard basis. Suppose instead I give you a differential 1-form $F = a \, dx + b \, dy + c \, dz$. Its exterior derivative is&#xA;$$dF = \left( \frac{\partial c}{\partial y} - \frac{\partial b}{\partial z} \right) \, dy \wedge dz + \left( \frac{\partial a}{\partial z} - \frac{\partial c}{\partial x} \right) \, dz \wedge dx + \left( \frac{\partial b}{\partial x} - \frac{\partial a}{\partial y} \right) \, dx \wedge dy$$&#xA;which, with a little squinting, is seen to be more-or-less the curl $\nabla \times F$. (This is basically the chain rule again, but you might be wondering why there aren't 9 terms in the exterior derivative and where all the minuses are coming from. This is because for any differential 1-form $\alpha$, $\alpha \wedge \alpha = 0$. Applying this to $(dx + dy) \wedge (dx + dy)$, we see that $dx \wedge dy = -dy \wedge dx$ — that is, the wedge product $\wedge$ of differential 1-forms is anticommutative, just like the cross product of two vectors.)&#xA;Finally, suppose I have a differential 2-form $A = a \, dy \wedge dz + b \, dz \wedge dx + c \, dx \wedge dy$. Its exterior derivative is&#xA;$$dA = \left( \frac{\partial a}{\partial x} + \frac{\partial b}{\partial y} + \frac{\partial c}{\partial z} \right) \, dx \wedge dy \wedge dz$$&#xA;which is quite obviously related to the divergence $\nabla \cdot A$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The generalised Stokes theorem is a vast generalisation of the fundamental theorem of calculus, the Kelvin—Stokes theorem, and the divergence theorem. Indeed, they are all special cases: again, working in 3-dimensional Euclidean space, we get&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;When $\omega$ is a 0-form, the fundamental theorem of calculus: $$\displaystyle \int_{P}^{Q} \nabla f \cdot d\vec{x} = f(Q) - f(P)$$&lt;/li&gt;&#xA;&lt;li&gt;When $\omega$ is a 1-form, the Kelvin—Stokes theorem: $$\displaystyle \int_S \nabla \times \vec{F} \cdot d\vec{S} = \oint_{\partial S} \vec{F} \cdot d\vec{x}$$&lt;/li&gt;&#xA;&lt;li&gt;When $\omega$ is a 2-form, the divergence theorem: $$\displaystyle \int_V \nabla \cdot \vec{A} \, dV = \int_{\partial V} \vec{A} \cdot d\vec{S}$$&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Of course, exactly as with div, grad, and curl, just by fiddling with the definitions, you can show that for &lt;em&gt;any&lt;/em&gt; differential $n$-form $\omega$, the exterior derivative of the exterior derivative is $d^2 \omega = 0$. But this doesn't explain why it &lt;em&gt;should&lt;/em&gt; happen, and I think the best explanation is through the generalised Stokes theorem:&#xA;$$\int_M d^2 \omega = \int_{\partial M} d\omega = \int_{\partial^2 M} \omega$$&#xA;This shows that the fact that $d^2 \omega = 0$ is intimately linked to the fact that $\partial^2 M = \emptyset$, i.e. that the boundary of a smooth manifold itself doesn't have a boundary. Translating this to the classical notation, we immediately have the identities&#xA;$$\nabla \times (\nabla f) = 0$$&#xA;$$\nabla \cdot (\nabla \times F) = 0$$&#xA;as promised.&lt;/p&gt;&#xA;" OwnerUserId="5191" LastActivityDate="2011-03-14T20:10:15.907" />
  <row Id="27014" PostTypeId="1" AcceptedAnswerId="27016" CreationDate="2011-03-14T20:17:43.037" Score="1" ViewCount="128" Body="&lt;p&gt;Hey guys,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need your help with this:&#xA;Let $f$, $p$, and $g$ be a polynomials in $F[x]$, assume that $f$, $g$, and $p$ are nononzero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $\gcd(f,p)=1$ and $fg$ is divisible by $p$ with no remainder,&#xA;I need to prove that $g$ is  divisible by $p$ with no remainder&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="6163" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-14T20:25:34.380" LastActivityDate="2011-03-14T20:31:01.113" Title="Euclid's Lemma for polynomials" Tags="&lt;abstract-algebra&gt;" AnswerCount="2" CommentCount="0" />
  <row Id="27015" PostTypeId="1" CreationDate="2011-03-14T20:23:57.547" Score="1" ViewCount="218" Body="&lt;p&gt;editing signal in frequency domain and converting back to time domain&#xA; the code is matlab / open source octave code&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Greetings All&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to be able to edit different aspects of a signal&#xA;(frequency, magnitude) in the frequency domain is this possible?&#xA;I do know about fft and ifft  which work great but I'm a little&#xA;confused as how to edit the data in the frequency domain and properly&#xA;get it back out in the time domain.  I can get it out using ifft but&#xA;when&#xA;I try and edit the signal in the frequency domain and plot it in the&#xA;time domain the values&#xA;are very different.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example: I create an array with the variables frequency, magnitude.&#xA;created from the frequency domain I then want to change the magnitude&#xA;of one of the frequencies&#xA;and then convert the signal back to the time domain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was reading up on it and it had mentioned using phase some how&#xA;phase = unwrap(angle(ya));  the problem is that the phase array&#xA;is twice as long as the array for the frequency and magnitude.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;code below:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;%signal from frequency domain to time domain&#xA;clear all,clf&#xA;%addpath(&quot;/home/transform/&quot;); %add path to location of functions&#xA;%create time domain&#xA;Fs = 1000;                    % Sampling frequency&#xA;t=linspace(0,1,1000);&#xA;&#xA;%1a create signal&#xA;ya = .5*sin(2*pi*10*t) + 1*sin(2*pi*50*t);&#xA;ya_fft = fft(ya);&#xA;&#xA;%Original data in frequency domain&#xA;[xfreqa,yampa]=rtplotfft(ya,Fs);&#xA;yampa1=(yampa(:,1)/max(abs(yampa(:,1)))*1); %keep at 1, amplitude&#xA;levels adjustied&#xA;phase = unwrap(angle(ya_fft));&#xA;&#xA;%Create array with Freq,magnitude&#xA;freq_mag_phase=[xfreqa yampa]; %can't add phase because length is to&#xA;large&#xA;&#xA;%Edit data in frequency domain&#xA;freq_mag_phase(52,2)=[.1] ; %changes amplitude to .1&#xA;[xfreqb,yampb]=rtplotfft(ya,Fs); %not sure what to replace ya with?&#xA;yampb1=(yampb(:,1)/max(abs(yampb(:,1)))*1); %keep at 1, amplitude&#xA;levels adjustied&#xA;&#xA;%3a frequency back to time domain&#xA;ya_ifft=real(ifft(ya_fft));&#xA;&#xA;%1b time domain plot&#xA;subplot(2,2,1),plot(t,ya)&#xA;title('1) Orginal Signal ')&#xA;ylabel('amplitude')&#xA;xlabel('time domain')&#xA;&#xA;%2b frequency domain plot.&#xA;subplot(2,2,2),plot(xfreqa,yampa)&#xA;title('2) Orginal signal in Frequency domain')&#xA;xlabel('Frequency (Hz)')&#xA;ylabel('amplitude')&#xA;&#xA;%3b rebuilt time domain from frequency (ifft)&#xA;subplot(2,2,3),plot(t,ya_ifft)&#xA;title('3) rebuild time domain using ifft but it hasnt been edited')&#xA;xlabel('time domain ')&#xA;ylabel('amplitude')&#xA;&#xA;%4b rebuilt of signal in frequency (ifft)&#xA;subplot(2,2,4),plot(xfreqb,yampb)&#xA;title('4) rebuild in frequency domain not edited')&#xA;xlabel('Frequency (Hz)')&#xA;ylabel('amplitude')&#xA;&#xA;fprintf('done');&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I've also included the function rtplotfft that is used&#xA;to create the frequency and magnitude plot:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;function [x,freq]=rtplotfft(vp_sig_orig,Fs)&#xA;       vp_sig_orig=vp_sig_orig';&#xA;       vp_sig_len=length(vp_sig_orig); %get sample rate from vp fs_rate&#xA;needs to be an even number?&#xA;&#xA;       % Use next highest power of 2 greater than or equal to length(x) to&#xA;calculate FFT.&#xA;       nfft= 2^(nextpow2(length(vp_sig_orig)));&#xA;&#xA;       % Take fft, padding with zeros so that length(fftx) is equal to nfft&#xA;       fftx = fft(vp_sig_orig,nfft);&#xA;&#xA;       % Calculate the number of unique points&#xA;       NumUniquePts = ceil((nfft+1)/2);&#xA;&#xA;       % FFT is symmetric, throw away second half&#xA;       fftx = fftx(1:NumUniquePts);&#xA;&#xA;       % Take the magnitude of fft of x and scale the fft so that it is not&#xA;a function of the length of x&#xA;       mx = abs(fftx)/length(vp_sig_orig);&#xA;&#xA;       % Take the square of the magnitude of fft of x.&#xA;       %mx = mx.^2; rem'd out to get amplitude to work&#xA;&#xA;       % Since we dropped half the FFT, we multiply mx by 2 to keep the same&#xA;energy.&#xA;       % The DC component and Nyquist component, if it exists, are unique&#xA;and should not be multiplied by 2.&#xA;       if rem(nfft, 2) % odd nfft excludes Nyquist point&#xA;       mx(2:end) = mx(2:end)*2;&#xA;       else&#xA;       mx(2:end -1) = mx(2:end -1)*2;&#xA;       end&#xA;&#xA;       freq=mx;&#xA;&#xA;       % This is an evenly spaced frequency vector with NumUniquePts&#xA;points.&#xA;       freq_vect = (0:NumUniquePts-1)*vp_sig_len/nfft;&#xA;       x=freq_vect';&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="5957" LastEditorUserId="536" LastEditorDisplayName="" LastEditDate="2011-03-21T21:44:25.857" LastActivityDate="2011-06-03T22:10:46.757" Title="editing signal in frequency domain and converting back to time domain " Tags="&lt;fourier-analysis&gt;&lt;matlab&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" />
  <row Id="27016" PostTypeId="2" ParentId="27014" CreationDate="2011-03-14T20:24:28.593" Score="2" ViewCount="" Body="&lt;p&gt;Polynomials over a field are an Euclidean domain relative to the degree function.  In particular, for every polynomials $a(x)$ and $b(x)$ in $F[x]$, with $b(x)\neq 0$, there exist unique polynomials $q(x),r(x)\in F[x]$ such that&#xA;$$ a(x) = q(x)b(x) + r(x),\quad\mbox{$r(x)=0$ or $\mathrm{deg}(r)\lt \mathrm{deg}(b)$}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the Euclidean algorithm also applies to polynomials. In particular, if $q(x) = \gcd(m(x),n(x))$, then there exist polynomial $\alpha(x),\beta(x)$ such that $q(x) = \alpha(x)m(x) + \beta(x)n(x)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now you can use the same proof as the proof of Euclid's Lemma for integers: if $a|bc$ and $\gcd(a,b)=1$, then $a|c$. &lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2011-03-14T20:24:28.593" CommentCount="5" />
  <row Id="27017" PostTypeId="2" ParentId="27014" CreationDate="2011-03-14T20:31:01.113" Score="3" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Proof&lt;/strong&gt; $\rm\ \ \ p\ |\ fg,\:pg\ \Rightarrow\ p\ |\ (fg,pg)\ =\ (f,p)\ g\ =\ g\ $ since $\rm\ (f,p)\ =\ 1\:.\quad\ $ &lt;strong&gt;QED&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This proof of &lt;strong&gt;Euclid's Lemma&lt;/strong&gt; works in any GCD domain, e.g. any domain like $\rm\:F[x]\:$ enjoying a Euclidean algorithm to compute the GCD.&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2011-03-14T20:31:01.113" CommentCount="4" />
  <row Id="27018" PostTypeId="1" AcceptedAnswerId="27031" CreationDate="2011-03-14T20:34:16.920" Score="0" ViewCount="79" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;If $M$ is Noetherian, any submodule $N \neq M$ can be written as a finite intersection of intersection-indecomposable ones.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I just don't know how to begin with the proof of this proposition. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Very kind of you for reading or answering.&lt;/p&gt;&#xA;" OwnerUserId="4736" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-15T12:05:48.110" LastActivityDate="2011-03-15T14:59:28.233" Title="Proper submodules of Noetherian modules are intersections of indecomposable modules" Tags="&lt;abstract-algebra&gt;&lt;modules&gt;" AnswerCount="2" CommentCount="6" />
  <row Id="27019" PostTypeId="2" ParentId="26952" CreationDate="2011-03-14T20:47:21.647" Score="5" ViewCount="" Body="&lt;p&gt;Got it.  The constant in Moron's answer is $C = \log A$, where $A$ is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Glaisher%E2%80%93Kinkelin_constant&quot; rel=&quot;nofollow&quot;&gt;Glaisher-Kinkelin constant&lt;/a&gt;.  Thus $C = \frac{1}{12} - \zeta&amp;#39;(-1)$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The expression $H(n) = \prod_{k=1}^n k^k$ is called the &lt;a href=&quot;http://mathworld.wolfram.com/Hyperfactorial.html&quot; rel=&quot;nofollow&quot;&gt;hyperfactorial&lt;/a&gt;, and it has the known asymptotic expansion&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$H(n) = A e^{-n^2/4} n^{n(n+1)/2+1/12} \left(1 + \frac{1}{720n^2} - \frac{1433}{7257600n^4} + \cdots \right).$$&#xA;Taking logs and using the fact that $\log (1 + x) = O(x)$ yields an asymptotic expression for the OP's sum&#xA;$$\sum_{k=1}^n k \log k = C - \frac{n^2}{4} + \frac{n(n+1)}{2} \log n + \frac{\log n}{12} + O \left(\frac{1}{n^2}\right),$$ &#xA;the same as the one Moron obtained with Euler-Maclaurin summation.&lt;/p&gt;&#xA;" OwnerUserId="2370" LastActivityDate="2011-03-14T20:47:21.647" CommentCount="7" />
  <row Id="27020" PostTypeId="2" ParentId="27018" CreationDate="2011-03-14T20:48:26.747" Score="2" ViewCount="" Body="&lt;p&gt;This is obviously false; perhaps you meant something else?  Consider a vector space of dimension 3 as the module M.  It is Noetherian, but its directly indecomposable submodules are all dimension 1.  If N has dimension 2, then N is not the intersection of one-dimensional submodules.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suspect this is just a translation problem.  There is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Lasker-Noether_theorem&quot; rel=&quot;nofollow&quot;&gt;Lasker–Noether theorem&lt;/a&gt; that expresses each proper submodule of a Noetherian module as an intersection of &lt;strong&gt;meet-irreducible&lt;/strong&gt; (or primary) submodules.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See page 84 and page 102 of Lam's Lectures on Modules and Rings for my understanding of it, or the proper chapter and exercises in Atiyah-MacDonald for a more commutative view.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The part you asked about I think just follows from the definition of Noetherian and should be true in any lattice with appropriate finiteness condition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In section 7.13 of Jacobson's Basic Algebra II (pagination differs a little, page 438 for me), Jacobson uses &lt;strong&gt;(intersection) indecomposable&lt;/strong&gt; to mean what I have called meet-irreducible.  His proof is by Noetherian induction and holds in any Noetherian lattice.&lt;/p&gt;&#xA;" OwnerUserId="583" LastEditorUserId="583" LastEditorDisplayName="" LastEditDate="2011-03-14T21:11:55.200" LastActivityDate="2011-03-14T21:11:55.200" CommentCount="5" />
  <row Id="27021" PostTypeId="2" ParentId="26926" CreationDate="2011-03-14T20:49:38.587" Score="1" ViewCount="" Body="&lt;p&gt;Works as expected in &lt;em&gt;Mathematica&lt;/em&gt; 7.0.1 for Windows. Try to create a new Notebook and copy-paste the code from your post (not from original Notebook). Should work.&lt;/p&gt;&#xA;" OwnerUserId="7617" LastActivityDate="2011-03-14T20:49:38.587" />
  <row Id="27022" PostTypeId="1" AcceptedAnswerId="27028" CreationDate="2011-03-14T20:51:57.057" Score="0" ViewCount="130" Body="&lt;p&gt;I have always been interested in cryptology, but struggled with Algebra as a kid. As a result I know the basics of how the Diffie-Hellman Key Exchange (DH from here on) works, but I don't know why it does. I know that by two similar algorithms with two private numbers, both parties come to the same conclusion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I have used &lt;a href=&quot;http://en.wikipedia.org/wiki/Diffie-Hellman_Key_Exchange&quot; rel=&quot;nofollow&quot;&gt;the wikipedia article&lt;/a&gt; to gain my understanding of the formulae and reproduce them in a Ruby script for my own learning.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My primary curiosity is over the two primary steps of deriving the public keys (which are exchanged) and then deriving the same, final encryption key. How do Alice and Bob actually get the same key without the sharing of their two private keys? I'm probably missing an obvious &quot;solve for ...&quot; or a simple rule, somewhere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;G: a small prime&lt;/li&gt;&#xA;&lt;li&gt;P: a very large prime (EG, 1024 bits)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Secret Keys (just a random number):&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$X_a$ Alice's secret key&lt;/li&gt;&#xA;&lt;li&gt;$X_b$ Bob's secret key&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Temporary (shared) Keys:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Alice: $Y_a = G^{X_a} \mod P$&lt;/li&gt;&#xA;&lt;li&gt;Bob: $Y_b = G^{X_b} \mod P$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Final Key:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Alice: ${Y_b}^{X_a} \mod P$&lt;/li&gt;&#xA;&lt;li&gt;Bob: ${Y_a}^{X_b} \mod P$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$G = 7$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$P = 11$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$X_a = 6$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$X_b = 9$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Should be:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$Y_a = 4$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$Y_b = 8$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By lhf's answer and a proper application of simplifying exponents, $G^{a \centerdot b} \mod P = 3$ should be correct. However, it is nowhere close. Manipulating it to $G^{a + b} \mod P = 3$ is still incorrect, but closer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is $(G^{a + b} \mod P) \mod P = 3$ correct?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit 2:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The calculator that I use on my computer incorrectly handles % as modulo, and therefore does not emit the correct answer. If you have a Mac, you can use Spotlight (cmd+space) and &lt;code&gt;(7^(6*9)) % 11&lt;/code&gt; (even if you use &quot;modulo&quot;) becomes &lt;code&gt;8&lt;/code&gt;—and no, I have no idea why.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So... chalk one problem up to trusting calculators. I understood exponents and their rules to begin with, but with a faulty calculator in the way it even made me question what I practically aced as a kid. Sheesh.&lt;/p&gt;&#xA;" OwnerUserId="8264" LastEditorUserId="8264" LastEditorDisplayName="" LastEditDate="2011-03-14T22:47:02.153" LastActivityDate="2011-03-14T22:47:02.153" Title="Why/How does the Diffie-Hellman Key Exchange work?" Tags="&lt;algebra&gt;&lt;cryptography&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="27023" PostTypeId="1" AcceptedAnswerId="27025" CreationDate="2011-03-14T20:54:30.520" Score="4" ViewCount="104" Body="&lt;p&gt;I came across an exercise in &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0486474178&quot; rel=&quot;nofollow&quot;&gt;this book&lt;/a&gt; where the question was to define a collection of sets and the union operator as a group. The two parts of the question were to (1) find the identity element and (2) find the inverse element.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming some collection of sets that's closed over set union $\mathcal{X}$ and some element $A \in \mathcal{X}$, my answer to the first question was that any set $A&amp;#39; \subseteq A$ and $\emptyset$ could be the identity element. But I run into trouble if I try to build an inverse out of this. For example, there's no way to obtain $\emptyset$ from a non-empty $A$ using union. And it's obvious that if uniqueness of identity and inverse is a requirement for defining groups, then for any $A$, $A$ is also its own identity and inverse.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Is it valid to have multiple non-unique identity elements in a group for each item in the collection?&lt;/li&gt;&#xA;&lt;li&gt;Is it valid to have a unique inverse element for each item in the collection even if the identity elements are not unique?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Judging by what I read on wikipedia, uniqueness doesn't seem to be a criterion.&lt;/p&gt;&#xA;" OwnerUserId="6690" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-14T21:23:13.783" LastActivityDate="2011-03-14T21:45:40.260" Title="Defining sets and union as a group" Tags="&lt;group-theory&gt;&lt;elementary-set-theory&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="27024" PostTypeId="1" AcceptedAnswerId="27049" CreationDate="2011-03-14T20:57:12.857" Score="1" ViewCount="196" Body="&lt;p&gt;How to prove that the only subgroup of the symmetric group $S_n$ of order $n!/2$ is $A_n$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why isn't there other possibility?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks :)&lt;/p&gt;&#xA;" OwnerUserId="4736" LastEditorUserId="360" LastEditorDisplayName="" LastEditDate="2011-03-14T23:42:46.307" LastActivityDate="2011-03-14T23:42:46.307" Title="$A_n$ is the only subgroup of $S_n$ of index $2$. " Tags="&lt;abstract-algebra&gt;&lt;group-theory&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="27025" PostTypeId="2" ParentId="27023" CreationDate="2011-03-14T21:05:46.640" Score="4" ViewCount="" Body="&lt;p&gt;If you have a group, then the identity element must be unique: if $e$ and $e&amp;#39;$ both satisfy the identity condition, $ex = xe = x$ for all $x$ and $e&amp;#39;x = xe&amp;#39; = x$ for all $x$, then you have $ee&amp;#39; = e$ (because $e&amp;#39;$ is an identity) and $ee&amp;#39; = e&amp;#39;$ (because $e$ is an identity), so $e=ee&amp;#39;=e&amp;#39;$, proving that in fact $e=e&amp;#39;$. So the identity element must be unique.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Likewise, if you have a group, then the inverse of an element must be unique: if both $y$ and $z$ are inverses of $x$, so $xy=yx = e$ and $xz=zx=e$, then&#xA;$$z = ze = z(xy) = (zx)y = ey = y$$&#xA;so $z=y$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note, however, that while any $A&amp;#39;\subseteq A$ satisfies $A&amp;#39;\cup A = A$, this does not prove by itself that you have more than one identity element: you would need the &lt;em&gt;same&lt;/em&gt; $A&amp;#39;$ to work for &lt;strong&gt;every&lt;/strong&gt; subset $B$ of $X$ for $A&amp;#39;$ to be the identity; if you need to &quot;change&quot; the $A&amp;#39;$, then you are not constructing an identity for this possible group.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the first step is to figure what subset(s) $E$, if any, satisfy that $E\cup A = A$ &lt;strong&gt;for all&lt;/strong&gt; $A\subseteq\mathcal{X}$. If you can find more than one that works for all $A$, then you know it's not a group (because in a group, the identity element must be unique).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if you can only find a single $E\subseteq\mathcal{X}$ that can work, then you know that one would have to be the identity, and now you can go ahead and see if you can find inverses for every element. If you can find an $A\subseteq \mathcal{X}$ that has at least two distinct inverses, then you would conclude that you cannot get a group (because inverses are unique in a group); or if you can find at least one $A\subseteq\mathcal{X}$ that has no inverse, then you would also know that this is not a group (because in a group, &lt;em&gt;every&lt;/em&gt; element must have an inverse). &lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2011-03-14T21:05:46.640" CommentCount="2" />
  <row Id="27026" PostTypeId="1" CreationDate="2011-03-14T21:07:15.563" Score="3" ViewCount="187" Body="&lt;p&gt;Suppose we have P, a string denoting a C program which takes in a natural number as input and outputs &lt;strong&gt;at most&lt;/strong&gt; one natural number. Prove that it is impossible (unsolvable) to determine that P has infinite range. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, show that it is impossible to determine whether $\mid\{y\in N | \exists x \in N.(P\ outputs\ y\ given\ input\ x)\}\mid$ is infinite.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My approach to this problem is by contradiction &amp;amp; reduction. Let's call this problem Infinite Range (IR). I first assume that it is solvable, and let A be such algorithm that solves IR. Then I want to create a algorithm B that will use A and will solve the Halting Problem, and prove this claim. Then this program uses A and solves the Halting problem, but since we know that the halting problem is unsolvable, this is a contradiction and by reduction we see that since Halting Problem is unsolvable, so is IR.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My problem lies in creating such program B to lead to the contradiction! I was thinking of creating a program D that takes in the program P as mentioned and covert it into program Q: run P in a loop that tries to find a number x such that P(x) = y, from y = 0 ... infinity. It will be ran in another loop that increments x until P(x) = y. And then if it fails to halt, as in it fails to find an x for y, or outputs no number, returns false. We insert this program into A, and then send the NOT of the answer A returns. IE: P halts, so Q sends false. A returns false, so B return true.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then this solves the halting problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However my program D seems awfully flawed and shaky and I feel rather insecure with it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance!&lt;/p&gt;&#xA;" OwnerUserId="6523" LastActivityDate="2011-04-15T13:40:29.733" Title="Proving problems are unsolvable - reducibility proof." Tags="&lt;homework&gt;&lt;computer-science&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" />
  <row Id="27027" PostTypeId="2" ParentId="27024" CreationDate="2011-03-14T21:08:37.340" Score="5" ViewCount="" Body="&lt;p&gt;subgroups of index two are normal (exercise).  $A_n$ is simple, $n\geq 5$ (exercise).  if there were another subgroup $H$ of index two, then $H\cap A_n$ would be normal in $A_n$, contradiction.&lt;/p&gt;&#xA;" OwnerUserId="6925" LastActivityDate="2011-03-14T21:08:37.340" CommentCount="1" />
  <row Id="27028" PostTypeId="2" ParentId="27022" CreationDate="2011-03-14T21:12:49.020" Score="3" ViewCount="" Body="&lt;p&gt;The wikipedia article says it plainly: &quot;Both Alice and Bob have arrived at the same value, because $(g^a)^b$ and $(g^b)^a$ are equal mod $p$.&quot; Note that they're both equal to $g^{ab}$ mod $p$.&lt;/p&gt;&#xA;" OwnerUserId="589" LastActivityDate="2011-03-14T21:12:49.020" />
  <row Id="27029" PostTypeId="1" CreationDate="2011-03-14T21:17:22.480" Score="3" ViewCount="81" Body="&lt;p&gt;I have a system of equations with two equations containing the Lambert W function as follows,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{cases} x = 1 - W_0(\frac{C_1 e^{y + 1}}{y + 1}) \\ y = 1 - W_0(\frac{C_2 e^{x + 1}}{x + 1}) \end{cases}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have already solved the system numerically using a simple iterative method. A gentelman told me that the answer to the system is $y = \frac{k_1 - k_2 x}{k_1 x - k_2}$ where $k_1 = C_2 - C_1$ and $k_2 = C_2 - C_1$ but I have no clue how can one compute the answer.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;How can I get the answer to the system (i.e the steps)?&lt;/li&gt;&#xA;&lt;li&gt;While we have two curves in the system of equation, the intersection (i.e the answer of the system) must be a point not a curve. But $y = \frac{k_1 - k_2 x}{k_1 x - k_2}$ is a hyperbola. How can I describe it?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="8184" LastEditorUserId="8184" LastEditorDisplayName="" LastEditDate="2011-03-14T21:44:22.517" LastActivityDate="2011-03-14T21:44:22.517" Title="System of equations with Lambert W function" Tags="&lt;functions&gt;" CommentCount="5" />
  <row Id="27031" PostTypeId="2" ParentId="27018" CreationDate="2011-03-14T21:21:24.627" Score="2" ViewCount="" Body="&lt;p&gt;The three equivalent definitions of &quot;noetherian module&quot; (in the presence of the Axiom of Choice) are:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;$M$ is noetheorian if and only if every submodule of $M$ is finitely generated.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$M$ is noetherian if and only if every ascending chain of submodules of $M$ stabilizes: $N_1\subseteq N_2\subseteq\cdots$ implies that there exists $m$ such that for all $k\geq m$, $N_m=N_k$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$M$ is noetherian if and only if every nonempty collection of submodules of $M$ has maximal elements.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;So look at the collection of all proper submodules that are not finite intersections of intersection-indecomposable submodules. If this collection is empty, you are done. If it is not empty, then by (3) above it has a maximal element $N$. Note that $N$ cannot be intersection-indecomposable, nor can $N$ be maximal in $M$ (if $N$ is maximal in $M$, then it is intersection indecomposable). So there must exist two proper submodules $N_1$ and $N_2$ of $M$ such that $N=N_1\cap N_2$, and $N\neq N_1$, $N\neq N_2$. Since $N$ is maximal among submodules that are not finite intersections of intersection-indecomposable submodules, then neither $N_1$ nor $N_2$ belong to the collection. So $N_1$ &lt;em&gt;is&lt;/em&gt; a finite intersection of intersection-indecomposable submodules, as is $N_2$, $N_1 = \cap_{j=1}^k M_j$, $N_2 = \cap_{j=k+1}^{\ell}M_j$. Then&#xA;$$N = N_1\cap N_2 = \cap_{j=1}^{\ell} M_j$$&#xA;is a finite intersection of intersection-indecomposable submodules, a contradiction. The contradiction arises from assuming that the collection of submodules that do not satisfy the condition is nonempty, so the collection &lt;em&gt;is&lt;/em&gt; empty and we are done.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Note.&lt;/strong&gt; Just before stating the lemma, Jacobson explicitly defines &quot;indecomposable&quot; as follows: &quot;We shall call a submodule $N$ of $M$ &lt;em&gt;(intersection) indecomposable&lt;/em&gt; if we cannot write $N$ as $N_1\cap N_2$ where $N_i\neq N$ for $i=1,2$.&quot; This is the definition I am using above, rather than the more common &quot;not expressible as a direct sum of two proper submodules&quot; (or even &quot;subdirectly irreducible&quot;). &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-15T14:59:28.233" LastActivityDate="2011-03-15T14:59:28.233" CommentCount="6" />
  <row Id="27032" PostTypeId="1" AcceptedAnswerId="27033" CreationDate="2011-03-14T21:30:26.883" Score="6" ViewCount="192" Body="&lt;p&gt;Is there any problem which is in NP but not in NP-Complete?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any possibility that this problem is analogous to P=NP problem, if so is there any problem which is in NP, but currently there is no proof or disproof about its NP-Completeness?&lt;/p&gt;&#xA;" OwnerUserId="7965" LastActivityDate="2011-03-15T14:44:55.350" Title="NP vs NP-Complete" Tags="&lt;computer-science&gt;&lt;np-complete&gt;" AnswerCount="3" />
  <row Id="27033" PostTypeId="2" ParentId="27032" CreationDate="2011-03-14T21:37:42.090" Score="8" ViewCount="" Body="&lt;p&gt;If there is some non-trivial* problem in $\text{NP}$, but it is not $\text{NP}$-Complete, it would imply that $\text{P} \neq \text{NP}$. This is because if $\text{P} = \text{NP}$, then &lt;em&gt;every&lt;/em&gt; non-trivial problem in $\text{P}$ (and so by assumption, in $\text{NP}$) is $\text{NP}$-Complete, by a polynomial time reduction which solves the problem as part of the reduction, and creates dummy input to the target problem (this is where we need the non-trivial assumption).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A classic example of a problem in $\text{NP}$, which is conjectured not to be in $\text{P}$ and not $\text{NP}$-Complete is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Graph_isomorphism_problem&quot; rel=&quot;nofollow&quot;&gt;Graph Isomorphism Problem&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, under the assumption that $\text{P} \neq \text{NP}$, it has already been &lt;em&gt;proven&lt;/em&gt; that there are an infinite number of hierarchies in $\text{NP}$: i.e. starting with $\text{P}$ the problems becoming increasingly harder (i.e. proper subset), ending with the $\text{NP}$-Complete problems. See this cstheory question: &lt;a href=&quot;http://cstheory.stackexchange.com/questions/52/hierarchies-in-np-under-the-assumption-that-p-np&quot;&gt;http://cstheory.stackexchange.com/questions/52/hierarchies-in-np-under-the-assumption-that-p-np&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;*non-trivial means here: any language other than the empty set or it's complement.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2011-03-15T14:44:55.350" LastActivityDate="2011-03-15T14:44:55.350" CommentCount="13" />
  <row Id="27034" PostTypeId="2" ParentId="27023" CreationDate="2011-03-14T21:45:40.260" Score="1" ViewCount="" Body="&lt;p&gt;Let $(\mathcal{X}, \cup, E, -)$ be a group. Take arbitrary $A\in \mathcal{X}$. $A\cup E = E$, then $A\supseteq E$. $A\cup (-A) = E$, then $A\subseteq E$. Then $A=E$. This group is trivial.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\cup$ is idempotent. Any idempotent group is trivial. Proof. Let $(G, +, 0, -)$ be a group. Take arbitrary $g\in G$. Then $g+g=g, -g+g+g=-g+g, 0+g=0, g=0$. Qed.&lt;/p&gt;&#xA;" OwnerUserId="7011" LastActivityDate="2011-03-14T21:45:40.260" />
  <row Id="27035" PostTypeId="2" ParentId="27032" CreationDate="2011-03-14T21:48:48.317" Score="1" ViewCount="" Body="&lt;p&gt;The class of NP problems contain all the problems in P as well. In fact, NP-complete problems are the hardest problem of the class NP. In addition, there are problems that are not known to be NP-complete (no proof yet) while we don't know any polynomial solution for them (yet).&lt;/p&gt;&#xA;" OwnerUserId="8184" LastActivityDate="2011-03-14T21:48:48.317" />
  <row Id="27036" PostTypeId="1" AcceptedAnswerId="27037" CreationDate="2011-03-14T22:00:52.737" Score="3" ViewCount="40" Body="&lt;p&gt;Consider the principal bundle $S^7$ with base space $\mathbb{C}P^3$ (3-dimensional complex projective space) and fiber $S^1\cong U(1)$.  Can someone write to me the bundle projection $\pi:S^7\rightarrow\mathbb{C}P^3$ explicitly?&lt;/p&gt;&#xA;" OwnerUserId="7831" LastActivityDate="2011-03-14T22:08:25.260" Title="Explicit bundle projection" Tags="&lt;projective-geometry&gt;&lt;fiber-bundles&gt;" AnswerCount="1" />
  <row Id="27037" PostTypeId="2" ParentId="27036" CreationDate="2011-03-14T22:08:25.260" Score="5" ViewCount="" Body="&lt;p&gt;Regard $S^7$ as the set of points in $\mathbb C^4$ of length $1$. If $p:\mathbb C^4\to\mathbb CP^3$ is the usual quotient map, then the restriction $p|_{S^7}:S^7\to\mathbb CP^3$ is the projection you are after.&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2011-03-14T22:08:25.260" CommentCount="3" />
  <row Id="27038" PostTypeId="5" CreationDate="2011-03-14T22:50:22.133" Score="0" ViewCount="" Body="&lt;p&gt;Homework help&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Questions regarding homework assignments are more than welcome, provided that they:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Briefly explain the problem you are trying to solve—do not post your entire assignment verbatim.&lt;/li&gt;&#xA;&lt;li&gt;Explain what you tried and where you're stuck (showing your work is a good idea).&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Don't&lt;/strong&gt; ask for complete solutions to the problem—we're not here to do your homework for you.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;This tag is intended for use along with other, more &quot;mathematical&quot; tags. Please do not use this as the only tag for your homework questions.&lt;/p&gt;&#xA;" OwnerUserId="-1" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2011-04-29T17:28:59.833" LastActivityDate="2011-04-29T17:28:59.833" />
  <row Id="27039" PostTypeId="4" CreationDate="2011-03-14T22:50:22.133" Score="0" ViewCount="" Body="Questions regarding homework assignments are welcome as long as they are asked honestly, explain the problem, and show sufficient effort. Please do not use this as the only tag for a question; this is intended to be used with other tags." OwnerUserId="-1" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2011-04-29T17:28:50.923" LastActivityDate="2011-04-29T17:28:50.923" />
  <row Id="27040" PostTypeId="5" CreationDate="2011-03-14T22:53:41.913" Score="0" ViewCount="" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditorDisplayName="" LastEditDate="2011-03-14T22:53:41.913" LastActivityDate="2011-03-14T22:53:41.913" />
  <row Id="27041" PostTypeId="4" CreationDate="2011-03-14T22:53:41.913" Score="0" ViewCount="" Body="linear, exponential, logarithmic, polynomial, rational, and trigonometric functions, conic sections, graphs and transformations of graphs, equation- and system-solving, and other symbolic-manipulation topics" OwnerUserId="-1" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2011-03-14T22:54:32.210" LastActivityDate="2011-03-14T22:54:32.210" />
  <row Id="27042" PostTypeId="5" CreationDate="2011-03-14T22:55:06.030" Score="0" ViewCount="" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditorDisplayName="" LastEditDate="2011-03-14T22:55:06.030" LastActivityDate="2011-03-14T22:55:06.030" />
  <row Id="27043" PostTypeId="4" CreationDate="2011-03-14T22:55:06.030" Score="0" ViewCount="" Body="trigonometric functions (both geometric and circular), relationships between lengths and angles in triangles, and other topics relating to measuring triangles" OwnerUserId="-1" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2011-03-14T22:55:35.360" LastActivityDate="2011-03-14T22:55:35.360" />
  <row Id="27044" PostTypeId="5" CreationDate="2011-03-14T22:57:57.103" Score="0" ViewCount="" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditorDisplayName="" LastEditDate="2011-03-14T22:57:57.103" LastActivityDate="2011-03-14T22:57:57.103" />
  <row Id="27045" PostTypeId="4" CreationDate="2011-03-14T22:57:57.103" Score="0" ViewCount="" Body="groups, rings, fields, vector spaces, modules, and other topics arising from the abstraction of properties of number systems." OwnerUserId="-1" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2011-03-14T23:03:34.450" LastActivityDate="2011-03-14T23:03:34.450" />
  <row Id="27046" PostTypeId="1" AcceptedAnswerId="27067" CreationDate="2011-03-14T22:58:24.920" Score="1" ViewCount="118" Body="&lt;p&gt;Let's say I've got a plane P, a point A and a vector V, how do I find the distance from A to P moving along the direction to which V points?&#xA;Also, once I've got that, how do I get the point of intersection between the line indicated by V and the plane?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S.: of course this is in an N-dimensional case&lt;/p&gt;&#xA;" OwnerUserId="5412" LastActivityDate="2011-03-15T01:40:42.610" Title="How do I find the distance of a point from a plane while moving along a line?" Tags="&lt;linear-algebra&gt;&lt;geometry&gt;" AnswerCount="2" />
  <row Id="27047" PostTypeId="5" CreationDate="2011-03-14T22:59:39.177" Score="0" ViewCount="" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditorDisplayName="" LastEditDate="2011-03-14T22:59:39.177" LastActivityDate="2011-03-14T22:59:39.177" />
  <row Id="27048" PostTypeId="4" CreationDate="2011-03-14T22:59:39.177" Score="0" ViewCount="" Body="use the (abstract-algebra) tag for topics relating to groups, rings, fields, etc.; use the (algebra-precalculus) tag for topics relating to symbolic manipulation, basic functions, and other algebra/advanced-algebra/precalculus topics." OwnerUserId="-1" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2011-03-14T23:03:44.213" LastActivityDate="2011-03-14T23:03:44.213" />
  <row Id="27049" PostTypeId="2" ParentId="27024" CreationDate="2011-03-14T23:01:33.980" Score="8" ViewCount="" Body="&lt;p&gt;As mentioned by yoyo: if $H\subset S_n$ is of index 2 then it is normal and $S_n/H$ is isomorphic to $C_2=\{1,-1\}$. We thus have a surjective homomorphism $f:S_n\to C_2$ with kernel $H$. All transpositions in $S_n$ are conjugate, hence $f(t)\in C_2$ is the same element for every transposition $t\in S_n$ (this uses the fact that $C_2$ is commutative). $S_n$ is generated by transpositions, therefore $C_2$ is generated by $f(t)$ (for any transposition $t\in S_2$), therefore $f(t)=-1$, therefore ker $f=A_n$.&lt;/p&gt;&#xA;" OwnerUserId="8268" LastActivityDate="2011-03-14T23:01:33.980" CommentCount="1" />
  <row Id="27050" PostTypeId="1" AcceptedAnswerId="27051" CreationDate="2011-03-14T23:02:07.630" Score="11" ViewCount="251" Body="&lt;p&gt;$e^{i \pi} = -1$. I get why this works from a sum-of-series perspective and from an integration perspective, as in I can evaluate the integrals and find this result. However, I don't understand it intuitively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Intuitively, what this means to me is that if you rotate pi radians around a unit circle, you will end up exactly opposite of where you started.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Expanding upon this, for any theta, $e^{i\theta}$ is equivalent to rotating $\theta$ radians around a unit circle. So it's obvious (if not intuitive) to me that $e^{(\pi/2)i} = i$ and $e^{2\pi i} = 1$ and so on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I'm wondering is, intuitively, how is the natural logarithm related so closely to circles? My understanding of $e$ stems from exponential growth, and I don't see how that ties to rotation around a unit circle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know the formulas, but I'm looking for an intuitive explanation. For example, when I used to ask how sin and cos related to circles, people would show me taylor series or tables with a bunch of values or tell me to use a calculator, but it didn't click until somebody told me that sin is just a measure of the height of a point as you travel around the unit circle. Then it all makes sense.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm looking for that kind of explanation of $e$ - how is $e$ related to circles and why does $e^{ix} = \cos(x) + i\sin(x)$?&lt;/p&gt;&#xA;" OwnerUserId="8267" LastEditorUserId="9003" LastEditorDisplayName="" LastEditDate="2011-06-05T21:46:19.730" LastActivityDate="2011-06-05T21:46:19.730" Title="How does e, or the exponential function, relate to rotation?" Tags="&lt;trigonometry&gt;&lt;logarithms&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="27051" PostTypeId="2" ParentId="27050" CreationDate="2011-03-14T23:11:12.320" Score="9" ViewCount="" Body="&lt;p&gt;There is a closely related discussion at &lt;a href=&quot;http://math.stackexchange.com/questions/3510/how-to-prove-eulers-formula-expi-t-costi-sint&quot;&gt;this math.SE question&lt;/a&gt; with lots of details. Let me see if I can summarize it as concisely as I can:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Let $r(t) : \mathbb{R} \to \mathbb{R}^2$ parameterize a particle moving uniformly around the unit circle. Then $|r(t)|^2 = \langle r(t), r(t) \rangle = 1$ (where $\langle \cdot, \cdot \rangle$ denotes the dot product). Differentiating this relation gives $\langle r(t), r&amp;#39;(t) \rangle = 0$; in other words, the displacement is always orthogonal to the velocity. (This should make physical sense.)&lt;/li&gt;&#xA;&lt;li&gt;Since in addition $r(t)$ is moving uniformly we have $|r&amp;#39;(t)|^2$ is a constant, and we may as well assume $|r&amp;#39;(t)| = 1$ by a suitable change of units. Hence $r&amp;#39;(t)$ is either a $90^{\circ}$ clockwise or counterclockwise rotation from $r(t)$. &lt;/li&gt;&#xA;&lt;li&gt;Now identify $\mathbb{R}^2$ with $\mathbb{C}$ and identify $r$ with a function $z(t) : \mathbb{R} \to \mathbb{C}^2$. If the particle is moving counterclockwise, then the above implies that $z&amp;#39;(t) = i z(t)$.&lt;/li&gt;&#xA;&lt;li&gt;But this differential equation clearly has unique solution $z(t) = e^{it} z(0)$. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So the fact that multiplication by $i$ is the same as rotation by $90^{\circ}$ actually immediately implies the more general fact about arbitrary rotations through Euler's formula (although one does not need Euler's formula to see this, of course). The other lesson to keep in mind here is that $e$ shows up whenever you solve linear ODEs. Abstractly this is because $e^{\lambda z}$ is an &lt;a href=&quot;http://en.wikipedia.org/wiki/Eigenvalues_and_eigenvectors&quot; rel=&quot;nofollow&quot;&gt;eigenvector&lt;/a&gt; for the derivative operator of eigenvalue $\lambda$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think these are important and fundamental questions, and it's a pity they aren't more clearly addressed anywhere in the undergraduate curriculum. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;So, to summarize: $e^{it}$ is a complex number $\cos t + i \sin t$ which describes counterclockwise rotation by $t$ radians. It follows that if $z$ is a complex number of absolute value $1$, then the possible values of $\log z$ are the purely imaginary numbers $it$ such that $e^{it} = z$; in other words, they're the possible values of $t$ such that $z$ describes rotation by $t$ radians. So taking the logarithm of a rotation gives you an angle. &lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2011-03-15T02:00:39.207" LastActivityDate="2011-03-15T02:00:39.207" CommentCount="4" />
  <row Id="27052" PostTypeId="2" ParentId="27026" CreationDate="2011-03-14T23:14:10.010" Score="1" ViewCount="" Body="&lt;p&gt;Maybe I missed something, but I don't see how that program B would solve the halting problem for &quot;every&quot; possible program, it would work on those type of programs you construct from program like P, but for every arbitrary program I don't see how B would behave.&lt;br/&gt;&#xA;But you can prove in a similar way to the halting problem, if you suppose that a program F solves IR (return true if the input program has infinite range, or false otherwise), you can construct this program:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;program Z(integer n)&#xA;    if(F(Z))&#xA;        return 0&#xA;    else&#xA;        return n&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Which is a program that takes a natural number and produces at most one natural number. The thing is that Z could either have an infinite range or not, if Z has an infinite range, F(Z) would evaluate true, so Z would produce just 0, which is a contradiction, if Z doesn't have an infinite range F(Z) would evaluate false, which means that Z(n) = n, so that Z would produce all natural numbers, another contradiction.&lt;br/&gt;&#xA;So a program like F can't exist.&lt;/p&gt;&#xA;" OwnerUserId="1318" LastActivityDate="2011-03-14T23:14:10.010" />
  <row Id="27053" PostTypeId="2" ParentId="27046" CreationDate="2011-03-14T23:17:22.033" Score="1" ViewCount="" Body="&lt;p&gt;Parametrize the straight line from $A$ pointing in the direction of $v$. Plug that into the equation for the plane and solve for $t$ (parametrization variable). Then you have the point of intersection. &lt;/p&gt;&#xA;" OwnerUserId="2900" LastActivityDate="2011-03-14T23:17:22.033" CommentCount="1" />
  <row Id="27054" PostTypeId="1" AcceptedAnswerId="27064" CreationDate="2011-03-14T23:19:19.097" Score="2" ViewCount="82" Body="&lt;p&gt;I'm interested in finding sums of the form&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sum_i{\sum_j{(-1)^{N-i+j}} }$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $N$ can be supposed to be constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm using the difference calculus to help solve these sums efficiently.  I believe that I can simulate this sum as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sum_i{\sum_j{cos(\pi(N-i+j))}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm attempting to simulate the sum, since I'm having trouble directly evaluating it any other way.  I'm wondering if there are alternatives that may be easier to work with.  So my question is basically what's the best way to evaluate or simulate this?  Alternatives are welcome, just as any good method is!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;*&lt;em&gt;EDIT&lt;/em&gt;*&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is actually part of a larger formula.  In other words, I'm working with something like&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sum_a{f_1 \sum_b{f_2\sum_i{\sum_j{cos(\pi(N-i+j))}}}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I'm especially interested in something that I can easily work with inside of a larger difference equation.&lt;/p&gt;&#xA;" OwnerUserId="2626" LastEditorUserId="2626" LastEditorDisplayName="" LastEditDate="2011-03-14T23:51:11.147" LastActivityDate="2011-03-15T05:37:09.883" Title="$(-1)^t$ in Difference Calculus" Tags="&lt;finite-differences&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="27055" PostTypeId="1" AcceptedAnswerId="27065" CreationDate="2011-03-14T23:26:20.430" Score="3" ViewCount="50" Body="&lt;p&gt;One property of Euler graphs is that if a graph with $n \geq 3$ vertices and more than $n$ edges has an Euler circuit, then it has more than two simple cycles.&#xA;How can we prove it?&lt;/p&gt;&#xA;" OwnerUserId="7843" LastEditorUserId="7003" LastEditorDisplayName="" LastEditDate="2011-03-14T23:35:58.990" LastActivityDate="2011-03-15T00:12:48.937" Title="Euler graph having simple cycles" Tags="&lt;graph-theory&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="27056" PostTypeId="5" CreationDate="2011-03-14T23:32:55.170" Score="0" ViewCount="" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditorDisplayName="" LastEditDate="2011-03-14T23:32:55.170" LastActivityDate="2011-03-14T23:32:55.170" />
  <row Id="27057" PostTypeId="4" CreationDate="2011-03-14T23:32:55.170" Score="0" ViewCount="" Body="geometry assuming the parallel postulate of Euclid: in a plane, given a line and a point not on that line, there is exactly one line parallel to the given line through the given point." OwnerUserId="-1" LastEditorUserId="1543" LastEditorDisplayName="" LastEditDate="2011-03-15T01:07:00.787" LastActivityDate="2011-03-15T01:07:00.787" />
  <row Id="27058" PostTypeId="5" CreationDate="2011-03-14T23:33:44.280" Score="0" ViewCount="" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditorDisplayName="" LastEditDate="2011-03-14T23:33:44.280" LastActivityDate="2011-03-14T23:33:44.280" />
  <row Id="27059" PostTypeId="4" CreationDate="2011-03-14T23:33:44.280" Score="0" ViewCount="" Body="geometry with coordinatization" OwnerUserId="-1" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2011-03-15T01:05:45.950" LastActivityDate="2011-03-15T01:05:45.950" />
  <row Id="27060" PostTypeId="3" CreationDate="2011-03-14T23:34:18.057" Score="0" ViewCount="" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditorDisplayName="" LastEditDate="2011-03-14T23:34:18.057" LastActivityDate="2011-03-14T23:34:18.057" />
  <row Id="27061" PostTypeId="3" CreationDate="2011-03-14T23:34:18.057" Score="0" ViewCount="" Body="DO NOT EVER USE THIS TAG" OwnerUserId="-1" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2011-03-14T23:47:51.697" LastActivityDate="2011-03-14T23:47:51.697" />
  <row Id="27062" PostTypeId="5" CreationDate="2011-03-14T23:35:35.510" Score="0" ViewCount="" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditorDisplayName="" LastEditDate="2011-03-14T23:35:35.510" LastActivityDate="2011-03-14T23:35:35.510" />
  <row Id="27063" PostTypeId="4" CreationDate="2011-03-14T23:35:35.510" Score="0" ViewCount="" Body="curves of intersection of a plane and a cone: parabolas, hyperbolas, ellipses, circles, lines, pairs of intersecting lines, pairs of parallel lines" OwnerUserId="-1" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2011-03-14T23:47:45.630" LastActivityDate="2011-03-14T23:47:45.630" />
  <row Id="27064" PostTypeId="2" ParentId="27054" CreationDate="2011-03-14T23:43:27.533" Score="4" ViewCount="" Body="&lt;p&gt;For the general sum, Mathematica gives the closed-form expression&lt;br&gt;&#xA;$$\sum_{i=a}^b \sum_{j=c}^d (-1)^{N-i+j} = \frac{(-1)^{N-a-b}}{4} \left( (-1)^a + (-1)^b \right) \left( (-1)^c + (-1)^d \right).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or, if you prefer a simpler answer but in piecewise form, write&#xA;$$\sum_{i=a}^b \sum_{j=c}^d (-1)^{N-i+j} = (-1)^N \left(\sum_{i=a}^b (-1)^i\right) \left(\sum_{j=c}^d (-1)^j\right),$$&#xA;which is&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$0$, if $b-a$ is odd or $d-c$ is odd,&lt;/li&gt;&#xA;&lt;li&gt;$-1$, if $b-a$ and $d-c$ are even, and $N+a+c$ is odd,&lt;/li&gt;&#xA;&lt;li&gt;$1$, if $b-a$ and $d-c$ are even, and $N+a+c$ is even.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;HR&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the sum $\sum_{i=1}^N \sum_{j=1}^N (-1)^{N-i+j}$, which appears to be what you want, this reduces to &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{i=1}^N \sum_{j=1}^N (-1)^{N-i+j} = \frac{-((-1)^N - 1)^2}{4},$$&#xA;or, more succinctly,&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$0$, if $N$ is even,  &lt;/li&gt;&#xA;&lt;li&gt;$-1$, if $N$ is odd.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="2370" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2011-03-15T05:37:09.883" LastActivityDate="2011-03-15T05:37:09.883" />
  <row Id="27065" PostTypeId="2" ParentId="27055" CreationDate="2011-03-14T23:44:31.483" Score="0" ViewCount="" Body="&lt;p&gt;Hint (for &lt;em&gt;atleast&lt;/em&gt; two, see Henry's comment): Consider the vertex with the maximum degree.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2011-03-14T23:49:34.317" LastActivityDate="2011-03-14T23:49:34.317" />
  <row Id="27066" PostTypeId="1" AcceptedAnswerId="27104" CreationDate="2011-03-14T23:45:34.697" Score="0" ViewCount="72" Body="&lt;p&gt;I have come across the terminology $\mathbb{Z}[X] \stackrel{\mathrm{quot.}}{\to} \mathbb{Z}$ where $\mathbb{Z}[X]$ is the ring of integer polynomials in &#xA;$\mathbb{Z}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am wondering what this quotient map is. From &lt;a href=&quot;http://en.wikipedia.org/wiki/Polynomial_ring&quot; rel=&quot;nofollow&quot;&gt;wikipedia&lt;/a&gt;:&#xA;&quot;..given a (not necessarily commutative) ring $A$ containing $K$ and an element $a$ of $A$ that commutes with all elements of $K$, there is a unique ring homomorphism from the polynomial ring $K[X] \to A$ that maps $X$ to $a$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\phi: K[X]\to A, \quad \phi(X)=a.$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; So perhaps there is no 'cannoical' map for the question? For what it is worth, I am trying to calculate homology of $0\to R \stackrel{x}{\to} R \stackrel{\mathrm{quot.}}{\to} \mathbb{Z} \to 0$ where $R = \mathbb{Z}[x]$&lt;/p&gt;&#xA;" OwnerUserId="2219" LastEditorUserId="2219" LastEditorDisplayName="" LastEditDate="2011-03-15T00:13:44.147" LastActivityDate="2011-03-15T03:59:06.130" Title="Quotient map in polynomial ring" Tags="&lt;ring-theory&gt;" AnswerCount="3" CommentCount="3" />
  <row Id="27067" PostTypeId="2" ParentId="27046" CreationDate="2011-03-14T23:51:39.943" Score="1" ViewCount="" Body="&lt;p&gt;I don't know about the $N$-dimensional space, but in the &lt;em&gt;real world&lt;/em&gt; of Homogeneous coordinates and and projective geometry your quantities have coordinates&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$P = \left[\vec{n},-d\right]$&lt;/li&gt;&#xA;&lt;li&gt;$A = \left(\vec{a},\delta\right)$&lt;/li&gt;&#xA;&lt;li&gt;$V = \left(\vec{v}, 0\right)$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The parametrized point location is $A+t V$ and the distance to the plane is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \rho = (A+t V)^T (P\,) = (\vec{a}+t\,\vec{v})\cdot\vec{n}-\delta d $$ when $\vec{n}$ and $\vec{v}$ are unit vectors, and  and $\delta=1$.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Example&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;A plane along the YZ axes at a distance of 5 from the origin is&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$ P =\left[\vec{n},-d\right] = [ (1,0,0), -5 ] $&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;A starting point $\vec{a} = (0,2,0)$ moves along the $\vec{v}=(1,-2,0)$ axis. The point is located at&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$ A = \left(\vec{a} + t \vec{v},\delta\right) = \left((t,2-2\,t,0), 1\right)$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;and the distance is&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$ \rho = (t,2-2 t,0)\cdot(1,0,0)+(-5)*1 = t-5 $&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So the intersection is when $t=5$ when $Q = A + t V = \left((5,-8,0),1\right) $ or at $(5,-8,0)$ coordinates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suppose for the $N$-dimensional case the homogeneous coordinates are of $N+1$ size with the appropriate dot products and such defined.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a way to construct a homogeneous line using &lt;a href=&quot;http://www.flipcode.com/archives/Introduction_To_Plcker_Coordinates.shtml&quot; rel=&quot;nofollow&quot;&gt;pluecker&lt;/a&gt; coordinates from the point and direction, and then directly intersect the line with the plane to yield point Q.&lt;/p&gt;&#xA;" OwnerUserId="3301" LastEditorUserId="3301" LastEditorDisplayName="" LastEditDate="2011-03-15T01:40:42.610" LastActivityDate="2011-03-15T01:40:42.610" />
  <row Id="27068" PostTypeId="2" ParentId="27066" CreationDate="2011-03-15T00:04:30.277" Score="2" ViewCount="" Body="&lt;p&gt;The map $\mathbb Z[x] \to \mathbb Z$ sending a polynomial $p$ to $p(0)$ is a surjective ring homorphism.  The kernel of this is the ideal generated by $x$ so we have the projection $\mathbb Z[x] \to \mathbb Z[x]/(x) \simeq \mathbb Z$.&lt;/p&gt;&#xA;" OwnerUserId="9" LastActivityDate="2011-03-15T00:04:30.277" />
  <row Id="27069" PostTypeId="2" ParentId="27055" CreationDate="2011-03-15T00:12:48.937" Score="0" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Hints:&lt;/strong&gt; Consider a Euler circuit as you know there is at least one.  &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Can you split the circuit into cycles?  &lt;/li&gt;&#xA;&lt;li&gt;Where?  &lt;/li&gt;&#xA;&lt;li&gt;Why?  &lt;/li&gt;&#xA;&lt;li&gt;Does that leave you with simple cycles?  &lt;/li&gt;&#xA;&lt;li&gt;If not, what can you do to create simple cycles?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="6460" LastActivityDate="2011-03-15T00:12:48.937" />
  <row Id="27070" PostTypeId="2" ParentId="27066" CreationDate="2011-03-15T00:12:51.970" Score="5" ViewCount="" Body="&lt;p&gt;The map isn't unique, so I don't know what this notation means without further context. A homomorphism $\mathbb{Z}[x] \to \mathbb{Z}$ is uniquely determined by where $x$ is sent, and $x$ can be chosen freely. In other words, evaluation $p \mapsto p(a)$ is a homomorphism for any $a \in \mathbb{Z}$ and every homomorphism has this form. &lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2011-03-15T00:12:51.970" />
  <row Id="27071" PostTypeId="2" ParentId="27050" CreationDate="2011-03-15T00:14:16.313" Score="3" ViewCount="" Body="&lt;p&gt;The complex logarithm is quite different from the real-valued logarithm.  In fact, it is multi-valued.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Complex_logarithm&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Complex_logarithm&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While it agrees with the real-valued logarithm on the real axis, if you take $R&amp;gt;0$, then as you traverse the circle $|z| = R$, the value of the complex logarithm increases linearly with the $\arg z$ until you return to the same place.  For example, if you take the unit circle, start at $1$, pass through $i$, $-1$, and $-i$, then when you return to $1$ the logarithm will have increased by $2\pi i$.  This comes from the fact that $\exp$ is periodic, as you have noted by realizing its relationship to Euler's identity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Typically, we take one branch of the complex logarithm (which has discontinuities along a ray from the origin, whose location we can choose, and we typically choose either the positive or negative real axis), but the actual function looks a bit like a helix or a spiral.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, the relationship between $\log$ and the circle is that, each time you go once around the circle, you gain $2\pi i$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are interested in more of this, you should study holomorphic functions and residues.  In a region where a holomorphic function is holomorphic (with certain nice properties like being simply connected), the integral around any closed curve is zero.  Simple poles of a holomorphic function $f$ are points where, locally, $f$ looks like $1/z$.  You will recall that, in real analysis, the anti-derivative of $1/z$ is $\log(z)$.  Since $\log$ has a branch cut, this fails to be a holomorphic anti-derivative of $1/z$, so you need some more power to integrate $1/z$, but it turns out that the integral along a circle around a pole like that will be exactly $2\pi i$, and this is precisely because, by taking one branch of $\log$, you get a jump of $2\pi i$ when you return to your starting point.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;More simply, recall that multiplication is the same as addition of logarithms.  In the complex numbers, multiplication corresponds to addition of arguments (angles).  Therefore, you can add something ($2\pi i$ it turns out) to your logarithm, which corresponds to multiplying by something that takes you once around the circle.  This is maybe an easier to see, more intuitive relationship.&lt;/p&gt;&#xA;" OwnerUserId="8137" LastActivityDate="2011-03-15T00:14:16.313" />
  <row Id="27072" PostTypeId="1" AcceptedAnswerId="27075" CreationDate="2011-03-15T00:20:54.833" Score="4" ViewCount="108" Body="&lt;p&gt;Hello. Suppose $R$ is a local ring with maximal ideal $m$ and suppose $M$ and $N$ are $R$-finitely generated modules. Let $f,g: M \rightarrow N$ be an $R$-module homomorphism. If $f$ is an isomorphism and $g(M) \subset mN$ why is $f+g$ in fact an isomorphism?  I think this might follow by Nakayama's lemma but I don't see this. Can you please help?&lt;/p&gt;&#xA;" OwnerUserId="6495" LastEditorUserId="6495" LastEditorDisplayName="" LastEditDate="2011-03-15T00:26:27.887" LastActivityDate="2011-03-15T00:48:03.390" Title="Sum is an isomorphism, finitely generated modules" Tags="&lt;modules&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="27074" PostTypeId="2" ParentId="27072" CreationDate="2011-03-15T00:44:04.343" Score="2" ViewCount="" Body="&lt;p&gt;Composing with $f^{-1}$ allows us to assume $M=N$, $f=\mathrm{Id}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $M&amp;#39;=(\mathrm{Id}+g)(M)$.&#xA;Then $P=M/M&amp;#39;$ satisfies $P=mP$ (if $x \in M$, $x=x+g(x)-g(x) \in M' + mM$) and is finitely generated, so that $P=0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moreover, for all $x \in K = \ker (\mathrm{Id}+g)$, $x=-g(x)=g^2(x)=\ldots=(-1)^n g^n(x)$ for any integer $n$, so that $K \subset \bigcap_{n \geq 0} m^n M$.&#xA;Now &lt;strong&gt;if $R$ is noetherian&lt;/strong&gt;, this last module is finitely generated and satisfies the same condition as $P$, so Nakayama's lemma can be applied again and $K=0$.&lt;/p&gt;&#xA;" OwnerUserId="2660" LastActivityDate="2011-03-15T00:44:04.343" CommentCount="1" />
  <row Id="27075" PostTypeId="2" ParentId="27072" CreationDate="2011-03-15T00:44:38.713" Score="5" ViewCount="" Body="&lt;p&gt;The hypotheses ensure that $g \otimes  R/\mathfrak{m}: M/\mathfrak{m}M \rightarrow N/\mathfrak{m} N$ is zero and thus that $(f+g) \otimes R/\mathfrak{m}$ is an isomorphism.  It is a standard consequence of Nakayama's Lemma that this implies that $f+g$ is surjective: see e.g. the end of Section 3.8.1 of &lt;a href=&quot;http://math.uga.edu/~pete/integral.pdf&quot; rel=&quot;nofollow&quot;&gt;my commutative algebra notes&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We still need to show that $f+g: M \rightarrow N$ is injective.  Let $K$ be the kernel of this map.  &lt;strong&gt;If&lt;/strong&gt; we may assume that $K$ is a finitely generated $R$-module, then again this follows from Nakayama's Lemma since the hypotheses give $K/\mathfrak{m} K = &#xD;&#xA;\ker ((f+g) \otimes R/\mathfrak{m}) = 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the assumption of the previous paragraph does seem to be an additional assumption, albeit a mild one.  It is automatic if $M$ is not just finitely generated but &lt;strong&gt;finitely presented&lt;/strong&gt; and this in turn is automatic if the ring $R$ is Noetherian.  (In the olden days, &quot;local rings&quot; were required to be Noetherian.  I wonder if that's what you meant?)  Whether the result is still true without any additional assumptions I'm not sure off the top of my head.  Perhaps some context would be helpful: why do you think this result is true?&lt;/p&gt;&#xA;" OwnerUserId="299" LastActivityDate="2011-03-15T00:44:38.713" CommentCount="3" />
  <row Id="27076" PostTypeId="2" ParentId="27072" CreationDate="2011-03-15T00:44:49.353" Score="3" ViewCount="" Body="&lt;p&gt;You need to apply Nakayama to both the kernel of $f+g$ and to the cokernel of $f+g$.  I suspect that you need to suppose $R$ to be Noetherian local, in order to know that the kernel is again finitely generated. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now if $K$ is either the kernel or the cokernel, you must argue that $K = mK$.&#xA;Give it a try...&lt;/p&gt;&#xA;" OwnerUserId="5020" LastActivityDate="2011-03-15T00:44:49.353" />
  <row Id="27077" PostTypeId="1" AcceptedAnswerId="27086" CreationDate="2011-03-15T00:48:02.093" Score="2" ViewCount="112" Body="&lt;p&gt;The upper Minkowski dimension of a compact set $A$ in $\mathbb{R}$ is defined as&#xA;$$&#xD;&#xA;\overline{\dim}_M = \inf \{ \epsilon &amp;gt; 0 : \text{ there is a constant } C(\epsilon) \text{ such that } \mu(A_\delta) \leq C(\epsilon) \delta^{1-\epsilon}  \text{ for all } \delta &amp;gt; 0       \}&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here $\mu$ is the Lebesgue measure and &#xA;$$&#xD;&#xA;A_\delta = \{x : \text{dist}(x,A) &amp;lt; \delta \}&#xD;&#xA;$$ &#xA;is the $\delta$-neighborhood of $A$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do I compute the upper Minkowski dimension of &#xA;$$&#xD;&#xA;A = \{ 0,1,\frac{1}{2}, \frac{1}{3}, \ldots \}&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I see from Wikipedia that the answer should be $\frac{1}{2}$, but I want to know how this is calculated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I see that $A_\delta$ is the union of the open intervals $(\frac{1}{n} - \delta, \frac{1}{n} + \delta)$ for $n \geq 1$.  For any fixed $\delta &amp;gt; 0$, all but finitely many points of $A$ belong to $(-\delta,\delta) \subseteq A_\delta$, so that $\mu(A_\delta)$ is bounded by a constant multiple of $\delta$.  However, this constant will depend on $\delta$.  I suppose my aim is to show that the constant behaves like $\delta^{-1/2}$ as $\delta$ varies, but I don't see how to do that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can someone please help me?&lt;/p&gt;&#xA;" OwnerUserId="6887" LastActivityDate="2011-03-28T16:55:18.513" Title="Calculating the upper Minkowski dimension of the set $\{0,1,\frac{1}{2}, \frac{1}{3}, \ldots \}$" Tags="&lt;geometry&gt;&lt;real-analysis&gt;&lt;measure-theory&gt;" AnswerCount="1" />
  <row Id="27079" PostTypeId="2" ParentId="26949" CreationDate="2011-03-15T01:04:09.187" Score="0" ViewCount="" Body="&lt;h3&gt;Given:&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;$2n+1, 2n+3, 2n+5, 2n+7$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$5((2n+1) + (2n+5)) = 5(2n+7) + 15$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;$5(4n + 6) = 10n + 35 + 15$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$20n + 30 = 10n + 35 + 15$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$10n = 20$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$n = 2$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;$2n+1, 2n+3, 2n+5, 2n+7$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$2(2)+1, 2(2)+3, 2(2)+5, 2(2)+7$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$5, 7, 9, 11$&lt;/p&gt;&#xA;" OwnerUserId="2736" LastEditorUserId="2736" LastEditorDisplayName="" LastEditDate="2011-03-15T01:35:30.740" LastActivityDate="2011-03-15T01:35:30.740" CommentCount="3" />
  <row Id="27080" PostTypeId="1" AcceptedAnswerId="27090" CreationDate="2011-03-15T01:05:08.170" Score="5" ViewCount="134" Body="&lt;p&gt;I have some homework problems from Greene and Krantz' &lt;em&gt;Function Theory of One Complex Variable&lt;/em&gt;.  They come from Chapter 5.  I definitely do not want answers, just light prodding in the right direction.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Let $f_j: D(0, 1)\to\mathbb C$ be&#xA;  holomorphic and suppose that each&#xA;  $f_j$ has at least $k$ roots in $D(0,1)$, counting multiplicities.  Suppose&#xA;  that $f_j\to f$ uniformly on compact&#xA;  sets.  Show by example that it does&#xA;  &lt;em&gt;not&lt;/em&gt; follow that $f$ has at least $k$ roots counting multiplicities.  In&#xA;  particular, construct examples, for&#xA;  each fixed $k$ and each $\ell$,&#xA;  $0\le\ell\le k$, where $f$ has exactly&#xA;  $\ell$ roots.  What simple hypothesis&#xA;  can you add that will guarantee that&#xA;  $f$ &lt;em&gt;does&lt;/em&gt; have at least $k$ roots?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I know we require continuity of the $f_j$ on the boundary for the number of zeroes to be the same in the limit, but I'm not clear on why this is.  Presumably, that's the purpose of the question.  In another question, the goal is to prove this when the disk and its boundary are in the region of holomorticity of the $f_j$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm tempted to think that the problem we run in to is when the zeroes move to the boundary in the limit, but maybe I'm just not familiar enough to construct an actual example of this.  I'm also confused by their use of &lt;em&gt;at least&lt;/em&gt; $k$ roots.  It seems simplest to start with a sequence of functions that all have exactly $k$ roots, but I'm afraid I'm missing something, and that it will only work if the functions have different numbers of zeroes, for some reason.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically, I would like some intuition about how things can go wrong when we only have holomorticity on the interior of the disk, and maybe a (small) clue about the form the sequence will take.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Prove:&lt;/strong&gt; If $f$ is a polynomial on $\mathbb C$, then the zeroes of $f^\prime$ are contained in the closed convex hull of the zeroes of $f$.  (Here the &lt;em&gt;closed convex hull&lt;/em&gt; of a set $S$ is the intersection of all closed convex sets that contain $S$.) [&lt;em&gt;Hint:&lt;/em&gt; If the zeroes of $f$ are contained in a halfplane $V$, then so are the zeroes of $f^\prime$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I would really like to use the maximum modulus theorem to say that the zeroes of $f^\prime$ occur at maxima (minima) of $f$, and therefore that these only happen on the boundary of some set $U$ (where $f$ is continuous on $\overline U$ and holomorphic on $U$), but I can't see a way to relate this statement about general bounded domains and convex hulls.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I could be looking at these entirely wrong.  If I need to clarify my thoughts or say more, please let me know, and again, I definitely don't want more than little hints.  Thanks all.&lt;/p&gt;&#xA;" OwnerUserId="8137" LastEditorUserId="8137" LastEditorDisplayName="" LastEditDate="2011-03-15T03:53:59.440" LastActivityDate="2011-03-15T03:53:59.440" Title="Two basic complex analysis homework questions" Tags="&lt;homework&gt;&lt;complex-analysis&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="27081" PostTypeId="1" AcceptedAnswerId="27094" CreationDate="2011-03-15T01:06:22.353" Score="1" ViewCount="110" Body="&lt;p&gt;I have a hypothesis about regular polygons, but in order to prove or disprove it I need a way to determine whether an expression is rational. Once I boil down my expression the only part that could be irrational is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$S_N = cot \frac{\pi}{N} \text{for } N\inℕ_1 ∖ \left\{1, 2, 4\right\}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there at least one such $N$ for which $S_N$ is rational? Can it be proven that $S_N$ is never rational for any such $N$? How would I go about proving one or the other?&lt;/p&gt;&#xA;" OwnerUserId="8240" LastActivityDate="2011-03-15T02:45:09.870" Title="How can I determine a number is irrational?" Tags="&lt;trigonometry&gt;&lt;irrational-numbers&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="27083" PostTypeId="1" AcceptedAnswerId="28344" CreationDate="2011-03-15T01:20:34.873" Score="1" ViewCount="93" Body="&lt;p&gt;I am wondering if there is a strict argument about the probabilities of Christmas (Dec. 25) on Monday, Tuesday, ..., Sunday. My experiments give:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Sunday 0.145&#xA;Monday  0.14&#xA;Tuesday 0.145&#xA;Wednesday 0.1425&#xA;Thursday  0.1425&#xA;Saturday 0.14&#xA;Friday 0.145&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;It looks that they are not equal. :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is:&#xA;1) How to obtain the probabilities without restricting the counts over certain range of years?&#xA;2) why Sunday is more probable than Wednesday, which is more probable than Monday, if it is true? &lt;/p&gt;&#xA;" OwnerUserId="6016" LastEditorUserId="6016" LastEditorDisplayName="" LastEditDate="2011-03-15T01:38:18.360" LastActivityDate="2011-03-21T18:48:13.767" Title="distribution of days in a week on Christmas" Tags="&lt;probability&gt;" AnswerCount="1" CommentCount="9" FavoriteCount="1" />
  <row Id="27084" PostTypeId="1" AcceptedAnswerId="27128" CreationDate="2011-03-15T01:21:42.410" Score="0" ViewCount="130" Body="&lt;p&gt;A right angled triangle has side lengths labeled as so.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://upload.wikimedia.org/wikipedia/commons/d/d2/Pythagorean.svg&quot; rel=&quot;nofollow&quot;&gt;A common geometric construction that shows three squares sitting upon the sides of a right triangle with lengths A, B, and C&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However unlike in this diagram $a = b$. How can $a$ be calculated given $c$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would $a = d \cdot c$ where $n$ is a constant?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: Sorry PEV to be a pain and change the question, I mistyped it originally. Also I've changed N to D as per the comment about natural numbers. I had to edit the image link due to spam prevention for new users, I can't comment much below for the same reasons.&lt;/p&gt;&#xA;" OwnerUserId="6982" LastEditorUserId="6982" LastEditorDisplayName="" LastEditDate="2011-03-16T16:21:27.380" LastActivityDate="2011-03-16T16:21:27.380" Title="How long is side $c$ in this right triangle if $a = b$?" Tags="&lt;algebra-precalculus&gt;&lt;triangle&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="27085" PostTypeId="2" ParentId="27084" CreationDate="2011-03-15T01:24:17.227" Score="5" ViewCount="" Body="&lt;p&gt;We know that $c^2 = a^2+b^2$ from Pythagorean Theorem or $c^2 = 2a^2$. Thus $c = a \sqrt{2}$.&lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2011-03-15T01:24:17.227" CommentCount="1" />
  <row Id="27086" PostTypeId="2" ParentId="27077" CreationDate="2011-03-15T01:26:32.637" Score="1" ViewCount="" Body="&lt;p&gt;In practice, I prefer the (equivalent) following definition of upper Minkowski dimension:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\overline{\dim}(E) = \limsup_{\delta \to 0} \frac{\log N(E,\delta)}{\log(1/\delta)}.&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$N(E, \delta)$ is the number of $\delta$-balls needed to cover the set $E$. (We need $E$ to be compact).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Throughout, we write $X \approx Y$ if there is a constant $c\geq 1$ such that $\frac{1}{c}X \leq Y \leq c X$.  We also write $X \sim Y$, if $\lim (X/Y) = 1$ with respect to some common paramater of $X$ and $Y$.)&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt; To see that these notions are equivalent, just notice that $\mu(E_{\delta}) \approx \delta^n N(E,\delta)$.  Formally, for a compact set $E \subset \mathbb{R}^d$, we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xD;&#xA;\overline{\dim}_M(E) &amp;amp;= \inf\left\{\alpha \in [0,d] : \delta^{-n}\mu(E_{\delta}) \lesssim \delta^{-\alpha}\right\}&#xD;&#xA;\\&#xD;&#xA;&amp;amp;= \inf \left\{\alpha \in [0,d] : \limsup_{\delta \to 0} N(E,\delta) \cdot \delta^{\alpha} = 0\right\}&#xD;&#xA;\\&#xD;&#xA;&amp;amp;= \limsup_{\delta \to 0} \frac{\log N(E,\delta)}{\log (1/\delta)}&#xD;&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;With this in mind, let's consider how many $\delta$-balls are necessary to cover $E = \{0,1, 1/2, \dots\}$.  First, break up the set $E$ into two pieces:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left\{1, \frac{1}{2}, \dots , \frac{1}{n}\right\} \cup \left( \left\{\frac{1}{n+1}, \dots \right\} \cup \{ 0 \} \right) = A \cup B$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, we need $n$ $\delta$-balls to cover $A$ (There are $n$ distinct points), and $\approx n$ $\delta$-balls to cover $B$ (since $| \frac{1}{n+1} - 0| \cdot n \approx 1$). (We needn't be exact with the precise value of $N(E, \delta)$.  We really only need an asymptotic value).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, $N(E, \delta) = n + n \approx n$.  However, what is $n$ in terms of $\delta$?  As long as $\delta \ll \frac{1}{n} - \frac{1}{n+1} \approx \frac{1}{n^2}$, we are not overlapping.  Therefore, we can choose $\delta \approx \frac{1}{n^2}$ or $n \approx 1/\sqrt{\delta}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;\overline{\dim(E)} = \limsup_{\delta \to 0} \frac{\log(1/\sqrt{\delta})}{\log(1 /\delta)} = \frac{1}{2}.&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This argument can be made as formal as you like, though we really only need an asymptotic value for $N(E, \delta)$, and $\delta$ (since $\log cx \sim \log x$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The trick with calculating the Minkowski dimension of a set is to let $\delta \to 0$ and $n \to \infty$ in a controlled way.  Here, setting $\delta \approx \frac{1}{n^2}$ did the trick.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would try adapting the argument to find the (upper) Minkwoski dimension of the set $E_{\alpha} = \{1, \frac{1}{2^{\alpha}}, \frac{1}{3^{\alpha}}, \dots \} \cup \{ 0 \}$ (here, $\alpha &gt; 0$).  I find these examples really fun to play with!  (This also let's one construct a set with Minkowski dimension $\beta$ for any $0 &amp;lt; \beta &amp;lt; 1$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore, if you are ever in need of a good textbook or reference on harmonic analysis and Hausdorff or Minkowski dimensions, I would strongly recommend Tom Wolff's &lt;a href=&quot;http://www.math.ubc.ca/~ilaba/wolff/&quot; rel=&quot;nofollow&quot;&gt;&quot;Lectures on harmonic analysis&quot;&lt;/a&gt; and Pertti Mattila's &quot;Geometry of Sets and Measures in Euclidean Spaces&quot;(which is basically the gold standrad for geometric measure theory!).&lt;/p&gt;&#xA;" OwnerUserId="6491" LastEditorUserId="6491" LastEditorDisplayName="" LastEditDate="2011-03-28T16:55:18.513" LastActivityDate="2011-03-28T16:55:18.513" CommentCount="2" />
  <row Id="27087" PostTypeId="2" ParentId="27050" CreationDate="2011-03-15T01:34:18.153" Score="1" ViewCount="" Body="&lt;p&gt;$ \ln{\left( x+ \hat{i}\,y \right)} = &#xD;&#xA;  \ln{\sqrt{\left( x^2+y^2 \right)}+ \hat{i}\,\tan^{-1}\left(\dfrac{y}{x}\right)}&#xD;&#xA; $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;when polar coordinates are used $x=r\,\cos\theta$ , $y=r\,\sin\theta$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ \ln{\left( r\,\cos\theta+ \hat{i}\, r\,\sin\theta \right)} = &#xD;&#xA;  \ln{r} + \hat{i}\,\theta &#xD;&#xA; $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the logarithm of an complex number represents the rotation angle in the imaginary units.&lt;/p&gt;&#xA;" OwnerUserId="3301" LastActivityDate="2011-03-15T01:34:18.153" />
  <row Id="27088" PostTypeId="1" CreationDate="2011-03-15T01:59:13.610" Score="2" ViewCount="62" Body="&lt;p&gt;All the definitions I can find of a limit (with functions from R to R) define something like:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;as x approaches a, f(x) approaches L&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where x is treated as a variable that is quantified over in the definition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whereas many of these books then go on to use expressions of the form:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;as g(x) approaches a, f(x) approaches L&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;without generalizing the definition appropriately.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Two questions:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;what on earth makes this seem unproblematic to the authors?  I'm guessing that the way I view things makes this use of notation seem more problematic than it is.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;What is the appropriate formal defintion of the limit of f(x) as g(x) approaches a, where f:S-&gt;R and S is a subset of R.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="7546" LastActivityDate="2011-03-15T02:30:29.403" Title="Extending &quot;as x approaches a&quot; to &quot;as g(x) approaches a&quot;" Tags="&lt;calculus&gt;&lt;definition&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="27089" PostTypeId="2" ParentId="27032" CreationDate="2011-03-15T01:59:34.560" Score="0" ViewCount="" Body="&lt;p&gt;Yes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's a whole lattice of complexity classes under NP, most of the ones defined by name are not known but highly suspected to be proper subsets of NP, most famously P. That is, it is not known if any particular problem in P is not NP-complete (though this is expected).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But P is not the only subclass of NP. There are the classes NL, L, AC_n, NC_n, LOGCFL, DET, etc. presumably much smaller than NP. (see the &lt;a href=&quot;http://complexityzoo.org&quot; rel=&quot;nofollow&quot;&gt;Complexity Zoo&lt;/a&gt; for class descriptions). There's lots of open problems about proper inclusion here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But it turns out that NLTIME, a very restricted class, the class of problems solvable in nondeterministic log -time- (not in the Complexity Zoo)), is -provably- a proper subset of NP. That is, there are problems in NP that provably cannot be reduced to a problem in NLTIME.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The following loose inclusion order is known:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$NLTIME \subseteq L \subseteq NL \subseteq P \subseteq NP \subseteq PSPACE$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice that NLTIME is a subset of NP. The only known proper inclusions in this particular list are between NLTIME and NP, and between L and PSPACE. But no one (yet) knows where (or if) there's an adjacent break in this particular linear order.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the real question you are asking is not about complexity classes but about problems. THere are many real world problems that have complexity class analogs in the 'Zoo' (e.g. Reachability in undirected graphs is in NL).  I can't seem to find a reference for the class NLTIME (it is easily definable by NTIME($\log(n)$) ), much less a given problem for it. Without thinking too hard, Im guessing some kind of decision version of &lt;strong&gt;Binary Search&lt;/strong&gt; would fit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(as an aside, there very well may be other more prominent classes, like $AC_0$, having well known problems, that are known to be properly included in NP, but I just happen not to know)&lt;/p&gt;&#xA;" OwnerUserId="1919" LastActivityDate="2011-03-15T01:59:34.560" />
  <row Id="27090" PostTypeId="2" ParentId="27080" CreationDate="2011-03-15T02:10:10.397" Score="3" ViewCount="" Body="&lt;p&gt;It seems to me that your intuition of zeroes wandering off towards the boundary is the correct one and you don't have to look for nastiness of the behavior on the boundary. In fact, Hurwitz's theorem can be stated as follows:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Let $G$ be open and connected and let $f_{n},f: G \to \mathbb{C}$ be holomorphic such that $f_{n} \to f$ uniformly on compact sets. Assume that $U \subset G$ is bounded and open, that $\overline{U} \subset G$ and that $f$ has no zeroes on $\partial U$ then there exists an index $n_{0} = n_{0}(U)$ such that $f_{n}$ and $f$ have the same number of zeroes (counted with multiplicity) in $\overline{U}$ for all $n \geq n_{0}$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you want to produce a sequence of functions with a zero wandering towards $\zeta \in \partial D$, simply put $f_{n} = (z - (1 - \frac{1}{n})\zeta)g$ where $g$ is an arbitrary non-constant holomorphic function on $D$.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Concerning the second question, I don't quite understand why one would want to use the maximum principle. It can be done completely explicitly:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The logarithmic derivative of a polynomial $f$ of degree $n$ is&#xA;\[&#xA;\frac{f'(z)}{f(z)} = \frac{1}{z-z_{1}} + \cdots + \frac{1}{z-z_{n}} = \overline{\sum_{k=1}^{n} \frac{z - z_{k}}{\vert z-z_{k}\vert^2}}&#xA;\]&#xA;where the $z_{1},\ldots,z_{n}$ are the (not necessarily distinct) zeroes of $f$. This is easily proved by induction on the degree of $f$. Note that for $g(z) = (z-z_{0})f(z)$ we have $\frac{g&amp;#39;}{g} = \frac{1}{z-z_{0}} + \frac{f&amp;#39;}{f}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $f&amp;#39;(c) = 0$ we want to show that $c$ is in the convex hull of the $z_{k}$. Now if $c \in \{z_{1},\ldots,z_{n}\}$ this is trivial, so we may assume that this is not the case. But $0 = \overline{\frac{f&amp;#39;(c)}{f(c)}}$ then gives&#xA;\[&#xA;\left(\sum_{k=1}^{n} \frac{1}{|c-z_{k}|^2} \right) \cdot c = \sum_{k=1}^{n} \frac{1}{|c-z_{k}|^2} z_{k}&#xA;\]&#xA;and this easily gives $c$ as a convex combination of the $z_{k}$. Recall that a convex combination is a sum of the form $\sum_{k=1}^{n} \lambda_{k} z_{k}$ with $\lambda_{k} \geq 0$ and $\sum \lambda_{k}= 1$.&lt;/p&gt;&#xA;" OwnerUserId="5363" LastActivityDate="2011-03-15T02:10:10.397" CommentCount="4" />
  <row Id="27091" PostTypeId="2" ParentId="27088" CreationDate="2011-03-15T02:30:17.013" Score="1" ViewCount="" Body="&lt;p&gt;The delta-epsilon definition of the limit generalizes in a straightforward way.  Suppose that for each $\epsilon &amp;gt; 0$, there exists $\delta(\epsilon) &amp;gt; 0$ such that $|g(x)-a|&amp;lt;\delta(\epsilon) \implies |f(x)-L|&amp;lt;\epsilon$.  Then we say that $f(x)$ approaches $L$ as $g(x)$ approaches $a$.  Though it might be more appropriate to say &quot;whenever $g(x)$ approaches $a$&quot;, since this could happen for more than one value of $x$.  For instance, $\sin(x)$ approaches $0$ whenever $\cos(x)$ approaches $1$.&lt;/p&gt;&#xA;" OwnerUserId="5546" LastActivityDate="2011-03-15T02:30:17.013" />
  <row Id="27092" PostTypeId="2" ParentId="27088" CreationDate="2011-03-15T02:30:29.403" Score="0" ViewCount="" Body="&lt;p&gt;I would interpret the statement that f(x) approaches L as g(x) approaches a to mean that for every $\epsilon &amp;gt; 0$ there exists $\delta &amp;gt; 0$ such that for all x, $0 &amp;lt; |g(x) - a| &amp;lt; \delta$ implies $|f(x) - L| &amp;lt; \epsilon$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that x need not be a real variable in order to apply this definition, though doubtless examples of that type come to mind.  For example, sin(x) tends to 0 as cos(x) approaches 1, but the converse is not true.&lt;/p&gt;&#xA;" OwnerUserId="3111" LastActivityDate="2011-03-15T02:30:29.403" CommentCount="7" />
  <row Id="27093" PostTypeId="1" CreationDate="2011-03-15T02:35:25.777" Score="0" ViewCount="54" Body="&lt;p&gt;I am trying to write a physics simulator and have had a problem for weeks now.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have two oriented bounding boxes in 3D space. An oriented bounding box is just a box of any dimension, location, and orientation. I am able to tell when these two boxes are intersecting by using the separating axis theorem. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However once they have intersected I want them to respond in a physically realistic way. In order for them to do so I need to know the point of intersection between the boxes. Before I can explain what this point is I need to explain what the normal of the collision is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The separating axis theorem tests if the boxes are overlapping. If they are then there is no axis (x,y,z,u,w,etc) in 3D space that can separate them, ie you can't draw a line between them that would separate the boxes. The collision normal would be the axis on which the two bounding boxes overlap the least. For instance imagine that two identical cubes were lined up perfectly on the y and z axes but separated on the x axis so that they weren't touching. If they started to move towards each other on the x axis and finally came to overlap one another a bit the x axis would be the collision normal since it is the axis on which there is the least amount of overlap between the two cubes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now getting back to the point of intersection. This point would be the point that is on the collision normal and in between the depth of overlap between the two boxes on the axis of the collision normal. Going back to the cube example it would be the point found by averaging their x positions. Of course this point is not so easy to find when it comes to oriented bounding boxes as they can be unequal in size, rotated in any which way, and probably won't be lined up perfectly on any specific axis. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to know any general method for finding this point. All I know is the location of each boxes center, the locations of their vertices, and their orientation (ie how they are rotated) which I am storing as a 3x3 matrix. Each row/column of the matrix represents one arbitrary axis which describes the rotation of the box. Finally I know a unit vector describing the normal of the collision.&lt;/p&gt;&#xA;" OwnerUserId="8272" LastActivityDate="2011-03-15T03:18:07.907" Title="How do I find the average point of contact between overlapping boxes?" Tags="&lt;computational-geometry&gt;" AnswerCount="1" />
  <row Id="27094" PostTypeId="2" ParentId="27081" CreationDate="2011-03-15T02:37:49.207" Score="3" ViewCount="" Body="&lt;p&gt;A simple, complete proof can be found in Olmsted, J. M. H., &lt;a href=&quot;http://dx.doi.org/10.2307/2304540&quot; rel=&quot;nofollow&quot;&gt;Rational Values of Trigonometric Functions&lt;/a&gt;, Amer. Math. Monthly 52 (1945), no. 9, 507–508.&lt;/p&gt;&#xA;" OwnerUserId="589" LastEditorUserId="589" LastEditorDisplayName="" LastEditDate="2011-03-15T02:45:09.870" LastActivityDate="2011-03-15T02:45:09.870" />
  <row Id="27095" PostTypeId="1" CreationDate="2011-03-15T02:42:45.787" Score="3" ViewCount="153" Body="&lt;p&gt;I want to ask if there is a non-zero function like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f:D^{n}\rightarrow [-a,a]$ for some $a\in \mathbb{R}$. In other words $f$ is bounded. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;such that $$f(\frac{Ax+By}{A+B})=\frac{Af(x)+Bf(y)}{A+B}$$ for some fixed $A,B\in \mathbb{R}-\mathbb{Q}$. And $$f(\partial D^{n})=0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This question originated from one of my mid term problems. I solved the mid term problem and submitted my work, so now I can ask for a more general question online. The original one assume $A=B$. I do not know how to write such a function explicitly. &lt;/p&gt;&#xA;" OwnerUserId="7887" LastEditorUserId="7887" LastEditorDisplayName="" LastEditDate="2011-03-15T03:17:53.293" LastActivityDate="2011-03-15T12:38:21.107" Title="Does such a function exist?" Tags="&lt;real-analysis&gt;&lt;measure-theory&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="27096" PostTypeId="1" AcceptedAnswerId="27098" CreationDate="2011-03-15T03:13:58.837" Score="2" ViewCount="142" Body="&lt;p&gt;Let $X$ be an infinite set of cardinality $|X|$, and let $S$ be the set of all finite subests of $X$. How can we show that Card($S$)$=|X|$? Can anyone help, please?&lt;/p&gt;&#xA;" OwnerUserId="8209" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-15T03:16:33.513" LastActivityDate="2011-03-15T03:19:34.493" Title="The cardinality of the set of all finite subsets of an infinite set" Tags="&lt;elementary-set-theory&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" />
  <row Id="27097" PostTypeId="2" ParentId="27093" CreationDate="2011-03-15T03:18:07.907" Score="1" ViewCount="" Body="&lt;p&gt;In 3D, you (except for incredible coincidences) have only two cases:  a corner of one box hits a face of the other or the edges of two boxes hit.  The first is pretty simple:  some linear equations will give the time of impact and since you know the orientation of the struck box you know the normal.  The second also will yield a time of impact to linear equations, but defining the change to the motion of each box is harder.  I would transform to a frame where the point of impact is stationary, then make the recoil a given fraction of the velocity of impact, then transform back to your usual coordinates.  Some one-time algebra will save the recurrent transformations, but if it is fast enough for your purposes I would suggest transforming.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-03-15T03:18:07.907" CommentCount="1" />
  <row Id="27098" PostTypeId="2" ParentId="27096" CreationDate="2011-03-15T03:19:34.493" Score="8" ViewCount="" Body="&lt;p&gt;The cardinality is at least $|X|$, since $S$ contains all singletons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $S_n$ be the subset of $S$ consisting of all subsets of cardinality exactly $n$. Then $S$ is the disjoint union of the $S_n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, for any positive integer $n$, the number of subsets of $X$ of cardinality $n$ is at most $|X|^n = |X|$ (equality since $|X|$ is infinite); because an $n$-tuple of elements of $X$ determines a subset of $X$ of cardinality at most $n$; and every subset with $n$ elements determines only finitely many distinct $n$-tuples of elements of $X$ (namely, $n!$). So $|S_n|\leq n!|X|^n = |X|$ for all $n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore:&#xA;\begin{align*}&#xA;|X| &amp;amp;\leq |S| = \left|\bigcup_{n=0}^{\infty} S_n\right| = \sum_{n=0}^{\infty}|S_n|\\&#xA;&amp;amp;= \sum_{n=1}^{\infty}|S_n| \leq \sum_{n=1}^{\infty}|X|\\&#xA;&amp;amp;= \aleph_0|X| = |X|,&#xA;\end{align*}&#xA;with the last equality since $|X|\geq\aleph_0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, $|X|\leq |S|\leq |X|$, so  $|S|=|X|$.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2011-03-15T03:19:34.493" CommentCount="17" />
  <row Id="27099" PostTypeId="1" CreationDate="2011-03-15T03:23:24.830" Score="1" ViewCount="43" Body="&lt;p&gt;Good day,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anybody know where I could find a recursive contour plotting algorithm which is compliant with breaks and discontinues in the objective function? Recursive subdivision should be done only in those places where more sample points are needed to get a plot with smooth contours.&lt;/p&gt;&#xA;" OwnerUserId="7617" LastActivityDate="2011-03-15T03:54:26.520" Title="Contour plotting algorithm compliant with breaks and discontinues" Tags="&lt;algorithms&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="27100" PostTypeId="1" CreationDate="2011-03-15T03:27:09.740" Score="2" ViewCount="95" Body="&lt;p&gt;Hello,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am trying to prove a set is convex. This is a problem from Boyd's Convex Optimization text.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The set is $\{\hat{x} + tv | \alpha t^2 + \beta t + \gamma \leq 0\}$. This is actually an intersection of a line and a set, the line can be thought of as the part before the | symbol, and the original set contributed the part after the | symbol.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To prove convexity, I must show: $x,y \in C \Rightarrow \theta x + (1-\theta)y \in C$, with $0 \leq \theta \leq 1$. For this particular set, I tried:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\theta (\hat{x}+t_1 v) + (1-\theta)(\hat{x}+t_2 v) = \theta \hat{x} + \theta t_1 v + \hat{x} + t_2 v - \theta \hat{x} - \theta t_2 v = \hat{x} + v(\theta t_1 + (1-\theta)t_2)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I must show: $\hat{x} + v(\theta t_1 + (1-\theta)t_2) \in \{\hat{x} + tv | \alpha t^2 + \beta t + \gamma \leq 0\}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the same as showing: $\alpha (\theta t_1 + (1-\theta)t_2)^2 + \beta(\theta t_1 + (1-\theta)t_2) + \gamma \leq 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By setting $t = \theta t_1$, I can simplify a little.&#xA;Somehow, I'm supposed to reach the conclusion $\alpha \geq 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can someone tell me a hint? Thanks.&lt;/p&gt;&#xA;" OwnerUserId="8140" LastActivityDate="2011-03-15T04:48:46.257" Title="Proving a Set Convex" Tags="&lt;convex-optimization&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="27101" PostTypeId="2" ParentId="26872" CreationDate="2011-03-15T03:53:21.407" Score="5" ViewCount="" Body="&lt;p&gt;The previous examples are all perfectly good, of course, but I thought I would add a rather low-brow way to see that such examples must be easy to generate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $S$ be a finite set, and let's label its elements $x_1,\ldots,x_n$.  To define a binary operation $\star: S \times S \rightarrow S$ is equivalent to filling in all the entries of an $n \times n$ matrix $M = \{m_{ij} \}$ with elements of $S$.  In particular, since we have $n^2$ entries to fill out and $n$ possible choices for each entry, the total number of such matrices -- i.e., the total number of binary operations on $S$ -- is $n^{n^2}$.  This is a pretty big number even when $n$ is rather small.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose we wish to enforce that the first element $x_1$ is a two-sided identity for our operation $\star$.  Then what we need is that for all $i,j \in \{1,\ldots,n\}$, &#xA;$m_{1j} = x_j$ and $m_{i1} = x_i$.  In other words, the first row and first column of &#xA;the matrix are entirely determined, but this leaves us with $n^2 - 2n + 1 = (n-1)^2$ entries left to choose.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is clear that the operation is commutative iff the matrix is symmetric, i.e., $m_{ij} = m_{ji}$ for all $i,j$.  It is easy to give a precise count of the number &#xA;of symmetric matrices with first row and first column as determined above and even easier to see that, as long as $n \geq 3$, most such matrices are not of this form.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally we come to associativity.  This is harder to see directly from the matrix, but it is easy to see that we have an awful lot of choices that will result in non-associative operations.  For instance, let us say for the sake of argument that &#xA;$x_2 \star x_3 = x_4$ and $x_3 \star x_4 = x_5$ (all elements distinct): up to relabelling the elements, this is in some sense the generic situation.  Then we find&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(x_2 \star x_3) \star x_4) = x_4 \star x_4$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;whereas&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x_2 \star (x_3 \star x_4) = x_2 \star x_5$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now for a randomly defined binary operation, the chance that these two products are &#xA;equal is only $\frac{1}{n}$.  In any case, we can certainly define them to have different values even among $x_1,\ldots,x_5$ and get a nonassociative binary operation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you like counting things, it is a fun exercise to see what you can prove about &lt;strong&gt;how many&lt;/strong&gt; of the $n^{(n-1)^2}$ binary operations on $S$ having $x_1$ as an identity &#xA;are commutative, or associative, or both, or neither.  If $X_n$ is the number of &#xA;such operations (having $x_1$ as an identity) which are neither commutative nor associative, it should be easy to show that $\lim_{n \rightarrow \infty} \frac{X_n}{n^{(n-1)^2}} = 1$, i.e., that &quot;with probability $1$&quot; a randomly chosen such operation is neither commutative nor associative.  I have no doubt that such results appear in the literature, with explicit bounds.  If someone knows a nice argument leading to an asymptotic or exact formula for $X_n$, I would be interested to see it.&lt;/p&gt;&#xA;" OwnerUserId="299" LastActivityDate="2011-03-15T03:53:21.407" CommentCount="1" />
  <row Id="27102" PostTypeId="2" ParentId="27100" CreationDate="2011-03-15T03:53:59.047" Score="0" ViewCount="" Body="&lt;p&gt;$\alpha$ is a given and could be positive or negative.  The conclusion you are looking for is exactly what you said: $\alpha (\theta t_1 + (1-\theta)t_2)^2 + \beta(\theta t_1 + (1-\theta)t_2) + \gamma \leq 0$.  This shows that the point $\theta t_1 + (1-\theta)t_2 \in C$.  You have that the quadratic is $\leq 0$ for both $t_1$ and $t_2$.  Can you combine those two to get the result?&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-03-15T03:53:59.047" CommentCount="5" />
  <row Id="27103" PostTypeId="2" ParentId="27099" CreationDate="2011-03-15T03:54:26.520" Score="4" ViewCount="" Body="&lt;p&gt;I would suggest looking into &lt;a href=&quot;http://dx.doi.org/10.1145/383259.383267&quot; rel=&quot;nofollow&quot;&gt;Jeff Tupper's successive refinement scheme&lt;/a&gt;, which does a pretty good job (the best that can be done with inexact arithmetic). This is the very method he used for his program GrafEq.&lt;/p&gt;&#xA;" OwnerUserId="8273" LastActivityDate="2011-03-15T03:54:26.520" CommentCount="1" />
  <row Id="27104" PostTypeId="2" ParentId="27066" CreationDate="2011-03-15T03:59:06.130" Score="3" ViewCount="" Body="&lt;p&gt;Your edit makes clear that the quotient map you want is $\mathbb{Z}[x] \rightarrow \mathbb{Z}[x]/(x)$; the latter is indeed isomorphic to $\mathbb{Z}$ in a natural way: &#xA;each equivalence class has a unique representative which is a polynomial of degree $0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, to give yourself a quotient map you need to give yourself an ideal $I$ to &#xA;mod out by.  When -- as here -- you have an exact sequence -- looking at the image of &#xA;the previous term will tell you what the kernel of the quotient map is.&lt;/p&gt;&#xA;" OwnerUserId="299" LastActivityDate="2011-03-15T03:59:06.130" CommentCount="1" />
  <row Id="27105" PostTypeId="1" AcceptedAnswerId="27114" CreationDate="2011-03-15T04:06:38.317" Score="0" ViewCount="57" Body="&lt;p&gt;I have a question about the words &quot;acyclic&quot; and &quot;exact.&quot; Why does Brown use the term &quot;acyclic&quot; instead of &quot;exact&quot; in his book Cohomology of Groups? It seems to me that these two terms exactly coincide. Are there examples(or topics in math) in which being acyclic means being sth1 and being exact means being sth2, and when restricted to the homology theory sth1 and sth2  coincide? Thank you. &lt;/p&gt;&#xA;" OwnerUserId="7796" LastActivityDate="2011-03-15T04:41:04.173" Title="Acyclic vs Exact" Tags="&lt;algebra&gt;&lt;cohomology&gt;&lt;homology&gt;&lt;group-cohomology&gt;" AnswerCount="2" />
  <row Id="27106" PostTypeId="2" ParentId="27095" CreationDate="2011-03-15T04:08:33.327" Score="7" ViewCount="" Body="&lt;p&gt;A and B are non-zero since $0\in\mathbb{Q}$. This is the only point at which I'm going to use $A,B\notin \mathbb{Q}$, so that condition could be replaced by $A,B\neq 0$. (The given equation reduces to $f(x)=f(x)$ if $A=0$ or $B=0$, so in that case any $f$ would do.) Also $A+B$ must be non-zero since otherwise the given equation is not defined.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For each $x$ and $y$ the given equation determines $f$ to be linear for one particular linear combination of $x$ and $y$ with the proportions $A/(A+B)$ and $B/(A+B)$. If $A$ and $B$ have the same sign, this is a convex combination. If $A$ and $B$ have opposite signs, write&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$z=\frac{Ax+By}{A+B}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f(z)=\frac{Af(x)+Bf(y)}{A+B}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and solve for $x$ and $f(x)$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x=\frac{(A+B)z-By}{A}\;,$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f(x)=\frac{(A+B)f(z)-Bf(y)}{A}\;.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $A+B$ and $-B$ have the same sign, this is a convex combination with the proportions $(A+B)/A$ and $-B/A$. If not, then $A+B$ and $-A$ have the same sign, so we can do the same thing solving for $y$ instead of $x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, in any case the given condition fixes $f$ to be linear for a particular convex combination of any pair of arguments. Denote by $g$ the function that maps a pair of points to this convex combination and by $S$ the closure of $\partial D^n$ under $g$, i.e. the set of all points that can be reached in a finite number of steps by starting with all points on the boundary of the cube and in each step adding the given convex combination for each pair of points reached so far. Then there exists a non-zero $f$ iff $S\neq D^n$, since $f$ has to be zero on $S$ and we can choose $f$ to be any linear function on $D^n\backslash S$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $n=1$, $S$ is countable and $D^1$ is uncountable, so $S\neq D^n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $n&amp;gt;1$, for given w in the interior of $D^n$ each line through $w$ has two intersections with $\partial D^n$. Choose some continuous family of lines through $w$ such that the proportion at which $w$ lies between these intersections varies across the family. For each line, $S$ contains at least the closure of the set of the two intersections under $g$. The points in that closure lie at the same proportions between the intersections for each line, and these proportions form a (countable) dense subset $P$ of $[0,1]$. Now as the line is continuously varied in the continuous family of lines, the proportion at which $w$ lies on the line varies continuously, and hence sweeps through elements of $P$. Thus $w\in S$, and hence $S=D^n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In summary, for arbitrary $A$ and $B$ there is a non-zero $f$ satisfying the given equation iff $n=1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that I didn't use the boundedness condition.&lt;/p&gt;&#xA;" OwnerUserId="6622" LastEditorUserId="6622" LastEditorDisplayName="" LastEditDate="2011-03-15T12:38:21.107" LastActivityDate="2011-03-15T12:38:21.107" CommentCount="2" />
  <row Id="27107" PostTypeId="1" CreationDate="2011-03-15T04:09:19.317" Score="0" ViewCount="90" Body="&lt;p&gt;I have the first point is (0, -100) and the second point is (7500, -250), and the maximum point is at (x, 210). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it possible to find X or the equation of the parabola using this information alone?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If so how?&lt;/p&gt;&#xA;" OwnerUserId="6127" LastEditorUserId="6127" LastEditorDisplayName="" LastEditDate="2011-03-15T04:43:51.910" LastActivityDate="2011-03-15T04:43:51.910" Title="parabola equation given only two points" Tags="&lt;conic-sections&gt;" AnswerCount="2" CommentCount="6" />
  <row Id="27108" PostTypeId="2" ParentId="27107" CreationDate="2011-03-15T04:13:46.877" Score="0" ViewCount="" Body="&lt;p&gt;Three points determine a parabola, since you have three constants to account for in $y=ax^2+bx+c$ . Construct the appropriate system of linear equations and you're golden.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Otherwise, if you're too cool for solving linear equations, there is a determinant expression for the parabola passing through three points $(x_1,y_1)$, $(x_2,y_2)$, and $(x_3,y_3)$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{vmatrix}x^2&amp;amp;x&amp;amp;y&amp;amp;1\\x_1^2&amp;amp;x_1&amp;amp;y_1&amp;amp;1\\x_2^2&amp;amp;x_2&amp;amp;y_2&amp;amp;1\\x_3^2&amp;amp;x_3&amp;amp;y_3&amp;amp;1\end{vmatrix}=0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;whose verification is left as an exercise.&lt;/p&gt;&#xA;" OwnerUserId="8276" LastEditorUserId="8276" LastEditorDisplayName="" LastEditDate="2011-03-15T04:30:44.850" LastActivityDate="2011-03-15T04:30:44.850" CommentCount="3" />
  <row Id="27109" PostTypeId="1" AcceptedAnswerId="27116" CreationDate="2011-03-15T04:14:42.727" Score="1" ViewCount="69" Body="&lt;p&gt;Hey,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First time here, more frequent over at StackOverflow. So not sure if calculator questions are accepted here, but I'll ask. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a quick way to solve problems like this in the ti-83:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;82% shot success rate out of 7 shots taken, find odds of exactly 3 shots going in, more than 5 shots going in. (I know how to do the written out way using nCr but it would save time on the exam if I knew a faster way).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Lets say 10% of people pay $\$$30, 70% pay $\$$52, and 20% pay $\$$60. Is there a way to get the mean and standard deviation of that by modifying the stats on ti-83 (I know, very easy mean is (.1*30)+... just wondering if there a way without entering 30 1 time and 52 7 times and so on in L1 then calculating the 1 var stats. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="8277" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2011-03-15T05:15:57.513" LastActivityDate="2011-03-15T05:15:57.513" Title="Ti-83 Probability of events occurring" Tags="&lt;probability&gt;&lt;statistics&gt;&lt;calculator&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="27110" PostTypeId="2" ParentId="27105" CreationDate="2011-03-15T04:18:47.823" Score="1" ViewCount="" Body="&lt;p&gt;Indeed, &quot;acyclicity&quot; of a complex means mostly the same thing as exactness (except in degree zero, as Mariano notes). I think they are just used in different contexts: people often call sequences (such as &quot;short exact sequences&quot; $0 \to A \to B \to C \to 0$ or also &quot;long exact sequences&quot; that are derived from complexes) exact, while a complex by itself is often called acyclic if it has the same property.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, one would say that the long exact sequence for group cohomology is exact (probably not acyclic), but that the  resolution used to compute it was acyclic (though here &quot;exact&quot; is probably used more often).&lt;/p&gt;&#xA;" OwnerUserId="536" LastEditorUserId="274" LastEditorDisplayName="" LastEditDate="2011-03-15T04:41:04.173" LastActivityDate="2011-03-15T04:41:04.173" CommentCount="3" />
  <row Id="27111" PostTypeId="1" AcceptedAnswerId="27113" CreationDate="2011-03-15T04:30:23.670" Score="1" ViewCount="60" Body="&lt;p&gt;Is  $ax^3 + bx^2 +cx + d$, for which $a$, $b$, $c$, $d$ are integers,  subset of P3. I got that it was a subspace but the book got that it wasn't a subspace and now I'm confused. &lt;/p&gt;&#xA;" OwnerUserId="8278" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-15T04:43:15.700" LastActivityDate="2011-03-15T04:43:15.700" Title="General Vector subspaces and how to find them?" Tags="&lt;linear-algebra&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="27112" PostTypeId="2" ParentId="27107" CreationDate="2011-03-15T04:35:52.900" Score="4" ViewCount="" Body="&lt;p&gt;Assume you have $y=ax^2 + bx + c$. The maximum is achieved at $x = -\frac{b}{2a}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The values you have give you the value of $c$ ($-100$), and a relation between $a$ and $b$ obtained by plugging in $(7500,-250)$. You also know that the maximum is achieved at $-\frac{b}{2a}$, so plugging that will give you another relation between $a$ and $b$. Putting them together will give you a quadratic equation that $b$ must satisfied.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2011-03-15T04:35:52.900" />
  <row Id="27113" PostTypeId="2" ParentId="27111" CreationDate="2011-03-15T04:36:56.160" Score="0" ViewCount="" Body="&lt;p&gt;What field is your vector space over? Is the set of all such polynomials closed under scalar multiplication?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer to this question will show why the set of these polynomials do not form a subspace.&lt;/p&gt;&#xA;" OwnerUserId="1583" LastActivityDate="2011-03-15T04:36:56.160" />
  <row Id="27114" PostTypeId="2" ParentId="27105" CreationDate="2011-03-15T04:39:36.687" Score="3" ViewCount="" Body="&lt;p&gt;Acyclic and exact are not the same. As Akhil says in his answer, the long exact for group cohomology is indeed exact, but a projective resolution of a module is acyclic because it is not exact in degree zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Originally, one used to say that a projective resolution $P_\bullet$ of a module $M$ was &quot;acyclic over $M$&quot;, and that means that there is a map $\varepsilon:P_0\to M$, called an augmentation, such that if one extends the complex $P_\bullet$ so as to put $M$ in degree $-1$ with $\varepsilon$ as the last differential, &lt;em&gt;then&lt;/em&gt; the resulting complex is exact.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, an acyclic space is not one whose singular complex is exact (there are very few such spaces!) but one whose singular complex is acyclic over $\mathbb Z$.&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2011-03-15T04:39:36.687" CommentCount="3" />
  <row Id="27115" PostTypeId="2" ParentId="27111" CreationDate="2011-03-15T04:41:06.730" Score="1" ViewCount="" Body="&lt;p&gt;It's not a subspace because when you multiply by a scalar (I assume $P_3$ is the vector space of polynomials of degree at most 3, with real coefficients), you get polynomials that are not necessarily in that subset, i.e., polynomials whose coefficients are not integers. For instance, $\sqrt{2}(x^3+x^2+x+1)=\sqrt{2}x^3+\sqrt{2}x^2+\sqrt{2}x+\sqrt{2}$, the coefficients of this polynomial are no longer integers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When checking for subspaces, make sure you don't forget multplication by scalars (you only checked that the subset was closed under vector addition).&lt;/p&gt;&#xA;" OwnerUserId="1313" LastActivityDate="2011-03-15T04:41:06.730" />
  <row Id="27116" PostTypeId="2" ParentId="27109" CreationDate="2011-03-15T04:44:02.140" Score="2" ViewCount="" Body="&lt;p&gt;Your first question,  my first answer :-)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1a.  2nd---&gt;distr---&gt;binompdf---&gt;(7, 0.82, 3)&#xA;1b.  1 - 2nd---&gt;distr---&gt;binomcdf---&gt;(7, 0.82, 5) Note thats CDF, not pdf&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Enetering 7 52's can be done easily, but I don't rememebr how.  Sorry.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="8279" LastActivityDate="2011-03-15T04:44:02.140" CommentCount="1" />
  <row Id="27117" PostTypeId="2" ParentId="19738" CreationDate="2011-03-15T04:44:28.707" Score="0" ViewCount="" Body="&lt;p&gt;First of all dear @Math Student as you say applied force is 4 pounds which (though i seldom work in such units) means that fore applied is 4 * g (acceleration due to gravity).&lt;/p&gt;&#xA;" OwnerUserId="8000" LastActivityDate="2011-03-15T04:44:28.707" CommentCount="1" />
  <row Id="27118" PostTypeId="2" ParentId="27100" CreationDate="2011-03-15T04:48:46.257" Score="1" ViewCount="" Body="&lt;p&gt;Think geometrically.  Assuming $\alpha \ge 0$ (which is necessary, as Ross remarked) $\alpha t^2 + \beta t + \gamma \le 0$ for $t$ in an interval of the real line.  Then $x + t v$ for such $t$ forms what kind of object?&lt;/p&gt;&#xA;" OwnerUserId="8508" LastActivityDate="2011-03-15T04:48:46.257" CommentCount="2" />
  <row Id="27119" PostTypeId="1" AcceptedAnswerId="27325" CreationDate="2011-03-15T04:53:45.450" Score="1" ViewCount="338" Body="&lt;p&gt;$F$ is an extension field of a field $K$.&lt;br&gt;&#xA;Let $F$  be an algebraic closure of $\mathbb{Z}_p $ ($p$ prime). Show that&lt;br&gt;&#xA;$(i)$ $F$ is algebraic Galois over $\mathbb{Z}_p$&lt;br&gt;&#xA;$(ii)$  The map $\alpha:F\rightarrow F$ given by $u\mapsto u^p$ is a nonidentiy $\mathbb{Z}_p$-automorphism of $F$.&lt;br&gt;&#xA;$(iii)$  The subgroup $H=\langle \alpha \rangle$ is a proper subgroup of Aut$(F/\mathbb{Z}_p)$ where the fixed field is $\mathbb{Z}_p$, which is also the fixed field of Aut$(F/\mathbb{Z}_p)$ by $(i).$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, here is my attempt for (i). Let  $S \subset \mathbb{Z}_p[x])$ of monic polynomials of the form $x^{p^n}-x$.  Then for all $f\in S$, gcd$(f,f^{\prime})=1$(i.e, the polynomials are separble) and $F=\mathbb{Z}_p(a\in F:f(a)=0)$. So $(F/\mathbb{Z}_p)$ is Galois. I'd be glad if I could get assistance for (ii) and (iii) as well.  Thanks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;ADDED:&lt;/strong&gt;  Attempt at (iii).   &lt;strike&gt;but I know that since $F/\mathbb{Z}_p$ is a finite Galois extension, its fixed field is $\mathbb{Z}_p$&lt;/strike&gt; hints...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The field $\mathbb{Z}_p$ must be contained in $F$. For $a\in \mathbb{Z}_p$, $\alpha(a)=a^p=a$. Thus the polynomial $x^p-x$ has $p$ zeros in $F$, namely, the elements of $\mathbb{Z}_p$. But the elements fixed under $\alpha$ are precisely the zeros in $F$ of $x^p-x$. Hence the fixed field of $\alpha$ is $\mathbb{Z}_p$ which is also the fixed field of Aut$(F/\mathbb{Z}_p)$.&lt;br&gt;&#xA;Left to show that $H$ is a proper subgroup... If it helps I know that the order of $\langle \alpha \rangle $ is $n$...&lt;/p&gt;&#xA;" OwnerUserId="8280" LastEditorUserId="8280" LastEditorDisplayName="" LastEditDate="2011-03-16T13:00:48.777" LastActivityDate="2011-03-16T19:51:13.547" Title="The algebraic closure of a finite field and its Galois group" Tags="&lt;abstract-algebra&gt;&lt;galois-theory&gt;" AnswerCount="2" CommentCount="8" />
  <row Id="27121" PostTypeId="2" ParentId="27119" CreationDate="2011-03-15T05:12:24.860" Score="0" ViewCount="" Body="&lt;p&gt;For (ii) we have the following: Let $a,b \in F$. Then $$\alpha(a+b) = (a+b)^{p} = a^p+b^p = \alpha(a)+\alpha(b)$$ Also $$\alpha(ab) = \alpha(a) \alpha(b)$$ Now $\alpha(a) = 0 \implies a = 0$ so that $\alpha$ is injective (i.e. the kernel of $\alpha$ is ${0}$). We also know that $\alpha$ is surjective since $F$ is finite. This is known as the &lt;a href=&quot;http://mathworld.wolfram.com/FrobeniusAutomorphism.html&quot; rel=&quot;nofollow&quot;&gt;Frobenius automorphism&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2548" LastEditorUserId="2548" LastEditorDisplayName="" LastEditDate="2011-03-15T05:17:42.250" LastActivityDate="2011-03-15T05:17:42.250" CommentCount="2" />
  <row Id="27122" PostTypeId="1" AcceptedAnswerId="27149" CreationDate="2011-03-15T05:16:43.440" Score="11" ViewCount="177" Body="&lt;p&gt;I know that there are isomorphic projective varieties which have nonisomorphic coordinate rings, but I'm a little mystified as to &quot;why&quot; this is the case. Why doesn't a usual functoriality proof go through to prove this, and is there any insight into this other than that the definitions just work out this way? &lt;/p&gt;&#xA;" OwnerUserId="1491" LastActivityDate="2011-03-16T12:46:51.120" Title="Why is the coordinate ring of a projective variety not determined by the isomorphism class of the variety?" Tags="&lt;algebraic-geometry&gt;" AnswerCount="3" CommentCount="4" FavoriteCount="3" />
  <row Id="27123" PostTypeId="1" CreationDate="2011-03-15T05:22:58.320" Score="6" ViewCount="262" Body="&lt;p&gt;I am having trouble in proving the pointwise convergence of $g(x)=\sum_{n=1}^\infty \frac{\sin(\sqrt{n}x)}{n}$ for all real numbers $x$ using elementary methods (e.g. integral test, Weierstrass M-test, Abel's Test, Dirichlet's Test, Comparison with Riemann sum, etc.).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can someone help me on this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="8252" LastActivityDate="2011-03-15T21:55:27.893" Title="Pointwise Convergence of $\sum \frac{\sin(\sqrt{n}x)}{n}$" Tags="&lt;real-analysis&gt;&lt;sequences-and-series&gt;&lt;trigonometry&gt;&lt;convergence&gt;&lt;classical-analysis&gt;" AnswerCount="4" CommentCount="0" FavoriteCount="1" />
  <row Id="27124" PostTypeId="1" AcceptedAnswerId="27143" CreationDate="2011-03-15T05:38:48.093" Score="0" ViewCount="109" Body="&lt;p&gt;Let's say I'm specifying a circular arc with the following information:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;x1,y1,x2,y2,radius,direction&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and that I have 2D boundaries of:&#xA;minX = -5, minY = -5, maxX = 5, maxY = 5&lt;/p&gt;&#xA;&#xA;&lt;p&gt;how can i tell if the arc will go outside the boundaries?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so, for example, an arc with these values &lt;em&gt;would&lt;/em&gt; go outside the bounds:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-4.5,0,-4.5,4,2,clockwise&lt;/p&gt;&#xA;&#xA;&lt;p&gt;in fact, it's almost entirely outside the bounds.  but if I change the radius to 6, it won't.  (or if I change the direction to ccw of course)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;thanks for any help!!&lt;/p&gt;&#xA;" OwnerUserId="7279" LastEditorUserId="7279" LastEditorDisplayName="" LastEditDate="2011-03-15T05:56:38.897" LastActivityDate="2011-03-15T08:40:57.410" Title="how to tell if an arc is within bounds?" Tags="&lt;algebra-precalculus&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="27126" PostTypeId="2" ParentId="27123" CreationDate="2011-03-15T06:29:24.877" Score="2" ViewCount="" Body="&lt;p&gt;I would try approximating $\sin(\sqrt{n} x)/n$ by $\int_{\sqrt{n-1/2}}^{\sqrt{n+1/2}} \sin(t x)/(2 t) \, dt$, noting that $\int_{0}^\infty \sin(t)/t \, dt$ converges.&lt;/p&gt;&#xA;" OwnerUserId="8508" LastActivityDate="2011-03-15T06:29:24.877" />
  <row Id="27127" PostTypeId="2" ParentId="27123" CreationDate="2011-03-15T06:34:37.160" Score="3" ViewCount="" Body="&lt;p&gt;Look at the sequence of partial sums up to each sign change in the sine, i.e. combine all consecutive terms with the same sign. Since the length of these groups of terms grows as $\sqrt{n}$, their sum decays as $1/\sqrt{n}$, so you get an alternating series with terms converging to zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: Didier rightly pointed out that I didn't show that the absolute values of the terms strictly decrease. To show this rigorously would probably be more hassle than pursuing one of the approaches in the other answers, but here's a not-so-rigorous argument: Each group of terms can be considered as an approximation with the trapezoidal rule of the integral over $\sin \sqrt{ux}/u$ over an interval of $\pi/x$, scaled in proportion to the number of terms in the group. The scaling by the number of terms grows as $\sqrt{n}$ and the integrand falls off as $1/n$, for a decay of $1/\sqrt{n}$. There are three sources of approximation error. Two occur at the endpoints, one from the fact that the trapezoidal rule specifies a weight of $1/2$ there, and another from the fact that the ends of the sum and the integral don't coincide. Both of these errors are proportional to the outermost terms in the group, which are of order $1/\sqrt{n}^3$, and to the width of the integration interval they represent, which is of order $1/\sqrt{n}$, for a total decay of $1/n^2$. The third approximation error is the error in the trapezoidal rule itself, which is cubic in the width of the integration intervals for a factor of $1/\sqrt{n}^3$, and proportional to the second derivative of the integrand, which falls off with $1/n$, for a total decay of $1/\sqrt{n}^5$. So both approximation errors decay faster than the difference of order $1/\sqrt{n}^3$ in the contributions of order $1/\sqrt{n}$.&lt;/p&gt;&#xA;" OwnerUserId="6622" LastEditorUserId="6622" LastEditorDisplayName="" LastEditDate="2011-03-15T21:19:55.583" LastActivityDate="2011-03-15T21:19:55.583" CommentCount="4" />
  <row Id="27128" PostTypeId="2" ParentId="27084" CreationDate="2011-03-15T06:46:16.920" Score="0" ViewCount="" Body="&lt;p&gt;If A=B , then the angles of the triangle are 45:45:90,&#xA;And as per the 45:45:90 theorem, the side opposite the 45 degree angle is &lt;code&gt;hypotenuse/√2&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in this case value of &lt;code&gt;A&lt;/code&gt; will be &lt;code&gt;C/√2&lt;/code&gt;&lt;/p&gt;&#xA;" OwnerUserId="6305" LastActivityDate="2011-03-15T06:46:16.920" />
  <row Id="27129" PostTypeId="1" AcceptedAnswerId="27130" CreationDate="2011-03-15T06:48:51.787" Score="0" ViewCount="76" Body="&lt;p&gt;I know the order of a &lt;em&gt;group&lt;/em&gt; is the size of the group, ie the number of elements. &#xA;But what does it mean for an element of that group to have order? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, what are the precise definitions for&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;1) &quot;element of a finite order of a&#xA;  group&quot; and &#xA;  2) order of an element of&#xA;  a group (assuming that the element has&#xA;  finite order)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If I remember correctly, the order of $\mathbb{Z}$ is one, however the order of the elements in this group have order infinity. Why is that? (Also I dont think $\mathbb{Z}$ is a group in the first place, is it? )&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would also like to ask another question if thats okay:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What is a cyclic group (and a precise&#xA;  definition for it as well) ?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="287" LastActivityDate="2011-03-15T06:55:05.800" Title="What is the difference between the order of a group and the order of the elements of the group" Tags="&lt;group-theory&gt;" AnswerCount="2" CommentCount="5" />
  <row Id="27130" PostTypeId="2" ParentId="27129" CreationDate="2011-03-15T06:51:34.870" Score="1" ViewCount="" Body="&lt;p&gt;An element $g \in G$ has order $n$ if $g^n = e$ ($n$ is the smallest positive integer for which this is true). Where $e$ is the identity. See this wikipedia &lt;a href=&quot;http://en.wikipedia.org/wiki/Order_%28group_theory%29&quot; rel=&quot;nofollow&quot;&gt;articles&lt;/a&gt; for more detail.&lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2011-03-15T06:51:34.870" />
  <row Id="27131" PostTypeId="2" ParentId="27129" CreationDate="2011-03-15T06:55:05.800" Score="0" ViewCount="" Body="&lt;p&gt;You can usually find out about this sort of thing on Wikipedia. In this case, the answers to your questions are here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Order_%28group_theory%29&quot; rel=&quot;nofollow&quot;&gt;Order (group theory)&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Cyclic_group&quot; rel=&quot;nofollow&quot;&gt;Cyclic group&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Integers#Algebraic_properties&quot; rel=&quot;nofollow&quot;&gt;Integers#Algebraic_properties&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="6622" LastActivityDate="2011-03-15T06:55:05.800" />
  <row Id="27132" PostTypeId="1" AcceptedAnswerId="27135" CreationDate="2011-03-15T07:03:24.953" Score="3" ViewCount="78" Body="&lt;p&gt;It is well known that any finite group can be embedded in Symmetric group $S_n$, $GL(n,q)$ ($q=p^m$) for some $m,n,q\in \mathbb{N}$. Can we embed any finite group in $A_n$, or $SL(n,q)$ for some $n,q\in \mathbb{N}$?&lt;/p&gt;&#xA;" OwnerUserId="8186" LastEditorUserId="8186" LastEditorDisplayName="" LastEditDate="2011-03-15T07:12:51.013" LastActivityDate="2011-03-15T07:31:15.927" Title="Embedding of finite groups" Tags="&lt;group-theory&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="27133" PostTypeId="2" ParentId="27132" CreationDate="2011-03-15T07:22:43.563" Score="3" ViewCount="" Body="&lt;p&gt;Yes we can.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $A_n$, we can embed the given group in some $S_{n-2}$ and then for the additional two elements choose the identity or the transposition according as the element of $S_{n-2}$ is even or odd.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $SL(n,q)$, we can embed the given group in some $GL(n-1,q)$ and then choose the diagonal element in the additional row and column as the reciprocal of the determinant of the element of $GL(n-1,q)$.&lt;/p&gt;&#xA;" OwnerUserId="6622" LastEditorUserId="6622" LastEditorDisplayName="" LastEditDate="2011-03-15T07:31:15.927" LastActivityDate="2011-03-15T07:31:15.927" CommentCount="3" />
  <row Id="27134" PostTypeId="2" ParentId="27132" CreationDate="2011-03-15T07:23:02.437" Score="2" ViewCount="" Body="&lt;p&gt;To your first question (embeddable within $A_n$) I think the answer is yes for obvious reasons: one can embed $S_n$ within $A_{n+2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Consider the subset of $A_{n+2}$ that stabilizes the first $n$ elements (as a set), it's obvious that this set will consist of all permutations of these elements, where it may or may not interchange the last two points, depending on the sign. For instance for $n=3$ we retrieve $S_3$ as $(1\ 2\ 3)$, $(1\ 2)(4\ 5)$, etc...)&lt;/p&gt;&#xA;" OwnerUserId="5189" LastActivityDate="2011-03-15T07:23:02.437" />
  <row Id="27135" PostTypeId="2" ParentId="27132" CreationDate="2011-03-15T07:26:50.303" Score="5" ViewCount="" Body="&lt;p&gt;Yes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The symmetric group Sym(n) is generated by { (1,2), (2,3), …, (n−1,n) }.  You can embed Sym(n) into Alt(n+2) as the group generated by { (1,2)(n+1,n+2), (2,3)(n+1,n+2), …, (n−1,n)(n+1,n+2) }.  This embedding takes a permution π in Sym(n) and sends it to π⋅(n+1,n+2)&lt;sup&gt;sgn(π)&lt;/sup&gt;, where sgn(π)=±1 is the sign of the permutation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, G ≤ Sym(n) ≤ Alt(n+2) embeds any group into a (slightly larger) alternating group.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The general linear group GL(n,q) embeds in the special linear group SL(n+1,q) using a determinant trick.  We just add a new coordinate to cancel out the determinant of the matrix from GL(n,q) so the result lands in SL(n+1,q).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\operatorname{GL}(n,q) \cong \left\{ \begin{bmatrix} A &amp;amp; 0 \\ 0 &amp;amp; 1/\det(A) \end{bmatrix} : A \in \operatorname{GL}(n,q) \right\} ≤ \operatorname{SL}(n+1,q)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, G ≤ GL(n,q) ≤ SL(n+1,q) embeds any group into a (slightly larger) special linear group.&lt;/p&gt;&#xA;" OwnerUserId="583" LastActivityDate="2011-03-15T07:26:50.303" />
  <row Id="27136" PostTypeId="1" CreationDate="2011-03-15T07:48:32.093" Score="2" ViewCount="232" Body="&lt;p&gt;Hi There, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a question to ask down below, that I have been having some trouble with and would like some help and clarification on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose &lt;strong&gt;A&lt;/strong&gt;  is an $n \times n$ matrix with (not necessarily distinct) eigenvalues $\lambda_{1}, \lambda_{2}, \ldots, \lambda_{n}$. Can it be shown that:   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(a) The sum of the main diagonal entries of &lt;strong&gt;A&lt;/strong&gt;, called the &lt;em&gt;trace&lt;/em&gt; of &lt;strong&gt;A&lt;/strong&gt;, equals the sum of the eigenvalues of &lt;strong&gt;A&lt;/strong&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(b) &lt;strong&gt;A&lt;/strong&gt; $- ~ k$ &lt;strong&gt;I&lt;/strong&gt; has the eigenvalues $\lambda_{1}-k, \lambda_{2}-k, \ldots, \lambda_{n}-k$ and the same eigenvectors as &lt;strong&gt;A&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank You very much.&lt;/p&gt;&#xA;" LastActivityDate="2011-03-19T04:55:03.990" Title="Can it be determined that the sum of the diagonal entries, of matrix A, equals the sum of eigenvalues of A." Tags="&lt;linear-algebra&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="27138" PostTypeId="1" AcceptedAnswerId="27139" CreationDate="2011-03-15T07:58:59.210" Score="4" ViewCount="123" Body="&lt;p&gt;Hi, I am trying to understand if there is an error in this question, or if the answer is trivial.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Define a topology on $[0,\infty)$ with open sets&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;$(a,\infty), a \in (0,\infty)$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$[0,\infty)$ and $\emptyset$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Show $[0,\infty)$ is compact in this topology.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since we can only build an open cover out of open sets, surely we can only take $\cup_{n &amp;gt; 0} (n,\infty)$ which does not cover $[0,\infty)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can we either:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Trivially take $[0,\infty)$ as an open finite cover, meaning whenever we take the whole space of any topology it is automatically compact.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The question should read $[a,\infty), a \in [0,\infty)$ are open in this topology&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Any help would be greatly appreciated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many thanks,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ash&lt;/p&gt;&#xA;" OwnerUserId="1763" LastActivityDate="2011-03-16T10:26:28.763" Title="Compactness of topology on $[0,\infty)$" Tags="&lt;topology&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="27139" PostTypeId="2" ParentId="27138" CreationDate="2011-03-15T08:06:27.270" Score="5" ViewCount="" Body="&lt;p&gt;The first variant is correct one. The only possible cover of $[0,\infty)$ in this topology is a cover which has $[0,\infty)$ in it. This follows since $\cup_{a}(a,\infty)=(0,\infty)$, so it is not possible to construct cover of $[0,\infty)$ using only sets $(a,\infty)$ with $a&amp;gt;0$.&lt;/p&gt;&#xA;" OwnerUserId="4742" LastEditorUserId="4742" LastEditorDisplayName="" LastEditDate="2011-03-16T10:26:28.763" LastActivityDate="2011-03-16T10:26:28.763" CommentCount="11" />
  <row Id="27140" PostTypeId="2" ParentId="27136" CreationDate="2011-03-15T08:23:25.230" Score="3" ViewCount="" Body="&lt;p&gt;For the first, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$A = P^{-1} M P$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where M is a (upper triangular) matrix with eigenvalues of A as diagonal elements. This is what it means to say that  A is always similar to its Jordan form.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Use $Tr(AB)=Tr(BA)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$Tr(A)= Tr( P^{-1} M P) = Tr(MPP^{-1})=Tr(M)=\sum_n\lambda_n$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;b) Let $B=A-kI$ with eigenvalues be $\chi_n$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Eigenvalues are determined by solutions of &#xA;$$|B-\chi I|=0$$&#xA;or,&#xA;$$|A-(\chi+k)I|=0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but since you know $$|A-\lambda I|=0$$ you get $\chi_n = \lambda_n-k$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $Y$ be an eigenvector of $B$. So $BY=\chi Y$. Now plug stuff in for $B$ and $\chi$ and see what you'd get.&lt;/p&gt;&#xA;" LastActivityDate="2011-03-15T08:50:19.030" CommentCount="3" />
  <row Id="27141" PostTypeId="1" AcceptedAnswerId="27197" CreationDate="2011-03-15T08:30:35.283" Score="3" ViewCount="69" Body="&lt;p&gt;Hello everyone.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have come across this question in a course about Cayley graphs recently. I don't really have a clue how to answer this question. Here it is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider two groups $A,B$. Let $K&amp;lt; A*B$ be the kernel of the homomorphism $\phi: A * B \to A \times B$ that extends the inclusions $A \hookrightarrow A \times B$ and $B \hookrightarrow A\times B$.&#xA;Show that $K$ is free.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As usual, thanks in advance :)&lt;/p&gt;&#xA;" OwnerUserId="5465" LastActivityDate="2011-03-15T14:53:23.960" Title="Kernel of $\phi: A * B \to A \times B$ is free" Tags="&lt;homework&gt;&lt;group-theory&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="27142" PostTypeId="1" CreationDate="2011-03-15T08:31:29.093" Score="0" ViewCount="50" Body="&lt;p&gt;Let K be an algebraically closed field of arbitrary characteristic. Then there is a Grassmannian of subspaces of fixed dimension say m of n-dimentional space and it is a projective variety. If you have an algebra A over field K you can vew say left modules over this algebra and for a given module of dimension n you can vew a Grassmannian of m-dimentional submodules. I want to know why the Grassmannian of submodules is closed in the Grassmannian of subspaces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a proof for characteristic 0. Take a subspace U, U is a submodule iff for any  $a \in A$ $a(U) \subseteq U$, for any $a \in A$ there is $c \in K$ (that is true only if charK=0) such that multipaing the whole module by a-c1 is an isomorphism. But $a(U) \subseteq U$ iff  for any c $(a-c1)(U) \subseteq U$ or also there exists c such that $(a-c1)(U) \subseteq U$. So you can chose c such that (a-c1) induces an isomorphism and have $(a-c1)(U) = U$. If you have an iso induced by multiplication by b on the whole module than you have a morphism on the Grassmannian of submodules say b'. So the Grassmannian of submodules is {subspaces U such that b'(U)=U for any iso b} and it is a set of equasions and so it is closed and the Grassmannian is projective.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if charK=p than it does't work. And I don't know what to do.&lt;/p&gt;&#xA;" OwnerUserId="8283" LastActivityDate="2011-03-15T20:08:18.607" Title="How to show that a Grassmannian of submodules is a projective variety?" Tags="&lt;algebraic-geometry&gt;" AnswerCount="2" />
  <row Id="27143" PostTypeId="2" ParentId="27124" CreationDate="2011-03-15T08:40:57.410" Score="1" ViewCount="" Body="&lt;p&gt;This attempts a systematic approach:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Check your two points are inside the bounds.  In your example they are.&lt;/li&gt;&#xA;&lt;li&gt;Find the centre of the circular arc, which lies on the perpendicular bisector between the two points. &lt;strong&gt;Note that if the distance between the two points is not twice the radius then there will be two such points meaning two such centres and so two different arcs.&lt;/strong&gt;  In your example the distance is twice the radius and the centre is $(-4.5,2)$.&lt;/li&gt;&#xA;&lt;li&gt;Find the (real) points where the circle meets the extended bounds. There will be up to 2 for each bound.  If there is just one then the circle touches the bound rather than going outside, so this can be ignored.  In your example it meets the bounds at about $(-5,3.93649)$ and $(-5,0.0635)$ where the decimals are $(4\pm \sqrt{15})/2$. &lt;/li&gt;&#xA;&lt;li&gt;Drop any of these points which are outside the other bounds.  In your example you do not drop anything so still have the two points where the circle meets the bounds.&lt;/li&gt;&#xA;&lt;li&gt;Decide on which side of the line which joins the two initial points the points on the circle and bounds lie. You could do this by looking at angle bearings from the centre of the various points. In your example, both of the two points where the circle meets the bounds lie on the clockwise side of the line.&lt;/li&gt;&#xA;&lt;li&gt;You have your answer.  In your example, the clockwise arc crosses the bounds.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="6460" LastActivityDate="2011-03-15T08:40:57.410" CommentCount="7" />
  <row Id="27144" PostTypeId="1" CreationDate="2011-03-15T08:58:22.637" Score="1" ViewCount="44" Body="&lt;p&gt;Are there two matrices $A_{m\times n}$, and $B_{n\times m}$ such that $A.B=I_{m}$ and $B.A=I_{n}$ (here $m\neq n$)&lt;/p&gt;&#xA;" OwnerUserId="8186" LastActivityDate="2011-03-15T09:03:10.017" Title="Product of Matrices" Tags="&lt;linear-algebra&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="27145" PostTypeId="1" AcceptedAnswerId="27147" CreationDate="2011-03-15T08:59:32.063" Score="3" ViewCount="138" Body="&lt;p&gt;Why are Artinian rings of Krull dimension 0?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As in the example of $\mathbb{Z}/(6)$, the ideal $\mathbb{Z}/(2)$ is prime, I think. So, Artinian rings may contain prime ideals. But why does not the primes ideal contain other prime ideals properly?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another question is that, when a ring is of Krull dimension 0, is it necessarily Artinian?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="4736" LastActivityDate="2011-03-15T23:06:36.177" Title="Why are Artinian rings of Krull dimension 0?" Tags="&lt;abstract-algebra&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="27146" PostTypeId="2" ParentId="27144" CreationDate="2011-03-15T09:03:10.017" Score="5" ViewCount="" Body="&lt;p&gt;No. This follows from the facts that $rk(AB) \leq \min(rk(A),rk(B))$, that $rk(I_n) = n$, and that for an $n \times m$ matrix $A$ we have $rk(A) \leq \min(m,n)$. Here $rk(\cdot)$ denotes the rank of a matrix.&lt;/p&gt;&#xA;" OwnerUserId="677" LastActivityDate="2011-03-15T09:03:10.017" />
  <row Id="27147" PostTypeId="2" ParentId="27145" CreationDate="2011-03-15T09:09:59.803" Score="6" ViewCount="" Body="&lt;p&gt;The following proof that any artinian ring $R$ has $\text{dim}(R)=0$ is from Theorem 8.1 of Atiyah Macdonald:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $\frak{p}$ be a prime ideal of $R$, so that $S=R/\frak{p}$ is an artinian domain (this is because any infinite descending chain of ideals in $S$ could be lifted to an infinite descending chain of ideals in $R$, so because $R$ is artinian, $S$ must be artinian too). For any non-zero $x\in S$, we must have that $(x^n)=(x^{n+1})$ for some $n$ (this is  because $S$ is artinian, so we can't have an infinite descending chain $S\supset (x)\supset (x^2)\supset\cdots$), hence $x^n=x^{n+1}y$ for some $y\in S$, but because $S$ is a domain, we can cancel to get $1=xy$, hence $x$ is invertible. Any non-zero element of $S$ is invertible, hence $S=R/\frak{p}$ is a field, hence $\frak{p}$ is maximal. Because any prime ideal of $R$ is maximal, we must have that $\text{dim}(R)=0$.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Theorem 8.5 of Atiyah Macdonald says that $R$ is artinian $\iff$ $R$ is noetherian and $\text{dim}(R)=0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So here is an example of a ring $R$ with $\text{dim}(R)=0$ but $R$ not noetherian, and hence not artinian:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$R=k[x_1,x_2,\ldots]/(x_1,x_2,\ldots)^2\cong k[\epsilon_1,\epsilon_2,\ldots]$$&#xA;which is a field $k$ with infinitely many nilpotent elements $\epsilon_i$ added in. There's only one prime ideal of $k$, namely the 0 ideal, and nilpotents won't change that, so the 0 ideal is still the only prime ideal of $R$, but $R$ is certainly not noetherian - the chain of ideals&#xA;$$(0)\subset (\epsilon_1)\subset(\epsilon_1,\epsilon_2)\subset\cdots$$&#xA;is an infinite ascending chain.&lt;/p&gt;&#xA;" OwnerUserId="264" LastEditorUserId="264" LastEditorDisplayName="" LastEditDate="2011-03-15T09:23:41.057" LastActivityDate="2011-03-15T09:23:41.057" CommentCount="1" />
  <row Id="27148" PostTypeId="1" AcceptedAnswerId="27152" CreationDate="2011-03-15T09:20:32.680" Score="0" ViewCount="109" Body="&lt;p&gt;The quaternion group is non-abelian p-group with every proper subgroup cyclic. In general, if $G$ is a non-abelian $p$-group with every proper subgroup cyclic, is it necessarily generalized quaternion?&lt;/p&gt;&#xA;" OwnerUserId="8186" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2011-03-15T16:10:07.393" LastActivityDate="2011-03-15T16:10:07.393" Title="p-Groups with every proper subgroup cyclic" Tags="&lt;group-theory&gt;" AnswerCount="1" FavoriteCount="2" />
  <row Id="27149" PostTypeId="2" ParentId="27122" CreationDate="2011-03-15T09:38:51.370" Score="7" ViewCount="" Body="&lt;p&gt;The homogeneous coordinate ring of a projective variety embedded in $\mathbb{P}^n$ is not the &quot;ring of functions&quot; on the variety but actually the direct sum of sections of symmetric powers of some &lt;a href=&quot;http://en.wikipedia.org/wiki/Line_bundle&quot; rel=&quot;nofollow&quot;&gt;line bundle&lt;/a&gt;. The point is that this construction actually depends on the line bundle, and the same variety can have many line bundles which give rise to non-isomorphic rings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure what &quot;a usual functoriality proof&quot; is, but you should find either that you end up proving something about the category of projective varieties together with an embedding or that you can't define the homogeneous coordinate ring at &lt;em&gt;all&lt;/em&gt; without a choice of embedding.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I'm not totally comfortable with this answer since I can't claim to have much experience with line bundles, so here's a more straightforward version. First, the reason this issue doesn't come up for affine varieties is that once you have a definition of a morphism of affine varieties, the ring of functions $k[V]$ on an affine variety $V$ really is just the set of morphisms $V \to \mathbb{A}^1(k)$, and this definition is clearly functorial.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same cannot be said for the homogeneous coordinate ring of a projective variety $V \subset \mathbb{P}^n$. First of all, its elements are not functions on the variety. Quotients of two elements of the same degree do define &lt;strike&gt;morphisms&lt;/strike&gt; rational maps $V \dashrightarrow \mathbb{P}^1$, but you're considering much more than just such quotients. When $V$ is irreducible you might be tempted to replace the ring of functions with the field $k(V)$ of rational functions, but $k(V)$ does not completely capture $V$ (for example it ignores the removal of finitely many points).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After you think about this problem for awhile you should realize that the problem here is that the homogeneous coordinate ring really is defined in terms of the embedding, and so there's no reason to expect it to be independent of the choice of embedding. So next you might look for a definition of projective variety that is independent of the choice of embedding, and then sooner or later you have to start studying schemes. &lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2011-03-16T12:46:51.120" LastActivityDate="2011-03-16T12:46:51.120" CommentCount="3" />
  <row Id="27150" PostTypeId="1" AcceptedAnswerId="27158" CreationDate="2011-03-15T09:40:19.953" Score="0" ViewCount="234" Body="&lt;p&gt;Free modules are projective, and projective modules are direct summand of free modules. Is there any example of projective modules that are not free? (I know this is not possible for modules of fields.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Free modules are torsion-free. But is the inverse true? Are there examples of torsion-free modules that are not free?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you~&lt;/p&gt;&#xA;" OwnerUserId="4736" LastActivityDate="2011-03-15T20:34:38.010" Title="Example of modules that are projective but not free; torsion-free but not free." Tags="&lt;abstract-algebra&gt;" AnswerCount="4" />
  <row Id="27151" PostTypeId="1" CreationDate="2011-03-15T09:41:33.883" Score="2" ViewCount="183" Body="&lt;p&gt;Hello Everyone, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I had been working on this problem here below, but seem to not know a precise and clean way to show the proof to the question below. I had about a few ways of doing it, but the statements/operations were pretty loosely used. The problem is as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose $\mathbf{A}$ is an $n \times n$ matrix with (not necessarily distinct) eigenvalues $\lambda_{1}, \lambda_{2}, \ldots, \lambda_{n}$. Can it be shown that the &lt;em&gt;polynomial matrix&lt;/em&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$p( \mathbf{A} ) = k_{m} \mathbf{A}^{m}+k_{m-1} \mathbf{A}^{m-1}+\ldots+k_{1} \mathbf{A} +k_{0} \mathbf{I} $  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;has the eigenvalues  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$p(\lambda_{j}) = k_{m}{\lambda_{j}}^{m}+k_{m-1}{\lambda_{j}}^{m-1}+\ldots+k_{1}\lambda_{j}+k_{0}$   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $j = 1,2,\ldots,n$ and the same eigenvectors as $\mathbf{A}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" LastActivityDate="2011-03-19T08:36:16.200" Title="Could it be shown that the polynomial matrix $p(\mathbf{A})$ has eigenvalues and same eigenvectors as $\mathbf{A}$?" Tags="&lt;polynomials&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="1" />
  <row Id="27152" PostTypeId="2" ParentId="27148" CreationDate="2011-03-15T09:46:59.697" Score="3" ViewCount="" Body="&lt;p&gt;Yes. If all proper subroups are cyclic, the group must have a unique subgroup of order $p$ (if it contains two subgroups of order $p$, one of them can be taken to lie in the center and thus their product is a non-cyclic subgroup). It is a basic result in the theory of $p$-groups that a $p$-group with a unique subgroup of order $p$ is either cyclic or generalized quaternion (see for instance Proposisition 1.3 of Berkovich's &quot;Groups of Prime Power Order&quot;).&#xA;However, if $G$ is generalized quaternion of order $&amp;gt;8$ then $G/Z(G)$ is a dihedral group which contains a proper non-cyclic subgroup, and this subgroup then corresponds to a non-cyclic subgroup of $G$.&#xA;So the quaternion group of order 8 really is the only non-abelian $p$-group with this property.&lt;/p&gt;&#xA;" OwnerUserId="2538" LastActivityDate="2011-03-15T09:46:59.697" CommentCount="2" />
  <row Id="27153" PostTypeId="2" ParentId="26569" CreationDate="2011-03-15T09:47:25.137" Score="2" ViewCount="" Body="&lt;p&gt;One characterization of the generalized quaternion group is:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If $G$ is a non-abelian $p$-group which contains only-one subgroup of order $p$, then $G$ is generalized quaternion group &lt;strong&gt;[Hall-Theory of groups; Theorem 12.5.2]&lt;/strong&gt;. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So if we try to write the generalized quaternion group $Q_n$ as semi-direct product, then we should have a normal  subgroup $N$, a subgroup $H$, with one necessary condition that $N\cap H=1$; which is not possible because of uniqueness of subgroup of order $p$; it will contained in all subgroups of $Q_n$. Hence the generalized quaternion group is not semi-direct product of smaller $p$-groups.&lt;/p&gt;&#xA;" OwnerUserId="8186" LastEditorUserId="8186" LastEditorDisplayName="" LastEditDate="2011-03-16T08:13:40.623" LastActivityDate="2011-03-16T08:13:40.623" CommentCount="3" />
  <row Id="27155" PostTypeId="2" ParentId="27151" CreationDate="2011-03-15T09:48:56.093" Score="1" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Hint:&lt;/strong&gt; write $A$ in upper triangular form. Locate the eigenvalues of $A$. Show that $p(A)$ is upper triangular as well and compute its diagonal. Conclude.&lt;/p&gt;&#xA;" OwnerUserId="6179" LastEditorUserId="6179" LastEditorDisplayName="" LastEditDate="2011-03-15T11:13:55.330" LastActivityDate="2011-03-15T11:13:55.330" />
  <row Id="27156" PostTypeId="1" CreationDate="2011-03-15T09:51:44.930" Score="9" ViewCount="216" Body="&lt;p&gt;If $G$ is a finite group, then it is well known that there are finitely many inequivalent irreducible representations of $G$ over $\mathbb{C}$; moreover the sum of squares of dimensions of the representations is equal to $|G|$. Also, the dimension of representation divides $|G|$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we consider representations over $\mathbb{Q}$, are the dimensions of irreducible representations related to $|G|$ in such a nice way?&lt;/p&gt;&#xA;" OwnerUserId="8186" LastEditorUserId="5465" LastEditorDisplayName="" LastEditDate="2011-05-21T13:26:25.037" LastActivityDate="2011-05-21T13:26:25.037" Title="Representations of finite groups" Tags="&lt;representation-theory&gt;&lt;finite-groups&gt;" AnswerCount="2" FavoriteCount="2" />
  <row Id="27157" PostTypeId="1" AcceptedAnswerId="27161" CreationDate="2011-03-15T09:52:19.050" Score="0" ViewCount="59" Body="&lt;p&gt;My Maths notes don't explain this, but is there a saddle point or a local maximum?&lt;/p&gt;&#xA;" OwnerUserId="7087" LastActivityDate="2011-03-15T10:47:20.747" Title="If a matrix is negative semidefinite, does it have a local maximum?" Tags="&lt;eigenvalues&gt;" AnswerCount="1" CommentCount="0" />
  <row Id="27158" PostTypeId="2" ParentId="27150" CreationDate="2011-03-15T09:54:45.023" Score="3" ViewCount="" Body="&lt;p&gt;Let $R=\mathbb{Z}/6\mathbb{Z}$. Obviously, $R$ is a free module over itself. Because $\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/3\mathbb{Z}\cong R$, we have that $\mathbb{Z}/2\mathbb{Z}$, considered as an $R$-module, is projective, but it cannot be free - any non-trivial direct sum of $R$'s would have at least 6 elements. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Baer%E2%80%93Specker_group&quot; rel=&quot;nofollow&quot;&gt;Baer-Specker group&lt;/a&gt; is an example of a torsion-free $\mathbb{Z}$-module which is not free.&lt;/p&gt;&#xA;" OwnerUserId="264" LastActivityDate="2011-03-15T09:54:45.023" CommentCount="4" />
  <row Id="27160" PostTypeId="2" ParentId="27123" CreationDate="2011-03-15T09:59:15.143" Score="4" ViewCount="" Body="&lt;p&gt;This may be very closely related to Robert's suggestion. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Elaborated answer&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the following inequality (with $n,m \to \infty$):&#xA;$$&#xD;&#xA;\Bigg|\sum\limits_{k = n}^m {\frac{{\sin (\sqrt k x)}}{k}}  - \sum\limits_{k = n}^m {\int_k^{k + 1} {\frac{{\sin (\sqrt u x)}}{u}du} } \Bigg| \le \sum\limits_{k = n}^m {\int_k^{k + 1} \Bigg| \frac{{\sin (\sqrt k x)}}{k} - \frac{{\sin (\sqrt u x)}}{u}\Bigg|du} .&#xD;&#xA;$$&#xA;For fixed $x &amp;gt; 0$, show that, for any $u \in [k,k+1]$,&#xA;$$&#xD;&#xA;{\Bigg|\frac{{\sin (\sqrt k x)}}{k} - \frac{{\sin (\sqrt u x)}}{u}\Bigg|} \leq \frac{x}{{2k\sqrt k }} + \frac{1}{{k^2 }}.&#xD;&#xA;$$&#xA;(Thus the same inequality holds for the integral from $k$ to $k+1$ of the left-hand side.)&#xA;For this purpose, first write&#xA;$$&#xD;&#xA;\Bigg|\frac{{\sin (\sqrt k x)}}{k} - \frac{{\sin (\sqrt u x)}}{u}\Bigg| = \Bigg|\frac{{\sin (\sqrt k x)k - \sin (\sqrt u x)k + \sin (\sqrt k x)(u - k)}}{{ku}}\Bigg|.&#xD;&#xA;$$&#xA;Then apply the triangle inequality, and use the mean value theorem (twice). &#xA;Further note that $\int_1^\infty  {\frac{{\sin (\sqrt u x)}}{u}} du$ converges (for this purpose, make a change of variable $y=\sqrt u x$). The rest is straightforward.&lt;/p&gt;&#xA;" OwnerUserId="2810" LastEditorUserId="2810" LastEditorDisplayName="" LastEditDate="2011-03-15T15:03:52.690" LastActivityDate="2011-03-15T15:03:52.690" />
  <row Id="27161" PostTypeId="2" ParentId="27157" CreationDate="2011-03-15T10:47:20.747" Score="1" ViewCount="" Body="&lt;p&gt;I assume you're trying to classify a stationary point by looking at the matrix of second derivatives? In that case, you can't draw any conclusion if the matrix is semidefinite, because what happens can be influenced by higher-order terms in the Taylor expansion of the function at the critical point. Investigation of what really happens must be done on a case-by-case basis, using whatever methods one might come up with, and since this goes beyond the simple standard methods, it's usually swept under the rug in introductory calculus courses.&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2011-03-15T10:47:20.747" />
  <row Id="27162" PostTypeId="2" ParentId="27151" CreationDate="2011-03-15T10:53:15.590" Score="5" ViewCount="" Body="&lt;p&gt;Here's another hint: If $X$ is an eigenvector of $A$, say $AX=\lambda X$, then you can use that to simplify $p(A)X$ into $(\text{some scalar value})X$, and that thing in front of $X$ is then an eigenvalue of $p(A)$, corresponding to the eigenvector $X$.&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2011-03-15T10:53:15.590" CommentCount="2" />
  <row Id="27163" PostTypeId="1" AcceptedAnswerId="27184" CreationDate="2011-03-15T11:07:18.950" Score="3" ViewCount="63" Body="&lt;p&gt;How the the monodromy group and deck transformation group of a covering $p\colon Y\rightarrow X$ are related in the cases when covering is branched and covering is unbranched?&lt;/p&gt;&#xA;" OwnerUserId="8186" LastActivityDate="2011-04-14T14:40:24.337" Title="Connection between two groups related to covering" Tags="&lt;algebraic-topology&gt;" AnswerCount="1" FavoriteCount="4" />
  <row Id="27164" PostTypeId="5" CreationDate="2011-03-15T11:10:47.260" Score="0" ViewCount="" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditorDisplayName="" LastEditDate="2011-03-15T11:10:47.260" LastActivityDate="2011-03-15T11:10:47.260" />
  <row Id="27165" PostTypeId="4" CreationDate="2011-03-15T11:10:47.260" Score="0" ViewCount="" Body="geometry as on the surface of a sphere, where &quot;lines&quot; are great circles and any pair of lines must intersect" OwnerUserId="-1" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2011-03-15T11:20:07.707" LastActivityDate="2011-03-15T11:20:07.707" />
  <row Id="27166" PostTypeId="5" CreationDate="2011-03-15T11:12:21.143" Score="0" ViewCount="" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditorDisplayName="" LastEditDate="2011-03-15T11:12:21.143" LastActivityDate="2011-03-15T11:12:21.143" />
  <row Id="27167" PostTypeId="4" CreationDate="2011-03-15T11:12:21.143" Score="0" ViewCount="" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditorDisplayName="" LastEditDate="2011-03-15T11:12:21.143" LastActivityDate="2011-03-15T11:12:21.143" />
  <row Id="27168" PostTypeId="5" CreationDate="2011-03-15T11:15:32.157" Score="0" ViewCount="" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditorDisplayName="" LastEditDate="2011-03-15T11:15:32.157" LastActivityDate="2011-03-15T11:15:32.157" />
  <row Id="27169" PostTypeId="4" CreationDate="2011-03-15T11:15:32.157" Score="0" ViewCount="" Body="shape, congruence, similarity, transformations, properties of classes of figures, points, lines, angles" OwnerUserId="-1" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2011-03-15T11:19:44.693" LastActivityDate="2011-03-15T11:19:44.693" />
  <row Id="27170" PostTypeId="5" CreationDate="2011-03-15T11:18:02.497" Score="0" ViewCount="" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditorDisplayName="" LastEditDate="2011-03-15T11:18:02.497" LastActivityDate="2011-03-15T11:18:02.497" />
  <row Id="27171" PostTypeId="4" CreationDate="2011-03-15T11:18:02.497" Score="0" ViewCount="" Body="a graph-theory graph is a collection of vertices and a collection of edges that connect pairs of vertices" OwnerUserId="-1" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2011-03-15T11:19:48.283" LastActivityDate="2011-03-15T11:19:48.283" />
  <row Id="27172" PostTypeId="5" CreationDate="2011-03-15T11:19:43.493" Score="0" ViewCount="" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditorDisplayName="" LastEditDate="2011-03-15T11:19:43.493" LastActivityDate="2011-03-15T11:19:43.493" />
  <row Id="27173" PostTypeId="4" CreationDate="2011-03-15T11:19:43.493" Score="0" ViewCount="" Body="calculus and analysis with functions of a complex variable" OwnerUserId="-1" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2011-03-15T11:32:22.173" LastActivityDate="2011-03-15T11:32:22.173" />
  <row Id="27174" PostTypeId="5" CreationDate="2011-03-15T11:21:24.147" Score="0" ViewCount="" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditorDisplayName="" LastEditDate="2011-03-15T11:21:24.147" LastActivityDate="2011-03-15T11:21:24.147" />
  <row Id="27175" PostTypeId="4" CreationDate="2011-03-15T11:21:24.147" Score="0" ViewCount="" Body="theoretical foundations of calculus: limits, convergence of sequences, construction of the real numbers, and other analysis topics on real numbers" OwnerUserId="-1" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2011-03-15T11:32:24.107" LastActivityDate="2011-03-15T11:32:24.107" />
  <row Id="27176" PostTypeId="2" ParentId="27136" CreationDate="2011-03-15T11:23:37.003" Score="2" ViewCount="" Body="&lt;p&gt;A more direct way of showing (a) (which doesn't involve the Jordan normal form) is to look at the second highest term in the characteristic polynomial&#xA;$$\det(\lambda I - A) = (\lambda - \lambda_1)(\lambda - \lambda_2) \dots (\lambda - \lambda_n).$$&#xA;When you expand the left-hand side using permutations and products of entries of $A$, you will get minus the sum of the diagonal entries of $A$ as the coefficient of $\lambda^{n-1}$, and when you multiply out the right-hand side, you will get minus the sum of the eigenvalues as the coefficient of $\lambda^{n-1}$.&lt;/p&gt;&#xA;" OwnerUserId="1242" LastEditorUserId="1242" LastEditorDisplayName="" LastEditDate="2011-03-15T12:27:43.977" LastActivityDate="2011-03-15T12:27:43.977" CommentCount="1" />
  <row Id="27177" PostTypeId="2" ParentId="26925" CreationDate="2011-03-15T11:49:44.077" Score="7" ViewCount="" Body="&lt;p&gt;Let's suppose that, one way or another, you know that $\int_{-\infty}^{\infty} e^{-x^2} \, dx = \sqrt{\pi}$. Then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_{-\infty}^{\infty} e^{2tx - x^2} = e^{t^2} \int_{-\infty}^{\infty} e^{-(t - x)^2} \, dx = e^{t^2} \sqrt{\pi}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_{-\infty}^{\infty} e^{2tx - x^2} = \int_{-\infty}^{\infty} \left( \sum_{n \ge 0} \frac{2^n t^n x^n}{n!} \right) e^{-x^2} = \sum_{n \ge 0} \frac{2^n t^n}{n!} \int_{-\infty}^{\infty} x^n e^{-x^2} \, dx.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, note that by evenness, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_{-\infty}^{\infty} x^{2n} e^{-x^2} \, dx = 2 \int_0^{\infty} x^{2n} e^{-x^2} \, dx.$$ &lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2011-03-15T12:50:50.170" LastActivityDate="2011-03-15T12:50:50.170" CommentCount="2" />
  <row Id="27178" PostTypeId="1" CreationDate="2011-03-15T11:52:53.067" Score="0" ViewCount="28" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have the following situation: $R,H$ schemes over a field $k$ which we can assume to be algebraically closed, with $H$ reduced, $Y\subset R\times \mathbb{P}_k^n$ an open subset, $p:Y\rightarrow R$ the restriction of the projection onto the first factor and $w: Y\rightarrow H$ a surjective formally smooth morphism. How can I show that $R$ is reduced?&#xA;Thank you&lt;/p&gt;&#xA;" OwnerUserId="6583" LastActivityDate="2011-03-15T11:52:53.067" ClosedDate="2011-03-15T17:08:14.820" Title="formal smoothness versus reductness" Tags="&lt;algebraic-geometry&gt;&lt;arithmetic-geometry&gt;" CommentCount="2" />
  <row Id="27179" PostTypeId="1" CreationDate="2011-03-15T12:06:49.530" Score="0" ViewCount="78" Body="&lt;p&gt;I'm trying to derive a result for the number of possible combinations of r objects from n, when we have unlimited numbers of objects to select from. (Wikipedia suggests that these are called multi-combinations). My current line of reasoning is like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose we can choose 4 letters from an unlimited set containing {A, B, C, D, E}. Then we can partition this into choices where 4 are identical, 3 are identical, 2 are identical and all are different, like so:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;combs like AAAA + combs like AAAB + combs like AABB + combs like AABC + combs like ABCD&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, it seems to me that to go this route, I'll have to start enumerating the partitions of r (here 4) to ensure that I'll get all possible combinations with r, r-1, r-2, ... objects. This looks tricky to me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can someone tell me a) if this is not a fruitful approach and b) suggest a hint (and no more than a hint please) as to a more straightforward method, as at the moment I don't see one ?&lt;/p&gt;&#xA;" OwnerUserId="7597" LastActivityDate="2011-04-14T12:40:23.060" Title="counting multi-combinations ?" Tags="&lt;combinatorics&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="27180" PostTypeId="2" ParentId="27179" CreationDate="2011-03-15T12:23:21.153" Score="1" ViewCount="" Body="&lt;p&gt;It makes a difference as to whether you are counting AAAB as distinct from BAAA, or as the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If they are distinct, the easiest way is to count how many ways you can choose the first element, how many the second, ... etc. and then produce a number for all the possible choices combined.  You can forget the patterns.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to treat AAAB and BAAA as being the same, then it gests slightly more complicated, and you have to identify each type of pattern AAAA, AAAB, AABB, AABC and ABCD (corresponding to partitions of 4), and then work out how many patterns fit each type.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an illustration of calculating how many ways of getting a pattern like AABB, you are choosing 0 letters four times, 0 letters three times, 2 letters twice, 0 letter one time and 3 letters 0 times so the number of ways is &#xA;$$ \frac{5!}{0! \, 0! \, 2! \, 0! \, 3!} = 10$$ &#xA;while if different orders make distinct patterns then you getter a larger number &#xA;$$ \frac{5!}{0! \, 0! \, 2! \, 0! \, 3!} \,  \frac{4!}{2! \, 2! \, 0! \, 0! \, 0!} = 60$$ &#xA;$$ &lt;/p&gt;&#xA;" OwnerUserId="6460" LastEditorUserId="6460" LastEditorDisplayName="" LastEditDate="2011-03-15T12:31:19.697" LastActivityDate="2011-03-15T12:31:19.697" CommentCount="3" />
  <row Id="27181" PostTypeId="1" AcceptedAnswerId="27187" CreationDate="2011-03-15T13:14:44.790" Score="1" ViewCount="58" Body="&lt;p&gt;Fourier series of  function f: &#xA; $$f(x)=\sum_{s=-\infty}^{\infty}f_{s}\exp(2\pi isx)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose $f_{s}\sim\frac{1}{s^{p}}$. What can we say about $f(x)$? Can we find some bounds for $f(x)$ like $f(x)&amp;lt;c\sum_{s}\frac{\exp(-2\pi isx)}{s^{p}}$? &lt;/p&gt;&#xA;" OwnerUserId="6052" LastActivityDate="2011-03-15T14:06:38.217" Title="Bounds for Fourier series" Tags="&lt;fourier-series&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="27182" PostTypeId="1" CreationDate="2011-03-15T13:37:17.970" Score="1" ViewCount="125" Body="&lt;p&gt;Hello,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When trying to reproduce the plot of &lt;a href=&quot;http://en.wikipedia.org/wiki/Tupper%27s_self-referential_formula&quot; rel=&quot;nofollow&quot;&gt;Tupper's self-referential formula&lt;/a&gt; (original paper is available &lt;a href=&quot;http://www.dgp.toronto.edu/people/mooncake/papers/SIGGRAPH2001_Tupper.pdf&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;) with &lt;a href=&quot;http://www.wolfram.com/mathematica/&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;Mathematica&lt;/em&gt;&lt;/a&gt; I have faced unexpected difficulties. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;First of all, &lt;a href=&quot;http://en.wikipedia.org/wiki/Tupper%27s_self-referential_formula&quot; rel=&quot;nofollow&quot;&gt;the algorithm&lt;/a&gt; does not give the image shown and I was forced to find the right way by myself (code for &lt;em&gt;Mathematica&lt;/em&gt;):&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;k = 960939379918958884971672962127852754715004339660129306651505519271\&#xA;7028023952664246896428421743507181212671537827706233559932372808741443\&#xA;0789132596394133772348785773574982392662971551717371699516523289053822\&#xA;1612403238855866184013235585136048828693337902491454229288667081096184\&#xA;4960917051834540678277315517054053816273809676025656250169814820834187\&#xA;8316384911559022561000365235137034387446184837873723819822484986346503\&#xA;3159410054974700593138339226497249461751545728366702369745461014655997\&#xA;933798537483143786841806593422227898388722980000748404719;&#xA;tb = Table[&#xA;   1/2 &amp;lt; Floor[&#xA;     Mod[Floor[y/17]*2^(-17 Floor[x] - Mod[Floor[y], 17]), 2]], {y, &#xA;    k + 17, k, -1}, {x, 106, 0, -1}];&#xA;g = Graphics[Raster[tb /. {True -&amp;gt; 0, False -&amp;gt; 1}], &#xA;  ImagePadding -&amp;gt; None, PlotRangePadding -&amp;gt; None]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;But even after this I get artefacts on the bottom of the image produced. Why this happens? Is this an error in the original description or in my mind? The correct image can be produced by the following code (where &lt;code&gt;sh&lt;/code&gt; may be any rational number of the form &lt;code&gt;1/n&lt;/code&gt;):&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;k = 960939379918958884971672962127852754715004339660129306651505519271\&#xA;7028023952664246896428421743507181212671537827706233559932372808741443\&#xA;0789132596394133772348785773574982392662971551717371699516523289053822\&#xA;1612403238855866184013235585136048828693337902491454229288667081096184\&#xA;4960917051834540678277315517054053816273809676025656250169814820834187\&#xA;8316384911559022561000365235137034387446184837873723819822484986346503\&#xA;3159410054974700593138339226497249461751545728366702369745461014655997\&#xA;933798537483143786841806593422227898388722980000748404719;&#xA;sh = 1;&#xA;tb = Table[&#xA;   1/2 &amp;lt; Floor[&#xA;     Mod[Floor[y/17]*2^(-17 Floor[x] - Mod[Floor[y], 17]), 2]], {y, &#xA;    k + 17 - sh, k, -sh}, {x, 106 - sh, 0, -sh}];&#xA;g = Graphics[Raster[tb /. {True -&amp;gt; 0, False -&amp;gt; 1}], &#xA;  ImagePadding -&amp;gt; None, PlotRangePadding -&amp;gt; None]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;And I also cannot decode the &lt;code&gt;k&lt;/code&gt; directly into an image in the right way. Here is my code:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;g=Graphics[Raster[Transpose@&#xA;(Partition[IntegerDigits[k/17,2]/.{1-&amp;gt;0,0-&amp;gt;1},17])],&#xA;ImagePadding-&amp;gt;None,PlotRangePadding-&amp;gt;None]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;What am I doing wrong?&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;EDIT:&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;I have found a way to decode the constant &lt;code&gt;k&lt;/code&gt;. The problem was that the original binary representation of the encoded image was truncated when converting to the number due to dropping leading zeros. We should also take into account that Tupper encoded black pixels as &quot;1&quot; and white pixels as &quot;0&quot;. So we need to pad the binary representation of &lt;code&gt;k&lt;/code&gt; with &quot;1&quot; at the start. Here is the solution (for &lt;em&gt;Mathematica&lt;/em&gt; 7+):&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Image[Transpose[&#xA;  Reverse@Partition[Reverse[1 - IntegerDigits[k/17, 2]], 17, 17, 1, &#xA;    1]], Magnification -&amp;gt; 4]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Here is also more elegant code for plotting Tupper's function:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Image[Table[&#xA;  1 - Boole[&#xA;    1/2 &amp;lt; Floor[&#xA;      Mod[Floor[y/17] 2^(-17 Floor[x] - Mod[Floor[y], 17]), 2]]], {y, &#xA;   k, k + 16}, {x, 105, 0, -1}], Magnification -&amp;gt; 4]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="7617" LastEditorUserId="7617" LastEditorDisplayName="" LastEditDate="2011-03-16T05:02:14.067" LastActivityDate="2011-03-16T05:02:14.067" Title="Tupper's self-referential formula with Mathematica" Tags="&lt;mathematica&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="27183" PostTypeId="2" ParentId="27156" CreationDate="2011-03-15T13:47:59.897" Score="3" ViewCount="" Body="&lt;p&gt;No. That's not even true over $\mathbb{R}$. For example the irreducible real representations of $C_3$ are the trivial one and a two-dimensional one, the sum of the conjugate 1-dimensional non-trivial complex representations. This two-dimensional representation is also defined over $\mathbb{Q}$.&lt;/p&gt;&#xA;" OwnerUserId="3212" LastActivityDate="2011-03-15T13:47:59.897" />
  <row Id="27184" PostTypeId="2" ParentId="27163" CreationDate="2011-03-15T13:48:45.670" Score="1" ViewCount="" Body="&lt;p&gt;I'm referring to &lt;a href=&quot;http://en.wikipedia.org/wiki/Monodromy&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Monodromy&lt;/a&gt; for the definition of monodromy group.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems to me, that the two coincides in the case of unbranchend Galois covering.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $p:\widetilde X \to X$ be a covering map, $p$ is Galois if $p_*(\pi_1(\widetilde X, \widetilde x_0))$ is a normal subgroup of $\pi_1(X,x_0)$. In this case we have the following description of the deck transformations:&#xA;$$&#xD;&#xA;\mbox{Aut}(p) = \pi(X, x_0)/p_*(\pi_1(\widetilde X, \widetilde x_0))&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To prove it, let the fundamental group act on every fiber. This way you obtain a map $\Phi_{[\gamma]}:\widetilde X \to \widetilde X$ for every $[\gamma] \in \pi_1(X,x_0)$, which turns out to be an automorphism of the covering.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This construction defines a surjective homomorphism $\pi(X, x_0) \to \mbox{Aut}(p)$ whose kernel is exactly $p_*(\pi_1(\widetilde X, \widetilde x_0))$. According to the wikipedia article, the same quotient is isomorph to the monodromy group.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe you can also prove it directly. The group of deck transformation preserves fibers, therefore you have a restriction homomorphism $\mbox{Aut}(p) \to \mbox{Mon}(p)$. You should be able to prove that this morphism is an isomorphism.&lt;/p&gt;&#xA;" OwnerUserId="8287" LastActivityDate="2011-03-15T13:48:45.670" />
  <row Id="27185" PostTypeId="2" ParentId="13662" CreationDate="2011-03-15T13:49:26.843" Score="0" ViewCount="" Body="&lt;p&gt;The probability of dying by 90 is the same for all four.  The probability that A dies first is simply 1/4 since we are give no more information.  Since the two events are independent, the probability of their conjunction, i.e., that A dies and is the first to die, is simply the product of the two probabilities, or 1/12.&lt;/p&gt;&#xA;" OwnerUserId="8288" LastActivityDate="2011-03-15T13:49:26.843" CommentCount="1" />
  <row Id="27186" PostTypeId="2" ParentId="27182" CreationDate="2011-03-15T13:53:56.817" Score="2" ViewCount="" Body="&lt;p&gt;At the risk of sounding gauche, I presume you've tried out the &lt;em&gt;Mathematica&lt;/em&gt; notebook &lt;a href=&quot;http://mathworld.wolfram.com/notebooks/RecreationalMath/TuppersSelf-ReferentialFormula.nb&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; and found it wanting?&lt;/p&gt;&#xA;" OwnerUserId="8289" LastActivityDate="2011-03-15T13:53:56.817" CommentCount="2" />
  <row Id="27187" PostTypeId="2" ParentId="27181" CreationDate="2011-03-15T14:06:38.217" Score="2" ViewCount="" Body="&lt;p&gt;The size of the Fourier coefficients of a function is related to the smoothness of the function, not to its size. The smoother the function (i.e. the more derivatives it has), the faster the convergence to $0$ of its Fourier coefficients.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $f$ is a periodic function of period $2\pi$, integrable on an interval of length $2\pi$, then, by the Riemann-Lebesgue theorem, $\hat f_s\to0$, where $\hat f_s$ is the $s$-th Fourier coefficient of $f$. If moreover $f$ is $k$ times differentiable and $f^{(k)}$ is integrable, then $|\hat f_s|\le C|s|^{-k}$ (even more, $|\hat f_s||s|^{k}\to0$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conversely, if $|\hat f_s|\le C|s|^{-k}$ for some $k\ge2$, then $f$ is continuous with $k-2$ continuous derivatives. But there is nothing you can say about the size of $f$. For instance, if $f=M$ is constant, then  $\hat f_s=0$ for all $s\ne0$.&lt;/p&gt;&#xA;" OwnerUserId="4791" LastActivityDate="2011-03-15T14:06:38.217" CommentCount="2" />
  <row Id="27188" PostTypeId="1" CreationDate="2011-03-15T14:07:25.910" Score="3" ViewCount="163" Body="&lt;p&gt;Hi, im doing a question that involves a shape with 8 faces, 10 vertices and 16 edges. Can anyone enlighten me as to what this shape is called?&#xA;Many Thanks&lt;/p&gt;&#xA;" OwnerUserId="8291" LastEditorUserId="2937" LastEditorDisplayName="" LastEditDate="2011-03-15T19:14:50.683" LastActivityDate="2011-03-15T21:07:11.693" Title="What shape is this?" Tags="&lt;homework&gt;&lt;polyhedra&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="1" />
  <row Id="27189" PostTypeId="2" ParentId="26925" CreationDate="2011-03-15T14:21:48.290" Score="11" ViewCount="" Body="&lt;p&gt;Alternatively, integration by parts works immediately.&lt;br&gt;&#xA;Let $$a_n=\int_0^\infty x^{2n}e^{-x^2}.$$  Consider $U=x^{2n-1}$ so that $du=(2n-1)x^{2n-2}$, and $dv=xe^{-x^{2}}$ so that $V=-\frac{1}{2}e^{-x^{2}}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then&#xA;$$\int_{0}^{\infty}x^{2n}e^{-x^{2}}dx=\frac{1}{2}e^{-x^{2}}x^{2n-1}\biggr|_{0}^{\infty}-\int_{0}^{\infty}(2n-1)x^{2n-2}\frac{-1}{2}e^{-x^{2}}dx$$ &#xA;$$=\frac{(2n-1)}{2}\int_{0}^{\infty}x^{2n-2}e^{-x^{2}}dx=\frac{(2n-1)2n}{2^{2}n}\int_{0}^{\infty}x^{2n-2}e^{-x^{2}}dx$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence $$a_n=\frac{(2n)(2n-1)}{2^2n}a_{n-1}$$ and since $a_0=\frac{\sqrt{\pi}}{2}$ we conclude $$a_n=\int_{0}^{\infty}x^{2n}e^{-x^{2}}dx=\frac{(2n)!}{2^{2n}n!}\frac{\sqrt{\pi}}{2}$$ by induction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope that helps,&lt;/p&gt;&#xA;" OwnerUserId="6075" LastActivityDate="2011-03-15T14:21:48.290" />
  <row Id="27190" PostTypeId="2" ParentId="27188" CreationDate="2011-03-15T14:24:57.447" Score="2" ViewCount="" Body="&lt;p&gt;The &lt;a href=&quot;http://mathworld.wolfram.com/TetragonalTrapezohedron.html&quot; rel=&quot;nofollow&quot;&gt;tetragonal trapezohedron&lt;/a&gt; seems to fit your bill.&lt;/p&gt;&#xA;" OwnerUserId="8289" LastActivityDate="2011-03-15T14:24:57.447" />
  <row Id="27191" PostTypeId="1" AcceptedAnswerId="27194" CreationDate="2011-03-15T14:25:57.257" Score="1" ViewCount="77" Body="&lt;p&gt;Hey,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I prove by definition(with epsilon and N) that the nth root of c (c&gt;0), when n tends to infinity, is 1?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="6163" LastEditorUserId="6163" LastEditorDisplayName="" LastEditDate="2011-03-15T14:37:05.063" LastActivityDate="2011-03-15T14:48:50.520" Title="Limits by definition" Tags="&lt;real-analysis&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="27192" PostTypeId="2" ParentId="27191" CreationDate="2011-03-15T14:31:17.097" Score="-1" ViewCount="" Body="&lt;p&gt;$$\lim_{n\to\infty}\exp\left(\frac{\ln(c)}{n}\right)=\exp\left(\ln(c)\lim_{n\to\infty}\frac1{n}\right)=\dots$$&lt;/p&gt;&#xA;" OwnerUserId="8289" LastActivityDate="2011-03-15T14:31:17.097" CommentCount="8" />
  <row Id="27194" PostTypeId="2" ParentId="27191" CreationDate="2011-03-15T14:40:47.907" Score="5" ViewCount="" Body="&lt;p&gt;Assume first $c&amp;gt;1$ and write $c^{1/n}=1+h_n$ with $h_n&amp;gt;0$. Taking $n$-th powers and using the binomial theorem get&#xA;$$&#xD;&#xA;c=(1+h_n)^n=1+nh_n+\dots&amp;gt;1+nh_n\implies 0&amp;lt;h_n&amp;lt;\frac{c-1}{n}.&#xD;&#xA;$$&#xA;Then&#xA;$$&#xD;&#xA;0&amp;lt;c^{1/n}-1&amp;lt;\frac{c-1}{n}.&#xD;&#xA;$$&#xA;You can use this inequality to apply the standard $\epsilon$-$N$ argument.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $0&amp;lt;c&amp;lt;1$, use $c^{1/n}=1/(1/c)^{1/n}$.&lt;/p&gt;&#xA;" OwnerUserId="4791" LastEditorUserId="4791" LastEditorDisplayName="" LastEditDate="2011-03-15T14:48:50.520" LastActivityDate="2011-03-15T14:48:50.520" CommentCount="4" />
  <row Id="27195" PostTypeId="1" CreationDate="2011-03-15T14:44:33.680" Score="0" ViewCount="48" Body="&lt;p&gt;I noticed that on a road, if the number of cars reaching a certain critical stage, all the cars would slow down immediately.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any theory/model that connects the number of cars $N$ with the average car speed $V$, and maybe with road width $R$  ( and other variables)?&lt;/p&gt;&#xA;" OwnerUserId="550" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2011-03-15T16:04:58.600" LastActivityDate="2011-03-15T16:04:58.600" Title="Modelling Traffic Congestion" Tags="&lt;mathematical-modeling&gt;&lt;traffic&gt;" CommentCount="2" />
  <row Id="27196" PostTypeId="1" CreationDate="2011-03-15T14:52:46.780" Score="0" ViewCount="91" Body="&lt;p&gt;If I know the following parameters how to find h co-ordinate of ellipse center &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1.Circle : center (0,0), radius = r &#xA;2.Ellipse : center (h,k), semi-major axis = a and semi-minor axis = b &#xA;3.I know that ellipse is tangent to circle. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I know all parameters, except h , what is the formula /equation find h ? &lt;/p&gt;&#xA;" OwnerUserId="8293" LastActivityDate="2011-05-14T15:57:30.570" Title="Ellipse : Equation to find &quot;h&quot; X co-ordinate of ellipse tangent to circle." Tags="&lt;conic-sections&gt;" AnswerCount="1" />
  <row Id="27197" PostTypeId="2" ParentId="27141" CreationDate="2011-03-15T14:53:23.960" Score="3" ViewCount="" Body="&lt;p&gt;I don't know if Bass--Serre theory is covered in your course, but it gives a slick proof.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $T$ be the Bass--Serre tree of the splitting $A*B$ and consider the action of $K$ on $T$.  Then $K$ acts freely.  For, if $g\in K$ fixes a point in $T$ then, by definition, $g$ is conjugate into $A$ or $B$, and so $\phi(g)\neq 1$ unless $g=1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, we have constructed a free action of $K$ on a tree $T$, so $K$ is the fundamental group of $T/K$, which is a graph.  Therefore $K$ is free.&lt;/p&gt;&#xA;" OwnerUserId="2287" LastActivityDate="2011-03-15T14:53:23.960" />
  <row Id="27198" PostTypeId="2" ParentId="27145" CreationDate="2011-03-15T14:56:35.210" Score="2" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\ $ By &lt;a href=&quot;http://math.stackexchange.com/questions/3698/why-doesnt-0-being-a-prime-ideal-in-z-imply-that-0-is-a-prime-number/3716#3716&quot;&gt;factoring out a prime ideal&lt;/a&gt; it reduces to showing that an Artinian &lt;em&gt;domain&lt;/em&gt; is a field, which follows immediately by DCC.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This method of factoring out by a prime ideal to reduce from rings to domains is a ubiquitous algebraic problem solving technique. Indeed, the eminent algebraist Irving Kaplansky explicitly pauses to mention this method (in a less trivial context) in his classic textbook &lt;em&gt;Commutative Rings&lt;/em&gt;. Follow the above link for an excerpt. Not only was Kap a great algebraist but also a great expositor - a rare combination. I wholeheartedly recommend his expositions, where I recall learning many beautiful mathematical &lt;em&gt;ideas&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2011-03-15T23:06:36.177" LastActivityDate="2011-03-15T23:06:36.177" CommentCount="2" />
  <row Id="27199" PostTypeId="1" AcceptedAnswerId="27207" CreationDate="2011-03-15T15:17:20.823" Score="6" ViewCount="170" Body="&lt;p&gt;This question requires finding the Cartesian equation for the locus: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$|z-1| = 2|z+1|$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;that is, where the modulus of $z -1$ is twice the modulus of $z-2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've solved this problem &lt;em&gt;algebraically&lt;/em&gt; (by letting z=x+iy) as follows:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$\sqrt{(x-1)^2 + y^2} = 2\sqrt{(x+1)^2 + y^2}$  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$(x-1)^2 + y^2 = 4\big((x+1)^2 + y^2\big)$  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$x^2 - 2x + 1 + y^2 = 4x^2 + 8x + 4 + 4y^2$ &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$3x^2 + 10x + 3y^2 = -3$  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$x^2 + \frac{10}{3}x + y^2 = -1$  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$(x + \frac{5}{3})^2 +y^2 = -1 + \frac{25}{9}$  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;therefore, $(x+\frac{5}{3})^2 + y^2 = \frac{16}{9}$, which is a circle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;However&lt;/em&gt;, I was wondering if there is a method, simply by inspection, of immediately concluding that the locus is a circle, based on some relation between the distance from $z$ to $(1,0)$ on the plane being twice the distance from $z$ to $(-1,0)$?&lt;/p&gt;&#xA;" OwnerUserId="6520" LastEditorUserId="6520" LastEditorDisplayName="" LastEditDate="2011-03-15T15:32:08.550" LastActivityDate="2011-03-15T23:17:57.480" Title="Equation of the complex locus: |z-1|=2|z +1|" Tags="&lt;algebra&gt;&lt;complex-numbers&gt;&lt;complex-geometry&gt;" AnswerCount="3" />
  <row Id="27200" PostTypeId="1" CreationDate="2011-03-15T15:18:43.327" Score="5" ViewCount="161" Body="&lt;p&gt;Suppose I have $G = Z_{p^m} \times Z_{p^n}$ for $m, n$ distinct natural numbers and $p$ a prime. Is there a combinatorial way to determine the number of automorphisms of $G$? &lt;/p&gt;&#xA;" OwnerUserId="8294" LastEditorUserId="8294" LastEditorDisplayName="" LastEditDate="2011-03-15T18:03:15.360" LastActivityDate="2011-03-30T13:35:12.580" Title="Number of automorphisms of a direct product of two cyclic $p$-groups" Tags="&lt;group-theory&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="27201" PostTypeId="2" ParentId="27196" CreationDate="2011-03-15T15:22:06.380" Score="0" ViewCount="" Body="&lt;p&gt;The circle is $x^2+y^2=r^2$, the ellipse $\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1$  Then taking implicit derivatives $y&amp;#39;=-\frac{x}{y}$ for the circle and $y&amp;#39;=-\frac{b^2(x-h)}{a^2(y-k)}$ for the ellipse.  You get another equation by equating them $\frac{x}{y}=\frac{b^2(x-h)}{a^2(y-k)}$  Pick your favorite pair to solve simultaneously, but it is a mess.  You expect four solutions (some may be complex) and tangency will happen when there is a double root.  &lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-03-15T15:22:06.380" CommentCount="1" />
  <row Id="27202" PostTypeId="1" CreationDate="2011-03-15T15:27:13.797" Score="2" ViewCount="127" Body="&lt;p&gt;I was reading &lt;a href=&quot;http://www.brentozar.com/archive/2011/03/consulting-lines-do-want-faster-faster-or-faster/?utm_source=feedburner&amp;amp;utm_medium=feed&amp;amp;utm_campaign=Feed%3a+BrentOzar-SqlServerDba+%28Brent+Ozar+-+SQL+Server+DBA%29&amp;amp;utm_content=Google+Feedfetcher&quot; rel=&quot;nofollow&quot;&gt;something&lt;/a&gt; today that was talking in terms of 10%, 100% and 1000% faster.  I assumed that 10% faster means it takes 10% less time (60 seconds down to 54 seconds).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If that is correct wouldn't 100% faster mean 0 time and 1000% mean traveling back in time?&lt;/p&gt;&#xA;" OwnerUserId="2490" LastActivityDate="2011-03-16T01:45:44.780" Title="What does &quot;X% faster&quot; mean?" Tags="&lt;logic&gt;" AnswerCount="3" CommentCount="3" />
  <row Id="27203" PostTypeId="1" CreationDate="2011-03-15T15:31:15.197" Score="0" ViewCount="107" Body="&lt;p&gt;&lt;a href=&quot;http://i.imgur.com/o4AAt.png&quot; rel=&quot;nofollow&quot;&gt;http://i.imgur.com/o4AAt.png&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could someone walk me through this derivative?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="8295" LastActivityDate="2011-03-15T16:36:36.200" Title="Need help understanding differentiation. Question and Answer provided." Tags="&lt;calculus&gt;" AnswerCount="1" />
  <row Id="27204" PostTypeId="2" ParentId="27203" CreationDate="2011-03-15T15:40:08.497" Score="2" ViewCount="" Body="&lt;p&gt;The solution you linked corresponds to a simple case of &lt;a href=&quot;http://www.cis.udel.edu/~breech/physics/physics-notes/node10.html&quot; rel=&quot;nofollow&quot;&gt;Leibnitz's Rule for Differentiating Integrals&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note, however, that there is an obvious typo in that solution: there should be $ - \frac{1}{k}Qf(\frac{Q}{k})$ instead of $ - \frac{1}{k}Q (\frac{Q}{k})$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: While the solution you linked is instructive, it is by far not the simplest one. In particular, note that&#xA;$$&#xD;&#xA;\frac{d}{{dQ}}\int_{Q/k}^\infty  {Qf(x)dx}  = \frac{d}{{dQ}}\bigg[Q\int_{Q/k}^\infty  {f(x)dx} \bigg] = \bigg[1 - F\bigg(\frac{Q}{k}\bigg)\bigg] + Q\frac{d}{{dQ}}\bigg[1 - F\bigg(\frac{Q}{k}\bigg)\bigg].&#xD;&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="2810" LastEditorUserId="2810" LastEditorDisplayName="" LastEditDate="2011-03-15T16:36:36.200" LastActivityDate="2011-03-15T16:36:36.200" />
  <row Id="27205" PostTypeId="2" ParentId="27202" CreationDate="2011-03-15T15:47:20.563" Score="8" ViewCount="" Body="&lt;p&gt;When I see numbers like that I don't believe they are honest numbers-they really mean &quot;faster&quot;, &quot;much faster&quot; and &quot;much much faster&quot;.  But taking them seriously, if you were traveling 60 mph, 10% faster would be 66 mph (taking 9.09% less time), 100% faster would be 120 mph (taking 50% less time) and 1000% faster would be 660 mph (taking 91.91% less time).&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-03-15T15:47:20.563" CommentCount="2" />
  <row Id="27206" PostTypeId="2" ParentId="27202" CreationDate="2011-03-15T15:50:27.253" Score="1" ViewCount="" Body="&lt;p&gt;Faster here is referring to the speed, it means that if you normalize $X$ as a number between $0$ and $1$, you will have that $$newSpeed = oldSpeed(1+X)$$ But faster doesn't mean that the time taken for the activity (this case is data processing or something like that) to complete is 0, or even negative, because the time is always some ratio of the form $t = \frac{d}{s}$, where $d$ would be the number of activities or data to process, an $s$ is the speed to complete one activity or process one data, so $t$ is always positive. But of course as X increases, the time tends to 0.&lt;/p&gt;&#xA;" OwnerUserId="1318" LastActivityDate="2011-03-15T15:50:27.253" />
  <row Id="27207" PostTypeId="2" ParentId="27199" CreationDate="2011-03-15T15:50:28.923" Score="5" ViewCount="" Body="&lt;p&gt;It is a classical theorem due to Apollonius. See &lt;a href=&quot;http://en.wikipedia.org/wiki/Circle#Apollonius_circle&quot; rel=&quot;nofollow&quot;&gt;Apollonius circle&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="8268" LastEditorUserId="8268" LastEditorDisplayName="" LastEditDate="2011-03-15T15:55:36.720" LastActivityDate="2011-03-15T15:55:36.720" CommentCount="1" />
  <row Id="27208" PostTypeId="2" ParentId="27200" CreationDate="2011-03-15T15:50:55.787" Score="7" ViewCount="" Body="&lt;p&gt;I'm assuming that $p$ is a prime. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume $m\geq n$, let $x$ be a generator of the first cyclic group, and $y$ a generator of the second cyclic group.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Method 1.&lt;/strong&gt; An endomorphism of $G$ is completely determined by the images of $x$ and $y$:&#xA;\begin{align*}&#xA;x &amp;amp;\mapsto x^a y^b\\&#xA;y &amp;amp;\mapsto x^c y^d&#xA;\end{align*}&#xA;Since the image of $y$ must have order dividing $p^n$, the order of $x^c$ must divide $p^n$, so $c$ must be a multiple of $x^{p^{m-n}}$. So in fact, we can write&#xA;\begin{align*}&#xA;x &amp;amp;\mapsto x^a y^b\\&#xA;y &amp;amp;\mapsto x^{cp^{m-n}} y^d&#xA;\end{align*}&#xA;$a$ is determined up to congruence modulo $p^m$; $b$ and $d$ are determined modulo $p^n$; and $c$ is determined up to congruence modulo $p^{n}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This can be &quot;coded&quot; with a matrix:&#xA;$$\left(\begin{array}{cc}&#xD;&#xA;a &amp;amp; b\\&#xD;&#xA;cp^{m-n} &amp;amp; d&#xD;&#xA;\end{array}\right)$$&#xA;where the first column is taken modulo $p^n$ and the second column is taken modulo $p^n$. An easy computation shows that composition of these endomorphisms corresponds to matrix multiplication, and that matrix multiplication is well-defined (thanks to that factor of $p^{m-n}$ in the bottom left).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The matrix is invertible if and only if $ad - cdp^{m-n}$ is relatively prime to $p$. So the endomorphism is an automorphism if and only if $\gcd(ad - cdp^{m-n},p) = 1$. Note again that $a$ is determined modulo $p^m$, while $b$, $c$, and $d$ are determined modulo $p^n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $m\gt n$, then the endomorphism is an automorphism if and only if both $a$ and $d$ are relatively prime to $p$, so the number of endomorphisms is $\phi(p^n)\phi(p^m)p^{2n}$: $\phi(p^m)$ choices for $a$, $\phi(p^n)$ choices for $d$, and arbitrary choices for $c$ and $d$ modulo $p^n$. This does not seem to match your computations. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $m=n$, then you are simply looking for invertible matrices over $\mathbb{Z}_{p^m}$, and there are well-known formulas for that, but you say you are not interested, so I don't need to count them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Method 2.&lt;/strong&gt; Every map into $\mathbb{Z}_{p^m}\times\mathbb{Z}_{p^n}$ is determined by a pair of maps, one into $\mathbb{Z}_{p^m}$ and one into $\mathbb{Z}_{p^n}$. Every map from $\mathbb{Z}_{p^{m}}\times\mathbb{Z}_{p^n}$ is determined by a pair of maps, one from $\mathbb{Z}_{p^m}$ and one from $\mathbb{Z}_{p^n}$. So every endomorphism of $\mathbb{Z}_{p^m}\times\mathbb{Z}_{p^n}$ is determined by a $4$-tuple of maps, $(h_{mm},h_{mn},h_{nm},h_{nn})$, with $h_{ij}\colon\mathbb{Z}_{p^i}\to\mathbb{Z}_{p^j}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The number of possible $4$-tuples is then $(p^m)(p^n)^3 = p^{m+3n}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The map corresponding to the $4$-tuple $(h_{mm},h_{mn},h_{nm},h_{nn})$ is an automorphism if and only if $h_{mm}$ and $h_{nn}$ are both automorphisms (here we are using the fact that $m\gt n$; if $m=n$, then this claim does not hold). Thus, the number of choices is reduced: there are $\phi(p^m)$ choices for $h_{mm}$ and $\phi(p^n)$ choices for $h_{nn}$. However, $h_{mn}$ and $h_{nm}$ are free, each of them with $p^n$ choices. So the total number is $\phi(p^m)\phi(p^n)p^{2n}$, as before.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Summary.&lt;/strong&gt; If $n\neq m$, the number of automorphisms is $\phi(p^m)\phi(p^n)p^{2\min(m,n)}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Coda.&lt;/strong&gt; The general case of a direct product of finite groups is treated in&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;J.N.S. Bidwell, M.J. Curran, and D.J. McCaughan, &lt;em&gt;Automorphisms of direct products of finite groups&lt;/em&gt;, Arch. Math. (Basel) &lt;strong&gt;86&lt;/strong&gt; (2006) no. 6, 481-489, MR 2241597 (2007e:20047)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;(I had figured out the case of cyclic $p$-groups for a paper from scratch before I was given this reference). The main theorem is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Theorem.&lt;/strong&gt; Let $H$ and $K$ be finite groups with no common direct factor. Then $\mathrm{Aut}(H\times K) \cong\mathcal{A}$, where $\mathcal{A}$ is the group of all $2\times 2$ matrices&#xA;$$\left(\begin{array}{cc}&#xD;&#xA;\alpha &amp;amp; \beta\\&#xD;&#xA;\gamma &amp;amp; \delta&#xD;&#xA;\end{array}\right)$$&#xA;with $\alpha\in\mathrm{Aut}(H)$, $\delta\in\mathrm{Aut}(K)$, $\beta\in\mathrm{Hom}(H,Z(K))$, $\gamma\in\mathrm{Hom}(K,Z(H))$, and in particular&#xA;$$|\mathrm{Aut}(H,K)| = |\mathrm{Aut}(H)||\mathrm{Aut}(K)||\mathrm{Hom}(H,Z(K))||\mathrm{Hom}(K,Z(H))|.$$&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-15T18:13:58.597" LastActivityDate="2011-03-15T18:13:58.597" CommentCount="4" />
  <row Id="27209" PostTypeId="1" AcceptedAnswerId="27211" CreationDate="2011-03-15T15:53:37.517" Score="1" ViewCount="43" Body="&lt;p&gt;I'm intested in learning how to get a sum in difference calculus.  I'm willing to settle, at least for now, in learning how to integrate the following integral in calculus:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\int(a^x(1-a^x)(1-a^x))dx$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sum that I'm really interested in learning how to get, hopefully using difference calculus, is:&#xA;$\sum_x{a^x(1-a^x)(1-a^x)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even better still, extend this technique to include more $(1-a^x)$, and add limits of integration or summation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;_&lt;em&gt;EDIT&lt;/em&gt;_&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd like to do so without multiplying everything out, if possible&lt;/p&gt;&#xA;" OwnerUserId="2626" LastEditorUserId="2626" LastEditorDisplayName="" LastEditDate="2011-03-15T16:08:09.543" LastActivityDate="2011-03-15T16:08:09.543" Title="Integral or sum of $a^x(1-a^x)(1-a^x)$" Tags="&lt;calculus&gt;&lt;finite-differences&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="27210" PostTypeId="2" ParentId="27150" CreationDate="2011-03-15T15:58:29.963" Score="4" ViewCount="" Body="&lt;p&gt;The issue of when projective modules are free is discussed in $\S 3.5.4$ of &lt;a href=&quot;http://www.math.uga.edu/~pete/integral.pdf&quot; rel=&quot;nofollow&quot;&gt;my commutative algebra notes&lt;/a&gt;.  In particular one gets very easy (but not very satisfying) examples by looking at disconnected rings: e.g. $\mathbb{C} \times \{0\}$ is quite clearly projective but not free over $\mathbb{C} \times \mathbb{C}$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is more interesting to ask for examples over domains.  One huge class of examples comes from invertible fractional ideals which are not principal.  In particular a Dedekind domain possesses such ideals iff it is not a PID.  This is worked out much later in my notes in the section on Dedekind domains (currently $\S 22$, but this is subject to change).  However, in $\S 3.5.4$ I take some time to exhibit from scratch a specific projective, nonfree module over $R = \mathbb{Z}[\sqrt{-5}]$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that free implies torsionfree holds for modules over a &lt;em&gt;domain&lt;/em&gt; $R$.  (In fact the usual definition of &quot;torsionfree module&quot; is only useful over domains, so far as I know.  I seem to recall that Lam's book &lt;em&gt;Lectures on modules and rings&lt;/em&gt; gives a more sophisticated definition for modules over a general ring...)  To be more precise, over any commutative ring $R$, free $\implies$ projective $\implies$ flat, and over a domain flat $\implies$ torsionfree.  So any example of a projective nonfree module over a domain is also an example of a torsionfree nonfree module.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Over a general domain there is a vast gap between torsion free modules and flat modules.  For instance, a prime ideal $\mathfrak{p}$ in a Noetherian domain is always a torsionfree module, but it is flat only if it has height at most one.  However, a torsionfree module over a PID is flat -- see e.g. $\S 3.11$ of my notes -- but need not be free.  For instance, let $R$ be any PID which is not a field and let $K$ be its field of fractions.  Then $K$ is a nontrivial torsionfree $R$-module which is divisible, hence not free.&lt;/p&gt;&#xA;" OwnerUserId="299" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2011-03-15T16:08:06.537" LastActivityDate="2011-03-15T16:08:06.537" CommentCount="1" />
  <row Id="27211" PostTypeId="2" ParentId="27209" CreationDate="2011-03-15T15:58:50.007" Score="2" ViewCount="" Body="&lt;p&gt;Well, integration is easy: just multiply everything out to get a sum where each term has the form $a^{kx} = e^{(k \ln a)x}$, then integrate term by term.&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2011-03-15T15:58:50.007" CommentCount="1" />
  <row Id="27212" PostTypeId="2" ParentId="2949" CreationDate="2011-03-15T16:04:52.637" Score="4" ViewCount="" Body="&lt;p&gt;The one result that puzzled me most is from the ACM's communication ... &quot;Puzzled&quot; section by Peter Winkler:&#xA;&quot;We are in a large rectangular room with mirrored walls, while elsewhere in the same room is our mortal enemy, armed with a laser gun. Our only defense is our ability to summon graduate students to stand at designated spots in the room blocking all possible shots by the enemy. How many students would we need, assuming for the purposes of the problem that we, our enemy, and the students are all thin enough to be considered points?&quot;&#xA;The answer is 16. I still didn't do the calculation end-to-end, though a buddy of mine did it and got the result. What i find most puzzling is that the trajectory may turn dense, but still has 16 such points.&lt;/p&gt;&#xA;" OwnerUserId="5878" LastActivityDate="2011-03-15T16:04:52.637" CommunityOwnedDate="2011-03-15T16:04:52.637" />
  <row Id="27213" PostTypeId="2" ParentId="27150" CreationDate="2011-03-15T16:08:49.220" Score="1" ViewCount="" Body="&lt;p&gt;An ideal $I$ of an integral domain $R$ is a free $R$-module iff it is generated by one element.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a Dedekind domain every non-zero ideal is invertible, thus projective. However Dedekind domains are usually not principal ideal domains - algebraic number theory yields many examples.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This way one gets torsion-free, finitely generated, projective modules that are not free.&lt;/p&gt;&#xA;" OwnerUserId="2479" LastActivityDate="2011-03-15T16:08:49.220" />
  <row Id="27214" PostTypeId="2" ParentId="1817" CreationDate="2011-03-15T16:11:57.970" Score="0" ViewCount="" Body="&lt;p&gt;Every Tychonoff space (and consequently every compact metric space) X can be embedded into the topological power $[0,1]^C$ (the functions from C to [0,1] with the usual product topology, i.e., the pointwise convergence).&#xA;If you work in $[0,1]^C$, then you can define the &quot;limit&quot; of a sequence $(f_n)$ by putting $L(f_n)(c)=\mathcal{U}-lim f_n(c)$, i.e. you can work pointwise. This limit is obviously shift-invariant and every convergent sequence is convergent in this sense, too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I do not like calling such a thing Banach limit, since it is different from what you have seen in the reals. The usual Banach limit respects the usual linear structure of reals. If you use this construction, then this structure is lost by arbitrarily choosing the embedding into the cube $[0,1]^C$.&lt;/p&gt;&#xA;" OwnerUserId="8297" LastActivityDate="2011-03-15T16:11:57.970" />
  <row Id="27215" PostTypeId="2" ParentId="27188" CreationDate="2011-03-15T16:13:43.087" Score="5" ViewCount="" Body="&lt;p&gt;Additionally, any polyhedron represented at this link could also be the shape you're looking for:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.uwgb.edu/dutchs/symmetry/poly8f2.htm&quot; rel=&quot;nofollow&quot;&gt;http://www.uwgb.edu/dutchs/symmetry/poly8f2.htm&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1782" LastActivityDate="2011-03-15T16:13:43.087" CommentCount="1" />
  <row Id="27216" PostTypeId="1" AcceptedAnswerId="27221" CreationDate="2011-03-15T16:17:04.720" Score="1" ViewCount="58" Body="&lt;p&gt;Let $k$ be a commutative ring, and $A$ be a $k$-module. Does $\mathrm{End}(A) \cong A \otimes_k A^*$ hold in general? It holds for finite dimensional vector spaces, but I'm not sure how to construct an element of $A \otimes_k A^*$ that corresponds to $\mathrm{id}_A$ in general, without picking a basis.&lt;/p&gt;&#xA;" OwnerUserId="3793" LastEditorUserId="3793" LastEditorDisplayName="" LastEditDate="2011-03-15T16:42:33.293" LastActivityDate="2011-03-15T17:12:49.930" Title="Is $\mathrm{End}(A) \cong A \otimes_k A^*$ for module $A$ over $k \in \mathsf{Comm}$" Tags="&lt;linear-algebra&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="27217" PostTypeId="1" AcceptedAnswerId="27220" CreationDate="2011-03-15T16:18:47.057" Score="0" ViewCount="74" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;here is a fomula: angle=arctan(dy/dx)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can find an angle with my calculator for any value except dx=0,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;my question is: Is there no angle or , there is something that says when dx==0 the angle=?...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm really sorry for my bad english.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;thank you very much.&lt;/p&gt;&#xA;" OwnerUserId="8009" LastActivityDate="2011-03-15T16:29:01.827" Title="angle=arctan(dy/dx), what happens when dx=0" Tags="&lt;trigonometry&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" />
  <row Id="27218" PostTypeId="1" AcceptedAnswerId="27293" CreationDate="2011-03-15T16:19:58.743" Score="4" ViewCount="320" Body="&lt;p&gt;Hey again,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using only the delta definition of a limit, how can we prove that the sequence $\{a_n\}$, where $a_n = \sin n$, as $n$ tends to infinity does not have a limit?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="6163" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-16T03:06:28.823" LastActivityDate="2011-03-16T03:06:28.823" Title="Prove that the limit of $\sin n$ as $n \rightarrow \infty$ does not exist." Tags="&lt;calculus&gt;&lt;real-analysis&gt;&lt;limit&gt;" AnswerCount="3" CommentCount="5" FavoriteCount="1" />
  <row Id="27219" PostTypeId="2" ParentId="26888" CreationDate="2011-03-15T16:23:20.267" Score="2" ViewCount="" Body="&lt;p&gt;The idea is to find a sequence of pairwise disjoint sets $F_n$ in $\mathcal F_{n + 1} \setminus \mathcal F_n$ and then prove that $F_n$ must actually be an element of $\mathcal F_n$ if $\mathcal F = \bigcup_n \mathcal F_n$ were a $\sigma$-algebra.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First pick a partition $N_1, N_2,\ldots$ of infinite sets of $\mathbb N$ and set&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$A_i = \bigcup_{n \in N_i} F_n.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, by assumption $A_i$ is an element of $\mathcal F$. So, for every $i$ we have an $n_i$ such that $A_i \in \mathcal F_{n_i}$. Now, note that since $(\mathcal F_n)$ is strictly increasing, so we can choose $(n_i)$ to be strictly increasing as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Okay, nice. So now for every $j$ we can pick an integer $m_j \in N_j$ such that $m_j &amp;gt; n_j$. So now we set&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$B = \bigcup_n \, F_{m_n} \in \mathcal F.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, eventually (for large $n$) $B$ will belong to all $\mathcal F_n$. So, eventually $B$ is in some $\mathcal F_{n_k}$(since $n_i$ is strictly increasing). By construction we now have that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$A_k \cap B = A_{m_k} \in \mathcal F_{n_k}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But! $n_k &amp;lt; m_k$ so $A_{m_k} \notin \mathcal F_{n_k}$. Contradiction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, now what is left is to show that such a disjoint sequence exists. Well, that is quite messy, so I will only add it if you want.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to add that this is not my idea but that I have got the idea from somewhere in the past, but I forgot exactly where from.&lt;/p&gt;&#xA;" OwnerUserId="1120" LastEditorUserId="5593" LastEditorDisplayName="" LastEditDate="2011-03-16T12:51:34.517" LastActivityDate="2011-03-16T12:51:34.517" CommentCount="13" />
  <row Id="27220" PostTypeId="2" ParentId="27217" CreationDate="2011-03-15T16:25:19.923" Score="1" ViewCount="" Body="&lt;p&gt;If $dx$ is $0$, if you look in the Cartesian plane, you are standing in the y-axis, so the angle would be $\frac{\pi}{2}$ or $\frac{3\pi}{2}$, depending if $dy &amp;gt; 0$ or $dy &amp;lt; 0$.&lt;/p&gt;&#xA;" OwnerUserId="1318" LastActivityDate="2011-03-15T16:25:19.923" />
  <row Id="27221" PostTypeId="2" ParentId="27216" CreationDate="2011-03-15T16:25:34.573" Score="5" ViewCount="" Body="&lt;p&gt;Notice that $\operatorname{End}(A)$ is never zero, unless $A$ is zero: the identity map is then a not-zero element. So one way to break the isomorphism is to pick a $k$ and $A$ so that the right hand side is zero. One way to do that is to have $A^*$ zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example: let $k=\mathbb Z$ and let $A=\mathbb Z/2\mathbb Z$. As there is no linear map $\mathbb Z/2\mathbb Z\to\mathbb Z$, we have $A^*=0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The isomorphism &lt;em&gt;does&lt;/em&gt; work when $A$ is a finitely generated projective, though.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Later&lt;/strong&gt;. This is treated, for example, in F. W. Anderson and K. R. Fuller's &lt;em&gt;Rings and categories of modules&lt;/em&gt;, where it is Proposition 20.10. The general statement is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Let $R$, $S$ be two rings and let ${}_SP$, ${}_SU_T$ ${}_TN$ be modules. Then there is a $\mathbb Z$-linear map $$\phi_{P,T,N}:\hom_S(P,U)\otimes_TN\to\hom_S(P,U\otimes_TN)$$ which is natural in $P$, $U$ and $N$ (and in $R$ and $S$, in the appropriate sense!) which is an isomorphism if $P$ is a finitely generated projective. The map $\phi$ is uniquely determined by the condition that $$\phi_{P,T,N}(f\otimes n)(p)=f(p)\otimes n$$ for all $f\in\hom_S(P,U)$, $n\in N$ and $p\in P$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You obtain the map in the question as the inverse of what you get from this statement by taking $S=T=k$, $P=N=A$ and $U=k$. The proof of the statement is easy. One first checks that for each choice of $P$, $U$, $N$ there is indeed a map $_{P,T,N}$ satisfying the condition mentioned in the last sentence, and that it is in fact natural; this is straightforward but a bit boring. If now $P$ is finitely generated and projective, there exists a ${}_SP&amp;#39;$ such that $P\oplus P&amp;#39;$ is finitely generated and free, isomorphic to $S^n$ for some $n\geq0$. Since the map $\phi$ is a natural transformation between &lt;em&gt;additive&lt;/em&gt; functors of $P$, so show that $\phi_{P,T,N}$ is an isomorphism it is enough to show that $\phi_{P\oplus P&amp;#39;,T,N}$, which is conjugate to $\phi_{S^n,T,N}$. Again, additivity implies that to show that $\phi_{S^n,T,N}$ is an isomorphism it is enough to show that the map $\phi_{S,T,N}$ is an isomorphism. And the latter is trivial.&lt;/p&gt;&#xA;" OwnerUserId="274" LastEditorUserId="274" LastEditorDisplayName="" LastEditDate="2011-03-15T17:12:49.930" LastActivityDate="2011-03-15T17:12:49.930" CommentCount="2" />
  <row Id="27222" PostTypeId="2" ParentId="27217" CreationDate="2011-03-15T16:29:01.827" Score="1" ViewCount="" Body="&lt;p&gt;Actually, the $angle$ that a point in $R^2$ with cartesian coordinates $(x,y)\neq (0,0)$ makes with the positive x-axis is defined as the value $\theta (x,y)$ such that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\cos \theta (x,y) = \frac{x}{\sqrt{x^2+y^2}}$ and &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sin \theta (x,y) = \frac{y}{\sqrt{x^2+y^2}}$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with $\theta\in [0, 2\pi)$. Note that, defined in this way, there is a unique $\theta$ for each point different from $(0,0)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $x&amp;gt;0$ and $y&amp;gt;0$, these formulas combined give $\theta (x,y) = \arctan(y/x)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $x=0$ and $y&amp;gt;0$, it follows from the definition that $\theta=\pi/2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can also consider $\theta \in[-\pi,\pi)$, and in this case the formula $\theta (x,y) = \arctan(y/x)$ is valid for all $y$ provided $x&amp;gt;0$.&lt;/p&gt;&#xA;" OwnerUserId="1178" LastActivityDate="2011-03-15T16:29:01.827" />
  <row Id="27223" PostTypeId="2" ParentId="27122" CreationDate="2011-03-15T16:51:51.163" Score="6" ViewCount="" Body="&lt;p&gt;Just to elaborate slightly on Qiaochu's answer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $X \subset \mathbb P^n$ is a projective variety, then the given embedding of $X$ in $\mathbb P^n$ equips $X$ with extra structure: a line bundle $\mathcal L$ (the restriction of $\mathcal O(1)$), and sections $x_0,\ldots,x_n \in \Gamma(&#xD;&#xA;X,\mathcal L)$ (the restriction of the sections $x_i$ of $\mathcal O(1)$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conversely, given an abstract variety $X$, equipped with a line bundle $\mathcal L$ and sections $x_0,\ldots,x_n$ of $\Gamma(X,\mathcal L)$ satisfying appropriate axioms (see e.g. Chapter II, Section 7 of Hartshorne), one obtains a projective embedding.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The homogeneous coordinate ring of $X$ depends functorially on the data&#xA;$\bigr(X,\mathcal L, (x_i)_{i= 0,\ldots,n}\bigr).$  But of course, this is more data than just $X$ alone.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a concrete example, think about the difference between the homogeneous coordiante ring of $\mathbb P^1$ embedded via the identity map into itself,&#xA;and embedded as a smooth conic in $\mathbb P^2$.&lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2011-03-15T16:51:51.163" />
  <row Id="27224" PostTypeId="2" ParentId="27218" CreationDate="2011-03-15T16:52:07.030" Score="5" ViewCount="" Body="&lt;p&gt;Assume $\lim \sin(n) = l$. Then so is $\lim \sin(2n) = l$. So $\lim \cos(2n) = 1 - 2l^2$, but so does the limit of $\cos(2(n + 1))$. Now apply the sum-formula to $\sin(2(n + 1) - 2n)$.&lt;/p&gt;&#xA;" OwnerUserId="1120" LastEditorUserId="1120" LastEditorDisplayName="" LastEditDate="2011-03-15T23:11:44.283" LastActivityDate="2011-03-15T23:11:44.283" CommentCount="4" />
  <row Id="27225" PostTypeId="2" ParentId="27142" CreationDate="2011-03-15T16:53:14.737" Score="1" ViewCount="" Body="&lt;p&gt;I will suppose in the following that you have a finitely generated $k$-algebra $A$ over a field $k$ and an $A$-module $M$ which is finite dimensional over $k$. If this is indeed the situation you have in mind, please do edit the question to make it explicit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $m\geq0$ We want to show that the set $G^A_m(M)$ of $A$-submodules of $M$ is a closed subset of the grassmanian $G^k_m(M)$ of $m$-dimensional vector subspaces.&#xA;If $a_1$, $\dots$, $a_n$ generate $A$ over $k$ as an algebra, then an element $U\in G^k_m(M)$ is in $G^A_m(M)$ iff $a_i(U)\subseteq U$ for all $i\in\{1,\dots,n\}$. Since the intersection of closed sets is closed, it follows immediately that it is enough to show the following:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If $M$ is a $k$-vector space and $f:M\to M$ is a linear map, then the  $$\{U\in G^k_m(M):f(U)\subseteq U\}$$ is a closed subset of $G^k_m(M)$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Can you see how to show this last statement?&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2011-03-15T16:53:14.737" CommentCount="2" />
  <row Id="27226" PostTypeId="2" ParentId="27122" CreationDate="2011-03-15T17:04:31.597" Score="2" ViewCount="" Body="&lt;p&gt;Here's a slightly tangential elaboration on the above two answers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $X = \mathrm{Proj} A$ for $A$ a noetherian ring. So here $A$ should fade into the background, but $X$ is the important thing; it turns out that the $\mathrm{Proj}$ depends only on the asymptotic part of $A$ in high dimensions. Then we can consider the category of coherent sheaves on $\mathrm{Proj} A$. The sheaf of regular functions is the canonical example.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a functor from finitely graded $A$-modules to coherent sheaves on $\mathrm{Proj} A$, denoted by a tilde. (This is as in Hartshorne II.5.) Unlike the case for an affine scheme, however, it is &lt;em&gt;not&lt;/em&gt; an equivalence of categories. It is not too hard to show that the functor is essentially surjective, since given a coherent sheaf $\mathcal{F}$ on $\mathrm{Proj} A$ you can construct the graded $A$-module $\bigoplus \Gamma(X, \mathcal{F}(n))$ of the global sections of all the twists. Then one can show using elementary methods that this recovers $\mathcal{F}$. (In fact, this is true even if $A$ is not noetherian and $\mathcal{F}$ is simply assumed quasi-coherent.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;However,&lt;/em&gt; if we have a graded $A$-module $M$, we cannot recover $M$ from the associated sheaf $\tilde{M}$. For instance, if $M$ has finitely many nonzero components, then the homogeneous localizations $M_{(f)}$ vanish (for any homogeneous element $f$), so the associated sheaf is zero. But $M$ may not be zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This suggests that if we consider the abelian category of finitely generated graded $A$-modules, and quotient out by the Serre class of modules that are &quot;asymptotically zero,&quot; then we will get the category of coherent sheaves on $\mathrm{Proj} A$. This is in fact true, though the proof uses a bit of cohomological machinery (see EGA III.2).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the moral of the story is that the $\mathrm{Proj}$ sees only asymptotic data. If we have  a morphism $A \to A&amp;#39;$ that induces an isomorphism in large enough degrees, then $\mathrm{Proj} A, \mathrm{Proj} A&amp;#39;$ will be isomorphic (as one example).&lt;/p&gt;&#xA;" OwnerUserId="536" LastActivityDate="2011-03-15T17:04:31.597" />
  <row Id="27227" PostTypeId="1" AcceptedAnswerId="27229" CreationDate="2011-03-15T17:05:04.990" Score="1" ViewCount="89" Body="&lt;pre&gt;&#xA;Given X and Y=a+bX, I have to prove that:&#xA;&#xA;if b &amp;lt; 0&#xA;  rho = -1&#xA;if b &amp;gt; 0&#xA;  rho = 1&#xA;&#xA;I've gotten to the point where I have:&#xA;&#xA;rho = b * sigma_x / sqrt(sigma_y^2)&#xA;&#xA;I've need to find why sigma_y^2 = b^2 * sigma_x^2.&#xA;&#xA;Can anyone please explain to me why that is the case?&#xA;&#xA;I tried rewriting sigma_y^2 as: E[(a+bX)(a+bX)] - E^2(a+bX)&#xA;And got: a^2 + 2abE(x) + b^2E(x^2) - a^2 - b^2E^2(X)&#xA;Which became: 2abE(x) + b^2 * sigma_x^2&#xA;&#xA;What is the extra 2abE(x) term?  Is that supposed to go away somehow?&#xA;&#xA;Thanks.&#xA;&lt;/pre&gt;&#xA;" OwnerUserId="8299" LastActivityDate="2011-03-15T17:15:06.277" Title="Proving correlation coefficient = 1 or -1 given X and Y=a + bX" Tags="&lt;statistics&gt;&lt;average&gt;&lt;correlation&gt;" AnswerCount="1" />
  <row Id="27228" PostTypeId="2" ParentId="21808" CreationDate="2011-03-15T17:11:05.063" Score="1" ViewCount="" Body="&lt;p&gt;The important identity that is missing from the Google Books preview is $(a \otimes b) v = (b \cdot v)a$.&#xA;Using this, and looking at the image of a vector $v$ under $S(a \otimes b)$ you get&#xA;$$\begin{aligned}&#xD;&#xA;S(a \otimes b) v &amp;amp;= S((b \cdot v)a)\\&#xD;&#xA;&amp;amp;= (b \cdot v) (Sa)\\&#xD;&#xA;&amp;amp;= ((Sa) \otimes b) v&#xD;&#xA;\end{aligned}$$&#xA;Hence $S(a \otimes b) = (Sa) \otimes b$.&lt;/p&gt;&#xA;" OwnerUserId="1016" LastActivityDate="2011-03-15T17:11:05.063" />
  <row Id="27229" PostTypeId="2" ParentId="27227" CreationDate="2011-03-15T17:15:06.277" Score="1" ViewCount="" Body="&lt;p&gt;${\rm Var}(Y) = {\rm Var}(a + bX) = {\rm Var}(bX) = b^2 {\rm Var(X)}$.&lt;/p&gt;&#xA;" OwnerUserId="2810" LastActivityDate="2011-03-15T17:15:06.277" />
  <row Id="27230" PostTypeId="1" CreationDate="2011-03-15T17:19:24.557" Score="0" ViewCount="52" Body="&lt;p&gt;I am working on a project regarding the number of zeros of a harmonic polynomial and am stuck with the proof of this: The zero set of the harmonic polynomial  h(z)= $z^n$ - $\bar{z}^n$ consists of n equally spaced lines through the origin. &#xA;Any suggestion is appreciated. Thanks.&lt;/p&gt;&#xA;" OwnerUserId="8301" LastActivityDate="2011-03-15T17:24:18.593" Title="Harmonic polynomial" Tags="&lt;analysis&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="27231" PostTypeId="2" ParentId="27230" CreationDate="2011-03-15T17:24:18.593" Score="1" ViewCount="" Body="&lt;p&gt;if $z=re^{i\theta}$ then this reduces to $e^{2ni\theta}=1$.  the solutions to this are $\theta=\pi k/n$.  so the lines $\{re^{k\pi i/n} : r\in(-\infty,\infty)\}, k=0,1,...n-1$ are the zero set.&lt;/p&gt;&#xA;" OwnerUserId="6925" LastActivityDate="2011-03-15T17:24:18.593" CommentCount="1" />
  <row Id="27232" PostTypeId="1" AcceptedAnswerId="27235" CreationDate="2011-03-15T17:24:26.190" Score="0" ViewCount="100" Body="&lt;p&gt;I am trying to integrate $x^3(x^2+1)^5$ and I believe the starting point is to utilize long division to break up the polynomial, however I am unsure how to divide this out.  Could someone give me what exactly I'm suppose to be dividing?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: Well the reason I ask is if you attempt to use wolframalpha.com, they state to use long division to simplify the problem and I don't see how that's possible.  My first thought was it was a parts problem.&lt;/p&gt;&#xA;" OwnerUserId="8302" LastEditorUserId="8302" LastEditorDisplayName="" LastEditDate="2011-03-15T17:34:33.850" LastActivityDate="2011-03-15T18:22:30.680" Title="Polynomial Long Division" Tags="&lt;integral&gt;" AnswerCount="3" CommentCount="7" />
  <row Id="27233" PostTypeId="2" ParentId="27232" CreationDate="2011-03-15T17:39:30.310" Score="0" ViewCount="" Body="&lt;p&gt;I would use integration by parts.  Technically you need to figure out the factors of this polynomial to know what possible parts are, though you don't actually need polynomial division to do that.&lt;/p&gt;&#xA;" OwnerUserId="7845" LastActivityDate="2011-03-15T17:39:30.310" />
  <row Id="27234" PostTypeId="2" ParentId="27218" CreationDate="2011-03-15T17:44:43.163" Score="2" ViewCount="" Body="&lt;p&gt;The following are true, based on standard trigonometric identities and $\sin(1) \approx 0.84147$ and $\sin(3) \approx 0.14112$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\textrm{if } \sin(n) \le -0.4  \textrm{ then } 0 &amp;lt; \sin(n+3) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\textrm{if } -0.4 \le \sin(n) \le 0.4  \textrm{ then } \sin(n+1) &amp;lt; -0.4 \textrm{ or} 0.4 &amp;lt; \sin(n+1) $$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\textrm{if } 0.4 \le \sin(n) \textrm{ then } \sin(n+3) &amp;lt; 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so there is no value $L$ where for any positive $\epsilon &amp;lt; 0.2$ you have all of $\sin(n)$, $\sin(n+1)$, $\sin(n+3)$, and $\sin(n+4)$ within $\epsilon$ of $L$.     &lt;/p&gt;&#xA;" OwnerUserId="6460" LastEditorUserId="1120" LastEditorDisplayName="" LastEditDate="2011-03-15T17:49:13.937" LastActivityDate="2011-03-15T17:49:13.937" />
  <row Id="27235" PostTypeId="2" ParentId="27232" CreationDate="2011-03-15T17:47:02.803" Score="2" ViewCount="" Body="&lt;p&gt;It's actually not that bad if we just expand the binomial and multiply through by $x^3$. &lt;br&gt;&lt;br&gt;&#xA;$\int {x^3(x^2+1)^5dx} = \int {x^3(x^{10} + 5x^8 + 10x^6 + 10x^4 + 5x^2 + 1)dx} =$&lt;br&gt;&lt;br&gt;$ \int {(x^{13} + 5x^{11}+10x^9+10x^7+5x^5+x^3)dx}$&lt;br&gt;&lt;br&gt;&#xA;And now you can integrate term by term. &lt;/p&gt;&#xA;" OwnerUserId="7850" LastActivityDate="2011-03-15T17:47:02.803" CommentCount="2" />
  <row Id="27237" PostTypeId="2" ParentId="13026" CreationDate="2011-03-15T18:12:46.570" Score="1" ViewCount="" Body="&lt;p&gt;What is a language for you?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The canonical definition is that any set $L \subseteq \Sigma^*$ with $\Sigma$ an alphabet (i.e. finite set of symbols) is a language. There are extensions to infinite, but countable $\Sigma$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we assume either of those definitions, any problem on real numbers can not be represented as language, since plenty of real numbers do not have finite representations (in countable universes). That means you can not even encode your inputs properly. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general terms, any problem that is defined on a non-countable is not representable as language (in the above terms).&lt;/p&gt;&#xA;" OwnerUserId="3330" LastActivityDate="2011-03-15T18:12:46.570" />
  <row Id="27238" PostTypeId="2" ParentId="27232" CreationDate="2011-03-15T18:22:30.680" Score="1" ViewCount="" Body="&lt;p&gt;You can also use a trig substitution here. Let $x=\tan(\theta)$. Then $x^2 + 1 = \tan^2\theta + 1 = \sec^2\theta$, $x^3 = \tan^3\theta$, and $dx = \sec^2\theta\,d\theta$. You get&#xA;\begin{align*}&#xA;\int x^3(x^2+1)^5\,dx &amp;amp;= \int \tan^3\theta\sec^{12}\theta\,d\theta\\&#xA;&amp;amp;= \int \tan\theta(\sec^2\theta-1)\sec^{12}\theta\,d\theta\\&#xA;&amp;amp;= \int\tan\theta\sec\theta \sec^{13}\theta\,d\theta - \int\tan\theta\sec\theta\sec^{11}\theta\,d\theta\\&#xA;&amp;amp;= \frac{1}{14}\sec^{14}\theta - \frac{1}{12}\sec^{12}\theta + C\\&#xA;&amp;amp;= \frac{1}{14}(1+x^2)^7 - \frac{1}{12}(1+x^2)^6 + C.&#xA;\end{align*}&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2011-03-15T18:22:30.680" CommentCount="3" />
  <row Id="27239" PostTypeId="1" AcceptedAnswerId="27255" CreationDate="2011-03-15T18:40:16.813" Score="1" ViewCount="92" Body="&lt;p&gt;It seems to be very simple question, but I'm afriad that I am missing something.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ok, so you have a basket with 1 ball- which can be black or white with an equal proability.&#xA;Then you insert to the basket white ball, and by coincidental selection you take out white ball out of the basket.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the probability that the remaining ball is white?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="6163" LastEditorUserId="6163" LastEditorDisplayName="" LastEditDate="2011-03-15T19:01:22.770" LastActivityDate="2011-03-15T19:46:38.830" Title="probability question" Tags="&lt;probability&gt;" AnswerCount="3" FavoriteCount="1" />
  <row Id="27241" PostTypeId="1" CreationDate="2011-03-15T18:46:58.240" Score="2" ViewCount="68" Body="&lt;p&gt;In the polynomial algebra $\mathbb{C}[X_1, X_2,\ldots, X_n]$, we define a set of symmetric polynomials as follows $h_i(X_k, X_{k+1}, \ldots, X_n)$ = Sum of all monomials of total degree $i$ in the set of variables $X_k, X_{k+1}, \ldots X_n$; $\forall i,k = 1,2, \ldots n$. Let $\sigma_i$ be elementary symmetric polynomial of degree $i$ in the variables $X_i$'s ie $\sigma_i = \Sigma_{t_1 &amp;lt; t_2 \ldots &amp;lt; t_i} X_{t_1}X_{t_2}\ldots X_{t_i}$. Then it is claimed in the book 'Algorithms of invariant theory' of Bernd Strurmfels page No 12 that $h_k(X_k, X_{k + 1}, \ldots X_n) + \Sigma_{i = 1}^k (-1)^i h_{k-i}(X_k, X_{k + 1}, \ldots X_n) \sigma_i = 0$. I am not able to see its proof. Please help me.&lt;/p&gt;&#xA;" OwnerUserId="7565" LastEditorUserId="7565" LastEditorDisplayName="" LastEditDate="2011-03-15T19:16:47.600" LastActivityDate="2011-03-15T19:40:28.117" Title="An identity on symmetric polynomial" Tags="&lt;combinatorics&gt;&lt;commutative-algebra&gt;" AnswerCount="1" />
  <row Id="27242" PostTypeId="2" ParentId="27239" CreationDate="2011-03-15T18:50:17.397" Score="3" ViewCount="" Body="&lt;p&gt;There are four equally likely possibilities:&#xA;&lt;br&gt;a) Ball 1 is white, and you remove ball 1 from the basket.&#xA;&lt;br&gt;b) Ball 1 is black, and you remove ball 1 from the basket.&#xA;&lt;br&gt;c) Ball 1 is white, and you remove ball 2 from the basket.&#xA;&lt;br&gt;d) Ball 1 is black, and you remove ball 2 from the basket.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We rule out case b), since the removed ball was white. The other three cases are equally likely, and in two of them (a) and c)) the remaining ball is white. The answer is thus 2/3.&lt;/p&gt;&#xA;" OwnerUserId="5656" LastActivityDate="2011-03-15T18:50:17.397" />
  <row Id="27243" PostTypeId="2" ParentId="27200" CreationDate="2011-03-15T18:50:33.463" Score="5" ViewCount="" Body="&lt;p&gt;$\def\ZZ{\mathbb{Z}}$&#xA;It's worth pointing out that Arturo's answer 2 generalizes very nicely to all abelian $p$-groups: Consider $G = \bigoplus \ZZ/p^{\lambda_i}$; let $\mu_{\ell}$ be the number of $\lambda$'s which are equal to $\ell$; let $r$ be the number of summands of $G$ (so $r = \sum \mu_i$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then $\mathrm{End}(G) \cong \bigoplus \mathrm{Hom}(\ZZ/p^{\lambda_i}, \ZZ/p^{\lambda_j}) \cong \bigoplus \ZZ/p^{\min(\lambda_i, \lambda_j)}$ as abelian groups.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An element of $\mathrm{End}(G)$ is invertible if and only if its image in $\mathrm{End}(G/pG)$ is invertible. (Nakayama's lemma.) Now, $G/pG \cong (\ZZ/p)^r$, so its endomorphism ring is $\mathrm{Mat}_{r \times r}(\ZZ/p)$. The map $\mathrm{End}(G) \to \mathrm{End}(G/pG)$ is NOT surjective. Rather, the image is the block-upper-triangular matrices, where the sizes of the blocks are the $\mu_i$. Such a matrix is invertible iff and only if the diagonal blocks are invertible. So the fraction of $\mathrm{End}(G)$ which is made up of invertible elements is &#xA;$$\prod_{\ell} \frac{|\mathrm{GL}_{\mu_{\ell}}(\ZZ/p)|}{|\mathrm{Mat}_{\mu_{\ell} \times \mu_{\ell}}(\ZZ/p)|} = \prod_{\ell} \frac{(p^{\mu_\ell}-1)(p^{\mu_\ell}-p) \cdots (p^{\mu_\ell} - p^{\mu_\ell -1})}{p^{\mu_{\ell}^2}}$$&#xA;$$=\prod_{\ell} \left( 1- p^{-1} \right) \left( 1-p^{-2} \right) \cdots \left( 1-p^{-\mu_{\ell} +1} \right).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Putting it all together, $|\mathrm{Aut}(G)|$ is found by multiplying the above formula by $|\mathrm{End}(G)|$ (computed in the second paragraph), and we get&#xA;$$|\mathrm{Aut}(G)| = p^{\sum_{i,j} \min(\lambda_i, \lambda_j)} \prod_{\ell} \left( 1- p^{-1} \right) \left( 1-p^{-2} \right) \cdots \left( 1-p^{-\mu_{\ell} +1} \right).$$&lt;/p&gt;&#xA;" OwnerUserId="448" LastEditorUserId="448" LastEditorDisplayName="" LastEditDate="2011-03-30T13:35:12.580" LastActivityDate="2011-03-30T13:35:12.580" CommentCount="1" />
  <row Id="27244" PostTypeId="2" ParentId="26985" CreationDate="2011-03-15T18:54:43.397" Score="3" ViewCount="" Body="&lt;p&gt;So the question can be marked off as answered...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't think there can be a &quot;purely functorial&quot; proof that a subgroup of a free group is free (that is, a proof using just the universal property of the free group). If there were such a proof, one would naturally expect that it can be applied to work for the relatively free groups in any variety of groups. But the only varieties of groups in which subgroups of free groups are always free are the variety of all groups, the variety of all abelian groups, and the varieties of abelian groups of prime exponent. So this argues strongly against the existence of such a proof.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As long as I'm writing this as an answer, I'll note that the technical name for such varieties is &lt;em&gt;Schreier varieties.&lt;/em&gt; That is, a variety $\mathfrak{V}$ of algebras (in the sense of universal algebra) is said to be a &lt;em&gt;Schreier variety&lt;/em&gt; if and only if every subalgebra of a free $\mathfrak{V}$-algebra is itself a free $\mathfrak{V}$-algebra. The proof that the only Schreier varieties of groups are the ones listed above is due to Peter Neumann and James Wiegold in &lt;em&gt;Schreier varieties of groups&lt;/em&gt;, Math. Z. &lt;strong&gt;85&lt;/strong&gt; (1964) 392-400. An alternate proof was given by Peter Neumann and Mike Newman, &lt;em&gt;On Schreier varieties of groups&lt;/em&gt;, Math. Z. &lt;strong&gt;98&lt;/strong&gt; (1967) 196-199. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A proof can also be found in Hanna Neumann's book &lt;strong&gt;Varieties of Groups&lt;/strong&gt;, in section 4.3.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-15T19:06:11.500" LastActivityDate="2011-03-15T19:06:11.500" CommunityOwnedDate="2011-03-15T18:54:43.397" />
  <row Id="27245" PostTypeId="1" AcceptedAnswerId="27263" CreationDate="2011-03-15T18:55:04.750" Score="1" ViewCount="70" Body="&lt;p&gt;Let's define the primarithm function, $pog : \mathbb{N} \rightarrow \mathbb{N}$, where $pog(n)$ is the largest number of distinct primes that can divide a natural number $k$, $k \leq n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does this function have a real name and, more importantly, is there a good estimating function that approaches $pog(n)$ as $n$ gets large?&lt;/p&gt;&#xA;" OwnerUserId="6272" LastActivityDate="2011-03-15T20:34:17.357" Title="Estimating the Primarithm" Tags="&lt;number-theory&gt;&lt;prime-numbers&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="27246" PostTypeId="1" CreationDate="2011-03-15T18:56:25.940" Score="2" ViewCount="86" Body="&lt;p&gt;Let $\mathbb{P}_{\mathbb{C}}$ be the set of &lt;a href=&quot;http://en.wikipedia.org/wiki/Gaussian_prime#As_a_unique_factorization_domain&quot; rel=&quot;nofollow&quot;&gt;Gaussian primes&lt;/a&gt; and $\mathbb{P}_{\mathbb{N}}$ the set of primes in $\mathbb{N}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $\pi_{\mathbf{C}}(\sqrt{n})$ be the number of Gaussian primes with norm $\leq \sqrt{n}$ and $\pi_{\mathbf{N}}(n)$ be, as usual, the number of primes $\leq n$ in $\mathbb{N}$. Recall that norm($x+iy$)=N($x+iy$)=$x^2 +y^2$; hence, my taking of a square root above.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am interested to know what the order of magnitude is for $$\frac{\pi_{\mathbf{N}}(n)}{\pi_{\mathbf{C}}(\sqrt{n})}$$i.e. Has the extension of the definition of primes increased/decreased the relative density of primes with respect to their set of definition? A rather quixotic question could be &quot; Is there a general asymptotic for the number of primes in an arbitrary infinite field with the definition of being a prime as usual?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Fact&lt;/strong&gt;: &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Prime numbers of the form $4n + 3$ are also Gaussian primes.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Gauss%27s_circle_problem&quot; rel=&quot;nofollow&quot;&gt;Gauss's circle problem&lt;/a&gt; which asks for the number of Gaussian integers with norm less than a given value is presently unresolved. I think this is tangentially related to the asymptotic I am looking for. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="5595" LastActivityDate="2011-03-15T20:45:50.323" Title="Relative density of primes under extension" Tags="&lt;number-theory&gt;&lt;complex-analysis&gt;&lt;prime-numbers&gt;" AnswerCount="2" CommentCount="5" />
  <row Id="27247" PostTypeId="2" ParentId="27245" CreationDate="2011-03-15T19:04:24.847" Score="3" ViewCount="" Body="&lt;p&gt;The &lt;a href=&quot;http://oeis.org/A002110&quot; rel=&quot;nofollow&quot;&gt;primorial function&lt;/a&gt; gives the product of the first $n$ primes.  Yours is the inverse of this.  Primorial grows as about $n^n$.  There are some references in OEIS.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-03-15T19:04:24.847" CommentCount="1" />
  <row Id="27248" PostTypeId="2" ParentId="27246" CreationDate="2011-03-15T19:06:18.567" Score="2" ViewCount="" Body="&lt;p&gt;I don't think you're asking the question you wanted to ask. First, if you're using $\sqrt{n}$ then it seems more natural to define the norm as the square root of what you've defined it. Second, either way the contribution from primes congruent to $3 \bmod 4$ is negligible, so you're basically only counting the contributions from the primes congruent to $1 \bmod 4$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you should be asking for is the relative density of primes congruent to $1 \bmod 4$ and primes congruent to $3 \bmod 4$, and it is known that both of these are $\frac{1}{2}$ by &lt;a href=&quot;http://en.wikipedia.org/wiki/Dirichlet%27s_theorem_on_arithmetic_progressions&quot; rel=&quot;nofollow&quot;&gt;Dirichlet's theorem&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A very general statement which might answer your follow-up question (in what sense is this quixotic? This is an extremely natural question to ask) is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Chebotarev%27s_density_theorem&quot; rel=&quot;nofollow&quot;&gt;Chebotarev density theorem&lt;/a&gt;. &lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2011-03-15T19:06:18.567" CommentCount="6" />
  <row Id="27249" PostTypeId="2" ParentId="27246" CreationDate="2011-03-15T19:08:01.687" Score="4" ViewCount="" Body="&lt;p&gt;I'm going to answer a different question, when $\sqrt{n}$ is replaced by $n$. The answer is that it goes to one!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can see this by noting that rational primes are evenly split (in terms of asymptotic density) between $4n+1$ and $4n+3$; this is a slight strengthening of the usual Dirichlet density theorem. Primes of the former type split into two primes in the Gaussian integers, while primes of the latter remain prime. The norms of the former are the same as the prime $4n+1$; the norms of the latter are $(4n+3)^2$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you take asymptotics by considering Gaussian primes whose norm is less than $N$ for $N$ large, only the former case matters ($(4n+3)^2 \gg 4n+3$ for $n$ large), and since there are roughly $N/(2 \log N)$ rational primes of the form $4n+1$ in this range, each of which splits into two distinct Gaussian primes, we get that the number of Gaussian primes of norm less than $N$ is roughly $N/\log N$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, though, this is true for general number fields. This is an &lt;a href=&quot;http://en.wikipedia.org/wiki/Landau_prime_ideal_theorem&quot; rel=&quot;nofollow&quot;&gt;extension&lt;/a&gt; of the usual prime number theorem, and follows by essentially similar arguments: the point is that one can define the &lt;a href=&quot;http://en.wikipedia.org/wiki/Dedekind_zeta_function&quot; rel=&quot;nofollow&quot;&gt;Dedekind zeta function&lt;/a&gt; which has the analogous properties of the Riemann-zeta function, and following the same proof of the prime number theorem, one can show that in a number field, the number of prime ideals of norm at most $N$ is asymptotically $N/\log N$. (Note that the primes that completely split over $\mathbb{Q}$ are the only ones that count asymptotically in the zeta-function; this is the generalization of the statement I made about the primes $4n+3$ not contributing much.)&lt;/p&gt;&#xA;" OwnerUserId="536" LastEditorUserId="536" LastEditorDisplayName="" LastEditDate="2011-03-15T20:45:50.323" LastActivityDate="2011-03-15T20:45:50.323" CommentCount="6" />
  <row Id="27250" PostTypeId="1" CreationDate="2011-03-15T19:15:12.637" Score="5" ViewCount="70" Body="&lt;p&gt;Let $1&amp;lt;p&amp;lt;\infty$, and define an isometry of normed linear spaces to be a norm-preserving surjection. Then all isometries from $\ell^p_n(\mathbb{R})$ to itself are given by linear transformations $T$ such that Mat$(T)$ is product of a permutation matrix and a diagonal matrix with $1$'s and $-1$'s along the diagonal. However it is slightly more complicated to classify all isometries from $\ell^p_n(\mathbb{C})$ to itself since $z\mapsto \overline{z}$ is an isometry of $\mathbb{C}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Question: What are all isometries from $\ell^p_n(\mathbb{C})$ to itself?&lt;/p&gt;&#xA;" OwnerUserId="8305" LastActivityDate="2011-03-15T19:15:12.637" Title="Isometries of $\ell^p_n(\mathbb{C})$" Tags="&lt;functional-analysis&gt;" FavoriteCount="2" />
  <row Id="27251" PostTypeId="2" ParentId="27239" CreationDate="2011-03-15T19:18:33.737" Score="2" ViewCount="" Body="&lt;p&gt;One can also use that fact that the probability the remaining ball is white is $$P(\text{remaining ball is white}) = 1-P(\text{remaining ball is black})$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ = 1- \frac{1}{3} = \frac{2}{3}$$ &lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2011-03-15T19:18:33.737" />
  <row Id="27252" PostTypeId="1" AcceptedAnswerId="27368" CreationDate="2011-03-15T19:21:09.553" Score="1" ViewCount="95" Body="&lt;p&gt;A linear system of equations of the form&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\begin{cases} ux+(u+d)y=u+2d\\&#xD;&#xA;ax+(a+i)y=a+2i\end{cases}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Will always have the solution $x=-1, y=2$ (easily proven through matrix algebra). How can I prove this by &lt;em&gt;induction&lt;/em&gt;?&lt;/p&gt;&#xA;" OwnerUserId="5877" LastActivityDate="2011-03-16T09:07:10.027" Title="Proof by induction for systems of equations with arithmetic progression" Tags="&lt;induction&gt;" AnswerCount="2" CommentCount="8" />
  <row Id="27253" PostTypeId="2" ParentId="27241" CreationDate="2011-03-15T19:22:21.843" Score="1" ViewCount="" Body="&lt;p&gt;(Edit:) This follows from the generating function identities&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$E(t) = \sum_{i \ge 0} \sigma_i t^i = (1 + t X_1)(1 + t X_2)...(1 + t X_n)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$H(t) = \sum_{i \ge 0} h_i t^i = \frac{1}{(1 - t X_k)(1 - t X_{k+1})...(1 - t X_n)}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The desired identity follows from comparing the coefficient of $t^k$ of both sides of the identity &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$H(t) E(-t) = (1 - t X_1)...(1 - t X_{k-1}).$$&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2011-03-15T19:40:28.117" LastActivityDate="2011-03-15T19:40:28.117" CommentCount="3" />
  <row Id="27254" PostTypeId="2" ParentId="27252" CreationDate="2011-03-15T19:28:02.373" Score="1" ViewCount="" Body="&lt;p&gt;We can prove this using the induction scheme that says it's true for all $u,d,a,i$ if it's true for $u=d=a=i=1$ and also the (four) induction step(s) given it is true for $(u,d,a,i)$ then it also holds for $(u+1,d,a,i)$,  $(u,d+1,a,i)$, $(u,d,a+1,i)$ and $(u,d,a,i+1)$.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The base case will be $a,u,d,i = 1$ $$\begin{cases} x+2y=3\\&#xD;&#xA;x+2y=3\end{cases}$$ clearly the solution holds for this case.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;In all four cases the induction hypothesis will be $$\begin{cases} ux+(u+d)y=u+2d\\&#xD;&#xA;ax+(a+i)y=a+2i\end{cases}$$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I will only show one induction step because the other three are so similar&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We want to prove&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{cases} (u+1)x+(u+1+d)y=u+1+2d\\&#xD;&#xA;ax+(a+i)y=a+2i\end{cases}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;by algebra that's the same as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{cases} ux+(u+d)y+x+y=u+2d+1\\&#xD;&#xA;ax+(a+i)y=a+2i\end{cases}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and since $x+y = 1$ this reduces to our hypothesis.&lt;/p&gt;&#xA;" OwnerUserId="4290" LastActivityDate="2011-03-15T19:28:02.373" />
  <row Id="27255" PostTypeId="2" ParentId="27239" CreationDate="2011-03-15T19:46:38.830" Score="3" ViewCount="" Body="&lt;p&gt;Excellent question. A quintessence of a problem on Bayes posterior probability.&#xA;Total probability for evidence to happen (for white ball to be removed):&#xA;$$P(E)=\frac{1}{2}\cdot 1+\frac{1}{2}\cdot \frac{1}{2}=\frac{3}{4}$$&#xA;Prior hypothesis probability (for the first ball to be white):&#xA;$$P(H)=\frac{1}{2}$$&#xA;The probability for the evidence to happen if the hypothesis is true (for white ball to be removed if the first one was white):&#xA;$$P(E|H)=1$$&#xA;Posterior probability for the hypothesis $H$ in the presence of the evidence $E$:&#xA;$$P(H|E)=\frac{P(E|H)\cdot P(H)}{P(E)}=\frac{1 \cdot \frac{1}{2}}{\frac{3}{4}}=2/3$$&lt;/p&gt;&#xA;" OwnerUserId="6554" LastActivityDate="2011-03-15T19:46:38.830" />
  <row Id="27256" PostTypeId="2" ParentId="27150" CreationDate="2011-03-15T20:00:55.790" Score="2" ViewCount="" Body="&lt;p&gt;If $k$ is a field and $A = \operatorname{Mat}_n(k) = \operatorname{End}_k(V)$ is the ring of $n\times n$ matrices with coefficients in $k$, then the (simple, left) $A$-module $V = k^n$ of &quot;column vectors&quot; is a projective $A$-module, but it is not free as an $A$-module as soon as $n&amp;gt;1$ (this non-freeness is easy to see by comparing the dimension of $V$ and of $A$ as $k$-vector spaces).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To see the fact that $V$ is projective $A$-module, fix a basis $b_1,\dots,b_n$ for $V$ as a vector space, and for each $j=1,\dots,n$, let $e_j \in A$ be the &quot;idempotent&quot; given by&#xA;projection on the 1 dimensional linear subspace of $V$ spanned by $b_j$. Form the left ideal $I_j = Ae_j$. The mapping&#xA;$$X \mapsto Xb_j$$&#xA;defines an $A$-module isomorphism $I_j \to V$, and it is straightforward to check that&#xA;$$A = \bigoplus_{j=1}^n I_j,$$&#xA;so indeed $V \simeq I_1$ is a direct summand of a free $A$-module, hence projective.&lt;/p&gt;&#xA;" OwnerUserId="5020" LastEditorUserId="5020" LastEditorDisplayName="" LastEditDate="2011-03-15T20:34:38.010" LastActivityDate="2011-03-15T20:34:38.010" />
  <row Id="27257" PostTypeId="2" ParentId="27142" CreationDate="2011-03-15T20:08:18.607" Score="0" ViewCount="" Body="&lt;p&gt;Well, this was exactly my question: how to get a set of equations, if I have a condition that $f(U) \subseteq U,$ the only idea I can think of is that the matrix of f must have some zero blocks, but this is not an equation in terms of points of Grassmannian. So my question was exactly in the last statement.&lt;/p&gt;&#xA;" OwnerUserId="8307" LastActivityDate="2011-03-15T20:08:18.607" CommentCount="2" />
  <row Id="27258" PostTypeId="1" AcceptedAnswerId="27261" CreationDate="2011-03-15T20:11:35.653" Score="2" ViewCount="77" Body="&lt;p&gt;How can one tell that a given sequence is perodic modulo m?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example its easy to see that the sequence $1^1, 2^2, 3^3, \dots$ is periodic modulo 10. But how can we prove this?&lt;/p&gt;&#xA;" OwnerUserId="8308" LastActivityDate="2011-03-15T20:57:44.760" Title="Prove that a given sequence is periodic modulo m" Tags="&lt;sequences-and-series&gt;&lt;modular-arithmetic&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" />
  <row Id="27259" PostTypeId="1" AcceptedAnswerId="27260" CreationDate="2011-03-15T20:14:37.650" Score="0" ViewCount="311" Body="&lt;p&gt;How can I prove that a 3x3 system of linear equations of the form:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\begin{pmatrix}&#xD;&#xA;a&amp;amp;a+b&amp;amp;a+2b\\&#xD;&#xA;c&amp;amp;c+d&amp;amp;c+2d\\&#xD;&#xA;e&amp;amp;e+f&amp;amp;e+2f&#xD;&#xA;\end{pmatrix}&#xD;&#xA;\begin{pmatrix}&#xD;&#xA;x\\ y\\ z&#xD;&#xA;\end{pmatrix}&#xD;&#xA;=\begin{pmatrix}&#xD;&#xA;a+3b\\&#xD;&#xA;c+3d\\&#xD;&#xA;e+3f&#xD;&#xA;\end{pmatrix}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for $a,b,c,d,e,f \in \mathbb Z$ will always have infinite solutions and will intersect along the line &#xA;$ r=&#xD;&#xA;\begin{pmatrix}&#xD;&#xA; -2\\3\\0&#xD;&#xA;\end{pmatrix}&#xD;&#xA;+\lambda&#xD;&#xA;\begin{pmatrix}&#xD;&#xA; 1\\-2\\1&#xD;&#xA;\end{pmatrix}$&lt;/p&gt;&#xA;" OwnerUserId="5877" LastEditorUserId="1543" LastEditorDisplayName="" LastEditDate="2011-05-03T10:02:50.330" LastActivityDate="2011-05-03T10:02:50.330" Title="Prove a 3x3 system of linear equations with arithmetic progression coefficients has infinitely many solutions" Tags="&lt;linear-algebra&gt;" AnswerCount="2" />
  <row Id="27260" PostTypeId="2" ParentId="27259" CreationDate="2011-03-15T20:28:54.907" Score="2" ViewCount="" Body="&lt;p&gt;First, consider the homogeneous system&#xA;$$\left(\begin{array}{ccc}&#xD;&#xA;a &amp;amp; a+b &amp;amp; a+2b\\\&#xD;&#xA;c &amp;amp; c+d &amp;amp; c+2d\\\&#xD;&#xA;e &amp;amp; e+f &amp;amp; e+2f&#xD;&#xA;\end{array}\right)\left(\begin{array}{c}x\\y\\z\end{array}\right) = \left(\begin{array}{c}0\\0\\0\end{array}\right).$$&#xA;If $(a,c,e)$ and $(b,d,f)$ are not scalar multiples of each other, then the coefficient matrix has rank $2$, so the solution space has dimension $1$. The vector $(1,-2,1)^T$ is clearly a solution, so the solutions are all multiples of $(1,-2,1)^T$. That is, the solutions to the homogeneous system are $\lambda(1,-2,1)^T$ for arbitrary $\lambda$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, the solutions to the inhomogeneous system are all of the form $\mathbf{x}_0 + \lambda(1,-2,1)^T$, where $\mathbf{x}_0$ is a particular solution to this system. Since $(-2,3,0)$ is a particular solution always, then all solutions have the described form.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If one of $(a,c,e)$ and $(b,d,f)$ is a multiple of the other, though, then there are other solutions: the matrix has rank $1$, so the nullspace has dimension $2$. Say $(a,c,e) = k(b,d,f)$ with $k\neq 0$, then there is another solution: $(-1-\frac{1}{k},1,0)$ would also be a solution to the system, so that the solutions to the inhomogeneous system would be of the form&#xA;$$r = \left(\begin{array}{r}-2\\3\\0\end{array}\right) + \lambda\left(\begin{array}{r}1\\-2\\1\end{array}\right) + \mu\left(\begin{array}{r}-1-\frac{1}{k}\\1\\0\end{array}\right).$$&#xA;This includes the solutions you have above, but also others. (If $k=0$, then you can use $(0,-2,1)$ instead of $(-1-\frac{1}{k},1,0)$) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $(b,d,f)=(0,0,0)\neq (a,c,e)$, then $(1,0,-1)$ can be used instead of $(-1-\frac{1}{k},1,0)$ to generate all solutions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And of course, if $(a,b,c)=(b,d,f)=(0,0,0)$, then every vector is a solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In all cases, you have an infinite number of solutions that &lt;em&gt;includes&lt;/em&gt; all the solutions you give (but there may be solutions that are not in that line).&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2011-03-15T20:28:54.907" CommentCount="2" />
  <row Id="27261" PostTypeId="2" ParentId="27258" CreationDate="2011-03-15T20:30:03.377" Score="3" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\rm\ \ \ \ \ \ \ \:mod\ 5\::\ (N+20)^{N+20}\ \equiv\ N^{20}\ N^N\ \equiv\ N^N\ $ for $\rm\ N\not\equiv 0\ $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Generally $\rm\ \ mod\ P\::\ (N+P(P-1))^{N+P(P-1)}\ \equiv\ N^N\ N^{P(P-1)}\ N^N\ \equiv\ N^N\ $ for $\rm\ N\not\equiv 0$&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2011-03-15T20:38:55.087" LastActivityDate="2011-03-15T20:38:55.087" CommentCount="0" />
  <row Id="27262" PostTypeId="1" AcceptedAnswerId="27268" CreationDate="2011-03-15T20:32:45.347" Score="1" ViewCount="217" Body="&lt;p&gt;I have some functionals that have lots of parameters, but all are volume integrals (not really important...)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's a variational function I'm minimizing and I'm only keeping the real solution (because it's my variational wavefunction and I demand only a real solution :) ).&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Solve[D[Subscript[E, HO][\[CapitalPsi]] + Subscript[E, Int][\[CapitalPsi]], b] == 0, b]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Basically, a &lt;code&gt;Solve&lt;/code&gt; for a parameter b. I get 5 roots, which is correct (fifth degree polynomial). Even when I add &lt;code&gt;, Reals&lt;/code&gt; (no change, still five solutions, of which 4 complex) and try &lt;code&gt;Reduce&lt;/code&gt; (some evil error message and none of the five answers). I even tried:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Simplify[Solve[D[Subscript[E, HO][\[CapitalPsi]] + Subscript[E, Int][\[CapitalPsi]], b] == 0, b], Element[b, Reals]]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Which apparently doesn't seem to work. The four complex solutions are of the form (-1)^(1/5) times a bunch of numbers/parameters. What can I do to filter out the rest (or only keep the first solution)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="939" LastActivityDate="2011-03-21T06:13:09.033" Title="Mathematica's &quot;Solve&quot; and filtering out the one real (non-complex) solution" Tags="&lt;mathematica&gt;" AnswerCount="3" CommentCount="0" />
  <row Id="27263" PostTypeId="2" ParentId="27245" CreationDate="2011-03-15T20:34:17.357" Score="1" ViewCount="" Body="&lt;p&gt;Let $\omega(n)$ be the number of distinct prime divisors of $n$, and let $\tau(n)$ be the number of divisors of $n$.  Then, one can show that&#xA;$$ \tau(n) \le \exp \left( \frac{ \log 2 \log n}{\log \log n} \left( 1 + O\left( \frac{\log \log \log n}{\log \log n} \right) \right) \right)$$&#xA;(see, for instance, G. Tenenbaum's Introduction to Analytic and Probabilistic Number Theory, chapter 5).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, since $2^\omega(n) \le \tau(n)$, we may conclude that&#xA;$$ \omega(n) \le (1+ o(1)) \frac{\log n}{\log \log n}$$&#xA;as $n \rightarrow \infty$.  By directly working with numbers which are the product of the first $k$ primes, a lower bound for $\tau$ is attainable which yields a lower bound for $\omega$ of the same order.  In short, one has&#xA;$$\limsup_{n \rightarrow \infty} \frac{\omega(n)}{(\log n)/(\log \log n)} = 1.$$&lt;/p&gt;&#xA;" OwnerUserId="2937" LastActivityDate="2011-03-15T20:34:17.357" CommentCount="2" />
  <row Id="27264" PostTypeId="1" CreationDate="2011-03-15T20:37:19.707" Score="1" ViewCount="28" Body="&lt;p&gt;How can I find a Chevalley basis of a type $B_2$ when the related lie algebra is defined as a linear Lie algebra of elements of the form $x= \begin{pmatrix} 0 &amp;amp; b_1 &amp;amp; b_2 \\ c_1 &amp;amp; m &amp;amp; n \\ c_2 &amp;amp; p &amp;amp; q \end{pmatrix}$, where $c_1=-b_2^t$, $c_2=-b_1^t$, $q=-m^t$, $n^t=-n$, $p^t=-p$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When trying to find such a base, the constraints, especially $[x_{\alpha}x_{\beta}]=c_{\alpha,\beta}x_{\alpha+\beta}$ for $\alpha,\beta$ independent, and $\alpha+\beta$ being a root, turn out to be very hard to follow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you~ :)&lt;/p&gt;&#xA;" OwnerUserId="4736" LastActivityDate="2011-03-15T20:37:19.707" Title="How can I find a Chevalley basis of $B_2$?" Tags="&lt;lie-algebra&gt;" />
  <row Id="27265" PostTypeId="1" AcceptedAnswerId="27288" CreationDate="2011-03-15T20:38:20.673" Score="0" ViewCount="53" Body="&lt;p&gt;Let $A$ and $B$ be $n \times n$ matrices. Strassen's algorithm for multiplication works on a partitioning of $A$ and $B$ into $2^2$ submatrices each. Is there any generalization of Strassen's algorithm where the partitioning is done into, say, $k^2$ submatrices?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: What I mean by a $3^2$ partitioning of $A$ is:&#xA;$$A = \left(\begin{array}{ccc}&#xD;&#xA;A_{1,1} &amp;amp; A_{1,2} &amp;amp; A_{1,3} \\&#xD;&#xA;A_{2,1} &amp;amp; A_{2,2} &amp;amp; A_{2,3} \\&#xD;&#xA;A_{3,1} &amp;amp; A_{3,2} &amp;amp; A_{3,3} &#xD;&#xA;\end{array} \right) $$&#xA;where each of $A_{i,j}$ is an $\frac{n}{3} \times \frac{n}{3}$ matrix.&lt;/p&gt;&#xA;" OwnerUserId="2468" LastActivityDate="2011-03-15T22:52:01.150" Title="Is there a generalization of Strassen algorithm?" Tags="&lt;linear-algebra&gt;&lt;matrices&gt;&lt;algorithms&gt;" AnswerCount="1" />
  <row Id="27266" PostTypeId="1" AcceptedAnswerId="27287" CreationDate="2011-03-15T20:48:22.833" Score="3" ViewCount="73" Body="&lt;p&gt;I was trying to solve the following exercise:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Suppose that the sequence of independent events $\{A_i\}$ satisfies $$\phi(n) = \sum_{i=1}^{n} P(A_i) \rightarrow \infty \text{ as } n \rightarrow \infty$$ &#xA;  and let $\tau_k :=\min\{n | \sum_{i=1}^{n} 1(A_i) = k\}$. By applying Doob's stopping time theorem to an appropriate martingale, prove the more quantitative result that $$E[\phi(\tau_k)] = k \text{ for all } k \geq 1$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here, $1(\cdot)$ is the indicator function. By Doob's stopping theorem, I think the book is referring to the following theorem (which is just named &quot;stopping time theorem&quot;):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If $(M_n)$ is a martingale with respect to $(\mathcal{F}_n)$, then the stopped process $(M_{\tau \wedge n})$  is also a martingale with respect to $(\mathcal{F}_n)$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;First of all, let me say that I think my proof is wrong. I don't use the stopping time theorem or the fact that $\phi(n)$ diverges for $n \rightarrow \infty$ and it just looks too easy to be true. In any case, I'd welcome hints and constructive criticism.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My attempt at a proof:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $P(A) = E[1(A)]$, we can write $$\phi(\tau_k) = \sum_{i=1}^{\tau_k} P(A_i) = \sum_{i=1}^{\tau_k} E[1(A_i)]$$ Now, $\tau_k$ is the smallest $n$ such that from the sequence $\{A_1, A_2, \ldots, A_n\}$, exactly $k$ of them occur. So $$\sum_{i=1}^{\tau_k} E[1(A_i)] = \sum_{j=1}^{k} E[1(A_{i_j})]$$ for indices $i_j$. Since, by definition of $\tau_k$, we know that the events $A_{i_j}$ occur for $j = 1, \ldots, k$, we have that $$\phi(\tau_k) = \underbrace{1 + 1 + \ldots + 1}_{k} = k$$ and taking the expectation of both sides gives $$E[\phi(\tau_k)] = E[k] = k$$&lt;/p&gt;&#xA;" OwnerUserId="7155" LastActivityDate="2011-03-15T22:37:36.703" Title="Expected time to $k$-th occurence" Tags="&lt;stochastic-processes&gt;&lt;probability&gt;" AnswerCount="1" CommentCount="7" FavoriteCount="1" />
  <row Id="27267" PostTypeId="2" ParentId="27259" CreationDate="2011-03-15T20:48:34.690" Score="1" ViewCount="" Body="&lt;p&gt;This follows immediately from a basic theorem of linear algebra. Namely, the solution of a nonhomogeneous linear system can be expressed as the sum of any particular solution plus the general solution of the homogeneous system. Since your exhibited solution has this form it is necessarily a solution, i.e. $\rm\ \ L(x_0) = a_0,\ L(x) = 0\ \Rightarrow\ L(x_0+\lambda\ x)\ =\ L(x_0)+\lambda\ L(x)\ =\ a_0\:.\ \ $ This sort of &quot;shifted&quot; vector space structure is known as an &lt;a href=&quot;http://en.wikipedia.org/wiki/Affine_space&quot; rel=&quot;nofollow&quot;&gt;affine space.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2011-03-15T20:53:53.233" LastActivityDate="2011-03-15T20:53:53.233" />
  <row Id="27268" PostTypeId="2" ParentId="27262" CreationDate="2011-03-15T20:50:19.353" Score="1" ViewCount="" Body="&lt;p&gt;Try&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Assuming[Element[b,Reals],~rest of your code~]&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt; then how about&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;b /. Solve[expr == 0, b] &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;N[b]&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Select[%,Element[#,Reals] &amp;amp; ] &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I tried this using a regular polynomial with complex solutions and it worked in my case, should work in yours. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;i.e, Create a list of solutions sols=Solve[expr==0,b]. Type sol// N to convert the fractional powers of 1 to &quot;numerical&quot; complex numbers Then use either Select or DeleteCases, which is a bit twisted and venture there only if Select didnt work. Or just numerically select sol[[n]] where n is the index of your real solution if you are willing to work the &quot;shortcut&quot; way.&lt;/p&gt;&#xA;" LastActivityDate="2011-03-15T21:33:54.243" CommentCount="1" />
  <row Id="27269" PostTypeId="2" ParentId="27156" CreationDate="2011-03-15T20:55:56.430" Score="5" ViewCount="" Body="&lt;p&gt;The best you can get if you take the character theory done over $\mathbb C$ and attempt to redo it over an arbitrary field $\Bbbk$ of characteristic $0$ (the usual character theory does not work in positive characteristic because $p$-dimensional representations could end up with the $0$ character), is that the group ring $\Bbbk G$ as a left module over itself is isomorphic to $\bigoplus_V \frac{\dim V}{\dim Hom_\Bbbk(V,V)^G} V$ where $V$ ranges over all irreducible representations and $Hom_\Bbbk(V,V)^G$ is the space of $G$-linear endomorphisms of $V$. This gives you $|G|=\sum_V\frac{(\dim V)^2}{\dim Hom_\Bbbk(V,V)^G}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All of the above follows immediately from noticing that the projection formula $\dim V^G=\frac1{|G|}\sum_{g\in G}\chi_V(g)$ still holds (because that formula says nothing more than that $\frac1{|G|}\chi_\Bbbk^*=\sum_{g\in G}g$ is an idempotent with $1$-dimensional image in the ring $\Bbbk G$), that the characters are additive and multiplicative on direct sums and tensor products, and that if $V$ is a representation with character $\chi_V$, then the dual $V^*$ is a representation with character $\chi_{V^*}(g)=\chi_V(g^{-1})$. Then the 'product' $(\alpha,\beta)=\frac1{|G|}\sum_{g\in G}\alpha(g)\beta^{-1}(g)$ is linear in the first variable, so for irreducible representations $V$ and $W$ we get $(\chi_V,\chi_W)=\frac1{|G|}\sum_{g\in G}\chi_V(g)\chi_W(g^{-1})=\frac1{|G|}\sum_{g\in G}\chi_{W^*\otimes V}(g)=\dim(W^*\otimes_{\Bbbk} V)^G=\dim Hom_\Bbbk(V,W)^G$. Since we still have that a $G$-linear map between irreducibles $V$ and $W$ either has trivial kernel or trivial image (as they are subrepresentations), it follows that if $V\neq W$, then $(\chi_V,\chi_W)=0$.&#xA;Then linearity in the first variable tells us that for an arbitrary representation $R$ we have a unique decomposition $\chi_R=\sum_V\frac1{\dim Hom_\Bbbk(V,W)^G}(\chi_R,\chi_V)\chi_V$, which implies that characters of irreducible representations are linearly independent. This establishes that the decomposition of a representation into irreducibles corresponds to the represenation of the character as the sum of irreducible characters. Also, because characters are class functions, and the space of class function is generatd by the linearly independent projections of conjugacy classes, we get that there are at most as many irreducible representations as there are conjugacy classes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then the usual observation that for the regular representation $R$ of the group ring $\Bbbk G$ as a left module over itself we have $\chi_R(e)=\dim\Bbbk G=|G|$ and $\chi_R(g)=0$ otherwise (look at the traces of the matrices with basisi $G$ for $\Bbbk G$ over $\Bbbk$) implies that $(\chi_R,\chi_V)=\frac1{|G|}|G|\chi_V^*(e)=\dim V$ as usual and hence&#xA;$\chi_R=\sum_V\frac{\dim V}{\dim Hom_\Bbbk(V,W)^G}\chi_V$ which gives us the direct sum decomposition of $\Bbbk G$ that I claimed in the beginning.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that when $\Bbbk$ is algebraically closed, then surely $\dim_\Bbbk(V,V)^G=1$ for all irreducible representations $V$ since any endomorphism of $V$ must have an eigenvalue and the eigenspace for that eigenvalue for $G$-linear endomorphism must be all of $V$, i.e. the $G$-linear endomorphism must be scalar multiplication, which gives you $\Bbbk G=\bigoplus_V(\frac{\dim V}1)V$ and hence $|G|=\sum_V(\dim V)^2$.&lt;/p&gt;&#xA;" OwnerUserId="400" LastEditorUserId="400" LastEditorDisplayName="" LastEditDate="2011-03-17T12:28:17.950" LastActivityDate="2011-03-17T12:28:17.950" />
  <row Id="27271" PostTypeId="1" AcceptedAnswerId="27274" CreationDate="2011-03-15T20:59:59.790" Score="3" ViewCount="189" Body="&lt;p&gt;So the question asks me to prove that an entire function with positive real parts is constant, and I was thinking that this might somehow be related to showing an entire bounded function is constant (Liouville's theorem), but are there any other theorems that might help me prove this fact? &lt;/p&gt;&#xA;" OwnerUserId="2919" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-16T03:03:39.200" LastActivityDate="2011-03-16T03:03:39.200" Title="What might I use to show that an entire function with positive real parts is constant?" Tags="&lt;homework&gt;&lt;complex-analysis&gt;" AnswerCount="4" CommentCount="2" />
  <row Id="27272" PostTypeId="2" ParentId="27188" CreationDate="2011-03-15T21:07:11.693" Score="0" ViewCount="" Body="&lt;p&gt;This paper, that enumerates and shows plane 3-connected graphs for the convex polyhedra with 4 to 8 faces, may be of interest: Federico, P. J. Polyhedra with 4 to 8 faces. Geometriae Dedicata 3 (1974/75), 469–481.&lt;/p&gt;&#xA;" OwnerUserId="1369" LastActivityDate="2011-03-15T21:07:11.693" />
  <row Id="27273" PostTypeId="2" ParentId="27271" CreationDate="2011-03-15T21:15:54.183" Score="1" ViewCount="" Body="&lt;p&gt;Have you learned the Riemann Mapping theorem?  If so, what can you do with the image of this entire function?  Remember, the composition of analytic functions is analytic.  Liousville's theorem should then finish the problem off.&lt;/p&gt;&#xA;" OwnerUserId="8309" LastActivityDate="2011-03-15T21:15:54.183" CommentCount="1" />
  <row Id="27274" PostTypeId="2" ParentId="27271" CreationDate="2011-03-15T21:21:20.557" Score="5" ViewCount="" Body="&lt;p&gt;It isn't a nonconstant polynomial, by the fundamental theorem of algebra.  It doesn't have an essential singularity at infinity, by the Casorati-Weierstrass theorem.  What other possibilities are there?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively, if you add $1$, you get a function satisfying $|g(z)|\geq 1$ for all $z$.  What can you say about the reciprocal of $g$?&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2011-03-15T21:21:20.557" />
  <row Id="27275" PostTypeId="1" CreationDate="2011-03-15T21:25:22.753" Score="3" ViewCount="48" Body="&lt;p&gt;For an arbitrary ring $R$ and a positive integer $n &amp;gt;1$, are the category of $R$-modules and the category of $M_n(R)$-modules isomorphic?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here, $M_n(R)$ denotes the $n$ plus $n$ matrices over the ring $R$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know these two categories are equivalent, and I guess they are not necessarily isomorphic, but I don't know how to prove it...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many thanks :)&lt;/p&gt;&#xA;" OwnerUserId="4736" LastEditorUserId="4736" LastEditorDisplayName="" LastEditDate="2011-03-15T21:40:31.157" LastActivityDate="2011-03-15T21:40:31.157" Title="For an arbitrary ring $R$ and a positive integer $n &gt;1$, are the category of $R$-modules and the category of $M_n(R)$-modules isomorphic?" Tags="&lt;category-theory&gt;" CommentCount="3" FavoriteCount="1" />
  <row Id="27277" PostTypeId="1" CreationDate="2011-03-15T21:29:47.130" Score="0" ViewCount="67" Body="&lt;p&gt;Let E be a set of finite measure. Let $\{f_n\}$ be a sequence of uniformly bounded non-measurable functions. When (state exact conditions if possible) would it be possible to assign a unique numerical value to $\lim_{n\to\infty} \int_E^* f_n$? (Here $\int_E^* f_n$ denotes a set of possible values for $\int_E f_n$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: It seems that we should require that $\{f_n\} \to f$ point-wise almost everywhere, otherwise we could get some alternating sequence when we take the limit. Furthermore since $\liminf_{n\to\infty} \int_E f_n \leq \lim_{n\to\infty} \int_E f_n \leq \limsup_{n\to\infty} \int_E f_n$ we should also probably require that the above inequalities are equalities. I am unsure if requiring $f$ to be measurable is necessary and also if all of these conditions together are sufficient.&lt;/p&gt;&#xA;" OwnerUserId="8043" LastEditorUserId="8043" LastEditorDisplayName="" LastEditDate="2011-03-15T22:03:08.557" LastActivityDate="2011-03-15T22:03:08.557" Title="Existence of Unique Value for Integral of Non-Measurable Sequence of Functions" Tags="&lt;homework&gt;&lt;measure-theory&gt;" CommentCount="2" />
  <row Id="27278" PostTypeId="2" ParentId="27123" CreationDate="2011-03-15T21:37:02.410" Score="2" ViewCount="" Body="&lt;p&gt;You can use summation by parts here: If $(a_n)_{n \geq 1}$ is a sequence of complex numbers, and $f : [1, \infty] \longrightarrow {\bf C}$ is a $C^1$ &#xA;function then&#xA;$$&#xD;&#xA;\sum_{n \leq N} a_n f(n) =&#xD;&#xA;(\sum_{n \leq N} a_n) f(N)&#xD;&#xA;-&#xD;&#xA;\int_{1}^{N} (\sum_{n \leq t} a_n) f&amp;#39;(t) \; d \, t.$$&#xA;Applying this to $a_n = {1 \over n}$ and $f(t) = \sin(\sqrt{t}x)$, one gets&#xA;$$\sum_{n \leq N} {\sin(\sqrt{n}x) \over n} = H(N)\sin(\sqrt{N}x) - \int_{1}^{N} H(\lfloor t \rfloor) {x\cos(\sqrt{t}x) \over 2\sqrt{t}} \; d \, t$$&#xA;Here $H(n)$ is the $n$th harmonic number $1 + {1 \over 2} + ... + {1 \over n}$. Fortunately, there are good asymptotics for $H(n)$, namely that $H(n) = \ln(n) + \gamma + O(1/n)$. As a result, the $H(\lfloor t \rfloor)$ appearing in the above integral differs from $\ln(t) + \gamma$ by at most ${C \over t}$. Since ${\displaystyle \int_1^{\infty} {1 \over t} {x\cos(\sqrt{t}x) \over 2\sqrt{t}}\,dt}$ is absolutely integrable, one can replace $H(\lfloor t \rfloor)$ by $\ln(t) + \gamma$ in that integral when trying to prove convergence. One can similarly disregard the $O({1 \over N})$ term in the $H(N)$ on the left. So our goal is to show that the following converges as $N$ goes to infinity.&#xA;$$(\ln(N) + \gamma) \sin(\sqrt{N}x) - \int_{1}^{N}  (\ln(t) + \gamma){x\cos(\sqrt{t}x) \over 2\sqrt{t}} \; d \, t.$$&#xA;We now integrate by parts again, in the opposite direction of our original summation by parts; we differentiate $(\ln(t) + \gamma)$ and integrate ${x\cos(\sqrt{t}x) \over 2\sqrt{t}}$ back to $\sin(\sqrt{t} x)$. Our goal now becomes to show the finiteness of &#xA;$$\int_1^{\infty}{\sin(\sqrt{t} x) \over t}\,dt$$&#xA;As others have said above, by changing $t$ to $t^2$, this is equivalent to showing the convergence of ${\displaystyle \int_1^{\infty} {\sin(t x) \over t}\,dt}$, which in turn by replacing $t$ by $tx$ is equivlalent to showing the convergence of ${\displaystyle \int_x^{\infty} {\sin(t) \over t}\,dt}$ which is well-known.&lt;/p&gt;&#xA;" OwnerUserId="3035" LastEditorUserId="6622" LastEditorDisplayName="" LastEditDate="2011-03-15T21:55:27.893" LastActivityDate="2011-03-15T21:55:27.893" CommentCount="1" />
  <row Id="27279" PostTypeId="1" CreationDate="2011-03-15T21:52:56.353" Score="3" ViewCount="60" Body="&lt;p&gt;I would like to tell me a good program to draw graphs and anylize them (finding euler,hamilton,etc)and export the drawing to word.&lt;/p&gt;&#xA;" OwnerUserId="7843" LastActivityDate="2011-03-15T22:14:23.177" Title="Graph draw software" Tags="&lt;math-software&gt;" AnswerCount="1" FavoriteCount="3" />
  <row Id="27280" PostTypeId="2" ParentId="27199" CreationDate="2011-03-15T21:57:17.470" Score="2" ViewCount="" Body="&lt;p&gt;View the point $z$ as a vector in $\mathbb R^2$. Without doing any calculations, we know that $\lVert z - (1,0) \rVert^2$ and $\lVert z - (-1,0) \rVert^2$ are of the form $z^Tz + b^T z + c$ for some $b$ and $c$. So the equation of the locus is $$(z^Tz + b_1^Tz + c_1) = 4(z^Tz + b_2^Tz + c_2),$$ or $$3z^Tz + b_3^Tz + c_3 = 0,$$ which is the equation of a circle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, the form $x^TAx + b^Tx + c$ defines a quadric, and if you know something about the properties of the matrix $A$, it can tell you what the shape of the quadric is. In this case, $A$ is a multiple of the identity, so it's a circle/sphere/hypersphere.&lt;/p&gt;&#xA;" OwnerUserId="856" LastEditorUserId="856" LastEditorDisplayName="" LastEditDate="2011-03-15T23:17:57.480" LastActivityDate="2011-03-15T23:17:57.480" />
  <row Id="27281" PostTypeId="1" AcceptedAnswerId="27283" CreationDate="2011-03-15T21:59:18.793" Score="3" ViewCount="77" Body="&lt;p&gt;From Richard Stanley's &lt;em&gt;Enumerative Combinatorics&lt;/em&gt;,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;2(j)&lt;/strong&gt; How many sequences $(a_1, a_2, \ldots, a_{12})$ are there consisting of four 0's and eight 1's, if no two consecutive terms are both 0's?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can brute force this, but Stanley provides the (mysterious) reasoning ${8 + 1 \choose 4} = 126$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any elegant argument is appreciated.&lt;/p&gt;&#xA;" OwnerUserId="3537" LastActivityDate="2011-03-15T22:02:01.833" Title="Counting Binary Sequences" Tags="&lt;combinatorics&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="27282" PostTypeId="1" CreationDate="2011-03-15T21:59:39.793" Score="-1" ViewCount="72" Body="&lt;p&gt;Can someone show that the collection of all simple functions is a function space? Also, is the collection an algebra?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you very much for your help...&lt;/p&gt;&#xA;" OwnerUserId="8245" LastActivityDate="2011-05-16T02:58:00.853" Title="Dealing with collection of simple or step functions as a function space." Tags="&lt;real-analysis&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="1" />
  <row Id="27283" PostTypeId="2" ParentId="27281" CreationDate="2011-03-15T22:02:01.833" Score="6" ViewCount="" Body="&lt;p&gt;From the answer here: &lt;a href=&quot;http://math.stackexchange.com/questions/18268/consecutive-birthdays-probability/18363#18363&quot;&gt;Consecutive birthdays probability&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We need to select $\displaystyle d$ numbers from $\displaystyle 1,2, \dots, k$ such that no two are consecutive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now if $\displaystyle b_1 \lt  b_2 \lt \dots \lt b_d$ were such numbers, then notice that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle 1 \le b_1 \lt b_2 - 1 \lt b_3 - 2 \dots \lt b_d - (d-1) \le k-(d-1)$ gives us a way to select numbers from $\displaystyle 1, 2, \dots, k-(d-1)$ without having to bother about the consecutive issue.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This can be done in $\displaystyle {k-d+1 \choose d}$ ways.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your case, $k=12, \ d=4$.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2011-03-15T22:02:01.833" CommentCount="3" />
  <row Id="27284" PostTypeId="2" ParentId="27279" CreationDate="2011-03-15T22:14:23.177" Score="3" ViewCount="" Body="&lt;p&gt;Why don't you try &lt;a href=&quot;http://www.sagemath.org/&quot; rel=&quot;nofollow&quot;&gt;Sage&lt;/a&gt; -- it's free and should do as good a job anything else. You just type in &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;G = Graph (M)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;for M an adjacency matrix to create your graph, then you can do &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;G.eulerian_circuit()&#xA;G.hamiltonian_cycle()&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;to find these things, and &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;plot(G)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;draws your graph as a .png file which shouldn't be too hard to open in Word.&lt;/p&gt;&#xA;" OwnerUserId="668" LastActivityDate="2011-03-15T22:14:23.177" />
  <row Id="27285" PostTypeId="1" CreationDate="2011-03-15T22:15:48.280" Score="1" ViewCount="58" Body="&lt;p&gt;Hi all,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Exercise 2.27 in Boyd and Vanderberghe:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose the set C is closed, has nonempty interior, and has a supporting hyperplane at every point in its boundary. Show that C is convex.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Seems to me one approach is to prove that the intersection of all the supporting hyperplanes is exactly C. Clearly this intersection contains C. Geometrically the other direction seems obvious, but any hint how to argue it rigorously?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="8130" LastActivityDate="2011-04-15T07:18:30.227" Title="Converse supporting hyperplane theorem" Tags="&lt;convex-analysis&gt;&lt;convex-optimization&gt;&lt;convex-sets&gt;" AnswerCount="2" CommentCount="0" />
  <row Id="27286" PostTypeId="2" ParentId="27199" CreationDate="2011-03-15T22:28:13.603" Score="1" ViewCount="" Body="&lt;p&gt;A perhaps easier way to see that it is a circle is to rewrite as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle |z&amp;#39;| = 2|z&amp;#39; + 2|, \text{where } z&amp;#39; = z -1$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And so&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle 1/4 = |1/2 + 1/z&amp;#39;|$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus $\displaystyle \frac{1}{z&amp;#39;}$ gives a circle centred at $(-1/2,0)$ and radius $1/4$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As we can see from the answers and comments to this recent question: &lt;a href=&quot;http://math.stackexchange.com/questions/26894/complex-number-property&quot;&gt;Complex number property&lt;/a&gt; that, the map $\displaystyle z \to \frac{1}{z}$ maps circles to circles, if the circle does not pass through the origin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case, it does not pass through the origin, and so the original curve you seek would be a circle too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course the fact that $\displaystyle z \to \frac{1}{\bar{z}}$ is inversion with respect to the unit circle probably uses the Appollonius theorem, but it looks like a useful fact to keep in mind.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2011-03-15T22:28:13.603" />
  <row Id="27287" PostTypeId="2" ParentId="27266" CreationDate="2011-03-15T22:37:36.703" Score="1" ViewCount="" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;em&gt;Doob's stopping theorem&lt;/em&gt; is probably intended to mean the &lt;a href=&quot;http://en.wikipedia.org/wiki/Optional_stopping_theorem&quot; rel=&quot;nofollow&quot;&gt;optional stopping theorem&lt;/a&gt;, named in honour of &lt;a href=&quot;http://en.wikipedia.org/wiki/Joseph_Leo_Doob&quot; rel=&quot;nofollow&quot;&gt;Joseph Leo Doob&lt;/a&gt;.  &lt;/li&gt;&#xA;&lt;li&gt;You implicitly use $\phi(n) \rightarrow \infty$ since if it had some finite upper bound $L$ then for $k&amp;gt;L$ you would not be able to have $\phi(\tau_k) = k$.&lt;/li&gt;&#xA;&lt;li&gt;When you say $\sum_{i=1}^{\tau_k} E[1(A_i)] = \sum_{j=1}^{k} E[1(A_{i_j})]$, it is unclear whether you are using $A_i$ to mean unknown future events at the start or known past events at the stopping time.  It does not matter which because of the optional stopping theorem as applied to martingales (meeting some additional conditions which are satisfied in this case), namely that the expected value of such a martingale at a stopping time is equal to its initial value. I would expect you to find something like $E[X_{\tau \wedge n}]=E[X_0]$ or $E[X_{\tau}]=E[X_0]$ somewhere in your book.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="6460" LastActivityDate="2011-03-15T22:37:36.703" CommentCount="5" />
  <row Id="27288" PostTypeId="2" ParentId="27265" CreationDate="2011-03-15T22:52:01.150" Score="3" ViewCount="" Body="&lt;p&gt;Yes, in some cases.  See e.g. &lt;a href=&quot;http://www.csd.uwo.ca/~mislam63/ms_thesis.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.csd.uwo.ca/~mislam63/ms_thesis.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="8508" LastActivityDate="2011-03-15T22:52:01.150" />
  <row Id="27289" PostTypeId="1" AcceptedAnswerId="27314" CreationDate="2011-03-15T22:55:52.410" Score="1" ViewCount="84" Body="&lt;p&gt;Let $0&amp;lt;x_1&amp;lt;x_2&amp;lt;...&amp;lt;x_n&amp;lt;1$. Let us consider the matrix M defined such that:&#xA;$$M_{i,j} = \min(x_i, x_j)^2 \left( 3 \max(x_i, x_j) - \min(x_i, x_j) \right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I &lt;em&gt;believe&lt;/em&gt; the determinant of M is different from zero. Any idea on how to prove it (&lt;em&gt;if true&lt;/em&gt;)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, any reason why M would be invertible would suit me as well, no real need to get the formula of the determinant.&lt;/p&gt;&#xA;" OwnerUserId="2380" LastEditorUserId="2641" LastEditorDisplayName="" LastEditDate="2011-03-15T23:45:59.060" LastActivityDate="2011-03-16T01:49:41.423" Title="Is $\det$ of $M_{i,j} = \min(x_i, x_j)^2 \left( 3 \max(x_i, x_j) - \min(x_i, x_j) \right)$ non zero?" Tags="&lt;linear-algebra&gt;&lt;matrices&gt;&lt;determinant&gt;" AnswerCount="3" />
  <row Id="27290" PostTypeId="2" ParentId="27285" CreationDate="2011-03-15T22:59:11.987" Score="1" ViewCount="" Body="&lt;p&gt;Hint: Suppose wlog 0 is in C.  Suppose $p$ is not in C.  Take $0 &amp;lt; t &amp;lt; 1$ such that $t p$ is on the boundary of C, and consider a supporting hyperplane at $t p$.&lt;/p&gt;&#xA;" OwnerUserId="8508" LastActivityDate="2011-03-15T22:59:11.987" CommentCount="1" />
  <row Id="27292" PostTypeId="1" CreationDate="2011-03-15T23:09:14.287" Score="0" ViewCount="47" Body="&lt;p&gt;$$.670=\frac{x}{(1-x)(2-x)}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Need to change into the quadratic equation.&lt;/p&gt;&#xA;" OwnerUserId="8310" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-16T03:03:00.387" LastActivityDate="2011-03-16T03:03:00.387" Title="changing formula into quadratic" Tags="&lt;algebra-precalculus&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="27293" PostTypeId="2" ParentId="27218" CreationDate="2011-03-15T23:16:23.080" Score="6" ViewCount="" Body="&lt;p&gt;No need for $\epsilon$ actually.&#xA;If $\sin(n) \rightarrow l$, then $\sin(n+1)$ also, and $\sin(n+1)=\sin(n)\cos(1)+\sin(1)\cos(n)$.&#xA;Since both $\sin(n)$ and $\sin(n+1)$ have limit $l$ and $sin(1) \neq 0$, $\cos(n) \rightarrow \frac{l(1-\cos(1))}{\sin(1)}$, and so $e^{in}=\cos(n)+i \sin(n)$ has a limit.&#xA;But $e^{i(n+1)}$ must then have the same limit (call it $x$), which implies $x=e^{i} x$, and since $e^{i} \neq 1$, $x$ has to be zero, which is a contradiction with the fact that $|e^{in}|=1$.&lt;/p&gt;&#xA;" OwnerUserId="2660" LastActivityDate="2011-03-15T23:16:23.080" />
  <row Id="27294" PostTypeId="1" CreationDate="2011-03-15T23:17:19.150" Score="9" ViewCount="267" Body="&lt;p&gt;The following is an identity concerning the Bessel functions of the first kind $J_n(x)$ for integers $n$ and $m$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{n=-\infty}^\infty J_n(x) J_{n+m}(x) = \delta(m)$$&#xA;where $\delta(x)$ is the Kronecker delta function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This can be &lt;a href=&quot;http://physics.stackexchange.com/questions/6778/how-does-phase-modulation-conspire-to-eliminate-power-variations/6850#6850&quot;&gt;derived&lt;/a&gt; from the &lt;a href=&quot;http://en.wikipedia.org/wiki/Jacobi%E2%80%93Anger_expansion&quot; rel=&quot;nofollow&quot;&gt;Jacobi-Anger identity&lt;/a&gt;, but is there a simpler way to derive it, for instance using well-known recurrence relations of the Bessel functions?&lt;/p&gt;&#xA;" OwnerUserId="2191" LastActivityDate="2011-03-24T20:49:15.753" Title="How to show $\sum_{n=-\infty}^\infty J_n J_{n+m} = \delta(m)$?" Tags="&lt;special-functions&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="27295" PostTypeId="2" ParentId="27292" CreationDate="2011-03-15T23:17:48.480" Score="4" ViewCount="" Body="&lt;p&gt;It is unclear whether .670=x/(1-x)*(2-x) is supposed to mean $0.670 = \frac{x}{(1-x)(2-x)}$ or $0.670 = \frac{x}{(1-x)}(2-x)$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In either case, Yuval Filmus's advice of multiplying by the denominator works.  In the first you get $0.670(1-x)(2-x) = x$ and you can expand and move all the terms to the left hand side.  In the second you get $0.670(1-x) = x(2-x)$ and you can again expand and move all the terms to the left hand side. &lt;/p&gt;&#xA;" OwnerUserId="6460" LastActivityDate="2011-03-15T23:17:48.480" />
  <row Id="27296" PostTypeId="2" ParentId="27289" CreationDate="2011-03-15T23:20:56.977" Score="3" ViewCount="" Body="&lt;p&gt;This checks out for $n=2$. The determinant is then [thanks, Paul Z] $$4x_1^3x_2^3 - x_1^4 (3x_2-x_1)^2 = -x_1^3(x_1-4x_2)(x_1-x_2)^2.$$ If $0 &amp;lt; x_1 = 4x_2$ then clearly $x_2 &amp;lt; x_1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the order or the $x_i$ doesn't matter (as long as they're all different), and the upper bound $1$ makes no difference since everything is homogeneous (if you multiply everything by some constant then regularity/singularity of your matrix remains the same).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The next step would be to repeat the explicit calculation for $n=3$, and try to either factor it or numerically find a counterexample.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2011-03-16T01:48:54.747" LastActivityDate="2011-03-16T01:48:54.747" />
  <row Id="27297" PostTypeId="1" AcceptedAnswerId="27319" CreationDate="2011-03-15T23:47:18.700" Score="0" ViewCount="52" Body="&lt;p&gt;If I have a know point cloud before and after a transformation (it could squish, &#xA;deform, rotate, etc.) and a known center of the point cloud before and after a transformation I'd like to see how a vector is transformed with that system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd like to have each point in the point cloud contribute to the vectors transformation. It seems like I'd like to use the positions of the points in my cloud as basis vectors and then project my vector onto the basis vectors to get the vectors coefficients in the initial system and then use the coefficients in the final system to reconstruct my vector in the new point cloud system. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm a bit confused as how to use a basis thats non-orthogonal and overly spans the r^3 basis. I do want each point in the cloud to contribute so I don't want to just use 3 basis vectors. Any idea how to do the projection and reconstruction?&lt;/p&gt;&#xA;" OwnerUserId="6732" LastActivityDate="2011-03-16T02:07:56.253" Title="Using a point cloud as a basis" Tags="&lt;linear-algebra&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="27298" PostTypeId="1" CreationDate="2011-03-15T23:47:57.600" Score="1" ViewCount="98" Body="&lt;p&gt;Suppose that $T: \mathbb{R}^m \to \mathbb{R}^n$ is a linear transformation and that $v_1, v_2, \ldots, v_p$ are vectors in $\mathbb{R}^m$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $W = \{T(v_1), T(v_2), \ldots, T(v_p)\}$ is linearly independent in $\mathbb{R}^n$, does it follow that $S = \{v_1, v_2, \ldots, v_p\}$ is linearly independent in $\mathbb{R}^m$? Justify your answer with either a proof or a counterexample.&lt;/p&gt;&#xA;" OwnerUserId="8312" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-16T03:04:19.980" LastActivityDate="2011-03-16T03:04:19.980" Title="If the image of a set is linearly independent, is the set linearly independent?" Tags="&lt;linear-algebra&gt;&lt;transformation&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="27299" PostTypeId="2" ParentId="27271" CreationDate="2011-03-15T23:48:34.393" Score="1" ViewCount="" Body="&lt;p&gt;Both of the two other answers are already excellent, but if you really want to bring on the big guns, use &lt;a href=&quot;http://en.wikipedia.org/wiki/Picard_theorem#Little_Picard&quot; rel=&quot;nofollow&quot; title=&quot;Picard's little theorem&quot;&gt;Picard's Little Theorem&lt;/a&gt; - noting that the half plane consists of more than two points.&lt;/p&gt;&#xA;" OwnerUserId="8246" LastActivityDate="2011-03-15T23:48:34.393" />
  <row Id="27300" PostTypeId="2" ParentId="26846" CreationDate="2011-03-15T23:54:33.780" Score="0" ViewCount="" Body="&lt;p&gt;I tried the two solutions I've linked to the original question (&lt;a href=&quot;http://www.gamedev.net/topic/545021-hermite-splines/page__view__findpost__p__4514299&quot; rel=&quot;nofollow&quot;&gt;Solution 1&lt;/a&gt; and &lt;a href=&quot;https://collada.org/public_forum/viewtopic.php?p=4016&amp;amp;sid=ae21755569bd1c1df6cff58061cd6ddb#p4016&quot; rel=&quot;nofollow&quot;&gt;Solution 2&lt;/a&gt;). Here are my results.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Solution 1&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The idea is to recreate the curve with a third degree polynomial, by matching the value at P1 and P4 and their respective slope. This polynomial can be solve using a system of linear equations. The only problem is that it's possible to create an infinite slope at P1 or P4 with the Bezier curve, which is impossible to recreate with the polynomial. I've also tried using a 5th degree polynomial (matching value, first derivative and second derivative) without success, because it has the same problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Solution 2&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This solution is pretty simple, but a bit &lt;em&gt;hacky&lt;/em&gt;. The idea is to use the De Casteljau algorithm to search for a &lt;code&gt;t&lt;/code&gt; (parameter of the Bézier curve) that matches a given &lt;code&gt;x&lt;/code&gt; variable (x-axis). The algorithm is simple :&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Split the curve in half&lt;/li&gt;&#xA;&lt;li&gt;Is the &lt;code&gt;x&lt;/code&gt; of the split point is approximatively equal to the searched &lt;code&gt;x&lt;/code&gt;? If yes, return &lt;code&gt;t&lt;/code&gt;  &lt;/li&gt;&#xA;&lt;li&gt;Else, if the searched &lt;code&gt;x&lt;/code&gt; is greater that the &lt;code&gt;x&lt;/code&gt; of the split point, repeat the algorithm with the segment on the right, else on the left.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Once you have the &lt;code&gt;t&lt;/code&gt;, you can obtain the &lt;code&gt;y&lt;/code&gt; value using the Bézier function $f_y(t)$ in my original post.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, yeah, working, but a bit &lt;em&gt;hacky&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="8222" LastActivityDate="2011-03-15T23:54:33.780" />
  <row Id="27301" PostTypeId="2" ParentId="27298" CreationDate="2011-03-16T00:06:08.440" Score="2" ViewCount="" Body="&lt;p&gt;Assuming $T: \mathbb R^m \to \mathbb R^n$ is linear and $v_1, v_2, \dots, v_p$ are vectors in $\mathbb R^m$ and the vectors in $W = \{ T(v_1), T(v_2), \dots, T(v_p)\}$ are linearly independent in $\mathbb R^m$. Showing that $S = \{v_1, v_2, \dots, v_p\}$ is linearly independent in $\mathbb R^n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume that the vectors in $S$ are linearly dependent, say &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\alpha_1 v_1 + \alpha_2 v_2 + \dots + \alpha_p v_p = 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where at least one $\alpha_i \neq 0$. Then we get: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$T(\alpha_1 v_1 + \alpha_2 v_2 + \dots + \alpha_p v_p) =&#xD;&#xA;\alpha_1 T(v_1) + \alpha_2 T(v_2) + \dots + \alpha_p T(v_p) = 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and the vectors in $W$ are thus linearly dependent, a contradiction. Therefore the vectors in $S$ have to be linearly independent.&lt;/p&gt;&#xA;" OwnerUserId="5966" LastActivityDate="2011-03-16T00:06:08.440" />
  <row Id="27302" PostTypeId="2" ParentId="27271" CreationDate="2011-03-16T00:13:37.380" Score="5" ViewCount="" Body="&lt;p&gt;The other three answers are overkill to me.. Simply consider $e^{-f}$ if $f$ is your function. Is it bounded?&lt;/p&gt;&#xA;" OwnerUserId="325" LastActivityDate="2011-03-16T00:13:37.380" CommentCount="2" />
  <row Id="27303" PostTypeId="2" ParentId="27289" CreationDate="2011-03-16T00:25:11.567" Score="4" ViewCount="" Body="&lt;p&gt;I've verified for n &amp;lt;= 8 that the matrices are all positive definite (and in particular nonsingular). &lt;/p&gt;&#xA;" OwnerUserId="8508" LastActivityDate="2011-03-16T00:25:11.567" CommentCount="1" />
  <row Id="27304" PostTypeId="1" CreationDate="2011-03-16T00:28:02.820" Score="2" ViewCount="83" Body="&lt;p&gt;For complex numbers $a_{n}, b_{n}$, what would be the next step to simplify this expression:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left \|\sum_{n=1}^\infty a_{n} b_{n}\right \|^{2}=\left \langle \sum_{n=1}^\infty a_{n}b_{n},\sum_{m=1}^\infty a_{m}b_{m}\right \rangle$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it $=\sum_{n=1}^\infty \sum_{m=1}^\infty |a_{n}|^{2}|b_{m}|^{2}$ !? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $|a_{n}|^{2}=a_{n}\overline{a_{n}}$&lt;/p&gt;&#xA;" OwnerUserId="8315" LastEditorUserId="5203" LastEditorDisplayName="" LastEditDate="2011-03-16T00:57:10.143" LastActivityDate="2011-03-16T02:49:54.793" Title="Inner product of an infinite sum" Tags="&lt;complex-analysis&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="27305" PostTypeId="2" ParentId="27298" CreationDate="2011-03-16T00:31:42.657" Score="1" ViewCount="" Body="&lt;p&gt;You need not argue to contraction.  Pick vectors $v_1, \cdots , v_n$ and constants&#xA;$\alpha_1, \cdots ,\alpha_n$ are constants with $\sum_k\alpha_kv_k = 0.$  Now apply $T$; $0 = T(\sum_k \alpha_k v_k) = \sum_k \alpha_k T(v_k)$. By the linear independence of the $T(v_k)$, $\alpha_k = 0$, $1\le k \le n$. &lt;/p&gt;&#xA;" OwnerUserId="4154" LastActivityDate="2011-03-16T00:31:42.657" />
  <row Id="27306" PostTypeId="2" ParentId="16160" CreationDate="2011-03-16T00:44:39.557" Score="4" ViewCount="" Body="&lt;p&gt;The steady state distribution is not enough to determine the mean hitting time $E(T)$ of a given subset $A$. To see this in a simple case, assume there are $2n$ states and consider the simple random walks on the discrete circle $C_{2n}$ and on the complete graph $K_{2n}$. By symmetry, the steady state distribution is uniform in both cases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Choose $A=\{j\}$ and $i$ at distance $n$ from $j$ in $C_{2n}$. Then, $T$ for $C_{2n}$ is distributed like the first hitting time of $\pm n$ by a standard symmetric random walk on the discrete line starting from $0$, hence $E_{C_{2n}}(T)=n^2$. On the other hand, $T$ for $K_{2n}$ is distributed like the time of first success in an i.i.d. sequence of trials with probability of success $1/(2n-1)$ at each trial, hence $E_{K_{2n}}(T)=2n-1$. For every $n\ge2$, $E_{C_{2n}}(T)\ne E_{K_{2n}}(T)$.&lt;/p&gt;&#xA;" OwnerUserId="6179" LastEditorUserId="6179" LastEditorDisplayName="" LastEditDate="2011-03-16T06:27:14.617" LastActivityDate="2011-03-16T06:27:14.617" CommentCount="2" />
  <row Id="27307" PostTypeId="1" AcceptedAnswerId="27312" CreationDate="2011-03-16T00:47:46.447" Score="4" ViewCount="86" Body="&lt;p&gt;Suppose that we have a sequence of finite sets $A_1, A_2, \ldots$, which partition $\mathbb{N}$.  I am making no other assumptions on the $A_n$ - i.e. there could be any amount of interleaving between them.  Now suppose we have $S\subset\mathbb{N}$.  If $\lim_{n\rightarrow\infty} \frac{|S\cap A_n|}{|A_n|}=0$, does it follow that $S$ has a natural density of 0?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(And if so, while I'm at it, can 0 be replaced by other numbers?  Natural density be replaced by upper, lower density?  I mostly just care about the density 0 (equivalently density 1) case, though.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: And if the statement is false, is there some sufficiently-little-interleaving condition on the $A_n$ I could assume that would make it true?&lt;/p&gt;&#xA;" OwnerUserId="2884" LastEditorUserId="2884" LastEditorDisplayName="" LastEditDate="2011-03-16T00:55:39.670" LastActivityDate="2011-03-16T09:23:24.493" Title="Basic question about natural density" Tags="&lt;number-theory&gt;&lt;real-analysis&gt;&lt;asymptotics&gt;&lt;average&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="27308" PostTypeId="1" CreationDate="2011-03-16T00:58:42.450" Score="5" ViewCount="99" Body="&lt;p&gt;Dear everyone,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose you have a number field $L$, and a non-zero ideal $I$ of the ring of integers $O$ of $L$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question part A:&lt;/strong&gt; Is there prime ideal $\mathcal{P} \subseteq O$ in the ideal class of $I$ such that $p =\mathcal{P} \cap \mathbb{Z}$ splits completely in $O$?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the extension $L/\mathbb{Q}$ is Galois I think one can show that the answer is yes. In fact I think it is possible to show that for every $I$ there are infinitely many such $\mathcal{P}$'s. The necessary tools for this are class field theory and Chebotarev's density theorem. On the other hand I'm not asking for infinitely many primes, only for one. So to be concrete:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question part B:&lt;/strong&gt; If the answer to A is yes, is it possible to give a proof that avoids showing that there are infinitely many $\mathcal{P}$'s?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you.&lt;/p&gt;&#xA;" OwnerUserId="8317" LastActivityDate="2011-03-16T05:23:05.773" Title="Does every ideal class contains a prime ideal that splits?" Tags="&lt;number-theory&gt;&lt;reference-request&gt;&lt;algebraic-number-theory&gt;" AnswerCount="1" />
  <row Id="27309" PostTypeId="1" AcceptedAnswerId="27317" CreationDate="2011-03-16T01:03:54.200" Score="5" ViewCount="127" Body="&lt;p&gt;I am trying to understand the proof in Carmichaels book &lt;em&gt;Diophantine Analysis&lt;/em&gt; but I have got stuck at one point in the proof where $w_1$ and $w_2$ are introduced.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The theorem it is proving is that the system of diophantine equations:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$$x^2 + y^2 = z^2$$&lt;/li&gt;&#xA;&lt;li&gt;$$y^2 + z^2 = t^2$$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;cannot simultaneously be satisfied.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The system is algebraically seen equivalent to&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$$t^2 + x^2 = 2z^2$$&lt;/li&gt;&#xA;&lt;li&gt;$$t^2 - x^2 = 2y^2$$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;and this is what will be worked on. We are just considering the case where the numbers are pairwise relatively prime. That implies that $t,x$ are both odd (they cannot be both even). Furthermore $t &amp;gt; x$ so define $t = x + 2 \alpha$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clearly the first equation $(x + 2\alpha)^2 + x^2 = 2 z^2$ is equivalent to $(x + \alpha)^2 + \alpha^2 = z^2$ so by the characterization of primitive Pythagorean triples there exist relatively prime $m,n$ such that $$\{x+\alpha,\alpha\} = \{2mn,m^2-n^2\}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the second equation $t^2 - x^2 = 4 \alpha (x + \alpha) = 8 m n (m^2 - n^2) = 2 y^2$ tells us that $y^2 = 2^2 m n (m^2 - n^2)$ by coprimality and unique factorization it follows that each of those terms are squares so define $u^2 = m$, $v^2 = n$ and $w^2 = m^2 - n^2 = (u^2 - v^2)(u^2 + v^2)$.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;It is now said that from the previous equation either&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$u^2 + v^2 = 2 {w_1}^2$,  $u^2 - v^2 = 2 {w_2}^2$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;or&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$u^2 + v^2 = w_1^2$,  $u^2 - v^2 = w_2^2$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;but $w_1$ and $w_2$ have not been defined and I cannot figure out what they are supposed to be. Any ideas what this last part could mean? Thanks a lot!&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;for completeness, if the first case occurs we have our descent and if the second case occurs $w_1^2 + w_2^2 = 2 u^2$, $w_1^2 - w_2^2 = 2 v^2$ gives the descent. Which finishes the proof.&lt;/p&gt;&#xA;" OwnerUserId="4290" LastEditorUserId="4290" LastEditorDisplayName="" LastEditDate="2011-03-16T01:09:16.757" LastActivityDate="2011-03-28T18:13:59.347" Title="Help to understand a proof by descent?" Tags="&lt;number-theory&gt;" AnswerCount="2" />
  <row Id="27310" PostTypeId="1" AcceptedAnswerId="27316" CreationDate="2011-03-16T01:06:48.637" Score="0" ViewCount="73" Body="&lt;p&gt;hello. I am having trouble showing the following and i was wondering if anyone can point out my mistake.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;let f be a 2pi periodic function and &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f_m(t)=f(mt)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;show the n-th fourier coefficient &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\hat{f_m}(n)=\hat{f}(n/m)$$ if m divides n&#xA;$$\hat{f_m}(n)=0$$ if m does not divides n&lt;/p&gt;&#xA;&#xA;&lt;p&gt;my attempt:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\hat{f_m}(n)=\frac{1}{2\pi}\int_0^{2\pi}f(mt)e^{-int}dt$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;substitute x=mt we get &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$=\frac{1}{2\pi}\int_0^{2\pi}f(x)e^{-i\frac{n}{m}x}\frac{dx}{m}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;as you can see, im having a extra m on the denominator. Furthermore, i do not see why it should equal 0 if m does not divide n&lt;/p&gt;&#xA;" OwnerUserId="6556" LastEditorUserId="6556" LastEditorDisplayName="" LastEditDate="2011-03-16T01:49:27.040" LastActivityDate="2011-03-16T01:49:27.040" Title="question about fourier coefficients" Tags="&lt;fourier-analysis&gt;" AnswerCount="1" />
  <row Id="27311" PostTypeId="2" ParentId="27307" CreationDate="2011-03-16T01:12:48.610" Score="6" ViewCount="" Body="&lt;p&gt;This is false. Let $S$ be any infinite coinfinite subset of $\mathbb{N}$, and let $A_n$ contain $1$ element of $S$ and $n$ elements of $\mathbb{N} \setminus S$ (it's clear we can always find such a partition, for example by recursively letting $A_n$ contain the smallest suitable numbers not already in $A_i$ for some $i &amp;lt; n$). Then $\lim_{n \to \infty} \frac{|S \cap A_n|}{|A_n|}=\lim_{n \to \infty} \frac{1}{n+1}=0$, but the density of $S$ can be anything in $[0,1]$, or not exist at all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words: your condition on $S$ and the $A_n$s makes no use of the order structure on $\mathbb{N}$, and just treats $\mathbb{N}$ as an unstructured set. It thus can't imply anything about $S$ that isn't already implied by its cardinality and cocardinality.&lt;/p&gt;&#xA;" OwnerUserId="5203" LastEditorUserId="5203" LastEditorDisplayName="" LastEditDate="2011-03-16T01:18:39.680" LastActivityDate="2011-03-16T01:18:39.680" CommentCount="3" />
  <row Id="27312" PostTypeId="2" ParentId="27307" CreationDate="2011-03-16T01:20:29.150" Score="6" ViewCount="" Body="&lt;p&gt;Fix an increasing sequence of positive integers $(k(n))_n$ and let $A_n=[k(n),k(n+1))$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fix a sequence of positive integers $(\ell(n))_n$ such that $\ell(n)\le k(n+1)-k(n)$ for every $n$ and define $S$ as the union of the intervals $[k(n),k(n)+\ell(n))$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume that $\ell(n)\ll k(n+1)-k(n)$. Then $|S\cap A_n|\ll|A_n|$ hence your hypothesis is met. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume furthermore that $\ell(n)\ge uk(n)$ for a given positive $u$, for every $n$. Then, for every $N=k(n)+\ell(n)$, &#xA;$$&#xD;&#xA;\frac1N|S\cap\{1,2,\ldots,N\}|\ge\frac{\ell(n)}{k(n)+\ell(n)}\ge\frac{u}{1+u}&#xD;&#xA;$$ &#xA;hence the natural density of $S$ cannot be $0$ and your conclusion does not hold. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Every condition above on the sequences $(k(n))$ and $(\ell(n))$ is met if, for example, $k(n)=2^{n^2}$ and $\ell(n)=2^{n^2+n}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this specific example, the sequence of densities of general term $N^{-1}|S\cap\{1,2,\ldots,N\}|$ converges to $0$ when restricted to the integers $N=k(n)$ and converges to $1$ when restricted to the integers $N=k(n)+\ell(n)$, hence the set of limit points of the whole sequence of densities is the interval $[0,1]$.&lt;/p&gt;&#xA;" OwnerUserId="6179" LastEditorUserId="6179" LastEditorDisplayName="" LastEditDate="2011-03-16T09:23:24.493" LastActivityDate="2011-03-16T09:23:24.493" CommentCount="6" />
  <row Id="27313" PostTypeId="2" ParentId="27309" CreationDate="2011-03-16T01:30:09.090" Score="2" ViewCount="" Body="&lt;p&gt;$u^2$ and $v^2$ are $m$ and $n$, respectively, which are coprime. Then since $(u^2+v^2)+(u^2-v^2)=2u^2$ and $(u^2+v^2)-(u^2-v^2)=2v^2$, the only factor that $u^2+v^2$ and $u^2-v^2$ can have in common is a single factor of $2$. Since their product is the square $w^2$, that leaves the two possibilities given.&lt;/p&gt;&#xA;" OwnerUserId="6622" LastActivityDate="2011-03-16T01:30:09.090" />
  <row Id="27314" PostTypeId="2" ParentId="27289" CreationDate="2011-03-16T01:31:01.930" Score="3" ViewCount="" Body="&lt;p&gt;Yuval's solution dropped an exponent; this is easily seen since his first expression is 6th degree and the later one is only 5th. It should be $$x_i^3(4x_2 - x_1) (x_2 - x_1)^2\;.$$ But his conclusion still holds; the claim is clearly true for $n = 2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the general case, it may help to remember that this is a symmetric matrix, and all symmetric matrices can be diagonalized. Furthermore, since you restricted the $x_i$ to be in strictly increasing order, all the $\min$ and $\max$ stuff devolves to $M_{i,j} = x_i^2(3x_j - x_i) = 3x_i^2x_j - x_i^3$ whenever $i \leq j$. This is immediately recognizable as $(D_{x_i} - D_{x_j})(x_i^3x_j)$, where $D$ is the differential operator. I don't know if either of these facts are helpful, but it sure seems suspicious. You might try showing that all the eigenvalues are always positive, which would mean the matrix is positive definite (implying nonsingular).&lt;/p&gt;&#xA;" OwnerUserId="8311" LastEditorUserId="8311" LastEditorDisplayName="" LastEditDate="2011-03-16T01:49:41.423" LastActivityDate="2011-03-16T01:49:41.423" CommentCount="3" />
  <row Id="27315" PostTypeId="2" ParentId="27304" CreationDate="2011-03-16T01:33:58.683" Score="0" ViewCount="" Body="&lt;p&gt;I don't think you can simplify it any further.  Your simplification does not work because $\sum_{n=1}^\infty \sum_{m=1}^\infty |a_{n}|^{2}|b_{m}|^{2}$ is actually an upper bound for your expression.  To see this, test the example $1*1 + 1*(-1) + \cdots = 0$.&lt;/p&gt;&#xA;" OwnerUserId="8173" LastActivityDate="2011-03-16T01:33:58.683" CommentCount="2" />
  <row Id="27316" PostTypeId="2" ParentId="27310" CreationDate="2011-03-16T01:36:59.067" Score="1" ViewCount="" Body="&lt;p&gt;You forgot to substitute the integral limits. Also I think your description of the problem is missing some premise -- is $f$ periodic with period $2\pi$?&lt;/p&gt;&#xA;" OwnerUserId="6622" LastActivityDate="2011-03-16T01:36:59.067" CommentCount="15" />
  <row Id="27317" PostTypeId="2" ParentId="27309" CreationDate="2011-03-16T01:38:40.537" Score="6" ViewCount="" Body="&lt;p&gt;This descent has a very beautiful presentation based upon ideas going back to Fibonacci.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Fibonacci's Lost Theorem&lt;/strong&gt; $\: $ The area of an integral &#xA;pythagorean triangle is not a square integer. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Over 400 years before Fermat's celebrated proof by infinite descent of the essentially equivalent $\rm\:FLT_4\:$ (Fermat's Last theorem for exponent $4$), Fibonacci claimed &#xA;to have a proof of this in his &lt;em&gt;Liber Quadratorum&lt;/em&gt; (Book of Squares). But, alas, to this day, his proof has never been found.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Below is my speculative reconstruction of Fibonacci's proof of this theorem, based upon similar ideas that survived in his extensive studies on squares and related topics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A &lt;em&gt;square&lt;/em&gt; arithmetic progression (SAP) is an AP  $\rm\ x^2,\ y^2,\ z^2\ $ with a square stepsize $\rm\ s^2\:,\ $ viz. $$\rm\ x^2\ \ \xrightarrow{s^2}\ \ y^2\ \ \xrightarrow{s^2}\ \ z^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Naturally associated with every SAP is a &quot;half square triangle&quot;, &#xA;ie. doubling  $\rm\ z^2 + x^2\ $  produces a triangle of square area $\rm\ s^2\ $ viz. &#xA;   $\rm\ (z + x)^2 + (z - x)^2\  =\  2\ (z^2 + x^2)\  =\  4\ y^2\ $ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;which indeed has $\ \ $ area  $\rm\ =\  (z + x)\ (z - x)/2\  = \ (z^2 - x^2)/2\  =\  s^2\ $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With these concepts in mind, the proof is very easy: &#xA;If there exists a pythagorean triangle with square area then it &#xA;may be primitivized and its area remains square. Let its primitive &#xA;parametrization be $\rm\:(a,b)\:$ and let its area be $\rm\:c^2\:,\:$ namely &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\rm\ \frac{1}2\ leg_1\ leg_2\ =\ \frac{1}2\ (2\:a\:b)\ (a^2-b^2)\ =\ (a-b)\ a\ (a+b)\ b\  =\  c^2\:.\ $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $\rm\:a\:$ and $\rm\:b\:$ are coprime of opposite parity,  $\rm\ a-b,\ a,\ a+b,\ b\ $ &#xA;are coprime factors of a square, hence must all be squares. &#xA;Then  $\rm\ a-b,\ a,\ a+b\ $  form a SAP; doubling its half square triangle &#xA;yields a triangle with smaller square area $\rm\ b &amp;lt; c^2\:,\ $ hence descent. $\ \ $ &lt;strong&gt;QED&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: the doubling construction was already in Euclid; it may be &#xA;  viewed as a composition of forms  $\rm\ (z^2 + x^2)\ (1^2 + 1^2)\:. $ &lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2011-03-28T18:13:59.347" LastActivityDate="2011-03-28T18:13:59.347" CommentCount="2" />
  <row Id="27318" PostTypeId="2" ParentId="27202" CreationDate="2011-03-16T01:45:44.780" Score="0" ViewCount="" Body="&lt;p&gt;I second Peláez's answer. In addition, I want to explain this mistake of yours: “I assumed that 10% faster means it takes 10% less time (60 seconds down to 54 seconds).” 10% normalized = 0.1 , oldTime=60 , and the correct result is not 54:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$newTime = oldTime/(1+0.1) = 54.54545454545454\dots$&#xA;$\neq 54 = oldTime\cdot(1-0.1)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $X\to 0$, then the difference between formulas $\to 0$, so in calculations $\cdot(1-X)$ is often used. In words, 10% faster means that something moves 10% further.&lt;/p&gt;&#xA;" OwnerUserId="7011" LastActivityDate="2011-03-16T01:45:44.780" />
  <row Id="27319" PostTypeId="2" ParentId="27297" CreationDate="2011-03-16T01:46:51.543" Score="0" ViewCount="" Body="&lt;p&gt;Let's say you have three vectors $u_{1,1}, u_{1,2}, u_{1,3}$ that span $\mathbb R^3$ and you have a fourth vector $v_1$, so you can write:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$v_1 = \tau_1 u_{1,1} + \tau_2 u_{1,2} + \tau_3 u_{1,3}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finding the $\tau$:s is just solving a linear system. Now, let's say that also have vectors $v_2$, $v_3$ such that $v_1, v_2, v_3$ span $\mathbb R^3$. And that you can decompose $v_2$ and $v_3$ as you decomposed $u_3$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$v_2 = \tau_1 u_{2,1} + \tau_2 u_{2,2} + \tau_3 u_{2,3}$$&#xA;$$v_3 = \tau_1 u_{3,1} + \tau_2 u_{3,2} + \tau_3 u_{3,3}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and you have an additional vector $w = \alpha_1 v_1 + \alpha_2 v_2 + \alpha_3 v_3$. Now you can write:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xD;&#xA;w &amp;amp;= \frac{1}{2} \left( \alpha_1 v_1 + \alpha_2 v_2 + \alpha_3 v_3 \right) +  \frac{1}{2} \left( \alpha_1 v_1 + \alpha_2 v_2 + \alpha_3 v_3 \right) = \\&#xD;&#xA;&amp;amp;= \frac{1}{2} \left( \alpha_1 v_1 + \alpha_2 v_2 + \alpha_3 v_3 \right) +&#xD;&#xA;\frac{1}{2}\alpha_1 \left(\tau_1 u_{1,1} + \tau_2 u_{1,2} + \tau_3 u_{1,3} \right) + \\&#xD;&#xA;&amp;amp;+ \frac{1}{2}\alpha_2 \left( \tau_1 u_{2,1} + \tau_2 u_{2,2} + \tau_3 u_{2,3} \right) +&#xD;&#xA;\frac{1}{2}\alpha_3 \left( \tau_1 u_{3,1} + \tau_2 u_{3,2} + \tau_3 u_{3,3} \right)&#xD;&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can continue this process as long as you need (decomposing the $u_{i,j}$ and so on). The $v$ and $u$-vectors are your vectors that you have got from the points. Of course, all the decompositions of these can be calculated beforehand, if you want to apply this to more than one vector, and you just calculate the $\alpha$:s for every vector. You will get zeros if one of the $v$:s is not part of $w$, and this zero will &quot;trickle down&quot; to produce a lot of zeros. You might want to have &quot;backup decompositions&quot; if you get a zero $\alpha$ (if this is an option). Of course, when you choose which vectors should go where, you should make sure that all coefficients are non-zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You don't have to decompose in halves, as I have done. You can use any other constants as long as they sum to 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, this decomposition is not unique. And technically, this is not a basis. A basis is linearly independent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I am curious, why do you want this? Can't you just calculate the image of every point?&lt;/p&gt;&#xA;" OwnerUserId="5966" LastEditorUserId="5966" LastEditorDisplayName="" LastEditDate="2011-03-16T02:07:56.253" LastActivityDate="2011-03-16T02:07:56.253" CommentCount="3" />
  <row Id="27320" PostTypeId="2" ParentId="27304" CreationDate="2011-03-16T02:30:03.323" Score="1" ViewCount="" Body="&lt;p&gt;You could simplify it some this way: &#xA;$$\left \langle \sum_{n=1}^\infty a_{n}b_{n},\sum_{m=1}^\infty a_{m}b_{m}\right \rangle=\lim_{N \to \infty} \lim_{M\to \infty} \left \langle \sum_{n=1}^N a_{n}b_{n},\sum_{m=1}^M a_{m}b_{m}\right \rangle = \lim_{N \to \infty} \lim_{M\to \infty} \sum_{n=1}^N \sum_{m=1}^M \left \langle a_{n}b_{n}, a_{m}b_{m}\right \rangle$$ $$ = \sum_{n=1}^\infty \sum_{m=1}^\infty \left \langle a_{n}b_{n}, a_{m}b_{m}\right \rangle=\sum_{n=1}^\infty \sum_{m=1}^\infty a_{n}b_{n} \overline{a_{m}b_{m}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are some details in between that I didn't want to type, and I'm a little iffy about the convergence, but I think it works.  You can pass the limit outside the inner product because inner products are continuous.  Your expression is missing the cross terms.&lt;/p&gt;&#xA;" OwnerUserId="4266" LastEditorUserId="4266" LastEditorDisplayName="" LastEditDate="2011-03-16T02:49:54.793" LastActivityDate="2011-03-16T02:49:54.793" />
  <row Id="27321" PostTypeId="2" ParentId="27308" CreationDate="2011-03-16T02:37:18.913" Score="5" ViewCount="" Body="&lt;p&gt;Yes, every ideal class contains a split prime.  Here is one possible proof, which however doesn't satisfy the conditions of part B of your question:&#xA;the analogue of Dirichlet's argument, using ideal class characters, shows that&#xA;$\sum_{\mathfrak p \in [I]} N\mathfrak p^{-1}$ (the sum being taken over prime ideals in the class of $I$) diverges.  On the other hand, the sum over non-split primes converges (because a non-split prime lying over $p$ has norm at least&#xA;$p^2$, and so the sum over non-split primes is majorized by (some constant times)&#xA;$\sum_p p^{-2}$, which converges).  Thus there must be infinitely many split primes contributing to this sum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: As the OP points out in a comment below, this argument applies only in the Galois case, and so doesn't actually address the question. &lt;/p&gt;&#xA;" OwnerUserId="221" LastEditorUserId="221" LastEditorDisplayName="" LastEditDate="2011-03-16T05:23:05.773" LastActivityDate="2011-03-16T05:23:05.773" CommentCount="2" />
  <row Id="27322" PostTypeId="2" ParentId="27294" CreationDate="2011-03-16T02:40:57.007" Score="3" ViewCount="" Body="&lt;p&gt;If $m = 2k+1$, this sum always contains the two different terms $J_n(x)J_{n+m}(x)$ and $J_{-(n+m)}(x)J_{-n}(x)$. I believe we can rewrite this sum as $$\sum_{n=-k}^\infty (J_n(x)J_{n+m}(x) + J_{-(n+m)}(x)J_{-n}(x))$$ and then by using the identity $J_{-r}(x) = (-1)^{r}J_r(x)$ we can rewrite this as $$\sum_{n=-k}^\infty (J_n(x)J_{n+m}(x) + (-1)^{2n+m} J_{n+m}(x)J_n(x) = \sum_{n=-k}^\infty (J_n(x)J_{n+m}(x) - J_{n+m}(x)J_n(x)) = 0.$$ So that does it for odd $m$, assuming that reordering the terms is legal in this case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For even $m = 2k \neq 0$ I haven't a clue. You can try this same trick but it doesn't work, because all the $-1$s cancel out completely rather than leaving one behind to cancel out the whole sum. Plus there's a single awkward $J_{-k}(x)J_{k}(x)$ term in the middle. I'm betting the trick is to use the recurrence $${2r \over x}J_r(x) = J_{r-1}(x) + J_{r+1}(x)$$ to unzip the whole thing starting from the awkward term into two similar sums with $m&amp;#39; = m \pm 1$ (which obviously won't work if $m = 0$, as expected), and then use the above to show both parts are zero. But I have no concrete reason to believe that this will actually work, just a gut instinct.&lt;/p&gt;&#xA;" OwnerUserId="8311" LastEditorUserId="8311" LastEditorDisplayName="" LastEditDate="2011-03-17T20:39:52.733" LastActivityDate="2011-03-17T20:39:52.733" CommentCount="4" />
  <row Id="27324" PostTypeId="1" AcceptedAnswerId="27335" CreationDate="2011-03-16T03:13:50.443" Score="19" ViewCount="394" Body="&lt;p&gt;I've seen the integral $\displaystyle \int_0^1\ln(1-x^2)\;{dx}$ on a thread in this forum and I tried to calculate it by using power series. I wrote the integral as a sum then again as an integral. Here is my calculation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle \begin{aligned} \int_0^1 \ln\left(1-x^2\right)\;{dx}  &amp;amp; = -\int_0^1\sum_{k \ge 0}\frac{x^{2k+2}}{k+1}\;{dx}  = -\sum_{k \ge 0}\int_{0}^{1}\frac{x^{2k+2}}{k+1}\;{dx} \\&amp;amp; = -\sum_{k \ge 0}\bigg[\frac{x^{2k+3}}{(k+1)(2k+3)}\bigg]_0^1  = -\sum_{k \ge 0}\frac{1}{(k+1)(2k+3)} \\&amp;amp; = -\sum_{k \ge 0}\frac{(2k+3)-2(k+1)}{(k+1)(2k+3)} = -\sum_{k\ge 0}\bigg(\frac{1}{k+1}-\frac{2}{2k+3}\bigg) \\&amp;amp; \color{blue}{= -\sum_{k\ge 0}\int_0^1\bigg(x^{k}-2x^{2k+2}\bigg)\;{dx} = -\int_0^1\sum_{k\ge 0}\bigg( x^{k}-2x^{2k+2}\bigg)\;{dx}} \\&amp;amp;= \int_0^1\frac{1}{1+x}-2\;{dx}  = \bigg[\ln(1+x)-2x\bigg]_0^1 = \ln{2}-2.\end{aligned} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I should have got $2\ln{2}-2$, not $\ln{2}-2$. I'm thinking that either I made a very silly mistake, or the blue step is wrong (probably the order cannot be switched, although I don't know why)!&lt;/p&gt;&#xA;" OwnerUserId="8164" LastEditorUserId="8164" LastEditorDisplayName="" LastEditDate="2011-03-16T05:46:45.683" LastActivityDate="2011-03-19T19:46:25.660" Title="On calculating $\displaystyle \int_0^1\ln(1-x^2)\;{dx}$ -- where is the mistake?" Tags="&lt;analysis&gt;&lt;sequences-and-series&gt;&lt;integral&gt;&lt;power-series&gt;" AnswerCount="2" CommentCount="8" FavoriteCount="3" />
  <row Id="27325" PostTypeId="2" ParentId="27119" CreationDate="2011-03-16T03:24:14.003" Score="2" ViewCount="" Body="&lt;p&gt;I will use $\mathbf{F}_p$ to denote the field of $p$ elements (what you denote as $\mathbb{Z}_p$), $\mathbf{F}_{p^k}$ to denote the field of $p^k$ elements, and $\mathbf{F}$ for the algebraic closure of $\mathbf{F}_p$ (what the OP calls $F$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your work in (i) is very rough (and at point unintelligible). For instance, you write &quot;$\gcd(f,f')$ (i.e., the polynomials are separable)&quot;. That is, &lt;em&gt;prima facie&lt;/em&gt;, nonsensical. (Did you forget to say &lt;em&gt;what&lt;/em&gt; the greatest common divisor was equal to?). Also, you are only dealing with certain kinds of polynomials. Finally, you did not say what you were going to do with $S$, it was just left hanging there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Presumably, you were trying to say: &quot;take $S$ to be the set of all polynomials of the form $x^{p^n}-x$, with $n$ a positive integer. These polynomials are separable, and $F$ is the splitting field of $S$. Thus, $F$ is Galois over $\mathbf{F}_p$, since it is the splitting field of a set of separable polynomials.&quot; That is okay as far as it goes, but how do you know that $\mathbf{F}$ &lt;em&gt;is&lt;/em&gt; the splitting field of $S$? It is clear that the splitting field of $S$ is contained in $\mathbf{F}$, but you have provided not argument (or ghost of one) for showing that every element of $\mathbf{F}$ must be in the splitting field of $S$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For that, you need to argue that given any $a\in\mathbf{F}$, you know that $[\mathbf{F}_p(a):\mathbf{F}]$ is finite (why?), hence $\mathbf{F}_p(a) = \mathbf{F}_{p^k}$ for some $k$ (why?), and so $a$ is a root of $x^{p^k}-x$ (why?), hence $a$ is in the splitting field of $S$. This will establish (i) correctly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For (ii), proving that $u\mapsto u^p$ is indeed an automorphism is straightforward. To prove that it is not the identity... Think about $\mathbf{F}_{p^2}$. What do you know about its multiplicative subgroup? Can exponentiation by $p$ be the identity map on all those elements? Are they in $\mathbf{F}$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For (iii): the point here is to show that in the infinite extension case, the correspondence between subgroups of the Galois group and subfields of the extension no longer holds. You are being asked to prove that there is more than one subgroup whose fixed field is precisely the ground field (if you think about the finite Galois case, that &lt;em&gt;never&lt;/em&gt; happens). Explicitly, you are being asked to show two things: that the fixed field of the subgroup generated by $\alpha$ is precisely the ground field; and that the subgroup generated by $\alpha$ is &lt;strong&gt;not&lt;/strong&gt; all of the Galois group. This will show that there are at least two distinct subgroups (both $H$ and all of $\mathrm{Aut}(\mathbf{F}/\mathbf{F}_p)$) that have the same fixed field, which shows the correspondence you know from the finite extension case no longer holds here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first part should be easy if you have managed to understand part (ii).&#xA;First, pick $a\in \mathbf{F}$, $a\notin \mathbf{F}_p$. Try thinking about $\mathbf{F}(a)$, and combining the ideas of (i) and (ii), use them to show that $\alpha(a)\neq a$, so that the fixed field of $\alpha$ is contained in $\mathbf{F}_p$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The trickier part is to show that the powers of $\alpha$ are &lt;em&gt;not&lt;/em&gt; the only automorphisms of the algebraic closure. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Added.&lt;/em&gt; Exhibing an explicit element of the Galois group that is not a power of $\alpha$ is fairly straightforward if you know what is going on behind the scenes, but it would likely be challenging when doing this problem, since it would seem to me that the point of the problem is to show you that there is &lt;em&gt;something&lt;/em&gt; going on behind the scenes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can show that the automorphism group is not cyclic by exhibiting two elements of infinite order such that the subgroups they generate intersection trivially. One possibility: for distinct primes $q$ and $\ell$, consider the two towers&#xA;\begin{align*}&#xA;&amp;amp;\mathbf{F}_p \subseteq \mathbf{F}_{p^q}\subseteq \mathbf{F}_{p^{q^2}}\subseteq\cdots\\&#xA;&amp;amp;\mathbf{F}_p \subseteq \mathbf{F}_{p^{\ell}} \subseteq \mathbf{F}_{p^{\ell^2}}\subseteq\cdots&#xA;\end{align*}&#xA;The two towers are dijsoint: since $\mathbf{F}_{p^n}\subseteq \mathbf{F}_{p^m}$ if and only if $n|m$, any field from one tower intersects any field in the other tower at just $\mathbf{F}_p$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Use the usual properties of extension of isomorphisms to show that there is an element in $\mathrm{Aut}(\mathbf{F}/\mathbf{F}_p)$ that acts like $\alpha$ on the first tower, but like the identity on the second tower. Then show that there is an element in $\mathrm{Aut}(\mathbf{F}/\mathbf{F}_p)$ that acts like $\alpha$ on the second tower and like the identity on the first tower. Call these $\beta_1$ and $\beta_2$. Now note that $\langle\beta_1\rangle\cap\langle\beta_2\rangle = \{1\}$. But if $\mathrm{Aut}(\mathbf{F}/\mathbf{F}_p) = \langle\alpha\rangle$, then any two nontrivial subgroups have a nontrivial intersection. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added.&lt;/strong&gt; I'm really at a loss on trying to give you enough without entirely spoiling the problem; based on the confusion that this (nontrivial) problem has engendered, I'm guessing that you are still very much a beginner with Galois groups in general and finite fields in particular...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another thing you can do: since $\alpha$ has infinite order, it would be enough to exhibit an element of $\mathrm{Aut}(\mathbf{F}/\mathbf{F}_p)$ that has &lt;strong&gt;finite&lt;/strong&gt; order. Again, you can try to do this by using extension properties of automorphisms. &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-16T19:51:13.547" LastActivityDate="2011-03-16T19:51:13.547" CommentCount="5" />
  <row Id="27326" PostTypeId="2" ParentId="27324" CreationDate="2011-03-16T03:25:39.850" Score="0" ViewCount="" Body="&lt;p&gt;Why not try splitting $\ln(1-x^2)$ as $\ln(1-x)+\ln(1+x)$ before applying the Mercator series?&lt;/p&gt;&#xA;" OwnerUserId="8321" LastActivityDate="2011-03-16T03:25:39.850" CommentCount="1" />
  <row Id="27328" PostTypeId="1" CreationDate="2011-03-16T03:47:04.953" Score="5" ViewCount="197" Body="&lt;p&gt;This was a mathematical induction question proposed in a textbook, and I've exhausted multiple approaches (proving RHS - LHS &gt; 0, splitting the fraction, fractional exponents, etc.)&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The geometric mean of $n$ positive numbers $a_1, a_2,\ldots,a_n$ is $\sqrt[n]{a_1a_2 \ldots a_n}$ and their arithmetic mean is $\frac{a_1+a_2+\ldots+a_n}{n}$. If $a_1, a_2,\ldots,a_n$ are $n$ positive real numbers, prove by induction that their geometric mean is always smaller than or equal to their arithmetic mean, i.e. $\sqrt[n]{a_1a_2\ldots a_n} \leq \frac{a_1+a_2+\ldots+a_n}{n}$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="6520" LastEditorUserId="6520" LastEditorDisplayName="" LastEditDate="2011-03-16T05:20:37.387" LastActivityDate="2011-05-01T20:47:52.400" Title="Geometric mean never exceeds arithmetic mean" Tags="&lt;sequences-and-series&gt;&lt;algebra&gt;&lt;statistics&gt;&lt;inequality&gt;&lt;induction&gt;" AnswerCount="4" CommentCount="1" />
  <row Id="27329" PostTypeId="1" CreationDate="2011-03-16T03:48:49.240" Score="1" ViewCount="78" Body="&lt;p&gt;I want to show through Boolean algebra that the statement &#xA;$$\neg Q\Rightarrow (R\Rightarrow \neg (P\land Q))$$&#xA;is equivalent to the tautology $Q\lor(\neg Q)$?&lt;/p&gt;&#xA;" OwnerUserId="8322" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-16T03:53:38.783" LastActivityDate="2011-03-16T14:39:17.467" Title="How do I show a statement is a tautology in the Booleans algebra" Tags="&lt;discrete-mathematics&gt;" AnswerCount="1" CommentCount="0" />
  <row Id="27330" PostTypeId="2" ParentId="11261" CreationDate="2011-03-16T03:50:17.520" Score="1" ViewCount="" Body="&lt;p&gt;I think &lt;a href=&quot;http://www.hicalc.com&quot; rel=&quot;nofollow&quot;&gt;http://www.hicalc.com&lt;/a&gt; is good. This is a smple &lt;a href=&quot;http://www.hicalc.com/program/24&quot; rel=&quot;nofollow&quot;&gt;http://www.hicalc.com/program/24&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hicalc is online platform to calculate expression quickly or create program with a simple, intuitive interface and share them with other people.&#xA;You can create program with almost all C# 4.0 syntax such as if, switch, while, for, foreach, etc., store and calculate your program anywhere at any time, share and Search Programs in public library. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;With Hicalc, you can customize any converter and share with other people&lt;/p&gt;&#xA;" OwnerUserId="8323" LastActivityDate="2011-03-16T03:50:17.520" CommunityOwnedDate="2011-03-16T03:50:17.520" />
  <row Id="27331" PostTypeId="1" AcceptedAnswerId="27339" CreationDate="2011-03-16T03:50:23.020" Score="1" ViewCount="66" Body="&lt;p&gt;Let I(x) be a function such that I(x)=0 if x$\leq$0, and I(x)=1 if x$&gt;0$.  Let $c_{n}$  be a sequence of numbers such that $\sum$ c$_n$ converges absolutely.  Let $x_n$ be some sequence of distinct real numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $f_n(x)=\sum_{i=1}^{n} c_i I(x-x_i)$ and $f(x)=\sum_{i=1}^{\infty}c_iI(x-x_i)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm kind of stumped as to how to demonstrate both point-wise convergence and uniform convergence.&lt;/p&gt;&#xA;" OwnerUserId="8324" LastEditorUserId="6491" LastEditorDisplayName="" LastEditDate="2011-03-16T03:57:20.543" LastActivityDate="2011-03-16T04:21:37.743" Title="Analysis, determing pointwise and uniform convergence." Tags="&lt;real-analysis&gt;&lt;analysis&gt;" AnswerCount="2" />
  <row Id="27332" PostTypeId="2" ParentId="27328" CreationDate="2011-03-16T03:55:45.210" Score="2" ViewCount="" Body="&lt;p&gt;Hint:  For induction, you need to prove the base case: $\sqrt{a_1+a_2} \le \frac{a_1+a_2}{2}$.  If you square both sides...  Then you need to prove that if it works for $n$ numbers, it works for $n+1$, so put the averages for $a_1 \ldots a_n$ in for $a_1$.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-03-16T03:55:45.210" />
  <row Id="27333" PostTypeId="2" ParentId="27329" CreationDate="2011-03-16T03:56:35.160" Score="3" ViewCount="" Body="&lt;p&gt;Don't know exactly what you mean by &quot;Boolean algebra&quot;, given that implications are not usually part of the language of Boolean algebras. If I translate $A\Rightarrow B$ to the logically equivalent $\neg A \lor B$, then it is straightforward:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align*}&#xA;\neg Q\Rightarrow (R\Rightarrow (\neg (P\land Q)) &amp;amp;\equiv \neg(\neg Q)\lor (R\Rightarrow (\neg (P\land Q)))\\&#xA;&amp;amp;\equiv Q \lor (\neg R \lor (\neg(P\land Q)))\\&#xA;&amp;amp;\equiv Q\lor (\neg R \lor (\neg P \lor \neg Q))\\&#xA;&amp;amp;\equiv Q \lor \neg R \lor \neg P \lor \neg Q\\&#xA;&amp;amp;\equiv (Q\lor \neg Q) \lor (R\lor \neg P)\\&#xA;&amp;amp;\equiv Q\lor \neg Q.&#xA;\end{align*}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added.&lt;/strong&gt; In the comments you ask how to prove it by contradiction: here we use the fact that $\neg(A\Rightarrow B) \equiv A\land \neg B$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align*}&#xA;\neg\Bigl(\neg Q\Rightarrow (R\Rightarrow (\neg (P\land Q))\Bigr) &amp;amp;\equiv&#xA;\neg Q \land \neg\Bigl(R\Rightarrow (\neg (P\land Q))\Bigr)\\&#xA;&amp;amp;\equiv \neg Q \land \Bigl( R \land \neg(\neg(P\land Q))\Bigr)\\&#xA;&amp;amp;\equiv \neg Q \land (R \land (P\land Q))\\&#xA;&amp;amp;\equiv \neg Q \land R \land P \land Q\\&#xA;&amp;amp;\equiv (\neg Q\land Q) \land R \land P\\&#xA;&amp;amp;\equiv (\neg Q\land Q).&#xA;\end{align*}&#xA;Since the negation of the proposition is equivalent to a contradiction, the original proposition is a tautology, i.e., true.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-16T04:30:02.387" LastActivityDate="2011-03-16T04:30:02.387" CommentCount="11" />
  <row Id="27334" PostTypeId="2" ParentId="27328" CreationDate="2011-03-16T03:58:04.180" Score="4" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Hint:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using induction:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Show the base case $n = 2$. &lt;/li&gt;&#xA;&lt;li&gt;Show that if the statement is true for $2^n$, then it is true for $2^{n+1}$.&lt;/li&gt;&#xA;&lt;li&gt;Show that if the statement is true for $n$, then it is true for $n - 1$.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;(This method of induction is sometimes called &lt;a href=&quot;http://www.artofproblemsolving.com/Wiki/index.php/Cauchy_Induction&quot; rel=&quot;nofollow&quot;&gt;Cauchy induction&lt;/a&gt;).&lt;/p&gt;&#xA;" OwnerUserId="6491" LastActivityDate="2011-03-16T03:58:04.180" CommentCount="5" />
  <row Id="27335" PostTypeId="2" ParentId="27324" CreationDate="2011-03-16T03:59:16.717" Score="12" ViewCount="" Body="&lt;p&gt;This is quite interesting. I think the problem does lie in the blue line. The sum over $\frac{1}{k+1}-\frac{2}{2k+3}$ converges absolutely, but the sums over the two terms individually diverge. What you're doing when you combine the two series as you do is effectively to move the terms $\frac{2}{2k+3}$ back to twice the $k$ value from where they belong, and that's a different series; you can only do that when the two terms absolutely converge individually.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you're effectively calculating by &quot;slowing down&quot; $k$ in the second term is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$-\sum_{k\ge0}\left(\frac{1}{2k+1}+\frac{1}{2k+2}-\frac{2}{2k+3}\right)\;,$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and that is indeed $\ln 2 - 2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[Update]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The above explanation refers to the series above the blue line to show that exchanging the sum and integral in the blue line can't be valid. But of course that should also be decidable from just the blue line itself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First off, note that actually the first step after the blue line isn't quite right, either: At $x=1$, the sum in the integrand actually diverges to $-\infty$, so the value $-3/2$ of the integrand in the last line for $x=1$ is obtained by rearranging a divergent series. But that problem only occurs at $x=1$; for all other values of $x$, the series converges absolutely and may validly be arranged to obtain the integrand in the last line. But this discontinuity at $x=1$ already shows that the partial sums don't converge uniformly. Indeed&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$-\sum_{k=0}^{n-1}\left(x^k-2x^{2k+2}\right)=-\frac{1-x^n}{1-x}+2x^2\frac{1-x^{2n}}{1-x^2}=\frac{1-x^n}{1-x}\left(-1+2x^2\frac{1+x^n}{1+x}\right)\;,$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which for large $n$ is close to the integrand in the last line for most $x$ but close to $x=1$ suddenly becomes positive and ends up at $n$ for $x=1$, as it must. &lt;a href=&quot;http://www.wolframalpha.com/input/?i=plot+%28%28%281-x%5E100%29%2F%281-x%29%29+%2a+%28-1+%2B+2+%2a+x%5E2+%2a+%281+%2B+x%5E100%29+%2F+%281+%2B+x%29%29%29+from+x+%3D+0+to+1+from+y+%3D+-2+to+2&quot; rel=&quot;nofollow&quot;&gt;Here&lt;/a&gt;'s a plot for $n=100$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://math.asu.edu/~jss/courses/fall06/mat472/limit_interchange.pdf&quot; rel=&quot;nofollow&quot;&gt;Here&lt;/a&gt;'s a very nice general treatment of the interchange of limits based on uniform convergence, which can be applied to integrals, derivatives and sums alike. Since the series in the integrand diverges at $x=1$, we have to treat the integral as a limit taking the upper integration limit to $1$, and this limiting operation and the one in the series can only be interchanged if the uniform convergence criterion is fulfilled.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(&lt;a href=&quot;http://www.mathreference.com/lc-ser,intlim.html&quot; rel=&quot;nofollow&quot;&gt;Here&lt;/a&gt;'s another example where an integral and a function limit can't be interchanged due to lack of uniform convergence; this one is a bit simpler because there's no sum involved.)&lt;/p&gt;&#xA;" OwnerUserId="6622" LastEditorUserId="6622" LastEditorDisplayName="" LastEditDate="2011-03-19T19:46:25.660" LastActivityDate="2011-03-19T19:46:25.660" CommentCount="3" />
  <row Id="27336" PostTypeId="1" AcceptedAnswerId="27337" CreationDate="2011-03-16T04:06:30.940" Score="1" ViewCount="192" Body="&lt;p&gt;Textbooks usually state &quot;&lt;em&gt;it is not hard to check that in modular arithmetic the usual associative, commutative and distributive properties continue to apply&lt;/em&gt;&quot;. Is there a way other than tedious proof by case analysis? For example, perhaps it is more convincing to notice that, say Z/13, is cyclic group over addition and multiplication and, therefore, commutativity and associativity follows? What about distributivity?&lt;/p&gt;&#xA;" OwnerUserId="5754" LastActivityDate="2011-03-16T05:27:27.677" Title="associativity, commutativity and distributivity of modulo arithmetic" Tags="&lt;algebra&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" />
  <row Id="27337" PostTypeId="2" ParentId="27336" CreationDate="2011-03-16T04:15:16.877" Score="6" ViewCount="" Body="&lt;p&gt;To prove (or &quot;notice&quot;) that $\mathbb{Z}/13\mathbb{Z}$ is a (cyclic) group under addition you would need to know first of all that it is a group under addition; and in order to know that it is a group under addition, you would need to know that addition is associative, has an identity element, and has inverses; so you have to show it is associative. You cannot save yourself knowing/showing it is associative by first showing it's a group: showing/knowing associativity is a prerequisite of it being a group. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You don't have to prove it by &quot;case analysis&quot;, but only through the basic properties of congruences. The key is really to show that addition and multiplication &lt;em&gt;are well defined&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is, we define addition and multiplication in $\mathbb{Z}/n\mathbb{Z}$ by &quot;class representatives&quot;: if $[a]$ is the modular class of $a$, and $[b]$ is the modular class of $b$, we define $[a]+[b]$ as $[a+b]$ and $[a][b]$ as $[ab]$. So long as these operations are &lt;em&gt;well-defined&lt;/em&gt;, the properties will be inherited from the properties of the usual addition and multiplication of integers (just like in any ring modulo an ideal).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To show it is well-defined, you need to show that if $[a]=[x]$ and $[b]=[y]$, then $[a+b]=[x+y]$ and $[ax]=[by]$. This is a direct consequence of the properties of congruences, namely that if $a\equiv x\pmod{n}$ and $b\equiv y\pmod{n}$, then $a+b\equiv x+y\pmod{n}$ and $ab\equiv xy\pmod{n}$. Verifying these through the definition of congruence is indeed trivial: If $a-x$ and $b-y$ are multiples of $n$, then so is $(a-x)+(b-y) = (a+b)-(x+y)$; and likewise, $(a-x)b + (b-y)x = ab - xy$ is also a multiple of $n$, being a sum of multiples of $n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The properties of the addition and multiplication of congruence classes now follows from the same properties for integers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added.&lt;/strong&gt; It's important to note that it is not merely the fact that we are dealing with equivalence classes that makes this work. The key is really that the operations on equivalence classes defined via representatives are &lt;strong&gt;well defined&lt;/strong&gt;.  We have a set $S$ on which we already have a (binary) operation $*$, and an equivalence relation $\sim$ on $S$. If we want to define an operation on the congruence classes $S/\sim$ by using $*$, then the natural definition is to try $[a]*[b]=[a*b]$. That is: the &quot;product&quot; of the classes is the class of the &quot;product&quot;. However, this does not always work correctly: it will depend on the operation and how $\sim$ relates to the operation. In the case of groups, for example, the &lt;strong&gt;only&lt;/strong&gt; equivalence relations for which this definition works are the equivalence relations induced by normal subgroups (see &lt;a href=&quot;http://math.stackexchange.com/questions/14282/why-do-we-define-quotient-groups-for-normal-subgroups-only/14315#14315&quot;&gt;my answer to this previous question&lt;/a&gt; for an ample discussion of trying to define an operation on equivalence classes in a group). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So it is not the fact that we have an equivalence relation/partition, but really &lt;em&gt;how&lt;/em&gt; the partition behaves relative to the operation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What we have is the following general result from Universal Algebra:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Theorem.&lt;/strong&gt; Let $S$ be a set, and let $*$ be an $n$-ary operation on $S$. If $\sim$ is an equivalence relation on $S$, then the operation on $S/\sim$ defined by&#xA;$$*([s_1],\ldots,[s_n]) = [*(s_1,\ldots,s_n)]$$&#xA;is well-defined if and only if $\sim$ is closed under $*\times*$ as a subset of $S\times S$, where &#xA;$$*\times*\Bigl((a_1,b_1),\ldots,(a_n,b_n)\Bigr) = \Bigl(*(a_1,\ldots,a_n),*(b_1,\ldots,b_n)\Bigr).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Proof.&lt;/em&gt; Suppose $*$ is well defined. If $(a_1,b_1),\ldots,(a_n,b_n)\in \sim$, then $[a_i]=[b_i]$, so by assumption we have that &#xA;$$[*(a_1,\ldots,a_n)] = *([a_1],\ldots,[a_n]) = *([b_1],\ldots,[b_n])=[*(b_1,\ldots,b_n)].$$&#xA;Therefore, &#xA;$$*\times *\Bigl( (a_1,b_1),\ldots,(a_n,b_n)\Bigr) = \bigl( *(a_1,\ldots,a_n),*(b_1,\ldots,b_n)\bigr)\in \sim,$$&#xA;so $\sim$ is closed under $*\times *$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conversely, suppose that $\sim$ is closed under $*\times *$, and that $[a_1]=[b_1],\ldots,[a_n]=[b_n]$. We need to show that $*([a_1],\ldots,[a_n]) = *([b_1],\ldots,[b_n])$. Since $(a_1,b_1),\ldots,(a_n,b_n)\in \sim$, then &#xA;$$*\times *\Bigl( (a_1,b_1),\ldots,(a_n,b_n)\Bigr) = \Bigl(*(a_1,\ldots,a_n),*(b_1,\ldots,b_n)\Bigr)\in \sim,$$&#xA;hence $[*(a_1,\ldots,a_n)] = [*(b_1,\ldots,b_n)]$. But this is exactly what we need to show $*$ is well defined on $S/\sim$. &lt;strong&gt;QED&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So &lt;em&gt;if&lt;/em&gt; the equivalence relation is a substructure of $S\times S$, &lt;em&gt;then&lt;/em&gt; we can define the induced operation on $S/\sim$ via representatives, and this operation on $S/\sim$ will inherit the properties of the operation from $S$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, to show that sums and products of congruence classes modulo $m$ in the integers inherit the properties of the sum and product of integers is equivalent to showing that the equivalence relation &quot;congruent modulo $m$&quot;, $\equiv_m$, interpreted as a subset of $\mathbb{Z}\times\mathbb{Z}$, is a subring of $\mathbb{Z}\times\mathbb{Z}$ (the latter with coordinate-wise operations). This is &lt;em&gt;precisely&lt;/em&gt; showing that if $a\equiv x\pmod{m}$ (i.e., $(a,x)\in\equiv_m$) and $b\equiv y\pmod{m}$ (i.e., $(b,y)\in\equiv_m$) , then $a+b\equiv x+y\pmod{m}$ (i.e., $(a,x)+(b,y)=(a+b,x+y)\in\equiv_m$) and $ab\equiv xy\pmod{m}$ (i.e., $(a,x)(b,y) = (ab,xy)\in\equiv_m$). So it still comes down to the &lt;em&gt;specific&lt;/em&gt; properties of the equivalence relation, and not merely to the fact that it &lt;em&gt;is&lt;/em&gt; an equivalence relation.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-16T05:12:40.010" LastActivityDate="2011-03-16T05:12:40.010" CommentCount="17" />
  <row Id="27338" PostTypeId="1" AcceptedAnswerId="27342" CreationDate="2011-03-16T04:15:34.347" Score="4" ViewCount="76" Body="&lt;p&gt;Hello, &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Prove that $x^3 \equiv a \pmod{p}$ has a solution where $p \equiv 2 \pmod{3}$?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;How can I prove a congruence equation has a solution? I tried to link Fermat's little theorem with this problem, but I couldn't find a way to solve it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My attempt was:&#xA;$$x^3 \equiv 1 \pmod{2}$$&#xA;$$x^3 \equiv a \pmod{p}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $p \equiv 2 \pmod{3}$, I have $p = 3k + 2$, for some integers $k$. But I was stuck here :(. Any idea?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another question is, is there are infinitely many primes of the form 3k + 2?&#xA;A hint would be sufficient.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks,&#xA;Thanks,&lt;/p&gt;&#xA;" OwnerUserId="5711" LastEditorUserId="5711" LastEditorDisplayName="" LastEditDate="2011-03-16T04:51:59.527" LastActivityDate="2011-03-16T06:05:03.480" Title="Prove that $x^3 \equiv a \pmod{p}$ has s solution where $p \equiv 2 \pmod{3}$?" Tags="&lt;elementary-number-theory&gt;" AnswerCount="2" CommentCount="6" />
  <row Id="27339" PostTypeId="2" ParentId="27331" CreationDate="2011-03-16T04:16:42.600" Score="2" ViewCount="" Body="&lt;p&gt;To show point-wise convergence, we must show that the sequence $\{f_n(x)=\sum_{i=1}^{n} c_i I(x-x_i)\}$ converges to $f(x)=\sum_{i=1}^{\infty}c_iI(x-x_i)$ for every value of $x$. This will follow trivially from the definition of an infinite series if we can show that the sequence converges at all. Since the space of real numbers is complete, it suffices to show that the sequence $\{f_n(x)=\sum_{i=1}^{n} c_i I(x-x_i)\}$ is Cauchy, that is for each $e &amp;gt; 0$ we can find an $N$ such that $\sum_{i=N}^{\infty} c_i I(x-x_i) &amp;lt; e$. But we can certainly find such an $N$ for $\sum_{i=N}^{\infty} c_i &amp;gt; \sum_{i=N}^{\infty} c_i I(x-x_i)$, thus we have that the sequence is point-wise convergent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Uniform convergence, on the other hand, requires us to be able to choose $N$ independently of the value of $x$. However, we did not use the value of $x$ in selecting $N$ above, so uniform convergence follows as well.&lt;/p&gt;&#xA;" OwnerUserId="8173" LastActivityDate="2011-03-16T04:16:42.600" />
  <row Id="27340" PostTypeId="2" ParentId="27331" CreationDate="2011-03-16T04:21:37.743" Score="2" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Hint:&lt;/strong&gt;  Ignore everything about the the way your function $I(x-x_i)$ is defined, except the fact that $|I(x-x_i)|\leq 1$ for all $x$.  Everything then follows since the sum of the $c_i$ converges absolutely.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $C_n=\sum_{i=1}^n c_i$.  Since $C_n$ is a Cauchy Sequence, we are able to conclude that $f_n (x)$ will be a Cauchy sequence at every $x$.  But the fact that $f_n$ is Cauchy is independent of $x$ since it only depends on the convergence of $\sum c_i$, and hence it is uniformly convergent, as desired.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope that helps,&lt;/p&gt;&#xA;" OwnerUserId="6075" LastActivityDate="2011-03-16T04:21:37.743" />
  <row Id="27341" PostTypeId="2" ParentId="27328" CreationDate="2011-03-16T04:24:43.523" Score="1" ViewCount="" Body="&lt;p&gt;An induction proof which I like, is to prove that&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If $\displaystyle \prod_{j=1}^{n} a_j = 1$ then $\displaystyle \sum_{j=1}^{n} a_j \ge n$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now for the induction step, if $\displaystyle \prod_{j=1}^{n+1} a_j = 1$, pick $\displaystyle a_r$ and $\displaystyle a_s$ such that $\displaystyle  0 \ge (1-a_r)(1-a_s)$ and...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a different proof, one could claim that using the concavity of $\displaystyle \log x$ is an induction proof too (going from $\displaystyle t + (1-t) = 1$ to $\displaystyle t_1 + t_2 + \dots + t_n = 1$).&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2011-03-16T04:24:43.523" />
  <row Id="27342" PostTypeId="2" ParentId="27338" CreationDate="2011-03-16T04:26:05.467" Score="4" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\:$ Show $\rm\: x\to x^3\: $ is a bijection using Fermat's little theorem and $\rm\: 3\ (2\:K+1) = 1 + 2\ (3\:K+1)$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In further detail: $\rm\ \ (x^{2\:K+1})^3\ =\ x\ (x^{3\:K+1})^2\ \equiv\ x\ \ (mod\ P)\ \ $ for $\rm\ x\not\equiv 0\:,\ $ prime $\rm\ P = 3\:K+2\:.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus $\rm\ x\to x^3\ $ is onto on the finite set $\rm\:\mathbb Z/P\:,\:$ so it is also 1-1, i.e. $\rm\ x^3 \equiv y^3\ \Rightarrow\ x\equiv y\:.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: this answers the original version of your question (existence &lt;em&gt;and&lt;/em&gt; uniqueness of cube roots).&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2011-03-16T06:05:03.480" LastActivityDate="2011-03-16T06:05:03.480" CommentCount="2" />
  <row Id="27343" PostTypeId="1" CreationDate="2011-03-16T04:29:22.937" Score="2" ViewCount="55" Body="&lt;p&gt;Explain why the following are true for arbitrary statements P and Q:&#xA;If P is a tautology then (P or Q) ≡ P&lt;/p&gt;&#xA;" OwnerUserId="8322" LastActivityDate="2011-03-16T04:38:42.940" Title="Explain why the following are true for arbitrary statements P and Q:" Tags="&lt;discrete-mathematics&gt;" AnswerCount="1" CommentCount="0" />
  <row Id="27344" PostTypeId="2" ParentId="27343" CreationDate="2011-03-16T04:38:42.940" Score="3" ViewCount="" Body="&lt;p&gt;$P\rightarrow (P\lor Q)$ is a tautology. Truth tables show it immediately; thinking about it shows it in about two and a half instants. And equivalences give it more slowly:&#xA;$$P\Rightarrow (P\lor Q) \equiv \neg P\lor (P\lor Q) \equiv (\neg P\lor P)\lor Q \equiv \neg P\lor P.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So whenver $P$ is true, $P\lor Q$ is also true.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if $P$ is &lt;strong&gt;always&lt;/strong&gt; true, then $P\lor Q$ is &lt;strong&gt;always&lt;/strong&gt; true.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if $P$ is a tautology, then $P\lor Q$ is a tautology; and two tautologies are always equivalent.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2011-03-16T04:38:42.940" />
  <row Id="27345" PostTypeId="1" CreationDate="2011-03-16T04:53:34.650" Score="1" ViewCount="103" Body="&lt;p&gt;Given a topological space $X$ and a finite group $G$, can we construct a covering space $Y$ of $X$, such that the full covering transformation group is isomorphic to $G$. (We may assume some conditions, if required, such as connectedness, path-connectedness etc.)&lt;/p&gt;&#xA;" OwnerUserId="8186" LastActivityDate="2011-04-25T07:36:49.440" Title="Spaces with given covering group" Tags="&lt;algebraic-topology&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="2" />
  <row Id="27346" PostTypeId="2" ParentId="27338" CreationDate="2011-03-16T05:00:43.140" Score="1" ViewCount="" Body="&lt;p&gt;My intuition tells me to attempt a proof by contradiction via factoring:&#xA;Assume $x^3 \equiv y^3 \equiv a \pmod p$ thus $(x-y)(x^2+xy+y^2) = x^3-y^3 \equiv 0$. Since $\mathbb{Z}_p$ is an integral domain and we are assuming $x$ and $y$ are distinct (as elements of $\mathbb{Z}_p$, we must have $x^2+xy+y^2 \equiv 0$, thus $x^2+xy+y^2 = n(3k+2)$ for some $n$. If $n$ is even we have that every term is even, which allows us to factor out $2$ from $x$ and $y$. We can assume this is true since if $x$ is an odd solution, $x+3k+2$ is an even solution as operations on $\mathbb{Z}_p$ are well-defined, and similarly for $y$. Thus we have $x/2$ or $(x+3k+2)/2$ and $y/2$ or $(y+3k+2)/2$ are solutions. You should be able to derive a contradiction from there.&lt;/p&gt;&#xA;" OwnerUserId="8173" LastActivityDate="2011-03-16T05:00:43.140" CommentCount="1" />
  <row Id="27349" PostTypeId="1" CreationDate="2011-03-16T05:14:52.213" Score="6" ViewCount="177" Body="&lt;p&gt;Is $\mathbb Q(\pi,i\pi):\mathbb Q$ a simple extension?&lt;/p&gt;&#xA;" OwnerUserId="8326" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-16T05:17:39.677" LastActivityDate="2011-03-16T05:44:48.903" Title="$\mathbb{Q}(\pi, i\pi)$ over $\mathbb{Q}$" Tags="&lt;abstract-algebra&gt;&lt;field-theory&gt;" AnswerCount="3" CommentCount="3" />
  <row Id="27350" PostTypeId="2" ParentId="27336" CreationDate="2011-03-16T05:27:27.677" Score="1" ViewCount="" Body="&lt;p&gt;That ring axioms are preserved in congruence rings is simply a &lt;em&gt;derived&lt;/em&gt; consequence of the fact that the map to the congruence class $\rm\ n\to [n]\ $ is a &lt;em&gt;homomorphism&lt;/em&gt;, i.e. it preserves all the fundamental operations of the structure. Thus it necessarily preserves expressions &lt;em&gt;composed&lt;/em&gt; of these operations, and hence it preserves all identities (laws) expressed in terms of these operations. For example, let's explicitly derive the preservation of the distributive law in a congruence ring:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\quad\quad\quad [a(b+c)]\ =\ [a][b+c]\ =\ [a]([b]+[c])$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\quad\quad\quad [ab+ac]\ =\ [ab]+[ac]\ =\ [a][b]+ [a][c]$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence $\rm\ [a]([b]+[c])\ =\ [a][b]+[a][c]\ $ i.e. the distributive law holds too in the congruence ring. Precisely the same proof works also for all the other ring laws, e.g. associative, commutative, etc. One is simply lifting a preservation property from the generating operations to expressions composed of those fundamental generating operations.&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2011-03-16T05:27:27.677" />
  <row Id="27351" PostTypeId="2" ParentId="27349" CreationDate="2011-03-16T05:32:46.993" Score="9" ViewCount="" Body="&lt;p&gt;No.  Let $x \in \mathbb Q(\pi,i\pi) = \mathbb Q(i,\pi).$  If $x$ is algebraic, then $\mathbb Q(x)$ is finite over $\mathbb Q$, hence an algebraic extension, and so does not contain the transcendental element $\pi.$  On the other hand, if $x$ is transcendental over $\mathbb Q$, then $\mathbb Q(x)$ is isomorphic to the field&#xA;of rational functions in one variable over $\mathbb Q$.  The algebraic closure of&#xA;$\mathbb Q$ in this extension is equal to $\mathbb Q$ itself, and so this field does not contain $i$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus $\mathbb Q(\pi,i\pi)$ is not of the form $\mathbb Q(x)$ for any of its elements $x$, and so is not a simple extension of $\mathbb Q$.&lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2011-03-16T05:32:46.993" />
  <row Id="27352" PostTypeId="2" ParentId="27349" CreationDate="2011-03-16T05:35:59.053" Score="2" ViewCount="" Body="&lt;p&gt;Let's see. Assume otherwise, thus $\mathbb Q(\pi,i\pi) = \mathbb Q(\pi,i) = \mathbb Q(x)$ for some x. This means that they are equivalent as modules over $\mathbb Q$, and thus $i = a_nx^n + \cdots + a_0$ for some polynomial with $n \geq 1$, and so we have that $(a_nx^n + \cdots + a_0)^2 + 1 = 0$ is a rational polynomial of nonzero degree with $x$ as a root, thus $x$ is algebraic. But $\pi$ is transcendental, and since the algebraic numbers are a field we cannot write $\pi$ as a polynomial in $x$, contradicting the assumption that $\mathbb Q(\pi,i) = \mathbb Q(x)$. So the answer is no.&lt;/p&gt;&#xA;" OwnerUserId="8173" LastActivityDate="2011-03-16T05:35:59.053" CommentCount="2" />
  <row Id="27353" PostTypeId="2" ParentId="27349" CreationDate="2011-03-16T05:44:48.903" Score="4" ViewCount="" Body="&lt;p&gt;Suppose there exists a $x\in\mathbb Q(\pi,i)$ such that $\mathbb Q(\pi,i)=\mathbb Q(x)$. Since $\mathbb Q(\pi,i)$ is infinite over $\mathbb Q$, $x$ must be trascendental. Lüroth's theorem, then, tells us that every subfield of $\mathbb Q(x)$ properly containing $\mathbb Q$ is itself a simple trascendental extension of $\mathbb Q$. But this then applies to $\mathbb Q(i)\subseteq\mathbb Q(x)$. Of course, this is absurd.&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2011-03-16T05:44:48.903" CommentCount="2" />
  <row Id="27354" PostTypeId="2" ParentId="13685" CreationDate="2011-03-16T05:52:08.047" Score="1" ViewCount="" Body="&lt;p&gt;if $z=i\pi$ we have problems with denominator because $e^z+1=e^{i\pi}+1=-1+1$ and remember thqt the exponencial function is a periodic function, then $\frac{e^z-1}{e^z+1}$ is not analytic in the point where $e^z+1=0$&lt;/p&gt;&#xA;" OwnerUserId="8326" LastActivityDate="2011-03-16T05:52:08.047" />
  <row Id="27355" PostTypeId="1" AcceptedAnswerId="27372" CreationDate="2011-03-16T05:54:28.153" Score="6" ViewCount="130" Body="&lt;p&gt;How does W behave near +∞ compared to log?  In particular, I'm interested in the asymptotic expansion of&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{W(x)}{\ln(x)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;near ∞ (but along the positive real line, if that matters).  Clearly $W(x)\sim\ln(x)$ as $x$ increases, and the next term looks hyperbolic.&lt;/p&gt;&#xA;" OwnerUserId="1778" LastActivityDate="2011-03-16T14:17:36.313" Title="How does Lambert's W behave near ∞?" Tags="&lt;analysis&gt;&lt;special-functions&gt;" AnswerCount="2" CommentCount="5" />
  <row Id="27356" PostTypeId="1" CreationDate="2011-03-16T06:17:20.943" Score="-1" ViewCount="83" Body="&lt;p&gt;Here are two expressions of the unemployed in the places $x$ and $y$&#xA;\[&#xA;\sim(\text{Employed} \cup \text{Children} \; \cup \sim (\text{PlaceX} \cup \text{PlaceY}))&#xA;\]&#xA;\[&#xA;(\text{PlaceX} \cup \text{PlaceY}) - (\text{Employed} \cup \text{Children}) &#xA;\]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Show that they denote the same quantity&lt;/p&gt;&#xA;" OwnerUserId="8322" LastEditorUserId="5363" LastEditorDisplayName="" LastEditDate="2011-03-16T06:24:32.807" LastActivityDate="2011-03-16T07:08:19.323" Title="Show that they denote the same quantity" Tags="&lt;discrete-mathematics&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="27358" PostTypeId="1" AcceptedAnswerId="27361" CreationDate="2011-03-16T06:42:48.933" Score="5" ViewCount="142" Body="&lt;p&gt;What's the difference between a vector and an hypercomplex number? For instance a 4-vector and a quaternion. They seem to share many properties.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps this question could be put more generally as: what's the difference between a vector space and a field?&lt;/p&gt;&#xA;" OwnerUserId="132" LastActivityDate="2011-04-27T16:13:10.810" Title="What's the Difference Between a Vector and an Hypercomplex Number?" Tags="&lt;linear-algebra&gt;&lt;field-theory&gt;&lt;vector-space&gt;&lt;complex-numbers&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="1" />
  <row Id="27359" PostTypeId="1" CreationDate="2011-03-16T06:50:27.233" Score="6" ViewCount="74" Body="&lt;p&gt;Let $t \geq 1$ and $\pi$ be an integer partition of $t$.  Then the number of set partitions $Q$ of $\{1,2,\ldots,t\}$ for which the multiset $\{|q|:q \in Q\}=\pi$ is given by \[\frac{t!}{\prod_{i \geq 1} \big(i!^{s_i(\pi)} s_i(\pi)!\big)},\] where $s_i(\pi)$ denotes the number of parts $i$ in $\pi$.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Question: Is there a book that contains a proof of this?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm looking to cite it in a paper and would prefer not to include a proof.  I attempted a search in Google books, but that didn't help too much.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A similar result is proved in &quot;Combinatorics: topics, techniques, algorithms&quot; by Peter Cameron (page 212), but has &quot;permutation&quot; instead of &quot;set partition&quot; and &quot;cycle structure&quot; instead of &quot;integer partition&quot;.&lt;/p&gt;&#xA;" OwnerUserId="139" LastEditorUserId="7676" LastEditorDisplayName="" LastEditDate="2011-03-16T07:54:11.250" LastActivityDate="2011-03-16T07:54:11.250" Title="Seeking a textbook proof of a formula for the number of set partitions whose parts induce a given integer partition." Tags="&lt;combinatorics&gt;&lt;reference-request&gt;&lt;partitions&gt;" />
  <row Id="27360" PostTypeId="2" ParentId="27358" CreationDate="2011-03-16T07:00:03.317" Score="4" ViewCount="" Body="&lt;p&gt;They are separate concepts, it's just that hypercomplex numbers can be used to model the behavior of vectors in $\mathbb{R}^n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A &lt;strong&gt;vector&lt;/strong&gt; is an element of a &lt;strong&gt;vector space&lt;/strong&gt;. Vector spaces are (up to isomorphism) direct sums of the field $k$, in that a vector is an ordered $n$-tuple $(a_1,\ldots,a_n)$ with each $a_i \in k$ which can be multiplied by an element $b \in k$, which is called a &lt;strong&gt;scalar&lt;/strong&gt;, as follows: $k*(a_1,\ldots,a_n) = (k*a_1,\ldots,k*a_n)$. In the case of real vectors, this field is $\mathbb{R}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hypercomplex numbers, such as the quaternions, are different in that they are elements of certain &lt;strong&gt;algebras&lt;/strong&gt; over the real numbers. Algebras are defined similarly to vector spaces, only they also allow you to multiply elements of the algebra by other elements of the algebra.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Essentially, vectors and hypercomplex numbers are different concepts and cannot be used interchangeably. However, they are similar in that often the same problem can be solved using either.&lt;/p&gt;&#xA;" OwnerUserId="8173" LastEditorUserId="8173" LastEditorDisplayName="" LastEditDate="2011-03-16T14:13:28.457" LastActivityDate="2011-03-16T14:13:28.457" CommentCount="2" />
  <row Id="27361" PostTypeId="2" ParentId="27358" CreationDate="2011-03-16T07:00:08.727" Score="8" ViewCount="" Body="&lt;p&gt;In a field you can multiply any two elements, as well as add them, while in a vector space you can add vectors, but you can only multiply by scalars; you can't multiply two vectors.  (Multiplication in a field has to satisfy some axioms, but I think this is the essential point for your question.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A field containing $\mathbb R$, or more generally a system of hypercomplex numbers containing $\mathbb R$, will in particular be a vector space (since you can add, and you can multiply by elements of $\mathbb R$, i.e. by scalars), but it will have extra structure.  (E.g. you can multiply quaternions, and this is extra structure which is not at all obvious if you just know about $4$-vectors.)&lt;/p&gt;&#xA;" OwnerUserId="221" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2011-04-27T16:13:10.810" LastActivityDate="2011-04-27T16:13:10.810" />
  <row Id="27362" PostTypeId="2" ParentId="27356" CreationDate="2011-03-16T07:08:19.323" Score="0" ViewCount="" Body="&lt;p&gt;$$\begin{align}&#xD;&#xA;&amp;amp;\neg (\text{Employed} \cup \text{Children} \cup \neg (\text{PlaceX} \cup \text{PlaceY})) \\&#xD;&#xA;= &amp;amp;\neg \text{Employed} \cap \neg \text{Children} \cap (\text{PlaceX} \cup \text{PlaceY}) \qquad \text{De Morgans&amp;#39;s law}\\&#xD;&#xA;=&amp;amp;(\text{PlaceX} \cup \text{PlaceY}) \cap \neg \text{Employed} \cap \neg \text{Children} \\&#xD;&#xA;= &amp;amp;(\text{PlaceX} \cup \text{PlaceY}) \cap \neg (\text{Employed} \cup \text{Children})\\&#xD;&#xA;= &amp;amp;(\text{PlaceX} \cup \text{PlaceY}) - (\text{Employed} \cup \text{Children})&#xD;&#xA;\end{align}$$&#xA;In the last step I used the relationship between absolute and relative complement: $A-B=A\cap \neg B$.&lt;/p&gt;&#xA;" OwnerUserId="7044" LastActivityDate="2011-03-16T07:08:19.323" />
  <row Id="27363" PostTypeId="1" AcceptedAnswerId="27365" CreationDate="2011-03-16T07:56:37.137" Score="0" ViewCount="128" Body="&lt;p&gt;Our book gives this problem:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Find the $\mathcal{B}$-matrix for the transformation $\vec{x} \rightarrow A\vec{x}$ when the basis $\mathcal{B} = \{ \vec{b}_1, \vec{b}_2 \}$, where $A = \left[\begin{array}{cc}&#xD;&#xA;3 &amp;amp; 4 \\&#xD;&#xA;-1 &amp;amp; -1 \\&#xD;&#xA;\end{array} \right]$, $\vec{b}_1 = \left[\begin{array}{cc}&#xD;&#xA;2 \\&#xD;&#xA;-1 \\&#xD;&#xA;\end{array} \right]$, and $\vec{b}_2 = \left[\begin{array}{cc}&#xD;&#xA;1 \\&#xD;&#xA;2 \\&#xD;&#xA;\end{array} \right]$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;From what I understand, it's asking us to find the matrix for the same exact transformation as $A$, except relative to to the given bases. I can't figure out where to go from here, though... any thoughts?&lt;/p&gt;&#xA;" OwnerUserId="4890" LastActivityDate="2011-03-16T08:10:05.097" Title="Find the Matrix of a Linear Transformation Relative to a Basis" Tags="&lt;homework&gt;&lt;linear-algebra&gt;&lt;matrices&gt;" AnswerCount="1" />
  <row Id="27364" PostTypeId="1" CreationDate="2011-03-16T08:08:58.193" Score="1" ViewCount="115" Body="&lt;p&gt;Let $A_i=K_1 \times K_2 \times K_3 \times \cdots \times K_{i-1}$ and $B_i=(K_1+1) \times (K_2+1) \times (K_3+1) \times \cdots \times (K_{i-1}+1)$, where $K_{i-1}=(n-1)2^{n-(i-1)}-1$, $i=2,3,\cdots$. Find $a_i=\frac{A_i}{B_i}$ and $\sum_{i=1}^n a_i$.&lt;/p&gt;&#xA;" OwnerUserId="8201" LastEditorUserId="8201" LastEditorDisplayName="" LastEditDate="2011-03-16T12:24:34.807" LastActivityDate="2011-03-16T12:24:34.807" Title="compute the fraction of products" Tags="&lt;combinatorics&gt;" CommentCount="16" FavoriteCount="1" />
  <row Id="27365" PostTypeId="2" ParentId="27363" CreationDate="2011-03-16T08:10:05.097" Score="1" ViewCount="" Body="&lt;p&gt;What you need to do is form the matrix $B = (\vec{b}_1|\vec{b}_2)$, where $\vec{b}_i$ is the $i$th column of B, and note that this matrix converts vectors from the standard basis into the basis $\mathcal{B}$, while the inverse $B^{-1}$ will convert vectors in the basis $\mathcal{B}$ into the standard basis. Thus if you have a vector already in the basis $\mathcal{B}$, you can convert it to standard basis by multiplying by $B^{-1}$, multiply it by $A$, and finally convert back to $\mathcal{B}$ by multiplying by $B$, so your overall matrix is $BAB^{-1}$.&lt;/p&gt;&#xA;" OwnerUserId="8173" LastActivityDate="2011-03-16T08:10:05.097" CommentCount="1" />
  <row Id="27366" PostTypeId="1" CreationDate="2011-03-16T08:24:54.600" Score="2" ViewCount="59" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a question about how well one can choose a partition of unity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So suppose $(M,g)$ is an open Riemannian manifold. I would like to have the following statement to be true:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;There exists an open cover $\{U_n\}$ of $M$ which is uniformly locally finite (i.e. each point of $p \in M$ lies in at most $q$ sets) and uniformly bounded (i.e. $\mathcal{U} := \bigcup U_n \times U_n$ is a controlled subset of $M \times M$, i.e. $\sup_{p \in \mathcal{U}} d(\pi_1(p), \pi_2(p)) &amp;lt; \infty$, where the $\pi_i$ are the projections on the first and second coordinate).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore, there exists an subordinate partition of unity $\{g_n\}$ with the properties that the functions $\{g_n^{1/2}\}$ are also smooth and for every $l \in \mathbb{N}$ the $i$-th derivatives $(g_n^{1/2})^{(i)}$ are uniformly bounded for all $i \le l$, i.e. $\|(g_n^{1/2})^{(i)}\|_\infty := \sup_{p \in M} |\nabla_{v_1, \ldots, v_i} g_n^{1/2}(p)| \le G_l$, where the $v_i$ are unit vectors at the point $p \in M$.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the most critical point is the uniform boundedness of the derivatives. Maybe someone knows a reference where it is proven that such a partition of unity always exists (or maybe just one, which has only the property of uniform boundedness of the derivatives)? Is this statement even true?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks,&#xA;Alex&lt;/p&gt;&#xA;" OwnerUserId="7110" LastActivityDate="2011-03-16T08:24:54.600" Title="Existence of a partition of unity with uniformly bounded derivatives." Tags="&lt;reference-request&gt;&lt;differential-geometry&gt;" CommentCount="3" FavoriteCount="1" />
  <row Id="27367" PostTypeId="1" CreationDate="2011-03-16T08:38:31.210" Score="0" ViewCount="54" Body="&lt;p&gt;Hey everybody! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone check my proof and let me know if I made some mistakes, please?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $R$ be an integral domain, let $K$ be its field of fraction Let $M=R^{\oplus A}$ be a free module on $R$ with&#xA;basis $A$.  View $M$ as a subset of $K^{\oplus A}$.  Let $S$ be a subset of M.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Q : Show that if $S$ is linearly independent on $R$, then so it is on $K$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried the following:[contrapositive]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume  $S$ is NOT linearly independent on $K$, then there exist $s_1,\ldots,s_n$ in $S$, and $a_1,\ldots, a_n$ in $K$ not all of the $a_i$'s are zeros&#xA;such that  $a_1s_1+\cdots+a_ns_n=0$. Write every $a_i$ as a fraction i.e. $ \displaystyle a_i= \frac{u_i }{v_i}$ where $u_i$ and $v_i$ are in $R$ and $v_i\neq 0$ .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $d= v_1v_2\ldots v_n$, then $d\neq 0$ because $R$ is integral domain.&#xA; Multiply by $d$&#xA; $$d (a_1 s_1+\cdots  +a_n s_n)=0$$ and then get $$c_1s_1+\cdots+c_ns_n=0$$&#xA;where $\displaystyle c_i=d \frac{u_i}{v_i}\in R$ and not all of the $c_i$'s are zeros.&#xA;This will imply $S$ is linearly independent on $R$. Hence the statement is true.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you.&lt;/p&gt;&#xA;" OwnerUserId="8209" LastEditorUserId="8209" LastEditorDisplayName="" LastEditDate="2011-03-16T08:43:55.010" LastActivityDate="2011-04-15T14:40:30.313" Title="Linearly independent set of a free module on a field of fractions of an integral domain." Tags="&lt;abstract-algebra&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="27368" PostTypeId="2" ParentId="27252" CreationDate="2011-03-16T08:48:57.467" Score="0" ViewCount="" Body="&lt;p&gt;As proposed by Bill Dubuque, a simple proof can be formulated like this &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We substitue $x=-1$ and $y=2$ into $ux+(u+d)y=u+2d$&#xA;$\Rightarrow -u+2(u+d)=u+2d$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To prove the above we must show that for $P(u,d)$, every of the following is true: $P(0,0)$, $P(u+1,d)$ and $P(u,d+1)$.&lt;/p&gt;&#xA;" OwnerUserId="5877" LastEditorUserId="5877" LastEditorDisplayName="" LastEditDate="2011-03-16T09:07:10.027" LastActivityDate="2011-03-16T09:07:10.027" CommunityOwnedDate="2011-03-16T08:48:57.467" CommentCount="1" />
  <row Id="27369" PostTypeId="1" AcceptedAnswerId="27380" CreationDate="2011-03-16T08:49:46.893" Score="2" ViewCount="72" Body="&lt;p&gt;Is it possible to prove that for all kinds of &lt;a href=&quot;http://en.wikipedia.org/wiki/Simple_polygon&quot; rel=&quot;nofollow&quot;&gt;simple polygon&lt;/a&gt;, regardless of whether it is convex or concave and with &lt;em&gt;no opening&lt;/em&gt;, the &lt;a href=&quot;http://en.wikipedia.org/wiki/Centroid&quot; rel=&quot;nofollow&quot;&gt;centroid of the polygon&lt;/a&gt; must ( or may not) lie inside the polygon?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The wiki link above gives example of polygon which has the centroid lying outside the polygon:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A non-convex object might have a&#xA;  centroid that is outside the figure&#xA;  itself. The centroid of a ring or a&#xA;  bowl, for example, lies in the&#xA;  object's central void.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But let's say my polygon has no opening, can it be proved that the centroid must lie inside the polygon?&lt;/p&gt;&#xA;" OwnerUserId="550" LastActivityDate="2011-03-16T10:00:11.060" Title="Prove ( or disprove) that for all kinds of simple polygon, the centroid lies inside the polygon" Tags="&lt;geometry&gt;&lt;computational-geometry&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="27370" PostTypeId="2" ParentId="26150" CreationDate="2011-03-16T09:19:29.583" Score="2" ViewCount="" Body="&lt;p&gt;You can bound $\lvert N(S)\rvert$ from below by considering the worst case that all neighbours have the maximal degree, $n-d+1$. Then each vertex in $S$ of degree $n-d+1$ has as many outgoing edges as a worst-case neighbour has incoming edges, so the &quot;cost balance&quot; of such a vertex is neutral. Each vertex in $S$ of degree $n-d$ &quot;saves&quot; one edge, but there can be at most $n-d$ of these in $S$, so only $n-d$ edges can be &quot;saved&quot;, whereas $n-d+1$ edges would have to be &quot;saved&quot; to &quot;save&quot; one worst-case neighbour. So no neighbour can be &quot;saved&quot; even in the worst case, and thus there must be as many neighbours as vertices in $S$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can write that out in formulas if you want, but you said you wanted a hint.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[Edit in response to request for formulas:]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since all edges incident at a vertex in $S$ are also incident at a vertex of $N(S)$, we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{w\in N(S)}\deg(w)\ge\sum_{v\in S}\deg(v)\;.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $S$ contains $n_-$ vertices of degree $n-d$ and $n_+$ vertices of degree $n-d+1$, then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{eqnarray}&#xD;&#xA;\sum_{v\in S}\deg(v)&amp;amp;=&amp;amp;n_-(n-d)+n_+(n-d+1)\\\&#xD;&#xA;&amp;amp;=&amp;amp;(n_-+n_+)(n-d+1)-n_-\\\&#xD;&#xA;&amp;amp;=&amp;amp;\lvert S \rvert (n-d+1)-n_-\\\&#xD;&#xA;&amp;amp;\ge&amp;amp;\lvert S \rvert (n-d+1)-(n-d)\;,&#xD;&#xA;\end{eqnarray}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;since there are only $n-d$ vertices of degree $n-d$. On the other hand, since no vertex has degree more than $n-d+1$, we also have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{w\in N(S)}\deg(w)\le \lvert N(S) \rvert (n-d+1)\;.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Putting this all together gives&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\lvert N(S) \rvert (n-d+1) \ge \lvert S \rvert (n-d+1)-(n-d)\;,$$&#xA;$$\lvert N(S) \rvert \ge \lvert S \rvert -\frac{n-d}{n-d+1}\;.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But $\lvert N(S) \rvert$ and $\lvert S \rvert$ are integers and the last term is less than $1$, so we can drop the last term to conclude that $\lvert N(S) \rvert\ge\lvert S \rvert$. Linking this back to the original answer, each vertex in $S$ of degree $n-d$ can change the balance in the penultimate inequality by at most $1$, and there are at most $n-d$ such vertices, so the balance can only be changed by $n-d$, but it would have to change by $n-d+1$ to make a difference in the last inequality, where the division by $n-d+1$ corresponds to the fact that we need to &quot;save&quot; $n-d+1$ edges in order to &quot;save&quot; one worst-case neighbour.&lt;/p&gt;&#xA;" OwnerUserId="6622" LastEditorUserId="6622" LastEditorDisplayName="" LastEditDate="2011-03-19T08:21:10.397" LastActivityDate="2011-03-19T08:21:10.397" CommentCount="3" />
  <row Id="27371" PostTypeId="1" CreationDate="2011-03-16T09:21:49.253" Score="0" ViewCount="64" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a question about complex analysis. I am trying to get some intuition about residues and integration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let c(t) = exp(2i.pi.t) be the unit circle of the complex plane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I consider the following functions :&lt;br&gt;&#xA;f(z) = z&lt;br&gt;&#xA;g(z) = 1/z&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Integrating f around the c(t) circle yields 0 because f has no pole inside the unit disk.&lt;br&gt;&#xA;Integrating g around the c(t) circle yields 2.i.pi (residues theorem).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But g(c(t)) = f(c(-t)) : intuitively, I should be integrating the same function around the same circle, but in opposite directions. I would expect the integral values to be opposite, but I don't understand why one is null and the other not!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is surely an obvious reasoning or calculus error, and I can't see where.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&#xA;Yuufo&lt;/p&gt;&#xA;" OwnerUserId="8329" LastActivityDate="2011-03-16T09:44:26.247" Title="A question about complex integration using the residues theorem" Tags="&lt;integral&gt;&lt;complex-analysis&gt;" AnswerCount="2" />
  <row Id="27372" PostTypeId="2" ParentId="27355" CreationDate="2011-03-16T09:22:39.940" Score="5" ViewCount="" Body="&lt;p&gt;According to Theorem 2.7 of &lt;a href=&quot;http://emis.impa.br/EMIS/journals/JIPAM/images/107_07_JIPAM/107_07.pdf&quot; rel=&quot;nofollow&quot;&gt;this paper&lt;/a&gt;, for every $x \geq e$,&#xA;$$&#xD;&#xA;\log x - \log \log x + \frac{1}{2}\frac{{\log \log x}}{{\log x}} \le W(x) \le \log x - \log \log x + \frac{e}{{e - 1}}\frac{{\log \log x}}{{\log x}}&#xD;&#xA;$$&#xA;(with equality only for $x=e$). Note that $\frac{e}{{e - 1}} \approx 1.582$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT (cf. Juan's answer). According to &lt;a href=&quot;http://mathworld.wolfram.com/LambertW-Function.html&quot; rel=&quot;nofollow&quot;&gt;Wolfram MathWorld&lt;/a&gt;, an asymptotic formula which yields reasonably accurate results for sufficiently large $x$ is &#xA;$$&#xD;&#xA;W(x) = L_1  - L_2  + \frac{{L_2 }}{{L_1 }} + \frac{{L_2 ( - 2 + L_2 )}}{{2L_1^2 }} + \frac{{L_2 (6 - 9L_2  + 2L_2^2 )}}{{6L_1^3 }}&#xD;&#xA;$$&#xA;$$&#xD;&#xA;+ \frac{{L_2 ( - 12 + 36L_2  - 22L_2^2  + 3L_2^3 )}}{{12L_1^4 }} + \frac{{L_2 (60 - 300L_2  + 350L_2^2  - 125L_2^3  + 12L_2^4 )}}{{60L_1^5 }} + O\bigg[\bigg(\frac{{L_2 }}{{L_1 }}\bigg)^6 \bigg],&#xD;&#xA;$$&#xA;where&#xA;$$&#xD;&#xA;L_1 = \log x&#xD;&#xA;$$&#xA;$$&#xD;&#xA;L_2 = \log \log x.&#xD;&#xA;$$&#xA;Useful references can be found in that link.&lt;/p&gt;&#xA;" OwnerUserId="2810" LastEditorUserId="2810" LastEditorDisplayName="" LastEditDate="2011-03-16T14:17:36.313" LastActivityDate="2011-03-16T14:17:36.313" CommentCount="3" />
  <row Id="27373" PostTypeId="1" CreationDate="2011-03-16T09:29:52.727" Score="7" ViewCount="297" Body="&lt;p&gt;Suppose &lt;strong&gt;B&lt;/strong&gt; represents the matrix of orthogonal (perpendicular) projection of $\mathbb{R}^{3}$ onto the plane $x_{2} = x_{1}$. Compute the eigenvalues and eigenvectors of &lt;strong&gt;B&lt;/strong&gt; and explain their geometric meaning.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I have I come up so far as an attempt to deduce this question down is, for instance. If we pick an arbitrary point in space ($\mathbb{R}^{3}$), then we must project this point onto a plane (in particular $x_{2} = x_{1}$) which I imagine on a three dimensional axis of ($x,y,z$), if we choose $x_{1}$ to represent the $x$-axis, $x_{2}$ to represent the $y$-axis, and $x_{3}$ to represent the $z$-axis then we would have a plane of the equation that looks like $y=z$ or conversely ($x_{2}=x_{1}$) which is its equivalent. Once you project this point onto the plane, I see that it is true to be perpendicular and its vector is coming out of the plane. My troubles are finding the new coordinates of the new point that is projected onto the plane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a skeleton sketch of what I had in math-ese.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\left[\begin{array}{c}&#xD;&#xA;?\\&#xD;&#xA;?\\&#xD;&#xA;? \end{array} \right]  = &#xD;&#xA;\left[\begin{array}{ccc}&#xD;&#xA;\Box &amp;amp; \Box &amp;amp; \Box \\&#xD;&#xA;\Box &amp;amp; \Box &amp;amp; \Box \\&#xD;&#xA;\Box &amp;amp; \Box &amp;amp; \Box              &#xD;&#xA;\end{array} \right]  &#xD;&#xA;\left[\begin{array}{c}&#xD;&#xA;x_{1} \\&#xD;&#xA;x_{2} \\&#xD;&#xA;x_{3}&#xD;&#xA;\end{array} \right] $, $~~$ where &lt;strong&gt;B&lt;/strong&gt; is the matrix with empty boxes for elements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These question marks inside of the first matrix represent the coordinates in which I am trying to find. Once these our found, making some appropriate choices for the entries in the coefficient matrix labeled &lt;strong&gt;B&lt;/strong&gt; in the question can be found, so that when B is multiplied by the last matrix $\left(\left[\begin{array}{c} x_{1} \\ x_{2} \\ x_{3} \end{array} \right]~\right) $, we will get back the matrix with the ? marks in the entries. I believe this is known as doing a linear transformation. I didn't know how to include graphics, but I hope the words was enough detail to be able to duplicate what I am saying on paper in a graphical meaning. If not, please let me know how I can clarify anything up.&#xA;Some help would be very appreciated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-19T01:34:52.040" LastActivityDate="2011-03-19T01:34:52.040" Title="Find eigenvalues of a projection and explain what they mean" Tags="&lt;linear-algebra&gt;&lt;geometry&gt;&lt;transformation&gt;&lt;eigenvalues&gt;&lt;eigenvectors&gt;" AnswerCount="6" CommentCount="2" FavoriteCount="2" />
  <row Id="27374" PostTypeId="2" ParentId="27371" CreationDate="2011-03-16T09:38:35.167" Score="1" ViewCount="" Body="&lt;p&gt;I think what you're missing here is the essential difference between real and complex integration. It seems you're thinking of this as just &quot;adding up&quot; all the function values along the way, so it shouldn't make a difference in which order you add them up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, complex integration doesn't work that way, because you integrate along a path and the &quot;path element&quot; $\mathrm{d}z$ is itself complex and its &quot;direction&quot; enters into the result. So it matters not only which function values you encounter along the way, but also which &quot;path element&quot; they get multiplied by. In the two different integrations you describe, the same function values get multiplied by different path elements. If you draw a diagram with little arrows showing the &quot;direction&quot; of the integrand and the path element at some points, you'll probably realize why they add up to $0$ in one case and not in the other.&lt;/p&gt;&#xA;" OwnerUserId="6622" LastActivityDate="2011-03-16T09:38:35.167" />
  <row Id="27375" PostTypeId="1" AcceptedAnswerId="27727" CreationDate="2011-03-16T09:41:28.957" Score="0" ViewCount="416" Body="&lt;p&gt;Hello,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm wondering does anyone know of a scientific calculator with a mod function?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In C# this is shown as follows (just in case there are any other mods that a mathematical non-savant such as myself wouldn't be aware of)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;5 % 4 = 1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've looked through the most common ones on the web but the function is either not in the list or not present.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a little bit of a bias towards Casio calculators as that's what I'm familiar with but will accept any suggestions that can be currently purchased.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Derek&lt;/p&gt;&#xA;" OwnerUserId="8330" LastActivityDate="2011-03-18T02:56:43.097" Title="Calculator model with mod function?" Tags="&lt;modular-arithmetic&gt;&lt;cryptography&gt;&lt;calculator&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="27376" PostTypeId="2" ParentId="27371" CreationDate="2011-03-16T09:44:26.247" Score="0" ViewCount="" Body="&lt;p&gt;Remember that for complex contour integration on a smooth curve $\gamma$ with endpoints $a$ and $b$, you have:$$\int_\gamma f(z)\,dz=\int_a^b f(\gamma (t))\gamma^\prime (t)\,dt.$$ The $\gamma^\prime (t)\,dt$ is the &quot;path element&quot; that joriki alludes to. So if you parametrize the unit circle by $\gamma (t)=e^{it}$ from 0 to $2\pi$, you get $\gamma^\prime (t)=ie^{it}$, and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align*}&#xA;&amp;amp;\int_\gamma z\,dz=\int_0^{2\pi} ie^{2it}\,dt=0\text{, and}\\&#xA;&amp;amp;\int_\gamma \frac{1}{z}\,dz=\int_0^{2\pi} \frac{ie^{it}}{e^{it}}\,dt=2\pi i.&#xA;\end{align*}&lt;/p&gt;&#xA;" OwnerUserId="7676" LastActivityDate="2011-03-16T09:44:26.247" CommentCount="2" />
  <row Id="27377" PostTypeId="2" ParentId="27373" CreationDate="2011-03-16T09:46:26.317" Score="3" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Hint:&lt;/strong&gt; First solve the same problem for the orthogonal projection on the plane $P$ of equation $z=0$. Surely you know what is the projection on $P$ of a point $(x,y,z)$, right? So, what is the matrix translation of the formulae you just wrote? Now, the eigenvectors and the eigenvalues of &lt;em&gt;this&lt;/em&gt; matrix should be obvious.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you write a complete solution of this case, the solution of the case you ask for should be within your reach.&lt;/p&gt;&#xA;" OwnerUserId="6179" LastActivityDate="2011-03-16T09:46:26.317" />
  <row Id="27378" PostTypeId="1" AcceptedAnswerId="27395" CreationDate="2011-03-16T09:49:31.373" Score="1" ViewCount="82" Body="&lt;p&gt;&lt;a href=&quot;http://math.stackexchange.com/questions/10848/does-rapid-decay-of-fourier-coefficients-imply-smoothness&quot;&gt;In this question&lt;/a&gt; the term rapidly decaying sequence is used. What is the definition of a rapidly decaying sequence.(in cases of terms being purely real or complex.) How to prove that the sequence of Fourier series coefficients of a compactly supported, real valued, smooth function is rapidly decaying ? Please suggest a reference if the proof is lengthy. Also for the converse statement.&lt;/p&gt;&#xA;" OwnerUserId="2987" LastActivityDate="2011-03-16T13:49:00.940" Title="rapidly decaying sequence and fourier series coefficients of a compactly supported smooth function." Tags="&lt;real-analysis&gt;&lt;fourier-series&gt;" AnswerCount="1" />
  <row Id="27379" PostTypeId="2" ParentId="27373" CreationDate="2011-03-16T09:54:45.170" Score="1" ViewCount="" Body="&lt;p&gt;You really should think of this geometrically:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The plane $P = \{x_{1}=x_{2}\}$ is defined by its normal vector $n = \frac{1}{\sqrt{2}}\begin{bmatrix} 1 \\\ -1 \\\ 0 \end{bmatrix}$ (the square root appears in order to achieve $\langle n, n \rangle = 1$). A vector $v$ lies in the plane $P$ if and only if $\langle n,v \rangle = 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Projecting orthogonally onto the plane by $p: \mathbb{R}^3 \to \mathbb{R}^{3}$ means that $p(v) \in P$ and $v - p(v) = \lambda(v) n$ for some $\lambda(v) \in \mathbb{R}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The condition $p(v) \in P$ is the equation $\langle p(v), n \rangle = 0$ and we can determine $\lambda(v)$ by&#xA;\[&#xA;\lambda(v) = \lambda(v) \langle n, n \rangle = \langle \lambda(v) n, n \rangle = \langle v - p(v), n \rangle = \langle v,n \rangle - \langle p(v), n \rangle = \langle v,n \rangle.&#xA;\]&#xA;Solving the equation $v - p(v) = \lambda(v) n$ for $p(v)$ and plugging in $\lambda(v) = \langle v,n \rangle$ gives us the formula&#xA;\[&#xA;p(v) = v - \langle v,n \rangle \cdot n = v - \frac{1}{2} \langle v, \begin{bmatrix} 1 \\ -1 \\ 0 \end{bmatrix}\rangle \, \begin{bmatrix} 1 \\ -1 \\ 0 \end{bmatrix}.&#xA;\]&#xA;This last expression is easily converted into matrix form by writing $v = \begin{bmatrix} x_{1} \\\ x_{2} \\\ x_{3} \end{bmatrix}$ and computing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the eigenvectors, simply observe that $p(n) = 0$ and $p(v) = v$ for all $v \in P$.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As Prof. Blatter gave away the solution, there is no harm in showing what I left for you:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have&#xA;\begin{align*}&#xA;p\left(\begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix}\right) &amp;amp; = &#xA;\begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} - \frac{1}{2} (x_1 - x_2) \begin{bmatrix} 1 \\ -1 \\ 0 \end{bmatrix} =&#xA;\begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} - \frac{1}{2}&#xA;\begin{bmatrix} x_1 - x_2 \\ x_{2} - x_{1} \\ 0 \end{bmatrix} \\ &amp;amp; = &#xA;\begin{bmatrix} \frac{1}{2} (x_{1} + x_{2}) \\ \frac{1}{2} (x_{1} + x_{2}) \\ x_{3} \end{bmatrix} =&#xA;\begin{bmatrix} 1/2 &amp;amp; 1/2 &amp;amp; 0 \\ 1/2  &amp;amp; 1/2 &amp;amp; 0 \\ 0 &amp;amp; 0 &amp;amp; 1\end{bmatrix}&#xA;\begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix}.&#xA;\end{align*}&#xA;You can also compute $p$ directly on the basis vectors to confirm what is written in other answers. However, the method I described above always works (since it is coordinate-free up to the end), and it is very rare that you can read the matrix off as easily as in Prof. Blatter's approach.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, the vector $n = \frac{1}{\sqrt{2}} \begin{bmatrix} 1 \\\ -1 \\\ 0\end{bmatrix}$ is an Eigenvector to the eigenvalue $0$ (geometrically it is a normal vector to the plane and hence it must be sent to zero). Every vector of the plane $P$ is an eigenvector to the eigenvalue $1$: Geometrically, this means that &lt;em&gt;because&lt;/em&gt; it lies in the plane onto which you project, it remains unaffected by the projection. If you want a basis of eigenvectors, take $n$ and  any two vectors that span the plane, e.g. $\frac{1}{\sqrt{2}} \begin{bmatrix} 1 \\\ 1 \\\ 0 \end{bmatrix}$ and $\begin{bmatrix} 0 \\\ 0 \\\ 1 \end{bmatrix}$.&lt;/p&gt;&#xA;" OwnerUserId="5363" LastEditorUserId="-1" LastEditorDisplayName="" LastEditDate="2011-03-18T01:28:48.940" LastActivityDate="2011-03-18T01:28:48.940" CommentCount="4" />
  <row Id="27380" PostTypeId="2" ParentId="27369" CreationDate="2011-03-16T10:00:11.060" Score="3" ViewCount="" Body="&lt;p&gt;Comment as answer as requested:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Think of a boomerang.   &lt;/li&gt;&#xA;&lt;li&gt;Alternatively, since you already understand the example of the polyon with an opening, imagine an arbitrarily narrow passage drilled to connect the opening to the outside and &quot;simplify&quot; the polygon. Since you can make the passage as narrow as you want, you can move the centroid as little as you want, so you can see it doesn't have to move inside the polygon.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="6622" LastActivityDate="2011-03-16T10:00:11.060" CommentCount="1" />
  <row Id="27381" PostTypeId="2" ParentId="27355" CreationDate="2011-03-16T10:02:31.200" Score="5" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://www.apmaths.uwo.ca/~djeffrey/Offprints/CorlessJeffreyKnuth.pdf&quot; rel=&quot;nofollow&quot;&gt;One of the original papers by Corless &lt;em&gt;et al&lt;/em&gt;.&lt;/a&gt; on the Lambert function gives the following series expansion at $\infty$ for the (principal branch of the) Lambert function:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$W_0(z)=\ln\;z-\ln\ln\;z+\sum_{n=1}^\infty\left(\frac{-1}{\ln\;z}\right)^n \sum_{m=1}^n (-1)^m \left[n\atop{n-m+1}\right]\frac{(\ln\ln\;z)^m}{m!}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\left[n\atop m\right]$ is a Stirling cycle number.&lt;/p&gt;&#xA;" OwnerUserId="8332" LastActivityDate="2011-03-16T10:02:31.200" CommentCount="1" />
  <row Id="27382" PostTypeId="1" AcceptedAnswerId="27392" CreationDate="2011-03-16T10:06:32.523" Score="2" ViewCount="81" Body="&lt;p&gt;A rather tight lower bound $c(n)$ of the number of unlabeled digraphs of order $n$ (loops allowed) seems to be&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$c(n) = 2^{n^2}/n!$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;because there are $2^{n^2}$ labeled graphs, almost all of them are asymmetric, thus division by $n!$ is almost always &quot;correct&quot;, but yields a slightly too small result, because it is sometimes &lt;em&gt;not&lt;/em&gt; correct. How can this error be estimated, of which order is it? But most of all I'd like to know:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Is the growth rate of $c(n)$ super- or supra-exponential?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I know Stirling's formula, but I didn't manage to simplify the resulting expression.&lt;/p&gt;&#xA;" OwnerUserId="1792" LastEditorUserId="1792" LastEditorDisplayName="" LastEditDate="2011-03-16T12:30:49.997" LastActivityDate="2011-03-16T17:36:58.547" Title="Asymptotic number of unlabeled graphs" Tags="&lt;combinatorics&gt;&lt;graph-theory&gt;&lt;approximation&gt;" AnswerCount="2" />
  <row Id="27383" PostTypeId="1" AcceptedAnswerId="27384" CreationDate="2011-03-16T10:06:49.713" Score="3" ViewCount="108" Body="&lt;p&gt;According to my math book, in order to find the intersecting line between two planes we need to:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Find the vector product of the direction normals of the two planes&lt;/li&gt;&#xA;&lt;li&gt;Write the equations of the planes in Cartesian form.&lt;/li&gt;&#xA;&lt;li&gt;&lt;em&gt;Assume that $z=0$ since the line has to intersect this plane.&lt;/em&gt;&lt;/li&gt;&#xA;&lt;li&gt;Solve simultaneously for a point on the line and write down a vector equation of the line&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;What is meant by point 3? I'm having problems visualizing it. Please be explicit.&lt;/p&gt;&#xA;" OwnerUserId="5877" LastActivityDate="2011-03-16T10:27:53.190" Title="A question on finding the intersecting line between two planes" Tags="&lt;vector-space&gt;&lt;intuition&gt;" AnswerCount="1" />
  <row Id="27384" PostTypeId="2" ParentId="27383" CreationDate="2011-03-16T10:27:53.190" Score="4" ViewCount="" Body="&lt;p&gt;To find the line of intersection, you need two pieces of information. The direction of the line as a vector and a point on the line. Step 3 and the first part of step 4 is about locating a single point on the line of intersection. Since you are only looking for a point, you might as well assume $z=0$, to reduce your two plane equations to two equations in two unknowns which can be solved simultaneously. Now this might not always work. Sometimes the line of intersection happens to be parallel to the $z=0$ plane. In that case you could try $y=0$ or $x=0$. (One of these is sure to work.)&lt;/p&gt;&#xA;" OwnerUserId="2941" LastActivityDate="2011-03-16T10:27:53.190" CommentCount="2" />
  <row Id="27385" PostTypeId="1" AcceptedAnswerId="27389" CreationDate="2011-03-16T10:39:48.063" Score="3" ViewCount="102" Body="&lt;p&gt;Q1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sum\limits_{j=0}^{N-1} \sum\limits_{i=0}^{N-1} \sum\limits_{y=0}^{j} \sum\limits_{x=0}^{i} a(x)a(y)b(|x-y|) $ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given that $b(0) = 0$, Find the coefficient of $b(k)$ , $0\le k \le N-1$ in the above expansion. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Q2. $\sum\limits_{j=0}^{N-1} \sum\limits_{i=0}^{N-1} \sum\limits_{x=0}^{min(i,j)} a(x) $.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Find the coefficient of $a(x)$ in the above expansion ?&lt;/p&gt;&#xA;" OwnerUserId="2987" LastEditorUserId="2987" LastEditorDisplayName="" LastEditDate="2011-03-16T11:41:53.927" LastActivityDate="2011-03-16T12:51:38.480" Title="a question on summation expansion" Tags="&lt;combinatorics&gt;&lt;algebra&gt;&lt;algebra-precalculus&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="27386" PostTypeId="2" ParentId="27375" CreationDate="2011-03-16T10:56:37.353" Score="0" ViewCount="" Body="&lt;p&gt;HP : buy a hewlett packard calculator. the model HP-35S contains hundreds of functions. &#xA;there is a description here ; &lt;a href=&quot;http://www.shopping.hp.com/product/calculator/Scientific/1/storefronts/F2215AA%2523ABA;HHOJSID=FW7TNQVJMqHh075qvYbPbRvjvx2K1ybHFyYXGDY9Vr8kpqFlknwg!492383249&quot; rel=&quot;nofollow&quot;&gt;http://www.shopping.hp.com/product/calculator/Scientific/1/storefronts/F2215AA%2523ABA;HHOJSID=FW7TNQVJMqHh075qvYbPbRvjvx2K1ybHFyYXGDY9Vr8kpqFlknwg!492383249&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The 32Sii used to have a button just for that, on the other hand the&#xA;HP35s is programmable and you can most probably do it easily. &lt;/p&gt;&#xA;" OwnerUserId="7431" LastActivityDate="2011-03-16T10:56:37.353" CommentCount="5" />
  <row Id="27387" PostTypeId="1" CreationDate="2011-03-16T11:17:04.817" Score="0" ViewCount="25" Body="&lt;p&gt;In &quot;On Fusion Categories&quot; by Etingof, Nikshych and Ostrik Proposition 2.1 there is used the following characterisation of a right dual in a rigid monoidal category (let us restrict to strict monoidal categories for simplicity):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $V$ be a simple object in a rigid monoidal category. A simple object $Y$ is the left dual of $V$ if and only if there is an epimorphism $V \otimes Y \rightarrow 1$. Let us call this map $\phi$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that left/right duals are unique, but the prove for $Y \simeq ^*X$ that I know, needs in addition (for the case of left duals) that there is a map $\psi: 1 \rightarrow Y \otimes X$ such that $X \stackrel{id \otimes \psi}{\rightarrow} X \otimes Y \otimes X \stackrel{\phi \otimes id}{\rightarrow} X$ is the identity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can somebody give me a hint or a source where the above uniqueness is proved? Thank you very much!&lt;/p&gt;&#xA;" OwnerUserId="8085" LastEditorUserId="8085" LastEditorDisplayName="" LastEditDate="2011-03-16T12:30:56.923" LastActivityDate="2011-03-16T12:30:56.923" Title="Characterisation of duals in rigid monoidal categories" Tags="&lt;monoidal-categories&gt;" CommentCount="2" />
  <row Id="27388" PostTypeId="1" AcceptedAnswerId="27390" CreationDate="2011-03-16T12:37:07.900" Score="1" ViewCount="54" Body="&lt;p&gt;Hi everybody ::- ). Recently, I asked you how to find the &quot;Intersection of 2 Lines in 2D&quot; and the answers revolved around the Determinants ( &lt;a href=&quot;http://en.wikipedia.org/wiki/Line-line_intersection&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Line-line_intersection&lt;/a&gt; ) or Systems ( &lt;a href=&quot;http://math.stackexchange.com/questions/25171/intersection-of-two-lines-in-2d/25173#25173&quot;&gt;Intersection Of Two Lines In 2D&lt;/a&gt; ).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, based on what I learned here from Isaac about how to get the General Form Linear Equation from 2 points ( &lt;a href=&quot;http://math.stackexchange.com/questions/25190/how-to-obtain-line-equation-of-the-form-ax-bx-c-0&quot;&gt;How to obtain Line Equation of the form ax + bx + c = 0&lt;/a&gt; ), isn't it simpler to find the intersection like this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where x1 to x4 and y1 to y4 are the 4 points defining the 2 lines.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;  //First line's Equation.&#xA;  var a: Number = y1 - y2;&#xA;  var b: Number = x2 - x1;&#xA;  var c: Number = (y2 - y1) * x1 - (x2 - x1) * y1;&#xA;&#xA;  //Second line's Equation.&#xA;  var d: Number = y3 - y4;&#xA;  var e: Number = x4 - x3;&#xA;  var f: Number = (y4 - y3) * x3 - (x4 - x3) * y3;&#xA;&#xA;  //Below formulas obtained via System.&#xA;  intersection.y = (c*d - f*a) / (-d*b + a*e);&#xA;  intersection.x = (-b * intersection.y - c) / a;&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;It seems to me to be less mathematical operations here compared to the determinants solution. Does this method seem ok to you, mathematically speaking? I like that it's slope-independent (vertical lines result in useless checks in those cases).&lt;/p&gt;&#xA;" OwnerUserId="7745" LastEditorUserId="7745" LastEditorDisplayName="" LastEditDate="2011-03-16T13:09:01.963" LastActivityDate="2011-03-16T13:09:01.963" Title="Intersection of 2 lines in 2D via General Form Linear Equations" Tags="&lt;geometry&gt;&lt;algebraic-geometry&gt;" AnswerCount="1" />
  <row Id="27389" PostTypeId="2" ParentId="27385" CreationDate="2011-03-16T12:51:38.480" Score="2" ViewCount="" Body="&lt;p&gt;Let us use a general approach. A first principle to evaluate multiple sums is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;When dealing with multiple sums, see what happens when exchanging the order of the summations.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Re Q2, this means writing the sum $S$ as a sum over $x$ of $a(x)$ times a sum over $i$ and $j$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which brings us to a second general principle:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Sums over a fixed set of integers are easier.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And to the most useful tool to apply this principle:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Write the spans of sums as indicator functions.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In this context, some people use the notation $[a]$ to mean $1$ is $a$ is true and $0$ otherwise and I will use these conventions. For instance, still Re Q2,&#xA;$$&#xD;&#xA;S=\sum_{j=0}^{N-1}\sum_{i=0}^{N-1}\sum_{x=0}^{N-1}[x\le \min\{i,j\}]a(x).&#xD;&#xA;$$&#xA;But $[x\le \min\{i,j\}]$ is the product $[x\le i][x\le j]$, hence exchanging the order of the summations yields&#xA;$$&#xD;&#xA;S=\sum_{x=0}^{N-1}a(x)\sum_{i=0}^{N-1}[x\le i]\sum_{j=0}^{N-1}[x\le j].&#xD;&#xA;$$&#xA;The sum over $i$ and the sum over $j$ coincide and their common value is&#xA;$$&#xD;&#xA;\sum_{j=0}^{N-1}[x\le j]=\sum_{j=x}^{N-1}[x\le j]=N-x,&#xD;&#xA;$$&#xA;hence&#xA;$$&#xD;&#xA;S=\sum_{x=0}^{N-1}(N-x)^2a(x).&#xD;&#xA;$$&#xA;This solves Q2. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The solution for Q1 is somewhat more involved but I suggest that you now see how far you can go with these principles to try to solve it.&lt;/p&gt;&#xA;" OwnerUserId="6179" LastActivityDate="2011-03-16T12:51:38.480" />
  <row Id="27390" PostTypeId="2" ParentId="27388" CreationDate="2011-03-16T12:56:55.637" Score="1" ViewCount="" Body="&lt;p&gt;Assuming your &lt;code&gt;ret.y&lt;/code&gt; is &lt;code&gt;intersection.y&lt;/code&gt;, your results look good to me.  If you want to make sure that you are safe for all lines, you probably should change the &lt;code&gt;intersection.x&lt;/code&gt; formula so that it doesn't depend on &lt;code&gt;a&lt;/code&gt; being nonzero: $$x=\frac{bf-ce}{-db+ae}$$ (If $-db+ae=0$, which both $x$ and $y$ have in the denominator, the two lines do not intersect.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These formulae for $x$ and $y$ in terms of $a$, $b$, $c$, $d$, $e$, and $f$ are equivalent to applying &lt;a href=&quot;http://en.wikipedia.org/wiki/Cramer%27s_rule#Explicit_formulas_for_small_systems&quot; rel=&quot;nofollow&quot;&gt;Cramer's rule&lt;/a&gt; (beware that the use of $a$, $b$, $c$, $d$, $e$, and $f$ there are not quite the same), which uses determinants.&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2011-03-16T12:56:55.637" CommentCount="3" />
  <row Id="27391" PostTypeId="2" ParentId="27373" CreationDate="2011-03-16T13:04:35.493" Score="2" ViewCount="" Body="&lt;p&gt;An orthogonal projection of $3$-dimensional space to whatever plane in it has the property that when repeated it is still equal to itself. Symbolically, if &#xA;$p:{\Bbb R}^3\rightarrow\pi$ denotes the orthogonal projection onto the plane $\pi$ through te origin, then $p^2=p\circ p=p$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means that if $v\in{\Bbb R}^3$ is an eigenvector, i.e. $p(v)=\lambda v$, then we must have $p^2(v)=p(p(v))=p(\lambda v)=\lambda^2 v$ but also $p^2(v)=p(v)=\lambda v$, thus $\lambda^2=\lambda$, which says that $\lambda$ is either $0$ or $1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With this intuition we get immediately that $\pi$ itsel must be the $1$-eigenspace and the line perpendicular to $\pi$ is the $0$-eigenspace.&lt;/p&gt;&#xA;" OwnerUserId="688" LastActivityDate="2011-03-16T13:04:35.493" />
  <row Id="27392" PostTypeId="2" ParentId="27382" CreationDate="2011-03-16T13:31:54.463" Score="2" ViewCount="" Body="&lt;p&gt;If you apply Stirling's formula, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$c(n)\approx \frac{2^{n^2}e^{-n}}{n^n}$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\log c(n)\approx n^2 \log 2 - n \log n + n$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the log increases faster than $n$, the growth is faster than any exponential.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-03-16T13:31:54.463" />
  <row Id="27394" PostTypeId="1" AcceptedAnswerId="27417" CreationDate="2011-03-16T13:46:29.603" Score="3" ViewCount="101" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;$x + y + z = 4$&lt;br&gt;&#xA;  $x^2 + y^2 + z^2 = 4$&lt;br&gt;&#xA;  $x^3 + y^3 + z^3 = 4$  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Any ideas on how to solve for $(x,y,z)$ satisfying the three simultaneous equations, provided there can be both real and complex solutions?&lt;/p&gt;&#xA;" OwnerUserId="6520" LastActivityDate="2011-03-16T16:20:31.530" Title="Three-variable system of simultaneous equations" Tags="&lt;algebra&gt;" AnswerCount="3" />
  <row Id="27395" PostTypeId="2" ParentId="27378" CreationDate="2011-03-16T13:49:00.940" Score="3" ViewCount="" Body="&lt;p&gt;A sequence of complex numbers $\{a_n\}$ is said to be rapidly decaying if it converges to $0$ at a faster rate than any negative power of $n$:&#xA;$$&#xD;&#xA;\sup_n|a_n|n^k&amp;lt;\infty\qquad\forall k\in\mathbb{N}.&#xD;&#xA;$$&#xA;Let $\mathbb{T}$ denote the unit circle. If $f\in C^\infty(\mathbb{T})$, that is, $f$ is a periodic function of period $2\pi$ that admits derivatives of all orders, then the (doubly infinte) sequence of its Fourier coefficients is rapidly decaying. Indeed, integration by parts shows that $|\hat f_n|\le n^{-1}\|f&amp;#39;\|_\infty$; iterating the argument one gets that for all $k\in\mathbb{N}$, $|\hat f_n|\le n^{-k}\|f^{(k)}\|_\infty$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conversely, if the sequence of Fourier coefficients of a function $f\in L^1(\mathbb{T})$ is rapidly decaying, then the corresponding Fourier series and all the series derived from it by term by term differentiation are uniformly convergent. It follows that $f$ is equal almost everywhwre to a $C^\infty$ function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Faster decay gives more smoothnes of $f$. See for instance Section 2 of chapter V in AN INTRODUCTION TO HARMONIC ANALYSIS by Yitzhak Katznelson.&lt;/p&gt;&#xA;" OwnerUserId="4791" LastActivityDate="2011-03-16T13:49:00.940" />
  <row Id="27396" PostTypeId="2" ParentId="27367" CreationDate="2011-03-16T13:49:33.220" Score="2" ViewCount="" Body="&lt;p&gt;Yes, your argument is correct.  In fact it works more generally to show that if $S$ is an $R$-linearly independent subset of any $R$-module $M$, then $S \otimes 1$ is a $K$-linearly independent subset of the $K$-module $M \otimes_R K$.  That is, you do not need $M$ to be a free $R$-module.  &lt;/p&gt;&#xA;" OwnerUserId="299" LastActivityDate="2011-03-16T13:49:33.220" />
  <row Id="27397" PostTypeId="2" ParentId="27394" CreationDate="2011-03-16T13:57:43.850" Score="1" ViewCount="" Body="&lt;p&gt;You can write the elementary functions in terms of these sums of powers, thus getting the coefficients for a polynomial equation whose roots are $x$, $y$, and $z$. For instance, squaring the first equation and subtracting the second, you get the value of $xy+yz+xz$. Cubing the first equation and using what you now know, you get the value of $xyz$.&#xA;In general you can use &lt;a href=&quot;http://en.wikipedia.org/wiki/Newton%27s_identities#Formulation_in_terms_of_symmetric_polynomials&quot; rel=&quot;nofollow&quot;&gt;Newton's identities&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="589" LastEditorUserId="589" LastEditorDisplayName="" LastEditDate="2011-03-16T14:08:41.237" LastActivityDate="2011-03-16T14:08:41.237" />
  <row Id="27398" PostTypeId="1" AcceptedAnswerId="27414" CreationDate="2011-03-16T14:01:34.877" Score="4" ViewCount="277" Body="&lt;p&gt;Even though there are uncountably many subsets of $\mathbb{N}$ there are only countably many isomorphism classes of countably infinite - or countable, for short - models of the empty theory (with no axioms) over one &lt;em&gt;unary&lt;/em&gt; relation.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;How many isomorphism classes of&#xA;  countable models of the empty theory&#xA;  over one &lt;em&gt;binary&lt;/em&gt; relation (a.k.a.&#xA;  graph theory) are there? I.e.: How many countable unlabeled graphs are there?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A handwaving argument might be: Since the number of unlabeled graphs with $n$ nodes &lt;a href=&quot;http://math.stackexchange.com/questions/27382/asymptotic-number-of-unlabeled-graphs/27392#27392&quot;&gt;grows (faster than) exponentially&lt;/a&gt; (as opposed to growing linearly in the case of a unary relation), there must be uncountably many countable unlabeled graphs. (Analogously to the case of subsets: the number of subsets of finite sets grows exponentially, &lt;em&gt;thus&lt;/em&gt; (?) there are uncountably many subsets of a countably infinite set.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How is this argument to be made rigorous?&lt;/p&gt;&#xA;" OwnerUserId="1792" LastEditorUserId="1792" LastEditorDisplayName="" LastEditDate="2011-03-16T14:55:27.093" LastActivityDate="2011-03-17T11:57:44.427" Title="How many countable graphs are there?" Tags="&lt;combinatorics&gt;&lt;graph-theory&gt;" AnswerCount="7" CommentCount="8" FavoriteCount="2" />
  <row Id="27399" PostTypeId="1" AcceptedAnswerId="38486" CreationDate="2011-03-16T14:07:21.580" Score="1" ViewCount="49" Body="&lt;p&gt;Please suggest a good reference for introduction to Kernel Density Estimation. I am an electrical engineer and i want some basic introduction so that i can get to start using it and also understand its working. I have basic knowledge of Probability and random variables as taught in engineering courses.&lt;/p&gt;&#xA;" OwnerUserId="2987" LastEditorUserId="2987" LastEditorDisplayName="" LastEditDate="2011-03-16T14:42:48.913" LastActivityDate="2011-05-11T16:08:48.873" Title="good reference to kernel density estimation." Tags="&lt;probability-theory&gt;&lt;reference-request&gt;&lt;probability-distributions&gt;" AnswerCount="1" />
  <row Id="27400" PostTypeId="2" ParentId="27358" CreationDate="2011-03-16T14:20:46.373" Score="3" ViewCount="" Body="&lt;p&gt;A hypercomplex number is an element of a certain kind of distributive unital algebra over $\mathbb{R}$. An algebra over $\mathbb{R}$ is, by definition, a vector space over $\mathbb{R}$ with additional structure so that you can multiply elements together. It's not very hard to turn a given finite-dimensional vector space into an algebra of hypercomplex numbers, but the question is whether you can do so in a natural, useful way. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, there are vector spaces over any field at all, so it's also easy to come up with a vector space which cannot be turned into an algebra of hypercomplex numbers. For instance, a finite dimensional vector space over a finite field has only finitely many elements, so obviously cannot be turned into an algebra over $\mathbb{R}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The interesting thing is that there are objects which are simultaneously all of these things, and in a recursive way: for example, $\mathbb{C}$ is both a field and an algebra over $\mathbb{R}$, and &lt;em&gt;a fortiori&lt;/em&gt; a vector space over $\mathbb{R}$. But $\mathbb{R}$ is in turn both a field and an algebra over $\mathbb{Q}$... The study of fields which are algebras over other fields is called Galois theory, which also finds applications in algebraic geometry (though probably not in the way you expect!).&lt;/p&gt;&#xA;" OwnerUserId="5191" LastActivityDate="2011-03-16T14:20:46.373" CommentCount="1" />
  <row Id="27402" PostTypeId="2" ParentId="27398" CreationDate="2011-03-16T14:22:46.303" Score="3" ViewCount="" Body="&lt;p&gt;One way is to construct an injection from $2^{\mathbb{N}}$ (or from some other uncountable set) to the set of countable graphs.  There are many, many ways to do this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's one way:  I will construct a countable graph corresponding to a function $f: \mathbb{N} \rightarrow \{0,1\}$.  The graph has vertices $y$ and $x_{n,i}$ for $n \in \mathbb{N}$ and $1 \leq i \leq n+1$.  For each fixed $n$, we connect $x_{n,i}$ for all $i$ (in a chain as Alex suggests, or a loop, or a complete graph-- any will work).  Then we add an edge from $y$ to $x_{n,1}$ if $f(n)=1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is indeed an injection; we can recover $f$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; At first I had a multigraph construction, with multiple edges to each vertex, which is not what the question wanted.  I have modified the construction to make a simple graph.&lt;/p&gt;&#xA;" OwnerUserId="1461" LastEditorUserId="1461" LastEditorDisplayName="" LastEditDate="2011-03-16T14:46:47.300" LastActivityDate="2011-03-16T14:46:47.300" CommentCount="3" />
  <row Id="27403" PostTypeId="2" ParentId="27398" CreationDate="2011-03-16T14:29:03.093" Score="2" ViewCount="" Body="&lt;p&gt;I assume you mean by countable graph one that is countably infinite. I will also assume that your relation can be an arbitrary binary relation and not just symmetric since you seem to be interested in that case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case there are uncountably many. For, a special case of a binary relation is a total order. We do not need to add anything to the theory to have a total order; it's just a special case and ordering will be preserved by isomorphism. There are uncountably many nonisomorphic orders on a totally ordered set.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Indeed let $N$ be the natural numbers. Let $S$ be a subset of $N$. Replace each $s\in S$ by a copy of $\mathbb{Q}\cap [0,1]$ where $\mathbb{Q}$ are the rationals. It's easy to show that if $S\not = T$ then you will get nonisomorphic orders this way. But there are uncountably many subsets of $N$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So there are uncountably many orders on a countable set.&lt;/p&gt;&#xA;" OwnerUserId="2055" LastActivityDate="2011-03-16T14:29:03.093" CommentCount="2" />
  <row Id="27404" PostTypeId="2" ParentId="27394" CreationDate="2011-03-16T14:38:47.467" Score="1" ViewCount="" Body="&lt;p&gt;If you just want an answer, Wolfram Alpha makes short work of it, producing $(2,1+i,1-i)$ and all permutations of that.  You expect six solutions from the product of the degrees.  You can easily try x=y=z and y=z and find a contradiction, proving that the values of x,y, and z are all distinct.  Then there is really only one solution, with permutations giving the six.  If the solution is real, substitution will yield a sixth degree polynomial.  If the solution is not all real, you must have one real value and two conjugate complex values, so set $y=a+bi, z=a-bi$ with $x, a$ and $b$ real.  This gives&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\begin{align} &amp;amp;x+2a=4 \\ &amp;amp;x^2+2a^2-2b^2=4 \\ &amp;amp;x^3+2a^3-6ab^2=4 \end{align}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is actually easier than the real version.  You can solve the second for $b^2$, insert it into the third along with $x=4-2a$ and get a cubic that will have $a=1$ as a solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried playing with the symmetric polynomials, but didn't find any slick approach.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-03-16T14:38:47.467" CommentCount="0" />
  <row Id="27406" PostTypeId="1" CreationDate="2011-03-16T14:58:14.157" Score="0" ViewCount="240" Body="&lt;p&gt;What is the volume of spheres in higher dimensions?&lt;/p&gt;&#xA;" OwnerUserId="1754" LastActivityDate="2011-03-17T09:43:21.670" Title="Volume of spheres in higher dimensions?" Tags="&lt;geometry&gt;&lt;differential-geometry&gt;" AnswerCount="5" CommentCount="2" FavoriteCount="1" />
  <row Id="27407" PostTypeId="2" ParentId="27406" CreationDate="2011-03-16T15:00:25.840" Score="11" ViewCount="" Body="&lt;p&gt;Did you try &lt;a href=&quot;http://en.wikipedia.org/wiki/N-sphere&quot; rel=&quot;nofollow&quot;&gt;the obvious&lt;/a&gt;?&lt;/p&gt;&#xA;" OwnerUserId="6179" LastActivityDate="2011-03-16T15:00:25.840" CommentCount="0" />
  <row Id="27408" PostTypeId="1" CreationDate="2011-03-16T15:02:05.950" Score="2" ViewCount="155" Body="&lt;p&gt;I am reading Kai Lai Chung's A Course in Probability Theory. I understood the concept of vague convergence of a sequence of probability measures in the book. But it seems to me the book uses the concept of vague convergence of a sequence of distribution functions without defining it first. But I could miss the definition, and if you can point me where the book defines it, it will be nice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I was wondering &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;how the vague convergence of a&#xA;sequence of distribution functions&#xA;is defined?&lt;/li&gt;&#xA;&lt;li&gt;how it is related to the vague&#xA;convergence of the corresponding&#xA;probability measures?&lt;/li&gt;&#xA;&lt;li&gt;how it is related to the point-wise&#xA;convergence of the distribution&#xA;functions?&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;ADDED&lt;/strong&gt;: for a sequence of&#xA;(sub)probability measures or a&#xA;sequence of distribution functions,&#xA;how weak convergence is defined? How it and vague&#xA;convergence are different from each&#xA;other?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Thanks and regards!&lt;/p&gt;&#xA;" OwnerUserId="1281" LastEditorUserId="6179" LastEditorDisplayName="" LastEditDate="2011-03-20T16:25:07.357" LastActivityDate="2011-03-20T16:25:07.357" Title="Definition of vague convergence of distibution functions" Tags="&lt;probability-theory&gt;&lt;measure-theory&gt;" CommentCount="4" />
  <row Id="27409" PostTypeId="2" ParentId="27398" CreationDate="2011-03-16T15:03:28.803" Score="4" ViewCount="" Body="&lt;p&gt;Here is one explicit way of producing continuum-many pairwise nonisomorphic (simple, undirected, loopless) graphs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Start with a $3$-cycle among vertices labelled $-2,-1,0$.  To the vertex $-1$ attach a tail (= path) of length $1$ and to the vertex $-2$ attach a tail of length $2$.  Now, for each positive integer $n$, attach a tail of length $n$ to the vertex $0$, and label the terminal vertex in this tail $n$.  (So we have some vertices that we have not labelled, but certainly only countably many of them.)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now focus on the vertices marked $1,2,3,\ldots$: color the odd numbered ones red and the even numbered ones green.  Consider the set of all possible bipartite graphs on this vertex set -- i.e., in which all edges connect red to green.  There are clearly continuum-many: for instance even the independent choices of including / not including the edges $1--2$, $1--4$, $1--6$, and so forth gives a set of continuum cardinality.  I claim that all of these graphs are pairwise nonisomorphic.  Indeed, the only three cycle in any of them is formed by the vertices $-2,-1,0$, and it follows easily from this that any isomorphism between any two of these graphs carries $0$ to $0$, $-1$ to $-1$ and $-2$ to $-2$.  From this it follows &#xA;that for all $n \in \mathbb{Z}^+$, an isomorphism takes the vertex labelled $n$ to the vertex labelled $n$ and thus every vertex is fixed.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(By the way, I don't find the argument you suggest to be at all valid.  I strongly suspect there are first order theories such that the number of nonisomorphic models of finite size $n$ grows at least exponentially with $n$ but for which there are only countably many isomorphism classes of countable models.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Added&lt;/b&gt;: I want to make sure that the paths from $0$ to $n$ stay pairwise nonisomorphic when edges are added, so it's best to modify the construction slightly.  For instance, call the penultimate vertex in the path $p_n$ and attach one more length one tail at $p_n$, so that now $p_n$ has degree $3$.  &lt;/p&gt;&#xA;" OwnerUserId="299" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2011-03-16T15:16:07.577" LastActivityDate="2011-03-16T15:16:07.577" CommentCount="6" />
  <row Id="27410" PostTypeId="1" CreationDate="2011-03-16T15:18:40.070" Score="5" ViewCount="138" Body="&lt;p&gt;Let $A$ be a subspace of $\mathbb{R}^{n-1}$ with $n \geq 4$.  And if we know $\pi_1(\mathbb{R}^n -A) \cong \pi_1(\mathbb{R}^{n+1} -A)$, then can we say that $\pi_1(\mathbb{R}^n -A) \cong \pi_1(\mathbb{R}^{n+k} -A)$   for all $k \geq 1, k \in \mathbb{Z}$ ? I have tried to prove it or find an counter example, but failed. Any suggestions and ideas are appreciated. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, we are given that $\mathbb{R}^n -A$ is path-connected. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sorry, I shall be more specific. Here $n$ is a fixed number. And we may want to do induction on $k$. I am sorry for any confusion.    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And I am not sure whether this statement is correct or not. This problem arose while my friends and I were discussing a homework problem. &lt;/p&gt;&#xA;" OwnerUserId="5980" LastEditorUserId="5980" LastEditorDisplayName="" LastEditDate="2011-03-17T12:44:57.163" LastActivityDate="2011-03-17T12:44:57.163" Title="$\pi_1(\mathbb{R}^n -A) \cong \pi_1(\mathbb{R}^{n+1} -A) \Rightarrow \pi_1(\mathbb{R}^n -A)  \cong \pi_1(\mathbb{R}^{n+k} -A) \ (k \geq 1)$ ?" Tags="&lt;homework&gt;&lt;algebraic-topology&gt;" CommentCount="7" FavoriteCount="1" />
  <row Id="27411" PostTypeId="1" AcceptedAnswerId="27498" CreationDate="2011-03-16T15:44:39.953" Score="2" ViewCount="59" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As stated in the title, I need a decision procedure for the problem of regular expression equivalence with not.  Wikipedia states the problem is in the &lt;a href=&quot;http://en.wikipedia.org/wiki/NONELEMENTARY&quot; rel=&quot;nofollow&quot;&gt;NONELEMENTARY&lt;/a&gt; complexity class.  All I really need is link to the paper that describes the problem and its solution.  This is NOT a homework problem.  I need the procedure for a program static analysis tool.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="6922" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2011-03-16T16:43:38.310" LastActivityDate="2011-03-17T01:25:48.973" Title="Decision procedure for the problem of regular expression equivalence with not." Tags="&lt;reference-request&gt;&lt;computational-mathematics&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="27412" PostTypeId="1" CreationDate="2011-03-16T15:52:51.490" Score="0" ViewCount="68" Body="&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/UHCc6.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to like to be able to find the point p3 given point p1 and p2. The length between point p1 and p2 is set to L and the length between point p2 and p3 is set to L. The angle between point p1, p2 and p3 is set to theta. I'm pretty sure the dot product will play a roll but I'm not sure how. Any help would be great. Thanks&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;&#xA;p0 should be p1.  This is a 3D problem and I know I'm basically looking for points on a circle.  I think I need to figure out how to define the circle such that it is perpendicular to the vector p1-p2 but I'm not really sure how to do that.  I will be using this to generate polymer chains with constrained bond angles. &lt;/p&gt;&#xA;" OwnerUserId="8341" LastEditorUserId="8341" LastEditorDisplayName="" LastEditDate="2011-03-16T17:28:34.637" LastActivityDate="2011-03-16T22:31:56.060" Title="Vector constrained to an angle and length?" Tags="&lt;trigonometry&gt;&lt;vector-analysis&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="27413" PostTypeId="1" CreationDate="2011-03-16T16:05:21.573" Score="2" ViewCount="47" Body="&lt;p&gt;&lt;em&gt;This is one of Do Carmo's excersices and I got it as homework. Part (a) is easy and I include it here for the sake of completness. But I am entirely lost on part (b).&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A function $g:\mathbb{R} \rightarrow \mathbb{R}$ given by $g(t)=yt+x$, $t,x,y \in \mathbb{R}, y&amp;gt;0$ is called an affine proper function. The subset of all such functions with respect to the usual composition law forms a Lie group $G$.As a differentiable manifold $G$ is just the half upper plane with the differentiable structure induced from $\mathbb{R}^{2}$. Prove that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(a) The left-invariant Riemannian metric of $G$ which at the neutral element $e=(0,1)$ coincides with the euclidean metric ($g_{11}=g_{22}=1$, $g_{12}=0$) is given by $g_{11}=g_{22}=\frac{1}{y^{2}}$, $g_{12}=0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(b) Putting $(x,y)= z=x + iy$, $i=\sqrt{-1}$, the transformation &#xA;         $z\rightarrow z&amp;#39;= \frac{az+b}{cz+d}$, $a,b,c,d \in \mathbb{R}$, $ad-bc=1$ is an isometry of $G$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Hint:&lt;/em&gt; Observe that the first fundamental form can be written as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ds^{2}= \frac{dx^{2} + dy^{2}}{y^{2}} = \frac{4dzd\overline{z}}{(z- \overline{z})^{2}}$.  &lt;/p&gt;&#xA;" OwnerUserId="5209" LastActivityDate="2011-03-16T16:05:21.573" Title="Excercise in isometries of the half upper plane." Tags="&lt;homework&gt;&lt;riemannian-geometry&gt;" CommentCount="3" />
  <row Id="27414" PostTypeId="2" ParentId="27398" CreationDate="2011-03-16T16:12:10.863" Score="2" ViewCount="" Body="&lt;p&gt;Another very simple way to show that they are uncountable is with well-orderings: By definition the well-orderings of a countable set are $\aleph_1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;EDIT:&lt;/em&gt; Here's another simple way to show that they are $2^{\aleph_0}$ many. Take all but one (let's call it $p$) of the countable nodes and order them with the standard order of the natural numbers. This essentially assigns to every of these nodes a natural number. Now use the edgeless node $p$ that we kept to describe characteristic functions: For every subset $X$ of the natural numbers add an edge from the node that denotes $n$ to $p$ if and only if $n\in X$. It's trivial to see that all these graphs are not isomorphic: If two of them were then two different sets of natural numbers would coincide.&lt;/p&gt;&#xA;" OwnerUserId="1772" LastEditorUserId="1772" LastEditorDisplayName="" LastEditDate="2011-03-17T11:57:44.427" LastActivityDate="2011-03-17T11:57:44.427" />
  <row Id="27415" PostTypeId="2" ParentId="27398" CreationDate="2011-03-16T16:18:48.177" Score="0" ViewCount="" Body="&lt;p&gt;In Wilfrid Hodges &quot;Model Theory&quot;, p. 355, I found as an exercise:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;(c) if L is the signature with one&#xA;  binary relation symbol, there are&#xA;  continuum many non-isomorphic&#xA;  ultrahomogeneous $\omega$-categorical&#xA;  structures of signature L.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;which seems a little bit stronger: ultrahomogeneous and $\omega$-categorical!&lt;/p&gt;&#xA;" OwnerUserId="1792" LastActivityDate="2011-03-16T16:18:48.177" />
  <row Id="27416" PostTypeId="2" ParentId="27412" CreationDate="2011-03-16T16:19:44.783" Score="0" ViewCount="" Body="&lt;p&gt;I think first you need coordinates, otherwise the question seems to be not well formulated.&#xA;And it should be $ p = (x, y) $ instead of $ p = (x, y, z) $, since it is a 2D problem.&#xA;Lets say in a Cartesian coordinate, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ p_{0} = (x_{0}, y_{0}) $(known),&#xD;&#xA;$ p_{2} = (x_{2}, y_{2}) $(known),&#xD;&#xA;$ p_{3} = (x_{3},y_{3}) $(unkown).&#xD;&#xA;&#xD;&#xA;&#xD;&#xA;Then there are two vectors:&#xD;&#xA;&#xD;&#xA;    $ \vec{u} = (x_{0} - x_{2}, y_{0} - y_{2}) $,&#xD;&#xA;    $ \vec{v} = (x_{3} - x_{2}, y_{3} - y_{2}) $.&#xD;&#xA;&#xD;&#xA;&#xD;&#xA;&#xD;&#xA;So we have 2 equations:&#xD;&#xA;&#xD;&#xA;    $ \vec{u} \cdot \vec{v} =(x_{0} - x_{2}) \cdot (x_{3} - x_{2}) + (y_{0} - y_{2}) \cdot (y_{3} - y_{2}) = \parallel u \parallel \parallel v \parallel \cos \theta = L^{2} \cos \theta $&#xD;&#xA;    $ \parallel \vec{v} \parallel ^{2} = (x_{3} - x_{2})^{2} + (y_{3} - y_{2})^{2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After solving these two equations, you can find $ x_{3} $ and $ y_{3} $.&lt;/p&gt;&#xA;" OwnerUserId="7920" LastActivityDate="2011-03-16T16:19:44.783" />
  <row Id="27417" PostTypeId="2" ParentId="27394" CreationDate="2011-03-16T16:20:31.530" Score="2" ViewCount="" Body="&lt;p&gt;For a fixed number of variables and a fixed power $n$ the sum of powers $$x^n + y^n + z^n + ... + w^n$$ is a symmetric polynomial.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is expressible in terms of elementary symmetric polynomials. The elementary symmetric polynomials for three variables are&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$e_1 = x + y + z$&lt;/li&gt;&#xA;&lt;li&gt;$e_2 = x y + x z + y z$&lt;/li&gt;&#xA;&lt;li&gt;$e_3 = x y z$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;and your polynomials expressed in terms of them are&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$x + y + z = e_1$&lt;/li&gt;&#xA;&lt;li&gt;$x^2 + y^2 + z^2 = e_1^2 - 2 e_2$&lt;/li&gt;&#xA;&lt;li&gt;$x^3 + y^3 + z^3 = e_1^3 - 3(e_1 e_2 - e_3)$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Now we can find the values of $e_1,e_2,e_3$ evaluated at the given $x,y,z$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$e_1 = 4$, $e_2 = 6$, $e_3 = 4$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now consider the polynomial $(t - x)(t - y)(t - z) = t^3 - e_1 t^2 + e_2 t - e_3 = t^3 - 4 t^2 + 6 t - 4$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It has the solutions $t = 2, 1 + i$ and $1 - i$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So now we can check if these are correct:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$(2) + (1+i) + (1-i) = 4$&lt;/li&gt;&#xA;&lt;li&gt;$(2)^2 + (1+i)^2 + (1-i)^2 = 4 + 2i - 2i = 4$&lt;/li&gt;&#xA;&lt;li&gt;$(2)^3 + (1+i)^3 + (1-i)^3 = 8 + -2 + 2i -2 - 2i = 4$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="4290" LastActivityDate="2011-03-16T16:20:31.530" />
  <row Id="27418" PostTypeId="1" AcceptedAnswerId="27430" CreationDate="2011-03-16T16:40:40.577" Score="1" ViewCount="136" Body="&lt;p&gt;&lt;em&gt;Got this as homework and I don't know how to tackle this. Help please!&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Prove that the isometries of $\mathbb{S}^{n} \subset \mathbb{R}^{n+1}$, with the induced metric, are restrictions to $\mathbb{S}^{n}$ of the linear orthogonal maps&lt;/p&gt;&#xA;" OwnerUserId="5209" LastActivityDate="2011-03-17T12:46:29.470" Title="Isometries of the sphere $\mathbb{S}^{n}$" Tags="&lt;homework&gt;&lt;riemannian-geometry&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="27419" PostTypeId="2" ParentId="27406" CreationDate="2011-03-16T16:53:37.820" Score="0" ViewCount="" Body="&lt;p&gt;Perhaps something like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine a 1D curve that follows the curved edge of a semi-circle, lets call a constant number determined by this curve $c$ and the curve $v$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For each additional dimention in the $n$ dimensional sphere generalization the internal space would increase relative to $c$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example a sphere could be thought of as neighbouring circles of varying sizes, the sizes would vary through the sphere to follow $v$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To put it a more generic way:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An $n$ sphere could be thought of as neighbouring $n-1$ spheres of varying sizes, the sizes would vary through the $n$ sphere to follow $v$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$space_1 = length$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$space_n = space_{n-1}*c$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, possibly, the internal space of the n-sphere could be defined as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$units^n = c^n$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;?&lt;/p&gt;&#xA;" OwnerUserId="6982" LastEditorUserId="6982" LastEditorDisplayName="" LastEditDate="2011-03-16T17:11:00.020" LastActivityDate="2011-03-16T17:11:00.020" />
  <row Id="27420" PostTypeId="1" AcceptedAnswerId="27497" CreationDate="2011-03-16T16:53:46.260" Score="1" ViewCount="64" Body="&lt;p&gt;I am having a little trouble understanding the notion of non-degeneracy for bilinear maps.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A bilinear map $\psi : V \times V \mapsto \mathbb{R}$ always has a matrix representation, so $\psi(x,y)=x^TAy$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A bilinear map is non-degenerate if and only if $A$ is full rank.  This is also equivalent to stating that $\psi(x,y)=0 \; \forall y \in V \Rightarrow x = 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider now the following symmetric bilinear form $r^TQ^{-1}q = 0$ ($Q$ is symmetric).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Define $W_q = \{ q \}$ and $W_r = \{ r \}$.  We can define the b-orthogonal subspaces as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$W_q^{\perp} = \{x \; | \; x=Qy, \; y \in \mathcal{N}(q^T)\}$, where $\mathcal{N}(q^T)$ is the null-space of $q$.  In particular, we can express $y$ as $y=T_q\alpha$, with columns of $T_q$ spanning the null-space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly,  $W_r^{\perp} = \{x \; | \; x=Qy, \; y \in \mathcal{N}(r^T)\}$, with $y=T_r\beta$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $r \in W_q^{\perp}$ and $q \in W_r^{\perp}$, then &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$0 = r^TQ^{-1}q = \alpha^TT_q^TQT_r\beta$, and this should hold for all $\alpha$ and $\beta$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the matrix $T_q^TQT_r$ is full-rank, and the transformed bilinear map is non-degenerate suggesting this equality can not hold.  I feel like I am missing an important argumentation step.  Any help would be much appreciated.&lt;/p&gt;&#xA;" OwnerUserId="7683" LastActivityDate="2011-03-17T01:25:31.953" Title="bi-linear forms and non-degeneracy" Tags="&lt;linear-algebra&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="27421" PostTypeId="2" ParentId="27406" CreationDate="2011-03-16T16:57:22.590" Score="6" ViewCount="" Body="&lt;p&gt;Let $\kappa_n$ be the volume of the unit $n$-sphere $B_n$ and write the points of $B_n$ in the form $(x,y,{\bf z})$ with ${\bf z}\in{\mathbb R}^{n-2}$. For given $(x,y)\in B_2$ one has $|{\bf z}|^2\leq 1-r^2$, where $r:=\sqrt{x^2+y^2}$. These ${\bf z}$ fill an $(n-2)$-sphere of radius $\sqrt{1-r^2}$, and the $(n-2)$-dimensional volume of this sphere amounts to $\kappa_{n-2}(1-r^2)^{(n-2)/2}$. Therefore we get $$\kappa_n=\int_{B_2}\kappa_{n-2}(1-r^2)^{(n-2)/2}{\rm d}(x,y)=2\pi \kappa_{n-2}\int_0^1 (1-r^2)^{(n-2)/2}\&amp;gt;r\&amp;gt;dr={2\pi\over n}\&amp;gt;\kappa_{n-2}\ .$$&#xA;By means of this this recursion formula and using the known values $\kappa_1=2$, $\kappa_2=\pi$ one easily obtains the $\Gamma$-formula quoted in other answers.&lt;/p&gt;&#xA;" OwnerUserId="1303" LastEditorUserId="1303" LastEditorDisplayName="" LastEditDate="2011-03-17T09:43:21.670" LastActivityDate="2011-03-17T09:43:21.670" />
  <row Id="27422" PostTypeId="1" CreationDate="2011-03-16T16:58:30.827" Score="1" ViewCount="94" Body="&lt;p&gt;Hi. I'm trying to figure out how to compute a particular integral using Lebesgue integration.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;For a number $a$, define $f(x) = x^a$ for $0 \lt x \leq 1$, and $f(0) = 0$. Compute $\int_0^1 f$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here is what I have so far:&lt;br&gt;&#xA;$$\int_0^1 f = \int_{(0,1]}f + 0 = (x^a)*m((0,1]) = x^a$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure if I'm doing this correct. I'd appreciate some help, thanks in advance. This problem appears in Real Analysis by Royden (4th Edition) on p. 84, Exercise 19. &lt;/p&gt;&#xA;" OwnerUserId="8344" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2011-03-16T17:04:32.730" LastActivityDate="2011-03-16T17:29:03.763" Title="Computing lebesgue integral of a piecewise function" Tags="&lt;real-analysis&gt;" AnswerCount="1" CommentCount="6" />
  <row Id="27423" PostTypeId="1" AcceptedAnswerId="27441" CreationDate="2011-03-16T16:59:34.570" Score="1" ViewCount="175" Body="&lt;p&gt;Let $X$ be a Banach space, and let $X^*$ denote its continuous dual space.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Under the weak* topology, do compactness and sequential compactness coincide?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That is, is a subset of $X^*$ weakly* compact if and only if it is weakly* sequentially compact?  Does one imply the other?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Perhaps I should make this next one a separate question, but I'd prefer to keep all of this in one place.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Is the weak* topology on $X^*$ Hausdorff?  Is the weak topology on $X$ Hausdorff?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Motivation:&lt;/strong&gt; I would like to say that if a subset of $X^*$ is weakly* compact, then it is weakly* closed, and that if a subset of $X$ is weakly compact, then it is weakly closed.&lt;/p&gt;&#xA;" OwnerUserId="640" LastActivityDate="2011-04-05T15:58:01.120" Title="Compactness in the weak* topology" Tags="&lt;analysis&gt;&lt;functional-analysis&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" />
  <row Id="27424" PostTypeId="1" CreationDate="2011-03-16T17:02:59.597" Score="2" ViewCount="82" Body="&lt;p&gt;Show that if a 3-D surface is tangent to a plane along a curve, then the points of this curve are either parabolic or planar.&lt;/p&gt;&#xA;" OwnerUserId="8252" LastActivityDate="2011-05-31T00:06:54.237" Title="parabolic or planar points" Tags="&lt;geometry&gt;&lt;differential-geometry&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="27425" PostTypeId="1" AcceptedAnswerId="27433" CreationDate="2011-03-16T17:03:13.380" Score="12" ViewCount="246" Body="&lt;p&gt;I see an equation like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$y\frac{\textrm{d}y}{\textrm{d}x} = e^x$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and solve it by &quot;separating variables&quot; like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$y\textrm{d}y = e^x\textrm{d}x$$&#xA;$$\int y\textrm{d}y = \int e^x\textrm{d}x$$&#xA;$$y^2/2  = e^x + c$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What am I doing when I solve an equation this way?  Because $\textrm{d}y/\textrm{d}x$ actually means&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\lim_{\Delta x \to 0} \frac{\Delta y}{\Delta x}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;they are not really separate entities I can multiply around algebraically.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can check the solution when I'm done this procedure, and I've never run into problems with it.  Nonetheless, what is the justification behind it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I thought of to do in this particular case is write&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int y \frac{\textrm{d}y}{\textrm{d}x}\textrm{d}x = \int e^x\textrm{d}x$$&#xA;$$\int \frac{\textrm{d}}{\textrm{d}x}(y^2/2)\textrm{d}x = e^x + c$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;then by the fundamental theorem of calculus&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$y^2/2 = e^x + c$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this correct?  Will such a procedure work every time I can find a way to separate variables?&lt;/p&gt;&#xA;" OwnerUserId="3305" LastActivityDate="2011-03-17T09:41:28.050" Title="What am I doing when I separate the variables of a differential equation?" Tags="&lt;calculus&gt;&lt;differential-equations&gt;" AnswerCount="4" FavoriteCount="1" />
  <row Id="27426" PostTypeId="2" ParentId="27425" CreationDate="2011-03-16T17:07:29.103" Score="1" ViewCount="" Body="&lt;p&gt;maybe its better to think of it as $y\frac{dy}{dx}=e^x$. the two functions of $x$ are equal, so their indefinite integrals (with respect to $x$) are equal (i.e. the way you talked about it at the end).  moving the &quot;differentials&quot; around is more of a convenience.&lt;/p&gt;&#xA;" OwnerUserId="6925" LastActivityDate="2011-03-16T17:07:29.103" />
  <row Id="27427" PostTypeId="2" ParentId="27418" CreationDate="2011-03-16T17:11:41.687" Score="1" ViewCount="" Body="&lt;p&gt;Any isometry on $\mathbb{S}^n$ is the composition of atmost $n + 1$ reflections. I think it is followed from &lt;a href=&quot;http://en.wikipedia.org/wiki/Cartan%E2%80%93Dieudonn%C3%A9_theorem&quot; rel=&quot;nofollow&quot;&gt;Cartan–Dieudonné theorem&lt;/a&gt;.Please tell me if I am wrong.&lt;/p&gt;&#xA;" OwnerUserId="7565" LastEditorUserId="7565" LastEditorDisplayName="" LastEditDate="2011-03-17T12:46:29.470" LastActivityDate="2011-03-17T12:46:29.470" CommentCount="4" />
  <row Id="27428" PostTypeId="2" ParentId="27425" CreationDate="2011-03-16T17:12:25.430" Score="1" ViewCount="" Body="&lt;p&gt;Seperation of variables involves manipulating the &lt;em&gt;differentials&lt;/em&gt; (the $dx$'s and $dy$'s in your equation). A differential is the infinitesimal change in a variable, and can be treated as a variable in its own right in many applications. With this perspective, $dy$ is a function of $x$ and $dx$, and the derivative $dy/dx$ is the ratio of these two differentials, which is a function of $x$. What you are doing is simply performing an algebraic manipulation of these variables and then using calculus to remove the differential terms.&lt;/p&gt;&#xA;" OwnerUserId="8173" LastActivityDate="2011-03-16T17:12:25.430" CommentCount="6" />
  <row Id="27429" PostTypeId="1" CreationDate="2011-03-16T17:16:10.063" Score="4" ViewCount="179" Body="&lt;p&gt;If I understand correctly, most definitions of 'limit' require that the function either a) be defined in an open neighborhood around the relevant point or b) more permissively, that the relevant point is a limit point--the definition of 'continuity' is then given a special case so that functions are continous at isolated points--why not extend the notion of 'limit' so that the limit of a function at an isolated point is just whatever the function's value is there?  Is there some good reason not to?&lt;/p&gt;&#xA;" OwnerUserId="7546" LastActivityDate="2011-03-16T18:08:15.587" Title="Why not define 'limit' to include isolated points?" Tags="&lt;calculus&gt;&lt;real-analysis&gt;&lt;definition&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="1" />
  <row Id="27430" PostTypeId="2" ParentId="27418" CreationDate="2011-03-16T17:22:36.000" Score="3" ViewCount="" Body="&lt;p&gt;Here's a different approach than the one Qiaochu is suggesting, assuming you have proved that every point in a Riemannian manifold has a normal neighborhood: A neighborhood for which any two points in the neighborhood can be connected by a unique minimizing geodesic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, a lemma:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose $M$ is a (connected) Riemannian manifold and $f:M\rightarrow M$ is an isometry.  Suppose further that for some $p\in M$, that $f(p) = p$ and $d_p f = Id$.  Then $f = Id$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proof (sketch):  Let $X \subseteq M$ with $X = \{q\in M|$ $f(q) = q$ and $d_q f = Id \}$.  Then $p\in X$ by definition so $X$ is nonempty.  We will show $X$ is both open and closed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$X$ is closed because it's defined by equations.  $X$ is open becuase if $q\in X$, using a normal neighborhood $U$ and the fact that $f(exp_q(tv)) = exp_{f(q)}(t$ $d_q f$ $v)$ shows all points $r\in U$ satisfy $f(r) = r$.  If $d_r f \neq Id$, draw a small triangle connecting $r, q$ and $exp(tv)$ where $d_r f v \neq v$ and $t$ is very small.  Appealing to the fact that vertices of the triangle have unique minimizing geodesics between them will give a contradiction, showing $d_r f = Id$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, $X$ is both open and closed and so, since $M$ is connected, $X = \emptyset $ or $X = M$.  But $p\in X$, so $X = M$.  But this says that $f(q) = q$ for all $q\in M$ (and that $d_q f = Id$, but we don't really care about that), so $f$ is the identity map.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With this lemma proved, we get a pretty easy corollary:  An isometry is determined by where it sends one point and what its differential does at that point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This in turn gives a potential solution to our problem:  If you can show&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Every linear orthogonal map is an isometry of $S^n$ and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Given any $p, q \in S^n$, and any o.n. bases $\{v_1,...,v_n\}$ and $\{w_1,...,w_n\}$ of $T_p S^n$ and $T_q S^n$ respectively, there is a linear orthogonal map taking $p$ to $q$ and $\{v_1,...,v_n\}$ to $\{w_1,...,w_n\}$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, it follows that you've found all of the isometries of $S^n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The proof of 1) is straightforward owing to the fact that the linear orthogonal matrices preserve the metric on $\mathbb{R}^{n+1}$ and $S^n$ inherits it's metric from $\mathbb{R}^{n+1}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The proof of 2) is slightly more involved, but not so bad:  First, prove that you can assume wlog that $p = q = $ north pole and that $\{v_1,...,v_n\} = \{e_2,...,e_{n+1}\}$ using the standard o.n. basis $\{e_1,...,e_{n+1}\}$ of $\mathbb{R}^{n+1}$.  From here, if you think about it, given $\{ w_1,...,w_n\}$, you can explicitly write down the linear orthgonal matrix you want.&lt;/p&gt;&#xA;" OwnerUserId="331" LastActivityDate="2011-03-16T17:22:36.000" />
  <row Id="27431" PostTypeId="2" ParentId="27429" CreationDate="2011-03-16T17:23:18.043" Score="3" ViewCount="" Body="&lt;p&gt;The entire point of the notion of a 'limit' is to capture the behavior of a function as it gets &quot;close&quot; to a point, which is not possible for isolated points, thus there is no utility to extending the notion in the way you described. Your definition would make &quot;limit&quot; the same as &quot;value&quot; at these points, which is not very useful.&lt;/p&gt;&#xA;" OwnerUserId="8173" LastActivityDate="2011-03-16T17:23:18.043" CommentCount="4" />
  <row Id="27432" PostTypeId="2" ParentId="27422" CreationDate="2011-03-16T17:23:56.140" Score="1" ViewCount="" Body="&lt;p&gt;Your problem seems to a very basic misunderstanding of the difference between definite integrals and functions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\int_{(0,1]} f$ is a specific number, not a function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You cannot just do $\int_{(0,1]} f = f*m((0,1])$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The statement that $\int_{(0,1]} f = x^a * m((0,1])$ is meaningless.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2011-03-16T17:29:03.763" LastActivityDate="2011-03-16T17:29:03.763" CommentCount="2" />
  <row Id="27433" PostTypeId="2" ParentId="27425" CreationDate="2011-03-16T17:26:41.580" Score="11" ViewCount="" Body="&lt;p&gt;The basic justification is that integration by substitution works, which in turn is &lt;a href=&quot;http://en.wikipedia.org/wiki/Integration_by_substitution#Relation_to_the_fundamental_theorem_of_calculus&quot; rel=&quot;nofollow&quot;&gt;justified by the chain rule and the fundamental theorem of calculus&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More specifically, suppose you have $\frac{dy}{dx} = g(x) h(y)$.  Rewrite as $\frac{1}{h(y)} \frac{dy}{dx} = g(x).$  Add the implicit dependency of $y$ on $x$ to obtain $\frac{1}{h(y(x))} \frac{dy}{dx} = g(x).$  Now, integrate both sides with respect to $x$: $\int \frac{1}{h(y(x))} \frac{dy}{dx} \, dx = \int g(x) \, dx. $ If we do a variable substitution of $y$ for $x$ on the left-hand side (i.e., use the integration by substitution technique), we replace $\frac{dy}{dx} dx$ with $dy$.  Thus we have $$\int \frac{1}{h(y)} \, dy = \int g(x) \, dx,$$&#xA;which is the separation of variables formula.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if you believe integration by substitution, then separation of variables is valid.  &lt;/p&gt;&#xA;" OwnerUserId="2370" LastActivityDate="2011-03-16T17:26:41.580" CommentCount="1" />
  <row Id="27434" PostTypeId="2" ParentId="27398" CreationDate="2011-03-16T17:29:52.157" Score="2" ViewCount="" Body="&lt;p&gt;Here is a bijection between $[0,1) \subset \mathbb{R}$, which is uncountable, and a pairwise non-isomorphic collection of countably infinite graphs:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $x \in [0,1)$, let $0.x_1 x_2 x_3 ....$ be its unique decimal expansion (so $0\le x_i&amp;lt;10$ for each $i$, and no expansion ends in infinitely many 9's).  Construct the graph $G_x$ as follows: start with the 3-cycle on $\{a,b,c\}$; attach the infinite chain $(a,1,2,3,...)$ to node $a$; and attach a chain of length $x_i$ to each node $i \ge 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A minimal modification of this (use chain length $x_1 x_2 ... x_i$ in place of $x_i$) gives a bijection that preserves the order relation: $G_x$ is a subgraph of $G_y$ if and only if $x \le y$.&lt;/p&gt;&#xA;" OwnerUserId="5546" LastEditorUserId="5546" LastEditorDisplayName="" LastEditDate="2011-03-16T17:39:06.627" LastActivityDate="2011-03-16T17:39:06.627" />
  <row Id="27435" PostTypeId="2" ParentId="27429" CreationDate="2011-03-16T17:31:29.470" Score="3" ViewCount="" Body="&lt;p&gt;You &lt;em&gt;could&lt;/em&gt; make that definition I suppose, but what use would it have, and how would it relate to the usual notion of limit?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's look at what a limit of a function $f$ at a point $x$ should mean (let's say for a real-valued function on a metric space). You want the limit of $f$ at $x$ to be a real number $L$ such that for all $\varepsilon&amp;gt;0$ there exists a $\delta&amp;gt;0$ such that $0&amp;lt;d(x,y)&amp;lt;\delta$ implies that $|f(y)-L|&amp;lt;\varepsilon$.  Now if $x$ is an isolated point, then you could take any $L$ you want, because whatever $\varepsilon$ is, you can choose $\delta$ sufficiently small so that $|f(y)-L|&amp;lt;\varepsilon$ is vacuously true for all $y$ with $0&amp;lt;d(x,y)&amp;lt;\delta$; that is, just make sure that there are no $y$ satisfying the latter condition. For non-isolated points, limits are unique.  For isolated points, trying to extend the usual definition leads to making every real number a limit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For one of the applications of limits, namely continuity, not defining the limit at an isolated point causes no problem.  You can say that $f$ is continuous at a point $x$ in the domain if for all $\varepsilon&amp;gt;0$ there exists a $\delta&amp;gt;0$ such that $d(x,y)&amp;lt;\delta$ implies $|f(y)-f(x)|&amp;lt;\varepsilon$.  If $x$ is an isolated point, then this will always be true, because for sufficiently small $\delta$ the only $y$ with $d(x,y)&amp;lt;\delta$ is $x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added&lt;/strong&gt;: I was writing when Alex posted, and part of my post makes a similar point to his.  Qiaochu's comment on Alex's post gives an answer to my question at the beginning of my post.  Making this definition allows continuity to be defined in terms of respecting limits without making isolated points a special case, something I had overlooked.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nonetheless, continuity can be defined in terms of respecting limits without actually defining the limit of a function at a point.  A function $f$ between metric spaces [resp. topological spaces] is continuous at $x$ if for every sequence [resp. net] $(x_n)_n$ in the domain converging to $x$, $\lim_n f(x_n)=f(x)$.  In case $x$ is an isolated point, a sequence converging to $x$ is eventually constantly equal to $x$, so this will be satisfied.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2011-03-16T17:45:52.767" LastActivityDate="2011-03-16T17:45:52.767" CommentCount="7" />
  <row Id="27436" PostTypeId="2" ParentId="27382" CreationDate="2011-03-16T17:36:58.547" Score="2" ViewCount="" Body="&lt;p&gt;See the &lt;a href=&quot;http://oeis.org/A000088&quot; rel=&quot;nofollow&quot;&gt;OEIS, sequence A000088&lt;/a&gt; for an asymptotic formula for the error. In particular, if $G(n)$ is the number of unlabeled graphs on n nodes, then &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ {G(n) \over c(n)} = 1 + {n^2 - n \over 2^{n-1}} + O(n^2 2^{-2n}) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;as $n \to \infty$. I suspect there is a simple combinatorial explanation for that first term but I don't know what it is; the source is Harary and Palmer, &lt;em&gt;Graphical Enumeration&lt;/em&gt;, which may explain that.&lt;/p&gt;&#xA;" OwnerUserId="173" LastActivityDate="2011-03-16T17:36:58.547" />
  <row Id="27437" PostTypeId="1" CreationDate="2011-03-16T17:43:09.920" Score="5" ViewCount="105" Body="&lt;p&gt;I stumbled upon this number theory problem while I was solving another problem. Here is the equation: $$3^kn + 3^{k-1} + 2^m(3^{k-1} + 2h) = 2^{m+l}n$$ where $k \geq 3, h,l,m,n\in\mathbb{N}$, $n$ is odd and $n$ is not a multiple of $3$. My impression is that it does not have a solution. However, I have not progressed on the problem anymore than that. Could you please help?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="5595" LastEditorUserId="2937" LastEditorDisplayName="" LastEditDate="2011-04-15T23:46:06.817" LastActivityDate="2011-04-15T23:46:06.817" Title="Solving in positive integers an equation containing exponentials" Tags="&lt;number-theory&gt;&lt;diophantine-equations&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="27438" PostTypeId="2" ParentId="27429" CreationDate="2011-03-16T17:46:29.443" Score="3" ViewCount="" Body="&lt;p&gt;Okay, now that I've checked Rudin to see that this really is the definition he gives of a limit of a function, I have an answer, but I don't like it. The motivation behind the definition of $\lim_{x \to a} f(x)$ is that you want to understand what $f$ is doing in a neighborhood of $a$ in order to compare it to what is happening at at $a$. If $a$ is an isolated point, there's nothing to compare.&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2011-03-16T18:08:15.587" LastActivityDate="2011-03-16T18:08:15.587" CommentCount="3" />
  <row Id="27439" PostTypeId="2" ParentId="27423" CreationDate="2011-03-16T17:46:46.913" Score="4" ViewCount="" Body="&lt;p&gt;(i) No. Consider $\ell^\infty$ (bounded sequences). The unit ball of ${\ell^\infty} ^*$ is compact by Alaoglu's theorem, but not sequencially compact: the sequence of functionals $a\mapsto a(n)$ (picking the nth element of the sequence $a\in \ell^\infty$) is bounded, but does not have a *-weakly convergent subsequence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(ii) Yes. For the weak* topology this follows directly from the definition; for the weak topology this follows from the Hahn-Banach theorem.&lt;/p&gt;&#xA;" OwnerUserId="1609" LastActivityDate="2011-03-16T17:46:46.913" />
  <row Id="27440" PostTypeId="2" ParentId="27437" CreationDate="2011-03-16T17:57:05.707" Score="1" ViewCount="" Body="&lt;p&gt;I think you can use &lt;a href=&quot;http://en.wikipedia.org/wiki/Infinite_descent&quot; rel=&quot;nofollow&quot;&gt;infinite descent&lt;/a&gt; to show that the equation does not have a solution. We know that $n= 3k+1$ for $k$ an odd integer or $n =3k+2$ for $k$ an even integer. If $(k,h,l,m,n)$ is a solution, then $2^{m+1}$ divides the LHS and $n$ divides the LHS.&lt;/p&gt;&#xA;" OwnerUserId="2548" LastEditorUserId="2548" LastEditorDisplayName="" LastEditDate="2011-03-16T18:36:01.740" LastActivityDate="2011-03-16T18:36:01.740" CommentCount="3" />
  <row Id="27441" PostTypeId="2" ParentId="27423" CreationDate="2011-03-16T17:58:04.423" Score="2" ViewCount="" Body="&lt;p&gt;Let me answer your second question first.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The weak$^{\ast}$-topology is Hausdorff (let me treat the real case, the complex case is similar): If $\phi \neq \psi$ are two linear functionals then there is $x \in X$ such that $\phi(x) \lt r \lt \psi(x)$. The sets $U = \{f \in X^{\ast} \,:\,f(x) \lt r\}$ and $V = \{f \in X^{\ast}\,:\,f(x) \gt r\}$ are weak$^{\ast}$-open (since evaluation at $x$ is weak$^{\ast}$-continuous) and disjoint neighborhoods of $\phi$ and $\psi$, respectively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That the weak topology is Hausdorff is shown similarly, using Hahn-Banach.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Next, if $X$ is separable, then the unit ball in the dual space is metrizable with respect to the weak$^{\ast}$-topology: pick a countable dense set $\{x_{n}\}_{n \in \mathbb{N}}$ of the unit ball of $X$ and verify that&#xA;\[&#xA;d(\phi,\psi) = \sum_{n=1}^{\infty} 2^{-n} \frac{|\phi(x_n) - \psi(x_n)|}{1+|\phi(x_n) - \psi(x_n)|}&#xA;\]&#xA;defines a metric compatible with the weak$^{\ast}$-topology. Hence the unit ball is sequentially compact in the weak$^{\ast}$-topology (this can be shown directly using Arzel&amp;agrave;-Ascoli, by the way).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using a standard Baire category argument, one can show that weak$^{\ast}$-compact sets are norm-bounded: Indeed, if $K$ is weak$^{\ast}$-compact, it is a Baire space. Write $B^{\ast}$ for the closed unit ball in $X^{\ast}$. Clearly $K = \bigcup_{n = 1}^{\infty} (K \cap n \cdot B^{\ast})$, so at least one of the closed subsets $K \cap n \cdot B^{\ast}$ of $K$ must have non-empty interior. By compactness finitely many translates of $n\cdot B^{\ast}$ must cover $K$, thus $K$ is bounded in norm and hence $K$ is a closed subset of a large enough ball.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Conclusion:&lt;/strong&gt; If $X$ is separable then every weak$^{\ast}$-compact subset of $X^{\ast}$ is sequentially compact.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I don't know if the converse is true.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $X$ is not separable, then weak$^{\ast}$-compactness does not imply weak$^{\ast}$-sequential compactness, the standard example is mentioned in Florian's post.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Since you might be interested in the weak topology as well, there's a rather difficult result due to Eberlein:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Recall that a space is countably compact if every &lt;em&gt;countable&lt;/em&gt; open cover has a finite subcover. A sequentially compact space is countably compact.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Theorem (Eberlein)&lt;/strong&gt; If a subset of a Banach space is weakly countably compact then it is weakly compact and weakly sequentially compact.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and finally:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Theorem (Eberlein-&amp;#352;mulian)&lt;/strong&gt; A &lt;s&gt;bounded&lt;/s&gt; subset of a Banach space is weakly sequentially compact if and only if it is weakly compact. In particular, if the unit ball is weakly sequentially compact then $X$ is reflexive.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5363" LastEditorUserId="5363" LastEditorDisplayName="" LastEditDate="2011-04-05T15:58:01.120" LastActivityDate="2011-04-05T15:58:01.120" CommentCount="2" />
  <row Id="27443" PostTypeId="1" CreationDate="2011-03-16T18:41:49.147" Score="4" ViewCount="131" Body="&lt;h2&gt;Background&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;I would appreciate help in identifying / explaining this operation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To calculate each of the $n$ values of $f(\Phi)$:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;sample from the distribution of each of $i$ parameters, $\phi_i$&lt;/li&gt;&#xA;&lt;li&gt;calculate the $i$ values of $g(\phi_{i})$&lt;/li&gt;&#xA;&lt;li&gt;subtract each $g_i(\phi_{i,n})$ from $g_i(\hat{\phi_i})$ (these are deviation)&lt;/li&gt;&#xA;&lt;li&gt;take the sum of these deviations    &lt;/li&gt;&#xA;&lt;li&gt;add the sum of these deviations to the median &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;in summary, this is the calculation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f(\Phi_n)=g(\hat{\phi}) + \sum_i(g_i(\phi_{i,n})-g_i(\hat{\phi_i}))$$&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\phi_i$ is the distribution of each of $i$ parameters &lt;/li&gt;&#xA;&lt;li&gt;$\hat{\phi}$ is a vector of the parameter medians &lt;/li&gt;&#xA;&lt;li&gt;$g$ is a vector of $i$ univariate splines, one for each parameter estimated by evaluating a multivariate model across the range of $\phi_i$ while all $\phi_{\text{not}i}$ held at their medians (a univariate sensitivity analysis of a computationally intensive prognostic model)&lt;/li&gt;&#xA;&lt;li&gt;$g_i(\hat{\phi_i})$ is the $i^{th}$ spline evaluated at the median of $\phi_i$  &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;h2&gt;Questions:&lt;/h2&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Is there a name or simple way to describe this calculation?&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;My first attempt to describe the above operation:&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;The spline ensemble is calculated as the sum of deviations from the median for for each parameter added to the median.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;An alternative suggestion:&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;The spline ensemble is calculated based on the univariate anomalies for each parameter.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Is there a simplified form of this computation, or expression of the equation?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="3733" LastEditorUserId="3733" LastEditorDisplayName="" LastEditDate="2011-03-17T02:52:08.397" LastActivityDate="2011-03-17T02:52:08.397" Title="Does this calculation have a name, or a generic formulation?" Tags="&lt;statistics&gt;&lt;algorithms&gt;&lt;functions&gt;&lt;probability&gt;" FavoriteCount="2" />
  <row Id="27444" PostTypeId="1" AcceptedAnswerId="27536" CreationDate="2011-03-16T18:48:30.263" Score="3" ViewCount="96" Body="&lt;p&gt;I've some trouble in understanding integer multiplication using FFT.&lt;br&gt;&#xA;I'm using the algorithm described on &lt;a href=&quot;http://en.wikipedia.org/wiki/Multiplication_algorithm#Fourier_transform_methods&quot; rel=&quot;nofollow&quot;&gt;wikipedia&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is an example of how I understand this algorithm:&lt;br&gt;&#xA;$$a=173$$&lt;br&gt;&#xA;$$b=95$$&lt;br&gt;&#xA;Lets take $w=4$, then we have&lt;br&gt;&#xA;$$a=13*2^{4\cdot 0}+10*2^{4\cdot1}$$&lt;br&gt;&#xA;$$b=15*2^{4\cdot 0}+5*2^{4\cdot1}$$&lt;br&gt;&#xA;In the vector notation it looks like this:&lt;br&gt;&#xA;$$a = \left[ \begin{array}{cc} 13 \\ 10 \end{array} \right]$$&lt;br&gt;&#xA;$$b = \left[ \begin{array}{cc} 15 \\ 5 \end{array} \right]$$&lt;br&gt;&#xA;The FFT of those vectors are:&lt;br&gt;&#xA;$$f(a) = \left[ \begin{array}{cc} \frac{23}{\sqrt{2}} \\ \frac{3}{\sqrt{2}} \end{array} \right]$$&#xA;$$f(b) = \left[ \begin{array}{cc} \frac{20}{\sqrt{2}} \\ \frac{10}{\sqrt{2}} \end{array} \right]$$&lt;br&gt;&#xA;The product of those results entry by entry is:&lt;br&gt;&#xA;$$c = \left[ \begin{array}{cc} 230 \\ 15 \end{array} \right]$$&lt;br&gt;&#xA;The inverse FFT of $c$ is:&lt;br&gt;&#xA;$$f^{-1}(c)= \left[ \begin{array}{cc} \frac{245}{\sqrt{2}} \\ \frac{215}{\sqrt{2}} \end{array} \right]$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And now what should be done?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;&lt;br&gt;&#xA;As Myself noticed those vectors should be 4dimensional instead of two dimensional. Here are correct calculations (in case anyone has similar problem): &#xA;$$a = \left[ \begin{array}{cc} 13 \\ 10 \\ 0 \\ 0 \end{array} \right]$$&lt;br&gt;&#xA;$$b = \left[ \begin{array}{cc} 15 \\ 5 \\ 0 \\ 0 \end{array} \right]$$&lt;br&gt;&#xA;The FFT of those vectors are:&lt;br&gt;&#xA;$$f(a) = \left[ \begin{array}{cc} 11.5 \\ 6.5+5i \\ 1.5 \\ 6.5-5i \end{array} \right]$$&#xA;$$f(b) = \left[ \begin{array}{cc} 10 \\ 7.5+2.5i \\ 5 \\ 7.5-2.5i \end{array} \right]$$&lt;br&gt;&#xA;The product of those results entry by entry is:&lt;br&gt;&#xA;$$c = \left[ \begin{array}{cc} 115 \\ 36.25+53.75i \\ 7.5 \\ 36.25-53.75i \end{array} \right]$$&lt;br&gt;&#xA;The inverse FFT of $c$ is:&lt;br&gt;&#xA;$$f^{-1}(c)= \left[ \begin{array}{cc} 195 \\ 215 \\ 50 \\ 0 \end{array} \right]$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the final result is $ab=195\cdot 2^{0} + 215 \cdot 2^{4} + 50 \cdot 2^{8} = 16435$&lt;/p&gt;&#xA;" OwnerUserId="4811" LastEditorUserId="4811" LastEditorDisplayName="" LastEditDate="2011-03-16T19:56:19.347" LastActivityDate="2011-03-17T06:17:32.060" Title="Integer multiplication using FFT" Tags="&lt;fourier-analysis&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="2" />
  <row Id="27445" PostTypeId="2" ParentId="27406" CreationDate="2011-03-16T18:52:49.463" Score="4" ViewCount="" Body="&lt;p&gt;$$V_n = \frac{\pi^{n/2}}{\Gamma(n/2+1)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As posted earlier &lt;a href=&quot;http://math.stackexchange.com/questions/15656/volumes-of-n-balls-what-is-so-special-about-n-5&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="3787" LastActivityDate="2011-03-16T18:52:49.463" CommentCount="1" />
  <row Id="27446" PostTypeId="1" AcceptedAnswerId="27451" CreationDate="2011-03-16T19:00:39.440" Score="1" ViewCount="304" Body="&lt;p&gt;I have a linear operator $T_1$ which acts on the vector space of polynomials in this way:&#xA;$$T_1(p(x))=p&amp;#39;(x).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I find its eigenvalues and how can I know whether it is diagonalizable or not?&lt;/p&gt;&#xA;" OwnerUserId="6163" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-16T19:14:17.727" LastActivityDate="2011-03-18T14:17:34.803" Title="Eigenvalues of the differentiation operator" Tags="&lt;linear-algebra&gt;&lt;eigenvalues&gt;&lt;eigenvectors&gt;" AnswerCount="4" CommentCount="4" FavoriteCount="1" />
  <row Id="27447" PostTypeId="2" ParentId="26954" CreationDate="2011-03-16T19:02:45.883" Score="4" ViewCount="" Body="&lt;p&gt;The claim is false as stated, but I think a refinement can fix the problem.  (Not sure what the correct refinement is yet.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My counterexample is essentially a generalization of taking $\Phi: \mathbb R \to \mathbb R, x \mapsto x^3$.  For $n=2, m=1$, we will take $M=M&amp;#39; = \mathbb R^2 = \mathbb C$.  Using complex coordinates $z = x+iy$, let $\Phi(z) = z |z|^2$.  This is a diffeomorphism of $\mathbb C \setminus \{ 0 \}$, and is a homeomorphism of $\mathbb C$.  The inverse map is $\Phi^{-1}(z) = z |z|^{-2/3}$ (with $\Phi^{-1}(0) = 0$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I will show that if $S$ is embedded in $\mathbb C$ then $\Phi(S)$ is embedded.  The other direction follows the same argument, but using the inverse function instead.&#xA;If $S$ avoids $0$, there is nothing to check.  If $S$ intersects $0$, we reduce to the local problem of a smooth curve $\gamma: [-1,1] \to \mathbb{C}$, $\gamma(0) = 0$, $\gamma&amp;#39;(t) \ne 0$.  We then check that $\gamma( t^{1/3})$ satisfies that $\Phi( \gamma( t^{1/3}))$ is smooth.  (The easy way to see this is to note that $\gamma(t) = ut + O(t^2)$ for a non-zero constant $u$, and thus $\Phi( \gamma(t) ) = u|u|^2 t^3 + O(t^4).)&lt;/p&gt;&#xA;" OwnerUserId="8343" LastActivityDate="2011-03-16T19:02:45.883" CommentCount="8" />
  <row Id="27449" PostTypeId="2" ParentId="27446" CreationDate="2011-03-16T19:08:54.800" Score="5" ViewCount="" Body="&lt;p&gt;Take the derivative of $a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0$ (with $a_n\neq 0$), and set it equal to $\lambda a_nx^n+\cdots+\lambda a_0$.  Look particularly at the equality of the coefficients of $x^n$ to determine what $\lambda$ must be.  Once you know what the eigenvalues are, consider which possible diagonalized linear transformations have that eigenvalue set, and whether such linear transformations can be similar to differentiation.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2011-03-16T19:19:09.127" LastActivityDate="2011-03-16T19:19:09.127" CommentCount="11" />
  <row Id="27450" PostTypeId="2" ParentId="27446" CreationDate="2011-03-16T19:10:58.683" Score="3" ViewCount="" Body="&lt;p&gt;Differentiating lowers the degree, so the only case where you get out a scalar multiple of what you put in is when you differentiate a constant.&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2011-03-16T19:10:58.683" />
  <row Id="27451" PostTypeId="2" ParentId="27446" CreationDate="2011-03-16T19:22:19.127" Score="6" ViewCount="" Body="&lt;p&gt;In the finite dimensional case, finding the eigenvalues can be done by considering the matrix of the operator, computing the characteristic polynomial, and finding the roots. This is not possible in the infinite dimensional case (as occurs in the case of the vector space of all polynomials with coefficients in $F$), because there is no matrix for the operator and no characteristic polynomial.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instead, you have to go back to the definitions. An eigenvalue of $T$ is a scalar $\lambda$ for which there exists a nonzero vector $\mathbf{x}$ with $T(\mathbf{x}) = \lambda\mathbf{x}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose $\mathbf{x}$ is an eigenvector. What can we say about $\mathbf{x}$ and $\lambda$? As usual in this kind of cases, we write down what everything means, and see what this entails/implies. Often, we can gain enough information to figure out who $\mathbf{x}$ and $\lambda$ have to be.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's write $\mathbf{x} = a_nx^n + \cdots + a_0$, with $a_n\neq 0$ (we know at least one coefficient has to be nonzero for $\mathbf{x}$ to be nonzero, a precondition for being an eigenvector; and so we may as well write it going up to just the degree we need; so we are going to write $x^2+0x+1$, but not $0x^4+0x^3+x^2+0x+1$, in order to make our life easier). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then the equation  $T(\mathbf{x}) = \lambda\mathbf{x}$ becomes&#xA;$$ na_nx^{n-1}+\cdots + a_1 = \lambda a_nx^n + \lambda a_{n-1}x^{n-1}+\cdots + \lambda a_0.$$&#xA;This gives you a system of equations&#xA;\begin{align*}&#xA;\lambda a_n &amp;amp;= 0\\&#xA;\lambda a_{n-1} - na_n &amp;amp;= 0\\&#xA;\lambda a_{n-2} - (n-1)a_{n-1} &amp;amp;= 0\\&#xA;&amp;amp;\vdots\\&#xA;\lambda a_1 - 2a_2 &amp;amp;= 0\\&#xA;\lambda a_0 - a_1 &amp;amp;=0.&#xA;\end{align*}&#xA;Since we are assuming $a_n\neq 0$, it should be an easy matter to determine all eigenvalues, and all corresponding eigenvectors from this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, a linear transformation is diagonalizable if and only if there is a basis for the vector space that consists entirely of eigenvectors. Since you know by now what all the eigenvectors of $T$ are, you can figure whether you can find a linearly independent set of eigenvectors that spans the vector space of all polynomials. If you can, then $T$ is diagonalizable. If you cannot, then $T$ is not diagonalizable.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-16T19:40:23.790" LastActivityDate="2011-03-16T19:40:23.790" CommentCount="5" />
  <row Id="27452" PostTypeId="1" CreationDate="2011-03-16T19:42:46.347" Score="1" ViewCount="76" Body="&lt;p&gt;I have a waveform which I would like to take the fourier transform of. However, the simulator which generated the waveform uses an adaptive algorithm to determine the timestep for each calculation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, I figure I can regularize the timestep to first order by creating a new timebase and for each point in the new base calculating a linear interpolation from the two nearest points in the original sequence, but this seems inelegant, and potentially incorrect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also considered expanding the original sequence with whatever the smallest time-step is, replicating data points to fill-in the missing data (effectively creating steps), and then filtering the resulting fourier transform above some frequency (or simply ignoring data above some frequency). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Either of these will probably work with the data I have, since it is heavily oversampled for the frequency range I am interested in, but I worry that both would cause distortions if that were not the case. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a better method?&lt;/p&gt;&#xA;" OwnerUserId="8347" LastActivityDate="2011-03-17T00:43:56.197" Title="FFT of waveform with non-constant timestep" Tags="&lt;algorithms&gt;&lt;fourier-analysis&gt;" AnswerCount="2" FavoriteCount="2" />
  <row Id="27453" PostTypeId="1" AcceptedAnswerId="27466" CreationDate="2011-03-16T19:42:54.240" Score="1" ViewCount="59" Body="&lt;p&gt;Forgive me for my notation and lack of formatting prowess, it's been a while since I've done any of this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have flipped a biased coin (unknown weight) $n$ times and received $k$ successes (or heads).  I am trying to calculate the probability that there be a success on flip $N+1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we assume the bias $p$ is sampled uniformly (and is constant throughout the flip), then is the correct way to calculate this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\int P(p| X)*p \partial p$  where $X$ is n successes and k failures&lt;/p&gt;&#xA;&#xA;&lt;p&gt;thus&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$P(p| X) = \frac{P(X|p)*f(p)}{\int_{0}^{1} P(X|p) \partial p}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming a uniform prior for $p$ and that ${\int_{0}^{1} P(X|p)} = \frac{1}{n+1}$, we get &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\int_{0}^{1} P(X|p)*(n+1)*p \partial p $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $P(X|p)$ is the binomial probability.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this correct?       &lt;/p&gt;&#xA;" OwnerUserId="8262" LastEditorUserId="6460" LastEditorDisplayName="" LastEditDate="2011-03-16T20:14:46.630" LastActivityDate="2011-03-16T21:30:26.473" Title="Likelihood of next trial given N trials" Tags="&lt;probability&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="1" />
  <row Id="27454" PostTypeId="2" ParentId="27425" CreationDate="2011-03-16T19:50:15.920" Score="2" ViewCount="" Body="&lt;p&gt;&quot;Separation of variables&quot; in ODE (which has nothing to do with separation of variables in PDE) is a kind of magic that is easy to perform but difficult to justify. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume that in the given differential equation the quantities $x$ and $y$ are functions of a hidden variable $t$ (time). Then the equation $y\&amp;gt;y&amp;#39;=e^x$ is equivalent to $y(t){\dot y(t)\over \dot x(t)}\equiv e^{x(t)}$, resp.&#xA;$$y(t)\dot y(t)\equiv e^{x(t)}\dot x(t).$$&#xA;Integrating this from $t=0$ to $t=T$ one gets&#xA;$${1\over2}(y^2(T)-y_0^2)=e^{x(T)}-e^{x_0},$$&#xA;where $(x_0,y_0)$ is the initial condition and $T$ is arbitrary. This means: At any given time the quantities $x$ and $y$ are related by the equation&#xA;$${1\over2}(y^2-y_0^2)=e^x-e^{x_0}.$$&#xA;Looking back, one can see that the relation between $x$ and $y$ obtained in this way is exactly the equation obtained by following the recipe given in the books. &lt;/p&gt;&#xA;" OwnerUserId="1303" LastEditorUserId="1303" LastEditorDisplayName="" LastEditDate="2011-03-17T09:41:28.050" LastActivityDate="2011-03-17T09:41:28.050" />
  <row Id="27455" PostTypeId="1" AcceptedAnswerId="27488" CreationDate="2011-03-16T20:03:09.947" Score="1" ViewCount="85" Body="&lt;p&gt;Let $(f_n)_{n \geq 1}$ be disjointly supported sequence of functions in $L^\infty(0,1)$. Is the space $\overline{\mathrm{span}(f_n)}$ (the closure of linear span) complemented in $L^\infty(0,1)$? By complemented we mean that $L^\infty(0,1) = \overline{\mathrm{span}(f_n)} \oplus X$, where $X$ is a subspace of $L^\infty$ and $\oplus$ is direct sum. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Equivalently, we can ask if there exists a projection $P\colon L^\infty(0,1) \to \overline{\mathrm{span}(f_n)}$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is quite easy to prove this in $C[0,1]$. Indeed, let $(f_n)$ be disjointly supported sequence in $C[0,1]$ and fix $x_n \in \mathrm{supp}(f_n)$, $n \in \mathbb{N}$. Then the space $C[0,1]$ can be written as&#xA;$$&#xD;&#xA;C[0,1] = \overline{\mathrm{span}(f_n)} \oplus \{f \in C[0,1]\colon f(x_n) = 0, n = 1,2,\dots \}.&#xD;&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="8229" LastEditorUserId="8229" LastEditorDisplayName="" LastEditDate="2011-03-16T22:51:24.313" LastActivityDate="2011-05-30T16:45:42.393" Title="Is a sequence of disjointly supported functions in $L^\infty$ complemented?" Tags="&lt;functional-analysis&gt;&lt;banach-spaces&gt;" AnswerCount="2" />
  <row Id="27456" PostTypeId="1" AcceptedAnswerId="36132" CreationDate="2011-03-16T20:20:17.267" Score="3" ViewCount="98" Body="&lt;p&gt;Given a pair of positive coprime integers $a_1$ and $a_2$, I'd like to determine the set of positive integers $x$ which satisfy the following system of $\text{gcd}$ equations:&#xA;\begin{eqnarray}&#xA;\gcd(x,a_1) = 1, \gcd(x - a_1,a_2) = 1, \dots, \gcd(x - a_1 \lfloor \tfrac{x}{a_1} \rfloor, a_2) = 1,&#xA;\end{eqnarray}&#xA;where $x \ \text{mod} \ a_1 = x - a_1 \lfloor \tfrac{x}{a_1} \rfloor$ (and we ignore those equations if any of the arguments are zero or negative). Are there methods to do so? More generally, given a set of coprime integers $a_1, \dots, a_n$, I'd like to determine the positive integer solutions of the following system&#xA;\begin{eqnarray}&#xA;\gcd(x,a_1) = 1, \gcd(x - a_1 i_1 ,a_2) = 1, \dots, \gcd(x - \sum_{k = 1}^{n-1} a_k i_k, a_n) = 1&#xA;\end{eqnarray}&#xA;where $i_j \in \{ 0 , \dots, \lfloor \frac{1}{a_j} (x - \sum_{k = 1}^{j-1} a_k i_k) \rfloor \}$ and  $j = 1, \dots, n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Update:&lt;/em&gt; Perhaps it is easier to try to solve the following related question in the two coefficient case. Given a pair of positive coprime integers $a_1$ and $a_2$, determine the set of positive integers $x$ which satisfy the following system of congruences:&#xA;\begin{eqnarray}&#xA;x \not \equiv 0 \ \text{mod} \ a_1, \quad x - a_1 \not \equiv  0 \ \text{mod} \ a_2, \quad \dots, \quad x - a_1 \lfloor \tfrac{x}{a_1} \rfloor \not \equiv 0 \ \text{mod} \ a_2,&#xA;\end{eqnarray}&#xA;where $x \ \text{mod} \ a_1 = x - a_1 \lfloor \tfrac{x}{a_1} \rfloor$.&lt;/p&gt;&#xA;" OwnerUserId="2720" LastEditorUserId="2720" LastEditorDisplayName="" LastEditDate="2011-03-16T23:05:49.937" LastActivityDate="2011-05-01T04:53:39.247" Title="Solving a System of GCD Equations" Tags="&lt;number-theory&gt;&lt;linear-algebra&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" />
  <row Id="27457" PostTypeId="1" AcceptedAnswerId="27505" CreationDate="2011-03-16T20:30:39.067" Score="1" ViewCount="175" Body="&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/Y3r33.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If anyone has some insight on how to do this it would be very much appreciated.&lt;/p&gt;&#xA;" OwnerUserId="3691" LastActivityDate="2011-03-17T23:52:52.083" Title="Floyd's algorithm for the shortest paths....challenging" Tags="&lt;algorithms&gt;&lt;graph-theory&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="27458" PostTypeId="2" ParentId="26160" CreationDate="2011-03-16T20:36:19.890" Score="0" ViewCount="" Body="&lt;p&gt;As Carl and Alex have mentioned, it is possible to prove the existence of an object satisfying given properties (e.g., is a counterexample to a given statement) without explicitly constructing the required object.  Going along a slightly different route, consider the logical law of &lt;em&gt;Tertium Non Datur&lt;/em&gt;, or &lt;em&gt;Law of Excluded Middle&lt;/em&gt;, which essentially states that given any sentence either it or its negation is true.  (By a &lt;em&gt;sentence&lt;/em&gt; I mean any statement that is unambiguous; the statement $x=2$ is not a sentence, since the meaning of $x$ is not fixed, however the statements $(\forall x) (x = 2)$ (&quot;every $x$ is equal to $2$&quot;) and $(\exists x) ( x = 2 )$ (&quot;there is an $x$ which is equal to $2$&quot;) are sentences.)  With &lt;em&gt;Tertium Non Datur&lt;/em&gt; one can prove that sentences $( \exists x ) \phi$ (&quot;there is an $x$ for which $\phi$ holds&quot;) and $\neg ( (\forall x ) \neg \phi)$ (&quot;it is not the case that every $x$ fails to satisfy $\phi$&quot;) are logically equivalent, and so instead of showing that such an object exists, you can show that the non-existence of such an object is contradictory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps one of the earliest and most (in)famous examples of such a proof was David Hilbert's proof of a special case of (what is now called) &lt;em&gt;Hilbert's Basis Theorem&lt;/em&gt;.  At the time this method was quite controversial, and the referee his proof was sent to replied, &quot;This is not mathematics. This is theology.&quot;  (Well, he actually replied in German.)&lt;/p&gt;&#xA;" OwnerUserId="8348" LastActivityDate="2011-03-16T20:36:19.890" />
  <row Id="27459" PostTypeId="2" ParentId="27406" CreationDate="2011-03-16T20:40:40.913" Score="4" ViewCount="" Body="&lt;p&gt;Start with an unrelated integral&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$I = \int_{-\infty}^\infty \cdots \int_{-\infty}^\infty e^{(-x_1^2+x_1^2+\cdots x_n^2) } dx_1 dx_2 \cdots dx_n = \pi^{n/2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This should be easy to easy if you notice that this can be factored as a product of n integrals each in the variable $x_i$ and that $\int_{-\infty}^\infty e^{-x^2}dx =\sqrt{\pi}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now try to calculate the same integrate by making a change to polar coordinates. These new coordinates will have one $r$ coordinate and $n-1$ angular coordinates $\{\theta_i\}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$I = \int e^{(-x_1^2+x_1^2+\cdots x_n^2) } dV_n =$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To express the volume element in the new coordinates you can either use inference from lower dimensions (or if you're a physicist, a dimensional argument) and say that $$V = C_n r^n \Rightarrow dV_n = nC_n r^{n-1}dr$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or you can explicitly compute the Jacobian to show that $|J(r,\theta_1,\cdots,\theta_n)|= r^{n-1} \cos^{n-2}\theta_1\cos^{n-3}\theta_2 \cdots \cos \theta_{n-2} $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$I = \int_0^\infty e^{-r^2} nC_nr^{n-1} dr = C_n\frac{n}{2} \Gamma (\frac{n}{2}) = C_n\left(\frac{n}{2}\right)! $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ C_n = \frac{\pi^{n/2}}{\left(\frac{n}{2}\right)!} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$V_n =  \frac{\pi^{n/2}}{\left(\frac{n}{2}\right)!}  r^n$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;on a sidenote, you also get the surface area from $dV_n = S_n(r) dr$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$S_n(r)= \frac{(2\pi)^{n/2}}{\Gamma (\frac{n}{2})} r^{n-1}$$&lt;/p&gt;&#xA;" LastActivityDate="2011-03-16T20:55:35.130" />
  <row Id="27460" PostTypeId="1" CreationDate="2011-03-16T21:03:40.927" Score="0" ViewCount="73" Body="&lt;p&gt;Let $k$ be an algebraically closed field, and let $\mathbb{A}^n(k)$ denote the linear $n$-dimensional vector space over $k$. Suppose $X \subset \mathbb{A}^n$, $X&amp;#39; \subset \mathbb{A}^m$ are two varieties. We say that $X, X&amp;#39;$ are isomorphic if there exists a polynomial map $\phi : X \rightarrow X&amp;#39;$ which is bijective and whose inverse is a polynomial map. Note that $\phi$ is polynomial if $\phi(x) = (f_1(x), \cdots, f_m(x))$ where each $f_j$ is a polynomial. I want to show that if $p \in X$ is smooth (the tangent space of $X$ at $p$ has the same dimension as $X$) if and only if $\phi(p)$ is a smooth point of $X&amp;#39;$. It is not clear to me why if $(x_1, \cdots, x_n)$ is an eigenvector of the Jacobian of the generators of $I(X)$ at $p$, that $\phi((x_1, \cdots, x_n))$ must be an eigenvector of the Jacobian of the generators of $I(X&amp;#39;)$, which I presume to be the natural approach to show that the two tangent spaces have equal dimension. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any suggestions?&lt;/p&gt;&#xA;" OwnerUserId="7871" LastActivityDate="2011-03-16T21:24:11.203" Title="Smoothness of an affine variety is preserved under isomorphism" Tags="&lt;algebraic-geometry&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="27461" PostTypeId="2" ParentId="27412" CreationDate="2011-03-16T21:06:57.133" Score="1" ViewCount="" Body="&lt;p&gt;Solve&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(\vec{p_3}-\vec{p_2})\cdot(\vec{p_2}-\vec{p_0}) = L^2 \cos\theta$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you get is an equation of a plane. Your locus is a circle on that plane centered at $\vec{p_2}$ and of radius $L$. &lt;/p&gt;&#xA;" LastActivityDate="2011-03-16T22:31:56.060" />
  <row Id="27462" PostTypeId="1" CreationDate="2011-03-16T21:11:13.447" Score="3" ViewCount="56" Body="&lt;p&gt;Some standard examples on various quals seems to be computing units/class numbers etc. of the ring $\mathbb{Q}(\alpha)$, where $\alpha$ is a root of either $X^3+aX+b$ or $X^5+aX+b$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My questions is the following: What are some standard tricks that can be used to deduce that a particular unit is actually a fundamental unit?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm familiar with class field theory and L-series, so I don't mind higher-level methods. I'm sort of trying to figure out what's usable during an exam. Books on number theoretic algorithms only cover pretty general cases and these algorithms are too computationally intensive to be done manually.&lt;/p&gt;&#xA;" OwnerUserId="8351" LastActivityDate="2011-03-16T21:11:13.447" Title="Computing units of certain number fields" Tags="&lt;algebraic-number-theory&gt;" CommentCount="3" FavoriteCount="2" />
  <row Id="27463" PostTypeId="1" CreationDate="2011-03-16T21:13:50.870" Score="0" ViewCount="103" Body="&lt;p&gt;I'm preparing for an exam in the signals and systems class I'm taking. One of the practice exams has a problem that requires you to take the Fourier transform of $\text{sinc}(4t)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From a table of fourier transform pairs I found: $\frac{\omega_b}{\pi}\text{sinc}(\frac{\omega_b t}{\pi})\Rightarrow \text{rect}(\omega/2\omega_b)$. Using this I tried to match the given $\text{sinc}(4t)$ by rewriting it as $\frac{1}{4}\frac{4\pi}{\pi}\text{sinc}(\frac{4\pi t}{\pi})$. From this I get that $\omega_b = 4\pi$ and thus the fourier transform should yield $\frac{1}{4}\text{rect}(\omega/8\pi)$. But, in the exam solutions they show the fourier transform to yield $\frac{\pi}{4}\text{rect}(\omega/8)$. Any ideas where I'm going wrong?&lt;/p&gt;&#xA;" OwnerUserId="8350" LastEditorUserId="1120" LastEditorDisplayName="" LastEditDate="2011-03-16T21:30:22.337" LastActivityDate="2011-03-16T21:30:22.337" Title="Fourier transform of sinc(4t)" Tags="&lt;fourier-analysis&gt;&lt;fourier-series&gt;" CommentCount="3" />
  <row Id="27464" PostTypeId="2" ParentId="27460" CreationDate="2011-03-16T21:24:11.203" Score="2" ViewCount="" Body="&lt;p&gt;A more intrinsic way to express smoothness is via the (sheaf of) ring(s) of functions: in your case, $X$ is smooth at $x$ iff the local ring at $x$ is regular.&#xA;The isomorphism between $X$ and $X&amp;#39;$ gives an isomorphism between (sheaves of) rings of functions.&lt;/p&gt;&#xA;" OwnerUserId="2660" LastActivityDate="2011-03-16T21:24:11.203" CommentCount="3" />
  <row Id="27465" PostTypeId="1" CreationDate="2011-03-16T21:28:47.660" Score="1" ViewCount="45" Body="&lt;p&gt;If $R \subseteq A \times A$ is it true that $R$ is symmetrical since $xRy$ then $yRx$ ? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have written that this is also antisymmetrical if both $x\leq y$ and $y\leq x$ if $x =y$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How does I then make $R \subseteq A \times A$ transitive?&lt;/p&gt;&#xA;" OwnerUserId="8322" LastEditorUserId="8322" LastEditorDisplayName="" LastEditDate="2011-03-16T21:50:26.517" LastActivityDate="2011-03-16T21:50:26.517" Title="Symmetrical, Transitional and Antisymmetrical question!" Tags="&lt;discrete-mathematics&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="27466" PostTypeId="2" ParentId="27453" CreationDate="2011-03-16T21:30:26.473" Score="1" ViewCount="" Body="&lt;p&gt;You have a problem using $p$ both as a probability or bias and as a density function, so let's use $f(p)$ as the prior density instead.  Then $f(p)=1$ since you have assumed it is uniform on $[0,1]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is useful to know that $(n+1) {n \choose k} \int_{x=0}^1 x^k (1-x)^{n-k} \; dx = 1$, related to the Beta distribution&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the posterior density is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f(p|k \text{ heads out of } n) = \frac{Pr(k \text{ heads out of } n|p) f(p)}{\int_{q=0}^1 Pr(k \text{ heads out of } n|q) f(q) \, dq} $$ $$ = \frac{ {n \choose k} p^k (1-p)^{n-k}  1}{\int_{q=0}^1 {n \choose k} q^k (1-q)^{n-k} 1 \; dq} = (n+1){n \choose k} p^k (1-p)^{n-p}$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;and the answer to the question is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$Pr({n+1}^\text{th} \text{ trial is heads }| k \text{ heads out of } n) = E[p|k \text{ heads out of } n]$$ &#xA;$$= \int_{p=0}^1 p \; f(p|k \text{ heads out of } n) \; dp = \int_{p=0}^1 (n+1){n \choose k} p^{k+1} (1-p)^{n-k} \; dp =  \frac{k+1}{n+2}$$&lt;/p&gt;&#xA;" OwnerUserId="6460" LastActivityDate="2011-03-16T21:30:26.473" />
  <row Id="27467" PostTypeId="2" ParentId="27465" CreationDate="2011-03-16T21:41:06.213" Score="2" ViewCount="" Body="&lt;p&gt;You are either confused, or we are running against a language barrier (both are certainly possible).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is not true that simply by virtue of being a subset of $A\times A$, a subset $R$ will be &quot;symmetrical.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rather: we define $R$ to be &quot;symmetrical&quot; &lt;strong&gt;if and only if&lt;/strong&gt; for every $x$ and $y$ in $A$, &lt;em&gt;if&lt;/em&gt; $(x,y)\in R$ (that is, if $xRy$), &lt;em&gt;then&lt;/em&gt; $(y,x)\in R$ (that is, $yRx$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An example of a symmetric relation on $\mathbb{R}$ is&#xA;$$R = \{ (a,b)\in\mathbb{R}\times\mathbb{R} \mid |a|=|b|\}$$&#xA;since, if $aRb$, then $|a|=|b|$, hence $|b|=|a|$, hence $bRa$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An example of a non-symmetric relation on $\mathbb{R}$ is&#xA;$$S = \{(a,b)\in\mathbb{R}\times\mathbb{R} \mid a\geq 0\}.$$&#xA;It is not symmetric, because $(1,-1)\in S$, but $(-1,1)\notin S$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We define $R$ to be &quot;antisymmetric&quot; &lt;strong&gt;if and only if&lt;/strong&gt; for every $x$ and $y$ in $A$, &lt;strong&gt;if&lt;/strong&gt; $xRy$ and $yRx$ are both true, &lt;strong&gt;then&lt;/strong&gt; $x=y$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(You have the implication reversed).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the example $R$ above, $R$ is &lt;em&gt;not&lt;/em&gt; antisymmetric, because $(1,-1)\in R$, $(-1,1)\in R$, but $-1\neq 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, an example of an antisymmetric relation on $\mathbb{R}$ is:&#xA;$$T = \{(a,b)\in\mathbb{R}\times\mathbb{R}\mid a\leq b\}.$$&#xA;This is antisymmetric because if $(a,b)\in T$ and $(b,a)\in T$, that means that $a\leq b$ and $b\leq a$, so then we conclude that $a=b$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, a relation $R\subseteq A\times A$ is said to be &quot;transitive&quot; &lt;strong&gt;if and only if&lt;/strong&gt; for every $a,b,c\in A$, &lt;em&gt;if&lt;/em&gt; $aRb$ and $bRc$ both hold, then $aRc$ holds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both $R$, $S$, and $T$ above are transitive. To see an example of a relation on $\mathbb{R}$ that is not transitive, let&#xA;$$U = \{(a,b)\in\mathbb{R}\times\mathbb{R}\mid a\neq b\}.$$&#xA;Then $0U1$ holds, and $1U0$ holds, but $0U0$ does not hold. So $U$ is not transitive.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2011-03-16T21:41:06.213" CommentCount="1" />
  <row Id="27468" PostTypeId="1" AcceptedAnswerId="27476" CreationDate="2011-03-16T21:58:04.857" Score="3" ViewCount="119" Body="&lt;p&gt;Let&#xA;$$&#xD;&#xA;f(x, y) = \frac{x + y}{2e}&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;g(x, y) = \frac{(x+y)+\sqrt{x^2 + y^2 + 6xy}}{2e^{\frac{\sqrt{x^2 + y^2 + 6xy}}{2(x + y)}}} &#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am looking for the largest $K$ so that,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \forall x, y &amp;gt; 0, \frac{g(x, y)}{f(x, y)} &amp;gt; K$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By numerical computation I found that $K$ must be around $3.3$. But the best thing I could prove is about 2.28 for $K$ using the following method,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;x^2 + y^2 + 2xy &amp;lt; x^2 + y^2 + 6xy &amp;lt; (\sqrt{3}x)^2 + (\sqrt{3}y)^2 + 2(\sqrt{3}x)(\sqrt{3}y)&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which results in,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;(x+y) &amp;lt; \sqrt{x^2 + y^2 + 6xy} &amp;lt; \sqrt{3}(x+y)&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;\frac{g(x, y)}{f(x, y)} &amp;gt; \frac{\frac{x+y}{e^{\sqrt{3}/2}}}{\frac{x+y}{2e}} &amp;gt; 2.28&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm wondered if anyone can propose a way to prove a larger K?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: This is the improvement of the network throughput based on the method I proposed (g(.)) over an old method (f(.)). So, this is important for me to show that how much better my proposed method works.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance :)&lt;/p&gt;&#xA;" OwnerUserId="8184" LastEditorUserId="8184" LastEditorDisplayName="" LastEditDate="2011-03-16T22:03:24.977" LastActivityDate="2011-03-17T10:21:19.110" Title="Finding a Lower Bound of a Function" Tags="&lt;calculus&gt;&lt;functions&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="27470" PostTypeId="1" CreationDate="2011-03-16T22:20:33.213" Score="-1" ViewCount="65" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am implementing a program in Java that needs the &lt;strong&gt;Bank of England base rate&lt;/strong&gt;. Rather than the user inputting this into the system, I have heard that there is a way to get a &lt;strong&gt;live feed&lt;/strong&gt; of the base rate directly from (I am guessing) the Bank of England. Has anyone heard of or used something similar? - any advice would be helpful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="8352" LastActivityDate="2011-03-16T22:39:47.927" ClosedDate="2011-03-16T22:40:49.877" Title="Bank of England base rate feed" Tags="&lt;finance&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="27471" PostTypeId="1" CreationDate="2011-03-16T22:30:31.413" Score="5" ViewCount="146" Body="&lt;p&gt;&lt;strong&gt;Theorem&lt;/strong&gt; Legendre: let $a,b,c$ coprime positive integers, then $ax^2 + by^2 = cz^2$ has a nontrivial solution in rationals $x,y,z$ iff $\left(\frac{-bc}{a}\right)=\left(\frac{-ac}{b}\right)=\left(\frac{ab}{c}\right)=1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm trying to prove this theorem using&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Lemma&lt;/strong&gt; Modified global square theorem: The rational number $cz$ is a $c$ times a square iff it is a $c$ times a square in $\mathbb{Q}_p$ for every prime $p$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So far, it is possible to show that the equation can be solved locally for all primes powers except $2^r$:&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;To solve this equation $\pmod {abc}$ just put $x = bc$, $y = ac$, $z = ab$. Reducing this gives a solution for every odd prime $p | abc$. Hensel lifts it to $\mathbb{Q}_p$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If an odd prime $p \not| abc$ then put $z = 1$ so that we have $ax^2 \equiv c - by^2 \pmod p$, $x$ and $y$ can take on $(p+1)/2$ values each so they must have an intersection which solves this congruence. Hensel lifts it to $\mathbb{Q}_p$ again.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's easy to solve when $p=2$ but I think it $p=4$ needs to be done for Hensel to apply.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;My question is how to get a rational number out of this, so that I can apply the global square theorem and conclude? It seems like the $p$-adic numbers have to have a finite expansion but that seems just as hard to prove as anything. Also if anyone has a hint for the $2^r$ case that would be great too! Thanks a lot.&lt;/p&gt;&#xA;" OwnerUserId="4290" LastEditorUserId="4290" LastEditorDisplayName="" LastEditDate="2011-03-16T22:43:13.493" LastActivityDate="2011-04-02T08:05:04.607" Title="Proof of Legendre's theorem on the terniary quadratic form." Tags="&lt;number-theory&gt;&lt;quadratic-forms&gt;" CommentCount="4" FavoriteCount="1" />
  <row Id="27472" PostTypeId="2" ParentId="27470" CreationDate="2011-03-16T22:39:47.927" Score="0" ViewCount="" Body="&lt;p&gt;Not sure if this belongs here... but:&#xA;&lt;a href=&quot;http://www.bankofengland.co.uk/&quot; rel=&quot;nofollow&quot;&gt;http://www.bankofengland.co.uk/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Current Bank Rate is shown in the top of the box in the bottom right corner. &lt;/p&gt;&#xA;" OwnerUserId="8171" LastActivityDate="2011-03-16T22:39:47.927" CommentCount="1" />
  <row Id="27473" PostTypeId="1" CreationDate="2011-03-16T22:45:05.290" Score="1" ViewCount="192" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am using matlab to solve the following ODE&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$0.02K^2 F&amp;#39;&amp;#39;(K)-3.45 F&amp;#39;(K)-3.45=0.08 F(K) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;subject to boundary conditions&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$F(K=0)=63.09$$ and $$ F(K=700)=0.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found that I had to give &lt;code&gt;bvpinit&lt;/code&gt; an offset such as &#xA;&lt;code&gt;bvpinit(linspace(1e-3,700,10000),[1 0.1])&lt;/code&gt;&#xA;from the left boundary &lt;code&gt;0&lt;/code&gt; because otherwise, I would get an error&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;Error using ==&amp;gt; bvp4c&#xA;Unable to solve the collocation equations -- a singular Jacobian encountered&#xA;&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But this isn't really elegant, because the offset seems arbitary. This also did not extend the solution all the way to the left end point &lt;code&gt;0&lt;/code&gt;. Would appreciate if someone can provide a better solution to this problem. Many thanks.&lt;/p&gt;&#xA;" OwnerUserId="6016" LastActivityDate="2011-03-22T08:33:27.053" Title="solving Boundary-value problem ODE in matlab numerically" Tags="&lt;numerical-methods&gt;&lt;matlab&gt;" AnswerCount="1" />
  <row Id="27474" PostTypeId="2" ParentId="27452" CreationDate="2011-03-16T22:55:01.330" Score="0" ViewCount="" Body="&lt;p&gt;If the you know a priori that the signal is sparse in the Fourier Domain ( sum of a few sinusoids), you may want to try compressive sensing techniques (e.g. Basis Pursuit) to determine the Fourier transform&lt;/p&gt;&#xA;" OwnerUserId="832" LastActivityDate="2011-03-16T22:55:01.330" />
  <row Id="27475" PostTypeId="2" ParentId="27468" CreationDate="2011-03-16T23:06:22.640" Score="3" ViewCount="" Body="&lt;p&gt;The minimum of $g(x,y)/f(x,y)$ occurs at the points $(x,y)$ where $x=y$, and is given by&#xA;$$&#xD;&#xA;K = (1 + \sqrt 2 )e^{1 - \sqrt 2 /2} \approx 3.235769626477154.&#xD;&#xA;$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sketch of proof. Using the transformations $s=x+y$, $t=xy$, one finds that&#xA;$$&#xD;&#xA;\frac{{g(x,y)}}{{f(x,y)}} = e\frac{{1 + \xi }}{{e^{\xi /2} }},&#xD;&#xA;$$&#xA;where&#xA;$$&#xD;&#xA;\xi  = \sqrt {1 + \frac{{4xy}}{{(x + y)^2 }}} .&#xD;&#xA;$$&#xA;The maximum of $\xi$ occurs at the points $(x,y)$ where $x=y$ (straightforward exercise), hence equal to $\sqrt{2}$.&#xA;The result then follows from the fact that the function $(1 + \xi )/e^{\xi /2}$&#xA;is decreasing (which is verified upon differentiation).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;: It is interesting (and maybe important) to note that the function $g(x,y)/f(x,y)$, $x,y &amp;gt; 0$, though quite complicated in appearance, is not far from being a constant. Indeed, by considering the infimum of $\xi$ (which is equal to $1$), we get the (tight) upper bound of $2\sqrt{e} \approx 3.297442541400256$, so&#xA;$$&#xD;&#xA;3.2357 &amp;lt; (1 + \sqrt 2 )e^{1 - 2/\sqrt 2 }  \le \frac{{g(x,y)}}{{f(x,y)}} &amp;lt; 2\sqrt e  &amp;lt; 3.2975,&#xD;&#xA;$$&#xA;for any $x, y &amp;gt; 0$.&lt;/p&gt;&#xA;" OwnerUserId="2810" LastEditorUserId="2810" LastEditorDisplayName="" LastEditDate="2011-03-17T10:21:19.110" LastActivityDate="2011-03-17T10:21:19.110" CommentCount="6" />
  <row Id="27476" PostTypeId="2" ParentId="27468" CreationDate="2011-03-16T23:06:39.883" Score="6" ViewCount="" Body="&lt;p&gt;$K=(1+\sqrt{2})\mathrm{e}^{1-\sqrt{2}/2}=3.23576963$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In short, this is because $x^2+y^2+6xy\le2(x+y)^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a more detailed proof, write the ratio $g(x,y)/f(x,y)$ as $h(z)=\mathrm{e}^{1-z/2}(1+z)$ with $z^2=(x^2+y^2+6xy)/(x+y)^2$. Now, $z\mapsto h(z)$ is a decreasing function on $z\ge1$ (differentiate the logarithm) and, for every positive $x$ and $y$, $z^2-1=4xy/(x+y)^2\le1$ hence $1\le z^2\le 2$. This means that $h(z)$ is minimal when $z^2=2$, that is, when $x=y$. This yields the value of $K$ given above, if one replaces the strict inequality sign $&amp;gt;$ in your post by a $\ge$ sign.&lt;/p&gt;&#xA;" OwnerUserId="6179" LastActivityDate="2011-03-16T23:06:39.883" CommentCount="1" />
  <row Id="27477" PostTypeId="1" AcceptedAnswerId="27479" CreationDate="2011-03-16T23:06:58.807" Score="4" ViewCount="104" Body="&lt;p&gt;Coonsider the polynomial $p(x) = x^2(4x^2-3)^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This polynomial is special because:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;All of the local maxima $M_i$ are of the form $p(M_i)=1$ &lt;/li&gt;&#xA;&lt;li&gt;All of the local minima $m_i$ are of the form $p(m_i)=0$&lt;/li&gt;&#xA;&lt;li&gt;and $p(-1)=p(1)=1$&lt;/li&gt;&#xA;&lt;li&gt;The polynomial has no other zeros, local maxima or local minima.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;My questions are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;what conditions must a polynomial satisfy in order to have the above characteristics?&lt;/li&gt;&#xA;&lt;li&gt;do they have a specific name?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These are the elements of the family I have so far:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$p_1(x)=x^2 (4x^2-3)^2$$&#xA;$$p_2(x)=x^2 (16 x^4-20 x^2+5)^2$$&#xA;$$p_3(x)=x^2 (64 x^6-112 x^4+56 x^2-7)^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can someone find the pattern? What is $p_4(x)$?&lt;/p&gt;&#xA;" OwnerUserId="1747" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-17T04:18:10.110" LastActivityDate="2011-03-17T04:21:08.093" Title="A special kind of polynomial: $p(x)=x^{2}(4x^{2}-3)^2$" Tags="&lt;polynomials&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" />
  <row Id="27478" PostTypeId="1" CreationDate="2011-03-16T23:12:53.513" Score="3" ViewCount="129" Body="&lt;p&gt;This question is from a math-essentials booklet for physicists. The function is not analytic from the C-R conditions but that is all I know. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ \rlap{\textbf{-------------------------------------------------------------------------------------------------}}{\mbox{Its derivative should be}}$ $2z + 2\bar{z} =4Re[z]$ So are its critical points all over the real axis? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question also asks for a roiugh sketch of the graph. How would I sketch $(x,y)\mapsto (x+x^2-y^2,2xy-y)$ ? &lt;/p&gt;&#xA;" LastEditorUserId="1303" LastEditorDisplayName="" LastEditDate="2011-03-17T10:14:43.937" LastActivityDate="2011-03-17T10:30:07.963" Title="Critical values and critical points of the mapping $z\mapsto z^2 + \bar{z}$" Tags="&lt;complex-analysis&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="2" />
  <row Id="27479" PostTypeId="2" ParentId="27477" CreationDate="2011-03-16T23:16:08.133" Score="3" ViewCount="" Body="&lt;p&gt;I think you will find the results and techniques of the following paper useful in this regard.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;MR1752251 (2001c:11035) 11D25 (11D41 11G05 11G30)&lt;br&gt;&#xA;Buchholz, Ralph H.; MacDougall, James A.(5-NEWC)&lt;br&gt;&#xA;&lt;strong&gt;When Newton met Diophantus: a study of rational-derived polynomials&lt;br&gt;&#xA;and their extension to quadratic fields.&lt;/strong&gt;&lt;br&gt;&#xA;J. Number Theory 81 (2000), no. 2, 210–233.&lt;br&gt;&#xA;&lt;a href=&quot;http://dx.doi.org/10.1006/jnth.1999.2473&quot; rel=&quot;nofollow&quot;&gt;http://dx.doi.org/10.1006/jnth.1999.2473&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is an interesting paper, which surveys the problem of determining the set $D(n)$ of all &quot;$k$-derived'' univariate polynomials of degree $n$ (where a polynomial $f \in k[x]$ is $k$-derived if $f$ and each of its successive derivatives has all roots in the ground field $k$). Define two polynomials $f_1,f_2\in k[x]$ to be equivalent if $f_1(x)=r f_2(s x+t)$ for $r,s,t\in k$, $r,s \neq 0$. Then up to equivalence, the following is known about $\mathbb Q$-derived polynomials: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$D(1)=\{x\};\quad D(2)=\{x^2,x(x-1)\};$$&#xA;$$D(3)=\{x^3\}\cup\bigg\{x(x-1)(x-a)\ :\ a=\frac{w(w-2)}{w^2-1},w\in \mathbb Q\bigg\};$$&#xA;$$ D(4)\supseteq \{x^4\}\cup\left\{x^2(x-1)(x-a)\ \left|\ \begin{array}{l}a=\frac{9(2w+z-12)(w+z)}{(z-w-18)(8w+z)},&#xD;&#xA; (w,z)\in E(\mathbb Q),\\&#xD;&#xA; E\colon z^2=w(w-6)(w+18)\end{array}\right.\right\};$$&#xA;$$ D(n)\supseteq \{x^n, x^{n-1}(x-1)\}\ {\rm for}\ n\geq 5.$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The authors prove that determining $D(n)$ in general devolves into two conjectures: (1) that no quartic with four distinct roots is $\mathbb Q$-derived; (2) that no quintic of type $x^3(x-a)(x-b)$, $a\neq b,\ a,b\neq0$, is $\mathbb Q$-derived. The first conjecture can be solved by determining all rational points on a hyperelliptic surface of degree 10. The second conjecture can be solved by determining all rational points on a curve of genus 2 (E. V. Flynn [&quot;On $\mathbb Q$-derived polynomials'', Preprint; per revr.] has now proved this second conjecture). The authors also discuss briefly the situation of $K$-derived polynomials for quadratic extensions $K$ of $\mathbb Q$; there is, for example, the quartic $y^2=x^2(x-1)(x-\frac{37-20\sqrt{3}}{13})$ which is a ${\mathbb Q}(\sqrt{3})$-derived polynomial.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reviewed by Andrew Bremner&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-17T04:21:08.093" LastActivityDate="2011-03-17T04:21:08.093" CommentCount="1" />
  <row Id="27480" PostTypeId="1" AcceptedAnswerId="27481" CreationDate="2011-03-16T23:49:00.830" Score="7" ViewCount="220" Body="&lt;p&gt;What are some of the best books on graph theory, particularly directed towards an upper division undergraduate student who has taken most the standard undergraduate courses? I'm learning graph theory as part of a combinatorics course, and would like to look deeper into it on my own. Thank you.&lt;/p&gt;&#xA;" OwnerUserId="8354" LastActivityDate="2011-04-05T23:15:05.687" Title="What are good books to learn graph theory?" Tags="&lt;reference-request&gt;&lt;graph-theory&gt;&lt;books&gt;" AnswerCount="6" FavoriteCount="1" />
  <row Id="27481" PostTypeId="2" ParentId="27480" CreationDate="2011-03-17T00:01:50.463" Score="5" ViewCount="" Body="&lt;p&gt;Try Bondy and Murthy, &lt;a href=&quot;http://www.springer.com/new+&amp;amp;+forthcoming+titles+%28default%29/book/978-1-84628-969-9&quot; rel=&quot;nofollow&quot;&gt;Graph Theory&lt;/a&gt;. The previous version, &lt;a href=&quot;http://www.math.jussieu.fr/~jabondy/books/gtwa/gtwa.html&quot; rel=&quot;nofollow&quot;&gt;Graph Theory with Applications&lt;/a&gt;, is available online.&lt;/p&gt;&#xA;" OwnerUserId="589" LastEditorUserId="589" LastEditorDisplayName="" LastEditDate="2011-04-05T23:14:49.993" LastActivityDate="2011-04-05T23:14:49.993" CommentCount="2" />
  <row Id="27482" PostTypeId="1" AcceptedAnswerId="27484" CreationDate="2011-03-17T00:04:06.093" Score="2" ViewCount="119" Body="&lt;p&gt;The textbook I'm using has a somewhat confusing proof of one of the directions of Heine-Borel's Theorem, namely the one that states that a closed and bounded subset of $\mathbb R^n$ is compact.  It uses sequences and &quot;Changing the centers&quot; to 'move' the balls to $\mathbb Q^n$ and then constructs a sequence which leads to a contradiction, the latter of which I'm confused about.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could someone give me a more coherent proof in the same style or point me in the direction of a fairly understandable proof?&lt;/p&gt;&#xA;" OwnerUserId="7354" LastActivityDate="2011-03-17T00:37:50.433" Title="Proof of Heine-Borel in $\mathbb R^n$" Tags="&lt;real-analysis&gt;" AnswerCount="3" CommentCount="3" />
  <row Id="27483" PostTypeId="2" ParentId="27482" CreationDate="2011-03-17T00:20:57.890" Score="0" ViewCount="" Body="&lt;p&gt;One can use the following lemma to prove it: Let $(\mathbb{R}^n, d)$ be a Euclidean space with either the Euclidean metric or the taxicab metric. Then every bounded sequence in $(\mathbb{R}^n, d)$ has at least one convergent subsequence.&lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2011-03-17T00:20:57.890" />
  <row Id="27484" PostTypeId="2" ParentId="27482" CreationDate="2011-03-17T00:32:30.733" Score="1" ViewCount="" Body="&lt;p&gt;By Bolzano-Weierstrass, any bounded sequence $(x_k)$ in $\mathbb R^n$ has a convergent subsequence. Now try to construct a sequence sequence that does not converge, assuming a set is closed and bounded, yet there exists an open covering that has no finite subcover.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Spoiler:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $A \subset \mathbb R^n$ be closed and bounded, s.t. there exists an open covering which has no finite subcovering. So let $U_i$ be such an open covering. Pick an open set, which we denote by $U_0$. Let &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x_0 \in A - U_0$ , which exists by assumption.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Choose $U_1 \in (U_i)$ with $x_0 \in U_1$. This exists by assumption.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we proceed in that manner and choose a sequence of open sets from the covering and a sequence in $A$ that is always one step before the covering sequence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then the constructed sequence $(x_k)$ is bounded, as it is contained in $A$. It has a convergent subsequence, say, to $x$. Choose such a sequence and keep the label $x_k$. As $A$ is closed, $x \in A$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By assumption, we can choose some $V$ which covers $x$. We do so at any point of the sequence, say after some fixed number of steps. We strip off all elements of the sequence which are in $V$, and construct the two sequences again in the same fashion, with $V$ being included in the sequence of open sets that we used to define the sequence $x_k$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We repeat this. Thus we construct a bounded sequence that does not converge - because otherwise, it would be covered by an open set in the sequence we have choosen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, if $A$ is closed and bounded, every open covering has a finite subcover. $qed$&lt;/p&gt;&#xA;" OwnerUserId="3557" LastEditorUserId="3557" LastEditorDisplayName="" LastEditDate="2011-03-17T00:37:50.433" LastActivityDate="2011-03-17T00:37:50.433" CommentCount="1" />
  <row Id="27485" PostTypeId="2" ParentId="15201" CreationDate="2011-03-17T00:34:10.003" Score="2" ViewCount="" Body="&lt;p&gt;Something that often gets ignored is Schaum's Outlines: Combinatorics by V. K. Balakrishnan. Lots of small examples that are manageable by a beginner, shows you 'how to do it' in a straightforward manner.&lt;/p&gt;&#xA;" OwnerUserId="1919" LastActivityDate="2011-03-17T00:34:10.003" />
  <row Id="27486" PostTypeId="2" ParentId="27482" CreationDate="2011-03-17T00:37:13.163" Score="1" ViewCount="" Body="&lt;p&gt;The best proof I've seen is a topological one, using a few theorems about different topologies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Namely:&#xA;1) In an order topology, $\left[ a,b \right]$ is a compact set.&#xA;2) The order topology on $\mathbb{R}$ is equivalent to the metric induced topology on $\mathbb{R}$.&#xA;3) Closed subsets of compact sets are compact.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Compact subsets of Hausdorff spaces are closed, and in the metric topology we can take $B_{n}(0)$, the ball of radius $n$ around 0, and since $\cup_{n=1}^{\infty} B_{n}(0) = \mathbb{R^{m}}$, we can choose a finite subcover, and so it is a bounded set.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That proves necessity. To prove sufficiency we simply note the following.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $K$ be a closed, bounded subset of $\mathbb{R}^n$. Then $K \subseteq \left[ -M, M\right]^n$ is actually compact, as $\left[ -M, M\right]$ is compact by 1), and since the order and metric topologies are equivalent, it is compact in the metric topology. By Tychonoff's Theorem(not even necessary, finite products of compact sets are compact), we have $\left[ -M, M\right]^n$ is also compact. Closed subsets of compact sets are compact, and $K$ is a closed subset of $\left[-M,M\right]$, so $K$ is compact, proving sufficiency.&lt;/p&gt;&#xA;" OwnerUserId="8356" LastActivityDate="2011-03-17T00:37:13.163" />
  <row Id="27487" PostTypeId="2" ParentId="27452" CreationDate="2011-03-17T00:38:34.313" Score="1" ViewCount="" Body="&lt;p&gt;There may be literature on this, and I'm no expert, but I believe there is a simple method to directly compute what you want:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, FFT is just a fast algorithm for the DFT. For the following method, you will need to compute the DFT coefficients directly. Recall the formula for a single coefficient (slightly modified from Wikipedia to make it match what I say below):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$X_k := \frac{1}{N} \sum_{n=0}^{N-1} x_n e^{-\pi i k \frac{2n+1}{N}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can interpret this as a discretization of the integral (hope I'm getting this right):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$Y_k := \int_0^1 y(t) e^{-2 \pi i k t} dt$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While $x$ is an $N$-tuple, $y$ is supposed to be the continuous function on $[0,1]$ that $x$ approximates. Each $n=0,\dots,N-1$ is associated with the interval $[a_n,b_n]:=[\frac{n}{N},\frac{n+1}{N}]$, so the union of all intervals is $[0,1]$. It also makes sense to assume there is a specific $t_n\in[0,1]$ such that $x_n = y(t_n)$, say $t_n := \frac{a_n + b_n}{2}$. (This is what I implicitly used in the first formula, making it a midpoint sum approximation. Please ask if this is not clear.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the input you have is basically a different discretization $z$ of $y$, which is an $M$-tuple ($M$ possibly being different from $N$), and where $a_m$, $b_m$, and $t_m$ for $m=0,\dots,M-1$ are given by your variable timestep. (You have to decide for yourself whether the interval boundaries or the measurement points are given; however, you have to make sure that $a_0=0$ and $b_{M-1}=1$.) What you need is a sum that appropriately discretizes the above integral, which would (hopefully, again) be:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$Z_k := \sum_{m=0}^{M-1} (b_m-a_m) z_m e^{-2 \pi i k t_m}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the timestep is completely variable, you can get nonzero coefficients for arbitrarily large $k$, but you can stop calculating wherever you want, of course.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, computing these sums is not nearly as fast as an FFT, but there are several things you can do to make it computationally feasible. The most important thing is to calculate the sines and cosines just once for each $m$, i.e. iterate over $m$ at the outer level, not $k$. You should pre-calculate sine/cosine tables if you have a fixed &quot;time resolution&quot; that every timestep is a multiple of; otherwise use sine/cosine implementations that favor speed over accuracy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want overlap between consecutive DFT &quot;windows&quot; (not sure if this is the correct word here), there is a way to reuse the part of each sum that is in both windows. Another way to reduce the time overhead is to use different window lengths for each coefficient, which gives you a variable (e.g. logarithmical) frequency resolution. I don't have time to explain this now, but you can mail me for details at SebastianR@gmx.de if you really want to implement this.&lt;/p&gt;&#xA;" OwnerUserId="1386" LastEditorUserId="1386" LastEditorDisplayName="" LastEditDate="2011-03-17T00:43:56.197" LastActivityDate="2011-03-17T00:43:56.197" />
  <row Id="27488" PostTypeId="2" ParentId="27455" CreationDate="2011-03-17T00:40:03.520" Score="0" ViewCount="" Body="&lt;p&gt;$X$ can be taken to be $\left\{ f \in L^{\infty} | \forall n,\  \int_{[0,1]} f f_n =0 \right\}$&lt;/p&gt;&#xA;" OwnerUserId="2660" LastActivityDate="2011-03-17T00:40:03.520" CommentCount="4" />
  <row Id="27489" PostTypeId="1" CreationDate="2011-03-17T00:41:25.533" Score="2" ViewCount="57" Body="&lt;p&gt;I usually work in number theory so I am not familiar with fourier transforms, I have read up on them and know the basics but it never seems to be in number theory language.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am trying to find the transform of a primitive dirichlet character $\chi(n)$ mod q. I know this is a periodic function and $\chi(n)=exp(\frac{Kv(n)}{\phi(p^\alpha)})$ but I have no idea have to find its transform or the transform of $f(n)\chi(n)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;yes you are right, say how would you calculate $\sum_(n\epsilon Z) f(n)\chi(n)$&lt;/p&gt;&#xA;" OwnerUserId="8355" LastEditorUserId="8355" LastEditorDisplayName="" LastEditDate="2011-03-17T01:23:06.263" LastActivityDate="2011-03-17T01:23:06.263" Title="The discrete fourier transform of a dirichlet charachter" Tags="&lt;number-theory&gt;&lt;fourier-analysis&gt;" CommentCount="2" FavoriteCount="3" />
  <row Id="27490" PostTypeId="1" AcceptedAnswerId="27519" CreationDate="2011-03-17T00:45:24.607" Score="4" ViewCount="177" Body="&lt;p&gt;How to prove the following theorem? &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Let $I \in R$ be an open interval, let $c \in I$, and let $f: I-{c} \rightarrow R$ be a function. Then $\lim \limits_{x \rightarrow c} {f(x)}$ exists if for each $\epsilon &amp;gt; 0$, there is some $\delta &amp;gt; 0$ such that $x,y \in I$ and $\vert x-c \vert &amp;lt; \delta$ and $\vert y-c \vert &amp;lt; \delta$ implies $\vert f(x)-f(y) \vert &amp;lt; \epsilon$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I think this needs standard limit definition, sup and inf properties to prove. And I came up with a following scratch of proof:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) Suppose for each $\epsilon &amp;gt;0$, there is some $\delta &amp;gt;0$ such that $\vert x-c \vert &amp;lt; \delta$ and $\vert y-c \vert &amp;lt; \delta$ implies $\vert f(x)-f(y) \vert &amp;lt; \epsilon$. For each $r&amp;gt;0$, let $A_r$ = $I \bigcap (c-r,c+r)$. Then, for each $\epsilon&amp;gt;0$, there is some $\delta&amp;gt;0$, such that $x,y \in A_\delta$ implies $\vert f(x)-f(y) \vert &amp;lt; \epsilon$. Then I want to show there is some $a&amp;gt;0$ such that $f(A_a)$ is bounded.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) If $f(A_a)$ is bounded, then for each $s \in (0,a)$, $f(A_s) \subseteq f(A_a)$, and thus $f(A_s)$ is bounded. Define $a_s = \mathrm{glb}~f(A_s)$ and $b_s = \mathrm{lub}~f(A_s)$. Let $A = \{a_s \mid s \in (0,a)\}$ and $B=\{b_s \mid s \in (0,a)\}$. Then we know that A has a least upper bound and B has a greatest lower bound, and $\mathrm{lub}(A) \leq \mathrm{glb}(B)$. Now I want to show $\mathrm{lub}(A) = \mathrm{glb}(B)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(3) If I could show $\mathrm{lub}(A) = \mathrm{glb}(B)$, let $M=\mathrm{lub}(A) = \mathrm{glb}(B)$, I want to show $\lim \limits_{x \rightarrow c} {f(x)} = M$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can someone give me some help on how to prove (1) (2) (3), $f(A_a)$ is bounded, $\mathrm{lub}(A) = \mathrm{glb}(B)$, and $\lim \limits_{x \rightarrow c} {f(x)} = M$? Thanks!&lt;/p&gt;&#xA;" OwnerUserId="7429" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-17T04:14:29.410" LastActivityDate="2011-03-17T04:14:29.410" Title="proof that an alternative definition of limit is equivalent to the usual one." Tags="&lt;real-analysis&gt;" AnswerCount="1" CommentCount="6" FavoriteCount="1" />
  <row Id="27491" PostTypeId="1" AcceptedAnswerId="27515" CreationDate="2011-03-17T01:05:05.880" Score="2" ViewCount="62" Body="&lt;p&gt;If N is characteristic in a group G and if we have that K/N is characteristic in G/N why do we have that K is characteristic in G. &lt;/p&gt;&#xA;" OwnerUserId="1127" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-17T03:49:27.137" LastActivityDate="2011-03-17T03:49:27.137" Title="Question about characteristic subgroups" Tags="&lt;group-theory&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="27492" PostTypeId="2" ParentId="27480" CreationDate="2011-03-17T01:11:51.847" Score="0" ViewCount="" Body="&lt;p&gt;Doug West, Introduction to Graph Theory. Rigorous but readable, proof based rather than simply descriptive, but the proofs are explanatory rather than simply justification of truth (by any arbitrary means). Even so may not be the best total beginner's book.&lt;/p&gt;&#xA;" OwnerUserId="1919" LastActivityDate="2011-03-17T01:11:51.847" />
  <row Id="27493" PostTypeId="2" ParentId="27491" CreationDate="2011-03-17T01:14:42.113" Score="1" ViewCount="" Body="&lt;p&gt;This is false. $N/N$ is always characteristic in $G/N$, so if this were true, then normal subgroups would always be characteristic.&lt;/p&gt;&#xA;" OwnerUserId="5203" LastActivityDate="2011-03-17T01:14:42.113" CommentCount="1" />
  <row Id="27495" PostTypeId="2" ParentId="23765" CreationDate="2011-03-17T01:23:48.507" Score="0" ViewCount="" Body="&lt;p&gt;ATM is mapping reducable to ETM's compliment. Proving undecidablitiy can be done with either the original language, or it's compliment. This is not the case, for proving unrecognivability, however.&lt;/p&gt;&#xA;" OwnerUserId="8358" LastActivityDate="2011-03-17T01:23:48.507" />
  <row Id="27496" PostTypeId="1" CreationDate="2011-03-17T01:24:07.150" Score="7" ViewCount="168" Body="&lt;p&gt;This might be a stupid question but what is the integral of $\frac{1}{1 + 9x^2}$? I want to think it's $\tan^{-1}(3x) + c$ but the book I'm working in says $\frac{1}{3}\tan^{-1}(3x) + c$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How did they get $1/3$? &lt;/p&gt;&#xA;" OwnerUserId="8359" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2011-03-17T04:00:15.490" LastActivityDate="2011-04-17T07:35:35.037" Title="Question about the integral of $1/(1+9x^2)$" Tags="&lt;calculus&gt;&lt;integral&gt;" AnswerCount="3" CommentCount="4" />
  <row Id="27497" PostTypeId="2" ParentId="27420" CreationDate="2011-03-17T01:25:31.953" Score="1" ViewCount="" Body="&lt;p&gt;Hmm I think the problem is this: $T_r\beta$ is orthogonal to $r$ for all $\beta$, and $T_q \alpha$ is orthogonal to $q$ for all $\alpha$, but that doesn't mean that $T_q\alpha$ is orthogonal to $T_r \beta$. For example think of $\mathbb{R}^3$ with the usual dot product. Then &lt;strong&gt;k&lt;/strong&gt; is orthogonal to &lt;strong&gt;j&lt;/strong&gt; and it is also orthogonal to &lt;strong&gt;i&lt;/strong&gt; but it is not orthogonal to itself!&lt;/p&gt;&#xA;" OwnerUserId="668" LastActivityDate="2011-03-17T01:25:31.953" CommentCount="1" />
  <row Id="27498" PostTypeId="2" ParentId="27411" CreationDate="2011-03-17T01:25:48.973" Score="1" ViewCount="" Body="&lt;p&gt;Stockmeyer and Meyer, &quot;Word problems requiring exponential time&quot;, Proc. 5th ACM Symposium on the Theory of Computing, pp 1-9, 1973.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The derivation is fairly intuitive: every negation in the regular expression 'somehow' requires conversion to a DFA, and a lower bound of $\Theta(2^n)$ states guaranteed for conversion. Nest your $\neg$'s and you get a stack of exponentials.&lt;/p&gt;&#xA;" OwnerUserId="1919" LastActivityDate="2011-03-17T01:25:48.973" />
  <row Id="27499" PostTypeId="2" ParentId="27496" CreationDate="2011-03-17T01:27:50.267" Score="10" ViewCount="" Body="&lt;p&gt;Set $u=3x$. Then the integrand is $1/(1+u^2)$ which integrates to $\tan^{-1} u$ but $du=3dx$. This gives you the $1/3$.&lt;/p&gt;&#xA;" OwnerUserId="589" LastActivityDate="2011-03-17T01:27:50.267" />
  <row Id="27501" PostTypeId="2" ParentId="27446" CreationDate="2011-03-17T01:53:38.633" Score="0" ViewCount="" Body="&lt;p&gt;When talking about linear operators on Banach spaces, the problem of eigenvalues is always examined in the presence of certain boundary conditions. So, let me restate the above problem as follows: We are looking for the eigenvalues of the differential operator $T:A \rightarrow C^{\infty}(\Re)$ where $A \subset C^{\infty}(\Re)$ with $f \in A \Rightarrow f(0) = 1$. Then we need to find $\lambda \in \Re$ such that :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align*}&#xA;T(f) &amp;amp;= \lambda f \\ &#xA;f(0) &amp;amp;= 1&#xA;\end{align*}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $\lambda = 0$ we have the case where the function $f(t)=0 \forall t \in \Re$ is the corresponding eigenfunction (trivial case). If $\lambda \neq 0$, then $f(t)=e^{\lambda t}$ with $t \in \Re$. So my main remarks here are:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Eigenvectors in this case are substituted by eigenfunctions&lt;/li&gt;&#xA;&lt;li&gt;Differential operators have lots of eigenvalues (usually infinite) followed by certain eigenfunctions. One cannot refer to the eigenvalues of a differential operator (or in general, of an operator on an infinite-dimensional space) regardless of its corresponding eigenfunctions.&lt;/li&gt;&#xA;&lt;li&gt;Boundary conditions should be stated in the problem formulation. &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update 1:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regarding the boundary conditions involved in the eigenvalue problem of operators on infinite-dimensional spaces let me give a few references from the literature:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Albrecht Böttcher and Harold Widom, &quot;On the eigenvalues of certain canonical higher-order ordinary differential operators&quot;, J. Math. Anal. Appl. 322 (2006) 990–1000 : The authors study an eigenvalue problem with respect to certain boundary conditions. Of course the problem has infinitely many eigenvalues.&lt;/li&gt;&#xA;&lt;li&gt;Qiaoling Wang and Changyu Xia, &quot;Universal bounds for eigenvalues of the biharmonic operator&quot;, J. Math. Anal. Appl. 364 (2010) 1–17 : Here the authors study the eigenvalue problem for the well known operators $\Delta$ and $\Delta^2$.&lt;/li&gt;&#xA;&lt;li&gt;Strum-Liouville operators, which are of high importance (e.g. for the solution of PDEs) are studied by Aceto et al. in L. Aceto, P. Ghelardoni and M. Marletta, &quot;Numerical computation of eigenvalues in spectral gaps of Sturm–Liouville operators&quot;, J. Comput. Appl. Math. 189 (2006) 453 – 470. Again with respect to certain boundary conditions.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="8357" LastEditorUserId="8357" LastEditorDisplayName="" LastEditDate="2011-03-18T14:17:34.803" LastActivityDate="2011-03-18T14:17:34.803" CommentCount="2" />
  <row Id="27502" PostTypeId="1" CreationDate="2011-03-17T01:55:37.610" Score="2" ViewCount="183" Body="&lt;p&gt;Since $X$ is a compact metric space, it is sequentially compact so every sequence in $X$ has a subsequence which converges. Now, I want to show that $X^\mathbb{N}$ is sequentially compact, but I'm getting confused because we must show that the sequence of a sequence has a convergent subsequence. I know that we must use the fact that $X$ is sequentially compact. Thanks!&lt;/p&gt;&#xA;" OwnerUserId="8361" LastEditorUserId="6622" LastEditorDisplayName="" LastEditDate="2011-03-17T10:01:40.440" LastActivityDate="2011-03-17T17:23:43.493" Title="If $X$ is a compact metric space, then the set of all sequences in $X$, $X^\mathbb{N}$ is sequentially compact." Tags="&lt;general-topology&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="1" />
  <row Id="27503" PostTypeId="1" AcceptedAnswerId="27510" CreationDate="2011-03-17T01:57:59.323" Score="1" ViewCount="73" Body="&lt;p&gt;You're computing an improper integral (with limits on it) to see if something converges, and you're at the part where you have something like: &#xA;$$\lim_{b\to\infty}\left(\frac{\tan^{-1}(b)^2}{2} - \frac{\tan^{-1}(1)^2}{2}\right).$$ &#xA;Wouldn't it diverge because you have infinity in the equation? &lt;/p&gt;&#xA;" OwnerUserId="8278" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-17T03:52:35.997" LastActivityDate="2011-03-17T03:52:35.997" Title="Question about using an integral to see if something converges." Tags="&lt;calculus&gt;" AnswerCount="3" />
  <row Id="27504" PostTypeId="2" ParentId="27478" CreationDate="2011-03-17T02:04:23.480" Score="0" ViewCount="" Body="&lt;p&gt;About the second question: here are some nice ways to plot complex functions: &lt;a href=&quot;http://www.maa.org/pubs/amm_complements/complex.html&quot; rel=&quot;nofollow&quot;&gt;http://www.maa.org/pubs/amm_complements/complex.html&lt;/a&gt; , &#xA;&lt;a href=&quot;http://www.mai.liu.se/~halun/complex/&quot; rel=&quot;nofollow&quot;&gt;http://www.mai.liu.se/~halun/complex/&lt;/a&gt; , &#xA;&lt;a href=&quot;http://www1.math.american.edu/People/lcrone/ComplexPlot.html&quot; rel=&quot;nofollow&quot;&gt;http://www1.math.american.edu/People/lcrone/ComplexPlot.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or &lt;a href=&quot;http://www.wolframalpha.com/input/?i=+z%5E2+%2B+Conj%5Bz%5D&quot; rel=&quot;nofollow&quot;&gt;http://www.wolframalpha.com/input/?i=+z%5E2+%2B+Conj%5Bz%5D&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="589" LastActivityDate="2011-03-17T02:04:23.480" CommentCount="1" />
  <row Id="27505" PostTypeId="2" ParentId="27457" CreationDate="2011-03-17T02:18:34.010" Score="1" ViewCount="" Body="&lt;p&gt;One way, as others have said, is to just follow the book.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But what the book probably says is: you'll have at the end for your answer seven 7x7 matrices, each entry in a matrix $A_{i, j}$ will be the pair (shortest path from $i$ to $j$, max labelled vertex on that path), the max or shortest known so far.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first matrix will be about the paths that allow a path through $v_1$, the second paths through any of $v_1, v_2$, ..., the seventh through any of them. At the $k$th matrix you update all entries involving vertices up to $v_k$ using the formula most likely given in the book: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$a_{ij} = a_{ij} + a_{ik}\cdot a_{kk}^{*} \cdot a_{kj}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where '$+$', '$*$', and even multiplication ('$\cdot$') have their special meanings for shortest paths and other special meanings for max vertex label, and the LHS is the new value (value in matrix $n+1$) and the RHS uses old values (values from matrix $n$. In practice you can do all this in-place in just one $n^2$ matrix).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The 0th matrix, the base case, where the paths don't go through any other vertex, is just for the edges. For shortest path, it is either the values of the edge weight, or $\infty$ if no edge because any finite value is less then/shorter than $\infty$). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The next exercise in the book will be to use the same method to convert a finite automaton into a regular expression. (Really!) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Extra commentary&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;'+' is usually taken to mean 'or', min (to get the shortest or least), or max (to get the largest vertex label&lt;/li&gt;&#xA;&lt;li&gt;'multiplication' (or '$\cdot$' is taken to mean 'plus' (to get the sum of weights along a path) or max again (yes, 'multiplication' is the same as 'addition' here)&lt;/li&gt;&#xA;&lt;li&gt;'*' is taken to mean 'for any number of cycles'. Since for both interpretations (shortest path, or max vertex label) a cycle isn't involved in any shortest path/max vertex, it can be ignored here.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The informative thing about this multiple interpretation, is that it is the first practical instance where abstract algebra (overloading of the interpretation of the +, $\cdot$, and *, but provable facts knowing only a few properties of those interpreted operations) in the CS curriculum. &lt;/p&gt;&#xA;" OwnerUserId="1919" LastEditorUserId="1919" LastEditorDisplayName="" LastEditDate="2011-03-17T23:52:52.083" LastActivityDate="2011-03-17T23:52:52.083" CommentCount="1" />
  <row Id="27506" PostTypeId="1" AcceptedAnswerId="27628" CreationDate="2011-03-17T02:22:16.487" Score="8" ViewCount="98" Body="&lt;p&gt;I'd like a closed form expression x(n) for the number of orbits of the symmetric group on 3 points acting on the triples in { (a,b,c) : a,b,c in &lt;strong&gt;Z&lt;/strong&gt;, 1 ≤ a,b,c ≤ n, c = 2n−a−b }.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I feel like this should be a really basic problem, but my standard method of attack fails: look it up in OEIS and prove the known formula.  My backup plan of &quot;think about it&quot; has failed: I don't know how to deal with the restriction on c. &lt;sup&gt;(Without the restriction, I happen to have learned this is the same thing as counting 1 ≤ a ≤ b ≤ c ≤ n, and I happen to have learned this is Binomial(n+2,3) because you need to place two bars between n stars, but I have no general context for this.)&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suspect this is a pretty standard counting problem even with the restriction, but I never really learned how to count (fish, fish, fish, …, fish, fish, …, fish).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The counts x(n) for n=1 to 30 are: 0, 1, 2, 3, 4, 6, 7, 9, 11, 13, 15, 18, 20, 23, 26, 29, 32, 36, 39, 43, 47, 51, 55, 60, 64, 69, 74, 79, 84, 90.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think there are (n-1)⋅(n+4)/2 triples, but even that is a little fuzzy (increase increases by 1 each time).  I have no idea how many of them have 2 equal components.&lt;/p&gt;&#xA;" OwnerUserId="583" LastActivityDate="2011-03-17T16:43:37.493" Title="Counting orbits under Sym(3)" Tags="&lt;combinatorics&gt;&lt;counting&gt;" AnswerCount="2" />
  <row Id="27507" PostTypeId="1" AcceptedAnswerId="27538" CreationDate="2011-03-17T02:22:58.860" Score="1" ViewCount="112" Body="&lt;p&gt;I am solving a problem in &lt;a href=&quot;http://books.google.com/books?id=cyqTPotl7QcC&amp;amp;pg=PA84&amp;amp;dq=one+to+one+borel+markov+chain+shao+jun&amp;amp;hl=en&amp;amp;ei=nGqBTf_tD8650QGbnIHxCA&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=1&amp;amp;ved=0CCgQ6AEwAA#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;Mathematical Statistics&lt;/a&gt; by Jun Shao&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Let $\{X_n \}$ be a Markov chain. Show&#xA;  that if $g$ is a one-to-one Borel&#xA;  function, then $\{g(X_n )\}$ is also a&#xA;  Markov chain. Give an example to show&#xA;  that $\{g(X_n )\}$ may not be a Markov&#xA;  chain in general.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have a hard time on solving it, even though I have been staring it and thinking about it for a whole day. &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;For the first part, which is to show that any&#xA;one-to-one Borel $g$ preserves&#xA;Markov property of a Markov chain, I&#xA;guess using the formula for density function under the change&#xA;of random variables by $g$, learned in elementary probability course, might&#xA;help, but I am not sure how to use it, or maybe the tools needed to solve the problem are not that simple?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;For the second part, I really have no idea of how to construct some $\{ X_n\}$ so that $\{g(X_n )\}$ is not a Markov chain, for example, when $g(x)=x^2$?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Here are also some extended thoughts and questions:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;If $g$ is not one-to-one, is&#xA;$\{g(X_n )\}$ always not a Markov chain for any Markov chain $\{ X_n\}$?&lt;/li&gt;&#xA;&lt;li&gt;How about if $\{X_t \}, t \in&#xD;&#xA;    \mathbb{R}$? Does any one-to-one $g$&#xA;also preserve Markov property of&#xA;continuous-time stochastic&#xA;processes?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Thanks a lot!&lt;/p&gt;&#xA;" OwnerUserId="3616" LastActivityDate="2011-03-17T07:27:10.290" Title="Transformation on state-space that preserves Markov property" Tags="&lt;stochastic-processes&gt;" AnswerCount="2" FavoriteCount="2" />
  <row Id="27508" PostTypeId="2" ParentId="27503" CreationDate="2011-03-17T02:23:24.287" Score="0" ViewCount="" Body="&lt;p&gt;You should show us the sequence you are working with. $tan^{-1}(\infty)$ is not well-defined because $\infty$ is not a real number; however, the limit of $tan^{-1}(x)$ as $x$ goes to $\infty$ is $\pi/2$.&lt;/p&gt;&#xA;" OwnerUserId="8173" LastActivityDate="2011-03-17T02:23:24.287" />
  <row Id="27509" PostTypeId="2" ParentId="27503" CreationDate="2011-03-17T02:25:02.130" Score="1" ViewCount="" Body="&lt;p&gt;Not in this case, $\lim_{b\to\infty}\tan^{-1}(b)=\frac{\pi}{2}$. To see this, it helps to think of $\tan(x)$ as the slope of a ray from the origin rotating $x$ radians counterclockwise around the unit circle. So after travelling $\frac{\pi}{2}$ radians, the line from the origin is pointing straight up in the positive $y$ direction. The slope of this line is undefined, or $\infty$. So it makes sense that the limit above evaluates to what it does.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in this particular case, if I'm interpreting your powers correctly, your expression evaluates to &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{(\pi/2)^2}{2}-\frac{\tan^{-1}(1)^2}{2}=\frac{\pi^2}{8}-\frac{\pi^2}{32}=\frac{3\pi^2}{32}$$&#xA;since $\tan^{-1}(1)=\frac{\pi}{4}$.&lt;/p&gt;&#xA;" OwnerUserId="1583" LastActivityDate="2011-03-17T02:25:02.130" />
  <row Id="27510" PostTypeId="2" ParentId="27503" CreationDate="2011-03-17T02:26:50.570" Score="4" ViewCount="" Body="&lt;p&gt;Basically, i understand your question as &quot;What is the limit $\lim_{b \to \infty} \arctan(b)$?&quot; ($\arctan$ is just another notation for $\tan^{-1}$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, $$\lim_{b \to \infty} \arctan(b) = \frac{\pi}{2}$$ How do know this? What does this mean?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To get a start, look at &lt;a href=&quot;http://File%3aArctangent%20Arccotangent.svg&quot; rel=&quot;nofollow&quot;&gt;this plot&lt;/a&gt; of $\arctan(x)$. Doesn't it look like it tends to $\frac{\pi}{2}$ as $x$ gets larger? The mathematical reason for the limit being what it is, is that we can show, as $x$ gets larger and larger, $\arctan(x)$ gets closer and closer to $\frac{\pi}{2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another way to view this, in the case of $\arctan$ is considering what $\arctan$ actually is: the inverse of $\tan$ in the open intervall $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$. What does $\tan(x)$ tend to as $x$ gets closer to $\frac{\pi}{2}$? That's right, infinity. (What is $\lim_{x \to -\infty} \arctan(x)$?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a simpler example of a function that does not tend to infinity as it's argument gets larger and larger, you can consider a constant function $f(x) = b$. What is $\lim_{x \to \infty} f(x)$? It is, of course, $b$, since $f(x)$ doesn't actually depend on $x$. &lt;/p&gt;&#xA;" OwnerUserId="5966" LastActivityDate="2011-03-17T02:26:50.570" CommentCount="2" />
  <row Id="27511" PostTypeId="2" ParentId="9758" CreationDate="2011-03-17T02:28:31.447" Score="0" ViewCount="" Body="&lt;p&gt;Intuitively, it's a matrix that's &quot;like&quot; a number $\geq 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Relatedly you can have a positive semidefinite function that's also &quot;like&quot; a number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both matrices and functions in general have many (, many, many) more degrees of freedom than members of $\mathbb R$. But these classes of matrices and functions &quot;boil down to&quot; something much simpler.&lt;/p&gt;&#xA;" OwnerUserId="1457" LastEditorUserId="1457" LastEditorDisplayName="" LastEditDate="2011-03-17T02:34:32.563" LastActivityDate="2011-03-17T02:34:32.563" />
  <row Id="27512" PostTypeId="2" ParentId="27282" CreationDate="2011-03-17T02:32:34.297" Score="1" ViewCount="" Body="&lt;p&gt;If I understand right:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A function space is a vector space $W$ of real-valued functions on some set $X$ with scalars in $\mathbb{R}$(since you're taking min, max), with the additional properties that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\min(f,g) \in W$ and also $\max(f,g) \in W$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We must simply need to show that $\max(\sum_{i=1}^{L1} r_i \chi_{I_i}, \sum_{j=1}^{L2} \rho_j \chi_{J_j}) = \sum_{k=1}^{L3} p_k \chi_{K_k}$. Well, we can even assume that the $I_i$'s are disjoint(if not, we can split up the sum further into a finite sum of characteristic functions on disjoint sets), so assume they are disjoint. then let $L3=L1 \cdot L2$, and define for $ 1 \leq m \leq L3$( with $m = qj + t$ for $0 \leq q &amp;lt; L1$ and $1 \leq t \leq L2$) the interval $K_m=I_q \cap I_t$ and define $p_k = \max(r_i,\rho_j)$, similarly to define $\min(\sum_{i=1}^{L1} r_i \chi_{I_i}, \sum_{j=1}^{L2} \rho_j \chi_{J_j}) = \sum_{k=1}^{L3} p_k \chi_{K_k}$ we would simply define $p_k = \min(r_i,\rho_j)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This should be a more or less trivial verification away, pick a random point and verify this construction gives the correct value of $\max(f,g)$ and $\min(f,g)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What do you mean by algebra, an algebra over a field(vector space with bilinear product)? If so yes, just take pointwise multiplication.&lt;/p&gt;&#xA;" OwnerUserId="8356" LastActivityDate="2011-03-17T02:32:34.297" />
  <row Id="27513" PostTypeId="2" ParentId="27480" CreationDate="2011-03-17T02:41:12.897" Score="3" ViewCount="" Body="&lt;p&gt;Diestel's &lt;a href=&quot;http://diestel-graph-theory.com/&quot; rel=&quot;nofollow&quot;&gt;Graph Theory&lt;/a&gt; (which has a &quot;free preview&quot; online) is presented as a graduate textbook, but it does not really have any prerequisites.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It goes quite deep in some parts, and includes material (such as the chapter on the graph minor theorem) that you won't find in other textbooks. Some proofs have a sort of &quot;how would someone ever think of that?&quot; feel, but this may have been remedied in the fourth edition (I have the third).&lt;/p&gt;&#xA;" OwnerUserId="2954" LastEditorUserId="2954" LastEditorDisplayName="" LastEditDate="2011-04-05T22:32:41.280" LastActivityDate="2011-04-05T22:32:41.280" CommentCount="2" />
  <row Id="27514" PostTypeId="1" AcceptedAnswerId="27516" CreationDate="2011-03-17T02:54:40.393" Score="4" ViewCount="74" Body="&lt;p&gt;I am currently learning about Markov chains and Markov processes, as part of my study on stochastic processes. I feel there are so many properties about Markov chain, but the book that I have  makes me miss the big picture, and I might better look at some other references.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I would like to know if there are some references with nice treatment on Markov chains/processes? They can be a whole book, some chapters of a book, some lecture notes, some webpages, .... Their approaches can be different, either with or without referring to measure theory (I think a measure-theory approach will be helpful in clarifying things, but I hope it also has some intuitive interpretation),  as long as they have clear organization, and provide both a big picture and cover most of the important topics of Markov chains/processes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I really appreciate your contribution!&lt;/p&gt;&#xA;" OwnerUserId="3616" LastEditorUserId="3616" LastEditorDisplayName="" LastEditDate="2011-03-17T03:14:01.687" LastActivityDate="2011-03-17T03:35:28.227" Title="Nice references on Markov chains/processes?" Tags="&lt;reference-request&gt;&lt;stochastic-processes&gt;" AnswerCount="1" FavoriteCount="3" />
  <row Id="27515" PostTypeId="2" ParentId="27491" CreationDate="2011-03-17T02:58:06.813" Score="4" ViewCount="" Body="&lt;p&gt;Let $\varphi\colon G\to G$ be a homomorphism of $G$. Consider $K$ and $\varphi(K)$; since $N\subseteq K$, then $N=\varphi(N)\subseteq\varphi(K)$, so then by the lattice isomorphism theorem, $K$ and $\varphi(K)$ correspond to subgroups of $G/N$, and $K=\varphi(K)$ if and only if $K/N = \varphi(K)/N$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But since $\varphi(N)=N$, then $\varphi$ induces an automorphism $\overline{\varphi}$ on $G/N$ by $\overline{\varphi}(gN) = \varphi(g)N$. This is well defined, since $xN=yN$ if and only if $y^{-1}x\in N$, if and only if $\varphi(y)^{-1}\varphi(x)\in\varphi(N)=N$, so $\varphi(x)N = \varphi(y)N$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, $\varphi(K)/N = \overline{\varphi}(K/N) = K/N$ (since $K/N$ is characteristic in $G/N$), so $K=\varphi(K)$. Thus, $K$ is characteristic in $G$.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-17T03:46:08.687" LastActivityDate="2011-03-17T03:46:08.687" />
  <row Id="27516" PostTypeId="2" ParentId="27514" CreationDate="2011-03-17T03:16:55.263" Score="3" ViewCount="" Body="&lt;p&gt;Two excellent introductions are James Norris's &lt;a href=&quot;http://www.statslab.cam.ac.uk/~james/Markov/&quot; rel=&quot;nofollow&quot;&gt;&quot;Markov Chains&quot;&lt;/a&gt; and Pierre Bremaud's &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0387985093&quot; rel=&quot;nofollow&quot;&gt;&quot;Markov Chains: Gibbs fields, Monte Carlo simulation, and queues&quot;&lt;/a&gt;. Both books assume a motivated student who is somewhat mathematically mature, though Bremaud reviews basic probability before he gets going. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also the wonderful book &quot;Markov Chains and Mixing Times&quot; by Levin, Peres, and Wilmer is available &lt;a href=&quot;http://pages.uoregon.edu/dlevin/MARKOV/&quot; rel=&quot;nofollow&quot;&gt;online here&lt;/a&gt;.  It starts right with the definition of Markov Chains, but eventually touches on topics in current research. So it is pretty advanced, but also well worth a look.&lt;/p&gt;&#xA;" OwnerUserId="940" LastEditorUserId="940" LastEditorDisplayName="" LastEditDate="2011-03-17T03:35:28.227" LastActivityDate="2011-03-17T03:35:28.227" CommunityOwnedDate="2011-03-17T03:16:55.263" CommentCount="4" />
  <row Id="27517" PostTypeId="1" AcceptedAnswerId="27563" CreationDate="2011-03-17T03:34:54.427" Score="1" ViewCount="141" Body="&lt;p&gt;Hello.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm having trouble showing that the $L^p$ norm of the $n$-th order Dirichlet kernel is proportional to $n^{1-1/p}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've tried brute force integration and it didn't work out. I would be grateful for any hints.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="6556" LastEditorUserId="6296" LastEditorDisplayName="" LastEditDate="2011-03-17T19:16:49.060" LastActivityDate="2011-03-17T19:16:49.060" Title="$L^p$ norm of Dirichlet Kernel" Tags="&lt;fourier-analysis&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" />
  <row Id="27518" PostTypeId="1" AcceptedAnswerId="29062" CreationDate="2011-03-17T03:43:46.467" Score="0" ViewCount="114" Body="&lt;p&gt;I'm trying to come up with a 95% confidence interval for the click-through-rate of particular advertisement.  It has $x$ clicks out of $n$ impressions so far.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What's the best way to compute this, given that I expect the click-through-rate to be small? I've been told that the &quot;usual&quot; methods of computing a confidence interval don't do well when the true probability $p$ is near 0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For advertisements, the true click probability is typically in the (0, 0.02) range.  I don't have an exact formula for the prior, but any reasonable approximation centered in the (0, 0.02) range would do.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a nice formula of something like&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;(lower, upper) = confidence_interval($x$, $n$, $prior_p$, 0.95)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;out there?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or alternately, has anyone out there used one of the &quot;usual&quot; confidence interval formulas in this situation, and can confirm that it produces &quot;close enough&quot; results?&lt;/p&gt;&#xA;" OwnerUserId="8363" LastActivityDate="2011-03-25T20:41:45.180" Title="How to calculate a confidence interval for a binomial, given a specific prior" Tags="&lt;probability&gt;&lt;statistics&gt;" AnswerCount="2" />
  <row Id="27519" PostTypeId="2" ParentId="27490" CreationDate="2011-03-17T03:44:09.640" Score="4" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Regarding your &quot;scratch of proof&quot;.&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) To show there is an $a$ such that $f(A_a)$ is bounded (and therefore, $f(A_b)$ is bounded for all $b\leq a$), simply pick $\delta$ that &quot;works&quot; for $\epsilon=1$. Now fix $x\in A_{\delta}$, and let $M = f(x)$. For all $y\in A_{\delta}$, you know that $|f(x)-f(y)|\lt 1$, hence &#xA;$$|f(y)| - |f(x)| \leq |f(y)-f(x)| \lt 1$$&#xA;so $|f(y)|\lt 1 + |f(x)| = 1+M$. This shows that $f(A_{\delta})\subseteq [-1-M, 1+M]$, hence $f(A_{\delta})$ is bounded.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I confess that I don't really see how to handle (2) easily.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Alternative way.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a possible way of attacking the problem which is closely related to what you are trying to do, but perhaps a bit easier: try constructing a sequence of points $x_1,x_2,\ldots$, with $x_1\to c$, and such that $f(x_1),f(x_2),\ldots$ is a Cauchy sequence. This will give you a &quot;target&quot; value for the limit, and then you can prove that the limit equals that target.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So: let $\epsilon_1 = \frac{1}{2}$. Then there exists a $\delta_1$, and we may assume $\delta_1\lt \frac{1}{2}$, such that $|x-c|\lt \delta_1$ and $|y-c|\lt \delta_1$ implies $|f(x)-f(y)|\lt \epsilon_1$. Take $x_1 = c-(\delta_1/2)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let $\epsilon_2 = \frac{1}{4}$. There exists a $\delta_2$, and we may assume $\delta_2\lt \min(\frac{1}{4},\delta_1)$, such that $|x-c|\lt \delta_2$ and $|y-c|\lt\delta_2$ implies $|f(x)-f(y)|\lt \epsilon_2$. Take $x_2 = c-(\delta_2/2)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $\epsilon_3 = \frac{1}{8}$. Then there exists a $\delta_3$, which we may assume satisfies $\delta_3\lt \min(\frac{1}{8},\delta_2)$, such that $|x-c|\lt\delta_3$ and $|y-c|\lt \delta_3$ implies $|f(x)-f(y)|\lt \epsilon_3$. Take $x_3 = c-(\delta_3/2)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Continuing this way, let $\epsilon_{n+1} = \frac{1}{2^{n+1}}$, and let $\delta_{n+1}\lt \min(\frac{1}{2^{n+1}},\delta_{n})$ be such that if $|x-c|\lt\delta_{n+1}$ and $|y-c|\lt\delta_{n+1}$ then $|f(x)-f(y)|\lt \epsilon_{n+1}$. Let $x_{n+1} = c - (\delta_{n+1}/2)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we have a sequence of points $\{x_n\}$ in $I$, with $x_n\to c$, and such that $|x_n - c| \lt \frac{1}{2^n}$ for all $n$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider now the sequence $\{ f(x_m)\mid m=1,2,\ldots\}$. I claim this is a Cauchy sequence. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Indeed: let $\epsilon\gt 0$. Then there exists a natural number $N\gt 0$ such that $1/2^N \lt \epsilon$. Let $n,m\geq N$. Then $|x_n-c| \lt \delta_{N}$ and $|x_m-c|\leq \delta_N$, so we know that $$|f(x_n)-f(x_m)|\leq \epsilon_N = \frac{1}{2^N}\lt \epsilon.$$&#xA;Thus, for every $\epsilon\gt 0$ there exists $N\gt 0$ such that for all $n,m\geq N$, $|f(x_n)-f(x_m)|\lt \epsilon$. Hence, the sequence $\{f(x_n)\}$ is a Cauchy sequence, and therefore has a limit, $L$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, since $x_n\to c$, then &lt;em&gt;if&lt;/em&gt; there is a limit for $f(x)$ as $x\to c$, then it will have to be equal to $L$. Prove that this is indeed the limit.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-17T04:12:13.407" LastActivityDate="2011-03-17T04:12:13.407" CommentCount="5" />
  <row Id="27520" PostTypeId="2" ParentId="27496" CreationDate="2011-03-17T03:52:58.707" Score="3" ViewCount="" Body="&lt;p&gt;Another way to think about this question is to differentiate $\tan^{-1}(3x) + C$ and see what you get.  Using the chain rule, &#xA;$$&#xD;&#xA;\frac{d}{dx} \left ( \tan^{-1}(3x) + C \right ) = \frac{3}{1 + (3x)^2} = \frac{3}{1 + 9x^2}.&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus the correct antiderivative of &#xA;$$\displaystyle \frac{1}{1+9x^2}&#xD;&#xA;$$ &#xA;must include a factor of $1/3$ to compensate: &#xA;$$&#xD;&#xA;\displaystyle \frac{1}{3} \tan^{-1}(3x) + C.&#xD;&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="4229" LastActivityDate="2011-03-17T03:52:58.707" />
  <row Id="27521" PostTypeId="2" ParentId="27506" CreationDate="2011-03-17T04:05:51.690" Score="4" ViewCount="" Body="&lt;p&gt;It is $x(n)=\lfloor \frac{n^2+4n+1}{12} \rfloor + \lfloor \frac {n+3}{6} \rfloor$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;No proof, just used your data, took two differences, and played.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2011-03-17T15:13:29.367" LastActivityDate="2011-03-17T15:13:29.367" CommentCount="3" />
  <row Id="27522" PostTypeId="2" ParentId="27507" CreationDate="2011-03-17T04:08:36.570" Score="2" ViewCount="" Body="&lt;p&gt;The key is that the Markov property is fundamentally about information: if you want to know something about the process's behavior after some time $t$, and you know its location at time $t$, then any information about the process's behavior before time $t$ is irrelevant.  All the relevant information from the history is packed up in the value of $X_t$.  You can express this in terms of conditional probabilities or conditional independence, but the idea is the same, in either discrete or continuous time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if $g$ is a one-to-one function, then $g(X)$ contains exactly the same information as $X$.  This is easy to understand, because $g$ has an inverse.  Thus you should be able to replace $X$ by $g(X)$ and not change the truth of any statement that is only about information; in particular the Markov property should be preserved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A more formal statement would be: for any $\sigma$-field $\mathcal{G}$, $X \in \mathcal{G}$ iff $g(X) \in \mathcal{G}$.  Now, if you express the Markov property formally in terms of $\sigma$-fields, it should become clear how to prove the first part, in either discrete or continuous time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a counterexample to the second part, note that if $g$ is not one-to-one, then $g(X)$ may contain less information than $X$ (but never more).  If we think about discrete time, in deciding what state to occupy at time $n+1$, the only information we can use is what state we are in at time $n$.  But the function $g$ can merge distinct states into a single state, so that we can no longer tell them apart.  If the original distinct states influenced $X_{n+1}$ in an important way, then the process can cease to be Markov.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This doesn't have to happen and you can certainly find particular examples of Markov chains $X_n$ and non-one-to-one functions $g$ so that $g(X_n)$ is not Markov.  Just try!&lt;/p&gt;&#xA;" OwnerUserId="822" LastActivityDate="2011-03-17T04:08:36.570" CommentCount="3" />
  <row Id="27523" PostTypeId="2" ParentId="15201" CreationDate="2011-03-17T04:25:54.590" Score="3" ViewCount="" Body="&lt;p&gt;I'm quite fond of Benjamin and Quinn's &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0883853337&quot; rel=&quot;nofollow&quot;&gt;Proofs that Really Count: The Art of Combinatorial Proof&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1930" LastActivityDate="2011-03-17T04:25:54.590" />
  <row Id="27524" PostTypeId="1" CreationDate="2011-03-17T04:38:51.777" Score="6" ViewCount="158" Body="&lt;p&gt;I've been going through a problem solving book, and I'm a little stumped on the following question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At each round, draw a number 1-100 out of a hat (and replace the number after you draw). You can play as many rounds as you want, and the last number you draw is the number of dollars you win, but each round costs an extra $1. What is a fair value to charge for entering this game?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One thought I had was to suppose I only have N rounds, instead of an unlimited number. (I'd then let N approach infinity.) Then my expected payoff at the Nth round is (Expected number I draw - N) = 50.5 - N. So if I draw a number d at the (N-1)th round, my current payoff would be d - (N-1), so I should redraw if d - (N-1) &amp;lt; 50.5 - N, i.e., if d &amp;lt; 49.5. So my expected payoff at the (N-1)th round is 49(50.5-N) + 1/100*[(50 - (N-1)) + (51 - (N-1)) + ... + (100 - (N-1))] = 62.995 - N (if I did my calculations correctly), and so on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is that this gets messy, so I think I'm doing something wrong. Any hints/suggestions to the right approach?&lt;/p&gt;&#xA;" OwnerUserId="8364" LastActivityDate="2011-03-17T17:42:59.960" Title="fair value of a hat-drawing game" Tags="&lt;probability&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="27526" PostTypeId="1" AcceptedAnswerId="27530" CreationDate="2011-03-17T05:11:53.807" Score="8" ViewCount="427" Body="&lt;p&gt;its written that if you multiply the sum above by -9 and use the distributive law, all terms except &quot;1&quot; will cancel. I can't see that. I know that this is a divergent series. (the article I was reading was a layman's introduction to zeta regularization)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, how is $S=1+2+3+4... = -1/12$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even if the series were to terminate somewhere, I dont see these value. This is completely baffling me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using $\frac{1}{1-x}$ I can put x=10 but that is not fair,isn't it? &lt;/p&gt;&#xA;" OwnerUserId="8365" LastEditorUserId="8365" LastEditorDisplayName="" LastEditDate="2011-03-17T05:18:30.587" LastActivityDate="2011-03-17T16:05:12.097" Title="$S=1+10+100+100+10000+... = -1/9$? How is that" Tags="&lt;sequences-and-series&gt;" AnswerCount="5" CommentCount="6" />
  <row Id="27527" PostTypeId="1" CreationDate="2011-03-17T05:20:12.193" Score="0" ViewCount="71" Body="&lt;p&gt;Hi, I posted &lt;a href=&quot;http://electronics.stackexchange.com/q/10632/2939&quot;&gt;this&lt;/a&gt; question on Electronics.Stackexchange and someone told me I'll be better off posting it here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Its an implementation of the Particle Filter using MATLAB but the results never follow the observations. I changed the weighting function to be Gaussian but still no avail.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;NOTE: I haven't taken into account that the process/measurement noise is time-correlated. Will this significantly change the accuracy of my program? If so, how do I correct for it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imelza&lt;/p&gt;&#xA;" OwnerUserId="7101" LastActivityDate="2011-03-17T05:20:12.193" Title="MATLAB programming of Particle Filter - what is going wrong?" Tags="&lt;signal-processing&gt;&lt;matlab&gt;&lt;monte-carlo&gt;&lt;correlation&gt;" CommentCount="1" FavoriteCount="2" />
  <row Id="27528" PostTypeId="2" ParentId="15201" CreationDate="2011-03-17T05:35:55.113" Score="1" ViewCount="" Body="&lt;p&gt;This is not precisely what you want, but you could look at &quot;Generatingfunctionology&quot;. The second edition is free, and can be downloaded here &lt;a href=&quot;http://www.math.upenn.edu/~wilf/DownldGF.html&quot; rel=&quot;nofollow&quot;&gt;http://www.math.upenn.edu/~wilf/DownldGF.html&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The book is about generating functions, which are helpful in combinatorial arguments.&lt;/p&gt;&#xA;" OwnerUserId="4284" LastActivityDate="2011-03-17T05:35:55.113" />
  <row Id="27529" PostTypeId="2" ParentId="27524" CreationDate="2011-03-17T05:42:42.510" Score="5" ViewCount="" Body="&lt;p&gt;Your expected return if you draw a number on the last round is 49.5 (because it costs a dollar to make the draw).  On round N-1, you should keep what you have if it is greater than 49.5, or take your chances if it is less.  The expected value if N=2 is then $\frac {51}{100}\frac {100+50}{2} -1 + \frac {49}{100}49.5=61.505$ where the first term is the chance that you keep the first draw times the expectation of that draw (assuming you will keep it), the second is the cost of the first draw, and the third is the chance that you will decline the first draw and take your chances on the second times the expectation of the second draw.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Added:  As Yuval makes more explicit, your strategy will be to quit when you get a number at least $X$.  The gain then is $\frac{100+X}{2}-\frac{100}{101-X}$ where the first is the payoff and the second is the cost of the expected number of plays.  As he says, this is maximized at X=87 with value $\frac{1209}{14}=86.3571$.  I'll have to think where I was a bit off.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2011-03-17T06:57:15.753" LastActivityDate="2011-03-17T06:57:15.753" />
  <row Id="27530" PostTypeId="2" ParentId="27526" CreationDate="2011-03-17T05:46:35.460" Score="15" ViewCount="" Body="&lt;p&gt;This does not hold for real numbers, the numbers you are probably used to working with. Rather it works with a different type of number, called a p-adic number. I will get to what these are in a moment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But first, what is a number? One might be tempted to give examples, such as $1$ or $-3/4$. But these are just symbols, which have no meaning on their own. Rather, mathematicians have &lt;em&gt;defined&lt;/em&gt; them in a way that they all agree upon, which allows them to be added, multiplied, etc. It is by defining them that we give them meaning.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But why do we define them the way we do? We could just as easily define them in other ways; it just so happens that the typical definition is often the most useful. But this is not always the case. The p-adic numbers are a different way of defining &quot;number&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To explain, I'll take the special case of 10-adic integers, which is where the equation you are asking about applies. The normal integers are what we are used to using daily, and if we take two integers $a$ and $b$ we can talk about how &quot;close&quot; $a$ is to $b$. Specifically, we say that $a$ is &quot;close&quot; to $b$ if they differ by a &lt;strong&gt;small&lt;/strong&gt; power of ten, i.e. $40$ is sort of close to $30$ because $40-30=10=10^1$ while $4$ is very close to $3$ because $4-3=1=10^0$ (the same process works for numbers which do not differ by an integer power of ten if we allow non-integer powers). But why do we define &quot;closeness&quot; this way? What if we said $a$ is close to $b$ if they differ by a &lt;strong&gt;large&lt;/strong&gt; power of ten? For example, $40,000$ would be sort of close to $30,000$ because $40,000-30,000=10,000=10^4$, while $4,000,000$ would be very close to $3,000,000$ because $4,000,000-3,000,000=1,000,000=10^6$. If we define &quot;closeness&quot; in this way, we get the 10-adic integers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now for the series you posted. Multiplying on both sides by $9$ gives $9+90+900+9000+90000+… = -1$. Well, we can see that this is giving us $\cdots999999 = -1$, with an infinite number of $9$'s to the left. But $9-(-1)=10=10^1$, $99-(-1)=100=10^2$, etc. so as we add more $9$'s onto the left we are getting closer and closer to $-1$, using the way we defined &quot;closeness&quot; for the 10-adic integers. In fact, we get &lt;em&gt;infinitely&lt;/em&gt; close to $-1$, so this series must in fact be equal to $-1$ in the 10-adic integers. While this is not a rigorous argument, I hope it provides you with a rough sense of what's going.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: Tying in what Ross said in his answer, the reason the formula $1/(1-10) = 1 + 10 + \cdots$ works in the 10-adic integers is that each successive term is getting closer to zero using the 10-adic notion of &quot;closeness&quot;, so we &lt;em&gt;can&lt;/em&gt; in fact ignore sufficiently late terms.&lt;/p&gt;&#xA;" OwnerUserId="8173" LastEditorUserId="8173" LastEditorDisplayName="" LastEditDate="2011-03-17T06:13:11.257" LastActivityDate="2011-03-17T06:13:11.257" CommentCount="7" />
  <row Id="27531" PostTypeId="2" ParentId="27524" CreationDate="2011-03-17T05:50:32.127" Score="8" ViewCount="" Body="&lt;p&gt;Consider what happens in the first stage. You get a number out of the hat, and either stop or go on. And you do it in some &quot;optimal&quot; way (maximizes expectation). If you do go on, the rule that will ensure you the maximal expectation is the same rule you used before - your original optimal rule.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This shows that your stopping rule can be characterized by some threshold $X$, which is the minimal number for which you'd stop (alternatively we could have chosen $X-1$, which is the maximal number for which you'd go on). Denote the expected gain by $E$. We have&#xA;$$ E = -1 + \frac{X-1}{100}E + \frac{\sum_{t=X}^{100} t}{100} = -1 + \frac{X-1}{100} + \frac{(100-X+1)(100+X)}{200}. $$&#xA;Multiplying by $200$ and rearranging, we get&#xA;$$ 2(101-X)E = (101-X)(100+X) - 200. $$&#xA;Therefore&#xA;$$ E = \frac{(101-X)(100+X) - 200}{2(101-X)}. $$&#xA;This is maximized by $101 - 10\sqrt{2} \approx 86.86$. Thus the best value for $X$ is either $86$ or $87$. These values give $E \approx 86.33$ and $E \approx 86.37$, so $X = 87$ is better, and the value of the game is $1209/14$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My calculations don't agree with Ross's, so there's probably a mistake somewhere; this doesn't invalidate the method.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2011-03-17T05:50:32.127" CommentCount="1" />
  <row Id="27532" PostTypeId="2" ParentId="27518" CreationDate="2011-03-17T05:53:56.770" Score="0" ViewCount="" Body="&lt;p&gt;You can always calculate lower and upper numerically. I'm not sure how the confidence interval is usually defined (Bayesian? likelihood?), but surely you can scan over $p$ and determine what you want numerically.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively, you can try the Poisson approximation, which works if $np$ is small. If $np$ is large then the normal approximation should work, and this probably the assumption underlying the usual formula, whatever it is.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2011-03-17T05:53:56.770" CommentCount="1" />
  <row Id="27533" PostTypeId="2" ParentId="27526" CreationDate="2011-03-17T05:54:00.067" Score="6" ViewCount="" Body="&lt;p&gt;There are many formulas that fall in this category.  The sum is not convergent, but you can do a formal manipulation that seems to sum it.  Maybe the simplest is 1-1+1-1+1-1 ...  If you represent it as $\sum (-1)^i$ you can sum it as $\frac {1}{1-(-1)}=\frac{1}{2}$.  This is exactly in the spirit of your original comment &quot;Using $\frac{1}{1−x}$ I can put $x=10$ but that is not fair,isn't it?&quot;  No, it's not fair, as the series is not convergent in the usual real number system, and as the p-adic numbers are not &quot;normal&quot; you should alert your readers when you use them.  The formal power series manipulation that leads to the sum of the geometric series cancels all the intermediate terms just fine, but the last term doesn't go to zero so you can't ignore it.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-03-17T05:54:00.067" />
  <row Id="27534" PostTypeId="1" AcceptedAnswerId="27537" CreationDate="2011-03-17T06:06:07.113" Score="5" ViewCount="142" Body="&lt;p&gt;Apologies for a soft question. I do not have a lot of time because of my job (programmer) and my wife cannot work so I cannot quit. I also did not graduate from a very prestigious school with good grades. So grad school is out of the question for me. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am 23 years old, I find that because I did not learn many topics well when I was younger, I have accumulated a lot of half baked knowledge about stuff. For example, I can spell out some theorems, results in fancy notation, but I don't really understand &lt;em&gt;those&lt;/em&gt; things. So when I try to restart and re-educate myself I find that I have to unlearn a lot of gibberish that I put in my head just to pass through examinations. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I visited this site and saw some solutions people posted. I was so exhilerated after getting some of them that I wish I had the same understanding as the person who could think of such solutions. So in my private time, I have embarked on a project to learn math from scratch. I do not aspire to make meaningful contributions, because of the late start, but to discover mathematics and get some personal peace.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Before beginning, I decided that I am most interested in stuff like graph theory, and perhaps I would like to explore geometry as well (algebraic, differential)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I planned a halfway route and here is my question... what is your advise for a person like me to learn mathematics and reach his goal (with literature recommendations possibly online)&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Start from basics, do good problems in pre-calculus algebra (inequalities, permutations, combinations, sequences and series etc?) to get some mathematical thinking&lt;/li&gt;&#xA;&lt;li&gt;Calculus from Rudin, to get rid of most of the algorithmic procedures in my minds (is this too tough for me)&lt;/li&gt;&#xA;&lt;li&gt;Where do I go from here? Follow some college's standard curriculum? Is there a curriculum for undergrads designed to specialize in graph theory/topology?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="8365" LastEditorUserId="8365" LastEditorDisplayName="" LastEditDate="2011-03-17T07:11:02.080" LastActivityDate="2011-03-17T07:21:53.143" Title="Math route for someone of this background" Tags="&lt;soft-question&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="3" />
  <row Id="27535" PostTypeId="1" CreationDate="2011-03-17T06:15:29.270" Score="2" ViewCount="153" Body="&lt;p&gt;given an arbitrary arc, where you know the following values: start point (x0,y0), end point (x1,y1), radius (r) and arc direction (e.g. clockwise or counterclockwise from start to end), how can I calculate the arc's center?  i know from &lt;a href=&quot;http://math.stackexchange.com/questions/27124/how-to-tell-if-an-arc-is-within-bounds&quot;&gt;this previous post&lt;/a&gt; (thanks!) that the center lies on the perpendicular bisector between the two points, but don't know how to calculate it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;thanks!&lt;/p&gt;&#xA;" OwnerUserId="7279" LastActivityDate="2011-03-17T14:22:40.387" Title="how to find center of an arc given start point, end point, radius, and arc direction?" Tags="&lt;geometry&gt;&lt;algebra-precalculus&gt;" AnswerCount="2" />
  <row Id="27536" PostTypeId="2" ParentId="27444" CreationDate="2011-03-17T06:17:32.060" Score="1" ViewCount="" Body="&lt;p&gt;-- answer first posted as comment -- &lt;/p&gt;&#xA;&#xA;&lt;p&gt;At this point I think you're supposed to reinterpret this result as a natural number that's the product of the numbers you started with. Wait, $\sqrt{2}$? Have you double-checked the calculations? Also I think your vectors should have 4 positions because now the result will be cropped (i.e. the circular convolution gives a different result than the real convolution).&lt;/p&gt;&#xA;" OwnerUserId="5189" LastActivityDate="2011-03-17T06:17:32.060" CommentCount="2" />
  <row Id="27537" PostTypeId="2" ParentId="27534" CreationDate="2011-03-17T06:26:56.877" Score="2" ViewCount="" Body="&lt;p&gt;I can speak from experience in saying that teaching oneself mathematics, while difficult, is possible. However, I cannot stress enough the importance of being able to talk with knowledgeable people in order to check one's understanding. I am fortunate enough to live in a university town with a mathematics faculty eager to talk math with anyone interested. This may well not be the case for you, in which case I suspect you will be making frequent use of this site.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for your &quot;curriculum&quot;, to some extend you need to find what works best for you. However, if you really are interested in learning higher math, I suggest you not spend much time memorizing formulas or doing algebraic busywork. Rather, try to reason your way through elementary problems in fields such as combinatorics or probability. Once you feel safe working with these, I suggest you either move on to basic calculus or an introductory text in abstract algebra. You didn't mention algebra in your list, but I really think that an introductory abstract algebra text will do the best job of introducing the fundamental concepts of abstract mathematics (which you will certainly need for graph theory or other branches of higher mathematics). Personally, I recommend &quot;Modern Algebra: An Introduction&quot; by John Durbin, which I found both intuitive and eye-opening. Of course, these recommendations are based off my own experiences and may or may not be well-suited to you.&lt;/p&gt;&#xA;" OwnerUserId="8173" LastActivityDate="2011-03-17T06:26:56.877" />
  <row Id="27538" PostTypeId="2" ParentId="27507" CreationDate="2011-03-17T06:49:05.220" Score="4" ViewCount="" Body="&lt;p&gt;The transformation you are interested in is often called &lt;em&gt;lumping&lt;/em&gt; some states of the process and the resulting process a &lt;em&gt;lumped chain&lt;/em&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a given denumerable Markov chain $X$ on a state space $E$ with transitions $p$ and a given function $g$, whether $g(X)$ is still a Markov chain or not may depend on the initial distribution of $X$, but a necessary and sufficient condition for $g(X)$ to be Markov for any initial distribution of $X$ (condition known at least since C. J. Burke and M. Rosenblatt &lt;a href=&quot;http://projecteuclid.org/euclid.aoms/1177706444&quot; rel=&quot;nofollow&quot;&gt;A Markovian Function of a Markov Chain&lt;/a&gt; in 1958 and widely used in the applications) is that for any $x$ and $x&amp;#39;$ in $E$ such that $g(x)=g(x&amp;#39;)$ and any $z$ in $g(E)$,&#xA;$$&#xD;&#xA;\sum_{y:g(y)=z}p(x,y)=\sum_{y:g(y)=z}p(x&amp;#39;,y).&#xD;&#xA;$$&#xA;In words, being at a state $x$, the probability to jump to a lumped state $z$ depends on $x$, only through the lumped state $g(x)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A simple but inspiring example is a Markov chain $X$ on three states $0$, $1$ and $2$ with transitions from $0$ to $1$, from $1$ to $2$, from $2$ to $1$ and from $2$ to $0$ and no other transition, thus, for a given $u$ in $(0,1)$, $$&#xD;&#xA;p(0,1)=p(1,2)=1,\quad&#xD;&#xA;p(2,0)=u=1-p(2,1).&#xD;&#xA;$$ &#xA;The chain $X$ is kosher: $X$ is irreducible and aperiodic and converges in distribution to the unique stationary distribution $\pi$ given by $\pi(0)=u/(2+u)$ and $\pi(1)=\pi(2)=1/(2+u)$. Now, lump together states $1$ and $2$, for example by a function $g$ to a two-state space $\{a,b\}$ such that $g(0)=a$ and $g(1)=g(2)=b$. Then,&#xA;$$&#xD;&#xA;P(g(X_{n})=a|g(X_{n-1})=\cdots=g(X_{n-k})=b,g(X_{n-k-1})=a)&#xD;&#xA;$$&#xA;is $0$ if $k$ is odd and $u$ if $k$ is even. This shows that $g(X)$ is not Markov and even that $g(X)$ is not a Markov chain of any order.&lt;/p&gt;&#xA;" OwnerUserId="6179" LastEditorUserId="6179" LastEditorDisplayName="" LastEditDate="2011-03-17T07:27:10.290" LastActivityDate="2011-03-17T07:27:10.290" />
  <row Id="27539" PostTypeId="1" CreationDate="2011-03-17T06:53:34.820" Score="1" ViewCount="89" Body="&lt;p&gt;Hello, I was wanting to know if I can get some help with this proof. I tried but i just cannot seem to get $2^{k}$. It states that,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $k \in \mathbb{Z}_{\ge} 0$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sum^{k}_{m=0}$ &#xA;$\left( \begin{array}{c} &#xD;&#xA;k \\ &#xD;&#xA;m \\ &#xD;&#xA;\end{array} \right)$ = $2^{k}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank You&lt;/p&gt;&#xA;" OwnerUserId="8367" LastActivityDate="2011-03-17T07:48:12.113" Title="Binomial Thereom" Tags="&lt;binomial-coefficients&gt;&lt;sequences-and-series&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="27540" PostTypeId="2" ParentId="27539" CreationDate="2011-03-17T07:11:05.560" Score="3" ViewCount="" Body="&lt;p&gt;The right-hand side is the number of subsets of a set with $k$ elements.  The left hand side gives a sum of the number of subsets of a set with $k$ elements having $0$ elements, then $1$ element, then $2$ elements, etc., up to the subsets with $k$ elements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Leaving that aside, this follows from applying the binomial theorem in a way that might seem like a trick.  Here's another example to motivate this.  If $k\geq 1$, then&#xA;$${k \choose 0}-{k \choose 1}+{k \choose 2}\mp\cdots +(-1)^k{k \choose k}=0.$$&#xA;Proof Sketch: Expand $(1-1)^k$ using the binomial theorem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could use induction and the factorial formula for the binomial coefficients, but I do not recommend doing so.  However, using induction and Pascal's identity ${k\choose {m-1}}+{k\choose m}={{k+1}\choose m}$ would work well.  Consider moving down a row in Pascal's triangle to motivate the proof.  Each entry from row $k$ is added twice to obtain the entries in row $k+1$, so the sum of the entries doubles.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2011-03-17T07:26:09.183" LastActivityDate="2011-03-17T07:26:09.183" />
  <row Id="27541" PostTypeId="1" AcceptedAnswerId="27550" CreationDate="2011-03-17T07:18:42.507" Score="3" ViewCount="90" Body="&lt;p&gt;Hello, I am an undergraduate of CS and I participate in TCS.SE. I am having trouble finding the error in a proof about the polynomial hierarchy collapsing to the first level ($NP = coNP$). I believe the reasoning to be false, because it is very simple and it would be common knowledge, however it is not. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $M$ be any oracle TM in $NP^{NP}$ (i.e. $ \Sigma^{P}_{2} $ ) . Obviously, by ignoring its oracle, $M$ can solve any problem in NP.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other (and most important) direction, let $M&amp;#39;$ be a TM where instead of using the oracle, we encode it into a simple non-deterministic machine with a running time of $NTIME( k(n) )$. If $M$ had a running time of $NTIME( p(n) )$ , $M&amp;#39;$ will have a running time of $NTIME( p(n)*k(n) )$ which is polynomial. Therefore, for any oracle TM in $NP^{NP}$, there is a non-deterministic TM that can solve the same problem in polynomial time. It follows that $NP^{NP} /subset NP$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By combining the two statements we have that $NP=NP^{NP}$ and extending this,&#xA;$NP=coNP=P^{NP}=NP^{NP}$ and the polynomial hierarchy collapses to the 1st level.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where is the error in the above proof?&lt;/p&gt;&#xA;" OwnerUserId="8327" LastEditorUserId="8327" LastEditorDisplayName="" LastEditDate="2011-03-17T11:27:41.427" LastActivityDate="2011-03-17T11:27:41.427" Title="Faulty proof of polynomial hierarchy first level collapse?" Tags="&lt;computer-science&gt;&lt;fake-proofs&gt;" AnswerCount="2" />
  <row Id="27542" PostTypeId="2" ParentId="27534" CreationDate="2011-03-17T07:21:53.143" Score="1" ViewCount="" Body="&lt;p&gt;First, I really admire your courage and perseverance. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second, pick up a &lt;em&gt;Schaum's outline&lt;/em&gt; on secondary school mathematics and make sure you understand every chapter. If you cannot answer a question by looking at it, then take out a pencil and paper and go to work on solving it. Make sure you understand the techniques involved and the reasoning behind each problem. Reread the chapter if you need to and make sure you check your answers with those in the book. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suggest this approach rather than starting with a standard math text, because Schaum's outlines are specifically geared toward self-study. Remember, if your foundation is not strong, succeeding in mathematics --- even elementary mathematics --- will be nearly impossible. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have any math-related questions, don't be afraid to share it on Math.SE. Once you're done with a thorough evaluation of the fundamentals, then you're ready for something more. In the meantime, get to work!&lt;/p&gt;&#xA;" OwnerUserId="2720" LastActivityDate="2011-03-17T07:21:53.143" />
  <row Id="27544" PostTypeId="2" ParentId="27502" CreationDate="2011-03-17T07:44:21.577" Score="5" ViewCount="" Body="&lt;p&gt;The answers in the comments are somewhat misleading in two respects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For one, they don't mention that they use the facts that a) &lt;a href=&quot;http://www.mathreference.com/top-ms,cprod.html&quot; rel=&quot;nofollow&quot;&gt;the space of sequences in a metric space is metrizable&lt;/a&gt; and b) compactness and sequential compactness are equivalent in metrizable spaces. The desired result can only be proved from Tychonoff's theorem using these facts; the product of an uncountable number of compact metric spaces is in general not metrizable and not sequentially compact.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But more importantly, they obscure the set-theoretic premises required for the desired result. &lt;a href=&quot;http://en.wikipedia.org/wiki/Tychonoff%27s_theorem#Tychonoff.27s_theorem_and_the_axiom_of_choice&quot; rel=&quot;nofollow&quot;&gt;Tychonoff's theorem in full generality is equivalent to the axiom of choice&lt;/a&gt;. We're only dealing with a countable product here, but I believe Tychonoff's theorem for countable products still requires the boolean prime ideal theorem and the axiom of countable choice (see &lt;a href=&quot;http://www.math.purdue.edu/~hrubin/JeanRubin/Papers/BPI.pdf&quot; rel=&quot;nofollow&quot;&gt;this article&lt;/a&gt;, p. 2). By contrast, the diagonal argument the OP was looking for requires no choice (not even countable choice):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given a sequence $s$ of sequences in $X$, denote by $s_i(n)$ the $n$-th element of the $i$-th sequence of $s$. (If you tend to get as confused about such things as I do, it helps to keep in mind the concrete example where $s_i(n)$ is the $n$-th digit in the binary representation of $i$.) Consider the sequence $s_i(1)$ in $X$. Since $X$ is a compact metric space and hence sequentially compact, this sequence contains a convergent subsequence $s_{i_1(k)}(1)$. Denote its limit by $a (1)$. Then consider the sequence $s_{i_1(k)}(2)$ (a subsequence of the sequence $s_i(2)$ in $X$). This sequence, too, has a convergent subsequence $s_{i_2(k)}(2)$ with limit $a (2)$. We can continue this construction for all $n$. Now consider $s_{i_l(l)}$, which is a subsequence of the sequence $s$ of sequences. We can show that it converges against the sequence $a$ of the limits $a (n)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An element of the basis of the product topology containing $a$ is a product of factors, only a finite number of which differ from $X$. For each $n$, apart from the first $n$ elements the sequence $s_{i_l(l)}(n)$ is a subsequence of $s_{i_n(l)}(n)$, and thus it is a convergent subsequence of $s_i(n)$ with limit $a (n)$, which will eventually end up in the $n$-th factor of the basis element. Since there are finitely many of these factors, eventually all these convergent subsequences will have ended up in their corresponding factors, so that the entire sequence eventually ends up in the given basis element.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[Edit in response to comments by Nate and Theo:]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a construction that picks a particular convergent subsequence of a sequence having at least one convergent subsequence in a metric space without making arbitrary choices (in case Nate and Theo are right that that requires the axiom of choice):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a given sequence $b_n$, consider the (non-empty) set of all convergent subsequences for which $\lvert b_n - b \rvert &amp;lt; 2^{-n}$ for all $n$, where $b$ is the limit of the subsequence, and take the least of these subsequences in lexicographical order. We can't just take the lexicographically least convergent subsequence because there isn't one, but we can explicitly construct the lexicographically least convergent subsequence that fulfills the given condition: For each element, just pick the earliest element in the sequence such that the resulting partial subsequence is a prefix of at least one subsequence that fulfills the condition. The result is a Cauchy sequence, and hence, since $X$ is compact, converges. (The same construction doesn't work without the additional condition, since the resulting subsequence would just be the sequence itself.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So it appears that we can avoid choice. However, that would seem to allow us to reason as follows: The fact that all factors in the product are $X$ was not used. Thus, the result generalizes to arbitrary countable products of compact metric spaces. A countable product of  metric spaces is metrizable, and in a metrizable space compactness and sequential compactness are equivalent; thus we've proved that the countable product of compact metric spaces is compact. But now we can plug that into the proof of the equivalence of the axiom of choice and Tychonoff's theorem (linked to above). Since to any compact metric space we can add an isolated point and still have a compact metric space (by limiting the metric to $1$ and choosing a uniform distance to the new point $&gt;1$), this seems to prove the possibility of choice from a countable product of compact metric spaces. This is not the full axiom of countable choice, but it does allow considerable choices that I would have expected to be independent of ZF. Is this true, or is there another fallacy somewhere?&lt;/p&gt;&#xA;" OwnerUserId="6622" LastEditorUserId="6622" LastEditorDisplayName="" LastEditDate="2011-03-17T17:23:43.493" LastActivityDate="2011-03-17T17:23:43.493" CommentCount="18" />
  <row Id="27545" PostTypeId="2" ParentId="27539" CreationDate="2011-03-17T07:48:12.113" Score="1" ViewCount="" Body="&lt;p&gt;Take the binomial expansion of $(1+x)^k$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;(1+x)^k = \sum_{m=0}^k \binom{k}{m} x^{k-m}&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The above expansion holds for all values of $x$ and for all $k \geq 0$. Substitute $x=1$ and you have the result.&lt;/p&gt;&#xA;" OwnerUserId="2641" LastActivityDate="2011-03-17T07:48:12.113" />
  <row Id="27546" PostTypeId="1" CreationDate="2011-03-17T07:57:05.020" Score="1" ViewCount="141" Body="&lt;p&gt;I know that zeroes of holomorphic functions are isolated,and I know that if a holomorphic function has zero set whic has a limit point then it is identically zero function,i know a holomorphic function can have countable zero set, does there exixt a holomorphic function which is not identically zero, and has uncountable number of zeroes?&lt;/p&gt;&#xA;" OwnerUserId="6990" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2011-03-17T08:13:54.927" LastActivityDate="2011-03-17T12:28:45.813" Title="zeroes of holomorphic function" Tags="&lt;complex-analysis&gt;" AnswerCount="4" />
  <row Id="27547" PostTypeId="2" ParentId="27546" CreationDate="2011-03-17T08:01:26.267" Score="0" ViewCount="" Body="&lt;p&gt;Your question needs some clarification, as it is not internally consistent. If I understand correctly, you're asking if all holomorphic functions have uncountably many zeros? In that case, the answer is definitely no, for example $f(z) = 1$.&lt;/p&gt;&#xA;" OwnerUserId="8173" LastActivityDate="2011-03-17T08:01:26.267" CommentCount="2" />
  <row Id="27548" PostTypeId="2" ParentId="27398" CreationDate="2011-03-17T08:02:43.833" Score="1" ViewCount="" Body="&lt;p&gt;You might be interested in a related, but quite different, question. Suppose we choose the edges of a countable graph &lt;em&gt;randomly&lt;/em&gt;, according to some probability $p$. How many different graphs do we expect to get, if we repeat the experiment several times?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might think that every time we'll get a different graph, but in fact, with probability $1$ the graph generated will be isomorphic to the so-called &lt;a href=&quot;http://en.wikipedia.org/wiki/Rado_graph&quot; rel=&quot;nofollow&quot;&gt;Rado graph&lt;/a&gt;. This is the same graph immaterial of the probability $p$, and you even get the same graphs if the probabilities decay not too fast (under some arbitrary $\omega$-ordering of the edges).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The charming proof is an illustration of the back-and-forth argument - check it out.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2011-03-17T08:02:43.833" />
  <row Id="27549" PostTypeId="1" AcceptedAnswerId="27551" CreationDate="2011-03-17T08:03:49.603" Score="1" ViewCount="66" Body="&lt;p&gt;Let $X$ be non-empty set and $S$ be the set of all subsets of $X$, which will be a poset, under subset relation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $\phi \colon S\rightarrow \mathbb{Z}$ be any function and for each $H\in S$, define $\sigma(H)=\sum_{K\leq H}\phi(K)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\mu$ be the Mobius function defined by $\sum_{K\leq H} \mu(H)=\delta_{K,X}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How to show that $\sum_{H\leq X} \sigma(H)\mu(H)=\phi(X)$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I considered LHS, and substituted for \sigma(H), interchanged the sum, but I couldn't proceed further...)  &lt;/p&gt;&#xA;" OwnerUserId="8186" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2011-03-17T08:12:40.387" LastActivityDate="2011-03-17T08:24:45.467" Title="Mobius inversion formula" Tags="&lt;order-theory&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" />
  <row Id="27550" PostTypeId="2" ParentId="27541" CreationDate="2011-03-17T08:11:51.667" Score="3" ViewCount="" Body="&lt;p&gt;If you combine an NP oracle with an NP machine you don't get another NP machine. The easiest way to see that is to query the oracle once and answer the negation - this machine is clearly co-NP rather than NP.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you take a general $NP^{NP}$ machine, you need to quantify both existentially and universally. This is very lucidly explained in &lt;a href=&quot;http://www.cs.cornell.edu/courses/cs6810/2009sp/scribe/lecture5.pdf&quot; rel=&quot;nofollow&quot;&gt;these lecture notes&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2011-03-17T08:11:51.667" CommentCount="1" />
  <row Id="27551" PostTypeId="2" ParentId="27549" CreationDate="2011-03-17T08:17:50.963" Score="2" ViewCount="" Body="&lt;p&gt;As you said,&#xA;$$\sum_{H\le X}\sigma(H)\mu(H)=\sum_{H\le X}\sum_{K\le H}\phi(K)\mu(H)=\sum_{K\le X}\phi(K)\tau_X(K)&#xD;&#xA;$$ with $$\tau_X(K)=\sum_{K\le H\le X}\mu(H),&#xD;&#xA;$$&#xA;hence all there is to show is that $\tau_X(X)=1$ and $\tau_X(K)=0$ for every $K\subset X$, $K\ne X$. This is how you &lt;em&gt;defined&lt;/em&gt; the Möbius function $\mu$, hence the proof is complete.&lt;/p&gt;&#xA;" OwnerUserId="6179" LastEditorUserId="6179" LastEditorDisplayName="" LastEditDate="2011-03-17T08:24:45.467" LastActivityDate="2011-03-17T08:24:45.467" CommentCount="5" />
  <row Id="27552" PostTypeId="2" ParentId="27546" CreationDate="2011-03-17T08:17:58.967" Score="2" ViewCount="" Body="&lt;p&gt;A holomorphic function on a connected open set that is not identically zero cannot have uncountably many zeros.  Open subsets of $\mathbb{C}$ are $\sigma$-compact, so if $G$ is the domain, then there is a sequence $K_1,K_2,\ldots$ of compact subsets of $G$ such that $G=K_1\cup K_2\cup\cdots$.  (It is not hard to construct $K_n$; e.g., if $G$ is the whole plane, you can take $K_n=\{z:|z|\leq n\}$.  Otherwise you can take $\{z\in G:|z|\leq n\text{ and }d(z,\partial G)\geq \frac{1}{n}\}$.) An uncountable subset of $G$ must have uncountable intersection with one of the $K_n$s, because a countable union of countable sets is countable.  An infinite subset of $K_n$ has a limit point in $K_n$ by compactness.  The rest follows from the result you mentioned that a holomorphic function that is not identically zero cannot have a limit point of its zero set in the connected open set on which it is defined.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2011-03-17T08:17:58.967" CommentCount="3" />
  <row Id="27553" PostTypeId="2" ParentId="27526" CreationDate="2011-03-17T08:20:58.120" Score="1" ViewCount="" Body="&lt;p&gt;This formula is justified by analytic continuation (so you can, in fact plug in x=10 in $\frac{1}{1-x}$). See&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/1_%2B_2_%2B_4_%2B_8_%2B_%C2%B7_%C2%B7_%C2%B7&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/1_%2B_2_%2B_4_%2B_8_%2B_%C2%B7_%C2%B7_%C2%B7&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/1_%2B_2_%2B_3_%2B_4_%2B_%C2%B7%C2%B7%C2%B7&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/1_%2B_2_%2B_3_%2B_4_%2B_%C2%B7%C2%B7%C2%B7&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The relation to p-adics is simple enough: $\frac{1}{1-x} - (1+x+x^2+\ldots +x^n)= x^{n+1}(1+x+\ldots)$, so to equate the two you want the difference to get small, which is true either for real numbers when $x&amp;lt;1$, or for p-apics when $x=p$.&lt;/p&gt;&#xA;" OwnerUserId="2633" LastActivityDate="2011-03-17T08:20:58.120" />
  <row Id="27554" PostTypeId="2" ParentId="27546" CreationDate="2011-03-17T08:32:50.967" Score="2" ViewCount="" Body="&lt;p&gt;Existence of such a function would imply that its zero locus were an uncountable closed discrete subset of the complex plane, contradicting that the complex plane is separable and normal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT in response to question: (For simplicity's sake $f$ is assumed to be defined on all of $\mathbb{C}$ — this isn't really a problem because $\mathbb{C}$ is hereditarily normal, so if $f$ is defined on an open set $U$ we also have that $U$ is separable and normal)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The zero locus of $f$ means $f^{-1}(\{0\})$ (i.e. the set of points where $f$ is $0$ — call it $Z$) and it's the pre-image of a closed set under a continuous function so it's closed. Assume that $Z$ is uncountable and discrete. Then every function from $Z$ to $\mathbb{R}$ is continuous and thus there are $2^{2^{\aleph_0}}$ many continuous functions from $Z$ to $\mathbb{R}$. Since $\mathbb{Q}(i)$ is a countable dense subset of $\mathbb{C}$ any continuous real-valued function on $\mathbb{C}$ is determined by its values on $\mathbb{Q}(i)$, i.e. there are $2^{\aleph_0}$ many continuous real-valued functions on $\mathbb{C}$. Hence there is some continuous real-valued function on $Z$ that cannot be extended to the whole space, but $\mathbb{C}$ is normal so any continuous function on a closed subspace is extendable to a continuous function on the whole space by the Tietze extension theorem; contradicting our assumption of $Z$ being uncountable and discrete.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is essentially the same argument as why the &lt;a href=&quot;http://en.wikipedia.org/wiki/Moore_plane&quot; rel=&quot;nofollow&quot;&gt;Moore plane&lt;/a&gt; isn't normal.&lt;/p&gt;&#xA;" OwnerUserId="1269" LastEditorUserId="1269" LastEditorDisplayName="" LastEditDate="2011-03-17T09:27:56.973" LastActivityDate="2011-03-17T09:27:56.973" CommentCount="4" />
  <row Id="27555" PostTypeId="1" AcceptedAnswerId="27581" CreationDate="2011-03-17T08:37:13.057" Score="5" ViewCount="103" Body="&lt;p&gt;I'm having problems with exercise 1 of chapter 3 of do Carmo's &quot;Riemannian Geometry&quot;.  Here is the background:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $(u,v)$ be the coordinates on $\mathbb{R}^2$.  Let $f,g\in C^\infty(\mathbb{R})$, and observe that $\varphi:\mathbb{R}^2\rightarrow \mathbb{R}^3$ given by $\varphi(u,v)=(f(v)\cos u,f(v)\sin u,g(v))$ is an immersion assuming $f&amp;#39;(v)^2+g&amp;#39;(v)^2\not= 0$ and $f(v)\not= 0$.  The image is the surface of revolution generated by the curve $(f(v),g(v))$ being rotated about the $z$-axis.  The induced metric is&#xA;$$ (g_{ij})=\left( \begin{array}{cc} f^2 &amp;amp; 0 \\ 0 &amp;amp; f&amp;#39;^2+g&amp;#39;^2 \end{array} \right), $$&#xA;and the local equations of a geodesic $\gamma$ are&#xA;$$ \left\{ \begin{array}{l} \frac{d^2 u}{dt^2} + \frac{2ff&amp;#39;}{f^2}\frac{du}{dt}\frac{dv}{dt}=0 \\ \frac{d^2 v}{dt^2}-\frac{ff&amp;#39;}{f&amp;#39;^2+g&amp;#39;^2}\left( \frac{du}{dt} \right)^2 + \frac{f&amp;#39;f&amp;#39;&amp;#39; + g&amp;#39;g&amp;#39;&amp;#39;}{f&amp;#39;^2+g&amp;#39;^2} \left( \frac{dv}{dt} \right)^2 = 0. \end{array} \right. $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, do Carmo says: Obtain the following geometric meaning of the equations above: the second equation is, except for meridians ($u=u_0$) and parallels ($v=v_0$), equivalent to the fact that the &quot;energy&quot; $|\gamma&amp;#39;(t)|^2$ of a geodesic is constant along $\gamma$; the first equation signifies that if $\beta(t)$ is the oriented angle, $\beta(t)&amp;lt;\pi$, of $\gamma$ with a parallel $P$ intersecting $\gamma$ at $\gamma(t)$, then $r\cos \beta$ is constant, where $r$ is the radius of $P$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This last paragraph is what's confusing me.  First of all, I've seen the energy of a path as $\int_a^b |\gamma&amp;#39;(t)|^2 dt$, so maybe that's why he put &quot;energy&quot; in quotes.  But also, geodesics have constant speed!  So this should be constant along all geodesics too (regardless of whether they're meridians or parallels).  But I figured that maybe I should just blindly plug and chug since that seems to work scarily often in Riemannian geometry, so I found $|\gamma&amp;#39;(t)|^2$, took its $t$-derivative, and substituted in the second equation in the system for geodesics.  Here it is:&#xA;$$ |\gamma&amp;#39;(t)|^2 = \left\langle \frac{d \gamma}{dt} , \frac{d \gamma}{dt} \right\rangle = u&amp;#39;^2 g_{11} + 2u&amp;#39;v&amp;#39;g_{12} + v&amp;#39;^2 g_{22} = u&amp;#39;^2f^2+v&amp;#39;^2(f&amp;#39;^2+g&amp;#39;^2)$$&#xA;so&#xA;$$ \frac{d}{dt} |\gamma&amp;#39;(t)|^2 = 2u&amp;#39;u&amp;#39;&amp;#39;f^2+2u&amp;#39;^2ff&amp;#39; + 2v&amp;#39;v&amp;#39;&amp;#39;(f&amp;#39;^2+g&amp;#39;^2)+v&amp;#39;^2(2f&amp;#39;f&amp;#39;&amp;#39;+2g&amp;#39;g&amp;#39;&amp;#39;) $$&#xA;$$ = 2u&amp;#39;u&amp;#39;&amp;#39;f^2+2u&amp;#39;^2ff&amp;#39; + 2v&amp;#39;(ff&amp;#39;u&amp;#39;^2-(f&amp;#39;f&amp;#39;&amp;#39;+g&amp;#39;g&amp;#39;&amp;#39;)v&amp;#39;^2)) + 2v&amp;#39;v&amp;#39;&amp;#39;(f&amp;#39;^2+g&amp;#39;^2)+2v&amp;#39;^2(f&amp;#39;f&amp;#39;&amp;#39;+g&amp;#39;g&amp;#39;&amp;#39;)$$&#xA;$$ = 2u&amp;#39;u&amp;#39;&amp;#39;f^2+2u&amp;#39;^2ff&amp;#39;(1+v&amp;#39;)+2v&amp;#39;v&amp;#39;&amp;#39;(f&amp;#39;^2+g&amp;#39;^2)+2v&amp;#39;^2(f&amp;#39;f&amp;#39;&amp;#39;+g&amp;#39;g&amp;#39;&amp;#39;)(1-v&amp;#39;)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and this looks hopeless.&lt;/p&gt;&#xA;" OwnerUserId="2159" LastActivityDate="2011-03-17T12:07:00.407" Title="geodesics on a surface of revolution" Tags="&lt;differential-geometry&gt;&lt;riemannian-geometry&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" />
  <row Id="27556" PostTypeId="1" AcceptedAnswerId="27560" CreationDate="2011-03-17T08:42:23.530" Score="3" ViewCount="69" Body="&lt;p&gt;I was given this question on a college assessment pre-test.  I got the correct answer in a reasonable amount of time, but mostly because I worked backwards and double checked my answer.  After I was done, I tried to find math on the net to solve it correctly, but I feel there is probably a simpler solution than this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Is there an easier way to solve problems in this template?&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Given the first and &lt;em&gt;n&lt;/em&gt;-th values in an arithmetic progression, and the sum of the progression up to &lt;em&gt;n&lt;/em&gt; (inclusive), give the first &lt;em&gt;x&lt;/em&gt; terms of the series.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;The actual question on the quiz&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In an arithmetic series, the terms of the series are equally spread out. For example, in&#xA;  1 + 5 + 9 + 13 + 17, consecutive terms are 4 apart. If the first term in an arithmetic series is 3, the last term is 136, and the sum is 1,390, what are the first 3 terms?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This one lended itself to intuition and backwards-work because I was reasonably certain that &lt;em&gt;d&lt;/em&gt; would be an integer, and 133 (&lt;code&gt;136 - 3&lt;/code&gt;) is evenly divisible by 7.  But I would like to find a simpler way to solve problems like this in the future.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;The work I did to solve the question&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I used the formulas:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;S&lt;sub&gt;n&lt;/sub&gt; = &amp;frac12; n(a&lt;sub&gt;1&lt;/sub&gt; + a&lt;sub&gt;n&lt;/sub&gt;)&lt;/li&gt;&#xA;&lt;li&gt;a&lt;sub&gt;n&lt;/sub&gt; = dn + c&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Where &lt;em&gt;n&lt;/em&gt; is the count of values in the sequence, &lt;em&gt;d&lt;/em&gt; is the common difference (distance between values in the sequence), and &lt;em&gt;c&lt;/em&gt; is an unknown constant&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the first equation, I solved for &lt;em&gt;n&lt;/em&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;1390 = &amp;frac12; n(3 + 136)&lt;br&gt;&#xA;  1390 = &amp;frac12; n(139)&lt;br&gt;&#xA;  10 = &amp;frac12; n&lt;br&gt;&#xA;  &lt;strong&gt;20 = n&lt;/strong&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For the second equation, I first solved for &lt;em&gt;c&lt;/em&gt;, by making two equations out of the values I had for &lt;em&gt;a&lt;sub&gt;n&lt;/sub&gt;&lt;/em&gt;, and the previously calculated value for &lt;em&gt;n&lt;/em&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;3 = d + c,&lt;br&gt;&#xA;  136 = 20d + c  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;(with the first): d = 3 - c&lt;br&gt;&#xA;  (substitute): 136 = 20(3 - c) + c&lt;br&gt;&#xA;  136 = 60 - 19c&lt;br&gt;&#xA;  76 = -19c&lt;br&gt;&#xA;  &lt;strong&gt;-4 = c&lt;/strong&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Then I solved for &lt;em&gt;d&lt;/em&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;(substitute): 3 = d - 4&lt;br&gt;&#xA;  &lt;strong&gt;d = 7&lt;/strong&gt;&lt;br&gt;&#xA;  (double-check): 136 = 20d - 4&lt;br&gt;&#xA;  140 = 20d&lt;br&gt;&#xA;  &lt;strong&gt;7 = d&lt;/strong&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The final answer was 3, 3 + d, 3 + 2d:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;3, 10, 17&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Which I did a brute force double-check on.&lt;/p&gt;&#xA;" OwnerUserId="8371" LastEditorUserId="8371" LastEditorDisplayName="" LastEditDate="2011-03-17T10:01:26.223" LastActivityDate="2011-03-17T10:01:26.223" Title="Simpler way to determine terms in arithmetic progression" Tags="&lt;homework&gt;&lt;sequences-and-series&gt;" AnswerCount="1" />
  <row Id="27557" PostTypeId="1" CreationDate="2011-03-17T08:45:07.993" Score="3" ViewCount="137" Body="&lt;p&gt;Find two pairs of relatively prime positive integers $(a,c)$ so that $a^2 + 5929 = c^2$.&#xA;Can you find additional pairs with $gcd(a,c) &amp;gt; 1$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&lt;strong&gt;What I know:&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$gcd(a,c) = 1$ implies that there are some $x$ and $y$ such that $ax + cy = 1$. Since $a d$oes not divide $c$, I'm guessing that $a^2$ does not divide $c^2$ as well (need confirmation). In that case we then have $gcd(a^2, c^2) = 1$ so there are some x and y such that $a^2 x + c^2 y = 1$. I'm not 100% sure if that leads us anywhere but it does give an equation that is somewhat matching the question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $c^2 = d$ We know that a number $d$ can be written as a sum of two squares if all its prime factors are either 2 or congruent to $1 (mod  4)$. We have $\sqrt(5929) = 77$. So we have that if $d = a^2 + 5929$, $d$ must be a product of distinct primes that are congruent to $1 (mod  4)$ or 2. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What am I missing from here that is keeping me back from answering this? It doesn't seem like a very difficult question yet I'm having trouble with it. Maybe its midnight speaking :(.&lt;/p&gt;&#xA;" OwnerUserId="6523" LastActivityDate="2011-03-17T08:56:12.490" Title="Sum of two squares proof" Tags="&lt;number-theory&gt;&lt;homework&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="0" />
  <row Id="27558" PostTypeId="2" ParentId="27557" CreationDate="2011-03-17T08:56:12.490" Score="5" ViewCount="" Body="&lt;p&gt;This is a naïve approach that comes to mind.  Seeing squares on opposite sides of an equation makes me want to move one of them and use the difference of squares formula.  That is, I can't resist rewriting that as $5929=(c-a)(c+a)$.  This then leads to the approach of factoring $5929=7^211^2$, leading to all possible choices of $c-a$ and $c+a$.  E.g., choosing $c+a=11^27$ and $c-a=7$ leads to $c=\frac{7}{2}(11^2+1)$ and $a=\frac{7}{2}(11^2-1)$.  Analyzing all such possibilities is one way to find your answer.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2011-03-17T08:56:12.490" CommentCount="3" />
  <row Id="27559" PostTypeId="1" AcceptedAnswerId="27568" CreationDate="2011-03-17T09:34:35.533" Score="2" ViewCount="64" Body="&lt;p&gt;Hey,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have two linear, skew, 3D lines, and I was wondering how I could find a points on each of the lines whereby the distance between the two points are a particular distance apart?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not after the points where the lines are the closest distance apart, nor the furthest distance apart, nor the point at which the lines cross! I'm after the points on the lines where the two lines are a &lt;em&gt;particular&lt;/em&gt; distance apart. I'd like to also be able to change this distance and find the new points.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks very much in advance! :)&lt;/p&gt;&#xA;" OwnerUserId="8372" LastEditorUserId="6296" LastEditorDisplayName="" LastEditDate="2011-03-17T10:47:14.703" LastActivityDate="2011-03-17T11:21:57.903" Title="Finding points on two linear lines which are a particular distance apart." Tags="&lt;linear-algebra&gt;&lt;geometry&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="27560" PostTypeId="2" ParentId="27556" CreationDate="2011-03-17T09:35:26.213" Score="2" ViewCount="" Body="&lt;p&gt;Your approach is good. The only simplification that I can see is to get rid of that rather uninteresting variable $c=-4$ and compute directly in terms of $a_1=3$, by using $a_n = a_1 + (n-1)d$ instead of $a_n = c+nd$.&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2011-03-17T09:35:26.213" CommentCount="1" />
  <row Id="27561" PostTypeId="1" AcceptedAnswerId="27569" CreationDate="2011-03-17T09:36:46.870" Score="0" ViewCount="110" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm a uni student and having a bit of trouble with a homework question in an engineering &quot;Optimisation&quot; class. I have a matrix $N$, where they ask you to find the eigenvalues and eigenvectors, which I have done.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then they ask for you to plot the eigenvalue ($\lambda$), the eigenvector ($z$) and then $\lambda * z$ on an Argand diagram.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know how to plot the eigenvalue, but am having trouble with plotting the eigenvector (being a $2 \times 1$ matrix).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Say the eigenvector is $[(2 + 3i)/5 \, , 1]^{T}$, how would I plot this on an Argand diagram? What does it even mean to plot a matrix on an Argand diagram? Sorry for my lack of formatting, I am not very familiar with LaTeX. Thanks for any help you can provide! :) &lt;/p&gt;&#xA;" OwnerUserId="8373" LastEditorUserId="6296" LastEditorDisplayName="" LastEditDate="2011-03-17T11:39:49.357" LastActivityDate="2011-03-17T11:39:49.357" Title="Plotting eigenvectors on an Argand diagram" Tags="&lt;complex-analysis&gt;&lt;eigenvalues&gt;&lt;eigenvectors&gt;" AnswerCount="1" CommentCount="0" />
  <row Id="27562" PostTypeId="2" ParentId="27541" CreationDate="2011-03-17T09:39:45.167" Score="0" ViewCount="" Body="&lt;p&gt;If you want to convert the $NP^{NP}$ TM to a $NP$ TM you have to check every oracle answer in a $NP$ machine. For 'Yes' answers this is easy because it is a $NP$ oracle. But if you want to check a 'No' answer you have to check that for every nondeterministic choice within the oracle machine the answer is 'No'. This can't work unless $NP = coNP$.&lt;/p&gt;&#xA;" OwnerUserId="4857" LastActivityDate="2011-03-17T09:39:45.167" />
  <row Id="27563" PostTypeId="2" ParentId="27517" CreationDate="2011-03-17T09:52:33.847" Score="2" ViewCount="" Body="&lt;p&gt;Assuming the notation of DJC, apply the transformation $y = n x$ to obtain&#xA;$$&#xD;&#xA;\frac{1}{n^{p-1}} \left\| D_n \right\|^p_p = \int_{- n \pi}^{n \pi} \left| \frac{\sin(y + \frac{y}{2 n})}{n \sin(\frac{y}{2 n})} \right|^p dy .&#xD;&#xA;$$&#xA;Fatou's Lemma now shows that&#xA;$$&#xD;&#xA;\frac{1}{n^{p-1}} \left\| D_n \right\|^p_p \leq \int_{\mathbb{R}} \left| \frac{2 \sin(y)}{y} \right|^p dy .&#xD;&#xA;$$&#xA;For the other side of the inequality, note, that&#xA;$$&#xD;&#xA;\frac{1}{n^{p-1}} \left\| D_n \right\|^p_p \geq \int_{- n \pi}^{n \pi} \left| \frac{\sin(y + \frac{y}{2 n})}{\frac{y}{2}} \right|^p dy,&#xD;&#xA;$$&#xA;so using the triangular inequality of the $L^p$ norm we obtain&#xA;$$&#xD;&#xA;\frac{1}{n^{p-1}} \left\| D_n \right\|^p_p \geq \left( \left( \int_{\mathbb{R}} \left| \frac{2 \sin(y)}{y} \right|^p dy \right)^{1/p} - \left( \int_{- n \pi}^{n \pi} \left| \frac{\sin(y + \frac{y}{2 n}) - \sin y}{\frac{y}{2}} \right|^p dy \right)^{1/p} \right)^p .&#xD;&#xA;$$&#xA;The inequality now follows from&#xA;$$&#xD;&#xA;\int_{- n \pi}^{n \pi} \left| \frac{\sin(y + \frac{y}{2 n}) - \sin y}{\frac{y}{2}} \right|^p dy \leq \int_{- n \pi}^{n \pi} \left| \frac{\frac{y}{2 n}}{\frac{y}{2}} \right|^p dy = \frac{2 \pi}{n^{p - 1}} = o(n) .&#xD;&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="960" LastActivityDate="2011-03-17T09:52:33.847" />
  <row Id="27564" PostTypeId="1" CreationDate="2011-03-17T10:07:07.610" Score="4" ViewCount="75" Body="&lt;p&gt;I don't know if this is the right kind of question for here. &#xA;But, can someone help me find an english translation (a link to a pdf would be nice) of: B. Riemann, &quot;Theorie der Abelschen Funktionen&quot;, J. Reine Angew. Math., 54 (1857)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many Thanks.&lt;/p&gt;&#xA;" OwnerUserId="4806" LastActivityDate="2011-03-17T12:11:35.317" Title="Reference request: Riemann's paper on abelian functions" Tags="&lt;math-history&gt;&lt;riemann-surface&gt;" AnswerCount="1" />
  <row Id="27565" PostTypeId="1" CreationDate="2011-03-17T10:23:08.233" Score="0" ViewCount="76" Body="&lt;p&gt;Suppose that a 3-D surface has the property that $|k_1|\leq 1$ and $|k_2|\leq 1$ everywhere, where $k_1$ and $k_2$ are the principal curvatures. Prove or disprove that the curvature $k$ of a curve on that surface also satisfies $|k|\leq 1$.&lt;/p&gt;&#xA;" OwnerUserId="8252" LastActivityDate="2011-05-24T17:53:25.903" Title="Principal Curvatures of a surface" Tags="&lt;geometry&gt;&lt;differential-geometry&gt;&lt;curvature&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="27566" PostTypeId="1" AcceptedAnswerId="27601" CreationDate="2011-03-17T10:24:06.620" Score="1" ViewCount="50" Body="&lt;p&gt;Hello!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $f(x)$ be a scalar function of a vector $x\in R^n$. If we want to find $x$ such that $f(x)$ is maximized or minimized. I think we can compute the differential of $f(x)$ with respect to $x$ and then let the coefficient of $dx$ be zero. For example, I get the differential of $f(x)$ as&#xA;$$df(x)=g(x)^T dx$$&#xA;where $g(x)$ is a vector function of $x$. Then $g(x)=0$ can give the extrema.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is: if in some context the vector $x$ is a unit vector ($||x||=1$). At this time, $dx$ is constrained. Can we still get the extrema by solving $g(x)=0$? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you very much.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Shiyu&lt;/p&gt;&#xA;" OwnerUserId="7156" LastEditorUserId="1242" LastEditorDisplayName="" LastEditDate="2011-03-17T10:48:11.380" LastActivityDate="2011-03-17T14:08:19.403" Title="Differential of a scalar function with respect to a 'unit-length' vector" Tags="&lt;multivariable-calculus&gt;&lt;differential&gt;" AnswerCount="3" />
  <row Id="27567" PostTypeId="2" ParentId="27478" CreationDate="2011-03-17T10:30:07.963" Score="1" ViewCount="" Body="&lt;p&gt;Hint: This problem has nothing to do with complex analysis. We just have a map&#xA;$$f:\  {\mathbb R}^2\to{\mathbb R}^2, \quad (x,y)\mapsto (u,v):=(x+x^2-y^2, 2xy-y).$$&#xA;The graph of this map lives in ${\mathbb R}^4$ and so is of no help in visualizing $f$. I suggest drawing images of some lines $x=$const. (horizontal parabolas) and $y=$const. (again horizontal parabolas). Along the circle $x^2+y^2={1\over4}$ the Jacobian of $f$ vanishes; this will lead to special effects in the picture.&lt;/p&gt;&#xA;" OwnerUserId="1303" LastActivityDate="2011-03-17T10:30:07.963" />
  <row Id="27568" PostTypeId="2" ParentId="27559" CreationDate="2011-03-17T10:35:07.837" Score="2" ViewCount="" Body="&lt;p&gt;Here is one way to do it. Assuming you have (or are able to find) a parametric equation for the lines on the form&#xA;$$\begin{align}&#xD;&#xA;  l:&#xD;&#xA;  \begin{cases}&#xD;&#xA;    x=x_{l0}+at \\&#xD;&#xA;    y=y_{l0}+bt \\&#xD;&#xA;    z=z_{l0}+ct&#xD;&#xA;  \end{cases},&#xD;&#xA;    \qquad m:&#xD;&#xA;  \begin{cases}&#xD;&#xA;    x=x_{m0}+ds \\&#xD;&#xA;    y=y_{m0}+es \\&#xD;&#xA;    z=z_{m0}+fs&#xD;&#xA;  \end{cases}.&#xD;&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, $P_0=(x_{l0}+at,y_{l0}+bt,z_{l0}+ct)$ is a general point on the line $l$ and $Q_0=(x_{m0}+ds,y_{m0}+es,z_{m0}+fs)$ a point on $m$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can find a vector $\overrightarrow{P_0Q_0}$ from a point on $l$ to a point on $m$:&#xA;$$\overrightarrow{P_0Q_0} = (x_{m0}+ds-x_{l0}-at,y_{m0}+es-y_{l0}-bt,z_{m0}+fs - z_{l0}-at)$$&#xA;The distance of the vector is given by&#xA;$$||\overrightarrow{P_0Q_0}|| = \sqrt{(x_{m0}+ds-x_{l0}-at)^2 +(y_{m0}+es-y_{l0}-bt)^2 + (z_{m0}+fs - z_{l0}-at)^2}.$$&#xA;In other words, if we let $D$ denote your distance, then &#xA;$$D^2=(x_{m0}+ds-x_{l0}-at)^2 +(y_{m0}+es-y_{l0}-bt)^2 + (z_{m0}+fs - z_{l0}-at)^2 .$$&#xA;This may not be the best way to do it. It could be difficult to find a $t$ and an $s$ that fits, and there will probably be several. If you are able to first choose a point on one line, and then find a point on the other line with desired distance, it will be easier and you will only get two solutions (at most).&lt;/p&gt;&#xA;" OwnerUserId="7044" LastActivityDate="2011-03-17T10:35:07.837" CommentCount="7" />
  <row Id="27569" PostTypeId="2" ParentId="27561" CreationDate="2011-03-17T10:40:53.850" Score="0" ViewCount="" Body="&lt;p&gt;What they probably mean is to take the two elements of your vector&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathbf v = \begin{pmatrix} v_1 \\ v_2 \end{pmatrix} = \begin{pmatrix} \frac{2 + 3i}{5} \\ 1 \end{pmatrix}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and plot the elements individually as vectors in the complex plane. That is, you will have one vector to the point $v_1 = \frac{2+3i}{5}$ and one to  $v_2 = 1$. Then mark where your eigenvalue (lets say it's $\lambda = \lambda_x + i\lambda_y$) is. Then multiply your vector by the eigenvalue, which yields a new vector with elements $w_1 = (\lambda_x + i\lambda_y) \frac{2 + 3i}{5}$ and $w_2 = (\lambda_x + i\lambda_y)1$ and mark these in the complex plane as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It will probably be easier if you use polar form (at least if your eigenvalue is not real).&lt;/p&gt;&#xA;" OwnerUserId="5966" LastActivityDate="2011-03-17T10:40:53.850" CommentCount="1" />
  <row Id="27570" PostTypeId="2" ParentId="27566" CreationDate="2011-03-17T10:47:06.077" Score="2" ViewCount="" Body="&lt;p&gt;See &lt;a href=&quot;http://en.wikipedia.org/wiki/Lagrange_multipliers&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Lagrange_multipliers&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2011-03-17T10:47:06.077" CommentCount="1" />
  <row Id="27571" PostTypeId="1" AcceptedAnswerId="27575" CreationDate="2011-03-17T10:47:56.620" Score="1" ViewCount="100" Body="&lt;p&gt;Exercise 3.32 page 119 of &lt;a href=&quot;http://www.stanford.edu/~boyd/cvxbook/bv_cvxbook.pdf&quot; rel=&quot;nofollow&quot;&gt;Convex Optimization&lt;/a&gt; is concerned with the proof that if $f:\mathbb{R}\rightarrow\mathbb{R}:x\mapsto f(x)$ and $g:\mathbb{R}\rightarrow\mathbb{R}:x\mapsto g(x)$ are both convex, nondecreasing (or nonincreasing) and positive, then $h:\mathbb{R}\rightarrow\mathbb{R}:x\mapsto h(x)=f(x)g(x)$ is also convex.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there any generalisations or analogies of this claim to the case where both $f$ and $g$ are convex functions mapping elements from $\mathbb{R}^n$ to $\mathbb{R}$, for any $n&amp;gt;1$?&lt;/p&gt;&#xA;" OwnerUserId="689" LastActivityDate="2011-03-17T13:15:33.913" Title="Convexity of the product of two functions in higher dimensions" Tags="&lt;convex-analysis&gt;&lt;convex-optimization&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="27572" PostTypeId="1" CreationDate="2011-03-17T10:51:31.470" Score="0" ViewCount="96" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Possible Duplicate:&lt;/strong&gt;&lt;br&gt;&#xA;  &lt;a href=&quot;http://math.stackexchange.com/questions/13575/teaching-myself-differential-topology-and-differential-geometry&quot;&gt;Teaching myself differential topology and differential geometry.&lt;/a&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;p&gt;Dear all,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I really need some reference books about differential geometry.&#xA;Recently I am taking a course of representation theory and symmetry method for DE. I feel too difficult to read the materials without the knowledge of differential geometry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to have a reference book on elementary level treating manifolds relevant to differential equations. And it's better with some concrete examples.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you very much.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cheers,&#xA;Jan&lt;/p&gt;&#xA;" OwnerUserId="7920" LastActivityDate="2011-03-17T10:51:31.470" ClosedDate="2011-03-17T12:34:17.017" Title="Books of differential geometry" Tags="&lt;differential-geometry&gt;&lt;books&gt;" CommentCount="3" />
  <row Id="27574" PostTypeId="2" ParentId="11031" CreationDate="2011-03-17T11:06:50.930" Score="1" ViewCount="" Body="&lt;p&gt;Hi,&lt;br&gt;&#xA;To finish the job initiated by  Sivaram Ambikasaran, first you can say that $\mu(x)=\alpha x+ \beta 1/x$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The stochastic integral term with $\beta=0$ case is answered by Sivaram Ambikasaran. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the $\alpha=0$ case you have using Itô's lemma :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$dLnB_t= \frac{1}{B_t}dB_t-\frac{1}{2.B_t^2}dt$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So integrating this gives you :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\beta.\int_0^T \frac{dB_s}{B_s}=\beta.(Ln(\frac{B_T}{B_0})+1/2\int_0^T \frac{1}{B_s^2}ds)$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you start your Brownian Motion at 0, the problem is not well defined. If $B_0\not=0$ then returning to the intial problem (still with $\alpha=0$) you have :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ \int_0^T \mu(B_s) dB_s - \frac{\int_0^T (\mu(B_s))^2 ds}{2} =&#xD;&#xA;Ln(\frac{B_T}{B_0})+\frac{\beta.(1-\beta)}{2}.\int_0^T \frac{1}{B_s^2}ds$ if I am not mistaken.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the general case ($\alpha\not= 0$ and $\beta\not= 0$) just adds the solutions ( i.e. respectively for $\alpha=0$ and $\beta=0$) and also add the cross product term from $\int_0^T (\mu(B_s))^2 ds$ (which should be something like $-T.\alpha.\beta$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regards&lt;/p&gt;&#xA;" OwnerUserId="4437" LastActivityDate="2011-03-17T11:06:50.930" CommentCount="1" />
  <row Id="27575" PostTypeId="2" ParentId="27571" CreationDate="2011-03-17T11:12:51.343" Score="4" ViewCount="" Body="&lt;p&gt;Yes a generalization is possible. Here is an elementary approach to the convexity of the product of two nonnegative convex functions defined on a convex domain of $\mathbb{R}^n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Choose $x$ and $y$ in the domain and $t$ in $[0,1]$. Your aim is to prove that $\Delta\ge0$ with&#xA;$$\Delta=t(fg)(x)+(1-t)(fg)(y)-(fg)(tx+(1-t)y).&#xD;&#xA;$$&#xA;But $f$ and $g$ are nonnegative and convex, hence&#xA;$$&#xD;&#xA;(fg)(tx+(1-t)y)\le(tf(x)+(1-t)f(y))(tg(x)+(1-t)g(y)).&#xD;&#xA;$$&#xA;Using this and some easy algebraic manipulations, one sees that $\Delta\ge t(1-t)D(x,y)$ with&#xA;$$&#xD;&#xA;D(x,y)=f(x)g(x)+f(y)g(y)-f(x)g(y)-f(y)g(x),&#xD;&#xA;$$&#xA;that is,&#xA;$$&#xD;&#xA;D(x,y)=(f(x)-f(y))(g(x)-g(y)).&#xD;&#xA;$$&#xA;This proves a generalization of the result you quoted to any product of convex nonnegative functions $f$ and $g$ such that $D(x,y)\ge0$ for every $x$ and $y$ in the domain of $f$ and $g$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, if $f$ and $g$ are differentiable at a point $x$ in the domain, one asks that their gradients $\nabla f(x)$ and $\nabla g(x)$ are such that $z^*M(x)z\ge0$ for every $n\times 1$ vector $z$, where $M(x)$ is the $n\times n$ matrix&#xA;$$&#xD;&#xA;M(x)=\nabla f(x)\cdot(\nabla g(x))^*.&#xD;&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="6179" LastEditorUserId="6179" LastEditorDisplayName="" LastEditDate="2011-03-17T13:15:33.913" LastActivityDate="2011-03-17T13:15:33.913" CommentCount="3" />
  <row Id="27576" PostTypeId="2" ParentId="27559" CreationDate="2011-03-17T11:21:57.903" Score="1" ViewCount="" Body="&lt;p&gt;In general, this is not possible for all distances. If the lines do not cross, there willl be a minimum distance between the lines which you can never reach below, but you should be able to solve it for all distances equal to or larger than this minimum distance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the general case has been shown by Eivind, I will take the special case where the lines do cross, so a solution is guaranteed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the lines, say $l$ and $m$, do cross, you can write them as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xD;&#xA;l(t)&amp;amp;= \mathbf a + t \mathbf b \\&#xD;&#xA;m(s)&amp;amp;= \mathbf a + s \mathbf c&#xD;&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The distance between two lines will be $\|l(t) - m(s)\| = \| t \mathbf b - s \mathbf c\|$. Say you want this to be equal to $d$, then you can solve for $d^2$ (I am here assuming that you are working in $\mathbb R^3$):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xD;&#xA;d^2 &amp;amp;= \langle \mathbf b t - \mathbf c s,  \mathbf b t - \mathbf c s \rangle = \langle \mathbf b t, \mathbf b t - \mathbf c s \rangle - \langle \mathbf c s,  \mathbf b t - \mathbf c s \rangle \\&#xD;&#xA;&amp;amp;= \langle \mathbf b t, \mathbf b t \rangle - 2 \langle \mathbf b t, \mathbf c s \rangle + \langle \mathbf c s, \mathbf c s \rangle = \\&#xD;&#xA;&amp;amp;= t^2\|b\|^2 + s^2 \|c\|^2 - 2 \langle \mathbf b t, \mathbf c s \rangle&#xD;&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have normalized $\mathbf b$ and $\mathbf c$, this will reduce to:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$d^2 = t^2 + s^2 - 2 (b_1c_1 + b_2c_2 + b_3c_3) ts$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which you can solve. You can here choose a value for $t$ (which will tell you how far the crossing point the point on line $l$ is) and then solve for $s$. If you want the points to be on equal distance from the crossing point, take $t = s$.&lt;/p&gt;&#xA;" OwnerUserId="5966" LastActivityDate="2011-03-17T11:21:57.903" />
  <row Id="27577" PostTypeId="1" AcceptedAnswerId="27590" CreationDate="2011-03-17T11:38:18.013" Score="6" ViewCount="175" Body="&lt;p&gt;Hello:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I ran across this confounding limit I am wondering about.  It is as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle \lim_{n\to \infty}\frac{1-(1-c^{n})^{2n}}{(1-c)^{2n}}, \;\ 0&amp;lt;c&amp;lt;1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I played around with this on Maple and found that if c is less than approximately .382 (but &lt;/p&gt;&#xA;&#xA;&lt;p&gt;greater than 0), it converges to 0. If c is greater than .382 (but less than 1), it &lt;/p&gt;&#xA;&#xA;&lt;p&gt;diverges. What is it about .382?.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;.382 is an approximation. By playing around more I could have taken it out to more decimal places.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The actual problem asks to prove that the above limit is &amp;lt; $\frac{1}{p(n)}$, where p(n) is a polynomial. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was mainly wondering how to solve the limit and why .382 is so significant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you all very much. You are always a big help.&lt;/p&gt;&#xA;" OwnerUserId="6358" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2011-03-17T16:58:28.143" LastActivityDate="2011-03-17T16:58:28.143" Title="&quot;Boundary&quot; of convergence of  $\frac{1-(1-c^{n})^{2n}}{(1-c)^{2n}}$" Tags="&lt;real-analysis&gt;&lt;limit&gt;" AnswerCount="3" CommentCount="4" />
  <row Id="27578" PostTypeId="1" CreationDate="2011-03-17T11:42:33.687" Score="2" ViewCount="147" Body="&lt;p&gt;If A has eigenvector $\mathbf{v}_1$ so that $A\mathbf{v}_1=\lambda_1\mathbf{v}_1$and B has eignenvector  $\mathbf{v}_2$ so that $B\mathbf{v}_2=\lambda_2\mathbf{v}_2$, then what can you say about AB? can you say $AB\mathbf{v_3}=\lambda_3\mathbf{v}_3$? and what would be the relationship between $\mathbf{v}_1,\mathbf{v}_2,\mathbf{v}_3$ and what would be the relationship between $\lambda_1,\lambda_2,\lambda_3$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;&#xA;$A,B$ are 3 by 3 matrices and $\lambda_1,\lambda_2,\lambda_3$ can be real or complex numbers and $\mathbf{v}_1,\mathbf{v}_2,\mathbf{v}_3$ is a triple.&lt;/p&gt;&#xA;" OwnerUserId="7340" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-17T13:18:12.313" LastActivityDate="2011-03-17T15:22:03.237" Title="the relationship between eigenvectors and matrix multiplication" Tags="&lt;homework&gt;&lt;linear-algebra&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="27579" PostTypeId="1" AcceptedAnswerId="27818" CreationDate="2011-03-17T11:45:28.990" Score="4" ViewCount="85" Body="&lt;p&gt;Ciao!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $A$ be a finite abelian group, and let &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ \psi : A \times A \to \mathbb{Q}/\mathbb{Z} $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;be an alternating, non-degenerate bilinear form on $A$. Maybe I should say what I mean by these words; bilinear means it is linear in each argument separately; alternating means that $\psi(a,a) = 0$ for all $a$; non-degenerate means that, if $\psi(a,b) = 0$ for all $b$, then $a$ must be $0$.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Why must $A$ have square cardinality?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I believe it will follow from the following theorem in Linear algebra:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Theorem. Let $V$ be a finite dimensional vector space over a field $K$ that has an alternating, non-degenerate bilinear form on it (from $V \times V \to K$). Then dim $V$ is even. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My idea was to proceed as follows: If the size of $A$ is not square, then for some prime $p$, $A(p)$ is not square, where $A(p)$ means the $p$-primary part of $A$. The original $\psi$ induces a map on $A(p)$ that is non-degenerate, alternating and bilinear. I then wanted to say that $A(p)$ is an $\mathbb{F}_p$-vector space, and then applying the theorem I am done, but this is not true, e.g, $\mathbb{Z}/25\mathbb{Z}$ is not an $\mathbb{F}_5$-vector space. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any pointers anyone? &lt;/p&gt;&#xA;" OwnerUserId="8374" LastActivityDate="2011-03-18T15:49:57.750" Title="Non-degenerate alternating bilinear form on a finite abelian group" Tags="&lt;linear-algebra&gt;&lt;abstract-algebra&gt;&lt;abelian-groups&gt;" AnswerCount="1" FavoriteCount="2" />
  <row Id="27580" PostTypeId="1" CreationDate="2011-03-17T11:46:14.413" Score="0" ViewCount="40" Body="&lt;p&gt;I have a heartrate signal which is a few hours long. I need to preform a spectral analysis to get the ratio high frequency (0.15Hz-0.4Hz) to low frequency (0.04Hz - 0.15Hz) changes in heartrate within a window.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maths is not my natrual habitat, but am I right using an FFT and and summing the absolute values of the bins that lie inside my frequency range?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't expect a full blown explanation of how fft works, but links to any tuts would be nice. I know what an FFT does, I just don't know how to use it.&lt;/p&gt;&#xA;" OwnerUserId="8375" LastActivityDate="2011-03-17T11:46:14.413" Title="Extracting the power of a frequency range in a signal." Tags="&lt;fourier-analysis&gt;&lt;signal-processing&gt;" CommentCount="6" FavoriteCount="2" />
  <row Id="27581" PostTypeId="2" ParentId="27555" CreationDate="2011-03-17T12:07:00.407" Score="1" ViewCount="" Body="&lt;p&gt;What you did wrong: the function $f$ should be treated as $f = f(\gamma(t)) = f(u(t),v(t))$. So $\frac{d}{dt}f = \partial_u f \frac{d}{dt}u + \partial_vf\frac{d}{dt}v$ using the chain rule. By the parametrization, you $\partial_uf = 0$, while $\partial_v f$ is what you wrote $f&amp;#39;$ originally. (Whereas in the above you implicitly wrote $\frac{d}{dt}f^2 = 2f f&amp;#39;$, which is wrong.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is an instance where you let notation get in your way. If would be clearer if you reserve the $\prime$ for $f&amp;#39;$, the derivative of the function $f$ relative to the parameter $v$ and $g&amp;#39;$ for the derivative of the function $f$ relative to parameter $v$, and use explicitly either $\frac{d}{dt}$ or $\cdot$ to denote time derivatives along the geodesic (as the geodesic equations that you originally copied down does). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For deriving the &quot;conservation of energy&quot;, it may help if instead of the second equation in the two geodesic equations (also, I think there may be a sign error in it, but I am not 100% certain), you look at that equation multiplied by $(f&amp;#39;^2 + g&amp;#39;^2)\frac{dv}{dt}$, that is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ (f&amp;#39;^2 + g&amp;#39;^2) \dot{v} \ddot{v} + ff&amp;#39;\dot{v}(\dot{u})^2 + (f&amp;#39;f&amp;#39;&amp;#39; + g&amp;#39;g&amp;#39;&amp;#39;)(\dot{v})^3 = 0 $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Incidentally, the second thing about angle against the parallels is called &quot;Clairaut's relation&quot;. &lt;/p&gt;&#xA;" OwnerUserId="1543" LastActivityDate="2011-03-17T12:07:00.407" CommentCount="1" />
  <row Id="27582" PostTypeId="2" ParentId="27564" CreationDate="2011-03-17T12:11:35.317" Score="1" ViewCount="" Body="&lt;p&gt;I doubt you will find it as a free pdf.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other possibilities include purchasing an English translation of &lt;a href=&quot;http://www.kendrickpress.com/Riemann.htm&quot; rel=&quot;nofollow&quot;&gt;Bernhard Riemann, Collected Papers&lt;/a&gt; which includes it.  Or you can double read the &lt;a href=&quot;http://www.maths.tcd.ie/pub/HistMath/People/Riemann/AbelFn/AbelFn.pdf&quot; rel=&quot;nofollow&quot;&gt;original in German&lt;/a&gt; and a &lt;a href=&quot;http://translate.google.co.uk/translate?js=n&amp;amp;prev=_t&amp;amp;hl=en&amp;amp;ie=UTF-8&amp;amp;layout=2&amp;amp;eotf=1&amp;amp;sl=de&amp;amp;tl=en&amp;amp;u=http://www.maths.tcd.ie/pub/HistMath/People/Riemann/AbelFn/AbelFn.pdf&amp;amp;act=url&quot; rel=&quot;nofollow&quot;&gt;machine translation into English&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="6460" LastActivityDate="2011-03-17T12:11:35.317" CommentCount="3" />
  <row Id="27583" PostTypeId="2" ParentId="27524" CreationDate="2011-03-17T12:22:42.020" Score="3" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; I've added some information about the game with a restricted number $N$ of rounds, since the OP mentioned this. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Just a comment to complement the nice answers we have already.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is an algorithm that calculates the optimal strategy and &#xA;the value of each state for such an optimal stopping problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $f(x)$ be the payout&#xA;at each state $x\in {\cal S}$ for the underlying Markov chain $(X_n)$, and let &#xA;$g(x)$ be the cost of leaving state $x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Set $u_0=f$ and for $N\geq 1$ put $u_N=\max(Pu_{N-1}-g,f)$. Here $P$ &#xA;is the transition matrix of the Markov chain. &#xA;Then $u_N$ is the value function for the restricted game with at most $N$ rounds; that is, &#xA;$$u_N(x)=\sup_{T\leq N}\ \mathbb{E}\left[ f(X_T)\, | \, X_0=x\right],$$&#xA;where the supremum is over all stopping times $T$ that satisfy $T\leq N$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As $N\to\infty$, we get $u_N\uparrow v$&#xA;where $v$ is the value function for the unrestricted game. The optimal strategy &#xA;is to stop when the chain hits the set $\lbrace x: f(x)=v(x)\rbrace$.&#xA;Here $v(x)$ means the &lt;em&gt;value&lt;/em&gt; of state $x$, i.e.,&#xA;$v(x)$ is the maximum expected payout starting in state $x$ and using&#xA;&lt;em&gt;any&lt;/em&gt; finite stopping time $T$ as a strategy: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$v(x)=\sup_T\ \mathbb{E}\left[ f(X_T)\, | \, X_0=x\right].$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is all nicely explained in the section on Optimal Stopping in &#xA;Gregory Lawler's book &quot;Introduction to Stochastic Processes&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your problem we have $f(x)=x$ for $1\leq x\leq 100$, and $g(x)\equiv 1$.&#xA;Your Markov chain $(X_n)$ is a sequence of independent, uniform $\lbrace 1,2,\dots,  100\rbrace $ random variables. &#xA;Thus the $P$ operator applied to $h$ gives the constant function whose value is the average value of $h$, that is, $Ph(x)\equiv \sum_{i=1}^{100} h(i)/100$. So we calculate &#xA;$$u_0(x)=x,\quad u_1(x)=\max\left(x,{99\over 2}\right), \quad u_2(x)=\max\left(x,{12301\over 200}\right). $$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Taking very large $N$ gives $$v(x)\approx \max\left(x,{86.35714}\right),$$&#xA;which shows that the optimal strategy (over all finite stopping times!) is to quit&#xA;as soon as we get $87$ or higher, and the value of the game is $86.35714$ dollars.    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your problem it is pretty straightforward to calculate the exact answer, &#xA;but this algorithm also gives the answer for more complicated games where &#xA;exact calculations are not so easy.        &lt;/p&gt;&#xA;" OwnerUserId="940" LastEditorUserId="940" LastEditorDisplayName="" LastEditDate="2011-03-17T17:42:59.960" LastActivityDate="2011-03-17T17:42:59.960" />
  <row Id="27584" PostTypeId="1" CreationDate="2011-03-17T04:58:47.257" Score="1" ViewCount="54" Body="&lt;p&gt;I met the question in my exam:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;flip one coin, the probability of head is 1/3, Pr{tail} = 2/3. Flip the coin twice, what are the four possible outcomes and their probabilities?&lt;/p&gt;&#xA;" LastActivityDate="2011-03-17T12:25:24.223" Title="basic probability question" Tags="&lt;probability&gt;" AnswerCount="2" CommentCount="0" />
  <row Id="27585" PostTypeId="2" ParentId="27584" CreationDate="2011-03-17T05:02:33.100" Score="3" ViewCount="" Body="&lt;pre&gt;&lt;code&gt;Head Head 1/3 * 1/3 = 1/9&#xA;Head Tail 1/3 * 2/3 = 2/9&#xA;Tail Head 2/3 * 1/3 = 2/9&#xA;Tail Tail 2/3 * 2/3 = 4/9&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;By the way, the results of 1st and 2nd flips are &lt;a href=&quot;http://en.wikipedia.org/wiki/Independence_%28probability_theory%29&quot; rel=&quot;nofollow&quot;&gt;independent&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="6023" LastActivityDate="2011-03-17T05:02:33.100" />
  <row Id="27586" PostTypeId="2" ParentId="27584" CreationDate="2011-03-17T05:02:37.860" Score="2" ViewCount="" Body="&lt;pre&gt;&lt;code&gt;Head Head: 1/3 * 1/3 = 1/9&#xA;Head Tail: 1/3 * 2/3 = 2/9&#xA;Tail Head: 2/3 * 1/3 = 2/9&#xA;Tail Tail: 2/3 * 2/3 = 4/9&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" LastActivityDate="2011-03-17T05:02:37.860" CommentCount="0" />
  <row Id="27587" PostTypeId="2" ParentId="27546" CreationDate="2011-03-17T12:28:45.813" Score="3" ViewCount="" Body="&lt;p&gt;As others have said, it comes down to the nonexistence of an uncountable discrete subset $Z$ of an open subset $U$ of the complex plane $\mathbb{C}$.  The point of this answer is to record a proof of this that I find simplest.  Namely, the space $\mathbb{C}$ is &lt;a href=&quot;http://en.wikipedia.org/wiki/Second-countable_space&quot; rel=&quot;nofollow&quot;&gt;second countable&lt;/a&gt;: there is a countable base for the topology (take, e.g., open balls with rational radii centered at points $x+yi$ with $x,y \in \mathbb{Q})$).  But every subspace of a second countable space is second countable: just restrict a countable base to the subspace.  In particular $U$ is second countable and so is the putative uncountable discrete subset $Z$ of $U$.  But this is absurd: since the only base of a discrete space consists of all the singleton sets, an uncountable discrete space is not second-countable!&lt;/p&gt;&#xA;" OwnerUserId="299" LastActivityDate="2011-03-17T12:28:45.813" />
  <row Id="27588" PostTypeId="1" CreationDate="2011-03-17T12:49:25.447" Score="2" ViewCount="61" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Possible Duplicate:&lt;/strong&gt;&lt;br&gt;&#xA;  &lt;a href=&quot;http://math.stackexchange.com/questions/24429/on-the-boundary-of-analytic-functions&quot;&gt;on the boundary of analytic functions&lt;/a&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;p&gt;I was thinking of a problem I will feel happy if you&#xA;give hint to solve this. Suppose I have a function $f$ that is analytic&#xA;on the unit disk $D=\{z \in \mathbb{C}:|z|&amp;lt;1\}$ that extends continuously to $\overline{D}$. If $f$ is identically zero on some segment of of the boundary,&#xA;is it then true that $f$ is identically zero on the entire boundary?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that analytic functions that are zero at an&#xA;accumulation point inside the domain of analyticity are identically&#xA;zero throughout the entire domain, but I don't know what (if anything)&#xA;can be said if something similar occurs on the boundary of the domain.&lt;/p&gt;&#xA;" OwnerUserId="8378" LastEditorUserId="5363" LastEditorDisplayName="" LastEditDate="2011-03-17T12:59:47.577" LastActivityDate="2011-03-17T12:59:47.577" ClosedDate="2011-03-17T13:31:20.663" Title="analytic function which is zero on a segment of a circle" Tags="&lt;complex-analysis&gt;" CommentCount="2" />
  <row Id="27590" PostTypeId="2" ParentId="27577" CreationDate="2011-03-17T13:04:08.510" Score="4" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Edited&lt;/strong&gt; to use Christian Blatter's simpler formula&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$0.38196...$ is $(3-\sqrt{5})/2$, which is a root of the equation $x^2-3x+1=0$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When $n$ is large, your numerator is about $2nc^n$, so your expression is about  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$2n(c/(1-c)^2)^n$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $c \in (0,1)$, this tends to zero if and only if $c/(1-c)^2 &amp;lt; 1$, giving:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$c &amp;lt; (1-c)^2$&lt;br&gt;&#xA;$c^2-3c+1&amp;gt;0$&lt;br&gt;&#xA;$c &amp;lt; (3-\sqrt{5})/2$.&lt;/p&gt;&#xA;" OwnerUserId="1508" LastEditorUserId="1508" LastEditorDisplayName="" LastEditDate="2011-03-17T14:39:18.800" LastActivityDate="2011-03-17T14:39:18.800" CommentCount="4" />
  <row Id="27591" PostTypeId="1" AcceptedAnswerId="27599" CreationDate="2011-03-17T13:04:48.850" Score="3" ViewCount="63" Body="&lt;p&gt;I have the following equation as a question for homework: $\lfloor 2x \rfloor = 2\lfloor x\rfloor$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know what the solution is by deducting to the possibilities.&#xA;Meaning this equation is true for any x which is between n (an integer) and $n+y$ where $0\leq y &amp;lt; \frac{1}{2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;putting it simple:&#xA;\begin{align}&#xA;x \in \lbrace n + y | n \in \mathbb{Z} , y \in [0, 0.5) \rbrace&#xA;\end{align}&#xA;I just don't know how to algebraically get to this solution.&#xA;Help will be appreciated!&lt;/p&gt;&#xA;" OwnerUserId="7295" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-17T13:46:59.340" LastActivityDate="2011-03-17T13:53:18.447" Title="Algebraic notation of floor function in an equation" Tags="&lt;homework&gt;&lt;algebra-precalculus&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="27592" PostTypeId="1" CreationDate="2011-03-17T13:08:07.060" Score="0" ViewCount="67" Body="&lt;p&gt;I am wondering is xy/7 a group. I think it may be and I found the identity to be 7 and found inverses for each element. Mark&lt;/p&gt;&#xA;" OwnerUserId="8379" LastActivityDate="2011-03-17T13:08:07.060" Title="groups and checking group properties" Tags="&lt;group-theory&gt;" CommentCount="3" />
  <row Id="27593" PostTypeId="1" AcceptedAnswerId="27597" CreationDate="2011-03-17T13:09:44.677" Score="2" ViewCount="175" Body="&lt;p&gt;What's the numerical method? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't get &lt;a href=&quot;http://www.wolframalpha.com/input/?i=n%5En+=+2%5E1000&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For c=1000, n should be aprox. 140, right?&lt;/p&gt;&#xA;" OwnerUserId="4474" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-17T13:16:22.253" LastActivityDate="2011-03-17T16:38:40.133" Title="How to solve $n$ for $n^n = 2^c$?" Tags="&lt;algebra&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="27594" PostTypeId="2" ParentId="27577" CreationDate="2011-03-17T13:11:55.787" Score="3" ViewCount="" Body="&lt;p&gt;Hint: Use Bernoulli's inequality to estimate $(1-c^n)^{2n}$. You will then find out that the  critical value is $c_0:=(3-\sqrt{5})/2\doteq 0.382$. For $c_0\leq c&amp;lt;1$ an estimate in the other direction is required. Maybe a suitable Taylor approximation of $(1+x)^{2n}$ for small $|x|$ helps.&lt;/p&gt;&#xA;" OwnerUserId="1303" LastActivityDate="2011-03-17T13:11:55.787" CommentCount="1" />
  <row Id="27595" PostTypeId="2" ParentId="27566" CreationDate="2011-03-17T13:14:41.400" Score="0" ViewCount="" Body="&lt;p&gt;It seems a quite simple problem if we use Lagrange multiplier. But I have some other methods which seem very weird. Can anybody help me to fix them?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Weird solution 1: Using a rotation matrix.~~~~~~~~~~~~~~~~~~~~~~~~~&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the length of $x$ is fixed, $x$ can vary on a sphere. That means any variance of $x$ is caused by a rotation matrix $R$.&#xA;$$\delta x = Rx-x=(R-I)x$$&#xA;According to rotation axis-angle formula, a rotation matrix can be written as&#xA;$$R=e^{[w]_{\times}}$$&#xA;where $w$ is the product of the rotation axis and rotation angle. $[w]_{\times}$ is the associating skew-symmetric matrix of the vector $w$. The first order Taylor expansion of $R$ is&#xA;$$R=I+[w]_{\times}$$&#xA;Hence we have&#xA;$$dx=(R-I)x=[dw]_{\times}x=[x]_{\times}dw$$&#xA;Now&#xA;$$df(x)=g(x)^Tdx=g(x)^T[x]_{\times}dw$$&#xA;Since $dw$ has no constraints, solving the following equation can give the extrema. &#xA;$$g(x)^T [x]_{\times}=0$$&#xA;That is the cross product&#xA;$$g(x) \times x=0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Weird solution 2:~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$||x||=1$, $dx$ is not free. But we can find a free vector $y$ such that $x=\frac{y}{||y||}$. Then $dy$ should be free. And we have&#xA;$$dx=\frac{dy||y||-yd||y||}{||y||^2}=\frac{1}{||y||}\left( I-xx^T \right) dy$$&#xA;Therefore&#xA;$$df(x)=g(x)^Tdx=\frac{1}{||y||}g(x)^T\left( I-xx^T \right) dy$$&#xA;Since $dy$ has no constraints, solving the following equation can give the extrema. &#xA;$$\frac{1}{||y||}g(x)^T\left( I-xx^T \right)=0 $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Normal Solution 3: using Lagrange multiplier.~~~~~~~~~~~~~~~~~~~~~~~~~&#xA;$$d[f(x)+\lambda (x^T x-1)]=\left( g(x)^T+2\lambda x^T \right)dx$$&#xA;So we have&#xA;$$g(x)+2\lambda x=0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remark: Note the solution 1 is the same as the solution 3, because both of them just mean $g(x)$ is collinear with $x$.&lt;/p&gt;&#xA;" OwnerUserId="7156" LastActivityDate="2011-03-17T13:14:41.400" />
  <row Id="27596" PostTypeId="2" ParentId="27593" CreationDate="2011-03-17T13:20:21.310" Score="5" ViewCount="" Body="&lt;p&gt;Hint: Consider &lt;a href=&quot;http://en.wikipedia.org/wiki/Lambert_W_function&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint 2: First take $\log$ on both sides.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And explicitly: The solution to your question is given by&#xA;$$&#xD;&#xA;n = e^{W(c\log 2)}  = \frac{{c\log 2}}{{W(c\log 2)}}.&#xD;&#xA;$$&#xA;For $c=1000$, this gives $n \approx 140.2217$.&#xA;The function $W$ is standard (&lt;a href=&quot;http://reference.wolfram.com/mathematica/ref/ProductLog.html&quot; rel=&quot;nofollow&quot;&gt;ProductLog&lt;/a&gt; in Wolfram Mathematica).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: For large $c$, a rough but very simple approximation to the solution $n$ of $n^n = 2^c$ can be obtained as follows (cf. &lt;a href=&quot;http://math.stackexchange.com/questions/27355/how-does-lamberts-w-behave-near/27372#27372&quot;&gt;this&lt;/a&gt;, also for improvement of the approximation):&#xA;$$&#xD;&#xA;n \approx (c\log 2)[\log (c\log 2)]^{1/\log (c\log 2) - 1} .&#xD;&#xA;$$&#xA;For example, for $c=1000$ this gives $n \approx 141.2083$, not far from the exact value of about $140.2217$.&lt;/p&gt;&#xA;" OwnerUserId="2810" LastEditorUserId="2810" LastEditorDisplayName="" LastEditDate="2011-03-17T15:05:27.923" LastActivityDate="2011-03-17T15:05:27.923" CommentCount="2" />
  <row Id="27597" PostTypeId="2" ParentId="27593" CreationDate="2011-03-17T13:26:48.610" Score="3" ViewCount="" Body="&lt;p&gt;Alpha sometimes goes off into the complex plane when what you want is only the reals.  I agree with you and get about 140.222.  If you ask to solve n(ln(n))=1000 ln(2) you get what you want.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-03-17T13:26:48.610" />
  <row Id="27598" PostTypeId="1" CreationDate="2011-03-17T13:38:02.557" Score="2" ViewCount="53" Body="&lt;p&gt;To iteratively solve large linear systems, many current state-of-the-art methods work by finding approximate solutions in successively larger (Krylov) subspaces. Are there similar iterative methods for solving systems of polynomial equations by finding approximate solutions on successively larger algebraic sets?&lt;/p&gt;&#xA;" OwnerUserId="3060" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2011-04-18T12:41:43.553" LastActivityDate="2011-04-18T12:41:43.553" Title="Krylov-like method for solving systems of polynomials?" Tags="&lt;algebraic-geometry&gt;&lt;polynomials&gt;&lt;numerical-methods&gt;&lt;computational-mathematics&gt;" FavoriteCount="1" />
  <row Id="27599" PostTypeId="2" ParentId="27591" CreationDate="2011-03-17T13:53:18.447" Score="5" ViewCount="" Body="&lt;p&gt;The best I can come up with is to replace manipulation of the floor function with manipulation of the fractional part function:&#xA;$$\{x\} = x-\lfloor x\rfloor.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then you have $\lfloor 2x\rfloor = 2\lfloor x \rfloor$ if and only if $2x- \{2x\}  = 2(x - \{x\})$, which holds if and only if $2\{x\} = \{2x\}$, which holds if and only if $0\leq 2\{x\}\lt 1$, which holds if and only if $0\leq \{x\}\leq \frac{1}{2}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would that do?&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2011-03-17T13:53:18.447" CommentCount="2" />
  <row Id="27600" PostTypeId="2" ParentId="27535" CreationDate="2011-03-17T14:00:28.190" Score="1" ViewCount="" Body="&lt;p&gt;Let $(x,y)$ be the center of the arc. For $i=1,2$ write the equation&#xA;$$&#xA;\text{distance from $(x,y)$ to $(x_i,y_i)$}=r.&#xA;$$&#xA;You get a system of two quadratic equations in the two unknowns $x$ and $y$. Solve it to find two solutions, each on a different side of the segment from $(x_1,y_1)$ to $(x_2,y_2)$. Choose the one that corresponds to the direction of the arc.&lt;/p&gt;&#xA;" OwnerUserId="4791" LastActivityDate="2011-03-17T14:00:28.190" />
  <row Id="27601" PostTypeId="2" ParentId="27566" CreationDate="2011-03-17T14:08:19.403" Score="3" ViewCount="" Body="&lt;p&gt;This really is a comment to &lt;a href=&quot;http://math.stackexchange.com/questions/27566/differential-of-a-scalar-function-with-respect-to-a-unit-length-vector/27595#27595&quot;&gt;Shiyu's &quot;weird methods&quot;&lt;/a&gt;, but is too long, and contains too much mathematics to fit into the comment box. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Diagrammatically, the three different methods can be captured by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \mathbb{S}^n \overset{i}{\longrightarrow} \mathbb{R}^{n+1}\setminus \{0\} \overset{j}{\longrightarrow} \mathbb{S}^n $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $i$ is the inclusion map of the unit sphere into Euclidean space, and $j$ is the projection map $j(y) = y / \|y\|$. Note that $j\circ i$ is the identity map on the sphere. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given a function $f:\mathbb{R}^{n+1}\to\mathbb{R}$. You want to find its maximum on the unit sphere. Geometrically your methods correspond to&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Method 1: consider instead the function $\tilde{f} = f \circ i : \mathbb{S}^n \to \mathbb{R}$. Now $\tilde{f}$ is a function on the sphere, which is a smooth manifold. So the critical points of $\tilde{f}$ corresponds to $d\tilde{f} = 0$ where the derivative is taken on the smooth manifold. (As it happens, the sphere is extra nice, in the sense that the set of rotations gives rise to a set of vectors that span the tangent space at every point. Since $d\tilde{f} = 0$ is equivalent to the directional derivatives of $\tilde{f}$ relative to any tangent vector being 0.) Now, by the chain rule, $d\tilde{f} = df|_{i(\mathbb{S}^{n})} \cdot di$. $di$ being the differential of the inclusion map, it spans precisely the tangent plane to the unit sphere. So this says that $df$, when restricted to the tangent plane to the unit sphere, must be 0. And this is precisely $g(x)\times x = 0$. In general, for an arbitrary submanifold $S$ of $\mathbb{R}^{n+1}$. You can do the same canonical inclusion and get that $f|_{S}$ has a critical point if and only if $df$ is normal to the surface $S$ at the point. (Unlike the sphere, the normals are not so nicely parametrized in the general case.)&lt;/li&gt;&#xA;&lt;li&gt;Method 2: you are looking at the map $\bar{f} :\mathbb{R}^{n+1}\setminus\{0\}\to \mathbb{R}$ given by $\bar{f} = f \circ i \circ j$. At any critical point, $d(f\circ i) = 0$ as shown above, and hence $d\bar{f}$ will also be zero. But doing the expansion a different way, you have&#xA;$$ d\bar{f} = df \cdot d(i\circ j) = 0$$&#xA;(where $i\circ j:\mathbb{R}^{n+1}\setminus\{0\} \to \mathbb{R}^{n+1}\setminus\{0\}$ is given by $i\circ j(y) = y / \|y\|$). The computation give what you wrote. Notice that &#xA;$$ I - xx^t $$ &#xA;for $\|x\| = 1$ is a projection matrix to the orthogonal complement of $x$. So indeed, the second method also gives rise to the statement that $df$ restricted to the tangent space of the sphere (remember that for the sphere, its tangent is orthogonal to the radial vector) is 0. Which is the same conclusion as method 1. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Geometrically the method of Lagrange multipliers and the two methods described above are completely identical. They just represent different ways to compute the same thing. &lt;/p&gt;&#xA;" OwnerUserId="1543" LastActivityDate="2011-03-17T14:08:19.403" CommentCount="1" />
  <row Id="27602" PostTypeId="1" CreationDate="2011-03-17T14:12:33.707" Score="-1" ViewCount="164" Body="&lt;p&gt;This problem is Proposition 3.13 c. ( or Exercise 18) of Folland's &quot;real analysis: modern techniques and their applications&quot; in page 94. I just have no idea how to prove it, can you help me? Please use notions and definitions in this book. Thanks!&#xA;The following images are section 3.3 that contains this problem and related notions and definitions.&#xA;&lt;img src=&quot;http://i.imgur.com/8mke0.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/NcBiR.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="7321" LastActivityDate="2011-03-17T15:39:29.263" Title="A problem about total variation of complex measure" Tags="&lt;real-analysis&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="27603" PostTypeId="2" ParentId="27577" CreationDate="2011-03-17T14:16:01.987" Score="3" ViewCount="" Body="&lt;p&gt;My previous comment was wrong of course. One can solve the problem using L'Hopital's rule. $0.38196$ appears as the solution to the equation $x^2-3x+1=0$ and equals $\frac{3-\sqrt{5}}{2}$ to be precise. &lt;/p&gt;&#xA;" OwnerUserId="4623" LastActivityDate="2011-03-17T14:16:01.987" CommentCount="3" />
  <row Id="27605" PostTypeId="2" ParentId="27535" CreationDate="2011-03-17T14:22:40.387" Score="0" ViewCount="" Body="&lt;p&gt;To be explicit, your equations are $r^2=(x-x_0)^2+(y-y_0)^2$ and $r^2=(x-x_1)^2+(y-y_1)^2$  Subtracting them gives the bisector line:  $0=2x(x_1-x_0)+x_0^2-x_1^2+2y(y_1-y_0)+y_0^2-y_1^2$.  You can solve any two of these simultaneously to get the two candidate centers.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-03-17T14:22:40.387" CommentCount="4" />
  <row Id="27606" PostTypeId="1" AcceptedAnswerId="27607" CreationDate="2011-03-17T14:22:41.323" Score="2" ViewCount="85" Body="&lt;p&gt;Let $M$ be an off-line Turing machine over the input alphabet $\{0,1\}^{*}$, that uses only one working tape in addition to the input tape. Construct a Turing machine $M&amp;#39;$, such that:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$L(M) = L(M&amp;#39;)$&lt;/li&gt;&#xA;&lt;li&gt;$M&amp;#39;$ never loops in a bounded space&#xA;(that is, $M&amp;#39;(w)\uparrow$ may happen&#xA;only if $M&amp;#39;$ visits infinitely many&#xA;cells in the computation on $w$)&lt;/li&gt;&#xA;&lt;li&gt;for each input word $w$, the number&#xA;of cells visited by $M&amp;#39;$ in the&#xA;computation on $w$ is the same as the&#xA;analogical number for $M$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;$M&amp;#39;$ may use larger working alphabet than $M$.&lt;/p&gt;&#xA;" OwnerUserId="8380" LastEditorUserId="8380" LastEditorDisplayName="" LastEditDate="2011-03-18T21:37:12.660" LastActivityDate="2011-03-18T21:37:12.660" Title="Detecting cycles in off-line Turing machines" Tags="&lt;algorithms&gt;&lt;computer-science&gt;&lt;formal-systems&gt;" AnswerCount="1" />
  <row Id="27607" PostTypeId="2" ParentId="27606" CreationDate="2011-03-17T14:32:41.520" Score="0" ViewCount="" Body="&lt;p&gt;A standard way to detect cycles in a TM is by counting configurations. For a machine with one input tape and one work tape, a configuration (of M's run on the input x) consists of the current state, the location of the heads in both tapes, and the content of the work tape.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the location of the heads is known to be bounded, the number of configurations itself is bounded. So if the machine runs more steps than the possible number of configuration, by the pigeonhole principle it entered the same configuration twice, hence it is in a loop, hence we can safely reject.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only problem in our case is that we don't know a-priory a bound on the location of the head in the work tape. So we adept dynamically - denote by $k$ the leftmost cell reached so far by the head, and count configurations according to $k$. If the head passes the $k$th cell, update $k$ accordingly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only remaining problem is counting configurations with only limited space; however, this can be fixed using extended alphabet. Think of every letter in the new alphabet as a pair - one letter from the old alphabet, representing what M sees, and the other letter is in an alphabet big enough so that $k$ digits are enough to represent a number as large as the possible number of configurations.&lt;/p&gt;&#xA;" OwnerUserId="1818" LastEditorUserId="1818" LastEditorDisplayName="" LastEditDate="2011-03-18T21:29:42.657" LastActivityDate="2011-03-18T21:29:42.657" CommentCount="0" />
  <row Id="27608" PostTypeId="2" ParentId="24081" CreationDate="2011-03-17T14:35:57.510" Score="2" ViewCount="" Body="&lt;p&gt;Rice's theorem cannot apply to machines, only to properties of languages (since the proof of Rice's theorem relies heavily on us being able to control the language which a machine accepts, although we have very limited control on what the machine does &lt;strong&gt;until&lt;/strong&gt; it accepts).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So yes, as you've answered yourself, the language is decidable.&lt;/p&gt;&#xA;" OwnerUserId="1818" LastActivityDate="2011-03-17T14:35:57.510" />
  <row Id="27609" PostTypeId="1" AcceptedAnswerId="27613" CreationDate="2011-03-17T14:39:12.510" Score="10" ViewCount="298" Body="&lt;p&gt;I'm dealing with a time-complexity problem in which I know the running time of an algorithm:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$t = 1000ms$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also know that the algorithm is upper bounded by $O(n!)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to know the approximate size of the input $n$ based on this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f(n) = n! = t$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f^{-1}(t) = n = ?$$&lt;/p&gt;&#xA;" OwnerUserId="4474" LastEditorUserId="2736" LastEditorDisplayName="" LastEditDate="2011-03-18T02:55:54.260" LastActivityDate="2011-03-18T02:55:54.260" Title="n! = c, find n?" Tags="&lt;algebra&gt;" AnswerCount="2" CommentCount="11" FavoriteCount="4" />
  <row Id="27610" PostTypeId="1" AcceptedAnswerId="27733" CreationDate="2011-03-17T14:40:12.327" Score="2" ViewCount="119" Body="&lt;p&gt;I have a linear operator $T$ which acts on the vector space of square $N\times N$ matrices in this way:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$T(A)=0.5(A-A^\mathrm{t})$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;($A^\mathrm{t}$: the transpose of a matrix $A$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need to prove that this operator is diagonalizable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that a linear transformation is diagonalizable if and only if there is a basis for the vector space that consists entirely of eigenvectors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to find the possible eigenvalues for this operator, so by definition I need to see what are the $\lambda$s that imply $t(v)=\lambda v$,and by assignment I get: $A(1-2\lambda)=A^\mathrm{t}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I go on from here?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you guys.&lt;/p&gt;&#xA;" OwnerUserId="6163" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-17T15:09:09.427" LastActivityDate="2011-03-18T03:42:44.377" Title="Eigenvalues of operators" Tags="&lt;linear-algebra&gt;&lt;eigenvalues&gt;&lt;operator-algebras&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="3" />
  <row Id="27611" PostTypeId="2" ParentId="27526" CreationDate="2011-03-17T14:53:45.883" Score="1" ViewCount="" Body="&lt;p&gt;I recommend &lt;a href=&quot;http://terrytao.wordpress.com/2010/04/10/the-euler-maclaurin-formula-bernoulli-numbers-the-zeta-function-and-real-variable-analytic-continuation/&quot; rel=&quot;nofollow&quot;&gt;this post&lt;/a&gt; in Terry Tao's blog. It is not elementary, but can be understood by someone with knowledge of the calculus of one real varable.&lt;/p&gt;&#xA;" OwnerUserId="4791" LastActivityDate="2011-03-17T14:53:45.883" />
  <row Id="27612" PostTypeId="2" ParentId="27610" CreationDate="2011-03-17T14:58:01.963" Score="1" ViewCount="" Body="&lt;p&gt;This is a funny problem because although $T(A)$ is the skew symmetric part of $A$, $T$ itself has a matrix representation that is symmetric.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, let $E_{nm}= \left (\delta_{nm} \right)$, i.e. the matrix of 0s everywhere except a 1 at the $(n,m)$ entry. Then&#xA;$T(E_{nm}) =\frac{1}{2} \left(E_{nm}-E_{mn}\right)$. With the usual inner product on $N\times N$ matrices, we have that $$T=\frac{1}{2} \sum_{n,m=1}^{N} (E_{nm}\otimes E_{nm} -E_{mn}\otimes E_{nm}).$$ So $$T^{\top} = \frac{1}{2} \sum_{n,m=1}^{N} (E_{nm}\otimes E_{nm}-E_{nm}\otimes E_{mn})=T$$ (by invariance under change of notation ;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Added: Since real symmetric matrices are diagonalizable, $T$ is.&lt;/p&gt;&#xA;" OwnerUserId="5253" LastEditorUserId="5253" LastEditorDisplayName="" LastEditDate="2011-03-17T16:40:24.003" LastActivityDate="2011-03-17T16:40:24.003" CommentCount="2" />
  <row Id="27613" PostTypeId="2" ParentId="27609" CreationDate="2011-03-17T14:59:12.097" Score="5" ViewCount="" Body="&lt;p&gt;You can use Stirling' approximation formula:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\log n! = n\log n - n + \log (2\pi n)/2 + \mathcal{O}(1/n)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you can take the approximation as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\log n! \approx n\log n - n + \log (2\pi n)/2 $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now given $n! = c$, you can compute an approximate value for $n$ by doing a simple binary search on the above formula.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want a direct formula to compute and approximate value for $n$, Shai Covo has given you a link to a mathoverflow thread.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2011-03-17T14:59:12.097" CommentCount="5" />
  <row Id="27614" PostTypeId="2" ParentId="27610" CreationDate="2011-03-17T15:07:47.837" Score="8" ViewCount="" Body="&lt;p&gt;$T$ is the projection onto to the space of antisymmetric matrices along the symmetric matrices. Thus the spectrum is $\{0;1\}$, and those are the eigenspaces. Obviously they span the whole space, so $T$ is diagonalizable.&lt;/p&gt;&#xA;" OwnerUserId="7598" LastActivityDate="2011-03-17T15:07:47.837" CommentCount="2" />
  <row Id="27615" PostTypeId="2" ParentId="27578" CreationDate="2011-03-17T15:22:03.237" Score="2" ViewCount="" Body="&lt;p&gt;It is possible to have $2 \times 2$ diagonal matrices $A$ and $B$ so that $A$ has eigenvalues $0$ and $\lambda_1$ and $B$ has eigenvalues $0$ and $\lambda_2$, but $AB$ is the $0$ matrix. This can be extended to $3\times 3$ matrices. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It can also be the case for $2 \times 2$ matrices or larger that the only eigenvalues of $A$ and $B$ are $0$, but $AB$ has an arbitrary nonzero eigenvalue. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If either of $A$ or $B$ is not invertible, then $AB$ can't be, so it must have $0$ as an eigenvalue.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the eigenvectors $v_1$ and $v_2$ are parallel, then $v_1$ must be an eigenvector of $AB$ of eigenvalue $\lambda_1 \lambda_2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, my guess is that if $v_1$ and $v_2$ are not parallel, and $\lambda_1$ and $\lambda_2$ are not $0$, then there is no restriction on the set of eigenvalues of $AB$.&lt;/p&gt;&#xA;" OwnerUserId="8345" LastActivityDate="2011-03-17T15:22:03.237" />
  <row Id="27616" PostTypeId="1" AcceptedAnswerId="27619" CreationDate="2011-03-17T15:37:25.723" Score="3" ViewCount="95" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;If $p$ is prime, determine the number of abelian groups of order $p^n$ for each $1\leq n\leq8$&lt;br&gt;&lt;br&gt;&#xA;  (I assume that &quot;up to isomorphism&quot; should be included somewhere in the question for the sake of precision...)&lt;br&gt;&lt;br&gt;Could someone please review/confirm my work?&lt;br&gt;&#xA;  n = 1: $\mathbb{Z}_p $ &lt;br&gt;&#xA;  n = 2: $\mathbb{Z}_{p^2}$ and $\mathbb{Z}_p\times \mathbb{Z}_p$ &lt;br&gt;&#xA;  n = 3: $\mathbb{Z}_{p^3}$,  $\mathbb{Z}_{p^2}\times \mathbb{Z}_p$, and $\mathbb{Z}_p\times \mathbb{Z}_p \times\mathbb{Z}_p$&lt;br&gt;&#xA;  n = 4: $\mathbb{Z}_{p^4}$, &#xA;  $\mathbb{Z}_{p^3} \times \mathbb{Z}_p$, $\mathbb{Z}_{p^2}\times \mathbb{Z}_{p^2}$,  $\mathbb{Z}_{p^2}\times \mathbb{Z}_p \times \mathbb{Z}_p$, and $\mathbb{Z}_p\times \mathbb{Z}_p \times\mathbb{Z}_p \times \mathbb{Z}_p$&lt;br&gt;&lt;br&gt;&#xA;  &lt;em&gt;et cetera&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am simply considering all the options for when the largest exponent of $p$ is $n$, then $n-1$, and so on. How does this look? Thanks!&lt;br&gt;&#xA;(Apparently I don't know how to &quot;end a quote&quot;...)&lt;/p&gt;&#xA;" OwnerUserId="7850" LastActivityDate="2011-03-17T15:41:06.747" Title="Number of abelian groups of order $p^n$" Tags="&lt;homework&gt;&lt;abstract-algebra&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="27617" PostTypeId="2" ParentId="27602" CreationDate="2011-03-17T15:39:29.263" Score="0" ViewCount="" Body="&lt;p&gt;By part (b) you have a measurable $g$ such that $d\nu = gd|\nu|$ and $|g(x)| = 1$ for all $x$. Use this and apply $|\int fd\nu| \leq \int|f|d\nu$, replacing $d\nu$ with $gd\nu$ first.&lt;/p&gt;&#xA;" OwnerUserId="2055" LastActivityDate="2011-03-17T15:39:29.263" CommentCount="3" />
  <row Id="27618" PostTypeId="2" ParentId="27616" CreationDate="2011-03-17T15:40:34.603" Score="4" ViewCount="" Body="&lt;p&gt;It's equal to the number of &lt;a href=&quot;http://en.wikipedia.org/wiki/Partition_%28number_theory%29&quot; rel=&quot;nofollow&quot;&gt;partitions&lt;/a&gt; of $n$. See &lt;a href=&quot;http://mathworld.wolfram.com/AbelianGroup.html&quot; rel=&quot;nofollow&quot;&gt;http://mathworld.wolfram.com/AbelianGroup.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="589" LastActivityDate="2011-03-17T15:40:34.603" CommentCount="1" />
  <row Id="27619" PostTypeId="2" ParentId="27616" CreationDate="2011-03-17T15:41:06.747" Score="8" ViewCount="" Body="&lt;p&gt;Your work is correct, except that you aren't answering the question asked (they asked you for the &lt;em&gt;number&lt;/em&gt; of (nonisomorphic) groups, not for a list of the groups). So for $n=1$, the answer should be &quot;1&quot;; for $n=2$ the answer should be &quot;2&quot;; for $n=3$ the answer should be &quot;3&quot;; for $n=4$ the answer should be &quot;5&quot;, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The magic words  you are looking for are &lt;a href=&quot;http://en.wikipedia.org/wiki/Partition_%28number_theory%29&quot; rel=&quot;nofollow&quot;&gt;&quot;partitions of $n$&lt;/a&gt;.&quot; You should verify that there is a bijection between the isomorphism types of abelian groups of order $p^n$ and the partitions of $n$.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2011-03-17T15:41:06.747" CommentCount="6" />
  <row Id="27620" PostTypeId="2" ParentId="27593" CreationDate="2011-03-17T15:42:05.647" Score="3" ViewCount="" Body="&lt;p&gt;Yes, take the $log_2$ of both sides gives you:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$n log_2(n) = c$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can use Newton's method to solve this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x_0 = c$$&#xA;$$x_{k+1} = x_k - (x_k log(x_k) - c log(2))/(1 + log(x_k))$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where now &quot;log&quot; is the natural logarithm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This gives the solution $n \sim x_4 = 140.221667$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Starting with a better $x_0$, like $x_0=c / log(c)$ gives you even faster convergence.  With c=1000 or c=1000000, the value $x_3$ is correct with an error of $10^{-8}$.&lt;/p&gt;&#xA;" OwnerUserId="7933" LastEditorUserId="7933" LastEditorDisplayName="" LastEditDate="2011-03-17T16:38:40.133" LastActivityDate="2011-03-17T16:38:40.133" />
  <row Id="27621" PostTypeId="1" AcceptedAnswerId="27622" CreationDate="2011-03-17T15:44:00.663" Score="6" ViewCount="198" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Let $x = (a, b), y = (c, d) \in \mathbb{Z}^2$. What is the condition on $a, b, c, d$ so that ${x, y}$ is a basis?&lt;br&gt;&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My answer: $ad\neq bc$ and $gcd(a, c) = gcd(b, d) = 1$. &lt;br&gt;&#xA;The first condition ensures that they aren't the same vector; the second ensures that we can actually &lt;em&gt;&quot;get&quot;&lt;/em&gt; all of the integer values/lattice points. &lt;br&gt;&#xA;Is this correct? &lt;br&gt;&lt;br&gt;&#xA;Thanks.  &lt;/p&gt;&#xA;" OwnerUserId="7850" LastActivityDate="2011-03-17T19:58:47.807" Title="Basis for $\mathbb{Z}^2$" Tags="&lt;homework&gt;&lt;abstract-algebra&gt;" AnswerCount="3" CommentCount="8" />
  <row Id="27622" PostTypeId="2" ParentId="27621" CreationDate="2011-03-17T15:48:29.260" Score="6" ViewCount="" Body="&lt;p&gt;If that is a basis, then you can write $e_1=(1,0)$ and $e_2=(0,1)$ in terms of the basis using integer coefficients. This implies that the matrix determined by the basis in invertible in $\mathbb Z$. That is your condition. It can be expressed neatly using determinants.&lt;/p&gt;&#xA;" OwnerUserId="589" LastActivityDate="2011-03-17T15:48:29.260" CommentCount="6" />
  <row Id="27624" PostTypeId="2" ParentId="27526" CreationDate="2011-03-17T16:05:12.097" Score="2" ViewCount="" Body="&lt;p&gt;for your $1+2+3+4+...=-1/12$ this is the value of $\zeta(-1)$, the riemann zeta function at $-1$. so while it is not an equality as written, the definition is usually given as $\zeta(s)=\sum_{n=1}^{\infty}n^{-s}$ which only converges for real $s$ greater than 1.  however it can be continued to the rest of the plane (with a pole at 1).&lt;/p&gt;&#xA;" OwnerUserId="6925" LastActivityDate="2011-03-17T16:05:12.097" />
  <row Id="27625" PostTypeId="1" CreationDate="2011-03-17T16:12:50.997" Score="2" ViewCount="61" Body="&lt;p&gt;Let $G=(V,E)$ be a graph, and $A$ be its adjacency matrix. Define $n = |V|$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given $A$ and a natural number $m \le n$, I'm interested in the following problem:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;How many &lt;strong&gt;simple&lt;/strong&gt; cycles of length $m$ exist in $G$?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;By &lt;strong&gt;simple&lt;/strong&gt; cycle, I mean no repeated vertices along the cycle is allowed (other than the starting and ending vertices, which coincide).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is NP-hard. However, I'm not asking its complexity; I'm merely interested in whether there is a closed-form expression for computing it. (Thus, computing the expression can be NP-hard.)&lt;/p&gt;&#xA;" OwnerUserId="2420" LastEditorUserId="2420" LastEditorDisplayName="" LastEditDate="2011-03-20T22:17:57.133" LastActivityDate="2011-03-20T22:17:57.133" Title="Cycles of Specified Length in a Graph" Tags="&lt;combinatorics&gt;&lt;graph-theory&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" />
  <row Id="27626" PostTypeId="1" CreationDate="2011-03-17T16:35:45.103" Score="1" ViewCount="61" Body="&lt;p&gt;Hi all, I am studing if I could modell a function using LUTs (ROMs) for a electronic digital design.&#xA;I have two functions, a FFT and other function, S.&#xA;S function system recive as input the output of the FFT system. Due there are 2000*2000 possibles values as input for my FFT system if my FFT is a inyective function then is not a good a idea to try modell S through a LUT because the set of the input of S system would be 2000*2000 and these are so many entries for a ROM in a FPGA, I think so...&#xA;If FFT is surjective, maybe the 2000*2000 values of the input set of FFT system might be a smaller set of the FFT image, small enough to consider the save the values in a ROM.&#xA;Thank you! &lt;/p&gt;&#xA;" OwnerUserId="4255" LastEditorUserId="1543" LastEditorDisplayName="" LastEditDate="2011-04-17T20:30:43.287" LastActivityDate="2011-04-17T20:30:43.287" Title="How to know if the Fourier Transform is surjective" Tags="&lt;homework&gt;&lt;fourier-analysis&gt;&lt;mathematical-modeling&gt;" CommentCount="1" FavoriteCount="1" />
  <row Id="27627" PostTypeId="1" CreationDate="2011-03-17T16:41:36.493" Score="0" ViewCount="45" Body="&lt;p&gt;Hi, &#xA;simple question.&#xA;I am looking at a presentation of the ring $k[x,yx^{-1}]$ in the following form $k[a,b]/I$ where $I$ is an ideal.&lt;/p&gt;&#xA;" OwnerUserId="6583" LastActivityDate="2011-03-17T18:45:41.193" Title="simple toric variety" Tags="&lt;algebraic-geometry&gt;" AnswerCount="1" />
  <row Id="27628" PostTypeId="2" ParentId="27506" CreationDate="2011-03-17T16:43:37.493" Score="2" ViewCount="" Body="&lt;p&gt;For each $n$ in $\mathbb{N}$, let $A_n \subset [n]^3$ be the set of triples $(a,b,c)$ such that $a+b+c=2n$.  We want to count the number of orbits of $S_3$ acting on the triples of $A_n$.  This is given by &lt;a href=&quot;http://en.wikipedia.org/wiki/Burnside%27s_lemma&quot; rel=&quot;nofollow&quot;&gt;Burnside's lemma&lt;/a&gt; as&#xA;$$&#xD;&#xA;|A_n/S_3| = \frac{1}{6}\sum_{g \in S_3}|A_n^{g}|,&#xD;&#xA;$$&#xA;where for each $g \in S_3$, $A_n^g$ is the set of triples fixed by $g$.  So we need to count the elements of $A_n$ (since these are preserved by the identity operation), the elements of $A_n$ that are of the form $(a,a,c)$ (since these are preserved under a two-component swap), and the elements of $A_n$ that are of the form $(a,a,a)$ (since these are preserved under the two rotations).  Calling these three counts $a_n$, $b_n$, and $c_n$ respectively, the result will be&#xA;$$&#xD;&#xA;|A_n/S_3| = \frac{1}{6}\left(a_n + 3b_n + 2c_n\right).&#xD;&#xA;$$&#xA;The number of triples of the form $(a,a,a)\in A_n$ is $c_n=1$ if $n$ is a multiple of $3$, and $0$ otherwise.  The number of triples of the form $(a,a,c)\in A_n$ is $b_n = \lfloor{n/2}\rfloor$, since the doubled element can take any value from $\lceil{n/2}\rceil$ to $n-1$, inclusive.  Finally, we must compute the total number of triples $(a,b,c)\in A_n$.  If $(a,b,c)$ were chosen from all of $\mathbb{N}^3$, the count would be $(n-1)(2n-1)$, since this is the number of ways to break a line of $2n$ elements in two distinct places.  Here we must remove the cases where one of the components is greater than $n$.  The number to be removed is&#xA;$$&#xD;&#xA;\begin{eqnarray}&#xD;&#xA;3 \sum_{i=n+1}^{2n-2} (2n-i-1) &amp;amp;=&amp;amp; 3\sum_{i=1}^{n-2} (i) \\&#xD;&#xA;&amp;amp;=&amp;amp; \frac{3}{2}(n-1)(n-2),&#xD;&#xA;\end{eqnarray}&#xD;&#xA;$$&#xA;giving us&#xA;$$&#xD;&#xA;a_n = (n-1)(2n-1) - \frac{3}{2}(n-1)(n-2) = \frac{1}{2}(n-1)(n+4).&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the final result is&#xA;$$&#xD;&#xA;\begin{eqnarray}&#xD;&#xA;|A_n/S_3| &amp;amp;=&amp;amp; \frac{1}{6}\left(\frac{1}{2}(n-1)(n+4) + 3\left\lfloor{\frac{1}{2}n}\right\rfloor + 2c_n\right) \\&#xD;&#xA;&amp;amp;=&amp;amp; \frac{1}{12}\left(n^2 + 6n - 4\right) + \epsilon_n,&#xD;&#xA;\end{eqnarray}&#xD;&#xA;$$&#xA;where $|\epsilon_n| &amp;lt; 1/2$; i.e., the number of orbits is the closest integer to $(n^2+6n-4)/12$.&lt;/p&gt;&#xA;" OwnerUserId="5546" LastActivityDate="2011-03-17T16:43:37.493" CommentCount="1" />
  <row Id="27629" PostTypeId="1" AcceptedAnswerId="27632" CreationDate="2011-03-17T16:44:13.560" Score="3" ViewCount="184" Body="&lt;p&gt;Can anyone suggest a collection of (solved) exercises in topology? Undergrad level, as a companion to Dugundji's Topology (although excellent it doesn't provide the solutions to the problems).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="8383" LastActivityDate="2011-03-17T17:01:15.183" Title="Topology exercises" Tags="&lt;topology&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="4" />
  <row Id="27630" PostTypeId="2" ParentId="27629" CreationDate="2011-03-17T16:48:57.940" Score="1" ViewCount="" Body="&lt;p&gt;&quot;Fundamentals of General Topology: Problems and Exercises&quot; by A. V. Arkhangel'skii; V.I. Ponomarev is a fun book, though it might be hard to find and the level might be higher than you'd like.&lt;/p&gt;&#xA;" OwnerUserId="3852" LastActivityDate="2011-03-17T16:48:57.940" CommentCount="1" />
  <row Id="27631" PostTypeId="2" ParentId="27629" CreationDate="2011-03-17T16:59:12.640" Score="0" ViewCount="" Body="&lt;p&gt;I'd recommend &lt;a href=&quot;http://pages.cs.wisc.edu/~dluu/ps/Viro,%20Ivanov,%20Kharlamov%20and%20Netsvetaev%20-%20Topology%20with%20problems%20and%20solutions.pdf&quot; rel=&quot;nofollow&quot;&gt;&quot;Elementary Topology Problem Textbook&quot;&lt;/a&gt; by Viro, Ivanov, Netsvetaev, Kharlamov.&lt;/p&gt;&#xA;" OwnerUserId="264" LastActivityDate="2011-03-17T16:59:12.640" />
  <row Id="27632" PostTypeId="2" ParentId="27629" CreationDate="2011-03-17T17:01:15.183" Score="1" ViewCount="" Body="&lt;p&gt;Elementary Topology Problem Textbook&#xA;O. Ya. Viro, O. A. Ivanov,&#xA;N. Yu. Netsvetaev, V. M. Kharlamov, is available freely online at &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.pdmi.ras.ru/~olegviro/topoman/eng-book-nopfs.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.pdmi.ras.ru/~olegviro/topoman/eng-book-nopfs.pdf&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This book is published by &lt;strong&gt;AMS&lt;/strong&gt; and that comes with solutions.&lt;/p&gt;&#xA;" LastActivityDate="2011-03-17T17:01:15.183" CommentCount="1" />
  <row Id="27633" PostTypeId="1" CreationDate="2011-03-17T17:07:20.030" Score="5" ViewCount="200" Body="&lt;p&gt;Is there an elementary(or not) proof that there are at least two consecutive primes which have difference $2n$ for every natural number $n$? i remind that Polignac's conjecture states that there should be infinite such pairs for every $n$ so it is a much more easy question to ask and of course should be valid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In addition, for every $n\in\mathbb N$, is there any $m_n\in\mathbb N$ such that there are  two natural numbers $k_1,k_2$ so that $k_2-k_1=2n$, $k_1, k_2$ are not divisible by $p_1,\dots p_{m_n}$, where $p_i$ is the $i$-th prime, and for every $k\in\{k_1+1,\dots k_2-1\}$ there exists a $p\in\{p_1,\dots p_{m_n}\}$ such that $p|k$?(elementary proof)&lt;/p&gt;&#xA;" OwnerUserId="5010" LastEditorUserId="5010" LastEditorDisplayName="" LastEditDate="2011-03-18T19:55:59.760" LastActivityDate="2011-03-19T00:02:41.567" Title="distance between consecutive primes (related  to Polignac's conjecture)" Tags="&lt;number-theory&gt;" AnswerCount="2" CommentCount="7" />
  <row Id="27634" PostTypeId="2" ParentId="27627" CreationDate="2011-03-17T17:19:39.223" Score="1" ViewCount="" Body="&lt;p&gt;Note that $x$ and $y/x$ are algebrically independent since the field of fraction of $k[x.y/x]$ is $k(x.y)$. If there is map from $k[a,b]$ to $k[x,y/x]$ sending $a$ to $x$ and $b$ to $y/x$ then the kernel $I$ is definitely a prime ideal as $k[x.y/x]$ is a domain. Now equating transcendence degree of $k[a.b]/I$ and $k[x.y/x]$ over $k$ we can conclude that the kernel is zero.&lt;/p&gt;&#xA;" OwnerUserId="7565" LastEditorUserId="7565" LastEditorDisplayName="" LastEditDate="2011-03-17T18:45:41.193" LastActivityDate="2011-03-17T18:45:41.193" CommentCount="2" />
  <row Id="27635" PostTypeId="1" AcceptedAnswerId="27636" CreationDate="2011-03-17T17:21:35.117" Score="3" ViewCount="115" Body="&lt;p&gt;I cannot figure out the solution to this exercise in Marker. Can someone help me?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(Z \oplus Z, +, 0) \not\equiv (Z, +, 0)$&lt;/p&gt;&#xA;" OwnerUserId="8121" LastActivityDate="2011-03-17T18:07:56.030" Title="[Model Theory] Problem" Tags="&lt;logic&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="27636" PostTypeId="2" ParentId="27635" CreationDate="2011-03-17T17:27:48.033" Score="9" ViewCount="" Body="&lt;p&gt;EDIT:&#xA;The sentence $\exists z \forall y \exists x (x+x=y \vee x+x+z=y)$ is true in $\mathbb{Z}$ but not in $\mathbb{Z} \oplus \mathbb{Z}$.&lt;/p&gt;&#xA;" OwnerUserId="5203" LastEditorUserId="5203" LastEditorDisplayName="" LastEditDate="2011-03-17T18:07:56.030" LastActivityDate="2011-03-17T18:07:56.030" CommentCount="4" />
  <row Id="27637" PostTypeId="2" ParentId="27621" CreationDate="2011-03-17T17:32:19.910" Score="9" ViewCount="" Body="&lt;p&gt;This has a beautiful geometrical interpretation. Note $\rm\: x,y\: $ is a $\rm\:\mathbb Z$-basis of $\rm\: \mathbb Z^2\: $ iff $\rm\: \mathbb Z^2\: $ is tiled by the fundamental parallelogram $\rm\:P\:$ with sides $\rm\:x,y\:.\ $ But this is true iff the only lattice points that are inside $\rm\:P\:$ or on the boundary of $\rm\:P\:$ are its vertices. However, by &lt;a href=&quot;http://en.wikipedia.org/wiki/Pick%27s_theorem&quot; rel=&quot;nofollow&quot;&gt;Pick's area formula,&lt;/a&gt; this is true iff $\rm\ area\ P\: =\: $  #interior_points $\ +\ 1/2\ $ #boundary_points $\:-\ 1\ =\ 0 + 4/2 - 1 = 1\:.\ $ But by basic analytic geometry $\rm\: area\ P\: =\: |det(x,y)|\:.\: $ Therefore $\rm\: x,y\:$ is a $\rm\:\mathbb Z$-basis of $\rm\: \mathbb Z^2 \iff  |det(x,y)| = 1\:.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact it deserves to be much better known that Pick originally applied his area formula in a similar way to give a beautiful geometric proof of the Bezout linear representation of the gcd.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2011-03-17T19:28:13.183" LastActivityDate="2011-03-17T19:28:13.183" CommentCount="7" />
  <row Id="27638" PostTypeId="2" ParentId="27473" CreationDate="2011-03-17T17:38:40.503" Score="1" ViewCount="" Body="&lt;p&gt;This is called a &lt;em&gt;singular&lt;/em&gt; boundary-value problem. Matlab can handle some singular BVPs (look at the documentation for &lt;code&gt;bvp4c&lt;/code&gt; and the &lt;code&gt;SingularTerm&lt;/code&gt; option in bvpset) so you need to bring your equation in the form that Matlab can handle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added later:&lt;/strong&gt; I've never used this option before, but here is how I would start. When bringing the equation to first-order form, one normally introduces a vector $(F_1,F_2)$ with $F_1(K) = F(K)$ and $F_2(K) = F&amp;#39;(K)$ so that the equation becomes&#xA;$$&#xD;&#xA;\begin{align}&#xD;&#xA;F_1&amp;#39; &amp;amp;= F_2, \\&#xD;&#xA;F_2&amp;#39; &amp;amp;= \frac{3.45}{0.02K^2} F_2 + \frac{0.08}{0.02K^2} F_1 + \frac{3.45}{0.02K^2}.&#xD;&#xA;\end{align}&#xD;&#xA;$$&#xA;As you noticed, Matlab requires the equation to be of the form $y&amp;#39; = \frac1x Sy + f(x,y)$  ...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added yet later:&lt;/strong&gt; I initially thought that one could get rid of the $1/K^2$ terms by using $F_1(K) = F(K)$ and $F_2(K) = KF&amp;#39;(K)$ but as Qiang Li noted (in more polite terms) I made a stupid mistake. In fact, in the original equation the boundary point $K=0$ is an irregular singular point, while in the equations that Matlab can handle the boundary point is a regular singular point. My uneducated guess would be that the algorithm implemented in Matlab won't be able to handle your equation. Sorry.&lt;/p&gt;&#xA;" OwnerUserId="520" LastEditorUserId="520" LastEditorDisplayName="" LastEditDate="2011-03-22T08:33:27.053" LastActivityDate="2011-03-22T08:33:27.053" CommentCount="5" />
  <row Id="27639" PostTypeId="2" ParentId="27625" CreationDate="2011-03-17T17:42:26.507" Score="1" ViewCount="" Body="&lt;p&gt;I think the trace of $A^m$ gives the number of cycles of length $m$ of a graph $G$.&lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2011-03-17T17:42:26.507" CommentCount="3" />
  <row Id="27640" PostTypeId="1" CreationDate="2011-03-17T17:48:25.320" Score="1" ViewCount="89" Body="&lt;p&gt;I am trying to do contour integration on $\int_c \zeta(s)\zeta(2s) \frac{x^s}{s} ds$ and &#xA;$\int_c \frac{x^s}{s^ks}ds$&#xA;where c is the line segment joining c-iT c+iT &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand the basic theory behind it, how to find residues and Cauchy's theorem. I am just having trouble with bounding these integrals when using Cauchy's theorem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I mean to say here is let's say I am doing contour integration for x&gt;1 on a square say for $\int_c \frac{x^s}{s}ds$ I can bound the horizontal integral by $\frac{1}{2\pi T}\frac{y^c}{log(y)}$  can I bound the horizontal integral for  $\int_c \zeta(s)\zeta(2s) \frac{x^s}{s^ks} ds$  by $\frac{1}{2\pi T^kT}\frac{y^c}{log(y)}$?&lt;/p&gt;&#xA;" OwnerUserId="8355" LastEditorUserId="8355" LastEditorDisplayName="" LastEditDate="2011-03-17T19:13:44.760" LastActivityDate="2011-03-17T22:53:18.320" Title="contour integration $\zeta(s)\zeta(2s)$ and $x^s/s^(k+1)ds$" Tags="&lt;number-theory&gt;&lt;complex-analysis&gt;" AnswerCount="1" />
  <row Id="27641" PostTypeId="1" AcceptedAnswerId="27645" CreationDate="2011-03-17T18:17:52.287" Score="0" ViewCount="58" Body="&lt;p&gt;I want to calculate how much I would pay monthly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$X$ = Amount borrowed. $I$= Interest Rate. $Y$ = number of years&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is what I have so far, but it's wrong; I'm getting a larger value.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{X\left(1 + \frac{I}{100}\right)^{Y}\left(\frac{I}{Y}\right)}{12\left(\left(1 + \frac{I}{100}\right)^Y - 1\right)}.$$&lt;/p&gt;&#xA;" OwnerUserId="8385" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2011-03-17T18:39:52.290" LastActivityDate="2011-03-17T19:20:10.103" Title="Interest Calculation Problem" Tags="&lt;finance&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="27642" PostTypeId="2" ParentId="27640" CreationDate="2011-03-17T18:31:50.160" Score="3" ViewCount="" Body="&lt;p&gt;First multiply out $\zeta(s)\zeta(2s)$ as a Dirichlet series $\sum_{n = 1}^{\infty} \frac{a_n}{n^s}$.  (I'll let you compute the values of $a_n$.) Your integral is then equal to&#xA;$$\sum_{n \geq 1} a_n \int_c \bigl(\frac{x}{n}\bigr)^s \frac{ds}{s}.$$&#xA;Now if we let $T \to \infty$,&#xA;then $\int_c x^s\frac{dx}{s}$ is equal to $0$, $\pi i$, or $2\pi i$ depending &#xA;on whether $0  &amp;lt; x &amp;lt; 1$, $x = 1$, or $x &amp;gt; 1$ (assuming that the real part of the vertical line $c$ is positive).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus your integral is equal to $2\pi i \sum_{n \leq x} a_n,$ with the convention&#xA;that you weight the final summand by $1/2$ if $x$ is an integer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you really want $T$ to be a finite value, rather than infinite, you will have&#xA;to compute more carefully.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first few chapters of Edward's book &lt;em&gt;Riemann's zeta function&lt;/em&gt; are a good reference for these sorts of integrals, including the precise issues related to convergence, interchanging the summation and integration, and so on.&lt;/p&gt;&#xA;" OwnerUserId="221" LastEditorUserId="221" LastEditorDisplayName="" LastEditDate="2011-03-17T22:53:18.320" LastActivityDate="2011-03-17T22:53:18.320" CommentCount="1" />
  <row Id="27643" PostTypeId="2" ParentId="27641" CreationDate="2011-03-17T18:40:40.637" Score="2" ViewCount="" Body="&lt;p&gt;The present value is $X$. The future value is $0$. We know $I/Y$. So you can compute the payment (which would be negative assuming that positive amounts imply money is coming to you and negative amounts mean that you are paying money). In particular, suppose $$a_{\overline{12Y}|i^{(12)}} = \frac{1-v^{12y}}{i^{(12)}}$$ where $v = \frac{1}{1+i^{(12)}}$ and $i^{(12)}$ is the monthly effective interest rate. So $$1+i = \left(1+i^{(12)} \right)^{12}$$ Then the level monthly payment would be $$P = \frac{X}{a_{\overline{12Y}|i^{(12)}}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can do this easily on a BA-II-PLUS calculator as well.&lt;/p&gt;&#xA;" OwnerUserId="2548" LastEditorUserId="2548" LastEditorDisplayName="" LastEditDate="2011-03-17T19:20:10.103" LastActivityDate="2011-03-17T19:20:10.103" CommentCount="1" />
  <row Id="27644" PostTypeId="2" ParentId="7512" CreationDate="2011-03-17T18:55:39.297" Score="0" ViewCount="" Body="&lt;p&gt;If you are EE you should read Kreyszig. You don't need the rigor of Rudin or anything similar, you want to be able to apply it. Every physicist reads Kreyszig. You might also try the Dover reprint of Griffel, Applied Functional Analysis, under $20. It is a nice read for someone with only an undergrad analysis course. My favorite, although you might have trouble with your background, is Applications of Functional Analysis and Operator Theory by Hutson and Pym, if you can find a copy. I learned functional analysis from doing quantum mechanics and then read all of the above books. If you take a grad level pure functional analysis course in your math department without the requisite background you may regret it.&lt;/p&gt;&#xA;" OwnerUserId="8388" LastActivityDate="2011-03-17T18:55:39.297" CommentCount="1" />
  <row Id="27645" PostTypeId="2" ParentId="27641" CreationDate="2011-03-17T18:57:03.617" Score="0" ViewCount="" Body="&lt;p&gt;Well, we know the amount borrowed, the interest (I'm going to assume it is the effective rate per year), and the amount of time we have to pay it off. Simplifying some things first off, we must be able to find the effective rate of interest per month, $i$, which can be gleaned from the equation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(1+i)^{12} = 1+I$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, in the language of annuities, we are trying to solve an equation for the present value of $X$, which is given by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$X = Pa_{\overline{12Y}|i}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $a_{\overline{12Y}|i} = \left[\frac{1-\left(1+i\right)^{-12Y}}{i}\right]$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, it is quite simple to solve for the amount of each payment, $P$.&lt;/p&gt;&#xA;" OwnerUserId="2465" LastEditorUserId="2465" LastEditorDisplayName="" LastEditDate="2011-03-17T19:15:30.107" LastActivityDate="2011-03-17T19:15:30.107" />
  <row Id="27646" PostTypeId="1" AcceptedAnswerId="27649" CreationDate="2011-03-17T18:58:21.133" Score="0" ViewCount="26" Body="&lt;p&gt;I'm trying to figure out the fee-adjusted interest rate paid when incurring a balance transfer fee on a loan, not accounting for inflation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the APR is 4%, the loan 5000, the transfer 2% and you pay it off in 200 days (54.8% of the year when rounded to 2 decimals), I see it this way:&lt;br&gt;&#xA;(APR: 4%  Loan: 5,000  Transfer fee: 2%  Days: 200)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Present Value&lt;/strong&gt;: 4,900  (Loan - transfer fee %)&lt;br&gt;&#xA;&lt;strong&gt;Future Value&lt;/strong&gt;: 5,109.589041095890411 (loan * ( APR * (200/365) ) )&lt;br&gt;&#xA;&lt;strong&gt;Real Rate Over 200 Days&lt;/strong&gt;: 4.2773273693039%  ( (FV - PV) / PV )&lt;br&gt;&#xA;&lt;strong&gt;Real Rate Over 365 days&lt;/strong&gt;: 7.806122448979618% ( ( (FV - PV) / PV ) * (365/200) )  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there something missing and if so, what?&lt;/p&gt;&#xA;" OwnerUserId="8386" LastEditorUserId="8386" LastEditorDisplayName="" LastEditDate="2011-03-17T19:16:33.287" LastActivityDate="2011-03-17T19:23:14.263" Title="How Can I Arrive at the Fee-Adjusted APR, Accounting for Balance Transfer Fee (But Not Inflation)?" Tags="&lt;finance&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="27647" PostTypeId="1" AcceptedAnswerId="27648" CreationDate="2011-03-17T19:04:42.863" Score="1" ViewCount="168" Body="&lt;p&gt;If I have a field $K$ and an extension $L$ of $K$ such that all (non-constant)&#xA;polynomials in $K[X]$ have a root in $L$, is the set of algebraic elements &#xA;of $L$ over $K$ (the sub-field of all the elements of $L$ which are roots of a polynomial in $K[X]$) algebraically closed ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do you have a counterexample ?&lt;/p&gt;&#xA;" OwnerUserId="8387" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-18T17:09:51.487" LastActivityDate="2011-03-18T17:09:51.487" Title="Is the sub-field of algebraic elements of a field extension of K containing roots of polynomials over K algebraically closed ?" Tags="&lt;abstract-algebra&gt;&lt;field-theory&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="27648" PostTypeId="2" ParentId="27647" CreationDate="2011-03-17T19:22:59.567" Score="4" ViewCount="" Body="&lt;p&gt;I'm guessing that what you wanted to ask is the following:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Let $K$ be a field, and let $L$ be an extension of $K$ (that is, $K\subseteq L$). Suppose futher that $L$ is algebraically closed (all non-constant polynomials in $L[x]$ have a root in $L$). Is the set $$\{a\in L\mid a\text{ is algebraic over }K\}$$algebraically closed?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The answer is &quot;yes.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $F = \{ a\in L\mid a\text{ is algebraic over }K\}$, and let $f(x)\in F[x]$ be a nonconstant polynomial. We may assume $f(x)$ is monic,&#xA;$$f(x) = x^n + a_{n-1}x^{n-1} + \cdots + a_0,\quad a_i\in F,\quad n\gt 0.$$&#xA;Since $a_i\in F$, then $a_i$ is algebraic over $K$ for all $i$, so the field $E=K(a_0,\ldots,a_{n-1})$ is a finite extension of $K$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now $f(x)\in F[x]\subseteq L[x]$, so we know that $L$ contains a root $r$ of $f(x)$, since $L$ is algebraically closed. Therefore, $[E(r):E] \leq n$ is finite, so $[E(r):K] = [E(r):E][E:K]$ is also finite. Therefore, $[K(r):K] \leq [E(r):K]$ is finite, so $r$ is algebraic over $K$. Since $r$ is algebraic over $K$, then $r\in F$. Thus, $f(x)$ has a root in $F$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This shows that every nonconstant polynomial in $F(x)$ has a root in $F$, so $F$ is algebraically closed.&lt;/p&gt;&#xA;&#xA;&lt;hr/&gt;&#xA;&#xA;&lt;p&gt;Apparently not. Rather, you want:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Let $K$ be a field, and let $L$ be an extension of $K$ (that is, $K\subseteq L$). Suppose further that  all non-constant polynomials in $K[x]$ have a root in $L$. Is the set $$\{a\in L\mid a\text{ is algebraic over }K\}$$algebraically closed?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strike&gt;This is equivalent to asking whether an algebraic extension of $K$ in which every polynomial in $K$ has a root is algebraically closed (since the subfield of $L$ of all elements that are algebraic over $K$ contains all the roots of polynomials in $K[x]$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again, the answer is yes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take $L$, $F$, and $f$ as above. Since $K(a_0,\ldots,a_{n-1})$ is algebraic over $K$, and $f(x)$ is a polynomial in $K(a_0,\ldots,a_{n-1})$, then the roots of $f(x)$ are algebraic over $K$. Thus, if $r$ is a root of $f(x)$ in some algebraic closure of $F$, then $r$ has an irreducible polynomial $g(x)\in K[x]$. Let $h(x)$ be the irreducible polynomial of $r$ in $F[x]$. In particular, $h(x)$ divides both $f(x)$ and $g(x)$ in $F[x]$. Since $g(x)$ has all its roots in $L$, and all its roots lie in $F$, then $g(x)$ has all its roots in $F$. Moreover, $h(x)|g(x)$ in $F$, so every root of $h(x)$ is a root of $g(x)$, hence every root of $h(x)$ lies in $F$. But $h(x)$ divides $f(x)$, so $f(x)$ has a root in $F$. Therefore, $F$ is algebraically closed.&lt;/strike&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added.&lt;/strong&gt; Hmph. Wait a second here. Every polynomial in $K[x]$ has at least one root in $L$, but we're not asserting they split. &quot;Every polynomial in $L$ has at least one root in $L$&quot; &lt;em&gt;is&lt;/em&gt; equivalent to &quot;every polynomial in $L$ splits&quot;, but this is not what we have here. We would need to show that&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If $L$ is an algebraic extension of $K$ such that every every nonconstant polynomial in $K[x]$ has a at least one root in $L$, then every nonconstant polynomial in $K[x]$ splits over $L$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;On reflection, I'm not so sure this is clear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strike&gt;Let $K=\mathbb{Q}$, and let $f(x)=x^3-2$. Consider the collection $\mathfrak{F}$ of all subfield of $\mathbb{C}$ that contain $\sqrt[3]{2}$, but do not contain the complex roots of $f(x)$. The collection is nonempty (it contains $\mathbb{Q}(\sqrt[3]{2})$). Partially order $\mathfrak{F}$ by inclusion. If we have a chain $\mathcal{C}$ of elements of $\mathfrak{F}$, $\mathcal{C}=\{F_i\}_{i\in I}$, then $\cup\mathcal{C}$ is a field, contained in $\mathbb{C}$, and does not contain the complex roots of $f(x)$. By Zorn's Lemma, $\mathfrak{F}$ contains maximal elements. Let $E$ be a maximal element of $\mathfrak{F}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does every irreducible nonconstant polynomial in $\mathbb{Q}[x]$ have at least one root in $E$?&lt;/strike&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On further reflection, the result is definitely true if $K$ is perfect (in particular, in characteristic zero, which takes care of my stricken attempt at an example above). I think it is true in general (by first closing under adjunction of $p$th roots and then using the result for perfect fields to get the conclusion), but I haven't quite worked out all the details yet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Theorem.&lt;/strong&gt; Let $L$ be an extension of $K$, and assume that every &lt;em&gt;separable&lt;/em&gt; nonconstant polynomial in $K[x]$ has at least one root in $L$. Then every nonconstant separable polynomial in $K[x]$ splits over $L$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Proof.&lt;/em&gt; Let $f(x)$ be a separable nonconstant polynomial in $K[x]$, and let $r\in L$ be a root of $f(x)$. Let $\overline{K}$ be an algebraic closure of $K$. Then there is an embedding of the subfield $K(r)$ of $L$ into $\overline{K}$, so we may assume $K(r)\subseteq \overline{K}$. Let $F$ be the splitting field of $f(x)$ over $K$. Since $F$ is separable and of finite degree over $K$, by &lt;a href=&quot;http://en.wikipedia.org/wiki/Primitive_element_theorem#Existence_statement&quot; rel=&quot;nofollow&quot;&gt;the Primitive Element Theorem&lt;/a&gt;, there exists $a\in\overline{K}$ such that $F=K(a)$. Let $g(x)$ be the irreducible polynomial of $a$ over $K$. Then we know that $g(x)$ has a root $\alpha\in L$, and that the subfield $K(\alpha)$ of $L$ embeds into $\overline{K}$. Since $K(a)$ is normal, we must have that the image of $K(\alpha)$ in $\overline{K}$ is $K(a)$; that is, $K(\alpha)\cong K(a)$. But $f(x)$ splits in $K(a)$, hence $f(x)$ splits in $K(\alpha)\subseteq L$ Thus, $f(x)$ splits in $L$, as claimed. &lt;strong&gt;QED&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Corollary.&lt;/strong&gt; Let $K$ be a field, and let $L$ be an extension of $K$ such that every nonconstant polynomial in $K[x]$ has at least one root in $L$. If $K$ is perfect, then the subfield $E=\{a\in L\mid a\text{ is algebraic over }K\}$ is algebraically closed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Proof.&lt;/em&gt; If $f(x)$ is any nonconstant polynomial in $K[x]$, then $f(x)$ is separable (since $K$ is perfect). Since $L$ satisfies the hypothesis of the theorem, then $f(x)$ splits over $L$, hence over $E$. Thus, $E$ is a splitting field of all nonconstant polynomials in $K[x]$, and algebraic over $K$, hence $E$ is algebraically closed. &lt;em&gt;QED&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;hr/&gt;&#xA;&lt;em&gt;Final addition:&lt;/em&gt; And Keith Conrad has been kind enough to point us in the comments to his notes on the subject, which shows that, indeed, first adjoining all the $p^k$th roots to all elements of $K$ will work to give the result in general. So the result &lt;em&gt;does&lt;/em&gt; hold: if every polynomial in $K[x]$ has at least one root in $L$, then $L$ contains an algebraic closure for $K$. &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-18T15:17:25.577" LastActivityDate="2011-03-18T15:17:25.577" CommentCount="8" />
  <row Id="27649" PostTypeId="2" ParentId="27646" CreationDate="2011-03-17T19:23:14.263" Score="1" ViewCount="" Body="&lt;p&gt;I would expect the amount to be paid after 200 days to be $$5000 \times 1.04^{200/365} \approx 5108.617$$ so the effective interest rate including fees over 200 days to be $$\frac{5108.617}{5000\times 0.98}-1 \approx  4.2575\%$$ and the annualised effective interest rate including fees to be $$\left(\frac{5108.617}{5000\times 0.98}\right)^{365/200}-1 \approx 7.9060\%$$&#xA;and these numbers are close to but not the same as yours.&lt;/p&gt;&#xA;" OwnerUserId="6460" LastActivityDate="2011-03-17T19:23:14.263" />
  <row Id="27650" PostTypeId="1" CreationDate="2011-03-17T19:36:43.140" Score="1" ViewCount="95" Body="&lt;p&gt;I've been wrestling with this question for quite some time now, and the result was like 20 leaves of paper packed with scribbling...anyway, here's the question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need to find the solution to the following source excited Helmholtz equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(\nabla^2 + k^2)G_1(r,a) = -\delta(r-a)$$&#xA;With the boundary condition:&#xA;$$\left. G_1(r,a)\right|_{z=0} = 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the solution is given by the integral:&#xA;$$G_1(r,a)=\iiint G(r,r&amp;#39;) \delta(r&amp;#39;-a) dV&amp;#39; + \iint [G(r,r&amp;#39;)\nabla&amp;#39;G_1(r&amp;#39;,a)-G_1(r&amp;#39;,a)\nabla&amp;#39;G(r,r&amp;#39;)]\cdot \hat{n} dS&amp;#39;$$&#xA;(The surface integral is on a closed surface, for some reason MathJaX doesn't understand what \oiint is)&#xA;The second term in the surface integral drops since it's 0 on all the boundaries of the volume that interests us (which is $z \geq 0$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;HOWEVER, I do not know what the gradient of G_1 is on $z=0$, and I understand that I'm free to chose it however I see fit as long as it only affects the $z&amp;lt;0$ region.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even simplifying the problem to a one dimensional problem I still can't solve it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I do know what the solution is, because I've been thought many years ago in introduction to Electromagnetics that when encountering a boundary over which the electrical potential is 0, I need to mirror all my sources and give them a negative sign, and if I plug that solution back into the equation is does satisfy it and the boundary condition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But how THAT solution is supposed to be inferred from the mathematical problem is beyond me. Any help/hints/insights would be most welcome.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Sorry for the paucity of tags, I wanted to use &quot;Helmholtz&quot; and &quot;Green&quot; but I'm still a newbie so I can't create new tags...)&lt;/p&gt;&#xA;" OwnerUserId="7070" LastEditorUserId="1543" LastEditorDisplayName="" LastEditDate="2011-03-17T19:50:00.233" LastActivityDate="2011-03-17T19:50:00.233" Title="Finding the modified Green function for the Helmholtz equation" Tags="&lt;differential-equations&gt;&lt;pde&gt;" CommentCount="2" FavoriteCount="1" />
  <row Id="27651" PostTypeId="2" ParentId="27621" CreationDate="2011-03-17T19:43:43.187" Score="0" ViewCount="" Body="&lt;p&gt;For $ (a, b) $ and $ (c, d) $ to be a basis, they must be linearly independent. In other words, &lt;br&gt;&#xA;det $ \begin{bmatrix}&#xD;&#xA;a &amp;amp;b \\ &#xD;&#xA; c&amp;amp; d&#xD;&#xA;\end{bmatrix} $&#xA;must be invertible. Over a field, this would imply that $det(A) \neq 0 $, but since we are in $\mathbb{Z}$, we require $det(A) = \pm 1$ so that each entry in $A^{-1} = $&lt;br&gt;&#xA;$\frac{1}{det(A)} \begin{bmatrix}&#xD;&#xA;d &amp;amp;-b \\ &#xD;&#xA; -c&amp;amp; a&#xD;&#xA;\end{bmatrix}$&#xA;is an integer. &lt;br&gt;&lt;br&gt;&#xA;Note that $ad - bc = 1$ implies that $gcd(a,c) = gcd(b,d) = 1$&lt;br&gt;&#xA;..&lt;/p&gt;&#xA;" OwnerUserId="7850" LastEditorUserId="7850" LastEditorDisplayName="" LastEditDate="2011-03-17T19:58:47.807" LastActivityDate="2011-03-17T19:58:47.807" CommentCount="5" />
  <row Id="27652" PostTypeId="1" AcceptedAnswerId="27663" CreationDate="2011-03-17T19:54:26.877" Score="4" ViewCount="240" Body="&lt;p&gt;$$f(x)=\frac{x^2+1}{x^3-x}$$&#xA;$$f^{(100)}(x)=?$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried differnetiating once and twice, but did not see any pattern emerging and can't guess what the 100th derivative should be. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;&#xA;so decomposing this as $$f(x)=-\frac{1}{x}+\frac{1}{x+1}+\frac{1}{x-1}&#xD;&#xA;$$  does the job. Thanks for the hints! (edit: Sivaram has a complete calculation) Although a similar approach would greatly simplify this (next) problem can someone tell me what is wrong with my approach&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;My usual line of attack is to use talor expansion. For example the next problem in the same list asks for the $100^{th}$ derivative of $$\frac{1}{x^2-3x+2}$$ at $x=0$ within 10% relative error. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;NOTE&lt;/strong&gt;:The above is a mistype, the following attempt is for $\frac{1}{x^2+3x+2}$. A better general approach, which is what I was looking for is described in the answer posted below.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I know I can expand in a Maclaurin series $$\frac{1}{x^2+3x+2}=\frac{1}{2} (1+\frac{x^2+3x}{2} + (\frac{x^2+3x}{2})^2 +\cdots)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After taking 100 derivatives I would be left to differentiate the following. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{1}{2}((\frac{x^2+3x}{2})^{50}+(\frac{x^2+3x}{2})^{51}+\cdots)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$=\frac{1}{2}\left(\frac{\sum_{k=0}^{50}{{50}\choose{k}}3^{50-k} x^{50+k}&#xD;&#xA;}{2^{50}}+\frac{\sum_{k=0}^{51}{{51}\choose{k}}3^{51-k} x^{51+k}&#xD;&#xA;}{2^{51}}+\cdots\frac{\sum_{k=0}^{100}{{100}\choose{k}}3^{100-k} x^{100+k}&#xD;&#xA;}{2^{100}}\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because anything on either side of these values would disappear when i take the hundreth derivative at $x=0$ . And it is also easy to sea that I will get exactly one term from each of the sums, so I get an answer, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$=100!\sum_{k=0}^{50}\frac{3^{2k}}{2^{50+k}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which is wrong, well because the answer is too huge and Im to find a number within 10%. Can someone tell me where I went wrong, and if there is a cleaner way to approach these problems. &lt;/p&gt;&#xA;" LastActivityDate="2011-03-17T22:06:36.803" Title="How to find the $100^{th}$ derivative of $\frac{x^2+1}{x^3-x}$" Tags="&lt;calculus&gt;" AnswerCount="3" CommentCount="12" />
  <row Id="27653" PostTypeId="5" CreationDate="2011-03-17T19:57:36.937" Score="0" ViewCount="" Body="&lt;p&gt;A triangle is one of the basic shapes of geometry: a polygon with three corners or vertices and three sides or edges which are line segments.&lt;/p&gt;&#xA;" OwnerUserId="-1" LastEditorUserId="7149" LastEditorDisplayName="" LastEditDate="2011-03-17T19:59:12.750" LastActivityDate="2011-03-17T19:59:12.750" />
  <row Id="27654" PostTypeId="4" CreationDate="2011-03-17T19:57:36.937" Score="0" ViewCount="" Body="For questions about triangles" OwnerUserId="-1" LastEditorUserId="7149" LastEditorDisplayName="" LastEditDate="2011-03-17T19:59:15.467" LastActivityDate="2011-03-17T19:59:15.467" />
  <row Id="27656" PostTypeId="1" AcceptedAnswerId="30097" CreationDate="2011-03-17T19:59:32.177" Score="4" ViewCount="254" Body="&lt;p&gt;If $\mathbf{A}_{n\times n}$ is a positive semi-definite matrix with eigenvalues $\{\alpha_k\},\ k\in\{1,...,n\}$, and $\mathbf{B}_{m\times n}$ is an arbitrary matrix with singular values $\{\beta_k\},\ k\in\{1,...,\min(m,n)\}$, can anything be said about the singular values $\{\gamma_k\},\ k\in\{1,...,\min(m,n)\}$ of the matrix $\mathbf{\Gamma}=\mathbf{BA}$? Is there a way I can relate $\gamma_k$ to $\alpha_k$ and $\beta_k$?&lt;/p&gt;&#xA;" OwnerUserId="7815" LastActivityDate="2011-03-31T07:46:59.870" Title="Eigenvalues of product of matrices" Tags="&lt;matrices&gt;&lt;eigenvalues&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="3" />
  <row Id="27658" PostTypeId="2" ParentId="27652" CreationDate="2011-03-17T20:03:18.737" Score="5" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt;: Try using &lt;a href=&quot;http://en.wikipedia.org/wiki/Partial_fraction#Procedure&quot; rel=&quot;nofollow&quot;&gt;partial fraction decomposition&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2465" LastEditorUserId="2465" LastEditorDisplayName="" LastEditDate="2011-03-17T20:09:07.453" LastActivityDate="2011-03-17T20:09:07.453" CommentCount="3" />
  <row Id="27660" PostTypeId="1" AcceptedAnswerId="27676" CreationDate="2011-03-17T20:15:56.157" Score="1" ViewCount="190" Body="&lt;p&gt;This question may be a bit vague, but neverthless, i would like to see an answer. Wikipedia tells me that:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;em&gt;In abstract algebra, localization is a systematic method of adding multiplicative inverses to a ring.&lt;/em&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;It is clear that for &lt;em&gt;integral domains&lt;/em&gt;, we have the &lt;em&gt;Field of Fractions&lt;/em&gt;, and we work on it. It's obvious that a Field has multiplicative inverses. Now if we consider any arbitrary ring $R$, by the definition of localization, it means that we are adding multiplicative inverses to $R$ thereby wanting $R$ to be a &lt;em&gt;division ring&lt;/em&gt; or a &lt;em&gt;field&lt;/em&gt;. My question, why is it so important to look at this concept of Localization. Or how would the theory look like if we never had the concept of Localization. &lt;/p&gt;&#xA;" LastActivityDate="2011-03-18T06:51:00.263" Title="Why do the Localization of a Ring" Tags="&lt;abstract-algebra&gt;&lt;ring-theory&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="27661" PostTypeId="2" ParentId="27652" CreationDate="2011-03-17T20:18:19.290" Score="0" ViewCount="" Body="&lt;p&gt;Doesn't look like there's a pattern to it. Mathematica gives me an answer that's like 4 pages long. And I don't think your answer is too large, because plotting it out gave values of the order $10^{200}$ and $100!\approx 10^{158}$, and the sum is approx $10^{17}$. So, the numbers aren't all that off. However, the 100th derivative evaluated at 0, was 0.&lt;/p&gt;&#xA;" OwnerUserId="7815" LastActivityDate="2011-03-17T20:18:19.290" CommentCount="2" />
  <row Id="27662" PostTypeId="1" AcceptedAnswerId="27671" CreationDate="2011-03-17T20:18:19.753" Score="4" ViewCount="95" Body="&lt;p&gt;The following theorem is an alternative definition of limits. In this theorem, you don't need to know the value of $\lim \limits_{x \rightarrow c} {f(x)}$ in order to prove the limit exists.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Let $I \in R$ be an open interval, let $c \in I$, and let $f: I-{c} \rightarrow R$ be a function. Then $\lim \limits_{x \rightarrow c} {f(x)}$ exists if for each $\epsilon &amp;gt; 0$, there is some $\delta &amp;gt; 0$ such that $x,y \in I-{c}$ and $\vert x-c \vert &amp;lt; \delta$ and $\vert y-c \vert &amp;lt; \delta$ implies $\vert f(x)-f(y) \vert &amp;lt; \epsilon$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have extended this to a theorem for one-sided limits: $\lim \limits_{x \rightarrow c+} {f(x)}$ exists if or each $\epsilon &amp;gt; 0$, there is some $\delta &amp;gt; 0$ such that $c&amp;lt;x&amp;lt;c+ \delta$, and $c&amp;lt;y&amp;lt;c+ \delta$ implies $\vert f(x)-f(y) \vert &amp;lt; \epsilon$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is: how to prove this extension to one-sided limits? I'm not familiar with Cauchy sequences, so I would prefer $\epsilon$-$\delta$ proof. &lt;/p&gt;&#xA;" OwnerUserId="7429" LastEditorUserId="7429" LastEditorDisplayName="" LastEditDate="2011-03-17T20:30:10.633" LastActivityDate="2011-03-17T21:27:47.097" Title="extend an alternative definition of limits to one-sided limits" Tags="&lt;real-analysis&gt;" AnswerCount="3" CommentCount="4" FavoriteCount="1" />
  <row Id="27663" PostTypeId="2" ParentId="27652" CreationDate="2011-03-17T20:19:00.533" Score="5" ViewCount="" Body="&lt;p&gt;$$\frac{x^2+1}{x^3-x} = -\frac{1}{x} + \frac{1}{x+1} + \frac{1}{x-1} = -x^{-1} + (x+1)^{-1} + (x-1)^{-1}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{d^n(y^{-1})}{dy^n} = \frac{(-1)^n n!}{y^{n+1}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, the $n^{th}$ derivative is&#xA;$$\frac{(-1)^{n+1} n!}{x^{n+1}} +  \frac{(-1)^n n!}{(x-1)^{n+1}} +  \frac{(-1)^n n!}{(x+1)^{n+1}} = (-1)^n n! \times \left( \frac{-1}{x^{n+1}} + \frac{1}{(x-1)^{n+1}} + \frac{1}{(x+1)^{n+1}}\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly,&#xA;$$\frac{1}{x^2-3x+2} = \frac{1}{x-2} - \frac{1}{x-1}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, the $n^{th}$ derivative is&#xA;$$\frac{(-1)^n n!}{(x-2)^{n+1}} -  \frac{(-1)^n n!}{(x-1)^{n+1}} = (-1)^n n! \times \left( \frac{1}{(x-2)^{n+1}} - \frac{1}{(x-1)^{n+1}}\right)$$&lt;/p&gt;&#xA;" OwnerUserId="3302" LastActivityDate="2011-03-17T20:19:00.533" CommentCount="1" />
  <row Id="27664" PostTypeId="2" ParentId="27373" CreationDate="2011-03-17T20:21:58.630" Score="1" ViewCount="" Body="&lt;p&gt;I think that Kristi is just looking for the new coordinates that she would get after projecting her arbitrary point ($x_1,x_2,x_3$) onto the plane $x_2=x_1$. I don't think she was looking for the other responses in particular, but they could have been some help maybe. Once the new coordinates are found, which she states as the question marks in the the first matrix of elements, she would be able to find the values that she could plug in to the coefficient matrix &lt;strong&gt;B&lt;/strong&gt; to find the eigenvalues and eigenvectors. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So restating, I think she is just having a troublesome time finding those new coordinates after projecting the arbitrary point onto the plane. I would answer but I am not sure I know how to do this. I just thought I should clear up the question for everyone else who views and responds.&lt;/p&gt;&#xA;" LastActivityDate="2011-03-17T20:21:58.630" />
  <row Id="27665" PostTypeId="1" CreationDate="2011-03-17T20:29:02.950" Score="2" ViewCount="53" Body="&lt;p&gt;Let $k$ be a field, not necessarily algebraically closed, not necessarily of characteristic 0 (actually, the example I have in mind is $k=F_2$).  Let $V,W$ be varieties over $k$, and $W\to V$ a morphism defined over $k$, such that every fiber is isomorphic to an affine space of the same dimension.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What conditions imply that $W$ is actually $V\times A^n$?&lt;/p&gt;&#xA;" OwnerUserId="176" LastActivityDate="2011-03-17T20:29:02.950" Title="When is an affine bundle trivial?" Tags="&lt;algebraic-geometry&gt;" CommentCount="2" />
  <row Id="27666" PostTypeId="2" ParentId="27662" CreationDate="2011-03-17T20:29:40.183" Score="2" ViewCount="" Body="&lt;p&gt;Answer: For each positive $\epsilon$, there is some positive $\delta$ such that  $\vert f(x)-f(y) \vert \le \epsilon$ for every $x$ and $y$ in $I$ such that $c&amp;lt;x\le c+\delta$ and $c&amp;lt;y\le c+\delta$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;About the question added later on in your post: you could try to mimick the proof given in the two sided case.&lt;/p&gt;&#xA;" OwnerUserId="6179" LastActivityDate="2011-03-17T20:29:40.183" CommentCount="0" />
  <row Id="27667" PostTypeId="1" AcceptedAnswerId="27668" CreationDate="2011-03-17T20:29:59.920" Score="4" ViewCount="127" Body="&lt;p&gt;Let $G$ and $H$ be finite abelian groups such that $G \times G \cong H \times H$ &lt;br&gt;&#xA;Then $G \cong H$&lt;br&gt;&lt;br&gt;&#xA;I &lt;em&gt;was&lt;/em&gt; going to just write the hypothesis as $G^2 \cong H^2$ and take square roots on both sides, but I don't think that would suffice (and neither would saying &quot;true&quot; &lt;em&gt;a la&lt;/em&gt; &lt;strong&gt;Myself&lt;/strong&gt;!)...&lt;br&gt;&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The hypothesis tells us that there is an isomorphism, say $f: G \times G\to H \times H$.&lt;br&gt;&#xA;I would like to use this to come to the conclusion that there is a bijective homomorphism &lt;br&gt;$g: G \to H$.  (I will be using additive notation...)&lt;br&gt; &lt;br&gt;&#xA;Since $f((a, b) + (c, d)) = f(a,b) + f(c, d)$ for all $a,b \in G$ and $c, d \in H$, I was thinking about choosing an arbitrary $a, b \in G$ and calculating:&lt;br&gt;&#xA;$f((a, 0) + (b, 0)) = f(a, 0) + f(b, 0) \Rightarrow$&lt;br&gt;$&#xA;f(a + b, 0) = f(a, 0) + f(b, 0)$.&lt;br&gt;&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I basically need to define g in such a way that it extracts the first dimension from the equation. Clearly (I think!), g will invoke f in some way. Can I have a tip on this? It's probably very simple, but I'm not sure how to express it symbolically.&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I &lt;em&gt;might&lt;/em&gt; not need to show that g is bijective if I can say something to the effect of &quot;this routine verification is straightforward and left to the reader&quot;, but I'm afraid I might not be able to perform such a verification if put on the spot! Here's a stab:&lt;br&gt;&#xA;f injective $\Leftrightarrow f(a,b) = f(c,d) \Rightarrow (a, b) = (c, d) \Rightarrow a = c \wedge b = d$&lt;br&gt;&#xA;So by definition of g (forthcoming...), g(a) = g(c) implies that a = c. &lt;br&gt;&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Surjective. For all $x, y \in H$, there exists $a, b \in G : f(a,b) = (x, y)$&lt;br&gt;&#xA;Man, this really seems trivial, but without my definition of g, I feel like I'm handwaving... &#xA;&lt;br&gt;&lt;br&gt;&#xA;Thanks again, guys! &lt;/p&gt;&#xA;" OwnerUserId="7850" LastEditorUserId="5189" LastEditorDisplayName="" LastEditDate="2011-03-18T12:19:29.663" LastActivityDate="2011-03-31T18:49:32.677" Title="Finite abelian groups and isomorphism" Tags="&lt;homework&gt;&lt;abstract-algebra&gt;&lt;group-theory&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="27668" PostTypeId="2" ParentId="27667" CreationDate="2011-03-17T20:45:41.230" Score="4" ViewCount="" Body="&lt;p&gt;Note that it is enough to assume that both $G$ and $H$ are finite $p$ torsion groups for some prime $p$ because under any isomorphism $p$ torsion elements go to $p$ torsion elements and any finite abelian group is direct sum of $p$ torsion groups for finitely many distinct prime $p$. So both $G$ and $H$ have the form $\mathbb{Z}/p^{n_1}\mathbb{Z} \times \mathbb{Z}/p^{n_2}\mathbb{Z} \times \ldots \mathbb{Z}/p^{n_m}\mathbb{Z}$ for some prime $p$. Now you may use structure theorem.&lt;/p&gt;&#xA;" OwnerUserId="7565" LastEditorUserId="7565" LastEditorDisplayName="" LastEditDate="2011-03-31T18:49:32.677" LastActivityDate="2011-03-31T18:49:32.677" CommentCount="9" />
  <row Id="27669" PostTypeId="1" CreationDate="2011-03-17T20:49:39.897" Score="-1" ViewCount="69" Body="&lt;p&gt;Consider the category of all open sets of a given topological space where the morphism are inclusions,why one can see a Presheaf as a contravariant functor?&lt;/p&gt;&#xA;" OwnerUserId="8390" LastActivityDate="2011-03-17T21:04:13.427" Title="Presheaf on a topological space" Tags="&lt;category-theory&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="27670" PostTypeId="1" CreationDate="2011-03-17T20:50:19.027" Score="-2" ViewCount="95" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Possible Duplicate:&lt;/strong&gt;&lt;br&gt;&#xA;  &lt;a href=&quot;http://math.stackexchange.com/questions/27609/n-c-find-n&quot;&gt;n! = c, find n?&lt;/a&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;p&gt;I want to find the approximate n from &lt;a href=&quot;http://hyperphysics.phy-astr.gsu.edu/hbase/math/stirling.html&quot; rel=&quot;nofollow&quot;&gt;Stirling's approximation&lt;/a&gt; with a known n!. How do I deal with this equation?&lt;/p&gt;&#xA;" OwnerUserId="4474" LastActivityDate="2011-03-17T21:23:07.190" ClosedDate="2011-03-18T00:41:12.070" Title="1000 = ((n/e)^n) * sqrt(2pi*n), Solve n" Tags="&lt;algebra&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" />
  <row Id="27671" PostTypeId="2" ParentId="27662" CreationDate="2011-03-17T20:53:14.030" Score="3" ViewCount="" Body="&lt;p&gt;You have essentially the same difficulty as in the &lt;a href=&quot;http://math.stackexchange.com/questions/27490/proof-that-an-alternative-definition-of-limit-is-equivalent-to-the-usual-one&quot;&gt;previous question&lt;/a&gt;: figuring out a &quot;target&quot; number $L$ for the limit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(By the way: if you don't know what Cauchy sequences are, you should have said something in that previous problem!)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can proceed in a similar manner. First, you want to find a potential &quot;target.&quot; For $\epsilon_n = \frac{1}{n}$, you know there is a $\delta_n$ (which you may assume is less than or equal to $\frac{1}{n}$ and less than $\delta_{n-1}$) such that for all $x$ and $y$, if $c\lt x\lt c+\delta_n$ and $c\lt y \lt c+\delta_n$, then $|f(x)-f(y)|\lt \frac{1}{n}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that this means that the values of $f$ on $(c,c+\delta_n)$ are bounded: for any $x\in (c,c+\delta_n)$, you know that $|f(x) - f(c+\delta_n/2)|\lt \frac{1}{n}$, so $|f(x)| \lt \frac{1}{n}+|f(c+\delta_n/2)|$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, there is a greatest lower bound $a_n$ to $f(c,c+\delta_n)$. Since $(c,c+\delta_{n+1})\subseteq (c,c+\delta_n)$, we have that $a_n\leq a_{n+1}$. So the sequence $a_1,a_2,\ldots$ is an increasing function. The sequence is bounded above, so the sequence converges to some $L$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The obvious &quot;target&quot; for the limit is $L$. So, let $\epsilon\gt 0$, and you want to show that there is a $\delta\gt 0$ such that for all $x$, if $c\lt x\lt c+\delta$, then $|f(x)-L|\lt \epsilon$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $L$ is the least upper bound of $a_1,a_2,\ldots$, there exists $N$ such that $L-\frac{\epsilon}{2} \lt a_n \leq L$ for all $n\geq N$. And there exists an $M$ such that $\frac{1}{M}\lt \frac{\epsilon}{4}$. Let $K=\max\{M,N\}$, and consider $\delta=\delta_K$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know that since $a_K$ is the greatest lower bound of $f(c,c+\delta_K)$, there exists $y\in (c,c+\delta_K)$ such that $a_K\leq f(y)\lt a_K+\frac{\epsilon}{4}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now suppose that $c\lt x \lt c+\delta_K$. Then&#xA;\begin{align*}&#xA;|f(x)-L| &amp;amp;\leq |f(x)-f(y)|+|f(y)-a_K| + |a_K-L|  &amp;amp;&amp;amp;\mbox{(triangle inequality)}\\&#xA;&amp;amp;\lt \frac{1}{K} + |f(y)-a_K| + |a_K-L| &amp;amp;&amp;amp;\mbox{(choice of $\delta_K$)}\\&#xA;&amp;amp;\leq \frac{1}{K} + \frac{\epsilon}{4} + |a_K-L| &amp;amp;&amp;amp;\mbox{(choice of $y$)}\\&#xA;&amp;amp;\leq \frac{1}{K} + \frac{\epsilon}{4} + \frac{\epsilon}{2} &amp;amp;&amp;amp;\mbox{(since $K\geq N$)}\\&#xA;&amp;amp;\lt \frac{\epsilon}{4} + \frac{\epsilon}{4} + \frac{\epsilon}{2} &amp;amp;&amp;amp;\mbox{(since $K\geq M$)}\\&#xA;&amp;amp;=\epsilon.&#xA;\end{align*}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(With this in mind, you can try doing the two-sided version of the problem without invoking Cauchy sequences; the idea is very similar, and you don't need to consider both least upper bounds and greatest lower bounds for $f(c-\delta_n,c+\delta_n)$, just one of the two and use their limit as a &quot;target&quot;. Once you know what to aim for, hitting it becomes much easier).&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2011-03-17T20:53:14.030" CommentCount="1" />
  <row Id="27672" PostTypeId="2" ParentId="27669" CreationDate="2011-03-17T21:04:04.270" Score="0" ViewCount="" Body="&lt;p&gt;Because presheaf deals with restriction. Please read 1st chapter of Red book by Mumford for more details. It is explained very well there.&lt;/p&gt;&#xA;" OwnerUserId="7565" LastActivityDate="2011-03-17T21:04:04.270" />
  <row Id="27673" PostTypeId="2" ParentId="27669" CreationDate="2011-03-17T21:04:13.427" Score="2" ViewCount="" Body="&lt;p&gt;Suppose your presheaf $F$ has values in some category, say rings. Then to each open $U $ there corresponds a ring $F(U)$, and to the inclusion map $U\to V$ where $U\subseteq V$ there corresponds map $V\to U$ in the opposite direction, sometimes called the restriction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then you should verify that the axioms of a functor hold. The domain of your functor is just the poset category where the objects are open sets and the morphisms are inclusion maps&lt;/p&gt;&#xA;" OwnerUserId="2055" LastActivityDate="2011-03-17T21:04:13.427" CommentCount="1" />
  <row Id="27674" PostTypeId="2" ParentId="27660" CreationDate="2011-03-17T21:13:37.977" Score="6" ViewCount="" Body="&lt;p&gt;In Algebraic Geometry, if you have a variety $\mathbf{V}$, then you have the corresponding coordianate ring $A(\mathbf{V})$ to study the variety. To each point $p\in\mathbf{V}$ there corresponds a maximal ideal $\mathfrak{M}$ of $A(\mathbf{V})$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The ring $A(\mathbf{V})$ carries information about what is going on in all of $\mathbf{V}$. If you are only interested in understanding what is happening at (or near) the point $p$, then a lot of the information in $A(\mathbf{V})$ doesn't really matter. If you localize $A(\mathbf{V})$ at $\mathfrak{M}$, what you get is precisely the information about what is going on near $p$.  For example, there is a one-to-one correspondence between the prime ideals of this localization and the subvarieties of $\mathbf{V}$ that go through the point $p$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The coordinate ring is often not an integral domain (e.g., if the variety is not irreducible). So you want to be able to localize in rings that are not integral domains. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;More generally, localizing gives you a way to focus on particular parts of the ideal structure of $R$, even when $R$ is not an integral domain (so that there is no ring of fractions to embed into); if the ideal you are looking at is a prime ideal, then you can &quot;strip away&quot; all the stuff that is not inside of $\mathfrak{p}$ and make it &quot;not matter&quot;, so that you can just focus on the stuff what is going on inside of $\mathfrak{p}$. This is important and useful even when we don't have a division ring or a field to embed into. &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-17T21:22:45.080" LastActivityDate="2011-03-17T21:22:45.080" />
  <row Id="27675" PostTypeId="1" AcceptedAnswerId="27680" CreationDate="2011-03-17T21:18:53.913" Score="2" ViewCount="94" Body="&lt;p&gt;This is not really significant to what I am trying to do right now but I'm just wondering...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is it that the discriminant for the cyclotomic polynomials $\Phi_{1}(x)$ and $\Phi_{2}(x)$ is 1? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using WolframAlpha, the discriminant for other linear polynomials is also 1. Why is that?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you.&lt;/p&gt;&#xA;" OwnerUserId="6838" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-17T21:29:40.503" LastActivityDate="2011-03-18T00:44:46.810" Title="Why is the discriminant of a linear polynomial equal to  1?" Tags="&lt;polynomials&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="27676" PostTypeId="2" ParentId="27660" CreationDate="2011-03-17T21:20:38.720" Score="9" ViewCount="" Body="&lt;p&gt;Localization is a technique which allows one to concentrate attention to what is happening &lt;em&gt;near&lt;/em&gt; a prime, for example. When you localize at a prime, you have simplified abruptly the behavior of your ring &lt;em&gt;outside&lt;/em&gt; that prime but you have more or less kept everything &lt;em&gt;inside&lt;/em&gt; it intact.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For lots of questions, this significantly simplifies things.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Indeed, there are very general procedures, in lots of contexts, which go by the name of &lt;em&gt;localization&lt;/em&gt;, and their purpose is usually the same: if you are lucky, the problems you are interested in can be solved &lt;em&gt;locally&lt;/em&gt; and then the &quot;local solutions&quot; can be glued together to obtain a solution to your original problem. Moreover, an immense deal of effort has been done in order to extent the meaning of &quot;local&quot; so as to be able to apply this strategy in more contexts: I have always loved the way the proofs of some huge theorems of algebraic geometry consist more or less in setting up an elaborate technology in order to be able to say the magical &quot;It is enough to prove this locally&quot;, and then, thanks to the fact that we worked so much in that technology, immediately conclude the proof with a &quot;where it is obvious&quot; :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, all sort of bad things can happen. For example, sometimes the &quot;local solutions&quot; cannot be glued together into a &quot;global solution&quot;, &amp;amp;c. (Incidentally, when this happens, so that you can do something locally but not glue the result, you end up with a cohomology theory which, more or less, is the art of dealing with that problem)&lt;/p&gt;&#xA;" OwnerUserId="274" LastEditorUserId="274" LastEditorDisplayName="" LastEditDate="2011-03-17T22:38:58.870" LastActivityDate="2011-03-17T22:38:58.870" />
  <row Id="27677" PostTypeId="2" ParentId="27373" CreationDate="2011-03-17T21:21:07.187" Score="1" ViewCount="" Body="&lt;p&gt;To sum it all up:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First the matrix: Its columns are the images of the basis vectors. As $p({\bf e}_1) =p({\bf e}_2)=({1\over2},{1\over2},0)$ and $p({\bf e}_3)=(0,0,1)$ we obtain the matrix&#xA;$$[p]=\left[\matrix{1/2 &amp;amp; 1/2 &amp;amp; 0 \cr&#xD;&#xA;1/2 &amp;amp; 1/2 &amp;amp;0 \cr 0&amp;amp;0&amp;amp;1\cr}\right]\&amp;gt;.$$&#xA;Since $p$ is a projection its eigenvalues are $0$ and $1$. The eigenspace $E_0$ consists of the vectors which are parallel to the direction of the projection, i.e. the scalar multiples of $(1, -1,0)$, and the eigenspace $E_1$ consists of all vectors that stay fixed under $p$, i.e., $E_1=P$. A basis of $E_1$ is given by, e.g., $\bigl((1,1,0),(0,0,1)\bigr)$.&lt;/p&gt;&#xA;" OwnerUserId="1303" LastEditorUserId="5363" LastEditorDisplayName="" LastEditDate="2011-03-18T00:46:02.440" LastActivityDate="2011-03-18T00:46:02.440" CommentCount="2" />
  <row Id="27678" PostTypeId="2" ParentId="27670" CreationDate="2011-03-17T21:23:07.190" Score="2" ViewCount="" Body="&lt;p&gt;If you take the log, you have $5.07\approx \ln 1000-\ln \sqrt{2\pi}=(n+\frac{1}{2})\ln n-n$.  This is slowly varying, so any numerical method will work.  I would just put it into a spreadsheet and use the goal seek function, or calculate for 3 through 7 and use bisection or the secant method.  For 1000, I would know 6!=720 and 7!=5040, so it is a bit more than 6, but maybe you want it for much larger numbers.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-03-17T21:23:07.190" />
  <row Id="27679" PostTypeId="2" ParentId="27662" CreationDate="2011-03-17T21:27:47.097" Score="2" ViewCount="" Body="&lt;p&gt;Assume that $\lim_{x \to c^+} f(x) = g$ exists. Then for any $\varepsilon &amp;gt; 0$ there exists $\delta &amp;gt; 0$ such that if $0 &amp;lt; x - c &amp;lt; \delta$ then $|f(x) - g| &amp;lt; \varepsilon$ and if $0 &amp;lt; y - c &amp;lt; \delta$ then $|f(y) - g| &amp;lt; \varepsilon$. Hence&#xA;$$&#xD;&#xA;|f(x) - f(y)| \leq |f(x) - g| + |f(y) - g| &amp;lt; 2\varepsilon&#xD;&#xA;$$&#xA;for $0 &amp;lt; x - c &amp;lt; \delta$ and $0 &amp;lt; y - c &amp;lt; \delta$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, assume that your definition holds. Then there exists a strictly decreasing sequence $(c_n)_{n \geq 1}$ such that $c_n \to c$. Thus for a given $\varepsilon &amp;gt; 0$ there exists $\delta &amp;gt; 0$ such that there exists positive integer $N$ such that $|f(c_m)-f(c_n)| &amp;lt; \varepsilon$ for $m,n &amp;gt; N$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It means that $(f(c_n))$ is Cauchy sequence and converges to, say, $g$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(If you are not familiar with Cauchy sequences, it is easy to prove that the sequence $(a_n)$ such that for $\varepsilon &amp;gt; 0$ there exists $N &amp;gt; 0$ such that $|a_m - a_n| &amp;lt; \varepsilon$ for $m,n &amp;gt; N$ converges. To prove it, notice first that $(a_n)$ is bounded. Then, by Bolzano-Weierstrass theorem there is a subsequence $(a_{n_k})$ which converges to some $a$. Then, by Cauchy condition and triangle inequality $a_n \to a$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume now that there exists a sequence $(c&amp;#39;_n)_{n \geq 1}$ such that $f(c&amp;#39;_n) \to g&amp;#39; \neq g$. Let $C_n$ equals $c_n$ if $n$ is even, and equals $c&amp;#39;_n$ if $n$ is odd. Hence $(f(C_n))$ diverges but on the other hand we have $|f(C_m) - f(C_n)| &amp;lt; \varepsilon$ for sufficiently large $m,n$ and this means that $(f(C_n))$ is Cauchy, hence converges. Contradiction.&lt;/p&gt;&#xA;" OwnerUserId="8229" LastActivityDate="2011-03-17T21:27:47.097" />
  <row Id="27680" PostTypeId="2" ParentId="27675" CreationDate="2011-03-17T21:27:58.133" Score="7" ViewCount="" Body="&lt;p&gt;If $f(x)$ is a polynomial with complex coefficients, and $a_1\ldots,a_n$ are all its roots (with multiplicity) in $\mathbb{C}$, then by definition the discriminant of $f(x)$ is the product of the squares of the differences between any two roots:&#xA;$$f_n^{2n-2}\prod_{1\leq i\lt j\leq n} (a_i-a_j)^2,$$&#xA;where $f_n$ is the leading coefficient of $f(x)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $f(x)$ is linear, then $n=1$. So there is no way to pick $i$ and $j$, with $1\leq i\lt j\leq 1$. So the question then becomes: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;How much is a product with no factors?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Well, if we want the associativity law to hold in general, then we need the product with no factors to not affect multiplication. Because if we have $x_1\cdot x_2 \cdots x_n$, we want to be able to associate any which way and get the same answer; but one way to associate would be, for example, $(x_1\cdot x_2)()(x_3\cdots x_n)$ (with a product with no factors in the middle, or really anywhere for that matter). So we need the product with no factors to equal $1$ for everything to work out well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we &lt;strong&gt;define&lt;/strong&gt; a product with no factors to be equal to $1$. (For the same reason, a sum with no summands is defined to be equal to $0$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the discriminant of a linear polynomial is a product with no factors, and a product with no factors is equal to $1$, the discriminant of a linear polynomial is equal to $1$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $\Phi_1(x) = x-1$, and $\Phi_2(x) = x+1$, both are linear, so their discriminant is $1$. &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="6622" LastEditorDisplayName="" LastEditDate="2011-03-18T00:44:46.810" LastActivityDate="2011-03-18T00:44:46.810" CommentCount="3" />
  <row Id="27681" PostTypeId="2" ParentId="27633" CreationDate="2011-03-17T21:31:13.130" Score="7" ViewCount="" Body="&lt;p&gt;I don't know the state of the art on this question, but some searches indicate that it is not known whether every even number is a difference of two primes, let alone a difference of consecutive primes.  Perhaps a little old, but this is mentioned in &lt;em&gt;The new book of prime number records&lt;/em&gt; by Paulo Ribenboim, 1996, on &lt;a href=&quot;http://books.google.com/books?id=72eg8bFw40kC&amp;amp;lpg=PP1&amp;amp;pg=PA250#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;page 250&lt;/a&gt;.  (This problem is called the &quot;Goldbach Variation&quot; in Hofstadter's &lt;em&gt;Gödel, Escher, Bach&lt;/em&gt;.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The following source seems more recent, and mentions that Chen's work related to the Goldbach conjecture showed that every even number is a difference of a prime and a product of at most two primes: &lt;a href=&quot;http://primes.utm.edu/notes/conjectures/&quot; rel=&quot;nofollow&quot;&gt;http://primes.utm.edu/notes/conjectures/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2011-03-17T21:31:13.130" CommunityOwnedDate="2011-03-17T21:31:13.130" CommentCount="4" />
  <row Id="27682" PostTypeId="1" AcceptedAnswerId="27685" CreationDate="2011-03-17T21:35:12.417" Score="1" ViewCount="103" Body="&lt;p&gt;if $f$ is integrable, i wish to show&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{n}{2} \int_{-1/n}^{1/n} (f(x+y) - f(x))dy \to 0 $ as $ n \to \infty $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;this looks to be very intuitive, but im having trouble proving it formally. any tips?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: added the n infront of the integral, thanks Jonas&lt;/p&gt;&#xA;" OwnerUserId="6556" LastEditorUserId="6556" LastEditorDisplayName="" LastEditDate="2011-03-17T22:22:21.610" LastActivityDate="2011-03-17T22:26:31.297" Title="convergence of integrals" Tags="&lt;measure-theory&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" />
  <row Id="27683" PostTypeId="1" AcceptedAnswerId="27780" CreationDate="2011-03-17T21:57:15.243" Score="3" ViewCount="109" Body="&lt;p&gt;I have begun working with Mathematica for some Graph Theory, and I want to compute the number of spanning trees of all cubic graphs up to 12 vertices. I have found that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;GraphData[&quot;Cubic&quot;, 12]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;will give me the cubic graphs with 12 vertices. But how do I compute the number of spanning trees in each of these graphs. A good tutorial for Graph Theory with Mathematica seems impossible to find (I have found many, but none are actually readable). I can find this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;GraphData[8];&#xA;Union[GraphData[#, &quot;VertexCount&quot;] &amp;amp; /@ %%]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;to compute the number of vertices in all graphs on 8 vertices, but the #-tag is not defined, and &amp;amp;/@%% seems pretty magical.&lt;/p&gt;&#xA;" OwnerUserId="3317" LastActivityDate="2011-05-08T13:02:59.950" Title="Union on Graph Data in Mathematica" Tags="&lt;graph-theory&gt;&lt;mathematica&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="27684" PostTypeId="1" AcceptedAnswerId="27687" CreationDate="2011-03-17T22:03:49.210" Score="3" ViewCount="146" Body="&lt;p&gt;Hey, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would really appreciate if you could help out with the following question&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Let $W_1,W_2,\ldots$ be the event times in a Poisson process $\{X(t);t\ge0\}$ of rate $\lambda$, and let $f(w)$ be an arbitrary function. Verify that&#xA;  $$&#xD;&#xA;\mathbb{E}\left[\sum_{i=1}^{X(t)}f(W_i)\right]=\lambda\int_0^tf(w)\,dw.&#xD;&#xA;$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(originally posted to this link: &lt;a href=&quot;http://i.imgur.com/3e6mf.png&quot; rel=&quot;nofollow&quot;&gt;http://i.imgur.com/3e6mf.png&lt;/a&gt;). I tried conditioning on X(t) = n but I am unsure how to proceed after that... Also I am not exactly sure how to show this is true since f(w) is an arbitrary function and I do not understand what that means.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks a bunch for the help! I really &lt;/p&gt;&#xA;" OwnerUserId="8393" LastEditorUserId="1321" LastEditorDisplayName="" LastEditDate="2011-03-18T01:10:34.727" LastActivityDate="2011-03-18T01:10:34.727" Title="Poisson Stochastic Process Question" Tags="&lt;stochastic-processes&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="27685" PostTypeId="2" ParentId="27682" CreationDate="2011-03-17T22:07:50.623" Score="3" ViewCount="" Body="&lt;p&gt;A plausible interpretation of the question is that you're asking whether $\int_{-1/n}^{1/n}f(x+y)dy$ is asymptotically equivalent to $\frac{2}{n}f(x)$ as $n\to\infty$, which amounts to the same as asking whether $\frac{n}{2}\int_{-1/n}^{1/n}f(x+y)dy\to f(x)$ as $n\to\infty$.  If $f$ is locally integrable, then this holds almost everywhere by &lt;a href=&quot;http://en.wikipedia.org/wiki/Lebesgue_differentiation_theorem&quot; rel=&quot;nofollow&quot;&gt;Lebesgue's differentiation theorem&lt;/a&gt;.  You could not hope for more than this, because changing the value of $f$ at $x$ (or on any set of measure $0$) will not affect the integrals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: Since I first posted, the question was edited to what I answered.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2011-03-17T22:26:31.297" LastActivityDate="2011-03-17T22:26:31.297" />
  <row Id="27686" PostTypeId="1" AcceptedAnswerId="27688" CreationDate="2011-03-17T22:22:40.643" Score="2" ViewCount="116" Body="&lt;p&gt;From what I understand, Black-Scholes equation in finance is used to price options which are a contract between a potential buyer and a seller. Can I use this mathematical framework to &quot;buy&quot; a stock? I do not have the choice using options in the market I am dealing with -- I either buy something or I don't. So I was wondering if B-S be used to decide to buy a stock, the next day, taking its last price, volatility and other necessary variables into account.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks, &lt;/p&gt;&#xA;" OwnerUserId="6786" LastEditorUserId="5363" LastEditorDisplayName="" LastEditDate="2011-03-17T23:23:49.937" LastActivityDate="2011-03-17T23:23:49.937" Title="Using Black-Scholes Equation to &quot;buy&quot; stocks" Tags="&lt;probability&gt;&lt;stochastic-processes&gt;&lt;finance&gt;&lt;applications&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="27687" PostTypeId="2" ParentId="27684" CreationDate="2011-03-17T22:27:49.687" Score="2" ViewCount="" Body="&lt;p&gt;Hint: Indeed, condition on $X(t)=n$. Then, note that given $X(t)=n$, the locations of the $n$ points are distributed as $n$ i.i.d. uniform$[0,t]$ rv's. This should lead you easily to the result, noting that ${\rm E}f(U)=\frac{1}{t}\int_0^t {f(w)dw} $, where $U$ is a uniform$[0,t]$ rv.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: More precisely (cf. cardinal's comment below), given $X(t)=n$, the points $W_1,\ldots,W_n$ are distributed as $n$ order statistics from the uniform distribution on $[0,t]$. However, the sum $f(W_1)+\cdots+f(W_n)$ (given $X(t)=n$) is equal in distribution to the sum $f(U_1)+\cdots+f(U_n)$ where the $U_i$ are i.i.d. uniform$[0,t]$ rv's, and thus the result follows straightforwardly using the hint.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT (additional hint, in response to the OP's request):&#xA;$$&#xD;&#xA;\sum\nolimits_{n = 0}^\infty  {n{\rm P}(X(t) = n)}  = {\rm E}[X(t)] = ?&#xD;&#xA;$$&#xA;The left-hand side is given by&#xA;$$&#xD;&#xA;\sum\nolimits_{n = 0}^\infty  {n{\rm E}[f(U)]{\rm P}(X(t) = n)}, &#xD;&#xA;$$&#xA;where $U$ is a uniform$[0,t]$ random variable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT (in response to the OP's request): &#xA;In general, if $Y$ is a random variable with density function $h$, then ${\rm E}[f(Y)] = \int {f(y)h(y)dy} $. A uniform$[0,t]$ random variable, $U$, has constant density function $h(y)=1/t$ for $y \in [0,t]$ (and $0$ otherwise). From this it follows that ${\rm E}[f(U)] = \frac{1}{t}\int_0^t {f(y)dy} $.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the summation $\sum\nolimits_{n = 0}^\infty  {n{\rm E}[f(U)]{\rm P}(X(t) = n)}$, first note that&#xA;$$&#xD;&#xA;{\rm E}\Big[\sum\nolimits_{i = 1}^{X(t)} {f(W_i )} \Big] = \sum\limits_{n = 0}^\infty  {{\rm E}\Big[} \sum\nolimits_{i = 1}^n {f(W_i )} |X(t) = n \Big]{\rm P}(X(t) = n).&#xD;&#xA;$$&#xA;Now, recall that given $X(t)=n$, the sum $f(W_1)+\cdots+f(W_n)$ is equal in distribution to the sum $f(U_1)+\cdots+f(U_n)$, where the $U_i$ are i.i.d. uniform$[0,t]$ rv's. This accounts for the $n {\rm E}[f(U)]$ in the sum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Further questions?&lt;/p&gt;&#xA;" OwnerUserId="2810" LastEditorUserId="2810" LastEditorDisplayName="" LastEditDate="2011-03-18T01:01:09.383" LastActivityDate="2011-03-18T01:01:09.383" CommentCount="10" />
  <row Id="27688" PostTypeId="2" ParentId="27686" CreationDate="2011-03-17T22:28:23.057" Score="4" ViewCount="" Body="&lt;p&gt;The Black-Scholes models assumes the price of the underlying asset (stock price) is given.  It therefore could not tell you if the stock price is over-/under-priced.  Risk-neutral pricing also won't give you any information about the likely drift of the stock in the future --- by definition, under the risk-neutral measure, the expected value of &lt;em&gt;any&lt;/em&gt; tradeable asset is a martingale process; thus the expected value of any stock (under the martingale measure) is just the current price discounted by the risk-free rate.&lt;/p&gt;&#xA;" OwnerUserId="3852" LastActivityDate="2011-03-17T22:28:23.057" CommentCount="3" />
  <row Id="27689" PostTypeId="1" AcceptedAnswerId="27691" CreationDate="2011-03-17T22:30:53.263" Score="1" ViewCount="60" Body="&lt;p&gt;Suppose that V and W are vector spaces (say over the reals) and that T is a linear surjection from V onto W with kernel K. By picking a basis for W (using choice) and pulling it back through T to a linearly independent set in V (again using choice) it is not difficult to get a linear embedding i of W into V satisfying T(i(w)) = w. It is then easily checked that the map $(k,w) \mapsto k + i(w)$ gives an isomorphism from $K \oplus W$ onto V. This all seems a little hamfisted though for such an algebraic result. I thought I would check whether there is some way to get such an isomorphism which avoids choice. I would guess there is no chance since there this isomorphism is so devastatingly nonunique... In fact I'm not sure I even want to ask this question anymore now that I think about it...&lt;/p&gt;&#xA;" OwnerUserId="6608" LastActivityDate="2011-03-17T22:39:58.193" Title="Rank-nullity depends on the existence of a Hamel basis?" Tags="&lt;linear-algebra&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="27690" PostTypeId="1" AcceptedAnswerId="27703" CreationDate="2011-03-17T22:33:44.460" Score="5" ViewCount="119" Body="&lt;p&gt;My friend and I have been stumped on this problem for a little while and I thought asking for tips couldn't hurt (we did ask the teacher, but we got other problems after)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is the question :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $\{X_n\}_{n \geq 1}$ be a sequence of random variables defined on the same probability space $(\Omega, F, \mathbb{P})$ with the same law of finite expected value (E(|X_1|)&amp;lt;\infty ). Let&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$Y_n = n^{-1} \max_{1 \leq i \leq n} |X_i|$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Show that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\lim_{n\rightarrow \infty} E(Y_n) = 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$Y_n \rightarrow 0$ almost surely.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have ideas of many parts of the proof, for example for the first one it would suffice to show that the expected value of the max of all $|X_i|$ is finite... and since the max is one of the $|X_i|$ for each $\omega \in \Omega$ it seems reasonable but we're not sure how to show it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We also tried splitting the integral for the expected value into a partition of $\Omega$ considering the sets on which $X_i$ is the max, but didn't get too far with that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the second part, I think we could show it if we knew that $X_i(\omega)$ diverges for only a measure 0 set, but it's not that obvious (I think).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any pointers to the right direction appreciated!&lt;/p&gt;&#xA;" OwnerUserId="887" LastActivityDate="2011-03-18T06:48:00.983" Title="convergence of the maximum of a series of identically distributed variables" Tags="&lt;homework&gt;&lt;measure-theory&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="2" />
  <row Id="27691" PostTypeId="2" ParentId="27689" CreationDate="2011-03-17T22:39:58.193" Score="3" ViewCount="" Body="&lt;p&gt;You do need choice in general:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;without it, there may exist vector spaces none of whose nontrivial subspaces has a complement (Herrlich, Axiom of Choice, LNM 1876, Disaster 4.43). &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;-&lt;a href=&quot;http://mathoverflow.net/questions/7378/exact-short-sequences-of-vector-spaces/7386#7386&quot; rel=&quot;nofollow&quot;&gt;JS Milne, MathOverflow, Dec 1, 2009&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2011-03-17T22:39:58.193" CommunityOwnedDate="2011-03-17T22:39:58.193" CommentCount="1" />
  <row Id="27692" PostTypeId="1" AcceptedAnswerId="27696" CreationDate="2011-03-17T22:50:46.923" Score="2" ViewCount="134" Body="&lt;p&gt;Hey guys, simple question in linear algebra.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to find the nth exponent of this matrix:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;\left[&#xD;&#xA;\begin{array}{cc}&#xD;&#xA; 1 &amp;amp; 1 \\&#xD;&#xA;0 &amp;amp; 2&#xD;&#xA;\end{array}&#xD;&#xA;\right]&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm trying to use the diagonal algorithm, first by finding the eigenvalues, so I get $t_1=1$, and $t_2=2$, and than I realize that the matrix is not diagonalizable... So how can I still calculate the nth exponent?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="6163" LastEditorUserId="3731" LastEditorDisplayName="" LastEditDate="2011-03-18T00:20:42.570" LastActivityDate="2011-03-18T00:20:42.570" Title="How can I find the nth exponent of the matrix using the diagonalization algorithm?" Tags="&lt;linear-algebra&gt;&lt;eigenvalues&gt;&lt;eigenvectors&gt;" AnswerCount="3" CommentCount="3" />
  <row Id="27693" PostTypeId="1" CreationDate="2011-03-17T23:00:47.453" Score="0" ViewCount="37" Body="&lt;p&gt;If $\sum_n \frac{c_n}{n^s} = \pi_{k&amp;gt;3}\zeta(ks)$ then how do I find the partial sum $\sum_{n&amp;lt;x} \frac{c_n}{n^s}$&lt;/p&gt;&#xA;" OwnerUserId="8355" LastActivityDate="2011-03-17T23:00:47.453" Title="Partial summation- dirichlet series" Tags="&lt;number-theory&gt;" CommentCount="1" />
  <row Id="27694" PostTypeId="2" ParentId="27692" CreationDate="2011-03-17T23:02:05.480" Score="3" ViewCount="" Body="&lt;p&gt;In general, you can use the &lt;a href=&quot;http://en.wikipedia.org/wiki/Cayley-Hamilton_theorem&quot; rel=&quot;nofollow&quot;&gt;Cayley–Hamilton theorem&lt;/a&gt; to reduce all powers of a matrix to a low degree polynomial in the matrix.&lt;/p&gt;&#xA;" OwnerUserId="589" LastActivityDate="2011-03-17T23:02:05.480" />
  <row Id="27695" PostTypeId="1" AcceptedAnswerId="27702" CreationDate="2011-03-17T23:16:24.113" Score="1" ViewCount="100" Body="&lt;p&gt;I was skimming through the topology book recommended through &lt;a href=&quot;http://math.stackexchange.com/questions/27629/topology-exercises&quot;&gt;this question&lt;/a&gt; and I came across a question that I apparently solved incorrectly. It's question 2.5 on page 12.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Repeated here:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Let $X$ be $\mathbb{R}$, and let $\Omega$ consists [sic] of the empty set and complements of all finite subsets of $\mathbb{R}$. Is $\Omega$ a topological structure?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And my answer is &quot;no&quot;, because I can posit some finite intersection of infinite sets and come up with a finite set, which is no longer in $\Omega$. For example, for some $a,b \in \mathbb{R}$ and $a&amp;lt;b$, two infinite sets are $X_1 = (-\infty,b)$ and $X_2 = (a,\infty)$ and obviously, $X_1, X_2 \in \Omega$ but $X_1 \cap X_2 \notin \Omega$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But it seems it is a topology, because in the line following the question, they state:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The space of Problem 2.5 is denoted by $\mathbb{R}_{T_1}$ and called the line with $T_1$-topology.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So where am I going wrong?&lt;/p&gt;&#xA;" OwnerUserId="6690" LastActivityDate="2011-03-17T23:26:11.760" Title="Complement of all finite subsets of $\mathbb{R}$" Tags="&lt;topology&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="27696" PostTypeId="2" ParentId="27692" CreationDate="2011-03-17T23:18:18.680" Score="10" ViewCount="" Body="&lt;p&gt;Nir, it is impossible to have less linearly independent eigenvectors than distinct eigenvalues; each distinct eigenvalue corresponds to the span of at least one eigenvector. Eigenvectors for your values of 1 and 2 are $\begin{bmatrix}&#xD;&#xA;1\\ &#xD;&#xA;0&#xD;&#xA;\end{bmatrix}$ and $\begin{bmatrix}&#xD;&#xA;1\\ &#xD;&#xA;1&#xD;&#xA;\end{bmatrix}$ respectively. By the definition of diagonalization, if you have an eigenbasis for $\mathbb{R^{2}}$, your matrix is diagonalizable. Since you have a basis consisting of two eigenvectors (and we know that for any vector space of dimension n, n linearly independent vectors form a basis, in addition to knowing that any eigenvectors that correspond to different eigenvalues are linearly independent), you can diagonalize your matrix. From here on, you can decompose your matrix into the product of an eigenbasis matrix, the eigenvalue matrix, and the inverse of the eigenbasis matrix, and compute the powers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am assuming this is not homework, since you did not tag it as such. Here is the full plan of action: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now that you know the corresponding eigenvectors for the eigenvalues, you can represent your matrix $A$ as &#xA;$\left[&#xD;&#xA;\begin{array}{cc}&#xD;&#xA; 1 &amp;amp; 1 \\&#xD;&#xA;0 &amp;amp; 2&#xD;&#xA;\end{array}&#xD;&#xA;\right]$&#xA;$=$&#xA;$\left[&#xD;&#xA;\begin{array}{cc}&#xD;&#xA; 1 &amp;amp; 1 \\&#xD;&#xA;0 &amp;amp; 1&#xD;&#xA;\end{array}&#xD;&#xA;\right]$&#xA;$\left[&#xD;&#xA;\begin{array}{cc}&#xD;&#xA; 1 &amp;amp; 0 \\&#xD;&#xA;0 &amp;amp; 2&#xD;&#xA;\end{array}&#xD;&#xA;\right]$&#xA;$\left[&#xD;&#xA;\begin{array}{cc}&#xD;&#xA; 1 &amp;amp; -1 \\&#xD;&#xA;0 &amp;amp; 1&#xD;&#xA;\end{array}&#xD;&#xA;\right]$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now that you have $A=SDS^{-1}$ where $S$ is the matrix with your eigenbasis as columns, and $D$ is the diagonal matrix consisting of the eigenvalues, you can find an $A^{k}$ by representing $A^{k}$ in the following way: $A^{k}=(SDS^{-1})^{k}=\underset{k}{\underbrace{(SDS^{-1})(SDS^{-1})...(SDS^{-1})}}$. As you can see, all $S$ and $S^{-1}$ besides the ones on the very left and very right will cancel, leaving you with $A^{k}=SD^{k}S^{-1}$. Now, you are left with an equation &#xA;$\left[&#xD;&#xA;\begin{array}{cc}&#xD;&#xA; 1 &amp;amp; 1 \\&#xD;&#xA;0 &amp;amp; 2&#xD;&#xA;\end{array}&#xD;&#xA;\right]^{k}$&#xA;$=$&#xA;$\left[&#xD;&#xA;\begin{array}{cc}&#xD;&#xA; 1 &amp;amp; 1 \\&#xD;&#xA;0 &amp;amp; 1&#xD;&#xA;\end{array}&#xD;&#xA;\right]$&#xA;$\left[&#xD;&#xA;\begin{array}{cc}&#xD;&#xA; 1 &amp;amp; 0 \\&#xD;&#xA;0 &amp;amp; 2&#xD;&#xA;\end{array}&#xD;&#xA;\right]^{k}$&#xA;$\left[&#xD;&#xA;\begin{array}{cc}&#xD;&#xA; 1 &amp;amp; -1 \\&#xD;&#xA;0 &amp;amp; 1&#xD;&#xA;\end{array}&#xD;&#xA;\right]$ and you can simply exponentiate the diagonal values and multiply the matrices out to get the result.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You mentioned that you realized that the matrix is not diagonalizable, so we can try finding eigenvectors by hand, to make sure that the process is clear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the eigenvalue 1, you need to find when $(A-1I)\vec{x}$ is equal to zero. So, you compute the kernel of $(A-I)$: $\left[&#xD;&#xA;\begin{array}{cc}&#xD;&#xA; 0 &amp;amp; 1 \\&#xD;&#xA;0 &amp;amp; 1&#xD;&#xA;\end{array}&#xD;&#xA;\right]$. Row-reducing, you get $\left[&#xD;&#xA;\begin{array}{cc}&#xD;&#xA; 0 &amp;amp; 1 \\&#xD;&#xA;0 &amp;amp; 0&#xD;&#xA;\end{array}&#xD;&#xA;\right]$, so the kernel is represented by the span of $\begin{bmatrix}&#xD;&#xA;1\\ &#xD;&#xA;0&#xD;&#xA;\end{bmatrix}$, meaning that a basis for $\xi_{1}$ is $\begin{bmatrix}&#xD;&#xA;1\\ &#xD;&#xA;0&#xD;&#xA;\end{bmatrix}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Same process for the eigenvalue 2: Find the kernel of $(A-2I)$, meaning the kernel of $\left[&#xD;&#xA;\begin{array}{cc}&#xD;&#xA; -1 &amp;amp; 1 \\&#xD;&#xA;0 &amp;amp; 0&#xD;&#xA;\end{array}&#xD;&#xA;\right]$. As you can see, the kernel consists of the span of $\begin{bmatrix}&#xD;&#xA;1\\ &#xD;&#xA;1&#xD;&#xA;\end{bmatrix}$, meaning that a basis for $\xi_{2}$ is $\begin{bmatrix}&#xD;&#xA;1\\ &#xD;&#xA;1&#xD;&#xA;\end{bmatrix}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Generally, an $n\times n$ matrix is always diagonalizable if it has $n$ distinct eigenvalues, which your matrix does.&lt;/p&gt;&#xA;" OwnerUserId="3731" LastEditorUserId="3731" LastEditorDisplayName="" LastEditDate="2011-03-18T00:12:00.953" LastActivityDate="2011-03-18T00:12:00.953" CommentCount="1" />
  <row Id="27697" PostTypeId="2" ParentId="27692" CreationDate="2011-03-17T23:18:26.977" Score="6" ViewCount="" Body="&lt;p&gt;This matrix &lt;em&gt;is&lt;/em&gt; diagonalizable. As you said the eigenvalues are $\lambda = 1$ and $\lambda = 2$ with corresponding eigenvectors $\begin{bmatrix} 1 &amp;amp; 0\end{bmatrix}^T$ and $\begin{bmatrix} 1 &amp;amp; 1\end{bmatrix}^T$ for example. Hence&#xA;$$&#xD;&#xA;\begin{bmatrix}&#xD;&#xA;1 &amp;amp; 0\\&#xD;&#xA;0 &amp;amp; 2&#xD;&#xA;\end{bmatrix} =&#xD;&#xA;\begin{bmatrix}&#xD;&#xA;1 &amp;amp; 1\\&#xD;&#xA;0 &amp;amp; 1&#xD;&#xA;\end{bmatrix}^{-1}&#xD;&#xA;\begin{bmatrix}&#xD;&#xA;1 &amp;amp; 1\\&#xD;&#xA;0 &amp;amp; 2&#xD;&#xA;\end{bmatrix}&#xD;&#xA;\begin{bmatrix}&#xD;&#xA;1 &amp;amp; 1\\&#xD;&#xA;0 &amp;amp; 1&#xD;&#xA;\end{bmatrix}.&#xD;&#xA;$$&#xA;Since $\begin{bmatrix}&#xD;&#xA;1 &amp;amp; 1\\&#xD;&#xA;0 &amp;amp; 1&#xD;&#xA;\end{bmatrix}^{-1} = \begin{bmatrix}&#xD;&#xA;1 &amp;amp; -1\\&#xD;&#xA;0 &amp;amp; 1&#xD;&#xA;\end{bmatrix}$ we obtain&#xA;$$&#xD;&#xA;\begin{bmatrix}&#xD;&#xA;1 &amp;amp; 1\\&#xD;&#xA;0 &amp;amp; 2&#xD;&#xA;\end{bmatrix}^n =&#xD;&#xA;\begin{bmatrix}&#xD;&#xA;1 &amp;amp; 1\\&#xD;&#xA;0 &amp;amp; 1&#xD;&#xA;\end{bmatrix}&#xD;&#xA;\begin{bmatrix}&#xD;&#xA;1 &amp;amp; 0\\&#xD;&#xA;0 &amp;amp; 2^n&#xD;&#xA;\end{bmatrix}&#xD;&#xA;\begin{bmatrix}&#xD;&#xA;1 &amp;amp; -1\\&#xD;&#xA;0 &amp;amp; 1&#xD;&#xA;\end{bmatrix}.&#xD;&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="8229" LastActivityDate="2011-03-17T23:18:26.977" />
  <row Id="27698" PostTypeId="2" ParentId="27695" CreationDate="2011-03-17T23:20:28.287" Score="1" ViewCount="" Body="&lt;p&gt;Not all infinite subsets of $\mathbb{R}$ are complements of finite subsets of $\mathbb{R}$. For example, $X_1=(-\infty,b)$ is the complement of $[b,\infty)$, which is infinite. Thus $X_1$ is not open in this topology.&lt;/p&gt;&#xA;" OwnerUserId="264" LastActivityDate="2011-03-17T23:20:28.287" CommentCount="1" />
  <row Id="27699" PostTypeId="1" CreationDate="2011-03-17T23:22:40.970" Score="0" ViewCount="230" Body="&lt;p&gt;Hi,&#xA;I would appreciate any possible help for this question because I have no clue what to do! Thanks so much!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider a population made of a fixed number (N) of people. At time t=0 there is only one infected individual and N-1 susceptible people in the population. When you get infected, you remain in the infected state forever. In any short time interval that is h long, any given infected person will transmit the disease to any susceptible person with a probability of alpha * h + o(h) where o(h) is an error term and alpha is the individual infection rate. Let X(t) denote the number of infected individuals in the population at time t &gt;= 0. So X(t) is a pure birth process on states 0, 1, ... N. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are the birth parameters?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some issues I am having are: 1) What exactly does the question mean by birth parameters? 2) How would I set up this problem in the form of a probability statement? 3) Any hints to help me out would be great!&lt;/p&gt;&#xA;" OwnerUserId="8393" LastEditorUserId="8393" LastEditorDisplayName="" LastEditDate="2011-03-19T02:56:12.960" LastActivityDate="2011-04-18T03:42:08.093" Title="Pure Birth Process Question" Tags="&lt;stochastic-processes&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="27700" PostTypeId="2" ParentId="27686" CreationDate="2011-03-17T23:23:15.350" Score="2" ViewCount="" Body="&lt;p&gt;I'm not an expert, though I'm studying stochastic calculus right now in one of my classes. The Black-Scholes price gives an exact solution to European Call option prices, subject to fixed risk free rates, constant volatilities, and other assumptions. It can be used as one tool (out of many) to &lt;strong&gt;hedge&lt;/strong&gt; a portfolio, such that, at any time, regardless of the change in price of the underlying, the change in value of the portfolio will be positive, subject to other assumptions like, for example, the option price vs underlying price curve doesn't straighten out (it maintains convexity as time goes on). So, you can use the Black-Scholes equation to create a delta-neutral portfolio, in theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In actuality, it's much much more difficult than this. Everyone already knows the pricing formulae, so in theory, all of the profit has been arbitraged away. Any advantage will last for a few seconds (imagine 100 other people just as smart or smarter than you trying to find a pricing mistake). Secondly, even if an investor did have a strategy, the execution of that strategy requires considerable skill. Simply buying or selling a stock incurs price slippage, trading costs, feedback, etc. There are entire departments in the banks devoted to trade execution alone.&lt;/p&gt;&#xA;" OwnerUserId="8140" LastActivityDate="2011-03-17T23:23:15.350" />
  <row Id="27701" PostTypeId="2" ParentId="6920" CreationDate="2011-03-17T23:24:27.627" Score="0" ViewCount="" Body="&lt;p&gt;I liked the geometriYou can use integration by parts, following the well known formula:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align*}&#xA;\int_a^b f(x) \frac{dg(x)}{dx} dx = [f(x)g(x)]_a^b - \int_a^b \frac{df(x)}{dx} g(x) dx&#xA;\end{align*}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;setting $g(x)=x$ and $f(x)=\int_a^x f(\xi) d\xi$ you have your result :)&lt;/p&gt;&#xA;" OwnerUserId="8357" LastActivityDate="2011-03-17T23:24:27.627" />
  <row Id="27702" PostTypeId="2" ParentId="27695" CreationDate="2011-03-17T23:26:11.760" Score="3" ViewCount="" Body="&lt;p&gt;The problem with your argument is that not all infinite subsets of $\mathbb R$ are complements of finite sets. To see that we in fact have a topology, let $X_1$ and $X_2$ be finite subsets of $\mathbb R$ and observe that $(\mathbb R - X_1) \cap (\mathbb R - X_2) = \mathbb R - (X_1 \cup X_2)$ which is again the complement of a finite set. The union property should be obvious.&lt;/p&gt;&#xA;" OwnerUserId="8173" LastActivityDate="2011-03-17T23:26:11.760" CommentCount="1" />
  <row Id="27703" PostTypeId="2" ParentId="27690" CreationDate="2011-03-17T23:39:00.750" Score="3" ViewCount="" Body="&lt;p&gt;Assume without loss of generality that $X_1$ is almost surely nonnegative.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Almost sure convergence&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a consequence of the first Borel-Cantelli lemma. To see this, fix any positive $x$. Then $P(X_n\ge nx)=P(X_1\ge nx)$ for every $n$ and &#xA;$$\sum_{n\ge1}P(X_1\ge nx)\le x^{-1}E(X_1),&#xD;&#xA;$$ &#xA;hence the series of general term $P(X_n\ge nx)$ converges. By the first Borel-Cantelli lemma, the limsup of the events $[X_n\ge nx]$ has probability $0$. This means that $X_n&amp;lt;nx$ for every $n$ large enough and can be translated as $X_n\le Z+nx$ for every $n$, for an almost surely finite $Z$. Hence $nY_n\le Z+nx$ for every $n$, and $\limsup Y_n\le x$ almost surely. This holds for every positive $x$, hence $Y_n\to0$ almost surely.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Convergence of the expectations&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Two ingredients are useful here: the fact that, for every nonnegative $Z$, $E(Z)$ is the integral of the function $x\mapsto P(Z\ge x)$ over $x\ge0$, and the fact that if $nY_n\ge x$, then $X_k\ge x$ for at least one integer $k$ between $1$ and $n$, hence $P(nY_n\ge x)\le nP(X_1\ge x)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks to the first ingredient, $E(Y_n)$ is the integral of $g_n$ with $g_n(x)=P(nY_n\ge x)/n$. Thanks to the second ingredient, $g_n(x)\le P(X_1\ge x)=g_1(x)$. Now, $g_n(x)\le1/n$ hence $g_n(x)\to0$, and since $E(X_1)$ is finite, $g_1$ is integrable. By dominated convergence, the integral of $g_n$ converges to $0$, that is, $E(Y_n)\to0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Remark&lt;/em&gt; You do not say where you found the exercise but your source is to be complimented because many people add the hypothesis that the sequence $(X_n)$ is independent although it is not necessary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added later on&lt;/strong&gt; The upper bound of a series by $x^{-1}E(X_1)$ used above can be proved as follows. First assume that $x=1$ and note that for any nonnegative $Z$ (random or deterministic),&#xA;$$&#xD;&#xA;\sum_{n\ge1}\mathbf{1}_{Z\ge n}=\lfloor Z\rfloor\le Z,&#xD;&#xA;$$&#xA;where $\lfloor \ \rfloor$ denotes the integer part. Integrating both sides of the inequality with respect to $P$ yields, for any nonnegative random variable $Z$,&#xA;$$&#xD;&#xA;\sum_{n\ge1}P(Z\ge n)=E(\lfloor Z\rfloor)\le E(Z).&#xD;&#xA;$$&#xA;For the case at hand, apply this inequality to the random variable $Z=x^{-1}X_1$, using&#xA;$$&#xD;&#xA;\sum_{n\ge1}\mathbf{1}_{X_1\ge nx}=\lfloor x^{-1}X_1\rfloor\le x^{-1}X_1.&#xD;&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="6179" LastEditorUserId="6179" LastEditorDisplayName="" LastEditDate="2011-03-18T06:48:00.983" LastActivityDate="2011-03-18T06:48:00.983" CommentCount="3" />
  <row Id="27704" PostTypeId="2" ParentId="27690" CreationDate="2011-03-17T23:39:18.583" Score="1" ViewCount="" Body="&lt;p&gt;Note first that $E \max_{i \ge 0} |X_i|$ (one should really write $\sup$ instead of $\max$) need not be finite.  Indeed, if the $X_i$ are, say, iid normal, then one can show $\sup_{i \ge 0} |X_i| = +\infty$ almost surely.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To get $L^1$ convergence, the trick is to split into the events where the $X_i$ are small and where they are large.  When they are small they do not contribute much to $Y_n$, and they can be large only with small probability.  So fix $M$ and let $U_i = |X_i| 1_{\{|X_i| \le M\}}$, $V_i = |X_i| 1_{\{|X_i| &amp;gt; M\}}$.  Then $Y_n \le \frac{1}{n} (\max_{i \le n} U_i + \max_{i \le n} V_i)$.  The first term is bounded by $M$, and the second by $V_1 + \dots + V_n$.  Taking expectations, $E Y_n \le \frac{M}{n} + E V_1$, so $\limsup_{n \to \infty} E Y_n \le E V_1$.  By choosing $M$ large enough, $E V_1$ can be made as small as desired (think dominated convergence).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For almost sure convergence, the argument that went here previously was wrong.&lt;/p&gt;&#xA;" OwnerUserId="822" LastEditorUserId="822" LastEditorDisplayName="" LastEditDate="2011-03-17T23:51:30.787" LastActivityDate="2011-03-17T23:51:30.787" />
  <row Id="27705" PostTypeId="1" CreationDate="2011-03-18T00:08:30.613" Score="0" ViewCount="74" Body="&lt;p&gt;It's about the first step to find an upper bound to the order of a finite group with h conjugacy classes (right or left) that depends only on h. (h a natural non nul integer).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have some doubts about the rigor of my proof that I am sharing with you so that you can help me find a likely error or an omited step.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have attached the link to the scan of my proof to this post.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many thanks&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://i.imgur.com/EI2mL.jpg&quot; rel=&quot;nofollow&quot;&gt;http://i.imgur.com/EI2mL.jpg&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="8397" LastEditorUserId="2548" LastEditorDisplayName="" LastEditDate="2011-03-18T00:18:21.247" LastActivityDate="2011-03-18T00:18:21.247" Title="conjugacy classes" Tags="&lt;group-theory&gt;" CommentCount="5" />
  <row Id="27706" PostTypeId="2" ParentId="27699" CreationDate="2011-03-18T00:26:08.743" Score="1" ViewCount="" Body="&lt;p&gt;Hint: Suppose you have k infected people and N-k non-infected ones at the start of your very small interval of time h.  Let $ A_{ij}$ be the event that infected person #i transmits the disease to &#xA;non-infected person #j in this time interval.  The assumption should be that each $P(A_{ij}) = \alpha h + O(h^2)$&#xA;and the $A_{ij}$ are independent.&#xA;So what about the probability (up to $O(h^2)$) of the union of all the $A_{ij}$?  &lt;/p&gt;&#xA;" OwnerUserId="8508" LastActivityDate="2011-03-18T00:26:08.743" CommentCount="2" />
  <row Id="27707" PostTypeId="1" AcceptedAnswerId="27708" CreationDate="2011-03-18T00:41:03.850" Score="2" ViewCount="58" Body="&lt;p&gt;If I put 4 green rabbits and 1 red rabbit into a hat, that's 5 rabbits!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I then make 5 attempts at pulling the red rabbit out of the hat.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I get a green rabbit I put it back in the hat.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the first attempt I have a 1 in 5 chance of red rabbit success.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the first attempt fails, on the second attempt I still have a 1 in 5 chance of red rabbit success.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the unlikelihood that I will fail to pull out the red rabbit in 5 attempts?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Although I may never succeed, Is there some magic that makes it more likely that I will get the red rabbit over time?&lt;/p&gt;&#xA;" OwnerUserId="7497" LastActivityDate="2011-03-18T00:44:42.470" Title="Simple pondering of probability." Tags="&lt;probability&gt;" AnswerCount="1" />
  <row Id="27708" PostTypeId="2" ParentId="27707" CreationDate="2011-03-18T00:43:48.153" Score="3" ViewCount="" Body="&lt;p&gt;Each time you have $\frac{4}{5}$ chance of getting a green rabbit.  As the events are independent, for $n$ tries you have $\left(\frac{4}{5}\right)^n$ chance of never getting a red rabbit.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-03-18T00:43:48.153" CommentCount="2" />
  <row Id="27710" PostTypeId="2" ParentId="27684" CreationDate="2011-03-18T01:00:08.637" Score="2" ViewCount="" Body="&lt;p&gt;One way of solving this problem is to rewrite the sum as a &lt;a href=&quot;http://en.wikipedia.org/wiki/Riemann%E2%80%93Stieltjes_integral&quot; rel=&quot;nofollow&quot;&gt;Riemann-Stieltjes&lt;/a&gt; integral,&#xA;$$&#xD;&#xA;\sum_{i=1}^{X(t)}f(W_i)=\int_0^tf(s)\,dX_s.&#xD;&#xA;$$&#xA;Using the fact that $X(s)$ has expectation $\mathbb{E}[X(s)]=\lambda s$, you can take expectations* of this to get&#xA;$$&#xD;&#xA;\mathbb{E}\left[\sum_{i=1}^{X(t)}f(W_i)\right]=\lambda\int_0^tf(s)ds.&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(*) To be rigorous, you should show that the expectation does indeed commute with the integral in this way. However, this is a general result which can be used in lots of situations, and is a very special case of the fact that stochastic integrals with respect to martingales are martingales ($X(s)-\lambda s$ is actually a martingale). In the case where $f$ is piecewise linear, the integral reduces to a sum, and expectations commute with summation by linearity. The fact that it extends to the general case of locally integrable $f$ is a consequence of the monotone class theorem.&lt;/p&gt;&#xA;" OwnerUserId="1321" LastActivityDate="2011-03-18T01:00:08.637" />
  <row Id="27711" PostTypeId="2" ParentId="27609" CreationDate="2011-03-18T01:31:58.237" Score="1" ViewCount="" Body="&lt;p&gt;It also depends on the algorithm. You would need an algorithm that can translate to an &quot;oblivious TM&quot;, a Turing Machine where the head movements depend only on the time step we are currently at. If your algorithm running time is different for different inputs of the same size, that is if you use loops that can be terminated abruptly (like when we search for one or more occurences of an item) or conditionals where each case has different running times (e.g. some heuristics), I believe approximating the input size is even more difficult.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am however curious about your motive. Why isn't the input size known beforehand?&lt;/p&gt;&#xA;" OwnerUserId="8327" LastActivityDate="2011-03-18T01:31:58.237" CommentCount="1" />
  <row Id="27712" PostTypeId="1" AcceptedAnswerId="27716" CreationDate="2011-03-18T01:39:30.430" Score="1" ViewCount="67" Body="&lt;p&gt;Can anyone please tell me if there is any other proof for the cardinality of all mappings, that is not by induction, i.e., not this one (http://www.proofwiki.org/wiki/Cardinality_of_Set_of_All_Mappings) ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="8398" LastActivityDate="2011-03-18T02:18:28.163" Title="Alternative proof of the cardinality of the set of all mappings" Tags="&lt;set-theory&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="27713" PostTypeId="1" CreationDate="2011-03-18T01:41:46.603" Score="3" ViewCount="153" Body="&lt;p&gt;How to prove that any vector bundle with a fiber metric g admits a metric connection? It seems I should use partition of unity, but I have no idea how to proceed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also it seems there are two definitions of connection on a vector bundle E, one is that &#xA;$\nabla: \Gamma(TM) \times \Gamma(E) \to \Gamma$(E),&#xA;and the other one is that $\nabla: \Gamma(E) \to \Gamma(E) \otimes\Gamma(T^\ast M)$, and they should be equivalent. But I don't see how to use this to show the following equations are equivalent:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there is a fiber metric g on vector bundle E, then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$d(g(u,v))=g(\nabla u,v)+g(u,\nabla v)$ for all $u,v\in\Gamma(E)$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;is equivalent to &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$X(g(u,v))=g(\nabla_X u,v)+g(u,\nabla_X v)$ for all $X\in TM$, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;is it because that&#xA;$d(g(u,v))(X)=X(g(u,v))$? &#xA;I am just not sure about how vector fields act on metrics.&lt;/p&gt;&#xA;" OwnerUserId="8400" LastEditorUserId="8246" LastEditorDisplayName="" LastEditDate="2011-05-17T14:13:37.743" LastActivityDate="2011-05-17T14:13:37.743" Title="Metric connection" Tags="&lt;homework&gt;&lt;differential-geometry&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="27714" PostTypeId="1" AcceptedAnswerId="27723" CreationDate="2011-03-18T02:13:02.277" Score="0" ViewCount="81" Body="&lt;p&gt;I have no clue how to proceed with this question. Please help me in deriving the differential equations! Thanks very much... I really appreciate it :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/QJQUb.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="8393" LastActivityDate="2011-03-18T05:39:41.823" Title="Pure Birth Process Probabilities" Tags="&lt;stochastic-processes&gt;" AnswerCount="1" />
  <row Id="27715" PostTypeId="2" ParentId="27151" CreationDate="2011-03-18T02:14:55.363" Score="0" ViewCount="" Body="&lt;p&gt;You can start by writing $\mathbf{A} = P \mathbf{\Lambda} P^{-1}$, where $P$ is the matrix of eigenvectors and $\mathbf{\Lambda}$ is the diagonal matrix of eigenvalues. Therefore, you can write $\mathbf{A}^k$ as $P\mathbf{\Lambda}^k P^{-1}$, where $\mathbf{\Lambda}^k$ is simply the diagonal matrix made up of the $n^{th}$ powers of the eigenvalues of $\mathbf{A}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using this fact, you can write&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;p( \mathbf{A} ) =  P(k_{m}\mathbf{\Lambda}^{m}+k_{m-1} \mathbf{\Lambda}^{m-1}+\ldots+k_{1} \mathbf{\Lambda} +k_{0} \mathbf{I})P^{-1} \&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This shows that $P(\mathbf{A})$ has eigenvalues given by $p(\lambda_i)$ for $i = 1,2,\ldots,n$ and the same eigenvectors as $\mathbf{A}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: My proof works only for diagonalizable matrices as Theo Buehler has commented below. &lt;/p&gt;&#xA;" OwnerUserId="2641" LastEditorUserId="2641" LastEditorDisplayName="" LastEditDate="2011-03-18T03:35:29.723" LastActivityDate="2011-03-18T03:35:29.723" CommentCount="7" />
  <row Id="27716" PostTypeId="2" ParentId="27712" CreationDate="2011-03-18T02:18:28.163" Score="2" ViewCount="" Body="&lt;p&gt;If $A=\{a_1, a_2, \ldots, a_n\}$ has $n$ elements and $B$ has $m$ elements, a mapping $f$ from $A$ to $B$ is defined uniquely by choosing $f(a_1)$ ($m$ options), then $f(a_2)$ ($m$ options again), and so on until you choose $f(a_n)$ (as always, $m$ options). By the product rule, there are $m \times m \times \cdots \times m$ ($n$ factors) overall, namely, $m^n$.&lt;/p&gt;&#xA;" OwnerUserId="308" LastActivityDate="2011-03-18T02:18:28.163" />
  <row Id="27717" PostTypeId="1" AcceptedAnswerId="27725" CreationDate="2011-03-18T02:30:25.617" Score="1" ViewCount="136" Body="&lt;p&gt;define $$h(x)=\int_0^{2\pi}f(x-y)g(y)dy=f*g(x)$$&#xA;if $f,g \in L^2$ are $2\pi$ periodic, show that h is continuous on $[0,2\pi)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so let $x_n \to x$, then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$|h(x)-h(x_n)|=|\int f(x-y)g(y)-\int f(x_n-y)g(y)|$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i tried to manipulate this but couldnt get anywhere. I was also thinking of using Holder's inequality somehow but im not sure...&lt;/p&gt;&#xA;" OwnerUserId="8401" LastActivityDate="2011-03-18T03:32:26.757" Title="convolution a continuous function?" Tags="&lt;fourier-analysis&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="27718" PostTypeId="1" AcceptedAnswerId="27721" CreationDate="2011-03-18T02:33:01.447" Score="1" ViewCount="110" Body="&lt;p&gt;How does one find the center of the group G/Z(G)?&lt;/p&gt;&#xA;" OwnerUserId="1127" LastActivityDate="2011-03-18T03:16:37.520" Title="Center of G/Z(G)" Tags="&lt;group-theory&gt;" AnswerCount="1" />
  <row Id="27719" PostTypeId="1" CreationDate="2011-03-18T02:34:16.090" Score="2" ViewCount="221" Body="&lt;p&gt;I have tried $\gcd(0,8)$ in a lot of online  gcd (or hcf) calculators, but some say $\gcd(0,8)=0$, some other gives $\gcd(0,8)=8$ and some others give $\gcd(0,8)=1$. So really which one of these is correct and why there are different conventions?&lt;/p&gt;&#xA;" OwnerUserId="7340" LastEditorUserId="7340" LastEditorDisplayName="" LastEditDate="2011-03-18T04:10:34.930" LastActivityDate="2011-03-18T19:30:37.777" Title="What is $\gcd(0,a)$, where a is a positive integer?" Tags="&lt;number-theory&gt;&lt;abstract-algebra&gt;&lt;group-theory&gt;" AnswerCount="3" CommentCount="3" />
  <row Id="27720" PostTypeId="2" ParentId="27719" CreationDate="2011-03-18T02:38:35.450" Score="12" ViewCount="" Body="&lt;p&gt;First recall the definition of $\rm\:a\:$ divides $\rm\:b\:$ in a ring $\rm\:Z\:,\: $ often written as $\rm\ a\ |\ b\ \ in\ Z\:.$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad \rm\ a\ |\ b\ \ in\ Z\ \iff\ a\:c = b\ \ $ for some $\rm\ c\in Z$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Recall also the definition of $\rm\ gcd(a,b)\:,\ $ namely   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(1)\rm\quad\quad\quad\quad\quad\ \rm gcd(a,b)\ |\ a,b\quad\quad\quad\quad\quad\quad\quad\ $  the gcd is a &lt;em&gt;common&lt;/em&gt; divisor&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(2)\rm\quad\quad\quad\quad\quad\ \rm c\ |\ a,b\ \Rightarrow\ c\ |\ gcd(a,b)\quad\quad$  the gcd is the &lt;em&gt;greatest&lt;/em&gt; common divisor&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ \ \ \ $ i.e. $\rm\quad\ c\ |\ a,b\ \iff\ c\ |\ gcd(a,b)\quad\quad$  expressed in $\iff$ form (put $\rm\ c = gcd(a,b)\ $ for $(1)$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice  $\rm\quad\: c\ |\ a,0\ \iff\ c\ |\ a\:,\ $ so $\rm\ gcd(a,0)\ =\ a\ $ by the above $\iff$ form of the gcd definition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that $\rm\ gcd(0,8) \ne 0\:,\ $ since $\rm\ gcd(0,8) = 0\ \Rightarrow\ 0\ |\ 8\ $ contra $\rm\ 0\ |\ x\ \iff\ x = 0\:.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that $\rm\ gcd(0,8) \ne 1\:,\ $ else $\rm\ 8\ |\ 0,8\ \Rightarrow\ 8\ |\ gcd(0,8) = 1\ \Rightarrow\ 1/8 \in \mathbb Z\:. $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore it makes no sense to define $\rm\ gcd(0,8)\ $to be $\:0\:$ or $\:1\:$ since $\:0\:$ is not a common divisor of $\:0,8\:$ and $\:1\:$ is not the &lt;em&gt;greatest&lt;/em&gt; common divisor. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The $\iff$ gcd definition is &lt;em&gt;universal&lt;/em&gt; - it may be employed in any domain or cancellative monoid, with the convention that the gcd is defined only up to a unit factor. This $\iff$ definition is very convenient in proofs since it enables efficient simultaneous proof of both implication directions. $\ $ For example, below is a proof of this particular form for the fundamental GCD &lt;em&gt;distributive law&lt;/em&gt; $\rm\ (ab,ac)\  =\ a\ (b,c)\ $ slightly generalized (your problem is simply $\rm\ c=0\ $ in the special case $\rm\ (a,\ \ ac)\ =\:\ a\ &#xA; (1,c)\ =\ a\: $).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;THEOREM&lt;/strong&gt; $\rm\quad (a,b)\ =\ (ac,bc)/c\quad$  if $\rm\ (ac,bc)\ $ exists.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof&lt;/strong&gt; $\rm\quad d\ |\ a,b\ \iff\ dc\ |\ ac,bc\ \iff\ dc\ |\ (ac,bc)\ \iff\ d|(ac,bc)/c$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See my &lt;a href=&quot;http://math.stackexchange.com/questions/16042/about-greatest-common-divisor/16043#16043&quot;&gt;post here&lt;/a&gt; for further discussion of this property and its relationship with &lt;strong&gt;Euclid's Lemma&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://math.stackexchange.com/questions/16042/about-greatest-common-divisor/16043#16043&quot;&gt;Recall also&lt;/a&gt; how this universal approach simplifies the proof of the basic GCD * LCM law:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;THEOREM&lt;/strong&gt;  $\rm\;\; \ (a,b) = ab/[a,b] \;\;$  if $\;\rm\ [a,b] \;$  exists.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof&lt;/strong&gt;  $\rm\quad d|\:a,b \;\iff\; a,b\:|\:ab/d \;\iff\; [a,b]\:|\:ab/d \;\iff\; d\:|\:ab/[a,b] \quad\;\;$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For much further discussion see my &lt;a href=&quot;http://math.stackexchange.com/search?q=user:242+gcd&quot;&gt;many posts on GCDs.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2011-03-18T19:30:37.777" LastActivityDate="2011-03-18T19:30:37.777" CommentCount="0" />
  <row Id="27721" PostTypeId="2" ParentId="27718" CreationDate="2011-03-18T02:47:14.643" Score="6" ViewCount="" Body="&lt;p&gt;The same way one finds the center in any group: you find all elements that commute with everything in $G/Z(G)$, that's $Z(G/Z(G))$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The subgroup of $G$ that contains $Z(G)$ and corresponds to $Z(G/Z(G))$ is called the &quot;second center of $G$&quot;, $Z_2(G)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, recall that the &lt;em&gt;commutator&lt;/em&gt; of $x$ and $y$ is defined to be $[x,y] = x^{-1}y^{-1}xy$. The commutator is the unique element of $G$ such that&#xA;$$xy = yx[x,y]$$&#xA;and $x$ and $y$ commute if and only if $[x,y]=1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is not hard to see that if $N\triangleleft G$, then $gN\in Z(G/N)$ if and only if for every $x\in G$, $[g,x]\in N$ (this, because $gN$ commutes with $xN$ if and only if $1 = [gN,xN] = [g,x]N$, if and only if $[g,x]\in N$). So one can also describe the second center of $G$ as&#xA;$$Z_2(G) = \{ g\in G\mid [g,x]\in Z(G)\text{ for all }x\in G\}$$&#xA;and then the center of $G/Z(G)$ equals $Z_2(G)/Z(G)$.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-18T03:16:37.520" LastActivityDate="2011-03-18T03:16:37.520" CommentCount="1" />
  <row Id="27722" PostTypeId="2" ParentId="27683" CreationDate="2011-03-18T02:50:28.513" Score="1" ViewCount="" Body="&lt;p&gt;First, let's pick apart the command you mention.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Union[GraphData[#, &quot;VertexCount&quot;] &amp;amp; /@ %%]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;GraphData[#, &quot;VertexCount&quot;]&amp;amp;&lt;/code&gt; is the same as &lt;code&gt;Function[x, GraphData[x, &quot;VertexCount&quot;]]&lt;/code&gt;.  That is, the &lt;code&gt;#&lt;/code&gt; is an input parameter and the &lt;code&gt;&amp;amp;&lt;/code&gt; terminates the expression of the &lt;code&gt;#&lt;/code&gt; parameter to be considered a function.  &lt;code&gt;%%&lt;/code&gt; is the result-before-last (&lt;code&gt;%&lt;/code&gt; is the most recent result, &lt;code&gt;%%&lt;/code&gt; is the one before that, &lt;code&gt;%%%&lt;/code&gt; is the one before that one, etc.).  &lt;code&gt;/@&lt;/code&gt; applies a function to each item of a list.  So, &lt;code&gt;GraphData[#, &quot;VertexCount&quot;] &amp;amp; /@ %%&lt;/code&gt; returns the list that results from applying the function &lt;code&gt;GraphData[#, &quot;VertexCount&quot;]&amp;amp;&lt;/code&gt; to each element of the list in &lt;code&gt;%%&lt;/code&gt;.  Because of the &lt;code&gt;%%&lt;/code&gt;, the line&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;GraphData[8]; Union[GraphData[#, &quot;VertexCount&quot;] &amp;amp; /@ %%]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;is probably out of context (since &lt;code&gt;%%&lt;/code&gt; should refer to the result of the statement before &lt;code&gt;GraphData[8]&lt;/code&gt;), and beyond saying that &lt;code&gt;Union[]&lt;/code&gt; gives the union of sets, I can't tell what it's doing here because I don't know to what &lt;code&gt;%%&lt;/code&gt; refers.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Now, back to what you're trying to do.  &lt;code&gt;GraphData[some_graph, &quot;SpanningTreeCount&quot;]&lt;/code&gt; will give the number of spanning trees for *some_graph* (I think, based on the documentation for &lt;code&gt;GraphData&lt;/code&gt;), so &lt;code&gt;GraphData[#, &quot;SpanningTreeCount&quot;] &amp;amp; /@ GraphData[&quot;Cubic&quot;, 12]&lt;/code&gt; will give a list containing the number of spanning trees for each of the cubic graphs with 12 vertices, and &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Total[GraphData[#, &quot;SpanningTreeCount&quot;] &amp;amp; /@ GraphData[&quot;Cubic&quot;, 12]]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;will give the total of the numbers in that list.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, since &lt;code&gt;GraphData[#, &quot;Connected&quot;]&amp;amp;&lt;/code&gt; is a function that returns &lt;code&gt;True&lt;/code&gt; for connected graphs and &lt;code&gt;False&lt;/code&gt; otherwise,&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Select[GraphData[&quot;Cubic&quot;, 12], GraphData[#, &quot;Connected&quot;] &amp;amp;]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;will give you a list of only the connected cubic graphs with 12 vertices.&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2011-03-18T02:50:28.513" />
  <row Id="27723" PostTypeId="2" ParentId="27714" CreationDate="2011-03-18T02:53:02.563" Score="0" ViewCount="" Body="&lt;p&gt;Complete derivation of the differential equations. [Suggestion: After each step below, try completing the derivation on your own.]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;P_0 (t + h) = {\rm P}(N(t + h)\;{\rm is} \; {\rm even}|N(t)\;{\rm is} \; {\rm even})P_0 (t) +  {\rm P}(N(t + h)\;{\rm is} \; {\rm even}|N(t)\;{\rm is} \; {\rm odd})P_1 (t).&#xD;&#xA;$$&#xA;Put $\Delta_t(h) = N(t+h)-N(t)$.&#xA;Then, as $h \downarrow 0$,&#xA;$$&#xD;&#xA;P_0 (t + h) = {\rm P}(\Delta_t(h)=0|N(t)\;{\rm is} \; {\rm even})P_0 (t) + {\rm P}(\Delta_t(h)=1|N(t)\;{\rm is} \; {\rm odd})P_1 (t) + o(h).&#xD;&#xA;$$&#xA;Hence,&#xA;$$&#xD;&#xA;P_0 (t + h) = [1 - {\rm P}(\Delta_t(h)=1|N(t)\;{\rm is} \; {\rm even})]P_0 (t) + {\rm P}(\Delta_t(h)=1|N(t)\;{\rm is} \; {\rm odd})P_1 (t) + o(h),&#xD;&#xA;$$&#xA;leading to&#xA;$$&#xD;&#xA;P_0 (t + h) = (1-\beta h)P_0 (t) + \alpha h P_1 (t) + o(h).&#xD;&#xA;$$&#xA;Rearranging gives&#xA;$$&#xD;&#xA;\frac{{P_0 (t + h) - P_0 (t)}}{h} = \alpha P_1 (t) - \beta P_0 (t) + \frac{{o(h)}}{h}.&#xD;&#xA;$$&#xA;Letting $h \downarrow 0$ thus gives&#xA;$$&#xD;&#xA;P&amp;#39;_0 (t) = \alpha P_1 (t) - \beta P_0 (t) .&#xD;&#xA;$$&#xA;Similarly,&#xA;$$&#xD;&#xA;P&amp;#39;_1 (t) = \beta P_0 (t) - \alpha P_1 (t) ,&#xD;&#xA;$$&#xA;which also follows trivially from $P_0 (t) + P_1 (t) = 1$.&lt;/p&gt;&#xA;" OwnerUserId="2810" LastEditorUserId="2810" LastEditorDisplayName="" LastEditDate="2011-03-18T05:39:41.823" LastActivityDate="2011-03-18T05:39:41.823" CommentCount="0" />
  <row Id="27724" PostTypeId="2" ParentId="27373" CreationDate="2011-03-18T02:55:02.590" Score="8" ViewCount="" Body="&lt;p&gt;Kristi, first of all if you are projecting $\begin{bmatrix}&#xD;&#xA;x_1\\ &#xD;&#xA;x_2\\ &#xD;&#xA;x_3&#xD;&#xA;\end{bmatrix}$ onto the $x_{1}=x_{2}$ plane, then you are projecting onto the plane $x=y$, not $y=z$ (since you defined $x=x_{1}$, $y=x_{2}$, and $z=x_{3}$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, as you are trying to find the coordinates of the projection vector, imagine the geometric meaning -- $z$, the 'height' of the vector will not ever change, as it is not relevant to the equation, but $x$ and $y$ will, depending on where the vector lies. When we are trying to find a projection on an n-dimensional subspace $W$, we can use a formula of ${proj_{W}}{\vec{x}}$=$(\vec{u_1}\cdot \vec{x})$$\vec{u_1}$+$(\vec{u_2}\cdot \vec{x})$$\vec{u_2}$$+\cdots +$$(\vec{u_n}\cdot \vec{x})$$\vec{u_n}$, where $\vec{u_1}, \vec{u_2}\dots \vec{u_n}$ form an &lt;a href=&quot;http://en.wikipedia.org/wiki/Orthonormal_basis&quot; rel=&quot;nofollow&quot;&gt;orthonormal basis&lt;/a&gt; of the subspace $W$. Here, $W$ is defined as $x=y$, meaning it can be spanned by vectors $\vec{v_1}$ = $&#xD;&#xA;\begin{bmatrix}&#xD;&#xA;1\\&#xD;&#xA;1\\&#xD;&#xA;0&#xD;&#xA;\end{bmatrix}$ and $\vec{v_2}$ =&#xA;$&#xD;&#xA;\begin{bmatrix}&#xD;&#xA;1\\&#xD;&#xA;1\\&#xD;&#xA;2&#xD;&#xA;\end{bmatrix}$, for example. To find an orthonormal basis of our space (meaning that all vectors in it will be mutually orthogonal/perpendicular, as well as of a length one), let's  use the &lt;a href=&quot;http://en.wikipedia.org/wiki/Gram-schmidt&quot; rel=&quot;nofollow&quot;&gt;Gram-Schmidt&lt;/a&gt; process. An orthonormalized version of the vector $&#xD;&#xA;\begin{bmatrix}&#xD;&#xA;1\\&#xD;&#xA;1\\&#xD;&#xA;0&#xD;&#xA;\end{bmatrix}$ would be $\vec{u_1}$ = $\frac{1}{\sqrt{2}}$$&#xA;\begin{bmatrix}&#xA;1\\&#xA;1\\&#xA;0&#xA;\end{bmatrix}$, as that will make it of length one. Now, by Gram-Schmidt, $\vec{u_2}=\vec{v_2}-\frac{\vec{v_2}\cdot\vec{u_1}}{\vec{u_1}\cdot\vec{u_1}} \vec{u_1}$, since we are basically subtracting the $\vec{u_1}$ component from our second vector, in order to get a vector perpendicular to $\vec{u_1}$ as a result. Calculations result into the following: $\vec{u_2}$=&#xA;$\begin{bmatrix}&#xD;&#xA;1\\ &#xD;&#xA;1\\ &#xD;&#xA;2&#xD;&#xA;\end{bmatrix}$&#xA;$-$$\frac{\begin{bmatrix}&#xA;1\\ &#xA;1\\ &#xA;2&#xA;\end{bmatrix} \cdot \frac{1}{\sqrt{2}}&#xA;\begin{bmatrix}&#xA;1\\&#xA;1\\&#xA;0&#xA;\end{bmatrix}}{\frac{1}{\sqrt{2}}&#xA;\begin{bmatrix}&#xA;1\\&#xA;1\\&#xA;0&#xA;\end{bmatrix} \cdot \frac{1}{\sqrt{2}}&#xA;\begin{bmatrix}&#xA;1\\&#xA;1\\&#xA;0&#xA;\end{bmatrix}} \frac{1}{\sqrt{2}}&#xA;\begin{bmatrix}&#xA;1\\&#xA;1\\&#xA;0&#xA;\end{bmatrix}$ = $\begin{bmatrix}&#xD;&#xA;1\\ &#xD;&#xA;1\\ &#xD;&#xA;2&#xD;&#xA;\end{bmatrix}$ - $\begin{bmatrix}&#xA;1\\ &#xA;1\\ &#xA;0&#xA;\end{bmatrix}$=$\begin{bmatrix}&#xA;0\\ &#xA;0\\ &#xA;2&#xA;\end{bmatrix}$. Normalizing the resulting vector, we get $\vec{u_2}$ = $\begin{bmatrix}&#xD;&#xA;0\\&#xD;&#xA;0\\&#xD;&#xA;1&#xD;&#xA;\end{bmatrix}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now that we have an orthogonal basis $\vec{u_1}$ = $\frac{1}{\sqrt{2}}$$\begin{bmatrix}&#xA;1\\&#xA;1\\&#xA;0&#xA;\end{bmatrix}$ and $\vec{u_2}$ = $\begin{bmatrix}&#xD;&#xA;0\\&#xD;&#xA;0\\&#xD;&#xA;1&#xD;&#xA;\end{bmatrix}$, we can calculate the projection.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, to find the projection of your vector $\begin{bmatrix}&#xD;&#xA;x_1\\\ &#xD;&#xA;x_2\\\ &#xD;&#xA;x_3&#xD;&#xA;\end{bmatrix}$ we use our orthonormal basis and the projection formula: ${proj_{W}}\begin{bmatrix}&#xA;x_1\\ &#xA;x_2\\ &#xA;x_3&#xA;\end{bmatrix}$=($\frac{1}{\sqrt{2}}$$\begin{bmatrix}&#xA;1\\&#xA;1\\&#xA;0&#xA;\end{bmatrix} \cdot \begin{bmatrix}&#xA;x_1\\ &#xA;x_2\\ &#xA;x_3&#xA;\end{bmatrix})(\frac{1}{\sqrt{2}}$$\begin{bmatrix}&#xA;1\\&#xA;1\\&#xA;0&#xA;\end{bmatrix})$$+$$(\begin{bmatrix}&#xA;0\\&#xA;0\\&#xA;1&#xA;\end{bmatrix} \cdot \begin{bmatrix}&#xA;x_1\\ &#xA;x_2\\ &#xA;x_3&#xA;\end{bmatrix})(\begin{bmatrix}&#xA;0\\&#xA;0\\&#xA;1&#xA;\end{bmatrix})$. After arithmetic, this results into ${proj_{W}}\begin{bmatrix}&#xA;x_1\\ &#xA;x_2\\ &#xA;x_3&#xA;\end{bmatrix}$=$(\frac{x_1+x_2}{\sqrt{2}})$$(\frac{1}{\sqrt{2}}\begin{bmatrix}&#xA;1\\&#xA;1\\&#xA;0&#xA;\end{bmatrix})$+$\begin{bmatrix}&#xA;0\\&#xA;0\\&#xA;x_3&#xA;\end{bmatrix}$=$\begin{bmatrix}&#xA;\frac{x_1+x_2}{2}\\&#xA;\frac{x_1+x_2}{2}\\&#xA;x_3&#xA;\end{bmatrix}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, now you have your coordinates. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To find out the eigenvalues, think of the nature of the transformation -- the projection  will not do anything to a vector if it is within the plane onto which you are projecting, and it will crash it if the vector is perpendicular to the plane. So, your eigenvalues are 1 and 0. A basis of eigenspace of 1 $\xi_{1}$ will have two vectors, as the plane is spanned by two of them. You could choose them to be your original $v_1$ and $v_2$, which were $&#xD;&#xA;\begin{bmatrix}&#xD;&#xA;1\\&#xD;&#xA;1\\&#xD;&#xA;0&#xD;&#xA;\end{bmatrix}$ and $\begin{bmatrix}&#xD;&#xA;1\\&#xD;&#xA;1\\&#xD;&#xA;2&#xD;&#xA;\end{bmatrix}$. To find a basis of eigenspace of 0 $\xi_{0}$, you need to find a vector perpendicular to this plane. You could use a property of cross-product, which states that $\vec{v_1} \times \vec{v_2}$ produces a vector $\vec{v_3}$ perpendicular to both. Crossing the aforementioned vectors, you get $\vec{v_3}=\begin{bmatrix}&#xD;&#xA;2\\&#xD;&#xA;-2\\&#xD;&#xA;0&#xD;&#xA;\end{bmatrix}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now that you know all of this, finding the matrix &lt;strong&gt;B&lt;/strong&gt; is very easy by inspection; consider $\begin{bmatrix}&#xD;&#xA;\frac{1}{2} &amp;amp; \frac{1}{2} &amp;amp; 0 \\ &#xD;&#xA;\frac{1}{2} &amp;amp; \frac{1}{2} &amp;amp; 0 \\ &#xD;&#xA;0 &amp;amp; 0 &amp;amp; 1&#xD;&#xA;\end{bmatrix}$.&lt;/p&gt;&#xA;" OwnerUserId="3731" LastEditorUserId="3731" LastEditorDisplayName="" LastEditDate="2011-03-18T05:43:27.670" LastActivityDate="2011-03-18T05:43:27.670" CommentCount="4" />
  <row Id="27725" PostTypeId="2" ParentId="27717" CreationDate="2011-03-18T02:55:36.180" Score="2" ViewCount="" Body="&lt;p&gt;Note first that $|(f \ast g)(x)| \leq \Vert f \Vert_2 \, \Vert g\Vert_2$ by H&amp;ouml;lder's inequality. Therefore convolution is a continous bilinear map from $L^2([0,2\pi])$ to the space of bounded functions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the space of continous functions is dense in $L^2$, we may find a sequence of continuous $2\pi$-periodic functions such that $f_{n} \to f$ in $L^2$. Then the above inequality gives $f_{n} \ast g \to f \ast g$ &lt;em&gt;uniformly&lt;/em&gt; on $[0,2\pi]$. Since a uniform limit of continuous functions is continous, it suffices to prove that $f \ast g$ is continuous if $f$ is continuous.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $f$ is continuous then $f$ is &lt;em&gt;uniformly continuous&lt;/em&gt; (because $[0,2\pi]$ is compact), that is: for each $\varepsilon \gt 0$ there is $\delta \gt 0$ such that for all $|z - z&amp;#39;| \lt \delta$ we have $|f(z) - f(z&amp;#39;)| \lt \varepsilon$. In other words, $\|f(\cdot - x) - f(\cdot - x_{n})\|_{\infty} \lt \varepsilon$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $\varepsilon \gt 0$. For $n$ so large that $|x_{n} - x| \lt \delta$ we then have&#xA;\[&#xA;|(f \ast g)(x_{n}) - (f \ast g)(x)| \leq \Vert f(\cdot - x) - f(\cdot - x_{n})\Vert_{\infty} \Vert g \Vert_{1} \lt \varepsilon \cdot \Vert g \Vert_1&#xA;\]&#xA;again by H&amp;ouml;lder (note that $L^2([0,2\pi)] \subset L^1([0,2\pi])$ so that $\|g\|_{1}$ is finite). This shows that $f \ast g$ is continuous if $f$ is continuous and as argued in the second paragraph this shows that $f \ast g$ is continuous for all $f,g \in L^2([0,2\pi])$.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added:&lt;/strong&gt; To avoid confusion, let me mention that this argument is &lt;em&gt;not&lt;/em&gt; circular. In order to show that the space of continuous functions is dense in $L^2$ one often uses convolution with a mollifying sequence. Of course, we should avoid that and this can be done without too much pain (the crux of the proof lies in showing that the step functions are dense in $L^p$ for $1 \leq p \lt \infty$), see e.g. Royden, &lt;em&gt;Real Analysis&lt;/em&gt;, Proposition 8 of Chapter 4, on p.128 of the third edition.&lt;/p&gt;&#xA;" OwnerUserId="5363" LastEditorUserId="5363" LastEditorDisplayName="" LastEditDate="2011-03-18T03:32:26.757" LastActivityDate="2011-03-18T03:32:26.757" />
  <row Id="27726" PostTypeId="1" AcceptedAnswerId="27729" CreationDate="2011-03-18T02:56:05.290" Score="3" ViewCount="96" Body="&lt;p&gt;I have known for a while that $\mathbb{R}P^1 \simeq S^1$. Indeed, visually it is easy to see why this is the case (any two antipodal points define a line through the origin)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Say we now forget everything we know, and let $F$ be a division ring and let $n \ge 0$. We definie an equivalence relation of $F - \{ 0 \}$ by $x \sim y$ if there exists $\lambda \in F - \{ 0 \}$ with $x = \lambda y$. The quotient set $F^{n+1}-\{ 0 \} / \sim$ is called the F-projective n-space. The class of $x = (x_0,\ldots ,x_n)$ is denoted by $[x] = [x_0, \ldots, x_n] \in FP^n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Based on this definition I would like to show $\mathbb{R}P^1 \simeq S^1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So how do we go from this definition to prove the isomorphism? I am sure it is simple...but I am not really sure where to go. Do I just use the antipodal max $f(x)=-x$? (Note the next part is to extent to show $\mathbb{H}P^1 \simeq S^4$ which is a bit harder to visualise!)&lt;/p&gt;&#xA;" OwnerUserId="2219" LastEditorUserId="2219" LastEditorDisplayName="" LastEditDate="2011-03-18T03:35:02.637" LastActivityDate="2011-03-18T04:01:32.660" Title="Show that $\mathbb{R}P^1 \simeq S^1$" Tags="&lt;topology&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="27727" PostTypeId="2" ParentId="27375" CreationDate="2011-03-18T02:56:43.097" Score="0" ViewCount="" Body="&lt;p&gt;Virtually every graphing calculator made in the recent past has a mod function.  Many scientific calculators do as well, though the documentation is often lacking.&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2011-03-18T02:56:43.097" CommentCount="4" />
  <row Id="27728" PostTypeId="2" ParentId="27726" CreationDate="2011-03-18T03:11:44.763" Score="1" ViewCount="" Body="&lt;p&gt;I suppose you want a &lt;em&gt;homeomorphism&lt;/em&gt; between $\mathbb R\mathbb P^1$ and $S^1$. ($S^1$ does not have an algebraic structure, so isomorphism in that sense would not make sense.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Define $\mathbb R\mathbb P^1\to S^1$ by $[a:b]\mapsto (a/\sqrt{a^2+b^2}, b/\sqrt{a^2+b^2})$ and $S^1\to \mathbb R\mathbb P^1$ by $(a,b)\mapsto [a:b]$. These are inverses to each other and hence bijective. They are both clearly continuous, so we're done.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ps: Why do you skip $\mathbb C\mathbb P^1\sim S^2$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;pps: the proof is the same in all cases including $\mathbb C$ and $\mathbb H$. You just have to figure out the dimension of each division algebra over $\mathbb R$, but it seems you already know that.&lt;/p&gt;&#xA;" OwnerUserId="4654" LastActivityDate="2011-03-18T03:11:44.763" CommentCount="1" />
  <row Id="27729" PostTypeId="2" ParentId="27726" CreationDate="2011-03-18T03:16:10.623" Score="4" ViewCount="" Body="&lt;p&gt;First, consider the mapping&#xA;$$\varphi\colon (x,y) \mapsto \left(\frac{x}{\sqrt{x^2+y^2}}, \frac{y}{\sqrt{x^2+y^2}}\right)$$&#xA;which maps $\mathbb{R}^2-\{0\}$ to the unit circle $\mathcal{C}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $(x,y)$ and $(z,w)$ map to the same point, then it follows that $x = \lambda z$ and $y=\lambda w$, where &#xA;$$\lambda = \frac{\sqrt{x^2+y^2}}{\sqrt{z^2+w^2}}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, it is possible for $(x,y)$ and $(z,w)$ to satisfy $(x,y)=\lambda(z,w)$ and &lt;em&gt;not&lt;/em&gt; to map to the same point: indeed, if $(x,y)=\lambda(z,w)$, then it is easy to see that $(x,y)$ maps to&#xA;$$\frac{\lambda}{|\lambda|}\left(\frac{z}{\sqrt{z^2+w^2}},\frac{w}{\sqrt{z^2+w^2}}\right).$$&#xA;So we define an equivalence relation on the unit circle by $\mathbf{x}\sim\mathbf{y}$ if and only if $\mathbf{x}=\pm\mathbf{y}$, and consider $\mathcal{C}^1/\sim$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we consider the composition of $\varphi$ and $\pi\colon \mathcal{C}^1\to\mathcal{C}^1/\sim$. If $(x,y) =\lambda (z,w)$, then $\pi\circ\varphi(x,y) = \pi\circ\varphi(z,w)$ by the discussion above. Conversely, if $\pi\circ\varphi(x,y) = \pi\circ\varphi(z,w)$, then either $\varphi(x,y)=\varphi(z,w)$, in which case $(x,y) = \lambda(z,w)$ with $\lambda$ as above; or else $\varphi(x,y) = -\varphi(z,w)$, hence&#xA;$$(x,y) = -\frac{\sqrt{x^2+y^2}}{\sqrt{z^2+w^2}}(z,w).$$&#xA;So we conclude that $\pi\circ\varphi$ factors through $\mathbb{R}^2-\{(0,0)\}/\equiv$, where $\equiv$ is the equivalence relation on $\mathbb{R}^2=\{(0,0)\}$ given by $\mathbf{x}\equiv \mathbf{z}\Longleftrightarrow \mathbf{x}=\lambda\mathbf{z}$ for some scalar $\lambda$, and induces a bijection&#xA;$$\frac{\mathbb{R}^2-\{(0,0)\}}{\equiv} \to \mathcal{C}^1/\sim \cong S^1.$$&#xA;The maps are all continuous.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-18T04:01:32.660" LastActivityDate="2011-03-18T04:01:32.660" CommentCount="5" />
  <row Id="27730" PostTypeId="1" CreationDate="2011-03-18T03:23:14.540" Score="3" ViewCount="87" Body="&lt;p&gt;I posted this on cs theory &lt;a href=&quot;http://cstheory.stackexchange.com/questions/5534/finding-svd-efficiently-for-abt&quot;&gt;yesterday&lt;/a&gt; but did not get an answer and hence I am posting here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a low rank matrix given as $AB^T$ where $A,B \in \mathbb{R}^{n \times p}$ and $p \ll n$. (I know $A$ and $B$ separately)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One efficient way to get the Singular Value Decomposition is to do a reduced QR on $A$ and $B$ i.e. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$A = Q_A R_A$ and $B = Q_B R_B$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The above can be done in $\mathcal{O}(p^2n)$ cost.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I could then compute the svd of $R_A R_B^T = U_1 \Sigma V_1^T$ which is $\mathcal{O}(p^3)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence this will give me $$AB^T = (Q_A U_1) \Sigma (Q_B V_1)^T$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The total cost is $\mathcal{O}(np^2)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I am wondering if there are other efficient ways to go about doing this to reduce the coefficient infront of $p^2n$. If I am right, the coefficient infront of $np^2$ is $3$ if we go about doing QR. The reason why I am interested in minimizing the coefficient is that my $n$ is really ginormous. So if I were to implement it, a cost cutting on the coefficient of the $\mathcal{O}(n)$ could be significant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am wondering if we could exploit the fact that the left singular vectors must span $A$ and the right singular vectors span $B$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i.e. we know that if $AB^T = U \Sigma V^T$, then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$A = U \alpha \text{ and } B = V \beta$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and we want $\alpha \beta^T$ to be diagonal and $U$,$V$ needs to be unitary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt; What I essentially want is an exact rank $r&amp;lt;p$ approximation to the product $AB^T$ in the form $U_r V_r^T$. I am not looking for approximate or probabilistic algorithms.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2011-03-18T05:47:19.627" LastActivityDate="2011-05-17T12:58:19.687" Title="Finding SVD efficiently for $AB^T$" Tags="&lt;matrices&gt;&lt;algorithms&gt;&lt;numerical-linear-algebra&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="1" />
  <row Id="27731" PostTypeId="1" AcceptedAnswerId="27737" CreationDate="2011-03-18T03:34:23.530" Score="4" ViewCount="61" Body="&lt;p&gt;How do you transform isoperimetric inequalities of a group to the of Riemann  integrals of functions of the form $f\colon \mathbb{R}\rightarrow G$ where $G$ is a metric group so that being $\delta-$hyperbolic in the sense of Gromov is expressible via Riemann integration? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, how do you define &quot;being $\delta-$hyperbolic group&quot; by using  integrals in metric groups? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Note: I am not interested in the &quot;Riemann&quot; part, so you are free to take commutative groups with lebesgue integration etc.)   &lt;/p&gt;&#xA;" OwnerUserId="7796" LastActivityDate="2011-03-18T04:40:02.210" Title="Isoperimetric inequalities of a group" Tags="&lt;group-theory&gt;&lt;integral&gt;&lt;metric-spaces&gt;&lt;hyperbolic-geometry&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="27732" PostTypeId="1" AcceptedAnswerId="27735" CreationDate="2011-03-18T03:37:13.697" Score="3" ViewCount="148" Body="&lt;p&gt;Hi Everyone!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am a total newbie to the world of math and was interesting in learning.  I just finished my degree(non-math) and am going to study a few math books to see if it interests me to apply for something more quantitative but I want to study something interesting with interesting problems that won't bore me.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I thought about it, and thought of the type of problems that intrest me.  One is predicting the future and the other is predicting the past.  Here's a problem that I think would be cool. Say you have a list&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;calories 89, 34, 67, 43, 54, 232, 623&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;and someone tells you that someone had a total of &quot;6553&quot; calories in a day.  What type of math would try to figure this out? Is it algebra? (by the way to get this question all I did was take each value above and multiple first one by 1, second one by two, etc.up to 7.)&lt;/p&gt;&#xA;" OwnerUserId="8402" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-18T03:46:28.163" LastActivityDate="2011-03-18T04:57:35.077" Title="what type of math is this?" Tags="&lt;elementary-number-theory&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" />
  <row Id="27733" PostTypeId="2" ParentId="27610" CreationDate="2011-03-18T03:42:44.377" Score="2" ViewCount="" Body="&lt;p&gt;An alternative way to proceed if you don't know about symmetric and skew-symmetric matrices (or projections) and you don't want to try to find the characteristic polynomial for arbitrary $N$ is the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Start with the general observation: if $\mathbf{x}$ is an eigenvector of $T$ corresponding to the eigenvalue $\lambda$, then $\mathbf{x}$ is also an eigenvector of $T^2$ ($T$ composed with itself) corresponding to $\lambda^2$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Indeed, we have that $\mathbf{x}\neq\mathbf{0}$ (since $\mathbf{x}$ is an eigenvector of $T$), and &#xA;\begin{align*}&#xA;T^2(\mathbf{x}) &amp;amp;= T(T(\mathbf{x})) = T(\lambda\mathbf{x})\\&#xA;&amp;amp;= \lambda T(\mathbf{x}) = \lambda(\lambda\mathbf{x}) = \lambda^2\mathbf{x}.&#xA;\end{align*}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Caveat.&lt;/em&gt; The converse does not hold: you can have $\mathbf{x}$ be an eigenvector of $T^2$, but not of $T$: consider the rotation of the real plane by ninety degrees; then ever vector is an eigenvalue of $T^2$, but $T$ has no eigenvectors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, consider your $T$. If you apply $T$ twice, you get:&#xA;$$T(T(A)) = T\left(\frac{1}{2}(A-A^t)\right) = \frac{1}{2}\left(\frac{1}{2}(A-A^t) - \frac{1}{2}(A^t-A)\right) = \frac{1}{2}(A-A^t) = T(A).$$&#xA;So if $A$ is an eigenvector if corresponding eigenvalue $\lambda$, then you have that $T(A)=\lambda A$ and $T^2(A)=\lambda^2A$. But since $T(A)=T^2(A)$, then $\lambda A = \lambda^2 A$, hence $(\lambda-\lambda^2)A=\mathbf{0}$. Since $A\neq\mathbf{0}$, then $\lambda=\lambda^2$, so either $\lambda=0$ or $\lambda=1$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the only possible eigenvalues of $T$ are $\lambda=0$ and $\lambda=1$. (Note: This is true of any linear transformation $T$ such that $T=T^2$; these are called &quot;projections&quot;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now that we know the &lt;em&gt;possible&lt;/em&gt; eigenvalues, it's a bit easier to check. If $A$ is an eigenvector corresponding to $0$, then we must have&#xA;$ 0 = \frac{1}{2}(A-A^t)$, so $A=A^t$; that is, $A$ must be equal to its transpose (symmetric)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $\lambda=1$, then we must have $A = \frac{1}{2}(A-A^t)$, so $A=-A^t$; that is, $A$ must be skew-symmetric. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So this determines all the eigenvectors and eigenvalues of $T$. It is now straightforward to verify that you can find enough linearly independent symmetric and skew-symmetric matrices (to wit, $n^2$ of them) to get a basis for the vector space, so that $T$ is diagonalizable.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2011-03-18T03:42:44.377" CommentCount="1" />
  <row Id="27734" PostTypeId="2" ParentId="518" CreationDate="2011-03-18T03:52:04.093" Score="3" ViewCount="" Body="&lt;p&gt;Denote set complements in $\rm\:A\:$ by $\rm\:\bar T = A - T\:.\ $ Consider the monoid $\rm\:S\:$ generated by $\rm\:\bar P\:$ and $\rm\:\bar Z\:,\ $ with $\rm\:Z = $ all zero-divisors in $\rm\:A\:$ (including $0$). Since $\rm\:0\not\in S\ $ (else $\rm 0 = a\:b\:,\ a\in \bar P,\ b\in \bar Z\ \Rightarrow\ b\in Z\:$) we can enlarge $0$ to an ideal $\rm\:Q\:$ maximally disjoint from $\rm\:S\:.\ $ Since $\rm\:S\:$ is a monoid, $\rm\:Q\:$ is prime.  $\rm\: S\:\supset\: \bar P \cup \bar Z\ \Rightarrow\ Q \subset \bar S \subset P\cap Z\ $ so by minimality of $\rm\:P\:$ we infer $\rm\ P = Q \subset Z\:.\quad$ &lt;strong&gt;QED&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2011-03-18T04:04:03.567" LastActivityDate="2011-03-18T04:04:03.567" CommentCount="4" />
  <row Id="27735" PostTypeId="2" ParentId="27732" CreationDate="2011-03-18T03:59:23.350" Score="5" ViewCount="" Body="&lt;p&gt;To cast the problem a little more clearly, you have a number of &quot;weights&quot;, $w_1,\ldots,w_n$, in this case:&#xA;\begin{align*}&#xA;w_1 &amp;amp;= 89\\&#xA;w_2 &amp;amp;= 34\\&#xA;w_3 &amp;amp;= 67\\&#xA;w_4 &amp;amp;= 43\\&#xA;w_5 &amp;amp;= 54\\&#xA;w_6 &amp;amp;= 232\\&#xA;w_7 &amp;amp;= 623,&#xA;\end{align*}&#xA;and a &quot;target total&quot; $T$, in this case $T=6553$. You want to find nonnegative integers $a_1,\ldots,a_n$ such that&#xA;$$a_1w_1 + \cdots + a_nw_n = T.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In its broadest sense, this is an example of what is called a &lt;a href=&quot;http://en.wikipedia.org/wiki/Diophantine_equation&quot; rel=&quot;nofollow&quot;&gt;Diophantine equation&lt;/a&gt; (an equation in which we require the solutions to be nonnegative integers, or more generally rational numbers). They are studied in the branch of mathematics called &lt;a href=&quot;http://en.wikipedia.org/wiki/Number_theory&quot; rel=&quot;nofollow&quot;&gt;Number theory.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2011-03-18T03:59:23.350" CommentCount="12" />
  <row Id="27736" PostTypeId="2" ParentId="27732" CreationDate="2011-03-18T03:59:40.040" Score="1" ViewCount="" Body="&lt;p&gt;For your question you need to find positive integer solutions to  $89a +34b +67c +43d+ 54d +232d+ 623e=6553$, $\{a,...,e\}$ are the number of items you eat of each to get 6553 calories. Problems like this where you need to find integer solutions are called linear Diophantine equations. But you can treat them as puzzles and try to use some hit and trial methods/computations. Sometimes, though not always, &lt;a href=&quot;http://www.wolframalpha.com&quot; rel=&quot;nofollow&quot;&gt;Wolfram alpha&lt;/a&gt; might help.&lt;/p&gt;&#xA;" LastActivityDate="2011-03-18T04:57:35.077" />
  <row Id="27737" PostTypeId="2" ParentId="27731" CreationDate="2011-03-18T04:14:58.327" Score="4" ViewCount="" Body="&lt;p&gt;You can do this using &lt;em&gt;metric currents&lt;/em&gt; in the sense of Ambrosio-Kirchheim. This is a rather new development of geometric measure theory, triggered by Gromov and really worked out only in the last decade. I should warn you that this is rather technical stuff and nothing for the faint-hearted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Urs Lang has a set of nice lecture notes, where you can find most of the relevant references, see &lt;a href=&quot;http://www.math.ethz.ch/~lang/loc.pdf&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My friend Stefan Wenger has done quite a bit of work on Gromov hyperbolic spaces and isoperimetric inequalities, his Inventiones paper &lt;em&gt;Gromov hyperbolic spaces and the sharp isoperimetric constant&lt;/em&gt; seems most relevant. You can find a link to the published paper and his other work on &lt;a href=&quot;http://www.math.uic.edu/~wenger/&quot; rel=&quot;nofollow&quot;&gt;his home page&lt;/a&gt;, the ArXiV-preprint is &lt;a href=&quot;http://arxiv.org/abs/math/0609310&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I should add that I actually prefer to prove that a linear (or subquadratic) isoperimetric inequality implies $\delta$-hyperbolicity using a coarse notion of area (see e.g. Bridson-Haefliger's book) or using Dehn functions, the latter can be found in Bridson's beautiful paper &lt;a href=&quot;http://people.maths.ox.ac.uk/bridson/papers/bfs/bfs.pdf&quot; rel=&quot;nofollow&quot;&gt;The geometry of the word problem&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="5363" LastEditorUserId="5363" LastEditorDisplayName="" LastEditDate="2011-03-18T04:40:02.210" LastActivityDate="2011-03-18T04:40:02.210" CommentCount="2" />
  <row Id="27740" PostTypeId="1" CreationDate="2011-03-18T04:51:46.393" Score="0" ViewCount="51" Body="&lt;p&gt;I'm massively out of my depths when it comes to the kind of maths that is posted here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I have a calculation related to a shopping cart i'm programming that was originally posted on Stack Overflow, however, since it boils down to a maths problem, I thought i'd give it a go here.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Setup&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;The cart allows items to be added with a different tax rate per item.&lt;br&gt;&#xA;As item tax rates can vary, the cart tax total cannot simply be calculated by for example doing:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Tax rate is 20%, so if sub-total is £100, we add 20% tax, the tax total equals £20.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Instead, the variable tax rate of each item is calculated against the items price and added to a running tax total.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So far, everything is running fine.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;The Problem&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Here's a quick example of the cart. Note that item prices include tax.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Item #1 price = £100, tax rate = 20%, tax total = £16.67&lt;br&gt;&#xA;  Item #2 price = £50, tax rate = 0%, tax total = £0&lt;br&gt;&#xA;  Cart Total = £150&lt;br&gt;&#xA;  Cart Tax-Total = £16.67 &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The problem occurs if a user applies a discount coupon, for example: £10 off grand total.&lt;br&gt;&#xA;If the grand total is £150 and I discount £10, the grand total is of course £140.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do I now work out what the tax total is of the new grand total, whilst still bearing in mind that not all items have tax applied? &lt;/p&gt;&#xA;&#xA;&lt;h3&gt;My own Possible Answer&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;If I divide the pre-discount cart total (£150), against the pre-discount tax total (£16.67), I get a factor of 8.998.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I then simply divide the discounted cart total (£140) by 8.998, I get a new discounted tax total of £15.56 - which seems to me about right.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, can someone please confirm whether my calculations are correct?&lt;/p&gt;&#xA;" OwnerUserId="8405" LastActivityDate="2011-03-18T04:51:46.393" Title="Calculate total of a variable tax rate when a flat rate discount is applied" Tags="&lt;soft-question&gt;" CommentCount="1" />
  <row Id="27741" PostTypeId="1" CreationDate="2011-03-18T04:57:38.733" Score="8" ViewCount="143" Body="&lt;p&gt;If $m$ is an outer measure on a set $X$, a subset $E$ of $X$ is called $m$-measurable iff &#xA;$$&#xD;&#xA;m(A) = m(A \cap E) + m(A \cap E^c)&#xD;&#xA;$$&#xA;for all subsets $A$ of $X$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The collection $M$ of all $m$-measurable subsets of $X$ forms a $\sigma$-algebra and $m$ is a complete measure when restricted to $M$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is $M$ the largest $\sigma$-algebra on $X$ on which $m$ is a measure (i.e., on which $m$ is countably additive)?  If not, what is?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is $M$ the largest $\sigma$-algebra on $X$ on which $m$ is a complete measure?  If not, what is?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am especially interested in the case when $X$ is $\mathbb{R}$ or $\mathbb{R}^n$ and $m$ is the Lebesgue outer measure.  In this case $M$ is the Lebesgue $\sigma$-algebra.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ADDED:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Julián Aguirre (thanks!) has shown in his response below that the answer to the first question is yes when $X$ is $\mathbb{R}^n$ and $m$ is the Lebesgue outer measure.  Hence the answer to the second question in this situation is also yes.&lt;/p&gt;&#xA;" OwnerUserId="6887" LastEditorUserId="6887" LastEditorDisplayName="" LastEditDate="2011-03-28T08:30:09.057" LastActivityDate="2011-03-28T08:30:09.057" Title="Largest $\sigma$-algebra on which an outer measure is countably additive" Tags="&lt;real-analysis&gt;&lt;analysis&gt;&lt;measure-theory&gt;" AnswerCount="1" FavoriteCount="2" />
  <row Id="27742" PostTypeId="1" CreationDate="2011-03-18T05:20:21.100" Score="1" ViewCount="88" Body="&lt;p&gt;I want to ask how to solve the equation $$-\frac{{\hbar}^{2}}{2m}\frac{\partial}{\partial_{r}^{2}}u(r)=(E-\frac{1}{2}Kr^{2})u(r)$$ with $K$ being a constant. &lt;/p&gt;&#xA;" OwnerUserId="7887" LastActivityDate="2011-03-18T05:26:48.753" Title="The quantum harmonic oscillator" Tags="&lt;homework&gt;" CommentCount="3" />
  <row Id="27743" PostTypeId="1" CreationDate="2011-03-18T05:34:59.710" Score="3" ViewCount="75" Body="&lt;p&gt;One reason why the iteration&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x_{n+1}=x_n-\tan\;x_n$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;converges quickly for appropriate starting values is that this is nothing more than the Newton-Raphson iteration for $\sin\;x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This got me thinking: given some arbitrary function $g(x)$, is there always a function $f(x)$ such that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{f(x)}{f^\prime(x)}=g(x)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or are there restrictions on the nature of $g(x)$ so that the differential equation has a solution?&lt;/p&gt;&#xA;" OwnerUserId="8406" LastActivityDate="2011-03-18T06:10:15.280" Title="Does a &quot;Newton differential equation&quot; always have a solution?" Tags="&lt;differential-equations&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="27744" PostTypeId="1" AcceptedAnswerId="27823" CreationDate="2011-03-18T05:56:23.207" Score="10" ViewCount="235" Body="&lt;p&gt;In &lt;a href=&quot;http://math.stackexchange.com/q/27667/5189&quot;&gt;this question&lt;/a&gt;, The Chaz asks whether $G\times G\cong H\times H$ implies that $G\cong H$, where $G$ and $H$ are finite abelian groups. The answer is to his question is yes, by the structure theorem for finite abelian groups, as noted in the answer by Anjan Gupta.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even though I don't know the first thing about categories -- except for the things that I do know -- I'm wondering if and how such property could be expressed and proven in terms of universal properties, without actually manoevring inside the objects. For instance one may attempt to create a morphism $G\to H$ somehow appealing to the universal property of $\oplus$, and subsequently show this morphism is an isomorphism by chasing diagrams. But it seems likely that the existence of a structure theorem of some sort will be required.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This question may be considered trivial or weird for someone who's fluent with categories, I don't know. &lt;a href=&quot;http://math.stackexchange.com/q/370/5189&quot;&gt;This topic&lt;/a&gt; contains quite a few references. I haven't really worked through any of them (yet) but I couldn't find anything helpful at first sight.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-- edit, So a more precise title would have been &quot;Under what conditions that can be expressed in a universal way does $G\oplus G\cong H\oplus H$ imply $G\cong H$ ?&quot; but I don't like long titles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-- edit2, By the comment by Alexei Averchenko, maybe It's more natural to ask this question with the product instead of the coproduct. An answer to my question with the 'real' product would be appreciated too, obviously.&lt;/p&gt;&#xA;" OwnerUserId="5189" LastEditorUserId="5189" LastEditorDisplayName="" LastEditDate="2011-03-18T06:12:14.423" LastActivityDate="2011-03-19T01:37:51.103" Title="Does $G\oplus G \cong H\oplus H$ imply $G\cong H$ in general?" Tags="&lt;category-theory&gt;" AnswerCount="1" CommentCount="10" FavoriteCount="2" />
  <row Id="27745" PostTypeId="2" ParentId="27743" CreationDate="2011-03-18T05:59:09.100" Score="2" ViewCount="" Body="&lt;p&gt;Certainly not for &quot;arbitrary&quot; functions, as functions of the form $\frac{f}{f&amp;#39;}$ satisfy properties that not all functions satisfy.  In fact, in any open set where $g$ is nonzero, $\frac{1}{g}=\log(|f|)&amp;#39;$, so $g$ must be the reciprocal of a derivative of a function.  This also leads to a way to find such $f$ in many cases, namely by integrating the last equation to get $\log(|f(x)|)=\int_a^x\frac{dt}{g(t)}+C$ provided the integral exists, in which case $f(x)=c\cdot\exp\left(\int_a^x\frac{dt}{g(t)}\right)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, attempting this with $g(x)=\tan(x)$ leads to integrating $\cot(x)$ and ending up with $f(x)=c\cdot\sin(x)$.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2011-03-18T06:10:15.280" LastActivityDate="2011-03-18T06:10:15.280" />
  <row Id="27746" PostTypeId="2" ParentId="27743" CreationDate="2011-03-18T06:01:19.677" Score="4" ViewCount="" Body="&lt;p&gt;The differential equation you have is an example of a first-order separable differential equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you write $y = f(x)$, then you have &#xA;$$&#xD;&#xA;\frac{y}{\frac{dy}{dx}} = g(x)&#xD;&#xA;$$&#xA;Writing this in terms of differentials (treat $\frac{dy}{dx}$ as a fraction) so that all the $y$'s are on one side and all the $x$'s are on the other gives&#xA;$$&#xD;&#xA;\frac{dx}{g(x)} = \frac{dy}{y}&#xD;&#xA;$$&#xA;Integrating both sides yields&#xA;$$&#xD;&#xA;\int \frac{dx}{g(x)} + C = \ln |y|&#xD;&#xA;$$&#xA;where $C$ is an arbitrary constant.  Then&#xA;$$&#xD;&#xA;\exp(\int \frac{dx}{g(x)} + C) = |y|&#xD;&#xA;$$&#xA;and so&#xA;$$&#xD;&#xA;y = \pm \exp(\int \frac{dx}{g(x)} + C)&#xD;&#xA;$$&#xA;This works as long as $\frac{1}{g(x)}$ has an antiderivative.  For this purpose, it is sufficient to have $g$ continuous and not everywhere zero.&lt;/p&gt;&#xA;" OwnerUserId="6887" LastActivityDate="2011-03-18T06:01:19.677" />
  <row Id="27747" PostTypeId="2" ParentId="21231" CreationDate="2011-03-18T06:06:16.173" Score="3" ViewCount="" Body="&lt;p&gt;f(n) is both time- and space-constructable. $M$ converts the input to binary, performs the necessary computations, and then spins for the appropriate number of steps.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Constructable functions are primarily considered in order to simplify technical issues in proofs. For example, consider the Time Hierarchy Theorem, which essentially says that having more time allows you to solve more stuff.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The formal statement is: Given a time-constructable function $f(n)$ and a function $g(n)=\omega(f(n)\log f(n))$, there exists a language decidable in $TIME(g(n))$ but not in $TIME(f(n))$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The proof is by diagonalization. Construct a machine $M$ which, on input $x$, simulates machine $x$ on input $x$ for $f(n)$ steps. If $x(x)$ halts within that time bound, $M$ outputs the opposite. If $x(x)$ hasn't halted, $M$ rejects (arbitrarily). The simulation can be done using $\log f(n)$ overhead, and the theorem follows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order for the proof to work, $M$ needs to be able to count $f(n)$ steps, and it needs to do so efficiently. Otherwise, the running time of $M$ won't be bounded by $g(n)$. Time-constructability is exactly the criteria needed to allow the proof to work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since basically every function of interest is time-constructable, there isn't much point in removing the technicality. Actually, in the case of the time hierarchy theorem, removing the constructability requirement makes the theorem spectacularly false. The Blum Gap Theorem says that given any function $r(n)$, such as $2^n$, there exists a function $f(n)$ such that $TIME(f(n))=TIME(f(r(n)))$. Obviously, this $f$ is not time-constructable.&lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2011-03-18T06:06:16.173" CommentCount="1" />
  <row Id="27748" PostTypeId="1" CreationDate="2011-03-18T06:18:07.037" Score="2" ViewCount="89" Body="&lt;p&gt;I am reading a section in a book that talks about normal matrices, and I see the following:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;A normal matrix is a matrix that commutes with its adjoint,&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/r9gD2.png&quot; alt=&quot;huh?&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Eh? What do the brackets and the comma inside mean?&lt;/p&gt;&#xA;" OwnerUserId="3143" LastActivityDate="2011-03-18T06:25:34.247" Title="Matrix bracket notation" Tags="&lt;matrices&gt;&lt;notation&gt;" AnswerCount="1" />
  <row Id="27749" PostTypeId="2" ParentId="27748" CreationDate="2011-03-18T06:25:34.247" Score="7" ViewCount="" Body="&lt;p&gt;The notation means that $AA^{T}-A^{T}A$=0. &lt;/p&gt;&#xA;" OwnerUserId="3731" LastActivityDate="2011-03-18T06:25:34.247" CommentCount="1" />
  <row Id="27750" PostTypeId="2" ParentId="27730" CreationDate="2011-03-18T06:41:18.423" Score="1" ViewCount="" Body="&lt;p&gt;I presume you're already aware that just the mere fact of multiplying $\mathbf A$ and $\mathbf B^T$ already exposes you to numerical instability, similar to the reasons why one is better off computing the singular value decomposition of a matrix $\mathbf A$ instead of the eigendecomposition of $\mathbf A^T \mathbf A$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Luckily, you don't have to go through that trouble: there is such a thing called &lt;a href=&quot;http://dx.doi.org/10.1137/S0036142995292633&quot; rel=&quot;nofollow&quot;&gt;&quot;product-induced singular value decomposition&quot;&lt;/a&gt; that allows you to formally decompose the product without having to multiply the two matrices. See the paper and the references for details.&lt;/p&gt;&#xA;" OwnerUserId="8407" LastActivityDate="2011-03-18T06:41:18.423" CommentCount="2" />
  <row Id="27751" PostTypeId="1" AcceptedAnswerId="27762" CreationDate="2011-03-18T06:43:58.627" Score="2" ViewCount="132" Body="&lt;p&gt;I am unable to see how to eliminate $t$. Wolfram Alpha fails at it too. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x=2t-4t^3$$ $$y=t^2-3t^4$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can guess that the curve is a polynomial equation so in principle I can write this as &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$w_1 x^4 +w_2 x^3 +w_3 x^2 +w_4 x +w_5 y^3 +w_6 y^2 +w_7 y +w_8=0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for the powers of $t$ to cancel when plugged in I am determine all these $w&amp;#39;s$ by substituting the $t&amp;#39;s$ for $x(t)$ and $y(t)$ and setting each of the new coefficients of the $t$'s $C_i(w_1,...w_8)$ to zero individually. I have my curve determined if I put each of the $C_i(w_1,...w_8)=0$ and solve for the $w$'s. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;All this was in principle, but this is like an examination question so there must be a clever manipulation/way, which I have been unable to find.&lt;/p&gt;&#xA;" LastActivityDate="2011-03-18T08:46:31.887" Title="Sketch a curve given parametrically by $x=2t-4t^3$ and $t^2-3t^4$" Tags="&lt;plane-curves&gt;&lt;parametric&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="1" />
  <row Id="27752" PostTypeId="2" ParentId="27751" CreationDate="2011-03-18T06:53:45.443" Score="3" ViewCount="" Body="&lt;p&gt;I wouldn't know how to do it in an exam, but using a Gröbner basis routine such as that in &lt;em&gt;Mathematica&lt;/em&gt; easily gives the implicit Cartesian equation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$27x^4-4x^2(36y+1)+16y(4y+1)^2=0$$&lt;/p&gt;&#xA;" OwnerUserId="8407" LastActivityDate="2011-03-18T06:53:45.443" CommentCount="4" />
  <row Id="27753" PostTypeId="1" AcceptedAnswerId="27756" CreationDate="2011-03-18T07:04:53.070" Score="4" ViewCount="236" Body="&lt;p&gt;This is a question I had on a test, we were told not to use brute-force and figure out a smart way to solve the problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have a matrix $A$ &#xA;$\begin{pmatrix}&#xD;&#xA;2 &amp;amp; 3\\ &#xD;&#xA;3 &amp;amp; 2&#xD;&#xA;\end{pmatrix}$. Find a matrix $C$ such that $C^{3}$=$A$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the 'smart not brute-force' way to solve this, without picking numbers, looking for patterns and so on?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;it was in eigenvalues section&quot; in the end?&lt;/p&gt;&#xA;" OwnerUserId="8408" LastEditorUserId="8408" LastEditorDisplayName="" LastEditDate="2011-03-18T07:15:33.887" LastActivityDate="2011-03-29T04:19:41.563" Title="Given a matrix $A$ find a matrix $C$ such that $C^3$=$A$" Tags="&lt;linear-algebra&gt;" AnswerCount="5" FavoriteCount="1" />
  <row Id="27754" PostTypeId="2" ParentId="27753" CreationDate="2011-03-18T07:13:37.177" Score="1" ViewCount="" Body="&lt;p&gt;Eigendecompose $\mathbf A$ (easily done since you have a symmetric matrix), take the cube root of the eigenvalues, and multiply back the matrix of eigenvectors appropriately.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac12\begin{pmatrix}\sqrt[3]{5}-1&amp;amp;\sqrt[3]{5}+1\\\sqrt[3]{5}+1&amp;amp;\sqrt[3]{5}-1\end{pmatrix}$$&lt;/p&gt;&#xA;" OwnerUserId="8407" LastActivityDate="2011-03-18T07:13:37.177" CommentCount="2" />
  <row Id="27755" PostTypeId="2" ParentId="27753" CreationDate="2011-03-18T07:14:31.397" Score="5" ViewCount="" Body="&lt;p&gt;Remember that:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Matrices act on vectors.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here the linearly independent vectors $u=\begin{pmatrix} 1\\ 1\end{pmatrix}$ and $v=\begin{pmatrix} 1\\ -1\end{pmatrix}$ are such that $Au=5u$ and $Av=-v$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence a suitable $C=\begin{pmatrix} a &amp;amp; b\\ c&amp;amp; d\end{pmatrix}$ could be defined by the conditions that $Cu=5^{1/3}u$ and $Cv=-v$. This gives you a linear system of two equations for $(a,b)$ and another linear system of two equations for $(c,d)$, and the matrix $C$ follows.&lt;/p&gt;&#xA;" OwnerUserId="6179" LastEditorUserId="6179" LastEditorDisplayName="" LastEditDate="2011-03-18T07:58:15.387" LastActivityDate="2011-03-18T07:58:15.387" CommentCount="2" />
  <row Id="27756" PostTypeId="2" ParentId="27753" CreationDate="2011-03-18T07:15:21.230" Score="8" ViewCount="" Body="&lt;p&gt;One thing that jumps out at me is that this matrix has two eigenvalues. Find them: det&#xA;$&#xA;(\begin{bmatrix}&#xA;2-\lambda &amp;amp; 3\\ &#xA;3 &amp;amp; 2-\lambda&#xA;\end{bmatrix})$=&#xA;$(2-\lambda)^{2}-9$=$-5-4\lambda+\lambda^{2}$. Factoring this you get $(\lambda+1)(\lambda-5)=0$, so your eigenvalues are -1 and 5. Now, find bases for their respective eigenspaces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basis $\xi_{-1}$ = ker$(A+I)$=ker $(\begin{bmatrix}&#xD;&#xA;3 &amp;amp; 3\\\ &#xD;&#xA;3 &amp;amp; 3&#xD;&#xA;\end{bmatrix})$. Row-reducing, you get ker $(\begin{bmatrix}&#xD;&#xA;1 &amp;amp; 1\\\ &#xD;&#xA;0 &amp;amp; 0&#xD;&#xA;\end{bmatrix})$, which is equal to the span of $\begin{bmatrix}&#xD;&#xA;-1\\ &#xD;&#xA;1&#xD;&#xA;\end{bmatrix}$. So a basis of $\xi_{-1}$ = $\begin{bmatrix}&#xD;&#xA;-1\\ &#xD;&#xA;1&#xD;&#xA;\end{bmatrix}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Same for $\xi_{5}$ -- ker $(\begin{bmatrix}&#xD;&#xA;-3 &amp;amp; 3\\\ &#xD;&#xA;3 &amp;amp; -3&#xD;&#xA;\end{bmatrix})$, row-reducing we get ker $(\begin{bmatrix}&#xD;&#xA;1 &amp;amp; -1\\\ &#xD;&#xA;0 &amp;amp; 0&#xD;&#xA;\end{bmatrix})$, so ker = span $\begin{bmatrix}&#xD;&#xA;1\\ &#xD;&#xA;1&#xD;&#xA;\end{bmatrix}$. So a basis of $\xi_{5}$ = $\begin{bmatrix}&#xD;&#xA;1\\ &#xD;&#xA;1&#xD;&#xA;\end{bmatrix}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now you know that $A$=$C^{3}$=$CCC$. So if $\vec{x}$ is an eigenvector of $A$, it is clearly an eigenvector of $C$, but perhaps with a different eigenvalue. You know that $A$$&#xA;\begin{bmatrix}&#xA;1\\ &#xA;1&#xA;\end{bmatrix}$=$CCC$$&#xA;\begin{bmatrix}&#xA;1\\ &#xA;1&#xA;\end{bmatrix}$=$\begin{bmatrix}&#xA;5\\ &#xA;5&#xA;\end{bmatrix}$. So, $\sqrt[3]{5}$ must be an eigenvalue of $C$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Same goes for $A$$\begin{bmatrix}&#xA;-1\\ &#xA;1&#xA;\end{bmatrix}$=$CCC$$\begin{bmatrix}&#xA;-1\\ &#xA;1&#xA;\end{bmatrix}$=$\begin{bmatrix}&#xA;1\\ &#xA;-1&#xA;\end{bmatrix}$. This would be possible only if -1 was an eigenvalue of $C$, since $(-1)^{3}$=$-1$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So now you can construct a system of equations -- you know that $C$ is of some form $(\begin{bmatrix}&#xD;&#xA;a &amp;amp; b\\\ &#xD;&#xA;c &amp;amp; d&#xD;&#xA;\end{bmatrix})$, and based on the eigenvectors and eigenvalues of $C$ that we just found out, you can devise the following system:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\left\{\begin{matrix}&#xD;&#xA;a &amp;amp; + &amp;amp; b &amp;amp; = &amp;amp; \sqrt[3]{5}\\ &#xD;&#xA;c &amp;amp; + &amp;amp; d &amp;amp; = &amp;amp; \sqrt[3]{5}\\ &#xD;&#xA;-a &amp;amp; + &amp;amp; b &amp;amp; = &amp;amp; 1\\ &#xD;&#xA;-c &amp;amp; + &amp;amp; d &amp;amp; = &amp;amp; -1&#xD;&#xA;\end{matrix}\right.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now solve for the variables -- $b=a+1$, substitute into the other $a, b$ equation to get $2a+1=\sqrt[3]{5}$, so $a=\frac{\sqrt[3]{5}-1}{2}$. $b$, then, is equal to $\frac{\sqrt[3]{5}-1}{2}+1$, as obvious from the third equation. Same for $c$ and $d$ -- $c=d+1$, so substituting, we get $2d+1=\sqrt[3]{5}$, so $d=\frac{\sqrt[3]{5}-1}{2}$, and hence $c=\frac{\sqrt[3]{5}-1}{2}+1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This results into $C$=$&#xA;\begin{bmatrix}&#xA;\frac{\sqrt[3]{5}-1}{2} &amp;amp; \frac{\sqrt[3]{5}-1}{2}+1 \\ &#xA;\frac{\sqrt[3]{5}-1}{2}+1 &amp;amp; \frac{\sqrt[3]{5}-1}{2}&#xA;\end{bmatrix}$.&lt;/p&gt;&#xA;" OwnerUserId="3731" LastActivityDate="2011-03-18T07:15:21.230" CommentCount="4" />
  <row Id="27757" PostTypeId="2" ParentId="27753" CreationDate="2011-03-18T07:16:55.413" Score="5" ViewCount="" Body="&lt;p&gt;This would be easy for a diagonal matrix, because $\begin{bmatrix}a &amp;amp; 0\\ 0 &amp;amp; b\end{bmatrix}^3=\begin{bmatrix}a^3 &amp;amp; 0\\ 0 &amp;amp; b^3\end{bmatrix}$, which means that you could just take the cube root of each diagonal entry to solve the problem.  While $A$ is not diagonal, it is symmetric and therefore diagonalizable.  If you're comfortable with diagonalizing, find $S$ such that $SAS^{-1}=\begin{bmatrix}a &amp;amp; 0 \\ 0 &amp;amp; b \end{bmatrix}$.  You know how to find a matrix whose cube gives the right hand side.  Then notice how conjugation behaves with cubing: $(S^{-1}C&amp;#39;S)^3=S^{-1}C&amp;#39;^3S$.  Therefore, you can take $C=S^{-1}\begin{bmatrix}\sqrt[3]{a} &amp;amp; 0 \\ 0 &amp;amp; \sqrt[3]{b} \end{bmatrix}S$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I was going to write more involving eigenvectors, but then other answers were posted covering this.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An alternative approach using polynomial interpolation will work for all diagonalizable matrices having eigenvalues $-1$ and $5$, and does not require finding eigenvectors.  For more on this and generalizations, see Chapter 1 of Higham's &lt;em&gt;&lt;a href=&quot;http://books.google.com/books?id=S6gpNn1JmbgC&amp;amp;dq&quot; rel=&quot;nofollow&quot;&gt;Functions of matrices&lt;/a&gt;&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case, the &lt;a href=&quot;http://en.wikipedia.org/wiki/Lagrange_polynomial&quot; rel=&quot;nofollow&quot;&gt;Lagrange interpolating polynomial&lt;/a&gt; of the cube root function on the spectrum $\{5,-1\}$ of $A$ is $$p(t)=\sqrt[3]{5}\cdot\frac{t+1}{5+1}+\sqrt[3]{-1}\cdot\frac{t-5}{-1-5}=\frac{\sqrt[3]{5}+1}{6}\cdot t +\frac{\sqrt[3]{5}-5}{6},$$ so that a matrix cube root for $A$ can be obtained as $$p(A)=\frac{\sqrt[3]{5}+1}{6}\cdot A +\frac{\sqrt[3]{5}-5}{6}\cdot I=\frac{1}{2}\begin{bmatrix}\sqrt[3]5-1&amp;amp;\sqrt[3]5+1\\\sqrt[3]5+1&amp;amp;\sqrt[3]5-1\end{bmatrix}.$$&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2011-03-29T02:33:24.073" LastActivityDate="2011-03-29T02:33:24.073" CommentCount="9" />
  <row Id="27758" PostTypeId="1" CreationDate="2011-03-18T07:29:44.600" Score="2" ViewCount="231" Body="&lt;p&gt;&lt;a href=&quot;http://www.scribd.com/doc/50636877/V-I-Arnold-A-mathematical-trivium&quot; rel=&quot;nofollow&quot;&gt;Here&lt;/a&gt; is the source of most of the problems I ask here, a majority of which I am currently unable to solve. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have asked three questions related to it on this site$^{(&lt;a href=&quot;http://math.stackexchange.com/questions/24533/find-the-average-of-sin100-x-in-5-minutes&quot;&gt;1&lt;/a&gt;), (&lt;a href=&quot;http://math.stackexchange.com/questions/27478/critical-values-and-critical-points-of-the-mapping-z-mapsto-z2-barz&quot;&gt;2&lt;/a&gt;), (&lt;a href=&quot;http://math.stackexchange.com/questions/27751/sketch-a-curve-given-parametrically-by-x-2t-4t3-and-t2-3t4&quot;&gt;3&lt;/a&gt;)}$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From responses to one of these questions I get the idea that there might not be a solution in the ordinary sense to some of these problems. For example, for (3) someone posted a helpful answer of how to get mathematica to solve it, and after staring at the solution I realized it was not in my capacity to have arrived at that answer using my method, or probably any method without a computer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I would ask if someone more experienced could give a quick look at the problems and tell me what analytical virtues the author demands and is trying to test, and hence deems essential? So that I can start working on those skill areas so that I know what mathphy minimum expects me to know.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also a reference request at similar stuff would be good. It is aimed at physicists and as someone hoping to be a mathematical physicist, the premise of the question bank provoked me a little. &lt;/p&gt;&#xA;" LastActivityDate="2011-03-18T15:16:07.503" Title="What is the idea of the problems in V. I. Arnold's problem book &quot;A mathematical trivium&quot;?&quot;" Tags="&lt;soft-question&gt;" CommentCount="5" FavoriteCount="1" />
  <row Id="27759" PostTypeId="1" AcceptedAnswerId="28678" CreationDate="2011-03-18T07:34:33.293" Score="2" ViewCount="134" Body="&lt;p&gt;Suppose $N$ is an RSA modulus (ie, it's the product of two distinct primes), 256 bits long.  What is the best method to factor it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Trial division is out of the question, Pollard's Rho is probably out as well (without significant parallelization).  I doubt there are any online tools or math libraries that can handle this number (I think Wolfram Alpha uses Pollard's Rho algorithm).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moduli up to 768 bits have been factored, and RSA Corp's (now defunct) challenge list doesn't even address numbers as small as 256 bits, so it must be pretty easy... but how?&lt;/p&gt;&#xA;" OwnerUserId="7162" LastEditorUserId="7162" LastEditorDisplayName="" LastEditDate="2011-03-18T15:57:07.213" LastActivityDate="2011-03-23T17:01:40.697" Title="What's the best way to factor a 256-bit number?" Tags="&lt;math-software&gt;&lt;factoring&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" />
  <row Id="27760" PostTypeId="2" ParentId="27713" CreationDate="2011-03-18T07:42:38.097" Score="2" ViewCount="" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Yes, if you put an arbitrary metric on each trivialized restriction and then glue them together using a partition of unity, you'll get a metric on the whole thing.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;To go from a map $\Gamma(E) \rightarrow \Gamma(E)\otimes \Gamma(T^*M)$ to a map $\Gamma(E)\otimes \Gamma(TM)\rightarrow \Gamma(E)$, if $s_0\mapsto s_1\otimes x$, then you can put $s \otimes v \mapsto s_1 \cdot x(v)$ -- basically, this all comes from the fact that $Hom_{\mathbb{R}}(V,W)\cong V^*\otimes W$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;It depends which version of $\nabla$ you're using.  From your notation it looks like either you're using the definition $\nabla:\Gamma(E)\rightarrow \Gamma(E)\otimes \Gamma(T^*M)$ or you're using the definition $\nabla:\Gamma(E)\otimes \Gamma(TM) \rightarrow \Gamma(E)$ and suppressing the vector field.  My guess is that for $s_0\otimes x\in \Gamma(E)\otimes \Gamma(T^*M)$ you're extending the metric by $g(s_0\otimes x,s_1)=g(s_0,s_1)\cdot x \in \Gamma(T^*M)$; in this case, yes the fact is true by what you said, which can be taken as the definition of applying a 1-form to a vector.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="2159" LastActivityDate="2011-03-18T07:42:38.097" />
  <row Id="27761" PostTypeId="1" CreationDate="2011-03-18T07:51:05.183" Score="4" ViewCount="65" Body="&lt;p&gt;While studying the use of the trapezoidal rule for numerically evaluating the complementary error function $\mathrm{erfc}(z)$, the following integrals showed up when I was trying to derive expressions for the truncation error:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_0^\pi \exp\left(-z^2\tan^2\frac{u}{2}\right)\cos(2mu) \mathrm du$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $z$ is positive and $m$ is a positive integer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Evaluating a bunch of these integrals in &lt;em&gt;Mathematica&lt;/em&gt;, I gather that these integrals follow the pattern&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\pi z^2\exp(z^2)\mathrm{erfc}(z)R_n(z)-2\sqrt{\pi}z S_n(z)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $R_n(z)$ and $S_n(z)$ are polynomials.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there any closed forms for these two polynomials?&lt;/p&gt;&#xA;" OwnerUserId="8407" LastActivityDate="2011-05-19T21:59:20.037" Title="Closed form for some integrals related to the complementary error function" Tags="&lt;special-functions&gt;" AnswerCount="1" />
  <row Id="27762" PostTypeId="2" ParentId="27751" CreationDate="2011-03-18T07:53:08.023" Score="7" ViewCount="" Body="&lt;p&gt;My opinion is that you should not try to eliminate $t$ at all. Instead, you should think about how $x$ and $y$ behave as $t$ varies, find some important points such as the critical points with respect to $x$ and $y$, and try to sketch the graph from that information.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x=2t-4t^3$ means that as $t$ goes from $-\infty$ to $\infty$, $x$ decreases from $\infty$ to a local minimum of $-\sqrt{8/27}$ at $t = -1/\sqrt{6}$, then rises to $\sqrt{8/27}$ at $t = 1/\sqrt{6}$, then falls again to $-\infty$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$y=t^2-3t^4$ means that $y$ starts and ends at $-\infty$, with local maxima of $1/12$ when $t = \pm 1/\sqrt{6}$, and a local minimum of $0$ at $t = 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Interestingly, $t = \pm 1/\sqrt{6}$ is a critical point for both $x$ and $y$, both of whose derivatives change sign, so the curve forms a cusp at that point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's enough information to follow the curve as $t$ goes from $-\infty$ to $\infty$: it comes in from the bottom right quadrant $(\infty,-\infty)$, goes upward and leftward till it hits the cusp $(-\sqrt{8/27},1/12)$, upon which it turns around, passing smoothly through $(0,0)$ where its tangent is horizontal, until it reaches the second cusp $(\sqrt{8/27},1/12)$, and then turns around a second time and exits at the bottom left towards $(-\infty,-\infty)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is consistent with &lt;a href=&quot;http://www.wolframalpha.com/input/?i=plot+27x%5E4-4x%5E2%2836y+1%29+16y%284y+1%29%5E2=0&quot; rel=&quot;nofollow&quot;&gt;the plot arising from Juan Joder's answer&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="856" LastEditorUserId="856" LastEditorDisplayName="" LastEditDate="2011-03-18T07:59:10.573" LastActivityDate="2011-03-18T07:59:10.573" />
  <row Id="27763" PostTypeId="2" ParentId="27759" CreationDate="2011-03-18T07:58:35.613" Score="1" ViewCount="" Body="&lt;p&gt;Charles's nice answer to &lt;a href=&quot;http://math.stackexchange.com/questions/13387&quot;&gt;this question&lt;/a&gt; might be of interest. Briefly: look into ECM, or GNFS if ECM chokes.&lt;/p&gt;&#xA;" OwnerUserId="8407" LastActivityDate="2011-03-18T07:58:35.613" />
  <row Id="27764" PostTypeId="1" CreationDate="2011-03-18T08:08:58.797" Score="-1" ViewCount="142" Body="&lt;p&gt;Greetings All&#xA;I have a function that creates an affine matrix that works for &#xA;2D rotation around an arbitrary point but now I would like&#xA;it to work for 3D rotations also.  The working function example is below&#xA;with it's example function:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;%affine matrix rotation about a point rtaffine[theta,rpx,rpy,sigx,sigy,sigz]&#xA;%crx cry rpx rpy represent center of rotation&#xA;function [rotsig,theta,crpx,crpy,sigx,sigy,sigz] = rtaffine(theta,rpx,rpy,sigx,sigy,sigz) %format in rrtaffine[theta,rpx,rpy,sigx,sigy,sigz]&#xA;    rotsig=[];&#xA;    %affinematrix=[];&#xA;    siga=[sigx;sigy;sigz];&#xA;&#xA;    r00 = cosd(theta); r01 = -sind(theta); r10 = sind(theta); r11 = cosd(theta);&#xA;    affinematrix = [r00, r01, rpx(1,1) - r00*rpx(1,1) - r01*rpy(1,1);...&#xA;    r10, r11, rpy(1,1) - r10*rpx(1,1) - r11*rpy(1,1);...&#xA;    0, 0, 1];   &#xA;    rotsig=affinematrix*siga; %new affine matrix&#xA;endfunction&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Example code to use with function below:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;clear all, clc,clf,tic&#xA;t= linspace(0,2*pi,1000); %from 0 to 1 with 100 equal spaces between&#xA;freq=1;&#xA;y=sin(freq*(t)); %pi will change phase of angle (t+pi)&#xA;y2=sin(freq*(t)); %pi will change phase of angle (t+pi)&#xA;z = ones(size(y));&#xA;siga=[t;y;z];&#xA;siga2=[t;y2;z];&#xA;rpcalc=[t(1,size(t,2)/2),0]; %size(b,2) %will count all the rows with a 1, count all colmns with 2&#xA;rp=[t(1,size(t,2)/2),0]; %rotation point half way on x/t axis&#xA;rotsigt=[];&#xA;for theta=0:1:360&#xA;    rotsig=rtaffine(theta,rp(1,1),rp(1,2),siga(1,:),siga(2,:),siga(3,:));&#xA;    rotsig2=rtaffine((theta*2),rp(1,1),rp(1,2),siga(1,:),siga(2,:),siga(3,:)); &#xA;    rotsigcomb=rotsig+rotsig2;&#xA;    axis([-10 15 -10 10])&#xA;    plot(0,0,'g*',rp(1,1),rp(1,2),'r*',rotsig(1,:),rotsig(2,:),'b-',rotsig2(1,:),rotsig2(2,:),'r',rotsigcomb(1,:),rotsigcomb(2,:),'g')&#xA;    %plot(0,0,'g*',rp(1,1),rp(1,2),'r*',rotsigcomb(1,:),rotsigcomb(2,:),rotsig2(1,:),rotsig2(2,:))&#xA;    axis([-10 15 -10 10])&#xA;    title(['Phase Shift is ',num2str(theta),' deg']);&#xA;    grid on&#xA;    pause(.001)&#xA;    %rotsigt=[rotsigt(:,:),rotsig(:,:)];&#xA;end&#xA;fprintf('- Complete re-import test in %4.4fsec or %4.4fmins\n',toc,toc/60);&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Any suggestions?&lt;/p&gt;&#xA;" OwnerUserId="5957" LastEditorUserId="5957" LastEditorDisplayName="" LastEditDate="2011-03-21T21:45:27.337" LastActivityDate="2011-05-25T01:01:13.223" Title="help converting 2D rotating affine matrix to 3D rotating (matlab / octave code) " Tags="&lt;matrices&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="27765" PostTypeId="2" ParentId="136" CreationDate="2011-03-18T08:37:41.337" Score="0" ViewCount="" Body="&lt;p&gt;to extend this idea, there is something i came up with in high-school (now i suck at even elementary math)...&#xA;I'm sure someone had come up with this much earlier ..&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;the Nth difference between Nth consecutive powers is equal to N!&quot;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;for N=2&#xA;1  4  9  16  &amp;lt;- Nth powers&#xA;  3  5  7     &amp;lt;- 1st Difference&#xA;   2  2       &amp;lt;- Nth Difference&#xA;&#xA;for N=3&#xA;&#xA;1  8  27  64  125&#xA; 7  19  37  61&#xA;  12  18  24&#xA;     6   6&#xA;&#xA;for N=4&#xA;&#xA;1  16  81  256  625  1296  2401&#xA; 15  65  175  369  671&#xA;   50  110  194  302&#xA;     60   84   108&#xA;       24    24&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;this can also be used to calculate the next number in the series&#xA;eg: 7^4 = 24+108+302+671+1296 = 2401&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i wish i could remember the proof too ... Yikes&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so for N=2 difference is 2 hence leading to a sequence of odd numbers &lt;/p&gt;&#xA;" OwnerUserId="8410" LastEditorUserId="8410" LastEditorDisplayName="" LastEditDate="2011-03-18T08:53:26.810" LastActivityDate="2011-03-18T08:53:26.810" CommentCount="1" />
  <row Id="27767" PostTypeId="2" ParentId="27751" CreationDate="2011-03-18T08:46:31.887" Score="2" ViewCount="" Body="&lt;p&gt;Natural points for sketching include: when $x=0$, giving $t = 0 \text{ or } \pm 1/\sqrt{2}$ and so $y = 0 \text{ or } -1/4$; and when $y=0$, giving $t = 0 \text{ or } \pm 1/\sqrt{3}$ and so $x = 0 \text{ or } \pm 2/\sqrt{27}$.  So these four points suggest something interesting is going on near the origin.     &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just drawing the curve parametrically using Excel gives something like &lt;img src=&quot;http://i.imgur.com/t9TCB.png&quot; alt=&quot;large scale curve&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but zooming in gives the more interesting &lt;img src=&quot;http://i.imgur.com/vFz3I.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;largely matching Rahul Narain's answer and &lt;a href=&quot;http://www.wolframalpha.com/input/?i=27+x%5E4-144+x%5E2+y-4+x%5E2+256+y%5E3+128+y%5E2+16+y+=0&quot; rel=&quot;nofollow&quot;&gt;Wolfram Alpha&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="6460" LastActivityDate="2011-03-18T08:46:31.887" />
  <row Id="27768" PostTypeId="1" CreationDate="2011-03-18T08:54:53.113" Score="2" ViewCount="81" Body="&lt;p&gt;Hello,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can one show the following identity &#xA;$ne^n = \sum_{k=0}^{\infty}n^k(n - k)^2/k!$ ?&lt;/p&gt;&#xA;" OwnerUserId="8412" LastActivityDate="2011-03-18T11:32:56.297" Title="Exponential identity" Tags="&lt;sequences-and-series&gt;" AnswerCount="2" />
  <row Id="27769" PostTypeId="2" ParentId="27768" CreationDate="2011-03-18T09:02:05.533" Score="2" ViewCount="" Body="&lt;p&gt;By developing the square as $(n-k)^2=n^2-(2n-1)k+k(k-1)$ and evaluating each of the three resulting sums over $k$.&lt;/p&gt;&#xA;" OwnerUserId="6179" LastActivityDate="2011-03-18T09:02:05.533" CommentCount="1" />
  <row Id="27770" PostTypeId="1" CreationDate="2011-03-18T09:06:30.303" Score="0" ViewCount="72" Body="&lt;p&gt;I want to show, Image of  $unit~ disc$ under a discontinuous linear functional from a Normed Linear Space is $\mathbb{C}$ . I know the Image is Unbounded&lt;/p&gt;&#xA;" OwnerUserId="8413" LastActivityDate="2011-03-18T12:20:19.223" Title="Image Of a Discontinuous linear functional" Tags="&lt;functional-analysis&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="27771" PostTypeId="2" ParentId="11780" CreationDate="2011-03-18T09:13:14.370" Score="1" ViewCount="" Body="&lt;p&gt;Some books that might help:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://books.google.co.in/books?id=FhYGYJtMwcUC&amp;amp;printsec=frontcover&amp;amp;dq=acourse+in+borel+sets&amp;amp;hl=en&amp;amp;ei=4iCDTZDqIozrrQeQ79S-CA&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=1&amp;amp;ved=0CDIQ6AEwAA#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;A Course in Borel Sets&lt;/a&gt; by S.M. Srivastava&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://books.google.co.in/books?id=KlmqW2uuGOkC&amp;amp;printsec=frontcover&amp;amp;source=gbs_atb#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;Selectors&lt;/a&gt; by J.E. Jayne and C.A. Rogers&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, as a suggestion, if you tag your questions better (this one with say 'topology' or 'set theory'), it'll help you get better, quicker answers from the experts here.&lt;/p&gt;&#xA;" OwnerUserId="7835" LastActivityDate="2011-03-18T09:13:14.370" />
  <row Id="27772" PostTypeId="1" AcceptedAnswerId="27793" CreationDate="2011-03-18T09:43:43.687" Score="0" ViewCount="72" Body="&lt;p&gt;Is this true that: &quot;The order of wreath product of $\mathbf{Z}_{p^n}$ and $\mathbf{Z}_{p^m}$ is $p^{m+n}$&quot; ?&lt;/p&gt;&#xA;" OwnerUserId="8581" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-18T13:19:54.743" LastActivityDate="2011-03-18T13:29:39.367" Title="Wreath Product of $\mathbf{Z}_{p^n}$ and $\mathbf{Z}_{p^m}$" Tags="&lt;abstract-algebra&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="27773" PostTypeId="2" ParentId="27772" CreationDate="2011-03-18T09:59:10.017" Score="2" ViewCount="" Body="&lt;p&gt;No. I assume you mean $Z_{p^m}$ as a permutation group on $p^m$ elements.&#xA;In this case, the wreath product has order $(p^n)^{p^m}p^m$.&#xA;In general, if $H$ is a permutation group on $n$ elements and $G$ is some group then the wreath product of $G$ and $H$ has order $|G|^n|H|$.&lt;/p&gt;&#xA;" OwnerUserId="2538" LastActivityDate="2011-03-18T09:59:10.017" />
  <row Id="27774" PostTypeId="1" AcceptedAnswerId="27786" CreationDate="2011-03-18T10:03:01.323" Score="0" ViewCount="79" Body="&lt;p&gt;Dear everyone,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd like to understand a step in a counterexample of Reichardt and Lind to the Hasse principle. The example is given by the equation $2y^2=x^4-17z^4$ (1).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) has no rational solutions ($\neq (0,0,0)$), but it has non-zero solutions in $\mathbb{Q}_{p}$ for all $p$. The part that I don't get is the justification that the equation has solutions in $\mathbb{Q}_{p}$ for all $p$. First one can find explicit solutions for $p=2,17,\infty$. For the other values of $p$ the idea is to first show the existence of non-zero solutions over $\mathbb{F}_{p}$, and then use Hensel's lemma to lift these solutions. To show that (1) has solutions over $\mathbb{F}_{p}$ what I've seen people doing is using Hasse's bound on number of points over $\mathbb{F}_p$ of an elliptic curve. (see &lt;a href=&quot;http://www.uni-math.gwdg.de/aufzeichnungen/SummerSchool/SummerSchool_20060717-1430_Kresch/avi/SummerSchool_20060717-1430_Kresch_xvid.avi&quot; rel=&quot;nofollow&quot;&gt;http://www.uni-math.gwdg.de/aufzeichnungen/SummerSchool/SummerSchool_20060717-1430_Kresch/avi/SummerSchool_20060717-1430_Kresch_xvid.avi&lt;/a&gt; between minutes 15 and 30) or &lt;a href=&quot;http://www.warwick.ac.uk/~maseap/arith/notes/elementary.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.warwick.ac.uk/~maseap/arith/notes/elementary.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The key point is the following observation:&#xA;&quot;The curve defined by the equation (1) has genus 1&quot;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;and thats precisely what I don't get, so my question is: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; Why is that a curve? and moreover why does it have genus 1?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you. &lt;/p&gt;&#xA;" OwnerUserId="8317" LastEditorUserId="8317" LastEditorDisplayName="" LastEditDate="2011-03-18T10:16:33.777" LastActivityDate="2011-03-18T12:28:49.177" Title="Understanding a counterexample to Hasse principle." Tags="&lt;number-theory&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="27775" PostTypeId="1" AcceptedAnswerId="27782" CreationDate="2011-03-18T10:45:51.623" Score="14" ViewCount="389" Body="&lt;p&gt;The function $f(x) = sin(x)/x$ is Riemann Integrable from $0$ to $\infty$, but it is not Lebesgue Integrable on that same interval.  (Note, it is not absolutely Riemann Integrable.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is it we restrict our definition of Lebesgue Integrability to absolutely integrable?  Wouldn't it be better to extend our definition to include ALL cases where Riemann Integrability holds, and use the current definition as a corollary for when the improper integral is absolutely integrable?&lt;/p&gt;&#xA;" OwnerUserId="7051" LastActivityDate="2011-03-25T01:40:15.897" Title="Why do we restrict the definition of Lebesgue Integrability?" Tags="&lt;measure-theory&gt;&lt;improper-integrals&gt;" AnswerCount="4" FavoriteCount="2" />
  <row Id="27776" PostTypeId="1" AcceptedAnswerId="27785" CreationDate="2011-03-18T10:50:42.820" Score="0" ViewCount="36" Body="&lt;p&gt;Hey everyone,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm currently working on some graph theory, and I'm checking the robustness of graphs, by removing random vertices from a graph and then check how many vertices are outside the giant component. Then I'll do that for a lot of vertices, and plot them using ListPlot.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This takes along time to calculate, is there a way I can visualise it, that it plots something everytime a value is calculated.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;thisFunctionTakesVeryLongAndReturnsAList := ....&#xA;ListPlot[thisFunctionTakesVeryLongAndReturnsAList [graph]]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;thisFunctionTakesVeryLongAndReturnsAList  calculates a value that can be plotted all the time&lt;/p&gt;&#xA;" OwnerUserId="3605" LastActivityDate="2011-03-18T12:04:16.697" Title="Mathematica asycn rendering of ListPlot" Tags="&lt;mathematica&gt;" AnswerCount="1" />
  <row Id="27777" PostTypeId="2" ParentId="22320" CreationDate="2011-03-18T10:57:00.907" Score="1" ViewCount="" Body="&lt;p&gt;First, assume not all digits are the same (since then this problem is trivial).&#xA;Now note what happens if you exahcnge the two least significant digits. You either add or subtract a certain number a 2-digit one). Now, if you have a divisor in both the number and the one obtained by exchanging these digits, then this is also a divisor in this 2-digit number, so you have severely reduced the problem. You can do this for any two distinct digits in the number, thus getting a set of -digit numbers, the gcd of the permutations of the original number will divide the gcd of these numbers. (note that all these 2-digit number will be divisible by 9, so you need to check if the original number is divisible by 9, but this property is not changed by permuting the number anyway).&#xA;Now the problem has been reduced to checking among some 2-digit numbers, but we might not be done yet, since the gcd of the permutations of the original numbers might be even smaller.&#xA;I will think a bit about how one can easily make sure to get the actual gcd.&lt;/p&gt;&#xA;" OwnerUserId="2538" LastActivityDate="2011-03-18T10:57:00.907" />
  <row Id="27778" PostTypeId="1" AcceptedAnswerId="27783" CreationDate="2011-03-18T11:00:13.763" Score="0" ViewCount="79" Body="&lt;p&gt;I have two linear, skew, 3D lines, and I was wondering how I could find a point on each of the lines whereby the distance between the two points are a particular distance apart, and that the vector formed between these points is perpendicular to both original lines?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not after the points where the lines are the closest distance apart, nor the furthest distance apart! I'm after the points on the lines where the two lines are a particular distance apart and the vector between them is perpendicular. I'd like to also be able to change this distance and find the new points.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please note that this is a different question to the one &lt;a href=&quot;http://math.stackexchange.com/questions/27559/finding-points-on-two-linear-lines-which-are-a-particular-distance-apart&quot;&gt;I posted here&lt;/a&gt;. I wanted to rewrite the question with the necessary changes because I got some really good answers to the last question I asked.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks very much in advance for any help! :)&lt;/p&gt;&#xA;" OwnerUserId="8372" LastActivityDate="2011-03-18T11:30:08.213" Title="Finding a perpendicular line that connects two skew 3D lines at a particular distance?" Tags="&lt;linear-algebra&gt;&lt;vector-calculus&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="27779" PostTypeId="1" AcceptedAnswerId="27790" CreationDate="2011-03-18T11:02:40.083" Score="0" ViewCount="66" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm trying to get a value in a web application, using some informations. I succeeded on creating this system (I need $x$):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$&#xD;&#xA;&#xD;&#xA;\begin{equation}&#xD;&#xA;\left\{&#xD;&#xA;\begin{matrix} &#xD;&#xA;x &amp;amp; = &amp;amp; \dot a + b - c - (\dot d \cdot \dot e)\\&#xD;&#xA;b &amp;amp; = &amp;amp; f \cdot \dot g\\&#xD;&#xA;f &amp;amp; = &amp;amp; \frac{x - \dot d - h}{1 + \dot i}\\&#xD;&#xA;c &amp;amp; = &amp;amp; \dot \alpha(b)\\&#xD;&#xA;h &amp;amp; = &amp;amp; x \cdot \dot e&#xD;&#xA;\end{matrix}&#xD;&#xA;\right.&#xD;&#xA;\end{equation}&#xD;&#xA;&#xD;&#xA;$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But it seems I can't solve it. Please, note that I know the value of the dotted variables (like $\dot p$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My fear is that I can't solve this system. If so, is there a way to approximate the value of $x$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any help would be very appreciated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Edit&lt;/em&gt;: I added two equations. I don't know it this can help, anyway I added them. Note that I know the value of $\alpha()$, but I did not put that here because it does not contain any of the variables of the system. It's there as a placeholder for an if statement.&lt;/p&gt;&#xA;" OwnerUserId="8416" LastEditorUserId="8416" LastEditorDisplayName="" LastEditDate="2011-03-18T13:20:44.663" LastActivityDate="2011-03-18T13:20:44.663" Title="Solving a complicated equation system" Tags="&lt;linear-algebra&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="27780" PostTypeId="2" ParentId="27683" CreationDate="2011-03-18T11:05:28.140" Score="2" ViewCount="" Body="&lt;p&gt;To help figure out a Mathematica expression, following is useful&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;GraphData[8]; Union[GraphData[#, &quot;VertexCount&quot;] &amp;amp; /@ %%] // Hold // TreeForm&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Also, try &lt;code&gt;FullForm&lt;/code&gt; instead of &lt;code&gt;TreeForm&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since &lt;code&gt;GraphData&lt;/code&gt; has property &lt;code&gt;SpanningTreeCount&lt;/code&gt;, you could do&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;graphs = GraphData[{&quot;Cubic&quot;, &quot;Connected&quot;}, 12];&#xA;counts = GraphData[#, &quot;SpanningTreeCount&quot;] &amp;amp; /@ graphs;&#xA;Column@Thread@{counts, graphs}&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Alternatively, you could get number of spanning trees from Laplacian matrix of the graph using &lt;a href=&quot;http://en.wikipedia.org/wiki/Kirchhoff%27s_theorem&quot; rel=&quot;nofollow&quot;&gt;Kirchoff's matrix tree theorem&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;kirk[L_] := Det@Drop[L, {1}, {1}];&#xA;counts = kirk[GraphData[#, &quot;LaplacianMatrix&quot;]] &amp;amp; /@ graphs;&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The &lt;code&gt;Drop&lt;/code&gt; command drops 1st row and 1st column of the matrix, but you could drop any other row/column.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To display 12 vertex cubic graphs nicely, with graph name in tooltip and sorted by number of spanning trees, you could do something like this&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;count[name_] := GraphData[name, &quot;SpanningTreeCount&quot;];&#xA;showCount[name_] := Show[GraphData@name, PlotLabel -&amp;gt; count@name];&#xA;showLabeledCount[name_] := Tooltip[showCount@name, name];&#xA;graphs = GraphData[&quot;Cubic&quot;, 12];&#xA;pics = showLabeledCount /@ Sort[graphs, count[#1] &amp;lt; count[#2] &amp;amp;];&#xA;pics = ArrayPad[pics, {0, 2}, Graphics[]];&#xA;GraphicsGrid@Partition[pics, 10]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://yaroslavvb.com/upload/save/spanning-trees.png&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To get all cubic, connected graphs with &lt;strong&gt;up to&lt;/strong&gt; 12 vertices (disconnected graphs are not interesting since they have 0 spanning trees), do &lt;code&gt;GraphData[{&quot;Cubic&quot;, &quot;Connected&quot;}, ;; 12]&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are doing graph theory with Mathematica, you could find my graph utilities &lt;a href=&quot;http://yaroslavvb.com/blog/Bulatov-sdp.zip&quot; rel=&quot;nofollow&quot;&gt;packages&lt;/a&gt; helpful, documented &lt;a href=&quot;http://mathematica-bits.blogspot.com/2011/01/listing-all-hamiltonian-cycles-of-graph.html&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;, &lt;a href=&quot;http://mathematica-bits.blogspot.com/2011/02/chordal-graph-usage.html&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;, and &lt;a href=&quot;http://mathematica-bits.blogspot.com/2011/01/tree-decomposition-package.html&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;. For instance, with &lt;code&gt;showGraphs&lt;/code&gt; package, you could display all Cubic graphs over 12 vertices that are not connected with &lt;code&gt;showGraphs[12,&quot;Cubic&quot;,!&quot;Connected&quot;]&lt;/code&gt;&lt;/p&gt;&#xA;" OwnerUserId="998" LastEditorUserId="9909" LastEditorDisplayName="" LastEditDate="2011-05-08T13:02:59.950" LastActivityDate="2011-05-08T13:02:59.950" CommentCount="1" />
  <row Id="27781" PostTypeId="2" ParentId="27768" CreationDate="2011-03-18T11:11:29.523" Score="4" ViewCount="" Body="&lt;p&gt;Expanding on Didier's answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is convenient to replace $n$ by $\lambda$. Then, you want to show that&#xA;$$&#xD;&#xA;\sum\limits_{k = 0}^\infty  {\frac{{\lambda ^k (\lambda  - k)^2 }}{{k!}}}  = \lambda e^\lambda . &#xD;&#xA;$$&#xA;Starting as Didier suggested, write the left-hand side as&#xA;$$&#xD;&#xA;\sum\limits_{k = 0}^\infty  {\frac{{\lambda ^k (\lambda  - k)^2 }}{{k!}}}  = \lambda ^2 e^\lambda  \sum\limits_{k = 0}^\infty  {e^{ - \lambda } \frac{{\lambda ^k }}{{k!}}}  - 2\lambda e^\lambda  \sum\limits_{k = 0}^\infty  {ke^{ - \lambda } \frac{{\lambda ^k }}{{k!}}}  + e^\lambda  \sum\limits_{k = 0}^\infty  {k^2 e^{ - \lambda } \frac{{\lambda ^k }}{{k!}}}.&#xD;&#xA;$$&#xA;Now, if $X$ is a Poisson random variable with mean $\lambda$, then&#xA;$$&#xD;&#xA;{\rm E}(X) = \sum\limits_{k = 0}^\infty  {ke^{ - \lambda } \frac{{\lambda ^k }}{{k!}}}  &#xD;&#xA;$$&#xA;and&#xA;$$&#xD;&#xA;{\rm E}(X^2) = \sum\limits_{k = 0}^\infty  {k^2 e^{ - \lambda } \frac{{\lambda ^k }}{{k!}}} .&#xD;&#xA;$$&#xA;From the facts that ${\rm E}(X) = \lambda$ and ${\rm E}(X^2) = {\rm Var}(X) + {\rm E}^2 (X) = \lambda + \lambda^2$ (and noting that $\sum\nolimits_{k = 0}^\infty  {e^{ - \lambda } \frac{{\lambda ^k }}{{k!}}} = 1$), it follows that&#xA;$$&#xD;&#xA;\sum\limits_{k = 0}^\infty  {\frac{{\lambda ^k (\lambda  - k)^2 }}{{k!}}} = e^\lambda  (\lambda ^2  - 2\lambda ^2  + \lambda  + \lambda ^2 ) = \lambda e^\lambda .&#xD;&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="2810" LastEditorUserId="2810" LastEditorDisplayName="" LastEditDate="2011-03-18T11:32:56.297" LastActivityDate="2011-03-18T11:32:56.297" />
  <row Id="27782" PostTypeId="2" ParentId="27775" CreationDate="2011-03-18T11:21:38.263" Score="11" ViewCount="" Body="&lt;p&gt;You could just as well ask the opposite question: why do we define Riemann integration in such a way that an integral can be convergent without being absolutely convergent? The definition of each type of integral &quot;is what it is,&quot; and the way the Lebesgue definition is defined there is no need for improper integrals like in Riemann integration. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We could simulate an improper integral with Lebesgue integration by taking a limit of Lebesgue integrals over bounded regions.  But that's not something that's usually of interest in the Lebesgue theory. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The things that are of interest are convergence theorems like the dominated convergence theorem. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;em&gt;&lt;strong&gt;Dominated convergence theorem:&lt;/em&gt;&lt;/strong&gt; If $(f_n)$ is sequence of measurable functions, $|f_n|&amp;lt;|g|$ for $n \in \mathbb{N}$, $\int |g| &amp;lt; \infty$, and $f_n\to f$ pointwise then $\int f &amp;lt; \infty$ and $\int f_n \to \int f$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In that theorem, the dominating function $g$ needs to be absolutely integrable. Your example, in fact, can be modified to give a counterexample to this statement:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;em&gt;&lt;strong&gt;False:&lt;/em&gt;&lt;/strong&gt; if $(f_n)$ is sequence of measurable functions, $|f_n|&amp;lt;|g|$ for $n \in \mathbb{N}$, $\int g &amp;lt; \infty$ when the integral is computed in the improper sense, and $f_n\to f$ pointwise then $\int f &amp;lt; \infty$ (again in the improper sense) and $\int f_n \to \int f$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The actual theorem requires $\int |g|$ to be finite. Since that is the sort of condition that we work with most of the time, we use the word &quot;integrable&quot; for it to save space. We can still recapture improper integrals if we have to, but they're rarely of interest in the context of Lebesgue integration, so we don't want to spend a good word like &quot;integrable&quot; on them.  &lt;/p&gt;&#xA;" OwnerUserId="630" LastEditorUserId="630" LastEditorDisplayName="" LastEditDate="2011-03-20T21:48:12.753" LastActivityDate="2011-03-20T21:48:12.753" CommentCount="2" />
  <row Id="27783" PostTypeId="2" ParentId="27778" CreationDate="2011-03-18T11:30:08.213" Score="3" ViewCount="" Body="&lt;p&gt;You can't do this. If the lines are skew, there is only one line between them that is perpendicular to them both. (And if the lines are parallel, there are many such lines, but they all have the same length.)&lt;/p&gt;&#xA;" OwnerUserId="1508" LastActivityDate="2011-03-18T11:30:08.213" CommentCount="1" />
  <row Id="27784" PostTypeId="1" AcceptedAnswerId="27801" CreationDate="2011-03-18T11:51:59.803" Score="1" ViewCount="72" Body="&lt;p&gt;An exercise says: if the real sequence $(x_n)$ is defined by $x_{n+1}=F(x_n)$ for some $F$, $x_n\to x$ and $F&amp;#39;(x)=0$, prove that $x_{n+2}-x_{n+1}=o(x_{n+1}-x_n)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The textbook says assume $F$ is continuously differentiable and apply the mean-value theorem. It's straightforward: $$\frac{x_{n+2}-x_{n+1}}{x_{n+1}-x_n}=\frac{F(x_{n+1})-F(x_n)}{x_{n+1}-x_n}=F&amp;#39;(y_n)$$ for $x_n&amp;lt;y_n&amp;lt;x_{n+1}$ or $x_{n+1}&amp;lt;y_n&amp;lt;x_n$, so the continuous derivative at $x$ taking on the value $0$ ensures the limit of this quotient is $0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can it be proved without assuming the continuity of the derivative? It holds (by differentiability and continuity of $F$ at $x$) that $$0\gets \frac{F(x_n)-F(x)}{x_n-x} = \frac{x_{n+1}-x}{x_n-x}$$ Can the limit of $$\frac{x_{n+2}-x_{n+1}}{x_{n+1}-x_n}$$ be obtained from here?&lt;/p&gt;&#xA;" OwnerUserId="1313" LastEditorUserId="1508" LastEditorDisplayName="" LastEditDate="2011-03-18T13:25:37.537" LastActivityDate="2011-03-18T13:46:59.250" Title="Proving that $x_{n+2}-x_{n+1}=o(x_{n+1}-x_n)$ in a certain case." Tags="&lt;calculus&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="27785" PostTypeId="2" ParentId="27776" CreationDate="2011-03-18T12:04:16.697" Score="1" ViewCount="" Body="&lt;p&gt;Timo,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have &lt;em&gt;Mathematica&lt;/em&gt; 6 or later, you can use &lt;code&gt;Dynamic&lt;/code&gt; for this. You should rewrite your &lt;code&gt;thisFunctionTakesVeryLongAndReturnsAList&lt;/code&gt; like this:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;list = {};&#xA;calculateOnePointVeryLong[x_] := (Pause[1]; result = {x, x^2}; &#xA;   AppendTo[list, result]; result);&#xA;Graphics[Point[Dynamic[list]], Axes -&amp;gt; True, &#xA; AspectRatio -&amp;gt; 1/GoldenRatio]&#xA;makeList = Table[calculateOnePointVeryLong[x], {x, 0, 10}];&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="7617" LastActivityDate="2011-03-18T12:04:16.697" CommentCount="1" />
  <row Id="27786" PostTypeId="2" ParentId="27774" CreationDate="2011-03-18T12:15:17.473" Score="3" ViewCount="" Body="&lt;p&gt;I don't really know what you mean when you ask &quot;Why is that a curve?&quot;. The question can be taken to mean either &quot;I don't know what an algebraic curve is&quot;, or &quot;Since there are three unknowns and two equations, it looks more like a surface to me&quot;. I am going to assume the latter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So recall that a curve is most often defined either by a set of homogeneous equations in $n+1$ variables, in which case their solutions form a well defined subvariety of the projective space $\mathbf P^n$, or a set of arbitrary equations in $n$ variables, in which case their solutions are taken inside of affine space $\mathbf A^n$. We can go from one to the other by setting one of the variables equal to one and homogenizing respectively (but note that this process is not bijective).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So note first that our equation is not homogeneous. Based on the previous paragraph it would make most sense to interpret it as a subvariety of $\mathbf A^3$, but this is clearly not what they mean since that would give us an algebraic surface and not an algebraic curve. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are two ways to make sense of this:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Set $z=1$, for example. Then we get the equation $2y^2 = x^4 - 17$ in two variables, &lt;em&gt;and whenever we have a solution to this equation we also have a solution to the original one&lt;/em&gt;. This equation defines a plane curve in $\mathbf A^2$. By embedding $\mathbf A^2$ in a projective space and taking the closure we get a projective curve, and by desingularizing it we also get a smooth projective curve.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Give the variable $y$ degree $2$, so that the equation is homogeneous with respect to a shifted degree. Then we can talk about the set of solutions inside of a &lt;em&gt;weighted projective space&lt;/em&gt;. (Over $\mathbf C$ this can be defined as $\mathbf C^3 / \mathbf C^\ast$ where $\mathbf C^\ast$ acts by $\lambda \cdot (x_0,x_1,x_2) = (\lambda^2 x_0, \lambda x_1, \lambda x_2)$.) &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Your second question - how do we see that this curve has genus one - can be determined in several ways. I am not sure what is the simplest way but here is one, using model (1) above and calculations in affine coordinates. Your curve has a degree 2 map to $\mathbf A^1$ given by projection to the $x$-coordinate, and this map will have branch points wherever $x^4-17 = 0$. The map can be extended to a map from the smooth projective version of the curve to $\mathbf P^1$ (this is a general fact about smooth curves, but you can also prove this by computing in explicit coordinates, see the Wikipedia article on hyperelliptic curves). Using the Riemann-Hurwitz formula one sees that:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;a map of degree two must have an even number of branch points, so the point $x = \infty$ can not be a branch point;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;a curve which has a degree two map to $\mathbf P^1$ with four branch points has genus one.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Oh, and please tell me which parts of this need elaborating.&lt;/p&gt;&#xA;" OwnerUserId="677" LastEditorUserId="677" LastEditorDisplayName="" LastEditDate="2011-03-18T12:28:49.177" LastActivityDate="2011-03-18T12:28:49.177" CommentCount="8" />
  <row Id="27787" PostTypeId="2" ParentId="27770" CreationDate="2011-03-18T12:20:19.223" Score="2" ViewCount="" Body="&lt;p&gt;Hint: a discontinuous linear functional is still linear!  So if $f(x)=z$, consider $f(wx)$, where $|w|=1$.&lt;/p&gt;&#xA;" OwnerUserId="822" LastActivityDate="2011-03-18T12:20:19.223" />
  <row Id="27788" PostTypeId="2" ParentId="24660" CreationDate="2011-03-18T12:31:03.137" Score="2" ViewCount="" Body="&lt;p&gt;Yes, surjectivity of dg(x) is sufficient.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The proof is not very complicated, but takes some work.  The reference cited by Planetmath is:&#xA;Eberhard Zeidler. &lt;i&gt;Applied functional analysis: main principles and their applications.&lt;/i&gt; Springer-Verlag, 1995.&#xA;Take a look at page 268 and following.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The key ingredient is a weaker formulation of the implicit function theorem:&#xA;Suppose $F: X \times Y \to Z$ is $C^1$ in a neighbourhood of $0$, with $F(0,0) = 0$&#xA;and $D_Y F(0,0) : Y \to Z$ is surjective.&#xA;Then, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) For each $r &amp;gt; 0$, there exists $\rho &amp;gt; 0$ so for every $|| u || &amp;lt; \rho$, there exists $v = v(u)$ with $F(u, v(u) ) = 0$ and $|| v || &amp;lt; r$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) There exists $d &amp;gt; 0$ so that $|| v(u) || \le d || D_Y F(0,0) v(u) ||$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The proof of this is a bit more delicate than the standard implicit function theorem, but you use the closed range theorem to construct a surrogate for the inverse, and then use this to construct your iteration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You then prove the Lagrange multiplier theorem from this, essentially following the standard proof (presumably the one you had in mind above).&lt;/p&gt;&#xA;" OwnerUserId="8343" LastEditorUserId="8343" LastEditorDisplayName="" LastEditDate="2011-03-18T14:15:01.793" LastActivityDate="2011-03-18T14:15:01.793" />
  <row Id="27789" PostTypeId="1" AcceptedAnswerId="27795" CreationDate="2011-03-18T12:50:16.717" Score="1" ViewCount="261" Body="&lt;p&gt;According to &lt;a href=&quot;http://en.wikipedia.org/wiki/Markov_chain#Reducibility&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt; with a little rephrasing:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A state $i$ is transient if and only&#xA;  if $P(T_i &amp;lt; \infty) &amp;lt;1$, recurrent if&#xA;  and only if $P(T_i &amp;lt; \infty) =1$,&#xA;  where $T_i$ is the first hitting time&#xA;  to i, i.e. $T_i=\inf\{n \in \mathbb{N} \cup \{ \infty \}: X_n=i \mid X_0=i \}.$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If I understand correctly, this can be used as the definition of transient/recurrent state. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Usually $P(T_i &amp;lt; \infty)$ is written as a series $\sum_{n \in \mathbb{N}} P(T_i = n)$. But I would like to learn other ways to tell if a state is recurrent/transient, which might be easier in some cases.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;For example, can a&#xA;transient/recurrent state be&#xA;completely characterized in terms of&#xA;closed subsets of states (defined similarly as an absorbing state), as follows&#xA;(my own quote)?&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;State $i$ is transient if and only if&#xA;  there exists a closed subset $S$ of&#xA;  states, s.t. $i \notin S$ and there exists $s \in S$&#xA;  and $n \in \mathbb{N}$ and the&#xA;  $n$-step transition probability&#xA;  $p_{is}^{(n)} &amp;gt; 0$.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Similarly, State $i$ is recurrent if&#xA;  and only if there does not exist such a&#xA;  closed subset of states as described above?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Can we also characterize positive/null&#xA;recurrence in terms of closed&#xA;subsets of states?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;Off the top of your head, what are some other necessary and/or sufficient&#xA;conditions for recurrent/transient&#xA;and positive/null recurrent state?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Thanks and regards!&lt;/p&gt;&#xA;" OwnerUserId="1281" LastEditorUserId="1281" LastEditorDisplayName="" LastEditDate="2011-03-18T13:02:24.470" LastActivityDate="2011-03-18T15:44:12.367" Title="How to characterize recurrent and transient states of Markov chain" Tags="&lt;stochastic-processes&gt;" AnswerCount="2" FavoriteCount="2" />
  <row Id="27790" PostTypeId="2" ParentId="27779" CreationDate="2011-03-18T12:51:19.760" Score="1" ViewCount="" Body="&lt;p&gt;If you don't know $\alpha(b)$ (I don't see it dotted) you still have six unknowns and five equations. If you do know $\alpha(b)$ you have a linear system.  If you insert $x=he$ for $x$ in the third equation, then substitute all the rest into the first you have &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x=a+\frac{x-d-he}{1+i}-\alpha(b)-de$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x\left(1-\frac{1}{1+i}\right)=x\frac{i}{1+i}=a+\frac{-d-he}{1+i}-\alpha(b)-de$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x=\frac{i+1}{i}(a-\alpha(b)-de)-i(d+he)$&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-03-18T12:51:19.760" CommentCount="1" />
  <row Id="27791" PostTypeId="2" ParentId="27775" CreationDate="2011-03-18T13:13:12.913" Score="10" ViewCount="" Body="&lt;p&gt;Technically speaking, the function $\displaystyle{f(x) = \frac{\sin x}{x}}$ is &lt;strong&gt;not&lt;/strong&gt; Riemann integrable on $(0, \infty)$, but rather &lt;em&gt;improperly&lt;/em&gt; Riemann integrable on $(0, \infty)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The construction of the Riemann integral only works for bounded intervals.  We can extend this construction to unbounded intervals like $(0, \infty)$, but that requires an additional limiting process.  It is the first construction (Riemann integrals for bounded integrals) that the Lebesgue integral generalizes.&lt;/p&gt;&#xA;" OwnerUserId="640" LastActivityDate="2011-03-18T13:13:12.913" CommentCount="3" />
  <row Id="27792" PostTypeId="1" AcceptedAnswerId="27796" CreationDate="2011-03-18T13:19:29.507" Score="2" ViewCount="86" Body="&lt;p&gt;There are three circles, all of them touching each other. The bottom two circles are laying on an imaginary floor, such that they touch the line g=-r as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given are all three radii, r1 (A), r2 (B) and r3 (C).&#xA;Assuming circle A has its center on (0/0), B has its center on (2 sqrt(r1 * r2), r2 - r1). I am now supposed to find the coordinates of C.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this a know problem and has an easy/straighforward solution? I can't seem to find a nice approach.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/3z9OW.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="8417" LastActivityDate="2011-03-18T14:39:07.020" Title="Positioning three circles, all of them touching each other" Tags="&lt;trigonometry&gt;&lt;circle&gt;&lt;triangle&gt;" AnswerCount="1" />
  <row Id="27793" PostTypeId="2" ParentId="27772" CreationDate="2011-03-18T13:29:39.367" Score="1" ViewCount="" Body="&lt;p&gt;If $K$ acts on $\Omega$, then we can define an unrestricted wreath product of $G$ by $K$ (relative to $\Omega$) as follows: first, $G^{\Omega} = \{\mathbf{g}\colon\Omega\to G\}$ has the pointwise multiplication (it's the set of $\Omega$-tuples of elements of $G$). We define an action of $K$ on $G^{\Omega}$ by&#xA;$$\mathbf{g}^k(\omega) = \mathbf{g}(k\cdot\omega).$$&#xA;This action gives a semidirect product $G^{\Omega}\rtimes K$; this semidirect product is called the (unrestricted) wreath product of $G$ by $K$, $G\wr_{\Omega}K$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(The &quot;restricted&quot; wreath product is the subgroup in which $\mathbf{g}(\omega)=e$ for almost all $\omega\in\Omega$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The order of $G^{\Omega}$ is $|G|^{|\Omega|}$, so the order of $G\wr_{\Omega}K$ is $|G|^{|\Omega|}|K|$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When we don't specify $\Omega$ and there is no natural action of $K$, then we take $\Omega=K$ and let $K$ act on itself by left multiplication. Then we just write $G\wr K$, and the order of $G\wr K$ is then $|G|^{|K|}|K|$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $\mathbf{Z}_{p^n}\wr\mathbf{Z}_{p^m}$, the order is $|p^n|^{|p^m|}|p^m| = p^{m+np^m}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the same as the order of $(\mathbf{Z}_p)^n\wr(\mathbf{Z}_p)^m$, since the orders of the groups are the same, though the structure of the group is very different.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2011-03-18T13:29:39.367" />
  <row Id="27794" PostTypeId="1" CreationDate="2011-03-18T13:30:50.317" Score="3" ViewCount="82" Body="&lt;p&gt;Can you use noninteger powers Like is $x^{8.3} / x^{2.2} = x^{6.1}$?&lt;/p&gt;&#xA;" OwnerUserId="8418" LastEditorUserId="3731" LastEditorDisplayName="" LastEditDate="2011-03-18T13:39:28.620" LastActivityDate="2011-03-18T13:41:56.597" Title="Non-integer exponents" Tags="&lt;homework&gt;&lt;algebra&gt;&lt;algebra-precalculus&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="27795" PostTypeId="2" ParentId="27789" CreationDate="2011-03-18T13:33:35.023" Score="4" ViewCount="" Body="&lt;p&gt;(0) The definition of $T_i$ on Wikipedia is awful on at least three counts. One should define $T_i=\inf A_i$ with $A_i=\{n\ge1:X_n=i\}\cup\{+\infty\}$ and one should say that $i$ is transient if and only if $P(T_i&amp;lt;+\infty|X_0=i)&amp;lt;1$ and recurrent otherwise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) In some cases there exists no closed subset at all and the existence of closed sets is not related to recurrence or transience. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;First example: consider a homogenous random walk on $\mathbb{Z}$. Thus $p_{n,n+1}=a$ and $p_{n,n-1}=1-a$ for every $n$, for a given $a$ in $(0,1)$. Then there exists no closed set except $\mathbb{Z}$ and the chain is recurrent if $a=\frac12$ and transient otherwise. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second example: consider a homogenous birth-and-death chain such that $0$ is absorbing. Thus $p_{0,0}=1$ and $p_{n,n+1}=a$ and $p_{n,n-1}=1-a$ for every $n\ge1$, for a given $a$ in $(0,1)$. Then the set $S=\{0\}$ is closed and the chain is recurrent if $a\le\frac12$ and transient otherwise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) You could (should?) read the beautiful small book &lt;a href=&quot;http://arxiv.org/abs/math/0001057&quot; rel=&quot;nofollow&quot;&gt;Random Walks and Electric Networks&lt;/a&gt; by Peter G. Doyle and J. Laurie Snell, which explains this and a lot of related stuff in a very accessible way.&lt;/p&gt;&#xA;" OwnerUserId="6179" LastEditorUserId="6179" LastEditorDisplayName="" LastEditDate="2011-03-18T14:16:17.000" LastActivityDate="2011-03-18T14:16:17.000" CommentCount="9" />
  <row Id="27796" PostTypeId="2" ParentId="27792" CreationDate="2011-03-18T13:35:52.140" Score="4" ViewCount="" Body="&lt;p&gt;Since you know the radii $r_1$, $r_2$, and $r_3$ and the centers $A$ and $B$, the center $C(x_c,y_c)$ must be a distance $r_1+r_3$ from $A$ and a distance $r_2+r_3$ from $B$, which gives the two equations $$x_c^2+y_c^2=(r_1+r_3)^2$$ and $$(x_c-2\sqrt{r_1r_2})^2+(y_c-(r_2-r_1))^2=(r_2+r_3)^2$$  Solving the system for $(x_c,y_c)$ should give the two possible coordinate pairs for $C$ (one as you've pictured, the other &quot;below&quot; the first two circles).&lt;/p&gt;&#xA;" OwnerUserId="72" LastEditorUserId="1508" LastEditorDisplayName="" LastEditDate="2011-03-18T14:39:07.020" LastActivityDate="2011-03-18T14:39:07.020" CommentCount="3" />
  <row Id="27797" PostTypeId="2" ParentId="27794" CreationDate="2011-03-18T13:37:28.353" Score="0" ViewCount="" Body="&lt;p&gt;Yes you can. In fact, $$\frac{1}{x^y}=x^{-y}$$ for $x,y \in \mathbb{R}, x \neq 0$. So $$\frac{x^y}{x^z}=x^{y-z}$$ for $x,y,z \in \mathbb{R}, x \neq 0$.&lt;/p&gt;&#xA;" OwnerUserId="5465" LastActivityDate="2011-03-18T13:37:28.353" />
  <row Id="27799" PostTypeId="1" AcceptedAnswerId="27803" CreationDate="2011-03-18T13:40:46.323" Score="1" ViewCount="58" Body="&lt;p&gt;Let $$f(x)=\sum_{s=-\infty}^{\infty}e^{-2\pi ixsk}$$&#xA;$k$ integer.&#xA;Does this integral exist?&#xA;$$\int_{0}^{1}(f(x))^{2}dx$$&lt;/p&gt;&#xA;" OwnerUserId="6052" LastActivityDate="2011-03-18T13:55:18.507" Title="Does the  integral of squared Shah function exist?" Tags="&lt;analysis&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="27800" PostTypeId="2" ParentId="27794" CreationDate="2011-03-18T13:41:56.597" Score="4" ViewCount="" Body="&lt;p&gt;Yes, you can. The definitions grow every more complicated depending on what kinds of exponents you allow, but the laws of exponents hold in general.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For positive integer powers, you can define exponentials (&quot;powers&quot;) for any base $x$:&#xA;\begin{align*}&#xA;x^1&amp;amp; = x\\&#xA;x^{n+1} &amp;amp;= x^nx&#xA;\end{align*}&#xA;which is equivalent to&#xA;$$x^n = \underbrace{x\times x\times\cdots\times x}_{n\text{ factors}}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This satisfies the usual &quot;exponent rules&quot;: $x^nx^m = x^{n+m}$, and $(xy)^n = x^ny^n$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to extend it to exponent $0$, you now have to exclude $x=0$ ($0^0$ is usually left undefined in general; see &lt;a href=&quot;http://math.stackexchange.com/questions/11150/zero-to-zero-power/&quot;&gt;this previous question&lt;/a&gt;). You define $x^0 = 1$ for all $x\neq 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then you can also extend to &lt;em&gt;negative&lt;/em&gt; integer exponents: if $n\gt 0$, then we define $x^{-n} = \frac{1}{x^n}$. This adds the third &quot;exponent rule&quot;: $x^n/x^m = x^{n-m}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can then extend the exponents to &lt;em&gt;fractions&lt;/em&gt;, but now you have to restrict the base to positive real numbers in general. If $p$ is an integer, we define $x^{1/p} = \sqrt[p]{x}$, the $p$th root of $x$ (the unique positive real number $y$ such that $y^p= x$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, if $p$ and $q$ are integers, with $q\gt 0$, we define $x^{p/q} = (x^p)^{1/q} = \sqrt[q]{x^p}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;With this definition, the exponentiation rules still hold; this applies to the case you have above, since $8.3 = \frac{83}{10}$, $2.2 = \frac{22}{10}$, and $6.1 = \frac{61}{10}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can extend the definition further, to &lt;em&gt;any&lt;/em&gt; real number as an exponent, through the use of either limits or logarithms. We define $a^b = e^{b\ln(a)}$, provided $a\gt 0$. This still satisfies the exponent laws.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(You can extend the definition &lt;em&gt;even further&lt;/em&gt; by allowing complex numbers as answers; then you can remove the restriction on the base being positive).&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2011-03-18T13:41:56.597" />
  <row Id="27801" PostTypeId="2" ParentId="27784" CreationDate="2011-03-18T13:46:05.273" Score="1" ViewCount="" Body="&lt;p&gt;We may suppose $x=0$. So the question is: given a sequence $(x_n)$ such that $x_{n+1}/x_n \to 0$, does this imply $$\frac{x_{n+1}-x_{n+2}}{x_{n}-x_{n+1}} \to 0 ?$$ The answer is yes. Take any $\epsilon \in (0,\frac{1}{2})$. Then there exists $N$ such that$|x_{n+1}/x_n| &amp;lt; \epsilon$ for $n&amp;gt;N$. So if $n&amp;gt;N$, we have $|x_n-x_{n+1}| &amp;gt; |(1-\epsilon)x_n|$, and $|x_{n+1}-x_{n+2}| &amp;lt; |(1+\epsilon)x_{n+1}|$. This gives $$|\frac{x_{n+1}-x_{n+2}}{x_{n}-x_{n+1}}| &amp;lt; |\frac{1+\epsilon}{1-\epsilon}\frac{x_{n+1}}{x_n}| &amp;lt; |3\frac{x_{n+1}}{x_n}| &amp;lt; 3\epsilon$$&lt;/p&gt;&#xA;" OwnerUserId="1508" LastActivityDate="2011-03-18T13:46:05.273" />
  <row Id="27802" PostTypeId="2" ParentId="27784" CreationDate="2011-03-18T13:46:59.250" Score="3" ViewCount="" Body="&lt;p&gt;Consider the ratios&#xA;$$&#xD;&#xA;w_n=\frac{x_{n+2}-x_{n+1}}{x_{n+1}-x_n},\quad&#xD;&#xA;u_n=\frac{x_{n+1}-x}{x_{n}-x}.&#xD;&#xA;$$&#xA;Then&#xA;$$&#xD;&#xA;w_n=u_n\frac{u_{n+1}-1}{u_n-1},&#xD;&#xA;$$&#xA;your hypothesis is that $u_n\to0$ and your aim is to show that $w_n\to0$. Well...&lt;/p&gt;&#xA;" OwnerUserId="6179" LastActivityDate="2011-03-18T13:46:59.250" CommentCount="4" />
  <row Id="27803" PostTypeId="2" ParentId="27799" CreationDate="2011-03-18T13:55:18.507" Score="2" ViewCount="" Body="&lt;p&gt;I think the problem is, to have a meaning for $f(x)^2$, having in mind that the &quot;Shah function&quot; is merely a distribution (also known as &lt;a href=&quot;http://en.wikipedia.org/wiki/Dirac_comb&quot; rel=&quot;nofollow&quot;&gt;Dirac comb&lt;/a&gt;). It is known that it is notoriously difficult to have a proper notion for the &lt;a href=&quot;http://en.wikipedia.org/wiki/Distribution_%28mathematics%29#Problem_of_multiplication&quot; rel=&quot;nofollow&quot;&gt;multiplication of distributions&lt;/a&gt; and I do not know a proper way of multiplying a $\delta$ with itself.&lt;/p&gt;&#xA;" OwnerUserId="3148" LastActivityDate="2011-03-18T13:55:18.507" />
  <row Id="27804" PostTypeId="1" CreationDate="2011-03-18T14:06:14.313" Score="2" ViewCount="136" Body="&lt;p&gt;We recently went through the wreath product in my group theory class, but the definition still seems a bit unmotivated to me. The two reasons I can see for it are 1) it allows us to construct new groups, and 2) we can use it to reconstruct imprimitive group actions. Are there any applications of the wreath product outside of pure group theory?&lt;/p&gt;&#xA;" OwnerUserId="8420" LastEditorUserId="8420" LastEditorDisplayName="" LastEditDate="2011-03-18T14:13:58.873" LastActivityDate="2011-03-18T18:37:07.127" Title="Applications of the wreath product?" Tags="&lt;group-theory&gt;" AnswerCount="4" CommentCount="1" FavoriteCount="1" />
  <row Id="27805" PostTypeId="1" AcceptedAnswerId="27806" CreationDate="2011-03-18T14:30:03.957" Score="3" ViewCount="122" Body="&lt;p&gt;Is it possible to simplify this expression? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{2^{2x-1} - 2^{x-1}}{2^{2x-1}}$ &lt;/p&gt;&#xA;" OwnerUserId="7403" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2011-03-18T18:30:38.953" LastActivityDate="2011-03-18T18:32:32.837" Title="Is it possible to simplify $\frac{2^{2x-1} - 2^{x-1}}{2^{2x-1}}$ ?" Tags="&lt;algebra&gt;&lt;algebra-precalculus&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="27806" PostTypeId="2" ParentId="27805" CreationDate="2011-03-18T14:39:11.047" Score="6" ViewCount="" Body="&lt;p&gt;Yeah you can do this: $$\frac{ 2^{x-1} \Bigl[ 2^{x}-1\Bigr]}{2^{2x-1}} = \frac{2^{x}-1}{2^{x}}$$&lt;/p&gt;&#xA;" LastActivityDate="2011-03-18T14:39:11.047" CommentCount="1" />
  <row Id="27807" PostTypeId="2" ParentId="27804" CreationDate="2011-03-18T14:52:17.213" Score="1" ViewCount="" Body="&lt;p&gt;Hi-- &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't have much knowledge about Automata theory, but here is a paper, which presents, application of Wreath Products to Rational Languages.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.springerlink.com/content/5678n51v657t21l6/fulltext.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.springerlink.com/content/5678n51v657t21l6/fulltext.pdf&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" LastActivityDate="2011-03-18T14:52:17.213" />
  <row Id="27808" PostTypeId="1" AcceptedAnswerId="27810" CreationDate="2011-03-18T15:04:05.440" Score="5" ViewCount="199" Body="&lt;p&gt;The question is from Axler's &quot;&lt;em&gt;Linear Algebra Done Right&lt;/em&gt;&quot;, which I'm using for self-study.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We are given a linear operator $T$ over a finite dimensional vector space $V$. We have to show that $T$ is a scalar multiple of the identity iff $\forall S \in L(V), TS = ST$. Here, $L(V)$ denotes the set of all linear operators over $V$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One direction is easy to prove. If $T$ is a scalar multiple of the identity, then there exists a scalar $a$ such that $Tv = av$, $\forall v \in V$. Hence, given an arbitrary vector $w$, $$TS(w) = T(Sw) = a(Sw) = S(aw) = S(Tw) = ST(w)$$ where the third equality is possible because $S$ is a linear operator. Then, it follows that $TS = ST$, as required.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am, however, at a loss as to how to tackle the other direction. I thought that a proof by contradiction,  ultimately constructing a linear operator $S$ for which $TS \neq ST$, might be the way to go, but haven't made much progress. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance!&lt;/p&gt;&#xA;" OwnerUserId="8421" LastEditorUserId="8421" LastEditorDisplayName="" LastEditDate="2011-03-19T00:38:26.407" LastActivityDate="2011-03-19T00:38:26.407" Title="Simple fact on linear operators" Tags="&lt;linear-algebra&gt;&lt;vector-space&gt;&lt;self-learning&gt;" AnswerCount="7" CommentCount="4" FavoriteCount="2" />
  <row Id="27809" PostTypeId="2" ParentId="27804" CreationDate="2011-03-18T15:10:46.317" Score="7" ViewCount="" Body="&lt;p&gt;Even within Group Theory, wreath products have more interests than you note; I'll give a couple below. But to answer your question, though perhaps not very satisfactorily, you can define wreath products of semigroups in precisely the analogous way as you do for groups. Semigroups are closely related to (and key to understanding) automata theory (which itself has many applications), and wreath products can play an important role in the study and construction of come automata.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I say it may not be very satisfying, because it sounds as if I'm saying &quot;Sure! It has lots of applications in &quot;pure &lt;strong&gt;semi&lt;/strong&gt; group theory!&quot;...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But within Group Theory, one very important property of wreath products is the theorem of Kaloujnine and Krasner:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Theorem.&lt;/strong&gt; &lt;em&gt;Let&lt;/em&gt; $H$ &lt;em&gt;and&lt;/em&gt; $K$ &lt;em&gt;be any groups. If&lt;/em&gt; $G$ &lt;em&gt;is an extension of&lt;/em&gt; $H$ &lt;em&gt;by&lt;/em&gt; $K$ &lt;em&gt;(that is,&lt;/em&gt; $G$ &lt;em&gt;contains a normal subgroup&lt;/em&gt; $N$ &lt;em&gt;such that&lt;/em&gt; $N\cong H$ &lt;em&gt;and&lt;/em&gt; $G/N\cong K$), &lt;em&gt;then&lt;/em&gt; $G$ &lt;em&gt;is isomorphic to a subgroup of the wreath product&lt;/em&gt; $H\wr K$. &lt;em&gt;That is,&lt;/em&gt; $H\wr K$ &lt;em&gt;contains isomorphic copies of every extension of&lt;/em&gt; $H$ &lt;em&gt;by&lt;/em&gt; $K$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In principle, if you understand all simple groups and you understand all possible extensions of two given groups (in terms of the groups, perhaps), then you understand all finite groups. Though this &quot;in principle&quot; is hopeless in practice, it can be useful in specific circumstances.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, the iterated wreath product of $\mathbf{Z}_p$ with itself plays an important role in the study of $p$-groups, and is associated to the $p$-Sylow subgroups of symmetric groups.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2011-03-18T15:10:46.317" CommentCount="1" />
  <row Id="27810" PostTypeId="2" ParentId="27808" CreationDate="2011-03-18T15:19:26.530" Score="2" ViewCount="" Body="&lt;p&gt;Suppose $TS = ST$ for every $S$. Show that $Tv = a_{d}v$ for every $v$ where $a_d$ could depend on $v$. In other words, show that $v$ and $Tv$ are linearly dependent for each $v \in V$. Suppose for contradiction that they are linearly independent. Since $(v, Tv)$ is linearly independent, it can be extended to a basis $(v,Tv, u_1, \dots, u_n)$ of $V$. So define $S$ as following: $Sv = v$, $S(Tv) = v$ and $S(u_1) = 0, \dots, S(u_n) = 0$. Then, $Tv = TSv = STv = v$. Hence $v$ and $Tv$ are linearly dependent, which is a contradiction. Then you have to show uniqueness. &lt;/p&gt;&#xA;" OwnerUserId="2548" LastEditorUserId="8421" LastEditorDisplayName="" LastEditDate="2011-03-18T16:54:56.423" LastActivityDate="2011-03-18T16:54:56.423" CommentCount="3" />
  <row Id="27811" PostTypeId="2" ParentId="27808" CreationDate="2011-03-18T15:24:02.350" Score="0" ViewCount="" Body="&lt;p&gt;If $v$ is an eigenvector of T with eigenvalue $\lambda$, then $Sv$ is also an eigenvector with the same eigenvalue.  For any two $v$ and $w$ in $V$, there exists a transformation $S$ mapping $v$ to $w$; so all elements of $V$ are eigenvectors with eigenvalue $\lambda$.&lt;/p&gt;&#xA;" OwnerUserId="8423" LastActivityDate="2011-03-18T15:24:02.350" CommentCount="1" />
  <row Id="27812" PostTypeId="1" AcceptedAnswerId="27855" CreationDate="2011-03-18T15:30:22.367" Score="1" ViewCount="128" Body="&lt;p&gt;Given integers $a, b, c$ construct a single-tape Turing machine recognizing the language&#xA;$\{w \in \{0,1\}^{*}: a*\#_{0}w+b*\#_{1}w+c=0\}$ in time $O(n*logn)$, where $n=|w|$. &#xA;$\#_{x}w$ denotes the number of occurrences of the symbol $x$ in $w$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Updating counters somewhere on the tape would result in $O(n^2)$ complexity. I suspect extended euclidean algorithm has to be used during construction of this machine.&lt;/p&gt;&#xA;" OwnerUserId="8380" LastEditorUserId="6296" LastEditorDisplayName="" LastEditDate="2011-03-18T17:02:30.360" LastActivityDate="2011-03-18T21:09:18.090" Title="Recognizing language using Turing machine" Tags="&lt;computer-science&gt;&lt;formal-systems&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="1" />
  <row Id="27813" PostTypeId="2" ParentId="27808" CreationDate="2011-03-18T15:34:15.720" Score="4" ViewCount="" Body="&lt;p&gt;In general, when one has a condition of the form &quot;$A$ is a &lt;code&gt;blah&lt;/code&gt; if and only if for every $B$ &lt;code&gt;this happens&lt;/code&gt;&quot;, the &quot;if&quot; direction can often be established by selecting suitably/cleverly chosen $B$ that show everything works.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is just such a situation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $\beta = \{\mathbf{v}_i\}_{i\in I}$ be a basis for $\mathbf{V}$. For each $i,j\in I$, let $S_{ij}$ be the linear operator on $\mathbf{V}$ given by&#xA;$$S_{ij}(\mathbf{v}_k) = \left\{\begin{array}{ll}&#xA;\mathbf{v}_j &amp;amp;\mbox{if $k=i$,}\\&#xA;\mathbf{v}_i &amp;amp;\mbox{if $k=j$,}\\&#xA;\mathbf{0} &amp;amp;\mbox{if $k\neq i$ and $k\neq j$.}&#xA;\end{array}\right.$$&#xA;That is: for $i\neq j$, $S_{ij}$ exchanges $\mathbf{v}_i$ and $\mathbf{v}_j$, and maps all other basis elements to $\mathbf{0}$. And $S_{ii}$ maps $\mathbf{v}_i$ to itself, and all other basis elements to $\mathbf{0}$. These are our &quot;suitably chosen&quot; $S$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now consider $S_{ii}T(\mathbf{v}_j)$ and $TS_{ii}(\mathbf{v}_j)$ first to get information about what $T$ does to $\beta$; then consider $S_{ij}T(\mathbf{v}_j)$ and $TS_{ij}(\mathbf{v}_j)$ for $i\neq j$.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2011-03-18T15:34:15.720" CommentCount="1" />
  <row Id="27814" PostTypeId="2" ParentId="27647" CreationDate="2011-03-18T15:42:00.533" Score="1" ViewCount="" Body="&lt;p&gt;This is a standard result that follows by factorization into separable and inseparable extensions. See the post below from Ask an Algebraist. Alas, the thread has since been deleted (due to vandals), but the references below to Isaacs will help you to locate the standard &lt;em&gt;conceptual&lt;/em&gt; proof (to be contrasted with the non-conceptual elementary form given below by a student - cf. my remarks following it). See Isaacs: &lt;em&gt;Algebra: a graduate course. Ch. 19: Separability and Inseparability&lt;/em&gt; and see esp. the remark preceding Theorem 19.18.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;From: Bill Dubuque; Date: July 6, 2010&lt;br&gt;&#xA;In reply to: &quot;Thanks &amp;amp; Summary&quot;, posted by student on July 6, 2010:&lt;br&gt;&#xA;student wrote:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Thanks for your help, it's all clear now. The proof you gave provides good intuition!&#xA;  For my own understanding I have written it up in total and tried to make the paralles &#xA;  between the separable and the inseparable case clear. I post the proof below,&#xA;  maybe someone reading this will benefit from it.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;BTW you said that this theorem occurs in many textbooks.&#xA;  Actually I could only locate it in Isaacs book (pp. 303-304),&#xA;  however with a different proof (as you suggested).&#xA;  Are there any specific books about algebra in general&#xA;  or field theory in particular you would recommend?&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;OK here comes the proof (hopefully without typos),&#xA;  which is basically a summary of the previous posts in this thread:&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;THEOREM&#xA;  Let L and K be fields s.t. L is an algebraic extension of K.&#xA;  Assume every irreducible polynomial in K[X] has at least one zero in L.&#xA;  Then L is an algebraic closure of K.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;REMARK&#xA;  If in the situation of the theorem char(L) = p, then for every integer k &gt;= 1&#xA;  every element c in K has a unique p^k-th root d in L (i.e. d^p^k = c).&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;PROOF (REMARK)&#xA;  Existence: The polynomial X^p^k - c is in K[X] and hence has a root d in L.&#xA;  Uniqueness: Assume d^p^k = c = e^p^k. Then (d - e)^p^k = 0 [freshman's dream]&#xA;  and hence d = e.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;PROOF (THEOREM)&#xA;  Suppose f is an irreducible polynomial with coefficients in K.&#xA;  To prove the theorem we have to show that f splits over L.&#xA;  The proof is divided into two cases: f separable and f inseparable.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;*) 1st case: f separable.&#xA;  Let S denote a splitting field for f over K&#xA;  lying inside a splitting field for f over L.&#xA;  Being the splitting field for a separable polynomial over K,&#xA;  S has a primitive element: S = K(a).&#xA;  Let g be the minimal polynomial of a over K.&#xA;  Then g has a root, say b, in L.&#xA;  Since g is irreducible and both a and b are roots of g,&#xA;  K(a) is isomorphic to K(b).&#xA;  [In fact K(a) = K(b):&#xA;  Since S (being the splitting field for a polynomial in K[X])&#xA;  is normal over K, and since g is irreducible and&#xA;  has a root in S (namely a), g must split over S, hence b is in K(a).&#xA;  Since K(a) and K(b) both have the same degree (namely deg(g)) over K,&#xA;  K(a) = K(b).]&#xA;  Thus L &gt;= K(b) = K(a) = S contains all roots of f.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;*) 2nd case: f inseparable.&#xA;  Then char(K) = p &gt; 0,&#xA;  and f = g(X^p^n) for a separable irreducible polynomial g in K[X] and some n &gt; 0.&#xA;  By the 1st case g splits in L, and we may let a_1,...,a_m be the roots of g in L.&#xA;  Then S = K(a_1,...,a_m) (being the splitting field for a separable polynomial over K)&#xA;  has a primitive element: S = K(a).&#xA;  Let h be the minimal polynomial of a over K.&#xA;  The polynomial h(X^p^n) is in K[X] and hence has a root in L, say b.&#xA;  Then b^p^n is a root of h in L.&#xA;  Since h is irreducible and both a and b^p^n are roots of h,&#xA;  K(a) is isomorphic to K(b^p^n).&#xA;  In fact K(a) = K(b^p^n) [same argument as in the 1st case].&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Now let r denote a root of f lying inside a splitting field for f over L.&#xA;  Then r^p^n is a root of g, hence r^p^n is in S = K(a) = K(b^p^n).&#xA;  Thus r^p^n = F(b^p^n) for some polynomial F in K[X].&#xA;  By the remark, every coefficient of F has a p^n-th root in L,&#xA;  so F(b^p^n) = G(b)^p^n for some G in L[X] [freshman's dream].&#xA;  Thus (r - G(b))^p^n = 0, hence r = G(b) is in L(b) = L.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Q.E.D.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is just an &quot;elementary&quot; form of the standard proof (e.g. cf. Isaacs ref. above).&#xA;But, alas, by eliminating the innate structural elements (factorization into separable&#xA;and inseparable extensions) in favor of &quot;simpler&quot; calculations with elements,&#xA;the innate structure of the proof is greatly obfuscated. The elementary proof&#xA;has been constructed by directly inlining the lemmas leading up to Isaacs proof&#xA;vs. invoking them by name. This is not good pedagogically. Decomposing algebraic &#xA;extensions into their separable and inseparable parts is an essential tool required &#xA;to study general algebraic extensions. Such key ideas should not be obscured such&#xA;as above - esp. when, as here, the structural form is just as short and simple as is&#xA;the elementary form. Please read the section of Isaacs leading up to the proof to&#xA;better understand the innate structure. You've a much better chance of remembering&#xA;the structural proof years later because it has been abstracted into simple steps&#xA;using key structures/lemmas that will often be reused. But few if any students&#xA;would probably remember the non-conceptual elementary form as presented above.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2011-03-18T15:52:24.237" LastActivityDate="2011-03-18T15:52:24.237" />
  <row Id="27815" PostTypeId="2" ParentId="27804" CreationDate="2011-03-18T15:43:43.967" Score="4" ViewCount="" Body="&lt;p&gt;It's not exactly an application, but the Rubik Cube group provides some insight into the reason why wreath products are interesting and natural objects to study.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The cube has 8 corner cubies each with three faces, and 12 edge corner cubies each with 2 faces. If you imagine all permutations of the faces of the corner cubies that permute the cubies and may also rotate them through 120 or 240 degrees, then you get a group $C$ which is the (permutation) wreath product of a cyclic group of order 3 and the symmetric group $S_8$, and $|C| = 3^88!$. Simiarly the group $E$ of permutations of the faces of the edge cubies that permute the cubies and may also flip them through 180 degrees is the wreath product of a cyclic group of order 2 and $S_{12}$ and has order $2^{12}12!$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Rubik Cube group itself is a subgroup $G$ of the direct product $C \times E$. It turns out that only $1/12$ of the possible permutations in $C \times E$ are attainable without taking the cube to pieces and reconstructing it, so $G$ has index 12 in $C \times E$.&lt;/p&gt;&#xA;" OwnerUserId="2820" LastActivityDate="2011-03-18T15:43:43.967" />
  <row Id="27816" PostTypeId="2" ParentId="27789" CreationDate="2011-03-18T15:44:12.367" Score="3" ViewCount="" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Tim's characterization of states in terms of closed sets is correct &#xA;for &lt;em&gt;finite&lt;/em&gt; state space Markov chains. Partition the state&#xA;space into communicating classes. Every recurrent class is closed, &#xA;but no transient class is closed (because the chain must eventually&#xA;get &quot;stuck&quot; in some recurrent class). The part in parentheses is false&#xA;for infinite state space chains, as Didier's answer shows. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Another well-known characterization is that a state $i$ is transient&#xA;if and only if $$\sum_{n=1}^\infty P(X_n=i | X_0=i)&amp;lt;\infty.$$ This criterion is used,&#xA;for example, to prove Polya's result that the symmetric random walk on  $\mathbb{Z}^d$ is recurrent if $d=1,2$, but transient when $d\geq 3$.   &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Similarly, the probability&lt;br&gt;&#xA;$$P(X_n=i\mbox{ for infinitely many } n | X_0=i)$$ &#xA;is equal to zero or one, depending on whether the state $i$ is transient or recurrent.  &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="940" LastActivityDate="2011-03-18T15:44:12.367" CommentCount="4" />
  <row Id="27817" PostTypeId="2" ParentId="27775" CreationDate="2011-03-18T15:45:53.190" Score="6" ViewCount="" Body="&lt;p&gt;You might be interested in &lt;a href=&quot;http://en.wikipedia.org/wiki/Henstock%E2%80%93Kurzweil_integral&quot; rel=&quot;nofollow&quot;&gt;Henstock-Kurzweil&lt;/a&gt; (HK) integral. Its definition is an easy modification of Riemann integral. All Lebesgue-integrable functions are integrable in the HK-sense, and so is your function $\sin(x)/x$. The usual theorems from Lebesgue theory (such as the dominated convergence theorem) have extensions to the HK theory. On top of that, Newton-Leibniz formula holds for any fuction admitting an anti-derivative (which is not true in Lebesgue theory).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That being said, you definitely need Lebesgue theory for spaces different from $\mathbb{R}^n$. Notice also that $\mathbb{R}^2$ is isomorphic to $\mathbb{R}$ as a measure space (and that any sensible measure space is isomorphic to an interval); you would lose this unity if you wanted e.g. $\sin(x)/x$ to be integrable. &lt;/p&gt;&#xA;" OwnerUserId="8268" LastActivityDate="2011-03-18T15:45:53.190" CommentCount="2" />
  <row Id="27818" PostTypeId="2" ParentId="27579" CreationDate="2011-03-18T15:49:57.750" Score="1" ViewCount="" Body="&lt;p&gt;After not receiving a response here, Giuseppe asked in Mathoverflow, where &lt;a href=&quot;http://mathoverflow.net/questions/58825/non-degenerate-alternating-bilinear-form-on-a-finite-abelian-group&quot; rel=&quot;nofollow&quot;&gt;an answer has been given.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I post this in the hope that Giuseppe will &quot;accept&quot; so the question does not appear un-answered here.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2011-03-18T15:49:57.750" CommunityOwnedDate="2011-03-18T15:49:57.750" CommentCount="1" />
  <row Id="27819" PostTypeId="2" ParentId="27808" CreationDate="2011-03-18T15:55:13.990" Score="0" ViewCount="" Body="&lt;p&gt;Let $\{e_1,\dots,e_n\}$ be a basis for the space $V$. Then you need to show that there exists $a \in \mathbb{R}$ such that $Te_i = a e_i$ for $i = 1,\dots,n$ (every linear operator on finite dimensional space is determined by its values on basis vectors).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $S_i \in L(V)$ be defined as $S(a_1 e_1 + \dots + a_n e_n) = a_i e_i$. Assume that $Te_i = b_{i,1} e_1 + \dots + b_{i,n} e_n$. Then&#xA;$$&#xD;&#xA;T e_i = TS_i e_i = S_i Te_i = b_{i,1} S_i e_1 + \dots + b_{i,n} S_i e_n = b_{i,i} e_i.&#xD;&#xA;$$&#xA;Now we need to show that $b_{i,i} = b_{j,j}$ for all $i,j=1,\dots,n$. For a given $i,j$ let $S \in L(V)$ be defined by&#xA;$$&#xD;&#xA;S(a_1 e_1 + \dots + a_n e_n) = a_j e_i + a_i e_j.&#xD;&#xA;$$&#xA;Then&#xA;$$&#xD;&#xA;b_{i,i} e_i + b_{j,j} e_j = T(e_i + e_j) = TS(e_i + e_j) = ST(e_i + e_j) = S(b_{i,i} e_i + b_{j,j} e_j) = b_{j,j} e_i + b_{i,i} e_j.&#xD;&#xA;$$&#xA;Hence, because $(e_k)$ form basis, we obtain $b_{i,i} = b_{j,j}$ which completes the proof.&lt;/p&gt;&#xA;" OwnerUserId="8229" LastActivityDate="2011-03-18T15:55:13.990" />
  <row Id="27820" PostTypeId="2" ParentId="18382" CreationDate="2011-03-18T16:13:40.980" Score="0" ViewCount="" Body="&lt;p&gt;Rotation happens within a plane.  In three dimensions, an axis is enough to specify the plane of rotation, but a point is not enough (as Steven said).  As I understand it, rotations with quaternions happen with respect to the origin.  So if you want to apply a transformation to a point that produces a rotation around another point, you have to sandwich your rotation with translations in order to move the reference frame to the origin and then back.  That is, to use quaternion &lt;strong&gt;Q&lt;/strong&gt; to rotate point &lt;strong&gt;P&lt;/strong&gt; with respect to translated origin &lt;strong&gt;G&lt;/strong&gt;: &#xA;&lt;strong&gt;P' = Q(P-G)Q' + G&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where &lt;strong&gt;P'&lt;/strong&gt; is the transformed point, and &lt;strong&gt;Q'&lt;/strong&gt; is the quaternion inverse.&lt;/p&gt;&#xA;" OwnerUserId="8424" LastActivityDate="2011-03-18T16:13:40.980" CommentCount="1" />
  <row Id="27821" PostTypeId="2" ParentId="27719" CreationDate="2011-03-18T16:14:47.273" Score="6" ViewCount="" Body="&lt;p&gt;Another way to look at it is by the divisibility lattice, where gcd is the &lt;strong&gt;greatest lower bound&lt;/strong&gt;. So 5 is the greatest lower bound of 10 and 15 in the lattice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The counter-intuitive thing about this lattice is that the 'bottom' (the absolute lowest element) is 1 (1 divides everything), but the highest element, the one above everybody, is 0 (everybody divides 0).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So $\gcd(0, x)$ is the same as ${\rm glb}(0, x)$ and should be $x$, because $x$ is the lower bound of the two: they are not 'apart' and 0 is '$&gt;'$ $x$ (that is the counter-intuitive part).&lt;/p&gt;&#xA;" OwnerUserId="1919" LastActivityDate="2011-03-18T16:14:47.273" CommentCount="0" />
  <row Id="27822" PostTypeId="2" ParentId="15143" CreationDate="2011-03-18T16:28:40.447" Score="6" ViewCount="" Body="&lt;p&gt;The idea is essentially correct, but what you wrote needs some editing.  You have independently rediscovered some beautiful mathematics!  Bravo!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your Theorem (as stated) has a much easier proof, but in your proof, you address an important point not mentioned in the theorem statement.  There are two key ideas you really need to separate.  Once you do that, I think you will see how to clean up your arguments.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the following, I assume that you are familiar with differential forms and the exterior derivative, so that expressions like &#xA;$$ df = \sum \frac{\partial f}{\partial x_i} d x_i$$ seem reasonable to you, but also that you know something about what $d( Adx + Bdy + Cdz)$ is.  If not, let me know in a comment, and I will gladly rewrite this in $\mathbb R^3$ using the curl operator ($\nabla \times$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, about the proof of the theorem you stated.  You need to think about what you mean by the differential equation $$A dx + Bdy = 0.$$  Obviously, you can &quot;divide through&quot; by $dx$ and then get an ordinary differential equation written in the normal way.  Another way of seeing it is to say that you are looking for a parametrized curve $( x(t), y(t) )$ such that the tangent vector $(x&amp;#39;(t), y&amp;#39;(t))$ satisfies&#xA;$$ 0 = A d x + B dy = A x&amp;#39;(t) dt + B y&amp;#39;(t) dt = (Ax&amp;#39;(t) + By&amp;#39;(t)) dt.$$&#xA;In other words, we want a curve $(x(t), y(t))$ so that its tangent vector is orthogonal to the vector $(A(x,y), B(x,y))$.  If you happen to know that $(A, B) = \nabla f$ for some function $f$ (i.e. $df = Adx + Bdy$) then you have implicitly defined solutions given by $f = c$ for any constant $c$, since &lt;i&gt;$\nabla f$ is orthogonal to level sets of $f$&lt;/i&gt;!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In higher dimensions, this picture generalizes perfectly.  The solution to an equation given by $$ Adx + B dy + Cdz = 0$$ is going to be a surface in $\mathbb R^3$.  If the differential form $Adx + Bdy + Cdz = df$ then by the same reasoning as above, level sets of $f$ will give you surfaces that satisfy this equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now for the important point you address in your proof, but which you do not mention in the theorem statement (but you do mention in the text above the theorem statement).  This is something called the Poincaré Lemma.  A special case of this says that in $\mathbb R^n$, if you have a differential 1-form $\alpha$ and $d\alpha = 0$ then $\alpha = df$ for some function $f$.  You have all the key ideas of the proof of this &quot;lemma&quot; lurking in your proof.  (It's called a lemma, but it is a key result in differential geometry and differential topology.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(The Poincaré Lemma does not hold if you work in a space with &quot;holes&quot;.  E.g. consider the 1-form given by $$\alpha := -\frac{y}{x^2+y^2} dx + \frac{x}{x^2+y^2} dy.$$&#xA;This has $d \alpha = 0$, but you can't find a function so that $\alpha = df$.)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;More generally, you can define a (partial) differential equation in $\mathbb R^n$ by taking $k$ differential 1-forms $\alpha_i, i=1\dots k$, and asking to find a surface $S$ with the property that every vector $v$ tangent to $S$ satisfies $\alpha_i(v) = 0$ for $i=1 \dots k$.  The condition to find a solution is called the &lt;i&gt;Frobenius integrability theorem&lt;/i&gt; and the special case when $k=1$ is the requirement that $d\alpha_1 = 0$.  This is generally a topic you will see in a differential geometry class.&lt;/p&gt;&#xA;" OwnerUserId="8343" LastActivityDate="2011-03-18T16:28:40.447" CommentCount="1" />
  <row Id="27823" PostTypeId="2" ParentId="27744" CreationDate="2011-03-18T16:34:44.870" Score="15" ViewCount="" Body="&lt;p&gt;Even in the category of &lt;em&gt;all&lt;/em&gt; abelian groups, $G\oplus G\cong H\oplus H$ does not imply $G\cong H$: there are countably generated torsion free abelian groups $G$ such that $G\not\cong G\oplus G$, but $G\cong G\oplus G\oplus G$. Setting $H=G\oplus G$ gives an example where the implication does not hold. (The first examples were constructed by A.L.S. Corner, &lt;em&gt;On a conjecture of Pierce concerning direct decompositions of abelian groups&lt;/em&gt;, Proc. Colloq. Abelian Groups (Tihany, 1963), Akadémiai Kiadó, Budapest, 1964, pp. 43-48; he proves that for any positive integer $r$ there exists a countable torsion free abelian group $G$ such that the direct sum of $m$ copies of $G$ is isomorphic to the direct sum of $n$ copies of $G$ if and only if $m\equiv n\pmod{r}$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, the analogous result for direct products fails for groups, though it holds for groups having both chain conditions (by the Krull-Schmidt theorem). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fact that you have categories with arbitrary coproducts/products (like the category of all groups and the category of all abelian groups) in which the result fails means that no proof via universal properties alone can exist. A proof that depended only the universal properties would translate into any category in which products/coproducts exist, but the result is false in general even for very nice categories. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whether the conditions for the implication to hold can be expressed via universal conditions also seems to me to be doubtful. Universal conditions don't lend themselves easily to statements of the form &quot;If something happens for $A$ and $B$, then it happens for $f(A)$ and $f(B)$&quot; (where by &quot;$f(-)$&quot; I just mean &quot;this other object cooked up or related to $A$ in some way&quot;) unless the construction/relation happens to be categorical.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-03-19T01:37:51.103" LastActivityDate="2011-03-19T01:37:51.103" CommentCount="8" />
  <row Id="27824" PostTypeId="2" ParentId="27808" CreationDate="2011-03-18T17:05:09.780" Score="1" ViewCount="" Body="&lt;p&gt;One demonstration of your &quot;other direction&quot;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $T$ and $S$ are two operators that comute, then $S(\mathrm{Ker}T)\leqslant \mathrm{Ker}T$. In fact,&#xA;$$v \in \mathrm{Ker}T \Rightarrow T(Sv)=S(Tv)=S(0)=0$$&#xA;In words, $\mathrm{Ker}T$ is invariant under $S$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, in our case we have that $\mathrm{Ker}T$ is invariant under any linear transformation in $L(V)$. This implies that $\mathrm{Ker}T = V$ or $0$. In fact, in other cases we would have $S\in L(V)$ such that $S(\mathrm{Ker}T)\nsubseteq\mathrm{Ker}T$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We now show that $T$ has an eigenvalue. In fact, let $S \in L(V)$ be a projection on a non-zero one-dimensional subspace $\langle v \rangle$. Since,&#xA;$$Tv=T(Sv)=S(Tv)$$&#xA;we have that $Tv \in \langle v\rangle$. Equivalently, $Tv=\lambda v$ for some $\lambda \in \mathbb K$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since&#xA;$$TS=ST \iff (T-\lambda I)S=S(T-\lambda I)$$&#xA;for any $S \in L(V)$, we have as above that&#xA;$$\mathrm{Ker}(T-\lambda I)=V$$&#xA;or&#xA;$$\mathrm{Ker}(T-\lambda I)=0.$$&#xA;But, since $\lambda$ is an eigenvalue, $\mathrm{Ker}(T-\lambda I)\neq 0$. Therefore, $T=\lambda I$. QED.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This result is a special case of the Schur Lemma wich states: &quot;If $T$ is an operator in $V$ with an eigenvalue $\lambda \in \mathbb K$ and $C \subseteq L(V)$ is a set of operators such that&#xA;$$TS=ST \forall S \in C$$&#xA;and for each non-trivial subspace $W$ there is some $S \in C$ such that&#xA;$$S(W) \nsubseteq W,$$&#xA;then we must have $T=\lambda I$&quot;. And whose demonstration is essentialy as above.  &lt;/p&gt;&#xA;" OwnerUserId="370" LastEditorUserId="370" LastEditorDisplayName="" LastEditDate="2011-03-18T19:51:41.597" LastActivityDate="2011-03-18T19:51:41.597" CommentCount="0" />
  <row Id="27825" PostTypeId="2" ParentId="27719" CreationDate="2011-03-18T17:09:33.210" Score="1" ViewCount="" Body="&lt;p&gt;In fact, the top answer can be generalized slightly (sorry, I don't have enough points to post comments, yet):  if $a \vert b$, then $gcd(a,b)=a$ (and this holds in any algebraic structure where divisibility makes sense (eg a commutative, cancellative monoid)).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To see why, well, it's clear that $a$ is a common divisor of $a$ and $b$, and if $\alpha$ is any common divisor of $a$ and $b$, then, of course, $\alpha \vert a$.  Thus, $a=gcd(a,b)$.&lt;/p&gt;&#xA;" OwnerUserId="5137" LastEditorUserId="5137" LastEditorDisplayName="" LastEditDate="2011-03-18T18:40:17.953" LastActivityDate="2011-03-18T18:40:17.953" CommentCount="1" />
  <row Id="27827" PostTypeId="2" ParentId="27808" CreationDate="2011-03-18T17:55:51.593" Score="1" ViewCount="" Body="&lt;p&gt;Perhaps somewhat against the spirit of Axler's book &quot;linear maps over matrices&quot; (although quite conceptual):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose T commutes with all S. Then in particular it commutes with all invertible S: so&#xA;$T=STS^{-1}$ for all invertible S. But this means the matrix of T is the same, no matter what basis we choose! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then it must be diagonal: for fixed $j$, replace basis vector $e_j$ with $2e_j$; then if $i\neq j$, we get $t_{ij}=2t_{ij}$, so $t_{ij}=0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;[elaboration on the previous line]: Suppose $T$ has matrix $(t_{ij})_{ij}$ w.r.t. the basis $\{e_1,...,e_n\}$. Fix $k$, and consider the basis $B_k=\{v_1,..,v_n\}$ where $v_i=e_i$ if $i\neq k$ and $v_k=2e_k$. Then, for $i\neq k$, the matrix of $T$ w.r.t. $B_k$ has $2t_{ik}$ at entry $i,k$. Hence $t_{ik}=2t_{ik}$ and consequently $2t_{ik}=0$. [End edit.]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then also all diagonal entries are the same: for fixed $i$ and $j$, interchange $e_j$ and $e_i$, and get $t_{ii}=t_{jj}$.&lt;/p&gt;&#xA;" OwnerUserId="6490" LastEditorUserId="6490" LastEditorDisplayName="" LastEditDate="2011-03-19T00:23:37.233" LastActivityDate="2011-03-19T00:23:37.233" CommentCount="4" />
  <row Id="27828" PostTypeId="1" AcceptedAnswerId="28078" CreationDate="2011-03-18T18:10:32.173" Score="1" ViewCount="131" Body="&lt;p&gt;I've kind of forgotten the name of the following statements:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;9x = 11&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;10x = 9y&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What &lt;strong&gt;are&lt;/strong&gt; they called? And how do you solve them?&lt;/p&gt;&#xA;" OwnerUserId="8431" LastEditorUserId="6296" LastEditorDisplayName="" LastEditDate="2011-03-18T20:25:39.890" LastActivityDate="2011-03-20T07:10:06.997" Title="Calculate the unknown" Tags="&lt;homework&gt;&lt;algebra-precalculus&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="27829" PostTypeId="2" ParentId="27828" CreationDate="2011-03-18T18:16:54.960" Score="2" ViewCount="" Body="&lt;p&gt;Dividing both sides of the first by $9$ gives $x=\frac{11}{9}$.  You can then insert this into the second, giving $9y=\frac{110}{9}$, then divide both sides by $9$ to get $y=\frac{110}{81}$&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-03-18T18:16:54.960" CommentCount="4" />
  <row Id="27830" PostTypeId="1" CreationDate="2011-03-18T18:18:06.117" Score="0" ViewCount="53" Body="&lt;ol&gt;&#xA;&lt;li&gt;Is &lt;a href=&quot;http://en.wikipedia.org/wiki/Karush%E2%80%93Kuhn%E2%80%93Tucker_conditions#Necessary_conditions&quot; rel=&quot;nofollow&quot;&gt;the set of KKT conditions&lt;/a&gt;&#xA;necessary for local min, only under&#xA;some &lt;a href=&quot;http://en.wikipedia.org/wiki/Karush%E2%80%93Kuhn%E2%80%93Tucker_conditions#Regularity_conditions_.28or_constraint_qualifications.29&quot; rel=&quot;nofollow&quot;&gt;regularity conditions (ie&#xA;constraint qualifications)&lt;/a&gt;?&lt;/li&gt;&#xA;&lt;li&gt;So does this mean that when the cost&#xA;functions and the functions in&#xA;constraints are continuously&#xA;differentiable at a point, it is&#xA;possible that the point is a local&#xA;min but is not regular?&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I am more interested in how to apply&#xA;KKT to find candidates of local mins&#xA;and even global mins.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If we use KKT conditions to find candidates of local mins, does it mean that we make a mistake of not checking if those non-regular points are local mins?&lt;/li&gt;&#xA;&lt;li&gt;When trying to compute candidates for local mins by solving the KKT&#xA;conditions, is it necessary to check&#xA;if the computed points are regular? How to&#xA;deal    with those that you got but turn out to&#xA;be    non-regular? Shall we discard&#xA;them?    Does this also mean that we&#xA;are    making the same mistake of&#xA;not    checking non-regular points?&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Thanks and regards!&lt;/p&gt;&#xA;" OwnerUserId="1281" LastEditorUserId="1281" LastEditorDisplayName="" LastEditDate="2011-03-19T01:56:00.853" LastActivityDate="2011-03-19T01:56:00.853" Title="Is it necessary to check for regularity when applying KKT thoerem?" Tags="&lt;linear-algebra&gt;&lt;real-analysis&gt;&lt;optimization&gt;" CommentCount="3" FavoriteCount="1" />
  <row Id="27832" PostTypeId="2" ParentId="27808" CreationDate="2011-03-18T18:28:58.320" Score="6" ViewCount="" Body="&lt;p&gt;For a basis-free answer, consider $S \in L(V)$ given by $S x = f(x) v$ for some vector $v$ and some linear functional $f$ on V.  Then $T S x = f(x) T v = S T x = f(T x) v$ for any x.  In particular, as long as a nontrivial linear functional $f$ on $V$ exists, there is $x$&#xA;such that $f(x) \ne 0$, and then $T v = \alpha v$ for all $v$, where $\alpha = f(T x)/f(x)$.&#xA;This works even for infinite-dimensional spaces, although I think in general you need the Axiom of Choice to get a nontrivial linear functional on a vector space.&lt;/p&gt;&#xA;" OwnerUserId="8508" LastActivityDate="2011-03-18T18:28:58.320" CommentCount="2" />
  <row Id="27833" PostTypeId="1" AcceptedAnswerId="27838" CreationDate="2011-03-18T18:29:50.017" Score="2" ViewCount="118" Body="&lt;p&gt;I'm in a class titled &quot;Puzzle Based Learning&quot; and we were given this problem:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;There is a new game show and you are&#xA;  the participant. There are two doors,&#xA;  each has a suitcase with gold coins&#xA;  behind it. You know that these&#xA;  suitcases contain amounts from the&#xA;  set: 25, 50, 100, 200, 400, 800,&#xA;  1,600, 3,200, and 6,400. You also know&#xA;  that one suitcase has twice as much&#xA;  coins as the other. You select the&#xA;  door and open the suitcase. You find&#xA;  1,600 gold coins. Then the game show&#xA;  host offers you the opportunity to&#xA;  switch doors. Do you do it? Justify&#xA;  your answer.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In class, the professor tells us:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The two possibilities of the suitcase&#xA;  are 800 and 3200.  The expected&#xA;  outcome of choosing one suitcase would&#xA;  be 800*.5 + 3200*.5 = 2000.  Since&#xA;  2000&gt;1600, you should choose again&#xA;  because the expected outcome is&#xA;  greater than 1,600.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I was hoping someone could explain this a little more clearly.  In my mind, it seems there is a 50/50 chance of getting 800 and 3200 and the &quot;expected outcome&quot; is meaningless because we shouldn't care about the payoff of choosing again. You have a 50% chance of losing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Am I missing something? Is this a trick question? Is my professor pulling my leg?&lt;/p&gt;&#xA;" OwnerUserId="6599" LastEditorUserId="6599" LastEditorDisplayName="" LastEditDate="2011-03-18T18:48:49.277" LastActivityDate="2011-03-19T10:59:52.700" Title="Probability of an expected outcome" Tags="&lt;puzzle&gt;&lt;probability&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="27835" PostTypeId="2" ParentId="27805" CreationDate="2011-03-18T18:32:32.837" Score="5" ViewCount="" Body="&lt;p&gt;$$\begin{align}\frac{2^{2x-1} - 2^{x-1}}{2^{2x-1}}&#xD;&#xA;&amp;amp;=\frac{2^{2x-1}}{2^{2x-1}}-\frac{2^{x-1}}{2^{2x-1}}&#xD;&#xA;\\&#xD;&#xA;&amp;amp;=1-2^{(x-1)-(2x-1)}&#xD;&#xA;\\&#xD;&#xA;&amp;amp;=1-2^{-x}&#xD;&#xA;\end{align}$$&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2011-03-18T18:32:32.837" />
  <row Id="27836" PostTypeId="2" ParentId="27804" CreationDate="2011-03-18T18:37:07.127" Score="3" ViewCount="" Body="&lt;p&gt;The Lamplightergroup &lt;a href=&quot;http://en.wikipedia.org/wiki/Lamplighter_group&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Lamplighter_group&lt;/a&gt;&#xA;is a nice group constructed via the wreathproduct.&#xA;It is an example of a group of exponential growth which is still amenable and &#xA;the notion of amenability is not pure group theory anymore.&lt;/p&gt;&#xA;" OwnerUserId="8432" LastActivityDate="2011-03-18T18:37:07.127" />
  <row Id="27838" PostTypeId="2" ParentId="27833" CreationDate="2011-03-18T18:49:20.253" Score="3" ViewCount="" Body="&lt;p&gt;There is little typo in your question. There should be&#xA;$$ 800 \cdot \frac{1}{2} + 3200 \cdot \frac{1}{2} = 2000. $$&#xA;What your professor said is that the &lt;em&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Expected_value&quot; rel=&quot;nofollow&quot;&gt;expected value&lt;/a&gt;&lt;/em&gt; of your winning is 2000 when you change the doors. In some way it means that when you play this game many times then your average winning is going to be closer and closer to 2000 (assuming that you always chose the door with 1600 first).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Formally, the &lt;a href=&quot;http://en.wikipedia.org/wiki/Law_of_large_numbers&quot; rel=&quot;nofollow&quot;&gt;Law of large numbers&lt;/a&gt; plays role here. If you take $X_i$ to be your winning in $i$'th game and $S_n$ be the sum of your winnings in $n$ games then&#xA;$$\frac{S_n}{n} \to 2000.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence if you change the door then your &lt;em&gt;average&lt;/em&gt; win will be greater then 1600. Here it is important that you either win 1600 more or 800 less then in your first choice with equal probability.&lt;/p&gt;&#xA;" OwnerUserId="8229" LastEditorUserId="8229" LastEditorDisplayName="" LastEditDate="2011-03-19T10:59:52.700" LastActivityDate="2011-03-19T10:59:52.700" CommentCount="3" />
  <row Id="27840" PostTypeId="2" ParentId="27741" CreationDate="2011-03-18T19:20:09.830" Score="1" ViewCount="" Body="&lt;p&gt;The answer to the first question (and the one in the title) is yes when $M$ is the $\sigma$-algebra of Lebesgue measurable subsets of $\mathbb{R}^n$. Suppose $N$ is another $\sigma$-algebra such that $M\subsetneq N$. Then there exists $E\in N$ non measurable. Since $E=\cup_{k=1}^\infty (E\cap\{x\in\mathbb{R}^n:|x|\le k\})$, for at least one $k$ the set $E\cap\{x\in\mathbb{R}^n:|x|\le k\}$ is not measurable. Thus, we may assume without loss of generality that $E$ is bounded, and in particular, $m(E)&amp;lt;+\infty$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $E$ is not measurable, there exists $\epsilon&amp;gt;0$ such that if $G$ is an open set and $E\subset G$, then $m(G\setminus E)\ge \epsilon$. (This follows from an equivalent definition of measurable set; cf. Proposition 15 on p.63 of Royden's &lt;em&gt;Real Analysis, 3 ed.&lt;/em&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, since $m(E)=\inf\{m(G):G\text{ open, }E\subset G\}$, there exists an open set $O$ such that $E\subset O$ and $m(E)\ge m(O)-\epsilon/2$. Then $O=(O\setminus E)\cup E$ but&#xA;$$&#xD;&#xA;m(O\setminus E)+m(E)\ge\epsilon+ m(O)-\epsilon/2&amp;gt;m(O),&#xD;&#xA;$$&#xA;and hence $m$ is not additive on $N$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the general case it is easy to see that if $N$ is another $\sigma$-algebra such that $M\subsetneq N$, then there exists $A\subset X$ such that $m$ is not additive on the $\sigma$-algebra generated by $N\cup\{A\}$, but I have not been able to show that it is not additive on $N$.&lt;/p&gt;&#xA;" OwnerUserId="4791" LastEditorUserId="6887" LastEditorDisplayName="" LastEditDate="2011-03-19T01:23:40.833" LastActivityDate="2011-03-19T01:23:40.833" />
  <row Id="27841" PostTypeId="1" CreationDate="2011-03-18T19:25:51.087" Score="3" ViewCount="129" Body="&lt;p&gt;Let $W_t$ be a standard one dimension Brownian Motion with $W_0=0$ and $X_t=\int_0^t{W_sds}$.&#xA;With the help of ito formula, we could get&#xA;$$E[(X_t)^2]=\frac{1}{3}t^3$$ &#xA;$$E[(X_t)^3]=0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I try to employ the same method to calculate the general case $E[(X_t)^n]$, I got stuck.&#xA;I guess $X_t$ should be normal distribution since it could be the limit of the following&#xA;$$\lim_{n\rightarrow \infty}{\sum_{i=0}^{n-1}{W_{t_i}(t_{i+1}-t_i)}},$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $ W_{t_i}\sim norm(0,\sqrt{\frac{t_i}{n}}).$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it is true, the problem would be trivial.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update:&lt;/strong&gt; Thanks for all the suggestions. Now I believe $X_t$ is a Gaussian process.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How about for this integral&#xA;$$Y_t=\int_0^t{f(W_s)ds}$$&#xA;if we assume that $f$ is some good function, say polynomial or exponential, i.e &#xA;$$Y_t=\int_0^t{e^{W_s}ds}$$&#xA;$$Y_t=\int_0^t{[a_n(W_s)^n+a_{n-1}(W_s)^{n-1}+...+a_0]ds}$$&lt;/p&gt;&#xA;" OwnerUserId="8183" LastEditorUserId="8183" LastEditorDisplayName="" LastEditDate="2011-03-18T21:33:45.827" LastActivityDate="2011-03-18T21:33:45.827" Title="How to calculate $E[(\int_0^t{W_sds})^n], n \geq 2$" Tags="&lt;probability&gt;&lt;stochastic-processes&gt;&lt;probability-distributions&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="1" />
  <row Id="27842" PostTypeId="1" AcceptedAnswerId="27846" CreationDate="2011-03-18T19:30:29.243" Score="2" ViewCount="48" Body="&lt;p&gt;I am reading a document it says &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A group &lt;strong&gt;realization&lt;/strong&gt; is a map from elements of G to transformations of a space M that is a &lt;strong&gt;group homomorphism&lt;/strong&gt;, i.e. it preserves the group multiplication law. Thus if $T:G \rightarrow T(M): g \mapsto T(g)$ where $T(g)$ is some transformation on M, then T is a homomorphism if $T(g_1 g_2) = T(g_1)T(g_2)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have got a bad feeling about this definition. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If T(M) is transformations on M then how can $T$ also be a map from $G$ to $T(M)$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.physics.harvard.edu/~dtlarson/tutorial05/lecture1.pdf&quot; rel=&quot;nofollow&quot;&gt;This document page 4.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="705" LastActivityDate="2011-03-18T19:51:12.220" Title="Help with definition of group realization" Tags="&lt;group-theory&gt;&lt;notation&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="27843" PostTypeId="2" ParentId="27833" CreationDate="2011-03-18T19:44:58.037" Score="3" ViewCount="" Body="&lt;p&gt;There is something missing, namely any statement that the 16 ways of putting the suitcases behind the doors are equally likely.  If they are equally likely then you should always switch after opening a door unless you see 6400, for the reason given. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But suppose that you know each pair of suitcases $(n,2n)$ is three times as likely as the next pair up $(2n, 4n)$.  In your particular example the choice would be between $1600$ and an expected $800\times \frac{3}{4} + 3200\times \frac{1}{4} = 1400$ so you would stay with what you see first; similarly for any amount you would keep what you see, unless you see 25 in which case you would switch doors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you simply don't know - it is a &lt;em&gt;new game show&lt;/em&gt; - then you have to judge the psychology of the game designers.  They may think more big winners will attract more viewers and so pay off in advertising; or they may just want to drive down prize costs.  Your call. &lt;/p&gt;&#xA;" OwnerUserId="6460" LastActivityDate="2011-03-18T19:44:58.037" />
  <row Id="27844" PostTypeId="2" ParentId="21869" CreationDate="2011-03-18T19:47:08.270" Score="0" ViewCount="" Body="&lt;p&gt;It makes more sense to look at it in terms of partials &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_{a}^{b}\dfrac{\partial y}{\partial x}\,{\rm d}x={\Delta y}_{{x\rightarrow a\ldots b}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with ${\Delta y}$ indicating the change in $y$ as $x$ varies from $a$ to $b$. If there are more variables involved ($y(x,z,w)$) then $z$ and $w$ are assumed to be constant during the variation in $x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is kind of the inverse of the chain rule.&lt;/p&gt;&#xA;" OwnerUserId="3301" LastActivityDate="2011-03-18T19:47:08.270" />
  <row Id="27845" PostTypeId="2" ParentId="27841" CreationDate="2011-03-18T19:49:16.003" Score="1" ViewCount="" Body="&lt;p&gt;Note that $X_T = \int_0^T W_t\,dt = T W_T - \int_0^T t\,dW_t$.  Thus $X_T$ is a linear combination of normally distributed random variables.  (The latter integral may be easier to show that it is normal.)&lt;/p&gt;&#xA;" OwnerUserId="3852" LastActivityDate="2011-03-18T19:49:16.003" />
  <row Id="27846" PostTypeId="2" ParentId="27842" CreationDate="2011-03-18T19:51:12.220" Score="1" ViewCount="" Body="&lt;p&gt;The problem here is really just the notation: $T$ is playing two roles, both the name of the group homomorphism from $G$ to the transformation group of $M$, and as part of the name of the group of transformations of $M$. Since the notes seem to often give special names to the group of transformations, depending on who $M$ is, a simple fix is just to change the generic name for this group.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instead of using $T(M)$ to denote &quot;a group of transformations $M$&quot;, call it something else, say $\mathrm{Tr}(M)$. Then $T\colon G\to \mathrm{Tr}(M)$ is a realization of $G$ if $T$ is a homomorphism from $G$ to a group of transformations of the a space $M$.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2011-03-18T19:51:12.220" CommunityOwnedDate="2011-03-18T19:51:12.220" CommentCount="3" />
  <row Id="27847" PostTypeId="2" ParentId="27812" CreationDate="2011-03-18T19:56:01.257" Score="0" ViewCount="" Body="&lt;p&gt;I don't think that O(n * logn) is your friend here, since this is basically a counting problem.  If the values for a, b and c result in a finite number of words in the language then this is &lt;em&gt;probably&lt;/em&gt; possible; the counting can be accomplished with a (potentially large, but finite) number of states.  But if the values for a, b and c result in an infinite number of words in the language then the counting will need to be done by traversing w multiple times.&lt;/p&gt;&#xA;" OwnerUserId="4181" LastActivityDate="2011-03-18T19:56:01.257" />
  <row Id="27848" PostTypeId="1" AcceptedAnswerId="27888" CreationDate="2011-03-18T20:15:10.640" Score="2" ViewCount="64" Body="&lt;p&gt;Does there exist a quaternion $q$ on the unit sphere such that, given the vanilla complex plane $\mathbb{C}$, $q\mathbb{C}q^{-1} = \bar{\mathbb{C}}$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Motivation: ordinarily, the plane is rotated by multiplying it by a complex number in the unit circle. Taking the conjugate amounts to flipping the plane along the real axis, which can be viewed as a rotation.&lt;/p&gt;&#xA;" OwnerUserId="890" LastActivityDate="2011-03-19T00:43:49.130" Title="$\bar{z}$ and rotation by quaternion multiplication" Tags="&lt;analysis&gt;&lt;quaternions&gt;&lt;rotations&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" />
  <row Id="27849" PostTypeId="2" ParentId="27841" CreationDate="2011-03-18T20:31:45.483" Score="4" ViewCount="" Body="&lt;p&gt;As a linear functional of the Gaussian process $(W_s)_{0\le s\le t}$, $X_t$ is a Gaussian random variable. You indicate yourself that $X_t$ is centered and has variance $\sigma^2_t=\frac13t^3$, hence $X_t$ is distributed like $\sigma_tN$ with $N$ standard Gaussian. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, for every $n$, $E((X_t)^{2n+1})=0$ and $E((X_t)^{2n})=(\sigma_t^2)^nE(N^{2n})$. If you know the moments of a standard Gaussian (and you should...), you are done.&lt;/p&gt;&#xA;" OwnerUserId="6179" LastActivityDate="2011-03-18T20:31:45.483" CommentCount="11" />
  <row Id="27850" PostTypeId="2" ParentId="27848" CreationDate="2011-03-18T20:32:34.263" Score="2" ViewCount="" Body="&lt;p&gt;I don't know if there's a single rotation that gives you what you want, but given any complex number $z$, the complex conjugate of $z$ is given by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\bar{z}=-\frac{1}{2}(z-izi^{-1}-jzj^{-1}-kzk^{-1}).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Complex conjugation is a linear combination of quaternion rotations, at least. Note that this equation actually works for all quaternions.&lt;/p&gt;&#xA;" OwnerUserId="7676" LastEditorUserId="7676" LastEditorDisplayName="" LastEditDate="2011-03-18T20:39:39.560" LastActivityDate="2011-03-18T20:39:39.560" />
  <row Id="27851" PostTypeId="1" AcceptedAnswerId="27993" CreationDate="2011-03-18T20:40:51.030" Score="0" ViewCount="145" Body="&lt;p&gt;Triggered by &lt;a href=&quot;http://math.stackexchange.com/questions/27719/what-is-gcd0-a-where-a-is-a-positive-integer&quot;&gt;previous question&lt;/a&gt;, can one prove GCD(0,8)≠1 purely by lattice laws?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Brute force Prover9/Mace4 assertions&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;x ^ y = y ^ x. &#xA;(x ^ y) ^ z = x ^ (y ^ z). &#xA;x ^ (x v y) = x. &#xA;x v y = y v x. &#xA;(x v y) v z = x v (y v z). &#xA;x v (x ^ y) = x.&#xA;&#xA;1 v x = x.&#xA;1 ^ x = 1.&#xA;0 ^ x = 1.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;exhibit no [finite] model, which is indication that the system is inconsistent. I have trouble, however, understanding how to elevate this intuition into a formal proof (there is no goal).  &lt;/p&gt;&#xA;" OwnerUserId="5754" LastEditorUserId="5754" LastEditorDisplayName="" LastEditDate="2011-03-19T17:52:23.967" LastActivityDate="2011-03-21T21:51:32.917" Title="can GCD(0,8)≠1 be proven purely by lattice laws?" Tags="&lt;lattices&gt;" AnswerCount="2" CommentCount="8" />
  <row Id="27852" PostTypeId="2" ParentId="11214" CreationDate="2011-03-18T20:44:29.710" Score="2" ViewCount="" Body="&lt;p&gt;The Laplace transform does exactly what you're talking about. The polynomials don't necessarily come in the form ${1 \over n!} x^n$ but can be substituted into that form.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Check out &lt;a href=&quot;http://academicearth.org/lectures/laplace-transform-odes-with-discontinuous-inputs&quot; rel=&quot;nofollow&quot;&gt;this lecture&lt;/a&gt; by Arthur Mattuck from MIT for the details.&lt;/p&gt;&#xA;" OwnerUserId="1457" LastActivityDate="2011-03-18T20:44:29.710" />
  <row Id="27853" PostTypeId="2" ParentId="27841" CreationDate="2011-03-18T20:49:52.057" Score="3" ViewCount="" Body="&lt;p&gt;The random variable $X_t$ is Gaussian for the reasons in Didier's answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can calculate the variance directly (without Ito's formula) as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathbb{E}\left[\left( \int^t_0 W_s ds \right)^2\right] &#xD;&#xA;= \int^t_0 \int^t_0  \mathbb{E}(W_r W_s)  dr ds &#xD;&#xA;= \int^t_0 \int^t_0  (r\wedge s) dr ds ={t^3\over 3}.$$ &lt;/p&gt;&#xA;" OwnerUserId="940" LastActivityDate="2011-03-18T20:49:52.057" CommentCount="5" />
  <row Id="27854" PostTypeId="2" ParentId="1002" CreationDate="2011-03-18T21:03:58.180" Score="2" ViewCount="" Body="&lt;p&gt;Think about light coming from stars. The light has color or &quot;spectrum&quot; but of course the data comes in a 1-D stream.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The Fourier transform gives you the spectrum of the time series.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can also think about the EQ on your stereo -- the 2kHz slider, the 5kHz slider, etc. Those sliders are adjusting the constants in the Fourier realm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;(Inverse Fourier just takes you back from spectrum to signal. So what does it mean that $\mathcal{F}^{-1} = \mathcal{F}$?)&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To get into the math of it, you have to remember that $\cos$ and $\sin$ a